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  <row Id="1" PostTypeId="1" AcceptedAnswerId="147" CreationDate="2014-05-14T20:17:34.103" Score="29" ViewCount="1844" Body="&lt;p&gt;Most American crosswords use rotational (or &quot;radial&quot;) symmetry so that squares directly opposite each other (through the center) are identical.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Times, I've only seen this &lt;strong&gt;not&lt;/strong&gt; be the case when the layout was directly linked to the theme.  Is that one of the &quot;rules&quot; you can count on, or are there (modern, Will Shortz era) examples of cases where you can't assume that non-radial symmetry will be somehow related to the theme?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example where the theme &lt;strong&gt;does&lt;/strong&gt; justify the deviation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CQ2Hy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theme in this example was celebrating the 50th anniversary of Frank Lloyd Wright's landmark Guggenheim building.&lt;/p&gt;&#xA;" OwnerUserId="5" LastEditorUserId="2383" LastEditDate="2014-12-07T17:02:14.893" LastActivityDate="2014-12-07T17:02:14.893" Title="Are there &quot;rules&quot; (or consistent conventions) that limit when the NY Times Crossword can deviate from rotational symmetry?" Tags="|puzzle-creation|crosswords|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2" PostTypeId="1" CreationDate="2014-05-14T20:23:29.700" Score="60" ViewCount="83832" Body="&lt;p&gt;I've seen Sudoku puzzles that can be solved two different ways. Referring to a traditional 9x9 grid:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Is it possible for a Sudoku puzzle to have more than two solutions? If so, what is the maximum number of solutions a single Sudoku puzzle can have?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For any given puzzle, is there a way to find out how many solutions it has (other than brute force)?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="10" LastEditorUserId="69582" LastEditDate="2021-09-04T19:11:41.127" LastActivityDate="2021-09-05T06:57:22.170" Title="What is the maximum number of solutions a Sudoku puzzle can have?" Tags="|logical-deduction|grid-deduction|sudoku|solvability|" AnswerCount="4" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4" PostTypeId="2" ParentId="2" CreationDate="2014-05-14T20:39:14.820" Score="28" Body="&lt;p&gt;I have seen two dissenting opinions on this subject (and in my opinion, the first option is right):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;By definition, a Sudoku has only one solution. Anything else is just a grid of numbers. Sometimes, there are errors in a publication, and a starting grid has multiple solutions, but, then the starting grid was not a Sudoku!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sudoku&quot; rel=&quot;noreferrer&quot;&gt;From Wikipedia&lt;/a&gt;: &lt;em&gt;The number of classic 9×9 Sudoku solution grids is 6,670,903,752,021,072,936,960, or approximately 6.67×10&lt;sup&gt;21&lt;/sup&gt; (ignoring rotations and other factors, there are many fewer solutions, just 5,472,730,538&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Assuming you define a sudoku to have just one possible solution, then the rest of your questions are moot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the sudoku is defined to have multiple solutions, then, brute-forcing is one way, but it is also possible to pre-compute all possible solutions for all inputs, and look up the result that way (a &lt;a href=&quot;http://en.wikipedia.org/wiki/Rainbow_table&quot; rel=&quot;noreferrer&quot;&gt;rainbow table&lt;/a&gt;). &lt;/p&gt;&#xA;" OwnerUserId="14" LastEditorUserId="-1" LastEditDate="2017-07-27T18:38:17.087" LastActivityDate="2017-07-27T18:38:17.087" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5" PostTypeId="2" ParentId="2" CreationDate="2014-05-14T20:39:38.710" Score="12" Body="&lt;p&gt;By definition, all valid Sudoku puzzles should have only one solution. In point of fact, many of the techniques used for solving puzzles &lt;em&gt;depend&lt;/em&gt; on there being only one solution. All of the Unique Rectangle techniques for example, only work if there is one, and only one, solution.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2014-05-14T20:39:38.710" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6" PostTypeId="2" ParentId="2" CreationDate="2014-05-14T20:39:50.493" Score="44" Body="&lt;p&gt;Most Sudoku puzzles published have only one solution. If there is more than one solution, it is probably a mistake. That said, puzzles with incomplete clues can have multiple solutions. In the extreme case, a puzzle with no clues has 6,670,903,752,021,072,936,960 solutions according to &lt;a href=&quot;http://en.wikipedia.org/wiki/Sudoku&quot;&gt;Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if it's possible to have exactly 3 solutions, but boards with 2 and 4 (and more) solutions are easy to find. In general, I think boards with an even number of solutions are easier to create.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding the number of solutions is a generalization of Sudoku solving algorithms, and there are Sudoku algorithms that do significantly better than brute forcing. Once the board has been filled out as far as possible, it can be brute forced the rest of the way.&lt;/p&gt;&#xA;" OwnerUserId="18" LastActivityDate="2014-05-14T20:39:50.493" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="7" PostTypeId="1" AcceptedAnswerId="8" CreationDate="2014-05-14T20:40:44.423" Score="29" ViewCount="5323" Body="&lt;p&gt;What is the maximum number of queens that can be placed on a standard 8x8 chessboard such that no one of them is capable of attacking any of the others in a single move?&lt;/p&gt;&#xA;" OwnerUserId="11" LastEditorUserId="11" LastEditDate="2014-05-27T23:39:46.320" LastActivityDate="2019-08-28T17:51:27.837" Title="How many queens can be on a chessboard without attacking each other?" Tags="|checkerboard|chess|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8" PostTypeId="2" ParentId="7" CreationDate="2014-05-14T20:40:44.423" Score="35" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 8.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Obviously, you can't have more than 8, since you would have a row (and column) with more than one queen on it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://mathworld.wolfram.com/QueensProblem.html&quot; rel=&quot;noreferrer&quot;&gt;Wolfram-Alpha&lt;/a&gt;, there are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 12 unique solutions, plus an additional 80 solutions from rotation/reflection.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One possible solution is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/npc8B.gif&quot; alt=&quot;8 Queens not attacking each other.  Image from codereview.stackexchange.com&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A list (and images!) of all &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 12 base solutions can be found both at the above Wolfram-Alpha link and &lt;a href=&quot;http://en.wikipedia.org/wiki/Eight_queens_puzzle&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;.  (Thank you &lt;a href=&quot;https://puzzling.stackexchange.com/users/4/kevin&quot;&gt;Kevin&lt;/a&gt; for the link). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11" LastEditorUserId="46903" LastEditDate="2018-09-16T07:09:19.513" LastActivityDate="2018-09-16T07:09:19.513" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="12" PostTypeId="1" CreationDate="2014-05-14T20:45:53.563" Score="57" ViewCount="101716" Body="&lt;p&gt;I believe in a number of books I have read and solved myself, the introduction stated that guessing should never be necessary for any of the puzzles in the book.  Instead, a handful of techniques could be used to solve the puzzle completely, including the hard puzzles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there published puzzles out there that do require guessing? Perhaps designed to challenge those who have &quot;mastered&quot; the more basic Sudoku puzzles?&lt;/p&gt;&#xA;" OwnerUserId="35" LastEditorUserId="30633" LastEditDate="2016-12-19T20:13:58.933" LastActivityDate="2019-01-18T17:52:15.980" Title="Are there published Sudoku puzzles that require guessing?" Tags="|sudoku|" AnswerCount="11" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="13" PostTypeId="1" CreationDate="2014-05-14T20:51:33.050" Score="34" ViewCount="5844" Body="&lt;p&gt;How can &lt;a href=&quot;http://en.wikipedia.org/wiki/Sokoban&quot;&gt;Sokoban&lt;/a&gt; puzzles be procedurally generated with a computer program, guaranteeing at least one solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple algorithm might be:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;{n} times&#xA;    place a &quot;target&quot; square and a box on the square&#xA;end&#xA;until {puzzle is done (?)}&#xA;    move a character to a location next to the box, and &quot;pull&quot; it towards the character&#xA;end&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here are my two main questions about this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How do you tell when the puzzle is &quot;done&quot; enough? (line 4)&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How would you move the character next to a box so that it could be pulled?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking about some kind of flood-fill to see if the character can reach a box, but how do I go to a random location from which I can pull a box?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="39" LastActivityDate="2014-12-06T03:04:55.467" Title="How can I procedurally generate Sokoban puzzles?" Tags="|puzzle-creation|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14" PostTypeId="2" ParentId="12" CreationDate="2014-05-14T20:53:20.593" Score="4" Body="&lt;p&gt;If by 'guessing' you mean writing down numbers that you are not sure about, then no, every proper puzzle is solvable (in theory) by looking at all the possibilities remaining. This is the brute force algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, some puzzles are complicated enough that you won't be able to remember all the possible values.  This is simply a case of humans having poor short-term memory, and you will require additional resources (like writing down numbers and trying them).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you get better and better at solving Sudoku puzzles, your brain recognizes more and more patterns, and when the puzzle you're solving matches one of those patterns, your brain can remember the solution. The more experience you have, the less difficult your reasoning becomes, and the less likely you are to find a puzzle that requires brute forcing.&lt;/p&gt;&#xA;" OwnerUserId="18" LastEditorUserId="18" LastEditDate="2014-05-14T22:12:01.373" LastActivityDate="2014-05-14T22:12:01.373" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15" PostTypeId="2" ParentId="12" CreationDate="2014-05-14T20:54:19.940" Score="10" Body="&lt;p&gt;TL;DR - short of a definitive mathematical proof, there is no way of knowing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do a web search for Sudoku solving techniques, you will find that there are a great many. Some of them are difficult and obscure, but handle cases that can't be solved any other way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My own experience, using the Sudoku app on my tablet, has been that every puzzle, no matter how difficult, can be solved by some technique, no matter how obscure. So, if you don't know a technique for a given situation, you may be reduced to guessing, but when asked for a hint, my tablet has always had a technique up it's sleeve for every possible situation.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2014-05-14T20:54:19.940" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16" PostTypeId="1" CreationDate="2014-05-14T20:56:06.477" Score="122" ViewCount="29664" Body="&lt;blockquote&gt;&#xA;&lt;h1&gt;Names in Boxes&lt;/h1&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The names of 100 prisoners are placed in 100 wooden boxes, one name to a box, and the boxes are lined up on a table in a room.&lt;/li&gt;&#xA;&lt;li&gt;One by one, the prisoners are led into the room; each may look in at most 50 boxes, but must leave the room exactly as he found it and is permitted no further communication with the others.&lt;/li&gt;&#xA;&lt;li&gt;The prisoners have a chance to plot their strategy in advance, and they are going to need it, because unless every single prisoner finds his own name all will subsequently be executed.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Find a strategy for them which has probability of success (all  prisoners' survival) exceeding 30%.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Comment:&lt;/strong&gt; If each prisoner examines a random set of 50&#xA;boxes, their probability of survival is an unenviable &lt;span class=&quot;math-container&quot;&gt;$1 /2^{100} \approx  0.0000000000000000000000000000008 $&lt;/span&gt;. They could do worse—if they all look in the same 50 boxes, their chances drop to zero. 30% seems&#xA;ridiculously out of reach—but yes, you heard the problem correctly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This problem was sourced from the excellent, &amp;quot;&lt;a href=&quot;https://www.math.dartmouth.edu/%7Epw/solutions.pdf&quot; rel=&quot;noreferrer&quot;&gt;Seven Puzzles You Think You Must Not Have Heard Correctly&lt;/a&gt;,&amp;quot; (Note - link includes solutions!)  Compiled by Peter Winkler&lt;/p&gt;&#xA;&lt;p&gt;This is the hardest brainteaser that I've ever actually been able to solve. (Eventually, anyway - I spent my spare time for a couple of &lt;em&gt;weeks&lt;/em&gt; on it, and at the time, even though I found the solution, I lacked the math skills to calculate the precise odds of the optimal solution working.) Can you explain the solution without too much mathematical background (less than what the linked solution assumes)?&lt;/p&gt;&#xA;" OwnerUserId="5" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2022-07-19T07:23:51.847" Title="100 Prisoners' Names in Boxes" Tags="|mathematics|word-problem|" AnswerCount="7" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="18" PostTypeId="1" AcceptedAnswerId="326" CreationDate="2014-05-14T20:59:29.090" Score="32" ViewCount="21543" Body="&lt;p&gt;I have seen many variations of the following puzzle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Once upon a time, a farmer went to the market and purchased a fox, a goose, and a bag of beans.  On his way home, the farmer came to the bank of a river and rented a small boat.  The boat was so small that the farmer could carry only himself and a single one of his purchases - the fox, the goose, or the bag of the beans.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The farmer knew that if he left the fox and the goose together, the fox would eat the goose.  Likewise, if left alone with the beans, the goose would eat them.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The farmer was able to get all of his purchases safely across the river.  How did he do it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know that the solution is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Take the goose across, come back empty.&lt;/li&gt;&#xA;&lt;li&gt;Take the beans across, bring the goose back.&lt;/li&gt;&#xA;&lt;li&gt;Leave the goose on the bank, take the fox across, come back empty.&lt;/li&gt;&#xA;&lt;li&gt;Take the goose across.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(The fox and beans are interchangeable).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is this:&lt;br&gt;&#xA;&lt;strong&gt;When given a problem in this format, what is the best way to begin solving it?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="11" LastEditorUserId="30633" LastEditDate="2016-12-19T23:54:22.200" LastActivityDate="2023-06-15T17:27:08.120" Title="Learning to Solve a River Crossing Puzzle" Tags="|strategy|river-crossing|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19" PostTypeId="1" AcceptedAnswerId="21" CreationDate="2014-05-14T21:01:28.680" Score="8" ViewCount="22972" Body="&lt;p&gt;Sometimes, when solving the &lt;a href=&quot;https://en.wikipedia.org/wiki/Void_Cube&quot;&gt;Void Rubik's Cube&lt;/a&gt; using the &lt;a href=&quot;http://www.youcandothecube.com/&quot;&gt;standard beginner method&lt;/a&gt;, it gets &quot;stuck.&quot; What I mean by this is that the top edge of one face and the top edge of an adjacent face's colors are swapped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, I've just turned random faces until the cube is sufficiently messed up and then started over. However, there must be a better way - what is it?&lt;/p&gt;&#xA;" OwnerUserId="39" LastEditorUserId="42" LastEditDate="2014-05-14T21:13:48.160" LastActivityDate="2017-10-01T22:58:18.260" Title="How can I fix &quot;stuck&quot; positions on the Void Rubik's Cube (using the standard beginner's method)?" Tags="|rubiks-cube|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20" PostTypeId="1" AcceptedAnswerId="76" CreationDate="2014-05-14T21:03:31.607" Score="35" ViewCount="25438" Body="&lt;p&gt;For those of you familiar with the solution to the Rubik's cube, you will know that a single-corner twist is not a valid position. It can neither be attained by scrambling, nor can it be solved if it occurs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that there's some fundamental group theory at work here; the corners on a Rubik's cube can only be rotated in groups of three. This logically implies that one can't have a single (or two, in the same direction) corners rotated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why, mathematically, is this the case?&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastActivityDate="2017-12-30T17:54:09.293" Title="Why is a single-corner twist not a valid position on a Rubik's cube?" Tags="|rubiks-cube|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21" PostTypeId="2" ParentId="19" CreationDate="2014-05-14T21:10:08.580" Score="8" Body="&lt;p&gt;When you work with a void cube, you've de facto lost a degree of visibility. The algorithm to take you out of this position is &lt;code&gt;M' U M' U' M U' M U2 M'&lt;/code&gt; - then you just solve the remaining edges as-is, and I'm going to attempt to briefly explain why this works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You, as you know, can no longer see the centers on the void cube. However, it's important to bear in mind that they're actually still there, just hidden. When the 'virtual' centers are rotated incorrectly - that is to say, you're attempting to solve the cube around an invalid position - the pieces won't line up. In the case you're seeing, the centers along one axis have been rotated 90 degrees out of position. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see for yourself why this might be, attempt to solve a regular Rubik's cube with one row of centers rotated 90 degrees. You'll find yourself in an equivalent position on the 3x3. (Note: a position which is mathematically equivalent to the one you are in is where two corners are swapped; if you end here, you're in literally the exact same position.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This algorithm moves them back into a legal position, and, though it disrupts a couple edges in the process, actually allows you to solve it regularly again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can also be done intuitively. Simply do a middle slice along any face, then keep one corner fixed in your mind relative to the three virtual centers (i.e., keep it in the same place relative to them), and solve the cube again.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastActivityDate="2014-05-14T21:10:08.580" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22" PostTypeId="1" AcceptedAnswerId="23" CreationDate="2014-05-14T21:11:34.290" Score="37" ViewCount="17516" Body="&lt;p&gt;Given an 8x8 chessboard, your goal is to &quot;cover&quot; each space on the board with the fewest possible number of pieces.  A space is &quot;covered&quot; if there is a piece on it, or if a piece on the board can be moved to that space in one move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A trivially easy solution would be that a board could be covered with 64 pieces.  If you place a piece on every square, every square is obviously covered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A less trivial solution is 8 - fill an entire row or column with rooks.  Obviously, each rook can cover all spaces in its row or column, so the board is covered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can this be done with less than 8 pieces?  If so, what is the minimum number of pieces required?&lt;/p&gt;&#xA;" OwnerUserId="11" LastEditorUserId="11" LastEditDate="2016-09-12T19:40:29.087" LastActivityDate="2020-01-28T05:37:22.910" Title="How many chess pieces does it take to &quot;cover&quot; all spaces on a chessboard?" Tags="|checkerboard|chess|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23" PostTypeId="2" ParentId="22" CreationDate="2014-05-14T21:11:34.290" Score="42" Body="&lt;p&gt;Yes.  The minimum number of pieces required is &lt;strong&gt;5&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5 queens can be places such that they cover every space on the board, as in the following example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3MsU5.gif&quot; alt=&quot;It only takes 5 queens to &amp;quot;cover&amp;quot; a full 8x8 chessboard.&quot;&gt; &#xA;&lt;img src=&quot;https://i.stack.imgur.com/EKhvj.png&quot; alt=&quot;color coded version&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 12 such arrangements, along with rotation and reflection of each of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: The above proves that 5 queens is enough, but it doesn't prove that 4 queens isn't enough.  According to &lt;a href=&quot;https://mathoverflow.net/q/30330/47866&quot;&gt;this MathOverflow question&lt;/a&gt; and its answers, there is no easy logical or mathematical proof, but it has been proven by completely evaluating all possible arrangements of queens on a board.  &lt;a href=&quot;http://oeis.org/A075458&quot; rel=&quot;noreferrer&quot;&gt;OEIS sequence A075458&lt;/a&gt; gives the minimum number of required queens for any square board from &lt;span class=&quot;math-container&quot;&gt;$1\times1$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$18\times18$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11" LastEditorUserId="3896" LastEditDate="2020-01-28T05:37:22.910" LastActivityDate="2020-01-28T05:37:22.910" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="24" PostTypeId="2" ParentId="20" CreationDate="2014-05-14T21:25:57.337" Score="13" Body="&lt;p&gt;This is not a rigorous mathematical proof, but I will try to explain it intuitively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of each corner having a property that is the amount that it's been rotated in 3 dimensions. The whole cube has a property that is the sum of all its corners. Initially everything is at 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you rotate one side of the cube, you rotate 4 corners a quarter-turn each. This adds up to a total of one full turn. Since rotation wraps around from 360° to 0°, you are back where you started. No matter what move you make, you will always end up with a whole number of full turns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since having a single corner rotated means that the total rotation is not 0, it is impossible to reach that state by making legal moves. However, when you have 3 corners rotated by a third of a turn each, the sum is once again a whole turn. Likewise it is legal to have two corners rotated, but in opposite directions, as the sum is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though it's hard to grasp 3D rotation and summing of angles, the concept should be enough to explain why corners cannot be rotated independently.&lt;/p&gt;&#xA;" OwnerUserId="18" LastEditorUserId="18" LastEditDate="2014-05-14T21:37:17.717" LastActivityDate="2014-05-14T21:37:17.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25" PostTypeId="1" AcceptedAnswerId="1804" CreationDate="2014-05-14T21:27:26.287" Score="25" ViewCount="1711" Body="&lt;p&gt;Quick summary of the Prisoner's Dilemma: Two criminals are charged with a crime. Each is given an opportunity by the police to confess.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If they &lt;strong&gt;both stay silent&lt;/strong&gt;, they don't have enough evidence to convict for much, and they both get &lt;strong&gt;1 year&lt;/strong&gt; in prison&lt;/li&gt;&#xA;&lt;li&gt;If they &lt;strong&gt;both confess&lt;/strong&gt;, they have a lot more evidence and they both get &lt;strong&gt;2 years&lt;/strong&gt; in prison.&lt;/li&gt;&#xA;&lt;li&gt;If &lt;strong&gt;one confesses and the other does not&lt;/strong&gt;, the one who confesses gets a plea bargain and &lt;strong&gt;gets to leave freely&lt;/strong&gt;. The one who doesn't confess gets all the liability and gets &lt;strong&gt;3 years in prison&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You can put these in a chart, called a &quot;payoff matrix&quot; in Game Theory. Here's a typical, standard payoff Matrix for Prisoner's Dilemma (&lt;a href=&quot;https://en.wikipedia.org/wiki/Prisoner%27s_dilemma&quot;&gt;adapted from Wikipedia&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;         | B coop | B defect&#xA;----------------------------&#xA;A coop   | R, R   | S, T&#xA;----------------------------&#xA;A defect | T, S   | P, P&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Where &lt;code&gt;T &amp;gt; R &amp;gt; P &amp;gt; S&lt;/code&gt;. It doesn't matter much if you're only playing one version of the game. But in iterated Prisoners dilemma, you want to maximize your score over many rounds, which has optimal strategies beyond &quot;always defect&quot;, e.g. &lt;a href=&quot;https://en.wikipedia.org/wiki/Tit_for_tat&quot;&gt;tit for tat&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I would like to know is, what happens if, for example, &lt;code&gt;R &amp;gt;&amp;gt;&amp;gt; P&lt;/code&gt;? Example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;T = 101&#xA;R = 100&#xA;P = 1&#xA;S = 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In this case, the penalty for having your opponent defect is so much worse, it dominates anything about your own decision. Your own decision can only move your score by 1 point, but your opponent can move your score by 100 points. To me, this implies that both players would be much more inclined to cooperate (and, of course, always defect in the last round).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Does this make sense? Would you adjust an algorithm to be more cooperative if this were the payout matrix?&lt;/li&gt;&#xA;&lt;li&gt;What about other extremes? e.g. (&lt;code&gt;P &amp;gt;&amp;gt;&amp;gt; S&lt;/code&gt;, or &lt;code&gt;T = 101, R = 100, P = 99, S = 0&lt;/code&gt;). Would you ever cooperate?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="42" LastEditorUserId="52046" LastEditDate="2019-12-15T18:12:23.473" LastActivityDate="2019-12-15T22:57:01.273" Title="In the iterated Prisoner's Dilemma, how would a change in the payoff matrix affect strategy?" Tags="|game-theory|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26" PostTypeId="2" ParentId="12" CreationDate="2014-05-14T21:31:18.097" Score="13" Body="&lt;p&gt;Every &lt;a href=&quot;http://en.wikipedia.org/wiki/Mathematics_of_Sudoku#Minimum_number_of_givens&quot;&gt;&quot;proper&quot; sudoku puzzle&lt;/a&gt; which only has one solution can be solved without guessing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only time you would need to guess is when the sudoku puzzle is improper, and thus has multiple possible solutions, and your guesses would be irrelevant to whether the puzzle could be solved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, just so people are aware, improper sudoku puzzles are actually not uncommon. I have a booklet of them on my shelf, so these puzzles are actually used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In these cases, the puzzles aren't more challenging necessarily (though they're typically written to be). While improper puzzles &lt;em&gt;can&lt;/em&gt; be harder to solve, they are not usually more difficult to solve &lt;em&gt;because&lt;/em&gt; they are improper, but rather because the design of the puzzle is more difficult.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastActivityDate="2014-05-14T21:31:18.097" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27" PostTypeId="1" AcceptedAnswerId="28" CreationDate="2014-05-14T21:31:53.123" Score="47" ViewCount="4708" Body="&lt;p&gt;You are in a room with three switches, labeled $A$, $B$, and $C$.  Initially, each switch is in an off state.  In an adjoining room, there are three incandescent light bulbs.  Your task is to correctly label which switch goes to which bulb.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The catch: you may only enter the bulb room once.  You have as long as you need in the switch room, and you may flip any switch as many times as you'd like, but once you leave the switch room, you may not enter it again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a way to know for certain which bulb goes to which switch.  How?&lt;/p&gt;&#xA;" OwnerUserId="11" LastEditorDisplayName="user88" LastEditDate="2016-03-04T02:26:55.277" LastActivityDate="2018-09-08T23:04:24.507" Title="Which switch goes to which bulb?" Tags="|lateral-thinking|strategy|" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="28" PostTypeId="2" ParentId="27" CreationDate="2014-05-14T21:31:53.123" Score="36" Body="&lt;p&gt;The answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Turn switches A and B on.  Wait about a minute.  Turn switch B off.  Go to the bulb room.  The bulb that is on corresponds to switch A.  The bulb that is off, but warm, corresponds to switch B.  The bulb that is off, but cold, corresponds to switch C.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Incandescent bulbs produce a significant amount of heat.  In the amount of time it takes you to walk from one room to the other, the bulb will have not had time to cool down, unless the rooms are unreasonably large.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11" LastEditorUserId="11" LastEditDate="2014-12-04T17:38:22.637" LastActivityDate="2014-12-04T17:38:22.637" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="29" PostTypeId="1" CreationDate="2014-05-14T21:34:12.920" Score="35" ViewCount="35884" Body="&lt;p&gt;What are the criteria for determining the difficulty of Sudoku puzzles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it a subjective estimation based on the author's experience, or are there any precise guidelines?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, there are some sudokus, where in each step there's always at least one digit that, if you look on some line, must be exactly in that position. Some, however, require to determine that field A can hold 2 values, but assuming it contains X, there's a contradiction in next step, so it must be Y... in other words, require either memorizing or making notices in the corner of the fields.&lt;/p&gt;&#xA;" OwnerDisplayName="user66" LastEditorUserId="5373" LastEditDate="2016-12-19T23:14:55.223" LastActivityDate="2018-01-04T00:34:05.270" Title="What are the criteria for determining the difficulty of Sudoku puzzle?" Tags="|sudoku|difficulty-rating|" AnswerCount="6" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30" PostTypeId="1" AcceptedAnswerId="47" CreationDate="2014-05-14T21:39:38.370" Score="42" ViewCount="24831" Body="&lt;p&gt;I have been told that it is possible to guarantee no worse than a tie in normal 3x3 tic-tac-toe, but that the first move is very important.  What is that important first move?&lt;/p&gt;&#xA;" OwnerUserId="11" LastEditorUserId="8874" LastEditDate="2015-02-01T17:53:26.050" LastActivityDate="2021-10-25T11:36:10.260" Title="What is the optimal first move in tic-tac-toe?" Tags="|tic-tac-toe|game|" AnswerCount="6" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31" PostTypeId="2" ParentId="30" CreationDate="2014-05-14T21:39:38.370" Score="8" Body="&lt;p&gt;If you're first: Go in a corner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're second, and the first person went in a corner: Go in the middle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From both of these positions, it is possible to guarantee no worse than a tie.&lt;/p&gt;&#xA;" OwnerUserId="11" LastActivityDate="2014-05-14T21:39:38.370" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32" PostTypeId="2" ParentId="29" CreationDate="2014-05-14T21:39:43.630" Score="9" Body="&lt;p&gt;In my experience, it's generally &lt;strong&gt;What is the hardest technique required to solve it.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Example hardest: &lt;strong&gt;Single Position&lt;/strong&gt; - Easy puzzle&lt;/li&gt;&#xA;&lt;li&gt;Ex. &lt;strong&gt;Candidate Line&lt;/strong&gt; - Medium Puzzle&lt;/li&gt;&#xA;&lt;li&gt;Ex. &lt;strong&gt;Naked Pair&lt;/strong&gt; - Hard Puzzle&lt;/li&gt;&#xA;&lt;li&gt;Ex. &lt;strong&gt;Swordfish&lt;/strong&gt; - Very Hard Puzzle&lt;/li&gt;&#xA;&lt;li&gt;Ex. &lt;strong&gt;Forcing Chains&lt;/strong&gt; - Master Puzzle&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Definitions can be found here: &lt;a href=&quot;http://www.sudokuoftheday.com/pages/techniques-overview.php&quot;&gt;http://www.sudokuoftheday.com/pages/techniques-overview.php&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="42" LastEditorUserId="42" LastEditDate="2014-05-14T21:46:38.957" LastActivityDate="2014-05-14T21:46:38.957" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33" PostTypeId="1" AcceptedAnswerId="37" CreationDate="2014-05-14T21:42:00.803" Score="16" ViewCount="6323" Body="&lt;p&gt;If an algo &lt;code&gt;X&lt;/code&gt; outputs the minimum &quot;number of moves&quot; (not the actual moves, just their count) to solve an input configuration, what's the maximum number &lt;code&gt;X&lt;/code&gt; can output for any configuration?&lt;/p&gt;&#xA;" OwnerUserId="62" LastActivityDate="2014-11-15T06:40:48.130" Title="What is the maximum &quot;minimum number of steps&quot; needed to solve a rubiks cube?" Tags="|rubiks-cube|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34" PostTypeId="2" ParentId="29" CreationDate="2014-05-14T21:43:40.707" Score="16" Body="&lt;p&gt;Modern sudoku puzzles are ranked according to the difficulty of the techniques required to generate a solution. It's important to note that there is no standardized system or metric, but rather, these are based on qualitative assessments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.playr.co.uk/sudoku/ratings.php&quot;&gt;playr.co.uk&lt;/a&gt; lists these requirements for its various ratings:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Level 1 puzzles can be solved simply by reducing the possibilities of a square to one, and filling it in.&lt;/li&gt;&#xA;&lt;li&gt;Level 2 puzzles use the occurrence of pairs of numbers within a block (cell, row, or column). This includes pairs of numbers, number chains, and hidden pairs and chains.&lt;/li&gt;&#xA;&lt;li&gt;Level 3 uses the &lt;a href=&quot;http://www.janko.at/Raetsel/Sudoku/Loesung/X-Wing1.jpg&quot;&gt;X-wing&lt;/a&gt; and &lt;a href=&quot;http://www.brainbashers.com/gifs/sudokuxywing2.gif&quot;&gt;Y-wing&lt;/a&gt; techniques, as well as &lt;/li&gt;&#xA;&lt;li&gt;Level 4 typically requires Nishio or Forcing Chains (which effectively involve making assumptions and tracing their results)&lt;/li&gt;&#xA;&lt;li&gt;Level 5 requires flat-out trial and error or lookup tables.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="user20" LastActivityDate="2014-05-14T21:43:40.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35" PostTypeId="2" ParentId="30" CreationDate="2014-05-14T21:44:05.120" Score="35" Body="&lt;p&gt;The first move can be made anywhere without sacrificing the game. If the opponent plays perfectly, any first move leads to a draw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the opponent does not play perfectly, then the optimal place to go is the corner, since that leaves only one spot (the center) for the opponent to go to get a draw, increasing their chance of making a mistake.&lt;/p&gt;&#xA;" OwnerUserId="18" LastActivityDate="2014-05-14T21:44:05.120" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="36" PostTypeId="2" ParentId="33" CreationDate="2014-05-14T21:46:00.467" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Optimal_solutions_for_Rubik%27s_Cube&quot; rel=&quot;nofollow&quot;&gt;It's well known that the diameter of the Cayley graph for Rubik's cube positions is 20.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that's the question you were asking. If not, please edit your question to clarify.&lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2014-05-14T21:46:00.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="37" PostTypeId="2" ParentId="33" CreationDate="2014-05-14T21:48:22.060" Score="14" Body="&lt;p&gt;The maximum number of positions a valid Rubik's cube configuration takes to solve is 20, and is called God's Number. No rigorous mathematical proof has been provided for the number; however, every configuration was tried. &lt;a href=&quot;http://www.cube20.org/&quot;&gt;Cube20&lt;/a&gt; gives an explanation of what they did, as well as provides source code for their project. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They achieved this through a few groups, discovered from other proofs of maximum move counts, which reduced the set of positions that actually needed to be tried significantly. They only needed to solve 55,882,296 positions as a result (they took quite a bit of advantage of symmetry). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, without brute-force, this has only been mathematically &lt;a href=&quot;http://link.springer.com/article/10.1007%2Fs00283-009-9105-3&quot;&gt;proven for 22 moves&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastActivityDate="2014-05-14T21:48:22.060" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="38" PostTypeId="1" AcceptedAnswerId="42" CreationDate="2014-05-14T21:51:27.490" Score="15" ViewCount="5838" Body="&lt;p&gt;The basic algorithm for procedurally generating mazes goes something like this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Choose a random available direction from the current square.&lt;/li&gt;&#xA;&lt;li&gt;If there is one, go there and repeat from step 1.&lt;/li&gt;&#xA;&lt;li&gt;Otherwise, backtrack to the previous square and repeat from step 1.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This is a fairly good algorithm, but the problem is that it typically generates easy mazes. They usually only have one option or path, and alternate paths don't appear very often and reach dead-ends quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could I alter the algorithm (or change it entirely) to create more branching?&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2014-08-11T04:16:35.613" Title="How can I generate mazes that often have multiple &quot;forks&quot; or choices?" Tags="|puzzle-creation|mazes|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="39" PostTypeId="1" CreationDate="2014-05-14T21:55:35.107" Score="21" ViewCount="1196" Body="&lt;p&gt;I, being a fan of 2048, was wondering if anyone has any proven strategies for playing &lt;a href=&quot;http://gabrielecirulli.github.io/2048&quot;&gt;2048&lt;/a&gt;? &lt;/p&gt;&#xA;" OwnerUserId="63" LastEditorUserId="8874" LastEditDate="2015-02-01T17:53:37.500" LastActivityDate="2015-03-26T18:36:56.373" Title="General strategy for 2048" Tags="|2048-game|game|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="40" PostTypeId="2" ParentId="39" CreationDate="2014-05-14T21:57:38.547" Score="3" Body="&lt;p&gt;The number one most important tip in my opinion is to keep your highest tile in one corner, and your next highest tiles adjacent to it. You should only move your top tile from the corner when absolutely necessary.&lt;/p&gt;&#xA;" OwnerUserId="18" LastActivityDate="2014-05-14T21:57:38.547" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="41" PostTypeId="2" ParentId="39" CreationDate="2014-05-14T21:57:45.937" Score="12" Body="&lt;p&gt;Try to line up numbers in descending order on a row of your choice (I prefer the top row), and avoid messing this up at all costs (especially the cell with the biggest number, which should be in a corner).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{|r|r|}&#xA;\hline&#xA;1024 &amp;amp; \phantom{0}512 &amp;amp; \phantom{00}16 &amp;amp; \phantom{000}2 \\&#xA;\hline&#xA;2 &amp;amp; 2 &amp;amp; 4 &amp;amp; 2 \\&#xA;\end{array}&#xA;\\&#xA;\text{(etc.)}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good method for achieving this is to avoid a certain key (i.e., the down key, if your large numbers are in the top row) so that you won't accidentally move a large tile away from the row. Also remember to not use the right key (in this example) unless the top row is full (which it usually is, but be careful after combining tiles on the top row).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reasons for this are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If you let the tile with your greatest number get out of a corner, it will &quot;clog&quot; the board and effectively take away one of your 16 available spaces.&lt;/li&gt;&#xA;&lt;li&gt;If you let tiles with large numbers get separated, such as $\begin{array}{|r|r|} \hline 1024 &amp;amp; \phantom{0}512 &amp;amp; \phantom{000}2 &amp;amp; \phantom{0}512 \end{array}$, they will stick around for much longer, get in the way, be hard to combine, and be generally pointless.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="39" LastEditorDisplayName="user20" LastEditDate="2014-06-22T02:44:11.113" LastActivityDate="2014-06-22T02:44:11.113" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="42" PostTypeId="2" ParentId="38" CreationDate="2014-05-14T21:58:16.690" Score="7" Body="&lt;p&gt;The algorithm you've mentioned is called depth-first generation, and actually, it's quite easy to intuitively demonstrate a way to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a maze generated via a depth-first algorithm, there is one and only one way to each square. DFS generates a &quot;perfect&quot; maze - one without any closed loops, without inaccessible areas, and is simply connected. There is one and only one path from any point to any other point. Thus, the maze has exactly one solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take adjacent - but walled - points A and B. There exists a path from the entrance to point A, and there exists a path from point B to the exit. However, because there is a wall between points A and B, there also exists a path which does not go through both A and B that leads to the exit. By removing the wall between A and B, you create a new (second) path through both A and B that leads to the exit, thus creating an alternate path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, if you remove &lt;em&gt;any&lt;/em&gt; wall in the maze, you create another path, or multiple valid paths. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this makes your maze imperfect by definition, as there are now two paths from the start to the exit.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorDisplayName="user20" LastEditDate="2014-05-14T22:05:01.687" LastActivityDate="2014-05-14T22:05:01.687" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="43" PostTypeId="1" AcceptedAnswerId="64" CreationDate="2014-05-14T22:01:12.957" Score="53" ViewCount="6894" Body="&lt;p&gt;You are in a room with two adjacent doors locked by four sliding bolts. The bolts are movable and block only one door at a time; however, you do not know which bolt currently is blocking which door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example configuration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Di9bU.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Di9bU.png&quot; alt=&quot;two doors blocked by four sliding bolts (example configuration)&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Luckily, there are three buttons &lt;kbd&gt;A&lt;/kbd&gt;, &lt;kbd&gt;B&lt;/kbd&gt;, &lt;kbd&gt;C&lt;/kbd&gt;, which cause some bolts to slide from the door they are curently blocking over to the other door. However, each button can trigger different actions, which one is used is determined at each push in a fully random way:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Button &lt;kbd&gt;A&lt;/kbd&gt; slides &lt;em&gt;at random&lt;/em&gt; either&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;bolt 1 &lt;em&gt;or&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;bolt 2 &lt;em&gt;or&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;bolt 3 &lt;em&gt;or&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;bolt 4.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Button &lt;kbd&gt;B&lt;/kbd&gt; slides at random either&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;bolt 1 and 2 &lt;em&gt;or&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;bolt 2 and 3 &lt;em&gt;or&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;bolt 3 and 4 &lt;em&gt;or&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;bolt 4 and 1.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Button &lt;kbd&gt;C&lt;/kbd&gt; slides at random either&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;bolt 1 and 3 &lt;em&gt;or&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;bolt 2 and 4.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Your task is to find a sequence of button activations to get out of the room (i.e., to move all the bolts to one door so the other one is open) regardless of the unknown initial position of the bolts. (You may try the doors after each touch of a button to see whether they are open.) The sequence should be as short as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Source:&lt;/em&gt; &lt;a href=&quot;https://groups.google.com/d/msg/de.rec.denksport/JlxxqX7-4Hs/WGeV4Boiw&quot; rel=&quot;noreferrer&quot;&gt;Newsgroup de.rec.denksport, &lt;em&gt;Zwei Tueren, vier Riegel&lt;/em&gt;, GJ Woeginger, 2011-10-15 (German)&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12" LastEditorUserId="19765" LastEditDate="2017-06-27T18:08:00.207" LastActivityDate="2020-12-28T22:05:38.230" Title="Sliding Bolt Puzzle" Tags="|strategy|logical-deduction|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="45" PostTypeId="2" ParentId="18" CreationDate="2014-05-14T22:01:46.367" Score="10" Body="&lt;p&gt;In this case, you can look at the first few moves.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can't take the fox or the beans, and if you go solo you will come back solo and be back at the start.  For the second move, if you take the goose back you are back where you started, so you come back solo.  You are a long way into the puzzle by then.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="73" LastEditorUserId="46903" LastEditDate="2018-09-18T23:59:27.177" LastActivityDate="2018-09-18T23:59:27.177" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="46" PostTypeId="2" ParentId="39" CreationDate="2014-05-14T22:07:47.210" Score="2" Body="&lt;p&gt;Up until 4096, @Doorknob's solution seems optimum, using a zig zag along one row then back along the next. Once you have a 4096 in one corner, the number of free tiles limits your options if you go for a single direction, though, so the technique I use once I have my bottom row filled (I like my top number at bottom left) is to add to both ends of the next row up as necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This often lets me sidestep the bottom right tile and double the next one in, and protects against an unlucky sequence of poor number tiles.&lt;/p&gt;&#xA;" OwnerUserId="53" LastActivityDate="2014-05-14T22:07:47.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="47" PostTypeId="2" ParentId="30" CreationDate="2014-05-14T22:15:32.833" Score="56" Body="&lt;p&gt;Tic-tac-toe has been solved. The optimal first move is to go in the corner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As always, there is a relevant &lt;a href=&quot;https://xkcd.com/832/&quot; rel=&quot;noreferrer&quot;&gt;xkcd&lt;/a&gt;.&#xA;&lt;img src=&quot;https://imgs.xkcd.com/comics/tic_tac_toe.png&quot; alt=&quot;xkcd tic tac toe&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditDate="2017-03-09T18:58:38.123" LastActivityDate="2017-03-09T18:58:38.123" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="48" PostTypeId="1" AcceptedAnswerId="49" CreationDate="2014-05-14T22:16:34.527" Score="57" ViewCount="76596" Body="&lt;p&gt;I can get to 4096 max. There are videos on youtube of guys making it to 8192. What is the theoretical maximum limit on this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: if you haven't played this, here is the &lt;a href=&quot;http://gabrielecirulli.github.io/2048&quot;&gt;link to addiction.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="62" LastEditorUserId="8874" LastEditDate="2015-02-01T17:53:47.070" LastActivityDate="2016-05-10T05:57:45.367" Title="What is the largest tile possible in 2048?" Tags="|game|2048-game|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="49" PostTypeId="2" ParentId="48" CreationDate="2014-05-14T22:16:34.527" Score="57" Body="&lt;p&gt;In the original game, sometimes the new tile that pops up is a 4. Let's ignore that possibility for the sake of simplicity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We've 16 squares in total and considering the new tile popped up at the position we chose, to make it to 4096, we need minimum these tiles: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;2048, 1024, 512, 256, 128, 64, 32, 16, 8, 4, 2, 2&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These can be combined from right to left to get to 4096. That's a total of 12 tiles. Since that's less than 16 (max number of tiles that can fit),  4096 is possible. Following along:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;8192: &lt;code&gt;4096, 2048, 1024, ..., 4, 2, 2&lt;/code&gt; Total 13&lt;/li&gt;&#xA;&lt;li&gt;16384: &lt;code&gt;8192, 4096, 2048, ..., 4, 2, 2&lt;/code&gt; Total 14&lt;/li&gt;&#xA;&lt;li&gt;32768: &lt;code&gt;16384, 8192, 4096, ..., 4, 2, 2&lt;/code&gt; Total 15&lt;/li&gt;&#xA;&lt;li&gt;65536: &lt;code&gt;32768, 16384, 8192, 4096, ..., 4, 2, 2&lt;/code&gt; Total 16&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Next is 131072 which will require at least 17 tiles which can't fit on the board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So, the largest possible tile is 65536.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Since the game pops up some 4's at times, &lt;strong&gt;131072 is still theoretically possible&lt;/strong&gt; if the last tile that pops up is a 4 making the board look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;131072: &lt;code&gt;65536, 32768, 16384, 8192, 4096, 2048, 1024, 512, 256, 128, 64, 32, 16, 8, 4, 4&lt;/code&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;that's 16 tiles in total.&lt;/p&gt;&#xA;" OwnerUserId="62" LastEditorUserId="62" LastEditDate="2016-05-10T05:57:45.367" LastActivityDate="2016-05-10T05:57:45.367" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="50" PostTypeId="2" ParentId="13" CreationDate="2014-05-14T22:20:26.403" Score="15" Body="&lt;p&gt;There is, indeed, a way! No way has yet been found (at least that I've found) to generate them infallibly, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.47.2303&amp;amp;rep=rep1&amp;amp;type=pdf&quot;&gt;a paper&lt;/a&gt; by Y. Murase, H. Matsubara, and Y. Higara, called &quot;Automatic Making of Sokoban Puzzles&quot;, they explain exactly how one makes unique puzzles. While their algorithm isn't perfect (about half of the puzzles they generated aren't solvable), it offers a significant statistical advantage over other possible methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In essence, they rely on three basic templates which are applied over a basic template which statistically are likely to generate solvable situations. Then, they mark specific spaces as invalid for both goals and boxes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Goals are generated first, as boxes' positions are restricted by the goals. Invalid positions for goals can be categorized by checking each direction; if there is no valid way to push a box into the goal from any direction, that position cannot be a goal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, crates are marked. The majority of invalid crate positions are those in corners (positions bounded by two walls) and those against walls with no goal access. These invalid positions are re-evaluated after placing each crate to insure that, for instance, two crates aren't placed adjacently against a wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then they literally run a solving algorithm on it and see if it works. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see a demonstration of this algorithm, &lt;a href=&quot;http://www.linusakesson.net/games/autosokoban/index.php&quot;&gt;called Autosokoban&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastActivityDate="2014-05-14T22:20:26.403" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="51" PostTypeId="1" AcceptedAnswerId="31429" CreationDate="2014-05-14T22:31:47.530" Score="20" ViewCount="5782" Body="&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Verbal_arithmetic&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, Alphametics, also known as verbal arithmetic, can be defined as &quot;a type of mathematical game consisting of a mathematical equation among unknown numbers, whose digits are represented by letters. The goal is to identify the value of each letter.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lqGeq.png&quot; alt=&quot;SEND + MORE = MONEY&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;has the solution (mouse over to see):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;O = 0, M = 1, Y = 2, E = 5, N = 6, D = 7, R = 8, S = 9.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Obviously, these can be solved by a &lt;a href=&quot;http://en.wikipedia.org/wiki/Brute-force_search&quot; rel=&quot;noreferrer&quot;&gt;brute force algorithm&lt;/a&gt; that involves attempting every possible variation of number/letter pairs.  However, I don't want to use brute-force - that's an easy programming hack.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I want to know is: &lt;strong&gt;What are some good general strategies for intelligently solving these verbal arithmetic problems by hand?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="11" LastEditorUserId="7586" LastEditDate="2016-04-28T23:35:05.443" LastActivityDate="2019-09-07T21:27:40.253" Title="Alphametic (Verbal Arithmetic) general strategy" Tags="|alphametic|solvability|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="52" PostTypeId="2" ParentId="51" CreationDate="2014-05-14T22:31:47.530" Score="13" Body="&lt;p&gt;Some simple strategies I've found are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Look for zeroes.  Any time you have $X + Y = X$ in the ones column, you know that $Y = 0$.  &lt;/li&gt;&#xA;&lt;li&gt;Look for doubles.  Any time you have numbers of the format $A + A = B$ in the ones column, you know that $B$ is even, and you know that either $0 &amp;lt; A &amp;lt; 5$ OR that there is a carrying one to the next column.&lt;/li&gt;&#xA;&lt;li&gt;Look for carrying ones.  Any time the answer has more columns in the answer than in the problem, you know that the value in the &quot;extra&quot; column is a $1$.  (In the example problem given, you know that $M = 1$.)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Tips 1 and 2 &lt;em&gt;can&lt;/em&gt; work for columns other than the ones column, but make sure you consider the possibility of carrying.  Also, these strategies must be modified if there are more than two &quot;words&quot; being added together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit to add: My answer only covers a very small subset of cases, and will not solve all scenarios.  &lt;a href=&quot;https://puzzling.stackexchange.com/a/71/11&quot;&gt;Emrakul's answer&lt;/a&gt; describes how to translate the alphametic problem into smaller algebraic equations, and gives a sample of how to start solving the problem.  &lt;a href=&quot;https://puzzling.stackexchange.com/a/31429/11&quot;&gt;Reti43's answer&lt;/a&gt; goes into more detail about how to find certain patterns in those algebraic equations that can help you solve it faster.  While both answers are a bit technical, they are both very good reads, and far more helpful than my answer.&lt;/p&gt;&#xA;" OwnerUserId="11" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2016-04-26T17:38:54.880" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="53" PostTypeId="1" CreationDate="2014-05-14T23:01:41.327" Score="8" ViewCount="725" Body="&lt;p&gt;Once one has mastered &lt;a href=&quot;https://puzzling.stackexchange.com/questions/39/general-strategy-for-2048&quot;&gt;winning 2048 in general&lt;/a&gt;, it's often an interesting exercise to try to do it as fast as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What strategies are there for speeding up 2048-solving?&lt;br/&gt;&#xA;For example, how can you pass the boring beginning game quickly (ex. mashing up and left), &lt;br/&gt;&#xA;and how do you know when to stop so you don't mess up your middlegame?&lt;/p&gt;&#xA;" OwnerUserId="39" LastEditorUserId="51627" LastEditDate="2018-08-29T13:20:55.930" LastActivityDate="2018-08-29T13:20:55.930" Title="Strategy for speedsolving 2048" Tags="|2048-game|speedsolving|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="56" PostTypeId="1" AcceptedAnswerId="57" CreationDate="2014-05-14T23:20:49.930" Score="14" ViewCount="9778" Body="&lt;p&gt;In a crossword, some clues have a question mark at the end. Is there any way to tell what these question marks mean based on the puzzle? I.E. rhetorical, compound word, etc.?&lt;/p&gt;&#xA;" OwnerUserId="83" LastEditorDisplayName="user88" LastEditDate="2015-04-25T02:17:44.953" LastActivityDate="2015-04-25T02:17:44.953" Title="What do question marks mean in crossword clues?" Tags="|crosswords|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="57" PostTypeId="2" ParentId="56" CreationDate="2014-05-14T23:28:28.367" Score="11" Body="&lt;p&gt;There's only one canonical example I can find of a crossword using question marks to mean anything special.&lt;/p&gt;&#xA;&lt;p&gt;Many American crossword puzzles (for instance, the New York Times) rely on a convention that, if a clue ends in a question mark, it's a play on words.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Mental Block?&lt;/p&gt;&#xA;&lt;p&gt;RUBIKSCUBE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;It can also be used to indicate something only loosely related to the topic:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Cause for a head slap?&lt;/p&gt;&#xA;&lt;p&gt;BONER&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;In essence, the clue shouldn't be taken too literally. The British use an exclamation mark for a similar purpose.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-05-14T23:28:28.367" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="59" PostTypeId="1" CreationDate="2014-05-15T00:03:24.913" Score="13" ViewCount="1506" Body="&lt;p&gt;When playing &lt;a href=&quot;http://en.wikipedia.org/wiki/Slitherlink&quot; rel=&quot;noreferrer&quot;&gt;SlitherLink&lt;/a&gt; on &lt;a href=&quot;http://www.puzzle-loop.com/?size=6&quot; rel=&quot;noreferrer&quot;&gt;moderate&lt;/a&gt; or &lt;a href=&quot;http://www.puzzle-loop.com/?size=12&quot; rel=&quot;noreferrer&quot;&gt;large&lt;/a&gt; boards I usually find that after using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Slitherlink#Exactly_2_or_0_lines_at_each_point&quot; rel=&quot;noreferrer&quot;&gt;basic techniques&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;filling out the values in the corners&lt;/li&gt;&#xA;&lt;li&gt;setting lines where there are adjacent 3s&lt;/li&gt;&#xA;&lt;li&gt;setting lines where there are 3s diagonally (including cases where they are only separated by 2s)&lt;/li&gt;&#xA;&lt;li&gt;clearing everything around 0s and filling in the lines (if there is only a unique way).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Between them I still only solved like 25% of the board. The rest of the board is usually either empty, or only contain 1s and 2s, from which I cannot really deduce anything. In these cases I usually resort to guessing a line, and trying to trace whether it results in a contradiction eventually, so I can cross it off, but this doesn't seem like an optimal solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Are there additional strategies that I can use to continue solving the puzzle?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="85" LastEditorUserId="5373" LastEditDate="2016-12-19T23:15:46.310" LastActivityDate="2016-12-20T08:47:21.953" Title="SlitherLink on moderate or large boards: going beyond 0s and 3s" Tags="|slitherlink|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="60" PostTypeId="2" ParentId="22" CreationDate="2014-05-15T00:17:35.343" Score="24" Body="&lt;p&gt;This type of chess puzzle is known as a &lt;a href=&quot;http://en.wikipedia.org/wiki/Mathematical_chess_problem#Domination_problems&quot; rel=&quot;noreferrer&quot;&gt;domination problem&lt;/a&gt;, and as @Xynariz points out, only five queens are needed for the 8x8 board. It's also interesting to note that five queens are also sufficient for the 9x9, 10x10, and 11x11 boards, as shown by the following diagram taken from a Russian chess-puzzle book found &lt;a href=&quot;http://golovolomka.hobby.ru/books/gik/04.shtml&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9sQqA.png&quot; alt=&quot;5 queens suffice&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="87" LastActivityDate="2014-05-15T00:17:35.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="61" PostTypeId="2" ParentId="59" CreationDate="2014-05-15T00:20:18.750" Score="1" Body="&lt;p&gt;There are many more configurations:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If you see a 1 and 3 next to each other on the edge, fill the wall of the 3 on the edge of the puzzle and blank the two sides of the 1 not connected to that.  &lt;/li&gt;&#xA;&lt;li&gt;Just like a 2 on the corner forces the two edges one step away from it to be filled, you can often find 2's in the middle that do the same.  &lt;/li&gt;&#xA;&lt;li&gt;It is very useful to identify what squares are inside the loop-they get divided from the outside ones by the path and often that can show you larger features of the path.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="73" LastEditorUserId="30903" LastEditDate="2016-12-20T08:47:21.953" LastActivityDate="2016-12-20T08:47:21.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="62" PostTypeId="1" AcceptedAnswerId="111" CreationDate="2014-05-15T00:22:06.507" Score="22" ViewCount="1705" Body="&lt;p&gt;The game 2048 can be thought as a two player game, where one player shifts the table around to get to the 2048 tile, while the other player tries to hinder them by putting down 2s and 4s on the table. While the &lt;a href=&quot;http://gabrielecirulli.github.io/2048/&quot; rel=&quot;nofollow noreferrer&quot;&gt;original game&lt;/a&gt; will put down the tiles randomly &lt;a href=&quot;http://sztupy.github.io/2048-Hard/&quot; rel=&quot;nofollow noreferrer&quot;&gt;there&lt;/a&gt; &lt;a href=&quot;http://aj-r.github.io/Evil-2048/&quot; rel=&quot;nofollow noreferrer&quot;&gt;are&lt;/a&gt; variants, such as some where this is not random, and also &lt;a href=&quot;http://lymia.github.io/Anti-2048/&quot; rel=&quot;nofollow noreferrer&quot;&gt;some&lt;/a&gt; where you are the one putting down the tiles, and the computer will shift the table around (not to mention the &lt;a href=&quot;http://akhilanb.github.io/Human2048/&quot; rel=&quot;nofollow noreferrer&quot;&gt;general&lt;/a&gt; one, where you can play both sides).&lt;/p&gt;&#xA;&lt;p&gt;Theoretically, the worst situation is the following table, but for this the player shifting the table has to play really badly.&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;If however both players play flawlessly against each other, then what is the highest tile that the shifting player can achieve?&lt;/p&gt;&#xA;" OwnerUserId="85" LastEditorUserId="69582" LastEditDate="2021-02-17T22:31:47.720" LastActivityDate="2021-02-17T22:31:47.720" Title="Worst case scenario in 2048" Tags="|optimization|game|2048-game|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="63" PostTypeId="2" ParentId="53" CreationDate="2014-05-15T00:24:58.270" Score="0" Body="&lt;p&gt;You can't speed it up by much. To get to the 2048 tile, you need 1024 2 tiles, 512 4 tiles or a combination, and that many moves is your basic minimum. There is no way to cross those limits. Because 4 shows up 1/4 times, it will take about 850 moves minimum to finish, regardless of skill or strategy.&lt;/p&gt;&#xA;" OwnerUserId="92" LastEditorUserId="29567" LastEditDate="2016-12-19T20:55:05.317" LastActivityDate="2016-12-19T20:55:05.317" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="64" PostTypeId="2" ParentId="43" CreationDate="2014-05-15T00:26:09.070" Score="38" Body="&lt;p&gt;&lt;kbd&gt;C&lt;/kbd&gt;&#xA;&lt;kbd&gt;B&lt;/kbd&gt;&#xA;&lt;kbd&gt;C&lt;/kbd&gt;&#xA;&lt;kbd&gt;A&lt;/kbd&gt;&#xA;&lt;kbd&gt;C&lt;/kbd&gt;&#xA;&lt;kbd&gt;B&lt;/kbd&gt;&#xA;&lt;kbd&gt;C&lt;/kbd&gt;, trying the doors at each step.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;There are 4 possible states of the bolts:&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;1&lt;/strong&gt;. All on one door. It is possible to get out.&lt;br /&gt;&#xA;&lt;strong&gt;2&lt;/strong&gt;. Exactly one bolt is on one side.&lt;br /&gt;&#xA;&lt;strong&gt;3&lt;/strong&gt;. Bolts {1, 2}, {2 ,3}, {3, 4} or {4, 1} are on one side.&lt;br /&gt;&#xA;&lt;strong&gt;4&lt;/strong&gt;. Bolts {1, 3} or {2, 4} are on one side.&lt;/p&gt;&#xA;&lt;p&gt;Initially we can be in states &lt;strong&gt;2&lt;/strong&gt;, &lt;strong&gt;3&lt;/strong&gt;, or &lt;strong&gt;4&lt;/strong&gt;. These are the possible states after each step. (We first try both doors to make sure we aren't in state &lt;strong&gt;1&lt;/strong&gt;.)&lt;/p&gt;&#xA;&lt;p&gt;&lt;kbd&gt;C&lt;/kbd&gt;: &lt;strong&gt;2&lt;/strong&gt; or &lt;strong&gt;3&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;kbd&gt;B&lt;/kbd&gt;: &lt;strong&gt;2&lt;/strong&gt; or &lt;strong&gt;4&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;kbd&gt;C&lt;/kbd&gt;: &lt;strong&gt;2&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;kbd&gt;A&lt;/kbd&gt;: &lt;strong&gt;3&lt;/strong&gt; or &lt;strong&gt;4&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;kbd&gt;C&lt;/kbd&gt;: &lt;strong&gt;3&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;kbd&gt;B&lt;/kbd&gt;: &lt;strong&gt;4&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;kbd&gt;C&lt;/kbd&gt;: &lt;strong&gt;1&lt;/strong&gt; (We will definitely have escaped by now)&lt;/p&gt;&#xA;&lt;p&gt;Here's the state transition matrix I used to calculate the answer:&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;A&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;B&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;C&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1/3/4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1/4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;" OwnerUserId="18" LastEditorUserId="69582" LastEditDate="2020-12-28T22:00:09.270" LastActivityDate="2020-12-28T22:00:09.270" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="65" PostTypeId="1" AcceptedAnswerId="1661" CreationDate="2014-05-15T00:35:11.300" Score="12" ViewCount="1269" Body="&lt;p&gt;Growing up on the cryptic crosswords I found in &lt;em&gt;Games&lt;/em&gt; magazine and &lt;em&gt;The New York Times&lt;/em&gt;, I'm used to a very rigid rule for clues: Every clue comprises two parts, one of which directly (if opaquely) clues the word as in a non-cryptic crossword and the other of which does so cryptically. (These may be separated by &quot;is&quot; or &quot;to yield&quot; or some similar copula. And so-called &amp;amp;lit clues are an exception.) The cryptic half, in turn, has some fairly rigid rules: If it uses an anagram, it must indicate as much (thus, e.g., &quot;turned handle&quot; for HANDEL, never just &quot;handle&quot; for HANDEL); and if it hides a word within another, it must indicate as much (thus, e.g., &quot;in the best&quot; for THEBES, never just &quot;the best&quot; for THEBES).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I've seen cryptics &amp;mdash; mostly, I think, from Britain &amp;mdash; whose clues don't follow these rules. They seem to allow anagrams and words hidden within other words even without signaling as much, and they even seem to allow cryptic-half-only clues, without a non-cryptic component of the clue. What rules, if any, do such crosswords follow for clue composition?&lt;/p&gt;&#xA;" OwnerUserId="93" LastActivityDate="2014-06-10T11:32:33.313" Title="What rules are there for clue composition in a British cryptic?" Tags="|cryptic-crosswords|crossword-clues|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="66" PostTypeId="2" ParentId="16" CreationDate="2014-05-15T00:46:38.223" Score="69" Body="&lt;p&gt;The solution I read when I heard about this problem (which is most likely the same one in your solution booklet) goes like this:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Before they begin, give each prisoner a fixed number from 1 to 100, and correlate this with their name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, when it is a prisoner's turn to open the boxes, he will start with the box of his own number. If the box does not contain his name in it, he will recall the number of the prisoner whose name &lt;em&gt;is&lt;/em&gt; in the box, and open that number's box. He then continues this cycle of recalling a number and opening that box until either he finds the box with his own name in it, or 50 boxes are opened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This strategy essentially maps one prisoner to another unique prisoner, or a number from 1 to 100 to another unique number from 1 to 100.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is known as a &lt;em&gt;permutation&lt;/em&gt;, and one of the things that can be shown about permutations is that they all consist of &lt;em&gt;cycles&lt;/em&gt;. This is because no two prisoners can have the same prisoner's name in &quot;their&quot; boxes, so eventually each prisoner &lt;em&gt;would&lt;/em&gt; come across their name by following the procedure above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, obviously, if there are no cycles of length 51 or longer, then the prisoners will go free. And if there is a cycle of length 51 or longer, then the prisoners will all die.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, we just need to find the probability that there &lt;em&gt;is&lt;/em&gt; a cycle of 51 or longer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do this, note first that there can only be one cycle of length 51 or more in a permutation of 100 elements - if there were more than one, there'd have to be at least 102 elements, which is impossible!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for a cycle of length exactly $n$, we select the elements that are going to be in the cycle, and the rest can be permuted whatever way they want. This results in $\binom {100} {n}$ possible sets of $n$ to be chosen, times $\frac {n!}{n}$ ways to permute the cycle in that set (because the cycle has $n$ states itself), times $(100-n)!$ ways to permute the rest of the elements. Conveniently, this equals exactly $\frac{100!}{n}$, so the probability that there is a cycle of length exactly $n$ is just $\frac 1n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, the probability that there is a cycle of length 51 or longer is just $$1/51 + 1/52 + 1/53 + ... + 1/100 \approx 0.688172$$, so the probability of the opposite is $0.311828$ which is above 30%.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="15544" LastEditDate="2015-09-11T22:53:11.257" LastActivityDate="2015-09-11T22:53:11.257" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="69" PostTypeId="2" ParentId="16" CreationDate="2014-05-15T02:16:15.947" Score="78" Body="&lt;p&gt;I've tried to explain the solution using as little math as possible, and to give some intuition as to what makes it tick. Nonetheless there will be a little mathematical notation at the end.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;First steps: going beyond the obvious solution in the simplest case (N=2)&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The statement of the puzzle as presented here doesn't make this very clear, but the puzzle relies on the prisoners not knowing anything about which name is located in which box (until they get into the room, after which they cannot communicate anymore). If every prisoner checks 50 boxes at random, then each prisoner has a ½ chance of finding his own name. If all the prisoners choose a set of boxes at random, independently, then the probability that they all find their own name is ½ × … × ½ = 1/2&lt;sup&gt;100&lt;/sup&gt; — infinitesimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Making independent choices is a waste, though. If anyone gets it wrong, the situation isn't worse than if everybody gets it wrong. Rather than make independent choices, they can make correlated choices; the idea is to try to arrange that either everybody gets it right, or many get it wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider the simpler case when there are two prisoners. If they both choose at random, then they have ½ × ½ = ¼ chance of surviving. But there's an obvious waste: suppose prisoner #1 opens the left-hand box and finds his name: then prisoner #2 will not find his name in the left-hand box. So the prisoners can decide that #1 will look at the box on the left and #2 looks at the box on the right: that way, either they both get it right or both get it wrong, and they have a ½ chance of survival.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, note that another assumption that wasn't clearly stated here is that the prisoners get to formulate their strategy in secret. If the warden knows which prisoner chose which box, he can arrange for the prisoners to fail by putting prisoner #1's name in the right-hand box.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;The next step: N=4&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The obvious way to generalize this to more prisoners is to assign each prisoner a fixed set of boxes that he'll open. However, I won't pursue this further, because it doesn't take advantage of an important ability: after a prisoner has opened a first box, he can base his decision on which box to open next on the content of the first box, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the case with 4 prisoners and 4 boxes. I'll use numbers for the prisoners' names, and assume that the boxes are numbered as well. Intuitively, it is preferable for each prisoner to pick a different box to open first, since otherwise some common choices are wasted. So prisoner #1 opens box #1 and finds a name (number). Now what? If he finds his own name (#1) (¼ chance), of course, he can stop. If he finds some different name (say 2) (¾ chance), what information does this provide? Well, since each box contains a different name, prisoner #1 now knows that box #2 does not contain 2, so prisoner #2 will not be lucky the first time either. Furthermore, the strategy should favor arranging for prisoner #2 to pick box #1 next.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To simplify the analysis, I'll only look at cases where all prisoners follow the same strategy. (I don't have an intuitive argument as to why breaking the symmetry wouldn't be advantageous.) Either they all open the box whose number they found in the first box, or they all open a different box.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If #1 opens box #2 and finds his name there, then either boxes #3 and #4 contain 3 and 4 respectively, or they contain 4 and 3. Either way, with the strategy of using the name in the first box, if one prisoner is lucky the second time then every prisoner is lucky!&lt;/li&gt;&#xA;&lt;li&gt;If #1 opens box #3 instead and finds his name there, then there is a ½ chance that prisoner #2 will find his name in box #2, and a ½ chance that he'll find his name in box #4. But what about prisoner #3? He finds the name of prisoner #1 in box #3, which doesn't give any clue as to where 3 might be instead.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So let's concentrate on the strategy where each prisoner opens the second box whose number is what he found in the first box. What arrangement of numbers in boxes make it work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 4 ways to choose which box contains the number 1. The number 2 can go into any of the 3 remaining boxes. The number 3 can go in either of the 2 remaining boxes, and the number 4 must go into the one remaining box. So there are 4×3×2 = 24 different arrangements. The following arrangements lead to success because each number is either in its own box or swapped with another number:&lt;br&gt;&#xA;1234 1243 1324 1432 2134 2143 3214 3412 4231 4321&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's 10 successful arrangements out of 24. The chance of success isn't very far from the theoretical maximum of ½, which is encouraging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in order for the chance of success to be 10/24, we need to know that the arrangements have an equal chance of being chosen. If the warden is nasty and arranges the numbers as 2341, the prisoners are sure to lose. This is where the fact that the prisoners choose a strategy in secret comes in. In my analysis, I used numbers for prisoners — but fact the prisoners are names, not numbers, and they can pick a random assignment of names to numbers as part of their secret strategy (in fact, this assignment is the only secret part, since the warden may have looked up the solution of the puzzle).&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;General analysis&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Let's explore a strategy that generalizes what we explored for 4 boxes: each prisoner opens the box with his own number, then the box whose number is contained in the first box, and so on. Consider the sequence of numbers that a certain prisoner encounters: $x_0$ (the inital box numbered with the prisoner's own number), $x_1$ (number contained in box $x_0$), $x_2$ (number contained in box $x_1$), … Since each number is contained in only one box, this sequence cannot contain any repeated element as long as it doesn't loop back to $x_0$. Eventually the sequence has to loop back to $x_0$ since it will run out of numbers. At that point, the prisoner has found his own name. The critical problem for the prisoner is whether the loop completes before or after the prisoner has opened the maximum of 50 boxes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From now on, let me use the proper mathematical vocabulary. A way to arrange distinct numbers into as many boxes is called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Permutation&quot;&gt;&lt;em&gt;permutation&lt;/em&gt;&lt;/a&gt;. Opening box number $k$ and looking at the number that it contains is called &lt;em&gt;applying&lt;/em&gt; that permutation. Repeated applications of a permutation eventually runs into a loop; such a loop is called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Cycle_(mathematics)&quot;&gt;&lt;em&gt;cycle&lt;/em&gt;&lt;/a&gt;. The prisoners succeed if all of the cycles for the permutation have a length of at most 50.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's call a cycle &lt;em&gt;long&lt;/em&gt; if it contains 51 or more elements. Observe that there cannot be more than one long cycle (if one cycle has at least 51 elements, then there are only 49 or fewer elements to share between the other cycles). So we can count the losing configurations by adding up the permutations of 100 elements that have a cycle of length 51, 52, …, 100.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma: there are $n! = 1 \cdot 2 \cdot 3 \cdot \ldots \cdot (n-1) \cdot n$ distinct permutations of $n$ elements. Proof: there are $n$ ways to pick the image of the first element, $n-1$ remaining ways to pick the image of the second elements, etc., down to a single way to pick the image of the last element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's count the number of permutations that have a cycle of length $c$ (for $c \ge 51$, so that there's a single such cycle). We're actually going to count each permutation $c$ times, once for each element of the cycle. Pick the first element in the cycle: there are 100 possibilities. There are 99 possibilities for the second element, and so on, until we've picked $c$ elements. So far, that's $100 \times 99 \times \ldots \times (100-c+1)$ possibilities. There are $100-c$ remaining elements, and they can be permuted in any way, so there are $(100-c)!$ possibilities as per the lemma above. That's a total of $(100 \times 99 \times \ldots \times (100-c+1)) \times ((100-c) \times \ldots \times 2 \times 1)$ possibilities, which nicely collapses to $100!$. Recall that we counted each permutation $c$ times, since we counted it once per element in the cycle. So the number of permutations with a cycle of length $c$ is $100! / c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of permutations with a long cycle is thus&#xA;$$ \frac{100!}{100} + \frac{100!}{99} + \ldots + \frac{100!}{51} $$&#xA;That's out of a total of $100!$ permutation. The proportion of failing permutations is thus&#xA;$$ \frac{1}{100} + \frac{1}{99} + \ldots + \frac{1}{51} $$&#xA;Numerically, this is about 0.6882, i.e. the chance of success is about 31.18%, a little over the requisite 30%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the proportion of failing permutations for $N$ prisoners is $H_N - H_{N/2}$ where&#xA;$$ H_n = 1 + \frac{1}{2} + \ldots + \frac{1}{n} $$&#xA;is called the $n$th &lt;a href=&quot;http://en.wikipedia.org/wiki/Harmonic_number&quot;&gt;harmonic number&lt;/a&gt;. For large values of $n$, $H_n \approx \ln n + C$ for some number C, and the series $H_N - H_{N/2}$ converges to $\ln 2 \approx 0.6931$ from below. (I will not provide an elementary proof of that). This gives a lower limit to the chance of success for large numbers of prisoners: 30.68% is achievable.&lt;/p&gt;&#xA;" OwnerUserId="71" LastEditorUserId="71" LastEditDate="2014-09-24T18:13:57.533" LastActivityDate="2014-09-24T18:13:57.533" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="70" PostTypeId="1" AcceptedAnswerId="557" CreationDate="2014-05-15T02:43:42.597" Score="24" ViewCount="7373" Body="&lt;p&gt;N logicians are wearing hats which can be of N different colors (each hat has one color; there can be multiple hats with the same color). Each logician can see the colors of all hats except his own. The logicians must simultaneously call out a color; they win if at least one logician calls the color of his own hat. They can agree on a strategy beforehand; the set of possible colors is known. There is a winning strategy (i.e. the logicians can arrange to win no matter what colors the hats end up being).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I definitely didn't invent this; I don't know who did.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know of a solution that requires a little knowledge of abstract algebra (you can read a &lt;a href=&quot;https://math.stackexchange.com/questions/79333/how-many-strategies-are-there-for-this-puzzle-where-one-of-n-logicians-must-call&quot;&gt;sketch of that solution on Math Stack Exchange&lt;/a&gt;). Is there a solution that doesn't require any mathematical background?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Specifically, how do you explain the solution to someone who isn't familiar with modulo arithmetic ($\mathbb{Z}/n\mathbb{Z}$)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt; (Hint, if you want to try to solve it on your own: start with N=2 — white hat and black hat. Then work on N=4 — red, pink, blue and cyan.) &lt;/sub&gt;  &lt;/p&gt;&#xA;" OwnerUserId="71" LastEditorUserId="-1" LastEditDate="2017-04-13T12:19:36.950" LastActivityDate="2018-03-01T15:55:13.783" Title="N logicians wearing hats of N colors" Tags="|logical-deduction|mathematics|hat-guessing|" AnswerCount="5" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="71" PostTypeId="2" ParentId="51" CreationDate="2014-05-15T03:05:55.813" Score="14" Body="&lt;p&gt;The optimal way to solve this is to create linear equations of the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll take the following problem:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;K Y O T O&#xA;O S A K A + &#xA;-----------&#xA;T O K Y O&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What you need to do to solve this is to set up a sequence of linear equations that represent the carryovers and equations. I'll denote $x_n$ as the single-digit carryover, and $C_n$ as the double-digit carryover. $x_n$ must be either 0 or 1, and $C_n$ must be either $0$ or $10$. (I'm going to replace $O$ with $P$ for readability.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What one ends up with is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{align}&#xA;P+A&amp;amp;=P+C_1\\&#xA;T+K&amp;amp;=Y+C_2-x_1\\&#xA;P+A&amp;amp;=K+C_3-x_2\\&#xA;Y+S&amp;amp;=P+C_4-x_3\\&#xA;K+P&amp;amp;=T-x_4\end{align}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And an important property: $C_n=10*x_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In essence, these equations represent the individual carryovers. Take the second column: $T + K = Y$, but if $Y$ has $1$ added from the carryover of $O + A$, then it needs to allow for that with $x_1$. Additionally, $10$ of the value of $T + K$ may have carried over into the next column, so one needs to add $C_2$. This is repeated for each column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next thing to do is set up a table of possible values, that looks something like:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A : 0 1 2 3 4 5 6 7 8 9&#xA;K : 0 1 2 3 4 5 6 7 8 9&#xA;P : 0 1 2 3 4 5 6 7 8 9&#xA;S : 0 1 2 3 4 5 6 7 8 9 &#xA;T : 0 1 2 3 4 5 6 7 8 9&#xA;Y : 0 1 2 3 4 5 6 7 8 9 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Additionally, make note of the values of $C_1$ through $C_4$ and $x_1$ through $x_4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a couple conditions to take note of, which will further restrict what the values of $A, K, P, S, T,$ and $Y$ (and therefore $C_n$ and $x_n$) can be: Primarily, no single value can be greater than nine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, it turns into a reduction of linear equations. &lt;strong&gt;This method is guaranteed to quickly solve any puzzle.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'll go a couple steps into the problem here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P + A = P + C_1$, so $A = C_1$. We therefore know that $A = C_1 = x_1 = 0$, since $C_1$ is either $0$ or $10$. First problem solved! We update our table and equations (the first equation is now useless):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;T + K = Y + C_2&#xA;P     = K + C_3 - x_2&#xA;Y + S = P + C_4 - x_3&#xA;K + P = T       - x_4&#xA;&#xA;A : 0&#xA;K :   1 2 3 4 5 6 7 8 9&#xA;P :   1 2 3 4 5 6 7 8 9&#xA;S :   1 2 3 4 5 6 7 8 9 &#xA;T :   1 2 3 4 5 6 7 8 9&#xA;Y :   1 2 3 4 5 6 7 8 9 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The rest is left as an exercise to the reader. It isn't an easy problem to solve, and takes some manual reduction. However, if you'd like to verify your solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $A=0, K=4, P/O=3, S=2, T=7, Y=1$; $C_2=10x_1=1$; all other $C_n = 0$&#xA; The final problem becomes $41373+32040=73413$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user20" LastEditorUserId="16717" LastEditDate="2016-04-25T18:00:41.527" LastActivityDate="2016-04-25T18:00:41.527" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="72" PostTypeId="1" CreationDate="2014-05-15T03:21:51.380" Score="5" ViewCount="287" Body="&lt;p&gt;The MPire coloring game is a game where several &quot;empires&quot; are laid out touching each other. Your task is to color each empire a different color, and not let two of the same color touch. You also have to use as few colors as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if anyone had specific strategies for this game? Is it better to start in the center or a corner? Plan a strategy around trick points in the map?&lt;/p&gt;&#xA;" OwnerUserId="63" LastEditorUserId="8874" LastEditDate="2016-03-21T10:41:08.553" LastActivityDate="2016-03-21T10:41:08.553" Title="MPire coloring game basic strategy" Tags="|combinatorics|game|graph-theory|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="73" PostTypeId="1" AcceptedAnswerId="74" CreationDate="2014-05-15T03:29:17.543" Score="8" ViewCount="754" Body="&lt;p&gt;I understand the basic method of generating sudoku puzzles, which is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Put a random number in the current square.&lt;/li&gt;&#xA;&lt;li&gt;If the puzzle is still valid, go to the next square. Otherwise, stay on the same square.&lt;/li&gt;&#xA;&lt;li&gt;Repeat from step 1.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;However, sometimes a sudoku puzzle gets &quot;stuck&quot; so that no matter what you put in the square, it still won't work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I avoid getting stuck in an infinite loop here?&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2014-05-30T17:33:27.160" Title="How to avoid getting &quot;stuck&quot; while generating sudoku puzzles?" Tags="|sudoku|puzzle-creation|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="74" PostTypeId="2" ParentId="73" CreationDate="2014-05-15T03:29:17.543" Score="5" Body="&lt;p&gt;You must keep track of all numbers that you have tried in a square. Then, on the random number generation step, if all numbers have been tried, go &lt;em&gt;back&lt;/em&gt; one square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When going back, remember to clear the &quot;numbers already tried&quot; list; otherwise you can't get past that point. This won't lose any information, because sudoku generation is like branching out on a &quot;tree&quot; - you don't care what's past that branch as long as you know it's a dead end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this same reason, you must keep a &quot;points tried&quot; list on &lt;em&gt;all&lt;/em&gt; squares, in case you have to go back two or more squares.&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2014-05-15T03:29:17.543" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="75" PostTypeId="1" AcceptedAnswerId="3185" CreationDate="2014-05-15T04:19:35.237" Score="5" ViewCount="1794" Body="&lt;p&gt;20 teams played in a round-robin tournament.&lt;br&gt;&#xA;The winner of each game got &lt;strong&gt;1&lt;/strong&gt; point, and the loser got &lt;strong&gt;0&lt;/strong&gt;; there were no draws.&lt;br&gt;&#xA;Each team's total score was a square number.&lt;br&gt;&#xA;At least one team had a unique score.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How many upsets were there&lt;/strong&gt;, defined as times a team with a lower total score beat a team with a higher total score?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Source:  &lt;em&gt;New Scientist&lt;/em&gt; &amp;nbsp;&lt;strong&gt;Enigma&lt;/strong&gt; column, long ago.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="30633" LastEditDate="2016-12-19T23:44:08.987" LastActivityDate="2016-12-19T23:44:08.987" Title="Round-Robin Tournament" Tags="|logical-deduction|calculation-puzzle|mathematics|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="76" PostTypeId="2" ParentId="20" CreationDate="2014-05-15T05:06:18.763" Score="39" Body="&lt;p&gt;Okay, I'll give a group-theoretic answer, since that was requested, which goes most of the way to a proof. I'm assuming you mean a 3x3x3 Rubik's cube. For larger cubes, the proof is just a little bit less convenient, because these cubes &lt;a href=&quot;https://math.stackexchange.com/q/144424/8473&quot;&gt;don't form groups&lt;/a&gt; (or more precisely, the action of the group of all moves on the cube on the states of the cube is not free). Basically everything still holds though for all cubes larger than 3x3x3 when talking specifically about the corner pieces, except that if the number of layers is even nothing fixes the orientation. Edge pieces and centers can get a bit more complicated though. Also, throughout, I'll use the notation $\mathbb Z_n = \mathbb Z / n \mathbb Z$ for the cyclic group of order n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a number of groups one could be referring to when using the term &quot;cube group&quot;. Here, by the &quot;full cube group&quot; $G$, I mean the set of all states of the cube that can be reached by disassembling it and reassembling it, keeping all the center pieces fixed in position. In order to define the product, we'll identify it as a subset of the permutation group on the stickers. Note that each corner/edge sticker is distinguishable from all 47 other stickers. This is because on the 3x3x3 cube, each cubie has a unique collection of sticker colors. Hence, each state $\Sigma$ corresponds uniquely to a permutation $\sigma_\Sigma \in S_{48}$, where $\sigma_\Sigma(s_i)$ is the sticker in the position that sticker $s_i$ would be in the state $\Sigma_0$ corresponding to the solved cube. The product structure on the full cube group is induced by this map. As a trivial consequence of this definition, the identity of the cube group is the solved cube position $\Sigma_0$. (If this multiplication looks a bit backwards to you, it's because I want my permutation group to act on the right, which is primarily since the standard notation of cube moves (e.g. R U R' U R U2 representing the &lt;a href=&quot;http://www.speedsolving.com/wiki/index.php/Sune&quot; rel=&quot;noreferrer&quot;&gt;Sune&lt;/a&gt;) is read left-to-right.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are $8! \; 3^8$ ways of putting in the 8 corner pieces (because each can be rotated any of 3 ways, and they can be permuted), and independently $12! \; 2^{12}$ ways of putting in the 12 edge pieces. Hence, the full cube group has order $8! \; 3^8 \; 12! \; 2^{12}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This group is too big for cubing purposes. When cubing, there are 6 basic moves you can make, specifically, U, D, L, R, F and B (as well as their inverses, which are typically denoted by primes e.g. U'). The cube group $H$ is the subgroup of the full cube group generated by these 6 permutations, i.e. which can be written as words in these 6 letters and their inverses. These represent the states of the cube which are reachable with a finite sequence of turns. On this subgroup, the product operation is especially easy to understand. Let $a_1 \cdots a_n$ be any word in the generators which takes you from $\Sigma_0$ to $\Sigma_1$, and $b_1 \cdots b_m$ any word taking you from $\Sigma_0$ to $\Sigma_2$. Their product $\Sigma_1 \Sigma_2$ is the state you reach from the sequence of moves $a_1 \cdots a_n b_1 \cdots b_m$. Now that we have this identification, I'll be a bit cavalier in equating moves with states, but it should always be clear in context what is meant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to prove that a particular state $\Sigma'$, that with a single corner turned by $120^\circ$, is not reachable, which is to say that it is not in the cube group (though it is in the full cube group). To do this, we'll define a homomorphism $\varphi: G \rightarrow \mathbb Z_3$. We'll then show that $\varphi(a_i) = 0$ for any one of the 6 generators $a_i$. Since any state $\Sigma$ in the cube group can be written as a word in the generators, that means that $\varphi(\Sigma) = 0$. That is to say that $\varphi$ is an invariant under the cube group. However, we can compute $\varphi(\Sigma') \ne 0$, which shows that $\Sigma' \not \in H$. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Aside&lt;/em&gt;: I'll just point out that this sort of approach of defining invariants that are preserved under a set of transformations is very common in mathematics; for instance, point-set topology could be characterized broadly as the study of invariants of topological spaces under homeomorphisms. Here, we're dealing with an algebraic invariant. To be a bit fancier, we're trying to define a surjective homomorphism $\varphi$ so that the composition $H \hookrightarrow G \overset{\varphi}{\twoheadrightarrow} \mathbb Z_3$ is the trivial homomorphism. Keep in mind though that this isn't an exact sequence-I haven't required that the kernel of $\varphi$ is exactly $H$. It isn't a chain complex either, at least not in standard parlance, since $H$ and $G$ are nonabelian groups.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To define the invariant, label each of the corner cubie stickers in the solved state with a number 0, 1, or 2, according to this diagram of a net of the cube:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qiiQJl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $\varphi(\Sigma)$ to be the total number on the stickers on the up and down faces mod 3 in the state $\Sigma$. This of course gives a (surjective) map from the full cube group to the group $\mathbb Z/3\mathbb Z$, which is not clearly a homomorphism. I won't give a full proof that it is, but I'll take you most of the way there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may wonder where this particular choice comes from. In fact, the only reason I chose it is because it makes it very easy to see that $\varphi(a_i) = 0$ for each generator $a_i$. For the up and down moves, these don't even change any of the numbers at all. For the other 4 moves, the numbers do change, but they change the same way for all 4. Hence, you only need to check one position, and it's easy to check $\varphi(F) = 0$. For the proof I give below, it's a bit more convenient to check that the inverses of the generators also preserve $\varphi$, which only requires checking one more state $\varphi(F')=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This ambiguity of choice of edge-labels for the solved state turns out to be important for proving that $\varphi$ is a homomorphism. Specifically, you'll want to prove the following lemma: Any choice of numbering will define the same function $\varphi$ provided is satisfies two criteria:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Each cubie contains each of the numbers 0, 1, 2, exactly once, and they are in that cyclic order when read counter-clockwise around the vertex.&lt;/li&gt;&#xA;&lt;li&gt;The solved state still satisfies $\varphi(\Sigma_0)=0$ for this choice of numbering.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This isn't too hard to prove, but I'll skip it here. Once you can prove that, the homomorphism property of $\varphi$ comes basically for free. For simplicity, I'll just prove that the restriction $\varphi |_H$ is a homomorphism (specifically the trivial homomorphism), which is all we need. The proof is by induction. The base case is the identity. Suppose we can show that for every word of length $n$ in the generators, $\varphi(a_1 \cdots a_n) = \varphi(a_1) \cdots \varphi(a_n) = 0$. Consider an arbitrary word $a_1 \cdots a_n a_{n+1}$. Now note that for $\Sigma = a_1 \cdots a_n$, we could define a different numbering on the cube, specifically, the one we would get by first doing $\Sigma$, then numbering the cube as in the above figure (and then of course returning it to the initial solved state). By inductive hypothesis, $\varphi(\Sigma^{-1}) = \varphi (a_n^{-1} \cdots a_1^{-1}) = 0$. But $\Sigma^{-1}$ has the same numbering in the original scheme as $\Sigma_0$ in the new one, because to get to $\Sigma_0$ from the above we just undo the moves we did before by applying $\Sigma^{-1}$. Thus, this alternate numbering pattern fulfills both requirements of the lemma (the second one is clear), so it gives a different method to compute the same function $\varphi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, then the state $a_1 \cdots a_n a_{n+1} = \Sigma a_{n+1}$, by construction in the new edge-numbering, will have the same numbering as if you started with the original edge-numbering in the figure above and did the single turn $a_{n+1}$, which we know will still give $\varphi(a_1 \cdots a_{n+1}) = 0$. Hence, for any reachable state $\Sigma \in H$, we'll have $\varphi(\Sigma) = 0$, so $\varphi |_{H} = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's all you need, but I'll just say that proving that $\varphi$ is a homomorphism on $G$ isn't too much harder. You have to choose a different generating set though, since the generators for the cube group don't generate the full cube group. The generators you want to choose are basically &quot;swap corners i and i+1&quot; and &quot;twist corner 1 by $120 ^\circ$&quot;. Those generate all possible configurations of the corner pieces (you should prove this). Of course, you also need to include things which generate the edge permutations, though these aren't important at all since we only care about the corners. The proof will proceed much the same as above, with a bit of special consideration taken for the &quot;twist corner 1&quot; generator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, once you know $\varphi(\Sigma) = 0$ whenever $\Sigma \in H$, you just have to compute $\varphi(\Sigma')$ for the state with a single corner turned. This is easily seen to be either 1 or 2 depending on which direction you twist the single corner. So $\Sigma' \not \in H$, and we're done with the proof. As you can see, it's not really that hard, though it is very detailed and long.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'll take a moment to mention some nice group theoretical stuff that we're actually not far off from proving. Note that as I said before, the sequence of homomorphisms $H \hookrightarrow G \twoheadrightarrow \mathbb Z_3$ is not exact. However, we can basically completely understand the group structure of the cube group. It is a normal subgroup of the full cube group, and the quotient is given by $H \hookrightarrow G \twoheadrightarrow \mathbb Z_3 \times \mathbb Z_2 \times \mathbb Z_2$ (which is exact). The latter factors of $\mathbb Z_2$ are also describable simply in terms of cube positions. The first represents edge orientation; a state with a singe edge flipped is not solvable. The second represents odd permutations. A state with 2 corner pieces (or 2 edge pieces, but not both) swapped, but with all other pieces in the right position, is not solvable. It's impossible to permute a single pair of pieces, though it is possible to permute 2 pairs simultaneously. For any cube state, when solving it, you'll either eventually come to a fully solved cube, or you'll reach something which has a single corner twisted and/or a single edge flipped and/or a single pair of either corners or edges swapped. In any of these latter cases, the cube is not solvable and will need to be disassembled to return it to the solved state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the above, the cube group $H$ is an index 12 subgroup of the full cube group $G$, which means that 1/12 of the positions will be solvable. The cube group $H$ thus has order $8! \; 3^7 \; 12! \; 2^{10}$. Furthermore, if you know a bit of group theory, it's pretty easy to see that the full cube group is isomorphic to $\mathbb Z_3 \wr S_8 \times \mathbb Z_2 \wr S_{12}$. The edge and corner orientations are just the diagonal homomorphisms of their respective wreath products (which incidentally is the easiest way to prove the above lemma by showing that this gives you the same homomorphism), while the permutation parity is a bit more complicated to understand. Once you've understood these fully, you can compute that the cube group is isomorphic to $(\mathbb Z_3^7 \times \mathbb Z_2^{11}) \rtimes ((A_8 \times A_{12}) \rtimes \mathbb Z_2)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, since it's relevant to my interests, I'll briefly note that much of what I said here still works for higher dimensional cubes. See, for instance, &lt;a href=&quot;https://mathoverflow.net/q/147172&quot;&gt;this MO question&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="94" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:31.357" LastActivityDate="2014-06-16T18:50:42.693" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="77" PostTypeId="1" AcceptedAnswerId="79" CreationDate="2014-05-15T07:10:48.480" Score="11" ViewCount="2291" Body="&lt;p&gt;I learned how to solve a Rubik’s cube with the normal beginner method. Now I wanted to know how to get from this to speedcubing. Is speedcubing just executing the steps really fast? I don't think so. What are the fastest ways to solve the cube?&lt;/p&gt;&#xA;" OwnerUserId="113" LastActivityDate="2020-07-17T16:05:27.387" Title="How to get started in speedcubing?" Tags="|rubiks-cube|speedsolving|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="78" PostTypeId="1" AcceptedAnswerId="125" CreationDate="2014-05-15T07:15:32.787" Score="16" ViewCount="17298" Body="&lt;p&gt;The &lt;a href=&quot;http://www.stanford.edu/~laurik/fsmbook/examples/Einstein%27sPuzzle.html&quot;&gt;Einstein's puzzle&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Zebra_Puzzle&quot;&gt;zebra puzzle&lt;/a&gt; is a well-known logic puzzle. Are there any very easy ways to solve it fast?&lt;/p&gt;&#xA;" OwnerUserId="113" LastEditorUserId="7586" LastEditDate="2016-04-26T06:42:37.080" LastActivityDate="2016-04-26T15:23:43.013" Title="Fast strategies for solving Einstein's puzzles?" Tags="|solvability|logic-grid|speedsolving|" AnswerCount="4" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="79" PostTypeId="2" ParentId="77" CreationDate="2014-05-15T07:38:26.447" Score="10" Body="&lt;p&gt;The answer quite honestly is: depends on how much you want to memorize. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd personally recommend exploring the algorithms and why they work before delving into speedcubing algorithms. The reason for this is simple: over the use of the cube, you'll see faster ways to do what you're trying to do. My solve time using the beginner's method is between 30 and 40 seconds (PB avg of 5 for me is about 24-6, if I recall?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, do remember that a lot of it &lt;em&gt;does&lt;/em&gt; actually have to do with how fast you move. Never underestimate the impact of training yourself to build up sheer speed in execution - but also never forget the benefits of muscle memory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, the ways to truly speed up your solves are with memorizations of specific steps. And to reiterate, this all depends on how much you intend to memorize. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Fridrich_Method&quot;&gt;Fridrich Method&lt;/a&gt; is widely regarded to be one of the fastest modern methods for solving the cube, though it is certainly not the fastest. In essence, it relies on the cross, then F2L, then OLL, then PLL, and takes a significant amount of memorization to execute in full. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The Cross is what you're familiar with from the basic solve. &lt;/li&gt;&#xA;&lt;li&gt;F2L stands for First Two Layers, and indicates a concurrent solve of both of the first two layers. This revolves, in essence, around pairing edges with corners and inserting both at once. There are a decent number of algorithms associated with specific conditions, all designed for the shortest path. &lt;/li&gt;&#xA;&lt;li&gt;OLL stands for Orientation of the Last Layer, and is, in essence, a dictionary method that orients all the pieces on the last layer. &lt;a href=&quot;http://www.cubewhiz.com/oll.php&quot;&gt;Lots of documentation is available&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;PLL stands for Permutation of the Last Layer, and is also effectively a dictionary attack on the permutation of pieces. This is &lt;a href=&quot;http://www.cubewhiz.com/pll.php&quot;&gt;slightly less intensive than OLL&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However, these are the extremes.&lt;/strong&gt; I'm using these to illustrate a point that the more you memorize, the faster you'll get. You can find a list of methods &lt;a href=&quot;http://www.speedsolving.com/wiki/index.php/3x3x3_speedsolving_methods&quot;&gt;on the speedcubing wiki&lt;/a&gt;. There are a &lt;em&gt;lot&lt;/em&gt; of intermediate steps, though, which can help you along your path. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;F2L doesn't necessarily need piles of algorithms. You can learn partial solutions, and intuitively guide yourself to cases you know. &lt;/li&gt;&#xA;&lt;li&gt;You can split OLL into two steps: one to orient edges, and one for corners. &lt;/li&gt;&#xA;&lt;li&gt;You can do the same for PLL. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In essence, the more shortcuts you discover and memorize, the faster you'll go. I definitely recommend looking through the &lt;a href=&quot;http://www.speedsolving.com/wiki/index.php/3x3x3_speedsolving_methods&quot;&gt;speedcubing methods&lt;/a&gt; wiki page, as it contains further useful information. &lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastActivityDate="2014-05-15T07:38:26.447" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="80" PostTypeId="1" AcceptedAnswerId="82" CreationDate="2014-05-15T08:34:56.247" Score="14" ViewCount="1591" Body="&lt;p&gt;Can you tell me what is the name of the game I used to play in school?&lt;/p&gt;&#xA;&lt;p&gt;It's a game for two players. You need grid paper and two pens.&lt;/p&gt;&#xA;&lt;p&gt;The game is played on a 10x10 arena, shifted by half a grid to one side.&lt;/p&gt;&#xA;&lt;p&gt;The players take turns, each drawing a one grid length line. One player plays &amp;quot;middles&amp;quot;, the other plays &amp;quot;edges&amp;quot;, that means one of the players draws lines over the lines of the grid, the other connects centers of adjacent boxes the grid forms. Lines of the two players can't cross - if there's a line drawn on the grid line between two adjacent boxes, the player playing middles can't connect them, and vice versa.&lt;/p&gt;&#xA;&lt;p&gt;The winner is the one to connect their &amp;quot;own&amp;quot; sides of the grid - &amp;quot;middle&amp;quot; - left and right, &amp;quot;edge&amp;quot; - top and bottom. Below is an example image of the game in progress.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gfute.png&quot; alt=&quot;Game in progress&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;What is the name of this game?&lt;/p&gt;&#xA;" OwnerUserId="117" LastEditorUserId="-1" LastEditorDisplayName="user121" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-02-02T09:24:35.227" Title="What is the name of this line-connecting game?" Tags="|puzzle-identification|game|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="81" PostTypeId="1" AcceptedAnswerId="3226" CreationDate="2014-05-15T09:01:30.610" Score="11" ViewCount="966" Body="&lt;p&gt;How, as a writer of riddles, can I rate their difficulty?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess, for me alone, knowing the answer even before the riddle existed, it's pretty much impossible, and I should need some external help. Are there some riddling communities or something like that, who could help rate the difficulty? How would I approach this problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: to clarify. I mean classic riddles based on metaphors - like the ones found in Tolkien's books. Not 'trick questions', math problems posed as riddles, or other weird riddle-like creatures.&lt;/p&gt;&#xA;" OwnerUserId="117" LastEditorUserId="8874" LastEditDate="2015-01-31T18:55:17.540" LastActivityDate="2015-01-31T18:55:17.540" Title="How to rate difficulty of a classic riddle?" Tags="|riddle|difficulty-rating|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="82" PostTypeId="2" ParentId="80" CreationDate="2014-05-15T09:02:18.477" Score="9" Body="&lt;p&gt;It's called &lt;a href=&quot;http://en.wikipedia.org/wiki/Bridg-It#Gale&quot;&gt;Bridg-It&lt;/a&gt;, or &lt;a href=&quot;http://www.papg.com/show?1TPI&quot;&gt;Game of Gale&lt;/a&gt; (as it was invented by David Gale). It is the rectangular grid equivalent of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Shannon_switching_game&quot;&gt;Shannon switching game&lt;/a&gt;, which is a generalized version of &lt;a href=&quot;http://en.wikipedia.org/wiki/Hex_%28board_game%29&quot;&gt;Hex&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="85" LastEditorUserId="10" LastEditDate="2014-05-17T08:52:07.220" LastActivityDate="2014-05-17T08:52:07.220" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="84" PostTypeId="2" ParentId="59" CreationDate="2014-05-15T09:16:31.860" Score="10" Body="&lt;p&gt;There are a couple of further strategies. To start, there are a lot more local patterns that allow marking some loop segments and non-loop segments. Bram de Laat wrote up an excellent &lt;a href=&quot;http://puzzleparasite.blogspot.de/2011/11/slitherlink-pattern-guide_23.html&quot;&gt;guide&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, there are some global arguments, which hinge on the requirement for a single loop: Some edges may result in a small closed loop, or some edges must be part of the loop in order for the loop to reach some remote area of the grid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you bisect the grid along any path, the loop must cross the cut an even number of times. (This is basically the &lt;a href=&quot;http://en.wikipedia.org/wiki/Jordan_curve_theorem&quot;&gt;Jordan Curve Theorem&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another set of techniques that I don't want to detail here exploit the fact that the puzzle has a unique solution.&lt;/p&gt;&#xA;" OwnerDisplayName="user121" LastActivityDate="2014-05-15T09:16:31.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="85" PostTypeId="2" ParentId="29" CreationDate="2014-05-15T09:47:47.450" Score="4" Body="&lt;p&gt;To be able to judge the difficulty of a Sudoku, you first need to define &quot;difficulty&quot;. The most useful definition seems to rely on the question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How long does it take an &quot;average&quot; solver to solve the puzzle?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Making this precise will involve fixing the population of solvers, and setting up some reference. E.g., take the 10 best Sudoku solvers at the last World Sudoku Championships and say a puzzle is of &quot;medium&quot; difficulty if they take 3 minutes on average.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, assembling a representative group of a few solvers and having them solve your puzzles will give a reliable view of their relative difficulty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the technical difficulty definitions, note that it is possible to make reasonably tricky puzzles with basic techniques (hidden singles can be hard to spot, but a skilled author can make them obvious). Similarly, X-wings can be in plain sight or hidden. So judging difficulty by required techniques alone is not particularly helpful.&lt;/p&gt;&#xA;" OwnerDisplayName="user121" LastActivityDate="2014-05-15T09:47:47.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="86" PostTypeId="1" AcceptedAnswerId="90" CreationDate="2014-05-15T10:21:20.207" Score="14" ViewCount="2046" Body="&lt;p&gt;I used to play &lt;a href=&quot;http://www.stanford.edu/~laurik/fsmbook/examples/Einstein%27sPuzzle.html&quot; rel=&quot;noreferrer&quot;&gt;Einstein's puzzles&lt;/a&gt; a lot when I was younger, and very quickly learned to assess quite rapidly whether I thought a puzzle would be a quick solve or more time-consuming, but I never worked out the detail on how many clues are required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a mathematical description of the minimum number of clues required to guarantee a solve, beyond which all further clues are unneeded but make life easier?&lt;/p&gt;&#xA;" OwnerUserId="53" LastEditorUserId="463" LastEditDate="2015-06-02T02:33:15.507" LastActivityDate="2021-11-14T09:47:57.590" Title="How many clues are required to ensure Einstein's Puzzle is solvable?" Tags="|logic-grid|minimum-solve|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="87" PostTypeId="2" ParentId="75" CreationDate="2014-05-15T10:55:42.007" Score="3" Body="&lt;p&gt;To solve this puzzle you first have to know the boundaries in whose you are working. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 20 teams, a total of 190 matches (and 190 points).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to divide the 190 points to all teams. The points you have to distribute are 0, 1, 4, 9 and 16 since a team can't get more than 19 points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So first we go to the team with 0 points. This can only be one because all the other teams have at least 1 win against this team. The second team then has 1 point. Again all other teams now have at least 2 points so the next few teams can have at least 4 points. If we follow this in the end I got 4 teams with 4, 5 teams with 9, 6 teams with 16 2 with 10 and 1 with 12. To fix those few teams with an non square number, we either have to get them to 16 or to 9. when we get them all to 9 we can get a 4 to 9 too (5 points in total). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the result then is the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 team with 0 points&lt;br&gt;&#xA; 1 team with 1 points&lt;br&gt;&#xA; 3 team with 4 points&lt;br&gt;&#xA; 9 team with 9 points&lt;br&gt;&#xA; 6 team with 16 points  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the way I can solve this puzzle, maybe there is a mathematical way I don't know of. &lt;/p&gt;&#xA;" OwnerUserId="113" LastEditorUserId="113" LastEditDate="2014-05-16T12:57:17.243" LastActivityDate="2014-05-16T12:57:17.243" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="88" PostTypeId="1" AcceptedAnswerId="96" CreationDate="2014-05-15T11:02:40.687" Score="15" ViewCount="2075" Body="&lt;p&gt;Once I solve a Rubik's Cube I obviously want to start again. I just scramble it until it seems to be enough; however, sometimes when I then solve it, I can skip entire steps from the &lt;a href=&quot;http://www.youcandothecube.com/&quot; rel=&quot;noreferrer&quot;&gt;beginner's method&lt;/a&gt;. Is this normal or was the cube not scrambled enough? How can one determine if the cube is scrambled enough?&lt;/p&gt;&#xA;" OwnerUserId="113" LastEditorUserId="30633" LastEditDate="2016-12-20T07:20:55.877" LastActivityDate="2019-12-24T11:12:28.900" Title="How to determine if a Rubik's Cube is sufficiently scrambled?" Tags="|rubiks-cube|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="90" PostTypeId="2" ParentId="86" CreationDate="2014-05-15T11:20:38.063" Score="12" Body="&lt;p&gt;I know this type of puzzle simply as a &quot;Logic Puzzle&quot; or sometimes &quot;Logic Grid Puzzle&quot; (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Logic_puzzle&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia entry here&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's impossible to come to a fixed number, because each clue may reveal a different amount of information. For example &quot;The Englishman lives in the red house but doesn't own a dog&quot; essentially has two clues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you break those down into separate pieces of information, there are different &lt;em&gt;types&lt;/em&gt; of information:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Straight truths&lt;/strong&gt;, e.g. &quot;The Englishman lives in the red house&quot;. This is more than one piece of information, since you also exclude multiple possibilities (i.e. No one else lives in the red house).&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Negatives&lt;/strong&gt;, e.g. &quot;The Swede doesn't live in the green house&quot;. Here you only gain one piece of information.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Spatial or relative clues&lt;/strong&gt;, e.g. &quot;The Dane's house is somewhere to the right of the Norwegian's&quot;. These can get more complex depending on the specific puzzle (e.g. &quot;Adam arrived at least 10 minutes later than Betty&quot;). It's impossible to determine how much information you extract from a clue here.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However, I do believe there is one metric that is required to ensure a puzzle is solvable: &lt;em&gt;all but one of each 'entity' must be mentioned in the puzzle&lt;/em&gt;. In other words, if you have 5 house colors then the clues must make reference to 4 of them (presuming all the colors are listed in the introduction or accompanying logic grid). If two colors are not mentioned, there is no way to determine what goes with each.&lt;/p&gt;&#xA;" OwnerUserId="125" LastEditorUserId="463" LastEditDate="2015-06-02T02:37:57.687" LastActivityDate="2015-06-02T02:37:57.687" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="91" PostTypeId="1" AcceptedAnswerId="93" CreationDate="2014-05-15T11:32:20.160" Score="10" ViewCount="789" Body="&lt;p&gt;A &lt;a href=&quot;https://en.wikipedia.org/wiki/River_crossing_puzzle&quot; rel=&quot;nofollow noreferrer&quot;&gt;River Crossing puzzle&lt;/a&gt; puts limits on the number of items that may cross a boundary, with limits on what may be left together. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Fox,_goose_and_bag_of_beans_puzzle&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Fox, goose and bag of beans&quot;&lt;/a&gt; is one of the earliest examples.  Of course, by increasing the capacity of the boat, or the number/type of barriers, infinity is the upper limit.  For this question, I am asking with the classical configuration of one barrier, with a limit of one transport per move. You can increase the number of items and their ability to coexist. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the largest number of items in a solvable river crossing puzzle that has been documented?  What is the maximum that could be created?&lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="30633" LastEditDate="2016-12-20T07:22:15.543" LastActivityDate="2016-12-20T07:22:15.543" Title="What is the maximum number of items in a river crossing puzzle?" Tags="|river-crossing|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="93" PostTypeId="2" ParentId="91" CreationDate="2014-05-15T12:00:33.453" Score="10" Body="&lt;p&gt;If you have just items to transport, infinity. But it's no riddle, then. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in 'classical' problem, every item except 1 is endangered by another item, every item except 1 is threatened by another item. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under that circumstances you can't even solve it with 4, because no matter what item you take, there will be at least one 'colission' pair:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A -&gt; B -&gt; C -&gt; D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's unsolvable even if there's no threat from B to C, because you can 'break' only 1 danger pair, unless they have similar element (which can take place only by 3 items, because, in case of 4 and more 1 element would have to be neutral:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A-&gt;B-&gt;C, D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So without neutral elements, it's not solvable under that model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take, that an element can be a danger for many, it's not going better:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A-&gt;B, A-&gt;C &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;this is solvable (removing A, than taking B and exchanging it for A, than taking C), but with any element more, it's unsolvable. Because you can either have A in boat, or have maximal 2 elements safe from A (one that you moved to the island in one step, the second in your boat). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So with 'simple' threatening rules (A can't be with B) it's solvable only for 3.&lt;/p&gt;&#xA;" OwnerDisplayName="user66" LastActivityDate="2014-05-15T12:00:33.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="94" PostTypeId="2" ParentId="91" CreationDate="2014-05-15T12:33:40.397" Score="7" Body="&lt;p&gt;For another viewpoint, let's take the &quot;classic&quot; version, of &lt;code&gt;A -&amp;gt; B, B -&amp;gt; C&lt;/code&gt; (that is, fox eats goose, goose eats beans) and add a &lt;code&gt;D&lt;/code&gt; such that it neither threatens anything nor is threatened by anything (a rock, say), as in the &quot;neutral&quot; element of @Lukasz's answer. Now is it solvable? Well, let's see (&lt;code&gt;\/&lt;/code&gt; = boat):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  ABCD\/~~~~~~~~~~&#xA;  A CD  ~~~~~~~~~~\/ B&#xA;* A CD\/~~~~~~~~~~   B&#xA;* A C   ~~~~~~~~~~\/ B D&#xA;  A C \/~~~~~~~~~~   B D&#xA;  A     ~~~~~~~~~~\/ BCD&#xA;  AB  \/~~~~~~~~~~    CD&#xA;   B    ~~~~~~~~~~\/A CD&#xA;   B  \/~~~~~~~~~~  A CD&#xA;        ~~~~~~~~~~\/ABCD&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As you can see, yes, it is solvable with two extra steps (marked with an &lt;code&gt;*&lt;/code&gt;). Similarly, by adding more &quot;neutral&quot; elements and adding more extra steps, there is no limit on how many items you can have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that there can only be two &quot;predators&quot; (objects that threaten other objects) because you can break a chain of two by removing one, but you can't break a chain of three by removing one:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A -&amp;gt; B -&amp;gt; C -&amp;gt; D            A -&amp;gt; B&#xA;                            C -&amp;gt; D&#xA;             !      (or)&#xA;A         C -&amp;gt; D            A  !&#xA;                            C -&amp;gt; D   &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is not true if they have the same &quot;prey,&quot; since removing that prey breaks all chains. Note that, however, you can't have &lt;em&gt;three&lt;/em&gt; predators for one prey, because you can only carry one item, and there will have to be an intermediate step with one of those predators with the prey (try it!).&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2014-05-15T12:33:40.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="95" PostTypeId="2" ParentId="48" CreationDate="2014-05-15T13:14:20.147" Score="3" Body="&lt;p&gt;We can simplify 2048 by discarding the layout and focusing on a multiset (or bag) of tiles. We can do this, if we assume that a new tile will always appear in a position that is most advantageous to us.&lt;br&gt;&#xA;We can then define two rules to make a new bag out of the current one (we also ignore the possibility of getting 4s for simplicity's sake):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Combine one or more pairs of tiles of equal value into tiles of double that value.&lt;/li&gt;&#xA;&lt;li&gt;Add a 2-tile if there's an empty slot.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Rule 1 is optional, rule 2 is mandatory. If rule 2 fails, we stop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this, we can see that for a bag of maximum size 3, we have the following states:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;{} → {2} → {2,2} → {2,2,2}&#xA;             ↓        ↓&#xA;           {2,4} → {2,2,4}&#xA;                      ↓&#xA;                   {2,4,4}&#xA;                      ↓&#xA;                   {2,2,8}&#xA;                      ↓&#xA;                   {2,4,8}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For a bag of maximum size 4, we build from this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;{} → {2} → {2,2} → {2,2,2} → {2,2,2,2}&#xA;             ↓        ↓          ↓&#xA;           {2,4} → {2,2,4} → {2,2,2,4}&#xA;                      ↓          ↓&#xA;                   {2,4,4} → {2,2,4,4}&#xA;                      ↓          ↓    ↘&#xA;                   {2,2,8} → {2,2,2,8}  {2,4,4,4}&#xA;                      ↓          ↓    ↙&#xA;                   {2,4,8} → {2,2,4,8}&#xA;                                 ↓&#xA;                             {2,2,8,8}&#xA;                                 ↓     ↘&#xA;                             {2,2,2,16}  {2,4,8,8}&#xA;                                 ↓     ↙&#xA;                  {2,4,16} → {2,2,4,16}&#xA;                                 ↓&#xA;                             {2,4,4,16}&#xA;                                 ↓&#xA;                             {2,2,8,16}&#xA;                                 ↓&#xA;                             {2,4,8,16}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;To allow for more than one pair of tiles combining, imagine a couple of diagonal arrows from &lt;code&gt;{2,2,2,2}&lt;/code&gt; to &lt;code&gt;{2,4,4}&lt;/code&gt;, from &lt;code&gt;{2,2,4,4}&lt;/code&gt; to &lt;code&gt;{2,4,8}&lt;/code&gt;, and from &lt;code&gt;{2,2,8,8}&lt;/code&gt; to &lt;code&gt;{2,4,16}&lt;/code&gt;.&lt;sup&gt;&amp;dagger;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, we can infer that for a table of size $n$, the maximum value attainable is $2^n$. For a standard 2048 game, &lt;em&gt;assuming only 2-tiles appear&lt;/em&gt;, this means the maximum tile value is &lt;strong&gt;65536&lt;/strong&gt;.&#xA;If we allow 4-tiles, 131072.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&#xA;&amp;dagger;: Drawing those arrows isn't easy and would've made for a less clear diagram.  &lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditDate="2015-06-18T20:42:03.813" LastActivityDate="2015-06-18T20:42:03.813" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="96" PostTypeId="2" ParentId="88" CreationDate="2014-05-15T13:18:09.547" Score="16" Body="&lt;p&gt;&lt;a href=&quot;http://dilbert.com/strip/2001-10-25&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/P7xj1.gif&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each set of moves is designed to correct a parity error in the cube's layout, but each group of parity it corrects is essentially independent and random. Each step has a chance of already having the right parity. For instance, the way I do it is solve the bottom two layers and then attack the parity errors on the top one at a time&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Make sure all the edges are right-side-up&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;3 possibilities &lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Make sure the corners are in the right order&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;3 possibilities&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Make sure the corners are right-side-up&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Not exactly sure. 5?&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Make sure the edges are in the right order.&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Not quite sure either. 5?&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;on a 4x4 there can be a flipped edge&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;2 possibilities&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;on a 5x5 there are a couple other funny edge things&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;5?&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now, the probabilities probably aren't equal for each parity state, but at each step there is a fair chance at already being in the right state. That's just how randomness works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd say if you need most of the steps most of the time, you're doing fine. After all, when you practice you are really just training yourself to recognize and correct the errors. As long as you're encountering all of them once in a while, that's all that really matters.&lt;/p&gt;&#xA;" OwnerUserId="4" LastEditorUserId="30903" LastEditDate="2017-03-09T07:33:18.690" LastActivityDate="2017-03-09T07:33:18.690" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="97" PostTypeId="2" ParentId="88" CreationDate="2014-05-15T13:18:19.293" Score="3" Body="&lt;p&gt;A good, scrambled Rubik's cube, in my opinion, has no lines of 3 of the same color on one side (green). It also has none of the same colors on a touching center and corner piece (red).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HnLg5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="63" LastEditorUserId="30633" LastEditDate="2017-03-09T15:12:31.307" LastActivityDate="2017-03-09T15:12:31.307" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="98" PostTypeId="2" ParentId="81" CreationDate="2014-05-15T13:26:44.890" Score="3" Body="&lt;p&gt;I would say that riddle difficulty is hard to rate. Some people will find one riddle easy, while another person will find it hard. It all depends how you approach the solution. Some riddles involve little-known facts, and therefore are difficult to solve.&lt;/p&gt;&#xA;" OwnerUserId="63" LastActivityDate="2014-05-15T13:26:44.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="99" PostTypeId="1" CreationDate="2014-05-15T13:32:13.850" Score="9" ViewCount="2419" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;This is a follow-up question for the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/43/sliding-bolt-puzzle&quot;&gt;Sliding Bolt Puzzle&lt;/a&gt;. If you have not solved it yet, you might want to head there first, as the extended discussion of its solution in this question will contain &lt;strong&gt;spoilers&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the original version of the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/43/sliding-bolt-puzzle&quot;&gt;Sliding Bolt Puzzle&lt;/a&gt;, I asked for the fastest sequence in terms of the maximum number of button presses it takes to definitely open the door, which was given in the accepted answer. However, I was wondering whether there might be a sequence which, while having more button presses, is still faster time-wise when you take into account the probability of the unknown initial position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To explain a bit more what I mean: As Kendall Frey explained, there are four possible states of the bolts, numbered 1 to 4. (We will neglect state 1 as an initial position, because it is specified that the door is locked at first.) However, these four states are not equally probable as an initial position: The probabilities are&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;8/14 for state 2,&lt;/li&gt;&#xA;&lt;li&gt;4/14 for state 3,&lt;/li&gt;&#xA;&lt;li&gt;2/14 for state 4.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(obtained by simply listing all 16-2=14 possible initial configurations and categorizing them accordingly).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we see that state 2 is by far the most probable one; the sequence &lt;strong&gt;CBCACBC&lt;/strong&gt; will open it after four touches of a button at the earliest, while the most improbable state 4 will be opened directly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hence, my question is: Which is the sequence with the smallest expected value E[t], where t is the number of button pushes until the door opens, taking into account the probability distribution of the initial configuration?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that there is also some randomness regarding the next state when you press button A while in state 2 or button B while in state 3 (see the transition state matrix in the accepted answer), which should also be considered.&lt;/p&gt;&#xA;" OwnerUserId="12" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2016-12-20T08:36:47.270" Title="Sliding Bolt Puzzle - fastest solution (time-wise)" Tags="|strategy|logical-deduction|mathematics|probability|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="100" PostTypeId="1" AcceptedAnswerId="104" CreationDate="2014-05-15T13:52:19.957" Score="13" ViewCount="8358" Body="&lt;p&gt;In a sudoku puzzle, what is the absolute minimum number of clues that must be given to specify a layout with a unique solution? Is one in each row, column, and box (9 total) necessary or sufficient?&lt;/p&gt;&#xA;" OwnerUserId="4" LastEditorUserId="4" LastEditDate="2014-05-16T15:09:23.247" LastActivityDate="2020-04-08T04:01:37.013" Title="Minimum number of clues for a sudoku puzzle" Tags="|sudoku|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="102" PostTypeId="2" ParentId="78" CreationDate="2014-05-15T14:05:52.257" Score="4" Body="&lt;p&gt;For small puzzles of this type, a grid is very useful.  If there are three types of item, with five of each type, you can use a grid like this.  Put an x in all the squares you know cannot hold, and fill the boxes you know are true.  So if you are told 1 is not a, you x the upper left box.  If you are told B is 2 you fill the corresponding box, then x all the boxes in the same row and column.  You can also copy x's from one grid to another when you fill a box.  For four types, put types 1,2,3 across the top, 4,3,2 down the left and you will have six grids.  It becomes unwieldy after five types&lt;img src=&quot;https://i.stack.imgur.com/Sg0fp.jpg&quot; alt=&quot;like this&quot;&gt; This is especially useful to see when you have eliminated all but one possibility.  To capture the left/right information, one of your types of item can be the house number.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditDate="2014-05-15T17:10:07.730" LastActivityDate="2014-05-15T17:10:07.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="103" PostTypeId="1" AcceptedAnswerId="279" CreationDate="2014-05-15T14:07:22.947" Score="0" ViewCount="1323" Body="&lt;p&gt;One of the classical puzzles is pouring water into the 5 gallon jug so there is 4 gallons of water remaining. This was popularized by Bruce Willis in &quot;Die Hard with a Vengeance&quot;. I was wondering however, if there was a possible way to complete the puzzle using a 3 gallon jug and a 7 gallon jug to still get the 4 gallons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, what other combinations of jugs could you use to still get the same four gallon requirement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: When doing the 4 gallons remaining from the 7 gallon jug, you must have 1 gallon of water left in the three gallon jug.&lt;/p&gt;&#xA;" OwnerUserId="83" LastEditorUserId="4050" LastEditDate="2016-01-02T12:22:13.443" LastActivityDate="2016-01-02T12:22:13.443" Title="Three Gallon Jug and 5 Gallon Jug" Tags="|calculation-puzzle|decanting-problems|" AnswerCount="4" CommentCount="5" ClosedDate="2016-01-02T22:08:54.833" ContentLicense="CC BY-SA 3.0" />
  <row Id="104" PostTypeId="2" ParentId="100" CreationDate="2014-05-15T14:09:10.157" Score="19" Body="&lt;p&gt;For a standard &lt;span class=&quot;math-container&quot;&gt;$9 \times 9$&lt;/span&gt; sudoku, the minimum is &lt;span class=&quot;math-container&quot;&gt;$17$&lt;/span&gt; squares.  It had long been known that at least &lt;span class=&quot;math-container&quot;&gt;$16$&lt;/span&gt; were required and that &lt;span class=&quot;math-container&quot;&gt;$17$&lt;/span&gt; was sufficient.  &lt;a href=&quot;https://www.technologyreview.com/2012/01/06/188520/mathematicians-solve-minimum-sudoku-problem/&quot; rel=&quot;nofollow noreferrer&quot;&gt;This article&lt;/a&gt; closes the gap.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="68054" LastEditDate="2020-04-08T04:01:37.013" LastActivityDate="2020-04-08T04:01:37.013" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="105" PostTypeId="1" AcceptedAnswerId="118" CreationDate="2014-05-15T14:19:37.507" Score="23" ViewCount="7184" Body="&lt;p&gt;I'm asking about classic riddles as they appear in books and kids' literature, based on metaphors - not maths problems or tasks, or deceitful questions but just an old good-natured rhyme describing the subject in a way that makes it tricky to guess. Let me give some examples from J.R.R. Tolkien for what I mean:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This thing all things devours:&lt;br&gt;&#xA;  Birds, beasts, trees, flowers;&lt;br&gt;&#xA;  Gnaws iron, bites steel;&lt;br&gt;&#xA;  Grinds hard stones to meal;&lt;br&gt;&#xA;  Slays king, ruins town,&lt;br&gt;&#xA;  And beats high mountain down.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Voiceless it cries,&lt;br&gt;&#xA;  Wingless flutters,&lt;br&gt;&#xA;  Toothless bites,&lt;br&gt;&#xA;  Mouthless mutters. &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wind&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Writing these riddles is a tricky business, because you must balance difficulty, come up with clues corrupted enough not to be trivial, and simultaneously not too obscure, and get the final result in rhyme form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are some techniques of coming up with such a riddle?&lt;/p&gt;&#xA;" OwnerUserId="117" LastEditorUserId="5373" LastEditDate="2018-11-10T10:35:30.590" LastActivityDate="2018-11-10T10:35:30.590" Title="How do you come up with a good classic rhyme riddle?" Tags="|riddle|rhyme|puzzle-creation|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="106" PostTypeId="2" ParentId="81" CreationDate="2014-05-15T14:22:14.177" Score="7" Body="&lt;p&gt;I think the problem with trying to rate these with one number is that &lt;em&gt;most&lt;/em&gt; of the most popular brainteasers present challenges on up to three different &quot;challenge spectrums&quot;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Some are highly dependent on actual skills, knowledge or tools,&lt;/strong&gt; like ones that can't easily be solved without a spreadsheet, or that are incredibly hard without knowledge of mathematical techniques like probability (trees, permutation odds, etc.) Monty Hall is one example that is &lt;em&gt;much&lt;/em&gt; harder to solve -even if you can get past how unintuitive the answer is -if you've never studied at least basic probability techniques.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Some require complex reasoning or creative (but not &quot;sneaky&quot;) thinking.&lt;/strong&gt; Complex reasoning (including some experimentation with different approaches) would include puzzles like &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16/100-prisoners-names-in-boxes&quot;&gt;100 boxes&lt;/a&gt;. the I'd toss out &quot;Everyone may have children until they have a boy, and then must stop - what will the eventual boy/girl ratio be?&quot; as an example of creative thinking - it's not a &quot;trick,&quot; but you have to approach it in a highly non-intuitive way.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Some require lateral thinking, of the variety that can seem kind of like a trick.&lt;/strong&gt; I've seen this referred to as the &quot;Aha!&quot; characteristic of a puzzle.  Classic examples end with &quot;The doctor is his &lt;em&gt;mother&lt;/em&gt; (you sexist!) or &quot;The victim is a fish!!&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2014-05-15T14:39:49.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="107" PostTypeId="2" ParentId="103" CreationDate="2014-05-15T14:22:28.177" Score="2" Body="&lt;p&gt;For 7 and 3 wanting 4, just fill the 7 and fill the 3 from it, leaving 4.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say you had 7 and 3 and want 2.&lt;br&gt;&#xA;Fill the 7&lt;br&gt;&#xA;fill the 3 from the 7, leaving 4&lt;br&gt;&#xA;empty the 3&lt;br&gt;&#xA;fill the 3 from the 7, leaving 1&lt;br&gt;&#xA;empty the 3&lt;br&gt;&#xA;put the 1 in the 3&lt;br&gt;&#xA;fill the 7&lt;br&gt;&#xA;fill the 3 from the 7, leaving 5&lt;br&gt;&#xA;empty the 3&lt;br&gt;&#xA;fill the 3 from the 7, leaving 2  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically you need the desired quantity to be less than the large jug and to have the desired quantity to be a multiple of the least common multiple of the jug volumes.  The algorithm is simple:  fill the large jug, keep filling the small jug from it, refill the large jug when you must, and you will get there.  You are finding a solution to &lt;a href=&quot;http://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity&quot; rel=&quot;nofollow&quot;&gt;Bézout's identity&lt;/a&gt;  My solution is 2*7-4*3=2  The solution in the comments is 3*3-1*7=2  Note that if you subtract these two equations you get 3*7-7*3=0&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditDate="2014-05-15T14:30:00.670" LastActivityDate="2014-05-15T14:30:00.670" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="108" PostTypeId="2" ParentId="75" CreationDate="2014-05-15T14:32:41.367" Score="2" Body="&lt;p&gt;Within a group of $n$ teams, there are $\frac{n (n - 1)}{2}$ points to share. But that goes for each subgroup as well. A subgroup of m teams, where $m &amp;lt; n$, will have $\frac{m (m - 1)}{2}$ points between them, plus (at most) $m (n - m)$ points from the teams outside the subgroup. If we want to give this group the same number of points ($x$), that total is equal to $m \cdot x$. So &#xA;$$\begin{align}&#xA;m \cdot x &amp;amp; \le m \cdot \frac{m - 1}{2} +  m \cdot (n - m) \\&#xA;x &amp;amp; \le \frac{m - 1}{2} + (n - m) \\&#xA;x &amp;amp; \le n - \frac{m - 1}{2}&#xA;\end{align}$$&lt;br&gt;&#xA;Plugging in the numbers, we get that we can have &lt;em&gt;at the most&lt;/em&gt; 7 teams with 16 points. That leaves us with 13 teams and 78 points between them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying the same principle again and again, we end up with&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 7 teams scored 16 points&lt;br&gt;&#xA; 7 teams scored  9 points&lt;br&gt;&#xA; 3 teams scored  4 points&lt;br&gt;&#xA; 3 teams scored  1 point&lt;br&gt;&#xA; 1 team  scored  0 points  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="68" LastEditorDisplayName="user20" LastEditDate="2014-06-22T02:47:45.793" LastActivityDate="2014-06-22T02:47:45.793" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="109" PostTypeId="2" ParentId="103" CreationDate="2014-05-15T14:37:41.087" Score="1" Body="&lt;p&gt;for 2 jugs that can contain $x$ and $y$ you can get any multiple of $GCD(x,y)$ between $0$ and $max(x,y)$ using the algorithm that @Ross described.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can understand this by creating a new unit so that $1 \text{ unit} = GCD(x,y)$ gallons, and now you have 2 jugs with $\frac x {GCD(x,y)} = x'$ and $\frac y{GCD(x,y)} = y'$ and because you can't take out fractional units it means that you alway have a multiple of $GCD(x,y)$ in your jugs.&lt;/p&gt;&#xA;" OwnerUserId="81" LastEditorUserId="68" LastEditDate="2014-07-02T09:45:09.813" LastActivityDate="2014-07-02T09:45:09.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="111" PostTypeId="2" ParentId="62" CreationDate="2014-05-15T14:57:13.600" Score="7" Body="&lt;p&gt;When playing &lt;a href=&quot;http://jmfork.github.io/2048/&quot; rel=&quot;noreferrer&quot;&gt;deterministic 2048&lt;/a&gt;, where a new tile is placed in the first available spot, you can get a pattern like this using left-down-left-down:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4FsNh.png&quot; alt=&quot;det 2048&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The worst I have been able to get in regular 2048 is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/miKCM.png&quot; alt=&quot;reg 2048&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It &lt;em&gt;is&lt;/em&gt; possible to get that worst-case scenario if your tile generator and human are working together to do so.&lt;/p&gt;&#xA;" OwnerUserId="63" LastActivityDate="2014-05-15T14:57:13.600" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="112" PostTypeId="1" AcceptedAnswerId="119" CreationDate="2014-05-15T15:02:09.310" Score="14" ViewCount="6696" Body="&lt;p&gt;Here's an old puzzle I found in a book once.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A box has some number of locks on it. It is guarded by a Grand Vizier and four slaves, each of which hold keys to the locks.&lt;/p&gt;&#xA;&lt;p&gt;The Grand Vizier and slaves each hold a different set of keys to the locks, and each key they hold opens a different lock.&lt;/p&gt;&#xA;&lt;p&gt;The keys they hold are in a certain combination such that as long as the Grand Vizier is with one slave, or if any three slaves are together, they will have all the keys to open the box; but if less than that are together, they cannot. How many locks did this box have at minimum?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2018-05-14T16:54:32.360" Title="The Puzzle of the Locked Box" Tags="|logical-deduction|story|keys-and-locks|" AnswerCount="4" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="114" PostTypeId="2" ParentId="12" CreationDate="2014-05-15T15:08:36.330" Score="39" Body="&lt;p&gt;The immediate answer is no. Any valid Sudoku can be solved without guessing, just by exhaustively trying all possibilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are two interesting variant interpretations of the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Are there Sudoku puzzles that can't be solved logically?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This one depends on what you accept as a &quot;logical solution.&quot; In the strictest sense, the answer is again &lt;em&gt;no.&lt;/em&gt; However, if you define a &quot;logical solution&quot; as excluding brute-force solving, the answer is probably &quot;yes.&quot; There is some imprecise agreement on what constitutes a logical solution under this definition in the puzzling community, but based on this, we will again find that yes, there are definitely Sudoku puzzles that have no logical solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thomas Snyder has given an insightful &lt;a href=&quot;http://www.gmpuzzles.com/blog/2013/03/ask-dr-sudoku-12-the-line-must-be-drawn-here/&quot;&gt;definition&lt;/a&gt; of when a puzzle is solvable logically:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When I can solve a puzzle in ink, without erasures, with all deductions either positive or negative coming from visualization in my head and not making scratch-work on the paper, the puzzle is solvable by logic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Are there Sudoku puzzles that are faster to solve by guessing?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, the answer is &quot;yes.&quot; In a competitive setting, most solvers would not use the more obscure techniques. After having narrowed possibilities somewhat with realistic techniques, some puzzles &quot;require&quot; guessing so as to not put the solver at a competitive disadvantage.&lt;/p&gt;&#xA;" OwnerDisplayName="user121" LastEditorUserId="29309" LastEditDate="2016-08-22T16:48:41.510" LastActivityDate="2016-08-22T16:48:41.510" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="118" PostTypeId="2" ParentId="105" CreationDate="2014-05-15T15:22:18.687" Score="24" Body="&lt;p&gt;Before you proceed, here's the riddle I came up with, through composing this answer. Obviously, the answer = spoiler follows close behind, so if you want to try to guess it, do it before proceeding below the line.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;Eight hours of death. Beneath your head -&lt;/em&gt;.&lt;br&gt;&#xA;&lt;em&gt;a colored box that's full of dreams.&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;It listens to your tales unsaid.&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;Soft gifts of birds escape its seams.&lt;/em&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Start from the end: the answer.&lt;/strong&gt; Think about your target audience - if it's a kids' riddle you won't give them a riddle with 'Euler's Constant' for an answer. Make sure it's not too obscure, unless you're aiming at a very specific audience.&lt;/p&gt;&#xA;&lt;p&gt;Let us try something nice and simple. Oh, out of top of my head, since I'm feeling sleepy, &lt;em&gt;a pillow&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Pick a set of associations&lt;/strong&gt; which can describe the subject, themes realated to it.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;feathers/down&lt;/li&gt;&#xA;&lt;li&gt;sleep&lt;/li&gt;&#xA;&lt;li&gt;soft&lt;/li&gt;&#xA;&lt;li&gt;cloth&lt;/li&gt;&#xA;&lt;li&gt;bed&lt;/li&gt;&#xA;&lt;li&gt;under head&lt;/li&gt;&#xA;&lt;li&gt;square&lt;/li&gt;&#xA;&lt;li&gt;pillowcase&lt;/li&gt;&#xA;&lt;li&gt;dreams&lt;/li&gt;&#xA;&lt;li&gt;container [contains feathers]&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;strong&gt;Turn some or that into either metaphors or unclear synonyms.&lt;/strong&gt; Think of ways of bundling some of them, possibly unrelated. Try to have the metaphors related.&lt;/p&gt;&#xA;&lt;p&gt;If you have difficulty coming up with metaphors, see &lt;a href=&quot;https://writers.stackexchange.com/a/7140/4291&quot;&gt;my crude recipe from Writers.SE&lt;/a&gt; although it's always better to craft these by hand.&lt;/p&gt;&#xA;&lt;p&gt;Craft roughly half the required number of them for now. I'm aiming for a four-liner, so two will do.&lt;/p&gt;&#xA;&lt;p&gt;square + container + dreams = &lt;em&gt;a box of dreams&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;feathers - fluffy; taken from birds - &lt;em&gt;fluffy gift of birds&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now is a good moment to &lt;strong&gt;start looking for rhymes.&lt;/strong&gt; Professional poets will wince, but I cheat with rhyming dictionaries. &lt;a href=&quot;http://www.rhymezone.com/&quot; rel=&quot;noreferrer&quot;&gt;RhymeZone&lt;/a&gt; has fewer hits but of higher quality. &lt;a href=&quot;http://www.rhymer.com/&quot; rel=&quot;noreferrer&quot;&gt;Rhymer&lt;/a&gt; finds much more but some of them are quite poor.&lt;/p&gt;&#xA;&lt;p&gt;Dreams:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;1 syllable:&lt;/p&gt;&#xA;&lt;p&gt;beams, beam's, creams, deems, eames, gleams, hiems, nemes, reames, reams, reims, schemes, screams, seams, seems, siems, steams, streams, teams, team's, teams', teems, themes, weems&lt;/p&gt;&#xA;&lt;p&gt;2 syllables:&lt;/p&gt;&#xA;&lt;p&gt;basim's, esteem's, extremes, hakeem's, hakim's, kareem's, pezim's, racemes, redeems, regimes, regime's, supremes&lt;/p&gt;&#xA;&lt;p&gt;3 syllables:&lt;/p&gt;&#xA;&lt;p&gt;joachims, joachim's&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Let's look for something promising. Oh, a pillow has &lt;em&gt;seams&lt;/em&gt;. Now, what can we write about seams? Sometimes the fall apart, sometimes feathers come out through these. We already have a line about feathers, so let's consume that one. Synonyms, extra adjectives, to &lt;strong&gt;pad for rhythm&lt;/strong&gt;, and we have two first very nice lines.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;Colorful box that's full of dreams&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;Soft gifts of birds escape its seams.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Now, since we consumed the 'gifts of birds' we need to resupply our store of metaphors to fit to half of the remaining lines. So, one. Similar noun, difference between them and the metaphor is ready.&lt;/p&gt;&#xA;&lt;p&gt;Sleep - death - 8 hours. &amp;quot;Eight hours of death.&amp;quot; Tell me if that's not a nice metaphor.&lt;/p&gt;&#xA;&lt;p&gt;Another visit to rhymezone and... nothing promising. &amp;quot;Breath-death&amp;quot; is totally ridden into ground. Let's pad with something at the end instead. Straight from the list above, 'head'.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;Eight hours of death under your head&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;1 syllable:&lt;/p&gt;&#xA;&lt;p&gt;bled, bread, bred, dread, dred, dredd, ed, fled, fread, fred, freda, pled, schwed, sffed, shred, sled, sledd, sped, spread, stead, swed, szwed, thread, tread&lt;/p&gt;&#xA;&lt;p&gt;2 syllables:&lt;/p&gt;&#xA;&lt;p&gt;abed, behead, embed, imbed, instead, misled, misread, numed, purebred, retread, sayed, unread, unsaid, unted, unwed, widespread&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;We have one cheap lockpick: 'instead', there's half the universe you can write ending with 'instead'. But let's pick something more ambitious. 'unsaid' has a nice poetic ring, and we should be able to come up with something neat and moody, since we have quite enough actual hints made.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Listens to your confessions unsaid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Nice, but we're one syllable too long. Another modern poet's friend comes in handy: &lt;a href=&quot;http://thesaurus.com&quot; rel=&quot;noreferrer&quot;&gt;Thesaurus.com&lt;/a&gt;. The wrong rhythm is after the caesura, and we don't want to touch the rhyming word for now, so 'confessions' gets the axe.&lt;/p&gt;&#xA;&lt;p&gt;Who'd think, &lt;a href=&quot;http://thesaurus.com/browse/confessions&quot; rel=&quot;noreferrer&quot;&gt;lots of results&lt;/a&gt; but only three 2-syllable synonyms. Still, 'stories' will be acceptable. Let's go with it.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Listens to your stories unsaid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;We can &lt;strong&gt;reshuffle some lines&lt;/strong&gt; - you're rarely at such liberty with other poems, and the new arrangement both gives a nicer rhyme scheme, and composes second and third verse into one sentence:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;Eight hours of death under your head&lt;/em&gt;.&lt;br&gt;&#xA;&lt;em&gt;Colorful box that's full of dreams&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;Listens to your stories unsaid.&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;Soft gifts of birds escape its seams.&lt;/em&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;There.&lt;/p&gt;&#xA;&lt;p&gt;edit:&lt;/p&gt;&#xA;&lt;p&gt;...and then you show it to someone and they suggest some nice corrections. Thanks, Joe Z.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;Eight hours of death. Beneath your head -&lt;/em&gt;.&lt;br&gt;&#xA;&lt;em&gt;a colored box that's full of dreams.&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;It listens to your tales unsaid.&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;Soft gifts of birds escape its seams.&lt;/em&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="117" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-03-10T03:45:13.597" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="119" PostTypeId="2" ParentId="112" CreationDate="2014-05-15T15:44:55.220" Score="15" Body="&lt;p&gt;The best solution I've found has 7 locks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Grand Vizier has keys 1, 2, 3, 4, 5, 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Slave 1 has keys 1, 2, 3, 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Slave 2 has keys 1, 4, 5, 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Slave 3 has keys 2, 4, 6, 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Slave 4 has keys 3, 5, 6, 7.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A couple of things simplify the deduction of this puzzle:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If the Grand Vizier has every key but one, and every slave has the missing key, that reduces the problem to 4 slaves and one less lock.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Every key must be held by at least two slaves.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The rest I just brute forced with a (digital) pencil and paper.&lt;/p&gt;&#xA;" OwnerUserId="18" LastEditorUserId="18" LastEditDate="2014-05-15T16:02:28.853" LastActivityDate="2014-05-15T16:02:28.853" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="121" PostTypeId="2" ParentId="99" CreationDate="2014-05-15T16:09:03.967" Score="2" Body="&lt;p&gt;I think Kendall Frey's solution is the best you can do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original CBC is needed to make sure you started in state 2 (and are still there).&lt;br&gt;&#xA;Then, after the A, you know you are in state 3 or 4 and don't need A again.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do the A earlier, it:&lt;br&gt;&#xA;- &lt;code&gt;25% chance&lt;/code&gt; Solves state 2&lt;br&gt;&#xA;- &lt;code&gt;75% chance&lt;/code&gt; Puts you in state 3 or 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, you didn't know you were in state 2, so it could take you from 3 or 4 to 2 instead.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="30903" LastEditDate="2016-12-20T08:36:47.270" LastActivityDate="2016-12-20T08:36:47.270" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="123" PostTypeId="1" AcceptedAnswerId="127" CreationDate="2014-05-15T16:40:22.120" Score="15" ViewCount="1789" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A magic square (of order 3) is a 3x3 matrix consisting of distinct numbers from 1 to 9, where the numbers in each row, column and diagonal add up to 15.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example, the following would be a magic square:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Xlrc1.png&quot; alt=&quot;A magic square (Source: Wikipedia)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The problem is to construct a magic square that has the position of the number 8 fixed.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mWm4y.png&quot; alt=&quot;A magic square with the position of 8 fixed&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="129" LastEditorUserId="463" LastEditDate="2015-06-05T12:12:49.233" LastActivityDate="2015-06-05T12:12:49.233" Title="Magic square with the position of 8 fixed" Tags="|magic-square|" AnswerCount="4" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="124" PostTypeId="2" ParentId="123" CreationDate="2014-05-15T16:44:51.420" Score="17" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Every normal magic square of order three is obtained from the Lo Shu by rotation or reflection.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From Wikipedia: &lt;a href=&quot;http://en.wikipedia.org/wiki/Lo_Shu_Square&quot;&gt;Lo Shu Square&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since rotation and reflection cannot move the 8 out of a corner, such a square is &lt;strong&gt;impossible&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="18" LastActivityDate="2014-05-15T16:44:51.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="125" PostTypeId="2" ParentId="78" CreationDate="2014-05-15T16:52:46.910" Score="15" Body="&lt;p&gt;The canonical Einstein's Puzzle can be solved using a grid of possibilities, and the easiest way to do so is to track each of the possible states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each letter in this grid stands for the associated entity:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Nationality stands for Enlgishman, Swede, Dane, Norwegian, and German&lt;/li&gt;&#xA;&lt;li&gt;Colors (house color) stand for Blue, Green, Red, White, Yellow &lt;/li&gt;&#xA;&lt;li&gt;Animals stand for Bird, Cat, Dog, Fish, Horse&lt;/li&gt;&#xA;&lt;li&gt;Drinks stand for Bier, Coffee, Milk, Tea, Water&lt;/li&gt;&#xA;&lt;li&gt;Cigarette stands for Blend, Blue Master (M), Dunhill, Pall Mall, Prince (R)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We therefore have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Natl. | ESDNG | ESDNG | ESDNG | ESDNG | ESDNG |&#xA;Color | BGRWY | BGRWY | BGRWY | BGRWY | BGRWY |&#xA;Animl | BCDFH | BCDFH | BCDFH | BCDFH | BCDFH |&#xA;Drink | BCMTW | BCMTW | BCMTW | BCMTW | BCMTW |&#xA;Cigar | BMDPR | BMDPR | BMDPR | BMDPR | BMDPR |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Thus, the clues help you immediately reduce the possibilities in this grid. I'll go through a couple steps to exemplify. I've pulled the rules from &lt;a href=&quot;http://www.stanford.edu/~laurik/fsmbook/examples/Einstein%27sPuzzle.html&quot;&gt;Stanford's page&lt;/a&gt;, and they look correct from memory.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(4) The green house is just to the left of the white one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, because the green house must be to the left of a house, and the white house must be to the right of a house, we know that the green house cannot be on the right, and the white house cannot be on the left.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Natl. | ESDNG | ESDNG | ESDNG | ESDNG | ESDNG |&#xA;Color | BGR Y | BGRWY | BGRWY | BGRWY | B RWY |&#xA;Animl | BCDFH | BCDFH | BCDFH | BCDFH | BCDFH |&#xA;Drink | BCMTW | BCMTW | BCMTW | BCMTW | BCMTW |&#xA;Cigar | BMDPR | BMDPR | BMDPR | BMDPR | BMDPR |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(5) The owner of the green house drinks coffee.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We know that, since the fifth house cannot be green, its owner cannot drink coffee.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Natl. | ESDNG | ESDNG | ESDNG | ESDNG | ESDNG |&#xA;Color | BGR Y | BGRWY | BGRWY | BGRWY | B RWY |&#xA;Animl | BCDFH | BCDFH | BCDFH | BCDFH | BCDFH |&#xA;Drink | BCMTW | BCMTW | BCMTW | BCMTW | B MTW |&#xA;Cigar | BMDPR | BMDPR | BMDPR | BMDPR | BMDPR |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(8) The man in the center house drinks milk.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This one should be obvious.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Natl. | ESDNG | ESDNG | ESDNG | ESDNG | ESDNG |&#xA;Color | BGR Y | BGRWY | BGRWY | BGRWY | B RWY |&#xA;Animl | BCDFH | BCDFH | BCDFH | BCDFH | BCDFH |&#xA;Drink | BC TW | BC TW |   M   | BC TW | B  TW |&#xA;Cigar | BMDPR | BMDPR | BMDPR | BMDPR | BMDPR |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(9) The Norwegian lives in the first house.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once again, this is obvious.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Natl. |    N  | ESD G | ESD G | ESD G | ESD G |&#xA;Color | BGR Y | BGRWY | BGRWY | BGRWY | B RWY |&#xA;Animl | BCDFH | BCDFH | BCDFH | BCDFH | BCDFH |&#xA;Drink | BC TW | BC TW |   M   | BC TW | B  TW |&#xA;Cigar | BMDPR | BMDPR | BMDPR | BMDPR | BMDPR |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(14) The Norwegian lives next to the blue house.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since the Norwegian lives in house 1, we know that the blue house must be house 2, because that is the only house the Norwegian is next to.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Natl. |    N  | ESD G | ESD G | ESD G | ESD G |&#xA;Color |  GR Y | B     |  GRWY |  GRWY |   RWY |&#xA;Animl | BCDFH | BCDFH | BCDFH | BCDFH | BCDFH |&#xA;Drink | BC TW | BC TW |   M   | BC TW | B  TW |&#xA;Cigar | BMDPR | BMDPR | BMDPR | BMDPR | BMDPR |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;It should be immediately obvious why this strategy works.&lt;/strong&gt; You can both reduce clues away as well as insert solutions into this grid, which is vitally important. Without tracking where pieces &lt;em&gt;can't&lt;/em&gt; be, you lose a significant amount of vital information.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorDisplayName="user20" LastEditDate="2014-05-15T17:19:03.997" LastActivityDate="2014-05-15T17:19:03.997" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="126" PostTypeId="2" ParentId="123" CreationDate="2014-05-15T17:07:16.647" Score="8" Body="&lt;p&gt;Here's an intuitive explanation: let's set the squares to the things we know:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;? 8 ?&#xA;? 5 ?&#xA;? ? ?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We can figure out the bottom one now, since &lt;code&gt;15-8-5=2&lt;/code&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;? 8 ?&#xA;? 5 ?&#xA;? 2 ?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The other two numbers on the bottom row now must add up to 13, which can only be 9/4 or 7/6. Now let's split this up into those two cases.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Case 1 (9/4)&lt;/h2&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;? 8 ?&#xA;? 5 ?&#xA;9 2 4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note that the top right square has to be 1, and the top left 6 (using the diagonals).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;6 8 1&#xA;? 5 ?&#xA;9 2 4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now let's use the columns to fill in the rest:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;6 8 1&#xA;1 5 10&#xA;9 2 4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;... notice a problem?&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Case 2 (7/6)&lt;/h2&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;? 8 ?&#xA;? 5 ?&#xA;7 2 6&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Fill in the corners as before:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;4 8 3&#xA;? 5 ?&#xA;7 2 6&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And the edges:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;4 8 3&#xA;4 5 6&#xA;7 2 6&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This one doesn't work either. So, we've proved that this is impossible!&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2014-05-15T17:07:16.647" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="127" PostTypeId="2" ParentId="123" CreationDate="2014-05-15T17:10:41.217" Score="15" Body="&lt;p&gt;As Kendall mentions, it is impossible to move pieces out of a 3x3 grid in this manner. This is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lo_Shu_Square&quot;&gt;Lo Shu square&lt;/a&gt;, and is the only unique solution to the 3x3 magic grid puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for this is actually rather simple. If you look at the Lo Shu square, you'll notice that the even numbers always appear in corners. To demonstrate why this would be, note that each row and column must sum to an odd number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a row contains an even number of odd numbers, then its solution is either $2(2N+1)+2N=6N+2$ or $3(2N)=6N$, both of which are divisible by two, and therefore cannot add to 15. If a row contains an odd number of odd numbers, then its solution is either $1(2N+1)+2(2N)=6N+1$ or $3(2N+1)=6N+3$, both of which are odd and therefore can equal 15.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that each row must contain and odd number of odd numbers. In a 3x3 grid, a row must contain either 1 or 3 odd numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the center contains an odd number, then two adjacent edges must contain an odd number. However, if two edges contain even numbers, then the corner between them must contain an even. However, if this is the case, the opposing diagonal must also contain two even numbers - but at this point, if it does, then a row contains three even numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the following grid, it is clear that the question marks indicate positions that must be, but can't be, even.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;----------    ----------    ----------&#xA;|  |  |  |    |  |EE|  |    |EE|EE|??|&#xA;----------    ----------    ----------&#xA;|  |EE|  | -&amp;gt; |EE|EE|  | -&amp;gt; |EE|EE|  |&#xA;----------    ----------    ----------&#xA;|  |  |  |    |  |  |  |    |??|  |  |&#xA;----------    ----------    ----------&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It is trivial from this point to demonstrate that if there is an even number on an edge, then the grid is unsolvable. It is also trivial to demonstrate that if one even number is in a corner, the rest must be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, every solution is the permutation of one solution: where the even numbers are in the corners; and if the even numbers are in the corners, then their positions are irrelevant, because a solution will always be a reflection of that position.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorDisplayName="user20" LastEditDate="2014-06-15T04:37:02.140" LastActivityDate="2014-06-15T04:37:02.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="128" PostTypeId="1" AcceptedAnswerId="294" CreationDate="2014-05-15T17:26:37.300" Score="39" ViewCount="6064" Body="&lt;p&gt;A group of N pirates has come by a chest containing 200 gold coins.  Their rules require that the coins be distributed by the following approach.  The pirates are ranked from fiercest to meekest and all pirates know the ranking.  The fiercest must propose a division, which is put to majority vote with the fiercest voting and winning ties.  If the proposal wins, it is implemented.  If not, the fiercest is fed to the sharks and the responsibility falls to the (newly promoted) fiercest.  The pirates are selfish and quite rational, with the following priorities:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Stay alive&lt;/li&gt;&#xA;&lt;li&gt;Get as much gold as possible, as long as 1 is satisfied&lt;/li&gt;&#xA;&lt;li&gt;If 1 is satisfied and two options are equivalent in terms of gold, feed somebody to the sharks by preference.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Depending on N, what happens?  To be complete, your answer should deal with large values of N.  Source:  I first saw it in a Martin Gardner column.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="11876" LastEditDate="2016-08-03T22:51:12.160" LastActivityDate="2021-01-05T19:12:47.617" Title="Pirates and gold coins" Tags="|mathematics|logical-deduction|" AnswerCount="3" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="130" PostTypeId="1" CreationDate="2014-05-15T17:57:41.363" Score="3" ViewCount="1011" Body="&lt;p&gt;When solving Sudoku on paper, I sometimes get to the point where I need to resort to guessing or bifurcation. At that point, I might guess one number, and continue solving, changing that guess if I arrive at a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've experimented with a couple of approaches, e.g., circling the guess and writing the consequences more lightly or small in a particular corner of each cell. But none of this works too well, particularly not if I need a second level of guessing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is anyone aware of a reliable notation in this situation? Do I need coloured pencils or transparencies?&lt;/p&gt;&#xA;" OwnerDisplayName="user121" LastEditorUserId="11110" LastEditDate="2018-11-27T19:35:21.747" LastActivityDate="2020-11-11T01:13:03.083" Title="What is a good notation for trial-and-error when solving Sudoku on paper?" Tags="|sudoku|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="131" PostTypeId="1" AcceptedAnswerId="137" CreationDate="2014-05-15T17:58:26.700" Score="17" ViewCount="47382" Body="&lt;p&gt;In the &lt;a href=&quot;http://en.wikipedia.org/wiki/Missionaries_and_cannibals_problem#The_problem&quot;&gt;Missionaries and Cannibals problem&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Three missionaries and three cannibals must cross a river using a boat which can carry at most two people, under the constraint that, for both banks and the boat, if there are missionaries present on the bank (or the boat), they cannot be outnumbered by cannibals (if they were, the cannibals would eat the missionaries). The boat cannot cross the river by itself with no people on board.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Missionaries_and_cannibals_problem#Solution&quot;&gt;The shortest solution&lt;/a&gt; for this puzzle has 11 one-way trips.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What should be the strategy to solve this puzzle for M missionaries and C cannibals&lt;/strong&gt;, given that M is not less than C, or else the puzzle wouldn't be solvable.&lt;/p&gt;&#xA;" OwnerUserId="129" LastActivityDate="2021-05-12T14:52:07.477" Title="Strategy to solve the Missionaries and Cannibals problem" Tags="|strategy|river-crossing|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="132" PostTypeId="2" ParentId="128" CreationDate="2014-05-15T18:20:35.570" Score="10" Body="&lt;p&gt;I believe pirate 1 (fiercest) should give every odd-numbered pirate (3rd fiercest, 5th, etc.) 1 gold, starting from the least-fierce end, and take the remainder (if any) for himself. This has the unintuitive effect that if there are more than 400 pirates, the only way for the fiercest to survive is to take none for himself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start with small N and use inference:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N = 2: 1 takes all gold (200) for himself. He votes yes, 2 votes no, 1, being fiercest, breaks tie and takes all the gold&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N = 3: 1 needs one vote for him. If he is killed, 3 gets nothing, so if he gives 3 1 piece, 3 votes for him, he wins 2-1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N = 4: 1 just needs one other vote (and he breaks the tie by being fiercest). If he is killed, 3 gets nothing (as he becomes 2 in the previous case)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N = 5: 1 needs two votes, so he gives 1g to the two who would get nothing if he were killed, the current 3 and 5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(somewhat) graphically:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;N = 2:&#xA;        1 = 200&#xA;        2 = 0&#xA;N = 3:&#xA;        1 = 199&#xA;        2 = 0&#xA;        3 = 1&#xA;N = 4:&#xA;        1 = 199&#xA;        2 = 0&#xA;        3 = 1&#xA;        4 = 0&#xA;N = 5:&#xA;        1 = 198&#xA;        2 = 0&#xA;        3 = 1&#xA;        4 = 0&#xA;        5 = 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So we see that the fiercest pirate gives 1 gold to those who would get nothing for N-1. These happen to always be the odd-numbered pirates. Once the money runs out (N &gt; 400), only the (odd) least fierce pirates get paid, since if they did not, enough would object to kill the &lt;/p&gt;&#xA;" OwnerUserId="4" LastActivityDate="2014-05-15T18:20:35.570" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="133" PostTypeId="1" AcceptedAnswerId="162" CreationDate="2014-05-15T18:34:58.380" Score="13" ViewCount="1091" Body="&lt;p&gt;In &lt;a href=&quot;https://puzzling.stackexchange.com/questions/112/the-puzzle-of-the-locked-box&quot;&gt;this question&lt;/a&gt;, I presented a puzzle involving seven locks, with a Grand Vizier and four slaves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I would like to use this style of puzzle in some of my own secret-unlocking riddles, in order to require that a person solve a specific number of problems in order to proceed, without regard for what order they solve them in or which ones they decide to solve. To do this, I make each riddle reveal a specific section of a secret key upon being solved, which is analogous to getting one person to use all his keys to unlock some of the locks on the box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I have two level A riddles, four level B riddles, and six level C&#xA;  riddles, and I want the challenger to pass if and only if he's solved&#xA;  at least one of the following sets:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;One level A riddle and two level B riddles.&lt;/li&gt;&#xA;  &lt;li&gt;One level A riddle, one level B riddle, and two level C riddles.&lt;/li&gt;&#xA;  &lt;li&gt;Two level B riddles and three level C riddles.&lt;/li&gt;&#xA;  &lt;li&gt;All six level C riddles.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How do I go about finding a configuration of keys that would make these conditions (or any other conditions I might care to devise) hold?&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2016-06-04T06:43:06.233" Title="How do you create a general locked-box problem?" Tags="|puzzle-creation|keys-and-locks|" AnswerCount="5" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="134" PostTypeId="2" ParentId="123" CreationDate="2014-05-15T19:10:54.350" Score="9" Body="&lt;p&gt;According to Kendall and others, the puzzle isn't solvable for integers. This can be proved using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lo_Shu_Square&quot;&gt;Lo Shu Square&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the problem statement says &lt;em&gt;&quot;A magic square (of order 3) is a 3x3 matrix consisting of &lt;strong&gt;distinct numbers&lt;/strong&gt; from 1 to 9&quot;&lt;/em&gt;, and doesn't explicitly specify integers, a solution with &lt;em&gt;&lt;strong&gt;real numbers&lt;/strong&gt;&lt;/em&gt; could be obtained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's one:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;+---+---+---+&#xA;|4.5| 8 |2.5|&#xA;+---+---+---+&#xA;| 3 | 7 | 5 |&#xA;+---+---+---+&#xA;|7.5| 2 |5.5|&#xA;+---+---+---+&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'd have used complex numbers, but the condition &lt;em&gt;&quot;numbers must be from 1 to 9&quot;&lt;/em&gt; restricts it to real numbers.&lt;/p&gt;&#xA;" OwnerUserId="129" LastActivityDate="2014-05-15T19:10:54.350" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="135" PostTypeId="2" ParentId="133" CreationDate="2014-05-15T19:56:31.260" Score="1" Body="&lt;p&gt;Each set of keys fundamentally represents a set of possible combinations of choices that need to be an allowed condition.  You add keys to deal with more possible situations, so as long as the number of situations that can occur is enumerable, then it should be possible to make a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in the original problem, you have the vizier with each of 4 slaves (4 cases) and 4 different cases of slaves (1,2,3), (1,2,4), (1,3,4) and (2,3,4).  It may be somewhat non-trivial to determine which keys go where, but if the possible cases that need to work can be enumerated, then the keys can be assigned.&lt;/p&gt;&#xA;" OwnerUserId="65" LastEditorUserId="65" LastEditDate="2014-05-16T01:49:39.730" LastActivityDate="2014-05-16T01:49:39.730" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="136" PostTypeId="2" ParentId="131" CreationDate="2014-05-15T20:01:17.243" Score="2" Body="&lt;p&gt;For the case of more M than C it becomes trivial as a canibal can run the ferry after some initial setup.  It starts with 1 M and 1C in the boat with C+1 M on the original shore.  The M takes the boat back and gets another M, leaving C Ms on the original shore, The Ms get to the other side and send the C back with the boat.  The C can now ferry alternating C's and Ms without issue.  The Ms will always outnumber the number left on either shore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If M is 2 or more higher than C, then M can actually run the boat as well since they can start with 2 Ms in the boat and keep a higher number on both shores simultaneously while also operating the boat.&lt;/p&gt;&#xA;" OwnerUserId="65" LastEditorUserId="65" LastEditDate="2014-05-15T20:20:46.187" LastActivityDate="2014-05-15T20:20:46.187" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="137" PostTypeId="2" ParentId="131" CreationDate="2014-05-15T20:15:46.377" Score="13" Body="&lt;p&gt;Let's define &lt;code&gt;a b * c d&lt;/code&gt; to represent the state of the river at any given time: &lt;code&gt;a&lt;/code&gt; missionaries and &lt;code&gt;b&lt;/code&gt; cannibals on the left, &lt;code&gt;c&lt;/code&gt; missionaries and &lt;code&gt;d&lt;/code&gt; cannibals on the right, and &lt;code&gt;*&lt;/code&gt; being &lt;code&gt;&amp;lt;&lt;/code&gt; if the boat is on the left and &lt;code&gt;&amp;gt;&lt;/code&gt; if the boat is on the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem starts out in the state &lt;code&gt;M C &amp;lt; 0 0&lt;/code&gt;, and we want to get &lt;code&gt;0 0 &amp;gt; M C&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the case of &lt;code&gt;M&lt;/code&gt; being more than &lt;code&gt;C&lt;/code&gt;, here's an algorithm to transfer 1 missionary and 1 cannibal at a time:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Bring 1 missionary and 1 cannibal over. (&lt;code&gt;M-1 C-1 &amp;gt; 1 1&lt;/code&gt;)&lt;/li&gt;&#xA;&lt;li&gt;Bring the &lt;em&gt;cannibal&lt;/em&gt; back. (&lt;code&gt;M-1 C &amp;lt; 1 0&lt;/code&gt;; since &lt;code&gt;M &amp;gt; C&lt;/code&gt;, &lt;code&gt;M-1 &amp;gt;= C&lt;/code&gt;, as required.) &lt;/li&gt;&#xA;&lt;li&gt;Bring 1 missionary and 1 cannibal over again. (&lt;code&gt;M-2 C-1 &amp;gt; 2 1&lt;/code&gt;)&lt;/li&gt;&#xA;&lt;li&gt;Bring a missionary back. (&lt;code&gt;M-1 C-1 &amp;lt; 1 1&lt;/code&gt;).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And then you're left with the case of &lt;code&gt;M-1&lt;/code&gt; missionaries and &lt;code&gt;C-1&lt;/code&gt; cannibals on one side of the river. Just keep repeating this procedure (it works by induction) until you have only missionaries on the left side of the river. Then you're home free.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &lt;code&gt;M&lt;/code&gt; equal to &lt;code&gt;C&lt;/code&gt;, I don't think you'll have a solution for &lt;code&gt;M &amp;gt; 3&lt;/code&gt;, because the solution for &lt;code&gt;M = 3&lt;/code&gt; already depends on the fact that you can have &lt;em&gt;only&lt;/em&gt; cannibals on one side and they won't be able to do anything to the missionaries no matter how many there are on one side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The part of the solution where this applies looks something like this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;code&gt;3 1 &amp;lt; 0 2&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;1 1 &amp;gt; 2 2&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;2 2 &amp;lt; 1 1&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;0 2 &amp;gt; 3 1&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There's no way to get a similar arrangement for anything greater than three, because you can't send enough missionaries at once to balance out the rest of the cannibals.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-09-03T19:35:37.163" LastActivityDate="2015-09-03T19:35:37.163" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="138" PostTypeId="2" ParentId="130" CreationDate="2014-05-15T20:20:14.953" Score="3" Body="&lt;p&gt;You could decide to partition the corners and use one for each level, writing even smaller numbers. Or tilting the numbers at different angles. But that sounds boring, why not use Arabic, Roman and Persian/Farsi (oops, there is some digit overlap) numerals at each level?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I consider sudoku puzzles that legitimately require trial and error unpure and stop solving them, but this is a topic for &lt;a href=&quot;https://puzzling.stackexchange.com/questions/12/are-there-sudoku-puzzles-published-that-do-require-guessing&quot;&gt;separate questions&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2014-05-15T20:58:52.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="139" PostTypeId="2" ParentId="128" CreationDate="2014-05-15T20:34:37.243" Score="5" Body="&lt;p&gt;Oh my. For large N ( &gt; 400) this gets an exponential structure of strings of carnage that can be stopped at certain points: a simple strategy exists for each pirate of fierceness 400+100*2^n (negative n allowed until we start getting fractional pirates). No optimal strategy exists for the remaining leaders; they die.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start with the first carnage at N&gt;800.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's at most 200 pirates who got any gold, and would prefer to keep it, but they have no say in it, for there's another 600 blood-thirsty jerks who believe they can murder the fiercest with full impunity and have nothing to lose (and either gain a chance for gold, or not gain anything).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The leaders are slaughtered until the total number drops to 800. Then the leader distributes gold to the bottom 600 at random (1 piece per pirate though).&#xA;Each of those on the bottom, who got a gold coin will vote to keep it, because they won't earn more anyway, and have a chance to lose it. Then the top 200 including the tie-breaking boss joins with approval and with 400+tie votes they outvote these out of bottom 600 who didn't get any gold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, assume one of the top 200 is an idiot and votes for sharks at the critical point. 200 satisfied with their coins and 199 currently-victims won't outvote the 400 who want blood (and money). The 400 keep killing until there's only 600 pirates left: 300 without gold, 200 with gold, and 100 potential victims if they don't come to agreement right now: gold distributed at random to 200 out of the bottom 500, the 100 come out alive, 300:300.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, assume an idiot botches the voting. 250 blood-thirsty pirates slaughter 50 leaders, then best 50 of them join the 200 satisfied to vote own survival.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see the pattern?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's try upwards. In our 800-pirate population we have 600 blood-thirsty jerks, who need to be outvoted by 200 with gold and a certain number who just wants to live. Well, that number appears to be 400, bringing the total number to 1200. But if the population is bigger, these 400 know they can kill with impunity (and won't get any gold anyway) so 1200 is the next population threshold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, 25, 50, 100, 200, 400...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;500 = 400+100, 600=400+200, 800=400+400, 1200=400+800&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each pirate of fierceness 400+100*2^n has the opportunity to come out of this penniless, alive, and stop the slaughter, in exchange for distributing all gold to the bottom 3/4 of the band, randomly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each other pirate above these specific ones, I think 400, correct me if I'm wrong, gets fed to the shark no matter what they offer.&lt;/p&gt;&#xA;" OwnerUserId="117" LastActivityDate="2014-05-15T20:34:37.243" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="141" PostTypeId="2" ParentId="78" CreationDate="2014-05-15T20:52:24.073" Score="1" Body="&lt;p&gt;This answer is more about choosing the puzzle than general strategies. Once you have the grid, a lot of reduction tactics are the same as with Sudoku.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Playing around with timed versions of the Einstein puzzle like &lt;a href=&quot;https://www.thinkpenguin.com/gnu-linux/einstein&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; can soon net you shortcuts for solving the grid. Granted, you get everything laid out and can start solving immediately, but our brain is more suited to certain types of cues. Some solved squares may help, but it all depends on its position and the other cues. Particularly the &quot;X is between A and B&quot; hints can be very helpful, especially when they are chained or you can place them immediately. Since most hints are relative, you get several new anchor points to try out. Perhaps their higher apparent benefit is also due to the way we process information, as they usually clear the most amount of possibilities upon placement and the same amount (2) as is-before when uncertain, reducing the problem space. In general, if you can solve any of the near-central squares, that can reduce the positioning options of the triplets significantly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During plays, I've noticed a particular pattern that I can solve fast (as opposed to 15+ minutes). All my best time (2 min) games started like this, but I don't know if the speed can be attributed to just the initial layout. It's when you start with a solved square one square from the border that can be expanded with an in-between cue. It's better than border squares, since you can then also discard some guesses via is-a-neighbour cues, but still gives only one possible orientation.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2014-05-15T20:52:24.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="142" PostTypeId="1" AcceptedAnswerId="146" CreationDate="2014-05-15T21:13:03.947" Score="13" ViewCount="10843" Body="&lt;p&gt;What is a simple and efficient way of removing numbers from a full sudoku puzzle to create one with a single unique solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have already figured out how to generate sudoku grids, like this one:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;5 6 7 2 3 8 1 4 9 &#xA;4 8 9 7 6 1 3 5 2 &#xA;2 3 1 9 5 4 6 8 7 &#xA;1 7 5 6 8 9 2 3 4 &#xA;8 4 3 1 2 5 9 7 6 &#xA;9 2 6 3 4 7 5 1 8 &#xA;3 9 4 8 1 6 7 2 5 &#xA;7 1 8 5 9 2 4 6 3 &#xA;6 5 2 4 7 3 8 9 1 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now, I need to make it an actual puzzle by removing some of the numbers. How can I do this in a simple and efficient way (that can be done by a computer)?&lt;/p&gt;&#xA;" OwnerUserId="39" LastEditorUserId="463" LastEditDate="2015-06-04T15:35:53.513" LastActivityDate="2015-06-06T05:49:47.080" Title="Removing numbers from a full sudoku puzzle to create one with a unique solution" Tags="|puzzle-creation|sudoku|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="143" PostTypeId="1" CreationDate="2014-05-15T21:52:59.997" Score="3" ViewCount="1729" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Leonhard_Euler&quot; rel=&quot;nofollow noreferrer&quot;&gt;Leonhard Euler&lt;/a&gt; proved that it is impossible to solve &lt;a href=&quot;http://en.wikipedia.org/wiki/Seven_Bridges_of_K%C3%B6nigsberg&quot; rel=&quot;nofollow noreferrer&quot;&gt;this puzzle&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/9/91/7_bridges.svg/500px-7_bridges.svg.png&quot; alt=&quot;wikimedia&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The challenge is to take a walk around the area depicted, and cross each bridge (yellow) &lt;em&gt;exactly&lt;/em&gt; once. Using topology, Euler proved that it is impossible to walk each bridge exactly once. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any similar challenges? I would like to try to solve a more complex one.&lt;/p&gt;&#xA;" OwnerUserId="63" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:48.607" LastActivityDate="2015-06-01T18:45:11.343" Title="Are there any topology puzzles similar to The Seven Bridges of Königsberg?" Tags="|topology|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="144" PostTypeId="2" ParentId="143" CreationDate="2014-05-15T22:02:27.050" Score="3" Body="&lt;p&gt;this is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Eulerian_path&quot; rel=&quot;nofollow&quot;&gt;Eulerian path problem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are more than 2 points with an odd number of bridges then the problem is impossible, when here are exactly 0 or 2 such points then it is possible, with 2 odd points you start at one and then end up at the other, with 0 you come back where you start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution for the path is simple, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start at a point with an odd number of bridges and add it to your path and remove it from the set, from there move further by picking any bridge forward, repeat until there are no more bridges at your current point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then at a point you passed and there are still bridges to cross split your path there and go over a bridge until you eventually come back to that point. Reconnect the path with the one you just followed.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2014-05-15T22:02:27.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="145" PostTypeId="2" ParentId="143" CreationDate="2014-05-15T22:02:36.307" Score="5" Body="&lt;p&gt;It would be easy to make an impossible puzzle like this. First, draw some rivers and temporarily number the areas that there are (not required, just used for reference in this answer):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mj05Z.png&quot; alt=&quot;rivers&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, draw some bridges and count how many of the areas have an odd amount of bridges going out:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8DlSK.png&quot; alt=&quot;bridges&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The red numbers are how many bridges there are that come out from that area. Since areas 1, 2, 4, and 6 have an odd number of bridges, this puzzle is complete! This is because &lt;a href=&quot;http://en.wikipedia.org/wiki/Seven_Bridges_of_K%C3%B6nigsberg&quot; rel=&quot;nofollow noreferrer&quot;&gt;there has to be exactly zero or two areas with an odd number of bridges for the walk to be possible&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your question literally, no, there is no &quot;list&quot; or &quot;collection&quot; of these puzzles, since each one would be so similar and solved the same way that it would be pointless. The real puzzle is figuring out &lt;em&gt;why&lt;/em&gt; there has to be zero or two nodes with an odd number of entry/exits (which is also discussed in the Wikipedia article linked above).&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2014-05-15T22:02:36.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="146" PostTypeId="2" ParentId="142" CreationDate="2014-05-15T23:27:14.380" Score="9" Body="&lt;p&gt;Here's how to remove numbers from a full Sudoku puzzle to create one with a single unique solution:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Remove a number,&lt;/li&gt;&#xA;&lt;li&gt;Run a &lt;a href=&quot;http://norvig.com/sudoku.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;solver&lt;/a&gt;, &lt;/li&gt;&#xA;&lt;li&gt;Check if the puzzle is still &lt;a href=&quot;https://stackoverflow.com/questions/5052866/how-to-show-a-sodoku-solution-is-unique&quot;&gt;unique&lt;/a&gt;, &lt;/li&gt;&#xA;&lt;li&gt;Repeat until solver is unable to find a solution, or solution is not unique.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="96" LastEditorUserId="-1" LastEditDate="2017-05-23T12:38:53.617" LastActivityDate="2014-05-15T23:27:14.380" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="147" PostTypeId="2" ParentId="1" CreationDate="2014-05-15T23:34:25.477" Score="18" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Nearly all the Times crossword grids have rotational symmetry: they&#xA;  can be rotated 180 degrees and remain identical. Rarely, puzzles with&#xA;  only vertical or horizontal symmetry can be found; yet rarer are&#xA;  asymmetrical puzzles, usually when an unusual theme requires breaking&#xA;  the symmetry rule. &lt;strong&gt;This rule has been part of the puzzle since the&#xA;  beginning; when asked why, initial editor Margaret Farrar is said to&#xA;  have responded, &quot;Because it is prettier.&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;SOURCE:&lt;/strong&gt; &lt;a href=&quot;http://www.nytimes.com/1992/02/16/magazine/crosswords-fiftyyears.html?ei=5070&amp;amp;en=eb226eeaee1ab061&amp;amp;ex=1237089600&amp;amp;pagewanted=all&quot;&gt;&quot;The Ins and Outs of Across and Down&quot; The New York Times Magazine, 1992-02-16.&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="96" LastActivityDate="2014-05-15T23:34:25.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="148" PostTypeId="2" ParentId="1" CreationDate="2014-05-16T00:28:28.737" Score="8" Body="&lt;p&gt;Quality crossword puzzles have had a &lt;a href=&quot;http://www.cruciverb.com/index.php?action=ezportal;sa=page;p=21&quot;&gt;rotational symmetry requirement&lt;/a&gt; for a very long time. It was determined early on in the development of the puzzle that the symmetry rule (along with the fully checked and all-over interlock rules) made the puzzles more interesting. (IMHO, that is because it puts interesting constraints on the beginnings of words towards the upper left corner, with mirrors of those constraints on the endings of words in the lower right.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The NYT does expressly give the editor freedom to break the rules, &lt;a href=&quot;http://www.cruciverb.com/index.php?action=ezportal;sa=page;p=16&quot;&gt;&quot;if the theme warrants&quot;&lt;/a&gt;. But notably none of my searches for submissions guidelines to the NYT made any more explicit statements than that.&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2014-05-16T00:28:28.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="149" PostTypeId="1" AcceptedAnswerId="150" CreationDate="2014-05-16T01:25:15.783" Score="9" ViewCount="1171" Body="&lt;p&gt;How can I solve 5-10 letter anagrams without a computer or electronics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was playing the Word Scramble category in the game &lt;em&gt;Quiz Up&lt;/em&gt;  and found it complicated, only in some parts. At home, I used an anagram solver, but elsewhere my friends have aced me in this section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I searched on the internet and found &lt;a href=&quot;http://www.wikihow.com/Solve-Anagrams-Effectively&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt;, but it requires either buying something, time, effort, or memorizing anagrams beforehand. &lt;/p&gt;&#xA;" OwnerUserId="135" LastEditorUserId="463" LastEditDate="2015-06-02T02:39:44.687" LastActivityDate="2018-06-22T07:30:58.503" Title="How to solve 5-10 letter anagrams (without a computer or electronics)?" Tags="|anagram|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="150" PostTypeId="2" ParentId="149" CreationDate="2014-05-16T01:47:57.480" Score="29" Body="&lt;ul&gt;&#xA;&lt;li&gt;Look for patterns, such as &lt;code&gt;ing&lt;/code&gt;, &lt;code&gt;tion&lt;/code&gt;, &lt;code&gt;re&lt;/code&gt;/&lt;code&gt;er&lt;/code&gt;, &lt;code&gt;de&lt;/code&gt;/&lt;code&gt;ed&lt;/code&gt;, &lt;code&gt;qu&lt;/code&gt;, &lt;code&gt;est&lt;/code&gt;, &lt;code&gt;un&lt;/code&gt;, &lt;code&gt;th&lt;/code&gt;, etc.&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Remove the letters in these patterns first, then see if you can use the prefix/suffix with the remaining letters. (If you can't, then try another prefix/suffix. Some words just don't have prefixes/suffixes, though.)&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Look at only the consonants, since th hmn brn cn stll ndrstnd mst wrds wtht vwls.&lt;/li&gt;&#xA;&lt;li&gt;Try compound words too (i.e. &lt;code&gt;nodbkoro&lt;/code&gt; -&gt; &lt;code&gt;knob doro&lt;/code&gt; -&gt; &lt;code&gt;knob door&lt;/code&gt; -&gt; &lt;code&gt;doorknob&lt;/code&gt;).&lt;/li&gt;&#xA;&lt;li&gt;If all else fails, consider each letter individually as the first letter of the word, and then quickly scan the letters for the last letter. The hamun biran is aslo sirpulgrisny good at rindaeg wdros wtih tiehr frist and lsat lteetr itnact.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="39" LastEditorUserId="463" LastEditDate="2015-06-02T02:41:37.167" LastActivityDate="2015-06-02T02:41:37.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="151" PostTypeId="2" ParentId="70" CreationDate="2014-05-16T02:21:06.907" Score="-1" Body="&lt;p&gt;Say there are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5 logicians &lt;code&gt;1, 2, 3, 4, 5&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;amp; 5 colors &lt;code&gt;A, B, C, D, E&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the logician, say A, is chosen the leader. Everyone will call out the leader's hat color. The leader will call out the color that is missing from the remaining N-1 hat colors he can see. Two cases are possible:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is no repeating color, then it's easy to see that the leader will call out his own hat's color and they'll win. If there is a repeating color, then further two cases are possible:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the leader's color is repeated, then the logician with the same color as leader will call out his own color and they'll win.&lt;/li&gt;&#xA;&lt;li&gt;If the leader's color is not repeated but someone else's color is, then idk...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Help me to build upon this? This was too long for a comment..&lt;/p&gt;&#xA;" OwnerUserId="62" LastActivityDate="2014-05-16T02:21:06.907" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="152" PostTypeId="1" AcceptedAnswerId="154" CreationDate="2014-05-16T03:01:10.620" Score="9" ViewCount="3999" Body="&lt;p&gt;Here's another oldie from a book, slightly paraphrased to make it more quantitative.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You're in a room with a ceiling exactly 100 feet high. Two thin but sturdy ropes are hanging from fixed hooks at the ceiling and just touch the ground, close enough that you can grab both of them at the same time comfortably. You can climb these ropes, but there is nothing else to climb on. Equipped with only a knife, how much of the rope can you cut down and still end up on the ground unharmed if you can only fall 20 feet without injuring yourself?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2014-05-16T14:36:06.330" LastActivityDate="2016-06-06T19:38:27.563" Title="A rope-cutting problem" Tags="|lateral-thinking|" AnswerCount="5" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="153" PostTypeId="2" ParentId="152" CreationDate="2014-05-16T03:06:55.147" Score="7" Body="&lt;p&gt;You can cut 199 feet of the rope. Climb to the top, cut one rope, and cut all but 1 foot off the other. You don't &lt;em&gt;need&lt;/em&gt; to fall... If you have 1 or 2 feet at the top, you can make a loop to hold on to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to get to the ground, you can cut &lt;strong&gt;&lt;em&gt;200&lt;/em&gt;&lt;/strong&gt; feet. Cut one rope and feed it through the hook, and tie one end around your waist while holding onto the other end. Cut the second rope, and tie it to the first. Then, holding one end, lower yourself down by letting it out.&lt;/p&gt;&#xA;" OwnerUserId="63" LastEditorUserId="63" LastEditDate="2014-05-16T03:20:39.630" LastActivityDate="2014-05-16T03:20:39.630" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="154" PostTypeId="2" ParentId="152" CreationDate="2014-05-16T03:20:59.233" Score="10" Body="&lt;p&gt;All 200 feet, as long as they're no more than 100 feet apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take one end of one rope. Climb up the other. Tighten the first rope as much as possible, cut (or unhook) the one you just climbed up (keep ahold of it) and swing on the intact rope. Climb to the top of that one. Put the cut rope through the hook. Tie yourself a harness of some sort on one end of it, keep hold of the other, and cut/unhook the rope originally there (again keeping hold of it).  Lower yourself down until you are near the end of the rope you are attached to. Tie the spare rope to the end of it, and keep lowering yourself. Once within your safe fall zone, undo the harness and let go that side of the rope, and once you're down pull it through the hook.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it has been clarified that &quot;the hooks are close enough that you can comfortably reach both of them,&quot; the description can be somewhat simplified. Climb up rope A, unhook B, tie it into a harness and re-hook it (such that you can lower yourself), unhook A and continue as above (lower halfway, tie A to the end of B, lower the rest of the way).&lt;/p&gt;&#xA;" OwnerUserId="4" LastEditorUserId="4" LastEditDate="2014-05-16T14:57:09.483" LastActivityDate="2014-05-16T14:57:09.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="155" PostTypeId="2" ParentId="142" CreationDate="2014-05-16T13:12:13.947" Score="5" Body="&lt;p&gt;Strictly speaking, you don't need to do anything because the puzzle already has a unique solution. Let's redefine your goal. Starting with a valid number grid, you want to eliminate numbers to arrive at a puzzle that &lt;em&gt;still&lt;/em&gt; has a unique solution, but also has a specified level of difficulty (e.g., &amp;quot;Easy&amp;quot;, &amp;quot;Medium&amp;quot;, &amp;quot;Hard&amp;quot;, etc.) and (optionally) has a symmetrical arrangement of numbers.&lt;/p&gt;&#xA;&lt;p&gt;To do this, it helps to have a sudoku solver that can also tell you (a) if a puzzle has more than one solution, and (b) how difficult it is to solve. I'm using &lt;a href=&quot;http://www.techfinesse.com/game/sudoku_solver.php&quot; rel=&quot;noreferrer&quot;&gt;Bill DuPree's sudoku solver&lt;/a&gt; here, although you will get faster results using a solver that aborts as soon as it finds more than one solution instead of going on to count them all.&lt;/p&gt;&#xA;&lt;p&gt;As others have suggested, a simple approach is to remove as many numbers as possible from the grid without losing the uniqueness of the solution. You will always get an empty top row if you clear the cells in strict sequential order, so you'll need to check them randomly. This is what I ended up with:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;. 6 7 2 . . . 4 .&#xA;. . 9 . . 1 3 . .&#xA;2 . . . . . . 8 .&#xA;. 7 . . 8 . 2 3 .&#xA;8 . 3 . . . . . .&#xA;. . 6 . 4 . . . .&#xA;. . . . 1 6 7 . 5&#xA;. . . 5 . . . . .  24 numbers&#xA;. . . . . . 8 9 .  Difficulty: ultra-diabolical&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;We can get a symmetrical puzzle by clearing two cells at a time (cells &lt;em&gt;n&lt;/em&gt; and 80-&lt;em&gt;n&lt;/em&gt;), resulting in the following grid:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;. 6 7 . . . . . .&#xA;4 . . . . 1 . 5 .&#xA;2 . . 9 5 4 . . 7&#xA;. . . . 8 . 2 . .&#xA;. . . 1 2 5 . . .&#xA;. . 6 . 4 . . . .&#xA;3 . . 8 1 6 . . 5&#xA;. 1 . 5 . . . . 3  27 numbers&#xA;. . . . . . 8 9 .  Difficulty: easy&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Or, with a different random seed:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;5 . . . 3 8 . . 9&#xA;4 . . . 6 . . . .&#xA;. 3 . . . . 6 . .&#xA;1 . 5 6 . 9 . . 4&#xA;. 4 . . . . . 7 .&#xA;9 . . 3 . 7 5 . 8&#xA;. . 4 . . . . 2 .&#xA;. . . . 9 . . . 3  28 numbers&#xA;6 . . 4 7 . . . 1  Difficulty: ultra-diabolical&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;As you can see, this approach results in puzzles with widely varying difficulty levels. To get a puzzle with a particular difficulty level, you can simply repeat this process with different random numbers until you get a result you like. (Using a &lt;a href=&quot;http://en.wikipedia.org/wiki/Hill_climbing&quot; rel=&quot;noreferrer&quot;&gt;hill climbing&lt;/a&gt; algorithm might make this process more efficient, although I haven't looked into this.)&lt;/p&gt;&#xA;&lt;h2&gt;Another approach&lt;/h2&gt;&#xA;&lt;p&gt;Instead of &lt;em&gt;removing&lt;/em&gt; numbers from a completed grid, you could try &lt;em&gt;adding&lt;/em&gt; numbers to a &lt;a href=&quot;http://web.archive.org/web/20130101205725/http://units.maths.uwa.edu.au/%7Egordon/sudokumin.php&quot; rel=&quot;noreferrer&quot;&gt;minimum sudoku puzzle&lt;/a&gt; (i.e., a puzzle with a unique solution but only 17 filled cells). Even after filling cells to make these puzzles symmetrical, they should still be fairly challenging to solve.&lt;/p&gt;&#xA;" OwnerUserId="40" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-05-16T13:12:13.947" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="156" PostTypeId="2" ParentId="152" CreationDate="2014-05-16T13:45:10.313" Score="3" Body="&lt;p&gt;So far as people have bested the solution I could come up with, I suppose I'll post the one that was in the book as well (again paraphrased to account for some changes I made):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose the ropes are called rope A and rope B. Climb up rope A and cut about 99 feet off of rope B, leaving just enough to tie a loop with the remaining top portion. Tie this loop. Slide back down rope A, and tie one end of the cut portion of rope B to the bottom of this rope. Now, taking the other end, climb back up to the top, and feed that end through the loop until it's close to touching the ground. Then, swing over to the double rope you've just made, and slide rope A off of its hook, letting that end fall to the ground as well. Climb down this double rope which is hanging about 6 inches off the ground, and then pull on one of the ends to get it all down, which is about 199 feet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of course, if you assume that you can feed the rope through the hook like a pulley, as other solutions have done, it also becomes possible to get all 200 feet, by eliminating the &quot;forming a loop&quot; step:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Climb up rope A and take rope B off of its hook. Slide down and tie rope B to rope A. Taking the other end, climb back up the the hook and slide the end in your hand through the hook until it touches the ground. Then, swing over to the double rope, and slide rope A off its hook, letting that end fall to the ground as well. Climb down your double rope, and pull on one of the ends to slide the other end off the hook, getting all 200 feet of the rope down.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I've since found the old puzzle book that this puzzle came from (&lt;a href=&quot;http://www.amazon.ca/Giant-Book-Mensa-Mind-Challenges/dp/0806920939&quot; rel=&quot;nofollow&quot;&gt;Giant Book of Mensa Mind Challenges&lt;/a&gt;, 1998 ed., section 1, puzzle 196). The solution in there is presented as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;First, tie the lower ends. Then climb up the first rope and cut the second one, close to the ceiling, leaving an end long enough to form a loop. (You can instead cut it off entirely if you slide the rope through the hook.) Hanging from the loop, cut the first rope at the hook. &lt;strong&gt;Be careful not to drop it.&lt;/strong&gt; Then slide the rope through the loop until it's even. Climb down the double rope, and once on the floor, pull one end to get it all.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This solution involves grabbing hold of the loop or hook and hanging from it while working with the other rope, which seems a bit dangerous. I prefer my strategy (which is the same except doesn't involve hanging 99.5 feet off the ground) which involves one more up/down trip but never puts you in a situation in which you might conceivably end up stuck hanging near the ceiling if you're not careful.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2014-12-06T18:44:25.683" LastActivityDate="2014-12-06T18:44:25.683" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="157" PostTypeId="1" CreationDate="2014-05-16T14:01:10.993" Score="8" ViewCount="938" Body="&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/100/minimum-number-of-squares-for-a-sudoku-puzzle&quot;&gt;It is known&lt;/a&gt; that the minimum number of clues a sudoku must have to have a unique solution is 17. But on every website I've seen for them, I haven't found any that are rotationally symmetrical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I once saw a puzzle in a book which only had 19 clues and was rotationally symmetrical, but I don't remember whether the solver I ran it through said that there was a unique solution or not. My question is, what is the minimum number of already-filled-in squares that a &lt;em&gt;rotationally symmetrical&lt;/em&gt; Sudoku must have in order to have a unique solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: by rotationally symmetrical, I'm referring to it in the standard sense that if a clue appears in one position, a clue will also appear in the position opposite it on the board.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="69582" LastEditDate="2021-05-11T19:40:27.647" LastActivityDate="2021-05-11T19:40:27.647" Title="What is the fewest number of clues on a rotationally symmetrical sudoku grid with a unique solution?" Tags="|logical-deduction|geometry|grid-deduction|optimization|sudoku|" AnswerCount="1" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="158" PostTypeId="1" CreationDate="2014-05-16T15:14:08.357" Score="11" ViewCount="2733" Body="&lt;p&gt;It is known that there are a minimum of 17 clues needed to create a &lt;a href=&quot;http://en.wikipedia.org/wiki/Mathematics_of_Sudoku#Minimum_number_of_givens&quot;&gt;proper sudoku puzzle&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For how many puzzles is 17 sufficient? From the history of the minimal proof, I gather that there are multiple puzzles for which 17 clues are enough for a unique solution. Is 17 enough to specify every proper sudoku puzzle? Just a class or a few classes of puzzles?&lt;/p&gt;&#xA;" OwnerUserId="4" LastEditorDisplayName="user20" LastEditDate="2014-05-16T17:36:07.850" LastActivityDate="2018-08-14T21:28:32.287" Title="How many minimal-clue sudoku puzzles are there?" Tags="|sudoku|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="161" PostTypeId="2" ParentId="157" CreationDate="2014-05-16T16:17:30.047" Score="10" Body="&lt;p&gt;&lt;a href=&quot;http://books.google.ca/books?id=U6RpAgAAQBAJ&amp;amp;pg=PA111&quot; rel=&quot;noreferrer&quot;&gt;This book&lt;/a&gt; appears to have a puzzle with only 18 clues that is rotationally symmetrical.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;7 2 . | . . . | . . .&#xA;. 5 . | . . 9 | . . .&#xA;. . . | . 3 8 | . . .&#xA;------+-------+-------&#xA;. . . | 4 . . | 5 . .&#xA;. . 3 | . . . | 9 . .&#xA;. . 1 | . . 3 | . . .&#xA;------+-------+-------&#xA;. . . | 2 5 . | . . .&#xA;. . . | 6 . . | . 3 .&#xA;. . . | . . . | . 1 9&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But I still don't know if a rotationally symmetrical 17 exists.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="15544" LastEditDate="2015-09-13T10:11:24.270" LastActivityDate="2015-09-13T10:11:24.270" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="162" PostTypeId="2" ParentId="133" CreationDate="2014-05-16T16:25:23.240" Score="6" Body="&lt;p&gt;In order to construct the solution to this problem in the general space, it's helpful to ignore the Grand Vizier. The Grand Vizier can hold all the keys except one, which all the slaves hold, and that answers that component minimally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have discovered a generic method for generating solutions to these puzzles given arbitrary values. The general method relies on some stuff which I develop in the beginning, but if you want to just go ahead and read that, it's below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;tl;dr? This is how you do it: list out all the possible combinations of ones and zeros for $(N-R+1)$ ones, filling in the rest with zeros. Assign each unique combination to its own key, and boom! Generic solution matrix!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To cover the even more generic solution you've requested would take an insanely long document, and would probably best just be written as a paper. I'm not going to do that here, but I will generally outline how this might work.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A Method for $N$ or $N-1$ slaves&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will demonstrate the method for $N$ of $N$ slaves, and $N-1$ of $N$ slaves, as well as a generic method below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is $N$, then trivially give each slave a key - then they must all be together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is $N-1$, then things become a little trickier. Construct a table with N rows and columns (including the first one, which is numbered):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;_______________&#xA;S1|  |  |  |  |&#xA;S2|  |  |  |  |&#xA;S3|  |  |  |  |&#xA;S4|  |  |  |  |&#xA;S5|  |  |  |  |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In this example, there are five slaves, numbered. Give the first two slaves a key:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;_______________&#xA;S1| 1|  |  |  |&#xA;S2| 1|  |  |  |&#xA;S3|  |  |  |  |&#xA;S4|  |  |  |  |&#xA;S5|  |  |  |  |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then, for each remaining space in the first column, create a new key and copy it into the corresponding space in the first row:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;_______________&#xA;S1| 1| 2| 3| 4|&#xA;S2| 1|  |  |  |&#xA;S3| 2|  |  |  |&#xA;S4| 3|  |  |  |&#xA;S5| 4|  |  |  |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then, insert new keys into the second column:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;_______________&#xA;S1| 1| 2| 3| 4|&#xA;S2| 1| 5|  |  |&#xA;S3| 2| 5|  |  |&#xA;S4| 3|  |  |  |&#xA;S5| 4|  |  |  |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Repeat the above until you have filled in the grid:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;_______________&#xA;S1| 1| 2| 3| 4|&#xA;S2| 1| 5| 6| 7|&#xA;S3| 2| 5| 8| 9|&#xA;S4| 3| 6| 8|10|&#xA;S5| 4| 7| 9|10|&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This grid requires $N-1$ slaves to open the box. You also need to remember to give them the Vizier's key.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your original problem's filled matrix is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;____________&#xA;S1| 1| 2| 3|&#xA;S2| 1| 4| 5|&#xA;S3| 2| 4| 6|&#xA;S4| 3| 5| 6|&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In general, if you want $N-1$ slaves to open the box, you must have $N*(N-1)/2$ keys. For four slaves, this is $4*3/2=12$; for $5$ slaves, this is $5*4/2=10$. Add one key if you want a Vizier.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Generic Method&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, let's go back and change the way we've been representing this problem. Let's instead create a matrix of keys and slaves, and use 1s and 0s to represent which slaves hold which keys.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;4/4 Required  3/4 Required      2/4 Required &#xA;------------  ----------------  ------------&#xA;Key|1|2|3|4|  Key|1|2|3|4|5|6|  Key|1|2|3|4|&#xA;------------  ----------------  ------------&#xA; S1|1|0|0|0|   S1|1|1|1|0|0|0|   S1|1|1|1|0|&#xA; S2|0|1|0|0|   S2|1|0|0|1|1|0|   S2|1|1|0|1|&#xA; S3|0|0|1|0|   S3|0|1|0|1|0|1|   S3|1|0|1|1|&#xA; S4|0|0|0|1|   S4|0|0|1|0|1|1|   S4|0|1|1|1|&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The question we have to ask is: why these particular combinations? If you notice, all columns in the 4/4 matrix have one 1; all in the 3/4 have 2; all in the 2/4 have 3; and trivially, all in the 1/4 have 4. This points to a formula: there exist $(N-R+1)$ copies of each key, where $N$ is the number of slaves, and R is the number required. I have yet to prove this, but it appears to be consistent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, as a curiosity, that 4/4 and 2/4 are effectively the &quot;inverses&quot; of each other (i.e. 1 swaps with 0). With $5x5$, I'll demonstrate (but again, unfortunately, not prove) that the solutions for problems whose distance from the center of $(2,N)$ (which is $(N+2)/2)$ is equivalent are inverses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the 5x5 matrices:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;5/5 Required    4/5 Required               3/5 Required               2/5 Required&#xA;--------------  -------------------------  -------------------------  -------------&#xA;Key|1|2|3|4|5|  Key|1|2|3|4|5|6|7|8|9|10|  Key|1|2|3|4|5|6|7|8|9|10|  Key|1|2|3|4|5|&#xA;--------------  -------------------------  -------------------------  --------------&#xA; S1|1|0|0|0|0|   S1|1|1|1|1|0|0|0|0|0| 0|   S1|1|1|1|1|1|1|0|0|0| 0|   S1|1|1|1|1|0|&#xA; S2|0|1|0|0|0|   S2|1|0|0|0|1|1|1|0|0| 0|   S2|1|1|1|0|0|0|1|1|1| 0|   S2|1|1|1|0|1|&#xA; S3|0|0|1|0|0|   S3|0|1|0|0|1|0|0|1|1| 0|   S3|1|0|0|1|1|0|1|1|0| 1|   S3|1|1|0|1|1|&#xA; S4|0|0|0|1|0|   S4|0|0|1|0|0|1|0|1|0| 1|   S4|0|1|0|1|0|1|1|0|1| 1|   S4|1|0|1|1|1|&#xA; S5|0|0|0|0|1|   S5|0|0|0|1|0|0|1|0|1| 1|   S5|0|0|1|0|1|1|0|1|1| 1|   S5|0|1|1|1|1|&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;No, I won't make you suffer through 6x6 matrices, but these are worth a good inspection. Note that 4/5 and 3/5 are equivalent. $(N+2)/2$ for these is 3.5; 4 and 3 are equidistant and are inverses; so are 5 and 2. But this is an aside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So, this is ridiculous and all - what's the pattern here?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, take a look again! The pattern is that the distribution of keys &lt;strong&gt;matches all possible combinations of ones and zeros for that number of keys.&lt;/strong&gt; For instance, the 4/5 matrix has two keys. All of the columns of that matrix are the unique combinations of those numbers in this space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of keys? That's just the factorial divided by duplicates! You have five entries in 4/5, so you have $5!$ combinations of keys - but you have to take out $3!$ for the three zeros, and $2!$ for the two ones, which gives you $5!/(3!*2!) = 10 \mbox{ keys}$. For 3/4, it's $4!/(2!*2!) = 6\mbox{ keys}$. For 3/11, there are $11!/(9!*2!) = 55\mbox{ keys}$. For 2/40, there are $40!/(38!*2!) = 780\mbox{ keys}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's now possible to prove why equidistant matrices would be pseudo-inverses of each other, but the proof for this part is left to the reader. $N$ slaves, $K$ keys - $N!/(K!(N-K)!)$ &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How do I make a solution matrix, then?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;List out all the possible combinations of ones and zeros for $(N-R+1)$ ones, filling in the rest with zeros. Assign each unique combination to its own key, and boom! Generic solution matrix!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But what about &lt;em&gt;my&lt;/em&gt; problem?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your problem gives three different keyrings: the A-class, B-class, and C-class keyrings. In order to specify each set of keyrings, you need to append an additional set of restricting keys. These keys will be literally equal to the matches required for keys in each of the combinations. You can create an additional matrix to append to the problem to represent these combinations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each class's matrix is generated based on the number of that class required; everything else gets keys for that class. Then, remove sets of keys which are equivalently distributed to each participant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The example you have provided is, quite frankly, too long to give a solution for in this answer. However, using this method, you can generate however many combinations and conditions you would like.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorUserId="16717" LastEditDate="2016-06-04T06:43:06.233" LastActivityDate="2016-06-04T06:43:06.233" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="163" PostTypeId="1" AcceptedAnswerId="164" CreationDate="2014-05-16T16:37:33.620" Score="14" ViewCount="99709" Body="&lt;p&gt;If you use the Beginners' Method to solve the Rubik's cube, and solve the equator layer (the layer between the U-layer and D-layer), it can happen that one of the edges is flipped, like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/U1npR.png&quot; alt=&quot;Flipped edge in Equator layer&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Screenshot generated using Cube Explorer&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A way to solve this is to replace the flipped edge by a random block and then put the edge at its right place. However, I would prefer a shorter way to do this. It does not matter what happens to the top layer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I do this?&lt;/p&gt;&#xA;" OwnerUserId="153" LastActivityDate="2020-01-26T15:08:50.197" Title="Flip edge in Equator layer" Tags="|rubiks-cube|" AnswerCount="9" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="164" PostTypeId="2" ParentId="163" CreationDate="2014-05-16T16:50:48.400" Score="6" Body="&lt;p&gt;An intuitive solution is to remove both the corner and edge, splitting them up, then move them back in an intelligent way. This can be done with:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;R U R' U' R U' R'&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;then by solving again. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, you could just memorize an algorithm. A couple easy solutions:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(R2 U2) (F R2 F') (U2 R' U R')&#xA;(R' F R F') (R U' R' U) (R U' R' U2) (R U' R')&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It's also a good idea to actively avoid being stuck in this situation whenever possible.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorDisplayName="user20" LastEditDate="2014-05-17T03:53:35.177" LastActivityDate="2014-05-17T03:53:35.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="166" PostTypeId="1" AcceptedAnswerId="167" CreationDate="2014-05-16T17:11:51.773" Score="44" ViewCount="7699" Body="&lt;p&gt;You have 2 identical ropes which burn at a specific rate, and an unlimited supply of matches. When you light one end of a rope, the fire will take exactly 1 hour to travel to the other end of the rope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to measure exactly 45 minutes. You must start by lighting one or both of the ropes. You can light or extinguish either end of either rope later, but you must only do this immediately after a rope has finished burning, as this is the only accurate way to measure elapsed time. You also must not light anywhere but the end of a rope, or any other form of guessing. You may finish with or without any remaining rope.&lt;/p&gt;&#xA;" OwnerUserId="18" LastEditorUserId="45524" LastEditDate="2018-10-16T02:58:01.970" LastActivityDate="2020-10-02T06:12:14.143" Title="Burning ropes as timers" Tags="|logical-deduction|time|" AnswerCount="6" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="167" PostTypeId="2" ParentId="166" CreationDate="2014-05-16T17:22:11.937" Score="59" Body="&lt;p&gt;Hint: You can light both sides of one rope. Solution: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Light rope $A$ on both sides so that the rope will be gone in $30$ minutes. You need to light rope $B$ at the same time you light rope $A$. Once rope $A$ is gone, light the other side of rope $B$. Rope $B$ will be gone after another $15$ minutes. That will add up to $45$ minutes. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="146" LastEditorUserId="146" LastEditDate="2014-06-29T08:49:53.360" LastActivityDate="2014-06-29T08:49:53.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="170" PostTypeId="1" CreationDate="2014-05-16T17:52:38.863" Score="13" ViewCount="3771" Body="&lt;p&gt;&lt;sup&gt;&lt;strong&gt;Note&lt;/strong&gt;: This is a follow-up question to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/166/burning-ropes-as-timers&quot;&gt;Burning ropes as timers&lt;/a&gt;. The following question and its answers may contain spoilers.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To sum up the puzzle, these are the rules (slightly modified):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You have some number &lt;em&gt;N&lt;/em&gt; of ropes, each with the following property.&lt;/li&gt;&#xA;&lt;li&gt;When you light one end of a rope, the fire will reach the other end after exactly one hour.&lt;/li&gt;&#xA;&lt;li&gt;You start the puzzle by lighting one or more ends of the ropes.&lt;/li&gt;&#xA;&lt;li&gt;You may light or extinguish any end of a rope after that.&lt;/li&gt;&#xA;&lt;li&gt;You can only light or extinguish a rope once another rope has completely burned out.&lt;/li&gt;&#xA;&lt;li&gt;You may only light the end of a rope, since lighting anywhere else is inaccurate.&lt;/li&gt;&#xA;&lt;li&gt;You may not organize the ropes so that they light each other or themselves.&lt;/li&gt;&#xA;&lt;li&gt;You must accurately measure some interval of time, between any two distinct times.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The question is, how many intervals of time (excluding 0) is it possible to measure with &lt;em&gt;N&lt;/em&gt; ropes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, when &lt;em&gt;N = 1&lt;/em&gt;, the answer is 2.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1 hour (by burning the length of the rope)&lt;/li&gt;&#xA;&lt;li&gt;1/2 hour (by lighting both ends of the rope at once)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="18" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-09-04T10:01:49.860" Title="Burning ropes as timers - How many time intervals can be measured?" Tags="|puzzle-creation|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="175" PostTypeId="5" CreationDate="2014-05-16T20:00:10.833" Score="0" Body="&lt;p&gt;2048 is a sliding block puzzle game which is played by combining adjacent matching number tiles to form a single tile bearing the sum. &#xA;2048 is played on a simple gray 4×4 grid, with numbered tiles that slide smoothly when a player moves them using the four arrow keys.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In every turn, a new tile randomly appears in an empty spot on the board with a value of either 2 or 4.&lt;/li&gt;&#xA;&lt;li&gt;Tiles slide as far as possible in the chosen direction until they are stopped by either another tile or the edge of the grid. &lt;/li&gt;&#xA;&lt;li&gt;If two tiles of the same number collide while moving, they will merge into a tile with the total value of the two tiles that collided. &#xA;The resulting tile cannot merge with another tile again in the same move. &lt;/li&gt;&#xA;&lt;li&gt;Creating a 2048 tile in this manner wins the game.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The original browser-based version was created by the 19-year-old Italian web developer &lt;em&gt;Gabriele Cirulli&lt;/em&gt; and can &lt;a href=&quot;http://gabrielecirulli.github.io/2048/&quot; rel=&quot;nofollow&quot;&gt;be played on the github page&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="52" LastEditorUserId="8874" LastEditDate="2015-02-06T18:11:35.943" LastActivityDate="2015-02-06T18:11:35.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="176" PostTypeId="4" CreationDate="2014-05-16T20:00:10.833" Score="0" Body="2048 is a sliding block puzzle game in which the object is to combine matching number tiles until a 2048 tile is created." OwnerUserId="52" LastEditorUserId="52" LastEditDate="2014-05-19T15:58:49.440" LastActivityDate="2014-05-19T15:58:49.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="177" PostTypeId="5" CreationDate="2014-05-16T20:01:11.113" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-05-16T20:01:11.113" LastActivityDate="2014-05-16T20:01:11.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="178" PostTypeId="4" CreationDate="2014-05-16T20:01:11.113" Score="1" Body="Questions asking for tips rather than specific solutions to a puzzle." OwnerUserId="52" LastEditorUserId="52" LastEditDate="2014-05-19T15:57:58.170" LastActivityDate="2014-05-19T15:57:58.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="180" PostTypeId="1" AcceptedAnswerId="181" CreationDate="2014-05-16T20:14:06.110" Score="9" ViewCount="621" Body="&lt;p&gt;Suppose you have a checkerboard with two opposite corner squares removed, like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AEqI3.gif&quot; alt=&quot;mutilated chessboard&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you move a pawn on the board only horizontally and vertically, one square at a time, and have it touch every square exactly once?&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="39" LastEditDate="2014-05-20T22:57:01.360" LastActivityDate="2014-05-20T22:57:01.360" Title="Moving a pawn around on a chessboard" Tags="|checkerboard|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="181" PostTypeId="2" ParentId="180" CreationDate="2014-05-16T21:03:10.800" Score="22" Body="&lt;h2&gt;No.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Every move the pawn makes it switches from a white to a black square or vice versa. Therefore it must either touch an equal number of white and black squares with an even number of moves, or one more square of either color with an odd number of moves. (A move including initially placing it on the board.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because there are two more white squares than black squares (32 and 30) there is no way it can touch each square exactly once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same argument applies to show that the Knight's Tour cannot be completed on this board either.&lt;/p&gt;&#xA;" OwnerUserId="18" LastActivityDate="2014-05-16T21:03:10.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="182" PostTypeId="2" ParentId="170" CreationDate="2014-05-16T22:27:28.717" Score="12" Body="&lt;p&gt;To get started on this problem, here is a general observation for any $n$: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You can only measure intervals of time of the form&#xA;  $$\frac{m}{2^n}\cdot 1\mbox{h}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where $m$ and $n$ are non-negative integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof (using the &lt;a href=&quot;http://www.artofproblemsolving.com/Wiki/index.php/Extreme_principle&quot; rel=&quot;nofollow&quot;&gt;extremal principle&lt;/a&gt;)&lt;/strong&gt;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose there was a number of ropes you could use to measure an interval which is not of the proposed form above. Take the minimum number of ropes $n$ for which this is possible, then $n≥2$ (as for $n=1$ it is not possible as shown in the question). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There must be a way of lighting the $n$ ropes so that the interval $\ t = t_2-t_1$ between two &quot;distinct times&quot;&lt;sup&gt;&lt;strong&gt;1&lt;/strong&gt;&lt;/sup&gt;  $\ t_1$ and $t_2$, where $t_1&amp;lt;t_2$, cannot be described by the above form. As $t_2&amp;gt;0$, there must be a rope $r$ that has completely burned down (otherwise, $t_2$ would not be a distinct time - see the footnote). Due to $t_1&amp;lt;t_2$, you can measure $t_1$ without using $r$ (so $n-1$ ropes suffice), which means that $t_1$ has the form above due to the minimal choice of $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the very last time $t_3$ rope $r$ is lit. By the same argument as above (you do not need $r$ to measure this time), it follows that $t_3$ has the above form. Accordingly, $\left|t_3-t_1\right|$ is also of that form. As $t = t_2-t_1 = (t_2-t_3)+(t_3-t_1)$ is by assumption not of the above form  (but $\left|t_3-t_1\right|$ is), it follows that $t_2-t_3$ is not of the form above either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this cannot be true: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at amounts of time $t_{r,1},\ldots,t_{r,l}$ that $r$ actually burns. These satisfy the relation $$n_1\cdot t_{r,1} + \ldots + n_1\cdot t_{r,l} = 1\mbox{h}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $n_i = 1,2$, depending on whether $r$ burns at one or both ends. All the amounts of time $t_{r,1},\ldots,t_{r,l-1}$ are of the above form, because they are measured without using $r$. So $t_{r,l}$ also has to be of the form above. But $t_{r,l} = t_2-t_3$, which is a contradiction to the conclusion of the last paragraph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the assumption that there is a number of ropes which can be used to measure an amount of time not of the form above has to be wrong.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt; 1. The start of the experiment or the time a rope has burned out, as there are no other possible events.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="12" LastEditorUserId="15544" LastEditDate="2015-09-04T10:01:49.860" LastActivityDate="2015-09-04T10:01:49.860" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="183" PostTypeId="1" AcceptedAnswerId="185" CreationDate="2014-05-16T22:58:13.827" Score="72" ViewCount="132455" Body="&lt;p&gt;You are given twelve identical-looking balls and a two-sided scale. One of the balls is of a different weight, although you don't know whether it's lighter or heavier. How can you use just three weighings of the scale to determine not only what the different ball is, but also whether it's lighter or heavier?&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="8874" LastEditDate="2015-01-31T12:27:32.677" LastActivityDate="2022-07-07T20:08:07.563" Title="Twelve balls and a scale" Tags="|strategy|combinatorics|weighing|" AnswerCount="7" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="184" PostTypeId="2" ParentId="163" CreationDate="2014-05-17T03:47:42.037" Score="0" Body="&lt;p&gt;If you have an open slot on the left like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TEX1Y.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TEX1Y.png&quot; alt=&quot;open slot on left&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can use this algorithm:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(D (R U R') d'(L' U L) D')&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This algorithm is especially versatile since it can be modified to accommodate an empty slot in any other position by changing the first and last &lt;code&gt;D&lt;/code&gt;s. Such as with &lt;code&gt;D2&lt;/code&gt; for the back-left slot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you want something that doesn't touch the F2L at all AND is quick to execute, then you'll have a harder time finding them. (Generally speaking, the heavier the constraints, the worse the algorithms will be.) My advice is that if you don't have any empty slots, then you might as well continue doing what you're doing. Or you can use the slightly longer algorithm given by @Emrakul.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way, once you learn a more advanced method, the solution to this problem would be to avoid being backed into a corner with this case at all. (:&lt;/p&gt;&#xA;" OwnerUserId="157" LastEditorUserId="5373" LastEditDate="2017-03-11T00:39:01.073" LastActivityDate="2017-03-11T00:39:01.073" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="185" PostTypeId="2" ParentId="183" CreationDate="2014-05-17T03:50:07.617" Score="43" Body="&lt;p&gt;Split this into three groups of four, A1, A2, A3, A4; B1, B2...; C1, C2... Each step here corresponds to one weighing.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Weigh A against B. &#xA;&lt;ul&gt;&#xA;&lt;li&gt;If A &gt; B, then weigh A1, B1, and B2 against B3, B4, and C1.&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the weights are equal, then one of A2...4 is heavier; weigh A2 and A3. If they are equal, A4 is heavier. If one is heavier, then that ball is heaviest.&lt;/li&gt;&#xA;&lt;li&gt;If the first group is heavier, then either A1 is heavier, or B3-4 is lighter. Compare B3 and B4; if they are equal, A1 is heavier; if they are different, the lightest is the lightest ball.&lt;/li&gt;&#xA;&lt;li&gt;If the first group is lighter, then either B1 or B2 is lighter. Weigh them and see.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;If A &amp;lt; B, renumber all A-balls to B-balls, and perform the above steps.&lt;/li&gt;&#xA;&lt;li&gt;If A = B, weigh A1, A2, A3 against C1, C2, C3&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If they are equal, then weigh A1 against C4.  If A1 is lighter, then C4 is the odd ball and it is heavy. If A1 is heavier, then C4 is the odd ball and it is light.&lt;/li&gt;&#xA;&lt;li&gt;If A is heavier than C, weigh C1 against C2. If they are equal, then C3 is the odd ball and it is lighter. If they are not equal, then the lighter of the two balls is the lightest ball&lt;/li&gt;&#xA;&lt;li&gt;If A is lighter than C, weigh C1 against C2. If they are equal, then C3 is the odd ball and it is heavier. If they are not equal, then the heavier of the two balls is the heaviest ball.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We can work backwards from the third step to see, approximately, why this works. At the third weighing, the options need to be reduced to either two or three balls. This means that the second weighing must reduce to either two or three possible balls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that the first step will remove either 1/3 or 2/3 of the possible solutions, no matter what you do. This means that, in the 1/3 case, you need to split the possibilities down from 8 into a group of 3, a group of 3, and a group of 2. From this, the third weighing points to the odd ball out. Because this case implies one set of balls is heavier, by virtue of finding the odd ball out, we know whether it's heavier or lighter, so we actually don't need to worry about this piece of information at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the 2/3 case, you need to reduce the possibilities into a group of 3 and a group of 1, which is easy enough to do intuitively. Because we actually don't know the relative weight of the odd ball in this case, the information from the third weighing must be used to determine whether the ball is heavier or lighter.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorUserId="5373" LastEditDate="2015-04-03T12:21:18.477" LastActivityDate="2015-04-03T12:21:18.477" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="186" PostTypeId="1" AcceptedAnswerId="189" CreationDate="2014-05-17T04:00:46.343" Score="23" ViewCount="35864" Body="&lt;p&gt;Suppose you want to create a set of weights so that any object with an integer weight from 1 to 40 pounds can be balanced on a two-sided scale by placing a certain combination of these weights onto that scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the fewest number of weights you need, and what are their weights?&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="8874" LastEditDate="2015-01-31T12:27:06.117" LastActivityDate="2023-03-03T06:42:11.517" Title="What's the fewest weights you need to balance any weight from 1 to 40 pounds?" Tags="|calculation-puzzle|combinatorics|weighing|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="187" PostTypeId="1" AcceptedAnswerId="333" CreationDate="2014-05-17T04:18:52.833" Score="5" ViewCount="266" Body="&lt;p&gt;I've tried to begin solving a cryptogram by making educated guesses from the context: position of punctuation, the presence of single-letter words, or assumptions about the frequency of the letter E, for a few examples. I often find 2-letter and 3-letter words in the puzzle and am looking for a more logical approach to such words than just blind guessing. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="158" LastEditorUserId="158" LastEditDate="2014-05-18T00:41:05.333" LastActivityDate="2015-06-02T03:01:14.237" Title="How do I choose between letter possibilities for 2- and 3-letter words in a cryptogram?" Tags="|strategy|cipher|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="189" PostTypeId="2" ParentId="186" CreationDate="2014-05-17T04:37:16.167" Score="31" Body="&lt;p&gt;It turns out that you need four weights, measuring 1, 3, 9, and 27 pounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trick here is that you can put weights on both sides of the scale. If you have an object of 2 pounds on the left side, you can place the 3-lb. weight on the right side and the 1-lb. weight on the &lt;em&gt;left&lt;/em&gt; side for the scale to balance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, if the object weighs 5 pounds, you put the 9-lb. weight on the right side and the 3-lb. and 1-lb. on the left side.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If you have a set of weights that can balance any weight from $1$ to $N$ pounds, then they can also balance any weight from $-1$ to $-N$ pounds just by switching the sides of the scale that you place them on. So the next weight you'd want to add would weigh $2N + 1$ pounds, and then you can balance any weight from $1$ to $3N + 1$ pounds on the scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, starting with the 1-lb. weight, which can balance any weight from $1$ to $1$ pounds, we get $2(1) + 1 = 3$ pounds for the next weight, and they can balance anything up to $1 + 3 = 4$ pounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, for those two weights, we get $2(4) + 1 = 9$, and they can balance up to $1 + 3 + 9 = 13$ pounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally, for all three of those weights, we get $2(13) + 1 = 27$, and they can balance up to $1 + 3 + 9 + 27 = 40$ pounds, as required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, to measure any weight up to $n$ pounds, you will need at least $\lceil\log_3(2n)\rceil$ weights due to the above formula.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="99" LastEditDate="2014-06-24T17:00:50.073" LastActivityDate="2014-06-24T17:00:50.073" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="190" PostTypeId="1" AcceptedAnswerId="191" CreationDate="2014-05-17T04:46:12.427" Score="14" ViewCount="1347" Body="&lt;p&gt;The magic square is a well-known grid of the numbers from 1 to 9 in which every row, column, and diagonal adds up to 15:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;4 9 2&#xA;3 5 7&#xA;8 1 6&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But it is also possible to create magic squares using other numbers:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;24 87 45&#xA;73 52 31&#xA;59 17 80&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It's also known that given just a few filled-in squares, you can determine the rest logically. For example, given the partially filled-in grid:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;8 9 .&#xA;. 6 .&#xA;. 3 .&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;you can immediately infer that the rows, columns, and diagonals add up to 18, and so the bottom-right square is 4 and the top-right square is 1:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;8 9 1&#xA;. 6 .&#xA;. 3 4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then the right square is 13 and the bottom-left square is 11:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 8 9  1&#xA; . 6 13&#xA;11 3  4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And finally, in a slightly uncouth twist, the left square is -1.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 8  9  1&#xA;-1  6 13&#xA;11  3  4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But in fact, it's possible to create a set of filled-in numbers that don't have any completed rows at all and still be able to fill the rest of the numbers in:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;12 .. 27&#xA;.. ..  6&#xA;.. 18 ..&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This, as it turns out, has a (unique) solution of:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;12 24 27&#xA;36 21  6&#xA;15 18 30&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So what is the fewest number of filled-in squares that are actually possible to derive the whole square from, and what arrangement are they in? And what about the case of higher-order magic squares?&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-06-02T03:11:52.127" Title="What is the fewest number of filled-in squares required to uniquely define a magic square?" Tags="|magic-square|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="191" PostTypeId="2" ParentId="190" CreationDate="2014-05-17T04:46:12.427" Score="18" Body="&lt;p&gt;It turns out that the magic squares form a vector space. You can add them (by adding each corresponding entry), multiply them by scalars (by multiplying every entry by the same number), and invert them (take the negative of each entry) around a zero element (the magic square where everything is zero), and they will still stay magic squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this vector space has a basis of size three, as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 0 -3  0     0  2  1     1  2  0&#xA;-1 -1 -1  ,  2  1  0  ,  0  1  2&#xA;-2  1 -2     1  0  2     2  0  1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This means that given any three constants &lt;code&gt;a&lt;/code&gt;, &lt;code&gt;b&lt;/code&gt; and &lt;code&gt;c&lt;/code&gt;, you can define the following magic square:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;          c,   2b + 2c - 3a,             b&#xA;     2b - a,      b + c - a,        2c - a&#xA;b + 2c - 2a,              a,   2b + c - 2a&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where the magic constant is &lt;code&gt;3b + 3c - 3a&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The classical (Lo Shu) magic square has &lt;code&gt;a = 1&lt;/code&gt;, &lt;code&gt;b = 2&lt;/code&gt;, and &lt;code&gt;c = 4&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you are given three numbers in the V shape (or its reflections and rotations):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;xx .. ..&#xA;.. .. xx&#xA;xx .. ..&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;then you can make a magic square out of it. In fact, as long as you're given any three positions so that for each position, when you take the number in that position from each square in the basis and form a vector out of it, you end up with three linearly independent vectors, then you can still uniquely determine a magic square out of it by solving a system of linear equations.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I don't know anything about the higher-order squares, though. That can be left for another answer.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2014-05-17T18:30:48.780" LastActivityDate="2014-05-17T18:30:48.780" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="192" PostTypeId="5" CreationDate="2014-05-17T05:05:29.057" Score="0" Body="&lt;p&gt;A &lt;strong&gt;lateral thinking puzzle&lt;/strong&gt; is a puzzle that involves thinking in ways outside of a challenger's preconception &amp;ndash; that is, &quot;thinking outside the box&quot;. The solution is usually obtained through an indirect and creative approach, using reasoning that is not immediately obvious and involving ideas that may not be obtainable by using only traditional step-by-step logic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When writing a lateral thinking puzzle (this goes for all puzzles on the site, but especially for this genre), one must take care to compose it in such a way that &lt;strong&gt;there is a unique correct answer&lt;/strong&gt;, otherwise the question is likely to be closed as &quot;too broad&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term &lt;strong&gt;lateral thinking&lt;/strong&gt; was coined in 1967 by Edward de Bono in his book &lt;em&gt;&quot;Lateral Thinking: Creativity Step by Step&quot;&lt;/em&gt; (Harper &amp;amp; Row publishers).&#xA;The following riddle (due to Edward de Bono) is typical for this puzzle type:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A boy and his father are driving on a road when suddenly their car crashes. They are both severely injured and taken to a hospital. When they arrive in the operating room, the surgeon says, &quot;I can't operate on this boy, he's my son!&quot; How is this possible?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The surgeon is the boy's mother. The difficulty of the riddle depends on the reader assuming that a surgeon is male, and therefore wondering how the boy's father could be in any condition to operate on anybody when he arrived severely injured in the same operating room.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="13796" LastEditDate="2015-07-06T15:57:36.143" LastActivityDate="2015-07-06T15:57:36.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="193" PostTypeId="4" CreationDate="2014-05-17T05:05:29.057" Score="0" Body="A puzzle solved through an indirect and creative approach, using reasoning that is not immediately obvious and involving ideas that may not be obtainable only using logic." OwnerDisplayName="user88" LastEditorUserId="5373" LastEditDate="2015-03-14T13:49:57.897" LastActivityDate="2015-03-14T13:49:57.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="196" PostTypeId="2" ParentId="190" CreationDate="2014-05-17T14:34:22.207" Score="4" Body="&lt;p&gt;I believe (and can almost prove) that for an $n \times n$ square you need $(n-1)^2-1=n^2-2n$ givens. First, let us ignore the requirement that the diagonal sums match the row/column sums.  Then you can fill the upper $(n-1) \times (n-1)$ grid however you want.  If you pick the upper right element, you can calculate the row sum $R.$  The last element of the next $n-2$ rows is now fixed, as is the bottom element in the first $n-1$ columns.  If $S$ is the sum of the original set of entries, the sum of the elements in the last column so far is $(n-1)R-S$, as is the sum of the elements in the last row, so you pick the lower right to be $R-((n-1)R-S)=S-(n-2)R$.  This shows there are $(n-1)^2+1$ free elements in a square where the rows and columns have a common sum.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diagonals introduce two constraints, so you would expect to subtract two from this count for fully magic squares.  This would give $(n-1)^2-1$ as the number of free elements, in agreement with Joe Z's calculation for $3 \times 3$.  I have not found a way to show these constraints are not redundant in the general case, but it seems unlikely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $n \times n$ matrices form a vector space, and I am counting the dimension of the subspace that is magic.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditDate="2015-06-02T03:11:52.127" LastActivityDate="2015-06-02T03:11:52.127" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="197" PostTypeId="1" CreationDate="2014-05-17T14:45:02.093" Score="12" ViewCount="1342" Body="&lt;p&gt;I know that in ancient times puzzles and riddles were as popular—if not more popular—than they are now. (Let us define &quot;ancient times&quot; as any time predating the fall of the Roman Empire, around A.D. 470).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many of these puzzles survive to this day, e.g. the Sphinx Riddle (&quot;What animal walks on four legs in the morning, two at noon, and three in the evening?&quot;), or variants of &quot;measure X amount of liquid, having only vessels of Y and Z sizes&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any riddles or puzzles left from these times for which the answer was either lost or never known? Can you provide any?&lt;/p&gt;&#xA;" OwnerUserId="117" LastEditorUserId="31484" LastEditDate="2017-01-30T03:44:55.867" LastActivityDate="2017-01-30T03:44:55.867" Title="Are there any puzzles/riddles left from ancient times, to which no answer is known nowadays?" Tags="|puzzle-history|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="199" PostTypeId="1" AcceptedAnswerId="200" CreationDate="2014-05-17T15:05:47.717" Score="11" ViewCount="3245" Body="&lt;p&gt;Is there a common/official name for visual puzzles that are based strictly on the observation skill - like the common &quot;Where is Waldo&quot; pictures, &quot;find all faces&quot;, &quot;count all X in the picture (or find a given number of them), &quot;Spot a thief&quot; and the likes - where the visual mess or tricky composition of the image obscures the location of the puzzle target?&lt;/p&gt;&#xA;" OwnerUserId="117" LastEditorUserId="25966" LastEditDate="2016-12-19T19:29:07.490" LastActivityDate="2016-12-19T19:29:07.490" Title="Is there an official name for visual puzzles based on observation skill?" Tags="|puzzle-identification|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="200" PostTypeId="2" ParentId="199" CreationDate="2014-05-17T16:45:11.770" Score="11" Body="&lt;p&gt;In the online realm, these are known as &quot;hidden object/picture&quot; games, and the definition also matches those which appear in books. &lt;a href=&quot;http://en.wikipedia.org/wiki/Puzzle_video_game#Hidden_object_game&quot;&gt;From Wikipedia (emph. not mine)&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A &lt;strong&gt;hidden object game&lt;/strong&gt; (sometimes called &lt;strong&gt;hidden picture&lt;/strong&gt;) is a genre of puzzle video game in which the player must find items from a list that are hidden within a picture.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In essence, &quot;Where's Waldo&quot; is just a simpler version of this game, in which the list of items you are looking for has a length of one (though if I recall, there are sometimes more objects to find in Waldo).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Physical copies of these games also, I believe, can go by the same name.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastActivityDate="2014-05-17T16:45:11.770" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="201" PostTypeId="1" AcceptedAnswerId="202" CreationDate="2014-05-17T16:52:00.250" Score="13" ViewCount="4267" Body="&lt;h1&gt;Red or Black&lt;/h1&gt;&#xA;&lt;p&gt;There are 3 children sitting on three chairs. The children can only look forward and all wear one hat. The hats are either black or red. They are a total of 5 hats, 3 are colored red and 2 are colored black. The child on chair A says: &amp;quot;I don't know what color of the hat I have.&amp;quot; Then the child on chair B says: &amp;quot;Also, I don't know the color of my hat.&amp;quot; Then the child on chair C says: &amp;quot;In that case, I know the color of my hat.&amp;quot;&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt; |        |        |&#xA; |        |        |&#xA; +---+    +---+    +---+&#xA; |   |    |   |    |   |&#xA;   A        B        C&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;The &lt;strong&gt;question&lt;/strong&gt; is: What color of hat does the child on chair C have and why?&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Bonus question&lt;/strong&gt;: Do we also know the color of the hat of the child on chair A and the color of the hat on chair B?&lt;/p&gt;&#xA;" OwnerUserId="146" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2020-12-14T07:47:16.960" Title="Guessing hat colors" Tags="|logical-deduction|meta-knowledge|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="202" PostTypeId="2" ParentId="201" CreationDate="2014-05-17T17:00:36.030" Score="17" Body="&lt;p&gt;If A doesn't know the colour of his hat, then at least one of B and C must be wearing a red hat, because if they were both wearing black, then A would know immediately that he was wearing a red one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B knows this as well. But if C were wearing a black hat, then B would know that she was wearing a red one because of that. So since B doesn't know, C must be wearing a red hat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C knows this as well. So C knows that his hat is red.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As for the bonus question, no, we cannot know what colour hat either A or B is wearing. All else we can deduce from the problem is that at least one of B and C is wearing a red hat, but C wearing a red hat fulfills that already, so B could be wearing either colour of hat. And in general, nobody can know what A's hat's colour is, since nobody can see it. So all four arrangements where C wears a red hat are possible.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2014-05-17T17:07:02.533" LastActivityDate="2014-05-17T17:07:02.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="203" PostTypeId="1" AcceptedAnswerId="205" CreationDate="2014-05-17T17:48:05.437" Score="9" ViewCount="4539" Body="&lt;p&gt;The following shape is has the proportions of a square attached to a similar square divided diagonally - A &lt;em&gt;square and a half&lt;/em&gt;, if you may.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/40knC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The puzzle is to &lt;strong&gt;dissect the shape into 4 &lt;a href=&quot;http://en.wikipedia.org/wiki/Congruence_%28geometry%29&quot;&gt;congruent&lt;/a&gt; pieces.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Two figures or objects are &lt;strong&gt;congruent&lt;/strong&gt; if they have the same shape and size, or if one has the same shape and size as the mirror image of the other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="129" LastEditorUserId="129" LastEditDate="2014-05-17T18:02:15.400" LastActivityDate="2014-05-17T18:21:00.553" Title="Dissect a square-and-a-half into 4 equal pieces" Tags="|geometry|dissection|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="204" PostTypeId="2" ParentId="203" CreationDate="2014-05-17T18:08:00.430" Score="8" Body="&lt;p&gt;The solution appears as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bCp9L.png&quot; alt=&quot;solution to the puzzle&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a bit rough, but it shows the right places to cut the shape.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2014-05-17T18:08:00.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="205" PostTypeId="2" ParentId="203" CreationDate="2014-05-17T18:08:53.653" Score="18" Body="&lt;p&gt;The first thing to do is divide the figure into a number of sections which is a multiple of four. The easiest way to do this is to split it up into smaller triangles:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zw2dL.png&quot; alt=&quot;triangle division&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that there are now 12 sections, which divides into sections of 3. Three triangles forms a  square and a smaller triangle. We know, as a result, that a square must go here (another way to do this is that the triangle in the corner must be part of a shape of three parts):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rrbWj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, it becomes clear where the rest of the divisions lie. The remaining triangle, as part of the triangular section, must be part of a set of three, so as a result, we know that it is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6O2m4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest of the divisions are simple:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UInIz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, as a side note, for future reference, &lt;a href=&quot;http://livegeometry.com/&quot;&gt;LiveGeometry&lt;/a&gt; is a wonderful little tool.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorDisplayName="user20" LastEditDate="2014-05-17T18:21:00.553" LastActivityDate="2014-05-17T18:21:00.553" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="206" PostTypeId="1" AcceptedAnswerId="207" CreationDate="2014-05-17T18:19:58.867" Score="2" ViewCount="485" Body="&lt;p&gt;A farmer is looking to hire some hired hands. His usual three hires are named Barry, Harry, and Larry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He knows from hiring them before that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Working together, Barry and Harry can plow an acre in two hours.&lt;/li&gt;&#xA;&lt;li&gt;Working together, Barry and Larry can plow an acre in three hours.&lt;/li&gt;&#xA;&lt;li&gt;Working together, Harry and Larry can plow an acre in five hours.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How many hours would it take for all three of them working together to plow an acre?&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2014-05-20T04:11:15.490" LastActivityDate="2015-06-24T17:20:04.527" Title="The problem of the hired hands' work rates" Tags="|calculation-puzzle|" AnswerCount="2" CommentCount="2" ClosedDate="2015-07-01T08:41:09.537" ContentLicense="CC BY-SA 3.0" />
  <row Id="207" PostTypeId="2" ParentId="206" CreationDate="2014-05-17T18:25:05.497" Score="5" Body="&lt;p&gt;Suppose we divide an acre into 30 sections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Barry and Harry can plow 15 sections an hour.&lt;/li&gt;&#xA;&lt;li&gt;Barry and Larry can plow 10 sections an hour.&lt;/li&gt;&#xA;&lt;li&gt;Larry and Harry can plow 6 sections an hour.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If we add all these three up and divide by 2, we get the number of sections all three of them can plow in one hour:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Barry, Harry, and Larry can plow 15.5 sections an hour.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(Looks like poor Larry works 10 times slower than either of the other two boys.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, it will take all three of them (30/15.5 hours) = about 1 hour and 56 minutes to plow an acre.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2014-05-17T18:25:05.497" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="208" PostTypeId="1" AcceptedAnswerId="209" CreationDate="2014-05-17T18:39:15.333" Score="22" ViewCount="3100" Body="&lt;p&gt;My friend once told me a logic puzzle along the lines of:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A student asks a teacher what the ages of the teacher's three kids are. The teacher said, &amp;quot;the product of their ages is 72, and the sum of their ages is your [the student's] house number.&amp;quot; The student thought about this for a while, then came back to the teacher and said, &amp;quot;I don't have enough information!&amp;quot; The teacher replied, &amp;quot;Oh, and the oldest plays piano.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;What are the ages of the kids?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;How do I even go about solving this problem?&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-11-07T10:37:25.543" Title="How do I tell the children's ages?" Tags="|logical-deduction|calculation-puzzle|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="209" PostTypeId="2" ParentId="208" CreationDate="2014-05-17T18:39:15.333" Score="22" Body="&lt;p&gt;This has since become one of my favorite logic puzzles! &lt;strong&gt;I highly recommend that you make an effort to solve this one first.&lt;/strong&gt; It's a lot of fun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know the kids' ages must multiply to 72. Therefore, we need to find all combinations of three numbers that multiply to 72. 72 has the prime factors 2*2*2*3*3. For convenience, I've added a list of their sums for the student's house number.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 + 1 + 72 = 74&#xA;1 + 2 + 36 = 39&#xA;1 + 3 + 24 = 28&#xA;1 + 4 + 18 = 23&#xA;1 + 6 + 12 = 19&#xA;1 + 8 + 9  = 18&#xA;2 + 2 + 18 = 22&#xA;2 + 3 + 12 = 17&#xA;2 + 4 + 9  = 15&#xA;2 + 6 + 6  = 14&#xA;3 + 3 + 8  = 14&#xA;3 + 4 + 6  = 13&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here we see a list of possible ages and house numbers. Now, of course, we know that the student comes back and tells the teacher &lt;strong&gt;&quot;I need more information.&quot;&lt;/strong&gt; If the student didn't need more information, then the student's house number would appear once in this list. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, because the student needs more information, we know that the student's house number must appear more than once in this list. The only combinations for which this is the case are &lt;code&gt;2+6+6=14&lt;/code&gt; and &lt;code&gt;3+3+8=14&lt;/code&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The teacher tells the student that the oldest plays piano, implying that there must be just one oldest child. Therefore, the answer can't be 2, 6, and 6, because there is no oldest. The answer must, therefore, be 3, 3, and 8.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorUserId="4204" LastEditDate="2014-11-27T04:47:51.320" LastActivityDate="2014-11-27T04:47:51.320" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="210" PostTypeId="1" AcceptedAnswerId="213" CreationDate="2014-05-17T18:48:06.817" Score="13" ViewCount="1933" Body="&lt;p&gt;If the end of the world was supposed to come on the first day of a new century, what would be the chances that it would happen on a Sunday?&lt;/p&gt;&#xA;" OwnerUserId="129" LastEditorDisplayName="user88" LastEditDate="2014-05-25T21:38:53.653" LastActivityDate="2016-05-12T23:09:27.703" Title="End of the world - on a Sunday?" Tags="|calculation-puzzle|mathematics|word-problem|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="211" PostTypeId="5" CreationDate="2014-05-17T18:51:48.740" Score="0" Body="&lt;p&gt;A calculation puzzle is one that involves numerical calculations of some sort, generally using the basic mathematical operations: addition, subtraction, multiplication, and division.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Although involving math, these questions &lt;em&gt;CANNOT&lt;/em&gt; be math homework/textbook style questions as these are &lt;em&gt;off-topic&lt;/em&gt; for this site.&lt;/strong&gt; See &#xA;&lt;a href=&quot;https://puzzling.meta.stackexchange.com/q/2783/11110&quot;&gt;Are math-textbook-style problems on topic?&lt;/a&gt; for more details.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="11110" LastEditDate="2020-04-26T09:13:45.903" LastActivityDate="2020-04-26T09:13:45.903" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="212" PostTypeId="4" CreationDate="2014-05-17T18:51:48.740" Score="0" Body="A puzzle that involves numerical calculations, such as multiplication and addition. Use with the [mathematics] tag." OwnerDisplayName="user88" LastEditorUserId="11110" LastEditDate="2021-03-09T21:37:00.887" LastActivityDate="2021-03-09T21:37:00.887" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="213" PostTypeId="2" ParentId="210" CreationDate="2014-05-17T19:14:13.810" Score="30" Body="&lt;p&gt;The answer is slightly trickier than you might think at first - it has to do with a quirk in the calendar we currently use, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Gregorian_calendar#Adoption&quot;&gt;Gregorian Calendar&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Gregorian Calendar was instated by Pope Gregory (hence the name) in 1582 to account for a slight disparity in the Julian calendar that made the days drift about 18.75 hours every century. The Julian Calendar had one leap day every 4 years, which accounted for the gap between 365 days and a year, but overshot it by just a little too much to be unnoticeable. By the time Jesus had been gone for over 1,500 years, the dates on which they were celebrating Easter (and the dates on which the seasons were occurring) were drifting much too far away from their original dates for the church's comfort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the Catholic Church under Pope Gregory decided to remove the leap years for three of the four century years, specifically the ones that weren't divisible by 400. So while the year 1600 had a February 29 as usual, the years 1700, 1800, and 1900 didn't have an extra day at all. This corrected the disparity by an average of 18 hours every century, to the Catholic Church's satisfaction. (The remaining 0.75 hours will only start to be noticeable in about 20,000 years' time.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what does this have to do with the question at hand? Well, the Gregorian calendar runs a 400-year cycle with its leap years in this way, and every 400 years there are exactly 97 occurrences of February 29. This is a total of (365 * 400 + 97) = 146097 days, which happens to be exactly divisible by 7. So every 400 years, the days of the week complete one cycle as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that ultimately, the beginning of a century can only have one of up to four possible days of the week, because after every fourth century, the cycle restarts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this in mind, let's check the dates for each of the first days of the century from 2101 to 2401 (as centuries begin with the -01 year; the -00 year is actually the &lt;em&gt;last&lt;/em&gt; year of the &lt;em&gt;previous&lt;/em&gt; century):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;January 1, 2101 falls on a &lt;em&gt;Saturday&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;li&gt;January 1, 2201 falls on a &lt;em&gt;Thursday&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;li&gt;January 1, 2301 falls on a &lt;em&gt;Tuesday&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;li&gt;January 1, 2401 falls on a &lt;em&gt;Monday&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And this cycle repeats for every set of 400 years onwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So tough luck, the first day of the century &lt;em&gt;never&lt;/em&gt; falls on a Sunday. The probability is &lt;strong&gt;zero&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;But wait, what if we &lt;em&gt;do&lt;/em&gt; count the -00 year as the beginning of the century? Then, the days of the week are as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;January 1, 2100 falls on a &lt;em&gt;Friday&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;li&gt;January 1, 2200 falls on a &lt;em&gt;Wednesday&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;li&gt;January 1, 2300 falls on a &lt;em&gt;Monday&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;li&gt;January 1, 2400 falls on a &lt;em&gt;Saturday&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As it happens, Sunday doesn't appear here either, and in fact, Sunday is the &lt;em&gt;only&lt;/em&gt; day of the week that doesn't appear in either list, and so is the only day that cannot be the beginning of a century, regardless of whether you consider -00 or -01 to be the first year.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2014-11-11T19:34:16.090" LastActivityDate="2014-11-11T19:34:16.090" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="215" PostTypeId="2" ParentId="210" CreationDate="2014-05-17T19:29:14.400" Score="10" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;0 percent.&lt;/strong&gt;&#xA; This is because the first day of a century can never fall on a Sunday.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are 365 days in a year (which is not a leap year) - which is exactly 52 weeks plus one day. Therefore, from a common year to the following year, New Year's Day advances by one day of the week.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a leap year (which has 366 days) to the year following, New Year's Day advances by two  days of the week.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;So, advancing from one century to another, we advance a 100 days plus a few extra days for leap years. Every fourth year, except the year divisible by 100 is a leap year making a total of 24 ($\frac{100}{4} - 1$) leap years in a century. Therefore, we advance a total of 124 days of the week in a century. 124 divided by 7 leaves a remainder of 5 - so we're only moving 5 days of the week ahead.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;After 2 centuries have passed&lt;/strong&gt;, we advance $5 + 5 = 10$ days - or &lt;strong&gt;3 days&lt;/strong&gt; from the start, because $10 =3 \mod 7$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;After 3 centuries&lt;/strong&gt;, we've advanced 15 days - or &lt;strong&gt;1 day&lt;/strong&gt; from the start.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;On the &lt;strong&gt;4th century&lt;/strong&gt;, however, we have to take into account the fact that &lt;em&gt;years divisible by 400 have a leap day&lt;/em&gt;. So, instead of advancing 20 days, we advance 21 days - or &lt;strong&gt;0 days&lt;/strong&gt;. &lt;strong&gt;We're back at the day we started 4 centuries ago.&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;We can now check the 4 days that a century can begin with by examining the days of 2100 to 2400:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The century 2100 starts on a &lt;em&gt;Friday&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;2200 subsequently starts 5 days later, a &lt;em&gt;Wednesday&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;2300 then starts on 3 days ahead of Friday, i.e. a &lt;em&gt;Monday&lt;/em&gt;,&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;and 2400 starts on &lt;em&gt;Saturday&lt;/em&gt;, i.e. 1 day from Friday.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We return to a Friday in the year 2500, and the cycle continues.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Because of this &lt;em&gt;Friday-Wednesday-Monday-Saturday cycle&lt;/em&gt;, we cannot have a century starting on Sunday (and a Tuesday and a Thursday) - which is a good thing because we get an extra holiday!&lt;/p&gt;&#xA;" OwnerUserId="129" LastEditorUserId="129" LastEditDate="2016-05-12T23:09:27.703" LastActivityDate="2016-05-12T23:09:27.703" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="217" PostTypeId="5" CreationDate="2014-05-17T23:02:24.010" Score="0" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Rubik%27s_Cube&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rubik's Cube&lt;/a&gt; is a puzzle invented by &lt;a href=&quot;https://en.wikipedia.org/wiki/Ern%C5%91_Rubik&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ernő Rubik&lt;/a&gt; in 1974.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Official Rubik site can be found &lt;a href=&quot;https://www.rubiks.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://ruwix.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ruwix wiki site&lt;/a&gt; covers all aspects of the cube, and also covers Rubik variations, such as the Pyraminx and MegaMinx.&lt;/p&gt;&#xA;" OwnerUserId="13796" LastEditorUserId="11497" LastEditDate="2019-03-31T11:54:16.743" LastActivityDate="2019-03-31T11:54:16.743" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="218" PostTypeId="4" CreationDate="2014-05-17T23:02:24.010" Score="0" Body="For questions about the Rubik's Cube (a 3-D combination puzzle) and its variations." OwnerDisplayName="user20" LastEditorUserId="70545" LastEditDate="2021-02-02T13:28:14.613" LastActivityDate="2021-02-02T13:28:14.613" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="219" PostTypeId="1" AcceptedAnswerId="223" CreationDate="2014-05-17T23:13:32.257" Score="27" ViewCount="10776" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Four_color_theorem&quot;&gt;Four color theorem&lt;/a&gt; states that no more than four colors are required for any map.  Can it be proved or disproved that 3 colors can be used for United States map?&lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="8874" LastEditDate="2016-03-23T18:22:22.437" LastActivityDate="2018-11-30T10:02:06.520" Title="Is there a proof that a map of the United States requires 4 colors?" Tags="|logical-deduction|geography|graph-theory|topology|" AnswerCount="5" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="220" PostTypeId="2" ParentId="219" CreationDate="2014-05-17T23:20:38.013" Score="6" Body="&lt;p&gt;3 colors are not sufficient for a map of the US.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at this 4-color US map I found online.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YqJ42.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at Oklahoma. It has a lot of edges and touches a lot of other states. If you only had 3 colors, there would be no way to color that section. Make Oklahoma color 1, and then start alternating colors for the surrounding states. Kansas is 2, Colorado is 3, New Mexico is 2, Texas is 3, Louisiana is 2, and here's where it breaks down: Arkansas is touching both Texas and Oklahoma. There's no way to color those so that they don't conflict with Oklahoma and Texas.  See this edited version of the map:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/k25Ny.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Arkansas can't be yellow, green, or pink. So you must have a fourth color.&lt;/p&gt;&#xA;" OwnerUserId="10" LastEditorUserId="10" LastEditDate="2014-05-17T23:51:30.507" LastActivityDate="2014-05-17T23:51:30.507" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="221" PostTypeId="1" AcceptedAnswerId="222" CreationDate="2014-05-17T23:25:47.727" Score="2" ViewCount="3536" Body="&lt;p&gt;Oskar van Deventer's Gear Shift is a somewhat difficult puzzle to figure out how to solve. The puzzle looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4R6kn.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see a &lt;a href=&quot;https://www.youtube.com/watch?v=yO6YCBKvhzI&quot; rel=&quot;nofollow noreferrer&quot;&gt;video of it here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a simple, intuitive method for solving this puzzle? I'm sure the optimal solution could be reached via matrix reduction, but I'm asking for something which is intuitive to understand.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorUserId="71" LastEditDate="2014-11-12T10:59:44.330" LastActivityDate="2014-11-12T10:59:44.330" Title="How do I solve Oskar van Deventer's Gear Shift puzzle?" Tags="|twisty-puzzles|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="222" PostTypeId="2" ParentId="221" CreationDate="2014-05-17T23:25:47.727" Score="5" Body="&lt;p&gt;Solving this puzzle is actually rather easy once you know how, but first, we need to establish a couple things:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Puzzle turning notation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm going to use the standard L/R, U/D, F/B notation for twisting the puzzle; however, you'll note that the puzzle pulls apart along the three axes. In order  to execute an L move, for instance, you will need to pull the cube apart so that you can easily make the move. This is implied by the move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I want to indicate a rotation of two ticks, I will write &lt;code&gt;[face](2)&lt;/code&gt;; for instance, &lt;code&gt;L(2)&lt;/code&gt; will say to turn the large gears on the left side clockwise by two ticks. I will always standardize the rotations to those of the large gears for convenience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, I will keep track of the number of turns in terms of &lt;code&gt;N&lt;/code&gt;, where &lt;code&gt;N&lt;/code&gt; is defined in the step as the number of ticks a particular piece is away from its goal. Rotating &lt;code&gt;2N&lt;/code&gt; means to rotate the specified face by twice the number of ticks of N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solution to the puzzle&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may happen while solving this puzzle that you encounter a position in which the puzzle is already solved and you need not go on. If so, congratulations!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first step to solving this puzzle is trivial. Pull it apart along the left/right axis, and turn the two sides until the top two pieces line up with each other. Then, push the puzzle back together. Once this is completed, the top four pieces should be solved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flip the puzzle over. Separate the four solved pieces from the four unsolved ones, and turn the unsolved side until at least two of the pieces are solved. Push the puzzle back together. You should now have six pieces of the puzzle solved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reorient the puzzle such that the two unsolved pieces face down and toward you. The larger of the two pieces will be an even number of turns (2N) from its solved state. Separate the puzzle such that the two unsolved pieces are on different faces, then:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the large piece is on the left, turn &lt;code&gt;L(2N) R(N)&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;If the large piece is on the right, turn &lt;code&gt;R(2N) L(N)&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In other words, for each two spaces you turn the large gear, turn the small gear one in the opposite direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, separate the back face and turn it until the back two bottom pieces are solved. Now, separate and turn the top face until the back two upper pieces are solved. You should now be in a position with one or two small gears unsolved. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Separate the face with the two unsolved small gears, and rotate it until one gear remains unsolved. Position that small gear such that it is on the bottom front right. This small gear will be &lt;code&gt;N&lt;/code&gt; ticks away from a solved state. Perform the following move sequence: &lt;code&gt;U(8N) F(-8N) R(-8N)&lt;/code&gt;; each 8N is a full revolution of the large gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have done that last move, the final corner should be rotated in place and the puzzle should be solved.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastActivityDate="2014-05-17T23:25:47.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="223" PostTypeId="2" ParentId="219" CreationDate="2014-05-17T23:34:44.253" Score="53" Body="&lt;p&gt;I started by assuming two adjacent states must be different colors. Therefore, I arbitrarily assigned blue to California and beige to Oregon. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Nevada must therefore be a third color: green.&lt;/li&gt;&#xA;&lt;li&gt;Arizona must therefore be neither green nor blue: beige.&lt;/li&gt;&#xA;&lt;li&gt;Utah must therefore neither be green nor beige: blue.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Idaho must therefore be neither beige, green, nor blue - which disproves the ability of the US map to be covered by three colors. Idaho, here highlighted in dark red, must be color number 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1sJpy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorUserId="4536" LastEditDate="2014-12-15T17:32:24.957" LastActivityDate="2014-12-15T17:32:24.957" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="224" PostTypeId="2" ParentId="183" CreationDate="2014-05-18T00:21:55.507" Score="70" Body="&lt;p&gt;There is another way to do this problem, that doesn't involve any sort of conditional branching at all. It is in fact possible to set a fixed weighing schedule beforehand and still determine which ball is lighter or heavier in just 3 weighings. I'll explain how below.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;The gist of problems like these is, how much information can you get from the procedure you're allowed to undertake? With each weighing, the scale can either tip to the left, tip to the right, or stay balanced. This gives you a total of 3&lt;sup&gt;3&lt;/sup&gt; = 27 possible outcomes, and in this case you need to discern 24 results from them (one of 12 balls is either light or heavy, which is 12 × 2 = 24).&lt;/p&gt;&#xA;&lt;p&gt;So, we need to begin the tedious task of mapping each result to an outcome.&lt;/p&gt;&#xA;&lt;p&gt;One of the things we can immediately notice is that there are also three states each ball can be in during each weighing - on the left side of the scale, on the right side of the scale, or off the scale. Naturally, this maps to the states of the scale in a way that's intuitively analogous:&lt;/p&gt;&#xA;&lt;p&gt;If the odd ball out is heavier...&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;and the ball is placed on the left side, the scale will tip to the left.&lt;/li&gt;&#xA;&lt;li&gt;and the ball is placed on the right side, the scale will tip to the right.&lt;/li&gt;&#xA;&lt;li&gt;and the ball is off the scale, the scale will remain balanced.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;If the ball is lighter, the first two cases are inverted.&lt;/p&gt;&#xA;&lt;p&gt;There are 27 possible ways to place each ball in all three weighings, each corresponding to a different outcome if that ball is the odd one out. We need to find an arrangement of balls where each possible set of placements and its inverse (for the heavy and light cases) is distinct - so no two balls are in the same place for all three weighings.&lt;/p&gt;&#xA;&lt;p&gt;Here's a preliminary arrangement that satisfies the distinctness property. Notice that no possible arrangement appears more than once in both tables:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;       Normal  Inverse&#xA;Ball   1 2 3    1 2 3&#xA; 1     L        R    &#xA; 2       L        R&#xA; 3         L        R&#xA; 4     L R      R L&#xA; 5     L   R    R   L&#xA; 6       L R      R L&#xA; 7     L L      R R&#xA; 8     L   L    R   R&#xA; 9       L L      R R&#xA;10     L L R    R R L&#xA;11     L R L    R L R&#xA;12     R L L    L R R&#xA;   &#xA;L = place it on the left&#xA;R = place it on the right&#xA;  = leave it off&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Right away, we run into the problem that we're not putting the same number of balls onto each scale. If you have seven balls on one side and one on the other, of course the scale is going to tip to the side with seven balls (unless your odd ball out is ridiculously heavy, but let's not entertain that scenario). So we need to invert a few of these configurations so that we're putting four on each side for each weighing. With some trial and error, we can get something like this:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;       Normal  Inverse&#xA;Ball   1 2 3    1 2 3&#xA; 1     L        R    &#xA; 2       L        R&#xA; 3         R        L&#xA; 4     L R      R L&#xA; 5     R   L    L   R&#xA; 6       L R      R L&#xA; 7     R R      L L&#xA; 8     L   L    R   R&#xA; 9       L L      R R&#xA;10     R R L    L L R&#xA;11     R L R    L R L&#xA;12     L R R    R L L&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;So our final weighing schedule of balls is as follows:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;Weighing 1:  1 4 8 12 / 5 7 10 11&#xA;Weighing 2:  2 6 9 11 / 4 7 10 12&#xA;Weighing 3:  5 8 9 10 / 3 6 11 12&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;And the outcomes are interpreted as such:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;==L : 3L    L== :  1H    R== :  1L&#xA;==R : 3H    L=L :  8H    R=L :  5H&#xA;=L= : 2H    L=R :  5L    R=R :  8L&#xA;=LL : 9H    LL= :  7L    RL= :  4L&#xA;=LR : 6H    LLR : 10L    RLL : 12L&#xA;=R= : 2L    LR= :  4H    RLR : 11H&#xA;=RL : 6L    LRL : 11L    RR= :  7H&#xA;=RR : 9L    LRR : 12H    RRL : 10H&#xA;   &#xA; = : scale balanced&#xA; L : scale tipped to the left&#xA; R : scale tipped to the right&#xA;nL : ball n is light&#xA;nH : ball n is heavy&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;And thus, we've created a weighing scheme where each weighing is completely predetermined beforehand, that still manages to determine which ball is the odd one out, and whether it's lighter or heavier.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;You might notice that we didn't use &lt;code&gt;LLL&lt;/code&gt;, &lt;code&gt;RRR&lt;/code&gt;, or &lt;code&gt;===&lt;/code&gt; in our arrangements.&lt;/p&gt;&#xA;&lt;p&gt;We can't use &lt;code&gt;LLL&lt;/code&gt; and &lt;code&gt;RRR&lt;/code&gt; as a 13th pair for a 13th ball, because then we'd end up having to put &lt;em&gt;nine&lt;/em&gt; balls onto the scale, and there's no way to do that since nine is odd. We could probably use it &lt;em&gt;in place of&lt;/em&gt; one of the &lt;code&gt;LLR/RRL&lt;/code&gt; pairs, but leaving &lt;code&gt;LLL&lt;/code&gt; and &lt;code&gt;RRR&lt;/code&gt; out makes for a symmetry in the result chart that I rather like.&lt;/p&gt;&#xA;&lt;p&gt;However, what's interesting is that you &lt;em&gt;can&lt;/em&gt; have a 13th ball that you &lt;em&gt;never&lt;/em&gt; place on any scale, and if your scales balance out in all three weighings, the 13th ball you never weighed is the odd ball out (although you obviously can't tell without a fourth weighing whether it's lighter or heavier).&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="463" LastEditDate="2022-07-07T20:08:07.563" LastActivityDate="2022-07-07T20:08:07.563" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="225" PostTypeId="2" ParentId="187" CreationDate="2014-05-18T03:28:49.730" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://www.scrabblefinder.com/two-letter-scrabble-words/&quot; rel=&quot;nofollow&quot;&gt;http://www.scrabblefinder.com/two-letter-scrabble-words/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.scrabblefinder.com/3-letter-words/&quot; rel=&quot;nofollow&quot;&gt;http://www.scrabblefinder.com/3-letter-words/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not a perfect solution, but there are only 101 official 2 letter words and 1015 3 letter words, so with one letter, or letter repetition, or even context it can be quite trivial to lower the possibilities to a small group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even in those 101 2-letter words and 1015 3-letter words, most of them are obscure or rarely used, so for most quotes you can eliminate all that just are too esoteric or abstruse for that quote.&lt;/p&gt;&#xA;" OwnerUserId="92" LastActivityDate="2014-05-18T03:28:49.730" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="226" PostTypeId="2" ParentId="197" CreationDate="2014-05-18T07:52:56.173" Score="-5" Body="&lt;p&gt;See Daniel &lt;a href=&quot;http://mechon-mamre.org/p/pt/pt3410.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;10&lt;/a&gt;:4 – &lt;a href=&quot;http://mechon-mamre.org/p/pt/pt3412.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;12&lt;/a&gt;:9:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;10&lt;/strong&gt; …&lt;/p&gt;&#xA;&lt;p&gt;&lt;sup&gt;14&lt;/sup&gt; Now I am come to make thee understand what shall befall thy people in the end of days…&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;12&lt;/strong&gt; …&lt;/p&gt;&#xA;&lt;p&gt;&lt;sup&gt;8&lt;/sup&gt; And I heard, but I understood not; then said I: 'O my lord, what shall be the latter end of these things?'&lt;/p&gt;&#xA;&lt;p&gt;&lt;sup&gt;9&lt;/sup&gt; And he said: 'Go thy way, Daniel; for the words are shut up and sealed till the time of the end.…'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="93" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-05-18T07:52:56.173" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="227" PostTypeId="1" AcceptedAnswerId="229" CreationDate="2014-05-18T08:51:56.103" Score="13" ViewCount="3727" Body="&lt;p&gt;I wonder how accurately I will have to work if I make a wooden burr puzzle out of square sticks as for example described in &lt;a href=&quot;http://en.wikipedia.org/wiki/Burr_puzzle&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Burr_puzzle&lt;/a&gt;. Should the notches be exactly the width of the square sticks so that everything fits tight together or should I sand down the sticks a little or widen the notches so the pieces will work easily together? Does anybody have experiences with that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/4/45/Six-Piece_Burr_-_Nut_Puzzle.svg/320px-Six-Piece_Burr_-_Nut_Puzzle.svg.png&quot; alt=&quot;Burr pieces&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="143" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:48.657" LastActivityDate="2015-02-04T09:51:33.440" Title="How exact must the pieces of a wooden burr puzzle be?" Tags="|puzzle-creation|mechanical-puzzles|burr-puzzle|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="228" PostTypeId="1" CreationDate="2014-05-18T14:14:30.640" Score="15" ViewCount="19314" Body="&lt;p&gt;I know there are several different standard methods to solve a Rubik's cube, some easier, some more straightforward. It is known that the minimum number of moves to guarantee a solution is 20.  Which method comes closest to the theoretical minimum in fewest average moves from a randomized cube to a solution? What about worst case? Are there any potential methods that are more efficient but more difficult, but still within reason for a human to solve?&lt;/p&gt;&#xA;" OwnerUserId="4" LastActivityDate="2018-08-29T10:09:30.243" Title="What is the most efficient method to solve a Rubik's cube?" Tags="|rubiks-cube|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="229" PostTypeId="2" ParentId="227" CreationDate="2014-05-18T15:14:01.760" Score="4" Body="&lt;p&gt;I have a copy of this particular wooden burr puzzle. The main thing to watch out for (and not just with this burr) is not actually the width of the internal pieces, but the precision of the angle of the cuts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're making this using high-precision tools, then just make sure everything is squared properly. And you'll probably be fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are, however, having problems assembling it, look for places that don't quite fit together well under tension and sand those down specifically. With this puzzle, since the pieces are large, it's hard to sand too much, but on future burrs be careful with this. Just don't take off much and you should be fine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd recommend against sanding everywhere, though. It risks making the puzzle too loose. &lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastActivityDate="2014-05-18T15:14:01.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="230" PostTypeId="1" AcceptedAnswerId="267" CreationDate="2014-05-18T21:31:41.280" Score="85" ViewCount="207555" Body="&lt;p&gt;A somewhat well-known puzzle is described as such:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You have a pile of 3,000 bananas. You wish to transport them to a place 1,000 miles away on the back of a camel; however, the camel can only carry a maximum of 1,000 bananas, and will eat one banana every mile it travels (and will not go anywhere if it does not have any bananas). However, you can load and unload as many bananas as you want anywhere. What is the most bananas you can bring over to your destination?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Obviously you can't just load up 1,000 bananas and go, because the camel will have eaten them all by the time you get there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd never figured out the answer by myself before. Does anybody have any insights into how to solve this problem and other problems like it?&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="8874" LastEditDate="2016-03-22T16:56:02.437" LastActivityDate="2022-12-29T06:30:37.357" Title="A camel transporting bananas" Tags="|mathematics|calculation-puzzle|optimization|" AnswerCount="16" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="231" PostTypeId="1" AcceptedAnswerId="359" CreationDate="2014-05-18T22:29:49.567" Score="13" ViewCount="36928" Body="&lt;p&gt;Of the following 5 strategies for getting started assembling a jigsaw puzzle:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Assembling the border and connecting other pieces to it&lt;/li&gt;&#xA;&lt;li&gt;Working your way out from one corner  &lt;/li&gt;&#xA;&lt;li&gt;Trying to assemble faces or other recognizable parts of the picture&lt;/li&gt;&#xA;&lt;li&gt;Picking a random piece and working your way out from it  &lt;/li&gt;&#xA;&lt;li&gt;Finding random pieces that fit together&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;which one or what combination results in the shortest solving time, on average, statistically speaking, according to the experts? Note that this is &lt;em&gt;not&lt;/em&gt; a matter of experience. What do the studies on jigsaw-puzzle solving show?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be sure to include the reasons why the strategy or combination of strategies is the fastest.&lt;/p&gt;&#xA;" OwnerUserId="158" LastEditorUserId="158" LastEditDate="2014-05-19T18:32:42.270" LastActivityDate="2019-01-19T06:37:51.480" Title="Of these five approaches to starting a jigsaw puzzle, which one is fastest?" Tags="|strategy|jigsaw-puzzle|" AnswerCount="5" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="232" PostTypeId="2" ParentId="230" CreationDate="2014-05-18T22:34:50.503" Score="41" Body="&lt;p&gt;First of all, the brute-force approach does not work. If the Camel starts by picking up the 1000 bananas and try to reach point B, then he will eat up all the 1000 bananas on the way and there will be no bananas left for him to return to point A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have to take an approach that the Camel drops the bananas in between and then returns to point A to pick up bananas again.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;lt;---p1---&amp;gt;&amp;lt;--------p2-----&amp;gt;&amp;lt;-----p3----&amp;gt;&#xA;A----------------------------------------&amp;gt;B&#xA;-----&amp;gt; ------&amp;gt; --------&amp;gt;&#xA;&amp;lt;----- &amp;lt;------&#xA;-----&amp;gt; ------&amp;gt;&#xA;&amp;lt;-----&#xA;-----&amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Since there are 3000 bananas and the Camel can only carry 1000 bananas, he will have to make 3 trips to carry them all to any point in between.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When bananas are reduced to 2000 then the Camel can shift them to another point in 2 trips and when the number of bananas left are &amp;lt;= 1000, then he should not return and only move forward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first part, P1, to shift the bananas by 1Km, the Camel will have to&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Move forward with 1000 bananas – Will eat up 1 banana in the way forward&lt;/li&gt;&#xA;&lt;li&gt;Leave 998 banana after 1 km and return with 1 banana – will eat up 1 banana in the way back&lt;/li&gt;&#xA;&lt;li&gt;Pick up the next 1000 bananas and move forward – Will eat up 1 banana in the way forward&lt;/li&gt;&#xA;&lt;li&gt;Leave 998 banana after 1 km and return with 1 banana - will eat up 1 banana in the way back&lt;/li&gt;&#xA;&lt;li&gt;Will carry the last 1000 bananas from point a and move forward – will eat up 1 banana&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note: After point 5 the Camel does not need to return to point A again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to shift 3000 bananas by 1km, the Camel will eat up 5 bananas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After moving to 200 km the Camel would have eaten up 1000 bananas and is now left with 2000 bananas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the length of part P1 is 200 Km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in the Part P2, the Camel needs to do the following to shift the Bananas by 1km.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Move forward with 1000 bananas - Will eat up 1 banana in the way forward&lt;/li&gt;&#xA;&lt;li&gt;Leave 998 banana after 1 km and return with 1 banana - will eat up this 1 banana in the way back&lt;/li&gt;&#xA;&lt;li&gt;Pick up the next 1000 bananas and move forward - Will eat up 1 banana in the way forward&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note: After point 3 the Camel does not need to return to the starting point of P2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to shift 2000 bananas by 1km, the Camel will eat up 3 bananas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After moving to 333 km the camel would have eaten up 1000 bananas and is now left with the last 1000 bananas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because it is a multiple of 3, after 333 times there is 1001 bananas left. Therefore, the camel must take 1000 bananas and then take 1 back with him to pick up the last banana, leaving him with 999 bananas. This would leave the camel at 334 km + 200 km, so the merchant only needs to travel 466 km, eating 1 banana for every km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;999 bananas - 466 bananas for each remaining km equals 533 bananas left at point B. Therefore, the merchant has 533 bananas to sell at point B.&lt;/p&gt;&#xA;" OwnerUserId="184" LastEditorUserId="-1" LastEditDate="2017-11-16T15:00:38.443" LastActivityDate="2017-11-16T15:00:38.443" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="233" PostTypeId="1" AcceptedAnswerId="244" CreationDate="2014-05-18T22:47:42.777" Score="9" ViewCount="2626" Body="&lt;p&gt;This is a problem I once found in a math textbook a tutor used with me.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A group of explorers is trying to get from their base camp to another camp which is eight days' walk through the desert.&lt;/p&gt;&#xA;&lt;p&gt;However, each explorer can only carry five days' worth of supplies. How many explorers need to venture out to make sure that at least two of the explorers make it to the second camp and the rest make it safely back to base camp before being stranded without supplies?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This problem is somewhat similar to the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/230/a-camel-transporting-bananas&quot;&gt;camel transporting bananas puzzle&lt;/a&gt;, but this time there are multiple people who can exchange supplies between each other.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2021-05-07T09:26:48.040" Title="Travellers across a desert" Tags="|calculation-puzzle|word-problem|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="234" PostTypeId="2" ParentId="228" CreationDate="2014-05-18T23:14:21.130" Score="8" Body="&lt;p&gt;I like the Lars Petrus method. it's intuitive up until the last layer which you can then solve by memorizing only 3 turn sequences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First you solve a 2x2x2 corner. That is 3 sides all adjacent to a corner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then without turning the faces that make up the already solved corner you expand it to a 2x2x3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that you orient the remaining edges (you should be able to make a double cross on the unsolved faces without moving the partially solved faces) using the move sequence &lt;code&gt;R U R'&lt;/code&gt; (that is Right face away from you, Up clockwise and then Right back again) which orients the edge on the front face on the left side and the edge on the top face at the back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then moving only the top and front faces you solve the bottom 2 layers. This is a bit trickier but easily doable with some experience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should have the top edges oriented correctly by this point. Now begins the sequence memorization:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First you permute the corners using &lt;code&gt;R' U L U' R U L U2&lt;/code&gt;, this flips the left hand corners (and permutes the edges). so they are in the correct location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you flip the corners with &lt;code&gt;R U R' U R U2 R'&lt;/code&gt; which flips the corners in the top left, top right and bottom right corners (when looking at the top face).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now all you need to do is permute the edges. The sequence for this is &lt;code&gt;F2 U L R' F2 L' R U F2&lt;/code&gt; which permutes the front, left and right edges in a clockwise fashion, for a counter clockwise rotation use U' instead of U.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a full tutorial available on &lt;a href=&quot;http://lar5.com/cube/&quot; rel=&quot;nofollow&quot;&gt;his website&lt;/a&gt; complete with animated 3D cubes&lt;/p&gt;&#xA;" OwnerUserId="81" LastEditorUserId="81" LastEditDate="2016-09-28T20:44:52.357" LastActivityDate="2016-09-28T20:44:52.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="235" PostTypeId="2" ParentId="233" CreationDate="2014-05-18T23:21:46.090" Score="2" Body="&lt;p&gt;The 2 travelers need 16DoS (Days of Supplies).&#xA;They can carry 10DoS. They need 6 more DoS. 2 people are needed to carry their DoS. We want the minimal amount of people, so let us just say that the 2 people use 3 DoS from the 2 other people, which means that the 3 DoS that they are using finishes in 3 days. They 2 other people need 12 more for the trip for 3 days and back, but they only have room for 4 DoS. They need other people to carry the 8 DoS. The number went from 6--&gt;8 so this is a futile attempt of logic. &lt;/p&gt;&#xA;" OwnerUserId="167" LastEditorUserId="167" LastEditDate="2015-09-19T17:57:19.537" LastActivityDate="2015-09-19T17:57:19.537" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="236" PostTypeId="1" AcceptedAnswerId="2890" CreationDate="2014-05-18T23:25:56.707" Score="196" ViewCount="95767" Body="&lt;h2&gt;The riddle&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Randall Munroe (of &lt;a href=&quot;https://xkcd.com&quot;&gt;xkcd&lt;/a&gt; fame) has, a bit hidden on his site, &lt;a href=&quot;https://xkcd.com/blue_eyes.html&quot;&gt;a logic puzzle&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A group of people with assorted eye colors live on an island. They are&#xA;  all perfect logicians -- if a conclusion can be logically deduced,&#xA;  they will do it instantly. No one knows the color of their eyes. Every&#xA;  night at midnight, a ferry stops at the island. Any islanders who have&#xA;  figured out the color of their own eyes then leave the island, and the&#xA;  rest stay. Everyone can see everyone else at all times and keeps a&#xA;  count of the number of people they see with each eye color (excluding&#xA;  themselves), but they cannot otherwise communicate. Everyone on the&#xA;  island knows all the rules in this paragraph.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;On this island there are 100 blue-eyed people, 100 brown-eyed people,&#xA;  and the Guru (she happens to have green eyes). So any given blue-eyed&#xA;  person can see 100 people with brown eyes and 99 people with blue eyes&#xA;  (and one with green), but that does not tell him his own eye color; as&#xA;  far as he knows the totals could be 101 brown and 99 blue. Or 100&#xA;  brown, 99 blue, and he could have red eyes.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The Guru is allowed to speak once (let's say at noon), on one day in&#xA;  all their endless years on the island. Standing before the islanders,&#xA;  she says the following:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;I can see someone who has blue eyes.&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Who leaves the island, and on what night?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;There are no mirrors or reflecting surfaces, nothing dumb. It is not a&#xA;  trick question, and the answer is logical. It doesn't depend on tricky&#xA;  wording or anyone lying or guessing, and it doesn't involve people&#xA;  doing something silly like creating a sign language or doing genetics.&#xA;  The Guru is not making eye contact with anyone in particular; she's&#xA;  simply saying &quot;I count at least one blue-eyed person on this island&#xA;  who isn't me.&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;And lastly, the answer is not &quot;no one leaves.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He admits the puzzle isn't his:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I didn't come up with the idea of this puzzle, but I've written and rewritten it over the the years to try to make a definitive version. The guy who told it to me originally was some dude on the street in Boston named Joel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;The answer&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;He gives &lt;a href=&quot;https://xkcd.com/solution.html&quot;&gt;his solution&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The answer is that on the 100th day, all 100 blue-eyed people will leave. It's pretty convoluted logic and it took me a while to believe the solution, but here's a rough guide to how to get there. Note -- while the text of the puzzle is very carefully worded to be as clear and unambiguous as possible (thanks to countless discussions with confused readers), this solution is pretty thrown-together. It's correct, but the explanation/wording might not be the best. If you're really confused by something, let me know.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If you consider the case of just one blue-eyed person on the island, you can show that he obviously leaves the first night, because he knows he's the only one the Guru could be talking about. He looks around and sees no one else, and knows he should leave. So: [THEOREM 1] If there is one blue-eyed person, he leaves the first night.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If there are two blue-eyed people, they will each look at the other. They will each realize that &quot;if I don't have blue eyes [HYPOTHESIS 1], then that guy is the only blue-eyed person. And if he's the only person, by THEOREM 1 he will leave tonight.&quot; They each wait and see, and when neither of them leave the first night, each realizes &quot;My HYPOTHESIS 1 was incorrect. I must have blue eyes.&quot; And each leaves the second night.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;So: [THEOREM 2]: If there are two blue-eyed people on the island, they will each leave the 2nd night.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If there are three blue-eyed people, each one will look at the other two and go through a process similar to the one above. Each considers the two possibilities -- &quot;I have blue eyes&quot; or &quot;I don't have blue eyes.&quot; He will know that if he doesn't have blue eyes, there are only two blue-eyed people on the island -- the two he sees. So he can wait two nights, and if no one leaves, he knows he must have blue eyes -- THEOREM 2 says that if he didn't, the other guys would have left. When he sees that they didn't, he knows his eyes are blue. All three of them are doing this same process, so they all figure it out on day 3 and leave.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This induction can continue all the way up to THEOREM 99, which each person on the island in the problem will of course know immediately. Then they'll each wait 99 days, see that the rest of the group hasn't gone anywhere, and on the 100th night, they all leave.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Before you email me to argue or question: This solution is correct. My explanation may not be the clearest, and it's very difficult to wrap your head around (at least, it was for me), but the facts of it are accurate. I've talked the problem over with many logic/math professors, worked through it with students, and analyzed from a number of different angles. The answer is correct and proven, even if my explanations aren't as clear as they could be.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;User lolbifrons on reddit posted &lt;a href=&quot;http://www.reddit.com/r/AskReddit/comments/khhpl/reddit_what_is_your_favorite_riddle/c2kdlr6&quot;&gt;an inductive proof&lt;/a&gt;.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If you're satisfied with this answer, here are a couple questions that may force you to further explore the structure of the puzzle: &lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;What is the quantified piece of information that the Guru provides that each person did not already have? &lt;/li&gt;&#xA;  &lt;li&gt;Each person knows, from the beginning, that there are no less than 99 blue-eyed people on the island. How, then, is considering the 1 and 2-person cases relevant, if they can all rule them out immediately as possibilities? &lt;/li&gt;&#xA;  &lt;li&gt;Why do they have to wait 99 nights if, on the first 98 or so of these nights, they're simply verifying something that they already know? &lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;These are just to give you something to think about if you enjoyed the main solution. They have answers, but please don't email me asking for them. They're meant to prompt thought on the solution, and each can be answered by considering the solution from the right angle, in the right terms. There's a different way to think of the solution involving hypotheticals inside hypotheticals, and it is much more concrete, if a little harder to discuss. But in it lies the key to answering the four questions above.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;h2&gt;The question&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Everybody on the island could have come to the conclusion that 'There is at least one person with blue eyes', simply by looking around, seeing 100 people with blue eyes, and realising that everybody can see at least one person with blue eyes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So why is it necessary for the Guru to say 'I see at least one person with blue eyes' to get the ball rolling?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="185" LastEditorUserId="36023" LastEditDate="2020-10-09T19:22:15.817" LastActivityDate="2023-02-15T22:04:30.543" Title="In the 100 blue eyes problem - why is the oracle necessary?" Tags="|logical-deduction|story|meta-knowledge|blue-eyes|" AnswerCount="26" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="237" PostTypeId="2" ParentId="236" CreationDate="2014-05-19T01:58:03.463" Score="16" Body="&lt;p&gt;There are a lot of explanations for this, and certainly also a lot of debate over this question as the problem is extremely counterintuitive. Therefore, no explanation I could give or anybody could give will come close to satisfying everyone, but I will try anyway.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Although every islander knows that there is at least one person in the island with blue eyes, the blue-eyed people do &lt;em&gt;not&lt;/em&gt; know whether there are 99 or 100 people with blue eyes on the island.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The guru coming and saying there is a person on the island with blue eyes allows them to start the chain of inferences alluded to in the solution and conclude that if everybody does not leave in 99 days, they are a person with blue eyes as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason they cannot start this chain of inferences themselves boils down to the fact that although they see somebody with blue eyes, they cannot determine how many days to wait (either 98 and I am not blue-eyed, or 99 and I am blue-eyed) because they do not know the total number of blue-eyed people on the island. You need somebody &lt;em&gt;outside&lt;/em&gt; their group to come and tell them that there is at least one person with blue eyes, so that you have the inductive base case of one blue-eyed person to build on top of and determine how many days to wait.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="2099" LastEditDate="2014-08-15T11:40:06.210" LastActivityDate="2014-08-15T11:40:06.210" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="238" PostTypeId="2" ParentId="236" CreationDate="2014-05-19T01:59:15.837" Score="35" Body="&lt;p&gt;The whole process is inductive, so it needs a starting point. If there were only one blue-eyed person, he would never know that there is &quot;at least one person with blue eyes,&quot; so he would not go the first night. If there are only two, neither of them can know whether the other doesn't go the first night because he only sees brown eyes, so they don't know if they should go the second night. A third wouldn't be able to know if the first two hadn't gone because they only see one each or two, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the oracle makes his statement, it ensures that a hypothetical lone blue-eyed person would know he is the one, which allows the induction to begin. &lt;/p&gt;&#xA;" OwnerUserId="4" LastActivityDate="2014-05-19T01:59:15.837" CommentCount="20" ContentLicense="CC BY-SA 3.0" />
  <row Id="239" PostTypeId="5" CreationDate="2014-05-19T02:30:12.460" Score="0" Body="&lt;p&gt;A Sudoku is a puzzle consisting of a grid of &lt;em&gt;cells&lt;/em&gt; to be filled with symbols, such that each constraining &lt;em&gt;unit&lt;/em&gt; of the grid contains each symbol exactly once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The typical Sudoku is a 9x9 square to be filled with the numbers 1 to 9, with the constraining units being &lt;em&gt;rows&lt;/em&gt;, &lt;em&gt;columns&lt;/em&gt;, and 3x3 &lt;em&gt;blocks&lt;/em&gt;.&#xA;Some cells already have a value - these are called &lt;em&gt;clues&lt;/em&gt;, or &lt;em&gt;givens&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;em&gt;proper&lt;/em&gt; Sudoku has a unique solution. A &lt;em&gt;satisfactory&lt;/em&gt; Sudoku is a proper Sudoku that needs no trial and error to reach that solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other popular variants include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Killer - there are &lt;em&gt;cages&lt;/em&gt; around cells that tell the sum of the cage&lt;/li&gt;&#xA;&lt;li&gt;X-factor - on each of the two main diagonals the symbols must also be distinct&lt;/li&gt;&#xA;&lt;li&gt;Shape - other shapes are used instead of blocks&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="68" LastEditorUserId="11497" LastEditDate="2019-03-10T10:45:17.283" LastActivityDate="2019-03-10T10:45:17.283" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="240" PostTypeId="4" CreationDate="2014-05-19T02:30:12.460" Score="0" Body="A Sudoku is a puzzle consisting of a grid to be filled with symbols, such that each constraining unit contains each symbol exactly once.&#xD;&#xA;&#xD;&#xA;The typical example is a 9x9 square to be filled with the numbers 1 to 9, where the constraining units are rows, columns, and 3x3 blocks." OwnerUserId="166" LastEditorUserId="8874" LastEditDate="2015-01-31T18:10:03.563" LastActivityDate="2015-01-31T18:10:03.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="241" PostTypeId="2" ParentId="236" CreationDate="2014-05-19T05:18:03.590" Score="15" Body="&lt;p&gt;As you did, let's reduce it to the case of three people for clarity's sake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aaron, Bob, and Charlie have blue eyes. No guru says anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aaron thinks: If Bob sees only Charlie with blue eyes, then Bob knows after the first night, viz after Charlie doesn't leave, that Bob has blue eyes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Er, no. That'd be true if the guru said someone has blue eyes. But that's not true now: Charlie's not leaving doesn't mean anything, as no one has told him he has blue eyes. So (in Aaron's mind) Bob doesn't, even if he sees only Charlie with blue eyes, know after Charlie doesn't leave the first night that Bob has blue eyes.&lt;/p&gt;&#xA;" OwnerUserId="93" LastActivityDate="2014-05-19T05:18:03.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="242" PostTypeId="2" ParentId="236" CreationDate="2014-05-19T06:57:31.670" Score="132" Body="&lt;p&gt;The knowledge of each islander consists of:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the color of the eyes of every other islander;&lt;/li&gt;&#xA;&lt;li&gt;any past pronouncement from the guru;&lt;/li&gt;&#xA;&lt;li&gt;the history of who left the island on previous days (including their eye color), which provides knowledge about others' knowledge (that either they did or did not know their own eye color on previous days).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;At the beginning of the story, nobody has ever left the island and there is no past pronouncement. So the only information everyone has is the color of everybody else's eyes, and the fact that nobody has figured out their own eye color. This is a stable situation, which lasts forever. It is in fact quite intuitive that since nobody has any information that involves in any way the color of their own eyes, nobody can be certain of the color of their own eyes.&lt;/p&gt;&#xA;&lt;p&gt;Let's say that the guru makes her pronouncement on day 0. Starting on day 0, each islander has extra information: up to n days after the pronouncement, nobody left, meaning that nobody could figure out the color of their own eyes.&lt;/p&gt;&#xA;&lt;p&gt;Suppose that only Alice has blue eyes. Before day 0, she never knew anyone with blue eyes. On day 0, she learns that someone has blue eyes; since nobody else does, it has to be her and only her, so she takes the ferry that night.&lt;/p&gt;&#xA;&lt;p&gt;Now suppose that only Alice and Bill have blue eyes. Before day 0, Bill already knew that there was someone with blue eyes, but &lt;strong&gt;he did not know that Alice knew&lt;/strong&gt;. If Bill had had green eyes, Alice would have been the only blue-eyed person and would not have known. On the first night after the guru, Alice doesn't leave; this tells Bill that Alice did not know the color of her eyes, so Bill learns that she was not the only blue-eyed person. Since Bill knows that either Alice is the only blue-eyed person or Bill and Alice are the only two, Bill now knows that both he and Alice have blue eyes.&lt;/p&gt;&#xA;&lt;p&gt;If Charlie also has blue eyes, then he follows the reasoning above. Since Alice and Bill do not leave on the second night, it follows that they are not the only two people with blue eyes, so Charlie figures out that he's the third and leaves the next night.&lt;/p&gt;&#xA;&lt;p&gt;The information that islander X learns from the guru is not just “someone has blue eyes”, but also “&lt;strong&gt;Y knows that X knows&lt;/strong&gt; that someone has blue eyes”, “Z knows that Y knows that X knows that someone has blue eyes”, etc. It's vital to the puzzle that &lt;strong&gt;the guru's declaration is public and known to be public&lt;/strong&gt;. If some of the islanders didn't hear the announcement, the chain of deduction wouldn't work anymore.&lt;/p&gt;&#xA;" OwnerUserId="71" LastEditorUserId="43053" LastEditorDisplayName="user9366" LastEditDate="2021-06-27T20:08:16.707" LastActivityDate="2021-06-27T20:08:16.707" CommentCount="30" ContentLicense="CC BY-SA 4.0" />
  <row Id="243" PostTypeId="2" ParentId="228" CreationDate="2014-05-19T07:28:50.467" Score="15" Body="&lt;p&gt;The World Cubing Association actually has an entire event dedicated to solving the cube with the fewest move count (FMC). You are given a scramble, paper, pencil, up to 3 cubes and an hour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The methods used vary widely and involve much more &quot;freestyle&quot; than regular speedsolving. But I'll explain one of the more common &quot;methods.&quot; (Which is more like a set of guidelines.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just a warning, this will likely take a few readings to understand, as it's a very deep topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, you try to preserve any blocks that might be present in the scramble and aim to build a 3x2x2 block. It helps to first build a 2x2x2 block, then extend it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here, you can extend your 3*2*2 block to have a cross, with your choice from 3 cross colours. This is usually a fork in the road, where a cuber can explore the different paths to see which one gives the more favorable outcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can solve one of the two empty F2L slots that this creates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now various means are used to orient and permute the last 5 edges. This often involves a combination of freestyle and algorithms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point there will be at most 5 corners left to solve. This is done with &quot;commutators.&quot; They're essentially 8-move algorithms that cycle any 3 corners and can be come up with on the fly. 1-3 commutators are needed to finish off the cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it doesn't stop there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An experienced FMCer would backtrack through their solution to find a good spot to insert a commutator. A good spot means that you can cancel some moves. (eg. if your commutator alg starts with R, you might find a place that ends with R' so that you can cancel both moves.) Thus the 1-3 commutators that would usually take anywhere between 8 and 24 moves to complete, can now be done in a fraction of the moves. Sometimes you can cancel a ridiculous number of moves. A friend of mine went from requiring 16 moves to solve the corners to only 3 because of move cancelation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the masters of FMC use all kinds of other special tricks for special cases in order to get even shorter solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tomoaki Okayama from Japan (who I've had the pleasure of meeting) set the world record by finding a 20 move solution at a competition in 2012.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to finish this of by mentioning that this is much more of a dynamic programming approach than a &quot;greedy&quot; one, since it involves backtracking. The most efficient greedy and speedsolving method that can also be insanely fast (ie. 7 second average) would probably be the Roux method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All in all, FMC is a very interesting problem that takes a tremendous amount of patience and skill.&lt;/p&gt;&#xA;" OwnerUserId="157" LastEditorUserId="157" LastEditDate="2014-05-19T07:38:07.137" LastActivityDate="2014-05-19T07:38:07.137" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="244" PostTypeId="2" ParentId="233" CreationDate="2014-05-19T10:03:02.920" Score="7" Body="&lt;p&gt;Six explorers, according to the following schedule:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Day 1: 6 explorers start with 28 rations and end with 22. Two return with 2 rations (1 per).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Day 2: 4 explorers start with 20 rations and end with 16. One returns with 2 rations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Day 3: 3 explorers start with 14 rations and end with 11. One returns with 3 rations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Day 4: 2 explorers start with  8 rations and end with  6.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Day 5: 2 explorers start with  6 rations and end with  4.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Day 6: 2 explorers start with  4 rations and end with  2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Day 7: 2 explorers start with  2 rations and end with  0.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Could &amp;quot;almost&amp;quot; do this with 5 and just let them get a little hungry but the six is needed just for one meal. Note that there's no need to load everyone to max at the beginning since 4 explorers can only carry 20 meals.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;As pointed out in comments, the solution above assumes that they eat at the end of the day (where day 8 should be &amp;quot;start with 0 and arrive hungry&amp;quot;). Below follows the rather similar solution for the more reasonable scenario of eat at the start or during the day.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Day 1: 8 explorers start with 40 rations and end with 32. Two return with 2 rations (1 per).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Day 2: 6 explorers start with 30 rations and end with 24. Two return with 4 rations (2 per).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Day 3: 4 explorers start with 20 rations and end with 16. Two return with 6 rations (3 per).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Day 4: 2 explorers start with 10 rations and end with  8.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Day 5: 2 explorers start with  8 rations and end with  6.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Day 6: 2 explorers start with  6 rations and end with  4.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Day 7: 2 explorers start with  4 rations and end with  2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Day 8: 2 explorers start with  2 rations and end with  0.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This solution is more symmetrical, which feels better somehow.&lt;/p&gt;&#xA;" OwnerUserId="115" LastEditorUserId="63414" LastEditDate="2021-05-07T09:25:38.460" LastActivityDate="2021-05-07T09:25:38.460" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="245" PostTypeId="2" ParentId="230" CreationDate="2014-05-19T10:29:12.163" Score="19" Body="&lt;p&gt;While the amount of bananas is larger than 2000, the camel will have to make 5 trips to shift them. At a cost of 1 banana per mile, this will cost 5 bananas per mile in total.&lt;br&gt;&#xA;While the amount of bananas is larger than 1000, it takes 3 trips to shift them, so a total cost of 3 bananas per mile.&lt;br&gt;&#xA;For the final stretch, it just takes a single banana per mile.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;3000 - 2000 bananas (numbers given for first mile)&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Move 1000 bananas, feed 1, 2999 left.&lt;/li&gt;&#xA;&lt;li&gt;Go back, feed 1, 2998 left.&lt;/li&gt;&#xA;&lt;li&gt;Move 1000 bananas, feed 1, 2997 left.&lt;/li&gt;&#xA;&lt;li&gt;Go back, feed 1, 2996 left.&lt;/li&gt;&#xA;&lt;li&gt;Move 998 bananas, feed 1, 2995 left.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;2000 - 1000 bananas (numbers given for first mile)&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Move 1000 bananas, feed 1, 1999 left.&lt;/li&gt;&#xA;&lt;li&gt;Go back, feed 1, 1998 left.&lt;/li&gt;&#xA;&lt;li&gt;Move 999 bananas, feed 1, 1997 left.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;1000 bananas (numbers given for first mile)&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Move 1000 bananas, feed 1, 999 left.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So at 5 bananas/mile you're down to 2000 bananas after 200 miles. Your total mileage then improves to 3 bananas/mile for the next 333 miles. After that, you're on the home stretch and move at 1 banana/mile for the remaining 467 miles (after which you'll have 533 bananas left).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few things to note:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It is not necessary to move in one mile increments. For the first stretch, you can load up with 1000 bananas, move 200 miles, unload 600, and use 200 for your return trip. You do the same thing again, and after the last trip, you can unload 800, or rather, load 200 to be at a full 1000 again.&lt;/li&gt;&#xA;&lt;li&gt;After completing the 333 mile stretch, you're 467 miles away from your destination, with exactly 1001 bananas. There are now two solutions:&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Eat a banana (or leave it to rot) to have a nice, integer solution.&lt;/li&gt;&#xA;&lt;li&gt;Use it to move for a total of 333&amp;#8531; miles. If the camel needs to be fed &lt;em&gt;after&lt;/em&gt; walking a full mile, you will now make it to your destination with &amp;#8531; of a mile to spare, which should save you a banana, leaving you with &lt;strong&gt;534&lt;/strong&gt; bananas.&lt;br&gt;&#xA;Congratulations, you're now banana king!&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditDate="2014-05-20T06:12:25.450" LastActivityDate="2014-05-20T06:12:25.450" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="247" PostTypeId="2" ParentId="227" CreationDate="2014-05-19T11:16:32.973" Score="5" Body="&lt;p&gt;For puzzle and precision longevity, I would also suggest using harder woods. I have a few of these puzzles from olive wood and they're too soft (feel like lime/linden). If you're not careful while solving them, it's easy to damage the vertices. After several runs, the starting precision matters little.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2014-05-19T11:16:32.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="250" PostTypeId="2" ParentId="197" CreationDate="2014-05-19T11:20:41.703" Score="4" Body="&lt;p&gt;Perhaps not in the spirit of the question, but all the &lt;a href=&quot;http://www.omniglot.com/writing/undeciphered.htm&quot; rel=&quot;nofollow&quot;&gt;scripts&lt;/a&gt; we haven't yet decoded from that period could be considered unsolved puzzles.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2014-05-19T11:20:41.703" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="251" PostTypeId="1" AcceptedAnswerId="257" CreationDate="2014-05-19T13:25:54.697" Score="49" ViewCount="15631" Body="&lt;p&gt;Two perfect logicians, Summer and Proctor, are told that integers  and  have been chosen such that 1 &amp;lt;  &amp;lt;  and  +  &amp;lt; 100.  Summer is given the value  +  and Proctor is given the value  ⋅ .  They then have the following conversation.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Proctor:&lt;/em&gt;  &amp;quot;I cannot determine the two numbers.&amp;quot;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Summer:&lt;/em&gt;  &amp;quot;I knew that.&amp;quot;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Proctor:&lt;/em&gt;  &amp;quot;Now I can determine them.&amp;quot;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Summer:&lt;/em&gt;  &amp;quot;So can I.&amp;quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Given that the above statements are true, what are the two numbers?&lt;/p&gt;&#xA;&lt;p&gt;You may want to use a computer to assist you.&lt;/p&gt;&#xA;" OwnerUserId="42" LastEditorUserId="69582" LastEditDate="2021-01-28T20:48:38.570" LastActivityDate="2021-01-28T20:58:49.833" Title="I don't know the two numbers... but now I do" Tags="|mathematics|logical-deduction|calculation-puzzle|meta-knowledge|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="252" PostTypeId="1" AcceptedAnswerId="8032" CreationDate="2014-05-19T14:01:46.107" Score="22" ViewCount="49643" Body="&lt;p&gt;This question is posted under the guidelines &lt;a href=&quot;https://puzzling.meta.stackexchange.com/questions/3/is-it-on-topic-to-present-specific-logic-puzzles-or-brainteasers-as-questions&quot;&gt;&lt;strong&gt;Don't worry too hard about restricting or regulating anything that isn't turning into a problem yet.&lt;/strong&gt;&lt;/a&gt; If you don't agree that this question is on topic, &lt;a href=&quot;https://puzzling.meta.stackexchange.com/questions/3/is-it-on-topic-to-present-specific-logic-puzzles-or-brainteasers-as-questions&quot;&gt;&lt;strong&gt;please go to that meta thread&lt;/strong&gt;&lt;/a&gt; and talk about why you feel that way!*&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solve this Sudoku. Post how you did it in your answer. Enjoy!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IA61i.jpg&quot; alt=&quot;8..........36......7..9.2...5...7.......457.....1...3...1....68..85...1..9....4..&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I put this program into the solver on &lt;a href=&quot;http://www.sudokuwiki.org/sudoku.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;sudokuwiki.org&lt;/a&gt; and it couldn't find any numbers. I then gave it cell H7 (the only cell with two possibilities) and still no luck. Then I gave it cell G7 (which became the only cell with two possibilities) and it was only able to solve one cell before it got stuck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://aisudoku.com/index_en.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here's the website of the mathematician who discovered this puzzle.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="42" LastEditorUserId="-1" LastEditDate="2017-03-16T16:42:07.097" LastActivityDate="2016-08-29T06:59:03.930" Title="How do I solve the world's hardest sudoku?" Tags="|sudoku|" AnswerCount="4" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="253" PostTypeId="2" ParentId="233" CreationDate="2014-05-19T14:18:15.813" Score="1" Body="&lt;p&gt;Assuming you have to feed the travelers on day 8:&#xA;Solution 1: 8 leave and go to camp 1.  Two refill the packs of the other six and return.  Six go to camp 2.  Four refill the packs of the others and return.  Four go to camp 3.  Two refill the packs of the two that will complete the trip and return.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solution 2:  Seven leave and go to camp 1.  Two refill the packs of the other five, hide one ration, and return.  Five go to camp 2.  Two refill the packs of the other three, hide one ration, and return.  Three go to camp 3.  One refills the packs of the other two, who will complete the journey.  He returns, using the hidden rations.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the first solution is the desired one.  It is tight: there is no food to spare.  The second has a spare ration.  The accepted answer has more slack.  Usually setters try to avoid this and pick the parameters of the problem this way.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2014-05-19T14:18:15.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="255" PostTypeId="1" AcceptedAnswerId="258" CreationDate="2014-05-19T14:47:08.213" Score="16" ViewCount="9939" Body="&lt;p&gt;Suppose you have a checkerboard with two opposite corner squares removed, like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ln6nc.gif&quot; alt=&quot;Mutilated chessboard&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to place 31 dominoes of size 2x1 so as to cover all of these squares?&lt;/p&gt;&#xA;" OwnerUserId="129" LastEditorUserId="57212" LastEditDate="2022-07-11T07:04:14.347" LastActivityDate="2022-07-11T07:04:14.347" Title="Placing 2x1 dominoes on a chessboard with two corners removed" Tags="|checkerboard|dissection|tiling|reachability|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="256" PostTypeId="2" ParentId="255" CreationDate="2014-05-19T14:47:45.330" Score="6" Body="&lt;h1&gt;No!&lt;/h1&gt;&#xA;&lt;p&gt;Each domino would occupy one black and one white cell. So, 31 dominoes would require exactly 31 white, and 31 black cells.&lt;/p&gt;&#xA;&lt;p&gt;The chessboard given, however, has 32 white and 30 black cells making it impossible to place the dominoes.&lt;/p&gt;&#xA;" OwnerUserId="129" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-05-19T14:47:45.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="257" PostTypeId="2" ParentId="251" CreationDate="2014-05-19T14:59:35.373" Score="41" Body="&lt;p&gt;Let  =  ⋅  and  =  + .&lt;/p&gt;&#xA;&lt;p&gt;If Proctor could not determine the two numbers offhand, then there must be at least two valid factorizations of , which means that  cannot be the product of two primes.&lt;/p&gt;&#xA;&lt;p&gt;And if Summer &lt;em&gt;knew&lt;/em&gt; that Proctor could not determine the two numbers, then &lt;em&gt;every possible pair of&lt;/em&gt;  &lt;em&gt;and&lt;/em&gt;  that add up to  must have a product that is not a product of two primes - i.e. no two primes can sum to  and . This gives &lt;strong&gt;11, 17, 23, 27, 29&lt;/strong&gt;, etc. (in general, every odd composite number plus 2) as possible values of , because odd prime numbers plus 2 have &lt;strong&gt;2&lt;/strong&gt; and  - 2 as the prime numbers that sum to it, and any even number greater than or equal to 4 can be represented as the sum of two primes, by the Goldbach conjecture which has been verified up to 100.&lt;/p&gt;&#xA;&lt;p&gt;Proctor knows this as well. If he &lt;em&gt;can&lt;/em&gt; now determine the two numbers, then out of all of 's factorizations, only &lt;em&gt;one&lt;/em&gt; of them add up to one of those possible values of . For example, if  = 24, then only &lt;strong&gt;3&lt;/strong&gt; and &lt;strong&gt;8&lt;/strong&gt; add up to a number in that list. (The other factorizations give 1 + 24 = 25, 2 + 12 = 14, and 4 + 6 = 10, none of which are in the list.)&lt;/p&gt;&#xA;&lt;p&gt;Summer knows this as well. But if &lt;em&gt;Summer&lt;/em&gt; can also figure out what  is, then the number of  must also have a unique  for which a unique  exists. Our value of  = 24 above gives  = 11, which is &lt;em&gt;not&lt;/em&gt; the right answer because  = 18 gives (2, 9) as another solution to  = 11, and so our value of  is not unique.&lt;/p&gt;&#xA;&lt;p&gt;With some brute-force searching by going through all possible values of  and checking their respective unique values of , we find that only  = 17 has only one corresponding  value, which is  = 52.&lt;/p&gt;&#xA;&lt;p&gt;So,  = 17 and  = 52, and  = 4 and  = 13.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="69582" LastEditDate="2021-01-28T20:58:49.833" LastActivityDate="2021-01-28T20:58:49.833" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="258" PostTypeId="2" ParentId="255" CreationDate="2014-05-19T15:27:41.977" Score="12" Body="&lt;p&gt;This puzzle is known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Mutilated_chessboard_problem&quot;&gt;mutilated chessboard problem&lt;/a&gt;. The other answer correctly explains that such a covering is &lt;strong&gt;impossible&lt;/strong&gt; because it would require an equal number of black and white squares (since each domino must cover one black and one white square), which the corner-cut board does not have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The link above contains a slightly more general result as well:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The same impossibility proof shows that no &lt;a href=&quot;http://en.wikipedia.org/wiki/Domino_tiling&quot;&gt;domino tiling&lt;/a&gt; exists whenever any two white squares are removed from the chessboard. However, if two squares of opposite colors are removed, then it is always possible to tile the remaining board with dominos; this result is called Gomory's theorem, and is named after mathematician &lt;a href=&quot;http://en.wikipedia.org/wiki/Ralph_E._Gomory&quot;&gt;Ralph E. Gomory&lt;/a&gt;, whose proof was published in 1973. &lt;strong&gt;Gomory's theorem&lt;/strong&gt; can be proven using a &lt;a href=&quot;http://en.wikipedia.org/wiki/Hamiltonian_cycle&quot;&gt;Hamiltonian cycle&lt;/a&gt; of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Grid_graph&quot;&gt;grid graph&lt;/a&gt; formed by the chessboard squares; the removal of two oppositely-colored squares splits this cycle into two paths with an even number of squares each, both of which are easy to partition into dominos.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="198" LastActivityDate="2014-05-19T15:27:41.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="259" PostTypeId="1" AcceptedAnswerId="272" CreationDate="2014-05-19T15:34:46.030" Score="5" ViewCount="818" Body="&lt;p&gt;I was just thinking about this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say that we have a card deck with 54 cards and the $n$th card from the top is flipped. We repeat a short process of cutting the deck directly in the middle  and then interleaving the cards from each pile (in other words, a perfect shuffle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if the cards were numbered from 1 to 54 from top to bottom, after interleaving the cards, the top of the deck would look like: 28, 1, 29, 2, 30, 3, ... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We repeat this process until we cut the deck in half and find the flipped card where we cut.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a function of the position of the flipped card from the top ($f(n)$) for how many times you have to do this before you find the flipped card upon cutting it?&lt;/p&gt;&#xA;" OwnerUserId="167" LastEditorUserId="167" LastEditDate="2015-09-19T17:59:52.873" LastActivityDate="2015-09-19T17:59:52.873" Title="Shuffling Cards" Tags="|calculation-puzzle|mathematics|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="260" PostTypeId="1" AcceptedAnswerId="1670" CreationDate="2014-05-19T15:55:44.747" Score="17" ViewCount="3984" Body="&lt;p&gt;The &lt;a href=&quot;http://blog.codinghorror.com/the-problem-of-the-unfinished-game/&quot; rel=&quot;noreferrer&quot;&gt;first time&lt;/a&gt; I encountered the &lt;a href=&quot;http://en.wikipedia.org/wiki/Boy_or_Girl_paradox&quot; rel=&quot;noreferrer&quot;&gt;Boy or Girl paradox&lt;/a&gt; I was struck by how ambiguous all of the formulations in English are. As Wikipedia notes:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Gardner argued that a &amp;quot;failure to specify the randomizing procedure&amp;quot; could lead readers to interpret the question in two distinct ways:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;From all families with two children, at least one of whom is a boy, a family is chosen at random. This would yield the answer of 1/3.&lt;/li&gt;&#xA;&lt;li&gt;From all families with two children, one child is selected at random, and the sex of that child is specified. This would yield an answer of 1/2.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Many formulations have a less technical problem: when a parent says that they have &amp;quot;two children and one of them is a boy&amp;quot;, it's nearly impossible to read that in English without thinking &amp;quot;therefore the other one is a girl&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;Can you rephrase the paradox so that the proper randomizing procedure is specified and the Bayesian logic is the crux of the problem?&lt;/p&gt;&#xA;" OwnerUserId="15" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2021-04-23T10:27:04.467" Title="How can the &quot;Boy or Girl&quot; paradox be worded unambiguously?" Tags="|probability|puzzle-creation|paradox|" AnswerCount="11" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="261" PostTypeId="1" AcceptedAnswerId="262" CreationDate="2014-05-19T16:12:07.507" Score="1" ViewCount="580" Body="&lt;p&gt;Many people are familiar with the story of &lt;a href=&quot;https://en.wikipedia.org/wiki/Archimedes#Discoveries_and_inventions&quot; rel=&quot;nofollow&quot;&gt;Archimedes submerging a crown in water&lt;/a&gt; to identify if the volume of the crown was consistent with the volume of the same amount of gold.  The test was to see if a lighter cheaper metal (like silver) was included in crown.  If the crown maker substitutes silver for some of the gold, they make a sinister profit at the risk of their life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Platinum is heavier than gold, silver is lighter.  Using the values as of January of 2013, could &lt;strong&gt;any combination of metals&lt;/strong&gt; be used to create a crown that would have the same volume as a solid gold crown, with a cost effective benefit for the crown maker?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the values of metals in 287 BCE – 212 BCE, could it be cost effective for the crown maker.&lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="39" LastEditDate="2014-06-11T23:48:35.487" LastActivityDate="2014-07-03T23:38:00.850" Title="Could a gold/{second metal} crown the be created that would pass the Archimedes test cost effectively?" Tags="|logical-deduction|" AnswerCount="1" CommentCount="10" ClosedDate="2014-05-19T22:25:46.353" ContentLicense="CC BY-SA 3.0" />
  <row Id="262" PostTypeId="2" ParentId="261" CreationDate="2014-05-19T16:57:59.297" Score="6" Body="&lt;p&gt;Nowadays, depleted uranium, with around \$5/gram and 19.1 g/cm3 density could be used to replace some of the gold (probably as insert/skeleton of the crown, as both hardness and appearance wouldn't fool anyone) with gold at \$45/gram at 19.3 g/cm^3 - especially that pure gold would be very rare and it would be hard to discover such minimal difference in density, and adding a little platinum could balance it out to exact density (especially that the price is only marginally higher than gold, so 1g platinum + 1g DU costs way less than 2g of gold, and weighs more).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then, back in the past, uranium was nearly unobtainable, and there was definitely no technology of purifying it sufficiently (nor producing depleted... using normal uranium as from uranium ore to make a crown would be one really sinister method of murder...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now then, back in the past, platinum was called &quot;young silver&quot;, sometimes mined alongside real silver. It was deemed worthless though, and buried in the ground &quot;to lie another couple hundred years till it matures into real silver&quot;. Therefore, while the 2013 values of using platinum + silver (for example) instead of gold wouldn't earn you any considerable benefit, taking the prices from Archimedes' times, this would be quite the method.&lt;/p&gt;&#xA;" OwnerUserId="117" LastEditorUserId="228" LastEditDate="2014-07-03T23:38:00.850" LastActivityDate="2014-07-03T23:38:00.850" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="263" PostTypeId="1" AcceptedAnswerId="268" CreationDate="2014-05-19T17:02:00.497" Score="4" ViewCount="3708" Body="&lt;p&gt;Or phrased differently, are non-symmetrical Sudoku puzzles considered legal?&lt;/p&gt;&#xA;" OwnerUserId="200" LastEditorUserId="463" LastEditDate="2015-06-04T15:39:10.847" LastActivityDate="2018-08-20T01:59:00.953" Title="Is a 'legit' Sudoku puzzle supposed to be symmetrical?" Tags="|sudoku|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="264" PostTypeId="2" ParentId="263" CreationDate="2014-05-19T17:06:46.967" Score="4" Body="&lt;p&gt;Yes, non-symmetric Sudoku puzzles are valid. For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2HPTD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a minimal Sudoku puzzle, with exactly 17 clues, and is not symmetric. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image courtesy of &lt;a href=&quot;http://en.m.wikipedia.org/wiki/File:Oceans_Sudoku17_Puzzle-39451.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Wikimedia Commons&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastActivityDate="2014-05-19T17:06:46.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="265" PostTypeId="2" ParentId="260" CreationDate="2014-05-19T17:17:56.347" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Mr. Smith has two children. &lt;strong&gt;Mr. Smith only tells you that&lt;/strong&gt; at least one of them is a boy, &lt;strong&gt;but does not tell you anything else about them. Ceteris paribus,&lt;/strong&gt; what is the probability that both children are boys?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have added the part in bold, and this rids the problem of ambiguity. Mr. Smith has told you that, for his family, at least one of the two children is male.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gets you the intended answer of 1/3 for the problem by removing the possible interpretation that Mr. Smith specified the gender of one child.   &lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorDisplayName="user20" LastEditDate="2014-05-19T19:24:16.053" LastActivityDate="2014-05-19T19:24:16.053" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="266" PostTypeId="2" ParentId="260" CreationDate="2014-05-19T17:21:04.883" Score="3" Body="&lt;p&gt;The classical boy/girl question is asked like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I have two kids, and one of them is a boy. What is the probability that I have two boys?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The problem, as you said, is that the statement itself is incredibly ambiguous. The interpretation that most people think of (and correctly too, I'd say, taking linguistic context into account) when they hear the boy/girl problem is as such:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I have two kids, and one of them is a boy. What is the probability that the other one is also a boy?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In this case, the answer is 1/2, because &quot;one of them is a boy&quot; refers to a specific boy, rather than just the fact that at least one of the two children is a boy. This is equivalent to the second randomizing procedure set out in Gardner's procedure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To word it in a way that evokes the Bayesian inference, you need to perform gymnastics with the language of the problem that give the trick away.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I have two kids, and at least one of them is a boy, with all scenarios where this is applicable being equally likely. What is the probability that I have two boys?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's much too dry, and the ambiguity of the language is what makes the problem seem so controversial and counter-intuitive in the first place.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2014-05-19T17:39:18.830" LastActivityDate="2014-05-19T17:39:18.830" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="267" PostTypeId="2" ParentId="230" CreationDate="2014-05-19T17:30:12.490" Score="46" Body="&lt;p&gt;Solution for &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; bananas, where &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; is the number of bananas you own, and &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; is the number of bananas the camel can carry:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - For bananas &lt;span class=&quot;math-container&quot;&gt;$0 \rightarrow c$&lt;/span&gt; the cost to move a banana is &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; banana per mile.&lt;br&gt;&#xA;  - For bananas &lt;span class=&quot;math-container&quot;&gt;$c+1 \rightarrow 2c$&lt;/span&gt;, the cost to move a banana is &lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt; bananas per mile.&lt;br&gt;&#xA;  - For bananas &lt;span class=&quot;math-container&quot;&gt;$2c+1 \rightarrow 3c$&lt;/span&gt;, the cost to move a banana is &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt; bananas per mile.&lt;br&gt;&#xA; - etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is because, if the camel moves the bananas 1 mile at a time, he needs to make two trips for each load beyond his current capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define &lt;span class=&quot;math-container&quot;&gt;$t = \lfloor\frac{n}{c}\rfloor$&lt;/span&gt; Therefore, the total number of miles the camel can reach is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\left(\sum_{k=1}^{t} \frac{c}{2k - 1}\right) + \frac{(n \bmod c)}{2t+1}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In particular, plugging in the given &lt;span class=&quot;math-container&quot;&gt;$n = 3000$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$c = 1000$&lt;/span&gt;, we have that the camel can travel&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$1000 + 333 + 200 = 1533 \text{ miles}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To figure out how many bananas &lt;strong&gt;remain&lt;/strong&gt; for a given distance, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Subtract the extra miles and multiply back at the rate given above.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the first &lt;span class=&quot;math-container&quot;&gt;$1000$&lt;/span&gt; miles, this number is just the distance beyond the total capacity:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$1533 - 1000 = 533$&lt;/span&gt;, or &lt;strong&gt;533 bananas&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="42" LastEditorUserId="42" LastEditDate="2018-10-08T22:40:19.920" LastActivityDate="2018-10-08T22:40:19.920" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="268" PostTypeId="2" ParentId="263" CreationDate="2014-05-19T17:36:10.097" Score="6" Body="&lt;p&gt;Any sudoku puzzle with enough numbers filled in that the solution is unique is considered a &quot;valid&quot; or &quot;legal&quot; sudoku. Before some time in 2005, the Toronto Metro newspaper published sudoku puzzles (although not named Sudoku at the time) where exactly four numbers in each row, column, and 3x3 square were filled in, and those were never rotationally symmetrical.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;However, the sudokus that newspapers tend to publish usually have some form of symmetry, usually for aesthetic reasons. Usually this symmetry is rotational.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="463" LastEditDate="2015-06-04T15:40:27.697" LastActivityDate="2015-06-04T15:40:27.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="269" PostTypeId="1" AcceptedAnswerId="273" CreationDate="2014-05-19T18:33:23.137" Score="16" ViewCount="3953" Body="&lt;h2&gt;The Magic Hexagon Problem&lt;/h2&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FMgOn.gif&quot; alt=&quot;Magic Hexagon&quot; /&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A magic hexagon of order &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; is an arrangement of close-packed hexagons containing the numbers &lt;span class=&quot;math-container&quot;&gt;$1, 2, ..., H_{n-1}$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$H_n$&lt;/span&gt; is the &lt;span class=&quot;math-container&quot;&gt;$n^{th}$&lt;/span&gt; hex number such that the numbers along each straight line add up to the same sum. (Here, the hex numbers are i.e., &lt;span class=&quot;math-container&quot;&gt;$1, 7, 19, 37, 61, 91, 127$&lt;/span&gt;, ...; &lt;a href=&quot;https://oeis.org/A003215%7D&quot; rel=&quot;noreferrer&quot;&gt;Sloane's A003215&lt;/a&gt;. In the above magic hexagon of order &lt;span class=&quot;math-container&quot;&gt;$n=3$&lt;/span&gt;, each line (those of lengths &lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt;) adds up to &lt;span class=&quot;math-container&quot;&gt;$38$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;From &lt;a href=&quot;http://mathworld.wolfram.com/MagicHexagon.html&quot; rel=&quot;noreferrer&quot;&gt;Wolfram's Magic Hexagon&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Here is the programming puzzles entry to the &lt;a href=&quot;https://codegolf.stackexchange.com/questions/6304/code-solution-for-the-magic-hexagon&quot;&gt;Magic Hexagon problem&lt;/a&gt;. As you can see, the solutions there use appropriate systems of linear equations, and then try all the possibilities until it finds one. Boring, but effective; and can't be done by a human in a reasonable amount of time.&lt;/p&gt;&#xA;&lt;p&gt;This is &lt;a href=&quot;http://britton.disted.camosun.bc.ca/magicsq/magic.html&quot; rel=&quot;noreferrer&quot;&gt;de la Loubere's method&lt;/a&gt; for magic squares. Much cleaner and more beautiful, and a human can do this process easily.&lt;/p&gt;&#xA;&lt;p&gt;I want to know, is there such a solution / methodology to the Magic Hexagon problem that has the elegance of de la Loubere's method? How would you begin the process of coming up with one?&lt;/p&gt;&#xA;" OwnerUserId="42" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-08-03T23:46:58.097" Title="Elegant solution to the Magic Hexagon problem" Tags="|strategy|magic-square|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="271" PostTypeId="2" ParentId="260" CreationDate="2014-05-19T19:04:59.257" Score="5" Body="&lt;p&gt;One solution is to reverse the subject of puzzle. Instead of focusing on the father who is withholding an important detail, make it about the strategy of teasing out information.  Here's my attempt:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Last weekend our Parents of Twins Club had a father-daughter night at the ballet. Since he wasn't invited, my twin boy stayed home with my wife. When we arrived, my daughter and I discovered our seats were isolated from the rest of the club. I didn't have a chance to see any of the other fathers or their daughters.&lt;/p&gt;&#xA;&lt;p&gt;During intermission, I got into a conversation with another father in the washroom. I recognized him from other club gatherings, but for the life of me I couldn't recall if he had a twin son or not. Taking a chance, I asked if he'd be at an upcoming father-&lt;strong&gt;son&lt;/strong&gt; event the club was sponsoring next month.&lt;/p&gt;&#xA;&lt;p&gt;Assuming the other father has only one set of twins and that non-twin siblings are not allowed at Twins Club events, what are the odds I've just made a fool of myself?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The important bit is that it puts the puzzle into a relatable, real-life situation rather than &amp;quot;puzzleland&amp;quot;. The selection method is perhaps old-fashioned (my sons would enjoy ballet too!), but clear. The other father isn't being coy about the information he holds. Instead, his mere presence at the event gives us the needed information. We'd get a very different answer if the chance encounter occurred at some location that didn't ensure he had a twin daughter.&lt;/p&gt;&#xA;&lt;p&gt;While we are at it, ambiguity in the selection criteria is also the problem with variations of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Monty_Hall_problem#Origin_and_full_assumptions&quot; rel=&quot;nofollow noreferrer&quot;&gt;Monty Hall problem&lt;/a&gt; that leave out important assumptions. The puzzle only works if the host:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;has access to information to be revealed,&lt;/li&gt;&#xA;&lt;li&gt;can not reveal that information explicitly until the guess is made, and&lt;/li&gt;&#xA;&lt;li&gt;always presents a puzzle.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Much the same criteria are needed to make the &amp;quot;boy or girl&amp;quot; paradox unambiguous.&lt;/p&gt;&#xA;" OwnerUserId="15" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-05-19T19:04:59.257" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="272" PostTypeId="2" ParentId="259" CreationDate="2014-05-19T19:58:04.513" Score="8" Body="&lt;p&gt;This is a problem in group theory. You are trying to find out how many times you can apply a specific permutation to a deck of $54$ cards (in this case, the permutation is equivalent to a perfect shuffle) until a card that was originally $n$th from the top turns up $28$th from the top (which is where you would &quot;find the flipped card if the deck were cut exactly in half&quot;).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Firstly, let's take a look at what the permutation is like. In the permutation specified in your question, you take a deck of cards arranged like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 2 3 4 5 6 ... 51 52 53 54&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;cut it into two piles of 27 cards each, and interleave the cards so that they end up like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;28 1 29 2 30 3 31 4 32 5 ... 53 26 54 27&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This provides a function $f : [1..54] \rightarrow [1..54]$ where each number $x$ is mapped to a different $f(x)$, which is known as a &lt;em&gt;bijection&lt;/em&gt; or a &lt;em&gt;permutation&lt;/em&gt;. The same principle from the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16/100-prisoners-names-in-boxes/66#66&quot;&gt;prisoners'-names-in-boxes problem&lt;/a&gt; applies, in which a permutation is a set of disjoint cycles. All it remains to do is to figure out what cycles the cards go through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The full table of values that each number maps to are as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   x: 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18&#xA;f(x): 28 01 29 02 30 03 31 04 32 05 33 06 34 07 35 08 36 09&#xA;&#xA;   x: 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36&#xA;f(x): 37 10 38 11 39 12 40 13 41 14 42 15 43 16 44 17 45 18&#xA;&#xA;   x: 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54&#xA;f(x): 46 19 47 20 48 21 49 22 50 23 51 24 52 25 53 26 54 27&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And by repeatedly taking the $f$ of each number, we can figure out the cycles that result:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;cycle 1: 14 07 31 43 49 52 26 13 34 17 36 18 09 32 16 08 04 02 01 28&#xA;cycle 2: 46 23 39 47 51 53 54 27 41 48 24 12 06 03 29 42 21 38 19 37&#xA;cycle 3: 05 30 15 35 45 50 25 40 20 10&#xA;cycle 4: 11 33 44 22&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There are 20 numbers in the same cycle with $28$ (the positions mentioned in the first line), so if the flipped card is $n$th on top where $n$ is some number in that first row of numbers, it'll take as many shuffles to get that card 28th from the top as the distance in numbers from $n$ to $28$ going forward in the cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if the card is $31$st from the top, it will take 17 shuffles to get the card to 28th from the top, because $31$ is 17 spaces to the left of $28$ in that table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But consequently, if the number is some $n$ in the other three cycles, then it will &lt;em&gt;never&lt;/em&gt; end up 28th from the top, and you will never get it to show up when you cut the deck in half.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2014-06-13T19:30:42.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="273" PostTypeId="2" ParentId="269" CreationDate="2014-05-19T20:50:24.530" Score="9" Body="&lt;p&gt;According to the article, the Magic Hexagon is the only one that exists (aside from n=1). There wouldn't be a simpler method for generating one, because no other hexagons of order 2 or &gt;3 would have a solution. Quoting from the last paragraph:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Trigg showed that the magic constant for an order n hexagon would be&#xA;  (9(n^4-2n^3+2n^2-n)+2)/(2(2n-1)), the first few of which are&#xA;  1, 28/3, 38, 703/7, 1891/9, 4186/11, ... (Sloane's A097361 and A097362),&#xA;  which requires 5/(2n-1) to be an integer for a solution to exist.&#xA;  But this is an integer for only n=1 (the trivial case of a single hexagon)&#xA;  and Adams's n=3 (Gardner 1984, p. 24).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I had to guess, I might claim the n=3 magic hexagon working at all is just a coincidence.&lt;/p&gt;&#xA;" OwnerUserId="202" LastEditorUserId="202" LastEditDate="2014-05-19T20:55:37.130" LastActivityDate="2014-05-19T20:55:37.130" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="274" PostTypeId="1" AcceptedAnswerId="308" CreationDate="2014-05-20T04:12:07.267" Score="5" ViewCount="574" Body="&lt;p&gt;Here's a third puzzle that I found in a book, slightly paraphrased because I don't entirely remember the format of the original.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A medical researcher is trying to test the effect of a certain drug on mice. He is supposed to collect data for all doses from 1 to 50 units, but only has the data collection resources for 25 tests. He has decided to only test dose amounts that are not multiples or factors of each other (for example, if he decided to use a dose of 15 units for a specific test, that would preclude tests for 1, 3, 5, 30, or 45 units). Moreover, because the drug is relatively expensive, he wants to use as little of the drug as he can manage to run these tests.&lt;/p&gt;&#xA;&lt;p&gt;How can he arrange the dosage amounts so that he ends up using all 25 test packages, &lt;strong&gt;and the total units of dosage used in the tests are as low as possible&lt;/strong&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The book had the answer, but one, it didn't explain how the answer was arrived at, and two, I don't remember what the answer was and no longer have that book with me.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2018-03-02T10:59:00.673" Title="Choosing units for drug testing" Tags="|calculation-puzzle|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="276" PostTypeId="2" ParentId="48" CreationDate="2014-05-20T05:08:24.460" Score="18" Body="&lt;p&gt;BlueFlame gives an essentially correct answer, but I wanted to add a rigorous proof (for anyone looking for it). [EDIT: user2357112's comment points out an annoying hole in the proof, under the &quot;proof that 65536 is theoretically possible&quot; section. While I believe the hole can be filled, I don't have time to do it right now.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will prove that 65536 is the maximum when 2 is the only tile which appears newly on the board.  It is easy to adapt the proof to show that 131072 is the maximum when 4 may also appear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1. Proof that 65536 is an upper bound.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At every step of the game, the &lt;em&gt;total value among all tiles&lt;/em&gt; goes up by 2, or stays the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose towards contradiction that at some point in time the 131072 tile is attained.  That at this time the sum of all the tiles is at least 131072.  Thus &lt;em&gt;at some point in time,&lt;/em&gt; the sum of all the tiles was &lt;em&gt;exactly&lt;/em&gt; 131070 (= 131072 - 2).  Then at this time there was a set S of at most 16 powers of two adding to get 131070.  Recursively adding together pairs of equal elements of S until all the elements are distinct (i.e., adding them as if they had combined in the game), we obtain a set R with at most 16 &lt;em&gt;distinct&lt;/em&gt; powers of two adding to get 131070.  But 131070 = 2^16 + 2^15 + 2^14 + ... + 2^1 cannot be written as a sum of anything less than &lt;em&gt;exactly&lt;/em&gt; 16 distinct powers of two (using unique representation in binary).  Therefore the set S must have initially consisted of one each of the numbers 2, 4, 8, ..., 2^15, 2^16, and the entire 4x4 grid must have been filled.  But then the position would have been a losing position, i.e. no two tiles could have combined and there was no space left on the board.  Contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2. Proof that 65536 is theoretically possible.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is to use a strategy like this:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/rWrgE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Label the 4 x 4 grid with the numbers 1 through 16, with the numbers in the following pattern:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;13 14 15 16&#xA;12 11 10  9&#xA; 5  6  7  8&#xA; 4  3  2  1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Call a board state &quot;ideal&quot; if it has the following properties: (1) the tiles that are filled are exactly 16, 15, 14, ..., k for some k, and (2) if two tiles a and b are on the board, with a located on a bigger label than b, then the value of tile a is greater than the value of tile b.  This matches the general pattern of the picture above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, suppose we are in an ideal state with all the tiles less than 65536 = 2^16.  Since there are only 15 distinct tile values less than 2^16, there must be an open space.  Then we can choose to spawn the 2 in the &lt;em&gt;open space with the highest label&lt;/em&gt;.  After this, we can recursively combine the lowest two tiles if they have equal value, by sliding left, right, or up, depending on the board state.  But due to the nature of our &quot;ideal state&quot; strategy, the left, right, and up slides will never modify any tiles but the two lowest ones which are combining.  Thus after the entire process we are left in another ideal state. [EDIT: This paragraph has a hole: it forgets to consider that a new 2 tile must spawn after each slide. One should probably be fine if the new 2 tile always spawns, as the first 2 tile, in the open space with the highest label. But the reasoning gets a bit more complicated.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will be able to follow this strategy continually as long as there is no 65536 tile on the board.  Moreover, the total value of all values on the board will do nothing but increase.  Therefore, at some point we will either reach a 65536 tile, in which case we will be done, or we will reach an ideal state with the total value of all tiles equaling at least 65536.  Since no set of unique tiles less than 65536 can add to 65536 or more, and since an ideal state contains unique tiles, this is a contradiction and we will necessarily reach the 65536 tile at some point.&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="203" LastEditDate="2016-01-24T15:51:39.650" LastActivityDate="2016-01-24T15:51:39.650" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="277" PostTypeId="2" ParentId="230" CreationDate="2014-05-20T09:06:51.890" Score="16" Body="&lt;p&gt;First leg of the journey: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Load 1000, &lt;/li&gt;&#xA;&lt;li&gt;Drop 4 bananas every mile for 200 miles, feeding the camel from own supply.&lt;/li&gt;&#xA;&lt;li&gt;The route is primed for 4 passes of the camel:&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;return&lt;/li&gt;&#xA;&lt;li&gt;take another 1000 bananas&lt;/li&gt;&#xA;&lt;li&gt;return&lt;/li&gt;&#xA;&lt;li&gt;take remaining 1000 bananas.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Second leg, starting with mile 200:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;load 1000&lt;/li&gt;&#xA;&lt;li&gt;drop 1 banana a mile away, then 2 bananas every next mile for another 333 miles, for 534 miles total.&lt;/li&gt;&#xA;&lt;li&gt;return&lt;/li&gt;&#xA;&lt;li&gt;pick 1000 bananas&lt;/li&gt;&#xA;&lt;li&gt;on the first mile eat a banana from own supply, then continue using the ones left.&lt;/li&gt;&#xA;&lt;li&gt;arrive at 534th mile with 999 bananas.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Third leg of the journey:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Feed the camel the last banana on the ground, then starting with 535th mile, keep using bananas from your supply of 999.&lt;/li&gt;&#xA;&lt;li&gt;use up 465 bananas from own supply for the remaining 465 miles.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You're left with 999-465 = 534 bananas.&lt;/p&gt;&#xA;" OwnerUserId="117" LastEditorUserId="117" LastEditDate="2016-03-22T17:55:28.937" LastActivityDate="2016-03-22T17:55:28.937" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="278" PostTypeId="2" ParentId="166" CreationDate="2014-05-20T11:35:13.130" Score="3" Body="&lt;p&gt;Alternative solution (?)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lay the ropes next to each other and light each rope at opposing ends. WHen the flames meet in the middle, 30mins has passed. Then move the ropes so the unburnt sections are next to each other, when the flames meet again, 15mins has passed - hence 45mins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="118" LastActivityDate="2014-05-20T11:35:13.130" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="279" PostTypeId="2" ParentId="103" CreationDate="2014-05-20T11:52:46.390" Score="7" Body="&lt;p&gt;Your added constraint of having 1 gallon left in the small jug makes the puzzle impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only available moves are&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;fill a jug from the well, filling it &lt;em&gt;completely&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;empty a jug into the well, emptying it &lt;em&gt;completely&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;pour one jug into another, until either one jug is &lt;em&gt;completely&lt;/em&gt; empty or the other jug is &lt;em&gt;completely&lt;/em&gt; full&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Since these moves leave you with at least one jug either &lt;em&gt;completely&lt;/em&gt; empty or &lt;em&gt;completely&lt;/em&gt; full, nothing among these moves lets you end up with both jugs partially filled.&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditDate="2014-05-23T13:17:43.653" LastActivityDate="2014-05-23T13:17:43.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="280" PostTypeId="2" ParentId="201" CreationDate="2014-05-20T13:14:34.823" Score="10" Body="&lt;p&gt;With three children (Alice, Bob, and Carol), three red hats and two black hats, there are 7 different configurations possible (with knowable hats boxed):&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{array}{cccl}&#xA;\text{Alice} &amp;amp; \text{Bob} &amp;amp; \text{Carol} \\&#xA;\\ \hline&#xA;\text{red}   &amp;amp; \text{red}   &amp;amp; \fbox{red}   &amp;amp; \text{(1)} \\&#xA;\text{black} &amp;amp; \text{red}   &amp;amp; \fbox{red}   &amp;amp; \text{(2)} \\&#xA;\text{red}   &amp;amp; \text{black} &amp;amp; \fbox{red}   &amp;amp; \text{(3)} \\&#xA;\text{black} &amp;amp; \text{black} &amp;amp; \fbox{red}   &amp;amp; \text{(4)} \\&#xA;\text{red}   &amp;amp; \fbox{red}   &amp;amp; \fbox{black} &amp;amp; \text{(5)} \\&#xA;\text{black} &amp;amp; \fbox{red}   &amp;amp; \fbox{black} &amp;amp; \text{(6)} \\&#xA;\fbox{red}   &amp;amp; \fbox{black} &amp;amp; \fbox{black} &amp;amp; \text{(7)} \\&#xA;\end{array}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;If Alice sees two black hats (7), she'll know the black hats have been exhausted and she is wearing a red one. Since Alice knows the colour of her hat, Bob and Carol will know that they are both wearing black hats, since this is the only configuration that allows Alice to know her hat's colour.&lt;/p&gt;&#xA;&lt;p&gt;If Alice sees at least one red hat, she doesn't know the colour of her own hat and will state so. Now if Carol is wearing a black hat (5, 6), Bob will know the colour of his hat, since Alice has seen at least one red hat and Carol's hat isn't it, so his must be. When Carol hears that Bob knows the colour of his hat, she can deduce that her hat is black, or else Bob wouldn't have known.&lt;/p&gt;&#xA;&lt;p&gt;If Carol is wearing a red hat (1, 2, 3, 4), Alice won't know the colour of hers and neither will Bob know the colour of his, because all he can see is that Carol is wearing a red hat, but he will not know whether Alice saw two red hats or just one.&lt;/p&gt;&#xA;&lt;p&gt;As we can see, Carol will &lt;em&gt;always&lt;/em&gt; know the colour of her hat, while Alice will almost never know.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;em&gt;Alternative table using the new Markdown table formatting. Knowable hats are in all caps.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;Alice&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;Bob&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;Carol&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;red&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;red&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;RED&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;(1)&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;black&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;red&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;RED&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;(2)&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;red&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;black&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;RED&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;(3)&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;black&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;black&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;RED&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;(4)&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;red&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;RED&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;BLACK&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;(5)&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;black&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;RED&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;BLACK&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;(6)&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;RED&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;BLACK&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;BLACK&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;(7)&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;" OwnerUserId="68" LastEditorUserId="68" LastEditDate="2020-12-14T07:47:16.960" LastActivityDate="2020-12-14T07:47:16.960" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="282" PostTypeId="1" AcceptedAnswerId="285" CreationDate="2014-05-20T14:49:16.953" Score="44" ViewCount="34167" Body="&lt;p&gt;Here's a neat little puzzle taken from a Google interview question:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;10 humans are abducted by aliens; each represents 10% of the entire human population. The aliens give each abductee either a purple hat or a green hat. The 10 are lined up in a single file line, each facing forward, such that the last person can see the remaining 9's hats, the second to last person can see the remaining 8's hats and so on. No one can see his or her own hat.&lt;/p&gt;&#xA;&lt;p&gt;The aliens then proceed, starting from the last person, to ask each of the abductees what the color of their hat is. If they guess correctly, they and the 10% of the human population they represent survives; if not, the opposite happens.&lt;/p&gt;&#xA;&lt;p&gt;Assuming the abductees are given a chance to develop a strategy before they are lined up and questioned: what is the optimal strategy they can utilize (i.e. the one with the highest expected number of survivals)?&lt;/p&gt;&#xA;&lt;p&gt;During the questioning, the abductees are not allowed to say &lt;em&gt;anything&lt;/em&gt; besides their guess for the color of their hat when it is their turn.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The optimal strategy will always ensure 9 survivals, and will have a 50% chance of the 10th survival.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="198" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2018-09-16T19:07:09.453" Title="Hats and Aliens" Tags="|logical-deduction|hat-guessing|" AnswerCount="6" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="284" PostTypeId="1" AcceptedAnswerId="10829" CreationDate="2014-05-20T15:12:59.550" Score="13" ViewCount="12884" Body="&lt;p&gt;The question of &lt;a href=&quot;https://puzzling.stackexchange.com/questions/183/twelve-balls-and-a-scale&quot;&gt;twelve balls and a scale&lt;/a&gt; is probably the best-known example of the &quot;find the ball of a different weight&quot; problem. But does it generalize? Is there a general way to find a weighing algorithm for $N$ balls on a scale where $N \ne 12$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the variant where the direction of deviation is not known (one ball is slightly lighter or slightly heavier than the others) and the solution must both pinpoint the odd ball and indicate whether it is lighter or heavier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that for $K$ weighings, there will be no more than $(3^K - 3)/ 2$ balls to weigh, as that has been proven to be optimum.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-05-18T19:30:43.370" Title="N balls and a scale" Tags="|mathematics|combinatorics|weighing|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="285" PostTypeId="2" ParentId="282" CreationDate="2014-05-20T15:25:20.017" Score="45" Body="&lt;p&gt;I think this would work? Sorry if it's confusing. I sort of confused myself when I came to this conclusion.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first guy would count the number of hats before him of a particular color... like, I dunno, I guess purple? And if the count of purple hats is odd, then he would say that his hat is purple. But if the count of purple hats is even, he'd say his hat was green.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the next guy should then be able to determine the color of his hat by: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The remaining hats. So if the questioning went:&lt;br&gt;&#xA;&#xA; Alien: What color is your hat?&lt;br&gt;&#xA;&#xA; 10th person: Purple.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then the 9th person would look ahead and: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Count the number of purple hats. If it's an odd number, he has a green hat. If it's an even number, his hat is purple. So on and so forth. They just need to remember the &quot;number&quot; of purple hats.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But the first guy has no way of knowing his own hat color, so he's got a 50/50 chance of dying either way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone has trouble understanding this, a visual explanation can be found here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=N5vJSNXPEwA&quot; rel=&quot;noreferrer&quot;&gt;https://www.youtube.com/watch?v=N5vJSNXPEwA&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19" LastEditorUserId="46903" LastEditDate="2018-09-16T19:07:09.453" LastActivityDate="2018-09-16T19:07:09.453" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="286" PostTypeId="2" ParentId="284" CreationDate="2014-05-20T15:31:19.613" Score="1" Body="&lt;p&gt;I don't think there is a crisp way to describe it, but basically you round up to the next $K$ and find some balls to leave out of the strategy for $(3^K-3)/2$  The type of problem you can have is for $N=8$.  If you weigh four vs four, they will not balance.  The next step calls for you to use one of the known good coins, but you don't have any of those.  If you had weighed three vs three to start (and they didn't balance) you have two known goods available.  So start by weighing about $1/3$ of the coins against the same number and keep going.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="71" LastEditDate="2014-06-13T19:40:28.997" LastActivityDate="2014-06-13T19:40:28.997" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="287" PostTypeId="1" AcceptedAnswerId="288" CreationDate="2014-05-20T16:22:11.620" Score="13" ViewCount="8878" Body="&lt;p&gt;This problem is commonly cited as a Microsoft interview problem, although I have paraphrased it here to involve a river and a boat instead of a bridge and a flashlight, and tweaked the crossing times a little bit.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There are four people standing on one side of a river, and they need to cross it as quickly as possible to the other side. However, there is only one rowboat docked at the river's bank, and only two people can fit in it. These four people can row to the other side in 3, 7, 12 and 15 minutes, and if two people cross, they must row at the rate of the slower person. How can these four people cross as quickly as possible, and how long will it take them?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="5373" LastEditDate="2017-02-06T11:24:13.583" LastActivityDate="2017-02-06T11:24:13.583" Title="Fastest way to cross a river: four people with different rowing speeds" Tags="|logical-deduction|river-crossing|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="288" PostTypeId="2" ParentId="287" CreationDate="2014-05-20T16:28:46.767" Score="17" Body="&lt;p&gt;You'll want the two slowest people to cross together, but you'll also want one fast person on either side to bring the boat back quickly. So I believe this will do it (I labeled the people from A (fastest) to D (slowest)):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A and B cross (7 mins)&lt;/li&gt;&#xA;&lt;li&gt;A returns (3 mins)&lt;/li&gt;&#xA;&lt;li&gt;C and D cross (15 mins)&lt;/li&gt;&#xA;&lt;li&gt;B returns (7 mins)&lt;/li&gt;&#xA;&lt;li&gt;A and B cross (7 mins)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This takes a total of 39 minutes. Note that steps 2 and 4 are interchangeable.&lt;/p&gt;&#xA;" OwnerUserId="198" LastEditorUserId="198" LastEditDate="2014-05-20T17:32:50.787" LastActivityDate="2014-05-20T17:32:50.787" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="289" PostTypeId="5" CreationDate="2014-05-20T19:14:46.150" Score="0" Body="&lt;p&gt;A paradox is any statement that appears to contradict itself, and yet is still presented as true. Many paradoxes require &lt;a href=&quot;/questions/tagged/lateral-thinking&quot;&gt;lateral thinking&lt;/a&gt; to solve.&lt;/p&gt;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;This statement is false.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If the above statement is true, then &amp;quot;this statement is false&amp;quot; is true, &lt;em&gt;and therefore it is false&lt;/em&gt;. But if the statement is false, then &amp;quot;this statement is false&amp;quot; is false, &lt;em&gt;and therefore it is true&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Either way, a logical contradiction is reached. Thus, &amp;quot;This statement is false&amp;quot; is a paradoxical statement.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="60644" LastEditorDisplayName="user88" LastEditDate="2022-12-12T23:08:27.967" LastActivityDate="2022-12-12T23:08:27.967" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="290" PostTypeId="4" CreationDate="2014-05-20T19:14:46.150" Score="0" Body="A puzzle in which a seemingly impossible situation is presented, or any sort of construction that results in a logical contradiction." OwnerUserId="135" LastEditorDisplayName="user88" LastEditDate="2014-05-28T14:43:46.277" LastActivityDate="2014-05-28T14:43:46.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="291" PostTypeId="2" ParentId="282" CreationDate="2014-05-20T19:43:51.947" Score="16" Body="&lt;p&gt;Generically, what do you do if there are $c$ colors, and $n$ people? Imagine further the aliens say the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You can plan your strategy now, but we are only going to tell you what the colors are going to be (including their hex representation) and how many there are after you are no longer allowed to communicate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How well can the humans do? It's the same as before, except the human who goes first has a $\frac{1}{c}$ chance of survival.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In advance, the humans use the following strategy:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Each color is assigned a number from $0 \rightarrow c-1$ (this is arbitrary, the humans just have to agree on some numbering).&lt;/li&gt;&#xA;&lt;li&gt;The human who goes first adds up the colors of all the hats that he sees, and computes the result $\bmod c$.&lt;/li&gt;&#xA;&lt;li&gt;He then calls out the color required to make the total sum equal to $0 \bmod c$.&lt;/li&gt;&#xA;&lt;li&gt;Each remaining human can then use the colors called out behind them and the colors they see in front of them to see what the total must be.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This strategy works no matter how many or how few humans there are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: 5 colors, 8 humans. Their hats have color numbers (assume the humans are all facing right, so the first human is the leftmost hat):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;32130343&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The first human sees &lt;code&gt;2130343&lt;/code&gt;. This adds to $16$. Note $16 \bmod 5 \equiv 1$, so to make the total come out to be $0$, he calls out $4$, because $0 \equiv 4 + 1 \bmod 5$.&lt;/li&gt;&#xA;&lt;li&gt;The second human sees &lt;code&gt;130343&lt;/code&gt;, and has heard $4$. This adds to $18 \equiv 2 (\bmod 5)$, so he knows he must have color $2$.&lt;/li&gt;&#xA;&lt;li&gt;The third human sees &lt;code&gt;30343&lt;/code&gt;, and has heard &lt;code&gt;42&lt;/code&gt;. This adds to $19 \equiv 1 (\bmod 5)$, so he must have color $1$.&lt;/li&gt;&#xA;&lt;li&gt;This process proceeds down the line, and only the first person dies.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="42" LastEditorUserId="71" LastEditDate="2014-06-13T20:08:05.777" LastActivityDate="2014-06-13T20:08:05.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="292" PostTypeId="5" CreationDate="2014-05-20T20:14:41.077" Score="0" Body="&lt;p&gt;A logic puzzle is one where a conclusion can be deduced solely using &lt;a href=&quot;https://en.wikipedia.org/wiki/Logic&quot; rel=&quot;nofollow noreferrer&quot;&gt;formal logical deduction&lt;/a&gt; without needing to guess. It often takes the form of a series of statements that you must use to find the solution to the question posed using logical inference. For example, the puzzle might ask you to figure out the order of five people standing in line from statements like &quot;Jamie is not to Sarah's left.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes, the truth of the statements themselves is unknown and must be deduced. Use &lt;a href=&quot;/questions/tagged/liars&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;liars&amp;#39;&quot; rel=&quot;tag&quot;&gt;liars&lt;/a&gt; in addition for these.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You are in a room with two wizards. One of these wizards always tells the truth, while the other always lies. How do you figure out which does what by asking both of them just one question?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is to ask them any question that has one irrefutably true answer and one irrefutably false answer, such as &quot;Am I speaking to you right now?&quot; The wizard that tells the truth will say &quot;Yes,&quot; proving that he speaks the truth, and the lying wizard will say &quot;No,&quot; showing that he is the liar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that puzzles requiring you to reason through hints or clues, rather than apply formal logic rules to supplied facts, are &lt;em&gt;not&lt;/em&gt; logical-deduction. You might want the &lt;a href=&quot;/questions/tagged/situation&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;situation&amp;#39;&quot; rel=&quot;tag&quot;&gt;situation&lt;/a&gt; tag for those.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="30633" LastEditDate="2019-03-12T17:34:11.020" LastActivityDate="2019-03-12T17:34:11.020" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="293" PostTypeId="4" CreationDate="2014-05-20T20:14:41.077" Score="0" Body="A puzzle that requires formal logical deduction to arrive at the solution. This suggests more than merely reasoning through clues to find an answer (you might want [situation] for that)." OwnerDisplayName="user88" LastEditorUserId="30633" LastEditDate="2019-03-12T17:34:11.020" LastActivityDate="2019-03-12T17:34:11.020" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="294" PostTypeId="2" ParentId="128" CreationDate="2014-05-20T20:19:58.910" Score="40" Body="&lt;p&gt;Let's put the number of coins at &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; and see what happens if we increase &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;. We'll number the pirates from meekest (&lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;) to fiercest (&lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;).&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: right;&quot;&gt;Nr. of pirates&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: right;&quot;&gt;Pirate&lt;br&gt;meekest to fiercest&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: right;&quot;&gt;Nr. of coins&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: left;&quot;&gt;Vote&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;c&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;yes&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;1&lt;br&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;0&lt;br&gt;c&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;no&lt;br&gt;yes &lt;strong&gt;breaks tie&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;1&lt;br&gt;2&lt;br&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;1&lt;br&gt;0&lt;br&gt;c-1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;yes&lt;br&gt;no&lt;br&gt;yes&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;1&lt;br&gt;2&lt;br&gt;3&lt;br&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;0&lt;br&gt;1&lt;br&gt;0&lt;br&gt;c-1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;no&lt;br&gt;yes&lt;br&gt;no&lt;br&gt;yes &lt;strong&gt;breaks tie&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;5&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;1&lt;br&gt;2&lt;br&gt;3&lt;br&gt;4&lt;br&gt;5&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;1&lt;br&gt;0&lt;br&gt;1&lt;br&gt;0&lt;br&gt;c-2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;yes&lt;br&gt;no&lt;br&gt;yes&lt;br&gt;no&lt;br&gt;yes&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;...&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;2c&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;1&lt;br&gt;2&lt;br&gt;3&lt;br&gt;...&lt;br&gt;n&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;0&lt;br&gt;1&lt;br&gt;0&lt;br&gt;&lt;br&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;no&lt;br&gt;yes&lt;br&gt;no&lt;br&gt;&lt;br&gt;yes &lt;strong&gt;breaks tie&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;If we define a solution as stable if nobody gets fed to the sharks, we can say that all solutions so far have been stable. This is because for each &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;, there are more or at least as much (in which case the tie is broken) pirates who would be worse off in the solution for &lt;span class=&quot;math-container&quot;&gt;$n-1$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;But now, we're at &lt;span class=&quot;math-container&quot;&gt;$n = 2c + 1$&lt;/span&gt;. Are we getting ready for carnage? Not quite:&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: right;&quot;&gt;Nr. of pirates&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: right;&quot;&gt;Pirate&lt;br&gt;meekest to fiercest&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: right;&quot;&gt;Nr. of coins&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: left;&quot;&gt;Vote&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;2c+1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;1&lt;br&gt;2&lt;br&gt;3&lt;br&gt;...&lt;br&gt;n-1&lt;br&gt;n&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;1&lt;br&gt;0&lt;br&gt;1&lt;br&gt;&lt;br&gt;0&lt;br&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;yes&lt;br&gt;no&lt;br&gt;yes&lt;br&gt;&lt;br&gt;no&lt;br&gt;yes&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;Here, the fiercest pirate can escape alive, but without gold.&lt;/p&gt;&#xA;&lt;p&gt;Okay, but now surely the sharks are getting fed? Let's see.&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: right;&quot;&gt;Nr. of pirates&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: right;&quot;&gt;Pirate&lt;br&gt;meekest to fiercest&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: right;&quot;&gt;Nr. of coins&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: left;&quot;&gt;Vote&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;2c+2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;1&lt;br&gt;2&lt;br&gt;3&lt;br&gt;...&lt;br&gt;n-2&lt;br&gt;n-1&lt;br&gt;n&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;0&lt;br&gt;1&lt;br&gt;0&lt;br&gt;&lt;br&gt;1&lt;br&gt;0&lt;br&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;no&lt;br&gt;yes&lt;br&gt;no&lt;br&gt;&lt;br&gt;yes&lt;br&gt;no&lt;br&gt;yes &lt;strong&gt;breaks tie&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;The fiercest pirate now broke the tie to stay alive.&lt;/p&gt;&#xA;&lt;p&gt;How about &lt;span class=&quot;math-container&quot;&gt;$n = 2c + 3$&lt;/span&gt; then?&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: right;&quot;&gt;Nr. of pirates&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: right;&quot;&gt;Pirate&lt;br&gt;meekest to fiercest&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: right;&quot;&gt;Nr. of coins&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: left;&quot;&gt;Vote&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;2c+3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;1&lt;br&gt;2&lt;br&gt;3&lt;br&gt;...&lt;br&gt;n-3&lt;br&gt;n-2&lt;br&gt;n-1&lt;br&gt;n&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;1&lt;br&gt;0&lt;br&gt;1&lt;br&gt;&lt;br&gt;0&lt;br&gt;0&lt;br&gt;0&lt;br&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;yes&lt;br&gt;no&lt;br&gt;yes&lt;br&gt;&lt;br&gt;no&lt;br&gt;no&lt;br&gt;no&lt;br&gt;yes &lt;strong&gt;shark bait&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;Here, finally, at &lt;span class=&quot;math-container&quot;&gt;$n = 2c + 3$&lt;/span&gt;, we reach our first unstable solution. The fiercest pirate can buy &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; votes, adds his own vote for a total of &lt;span class=&quot;math-container&quot;&gt;$c+1$&lt;/span&gt;, but the opposition has &lt;span class=&quot;math-container&quot;&gt;$c + 2$&lt;/span&gt; votes.&lt;br /&gt;&#xA;Note that here, pirate &lt;span class=&quot;math-container&quot;&gt;$n-3$&lt;/span&gt; is pirate &lt;span class=&quot;math-container&quot;&gt;$n-2$&lt;/span&gt; from the solution before it. Voting &lt;span class=&quot;math-container&quot;&gt;$\text{no}$&lt;/span&gt; yields a coin. Pirates &lt;span class=&quot;math-container&quot;&gt;$n-2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$n-1$&lt;/span&gt; get no gold by voting against this proposal, but as per the third rule, vote to feed the sharks.&lt;sup&gt;‡&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;p&gt;But does that mean that from now on, all pirates get their feet wet? Not at all.&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: right;&quot;&gt;Nr. of pirates&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: right;&quot;&gt;Pirate&lt;br&gt;meekest to fiercest&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: right;&quot;&gt;Nr. of coins&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: left;&quot;&gt;Vote&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;2c+4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;1&lt;br&gt;2&lt;br&gt;3&lt;br&gt;...&lt;br&gt;n-4&lt;br&gt;n-3&lt;br&gt;n-2&lt;br&gt;n-1&lt;br&gt;n&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;1&lt;br&gt;0&lt;br&gt;1&lt;br&gt;&lt;br&gt;0&lt;br&gt;0&lt;br&gt;0&lt;br&gt;0&lt;br&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;yes&lt;br&gt;no&lt;br&gt;yes&lt;br&gt;&lt;br&gt;no&lt;br&gt;no&lt;br&gt;no&lt;br&gt;yes &lt;strong&gt;potential shark bait&lt;/strong&gt;&lt;br&gt;yes &lt;strong&gt;breaks tie&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;Now both the fiercest pirate and the almost-as-fierce pirate are voting for their lives, since the solution with one pirate less is unstable and will see the almost-as-fierce pirate being fed to the sharks. The fiercest pirate breaks the tie and this solution is stable.&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: right;&quot;&gt;Nr. of pirates&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: right;&quot;&gt;Pirate&lt;br&gt;meekest to fiercest&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: right;&quot;&gt;Nr. of coins&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: left;&quot;&gt;Vote&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;2c+5&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;1&lt;br&gt;2&lt;br&gt;3&lt;br&gt;...&lt;br&gt;n-5&lt;br&gt;n-4&lt;br&gt;n-3&lt;br&gt;n-2&lt;br&gt;n-1&lt;br&gt;n&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;0&lt;br&gt;1&lt;br&gt;0&lt;br&gt;&lt;br&gt;1&lt;br&gt;0&lt;br&gt;0&lt;br&gt;0&lt;br&gt;0&lt;br&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;no&lt;br&gt;yes&lt;br&gt;no&lt;br&gt;&lt;br&gt;yes&lt;br&gt;no&lt;br&gt;no&lt;br&gt;no&lt;br&gt;no&lt;br&gt;yes &lt;strong&gt;shark bait&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;2c+6&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;1&lt;br&gt;2&lt;br&gt;3&lt;br&gt;...&lt;br&gt;n-6&lt;br&gt;n-5&lt;br&gt;n-4&lt;br&gt;n-3&lt;br&gt;n-2&lt;br&gt;n-1&lt;br&gt;n&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;0&lt;br&gt;1&lt;br&gt;0&lt;br&gt;&lt;br&gt;1&lt;br&gt;0&lt;br&gt;0&lt;br&gt;0&lt;br&gt;0&lt;br&gt;0&lt;br&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;no&lt;br&gt;yes&lt;br&gt;no&lt;br&gt;&lt;br&gt;yes&lt;br&gt;no&lt;br&gt;no&lt;br&gt;no&lt;br&gt;no&lt;br&gt;yes&lt;br&gt;yes &lt;strong&gt;shark bait&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;2c+7&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;1&lt;br&gt;2&lt;br&gt;3&lt;br&gt;...&lt;br&gt;n-7&lt;br&gt;n-6&lt;br&gt;n-5&lt;br&gt;n-4&lt;br&gt;n-3&lt;br&gt;n-2&lt;br&gt;n-1&lt;br&gt;n&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;0&lt;br&gt;1&lt;br&gt;0&lt;br&gt;&lt;br&gt;1&lt;br&gt;0&lt;br&gt;0&lt;br&gt;0&lt;br&gt;0&lt;br&gt;0&lt;br&gt;0&lt;br&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;no&lt;br&gt;yes&lt;br&gt;no&lt;br&gt;&lt;br&gt;yes&lt;br&gt;no&lt;br&gt;no&lt;br&gt;no&lt;br&gt;no&lt;br&gt;yes&lt;br&gt;yes&lt;br&gt;yes &lt;strong&gt;shark bait&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;2c+8&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;1&lt;br&gt;2&lt;br&gt;3&lt;br&gt;...&lt;br&gt;n-8&lt;br&gt;n-7&lt;br&gt;n-6&lt;br&gt;n-5&lt;br&gt;n-4&lt;br&gt;n-3&lt;br&gt;n-2&lt;br&gt;n-1&lt;br&gt;n&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;0&lt;br&gt;1&lt;br&gt;0&lt;br&gt;&lt;br&gt;1&lt;br&gt;0&lt;br&gt;0&lt;br&gt;0&lt;br&gt;0&lt;br&gt;0&lt;br&gt;0&lt;br&gt;0&lt;br&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;no&lt;br&gt;yes&lt;br&gt;no&lt;br&gt;&lt;br&gt;yes&lt;br&gt;no&lt;br&gt;no&lt;br&gt;no&lt;br&gt;no&lt;br&gt;yes &lt;strong&gt;potential shark bait&lt;/strong&gt;&lt;br&gt;yes &lt;strong&gt;potential shark bait&lt;/strong&gt;&lt;br&gt;yes &lt;strong&gt;potential shark bait&lt;/strong&gt;&lt;br&gt;yes &lt;strong&gt;breaks tie&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;And we've found a stable one again.&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: right;&quot;&gt;Nr. of pirates&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: right;&quot;&gt;Pirate&lt;br&gt;meekest to fiercest&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: right;&quot;&gt;Nr. of coins&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: left;&quot;&gt;Vote&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;2c+9&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;1&lt;br&gt;2&lt;br&gt;3&lt;br&gt;...&lt;br&gt;n-9&lt;br&gt;n-8&lt;br&gt;n-7&lt;br&gt;n-6&lt;br&gt;n-5&lt;br&gt;n-4&lt;br&gt;n-3&lt;br&gt;n-2&lt;br&gt;n-1&lt;br&gt;n&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;1&lt;br&gt;0&lt;br&gt;1&lt;br&gt;&lt;br&gt;0&lt;br&gt;0&lt;br&gt;0&lt;br&gt;0&lt;br&gt;0&lt;br&gt;0&lt;br&gt;0&lt;br&gt;0&lt;br&gt;0&lt;br&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;yes&lt;br&gt;no&lt;br&gt;yes&lt;br&gt;&lt;br&gt;no&lt;br&gt;no&lt;br&gt;no&lt;br&gt;no&lt;br&gt;no&lt;br&gt;no&lt;br&gt;no&lt;br&gt;no&lt;br&gt;no&lt;br&gt;yes &lt;strong&gt;shark bait&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;And so on.&lt;/p&gt;&#xA;&lt;p&gt;When &lt;span class=&quot;math-container&quot;&gt;$n &amp;gt; 2c$&lt;/span&gt;, we'll see more and more large runs of downvoting pirates, until the number of pirates upvoting to save their lives becomes equal to the number of downvoters and we find a stable solution again.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;This yields the following formula for stable solutions:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ n \leq 2c \lor n = 2c + 2^z\,|\,z \in \mathbb Z_{\ge 0}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The procedure by which to distribute the coins is based on the fact that a pirate can only buy votes if coins are given to the opposite pirates from the next stable solution.&lt;br /&gt;&#xA;It is as follows:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For &lt;span class=&quot;math-container&quot;&gt;$n \leq 2c$&lt;/span&gt; give a coin to every pirate with the same parity as &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;, with any left over coins going to the fiercest pirate.&lt;/li&gt;&#xA;&lt;li&gt;For &lt;span class=&quot;math-container&quot;&gt;$2c + 2^{z-1} &amp;lt; n \leq 2c + 2^z\,|\,z \in \mathbb Z_{\ge 0}$&lt;/span&gt; start with the meekest pirate with the opposite parity of &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt; and move up in ferocity.&lt;sup&gt;†&lt;/sup&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;hr /&gt;&#xA;&lt;sup&gt;&#xA;&amp;Dagger;: I prefer to think of the pirates as kindhearted people who &lt;em&gt;care&lt;/em&gt; for the sharks and don't want them to go hungry.&lt;br /&gt;&#xA;&amp;dagger;: It can be argued that if a solution is known to be unstable, &lt;em&gt;all&lt;/em&gt; other pirates will vote no, since it doesn't matter who gets the coins — they won't get to keep them. But for the stable solution, it &lt;em&gt;is&lt;/em&gt; necessary to distribute the coins like this.&#xA;&lt;/sup&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditDate="2021-01-05T19:12:47.617" LastActivityDate="2021-01-05T19:12:47.617" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="295" PostTypeId="1" AcceptedAnswerId="299" CreationDate="2014-05-20T21:14:25.943" Score="16" ViewCount="6206" Body="&lt;p&gt;There are three cowboys in a Mexican standoff against each other, named Juan, José, and Jorge.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Juan is the straightest shooter in the West, and can hit his target 100% of the time.&lt;/li&gt;&#xA;&lt;li&gt;José, who has a cataract and can't see clearly, can hit his target 70% of the time.&lt;/li&gt;&#xA;&lt;li&gt;Jorge, who has no aim but is very clever, can hit his target only 30% of the time.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;They proceed to shoot starting from Jorge and going to José and finally Juan, continuing in the same order until only one of them is left standing. Who has the highest chance of surviving, and what strategies do each of them use to maximize their chances?&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="69582" LastEditDate="2020-12-25T01:17:09.643" LastActivityDate="2023-08-22T12:57:55.057" Title="The Mexican Standoff" Tags="|mathematics|calculation-puzzle|strategy|probability|game|" AnswerCount="4" CommentCount="11" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="296" PostTypeId="1" CreationDate="2014-05-20T23:26:48.513" Score="6" ViewCount="5674" Body="&lt;p&gt;The puzzle below features each week in a TV guide. It's called &quot;Number Workout&quot; there but I've done a bit of research and it seems to be called &quot;Hexanum&quot; elsewhere. It bears some resemblance to Sudoku, in that it's a logic puzzle with a single solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bAA4J.jpg&quot; alt=&quot;Hexanum&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Transcription: &quot;Fit the numbers 1-6 once in every hexagon so that where the hexagons touch, the numbers are the same.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've done lots of these and I have been able to solve them using simple techniques. In terms of the &lt;a href=&quot;https://puzzling.stackexchange.com/a/34/125&quot;&gt;Sudoku ratings listed here&lt;/a&gt;, it would be &quot;Level 1&quot; techniques: choosing a number and seeing where it can fit in each hexagon. (Several times I have got stuck, but when I finally found a new number it was still by simple deduction.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is: &lt;strong&gt;Is it possible to construct one of these puzzles, with a unique solution, that requires more advanced techniques to solve?&lt;/strong&gt; Note it doesn't need to be the same size as the one below - a smaller or larger puzzle may be required to prove it.&lt;/p&gt;&#xA;" OwnerUserId="125" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-01-31T18:55:41.013" Title="Are there 'Hexanum' puzzles that require advanced methods to solve?" Tags="|logical-deduction|difficulty-rating|" AnswerCount="1" CommentCount="5" ClosedDate="2015-01-05T07:43:35.067" ContentLicense="CC BY-SA 3.0" />
  <row Id="297" PostTypeId="2" ParentId="274" CreationDate="2014-05-20T23:32:27.230" Score="2" Body="&lt;p&gt;The best I can get by hand is $14,16,17,18,19,20,21,22,23,24,25,26,27,29,30,31,33,35,37,39,41,43,45,47,49$ for a sum of $731\ $.   I started with $26$ through $50$, then replaced big numbers by small numbers wherever I could.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="28" LastEditDate="2014-07-06T04:32:57.747" LastActivityDate="2014-07-06T04:32:57.747" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="298" PostTypeId="2" ParentId="295" CreationDate="2014-05-20T23:47:18.063" Score="5" Body="&lt;p&gt;If Juan gets a chance to shoot, he will shoot Jose (because Jose has a higher chance of killing him) and he will kill Jose because he has 100% accuracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jose knows this, and he gets a chance to shoot first, so he will shoot at Juan. There is a 70% chance that he will manage to kill Juan if he gets to shoot him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jorge knows that the other two are going to focus on shooting each other. If he succeeds at shooting either one of them, the other will go next and shoot him instead! That is not what he wants. So Jorge will go first, and he will not shoot at &lt;em&gt;either&lt;/em&gt; Juan or Jose. He can shoot at the ground, a passing bird, whatever. But he will not aim at either of the others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then it's Jose's turn. As we said, he'll aim at Juan with a 70% chance of success. If he misses, it's Juan's turn, and Juan will shoot and kill Jose. Then Jorge will have a 30% chance to kill Juan. If he fails, Juan will shoot him and he will die (Juan is the last left standing). If Jorge succeeds, he will be the last one left standing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going back to the top: Jorge passes, and it's Jose's turn. If he &lt;em&gt;succeeds&lt;/em&gt; in killing Juan, it's now Jorge's turn. He has a 30% change of killing Jose. If he fails, Jose has a 70% chance to kill Jorge. They will keep exchanging shots until one of them succeeds.&lt;/p&gt;&#xA;" OwnerUserId="10" LastActivityDate="2014-05-20T23:47:18.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="299" PostTypeId="2" ParentId="295" CreationDate="2014-05-20T23:50:04.960" Score="11" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Juan's strategy is simple: if he has to choose between opponents, then his shot &lt;em&gt;will&lt;/em&gt; turn it into a two-man contest, and he will have to survive a shot before he can kill the second man. So of course he will shoot Jose first.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;José's strategy, similarly, is to shoot at Juan. Leaving Juan alive and Jorge dead cannot possibly benefit him.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Jorge's strategy is the interesting case.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Aiming at José is clearly wrong, since hitting him would leave Juan with no other target.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's say Jorge aims at Juan and kills him. His chance of surviving is the sum of the chances that he will both survive a round and kill José in the next. Since both are 30% chances, he has a 30%×30% chance of winning on his next shot, a 30%×30%×30%×30% chance of winning on his shot after that, and so on. We can &lt;a href=&quot;http://www.wolframalpha.com/input/?i=sum%280.3%5E%28n%2a2%29%29%2C+n%3D1..infinity&quot;&gt;calculate&lt;/a&gt; his chances at $\sum_1^\infty 0.3^\left(2n\right)$ = 9.89%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can he do better? Yes!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jorge's best strategy (assuming he's a good enough shot to implement it) is to &lt;em&gt;intentionally miss&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if José misses Juan (30% chance), then Juan will kill José. Then Jorge's chances are exactly 30%: he gets one shot and must make it count.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if José hits Juan (70% chance), then it's Jorge's turn and he has a 30% chance of winning immediately, plus the chances of surviving a duel with Jorge as calculated above. Total chance: 39.89%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means Jorge's total chances are 70%×39.89% + 30%×30% = 36.9%. Not bad for a guy who can't shoot!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now that we have all three strategies, we can calculate the chances for Juan and José.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;José will get the first meaningful shot. He must first kill Juan (70% chance) and then survive his battle with Jorge (60.11% chance: 100% minus the 39.89% chance calculated above). Total chance: 70%×60.11% = 42.1%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Juan has a 30% chance of surviving José's shot and then killing him, then a 70% chance of beating Jorge. Total chance: 30%×70% = 21%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting it all together:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Juan: 21%&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;José: 42.1%&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Jorge: 36.9%&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The moral of the story: &lt;a href=&quot;https://en.wikipedia.org/wiki/The_Gunfighter&quot;&gt;never be the best gunfighter around.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="52" LastEditorUserId="52" LastEditDate="2014-06-23T20:47:50.807" LastActivityDate="2014-06-23T20:47:50.807" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="301" PostTypeId="2" ParentId="295" CreationDate="2014-05-21T00:38:05.763" Score="4" Body="&lt;p&gt;Let's approach this for a purely Game-theory perspective (which means, everyone simply wants to get the biggest possible payout, to live, and no one does actions like shooting someone out of revenge for shooting at them, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can deduce two things logically from the beginning. The first is that since Juan is such an excellent marksman, at the end of the first &quot;round&quot; (where everyone has had a chance to shoot at least once, assuming they haven't died), the game will &lt;strong&gt;always&lt;/strong&gt; end with either  one winner left alive, or a &quot;two-man standoff&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second thing we can deduce is that since everyone has an infinite amount of bullets (though, only three would actually be required per person until the game ends), if two people are left standing, the best thing they can do is keep shooting at each other until they or the other person is dead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you enjoy graphs, I have created a little ASCII diagram which illustrates how the first three turns can play out:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[Jorge]---(Shoot José)--- 0.3 HIT  ---[DEAD]-------------------------------[Juan]--(Shoot Jorge)-- # Juan WINS&#xA; \                    `-- 0.7 MISS ---[José]---(Shoot Jorge)--- 0.7 HIT  --[Juan]--(Shoot José )-- # Juan WINS&#xA;  \                                       \                 `-- 0.3 MISS --[Juan]--(Shoot Jorge)-- # José  vs. Juan two-man standoff&#xA;   \                                       \                                     `-(Shoot José )-- # Jorge vs. Juan two-man standoff&#xA;    \                                       `--(Shoot Juan )--- 0.7 HIT  --[DEAD]----------------- # Jorge vs. José two-man standoff&#xA;     \                                                      `-- 0.3 MISS --[Juan]--(Shoot Jorge)-- # José  vs. Juan two-man standoff&#xA;      \                                                                          `-(Shoot José )-- # Jorge vs. Juan two-man standoff&#xA;       `--(Shoot Juan)--- 0.3 HIT  ---[José]---(Shoot Jorge)--- 0.7 HIT  --[DEAD]----------------- # José WINS&#xA;                     \                                      `-- 0.3 MISS --[DEAD]----------------- # Jorge vs. José two-man standoff&#xA;                      `-- 0.7 MISS ---[José]---(Shoot Jorge)--- 0.7 HIT  --[Juan]--(Shoot José )-- # Juan WINS&#xA;                                          \                 `-- 0.3 MISS --[Juan]--(Shoot Jorge)-- # José  vs. Juan two-man standoff&#xA;                                           \                                     `-(Shoot José )-- # Jorge vs. Juan two-man standoff&#xA;                                            `--(Shoot Juan )--- 0.7 HIT  --[DEAD]----------------- # Jorge vs. José two-man standoff&#xA;                                                            `-- 0.3 MISS --[Juan]--(Shoot Jorge)-- # José  vs. Juan two-man standoff&#xA;                                                                                 `-(Shoot José )-- # Jorge vs. Juan two-man standoff&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h2&gt;Two-man standoffs&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There are three possible two-man standoffs at the end of the first round. Note that since everyone has had a chance to fire once, the person who is the poorest shot will be allowed to shoot first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Jorge vs. Juan&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Jorge&lt;/em&gt; is first to fire:&lt;br&gt;&#xA;&lt;strong&gt;&gt;&lt;/strong&gt; If he hits Juan (&lt;code&gt;30%&lt;/code&gt; chance), he wins.&lt;br&gt;&#xA;&lt;strong&gt;&gt;&lt;/strong&gt; If he misses (&lt;code&gt;70%&lt;/code&gt; chance), Juan fires back, and being an excellent marksman, is guaranteed to win the standoff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;José vs. Juan&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Jorge&lt;/em&gt; is first to fire:&lt;br&gt;&#xA;&lt;strong&gt;&gt;&lt;/strong&gt; If he hits Juan (&lt;code&gt;70%&lt;/code&gt; chance), he wins.&lt;br&gt;&#xA;&lt;strong&gt;&gt;&lt;/strong&gt; If he misses (&lt;code&gt;30%&lt;/code&gt; chance), same as above, Juan guarantees a win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Jorge vs. José&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Jorge faces off against José, since no one can hit the other person with &lt;code&gt;100%&lt;/code&gt; accuracy, there is an infinitesimally small probability that the standoff will go on forever with each person missing their shot until they both die of old age.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's possible to calculate an &quot;exact&quot; probability out of this game using integrals, however, we are going to stick to high-school algebra and basic logic to solve this puzzle in order to make it understandable to as many people as possible. Let's just say since Jorge shoots first, he has a greater than &lt;code&gt;30%&lt;/code&gt; chance of winning, and José has a less than &lt;code&gt;%70&lt;/code&gt; chance of winning. This is actually all the information we need to get an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's summarize the two-man standoffs. Instead of using percentages, we will be using values from &lt;code&gt;0&lt;/code&gt; to &lt;code&gt;1&lt;/code&gt; to make the math we will be doing later easier. The following values represent the probabilities of each person surviving the encounter:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Jorge vs. Juan: { Jorge:  0.3, José:  0.0, Juan: 0.7 }&#xA;José  vs. Juan: { Jorge:  0.0, José:  0.7, Juan: 0.3 }&#xA;Jorge vs. José: { Jorge: &amp;gt;0.3, José: &amp;lt;0.7, Juan: 0.0 }&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h2&gt;Juan&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Let's start with Juan. If only one person remains after José and Jorge had their turns, he will obviously shoot the remaining person, hitting the target perfectly, and will win the game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the other two have gotten their allowed shots, and Juan is left standing, he has two options:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A.&lt;/strong&gt; Shoot José&lt;br&gt;&#xA;&lt;strong&gt;B.&lt;/strong&gt; Shoot Jorge&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since he will always hit his opponent, the person he shoots will not make it to the next &quot;round&quot;. That means, he will be left in a two-man standoff (see Deduction #2) with the person he didn't shoot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at our previous summary, we know that he would rather be in a two-man standoff with the worst shooter (Jorge), so he will &lt;strong&gt;always&lt;/strong&gt; shoot José to avoid a two-man standoff with him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can now update our graph with this information:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[Jorge]---(Shoot José)--- 0.3 HIT  ---[DEAD]------------------------------- # Juan WINS&#xA;   \                  `-- 0.7 MISS ---[José]---(Shoot Jorge)--- 0.7 HIT  -- # Juan WINS&#xA;    \                                      \                `-- 0.3 MISS -- # Jorge vs. Juan&#xA;     \                                      `--(Shoot Juan )--- 0.7 HIT  -- # Jorge vs. José&#xA;      \                                                     `-- 0.3 MISS -- # Jorge vs. Juan&#xA;       `--(Shoot Juan)--- 0.3 HIT  ---[José]---(Shoot Jorge)--- 0.7 HIT  -- # José WINS&#xA;                     \                                      `-- 0.3 MISS -- # Jorge vs. José&#xA;                      `-- 0.7 MISS ---[José]---(Shoot Jorge)--- 0.7 HIT  -- # Juan WINS&#xA;                                           \                `-- 0.3 MISS -- # Jorge vs. Juan&#xA;                                            `--(Shoot Juan )--- 0.7 HIT  -- # Jorge vs. José&#xA;                                                            `-- 0.3 MISS -- # Jorge vs. Juan&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h2&gt;José&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Now it's José's turn. Obviously, if only one person remains, José will shoot whoever is left! But what if three people remain?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will update the graph to include the payoffs, which will see who José is better off shooting:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   [José]---(Shoot Jorge)--- 0.7 * { Jorge:  0.0, José:  0.0, Juan: 1.0 }&#xA;        \                `-- 0.3 * { Jorge:  0.3, José:  0.0, Juan: 0.7 }&#xA;         `--(Shoot Juan )--- 0.7 * { Jorge: &amp;gt;0.3, José: &amp;lt;0.7, Juan: 0.0 }&#xA;                         `-- 0.3 * { Jorge:  0.3, José:  0.0, Juan: 0.7 }&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If we multiply &lt;em&gt;the probability of José hitting the target&lt;/em&gt; by &lt;em&gt;the probability that José will win after that round&lt;/em&gt;, and then get the average of the two, we get the following results:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   [José]---(Shoot Jorge)--- { Jorge:  0.09, José:  0.00, Juan: 0.91 }&#xA;         `--(Shoot Juan )--- { Jorge: &amp;gt;0.30, José: &amp;lt;0.49, Juan: 0.21 }&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We can now clearly see that if there are three people remaining when José is ready to shoot, if he shoots Jorge he has absolutely &lt;code&gt;0%&lt;/code&gt; chance of surviving. On the other hand, if he shoots Juan, he has a bit less than &lt;code&gt;49%&lt;/code&gt; chance of winning. Not being an idiot, he will choose the latter!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our tree now looks like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[Jorge]---(Shoot José)--- 0.3 HIT  ---[DEAD]------------------- { Jorge:  0.00, José:  0.00, Juan: 1.00 }&#xA;     \                `-- 0.7 MISS ---[José]---(Shoot Juan )--- { Jorge: &amp;gt;0.30, José: &amp;lt;0.49, Juan: 0.21 }&#xA;       `--(Shoot Juan)--- 0.3 HIT  ---[José]---(Shoot Jorge)--- { Jorge: &amp;gt;0.09, José: &amp;lt;0.91, Juan: 0.00 }&#xA;                      `-- 0.7 MISS ---[José]---(Shoot Juan )--- { Jorge: &amp;gt;0.30, José: &amp;lt;0.49, Juan: 0.21 }&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h2&gt;Jorge&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Finally, it's Jorge's turn (well, actually, he shot first, but thinking backwards logically, we &lt;em&gt;finally&lt;/em&gt; arrive at him).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take a final look at that tree, collapsing the tree down so we see what Jorge's probabilities of winning are (hiding José's move):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[Jorge]---(Shoot José)--- 0.3 * { Jorge:  0.00, José:  0.00, Juan: 1.00 }&#xA;     \                `-- 0.7 * { Jorge: &amp;gt;0.30, José: &amp;lt;0.49, Juan: 0.21 }&#xA;       `--(Shoot Juan)--- 0.3 * { Jorge: &amp;gt;0.09, José: &amp;lt;0.91, Juan: 0.00 }&#xA;                      `-- 0.7 * { Jorge: &amp;gt;0.30, José: &amp;lt;0.49, Juan: 0.21 }&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And if we multiply the payoffs by the odds of them happening:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[Jorge]---(Shoot José)--- { Jorge: &amp;gt;0.210, José: &amp;lt;0.343, Juan: 0.447 }&#xA;       `--(Shoot Juan)--- { Jorge: &amp;gt;0.237, José: &amp;lt;0.616, Juan: 0.147 }&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It's a close tie! Jose has a slightly larger than &lt;code&gt;21.0%&lt;/code&gt; chance of winning if he shoots Jorge, but a &lt;strong&gt;very&lt;/strong&gt; slight lead of slightly larger than &lt;code&gt;23.7%&lt;/code&gt; chance of surviving if he shoots Juan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There isn't a very large chance that he will survive anyway, but Jorge hopes for the best, and aims his sights at &lt;strong&gt;Juan&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Results&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Assuming everyone plays rationally, here is how the game will go:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[Jorge]---(Shoot Juan)--- 0.3 HIT  ---[José]---(Shoot Jorge)--- 0.7 HIT  --[DEAD]--------------------- { Jorge:  0.0, José:  1.0, Juan: 0.0 }&#xA;                     \                                      `-- 0.3 MISS --[DEAD]---[Jorge vs. José]-- { Jorge: &amp;gt;0.3, José: &amp;lt;0.7, Juan: 0.0 }&#xA;                      `-- 0.7 MISS ---[José]---(Shoot Juan )--- 0.7 HIT  --[DEAD]---[Jorge vs. José]-- { Jorge: &amp;gt;0.3, José: &amp;lt;0.7, Juan: 0.0 }&#xA;                                                            `-- 0.3 MISS --[Juan]---(Shoot José)------ { Jorge:  0.3, José:  0.0, Juan: 0.7 }&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In the end, these are the combined chances that the respective marksmen will win:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Jorge:&lt;/strong&gt; &gt;23.7%&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;José:&lt;/strong&gt; &amp;lt;61.6%&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Juan:&lt;/strong&gt; 14.7%&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So surprisingly, despite being a semi-poor shot, assuming Jorge acts rationally, the odds are in José's favor! Yet, one can never guarantee anything with randomness.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Addendum&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There is a &quot;bonus answer&quot; for Jorge getting a better payoff if you are willing to &quot;think outside the box&quot;, but I'll leave that one for you to decipher. ;)&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="115" LastEditDate="2014-05-21T09:14:47.633" LastActivityDate="2014-05-21T09:14:47.633" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="302" PostTypeId="1" AcceptedAnswerId="304" CreationDate="2014-05-21T01:01:45.367" Score="33" ViewCount="2237" Body="&lt;p&gt;Three world powers have just entered a nuclear war!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They have been waiting for this day for decades; everyone has their nuclear missiles ready to fire, and only need to choose a target.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;The Chinese&lt;/strong&gt; missiles hit their  target &lt;strong&gt;30%&lt;/strong&gt; of the time.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;The American&lt;/strong&gt; missiles hit their target &lt;strong&gt;60%&lt;/strong&gt; of the time.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;The German&lt;/strong&gt; missiles hit their target &lt;strong&gt;90%&lt;/strong&gt; of the time.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is similar to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/295/the-mexican-standoff&quot;&gt;the Mexican Standoff&lt;/a&gt;, but instead, every round the three countries fire their missiles &lt;strong&gt;simultaneously&lt;/strong&gt; at a chosen target. After everyone has fired their missile and found out which countries are left standing, the surviving nations begin a second round, firing at a chosen nation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This continues until either the last World Power remains, or everyone has wiped each other out.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What are the strategies for these three countries?&lt;/li&gt;&#xA;&lt;li&gt;What is the probability of each country winning if they follow their strategies?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Clarifications:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Countries don't know where their neighbors fired the missiles until &lt;em&gt;after&lt;/em&gt; the round is over.&lt;/li&gt;&#xA;&lt;li&gt;Two countries can unknowingly hit and destroy the same target in the same round.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="72" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2023-08-26T21:57:15.593" Title="The Nuclear Missile Standoff" Tags="|calculation-puzzle|word-problem|game-theory|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="303" PostTypeId="2" ParentId="284" CreationDate="2014-05-21T02:33:14.097" Score="9" Body="&lt;p&gt;Yes, there is an algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh, you wanted to know what the algorithm is? Here is one, which presents an algorithm but not all the reasoning behind it. For that, and a proof, and the variant where the odd ball must be identified but not the direction of its deviance, read &lt;a href=&quot;https://math.stackexchange.com/questions/15423/optimal-algorithm-for-finding-the-odd-spheres/336224#336224&quot;&gt;Thomas Pornin's excellent presentation on the same question on Math Stack Exchange&lt;/a&gt; (which I used heavily while writing my answer here).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The core idea is that this problem works in threes. Each weighing involves three sets of balls: the left pane, the right pane, and the balls not being weighed. Each weighing has three possible results: lighter, balanced or heavier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll start with the simpler variant where you know that the odd ball out is heavier than the rest (but not as heavy as two normal balls). The general idea is to weigh one third of the balls against another third, determine which set contains the odd ball, and repeat. But there's a big wrinkle: at the beginning, you don't know whether the odd ball is heavier or lighter, so unless the result of the weighing is “equal”, you're left with two possibilities. To compensate for that, we'll put slightly less than 1/3 of the balls on each pane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determine the number $w$ such that $(3^{w-1} - 3)/2 \lt N \le (3^w - 3)/2$, i.e. $w = \lceil \log_3(2N+3) \rceil$. This will be the number of required weighings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first stage, weigh $(3^{w-1}-1)/2$ balls against the same number of balls.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the outcome is “heavier” or “lighter”, all the left-over balls are standard; apply the second stage to all the balls that were weighed, and put a marking on each of these balls to indicate whether it was on the lighter or heavier side (this will be used to tell whether the odd ball is lighter or heavier).&lt;/li&gt;&#xA;&lt;li&gt;If the outcome is “balanced”, then the odd ball is among the left-over balls, and all the weighed balls are standard; apply the third stage to the left-over balls.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the second stage, we have a pool of standard balls. Set $(3^w-1)/2$ balls apart, and weigh half of the remainder against the other half, using a standard ball to complete the set if the size is odd.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the outcome is “balanced”, repeat the second stage with the left-over balls.&lt;/li&gt;&#xA;&lt;li&gt;If the outcome is “lighter” or “heavier”, then move on to the third stage (same principle as in the first stage).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the third stage, all the balls involved have a mark which indicates whether they are part of the heavy group or of the light group. Put $\lceil N/3 \rceil$ balls on each side of the scale, taking care to put the same number of light balls on both sides (and also the same number of heavy balls).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the outcome is “balanced”, the odd ball is one that wasn't weighed. Repeat the third stage with the left-over balls.&lt;/li&gt;&#xA;&lt;li&gt;If the outcome is “lighter” or “heavier”, then the odd ball is either among the light balls on the lighter side or among the heavy balls on the heavier side. Repeat the third stage with the potential odd balls.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The process stops when there are at most two potential odd balls:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If there is a single potential odd ball, it's the one.&lt;/li&gt;&#xA;&lt;li&gt;If there are two balls marked heavy, weigh one of them against a known-standard ball. (There will be one, courtesy of stage 2, except if you started with two balls, in which case the problem cannot be solved.)&lt;/li&gt;&#xA;&lt;li&gt;If there are two balls with different markings, weigh them against each other.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In all cases, the last weighing reveals the odd ball, and its marking indicates whether it is lighter or heavier.&lt;/p&gt;&#xA;" OwnerUserId="71" LastEditorUserId="-1" LastEditDate="2017-04-13T12:19:36.950" LastActivityDate="2014-06-13T19:38:26.863" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="304" PostTypeId="2" ParentId="302" CreationDate="2014-05-21T03:29:39.673" Score="29" Body="&lt;p&gt;We can see that as long as all the countries fire, at least two countries will be targeted. There are eight possible strategies to be played in the first round:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; Atkrs  P.O.D.&#xA; A C G  USA CHN GER&#xA; ------------------&#xA; C A A  93% 60%  0%  &#xA; C A C  30% 96%  0%&#xA; C G A  90% 60% 30%&#xA; C G C   0% 96% 30%&#xA; G A A  93%  0% 60%&#xA; G A C  30% 90% 60%&#xA; G G A  90%  0% 72%&#xA; G G C   0% 90% 72%&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If any one of them are destroyed, the strategies for any two remaining countries are just to fire at each other, in which case the probabilities become:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; A vs. G: A wins 6.25%, G wins 37.5%, they both get destroyed 56.25%.&#xA; A vs. C: A wins 58.33%, C wins 16.67%, they both get destroyed 25%.&#xA; C vs. G: G wins 67.74%, C wins 3.23%, they both get destroyed 29.03%.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;These probabilities were calculated by dividing the overall probability of each event by the probability that not everybody survives. Because in that case, nothing happens, and they just keep on trying until something does happen, and the probabilities of the other events stay in the same proportion anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If any two of them are destroyed, then the last one standing is the winner by default.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Since they all fire simultaneously, their strategies can be evaluated independently, since they can't depend on any other information. If America and Germany were to target each other, China wouldn't benefit at all by choosing not to fire, because just firing at Germany increases the chances that both of those countries will be destroyed, and China would emerge the sole survivor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems obvious, given the symmetry of this problem, that each country would simply fire at the country that has the highest hit rate besides them, since by analyzing the two-country standoffs, eliminating that country will always maximize their chances in the resulting duel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the first round, America and China both fire at Germany, and Germany fires at America. The probability table then looks as such:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;America and Germany both get destroyed: 66.67%&#xA;Only America gets destroyed:            25.92%&#xA;Only Germany gets destroyed:             7.41%&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In the first case, which has a 2/3 probability of happening under this naïve strategy, China is the sole survivor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second case, the standoff between Germany and China has a &lt;code&gt;67.74%&lt;/code&gt; chance of Germany winning, which represents a &lt;code&gt;17.56%&lt;/code&gt; chance overall. And China &lt;em&gt;still&lt;/em&gt; has a &lt;code&gt;3.23%&lt;/code&gt; chance of surviving, which is a &lt;code&gt;0.84%&lt;/code&gt; chance overall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the third case, the standoff between America and China has a &lt;code&gt;58.33%&lt;/code&gt; chance of America winning, which represents a &lt;code&gt;4.32%&lt;/code&gt; chance overall. And China has a &lt;code&gt;16.67%&lt;/code&gt; chance of winning, which represents a &lt;code&gt;1.23%&lt;/code&gt; chance overall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the total probabilities of surviving are as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;China:    68.74%&#xA;Germany:  17.56%&#xA;America:   4.32%&#xA;Nobody:    9.38%&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;An interesting quirk of this problem is that America and Germany are in a sort of prisoner's dilemma when it comes to deciding whether to target each other or China.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;China will target Germany on the first round, no matter what. But if America and Germany decide to collude and both target China, then the probability matrix looks something like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;China and Germany both get destroyed: 29.63%&#xA;Only China gets destroyed:            69.14%&#xA;Only Germany gets destroyed:           1.23%&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And in the resulting standoff if only China gets destroyed, Germany has a &lt;code&gt;37.5%&lt;/code&gt; chance of winning, which is a total survival ratio of &lt;code&gt;25.92%&lt;/code&gt;. The resulting survival table looks like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Nobody:   39.20%&#xA;America:  34.67%&#xA;Germany:  25.92%&#xA;China:     0.21%&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So America and Germany's survival ratios &lt;em&gt;both&lt;/em&gt; go up if they decide to mutually target China. But since they can't contact each other to make this decision, they both &quot;defect&quot;, try to target each other, and China ends up surviving over two-thirds of the time.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2014-05-21T03:43:45.120" LastActivityDate="2014-05-21T03:43:45.120" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="305" PostTypeId="1" AcceptedAnswerId="389" CreationDate="2014-05-21T04:53:33.390" Score="13" ViewCount="4085" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;NOTE:&lt;/strong&gt; This is in reference to &lt;a href=&quot;https://puzzling.stackexchange.com/q/252/72&quot;&gt;an existing question&lt;/a&gt; which I feel is poor and off-topic. In the spirit of &quot;shaping the site while in private beta&quot;, I wanted show how the question can be re-phrased and &quot;fixed&quot; in order to be more on-topic here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is a specific Sudoku puzzle presented by the mathematician who discovered it as &lt;a href=&quot;http://aisudoku.com/index_en.html&quot; rel=&quot;noreferrer&quot;&gt;&quot;the most difficult Sudoku in the world&quot;&lt;/a&gt;. It looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IA61i.jpg&quot; alt=&quot;8..........36......7..9.2...5...7.......457.....1...3...1....68..85...1..9....4..&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this Sudoku puzzle considered to be so difficult? Can it be solved through strategy alone, or is brute-forcing it required?&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2018-05-21T07:16:51.883" Title="Why is this considered to be &quot;The World's Hardest Sudoku&quot;?" Tags="|sudoku|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="306" PostTypeId="2" ParentId="282" CreationDate="2014-05-21T04:56:59.840" Score="10" Body="&lt;p&gt;This puzzle has a computer science interpretation¹.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The humans must provide 10 bits of information in total (each hat color is one bit of information). Participant number $k$ (with #0 at the back and #9 at the front) will hear $k$ bits of information from the humans behind him and will see $10-k-1$ bits of information on the heads in front of him. So there's one bit missing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The person at the back cannot do better than guess at random, since he has no information whatsoever that would indicate his own hat's color. Therefore the strategy should provide information that will help his successors optimally. The amount of information is precisely one bit, which is what the other partipants need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that all the participants except poor #0 succeed in calling their own hat's color, participant $k$ will have complete information except for #0's hat color (which is irrelevant) and his own hat color. In other words, he has a bit string with one bit missing. This calls for an &lt;a href=&quot;http://en.wikipedia.org/wiki/Error_correcting_code&quot;&gt;error correcting code&lt;/a&gt;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the position of the error will be known, a very simple error correcting code is sufficient: a &lt;a href=&quot;http://en.wikipedia.org/wiki/Parity_bit&quot;&gt;parity bit&lt;/a&gt;. Participant #0 calls out (for example) “purple” if he sees an even number of purple hats and “green” otherwise. Each other participant deduces from what he hears and sees whether the number of purple hats other than #0's and his is even or odd, and calls accordingly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;¹ &lt;sub&gt; Which, in as much as puzzles belong in a job interview, makes it a little relevant when interviewing for a programming job. The puzzle itself is older than Google. It may well be older than computers, but I don't know its origin. &lt;/sub&gt;  &lt;/p&gt;&#xA;" OwnerUserId="71" LastEditorUserId="71" LastEditDate="2014-06-13T19:34:52.253" LastActivityDate="2014-06-13T19:34:52.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="307" PostTypeId="2" ParentId="29" CreationDate="2014-05-21T11:28:20.247" Score="3" Body="&lt;p&gt;One approach would be doing a breadth-first brute force and noting the highest depth level visited before arriving at a solution. The higher the needed depth, the more difficult the puzzle is. &lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2014-05-21T11:28:20.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="308" PostTypeId="2" ParentId="274" CreationDate="2014-05-21T12:44:05.500" Score="4" Body="&lt;p&gt;By logically starting from 26-50 and trying to shrink them one by one you can easily show: $8,12,14,17,18,19,20,21,22,23,25,26,27,29,30,31,33,35,37,39,41,43,45,47,49$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which equals $711$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is performed by replacing every even number by itself divided by two if and only if the result is not a divisor of any other number (usually if that number times $3$ remains) from $50$ to the lowest number.  Then you repeat this process until you get the answer above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not convinced that this is definitely the smallest and am therefore worked on an alternative solution but this answer was also MY programming result so I a confident enough to include it as an answer.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="44915" LastEditDate="2018-03-02T10:59:00.673" LastActivityDate="2018-03-02T10:59:00.673" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="309" PostTypeId="1" AcceptedAnswerId="319" CreationDate="2014-05-21T13:05:38.960" Score="10" ViewCount="4359" Body="&lt;p&gt;It is possible to have a proper Sudoku with one empty block (consisting of 3x3 cells). The trivial (and very boring) example would be a puzzle where all other cells are given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's impossible to have a proper Sudoku with an empty chute (a band of 3 blocks), since units within the chute that are aligned with it, could be swapped around while the puzzle remained valid. So we need &lt;em&gt;at least&lt;/em&gt; three non-empty blocks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what is the maximum number of empty blocks a proper Sudoku can have? I don't think I've seen Sudoku's with more than one block empty; is this even possible?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&#xA;A standard Sudoku consists of 81 cells in 9 3x3 blocks. Those 9 blocks live in 6 chutes; 3 bands (horizontal) and 3 stacks (vertical).&lt;br&gt;&#xA;A block is a unit; other units are rows and columns.&lt;br&gt;&#xA;For &lt;em&gt;chutes&lt;/em&gt; and other terminology, see this &lt;a href=&quot;http://en.wikipedia.org/wiki/Glossary_of_Sudoku&quot; rel=&quot;noreferrer&quot;&gt;glossary&lt;/a&gt;.&#xA;&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditDate="2016-05-23T11:41:54.387" LastActivityDate="2016-08-07T09:32:13.797" Title="What is the maximum number of empty 3x3 blocks a proper Sudoku can have?" Tags="|sudoku|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="311" PostTypeId="2" ParentId="309" CreationDate="2014-05-21T13:35:23.630" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Mathematics_of_Sudoku&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia has a good discussion of this.&lt;/a&gt; However, given the sheer size of the number of possible puzzles, it's not yet known whether the below puzzles can do better.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The largest rectangular orthogonal &quot;hole&quot; (region with no clues) in a proper sudoku is believed to be a rectangle of 30 cells (a 5 x 6 rectangular area). One such example is the following with 22 clues:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/e8bjd.jpg&quot; alt=&quot;Sudoku with 30 cell (5 x 6) empty rectangle&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The largest total number of empty groups (rows, columns, and squares) in a sudoku is believed to be nine. One example is the following; a sudoku which has 3 empty squares, 3 empty rows, and 3 empty columns and gives 22 clues.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eNWU3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="42" LastEditorUserId="42" LastEditDate="2014-05-21T14:30:37.780" LastActivityDate="2014-05-21T14:30:37.780" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="313" PostTypeId="1" AcceptedAnswerId="317" CreationDate="2014-05-21T14:31:24.993" Score="78" ViewCount="19683" Body="&lt;p&gt;This is an interesting puzzle which was passed to me by a friend some time ago. I do know the answer, but will refrain from self-answering on this to see where it goes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking of a number: 1, 2, or 3. You may ask me one question, which I will answer to the best of my ability. I may not, however, tell you my number or any codified version of my number. I can only answer yes, no, maybe, I don't know, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you tell the number I'm thinking of?&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorDisplayName="user20" LastEditDate="2014-05-21T14:40:38.153" LastActivityDate="2021-03-25T16:21:49.690" Title="You have one question to tell whether the number I'm thinking of is 1, 2, or 3" Tags="|logical-deduction|" AnswerCount="28" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="314" PostTypeId="2" ParentId="313" CreationDate="2014-05-21T14:38:01.890" Score="7" Body="&lt;p&gt;You should ask:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We have 1 ball on the table and 0 or 1 balls under a concealed cup. Now I ask you is the amount of balls on the table is larger than the number on your mind? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;No, because 1 is not larger than the 1 or 2 balls on the table&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Maybe, because maybe it's larger than 1, but it's not larger than 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Yes, because 3 is larger than the 1 or 2 balls on the table. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="146" LastEditorUserId="46903" LastEditDate="2018-09-21T17:09:25.687" LastActivityDate="2018-09-21T17:09:25.687" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="316" PostTypeId="2" ParentId="313" CreationDate="2014-05-21T14:48:14.957" Score="18" Body="&lt;p&gt;The question I would ask is this (where $n$ is the number you are thinking of):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is $\lim_{x \to n} \frac{1}{x-2} &amp;gt; 0$ (where we are approaching $n$ from either the left or the right)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The possible answers are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $n = 1$: $\frac{1}{1-2} = -$, so the answer is &lt;strong&gt;&quot;No&quot;&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;If $n = 2$: $\lim_{x \to 2^+} \frac{1}{x-2} = +\infty$ but $\lim_{x \to 2^-} \frac{1}{x-2}$ = $-\infty$ so the answer is &lt;strong&gt;&quot;I don't know&quot;&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;If $n = 3$: $\frac{1}{3-2} = 1$, so the answer is &lt;strong&gt;&quot;Yes&quot;&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="198" LastEditorUserId="28" LastEditDate="2014-07-02T18:12:43.400" LastActivityDate="2014-07-02T18:12:43.400" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="317" PostTypeId="2" ParentId="313" CreationDate="2014-05-21T14:48:32.477" Score="125" Body="&lt;p&gt;Would it work if I ask: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I'm thinking of a number: either 0 or 1. Is the sum of our numbers greater than 2?&quot; I could figure out the answer if that's a question I'm allowed to ask.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If your number is 3, then you would say &quot;yes&quot; because no matter what you add, it would be true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your number is 2, you would say &quot;I don't know&quot;. If my number was 0, then the answer would be no, but if my number is 1 the answer is yes. So you don't know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your number is 1, then you would respond &quot;no&quot;.  No matter what number you add, the number will never be greater than 2.&lt;/p&gt;&#xA;" OwnerUserId="19" LastEditorUserId="48462" LastEditDate="2018-09-08T09:40:20.263" LastActivityDate="2018-09-08T09:40:20.263" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="318" PostTypeId="2" ParentId="313" CreationDate="2014-05-21T14:49:12.440" Score="26" Body="&lt;p&gt;Here's my thought process on this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The three answer possibilities must be &quot;yes,&quot; &quot;no,&quot; and... something else.&lt;/li&gt;&#xA;&lt;li&gt;Perhaps there is a way to get an answer of &quot;maybe&quot; or &quot;sometimes?&quot;&lt;/li&gt;&#xA;&lt;li&gt;Therefore, this would have to involve some kind of randomness.&lt;/li&gt;&#xA;&lt;li&gt;The question should be in the form of &quot;Either X or Y is true. Is your number {some condition}?&quot;&lt;/li&gt;&#xA;&lt;li&gt;A simple condition I can think of is &quot;even or odd.&quot;&lt;/li&gt;&#xA;&lt;li&gt;Is there a way to get a known even and known odd result for two of the numbers, and an unknown one for the other?&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let's make a little chart:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 2 3&#xA;O E O n^2&#xA;E O O floor(n/2)&#xA;O O E ceil(n/2)&#xA;E E O floor(n/3)&#xA;O O O ceil(n/3)&#xA;&lt;/code&gt;&lt;/pre&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Hey, look! For &lt;code&gt;floor(n/2)&lt;/code&gt; and &lt;code&gt;floor(n/3)&lt;/code&gt;, our conditions are satisfied! &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;So, my final question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I will divide your number by either 2 or 3 and round it down. Is the result even or odd?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the answer is &quot;even,&quot; the number was 1. If it's &quot;odd,&quot; the number was 3. If it's &quot;I don't know,&quot; the number was 2.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;Note that an alternate question could use &lt;code&gt;n^2&lt;/code&gt; and &lt;code&gt;floor(n/3)&lt;/code&gt; instead (I will either square your number or divide it by three and round down).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="39" LastEditorUserId="39" LastEditDate="2014-05-21T14:54:18.140" LastActivityDate="2014-05-21T14:54:18.140" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="319" PostTypeId="2" ParentId="309" CreationDate="2014-05-21T15:12:45.763" Score="12" Body="&lt;p&gt;&lt;strong&gt;Four is provably the least number of empty boxes. I will proceed to demonstrate this  here.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start by defining a 3x3 block on the Sudoku grid. This has no restrictions, as all the numbers are currently interchangeable without issue. From here out, I will refer to the purple boxes as &quot;odd&quot; boxes, and the green ones as &quot;even.&quot; The images were generated using &lt;a href=&quot;http://www.sudokuwiki.org/sudoku.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;a Sudoku solver&lt;/a&gt;, so I trust the listings of possibilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VmCsi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to create a sequence of squares which will fully define other squares. We know a couple important properties:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;That if three consecutive boxes are empty, then there is not a singular solution&lt;/li&gt;&#xA;&lt;li&gt;That if all the other boxes in a given row and column are not defined, then the target box is not defined.&lt;/li&gt;&#xA;&lt;li&gt;That it is trivially irrelevant what the permutation of this puzzle is, but rather that one exists.&lt;/li&gt;&#xA;&lt;li&gt;That the removal of clues does not help to solve the puzzle, so we can assume the four known boxes are fully solved.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I also assume the following vernacular:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A rigorously defined box is one for which all values are known.&lt;/li&gt;&#xA;&lt;li&gt;A rigorously defined box is one for which all other boxes in the same row and column are rigorously defined.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We want to demonstrate that it is impossible to fully define the remaining five boxes with four others. To do this, we need to show another couple properties:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Using only the numbers in two adjacent boxes, it is impossible to fully define any numbers in a given box.&lt;/li&gt;&#xA;&lt;li&gt;Using three boxes, it is impossible to define any more than three numbers in a given box, and that those three numbers must lie on a diagonal (or trivial permutation of a diagonal).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Because the order of the columns is trivially irrelevant, as is the second selected box, we fill in another box as such:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HX8jY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can see that no matter which three boxes we select, the numbers must fall along a diagonal. As a consequence of two boxes being collinear, the possibilities of the target box must be inherently limited to the diagonal. The possibilities in each column are such that, horizontally, they are unique, and vertically, they are distinct. This proves the first lemma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the maximum solution to this situation is to place two numbers in a row to cancel two possibilities. The sudoku puzzle we currently have is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/iLQjS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of the remaining numbers in the third box are relevant to those in the second. Therefore, as this situation has trivially covered all permutations, the second lemma is (admittedly, intuitively) proven.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can actually move on to demonstrating why five empty boxes is impossible. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To execute this proof, I will first prove that all four boxes must fall on odd squares. Once that is done, I will prove that under no configuration of boxes in odd squares can a functional Sudoku be achieved.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the four filled boxes are placed on the green squares above (the edge squares), then the center box is rigorously defined, and none of the corner boxes are. &lt;/li&gt;&#xA;&lt;li&gt;If three edge boxes are filled, and any corner is filled, then no boxes are rigorously defined.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As a result, we can conclude easily that at least two boxes must not be on the edge squares. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If two boxes are placed on the corners, then there exists no combination of boxes that can define any other box. (The proof of this is intuitive, but has a lot of cases, so is left to the reader.)&lt;/li&gt;&#xA;&lt;li&gt;If one box is placed on the center, and one is placed in the corner, then it is at most possible to place the even boxes such that the diagonals of two other even boxes are known. However, at this point, it is not possible to continue, as no other boxes will be defined.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is a point useful to demonstrate with an image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CYOXF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that even though we now have the diagonals on two boxes, we only have $1\frac{1}{3}$ of a definition for the surrounding boxes. Thus, we cannot continue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result, we now know that at least three boxes must be on the odd squares. Now, let us consider the possibilities where there are three boxes on odd squares.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If there are three boxes on odd squares, then if the three boxes are on corners, we have three cases:&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the remaining green box falls on the edge between two defined corners, then the box partially defines the green box opposite it, but the solutions end here - no other boxes have the 3 required to be solvable.&lt;/li&gt;&#xA;&lt;li&gt;If the green box is not between two solved corners, then it partially defines the remaining unsolved corner box, as well as the edge opposite it. However, no other values can be derived.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;If there are two boxes on adjacent corners, and one box in the center:&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the remaining box falls between the solved corners, it partially defines the opposite edge box, but that is all.&lt;/li&gt;&#xA;&lt;li&gt;If the remaining box is adjacent to a single solved corner, then it partially defines the corner next to it, as well as the edge across from it, but that is all.&lt;/li&gt;&#xA;&lt;li&gt;If the remaining box is placed on an edge opposite the solved corners, then the edge opposite it is rigorously defined, but no other boxes can be solved.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;If there are two boxes on opposite corners, and one in the center:&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Placing the green box anywhere on the grid can partially define two boxes: the opposite edge and the unsolved adjacent corner. Nothing else can be derived. &lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As a result of these three cases, it is clear that all four boxes must lie on the odd spaces - the corners and center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is trivial. If all four boxes lie on corners, then the puzzle cannot be solved, as no box can be defined. If three boxes lie on corners and one lies in the center, then two boxes are rigorously defined, but nothing* else can be derived.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*: In some cases, it may be possible to derive the diagonals of the opposite boxes by the &lt;a href=&quot;http://www.sudokuwiki.org/Unique_Rectangles&quot; rel=&quot;nofollow noreferrer&quot;&gt;Unique Rectangles&lt;/a&gt; or &lt;a href=&quot;http://www.sudokuwiki.org/Hidden_Unique_Rectangles&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hidden Unique Rectangles&lt;/a&gt; reduction methods, but it goes no farther than this (I swear!). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In summary&lt;/strong&gt;, the four solved boxes must be placed on odd boxes (corners and center), but no combination exists there for the boxes to be solved. Thus, the puzzle requires more than four boxes to be solved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see why it is possible to have four boxes unsolved, this puzzle is trivially solvable:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tp9Ko.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: complete.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorUserId="68" LastEditDate="2016-08-07T09:32:13.797" LastActivityDate="2016-08-07T09:32:13.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="320" PostTypeId="5" CreationDate="2014-05-21T15:30:46.627" Score="0" Body="&lt;p&gt;A checkerboard is a grid of squares with alternating black and white colors, most commonly an 8x8 board but not necessarily even rectangular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many checkerboard puzzles concern the adjacency relations between squares:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Two squares are said to be &lt;em&gt;horizontally or vertically&lt;/em&gt; adjacent (sometimes also called &lt;em&gt;orthogonally&lt;/em&gt; adjacent or &lt;em&gt;strongly&lt;/em&gt; adjacent), if they share a common edge.&#xA;Every corner/boundary/interior square has 2/3/4 horizontally or vertically adjacent neighbors.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Two squares are said to be &lt;em&gt;weakly&lt;/em&gt; adjacent (sometimes also called &lt;em&gt;King-adjacent&lt;/em&gt;), if they share a common edge or a common point.&#xA;Every square has at most eight weakly adjacent neighbors.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Example&lt;/strong&gt;: The &lt;em&gt;mutilated chessboard&lt;/em&gt; problem considers a standard 8x8 chessboard that has two diagonally opposite corners removed (hence leaving 62 squares). Is it possible to place 31 dominoes of size 2x1 so as to cover all of these squares?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11" LastEditorUserId="8874" LastEditDate="2015-10-04T13:58:40.807" LastActivityDate="2015-10-04T13:58:40.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="321" PostTypeId="4" CreationDate="2014-05-21T15:30:46.627" Score="0" Body="A puzzle involving checkerboards: grids of squares alternating black and white in color, most commonly an 8x8 board." OwnerUserId="11" LastEditorUserId="8874" LastEditDate="2015-10-04T13:59:33.463" LastActivityDate="2015-10-04T13:59:33.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="323" PostTypeId="1" AcceptedAnswerId="325" CreationDate="2014-05-21T16:30:09.583" Score="52" ViewCount="323185" Body="&lt;p&gt;If I wanted to solve a physical Rubik's cube multiple times, for practice, what is the best, most random way, to scramble the solved cube?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best way I can think is to hold it behind my back and turn randomly until, when I look at it, it &lt;em&gt;looks&lt;/em&gt; random enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a better way?&lt;/p&gt;&#xA;" OwnerUserId="220" LastEditorUserId="73193" LastEditorDisplayName="user20" LastEditDate="2021-10-20T07:38:40.120" LastActivityDate="2021-10-20T07:38:40.120" Title="What is the best method of scrambling a rubik's cube?" Tags="|rubiks-cube|twisty-puzzles|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="324" PostTypeId="2" ParentId="309" CreationDate="2014-05-21T16:40:24.797" Score="3" Body="&lt;p&gt;You can have four $3 \times 3$ blocks empty. I took the example from Wikipedia, blanked four blocks to get the below, and it solved easily. I would be surprised but not shocked at five.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pQilW.jpg&quot; alt=&quot;A sudoku with 4 empty blocks&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="68" LastEditDate="2014-07-02T13:13:22.167" LastActivityDate="2014-07-02T13:13:22.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="325" PostTypeId="2" ParentId="323" CreationDate="2014-05-21T16:40:50.750" Score="37" Body="&lt;p&gt;If you're willing to take the time: The &lt;a href=&quot;https://www.worldcubeassociation.org&quot;&gt;World Cube Association&lt;/a&gt; has a specific method for shuffling their cubes. They use a program called &lt;a href=&quot;https://github.com/cubing/tnoodle&quot;&gt;TNoodle&lt;/a&gt; to do the scrambling; the program generates a &quot;scramble sequence&quot; which can be followed to produce a scrambled cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their code actually features a nice interface to generate the scrambles:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rOj6o.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have yet to figure out what the &quot;Adjust Events&quot; options do, but producing the scramble sequences just comes down to clicking the &quot;Scramble!&quot; button.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the result of running their program:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bLOyh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;There is also an online interface &lt;a href=&quot;https://www.worldcubeassociation.org/regulations/history/files/scrambles/scramble_cube.htm&quot;&gt;here&lt;/a&gt;.&lt;/strong&gt; I used it to scramble my new 30x30x30 Rubik's cube:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZyqN4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="198" LastEditorUserId="198" LastEditDate="2014-05-21T17:11:10.303" LastActivityDate="2014-05-21T17:11:10.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="326" PostTypeId="2" ParentId="18" CreationDate="2014-05-21T17:21:52.273" Score="43" Body="&lt;p&gt;The quickest way may be by some form of intelligent trial and error, but there is one systematic method which will always work, and that is to translate the problem into the language of &lt;a href=&quot;http://en.wikipedia.org/wiki/Graph_theory&quot; rel=&quot;noreferrer&quot;&gt;graph theory&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start by making a complete list of all possible configurations of people and items. In this case, the boat will always be together with the farmer, so there are four objects, and \$2^4=16\$ possible configurations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cross out all configurations where something will be eaten, and number the remaining configurations. Here, six configurations are forbidden, leaving us with ten, which I describe by what is found on the left (original) river bank:&#xA;$$1 : (M,F,G,B) \qquad  2: (M,F;G)\qquad 3:(M,F,B)$$&#xA;$$4 : (M,G;B)\qquad 5:(M,G)  \qquad 6:(F,B)$$&#xA;$$7:(F)\qquad 8:(G)\qquad 9:(B)\qquad 10:(none)$$&#xA;I'm using the letter M for the farmer, who is described as &quot;he&quot;, in order to avoid confusion with the fox (F).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we will make a graph with 10 nodes, where each node is one of the configurations above. Two nodes will be connected by an edge if and only if it is possible to get from one configuration to the other with a single boat trip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, node 1 will be connected to node 6, and node 2 will be connected to node 7 and to node 8. We find that node 3 will be connected to three nodes; node 6, node 7, and node 9. Node 4 will be connected to node 8 and node 9, and node 5 will be connected to node 8 and node 10. Going through the remaining nodes 6 - 10, we find no new connections (the farmer has to return from the other river bank, resulting in one of the configurations 1 - 5).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step is to make a drawing of this graph. My computer abilities are very limited, but you should get something like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;        7--2&#xA;       /    \&#xA;1--6--3      8--5--10&#xA;       \    /&#xA;        9--4&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where you are allowed to go from node 3 to either 7 or 9, and from 2 or 4 to node 8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we return to the original question, which can now be interpreted as: Find a path in our graph starting at node 1 and ending at node 10. The two solutions are immediately apparent!&lt;/p&gt;&#xA;" OwnerUserId="219" LastEditorUserId="29870" LastEditDate="2016-10-17T16:31:10.670" LastActivityDate="2016-10-17T16:31:10.670" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="327" PostTypeId="1" AcceptedAnswerId="328" CreationDate="2014-05-21T18:31:46.217" Score="8" ViewCount="570" Body="&lt;p&gt;In single-solution regular Sudoku puzzles, sometimes you may encounter a combination of possible cell solutions such as:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;{2,7} in A1 and A7 (same row)&lt;/li&gt;&#xA;&lt;li&gt;{2,7} in A7 and B8 (same box)&lt;/li&gt;&#xA;&lt;li&gt;{2,7} in B8 and E8 (same column)&lt;/li&gt;&#xA;&lt;li&gt;{2,3,5,7} in E1 (same column as A1 and same row as E8)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I call patterns like the one above &lt;em&gt;conjugate-pair pseudocycles&lt;/em&gt; because they're made of conjugate pairs that link to each other by common cells to form a chain, but one node has more solutions possible than just that pair. How do the cells with just that pair influence which digits are removed from the possibilities for that cell?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have answered below:&lt;/p&gt;&#xA;" OwnerUserId="158" LastActivityDate="2014-05-21T23:23:07.077" Title="How to deal with conjugate-pair pseudocycles when solving Sudoku" Tags="|sudoku|strategy|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="328" PostTypeId="2" ParentId="327" CreationDate="2014-05-21T18:31:46.217" Score="6" Body="&lt;p&gt;As it turns out, if the Sudoku is a regular puzzle with one solution, if such a pattern is present, the two digits involved in the conjugate pairs should always be removed from the notes of the cell in question. This is true whether there is an odd or even number of such cells&amp;mdash;if it is one parity, i.e. odd, the combination may result in a multiple-solution impasse; if it is the other, i.e. even, the digits will be eliminated anyway as solutions are assigned to the other cells in the pseudocycle. In this case, {2,7} would be removed from E1, leaving only {3,5} as the remaining possibilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's examine the case above more closely: if A1 is 2, then A7 is 7, B8 is 2, and E8 is 7, which eliminates {2,7} from E1 because both digits are already assigned. If you pick 7 for A1, E8 must likewise be 2, also eliminating both digits. Therefore E1 must be reduced to {3,5}. As another example,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;{2,7} in A1 and A7&lt;/li&gt;&#xA;&lt;li&gt;{2,7} in A7 and E7&lt;/li&gt;&#xA;&lt;li&gt;{2,3,5,7} in E1&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If A1 is 2, then A7 is 7 and E7 is also 2, so E1 &lt;em&gt;could&lt;/em&gt; be 7. But if A1 is 7, E7 is likewise 7, so E1 &lt;em&gt;could&lt;/em&gt; be 2. Now, reducing E1 to {2,7} will not work because the puzzle has only one solution, and doing so would make 2 solutions possible! So we must reduce E1 to {3,5} in this case as well.&lt;/p&gt;&#xA;" OwnerUserId="158" LastEditorUserId="158" LastEditDate="2014-05-21T23:23:07.077" LastActivityDate="2014-05-21T23:23:07.077" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="329" PostTypeId="2" ParentId="70" CreationDate="2014-05-21T19:20:29.663" Score="8" Body="&lt;p&gt;This answer attempts to give a dumbed down version of the &lt;a href=&quot;https://math.stackexchange.com/questions/79333/how-many-strategies-are-there-for-this-puzzle-where-one-of-n-logicians-must-call&quot;&gt;answer on Math.SE&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say there are two logicians, A and B; and two colors, $0$ and $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mr. A looks at his friend and guesses that they are wearing the same hat.  Mr. B knows that Mr. A is predictable and will guess that they are the same, so he will guess they are different. Without knowing what either person saw we can see that all answers are covered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Them being the same is the same as saying that the sum of their colors is equal to $0 \bmod 2$ as $0+0=0$ and $1+1=2=0 \bmod 2$.  Them being different is, similarly, $1 \bmod 2$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The easiest solution, therefore, is for every person to guess that the sum of all the colors' numbers is equal to their seat number $S$ modulo the number of people.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way to see this is to note that the equation for 5 logicians and 5 colors is:&#xA;$a+b+c+d+e=S \bmod 5$ where each variable refers to the color number of the player with the same name.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Mr. A,  he knows what $b$, $c$, $d$, and $e$ add up to be and only doesn't know $a$ so he guesses $S=0 \bmod 5$ which means that $a=0-b-c-d-e \mod 5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mr. B, he knows what Mr. A will guess, so he guesses $S=1 \bmod 5$, so $b=1-a-c-d-e \mod 5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the time we get through 5 guessers we will guess:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mr. A: $S=0 \bmod 5$&lt;br&gt;&#xA;Mr. B: $S=1 \bmod 5$&lt;br&gt;&#xA;Mr. C: $S=2 \bmod 5$&lt;br&gt;&#xA;Mr. D: $S=3 \bmod 5$&lt;br&gt;&#xA;Mr. E: $S=4 \bmod 5$&lt;br&gt;&#xA;which is of course every possibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you make up colors (one person in a circle with only black and white hats can yell 'purple'), this means any number of players can guess at least one of their hats correctly for any number of colors less than or equal to the number of players.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="11110" LastEditDate="2018-03-01T15:45:23.673" LastActivityDate="2018-03-01T15:45:23.673" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="332" PostTypeId="2" ParentId="99" CreationDate="2014-05-21T20:54:54.783" Score="5" Body="&lt;p&gt;There are $2^4-2=14$ initial configurations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initial cases are described as: &#xA;$(13)$ means 1 and 3 are on the left door while 2 and 4 are on the right.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;By playing C first, you eliminate 2 cases: $(13)$ and $(24)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;B: removes 2 cases: $(23)$ and $(14)$ or $(12)$ and $(34)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This continues the whole way so that every move removes two of the initial states until all 7 moves remove 14 original states.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;One can easily show that there does not exist a single move that can remove more than two initial states at a time. There is also no move that can make initial states equivalent without removing them besides the initial states that are equivalent from the start. (note that because you have two doors, you are really manipulating linked pairs of initial states). This means that the fewest moves solution you already have is, unfortunately for my curiousity, the fastest solution as well.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="15544" LastEditDate="2015-09-04T09:50:47.547" LastActivityDate="2015-09-04T09:50:47.547" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="333" PostTypeId="2" ParentId="187" CreationDate="2014-05-21T21:06:01.080" Score="3" Body="&lt;p&gt;I would suggest compiling a list from this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.shabanali.com/upload/1000words.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.shabanali.com/upload/1000words.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a list of the 1000 most common words in the English language.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no use for such a list and will, therefore, not make it for you... sorry but I'm selfish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would suggest that you note the following: of the first 100 most common words, 5 are two letter words that end in &quot;o&quot; (do to go so no).  It may be worth focusing on that possibility first. &quot;e&quot; of course is the most common overall, so should be easier to find.  &quot;be&quot; and &quot;he&quot; are somewhat common.  Look for &quot;xyz&quot; and &quot;yz&quot; both appearing in longer cryptograms as that can commonly be &quot;the&quot; and &quot;he&quot; respectively. Once these are found, &quot;she&quot;, &quot;it&quot;, and &quot;on&quot; are easy.  You will never see &quot;ne&quot; so that can help differentiate &quot;one&quot; from &quot;the&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, one of the first things to do is look for &quot;xy&quot; and &quot;yx&quot; as these will quickly help find on and no with a relatively high degree of certainty that can easily be checked. &quot;ot&quot; &quot;eb&quot; &quot;eh&quot; and &quot;sa&quot; are fairly uncommon!&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="463" LastEditDate="2015-06-02T03:01:14.237" LastActivityDate="2015-06-02T03:01:14.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="334" PostTypeId="2" ParentId="158" CreationDate="2014-05-21T21:51:46.553" Score="5" Body="&lt;p&gt;While not a perfect answer the, following is from the same paper that proved the non-existence of 16 clue sudoku. The paper can be read at &lt;a href=&quot;http://arxiv.org/pdf/1201.0749v2.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://arxiv.org/pdf/1201.0749v2.pdf&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For several years now, whenever people send Gordon Royle (see the second item in Section 2.1) a list of 17-clue puzzles, there are usually not too many new puzzles. One correspondent sent 700 puzzles, and it turned out that only 33 were new. Assuming that both Royle and this correspondent had drawn their puzzles at random from the universe of all 17-clue puzzles, Ed Russell computed the maximum likelihood estimate for the size of the universe to be approximately 34,550 at a time at which Royle’s list contained around 33,000 puzzles [26]. In retrospect this is an underestimate; nevertheless we may infer that the 49,151 puzzles on the most recent version of Royle’s list must be almost all the 17-clue puzzles in existence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="99" LastEditorUserId="294" LastEditDate="2018-08-14T21:28:32.287" LastActivityDate="2018-08-14T21:28:32.287" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="335" PostTypeId="5" CreationDate="2014-05-21T22:10:01.763" Score="0" Body="&lt;p&gt;Puzzle creation questions are about the creation and development of puzzles.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can cover all aspects of puzzle creation, from &lt;a href=&quot;/q/323&quot;&gt;scrambling a&lt;/a&gt; &lt;a href=&quot;/questions/tagged/rubiks-cube&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;rubiks-cube&amp;#39;&quot; rel=&quot;tag&quot;&gt;rubiks-cube&lt;/a&gt; to &lt;a href=&quot;/q/260&quot;&gt;phrasing questions&lt;/a&gt; to &lt;a href=&quot;/q/170&quot;&gt;calculating all possible times a rope puzzle can produce&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please do &lt;strong&gt;not&lt;/strong&gt; use this tag to request the community to make a puzzle for you - the post will likely be closed as too broad.&lt;/p&gt;&#xA;" OwnerUserId="11" LastEditorUserId="18250" LastEditDate="2017-12-16T22:19:55.413" LastActivityDate="2017-12-16T22:19:55.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="336" PostTypeId="4" CreationDate="2014-05-21T22:10:01.763" Score="0" Body="A question about the creation and development of puzzles. Not to be used for requesting the creation of a puzzle." OwnerUserId="-1" LastEditorUserId="18250" LastEditDate="2017-12-16T22:19:55.413" LastActivityDate="2017-12-16T22:19:55.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="337" PostTypeId="5" CreationDate="2014-05-21T22:14:45.237" Score="0" Body="&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/River_crossing_puzzle&quot; rel=&quot;nofollow&quot;&gt;river crossing puzzle&lt;/a&gt; is a type of puzzle that involves moving multiple items across a river.  Different puzzles have different rules about how many boats there are, what can be carried in a boat, how many trips the boat makes, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This tag may also be used for logically equivalent puzzles (e.g. trips across a bridge). The most classical and well-known example is Alcuin's river crossing puzzle, as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A peasant must transport a wolf, a goat and a piece of cabbage from one side of a river to the other while using a boat which can only hold one item in addition to the peasant. &#xA;  The wolf cannot be left alone with the goat, and the goat cannot be left alone with the cabbage. How does the peasant organize his trips?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11" LastEditorUserId="5373" LastEditDate="2015-07-07T17:15:40.747" LastActivityDate="2015-07-07T17:15:40.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="338" PostTypeId="4" CreationDate="2014-05-21T22:14:45.237" Score="0" Body="A river crossing puzzle is a type of puzzle that involves moving multiple items across a river subject to certain constraints." OwnerUserId="11" LastEditorUserId="5373" LastEditDate="2015-03-14T13:58:25.680" LastActivityDate="2015-03-14T13:58:25.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="339" PostTypeId="1" CreationDate="2014-05-21T22:20:29.440" Score="4" ViewCount="501" Body="&lt;p&gt;Assuming the game is played without cheating, in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Three-card_Monte&quot; rel=&quot;nofollow&quot;&gt;three-card monte&lt;/a&gt;, can a logical approach increase the player's chance of beating the house?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first glance it would seem to be unlikely but a &lt;a href=&quot;http://www.bbc.com/news/science-environment-27228416&quot; rel=&quot;nofollow&quot;&gt;recent article in the BBC news&lt;/a&gt; indicates that in the somewhat similar two person game rock-paper-scissors, knowledge and application of a simple set of natural human habits can have a pretty drastic impact on the odds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming you know where the card started can you make a logical choice, that will identify the most likely current position (regardless of distractions)? &lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorDisplayName="user88" LastEditDate="2014-05-23T13:55:43.687" LastActivityDate="2017-07-30T07:34:51.727" Title="In the three-card monte, can a logical principle be applied to increase the odds?" Tags="|cards|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="340" PostTypeId="1" AcceptedAnswerId="1900" CreationDate="2014-05-21T22:27:30.213" Score="10" ViewCount="2558" Body="&lt;p&gt;Logic puzzles with a set of conditions and a &quot;logic grid&quot; are very popular.  More than one publisher is currently producing puzzle books with these types of puzzles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Wikipedia article claims &lt;a href=&quot;https://en.wikipedia.org/wiki/Logic_problem&quot; rel=&quot;noreferrer&quot;&gt;Charles Lutwidge Dodgson, (&lt;em&gt;Lewis Carroll, author of Alice's Adventures in Wonderland.&lt;/em&gt;)&lt;/a&gt; was the instigator of what later became these type of puzzles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When and where was the first logic puzzle using a logic grid published? Here is an example of a logic grid from Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cMSWp.png&quot; alt=&quot;Logic grid example&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="8874" LastEditDate="2016-03-23T16:51:44.843" LastActivityDate="2016-05-25T08:47:07.063" Title="When and where was the first logic-grid-based puzzle published?" Tags="|logical-deduction|logic-grid|puzzle-history|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="341" PostTypeId="2" ParentId="339" CreationDate="2014-05-21T22:33:31.343" Score="2" Body="&lt;p&gt;The three-card monte depends on the mark following the wrong card but being entirely certain that that is the card he wants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you can &quot;select a card&quot; the way a mark would normally do so, and the con man hasn't seen through your double bluff, you can get away with instinctively selecting one card at first, and then committing &lt;em&gt;not&lt;/em&gt; to select that card and selecting one of the other two cards completely randomly which gives you even odds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's very difficult to do so in practice, though, because the con man employs a lot of psychological tricks to make you think that the card you want really &lt;em&gt;is&lt;/em&gt; the one he threw down, and, as referenced on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Three-card_Monte&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt; for the three-card monte, if you &lt;em&gt;do&lt;/em&gt; select the right card, there'll be somebody to intervene to make sure you never win any money.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2014-05-21T22:33:31.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="342" PostTypeId="1" AcceptedAnswerId="343" CreationDate="2014-05-21T22:55:51.090" Score="7" ViewCount="6317" Body="&lt;p&gt;Single player games using a single deck of playing cards often called &lt;a href=&quot;https://en.wikipedia.org/wiki/Patience_%28game%29&quot; rel=&quot;nofollow&quot;&gt;patience or solitaire&lt;/a&gt;, generally involve a degree of skill and luck to solve, with card shuffling ultimately being the deciding factor on if the game can be solved/won.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any version of the game with a single player and a standard 52 card poker deck, where every solution can be ultimately positive through deductive reasoning alone?  &lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="463" LastEditDate="2015-06-02T03:10:23.740" LastActivityDate="2015-06-02T03:12:19.813" Title="Are there any forms of patience or solitaire that can be solved by logic alone?" Tags="|logical-deduction|cards|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="343" PostTypeId="2" ParentId="342" CreationDate="2014-05-21T23:06:22.030" Score="7" Body="&lt;p&gt;Most patience games, as you've noted, require logical deduction, but most still require a degree of luck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Spider_(solitaire)&quot; rel=&quot;nofollow&quot;&gt;Spider Solitaire&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Mrs._Mop&quot; rel=&quot;nofollow&quot;&gt;Mrs. Mopp&lt;/a&gt; come close to being solely skill-based, but are both still vulnerable. &lt;a href=&quot;http://en.wikipedia.org/wiki/FreeCell&quot; rel=&quot;nofollow&quot;&gt;FreeCell&lt;/a&gt; is another game for which almost all games are thought to be playable, though the proportion has not yet been calculated. A rough estimate of winnable games &lt;a href=&quot;http://en.wikipedia.org/wiki/FreeCell_(Windows)#Unsolvable_combinations&quot; rel=&quot;nofollow&quot;&gt;was calculated&lt;/a&gt; to be around 99.9987% of games.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are a number of games which do not rely on luck if one assumes the game is winnable (as opposed to games like &lt;a href=&quot;http://en.wikipedia.org/wiki/Aces_Up&quot; rel=&quot;nofollow&quot;&gt;Aces Up&lt;/a&gt;, which are almost solely winnable based on luck). &lt;a href=&quot;http://en.wikipedia.org/wiki/Scorpion_(solitaire)&quot; rel=&quot;nofollow&quot;&gt;Scorpion&lt;/a&gt; is an example of a solitaire game which takes significant skill to actually win, even if the game is winnable. &lt;a href=&quot;http://en.wikipedia.org/wiki/Calculation_(card_game)&quot; rel=&quot;nofollow&quot;&gt;Calculation&lt;/a&gt; additionally is almost always winnable, but takes significant amounts of strategy to actually claim a victory (in most cases). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Klondike actually does &lt;em&gt;not&lt;/em&gt; fall into this category, as you are blind to the cards much of the time, and frequently would actually be better off taking your chances as opposed to playing logically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could go on, and may feel like editing more examples in later, but the short answer is: there are card games for which mostly skill is required, and card games for which, if victory is possible, significant skill is often required to achieve it. Whether there are card games which are provably always solvable remains to be seen. &lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorUserId="463" LastEditDate="2015-06-02T03:12:19.813" LastActivityDate="2015-06-02T03:12:19.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="344" PostTypeId="5" CreationDate="2014-05-21T23:13:07.360" Score="0" Body="&lt;p&gt;A brain teaser is a puzzle that involves thought to solve. Often it requires thinking in unconventional ways, or disregarding what seems obvious.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Logic puzzles and riddles are specific types of brain teasers.&lt;/li&gt;&#xA;&lt;li&gt;Brain teasers sometimes involve lateral thinking.&lt;/li&gt;&#xA;&lt;li&gt;Brain teasers typically requires inductive rather than deductive reasoning.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You should only use the brain teaser tag for a question, if a single answer is demonstrably correct.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-01-30T15:57:03.390" LastActivityDate="2015-01-30T15:57:03.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="345" PostTypeId="4" CreationDate="2014-05-21T23:13:07.360" Score="0" Body="A puzzle that involves thinking in unconventional ways or disregarding what seems obvious. " OwnerDisplayName="user20" LastEditorUserId="4551" LastEditDate="2015-01-30T22:26:04.377" LastActivityDate="2015-01-30T22:26:04.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="346" PostTypeId="5" CreationDate="2014-05-21T23:20:10.320" Score="0" Body="&lt;p&gt;A challenge question is one in which you present a puzzle to which you know the answer (and generally of your own creation) as a challenge to others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if you wish to challenge the community, it is requested that you not answer the question immediately yourself, but instead give time for others to answer first.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2014-05-28T14:43:36.920" LastActivityDate="2014-05-28T14:43:36.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="347" PostTypeId="4" CreationDate="2014-05-21T23:20:10.320" Score="0" Body="A question posed as a challenge for other people to answer." OwnerDisplayName="user20" LastEditorDisplayName="user88" LastEditDate="2014-05-28T14:44:37.663" LastActivityDate="2014-05-28T14:44:37.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="348" PostTypeId="5" CreationDate="2014-05-21T23:39:46.650" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-05-21T23:39:46.650" LastActivityDate="2014-05-21T23:39:46.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="349" PostTypeId="4" CreationDate="2014-05-21T23:39:46.650" Score="0" Body="Speedsolving typically refers to solving a physical puzzle as quickly as possible. For instance, one might use this tag when asking about optimal Rubik's cube algorithms. This tag is not necessarily localized to the Rubik's cube, however." OwnerDisplayName="user20" LastEditorDisplayName="user20" LastEditDate="2014-05-23T17:08:41.180" LastActivityDate="2014-05-23T17:08:41.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="350" PostTypeId="1" AcceptedAnswerId="351" CreationDate="2014-05-21T23:55:32.757" Score="0" ViewCount="1166" Body="&lt;p&gt;The puzzle grid pictured below is a prime example of the classic rush to judgment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aELsO.png&quot; alt=&quot;Minesweeper, all blown up&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was so sure of my decision, but when I double-clicked to clear more cells, the cell to the left of the one I had just flagged (the red one next to the X) was the mine, and the game ended. I see that apparently I wasn't paying close enough attention to the numbers in the surrounding cells. What kinds of moves could I make in the future in a similar situation, in order to to avoid a costly mistake such as this one? Include reasons so this can be helpful to other users.&lt;/p&gt;&#xA;" OwnerUserId="158" LastEditorUserId="158" LastEditDate="2014-05-23T20:02:55.263" LastActivityDate="2017-06-02T13:19:55.283" Title="How could I have avoided this mistake in Minesweeper?" Tags="|strategy|minesweeper|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="351" PostTypeId="2" ParentId="350" CreationDate="2014-05-22T00:32:56.523" Score="11" Body="&lt;p&gt;The 2 down-right from your bad flag should have tipped you off; it is touching two bombs you had already flagged. That would show the one next to it (which you triggered) to be a bomb; the two under that one would clear the one left of it (above the 4), leaving just the four bombs around the 4. You could also have figured out the 4's surroundings just from it and the 2s right and down-left of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Graphically; blues are squares of consideration, red is conclusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{|c|c|c|}&#xA;\hline&#xA;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;2\\&#xA;\hline&#xA;&amp;amp;&amp;amp;&amp;amp;\color{red}{3}&amp;amp;1&amp;amp;1&amp;amp;2&amp;amp;F&amp;amp;2\\&#xA;\hline&#xA;&amp;amp;4&amp;amp;2&amp;amp;2&amp;amp;\color{blue}{2}&amp;amp;2&amp;amp;2&amp;amp;1&amp;amp;1\\&#xA;\hline&#xA;2&amp;amp;F&amp;amp;1&amp;amp;1&amp;amp;\color{blue}{F}&amp;amp;\color{blue}{F}&amp;amp;1 \\&#xA;\hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{|c|c|c|}&#xA;\hline&#xA;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;2\\&#xA;\hline&#xA;&amp;amp;&amp;amp;\color{red}{F}&amp;amp;3&amp;amp;1&amp;amp;1&amp;amp;2&amp;amp;F&amp;amp;2\\&#xA;\hline&#xA;&amp;amp;4&amp;amp;2&amp;amp;\color{blue}{2}&amp;amp;2&amp;amp;2&amp;amp;2&amp;amp;1&amp;amp;1\\&#xA;\hline&#xA;2&amp;amp;F&amp;amp;1&amp;amp;1&amp;amp;\color{blue}{F}&amp;amp;F&amp;amp;1 \\&#xA;\hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{|c|c|c|}&#xA;\hline&#xA;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;2\\&#xA;\hline&#xA;&amp;amp;\color{red}{4}&amp;amp;\color{blue}{F}&amp;amp;3&amp;amp;1&amp;amp;1&amp;amp;2&amp;amp;F&amp;amp;2\\&#xA;\hline&#xA;&amp;amp;4&amp;amp;\color{blue}{2}&amp;amp;2&amp;amp;2&amp;amp;2&amp;amp;2&amp;amp;1&amp;amp;1\\&#xA;\hline&#xA;2&amp;amp;\color{blue}{F}&amp;amp;1&amp;amp;1&amp;amp;F&amp;amp;F&amp;amp;1 \\&#xA;\hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{|c|c|c|}&#xA;\hline&#xA;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;2\\&#xA;\hline&#xA;\color{red}{F}&amp;amp;4&amp;amp;\color{blue}{F}&amp;amp;3&amp;amp;1&amp;amp;1&amp;amp;2&amp;amp;F&amp;amp;2\\&#xA;\hline&#xA;\color{red}{F}&amp;amp;\color{blue}{4}&amp;amp;2&amp;amp;2&amp;amp;2&amp;amp;2&amp;amp;2&amp;amp;1&amp;amp;1\\&#xA;\hline&#xA;2&amp;amp;\color{blue}{F}&amp;amp;1&amp;amp;1&amp;amp;F&amp;amp;F&amp;amp;1 \\&#xA;\hline&#xA;\end{array}&lt;/p&gt;&#xA;" OwnerUserId="4" LastEditorUserId="4" LastEditDate="2014-05-22T00:55:19.670" LastActivityDate="2014-05-22T00:55:19.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="352" PostTypeId="2" ParentId="72" CreationDate="2014-05-22T00:49:26.797" Score="5" Body="&lt;p&gt;As others have mentioned in the comments, this is covered by a well-known mathematical theorem called &lt;a href=&quot;http://en.wikipedia.org/wiki/Four_color_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Four Colour Theorem&lt;/a&gt;. Any map can be coloured as you describe using only four colours. The actual algorithm, though, sounds rather complex. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://stackoverflow.com/a/22116707/4794&quot;&gt;This answer&lt;/a&gt; on Stack Overflow gives an easier algorithm that will use at most five colours.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The four-color mapping algorithm is very complex, with 1476 special cases that you have to handle in your code. If you can spare one more color, the five color mapping algorithm will meet your requirements, is much simpler, and there is a nice write up on it &lt;a href=&quot;http://www.devx.com/dotnet/Article/32927&quot; rel=&quot;nofollow noreferrer&quot;&gt;at devx.com&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a brief summary of the algorithm, the strategy is to find a region whose colour would be obvious if the rest of the map were already coloured. Then &quot;remove&quot; that region from the map and repeat the process until the map is empty. If you kept track of the order you removed the regions, you can now add them back in the reverse order, choosing colours as you do so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see the two rules for choosing a region to remove, see the detailed write up.&lt;/p&gt;&#xA;" OwnerUserId="38" LastEditorUserId="-1" LastEditDate="2017-05-23T12:38:53.617" LastActivityDate="2014-05-22T05:22:14.423" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="353" PostTypeId="5" CreationDate="2014-05-22T01:02:58.313" Score="0" Body="&lt;p&gt;Questions about Minesweeper (the game or the [grid-deduction] puzzle genre). Some non-bomb cells display the number of adjacent bombs, and finding all bombs is the goal. In the puzzle some numbers are pre-given, while in the game cells must be clicked on to reveal their numbers.&lt;/p&gt;&#xA;&lt;p&gt;Some examples of the game:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/357/what-frequent-patterns-can-be-used-to-overcome-minesweeper-obstacles&quot;&gt;What frequent patterns can be used to overcome Minesweeper obstacles?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/10885/can-you-create-a-minesweeper-puzzle-with-information-crossover&quot;&gt;Can you create a Minesweeper puzzle with &amp;#39;information crossover&amp;#39;?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;See also the &lt;a href=&quot;https://en.wikipedia.org/wiki/Minesweeper_(video_game)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article on Minesweeper&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;An example of the puzzle:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/102828/wait-so-how-many-mines-are-there-a-tetromino-minesweeper&quot;&gt;Wait, so how many mines are there? A tetromino minesweeper&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;See also GM Puzzles &lt;a href=&quot;https://www.gmpuzzles.com/blog/category/objectplacement/minesweeper/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Minesweepers&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="158" LastEditorUserId="69582" LastEditDate="2021-02-21T09:12:47.650" LastActivityDate="2021-02-21T09:12:47.650" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="354" PostTypeId="4" CreationDate="2014-05-22T01:02:58.313" Score="0" Body="Questions about Minesweeper (the game or the [grid-deduction] puzzle genre). Some non-bomb cells display the number of adjacent bombs, and finding all bombs is the goal. In the puzzle some numbers are pre-given, while in the game cells must be clicked on to reveal their numbers." OwnerUserId="158" LastEditorUserId="69582" LastEditDate="2021-02-21T09:12:52.007" LastActivityDate="2021-02-21T09:12:52.007" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="355" PostTypeId="1" AcceptedAnswerId="356" CreationDate="2014-05-22T01:03:24.930" Score="7" ViewCount="4400" Body="&lt;p&gt;I always start a Minesweeper puzzle in Windows by clicking the cell in the lower left corner. Even though I have never seen a bomb there, how sound is this strategy? Is it possible that some time in the future there will be a bomb there? Or is it possible to click anywhere the first time?&lt;/p&gt;&#xA;" OwnerUserId="158" LastEditorUserId="158" LastEditDate="2014-05-22T02:19:08.877" LastActivityDate="2017-01-06T03:04:17.420" Title="Is it always safe to click first in the lower lefthand corner of a Minesweeper puzzle?" Tags="|strategy|minesweeper|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="356" PostTypeId="2" ParentId="355" CreationDate="2014-05-22T01:28:07.730" Score="21" Body="&lt;p&gt;Many Minesweeper games, including the one that ships with Windows, do not allow the first square you click to be a mine. The lower-left-hand corner is not necessarily safe, but if it's the first square you click, then it will be safe.&lt;/p&gt;&#xA;" OwnerUserId="4" LastActivityDate="2014-05-22T01:28:07.730" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="357" PostTypeId="1" AcceptedAnswerId="358" CreationDate="2014-05-22T01:30:15.460" Score="6" ViewCount="28845" Body="&lt;p&gt;Minesweeper, to those who do not know the strategies involved, is a difficult game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What patterns exist in the game which I can take advantage of? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Meta-context: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/357/how-do-i-win-minesweeper-every-time/358&quot;&gt;this question&lt;/a&gt; is an offshoot of the current original question, for which the scope has been changed to be simply pattern-based. Feedback indicated this &lt;em&gt;was&lt;/em&gt; &lt;a href=&quot;https://puzzling.meta.stackexchange.com/questions/159/how-broad-is-too-broad-for-how-do-i-solve-this-puzzle-questions/&quot;&gt;somewhat too broad&lt;/a&gt;. &lt;strong&gt;This question has now been edited to address these concerns.&lt;/strong&gt; Please contribute to discussion on Meta if you feel inclined.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user20" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2016-02-24T18:34:50.383" Title="What frequent patterns can be used to overcome Minesweeper obstacles?" Tags="|strategy|minesweeper|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="358" PostTypeId="2" ParentId="357" CreationDate="2014-05-22T01:30:15.460" Score="11" Body="&lt;p&gt;&lt;strong&gt;Small notes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Remember that the number of remaining mines is the clue number minus the number of surrounding mines. This is important in patterns like 1-2-1; 3-2-3, for instance, also works if there are two mines next to each 3, since they both need 1 mine to be solved.&lt;/li&gt;&#xA;&lt;li&gt;A one by itself on a corner always indicates that there is a mine&lt;/li&gt;&#xA;&lt;li&gt;A two by itself next to a convex corner always indicates two mines&lt;/li&gt;&#xA;&lt;li&gt;Patterns that work on horizontal walls obviously work on vertical walls, and vice versa.&lt;/li&gt;&#xA;&lt;li&gt;Patterns do not work when there is interference from adjacent clues or unknowns; i.e. when a pattern is in a corner or has another unknown near it.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The double-click trick&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the Windows version of Minesweeper, if you click on a clue with both your left and right mouse buttons, if all that clue's mines have been identified, then the game will remove all the other surrounding unknowns. You should &lt;em&gt;always&lt;/em&gt; use this to clear spaces, unless you are guessing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, you can use this to check visually if all the spaces surrounding a clue are mines. If you need two mines, and two squares are highlighted when you use this trick, you know they're both mines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will primarily use this trick to identify patterns quickly through visual recognition of shapes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: if you have a mine incorrectly marked, this will cause you to lose the game - but then again, you would have anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The 1-2 pattern&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever you see a 1 next to a 2 against a wall, you know that the space opposite the 2 is a mine. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LcImU.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/lPQZe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is because there can only be one mine in the two spaces adjacent to the 1, which means the second mine must be adjacent to the 2 in the final space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in the second case, since there is a mine next to the 3 and 2, they need 2 and 1 mines respectively, matching the 1-2 pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The 1-2-1 pattern&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the frequently-encountered patterns along walls in Minesweeper is the 1-2-1 pattern. This is an extension of the 1-2 pattern, in that two of the pattern exist side-by-side. In this case, as an extension of 1-2, the mines must be underneath the ones.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FWZKN.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/FFgB4.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/NGbA8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note again that in the second case, the 2-3-2 are all adjacent to a known mine, which means they need 1-2-1. In the third case, we know that there is a mine next to the two by the one and corner in olive-green. therefore, the 1-2-2 matches the 1-2-1 pattern, implying there are two mines in front of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first position was honestly just dumb luck. Yes, the 1-2-1 pattern does chain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that it is very rare to actually encounter a 1-2-1 specifically. Usually, these are by reduction from surrounding mines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The 1-2-2-1 pattern&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever you find a 1-2-2-1 pattern against a wall, you know that the two squares under the 2s are mines. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EoBcL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I found myself unable to generate a true 1-2-2-1 case in a reasonable amount of time, but here is a valid position anyway that demonstrates the point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The 1-1-X pattern&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a 1-1-X are sequential against a wall &lt;em&gt;and&lt;/em&gt; against the edge of the grid, the space under the third clue, marked here with a question mark:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IcclP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is because the first two spaces must contain a mine by the first clue, which means both the first and second ones are already guaranteed to be satisfied. If there were a mine in the third space, it would conflict with the second one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This pattern is, admittedly, a lot less useful. However, it has the possibility to save your game, especially when you're stuck.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorDisplayName="user20" LastEditDate="2014-05-22T04:36:07.077" LastActivityDate="2014-05-22T04:36:07.077" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="359" PostTypeId="2" ParentId="231" CreationDate="2014-05-22T01:32:17.590" Score="12" Body="&lt;p&gt;Let's take a 25 x 25 puzzle of a color gradient(also called color ramp and color progression). No two connections the same, even it is off by a millimeter. All the pieces are in a pile, with no way to identify the pieces. The fastest would be the first solution, assembling the border first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number 1, which is the fastest, would require you to sift through the pile and find edges and corners. Since it is a color gradient, the top edges would take little effort and the bottom edge just more so. After finishing that, just grab a piece and put it where you think it goes. If you make a wrong guess, the piece will move around the surface you are building on into the right place, thus solving the puzzle quickest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number 2 is the third fastest. Not as fast as number 1, you would have to go through the pile I mentioned in my introduction to find a corner. It also depends on how you assemble from the corner. Since number 1 is building a border, this would be a block progression, starting from the one corner piece to a corner, 2 edges, and one four sided piece to a corner, 4 edges, and 4 four sided pieces, and so on so forth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number 3 is the second fastest. You would have to find pieces of similar color and, since there are no faces, start off with that. A color gradient would have two identical colors near each other and you could just fit them together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number 4 would be the second slowest, because you would have to find 2, 3, or 4 pieces that fit with it. Since every piece has a certain connection with 2-4 others, you would most likely have to sift through all the pieces to find the right one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number 5 would not be the fastest, but actually the slowest, because you would be required to go through all of the pieces to see ones that fit with another. Unless you got lucky, saying you chose two pieces and they just happened to fit together, you would take the most time trying to fit pieces together rather than putting all of them together.&lt;/p&gt;&#xA;" OwnerUserId="135" LastActivityDate="2014-05-22T01:32:17.590" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="360" PostTypeId="1" CreationDate="2014-05-22T01:59:52.653" Score="9" ViewCount="33063" Body="&lt;p&gt;Is it better to start in a corner, edge, or any random middle square? If the first square isn't empty, when is it better to click a neighbor vs. another random square? How does the strategy vary by mine density?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &quot;optimal&quot; I mean maximizing the probability of winning on a game not guaranteed to be logically solvable.&lt;/p&gt;&#xA;" OwnerUserId="4" LastEditorUserId="4" LastEditDate="2018-04-04T17:56:48.573" LastActivityDate="2018-04-04T17:56:48.573" Title="What is the optimal starting strategy in Minesweeper?" Tags="|minesweeper|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="361" PostTypeId="2" ParentId="360" CreationDate="2014-05-22T02:03:13.217" Score="15" Body="&lt;p&gt;It is better to start in the corner in Minesweeper. The reason for this is simple: corners are the most likely place for logically unsolvable positions. The more corners you remove, the greater the chance you'll be able to solve the puzzle, regardless of mine density. Since your starting move can't be a mine, starting there reduces the chance of being stuck on a corner by 25%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason corners are the most likely place for unsolvable positions is simple: you have very few clues there. Note that this will also tend to generate smaller starting cleared areas when you start in the corner, but this isn't generally a hindrance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are additionally not much more likely to have a faster game if you start in the corner. The reason for this is simple: until you run out of options, you're always executing further options. If a region would have been cleared by clicking in the middle of the puzzle, it will be cleared when you reach it normally. In this sense, you wouldn't actually be saving time by clearing the middle region, because you'll inevitably clear it anyway.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorDisplayName="user20" LastEditDate="2014-05-22T03:10:14.627" LastActivityDate="2014-05-22T03:10:14.627" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="362" PostTypeId="2" ParentId="342" CreationDate="2014-05-22T02:20:29.520" Score="7" Body="&lt;p&gt;Solitaire games that are entirely deductive like you mentioned will usually have all their cards exposed at the beginning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Games like that are called &lt;em&gt;perfect information games&lt;/em&gt; &amp;ndash; a term usually used in two-player game theory, but in the case of Solitaire applies to games where the position of all the cards is known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it's not always the case that these games are winnable. Famously, deal #11,982 in Microsoft Freecell is the only unsolvable one out of all 32,768 deals available in the original version of the program packaged with Windows 95. What you &lt;em&gt;can&lt;/em&gt; determine just from looking at the initial deal is whether the game is winnable or not, purely through deductive logic.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Examples of games like this include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Freecell&lt;/strong&gt;, in any of its implementations.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Canfield_%28solitaire%29&quot;&gt;Canfield&lt;/a&gt;&lt;/strong&gt;, when all 13 cards in the secondary stockpile are exposed and you're given unlimited redeals. Even so, Canfield is very hard to win, because you're dealing cards from the stockpile three at a time and only allowed to use the top card of three.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Simple_Simon_%28solitaire%29&quot;&gt;Simple Simon&lt;/a&gt;&lt;/strong&gt;, a game packaged with KPatience in the KDE game suite. It's similar to Spider Solitaire, but it only uses one deck and it deals all 52 cards face-up to the 10 piles with 8, 8, 8, 7, 6, 5, 4, 3, 2, 1 cards in each pile. This game is also very hard to win, but when you do, it's entirely skill-based.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2014-05-22T02:20:29.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="363" PostTypeId="2" ParentId="360" CreationDate="2014-05-22T02:38:13.277" Score="3" Body="&lt;p&gt;Emrakul's answer is correct, for some definitions of &quot;optimal&quot;. If your goal is just &quot;complete the puzzle&quot;, you want to start in the corner for the reason they describe. If your goal is to solve the puzzle quickly, you want the opposite: Start in the middle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Selecting a cell in the middle gives you more options to actually start building off of. It's true that you're more likely to get a logically solvable position this way, but when that happens you can do what the pros do: Guess. If you guess right, you saved time. If you guessed wrong, click reset and try again.&lt;/p&gt;&#xA;" OwnerUserId="224" LastActivityDate="2014-05-22T02:38:13.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="364" PostTypeId="1" CreationDate="2014-05-22T03:31:24.323" Score="13" ViewCount="28154" Body="&lt;p&gt;I've played many a-game of minesweeper, and I frequently reach positions in which there is no logical, deterministic way to come to a conclusion about how I should solve the puzzle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How should I best approach solving these puzzles to guarantee the highest statistical victory?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Meta-context: this question is an offshoot of &lt;a href=&quot;https://puzzling.stackexchange.com/questions/357/how-do-i-win-minesweeper-every-time/358&quot;&gt;this original question&lt;/a&gt;, which feedback indicated was &lt;a href=&quot;https://puzzling.meta.stackexchange.com/questions/159/how-broad-is-too-broad-for-how-do-i-solve-this-puzzle-questions/&quot;&gt;somewhat too broad&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user20" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2020-12-04T16:57:27.137" Title="How do I approach a &quot;stuck&quot; game of minesweeper?" Tags="|strategy|minesweeper|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="365" PostTypeId="2" ParentId="364" CreationDate="2014-05-22T03:31:24.323" Score="7" Body="&lt;p&gt;In general, when looking at these situations, you want to click on the squares that are the &lt;em&gt;least&lt;/em&gt; likely to contain mines. This will invariably give you more information about your current situation.&lt;/p&gt;&#xA;&lt;p&gt;Here's an example of a game that can't be beaten without strict testing, which I will use to demonstrate what I mean.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pzZXz.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;There are two mines remaining. I apologize, but the 3 underneath the square labeled 2 actually has two adjacent flags, and needs one more flag; one of them was cut off in the screenshot.&lt;/p&gt;&#xA;&lt;p&gt;Because there is a 1 next to 1 and 2, I know that one of these must be a mine. If 1 is a mine, then 2 is not, which means one of 4, 5, 6 is a mine, and 3 is not. This gives three cases when 1 is a mine.&lt;/p&gt;&#xA;&lt;p&gt;If 1 is not a mine, then 2 is a mine. If 2 is a mine, then none of 4, 5, 6 is a mine, which means 3 must be a mine. This has only one case.&lt;/p&gt;&#xA;&lt;p&gt;Therefore, the number of cases where 1 is a mine is 3/4; the number of cases where it is not is 1/4. We can proceed assuming that 1 is a mine - and as it happens to turn out, it is.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9hnnZ.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Here's another example of a possible setup. The method I use in these cases effectively can be extended across the entire board (which makes it a bit more accurate), but I only execute it a few levels in my head. Here's how my thought process works:&lt;/p&gt;&#xA;&lt;p&gt;We can see that if 5 is a mine, then 4 can't be, which means 3 is. At this point, either 1 or 2 could be a mine, which gives two cases.&lt;/p&gt;&#xA;&lt;p&gt;If 5 isn't a mine, then 4 must be a mine, which means 3 isn't, which means both 1 and 2 are - which gives one case.&lt;/p&gt;&#xA;&lt;p&gt;Therefore, the situation in which 5 is a mine covers 2/3 of the cases, and the situations in which 5 isn't covers 1/3. Therefore, we proceed on the assumption that 5 is a mine.&lt;/p&gt;&#xA;&lt;p&gt;Of course, there are more than three cases for the entire grid - this is only a rough estimate. I could go into further detail about what the squares above 1 and 2 could be, as well as the surrounding squares, but I won't here. It all depends on how long you want to take and how accurate you want to be.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorUserId="69582" LastEditDate="2020-12-04T16:48:49.743" LastActivityDate="2020-12-04T16:48:49.743" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="366" PostTypeId="5" CreationDate="2014-05-22T03:41:41.840" Score="0" Body="&lt;p&gt;Verbal arithmetic problems express numbers in an elementary math problem as letters. The classic example is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; SEND&#xA; MORE +&#xA;-------&#xA;MONEY&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The aim is to discover which integers from 0 to 9 the letters stand for. For instance, here is a puzzle next to its (unique) solution:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;DONALD   | 526485&#xA;GERALD + | 197485 +&#xA;-------- | --------&#xA;ROBERT   | 723970&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Verbal_arithmetic&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; (caution: spoils the above puzzle) describes these puzzles as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Verbal arithmetic, also known as alphametics, cryptarithmetic, crypt-arithmetic, cryptarithm or word addition, is a type of mathematical game consisting of a mathematical equation among unknown numbers, whose digits are represented by letters. The goal is to identify the value of each letter. The name can be extended to puzzles that use non-alphabetic symbols instead of letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solutions often involve number-theoretic solutions, or, in some easy cases, pure logic.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorDisplayName="user20" LastEditDate="2015-01-10T06:58:34.783" LastActivityDate="2015-01-10T06:58:34.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="367" PostTypeId="4" CreationDate="2014-05-22T03:41:41.840" Score="0" Body="Verbal arithmetic problems (a.k.a. &quot;alphametics&quot;, &quot;cryptarithms&quot;) involve equations where numbers have been replaced with letters. The aim is to find which numbers represent which letters. Use this tag for questions concerning verbal arithmetic creation, solutions, and strategies. &#xD;&#xA;" OwnerDisplayName="user20" LastEditorDisplayName="user20" LastEditDate="2014-06-20T02:03:45.810" LastActivityDate="2014-06-20T02:03:45.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="368" PostTypeId="1" AcceptedAnswerId="369" CreationDate="2014-05-22T09:23:13.153" Score="5" ViewCount="903" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kBk1l.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Today I came across this puzzle:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;How many minutes is it until six o'clock if fifty minutes ago it was&#xA;four times as many minutes past three o'clock?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Any help?&lt;/p&gt;&#xA;" OwnerUserId="50" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-07-02T09:47:09.077" Title="Clock Puzzle problem" Tags="|logical-deduction|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="369" PostTypeId="2" ParentId="368" CreationDate="2014-05-22T09:41:08.743" Score="9" Body="&lt;p&gt;Right now, we have $x$ minutes left until 6 o'clock. 50 minutes ago, it was $4x$ minutes past 3 o'clock. That means that the time between 3 o'clock and 6 o'clock, 180 minutes, is equal to the sum of those times. In other words&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;  4x + 50 + x &amp;amp; = 180 \\&#xA;  x &amp;amp; = \frac{180 - 50}{5} \\&#xA;  x &amp;amp; = 26&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it's now 26 minutes until 6 o'clock, or 34 minutes past 5.&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditDate="2014-07-02T09:47:09.077" LastActivityDate="2014-07-02T09:47:09.077" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="370" PostTypeId="1" AcceptedAnswerId="373" CreationDate="2014-05-22T10:05:27.610" Score="16" ViewCount="4472" Body="&lt;p&gt;In Heinlein's &lt;em&gt;Space Cadet&lt;/em&gt;, one of many tests administered to the protagonist seems to have no way of scoring any points. See the excerpt below.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Late in the day he was ushered into a cubbyhole containing a chair, a gadget mounted on a desk, pencil and paper, and framed directions.&lt;br&gt;&#xA;  &quot;If any score from a previous test,&quot; Matt read, &quot;appears in the window &#xA;  marked SCORE, return the starting lever to the position marked NEUTRAL to &#xA;  clear the board for your test.&quot;&lt;br&gt;&#xA;  Matt found the window labeled &quot;SCORE&quot;; it had a score showing in it - &quot;37.&quot;&lt;br&gt;&#xA;  Well, he thought, that gives me a mark to shoot at. He decided not to clear &#xA;  the board until he had read the instructions.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;After the test starts,&quot; he read, &quot;a score of T will result each time you press &#xA;  the lefthand button except as otherwise provided here below. Press the &#xA;  lefthand button whenever the red light appears provided the green light is not &#xA;  lighted as well except that no button should be pressed when the righthand &#xA;  gate is open unless all lights are out. If the right-hand gate is open and the &#xA;  lefthand gate is closed, no score will result from pressing any button, but the &#xA;  lefthand button must nevertheless be pressed under these circumstances if &#xA;  all other conditions permit a button to be pressed before any score may be &#xA;  made in succeeding phases of the test. To put out the green light, press the &#xA;  righthand button. If the lefthand gate is not closed, no button may be &#xA;  pressed. If the lefthand gate is closed while the red light is lighted, do not &#xA;  press the lefthand button if the green light is out unless the righthand gate is &#xA;  open. To start the test move the starting lever from neutral all the way to the &#xA;  right. The test runs for two minutes from the time you move the starting lever &#xA;  to the right. Study these instructions, then select your own time for &#xA;  commencing the test. You are not permitted to ask questions of the &#xA;  examiner, so be sure that you understand the instructions. Make as high a &#xA;  score as possible.&quot;  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;Whew!&quot; said Matt.&lt;br&gt;&#xA;  Still, the test looked simple-one lever, two pushbuttons, two colored lights, &#xA;  two little gates. Once he mastered the instructions, it would be as easy as &#xA;  flying a kite, and a durn sight simpler than flying a copter! - Matt had had his &#xA;  copter license since he was twelve. He got to work.&lt;br&gt;&#xA;  First, he told himself, there seems to be just two ways to make a score, one with the red light on and one with both lights out and one gate open.&lt;br&gt;&#xA;  Now for the other instructions - Let's see, if the lefthand gate is not closed - no, if the lefthand gate is closed - he stopped and read them over again.&lt;br&gt;&#xA;  Some minutes later he had sixteen possible positions of gates and conditions of lights listed. He checked them against the instructions, seeking scoring combinations. When he was through he stared at the result, then checked everything over again.&lt;br&gt;&#xA;  After rechecking he stared at the paper, whistled tunelessly, and scratched his head. Then he picked up the paper, left the booth, and went to the examiner.&lt;br&gt;&#xA;  That official looked up. &quot;No questions, please.&quot;&lt;br&gt;&#xA;  &quot;I don't have a question,&quot; Matt said. &quot;I want to report something. There's &#xA;  something wrong with that test. Maybe the wrong instructions sheet was put in there. In any case, there is no possible way to make a score under the instructions that are in there.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Was Matt's conclusion correct?&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2021-01-05T21:36:18.620" Title="Is it indeed impossible to score any points in the test in Heinlein's &quot;Space Cadet&quot;?" Tags="|logical-deduction|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="371" PostTypeId="1" AcceptedAnswerId="372" CreationDate="2014-05-22T11:25:57.690" Score="16" ViewCount="4061" Body="&lt;p&gt;One classic puzzle has many different forms, but the basic strategy is the same.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Once upon a time there was a crazy king who had a very wise minister&#xA;with him. The king had a habit of playing a strange game on a circular table.&lt;/p&gt;&#xA;&lt;p&gt;The game was played as follows: The king told his minister a different number everyday -&#xA;let's say he says 5. Then minister would then arrange five people and ask&#xA;them to sit in a circular table with five chairs. Starting from a specific person, the king would then kill every&#xA;second person on the table, until only one remained at the end, to which he would give 1,000 gold coins.&lt;/p&gt;&#xA;&lt;p&gt;For example, if there were five people on the table, starting from&#xA;person 1, he would kill person 2, person 4, person 1, and person 5 (skipping over persons 2 and 4 since they are already dead),&#xA;leaving person 3 as the survivor and winner.&lt;/p&gt;&#xA;&lt;p&gt;Now the wise minister was so clever as to calculate the winner&#xA;beforehand as soon as king said the number, and asked his man to sit&#xA;at particular seat so as to not be killed and also win some gold.&lt;/p&gt;&#xA;&lt;p&gt;How did the minister decide the winner?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="189" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2023-12-27T04:57:38.263" Title="How do you solve the circular table problem?" Tags="|mathematics|josephus-problem|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="372" PostTypeId="2" ParentId="371" CreationDate="2014-05-22T13:04:36.410" Score="19" Body="&lt;p&gt;This is known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Josephus_problem&quot;&gt;Josephus problem&lt;/a&gt;.  If the number of people $n$ is expressed as $n=2^m+p$, with $m$ as large as possible, the survivor is in seat $2p+1$  Another way to express it is to write $n$ in binary and rotate left one position.  For example, let there be $19$ people at the table, so we write $19=2^4+3$ and the winner is $2\cdot 3+1=7$, or  $19_{10}=10011_2$  Left rotating one space gives $00111_2=7_{10}$&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="71" LastEditDate="2014-06-13T20:08:29.953" LastActivityDate="2014-06-13T20:08:29.953" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="373" PostTypeId="2" ParentId="370" CreationDate="2014-05-22T13:54:51.577" Score="4" Body="&lt;p&gt;It depends on the precise mechanics of the game, and the starting state.&lt;/p&gt;&#xA;&lt;p&gt;I can't find any rule forbidding pushing buttons when both gates are closed, except when the red light is on and the green is off (which has contradictory rules), so if the game starts in one of those positions, he should be able to score. If the game does &lt;em&gt;not&lt;/em&gt; start in one of those positions, and the board can only be changed by pressing a button, he's out of luck.&lt;/p&gt;&#xA;&lt;p&gt;The table of specified moves comes out to this:&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th&gt;red light&lt;/th&gt;&#xA;&lt;th&gt;green light&lt;/th&gt;&#xA;&lt;th&gt;left gate&lt;/th&gt;&#xA;&lt;th&gt;right gate&lt;/th&gt;&#xA;&lt;th&gt;button&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;on&lt;/td&gt;&#xA;&lt;td&gt;on&lt;/td&gt;&#xA;&lt;td&gt;open&lt;/td&gt;&#xA;&lt;td&gt;open&lt;/td&gt;&#xA;&lt;td&gt;None&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;on&lt;/td&gt;&#xA;&lt;td&gt;on&lt;/td&gt;&#xA;&lt;td&gt;open&lt;/td&gt;&#xA;&lt;td&gt;closed&lt;/td&gt;&#xA;&lt;td&gt;None&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;on&lt;/td&gt;&#xA;&lt;td&gt;on&lt;/td&gt;&#xA;&lt;td&gt;closed&lt;/td&gt;&#xA;&lt;td&gt;open&lt;/td&gt;&#xA;&lt;td&gt;None&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;on&lt;/td&gt;&#xA;&lt;td&gt;on&lt;/td&gt;&#xA;&lt;td&gt;closed&lt;/td&gt;&#xA;&lt;td&gt;closed&lt;/td&gt;&#xA;&lt;td&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;on&lt;/td&gt;&#xA;&lt;td&gt;off&lt;/td&gt;&#xA;&lt;td&gt;open&lt;/td&gt;&#xA;&lt;td&gt;open&lt;/td&gt;&#xA;&lt;td&gt;None&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;on&lt;/td&gt;&#xA;&lt;td&gt;off&lt;/td&gt;&#xA;&lt;td&gt;open&lt;/td&gt;&#xA;&lt;td&gt;closed&lt;/td&gt;&#xA;&lt;td&gt;None&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;on&lt;/td&gt;&#xA;&lt;td&gt;off&lt;/td&gt;&#xA;&lt;td&gt;closed&lt;/td&gt;&#xA;&lt;td&gt;open&lt;/td&gt;&#xA;&lt;td&gt;None&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;on&lt;/td&gt;&#xA;&lt;td&gt;off&lt;/td&gt;&#xA;&lt;td&gt;closed&lt;/td&gt;&#xA;&lt;td&gt;closed&lt;/td&gt;&#xA;&lt;td&gt;***&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;off&lt;/td&gt;&#xA;&lt;td&gt;on&lt;/td&gt;&#xA;&lt;td&gt;open&lt;/td&gt;&#xA;&lt;td&gt;open&lt;/td&gt;&#xA;&lt;td&gt;None&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;off&lt;/td&gt;&#xA;&lt;td&gt;on&lt;/td&gt;&#xA;&lt;td&gt;open&lt;/td&gt;&#xA;&lt;td&gt;closed&lt;/td&gt;&#xA;&lt;td&gt;None&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;off&lt;/td&gt;&#xA;&lt;td&gt;on&lt;/td&gt;&#xA;&lt;td&gt;closed&lt;/td&gt;&#xA;&lt;td&gt;open&lt;/td&gt;&#xA;&lt;td&gt;None&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;off&lt;/td&gt;&#xA;&lt;td&gt;on&lt;/td&gt;&#xA;&lt;td&gt;closed&lt;/td&gt;&#xA;&lt;td&gt;closed&lt;/td&gt;&#xA;&lt;td&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;off&lt;/td&gt;&#xA;&lt;td&gt;off&lt;/td&gt;&#xA;&lt;td&gt;open&lt;/td&gt;&#xA;&lt;td&gt;open&lt;/td&gt;&#xA;&lt;td&gt;None&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;off&lt;/td&gt;&#xA;&lt;td&gt;off&lt;/td&gt;&#xA;&lt;td&gt;open&lt;/td&gt;&#xA;&lt;td&gt;closed&lt;/td&gt;&#xA;&lt;td&gt;None&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;off&lt;/td&gt;&#xA;&lt;td&gt;off&lt;/td&gt;&#xA;&lt;td&gt;closed&lt;/td&gt;&#xA;&lt;td&gt;open&lt;/td&gt;&#xA;&lt;td&gt;[No Score]&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;off&lt;/td&gt;&#xA;&lt;td&gt;off&lt;/td&gt;&#xA;&lt;td&gt;closed&lt;/td&gt;&#xA;&lt;td&gt;closed&lt;/td&gt;&#xA;&lt;td&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;The asterisks (on/off/closed/closed) mark a contradictory row, with one direction specifying pressing the left button and the other prohibiting it.&lt;/p&gt;&#xA;&lt;p&gt;Text of the &amp;quot;No Score&amp;quot; row:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;the lefthand button must nevertheless be pressed under these circumstances if all other conditions permit a button to be pressed before any score may be made in succeeding phases of the test.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The rules don't seem to prohibit pushing a button, but he won't get a score from it.&lt;/p&gt;&#xA;" OwnerUserId="4" LastEditorUserId="69582" LastEditDate="2021-01-05T21:36:18.620" LastActivityDate="2021-01-05T21:36:18.620" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="374" PostTypeId="2" ParentId="260" CreationDate="2014-05-22T14:04:43.453" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Bob walked into the clothing store. He looked a bit lost. &quot;Hi, I'm Alice, can I help you?&quot; said Alice. &quot;Remember, you can get anything you want, at Alice's Restaurant.&quot; She looked at Bob again and swallowed her usual &quot;excepting Alice.&quot;  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Bob laughed. &quot;I was wondering about that name, but now I get the reference. Can you help me pick out a dress for a girl? My sister Carol is staying over with her two kids, and one of them spilled chocolate milk, ruining her favorite dress. I want to surprise her with a new one. You see, I don't have any kids myself, I'm single.&quot; Bob wondered why he had emphasized that.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;They talked, and looked at several dresses, but Bob just couldn't pick one. &quot;You know, maybe she should pick one out herself, I'm sorry,&quot; he said. Now Alice didn't want to lose a customer, and she didn't want to lose Bob. &quot;I tell you what,&quot; she said, &quot;bring both girls over to pick out a dress and I'll give you a discount.&quot;  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;To make a long story short, Bob did, the girls got dresses, Bob and Alice have been married for 7 years now, and they all lived happily ever after, proving that you &lt;em&gt;really&lt;/em&gt; can get anything you want, at Alice's Restaurant.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;But what are the chances that Alice would've made a wrong guess and the whole thing would've fallen apart?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Below is my first try, but I took Jon Ericson's advice and created the above.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Alice goes into a toy store and asks the clerk, Bob, to help her pick out a toy. Alice tells Bob she needs a toy for one of the two kids of Carol, a friend of hers. Since Alice has only girls herself, she needs his help picking out a toy for a boy. Bob helps her find something and they chat a while.&lt;br&gt;&#xA;  As a friendly gesture, Bob offers Alice a discount if she buys the same item twice, so she can give the other kid a present as well.&lt;br&gt;&#xA;  What is the chance that Alice has any use for Bob's offer?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditDate="2014-05-23T06:23:07.157" LastActivityDate="2014-05-23T06:23:07.157" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="375" PostTypeId="2" ParentId="370" CreationDate="2014-05-22T14:23:36.507" Score="12" Body="&lt;p&gt;&lt;strong&gt;The rules do not explicitly forbid pressing the button when not ordered to do so by the rules.&lt;/strong&gt; It is not possible to score at the times you are instructed to press the button, but it is possible to score at other times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the matrix I used to determine what's allowed.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;            Left/Right  OO OC CO CC&#xA;Red/Green&#xA;       00               11 11 11 11&#xA;       01               11 11 11 11&#xA;       10               11 11 11 11&#xA;       11               11 11 11 11&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The left digit of a value indicates whether the left button may be pressed in that state. The right digit is the right button.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I proceeded to eliminate my options like so:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the lefthand gate is not closed, no button may be pressed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;00 00 11 11&#xA;00 00 11 11&#xA;00 00 11 11&#xA;00 00 11 11&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;No button should be pressed when the righthand gate is open unless all lights are out.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;00 00 11 11&#xA;00 00 00 11&#xA;00 00 00 11&#xA;00 00 00 11&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the lefthand gate is closed while the red light is lighted, do not press the lefthand button if the green light is out unless the righthand gate is open.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;00 00 11 11&#xA;00 00 00 11&#xA;00 00 00 01&#xA;00 00 00 11&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;At this point, this command doesn't apply:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Press the lefthand button whenever the red light appears provided the green light is not lighted as well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since pressing the right button and turning off the green light will never send you from a 0 to a 1 for the left button, it can be ignored.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the right-hand gate is open and the lefthand gate is closed, no score will result from pressing any button, but the lefthand button must nevertheless be pressed under these circumstances if all other conditions permit a button to be pressed before any score may be made in succeeding phases of the test.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This command must be followed if both lights are off, like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;00 00 10 00&#xA;00 00 00 00&#xA;00 00 00 00&#xA;00 00 00 00&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Other than these possibilities, the only way you can score points by pressing the lefthand button are these 3 states:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;00 00 00 10&#xA;00 00 00 10&#xA;00 00 00 00&#xA;00 00 00 10&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If the rules specified that you cannot score points unless instructed to press the left button, then it would indeed be impossible.&lt;/p&gt;&#xA;" OwnerUserId="18" LastActivityDate="2014-05-22T14:23:36.507" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="376" PostTypeId="1" AcceptedAnswerId="378" CreationDate="2014-05-22T15:41:38.293" Score="21" ViewCount="6686" Body="&lt;p&gt;The following is a type of logic / math puzzle I've yet to see on this site. I feel it belongs because the kernel of this problem can be reworked into other puzzles.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; pirates find themselves marooned on an island with no food stores, no other people, and no animal life besides one monkey.  There are, however, many banana trees. The pirates spend the entire day gathering bananas to tide them over as they work on a raft.&lt;/p&gt;&#xA;&lt;p&gt;After they go to sleep, one pirate wakes up, decides he doesn't trust anyone else, and tries to take his even &lt;span class=&quot;math-container&quot;&gt;$(\frac{1}{N})^{\mathrm{th}}$&lt;/span&gt; share of the bananas from the big pile. Finding one extra banana over an even split, he gives one to the monkey and takes &lt;span class=&quot;math-container&quot;&gt;$(\frac{1}{N})^{\mathrm{th}}$&lt;/span&gt; of the remaining pile to hide.&lt;/p&gt;&#xA;&lt;p&gt;After the first pirate finally falls asleep, a second pirate wakes up and decides the same thing but does not realise the other pirate had taken his share.  Finding one extra banana over an even split, he gives one banana to the monkey and takes &lt;span class=&quot;math-container&quot;&gt;$(\frac{1}{N})^{\mathrm{th}}$&lt;/span&gt; of the remaining pile to hide it.&lt;/p&gt;&#xA;&lt;p&gt;All N pirates wake up over the night, do the same thing and every time they find one more banana than an even split, give one to the monkey, and take &lt;span class=&quot;math-container&quot;&gt;$(\frac{1}{N})^{\mathrm{th}}$&lt;/span&gt; of the remaining pile.&lt;/p&gt;&#xA;&lt;p&gt;When morning comes and all &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; pirates wake up, they see the greatly diminished pile and a happy fat monkey.&lt;/p&gt;&#xA;&lt;p&gt;The question is: how many bananas could there have been in the original pile for this to occur?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="99" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2020-04-13T07:16:58.843" Title="N pirates steal their share of bananas to the benefit of a monkey" Tags="|calculation-puzzle|mathematics|word-problem|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="378" PostTypeId="2" ParentId="376" CreationDate="2014-05-22T16:12:11.740" Score="22" Body="&lt;p&gt;The basic idea is to work backwards.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The last pirate must have found $N+1$ bananas, because he had to find enough bananas remaining for at there to be least 1 banana in the pile left for each pirate.  He took 1, fed 1 to the monkey, and left $N-1$. That $N+1$ means that the next to last pirate found  $ \frac N{N-1}\cdot (N+1)+1$ and so on.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A cute trick is to recognize that there could have been &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $-(N-1)$ bananas.  Each pirate gives the monkey $1$, takes $-1$, and leaves the pile the same size as before.  Since we need to divide it by $N$ for each pirate, the next solution is higher by $N^N$, so the minimal positive solution is $N^N-N+1$ bananas.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="73" LastEditorUserId="46903" LastEditDate="2018-09-16T07:42:13.773" LastActivityDate="2018-09-16T07:42:13.773" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="379" PostTypeId="2" ParentId="112" CreationDate="2014-05-22T17:19:30.300" Score="9" Body="&lt;p&gt;The question doesn't place any constraints on the design of the box, so we can exploit that to get a &lt;strong&gt;6-lock&lt;/strong&gt; solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can either handle the two use cases with two lids, each covering half of the top of the box, or we can have one lid with a set on locks on one side and a set of locks on the other side, such that the locks on the side we're not unlocking function as the hinge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We handle one of the two sides with a lock for which the Grand Vizier has the only key and a lock for which each of the slaves has a key.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the other side, we use a similar technique to &lt;a href=&quot;https://www.flickr.com/photos/65379344@N03/5954344148/&quot;&gt;this one&lt;/a&gt; to create a bolt which can slide only if three of four locks are removed. (The technique in the photo has a bolt which is constrained to slide in one direction, but is blocked by a disc. That disc has N holes, each of which has a padlock through it, and can rotate so as to put any of the holes above the bolt. In this case we want the bolt to end in a fork such that it has three ends which must pass through holes. There is more discussion on &lt;a href=&quot;https://www.schneier.com/blog/archives/2012/08/shared_lock.html&quot;&gt;complex lock configurations&lt;/a&gt; on Bruce Schneier's blog).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, it should be possible to design the bolt in such a way that we can reduce this to &lt;strong&gt;5 locks&lt;/strong&gt;. We use the same forked bolt and rotating disc approach, but the Vizier's lock allows us to fold back the fork such that the bolt only needs one hole in the disc to be free.&lt;/p&gt;&#xA;" OwnerUserId="228" LastEditorUserId="228" LastEditDate="2014-05-22T18:25:22.523" LastActivityDate="2014-05-22T18:25:22.523" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="380" PostTypeId="2" ParentId="260" CreationDate="2014-05-22T20:21:30.917" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Today was &quot;bring your kids to work&quot;-day at the office. Alice has no kids of her own, but she saw her co-worker Bob with a lovely girl.&lt;br&gt;&#xA;  &quot;That's weird,&quot; she tought, &quot;I thought he had two kids.&quot; So she approached them, asked the girl's name (Carol), and asked Bob if she was his only child.&#xA;  &quot;Nah, &quot; said Bob, &quot;my other kid is sick at home.&quot;&#xA;  Alice thought for a bit and said to Carol, &quot;wish your brother well when you get home.&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;What chance did Alice have in guessing correctly?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditDate="2014-05-23T06:23:50.810" LastActivityDate="2014-05-23T06:23:50.810" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="381" PostTypeId="2" ParentId="313" CreationDate="2014-05-22T21:27:31.013" Score="19" Body="&lt;p&gt;As is the case with all information puzzles, we have to find away to map each outcome (yes, no, I don't know) to a different number (1, 2, 3). To do this, we need some property that's unknown for exactly one of these three numbers.&lt;/p&gt;&#xA;&lt;p&gt;There are multiple ways to do this. The most basic (and most reliable) way would be to introduce a number that only &lt;em&gt;you&lt;/em&gt; know so I wouldn't know:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;I'm thinking of either 1.5 or 2.5. Is your number greater than mine?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;1 is smaller than either of these numbers, so the answer will always be no. 3 is greater, so yes. And 2 could be either greater or smaller, so I don't know.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Another way, which is less reliable, is to introduce an unknown is to use some unknown property in mathematics:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Is TREE(your number) divisible by your number?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Kruskal%27s_tree_theorem&quot; rel=&quot;noreferrer&quot;&gt;Kruskal tree theorem&lt;/a&gt; produces a sequence of numbers such that TREE(1) = 1 and TREE(2) = 3, but TREE(3) is so astronomically large that it makes Graham's number look like epsilon in comparison. Moreover, it's not even known how to calculate it, so we cannot even tell whether it's even or odd.&lt;/p&gt;&#xA;&lt;p&gt;In this case, if I'm thinking of 1, the answer is yes, if I'm thinking of 2, the answer is no, and if I'm thinking of 3, I'm certainly not going to know whether TREE(3) is divisible by 3. The reason this one is less reliable is because if we ever &lt;em&gt;do&lt;/em&gt; find a way to calculate TREE(3), the uncertainty will no longer be there.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Yet a third way, the least reliable of all, is to use some form of contingent unknown, like a form of obscure trivia or some event that has a relatively unpredictable or unknown occurrence.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Is the current world population greater than (your number + 5.3) billion?&lt;/p&gt;&#xA;&lt;p&gt;Will Easter Sunday ten years from now happen less than (your number * 32) days after February 1?&lt;/p&gt;&#xA;&lt;p&gt;Is the last bit of the SHA-256 hash of &amp;quot;entanglement&amp;quot; strictly smaller than (your number - 1)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The current world population is currently between 7.2 and 7.3 billion. It reached 6.3 billion ten years ago, and won't reach 8.3 billion for another ten years, but somebody who didn't actually look up the population tables wouldn't know this fact for sure.&lt;/p&gt;&#xA;&lt;p&gt;Easter Sunday is defined as the Sunday after the first full moon after the vernal equinox, so it can't happen before March 20 in any year (which is the earliest date the vernal equinox can happen), and it can't happen after April 25 (36 days after the vernal equinox) either. So it's not going to happen before March 3 or after May 8 either. But the date in the middle, April 5/6, is right in the middle of possible dates for Easter, so unless you've memorized the Computus table, it's not generally feasible to predict what day Easter is going to fall on.&lt;/p&gt;&#xA;&lt;p&gt;The SHA-256 hash of &amp;quot;entanglement&amp;quot; is generally impossible to calculate in a few seconds in your head, unless you looked it up ahead of time, but you &lt;em&gt;do&lt;/em&gt; know that its last bit (which is either 0 or 1) will always be less than (3 - 1), and never less than (1 - 1).&lt;/p&gt;&#xA;&lt;p&gt;Although this method isn't completely reliable in that they might &lt;em&gt;know&lt;/em&gt; that piece of trivia, the key is to provide two completely ridiculous values on each side and one that can be argued either way in the middle that the average person would answer &amp;quot;I don't know&amp;quot; to.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; For the record: &lt;br /&gt;&lt;br /&gt;&#xA; The current world population at the time of posting is about 7.23 billion, which is less than 7.3 billion. This makes the answer &amp;quot;no&amp;quot; when my number is 2.&lt;br /&gt;&lt;br /&gt;&#xA; Easter Sunday will happen on March 31 in 2024, which makes the answer &amp;quot;no&amp;quot;.&lt;br /&gt;&lt;br /&gt;&#xA; The SHA-256 hash of &amp;quot;entanglement&amp;quot; is &lt;code&gt;961B164F23EB33F8FDA12C95E8BD93F6 32A08A8B8A0A18B3DDE1CFE8926875FF&lt;/code&gt;, which has a last bit of &lt;code&gt;1&lt;/code&gt;. So the answer is &amp;quot;no&amp;quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-05-22T21:27:31.013" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="382" PostTypeId="1" AcceptedAnswerId="3297" CreationDate="2014-05-22T22:01:41.187" Score="29" ViewCount="4729" Body="&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/313/you-have-one-question-to-tell-whether-the-number-im-thinking-of-is-1-2-or-3&quot;&gt;This question&lt;/a&gt; requires a bit of logical distortion to get the information required to answer the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I've never seen it asked before what would happen if more modes of question unanswerability were introduced.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm thinking of a number from 1 to 5, and you're allowed to ask one&#xA;  yes-no question to which I can answer one of the following things:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Yes&lt;/li&gt;&#xA;  &lt;li&gt;No&lt;/li&gt;&#xA;  &lt;li&gt;Sometimes  (Both yes and no are known to be true with a nonzero frequency.)&lt;/li&gt;&#xA;  &lt;li&gt;I don't know  (There is a definite answer, but it is not known.)&lt;/li&gt;&#xA;  &lt;li&gt;Not applicable  (This question does not have an answer for my number.)&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;What question can you ask to determine what number I'm thinking of?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There may be many correct answers to this question, but try to make it as elegant as possible.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2015-10-21T22:23:14.450" Title="Differentiate between the numbers from 1 to 5 with one single yes/no question" Tags="|logical-deduction|lateral-thinking|" AnswerCount="3" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="383" PostTypeId="5" CreationDate="2014-05-22T22:58:37.273" Score="0" Body="&lt;p&gt;An &lt;a href=&quot;https://en.wikipedia.org/wiki/Anagram&quot; rel=&quot;nofollow noreferrer&quot;&gt;anagram&lt;/a&gt; is a type of word play, the result of rearranging the letters of a word or phrase to produce a new word or phrase, while using all the original letters exactly once. The original word or phrase is called the &lt;em&gt;subject&lt;/em&gt; of the anagram. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;William Shakespeare = I am a weakish speller&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="11497" LastEditDate="2020-03-06T10:40:44.493" LastActivityDate="2020-03-06T10:40:44.493" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="384" PostTypeId="4" CreationDate="2014-05-22T22:58:37.273" Score="0" Body="An anagram rearranges the letters of a word or phrase to produce a new word or phrase." OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-02T13:54:09.033" LastActivityDate="2015-02-02T13:54:09.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="385" PostTypeId="2" ParentId="252" CreationDate="2014-05-23T00:58:20.017" Score="13" Body="&lt;p&gt;For this puzzle, while it has one and only one solution, no known patterns work on it, other than a slightly more intelligent guess-and-check. The number of steps one has to look ahead in order to reduce away clues is the metric here, and this puzzle needs nine sequential guesses to reach a solvable state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solver on SudokuWiki can't get it because it would simply take too long to do in Javascript, and it's not programmed to guess numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution requires one to assume the values of squares, and then reduce the puzzle to see if you need more assumptions - if you do, make another one and continue. It is a depth-first-search of the possible solutions, in essence. The solver on &lt;a href=&quot;http://www.sudoku-solutions.com/&quot; rel=&quot;noreferrer&quot;&gt;sudoku-solutions&lt;/a&gt; does come up with the solution to this puzzle, but when asked to provide the steps, declares:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;This solver could not solve the puzzle completely by logic, this does not mean there is not a logical solution.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and then promptly fails to list &lt;em&gt;any&lt;/em&gt; of the steps it used to solve it. This only happens when the solver must use brute-force branching guessing to find the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result, there is no way I myself could reasonably provide a &quot;how to solve this puzzle&quot; answer, since doing so would involve finding these specific chains and explaining why the other vast quantity of chains don't work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that's how you do it: assume a square is a number, then another, then another, and keep checking until you've come to a sequence that still makes sense and allows you to solve the puzzle, or you've come to a contradiction and need to back up and try again. I'm afraid I think this is the best answer you can get to this question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you did ask for a solution to the puzzle, however, I can provide it (mouseover the spoiler block):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/11Rin.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user20" LastEditorDisplayName="user20" LastEditDate="2014-05-23T02:01:48.317" LastActivityDate="2014-05-23T02:01:48.317" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="386" PostTypeId="2" ParentId="382" CreationDate="2014-05-23T01:34:51.553" Score="27" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm thinking of a number, $s$, which equals either $0$ or $1$ with equal probability. If $s=1$, then $p$ is either $0$ or $1$ with an unknown probability distribution - otherwise $p=0$. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If your number is $n$, where $n\neq 4$, is the $\displaystyle\lim_{x\to n}\frac{1}{x-2-s-p}\geq0$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $x=1$, then the denominator is either $-1$, $-2$, or $-3$, for which the answer is &lt;strong&gt;no&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;If $x=2$, then the denominator is either $0$, $-1$, or $-2$, for which the answer is &lt;strong&gt;sometimes&lt;/strong&gt;. The $0$ case - the only one that matters - occurs with known probability of $50\%$.&lt;/li&gt;&#xA;&lt;li&gt;If $x=3$, then the denominator is $1$ if $s=0$ (known probability), $0$ if $s=1$ and $p=0$, or $-1$ if $s=1$ and $p=1$. Since the last two cases where $p$ changes with unknown probability produce a different result, the answer is &lt;strong&gt;I don't know&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;If $x=4$, then question does not apply, so the answer is &lt;strong&gt;not applicable&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;If $x=5$, then the denominator may be $1$, $2$, or $3$, for which the answer is &lt;strong&gt;yes&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="user20" LastEditorDisplayName="user20" LastEditDate="2014-06-15T04:33:49.980" LastActivityDate="2014-06-15T04:33:49.980" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="387" PostTypeId="5" CreationDate="2014-05-23T01:49:47.540" Score="0" Body="&lt;p&gt;For puzzles referencing or relating to historical events or knowledge. Sometimes centered around people, cultural heritage or legends. General History questions are off-topic but can be asked on our sister site &lt;a href=&quot;https://history.stackexchange.com/&quot;&gt;History Stack Exchange&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Note there is a separate &lt;a href=&quot;/questions/tagged/puzzle-history&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;puzzle-history&amp;#39;&quot; rel=&quot;tag&quot;&gt;puzzle-history&lt;/a&gt; tag to be used for questions about the history of puzzles.&lt;/p&gt;&#xA;" OwnerUserId="52483" LastEditorUserId="60644" LastEditDate="2020-11-24T23:04:27.077" LastActivityDate="2020-11-24T23:04:27.077" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="388" PostTypeId="4" CreationDate="2014-05-23T01:49:47.540" Score="0" Body="For puzzles referencing or relating to historical events or knowledge. Use [puzzle-history] for the history of puzzles." OwnerDisplayName="user20" LastEditorUserId="69582" LastEditDate="2021-02-03T07:12:37.490" LastActivityDate="2021-02-03T07:12:37.490" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="389" PostTypeId="2" ParentId="305" CreationDate="2014-05-23T01:56:14.860" Score="16" Body="&lt;p&gt;The claim that this is the world's hardest sudoku puzzle was actually made by the author, and popularized by the media. There is nothing to suggest that this is the &lt;em&gt;singular hardest puzzle in existence&lt;/em&gt;, however, it's still a pretty dern difficult puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To quote the author of the puzzle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Normal sudoku puzzle logic eliminates the possibilities for each box in two or three steps, but this one requires puzzlers to think ahead eight or nine steps at a time - making very long deductions to eliminate the possible candidates for each box.&quot; (&lt;a href=&quot;http://www.mirror.co.uk/news/weird-news/worlds-hardest-sudoku-puzzle-ever-942299&quot;&gt;source&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is what makes this puzzle difficult: the number of steps one has to look ahead in order to reduce away clues. The solver on SudokuWiki can't get it because it would simply take too long to do in Javascript, and it's not programmed to look to a depth of nine. Performing an exhaustive search at that depth would take an inconceivable amount of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As this is the furthest depth we know of, and the only puzzle of this depth known, it currently takes top spot as the computationally hardest Sudoku. &lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastActivityDate="2014-05-23T01:56:14.860" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="390" PostTypeId="5" CreationDate="2014-05-23T06:09:45.923" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-05-23T06:09:45.923" LastActivityDate="2014-05-23T06:09:45.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="391" PostTypeId="4" CreationDate="2014-05-23T06:09:45.923" Score="0" Body="A puzzle that is centered around scales, weights, and balancing problems. " OwnerDisplayName="user20" LastEditorUserId="8874" LastEditDate="2015-01-30T13:32:51.780" LastActivityDate="2015-01-30T13:32:51.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="392" PostTypeId="2" ParentId="133" CreationDate="2014-05-23T08:35:23.117" Score="1" Body="&lt;p&gt;There is a generic approach for the optimal solution to the &quot;Requires $M$ of $N$ slaves&quot; problem. I'm not sure whether it's the one Emrakul describes, because TL;DR. It certainly doesn't need to be that complicated. Take any $M-1$ subset of the slaves. There must be a key which none of them have, but which all of the other slaves have. So $\binom{N}{M-1}$ locks are necessary and sufficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be extended in a suboptimal way to a wider class of problem featuring people who count as an integral number of slaves. For example, in your desired approach to the Vizier + 4 slaves problem, the Vizier counts as 2 slaves, so that's a 3 of 6 problem. But this is a suboptimal way of solving the wider class, because in the generic construction 3 of 6 would take 15 locks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your example problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;One level A riddle and two level B riddles.&lt;/li&gt;&#xA;  &lt;li&gt;One level A riddle, one level B riddle, and two level C riddles.&lt;/li&gt;&#xA;  &lt;li&gt;Two level B riddles and three level C riddles.&lt;/li&gt;&#xA;  &lt;li&gt;All six level C riddles.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is trickier, because the third constraint means that B should be $\frac32$ C, so the second constraint means that A should be $\frac52$ C and the first constraint means that A should be 3 C. We have to double things up: C is two slaves, B is three slaves, A is six slaves, and you need 12 slaves to proceed. That means that you have one extra slave in the second case, but since there's no way of removing a single slave from the set that's not a problem. What is a problem is requiring $\binom{36}{11} = 600805296$ keys.&lt;/p&gt;&#xA;" OwnerUserId="228" LastActivityDate="2014-05-23T08:35:23.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="394" PostTypeId="1" AcceptedAnswerId="430" CreationDate="2014-05-23T17:15:24.293" Score="23" ViewCount="7625" Body="&lt;p&gt;There is a classic problem below for which &lt;a href=&quot;https://www.math.washington.edu/~morrow/336_11/papers/yisong.pdf&quot;&gt;Dr. Yisong Song wrote a very well written and thought out series of solutions&lt;/a&gt;. &#xA;The variation here I've never seen anywhere but am curious. I want to work it out myself but think others may also want to give it a go.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;One hundred prisoners have been newly ushered into prison. The warden tells them that starting tomorrow, each of them will be placed in an isolated cell, unable to communicate amongst each other. Each day, the warden will choose one of the prisoners uniformly at random with replacement, and place him in a central interrogation room containing only a light bulb with a toggle switch. The prisoner will be able to observe the current state of the light bulb. If he wishes, he can toggle the light bulb. He also has the option of announcing that he believes all prisoners have visited the interrogation room at some point in time. If this announcement is true, then all prisoners are set free, but if it is false, all prisoners are executed. The warden leaves, and the prisoners huddle together to&#xA;  discuss their fate. Can they agree on a protocol that will guarantee their freedom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For this variation, there is a set of such 100 prisoners who are all exactly 50 years old and reasonably expect to live to 80. The light starts off and they will be taken into room on average roughly once a day. The warden may, however, send an extra prisoner of his choosing to the cell on the same day (but they still never see each other) to disrupt strategies.  This excludes the use of binanary token strategy. Because the warden is not completely sadistic, a fair random number generator picks the first prisoners each day. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In their meeting, they initially agree to a simple one counter solution when one prisoner remembers (having once been a puzzle fan) that this should take about 30 years if they are lucky.  He argues that each year outside of prison has an equal value but each year in prison is no better or worse than being dead.  Being released on his 80th birthday is therefore worth about the same as never being released at all.  He would gladly accept a 50% chance of death to be released with 20 years left of life remaining compared to a 100% chance of survival but being 79 when that chance came.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if the value of making a guess is the number of years remaining of life multiplied by the probability of surviving the guess, what strategy should they employ to maximize this value?  &lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorDisplayName="user20" LastEditDate="2015-04-18T05:35:57.660" LastActivityDate="2018-04-02T13:20:55.557" Title="Variation of 100 prisoners light problem" Tags="|mathematics|logical-deduction|" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="396" PostTypeId="1" AcceptedAnswerId="397" CreationDate="2014-05-23T17:24:59.657" Score="6" ViewCount="4273" Body="&lt;p&gt;What is the lowest score you can get in 2048?  Creating high numbers is obviously bad, but what about not matching numbers?&#xA;What is the chance of getting this lowest score?&lt;/p&gt;&#xA;" OwnerUserId="95" LastEditorUserId="95" LastEditDate="2014-05-24T03:28:05.860" LastActivityDate="2014-05-24T03:28:05.860" Title="What is the lowest you can get in 2048?" Tags="|strategy|2048-game|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ClosedDate="2014-11-10T19:19:33.397" ContentLicense="CC BY-SA 3.0" />
  <row Id="397" PostTypeId="2" ParentId="396" CreationDate="2014-05-23T17:45:39.827" Score="13" Body="&lt;p&gt;If you were unlucky enough and were not trying...You could in theory end up with a square full of alternating 2s and 4s.  It should even be possible to do this without matching, which means a score of 0.&lt;/p&gt;&#xA;" OwnerUserId="220" LastActivityDate="2014-05-23T17:45:39.827" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="398" PostTypeId="2" ParentId="394" CreationDate="2014-05-23T18:27:56.020" Score="13" Body="&lt;p&gt;If you accept a probabilistic argument, they can get out (or die) much sooner.  After $d$ days, the chance that a given prisoner has seen the room is $1-\left(\frac {99}{100}\right)^d$.  It is not fair to consider these all independent, but it will not be far wrong.  The chance that all the prisoners have seen the room is then $\left(1-\left(\frac {99}{100}\right)^d\right)^{100}$.  If we want a probability of success of $p$, we need to solve &#xA;$$\displaystyle \left(1-\left(\frac {99}{100}\right)^d\right)^{100}=p\\1-\left(\frac {99}{100}\right)^d=p^{0.01}\\1-p^{0.01}=\left(\frac {99}{100}\right)^d\\d=\frac{\log(1-p^{0.01})}{\log 0.99} $$&lt;br&gt;&#xA;giving a result of $$ \begin {array} {r | r}p&amp;amp;d\\ \hline \\0.5 &amp;amp;495\\0.75&amp;amp;582\\0.9&amp;amp;682\\&#xA;0.95&amp;amp;754\\0.99&amp;amp;916\\0.999&amp;amp;1145 \end {array}$$  They can take their choice from this table.  Even $99\%$ probability take less than three years.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get an idea of the error of the independence assumption, at 582 days, the first two stages of an inclusion-exclusion argument give a chance of failure of $0.99^{582}\cdot 100-0.98^{582} {100 \choose 2}=0.2495$ and adding in the chance of $97$ takes it up to $0.2527$ so it is already very close.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="99" LastEditDate="2014-07-14T14:09:18.343" LastActivityDate="2014-07-14T14:09:18.343" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="399" PostTypeId="1" AcceptedAnswerId="401" CreationDate="2014-05-23T19:33:49.450" Score="16" ViewCount="2288" Body="&lt;p&gt;Number twenty-one cages containing mice, 1, 2, 3, etc., up to 21, and place them in the following order:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;1 14 15 17 3 7 13 6 20 2 19 21 5 18 8 11 16 9 12 4 10 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;You start from a cage of your choice, calling that cage &lt;em&gt;one&lt;/em&gt;, and count, &lt;em&gt;one, two, three, etc.,&lt;/em&gt; from left to right. Once you reach the end, you continue counting from the leftmost cage.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When your count matches with the number on the cage, you have made a &lt;em&gt;&lt;strong&gt;catch&lt;/strong&gt;&lt;/em&gt;, and you can eat that mouse &lt;sup&gt;&lt;sub&gt;(you're a cat)&lt;/sub&gt;&lt;/sup&gt;, and throw away the cage.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Then start at the cage next to it &lt;sup&gt;&lt;sub&gt;(or the leftmost cage, if you just caught the rightmost cage)&lt;/sub&gt;&lt;/sup&gt;, call that &lt;em&gt;one&lt;/em&gt;, and try to make another catch, and so on.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Let's say you start at &lt;strong&gt;18&lt;/strong&gt;, calling that cage &lt;em&gt;one&lt;/em&gt;, your first catch will be &lt;strong&gt;19&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;1 14 15 17 3 7 13 6 20 2 19 21 5 18 8 11 16 9 12 4 10 &#xA;                                 ^&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Remove &lt;strong&gt;19&lt;/strong&gt; and your next catch will be &lt;strong&gt;10&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;1 14 15 17 3 7 13 6 20 2    21 5 18 8 11 16 9 12 4 10 &#xA;                            ^  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Remove 10 and your next catch is &lt;strong&gt;1&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;1 14 15 17 3 7 13 6 20 2    21 5 18 8 11 16 9 12 4  &#xA;^&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Remove the 1, and you &lt;strong&gt;cannot make another catch, even if you count up to 21&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;  14 15 17 3 7 13 6 20 2    21 5 18 8 11 16 9 12 4  &#xA;  ^&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Since you're very hungry, &lt;strong&gt;you want to catch all the twenty-one mice&lt;/strong&gt;. But you'll soon figure out that this is not here possible - no matter which cage you start from.&lt;/p&gt;&#xA;&lt;p&gt;But here's what you can do:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;You can choose &lt;strong&gt;two&lt;/strong&gt; cages and swap them before you begin. Then, if you start at the right place, you might be able to catch all twenty-one mice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;You may never pass over a catch, i.e., you must always eat the mouse you caught, remove that cage and continue.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You can always count up to 21, even if fewer cages are left.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;h3&gt;What two cages would you swap, and which cage would you start counting from?&lt;/h3&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;sup&gt;It would be great if you could include the strategy you used to get to the answer, unless you used brute force.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="129" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2018-05-14T15:47:30.670" Title="You are a cat. How do you swap?" Tags="|logical-deduction|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="401" PostTypeId="2" ParentId="399" CreationDate="2014-05-23T21:29:45.067" Score="6" Body="&lt;p&gt;I employed a brute force strategy, testing every possible swap with every possible starting position. I used the following Python code (Python 2.7):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;cages = map(int, '1 14 15 17 3 7 13 6 20 2 19 21 5 18 8 11 16 9 12 4 10'.split())&#xA;&#xA;swaps = ((i,j) for i in xrange(len(cages)) for j in xrange(len(cages)) if i &amp;gt; j)&#xA;&#xA;for (i,j) in swaps:&#xA;    for start in xrange(len(cages)):&#xA;        l = list(cages)&#xA;        l[i], l[j] = l[j], l[i]  # perform the swap&#xA;        pos = start&#xA;        count = 1&#xA;&#xA;        while (count &amp;lt;= 21) and l:&#xA;            if l[pos] == count:&#xA;                del l[pos]&#xA;                count = 0&#xA;            else:&#xA;                pos += 1&#xA;&#xA;            if pos &amp;gt;= len(l):&#xA;                pos = 0&#xA;&#xA;            count += 1&#xA;&#xA;        if not l:  # if the list is empty, we've found one that works&#xA;            print 'swap', cages[i], 'with', cages[j], 'and start at', cages[start]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which produces the following output:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;swap 6 with 13 and start at 14&lt;br&gt;&#xA;swap 8 with 6 and start at 19&lt;br&gt;&#xA;swap 10 with 14 and start at 16&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="198" LastEditorUserId="198" LastEditDate="2014-05-23T21:45:00.593" LastActivityDate="2014-05-23T21:45:00.593" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="402" PostTypeId="1" AcceptedAnswerId="403" CreationDate="2014-05-24T03:43:48.317" Score="2" ViewCount="460" Body="&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/371/how-do-you-solve-the-circular-table-problem&quot;&gt;This question&lt;/a&gt; was asked about a group of people whom a king wishes to play a cruel game with. Going around in a circle, the king kills every other person. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question concerned how one could compute the answer without executing the sequence, and this answer comes out to $n=2^m+p$, where $n$ is the number of people, $2^m$ is the highest value less than $n$, and $p$ is the seat number of the person who lives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens if the king kills every third person? Fourth person? For convenience, let $s$ be the number of seats skipped. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've worked out the first few digits in the sequence where $s=3$, as $1, 2, 2, 1, 4, 1, 4, 7$, but I don't see a pattern here. I imagine I'd have to generate quite a number of these to see the pattern. The first few digits where $s=4$ are $1, 2, 3, 2, 1, 5, 2, 6, 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one generate this sequence in a generic way for any $s$?&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2014-06-13T19:58:45.840" Title="What is the solution to the generic round table problem?" Tags="|calculation-puzzle|josephus-problem|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="403" PostTypeId="2" ParentId="402" CreationDate="2014-05-24T04:44:43.353" Score="4" Body="&lt;p&gt;As mentioned in the previous thread, this is the (generalized) &lt;a href=&quot;http://en.wikipedia.org/wiki/Josephus_problem&quot; rel=&quot;nofollow&quot;&gt;Josephus problem&lt;/a&gt;, known as such because the oldest known reference (at least in Western history) is by the historian &lt;a href=&quot;http://en.wikipedia.org/wiki/Flavius_Josephus&quot; rel=&quot;nofollow&quot;&gt;Flavius Josephus&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no known general closed form. You can compute the position of the survivor by recurrence:&#xA;$$p_s^n = \begin{cases}&#xA;  1 + (p^{n-1}_s - 1 + s) \bmod n &amp;amp; \text{if \(n \gt 1\)} \\&#xA;  1 &amp;amp; \text{if \(n = 1\)} \\&#xA;\end{cases}$$&#xA;where $p_s^n$ is position of the survivor for the $s$-step Josephus problem with $n$ people.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://oeis.org/&quot; rel=&quot;nofollow&quot;&gt;Online Encyclopedia of Integer Sequences&lt;/a&gt; should always be your first hit when you have a sequence of integers. It has a &lt;a href=&quot;https://oeis.org/wiki/Index_to_OEIS%3a_Section_J&quot; rel=&quot;nofollow&quot;&gt;bunch of entries tagged “Josephus”&lt;/a&gt;, including&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://oeis.org/A006257&quot; rel=&quot;nofollow&quot;&gt;A006257&lt;/a&gt;: survivor for $s=2$&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://oeis.org/A054995&quot; rel=&quot;nofollow&quot;&gt;A054995&lt;/a&gt;: survivor for $s=3$&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://oeis.org/A088333&quot; rel=&quot;nofollow&quot;&gt;A088333&lt;/a&gt;: survivor for $s=4$&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://oeis.org/A181281&quot; rel=&quot;nofollow&quot;&gt;A181281&lt;/a&gt;: survivor for $s=5$&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://oeis.org/A032434&quot; rel=&quot;nofollow&quot;&gt;A032434&lt;/a&gt;: $p_s^n+1$ for $s \le n$, in lexicographic order of $(s,n)$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The reference for A054995 gives a formula for $s=3$:&#xA;$$p^3_n = 3n+1 - \lfloor K(3) \cdot (3/2)^{\lceil L(3) \rceil} \rfloor$$&#xA;where $ L(3) = \dfrac{\log((2n+1)/K(3))}{\log(3/2)} $&#xA;and $K(3) \approx 1.62227050288476731595695$ as seen in &lt;a href=&quot;https://oeis.org/A083286&quot; rel=&quot;nofollow&quot;&gt;A083286&lt;/a&gt;. See &lt;a href=&quot;ftp://ftp.cis.upenn.edu/pub/wilf/josephus.ps&quot; rel=&quot;nofollow&quot;&gt;“Functional iteration and the Josephus problem”&lt;/a&gt; by A. M. Odlyzko and H. S. Wilf for a proof and generalization. There is also a proof with didactic commentary in &lt;em&gt;Concrete Mathematics&lt;/em&gt; (Graham, Knuth and Patashnik, 1994 (2nd ed.)) §3.3. Note that this doesn't really help to compute the values, because computing a precise enough value of $K(3)$ is as hard as computing $p^3_n$.&lt;/p&gt;&#xA;" OwnerUserId="71" LastEditorUserId="71" LastEditDate="2014-06-13T19:40:40.350" LastActivityDate="2014-06-13T19:40:40.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="405" PostTypeId="2" ParentId="70" CreationDate="2014-05-24T09:02:14.103" Score="7" Body="&lt;p&gt;&lt;em&gt;The solution is &lt;strong&gt;very&lt;/strong&gt; simple. The solution is very simple ones you know it. But may be you would like to give it a try first?&lt;/em&gt;&#xA;&lt;em&gt;(Though my answer looks relatively complicated, this is because I was required to explain operation % with very basic operations)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Solution:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The logicians numerate themselves (L1, L2, ..., LN) and colours (C1, C2, ..., CN).&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let's introduce a new operation &lt;code&gt;%&lt;/code&gt;, it's meaning please see below*. There are N equations:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1) (C1 + C2 + ... + CN) % N = 0&#xA;2) (C1 + C2 + ... + CN) % N = 1&#xA;...&#xA;N) (C1 + C2 + ... + CN) % N = N-1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Logician &lt;code&gt;Li&lt;/code&gt; should suppose that equation number &lt;code&gt;i&lt;/code&gt; is true. Then he extracts &lt;code&gt;Ci&lt;/code&gt; from equation number &lt;code&gt;i&lt;/code&gt;: &lt;code&gt;Ci = (i - C1 - C2 - ... - C_(i-1) - C_(i+1) ... - CN) % N&lt;/code&gt;. He can do it because he knows all colours except &lt;code&gt;Ci&lt;/code&gt;. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Obviously exactly one equation is true. Therefore correspondent logician will be correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Definition of operation %:&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;&lt;code&gt;Z = X % Y&lt;/code&gt; operation is defined in the following way: One needs to find R and K integer number, which satisfy: &lt;code&gt;X = R + Y*K&lt;/code&gt; and &lt;code&gt;0 &amp;lt;= R &amp;lt; Y&lt;/code&gt;. Then &lt;code&gt;Z = R&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;li&gt;This means that &lt;code&gt;C1 + C2 + ... + Ci + ... + CN) % N = i&lt;/code&gt; is equivalent to &lt;code&gt;C1 + C2 + ... + Ci + ... + CN = i + N*k&lt;/code&gt;, where k is integer. &lt;code&gt;Ci&lt;/code&gt; is found as &lt;code&gt;Ci = i + N*k - C1 - C2 - ... - C_(i-1) - C_(i+1) ... - CN&lt;/code&gt; or, applying % operation to both parts, since &lt;code&gt;Ci&lt;/code&gt; is integer less than &lt;code&gt;N&lt;/code&gt;: &lt;code&gt;Ci = (i - C1 - C2 - ... - C_(i-1) - C_(i+1) ... - CN) % N&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="28" LastEditorUserId="44915" LastEditDate="2018-03-01T15:55:13.783" LastActivityDate="2018-03-01T15:55:13.783" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="407" PostTypeId="1" CreationDate="2014-05-24T19:08:22.187" Score="24" ViewCount="3329" Body="&lt;p&gt;I came across the following interesting problem today:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A game consists of a sequence of plays; on each play either you or your opponent scores a point, you with probability  &amp;lt; 1/2, he with probability (1 - ). The number of plays is to be even (2, 4, 6, ...). To win the game, you must score &lt;em&gt;more than&lt;/em&gt; half the points. You are allowed in advance to choose the number of plays. How many plays should you choose in terms of  to optimize your chances of winning?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The problem is much less trivial than I thought at first glance. I know the answer, but will refrain from posting it unless the problem goes unsolved for an extended time.&lt;/p&gt;&#xA;" OwnerUserId="198" LastEditorUserId="69582" LastEditDate="2021-01-03T22:03:32.537" LastActivityDate="2021-01-03T22:03:32.537" Title="Winning an unfair game" Tags="|logical-deduction|mathematics|probability|word-problem|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="408" PostTypeId="2" ParentId="407" CreationDate="2014-05-24T20:56:04.837" Score="0" Body="&lt;p&gt;Let's look at the number of plays vs the number of games he has to win:&lt;/p&gt;&#xA;&lt;p&gt;2: 2/2 + 1 = 2&lt;br /&gt;&#xA;4: 4/2 + 1 = 3&lt;br /&gt;&#xA;6: 6/2 + 1 = 4&lt;br /&gt;&#xA;n: n/2 + 1&lt;/p&gt;&#xA;&lt;p&gt;Or in relative terms: 1, 3/4, 2/3, 1/2 + 1/n. So the more they would play, the easier it would be for the first player to get more than half wins. This does not take into account the handicap though. For that, one would need to maximise the combined probability, but I'm not skilled enough to write that here or be certain of its correctness (binomial distribution of n over k for values of k from n/2 + 1 to n?).&lt;/p&gt;&#xA;&lt;p&gt;So I'll just cheat through induction. For p=1/4, there's a ~6%  chance of winning 2/2 games, ~5% of winning 3 or more of 4, ~4% for 4-6/6 ... So it seems the extra combinations do not offset the exponents and the best choice is the unintuitive &lt;strong&gt;2 games&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="69582" LastEditDate="2021-01-03T22:01:52.777" LastActivityDate="2021-01-03T22:01:52.777" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="409" PostTypeId="2" ParentId="407" CreationDate="2014-05-24T21:29:00.383" Score="10" Body="&lt;p&gt;If you play two games, your chance is &lt;span class=&quot;math-container&quot;&gt;$ p^2$&lt;/span&gt;.  If you play four, your chance is &lt;span class=&quot;math-container&quot;&gt;$p^4+4p^3(1-p)$&lt;/span&gt;, which becomes greater when  &lt;span class=&quot;math-container&quot;&gt;$p\gt \frac 13 $&lt;/span&gt;.  If you play six games, your chance is &lt;span class=&quot;math-container&quot;&gt;$p^6+6p^5(1-p)+15p^4(1-p)^2$&lt;/span&gt;, which is better than four games when &lt;span class=&quot;math-container&quot;&gt;$p \gt \frac 25$&lt;/span&gt;  It is tempting to conjecture that you should play &lt;span class=&quot;math-container&quot;&gt;$2n$&lt;/span&gt; games (instead of &lt;span class=&quot;math-container&quot;&gt;$2n-2$&lt;/span&gt;) when &lt;span class=&quot;math-container&quot;&gt;$p \gt \frac {n-1}{2n-1}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th&gt;p &amp;lt;&lt;/th&gt;&#xA;&lt;th&gt;n&lt;/th&gt;&#xA;&lt;th&gt;P (chances of winning the game)&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;&lt;span class=&quot;math-container&quot;&gt;$\frac{1}{3}$&lt;/span&gt;&lt;/td&gt;&#xA;&lt;td&gt;2&lt;/td&gt;&#xA;&lt;td&gt;&lt;span class=&quot;math-container&quot;&gt;$p^2$&lt;/span&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;&lt;span class=&quot;math-container&quot;&gt;$\frac{2}{5}$&lt;/span&gt;&lt;/td&gt;&#xA;&lt;td&gt;4&lt;/td&gt;&#xA;&lt;td&gt;&lt;span class=&quot;math-container&quot;&gt;$p^4+4p^3(1-p)$&lt;/span&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;&lt;span class=&quot;math-container&quot;&gt;$\frac{3}{7}$&lt;/span&gt;&lt;/td&gt;&#xA;&lt;td&gt;6&lt;/td&gt;&#xA;&lt;td&gt;&lt;span class=&quot;math-container&quot;&gt;$p^6+6p^5(1-p)+15p^4(1-p)^2$&lt;/span&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;&lt;span class=&quot;math-container&quot;&gt;$\frac{4}{9}?$&lt;/span&gt;&lt;/td&gt;&#xA;&lt;td&gt;8&lt;/td&gt;&#xA;&lt;td&gt;&lt;span class=&quot;math-container&quot;&gt;$p^8 + 8p^7(1-p) + 28p^6(1-p)^2 + 56p^5(1-p)^3$&lt;/span&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;" OwnerUserId="73" LastEditorUserId="69582" LastEditDate="2021-01-03T22:01:04.500" LastActivityDate="2021-01-03T22:01:04.500" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="410" PostTypeId="1" AcceptedAnswerId="421" CreationDate="2014-05-24T21:37:51.740" Score="27" ViewCount="8842" Body="&lt;p&gt;When you're looking at a logical word puzzle, what sorts of clues should you try to look for to solve them? There have been quite a few puzzles posted to this site so far that I spent a few minutes pondering, and I just had no idea where to start. Once I saw the answers they usually made sense, but I don't seem to have a good gameplan for trying to figure it out on my own (which is the fun part!) I'd like to get better at solving puzzles like this, and I imagine there are techniques I'm unaware of to finding the important information and where to begin (because part of the trick to puzzles like these is that they give you extra information you don't need to throw you off, right?)&lt;/p&gt;&#xA;&lt;p&gt;So here's a good example of one such puzzle. (&lt;a href=&quot;http://www.folj.com/puzzles/very-difficult-analytical-puzzles.htm&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;; solution also displays at link).&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;You are the ruler of a medieval empire and you are about to have a celebration tomorrow. The celebration is the most important party you have ever hosted. You've got 1000 bottles of wine you were planning to open for the celebration, but you find out that one of them is poisoned.&lt;/p&gt;&#xA;&lt;p&gt;The poison exhibits no symptoms until death. Death occurs within ten to twenty hours after consuming even the minutest amount of poison.&lt;/p&gt;&#xA;&lt;p&gt;You have over a thousand slaves at your disposal and just under 24 hours to determine which single bottle is poisoned.&lt;/p&gt;&#xA;&lt;p&gt;You have a handful of prisoners about to be executed, and it would mar your celebration to have anyone else killed.&lt;/p&gt;&#xA;&lt;p&gt;What is the smallest number of prisoners you must have to drink from the bottles to be absolutely sure to find the poisoned bottle within 24 hours?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I thought about it for a little while, and I just had no idea where to start. All I could think is that, since you have 24 hours and it takes less than 24 hours to die of the poisoning, each of the 1000 people should test one bottle of wine. But that's clearly not the best answer, because that's obvious; they're asking what's the &lt;em&gt;smallest&lt;/em&gt; number of testers. I reread the problem several times trying to pick out the piece of information that should be my starting point, but I was unsuccessful.&lt;/p&gt;&#xA;&lt;p&gt;Once I read the solution it made sense, but it would never have occurred to me. How do you figure out where to start solving a puzzle like this when you're completely flummoxed?&lt;/p&gt;&#xA;" OwnerUserId="10" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-11-01T15:06:19.910" Title="How do you solve a word puzzle when you have no idea where to start?" Tags="|logical-deduction|word-problem|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="411" PostTypeId="2" ParentId="410" CreationDate="2014-05-24T21:53:21.313" Score="5" Body="&lt;p&gt;There is an aha! moment when you realize that one person can sample more than one bottle of wine. Those are hard to describe how to come to.  Problems that involving find 1 out of N often come down to partitioning N in some useful way.  (The classic &lt;a href=&quot;https://puzzling.stackexchange.com/questions/183/twelve-balls-and-a-scale&quot;&gt;12 balls and a scale&lt;/a&gt;splits them in thirds).  Maybe you have seen the problem of guessing somebody's number where you are allowed a small number of guesses and are told higher/lower.  You want to split the possibilities in half, then in half again etc.  Here you can think of lining up the bottles in order and you are asked to find the chosen number.  However, because of the time it takes to die, you have to do all the tests simultaneously.  If you have played with binary numbers this problem is much easier.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2014-05-24T21:53:21.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="412" PostTypeId="2" ParentId="410" CreationDate="2014-05-24T23:03:32.220" Score="8" Body="&lt;p&gt;As Ross already mentioned, the first aha! moment should be when you realize that a prisoner can drink from multiple bottles. From this the next step is realizing that you can only perform the test once, so you cannot do separate tests based on the results of the previous test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key concept you should first realize after this is: this means that each prisoner can only signal you two states: he will die, or he will live.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the fact that you now know that you have tons of prisoners who can each signal you two different states you now need to think on something that can represent a number between 1 and 1000 in the least amount of died/lived states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what is the concept that resembles this?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; to use prisoners as binary digits! &#xA; &lt;br&gt;&#xA; &lt;br&gt;&#xA; So the solution: you have to label each bottle from 1 to 1000, and&#xA; label each prisoner with the powers of two. You can easily see, you need&#xA; &lt;code&gt;ceil(log_2(1000)) = 10&lt;/code&gt; prisoners for this. Then each prisoner should drink from&#xA; bottles which also have the specific power of two in its given representation.&#xA; &lt;br&gt;&#xA; &lt;br&gt;&#xA; So prisoner 1 will drink from the odd numbered bottles, prisoner 2 will drink from&#xA; bottles numbered 2,3,6,7,10,11,..., prisoner 3 (or more exactly 4) will drink from&#xA; bottles 4,5,6,7,12,13,14,15,..., prisoner 4 (8) will drink from&#xA; 8,9,10,11,12,13,14,15,24,25,..., etc.&#xA; &lt;br&gt; &#xA; &lt;br&gt;&#xA; After the drinking you have to wait for the prisoners to die. There will be exactly one&#xA; bottle, from which excactly those prisoners who have died drank.&#xA; Just add the prisoner's numbers&#xA; together and you'll get to the specific bottle you shouldn't drink from&#xA; &lt;br&gt;&#xA; &lt;br&gt;&#xA; The result again is 10.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="85" LastEditorUserId="4" LastEditDate="2014-05-24T23:53:10.720" LastActivityDate="2014-05-24T23:53:10.720" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="413" PostTypeId="1" AcceptedAnswerId="416" CreationDate="2014-05-25T01:04:55.223" Score="5" ViewCount="2525" Body="&lt;p&gt;With a 4x4x4 cube, it's possible to apply the same strategies as for a 3x3x3, either treating it as a 3x3x3 with a double-width middle or else as a 2x2x2 and just working on the corner &quot;quadrants&quot;. Centres of each face can also be handled as a 2x2x2 cube, ignoring the outer layers entirely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it better to apply those strategies in one order rather than the other?&lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2018-08-10T09:18:32.183" Title="4x4x4 Rubiks Cube strategy" Tags="|rubiks-cube|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="414" PostTypeId="2" ParentId="413" CreationDate="2014-05-25T02:06:20.787" Score="0" Body="&lt;p&gt;the strategies mostly used are to solve each of the 2x2 faces (and positioning them) then the edges and then you can treat the entire cube as a normal 3x3x3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the key algorithms are &lt;code&gt;2R U2 2R'&lt;/code&gt; that is move the right part of the front face to the top rotate it and then move the new part back, this will allow you to solve the center faces&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the other one is &lt;code&gt;2B R U R' 2B'&lt;/code&gt; or rotate the bottom 2 layers splitting up the edge, moving the top part of the split side up, rotating in a new edge and moving it all back again. This will allow you to solve the edges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;after that is it just a clunky 3x3x3.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2014-05-25T02:06:20.787" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="415" PostTypeId="1" CreationDate="2014-05-25T08:20:36.050" Score="10" ViewCount="3655" Body="&lt;p&gt;I have been solving the Rubik's Cube 2x2x2, 3x3x3, 4x4x4 and 5x5x5, and because things are very similar I thought a 6x6x6 wouldn't be a challenge much. So I bought the C4U 3x3x7 - see image. What I want to know is how to solve it, what algorithms do I need? I can't find any good YouTube tutorials and also not a good website with algorithms and explanation on how to solve it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5ozkA.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/KNXrY.jpg&quot; alt=&quot;Rubik&amp;#39;s Cube 3x3x7 scrambled&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I found in the tutorials are the next few steps, the first steps I can follow but steps 4 to the end not:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Start with a fully scrambled Rubik's cube. &lt;em&gt;(right image)&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Solve the middle layer with the normal blocks&lt;/li&gt;&#xA;&lt;li&gt;Make sure the upper and bottom layer pieces are aligned correctly so you can turn it. &lt;em&gt;(left image)&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;???&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Profit&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Where do I go from here?&lt;/p&gt;&#xA;" OwnerUserId="146" LastEditorUserId="71" LastEditDate="2014-11-12T11:02:22.813" LastActivityDate="2014-11-12T11:02:22.813" Title="How do I solve the C4Y 3x3x7?" Tags="|rubiks-cube|twisty-puzzles|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="416" PostTypeId="2" ParentId="413" CreationDate="2014-05-25T08:46:50.347" Score="5" Body="&lt;p&gt;&lt;strong&gt;Notation&lt;/strong&gt; &lt;br&gt;&#xA;We need to have some notation for the algorithms. We use:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$U$ for the upper layer and $u$ for the middle upper layer. &lt;/li&gt;&#xA;&lt;li&gt;$R$ for the right layer and $r$ for the middle right layer. &lt;/li&gt;&#xA;&lt;li&gt;$L$ for the right layer and $l$ for the middle left layer. &lt;/li&gt;&#xA;&lt;li&gt;$F$ for the front layer and $f$ for the middle front layer. &lt;/li&gt;&#xA;&lt;li&gt;$?^2$ for turning a layer twice&lt;/li&gt;&#xA;&lt;li&gt;$?'$ for turning counter clockwise&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Example: $(Uu)^2 ~ l'$ is turning the upper half of the cube twice and turning the middle left layer counter clockwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solving&lt;/strong&gt; &lt;br&gt;&#xA;Note that you need the same colours as on the 3x3x3, but with the 4x4x4 you can mix them up and later on while solving you have a problem when yours centers doesn't have the same colours as the 3x3x3. So it easier to reduce to a 3x3x3 than a 2x2x2. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Solve the centers, by bringing two times two pieces together and then bring those pieces together and you have a solved center: &lt;em&gt;(this can mostly done by intuition)&lt;/em&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;White Up, Red Front, Blue Right,&lt;/li&gt;&#xA;&lt;li&gt;Yellow opposite to White&lt;/li&gt;&#xA;&lt;li&gt;Red opposite to Orange&lt;/li&gt;&#xA;&lt;li&gt;Blue opposite to green&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Solve the edges &lt;em&gt;(watch the video linked below and this becomes intuition)&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Solve it as a 3x3x3&lt;/li&gt;&#xA;&lt;li&gt;Solve the parities:&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Edge parity: you want to flip one edge. $r' ~ U^2 ~ l ~ F^2 ~ l' ~ F^2 ~ r^2 ~ U^2 ~ r ~ U^2 ~ r' ~ U^2 ~ F^2 ~ r^2 ~ F^2$ &lt;/li&gt;&#xA;&lt;li&gt;Corner parity you want to swap two corners. $U^2 ~ r^2 ~ U^2 ~ r^2 ~ (Uu)^2 ~ r^2 ~  (Uu)^2$&lt;br&gt;&#xA;&lt;em&gt;Note: Actually you swap two edges, so you need to do some more solving after you have performed this algorithm.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Finish the cube and profit!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Corner parity and edge parity&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YhdbV.gif&quot; alt=&quot;Edge parity&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/jUGnH.gif&quot; alt=&quot;Corner parity&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I want to show you a good &lt;a href=&quot;https://www.youtube.com/watch?v=vArGJLLdWh8&quot; rel=&quot;nofollow noreferrer&quot;&gt;YouTube tutorial&lt;/a&gt;. &lt;em&gt;(in this video he never mentioned a 2x2x2, but he did find it necessary to master the 3x3x3)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="146" LastEditorUserId="146" LastEditDate="2014-06-29T08:46:03.860" LastActivityDate="2014-06-29T08:46:03.860" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="417" PostTypeId="1" AcceptedAnswerId="418" CreationDate="2014-05-25T11:26:28.513" Score="13" ViewCount="1035" Body="&lt;p&gt;In a building, there is an odd elevator, which has only two buttons: UP, which makes it go up 9 floors, and DOWN, which makes it go down 7 floors. (The ground floor is floor 0.) It is possible to reach each floor, but if the building was just &lt;em&gt;one&lt;/em&gt; floor less tall, this would not be possible anymore. &#xA;How many floors are there in this building?&lt;/p&gt;&#xA;" OwnerUserId="70" LastEditorUserId="8874" LastEditDate="2015-02-05T13:37:28.447" LastActivityDate="2015-02-05T13:37:28.447" Title="A building with an odd elevator" Tags="|mathematics|calculation-puzzle|" AnswerCount="1" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="418" PostTypeId="2" ParentId="417" CreationDate="2014-05-25T11:46:23.667" Score="11" Body="&lt;p&gt;The answer is 15:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;15             --v&#xA;14             . .           --v&#xA;13         --v . .           . .&#xA;12         . . . .       --v . .&#xA;11     --v . . . .       . . . .&#xA;10     . . . . . .   --v . . . .&#xA;9  --v . . . . . .   . . . . . .&#xA;8  . . . . . . . --v . . . . . .&#xA;7  . . . . . . .   . . . . . . --&#xA;6  . . . . . --^   . . . . . .&#xA;5  . . . . .       . . . . --^&#xA;4  . . . --^       . . . .&#xA;3  . . .           . . --^&#xA;2  . --^           . .&#xA;1  .               --^&#xA;0  ^&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;An intuitive explanation for this is that you can only go two floors &lt;em&gt;up&lt;/em&gt; at a time, and you have to get above floor 6 to go down and reach an odd numbered floor. But every time you go up by two, you also travel to the floor 7 floors above that one. Therefore, the minimum amount of floors for this to work is &lt;code&gt;[smallest even number &amp;gt; 6] + 7&lt;/code&gt;, which is &lt;code&gt;8+7=15&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the same logic, a general formula can be constructed, where &lt;code&gt;n&lt;/code&gt; = the greater number (9) and &lt;code&gt;m&lt;/code&gt; is the lesser:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{[smallest number &amp;gt; m divisible by n-m]} + m$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From which this slightly easier formula can be made:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m - (m \text{ mod } (n-m)) + (n-m) + m$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which can be simplified to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m + n - (m \text{ mod } (n-m))$$&lt;/p&gt;&#xA;" OwnerUserId="39" LastEditorUserId="39" LastEditDate="2014-05-25T12:21:03.417" LastActivityDate="2014-05-25T12:21:03.417" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="419" PostTypeId="2" ParentId="415" CreationDate="2014-05-25T17:27:24.447" Score="4" Body="&lt;p&gt;I used to have this puzzle, but now I don't. I apologize if these instructions have a few holes in them - they should be complete, though, as I tested them on a 3x3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general method for solving this puzzle is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Solve the middle layer pieces (as you've done)&lt;/li&gt;&#xA;&lt;li&gt;Rotate the top corners and edges so that the slices are turnable (as you've also done)&lt;/li&gt;&#xA;&lt;li&gt;Solve the closest two layers to the middle that are not yet solved&lt;/li&gt;&#xA;&lt;li&gt;If need be, reorient the middle layer so that it is in line with the outside unsolved layers.&lt;/li&gt;&#xA;&lt;li&gt;After they are solved, consider them to be part of the middle layer. This is important.&lt;/li&gt;&#xA;&lt;li&gt;Advance to the next closest outside layer.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The first point of note is that you should leave the cube oriented in one direction for the entire solve, unless you've actively identified that the middle layer is solved. If you don't, you might unintentionally mess it up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second point of note is that &lt;code&gt;U&lt;/code&gt; and &lt;code&gt;D&lt;/code&gt; apply only to the depth of layer that isn't solved. For instance, if you've solved the inside layers on both the top and bottom, &lt;code&gt;U&lt;/code&gt; and &lt;code&gt;D&lt;/code&gt; only tell you to rotate the top and bottom two layers. Or, phrased differently, once you're done solving the bottom and top layers from the inside out, don't rotate them again. So, if you're trying to solve pieces on the second layer, don't rotate the third layer as well.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Orient the cube so that the rotating sides are facing up and down. These form your bottom and top layers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bottom layer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First, solve the bottom-layer edges by intuition. This isn't difficult once you get used to it. &lt;/li&gt;&#xA;&lt;li&gt;Second, solve the corners by basic substitution with the corners on the top layer.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The first thing to do is solve the bottom edges by intuition. You can swap any edges with just the right-face rotations. For instance, if I wanted to swap the F and R edges, I would move &lt;code&gt;R2 U' R2 D R2 U R2 D' U' R2&lt;/code&gt; - this may look like a lot, but really all I'm doing here is moving an edge out of position, then moving it back into position over the correct edge. Go through it slowly and watch how the pieces move. The setup positions are actually fairly straightforward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, move the corners into place over the correct positions. Move the corners over the positions where they belong at the FR position, and execute the following sequence: &lt;code&gt;R2 U R2 U' R2&lt;/code&gt;; this will move the corner into place. If you're in a situation where corners are all on the white face but in the wrong spots, just remove a white piece with a yellow one, and use the floating white corner to solve the others. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Top layer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Solve the positions of the corners by swapping adjacent corners&lt;/li&gt;&#xA;&lt;li&gt;Solve the positions of the edges by swapping adjacent edges&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Next, we want to solve the top layer corners. This uses one algorithm, and is pretty simple. First, rotate the top layer so as many corners as possible are solved. Identify corners which need to be flipped. Use the following algorithm to swap the two corners on the top-right side: &lt;code&gt;(R2 U R2 U' R2) [U' D] (R2 U' R2 U R2) [D']&lt;/code&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If two corners need to be swapped diagonally, then execute this sequence:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;once, to place the two corners that need to be swapped next to each other&lt;/li&gt;&#xA;&lt;li&gt;once, to swap those two corners&lt;/li&gt;&#xA;&lt;li&gt;once, to undo the first swap&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;At this point, you need to solve the edges for the layers. The algorithm you need for this is: &lt;code&gt;(R2 U)2 (R2 U2)2 R2 U R2 U' R2&lt;/code&gt;, which swaps two edges that are in the F and R positions. You may also wish to use &lt;code&gt;(R2 U2)3&lt;/code&gt;, which swaps the F and B edges.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Middle layer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It happens \$50\%\$ of the time that two middle layer pieces may not be oriented correctly once you've solved a top and bottom layer. If this is the case, execute the following sequence: &lt;code&gt;(R2 U2)3 Sw U Sw U'&lt;/code&gt;, where &lt;code&gt;Sw = (R2 U)2 (R2 U2)2 R2 U R2 U' R2&lt;/code&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To explain why this sequence works, note first that the algorithms that swap two edges also swap two middle pieces. The above algorithm is a combination of the following: first, swapping the F and B edges; second, swapping the F and L edges; third, swapping the L and B edges. This actually solves the edges, but it does so with three swaps, flipping the middle layer into the correct orientation.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Follow these steps for all three layers and your puzzle is solved.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorDisplayName="user20" LastEditDate="2014-05-28T00:15:17.577" LastActivityDate="2014-05-28T00:15:17.577" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="420" PostTypeId="1" CreationDate="2014-05-25T17:55:38.253" Score="6" ViewCount="22680" Body="&lt;p&gt;The Ghost Cube is a 3x3 shape modification which is solved when two layers are rotated out of position. As a result, in order to solve the puzzles, you need to move pieces into positions where they don't necessarily line up with the other centers or nearby reference pieces, making it unclear as to when the pieces are solved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're unclear what I mean by this, &lt;a href=&quot;http://youtu.be/yomRUBvofb8?t=19s&quot; rel=&quot;nofollow noreferrer&quot;&gt;here is a video&lt;/a&gt; which should hopefully make this somewhat more clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/e2ICK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result of this strange cut pattern, each piece is of a different shape. This puzzle is notoriously difficult to solve, and is commonly considered the hardest 3x3 mod available. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What piece(s) should I focus on solving first so that I: a) don't end up trying to solve a confusing face, and b) can easily identify the pieces I need? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What shapes should I look for so that I can identify the pieces I need quickly and easily?&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorDisplayName="user20" LastEditDate="2014-06-04T18:00:08.570" LastActivityDate="2019-12-27T13:20:59.937" Title="Where should I start on the Ghost Cube?" Tags="|mechanical-puzzles|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="421" PostTypeId="2" ParentId="410" CreationDate="2014-05-25T21:04:40.957" Score="13" Body="&lt;p&gt;As I've said in many other questions, puzzles like the one you posted are what I call &lt;em&gt;information puzzles&lt;/em&gt; &amp;ndash; that is, puzzles where you perform some trial to get as much information as possible about a certain unknown fact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, you can notice that each slave you use can either die or not die, which is 2 possible outcomes. If you then use $n$ slaves, then there are a total of $2^n$ possible outcomes, and the smallest $n$ for which $2^n \ge 1000$ is $10$. If you think about it in this fashion from the beginning, you will probably realize that each prisoner can sample from more than one bottle in the process, because that's the only way this method can really work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a similar problem in which you have &lt;em&gt;two&lt;/em&gt; days to figure out which bottle out of 27 is poisoned, which only requires three slaves instead of the five it would require if you only had one day. Why is this?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because each slave now has &lt;em&gt;three&lt;/em&gt; possible states - not dead, dead in one day, dead in two days. So $3^3 = 27$, as required. But even knowing that a prisoner can sample from more than one bottle, it's still somewhat difficult to devise an algorithm for which bottles each slave should drink each day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;P.S. I don't personally like the formulation of &quot;anywhere between 10 to 20 hours, and you don't know how long&quot; to imply that you can't use a timing attack on the test. I prefer instead that &quot;anyone who drinks the poison dies at the stroke of midnight, regardless of when they drank it&quot;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As for the more general problem of how to solve word problems (which is what your question is about), there is only really a general set of principles you can follow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, identify the problem. This puzzle was an information puzzle. Generally, information puzzles will ask you to &quot;find out which [something] [has some property]&quot; &amp;ndash; in this case, figuring out which bottle of wine was poisoned. There might be some puzzles that are about direct calculation or traversing a graph or solving an equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, identify common steps to solve problems like these. In the case of information puzzles, you'll want to identify how many possible results you want, and how many possible outcomes the trial will give you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, try and solve the problem using those methods. In the case of information puzzles, this involves trying to map each desired result to an outcome of the trial. In the case of the prisoners having two days to sample wine bottles, you need some way of configuring the bottles on the first day, so that on the second day, there are still enough prisoners alive that you can carry on with the bottles you still don't know aren't poisoned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If these don't work, generally the most you can do is keep thinking about it until you do find some way to solve it. Sometimes, if you go looking for a solution, you won't even understand it or how it solves the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an exercise, try to devise the schedule that three slaves would follow for sampling wine if they had two days to test 27 bottles.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2014-10-21T00:31:34.067" LastActivityDate="2014-10-21T00:31:34.067" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="422" PostTypeId="5" CreationDate="2014-05-25T21:30:07.827" Score="0" Body="&lt;p&gt;A word problem is a mathematical problem stated using words, usually in terms of a story or a situation that dresses the problem up and makes extracting the actual question into a challenge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although generally all problems will use words, a word problem relies on the words to make the problem more interesting. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples of word problems are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/2444/the-tricky-tunnels&quot;&gt;The Tricky Tunnels!&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/80375/whats-bobs-age-again&quot;&gt;What&amp;#39;s Bob&amp;#39;s age again?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="11497" LastEditDate="2019-03-10T17:22:55.727" LastActivityDate="2019-03-10T17:22:55.727" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="423" PostTypeId="4" CreationDate="2014-05-25T21:30:07.827" Score="0" Body="A mathematical puzzle that's stated in words, usually in terms of a story or a situation that dresses the problem up." OwnerDisplayName="user88" LastEditorUserId="4421" LastEditDate="2016-08-22T17:42:39.507" LastActivityDate="2016-08-22T17:42:39.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="424" PostTypeId="2" ParentId="233" CreationDate="2014-05-26T09:19:30.660" Score="4" Body="&lt;p&gt;&lt;strong&gt;Minimum Number of Explorers = 2&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;If food supplies can be stored along the route, only these two explorers are needed. The rules don't offer this as an option but they don't seem to disallow it either. Assuming a liberal interpretation, here's an alternate solution:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The two explorers load their packs and walk two days. They leave two rations at a suitable location, &amp;quot;camp 2&amp;quot;, and return to the base on their remaining rations. This is repeated until the 8th trip where they instead walk three days to &amp;quot;camp 3&amp;quot;, i.e., where they would be at the end of day 3 of an uninterrupted journey. They leave two rations here and return to camp 2 (not back to base). They refill with 10 of the 14 rations stored here and walk to camp 3. They leave six rations and return to camp 2. They take the remaining four rations from camp 2, walk to camp 3 and refill with the eight rations stored there. With full packs, they can now complete their journey.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This solution takes twice as many supplies and dramatically more time than the non-storable solution. Which is better depends on resource management (time/supplies vs people) and whether the &lt;em&gt;five-days of rations&lt;/em&gt; limitation is due to perishability or weight ... if one were to care about practical issues in a puzzle.&lt;/p&gt;&#xA;" OwnerUserId="115" LastEditorUserId="63414" LastEditDate="2021-05-07T09:26:48.040" LastActivityDate="2021-05-07T09:26:48.040" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="426" PostTypeId="1" AcceptedAnswerId="428" CreationDate="2014-05-26T10:17:32.683" Score="1" ViewCount="2695" Body="&lt;p&gt;It seems like most word search puzzles have a quantity of words that match some theme, arranged haphazardly, with some random letters filling in the extra spaces.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I want to create word search puzzles what do I need to consider in my puzzle creation? &lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="4" LastEditDate="2014-05-26T23:25:46.097" LastActivityDate="2024-03-25T14:39:14.693" Title="If I want to create word search puzzles what approach is best?" Tags="|puzzle-creation|word-search|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="427" PostTypeId="2" ParentId="426" CreationDate="2014-05-26T13:32:24.557" Score="3" Body="&lt;p&gt;&lt;strong&gt;'Random' letters&lt;/strong&gt; &lt;br&gt;&#xA;Consider that the random letters you will end up with are not truly random and you don't want them to be! If they are random they can form words that need to be found, in this way you can the puzzle more easy. You could also make the 'random letters' such that they form a word at the end, that would be harder to make though. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theme&lt;/strong&gt; &lt;br&gt;&#xA;A theme is nice, but not entirely necessary. It is up to you to be creative and make the words themed. You can look for synonyms or look on the web for the words you already have made up. I think it's good to have more words and place them in the puzzle if they fit and else don't use them. The best way is to create or use software to create those puzzles. &lt;/p&gt;&#xA;" OwnerUserId="146" LastEditorUserId="146" LastEditDate="2014-05-27T09:08:49.160" LastActivityDate="2014-05-27T09:08:49.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="428" PostTypeId="2" ParentId="426" CreationDate="2014-05-27T13:25:29.123" Score="4" Body="&lt;p&gt;Here are a few technical factors to keep in consideration before starting your puzzle.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Grid size&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;A word search will tend to be more difficult on a larger grid than a smaller one, due to the larger search space. However, it will also be harder to fit all the letters onto whatever medium you're creating this word search on, so you need to find a balance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, a smaller word search is harder to &lt;em&gt;create&lt;/em&gt;, due to the fact that you need to fit so many words into such a small space and they might need to intersect. Be careful that you don't try and make a grid so small that the words become impossible to place.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Word length&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;As a rule, longer words are easier to find than smaller ones, simply because they're so big. However, they'll also be harder to place on a smaller grid, again because there are so many words they have to intersect with. At the same time, you need to be careful that you don't accidentally include a small word in the grid twice by virtue of the random letters coincidentally falling on that pattern.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For the actual puzzle creation process, you'll want to follow this general procedure:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Place big words first.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Big words are, as a rule, harder to place, because they have so many possible intersections and therefore requirements to fulfill. If you place them onto the grid first, there are more places to put them, and it's more likely the small words will fit onto the grid afterwards.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Look for clever combos.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If you notice that three words would work in a triangle or a group of four, put those three or four words together first. Making combos makes for more compact arrangements, which leave more room for other words to appear in the grid.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;If it fails, try again.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;You may not be able to get all the words you want onto the grid on the first try. If that happens, restart with the big words in a different position. If you've tried a few times and can't seem to get even half of the words onto the grid, try using a bigger grid or reducing your word list.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2014-06-13T19:39:49.250" LastActivityDate="2014-06-13T19:39:49.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="429" PostTypeId="1" AcceptedAnswerId="431" CreationDate="2014-05-27T14:25:41.133" Score="26" ViewCount="27582" Body="&lt;p&gt;My friend always insists on scrambling my Rubik's cube &amp;quot;perfectly&amp;quot; before giving it to me to solve. According to his definition, a &amp;quot;perfect&amp;quot; scramble must have no three of the same color touching (which takes an annoyingly long time to achieve). For example, this scramble is not a perfect one (showing the front face only):&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;RRG&#xA;BRB&#xA;YWO&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;because three red squares are touching, while this one is perfect:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;RRG&#xA;BBR&#xA;YWO&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Is it possible to achieve a scramble where no two same-colored squares are touching, on all sides?&lt;/p&gt;&#xA;" OwnerUserId="39" LastEditorUserId="69582" LastEditorDisplayName="user20" LastEditDate="2021-10-19T22:34:23.567" LastActivityDate="2024-03-03T03:21:42.450" Title="Is it possible to scramble a Rubik's cube such that no two squares of the same color are touching?" Tags="|rubiks-cube|twisty-puzzles|" AnswerCount="9" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="430" PostTypeId="2" ParentId="394" CreationDate="2014-05-27T14:36:15.990" Score="11" Body="&lt;p&gt;I still believe there is a better answer than this but that this is a significant improvement over the current best answer. I have used only the same math used by Mr. Millikan as I like how understandable it made the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On day 917, according to the &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \left(1-\left(\frac {99}{100}\right)^d\right)^{100}=p$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;used by Mr. Ross Milikan the value has a maximum with only a roughly 1% chance of everyone being executed horribly. This does not use the light bulb at all and the likelyhood of more than 1 person never having been in the room is less than .001%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strategize that every time a prisoner (assumed to be all male) walks into the room the first time, he switches the light. After the 100th person goes in the room for the first time the light will always be off.  On day 917, if the light is on: they know the 1% chance has happened and they must wait for it to go off before they say everyone has been there.  The likelyhood that there are 2 people who have not been there so the light was off but they die is very small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this strategy is used on day 495 (for instance).  The probability that everyone has been there is under 50%.  The probability that 2 people have not been there, however, is less than 1%. Therefore, on day 495 if the light is off they can guess with &gt;98% certainty that they can be freed; if the light is on, the person who has not been there yet, turns off the light and guesses that everyone is there and they win on the first day it is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless my calculations are too approximated, this strategy could yield a larger likelyhood of survival and will likely release prisonners a year ealier than Mr. Millikan's (depending on they day they choose).  This strategy has a large likelyhood of releasing the prisonners on the first day possible (but of course still has a non-zero chance of causing horrible death).&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditDate="2018-04-02T13:20:55.557" LastActivityDate="2018-04-02T13:20:55.557" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="431" PostTypeId="2" ParentId="429" CreationDate="2014-05-27T14:45:10.893" Score="40" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Yes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Examples&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Superflip&#xA;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/7EgQ7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7EgQ7.png&quot; alt=&quot;Superflip&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &amp;quot;Checkerboard&amp;quot;&#xA;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/sInIU.jpg&quot; alt=&quot;Checkerboard&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18" LastEditorUserId="63414" LastEditDate="2021-02-23T12:13:29.470" LastActivityDate="2021-02-23T12:13:29.470" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="432" PostTypeId="1" CreationDate="2014-05-27T15:03:47.797" Score="4" ViewCount="955" Body="&lt;p&gt;I've tried my hand at creating a crossword a few times, but generally there have been way too many interlinks between words for me to create even a solution grid with any ease. Often I get about a quarter of the way through a single 4x4 block, for example, when I realize that one of the words I'm about to create is total gibberish and I need to start over.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How do the professionals do it? What sort of tips and tricks are there to deciding which letters go into which boxes to form words?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="18250" LastEditDate="2016-04-09T11:51:27.933" LastActivityDate="2016-04-09T11:51:27.933" Title="How do people decide which words to use in crossword puzzles?" Tags="|puzzle-creation|crosswords|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="433" PostTypeId="1" CreationDate="2014-05-27T15:39:00.250" Score="15" ViewCount="2439" Body="&lt;p&gt;I was doing a cryptic crossword this morning and I came across the following clue:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Circus equipment has chimpanzee, say at the centre&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is &quot;TRAPEZE&quot; (confirmed by &quot;check answers&quot;). The APE is easy to see (chimpanzee), but I can't figure out where the TR___ZE comes into the clue. From my understanding, it &lt;em&gt;should&lt;/em&gt; be implied by the clue. Is there something I'm missing or is it just an omission we're supposed to be able to guess?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cross letters are T_A_E_E so the R and Z would have to be guessed entirely, based on TRAPEZE being the only word fitting the rest of the clues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;From Cryptic Crossword iPhone app, Pack 6, Puzzle 3, Clue 3d&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="4" LastActivityDate="2021-12-29T03:23:33.560" Title="Why is &quot;trapeze&quot; the solution to this cryptic crossword clue?" Tags="|cryptic-crosswords|crossword-clues|explanation|" AnswerCount="5" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="434" PostTypeId="2" ParentId="433" CreationDate="2014-05-27T15:58:20.750" Score="3" Body="&lt;p&gt;&quot;Chimpanzee&quot; is actually two clues in one &amp;ndash; the trick is that it's only the &quot;chimp&quot; part by itself that refers to APE, not the whole word. The &quot;zee&quot; refers to a Z that appears right afterwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the string that you get is actually &quot;APEZ&quot;. And of course, there's only one piece of circus equipment that has this &quot;chimp and Z&quot; at its centre.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2014-05-27T15:58:20.750" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="435" PostTypeId="1" AcceptedAnswerId="436" CreationDate="2014-05-27T16:00:32.420" Score="16" ViewCount="620" Body="&lt;p&gt;There are a number of conventions you can count on fairly consistently in the New York Times crossword.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One is that abbreviated answers are &quot;signaled&quot; somehow - an answer that is an acronym/initialism, or a shortened form of a longer word, has something to indicate it in the clue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my understanding, any abbreviated answer would always have one of the following in the clue:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;An explicit indicator&lt;/strong&gt; that it's shortened &quot;Abbreviated&quot; &quot;...for short&quot;, &quot;Casually&quot; etc. &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Abbreviations&lt;/strong&gt; in the clue itself: &quot;Low ranking &lt;strong&gt;GI&lt;/strong&gt;&quot; for &quot;pvt&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But consider the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HdDlI.jpg&quot; alt=&quot;crossword with &amp;quot;Capital of France&amp;#39;s Manche department&amp;quot; cluing &amp;quot;STLO&amp;quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The town the answer references is &lt;a href=&quot;https://en.wikipedia.org/wiki/Saint-L%C3%B4&quot; rel=&quot;noreferrer&quot;&gt;&quot;Saint-Lô;&quot;&lt;/a&gt; using &quot;St.&quot; is simply an abbreviation for &quot;Saint&quot;.  So why doesn't convention require that the abbreviation is telegraphed, in the same way it is in 9 Across? (&quot;Philatelists's abbr.&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the convention more complex than what I've described above?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I know that there aren't Crossword Courts, and that Will Shortz may do whatever he pleases, but crosswords are built on traditional expectations, so I'm trying to draw on the community's experience with why the indicator is expected on some clues and not others.&lt;/p&gt;&#xA;" OwnerUserId="5" LastEditorUserId="10" LastEditDate="2014-05-28T16:20:30.307" LastActivityDate="2015-06-03T16:16:15.910" Title="What are the conventions telegraphing when answers are abbreviated in New York Times crosswords?" Tags="|crosswords|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="436" PostTypeId="2" ParentId="435" CreationDate="2014-05-27T20:16:29.887" Score="12" Body="&lt;p&gt;The conventions for abbreviations in crosswords, particularly the New York Times crossword, are essentially what you described.  If there is any abbreviation in the clue itself, that dictates that the answer must be an abbreviation.  An indicator in the clue such as &quot;Abbreviated&quot; is also an obvious sign of an abbreviation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In regards to the clue that you used as an example, the proper convention concerning abbreviations is not necessary because Saint can be shortened to St. without being considered a specific abbreviation.  The same applies to mountains.  For example, the clue &quot;Sicilian volcano&quot; might garner the answer &quot;MTETNA&quot;.  The Mt. is not considered a formal abbreviation just like the St. in the example you provided.  I hope this clears things up.&lt;/p&gt;&#xA;" OwnerUserId="184" LastEditorUserId="463" LastEditDate="2015-06-03T16:16:15.910" LastActivityDate="2015-06-03T16:16:15.910" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="437" PostTypeId="1" AcceptedAnswerId="459" CreationDate="2014-05-27T20:28:14.053" Score="13" ViewCount="2503" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Rush_Hour_%28board_game%29&quot; rel=&quot;noreferrer&quot;&gt;Rush Hour&lt;/a&gt; is a well-known sliding-block puzzle in which you push cars around (each car moves horizontally or vertically) to let a red car escape the grid. The standard board is a 6x6 grid, with four 3x1 cars and twelve 2x1 cars available to place on the grid. Only the red car is ever placed horizontally on the third row, which is the escape row.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You cannot get the red car to escape with brute force; solving the gridlock sometimes requires some complicated and at other times counterintuitive manoeuvres. Generally, the more steps a solution has, the harder the puzzle is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Level 40 in the original set sold by Binary Arts has a solution with 50 steps. But there must certainly be harder puzzles than that. What is the Rush Hour puzzle with the most possible steps in its solution?&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2014-05-28T14:05:45.023" LastActivityDate="2014-05-28T17:14:47.657" Title="What is the Rush Hour puzzle with the most steps in its solution?" Tags="|rush-hour|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="438" PostTypeId="2" ParentId="432" CreationDate="2014-05-27T20:50:35.680" Score="3" Body="&lt;p&gt;Creating crosswords with easy to understand words can often be a challenge but there are many talented crossword constructors today.  Many of these constructors use software to help them fill the grid such as:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.crossword-compiler.com/&quot; rel=&quot;nofollow&quot;&gt;Crossword Compiler&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://beekeeperlabs.com/crossfire/&quot; rel=&quot;nofollow&quot;&gt;Crossfire&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Other than using these tools, it is also helpful to use short three or four letter words to fill in the grid around the longer words.  These easy words are collectively known as &quot;crosswordese&quot;.&lt;/p&gt;&#xA;" OwnerUserId="184" LastActivityDate="2014-05-27T20:50:35.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="439" PostTypeId="2" ParentId="433" CreationDate="2014-05-27T22:22:27.670" Score="7" Body="&lt;p&gt;This is still an incomplete answer, but I'm getting closer.   First, it's helpful to know that a cryptic clue has two components:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A traditional &quot;definition&quot; of the solution, much like a &quot;normal&quot; crossword clue.&lt;/li&gt;&#xA;&lt;li&gt;Another way to create the same answer, which can be another simple definition, or more complex ways to build the answer, like word rebuses, &quot;sounds like&quot; clues, etc.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The two clues are not mixed - either can come first, but one ends and then the other starts, so they're not mixed together.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back to Trapeze:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Obviously, the simple definition is &quot;circus equipment&quot;.&lt;/li&gt;&#xA;&lt;li&gt;So, the second clue that can be &quot;punnier&quot; is &quot;has chimpanzee, say at the centre&quot;, and for the most part, every word there should be relevant to the solution (or possibly a connector word, if the puzzle is from the UK).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&quot;APEZ&quot; almost surely comes from the phrase, &quot;chimpanzee, say&quot;.&lt;/strong&gt; &quot;Say&quot; is an indicator of a homophone here - it's a phrase, or manipulation of a phrase that &lt;em&gt;sounds like&lt;/em&gt; &quot;Chimpanzee&quot; - In this case, &quot;chimp AND zee&quot;, which gets us to &quot;APE Z&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's still unclear is how we get &quot;TR_E&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would work if &quot;has&quot; meant &quot;tre,&quot; because &quot;at the centre&quot; would just be the indicator to stick the first part inside the &quot;tre&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would also work if &quot;at the&quot; meant &quot;the end of&quot;, because taking the beginning or end of a word is typical, and you could (weakly) argue that &quot;has&quot; indicates the &quot;holding&quot; idea that gets part one stuck in the middle of TRE, and the tee could be coming from the end of &quot;centre&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But neither of those is the case.&lt;/p&gt;&#xA;" OwnerUserId="5" LastActivityDate="2014-05-27T22:22:27.670" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="442" PostTypeId="5" CreationDate="2014-05-28T02:13:14.477" Score="0" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Chess&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chess&lt;/a&gt; is a strategy game involving two players moving pieces on an 8×8 chessboard.&lt;/p&gt;&#xA;&lt;p&gt;There are six different pieces:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Pawn: can only move forward, except to capture pieces, in which case they move diagonally.&lt;/li&gt;&#xA;&lt;li&gt;Rook: can move horizontally or vertically anywhere.&lt;/li&gt;&#xA;&lt;li&gt;Knight: can move a &amp;quot;knight's move&amp;quot; all at once, which is two squares in one direction and one square in a perpendicular direction.&lt;/li&gt;&#xA;&lt;li&gt;Bishop: can move diagonally in any direction.&lt;/li&gt;&#xA;&lt;li&gt;Queen: can move horizontally, vertically, and diagonally.&lt;/li&gt;&#xA;&lt;li&gt;King: can move horizontally, vertically, and diagonally, but only one square at once.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;And there are a few more tricks, such as castling, promotion and en passant.&lt;/p&gt;&#xA;&lt;p&gt;The goal of the game is to &lt;em&gt;checkmate&lt;/em&gt; the other player's king - that is to position pieces in such a way so that no matter what move the opponent makes, their king can still be captured.&lt;/p&gt;&#xA;&lt;p&gt;To learn how to play chess, a good place to start is &lt;a href=&quot;https://www.chess.com/learn-how-to-play-chess&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chess.com&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Chess#Variants&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chess variants&lt;/a&gt; play a big part in chess puzzles, as do mathematical chess puzzles, such as &lt;a href=&quot;https://en.wikipedia.org/wiki/Knight%27s_tour&quot; rel=&quot;nofollow noreferrer&quot;&gt;Knights Tour&lt;/a&gt; or &lt;a href=&quot;https://en.wikipedia.org/wiki/Eight_queens_puzzle&quot; rel=&quot;nofollow noreferrer&quot;&gt;8-queens&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Questions that only involve the board itself and none of the pieces should be tagged &lt;a href=&quot;/questions/tagged/checkerboard&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;checkerboard&amp;#39;&quot; rel=&quot;tag&quot;&gt;checkerboard&lt;/a&gt; instead.&lt;/p&gt;&#xA;&lt;p&gt;Questions related to the game of chess itself should be posted to &lt;a href=&quot;https://chess.stackexchange.com&quot;&gt;Chess Stack Exchange&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="69582" LastEditDate="2021-01-28T15:48:00.983" LastActivityDate="2021-01-28T15:48:00.983" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="443" PostTypeId="4" CreationDate="2014-05-28T02:13:14.477" Score="0" Body="A puzzle related to chess, a two-player board game with white and black pieces of varying ranks." OwnerUserId="146" LastEditorUserId="69582" LastEditDate="2021-01-28T15:48:17.007" LastActivityDate="2021-01-28T15:48:17.007" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="444" PostTypeId="2" ParentId="407" CreationDate="2014-05-28T07:05:45.223" Score="13" Body="&lt;p&gt;This is really a comment on Ross Millikan's answer, but too cumbersome to enter that way. I have verified the $ p= \frac{n-1}{2n-1}$ threshold, although it really should be called $\frac{n/2}{n+1}$ (there is some confusion between $n$ and $2n$ in Ross's answer). The derivation is &lt;em&gt;roughly&lt;/em&gt; as follows.&#xA;For even $n$, let $P_n =$ the probability of winning a match when the probability of winning a game is $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ P_n = {n \choose 0} p^n (1-p)^{0} + {n \choose 1}p^{n-1}(1-p)^{1}+ \cdots + {n \choose m}p^{n-m}(1-p)^m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $m=n/2-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just a restatement of Ross's table in general form. If you want to solve for $P_8 &amp;gt;P_6$, say, where $0&amp;lt;p&amp;lt;1$, then we need to find the roots of $P_8-P_6$.&#xA;In general, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_{2k+2}-P_{2k} = (-1)^k C_k \,p^{k+1} (1-p)^{k}(p-\frac{k}{2k+1})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which has multiple roots at $0$ and $1$ and one root at $p=\frac{k}{2k+1}$ ($=\frac{n/2}{n+1}$ if we call $n=2k$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_k$ is the $k$th Catalan number.&lt;/p&gt;&#xA;" OwnerUserId="87" LastEditorUserId="4" LastEditDate="2015-02-09T04:00:21.387" LastActivityDate="2015-02-09T04:00:21.387" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="445" PostTypeId="2" ParentId="410" CreationDate="2014-05-28T07:10:16.727" Score="9" Body="&lt;p&gt;I'll give an answer that pertains to puzzles in general, not specifically this puzzle or information puzzles. You asked about &quot;word puzzles&quot;, but I don't think there is anything specific to word puzzles as opposed to other kinds. Anyway, most puzzles are word puzzles (one kind of a non-word puzzle is a &lt;a href=&quot;http://s4.zetaboards.com/science/topic/7693756/1/&quot; rel=&quot;nofollow noreferrer&quot;&gt;cryptarithmetic puzzle&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;em&gt;puzzle&lt;/em&gt;, by definition, doesn't have an algorithm for solving it. This is different from word &lt;em&gt;problems&lt;/em&gt; you were taught at school. (Although the first time you encounter a word problem of a certain kind, it's still a puzzle for you, because you don't know the algorithm for it yet.) Nevertheless, I can give you some tips.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, you write that you &quot;spent a few minutes pondering&quot;. Here's your problem. &lt;strong&gt;You need to think for a long while.&lt;/strong&gt; You try to think of one way of solving a puzzle. It doesn't lead you anywhere, so you think of another approach and so on. You might find certain environments stimulative. For example, I like to think while walking or showering, but some people like to think on the toilet. So instead of giving up after a few minutes, try to think of the puzzle at different times during the day, for at least one day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try to think of any other problems your puzzle resembles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One general approach is to try to reach a partial solution. There are several kinds of &quot;partial solutions&quot;. For example, if you are trying to solve an optimization problem (e.g., &quot;what is the &lt;em&gt;smallest&lt;/em&gt; number of prisoners?&quot;), usually there is an obviously-not-optimal answer (e.g., &quot;Make every prisoner taste a single bottle. I'll need 1000 prisoners.&quot;), then try to improve it (&quot;can I involve only 500 prisoners?&quot;). A partial solution can also be a solution to a simpler problem. For example, in the case of the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/230/a-camel-transporting-bananas&quot;&gt;camel transporting bananas&lt;/a&gt; problem, try first solving the problem for 2000 bananas instead of 3000. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make sure you understand the significance of every detail in the puzzle. For example, why is it important that death occurs after at most 20 hours? Because it means that you can't make a prisoner taste a bottle, wait to make sure he doesn't die, and only then test another bottle on the same prisoner. Why is it important that death occurs after at least 10 hours? No reason, because you always might be out of luck. Such cases of useless information are quite rare, so make sure you understand the significance of everything. Sometimes you have the opposite situation: It seems that you don't have enough info. If that happens, think why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck and have fun!&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2014-05-28T07:42:56.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="446" PostTypeId="1" AcceptedAnswerId="452" CreationDate="2014-05-28T07:13:33.220" Score="8" ViewCount="1472" Body="&lt;p&gt;Is there a rule/method to know how &quot;random&quot; should the cube be when solving it? in order for it to be a &quot;valid&quot; score?&lt;/p&gt;&#xA;" OwnerUserId="111" LastActivityDate="2015-11-16T22:09:27.190" Title="How Random does a 3x3x3 (or other rubik cubes) need to be in speed solving?" Tags="|rubiks-cube|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="447" PostTypeId="2" ParentId="446" CreationDate="2014-05-28T08:03:17.950" Score="5" Body="&lt;p&gt;A rubik's cube cannot be more than 20 moves ([called God's number][1]) away from its initial position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After about 16 moves, you should be random enough, since it's very hard to be in a specific 20-moves-away position.&lt;/p&gt;&#xA;" OwnerUserId="277" LastEditorUserId="463" LastEditDate="2015-06-03T16:12:37.873" LastActivityDate="2015-06-03T16:12:37.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="448" PostTypeId="2" ParentId="236" CreationDate="2014-05-28T09:29:47.103" Score="12" Body="&lt;p&gt;Let's take the case where there are 3 blue eyed people. each blue eyed person sees two blue eyed people but that is not enough for him/her to realize they have blue eyes. for that fact to be inferred he needs to observe the two blue eyed people he sees not leaving after two days.&#xA;and the only reason he would expect them to leave in two days is because he observed them listening to the remark that &quot;there is at least one blue eyed person&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the information was not shared to all at the same time there would be no reason for anyone to expect the group of blue eyed people to leave at any point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you see N blue eyed people around you expect them to all leave N days after the statement. if the information isn't shared there would be no reason for that expectation and therefore it would be impossible to infer your own eye color.&lt;/p&gt;&#xA;" OwnerUserId="280" LastActivityDate="2014-05-28T09:29:47.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="450" PostTypeId="5" CreationDate="2014-05-28T12:43:40.173" Score="0" Body="&lt;h1&gt;3x3x7&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;For Rubik's Cubes with a 3x3x7 layout. Example:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/1Runf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="146" LastEditorUserId="146" LastEditDate="2014-05-28T14:43:25.437" LastActivityDate="2014-05-28T14:43:25.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="451" PostTypeId="4" CreationDate="2014-05-28T12:43:40.173" Score="0" Body="For Rubik's Cubes with a 3x3x7 layout" OwnerUserId="146" LastEditorUserId="146" LastEditDate="2014-05-28T14:44:58.533" LastActivityDate="2014-05-28T14:44:58.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="452" PostTypeId="2" ParentId="446" CreationDate="2014-05-28T12:52:11.897" Score="6" Body="&lt;p&gt;The World Cube Association has regulations regarding the official scrambling algorithm used for their competitions.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An official scramble sequence must produce a random state from those that require at least 2 moves to solve.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;https://www.worldcubeassociation.org/regulations/#article-4-scrambling&quot;&gt;https://www.worldcubeassociation.org/regulations/#article-4-scrambling&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, as long as it takes at least 2 moves to solve, it's a valid scramble. All scrambles are computer-generated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most scrambles will require a minimum of 18 moves to solve. The number of possible permutations for a certain number of moves is given on &lt;a href=&quot;http://www.cube20.org&quot;&gt;http://www.cube20.org&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although a 2-move scramble is in theory possible, the chance of it happening is about 243 in 43 quintillion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I seem to recall that the standard for competition scrambles before this was 40 random turns, but I don't have a source for this.&lt;/p&gt;&#xA;" OwnerUserId="18" LastActivityDate="2014-05-28T12:52:11.897" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="454" PostTypeId="1" CreationDate="2014-05-28T13:45:37.360" Score="10" ViewCount="2238" Body="&lt;p&gt;If one has a very large puzzle of an image that one desires to mount and frame;  what kind of glues would damage the puzzle piece over time?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have heard of 'puzzle glues', but question if this is anything more than just a glue sold by a puzzle company.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have experience with puzzles that were mounted for long periods of time?  Anyone have experience from old / historic puzzles that they have seen mounted?  What about when the puzzle is very large (6' x 8' / 2 x 2.5m)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume the backing material is part of the 'solution', too.  &lt;/p&gt;&#xA;" OwnerUserId="283" LastEditorUserId="25966" LastEditDate="2016-12-01T10:10:43.970" LastActivityDate="2016-12-11T16:41:45.833" Title="What kind of glues are not conducive to long-term puzzle preservation?" Tags="|puzzle-history|jigsaw-puzzle|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="455" PostTypeId="1" AcceptedAnswerId="486" CreationDate="2014-05-28T14:55:24.840" Score="96" ViewCount="24782" Body="&lt;p&gt;I found this puzzle &lt;a href=&quot;http://www.folj.com/bb/viewtopic.php?t=12849&quot; rel=&quot;noreferrer&quot;&gt;online&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;On the top floor of a castle lives a princess. The floor has 17 bedrooms arranged in a row. Each bedroom has doors connecting to the adjoining bedrooms as well as to the outside corridor. The princess sleeps in a different bedroom each night by opening the door to an adjoining bedroom and spending the night and the next day in that room.&lt;/p&gt;&#xA;&lt;p&gt;One day a prince arrives at the castle and is desirous of marrying the princess. The guardian angel at the castle tells him of the princess' sleeping patterns and informs him that each morning he may knock on one of the outside doors. If the princess happens to be behind that door, she will open it and consent to marry him. The prince also has a return ticket to his kingdom in 30 days, so he can make at most 30 attempts. Can the prince win the hand of the princess, and if so, what is his strategy?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;An unstated assumption here is that the princess can move to &lt;em&gt;any&lt;/em&gt; adjacent room (she's not restricted to moving in one direction). So a possible sequence could be 3, 2, 3, 4, 5, 4, 3, 2, 1, 2...&lt;/p&gt;&#xA;&lt;p&gt;Further down the thread at the link, this solution is given:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The Prince should knock on the second door from one of the ends of the corridor (call it door #2), and knock on the next adjacent door each successive day until he reaches the second door from the opposite end of the corridor (door #16). The day after that, he should begin the same process in reverse order (meaning he will knock on the 16th door two days in a row). By the time he reaches his starting point (door #2 on the 30th day), he will have found the princess.&lt;/p&gt;&#xA;&lt;p&gt;Number the doors 1 thru 17.&lt;br /&gt;&#xA;If the princess occupies an even numbered room on the day the prince first knocks on a door (#2), then she will either occupy the same room he knocks on or she will be an even number of rooms away. Since both move to an adjacent room each day, this will hold true so she will never be in a room adjacent to the one he knocks on, and thus never be in a position to move past him the next day. She will have nowhere else to go by the time he reaches the 16th door, and will have by then been located. If she, on the other hand, was in an odd numbered room when he began, then she will be in an even numbered room when the prince starts the process again on day 16 (when he knocks on door #16 the second time).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I've been trying to follow the solution, but I don't quite get it. The main assumption the logic in the solution seems to hinge on is this: &lt;strong&gt;&amp;quot;she will never be in a room adjacent to the one he knocks on, and thus never be in a position to move past him the next day.&amp;quot;&lt;/strong&gt; I can't figure out what evidence from the puzzle supports that assumption. If we begin knocking at door #2 and the princess is in room #3, our next move (following this solution) will be to knock on door #3. But at this point the princess could have moved to room #2, and we will keep moving to higher numbered rooms, and the princess could easily remain in the lower-numbered rooms.&lt;/p&gt;&#xA;&lt;p&gt;So I'm having trouble understanding why this is the solution. Can someone explain it more clearly/in a different manner? Why will this process ensure the princess is found?&lt;/p&gt;&#xA;" OwnerUserId="10" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2019-06-19T17:17:02.207" Title="Why does this solution guarantee that the prince knocks on the right door to find the princess?" Tags="|mathematics|strategy|algorithm|" AnswerCount="7" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="456" PostTypeId="2" ParentId="455" CreationDate="2014-05-28T15:41:24.473" Score="8" Body="&lt;p&gt;Try a specific example.  Say she starts in 6.  He knocks on 2 and doesn't find her.  She can move to 5 or 7.  He knocks on 3 and doesn't find her.  She moves to 4,6, or 8.  He knocks on 4 and might find her.  The thing that can't happen (if she started in an even room) is for her to slip by him.  In this example, she must be in 6 or 8 if he doesn't find her, and now moves to 5,7, or 9.  He knocks on 5.  The rooms she might be in keep getting pushed toward high numbers.  If she started in an even room, he is guaranteed to find her by the time he knocks on 16.  If she started in an odd room, by the time he gets to 16 that is all he knows-that she started in an odd room.  If she is now in 17, she will move to 16 and he will find her.  Otherwise he knows that she is now in an even room and the series back to 2 will find her.  The critical point is that her room changes parity each move, so he takes advantage of that.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditDate="2018-06-16T03:01:10.740" LastActivityDate="2018-06-16T03:01:10.740" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="457" PostTypeId="2" ParentId="455" CreationDate="2014-05-28T15:43:34.227" Score="15" Body="&lt;p&gt;The solution assumes that she cannot ever stay in the the same room. Either on even days she is in even rooms or on even days she is in odd rooms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As he is going from room 2 to room 3 he is going from an odd day to an even day.  She can go from room 3 to 2 if she is in even rooms on even days. She can never pass him if she is in odd rooms on even days as he is moving to higher numbered rooms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When he moves from low numbers to high numbers, he eliminates one by one each possibility of her being in odd rooms on even days and then, as he goes backwards, he eliminates the possibilities of her being in even rooms on even days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;when he goes from 3 to 2 she cannot go from 2 to 3 as she cannot be in room 2 as that was already eliminated.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditDate="2014-05-28T15:49:47.523" LastActivityDate="2014-05-28T15:49:47.523" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="458" PostTypeId="2" ParentId="455" CreationDate="2014-05-28T16:07:31.090" Score="85" Body="&lt;p&gt;Maybe this diagram will help you visualize the solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pUezr.png&quot; alt=&quot;Diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this diagram, the vertical axis shows the room number (1&amp;ndash;17) and the horizontal axis shows the day (1&amp;ndash;30).  The $\color{red}{\text{red}}$ dots show the rooms on whose door the prince knocks on each day, and the $\color{darkgreen}{\text{green}}$ dots and arrows show the possible locations and movements available to the princess, assuming that she has not yet been found by the prince.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For the sake of clarity, I have omitted the green dots and arrows from the dark-colored squares on the left half of the diagram, representing the days 1&amp;ndash;15, even though the princess could also be in any of those rooms during those days.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the left side of the diagram shows, if the princess starts in an even-numbered room on day 1, then she will be restricted to the light-colored squares in the diagram (as she must move by exactly one room every day), and will be found by the prince by day 15 at the latest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, if she starts in an odd-numbered room on day 1, then she will not be found by the prince during the first 15 days.  However, this means that she's restricted to the dark-colored squares on the diagram, and so, on day 16, when the prince turns back, she must be in one of the even-numbered rooms, and will thus be found by the prince on his second pass through the rooms, during the days 16&amp;ndash;30.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ps.&lt;/strong&gt; In any case, the important thing to realize is that, if the princess is in an even-numbered room on day $n$, then she must move to an odd-numbered room on day $n+1$, and vice versa.  Thus, in the diagram above, she is restricted to always staying on squares of one color, like the bishop on a chessboard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, the diagram also shows that the given strategy is not the only possible one for the prince.  Besides the obvious mirror-image variant, the prince could also start as in the given solution, but decide on day 16 to move back to room #2 and work upwards from there, instead of staying at room #16 and working downwards. Either way, if he hasn't found the princess by day 15, she must be staying in the rooms corresponding to the dark-colored squares, and so, by systematically knocking on each room's door when it's dark, he will eventually find the princess by day 30.&lt;/p&gt;&#xA;" OwnerUserId="243" LastEditorUserId="243" LastEditDate="2014-07-23T01:17:29.610" LastActivityDate="2014-07-23T01:17:29.610" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="459" PostTypeId="2" ParentId="437" CreationDate="2014-05-28T17:14:47.657" Score="11" Body="&lt;p&gt;Frédéric Servais has apparently done a lot of study of this problem, particularly his Masters' thesis, &lt;a href=&quot;http://code.ulb.ac.be/dbfiles/Ser2005mastersthesis.pdf&quot;&gt;&lt;em&gt;Finding hard initial configurations of Rush Hour with Binary Decision Diagrams&lt;/em&gt;&lt;/a&gt;.  For the 6x6 case without 'trucks' (3x1 pieces), he finds a maximum of 65 steps (plus the final move of the red car out); for the &quot;full&quot; 6x6 case, the maximum he finds is 92 steps, but it's not as clear to me from the skimming I've done that he has a proof of maximality for that position.  You can see diagrams of his top positions (note that move counts are off by 1 for these, since the results on this page count the final move of the escaping car as well) at &lt;a href=&quot;http://cs.ulb.ac.be/~fservais/rushhour/&quot;&gt;http://cs.ulb.ac.be/~fservais/rushhour/&lt;/a&gt; .&lt;/p&gt;&#xA;" OwnerUserId="266" LastActivityDate="2014-05-28T17:14:47.657" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="460" PostTypeId="2" ParentId="415" CreationDate="2014-05-28T21:02:12.983" Score="0" Body="&lt;p&gt;&lt;strong&gt;Solving the top layer&lt;/strong&gt; &lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/users/20/emrakul&quot;&gt;Emrakul&lt;/a&gt; has a good &lt;a href=&quot;https://puzzling.stackexchange.com/a/419/146&quot;&gt;anwser&lt;/a&gt;, but the top layer takes a lot of time. To speed things up you can also use the following algorithm:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YgC87.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can come in the situation that you only need to swap the F and R edge then you need the algorithm &lt;a href=&quot;https://puzzling.stackexchange.com/users/20/emrakul&quot;&gt;Emrakul&lt;/a&gt; suggested in &lt;a href=&quot;https://puzzling.stackexchange.com/a/419/146&quot;&gt;his anwser&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(R2 U)2 (R2 U2)2 R2 U R2 U' R2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="146" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2014-05-29T07:50:22.103" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="461" PostTypeId="1" AcceptedAnswerId="463" CreationDate="2014-05-28T22:41:05.233" Score="10" ViewCount="740" Body="&lt;p&gt;I am making a puzzle based on a &lt;a href=&quot;http://projecteuler.net/&quot;&gt;Project Euler&lt;/a&gt; problem. Here is the puzzle at present:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What path in the triangle below, starting from the top number with each step moving to one adjacent number in the next row, gives the greatest possible sum?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;      35&#xA;    12  80&#xA;  88  79  96&#xA;73  56  20  55&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The intended solution to this problem is to, in the second-last row, draw an arrow from the greater of the two to the bottom. Then replace the number with the sum of those numbers, as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;       35&#xA;     12  80 &#xA;  161 135 151&#xA; ↗    ↗      ↖&#xA;73  56   20   55&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;When this process is repeated until the top row, it will give the best path. In this case, 35 to 80 to 96 to 55 is the best path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel that this puzzle is too easy, however. There are only so many paths and it's very easy to see which ones couldn't possibly be best. It's definitely possible to solve this without finding the &quot;correct&quot; solution. The obvious solution is to add more rows, but adding too many would require a lot of arithmetic. Already 6 additions must be performed to solve the puzzle, and puzzles get boring if there are too many additions. Five rows would require 10 additions and six rows would require 15.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I keep the puzzle at four rows, or increase its size? Is there general advice as to how large a puzzle should be?&lt;/p&gt;&#xA;" OwnerUserId="258" LastEditorUserId="258" LastEditDate="2014-05-28T22:51:11.117" LastActivityDate="2014-06-13T19:35:52.173" Title="What size of puzzle is appropriate?" Tags="|calculation-puzzle|puzzle-creation|mathematics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="462" PostTypeId="1" AcceptedAnswerId="468" CreationDate="2014-05-28T22:58:56.623" Score="32" ViewCount="4671" Body="&lt;p&gt;Though there is &lt;a href=&quot;https://mathoverflow.net/q/30330/51379&quot;&gt;no strict mathematical proof&lt;/a&gt;, it has been shown via brute-force that no arrangement of 4 or fewer queens can control every square on a standard 8x8 chessboard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, out of the 64 squares on an 8x8 chessboard, what is the &lt;strong&gt;maximum number of squares&lt;/strong&gt; that can be controlled by only 4 queens?&lt;/p&gt;&#xA;" OwnerUserId="291" LastEditorUserId="57521" LastEditDate="2021-02-04T17:54:33.953" LastActivityDate="2021-02-04T17:54:33.953" Title="How many squares can 4 queens control?" Tags="|chess|optimization|checkerboard|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="463" PostTypeId="2" ParentId="461" CreationDate="2014-05-28T22:59:20.427" Score="3" Body="&lt;p&gt;Determining an &quot;appropriate size&quot; for a puzzle is all about determining what constraints you consider &quot;appropriate&quot;, and then finding a puzzle that fits those constraints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, you are given a pyramid with $n$ levels, designed in such a way that the dynamic programming approach to the problem requires $\binom{n}{2}$ additions, but there are $2^n$ possible paths in total.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can make a table of values for this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{matrix}&#xA;n &amp;amp; \binom{n}{2} &amp;amp; 2^n\\&#xA;4 &amp;amp; 6 &amp;amp; 16 \\&#xA;5 &amp;amp; 10 &amp;amp; 32 \\&#xA;6 &amp;amp; 15 &amp;amp; 64 \\&#xA;7 &amp;amp; 21 &amp;amp; 128 \\&#xA;\end{matrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I personally feel that doing about 10 or 15 additions shouldn't be a problem, so you could get away with a six-row pyramid. However, you seem to feel that this is too much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes, the bounds you provide mean that there are no appropriate sizes for the puzzle because they don't require you to see the trick behind it. It just means that that &lt;em&gt;type&lt;/em&gt; of puzzle is probably off-limits to your audience.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2014-06-13T19:35:52.173" LastActivityDate="2014-06-13T19:35:52.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="464" PostTypeId="1" CreationDate="2014-05-28T23:01:19.797" Score="6" ViewCount="1087" Body="&lt;p&gt;There's a common category of mathematical puzzle which involves determining missing numbers in a &lt;a href=&quot;http://www.wikihow.com/Do-Long-Multiplication&quot; rel=&quot;nofollow&quot;&gt;long multiplication problem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example from &lt;a href=&quot;http://www.edhelper.com/math/missing_digits_14.htm&quot; rel=&quot;nofollow&quot;&gt;this site&lt;/a&gt; (problem 10):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;       1,7__ | (    1736)&#xA; x     _,_43 | (    5843)&#xA; ----------- | &#xA;       5,20_ | (    5208)&#xA;      69,440 | (   69440)&#xA;   1,38_,800 | ( 1388800)&#xA; + _,680,0_0 | ( 8680000)&#xA; ----------- | &#xA;  1_,14_,448 | (10143448)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It's possible without too much difficulty to deduce the original multiplication problem. However, it's fairly easy to see that not all of these numbers are required to achieve the same result. Trivially: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;       1,7__&#xA; x     _,_43&#xA; -----------&#xA;       5,20_&#xA;      _9,44_&#xA;   1,38_,8__&#xA; + _,680,___&#xA; -----------&#xA;  1_,14_,448&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;gives the same result. For a given multiplication problem, how many numbers can I maximally remove to still allow only one unique solution that can determined without brute-force guessing?&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorUserId="12408" LastEditDate="2016-07-31T16:29:26.847" LastActivityDate="2016-07-31T16:29:26.847" Title="How many digits can be removed from a multiplication puzzle and still give only one answer?" Tags="|mathematics|calculation-puzzle|puzzle-creation|missing-digits|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="465" PostTypeId="2" ParentId="461" CreationDate="2014-05-28T23:14:08.230" Score="7" Body="&lt;p&gt;I'm going to look at your question from a slightly different angle, and argue that there &lt;em&gt;is&lt;/em&gt; no correct size for this puzzle; the question isn't a function of the quantity of solving work that's done here, but its very nature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my mind, the core issue with the triangle-path problem as a 'human' puzzle is that it's a little too &lt;em&gt;mechanical&lt;/em&gt;: there's a relatively straightforward linear-time algorithm that gives the solution.  It may take a little while to discover the algorithm, but once you do, no puzzle of this form is going to give you any additional insight over the first; you just have to apply the solution steps that you've determined to the new instance.  (For an example of a released puzzle game with this problem, look at &lt;a href=&quot;https://itunes.apple.com/us/app/strata/id641702337?mt=8&quot;&gt;Strata&lt;/a&gt; on iOS or Mac; the game looks stunning, but the puzzles lose their long-term satisfaction once the algorithm is discovered.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By contrast, the puzzles that retain long-term appeal tend to be those that are &lt;em&gt;algorithmically&lt;/em&gt; hard; that is, those for which no (known) 'fast' algorithm exists, and where good algorithms are conjectured not to exist at all.  For instance, Sudoku is known to be NP-complete, and Rush Hour is known to be &lt;a href=&quot;http://en.wikipedia.org/wiki/PSPACE-complete&quot;&gt;PSPACE-complete&lt;/a&gt; &amp;mdash; while both puzzles are susceptible to 'smart' approaches that use knowledge to bound the search space, they're also both (presumably) hard enough that there's no straightforward mechanical approach to solving them that doesn't involve prohibitive amounts of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, making your puzzle larger would make it take longer to solve &amp;mdash; but it wouldn't make it any &lt;em&gt;harder&lt;/em&gt;, only force the Solver to spend more time on it repeating the same steps.  That's an issue inherent to the puzzle itself, and not one that making it larger can address.&lt;/p&gt;&#xA;" OwnerUserId="266" LastActivityDate="2014-05-28T23:14:08.230" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="467" PostTypeId="2" ParentId="464" CreationDate="2014-05-28T23:19:47.923" Score="1" Body="&lt;p&gt;In a $4 \times 4 $ multiply there are $ 8 $ single digit multiplies and $ 7 $ column additions.  In the sense of simultaneous equations, you would expect to be able to delete $ 15 $ digits and find your way home.  The fact that you know where digits belong, as well as pattern in arithmetic (like a product $5$ must come from $5$ times odd) gives you further information.  I have seen problems like this with &lt;em&gt;very&lt;/em&gt; few given digits that could be solved uniquely.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="153" LastEditDate="2014-06-15T17:59:41.540" LastActivityDate="2014-06-15T17:59:41.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="468" PostTypeId="2" ParentId="462" CreationDate="2014-05-28T23:47:38.997" Score="34" Body="&lt;p&gt;62 is the optimal result, I wrote an (admittedly somewhat hackneyed (and in retrospect not very good)) brute force program in Java which can be found &lt;a href=&quot;http://pastebin.com/mTKrrdq3&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;. It returns that 2 is the smallest number of uncovered squares possible using four queens. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, it places four queens on a board, scans the board for each position, and outputs the case with the maximum number of covered squares. Since we're looking at $(8^2)^4\approx 17\mbox{M}$ combinations, this is totally feasible for a regular computer to execute. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first case my program comes across is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/upOLo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here's a particular case example of this situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://puzzling.stackexchange.com/a/23/20&quot;&gt;Xynariz' question and answer&lt;/a&gt;, we know that the minimum number of queens is five, and that it occurs in this configuration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3MsU5.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, we want to count the number of squares uniquely covered by each queen (that is, how many squares one and only one queen covers for each queen.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do this, we draw a line through each queen. Warning: this gets nasty. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NSHxW.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/RgLag.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be clearly seen that any square with only one line through it is uniquely covered by a queen. Those squares are highlighted with blue on the right. We want to remove the queen that uniquely covers the fewest squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can either follow the lines out from the queens, or count how many blue squares each queen controls. Either way, you'll see that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Red covers one square&lt;/li&gt;&#xA;&lt;li&gt;Blue covers four squares&lt;/li&gt;&#xA;&lt;li&gt;Orange covers six squares&lt;/li&gt;&#xA;&lt;li&gt;Purple covers four squares &lt;/li&gt;&#xA;&lt;li&gt;Green covers four squares.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Therefore, we should remove the red queen from the grid. Removing the red queen uncovers red's uniquely covered square, as well as the queen's square. As a result, the greatest number of squares that can be covered in this case by four queens is $64-2=62$.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-01-27T22:58:18.597" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="469" PostTypeId="1" CreationDate="2014-05-28T23:58:13.227" Score="16" ViewCount="1778" Body="&lt;p&gt;First, a little background: as I noted in &lt;a href=&quot;https://puzzling.stackexchange.com/a/465/266&quot;&gt;my answer to another question here&lt;/a&gt;, most of the puzzles that seem to stay popular for extended periods of time are canonically 'hard' in an algorithmic sense: for instance, Rush Hour is known to be PSPACE-complete, Sudoku is known to be NP-complete, Nurikabe is (&lt;a href=&quot;https://math.stackexchange.com/questions/29992/what-is-the-complexity-of-succinct-binary-nurikabe&quot;&gt;at least&lt;/a&gt;) NP-complete, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is whether there are any good counterexamples to my assertion there; that is, whether there are any fairly popular puzzles that are still 'easy' in an algorithmic sense.  A few examples spring to mind, one that doesn't really fit the rather nebulous abstract-puzzle crtieria I had in mind for this question, but a couple that do:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Word search puzzles have a relatively trivial algorithm that takes O(m * n * s * |S|) time (specifically, comparisons) to find s words of total length |S| in an m&amp;times;n grid.  There are ways to make this somewhat better with string-matching algorithms like the Knuth-Morris-Pratt algorithm, but even the naive approach is usually 'fast enough'.  This isn't really the sort of puzzle I'm after, though &amp;mdash; it's more like a textual Where's Waldo than a deductive exercise.&lt;/li&gt;&#xA;&lt;li&gt;(Classical) Mazes can be solved efficiently by converting the grid into an abstract graph and then using basic pathfinding algorithms (e.g., breadth-first search) to find a route (the shortest route) from start to finish.  This is definitely an example of the sort of thing I'm after &amp;mdash; but I would argue that traditional mazes are rarer than they used to be for exactly this reason, and multistate mazes (which have gotten much harder) are much harder both for human and for algorithmic solvers; in the abstract, there can even be an exponential state-space blowup and I wouldn't be surprised to find hardness results for suitably general multistate mazes &amp;mdash; in fact, the PSPACE-completeness for Rush Hour or hardness for Sokoban could be thought of as examples of this.&lt;/li&gt;&#xA;&lt;li&gt;Rubik's Cube obviously has an algorithmic procedure for solving (in fact, several); on the other hand, it's hard to even speak about hardness results for a 'fixed-n' problem, and my understanding is that solving cube-style puzzles (though possibly not the cube itself) can get algorithmically hard for higher n and/or higher dimensions.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What other examples are there of puzzles that have maintained popularity despite having relatively mechanical solutions?&lt;/p&gt;&#xA;" OwnerUserId="266" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.923" LastActivityDate="2023-09-10T02:45:50.357" Title="Are there any popular puzzles with polynomial-time algorithms?" Tags="|solvability|puzzle-history|difficulty-rating|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="470" PostTypeId="2" ParentId="464" CreationDate="2014-05-29T00:10:09.877" Score="10" Body="&lt;p&gt;Kind of cheating, but...&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;       1,736 | (    1736)&#xA; x     5,843 | (    5843)&#xA; ----------- | &#xA;       _,___ | (    5208)&#xA;      __,___ | (   69440)&#xA;   _,___,___ | ( 1388800)&#xA; + _,___,___ | ( 8680000)&#xA; ----------- | &#xA;  __,___,___ | (10143448)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Answers the question in its literal form. ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, note that with lots of zeroes in the multiplicands this works:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;      _,___ | (   1000)&#xA;x     _,___ | (   1000)&#xA;-----------&#xA;  _,___,___ | (1000000)&#xA;-----------&#xA;  1,___,___ | (1000000)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="39" LastEditorUserId="39" LastEditDate="2014-05-29T02:48:47.793" LastActivityDate="2014-05-29T02:48:47.793" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="471" PostTypeId="2" ParentId="462" CreationDate="2014-05-29T01:43:28.883" Score="16" Body="&lt;p&gt;&lt;strong&gt;TL;DR:&lt;/strong&gt; The maximum number of squares that can be covered by 4 queens on a 8x8 board is &lt;strong&gt;62&lt;/strong&gt;, achievable through &lt;em&gt;64&lt;/em&gt; different board orientations. By &quot;different&quot; I mean that the mirror image/rotation of one of the orientations does not produce another that is counted.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I also used a brute-force approach, using the following Python 2.7 code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from itertools import combinations as combs&#xA;&#xA;def covered_set(row, col):&#xA;    controlled = set()&#xA;&#xA;    for i in range(1,9):&#xA;        controlled.add((row, i))&#xA;&#xA;    for i in range(1,9):&#xA;        controlled.add((i, col))&#xA;&#xA;    x = 1&#xA;    while row + x &amp;lt;= 8 and col + x &amp;lt;= 8:&#xA;        controlled.add((row + x, col + x))&#xA;        x += 1&#xA;&#xA;    x = 1&#xA;    while row - x &amp;gt;= 1 and col - x &amp;gt;= 1:&#xA;        controlled.add((row - x, col - x))&#xA;        x += 1&#xA;&#xA;    x = 1&#xA;    y = 1&#xA;    while row + x &amp;lt;= 8 and col - y &amp;gt;= 1:&#xA;        controlled.add((row + x, col - y))&#xA;        x += 1&#xA;        y += 1&#xA;&#xA;    x = 1&#xA;    y = 1&#xA;    while row - x &amp;gt;= 1 and col + y &amp;lt;= 8:&#xA;        controlled.add((row - x, col + y))&#xA;        x += 1&#xA;        y += 1&#xA;&#xA;    return controlled&#xA;&#xA;max_covered = 0&#xA;arrangement = []&#xA;for a,b,c,d in combs(range(64), 4):&#xA;    row1, col1 = divmod(a, 8)&#xA;    row2, col2 = divmod(b, 8)&#xA;    row3, col3 = divmod(c, 8)&#xA;    row4, col4 = divmod(d, 8)&#xA;&#xA;    controlled = len(covered_set(row1+1, col1+1) |&#xA;                     covered_set(row2+1, col2+1) |&#xA;                     covered_set(row3+1, col3+1) |&#xA;                     covered_set(row4+1, col4+1))&#xA;&#xA;    if controlled &amp;gt; max_covered:&#xA;        max_covered = controlled&#xA;        arrangement = [((row1+1, col1+1), (row2+1, col2+1), (row3+1, col3+1), (row4+1, col4+1))]&#xA;    elif controlled == max_covered:&#xA;        arrangement.append(((row1+1, col1+1), (row2+1, col2+1), (row3+1, col3+1), (row4+1, col4+1)))&#xA;&#xA;print max_covered&#xA;for v in arrangement:&#xA;    print v&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Output:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;62&#xA;((1, 1), (2, 5), (5, 8), (8, 2))&#xA;((1, 1), (2, 8), (5, 2), (8, 5))&#xA;((1, 1), (2, 8), (5, 4), (6, 5))&#xA;((1, 1), (3, 5), (5, 7), (7, 3))&#xA;((1, 1), (3, 7), (5, 3), (7, 5))&#xA;((1, 1), (4, 5), (5, 6), (6, 4))&#xA;((1, 1), (4, 5), (5, 6), (8, 2))&#xA;((1, 1), (4, 6), (5, 4), (6, 5))&#xA;((1, 2), (2, 6), (5, 1), (6, 5))&#xA;((1, 2), (4, 6), (5, 5), (8, 1))&#xA;((1, 2), (4, 8), (7, 5), (8, 1))&#xA;((1, 3), (2, 7), (5, 2), (6, 6))&#xA;((1, 3), (2, 7), (5, 2), (7, 5))&#xA;((1, 3), (3, 7), (5, 1), (7, 5))&#xA;((1, 4), (2, 8), (5, 3), (6, 7))&#xA;((1, 4), (3, 8), (5, 2), (7, 6))&#xA;((1, 4), (4, 7), (7, 1), (8, 8))&#xA;((1, 5), (2, 1), (5, 6), (6, 2))&#xA;((1, 5), (3, 1), (5, 7), (7, 3))&#xA;((1, 5), (4, 2), (7, 8), (8, 1))&#xA;((1, 6), (2, 2), (5, 7), (6, 3))&#xA;((1, 6), (2, 2), (5, 7), (7, 4))&#xA;((1, 6), (3, 2), (5, 8), (7, 4))&#xA;((1, 7), (2, 3), (5, 8), (6, 4))&#xA;((1, 7), (4, 1), (7, 4), (8, 8))&#xA;((1, 7), (4, 3), (5, 4), (8, 8))&#xA;((1, 8), (2, 1), (5, 5), (6, 4))&#xA;((1, 8), (2, 1), (5, 7), (8, 4))&#xA;((1, 8), (2, 4), (5, 1), (8, 7))&#xA;((1, 8), (3, 2), (5, 6), (7, 4))&#xA;((1, 8), (3, 4), (5, 2), (7, 6))&#xA;((1, 8), (4, 3), (5, 5), (6, 4))&#xA;((1, 8), (4, 4), (5, 3), (6, 5))&#xA;((1, 8), (4, 4), (5, 3), (8, 7))&#xA;((2, 2), (3, 6), (6, 1), (7, 5))&#xA;((2, 2), (4, 7), (6, 1), (7, 5))&#xA;((2, 3), (4, 7), (6, 1), (8, 5))&#xA;((2, 3), (4, 7), (6, 5), (8, 1))&#xA;((2, 4), (3, 8), (5, 2), (7, 7))&#xA;((2, 4), (3, 8), (6, 3), (7, 7))&#xA;((2, 4), (4, 6), (6, 2), (8, 8))&#xA;((2, 4), (4, 7), (7, 2), (8, 6))&#xA;((2, 4), (4, 8), (6, 2), (8, 6))&#xA;((2, 5), (3, 1), (5, 7), (7, 2))&#xA;((2, 5), (3, 1), (6, 6), (7, 2))&#xA;((2, 5), (4, 1), (6, 7), (8, 3))&#xA;((2, 5), (4, 2), (7, 7), (8, 3))&#xA;((2, 5), (4, 3), (6, 7), (8, 1))&#xA;((2, 6), (4, 2), (6, 4), (8, 8))&#xA;((2, 6), (4, 2), (6, 8), (8, 4))&#xA;((2, 7), (3, 3), (6, 8), (7, 4))&#xA;((2, 7), (4, 2), (6, 8), (7, 4))&#xA;((3, 2), (4, 6), (7, 1), (8, 5))&#xA;((3, 3), (4, 7), (7, 2), (8, 6))&#xA;((3, 4), (4, 5), (5, 3), (8, 8))&#xA;((3, 4), (4, 5), (7, 1), (8, 8))&#xA;((3, 4), (4, 6), (5, 5), (8, 1))&#xA;((3, 4), (4, 8), (7, 3), (8, 7))&#xA;((3, 5), (4, 1), (7, 6), (8, 2))&#xA;((3, 5), (4, 3), (5, 4), (8, 8))&#xA;((3, 5), (4, 4), (5, 6), (8, 1))&#xA;((3, 5), (4, 4), (7, 8), (8, 1))&#xA;((3, 6), (4, 2), (7, 7), (8, 3))&#xA;((3, 7), (4, 3), (7, 8), (8, 4))&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The first line is the maximum number of squares covered, the following lines are the 1-based &lt;code&gt;(row, col)&lt;/code&gt; positions of the four queens that achieve the optimal covering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the answer I received is &lt;strong&gt;62&lt;/strong&gt;, achieved as such (the two squares that are &lt;em&gt;not&lt;/em&gt; controlled I have highlighted with a red circle):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uulyP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="198" LastEditorUserId="-1" LastEditDate="2015-01-15T12:46:52.730" LastActivityDate="2015-01-15T12:46:52.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="472" PostTypeId="2" ParentId="469" CreationDate="2014-05-29T03:29:26.513" Score="9" Body="&lt;p&gt;I suspect the complexity class is not too important.  In the example of Sudoku, the NP-complete problem is &amp;quot;given a partially filled in &lt;span class=&quot;math-container&quot;&gt;$n \times n$&lt;/span&gt; grid, state whether it can be completed to a legal arrangement&amp;quot;.  The popular puzzle is &amp;quot;given a partially filled in &lt;span class=&quot;math-container&quot;&gt;$9 \times 9 $&lt;/span&gt; grid that is known to have a unique solution, find that solution&amp;quot;.  In the popular puzzle, we don't care how the complexity depends on &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;, just that the size we do has the correct difficulty.  Similarly, the substitution ciphers that are in the daily paper have &lt;span class=&quot;math-container&quot;&gt;$ 26! $&lt;/span&gt; possible keys, an enormous number, but the structure of English makes them soluble.&lt;/p&gt;&#xA;&lt;p&gt;Jumble, cryptarithms, and bridge problems are some that have stayed around and seem to have a lower complexity class.&lt;/p&gt;&#xA;&lt;p&gt;Added much later:  one of the things that makes Sudoku appealing is that if you find one thing you don't know where in the grid to look for the thing(s) you should now find.  This seems to be a common thread for the NP-hard problems, that you need to find a global solution because a problem in one area can propagate to affect the solution in other areas.  I think it makes puzzles more appealing.  In that sense there is a common thread between interesting puzzles and complexity class.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditDate="2023-09-10T02:45:50.357" LastActivityDate="2023-09-10T02:45:50.357" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="475" PostTypeId="1" AcceptedAnswerId="493" CreationDate="2014-05-29T06:50:54.980" Score="5" ViewCount="7222" Body="&lt;p&gt;Is there an algorithm to help me solve the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rubik%27s_360&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rubik's 360&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Sx70w.jpg&quot; alt=&quot;Image of Rubik&amp;#39;s 360&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually I just do it at random, but I rarely succeed.&lt;/p&gt;&#xA;" OwnerUserId="50" LastEditorUserId="25966" LastEditDate="2016-12-27T21:05:43.427" LastActivityDate="2016-12-27T23:39:51.110" Title="Is there an algorithm for the Rubik's 360?" Tags="|strategy|algorithm|twisty-puzzles|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="476" PostTypeId="1" CreationDate="2014-05-29T07:09:29.900" Score="14" ViewCount="1055" Body="&lt;p&gt;&lt;em&gt;First, I apologise for asking a question that's not strictly in the right Q&amp;amp;A format; I'm hoping there are quality answers anyway.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any puzzle or solitaire games based on chess? By this I mean&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Does not require more than one intelligent player (e.g. if there is a second player, their moves can be mechanical or based on simple rules)&lt;/li&gt;&#xA;&lt;li&gt;Based on the rules of chess (or chess variants such as xiangqi): either using real chess pieces, the rules of capture, or maybe even novelty pieces&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For example, the 8 queens puzzle might qualify, although in my opinion it's not very fun to play as a game. There's also a toy called &quot;Solitaire Chess&quot; based on a 4x4 board and the objective is to only use capturing moves until only one piece is left, just like solitaire. Are there any more like this?&lt;/p&gt;&#xA;" OwnerUserId="104" LastActivityDate="2019-12-02T20:11:27.703" Title="Are there any puzzle/solitaire games based on chess?" Tags="|chess|" AnswerCount="6" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="477" PostTypeId="5" CreationDate="2014-05-29T07:57:12.243" Score="0" Body="&lt;h2&gt;What is a Twisty Puzzle?&lt;/h2&gt;&#xA;&lt;p&gt;A twisty puzzle is another term for a &lt;a href=&quot;https://en.wikipedia.org/wiki/Combination_puzzle&quot; rel=&quot;nofollow noreferrer&quot;&gt;combination puzzle&lt;/a&gt; (also known as a sequential move puzzle). It is a puzzle that consists of a set of pieces which can be manipulated into different combinations by a group of operations. Many such puzzles are mechanical puzzles of polyhedral shape, consisting of multiple layers of pieces along each axis which can rotate independently of each other.&lt;/p&gt;&#xA;&lt;p&gt;The archetype of this kind of puzzle is the Rubik's Cube. Each rotating side is usually marked with different colors, intended to be scrambled, then 'solved' by a sequence of moves that sort the facets by color. As a generalization, combination puzzles also include mathematically defined examples either not yet or impossible to physically construct (e.g. a 4D twisty puzzle such as a tesseract version of the Rubik's cube).&lt;/p&gt;&#xA;&lt;h5&gt;How do I solve a twisty puzzle?&lt;/h5&gt;&#xA;&lt;p&gt;A combination puzzle is solved by achieving a particular combination, starting from a random (scrambled) combination. Often, the solution is required to be some recognizable pattern such as &amp;quot;all like colors together&amp;quot; or &amp;quot;all numbers in order&amp;quot;.&lt;/p&gt;&#xA;&lt;h5&gt;What are some examples of twisty puzzles?&lt;/h5&gt;&#xA;&lt;p&gt;The most common example of a twisty puzzle is the Rubik's cube, but there are many others, such as:&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th&gt;Picutre&lt;/th&gt;&#xA;&lt;th&gt;Name&lt;/th&gt;&#xA;&lt;th&gt;Comments&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;&lt;a href=&quot;https://i.stack.imgur.com/ESmSa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ESmSa.png&quot; alt=&quot;Picture of a 3x3x3 Rubik's cube.&quot; /&gt;&lt;/a&gt;&lt;/td&gt;&#xA;&lt;td&gt;Rubik's Cube&lt;/td&gt;&#xA;&lt;td&gt;The original Rubik's Cube.&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;&lt;a href=&quot;https://i.stack.imgur.com/Fnt1p.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fnt1p.png&quot; alt=&quot;Picture of a 4x4x4 Rubik's cube.&quot; /&gt;&lt;/a&gt;&lt;/td&gt;&#xA;&lt;td&gt;Rubik's Revenge&lt;/td&gt;&#xA;&lt;td&gt;Solution is much the same as 3×3×3 cube except additional (and relatively simple) algorithm(s) are required to unscramble the centre pieces and edges and additional parity not seen on the 3x3x3 Rubik's Cube.&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;&lt;a href=&quot;https://i.stack.imgur.com/nH4Ic.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nH4Ic.png&quot; alt=&quot;Picture of a computer rendering of a tesseract puzzle.&quot; /&gt;&lt;/a&gt;&lt;/td&gt;&#xA;&lt;td&gt;Tesseract&lt;/td&gt;&#xA;&lt;td&gt;This is the 4-dimensional analog of a cube and thus cannot actually be constructed. However, it can be drawn or represented by a computer. Significantly more difficult to solve than the standard cube, although the techniques follow much the same principles. There are many other sizes of virtual cuboid puzzles ranging from the trivial 3×3 to the 5-dimensional 7×7×7×7×7 which has only been solved twice so far.&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;&lt;a href=&quot;https://i.stack.imgur.com/KYtpy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KYtpy.png&quot; alt=&quot;Picture of a calendar cube.&quot; /&gt;&lt;/a&gt;&lt;/td&gt;&#xA;&lt;td&gt;Calendar Cube&lt;/td&gt;&#xA;&lt;td&gt;Mechanically identical to the standard 3×3×3 cube, but with specially printed stickers for displaying the date. Much easier to solve since five of the six faces are ignored. Ideal produced a commercial version during the initial cube craze. Sticker sets are also available for converting a normal cube into a calendar.&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;&lt;a href=&quot;https://i.stack.imgur.com/3XXaH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3XXaH.png&quot; alt=&quot;Picture of a sudoku cube.&quot; /&gt;&lt;/a&gt;&lt;/td&gt;&#xA;&lt;td&gt;Sudoku Cube&lt;/td&gt;&#xA;&lt;td&gt;Identical to the Rubik's Cube in mechanical function, it adds another layer of difficulty in that the numbers must all have the same orientation and there are no colors to follow. The name reflects its superficial resemblance to the two-dimensional Sudoku number puzzle.&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;&lt;a href=&quot;https://i.stack.imgur.com/UB0WE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UB0WE.png&quot; alt=&quot;Picture of a skewb.&quot; /&gt;&lt;/a&gt;&lt;/td&gt;&#xA;&lt;td&gt;Skewb&lt;/td&gt;&#xA;&lt;td&gt;Similar to the original Rubik's Cube, the Skewb differs in that its four axes of rotation pass through the corners of the cube rather than the centers of the faces. As a result, it is a deep-cut puzzle in which each twist scrambles all six faces.&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;&lt;a href=&quot;https://i.stack.imgur.com/ALYQ6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ALYQ6.png&quot; alt=&quot;Picture of a pyraminx.&quot; /&gt;&lt;/a&gt;&lt;/td&gt;&#xA;&lt;td&gt;Pyraminx&lt;/td&gt;&#xA;&lt;td&gt;Tetrahedral-shaped puzzle with axes on the corners and trivial tips. It was invented in 1970 by Uwe Mèffert.&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;&lt;a href=&quot;https://i.stack.imgur.com/Yni5L.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Yni5L.png&quot; alt=&quot;Picture of a megaminx.&quot; /&gt;&lt;/a&gt;&lt;/td&gt;&#xA;&lt;td&gt;Megaminx&lt;/td&gt;&#xA;&lt;td&gt;12-sided polyhedron puzzle similar to Rubik's Cube in operation and solution.&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;&lt;a href=&quot;https://i.stack.imgur.com/5Am6C.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5Am6C.png&quot; alt=&quot;Picture of an impossiball.&quot; /&gt;&lt;/a&gt;&lt;/td&gt;&#xA;&lt;td&gt;Impossiball&lt;/td&gt;&#xA;&lt;td&gt;Rounded icosahedron puzzle similar to Pocket Cube in operation and solution.&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;&lt;a href=&quot;https://i.stack.imgur.com/xx5e1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xx5e1.png&quot; alt=&quot;Picture of a Rubik's snake.&quot; /&gt;&lt;/a&gt;&lt;/td&gt;&#xA;&lt;td&gt;Rubik's Snake&lt;/td&gt;&#xA;&lt;td&gt;Some would not count this as a combinational puzzle though it bears the Rubik name. Also known as Rubik's Twist. There is no one solution to this puzzle but multiple different shapes can be made.&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;&lt;a href=&quot;https://i.stack.imgur.com/FklKF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FklKF.png&quot; alt=&quot;Picture of a traditional sliding puzzle.&quot; /&gt;&lt;/a&gt;&lt;/td&gt;&#xA;&lt;td&gt;Sliding Puzzle&lt;/td&gt;&#xA;&lt;td&gt;These ubiquitous puzzles come in many sizes and designs. The traditional design is with numbers and the solution forms a magic square. There have been many different designs, the example shown here uses graphic symbols instead of numbers. The solution requires that there are no repeated symbols in any row, column or diagonal. The picture shows the puzzle unsolved.&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;&lt;a href=&quot;https://i.stack.imgur.com/JGgVi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JGgVi.png&quot; alt=&quot;Picture of a fifteen puzzle.&quot; /&gt;&lt;/a&gt;&lt;/td&gt;&#xA;&lt;td&gt;Fifteen Puzzle&lt;/td&gt;&#xA;&lt;td&gt;The original sliding piece puzzle.&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;&lt;em&gt;A related tag that can additionally be used for these puzzles is &lt;a href=&quot;/questions/tagged/mechanical-puzzles&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;mechanical-puzzles&amp;#39;&quot; aria-label=&quot;show questions tagged &amp;#39;mechanical-puzzles&amp;#39;&quot; rel=&quot;tag&quot; aria-labelledby=&quot;mechanical-puzzles-container&quot;&gt;mechanical-puzzles&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;h5&gt;References&lt;/h5&gt;&#xA;&lt;p&gt;Wikipedia contributors. (2021, October 16). Combination puzzle. In Wikipedia, The Free Encyclopedia. Retrieved 14:17, October 18, 2021, from &lt;a href=&quot;https://en.wikipedia.org/w/index.php?title=Combination_puzzle&amp;amp;oldid=1050187792&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/w/index.php?title=Combination_puzzle&amp;amp;oldid=1050187792&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="146" LastEditorUserId="52483" LastEditDate="2023-01-31T19:25:33.833" LastActivityDate="2023-01-31T19:25:33.833" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="478" PostTypeId="4" CreationDate="2014-05-29T07:57:12.243" Score="0" Body="Twisty puzzles consist of a set of pieces which can be manipulated into different combinations by a group of operations. Use this tag for puzzles similar to the archetype of combination puzzles, the Rubik's cube." OwnerUserId="146" LastEditorUserId="51105" LastEditDate="2021-10-18T14:58:27.557" LastActivityDate="2021-10-18T14:58:27.557" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="479" PostTypeId="5" CreationDate="2014-05-29T07:59:28.547" Score="0" Body="&lt;p&gt;A puzzle whose solution is a methodological plan of action for realizing a specific goal as effectively as possible. The solution, rather than being an object or word or single action, specifies how to act in a multi-step interaction, basing each action off of previous actions not under your control. Often, this is in a turn-based game played against an adversary.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2014-11-22T06:49:53.567" LastActivityDate="2014-11-22T06:49:53.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="480" PostTypeId="4" CreationDate="2014-05-29T07:59:28.547" Score="0" Body="A puzzle whose solution is a methodological plan of action for realizing a specific goal." OwnerUserId="4219" LastEditorUserId="4551" LastEditDate="2014-11-22T06:59:16.157" LastActivityDate="2014-11-22T06:59:16.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="482" PostTypeId="2" ParentId="433" CreationDate="2014-05-29T11:27:10.297" Score="3" Body="&lt;p&gt;I think the &quot;chimp-and-zee&quot; idea posited in the other answers if possible. But it could also just be a coincidence, and all that's intended is that &quot;chimpanzee, say&quot; is indicating APE. This has the advantage that APE really is exactly at the centre of TRAPEZE. Certainly a 'good' British cryptic wouldn't use 'centre' to mean anything other than the exact middle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's certainly no rule that the 'wordplay' constituent of a clue must indicate every letter of the answer. It would be a matter of judgement for the setter whether the indication given is enough for the clue to be fair, taking into account the obscureness of the answer, the number of possible alternatives for the definition given, the helpfulness of the letters provided by crossing clues, etc. For what it's worth I'd say this clue is borderline but fair. It would be better reading &quot;...that has...&quot; to more clearly indicate that what follows is a description of the answer rather than a full instruction for arriving at it.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2014-05-29T11:27:10.297" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="483" PostTypeId="2" ParentId="455" CreationDate="2014-05-29T13:42:24.630" Score="11" Body="&lt;p&gt;The princess moves 1 room (never 0, never 2) each day. So does the prince. To meet, they have to be at distance 2, and move toward each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, their distance either stays constant, increases by 2, or decreases by 2, but the oddity is unchanged, except once on day 16.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the initial distance is even (first case discussed in the solution), it will always stay even, which is why the princess can never get past the prince, and therefore gets stuck in the final room. That also implies that the princess is not in room 1 (distance is 1), but only in rooms (2, 4, 6, 8, 10, 12, 14, 16).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the initial distance is odd, just move forward to the 15th day. At this point, the prince does not move but the princess still does, changing their distance by only 1, making it even, reducing to the initial problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back to your specific question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If we begin knocking at door #2 and the princess is in room #3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The distance in this case is 1, which is odd, meaning the prince will necessarily miss the princess in the first 15 days. (try it. the princess can skip the prince at any moment, and the prince can never get to here in the first part.). But as soon as day 16, the distance becomes even, and the prince is sure to catch up with the princess at some point&lt;/p&gt;&#xA;" OwnerUserId="310" LastActivityDate="2014-05-29T13:42:24.630" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="484" PostTypeId="1" AcceptedAnswerId="487" CreationDate="2014-05-29T14:56:13.633" Score="13" ViewCount="2124" Body="&lt;p&gt;One of Oskar van Deventer's puzzles is the gear maze 2x2.There are 4 gears connected together so if you move one gear all four gears rotate (two clockwise, the other two counter-clockwise).On the gears there is a maze and the object is to move a ball through this maze.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kCkCZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my question is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Is there an (easy) way to calculate the solution by  using an excel-based spreadsheet or …&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To make things not too complicated let us say:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;one gear consists of (only) 12 teeth&lt;/li&gt;&#xA;&lt;li&gt;so each gear is parted in 12 segments&lt;/li&gt;&#xA;&lt;li&gt;the connecting points of the gears are defined e. g.&lt;br&gt;&#xA;1(red) &amp;lt;-&gt; 1(blue); 4(blue) &amp;lt;-&gt; 4(green); 1(green) &amp;lt;-&gt;1(orange); 4(orange) &amp;lt;-&gt; 4(red)&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the paths of the maze start and end on a segment and do not cross each other&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tG0G3.png&quot; width=&quot;500&quot; /&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="143" LastEditorUserId="8874" LastEditDate="2015-02-06T07:54:48.347" LastActivityDate="2015-02-06T07:54:48.347" Title="How to solve Oskar van Deventer's gear maze 2x2" Tags="|mechanical-puzzles|mazes|geared-puzzles|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="485" PostTypeId="2" ParentId="484" CreationDate="2014-05-29T16:42:38.553" Score="5" Body="&lt;p&gt;I count (but you should check) that there are 23 teeth on each gear.  The exact number is not important, but the fact that there are the same number on each gear is.  That means that the paths will line up the same way every revolution and simplifies things considerably.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can label each end of a path through a gear like you are doing.  On the red gear I find 16 ends.  Most ends can connect with three other ends-one on the same gear and one on each neighboring gear.  Make a table with each end and the ones it can connect to.  Start and End can only connect to two other ends.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can do a &lt;a href=&quot;http://en.wikipedia.org/wiki/Breadth-first_search&quot; rel=&quot;nofollow&quot;&gt;breadth-first search&lt;/a&gt;.  I would start from both start and end.  From start, list the places you can go and color them red.  From end, list the places you can go and color them green.  Now from each red one, list where you can go and color those red and so on.  When you find a link between the red and green ends, you have a path.  You will probably find a path with about seven links, as the number of colored spots about triples each time.  After three links you have colored 1+2+6+18=27 vertices unless there are multiple paths to the same point.  Tow of them should touch and you are done.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditDate="2014-05-29T17:29:25.160" LastActivityDate="2014-05-29T17:29:25.160" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="486" PostTypeId="2" ParentId="455" CreationDate="2014-05-29T19:02:14.897" Score="119" Body="&lt;p&gt;Ilmari's answer covers the logic of the answer perfectly, but in case anyone's still confused, here's a version of the diagram I find more intuitive (I made it myself while solving the puzzle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Pink squares are rooms where the princess could be on the given day, the blue squares are where the prince knocks that day, and the black squares are rooms in which she logically cannot be. On day 30 all rooms but room 2 have been eliminated, meaning that if the prince has not found the princess already, he will find her there on day 30.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ybByw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="320" LastActivityDate="2014-05-29T19:02:14.897" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="487" PostTypeId="2" ParentId="484" CreationDate="2014-05-29T21:53:58.020" Score="11" Body="&lt;p&gt;Following &lt;a href=&quot;https://puzzling.stackexchange.com/a/485/12&quot;&gt;Ross Millikan's suggestion&lt;/a&gt;, I numbered the segments of the original puzzle as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pva8K.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pva8K.png&quot; alt=&quot;Gear Maze 2x2 with numbered sections&quot;&gt;&lt;/a&gt;&#xA;&lt;sub&gt;(Source: &lt;a href=&quot;http://www.laserexact.nl/images/stories/virtuemart/product/gear%20maze%202x2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.laserexact.nl/images/stories/virtuemart/product/gear%20maze%202x2.jpg&lt;/a&gt;, modified)&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The connections between the segments are given in the following table:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/niTOb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/niTOb.png&quot; alt=&quot;Table of all connections between segments in the Gear Maze 2x2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possible solution (there are several) is marked in bold. To follow it, start at 1(orange) and&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;for a vertical line, change to the accordingly coloured gear, or&lt;/li&gt;&#xA;&lt;li&gt;for a horizontal line, traverse the path on the same gear&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;until you reach the goal 19(green).&lt;/p&gt;&#xA;" OwnerUserId="12" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2014-05-29T21:53:58.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="488" PostTypeId="2" ParentId="455" CreationDate="2014-05-29T21:57:17.150" Score="7" Body="&lt;p&gt;I find it easier to understand the solution by remarking at the onset that there is an interesting parity invariant: the sum of the day number and the princess's room number always has the same parity (it's either always even or always odd). This holds because every time the day number increases by 1, the room number either decreases or increases by 1, so their sum increases by either 2 or 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that the sum is always odd. On day 1, the prince visits room 2; on day 2, he visits room 3, and so on. Notice how his visits have odd parity too, and the day+room sum of the prince increases by 2 every day. The princess's sum cannot have started smaller than the prince's, since on day 1 and with rooms numbered from 1 the smallest possible sum is 3. So the prince is pushing the princess towards higher-numbered rooms, until on day 15 he visits room 16, which is the highest possible odd sum. Thus, if the princess has an odd sum, the prince will flush her out no later than day 15. If the princess has an even sum, these 15 days are for nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On day 16, the prince starts exactly the same procedure, but with even sums. It doesn't actually matter whether he starts with door 2 or door 16. By the same phenomenon as before, after 15 days of going from the second door to the next-to-last door, the prince will flush out the princess if her sum is even, no matter where she started out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's important to realize that it doesn't matter if the princess starts in a high-numbered room and crosses the prince during the first phase. If the princess has an odd sum, then the two will meet in the middle. If she doesn't meet the prince somewhere, it means that she had an even sum, and he'll catch her during the second phase.&lt;/p&gt;&#xA;" OwnerUserId="71" LastActivityDate="2014-05-29T21:57:17.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="489" PostTypeId="2" ParentId="476" CreationDate="2014-05-29T22:04:35.337" Score="2" Body="&lt;p&gt;The ones I've seen mostly show a board position (often one that occurred in a real match) and the challenge is to figure out how to force checkmate in some given number of moves. These are often seen in one-a-day calendars. A quick &lt;a href=&quot;https://www.google.com/search?q=chess+puzzles&quot; rel=&quot;nofollow&quot;&gt;google search&lt;/a&gt; shows a number of online versions, though I'm having trouble finding a calendar version quickly.&lt;/p&gt;&#xA;" OwnerUserId="4" LastActivityDate="2014-05-29T22:04:35.337" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="490" PostTypeId="2" ParentId="476" CreationDate="2014-05-29T22:15:08.843" Score="8" Body="&lt;p&gt;Besides the classic &quot;Mate in 2&quot; or &quot;White to play and win&quot; puzzles there are &lt;a href=&quot;http://en.wikipedia.org/wiki/Retrograde_analysis&quot;&gt;retrograde analysis&lt;/a&gt; problems, where you are called upon to figure out what the last move was or how a position was achieved.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2014-05-29T22:15:08.843" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="491" PostTypeId="1" CreationDate="2014-05-29T23:28:09.173" Score="9" ViewCount="237" Body="&lt;p&gt;I was playing a &lt;a href=&quot;http://en.wikipedia.org/wiki/Set_(game)&quot; rel=&quot;nofollow&quot;&gt;game of Set&lt;/a&gt; with a group.  We had the usual twelve cards out and all agreed that there was no set.  We then increased the number of cards to fifteen and were still unable to find a set.  (Unfortunately I no longer have a picture of the board in question, so it's impossible to verify.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this even possible?  Were we all just really, really tired?  What are the odds of this occurring?&lt;/p&gt;&#xA;" OwnerUserId="67" LastEditorUserId="15544" LastEditDate="2015-09-11T07:35:02.723" LastActivityDate="2015-09-11T07:35:02.723" Title="What are the odds that there is no solution on a given Set board?" Tags="|probability|cards|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="492" PostTypeId="2" ParentId="491" CreationDate="2014-05-29T23:50:44.327" Score="3" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Set_%28game%29#Basic_combinatorics_of_Set&quot; rel=&quot;nofollow&quot;&gt;Wikipedia says&lt;/a&gt; the chance of no set being present in 15 cards when playing is $88:1$.  This assumes that the first twelve cards have no set.  It also says you can find twenty cards with no set.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="4204" LastEditDate="2014-12-04T03:45:46.647" LastActivityDate="2014-12-04T03:45:46.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="493" PostTypeId="2" ParentId="475" CreationDate="2014-05-29T23:52:25.337" Score="4" Body="&lt;p&gt;To get the balls to the ball layer so they can be dropped to the outer pockets, you need to flip the inner ball so the counterweight is pointing up after which it catches and will only rotate one way. This can be done by quickly rotating the entire ball around. I do this by quickly rotating the ball from where the black is horizontal to where the are vertical around 270° around 2 of the pockets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll know that it catches when the hole and counterweight of middle ball are horizontal when the black knobs are vertical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then guide one of the lower color balls into the hole and gently let it line up with the closest hole. Which will put it in the correct layer. After that it is easy to drop that ball into the correct pocket. Remember to not turn it around or you will lose the ball again.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2014-05-29T23:52:25.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="494" PostTypeId="5" CreationDate="2014-05-30T01:47:07.693" Score="0" Body="&lt;p&gt;For questions about the card game &lt;a href=&quot;http://en.wikipedia.org/wiki/Set_%28game%29&quot; rel=&quot;nofollow&quot;&gt;Set&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Set, cards have various colored symbols on them, with four different properties:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Shape: oval, squiggle, diamond&lt;/li&gt;&#xA;&lt;li&gt;Color: green, red, purple&lt;/li&gt;&#xA;&lt;li&gt;Shading: empty, lines, filled&lt;/li&gt;&#xA;&lt;li&gt;Quantity: one, two, or three shapes&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For each property, the cards must either share the same type, or be distinct types. For instance, you can have one oval, one squiggle, and one diamond; or, all three are ovals, squiggles, or diamonds. However, you could not have two ovals and one diamond.&lt;/p&gt;&#xA;" OwnerUserId="10" LastEditorDisplayName="user20" LastEditDate="2014-06-26T06:54:39.060" LastActivityDate="2014-06-26T06:54:39.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="495" PostTypeId="4" CreationDate="2014-05-30T01:47:07.693" Score="0" Body="Set is a game in which players form sets of matching cards to score points." OwnerUserId="10" LastEditorDisplayName="user20" LastEditDate="2014-06-26T06:54:39.060" LastActivityDate="2014-06-26T06:54:39.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="496" PostTypeId="1" CreationDate="2014-05-30T02:27:59.343" Score="29" ViewCount="9323" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Once upon a time, and old lady went to sell her vast quantity of eggs at the local market. When asked how many she had, she replied:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Son, I can't count past 100 but I know that:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If you divide the number of eggs by 2 there will be one egg left.&lt;br&gt;&#xA;  If you divide the number of eggs by 3 there will be one egg left.&lt;br&gt;&#xA;  If you divide the number of eggs by 4 there will be one egg left.&lt;br&gt;&#xA;  If you divide the number of eggs by 5 there will be one egg left.&lt;br&gt;&#xA;  If you divide the number of eggs by 6 there will be one egg left.&lt;br&gt;&#xA;  If you divide the number of eggs by 7 there will be one egg left.&lt;br&gt;&#xA;  If you divide the number of eggs by 8 there will be one egg left.&lt;br&gt;&#xA;  If you divide the number of eggs by 9 there will be one egg left.&lt;br&gt;&#xA;  If you divide the number of eggs by 10 there will be one egg left.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Finally. If you divide the number of eggs by 11 there will be NO EGGS left!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;How many eggs did the old lady have?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Source (with solution): &lt;a href=&quot;http://www.brainbashers.com/showanswer.asp?ref=ZFMU&quot;&gt;BrainBashers&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From reading the puzzle, some easily-indentifiable properties of the total number of eggs are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;it's a multiple of 11 (remainder of 0 if divided by 11)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;it's odd (if divided by 2 there's one egg remaining)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I haven't been able to come up with any other clues, so the only way I can see to solve this puzzle is to brute-force it (that is, start listing odd multiples of 11 and manually check if &lt;code&gt;x modulo 3-&amp;gt;10 = 1&lt;/code&gt;. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;11, 33, 55, 77, 99, 121, 143...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;At this point I got bored of brute-forcing it and decided there had to be a better way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there? Can I use the other clues to build more rules to make it easier to find the correct number? (I started to think that maybe combining two clues would lead to a nugget of information, but then I couldn't figure out how to do that.) What math can I use to solve this other than brute-forcing the set of odd multiples of 11?&lt;/p&gt;&#xA;" OwnerUserId="10" LastEditorDisplayName="user88" LastEditDate="2014-05-30T13:41:55.990" LastActivityDate="2021-07-19T17:39:59.290" Title="Aside from brute force, how can I solve this puzzle?" Tags="|calculation-puzzle|mathematics|word-problem|" AnswerCount="7" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="497" PostTypeId="2" ParentId="496" CreationDate="2014-05-30T02:45:57.447" Score="59" Body="&lt;p&gt;Remove one egg. Now the number of eggs is divisible by 2, 3, 4, 5, 6, 7, 8, 9, 10. Therefore it is divisible by their &lt;a href=&quot;http://en.wikipedia.org/wiki/Least_common_multiple&quot;&gt;least common multiple&lt;/a&gt;. There are algorithms to compute the LCM; for example you can write the prime factor decompositions and take the maximum for each factor. Here the result is 7*5*9*8 = 2520. So the original number of eggs is of the form 2520*x+1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2520 divided by 11 is 229 with a remainder of 1. The successive values of the remainder of 2520*x+1 modulo 11 for x=1,2,3,… are 1+1, 2+1, 3+1, … so x=10 gives the smallest possible number of eggs: 2520*10+1 = 25201.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;25201 is not the only answer: if you add 11 to x, you get another answer. The set of possible answers are all the numbers of the form 2520*(11*y+10)+1, i.e. 25201 plus any multiple of 27720.&lt;/p&gt;&#xA;" OwnerUserId="71" LastEditorUserId="71" LastEditDate="2014-05-30T18:19:54.433" LastActivityDate="2014-05-30T18:19:54.433" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="498" PostTypeId="1" AcceptedAnswerId="501" CreationDate="2014-05-30T02:54:44.640" Score="4" ViewCount="440" Body="&lt;p&gt;From the &lt;a href=&quot;http://www.theguardian.com/crosswords/cryptic/26271&quot; rel=&quot;nofollow&quot;&gt;Guardian Cryptic crossword No 26,271&lt;/a&gt;, set by &lt;a href=&quot;http://www.theguardian.com/profile/chifonie&quot; rel=&quot;nofollow&quot;&gt;Chifonie&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Mean to confine flower that's fibrous (7)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The confirmed answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; STRINGY&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can see the definition, and I think part of the wordplay, but not all of it. Here's my understanding:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; fibrous = def&lt;br&gt;&#xA; mean (to) = TRY&lt;br&gt;&#xA; flower = some river, presumably &lt;a href=&quot;http://en.wikipedia.org/wiki/Ing_River&quot; rel=&quot;nofollow&quot;&gt;ING&lt;/a&gt;&lt;br&gt;&#xA; So we have TRY confine ING = TRINGY.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But where does the first letter come from? (Or maybe I should phrase it this way: how does it get into first position?)&lt;/p&gt;&#xA;" OwnerUserId="71" LastActivityDate="2014-11-12T12:31:05.080" Title="Mean to confine flower that's fibrous (7)" Tags="|cryptic-crosswords|crossword-clues|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="499" PostTypeId="5" CreationDate="2014-05-30T03:02:57.920" Score="0" Body="&lt;p&gt;A &lt;a href=&quot;https://en.wikipedia.org/wiki/Cryptic_crossword&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cryptic crossword&lt;/a&gt; is a crossword puzzle where each clue is a word puzzle in and of itself.&#xA;Cryptic crosswords are particularly popular in the United Kingdom (where they originated) Ireland, the Netherlands, and in several Commonwealth nations, including Australia, Canada, India, and New Zealand.&#xA;In the United States, cryptics are sometimes known as &amp;quot;British-style&amp;quot; crosswords.&lt;/p&gt;&#xA;&lt;p&gt;Most clues are composed of two parts: a definition, and a wordplay. The wordplay describes how to arrange the letters to form the word; types of word plays include abbreviations, anagrams, hidden words, deleting and adding letters, assembling multiple parts, etc.&lt;/p&gt;&#xA;&lt;h2&gt;Example&lt;/h2&gt;&#xA;&lt;p&gt;Here is an example of a typical cryptic crossword clue (taken from The Guardian crossword of 6 August 2002, set by &amp;quot;Shed&amp;quot;).&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Very sad unfinished story about rising smoke (8)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;is a clue for TRAGICAL. This breaks down as follows.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&amp;quot;Very sad&amp;quot; is the definition&lt;/li&gt;&#xA;&lt;li&gt;&amp;quot;unfinished story&amp;quot; gives &amp;quot;tal&amp;quot; (&amp;quot;tale&amp;quot; with one letter missing; i.e., unfinished)&lt;/li&gt;&#xA;&lt;li&gt;&amp;quot;rising smoke&amp;quot; gives &amp;quot;ragic&amp;quot; (a &amp;quot;cigar&amp;quot; is a smoke and this is a down clue so &amp;quot;rising&amp;quot; indicates that &amp;quot;cigar&amp;quot; should be written up the page; i.e., backwards)&lt;/li&gt;&#xA;&lt;li&gt;&amp;quot;about&amp;quot; means that the letters of &amp;quot;tal&amp;quot; should be put either side of &amp;quot;ragic&amp;quot;, giving &amp;quot;tragical&amp;quot;&lt;/li&gt;&#xA;&lt;li&gt;&amp;quot;(8)&amp;quot; says that the answer is a single word of eight letters.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Use &lt;a href=&quot;/questions/tagged/cryptic-clues&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;cryptic-clues&amp;#39;&quot; aria-label=&quot;show questions tagged &amp;#39;cryptic-clues&amp;#39;&quot; rel=&quot;tag&quot; aria-labelledby=&quot;cryptic-clues-container&quot;&gt;cryptic-clues&lt;/a&gt; in preference to &lt;a href=&quot;/questions/tagged/cryptic-crosswords&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;cryptic-crosswords&amp;#39;&quot; aria-label=&quot;show questions tagged &amp;#39;cryptic-crosswords&amp;#39;&quot; rel=&quot;tag&quot; aria-labelledby=&quot;cryptic-crosswords-container&quot;&gt;cryptic-crosswords&lt;/a&gt; for cryptic clues that aren't part of a crossword puzzle. Use &lt;a href=&quot;/questions/tagged/crosswords&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;crosswords&amp;#39;&quot; aria-label=&quot;show questions tagged &amp;#39;crosswords&amp;#39;&quot; rel=&quot;tag&quot; aria-labelledby=&quot;crosswords-container&quot;&gt;crosswords&lt;/a&gt; for crosswords whose clues don't follow the conventions for cryptic clues.&lt;/p&gt;&#xA;&lt;h3&gt;Resources&lt;/h3&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chambers.co.uk/search/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chambers dictionary&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://sutherland-studios.com.au/apps/crypticguide.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cryptic abbreviations&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.crosswordunclued.com/2008/09/anagram-indicators.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Anagrinds&lt;/a&gt; (anagram indicators)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.alberichcrosswords.com/pages/id51.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ximenean clueing&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/45984/cryptic-clue-guide&quot;&gt;Cryptic clue guide&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="71" LastEditorUserId="52483" LastEditDate="2023-02-03T11:24:09.940" LastActivityDate="2023-02-03T11:24:09.940" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="500" PostTypeId="4" CreationDate="2014-05-30T03:02:57.920" Score="0" Body="Cryptic crosswords are a special type of crossword puzzle where clues combine a definition with wordplay. Use [crosswords] where the clues don't obey the conventions of cryptic crosswords; [cryptic-clues] or [crossword-clues] for clues not assembled into whole crosswords." OwnerUserId="71" LastEditorUserId="19114" LastEditDate="2017-11-10T17:32:09.343" LastActivityDate="2017-11-10T17:32:09.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="501" PostTypeId="2" ParentId="498" CreationDate="2014-05-30T03:06:00.590" Score="14" Body="&lt;p&gt;You're on the wrong path: there's a different breakdown where the S is part of the confining word.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I read “mean” as &lt;code&gt;STINGY&lt;/code&gt;, and “flower” as the &lt;code&gt;R&lt;/code&gt; that abbreviates “river”.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="341" LastEditorUserId="71" LastEditDate="2014-11-12T11:51:07.890" LastActivityDate="2014-11-12T11:51:07.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="502" PostTypeId="1" AcceptedAnswerId="505" CreationDate="2014-05-30T09:52:23.297" Score="0" ViewCount="295" Body="&lt;p&gt;I am interested in finding out if there is any specific way (algorithm) to play solitaire (not the card game)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bVz0V.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="50" LastEditorUserId="4421" LastEditDate="2015-09-02T18:38:33.877" LastActivityDate="2015-09-02T18:38:33.877" Title="Is there any specific way to play Solitaire?" Tags="|strategy|board-games|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="503" PostTypeId="1" AcceptedAnswerId="504" CreationDate="2014-05-30T11:37:00.457" Score="4" ViewCount="629" Body="&lt;p&gt;I'm stuck on the last clue of a cryptic crossword. The clue is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A non-PC American woman overseas? (6)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The cross letters are A_R_A_, confirmed by &quot;check answers.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Cryptic Crosswords iPhone app, Pack 1, Puzzle 16, Clue 16&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="4" LastEditorUserId="4" LastEditDate="2014-05-30T14:07:18.053" LastActivityDate="2014-05-30T14:07:18.053" Title="What is &quot;A non-PC American woman overseas?&quot;" Tags="|cryptic-crosswords|crossword-clues|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="504" PostTypeId="2" ParentId="503" CreationDate="2014-05-30T11:37:00.457" Score="7" Body="&lt;p&gt;As I was typing this up, I figured it out:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ABROAD&lt;br&gt;&#xA; [A] [non-PC American woman = broad] [overseas = abroad]&lt;br&gt;&#xA;&lt;br&gt;&#xA; I'm not quite sure if &quot;American&quot; is there for a particular reason besides just being filler. Is &quot;broad&quot; really restricted to American women?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4" LastActivityDate="2014-05-30T11:37:00.457" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="505" PostTypeId="2" ParentId="502" CreationDate="2014-05-30T13:42:56.637" Score="2" Body="&lt;p&gt;I'm not familiar with formal algorithms, but there are certainly effective strategies.  For the type of solitaire you picture there (along with all other &quot;hop to remove&quot; solitaires), my typical strategy is &quot;avoid the corners&quot;... Given a board like yours:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    o * o&#xA;    * * *&#xA;o * @ @ @ * o&#xA;* * @ @ @ * *&#xA;o * @ @ @ * o&#xA;    * * *&#xA;    o * o&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;... I will move the pieces out of the spots marked with &lt;code&gt;o&lt;/code&gt; as soon as possible, and will never move a piece back into them unless I have a plan to move it back out.  Additionally, the spots marked with &lt;code&gt;@&lt;/code&gt; are gold, if you aim to keep your pieces there then your chances are pretty good.  As a strategy it's fairly simplistic and doesn't guarantee a win (not to mention, it's hard to do), but I've won at least once or twice just trying to do that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's typically easier on the triangle layouts:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    o&#xA;   * *&#xA;  * @ *&#xA; * @ @ *&#xA;o * * * o&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;     o o&#xA;      *&#xA;     * *&#xA;    * @ *&#xA;   * @ @ *&#xA;o * * * * * o&#xA; o         o&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;... that last one I've only ever seen once, and it's where I learned this strategy to begin with.&lt;/p&gt;&#xA;" OwnerUserId="358" LastActivityDate="2014-05-30T13:42:56.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="506" PostTypeId="5" CreationDate="2014-05-30T13:53:51.417" Score="0" Body="&lt;p&gt;Puzzles that involve physical objects and have moving or rotating parts fall under this tag. For example, questions involving Rubik's cubes (or variants on a rubik's cubes) and trick boxes.&lt;/p&gt;&#xA;" OwnerUserId="52483" LastEditorUserId="52483" LastEditDate="2018-11-21T01:03:47.403" LastActivityDate="2018-11-21T01:03:47.403" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="507" PostTypeId="4" CreationDate="2014-05-30T13:53:51.417" Score="0" Body="Puzzles that are physically made and have some sort of mechanism for turning or moving them, such as Rubik's Cubes or trick boxes." OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2014-06-05T00:51:55.520" LastActivityDate="2014-06-05T00:51:55.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="508" PostTypeId="2" ParentId="496" CreationDate="2014-05-30T13:54:42.657" Score="18" Body="&lt;p&gt;Gilles certainly has the right idea, find the LCM of 2..10, +1. To take it a but further, here's the method I find easiest to find the LCM of relatively low (and/or easy-to-factor) numbers.&lt;/p&gt;&#xA;&lt;p&gt;First, find the prime factors of all the numbers:&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;number&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;factors&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2, 2&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;5&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;5&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;6&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2, 3&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;7&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;7&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;8&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2, 2, 2&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;9&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3, 3&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;10&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2, 5&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;Now go through that list and find the highest count of each factor in any of the numbers. It may be easiest to make a table&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;number&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;2&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;3&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;5&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;7&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2, 2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;5&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;5&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;6&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;7&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;7&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;8&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2, 2, 2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;9&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3, 3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;10&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;5&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;max&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2,2,2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3,3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;5&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;7&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;Now just multiply all the numbers in the &lt;span class=&quot;math-container&quot;&gt;$max$&lt;/span&gt; row and you have your number. If you rearrange and combine numbers in the right way it might be easier to multiply out in your head.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$2\times2\times2\times3\times3\times5\times7 = 7\times9\times4\times10 = 63\times4\times10 = 2520$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;So we need to find a number of the form &lt;span class=&quot;math-container&quot;&gt;$2520\times k+1 \equiv 0\mod11$&lt;/span&gt;.  Since &lt;span class=&quot;math-container&quot;&gt;$2520 \equiv 1 \mod 11$&lt;/span&gt;, we want the smallest &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; such that &lt;span class=&quot;math-container&quot;&gt;$1\times k + 1 \equiv 0\mod11$&lt;/span&gt;. Conveniently, we can easily see that that will be &lt;span class=&quot;math-container&quot;&gt;$k = 10$&lt;/span&gt;, for a final answer of &lt;span class=&quot;math-container&quot;&gt;$25201$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4" LastEditorUserId="69582" LastEditDate="2021-07-19T17:39:59.290" LastActivityDate="2021-07-19T17:39:59.290" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="509" PostTypeId="1" AcceptedAnswerId="510" CreationDate="2014-05-30T14:03:26.310" Score="5" ViewCount="723" Body="&lt;p&gt;I'm stumped on the last clue in another cryptic crossword. The clue is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Injured person's mostly easy-going &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The (confirmed) cross-clues give C_S_A_T_. Normally I can guess at a part or at least the structure, but I've got nothing here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Cryptic Crosswords iPhone app, Pack 3, Puzzle 5, Clue 12&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="4" LastActivityDate="2014-06-04T23:48:38.180" Title="Cryptic clue &quot;Injured person's mostly easy-going (8)&quot;" Tags="|cryptic-crosswords|crossword-clues|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="510" PostTypeId="2" ParentId="509" CreationDate="2014-05-30T14:08:32.570" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CASUALTY is mostly CASUAL&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The cryptic part of the clue doesn't seem to cover the whole word, as far as I can tell.&lt;/p&gt;&#xA;" OwnerUserId="18" LastActivityDate="2014-05-30T14:08:32.570" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="511" PostTypeId="2" ParentId="496" CreationDate="2014-05-30T14:43:27.267" Score="7" Body="&lt;p&gt;What do we know about your numbers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well it's 1 greater than a multiple of 2.&lt;br&gt;&#xA;1 greater than a multiple of 3.&lt;br&gt;&#xA;1 greater than a multiple of 4.&lt;br&gt;&#xA;...&lt;br&gt;&#xA;1 greater than a multiple of 10.&lt;br&gt;&#xA;a multiple of 11.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first 9 properties are easy. &lt;code&gt;10! + 1&lt;/code&gt; will cover that. The only question is, is it divisible by 11? Well, a quick check with a calculator reveals that it is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;therefore 3628801 is a possible answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question doesn't seem to ask for the smallest solution, and in fact this puzzle has an infinite number of solutions.&lt;/p&gt;&#xA;" OwnerUserId="361" LastActivityDate="2014-05-30T14:43:27.267" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="512" PostTypeId="1" AcceptedAnswerId="515" CreationDate="2014-05-30T15:23:29.070" Score="10" ViewCount="2752" Body="&lt;p&gt;Here's a relatively easy puzzle that's actually entirely my own creation.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;An old guru in an Asian monastery is showing his beloved student his gallery of items that he has either collected or kept from previous generations.&lt;/p&gt;&#xA;&lt;p&gt;One item that he is particularly fond of is an odd-looking walking stick. He shows it to his student, waving his hands over the evenly spaced notches in the stick. His student understands without talking that the stick served two purposes – as a walking stick and as a measuring device.&lt;/p&gt;&#xA;&lt;p&gt;He then proceeds to pop the handle off the stick. The student almost gasps but utters no sound. The guru continues to break the stick into pieces, while the student watches in silent horror at how inconsiderate the guru is being of an old relic.&lt;/p&gt;&#xA;&lt;p&gt;The student regains his calm, though, after noticing that the guru has not destroyed the stick after all, but is snapping pieces of the stick back together again, rearranging them into two smaller sticks. He then lines the two sticks on the ground, and the student sees that one piece is one notch longer than the other.&lt;/p&gt;&#xA;&lt;p&gt;He then proceeds to break the sticks into pieces again. The student watches curiously as he then rearranges the pieces into three sticks, each of them one notch longer than the next. The guru continues to do this for five sticks and seven sticks, and in each of the arrangements, each stick is one notch longer than the next.&lt;/p&gt;&#xA;&lt;p&gt;Finally, the guru snaps off all the pieces that he had been using to make the smaller sticks, and lines them up individually. The student knows what to expect. Each individual piece, ten of them in total, is one notch longer than the next.&lt;/p&gt;&#xA;&lt;p&gt;How long is this stick, in notches?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-06-23T21:37:16.097" Title="The Magic Stick" Tags="|calculation-puzzle|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="513" PostTypeId="2" ParentId="512" CreationDate="2014-05-30T16:06:56.973" Score="7" Body="&lt;p&gt;Here's my answer for the smallest possible solution, in number of notches:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$105$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's my logic (slightly mathematical, mostly intuitive).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The first case (splitting into $2$ sticks) is easy - any odd number satisfies that: $n + (n+1) = 2n+1$, which will always be odd.&lt;br&gt;&#xA; The second, third, and fourth cases (splitting into $3$, $5$, and $7$ sticks) are more interesting.  Consider the case of $3$ sticks.  You have sticks of length $n$, $n+1$, and $n+2$.  If you took one off of the $n+2$ stick, and put it on the $n$ stick, you have $3 \times (n+1)$, which means the final number has to be a multiple of $3$.  By similar logic, you can conclude that the number is also a multiple of $5$ and $7$.&lt;br&gt;&#xA; The final case was a bit more difficult, until I realized that $n + (n+1) + (n+2) + \dots + (n+9) = 5 * (2n+9)$.  I took the simplest case possible for the first for clues ($3 \times 5 \times 7 = 105$), and did some simple algebra to reduce to $n = 6$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is a breakdown of the length (in notches) of each stick, and how they can be combined as needed:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First case: $52 = 15+14+13+10$ and $53 = 12+11+9+8+7+6$.&lt;br&gt;&#xA; Second case: $34 = 15+13+6$, $35 = 14+12+9$, and $36 = 11+10+8+7$.&lt;br&gt;&#xA; Third case: $19 = 13+6$, $20 = 12+8$, $21 = 11+10$, $22 = 15+7$, and $23 = 14+9$.&lt;br&gt;&#xA; Fourth case: $12,13,14,15$, $16=10+6$, $17=8+9$, $18=7+11$&lt;br&gt;&#xA; Fifth case (trivially): $6,7,8,9,10,11,12,13,14,15$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/a/515/11&quot;&gt;Kaine's answer&lt;/a&gt; does prove that this is indeed the &lt;em&gt;only&lt;/em&gt; solution.&lt;/p&gt;&#xA;" OwnerUserId="11" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2014-06-23T21:37:16.097" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="515" PostTypeId="2" ParentId="512" CreationDate="2014-05-30T16:39:24.393" Score="5" Body="&lt;h3&gt;Part 1&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/512/the-magic-stick/513#513&quot;&gt;Xynariz's answer&lt;/a&gt; seems to be correct.  We can note all the infinite solutions if the stick has more than 10 breaks as follows:&#xA;$$5(2j+9)=3*5*7*k \quad \text{for some integers \(j\) and \(k\)} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $j = 21x+6$ and $k=2x+1$ for any integer $x$, then these equations are true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means the number of sticks is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $210x+105$ for any integer $x$. This number is odd and divisible by 3,5, and 7.  Also if you subtract 45, the answer is divisible by 10.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;Part 2&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;If there are only 9 breaks, each stick is $L+s$ long where $s$ is the stick number $0,1,2,3...$ while L is $21x+6$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we try to make 3 sticks out of the 10 parts, the minimum length of the stick that contains 4 parts (as 10 is not divisible by 3) is $(21x+6)*4+0+1+2+3=84x+30$.  The largest stick that can be made out of 3 small sticks is $(21x+6)*3+9+8+7 = 63x+42$.  The smalles difference between the 4 and 3 stick lengths is $21x-12$ while the guru's game requires at least one combination that differ in length by no more than $1$. As the 4-stick is more than $1$ greater than the 3-stick for any $x&amp;gt;0$, the only solution is $x=0$.  This means that the only solution is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $105$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="99" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2014-06-23T21:37:12.160" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="520" PostTypeId="2" ParentId="512" CreationDate="2014-05-30T17:41:52.413" Score="0" Body="&lt;p&gt;Well we can easily rephrase the question as follows (the subtractor being $k + (k - 1) / 2$ obviously for $\bmod k$):&#xA;$$\begin{align}&#xA;    (n - 1)  \equiv 0 &amp;amp; \mod 2  \\&#xA;    (n - 3)  \equiv 0 &amp;amp; \mod 3  \\&#xA;    (n - 10) \equiv 0 &amp;amp; \mod 5  \\&#xA;    (n - 21) \equiv 0 &amp;amp; \mod 7  \\&#xA;    (n - 45) \equiv 0 &amp;amp; \mod 10 \\&#xA;\end{align} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using congruence math we can prettify this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;    n \equiv 1 &amp;amp; \mod 2  \\&#xA;    n \equiv 0 &amp;amp; \mod 3  \\&#xA;    n \equiv 0 &amp;amp; \mod 5  \\&#xA;    n \equiv 0 &amp;amp; \mod 7  \\&#xA;    n \equiv 5 &amp;amp; \mod 10 \\&#xA;\end{align} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the number has to be divisible by $3$, $5$, $7$ which gives us as the minimum &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $105$, which also fulfills the other two requirements.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So is there really only one solution as claimed? Easy to disprove: $n \equiv 1 \mod 2$ is guaranteed if $n \equiv 5 \mod 10$ is fulfilled (I'm sure that's obvious although we could prove it with the CRT). So we just have to check when $105 n \equiv 5 \mod 10$ or rather $5n \equiv 5 \mod 10$ is true. Which is for all odd values of $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the correct solution is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\forall k \in \mathbb{N}, 105 (2k+1)$, i.e. $105, 315, 525, \dots$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="364" LastEditorUserId="71" LastEditDate="2014-06-22T16:38:30.653" LastActivityDate="2014-06-22T16:38:30.653" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="521" PostTypeId="1" CreationDate="2014-05-30T17:46:23.067" Score="7" ViewCount="3641" Body="&lt;p&gt;I was using the beginner method to deliberately solve a 3X3 cube with one edge piece flipped. But, by the time I got to the top I discovered that I couldn't solve it with only one edge flipped. I had to also flip an edge on the top layer. Why is that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cAylY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="130" LastActivityDate="2014-05-31T05:09:40.537" Title="Why can't I have only one edge piece flipped on a 3X3 Rubik's cube?" Tags="|rubiks-cube|" AnswerCount="1" CommentCount="2" ClosedDate="2014-06-03T15:27:01.570" ContentLicense="CC BY-SA 3.0" />
  <row Id="523" PostTypeId="1" AcceptedAnswerId="524" CreationDate="2014-05-30T18:49:03.767" Score="6" ViewCount="1225" Body="&lt;p&gt;I found the following problem in an number of places online:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;John once bet a fellow gambler that two of the first thirty persons they met and spoke to would prove to have the same birthday. Strong in the thought that he had 365 days running for him, the second hustler was pleased to accept. Suspecting, not unnaturally, a frame-up, he was careful to approach total strangers and chance passers-by, who could not be known to John. He lost the bet on the twenty-eighth question, when a duplicate birthday turned up.&lt;/p&gt;&#xA;&lt;p&gt;“To tell you the truth,” said John afterward, “on each of the last five guys we spoke to, the odds were already better than even money in my favor. I'll explain the mathematics to you some time.”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Why is this true?  At what point did John's chance of filling his bet pass 25%?  50%?  75%?  90%?  99%?  (For your calculations, assume there are only 365 unique birthdays.)&lt;/p&gt;&#xA;&lt;p&gt;N.B. John's statement about &amp;quot;the last five guys&amp;quot; assumes they will talk to all thirty people, regardless of whether the bet condition has been met or not, and the &amp;quot;better than even money&amp;quot; refers to the chance he will win the bet on &lt;em&gt;or before&lt;/em&gt; that person.&lt;/p&gt;&#xA;" OwnerUserId="11" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-10-15T13:43:17.610" Title="Odds of duplicate birthdays" Tags="|calculation-puzzle|mathematics|probability|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="524" PostTypeId="2" ParentId="523" CreationDate="2014-05-30T19:09:55.020" Score="8" Body="&lt;p&gt;Probability two people have the same birthday: $\frac{1}{365}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probability that (if the first two of people don't have the same birthday) the third person is a duplicate: $\frac{2}{365}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the probability that by the third person, there is a duplicate is: $\frac{1}{365}+(1-\frac{1}{365})(\frac{2}{365})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability that by the nth person they will have found a duplicate: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $P(n) = P(n-1)+(1-P(n-1))*\frac{n-1}{365}=\frac{n-1}{365}+P(n-1)\frac{366-n}{365}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't feel like calculating this exactly by hand so:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $P(15) \approx .255$&lt;br&gt;&#xA; $P(23) \approx .509$&lt;br&gt;&#xA; $P(32) \approx .754$&lt;br&gt;&#xA; $P(41) \approx .903$&lt;br&gt;&#xA; $P(47) \approx .955$&lt;br&gt;&#xA; $P(57) \approx .99$&lt;br&gt;&#xA; $P(366) = 1$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These answers give the probability that he would win the bet by this turn.  If, however, he were to say &quot;Given that I have not yet won on turn $n-1$ the probability I will win on turn $n$ is greater than 50%&quot; this does not occur so early. In this case, $P(n-1)=0$ so: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$P(n)=\frac{n-1}{365}$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In order for $P(n)&amp;gt;.5$ then &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$n&amp;gt;183$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="99" LastEditorUserId="68" LastEditDate="2014-10-15T13:43:17.610" LastActivityDate="2014-10-15T13:43:17.610" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="525" PostTypeId="1" AcceptedAnswerId="530" CreationDate="2014-05-30T20:02:07.660" Score="15" ViewCount="4745" Body="&lt;p&gt;I wanted to answer this question: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/521/why-cant-i-have-only-one-edge-piece-flipped-on-a-3x3-rubiks-cube&quot;&gt;Why can&amp;#39;t I have only one edge piece flipped on a 3X3 Rubik&amp;#39;s cube?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by saying that it is homomorphic to the illegal state in this question: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/20/why-is-a-single-corner-twist-not-a-valid-position-on-a-rubiks-cube&quot;&gt;Why is a single-corner twist not a valid position on a Rubik&amp;#39;s cube?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That does not appear to be nessesarily true.  A more interesting question than either of them is, therefore:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If I were to break apart, put back together randomly, and try to solve my Rubik's cube, what is the probability that the resulting cube is solvable?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If there are only 2 homomorphs, the answer should be $50\%$.  If there are more....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am inclined to believe there are more than 2 (at least 6) as if you rotate a corner, there are 3 different orientations that cannot be alternated.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2016-12-08T17:35:09.967" Title="How many distinct unsolvable Rubik's cubes exist?" Tags="|calculation-puzzle|rubiks-cube|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="526" PostTypeId="2" ParentId="525" CreationDate="2014-05-30T20:02:07.660" Score="7" Body="&lt;p&gt;Using Wikipedia I note the following facts: First, the number of permutations of a true $3\times 3\times 3$ cube is $43,252,003,274,489,856,000$. Second, the number of permutations of breaking and reassembling a cube is $519,024,039,293,878,272,000$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of non-interchangeable cubes, therefore, appears to be $ 12 $. This means that the probability of reassembling a solvable cube is roughly $8\%$.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="39" LastEditDate="2014-06-23T23:53:59.547" LastActivityDate="2014-06-23T23:53:59.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="527" PostTypeId="2" ParentId="433" CreationDate="2014-05-31T00:02:21.427" Score="12" Body="&lt;p&gt;It appears to be two clues in one:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;Chimpanzee, say&quot; (answer: APE)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;Circus equipment has &lt;em&gt;APE&lt;/em&gt; at the centre.&quot; (The letters &quot;APE&quot; are at the exact centre of the word &quot;TRAPEZE&quot;.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="381" LastActivityDate="2014-05-31T00:02:21.427" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="528" PostTypeId="2" ParentId="33" CreationDate="2014-05-31T02:25:33.763" Score="4" Body="&lt;p&gt;It depends on what you count as a move.  The &quot;God's number is 20&quot; result mentioned by earlier answerers is for the &lt;em&gt;face turn metric&lt;/em&gt;, where a 180 rotation/half turn of a face counts as one move.  (This &lt;a href=&quot;http://wdjoyner.wordpress.com/2012/07/06/the-man-who-found-gods-number/&quot; rel=&quot;nofollow&quot;&gt;blog entry&lt;/a&gt; by David Joyner gives a history of results toward this conclusion with a parallel human interest story on one of the primary researchers.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the &lt;em&gt;quarter turn metric&lt;/em&gt;, a 180 rotation counts as two moves so that &quot;God's number&quot; is presumably higher.  To date it is known to be at least 26 (&quot;superflip plus fourspot&quot;, &lt;a href=&quot;http://www.cube20.org/distance20s/&quot; rel=&quot;nofollow&quot;&gt;description here&lt;/a&gt;).  Similar to the long-known examples requiring 20 turns in the FTM that were eventually shown (with lots of computer power) to be optimal, the suspicion is that 26 is God's number in the QTM.&lt;/p&gt;&#xA;" OwnerUserId="383" LastEditorUserId="383" LastEditDate="2014-06-01T15:05:44.680" LastActivityDate="2014-06-01T15:05:44.680" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="529" PostTypeId="2" ParentId="521" CreationDate="2014-05-31T05:09:40.537" Score="3" Body="&lt;p&gt;12 edges on a cube. 24 positions an edge piece can be in once you consider flipping. If you consider one move to be a single quarter turn, then you can move an edge to exactly 12 of those 24 positions in an even number of moves, and to the other 12 only in an odd number of moves. You should be able to convince yourself of this by trial and error, but if you want proof you could come up with names for the 24 positions and check that each possible move only moves edges from a position in the even set to one in the odd set or vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider what a single move does. It moves each of four edge pieces from one set of positions to the other. This means if you started with (say) 6 in one position-set and 6 in the other, you now have 2-10 or 4-8 or 6-6 or 8-4 or 10-2, but you can't have an odd number of pieces that changed position-sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to flip a single edge piece, you need to move a single piece to the other position set without moving any of the others. But you can only switch the position-sets of an even number of pieces at a time.&lt;/p&gt;&#xA;" OwnerUserId="385" LastActivityDate="2014-05-31T05:09:40.537" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="530" PostTypeId="2" ParentId="525" CreationDate="2014-05-31T05:57:03.660" Score="25" Body="&lt;p&gt;My answer at &lt;a href=&quot;https://puzzling.stackexchange.com/questions/20/why-is-a-single-corner-twist-not-a-valid-position-on-a-rubiks-cube&quot;&gt;Why is a single-corner twist not a valid position on a Rubik&amp;#39;s cube?&lt;/a&gt; basically explains this, but in simpler terminology here are the facts presented without proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you randomly disassemble and reassemble a $3 \times 3 \times 3$ cube, the probability that it will be solvable with just face turns is exactly $1 \over 12$. Furthermore, this number has a simple interpretation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any random state of the cube can be reduced with only face turns to one of these 12 configurations. Furthermore, none of these configurations can be achieved from any other one with solely face turns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UrUZ1l.png&quot; alt=&quot;12 configurations image&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pattern here is pretty simple. Going from left to right is just twisting a corner (in this case the yellow-blue-red one). Going up or down by 2 rows is flipping an edge (in this case the yellow-green one). Going between the 1st and 2nd or 3rd and 4th rows is swapping 2 edges (in this case the yellow-orange and yellow-red ones).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set of all states which can be reduced to one of these 12 configurations is called an &quot;orbit&quot;, and each of these 12 orbits has the same number of states as the others (which is basically &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrange&amp;#39;s_theorem_(group_theory)&quot; rel=&quot;noreferrer&quot;&gt;Lagrange's theorem&lt;/a&gt;). Thus, each has exactly $1 \over 12$ of all states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the upper left corner is the solved cube. The orbit of this cube is exactly the set of states you can solve just with face turns. Since it has $1 \over 12 $ of all the possible states, the probability that a randomly reassembled cube will be solvable is exactly ${1 \over 12} \approx 8.3\% $. The other ${11 \over 12} \approx 91.7 \% $ of the time, you'll end up in a different orbit, and eventually reach one of the other states above, at which point you can't solve it any further without face turns.&lt;/p&gt;&#xA;" OwnerUserId="94" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2014-06-23T23:57:55.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="534" PostTypeId="2" ParentId="455" CreationDate="2014-05-31T20:26:00.930" Score="4" Body="&lt;p&gt;You can't just dissect each statement when following a proof. You have to keep the assumptions and the context in mind. The problematic statement you quote should be read like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;If&lt;/em&gt; the princess occupies an even numbered room on the day the prince first  knocks on a door &lt;em&gt;then&lt;/em&gt; she will never be in a room adjacent to the one he knocks on&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of course the assumption is not proven in any way. But the point is if the assumption doesn't hold then the second part of the proof applies.&lt;/p&gt;&#xA;" OwnerUserId="393" LastActivityDate="2014-05-31T20:26:00.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="540" PostTypeId="1" CreationDate="2014-06-01T08:06:45.497" Score="-1" ViewCount="249" Body="&lt;p&gt;To be able to solve a sudoku one usually has to first work out all the 'possible' numbers in an empty cell. How can one make a program to do this? In this case 'possible numbers' are ones which have not yet been provided elsewhere in the same line column or box and therefore this cell can possibly have those numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/B94wx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not meant to solve the puzzle.But helpful for those who want to do it by hand. &lt;a href=&quot;https://puzzling.stackexchange.com/users/365/alan-ross&quot;&gt;see also here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="365" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-04-16T18:58:30.387" Title="How does one make a program to provide all the 'possible' numbers in an empty cell" Tags="|sudoku|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="542" PostTypeId="2" ParentId="313" CreationDate="2014-06-01T13:13:39.680" Score="9" Body="&lt;p&gt;As you can see in this topic, there are a lot of possible solutions of course.&lt;br&gt;&#xA;But there is &lt;strong&gt;general solution&lt;/strong&gt;, which includes all of them. It is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are you thinking of 1 OR you are thinking of 2 AND [statement with unknown result]?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The interpretation of the answer is clear:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 -&amp;gt; yes&#xA;2 -&amp;gt; I do not know&#xA;3 -&amp;gt; no&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My most preferable solution is &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are you thinking of 1 OR you are thinking of 2 AND &lt;em&gt;tomorrow will be rain&lt;/em&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Just because it is simple and it is clear how it works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But any other solution will be equivalent to given general solution (up to a interchange of numbers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example accepted answer of &lt;a href=&quot;https://puzzling.stackexchange.com/a/317/28&quot;&gt;Ice-9&lt;/a&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;I'm thinking of a number: either 0 or 1. Is the sum of our numbers greater than 2?&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is equivalent to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are you thinking of 3 OR you are thinking of 2 AND &lt;em&gt;I thinking of 1&lt;/em&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or another solution of &lt;a href=&quot;https://puzzling.stackexchange.com/a/318/28&quot;&gt;Doorknob&lt;/a&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;I will divide your number by either 2 or 3 and round it down. Is the result even or odd?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is equivalent to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are you thinking of 1 OR you are thinking of 2 AND &lt;em&gt;I thinking of 3&lt;/em&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Even such sophisticated solution, which &lt;a href=&quot;https://puzzling.stackexchange.com/a/316/28&quot;&gt;arshajii&lt;/a&gt; gave:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Is $\lim_{x \to n} \frac{1}{x-2} &amp;gt; 0$ (where we are approaching $n$ from either the left or the right)?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is equivalent to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are you thinking of 3 OR you are thinking of 2 AND $\lim_{x \to 0} \frac{1}{x} &amp;gt; 0$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="28" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2014-07-17T08:51:16.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="544" PostTypeId="2" ParentId="25" CreationDate="2014-06-01T14:48:40.003" Score="-1" Body="&lt;p&gt;I would answer that the numbers are not important here unless you define &lt;strong&gt;full&lt;/strong&gt; model of the situation mathematically strictly and with exact numbers. The model includes both: prison rules and prisoners way of thinking. Currently you have defined only the first one.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;To illustrate this on a simple example what happens when a model is defined only partially:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the same way it is not important how big probability to lose in the game until you do not define how must you win when you win and how much you lose when you lose. Let's note probability to win as P, amount you win as AW, amount you lose as AL. Then in average you will win &lt;code&gt;A = AW*P+AL*(1-P)&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;If you take P = 50%, AW = 200, AL = 100. A = 50.&#xA;If you take P = 0.5%, AW = 20000, AL = 1. A ~= 99.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;One can see that looking on P one can not judge at all about decision.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Ones you define the prisoners strictly in mathematical sense I believe you will have no problems to predict they behaviour. For example superational prisoners will always cooperate (if R &gt; P AND S+T &amp;lt; 2R).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2014-06-01T14:56:21.543" LastActivityDate="2014-06-01T14:56:21.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="545" PostTypeId="2" ParentId="540" CreationDate="2014-06-01T15:23:06.353" Score="2" Body="&lt;p&gt;Algorithm:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;int[][] sudoku = new int[9][9];&#xA;sudoku = [input the numbers]; // 0 if not known&#xA;for-each cell (x,y) do:&#xA;  int[] O = {1,2,3,4,5,6,7,8,9};&#xA;  // remove from O all numbers that are in the same square&#xA;  (bX, bY) = getBlockXY( cell (x,y) ); // this can be done with some if-statements&#xA;  for i = 1 to 3 do:&#xA;    for j = 1 to 3 do:&#xA;      if sudoku[bX * 3 + i][bY * 3 + j] then : // evaluates to true if known&#xA;        O.remove(sudoku[bX * 3 + i][bY * 3 + j]); &#xA;      end if&#xA;    end for&#xA;  end for      &#xA;&#xA;  // remove from O all numbers that are in the same row&#xA;  for i = 1 to 9 do:&#xA;    if sudoku[i][y] then :&#xA;      O.remove(sudoku[i][y]);&#xA;    end if&#xA;  end for&#xA;&#xA;  // remove from O all numbers that are in the same column&#xA;  for i = 1 to 9 do:&#xA;    if sudoku[x][i] then :&#xA;      O.remove(sudoku[x][i]);&#xA;    end if&#xA;  end for&#xA;&#xA;  print O&#xA;end for-each&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I think you could do this more efficiently by using some kind of datastructure. I have no idea about that yet.&lt;/p&gt;&#xA;" OwnerUserId="146" LastEditorUserId="146" LastEditDate="2014-06-01T16:45:49.773" LastActivityDate="2014-06-01T16:45:49.773" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="546" PostTypeId="1" AcceptedAnswerId="547" CreationDate="2014-06-01T16:07:38.647" Score="30" ViewCount="197941" Body="&lt;p&gt;Is there any specific way to solve Mastermind?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rOeTu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apart from the first step that is pure chance, is there any way to continue based on the colors that you think are correct?&lt;/p&gt;&#xA;" OwnerUserId="50" LastEditorUserId="5373" LastEditDate="2015-04-28T08:13:24.853" LastActivityDate="2021-06-01T13:35:52.927" Title="Clever ways to solve Mastermind?" Tags="|strategy|mastermind|" AnswerCount="6" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="547" PostTypeId="2" ParentId="546" CreationDate="2014-06-01T17:13:48.607" Score="31" Body="&lt;p&gt;Wikipedia has the &lt;a href=&quot;http://en.wikipedia.org/wiki/Mastermind_(board_game)#Five-guess_algorithm&quot; rel=&quot;nofollow noreferrer&quot;&gt;nice section on optimal Mastermind strategies&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In 1977, Donald Knuth demonstrated that the codebreaker can solve the&#xA;pattern in five moves or fewer, using an algorithm that progressively&#xA;reduced the number of possible patterns. The algorithm works as&#xA;follows:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Create the set S of 1296 possible codes, 1111,1112,.., 6666.&lt;/li&gt;&#xA;&lt;li&gt;Start with initial guess 1122 (Knuth gives examples showing that some other first guesses such as 1123, 1234 do not win in five tries&#xA;on every code).&lt;/li&gt;&#xA;&lt;li&gt;Play the guess to get a response of colored and white pegs.&lt;/li&gt;&#xA;&lt;li&gt;If the response is four colored pegs the game is won, the algorithm terminates.&lt;/li&gt;&#xA;&lt;li&gt;Otherwise, remove from S any code that would not give the same response if it (the guess) were the code.&lt;/li&gt;&#xA;&lt;li&gt;Apply minimax technique to find a next guess as follows: For each possible guess, that is, any unused code of the 1296 not just those in&#xA;S, calculate how many possibilities in S would be eliminated for each&#xA;possible colored/white peg score. The score of a guess is the minimum&#xA;number of possibilities it might eliminate from S. From the set of&#xA;guesses with the maximum score select one as the next guess, choosing&#xA;a member of S whenever possible. (Knuth follows the convention of&#xA;choosing the guess with the least numeric value e.g. 2345 is lower&#xA;than 3456. Knuth also gives an example showing that in some cases no&#xA;member of S will be among the highest scoring guesses and thus the&#xA;guess cannot win on the next turn yet will be necessary to assure a&#xA;win in five.)&lt;/li&gt;&#xA;&lt;li&gt;Repeat from step 3.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Subsequent mathematicians have been finding various algorithms that&#xA;reduce the average number of turns needed to solve the pattern: in&#xA;1993, Kenji Koyama and Tony W. Lai found a method that required an&#xA;average of 5625/1296 = 4.340 turns to solve, with a worst-case&#xA;scenario of six turns. The minimax value in the sense of game theory&#xA;is 5600/1296 = 4.321.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathworld.wolfram.com/Mastermind.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;MathWorld's page on Mastermind&lt;/a&gt; also gives a nice synopsis and mention a few more strategies:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Knuth (1976-77) showed that the codebreaker can always succeed in five&#xA;or fewer moves (i.e., knows the code after four guesses). His&#xA;technique uses a greedy strategy that minimizes the number of&#xA;remaining possibilities at each step, and requires 4.478 guesses on&#xA;average, assuming equally likely code choice. Irving (1978-79)&#xA;subsequently found a strategy with slightly smaller average length.&#xA;Koyama and Lai (1993) described a strategy that minimizes the average&#xA;number of guesses, requiring on average 4.340 guesses, although may&#xA;require up to six in the worst case. A slight modification also&#xA;described by Koyama and Lai (1993) increases the average to 4.341, but&#xA;reduces the maximum number of guesses required to five.&lt;/p&gt;&#xA;&lt;p&gt;Swaszek (1999-2000) gives an analysis of practical strategies that do&#xA;not require complicated record-keeping or use of a computer. Making a&#xA;random guess from the set of remaining candidate code sequences gives&#xA;a surprisingly short average game length of 4.638, while interpreting&#xA;each guess as a number and using the next higher number consistent&#xA;with the known information gives a game of average length 4.758.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;In summary, there is a trade-off to make between the average length and the maximum length of the game. (length is expressed in the number of code guesses)&lt;/p&gt;&#xA;" OwnerUserId="410" LastEditorUserId="-1" LastEditDate="2021-06-01T13:35:52.927" LastActivityDate="2021-06-01T13:35:52.927" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="548" PostTypeId="1" CreationDate="2014-06-01T18:00:46.113" Score="13" ViewCount="3347" Body="&lt;p&gt;I am looking for solution of this puzzle:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;There is a kingdom with a square shape with sides of length 1. The castle is located at the center of the square. At the castle the king lives under the protection of one soldier. There are other soldiers, but each soldier is at his own house. Houses are spread throughout the area of the kingdom, and everyone is aware of where they are.&lt;/p&gt;&#xA;&lt;p&gt;A war has been declared with a neighboring kingdom, and the king needs to collect all the soldiers at the castle urgently. To do this, he sends the soldier at the castle with the news to gather troops from the other places. Each soldier who learned the news, if necessary, can participate in the gathering. They can travel at the speed of 1.&lt;/p&gt;&#xA;&lt;p&gt;What is the minimum time in which the king can collect all solders, &lt;strong&gt;regardless&lt;/strong&gt; of their number and initial placement? (Number of soldiers is finite).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I see that the time should satisfy:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$$T_{min} \geq 2+\sqrt{2} = 3.4142...$$&lt;/span&gt; This is the minimum time to gather solders from 4 houses in square's corners.&#xA;&#xA; &lt;span class=&quot;math-container&quot;&gt;$T = 2+\sqrt{2}$&lt;/span&gt; feels to be the minimum: somehow with help of new solders everyone in a middle zone should be gathered while the very first solder goes via 3 corners.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;But I have no idea how to find a generic algorithm for &lt;span class=&quot;math-container&quot;&gt;$T_{\mathrm{min}}$&lt;/span&gt;, which will achieve such a time for arbitrary number of houses.&lt;/p&gt;&#xA;&lt;p&gt;P.S. Please be aware when you formulate your algorithm and prove it that there are so many possibilities that it is easy to miss something. I tried many times already, feeling that I've almost got it. But my solutions always had some small unclearness, and when I investigated the &amp;quot;proof&amp;quot; more explicitly and carefully I would find a gap in it...&lt;/p&gt;&#xA;&lt;p&gt;P.P.S. I tried the following strategy:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;When solder decides to go to some house he &amp;quot;reserves&amp;quot; it.&lt;/li&gt;&#xA;&lt;li&gt;Once a solder is free (has delivered the news to another house) he takes the closest house from all unreserved and goes there. If there are several closest houses he chooses at random.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/605/does-this-strategy-works&quot;&gt;It doesn't work.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-12-24T18:32:52.657" Title="Fastest way to collect an arbitrary army" Tags="|mathematics|geometry|optimization|algorithm|" AnswerCount="7" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="549" PostTypeId="2" ParentId="546" CreationDate="2014-06-01T20:57:36.487" Score="11" Body="&lt;p&gt;I play Mastermind with numbers instead of colours, because I first learned it in the second grade as Bagel Pico Fermi which uses numbers. For the rest of my answer, I will refer to red pegs as &quot;bagels&quot;, and white pegs as &quot;picos&quot; (and holes without pegs as &quot;fermis&quot;).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The system I tend to use is suboptimal but very easy to follow. It goes as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Start with 0000. You can never get picos if all digits are the same, only bagels.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If the secret number was 0187, then you will get one bagel and three fermis.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the secret number was 2966, then you will get four fermis and you know that 0 is not in the secret number at all.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If there are any bagels with 0000, include that many 0's in your next answer, and replace the rest with 1's.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If the secret number was 0187, then you'd keep one of the 0's in your answer, and guess 0111 next, getting two bagels.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the secret number was 2966, then you'd guess 1111 next, getting four fermis again.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Keep increasing the extra digits by 1. Those digits are &quot;background digits&quot;, while the digits that you've kept the same should never change values and are &quot;foreground digits&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However many more pegs there are when you change the background digits, that many background digits then become foreground digits.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Eventually you'll get to a point where you have a total of four pegs. If you have four bagels, congratulations, you have the right answer. But if some of them are picos, then some of them are in the wrong order. At this point, just try rearranging them, paying attention to whether your arrangement matches the number of switched digits in each of your previous guesses.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;An example of this algorithm at work might be as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Secret number: 4034&#xA;Every round there are A bagels and B picos.&#xA;&#xA;[BG digit: 0] 1. 0000  1A0B  (so there's one 0) &#xA;[BG digit: 1] 2. 0111  0A1B  (so there's no 1's, and the 0 is in the wrong place)&#xA;[BG digit: 2] 3. 2022  1A0B  (so there's no 2's either, but we know where the 0 is now) &#xA;[BG digit: 3] 4. 3033  2A0B  (so there's one 3, because A+B increased by 1) &#xA;[BG digit: 4] 5. 3044  2A2B  (you now have all the numbers, just not in the right order)&#xA;[BG digit: -] 6. 4043  2A2B  (the switch didn't work, try another one)&#xA;[BG digit: -] 7. 4034  4A0B  (you win!)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This isn't as good as the algorithm that a computer would use, but it's very simple and systematic, and very easy to get the hang of once you understand what you're doing.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2014-06-01T20:57:36.487" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="550" PostTypeId="1" AcceptedAnswerId="555" CreationDate="2014-06-02T02:14:35.970" Score="9" ViewCount="3285" Body="&lt;p&gt;In chess puzzles, the objective of the puzzle is usually to checkmate the opponent using some form of novel strategy or tactic. For example, this puzzle &lt;a href=&quot;http://www.chess.com/forum/view/more-puzzles/mate-in-2-white-to-play&quot; rel=&quot;nofollow noreferrer&quot;&gt;from chess.com&lt;/a&gt; has white to play and mate in two.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6f5EP.gif&quot; alt=&quot;mate in two&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;But &lt;a href=&quot;https://pdb.dieschwalbe.de/P1236762&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;this&lt;/strong&gt;&lt;/a&gt; puzzle, created by Karl Fabel, is famous for having white to play and mate in &lt;strong&gt;sixty&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2HjDs.gif&quot; alt=&quot;mate in sixty&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;This puzzle got me thinking - what is the position in chess with the longest string of moves before mate?&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="69582" LastEditDate="2021-01-21T00:27:47.550" LastActivityDate="2021-01-21T00:27:47.550" Title="What chess position has the longest string of moves before mate?" Tags="|chess|construction|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="551" PostTypeId="1" CreationDate="2014-06-02T12:47:42.823" Score="0" ViewCount="4194" Body="&lt;p&gt;Why must a sudoku have a unique solution. In my books I don't guarantee it. But I am sure there are not many which don't. I doubt if one in a thousand dont have. And even if they don't its only a matter of interchanging about four numbers. Why is there this obsession that it has to be only one unique solution. This of course applies to other puzzles too. &lt;/p&gt;&#xA;" OwnerUserId="365" LastActivityDate="2014-06-02T18:15:49.740" Title="Why must a sudoku have a unique solution" Tags="|sudoku|" AnswerCount="3" CommentCount="14" ClosedDate="2014-06-02T21:51:24.873" ContentLicense="CC BY-SA 3.0" />
  <row Id="552" PostTypeId="2" ParentId="551" CreationDate="2014-06-02T13:52:09.197" Score="4" Body="&lt;p&gt;There is no law requiring that a published Sudoku have a unique solution.  When I see a puzzle of any type, I expect from experience that the setter has promised a unique solution (or occasionally will say there are some number to be found).  Some setters, Raymond Smullyan especially, create problems that challenge you to make use of the fact that there must be a unique solution.  Cryptarithms are especially prone to multiple solutions.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2014-06-02T13:52:09.197" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="553" PostTypeId="2" ParentId="548" CreationDate="2014-06-02T17:12:13.293" Score="2" Body="&lt;p&gt;Here's an algorithm that doesn't always work (see a counter-example at the end). A slight adjustment (if a knight has no houses in his region he can take a portion of the other knight at his house's region) seems like it would fix this but it is not in this answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 1: Determine the lower bound:&lt;br&gt;&#xA;If the knights are located at the 4 corners it will take: $\sqrt{(1/2)^2+(1/2)^2}+1+1+\sqrt{(1/2)^2+(1/2)^2} = 2+\sqrt{2}$.&lt;br&gt;&#xA;The first one goes to a corner, he and the corner liver goes to the other two closest corner, someone of those 4 go to the last corner, and then everyone goes to the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 2: Show the lower bound works for any pattern:&lt;br&gt;&#xA;Assume that there is a knight in a corner as this is the furthest point away from the king.  If no one is further than that it should make the solution faster but more complicated. Use the following strategy as it is easier to show but slowest. For this algorithm one knight goes to a corner, they progress to the other corner and when everyone is notified, go to the castle.  If every house is reached within 2 time after the first house is reached, they can all get back as they are at most $\sqrt{1/2}$ away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Send the first knight to a corner call the corner in this direction North (N) for simplicity. You will then have two knights.  Have them go to houses by this algorithm:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The knight divide the map into two regions by the East/West location of their current house. One knight goes to the northmost house which has not been visited, no other knight is headed towards, and is to the East of the current location. If multiple houses are the same distance north, he goes to the closer one. If there are no houses to the East he should go to the castle.  The other knight follows the same algorithm but to the West.  If there is a house directly South, it can count as either East or West but don't send both to the same house!.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each time you visit a house, repeat the same algorithm with the two knights then at the house except no knight may move into a region which was excluded by one of his predecessors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As no knight goes north again (until they are going to the castle) and the knights are interchangeable, the shortest path connecting the north and south corner houses via this algorithm will be taken.  The maximum distance for this shortest path is $2$.  This is because there is at least one path taken by which $\Delta S \ge |\Delta E|$ for every trip between houses.  Similarly paths will go to the East most and West most houses that will never go in the other direction and, therefore, will have a path length of at most $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are likely faster algorithms than this for a particular setup but this one goes to all within the lower bound.  Essentially each knight will take his map, draw two vertical lines based on his location and the history of his messenger, and go to the closest northmost house in his region.  As the regions get smaller, they can be filled faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yM9xq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the example above (which only shows the house covering part of the algorithm), purple is the paths knights take, blue the region dividers, and red the houses.  If the houses not at the corners are $0.05$ from the corners and $22.5^\circ$ degrees from the edge, it takes $1.42$ units of time to go from N to S and $1.004$ units to get to either other corner.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IiK9s.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This second example does not work by this algorithm. If the house near the south corner is slightly to the west of the house near the center (labeled). The limit of the time to complete this algorithm as d goes to zero is: $2\sqrt{.5}+1 \approx 2.4$. Several other knights are available to get to that last house before the knight that does but is not defined in the algorithm. &lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="71" LastEditDate="2014-06-22T16:35:51.037" LastActivityDate="2014-06-22T16:35:51.037" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="554" PostTypeId="2" ParentId="551" CreationDate="2014-06-02T17:31:03.567" Score="2" Body="&lt;p&gt;You can solve Sudoku with different approaches.  For instance, you can fill in all blank squares with random digits and see if the result meets the Sudoku 1-9 digit-set requirements; if not, try some other random digits. This is a trial-and-error approach. Most people, though, want to use deduction based on the digits previously established in puzzle squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a Sudoku has multiple solutions, deduction will fail to solve the puzzle at some (late) point.  The partial solution sequence will reach a point where no empty square can be deduced as having to be a specific digit.  For people who do not use guessing (trial-and-error), this can be a frustrating barrier (probably making them curse the puzzle maker).&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2014-06-02T17:31:03.567" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="555" PostTypeId="2" ParentId="550" CreationDate="2014-06-02T17:52:02.203" Score="11" Body="&lt;p&gt;A bit of digging around finds a&lt;a href=&quot;http://tb7.chessok.com/articles/Top8DTM_eng&quot; rel=&quot;nofollow noreferrer&quot;&gt; position&lt;/a&gt;, taken from the 7-piece endgame databases, that's claimed as  a mate in 549 moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YuSef.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this isn't a 'constructed' position; there's no theme to the mating sequence, and the moves are the classic 'no rhyme or reason' dance that shows up so much in these multi-piece endgames.  As far as actual chess problems go, the longest I was able to find, for a legal position, was a mate in 271 by Nenad Petrovic, seen &lt;a href=&quot;https://www.yacpdb.org/#97037&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LCA3p.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;White starts with 1. Bb1: the core idea here is that White wants to gain access to a6; Black shuttles his king back and forth between a8, b7 and c8, as White plays Ka5-a4-a3-a2-a1, Ba2, Ka1-b1-c1-d1-e1-f1-f2-e1-d1-c1-b1-a1, Bb1, Ka1-a2-a3-a4-a5.  Since White has 'lost a move' with the e1-f1-f2-e1 triangulation, Black now has to move a pawn in lieu of moving his King and giving White access to a6.  Black's available pawn moves are Pf6-f5-f4, Pf7-f6-f5, Ph4-h3-h2, Ph7-h6-h5-h4-h3 before he finally runs out of pawn moves and must allow White access to a6 to queen his lead b-pawn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that Bb1 must happen first because the Bishop has to be on b1 whenever the king isn't in the a1-f2 corridor, to prevent Black from winning by playing Pd3! and then queening either the d or c pawns.)&lt;/p&gt;&#xA;" OwnerUserId="266" LastEditorUserId="266" LastEditDate="2020-02-15T17:08:06.297" LastActivityDate="2020-02-15T17:08:06.297" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="556" PostTypeId="2" ParentId="551" CreationDate="2014-06-02T18:15:49.740" Score="4" Body="&lt;p&gt;Many of the advanced solving techniques rely on there being a unique solution. All of the &lt;a href=&quot;http://hodoku.sourceforge.net/en/tech_ur.php&quot; rel=&quot;nofollow&quot;&gt;Unique Rectangle&lt;/a&gt; techniques are predicated on this. To quote from that page:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Uniqueness Techniques are based on the fact that practically every Sudoku sudoku ever published has only one solution. More than that: Most Sudoku players consider sudokus with more than one solution invalid!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would say that uniqueness is the &lt;em&gt;default convention&lt;/em&gt;, in that unless you explicitly say otherwise, everyone will assume that any Sudoku puzzle has a unique solution, and failure to say that a given puzzle may not have a unique solution could result in unhappy customers.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2014-06-02T18:15:49.740" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="557" PostTypeId="2" ParentId="70" CreationDate="2014-06-02T18:34:48.047" Score="9" Body="&lt;p&gt;There are two secrets to this question. First is to realise that you want no player in the circle to guess that the circle looks the same.  If they guess the same circle, it is a wasted guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For these examples, the distinct colors are B (black), W (white), G (grey), N (neon), and F (fish).  There are $N$ players and $N$ color hats.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is never a chance you could both be correct in this one specific puzzle, there is never a chance you could both be wrong.  This has to do with the fact that there are $N \, N$ possible patterns of colors and each player can eliminate $N \, (N-1)$ possibilities by looking at the others.  The remaining $\frac{N \, N}{N\,(N-1)}=N$ possibilities must all be guessed somehow. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;$N=2$&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;For two players this is easy.  One of you simply says: you guess we have the same color while I guess that we have a different.  If the colors are BB or WW, the first person will guess correctly.  If they are BW or WB the second person will.  If they were to decide &quot;I pick B you pick W or we both pick B&quot; then there is a chance they could both be correct and an equal chance you could both be wrong.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;$N=3$&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If we have three players, each player needs a slightly different strategy to make this work! Decide, therefore, to elect an equalist (player 1) who believes that either every seat is the same (all the same color) OR every color is the same (all colors appear once).  If he sees BB he guesses B.  If he sees BW he guesses G.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The person in the chair to his right (player 2) knows what player 3 is wearing and what player 1 is wearing as well as player 1's strategy.  If player 3 is B and player 2 thinks he is B, player 1 will guess B.  There is no point in player 2 guessing that he is the same as the other two if they are equal, therefore, so he guesses (for this case) W.  If, however, the arrangement is B_G, picking W will backfire as player 1 will guess G.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easiest way to do this is to say that the colors W, G, and B are in a circle holding hands like the children you must be.  Player two figures out what color Player 1 would pick in his situation and picks the one to that color's right.  Player 3 does the same but picks the color to the left.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;$N \gt 3$&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The strategies for all players for any number of players and colors must meet only two criteria: (a) every player (A) must be able to tell by looking at each other players what color A's hat must be in order for that other character to guess correctly; (b) all player's guesses must be coordinated by (a) so that they are all different.  The previous examples show examples as to how this works.  Essentially, there are only $N$ different colored hats you could be wearing and if the other players take care of $N$-1 of them, you only have 1 left to cover...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One strategy to do this is to give every player a &quot;favorite&quot; color and then each color a proper fraction. This fraction is the seat (starting at 0) that has it as their favorite divided by $N$.  Aka if $N=5$: $\mathrm{B}=0$; $\mathrm{W}=\frac{1}{5}$; $\mathrm{G}=\frac{2}{5}$; $\mathrm{N}=\frac{3}{5}$; $\mathrm{F}=\frac{4}{5}$&#xA;You imagine that there is a circle in the center of the seats where every color is arranged in the same order as the favorite colors.  For every colored hat you see, you rotate that circle by whatever fraction of a circle that color is represented by.  When you are done, the color in front of each player is the color they would guess if they saw exactly what you see.  You should guess the color in front of you! As these are all different and your answer is dependent on every other players' hat color, this strategy works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are likely other strategies that work for every $N\gt2$ but there do not seem to be any that are easier to visualize or generalize for any $N$ than this!&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="11110" LastEditDate="2018-03-01T15:45:45.377" LastActivityDate="2018-03-01T15:45:45.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="558" PostTypeId="1" CreationDate="2014-06-02T18:41:40.850" Score="4" ViewCount="3832" Body="&lt;p&gt;On searching the internet for difficult crossword puzzles I find that the New York Times is often mentioned.  But is there a rating system? If so how does it work? &lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="8874" LastEditDate="2015-01-31T18:56:09.560" LastActivityDate="2015-03-17T15:18:39.080" Title="Is there a rating system for crossword puzzles?" Tags="|crosswords|difficulty-rating|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="559" PostTypeId="1" AcceptedAnswerId="615" CreationDate="2014-06-02T21:05:15.103" Score="11" ViewCount="2621" Body="&lt;p&gt;I did some turning on my 5x5x5 Rubik's Cube and I now have the cube like on the image below. 4 times a smiley &lt;code&gt;-_-&lt;/code&gt; and 2 times a solved plane. But I have no idea how I did this, does someone know the algorithm? The yellow plane is solved, the red plane has the smiley in orange and the green plane has the smiley in blue. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wZxJc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3tNpc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;br&gt;&#xA;The the smileys on the hidden faces are the same as in the opposite face. If you look straight at the smiley &lt;code&gt;-_-&lt;/code&gt; and turn the cube $180^\circ$ on the y-axis the smiley is orientated exactly the same &lt;code&gt;-_-&lt;/code&gt;. Example when turning the cube as mentioned with yellow on the table you see red/orange smileys oriented correctly and the blue/green smileys orientated $90^\circ$ wrong. &lt;/p&gt;&#xA;" OwnerUserId="146" LastEditorUserId="1477" LastEditDate="2014-12-17T21:23:07.570" LastActivityDate="2014-12-17T21:23:07.570" Title="5x5x5 Rubik's Cube Smiley" Tags="|rubiks-cube|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="560" PostTypeId="2" ParentId="559" CreationDate="2014-06-03T05:38:46.213" Score="6" Body="&lt;p&gt;This is brute-force approach, but it will work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose you can solve the cube entirely. Then you can make smiles in the same way, just imagine that they are the solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is too hard to imagine, you can use stickers:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Take solved cube.&lt;/li&gt;&#xA;&lt;li&gt;Put 20 stickers of 4 colours (5 red, 5 orange, 5 blue, 5 green) on 4 cube planes in a way they create smiley like on the photo.&lt;/li&gt;&#xA;&lt;li&gt;Solve the cube.&lt;/li&gt;&#xA;&lt;li&gt;Take stickers away.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2014-06-03T08:28:03.667" LastActivityDate="2014-06-03T08:28:03.667" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="561" PostTypeId="1" AcceptedAnswerId="562" CreationDate="2014-06-03T07:54:03.490" Score="5" ViewCount="7680" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XAvsu.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;I am trying to find out if there are any 'optimal' moves in the game, so as to trap my enemy to put his pawns in the place where &lt;strong&gt;I&lt;/strong&gt; need them&lt;/p&gt;&#xA;" OwnerUserId="50" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2021-12-28T05:06:58.153" Title="Is there any way to 'trap' my opponent on Connect Four?" Tags="|strategy|board-games|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="562" PostTypeId="2" ParentId="561" CreationDate="2014-06-03T08:37:17.350" Score="3" Body="&lt;p&gt;forcing only happens when you can either make a 4 in a row next turn (forcing him to put his piece there to remain in the game) or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a 2 in a row with 2 spots one way and 1 spot the other:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2  1  B  B  2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;you could place a piece in &lt;code&gt;1&lt;/code&gt; while the &lt;code&gt;2&lt;/code&gt;s are free then you have a guaranteed win, he needs to place in any of the free spots to be able to block you&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2014-06-03T08:37:17.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="563" PostTypeId="1" AcceptedAnswerId="624" CreationDate="2014-06-03T13:50:42.940" Score="7" ViewCount="32542" Body="&lt;p&gt;The &lt;a href=&quot;http://www.puzl.co.uk/gear-cube-extreme-level-p-534.html&quot; rel=&quot;nofollow&quot;&gt;Gear Cube Extreme&lt;/a&gt; is another Mechanical Puzzle. I ordered it not so long ago and a friend of mine &quot;accidentally&quot; scrambled it. Now I want to solve it again but I have no idea how or even where to start as it is my first non standard Rubik's cube to solve.&lt;/p&gt;&#xA;" OwnerUserId="113" LastEditorUserId="8874" LastEditDate="2015-02-06T14:46:33.297" LastActivityDate="2015-09-09T06:18:00.443" Title="How can I solve the Gear Cube Extreme?" Tags="|mechanical-puzzles|twisty-puzzles|geared-puzzles|" AnswerCount="3" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="564" PostTypeId="2" ParentId="546" CreationDate="2014-06-03T14:29:19.877" Score="3" Body="&lt;p&gt;It may not be the fastest technique, but I generally take the trivial solution of starting with a row with all one color.  This tells me how many of that color exist in the solution.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If none, I simply move on to another color.  If one or more match, I leave that many of the color and move on to the next color for the remaining spaces.  Using this technique it will determine the exact set of colors within 6 moves, but not the order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I am working on that technique, I also start working on determining position by swapping positions to rule out possible positions for each color.  This is where the technique I use gets a bit more complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I look carefully at the previous combinations and select combinations that will eliminate positions for certain colors based on whatever output I get, for example, placing colors I know are not present (or already know the position of) to blank portions of the board.  For each color I lock in, the number of possible guesses is reduced on future guesses.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If possible, I will also try solving more than one peg simultaneously by leaving one in place and moving the other.  If I don't lose a black peg, then I know that the one I didn't move was correct, if I gain a black peg, I know that both are correct.  If I lose a black peg, I know that the one I moved was correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this technique, I can reliably win pretty much any mastermind game, though I do sometimes use most of my guesses, so it isn't the most efficient solution out there.&lt;/p&gt;&#xA;" OwnerUserId="65" LastActivityDate="2014-06-03T14:29:19.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="565" PostTypeId="2" ParentId="540" CreationDate="2014-06-03T14:31:55.150" Score="1" Body="&lt;p&gt;I had my solver use a simple process of elimination.  I gave each cell an array of possible values and linked them by row, column and grid.  Then, whenever a number is selected, I removed it from the array of possible values in the grid, row and column for that number being selected.  I then iteratively looked for the space with the fewest possible options and selected it next.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are further efficiencies that can be gained, but this is one of the simplest ways to implement a solver.&lt;/p&gt;&#xA;" OwnerUserId="65" LastActivityDate="2014-06-03T14:31:55.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="566" PostTypeId="2" ParentId="561" CreationDate="2014-06-03T15:25:37.870" Score="2" Body="&lt;p&gt;In addition to @ratchetfreak's answer, it is possible to indirectly force a move in another way: when you can place a piece on your next turn such that there are multiple slots on which you can win (branching). This is a generalization of the &quot;two free slots&quot; case presented in his answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x&#xA;oxx&#xA;oxox&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If x moves on row 2 (counting from the left), he is guaranteed a win on the subsequent turn; therefore, o must place a piece to block this branching first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more complex example of this could be multi-stage branching, which could look something like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;oxoxo&#xA;oxxxo&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;o is forced to move in one of rows 1-5 in order to avoid losing; try playing it out to see why. (Needless to say, this would be very rare in a real game, but it's theoretically possible.)&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2014-06-03T15:25:37.870" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="567" PostTypeId="2" ParentId="313" CreationDate="2014-06-03T15:50:50.540" Score="7" Body="&lt;p&gt;Divide 2 by one less than the number you were thinking of. Is the result an even number?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Yes: You were thinking of 2 (because 2/(2-1) = 2)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;No: You were thinking of 3 (because 2/(3-1) = 1)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Unanswerable: You were thinking of 1 (because 2/(1-1) = ∞)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="40" LastActivityDate="2014-06-03T15:50:50.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="575" PostTypeId="1" AcceptedAnswerId="577" CreationDate="2014-06-03T22:26:26.933" Score="19" ViewCount="74503" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rmanH.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;If I'm not mistaken, this is a children's puzzle, but it got me thinking for a while.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;The bus is driving on the autobahn. Relative to you, which way is the bus going? Left or Right?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="50" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2020-05-30T15:15:18.910" Title="Which way is the bus going?" Tags="|visual|" AnswerCount="6" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="576" PostTypeId="2" ParentId="561" CreationDate="2014-06-04T00:36:23.100" Score="2" Body="&lt;p&gt;In addition to the answers above, another key concept for Connect-4 is the notion of &lt;em&gt;Zugzwang&lt;/em&gt;: having to make a move that one doesn't want to.  Often parity is a way to force this; eventually a player is forced to play in a column that they don't want to.  A very simple example of this concept is the following structure:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   x&#xA;  xo&#xA;  oo&#xA;__xx&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Suppose it's X's turn to move here (the underscores represent an empty column).  Then X can force a win by playing on the left hand side:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   x&#xA;  xo&#xA;  oo&#xA;x_xx&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now, o has to move into the second column to block the four-across - but doing so lets x win on the diagonal instead, by playing above o's piece.  This particular configuration is obviously highly contrived - but 'locking' columns so that it's suicidal for one player to play there is often easier than it might seen; once that's done, it becomes a question of leaving that player with no other good moves.&lt;/p&gt;&#xA;" OwnerUserId="266" LastActivityDate="2014-06-04T00:36:23.100" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="577" PostTypeId="2" ParentId="575" CreationDate="2014-06-04T01:50:37.963" Score="40" Body="&lt;p&gt;I saw this problem on a web page once.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The initial thought that most people have is that you can't tell, because the bus is totally symmetrical. However, this misses one important detail that apparently children will imply in their minds more often than adults.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The canonical answer is that &lt;em&gt;this picture of a bus is missing its door&lt;/em&gt;, and therefore the bus must be going &lt;em&gt;left&lt;/em&gt;, because the door is on the &lt;em&gt;right&lt;/em&gt; side of the bus and there's no door visible on the side that's facing you in the picture.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Of course, as Gilles pointed out, depending on which country you're in, the bus could be driving on either side of the road. If you're in Britain or some of the Commonwealth countries, then the bus must be going &lt;em&gt;right&lt;/em&gt;, because the door is on the &lt;em&gt;left&lt;/em&gt; side of the bus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And depending on where from the bus you're exiting, the door could be on the top or bottom as well. I've never seen a bus that had its wheels so far below the chassis like on this one, and maybe there's a stairwell that opens from the bottom to let people in and out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, it could be a maglev train given that the car appears to be floating above two round things. A maglev train has doors on &lt;em&gt;both&lt;/em&gt; sides. In both of the above cases, you can't tell which way the vehicle is going without further information.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The question was updated to indicate that the bus was driving on the Autobahn. However, a &lt;em&gt;bus&lt;/em&gt; on the &lt;em&gt;Autobahn&lt;/em&gt; sounds really unsafe, and so we might actually be viewing a tipped-over bus from a bird's eye view (it would certainly explain the detached wheels). In such a case, you still couldn't tell which way the bus was going unless you knew which side of the highway it was driving on.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-08-20T17:42:10.650" LastActivityDate="2015-08-20T17:42:10.650" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="579" PostTypeId="1" CreationDate="2014-06-04T05:48:55.963" Score="27" ViewCount="2085" Body="&lt;p&gt;I'm looking for a more efficient algorithm to rotate the center column of the Octo-Star puzzle 45 degrees.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FBiXj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FBiXj.jpg&quot; alt=&quot;Octo-Star&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The situation below occurs when four properly-oriented edges are next to each other, allowing for full freedom of movement.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rtCo4.jpg&quot; alt=&quot;Octo-Star face&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;As can clearly be seen, if the pieces highlighted in green are rotated 45° clockwise, the face is effectively solved. On this puzzle, you often want to rotate the centerpieces 45°, but the algorithm I currently have for doing so is slow.&lt;/p&gt;&#xA;&lt;p&gt;With &lt;code&gt;[face]&lt;/code&gt; representing a 45° turn, it is: &lt;code&gt;X = R2' D2 R2; [X U' X' U']4&lt;/code&gt;&lt;/p&gt;&#xA;&lt;p&gt;This algorithm takes 32 moves to complete, and I execute it frequently.&lt;/p&gt;&#xA;&lt;p&gt;Is there a faster way to rotate these pieces?&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorUserId="83061" LastEditDate="2023-04-06T01:30:12.147" LastActivityDate="2023-04-06T01:30:12.147" Title="Is there a move-efficient way to rotate the center pieces on the Octo-Star?" Tags="|logical-deduction|optimization|mechanical-puzzles|twisty-puzzles|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="580" PostTypeId="1" CreationDate="2014-06-04T10:05:53.737" Score="10" ViewCount="889" Body="&lt;p&gt;Does anybody know a nice chess puzzle, where the question states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Exactly one piece is drawn incorrectly on this diagram; another piece must be drawn on its position. Find the incorrectly drawn piece.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One needs to make retrograde analysis of the given position and find which of all the pieces is drawn incorrectly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really would like to find such a puzzle. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;P.S. I would rather like to &lt;em&gt;find&lt;/em&gt; an old puzzle like this than to see you &lt;em&gt;creating&lt;/em&gt; a new one, because:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I understand that it is easy to create a simple puzzle like &lt;a href=&quot;https://braingames.ru/chess/kw11pb22pb31pb33kb32rw21rw12.gif&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;. But this would be too simple.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When you try to create it, it is easy to miss some combination. Meanwhile old puzzles are checked &quot;by time&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;But if you really would like to try to create a nice puzzle like this - you are welcome.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorDisplayName="user88" LastEditDate="2016-05-08T15:36:03.877" LastActivityDate="2016-05-08T15:36:03.877" Title="Is there a retrograde analysis puzzle like &quot;Point on the incorrectly drawn chess piece&quot;?" Tags="|chess|retrograde-analysis|" AnswerCount="1" CommentCount="19" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="581" PostTypeId="1" CreationDate="2014-06-04T11:04:06.430" Score="18" ViewCount="1888" Body="&lt;p&gt;I am thinking how to reformulate a physical puzzle in a way that it becomes purely logic based (doesn't uses physical lows and can be solved by a person, who knows nothing about physics).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The puzzle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The electric circuit in the image is made from the infinite number of wire-frame-squares. The next square is exactly sqrt(2) smaller than the previous. The resistance of used wire is R per one side of the largest square. What is electrical resistance between A and B?&#xA;  &lt;img src=&quot;https://i.stack.imgur.com/8ofuC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Solution explores symmetry with respect to the diagonal AB line. Let's imagine one runs current from A to B. Potential of A is 1V, potential of B is 0V.&#xA; &#xA; 1. All symmetric points will have the same potential, therefore they can be connected together.&#xA; &#xA; 2. If we consider the second diagonal (let's call it CD, so square would be ACBD) we can note that all points on CD line have same way to point A and to point B. That means they all will have 0.5V potential and can be connected of disconnected freely.&#xA; &#xA; 3. Now we can make and equivalent circuit (see the picture)&#xA; &lt;img src=&quot;https://i.stack.imgur.com/8hh3z.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA; and find that Rc = (1 + sqrt(3) + sqrt(2))/2 R.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So can someone reformulate it in pure logic-based way?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My attempt:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Megamind created a micro robotic ants.&#xA;  Once upon a time he asked the ants to move a ton of sand through his anthill (from the entrance A to exit B) as quickly as possible. Anthill is a complex system of passes (see the figure, the side of the big square - 1 m). There are a lot of ants, they are very smart and always able to find the best approach to the problem. However, the throughput  of passes is very limited - one ant can't overtake the other one. How long will it take for ants to complete the task?&#xA;  It is known that for simple straight passage from A to B the ants could complete the task in sqrt(2) days?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here I am trying to simulate series and parallel resistors connections. If two passes with lengths L combined in series configuration it will be a path 2L, which will take twice more time to pass the sand through. If two passes with lengths L combined in parallel configuration it will be a 2 times wider path, which will take twice less time to pass the sand through. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are critical issues with this formulation:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I don't know how to express &quot;limited throughput&quot; strictly and naturally. If ants are points (case A) they can be as close as needed to each other and basically became one point and use only one path. If ants have some length (case B), then it will take time to fill all paths and free them. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If we consider parallel connection following a series connection of 3 resistors: 1 || 1 + 1 the resistance would be 0.5+1=1.5. Mean while in case (B) ants will wait at the correction, so &quot;resistance&quot; will be 1 + 1 = 2. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="28" LastEditorUserId="2484" LastEditDate="2014-11-25T22:05:58.900" LastActivityDate="2018-06-06T04:14:24.780" Title="How to create a pure mathematical-logical based puzzle equivalent to resistance based physical puzzle?" Tags="|logical-deduction|puzzle-creation|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="582" PostTypeId="1" CreationDate="2014-06-04T17:06:20.787" Score="7" ViewCount="829" Body="&lt;p&gt;I am attempting to create original Verbal Arithmetic puzzles such as the famous one from this question &lt;a href=&quot;https://puzzling.stackexchange.com/questions/51/verbal-arithmetic-general-strategy&quot;&gt;Alphametic (Verbal Arithmetic) general strategy&lt;/a&gt;&#xA;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;SEND+MORE=MONEY&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For these problems each letter represents a different one digit number but the equation still holds true.  The first letter (M or S in this case) cannot be 0 of course as that would mean they would be omitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My main issue with creating new ones is that most of the puzzles I develop have more than 1 solution.  For example, there are 3 correct and distinct solutions to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;TWO x SIX = TWELVE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In this case, I can reduce this to one answer by making it a long multiplication problem with marked blank areas for where digits are needed but this sort of ruins the problem.  My question, therefore, is what strategies can I use to develop original, unique verbal arithmetic problems that only have one possible solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Can anyone think of any strategies for designing the puzzles, rather than just creating them and testing for the existence of solutions and uniqueness, as suggested by Ross Millikan's answer below?&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="51989" LastEditDate="2019-09-08T06:53:16.907" LastActivityDate="2019-09-08T06:53:16.907" Title="Identify unique Verbal Arithmetic Puzzles" Tags="|puzzle-creation|alphametic|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="583" PostTypeId="2" ParentId="582" CreationDate="2014-06-04T19:59:27.970" Score="4" Body="&lt;p&gt;I would write a program that checks these for answers.  As $10!=3,628,800$ is small, it will be able to check any puzzle instantly.  You have to decide what forms you want it to do.  Then for ones like $TWO \times SIX = TWELVE$ you have have a calling program that goes through the multiplication table, or you can have one that does partitions checking $ONE+FOUR=FIVE$ or $ONE + FOUR+ FOUR=NINE$, first checking that the number of letters is acceptable, then checking for a solution.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="99" LastEditDate="2014-06-27T20:37:40.483" LastActivityDate="2014-06-27T20:37:40.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="587" PostTypeId="2" ParentId="509" CreationDate="2014-06-04T23:48:38.180" Score="2" Body="&lt;p&gt;You can look up possible combinations for a word like this one using a Regular Expression search at &lt;a href=&quot;http://www.dict.org&quot; rel=&quot;nofollow&quot;&gt;http://www.dict.org&lt;/a&gt;--select one of the two regular expression search options available, and type in the incomplete word, say &lt;code&gt;^c.s.a.t.$&lt;/code&gt;, in the search field. I find it &lt;strong&gt;most helpful indeed&lt;/strong&gt; when I'm stuck.&lt;/p&gt;&#xA;" OwnerUserId="158" LastActivityDate="2014-06-04T23:48:38.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="588" PostTypeId="7" CreationDate="2014-06-05T01:26:31.930" Score="0" Body="&lt;ul&gt;&#xA;&lt;li&gt;Specific questions about the creation and solving of puzzles&lt;/li&gt;&#xA;&lt;li&gt;Real problems or questions that you have encountered&lt;/li&gt;&#xA;&lt;li&gt;Puzzles you've created yourself&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="-1" LastEditorDisplayName="user20" LastEditDate="2016-05-13T20:15:03.733" LastActivityDate="2016-05-13T20:15:03.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="589" PostTypeId="7" CreationDate="2014-06-05T01:27:05.617" Score="0" Body="&lt;ul&gt;&#xA;&lt;li&gt;Anything not directly related to the creation and solving of puzzles&lt;/li&gt;&#xA;&lt;li&gt;Questions that are primarily opinion-based or could have multiple, subjectively-correct answers&lt;/li&gt;&#xA;&lt;li&gt;Questions with too many possible answers or that would require an extremely long answer&lt;/li&gt;&#xA;&lt;li&gt;Questions that are textbook math problems&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="-1" LastEditorDisplayName="user20" LastEditDate="2016-02-07T06:32:59.973" LastActivityDate="2016-02-07T06:32:59.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="590" PostTypeId="7" CreationDate="2014-06-05T01:27:49.370" Score="0" Body="&lt;p&gt;&lt;strong&gt;Puzzling Stack Exchange&lt;/strong&gt; is a question and answer site for those who create, solve, and study puzzles. It's built and run &lt;em&gt;by you&lt;/em&gt; as part of the &lt;a href=&quot;https://stackexchange.com&quot;&gt;Stack Exchange&lt;/a&gt; network of Q&amp;amp;A sites. With your help, we're working together to build a library of detailed answers to every question about creating, solving, and studying puzzles.&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="70029" LastEditorDisplayName="user20" LastEditDate="2021-07-13T11:26:45.033" LastActivityDate="2021-07-13T11:26:45.033" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="591" PostTypeId="2" ParentId="75" CreationDate="2014-06-05T03:19:09.200" Score="1" Body="&lt;ul&gt;&#xA;&lt;li&gt;We know there are $190$ points distributed among the teams, and the allowable scores are $16,9,4,1,0$.&lt;/li&gt;&#xA;&lt;li&gt;You can't have more than $7$ teams with $16$ because they would have to lose too many games in the round robin among themselves.&lt;/li&gt;&#xA;&lt;li&gt;You can have $ 7 \cdot 16, 7 \cdot 9, 3 \cdot 4, 3 \cdot 1$, but there is no team with a unique score.&lt;/li&gt;&#xA;&lt;li&gt;You can also have $ 6 \cdot 16, 9 \cdot 9, 3 \cdot 4, 1 \cdot 1, 1 \cdot 0$.  In that case the three teams that scored $4$ got two points each from the $0,1$ teams and an average of $1$ in their round robin, so they need three more points (which came at the expense of the $16$'s).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There were three upsets.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="39" LastEditDate="2014-06-23T23:56:38.837" LastActivityDate="2014-06-23T23:56:38.837" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="592" PostTypeId="2" ParentId="313" CreationDate="2014-06-05T12:39:53.600" Score="176" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you flip that many coins, will you get two the same?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As Doorknob put it... &quot;1 =&gt; no, because it's only one, 2 =&gt; maybe, because it could be one of TH, HT, TT, or HH, 3 =&gt; yes, because a coin flip has only two possible outcomes&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="385" LastEditorUserId="46903" LastEditDate="2018-09-17T01:11:55.107" LastActivityDate="2018-09-17T01:11:55.107" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="593" PostTypeId="1" AcceptedAnswerId="1676" CreationDate="2014-06-05T13:24:06.427" Score="18" ViewCount="3549" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;There are &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; men. &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; of them are knights, &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; of them are jokers.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; is known, &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; are unknown. You know that: &lt;span class=&quot;math-container&quot;&gt;$K + M = N$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$K \gt M$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$M \ge 1$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; is odd.&lt;/p&gt;&#xA;&lt;p&gt;Knights always tell the truth, jokers can tell whatever they want. All &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; people know who is who. You go to anyone of them and ask him a yes-or-no question. You can repeat this procedure any number of times with the same man or with different men. It is not allowed to ask a question which the man is not able to answer with &amp;quot;Yes&amp;quot; or &amp;quot;No&amp;quot;.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How to determine who is who if you can ask only &lt;span class=&quot;math-container&quot;&gt;$2N$&lt;/span&gt; questions?&lt;/li&gt;&#xA;&lt;li&gt;How to determine who is who with the minimum number of questions?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Explanations:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Each joker chooses what to tell according to unknown arbitrary algorithm. Examples of such algorithms: &amp;quot;Always tell Yes&amp;quot;, &amp;quot;Throw a dice and tell Yes only if you throw a one, otherwise tell No.&amp;quot;, &amp;quot;If it is raining tell Yes, otherwise tell No.&amp;quot;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;One question is asked to only one man and results in one answer (which provides 1 bit of information).&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;I do not know the author's solution. I have figured out a solution by my own, and now I would like to check whether it is correct. For this I need someone who figured out it independently. Also I don't have prove of optimality in b).&lt;/p&gt;&#xA;&lt;p&gt;To show that the puzzle is &amp;quot;objective&amp;quot; and result is achievable, here is the solution for the case &lt;span class=&quot;math-container&quot;&gt;$N = 3$&lt;/span&gt;:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; In this case there are 2 knights and 1 jokers.&lt;br /&gt;&#xA; 1. Ask the first one &amp;quot;Is the second man a knight?&amp;quot;. If he answers &amp;quot;Yes&amp;quot; then the 2nd is a knight. Indeed, let's suppose second is joker, then 1st is knight and he lied. If he answers &amp;quot;No&amp;quot; then 1st or 2nd is a joker and third is knight.&lt;br /&gt;&#xA; 2. Ask the found knight &amp;quot;Is the first one a knight?&amp;quot; and you will identify one more person, the identity of the last one is derived from the fact that there is exactly one joker.&lt;br /&gt;&#xA; 3. You need to identify 3 persons, which is 8 combination. &lt;span class=&quot;math-container&quot;&gt;$KKK, JJK, JKJ, KJJ, JJJ$&lt;/span&gt; are discarded by condition &lt;span class=&quot;math-container&quot;&gt;$K \gt M, M \ge 1$&lt;/span&gt;, that leaves us with 3 combinations. So you can't do it in less than 2 questions - this is the minimum.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="28" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-12-25T17:05:59.863" Title="Knights and jokers" Tags="|logical-deduction|liars|optimization|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="594" PostTypeId="1" AcceptedAnswerId="597" CreationDate="2014-06-05T14:47:23.357" Score="6" ViewCount="873" Body="&lt;p&gt;If a rod of unit length is broken into $n$ pieces, what is the expected length of the longest piece? The positions at which the rod is broken are chosen randomly uniformly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a generalization of a puzzle I recently encountered, the answer to which I do not know myself:&lt;/p&gt;&#xA;" OwnerUserId="198" LastEditorUserId="463" LastEditDate="2015-06-05T12:36:34.737" LastActivityDate="2015-06-05T12:36:34.737" Title="What is the expected length of the longest piece?" Tags="|mathematics|probability|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="597" PostTypeId="2" ParentId="594" CreationDate="2014-06-05T16:01:34.030" Score="7" Body="&lt;p&gt;This is known as the &lt;em&gt;broken stick rule&lt;/em&gt;. The &lt;a href=&quot;http://demonstrations.wolfram.com/BrokenStickRule/&quot; rel=&quot;nofollow&quot;&gt;Wolfram Demonstrations Project&lt;/a&gt; claims that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the unit interval is divided into  $n $ spaces by $ n-1$ random points, the expected size of the $i^{\text{th}}$ largest space is $$\frac 1n \sum_{j=0}^{n-i} \frac 1{n-j}$$  Setting $i=1$ gives the expected size of the largest space as $$\frac 1n H_{n-1}\approx \frac 1n(\ln(n-1)+\gamma)$$ where $H_{n-1}$ is the $n-1^{\text{st}}$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Harmonic_number&quot; rel=&quot;nofollow&quot;&gt;harmonic number&lt;/a&gt; and $\gamma \approx 0.5772$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant&quot; rel=&quot;nofollow&quot;&gt;Euler-Mascheroni constant&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="73" LastEditorUserId="71" LastEditDate="2014-06-22T16:48:07.430" LastActivityDate="2014-06-22T16:48:07.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="599" PostTypeId="2" ParentId="563" CreationDate="2014-06-06T03:05:18.903" Score="0" Body="&lt;p&gt;Jaap's Puzzle Page has lots of helpful twisty solutions. Here's the link for the Gear Cube Extreme:&#xA;&lt;a href=&quot;http://www.jaapsch.net/puzzles/gearcube2.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.jaapsch.net/puzzles/gearcube2.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This site also has steps to a solution, with some pictures to assist:&#xA;&lt;a href=&quot;http://ruwix.com/twisty-puzzles/gear-cube-extreme-ultimate-how-to-solve-gearcube-solution/&quot; rel=&quot;nofollow&quot;&gt;http://ruwix.com/twisty-puzzles/gear-cube-extreme-ultimate-how-to-solve-gearcube-solution/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="447" LastEditorUserId="447" LastEditDate="2014-06-06T03:19:05.070" LastActivityDate="2014-06-06T03:19:05.070" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="600" PostTypeId="2" ParentId="227" CreationDate="2014-06-06T03:15:41.710" Score="2" Body="&lt;p&gt;For tips on super-precise wooden puzzlemaking, check out this tutorial from Lee Krasnow, one of the masters of puzzlecrafting:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.instructables.com/id/Precision-Puzzlemaking-Primer----Volume-1/&quot; rel=&quot;nofollow&quot;&gt;http://www.instructables.com/id/Precision-Puzzlemaking-Primer----Volume-1/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="447" LastActivityDate="2014-06-06T03:15:41.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="603" PostTypeId="1" AcceptedAnswerId="611" CreationDate="2014-06-06T08:29:35.057" Score="11" ViewCount="757" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;The colors of the pieces in the diagram is unknown. The diagram can be obtained from a real game. One needs to find the color of each piece: Black or White.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9FqvN.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I do not know the full solution.&lt;/p&gt;&#xA;&lt;p&gt;It is easy to find the colors of most of the pieces:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 1. The kings should not be checkmated by pawns (otherwise what was the previous move?), so the upper one is white.&lt;br&gt;&#xA; 2. The knight on g4 gives a checkmate to some king, so this was the last move.&lt;br&gt;&#xA; 3. All other pieces should not give a check to any king, so we find the colors of them.&lt;br&gt;&#xA; 4. There are 12 pawns and 6 queens, so 4 queens were created, all other pieces are original.&lt;br&gt;&#xA; 5. There are 2 white knights on the top, so the knight on g4 is black.&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/iQjOk.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Then one needs to consider different possibilities of the pawns' placements, taking into account possible moves. It is hard for me to find a solid approach to it.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-06-04T16:39:21.117" Title="Find the color of the chess pieces" Tags="|chess|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="604" PostTypeId="1" CreationDate="2014-06-06T09:16:03.063" Score="9" ViewCount="2006" Body="&lt;p&gt;Given an 8x8 chessboard, your goal is to &quot;cover&quot; each space on the board. A space is &quot;covered&quot; if there is a piece on it, or if a piece on the board can be moved to that space in one move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have $n$ Queens, what is the minimum number of added Knights you have to have to do this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we solve this for $n = 1, 2, 3, 4$ ?  For $n = 5$ the answer is zero of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each Queen covers at most 29 new squares and each Knight covers at most 9.  So we know that we need at least $4, 2, 0, 0$  Knights $n = 1, 2, 3, 4$ .  However, we also know we need at least one Knight for $n &amp;lt; 5$.&lt;/p&gt;&#xA;" OwnerUserId="8998" LastEditorUserId="99" LastEditDate="2014-07-08T18:26:37.483" LastActivityDate="2018-04-03T14:01:47.307" Title="How many Queens and Knights does it take to cover a chessboard?" Tags="|chess|checkerboard|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="605" PostTypeId="1" AcceptedAnswerId="607" CreationDate="2014-06-06T13:08:57.577" Score="3" ViewCount="651" Body="&lt;p&gt;I'm thinking about the following strategy for &lt;a href=&quot;https://puzzling.stackexchange.com/questions/548/fastest-way-to-collect-an-arbitrary-army&quot;&gt;Fastest way to collect an arbitrary army&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;When a soldier decides to go to some house he &quot;reserves&quot; it.&lt;/li&gt;&#xA;&lt;li&gt;Once a soldier is free (has delivered the news to another house) he takes the closest house from all unreserved and goes there. If there are several closest houses he chooses at random.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My question is &quot;Is there a house placement example for which this strategy gives a time bigger than $2+\sqrt{2}$?&quot;.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-07-15T15:18:48.207" Title="Does this strategy work?" Tags="|mathematics|geometry|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="607" PostTypeId="2" ParentId="605" CreationDate="2014-06-06T13:38:32.657" Score="5" Body="&lt;p&gt;No I don't think that strategy works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ubXti.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we consider a clusters of very close houses such that every house will be reserved before they are reached, it is easy in imagine arbitrarily long paths like the one in the top left of my picture.  The arrangement in the diamond would fail yours but pass mine.  Each number represents a very dense arrangments of that number of houses.  At least 1 house will not be reached before your due date much less get back to the castle.&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2014-06-06T13:38:32.657" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="608" PostTypeId="2" ParentId="605" CreationDate="2014-06-06T13:40:35.043" Score="2" Body="&lt;p&gt;Yes, there is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider a spiral, the distance between neighbouring turns being 0.25. Houses are placed along the spiral in groups, the distance between groups being 0.125. Each group (starting from the centre) has twice as many houses as the previous one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to show that the soldiers will go consecutively from one group to another. So the path will take $2*0.25+2*0.5+2*0.75+3*1+\sqrt{0.5} = 6.7 &amp;gt; 2+\sqrt{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zXnTy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="5373" LastEditDate="2015-07-15T15:18:48.207" LastActivityDate="2015-07-15T15:18:48.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="611" PostTypeId="2" ParentId="603" CreationDate="2014-06-06T14:58:57.450" Score="9" Body="&lt;p&gt;I've made some progress by looking at the number of captures (7) vs the number of pawns that have changed file. The extra f pawns can't have come from the left in only seven captures, so we should have four captures bringing them from the right.&#xA;h-&gt;g x2&#xA;g-&gt;f x2&#xA;The extra d pawn must have come from the left, so we get at least one more capture.&#xA;c-&gt;d&#xA;And in order for the a and b pawns to no longer be there, a pawn must have moved &lt;em&gt;from&lt;/em&gt; both (the other must have been promoted), although this doesn't tell us where the b pawn went &lt;em&gt;to&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that's all captures accounted for, and no pawn has left d, e or f. That means one white pawn must be below one black pawn in each file. And there is only one way to fill those six pawns in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the extra f pawns came from the right, there can have been no promotions on the right side. So the f and g pawns all came from f, g and h. This means we need three white pawns and three black. We have two of each in f, so the g pawns must also be one of each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose there were white pawns at e2 and g2. They must have been there the whole game, so the bishop that started at f1 could not have moved away, nor could it have been captured because no promotions happened on the right side. Since the bishop is no longer at f1, and we know the e2 pawn is white, the g2 pawn must be black. And hence the g3 pawn must be white.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that black made two captures on the right side to get its pawns across, and one further capture in order to have the second d pawn. Since the second black d pawn came from c, if the c pawn is also black, it must have come from b, which would mean a fourth capture for black. But black has only captured three pieces. So the c pawn must be white.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We've now located three white queens and six white pawns in total, so all the other queens must be black.&lt;/p&gt;&#xA;" OwnerUserId="385" LastEditorUserId="385" LastEditDate="2014-06-07T15:58:05.220" LastActivityDate="2014-06-07T15:58:05.220" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="612" PostTypeId="2" ParentId="548" CreationDate="2014-06-06T15:48:45.443" Score="6" Body="&lt;p&gt;I've already shown how you can force my other answer to fail.  Hopefully this one will not have such an exploit.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Step 1: First knight goes to the extreme house on the map.&#xA;(Equivalent to just going the the north house if there is a house&#xA;there.) (Maximum time $\frac{\sqrt{2}}{2}$)  For a sometimes faster&#xA;but harder to visualize strategy, instead have him go to the closest house and&#xA;divide the entire map by the line along which he just travelled.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Step 2: Old knight updates his map by drawing two lines to the houses&#xA;on his left and right with a minimal angle between the such that all&#xA;other houses on his map are between these lines.  He then divides this map&#xA;into two halves by bisecting this angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 3: The old knight and the new knight proceed to the closest&#xA;houses that is on their halves of the map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 4: Each knight follows steps 2 and 3 until he is awarded a map&#xA;with no houses on it (and as a result no area). (Maximum time $2$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 5: Knight proceeds back to the castle. (Maximum time&#xA;$\frac{\sqrt{2}}{2}$)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A potentially simpler to state step 2 would be to merely divide the current map into two equal halves along a line which passes through the current location. Unfortunately, already emptied space would remain on one of these maps making them less useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 2 is easier than is sounds as it is usually trimming an emptied triangle off a triangle then bisecting the new triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to work for every answer so far.  One can imagine on map getting repeatedly divided into halves until nothing remains.  It is area filling so it resolves any issues from house bunches (More at &lt;a href=&quot;https://puzzling.stackexchange.com/questions/605/does-this-strategy-work/607?noredirect=1#comment1497_607&quot;&gt;Does this strategy work?&lt;/a&gt;). No knight can be eliminated without removing at least one house and significantly reducing the number of houses. (what killed my other answer).    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I've been completely unable to create a counter example or a proof.  I welcome any input.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6kItx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This (rather rough) diagram shows several moves of steps 2 and 3 for a random pattern of houses.  the original map is bounded by black.  They second map pink.  This continues through green and blue.  The houses are red and the path traveled is purple.  Original location is black.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2014-06-24T14:20:13.033" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="613" PostTypeId="2" ParentId="604" CreationDate="2014-06-06T20:26:31.343" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/CA6eq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You provided a lower bound. These were done by hand or by referencing other answers and should provide an upper bound.  I would love to see which ones can be reduced further.  I am fairly confident in 0 and 3 queens but less so on 1 and 2.  4 and 5 queens of course have already been pointed out on this site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I thought this was self explanitory in conjunction with the question i want to be clear: Kn stands for knights, Q stands for queens, Maroon squares are ones convered by the queens, Orange squares are ones covered by the knights.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've made a large mistake on this question.  I forgot to take into account the fact that queens cannot jump over knights to attack squares accross from them.  The next figure does better:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SGaum.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditDate="2014-06-12T19:33:03.133" LastActivityDate="2014-06-12T19:33:03.133" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="614" PostTypeId="1" AcceptedAnswerId="1740" CreationDate="2014-06-06T21:16:57.070" Score="8" ViewCount="1149" Body="&lt;p&gt;I want to create a Sudoku puzzle, and I want it to be of moderate difficulty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a specific pattern I want to include, called the X-Wing pattern:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LdynU.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/W4Rua.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This occurs when two numbers form locked pairs - the two squares are the only ones of a specific number in a row or column. As a result, the other in that row or column can be eliminated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to create Sudoku puzzles which require the use of this pattern to be solved, partially because I want to practice it, but also partially because my friends and I like to swap difficult puzzles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently, I use the &quot;generate a random Sudoku and remove numbers&quot; method to create new puzzles, then check in a solver to see whether it's valid and what patterns are used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I go about modifying or creating a new generation process to ensure I will create puzzles with the X-Wing pattern?&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastActivityDate="2014-06-18T20:03:06.107" Title="How do I necessitate the use of the X-Wing pattern during Sudoku creation?" Tags="|sudoku|puzzle-creation|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="615" PostTypeId="2" ParentId="559" CreationDate="2014-06-06T21:29:09.490" Score="8" Body="&lt;p&gt;&lt;strong&gt;Notation&lt;/strong&gt;&lt;br&gt;&#xA;From left layer to right layer:&lt;br&gt;&#xA;$L~l~m~r~R$&lt;br&gt;&#xA;From upper layer to down layer:&lt;br&gt;&#xA;$U~u~M~d~D$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Quick and Intuitive Solution&lt;/strong&gt;&lt;br&gt;&#xA;Take white as front:&lt;br&gt;&#xA;$l^2~(d~u)^2~l^2~(d~u)^2$&lt;br&gt;&#xA;Take red as front and do the same:&lt;br&gt;&#xA;$l^2~(d~u)^2~l^2~(d~u)^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take white as front and red top and do:&lt;br&gt;&#xA;$m~u^2~m'~u^2$&lt;/p&gt;&#xA;" OwnerUserId="146" LastEditorUserId="146" LastEditDate="2014-06-29T08:48:27.967" LastActivityDate="2014-06-29T08:48:27.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="616" PostTypeId="2" ParentId="558" CreationDate="2014-06-07T00:39:23.740" Score="3" Body="&lt;p&gt;As mgkrebbs stated, the most widely known system for ranking the difficulty of crossword puzzles is by the day of the week.  The New York Times is famous for doing this, with many other crosswords following in its footsteps.  Monday is the easiest and Saturday is the hardest, with Sunday puzzles being larger and having the difficulty of about a Thursday puzzle.  The biggest jump in difficulty is from Tuesday and Wednesday as the words get longer and longer and the clues get trickier.  Hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="184" LastActivityDate="2014-06-07T00:39:23.740" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="617" PostTypeId="1" AcceptedAnswerId="1630" CreationDate="2014-06-07T02:01:52.413" Score="8" ViewCount="406" Body="&lt;p&gt;In Wikipedia's bibliography of &lt;a href=&quot;https://en.wikipedia.org/wiki/Martin_David_Kruskal#Personal&quot; rel=&quot;nofollow&quot; title=&quot;Martin Kruskal&quot;&gt;Martin Kruskal&lt;/a&gt; it is said:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Martin, who had a great love of games, puzzles, and word play of all&#xA;  kinds, also invented several quite unusual origami models including an&#xA;  envelope for sending secret messages (anyone who unfolded the envelope&#xA;  to read the message would have great difficulty refolding it to&#xA;  conceal the deed).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, this is the only information in Wikipedia about Kruskal's secret message envelope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Does anyone know the history of this envelope, or where I can find more information about it?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This envelope seems like a puzzle, since there may have been a clever trick to refold it.&lt;/p&gt;&#xA;" OwnerUserId="270" LastEditorUserId="463" LastEditDate="2015-06-04T16:23:40.063" LastActivityDate="2015-06-04T16:27:50.547" Title="Kruskal's hard to refold envelope" Tags="|puzzle-identification|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="618" PostTypeId="2" ParentId="558" CreationDate="2014-06-07T04:34:39.373" Score="1" Body="&lt;p&gt;There is a big contrast with Sudoku rating.  For Sudoku, there is a list of techniques for solving and an agreed ranking of the difficulty of the techniques.  The further down the list you go, the harder the puzzle.  Between lines of puzzles, the number of stars may vary but the order of difficulty generally remains the same.  For crosswords, there is nothing of the sort.  Any difficulty rating is in the eyes of the constructor/editor.  Yes, generally as you go through the week they get harder, but what one person finds difficult somebody else may find easy.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2014-06-07T04:34:39.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="619" PostTypeId="1" AcceptedAnswerId="621" CreationDate="2014-06-07T12:10:04.743" Score="4" ViewCount="6514" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How do you write 23 using only the number 2?&lt;br /&gt; 34 using only the&#xA;  number 3?&lt;br /&gt; 56 using only the number 5?&lt;br /&gt; 100 using only the&#xA;  number 9?&lt;br /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can use only the numbers said, but any math you want in order to get the wanted result&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are allowed to use:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Fractions&lt;/li&gt;&#xA;&lt;li&gt;Addition&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="50" LastEditorUserId="4551" LastEditDate="2015-06-15T21:48:28.037" LastActivityDate="2018-08-30T10:55:54.713" Title="Writing numbers puzzle" Tags="|mathematics|formation-of-numbers|" AnswerCount="6" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="620" PostTypeId="2" ParentId="619" CreationDate="2014-06-07T12:18:31.657" Score="9" Body="&lt;p&gt;Trivially, I can write any integer using only the number N as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 = N/N&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 = N/N + N/N&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 = N/N + N/N + N/N&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you only want one occurrence of the number N, then you're going to have to define &quot;math&quot;. Otherwise I'll choose a math with an operator that multiplies by 11 and then adds 1. :)&lt;/p&gt;&#xA;" OwnerUserId="385" LastEditorUserId="385" LastEditDate="2014-06-07T12:27:41.557" LastActivityDate="2014-06-07T12:27:41.557" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="621" PostTypeId="2" ParentId="619" CreationDate="2014-06-07T13:09:38.523" Score="7" Body="&lt;p&gt;Greg m's answer is correct, but I'll attempt to do it using as few occurrences of the number as possible.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;We can see that for each of your numbers, using $NN + \frac NN$ (where $NN$ refers to stringing two of the digits together) leads to the desired number (e.g. $22 + \frac 22 = 23$, $33 + \frac 33 = 34$, etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if we don't allow the stringing of digits together, then the solutions become a little harder to come up with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $23$, we have $(2 + 2)! - \frac 22$, which also uses four instances of $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $34$, we have $3! \times 3! - \frac {3!}3$, which also uses four instances of $3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $56$, we have $5 \times (5 + 5) + 5 + \frac 55$, which uses six instances of $5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $100$, we have $(9 + \frac99)^{\frac{\sqrt{9}!}{\sqrt{9}}}$, which uses five instances of $9$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Thanks to greg m for all the improvements.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might be able to improve on these, but those are what I see for the moment.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2014-06-13T19:38:42.263" LastActivityDate="2014-06-13T19:38:42.263" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="623" PostTypeId="1" AcceptedAnswerId="1641" CreationDate="2014-06-07T15:35:54.210" Score="25" ViewCount="18777" Body="&lt;p&gt;Write 24 using the four &lt;strong&gt;numbers&lt;/strong&gt; 1, 3, 4, 6 and basic arithmetic.&lt;/p&gt;&#xA;&lt;p&gt;Explanations:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The given are numbers, not digits, so you can't group them using decimal notation. For example, you can't create the number 6431.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Only one occurrence of each number is allowed. For example, you can't write 24 = 6 + 6 + 6 + 6.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;All numbers have to be used.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Basic arithmetic means only the 4 operations: +, -, x, /. Also parentheses (...) are allowed to set the order of operations.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="28" LastEditorUserId="69582" LastEditDate="2020-12-02T15:08:27.270" LastActivityDate="2023-06-24T21:42:52.170" Title="Write twenty-four from four numbers" Tags="|mathematics|calculation-puzzle|formation-of-numbers|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="624" PostTypeId="2" ParentId="563" CreationDate="2014-06-08T06:20:55.677" Score="7" Body="&lt;p&gt;Here's how I solve it:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Solve the middle layer&lt;/li&gt;&#xA;&lt;li&gt;Solve the corners&lt;br&gt;&#xA;2.1 Resolve middle layer parity and centers (if there is a problem)&lt;/li&gt;&#xA;&lt;li&gt;Move the edges to the correct layers&lt;/li&gt;&#xA;&lt;li&gt;Move the edges into the correct positions&lt;/li&gt;&#xA;&lt;li&gt;Rotate the edges into the correct orientations&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;0. Notation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using &lt;code&gt;R, L, B, F&lt;/code&gt; to indicate a 180-degree turn of each side. This is because, as you know, you cannot rotate each side except in increments of 180. &lt;code&gt;U, D&lt;/code&gt;, however, will refer to 90 degree turns of the upper and lower layers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I refer to a corner using three letters, those letters indicate the faces that corner touches. For instance, the UFL corner is the upper front left corner.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1. Solve the middle layer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This step is pretty easy, honestly, and you should be able to do this intuitively using the side moves. However, a couple algorithms that can help if you're truly stuck on this: use &lt;code&gt;R B R&lt;/code&gt; to move the edges to a more understandable position, and &lt;code&gt;R2&lt;/code&gt; or &lt;code&gt;L2&lt;/code&gt; to rotate the middle slice 180 degrees to solve the centers.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2. Solve the corners&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For now, keep one hand firmly on two pieces of the center layer, and don't move it. If you do, you'll need to restore it, which isn't a difficult problem, but I'm not going to cover how to do it because it shouldn't happen. We'll restore the middle layer at the end of this step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I solve this in two steps: first, by solving the bottom layer corners; then, by solving the top layer corners. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The algorithm I use to solve bottom layer corners is pretty simple: &lt;code&gt;R U R U' R'&lt;/code&gt; This moves the UFL corner into the DFL position without disturbing the other pieces on the bottom layer. The process for solving the bottom layer pieces is simple: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Find a bottom-layer corner on the top layer, and find where it should go. Use &lt;code&gt;U&lt;/code&gt; to place the piece in the UFL position, and use &lt;code&gt;D&lt;/code&gt; to place its target location in the DFL position. Then, execute the algorithm.&lt;/li&gt;&#xA;&lt;li&gt;Here, 'target location' refers to a spot relative to other properly-placed pieces.&lt;/li&gt;&#xA;&lt;li&gt;If there are no top-layer pieces on the bottom layer, then just move a piece there for reference.&lt;/li&gt;&#xA;&lt;li&gt;If there are no bottom-layer pieces on the top layer, find an incorrectly-positioned piece on the bottom layer and swap it with any top-layer piece.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;After this, you're on to the top layer corners. One algorithm should be sufficient here: &lt;code&gt;R U R U' R' (U' D) R U' R U R'&lt;/code&gt;. This swaps the UFL and UBL corners. Executing this strategically should lead you to a solved upper layer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2.1 Resolve middle layer parity (if it exists)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the two middle layer pieces you've been holding onto are flipped, then execute the following algorithm: &lt;code&gt;D' R D R (D' R)2 D' R' U' D R'&lt;/code&gt;. At this point, there will be one yellow corner on the top, and a corresponding purple corner on the bottom. Proceed from the beginning of Step 2 - this shouldn't take long to complete. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there's a problem with a particular center, just rotate an adjacent &lt;code&gt;L/R&lt;/code&gt; or &#xA;&lt;code&gt;F/B&lt;/code&gt; slice 180 degrees and check again.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3. Move the edges to the correct layers&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This step takes only one algorithm: &lt;code&gt;(R U R' U')5&lt;/code&gt;. This swaps the UF and BR edges. Use this as many times as needed, placing an incorrect edge in each place and swapping them. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4. Move the edges into the correct positions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this, you need two algorithms. These images are courtesy of &lt;a href=&quot;http://ruwix.com/twisty-puzzles/gear-cube-extreme-ultimate-how-to-solve-gearcube-solution/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ruwix&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SUleU.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/IIuAJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Left: &lt;code&gt;R2 U R2 U&lt;/code&gt;; Right: &lt;code&gt;R D R U2 R2 U2 R D' R'&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should cover most cases intuitively. Use &lt;code&gt;U&lt;/code&gt; and &lt;code&gt;D&lt;/code&gt; to move the pieces in position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you end up with three pieces on one layer that need to be swapped, then do not solve the middle of the three first and you'll be alright. To do this, use the second algorithm twice - the first time will flip two edges on the other layer, but the second time will flip them back.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;5. Rotate the edges into the correct orientations&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, you'll need the following two algorithms:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;code&gt;(R U)2 (R )4 (U' R')2&lt;/code&gt; (CCW)&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;(R U)2 (R')4 (U' R')2&lt;/code&gt; (CW)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These two algorithms will do strange things. They both turn the bottom front edge piece in different directions, which is really the focus of them, but they also turn the UL, UB, and UR edges as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first step is to solve the bottom-layer edges. If the top layer has a solved edge, move it into the UF position. If it doesn't have one, do so immediately after you solve the first bottom-layer piece. For each unsolved edge on the bottom layer, execute the appropriate algorithm above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the cube is not solved, flip it over. Repeat this process for all the edges again. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the cube is still not solved, flip it over again. Once you repeat the process again, the edges are guaranteed to be solved.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastActivityDate="2014-06-08T06:20:55.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="625" PostTypeId="1" AcceptedAnswerId="1700" CreationDate="2014-06-08T07:39:48.533" Score="31" ViewCount="6300" Body="&lt;p&gt;There is a puzzle about two spies:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Two spies must pass each other two secret numbers (one number per spy), unnoticed by their enemies. They have agreed on a method for doing this using only 26 indistinguishable stones in advance.&lt;/p&gt;&#xA;&lt;p&gt;They meet at a river, where there is a pile of 26 stones. Starting with the first spy, they take turns throwing a group of stones into the river: the first spy throws some number of stones, then the second one, then the first one again...&lt;/p&gt;&#xA;&lt;p&gt;Each spy must throw at least one stone on his turn, until all the stones are gone.&lt;/p&gt;&#xA;&lt;p&gt;They observe all throws and diverge when there are no more stones. They keep silence all the time and no information is exchanged except number of stones thrown at each turn.&lt;/p&gt;&#xA;&lt;p&gt;How can they exchange the numbers successfully if the numbers can be from &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;To understand the idea you can read the following simple solution for &lt;span class=&quot;math-container&quot;&gt;$M = 42$&lt;/span&gt;:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 0. Let's call the spies spy 1 and spy 2. Spy 1 has number &lt;span class=&quot;math-container&quot;&gt;$N_1$&lt;/span&gt; to tell Spy 2, and Spy 2 has number &lt;span class=&quot;math-container&quot;&gt;$N_2$&lt;/span&gt; to tell spy 1.&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; 1. The spies agree that each number &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; from &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$42$&lt;/span&gt; shall be represented by a combination of two numbers &lt;span class=&quot;math-container&quot;&gt;$(m_1, m_2)$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$1 \le m_1 \le 7$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$1 \le m_2 \le 6$&lt;/span&gt;. Since there are exactly 42 possible combinations of &lt;span class=&quot;math-container&quot;&gt;$(m_1, m_2)$&lt;/span&gt;, it is possible for them to make this mapping.&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; 2. At the river each spy calculates his pair of numbers &lt;span class=&quot;math-container&quot;&gt;$(m_1, m_2)$&lt;/span&gt; that corresponds to the number he wants to send.&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; 3. Spy 1 throws stones in the river corresponding to his &lt;span class=&quot;math-container&quot;&gt;$m_1$&lt;/span&gt;, then spy 2 throws stones corresponding to his own &lt;span class=&quot;math-container&quot;&gt;$m_1$&lt;/span&gt;, then spy 1 throws &lt;span class=&quot;math-container&quot;&gt;$m_2$&lt;/span&gt; stones, then spy 2 throws &lt;span class=&quot;math-container&quot;&gt;$m_2$&lt;/span&gt; stones.&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; 4. Finally, spy 1 throws all the rest of the stones into the river if any left.&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; This solution works because neither spy throws more than 13 stones, so they can never throw more than 26 in total.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I know the algorithm for &lt;span class=&quot;math-container&quot;&gt;$M = 1716$&lt;/span&gt;. I know that an algorithm for &lt;span class=&quot;math-container&quot;&gt;$M = 2286$&lt;/span&gt; exists, but I do not know what it is. I wrote a program that seems to prove that it is impossible to do if &lt;span class=&quot;math-container&quot;&gt;$M &amp;gt; 2535$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;I would like to know, what is the algorithm for &lt;span class=&quot;math-container&quot;&gt;$M = 2286$&lt;/span&gt;? Is it possible to formulate an algorithm for &lt;span class=&quot;math-container&quot;&gt;$M &amp;gt; 2286$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Please note that people who attempt to solve this puzzle keep making the same mistakes:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;People forget that due to the alternation rule, if spy 1 needs to make &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; throws, then spy 2 can make only &lt;span class=&quot;math-container&quot;&gt;$K-1$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; throws.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;People do not take into account that spies must agree on everything in advance. Therefore it doesn't matter for the second spy how big the first spy's number is, and how many stones he actually needs to transmit this information; each spy's algorithm must work for any possible number (from &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt;), that the other spy can want to tell him.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="28" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2019-07-26T19:28:44.257" Title="Two spies throwing stones into a river" Tags="|logical-deduction|combinatorics|optimization|" AnswerCount="5" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="626" PostTypeId="1" AcceptedAnswerId="628" CreationDate="2014-06-08T09:50:35.680" Score="127" ViewCount="29841" Body="&lt;p&gt;You are on your way to visit your Grandma, who lives at the end of the valley. It's her birthday, and you want to give her the cakes you've made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Between your house and her house, you have to cross 7 bridges, and as it goes in the land of make believe, there is a troll under every bridge! Each troll, quite rightly, insists that you pay a troll toll. Before you can cross their bridge, you have to give them half of the cakes you are carrying, but as they are kind trolls, they each give you back a single cake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many cakes do you have to leave home with to make sure that you arrive at Grandma's with exactly 2 cakes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT :&lt;/strong&gt; If you go to your grandma's with a half eaten cake, she's gonna be pissed. The trolls can't give you half a cake back. It is unhygienic and disgusting.&lt;/p&gt;&#xA;" OwnerUserId="50" LastEditorUserId="50" LastEditDate="2014-06-09T14:07:53.093" LastActivityDate="2020-08-31T19:23:36.623" Title="Paying the Troll toll" Tags="|mathematics|" AnswerCount="12" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="627" PostTypeId="2" ParentId="626" CreationDate="2014-06-08T10:05:23.940" Score="5" Body="&lt;p&gt;&lt;strong&gt;Answer&lt;/strong&gt; is &lt;strong&gt;2&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Prove&lt;/strong&gt; is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have 2 cakes after last troll and X cakes after previous troll then 2=(X/2+1) and X = 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same we conclude that number of cakes after each troll wasn't changed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means you left with 2 cakes.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2014-06-08T20:24:41.853" LastActivityDate="2014-06-08T20:24:41.853" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="628" PostTypeId="2" ParentId="626" CreationDate="2014-06-08T10:39:39.377" Score="97" Body="&lt;p&gt;If you leave home with&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 2 cakes, you will never pay the troll toll. You give him half of your cakes (one) and he gives one cake back to you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So the answer is&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;em&gt;&lt;strong&gt;2&lt;/strong&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;The exact solution :&lt;/h2&gt;&#xA;&lt;p&gt;Assume we have &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; cakes,&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; - after the 1&lt;sup&gt;st&lt;/sup&gt; bridge, we have &lt;span class=&quot;math-container&quot;&gt;$\frac{x}{2}+1 = \frac{x+2}{2}$&lt;/span&gt; cakes&lt;br /&gt;&#xA; - after the 2&lt;sup&gt;nd&lt;/sup&gt; bridge, we have &lt;span class=&quot;math-container&quot;&gt;$\dfrac{\frac{x+2}{2}+2}{2} = \dfrac{x+2+4}{4}$&lt;/span&gt; cakes&lt;br /&gt;&#xA; - after the 3&lt;sup&gt;rd&lt;/sup&gt; bridge, we have &lt;span class=&quot;math-container&quot;&gt;$\dfrac{\frac{x+2+4}{4}+2}{2} = \dfrac{x+2+4+8}{8}$&lt;/span&gt; cakes&#xA; - ...&lt;br /&gt;&#xA; - after the n&lt;sup&gt;th&lt;/sup&gt; bridge, we have &lt;span class=&quot;math-container&quot;&gt;$\frac{x+2+4+\dots+2^n}{2^n}$&lt;/span&gt; cakes&lt;br /&gt;&#xA; - So, after the last bridge (&lt;span class=&quot;math-container&quot;&gt;$n=7$&lt;/span&gt;), we have &lt;span class=&quot;math-container&quot;&gt;$\frac{x+2+4+8+16+32+64+128}{128} = \frac{x+254}{128}$&lt;/span&gt; cakes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;According to the puzzle, we have&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 2 cakes at the end,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;because:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$$ \frac{x+254}{128} = 2 \\\implies \boxed{x = 2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;The answer is 2.&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="465" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2018-09-16T21:12:01.420" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="629" PostTypeId="2" ParentId="619" CreationDate="2014-06-08T11:09:52.563" Score="3" Body="&lt;p&gt;As the question is &lt;a href=&quot;https://puzzling.stackexchange.com/revisions/619/1&quot;&gt;currently written&lt;/a&gt;, there is a trivial answer. While this answer is boring in terms of algebra, it demonstrates the need for careful wording when writing mathematical reasoning. I expect that this is not what is meant, but it's what is written.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;any math you want&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;clarified in a &lt;a href=&quot;https://puzzling.stackexchange.com/questions/619/writing-numbers-puzzle?noredirect=1#comment1573_619&quot;&gt;comment&lt;/a&gt; to&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;any kind of mathematical operation is allowed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The function that takes no argument and always returns 23 is a mathematical operation. It isn't a common one, but it is one. You can define a notation for it if you want — let's call it $\mathrm{Bob}$ — but mathematical operations exist independently of notation. (There's a standard notation for it, which is $23$, but I don't want to risk any confusion with the use of the digit $2$). Here is the number 23 written with 0 uses of the number 2: $\mathrm{Bob}$. All other numbers can be written in this way.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It is typical in this kind of puzzles to restrict to a few operations. A typical set of allowed operations might be:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the binary operators on numbers addition, subtraction, multiplication, division, exponentiation;&lt;/li&gt;&#xA;&lt;li&gt;the unary operators square root and factorial;&lt;/li&gt;&#xA;&lt;li&gt;integers written in base 10 using allowed digits, typically also decimal numbers, sometimes including repeated decimals.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Using this set of operations as an example, I think that $23 = 22 + 2/2$ uses the fewest occurrences of the digits (a common goal in such puzzles — but not mentioned in the present question — is to minimize the number of digits that are used). Any integer $n$ can be written as a sum of $n$ occurrences of $2/2$.&lt;/p&gt;&#xA;" OwnerUserId="71" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2014-06-13T19:39:25.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="630" PostTypeId="2" ParentId="625" CreationDate="2014-06-08T13:20:10.320" Score="6" Body="&lt;p&gt;This is another example of an information puzzle. You are trying to find the most numbers a pair of spies can send to each other given that there are 26 stones in the river.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, the stones are all identical, and the only way the spies can communicate is by throwing a certain &lt;em&gt;number&lt;/em&gt; of stones into the pond at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are ultimately trying to devise an algorithm that can produce the most possible outcomes for this procedure of throwing 26 stones, which will them map to the greatest number of results.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The answer lies in the number of ways there are to divide 26 stones into groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although the stones themselves are identical, the order that they're thrown in isn't. So the groups that the stones are thrown in form an ordered partition, like such:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;o o o o|o o o|o o|o o o o o|o|o o|o o o|o o o o|o o&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Each &lt;code&gt;o&lt;/code&gt; represents a stone, and each &lt;code&gt;|&lt;/code&gt; represents a divider between a group of stones that were thrown. In the example above, there were 4, 3, 2, 5, 1, 2, 3, 4, 2 stones thrown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, notice that any two consecutive stones can have a divider between them or not. This is a total of 25 dividers that can either be present or not present, for a total of &lt;span class=&quot;math-container&quot;&gt;$2^{25} = 33~554~432$&lt;/span&gt; outcomes. So our first upper bound on the numbers the spies could exchange is &lt;span class=&quot;math-container&quot;&gt;$\lfloor2^{12.5}\rfloor = 5792$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Complicating this is the fact that each spy has to have some control over the stones they throw. If spy 1 throws all 26 stones (which is the case with no dividers), this leaves no choice in the matter for spy 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we decide instead to give each spy his own set of 13 stones, which again can either have dividers between them or not (which is a total of 12 positions where dividers can occur). In this case, the upper bound is &lt;span class=&quot;math-container&quot;&gt;$2^{12} = 4096$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Complicating this yet again is that each spy has to throw either the same number of groups of stones, or the first spy throws one more group of stones than the second spy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for each of the 13 stones, the spies need to decide on a way to divide them into groups of, say, 6 or 7. If each spy decides beforehand to divide the stones into exactly 7 groups to throw, this gives a total of &lt;span class=&quot;math-container&quot;&gt;$\binom{12}{6} = 924$&lt;/span&gt; choices in where to put the dividers, which is our first solution that actually works.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From here, we have to work up. Note that this first naïve solution doesn't take advantage of the fact that the first spy can throw one more group of stones than the second, or that either spy can throw less than 13 stones. So there's some information that we've ended up discarding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;hockey stick theorem&lt;/strong&gt; states that any number &lt;span class=&quot;math-container&quot;&gt;$\displaystyle \binom{n}{k}$&lt;/span&gt; is equal to &lt;span class=&quot;math-container&quot;&gt;$\displaystyle \sum_{m=0}^{k} \binom{n-1-m}{k-m}$&lt;/span&gt; (the theorem gets its name from the way those numbers form a hockey stick on Pascal's Triangle). So supposing we instead arrange that spy 2 can throw anywhere from 7 to 13 stones in 7 groups depending on how many spy 1 throws, he can still get a total of &lt;span class=&quot;math-container&quot;&gt;$\binom{13}{6} = 1716$&lt;/span&gt; combinations. The algorithm then becomes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Both spies determine which arrangement of stones to throw beforehand, with the restriction that they each throw exactly 7 groups of no more than 13 stones.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;They come to the river and alternate throwing groups of stones that correspond to their number.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Spy 1 then throws all the remaining stones into the river, and they depart.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This allows them to exchange two numbers up to &lt;span class=&quot;math-container&quot;&gt;$1716$&lt;/span&gt;, which is the same number you got up to.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, we note that in some of the above cases, we have some stones left over that Spy 1 has to throw away. Could we put these to better use?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;In &lt;span class=&quot;math-container&quot;&gt;$\binom{11}{5} = 462$&lt;/span&gt; cases, a spy will have 1 stone left.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In &lt;span class=&quot;math-container&quot;&gt;$\binom{10}{4} = 210$&lt;/span&gt; cases, a spy will have 2 stones left.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In &lt;span class=&quot;math-container&quot;&gt;$\binom{9}{3} = 84$&lt;/span&gt; cases, a spy will have 3 stones left, etc.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In each of these cases, the spy can throw any number from &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; stones, but Spy 2 cannot throw any stones if Spy 1 doesn't throw at least &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; first, so we consider the worst-case scenario where Spy 1 has thrown all his stones but Spy 2 still has &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; to throw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spy 1 throws a stone, leaving &lt;span class=&quot;math-container&quot;&gt;$n-1$&lt;/span&gt; for Spy 2. Spy 2 then throws any number from &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$n-1$&lt;/span&gt; and Spy 1 throws the rest away. This algorithm will still work if there are any other number of stones left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't really do anything in the case where there are only 1 or 2 stones left (in each of these cases, Spy 2 either has no stones to throw or must throw exactly 1 stone, which doesn't give any information). However, for 3 or more stones, we get the following improvements:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;For &lt;span class=&quot;math-container&quot;&gt;$s = 3$&lt;/span&gt;, we can express &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; cases per arrangement, for an improvement of &lt;span class=&quot;math-container&quot;&gt;$\binom{9}{3} \times (2-1) = 84$&lt;/span&gt; more cases.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For &lt;span class=&quot;math-container&quot;&gt;$s = 4$&lt;/span&gt;, we can express &lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt; cases per arrangement, for an improvement of &lt;span class=&quot;math-container&quot;&gt;$\binom{8}{2} \times (3-1) = 56$&lt;/span&gt; more cases.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For &lt;span class=&quot;math-container&quot;&gt;$s = 5$&lt;/span&gt;, we can express &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt; cases per arrangement, for an improvement of &lt;span class=&quot;math-container&quot;&gt;$\binom{7}{1} \times (4-1) = 21$&lt;/span&gt; more cases.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For &lt;span class=&quot;math-container&quot;&gt;$s = 6$&lt;/span&gt;, we can express &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt; cases per arrangement, for an improvement of &lt;span class=&quot;math-container&quot;&gt;$\binom{6}{0} \times (5-1) = 4$&lt;/span&gt; more cases.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All together, we get &lt;span class=&quot;math-container&quot;&gt;$84 + 56 + 21 + 4 = 145$&lt;/span&gt; extra cases from the extra stones, bringing the total up to &lt;span class=&quot;math-container&quot;&gt;$1861$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I can't see any elegant improvements to make on this algorithm past that, though. If you were to make a computer program to traverse all the possibilities, I suppose you could get to &lt;span class=&quot;math-container&quot;&gt;$2286$&lt;/span&gt;, but it would probably require a whole different approach, potentially involving the same sequence of throws from one spy representing different numbers depending on how the other spy threw &lt;em&gt;his&lt;/em&gt; stones.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2019-07-26T19:28:44.257" LastActivityDate="2019-07-26T19:28:44.257" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="631" PostTypeId="2" ParentId="626" CreationDate="2014-06-08T13:53:47.790" Score="8" Body="&lt;p&gt;It depends how you define half of cakes, for example for 3 cakes. Assuming round up:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;3 cakes -&amp;gt; half is 2 -&amp;gt; leaves 1 -&amp;gt; +1 back -&amp;gt; you have 2 cakes&#xA;4 cakes -&amp;gt; half is 2 -&amp;gt; leaves 2 -&amp;gt; +1 back -&amp;gt; you have 3 cakes&#xA;5 -&amp;gt; 3&#xA;6 -&amp;gt; 4&#xA;7 -&amp;gt; 4&#xA;8 -&amp;gt; 5&#xA;.. and so on&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Theoretically every number between 2 and N should work, figuring out the highest possible number N is left as homework exercise. ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint, for every number x, the highest one leading directly to it is 2x-1.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2014-06-08T13:53:47.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1630" PostTypeId="2" ParentId="617" CreationDate="2014-06-08T18:23:14.717" Score="1" Body="&lt;p&gt;It is really hard to find something on the Web about Martin Kruskal in relation to origami.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a link missing in Wikipedia, so maybe this information is just gossip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to see how to fold an envelope for sending secret messages, you can check out:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=R9wIbYjMLf8&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; method for a hard-to-open envelope.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=fAn5QFiKwFQ&quot; rel=&quot;nofollow&quot;&gt;This method&lt;/a&gt; and &lt;a href=&quot;http://berrylyn.blogspot.it/2012/10/letterfold-central-square-cross.html&quot; rel=&quot;nofollow&quot;&gt;this method&lt;/a&gt; for kind-of-hard-to-fold-back envelopes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="28" LastEditorUserId="463" LastEditDate="2015-06-04T16:27:50.547" LastActivityDate="2015-06-04T16:27:50.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1631" PostTypeId="2" ParentId="313" CreationDate="2014-06-08T20:29:29.503" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Does any user of puzzling.stackexchange have that many arms?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If your number is 3, the answer will be &quot;no&quot; (unless we have any aliens on the site)&lt;/li&gt;&#xA;&lt;li&gt;If your number is 2, the answer will be &quot;yes&quot;&lt;/li&gt;&#xA;&lt;li&gt;If your number is 1, the answer will most probably be &quot;maybe&quot;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="274" LastActivityDate="2014-06-08T20:29:29.503" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1633" PostTypeId="2" ParentId="626" CreationDate="2014-06-09T02:35:39.260" Score="59" Body="&lt;p&gt;The above answers are correct in a sense, but unfortunately incomplete.  There are three scenarios:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1) The trolls round down.&lt;/strong&gt;  That is, if you had 3 cakes, they take &lt;code&gt;floor(1.5) = 1&lt;/code&gt; cake from you.  In this case, you may leave home with, at minimum, &lt;strong&gt;zero (0)&lt;/strong&gt; cakes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why?  Consider each troll as below (the item number is the troll toll number):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Takes &lt;code&gt;1/2&lt;/code&gt; of your cakes (which is &lt;code&gt;0&lt;/code&gt;), and gives you &lt;code&gt;1&lt;/code&gt;.  Now you have &lt;code&gt;1&lt;/code&gt; cake.&lt;/li&gt;&#xA;&lt;li&gt;Takes &lt;code&gt;1/2&lt;/code&gt; of your cakes (which is &lt;code&gt;floor(0.5) = 0&lt;/code&gt;), and gives you &lt;code&gt;1&lt;/code&gt;.  Now you have &lt;code&gt;2&lt;/code&gt; cakes.&lt;/li&gt;&#xA;&lt;li&gt;(and all other trolls following) Takes &lt;code&gt;1/2&lt;/code&gt; of your cakes (which is &lt;code&gt;1&lt;/code&gt;), and gives you &lt;code&gt;1&lt;/code&gt;.  Now you have &lt;code&gt;2&lt;/code&gt; cakes.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thus, you may start with a minimum of &lt;code&gt;0&lt;/code&gt; cakes, and end up with &lt;code&gt;2&lt;/code&gt; at the end of the route. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2) The trolls round up.&lt;/strong&gt; This is a nearly identical process, so I will neglect the explanation.  In this case, you must &lt;strong&gt;leave home with a minimum of &lt;code&gt;2&lt;/code&gt; cakes and a maximum of &lt;code&gt;255&lt;/code&gt; cakes&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2) The trolls don't round, and deal with floating point numbers&lt;/strong&gt;: I don't really want to think about this case... &lt;code&gt;:)&lt;/code&gt;  I believe that a starting value of &lt;code&gt;2&lt;/code&gt; is minimal, but do not have time to prove so.&lt;/p&gt;&#xA;" OwnerUserId="1474" LastEditorUserId="1474" LastEditDate="2014-10-16T15:05:56.837" LastActivityDate="2014-10-16T15:05:56.837" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="1634" PostTypeId="2" ParentId="626" CreationDate="2014-06-09T08:28:16.093" Score="3" Body="&lt;p&gt;It depends on how forgiving your grandma is when you present her with fractional cakes.  If she accepts 1.99 cakes as being the same as 2 cakes, then you can start with 1 cake (because [1+254]/128&gt;1.99).  If she accepts 1.98 cakes as being the same as 2 cakes, you can start with 0 cakes (because [0+254]/128&gt;1.98).  &lt;/p&gt;&#xA;" OwnerUserId="1478" LastActivityDate="2014-06-09T08:28:16.093" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1635" PostTypeId="1" AcceptedAnswerId="1636" CreationDate="2014-06-09T09:36:02.463" Score="9" ViewCount="724" Body="&lt;p&gt;There are two friends, they are not related. Each has a daughter. Their wives are dead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each man marries his friend's daughter, and after a while, both couples will have a child (a daughter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the relationship between the two babies?&lt;/p&gt;&#xA;" OwnerUserId="465" LastEditorUserId="5373" LastEditDate="2014-12-30T14:47:46.000" LastActivityDate="2014-12-30T14:47:46.000" Title="Discover the relation of two babies of a awkward marriage" Tags="|family|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1636" PostTypeId="2" ParentId="1635" CreationDate="2014-06-09T09:48:26.930" Score="7" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The first baby would be a daughter of the daughter of the second man - his granddaughter.&#xA;The second baby would be a daughter of the second man.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Therefore first baby is &lt;strong&gt;niece&lt;/strong&gt; to the second baby. Similarly she is an &lt;strong&gt;aunt&lt;/strong&gt; for the second baby.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Starting from this point one can find out that first baby is step-granddaughter for herself. Therefore first baby for the second baby is &lt;strong&gt;step-great-granddaughter&lt;/strong&gt; and &lt;strong&gt;step-great-great-great-granddaughter&lt;/strong&gt;, and step-great-great-great-great-great-granddaughter, &lt;strong&gt;and so on&lt;/strong&gt;. The same is true for the first baby. Therefore first baby for the second baby is also &lt;strong&gt;step-great-grandmother&lt;/strong&gt;, &lt;strong&gt;step-great-great-great-grandmother&lt;/strong&gt; &lt;strong&gt;and so on&lt;/strong&gt;. (But I'm not a native language speaker and not sure how correct it is to use here prefix &quot;step&quot; only one time per word.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2014-06-09T17:00:34.717" LastActivityDate="2014-06-09T17:00:34.717" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1641" PostTypeId="2" ParentId="623" CreationDate="2014-06-09T17:14:09.380" Score="42" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\dfrac{6}{1 - \frac{3}{4}}$&lt;/span&gt;&lt;br /&gt;&#xA;&lt;br /&gt;&#xA;This simplifies to &lt;span class=&quot;math-container&quot;&gt;$\frac{6}{1/4}$&lt;/span&gt;, which becomes &lt;span class=&quot;math-container&quot;&gt;$6 \cdot 4$&lt;/span&gt;, which becomes &lt;span class=&quot;math-container&quot;&gt;$24$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1497" LastEditorUserId="80061" LastEditDate="2023-06-24T21:42:52.170" LastActivityDate="2023-06-24T21:42:52.170" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="1642" PostTypeId="2" ParentId="626" CreationDate="2014-06-09T18:55:35.090" Score="21" Body="&lt;p&gt;(presses the rewind button on reality)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You leave grandma's house with two cakes. Each time you cross a bridge, you give the troll one cake, and then he gives you back a number of cakes equal to how many you have left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to see that every time you leave a bridge, and thus when you return home, you still have two cakes.&lt;/p&gt;&#xA;" OwnerDisplayName="user1502" LastActivityDate="2014-06-09T18:55:35.090" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="1643" PostTypeId="5" CreationDate="2014-06-09T20:45:22.453" Score="0" Body="&lt;p&gt;From four given numbers, try to get to exactly 24, using all four numbers once and using only basic arithmetical operations: addition, subtraction, multiplication, division. Parentheses are allowed to change the order of operations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All numbers have to be used, they can be used only once, and they're numbers, not digits (so you can't group them together to form a number).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;(Classic) example&lt;/strong&gt;: Make 24 out of the four numbers 1, 3, 4, and 6. ---&#xA;  &lt;em&gt;Answer&lt;/em&gt;:  6/(1-3/4).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="68" LastEditorUserId="8874" LastEditDate="2015-02-06T18:14:02.863" LastActivityDate="2015-02-06T18:14:02.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1644" PostTypeId="4" CreationDate="2014-06-09T20:45:22.453" Score="0" Body="Get exactly 24 from four numbers, using only basic arithmetic." OwnerUserId="68" LastEditorUserId="68" LastEditDate="2014-06-09T22:56:41.217" LastActivityDate="2014-06-09T22:56:41.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1646" PostTypeId="5" CreationDate="2014-06-10T02:10:12.127" Score="0" Body="&lt;p&gt;The Octo-Star was originally created by David Pitcher, who called it the OctoCube, and released it in January 2013. It was subsequently mass-produced by Calvin Puzzles in January 2014.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original video from its creator is available &lt;a href=&quot;https://www.youtube.com/watch?v=xFN73NH5OBk&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, showcasing the original Shapeways print of the model. The Octo-Star is a shape modification of a fracture-sliced &lt;a href=&quot;http://en.wikipedia.org/wiki/Triakis_octahedron&quot; rel=&quot;nofollow noreferrer&quot;&gt;triakis octahedron&lt;/a&gt;. Here's what the puzzle looks like:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.robspuzzlepage.com/whatsnew.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DNWWtm.jpg&quot; alt=&quot;http://www.robspuzzlepage.com/whatsnew.htm&quot;&gt;&lt;/a&gt; &lt;img src=&quot;https://i.stack.imgur.com/i4vaem.jpg&quot; alt=&quot;source unknown&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user20" LastEditorDisplayName="user20" LastEditDate="2014-06-10T02:10:12.127" LastActivityDate="2014-06-10T02:10:12.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1647" PostTypeId="4" CreationDate="2014-06-10T02:10:12.127" Score="0" Body="The Octo-Star is a shape-shifting, jumbling mechanical puzzle created by David Pitcher, released in 2014." OwnerDisplayName="user20" LastEditorDisplayName="user20" LastEditDate="2014-06-10T02:10:12.127" LastActivityDate="2014-06-10T02:10:12.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1648" PostTypeId="1" AcceptedAnswerId="1654" CreationDate="2014-06-10T02:17:18.790" Score="5" ViewCount="1154" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;A lighthouse keeper decides to take a bath an hour before the next lighthouse keeper comes to relieve him of his shift.&lt;/p&gt;&#xA;&lt;p&gt;This lighthouse keeper is having a bad day. First he accidentally locks himself in the bathroom, so he's not going to be able to get out until the next lighthouse keeper arrives and is able to unlock the bathroom from the outside.&lt;/p&gt;&#xA;&lt;p&gt;He decides this isn't going to put him off having his bath, and so starts running the bath. Then the bath faucet's knob breaks off! He's unable to turn the bath water off. The walls and door of the bathroom are completely water tight. At this rate, the bathroom is going to be completely full after 55 minutes.&lt;/p&gt;&#xA;&lt;p&gt;How does our lighthouse keeper avoid drowning?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="185" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-06-10T21:00:38.747" Title="Lighthouse keeper stuck the bath" Tags="|lateral-thinking|" AnswerCount="2" CommentCount="10" ClosedDate="2014-06-10T22:35:38.603" ContentLicense="CC BY-SA 3.0" />
  <row Id="1650" PostTypeId="2" ParentId="1648" CreationDate="2014-06-10T05:02:16.103" Score="3" Body="&lt;p&gt;I see 4 options:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If there is some container (bath wash, pan), he upends it and use the air inside once the bath is full.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If &quot;at this rate&quot; means the current speed (litters per second), then he have more time, since when water surface will go up the water pressure will increase and compensate pressure in the bath faucet knob, therefore the speed will decrease with time, which can easily give him additional 5 minutes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If he has clothes or towel inside he can use it to plug the faucet hole, this will decrease the speed.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If he has phone he can call someone to help.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;P.S. If the water pressure is big enough there is no way to escape: he even can be smashed by it ones the bathroom is full.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2014-06-10T05:11:46.020" LastActivityDate="2014-06-10T05:11:46.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1652" PostTypeId="1" AcceptedAnswerId="1685" CreationDate="2014-06-10T05:49:33.747" Score="5" ViewCount="1242" Body="&lt;p&gt;There is a puzzle&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose a standard $8x8$ chessboard has two (arbitrary) squares removed. The only thing known is that these two squares have &lt;em&gt;different&lt;/em&gt; colours. Is it always possible to place 31 dominoes of size 2x1 so as to cover all of these squares?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; yes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and there is very clever and fast prove:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Consider a closed curve, which covers all cells on the chessboard. You can cover chess board by dominoes simply putting them along the curve.&lt;br&gt;&#xA; If two cells with different colours are missing the curve breaks to two curves, each has even number of cells, therefore they always can be covered with dominos.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would like to know is there a realistic brute-force solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is of course a brute-force solution, which considers all 32*32 combinations and solves them. But of course a man can't do it. Therefore, please understand, what I asking: is there a brute-force solution, which reduces all possible cells placements to several (let's say &lt;strong&gt;up to 10&lt;/strong&gt;) options and simply check them one by one? If yes - please, &lt;strong&gt;show me it&lt;/strong&gt;. If no - please prove it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is such an solution - how can one modify this puzzle to rid of it (that means to increase number of options one needs to consider to be bigger than 10)?&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="8874" LastEditDate="2015-10-18T16:03:15.193" LastActivityDate="2015-10-18T16:03:15.193" Title="Is there a brute-force solution of this puzzle?" Tags="|combinatorics|puzzle-creation|checkerboard|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1653" PostTypeId="1" CreationDate="2014-06-10T06:48:15.867" Score="3" ViewCount="721" Body="&lt;p&gt;There is well known task:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How to divide a pie between two people, so each will think that he got equal or more that the other one?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It has the following solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First person divide the pie on 2 equal (in his opinion) parts. Second one chooses the biggest one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know that it is possible to generalise in a way that N people will divide a pie on N pieces and each one will think that his part is not less than 1/N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to generalise this on N envy people? That means to divide a pie between N people in such a way that each one will think that his part &lt;strong&gt;is not less than part of any other man&lt;/strong&gt;. In this case, when a man has got a part bigger than 1/N, he will still be following the rest of the division between the others - may be some one will get even bigger part (in this case the man will not calm down until the situation changes).&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="11297" LastEditDate="2015-05-12T18:47:00.000" LastActivityDate="2015-05-12T18:47:00.000" Title="How to generalise Partition of the pie for N envy people?" Tags="|logical-deduction|game-theory|" AnswerCount="2" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="1654" PostTypeId="2" ParentId="1648" CreationDate="2014-06-10T07:36:14.123" Score="15" Body="&lt;p&gt;The walls and door may be water-tight, but the bathtub is not, there has to be an outflow. So he just has to pull the outflow-plug.&lt;/p&gt;&#xA;" OwnerUserId="1518" LastActivityDate="2014-06-10T07:36:14.123" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1656" PostTypeId="2" ParentId="626" CreationDate="2014-06-10T09:30:26.527" Score="11" Body="&lt;p&gt;This problem is really too easy, since a moment's reflection (or reading any of the previous answers) shows that travellers with 2 cakes can cross any bridge keeping all their cakes. To make it more interesting, one may generalise to what would be needed at departure if one wants to have 3 cakes (or any other number$~n$) at arrival. I'll assume cakes cannot be subdivided, and that although kind, these are real world trolls that won't let you pass taking any &lt;em&gt;less&lt;/em&gt; than half of your cakes (before returning one), so that (like for parking meters) if you ain't got exact change for what you want, you must  pay a bit more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only point I want to make is that this kind of problems becomes very easy working backwards. If you need$~n$ cakes after passing a bridge, one of those will be returned from the troll, and to have the remaining $n-1$ after paying the troll, you need to have had at least $2(n-1)$ before the bridge. So to cross $k$ bridges and have$~n$ you need to iterate $k$ times the operation $f: n\mapsto2(n-1)$ with starting value$~n$. For instance to cross 7 bridges and have 3 cakes left you need to start out with at least $f(f(f(f(f(f(f(3)))))))=130$ cakes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a nice formula for the result once you realise that $f(2+m)=2+2m$ for all non negative integers $m$, but that is not my main point.&lt;/p&gt;&#xA;" OwnerUserId="1521" LastEditorDisplayName="user20" LastEditDate="2014-06-15T04:38:34.137" LastActivityDate="2014-06-15T04:38:34.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1660" PostTypeId="1" AcceptedAnswerId="1662" CreationDate="2014-06-10T11:29:34.570" Score="11" ViewCount="20239" Body="&lt;p&gt;You are in the intersection of two roads, &#xA;there is a troll in the beginning of each road.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of them is liar (&lt;strong&gt;&lt;em&gt;always&lt;/em&gt;&lt;/strong&gt; lies), and one of them &lt;strong&gt;&lt;em&gt;always&lt;/em&gt;&lt;/strong&gt; tells the truth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the roads, ends to your destination and one of them ends to &lt;strong&gt;death&lt;/strong&gt; :) .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trolls know which way is the right one and which one ends to death.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can you find the way if you can ask only &lt;strong&gt;&lt;em&gt;one&lt;/em&gt;&lt;/strong&gt; question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it's your choice who to ask from and you don't know who tells the truth and who lies.&lt;/p&gt;&#xA;" OwnerUserId="465" LastEditorUserId="28" LastEditDate="2014-06-11T04:42:14.227" LastActivityDate="2019-11-21T18:18:25.487" Title="Survive from death by asking the right question from trolls" Tags="|logical-deduction|liars|" AnswerCount="9" CommentCount="3" ClosedDate="2021-06-05T23:13:05.250" ContentLicense="CC BY-SA 3.0" />
  <row Id="1661" PostTypeId="2" ParentId="65" CreationDate="2014-06-10T11:32:33.313" Score="12" Body="&lt;p&gt;Broadly speaking, &lt;a href=&quot;https://en.wikipedia.org/wiki/Cryptic_crossword#Clueing_technique_and_difficulty&quot;&gt;there are two styles of clues in British crosswords&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ximenean&lt;/strong&gt; clues (named after the pseudonym of &lt;a href=&quot;https://en.wikipedia.org/wiki/Derrick_Somerset_Macnutt&quot;&gt;Derrick Somerset Macnutt&lt;/a&gt;) adhere to a fairly strict recipe of [precise definition] + [cryptic indication], and are probably the sort of clues that American solvers would be most familiar with. For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;• &lt;em&gt;I feel wretched getting no Latin to penetrate (5)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;IMBUE&lt;/strong&gt; — (I'M BLUE, without L for Latin)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ximenes also favoured another type of clue called &lt;em&gt;Printer's Devilry&lt;/em&gt; (PD), where the clue is a phrase from which a word has been removed, and then the letters rearranged to disguise the fact, e.g.:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;• &lt;em&gt;She lay in the sunny meadow, chew in gab, utterly (5)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;CUPID&lt;/strong&gt; — (chewing a butterCUP IDly)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, PD clues are generally not mixed with other types of clue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Libertarian&lt;/strong&gt; clues break the rules in various ways. For example, they will use less precise language in the definition or cryptic parts as long as the clue remains sufficiently unambiguous. Solving them typically requires a leap of the imagination. The late lamented &lt;a href=&quot;https://en.wikipedia.org/wiki/John_Galbraith_Graham&quot;&gt;Araucaria&lt;/a&gt; (who sadly passed away last year) was one of the best setters of this type. Some examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;• &lt;em&gt;A forbidding place (7,4)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Auction Room&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;• &lt;em&gt;Of of of of of of of of of of (10)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Oftentimes&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;• &lt;em&gt;Yogdaws (3, 5, 2, 10, 4)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;God moves in mysterious ways&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By their very nature, Libertarian clues are difficult to define in terms of a set of rules. But they all make perfect sense once you know what the answer is.&lt;/p&gt;&#xA;" OwnerUserId="40" LastActivityDate="2014-06-10T11:32:33.313" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1662" PostTypeId="2" ParentId="1660" CreationDate="2014-06-10T11:44:40.553" Score="13" Body="&lt;p&gt;The idea is to ask question about question.&lt;br&gt;&#xA;In that case &lt;code&gt;Lie about Lie = Truth about Truth = Truth&lt;/code&gt; (if you ask person about question to himself)&lt;br&gt;&#xA;or &lt;code&gt;Lie about Truth = Truth about Lie = Lie&lt;/code&gt; (if you ask person about question to his antipode).&lt;br&gt;&#xA;Therefore it won't be matter who you are asking.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;An example of such a question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What would you tell me if I ask you whether this road ends to death?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If answer is &quot;yes&quot; you better do not take this road. If answer is &quot;No&quot; you should take this road.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof/explanation:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If we ask a liar whether this road ends in death and the road ends in death, he must answer &quot;No&quot;. Therefore to our question the liar must lie about his answer and say &quot;Yes&quot;.&lt;/li&gt;&#xA;&lt;li&gt;If we ask a liar whether this road ends in death and the road doesn't end in death, he must answer &quot;Yes&quot;. Therefore to our question the liar must lie about his answer and say &quot;No&quot;.&lt;/li&gt;&#xA;&lt;li&gt;If we ask a truth-teller whether this road ends in death, then he must answer truthfully; he will answer the same truth to our question.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="28" LastEditorUserId="20176" LastEditDate="2016-03-15T14:42:23.603" LastActivityDate="2016-03-15T14:42:23.603" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="1663" PostTypeId="5" CreationDate="2014-06-10T12:30:54.190" Score="0" Body="&lt;p&gt;The most common form of a puzzle with this tag focuses on getting information by asking questions to members of 2 or 3 of the following groups:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Truth-tellers (or Knights), who always tell the truth.&lt;/li&gt;&#xA;&lt;li&gt;Liars (or Knaves), who always lie.&lt;/li&gt;&#xA;&lt;li&gt;Jokers, who answer at random.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;The questioner doesn't know who they're asking — Knight, Knave or Joker — and often only yes/no-questions (or similarly restricted questions) are&#xA;allowed. Restrictions on the number of questions asked are also common for these questions.&lt;/p&gt;&#xA;&lt;p&gt;This tag can also be used for puzzles which have liars, truth-tellers, and/or jokers as an integral part. Some examples of this type:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/18448/murdered-down-the-rabbit-hole-a-long-logic-puzzle&quot;&gt;Murdered down the Rabbit Hole (a long logic puzzle)&lt;/a&gt; is a mystery logic puzzle where not all of the clues are truthful. Part of the puzzle is determining which clues are lying.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/49772/you-trolls-get-out-of-my-party&quot;&gt;You trolls, get out of my party!&lt;/a&gt; is a strategy puzzle where trolls (who answer at random) must be identified and expelled from a party that also contains friends (truth-tellers)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="28" LastEditorUserId="69582" LastEditDate="2021-02-06T18:12:53.203" LastActivityDate="2021-02-06T18:12:53.203" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="1664" PostTypeId="4" CreationDate="2014-06-10T12:30:54.190" Score="0" Body="Puzzles that focus on getting information from truth-tellers (who never lie), liars (who always lie), and/or Jokers (who answer at random), or which use such characters as an integral part of the puzzle." OwnerUserId="68" LastEditorUserId="69582" LastEditDate="2021-02-06T17:41:46.290" LastActivityDate="2021-02-06T17:41:46.290" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="1666" PostTypeId="2" ParentId="593" CreationDate="2014-06-10T16:03:18.890" Score="6" Body="&lt;p&gt;Here is the solution I found. Please tell, if you see any mistakes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's call $M'$ maximum number of jokers, which can be among group of chosen $N'$ man according to our knowledge.&lt;br&gt;&#xA;Our start group will include all men, $N' = N$. Our knowledge is that $K+M=N, K&gt;M$, therefore $M'= (N-1)/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal is to find a chain from $(M'+1)$ men, such that each men in the chain tells about the next one that he is a knight. The last man in the chain will be a knight.&lt;br&gt;&#xA;&lt;em&gt;Indeed, there are only $M'$ jokers, therefore at least one man in the chain is a knight. If one man is a knight, then he doesn't lie and men next to him is knight too. Therefore all men after very first knight in the chain are knights.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To achieve this goal we take an arbitrary man, he will be first in the chain, and ask him about random man: &quot;Is he a knight?&quot;. If answer is &quot;Yes&quot; we add that new man in the chain and continue.&lt;br&gt;&#xA;If at some point the answer is &quot;No&quot; we take that new man and the asked man and exclude them from the chain and from our group. In this case we set $N'=N'-2$ and $M'=M'-1$, since at least one from the excluded two men is joker. [Note, that here we obtain exactly the same puzzle, but with less number of men].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With each question we need one less men to finish the chain, also we can lose one man in the chain in case the answer is &quot;No&quot;. Therefore at most in $2\cdot(N-1)/2$ questions the chain will be completed and we will find a knight. [Even if $N'=1$ and $M'=0$, and the chain consists from only one man.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we should ask the knight about all other men. This will take another $(N-1)$ questions. In total $2\cdot(N-1)$ questions.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Some kind of proof of optimality [This given for an old answer, which was $3/2\cdot(N-1)$ and appeared to be incorrect. I hope I will find more optimal strategy later.]:&lt;br&gt;&#xA;1. Until we found a knight we can't get any concrete information.&lt;br&gt;&#xA;2. There are $(N-1)/2$ jokers, therefore if we unlucky all very first $(N-1)/2$ questions can be asked to jokers and will not give us any information about who is knight and who is not. Therefore we need at least $(N-1)/2$ questions before we found a knight.&lt;br&gt;&#xA;3. First $(N-1)/2$ questions haven't get us any concrete information and now to determine who is who we need at least one more question about other men: $(N-1)$ questions.&#xA;4. In total we need at least $3/2\cdot(N-1)$ questions.&#xA;I understand that there is difference between &quot;concrete&quot; and &quot;useful&quot; information. And this is a week place in the prove, that's why I call it &quot;kind of prove&quot;.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2014-06-22T16:59:50.083" LastActivityDate="2014-06-22T16:59:50.083" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1667" PostTypeId="2" ParentId="1653" CreationDate="2014-06-10T16:14:50.060" Score="1" Body="&lt;h2&gt;&lt;strong&gt;Example moving knife cake cutting method&lt;/strong&gt;&lt;/h2&gt;&#xA;&#xA;&lt;h2&gt;Assumptions:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;1) we assume that all players will accept that if they had as much opportunity to a given cake as every other player they are satisfied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) everyone will be fine with someone taking a small piece if that person is doing so under the impression that it is the same size&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) cakes are combinable...two small pieces can be pressed together to make one of the same size as them separately and everyone accepts that the new cake is the same as the old.  This would work better if we were splitting a beer or wine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) People can cut perfectly and don't try to cheat when performing step 3.  A referee might work for that.  This is by far the weakest part of the method.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Method:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Step 1) A particpant is chosen at random to cut his piece of cake.  He cuts off a piece he thinks is fair.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 2) Everyone is given the option to claim his piece is too big. The players who feel that way particpate in step 3.  If no one feels that way, he can keep his piece, skip step 3, and be no longer part of the game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 3) He holds his knife over the cake in an exact steady nature over the left hand side and slowly moves the knife to right above the cake.  When each person feels he would cut it accurately if he cuts there they yell &quot;STOP&quot;.  He cuts when the last person says stop and that person gets the remaining right hand side of the cake.  The player is no longer part of the game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 4) The cake and all scrap is pressed together to make one big cake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 5) Repeat all steps until there is only one player.  He gets all remaining cake.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Note:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;When N=2, this is equivalent to the standard answer you describe in your question as he will say &quot;STOP&quot; immediately.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Variation:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If cakes cannot be pressed together, assume that any scrap will be given to the person to be awarded cake on the next round. If any players don't say stop until after the knife has passes over the piece of cake they are playing for, the knife will proceed over the scrap pile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you must allow for people to want pieces that are produced after they are given their piece  the problem gets unsolvable for all N.  For N=3, however, there are solutions.  For example, if all players hold a knife over what they consider to be the center of the right hand side of a cake while a referee moves his knife from left to right, both cuts can effectively happen at the same time. When any player yells &quot;CUT&quot; the referee cuts and the yeller gets the cake to the left of the referee's knife.  The other cut (on the right hand side of the cake) is made at the knife in the middle of the other 3 players.  Besides the yeller the player with the leftmost knife gets the middle cake while the other player gets the rightmost cake. &lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditDate="2014-06-11T17:53:11.440" LastActivityDate="2014-06-11T17:53:11.440" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1668" PostTypeId="2" ParentId="593" CreationDate="2014-06-10T16:46:56.213" Score="3" Body="&lt;p&gt;I've figured out how to solve it with no more than $2(N-1)$ questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do a test, ask one man whether a second man is a Knight.  Then ask the second man whether the first one is a Knight.  This produces one of two results: They both say &quot;Yes&quot;, or at least one says &quot;No&quot;.  We'll call this $Q(n)$, defined as &lt;em&gt;&quot;Ask men $n$ and $n+1$ about each other&quot;&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In the scenario where at least one says &quot;No&quot;:&lt;/strong&gt; At least one of the two is a Joker. We'll call this $D$.&#xA;&lt;br&gt;&lt;strong&gt;In the scenario where both say &quot;Yes&quot;:&lt;/strong&gt; Either both are Knights or both are Jokers.  We'll call this $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, with $N=5$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Test the following pairs, for a total of $2(5-1) = 8$ questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;Q(1) = 1 \leftrightarrow 2 \\&#xA;Q(2) = 2 \leftrightarrow 3 \\&#xA;Q(3) = 3 \leftrightarrow 4 \\&#xA;Q(4) = 4 \leftrightarrow 5&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will produce a set of 4 scenarios, each of which is $B$ or $D$.  At that point, each possible combination (except ${B,B,B,B}$) will map to one of the 15 possible combinations of 3-4 Knights and 1-2 Jokers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&#xA;\begin{equation}&#xA;\left.&#xA;\left.&#xA;\begin{aligned}&#xA;Q(1) = B\\&#xA;Q(2) = B\\ &#xA;Q(3) = D\\&#xA;Q(4) = D&#xA;\end{aligned}&#xA;\right\} = {K,K,K,J,K}&#xA;\right.\quad\quad&#xA;\left.&#xA;\begin{aligned}&#xA;Q(1) = D\\&#xA;Q(2) = B\\ &#xA;Q(3) = D\\&#xA;Q(4) = D&#xA;\end{aligned}&#xA;\right\} = {J,K,K,J,K}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure there's a relationship between the results and where the Jokers are, but I'm not seeing it.  However, the numbers work out: The number of potential combinations is ${{K,J} \choose N} = {2 \choose N} = 2^N$, and the number of invalid combinations is ${{2^N}\over2}+1$ (because half of the combinations have more jokers than knights, and one combination has no jokers at all). So $2^N-\left({{2^N}\over2}+1\right) = {(2^N)\over2}-1 = 2^{N-1}-1 $.  This means that $N-1$ bits is enough to encode all possible cases (with one left over for the impossible &quot;No jokers&quot; case).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D$ and $B$ each represent one bit of data, so we need $N-1$ trials to encode one set of Jokers and Knights.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This solution will scale up for any $N$ - you just need to start with $Q(1)$ and continue until you're asking $Q(N-1)$.&lt;/p&gt;&#xA;" OwnerUserId="325" LastEditorUserId="71" LastEditDate="2014-06-22T17:49:06.883" LastActivityDate="2014-06-22T17:49:06.883" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="1670" PostTypeId="2" ParentId="260" CreationDate="2014-06-10T19:39:09.760" Score="20" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I was at a dinner party the other night and I spoke to Mr Smith. He mentioned having two kids, both with unisex names: Sam and Alex. I remember that he told a story about &quot;my son&quot;, but I don't recall which child he was talking about, and I don't know if he has a daughter or not. What are the odds that he has two sons?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This avoids selection entirely, and avoids bringing twins into the mix. It does, however, hinge on your ability to accept Sam and Alex as being short for Samantha and Alexander, Samantha and Alexandra, Samuel and Alexander, or Samuel and Alexandra. It makes it easier to enumerate the possibilities though, since you now have names -- I got lost in the weeds for a while discussing this problem with why having a boy first and then a girl was counted separately from having a girl first and then a boy and didn't quite realize I was talking about a different version of the question at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other version would read:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I was at a dinner party the other night and I spoke to Mr Smith. He mentioned having two kids, a boy named Jim and a child named Sam whose gender I don't recall. What are the odds that he has two sons?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which is obviously 50%. &lt;/p&gt;&#xA;" OwnerUserId="431" LastEditorUserId="431" LastEditDate="2014-06-10T19:44:21.543" LastActivityDate="2014-06-10T19:44:21.543" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1671" PostTypeId="1" AcceptedAnswerId="1679" CreationDate="2014-06-10T19:57:13.060" Score="16" ViewCount="37567" Body="&lt;p&gt;What is correct answer for this IQ task and &lt;strong&gt;why&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9fv2s.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="5373" LastEditDate="2017-03-24T15:14:02.547" LastActivityDate="2017-03-24T15:14:02.547" Title="How to solve this IQ test puzzle with arrows and pied squares?" Tags="|logical-deduction|pattern|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1672" PostTypeId="2" ParentId="1652" CreationDate="2014-06-10T19:57:23.890" Score="2" Body="&lt;p&gt;The answer to whether this could be brute forced is yes. However, would it be feasible to brute force the answer? Probably not. Let's look at why:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A chessboard has 64 squares — 32 white, and 32 black. Removing one of each color gives us 32^2, or 1024 possible ways to arrange it. However, there are 4 lines of symmetry on the board. Each line of symmetry divides the final number of layouts by 2. $1024 / 2 = 512$; $512 / 2 = 256$; $256 / 2 = 128$; and $128 / 2 = 64$ possibilities. This can go one of two ways:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Your friends don't realize this, so they start trying every possibility in the $32^2$ set. Depending on how they try to solve the domino layout, this could take a very long time. If they start noticing patterns, it will speed up, especially if they start noticing symmetry.&lt;/li&gt;&#xA;&lt;li&gt;Your friends already know &lt;em&gt;all&lt;/em&gt; the symmetry, so they try the 64 board layouts. They'll only try as many as they need to. Even then, solution time will vary based on how they decide to try laying out the dominoes.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the end, it's able to be brute-forced. However, even when you know there are only 64 unique ways to set up the board, the real brute force effort will be in arranging the dominoes, and this will be the most time consuming. Because your puzzle asks whether all of them can be filled with 2x1 dominoes, every board layout will have to be tried until they either find one that can't be 100% filled no matter what, or until they've tried all of them and found the condition to be true for each of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll give a little tip. Whenever the solution of a puzzle depends on combinations/permutations matching a certain condition, it can be brute-forced (think of it in the sense of passwords). In most cases, it only takes one combination/permutation that does or doesn't fit the conditions to solve the problem (the only situation I can think of where this is not the case is if it's asking how many meet the condition). Finding that will be a pain (like brute-forcing a bike lock).&lt;/p&gt;&#xA;" OwnerUserId="173" LastEditorUserId="71" LastEditDate="2014-06-23T21:37:24.200" LastActivityDate="2014-06-23T21:37:24.200" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1673" PostTypeId="1" AcceptedAnswerId="1674" CreationDate="2014-06-10T20:11:14.283" Score="11" ViewCount="3803" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DQaD7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been shown this &quot;Chinese puzzle from textbook for 2nd grade&quot; like a joke. But it is intriguing, so I would like to know the answer, but I am quite bad in associative puzzles and can't find the pattern. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The translation of the 3 lines of text should be the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;These creatures below are aliens:&lt;br&gt;&#xA;  These creatures below are not aliens:&lt;br&gt;&#xA;  Which of the following below is an alien? Circle it. (5 points)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="28" LastEditorUserId="39" LastEditDate="2015-04-27T22:03:50.153" LastActivityDate="2015-04-27T22:03:50.153" Title="How to solve this crazy Chinese puzzle for children?" Tags="|logical-deduction|pattern|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1674" PostTypeId="2" ParentId="1673" CreationDate="2014-06-10T20:20:10.860" Score="23" Body="&lt;p&gt;All the creatures in the top row have three dangly things and one triangle on their body. None of the creatures in the second row do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By that criterion, only the fourth creature in the bottom row is an alien.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2014-06-10T20:20:10.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1676" PostTypeId="2" ParentId="593" CreationDate="2014-06-10T20:49:06.603" Score="20" Body="&lt;p&gt;The minimum number of questions of the special form “Person $i$, is Person $j$ a knight?” that will guarantee to find everyone's identity is $3(N-1)/2$. This was first proved by P. Blecher in &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/0012365X83900262&quot;&gt;this paper&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we know for sure that at most $J$ jokers are present, where $J &amp;lt; N/2$. My &lt;a href=&quot;http://www.ma.rhul.ac.uk/~uvah099/Maths/ksDRev2.pdf&quot;&gt;paper on the problem&lt;/a&gt; shows that $N+J-1$ is the minimum number of questions that will guarantee to find everyone's identity. Taking $J$ to be $(N-1)/2$ recovers Blecher's result. My &lt;a href=&quot;http://www.ma.rhul.ac.uk/~uvah099/knights.html&quot;&gt;webpage on the problem&lt;/a&gt; gives an overview of the solution. These &lt;a href=&quot;http://www.ma.rhul.ac.uk/~uvah099/Talks/ksTalk.pdf&quot;&gt;slides&lt;/a&gt; might also be of interest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;The easier part is to show that $N+J-1$ questions suffice.&lt;/b&gt; The strategy used in my paper and in Blecher's is the same. Set $T$ equal to $J$. Pick one person (call him a candidate), and ask other people about him until either&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;for some $r$, the candidate has been accused $r$ times and supported $r-1$ times, &lt;em&gt;or&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;the candidate has been supported by at least $T$ people.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the first case either the candidate is a knight, and he has been falsely accused by $r$ jokers, or the candidate is a joker, who has been rightly accused by $r-1$ people (who might be knights or jokers). Either way, at least $r$ of the $2r$ people involved are jokers. We may therefore ignore all these people, and repeat with a fresh candidate, replacing $T$ with $T-r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually we end in the second case with a successful candidate, who must be a knight. Use this knight to identify the unsuccessful candidates (this will reveal some jokers), and finally use the knight to identify all the remaining people whose identity is still ambiguous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a special case of the argument that this strategy uses at most $N+J-1$ questions. Suppose that the first candidate is successful, after he is supported by $J$ people, and accused by $A$ people. These $A$ people must be jokers. We now ask the first candidate about the $N-J-A-1$ people not yet involved in proceedings, and the $J$ people who supported him. The total number of questions asked is $ J+A + (N-J-A-1) + J = N-1 +J $ as required. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;The hard part is to show that no questioning strategy, no matter how clever, can guarantee to use fewer than $N+J-1$ questions.&lt;/b&gt; For this I like to this of the problem as a two player game: the &lt;em&gt;Interrogator&lt;/em&gt; poses the questions, and the &lt;em&gt;Spy Master&lt;/em&gt; (or Joker Master in this setting) decides on the answers. The job of the Spy Master is to arrange matters so that the Interrogator faces the worst-case scenario for his strategy. The &lt;em&gt;Knight Hiding Strategy&lt;/em&gt; in my paper is an optimal strategy for the Spy Master that holds the Interrogator to the target of $N+J-1$ questions. Following this strategy, the first question is answered with an accusation, so at least one Joker is present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;More general questions.&lt;/b&gt; My guess is that allowing more general yes/no questions does not permit a quicker solution. But I do not have a proof of this.&lt;/p&gt;&#xA;" OwnerUserId="1539" LastEditorUserId="71" LastEditDate="2014-06-22T17:57:58.980" LastActivityDate="2014-06-22T17:57:58.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1677" PostTypeId="2" ParentId="1671" CreationDate="2014-06-10T20:50:55.210" Score="9" Body="&lt;p&gt;H.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Various categories of symbol appear on alternate rows and columns:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Odd row, odd column - rectangles shaded in two colours, with diagonal dividing line.&#xA;Odd row, even column - single headed arrows&#xA;Even row, odd column - double headed arrows&#xA;Even row, even column - cross&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each category of symbol appears to have its own type of transformation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Odd row, odd column - height increases moving right, which half is shaded swaps moving down (or 180 degree rotation moving down)&#xA;Odd row, even column - 45 degree rotation clockwise moving down&#xA;Even row, odd column - 45 degree rotation clockwise moving right&#xA;Even row, even column - ?  (only one member)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the absence of any obvious common rule for all symbols, I conclude that each set follows its own rules.  Thus the answer is H (column 3 = tall, row 3 = top left half black).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did wonder if the arrows represented transformations for the squares on either side - the arrow in row 1 column 2 could represent a height increase.  The double headed arrows suggest a flip to me but would have to represent a rotation for this to work.  But this system breaks down with row 3 column 2 - there are no answers which have been stretched along a diagonal (which would give a parallelogram).&lt;/p&gt;&#xA;" OwnerUserId="1541" LastActivityDate="2014-06-10T20:50:55.210" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1679" PostTypeId="2" ParentId="1671" CreationDate="2014-06-11T05:38:22.383" Score="23" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;F&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I googled and found the test this task is taken from: &lt;a href=&quot;http://www.iqtest.dk/testEnglish.swf&quot;&gt;http://www.iqtest.dk/testEnglish.swf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I answer 1-10 questions correctly and 38 H, I get IQ 92. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I answer 1-10 questions correctly and 38 F, I get IQ 93.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I conclude that correct answer is F.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;How I understand this puzzle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the corners (cells: 11,13,31,33) we have some objects. In the middle edges (12,21,23,32) we have operators on neighbouring two objects. The centre cell (22) is simply do not used here [this opinion agrees with tasks 4 and 9 at the same test, where it is used similarly].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look on (12) operator (one arrow), it leads to simple vertical stretching of object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look on (21) operator (two arrows), it leads to simple swap of object colours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we are looking on (23) and (32). (23), by our assumption, should swap colours of (13), creating the object from varian H, plus some additional operation, since (23) is different from (13). (32), by our assumption, should stretch (31), making variant H from it, plus some additional operation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are these additional operations? Since they make the same object from the same object (H) that must be the same. Looking on possible answers we see that all differences between objects are: size, swap of the colours, angle. We have notations for first two differences, so additional operation should be rotation.&lt;br&gt;&#xA;We see that (23) is rotated respectively to (21), by 45 degree clockwise. (32) is rotated respectively to (12) by 45 degree clockwise. So most probably both additional operations means rotation by 45 degree clockwise. This means that the right variant for (33) is F.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;What I do not like about this solution:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;One arrow and two arrows means completely different things. It is counter intuitive. And though double arrow associated with swap, no doubts, it is also associated with stretching.&lt;/li&gt;&#xA;&lt;li&gt;Arrows in (12) and (21) have different orientation, and this have no influence on the objects, meanwhile orientation of (32) and (23) is important. [The orientation of arrows can be explained by the fact that they also plays role of objects for meta operation, but this is lousy argument, based on &quot;how would author think creating this puzzle&quot; - the orientations of arrows when they play role operators still do not agree with each other.]&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I hope, someone else will find even better explanation, without mentioned two drawbacks.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="17438" LastEditDate="2015-12-12T16:23:40.573" LastActivityDate="2015-12-12T16:23:40.573" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="1685" PostTypeId="2" ParentId="1652" CreationDate="2014-06-11T13:24:46.970" Score="3" Body="&lt;p&gt;In fact, if you try it a few times for specific cases, you find it is easy.  There are lots of solutions.  This is not a proof, but will convince most novices that it can be done.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is another way:  Number the board $(0,0)$ through $(7,7)$ and let the removed cells be $(a,b)$, $(c,d)$ with $a \le b$ and $c \le d$ (rotate and mirror image if necessary to get this).  As the colors are different, one of $c-a$ or $d-b$ is odd, the other even.  If $c-a$ is odd, you can fill the rows above and below the holes with horizontal dominoes and the columns outside with vertical ones, leaving a rectangle $b-a$ times $d-b$ missing the corners.  Fill the top and bottom rows with horizontals and the intervening rows with verticals.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="71" LastEditDate="2014-06-22T16:37:07.060" LastActivityDate="2014-06-22T16:37:07.060" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1687" PostTypeId="1" AcceptedAnswerId="1688" CreationDate="2014-06-11T21:27:00.297" Score="6" ViewCount="11834" Body="&lt;p&gt;What is the maximal number of steps that a computer (using the fastest algorithm) would need to solve an arbitrary 2x2x2 Rubik's Cube configuration? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any other known algorithms that will perform similarly fast?&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="69582" LastEditDate="2021-08-10T21:36:51.623" LastActivityDate="2021-08-10T21:36:51.623" Title="Maximal number of steps needed to solve 2x2x2 Rubik's Cube" Tags="|strategy|optimization|rubiks-cube|mechanical-puzzles|twisty-puzzles|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1688" PostTypeId="2" ParentId="1687" CreationDate="2014-06-11T22:27:30.663" Score="8" Body="&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/Pocket_Cube&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Pocket_Cube&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The maximum number of turns required to solve the cube is up to 11 full turns, or up to 14 quarter turns.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As for an efficient algorithm, there are only ~3.7M configurations&lt;sup&gt;1&lt;/sup&gt; that the 2x2x2 cube can have, so it would not be unreasonable to store the solution for each of these positions in a look-up table. You can generate the look-up table by starting with a solved cube and &quot;scrambling&quot; it to arrive at each of the 3.7M configurations (keeping track of your turns along the way) and then &quot;backtrack&quot; to solve the cube from that configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also efficient algorithms for the 3x3x3 cube, such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Optimal_solutions_for_Rubik%27s_Cube#Korf.27s_Algorithm&quot; rel=&quot;nofollow&quot;&gt;Korf's Algorithm&lt;/a&gt;, which you may be able to employ when solving the 2x2x2 cube. However, note the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Although this algorithm will always find optimal solutions there is no worst case analysis. It is not known how many moves this algorithm might need.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; As opposed to the 3x3x3 cube for which there are 43,252,003,274,489,856,000 configurations.&lt;/p&gt;&#xA;" OwnerUserId="198" LastEditorUserId="198" LastEditDate="2014-06-12T13:16:22.830" LastActivityDate="2014-06-12T13:16:22.830" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="1689" PostTypeId="1" AcceptedAnswerId="1692" CreationDate="2014-06-12T12:02:09.923" Score="9" ViewCount="875" Body="&lt;p&gt;There is well known problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The picture is attached to a string and you want to hang it on N nails so that if one takes out any of the nails, the picture falls. How do you do it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;For those who have not heard this puzzle:&lt;br&gt;&#xA;No brainteaser tricks are here, this is basically topology problem: &quot;How to put N points and a closed curve on a plane in a way that curve can't be moved to infinity without crossing any point, but if one arbitrary point is removed - it can?&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to generalise in the following way?:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The picture is attached to a string and you want to hang it on N nails so that if one takes out any M of the N nails, the picture falls, but if one takes out any M-1 nails the picture will not be freed. How do you do it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Does this problem have a solution? Can it be shown that this is impossible to do if &lt;code&gt;M &amp;gt; 1&lt;/code&gt;?&lt;/p&gt;&#xA;" OwnerUserId="28" LastActivityDate="2017-06-26T11:33:05.077" Title="How to generalise hang picture puzzle?" Tags="|puzzle-creation|topology|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1690" PostTypeId="1" CreationDate="2014-06-12T13:49:40.547" Score="18" ViewCount="6542" Body="&lt;p&gt;My first question: how would one solve such problems (in general)? What should be the general technique?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uD7XC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My second question: When to be sure that we have counted all the squares in such problems?&lt;/p&gt;&#xA;" OwnerUserId="1553" LastEditorUserId="5208" LastEditDate="2015-05-18T03:44:13.947" LastActivityDate="2018-03-08T16:54:59.247" Title="When to be sure that we have counted all the squares in such problems" Tags="|logical-deduction|geometry|visual|explanation|" AnswerCount="8" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1691" PostTypeId="2" ParentId="1671" CreationDate="2014-06-12T14:46:25.143" Score="1" Body="&lt;p&gt;Like &lt;a href=&quot;https://puzzling.stackexchange.com/questions/1671/how-to-solve-this-puzzle-from-iq-test/1677#1677&quot;&gt;bmj&lt;/a&gt;, I also get &lt;strong&gt;H&lt;/strong&gt; as the answer. However, my logic is much more straightforward:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;We see:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Upper Left  $\rightarrow$  Upper Right $\Rightarrow$ Rectangle $\rightarrow$  Square&lt;/li&gt;&#xA;&lt;li&gt;Upper Left  $\rightarrow$  Lower Left $\Rightarrow$ Invert colors&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Therefore:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Lower Left  $\rightarrow$  Lower Right $\Rightarrow$ Rectangle $\rightarrow$  Square&lt;/li&gt;&#xA;&lt;li&gt;Upper Right $\rightarrow$  Lower Right $\Rightarrow$ Invert colors&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Resulting in a color-inverted square.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The arrows and $\times$ are just distractions.&lt;/p&gt;&#xA;" OwnerUserId="325" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.923" LastActivityDate="2014-06-22T17:49:23.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1692" PostTypeId="2" ParentId="1689" CreationDate="2014-06-12T15:09:24.910" Score="8" Body="&lt;p&gt;There's a paper &lt;a href=&quot;http://arxiv.org/abs/1203.3602&quot; rel=&quot;nofollow noreferrer&quot;&gt;available on the arXiv&lt;/a&gt;* which investigates both the &quot;remove one nail&quot; variant (which they call &quot;1-of-$n$&quot;) and a more generalized &quot;$k$-of-$n$&quot; version (remove $k$ out of $n$ nails to make the picture fall).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem 1, on page 18 addresses the $1$-of-$n$ version: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem 1&lt;/strong&gt; For any $n \geq 1$, there is a picture hanging on $n$ nails of length at most $2n^2$ that falls upon the removal of any one nail. For each $i = 1, 2, ..., n$, symbols $x_i$ and $x_i^{-1}$ appear at most $2n$ times.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not going to reproduce the whole proof here, because it involves more LaTeX than I want to try and transcribe, but the gist of it (if I'm understanding it correctly) is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;We have $n = 1$ and $n = 2$ solutions.&lt;/li&gt;&#xA;&lt;li&gt;We can show that any problem's solution is equivalent to splitting the problem in half and solving each half, then looping them together.&lt;/li&gt;&#xA;&lt;li&gt;All problems can be recursively reduced to a series of $n =1$ or $n = 2$ problems.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;* Paper source: &lt;a href=&quot;http://arxiv.org/abs/1203.3602&quot; rel=&quot;nofollow noreferrer&quot;&gt;arXiv:1203.3602v2 [cs.DS] 26 Apr 2014&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="325" LastEditorUserId="325" LastEditDate="2017-06-26T11:33:05.077" LastActivityDate="2017-06-26T11:33:05.077" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1693" PostTypeId="2" ParentId="1690" CreationDate="2014-06-12T15:14:08.467" Score="2" Body="&lt;p&gt;One potential strategy is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Call this example an square of dimentions 8x8&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Start from an extreme corner of the diagram.  (example: top left corner)&lt;/li&gt;&#xA;&lt;li&gt;Iterate across the board in such a way as to cover the entire board (example: move down 8 units, right 1 unit, up 8 units, right 1 unit and repeat)&lt;/li&gt;&#xA;&lt;li&gt;Each time you reach previously chosen corner type count the number of possible opposite corners and ensure that they all make a square. (exampe: Each time one sees a &quot;bottom left&quot; corner, draw a line segment of slope 1 towards the top right. Count the number of &quot;top right&quot; corners that this line intersects.  Eliminate from this count any corner pairs that have broken sides.)&lt;/li&gt;&#xA;&lt;li&gt;Continue until the entire map is covered and all squares counted.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;By this strategy I get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;40 squares&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="99" LastActivityDate="2014-06-12T15:14:08.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1694" PostTypeId="2" ParentId="1690" CreationDate="2014-06-12T15:41:56.093" Score="16" Body="&lt;p&gt;You need to find some systematic technique.  The technique will depend on the problem.  Usually there is a regularity to the problem that you can exploit.  In this case, I would note that the two squares that are not part of the main $4 \times 4$ do not interact with the lattice, so we can count them separately.  For the lattice, work by square size.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have four choices along each edge for the upper left corner of a $1 \times 1$ square for &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  $16$ in total.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Similarly there are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $9\  2 \times 2$ squares, $4\  3 \times 3$ squares and $1\  4 \times 4$ for a total of $30$.&lt;br&gt;&#xA; You might even know that in an $n \times n$ chessboard there are $\sum_{i=1}^ni^2=\frac 16n(n+1)(2n+1)$ squares.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Each of the stray squares gives us five, for a total &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $40$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="73" LastEditorUserId="35972" LastEditDate="2017-04-28T18:55:03.803" LastActivityDate="2017-04-28T18:55:03.803" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1695" PostTypeId="2" ParentId="1689" CreationDate="2014-06-12T15:49:23.867" Score="4" Body="&lt;p&gt;Yes. You can.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;First, let's reformulate the puzzle to mathematical language. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;We place all nails on one straight line L and take some point P, far away from line L and all nails on it.&lt;/li&gt;&#xA;&lt;li&gt;We number all nails from 1 to N.&lt;/li&gt;&#xA;&lt;li&gt;We take the string; place one end at P; move the other end around some nail number X clockwise; move the end back to point P. Let's note all this operations as &lt;code&gt;O(X)&lt;/code&gt;. Let's note the same operation, but done anticlockwise as &lt;code&gt;O*(X)&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Let's note several operations, which are done one after another (with nails A, B, ..., Z) as &lt;code&gt;O(A)O(B)..O(Z)&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;li&gt;If &lt;code&gt;S = O(A)O(B)..O(Z)&lt;/code&gt;, let's denote &lt;code&gt;S* = O*(Z)..O*(B)O*(A)&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Let's note operator, which does nothing as &lt;code&gt;E&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Now our task is reformulated like this: find sequence of operators, which is not &lt;code&gt;E&lt;/code&gt; when you remove any M-1 nails, and is &lt;code&gt;E&lt;/code&gt; when you remove M nails.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Second, let's investigate properties of the operator &lt;code&gt;O(...)&lt;/code&gt; and it's combinations &lt;code&gt;U&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;code&gt;E = E*&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;O**(X) = O(X)&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;li&gt;If we take some nail X out then &lt;code&gt;O(X) will become E&lt;/code&gt;. In sequence of operations we should remove all O(X) and O*(X). Since topologically these operations will do nothing.&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;O(X)O*(X) = E&lt;/code&gt;. If in sequence of operations we have &lt;code&gt;O(X)O*(X)&lt;/code&gt; we can remove it. Since topologically these operations will do nothing.&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;UU* = E&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Then, let's solve the puzzle for M = 1. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Let's note the sequence of operations, that solves the task for N nails 1,2,..,N as &lt;code&gt;F(1,2,..,N)&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;li&gt;If N = 1. Trivial solution is &lt;code&gt;F(1) = O(1)&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;li&gt;If N = K+1. Suppose that we solved N = K already (we can do it similarly to mathematical induction method). Solution is defined recursively from solution for K nails: &lt;code&gt;F(1,..,K,K+1) = F(1,...,K)O(K+1)F*(1,...,K)O*(K+1)&lt;/code&gt;. This works because if you take one nail X &amp;lt;= K out &lt;code&gt;F(1,...,K) will become E&lt;/code&gt; and &lt;code&gt;F(1,..,K,K+1) will became O(K+1)O*(K+1) = E&lt;/code&gt;, if you take nail K+1 out &lt;code&gt;O(K+1) will become E&lt;/code&gt;  and &lt;code&gt;F(1,..,K,K+1) will became F(1,..,K)F*(1,..,K) = E&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;li&gt;For example, solution for 3 nails is: &lt;code&gt;F(1,2,3) = O(1)O(2)O*(1)O*(2)O(3)O(2)O(1)O*(2)O*(1)O*(3)&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Now, M &gt; 1.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Let's note the sequence of operations, that solves the task with N and M as &lt;code&gt;S(N,M)&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Let's invent sequence &lt;code&gt;G(X1,X2,..,XM)&lt;/code&gt;, which is not &lt;code&gt;E&lt;/code&gt; until you remove all nails among &lt;code&gt;X1,X2,..,XM&lt;/code&gt;. It is quite trivial: &lt;code&gt;G(X1,X2,..,XM) = O(X1)O(X2)..O(XM)&lt;/code&gt; - we just must hang the picture on all nails as one.&lt;/li&gt;&#xA;&lt;li&gt;Let's consider all possible combinations of M nails out of N: &lt;code&gt;1,2,..,M-1,M; 1,2,..,M-1,M+1; ... N-M+1,N-M+2,..,N-1,N&lt;/code&gt;; Let's number them and call them &lt;code&gt;C1,C2,..,CR&lt;/code&gt;, where &lt;code&gt;R = N choose M&lt;/code&gt;. For each combination we can consider operator G: &lt;code&gt;G(Ck)&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Now we need to do the same procedure as we did with M=1, but instead of nails use combinations of nails. The solution is given recursively, in R steps: &lt;code&gt;S'(1,M) = G(C1); S'(k+1,M)=S'(k,M)G(Ck)S'*(k,M)G*(Ck)&lt;/code&gt;, here k is number of k-th combination, &lt;code&gt;S'(k,M)&lt;/code&gt; the sequience, which solves the task for first k combinations. Therefore &lt;code&gt;S'(k,M)&lt;/code&gt; will work when you remove groups of nails, which belong to any combinations from 1 to k.&lt;/li&gt;&#xA;&lt;li&gt;The required solution is &lt;code&gt;S(K,M) = S'(N choose M,M)&lt;/code&gt;. Indeed, if we remove one nail, neither combination from &lt;code&gt;Ck&lt;/code&gt; will not become &lt;code&gt;E&lt;/code&gt;, therefore &lt;code&gt;S(K,M)&lt;/code&gt; will not become &lt;code&gt;E&lt;/code&gt; and the picture will not be freed. If we remove any M nail, the removed nails are compose one combination from &lt;code&gt;Ck&lt;/code&gt;, then &lt;code&gt;S(K,M) will become E&lt;/code&gt; and the picture will fall.&lt;/li&gt;&#xA;&lt;li&gt;For example, solution for N=3, M=2 is: &lt;code&gt;S(3,2) = G(C1)G(C2)G*(C1)G*(C2)G(C3)G(C2)G(C1)G*(C2)G*(C1)G*(C3) = O(2)O(3)O(3)O(1)O*(3)O*(2)O*(1)O*(3)O(1)O(2)O(3)O(1)O(2)O(3)O*(3)O*(1)O*(3)O*(2)O*(2)O*(1)&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2014-06-12T17:29:17.060" LastActivityDate="2014-06-12T17:29:17.060" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1697" PostTypeId="2" ParentId="1690" CreationDate="2014-06-12T19:11:05.783" Score="1" Body="&lt;p&gt;Another strategy would be as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How many 1x1 squares can we have, ignoring the smaller squares? 4 rows of 4 --&gt; 16&lt;/li&gt;&#xA;&lt;li&gt;How many 2x2 squares can we have, ignoring the smaller squares? 3 rows of 3 --&gt; 9&lt;/li&gt;&#xA;&lt;li&gt;How many 3x3? 2 rows of 2 --&gt; 4&lt;/li&gt;&#xA;&lt;li&gt;How many 4x4? 1 --&gt; 1&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This gives us (16+9+4+1)=30 squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now add in the smaller squares:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How many 1x1 can we have? 8&lt;/li&gt;&#xA;&lt;li&gt;How many 2x2? 2&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This gives us 10 more squares. &#xA;Add those answers and we get 40.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2014-06-12T19:11:05.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1698" PostTypeId="2" ParentId="1690" CreationDate="2014-06-12T19:28:35.107" Score="0" Body="&lt;p&gt;Answer to question 1 and 2: Use brute force, that means test all possible combinations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will need a definition to check if a point exists and if a line exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Pseudocode that looks something like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;function get_square_count()&#xA;    counter = 0&#xA;    for each a in corners:&#xA;        for each b in corners:&#xA;            //same height level and b right of a&#xA;            if a.x=b.x and a.y&amp;lt;b.y:&#xA;                if is_square_top_line(a,b):&#xA;                    counter++&#xA;    return counter&#xA;&#xA;function is_square_top_line(a,b)&#xA;    // use square property to add height &#xA;    calculate virtual points c,d&#xA;    check if c and d exist&#xA;    check if all lines exists a,b b,c c,d, d,a&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1562" LastActivityDate="2014-06-12T19:28:35.107" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1700" PostTypeId="2" ParentId="625" CreationDate="2014-06-13T10:46:21.120" Score="14" Body="&lt;p&gt;I think that the answer lies in inductive construction of the decision tree. At each step there are $k$ stones remaining, and the current player has to throw somewhere from $1$ to $k$ stones. The full tree $D_k$ for a situation with an initial $k$ stones is thus built up with $D_k$ having $k$ children, one instance each of $D_i$ for $i$ from $0$ to $k-1$: e.g. $D_3$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BcJCL.png&quot; alt=&quot;Full decision tree with 3 stones&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the label on each edge is the number of stones thrown and every path ends with no stones remaining.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the spies must have pre-agreed an assignment of values to the leaves of this tree in such a way that the result is &lt;em&gt;decodable&lt;/em&gt;: i.e. that given their place on the tree, each spy knows what possible outputs there are, and can communicate his number regardless of his partner's number. For simplicity, we assume that they prune all branches which have no values assigned to their descendants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that we can construct all such decodable decision trees from a simple principle: the children of the current node must all be decodable in order for the current node to be decodable. Then we can build up the decodable trees starting with $D_0$. I'm going to use the notation $T_k$ to denote the set of decodable trees with $k$ stones, and I'm going to represent each such tree as the thing which we really care about: $(x, y) \in T_k$ means there is a decodable tree with $k$ stones which allows the player who goes next to say one of $x$ numbers, and the other player to say one of $y$ numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The base case is obviously $$T_0 = \{ (1, 1) \}$$ because neither player can do anything to distinguish between numbers. The recursive construction is $$T_k = \left\{ \left(\sum_{s \in S} \mathrm{snd}(s), \min_{s \in S} \mathrm{fst}(s) \right) \middle| S \in \prod_{i&amp;lt;k} \left(T_i \cup \{(\infty, 0)\}\right) \right\} \setminus \{(0, \infty)\}$$ where the $(\infty, 0)$ option corresponds to pruning the branch where the current player would throw $i$ stones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $k&amp;gt;1$ we can make a small transformation for ease of computation: $$\begin{eqnarray}\prod_{i&amp;lt;k} \left(T_i \cup \{(\infty, 0)\}\right) &amp;amp;=&amp;amp; \left(T_{k-1} \cup \{(\infty, 0)\}\right) \times \prod_{i&amp;lt;k-1} \left(T_i \cup \{(\infty, 0)\}\right)\\&#xA;&amp;amp;=&amp;amp; \left(T_{k-1} \times \prod_{i&amp;lt;k-1} \left(T_i \cup \{(\infty, 0)\}\right)\right) \cup \\&amp;amp;&amp;amp; \left(\{(\infty, 0)\} \times \prod_{i&amp;lt;k-1} \left(T_i \cup \{(\infty, 0)\}\right)\right)\\&#xA;\end{eqnarray}$$&#xA;so&#xA;$$\begin{eqnarray}T_k &amp;amp;=&amp;amp; \left\{ \left(\sum_{s \in S} \mathrm{snd}(s), \min_{s \in S} \mathrm{fst}(s) \right) \middle| S \in \left(T_{k-1} \times \prod_{i&amp;lt;k-1} \left(T_i \cup \{(\infty, 0)\}\right)\right) \right\} \cup \\&amp;amp;&amp;amp; \left\{ \left(\sum_{s \in S} \mathrm{snd}(s), \min_{s \in S} \mathrm{fst}(s) \right) \middle| S \in \prod_{i&amp;lt;k-1} \left(T_i \cup \{(\infty, 0)\}\right) \right\} \\&#xA;&amp;amp;=&amp;amp; \left\{ (x + y, \min(w, z)) \middle| (w, x) \in T_{k-1}, (y, z) \in T_{k-1} \cup \{(0, \infty)\} \right\} \cup T_{k-1} \\&#xA;&amp;amp;=&amp;amp; \left\{ (x, w) \middle| (w, x) \in T_{k-1} \right\} \cup \left\{ (x + y, \min(w, z)) \middle| (w, x) \in T_{k-1}, (y, z) \in T_{k-1} \right\} \cup T_{k-1} \\&#xA;\end{eqnarray}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also simplify the bookkeeping by removing &lt;em&gt;dominated&lt;/em&gt; pairs: if one pair is strictly worse than another, we forget about it. Denote the simplified sets as $T'_k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we get $$\begin{eqnarray}T'_0 &amp;amp;=&amp;amp; \{ (1, 1) \}\\&#xA;T'_1 &amp;amp;=&amp;amp; \{ (1, 1) \}\\&#xA;T'_2 &amp;amp;=&amp;amp; \{ (2, 1) \}\\&#xA;T'_3 &amp;amp;=&amp;amp; \{ (3, 1), (1, 2) \}\\&#xA;T'_4 &amp;amp;=&amp;amp; \{ (5, 1), (2, 2), (1, 3) \}\\&#xA;T'_5 &amp;amp;=&amp;amp; \{ (8, 1), (4, 2), (2, 3), (1, 5) \}\\&#xA;T'_6 &amp;amp;=&amp;amp; \{ (13, 1), (7, 2), (4, 3), (3, 4), (2, 5), (1, 8) \}\\&#xA;\vdots \\&#xA;&amp;amp;&amp;amp;(1782, 1782) \in T'_{25}\\&#xA;&amp;amp;&amp;amp;(2535, 2536) \in T'_{26}&#xA;\end{eqnarray}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the solution is $$M=2535$$&lt;/p&gt;&#xA;" OwnerUserId="228" LastEditorUserId="228" LastEditDate="2014-06-13T18:22:51.917" LastActivityDate="2014-06-13T18:22:51.917" CommentCount="20" ContentLicense="CC BY-SA 3.0" />
  <row Id="1701" PostTypeId="1" AcceptedAnswerId="1733" CreationDate="2014-06-13T13:47:55.790" Score="5" ViewCount="579" Body="&lt;p&gt;Imagine a very big number (2N) of people, who do not know each other, and play the following game:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;p&gt;Players are divided into two equal groups: A and B. In each group players are numbered from 1 to N.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Each A-player chooses a position for a white king on the chessboard. &lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Each B-player chooses a position for a black queen on the chessboard. &lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;For each pair of players with the same numbers, the king and queen are placed on a void chessboard. If the queen checks the king, the B-player wins. If not, the A-player wins. If they are on the same square, we call it even.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's call a player a 0-level player if he chooses his piece position by uniform random distribution over the whole board. Let's call a player an L-level player if he chooses his piece position randomly between the most winning positions on the assumption that he is playing with an (L-1)-level player.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that all 2N players are M-level players. What is the optimal strategy for A- and B-players to win as many games as possible? What will the wining percent in each group be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please, consider the following cases:  &lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;M = 2.  &lt;/li&gt;&#xA;&lt;li&gt;M = 3.  &lt;/li&gt;&#xA;&lt;li&gt;M = Infinity.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For example, for M = 1 &lt;em&gt;[I do not know the solution, so, please be careful, mistakes can be here]&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The B-player will suppose that king can be at any square with the same probability and will put queen at the positions where it covers as many cells as possible: &#xA;&lt;img src=&quot;https://i.stack.imgur.com/fi5E2.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;He will &lt;em&gt;expect&lt;/em&gt; a winning probability of &lt;code&gt;(6+7+7+7)/64 = 0.422&lt;/code&gt; and a draw probability of &lt;code&gt;1/64 = 0.016&lt;/code&gt;. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The A-player will suppose that the queen can be at any cell with the same probability, and will put king at the positions where the queen would cover as few cells as possible: &lt;img src=&quot;https://i.stack.imgur.com/MqrP9.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;He will &lt;em&gt;expect&lt;/em&gt; a winning probability of &lt;code&gt;(64-7-7-7)/64 = 0.672&lt;/code&gt; and a draw probability of &lt;code&gt;1/64 = 0.016&lt;/code&gt;.   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In reality, the winning probability for the A-player will be &lt;code&gt;(30-8)/30 = 0.733&lt;/code&gt;, the winning probability for the B-player will be &lt;code&gt;8/30 = 0.267&lt;/code&gt;, and the draw probability will be &lt;code&gt;0&lt;/code&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="28" LastEditorUserId="463" LastEditDate="2015-06-02T03:20:24.343" LastActivityDate="2015-06-02T03:27:00.707" Title="Optimal strategy against a random opponent" Tags="|chess|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1702" PostTypeId="2" ParentId="420" CreationDate="2014-06-13T18:04:02.457" Score="1" Body="&lt;p&gt;I never hold this awesome thing in my hands, but there is 2 very nice videos on youtube, that fully describes the approach:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=UBqfzmCcEeM&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=UBqfzmCcEeM&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=KWVVF4hC7p4&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=KWVVF4hC7p4&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are made in assumption that you know how to solve 3x3x3 Rubik's. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will do a short summary of those videos plus some basic advises, that should be exactly &lt;strong&gt;the answer on your question &quot;Where to start?&quot;&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;First, you need to understand, that the Ghost cube is basically a 3x3x3 Rubik's cube, but you need to turn two angles (moving parts) by 45 degrees (in opposite directions) to get it. In Rubik's cube you have colours to guide you, here you have shapes to do this.&lt;/li&gt;&#xA;&lt;li&gt;Second, you need to acknowledge yourself with each &quot;face&quot; of this pseudo Rubik's cube. Remember which shapes correspond to which colour, which shape correspond to which piece position (corner, edge, centre). Just turn the angles and learn, find the differences and patters, and remember (or use coloured stickers if you do not like remembering, but then you would lose most of the fun).&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Two corners of the two &quot;moving parts&quot; correspond to centers of opposite faces of Rubik cube. The other Rubik centres are in the middle of the edges in the part between the &quot;moving parts&quot;.&lt;/li&gt;&#xA;&lt;li&gt;I will not go to more details here to avoid the spoilers, patterns are not so hard to find and they are kinda subjective (I believe each person will prefer it's own patterns to use). But if would like patterns to be shown to you - they are very nicely explained in the video.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;When you solve it by (Rubik's-)layers (that means by shape layers, which correspond to Rubik 3x3 layers), the Ghost cube won't be that scary - once you solved first two (the easy ones) layers it is more or less structured and harder to be lost in.&lt;/li&gt;&#xA;&lt;li&gt;But be careful and pay attention to the smallest details. For example, the pieced can look like almost lined up, but not quite - then this is probably wrong position for them.&lt;/li&gt;&#xA;&lt;li&gt;If you forget differences between (Rubik's-)faces you may simply try different combinations and see how they line up at the end.&lt;/li&gt;&#xA;&lt;li&gt;You can use a Rubik's cube (if you have one) to help you to learn well-known Rubik's-combinations of moves in this shape-based cube.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="28" LastEditorUserId="1649" LastEditDate="2016-09-08T18:01:28.307" LastActivityDate="2016-09-08T18:01:28.307" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1703" PostTypeId="2" ParentId="623" CreationDate="2014-06-13T19:29:53.273" Score="15" Body="&lt;p&gt;This boils down to the same solution as &lt;a href=&quot;https://puzzling.stackexchange.com/users/1497/kevin&quot;&gt;Kevin&lt;/a&gt;'s, but uses &lt;a href=&quot;http://en.wikipedia.org/wiki/Reverse_Polish_notation&quot; rel=&quot;noreferrer&quot;&gt;Reverse Polish notation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 6 &amp;nbsp; 1 &amp;nbsp; 3 &amp;nbsp; 4 &amp;nbsp; / &amp;nbsp; - &amp;nbsp; / &amp;nbsp; =&gt; &amp;nbsp; 24&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Notice the elegance of the notation, the lack of brackets, the lack of different sized fonts, the lack of confusing latex markup to write it, the sheer incomprehensibility of what's actually going on, ... uhm, yeah, apart from that, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The contents of the stack at each step:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/lvEdw.png&quot; alt=&quot;The operations and the contents of the stack at each step&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="29" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2014-06-13T19:29:53.273" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1704" PostTypeId="2" ParentId="112" CreationDate="2014-06-13T19:44:46.970" Score="9" Body="&lt;p&gt;&lt;em&gt;Kendall Frey's answer gives an example of how to do this with 7 locks, this proves that minimum is not greater than 7.&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Unfortunately, Kendall Frey's answer does not give an answer to the question what exactly is the minimum number of locks. I met a similar puzzle, and can take the honour to prove that 7 is the minimum.&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;To do this I need to add one more condition to the puzzle: &quot;the locks are placed in such a way that to open the box one needs to open all locks&quot;. [According to reaction of the author (Joe Z.) to answer of Peter Taylor, that violates this condition, I can assume that this condition was meant to be in the puzzle formulation and it is ok to use it.]&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea of the prove of minimality is to count missing keys for different combinations of people. Here we go:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Each two slaves must have at least one key missing, since they are not able to open the box.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This missing key must be different for each pair of slaves. Otherwise combining two pairs we will have 3 or 4 slaves with one key missing - they won't be able to open the box.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Therefore we have at least as many keys as many pairs of slaves we can construct. There are ${4 \choose 2} = 4!/2!/2! = 6$ possible pairs of slaves. That means we have at least 6 different keys.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Each of those keys have at least one slave, which doesn't have it. Grand Vizier can't open the box by himself, therefore he has at least one key missing. But this can't be the key some of slaves missing, otherwise he won't be able to open the box with any slave. Therefore one more key exists.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Concluding, in total they must have at least 7 keys and 7 locks.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To make my answer complete I will reconstruct the possible keys distribution following my prove of minimality.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;There will be exactly 7 keys: 1,2,3,4,5,6,7.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We have 6 pairs of slaves: (1,2); (1,3); (1,4); (2,3); (2,4); (3,4). Let's assign one key to each pair (the missing key for a pair). (1,2) must miss key 1; (1,3) must miss key-2; ... (3,4) must miss key-6.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Take all pairs slave-1 belong to: (1,2); (1,3); (1,4). They must not have keys 1,2,3. So we will give slave-1 only keys 4,5,6,7. In the same way we decide, which key to give to all other slaves:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1 - 4,5,6,7.&lt;br&gt;&#xA;2 - 2,3,6,7.&lt;br&gt;&#xA;3 - 1,3,5,7.&lt;br&gt;&#xA;4 - 1,2,4,7.&lt;/strong&gt;  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The only thing that left to do is to assign missing to Grand Vizier. He will not have key-7:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Grand Vizier - 1,2,3,4,5,6.&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It is easy to check that Grand Vizier with any slave can open the box, and any 3 slaves can do it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="28" LastEditorUserId="2484" LastEditDate="2014-10-08T22:55:54.720" LastActivityDate="2014-10-08T22:55:54.720" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="1705" PostTypeId="1" AcceptedAnswerId="1714" CreationDate="2014-06-13T21:18:01.747" Score="8" ViewCount="1309" Body="&lt;p&gt;This &lt;a href=&quot;https://puzzling.stackexchange.com/a/379/28&quot;&gt;answer&lt;/a&gt; reminded me of a nice puzzle. I present it as a challenge for you.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;N thieves had stolen a magical artefact. They bought a lock with N keys and attached the artefact to a loop on a wall in their garage, so that each of them could use the artefact at any time. Immediately a big problem came up: if one of them decided to take the artefact from the garage, then they would not be able to find out who had done it. As they couldn't trust each-other, they were eager to always know who had taken the artefact whenever it was not in the garage. Meanwhile, any thief alone should be able to take it when he needed it.&lt;/p&gt;&#xA;&lt;p&gt;How could they achieve this goal, if they could buy any number of locks with any number of keys to each lock? How can this be done with the minimum number of locks?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;P.S. I do not know whether proof of minimality exists.&lt;/p&gt;&#xA;&lt;p&gt;Explanations:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Thieves can use only padlocks and keys, other things are forbidden.&lt;/li&gt;&#xA;&lt;li&gt;They can attach locks to the loop and to the artefact. They can attach the locks to each other. Shackles of the locks are wide enough to be attached to any number of other shackles simultaneously.&lt;/li&gt;&#xA;&lt;li&gt;They can distinguish locks by serial number.&lt;/li&gt;&#xA;&lt;li&gt;Thieves can't break locks. Also they can't open locks then do not have keys to.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;P.S. A common mistake is to not take into account that a thief, once the artefact is stolen, is free to leave his lock whenever he wants.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-10-18T15:57:28.043" Title="How to lock a shared artefact" Tags="|logical-deduction|keys-and-locks|locks|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1706" PostTypeId="5" CreationDate="2014-06-13T21:21:36.130" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-06-13T21:21:36.130" LastActivityDate="2014-06-13T21:21:36.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1707" PostTypeId="4" CreationDate="2014-06-13T21:21:36.130" Score="0" Body="For puzzles about locks or any other similar mechanisms." OwnerUserId="25957" LastEditorUserId="5373" LastEditDate="2016-12-02T13:36:47.260" LastActivityDate="2016-12-02T13:36:47.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1708" PostTypeId="2" ParentId="1705" CreationDate="2014-06-13T21:28:16.343" Score="0" Body="&lt;p&gt;Edit 3:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N Thieves can make N Chains of locks in the following fashion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Loop - Lock1 - Lock2 - Lock3 - Lock4 - ... - LockN-1 - LockN - Artefact&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Loop - Lock2 - Lock3 - Lock4 - Lock5 - ... - LockN - Lock1 - Artefact&#xA;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so on, shifting by 1 each time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to that, assuming that X steals the Artefact, he has to remove 1 Lock X attached to the artefact itself. At which point, assuming that the thieves write down the order that they assign themselves, the &lt;em&gt;longest&lt;/em&gt; chain will reveal the thief: the longest chain will necessarily be of length (N-1), ending in X-1. Or X could put his lock back at the end of the chain, but hey, he's just incriminating himself even more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If X tries to put his lock back on the second longest chain, such that the two longest chains have the same length, X will still be found out since the lock (X+1), missing at the end of the second (faked) longest chain, also exists at the beginning of the real longest one, thus making it impossible for (X+1) to have the artefact.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Unless I'm wrong, they can tie 2 chains of locks between the lop and the artefact, such that each chain is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Loop-Lock A-Lock B-Artefact&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Loop-Lock B-Lock A-Artefact&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where Lock A and Lock B denote a lock owned by the first or second thief, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That way, no matter who takes the artefact, there will remain a lock attached to the loop that belongs to the thief that &lt;em&gt;didn't&lt;/em&gt; borrow the artefact, making it very obvious who took it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, this can probably be extended to more thieves, but I can't think of it off the top of my head&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After mulling it out some more, N thieves just need to make chains in such a way so that every single one of N thieves has at least 1 lock attached directly to the loop, and has 1 lock somewhere in the other chains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's refer to a chain starting with (Loop - Lock X - ... - Artefact) as X's chain.&#xA;Let's refer to chains that look like (Loop - ... - LockX X - ... - Artefact) as other chains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way, any X can break all the other chains at some point in the middle, can break his own chains at the beginning, and it's pretty obvious that X took the artefact if X's chains are missing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit2: Which doesn't solve the problem of X leaving his lock back on the loop.&lt;/p&gt;&#xA;" OwnerUserId="1575" LastEditorUserId="1575" LastEditDate="2014-06-13T22:06:41.700" LastActivityDate="2014-06-13T22:06:41.700" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1709" PostTypeId="2" ParentId="1705" CreationDate="2014-06-13T22:42:49.227" Score="0" Body="&lt;p&gt;I can't tell if this answer is only true due to a misstatement in the rules, but I'll answer the question as written regardless:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&quot;Thieves can't break a lock they do not have a key to.&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which this answer assumes means thieves can break a lock they do have a key to.  The term breaking also implies a difference from &quot;unlocking&quot;, but again, this conclusion may stem from misstatements in the rules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under these assumptions-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To successfully protect this &quot;artefact&quot; N chain of locks must be procured, each comprised of a cycle of N - 1 locks which require a cycle of N - 1 keys.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example for N = 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chain 1 contains:&#xA;Lock 1.A which accepts key a&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chain 2 contains:&#xA;Lock 2.A which accepts key b&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example for N = 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chain 1 contains&quot;&#xA;Lock 1.A which accepts key a and b&#xA;Lock 1.B which accepts key b and c&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chain 2 contains&#xA;Lock 2.A which accepts key b and c&#xA;Lock 2.B which accepts key c and a&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works because a thief can break a unique series of chains to release the &quot;artefact&quot; but has no &lt;em&gt;written&lt;/em&gt; ability to repair the broken locks.  Leaving proof of their treachery behind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;:&#xA;For the sake of completion, each lock in the cycle is held in place by N locks with a cycle of single lock keys, thus preventing a perpetrator from stealing the proof of their crime along with the &quot;artefact&quot;.&lt;/p&gt;&#xA;" OwnerUserId="457" LastActivityDate="2014-06-13T22:42:49.227" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1712" PostTypeId="1" CreationDate="2014-06-14T11:57:54.823" Score="8" ViewCount="934" Body="&lt;p&gt;There are $N$ letters in an alphabet. There is a combination lock, the code to it consists from $M$ different letters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can input $M$ letters combination and try to open the lock. (But you can't check the lock unless you have inputted all $M$ letters).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you guess at least $M-1$ letters from the code correctly the lock will be opened. The order of letters does not matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How many&lt;/strong&gt; tries you need to guarantee that lock will be opened? How to find list of combinations you need to try?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;P.S. I am especially interested for the case $N=32, M=4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example for the case $N=4, M=3$. The alphabet is &quot;ABCD&quot;.&lt;br&gt;&#xA;The answer is 1, with working combination &quot;ABC&quot;. Indeed, if the code is different it can be different in $1$ letter only, therefore $M-1=2$ letters are always the same.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="8874" LastEditDate="2015-02-06T15:23:17.603" LastActivityDate="2015-02-06T15:23:17.603" Title="Number of tries to guess M-1 letters from M-letters-code" Tags="|logical-deduction|combinatorics|locks|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="1714" PostTypeId="2" ParentId="1705" CreationDate="2014-06-14T19:51:23.957" Score="9" Body="&lt;p&gt;You can do it with $2N$ locks.  Make one chain $123...N$.  If we see $123...p$ on the wall, we know the artifact was taken by $p$ or $p+1$.  Make another chain $246...N13...(N-1)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added:  you can do it with $N+3$ locks.  Make the $123...N$ chain as above.  Then make a three lock chain.  Give a key to the first lock to every even number, a key to the last lock to every odd number, and nobody a key to the middle lock.  The short chain will distinguish $p$ from $p+1$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even better: a $3 \log_2 N$ solution. Write each person's number in binary starting at $0$. Make a three lock chain for each bit, giving the key for the lock on the wall to those with a $0$ in that bit, the key for the lock on the artifact to those with a $1$, and nobody the key to the middle one.  So for $N=8$ you make three chains.  For the first, the wall lock can be unlocked by $0,1,2,3$ and the artifact lock by $4,5,6,7$.  For the second, the wall lock can be unlocked by $0,1,4,5$ and the artifact lock by $2,3,6,7$ and the third wall lock can be unlocked by $0,2,4,6$ and the artifact lock by $1,3,5,7$.  To see who took it, put a $1$ for each missing middle lock and a $0$ for each present middle lock.  Form a binary number and you have the number of the taker.  You can also do $5 \log_3 N$ with chains of $5$ and base $3$, but that is inferior unless $N=3$ or $65 \le N \le 81$&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditDate="2014-06-18T21:37:02.227" LastActivityDate="2014-06-18T21:37:02.227" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1715" PostTypeId="1" CreationDate="2014-06-14T20:36:49.567" Score="13" ViewCount="1995" Body="&lt;p&gt;Oskar van Deventer has created the “haunted vending machine puzzle”.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IaRaz.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;You have to insert coins into the slots 1, 2, 3 or 4. A coin that hits a green switch on its head rests there; a coin that hits a switch on its leg changes switch position. A starting position is shown on the left side. The goal is to have 7 coins in the machine (one on each switch). Partial solution is shown on right side (3 out of 7 coins are on the right places). You can insert as many coins as you need, your supply of coins is unlimited.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You can try this game &lt;a href=&quot;http://www.clickmazes.com/ovd/ixovend.htm&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; (a java applet).&lt;/p&gt;&#xA;&lt;p&gt;Is it always possible to solve the puzzle? Is there a way to calculate the solution?&lt;/p&gt;&#xA;" OwnerUserId="143" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2019-12-15T21:23:41.237" Title="Oskar van Deventer's “haunted vending machine puzzle”" Tags="|mechanical-puzzles|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1716" PostTypeId="2" ParentId="625" CreationDate="2014-06-14T23:11:05.587" Score="1" Body="&lt;p&gt;This to add to &lt;a href=&quot;https://puzzling.stackexchange.com/a/630/28&quot;&gt;Joe Z.'s answer&lt;/a&gt;, who achieved 1861. Hope this will help the other people to improve the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Joe Z.'s idea with some tuning allows to achieve 1891. The idea, which makes it possible to proceed further is to choose between two algorithms basing on the first turn result. &lt;em&gt;[Actually, this is a special case of more general idea, described in &lt;a href=&quot;https://puzzling.stackexchange.com/a/1700/28&quot;&gt;Peter Taylor's answer&lt;/a&gt;, what is good about below-described algorithm is that $M_{max}$ here can be found without help of computer.]&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, here is the algorithm for M = &lt;strong&gt;2106&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Each spy gets 13 stones as his &quot;property&quot; stones (let's call them PS). That means a spy can count on these stones to use them how he likes without worries about what needs the other spy.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Each spy also gets some number ($T$) of mandatory turns (let's call them MT) to distribute his 13 stones between. $i$-th spy guarantees to throw 13 or less stones in $T_i$ turns.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Spy 1 decides how many mandatory throws each spy will get. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If spy 1 trows 1 stone at turn one then $T_1 = 7, T_2 = 6$. &lt;/li&gt;&#xA;&lt;li&gt;If spy 1 trows 2 or more stones at turn one then $T_1 = 6, T_2 = 5$.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If some spy $i$ ($i$ = 1 or 2) has made $T_i$ turns but the other spy (spy $j$) needs more then one turn to finish, spy $j$ &quot;leaves&quot; one stone for the other spy to &quot;burn&quot;: he (spy $j$) subtracts one stone from his PS.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If spy $i$ sees that the other spy must left for him one stone to &quot;burn&quot; spy $i$ must &quot;burn&quot; this stone no matter what: he throws one more stone than he planed.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Spies agree on enumeration of all posible combinations each of them can do by this algorithm in advance.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Let's calculate the combinations each spy can throw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First lemma, what number of combinations a spy can do if he decides to throw $S$ stones in $T$ turns?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The possible map of throws looks like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;`Ao|o|o|o|o|o|o o o o o o oB` &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The notation is &lt;code&gt;o&lt;/code&gt; - a stone, and each &lt;code&gt;|&lt;/code&gt; - a finish of a turn (can be at any plase between &lt;code&gt;o&lt;/code&gt;'s), &lt;code&gt;A&lt;/code&gt; - start, spies just came to the river, &lt;code&gt;B&lt;/code&gt; - finish, spies diverge,  &lt;code&gt;&lt;/code&gt; - a free position, where &lt;code&gt;|&lt;/code&gt; can be put.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have $T-1$ &lt;code&gt;|&lt;/code&gt; to put in $S-1$ position. The number of combinations will be $C = \binom{S-1}{T-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second lemma, what number of combinations a spy can do if he decides to throw $S$ stones in $T$ or  $T+1$ turns? It would be $C = \binom{S-1}{T-1} + \binom{S-1}{T} = \binom{S}{T}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the mini decision tree:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;First spy:  &lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;Decides that he needs $T_1 = 7, T_2 = 6$. Then his first turn has been finished, 12 PS and 6 MT of the other spy left. He can make 6 &quot;free&quot; turns.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;He can leave 0 stones for spy 2 to burn and throw 12 stones in 6 turns, which will give him $C = \binom{11}{5}$ combinations.  &lt;/li&gt;&#xA;&lt;li&gt;He can leave 1 stone for spy 2 to burn and throw 11 stones in 6 or 7 turns, which will give him $C = \binom{11}{6}$ combinations.  &lt;/li&gt;&#xA;&lt;li&gt;... [full analogy].&lt;/li&gt;&#xA;&lt;li&gt;In total $C = \binom{11}{5} + \binom{11}{6} + \binom{10}{7} + \binom{9}{8} = 1053$.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Decides that he needs $T_1 = 6, T_2 = 5$. Then his first turn is not finished, 12 PS and 5 MT of the other spy left. He can make 6 &quot;free&quot; turns.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;... [exactly the same as the previous case]&lt;/li&gt;&#xA;&lt;li&gt;In total $C = 1053$.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;In total $C = 1053 + 1053 = 2106$.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Second spy:  &lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;$T_1 = 7, T_2 = 6$. He has 13 PS and 6 MT of the other spy left. &lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;He can leave 0 stones for spy 1 to burn and throw 13 stones in 6 or 7 turns, which will give him $C = \binom{13}{7}$ combinations.&lt;/li&gt;&#xA;&lt;li&gt;He can leave 0 stones for spy 1 to burn and throw 12 stones in 7 or 8 turns, which will give him $C = \binom{12}{8}$ combinations.&lt;/li&gt;&#xA;&lt;li&gt;... [full analogy].&lt;/li&gt;&#xA;&lt;li&gt;In total $C = \binom{13}{7} + \binom{12}{8} + \binom{11}{9} + \binom{10}{10} = 2267 $.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$T_1 = 6, T_2 = 5$. He has 13 PS and 5 MT of the other spy left. &lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;... [full analogy with the previous case].&lt;/li&gt;&#xA;&lt;li&gt;In total $C = \binom{13}{6} + \binom{12}{7} + \binom{11}{8} + \binom{10}{9} = 2683 $.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In total $C = min(2267, 2683) = 2267$. (This spy can't choose the case).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In totals both spies can do $C = min(2106, 2267) = 2106$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="28" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.923" LastActivityDate="2014-06-14T23:29:06.050" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1717" PostTypeId="2" ParentId="1715" CreationDate="2014-06-15T01:57:08.253" Score="0" Body="&lt;p&gt;The starting position shown cannot be solved in 7 coins. Look at the rightmost switch. You can't land a coin on its head without first flipping it. When you flip it, the coin will continue on through the bottom switch and out of the machine. Unless you've flipped the bottom switch already, which again means you've lost a coin out the bottom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: turns out I was misunderstanding the question, the wording was a bit confusing. You can use more than seven coins in order to get seven coins stuck.&lt;/p&gt;&#xA;" OwnerUserId="385" LastEditorUserId="385" LastEditDate="2014-06-16T13:40:56.283" LastActivityDate="2014-06-16T13:40:56.283" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1718" PostTypeId="2" ParentId="626" CreationDate="2014-06-15T17:24:16.380" Score="5" Body="&lt;p&gt;Any solution that is longer than three lines, not including any smug prefaces about other solutions, is overdoing things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start from the back: You want to reach grandma with two cakes, so you were left with one before the last troll gave you one. Thus, you reached that troll with two cakes. Thus, we find that reaching a troll with two cakes does the trick. And we extrapolate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we were to answer the question of finding &lt;em&gt;every&lt;/em&gt; solution, on the other hand...&lt;/p&gt;&#xA;" OwnerUserId="1581" LastActivityDate="2014-06-15T17:24:16.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1719" PostTypeId="1" AcceptedAnswerId="11550" CreationDate="2014-06-15T19:03:02.640" Score="7" ViewCount="1283" Body="&lt;p&gt;Similarly to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/464/how-many-digits-can-be-removed-from-a-multiplication-puzzle-and-still-give-only&quot;&gt;How many digits can be removed from a multiplication puzzle and still give only one answer?&lt;/a&gt; I am curious about division puzzles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this puzzle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NDuXn.jpg&quot; width=&quot;200&quot;&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has only 2 out of 42 digits $\approx$ 4.8% digits shown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a puzzle with less percent of digits shown?&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2022-08-21T20:45:49.183" Title="How many digits can be removed from a division puzzle?" Tags="|mathematics|calculation-puzzle|missing-digits|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1722" PostTypeId="2" ParentId="1715" CreationDate="2014-06-15T23:31:55.643" Score="1" Body="&lt;p&gt;This can be modelled as a graph transition problem. There are $(2 \times 2)^7 = 16~384$ possible states the machine can be in, with each switch being either on the left or the right, and the coin on the switch or off of it. This is represented as a directed graph with $16~384$ points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each point has four out edges, representing inserting a coin into one of the four slots, and a varying number of in edges, depending on which states get to that state by inserting a coin into them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your goal is to get from one of the $128$ states where none of the switches have coins on them to one of the $128$ states where all the switches have coins on them. This is equivalent to finding a path in the graph from the first state to the second state, which is doable with a computer using breadth-first search. The length of the path represents the total number of coins you need to insert.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2014-06-15T23:31:55.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1723" PostTypeId="2" ParentId="604" CreationDate="2014-06-16T01:35:05.013" Score="4" Body="&lt;p&gt;kaine's position for zero queens is optimal. $\:$ Proof:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Consider a 2-by-2 corner block such as {&lt;a href=&quot;http://en.wikipedia.org/wiki/Algebraic_notation_%28chess%29&quot; rel=&quot;nofollow&quot;&gt;a1,a2,b1,b2&lt;/a&gt;}. $\:$ Clearly, a knight that &lt;em&gt;occupies&lt;/em&gt; any of those squares doesn't &lt;em&gt;attack&lt;/em&gt; any of those squares. $\:$ Since the only squares from which a knight can attack a1 are b3 and c2, a knight that attacks a1 doesn't attack any of the other three squares in the 2-by-2 corner block. $\:$ Since a knight changes the color of its square with each of its moves, a knight that attacks b2 doesn't attack a2 or b1. $\:$ By the previous three sentences, that 2-by-2 corner block can't be covered by less than three knights. $\:$ By how knights move, a knight that is not in the lower-left &lt;a href=&quot;http://mathworld.wolfram.com/Quadrant.html&quot; rel=&quot;nofollow&quot;&gt;quadrant&lt;/a&gt; of the chess board (a1's 4-by-4 corner block) cannot help cover that 2-by-2 corner block. $\:$ Thus, if that 2-by-2 corner block is covered then there are at least three knights in the lower-left quadrant of the chess-board. $\:$ By rotational symmetry, that applies to &lt;em&gt;each&lt;/em&gt; 2-by-2 block and the corresponding quadrant. $\:$ Thus, if the entire board is covered then there are at least three knights in each quadrant. $\:$ Since there are four quadrants and the quadrants are disjoint, 12 knights are necessary to cover the board. $\:$ Therefore kaine's position for zero queens is optimal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for stuff involving computation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I show how to reduce the search space significantly below what was described by Steven Stadnicki,&#xA;&lt;br&gt;&#xA;and give the code for two Python 3 programs to brute-force the two-queen and three-queen cases.&#xA;&lt;br&gt;&#xA;The programs' outputs indicate that kaine's positions for two and three queens are optimal.&#xA;&lt;br&gt;&lt;br&gt;&lt;br&gt;&#xA;Consider the 8 squares {&lt;a href=&quot;http://en.wikipedia.org/wiki/Algebraic_notation_%28chess%29&quot; rel=&quot;nofollow&quot;&gt;a3,a6,c1,c8,f1,f8,h3,h6&lt;/a&gt;}. $\:$ (That set is preserved by rotation and reflection.) $\:$ Clearly, a knight which &lt;em&gt;occupies&lt;/em&gt; any of those squares doesn't &lt;em&gt;attack&lt;/em&gt; any of those squares. $\:$ Since knights change color on each of their moves, a knight can't attack two squares whose nearest corners of the board share exactly one side. $\:$ By looking at the knight's move, it also can't attack two squares that [are on a diagonal and exactly two apart on that diagonal] or [have more than three rows or columns between them]. $\:$ By the previous three sentences, a single knight covers at most one of those eight squares. $\:$ In particular, if the board could be covered by QQQNNN then at least five of those squares must be covered by queens. $\:$ Furthermore, the entire problem is preserved by rotation and reflection. $\:$ Thus, one only needs to check positions of the queens that cover at least five of those 8 squares&#xA;&lt;br&gt;&#xA;and are &quot;canonical&quot; with respect to rotation and reflection. $\:$ (There are exactly 4406 such positions.)&#xA;&lt;br&gt;&#xA;For any position of the queens and zero or more knights, one can pick any given square that's&#xA;&lt;br&gt;&#xA;not yet covered, and only have to deal with positions that have a knight covering that square.&#xA;&lt;br&gt;&#xA;As a final optimization, I apply the previous sentence to the not-yet-been-covered square&#xA;&lt;br&gt;&#xA;that would minimize the number of covering-squares if there were no queens on the board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, to make the coding somewhat easier, I observe that a knight can go from square_a to square_b if and only if a knight can go from square_b to square_a, and that queens &quot;might as well&quot; not block each other's movement, since one could just have the blocking piece move to the intended square.&#xA;&lt;br&gt;&#xA;The code itself is located at &lt;a href=&quot;http://pastebin.com/DaWta3T8&quot; rel=&quot;nofollow&quot;&gt;http://pastebin.com/DaWta3T8&lt;/a&gt;, since I can't figure out how to format large code blocks. $\:$ It took well over an hour to write, runs in under 20 seconds on my desktop computer, and outputs &lt;code&gt;False&lt;/code&gt;, which indicates that kaine's position for 3 queens is optimal. $\:$ Similarly, &lt;a href=&quot;http://pastebin.com/2fk3Viin&quot; rel=&quot;nofollow&quot;&gt;this modified code&lt;/a&gt; takes about 20 seconds to run and indicates that kaine's position for 2 queens is also optimal.&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user1579" LastEditorDisplayName="user1579" LastEditDate="2014-06-16T06:12:48.210" LastActivityDate="2014-06-16T06:12:48.210" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1724" PostTypeId="1" AcceptedAnswerId="3128" CreationDate="2014-06-16T02:54:47.040" Score="15" ViewCount="1359" Body="&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/1715/oskar-van-deventers-haunted-vending-machine-puzzle&quot;&gt;This question&lt;/a&gt; asks about a strange vending machine in which coins can fall out as well as being deposited in, and all the coins it asks for must be deposited before the machine will accept them.&lt;/p&gt;&#xA;&lt;p&gt;That got me thinking, what would happen if you couldn't see inside the machine to know which switches would be affected and which coins would fall out or stay in?&lt;/p&gt;&#xA;&lt;p&gt;Thus, consider the following problem:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Consider an &lt;a href=&quot;http://www.clickmazes.com/ovd/ixovend.htm&quot; rel=&quot;noreferrer&quot;&gt;Oskar van Deventer vending machine&lt;/a&gt; with seven switches. These seven switches are in some unknown position, and it's not known which position the switches are currently in, or which switches have coins in them. It's only known that at least one switch does not have a coin in it.&lt;/p&gt;&#xA;&lt;p&gt;This machine is also rather finicky in that it only accepts one denomination of coin, so all the coins are indistinguishable.&lt;/p&gt;&#xA;&lt;p&gt;Devise an algorithm of inserting coins that will be sure of inserting coins into all of the switches in the fewest number of moves, regardless of the starting position.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-10-21T22:54:27.917" Title="A haunted vending machine that you can't see inside" Tags="|mechanical-puzzles|" AnswerCount="1" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="1725" PostTypeId="1" AcceptedAnswerId="1735" CreationDate="2014-06-17T07:10:10.150" Score="13" ViewCount="2843" Body="&lt;p&gt;I heard Martin Gardner in one of his books presented a crossword, which looks like 3x3 square. &lt;strong&gt;Could you help me to find it?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, have been there a smaller crossword puzzle created?&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="8874" LastEditDate="2016-03-18T15:57:46.183" LastActivityDate="2016-03-18T15:57:46.183" Title="The smallest crossword" Tags="|crosswords|puzzle-history|" AnswerCount="7" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1726" PostTypeId="2" ParentId="1725" CreationDate="2014-06-17T08:48:38.180" Score="10" Body="&lt;p&gt;It's not hard to construct a 1x1 crossword:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;+---+&#xA;|1  |&#xA;|   |&#xA;+---+&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;  &#xA;  &lt;p&gt;Across:&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;A Roman one.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;Down:&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;The shortest pronoun.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's also a cheating one going around, with clues &quot;&lt;em&gt;What letter am I thinking of?&lt;/em&gt;&quot; and &quot;&lt;em&gt;What is the answer to 1 Across?&lt;/em&gt;&quot;, and claimed solution &quot;&lt;em&gt;I'm thinking of U&lt;/em&gt;&quot;.&lt;/p&gt;&#xA;" OwnerUserId="228" LastActivityDate="2014-06-17T08:48:38.180" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="1727" PostTypeId="2" ParentId="1725" CreationDate="2014-06-17T10:29:36.727" Score="4" Body="&lt;p&gt;What a about the null crossword, presented below. It has no clues. This is the smallest crossword possible and yet impossible to solve. (Or not solve.)&lt;/p&gt;&#xA;" OwnerUserId="1587" LastActivityDate="2014-06-17T10:29:36.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1728" PostTypeId="1" AcceptedAnswerId="1741" CreationDate="2014-06-17T13:22:32.410" Score="15" ViewCount="3124" Body="&lt;p&gt;Consider the classic pencil and paper game &lt;a href=&quot;http://en.wikipedia.org/wiki/Dots_and_Boxes&quot;&gt;Dots and Boxes&lt;/a&gt; . You can assume the 2 by 2 version (that is with 9 dots). &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Starting with an empty grid of dots, players take turns, adding a&#xA;  single horizontal or vertical line between two unjoined adjacent dots.&#xA;  A player who completes the fourth side of a 1×1 box earns one point&#xA;  and takes another turn. (The points are typically recorded by placing&#xA;  in the box an identifying mark of the player, such as an initial). The&#xA;  game ends when no more lines can be placed. The winner of the game is&#xA;  the player with the most points.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/blhGw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that there are at most 12 moves in the entire game and I think there will be between 9 and 11 turns.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The puzzle is to determine if the game is a win, lose or draw for the&#xA;  first player assuming optimal play by both sides.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clearly you could brute force it by computer, which is fine and perfectly interesting, but not as interesting as a human understandable proof.&lt;/p&gt;&#xA;" OwnerUserId="8998" LastEditorUserId="41973" LastEditDate="2019-07-08T07:56:20.293" LastActivityDate="2021-07-19T23:43:50.203" Title="Optimal play for 2 by 2 dots and boxes" Tags="|game|pencil-and-paper-games|dots-and-boxes|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1729" PostTypeId="1" CreationDate="2014-06-17T13:44:30.407" Score="2" ViewCount="1619" Body="&lt;p&gt;If  &quot;PSAKLJIUASP&quot; is &quot;PJIUAZAKLKP&quot;, then what is the word for &quot;LIOQWERRPOIL&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I noticed that the first 3 characters are reverse of the last 3 characters in both words in question. A is mapped to I and I is mapped to A. But the same logic doesn't work for the other cases. Any clues please?&lt;/p&gt;&#xA;" OwnerUserId="1590" LastEditorUserId="5373" LastEditDate="2017-02-06T11:58:21.157" LastActivityDate="2017-02-06T11:58:21.157" Title="character decoding problem at an interview" Tags="|cipher|pattern|" AnswerCount="4" CommentCount="8" ClosedDate="2015-04-20T03:09:49.013" ContentLicense="CC BY-SA 3.0" />
  <row Id="1730" PostTypeId="2" ParentId="1729" CreationDate="2014-06-17T14:10:22.457" Score="5" Body="&lt;p&gt;I don't see a nice answer, but there are some things you should notice.  First, the wording is ambiguous:  We are told &quot;PSAKLJIUASP&quot; is &quot;PJIUAZAKLKP&quot;, but it is not clear whether  &quot;LIOQWERRPOIL&quot; should be the source or destination.  I would notice the groups of three letters AKL and JIU which are maintained but shifted in position.  Unfortunately that leaves (besides the leading and trailing P's) S and AS, which might go to K and AZ, but it is not clear how.  If &quot;LIOQWERRPOIL&quot; is the source, that leads us to &quot;LERR??OQW?L&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an interview question, I would think this is coherent enough to show I wasn't flustered by nonsense and try to go on.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2014-06-17T14:10:22.457" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1731" PostTypeId="2" ParentId="1725" CreationDate="2014-06-17T14:42:18.127" Score="1" Body="&lt;p&gt;As Peter Taylor mentioned, it's possible to make a $1 \times 1$ crossword. Since crosswords don't necessarily have to use whole words in the English language like Scrabble does (abbreviations and proper nouns are totally fair game in crosswords), it's possible to use any of the $26$ letters in the single square, for a total of $26$ possible combinations.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If you get up to $2 \times 2$, there are four essentially distinct ways to block off squares that don't convert it into just a $1 \times 1$:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;+---+---+    +---+---+    +---+---+    +---+---+&#xA;|1  |2  |    |1  |   |    |1  |xxx|    |1  |xxx|&#xA;|   |   |    |   |   |    |   |xxx|    |   |xxx|&#xA;+---+---+    +---+---+    +---+---+    +---+---+&#xA;|3  |   |    |   |xxx|    |   |xxx|    |xxx|2  |&#xA;|   |   |    |   |xxx|    |   |xxx|    |xxx|   |&#xA;+---+---+    +---+---+    +---+---+    +---+---+&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The last one, however, is just two $1 \times 1$s superimposed on a $2 \times 2$ board, which is uninteresting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of these has multiple options for what words can go into them, which are hard to enumerate here.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If you are going by the rule that a crossword word must have at least three letters, the smallest grid you can make is $3 \times 3$, simply because a smaller grid can't fit a three-letter word into it. There are two ways such a grid can be made:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;+---+---+---+    +---+---+---+&#xA;|1  |2  |3  |    |1  |   |2  |&#xA;|   |   |   |    |   |   |   |&#xA;+---+---+---+    +---+---+---+&#xA;|4  |   |   |    |   |xxx|   |&#xA;|   |   |   |    |   |xxx|   |&#xA;+---+---+---+    +---+---+---+&#xA;|5  |   |   |    |3  |   |   |&#xA;|   |   |   |    |   |   |   |&#xA;+---+---+---+    +---+---+---+&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and again, a multitude of ways the letters can be filled in to make words.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2014-06-17T14:42:18.127" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1732" PostTypeId="2" ParentId="1715" CreationDate="2014-06-17T14:57:35.220" Score="7" Body="&lt;p&gt;First of all if either of the top two switches has a coin on it, drop another to clear it. Now, if either of them are leaning in (and by leaning I always mean the &lt;em&gt;top&lt;/em&gt; of the switch leaning that way), drop a coin on their outside so that lean out. From now on we will only ever hit a top-row switch with pairs of coins (one each side), so they will always stay clear and lean out. (until the final two or coins, anyway!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if the central switch has a coin on it, drop a pair of coins on either top-row switch. One of them will fall through and clear the central switch. We will also be keeping the central switch clear until right near the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observe that we can now drop a set of four coins one each into the four slots - in any order - and these three switches will all end up clear, and the top row leaning out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the outside switches on the middle row. After the first coin strikes them, they can only ever lean in. Let's drop a set of four coins so we know they're leaning in. The only possible states are with-coin and without.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we were to drop a pair of coins on the side away from which the central switch leans, we change the state of the outer-middle switch on that side while keeping the central switch clear and the top switches unchanged. If the two out-middle switches are in different states, do this to bring them into alignment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the outer-middle switches do not have coins on them, drop a set of four so that they will.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have the top and central switches clear, the outer-middle switches laden, the top switches leaning out and the outer-middle switches laden. This can be preserved by dropping two sets of four in a row, to make a set of eight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm going to ask that the very first coin of each set of eight be placed directly above the exposed leg of the central switch. This is to constrain the order in which the coins arrive at the lower switches. It doesn't actually make a difference to the &lt;em&gt;outcome&lt;/em&gt; in the next paragraph but does make it easier to calculate.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens to the lower switches when we drop a set of eight? They each get hit inside, then outside, then inside again. The effect this has depends on their initial state.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;out+laden -&gt; out+unladen&lt;/li&gt;&#xA;&lt;li&gt;out+unladen -&gt; in+unladen&lt;/li&gt;&#xA;&lt;li&gt;in+laden -&gt; in+unladen&lt;/li&gt;&#xA;&lt;li&gt;in+unladen -&gt; in+laden&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Drop two sets of eight. Now both bottom switches must leaning in, and will remain leaning in after any number of sets of eight after that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, we deal with the bottom switch toward which the central switch is leaning. We want it to be in+laden. If it is in+unladen, drop a set of eight. Now it is in+laden.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the other bottom switch is also in+laden then the bottom row is complete. Otherwise we complete is as such:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drop two pairs (preserving the state of all but the bottom row because the outer-middle switch gets toggled twice) on the side toward which the central switch is leaning (the side that is currently in+laden). The first pair will drop a coin on the outside of the near bottom switch, making it in+unladen just like the other bottom switch. The second pair will drop coins on the inside of both switches, loading them up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the bottom row is complete, we simply drop a coin from directly above the head of the central switch to load it up, then do the same for the two switches on the top row.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, now you can all have fun spotting the mistakes I no doubt made. :)&lt;/p&gt;&#xA;" OwnerUserId="385" LastEditorUserId="385" LastEditDate="2014-06-20T11:35:33.493" LastActivityDate="2014-06-20T11:35:33.493" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1733" PostTypeId="2" ParentId="1701" CreationDate="2014-06-17T15:04:32.297" Score="5" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KcrCm.png&quot; alt=&quot;grid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one were to work through the way you describe (each player taking into account only the previous players plan), then a 12-unit loop would quickly be observed.  Case $M=2$ is the same as case $M=14$ which is the same as case $M=26$ etc.  This doesn't really help anything.  This loop does not yield a perfect strategy for either player.  Essentially, no $L$ level player will be against an $L-1$ level player so why plan as if he were?  As there is a loop there isn't really an $M=\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xL0Cr.png&quot; alt=&quot;chart&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets pretend that a better player of level $L$ chooses a plan which assumes that the other player will randomly choose a strategy such that $ S &amp;lt; L $.  So $ M=3$ takes into account $M=1$ and $M=2$ (but not the totally random case).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After 10,000 iterations of this done (by making a matrix of what position has what probability of beating what position multiplied by vectors describing the probability of each player choosing each position) I get the answers below for which the queen wins about 35% of the time while the king wins 64% with both players playing optimally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DjtM4.png&quot; alt=&quot;another chart&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at it from the perspective of the queen trying to find a perfect strategy and the king reacting perfectly, this does not improve much over the queen just randomly picking any square!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also designed a pseudo-evolutionary algorithm where 20,000 games are played.  Each player chooses where he should go based on the third power of the percentage of times he won in that spot previously (plus a small mutation factor close to the value of 1 win).  This isn't worth much, but it claims the king should: prefer A, B, and C;  choose D, E, and F about 1/2 as much; and never choose H, I, and J.  The queen should chose about all randomly, but prefer H, I, J, and E. I think there is too much noise though in this program, and I don't want to focus on it anymore. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likely, the best strategy possible is the one that has no preferred placement of the kings, as everywhere is covered equally. I doubt that a strategy the would let the queen win more than 40% of the time if the king playing perfectly is possible.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="463" LastEditDate="2015-06-02T03:27:00.707" LastActivityDate="2015-06-02T03:27:00.707" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1734" PostTypeId="2" ParentId="1729" CreationDate="2014-06-17T16:54:11.593" Score="4" Body="&lt;p&gt;If you assign a value to each letter (a = 1, b = 2... z = 26) and add up each letter, the two sample words &quot;PSAKLJIUASP&quot; and &quot;PJIUAZAKLKP&quot; are both equal (137).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The test word adds up to 169, so it seems like any 12 character word with the same value should suffice (my favorite so far has been &quot;zzzsoeeaaasy&quot;).&lt;/p&gt;&#xA;" OwnerUserId="457" LastActivityDate="2014-06-17T16:54:11.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1735" PostTypeId="2" ParentId="1725" CreationDate="2014-06-17T18:33:05.143" Score="21" Body="&lt;p&gt;I found the book: &lt;em&gt;M.Gardner &quot;Aha. Insight.&quot;&lt;/em&gt;, p. 153.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the crossword, which was called there &quot;World's smallest crossword&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mXyrQ.png&quot; alt=&quot;crossword&quot;&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The clues are:&lt;br&gt;&#xA;  Horizontal:&#xA;  1. Insect&#xA;  4. To annoy&#xA;  5. Eavesdropper&lt;br&gt;&#xA;  Vertical:&#xA;  1. Stingers&#xA;  2. To employ&#xA;  3. Gigi has two&lt;br&gt;&#xA;  (There are typos in numbers 4 and 5).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solution:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/zJ0Jh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2014-06-18T16:25:41.113" LastActivityDate="2014-06-18T16:25:41.113" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="1736" PostTypeId="2" ParentId="1728" CreationDate="2014-06-17T20:43:07.353" Score="2" Body="&lt;p&gt;The key to perfect play in this game seems to be the number of lines in the center cross at the end of turn 6!  Warnig I&quot;ve not taken the time to put my 39 drawings into the computer so this is a map of the complex (largely brute force) proof.  This was all done by hand but I figured out which of the 3 out of 4 possible 7 line patterns with no squares or threes forced player 2 to give all the squares to player 1.  This made the problem much simpler&lt;/p&gt;&#xA;&#xA;&lt;p&gt;squares - the things that earn points&#xA;Threes - one line is left to make a square&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Turn seven study with no Threes or Squares&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Turn seven means there are 6 lines on the board and it is player 1's turn.  I also assumed no squares or threes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 1: Turn seven starts and there is one or fewer lines in the center cross and no squares or threes. Player 1 can force a win in any circumstance.  I proved this with brute force.  There are only 15 such truely unique cases due to symetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 2: Turn seven starts and the two center lines are at right angles to each other but there are no squares or threes.  Player 1 can force a win in any circumstance.  I proved this with brute force.  There are only 3 such truely unique cases due to symetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 3: Turn seven starts and there are two center lines in the same direction with no squares or threes. Only a draw can be forced.  AVOID There are only 7 such unique cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is impossible for there to be three in the center and no threes or squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Player 2 playing optimumly, therefore, needs to make a square or three on turn 6 or before! I also need to look for the possible force draw option for Player 2.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Early square study&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Player 1 does not want any squares or threes before turn seven so he will not make one on turn three.  The earliest possible Three will therefore be on turn four.  This will go to Player 1 on turn five.  This does not appear to give Player 2 an advantage and can only seem to force the same scenereos as him leaving one Three for turn seven.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Turn seven revisited with Threes or one square&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Assume that a Three is a Three and it's orientation does not matter and it is equivalent to a square already owned by that player (leaving an extra line on the board).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are more than two threes, they all go to player 1 and, therefore, He will win or draw.  If Player 2 wants a win, he can leave at most one three at the end of turn 6.  There are 10 such possibilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Player 1 can manage (after receiving his square on turn 7) to not create a three he wins.  This is because there will be 8 lines on the board after this turn and there is no way to place 9 such that only one square and no three's are formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is impossible if there are three lines in the center after he makes his square.  He can, however, chose in all cases to create the three in such a way that the other player can only get one square.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Conclusion&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I've walked through how to show by hand that Player 2 cannot force a win.  I have not, however, shown that he cannot force a tie.  I've also neglected to include the 39 drawings I made on paper to come to this conclusion.  I will include them if I find time to draw them on the computer on another date.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Player 2 can force a tie this requires him to either give the other player two squares or threes by turn seven or place exactly two parellel lines (and no perpendicular lines) in the center by turn seven.  Player 1 should be able to avoid this by placing more lines in the center that don't form three but I've not shown this yet for any series of moves.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditDate="2014-06-17T20:51:54.833" LastActivityDate="2014-06-17T20:51:54.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1737" PostTypeId="2" ParentId="1725" CreationDate="2014-06-18T08:23:55.540" Score="4" Body="&lt;p&gt;George Perec actually jokingly &lt;strong&gt;designed&lt;/strong&gt; a 1x1 grid, with the following clues:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Horizontalement : 1. Voyelle&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Verticalement : 1. Consonne &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;see &lt;a href=&quot;https://fr.wikipedia.org/wiki/Mots_crois%C3%A9s#Une_lettre&quot;&gt;French Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Italian, you may find a lot of examples in &lt;a href=&quot;http://www.aldospinelli.com/definizionario/&quot;&gt;the site of Aldo Spinelli&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="70" LastActivityDate="2014-06-18T08:23:55.540" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1738" PostTypeId="2" ParentId="1729" CreationDate="2014-06-18T12:15:10.193" Score="3" Body="&lt;p&gt;I suspect these words are deliberate gibberish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only letters in these words are A E I J K L O P Q R S U W and Z. Look at the positions of these letters on the keyboard:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[Q][W][E][R] T  Y [U][I][O][P]&#xA;&#xA; [A][S] D  F  G  H [J][K][L]&#xA;&#xA;  [Z] X  C  V  B  N  M&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This looks rather like the sort of distribution you would produce if you were mashing the keyboard randomly with both hands.&lt;/p&gt;&#xA;" OwnerUserId="40" LastActivityDate="2014-06-18T12:15:10.193" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1739" PostTypeId="2" ParentId="1728" CreationDate="2014-06-18T14:22:04.953" Score="3" Body="&lt;p&gt;Not as interesting as a manual analysis, but here is a brute-force solver, written in Python. (Warning, CPU intensive for about two minutes)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from collections import defaultdict&#xA;&#xA;VERTICAL, HORIZONTAL = 0, 1&#xA;&#xA;#represents a single line segment that can be drawn on the board.&#xA;class Line:&#xA;    def __init__(self, x, y, orientation):&#xA;        self.x = x&#xA;        self.y = y&#xA;        self.orientation = orientation&#xA;    def __hash__(self):&#xA;        return hash((self.x, self.y, self.orientation))&#xA;    def __eq__(self, other):&#xA;        if not isinstance(other, Line): return False&#xA;        return self.x == other.x and self.y == other.y and self.orientation == other.orientation&#xA;    def __repr__(self):&#xA;        return &quot;Line({}, {}, {})&quot;.format(self.x, self.y, &quot;HORIZONTAL&quot; if self.orientation == HORIZONTAL else &quot;VERTICAL&quot;)&#xA;&#xA;class State:&#xA;    def __init__(self, width, height):&#xA;        self.width = width&#xA;        self.height = height&#xA;        self.whose_turn = 0&#xA;        self.scores = {0:0, 1:0}&#xA;        self.lines = set()&#xA;    def copy(self):&#xA;        ret = State(self.width, self.height)&#xA;        ret.whose_turn = self.whose_turn&#xA;        ret.scores = self.scores.copy()&#xA;        ret.lines = self.lines.copy()&#xA;        return ret&#xA;    #iterate through all lines that can be placed on a blank board.&#xA;    def iter_all_lines(self):&#xA;        #horizontal lines&#xA;        for x in range(self.width):&#xA;            for y in range(self.height+1):&#xA;                yield Line(x, y, HORIZONTAL)&#xA;        #vertical lines&#xA;        for x in range(self.width+1):&#xA;            for y in range(self.height):&#xA;                yield Line(x, y, VERTICAL)&#xA;    #iterate through all lines that can be placed on this board, &#xA;    #that haven't already been placed.&#xA;    def iter_available_lines(self):&#xA;        for line in self.iter_all_lines():&#xA;            if line not in self.lines:&#xA;                yield line&#xA;&#xA;    #returns the number of points that would be earned by a player placing the line.&#xA;    def value(self, line):&#xA;        assert line not in self.lines&#xA;        all_placed = lambda seq: all(l in self.lines for l in seq)&#xA;        if line.orientation == HORIZONTAL:&#xA;            #lines composing the box above the line&#xA;            lines_above = [&#xA;                Line(line.x,   line.y+1, HORIZONTAL), #top&#xA;                Line(line.x,   line.y,   VERTICAL),   #left&#xA;                Line(line.x+1, line.y,   VERTICAL),   #right&#xA;            ]&#xA;            #lines composing the box below the line&#xA;            lines_below = [&#xA;                Line(line.x,   line.y-1, HORIZONTAL), #bottom&#xA;                Line(line.x,   line.y-1, VERTICAL),   #left&#xA;                Line(line.x+1, line.y-1, VERTICAL),   #right&#xA;            ]&#xA;            return all_placed(lines_above) + all_placed(lines_below)&#xA;        else:&#xA;            #lines composing the box to the left of the line&#xA;            lines_left = [&#xA;                Line(line.x-1, line.y+1, HORIZONTAL), #top&#xA;                Line(line.x-1, line.y,   HORIZONTAL), #bottom&#xA;                Line(line.x-1, line.y,   VERTICAL),   #left&#xA;            ]&#xA;            #lines composing the box to the right of the line&#xA;            lines_right = [&#xA;                Line(line.x,   line.y+1, HORIZONTAL), #top&#xA;                Line(line.x,   line.y,   HORIZONTAL), #bottom&#xA;                Line(line.x+1, line.y,   VERTICAL),   #right&#xA;            ]&#xA;            return all_placed(lines_left) + all_placed(lines_right)&#xA;&#xA;    def is_game_over(self):&#xA;        #the game is over when no more moves can be made.&#xA;        return len(list(self.iter_available_lines())) == 0&#xA;&#xA;    #iterates through all possible moves the current player could make.&#xA;    #Because scoring a point lets a player go again, a move can consist of a collection of multiple lines.&#xA;    def possible_moves(self):&#xA;        for line in self.iter_available_lines():&#xA;            if self.value(line) &amp;gt; 0:&#xA;                #this line would give us an extra turn.&#xA;                #so we create a hypothetical future state with this line already placed, and see what other moves can be made.&#xA;                future = self.copy()&#xA;                future.lines.add(line)&#xA;                if future.is_game_over(): &#xA;                    yield [line]&#xA;                else:&#xA;                    for future_move in future.possible_moves():&#xA;                        yield [line] + future_move&#xA;            else:&#xA;                yield [line]&#xA;&#xA;    def make_move(self, move):&#xA;        for line in move:&#xA;            self.scores[self.whose_turn] += self.value(line)&#xA;            self.lines.add(line)&#xA;        self.whose_turn = 1 - self.whose_turn&#xA;&#xA;    def tuple(self):&#xA;        return (tuple(self.lines), tuple(self.scores.items()), self.whose_turn)&#xA;    def __hash__(self):&#xA;        return hash(self.tuple())&#xA;    def __eq__(self, other):&#xA;        if not isinstance(other, State): return False&#xA;        return self.tuple() == other.tuple()&#xA;&#xA;#function decorator which memorizes previously calculated values.&#xA;def memoized(fn):&#xA;    answers = {}&#xA;    def mem_fn(*args):&#xA;        if args not in answers:&#xA;            answers[args] = fn(*args)&#xA;        return answers[args]&#xA;    return mem_fn&#xA;&#xA;#finds the best possible move for the current player.&#xA;#returns a (move, value) tuple.&#xA;@memoized&#xA;def get_best_move(state):&#xA;    cur_player = state.whose_turn&#xA;    next_player = 1 - state.whose_turn&#xA;    if state.is_game_over():&#xA;        return (None, state.scores[cur_player] - state.scores[next_player])&#xA;    best_move = None&#xA;    best_score = float(&quot;inf&quot;)&#xA;    #choose the move that gives our opponent the lowest score&#xA;    for move in state.possible_moves():&#xA;        future = state.copy()&#xA;        future.make_move(move)&#xA;        _, score = get_best_move(future)&#xA;        if score &amp;lt; best_score:&#xA;            best_move = move&#xA;            best_score = score&#xA;    return [best_move, -best_score]&#xA;&#xA;s = State(2,2)&#xA;best_move, relative_value = get_best_move(s)&#xA;print(&quot;The best first move is {}, with {} more points than the second player.&quot;.format(best_move, relative_value))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h1&gt;Result:&lt;/h1&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The best first move is [Line(0, 0, HORIZONTAL)], with 2 more points than the second player.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In English, this means the first player should place his first line horizontally in the lower left corner, and &lt;strong&gt;the final score will be 3-1 in his favor&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Interesting Observations&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Some trivia I found while writing the above program.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A full transcription of one possible optimal play:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Player 0 makes move [Line(0, 0, HORIZONTAL)]. &lt;br/&gt;&#xA;  Player 1 makes move [Line(0, 1, HORIZONTAL)]. &lt;br/&gt; &#xA;  Player 0 makes move [Line(1, 0, HORIZONTAL)].  &lt;br/&gt;&#xA;  Player 1 makes move [Line(1, 1, HORIZONTAL)].  &lt;br/&gt;&#xA;  Player 0 makes move [Line(1, 1, VERTICAL)].  &lt;br/&gt;&#xA;  Player 1 makes move [Line(0, 2, HORIZONTAL)].  &lt;br/&gt;&#xA;  Player 0 makes move [Line(0, 1, VERTICAL), Line(1, 2, HORIZONTAL)]. &lt;br/&gt;&#xA;  Player 1 makes move [Line(2, 1, VERTICAL), Line(0, 0, VERTICAL)]. &lt;br/&gt;&#xA;  Player 0 makes move [Line(1, 0, VERTICAL), Line(2, 0, VERTICAL)]. &lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the first player chooses an interior line as his first move (a sub-optimal play), the second player can force a tie if he chooses the opposite interior line, and loses 1-3 if he chooses anything else.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1497" LastEditorUserId="1497" LastEditDate="2014-06-18T14:38:08.530" LastActivityDate="2014-06-18T14:38:08.530" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1740" PostTypeId="2" ParentId="614" CreationDate="2014-06-18T20:03:06.107" Score="4" Body="&lt;p&gt;I don't think you can force someone to use the X-Wing technique. There are several techniques to eliminate clues. If they eliminate the 7s in your example by a different technique , then the X-Wing will be gone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eliminating clues in a different order will also cause the grid to look different.&#xA;This site has been around for years and may give some insight:&#xA;&lt;a href=&quot;http://www.setbb.com/sudoku/&quot; rel=&quot;nofollow&quot;&gt;http://www.setbb.com/sudoku/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1598" LastActivityDate="2014-06-18T20:03:06.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1741" PostTypeId="2" ParentId="1728" CreationDate="2014-06-19T14:45:06.893" Score="10" Body="&lt;p&gt;Going first, you should be able to force a win.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using @kaine's terminology:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;box - an area on the board that can be made into a square&lt;/li&gt;&#xA;&lt;li&gt;square - a box of 4 lines&lt;/li&gt;&#xA;&lt;li&gt;three - a box of three lines which can be made into a square on the next turn&lt;/li&gt;&#xA;&lt;li&gt;safe - a move that doesn't result in a three&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Moves are enumerated as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;* 1 * 2 *&#xA;3   4   5&#xA;* 6 * 7 *&#xA;8   9  10&#xA;*11 * 12*&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h1&gt;Winning Boards&lt;/h1&gt;&#xA;&#xA;&lt;h2&gt;Split Board&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;After 5 moves, the board is split down the middle into 2 1x2 half boards. One half contains 2 connected boxes with no threes and no safe moves, and the other contains 2 separate boxes with no threes and no safe moves.  For example, the following boards are split down the middle into left and right half boards.  Any move on either board will concede a square or two:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;o   o---o      o   o---o      o   o   o      o   o   o&#xA;    |              |              |   |          |   |&#xA;o---o   o      o---o   o      o---o   o      o---o   o&#xA;    |   |          |              |   |          |   &#xA;o   o   o      o   o---o      o   o   o      o   o---o&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If you can make any of these boards, or rotations/flips of them, you will win. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that there are three sections, one with 2 boxes (on the right in these examples), and 2 with 1 box (on the left).  Your opponent goes first, and must concede two of these sections to you.  Since you will have a choice of which to concede, you never have to concede the right half (with 2 squares).  Thus, you will win 3-1 every time.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Corner Board&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If a board can be divided in such a way after 6 turns that you are forced to concede a single square, but the other 3 squares are connected and will be conceded to you, then you will win.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below are some examples of such boards.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;o   o   o      o---o---o      o---o---o&#xA;|   | c            | c              c |    &#xA;o   o---o      o   o   o      o   o   o&#xA;|              |              |        &#xA;o---o---o      o---o---o      o---o---o&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For all three boards, concede the box 'c' in the upper right, and your opponent will be forced to concede the other 3 to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Corner Boards have the following properties:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the corner box has two out of four sides completed&lt;/li&gt;&#xA;&lt;li&gt;there are no threes&lt;/li&gt;&#xA;&lt;li&gt;the corner box is isolated so that making a square in it does not make a three elsewhere, but results in all boxes having 2 out of three sides completed&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h1&gt;General Strategy&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The general strategy to be used is as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If presented with a three, take the square and play again&lt;/li&gt;&#xA;&lt;li&gt;If no squares can be taken, play safe&lt;/li&gt;&#xA;&lt;li&gt;If you can't play safe, concede a single square&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This strategy is used in a number of situations and is referenced simply as &quot;General Strategy&quot; for the following scenarios.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Strategy&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;On your first move, you must take an outside edge.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; &lt;em&gt;If you don't and take one of the lines touching the middle dot, then your opponent can force a draw simply by splitting the board with their first move and then mirroring every move you make on the other half of the board.  When you are forced to make a three, he will take the square and make the same three on the other half board.  Repeat until the end of the game for a tie.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, here is an example of taking an outside edge:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;o   o---o&#xA;&#xA;o   o   o&#xA;&#xA;o   o   o&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Your goal will be to create the split board, but failing that, work towards the corner board.  Your opponent can now either take an interior line, or an edge.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Interior line&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If your opponent takes an interior line (4/6/7/9), then you can take another interior line 90 degrees to it without setting up a three.  Here are the moves you would make based on his move:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;4 --&amp;gt; 6&#xA;6 --&amp;gt; 4&#xA;7 --&amp;gt; 9&#xA;9 --&amp;gt; 7&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Thus, there are two possible boards (since either 4/6 and 7/9 result in the same board):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; Board 1        Board 2&#xA;o   o---o      o   o---o&#xA;    |                                  &#xA;o---o   o      o   o---o&#xA;                   |    &#xA;o   o   o      o   o   o&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Board 1&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your opponent has a few choices. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Moves 1/3/5/7: All concede a square, which you take on your first turn, and then make move 11.  Proceed with the General Strategy for the win.&lt;/li&gt;&#xA;&lt;li&gt;Moves 9/10/12: In one move, you can create a Split Board for a win.&lt;/li&gt;&#xA;&lt;li&gt;Move 8/11: Play safe with 10, and he will be forced to either concede a square, or make a Corner Board with move 12.  If he concedes, play the General Strategy for the win.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Board 2&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are the following options for your opponent:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Moves 1/3/6: You can make a Split Board on your next move.&lt;/li&gt;&#xA;&lt;li&gt;Moves 4/5/10/12: Take the square and make move 8. General Strategy results in a win.&lt;/li&gt;&#xA;&lt;li&gt;Move 8/11: Play safe with move 3.  He will be force to make a Corner Board with move 1 or concede a square.  General Strategy results in a win either way.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Exterior Line&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If your opponent takes an exterior line, then you can try to aim for a Split Board or Corner Board, if available.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Split Board Candidates&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following boards should be attempted to be made into a Split Board.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; Board 1        Board 2        Board 3&#xA;o   o   o      o   o---o      o   o---o&#xA;        |                               &#xA;o   o   o      o   o   o      o   o   o&#xA;        |              |               &#xA;o   o   o      o   o   o      o   o---o&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If the flipping/rotating the board matches any of these, then you can try for a Split Board by attempting to split the board vertically.  So, make move 4. To complete the Split Board, your opponent must make either 6 or 9, and when they do, you make the other for the guaranteed victory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they make a different move, then your goal is now to make a Corner Board.  Here are the break downs for each board above after you make move 4:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Boards 1 and 2&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;o   o   o      o   o---o&#xA;    |   |          |    &#xA;o   o   o      o   o   o&#xA;        |              | &#xA;o   o   o      o   o   o&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The only safe moves (other than 6/9 which can be made into a Split Board) are 1/3/8/11/12.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Move 2/7: He is conceding the upper right square early.  Take it and then play safe with move 11. General Strategy wins the day.&lt;/li&gt;&#xA;&lt;li&gt;Move 8/11/12: If you make move 6 (the bottom of the top left corner square) the only remaining safe move is 12 or 11, which makes a Corner Board.  If your opponent instead concedes a square, General Strategy wins.&lt;/li&gt;&#xA;&lt;li&gt;Move 1/3: Make move 11.  Your opponent can now complete the Corner Board by making move 12, or make the only other safe move - move 8.  By making move 12 your self, the next move by your opponent will concede all 4 squares to you.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Board 3&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;o   o---o&#xA;    |   &#xA;o   o   o&#xA;&#xA;o   o---o&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The only safe moves are 1/3/8/10/11.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Move 5/7: Similar to the previous case, you can take the square and then force a victory by making move 11 and using the General Strategy.&lt;/li&gt;&#xA;&lt;li&gt;Move 8/11/10: If you make move 6, the only remaining safe move is 10 or 11 which makes a Corner Board.  Anything else, and General Strategy will win for you.&lt;/li&gt;&#xA;&lt;li&gt;Move 1/3: &lt;em&gt;Concede the upper right square immediately by making move 7!&lt;/em&gt;  If your opponent takes the square, they will only have 2 possible safe moves for their second move - 8 or 11.  You take the other, and they will be forced to concede the other three squares.  Alternatively, your opponent may concede the one of the other squares, so take it, and then make move 11.  They will then be forced to concede the other two squares.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Corner Board Candidates&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After eliminating the Split Board candidates, and any rotations/flips, the following three boards remain:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;o   o---o      o   o---o      o   o---o&#xA;        |                              &#xA;o   o   o      o   o   o      o   o   o&#xA;                              |        &#xA;o   o   o      o---o   o      o   o   o&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;None of these are split board candidates.  So, you will aim for a Corner Board instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On your move, for the first and second boards, make move 8.  For the third, make move 11.  All three boards are now rotations/flips of one another, so they are all the same scenario.  Thus, without loss of generality, we can use the following board for your opponents move:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;o   o---o&#xA;        |&#xA;o   o   o&#xA;&#xA;o---o   o&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Your opponent can either play safe or concede the square.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Move 4/7:  Take the square on your first move of the turn and then make move 8.  General Strategy takes over from here.&lt;/li&gt;&#xA;&lt;li&gt;Moves 3/8/12: Try to make a Corner Board with the top right corner as the corner box.  Thus, you can play move 8 (or 3/12 is 8 is already taken), leaving your opponent move 12 or 3 as the only safe move, which results in a Corner Board.  If instead, they concede a square earlier, General Strategy can be used to win.&lt;/li&gt;&#xA;&lt;li&gt;Moves 1/10: Make move 8 and the board is now a rotation/flip of the previous case except the corner is the lower left.  Flip the board and use the same strategy.&lt;/li&gt;&#xA;&lt;li&gt;Move 6/9: Make move 10.  The only safe move left is 1 or 3, and if they take it, you have a Corner Board.  If they don't use General Strategy to win.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="1592" LastEditDate="2014-06-20T11:08:17.190" LastActivityDate="2014-06-20T11:08:17.190" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1742" PostTypeId="1" AcceptedAnswerId="1743" CreationDate="2014-06-19T18:06:09.617" Score="3" ViewCount="433" Body="&lt;p&gt;There are arithmetic puzzles with some numbers (typically four) and rules for combining them (perhaps just the four basic arithmetic operators or including things like combining digits 2 and 3 to make numbers like 23).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/623/write-twenty-four&quot;&gt;Write twenty-four from four numbers&lt;/a&gt; is an example which was recently posed on this site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the general name of this type of puzzle, or is it just an &quot;arithmetic puzzle&quot;?&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="69582" LastEditDate="2020-12-07T15:18:13.870" LastActivityDate="2020-12-07T15:18:13.870" Title="Is there a more specific name for this type of arithmetic puzzle?" Tags="|mathematics|calculation-puzzle|formation-of-numbers|puzzle-identification|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1743" PostTypeId="2" ParentId="1742" CreationDate="2014-06-19T18:32:25.647" Score="2" Body="&lt;p&gt;I have not heard any special name for them before.&lt;/p&gt;&#xA;&lt;p&gt;But &lt;a href=&quot;https://erich-friedman.github.io/puzzle/form/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, such puzzles are called &amp;quot;Number Formation Puzzles,&amp;quot; and if you google, you can find &amp;quot;About 26,300 results&amp;quot;, so it is more or less a common name.&lt;/p&gt;&#xA;&lt;p&gt;Meanwhile, if you google arithmetic puzzles you can find many different types of such a puzzles:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Verbal_arithmetic&quot; rel=&quot;nofollow noreferrer&quot;&gt;Verbal arithmetic&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.thepuzzleclub.com/mental-arithmetic.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mental Arithmetic&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.riddleriders.com/math-puzzles/arithmetic-puzzles&quot; rel=&quot;nofollow noreferrer&quot;&gt;Math puzzles&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;and others, practically any imaginary puzzle, which involves arithmetic is called with such a name by somebody.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="28" LastEditorUserId="4254" LastEditDate="2020-12-07T06:44:17.587" LastActivityDate="2020-12-07T06:44:17.587" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="1744" PostTypeId="1" CreationDate="2014-06-19T19:11:51.837" Score="6" ViewCount="145" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;You have N numbers written in one row. One needs to put any amount of the four symbols &lt;span class=&quot;math-container&quot;&gt;$($&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$)$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$+$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$\times$&lt;/span&gt; between them in such a way that the resulting expression is divisible by &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;What are &lt;span class=&quot;math-container&quot;&gt;$[N,M]$&lt;/span&gt; pairs it is always possible to do with (independently of the given sequence of numbers)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;For example, with &lt;span class=&quot;math-container&quot;&gt;$N=2, M=2$&lt;/span&gt; you can always do what is asked. Indeed, if you have two numbers &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; and at least one is even you write &lt;span class=&quot;math-container&quot;&gt;$a \times b$&lt;/span&gt;, if none is even you write &lt;span class=&quot;math-container&quot;&gt;$a+b$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-06-21T01:32:33.277" Title="When can one construct expression dividable by M from any N numbers?" Tags="|mathematics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1745" PostTypeId="2" ParentId="1744" CreationDate="2014-06-19T19:34:09.357" Score="5" Body="&lt;p&gt;You can always do it if $N \ge M$.  Reduce all the numbers $\pmod M$.  If any are zero, put multiply signs everywhere.  If not, imagine putting plus signs everywhere and keeping track of the running sum.  Since there are at least $M$, either one is $0 \pmod M$ or two are the same $\pmod M$.  If one of the running sums is zero, put parentheses around that addition and multiply all the rest of the numbers.  If two of the running sums are the same, start after the first one and go to the end of the second.  The sum of that interval will be $0 \pmod M$.  Multiply by all the other numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $M$ is composite and $N$ is at least the sum of the prime factors of $M$ (counting multiplicity) you can succeed.  For example, let $M=12, N=7$.  We have shown that the first three elements can contribute a factor $3$ and each of the pairs of the last four can contribute a factor $2$ so we can succeed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $N$ is at least the sum of the prime factors of $M$ you can succeed.  It may be possible to succeed with fewer-I thought I had a proof that you couldn't but it fails.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditDate="2014-06-21T01:32:33.277" LastActivityDate="2014-06-21T01:32:33.277" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="1746" PostTypeId="2" ParentId="432" CreationDate="2014-06-19T22:18:45.327" Score="5" Body="&lt;p&gt;While you can use software and premade corpuses (whatever the latin declination would be?), I think that's a bit of a cop out. The first creators had no such tools at their disposal!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first thing to do is to forget about trying to get the black cells in a pattern. I find it easiest to just start with a good long word or three and go from there, slowly weaving — horizontal, vertical, horizontal ... so the corner gets filled. It requires some sudoku-style trial-and-error, maybe talking to yourself aloud (to get more possible matches: jud-e, jud-f, jud-g[em], etc.) and backtracking, but it is quite doable and rewarding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The choice of the starting word will depend on the language and its crossword making rules. In Slovenian for example, most words have a roughly even distribution of vowels and consonants, plus you'll never find two of the first together. So I know I have to pick the initial word, so that it has at most two together, otherwise I'll usually have problems in the grid later and will either have to use more black cells to get the problematic parts out as 2/3-letter simpletons or resort to foreign words and names.&#xA;English has less regularity, but you could start the same way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can't get a 4x4 with what you've started with, try 5x3 or more in one of the favourable directions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if you still get stuck, going online for one word is still better than having most of the work done by a computer. There's always a 16th century Romanian poet, a Sudanese politician from 1922 or the wife of a Japanese olympian from 1984 to help you out.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2014-06-19T22:18:45.327" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1747" PostTypeId="2" ParentId="558" CreationDate="2014-06-19T22:23:16.233" Score="1" Body="&lt;p&gt;A simple metric I use for judging quality, could also be used for difficulty — but only somewhat. The lower the density of black cells, the longer the words are and the harder was for the puzzle to be constructed, so there's bound to be harder entries in it.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2014-06-19T22:23:16.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1748" PostTypeId="1" AcceptedAnswerId="1749" CreationDate="2014-06-19T22:23:57.240" Score="3" ViewCount="314" Body="&lt;p&gt;Is it common to see &lt;em&gt;non-words&lt;/em&gt; (like the homophone &lt;a href=&quot;https://puzzling.stackexchange.com/a/1735/139&quot;&gt;verticals in this answer&lt;/a&gt;)  used in English crosswords? Ignoring abbr., initials, odd names, various symbols, &quot;Xth and Yth letter&quot;, &quot;middle of X&quot; and other similar simple fillers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Was that particular puzzle just extra cryptic (for fun/art) or is this common?&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2014-06-19T22:55:43.133" Title="How common are non-words in crosswords?" Tags="|puzzle-creation|crosswords|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1749" PostTypeId="2" ParentId="1748" CreationDate="2014-06-19T22:55:43.133" Score="2" Body="&lt;p&gt;I would say they are rather rare.  This was a special puzzle-note the book title.  I have seen cases where there is a symbol like star in a square, but usually (if you spell it out) it makes words.  I have also seen cases where a few letters are deleted from theme answers.  Maybe the theme is POSTHOLE and the letters POST are deleted from the theme answers making non-words.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2014-06-19T22:55:43.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1750" PostTypeId="1" CreationDate="2014-06-20T02:21:08.513" Score="5" ViewCount="494" Body="&lt;p&gt;I see &lt;a href=&quot;https://puzzling.stackexchange.com/questions/582/identify-unique-verbal-arithmetic-puzzles&quot;&gt;Identify unique Verbal Arithmetic Puzzles&lt;/a&gt; deals with trying to deal with uniqueness when creating Verbal Arithmetic puzzles, but I have a question that deals with something earlier in the creation process, namely &lt;strong&gt;how does one go about picking words for the puzzle to begin with&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I could start with any phrase with word lengths roughly the right length and hope that it works, but that is obviously not very efficient.&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;So as an example, I know that &lt;code&gt;LYNDON x B = JOHNSON&lt;/code&gt; is one, and I could try other U.S. presidents (like &lt;code&gt;WARREN x G = HARDING&lt;/code&gt;), but it's basically like sifting through dirt trying to find gold--there's a low chance of success.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Another approach would be to start with the arithmetic equation (and then find words to fit the equations), but the problem there is that the search space there is extremely big. There might be patterns to look for, but I'm not aware of any that would narrow down the possibilities to something that isn't still overwhelming.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, is there some strategy that makes finding words for these puzzles easier, or am I stuck with the naive approaches?&lt;/p&gt;&#xA;" OwnerUserId="1600" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2016-03-05T17:35:08.513" Title="How does one go about picking words when creating Verbal Arithmetic puzzles?" Tags="|puzzle-creation|alphametic|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1751" PostTypeId="1" AcceptedAnswerId="2403" CreationDate="2014-06-20T07:06:24.013" Score="7" ViewCount="1579" Body="&lt;p&gt;I've been struggling for a while with the &lt;a href=&quot;https://play.google.com/store/apps/details?id=com.BitShiftInteractive.Games.BetweenWorlds&quot; rel=&quot;nofollow noreferrer&quot;&gt;Between Worlds&lt;/a&gt; Android game, and have not yet solved it. I've only noticed a couple vague properties that hold in localized cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yWdts.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To operate the puzzle, select the square, pentagon, or hexagon and move the lever. The spheres and rings on the right will move by some unknown conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would provide a summary of what I've done, but every example has at least two counterexamples. The only consistent pattern I've noticed is that, when a pair of pieces are in alignment, selecting the button counterclockwise of the circle that's in alignment doesn't move it out of place - but it doesn't hold in all cases, and I'm uncertain as to why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the logical pattern behind this puzzle? How do I solve it?&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastActivityDate="2014-10-11T12:35:14.973" Title="How do I solve the &quot;Between Worlds&quot; Android puzzle game?" Tags="|logical-deduction|calculation-puzzle|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1752" PostTypeId="2" ParentId="1750" CreationDate="2014-06-20T15:55:30.367" Score="1" Body="&lt;p&gt;If there is no worry about multiple solutions (which I feel is required for it to be a verbal arithmetic problem), find words is simple as long as one follows the following rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There are no more than 10 different letters.  7-9 is probably the&#xA;most useful in order to limit too many answers.&lt;/li&gt;&#xA;&lt;li&gt;Count the appropriate number of digits.  If you want $5$ letters&#xA;times $1$ letter, the result can only have $5$ or $6$ letters&#xA;as $99999*9=899991$ and $10000*1=10000$ which are the extreme&#xA;examples. Use those extremes to determine whether the equation make sense.&lt;/li&gt;&#xA;&lt;li&gt;Shorter words (3-5 letters) work better than longer ones.  You are setting up a&#xA;system of simultaneous equations.  The fewer equations, the more&#xA;likely they will work out but the more likely there will be duplicate&#xA;answers.&lt;/li&gt;&#xA;&lt;li&gt;Have a computer based solver ready to rapidly check whether a&#xA;solution exists.  If follow the above rules, most sets of words will yield&#xA;at least one solution.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="99" LastEditorUserId="12109" LastEditDate="2015-07-16T16:25:56.493" LastActivityDate="2015-07-16T16:25:56.493" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="1754" PostTypeId="1" AcceptedAnswerId="1759" CreationDate="2014-06-20T19:22:41.230" Score="6" ViewCount="1778" Body="&lt;p&gt;There is not winning strategy for 9cell 3x3 board.&#xA;There is a winning strategy for &lt;a href=&quot;http://all-r-math.blogspot.de/2012/11/4x4-tic-tac-toe-ii.html&quot;&gt;16cell 4x4 board&lt;/a&gt;, or even for 10cell &quot;3x3 + 1_cell_near_a_corner&quot; board (see M.Gardner &quot;Mathematical games&quot;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{|c|c|}&#xA;\hline&#xA;   &amp;amp;  &amp;amp;  &amp;amp;  - \\&#xA;\hline&#xA;    &amp;amp; &amp;amp;  &amp;amp; - \\&#xA;\hline&#xA;   \space\space &amp;amp; \space\space &amp;amp; X &amp;amp;  \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here &quot;$-$&quot; states for unused cells. &quot;$X$&quot; - for the first winning move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder is there a board with 9 or less cells where a player has a winning strategy? So the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the smallest tic-tac-toe board to have a winning strategy for first or second player?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. win means the usual thing: to put 3 signs (&quot;x&quot; or &quot;o&quot;) consecutively in a row (horizontally, vertically or diagonally). So one obviously needs at least 5 cells to perform the moves until the winning one.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="69582" LastEditDate="2021-07-19T18:13:36.147" LastActivityDate="2021-07-19T18:13:36.147" Title="What is the smallest tic-tac-toe board to have a winning strategy?" Tags="|optimization|game|tic-tac-toe|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1757" PostTypeId="2" ParentId="1754" CreationDate="2014-06-20T20:02:28.290" Score="9" Body="&lt;p&gt;What I found up to now is a 7 cell board:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{array}{|c|c|}&#xA;\hline&#xA;  - &amp;amp;  &amp;amp;  -&amp;amp;  -\\&#xA;\hline&#xA;   &amp;amp; X &amp;amp;  &amp;amp;  \\&#xA;\hline&#xA;   - &amp;amp; &amp;amp; - &amp;amp; - \\&#xA;\hline&#xA;   - &amp;amp; &amp;amp; - &amp;amp; - \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;(Two 4-cell runs, one vertical and one horizontal, which overlap at their second cells. The intersection cell has an X to indicate it is where to play first)&lt;/p&gt;&#xA;&lt;p&gt;But I do not have a proof of minimality.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="69582" LastEditDate="2021-07-19T17:57:53.143" LastActivityDate="2021-07-19T17:57:53.143" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="1758" PostTypeId="1" AcceptedAnswerId="1763" CreationDate="2014-06-21T06:08:34.197" Score="0" ViewCount="721" Body="&lt;p&gt;M.Gardner in his book &quot;Mathematical puzzles and diversions&quot; states that the following 2x4 map &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{|c|c|}&#xA;\hline&#xA;   1 &amp;amp; \phantom{1} &amp;amp; \phantom{1} &amp;amp; \phantom{1} \\&#xA;\hline&#xA;   \phantom{1} &amp;amp; \phantom{1} &amp;amp; \phantom{1} &amp;amp; \phantom{1} \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can be folded in 40 different ways (along 7 segments shown) that the cell with 1 on it will be the upper one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to count all these ways and prove that there are exactly 40 of them?&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="11876" LastEditDate="2016-08-03T22:50:47.500" LastActivityDate="2016-08-03T22:50:47.500" Title="Number of ways to fold a 2x4 map" Tags="|mathematics|topology|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="1759" PostTypeId="2" ParentId="1754" CreationDate="2014-06-21T07:47:18.960" Score="8" Body="&lt;p&gt;6 or fewer cells will never make a board in which either player can force a win. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a 6 cell board, each player moves exactly thrice. To win, $X$ must pick 3 cells which happen to lie adjacent to each other on a line, meaning that his only chance to win is on move 5. It's a standard strategy-stealing argument to show that $O$ can never force a win on any board (because making a move is always advantageous); either $X$ can force a win or the board is a draw with perfect play.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be able to win move 5 on a board of any size, he must have at least two potential winning cells after move 3, since $O$ will be able to block one with move 4. But he only has selected two cells after his second move. These two cells determine a line, so both of his winning cells must also be on that line. Hence, he must have picked adjacent cells along the same line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose there were only one line of 4 or more cells passing through $X$'s first move. $O$ knows $X$ can only force a win along this line with his second move (if at all). With move 2, she can pick either cell along this line adjacent to move 1. As we've already said, $X$'s only chance with move 3 is to play in the other square adjacent to move 1 along this line. But now $O$ already has one of the (at most) 2 cells $X$ could complete this line with blocked, and she can pick the other one with move 4. So with move 5, $X$ can't win since he's already committed to this line and $O$ has stopped him there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now we just have to show that a board with 6 cells or fewer can't possibly have two distinct lines of 4 or more. But this is actually obvious. If it did, by Pidgeonhole principle, those two lines must overlap on at least two cells. But if they overlap on two cells, they're each the line determined by those two cells; hence actually the same line. So 7 cells is required for a board to be a forced win for $X$. The proof also suggests the example given in &lt;a href=&quot;https://puzzling.stackexchange.com/questions/1754/what-is-the-smallest-tic-tac-toe-board-to-have-a-winning-strategy/1757#1757&quot;&gt;this answer&lt;/a&gt;, which indeed is a forced win for $X$ on move 5. Interestingly, although the board has 7 cells, $X$ still only needs 3 moves to win.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You'll note that the fact that two points on a plane uniquely determine a line was used repeatedly here. This is necessary; it's what tells you that the board is on Euclidean space as opposed to something else. You might ask what happens if we loosen this and allow boards on other sorts of spaces. This isn't exactly a precise question, though there are several ways one can make it precise. If you look at tic-tac-toe on a cube (where the faces are cells), for example, in some sense the &quot;diagonal&quot; lines correspond to picking 3 vertices which meet at a corner. So in fact, any collection of 3 faces of a cube are adjacent along a &quot;line&quot; of some sort. Thus, for either the full cube or a subset of 5 faces, $X$ automatically wins turn 5 regardless of the moves chosen by either player.&lt;/p&gt;&#xA;" OwnerUserId="94" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2014-06-21T07:47:18.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1760" PostTypeId="2" ParentId="1758" CreationDate="2014-06-23T13:57:48.270" Score="2" Body="&lt;p&gt;This is not the answer, but just one step forward it.&lt;br&gt;&#xA;I give here 29 ways to fold the map + idea for 4 more ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I number the cells according to positions they would take at the resulting 1x1 folder map, counting from top to bottom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notations: H - fold along horizontal line; VI - fold along I-th vertical line; in - fold in a way that front part (usually with numbers) will be hidden inside; out - fold in a way that front part (usually with numbers) will be outside; &quot;IJ&quot; - cell in I-th row and J-th column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Left-to-right order of vertical folds.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6 when fold horizontal first:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Hout, V1out, V2in, V3out&#xA;$$&#xA;\begin{array}{|c|c|}&#xA;\hline&#xA;   1 &amp;amp; 4 &amp;amp; 5 &amp;amp; 7 \\&#xA;\hline&#xA;   2 &amp;amp; 3 &amp;amp; 6 &amp;amp; 8 \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/li&gt;&#xA;&lt;li&gt;Hout, Vout, Vin, Vin&#xA;$$&#xA;\begin{array}{|c|c|}&#xA;\hline&#xA;   1 &amp;amp; 4 &amp;amp; 7 &amp;amp; 6 \\&#xA;\hline&#xA;   2 &amp;amp; 3 &amp;amp; 8 &amp;amp; 5 \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/li&gt;&#xA;&lt;li&gt;Hout, V1out, V2out, V3in&#xA;$$&#xA;\begin{array}{|c|c|}&#xA;\hline&#xA;   1 &amp;amp; 8 &amp;amp; 5 &amp;amp; 4 \\&#xA;\hline&#xA;   2 &amp;amp; 7 &amp;amp; 6 &amp;amp; 3 \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/li&gt;&#xA;&lt;li&gt;Hout, V1out, V2out, V3out&#xA;$$&#xA;\begin{array}{|c|c|}&#xA;\hline&#xA;   1 &amp;amp; 8 &amp;amp; 3 &amp;amp; 6 \\&#xA;\hline&#xA;   2 &amp;amp; 7 &amp;amp; 4 &amp;amp; 5 \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/li&gt;&#xA;&lt;li&gt;Hout, V1out, V2out, V3out (right end should go inside of left end)&#xA;$$&#xA;\begin{array}{|c|c|}&#xA;\hline&#xA;   1 &amp;amp; 8 &amp;amp; 2 &amp;amp; 7 \\&#xA;\hline&#xA;   4 &amp;amp; 5 &amp;amp; 3 &amp;amp; 6 \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/li&gt;&#xA;&lt;li&gt;Hout, V1out, V2out, V3out (folded half right end should go in-front of &quot;21&quot;)&#xA;$$&#xA;\begin{array}{|c|c|}&#xA;\hline&#xA;   1 &amp;amp; 8 &amp;amp; 2 &amp;amp; 5 \\&#xA;\hline&#xA;   6 &amp;amp; 7 &amp;amp; 3 &amp;amp; 4 \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;6 when fold horizontal 2-nd:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;V1out, Hout, V2in, V3out (folded half must be put behind &quot;1&quot;)&#xA;$$&#xA;\begin{array}{|c|c|}&#xA;\hline&#xA;   1 &amp;amp; 6 &amp;amp; 5 &amp;amp; 2 \\&#xA;\hline&#xA;   8 &amp;amp; 7 &amp;amp; 4 &amp;amp; 3 \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/li&gt;&#xA;&lt;li&gt;V1out, Hout, V2in, V3in (folded half must be put behind &quot;11&quot;)&#xA;$$&#xA;\begin{array}{|c|c|}&#xA;\hline&#xA;   1 &amp;amp; 6 &amp;amp; 3 &amp;amp; 4 \\&#xA;\hline&#xA;   8 &amp;amp; 7 &amp;amp; 2 &amp;amp; 5 \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/li&gt;&#xA;&lt;li&gt;V1out, Hout, V2out, V3in&#xA;$$&#xA;\begin{array}{|c|c|}&#xA;\hline&#xA;   1 &amp;amp; 2 &amp;amp; 6 &amp;amp; 7 \\&#xA;\hline&#xA;   4 &amp;amp; 3 &amp;amp; 5 &amp;amp; 8 \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/li&gt;&#xA;&lt;li&gt;V1out, Hout, V2out, V3in (folded half must be put in-front of &quot;21&quot;)&#xA;$$&#xA;\begin{array}{|c|c|}&#xA;\hline&#xA;   1 &amp;amp; 2 &amp;amp; 5 &amp;amp; 6 \\&#xA;\hline&#xA;   8 &amp;amp; 3 &amp;amp; 4 &amp;amp; 7 \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/li&gt;&#xA;&lt;li&gt;V1out, Hout, V2out, V3out&#xA;$$&#xA;\begin{array}{|c|c|}&#xA;\hline&#xA;   1 &amp;amp; 2 &amp;amp; 8 &amp;amp; 5 \\&#xA;\hline&#xA;   4 &amp;amp; 3 &amp;amp; 7 &amp;amp; 6 \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/li&gt;&#xA;&lt;li&gt;V1out, Hout, V2out, V3out (folded half must be put in-front of &quot;21&quot;)&#xA;$$&#xA;\begin{array}{|c|c|}&#xA;\hline&#xA;   1 &amp;amp; 2 &amp;amp; 7 &amp;amp; 4 \\&#xA;\hline&#xA;   7 &amp;amp; 3 &amp;amp; 6 &amp;amp; 5 \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;4 when fold horizontal 3-rd:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;V1out, V2in, Hout, V3out&#xA;$$&#xA;\begin{array}{|c|c|}&#xA;\hline&#xA;   1 &amp;amp; 6 &amp;amp; 3 &amp;amp; 8 \\&#xA;\hline&#xA;   8 &amp;amp; 5 &amp;amp; 4 &amp;amp; 7 \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/li&gt;&#xA;&lt;li&gt;V1out, V2in, Hout, V3out (folded half must be put in-front of &quot;21&quot; and &quot;22&quot;)&#xA;$$&#xA;\begin{array}{|c|c|}&#xA;\hline&#xA;   1 &amp;amp; 6 &amp;amp; 3 &amp;amp; 8 \\&#xA;\hline&#xA;   8 &amp;amp; 5 &amp;amp; 4 &amp;amp; 7 \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/li&gt;&#xA;&lt;li&gt;V1out, V2in, Hout, V3in (folded half must be put behind of &quot;11&quot; and &quot;12&quot;)&#xA;$$&#xA;\begin{array}{|c|c|}&#xA;\hline&#xA;   1 &amp;amp; 2 &amp;amp; 5 &amp;amp; 4 \\&#xA;\hline&#xA;   8 &amp;amp; 7 &amp;amp; 6 &amp;amp; 3 \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/li&gt;&#xA;&lt;li&gt;V1out, V2out, Hout, V3in (impossible)&lt;/li&gt;&#xA;&lt;li&gt;V1out, V2out, Hout, V3out (impossible)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;4 when fold horizontal 4-th:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;V1out, V2in, V3out, Hout&#xA;$$&#xA;\begin{array}{|c|c|}&#xA;\hline&#xA;   1 &amp;amp; 2 &amp;amp; 3 &amp;amp; 4 \\&#xA;\hline&#xA;   8 &amp;amp; 7 &amp;amp; 6 &amp;amp; 5 \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/li&gt;&#xA;&lt;li&gt;Vout, Vin, Vin, Hout &#xA;$$&#xA;\begin{array}{|c|c|}&#xA;\hline&#xA;   1 &amp;amp; 2 &amp;amp; 4 &amp;amp; 3 \\&#xA;\hline&#xA;   8 &amp;amp; 7 &amp;amp; 5 &amp;amp; 6 \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/li&gt;&#xA;&lt;li&gt;V1out, V2out, V3in, Hout&#xA;$$&#xA;\begin{array}{|c|c|}&#xA;\hline&#xA;   1 &amp;amp; 4 &amp;amp; 3 &amp;amp; 2 \\&#xA;\hline&#xA;   8 &amp;amp; 5 &amp;amp; 6 &amp;amp; 7 \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/li&gt;&#xA;&lt;li&gt;V1out, V2out, V3out, Hout&#xA;$$&#xA;\begin{array}{|c|c|}&#xA;\hline&#xA;   1 &amp;amp; 4 &amp;amp; 2 &amp;amp; 3 \\&#xA;\hline&#xA;   8 &amp;amp; 5 &amp;amp; 7 &amp;amp; 6 \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;V2 first.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 when fold horizontal 2-nd:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;V2out, Hout, V1out, V3out&#xA;$$&#xA;\begin{array}{|c|c|}&#xA;\hline&#xA;   1 &amp;amp; 8 &amp;amp; 7 &amp;amp; 2 \\&#xA;\hline&#xA;   4 &amp;amp; 5 &amp;amp; 6 &amp;amp; 3 \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/li&gt;&#xA;&lt;li&gt;V2out, Hout, V1out, V3out (&quot;14&quot; goes behind &quot;22&quot;)&#xA;$$&#xA;\begin{array}{|c|c|}&#xA;\hline&#xA;   1 &amp;amp; 8 &amp;amp; 7 &amp;amp; 4 \\&#xA;\hline&#xA;   2 &amp;amp; 3 &amp;amp; 6 &amp;amp; 5 \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/li&gt;&#xA;&lt;li&gt;V2out, Hout, V1out, V3in (&quot;14&quot; goes behind &quot;13&quot;)&#xA;$$&#xA;\begin{array}{|c|c|}&#xA;\hline&#xA;   1 &amp;amp; 8 &amp;amp; 5 &amp;amp; 7 \\&#xA;\hline&#xA;   2 &amp;amp; 3 &amp;amp; 4 &amp;amp; 6 \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;6 when fold horizontal 3-nd:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;V2out, V1***, Hout (is impossible)&lt;/li&gt;&#xA;&lt;li&gt;V2out, V3out, Hout, V1in&#xA;$$&#xA;\begin{array}{|c|c|}&#xA;\hline&#xA;   1 &amp;amp; 8 &amp;amp; 6 &amp;amp; 7 \\&#xA;\hline&#xA;   2 &amp;amp; 3 &amp;amp; 5 &amp;amp; 4 \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;V2out, V3in, Hout, V1in&#xA;$$&#xA;\begin{array}{|c|c|}&#xA;\hline&#xA;   1 &amp;amp; 8 &amp;amp; 7 &amp;amp; 6 \\&#xA;\hline&#xA;   2 &amp;amp; 3 &amp;amp; 4 &amp;amp; 5 \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;V2in, V1out, Hout, V1in (&quot;14&quot; in-front &quot;13)&#xA;$$&#xA;\begin{array}{|c|c|}&#xA;\hline&#xA;   1 &amp;amp; 8 &amp;amp; 5 &amp;amp; 4 \\&#xA;\hline&#xA;   2 &amp;amp; 7 &amp;amp; 6 &amp;amp; 3 \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;V2in, V1out, Hout, V1out (&quot;14&quot; behind &quot;13)&#xA;$$&#xA;\begin{array}{|c|c|}&#xA;\hline&#xA;   1 &amp;amp; 8 &amp;amp; 3 &amp;amp; 6 \\&#xA;\hline&#xA;   2 &amp;amp; 7 &amp;amp; 4 &amp;amp; 5 \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/li&gt;&#xA;&lt;li&gt;V2out, V3out, Hout, V1in&#xA;$$&#xA;\begin{array}{|c|c|}&#xA;\hline&#xA;   1 &amp;amp; 6 &amp;amp; 7 &amp;amp; 8 \\&#xA;\hline&#xA;   2 &amp;amp; 5 &amp;amp; 4 &amp;amp; 3 \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/li&gt;&#xA;&lt;li&gt;V2out, V3in, Hout, V1in&#xA;$$&#xA;\begin{array}{|c|c|}&#xA;\hline&#xA;   1 &amp;amp; 6 &amp;amp; 8 &amp;amp; 7 \\&#xA;\hline&#xA;   2 &amp;amp; 5 &amp;amp; 3 &amp;amp; 4 \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;V3 first. H 2-nd&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;V3in, Hout, V1out, V2out&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;V3in, Hout, V1out, V2in&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;V3out, Hout, V1out, V2out&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;V3out, Hout, V1out, V2in&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(to be continued)&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2014-06-23T15:23:37.603" LastActivityDate="2014-06-23T15:23:37.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1761" PostTypeId="1" AcceptedAnswerId="1762" CreationDate="2014-06-23T14:29:30.687" Score="33" ViewCount="7608" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oqwcY.jpg&quot; alt=&quot;Isaac Asimov's Science Fiction Magazine, spring 1977&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;I found the following puzzle by Martin Gardner in an archived copy of &lt;em&gt;&lt;a href=&quot;http://www.archive.org/stream/spgisaacasimovss1977asimrich/spgisaacasimovss1977asimrich_djvu.txt&quot; rel=&quot;nofollow noreferrer&quot;&gt;Isaac Asimov's Science Fiction Magazine&lt;/a&gt;&lt;/em&gt;. I've heard it before, but in a more vulgar setting, involving condoms instead of gloves. I think mr. Gardner knew this version too, hence the gender and the name of the patient.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The first earth colony on Mars has been swept by an epidemic of Barsoomian flu. The cause: a native Martian virus not yet isolated.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;There is no way to identify a newly infected person until the symptoms appear weeks later. The flu is highly contagious, but only by direct contact. The virus transfers readily from flesh to flesh, or from flesh to any object which in turn can contaminate any flesh it touches. Residents are going to extreme lengths to avoid touching one another, or touching objects that may be contaminated.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Ms. Hooker, director of the colony, has been seriously injured in a rocket accident.&lt;br /&gt;&#xA;Three immediate operations are required. The first will be performed by Dr. Xenophon, the second by Dr. Ypsilanti, the third by Dr. Zeno. Any of the surgeons may be infected with Barsoomian flu. Ms. Hooker, too, may have caught the disease.&lt;/p&gt;&#xA;&lt;p&gt;Just before the first operation it is discovered that the colony's hospital has only two pairs of sterile surgeon's gloves. No others are obtainable and there is no time for resterilizing. Each surgeon must operate with both hands.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;I don't see how we can avoid the risk of one of us becoming infected,&amp;quot; says Dr. Xenophon to Dr. Zeno. &amp;quot;When I operate, my hands may contaminate the insides of my gloves. Ms. Hooker's&#xA;body may contaminate the outsides. The same thing will happen to the gloves worn by Dr. Ypsilanti. When it's your turn, you'll have to wear gloves that could be contaminated on both sides.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;Au contraire,&amp;quot; says Dr. Zeno, who had taken a course in topology when he was a young medical student in Paris. &amp;quot;There's a simple procedure that will eliminate all risk of any of us catching the flu from one another or from Ms. Hooker.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;What does Dr. Zeno have in mind?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditDate="2021-06-04T07:49:12.547" LastActivityDate="2021-06-04T07:49:12.547" Title="Doctor's Dilemma" Tags="|lateral-thinking|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="1762" PostTypeId="2" ParentId="1761" CreationDate="2014-06-23T14:29:30.687" Score="32" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Let 1A stand for the insides of the first pair of gloves, 1B for the outsides. Let 2A stand for the insides of the second pair, 2B for the outsides.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dr. Xenophon wears both pairs, the second on top of the first. Sides 1A and 2B may become contaminated. Sides 1B and 2A remain sterile. Dr. Ypsilanti wears the second pair, with sterile sides 2A touching his hands. Dr. Zeno turns the first pair inside out before putting them on. Sterile sides 1B will then be touching his hands.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;After Dr. Zeno finished operating, his nurse, Ms. Frisbie, was furious. &quot;You boneheads ought to be ashamed! You protected yourselves, but forgot about poor Ms. Hooker. If Dr. Xenophon has the flu, Mrs. Hooker could catch it from the gloves you and Dr. Ypsilanti wore.&quot;  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;Are you suggesting, Ms. Frisbie,&quot; asked Dr. Zeno, &quot;that we could have prevented that?&quot;   &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;That's exactly what I'm suggesting.&quot;  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Then, to Dr. Zeno's amazement, Ms. Frisbie explained how they could have followed another procedure that would have eliminated not only the possibility of the surgeons catching the Barsoomian flu from one another or from Ms. Hooker, but also the possibility of Ms. Hooker catching it from the surgeons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dr. Xenophon wears both pairs of gloves. Sides 1A and 2B may become contaminated, while 1B and 2A remain sterile.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Dr. Ypsilanti wears the second pair, with sterile sides 2A against his hands.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Dr. Zeno turns the first pair inside out and then puts them on with sterile sides 1B against his hands. Then he puts on the second pair, with sides 2A over sides 1A and sides 2B outermost.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Because only sides 2B touch Ms. Hooker in all three operations, she runs no risk of catching the Barsoomian flu from any of the surgeons. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="68" LastActivityDate="2014-06-23T14:29:30.687" CommentCount="5" CommunityOwnedDate="2016-08-07T09:27:08.500" ContentLicense="CC BY-SA 3.0" />
  <row Id="1763" PostTypeId="2" ParentId="1758" CreationDate="2014-06-23T15:44:22.207" Score="3" Body="&lt;p&gt;For short hand: the horizontal line is $H$, the center vertical is $C$, the one to the left of $C$ is $L$, and the one to the right of $C$ is $R$. The cells are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{|c|c|}&#xA;\hline&#xA;   1 &amp;amp; 2 &amp;amp; 3 &amp;amp; 4 \\&#xA;\hline&#xA;   8 &amp;amp; 7 &amp;amp; 6 &amp;amp; 5 \\&#xA;\hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Fold $H$ last: 4 ways&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;$H$ and $L$ have 1 way to fold, so here can be ignored. As $C$ and $R$ have 2 ways to fold and their order doesn't matter (make dulplicate orientations) there are only $2\times2=4$ different ways to fold the sheet ignoring $H$ for the end.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Fold $H$ first: 7 ways&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If one were to fold $H$ first, there are 6 different ways to fold the sheet. 4 are simply folding $R$ and $C$ in their 2 possible ways.  The other 2 involve tucking cells $3-6$ between $1$ and $8$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The seventh was well decribed by klm123 and aschepler: Fold $Ho$ . Slide $4-5$  between $1,8$. Making it more like a cylinder than a triangle, continue sliding in until $4-5$  are between $2,7$  and $3,6$  are between $1,8$. Then you can fully flatten the LoCoRo  creases, making it a 1x1 square.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Fold $R$ then $H$ first: 6 ways&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If one folds $R$ then $H$ one gets 6 orientations in exactly the same way as the first 6 option from folding $H$ first.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Fold $R$ and $C$ then $H$ first: + 4 ways&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Neither $H$ last nor $L$ last scenereos can cover square $1$ as long as $H$ and $L$ are both folded out and no tucking possible as the first two folds quickly shorten the paper.  This means that there are 8 possibilitys: $RoCoHoLo,RiCoHoLo,RoCiHoLo,RiCiHoLo$ and the same ones with $LoHo$ (but these 4 are duplicates of $H$ last).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Fold $R$ and $L$ then $H$ first: see below (Fold $L$ and $R$ then $H$ then $C$)&lt;/h2&gt;&#xA;&#xA;&lt;h2&gt;Fold $C$ then $H$ first: 3 ways&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;$C$ can now only be folded &quot;out&quot; in the same direction as $H$ or cell $1$ is covered.&lt;br&gt;&#xA;Once this is done, however, $R$ can be folded into three different orientations.  The options are to place $4$ and $5$ between $1$ and $8$,$2$ and $3$ or $7$ and $6$.  Folding $L$ makes it a 1x1 but doesn't change the number of patterns.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Fold $C$ and $L$ then $H$ first: 3 ways&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;$C$ can now only be folded &quot;in&quot; and $L$ must be folded &quot;out&quot;.&lt;br&gt;&#xA;Similar to the previous case, there are 3 different options based on folding $R$ after $H$.  They are to place to place $4$ and $5$ at the bottom, between $6$ and $7$ or between $2$ and $3$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Fold $C$ and $R$ then $H$ first: see above (Fold $R$ and $C$ then $H$ first)&lt;/h2&gt;&#xA;&#xA;&lt;h2&gt;Fold $L$ then $H$ first: 7 ways&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;$L$ can now only be folded &quot;out&quot; and $H$ must be folded &quot;out&quot;.&lt;br&gt;&#xA;This yields 7 different possibilities based on where $3-6$ are placed.  There is: one with $4-5$ between $2$ and $7$; 2 ($Ri$ and $Ro$) with $3-6$ between $1$ and $2$; 2 with $3-6$ between $7$ and $8$; and 2 with $3-6$ on the bottom.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Fold $L$ and $R$ then $H$ then $C$: 6 ways&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;$L$ can now only be folded &quot;out&quot;.&lt;br&gt;&#xA;This yields 6 different possibilities based on where $3-6$ are placed.  There are: 2 ($Ri$ and $Ro$) with $3-6$ between $1$ and $2$; 2 with $3-6$ between $7$ and $8$; and 2 with $3-6$ on the bottom.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditDate="2014-06-30T13:04:17.853" LastActivityDate="2014-06-30T13:04:17.853" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="1764" PostTypeId="1" AcceptedAnswerId="1766" CreationDate="2014-06-24T02:44:05.267" Score="1" ViewCount="937" Body="&lt;p&gt;If four 2s make eight, and three 2s also make eight, how many 2s do you need to make three eights, when you only have less than eight 2s?&lt;/p&gt;&#xA;" OwnerUserId="1621" LastEditorUserId="2484" LastEditDate="2014-12-11T13:00:16.120" LastActivityDate="2014-12-11T18:51:00.263" Title="I Eight Two Much" Tags="|mathematics|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1766" PostTypeId="2" ParentId="1764" CreationDate="2014-06-24T03:00:31.013" Score="4" Body="&lt;p&gt;I take three eights as twenty-four, and can do $(2+2)(2+2+2)$ to get there with five twos.  One or two of the plus signs can become multiplies.  There are other solutions, depending on what operations are allowed.  If we add division, we can multiply the above by $\frac 22$ and have solutions with seven twos.  If concatenation and factorials are allowed, there are $22+2$ with only three, and $(2+2)!$ with just two.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorDisplayName="user88" LastEditDate="2014-12-11T18:51:00.263" LastActivityDate="2014-12-11T18:51:00.263" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="1769" PostTypeId="1" AcceptedAnswerId="1771" CreationDate="2014-06-24T17:21:22.560" Score="1" ViewCount="1973" Body="&lt;p&gt;Following this question &lt;a href=&quot;https://puzzling.stackexchange.com/questions/186/whats-the-fewest-weights-you-need-to-balance-any-weight-from-1-to-40-pounds&quot;&gt;What&amp;#39;s the fewest weights you need to balance any weight from 1 to 40 pounds?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested what is the minimum number M of weights you need to &lt;strong&gt;define&lt;/strong&gt; any integer weight from a) 1 to 40? b) 1 to N?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if N = 2, M = 1. Indeed, with one weight equal 2 you can define any weight X from 1 to 2 ( $1 \le X &amp;lt; 2 \implies X = 1$; $X = 2 \implies X = 2$ ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With two weights I can go up to N = 8. Take weights 2 and 6, then:&lt;br&gt;&#xA;$1 \le X &amp;lt; 2 \implies X = 1$;&lt;br&gt;&#xA; $X = 2 \implies X = 2$;&lt;br&gt;&#xA;$2 &amp;lt; X &amp;lt; 6-2 \implies X = 3$;&lt;br&gt;&#xA; $X = 6-2 \implies X = 4$;&lt;br&gt;&#xA; $6-2 &amp;lt; X &amp;lt; 6 \implies X = 5$;&lt;br&gt;&#xA; $X = 6 \implies X = 6$;&lt;br&gt;&#xA; $6 &amp;lt; X &amp;lt; 6+2 \implies X = 7$;&lt;br&gt;&#xA; $ X = 6+2 \implies X = 8$.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-01-31T12:26:13.263" Title="Minimum number of weights you need to define any integer weight from 1 to N" Tags="|mathematics|combinatorics|weighing|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1770" PostTypeId="2" ParentId="1761" CreationDate="2014-06-24T17:39:45.727" Score="8" Body="&lt;p&gt;Additional interesting question to consider is who and how should prepare the gloves and move them between doctors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basing on answer of SQB I give the answer that takes into account this problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let 1A stand for the insides of the first pair of gloves, 1B for the outsides. Let 2A stand for the insides of the second pair, 2B for the outsides.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0. There are 4 persons and 4 sides of gloves, therefore each person must contaminate exactly one side. Unfortunately, the operation must be split into 3 separate operations, because Ms. Hooker is the only one who can help to move 2-nd pair of gloves between doctors.&lt;br&gt;&#xA;&lt;br&gt;&#xA; 1. Dr. Xenophon wears 1-st pair of gloves side A in. He asks Ms. Hooker to take 2-nd pair and touching side 2B only put them on 1-st gloves.&lt;br&gt;&#xA; Side 1A may become contaminated by Dr. Xenophon.&lt;br&gt;&#xA; Side 2B may become contaminated by Ms. Hooker.&lt;br&gt;&#xA;&lt;br&gt;&#xA; 2. Ones Dr. Xenophon is finished he wakes the patient. Ms. Hooker takes 2-nd pair out from Dr. Xenophon hands and puts them on Dr. Ypsilanti.&lt;br&gt;&#xA; Side 2A may become contaminated by Dr. Ypsilanti.&lt;br&gt;&#xA;&lt;br&gt;&#xA; 3. Ones Dr. Ypsilanti is finished he wakes the patient. Ms. Hooker takes 2-nd pair out from Dr. Ypsilanti hands. Dr. Zeno takes 1-st pair out from Dr. Xenophon hands touching side 1B only, and (with help of Dr. Xenophon) turns them inside out and puts them on. Ms. Hooker takes 2-nd pair and puts them on Dr. Zeno.&lt;br&gt;&#xA; Side 1B may become contaminated by Dr. Zeno. But the operations are finished successfully! No one must touch more than one side and the Barsoomian flu will not be spread.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;P.S. I believe with the mentioned gloves movement problem, this algorithm becomes the only possible answer to the problem. One can switch doctors and gloves since they are equal, but the idea (1 side per person + first doctor put two pairs on) must stay the same.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="8822" LastEditDate="2017-02-26T11:22:10.417" LastActivityDate="2017-02-26T11:22:10.417" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1771" PostTypeId="2" ParentId="1769" CreationDate="2014-06-24T19:23:05.567" Score="6" Body="&lt;p&gt;If you can weigh all the even numbers, you can define all the odd numbers.  The question you linked to has the weights as powers of $3:  3^0=1, 3^1=3, 3^2=9, 3^3=27$ and can weigh all numbers up to $40$.  So doubling the weights (as you started) $2,6,18,54$ allows you to define all weights up to $81$.  More generally, the series of weights $2\cdot 3^i$, for $i=0,1,2,\dots n$ lets you define all weights up to $3^{n+1}$&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2014-06-24T19:23:05.567" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1772" PostTypeId="2" ParentId="1764" CreationDate="2014-06-24T22:29:59.457" Score="0" Body="&lt;p&gt;Using the two initial examples given:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(2*2*2*2) + (2*2*2)$$&lt;/p&gt;&#xA;" OwnerUserId="185" LastEditorUserId="39" LastEditDate="2014-06-25T14:53:48.607" LastActivityDate="2014-06-25T14:53:48.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1773" PostTypeId="2" ParentId="1764" CreationDate="2014-06-25T09:40:51.877" Score="12" Body="&lt;p&gt;Here's another way of producing three eights from five twos:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2 \times 2 \times {222} = 888$$&lt;/p&gt;&#xA;" OwnerUserId="40" LastActivityDate="2014-06-25T09:40:51.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1774" PostTypeId="2" ParentId="1764" CreationDate="2014-06-25T10:43:47.883" Score="0" Body="&lt;p&gt;Here is another way also assuming $3*8=24$ using $5*2$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(2+2+2)*2*2$$&lt;/p&gt;&#xA;" OwnerUserId="1570" LastEditorUserId="39" LastEditDate="2014-06-25T14:54:19.990" LastActivityDate="2014-06-25T14:54:19.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1775" PostTypeId="2" ParentId="548" CreationDate="2014-06-25T15:26:06.713" Score="1" Body="&lt;p&gt;&lt;em&gt;(Note: This answer is a work in progress. See &lt;a href=&quot;https://puzzling.meta.stackexchange.com/q/1283/1622&quot;&gt;relevant meta discussion&lt;/a&gt; if you would rather answers such as this did not exist.)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't like complicated things so I immediately chose to solve some simple variants of this problem before scaling up to the full deal. The simplest variant adds these assumptions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;No more than one house can occupy the same place&lt;/li&gt;&#xA;&lt;li&gt;All houses are located on a straight line&lt;/li&gt;&#xA;&lt;li&gt;The castle is located on the &quot;end&quot; of the line&lt;/li&gt;&#xA;&lt;li&gt;There is no size restriction for the kingdom&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This is a trivial variant of the problem and the solution is to send the original solider along the line until it reaches the last house on the line. Every other solider immediately returns to the castle. The time taken will be $t = 2x$ where $x$ is the distance between the castle and the house farthest away.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The next variant follows these assumptions instead:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;No more than one house can occupy the same place&lt;/li&gt;&#xA;&lt;li&gt;All houses are located on a straight line&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;The castle can be located anywhere on the line&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;There is no size restriction for the kingdom&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;To start digging for patterns, I used the following example (&lt;code&gt;X&lt;/code&gt; is the castle; other letters are houses):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A---B-X--C&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The optimal strategy will always be:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Send Solider 1 ($SX$) to $B$, generating Solider 2 ($SB$)&lt;/li&gt;&#xA;&lt;li&gt;Send $SX$ to $A$, send $SB$ to $C$&lt;/li&gt;&#xA;&lt;li&gt;Send all soldiers home&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The cost for this will be:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\overline{BX} + \max(\overline{AB}+\overline{AB}+\overline{BX}, \overline{BX}+\overline{CX}+\overline{CX})$&lt;/li&gt;&#xA;&lt;li&gt;$= \overline{BX} + \max(2*\overline{AB}+\overline{BX}, \overline{BX}+2*\overline{CX})$&lt;/li&gt;&#xA;&lt;li&gt;$= 2*\overline{BX} + \max(2*\overline{AB}+\overline{BX}, \overline{BX}+2*\overline{CX})$&lt;/li&gt;&#xA;&lt;li&gt;$= 2*(\overline{BX} + \max(\overline{AB}, \overline{CX}))$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The second example for this variant is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A---B-X--C--D&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The optimal strategy is no longer static and requires knowledge about the distance between the various houses. Specifically, there are two potential options.&lt;/p&gt;&#xA;&#xA;OPTION 1&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$S_X$ to $B$&lt;/li&gt;&#xA;&lt;li&gt;$S_X$ to $A$, $S_B$ to $C$&lt;/li&gt;&#xA;&lt;li&gt;$S_X$, $S_A$, $S_C$ to X, $S_B$ to $D$&lt;/li&gt;&#xA;&lt;li&gt;Send all soldiers home&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;OPTION 2&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$S_X$ to $C$&lt;/li&gt;&#xA;&lt;li&gt;$S_X$ to $D$, $S_C$ to $B$&lt;/li&gt;&#xA;&lt;li&gt;$S_X$, $S_D$, $S_B$ to X, $S_C$ to $A$&lt;/li&gt;&#xA;&lt;li&gt;Send all soldiers home&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Therefore, the optimal time will be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\min(\overline{XB}+\max(2*\overline{AB}+\overline{XB},\overline{XB}+2*\overline{XC}+2*\overline{CD}), \overline{XC}+\max(2*\overline{CD}+\overline{XC},\overline{XC}+2*\overline{XB}+2*\overline{AB}))$&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;$= \min(\max(2*\overline{AB}+2*\overline{XB},2*\overline{XB}+2*\overline{XC}+2*\overline{CD}), \max(2*\overline{CD}+2*\overline{XC},2*\overline{XC}+2*\overline{XB}+2*\overline{AB}))$&#xA;$= 2*\min(\max(\overline{AB}+\overline{XB},\overline{XB}+\overline{XC}+\overline{CD}), \max(\overline{CD}+\overline{XC},\overline{XC}+\overline{XB}+\overline{AB}))$&#xA;$= 2*\min(-\min(-\overline{AB}-\overline{XB},-\overline{XB}-\overline{XC}-\overline{CD}), -\min(-\overline{CD}-\overline{XC},-\overline{XC}-\overline{XB}-\overline{AB}))$&#xA;$= 2*(\min(\min(-\overline{CD}-\overline{XC},-\overline{XC}-\overline{XB}-\overline{AB}), \min(-\overline{AB}-\overline{XB},-\overline{XB}-\overline{XC}-\overline{CD}))-\min(-\overline{AB}-\overline{XB},-\overline{XB}-\overline{XC}-\overline{CD})-\min(-\overline{CD}-\overline{XC},-\overline{XC}-\overline{XB}-\overline{AB}))$&#xA;$= 2*(\min(-\overline{CD}-\overline{XC},-\overline{XC}-\overline{XB}-\overline{AB}, -\overline{AB}-\overline{XB},-\overline{XB}-\overline{XC}-\overline{CD})-\min(-\overline{AB}-\overline{XB},-\overline{XB}-\overline{XC}-\overline{CD})-\min(-\overline{CD}-\overline{XC},-\overline{XC}-\overline{XB}-\overline{AB}))$&#xA;$= 2*(\min(-\overline{XC}-\overline{XB}-\overline{AB},-\overline{XB}-\overline{XC}-\overline{CD})-\min(-\overline{AB}-\overline{XB},-\overline{XB}-\overline{XC}-\overline{CD})-\min(-\overline{CD}-\overline{XC},-\overline{XC}-\overline{XB}-\overline{AB}))$&#xA;$= 2*(-\max(\overline{XC}+\overline{XB}+\overline{AB},\overline{XB}+\overline{XC}+\overline{CD})+\max(\overline{AB}+\overline{XB},\overline{XB}+\overline{XC}+\overline{CD})+\max(\overline{CD}+\overline{XC},\overline{XC}+\overline{XB}+\overline{AB}))$&#xA;$= 2*(\overline{XB}+\overline{XC}+\max(\overline{AB},\overline{XC}+\overline{CD})+\max(\overline{CD},\overline{XB}+\overline{AB})-\overline{XC}-\overline{XB}-\max(\overline{AB},\overline{CD}))$&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;$= 2*(\max(\overline{AB},\overline{XC}+\overline{CD})+\max(\overline{CD},\overline{XB}+\overline{AB})-\max(\overline{AB},\overline{CD}))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using some quick numbers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\overline{AB}=3, \overline{BX}=1, \overline{XC}=2, \overline{CD}=2$ then:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$t=2*(\max(3,2+2)+\max(2,1+3)-\max(3,2))$&lt;/li&gt;&#xA;&lt;li&gt;$t=2*(4+4-3)=2*5=10$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Presumably this pattern can be extended to apply to &lt;em&gt;all&lt;/em&gt; scenarios where the houses fit on a line. The second example for this variant is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A---B-X--C--D---E&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Similarly to the second example, we really only have two potential options:&lt;/p&gt;&#xA;&#xA;OPTION 1&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$S_X$ to $B$&lt;/li&gt;&#xA;&lt;li&gt;$S_X$ to $A$, $S_B$ to $C$&lt;/li&gt;&#xA;&lt;li&gt;$S_X$, $S_A$, $S_C$ to X, $S_B$ to $D$&lt;/li&gt;&#xA;&lt;li&gt;$S_B$ to $E$&lt;/li&gt;&#xA;&lt;li&gt;Send all soldiers home&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;OPTION 2&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$S_X$ to $C$&lt;/li&gt;&#xA;&lt;li&gt;$S_X$ to $D$, $S_C$ to $B$&lt;/li&gt;&#xA;&lt;li&gt;$S_D$, $S_B$ to X, $S_X$ to $E$, $S_C$ to $A$&lt;/li&gt;&#xA;&lt;li&gt;Send all soldiers home&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$t = \min(\overline{XB}+\max(\overline{XB}+2*\overline{AB},\overline{XB}+2*\overline{XC}+2*\overline{CD}+2*DE), \overline{XC}+\max(\overline{XC}+2*\overline{CD}+2*DE,\overline{XC}+2*\overline{XB}+2*\overline{AB}))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we treat $2*\overline{CD}+2*DE$ as $2*CE$ we get this form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$t = \min(\overline{XB}+\max(\overline{XB}+2*\overline{AB},\overline{XB}+2*\overline{XC}+2*CE), \overline{XC}+\max(\overline{XC}+2*CE,\overline{XC}+2*\overline{XB}+2*\overline{AB}))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which matches our previous formula letting us jump to the reduced form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= 2*(\max(\overline{AB},\overline{XC}+CE)+\max(CE,\overline{XB}+\overline{AB})-\max(\overline{AB},CE))$&#xA;$= 2*(\max(\overline{AB},\overline{XC}+\overline{CD}+CE)+\max(\overline{CD}+CE,\overline{XB}+\overline{AB})-\max(\overline{AB},\overline{CD}+CE))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, one pattern does reveal itself. The only lines that &lt;em&gt;really&lt;/em&gt; matter are $XA = \overline{XB}+\overline{AB}$, $XE = \overline{XC}+\overline{CD}+DE$, $\overline{XB}$ and $\overline{XC}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= 2*(\max(\overline{AB},XE)+\max(\overline{CD}+CE,XA)-\max(\overline{AB},\overline{CD}+CE))$&#xA;$= 2*(\max(XA-\overline{XB},XE)+\max(XE-\overline{XC},XA)-\max(XA-\overline{XB},XE-\overline{XC}))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This form seems likely to apply to all scenarios where the houses fit on a line. If we redefine the points in our example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A1---A0-X--B0--B1---B2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We can write the formula as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= 2*(\max(XA-XA_0,\overline{XB})+\max(\overline{XB}-\overline{XB}_0,XA)-\max(XA-XA_0,\overline{XB}-\overline{XB}_0))$ where $XA=XA_0+A_0A_1+A_1A_2...$ and $\overline{XB}=\overline{XB}_0+B_0B_1+B_1B_2...$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This solution makes intuitive sense. The only real choice in a lined scenario is which direction to go first. Everything after that presents us with the trivial problem noted at the beginning of this post. The primary reason this formula works, however, is because we know that each $XA_0 ... XA_N$ is less than $B_0A_0 ... B_0A_N$ since any solider travelling from $B_0$ to $A_0$ has to pass through $X$. Therefore, this formula is only true if and only if $X$ perfectly splits two sections of the kingdom such that $X$ is always closer to each house in section $A$ compared to each house in section $B$ &lt;strong&gt;and&lt;/strong&gt; the houses in each section form a line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This last clause seems like a good place to extend our approach.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The next variant follows these assumptions instead:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;No more than one house can occupy the same place&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;All houses exist in either section A or section B such that there is no line between a house in section A and a house in section B that is shorter than travelling from that house to the castle&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;All house in a given section $S$ follow the rule that $S_N$ is closer to $S_{N-1}$ and $S_{N+1}$ than any other houses or the castle &lt;em&gt;unless&lt;/em&gt; the house is $S_0$&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;The castle can be located anywhere on the line&lt;/li&gt;&#xA;&lt;li&gt;There is no size restriction for the kingdom&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This set of assumptions removes the restriction that all houses exist on a line. It should make the math significantly more complicated but let's work it through.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A1   A0&#xA;           X   B0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This example moves $A_0$ and $A_1$ off of a line but preserves Assumption 3 which states that $A_1$ cannot be closer to $X$ than $A_0$. Let's see how this affects the math. The strategy is still the same:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Send $S_X$ to $A_0$&lt;/li&gt;&#xA;&lt;li&gt;Send $S_X$ to $A_1$, send $SA_0$ to $S_{B_0}$&lt;/li&gt;&#xA;&lt;li&gt;Send all soldiers home&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The two major differences are that $\overline{A_1X}$ no longer passes through $A_0$ and $\overline{A_0B_0}$ no longer passes through $X$. Here is our new formula:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\overline{XA_0} + \max(\overline{A_0A_1}+\overline{XA_1}, \overline{A_0B_0}+\overline{XB_0})$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Sadly, we don't get to reduce this at all since we don't get to replace $\overline{A_0B_0}$ with $\overline{XA_0} + \overline{XB_0}$. Let's extend section $A$ again and see what happens:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   A1   A0&#xA;              X   B0&#xA;A2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Is the optimal strategy still the same?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Send $S_X$ to $A_0$&lt;/li&gt;&#xA;&lt;li&gt;Send $S_X$ to $A_1$, send $S_{A_0}$ to $B_0$&lt;/li&gt;&#xA;&lt;li&gt;Send $S_X$ to $A_2$, send $S_{A_0}$, $S_{A_1}$, $S_{B_0}$ home&lt;/li&gt;&#xA;&lt;li&gt;Send all soldiers home&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I can see two alternative strategies and it may be tricky to determine if there is a case where one of those strategies outperforms this one. To make this easier to understand, I'll start using the notation $S_X \to A_0 \to A_1 \to A_2 \to X$ to describe one soldier's entire cycle. The above strategy expressed this way:&lt;/p&gt;&#xA;&#xA;OPTION 1&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$S_X \to A_0 \to A_1 \to A_2 \to X$&lt;/li&gt;&#xA;&lt;li&gt;$S_{A_0} \to {B_0} \to X$&lt;/li&gt;&#xA;&lt;li&gt;$S_{A_1} \to X$&lt;/li&gt;&#xA;&lt;li&gt;$S_{A_2} \to X$&lt;/li&gt;&#xA;&lt;li&gt;$S_{B_0} \to X$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;OPTION 2&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$S_X \to A_0 \to A_2 \to A_1 \to X$&lt;/li&gt;&#xA;&lt;li&gt;$S_{A_0} \to {B_0} \to X$&lt;/li&gt;&#xA;&lt;li&gt;$S_{A_1} \to X$&lt;/li&gt;&#xA;&lt;li&gt;$S_{A_2} \to X$&lt;/li&gt;&#xA;&lt;li&gt;$S_{B_0} \to X$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;OPTION 3&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$S_X \to A_1 \to A_2 \to X$&lt;/li&gt;&#xA;&lt;li&gt;$S_{A_1} \to A_0 \to B_0 \to X$&lt;/li&gt;&#xA;&lt;li&gt;$S_{A_0} \to X$&lt;/li&gt;&#xA;&lt;li&gt;$S_{A_2} \to X$&lt;/li&gt;&#xA;&lt;li&gt;$S_{B_0} \to X$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How do we confirm that OPTIONS 2 &amp;amp; 3 are suboptimal for all positions of $A_0...A_N$?&lt;/p&gt;&#xA;" OwnerUserId="1622" LastEditorUserId="-1" LastEditDate="2017-03-16T16:42:07.097" LastActivityDate="2014-06-25T22:35:36.263" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1776" PostTypeId="1" AcceptedAnswerId="1786" CreationDate="2014-06-25T17:39:47.633" Score="9" ViewCount="12297" Body="&lt;p&gt;This question is from a free aptitude test on naukri.com (&lt;a href=&quot;https://i.stack.imgur.com/VcgXc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;image&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;This question is based upon the figure shown below.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{array}{|ccc|c|ccc|c|ccc|}&#xA;\; &amp;amp; 41 &amp;amp; \; &amp;amp; \; &amp;amp; \; &amp;amp; 31 &amp;amp; \; &amp;amp; \; &amp;amp; \; &amp;amp; ?  &amp;amp; \; \\&#xA;5  &amp;amp; \; &amp;amp; 6  &amp;amp; \; &amp;amp; 9  &amp;amp; \; &amp;amp; 4  &amp;amp; \; &amp;amp; 8  &amp;amp; \; &amp;amp; 7  \\&#xA;7  &amp;amp; \; &amp;amp; 8  &amp;amp; \; &amp;amp; 2  &amp;amp; \; &amp;amp; 6  &amp;amp; \; &amp;amp; 2  &amp;amp; \; &amp;amp; 6  \\&#xA;\end{array}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a. 51&lt;/li&gt;&#xA;&lt;li&gt;b. 62&lt;/li&gt;&#xA;&lt;li&gt;c. 48&lt;/li&gt;&#xA;&lt;li&gt;d. 53&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;br /&gt;&#xA;Find the missing number on the basis of the pattern of figures shown above.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1637" LastEditorUserId="69582" LastEditDate="2021-07-19T18:17:37.657" LastActivityDate="2021-07-19T18:17:37.657" Title="Find the missing number: A Pentagonal pattern puzzle" Tags="|pattern|calculation-puzzle|" AnswerCount="10" CommentCount="13" ContentLicense="CC BY-SA 4.0" />
  <row Id="1777" PostTypeId="2" ParentId="1776" CreationDate="2014-06-25T18:13:33.937" Score="9" Body="&lt;p&gt;This answer somewhat works, but part of the logic seems a bit arbitrary, so maybe I am missing something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5/7 + 6/8 = 82/56 = 41/28.  The numerator 41 matches the top number in the first figure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;9/2 + 4/6 = 62/12 = 31/6.  The numerator 31 matches the top number in the second figure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8/2 + 7/6 = 62/12 = 31/6.  31 isn't a choice for the answer, but 62 is.  Why we don't reduce the fraction in this case I do not know.&lt;/p&gt;&#xA;" OwnerUserId="239" LastActivityDate="2014-06-25T18:13:33.937" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1778" PostTypeId="2" ParentId="548" CreationDate="2014-06-25T19:16:27.000" Score="0" Body="&lt;p&gt;I have written an algorithm to find minimal time for a specific kingdom.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;We need to start by creating a table/map/lookup etc. of distances between every single pair of knights' houses.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We then need to generate all the permutations of the ways that the &quot;news&quot; can spread. i.e. Knight0 can go first to Knight1, but could also start by going to Knight2 etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the Knight at each extreme point example mentioned in the question, I used the notation 0(1(4),2(),3()) to mean:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Knight0 travelled to Knight1's house ($\frac{\sqrt{2}}{2}$)&lt;/li&gt;&#xA;&lt;li&gt;Knight0 then travelled to Knight2's house ($1$)&lt;/li&gt;&#xA;&lt;li&gt;Knight0 then travelled to Knight3's house ($1$)&lt;/li&gt;&#xA;&lt;li&gt;Knight0 then travelled back to the castle ($\frac{\sqrt{2}}{2}$)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, after Knight0 had travelled to Knight1's house,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Knight1 then travelled to Knight4's house ($1$)&lt;/li&gt;&#xA;&lt;li&gt;Knight1 then travelled back to the castle ($\frac{\sqrt{2}}{2}$)&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Once we have every possibility (336 for the above example), we need to calculate the time taken for each of those possibilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the above example, this would be the distance travelled by Knight0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\frac{\sqrt{2}}{2}+1+1+\frac{\sqrt{2}}{2} = 2 + \sqrt{2}$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Once we have all the times of all the possibilities, we can then pick out the smallest time. (There is always more than one answer, as there is always a choice between two knights as to who visits the final knight before returning to the castle.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I have written this algorithm in C#. It doesn't take a long time to run, but it cannot handle more than 6 Knights in a kingdom at the moment due to memory problems, but am working on that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the algorithm, $2 + \sqrt{2}$ (or 3.414...) was output as the shortest distance in the above example. So far, randomly generating house positions for 6 Knights has produced no higher values.&lt;/p&gt;&#xA;" OwnerUserId="1616" LastEditorUserId="39" LastEditDate="2014-07-27T23:34:28.393" LastActivityDate="2014-07-27T23:34:28.393" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="1781" PostTypeId="2" ParentId="1776" CreationDate="2014-06-25T21:04:00.213" Score="6" Body="&lt;p&gt;a) The answer is obviously 51. The topmost number is decremented by 10 every step with an odd number, and incremented by 20 on every step with an even number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) The answer is obviously 62. The topmost number is decremented by 10 every step with an odd number, and doubled on every step with an even number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) The answer is obviously 48. The topmost number is decremented by 10 every step with an odd number, and incremented by 17 on every step with an even number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d) The answer is obviously 53. In every third image it is set to the constant 53.&lt;/p&gt;&#xA;" OwnerUserId="1643" LastEditorUserId="68" LastEditDate="2014-07-02T13:14:05.617" LastActivityDate="2014-07-02T13:14:05.617" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1782" PostTypeId="2" ParentId="1776" CreationDate="2014-06-25T21:16:53.273" Score="2" Body="&lt;p&gt;Wild guess here - but it still could be a pattern&lt;/p&gt;&#xA;&#xA;&lt;p&gt;41 // 4 + 1 = 5 (answer is on top-left)&#xA;31 // 3 + 1 = 4 (answer is on top-right)&#xA;51 // 5 + 1 = 6 (answer is on bottom-right)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe the pattern here is that the answer to adding each digit of the topmost number from each image is moving clockwise for each image. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;^Not sure if that makes sense... But I tried. Would be nice if you tell us the answer if you found it :D&lt;/p&gt;&#xA;" OwnerUserId="1645" LastActivityDate="2014-06-25T21:16:53.273" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1783" PostTypeId="2" ParentId="1776" CreationDate="2014-06-26T05:24:32.193" Score="3" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;(5+7+6+8)*2-11=41&lt;/p&gt;&#xA;&lt;p&gt;(9+2+4+6)*2-11=31&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;but&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;(8+2+7+6)*2-11=35&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Not the answer looking for but just can be another soln&lt;/p&gt;&#xA;&lt;p&gt;First I got the above two then I found this.&lt;/p&gt;&#xA;" OwnerUserId="444" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-06-26T05:29:51.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1784" PostTypeId="1" AcceptedAnswerId="1785" CreationDate="2014-06-26T09:14:42.657" Score="33" ViewCount="6522" Body="&lt;p&gt;Suppose there are $n$ ants on a stick which has length 1. At any time, the ants may be facing &lt;em&gt;left&lt;/em&gt; or &lt;em&gt;right&lt;/em&gt;, the initial directions of the ants are arbitrary. Each ant can be modeled as a point in the stick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At time 0, all ants start moving in the direction they're facing at speed 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each time an ant collides with another ant, both ants reverse direction instantly and continue moving in the new direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If an ant reaches the ends of the stick, it falls down (quite a strange species of ant).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you are observing this stick full of ants, starting at time 0. You are very interested of one ant, which appears in a distinct color from the other ants, so you focus your eyes just on that particular ant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the maximum time you'll be watching &lt;strong&gt;that particular ant&lt;/strong&gt; until it falls off the stick?&lt;/p&gt;&#xA;" OwnerUserId="1649" LastEditorUserId="4551" LastEditDate="2015-04-20T01:49:11.817" LastActivityDate="2020-06-23T10:49:48.380" Title="Maximum time for ants to fall off stick" Tags="|mathematics|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1785" PostTypeId="2" ParentId="1784" CreationDate="2014-06-26T09:14:42.657" Score="38" Body="&lt;p&gt;And the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Surprisingly, it's 1, that is, it's the same as if nothing happens when the ants collide, and this doesn't depend on $n$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The phrase that says you're watching a particular ant is just a distraction so that people will think that there is complexity in this riddle, while in fact it's very simple. When two ants collide and each reverses its direction, position-wise it's the same as if each ant simply continues moving in the original direction. So the maximum time is when there is an ant at on end of a stick facing the other end. Since the maximum time an ant can be in the stick is 1 (from one end to the other), therefore by time 1 all ants will already fall off the stick, including the particular ant that you're watching.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alternative explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Imagine that &lt;strong&gt;each ant holds a number&lt;/strong&gt;, and when two ants collide (and reverse their directions), they will &lt;strong&gt;exchange the number&lt;/strong&gt;. Now the crucial observation is: &lt;strong&gt;the numbers will never change its direction&lt;/strong&gt;. So the maximum time a number can be on the stick is 1 (the numbers are always moving on the same direction with the same speed). So by the time 1, all numbers will have fallen down. Since every number is held by an ant, when there is no number, there is no ant either. Therefore by the time 1, all ants will have already fallen from the stick, including the particular ant you are watching.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1649" LastEditorUserId="1649" LastEditDate="2015-04-22T04:22:11.920" LastActivityDate="2015-04-22T04:22:11.920" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="1786" PostTypeId="2" ParentId="1776" CreationDate="2014-06-26T10:06:32.373" Score="10" Body="&lt;p&gt;&lt;em&gt;I found a pattern, which I personally do not like because it uses two additional constants, but: 1. it is simple, 2. it takes into account all given numbers and 3. it provides expected result. I post this answer because there is not other working pattern was suggested up to now.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's name the numbers in pentagon like:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; E&#xA;a b  &#xA;c d&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then for each pentagon: $E = 10\times(a+b+c-d)-59$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case the answer is &lt;strong&gt;51&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2014-06-26T18:03:59.963" LastActivityDate="2014-06-26T18:03:59.963" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1789" PostTypeId="2" ParentId="1784" CreationDate="2014-06-26T14:26:57.293" Score="9" Body="&lt;p&gt;@justhalf is correct but here is a visual aid. The spots on the stick are labeled 0-9 and ants facing right are &lt;code&gt;&amp;gt;&lt;/code&gt; and ants facing left are &lt;code&gt;&amp;lt;&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0123456789&#xA;&amp;gt;        &amp;lt;&#xA; &amp;gt;      &amp;lt;&#xA;  &amp;gt;    &amp;lt;&#xA;   &amp;gt;  &amp;lt;&#xA;    &amp;gt;&amp;lt;&#xA;    &amp;lt;&amp;gt;&#xA;   &amp;lt;  &amp;gt;&#xA;  &amp;lt;    &amp;gt;&#xA; &amp;lt;      &amp;gt;&#xA;&amp;lt;        &amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note that the ants appear to &quot;pass through&quot; each other which was @justhalf's point. Two ants colliding mean they have each traveled half of the distance between them. They then turn around and walk back over that exact same distance. There will never be a case where an ant will have to walk past 5 to get to another ant and collide and, once it does, it will only have 5 more spaces to walk before it falls off of the stick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is another example with three ants:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0123456789&#xA;&amp;gt; &amp;gt;      &amp;lt;&#xA; &amp;gt; &amp;gt;    &amp;lt; &#xA;  &amp;gt; &amp;gt;  &amp;lt;   &#xA;   &amp;gt; &amp;gt;&amp;lt;    &#xA;    &amp;gt;&amp;lt;&amp;gt;    &#xA;    &amp;lt;&amp;gt; &amp;gt;  &#xA;   &amp;lt;  &amp;gt; &amp;gt; &#xA;  &amp;lt;    &amp;gt; &amp;gt;&#xA; &amp;lt;      &amp;gt; &#xA;&amp;lt;        &amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The ant that starts at 0 still turns around at the same spot and it still looks like the ants just walk through each other. It doesn't matter how many ants are added to the stick, the most one ant will ever have to travel is 10 spaces.&lt;/p&gt;&#xA;" OwnerUserId="1622" LastActivityDate="2014-06-26T14:26:57.293" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1790" PostTypeId="1" CreationDate="2014-06-26T14:32:00.703" Score="5" ViewCount="722" Body="&lt;p&gt;This question is purely a follow-up to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/1784/maximum-time-for-ants-to-fall-off-stick/&quot;&gt;Maximum time for ants to fall off stick&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose there are $n$ ants on a stick which has length 10. At any time, the ants may be facing left or right, the initial directions of the ants are arbitrary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At time 0, all ants start moving in the direction they're facing at speed 1. If an ant reaches the ends of the stick, it falls down (still quite a strange species of ant).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each time an ant collides with another ant, both ants reverse direction &lt;strong&gt;in one unit of time&lt;/strong&gt; and then continue moving in the new direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the maximum time an ant can stay on the stick before it falls off?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Clarification regarding ant collisions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $X$ encounters $Y$ and $Z$ in the middle of a colliding turn, then $X$ will turn around and leave in 1 time. $Y$ will finish its turn in 0.5 time and then turn around to face $Z$ again. This will take an additional 1 time. $Z$ will finish its turn in 0.5 time and continue on its way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes intuitive sense, by the way, when you remember that $X$ will still be turning when $Y$ finishes its first turn. It couldn't move forward because the path is blocked.&lt;/p&gt;&#xA;" OwnerUserId="1622" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2016-10-11T06:39:33.420" Title="Maximum time for ants to fall off stick (with a non-instant turning speed)" Tags="|mathematics|" AnswerCount="1" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="1791" PostTypeId="2" ParentId="1790" CreationDate="2014-06-26T14:50:18.747" Score="7" Body="&lt;p&gt;The ants-passing-each-other trick still works, with the additional condition that it takes a non-zero amount of time for two ants to pass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the passing delay is 1 time unit, the maximum time on the stick is equal to the time to traverse the stick plus the time to pass all the other ants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $t$ is time, $l$ is the length of the stick, $v$ is the speed of the ants, and $n$ is the number of ants:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$t_{max}=\frac{l}{v}+(n-1)t_{pass}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plug in some values (I'll take 10 for $n$ and assume 1 for $t_{pass}$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$19=\frac{10}{1}+(10-1)1$&lt;/p&gt;&#xA;" OwnerUserId="18" LastActivityDate="2014-06-26T14:50:18.747" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1792" PostTypeId="2" ParentId="1784" CreationDate="2014-06-26T15:20:15.817" Score="2" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;The answer is N, the length of the stick (which is 1 in this question).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The complexity is that you're being distracted by the fact that they're bumping into each other while in reality nothing happens.&lt;/p&gt;&#xA;&lt;p&gt;So you'd have to look for the maximum distance an ant has to travel for it to fall off, which would be on either sides of the stick.&lt;/p&gt;&#xA;" OwnerUserId="1659" LastEditorUserId="33177" LastEditDate="2020-06-23T10:49:48.380" LastActivityDate="2020-06-23T10:49:48.380" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="1795" PostTypeId="2" ParentId="548" CreationDate="2014-06-26T16:34:45.967" Score="-2" Body="&lt;p&gt;It could be &lt;code&gt;~1.707&lt;/code&gt; if the King helps.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;s4 ----- s1 --+&#xA; |       |    |&#xA; |       |    |&#xA; |   K   |    1&#xA; |       |    |&#xA; |       |    |&#xA;s3-------s2 --+&#xA;&#xA; |       |&#xA; +---1---+&#xA;&#xA; Route taken&#xA; K  -&amp;gt; S1 -&amp;gt; C  =&amp;gt; 2 * SQR(((1/2)^2)+((1/2)^2))&#xA; S0 -&amp;gt; S3 -&amp;gt; C  =&amp;gt; 2 * SQR(((1/2)^2)+((1/2)^2)) (Aka the Soldier at the castle.)&#xA; S1 -&amp;gt; S2 -&amp;gt; C  =&amp;gt; 1 + SQR(((1/2)^2)+((1/2)^2))&#xA; S2 -&amp;gt; C        =&amp;gt;     SQR(((1/2)^2)+((1/2)^2))&#xA; S3 -&amp;gt; S4 -&amp;gt; C  =&amp;gt; 1 + SQR(((1/2)^2)+((1/2)^2))&#xA; S4 -&amp;gt; C        =&amp;gt;     SQR(((1/2)^2)+((1/2)^2))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Think about it this way the King and S0 take half of the list of house.&#xA;   King:= S1, S2&#xA;     S0:= S3, S4&#xA;When they get to the first house.&#xA;They cross that house of the list and first divide the list.&#xA;If there is only one house on the list, the soldier who house it is go and fetches them, whilst the solider who originally brought the list heads back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is only 1 additional soldier then the answer is &lt;code&gt;2 x Distance To Solider&lt;/code&gt; Eg There and back. &lt;/p&gt;&#xA;" OwnerUserId="1656" LastEditorUserId="1656" LastEditDate="2014-06-26T17:18:35.770" LastActivityDate="2014-06-26T17:18:35.770" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="1797" PostTypeId="1" AcceptedAnswerId="1798" CreationDate="2014-06-26T22:25:22.040" Score="60" ViewCount="288405" Body="&lt;p&gt;One needs to find the number under the car.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CT56W.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CT56W.jpg&quot; alt=&quot;A picture of some parking spaces from the top down, appearing to be numbered &amp;quot;16, 06, 68, 88, ??, 98&amp;quot;&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;sub&gt;Source: unknown&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2023-02-07T20:04:33.897" LastActivityDate="2023-02-07T20:04:33.897" Title="What is the number of the parking spot?" Tags="|lateral-thinking|pattern|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="1798" PostTypeId="2" ParentId="1797" CreationDate="2014-06-26T22:28:54.777" Score="50" Body="&lt;p&gt;I believe the most obvious answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is a usual sequence of numbers, we're just looking at it in the wrong direction. The numbers of the parking spots must be viewed from the outside in, as a driver would see them, which is upside-down from the current view.&lt;br /&gt; &lt;br /&gt;&#xA; The shown numbers are: 86, ??, 88, 89, 90, 91. Therefore the answer is &lt;strong&gt;87&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="28" LastEditorDisplayName="user88" LastEditDate="2014-06-27T15:40:43.080" LastActivityDate="2014-06-27T15:40:43.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1800" PostTypeId="2" ParentId="1705" CreationDate="2014-06-27T03:14:20.437" Score="1" Body="&lt;p&gt;&lt;strong&gt;N+1 Locks&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 master lock attached to the artefact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N locks attached to the loop, each lock has a key to the master lock attached to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each thief has a key to one of the locks (except the master).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any thief can obtain a key to the master using their own key, and whichever lock is open/missing implicates the thief.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: This only works once - once a master key has been accessed, it could potentially be copied.&lt;/p&gt;&#xA;" OwnerUserId="1667" LastEditorUserId="1667" LastEditDate="2014-06-27T03:33:34.510" LastActivityDate="2014-06-27T03:33:34.510" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1801" PostTypeId="2" ParentId="1797" CreationDate="2014-06-27T03:58:59.407" Score="102" Body="&lt;p&gt;If you look at the numbers...&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; ...upside down, they read 86, __, 88, 89, 90, 91. The number that fits this pattern is, of course, 87.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Therefore, the space as the picture is currently viewed should be filled in with&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;strong&gt;L8&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="60644" LastEditorDisplayName="user88" LastEditDate="2023-02-07T14:48:46.317" LastActivityDate="2023-02-07T14:48:46.317" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="1803" PostTypeId="2" ParentId="1776" CreationDate="2014-06-27T14:08:32.113" Score="0" Body="&lt;p&gt;I may not be correct but just let me know if I have a mistake in my logic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thought is that I consider them as sequence with $15,10,25,10,35$ and so on...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$41-(5+6+7+8) = 15$$&#xA;$$31- (9+4+2+6) = 10$$&lt;br&gt;&#xA;$$X - (8+7+2+6) = 25$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, I think $48$ is my answer.&lt;/p&gt;&#xA;" OwnerUserId="1674" LastEditorDisplayName="user20" LastEditDate="2014-06-27T17:42:44.177" LastActivityDate="2014-06-27T17:42:44.177" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1804" PostTypeId="2" ParentId="25" CreationDate="2014-06-28T01:36:23.137" Score="6" Body="&lt;p&gt;The importance of the numbers depends on time discounting and what solution concepts you'd like to use. Your intuition is good. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You mention tit for tat. This and other punishment strategies exist to ensure cooperation. However, you may be interested in finding punishment strategies that are &lt;em&gt;credible&lt;/em&gt;, introducing refinements to the Nash equilibrium concept common in the economic literature. Credibility results don't necessarily hinge on the gain from defecting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your example, tit for tat is credible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my work showing why tit for tat is credible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let there be a discount factor $\delta\in(0,1)$. We can consider the average payoffs to player 1 in the four states given the tit for tat protocol: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When both players cooperate, player 1 earns $V_{1}(w_{CC})=100$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When both defect, $V_{1}(w_{DD})=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When only 1 defects, $V_{1}(w_{DC})=(1-\delta)101+\delta V_{1}(w_{CD})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When only 2 defects, $V_{1}(w_{CD})=(1- \delta )0+\delta V_{1}(w_{DC})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, $V_{1}(w_{DC})=\frac{101}{1+\delta}$ and $V_{1}(w_{CD})=\frac{\delta101}{1+\delta}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, CC is a Nash eq. in the state CC if&#xA;$100\geq(1-\delta)101+\delta\frac{\delta101}{1+\delta}$, requiring $\delta\geq\frac{1}{100}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we check that once one player has deviated, they will find it in their interest to continue playing tit for tat. We compare average payoffs from different strategies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in states CD and DC, cooperating is a best response if $\frac{\delta101}{1+\delta}&amp;gt;(1-\delta)1+\delta=1$. This is satisfied if $\delta\geq\frac{1}{100}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And defecting is a best response if $\frac{101}{1+\delta}\geq(1-\delta)3+\delta3=3,$ which is satisfied for any $\delta$ in $(0,1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following in this pattern, you would see that tit for tat is not credible for a game with payoffs T=4, R=3, S=-1, P=1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an academic reference, see&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mailath, G. J., &amp;amp; Samuelson, L. (2006). Repeated games and reputations: long-run relationships. OUP Catalogue.&lt;/p&gt;&#xA;" OwnerUserId="1611" LastEditorUserId="1611" LastEditDate="2014-06-28T22:12:27.340" LastActivityDate="2014-06-28T22:12:27.340" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1805" PostTypeId="2" ParentId="1776" CreationDate="2014-06-28T05:43:15.443" Score="2" Body="&lt;p&gt;Let's say $41$ is the top number and $a=5,b=6,c=7,d=8$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;cb-41 + ad &amp;amp;= 41\\&#xA;cb-31 + ad &amp;amp;= 31\\&#xA;cb-x + ad &amp;amp;= x\\&#xA;\implies 14-x+48 &amp;amp;= x\\&#xA;2x &amp;amp;= 62\\&#xA;x&amp;amp;=31&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;therefore answer is $31$.&lt;/p&gt;&#xA;" OwnerUserId="1674" LastEditorUserId="13494" LastEditDate="2017-09-03T13:54:55.290" LastActivityDate="2017-09-03T13:54:55.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1806" PostTypeId="2" ParentId="313" CreationDate="2014-06-28T19:42:57.997" Score="5" Body="&lt;p&gt;&lt;a href=&quot;http://www.fifa.com/worldcup/matches/&quot; rel=&quot;noreferrer&quot;&gt;Soccer World Cup&lt;/a&gt; is in progress, so that warrants a themed question (though just a bit late, as group stage is over now - oh well):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is your number both valid and not less than a number of points that a team got after finishing a group stage match in which they scored at least the same number of goals as their opponent?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A soccer team receives 1 point for a draw and 3 points for a win, 2 is not a valid number of points, so:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 =&amp;gt; yes&#xA;2 =&amp;gt; no&#xA;3 =&amp;gt; maybe&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I feel this doesn't bode well with the spirit of the posted question as some of the other answers do, though. ;)&lt;/p&gt;&#xA;" OwnerUserId="1693" LastActivityDate="2014-06-28T19:42:57.997" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1807" PostTypeId="2" ParentId="313" CreationDate="2014-06-28T19:53:53.773" Score="3" Body="&lt;p&gt;&lt;a href=&quot;http://www.wimbledon.com/index.html&quot; rel=&quot;nofollow&quot;&gt;Wimbledon&lt;/a&gt; is in progress, so let's put that to good use:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is your number less than a number of sets played in a straight-set win in a Wimbledon tennis match?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Women win in 2 sets, men in 3 sets.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 -&amp;gt; yes&#xA;2 -&amp;gt; maybe&#xA;3 -&amp;gt; no&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1693" LastActivityDate="2014-06-28T19:53:53.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1808" PostTypeId="1" CreationDate="2014-06-29T01:39:04.417" Score="15" ViewCount="3794" Body="&lt;p&gt;Suppose you have twelve identical-looking balls. Ten of them are the same weight, but one is slightly lighter and one slightly heavier, in such a way that the weight of the lighter and heavier ball add up to exactly the weight of two of the equal-weight balls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the minimum number of weighings on a two-sided scale that are needed to determine which ball is lighter and which one is heavier?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;What would the solution be if the weight of the lighter and heavier ball didn't &lt;em&gt;necessarily&lt;/em&gt; add up to the weight of two equal balls, but you didn't know whether it was lighter, heavier, or equal to that weight?&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="8874" LastEditDate="2015-01-31T12:25:38.090" LastActivityDate="2021-11-11T14:39:25.083" Title="Twelve balls on a scale, where one ball is lighter and another is heavier" Tags="|mathematics|strategy|weighing|" AnswerCount="7" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1810" PostTypeId="2" ParentId="1808" CreationDate="2014-06-29T09:19:41.443" Score="5" Body="&lt;p&gt;We divide the balls into two 6 balls groups, and compare them with scale &lt;kbd&gt;1&lt;/kbd&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Scenario 1&lt;/strong&gt; :&#xA;Assume the weights is &lt;strong&gt;not&lt;/strong&gt; equal, this means that the heavier and lighter ball are &lt;strong&gt;not&lt;/strong&gt; in the same side.(the weight of the lighter and heavier ball add up to exactly the weight of two of the equal-weight balls)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then, we divide heavier 6-balls groups into two 3-balls group and compare them with scale&lt;kbd&gt;2&lt;/kbd&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we know for sure that  the weight is &lt;strong&gt;not&lt;/strong&gt; equal (because there is only heavier ball in this 6-balls group)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then,we pick the heavier group and compare two of them with scale &lt;kbd&gt;3&lt;/kbd&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If they have equal weights,then the third one is the heavier one.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If they have different weights, we can pick the heavier one easily.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;by repeating weightings number &lt;kbd&gt;2&lt;/kbd&gt; and &lt;kbd&gt;3&lt;/kbd&gt; for the  lighter 6-balls groups, we can find the lighter one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in this scenario, we can find both of balls in &lt;kbd&gt;5&lt;/kbd&gt; weightings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this scenario is the shortest way.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Scenario 2&lt;/strong&gt; : the weight of two group is equal, this means the balls are in the same side...&#xA;            I'm working on this scenario :)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="465" LastActivityDate="2014-06-29T09:19:41.443" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1811" PostTypeId="1" AcceptedAnswerId="1812" CreationDate="2014-06-29T10:10:43.517" Score="5" ViewCount="2005" Body="&lt;p&gt;The question reads:&#xA;&lt;strong&gt;From the given pattern, What comes in place of '?'&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JImq8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1637" LastActivityDate="2017-11-22T14:32:10.867" Title="I have a pattern problem to solve!" Tags="|logical-deduction|pattern|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1812" PostTypeId="2" ParentId="1811" CreationDate="2014-06-29T10:56:52.810" Score="10" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;(4)&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It looks like:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0. Triangle from 4 figures rotates in the square anti-clockwise.&lt;BR&gt;&#xA; 1. &quot;S&quot; goes around anti-clockwise, from one corner to another corner.&lt;BR&gt;&#xA; 2. &quot;Square&quot; goes along main diagonal (from TopLeft to BottomRight and back).&lt;BR&gt;&#xA; 3. &quot;Triangle&quot; goes along second diagonal (from TopRight to BottomLeft and back).&lt;BR&gt;&#xA; 4. Other figures appear to chosen randomly. &lt;em&gt;May be there is some pattern I do not see, but it doesn't matter, since 2 and 4 leaves us&#xA; with only one possible answer, which perfectly agrees with all 5&#xA; patterns.&lt;/em&gt;&lt;BR&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="28" LastEditorUserId="4637" LastEditDate="2015-07-03T09:26:00.423" LastActivityDate="2015-07-03T09:26:00.423" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1813" PostTypeId="2" ParentId="1811" CreationDate="2014-06-29T14:13:27.947" Score="1" Body="&lt;p&gt;I also answer &lt;strong&gt;(4)&lt;/strong&gt; but &lt;strong&gt;I'd like to define the reason in another way&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears to be &lt;em&gt;a&lt;/em&gt; control structure in the patter, formed by:&#xA;a 'triangle', a 'square' and a 'S'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This 3 signs form a triangle where:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;'S' is in a corner and rotates anti-clockwise,&lt;/li&gt;&#xA;&lt;li&gt;the central vertex is 'square' or 'triangle', alternating,&lt;/li&gt;&#xA;&lt;li&gt;the third vertex of the triangle-structure is occupied by the&#xA;remaining sign.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Figure number (4) is the only one validated by this pattern. (1) and (3) don't show a 'S' and in particular the (1)'s 'triangle' is not where it were supposed to be. (2) shows two 'squares' and no 'triangle', quite strange.&lt;/p&gt;&#xA;" OwnerUserId="1703" LastActivityDate="2014-06-29T14:13:27.947" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1814" PostTypeId="1" CreationDate="2014-06-29T21:07:19.727" Score="20" ViewCount="1874" Body="&lt;p&gt;The following problem is found in the first Norwegian arithmetic, published in 1645 by Tyge Hansøn:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Three hundred oxen large and small,&lt;br&gt;&#xA;  A cattle owner wanted to buy:&lt;br&gt;&#xA;  3 for 63 daler he got.&lt;br&gt;&#xA;  Again, he let them sell,&lt;br&gt;&#xA;  3 for 63 daler they fetched,&lt;br&gt;&#xA;  both slim and fat.&lt;br&gt;&#xA;  787$1\over 2$ daler was his profit.&lt;br&gt;&#xA;  Tell me, how did that happen?&lt;br&gt;&#xA;  Whoever wants to calculate that, consider rightly&lt;br&gt;&#xA;  and argue the issue well.&lt;br&gt;&#xA;  Then it becomes quite simple.&lt;br&gt;&#xA;  to calculate for a farmer.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So apparently, the cattle owner bought and sold for the same price, but he still made a profit. There is no solution to this problem in the text. I have a puzzle-like solution, but would like to hear other suggestions before I reveal it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some notes:&lt;br&gt;&#xA;The original is written in verse, with rhymes. I have translated the text into modern English, but had to sacrifice some of the poetic effects in order to preserve the factual content.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the coinage system used in Denmark-Norway at the time, one daler could be divided into 96 skillings.&#xA;The problem is found in a chapter on the rule of three, so a solution should in some way be true to the principle of proportionality. Hence, any suggestion that the herd has increased in size, or similar ideas, would not be satisfactory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if anyone has seen a similar problem, I would be very interested in hearing about it.  &lt;/p&gt;&#xA;" OwnerUserId="219" LastEditorUserId="8874" LastEditDate="2016-03-21T14:02:48.767" LastActivityDate="2016-03-21T14:02:48.767" Title="An arithmetic puzzle from 1645" Tags="|mathematics|algebra|puzzle-history|" AnswerCount="3" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="1815" PostTypeId="1" AcceptedAnswerId="1816" CreationDate="2014-06-29T21:25:14.230" Score="3" ViewCount="587" Body="&lt;p&gt;This is a small generalization of well known &lt;a href=&quot;https://puzzling.stackexchange.com/q/183/28&quot;&gt;12 balls&lt;/a&gt; puzzle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You are given two-sided scale and some number of balls. All balls have the same weight but one. It is of a different weight, although you don't know whether it's lighter or heavier. You are allowed to do only 3 weighings to determine not only what the different ball is, but also whether it's lighter or heavier. You can distinguish balls, but you do not know a weight of some of them. How many balls with unknown weight you may have and still be able to find the different one?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="28" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-01-31T12:31:08.197" Title="$N_{max}$ balls in 3 weightings" Tags="|strategy|weighing|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1816" PostTypeId="2" ParentId="1815" CreationDate="2014-06-29T21:25:14.230" Score="2" Body="&lt;p&gt;This answer is a follow up of Joe Z.'s &lt;a href=&quot;https://puzzling.stackexchange.com/a/224/28&quot;&gt;answer&lt;/a&gt; on the similar puzzle. If you have problems with understanding this one, you can read Joe Z.'s answer, which explains used notations in much more details, which I would like to skip here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In the assumption of fixed weighing schedule the optimum (maximum) number of unknown balls are, so to say, 13.5, where optimum (minimum) number of known balls is 1.&lt;/strong&gt;&lt;br&gt;&#xA;The following is solution for tasks with 13 balls with unknown weight (they can be normal, light, heavy), 1 ball with partially unknown weight (it can be either normal or lighter, but it can't be heavier) and 1 ball, which is known to be normal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to prove that you can't add more unknown balls and you need at least one known ball.&lt;br&gt;&#xA;Indeed:&lt;br&gt;&#xA; 1. this has $13\times2+1 = 27$ possibilities to choose between, this is maximum having 3 weights and 3 states of them: $27=3^3$;&lt;br&gt;&#xA; 2. you must use all $27=3^3$ possible sets of ball placement: 2 sets per first 13 balls to distinguish between heavy and light and 1 set (which is &quot;always off the scale&quot;) for 14-th ball, but one can see that in this case exactly 27 balls must be on the scale in total during 3 weightings, that means at least one time there must be odd number of balls, this would lead to fault until we have one more ball is needed with known state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ball's arrangement:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Ball  1  2  3  4  5  6  7  8  9  10 11 12 13 14 15 -1 -2 -3 -4 -5 -6 -7 -8 -9 -10-11-12-13&#xA;W1    L        L  R     R  L     R  R  L  R     L    R        R  L     L  R     L  L  R  L&#xA;W2       L     R     L  R     L  R  L  R  R     L       R     L     R  L     R  L  R  L  L&#xA;W3          R     L  R     L  L  L  R  R  R     L          L     R  L     R  R  R  L  L  L&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Table of inverses here should not include 14th and 15th balls, since they are already different and you can distinguish between them using their properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weighing schedule of balls:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Weighing 1:  1  4  8 12 13 /  5  7 10 11 15&#xA;Weighing 2:  2  6  9 11 13 /  4  7 10 12 15&#xA;Weighing 3:  5  8  9 10 13 /  3  6 11 12 15&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The outcomes interpretation:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;= = L :  3L    L = = :  1H   R = = :  1L&#xA;= = R :  3H    L = L :  8H   R = L :  5H&#xA;= L = :  2H    L = R :  5L   R = R :  8L&#xA;= L L :  9H    L L = :  7L   R L = :  4L&#xA;= L R :  6H    L L R : 10L   R L L : 12L&#xA;= R = :  2L    L R = :  4H   R L R : 11H&#xA;= R L :  6L    L R L : 11L   R R = :  7H&#xA;= R R :  9L    L R R : 12H   R R L : 10H&#xA;= = = : 14L    L L L : 13L   R R R : 13H&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="28" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2014-06-30T09:06:59.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1817" PostTypeId="1" CreationDate="2014-06-29T21:41:28.753" Score="2" ViewCount="887" Body="&lt;p&gt;This question has been written in order to help answer &lt;a href=&quot;https://puzzling.stackexchange.com/questions/548/fastest-way-to-collect-an-arbitrary-army&quot;&gt;Fastest way to collect an arbitrary army&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a 1 x 1 square area with corners labelled clockwise A, B, C and D.&#xA;A finite number of stationary robots are spread out across the area.&#xA;Every robot knows where every other robot is at any time.&#xA;A robot may only pass a message to another robot by touch.&#xA;All robots travel with speed 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robot Zero is at point A and receives the following message from an off grid source:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Ensure that all robots leave the area as quickly as possible, through point C&quot;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that we do not know how many robots there are or where they are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the longest time it could take to clear the area of robots?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How can you prove it?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can assume the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Robots have negligible area, so when they touch in order to relay the message, they are occupying the same point on the square.&lt;/li&gt;&#xA;&lt;li&gt;Message is transferred in negligible time&lt;/li&gt;&#xA;&lt;li&gt;When faced with two options of equal efficiency a robot is able to make a decision (in collaboration with another robot where necessary) in negligible time.&#xA;So, if there are two robots at B and there is only one other robot on the grid, at D, the two robots at B are able to send one robot to D to relay the message, while the other robot goes straight from B to C.&lt;/li&gt;&#xA;&lt;li&gt;The robots are all able to calculate the most efficient way of relaying the message to all robots.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1616" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2014-07-31T16:50:05.507" Title="How long is needed to clear the area of robots?" Tags="|mathematics|geometry|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="1818" PostTypeId="2" ParentId="1814" CreationDate="2014-06-30T04:39:42.593" Score="17" Body="&lt;p&gt;The problem in modern English might be stated as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A farmer wanted to buy three hundred oxen. He bought them at a price of $3$ oxen for $\$63$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Afterwards, he sold the oxen at $3$ for $\$63$ as well, but managed to make a profit of $\$787.50$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;How did he manage this?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At first blush, as you said. it appears that the farmer should have left with exactly as much money as he'd come with - after all, he bought and sold the oxen at $\$21$ an ox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, let's find the price that $300$ oxen would have fetched. Since you mentioned the rule of three, we'll use that to calculate it. Let $P$ be the total amount of money the farmer spend to buy the cows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac {\$63}{3} = \frac{P}{300} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ P = \frac{$63 \times 300}{3} = \$6300 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $\$787.50$ is exactly one-eighth of $\$6300$. Somehow, when selling the oxen, he received 9 dollars for every 8 dollars he spent, even selling them at the same rate.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I can only imagine that the answer has something to do with currency exchange and the deceptive use of the word &quot;daler&quot; in the original riddle, and that another Nordic country that used dalers at the time divided them into $108$ skillings instead of $96$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this were the case and the farmer had done his entire bovine-selling operation in skillings, he could have bought the oxen for $21 \times 96$ skillings apiece in one country and sold them for $21 \times 108$ skillings in another country, making a profit of $21 \times 12$ skillings per ox for a total of $\frac{21 \times 12 \times 300}{96} = 787.5$ dalers in profit.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Doing some &lt;a href=&quot;https://en.wikipedia.org/wiki/Skilling_%28Scandinavian_monetary_unit%29&quot;&gt;cursory research on Wikipedia&lt;/a&gt; reveals that Norway/Denmark had two dalers - one worth 96 skillings and one worth 120. This still works as a solution as above &amp;ndash; if the farmer sold &lt;em&gt;half&lt;/em&gt; the cows for those 120-skilling dalers, he would have made the same profit of $\frac{21 \times 24 \times 150}{96} = 787.5$ dalers.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2014-06-30T04:51:58.973" LastActivityDate="2014-06-30T04:51:58.973" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1819" PostTypeId="2" ParentId="1764" CreationDate="2014-06-30T05:25:38.097" Score="0" Body="&lt;p&gt;Or, three eights is 8 × 8 × 8, which is 512.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8 × 8 × 8 = 2&lt;sup&gt;(2×2)&lt;/sup&gt; × 2&lt;sup&gt;(2×2)&lt;/sup&gt; × 2.&lt;/p&gt;&#xA;" OwnerUserId="455" LastActivityDate="2014-06-30T05:25:38.097" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1820" PostTypeId="1" AcceptedAnswerId="1823" CreationDate="2014-06-30T13:25:19.200" Score="3" ViewCount="1067" Body="&lt;p&gt;Here is more complicated &lt;a href=&quot;https://puzzling.stackexchange.com/questions/626/paying-the-troll-toll&quot;&gt;Paying the Troll toll&lt;/a&gt; like puzzle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Between you and your destination, you have 7 bridges, and there is a troll under every bridge. Each troll, quite rightly, insists that you pay a troll toll. Before you can cross the $K$-th bridge, you have to give $K+1$ cakes plus they are not kind trolls and each takes additionally exactly $1/(K+1)$ of the rest. Also trolls would not accept part of a cake - just a whole one. And you can't rid of cakes between bridges. This is a sad puzzle, because you can never reach your destination, but how many bridges can you cross if you try hard? How many cakes do you need to take with you?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;br&gt;&#xA;  1. If you have $X$ cakes you pay first troll: $Y = 2+(X-2)/2$, second troll: $3+(X-Y-3)/3$, etc.&lt;br&gt;&#xA;  2. That means that $Y = 2+(X-2)/2$ and $3+(X-Y-3)/3$ must be integer numbers. So if you start with 1001 cake you fail at the very first bridge.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2014-06-30T20:08:40.057" Title="Paying the unfair Troll toll" Tags="|mathematics|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1822" PostTypeId="1" AcceptedAnswerId="1824" CreationDate="2014-06-30T16:22:36.243" Score="14" ViewCount="2486" Body="&lt;p&gt;Suppose you've rigged the church charity lottery so that you cannot lose (what a swell guy you are).  There are 10 drawings, and you start off with 10 dollars exactly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At each drawing, you must chose an amount to bet, and that amount gets put into a pool (that money then can no longer be used to wager).  You then proceed with the next drawing, and you do this until you have no more money left.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total amount of money won is the product of the amounts wagered each round played that you won (and since it is rigged, everytime you wager you win).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if I wagered all 10 dollars in the first round, I will have won 1 drawing and earned exactly 10 dollars.  If I wager 2 in the first round and 8 in the second, I will have won 16 dollars.  Wagers &lt;em&gt;must&lt;/em&gt; be in whole dollar amounts.  Rounds where nothing is wagered are disregarded entirely, depending on how money is distributed amongst the various rounds.  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What is the best strategy to adopt to maximize winnings?&lt;/li&gt;&#xA;&lt;li&gt;For $n$ dollars and potentially up to $n$ drawings, what is the generalizable best strategy to maximize winnings?&lt;/li&gt;&#xA;&lt;li&gt;Let $\varphi(n)$ represent the maximum amount possible for start amount $n$ and rounds $n$ while observing the rules above.  What is the relationship between $\varphi(n)$ and $n$ (can a conclusion be made about its relationship)?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1672" LastEditorUserId="5373" LastEditDate="2018-05-23T20:29:15.297" LastActivityDate="2022-07-21T00:05:16.067" Title="Church charity lottery" Tags="|logical-deduction|strategy|" AnswerCount="6" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1823" PostTypeId="2" ParentId="1820" CreationDate="2014-06-30T16:23:09.093" Score="11" Body="&lt;p&gt;Say you start with $x$ cakes. After the first bridge, you're left with &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{x-2}{2}$ cakes, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{x-8}{3}$cakes,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{x-20}{4}$ cakes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{x-40}5$ cakes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{x-70}6$ cakes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{x-112}7$ cakes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want those to be integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus  $x-8 \equiv 0 \pmod 3$ ie $x\equiv 2 \pmod 3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $x\equiv 70 \equiv 4 \pmod 6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $x\equiv 2 \pmod 3$ and $x\equiv 4 \pmod 6$ is a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, you cannot get past the fifth bridge. To reach it, you can start with  $80$ cakes&lt;/p&gt;&#xA;" OwnerUserId="1636" LastEditorUserId="1636" LastEditDate="2014-06-30T16:53:37.100" LastActivityDate="2014-06-30T16:53:37.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1824" PostTypeId="2" ParentId="1822" CreationDate="2014-06-30T17:02:04.103" Score="14" Body="&lt;p&gt;This is basically a question of partitioning - which partition of $n$ generates the largest product?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's establish the rules that our partition should follow:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) It doesn't make sense to make a partition of 1, that won't increase the product. We'd rather make another partition one larger. Better to have $2,3$ than $2,2,1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) It also doesn't make sense to partition except into prime numbers. Since we're multiplying all the numbers together, then multiplying by a composite number is the same as multiplying by its prime factorization. So if we partitioned $10$ into $6,2,2$ then our winnings would be $6*2*2$, but since multiplying by $6$ is the same as multiplying by $2$ and $3$ (its prime factorization), we can partition into $3,2,2,2,1$ and get the same result. EDIT: The prime factorization won't always sum up to the same number, so we may get additional factors, which will only increase our end product, or leave it the same if the additional factor is just $1$ as above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) One additional rule is that one larger prime is not as good as two smaller primes. For example, a factor of $5$ is not as good as factors of $2$ and $3$. Similarly, factors of $7$ are not as good as factors of $3,2,2$. Lastly, a factor of $3$ is better than factors of $2$ and $1$. SO a partition that includes $11$ is beaten by a partition that includes $5,5,1$ instead, which is similarly beaten by a partition that includes $3,2,5,1$ which is beaten by a partition of $3,3,3,2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This fact can be &quot;proven&quot; by the simple fact that $2$ is less than half of any number greater than $4$. So if you take $c$ and split it into $2$ and $c-2$, then multiplying by $2$ and $c-2$ will yield a product larger than if you just multiplied by $c$. Repeat this process until your numbers are small enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, given the above rules, I think that the best partition would be to make as many 3's as possible, unless it would leave a partition of 1 left in which case you should fill in the rest with 2's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would leave the maximal partition of $10$ to be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3,3,2,2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximal partition of $11$ would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3,3,3,2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximal partition of $n$ would therefore be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n\mod3 \equiv 0$: $k$ $3$'s, where $k = \frac n3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n\mod3\equiv1$: $k$ $3$'s and two $2$'s, where $k = \lfloor {\frac n3}\rfloor -1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n\mod3\equiv2$: $k$ $3$'s and one $2$, where $k = \lfloor {\frac n3}\rfloor$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would love for someone to check my logic.&lt;/p&gt;&#xA;" OwnerUserId="1620" LastEditorUserId="1620" LastEditDate="2014-06-30T17:39:19.693" LastActivityDate="2014-06-30T17:39:19.693" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="1825" PostTypeId="2" ParentId="1822" CreationDate="2014-06-30T19:12:06.270" Score="4" Body="&lt;p&gt;If you take $k$ ($1 &amp;lt;= k &amp;lt;= 10$) as the number of wagers you want to join, then the optimal amount to bet for each wager is $10 / k$.&#xA;And your total winnings would then be $(10 / k)^k$.&#xA;So I calculated this for every $k$ from $1$ to $10$ and got this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;k    10/k   (10/k)^k&#xA;---------------------&#xA;1    10     10&#xA;2    5      25&#xA;3    3.33   37.04&#xA;4    2.5    39.06&#xA;5    2      32&#xA;6    1.67   21.43&#xA;7    1.43   12.14&#xA;8    1.25   5.96&#xA;9    1.11   2.58&#xA;10   1      1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And it clearly shows you that you should participate in 4 wagers of 2,5 dollars each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the extended case of $n$ dollars and up to $n$ drawings, you're looking for a $k$ $(1 &amp;lt;= k &amp;lt;= n)$, where $(n / k)^k$ is maximal.&#xA;I don't know yet how to find such a $k$ without brute forcing it.&lt;/p&gt;&#xA;" OwnerUserId="448" LastEditorUserId="448" LastEditDate="2014-06-30T19:21:45.330" LastActivityDate="2014-06-30T19:21:45.330" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1826" PostTypeId="1" AcceptedAnswerId="1827" CreationDate="2014-06-30T21:38:48.543" Score="2" ViewCount="1159" Body="&lt;p&gt;On HuDoKu's page on Uniqueness, under &lt;a href=&quot;http://hodoku.sourceforge.net/en/tech_ur.php#un2&quot; rel=&quot;nofollow&quot;&gt;Unique Rectangle&lt;/a&gt;, it says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A common mistake when looking for URs is to violate the &quot;2 blocks&quot; rule (see the &lt;a href=&quot;http://www.sudopedia.org/wiki/Unique_Rectangle&quot; rel=&quot;nofollow&quot;&gt;Sudopedia article&lt;/a&gt; for an example).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, the Sudopedia link they give is broken.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand the &quot;2 blocks rule&quot; from the section above that note as meaning that the rectangle must be in exactly two blocks, but what mistake was the example supposed to show, and how can it cause a problem if I make that mistake?&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditDate="2014-06-30T23:43:56.130" LastActivityDate="2014-07-01T17:45:17.850" Title="SuDoKu Unique Rectangles and the &quot;2 blocks&quot; rule" Tags="|sudoku|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1827" PostTypeId="2" ParentId="1826" CreationDate="2014-06-30T22:14:11.473" Score="4" Body="&lt;p&gt;Could this be an alternative link? &lt;a href=&quot;http://www.sudokuwiki.org/Unique_Rectangles&quot; rel=&quot;nofollow noreferrer&quot;&gt;Unique Rectangles&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The link is saying that if you end up with the 2 blocks like this (the four pink boxes can only have either 1 or 2 in them, &quot;MadOverlord&quot; named this Unique Rectangle pattern the Deadly Pattern apparently!):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ezy5u.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then either:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the setter has not created a sudoku with a unique solution (since there are two possible ways to populate the boxes)&lt;/li&gt;&#xA;&lt;li&gt;or you have gone wrong somewhere.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The writer from Sudoku Wiki goes on to say that if you know that there is a unique solution then you can use potential blocks like this to your advantage when solving the puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in the below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8OBrA.jpg&quot; alt=&quot;Potential Deadly Pattern&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will be able to see that you cannot place a 7 in any of the blue boxes as this will leave behind the deadly pattern, and so the 7 must be in the bottom left of that rectangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few variations on this example on that page - You can draw conclusions about the numbers in more than just the four rectangle boxes themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally, I am a bit wary of using this method as making any assumption in a sudoku makes me feel a little uneasy! It was an interesting read though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer you question, I don't think that finding that you have &quot;the deadly pattern&quot; is the mistake, but it shows you that maybe somewhere earlier in the solving, you've made a mistake. Or else, there are multiple solutions. As we can see from this question, not many people like multiple solutions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/551/why-must-a-sudoku-have-a-unique-solution/554&quot;&gt;Why must a sudoku have a unique solution&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that helps? Or at least that the sudokuwiki link helps!&lt;/p&gt;&#xA;" OwnerUserId="1616" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2014-07-01T17:45:17.850" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1829" PostTypeId="2" ParentId="1822" CreationDate="2014-07-01T09:52:51.280" Score="3" Body="&lt;p&gt;As long as there are enough drawings, the optimal strategy is to bet 3 dollars each time as long as you have more than 4 dollars left. Once you have 4 dollars or less left, you bet the rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any bet of 5 or more won't be optimal, because you could bet 2 of them and then the rest, and get a larger return. A bet of 1 obviously doesn't make sense either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strategy is optimal, because if you had spent more than 4 dollars on bets of sizes 2 or 4, the strategy could be improved as follows. Any bet of 4 dollars is split into two of 2 dollars (which doesn't change anything), and then you change three bets of 2 into two bets of 3, which is better because 3*3 &gt; 2*2*2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you started with less than 5 dollars, the above strategy will only pay back your initial bets, instead you just don't bet at all and win one dollar, because the product over the empty set is 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are too few drawings to use the above strategy, you split your bets evenly over the drawings. Round up or down enough times to exactly use up all your money.&lt;/p&gt;&#xA;" OwnerUserId="1619" LastActivityDate="2014-07-01T09:52:51.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1830" PostTypeId="2" ParentId="1822" CreationDate="2014-07-01T10:35:36.387" Score="3" Body="&lt;p&gt;I would like to combine answers of &lt;a href=&quot;https://puzzling.stackexchange.com/a/1824/28&quot;&gt;Duncan&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/a/1825/28&quot;&gt;PrisonMonkeys&lt;/a&gt; (my credits to them) and add some prove (not 100% strict, but it let to feel what is happening in this puzzle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To formulate problem on mathematical language we need to find $k&amp;lt;=N$ nonnegative integer numbers: $n_1,n_2,..n_k$, such that sum $S = n_1+n_2+..+n_k = N$ and product $P = n_1 \cdot n_2 \cdot .. \cdot n_k$ is maximal.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Let's first allow $n_i$ to be not integer.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If $k$ is fixed we need to find conditional extremum. I use &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrange_multiplier&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lagrange function&lt;/a&gt;, that would be:&#xA;$L = n_1 \cdot n_2 \cdot .. \cdot n_k + \lambda(n_1+n_2+..+n_k - N)$&lt;br&gt;&#xA;$\partial L/\partial n_i = P / n_i + \lambda$, for any $i = 1, .., k$&#xA;therefore $n_1 = n_2 = .. = n_k = P/\lambda$,&lt;br&gt;&#xA;$n_i = N/k$ for any $i = 1, .., k$,&lt;br&gt;&#xA;$P = (N/k)^k$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let's suppose $k$ is not integer too. Then maximum is achieved when&lt;br&gt;&#xA;$dP/dk = 0$ or&lt;br&gt;&#xA;$P(ln(N) - ln(k) - 1) = 0$, or&lt;br&gt;&#xA;$k = N/e$ therefore&lt;br&gt;&#xA;$n_i = e$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When we go from real $n_i = e \approx 2.7$ to integer we can &lt;em&gt;suppose&lt;/em&gt; that the closest possible integer number must be chosen. This is reasonable assumption, since function $P(k)=(N/k)^k$ has monotonic increase before $k&amp;lt;e$ and monotonic decrease with $k&amp;gt;e$.&lt;br&gt;&#xA;The closest numbers to $2.7$ are $2$ on the left and $3$ on the right, one can compare them directly and find out that $3$ is optimal.&lt;br&gt;&#xA;Since not all $N$ are dividable by $3$ we may need to use 2, which is also close to $2.7$, but not more that two $2$'s will be needed.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Summarising: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $N \bmod 3 = 0$ : we wager all $3$'s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $N\bmod3 = 1$ : we wager two $2$'s and rest are $3$'s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $N\bmod3 =2$ : we wager one $2$'s and rest are $3$'s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $N = 10 = 9+1$, we can win $2\cdot 2\cdot 3\cdot 3=36$.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2014-07-01T11:33:04.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1831" PostTypeId="2" ParentId="1822" CreationDate="2014-07-01T10:44:14.220" Score="3" Body="&lt;p&gt;Adding to PrisonMonkeys' answer, We need to find the value of &lt;em&gt;&lt;strong&gt;k&lt;/strong&gt;&lt;/em&gt; such that the value of (n/k)^k  is maximized. on differentiating and solving this, the value of &lt;em&gt;&lt;strong&gt;k&lt;/strong&gt;&lt;/em&gt; happens to be &lt;em&gt;&lt;strong&gt;n/e&lt;/strong&gt;&lt;/em&gt; , where &lt;em&gt;&lt;strong&gt;e&lt;/strong&gt;&lt;/em&gt; is the irrational 2.71828... . That means, we would have the most optimal product when we divide it into &lt;em&gt;&lt;strong&gt;n/e&lt;/strong&gt;&lt;/em&gt; parts with each part having the value of &lt;em&gt;&lt;strong&gt;e&lt;/strong&gt;&lt;/em&gt; . I think the accepted solution of Duncan makes sense because 3 is the closest integer value to &lt;em&gt;&lt;strong&gt;e&lt;/strong&gt;&lt;/em&gt; and 2 is the next closest. So when we are going for integer partitions, make maximum of the partitions equal to the integer nearest to &lt;em&gt;&lt;strong&gt;e&lt;/strong&gt;&lt;/em&gt; and the rest to the next nearest. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just my thought, not a rigorous proof as such. Feel free to comment and improve/ add more rigour to the answer. Lastly, I am new here, so sorry about the formatting.&lt;/p&gt;&#xA;" OwnerUserId="1725" LastActivityDate="2014-07-01T10:44:14.220" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1832" PostTypeId="2" ParentId="1826" CreationDate="2014-07-01T12:39:15.883" Score="3" Body="&lt;p&gt;Unique rectangles basically says that if you can find a group of four cells that form a rectangle, and such that two of them are in the same 9-block (which implies the other two also will be in the same block) and each have the same pair of numbers that can occur in them, then there can be two possible solutions that fit the Sudoku&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example if 2 and 7 are the numbers that can occur in the cells of the rectangle, then filling the Sudoku with&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    2---7        7---2&#xA;1.  |   |  or 2. |   |&#xA;    7---2        2---7&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;will not make any difference as long as one pair of cells are in one 9-block and the other pair in another 9-block ( That is they are in two 9-blocks)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If each of the cell of the rectangle is in a different 9-block (That is 4 9-blocks in total), then this rule wont apply because interchanging a pattern 1 into pattern 2 would not result in a valid solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where as if pairs of cells are in the same 9-block, then it would still make a valid solution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, applying the unique rectangle rule only makes sense when the four grid cells are spread over two 9-blocks , or simply two blocks. I think this is what the article might have meant by &quot;2 blocks&quot; rule.&lt;/p&gt;&#xA;" OwnerUserId="1725" LastActivityDate="2014-07-01T12:39:15.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1833" PostTypeId="1" AcceptedAnswerId="1834" CreationDate="2014-07-01T20:59:43.773" Score="25" ViewCount="3690" Body="&lt;p&gt;This is a puzzle I remember arguing with my sister about when we were little...&lt;br /&gt;&#xA;(Not quite as loud (or as public!) as the Monty Hall argument we had.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have two large beakers of the same size.&lt;br /&gt;&#xA;Beaker A is half filled with blue sand.&lt;br /&gt;&#xA;Beaker B is half filled with red sand.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You take a cup of blue sand from Beaker A and pour it into Beaker B.&lt;/li&gt;&#xA;&lt;li&gt;You mix up the contents of Beaker B until the colour is consistent.&lt;/li&gt;&#xA;&lt;li&gt;You take a cup of the contents of Beaker B and pour it into A.&lt;/li&gt;&#xA;&lt;li&gt;You mix up the contents of Beaker A until the colour is consistent.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Which is there now more of:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Blue sand in Beaker B or&lt;/li&gt;&#xA;&lt;li&gt;Red sand in Beaker A?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1616" LastActivityDate="2017-01-20T16:45:19.710" Title="How much red sand and how much blue sand?" Tags="|logical-deduction|" AnswerCount="5" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1834" PostTypeId="2" ParentId="1833" CreationDate="2014-07-01T21:33:22.977" Score="26" Body="&lt;p&gt;The answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are the same.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The solution is general:&lt;br&gt;&#xA;1. Beakers can have different size.&lt;br&gt;&#xA;2. You can either mix or not mix.&lt;br&gt;&#xA;3. You can repeat the procedure N times.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You have the same amount of sand in each beaker at the beginning and at the end.&lt;br&gt;&#xA; Therefore if &lt;strong&gt;&lt;em&gt;Beaker A&lt;/strong&gt; lacks X litres of blue sand&lt;/em&gt; then it has &lt;strong&gt;X litres of red sand&lt;/strong&gt; instead. This means &lt;strong&gt;&lt;em&gt;Beaker B&lt;/strong&gt; lacks this X litres of red sand&lt;/em&gt;, they are replaced by &lt;strong&gt;X litres of blue sand&lt;/strong&gt;, which we started our thinking from.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2014-07-02T06:26:45.410" LastActivityDate="2014-07-02T06:26:45.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1835" PostTypeId="2" ParentId="1833" CreationDate="2014-07-02T00:11:08.240" Score="0" Body="&lt;p&gt;A simple scenario to show that you can have equal parts of blue and red sand in both beaker:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the cup size is the same the the beaker size then you are&#xA;essentially combining equal parts of blue and red sand (by taking everything from one beaker and putting it into the other) and then splitting them apart&#xA;equally (by taking half of the evenly mixed sand in one beaker and putting it back in the other beaker).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To show that this holds regardless of the beaker and cup size (unless the cup volume is bigger than a beaker):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: Beaker A and B can hold 18 units of volume, and cup C holds 1 unit of volume, so a half full beaker will hold 9 units of volume.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Beaker A (9 units of blue sand), Beaker B (9 units of red sand)&lt;/li&gt;&#xA;&lt;li&gt;Beaker A (9 units of blue sand + 1 unit of red sand), Beaker B (8 units of red sand)&lt;/li&gt;&#xA;&lt;li&gt;Beaker A ((9 units of blue sand - 0.9 units of blue sand) + (1 unit of red sand - 0.1 unit of red sand), Beaker B (8 units of red sand + 0.1 units of red sand + 0.9 units of blue sand)&lt;/li&gt;&#xA;&lt;li&gt;Beaker A (8.1 units of blue sand, 0.9 units of red sand), Beaker B (8.1 units of red sand, 0.9 units of blue sand)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1730" LastEditorUserId="1730" LastEditDate="2014-07-02T01:16:36.330" LastActivityDate="2014-07-02T01:16:36.330" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1836" PostTypeId="2" ParentId="1833" CreationDate="2014-07-02T02:28:50.433" Score="7" Body="&lt;p&gt;Here's a mathematical proof, as requested by Michael Lai above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is $x$ amount of blue sand in beaker A and $x$ amount of red sand in beaker B. You take $y$ amount of blue sand from beaker A ($y \leq x$) and pour it into beaker B and mix it up. Now beaker A has $x-y$ blue sand in it and beaker B has $x$ red sand in it and $y$ blue sand in it, all mixed up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, beaker B is $\frac x{x+y}$% red sand and $\frac y{x+y}$% blue sand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You now take $y$ sand from beaker B. Because of the percentages above, this sand will contain: $$y*\frac x{x+y}$$ red sand and: $$y*\frac y{x+y}$$ blue sand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we add that to beaker A, it now has: $$y*\frac x{x+y}$$ red sand while beaker B has: $$y-y*\frac y{x+y}$$ blue sand. This second formula simplifies to: $$y*(1-\frac y{x+y})$$&#xA;$$y*(\frac {x+y}{x+y}-\frac y{x+y})$$&#xA;$$y*(\frac {x+y-y}{x+y}$$&#xA;$$y*(\frac x{x+y})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the same as the amount above for the amount of red sand in beaker A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;No Mixing&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the general, no mixing proof, let's take what klm123 says above and apply formal mathematical language to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a total of $x$ blue sand and $x$ red sand. After an unlimited amount of exchanges (with or without mixing), if the two beakers have the same amount of sand in them they have both returned to $x$ sand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beaker A has $a_R$ red sand and $a_B$ blue sand. $a_R + a_B = x$.&#xA;Similarly, Beaker B has $b_R$ red sand and $b_B$ blue sand. $b_R + b_B = x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the first equation above we can get $a_R = x-a_B$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the total amount of each color sand has not changed, we also know that $a_R + b_R = x$ and $a_B + b_B = x$. From the first of these two we get $a_R = x-b_R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $a_R = x-a_B = x-b_R$ then $a_B$ must equal $b_R$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course there's the philosophical question that if we mix the two sands together we no longer have red and blue sand but rather some type of purple sand, but that's silly.&lt;/p&gt;&#xA;" OwnerUserId="1620" LastEditorUserId="1620" LastEditDate="2014-07-02T07:39:10.207" LastActivityDate="2014-07-02T07:39:10.207" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1837" PostTypeId="2" ParentId="313" CreationDate="2014-07-02T10:27:50.587" Score="48" Body="&lt;p&gt;Here is my answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am thinking of an odd number , does your number divide the number I am thinking of?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    &lt;code&gt;1 -&amp;gt; yes&lt;/code&gt;&lt;br&gt;&#xA;    &lt;code&gt;2 -&amp;gt; no&lt;/code&gt;&lt;br&gt;&#xA;    &lt;code&gt;3 -&amp;gt; maybe&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1725" LastEditorUserId="46903" LastEditDate="2018-09-17T01:11:50.033" LastActivityDate="2018-09-17T01:11:50.033" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="1841" PostTypeId="2" ParentId="1690" CreationDate="2014-07-02T19:21:08.020" Score="10" Body="&lt;p&gt;Systematic technique is definitely required. I came across this particular puzzle a few years ago on facebook and got into arguments with some people about the answer, with them arguing that there were &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 41, 42 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or even more squares. Eventually I got bored, produced this animation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/YUw3e.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which shows 40 different squares, and challenged them to show me one that I'd not highlighted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general case would be similar - find the recurring pattern, and count them methodically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to knowing if you've &quot;found them all&quot;, that's just down to being rigorous and methodical in your search. This can get difficult if the puzzle is particularly complex or large, but if the answer is in the realms of &quot;easily countable&quot; as it is in this case then this is easily verified manually.&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="35972" LastEditDate="2017-04-28T18:54:21.090" LastActivityDate="2017-04-28T18:54:21.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1842" PostTypeId="1" AcceptedAnswerId="1846" CreationDate="2014-07-02T19:40:12.517" Score="6" ViewCount="3094" Body="&lt;p&gt;I have a complete set of encyclopaedias on my bookshelf, beautifully bound and neatly arranged with volume 1 on the left and volume 32 on the right. They were a talking point when I bought them 30-odd years ago and were regularly used, but as the internet took off so my use of them declined, preferring to use the online resources which were so conveniently at my fingertips.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it was that the books remained unexamined for a number of years, until I recently moved house, whereupon I discovered an awful fact: an insect larva - specifically, a bookworm (&lt;em&gt;Anobium punctatum&lt;/em&gt;, I've been told) - had found its way into the books and had munched its way through them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The entomologist who identified it for me was curious about it, and took it for study. One question she did ask, though, was just how far the creature had travelled. I looked back at the books and, as I'd previously noted, confirmed that it had started at the first page of volume 1 (evidence suggests that it hatched there, the parent having departed a different way) and had travelled until it had reached the final page of the final volume, which is where I discovered it, so I gave this information to her, along with the measurements of the books, which were all identical:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;each cover is 6 mm thick&lt;/li&gt;&#xA;&lt;li&gt;each book is 40 mm thick in total&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, how far had the bookworm gone?&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorDisplayName="user88" LastEditDate="2015-04-30T15:05:30.950" LastActivityDate="2015-04-30T21:15:20.080" Title="To be a bookworm" Tags="|calculation-puzzle|lateral-thinking|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1843" PostTypeId="2" ParentId="1842" CreationDate="2014-07-02T20:51:07.250" Score="-1" Body="&lt;p&gt;I feel like I'm missing something, but isn't this just $$(6*2)*32+(40-6*2)*32-6-6=1268 \text{ mm} = 1.268 \text{ m}$$&#xA;Which can be reduced to&#xA;$$&#xA;40*32-6-6 = 1268 \text{ mm}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The extra two $-6$ are for skipping the first cover of the first book (started on the first page) and the last cover of the last book (ended on the last page).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see the puzzle here.&lt;/p&gt;&#xA;" OwnerUserId="325" LastActivityDate="2014-07-02T20:51:07.250" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1844" PostTypeId="2" ParentId="219" CreationDate="2014-07-02T20:51:52.333" Score="11" Body="&lt;p&gt;That four colors will suffice may be easily shown by the fact that four-color maps exist.  That three colors will not suffice may readily be shown by pointing out particular places where it is insufficient, but it may be more enlightening to point out a required condition for three-coloring: if a region is encircled by other regions, and at most two of them touch it at any point, then none of those regions may be the same color as the encircled reason, and they must have three different colors unless the number of such regions is either one (in which case one color would obviously suffice for it) or there are an even number (in which case two would suffice).  Many land-locked states other than Colorado, Utah, Arizona, and New Mexico (the only states to meet more than two neighbors at a single point) have an odd number of neighbors; thus, any region of the map containing such a state and its neighbors would require four colors.&lt;/p&gt;&#xA;" OwnerUserId="1467" LastActivityDate="2014-07-02T20:51:52.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1846" PostTypeId="2" ParentId="1842" CreationDate="2014-07-02T22:40:22.330" Score="15" Body="&lt;p&gt;This riddle, like the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/575/which-way-is-the-bus-going/577#577&quot;&gt;question about the direction of a bus&lt;/a&gt;, is actually a trick riddle &amp;ndash; the answer is very simple, but most people won't get it on the first try (and won't understand why their answer is incorrect until you tell them) due to an inadvertent oversimplification on their part. And just like the direction-of-the-bus riddle, this riddle depends on an implied convention.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The intended naïve line of thought is that for the bookworm to have travelled from the first page of volume 1 to the last page of volume 32, that it would have to have chewed clean through all 32 volumes of the encyclopedia, excepting the two covers on the edge of the row of the bookshelf. This gives us a total distance of $40 \times 32 - 6 \times 2 = 1268\ \text{mm}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this line of reasoning is incorrect, because the front cover of a book is on the &lt;em&gt;right&lt;/em&gt; side of its spine. So the bookworm in fact didn't chew through the pages of volumes 1 or 32 at all &amp;ndash; just their covers. This means that the actual distance travelled is $40 \times 30 + 6 \times 2 = 1212\ \text{mm}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;However, even this answer is not necessarily correct. There's no information given about what language the encyclopedia is in. Certain languages like Chinese or Arabic are written from right-to-left, and the books in that language that are written in right-to-left generally have their front covers on the &lt;em&gt;left&lt;/em&gt; side of the spine. Supposing the encyclopedia was written in one of these languages, the initial answer of $1268\ \text{mm}$ would in fact be correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this answer is slightly unsatisfactory because volumes of an encyclopedia written in a right-to-left language are generally &lt;em&gt;also&lt;/em&gt; organized from right to left, so the fact that they were organized left-to-right on this particular shelf would seem somewhat unusual.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2014-07-02T22:40:22.330" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1847" PostTypeId="1" AcceptedAnswerId="1849" CreationDate="2014-07-02T22:47:34.727" Score="9" ViewCount="533" Body="&lt;p&gt;This was a puzzle that appeared in an old math contest as well as an unrelated puzzle book.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose we have a two-dimensional map with a lot of cities marked on it. For each city on the map, we draw a line segment to the city nearest to it (we assume that there is always a nearest city and that no two cities are exactly the same distance apart from any third city). Prove that no city will have more than five line segments connected to it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I seem to remember that the proof involved triangles, but to this day don't know the whole proof.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2014-07-03T06:56:40.280" Title="Drawing lines on a map" Tags="|mathematics|geometry|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1848" PostTypeId="2" ParentId="1847" CreationDate="2014-07-03T00:03:57.287" Score="3" Body="&lt;p&gt;I'm not sure this is rigorous per se, but I think it's accurate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider some number cities arranged in a circle of radius $r$ around a central city (the &quot;most connected&quot;).  The furthest these can be from each other is when they are distributed equally. When there are six cities around the circle, they are all at a distance $r$ from each other. Since each of the cities has exactly one closest city, there cannot be six cities around the circle. If there were more, they would be closer to each other than the central one, so there are strictly less than six cities closest to a single one; since we're working in integers, that means at most five.&lt;/p&gt;&#xA;" OwnerUserId="4" LastActivityDate="2014-07-03T00:03:57.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1849" PostTypeId="2" ParentId="1847" CreationDate="2014-07-03T00:11:09.523" Score="8" Body="&lt;p&gt;I'll assume the cities lie on a flat plane, which seems to be the most reasonable interpretation of the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume (for contradiction) that we had a city which is connected to at least 6 other cities. First, note that this implies that no two cities lie on the same ray extending outwards from the central city which I'll label as $0$. Picking the closest of these six cities to be $1$ and label the cities counterclockwise starting from the ray $\vec{01}$ as $2,3, \ldots, 6$. Now $0$ and $1$ are connected since $1$ is closest to $0$. For cities $2, \ldots, 6$, the only way they can be connected to $0$ is by being closer to $0$ than they are to any other city. Here's a conceptual diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2VNc0.png&quot; alt=&quot;Diagram of cities&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, let's deal with the two triangles $\Delta012$ and $\Delta 061$ which have city 1 as vertices. By construction, $d(0,1) &amp;lt; d(0,2)$, where $d(i,j)$ is the distance from $i$ to $j$. $1$ already has a segment connecting it to $0$ since it's the closest city to $0$. But for $2$, since it isn't the closest city to $0$, we need $0$ to be the closest city to it. So $d(0,2) &amp;lt; d(1,2)$. By transitivity, $d(0,1) &amp;lt; d(1,2)$, so the longest edge in $\Delta 012$ is $\overline{12}$. The same logic shows that the longest edge in $\Delta 061$ is $\overline{16}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for $\Delta 023$, we know that $d(0,2) &amp;lt; d(2,3)$ and similarly $d(0,3) &amp;lt; d(2,3)$, both since $2$ and $3$ each have $0$ as their closest city. Hence, the longest edge of the triangle $\Delta 023$ is $\overline{23}$. We apply the same to $\Delta 034, \Delta 045, \Delta 056$ to show that the side opposite the angle at city $0$ is always the longest. This is thus true for all 6 triangles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Law_of_sines&quot; rel=&quot;nofollow noreferrer&quot;&gt;law of sines&lt;/a&gt; tells us that either the largest side is opposite the largest angle, or else the &quot;triangle&quot; has a reflex angle (an angle greater than $180 ^\circ$). It's possible that $\angle 102$ is a reflex angle, but by construction it's not possible for either of the other angles to be a reflex angle, because we chose them to lie on &lt;em&gt;rays&lt;/em&gt; going outwards from $0$, and there's no way to create a reflex angle between the two rays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So either $\angle 102$ is the largest angle in the triangle, making it more than $60 ^\circ$, or it's more than $180 ^\circ$, which would of course still make it more than $60 ^\circ$. We can apply the same logic for $\Delta 023, \Delta034, \ldots, \Delta061$ and find that the central city has 6 angles each of over $60 ^\circ$, which means that the total angle is more than $360 ^\circ$. That's a contradiction since we know it has to be exactly $360 ^\circ$ by construction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, it's impossible for any city to be connected to 6 other cities this way.&lt;/p&gt;&#xA;" OwnerUserId="94" LastEditorUserId="94" LastEditDate="2014-07-03T06:56:40.280" LastActivityDate="2014-07-03T06:56:40.280" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="1850" PostTypeId="1" AcceptedAnswerId="1854" CreationDate="2014-07-03T05:36:38.023" Score="11" ViewCount="1698" Body="&lt;h3&gt;Background Story&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;A nomad is injured in the desert. And he can't move, but fortunately an oasis is nearby, and he can stay there for enough time until rescue arrives. But the water is a magical water, that it can't be stored. So it should be drunk immediately or it'll disappear. In other words, one cannot bring the water outside the oasis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other nomad friends, know the exact position of the injured nomad, which is $n$ days away from the base. The nomads are going to send delegations to save him, but the problem is that each nomad can only bring food supply for one nomad for $3$ days. The base has arbitrarily large number of nomads inside, so they decided to send a group of nomads to rescue the injured nomad, that is, to go to the place of injured nomad, and then go back to the base. All nomads must survive. But they have weird cultural rules, which is each time a nomad leaves the base, they have to kill a sheep as redemption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the injured nomad also need to eat while going back to the base.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Question&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;What is the number of sheep need to be slaughtered in order to save the injured nomad in case n=2? n=3? arbitrary n?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, what is the &lt;strong&gt;minimum&lt;/strong&gt; number of days required to save the nomad?&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Example&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;For example, if n=1, the nomad group can just send one nomad. It'll consume 1 day of supply, then arrives at the injured nomad place with 2 days of supply left, then each of them consume 1 day of supply while going back to the base, leaving no supply left when they arrived at the base. Since only one nomad leaves the base, only one sheep is killed.&#xA;So 1 sheep will be slaughtered, and the rescue time is 2 days.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Notes&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;This is generalisation to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/233/travellers-across-a-desert&quot;&gt;Travellers across a desert&lt;/a&gt; with a twist that everybody should come back to the base&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1649" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2014-07-04T06:39:48.553" Title="Stranded Nomad Riddle" Tags="|calculation-puzzle|word-problem|" AnswerCount="3" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="1851" PostTypeId="2" ParentId="1850" CreationDate="2014-07-03T07:00:38.747" Score="5" Body="&lt;p&gt;While I don't have a solution to the whole problem just yet, here's a solution for &lt;span class=&quot;math-container&quot;&gt;$n = 2$&lt;/span&gt; in 4 days:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;T+0 days: a single nomad (1) leaves and travels 1 day forward.&lt;/p&gt;&#xA;&lt;p&gt;T+1 day: a single nomad (2) leaves, and both nomads travel one day forward. Nomad 1 reaches the oasis, with one day of supplies left, which he shares with the stranded nomad (nomad 0).&lt;/p&gt;&#xA;&lt;p&gt;T+2 days: a single nomad (3) leaves, and nomads 0 and 1 go back for 1/2 day, running out of supplies. Nomad 2 meets them at 1.5 days out, with 1.5 days of supplies left, which he shares with the two other nomads.&lt;/p&gt;&#xA;&lt;p&gt;T+2.5 days: The group of three nomads go back for 1/2 day, meeting nomad 3 at 1 day out, with 2.0 days of supplies left, which he shares with the three other nomads.&lt;/p&gt;&#xA;&lt;p&gt;T+3.0 days: a single nomad (4) leaves, and the group of four nomads go back for 1/2 day, meeting nomad 4 at 0.5 day out, with 2.5 days of supplies left, which he shares with the four other nomads.&lt;/p&gt;&#xA;&lt;p&gt;T+3.5 days: The group of five nomads head back to base in 1/2 day with just enough supplies to make it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This solution is rather elegant, because it only requires four nomads to set out at T + 0, 1, 2, and 3 days (and thus four sheep are slaughtered), and all the resources are used up exactly when they're needed (every time the nomads run out of supplies, a nomad is there to provide just enough for yet another nomad to meet them in the middle). I'd be surprised if a better solution existed.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-07-03T07:00:38.747" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1852" PostTypeId="2" ParentId="1850" CreationDate="2014-07-03T11:08:01.553" Score="3" Body="&lt;p&gt;I have this answer for $n$, but in the end I lack the mathematical skills to come to a single formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general principle will be like this: You send a big group into the dessert. After each day, half of the nomads in the (remaining) group will give food to the other half and travel back one day. The nomads that are fully equiped travel one day further. This means that when the oasis is $n$ days away, the original group needs to be $2^{n-1}$ to make sure that one nomad arrives in the oasis with $2$ food left needed to take himself and the injured nomad to travel back one day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now all we need to do is send groups to resupply the parties that were starting their return. This can be seen as more of the same rescue operations. There are $2$ nomads at $n-1$ that need to be rescued, and furthermore, $1$ at $n-2$, 2 at $n-3$, $4$ at $n-4$ and so on. This can be written as $2^x$ people need to be rescued at $n-2-x$ for every $x$ from $0$ to $n-3$. It is $n-3$ since the last people that need rescuing will be at location $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that if $f(n)$ is the total amount of people that need to leave the base to rescue one person at $n$ then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(n) = 2^{n-1} + 2\cdot f(n-1) + \sum_0^{n-3} 2^x \cdot f(n-2-x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not able to solve the above formula, but I hope the general idea is clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the amount of time is $2n$, which can easily be seen since the ultimate rescuer travels there in one go and returns in one go as well.&lt;/p&gt;&#xA;" OwnerUserId="1745" LastEditorUserId="1649" LastEditDate="2014-07-04T06:39:48.553" LastActivityDate="2014-07-04T06:39:48.553" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1854" PostTypeId="2" ParentId="1850" CreationDate="2014-07-03T14:23:14.307" Score="7" Body="&lt;p&gt;Using Joe Z.'s method for &lt;span class=&quot;math-container&quot;&gt;$n=2$&lt;/span&gt;, I can think of a strategy for &lt;span class=&quot;math-container&quot;&gt;$n=3$&lt;/span&gt; in &lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt; days killing &lt;span class=&quot;math-container&quot;&gt;$13$&lt;/span&gt; sheep. Pretty sure this is the minimum amount of days because it's &lt;span class=&quot;math-container&quot;&gt;$2n$&lt;/span&gt;. Not sure if it is the minimum amount of sheep.&lt;/p&gt;&#xA;&lt;p&gt;I won't do a day by day guide like Joe but my explanation hopefully is clear.&lt;/p&gt;&#xA;&lt;p&gt;First start off with &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; nomad's.&#xA;After &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; day one nomad gives a day supply to the other and heads back.&#xA;So, the one returning has just enough to get back and the other has a full three-day supply.&lt;/p&gt;&#xA;&lt;p&gt;From his point of view it now is only an &lt;span class=&quot;math-container&quot;&gt;$n=2$&lt;/span&gt; problem because he is now at distance &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; with full supply.&lt;/p&gt;&#xA;&lt;p&gt;So use the &lt;span class=&quot;math-container&quot;&gt;$n=2$&lt;/span&gt; strategy here. You can do this by sending an extra nomad for each nomad that needs to be at &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; day travel with full supplies.&lt;/p&gt;&#xA;&lt;p&gt;The &lt;span class=&quot;math-container&quot;&gt;$n=2$&lt;/span&gt; strategy eventually also returns all these nomad's at this point with no food left so for each of them they need an extra nomad to supply them.&lt;/p&gt;&#xA;&lt;p&gt;So to sum it up: start sending &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; nomads every day for &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt; days. half of them return. the other half does the &lt;span class=&quot;math-container&quot;&gt;$n=2$&lt;/span&gt; strategy. One day before they are finished send &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt; nomads away so they meet up with the other five when they are finished with their &lt;span class=&quot;math-container&quot;&gt;$n=2$&lt;/span&gt; strategy. The now &lt;span class=&quot;math-container&quot;&gt;$10$&lt;/span&gt; nomads have exactly enough to return. counting the &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt; that returned in the beginning and not counting the stranded nomad, that's &lt;span class=&quot;math-container&quot;&gt;$13$&lt;/span&gt; sheep that were killed.&lt;/p&gt;&#xA;&lt;p&gt;ADDITION AND CONCLUSION:&lt;/p&gt;&#xA;&lt;p&gt;you can generalize strategy this by doing same thing for other &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;. You will have then:&#xA;sheep killed for &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; is three times the [number of sheep killed for strategy &lt;span class=&quot;math-container&quot;&gt;$n-1$&lt;/span&gt;] plus one. And days to take is always &lt;span class=&quot;math-container&quot;&gt;$2n$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Given this, this strategy is equally good as Joe's with going to &lt;span class=&quot;math-container&quot;&gt;$n=1$&lt;/span&gt; one to &lt;span class=&quot;math-container&quot;&gt;$n=2$&lt;/span&gt;. because &lt;span class=&quot;math-container&quot;&gt;$3*1+1 = 4$&lt;/span&gt; sheep.&lt;/p&gt;&#xA;&lt;p&gt;And &lt;span class=&quot;math-container&quot;&gt;$n=2$&lt;/span&gt; worked out to be:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;D0: 2 nomads leave (N1 and N2)&lt;/p&gt;&#xA;&lt;p&gt;D1: N1 gives 1 supply to N2 and heads back&lt;/p&gt;&#xA;&lt;p&gt;D2: N2 arrives at N0 and gives him 1 supply and start heading back. N3 and N4 leave.&lt;/p&gt;&#xA;&lt;p&gt;D3 N3 and N4 meat up with N0 and N1 and they give them each a supply.&lt;/p&gt;&#xA;&lt;p&gt;D4: they all arrive back with their supply used up.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So wrapping it up. If &lt;span class=&quot;math-container&quot;&gt;$F(n)$&lt;/span&gt; is the amount of sheep necessary or a nomad stranded at &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; days distance.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$F(1) = 1$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$F(n) = 3*F(n-1) + 1$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;And days it takes is always &lt;span class=&quot;math-container&quot;&gt;$2n$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;I'm not that good in math but I know there is a simple way to convert this recursive formula to a direct one.&lt;/p&gt;&#xA;" OwnerUserId="1748" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-07-04T05:36:55.897" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1855" PostTypeId="1" CreationDate="2014-07-03T14:26:12.710" Score="14" ViewCount="2741" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose you are allowed to use all 10 digits (0,1,2,...9) at most once each; 4 arithmetic operations ($-$,$+$,$\times$,$\div$), each any number of times; parenthesises to group operations; and you can create numbers from digits by writing them together.&lt;br&gt;&#xA;  What is the smallest natural number which you would not be able to write?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example, you can write:&lt;br&gt;&#xA;$135 = 12*3*(9+6)/4$&lt;br&gt;&#xA;and you can't write:&lt;br&gt;&#xA;$11 = 11$&lt;br&gt;&#xA;$3 = 2+3/3$&lt;br&gt;&#xA;$27 = 3+4!$&lt;br&gt;&#xA;$81 = 3^4$&lt;br&gt;&#xA;$1 = 5/3$&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="4551" LastEditDate="2015-06-22T22:59:56.787" LastActivityDate="2015-06-22T22:59:56.787" Title="What is the smallest positive integer, which can not be written without repetitions of digits and using arithmetics only?" Tags="|mathematics|formation-of-numbers|" AnswerCount="5" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="1856" PostTypeId="1" AcceptedAnswerId="1858" CreationDate="2014-07-03T20:57:19.807" Score="33" ViewCount="70265" Body="&lt;p&gt;I saw the following problem on 4chan and couldn't solve it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dnglT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's very likely to be some kind of troll (no solution).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm hoping to see some rigorous proofs that disprove the existence of such a line.&lt;/p&gt;&#xA;" OwnerUserId="1636" LastEditorUserId="5373" LastEditDate="2015-05-19T09:55:03.527" LastActivityDate="2019-02-19T13:40:29.203" Title="Draw a line through all doors" Tags="|combinatorics|graph-theory|" AnswerCount="18" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1857" PostTypeId="2" ParentId="1856" CreationDate="2014-07-03T21:35:21.727" Score="20" Body="&lt;p&gt;You can draw a graph with six vertices, one for each room and one for the outside.  Draw an edge to represent each door.  The puzzle asks for an &lt;a href=&quot;http://en.wikipedia.org/wiki/Eulerian_path&quot; rel=&quot;noreferrer&quot;&gt;Eulerian path&lt;/a&gt;, which can be done if no more than two vertices have an odd number of edges coming in.  In this graph the top two rooms, the middle bottom room, and the outside all have an odd number of edges coming in, so in the usual way there is no such path.  A sketch is below.  ABCDE are the rooms in the same locations as in the picture, F is outside, and the lines are connections through doors. ABD and F all have an odd number of edges coming in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3UztT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditDate="2014-07-03T21:40:56.937" LastActivityDate="2014-07-03T21:40:56.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1858" PostTypeId="2" ParentId="1856" CreationDate="2014-07-03T21:38:37.493" Score="42" Body="&lt;p&gt;It is impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quite the same problem is &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Seven_Bridges_of_K%C3%B6nigsberg&quot;&gt;Seven Bridges of Königsberg&lt;/a&gt;&quot;, it was solved (proven) by Euler.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Suppose you have drawn such a line and follow it from one room to another. Since you must use each door you must have a look at each room out of 5. What are these rooms?  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There will be 3 (at least) rooms you always go through - if you enter it you always exit it later.&lt;br&gt;&#xA;Indeed, 1 (at most) room you can start at, and another 1 (at most) room you can finish at, but others you must go through: $5-1-1 = 3$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Since you use each door exactly once, the mentioned 3 rooms must have an even number of doors, since you entry them the same number of times you exit them. But you have only 2 rooms with an even number of doors, the others have 5 doors. So you could not draw such a line.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2014-07-26T17:56:47.057" LastActivityDate="2014-07-26T17:56:47.057" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="1859" PostTypeId="2" ParentId="1856" CreationDate="2014-07-03T21:47:21.370" Score="35" Body="&lt;p&gt;&lt;em&gt;Just for fun:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, there is solution, which formally satisfies all the rules. You just need to walk through a wall!&lt;br&gt;&#xA;Hard, but possible! &lt;img src=&quot;https://i.stack.imgur.com/gpEsn.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="28" LastActivityDate="2014-07-03T21:47:21.370" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1860" PostTypeId="2" ParentId="296" CreationDate="2014-07-03T21:58:18.487" Score="-1" Body="&lt;p&gt;There is a standard reduction to/from 3SAT, so this problem is NP-complete. So the answer to your question is &quot;Yes, arbitrarily complex puzzles are possible.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2014-07-03T21:58:18.487" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1861" PostTypeId="2" ParentId="1855" CreationDate="2014-07-04T06:42:20.517" Score="0" Body="&lt;p&gt;I'm trying to approach the task solving simpler tasks with $N&amp;lt;9$ digits allowed.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;N = 3: 0,1,2.&lt;br&gt;&#xA;The Answer would be $A=4$.&lt;br&gt;&#xA;Obviously, you can't reach $(N-1)N/2+1 = 4$ if you use only $-,\div,+$ operations. $\times$ operation also would help here.&lt;br&gt;&#xA;So you need to use grouping operation. But then you will get a number $R \ge 10$ and the only way to decrease it drastically to $4$ is to use division. But $10/2 = 5$ - the division is not enough too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to check that you can make any number $\le4$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;N = 4: 0,1,2,3.&lt;br&gt;&#xA;The Answer is $A=25$.&lt;br&gt;&#xA;Similarly you can't reach $7$ if you use only $-,\div,+$ operations. The maximum you can get adding $\times$ operation is (2+1)*3 = 9 (you have to multiply maximum possible numbers to reach the maximum).&lt;br&gt;&#xA;So you need to use grouping operation.&lt;br&gt;&#xA;If you group 1 and other digit, you would need to use multiplication to reach 25, but 10*2 =20 and 10*3 = 30 leaves us with 3 or 2 to finish... 12*3 = 36 and 13*2 = 26 lives us with 0... so you can do nothing here.&lt;br&gt;&#xA;If you group 2 and other digit, for example, 20 you would need to add to it something and can't reach 25 with additional 1 and 3. The same for 21,23.&lt;br&gt;&#xA;If you group 3 and other digit, like 30, you would need to subtract from it something, similarly you can't reach 25.&lt;br&gt;&#xA;If you group three digits, then you would get $R&amp;gt;100$ and you would need to use division, but $100/3 &amp;gt; 25$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can check that you can make any number $\le24$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This really looks like there is no simple pattern, the $A$ is rapidly increases with $N$, the same for number of possible ways to achieve the result.&lt;br&gt;&#xA;The only patter I see is that $A &amp;lt; 10^{N-2}/(N-1)$, possibly this can be proven. And $10^{N-3} &amp;lt; A$, which is most probably will fail for big $N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most probably there is no pattern for $A$ at all, and one would need to brute force it for big $N$.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2014-07-04T06:48:11.763" LastActivityDate="2014-07-04T06:48:11.763" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1862" PostTypeId="2" ParentId="1856" CreationDate="2014-07-04T08:31:57.847" Score="9" Body="&lt;p&gt;It is actually impossible if you consider walking through walls impossible as stated in the other answers. That can be explained by graphs or in another  similar, but yet easier to understand way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ignoring the connections between the rooms, you have 5 rooms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 rooms with 5 doors and 2 rooms with 4 doors.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;if a room has an &lt;strong&gt;even&lt;/strong&gt; amount of doors, there are two ways it can be part of the solution: either the line &lt;strong&gt;starts and ends&lt;/strong&gt; within that room and all other doors are used to both enter and leave the room or the line passes through (amount of doors / 2) times.&lt;/li&gt;&#xA;&lt;li&gt;if a room has an &lt;strong&gt;uneven&lt;/strong&gt; amount of doors, there are another two ways it can be part of the solution: either the line &lt;strong&gt;starts&lt;/strong&gt; within that room and all the other doors are used to both enter and leave or the line &lt;strong&gt;ends&lt;/strong&gt; within that room and all the other doors are used to both enter and leave.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This gives us the insight that there are 3 rooms that &lt;strong&gt;need&lt;/strong&gt; to contain either the start or the end of the line and there are 2 rooms that &lt;strong&gt;might&lt;/strong&gt; contain both start and end. Following that logic, we have at least 2 of either start or end of the line and thus we can't use just a single line to solve this problem.&lt;/p&gt;&#xA;" OwnerUserId="1756" LastActivityDate="2014-07-04T08:31:57.847" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1863" PostTypeId="2" ParentId="1855" CreationDate="2014-07-04T10:45:38.650" Score="2" Body="&lt;p&gt;Here is my attempt at finding an upper bound:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have 10 atomic values ($0..9$) and five operations ($+$, $-$, $*$, $/$, $○$; according to rules concatenation may not be used on results of other operators, let's ignore this for now)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since all operations are binary, we need 9 operators to get a single result out of 10 values. &lt;a href=&quot;http://en.wikipedia.org/wiki/Catalan_numbers#Applications_in_combinatorics&quot; rel=&quot;nofollow&quot;&gt;There are 4862 binary trees with nine nodes&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are $5^9 = 1953125$ ways to choose operators for a tree and $10! = 3628800$ ways to arrange the digit leaves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the number must be smaller than $3628800*1953125*4862 = 34459425000000000$. Hey, this still fits into a signed 64-bit integer, so good look to the brute forcers :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to reduce the upper limit. The following rules were not considered by my approximation&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Concatenation may not be used on results of other operators&lt;/li&gt;&#xA;&lt;li&gt;$+$ and $*$ are commutative &lt;/li&gt;&#xA;&lt;li&gt;not all combinations result in integers&lt;/li&gt;&#xA;&lt;li&gt;some integers can be computed in more than one way&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1750" LastEditorUserId="1672" LastEditDate="2014-07-06T00:04:26.153" LastActivityDate="2014-07-06T00:04:26.153" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1864" PostTypeId="2" ParentId="1856" CreationDate="2014-07-04T14:38:12.067" Score="3" Body="&lt;p&gt;Even assuming that it is not allowed to draw the line through a &quot;wall&quot;, there is another kind of troll solution. You are forbidden to go through the same door twice, but you can go through that door four times. Then it is easy to do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you could maintain that there are no door at all in the drawing, just $12$ disconnected pieces of &quot;wall&quot;. This makes it a non-problem.&lt;/p&gt;&#xA;" OwnerUserId="1521" LastActivityDate="2014-07-04T14:38:12.067" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1866" PostTypeId="2" ParentId="1855" CreationDate="2014-07-04T16:57:24.590" Score="1" Body="&lt;p&gt;In the interest of crowd-sourcing the brute forcing, I've made a google spreadsheet with the list of the 4765 numbers from 1 to 9999 which can't be formed by simple concatenation.  Please pick a couple of them and work out solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://docs.google.com/spreadsheets/d/1oCv8zImAWJ4k-VOKRlVWQp5d58zA_C1BcJ_X7dfAwxY/edit?usp=sharing&quot; rel=&quot;nofollow&quot;&gt;Spreadsheet&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we finish this set with solutions for all numbers, I'll add more.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't &lt;em&gt;really&lt;/em&gt; expect it to be in this set, but we've got to start somewhere...&lt;/p&gt;&#xA;" OwnerUserId="325" LastActivityDate="2014-07-04T16:57:24.590" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1867" PostTypeId="1" CreationDate="2014-07-05T03:38:34.710" Score="6" ViewCount="819" Body="&lt;p&gt;There is a puzzle which I remember from many years ago which can be summarised as&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is easy to arrange 20 items in five rows of four items, but how could you arrange just 10 items in five rows of four items?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is this (spoiler-protected for those of you who want to try to solve it first):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/Pujkw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is all very clever, but what is the more general case of this lateral thinking puzzle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[edit]&lt;/strong&gt;: To clarify what I mean by &quot;general case&quot;, I'm not so much asking &lt;em&gt;whether&lt;/em&gt; there's a general case, but how to describe it, what values are possible, and how to find them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, for an $n$-sided polygon (n&gt;2), each vertex is in two rows of $2$ items, so the &lt;em&gt;&quot;trivial&quot;&lt;/em&gt; general case is that $n$ items can be arranged in $n$ rows of $2$ if you only count the edges. However, for n&gt;3, each vertex can be joined in $n-1$ rows of $2$ items, producing $\frac{1}{2}n(n-1)$ such rows in total.&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditDate="2014-07-05T22:59:27.340" LastActivityDate="2014-07-05T22:59:27.340" Title="What's the more general case of this puzzler?" Tags="|lateral-thinking|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1868" PostTypeId="2" ParentId="1856" CreationDate="2014-07-05T04:15:57.527" Score="24" Body="&lt;p&gt;Similar to klm's solution, this one requires a little bit of lateral thinking but doesn't actually involve walking &lt;em&gt;through&lt;/em&gt; a wall. Instead, you have to fold the corner of the piece of paper that the picture is drawn on, to form a bridge &lt;em&gt;over&lt;/em&gt; a wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mZXPu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2014-07-05T04:15:57.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1869" PostTypeId="2" ParentId="1867" CreationDate="2014-07-05T05:28:19.817" Score="9" Body="&lt;p&gt;Generally, for any two numbers $(n, k)$, where you want to arrange $n$ coins into rows of $k$, there is some maximum number $m$ of rows that you can arrange them into. We can define this to be $m = f(n, k).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $(10, 4)$, you showed that $m = 5$ in your question statement.  Another famous case is $(9, 3)$, in which case the answer is $m = 10$, as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9GBNp.gif&quot; alt=&quot;arranging nine coins into 10 rows of three&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way you could formulate the problem is to find the minimum number $n$ of coins it takes to form $m$ rows of $k$ coins. Then, we define $g(m, k)$ to equal $n$, and as above, we see that $g(5, 4) = 10$ and $g(10, 3) = 9$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In each of the above cases, the problem is generalizable in a different fashion, but the two examples given (of the 10 coins in 5 rows of 4 and the 9 coins in 10 rows of 3) are just some of the most well-known ones, probably because they give the most beautiful or symmetric patterns.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2014-07-05T05:28:19.817" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1870" PostTypeId="1" AcceptedAnswerId="1883" CreationDate="2014-07-05T17:37:54.683" Score="6" ViewCount="8784" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;The solution for this test is 1. But what is the pattern?&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QCBmB.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;P.S. I checked that 1 is the answer by simply trying different options in the test itself (Russian site for the test is here: &lt;a href=&quot;http://sly2m.com/sly2m/IQ&quot; rel=&quot;noreferrer&quot;&gt;http://sly2m.com/sly2m/IQ&lt;/a&gt; )&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2017-03-24T15:16:17.680" Title="What is the pattern in this IQ test puzzle with spotted triangles and arrowed circles?" Tags="|pattern|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1871" PostTypeId="2" ParentId="1870" CreationDate="2014-07-05T18:02:50.620" Score="3" Body="&lt;p&gt;I think that the arrow at the top of the circle, pointing to the bottom, refers to the semi-circle at the right side of the triangle, and the arrow at the bottom of the circle (pointing to the top), refers to the semi-circle at the left side of the triangle. So, the arrow at the left of the circle (pointing to the right) refers to the part of the circle at the top of the triangle, and then the arrow at the right side of the circle (pointing to the left) refers to the semi-circle at the bottom of the triangle, which gives #1 as solution.&lt;/p&gt;&#xA;" OwnerUserId="153" LastActivityDate="2014-07-05T18:02:50.620" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1872" PostTypeId="1" AcceptedAnswerId="27486" CreationDate="2014-07-05T18:16:58.633" Score="3" ViewCount="15157" Body="&lt;p&gt;What is the solution for this puzzle from this &lt;a href=&quot;http://iqtest.dk/main.swf&quot; rel=&quot;nofollow&quot;&gt;IQ test&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://www.braingames.ru/forum/uploads/post-13114-1262532170.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="28" LastEditorUserId="4" LastEditDate="2014-07-05T19:08:38.863" LastActivityDate="2017-07-19T11:41:22.260" Title="Task 39 from iqtest.dk" Tags="|pattern|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="1873" PostTypeId="2" ParentId="1872" CreationDate="2014-07-05T18:16:58.633" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The answer is &lt;strong&gt;B&lt;/strong&gt;.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Pictures must be considered consecutively from left to right, from top to bottom.&lt;br&gt;&#xA; Each picture is obtained from previous by several transformations:&lt;br&gt;&#xA;&lt;br&gt;&#xA; 1. Triangles are changed to crosses. Crosses -&gt; circles. Circles -&gt; triangles.&lt;br&gt;&#xA; 2. 1-st and 2-nd columns are shifted right (become 2-nd and 3-rd columns).&lt;br&gt;&#xA; 3. 3-rd column become 1-st column and circle-shifted down (1-st row become 2-nd, 2 -&gt; 3, 3 -&gt; 1).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="28" LastActivityDate="2014-07-05T18:16:58.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1874" PostTypeId="1" CreationDate="2014-07-05T20:44:38.690" Score="11" ViewCount="3192" Body="&lt;p&gt;I was doing a Sudoku puzzle (in my morning newspaper on the train to work) and got stuck at this point:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;_ _ _   _ 4 _   7 2 1 &#xA;4 _ 2   6 _ 7   _ 8 9&#xA;7 _ _   8 _ 2   _ _ _&#xA;&#xA;2 9 5   4 6 8   _ _ 7&#xA;3 4 6   5 7 1   8 9 2&#xA;8 _ _   _ _ 9   6 4 5&#xA;&#xA;_ 2 _   7 _ 4   _ _ 3&#xA;5 _ _   1 _ 6   2 7 _&#xA;1 _ _   _ _ _   _ _ _&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Having made no further progress by the time I got home, I plugged the numbers into HoDoKu for help, and got told that there was an X-Wing. I looked at the hints but couldn't find it, so eventually gave up and got it to tell me which number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's 9:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pUxok.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I've had it pointed out to me I can see it, but &lt;strong&gt;what strategies are there for &quot;seeing&quot; it?&lt;/strong&gt; Bear in mind I want to be able to do this on the train using paper &amp;amp; pen, without the aid of a computer.&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="5511" LastEditDate="2016-04-30T16:43:57.653" LastActivityDate="2018-11-22T18:36:39.693" Title="How can I spot an X-Wing?" Tags="|sudoku|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1875" PostTypeId="2" ParentId="1870" CreationDate="2014-07-06T00:10:02.003" Score="-1" Body="&lt;p&gt;Here's my opinion...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a container shape (triangle or circle), and then an object within it.&#xA;Treat the container object, separately from its content.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, imagine a line of symmetry on the first image (ie a mirror showing the content and its reflection)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second image shows the container after the content has been rotated 90 degrees, and the mirror removed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third image shows the reflection of the content&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is 1 because it is showing just the reflection of the circle, within the container triangle. &lt;/p&gt;&#xA;" OwnerUserId="1771" LastActivityDate="2014-07-06T00:10:02.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1876" PostTypeId="2" ParentId="1856" CreationDate="2014-07-06T05:07:31.123" Score="49" Body="&lt;p&gt;It works when you use a really wide line:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IYG5m.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1774" LastActivityDate="2014-07-06T05:07:31.123" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1877" PostTypeId="2" ParentId="1856" CreationDate="2014-07-06T08:27:12.547" Score="2" Body="&lt;p&gt;This &quot;puzzle&quot; can be described as an &lt;em&gt;Euler Trail&lt;/em&gt;, and was discussed in a video by James Grime in which he proves why the puzzle is impossible:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=CQNH2ifOAQE&quot; rel=&quot;nofollow&quot;&gt;YouTube: A Maths Puzzle: Euler Trail and Solution &lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The gist of it is, every time you enter a room, you have to leave it as well, which means all doors come in pairs (the door you entered the room with, and the door you left the room with). The only time you can have an odd number of doors is in &lt;strong&gt;two cases&lt;/strong&gt;: if you start the path in that room, or end in that room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look at the image, two of the rooms have 4 doors (an even number), while three of the rooms have 5 doors (an odd number). This means if we start in the top left room, make our path, then end in the top right room, the bottom center room will have been walked through an &lt;strong&gt;even&lt;/strong&gt; amount of times, meaning one of the doors will have remained unused and inaccessible. Therefore, it is indeed a &quot;troll puzzle&quot;.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditDate="2014-07-07T14:21:23.883" LastActivityDate="2014-07-07T14:21:23.883" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1880" PostTypeId="2" ParentId="1856" CreationDate="2014-07-06T12:42:34.103" Score="19" Body="&lt;p&gt;Alternatively, going &quot;through&quot; a door need not necessarily be interpreted in the same way as one would in a house.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;strong&gt;single, continuous line passes exactly once through each door&lt;/strong&gt;, which is the constraint in the original question. Of course, it also conveniently side-steps making any other assumptions and takes certain liberties with the walls. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kfk1F.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditDate="2014-07-07T22:41:11.027" LastActivityDate="2014-07-07T22:41:11.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1881" PostTypeId="2" ParentId="1856" CreationDate="2014-07-06T14:06:40.930" Score="6" Body="&lt;p&gt;As far as I can tell there are no doors in this image, therefore any line will cross every doors and no line will cross a door twice: just draw a line anywhere. Actually you might have already draw some lines of length 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: some people seem to think this is a troll, maybe they simply don't know what &lt;a href=&quot;http://en.wikipedia.org/wiki/Universal_quantification&quot;&gt;universal&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Existential_quantification&quot;&gt;existential quantification&lt;/a&gt; mean, or they disagree with the fact that there are no door on the image ?&lt;/p&gt;&#xA;" OwnerUserId="1779" LastEditorUserId="1779" LastEditDate="2014-07-07T15:44:39.117" LastActivityDate="2014-07-07T15:44:39.117" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1882" PostTypeId="2" ParentId="1855" CreationDate="2014-07-06T14:46:40.380" Score="8" Body="&lt;p&gt;Concatenation is the &quot;most efficient&quot; among the allowed operations, which is exmplified by this observation: If $x$ is a nonzero rational number obtained according to the rules using the $k$ digits $d_1&amp;gt;d_2&amp;gt;\ldots &amp;gt;d_k$, then &#xA;$$\tag1 \max\left\{|x|,\tfrac1{|x|}\right\}\le 10^k\cdot 0.d_1d_2\ldots d_k$$&#xA;This is clear for direct concatenations and otherwise follows by induction and using $(1)$ for the summands, factors, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can strengthen this: If the last operation is $\times$ or $\div$, then &#xA;$$\tag2\max\left\{|x|,\tfrac1{|x|}\right\}&amp;lt;10^k\cdot 0.85598232 $$&#xA;This follows because not both parts can use the digit $9$, hence the largest digit in on subterm is either $8$, leading to a bounding factor $0.97654321\cdot 0.87654321$; or the largest digit in one factor is $\le 7$, leading to a bounding factor $0.987654321\cdot 0.7654321$. If one takes these case distinctions a bit further, one readily finds that &#xA;$$\tag{2'} \max\left\{|x|,\tfrac1{|x|}\right\}\le10^k\cdot 0.843973902$$&#xA;if the last operation is $\times$ or $\div$ (with the extreme given by $9642\cdot 875310$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, if the last operation is $+$ or $-$, then &#xA;$$\tag3|x|\le (10^{k-1}+1)\cdot 0.987654321 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We thus are led to believe that $987654323$ is likely not expressible: Because of two difgits $3$ it cannot be obtained from concatenation; because of $(3)$ is cannot be obtained as sum or difference; because it is prime, it cannot be obtained as product of integers (unless one factor is $1$ and we are effectively using at most $9$ digits). Remains the case that the number is obtained as a product or quotient of fractions, but at first sight this seems unplausible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meanwhile I brute-forced all numbers that can be obtained with only $+,-,\times$ and digit concatenation (and with all intermediate resuls $\le 10^7$). The first number that cannot be expressed this way turns out to be&#xA;$$ 8480902.$$&#xA;So unless someone manages to express $8480902$ under the original rules (i.e. with division allowed), this is the answer to the original problem.&lt;/p&gt;&#xA;" OwnerUserId="1731" LastEditorUserId="1731" LastEditDate="2014-07-06T20:48:07.973" LastActivityDate="2014-07-06T20:48:07.973" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1883" PostTypeId="2" ParentId="1870" CreationDate="2014-07-07T06:51:29.433" Score="17" Body="&lt;p&gt;This one, I think, is a lot easier than it seems. The shapes of the objects aren't relevant, so ignore them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first set, we see a sequence: something is on the top and bottom, something is on the left, and something is on the right.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  |X|     | |     | |   &#xA; - - -   - - -   - - -  &#xA;  | |    X| |     | |X &#xA; - - -   - - -   - - -  &#xA;  |X|     | |     | |   &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Rotate this sequence 90 degrees, and you end up with the bottom row. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  | |     |X|     | |   &#xA; - - -   - - -   - - -  &#xA; X| |X    | |     | |  &#xA; - - -   - - -   - - -  &#xA;  | |     | |     |X|   &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now something is on the left and right, and something is on the top. This means that, in the third object, something is on the bottom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the bottom row object happens to be a triangle, the answer is #1: a triangle with something in the bottom.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastActivityDate="2014-07-07T06:51:29.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1885" PostTypeId="2" ParentId="1855" CreationDate="2014-07-07T16:40:50.660" Score="4" Body="&lt;p&gt;I have written a Delphi application to brute force a solution. Unfortunately I see no way of solving the problem: what expression(s) results in number N? That is I see no solution other than generating all expressions and then seeing what natural numbers they result in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I can't really help with Bobson's crowd-sourcing approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am following klm123's idea, and I have ran the application with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;N = 3 : 0,1,2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Execution time: 00:00:00&#xA;Expressions processed: 441 [Average: 27.562,5/sec]&#xA;Expressions calculated: 34 [Average: 2.125,0/sec]&#xA;Natural numbers calculated: 28 [Average: 1.750,0/sec]&#xA;Distinct natural numbers found: 14 [Average: 875,0/sec]&#xA;Highest natural found: 210&#xA;Lowest natural not found: 4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;N = 4 : 0,1,2,3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Execution time: 00:00:00&#xA;Expressions processed: 22.924 [Average: 363.873,0/sec]&#xA;Expressions calculated: 1.464 [Average: 23.238,1/sec]&#xA;Natural numbers calculated: 871 [Average: 13.825,4/sec]&#xA;Distinct natural numbers found: 110 [Average: 1.746,0/sec]&#xA;Highest natural found: 3.210&#xA;Lowest natural not found: 25&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;N = 5 : 0,1,2,3,4&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Execution time: 00:00:04&#xA;Expressions processed: 1.679.977 [Average: 401.812,2/sec]&#xA;Expressions calculated: 98.228 [Average: 23.493,9/sec]&#xA;Natural numbers calculated: 45.423 [Average: 10.864,1/sec]&#xA;Distinct natural numbers found: 884 [Average: 211,4/sec]&#xA;Highest natural found: 43.210&#xA;Lowest natural not found: 89&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;N = 6 : 0,1,2,3,4,5&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Execution time: 00:08:39&#xA;Expressions processed: 159.888.346 [Average: 307.590,0/sec]&#xA;Expressions calculated: 8.867.950 [Average: 17.060,0/sec]&#xA;Natural numbers calculated: 3.236.479 [Average: 6.226,3/sec]&#xA;Distinct natural numbers found: 8.661 [Average: 16,7/sec]&#xA;Highest natural found: 543.210&#xA;Lowest natural not found: 653&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;N = 7 : 0,1,2,3,4,5,6&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Execution time: 19:36:02&#xA;Expressions processed: 18.788.082.577 [Average: 266.262,5/sec]&#xA;Expressions calculated: 1.014.272.742 [Average: 14.374,2/sec]&#xA;Natural numbers calculated: 308.146.445 [Average: 4.367,0/sec]&#xA;Distinct natural numbers found: 93.219 [Average: 1,3/sec]&#xA;Highest natural found: 6.543.210&#xA;Lowest natural not found: 5.683&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now on to N=8. I got a few small optimizations in mind, but nothing that will cut this drastically, I presume.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Explanation of what the application is doing&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I have two input parameters: &lt;strong&gt;Digits&lt;/strong&gt; and &lt;strong&gt;Operators&lt;/strong&gt;. I also have some other parameters to handle output dump to disk, but they have nothing to do with the algo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The application starts processing the empty expression: 0 length string.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every time an expression &lt;strong&gt;E&lt;/strong&gt; is processed, I do the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1. check all Digits&lt;/strong&gt; - for any single digit &lt;strong&gt;D&lt;/strong&gt; not already included in &lt;strong&gt;E&lt;/strong&gt;, I process the expression &lt;strong&gt;E&lt;/strong&gt;+&lt;strong&gt;D&lt;/strong&gt; (concatenation, not sum). To optimize a bit I avoid this step if &lt;strong&gt;E&lt;/strong&gt; ends with a closed parenthesis. I also avoid this if &lt;strong&gt;D&lt;/strong&gt; is 0 and &lt;strong&gt;E&lt;/strong&gt; does not end with any other Digit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2. check all Operators&lt;/strong&gt; - for any single operator &lt;strong&gt;O&lt;/strong&gt;, I process the expression &lt;strong&gt;E&lt;/strong&gt;+&lt;strong&gt;O&lt;/strong&gt;. To optimize I avoid this step if &lt;strong&gt;E&lt;/strong&gt; is empty or ends with another operator or with an open parenthesis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3. check &quot;)&quot;&lt;/strong&gt; - if &lt;strong&gt;E&lt;/strong&gt; ends with a Digit or a closed parenthesis, and it has a pending unclosed parenthesis, and this contains at least one operatore, then I process the expression &lt;strong&gt;E&lt;/strong&gt;+&lt;strong&gt;)&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4. check &quot;(&quot;&lt;/strong&gt; - if &lt;strong&gt;E&lt;/strong&gt; is empty or ends with an Operator or an open parenthesis, then I process the expression &lt;strong&gt;E&lt;/strong&gt;+&lt;strong&gt;(&lt;/strong&gt;. To avoid infinite recursion there is some magic here... I do it only if the number of Digits still unused in &lt;strong&gt;E&lt;/strong&gt; is greater the number of pending unclosed parenthesis in &lt;strong&gt;E&lt;/strong&gt; plus 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;5. calculate the expression&lt;/strong&gt; - if &lt;strong&gt;E&lt;/strong&gt; ends with a Digit or a closed parenthesis, and all parenthesis are balanced and are useful (contain at least one operator), and it does not start with &lt;strong&gt;(&lt;/strong&gt; and end with &lt;strong&gt;)&lt;/strong&gt; then I calculate the expression. If it turns out to be a natural number, then I check it with the highest found natural and lowest unfound natural.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Processing of derived expressions is obviously done through recursion. The algo seems quite efficient, apart from the parenthesis part. I have made some changes to this code and it now validates parenthesis better, avoiding some recursion in useless branches and evaluation os redundant expressions. It's not perfect, but it does cut times as N goes up. &lt;em&gt;(I am using Artem V. Parlyuk's ArtFormula package of nonvisual Delphi component for symbolic expression parsing and evaluation).&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, I also keep track of the shortest expression that can generate each natural number found. The shortest is meaningful, as it is usually the most immediate and readable. It is also always the one with no useless parenthesis ;)&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Update&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Tweaking the parenthesis logic, I have stumbled upon a small bug. It was generating a lot of useless expressions (with redundant parenthesis) but it was not generating some useful expressions. With N&amp;lt;=5 this bug did not have any effects. With N=6 I did find 6 more natural numbers, though the lowest unfound natural is 653.&lt;/p&gt;&#xA;" OwnerUserId="1761" LastEditorUserId="1761" LastEditDate="2014-07-09T10:50:49.407" LastActivityDate="2014-07-09T10:50:49.407" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1886" PostTypeId="2" ParentId="433" CreationDate="2014-07-07T21:04:45.473" Score="0" Body="&lt;p&gt;Going to throw this out there. I like the &quot;APE&quot; explanation much better, but maybe someone can do something with this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Trap&quot; can be used as a synonym for equipment (though not a common one). If &quot;Circus&quot; is the general clue, leading to &quot;Trapeze&quot;, then the rest of it might be &quot;Equipment&quot; = trap and &quot;has chimpanzee, say at the centre&quot; would be &quot;eze&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like the other clues I've read, though, I can't figure out quite how the second part would work, unless maybe &quot;chimp&quot; = &quot;ee&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a real stretch, wiktionary.org says that &quot;chimp&quot; can mean:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(informal, often pejorative) To get very excited when showing images on a digital camera.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So maybe &quot;ee&quot; is the sound someone makes when chimping and &quot;zee&quot; goes in the centre? I know, it's a &lt;em&gt;real&lt;/em&gt; stretch. I much prefer the &quot;APE&quot; explanation, but can't figure out what leads to the &quot;TRE&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thought (again, that I don't think is right, but maybe it'll poke someone in the right direction): What if &quot;say at the centre&quot; means &quot;rap&quot;? &quot;Rap&quot; is a way of saying things. That would leave &quot;has chimpanzee&quot; or &quot;equipment has chimpanzee&quot; to be &quot;teze&quot;. I haven't been able to make that work yet, but it's another thought.&lt;/p&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2014-07-07T21:04:45.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1887" PostTypeId="1" AcceptedAnswerId="1908" CreationDate="2014-07-08T09:06:22.047" Score="7" ViewCount="1120" Body="&lt;p&gt;What would be a chess position, where a player (white or black) has a maximum possible number of different moves to make?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two options:&lt;br&gt;&#xA;1. The position must be possible to obtain in a chess game.&lt;br&gt;&#xA;2. Any position, involving any number of chess pieced are allowed. (Most probably it is optimal to fill the board with queens only, but what number of queens is optimal, they would get in a way of each other).&lt;/p&gt;&#xA;" OwnerUserId="28" LastActivityDate="2014-07-12T20:52:43.543" Title="Most moveable chess position" Tags="|chess|" AnswerCount="3" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="1888" PostTypeId="2" ParentId="1874" CreationDate="2014-07-08T11:30:56.140" Score="6" Body="&lt;p&gt;I've highlighted the X-Wing and the 9's involved in this X-Wing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rQqKf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See how's there's two 9's in row 3 and also two 9's in row 8? (highlighted in blue)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since every row must contain exactly one 9, either A and D must both be 9 OR B and C must both be 9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In either case, these 4 squares will provide the 9's for rows 3 and 8, and because of that, they will also provide the 9's for their columns (3 and 5).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that any other squares in the same column as these squares can NOT be 9. So any 9's in these squares can be deleted. (highlighted in red)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try making any of the red ones a 9, for instance row 1, column 3. If this were a 9, then A and C cannot be 9. But then, for rows 3 and 8 to both contain a 9, Both B and D would have to be 9 and this is not possible, because then column 5 would have two 9's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imo, X-Wings are not easy to spot without highlighting all occurrences of a number. I usually don't look for them when solving with pen &amp;amp; paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'd have to look for 2 numbers occurring in 2 rows or columns. It is even more difficult to find 3 numbers in 3 rows/columns, or 4, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There can also be an X-Wing involving a row and a 3x3 box, but they are even more difficult to see. Computers have no problem spotting them, of course.&lt;/p&gt;&#xA;" OwnerUserId="1598" LastEditorUserId="1598" LastEditDate="2014-07-08T12:11:36.613" LastActivityDate="2014-07-08T12:11:36.613" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1889" PostTypeId="2" ParentId="1808" CreationDate="2014-07-08T17:04:48.337" Score="1" Body="&lt;p&gt;I see from the previous reply this should be doable in 5 weighings. Though, it has been a week and no answer. So, i will post an answer with 6 weighings. I hope i got it right. If the weights are not necessarily equal, it might remove a weighing, but being they &lt;em&gt;can&lt;/em&gt; be equal, it shouldn't change much.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;hr /&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Split the balls into 4 groups of 3.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;hr /&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If a group is known to be heavier or lighter, it can be determined by weighing one ball against another.&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If they are equal, the third ball is the heavier or lighter ball.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;If they are not equal, the ball that matches the group as heavier or lighter, is the heavier or lighter ball.&lt;/li&gt;&#xA;&lt;li&gt;This will be referred to as determine(group).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;hr /&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Weigh Group 1 against Group 2. [1]&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If they are not equal, either Group 1 has the heavy ball or Group 2 has the light ball. (If it is the opposite, we will rename the lighter side Group 1 and the heavier Group 2)&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Weigh Group 1 against Group 3 [2]&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If they are equal, either Group 2 has the heavy ball and Group 4 has the light ball or Group 1 has the light ball and Group 3 has the heavy ball&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Weigh Group 1 against Group 4, they will be unequal. [3]&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If Group 1 is lighter, Group 1 has the light ball and Group 3 has the heavy ball&lt;/li&gt;&#xA;&lt;li&gt;If Group 4 is lighter, Group 2 has the heavy ball and Group 4 has the light ball&lt;/li&gt;&#xA;&lt;li&gt;Determine(lighter group) [4]&lt;/li&gt;&#xA;&lt;li&gt;Determine(heavier group) [5]&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;If Group 1 is lighter, Group 1 has the light ball and Group 4 has the heavy ball.&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Determine(Group 1) [3]&lt;/li&gt;&#xA;&lt;li&gt;Determine(Group 4) [4]&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;If Group 1 is heavier, Group 2 has the heavy ball and Group 3 has the light ball.&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Determine(Group 2) [3]&lt;/li&gt;&#xA;&lt;li&gt;Determine(Group 3) [4]&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;If they are equal, weigh Group 1 against Group 3 [2]&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If they are not equal, either Group 1 has the light ball and Group 2 has the heavy ball (see note above) or Group 3 has the heavy ball and Group 4 has the light ball.&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Weigh Group 1 against Group 4, they will be unequal. [3]&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Determine(lighter group) [4]&lt;/li&gt;&#xA;&lt;li&gt;Determine(heavier group) [5]&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;If they are equal, one of the groups has both balls.&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Weigh 1 ball from each group against another ball from each group, leaving the third from each on the side. [3] (Refer to these as Groups A, B, and C)&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If they are equal, either Group A is lighter and Group B is heavier or vice versa.&#xA;&lt;ul&gt;&#xA;&lt;li&gt;switch Group B for Group C and weigh again, they will be not equal. [4]&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Weigh two of the balls from the lighter group (either Group A or Group B) against each other. [5]&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If they are not equal, the lighter ball is the light ball and it's groupmate is the heavy ball.&lt;/li&gt;&#xA;&lt;li&gt;If they are equal, weigh the other two balls from the lighter group, they will be not equal. [6]&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The lighter ball is the light ball and it's mate in Group C is the heavy ball.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;If they are not equal, either Group A has the light ball and Group C has the heavy ball, or Group B has the heavy ball and Group C has the light ball. (See note above)&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Weigh 2 balls from Group A against the other 2 balls from Group A. [4]&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If they are not equal, weigh the 2 balls from the lighter side against the other. [5]&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The lighter ball is the light ball and its mate in Group C is the heavy ball.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;If they are equal, weigh 1 ball from Group B against another ball from Group B. [5]&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If they are not equal, the heavier ball is the heavy ball and its mate in Group C is the light ball.&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If they are equal, weigh the 2 remaining balls from Group B against each other. [6]&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The heavier ball is the heavy ball and its mate in Group C is the light ball.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1786" LastEditorUserId="75598" LastEditDate="2021-11-11T14:39:25.083" LastActivityDate="2021-11-11T14:39:25.083" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="1890" PostTypeId="2" ParentId="1887" CreationDate="2014-07-08T18:10:07.543" Score="4" Body="&lt;p&gt;The best result I can get &lt;strong&gt;for option #2 is 288&lt;/strong&gt;.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/fndoz.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not have a strict proof that this is a maximum, but I have something very close. Therefore I have feeling that this is it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is &lt;strong&gt;the &quot;proof&quot;&lt;/strong&gt;:&lt;br&gt;&#xA;Consider 1x8 board, you will easily see that the optimum value of 12 moves is achieved with 2 queens at the different ends.&lt;br&gt;&#xA;From this you can naturally suppose that optimally you need to put 2 queens on any possible line. You can't put exactly 2 queens on all lines, but the position I show you have maximum number of lines with 2 queens on it (6 vertical + 6 horizontal + 11*2 diagonal).&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2014-07-12T06:33:54.287" LastActivityDate="2014-07-12T06:33:54.287" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1891" PostTypeId="1" AcceptedAnswerId="1892" CreationDate="2014-07-08T18:27:38.437" Score="14" ViewCount="2969" Body="&lt;p&gt;This is from the current weekly math challenge from the newspaper &lt;em&gt;Le Monde&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Alice and Bob play the following game, in turn:&lt;/p&gt;&#xA;&lt;p&gt;A number between &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$10$&lt;/span&gt; is written on a blackboard.&lt;/p&gt;&#xA;&lt;p&gt;At each of their turns, they choose &lt;strong&gt;one&lt;/strong&gt; of these actions:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;multiply the current number by &lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;multiply the current number by &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;add &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; to the current number.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Alice begins.&lt;/p&gt;&#xA;&lt;p&gt;If after playing, a player gets a number greater or equal than &lt;span class=&quot;math-container&quot;&gt;$1000$&lt;/span&gt;, he wins.&lt;/p&gt;&#xA;&lt;p&gt;Given the initial number, who has a winning strategy ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;For example&lt;/em&gt;: the initial number is &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Multiplying by &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt;, Alice makes it &lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Then Bob adds &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; and gets &lt;span class=&quot;math-container&quot;&gt;$9$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;And so on.&lt;/p&gt;&#xA;" OwnerUserId="1636" LastEditorUserId="57212" LastEditDate="2022-03-07T09:29:31.970" LastActivityDate="2022-03-07T09:29:31.970" Title="A two player game: Player who gets to 1000 or more first, wins" Tags="|strategy|game|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="1892" PostTypeId="2" ParentId="1891" CreationDate="2014-07-08T19:56:00.880" Score="21" Body="&lt;p&gt;I suppose that both players plays optimally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since both players think similarly the win does not depend on player but on current number. Some numbers are clearly a wining ones and some are losing ones. Let's call first W-numbers (if a player gets one before his\her turn he will win), and others L-numbers. Let's find type of each number starting from biggest.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If a player gets (after a move of the opponent) a number  250 or bigger he multiplies by 4 and wins. So &lt;strong&gt;numbers $n \ge 250$ are W-numbers&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;If a player gets 249 he will lose next turn. 249 is L-number.&lt;/li&gt;&#xA;&lt;li&gt;If a player gets a number $249 &amp;gt; n &amp;gt; 83$ he will never multiply it, otherwise he will lose, so he will add 1. Since 249 is L-number, then all &lt;strong&gt;odd numbers $250 &amp;gt; n &amp;gt; 83$ are L-numbers&lt;/strong&gt; and all &lt;strong&gt;even $250 &amp;gt; n &amp;gt; 83$ are W-numbers&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;If a player gets an odd number $250/3 &amp;gt; n &amp;gt; 83/3$ he will multiply by 3 and get a L-number. So all &lt;strong&gt;odd numbers $83 &amp;gt;= n &amp;gt; 27$ are W-numbers&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;If a player gets an even number $83 &amp;gt; n &amp;gt; 27$ he can't add 1, he can't multiply by 3 or 4 will multiply by 4 - in any case he will get a W-number for opponent and lose. So &lt;strong&gt;&lt;em&gt;even&lt;/em&gt; numbers $83 &amp;gt;= n &amp;gt; 27$ are L-numbers&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;If a player gets an even number $83/3 &amp;gt;= n &amp;gt; 27/3$ he will multiply by 3 or 4 and get L-number for his opponent. So &lt;strong&gt;even numbers $26 &amp;gt;= n &amp;gt; 9$ are W-numbers&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;If a player gets an number $83/4 &amp;gt;= n &amp;gt; 27/4$ he will multiply by 4 and get L-number for his opponent. So &lt;strong&gt;numbers $20 &amp;gt;= n &amp;gt; 7$ are W-numbers&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;If a player gets an odd number $25 &amp;gt; n &amp;gt; 20$ he can't add 1, multiply by 4 or 3 - he will lose. So &lt;strong&gt;$23$ and $21$ are L-numbers&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;$7$ is clearly a W-number&lt;/strong&gt;, since a player can make $21$ from it.&lt;/li&gt;&#xA;&lt;li&gt;If a player get a number $26/4 &amp;gt;= n &amp;gt; 7/3$ from he can't multiply by 4 or 3 he will lose, so he forced to add 1. So &lt;strong&gt;$6$, $4$ are L-numbers&lt;/strong&gt; and &lt;strong&gt;$5$,$3$ are W-numbers&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;If a player get 2 or 1 he can get L-number (6 or 4) for the opponent. &lt;strong&gt;$1$ and $2$ are W-numbers&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Summarising:&lt;br&gt;&#xA;If initial number is $4$ or $6$ then the second player wins, otherwise the first player wins.&lt;/p&gt;&#xA;" OwnerUserId="28" LastActivityDate="2014-07-08T19:56:00.880" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="1893" PostTypeId="2" ParentId="1891" CreationDate="2014-07-08T20:04:53.813" Score="10" Body="&lt;p&gt;&lt;strong&gt;Note this answer only works if reaching 1000 does not warrent a win.  This was the original question but since it was changed to &quot;greater than or equal to 1000&quot; is a win it is incorrect.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets look at this for any $N$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $N&amp;gt;250$ player one can multiply by $4$ and win instantly so label all of those $F$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $251&amp;gt;N&amp;gt;82$ it will depend on if $N$ is even or odd.  Both players will just add one to prevent the other getting a $250+$.  If $N$ is even it is $S$ while if it is odd it is $F$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $83&amp;gt;N&amp;gt;63$ then if $N$ is even it is $F$ while if it is odd it is $S$. The reversal happens because when $N=82$ then $3N=246$ which is less than $250$ but even so by the previous paragraph the player awarded that will lose.  Simular strategys arrise by multiplying any $N$ by $3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $63&amp;gt;N&amp;gt;20$ then the first player always wins because he can multiply by $4$ to yield a $N&amp;gt;83$ which is even which means he wins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $21&amp;gt;N&amp;gt;7$ it is alternating again as if $N$ is even it is $S$ while if it is odd it is $F$.  This is because multiplying makes the other player win so they can only add.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all $N$ between $1$ and $10$ besides $1,8,10$, the first player can force the number into a region which is an $S$ by making sure it is even and between $20$ and $7$.  If he cannot, the second player can force a win.  This means if $N=1,8,10$ then the second player wins. If $N=2,3,4,5,6,7,9$ then the first player wins.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditDate="2014-07-08T20:30:17.750" LastActivityDate="2014-07-08T20:30:17.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1894" PostTypeId="2" ParentId="1887" CreationDate="2014-07-08T20:24:30.913" Score="4" Body="&lt;p&gt;Your answer shows a chess board with a white queen at each edge and corner location - 28 queens in total - and you claim that there are 228 possible moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I make it &lt;strong&gt;288&lt;/strong&gt; moves:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;each queen in a corner has six possible moves available along the diagonal&lt;br&gt;&#xA;$4 \times 6 = 24$&lt;/li&gt;&#xA;&lt;li&gt;each queen along an edge has 11 moves available, six straight and 5 diagonally&lt;br&gt;&#xA;$24 \times 11 = 264$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;making not &lt;strong&gt;228&lt;/strong&gt; but &lt;strong&gt;288&lt;/strong&gt; possible moves in total.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gygOl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether that is the maximum possible remains to be seen.&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditDate="2014-07-08T20:33:40.617" LastActivityDate="2014-07-08T20:33:40.617" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1895" PostTypeId="2" ParentId="1870" CreationDate="2014-07-09T00:38:31.687" Score="24" Body="&lt;p&gt;Rotate the circles clockwise 90 degrees so the arrows point to where you can see black circles through the triangle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JPGmY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1789" LastActivityDate="2014-07-09T00:38:31.687" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="1896" PostTypeId="2" ParentId="313" CreationDate="2014-07-09T14:40:41.510" Score="20" Body="&lt;p&gt;Slightly tongue-in-cheek, but a variation on the &quot;maybe&quot; option provided the other player interprets it literally enough.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Keep evaluating powers of your number until you reach one that exceeds 6. Is it an odd power? Answer yes = 2, no = 3, infinitely long pause = 1 …&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1801" LastEditorUserId="46903" LastEditDate="2018-09-19T00:06:23.413" LastActivityDate="2018-09-19T00:06:23.413" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="1897" PostTypeId="2" ParentId="1814" CreationDate="2014-07-10T17:56:59.013" Score="9" Body="&lt;p&gt;Following up on the suggestion by Dennis Meng, I'm posting my suggestion as an answer instead of as part of the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here are my thoughts on the problem. (1) As Joe Z comments, the cattle owner needs to increase prices by one eighth in order to make the stated profit. (2) For some reason, the price is given as 63 daler for 3 oxen, rather than the simpler 21 daler for one ox. (3) There seems to be two different kinds of oxen, where some are big and fat, and some are small and slim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suggestion: Let there be different prices for large and small oxen. The farmer buys 200 large oxen and 100 small oxen, where two large oxen and one small cost 63 daler together. He increases both prices by one eighth and sells the herd. The new prices happen to be such that one large and two small oxen cost 63 daler together!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Working out the details, large oxen were bought for $23 {1\over 3}$ daler each and sold for $26{1\over 4}$, whereas small oxen were bought for $16{1\over 3}$ daler each and sold for $18{3\over 8}$. Note that the fractions cause no problems, as there are 96 skilling in a daler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if this is the intended solution, but it seems consistent with both the statement of the problem and the principle of proportionality, which is present in all rule of three computations. I guess the only way to find out for sure is if the problem was taken from another source, with a fuller discussion.&lt;/p&gt;&#xA;" OwnerUserId="219" LastEditorUserId="219" LastEditDate="2014-10-30T09:45:51.637" LastActivityDate="2014-10-30T09:45:51.637" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1899" PostTypeId="2" ParentId="1725" CreationDate="2014-07-11T19:01:44.423" Score="1" Body="&lt;p&gt;The following is inspired by an old Swedish crossword puzzle, whose solution has the same structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Across: 1. Apex&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Down: 2. Burden&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;+---+---+   &#xA;|xxx|2  |   &#xA;|xxx|   |   &#xA;+---+---+  &#xA;|1  |   |  &#xA;|   |   |    &#xA;+---+---+    &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    X W&lt;br&gt;&#xA;    H 8    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="219" LastEditorUserId="219" LastEditDate="2014-07-11T19:19:17.940" LastActivityDate="2014-07-11T19:19:17.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1900" PostTypeId="2" ParentId="340" CreationDate="2014-07-11T22:30:38.760" Score="6" Body="&lt;p&gt;So while not explicitly a logic grid, Carrol's book &lt;a href=&quot;http://www.gutenberg.org/files/4763/4763-h/4763-h.htm&quot;&gt;The Game Of Logic&lt;/a&gt; popularized logic &quot;puzzles&quot; of this nature (his weren't exactly puzzles so much as complicated logical statements) with answers represented in a notational style of his own creation involving squares and counters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These problems were later refined in magazines and articles into the modern variety of what we call the logic puzzle: a single unique combination of facts which could be deduced through a provided set of statements. These became increasingly common during the 1940s and 1950s. When printed in magazines, some would suggest the form of an array or grid to assist with the problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first actual printed grid I can find is in Clarence Raymond Wylie, Jr's 1957 publication 101 Puzzles in Thought and Logic.&lt;/p&gt;&#xA;" OwnerUserId="1810" LastActivityDate="2014-07-11T22:30:38.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1901" PostTypeId="1" AcceptedAnswerId="1903" CreationDate="2014-07-12T05:51:49.890" Score="7" ViewCount="791" Body="&lt;p&gt;&lt;code&gt;This statement contains 3 0's, 1 1, 4 2's, 1 3, 5 4's 92 5's, 6 6's, 53 7's, 58 8's, and 9 9's.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly that statement is incorrect. It only has 1 occurrence of the character &quot;0&quot;, not 3; and there are definitely not 92 instances of &quot;5&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you rewrite it to make it true? You still must follow the same format, giving the exact number of times each digit has appeared in the statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically your statement must be in the format: &lt;code&gt;This statement contains A 0's, B 1, C 2's, D 3, E 4's F 5's, G 6's, H 7's, I 8's, and J 9's.&lt;/code&gt;, where A is the number of 0's, B is the number of 1's, etc. The number of 1's (or any digit) includes the 1 needed for &lt;code&gt;B 1's&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not can you prove that this is impossible?&lt;/p&gt;&#xA;" OwnerUserId="1811" LastEditorUserId="1811" LastEditDate="2014-07-12T07:06:30.427" LastActivityDate="2017-07-14T20:36:20.153" Title="Write a true self-reflective statement about the digits from 0 to 9 (or prove it can't be done)" Tags="|mathematics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1903" PostTypeId="2" ParentId="1901" CreationDate="2014-07-12T06:54:55.033" Score="7" Body="&lt;p&gt;Here we go:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This statement contains &lt;strong&gt;1&lt;/strong&gt; 0's, &lt;strong&gt;7&lt;/strong&gt; 1's, &lt;strong&gt;3&lt;/strong&gt; 2's, &lt;strong&gt;2&lt;/strong&gt; 3's, &lt;strong&gt;1&lt;/strong&gt; 4's, &lt;strong&gt;1&lt;/strong&gt; 5's, &lt;strong&gt;1&lt;/strong&gt; 6's, &lt;strong&gt;2&lt;/strong&gt; 7's, &lt;strong&gt;1&lt;/strong&gt; 8's, and &lt;strong&gt;1&lt;/strong&gt; 9's.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2014-07-12T07:12:17.280" LastActivityDate="2014-07-12T07:12:17.280" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="1904" PostTypeId="1" AcceptedAnswerId="1907" CreationDate="2014-07-12T07:25:07.337" Score="8" ViewCount="627" Body="&lt;p&gt;Following &lt;a href=&quot;https://puzzling.stackexchange.com/q/1901/28&quot;&gt;this question&lt;/a&gt;, consider the case where we change the number of digits and formulate the task like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Write the statement like &quot;This statement contains $M_0$ 0's, $M_1$ 1, ..., $M_N$ N's&quot;, which is true. You must write about all digits, which statement includes: that all $M_i (i = 0, .., N)$ must be written with $0, .., N$ digits only.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the smallest $N$ possible here?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Similarly to the given answer for $N=9$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;1&lt;/strong&gt; 0's, &lt;strong&gt;7&lt;/strong&gt; 1's, &lt;strong&gt;3&lt;/strong&gt; 2's, &lt;strong&gt;2&lt;/strong&gt; 3's, &lt;strong&gt;1&lt;/strong&gt; 4's, &lt;strong&gt;1&lt;/strong&gt; 5's, &lt;strong&gt;1&lt;/strong&gt; 6's, &lt;strong&gt;2&lt;/strong&gt; 7's, &lt;strong&gt;1&lt;/strong&gt; 8's, and &lt;strong&gt;1&lt;/strong&gt; 9's.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can construct answer for $N=6$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;1&lt;/strong&gt; 0's, &lt;strong&gt;4&lt;/strong&gt; 1's, &lt;strong&gt;3&lt;/strong&gt; 2's, &lt;strong&gt;2&lt;/strong&gt; 3's, &lt;strong&gt;2&lt;/strong&gt; 4's, &lt;strong&gt;1&lt;/strong&gt; 5's, &lt;strong&gt;1&lt;/strong&gt; 6's.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but the same approach fails for $N=5$. Meanwhile this doesn't prove that $N \le 5$ is not possible to achieve.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-07-16T16:38:10.623" Title="Find the smallest N to complete the task: &quot;Write a true self-reflective statement about the digits from 0 to N&quot;" Tags="|mathematics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1907" PostTypeId="2" ParentId="1904" CreationDate="2014-07-12T12:18:28.707" Score="5" Body="&lt;p&gt;I've seen this problem for numbers from $1$ to $N$ before, but not for $0$ to $N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the statements are from $1$ to $N$, the minimum you can have is $4$, as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This statement contains &lt;strong&gt;2&lt;/strong&gt; 1's, &lt;strong&gt;3&lt;/strong&gt; 2's, &lt;strong&gt;2&lt;/strong&gt; 3's, and &lt;strong&gt;1&lt;/strong&gt; 4.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now note that in the case from $0$ to $N$, you're guaranteed that there is only ever one $0$, because there is never $0$ of any other number. So, a trivial solution exists for $N = 0$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This statement contains &lt;strong&gt;1&lt;/strong&gt; 0.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And for $N = 1$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This statement contains &lt;strong&gt;1&lt;/strong&gt; 0 and &lt;strong&gt;2&lt;/strong&gt; 1's.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And for $N = 2$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This statement contains &lt;strong&gt;1&lt;/strong&gt; 0, &lt;strong&gt;3&lt;/strong&gt; 1's, and &lt;strong&gt;1&lt;/strong&gt; 2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, what these statements all have in common is that they involve writing a number greater than the maximum number being counted, which means that the sentence only describes &lt;em&gt;some&lt;/em&gt; of the numbers, not all of them. The smallest statement that fits your new criterion is indeed $N = 3$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This statement contains &lt;strong&gt;1&lt;/strong&gt; 0, &lt;strong&gt;2&lt;/strong&gt; 1's, &lt;strong&gt;3&lt;/strong&gt; 2's, &lt;strong&gt;2&lt;/strong&gt; 3's, &lt;strong&gt;and no other numbers&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2014-07-12T18:32:33.940" LastActivityDate="2014-07-12T18:32:33.940" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="1908" PostTypeId="2" ParentId="1887" CreationDate="2014-07-12T20:52:43.543" Score="9" Body="&lt;p&gt;For a legal position, this was (as of roughly 30 years ago, but I don't think it's been surpassed) the record for available moves for one side, &lt;strong&gt;218 moves&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/R7ohm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book &lt;em&gt;A Guide to Fairy Chess&lt;/em&gt; has a number more then-record positions, including confirming the queens-around-the-edge position from the other answers, as well as records for most moves available to both sides, most moves available without promoted pieces, most moves available using all 32 original men, etc.  Let me know if you'd like any of the other records and I'll add them to this answer.&lt;/p&gt;&#xA;" OwnerUserId="266" LastActivityDate="2014-07-12T20:52:43.543" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1909" PostTypeId="1" AcceptedAnswerId="1943" CreationDate="2014-07-13T20:04:37.217" Score="10" ViewCount="8219" Body="&lt;p&gt;This is probably my favorite puzzle. It uses logic, but it's not one of those &amp;quot;logic table&amp;quot; puzzles which i dislike immensely. What is this type of puzzle called and are there more like it that are not just variants?&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Three children are playing on the beach and all three have mud on their foreheads. An old man walks by and asks them, “After looking at both of your friends foreheads, raise your hand if one or both of them have mud on their foreheads.“ All three children look at both their friends and raise their hands.&lt;/p&gt;&#xA;&lt;p&gt;He then asks, “For a dollar, without touching your own forehead or looking in a mirror or anything similar, tell me if you have mud on your own forehead, and how you know that.” All three children are bewildered. Finally, one raises his hand and says, “i have mud on my forehead.” He explains how he knew this and receives the dollar.&lt;/p&gt;&#xA;&lt;p&gt;What was his reasoning?&lt;/p&gt;&#xA;&lt;p&gt;Note: The old man and the children are honest, able to employ logic, and want the dollar.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1786" LastEditorUserId="36023" LastEditDate="2020-10-09T19:25:36.857" LastActivityDate="2020-10-09T19:25:36.857" Title="Who has mud on their forehead?" Tags="|logical-deduction|meta-knowledge|" AnswerCount="6" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1910" PostTypeId="2" ParentId="1909" CreationDate="2014-07-13T20:31:36.797" Score="7" Body="&lt;p&gt;There is no special name for such type of puzzles as far as i know.&lt;br /&gt;&#xA;But usually they are formulated in terms of hats worn by logicians.&lt;/p&gt;&#xA;&lt;p&gt;If you google you can find a lot of puzzles of this type, for example, &lt;a href=&quot;http://www.mathpuzzle.com/LogicalHats.txt&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The Logical Hats Puzzle.  Logicians A, B and C each wear a hat with a positive integer on it such that the number on one hat is the sum of the numbers on the other two.  They can see the numbers on the other two hats but not their own.  They are given this information and asked in turn if they can identify their number.  In the first round A, B and C each in turn say they don't know.  In the second round A is first to go and states his number is 50.  What numbers are on B and C?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Or &lt;a href=&quot;http://en.wikipedia.org/wiki/Prisoners_and_hats_puzzle&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The jailer puts three of the men sitting in a line. The fourth man is put behind a screen (or in a separate room). He gives all four men party hats (as in diagram). The jailer explains that there are two red and two blue hats; that each prisoner is wearing one of the hats; and that each of the prisoners is only to see the hats in front of them but not on themselves or behind. The fourth man behind the screen can't see or be seen by any other prisoner. No communication between the prisoners is allowed.&lt;/p&gt;&#xA;&lt;p&gt;If any prisoner can figure out and say to the jailer what color hat he has on his head all four prisoners go free. If any prisoner suggests an incorrect answer, all four prisoners are executed. The puzzle is to find how the prisoners can escape, regardless of how the jailer distributes the hats.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Also there was a lot of them on this site:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/201/guessing-colours-hats&quot;&gt;Guessing hat colors&lt;/a&gt;&lt;br /&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/282/hats-and-aliens&quot;&gt;Hats and Aliens&lt;/a&gt;&lt;br /&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/70/n-logicians-wearing-hats-of-n-colors&quot;&gt;N logicians wearing hats of N colors&lt;/a&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/236/in-the-100-blue-eyes-problem-why-is-the-green-eyed-oracle-necessary&quot;&gt;In the 100 blue eyes problem - why is the oracle necessary?&lt;/a&gt; (suggested by Bobson)&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-07-17T13:22:01.640" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="1911" PostTypeId="1" AcceptedAnswerId="1912" CreationDate="2014-07-14T15:15:55.893" Score="6" ViewCount="8620" Body="&lt;p&gt;When I was a child I had a puzzle book.  One of the puzzles involved finding words where you can remove one letter from the beginning or end of the word, and you still have a valid word.  You can continue this process with these words until there is only one letter left.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, &quot;brandy&quot; is such a word, because after each letter is removed below, we get another word:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;brandy&#xA;brand&#xA;bran&#xA; ran&#xA;  an&#xA;  a&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I don't have an actual puzzle here. But instead I am looking for the name of this class of word, if it exists.&lt;/p&gt;&#xA;" OwnerUserId="1784" LastEditorUserId="5373" LastEditDate="2016-09-18T13:48:27.133" LastActivityDate="2016-09-18T13:48:27.133" Title="Remove first or last letter - What is the name of this word puzzle/type of word?" Tags="|wordplay|puzzle-identification|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1912" PostTypeId="2" ParentId="1911" CreationDate="2014-07-14T15:48:04.253" Score="6" Body="&lt;p&gt;In general, these puzzles are called &quot;Deletion Puzzles&quot; but the usually do not include the restriction to the beginning or end letters.  The most famous example is &quot;startling&quot; as it is the longest. It is also sometimes called &quot;Elision&quot;.  Please not that these terms also do not require it to be iterative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Terminal Deletion&quot; or &quot;Terminal Elision&quot; refer to removing both the first and last letters to make a work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Decapitation&quot; word puzzles refer to only removing the first letter to reveal a word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would most likely have luck calling a puzzle regarding these words an &quot;Iterated Amputation&quot; word puzzle.  Amputation usually has the same definition as Decapitation for this context but it less common.  It would be easy to draw contrast between them and describe removing either of the two &quot;arms&quot; of the word (right or left arm). Iterated, of course, refers to the fact you want to do this several times.  Does that mean you call a word &quot;amputatable&quot;? I don't know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately I don't think a single term for what you want exists and, if it did, I can't think of a context where you wouldn't have to define it for the audience anyways.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest you check out &quot;&lt;a href=&quot;http://books.google.com/books?id=1VhNAAAAYAAJ&amp;amp;pg=PA145&amp;amp;lpg=PA145&amp;amp;dq=terminal%20deletion%20word%20puzzle&amp;amp;source=bl&amp;amp;ots=KHwefbmvNi&amp;amp;sig=VjHopc2R8MkDp-l9c8IUx5UbF0U&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=hPfDU8GhA8qgyATs24L4BQ&amp;amp;ved=0CCUQ6AEwAQ#v=onepage&amp;amp;q&amp;amp;f=false&quot;&gt;A Key to Puzzledom: Or, Complete Handbook of the Enigmatic Art&lt;/a&gt;&quot; which is available on google books.&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2014-07-14T15:48:04.253" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1913" PostTypeId="1" AcceptedAnswerId="1914" CreationDate="2014-07-15T01:44:16.717" Score="-3" ViewCount="1277" Body="&lt;p&gt;Two workers A and B manufactured a batch of identical parts. A worked&#xA;for 2 hours and B worked for 5 hours and they did half the job. Then they&#xA;worked together for another 3 hours and they had to do (1/20)th of the&#xA;job. How much time does B take to complete the job, if he worked alone?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A) 24 hours&lt;/li&gt;&#xA;&lt;li&gt;B) 12 hours&lt;/li&gt;&#xA;&lt;li&gt;C) 15 hours&lt;/li&gt;&#xA;&lt;li&gt;D) 30 hours&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;HI can anyone please explain how to solve this.&#xA;Answer is given as option C&lt;/p&gt;&#xA;" OwnerUserId="1821" LastEditorUserId="39" LastEditDate="2014-07-15T14:32:09.883" LastActivityDate="2014-07-15T14:32:09.883" Title="Can anyone explain me how to solve this problem?" Tags="|calculation-puzzle|" AnswerCount="1" CommentCount="3" ClosedDate="2015-02-07T16:14:56.917" ContentLicense="CC BY-SA 3.0" />
  <row Id="1914" PostTypeId="2" ParentId="1913" CreationDate="2014-07-15T03:31:17.623" Score="2" Body="&lt;p&gt;Let's denote the speed of A (in working) as $x$ and the speed of B (in working) as $y$. Then we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2x+5y = \frac{1}{2}$$&#xA;$$3x+3y = \frac{1}{2} - \frac{1}{20} = \frac{9}{20}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just solve that &lt;sup&gt;(details left for exercise)&lt;/sup&gt; to get the final result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x = \frac{1}{12}, y = \frac{1}{15}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which means B will finish the job alone in 15 hours.&lt;/p&gt;&#xA;" OwnerUserId="1649" LastActivityDate="2014-07-15T03:31:17.623" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="1915" PostTypeId="2" ParentId="1909" CreationDate="2014-07-15T05:05:36.360" Score="12" Body="&lt;p&gt;For informational purposes, I'll give the solution to the quoted problem.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There are two possible scenarios that would cause all three children to raise their hands at the first question:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;All three children have mud on their foreheads.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Two children have mud on their foreheads and one doesn't.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Each of the children know this. Now, without loss of generality, let's pick a specific child; say, we'll name him Andy. Andy sees that the other two kids have mud on their foreheads. He can't deduce immediately whether he has mud or not, since both those two scenarios are part of the initial possibilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, Andy does see that neither of the other two kids are raising their hands and saying they have mud on their foreheads. If Andy's forehead was clean, each of the other two kids would only see one muddy forehead, and would therefore immediately know that their forehead was muddy. So he deduces that his forehead must also be muddy from this information, as well as the rather protracted pause after the old man asks the second question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually all three kids could deduce this conclusion, but Andy simply got to it first.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2014-07-15T05:05:36.360" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1916" PostTypeId="1" AcceptedAnswerId="1923" CreationDate="2014-07-15T10:17:32.830" Score="9" ViewCount="795" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;There are 100 white and 101 black balls. All the balls have different weights. Balls of the same colour look the same. White balls are numbered from lighter to heavier (from 1 to 100). The same is true for black balls (they are numbered from 1 to 101).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;What is the minimal number of weightings on a balance scale to find the ball with the middle weight?&lt;/strong&gt; (That is, the ball which will have the number 101 if all the balls would be sorted and numbered from lighter to heavier.)  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Additional explanations:&lt;br&gt;&#xA;   1. On a balance scale you can compare X balls on one arm to X balls on another arm and tell whether they are lighter, heavier or equal in relation to each other.&lt;br&gt;&#xA;  2. Since the balls are numbered, you don't have to compare balls of the same colour to each other using the scale as you know the result already.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know that there is a solution with 8 weightings (though I do not know it). &lt;strong&gt;I am especially interested in the proof of minimality&lt;/strong&gt;, since I want to be sure that 8 (or any other number) is the minimum.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="8874" LastEditDate="2015-01-31T12:29:48.503" LastActivityDate="2015-01-31T12:29:48.503" Title="Middle weight puzzle" Tags="|logical-deduction|strategy|weighing|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="1917" PostTypeId="1" AcceptedAnswerId="1918" CreationDate="2014-07-15T10:41:09.180" Score="5" ViewCount="485" Body="&lt;p&gt;It was traditional during the times of the Oregon Trail and the gold rush that men founding new towns were allowed to stake their claim on land upon arriving to the west coast.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose a man is given $n$ minutes to stake his claim on an area of land defined by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Convex_hull&quot; rel=&quot;nofollow&quot;&gt;convex hull&lt;/a&gt; of those stakes placed into the ground.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplicity, assume the man travels 1 kilometer in 1 minute and assume there are no obstructions of any kind and that the land is completely flat.  Also, assume an unlimited amount of stakes available.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Assuming that it takes no time to plant a stake into the ground, what is the optimal strategy for maximizing land area for stakes placed in this way?&lt;/li&gt;&#xA;&lt;li&gt;Supposing it takes $m$ minutes to plant a stake, how does this change the optimal strategy (where $0 &amp;lt; m &amp;lt; n$)?&lt;/li&gt;&#xA;&lt;li&gt;Assume the man is told he is given a fixed amount of time (you can assume that it takes no time to plant a stake into the ground as in scenario #1), but he is not told how long, how does this affect the optimal strategy for maximizing land area?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Edit:  Rephrasing scenario #3 for clarity.  What would be the best strategy such that $\sum\limits_{i=1}^n f(i)$ is optimized where $f(i)$ is the area given maximum amount of time $i$.&lt;/p&gt;&#xA;" OwnerUserId="1672" LastEditorUserId="1672" LastEditDate="2014-07-15T13:23:36.383" LastActivityDate="2014-07-16T13:14:08.130" Title="Land claim staking" Tags="|strategy|geometry|optimization|" AnswerCount="2" CommentCount="10" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1918" PostTypeId="2" ParentId="1917" CreationDate="2014-07-15T12:58:36.857" Score="2" Body="&lt;p&gt;It is clear that the area must be a polygon with stakes at corners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not have prove, but &lt;strong&gt;most probably&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The fastest way to put stake it each corner is to follow perimeter of the polygon. So we must optimise an area between the path and a line (which goes through first and last stake).   The maximum area is given by semicircle. (It is the shape of soap bubble put on a table, physical lows there optimise area of the bubble with given volume, which is virtually what we need). The aria would be $S = \pi(\frac{n}{\pi})^2 = n^2/\pi$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We should use regular polygon here with some number of corners $k$. But we must follow only half of it, similarly to semicircle (thanks &lt;a href=&quot;https://puzzling.stackexchange.com/a/1920/28&quot;&gt;kaine&lt;/a&gt; for insight on this). So side of polygon must be $A = (n-k/2\cdot m)/(k/2) = 2n/k-m$. The area of half polygon &lt;a href=&quot;http://www.mathwords.com/a/area_regular_polygon.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;would be&lt;/a&gt; $S = 1/8(2n/k-m)^2cot(180/k)$. One needs to find derivative and assign it to zero to find the maximum. Unfortunately I can't do it without computer, but what have to be done is quite clear, may be some one else would like to do it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Here the strategy must depend strongly on probability distribution of the given time $t$, which is not given &lt;em&gt;[Here I talk about initial formulation of the question, now it is changed].&lt;/em&gt;&lt;br&gt;&#xA;For standard uniform distribution from 0 to $T_{max}$, I don't know how to approach it strictly, but the wild guess would be to use logarithmic spiral, like in the task &quot;What is quickest way out from forest, if you lost?&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="28" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2014-07-15T19:41:46.893" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1920" PostTypeId="2" ParentId="1917" CreationDate="2014-07-15T18:59:03.933" Score="3" Body="&lt;p&gt;Klm123 is largely right but I am hoping to add to it. Obviously, the path he follows must remain on the perimeter of the shape or he is wasting running time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) A semicircle is correct.  Circles have the largest ratio of area to perimeter.  The semi-circle takes advantage of the longest straight line so he doesn't have to run it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) A regular polygon with $2x$ sides approximates a circle with the same radius as the polygons apothem.  $x$ is the number of stakes he places assuming there was already one at the start. $$A=\frac{(n-xm)^2}{4x \tan(\pi/2x)}$$ The derivative of this is too complex for wolfram to calculate the maximum for a general case but $x$ tends to be very small even if $m$ is low as they cause a relatively small improvement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) If he does not know how long he has, his strategy whilst constantly placing stakes is to maximize the derivative of the area with time.  Assume that he starts going in the $y$ direction on a parametric graph where $x$ is a function of $t$ describing his location perpendicular to his starting direction and $f(x)$ is his. (For ease f(x) is he same as y(x).) Lets say at any given time he is going in the direction $\phi$ relative to these axes at a given time: $$\frac{\partial A}{\partial t}=\frac{x(y+y')-y(x+x')}{2}=\frac{xy'-yx'}{2}=\frac{x \sin(\phi)-y \cos(\phi)}{2}$$ $$\frac{\partial^2 A}{\partial t \partial \phi}=\frac{x \cos(\phi)+y \sin(\phi)}{2}=0$$ $$\tan(\phi)=\frac{-x}{y}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This winds up making a spiral once we state that a zig zag is unacceptable. I am not satisfied with a spiral as that yields locations where he is wasting running time by covering the same area several times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As this is inherently an attempt to create a curve where the final line approaches being perpendicular to the first, a line along $f(x)=e^x$ seems to be very appropriate. He traps 227km after walking 100km.   Proving that is optimum...I can't seem to find a way to do that one way or he other.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditDate="2014-07-16T13:14:08.130" LastActivityDate="2014-07-16T13:14:08.130" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1922" PostTypeId="2" ParentId="1916" CreationDate="2014-07-15T19:26:24.610" Score="4" Body="&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; The first version contained a silly mistake (after the sixth weighing), which &lt;em&gt;justhalf&lt;/em&gt; has pointed out. I have corrected the solution, as well as the general comment at the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number the balls W1, W2, ..., W100 and B1, B2, ..., B101 in order of increasing weight. Bring out one table and two empty sacks. Put all the balls on the table in two ordered rows, white and black. Put one sack on the left side and one on the right side of the table. After each weighing, we will put all balls that we know are lighter than the middle ball in the left sack, and all balls that we know are heavier than the middle ball in the right sack. When we are done, there should be 100 balls in either sack and one ball remaining on the table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will always compare &lt;em&gt;one&lt;/em&gt; white ball against &lt;em&gt;one&lt;/em&gt; black ball, and we will pick each ball from the middle of its row. There will be an odd/even theme here, but let's just start and see what happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We begin by comparing W50 and B51. (The choice W51 and B51 gives similar results.) There are two possibilities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) If W50 &gt; B51 then the 51 black balls B1,...,B51 go into the left sack and the 50 white balls W51,...,W100 go into the right sack. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) If W50 &amp;lt; B51 then the 50 white balls W1,...,W50 go into the left sack and the 50 black balls B52,...,B101 go into the right sack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The worst case is (2), so we will focus on that. (In case (1), it is possible to leave B51 on the table and get the same situation as in case (2), if desired.) Rename the balls W1,...,W50 and B1,...,B51, and compare W25 with B26. The worst case is when W25 &amp;lt; B26, then W1,...,W25 go into the left sack and B27,...,B51 go into the right sack. We have 75 balls in each sack, and 25 white and 26 black balls on the table. (Note that this time it is the larger of these two numbers which is even!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, we rename the balls W1,...,W25 and B1,...,B26, and we compare W13 with B14. The worst case is when W13 &amp;lt; B14, then W1,...,W12 go into the left sack and B14,...,B26 go into the right sack. We now have 87 balls in the left sack, 88 balls in the right sack, and 13 white and 13 black balls on the table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rename tha balls W1,...,W13 and B1,...,B13, and compare W7 with B7. In either case, seven balls go into the left sack and six into the right sack. We have 94 balls in either sack, and 6 balls of one color (say white) and 7 balls of the other color (say black) on the table. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the next weighing (the fifth) we will in the worst case have 97 balls in each sack, and 3 white and 4 black balls on the table. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the sixth weighing we will in the worst case have 98 balls in the left sack, 99 balls in the right sack, and 2 balls of each color on the table. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rename the balls W1, W2 and B1, B2. Compare W2 with B2. If W2 &amp;lt; B2 then put W1 in the left sack and B2 in the right sack. If W2 &gt; B2 then put B1 in the left sack and W2 in the right sack. We are left with only two balls on the table, and in the eight and last weighing we compare them against each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Some general considerations&lt;/strong&gt;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this method, if there are 2k white balls and 2k+1 black balls on the table, with equally many balls in each sack, one will get k white balls and k+1 black balls on the table after one weighing (as always in the worst case). We can write this as $(2k,2k+1)\mapsto (k,k+1)$. Similarly, one can consider the other possibilities for even and odd numbers of balls, and get the following four possibilities for each weighing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(2k,2k+1)\mapsto(k,k+1), \qquad (2k+1,2k+2)\mapsto(k+1,k+1),$$&#xA;$$(2k+1,2k+1)\mapsto(k,k+1), \qquad (2k,2k)\mapsto (k,k).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If the number of white and black balls on the table is equal, then one sack has one ball more than the other, otherwise the sacks have the same number of balls.) &lt;/p&gt;&#xA;" OwnerUserId="219" LastEditorUserId="219" LastEditDate="2014-07-16T08:46:05.980" LastActivityDate="2014-07-16T08:46:05.980" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1923" PostTypeId="2" ParentId="1916" CreationDate="2014-07-16T03:40:41.850" Score="9" Body="&lt;h3&gt;Proof of minimality&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;8 is the minimum, with the following claims:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;In the worst case, the result of comparison can only be one of the two possibilities (left heavier or right heavier), since equality might not be present and hence we can't rely on that.&lt;/li&gt;&#xA;&lt;li&gt;Therefore one weighting gives 1 bit of information.&lt;/li&gt;&#xA;&lt;li&gt;There are 201 possible outcomes (each ball is a candidate of the median), which can't be encoded in 7 bits (7 bits give at most 128 possible outcomes)&lt;/li&gt;&#xA;&lt;li&gt;Therefore the minimum possible number of weightings is 8.&lt;/li&gt;&#xA;&lt;li&gt;Now it's left for us to prove that there is a solution with 8 weightings.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h3&gt;The 8 weightings&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Per Manne almost gets it correct, it's just that things get trickier after the sixth weighting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, assuming the same thing as what Per Manne has said (using left sack to store balls which are certainly lighter than the middle ball, and right sack to store those which are heavier), the procedure to find the $(k+1)$-th ball for $k$ white balls ($w_1, w_2, \ldots, w_k$) and $k'$ black balls ($b_1, b_2, \ldots, b_{k'}$), where $k'$ is either $k$ or $k+1$, is as follows.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Let $m = \lceil\frac{k+1}{2}\rceil$ and $n = \lceil\frac{k'+1}{2}\rceil$&lt;/li&gt;&#xA;&lt;li&gt;Compare $w_m$ with $b_n$&lt;/li&gt;&#xA;&lt;li&gt;&lt;ol&gt;&#xA;&lt;li&gt;If $w_m &amp;gt; b_n$ then put $b_1, b_2, \ldots, b_{n-1}$ on left sack and $w_m, w_{m+1}, \ldots, w_{k}$ on right sack&lt;/li&gt;&#xA;&lt;li&gt;If $w_m &amp;lt; b_n$ then put $w_1, w_2, \ldots, w_{m-1}$ on left sack and $b_n, w_{n+1}, \ldots, b_{k'}$ on right sack&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;Rename the balls, repeat to step 1 until we have 4 balls left (follow Per Manne's answer for more details) - you can actually continue this procedure until you get the result; the following steps are only for clarity.&lt;/li&gt;&#xA;&lt;li&gt;At 4 balls at two balls each (we have done 6 weightings up to this point), we have these 6 possible combinations:&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$w_1, w_2, b_1, b_2$&lt;/li&gt;&#xA;&lt;li&gt;$w_1, b_1, w_2, b_2$&lt;/li&gt;&#xA;&lt;li&gt;$b_1, w_1, w_2, b_2$&lt;/li&gt;&#xA;&lt;li&gt;$w_1, b_1, b_2, w_2$&lt;/li&gt;&#xA;&lt;li&gt;$b_1, w_1, b_2, w_2$&lt;/li&gt;&#xA;&lt;li&gt;$b_1, b_2, w_1, w_2$&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;Note that at this time we have one more ball in right sack compared to left sack, so we are looking for the third element here.&lt;/li&gt;&#xA;&lt;li&gt;Compare $w_2$ with $b_2$ (actually you would do this also if you follow step 1-4 above, this is just for clearer explanation)&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If $w_2 &amp;gt; b_2$ then we have either of the case (4), (5), or (6)&lt;/li&gt;&#xA;&lt;li&gt;If $w_2 &amp;lt; b_2$ then we have either of the case (1), (2), or (3)&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;At both case, we have three possibilities left. Since the two cases are analogous, we take the first three (1), (2), and (3).&lt;/li&gt;&#xA;&lt;li&gt;Compare $w_2$ and $b_1$.&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If $w_2 &amp;gt; b_1$ then the third element is $w_2$&lt;/li&gt;&#xA;&lt;li&gt;If $w_2 &amp;lt; b_1$ then the third element is $b_1$&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;Done&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note that after step 5 (that is, after 6 weightings), we only do two more weightings (step 6 and step 8), totaling to 8 weightings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore 8 is the minimum.&lt;/p&gt;&#xA;" OwnerUserId="1649" LastEditorUserId="1649" LastEditDate="2014-10-14T06:30:10.470" LastActivityDate="2014-10-14T06:30:10.470" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1924" PostTypeId="1" CreationDate="2014-07-16T07:07:10.837" Score="5" ViewCount="378" Body="&lt;p&gt;There is a problem from M.Gardner book &lt;em&gt;(&quot;Wheels, Life, and other Mathematical Amusements&quot;, p. 201)&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[Hallard T. Croft] asked if there existed a finite set of points on the plane such that the perpendicular bisector of the line segment joining any two points would always pass through at least two other points of the set.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The known solution to this problem is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You need 8 points. Take 4 points (A, B, C, D), put them in the corners of a square with side 1 (ABCD). Add 4 more points (E, F, G, H) outside of the square, each corresponds to one side of the square and creates an equilateral triangle with this side (ABE, BCF, CDG, DAH).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I wonder whether there is any other solution to this problem (with at least 2 points in the set)?&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="11876" LastEditDate="2016-08-03T22:50:19.577" LastActivityDate="2016-08-03T22:50:19.577" Title="Is there a second solution to points and perpendicular bisectors problem?" Tags="|geometry|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="1926" PostTypeId="1" CreationDate="2014-07-16T15:03:38.200" Score="4" ViewCount="362" Body="&lt;p&gt;Alice chooses a subset &lt;strong&gt;S&lt;/strong&gt; of {A,B,C,D,E}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bob makes a guess of any subset &lt;strong&gt;T&lt;/strong&gt; and Alice tells him the number of elements in &lt;strong&gt;S$\cap$T&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bob has to continue making guesses until he can exactly determine &lt;strong&gt;S&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that the minimum number of times Bob has to guess is 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: I have an algorithm that does it in 4 guesses but I do not have a proof that it is the lowest)&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="51105" LastEditDate="2018-09-19T14:55:50.593" LastActivityDate="2018-09-19T14:55:50.593" Title="How many subset intersections do you require?" Tags="|strategy|combinatorics|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="1928" PostTypeId="2" ParentId="1926" CreationDate="2014-07-16T17:36:34.260" Score="0" Body="&lt;p&gt;Here is the proof, if I understood the task correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subset $S$ either includes or doesn't each letter, so there are $2^5=32$ possible subsets. So you have $32$ combinations to chose between.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If guess $T$ includes $N$ letters it allows to chose between $N$ groups of combinations.&lt;br&gt;&#xA;&lt;strong&gt;One guess&lt;/strong&gt; may include all 5 letters (this is the best situation from information point of view). It gives nothing to guess $T=\{A,B,C,D,E\}$ two times, so &lt;strong&gt;another guess&lt;/strong&gt; must include 4 or less letters. In total you can chose between 20 or less combinations in two guesses. Therefore you will need &lt;strong&gt;3rd guess&lt;/strong&gt; to get additional information. Plus you will need &lt;strong&gt;4th one&lt;/strong&gt; $T=S$ to guess $S$ correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S Of course, you get guess the number correctly in 3 guesses, even in 1 guess, but you have to be lucky for this.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2014-07-17T16:42:48.657" LastActivityDate="2014-07-17T16:42:48.657" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="1929" PostTypeId="2" ParentId="1926" CreationDate="2014-07-16T19:02:07.073" Score="4" Body="&lt;p&gt;Guess the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{A,B,c} {A,D} {B,D} {C,E}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply the second answer by 5, the third by 25, and the fourth by 125.  Sum the answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each number will correspond to a different one of the 32 answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This requires 4 guess to determine the answer but doesn't state it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works by making inclusion omission patterns for each letter such that none are duplicated nor can be written as the sum of another 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only 4 non-interacting patterns exist for only 3 guesses. This means that if you want to know the answer after 3 guesses, there needs to be fewer letters to chose from.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Update&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I'd like to improve this slightly to make it easy to find the answer from the data without math skills.  Instead of $1,5,25,125$ multiply by $1,3,9,27$.  They work equally well but I prefer the former usually.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Call your numberic result $N_o$ and give me an number I can&#xA;manipulate $N$ that starts as $N=N_o$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As only $E$ uses an odd number of entry's, if you result is odd then &#xA;&quot;S&quot; contains an &quot;E&quot;. If it does, write it down and $N-27 \rightarrow &#xA;   N$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;No matter what number you have (it should be even now) perform $N/2  &#xA;   \rightarrow N$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $N$ is now odd, then you have a $B$. If it does, write it down and&#xA;$N-5 \rightarrow N$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;No matter what number you have (it should be even now) perform $N/2  &#xA;   \rightarrow N$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $N=0$, no more a present. If $N=1$, an $A$ is in $S$ but not $C$&lt;br&gt;&#xA;or $D$. If $N=3$, an $D$ is in $S$ but not $C$ or $A$. If $N=7$, an&lt;br&gt;&#xA;$C$ is in $S$ but not $A$ or $D$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The remaining $4$ options are the obvious combinations of these:&lt;br&gt;&#xA;$N=4$ means $AD$ no $C$. $N=8$ means $AC$ no $D$. $N=10$ means $CD$ no &#xA;$A$. $N=11$ means $AC$ and $D$ are all in $S$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditDate="2014-07-17T14:40:35.887" LastActivityDate="2014-07-17T14:40:35.887" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="1930" PostTypeId="2" ParentId="313" CreationDate="2014-07-16T22:20:23.273" Score="9" Body="&lt;p&gt;I wanted a solution that avoids uncertainty. So here's a good logical contradiction:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Is exactly one of these statements true?&lt;/p&gt;&#xA;&lt;p&gt;(a) Your number is not 2.&lt;/p&gt;&#xA;&lt;p&gt;(b) Your number is 3 and your answer to this question is Yes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Analysis&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;If the number is 1, then (a) is true and (b) is false, so the answer is &lt;code&gt;Yes&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;p&gt;If the number is 2, then (a) is false and (b) is false, so the answer is &lt;code&gt;No&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;p&gt;If the number is 3, then (a) is true. Now:&lt;/p&gt;&#xA;&lt;p&gt;If the answer is &lt;code&gt;Yes&lt;/code&gt;, then (b) is true and therefore the answer must be &lt;code&gt;No&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;p&gt;If the answer is &lt;code&gt;No&lt;/code&gt; then (b) is false and therefore the answer must be &lt;code&gt;Yes&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;p&gt;By contradiction, neither &lt;code&gt;Yes&lt;/code&gt; nor &lt;code&gt;No&lt;/code&gt; can be answered, and some other response must be used (e.g. &lt;code&gt;Unanswerable&lt;/code&gt;).&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Summary&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;If &lt;code&gt;Yes&lt;/code&gt;, then the number is 1.&lt;/p&gt;&#xA;&lt;p&gt;If &lt;code&gt;No&lt;/code&gt;, then the number is 2.&lt;/p&gt;&#xA;&lt;p&gt;If &lt;code&gt;Unanswerable&lt;/code&gt;, then the number is 3.&lt;/p&gt;&#xA;" OwnerUserId="1839" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-07-16T22:20:23.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1931" PostTypeId="1" AcceptedAnswerId="12163" CreationDate="2014-07-18T13:11:20.753" Score="7" ViewCount="1757" Body="&lt;p&gt;There are many &lt;strong&gt;Brainfuck&lt;/strong&gt; variations used in puzzles in geocaching, for example &lt;strong&gt;Ook!&lt;/strong&gt; They are usually not so hard to decode because they are 'standards' - therefore the substitution is googlable, there are often online converter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what is general strategy to dealing with such puzzles, if the only thing I know it's based on Brainfuck substitution? Or at least, I suppose it's a substitution? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've found that one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Buuga Uuga Uga Buga Búga Ùga Buuga Uuga Uga Ûga Úga Ùga Buuga Uuga Bùga Búga Bùga Bùga Bùga Ûga &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know (suppose) that the output is the code for a cipher lock. Judging for the number of the tokens, each Brainfuck token is likeyly to be substituted with separate uga token. Because brainfuck-encoded texts has very many repeating + on begin, and in that uga text there are little repetitions, I suppose that some uga symbols represent multiple +. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, encoding &lt;code&gt;1234&lt;/code&gt; gives the output:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;+++++ ++[-&gt; +++++ ++&amp;lt;]&gt; .+.+. +.&amp;lt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so any simple substitution would give visible repetitions. However, some symbol can encode &quot;+++&quot;, for example, or can be a &quot;special word&quot;, that encoded together with second one gives 2 up to 9 &quot;+&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any algorithms that would help to beat that beast, or the only way is to brute-force all possible substitutions? Taking into account, that the substitutions can be quite sophisticated, is it solvable at all?&lt;/p&gt;&#xA;" OwnerDisplayName="user66" LastEditorUserId="15544" LastEditDate="2015-09-04T10:57:56.577" LastActivityDate="2015-09-04T10:57:56.577" Title="Is there a general strategy for decoding Brainfuck mutations?" Tags="|cipher|calculation-puzzle|computer-science|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1932" PostTypeId="1" CreationDate="2014-07-18T19:42:46.380" Score="8" ViewCount="5019" Body="&lt;p&gt;The following puzzles were &lt;a href=&quot;http://news.generiq.net/LS24/ls24.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;created&lt;/a&gt; by Robert Lato to measure IQ in range between 160 and 190.&lt;/p&gt;&#xA;&lt;p&gt;Whats the solution to them?&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;You are supposed to draw in the dented boxes what the missing figure is for each of the five puzzles, the small symbols in top left are just the numbering of the test.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/V74Ev.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1885" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2018-02-03T04:03:19.713" Title="Spatial IQ test puzzles" Tags="|logical-deduction|pattern|" AnswerCount="2" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="1933" PostTypeId="2" ParentId="1926" CreationDate="2014-07-18T22:06:53.637" Score="1" Body="&lt;p&gt;Here's an alternative formulation that proves that it can't be done with three (&lt;em&gt;non-adaptive&lt;/em&gt;) guesses:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $U=\{A,B,C,D,E\}$ be the 'universe' here, with the subset we're guessing at then being $S\subseteq U$.  Label our three guesses as $T_0$, $T_1$, and $T_2$ (and note that this formulation assumes that we've made these choices beforehand; i.e., that $T_1$ is chosen without any knowledge of the value of $\left|T_0\cap S\right|$).  For each of the five elements $e\in U=\{A,B,C,D,E\}$ associate a 3-bit 'inclusion code' $I_e$ that says which of our three subsets of $U$ the element is in: for instance, if our guesses are $T_0=\{A,B,C,D\}$, $T_1=\{B,C\}$ and $T_2=\{C,D\}$ then the inclusion codes are $I_A=\langle1,0,0\rangle$ (since $A\in T_0$, $A\not\in T_1$, and $A\not\in T_2$), $I_B=\langle1,1,0\rangle$, $I_C=\langle1,1,1\rangle$, $I_D=\langle1,0,1\rangle$, and $I_E=\langle0,0,0\rangle$.  We'll call the number of 1s in an inclusion code its &lt;em&gt;weight&lt;/em&gt;; in this example, the weight of $I_A$ is 1, the weight of $I_C$ is 3, etc. (Note that for our guesses to have any chance of succeeding then every $I_e$ must have positive weight; otherwise we'll never be able to detect the presence/absence of $e$.  This leaves seven possible values of $I_e$, from which we'll be choosing five.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the 'winning condition' that we can successfully determine any subset of $U$ based on our guesses is exactly the condition that the 32 vector sums $\displaystyle\{\sum_{e\in E}I_e : E\subseteq U\}$ are distinct (since if any two of them are the same, then the corresponding subsets of $U$ are indistinguishable).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we can knock off the rest with some case-based analysis using the pigeonhole principle.  First, as noted above, we trivially can't have any of the $I_e=\langle0,0,0\rangle$.  Now, suppose we had e.g. $I_A=\langle1,1,1\rangle$.  Then we have four other $I_e$ to distribute amongst the weight-1 and weight-2 slots.  But any way you distribute those four, one of the weight-1 inclusion codes will be complementary to one of the weight-2 inclusion codes, and so they'll sum to $\langle 1,1,1\rangle$.  This means that there's no solution with a weight-3 inclusion code, and so the five inclusion codes must be distributed among the six possibilities of weight 1 or 2; but however this is done, either all the weight-one slots will be used (in which case some pair of them will add to one of the used weight-two slots) or all the weight-two slots will be used (in which case the pair of weight-one slots that are used will add to one of them).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may be a more direct way to show this based on linear algebra; another formulation of the 'all sums distinct' condition is that there's no linear combination $\sum_{e\in U}a_eI_e=\langle 0,0,0\rangle$ with the coefficients $a_e\in\{-1,0,1\}$.  Now, since we have five distinct vectors in $\mathbb{N}^3$ they're clearly linearly dependent, but it's not immediately clear to me that the usual proofs for linear dependence (e.g., Gram-Schmidt orthogonalization) won't lead to coefficients larger than 1.  If that hole can be patched, though, it's obviously a much cleaner proof.&lt;/p&gt;&#xA;" OwnerUserId="266" LastEditorUserId="266" LastEditDate="2014-07-18T23:12:02.040" LastActivityDate="2014-07-18T23:12:02.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1934" PostTypeId="1" AcceptedAnswerId="1936" CreationDate="2014-07-19T19:18:09.797" Score="4" ViewCount="806" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Countdown_%28game_show%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Countdown&lt;/a&gt; is a British game show in which people try and make the longest possible word out of nine given letters, reach a target number using six given numbers and basic arithmetic, and then solve a nine-letter anagram, each within 30 seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the numbers round, the players choose numbers from groups of &quot;large&quot; and &quot;small&quot; numbers. The large numbers contain one instance each of $25$, $50$, $75$, and $100$, while the small numbers contain two instances of each number from $1$ to $10$. They then use addition, subtraction, multiplication, division and bracketing with these numbers to try and reach a target number which can be anywhere between $100$ and $999$ inclusive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=pfa3MHLLSWI&quot; rel=&quot;nofollow noreferrer&quot;&gt;This video&lt;/a&gt; is famous for involving numbers as large as $23850$ in the solution process, where the starting numbers are $3,6,25,50,75,100$ and the target is: $$952 = \dfrac{(100+6)\cdot 3\cdot 75-50}{25} = \dfrac{23850-50}{25}$$&#xA;But is there a set of given numbers and a target number that requires even larger numbers as part of the solution? In particular, what set of numbers requires the largest intermediate number in the solution process?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that ideally the solution should be the &lt;strong&gt;only&lt;/strong&gt; solution.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2017-11-10T06:47:10.563" LastActivityDate="2017-11-10T13:07:01.470" Title="What's the largest number that can be obtained in the process of solving a numbers game on Countdown?" Tags="|calculation-puzzle|optimization|formation-of-numbers|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1935" PostTypeId="2" ParentId="1909" CreationDate="2014-07-20T05:40:30.720" Score="0" Body="&lt;p&gt;I would suspect the child knew he had mud on his forehead because he could feel it there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They don't know enough to to be able to employ other logic here.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The old man and the children are honest, able to employ logic, and want the dollar.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Each child must &lt;strong&gt;know&lt;/strong&gt; that the old man, and the other children are are honest, able to employ logic, and want the dollar.  The fact that the children are all honest does not imply that that know their fellow playmates are honest or that the old man is.  This knowledge is required for them to be able to make this logical leap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, what tends to hang me up about these puzzles (and I haven't yet seen mentioned) is that each child would have to know how fast the other children would react, if that child was the only one with mud on their forehead. The fact that only one wins (and there isn't a three-way draw) implies that they are not simply purely logical clones of each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I apologies if I am splitting hairs here.&lt;/p&gt;&#xA;" OwnerUserId="1893" LastActivityDate="2014-07-20T05:40:30.720" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1936" PostTypeId="2" ParentId="1934" CreationDate="2014-07-20T07:25:15.300" Score="8" Body="&lt;p&gt;Concerning the question &quot;is there a set of given numbers and a target number that requires even larger numbers as part of the solution?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, there is. Here we get $62550$ as an intermediate result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$834 = \frac{25\cdot 5\cdot 5\cdot 100+50}{75} = \frac{62550}{75}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;em&gt;Lopsy&lt;/em&gt; there is no other possible way to write $834$ using these numbers (see the first comment).&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2014-07-20T14:21:45.830" LastActivityDate="2014-07-20T14:21:45.830" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1941" PostTypeId="1" AcceptedAnswerId="1944" CreationDate="2014-07-21T14:36:56.570" Score="2" ViewCount="4723" Body="&lt;p&gt;One summer day John was jogging in the countryside wearing only his normal running clothes  (shorts, vest, sports shoes and socks).  After running for some time he saw that the lace of one of sports shoes was untied. While he was bending down to tie it up, a psychopath who was hidden nearby, approached him from behind and hit him on the head so hard that John became unconscious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When John recovered he found that he was inside a perfectly cylindrical room about 20 meters diameter and 5 meters high.  The walls, floor and ceiling of the room were perfectly smooth and symmetrical and had no visible feature so it was impossible to distinguish any particular direction inside the room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inside the room there was a cage. The cage itself also was in the shape of a perfect cylinder about 2 meters diameter and 2 meters high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cage was made of solid steel and it had only a very small spherical light bulb inside fixed exactly at the centre of the roof of the cage, and a solid door on one side. The light bulb was fixed to the roof, so that it rotated together with the cage. It was not possible to see anything outside the cage when someone was inside the cage and the door was closed (nor was it possible for someone outside the cage see inside after the door was closed)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cage was placed on a perfectly circular platform exactly the same diameter as the cage. The platform can rotate clockwise or anti-clockwise at the discretion of the psychopath, who uses a remote control to start and stop the rotation. The platform starts and stops rotating at a very slow rate of angular acceleration, so slow that someone inside the cage cannot even feel when the platform starts to rotate and when it stops. Also the maximum speed of rotation is so slow that a person inside the cage cannot feel any centrifugal force so he doesn’t even have any idea about how fast the cage is rotating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The psychopath then told John that he will challenge him to do a task. If he fails he will die, as have many others before. However, if he succeeds his psychopathic ego will be destroyed, John will be released and the psychopath will give himself up to the police.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The task consists of this. Surrounding the cage on the floor of the room about 10 meters from the center of the cylinder there are eight glasses each filled to the top with an orange liquid. The glasses are placed at the eight cardinal points: North, Northeast, East, Southeast, South, Southwest, West and Northwest. One of the glasses is filled up with pure orange juice. The other seven glasses have orange juice with a small amount of deadly poison mixed with the juice. It is not possible to distinguish between the glasses or their contents in any way at all except by tasting the liquid. John is not allowed to go anywhere near the glasses to examine them before the test starts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The psychopath then tells John that the only glass without poison is the glass placed in the north direction and tells him which one it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The psychopath does not allow John to leave any mark or object on the floor of the room by which to indicate a particular direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the psychopath tells John to get into the cage and the psychopath closes and locks the door. The psychopath remotely turns on the little light in the cage and gives John a couple of minutes to get used to the environment inside the cage. The psychopath then slowly starts to rotate the cage for about one minute then slowly stops the rotation  (which takes about another minute). The psychopath waits another couple of minutes then unlocks the door and tells John to get out and choose a glass from among the eight and to drink it all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before the psychopath unlocked the cage door he had already changed his own position in the room to make sure John was totally disoriented when he got out of the cage and had no possible way to tell from his surroundings in the room which direction he was facing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;John got out of the cage, quickly chose a glass of juice and drank it. The psychopath was amazed to see that John did not immediately fall down dead so he knew that John had chosen the right glass. He then told John to repeat task many more times until he realized that John had not chosen the right glass the first time just by luck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;John was released and the psychopath was put in prison for rest of his life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How did John manage to succeed in doing the task?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: Everything that is described could actually have happened in real life on this Earth.&lt;/p&gt;&#xA;" OwnerUserId="1885" LastEditorUserId="1885" LastEditDate="2014-07-21T14:45:29.773" LastActivityDate="2016-08-22T01:00:52.693" Title="The cage - a logic puzzle" Tags="|logical-deduction|" AnswerCount="10" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="1942" PostTypeId="2" ParentId="1911" CreationDate="2014-07-21T15:18:24.163" Score="2" Body="&lt;p&gt;The original game, perhaps, was &lt;a href=&quot;http://en.wikipedia.org/wiki/Word_ladder&quot; rel=&quot;nofollow&quot;&gt;Doublets&lt;/a&gt;, by Lewis Carroll. This had no addition or deletion, though variants allowed a variety of similar tasks. If so, wikipedia titles this form of game, &quot;Word Ladder&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1786" LastActivityDate="2014-07-21T15:18:24.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1943" PostTypeId="2" ParentId="1909" CreationDate="2014-07-21T16:05:25.980" Score="7" Body="&lt;p&gt;As with klm123, I don't know of a specific name for this type of problem (oracle riddle?).  However, I've usually seen this question in the context of &lt;a href=&quot;https://xkcd.com/blue_eyes.html&quot; rel=&quot;noreferrer&quot;&gt;an island with people of two eye colors and an oracle&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;In this case, the old man is the oracle, whether or not the children have mud on their face corresponds to whether the person has brown or blue eyes, and raising the hand is equivalent to leaving the island.&lt;/p&gt;&#xA;&lt;p&gt;This is a vaguer retelling in a more natural setting, however.  Only one child raises his hand, even though all of them should be able to reach the conclusion at the same time.  Likewise, the oracle problem has a defined amount of time that has to pass, rather than waiting for some arbitrary &amp;quot;Oh, he didn't react fast enough, which gives me information&amp;quot; point.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/a/242/325&quot;&gt;Gilles' explanation&lt;/a&gt; of the oracle problem is a great answer, and I'll quote a bit of it here:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Suppose that only Alice has blue eyes. Before day 0 [when the oracle speaks], she never knew anyone with blue eyes. On day 0, she learns that someone has blue eyes; since nobody else does, it has to be her and only her, so she takes the ferry that night.&lt;/p&gt;&#xA;&lt;p&gt;Now suppose that only Alice and Bill have blue eyes. Before day 0, Bill already knew that there was someone with blue eyes, but he did not know that Alice knew. If Bill had had green eyes, Alice would have been the only blue-eyed person and would not have known. On the first night after the guru, Alice doesn't leave; this tells Bill that Alice did not know the color of her eyes, so learn that she was the only blue-eyed person. Since Bill knows that either Alice is the only blue-eyed person or Bill and Alice are the only two, Bill now knows that both he and Alice have blue eyes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="325" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-07-21T16:05:25.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1944" PostTypeId="2" ParentId="1941" CreationDate="2014-07-21T16:47:33.527" Score="3" Body="&lt;p&gt;I'm not sure this would actually work but the implied solution sends to involve inertia.  Tie a shoelace from the light, dangle the heaviest thing you can from it,  and note how it deflects when the cage rotates.  The issue is how to make this system freely rotating/friction free to keep the (likely) shoe from rotating. &lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2014-07-21T16:47:33.527" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="1945" PostTypeId="2" ParentId="1941" CreationDate="2014-07-21T20:30:39.503" Score="10" Body="&lt;p&gt;Since we don't have any visual cues to measure the end orientation, John needs some way to judge how much he rotated. That'd require very good timekeeping and some way to measure the angular acceleration. We'll assume that John has perfect mental and physical skills and can track time, angular displacement, etc exactly. The first idea would be to hang one of his shoes from the lightbulb and measure the deflection from vertical. There's a couple of problems with this, though. One is that there's a lot of friction, and two is that since the lightbulb is in the center of the room the shoe won't actually displace. It will rotate, but it'll be very hard to make the shoe start out perfectly still, plus at some point the torque in the shoelace will mess with the rotation calculations so that John dies at the end. So what can we do instead?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can juggle!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;John starts juggling two socks in one hand. The socks are &quot;almost always&quot; in free motion, meaning they'll start revolving independently of John. If he walks a little to properly catch them, the angle he makes with the center of the room and the door will roughly equal the amount the cage rotated (although in the opposite direction). He can keep doing this until the rotation stops and calculate how much the cage rotated overall in order to find true north. A cool consequence of this solution is that if the cage remained still but the rest of the room rotated, then John would be out of luck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If John could also throw and catch objects perfectly, he can do this without juggling by throwing a shoe into the air with no angular momentum. The amount it rotates before he catches it, then, will be the amount the room rotates too.&lt;/p&gt;&#xA;" OwnerUserId="1906" LastActivityDate="2014-07-21T20:30:39.503" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1946" PostTypeId="1" AcceptedAnswerId="1947" CreationDate="2014-07-22T22:30:46.690" Score="1" ViewCount="677" Body="&lt;p&gt;Picture this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are in a hallway of a very large office building. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Near you are 50 light switches of which 3 are unlabled. These lightswitches are connected to 3 lights in 3 separate rooms. The rooms are not directly in your line of sight as there are 47 other rooms too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes it that walking up and down from the lightswitches to the rooms a tedious task that you want to avoid. Because the other 47 lightswitches do have lables, you do know which 3 of the 50 rooms have yet to be labled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your task is to efficiently find out which lightswitch is connected to which room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give an idea of the challenge, it means that you walk from the lightswitches to the 3 rooms then back and know which switch is connected to which room so you can label them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would you proceed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I do know the answer, but will refrain of answering my own question unless everyone gives up.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1915" LastEditorUserId="1915" LastEditDate="2014-07-22T22:54:19.037" LastActivityDate="2014-07-23T12:15:11.943" Title="3 Switches identification" Tags="|logical-deduction|lateral-thinking|" AnswerCount="1" CommentCount="7" ClosedDate="2014-07-26T17:52:24.103" ContentLicense="CC BY-SA 3.0" />
  <row Id="1947" PostTypeId="2" ParentId="1946" CreationDate="2014-07-23T12:15:11.943" Score="3" Body="&lt;ol&gt;&#xA;&lt;li&gt;On first switch among the three and keep it on for 10 mins.&lt;/li&gt;&#xA;&lt;li&gt;Off the first switch and on the second switch.&lt;/li&gt;&#xA;&lt;li&gt;Now proceed towards the room.&lt;/li&gt;&#xA;&lt;li&gt;The room in which light bulb is hot (you need to touch the light bulb) belongs to first switch&lt;/li&gt;&#xA;&lt;li&gt;The room in which light bulb is on belongs to second switch which we have left on before coming towards rooms.&lt;/li&gt;&#xA;&lt;li&gt;The room in which light is not on belongs to third switch, which we have not switched on yet.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1917" LastActivityDate="2014-07-23T12:15:11.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1948" PostTypeId="2" ParentId="1909" CreationDate="2014-07-23T13:39:21.787" Score="1" Body="&lt;p&gt;I've seen this puzzle with men and coloured hats.  The way it was framed was in stages, which led the reader (by the nose) to the solution of the most complex case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The puzzle was simple - there are three men, all wearing hats, and a diabolical king.  The hats are either white or black.  The men can see each other's hat, but not their own.  The king tells them that if they can guess the colour of their hat they need to simply raise their hand and (if they are right) they will live.  The king tells them that there is at least one white hat, and the men are not allowed to communicate in any way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each scenario, your task is to determine who can figure out what they are wearing.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;One man is wearing a white hat.&lt;br&gt;&#xA;Solution: The two men in black hats see one white and one black, but cannot determine their own hat colour, so they do not immediately raise their hands.  The man in the white hat sees two black hats, and since he knows at least one hat is white, he raises his hand immediately.  When the other two see this, they know he has seen two black hats, so then they know the colour of their own hats and are able to raise their hands as well.&lt;/li&gt;&#xA;&lt;li&gt;Two men are wearing white hats.&lt;br&gt;&#xA;Solution: The two men wearing white see one white and one black, and the man wearing black sees two white.  If only one man was wearing white, he would have raised his hand immediately as in the first scenario.  The men know this, and since no one raised their hand immediately, they determine that there is more than one white hat.  The two men who see one white and one black correctly deduce that they must be wearing white and raise their hand.  The last man in black sees that they hesitated before raising their hands, so is also able to rule out the first scenario.  Since they were able to figure out their hat colour relatively soon after, he then correctly guesses that his hat must be black.&lt;/li&gt;&#xA;&lt;li&gt;All three are wearing white hats.&lt;br&gt;&#xA;Solution: All three men see two white hats.  If it had been the previous scenario, then two of them would have raised their had after some initial hesitation.  Since this doesn't happen, they can all infer that they must be wearing white.  Thus after a second hesitation, they all raise their hand.  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In a twist, the question is sometimes asked as follows (same premise as above).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a lengthy pause, all three men finally raise their hands.  What colour were their hats?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the above logic, we can infer that they were all wearing white.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2014-07-23T13:39:21.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1949" PostTypeId="2" ParentId="1941" CreationDate="2014-07-23T21:38:54.033" Score="3" Body="&lt;ol&gt;&#xA;&lt;li&gt;John breaks off the zipper pull from his vest, which conveniently happens to be made of a ferrous metal.&lt;/li&gt;&#xA;&lt;li&gt;He then rubs it on his shorts in one direction. His shorts, conveniently made of silk, do the job of magnetizing the zipper pull.&lt;/li&gt;&#xA;&lt;li&gt;He rips a bit of buoyant foam from in-sole of one of his running shoes, making a circle as small as possible while still having enough buoyancy to hold up the zipper pull.&lt;/li&gt;&#xA;&lt;li&gt;John places his other shoe in the center of the cage, easily found by dropping something from just below the light bulb.&lt;/li&gt;&#xA;&lt;li&gt;Being a lifelong runner and very health-conscious, John is easily hydrated enough to urinate into his other running shoe, filling it up to the top and letting the liquid settle.&lt;/li&gt;&#xA;&lt;li&gt;He floats the magnetized zipper pull on the body of liquid with his circle of buoyant foam, noting its orientation relative to the cup he knows is North.&lt;/li&gt;&#xA;&lt;li&gt;The zipper pull will eventually return to that orientation, regardless of how the cage was rotated.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;He can repeat this process as many times as he wishes, since the would-be killer is content to keep giving him plenty of fluids throughout the process.&lt;/p&gt;&#xA;" OwnerUserId="1647" LastActivityDate="2014-07-23T21:38:54.033" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="1950" PostTypeId="1" AcceptedAnswerId="1952" CreationDate="2014-07-24T06:45:17.137" Score="7" ViewCount="538" Body="&lt;p&gt;You are given a sheet of paper with &lt;span class=&quot;math-container&quot;&gt;$100$&lt;/span&gt; sentences, in the form of: There are exactly &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; wrong sentences on this sheet.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;There is exactly &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; wrong sentence on this sheet.&lt;/p&gt;&#xA;&lt;p&gt;There are exactly &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; wrong sentences on this sheet.&lt;/p&gt;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&lt;p&gt;There are exactly &lt;span class=&quot;math-container&quot;&gt;$99$&lt;/span&gt; wrong sentences on this sheet.&lt;/p&gt;&#xA;&lt;p&gt;There are exactly &lt;span class=&quot;math-container&quot;&gt;$100$&lt;/span&gt; wrong sentences on this sheet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Your task is find out which sentence says the truth (there can be only one).&lt;/p&gt;&#xA;" OwnerUserId="1924" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-07-24T18:42:03.140" Title="Number of true sentences" Tags="|logical-deduction|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1952" PostTypeId="2" ParentId="1950" CreationDate="2014-07-24T08:17:22.753" Score="8" Body="&lt;p&gt;It depends on what you mean by the wording of the sentences: Does &quot;There is 1 wrong sentence on this sheet&quot; mean &quot;There is exactly 1 wrong sentence on this sheet,&quot; or does it mean &quot;There exists 1 wrong sentence on this sheet (but there may be more)&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is the first case, the answer is &lt;strong&gt;99&lt;/strong&gt;, because there are 100 sentences total on the sheet, and all but one of them are wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is the second case, the answer is &lt;strong&gt;50&lt;/strong&gt;. Sentences #1 to #50 inclusive are correct, and sentences #51 to #100 inclusive are incorrect.&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2014-07-24T08:17:22.753" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1953" PostTypeId="1" AcceptedAnswerId="1954" CreationDate="2014-07-24T09:44:03.977" Score="13" ViewCount="1197" Body="&lt;p&gt;You are kidnapped and your offender plays a game with you. In front of you are 2 boxes, containing a total of 50 white balls and 50 black balls. The kidnapper will pull out a ball from one of the two boxes, at random. If the ball is white you survive, if not, you are shot.&lt;/p&gt;&#xA;&lt;h1&gt;Rules:&lt;/h1&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Every ball must be in one of the boxes.&lt;/li&gt;&#xA;&lt;li&gt;No balls can be left out.&lt;/li&gt;&#xA;&lt;li&gt;Each box must have at least one ball.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;h1&gt;Example:&lt;/h1&gt;&#xA;&lt;p&gt;Box 1: 25 black balls and 25 white balls Box 2: 25 black balls and 25 white balls.&lt;/p&gt;&#xA;&lt;p&gt;In this example the chance to live is obviously &lt;span class=&quot;math-container&quot;&gt;$50 \% $&lt;/span&gt;. How can you distribute the balls so as to increase that chance?&lt;/p&gt;&#xA;" OwnerUserId="1924" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-02-11T23:50:05.527" Title="How to survive, with statistics" Tags="|strategy|probability|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1954" PostTypeId="2" ParentId="1953" CreationDate="2014-07-24T10:43:38.587" Score="22" Body="&lt;p&gt;Box 1: 50 black balls, 49 white balls Box 2: 0 black balls, 1 white ball.&lt;br&gt;&#xA;In this example the chance to live is obviously $50\%+50\%*49/99 \approx 75\%$. There is no way to increase that chance.&lt;/p&gt;&#xA;" OwnerUserId="28" LastActivityDate="2014-07-24T10:43:38.587" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1955" PostTypeId="1" AcceptedAnswerId="1956" CreationDate="2014-07-24T19:23:51.600" Score="8" ViewCount="744" Body="&lt;p&gt;Epimenides (a Cretan) once wrote a poem, in which he stated that all Cretans are liars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since he is a Cretan, and therefore a liar, Cretans are veracious.&lt;br&gt;&#xA;But then again he wouldn't be a liar!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This paradox can be solved, i.e. it can be shown that it's not an actual paradox. How?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: In this puzzle's &quot;universe&quot;, liars always lie. I don't know how to precicely explain it, but &quot;black and white logic&quot; applies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, there are at least 52 Cretans.&lt;/p&gt;&#xA;" OwnerUserId="1910" LastEditorUserId="1910" LastEditDate="2014-07-24T20:05:31.867" LastActivityDate="2014-07-29T08:25:41.203" Title="Epimenides paradox" Tags="|paradox|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1956" PostTypeId="2" ParentId="1955" CreationDate="2014-07-24T19:46:28.833" Score="11" Body="&lt;p&gt;The negation of &quot;All x are y&quot; is &quot;There is at least one x which is not y&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, Epimenides is a liar. Therefore his statement &quot;All Cretans are liars&quot; is false. This means that not all Cretans are liars. This means that at least one Cretan tells the truth. He can still be a liar, there just has to be at least one Cretan who's not a liar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if Epimenides is the only Cretan, we'd have a bigger problem.&lt;/p&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2014-07-24T19:46:28.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1957" PostTypeId="1" AcceptedAnswerId="1982" CreationDate="2014-07-25T22:19:42.327" Score="11" ViewCount="628390" Body="&lt;p&gt;In a 3x3 grid, I'd have to put numbers from 1 to 9 in a manner so that respective row, column and diagonal add up to 15.&lt;/p&gt;&#xA;&lt;p&gt;I have only been able to come up with one solution:&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th&gt;&lt;/th&gt;&#xA;&lt;th&gt;&lt;/th&gt;&#xA;&lt;th&gt;&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;6&lt;/td&gt;&#xA;&lt;td&gt;1&lt;/td&gt;&#xA;&lt;td&gt;8&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;7&lt;/td&gt;&#xA;&lt;td&gt;5&lt;/td&gt;&#xA;&lt;td&gt;3&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;2&lt;/td&gt;&#xA;&lt;td&gt;9&lt;/td&gt;&#xA;&lt;td&gt;4&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;Through some calculations and trial and error method.&lt;/p&gt;&#xA;&lt;p&gt;Is there any strategy or way of approach to this problem, or is trial and error method the solution to it?&lt;/p&gt;&#xA;" OwnerUserId="1935" LastEditorUserId="69582" LastEditDate="2020-12-30T19:13:24.890" LastActivityDate="2021-01-03T01:47:06.637" Title="Puzzle of putting numbers 1-9 in 3x3 Grid to add up to 15" Tags="|logical-deduction|calculation-puzzle|magic-square|" AnswerCount="5" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="1958" PostTypeId="2" ParentId="1957" CreationDate="2014-07-25T23:50:23.813" Score="10" Body="&lt;p&gt;There are 8 equations which must must add to $15$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a+b+c=15$$ $$d+e+f=15$$ $$ g+h+I=15$$ $$ a+d+g =15$$ $$b+e+h =15$$ $$c+f+I=15$$ $$ a+e+I =15$$ $$ c+e+g=15$$  I would place these in excel personally. So adding equations $2,5,7,8$ and then subtracting equations $1,2,3$ yields $3e=15$ so $ e=5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No others are solvable without guessing one but from there you just solve them one at a time. Guess $1$ is in a corner, requires $9$ in the other corner. There must be two pairs which add to $14$. The only one is $8+6$ so $1$ cannot be in the corner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Guess $1$ not in a corner, $9$ must still be opposite to it. $8$ and $6$ must be in the adjacent corners. The other corners can be solved by the diagonal equations ($2$ and $4$). Finally the last two squares can be easily solved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are only 8 different solutions and because this logic eliminates any others; they are all mirrors/rotations of the one you already posted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course there are some made from larger squares....&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="9346" LastEditDate="2015-09-22T17:19:04.597" LastActivityDate="2015-09-22T17:19:04.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1959" PostTypeId="1" CreationDate="2014-07-26T11:21:32.420" Score="0" ViewCount="47875" Body="&lt;p&gt;You are given two candles of equal size, which can burn 1 hour each. You have to measure 90 minutes with these candles. (There is no scale or clock). You are also given a lighter.&lt;/p&gt;&#xA;" OwnerUserId="1938" LastEditorUserId="39" LastEditDate="2014-07-27T23:32:20.213" LastActivityDate="2014-07-30T12:33:01.907" Title="Candle timing measurement" Tags="|calculation-puzzle|" AnswerCount="3" CommentCount="4" FavoriteCount="0" ClosedDate="2014-07-30T17:00:40.060" ContentLicense="CC BY-SA 3.0" />
  <row Id="1960" PostTypeId="2" ParentId="1959" CreationDate="2014-07-26T11:33:50.203" Score="3" Body="&lt;p&gt;OK. Old Infosys interview question.&#xA;As @klm123 said, you burn both ends of one candle. It will burn in 30 minutes and you burn the other one from one end only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initially, the problem was with &lt;code&gt;ropes / threads&lt;/code&gt; with each of &lt;code&gt;timespan 30 minutes&lt;/code&gt; and you have to &lt;code&gt;measure 45 minutes with 2 ropes&lt;/code&gt;; with which it was quite affirmative and little easy. But someone (NOT the asker here) become thinking himself/herself very smart changed the problem with candles and increased time to double. The smarty who changed it probably had no / little understanding of Physics / Chemistry.The candle will burn quicker if burned on both ends by placing it horizontal (burn a little faster from both sides) or vertical (burn quite faster from one side). So, we won't get exact 30 mins but have to approximate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rope problem on math exchange and glassdoor . Take a look : &lt;a href=&quot;https://math.stackexchange.com/questions/183282/puzzle-there-are-two-lengths-of-rope&quot;&gt;https://math.stackexchange.com/questions/183282/puzzle-there-are-two-lengths-of-rope&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.glassdoor.co.in/Interview/You-have-2-pieces-of-rope-each-of-which-burns-from-one-end-to-the-other-in-30-minutes-no-matter-which-end-is-lit-If-dif-QTN_108377.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.glassdoor.co.in/Interview/You-have-2-pieces-of-rope-each-of-which-burns-from-one-end-to-the-other-in-30-minutes-no-matter-which-end-is-lit-If-dif-QTN_108377.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1939" LastEditorUserId="-1" LastEditDate="2017-04-13T12:19:36.950" LastActivityDate="2014-07-30T12:33:01.907" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1961" PostTypeId="1" AcceptedAnswerId="1962" CreationDate="2014-07-26T13:57:06.520" Score="7" ViewCount="646" Body="&lt;p&gt;Fill in the missing number: 10, 30, 60, 01,__, 12&lt;br/&gt;&#xA;Use a bit of lateral thinking. &lt;br/&gt;&#xA;And math.&lt;br/&gt;&#xA;Also, I do know the answer. Just giving the question as a teaser.&lt;/p&gt;&#xA;" OwnerDisplayName="user360" LastEditorUserId="4050" LastEditDate="2015-01-03T17:56:29.393" LastActivityDate="2020-04-09T16:47:03.800" Title="Pattern 1 -- Numbers" Tags="|mathematics|pattern|number-sequence|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1962" PostTypeId="2" ParentId="1961" CreationDate="2014-07-26T16:06:42.500" Score="11" Body="&lt;p&gt;Let's try reversing the digits:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;01, 03, 06, 10, __, 21&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then I see this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;01 + &lt;strong&gt;2&lt;/strong&gt; = 03&lt;br&gt;&#xA;  03 + &lt;strong&gt;3&lt;/strong&gt; = 06&lt;br&gt;&#xA;  06 + &lt;strong&gt;4&lt;/strong&gt; = 10&lt;br&gt;&#xA;  10 + &lt;strong&gt;5&lt;/strong&gt; = &lt;strong&gt;15&lt;/strong&gt;&lt;br&gt;&#xA;  15 + &lt;strong&gt;6&lt;/strong&gt; = 21  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As 15 is the answer when the digits are reversed, the missing number in your sequence is &lt;strong&gt;51&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="153" LastActivityDate="2014-07-26T16:06:42.500" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1963" PostTypeId="1" AcceptedAnswerId="1964" CreationDate="2014-07-26T22:03:31.047" Score="48" ViewCount="288352" Body="&lt;p&gt;Here's a fun (albeit difficult) one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make these equations true using arithmetic operations:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 1 1 = 6&#xA;2 2 2 = 6&#xA;3 3 3 = 6&#xA;4 4 4 = 6&#xA;5 5 5 = 6&#xA;6 6 6 = 6&#xA;7 7 7 = 6&#xA;8 8 8 = 6&#xA;9 9 9 = 6&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For example: &lt;code&gt;6 + 6 - 6 = 6&lt;/code&gt; (I hope I did not spoil some of you :D)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Allowed operators are:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;+, -, *, /, ! , ^, %&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Setting parenthesis is also allowed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;code&gt;^&lt;/code&gt; operator is an exception as you are permitted to supply a second argument to it which may be any positive integer or the multiplicative inverse of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^{1/y}$ is &lt;strong&gt;always&lt;/strong&gt; positive and real.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you find an alternative solution using other operators you may post it but please also provide a solution using only these 7 operators.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For those of you who think this was easy, here is a bonus:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0 0 0 = 6&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1943" LastEditorUserId="1943" LastEditDate="2014-07-27T15:31:09.960" LastActivityDate="2019-04-30T19:06:28.153" Title="6, the magic number" Tags="|mathematics|" AnswerCount="11" CommentCount="20" ContentLicense="CC BY-SA 3.0" />
  <row Id="1964" PostTypeId="2" ParentId="1963" CreationDate="2014-07-26T23:26:11.823" Score="44" Body="&lt;p&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$(1+1+1)! = 6 $&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$2+2+2 = 6$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$3*3-3 = 6$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$\left(4-\frac 4 4\right)! = \sqrt 4+\sqrt 4+\sqrt 4=6$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$5+\frac 5 5 = 6$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$6*\frac 6 6 = 6 + 6 -6=6$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;7.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$7-\frac 7 7 = 6$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;8.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$\left(\sqrt{8+\frac 8 8}\right)! = 6$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;9.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$\left(\frac{\sqrt{9}\sqrt{9}}{\sqrt 9}\right)! = 6$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bonus: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$(0!+0!+0!)! = 6$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1944" LastEditorUserId="56016" LastEditDate="2019-04-30T19:06:28.153" LastActivityDate="2019-04-30T19:06:28.153" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="1965" PostTypeId="1" AcceptedAnswerId="1967" CreationDate="2014-07-26T23:59:25.407" Score="23" ViewCount="12587" Body="&lt;p&gt;You are given ten stacks of golden coins, each stack consisting of ten coins and a digital scale with arbitrary precision. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know that all stacks of coins are made from gold, weighing 10 grams per coin except one stack, which is made from silver but is painted golden. A silver coin weighs 1 gram less than a golden one, which makes its weight 9 grams. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With only being able to weigh exactly &lt;strong&gt;once&lt;/strong&gt;, how can you identify the stack containing the silver ones?&lt;/p&gt;&#xA;" OwnerUserId="1943" LastEditorUserId="8874" LastEditDate="2015-01-31T12:29:16.030" LastActivityDate="2020-07-10T14:37:50.480" Title="Coin weighing problem" Tags="|mathematics|logical-deduction|weighing|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1966" PostTypeId="2" ParentId="1963" CreationDate="2014-07-27T00:07:27.997" Score="8" Body="&lt;p&gt;The bottom five (0 through 4) can all be solved using the same construction:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;     &lt;code&gt;(0!+0!+0!)! = 6&lt;/code&gt;&lt;br&gt;&#xA;     &lt;code&gt;(1 +1 +1 )! = 6&lt;/code&gt;&lt;br&gt;&#xA;     &lt;code&gt;(2 +2 /2 )! = 6&lt;/code&gt;&lt;br&gt;&#xA;     &lt;code&gt;(3 +3 %3 )! = 6&lt;/code&gt;&lt;br&gt;&#xA;     &lt;code&gt;(4 -4 /4 )! = 6&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For 6 and 7, there are slightly more funky solutions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;     &lt;code&gt;(6!)%(6!-6)=6&lt;/code&gt;&lt;br&gt;&#xA;     &lt;code&gt;((7!)/7)%7=6&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(I haven't found an interesting solution for 5, nor any square-root-free solutions for 8 or 9.)&lt;/p&gt;&#xA;" OwnerUserId="1704" LastEditorUserId="1704" LastEditDate="2016-11-13T08:51:17.237" LastActivityDate="2016-11-13T08:51:17.237" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1967" PostTypeId="2" ParentId="1965" CreationDate="2014-07-27T06:28:01.487" Score="28" Body="&lt;ol&gt;&#xA;&lt;li&gt;You can number stacks from 1 to 10.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 2. From each stack take as many coins as the stack number. &lt;br&gt;&#xA; 3. Weigh them all. &lt;br&gt;&#xA; 4. Subtract the weight from 10+20+...+100=550. The result is the number of the silver stack.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;P.S. You can do the same even if you have 11 stacks of 10: just number them from 0 to 10.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="63414" LastEditDate="2020-07-10T14:36:46.190" LastActivityDate="2020-07-10T14:36:46.190" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="1970" PostTypeId="2" ParentId="1932" CreationDate="2014-07-27T10:00:13.533" Score="1" Body="&lt;p&gt;The third one seems to be rule based but can be interpreted in two ways:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The square binding prevents movement from the inner four tiles.&lt;/li&gt;&#xA;&lt;li&gt;The long line needs an arrow binding or it rotates.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Seeing as the first compares best to the fourth picture the most obvious choice would be the first one. Which makes me think it would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GSvHL.jpg&quot; alt=&quot;possible solution&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clarification&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shape can be displayed in a 4x4 grid in all examples. Seeing the size of the possible solution makes me believe the solution will be a 4x4 grid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The squares in picture one are in the inner 2x2 block and move away from the center in the next images. In the third picture it would be very likely for these squares to have the position (-1,-1)(5,5). Because of this it would be logical for this to happen in the solution as the fourth picture takes the same position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The outside half squares follow the rule; if you are not in the inner 2x2 grid then invert your position. If you are inside the 2x2 grid then hold your position for a single turn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last rule is that of the rotation through lack of squares in the field. When there are no squares in the field the grid rotates (either left or right) by 90 degrees and squares are created at (2,2)(3,3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for the actual solution the half squares would be inverted and moved to the inner 2x2 grid. The squares move to the outside and the center lines retain there position because of the presence of squares.&lt;/p&gt;&#xA;" OwnerUserId="1945" LastEditorUserId="1945" LastEditDate="2014-07-27T16:07:00.293" LastActivityDate="2014-07-27T16:07:00.293" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1971" PostTypeId="2" ParentId="1963" CreationDate="2014-07-27T21:05:19.800" Score="32" Body="&lt;p&gt;I insist on using all the digits!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(1 + 1^{1234567890} + 1)! = 6$&lt;br&gt;&#xA;&lt;br&gt;&#xA; $(2 + (2^{1234567890}\ \text{mod}\ 2)!)! = 6$&lt;br&gt;&#xA;&lt;br&gt;&#xA; $(3 + 3^{1234567890}\ \text{mod}\ 3)! = 6$&lt;br&gt;&#xA;&lt;br&gt;&#xA; $(4 - (4^{1234567890}\ \text{mod}\ 4)!)! = 6$&lt;br&gt;&#xA;&lt;br&gt;&#xA; $5 + (5^{1234567890}\ \text{mod}\ 5)! = 6$&lt;br&gt;&#xA;&lt;br&gt;&#xA; $6 + 6^{1234567890}\ \text{mod}\ 6 = 6$&lt;br&gt;&#xA;&lt;br&gt;&#xA; $7 - (7^{1234567890}\ \text{mod}\ 7)! = 6$&lt;br&gt;&#xA;&lt;br&gt;&#xA; $(\sqrt[3]8 + (8^{1234567890}\ \text{mod}\ 8)!)! = 6$&lt;br&gt;&#xA;&lt;br&gt;&#xA; $(\sqrt{9} + (9^{1234567890}\ \text{mod}\ 9))! = 6$&lt;br&gt;&#xA;&lt;br&gt;&#xA; $(0! + (0^{1234567890})! + 0!)! = 6$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, wait!  How about if we take subtraction out and put subfactorial in?  More exclamation points!!!!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $((!1)! + (!1)! + (!1)!)! = 6$&lt;br&gt;&#xA;&lt;br&gt;&#xA; $(!2 + !2 + !2)! = 6$&lt;br&gt;&#xA;&lt;br&gt;&#xA; $!3 + !3 + !3 = 6$&lt;br&gt;&#xA;&lt;br&gt;&#xA; $(\sqrt{!4} \times 4 \div 4)! = 6$&lt;br&gt;&#xA;&lt;br&gt;&#xA; $!(\sqrt{!5\ \text{mod}\ 5}) + 5 = 6$&lt;br&gt;&#xA;&lt;br&gt;&#xA; $!6\ \text{mod}\ 6 \times 6 = 6$&lt;br&gt;&#xA;&lt;br&gt;&#xA; $!7\ \text{mod}\ 7\ \text{mod}\ 7 = 6$&lt;br&gt;&#xA;&lt;br&gt;&#xA; $(!8\ \text{mod}\ 8 + \sqrt[3]8)! = 6$&lt;br&gt;&#xA;&lt;br&gt;&#xA; $\sqrt[3]{!9\ \text{mod}\ 9} \times \sqrt9 = 6$&lt;br&gt;&#xA;&lt;br&gt;&#xA; $(!0 + !0 + !0)! = 6$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1923" LastEditorUserId="4421" LastEditDate="2015-10-07T17:34:01.003" LastActivityDate="2015-10-07T17:34:01.003" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1973" PostTypeId="2" ParentId="236" CreationDate="2014-07-28T00:06:44.530" Score="58" Body="&lt;p&gt;Every blue-eyed person sees 99 blue-eyed people. Since they don't know that they have blue eyes, they suspect it might be the case that every other blue-eyed person can only see 98 blue-eyed people, and if those people only see 98 blue-eyed people, they might think that each of &lt;em&gt;them&lt;/em&gt; only see 97 blue-eyed people. And so it continues, until someone considers a hypothetical situation in which someone sees no blue-eyed people. Then the guru, in this hypothetical, really does make a difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the essential piece of information the Guru provides is that everyone knows that everyone knows that everyone knows that [... etc. ...] everyone knows that there is someone on the island with blue eyes. This enables everyone to discard that nested hypothetical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be easier if we assign everyone numbers. People 1 to 100 have blue eyes. Person 1 sees 99 people with blue eyes, so suspects that Person 2 might see only 98 people with blue eyes, in which case Person 2 would think that Person 3 might only be able to see 97 people with blue eyes, in which case they would think that Person 4 might only be able to see 96… all this speculating is unravelled when everyone finds out that if Person 100 could not see any blue eyes, then Person 100 would be able to leave, so if Person 99 could only see one set of blue eyes, Person 99 would be able to leave after they didn't, so… etc.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Perhaps this is enlightening: if the Guru went to each person individually, and told them each in secret that there was a person with blue eyes, then it would not help: they would truly have learned nothing. The Guru saying that someone has blue eyes does not change anyone's mind about whether or not anyone has blue eyes. But that's not all everyone gets from the situation: not only did everyone hear the announcement, everyone saw that everyone else heard the announcement, and everyone saw that everyone saw that etc. Everyone finds out something about other people's state of knowledge.&lt;/p&gt;&#xA;" OwnerUserId="1950" LastEditorUserId="1950" LastEditDate="2014-11-22T10:26:41.803" LastActivityDate="2014-11-22T10:26:41.803" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="1974" PostTypeId="2" ParentId="1909" CreationDate="2014-07-28T01:10:02.480" Score="3" Body="&lt;p&gt;Here a completely logical solution with no shortcuts involved :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 4 scenarios&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;None have mud.&lt;/li&gt;&#xA;&lt;li&gt;One has mud.&lt;/li&gt;&#xA;&lt;li&gt;2 have mud.&lt;/li&gt;&#xA;&lt;li&gt;All of them have mud.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Case 1: If no one of them have mud on their foreheads, then no one would raise their hands. However everyone raised their hands so this is not the case. (only 3 cases left)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 2: If one person had mud on his head, he would see that both of his friends raised their hands while they didn't have mud on their foreheads.  That must mean he has mud and he would immediately get the dollar.  However this is not the case because all three raised their hands. (only 2 cases left)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 3: Imagine people A, B, and C.  People &lt;strong&gt;A&lt;/strong&gt; and &lt;strong&gt;B&lt;/strong&gt; have mud on their foreheads.  Person &lt;strong&gt;A&lt;/strong&gt; would see that person &lt;strong&gt;B&lt;/strong&gt; raised his hand while person &lt;strong&gt;C&lt;/strong&gt; didn't have mud on their forehead.  That means person &lt;strong&gt;A&lt;/strong&gt; would know he is the only one person &lt;strong&gt;B&lt;/strong&gt; could be raising his hands for, and he would immediately know he had mud on his head and get the dollar.  The same logic goes for person &lt;strong&gt;B&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, &quot;All three children are bewildered.&quot;  so this cannot be the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That leaves us with case four being the only possible answer through process of elimination!!&lt;/p&gt;&#xA;" OwnerUserId="1819" LastActivityDate="2014-07-28T01:10:02.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1976" PostTypeId="1" CreationDate="2014-07-29T01:43:16.927" Score="9" ViewCount="369" Body="&lt;p&gt;There is a type of puzzle which can be modelled as finding a &lt;a href=&quot;https://en.wikipedia.org/wiki/Hamiltonian_path&quot; rel=&quot;noreferrer&quot;&gt;Hamiltonian path&lt;/a&gt; through a graph. One example is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Icosian_game&quot; rel=&quot;noreferrer&quot;&gt;Icosian game&lt;/a&gt;. What factors affect the difficulty of such puzzles, to be solved by humans? Let's assume the solver is unfamiliar with graph theory and Hamiltonian paths in particular, but can see the entire graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm partly motivated by the &lt;a href=&quot;http://www.astrolog.org/labyrnth/psych.htm&quot; rel=&quot;noreferrer&quot;&gt;Making Difficult Mazes&lt;/a&gt; page by Walter D. Pullen, which lists numerous factors affecting how difficult regular mazes are, such as solution length, the number of curves and loops. Do similar factors affect Hamiltonian puzzles? Different ones?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Full Disclosure:&lt;/em&gt; I recently made a &lt;a href=&quot;http://congusbongus.itch.io/drunken-viking&quot; rel=&quot;noreferrer&quot;&gt;puzzle game&lt;/a&gt; for a &lt;a href=&quot;http://www.ludumdare.com/compo/minild-53/?action=preview&amp;amp;uid=33168&quot; rel=&quot;noreferrer&quot;&gt;jam&lt;/a&gt;, whose mechanics could be modelled as a Hamiltonian path puzzle. Whilst making it, I realised I did not have a good idea of what made a puzzle easier or harder, and was wondering if there were studies or other examples of such puzzles.&lt;/p&gt;&#xA;" OwnerUserId="104" LastEditorUserId="8874" LastEditDate="2015-01-31T19:08:02.563" LastActivityDate="2015-01-31T19:08:02.563" Title="What factors affect the difficulty of Hamiltonian puzzles?" Tags="|puzzle-creation|difficulty-rating|graph-theory|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1977" PostTypeId="2" ParentId="1976" CreationDate="2014-07-29T03:53:00.877" Score="6" Body="&lt;p&gt;The critical factor is the connectivity of the graph.  The Hamiltonian path problem is known to be NP-complete.  Years ago I saw charts from a talk called &quot;Where are all the hard problems?&quot;  (seems no longer available) The point was that although Hamiltonian path is NP-complete, specific instances are easy.  Either (for high connectivity) you find a path quickly or (for low connectivity) you find there isn't one because you find where you get stuck.  There is a transition region, akin to a phase transition, where long-range correlations become important, and that is where the hard problems are.  A more recent paper is &lt;a href=&quot;http://www.dcs.gla.ac.uk/~pat/cpM/papers/cheeseman91where.pdf&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2014-07-29T03:53:00.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1978" PostTypeId="2" ParentId="1955" CreationDate="2014-07-29T08:25:41.203" Score="2" Body="&lt;p&gt;It is not written that the state of being a liar can't change in time. So we can find &lt;strong&gt;recursive&lt;/strong&gt; solution to the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stage 1: All Cretans are liars, including Epidemides. None of them has ever told a truth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stage 2: Epimenides says, that all Cretans are liars. This is a truth, but then, first time in his life, he has told the truth, therefore he isn't a liar anymore. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stage 3: If Epidemes would say that once again, it would &lt;strong&gt;not&lt;/strong&gt; be truth anymore, because Epidemes is no longer a liar. However, since he is already contaminated with truth, he is no liar anymore, so that statement is a lie. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is the consequence of start condition, that the liar is someone who have never told a truth. If you allow, that liar tells a truth from time to time, that statement could, as well, be correct from the beginning.&lt;/p&gt;&#xA;" OwnerDisplayName="user66" LastActivityDate="2014-07-29T08:25:41.203" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1980" PostTypeId="1" AcceptedAnswerId="52271" CreationDate="2014-07-29T14:14:50.173" Score="23" ViewCount="185310" Body="&lt;p&gt;There are four prisoners. All four prisoners will be freed, if at least one of them correctly guesses the color of the hat on his head.&lt;/p&gt;&#xA;&lt;p&gt;They can't speak to each other, and they can't touch each other.&lt;/p&gt;&#xA;&lt;p&gt;Number 1 sees number 2 and 3's hats.&lt;br /&gt;&#xA;Number 2 sees number 3's hat.&lt;br /&gt;&#xA;Number 3 sees only the wall.&lt;br /&gt;&#xA;Number 4 sees only the wall.&lt;/p&gt;&#xA;&lt;p&gt;There are no mirrors.&lt;/p&gt;&#xA;&lt;p&gt;They all know that there are 2 black hats and 2 white hats, and that there are four people.&lt;/p&gt;&#xA;&lt;p&gt;They know that their placement in this room is as follows:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wQdL9.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;All the prisoners are perfect logicians, and won't speak unless they are sure of their hat color.&lt;/p&gt;&#xA;&lt;p&gt;Can the four prisoners be freed? If yes, then how?&lt;/p&gt;&#xA;" OwnerUserId="1959" LastEditorUserId="65277" LastEditDate="2023-06-17T21:49:10.323" LastActivityDate="2023-06-17T21:49:10.323" Title="Four prisoners wearing black and white hats" Tags="|logical-deduction|meta-knowledge|hat-guessing|" AnswerCount="12" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="1981" PostTypeId="2" ParentId="1980" CreationDate="2014-07-29T14:37:44.753" Score="30" Body="&lt;p&gt;There are only 6 possible configuration of hats.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;wwbb&lt;/code&gt; &#xA;&lt;code&gt;wbwb&lt;/code&gt;&#xA;&lt;code&gt;bwwb&lt;/code&gt;&#xA;&lt;code&gt;wbbw&lt;/code&gt;&#xA;&lt;code&gt;bwbw&lt;/code&gt;&#xA;&lt;code&gt;bbww&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If $h(3)=h(2)$ then $1$ knows his.  This eliminates 2 configurations (wbbw,bwwb).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;When $2$ looks at $3$ and $1$ says nothing, then he knows his hat color is not the same as $3$.  He, therefore, knows he has the opposite color as $3$ and says it accordingly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This would be a better question if you specify that every player is killed if he guesses wrong (my answer) or they must all answer at the same time ($1$ and $2$ always guesses opposite of $3$).  &lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="46903" LastEditDate="2018-09-19T03:13:22.220" LastActivityDate="2018-09-19T03:13:22.220" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="1982" PostTypeId="2" ParentId="1957" CreationDate="2014-07-29T14:57:11.270" Score="18" Body="&lt;p&gt;There is a general, very simple, algorithm for generating any magic square which has an odd number of rows/columns as follows:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Start in the middle of the top row and enter 1.&lt;/li&gt;&#xA;&lt;li&gt;Move Up 1 and Right 1 wrapping both vertically and horizontally when you leave the grid *(see note below).&lt;/li&gt;&#xA;&lt;li&gt;If that square is empty enter the next number; if the square is not empty put the next number underneath the last number you entered.&lt;/li&gt;&#xA;&lt;li&gt;Repeat 2&amp;amp;3 until the grid is complete. All rows, columns and the two diagonals will sum to the same value.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Here is the 5x5 square:&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th&gt;&lt;/th&gt;&#xA;&lt;th&gt;&lt;/th&gt;&#xA;&lt;th&gt;&lt;/th&gt;&#xA;&lt;th&gt;&lt;/th&gt;&#xA;&lt;th&gt;&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;17&lt;/td&gt;&#xA;&lt;td&gt;24&lt;/td&gt;&#xA;&lt;td&gt;1&lt;/td&gt;&#xA;&lt;td&gt;8&lt;/td&gt;&#xA;&lt;td&gt;15&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;23&lt;/td&gt;&#xA;&lt;td&gt;5&lt;/td&gt;&#xA;&lt;td&gt;7&lt;/td&gt;&#xA;&lt;td&gt;14&lt;/td&gt;&#xA;&lt;td&gt;16&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;4&lt;/td&gt;&#xA;&lt;td&gt;6&lt;/td&gt;&#xA;&lt;td&gt;13&lt;/td&gt;&#xA;&lt;td&gt;20&lt;/td&gt;&#xA;&lt;td&gt;22&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;10&lt;/td&gt;&#xA;&lt;td&gt;12&lt;/td&gt;&#xA;&lt;td&gt;19&lt;/td&gt;&#xA;&lt;td&gt;21&lt;/td&gt;&#xA;&lt;td&gt;3&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;11&lt;/td&gt;&#xA;&lt;td&gt;18&lt;/td&gt;&#xA;&lt;td&gt;25&lt;/td&gt;&#xA;&lt;td&gt;2&lt;/td&gt;&#xA;&lt;td&gt;9&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;As the square square is symmetric there are eight symmetries. You can also get these symmetries by a simple variation of the start square and direction used in step 2.&lt;/p&gt;&#xA;&lt;p&gt;*Note:&#xA;So as you are in the middle of the top row on the first move you want to place the next number in the next column of the row above. The row above does not exist so move to the last row of the square in the same column. If you were in the last column you would move to the first column. If you look at the example of the 5x5 at number 15. The next position is the square up and to the right of 15 which wraps on both the row and column to point to the lower right square which has 11 in it. As that square is not empty we placed 16 underneath 15.&lt;/p&gt;&#xA;" OwnerUserId="1963" LastEditorUserId="69582" LastEditDate="2020-12-30T19:15:10.287" LastActivityDate="2020-12-30T19:15:10.287" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="1984" PostTypeId="2" ParentId="1959" CreationDate="2014-07-29T15:57:46.140" Score="1" Body="&lt;p&gt;One way that assumes the candle are exactly the same is to use the unlit candle to determine when the first candle has burned half-way.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Light the first candle and wait for it to burn to approximately half-way (must be before it reaches half-way).&lt;/li&gt;&#xA;&lt;li&gt;Place the unlit candle next to the lit candle and use your finger or thumb to mark where the lit candle is on the unlit candle.&lt;/li&gt;&#xA;&lt;li&gt;Invert the unlit candle keeping your finger/thumb on the position you measured in step 2 and compare it to the height of the lit candle again.&lt;/li&gt;&#xA;&lt;li&gt;Once the mark is the same on the first and inverted measurement you know the lit candle has burned exactly half-way. You can now light the unlit candle and wait for it to burn to the end.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1963" LastActivityDate="2014-07-29T15:57:46.140" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1985" PostTypeId="2" ParentId="1959" CreationDate="2014-07-29T16:38:33.593" Score="2" Body="&lt;p&gt;If the candles burn at the same rate when sideways vs. upright (which they probably won't, but let's make the assumption for the sake of having a solution), lie the candles horizontally next to each other, but with the wicks at opposite ends. Light both. When they meet, they'll have both burned halfway - 30 minutes gone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now blow one out. Let the other burn the rest of its 30 minute lifespan - now you've timed 60 minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it goes out, relight the other one for the remaining 30 minutes - total of 90 minutes.&lt;/p&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2014-07-29T16:38:33.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1987" PostTypeId="1" AcceptedAnswerId="1988" CreationDate="2014-07-29T22:34:50.120" Score="10" ViewCount="2008" Body="&lt;p&gt;Miss Danner, a primary school teacher, is on her way to school - it's the first day of school this year.&lt;br&gt;&#xA;In class, she sees two identically looking boys sitting in the first row.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Aw, twins, so cute!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;she exclaims.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We aren't twins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the boys respond unisono.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Miss Danner - astounded - doesn't comment it further. After class, she takes a look into their school files and finds the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Thomas Elmer&#xA;born 5.2.2008 &#xA;contact person(s): &#xA;Franz Elmer (father), telephone 013579  &#xA;Sophie Elmer (mother), telephone 012468&#xA;&#xA;Maximilian Elmer&#xA;born 5.2.2008 &#xA;contact person(s):&#xA;Franz Elmer (father), telephone 013579  &#xA;Sophie Elmer (mother), telephone 012468&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&lt;br&gt;&#xA;(I thought this was taken care of by using the word &lt;em&gt;identically&lt;/em&gt;, but here you go:&#xA;Miss Danner is &lt;em&gt;very&lt;/em&gt; curious, so that on the next parents meeting, she takes DNA samples from the parents. As Maximilian and Thomas attend her class, it is not a problem for her to acquire samples from the boys, too. She takes all four samples to the lab and the laboratory technician tells her that the two boys are undoubtedly the children of Franz and Sophie (and no one else). The lab technician did everything right and no one in the story is lying. Both the boys were naturally born by Sophie on the same day in the same hospital.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How is that possible?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1910" LastEditorUserId="1910" LastEditDate="2014-09-30T21:22:28.950" LastActivityDate="2017-02-22T17:37:58.143" Title="Miss Danner and the Elmer Boys" Tags="|lateral-thinking|" AnswerCount="6" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1988" PostTypeId="2" ParentId="1987" CreationDate="2014-07-29T23:49:31.473" Score="25" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;They're triplets, or have even more identical siblings than that, making them still identical siblings, but not twins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1968" LastEditorUserId="1968" LastEditDate="2014-07-30T00:04:06.320" LastActivityDate="2014-07-30T00:04:06.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1992" PostTypeId="1" AcceptedAnswerId="1995" CreationDate="2014-07-30T19:17:38.283" Score="6" ViewCount="535" Body="&lt;p&gt;My back yard forms a rectangular grid of squares except some of the squares are missing as they are covered by pipes or a small tree. The layout is as follows. A 'x' indicates a square that is free and '.' a covered square.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;. x x .&#xA;. x x x&#xA;. x x x&#xA;x x x x&#xA;x x x x&#xA;x x x x &#xA;x x x x&#xA;x x x x&#xA;. x x x&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I would like write distinct non-negative numbers in each one of the free squares.  However, to make it interesting I would like the sums of all the columns to be the same and the sum of the all the rows to be the same (although the row sum need not equal the column sum).  We can assume that the value of a covered square is zero.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is this possible with this layout? If not, why not?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8998" LastActivityDate="2014-08-02T18:42:37.690" Title="Magic back yards" Tags="|calculation-puzzle|magic-square|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1994" PostTypeId="2" ParentId="1987" CreationDate="2014-07-31T00:12:06.547" Score="3" Body="&lt;p&gt;Archetypes of alternate answers:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Atypical birth records:&lt;br&gt;&#xA;Mr. and Mrs. Elmer travel frequently, while traveling their first child was born in a region where the date is recorded month/day/year, and another was born ~12 weeks later in a region where it is recorded day/month/year.  The boys share the same DNA but do not consider themselves twins because of the disparity between their birth days.&lt;/li&gt;&#xA;&lt;li&gt;Premature births + Plastic Surgery:&lt;br&gt;&#xA;As a subset of the previous archetype, The Elmers got their birth certificates in different regions and conceived two children separately.  A horrible accident scared both children and they received plastic surgery to try and give them normal lives.  Unfortunately the doctor was a hack and only knew how to build one type of face.  Now they look identical.&lt;/li&gt;&#xA;&lt;li&gt;Not Technically lying:&lt;br&gt;&#xA;They aren't twins, they are &lt;em&gt;Super&lt;/em&gt; twins, a term sometimes used to refer to identical twins (or any number of other terms which a sarcastic 6 year old might enjoy using).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm sure there are more.&lt;/p&gt;&#xA;" OwnerUserId="457" LastEditorUserId="457" LastEditDate="2014-09-30T18:30:12.033" LastActivityDate="2014-09-30T18:30:12.033" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="1995" PostTypeId="2" ParentId="1992" CreationDate="2014-07-31T03:15:50.980" Score="7" Body="&lt;p&gt;Yes, this is possible:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;.   54  58  .&#xA;.   38  34  40&#xA;.   36  44  32&#xA;46   4   2  60&#xA;48  27  12  25&#xA;50  29  22  11&#xA;52  30  14  16&#xA;56  26  24   6&#xA;.    8  42  62&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Each column sums to 252; each row 112.&#xA;And this is certainly not the only solution, just try tweaking the odd numbers up and down a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To figure this out, I started with a spreadsheet with the grid, with equations for the sums around the border, and a list of numbers 1-31 (# of empty squares). I calculated the minimum total sum (504), and figured out how much each row and column should sum to given that total (56/126). I then placed numbers from the list into the empty spaces, shorter rows/cols first, until there were no more possibilities (for me, this left empty spaces where the 27-12-25/29-22-11 is). Then I doubled the value of each square and the expected totals. That gave me 6 even numbers and all the odds to work with; checking the missing sums showed that both the 3 missing numbers in each of the two rows and the two in each column must sum to even numbers, so the columns must be even-even or odd-odd and the rows must have two odd and one even. Looking over the available evens and the required sums, the 12-22 combination fit where it is, and it was trivial then to find four odd numbers to fit in the remaining spots.&lt;/p&gt;&#xA;" OwnerUserId="4" LastEditorUserId="4" LastEditDate="2014-08-01T01:53:16.083" LastActivityDate="2014-08-01T01:53:16.083" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1996" PostTypeId="2" ParentId="1992" CreationDate="2014-07-31T12:17:06.543" Score="-1" Body="&lt;p&gt;Assuming that 0 counts as &quot;non negative&quot; a solution with a much lower sum is possible:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; .  2  2  . &#xA; .  1  1  . &#xA; .  1  1  2 &#xA; 1  2  1  0 &#xA; 2  1  0  1 &#xA; 2  0  1  1 &#xA; 2  1  0  1 &#xA; 2  1  1  0 &#xA; .  0  2  2 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Each row adds up to 4, each column to 9.&#xA;I started by placing two numbers (2 and 2) in the top row and filling the left most column with 2s too and then filling each row that belongs to the leftmost column with 1 0 1//1 1 0//0 0 1.&#xA;However, the 3 rows left required 12 in sum to be satisfied, and the columns required 16. By shifting a sum of 1 from the left most column to the right columns, each row required 1 less to be satisfied, while 1 more was already placed in the middle rows. Therefore, colums and rows each needed 12 in sum to be satisfied. Then I just filled up the empty spaces, which was more a random process and quite easy.&lt;/p&gt;&#xA;" OwnerUserId="1976" LastActivityDate="2014-07-31T12:17:06.543" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1999" PostTypeId="1" AcceptedAnswerId="2006" CreationDate="2014-07-31T12:35:48.053" Score="29" ViewCount="5248" Body="&lt;p&gt;All numbers between 1 and 10&lt;sup&gt;10&lt;/sup&gt; are written out and ordered alphabetically into a dictionary (as the only entries).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spaces and hyphens are removed. 1024 would then be &quot;one&lt;b&gt;thousand&lt;/b&gt;twentyfour&quot;.&lt;br&gt;&#xA;Also, &quot;and&quot; is not being used as a link (I think that's American English).&lt;br&gt;&#xA;Furthermore, the American number scale is used: million, billion, trillion...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Myriad is not used, and numbers with (or like) 1984 are composed with &quot;onethousand&quot;,&lt;br&gt;&#xA;so &quot;nineteeneightyfour&quot; would not be allowed.&lt;br&gt;&#xA;Last but not least: the numbers are written out in, well, (American) English.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the first odd number?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Bonus question: What is the first prime number?&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1910" LastEditorUserId="1910" LastEditDate="2014-07-31T16:34:21.253" LastActivityDate="2018-08-01T12:50:59.587" Title="First odd number in a &quot;number dictionary&quot;" Tags="|logical-deduction|mathematics|" AnswerCount="5" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2000" PostTypeId="2" ParentId="1999" CreationDate="2014-07-31T13:05:11.400" Score="4" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;We know that the numbers go from &quot;one&quot; to &quot;tenbillion.&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We also know that the only phrases that can appear in this dictionary are: one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve, thirteen, fourteen, fifteen, sixteen, seventeen, eighteen, nineteen, twenty, thirty, forty, fifty, sixty, seventy, eighty, ninety, hundred, thousand, million, billion.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;This list was found by concatenating: The list of numbers 1-10, the list of &quot;special&quot; numbers 11-19, the list of &quot;tens place numbers&quot; 10-90, and the list of &quot;scales&quot; 100, 1000, ....&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Sorted alphabetically, that becomes: billion, eight, eighteen, eighty, eleven, fifteen, fifty, five, forty, four, fourteen, hundred, million, nine, nineteen, ninety, one, seven, seventeen, seventy, six, sixteen, sixty, ten, thirteen, thirty, thousand, three, twelve, twenty, two.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A number cannot start with &quot;billion,&quot; so that leaves &quot;eight&quot; as the first &quot;segment&quot; of the number we're looking for. (This still technically leaves the problem of what would happen if a shorter &quot;segment&quot; that appears later in the list had, for example, &quot;billion&quot; added to it, but the shortest segment that comes after &quot;eight&quot; is &quot;five,&quot; and &quot;fivebillion&quot; sorts after &quot;eight.&quot;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The number has to be odd, and eight is not odd, so something has to be appended. &quot;Billion&quot; is the obvious choice as it is the first item in the list.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Now, the number needs to have an odd ending. We can continue appending to it, though. &quot;Eightbillionbillion&quot; doesn't make sense, so it can become...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&quot;Eightbillioneight&quot;&lt;/li&gt;&#xA;&lt;li&gt;&quot;Eightbillioneighthundred&quot;&lt;/li&gt;&#xA;&lt;li&gt;&quot;Eightbillioneighthundredeight&quot;&lt;/li&gt;&#xA;&lt;li&gt;... uh oh, we can't append anything else while still having a valid number&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The last digit has to be odd, so change the &quot;eight&quot; to the next odd digit, &quot;five.&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Therefore, the first odd number in this dictionary is &quot;eightbillioneighthundredfive&quot; or 8,000,000,805.&lt;/p&gt;&#xA;" OwnerUserId="39" LastEditorUserId="39" LastEditDate="2014-07-31T13:17:56.797" LastActivityDate="2014-07-31T13:17:56.797" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="2001" PostTypeId="2" ParentId="1999" CreationDate="2014-07-31T13:10:27.377" Score="2" Body="&lt;p&gt;Using Doorknobs reasoning, I come to eightbillioneighteenmillioneighteenthousandeighthundredfive or, if that is possible in English eightbillioneighteenhundredfive&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edited to remove the 'ands' but the example in the question '1024 would then be &quot;onethousandfortytwo&quot;' is a bit misleading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edited again: last bit was my mistake by not reading thoroughly enough.&lt;/p&gt;&#xA;" OwnerUserId="1745" LastEditorUserId="1745" LastEditDate="2014-07-31T13:39:15.037" LastActivityDate="2014-07-31T13:39:15.037" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="2002" PostTypeId="2" ParentId="1999" CreationDate="2014-07-31T15:06:37.887" Score="16" Body="&lt;p&gt;Under the belief that Kevin has the correct answer with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8,018,018,885&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I decided to look at the bonus. Pulling the prime numbers from &lt;a href=&quot;http://www.primos.mat.br/&quot;&gt;http://www.primos.mat.br/&lt;/a&gt;, #37 (includes all primes that start with eightbillioneighteenmillion), there are 4 primes that start with eightbillioneighteenmillioneighteenthousandeighthundred:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8,018,018,819&lt;br&gt;&#xA;8,018,018,849&lt;br&gt;&#xA;&lt;strong&gt;8,018,018,851&lt;/strong&gt;&lt;br&gt;&#xA;8,018,018,893&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of those, the first alphabetically is in bold, and is written out as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;eightbillioneighteenmillioneighteenthousandeighthundredfiftyone&lt;/p&gt;&#xA;" OwnerUserId="375" LastEditorUserId="375" LastEditDate="2014-07-31T16:07:14.353" LastActivityDate="2014-07-31T16:07:14.353" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2003" PostTypeId="1" AcceptedAnswerId="2014" CreationDate="2014-07-31T15:27:50.127" Score="11" ViewCount="738" Body="&lt;p&gt;What is the appropriate answer to this question?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;What is the first natural number, which can not be defined with less than 15 words?&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Explanations:&lt;br&gt;&#xA;  1. Natural means integers greater than 0.&lt;br&gt;&#xA;  2. Word here is any group of letters separated by spaces or punctuation. &quot;a&quot;, &quot;of&quot;, for example, are words too.&lt;br&gt;&#xA;  3. You are not allowed to use digits. All numbers in the definition must be written with words. 1234 takes 6 words, for example.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="28" LastActivityDate="2014-08-03T14:34:52.303" Title="First number, which can not be defined" Tags="|logical-deduction|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2004" PostTypeId="2" ParentId="2003" CreationDate="2014-07-31T15:44:10.167" Score="11" Body="&lt;p&gt;This is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Richard%27s_paradox&quot;&gt;Richard's paradox&lt;/a&gt;. This is paradox because the phrase &quot;the first natural number, which can not be defined with less than 15 words&quot; has only 14 words and can be used as definition of a number.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="28" LastEditDate="2014-08-02T04:57:20.330" LastActivityDate="2014-08-02T04:57:20.330" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2005" PostTypeId="1" AcceptedAnswerId="2007" CreationDate="2014-07-31T16:13:15.307" Score="3" ViewCount="517" Body="&lt;p&gt;This is not a very difficult problem, but it seems that it hasn't been asked here.&#xA;It's one of my favorites, so here we go:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I am twice as old as you were when I was as old as you are.&lt;br /&gt;&#xA;  When you will be as old as I am, the sum of our ages will be 81.&lt;br /&gt;&#xA;  How old are we?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: I totally misremembered the second sentence, it's fixed now.&lt;/p&gt;&#xA;" OwnerUserId="1980" LastEditorUserId="5373" LastEditDate="2018-05-09T14:42:47.640" LastActivityDate="2018-05-09T14:42:47.640" Title="Your age and my age" Tags="|mathematics|calculation-puzzle|time|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ClosedDate="2015-04-15T16:01:44.953" ContentLicense="CC BY-SA 3.0" />
  <row Id="2006" PostTypeId="2" ParentId="1999" CreationDate="2014-07-31T16:27:26.883" Score="29" Body="&lt;p&gt;Number words in the given range are assembled from the following components:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The set $A$ of single digits: &lt;code&gt;eight&lt;/code&gt;, &lt;code&gt;five&lt;/code&gt;, &lt;code&gt;four&lt;/code&gt;, &lt;code&gt;nine&lt;/code&gt;, &lt;code&gt;one&lt;/code&gt;, &lt;code&gt;seven&lt;/code&gt;, &lt;code&gt;six&lt;/code&gt;, &lt;code&gt;three&lt;/code&gt;, &lt;code&gt;two&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;li&gt;The set $B$ of double-digit blocks: &lt;code&gt;eighteen&lt;/code&gt;, &lt;code&gt;eleven&lt;/code&gt;, &lt;code&gt;fifteen&lt;/code&gt;, &lt;code&gt;fourteen&lt;/code&gt;, &lt;code&gt;nineteen&lt;/code&gt;, &lt;code&gt;seventeen&lt;/code&gt;, &lt;code&gt;sixteen&lt;/code&gt;, &lt;code&gt;thirteen&lt;/code&gt;, &lt;code&gt;twelve&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;li&gt;The set $C$ of tens digits: &lt;code&gt;eighty&lt;/code&gt;, &lt;code&gt;fifty&lt;/code&gt;, &lt;code&gt;forty&lt;/code&gt;, &lt;code&gt;ninety&lt;/code&gt;, &lt;code&gt;seventy&lt;/code&gt;, &lt;code&gt;sixty&lt;/code&gt;, &lt;code&gt;thirty&lt;/code&gt;, &lt;code&gt;twenty&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;hundred&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;The powers of 1000: &lt;code&gt;billion&lt;/code&gt;, &lt;code&gt;million&lt;/code&gt;, &lt;code&gt;thousand&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Number words are generated by the following non-ambiguous grammar (I use the abbreviated notation $X^?$ to mean $X$ or nothing):&#xA;$$&#xA;\begin{array}{lllll}&#xA;  S \to &amp;amp;  &amp;amp;  &amp;amp; T \, \texttt{thousand} &amp;amp; T^? \\&#xA;  S \to &amp;amp;  &amp;amp; T \, \texttt{million} &amp;amp;  &amp;amp; T^? \\&#xA;  S \to &amp;amp;  &amp;amp; T \, \texttt{million} &amp;amp; T \, \texttt{thousand} &amp;amp; T^? \\&#xA;  S \to &amp;amp; A \, \texttt{billion} &amp;amp;  &amp;amp;  &amp;amp; T^? \\&#xA;  S \to &amp;amp; A \, \texttt{billion} &amp;amp;  &amp;amp; T \, \texttt{thousand} &amp;amp; T^? \\&#xA;  S \to &amp;amp; A \, \texttt{billion} &amp;amp; T \, \texttt{million} &amp;amp;  &amp;amp; T^? \\&#xA;  S \to &amp;amp; A \, \texttt{billion} &amp;amp; T \, \texttt{million} &amp;amp; T \, \texttt{thousand} &amp;amp; T^? \\&#xA;  T \to &amp;amp;  &amp;amp; A \\&#xA;  T \to &amp;amp;  &amp;amp; B \\&#xA;  T \to &amp;amp;  &amp;amp; C \, A^? \\&#xA;  T \to &amp;amp; A \, \texttt{hundred} &amp;amp; B \\&#xA;  T \to &amp;amp; A \, \texttt{hundred} &amp;amp; C^? \, A^? \\&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The constraint that the number is odd implies that the rightmost non-terminal will be $A$ or $B$ (otherwise the number is a multiple of $10$). Every production that has $A$ or $B$ at the rightmost position allows the rightmost non-terminals to be $A$, $B$, or $C \, A$. The first odd $A$ is &lt;code&gt;five&lt;/code&gt;, the first odd $B$ is &lt;code&gt;eleven&lt;/code&gt;, and the first odd $C \, A$ is &lt;code&gt;eightyfive&lt;/code&gt;. The winner is &lt;code&gt;eightyfive&lt;/code&gt;: that's the first odd number in the range $[1,99]$, and that's how the first odd number ends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going left, &lt;code&gt;eighthundred&lt;/code&gt; competes with leaving out the $A \, \texttt{hundred}$ part. Since &lt;code&gt;eighthundred&lt;/code&gt; comes before &lt;code&gt;eightyfive&lt;/code&gt;, the first number ends with &lt;code&gt;eighthundredeightyfive&lt;/code&gt; (which is the first odd number in the range $[1,999]$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the other occurrences of $T$ (in front of &lt;code&gt;thousand&lt;/code&gt; and &lt;code&gt;million&lt;/code&gt;), we don't have the constraint that the number must be odd. There &lt;code&gt;eight&lt;/code&gt; followed by a power of 1000 competes with &lt;code&gt;eighteen&lt;/code&gt; (which beats &lt;code&gt;eighthundred…&lt;/code&gt;); for both &lt;code&gt;thousand&lt;/code&gt; and &lt;code&gt;million&lt;/code&gt;, &lt;code&gt;eighteen&lt;/code&gt; wins. If &lt;code&gt;billion&lt;/code&gt; appears, then the first member of $A$ (&lt;code&gt;eight&lt;/code&gt;) is in front. Thus the following numbers compete:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                                              eighthundredeightyfive&#xA;                             eighteenthousand eighthundredeightyfive&#xA;             eighteenmillion                  eighthundredeightyfive&#xA;             eighteenmillion eighteenthousand eighthundredeightyfive&#xA;eightbillion                                  eighthundredeightyfive&#xA;eightbillion                 eighteenthousand eighthundredeightyfive&#xA;eightbillion eighteenmillion                  eighthundredeightyfive&#xA;eightbillion eighteenmillion eighteenthousand eighthundredeightyfive&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;eightbillion&lt;/code&gt; wins against &lt;code&gt;eighteen…&lt;/code&gt;. As for the powers of 1000 to include prefixed by &lt;code&gt;eighteen&lt;/code&gt; when they all start after &lt;code&gt;een&lt;/code&gt;, pick the first one in alphabetical order, then the first one that's numerically smaller, and so on. Or, with only four possibilities, you can check manually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The winner is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;eightbillion eighteenmillion eighteenthousand eighthundredeightyfive&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;or 8,018,018,885 for short. This would still be the first word for an odd number if the dictionary included all number words with conventionally-accepted names (in U.S. English, without &lt;code&gt;and&lt;/code&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. &lt;a href=&quot;https://oeis.org/A108067&quot;&gt;https://oeis.org/A108067&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="71" LastActivityDate="2014-07-31T16:27:26.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2007" PostTypeId="2" ParentId="2005" CreationDate="2014-07-31T16:30:16.437" Score="8" Body="&lt;p&gt;&lt;strong&gt;I am 36 years old, you are 27 years old.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I am $X$ years old and you are $Y$ years old. I am older than you by $(X-Y)$ years. Then when I was $Y$ you were $Y-(X-Y)$ years old. Therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X = 2 \cdot (Y-(X-Y)) = 4 \cdot Y-2 \cdot X$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X = 4/3 \cdot Y$ (1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you will be X years old I will be $X+(X-Y)$ years old. Therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X+(X+(X-Y)) = 81$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Y = 3 \cdot X-81$ (2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From (1) and (2) we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X = 4/3 \cdot (3 \cdot X-81) = 4 \cdot X-108$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $X = 36$. $Y = 3/4 \cdot X = 27$.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2014-07-31T16:36:01.227" LastActivityDate="2014-07-31T16:36:01.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2008" PostTypeId="2" ParentId="1817" CreationDate="2014-07-31T16:50:05.507" Score="1" Body="&lt;p&gt;Although the question is for one answer the comments are slightly confused about the minimum so below are the minimum and maximum times for all robots to exit the square:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Minimum time: √2&#xA;&lt;BR&gt;      &lt;em&gt;If all robots are on the line that joins A to C then the initial robot can travel directly to C in the shortest time/distance. Every other robot would join with the first and also travel directly to C.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maximum time: 2 + √2&#xA;&lt;BR&gt; &lt;em&gt;The worst scenario is where there are only three robots at A, B and D.The first robot has to travel to either B or D, then travel diagonally across the square to direct the final robot.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Justification:&lt;BR&gt;&#xA;The scenario presented in the maximum time represents the configuration where the robots and the exit are all as far away from each other as possible, hence each corner of the square has a robot or the exit. Adding more robots to the play-field cannot increase this maximum distance due to two reasons:&#xA;&lt;BR&gt;&lt;BR&gt;&#xA; 1. Each new robot can travel in a different direction to the first robot which can reduce the overall time needed for all robots to leave the square. For instance if there are four robots and the fifth is at the centre of the square the first robot would travel to the robot at the centre then it could travel to B and then directly to C. This would give a travel time of 1+√2. The robot at the centre would travel to D and then on to C. The travel to B and D happens at the same time and do reduces the overall cost.&#xA;&lt;BR&gt;&lt;BR&gt; 2. A mathematician could probably phrase this better so my apologies if it is not concise. If you imagine that there are at least three robots at A, B and D and lets imagine that the robot at A travels to B. From that point the longest distance to D and then C is diagonally across the square which divides the square into two triangles. Now assume that there are no robots in the triangle ABD but there is one somewhere in BCD. In order to increase the maximum length of robot A's journey this distance must be greater than the two legs B-&gt;D and D-&gt;C but there are no triangles that can fit within BCD with a greater total of side lengths. This means that the greatest distance the robot has to travel is BDC. If this is the case for a journey going from A to C through a triangle at BCD then it would also be true if we travelled in the opposite direction from C to A through triangle ABD.&#xA;&lt;BR&gt;&lt;BR&gt;&#xA;When there are robots at A, B and D then there is a better chance of achieving 1+√2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that explains it well enough but the following pictures describe the minimum, maximum and explanation of maximum in, hopefully, a clearer way:&lt;BR&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/mHXLc.png&quot; alt=&quot;Shortest&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EcXHl.png&quot; alt=&quot;longest&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KUoZf.png&quot; alt=&quot;Justification&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1963" LastActivityDate="2014-07-31T16:50:05.507" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2009" PostTypeId="2" ParentId="2005" CreationDate="2014-07-31T17:26:25.043" Score="6" Body="&lt;p&gt;Solution from the perspective of the asker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The difference $x$.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first sentence gives us two equations:&#xA;$$\text{I}~~~~ inow=2 \cdot youthen$$&#xA;$$&#xA;\begin{pmatrix}&#xA;{\scriptsize ithen\overset{\text{as old as}}= younow} \\&#xA;{\scriptsize younow=inow-x}\\&#xA;{\scriptsize \implies ithen=inow-x} \\ &#xA;{\scriptsize \implies ithen-x=inow-2x}\\&#xA;{\scriptsize  \text{and since}~~~~ youthen = ithen -x} \\&#xA;\end{pmatrix}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{II}~~~~ youthen = inow - 2x$$&#xA;$$ \Rightarrow inow=2 \cdot (inow-2x) \\&#xA;\Rightarrow inow=4x $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you become as old as I am now, I will be x years older (2nd sentence):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2 \cdot inow+x=81$$ &#xA;$$ inow=4x $$&#xA;$$ \Rightarrow x=9 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Our ages.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we know that&#xA;$$inow=36$$&#xA;$$inow-younow=9$$&#xA;$$ \Rightarrow younow =27$$&lt;/p&gt;&#xA;" OwnerUserId="1910" LastEditorUserId="1910" LastEditDate="2014-08-01T11:24:21.537" LastActivityDate="2014-08-01T11:24:21.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2010" PostTypeId="1" AcceptedAnswerId="2011" CreationDate="2014-07-31T18:12:44.690" Score="7" ViewCount="1092" Body="&lt;p&gt;The linguistic hardness ($LH$) of a natural number is the ratio of the amount of letters in the writing of this number in English to the amount of its digits.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, $LH(1234) = 7.75$, as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mbox{Number of letters}}{\mbox{Number of digits}} = \frac{\mbox{N(one thousand two hundred thirty four)}}{N(1234)}= 31/4 = 7.75$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the is $N: 0&amp;lt;N&amp;lt;10^9$ with biggest $LH(N)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. This puzzle appears to be much more interesting in Russian, but this is not Russian site to ask it here.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2015-01-05T09:11:30.860" LastActivityDate="2015-01-05T09:11:30.860" Title="What is the &quot;linguistically hardest&quot; number less than $10^9$?" Tags="|mathematics|word|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2011" PostTypeId="2" ParentId="2010" CreationDate="2014-07-31T18:39:40.913" Score="13" Body="&lt;p&gt;For any given digit range the numbers with the longest spelling have the highest LH.&lt;br&gt;&#xA;7-only-combinations are always among those numbers. Here's a table with their corresponding LH:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA; number from here | number segment | letters from here | LH&#xA;------------------+----------------+-------------------+-------&#xA;        777777777 | seven hundred  | 87                | 9 2/3 = 9 14/21&#xA;         77777777 | seventy        | 75                | 9 3/8&#xA;          7777777 | seven million  | 68                | 9 5/7 = 9 15/21&#xA;           777777 | seven hundred  | 56                | 9 1/3&#xA;            77777 | seventy        | 44                | 8 4/5&#xA;             7777 | seven thousand | 37                | 9 1/4&#xA;              777 | seven hundred  | 24                | 8&#xA;               77 | seventy        | 12                | 6&#xA;                7 | seven          | 5                 | 5&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Looks like 7777777 beat my previous suggestion. It worries me that LH(8878878) is the same (alongside others like 3878373).&lt;/p&gt;&#xA;" OwnerUserId="1910" LastEditorUserId="1910" LastEditDate="2014-08-01T00:33:12.607" LastActivityDate="2014-08-01T00:33:12.607" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2013" PostTypeId="1" CreationDate="2014-08-01T20:54:47.517" Score="4" ViewCount="530" Body="&lt;p&gt;The Tale of Genji is a classic japanese novel, probably written more than a thousand years ago. The 19th century artist Toyohara Kunichika painted a picture for every of the &lt;a href=&quot;http://www.man-pai.com/Genji/genji.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;54 chapters&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every &lt;a href=&quot;http://www.man-pai.com/Grandes_series/Kunichika_54sentimentos/kunichika_54sentimentos_52_e.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;picture&lt;/a&gt; is showing an open book, on the pages of which a glyph is repeated. For each chapter, a different glyph is used (also referred to as &lt;em&gt;crest&lt;/em&gt; or &lt;em&gt;Genjimon&lt;/em&gt;). There is a system to their creation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples from two different chapters:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rjhmS.jpg&quot; alt=&quot;picture example&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/UnMuM.jpg&quot; alt=&quot;glyph example&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here you can &lt;a href=&quot;http://www.man-pai.com/Grandes_series/Kunichika_54sentimentos/kunichika_54sentimentos_modernos1_e.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;browse&lt;/a&gt; through Kunichika's paintings (there's a previous/next navigation). Unfortunately the &lt;a href=&quot;http://www.man-pai.com/Grandes_series/Kunichika_54sentimentos/kunichika_54sentimentos_18_e.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;details&lt;/a&gt; under chapter 18 are messed up, but one can recognize the pattern anyhow.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can you make out the system and the 'mistakes'?&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: It is not necessary to know Japanese. Do not base your deductions on the summary linked to below, the representation of the brushstrokes is deceiving. Use it only to identify the indistinctly painted glyphs. The order of the glyphs is of no importance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you find it too hard making them out, view this &lt;a href=&quot;http://www.man-pai.com/Artigos/genjimon_e.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;glyph summary&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&#xA;Hint one:&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;sup&gt;Intersections should be interpreted like this:&lt;/sup&gt; &lt;img src=&quot;https://i.stack.imgur.com/2WvIn.png&quot; alt=&quot;interhint&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Hint two:&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;sup&gt;Certain glyphs are duplicates in terms of the system.&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Hint three:&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;sup&gt;My lucky number can be found &lt;a href=&quot;https://puzzling.stackexchange.com/questions/1955/epimenides-paradox&quot;&gt;here&lt;/a&gt;.&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1910" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2014-08-17T12:20:55.263" Title="What do the Glyphs from the Tale of Genji represent?" Tags="|pattern|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2014" PostTypeId="2" ParentId="2003" CreationDate="2014-08-02T05:33:45.220" Score="12" Body="&lt;p&gt;There is no such number. Like Ross mentioned in his answer, this is related to &lt;a href=&quot;http://en.wikipedia.org/wiki/Richard%27s_paradox&quot;&gt;Richard's paradox&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the heart of both this puzzle and Richard's paradox is (from the Wikipedia article) &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the resolution of Richard's paradox is that there is no way to unambiguously determine exactly which English sentences are definitions of real numbers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The resolution of this puzzle is that although it sure seems like we are defining a real number, we are actually not defining one. For example, consider the following statement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The first natural number that is greater than five and less than three.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This obviously does not describe a number as no matter which number you try one of the two requirements will always be left unsatisfied. Likewise, the statement&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The first natural number, which can not be defined with fewer than 15 words.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;has a requirement that cannot be met - it must not be able to be defined with fewer than 15 words, and yet we have done so. Thus, there is no such number. The only real difference between the two statements is in how obvious the unsatisfiable requirements are.&lt;/p&gt;&#xA;" OwnerUserId="1955" LastEditorUserId="1955" LastEditDate="2014-08-02T05:39:20.443" LastActivityDate="2014-08-02T05:39:20.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2015" PostTypeId="2" ParentId="1992" CreationDate="2014-08-02T14:56:22.337" Score="8" Body="&lt;p&gt;I have found a minimal solution.&lt;br&gt;&#xA;First of all, the puzzle asks for $31$ distinct non-negative numbers. If we take $0$ as a non-negative as well, the $31$ smallest integers are $0 - 30$ and they sum up to $(30*31)/2 = 465$. Now, it should be possible to divide this grand sum through both $4$ and $9$ and that is not possible with $465$. We need a multiple of $36$ and the first after $465$ is $468$, so this is the smallest possible grand total.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wrote a simple backtracking algorithm, which took a few hours to come to the following solution:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  . 21 31  .&#xA;  .  0 22 30&#xA;  .  1 24 27&#xA; 14 15 10 13&#xA; 23 20  2  7&#xA; 25 18  3  6&#xA; 26 12  5  9&#xA; 29 11  4  8&#xA;  . 19 16 17&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now, this solution is still very early in the search tree, and I already found two more solutions that look a lot like this, so I assume there will be plenty more solutions. I cannot tell you how many though, that will take a little rewriting the algorithm and a lot more time.&lt;/p&gt;&#xA;" OwnerUserId="1745" LastEditorUserId="1745" LastEditDate="2014-08-02T18:42:37.690" LastActivityDate="2014-08-02T18:42:37.690" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2017" PostTypeId="1" AcceptedAnswerId="2018" CreationDate="2014-08-04T03:44:54.553" Score="9" ViewCount="878" Body="&lt;p&gt;This question appeared in an old math contest, and I seem to remember that the answer I had back when I first saw it was unsatisfactory.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;You have a deck of cards numbered from &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$60$&lt;/span&gt;. First, you shuffle them, and deal them out into six rows of ten cards each. Then, you do the following two things:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;You rearrange the cards in each row so that they are in increasing order from left to right.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Then, you rearrange the cards in each column so that they are in increasing order from top to bottom.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Are the cards still in increasing order from left to right after the second step?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2022-02-04T19:14:44.230" Title="Arranging cards in rows" Tags="|mathematics|strategy|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2018" PostTypeId="2" ParentId="2017" CreationDate="2014-08-04T05:38:28.183" Score="7" Body="&lt;p&gt;Yes. They are.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Let's chose any 2 columns: $J$ and $I$, $1 \le J &amp;lt; I \le 10$.  &lt;/li&gt;&#xA;&lt;li&gt;After 1-st rearranging for any chosen card $C_{xI}$ in the column $I$ we have one card (more specifically: $C_{xJ}$) in the column $J$, which is smaller (has smaller number) than the chosen card. And each chosen card has it's own smaller card - different from the smaller card for the other cards at the column $I$.&lt;/li&gt;&#xA;&lt;li&gt;After 2-nd rearranging for any chosen card $C_{RI}$ in the column $I$ and row $R$ has exactly $R-1$ cards in the column $I$, which are smaller than the chosen card. As we remember from item 2, each of those $R-1$ cards has at least one smaller card in the column $J$. Plus the same is true for the chosen card itself. And all these $R$ cards are different. Therefore the chosen card $C_{RI}$ has at least $R$ cards in the column $J$, which are smaller then it is.&lt;/li&gt;&#xA;&lt;li&gt;Since cards in the column $J$ are sorted and $R$ smallest cards are in the rows $1,..,R$, then clearly $C_{RI}$ must be bigger then $C_{RJ}$.&lt;/li&gt;&#xA;&lt;li&gt;So we proved that for any $J&amp;lt;I$ and $R$: $C_{RJ} &amp;lt; C_{RI}$. This is exactly what we were required to do.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2014-08-04T05:49:30.610" LastActivityDate="2014-08-04T05:49:30.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2019" PostTypeId="2" ParentId="13" CreationDate="2014-08-04T12:36:15.993" Score="4" Body="&lt;p&gt;Couldn't you just use this easy approach:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply start with a solved Level and let a random generator play the level backwards!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the Figure can move in all directions and optionally pull crates while moving - just let a random monkey play the level for some time and if it looks pretty use it!&lt;/p&gt;&#xA;" OwnerUserId="1998" LastActivityDate="2014-08-04T12:36:15.993" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2020" PostTypeId="2" ParentId="1963" CreationDate="2014-08-04T13:11:54.427" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. $(1+1+1)! = 6 $&lt;br&gt;&#xA; 2. $2+2+2 = 6 $&lt;br&gt;&#xA; 3. $3*3-3 = 6$&lt;br&gt;&#xA; 4. $4^3/4^2+4^{1/2} = 6$&lt;br&gt;&#xA; 5. $5+(5/5) = 6$&lt;br&gt;&#xA; 6. $(6*6)/6 = 6$&lt;br&gt;&#xA; 7. $7-(7/7) = 6$&lt;br&gt;&#xA; 8. $8^3/8^2-8^{1/3} = 6$&lt;br&gt;&#xA; 9. $(9+9)/9^{1/2} = 6  $&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the bonus  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(0!+ 0! + 0!)! = 6 $&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For more info on the bonus take a look here: &#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Empty_product&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Empty_product&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1988" LastEditorUserId="42054" LastEditDate="2018-05-06T03:34:03.380" LastActivityDate="2018-05-06T03:34:03.380" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="2023" PostTypeId="2" ParentId="112" CreationDate="2014-08-04T16:20:34.913" Score="3" Body="&lt;p&gt;Another variant would be a Box where you have to open all Locks at once, or where you cannot pull out a key as long as the Lock is open!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you would only 3 Locks, all of them can be opened with the same type of key:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Vizir has 2 Standard Keys&#xA;The Slaves each have one 1 Standard Key&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If three Keys come together you open the box!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another way, which is also kind of trickey, but won't break any of the rules:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Box has 3 Locks: 1,2,3, The locks are attached in an even triangle, right next to each other and the locks are designed in a way, you can only remove the key when the lock is closed. As long as it is open the lock is stuck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each Key has a big nice cylindrical grip. So Big that you cannot put in two keys at once, if they have the same length, as their grips will hinder each other. So you can only insert 3 keys, if all 3 keys have different lengths: You insert the shortest first, then the second and the longest last.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each slave now gets 3 Keys (one for each lock) but each slave will get a set of 3 keys which are all of the same length (Slave 1 has 3 keys of length 1, Slave 2 three keys of length 2, ...) So although each slave has a key for all 3 locks, they can only use one of them at once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When 3 Slaves come together, they have 3 different lengths among them, so they can open the Box. The Vizir has 2 Keys: 1 and 2 with Length 5 and 6, so they can combine with a Key-3 form any Slave!&lt;/p&gt;&#xA;" OwnerUserId="1998" LastEditorUserId="1998" LastEditDate="2014-08-06T10:46:44.550" LastActivityDate="2014-08-06T10:46:44.550" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2024" PostTypeId="2" ParentId="38" CreationDate="2014-08-05T13:12:54.730" Score="1" Body="&lt;p&gt;Ok - you want your Algorithm to create more challenging mazes? Change it to a BFS - You start with your normal algorithm, but you always pick random where you continue and not just after you have reached the goal...&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1. Initialize path-list with one active square at the starting location.&#xA;2. Pick a random active squares: Referred to as 'current square'&#xA;3. Pick a random free square around the current square (if there are non skip to 5)&#xA;4. Connect the chosen free square to the current square and mark it NOT free and&#xA;   add it to the list of active squares.&#xA;5. If the current square has no empty squares around remove it from the list&#xA;6. If the new Square is next to the Goal, connect it to the goal&#xA;7. Goto 2 until there are no more active paths left.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This will give you a labyrinth with a single path to the goal, but many paths which look just as promising. This should be pretty hard. If you want more than a single path to the goal, you can add a switch in step 3, which can also chose an occupied square and open up a connection to it (but doesn't add it to the list of active squares)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you will get a labyrinth with round-trips and (possibly) multiple ways to the goal&lt;/p&gt;&#xA;" OwnerUserId="1998" LastActivityDate="2014-08-05T13:12:54.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2025" PostTypeId="1" AcceptedAnswerId="2027" CreationDate="2014-08-04T23:55:24.220" Score="9" ViewCount="453" Body="&lt;p&gt;The following is a paraphrasing of a problem from Martin Gardner called &quot;Tricky Track.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three schools $-$ $A,B,C$ $-$ participated in a track meet with several events. Each school was awarded some amount of points for each event (a positive integer). Each event was scored in the same way, and of course the first place winner of an event would receive more points than the second place one, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Susan, a student of school $B$, won the shot-put for her school. However, they were still beaten handily by school $A$, with the final result of the track meet being $22-9-9$, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many events took place at the track meet? Can you reconstruct the entire scoreboard?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no solution provided, just an answer.&lt;/p&gt;&#xA;" OwnerUserId="5778" OwnerDisplayName="ant11" LastEditorUserId="11876" LastEditDate="2016-08-03T22:50:09.280" LastActivityDate="2018-01-31T12:46:52.510" Title="A Martin Gardner problem about constructing a scoreboard given minimal information" Tags="|mathematics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2026" PostTypeId="2" ParentId="2025" CreationDate="2014-08-05T00:09:33.437" Score="1" Body="&lt;p&gt;$40$ total points were scored.  As the bottom two schools get at least $3$ points per event and $B$ won one, there were at most $5$ events.  So far, we could have $5$ events with a total score of $8$ per event or $4$ events with a total score of $10$ per event.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first thing I would try is five events with scores $5,2,1$.  Then $B$ can score $5,1,1,1,1, C$ can score $1,2,2,2,2$ and $A$ can score $ 2,5,5,5,5$  Assuming there is only one solution (and you can trust Gardner for that) here we are.&lt;/p&gt;&#xA;" OwnerUserId="73" OwnerDisplayName="Ross Millikan" LastActivityDate="2014-08-05T00:09:33.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2027" PostTypeId="2" ParentId="2025" CreationDate="2014-08-05T00:15:43.997" Score="16" Body="&lt;ol&gt;&#xA;&lt;li&gt;There are $22 + 9 + 9 = 40$ total points.  &lt;/li&gt;&#xA;&lt;li&gt;As each school receives a positive integer number of points for each event, and placing strictly better gives strictly more points, there must be at least $6 = 1 + 2 + 3$ points per event.  &lt;/li&gt;&#xA;&lt;li&gt;The divisors of $40$ that are greater than $6$ are $8,10,20,$ and $40$, and so there can be $5,4,2,$ or $1$ events.  &lt;/li&gt;&#xA;&lt;li&gt;As schools $B$ and $C$ are tied at the end, there must be more than one event.  &lt;/li&gt;&#xA;&lt;li&gt;As school $B$ won an event and there is more than one event, first place must give strictly fewer than $9$ points.  &lt;/li&gt;&#xA;&lt;li&gt;If there are exactly two events, then first place must give exactly $8$ points, as $8,7,5$ is the only possible distribution of points that gives first place fewer than $9$ points, sums to $20$ points per event, and gives strictly more points to schools placing strictly better.  &lt;/li&gt;&#xA;&lt;li&gt;Note, however, that this makes it impossible for school $B$ to have $9$ points.  &lt;/li&gt;&#xA;&lt;li&gt;Similarly, if there are exactly four events, then the only possible distribution of points is $1,3,6$.  &lt;/li&gt;&#xA;&lt;li&gt;Then, giving school $B$ 9 points can happen in only one way &amp;mdash; $1,1,1,6$, and now it is impossible for school $C$ to have only $9$ points.  &lt;/li&gt;&#xA;&lt;li&gt;Thus, there are five events. There are two possible distributions of points, now &amp;mdash; $1,2,5$ and $1,3,4$. The latter does not let school $B$ get exactly $9$ points, and so the answer is that there are $5$ events, with $5$ points for first place, $2$ points for second place, and $1$ point for third place.&lt;br&gt;&#xA;This works, with school $A$ getting $5,5,5,5,2$, school $B$ getting $1,1,1,1,5$, and school $C$ getting $2,2,2,2,1$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2007" OwnerDisplayName="qaphla" LastEditorUserId="36023" LastEditDate="2018-01-31T12:46:52.510" LastActivityDate="2018-01-31T12:46:52.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2028" PostTypeId="2" ParentId="1999" CreationDate="2014-08-05T14:31:56.197" Score="3" Body="&lt;p&gt;For the bonus question, the following python code can be used to find (or verify) the set of primes in the range 8018018800 to 8018018900 (a range found to be relevant by reasoning as in previous answers), the winning number being 8018018851.  If there had been no suitable primes in that first range, one could then have looked at ranges like 8018018500 to 8018018600 or  8018018400 to 8018018500, or even the whole range 8018018000 to 8018019000, to find candidates like 8018018581 or 8018018447 or 8018018081.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#!/usr/bin/env python&#xA;from gmpy import next_prime&#xA;v = next_prime(8018018800)&#xA;while 1:&#xA;    print v&#xA;    v = next_prime(v)&#xA;    if v &amp;gt; 8018018900: break&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Output:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;8018018819&#xA;8018018849&#xA;8018018851&#xA;8018018893&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2004" LastActivityDate="2014-08-05T14:31:56.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2029" PostTypeId="2" ParentId="496" CreationDate="2014-08-05T16:14:03.910" Score="8" Body="&lt;p&gt;As shown in Gilles' answer and others, it is straightforward to find answer E by noticing that E is divisible by 11, and E-1 is divisible by each of 2 through 10, hence by the LCM of 2 through 10, or 2520.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the remainders were not all 1, a more-general approach would be needed.  The  &lt;em&gt;Chinese remainder theorem&lt;/em&gt; (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Chinese_remainder_theorem&quot;&gt;Wikipedia&lt;/a&gt;) is good for such problems.  The calculations below show an application of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chinese_remainder_theorem#A_constructive_algorithm_to_find_the_solution&quot;&gt;constructive algorithm&lt;/a&gt; given in the Wikipedia article.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the method applies to pairwise coprime integers.  For the present problem, we can work with 11, 9, 8, 7, 5 and ignore the other numbers because if the remainder is 1 when dividing by 8, then it's 1 when dividing by 4 or by 2; similarly for 9 vs 3.  If it's 1 when dividing by 2 or 5, then it's 1 when dividing by 10; similarly for 2 and 3 vs 6.  So it's adequate to work with {11, 9, 8, 7, 5}, members of which are pairwise coprime.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, N = 11·9·8·7·5 = 27720; and dividing N by members of {11, 9, 8, 7, 5} gives {2520, 3080, 3465, 3960, 5544}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we apply the extended Euclidean algorithm to each nᵢ, N/nᵢ pair and obtain:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;gcd 11, 2520 = 1 =  -229 * 11 +  1 * 2520&#xA;gcd  9, 3080 = 1 =  1369 *  9 + -4 * 3080&#xA;gcd  8, 3465 = 1 =  -433 *  8 +  1 * 3465&#xA;gcd  7, 3960 = 1 = -1697 *  7 +  3 * 3960&#xA;gcd  5, 5544 = 1 =  1109 *  5 + -1 * 5544&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The products at the right are eᵢ; as a set, {2520, -12320, 3465, 11880, -5544}.  Now we compute x = Σ(aᵢ·eᵢ) where the aᵢ are from the set of desired remainders, {0,1,1,1,1} for respective members of {11, 9, 8, 7, 5}.  That is, x = 0*2520-12320+3465+11880-5544 = -2519, and we find that -2519 is indeed a solution, although negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, we have -2519 as a result of applying the Chinese Remainder Theorem; now we need to observe that adding the LCM of {11, 9, 8, 7, 5}, or 27720, produces another solution, giving us the solution -2519+27720 = 25201, the same as mentioned in previous answers.&lt;/p&gt;&#xA;" OwnerUserId="2004" LastActivityDate="2014-08-05T16:14:03.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2030" PostTypeId="1" AcceptedAnswerId="2031" CreationDate="2014-08-05T16:55:24.690" Score="5" ViewCount="2566" Body="&lt;p&gt;Which (if any) of these intervals is it possible to measure given two sand timers (which change from question to question)? There are no minute marks, halfway points, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A. 7 minutes with 4- and 5-minute timers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B. 15 minutes with 7- and 11-minute timers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C. 14 minutes with 5- and 8-minute timers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one (or more) is possible, specify how it can be done with the given timers.&lt;/p&gt;&#xA;" OwnerUserId="2001" LastEditorUserId="5373" LastEditDate="2014-11-29T14:48:03.540" LastActivityDate="2015-07-28T19:29:47.167" Title="A Sand Timer Challenge" Tags="|calculation-puzzle|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2031" PostTypeId="2" ParentId="2030" CreationDate="2014-08-05T17:09:18.377" Score="4" Body="&lt;p&gt;A is possible: Turn both the 5 and the 4. When the 4 runs out, flip it over.Now, when the 5 runs out, start timing. The 4 will run for three more minutes, after which, you can flip it over to reach 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B is possible: Turn both the 7 and the 11. When the 7 runs out, start timing. The 11 will run for 4 more minutes, after which it can be flipped to reach 15.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C is possible: Turn both the 5 and the 8. When the 5 runs out, flip it. The 8 will then run out after 3 minutes, leaving 2 minutes in the 5. Flip the 8 then. When the 5 runs out, start timing. There are now 6 minutes left in the 8, and flipping the 8 after those 6 minutes gives 6 + 8 = 14 minutes.&lt;/p&gt;&#xA;" OwnerUserId="2007" LastActivityDate="2014-08-05T17:09:18.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2032" PostTypeId="2" ParentId="1963" CreationDate="2014-08-05T17:32:23.600" Score="2" Body="&lt;p&gt;Having heard about this many times, I decided to give it a try. These are the answers that I came up with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(1+1+1)!=6$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2^2+2=6$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3*3-3=6$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4+(4/\sqrt4)=6$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(5-5)!+5=6$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$6*6/6=6$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$7-(7-7)!=6$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt[3]{8}+\sqrt[3]{8}+\sqrt[3]{8}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(9+9)/(\sqrt9)=6$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(0!+0!+0!)!=6$$ &lt;/p&gt;&#xA;" OwnerUserId="2001" LastEditorUserId="2001" LastEditDate="2016-04-27T03:12:23.880" LastActivityDate="2016-04-27T03:12:23.880" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2033" PostTypeId="2" ParentId="496" CreationDate="2014-08-05T18:36:50.713" Score="3" Body="&lt;p&gt;You can use the Least Common Multiple of all the numbers from 2 to 10, inclusive, which is 2520, then add 1, which would be 2521. Then take 2521 mod 11, which is congruent to 2 mod 11. Then you add another 2520, and take 5041 mod 11, which is congruent to 3 mod 11. Noticing that it increases by 1 mod 11 each time, and since you want 0 mod 11, or 11 mod 11, you'll need to add $2520*8$ to 5041, which gives us the desired $25201$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, 25201 eggs will weigh over 3000 pounds!!&lt;/p&gt;&#xA;" OwnerUserId="2001" LastActivityDate="2014-08-05T18:36:50.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2034" PostTypeId="1" AcceptedAnswerId="2037" CreationDate="2014-08-05T22:37:21.293" Score="5" ViewCount="4447" Body="&lt;p&gt;You've probably heard about challenges with water jugs, so why not try a few?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Use 3- and 5- liter jugs to get 4 liters. &lt;br/&gt; &lt;/li&gt;&#xA;&lt;li&gt;Use 5- and 7- liter jugs to get 4 liters. &lt;br/&gt;&lt;/li&gt;&#xA;&lt;li&gt;This one's different. How many different exact measurements can you get using only 8- and 11- liter jugs (not including 0 liters)? &lt;br/&gt;&lt;/li&gt;&#xA;&lt;li&gt;Use 3- and 11- liter jugs to get 7 liters.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;These numbers are made by me, so if any of these exactly match the popular ones out there (except for the first one), it is purely coincidental.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: accepted answer is the one that is first with all 4 correct answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Format to submit answer&#xA;&lt;hr/&gt;&#xA;(example with 5- and 8- liter jugs to get 6 liters) &lt;br/&gt;&#xA;0-0, 0-8, 5-3, 0-3, 3-0, 3-8, 5-&lt;strong&gt;6&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2001" LastEditorUserId="4050" LastEditDate="2015-01-27T13:26:45.253" LastActivityDate="2021-11-15T12:00:39.107" Title="A Set of Water Jug Challenges" Tags="|calculation-puzzle|decanting-problems|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2035" PostTypeId="2" ParentId="2034" CreationDate="2014-08-05T23:01:50.907" Score="4" Body="&lt;p&gt;For $1,2,4$ you can just fill one jug, fill the other from it, and repeat until solved.  This will work for any jug contents as long as they have no common factor.  If they have a common factor, you can only get multiples of the common factor.  One direction will be faster than the other for a given target.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 3, you can get any number from $1$ through $11$ in the $11$ jug, and have the $8$ full or not, so you can get anything $1$ through $19$, for $19$ values.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditDate="2014-08-06T19:41:42.223" LastActivityDate="2014-08-06T19:41:42.223" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2036" PostTypeId="2" ParentId="38" CreationDate="2014-08-06T04:29:35.513" Score="1" Body="&lt;p&gt;There are multiple maze generation algorithms. The one you describe the basic DFS algorithm. I'm aware of two other maze algorithms, both of which produce a lot more branching: &lt;a href=&quot;https://en.wikipedia.org/wiki/Prim%27s_algorithm&quot; rel=&quot;nofollow&quot;&gt;Prim's algorithm&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Kruskal%27s_algorithm&quot; rel=&quot;nofollow&quot;&gt;Kruskal's algorithm&lt;/a&gt;. Both of these algorithms are actually algorithms to find the minimum spanning tree of an undirected weighted graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have a grid of points, which are connected horizontally and vertically to their direct neighbours. Then, assign each edge a random weight, and run Prim's algorithm to find the minimum spanning tree of this grid. Ta-dah, you have a randomly generated maze.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the problem with mazes like this, from testing and experience, is that they branch a little &lt;em&gt;too&lt;/em&gt; much. Most of the branches are one-square to three-square ordeals, and it's really easy to determine which branches lead to dead ends most of the time. You want somebody to be able to follow the maze for a little while before reaching dead ends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what you can do is to tweak the algorithm a little bit. One of the easiest alternate algorithms you can make is a &quot;bounded DFS&quot;, in which you only go a certain distance inward on your search for a path before forcing a backtrack and branching again. This will create mazes with nice long stray paths most of the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for modifying Prim and Kruskal, I don't know enough about graph theory to ascertain whether this method will actually work, but consider making it so that when generating the random graph at the start, the weights of edges touching the same point have to be within a certain distance of each other.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You can read more about &lt;a href=&quot;https://en.wikipedia.org/wiki/Maze_generation_algorithm&quot; rel=&quot;nofollow&quot;&gt;maze generation algorithms&lt;/a&gt; on Wikipedia.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2014-08-06T04:29:35.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2037" PostTypeId="2" ParentId="2034" CreationDate="2014-08-06T05:47:06.663" Score="5" Body="&lt;p&gt;The answers are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;3-0, 0-3, 3-3, 1-5, 1-0, 0-1, 3-1, 0-&lt;strong&gt;4&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;5-0, 0-5, 5-5, 3-7, 3-0, 0-3, 5-3, 1-7, 1-0, 0-1, 5-1, 0-6, 5-6, &lt;strong&gt;4&lt;/strong&gt;-7.&lt;/li&gt;&#xA;&lt;li&gt;You can get 19 exact measurements (1 litre - 19 litres) not including 0 litre.&lt;/li&gt;&#xA;&lt;li&gt;3-0, 0-3, 3-3, 0-6, 3-6, 0-9, 3-9, 1-11, 1-0, 0-1, 3-1, 0-4, 3-4, 0-&lt;strong&gt;7&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2009" LastEditorUserId="2009" LastEditDate="2014-08-06T18:44:49.170" LastActivityDate="2014-08-06T18:44:49.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2038" PostTypeId="1" AcceptedAnswerId="2040" CreationDate="2014-08-06T21:09:14.017" Score="23" ViewCount="2539" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uO8xy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are given two of each from the array of 8 vegetables numbered 1 to 8 as shown above.&#xA;So in total you have 16 veggies(8 pairs). Your task is to make the longest kebab (sequence of vegetables arranged linearly) such that the number of vegetable between any pair of vegetables should match exactly to the number written on the pair itself.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For example a 3-dimensional (using 3 different veggies) kebab would be&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KusS4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the 3-d case, between the two red capsicums (n=1) there is exactly one vegetable (eggplant) as required and similar case holds for n=2 and n=3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Another example: For 4-d a valid sequence would be&lt;br&gt;&#xA;  &lt;strong&gt;41312432&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In addition to finding the longest kebab (8-d sequence) you can also try some related questions shown below.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1) Find one possible 8-d kebab.(main question)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;2) How many 8-d kebabs are possible?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;3) Are there any dimensions in which it is impossible to make a kebab?If yes then whats  the general formula for figuring out the allowed dimensions?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="101" LastEditorUserId="8874" LastEditDate="2016-03-21T14:01:09.733" LastActivityDate="2018-09-10T21:08:46.387" Title="The 8-dimensional vegetable kebab" Tags="|mathematics|combinatorics|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2039" PostTypeId="2" ParentId="2038" CreationDate="2014-08-07T01:50:13.360" Score="2" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;8 6 4 2 5 7 2 4 6 8 5 3 1 7 1 3&lt;/p&gt;&#xA;&lt;p&gt;8 6 4 2 7 5 2 4 6 8 3 5 7 1 3 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;should work. I started boxing the even numbers and then shuffled in the remaining odd numbers. If what I posted in the comment above is ok, then one could add 5 or 1.&lt;/p&gt;&#xA;&lt;p&gt;For the first two &amp;quot;dimensions&amp;quot;, there's obviously no such kebab. Thinking about &lt;span class=&quot;math-container&quot;&gt;$d&amp;gt;8$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;(Again, if my comment holds, kebabs for d=5, d=9  and d=10 are quite easily found.)&lt;/p&gt;&#xA;" OwnerUserId="1910" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-08-07T02:29:06.193" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2040" PostTypeId="2" ParentId="2038" CreationDate="2014-08-07T02:18:17.843" Score="17" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;1  3  1  6  7  3  8  5  2  4  6  2  7  5  4  8&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is probably the lexicographically-least 8D answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some solutions for various dimensions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;3: 2  3  1  2  1  3&lt;br&gt;&#xA;  4: 2  3  4  2  1  3  1  4&lt;br&gt;&#xA;  7: 1  4  1  5  6  7  4  2  3  5  2  6  3  7&lt;br&gt;&#xA;  8:   1  3  1  6  7  3  8  5  2  4  6  2  7  5  4  8&lt;br&gt;&#xA;  11:  1  2  1  4  2  8  9 10  4 11  6  3  7  5  8  3  9  6 10  5  7 11&lt;br&gt;&#xA;  12:  1  2  1  3  2  8  9  3 10 11 12  5  7  4  8  6  9  5  4 10  7 11  6 12&lt;br&gt;&#xA;  15:  1  2  1  3  2  4 10  3 11 12  4  9 14 15 13  7  8 10  5  6 11  9 12  7  5  8  6 14 13 15&lt;br&gt;&#xA;  16:  1  2  1  3  2  4  8  3 12 13  4 10 14 15 16  8  9  6 11  5  7 12 10 13  6  5  9 14  7 15 11 16&lt;br&gt;&#xA;  19:  1  2  1  3  2  4  5  3  9 15  4 14  5 13 16 17 18 19  9 10  7 12  6  8 11 15 14 13  7  6 10 16  8 17 12 18 11 19&lt;br&gt;&#xA;  20:  1  2  1  3  2  4  5  3  6 16  4 15  5 14 17  6 12 19 20 18  9 11 13  7 10  8 16 15 14 12  9  7 17 11  8 10 13 19 18 20  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There appear to be 300 8D solutions.  The reverse of a solution also is a solution.  I found no solutions for n in the set {1, 2, 5, 6, 9, 10, 13, 14} and found solutions for n in {3, 4, 7, 8, 11, 12, 15, 16, 19, 20}, suggesting there are  solutions only for n of the form 4k and 4k-1.  (I didn't test at n=17 or 18 or n&gt;20 as the program might take weeks to get done.  First-solutions at  n=19 and 20 took 20 minutes and 57 minutes to find.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a solution, let us write the occurrences of fruit i as i and i'.  Fruits i and i' are not distinguishable, so we can arrange that i always appears before i'.  Thus we can exhaust all possible orders of 2n items by generating and testing all permutations of n items, accepting those permutations such that an array of 2n cells&#xA;can be filled in via the following C code, with elements i and i' at distance i+1.  For a given n-permutation, there is one way or no way to fill in the array of 2n cells.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;void testPerm(int n) { // See if n-perm makes valid n-kabob&#xA;  int minplace, i, j;&#xA;  memset(dat, 0, sizeof(dat));&#xA;  minplace = 1;&#xA;  for (i=1; i&amp;lt;=n; ++i) {&#xA;    j = pi[i];&#xA;    while (dat[minplace])&#xA;      ++minplace;       // Get to next available spot in dat&#xA;    dat[minplace] = j;&#xA;    if (minplace+j+1 &amp;gt; 2*n) return; // Fail if out of bounds&#xA;    if (dat[minplace+j+1]) return;  // Fail if already in use&#xA;    dat[minplace+j+1] = j;&#xA;  }&#xA;  exitnow = 1;&#xA;  showKebab(n);&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note, the above routine uses some globally-declared variables, as below.  The current permutation is held in &lt;code&gt;pi[1]...pi[n]&lt;/code&gt;, and &lt;code&gt;dat&lt;/code&gt; is used in trying to generate an n-kabob from the n-permutation, or when printing a solution.  Var &lt;code&gt;exitnow&lt;/code&gt; tells the caller whether to exit (eg, if a solution is found).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#define NMAX 21&#xA;int pi[NMAX+1], dat[2*NMAX+2];&#xA;int exitnow;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The routine that calls &lt;code&gt;testPerm&lt;/code&gt; is like the following, which implements a permutation algorithm of Dijkstra's.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;void permuter(int n) {&#xA;  int i, j, r, s, temp;&#xA;  for (i=0; i &amp;lt;= n; i++) pi[i] = i; // Init to first perm, 1, 2...n&#xA;  i = 1;&#xA;  while (i &amp;amp;&amp;amp; !exitnow) {&#xA;    testPerm(n);&#xA;    // Generate next permutation&#xA;    i = n-1;&#xA;    while (pi[i] &amp;gt; pi[i+1]) i--;&#xA;    j = n;&#xA;    while (pi[i] &amp;gt; pi[j]) j--;&#xA;    temp = pi[i]; pi[i] = pi[j]; pi[j] = temp;&#xA;    r = n; s = i+1;&#xA;    while (r &amp;gt; s) {&#xA;      temp = pi[r]; pi[r] = pi[s]; pi[s] = temp;&#xA;      r--; s++;&#xA;    }&#xA;  }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2004" LastEditorUserId="2004" LastEditDate="2014-08-07T04:35:55.810" LastActivityDate="2014-08-07T04:35:55.810" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2041" PostTypeId="2" ParentId="2038" CreationDate="2014-08-07T02:51:42.697" Score="3" Body="&lt;p&gt;I wrote some code for this and brute forced it. I would greatly prefer a non-brute force solution or at very least more elegant code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 example solutions it found included but there are many:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - 4567348536271218 - 5673485364712182 - 6734583647512182 - 7345638475261218&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The code was:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import java.util.*;&#xA;import java.lang.*;&#xA;import java.io.*;&#xA;&#xA;/* Name of the class has to be &quot;Main&quot; only if the class is public. */&#xA;class Ideone&#xA;{&#xA;    public static void main (String[] args) throws java.lang.Exception&#xA;    {&#xA;        int[] desc = {1,2,3,4,5,6,7,8};&#xA;        int[] sep = {1,1,1,1,1,1,1,0};&#xA;        int c=0;&#xA;&#xA;        while (c&amp;lt;8*7*6*5*4*3*2){&#xA;            c++;&#xA;            boolean unique=false;&#xA;            while(!unique){&#xA;                int k=7;&#xA;                while(k&amp;gt;0&amp;amp;&amp;amp;sep[k]==7){sep[k]=1;k--;}&#xA;                sep[k]++;&#xA;                desc[0]=sep[0];&#xA;                for(int j=1;j&amp;lt;8;j++){desc[j]=(desc[j-1]+sep[j]-1)%8+1;}&#xA;                unique=true;&#xA;                for(int j=1;j&amp;lt;8;j++){&#xA;                    for(int x=0;x&amp;lt;j;x++){&#xA;                        if(desc[x]==desc[j]){unique=false;}&#xA;                    }&#xA;                }&#xA;            }&#xA;&#xA;            int[] list = {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0};&#xA;            int i=0;&#xA;            int j=0;&#xA;            while(j&amp;lt;8){&#xA;                while(list[i]&amp;gt;0){i++;}&#xA;                list[i]=desc[j];&#xA;                if(i+desc[j]+1&amp;gt;15||list[i+desc[j]+1]&amp;gt;0){&#xA;                    break;&#xA;                } else {&#xA;                    list[i+desc[j]+1]=desc[j];&#xA;                }&#xA;                j++;&#xA;            }&#xA;            boolean tester=true;&#xA;            for(int z=0; z&amp;lt;15; z++){if(list[z]&amp;lt;1||list[z]&amp;gt;8){tester=false;}}            &#xA;            if(j&amp;gt;7 &amp;amp;&amp;amp; tester){&#xA;            for(int z=0; z&amp;lt;15; z++){System.out.print(list[z]);}&#xA;            System.out.println(list[15]);}&#xA;        }&#xA;    }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditDate="2014-08-07T02:58:21.550" LastActivityDate="2014-08-07T02:58:21.550" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2042" PostTypeId="1" AcceptedAnswerId="2045" CreationDate="2014-08-07T05:51:50.860" Score="2" ViewCount="1756" Body="&lt;p&gt;A piece of fabric shaped like a regular hexagon with unit-length sides is given, with vertices A...F and center O.  Let a &lt;em&gt;fold&lt;/em&gt; correspond to folding some fabric along a straight line, without stretching.  For example, bringing A to E and B to D by folding along line FC is a fold.  Let a &lt;em&gt;uniform folding&lt;/em&gt; be a series of folds giving a result with uniform fabric thickness.  For example, if a uniform-folding result is 3 layers thick somewhere, it is 3 layers thick everywhere.&lt;/p&gt;&#xA;&lt;p&gt;Find a three-fold uniform folding giving a triangle.&lt;br /&gt;&#xA;Find a four-fold uniform folding giving a &lt;s&gt;square&lt;/s&gt; rectangle.&lt;br /&gt;&#xA;Find a &lt;s&gt;six-fold&lt;/s&gt; uniform folding giving a square, or prove that none exists.&lt;br /&gt;&#xA;Is there a five-fold uniform folding that gives a square?&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/S3u9t.jpg&quot; alt=&quot;Regular hexagon&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="2004" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-11-13T10:13:09.703" Title="Folding a hexagon to a rectangle or square, with uniform overlap" Tags="|geometry|optimization|paper-folding|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2043" PostTypeId="2" ParentId="2042" CreationDate="2014-08-07T07:55:02.493" Score="1" Body="&lt;p&gt;To get a triangle from a three-fold uniform folding, do the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Bring A to E and B to D by folding along FC.&lt;/li&gt;&#xA;&lt;li&gt;Bring F to D by folding along EO.&lt;/li&gt;&#xA;&lt;li&gt;Bring C to E by folding along DO.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now we have the triangle OED which is 6 layers thick everywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't found solutions to the other problems yet.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2014-08-07T07:55:02.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2044" PostTypeId="2" ParentId="2042" CreationDate="2014-08-07T13:29:27.813" Score="2" Body="&lt;p&gt;For a three fold giving a triangle of two thicknesses, fold along $AE, AC, CE$&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2014-08-07T13:29:27.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2045" PostTypeId="2" ParentId="2042" CreationDate="2014-08-07T16:04:52.290" Score="2" Body="&lt;p&gt;Here's a four fold uniform folding giving a &lt;s&gt;square&lt;/s&gt; rectangle that is slightly wider than it is tall:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;fold AB so that they lie on the line FC&lt;/li&gt;&#xA;&lt;li&gt;fold ED so that they lie on the line FC&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;After doing that, A and E are in the same position, and B and D are also in the same position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;fold so that F lies on top of A and E&lt;/li&gt;&#xA;&lt;li&gt;fold so that C lies on top of B and D&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thanks to PrisonMonkeys for correcting me on the square/rectangle nature of the result!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then obtain a 6-fold &quot;square&quot; by folding the 4-fold &quot;square&quot; above once vertically (down the middle), and then horizontally (down the middle), to obtain a &quot;square&quot; that is a quarter the size of the original.&lt;/p&gt;&#xA;" OwnerUserId="1608" LastEditorUserId="1608" LastEditDate="2014-08-07T17:50:55.903" LastActivityDate="2014-08-07T17:50:55.903" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2046" PostTypeId="1" CreationDate="2014-08-07T19:26:50.390" Score="1" ViewCount="422" Body="&lt;p&gt;Given $L$ units of wash water, we want to wash out a vessel (eg, a milk bottle, honey jar, chemistry flask...) using a series of rinses that minimize the amount of Stuff other than water remaining in the jar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $R$ units of liquid remain in the vessel each time we pour out the rinse; and suppose the original concentration of Stuff is $\alpha$, ie there is $R\cdot\alpha$ units of Stuff in the jar originally. Also suppose complete mixing of wash water with residue, so that when we add $x$ units of water to the $R$ in the jar, the new concentration is $\frac{R\cdot\alpha}{R+x}$. For example, if we put all $L$ units of wash water in at once, our final result is $R\cdot\frac{R\cdot\alpha}{R+L}$ units of Stuff left in the jar.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A better model would make $R$ a function of drainage time and viscosities, make new concentration a function of shake time, allow variable amounts of wash water, and account for differences in viscosity of water vs. Stuff, etc. In this simplified problem we ignore all that. However, if I have the physics all wrong even in the simplified scenario, please let me know.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 1:  Suppose $L=1$, $R=0.01$, $\alpha=1$. For each $k\in \{2...7\}$, where $k$ is a fixed number of rinses, what is an optimal division of wash water? (For example, when $k=2$, is $(L/2, L/2)$ the best division?  If not, what is?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 2:  For some $(L,R,\alpha)$, suppose the best result (ie, least residue) occurs in a $k$-way division. That is, if $B_k(L,R,\alpha)$ denotes the best result for a $k$-way division, we suppose $B_k(L,R,\alpha) \le  B_j(L,R,\alpha) \forall j \ne k$.]  With another value $L'$ in place of $L$, will it be that true $B_k(L',R,\alpha) \le  B_j(L',R,\alpha) \forall j \ne k$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other questions: [Like Q2, but with $R$ or $\alpha$ or combinations of $L,R,\alpha$ varying]&lt;/p&gt;&#xA;" OwnerUserId="2004" LastEditorUserId="2004" LastEditDate="2014-08-07T22:33:24.823" LastActivityDate="2014-08-08T00:01:32.147" Title="Bottle rinsing with fixed amount of wash water" Tags="|mathematics|calculation-puzzle|optimization|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2047" PostTypeId="2" ParentId="2046" CreationDate="2014-08-07T23:05:26.613" Score="3" Body="&lt;p&gt;Let's look at the case where $k$ is 2. We'll define $x_1, x_2$ such that $x_1+x_2=L$ to represent how much we use each time, with $0&amp;lt;=x_1,x_2&amp;lt;=L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you noted, after the first rinse, the concentration in the jar is $\alpha_1 = \frac{R\cdot\alpha}{R+x_1}$. Then, the concentration in the jar after the second rinse is $$R_2=\frac{R\cdot\alpha_1}{R+x_2}=\frac{R\cdot\frac{R\cdot\alpha}{R+x_1}}{R+x_2}=\alpha\cdot\frac{R}{R+x_1}\cdot\frac{R}{R+x_2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in general, we are trying to minimize $\prod_{n=1}^k{\frac{R}{R+x_n}}$, which is the same as maximizing $\prod_{n=1}^k{R+x_n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have time to do more than this right now, but hopefully this might be able to help other people who are looking at this problem.&lt;/p&gt;&#xA;" OwnerUserId="1955" LastActivityDate="2014-08-07T23:05:26.613" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2048" PostTypeId="2" ParentId="2046" CreationDate="2014-08-08T00:01:32.147" Score="6" Body="&lt;p&gt;More rinses are better, and equally divided is optimal.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show this, we first assume we will do only two rinses and show that evenly divided is best.  Let us use $a$ in the first rinse, then $L-a$ in the second.  After the two rinses, we have $\frac R{R+a}\cdot \frac R{R+L-a}$ left, so we want to maximize $(R+a)(R+L-a)=R^2+RL+La-a^2$, which occurs at $a=\frac L2$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we argue that given one rinse of a given size, we are better off splitting it in two equal halves. We do that until we have the maximum allowable number of rinses.  Given two unequal rinses, we are better evening them up.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2014-08-08T00:01:32.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2049" PostTypeId="1" AcceptedAnswerId="2050" CreationDate="2014-08-08T20:23:16.137" Score="2" ViewCount="386" Body="&lt;p&gt;To my knowledge, this puzzle is not previously published (except by me on google+ recently), but I would be interested to hear of any info otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This following graph was generated using a simple ruby program and gnuplot. It graphs a &quot;basic/fundamental&quot; number theoretic property of the natural numbers. Can you figure out what it is? Increasingly detailed hints below if you want more help. The answer and further background will be posted as a comment after some time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://vzn1.files.wordpress.com/2014/07/facta.png&quot; width=&quot;400&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;hint1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;fundamental&quot; as in &quot;fundamental theorem of arithmetic&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;hint2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; one word: &lt;em&gt;prime&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;hint3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; prime decomposition&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2017" LastEditorDisplayName="user31569" LastEditDate="2017-12-13T17:22:03.790" LastActivityDate="2019-03-29T00:48:32.520" Title="Guess the Function for Scatterplot of Number Theoretic Function" Tags="|mathematics|pattern|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2050" PostTypeId="2" ParentId="2049" CreationDate="2014-08-08T22:56:13.380" Score="4" Body="&lt;p&gt;This would appear to be a plot of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $x$ against its largest prime factor. The points on the top line are primes, the next line down are primes times two, then primes times three, and so on, with the scattering at the bottom being as a result of numbers that are a product of more than two primes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2007" LastEditorUserId="2007" LastEditDate="2014-08-08T23:04:11.850" LastActivityDate="2014-08-08T23:04:11.850" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2051" PostTypeId="1" AcceptedAnswerId="2052" CreationDate="2014-08-09T06:42:14.267" Score="6" ViewCount="842" Body="&lt;p&gt;obviously I understand the definition part &quot;One who will succeed&quot;. But can someone tell me how the word play part works in this clue?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;with melody, we hear&quot;&lt;/p&gt;&#xA;" OwnerUserId="2020" LastActivityDate="2014-08-09T09:20:24.547" Title="Cryptic Clue - &quot;One who will succeed with melody, we hear&quot; Answer - &quot;Heir&quot;" Tags="|crosswords|cryptic-crosswords|crossword-clues|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2052" PostTypeId="2" ParentId="2051" CreationDate="2014-08-09T09:20:24.547" Score="15" Body="&lt;p&gt;&quot;Heir&quot; is usually pronounced with a silent H, so sounds like &quot;air&quot;.&#xA;One meaning of the word air is a melody, such as &quot;air on a g string&quot;.&#xA;&quot;We hear&quot; is a common idiom in cryptic crossword clues for homophones ie &quot;sounds like&quot;.&#xA;So &quot;melody, we hear&quot; = &quot;sounds like another word for melody&quot; = &quot;sounds like air&quot; = &quot;heir&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2021" LastActivityDate="2014-08-09T09:20:24.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2053" PostTypeId="1" AcceptedAnswerId="2074" CreationDate="2014-08-09T23:40:07.280" Score="3" ViewCount="901" Body="&lt;p&gt;An $S$-&lt;em&gt;tileset&lt;/em&gt; is a collection  of $n$ oriented tiles, where no two tiles have the same size, each tile is one unit thick, and its non-zero-integer length and width add up to $n+1$.  (So, an $S$-tileset has $n$ tiles, of sizes $1\times n, 2\times{n-1}, ... n\times1$.)  You can make a tower of tiles by stacking up layers of tiles; tiles in a layer adjoin; rotation of tiles is not allowed; you don't have to use all tiles.  Overhang is not allowed:  all the area of each layer must be supported by an area of the layer below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How high is the tallest tower that can be made with one  $S$-tileset?  [If you can write the answer in terms of $n$, great; else exhibit solutions for $n\in\{6,9,12\}$]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;s&gt;How high can it be if given two $S$-tilesets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How high is the tallest tower that can be made with one  $S$-tileset if you break the rules and rotate some tiles?&lt;/s&gt;&lt;/p&gt;&#xA;" OwnerUserId="2004" LastEditorUserId="2004" LastEditDate="2014-08-10T03:28:24.197" LastActivityDate="2014-08-13T03:49:28.710" Title="How high a tower of tiles can be made?" Tags="|geometry|combinatorics|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2054" PostTypeId="1" AcceptedAnswerId="2059" CreationDate="2014-08-10T13:36:16.740" Score="10" ViewCount="6259" Body="&lt;p&gt;If you multiply $2178$ by $4$ you will get the same digits, but written in the back order, like if you look in a mirror: $2178 \cdot 4 = 8712$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to find all possible natural numbers, that are mirrored when multiplied by 4? &lt;strong&gt;And prove that these are all numbers.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What other factors $F \neq 4$ have mirrored numbers?&lt;/strong&gt;&lt;br&gt;&#xA;I see that $F = 1$ has a trivial mirror number $1$, but I would like to have help with other $F=2,..,9$.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2014-08-12T07:16:11.453" LastActivityDate="2017-07-16T19:58:41.800" Title="Mirrored numbers" Tags="|mathematics|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2055" PostTypeId="2" ParentId="2054" CreationDate="2014-08-10T14:26:47.203" Score="3" Body="&lt;p&gt;The numbers are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2178, 21978, 219978, ...$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ 21782178, 217821782178, 2178217821782178, ... $ with $F = 4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1089, 10989, 109989, ... $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$10891089, 108910891089, ... $ with $ F = 9 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F \in [2,9]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why I will have to check in more detail but AFAIK there are the only two series.&lt;/p&gt;&#xA;" OwnerUserId="1943" LastEditorUserId="2226" LastEditDate="2014-08-30T14:07:57.347" LastActivityDate="2014-08-30T14:07:57.347" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2056" PostTypeId="1" AcceptedAnswerId="2057" CreationDate="2014-08-10T16:50:10.903" Score="6" ViewCount="761" Body="&lt;p&gt;The image shows numbers on ring-dials on a curiously-geared combination lock, where the current setting is (3,3,3,3) and ring sizes are (5,7,8,11), inner to outer.  The setting (0,0,0,0) opens the lock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you rotate any ring, rings outside it move twice as fast in the same direction, while rings inside it move twice as fast in the opposite direction.  For example,&#xA;moving the 7-ring in the diagram 1 unit to 2 changes the 8- and 11-rings to 1 and the 5-ring to 0, for a new state of (0, 2, 1, 1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you find a minimal sequence to open this lock?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The cost of a sequence is the total number of places that selected dials are turned in one direction.  There is no cost for selecting a dial and direction.  For example, if you turn the 5-ring 18 places CCW and 11 places CW, and then the 11-ring 29 places CCW, the cost is 18+11+29 = 58.  At most one dial can be selected at a time.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tRC0A.jpg&quot; alt=&quot;CGCL S=(5,7,8,11), B=(3,3,3,3)&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2004" LastActivityDate="2014-08-13T14:08:05.897" Title="Unlocking a curiously-geared combination lock" Tags="|mathematics|mechanical-puzzles|combinatorics|twisty-puzzles|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2057" PostTypeId="2" ParentId="2056" CreationDate="2014-08-10T23:30:47.283" Score="10" Body="&lt;p&gt;First, what do we know about this puzzle?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There are 5*7*8*11 = 3080 possible positions.&lt;/li&gt;&#xA;&lt;li&gt;Moves do not affect each other, so if you have a good set of moves, you can do them in any order.&lt;/li&gt;&#xA;&lt;li&gt;There must be at least one move of the 8 ring, but not necessarily any of the others.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Given this, it is a relatively simple algorithm to find the shortest sequence of moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I did was to start with a list containing only (0,0,0,0), then for each item in that list, I did a +5,-5,+7,-7,+8,-8,+11,-11. For each item produced by this, I check if it is in the list and if not, add it to the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had it stop when it got to (3,3,3,3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(0,0,0,0) -&gt; +5&lt;br&gt;&#xA;(1,2,2,2) -&gt; +7&lt;br&gt;&#xA;(4,3,4,4) -&gt; -8&lt;br&gt;&#xA;(1,5,3,2) -&gt; -8&lt;br&gt;&#xA;(3,0,2,0) -&gt; -8&lt;br&gt;&#xA;(0,2,1,9) -&gt; -8&lt;br&gt;&#xA;(2,4,0,7) -&gt; -8&lt;br&gt;&#xA;(4,6,7,5) -&gt; -11&lt;br&gt;&#xA;(1,1,1,4) -&gt; -11&lt;br&gt;&#xA;(3,3,3,3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the minimum number of moves is 9. One example is(just reversing my list):&lt;br&gt;&#xA;&lt;strong&gt;+11,+11,+8,+8,+8,+8,+8,-7,-5&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&lt;br&gt;&#xA;The maximum distance from any position to any other position is 14. Here is a list of the distances and the number of points for each distance. This is done from (0,0,0,0), but it holds for any point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0: 1&lt;br&gt;&#xA;1: 8&lt;br&gt;&#xA;2: 32&lt;br&gt;&#xA;3: 84&lt;br&gt;&#xA;4: 168&lt;br&gt;&#xA;5: 284&lt;br&gt;&#xA;6: 420&lt;br&gt;&#xA;7: 397&lt;br&gt;&#xA;8: 344&lt;br&gt;&#xA;9: 330&lt;br&gt;&#xA;10: 324&lt;br&gt;&#xA;11: 324&lt;br&gt;&#xA;12: 250&lt;br&gt;&#xA;13: 100&lt;br&gt;&#xA;14: 14&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of an item that is 14 moves from (0,0,0,0) is (3,1,5,1), which happens to be the last item in my sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closest items with 3 0's are (0,0,0,3) and (0,0,0,8), which are 5 moves away.&lt;br&gt;&#xA;The closest items with 3 0's and a 1 are (1,0,0,0) and (0,0,1,0), which are 7 moves away.&lt;br&gt;&#xA;The furthest item with 3 0's and a 1 is (0,1,0,0) which is 12 moves away.&lt;br&gt;&#xA;(0,0,0,1) is 8 moves away (to complete the set).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update 2:&lt;/strong&gt;&lt;br&gt;&#xA;The question was asked how I knew the 8 ring must move at least once, but not necessarily any of the others. There are two different parts to this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, if you never move the 8 ring, moving other rings always causes it to move 2, which means it can only ever be at 1, 3, 5 or 7. So you must move it at least once to get it to an even number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, to never move another ring: I expected this to be true because all the ring sizes are relatively prime to each other (and to the moves, except for the 2-move and 8-ring), so if I move the 8 ring 3080 times, I should go through every permutation. If I move any other ring 1540 times, I'll go through half the permutations (because in that case, the 8 ring only stays on odd numbers). To verify this, I put together a quick spreadsheet to simulate moving each ring +1 3080 times. The spreadsheet showed this to be correct. It also showed that you can get to (0,0,0,0) by moving the 8 ring -1 only 51 times.&lt;/p&gt;&#xA;" OwnerUserId="375" LastEditorUserId="375" LastEditDate="2014-08-13T14:08:05.897" LastActivityDate="2014-08-13T14:08:05.897" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2058" PostTypeId="2" ParentId="38" CreationDate="2014-08-11T04:16:35.613" Score="2" Body="&lt;p&gt;First I want to question your premise a bit. Your &lt;strong&gt;&lt;em&gt;goal&lt;/em&gt;&lt;/strong&gt; is to create harder mazes, but you suggest that the &lt;strong&gt;&lt;em&gt;method&lt;/em&gt;&lt;/strong&gt; is to increase the branching factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are right up to a point; more branches is often associated with higher difficulty, but if you go too far you run into the problem that Joe Z. hinted at - you'll have lots of very short dead-ends, which to a human is very easy to spot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the ultimate maze/labyrinth creation resource, I often go to &lt;a href=&quot;http://www.astrolog.org/labyrnth.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Think Labyrinth by Walter D. Pullen&lt;/a&gt;. On his &lt;a href=&quot;http://www.astrolog.org/labyrnth/algrithm.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Maze Classification&lt;/a&gt; page, he lists a relevant metric for you, &lt;strong&gt;&lt;em&gt;river&lt;/em&gt;&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;River&lt;/strong&gt;: The &quot;river&quot; characteristic means that when creating the Maze, the algorithm will look for and clear out nearby cells (or walls) to the current one being created, i.e. it will flow (hence the term &quot;river&quot;) into uncreated portions of the Maze like water. A perfect Maze with less &quot;river&quot; will tend to have many short dead ends, while a Maze with more river will have fewer but longer dead ends. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think what you want is a moderate amount of &quot;river&quot;; lots of branches but not so many that will cause the dead ends to be too small and easy to spot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare the following mazes generated by different algorithms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Recursive backtracker (very high river)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DbqrB.gif&quot; alt=&quot;recursive backtracker&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Eller's Algorithm (moderate river)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VAi03.gif&quot; alt=&quot;eller&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Prim's Algorithm (very low river)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LOMwv.gif&quot; alt=&quot;prim&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that your algorithm is pretty much the recursive backtracker. Try a different algorithm!&lt;/p&gt;&#xA;" OwnerUserId="104" LastActivityDate="2014-08-11T04:16:35.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2059" PostTypeId="2" ParentId="2054" CreationDate="2014-08-11T08:54:22.467" Score="22" Body="&lt;h2&gt;The factors are 1, 4 and 9&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;We consider a number a factor iff there exist a non zero integer which it mirrors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F = 1$ is clearly a factor for all palindromes (numbers that are already mirrors of themselves). The remaining factors must be in $[2,9]$ because:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;negative $F$ produce numbers with opposite sign&lt;/li&gt;&#xA;&lt;li&gt;$F = 0$ is factor only for the number $0$&lt;/li&gt;&#xA;&lt;li&gt;$F \geq 10$ implies that the mirrored would have more digits than the original (we will stick to classical decimal notation: $F = 10$ is not a factor for $1 = 01$ and $9.999\ldots$ is not a valid notation for 10).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Let $\overline{a_n\ a_{n-1}\ \ldots a_0}$ be the decimal notation of a mirror $\sum\limits_{0 \leq i \leq n} a_i 10^{i}$ (mirrored in $\overline{a_0\ a_1\ \ldots a_n}$) for the factor $F$. Note that we will always use this notation with $n \geq 1$ and $a_n \neq 0$, it applies for all non-zero mirrors with $F &amp;gt; 1$ because they have a least two digits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $a_0 \times F \equiv a_n$ mod $10$, $a_n \times F \leq a_0$ and $a_n \times F &amp;lt; 10$ therefore a simple brute force like this&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;code&gt;for&lt;/code&gt;$a_0$, $a_n$, $F$ in $[1,9]^2 \times [2,9]$&lt;code&gt;:&#xA;    if&lt;/code&gt; $a_0 \times F \equiv a_n$ mod $10$, $a_n \times F \leq a_0$ and $a_n \times F &amp;lt; 10$&lt;code&gt;:&#xA;         print&lt;/code&gt;$a_0$, $a_n$, $F$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yields that the only possibilities with $F \in [2,9]$ are :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{|c|c|c|c|c|}&#xA;\hline&#xA;F &amp;amp; 2 &amp;amp; 3 &amp;amp; 4 &amp;amp; 9\\\hline&#xA;a_0 &amp;amp; 6 &amp;amp; 7 &amp;amp; 8 &amp;amp; 9\\\hline&#xA;a_n &amp;amp; 2 &amp;amp; 1 &amp;amp; 2 &amp;amp; 1\\\hline&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can eliminate $F = 2$ instantly because $\overline{2\ a_{n-1}\ \ldots 6} \times 2 = \overline{6\ a_{1}\ \ldots 2}$ implies that $(2 \times 10^n + \overline{a_{n-1}\ \ldots 6}) \times 2 \geq 6 \times 10^n$ hence that $\overline{a_{n-1}\ a_{n-2}\ \ldots 6} \geq (\frac{6}{2} - 2) 10^n \geq 10^n$ yet we know that $\overline{a_{n-1}\ a_{n-2}\ \ldots 6} &amp;lt; 10^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same argument can be used for $F = 3$ :  $\overline{1\ a_{n-1}\ \ldots 7} \times 3 = \overline{7\ a_{1}\ \ldots 3}$ therefore $\overline{a_{n-1}\ a_{n-2}\ \ldots 7} \times 3 \geq (7 - 1 \times 3) 10^n$ which is impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the possible factors are included in $\{1,4,9\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these factors have mirrors (for instance $4224$, $2178$ and $1089$ for respectively $1$, $4$ and $9$) so they are the only possible factors.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The mirrors are expressible with a context-free grammar&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Using the same constraints as before we can numerically compute the possible extremities ($a_0$ and $a_n$) of the admissible numbers, then we have a slightly altered problem for the number without its extremities: in $2178 \times 4= 8712$ it corresponds to $17 \times 4 + 3 = 71$ (this $+3$ comes from $2008 \times 4 - 8002 = 3 . 10^1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can show that a slightly more general class of mirror problems (with small offsets like this $+3$) can be solved using only very basic operators and other problems of the same class. This will allow us to express the mirrors numbers using context free grammar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the following $S_F(u,v)$ will be used to denote the solution of a mirror problem for factor $F$ with left offset $u$ and right offset $v$, more formally:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $F \in \{1,4,9\}$ and $u, v \in  [0,9]^2$, let $S_F(u,v)$ be the set of numbers such that $\overline{a_n\ a_{n-1}\ \ldots a_0} \in S_F(u,v)$ iff $\overline{a_n\ a_{n-1}\ \ldots a_0} \times F + v - u 10^{n+1} = \overline{a_0\ a_1\ \ldots a_n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are interested in $S_4(0,0) = \{\overline{a_n\ a_{n-1}\ \ldots a_0}\ |\ \overline{a_n\ a_{n-1}\ \ldots a_0} \times 4 = \overline{a_0\ a_1\ \ldots a_n} \}$ and $S_9(0,0)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have seen that the decimal notation of elements of $S_4(0,0)$ starts with 2 and ends 8, lets have a look at what is between:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{align}&#xA;\overline{a_{n}\ \ldots a_0} \in S_4(0,0) &amp;amp;\Leftrightarrow \overline{2\ a_{n-1}\ \ldots a_1, 8} \times 4 = \overline{8\ a_{1}\ \ldots a_n, 2}, a_0 = 8, a_n = 2\\&#xA;&amp;amp;\Leftrightarrow \overline{a_{n-1}\ \ldots a_1, 0} \times 4 + 32 = \overline{a_{1}\ \ldots a_n, 2}, a_0 = 8, a_n = 2\\&#xA;&amp;amp;\Leftrightarrow \overline{a_{n-1}\ \ldots a_1} \times 4 + 3 = \overline{a_{1}\ \ldots a_n}, a_0 = 8, a_n = 2\\&#xA;\overline{a_{n}\ \ldots a_0} \in S_4(0,0) &amp;amp;\Leftrightarrow \overline{a_{n-1}\ \ldots a_1} \in S_4(0,3), a_0 = 8, a_n = 2&#xA;\end{align}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be expressed as $S_4(0,0) = 2.S_4(0,3).8$ (plus potential mirors with less than 2 digits as we only considered element of $S_4(0,0)$ of the form $\overline{a_{n}\ \ldots a_0}_{n \geq 1}$) where $.$ is the concatenation of digits. The mirrors with less than two digits can only be $0$ (1 digit numbers don't work), so $S_4(0,0) = 0 + 2.S_4(0,3).8$. where $+$ is the set union and $0$ stand for the singleton containing $0$, i.e. $\{0\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly $S_9(0,0) = 0 + 1.S_9(0,8).9$ because we know that $a_0 = 9, a_n = 1$ is the only possibility and $v = (a_0 \times F - a_n) /10 = (9 \times 9 - 1)/10 = 8$ and $u = a_0 - a_n \times F = 9 - 1 \times 9 = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally any $S_F(u,v)$ can be expressed using digit concatenation, digits, union and &#xA;$\{S_F(i,j) | i,j \in [0,9]^2\}$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Explicitly the mirrors are:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;$\begin{align*}&#xA;S_1(0,0) &amp;amp;= 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 \\&#xA;&amp;amp;\ \ + 1.S_1'(0,0).1 + 2.S_1'(0,0).2 + 3.S_1'(0,0).3 + 4.S_1'(0,0).4 + 5.S_1'(0,0).5 + 6.S_1'(0,0).6 + 7.S_1'(0,0).7 + 8.S_1'(0,0).8 + 9.S_1'(0,0).9\\&#xA;&amp;amp;S_1'(0,0) = S_1(0,0) + \epsilon + 0.S_1'(0,0).0\\&#xA;S_4(0,0) &amp;amp;= 0 + 2.S_4(0,3).8\\&#xA;&amp;amp;S_4(0,3) = 17 + 1.S(3,3).7\\&#xA;&amp;amp;S_4(3,3) = 9 + 99 + 7.S(3,0).1 + 9.S_4(3,3).9\\&#xA;&amp;amp;S_4(3,0) = 82 + 8.S_4'(0,0).2\\&#xA;&amp;amp;S_4'(0,0) = S_4(0,0) + 00 + 0.S_4'(0,0).0\\&#xA;S_9(0,0) &amp;amp;= 0 + 1.S_9(0, 8). 9\\&#xA;&amp;amp;S_9(0, 8) = 08 + 0.S_9(8, 8).8\\&#xA;&amp;amp; S_9(8, 8) = 9 + 99 + 8.S_9(8, 0).0 + 9.S_9(8, 8).9\\&#xA;&amp;amp;S_9(8, 0) =  91 + 9.S_9'(0, 0).1\\&#xA;&amp;amp;S_9'(0, 0) = S_9(0, 0) + 00 + 0.S_9'(0, 0).0&#xA;\end{align*}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$.$ is the digit concatenation over sets: $a.b = \{\overline{u\ v} | u \in a, v \in b\}$&lt;/li&gt;&#xA;&lt;li&gt;$\epsilon$ is its neutral, i.e. $\epsilon.a = a.\epsilon = a$&lt;/li&gt;&#xA;&lt;li&gt;$+$ is the set union&lt;/li&gt;&#xA;&lt;li&gt;numbers stands for singleton containing only themselves&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We consider the minimal sets, for inclusion, satisfying the relations.&#xA;The primed set are an artifact to avoid numbers starting with zeros.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Some &lt;a href=&quot;http://pastebin.com/gVcD2vQm&quot; rel=&quot;noreferrer&quot;&gt;quick and dirty python code&lt;/a&gt; to get this result.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Some examples&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Let's construct some examples from the above description, we'll use some colors and parenthesis to highlight the construction. We will stick with $F = 4:$ and the following color code: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{|c|c|c|c|c|}&#xA;\hline&#xA;S_4(0,0) &amp;amp;&#xA; S_4(0,3) &amp;amp;&#xA; S_4(3,3) &amp;amp;&#xA; S_4(3,0) &amp;amp;&#xA; S_4'(0,0)\\\hline&#xA;\color{blue}{0}, &amp;amp;&#xA; \color{Orange}{17}, &amp;amp;&#xA; \color{Green}{9}, \color{Green}{7}.S_4(3,0).\color{Green}{1}, &amp;amp;&#xA; \color{Purple}{82}, &amp;amp;&#xA; S_4(0,0), \color{Red}{00}\\&#xA;\color{blue}{2}.S_4(0,3).\color{blue}{8} &amp;amp;&#xA; \color{Orange}{1}.S_4(3,3).\color{Orange}{7} &amp;amp;&#xA; \color{Green}{99}, \color{Green}{9}.S_4(3,3).\color{Green}{9} &amp;amp;&#xA; \color{Purple}{8}.S_4'(0,0).\color{Purple}{2} &amp;amp;&#xA; \color{Red}{0}.S_4'(0,0).\color{Red}{0}\\\hline&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This table says that to get an element from a set you have to take elements from its column. Let's say we want an element of $S_4(0,0)$ (it's the ones we are interested in: the mirrors for $F = 4$ without offsets) : so you can take $\color{blue}{0}$ for instance, and indeed $0 \times 4 = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only other possibility would be $\color{blue}{2}.S_4(0,3).\color{blue}{8}$, that means take an element of $S_4(0,3)$ put a $\color{blue}{2}$ before and a $\color{blue}{8}$ after and you are good.&#xA;For instance $\color{Orange}{17}$ is an element of $S_4(0,3)$ therefore $\color{blue}{2}\color{Orange}{17}\color{blue}{8}$ is an element of $S_4(0,0)$, and indeed $2178 \times 4 = 8712$. In the following I'm gonna add parenthesis -like $\color{blue}{2}(\color{Orange}{17})\color{blue}{8}$ - to make the construction clearer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We had to take an element from $S_4(0,3)$ and we took $\color{Orange}{17}$, we could also have chosen to use $\color{Orange}{1}.S_4(3,3).\color{Orange}{7}$, for instance $\color{Orange}{1}(\color{Green}{9})\color{Orange}{7}$ which leads to $\color{blue}{2}(\color{Orange}{1}(\color{Green}{9})\color{Orange}{7})\color{blue}{8}$ (and yes $21978 \times 4 = 87912$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some others examples are:&#xA;$$\begin{array}{c|c|c}&#xA;\color{blue}{2}(\color{Orange}{1}(\color{Green}{99})\color{Orange}{7})\color{blue}{8}&#xA;&amp;amp;&#xA; \color{blue}{2}(\color{Orange}{1}(\color{Green}{7}(\color{Purple}{82})\color{Green}{1})\color{Orange}{7})\color{blue}{8}&#xA;&amp;amp;&#xA; \color{blue}{2}(\color{Orange}{1}(\color{Green}{9}(\color{Green}{7}(\color{Purple}{82})\color{Green}{1})\color{Green}{9})\color{Orange}{7})\color{blue}{8}&#xA;\\\hline&#xA; \color{blue}{2}(\color{Orange}{1}(\color{Green}{9}(\color{Green}{99})\color{Green}{9})\color{Orange}{7})\color{blue}{8}&#xA;&amp;amp;&#xA; \color{blue}{2}(\color{Orange}{1}(\color{Green}{9}(\color{Green}{9}(\color{Green}{7}(\color{Purple}{82})\color{Green}{1})\color{Green}{9})\color{Green}{9})\color{Orange}{7})\color{blue}{8}&#xA;&amp;amp;&#xA; \color{blue}{2}(\color{Orange}{1}(\color{Green}{7}(\color{Purple}{8}(\color{Red}{00})\color{Purple}{2})\color{Green}{1})\color{Orange}{7})\color{blue}{8}&#xA;\\\hline&#xA; \color{blue}{2}(\color{Orange}{1}(\color{Green}{7}(\color{Purple}{8}(\color{blue}{0})\color{Purple}{2})\color{Green}{1})\color{Orange}{7})\color{blue}{8}&#xA;&amp;amp;&#xA; \color{blue}{2}(\color{Orange}{1}(\color{Green}{7}(\color{Purple}{8}(\color{blue}{2}(\color{Orange}{17})\color{blue}{8})\color{Purple}{2})\color{Green}{1})\color{Orange}{7})\color{blue}{8}&#xA;&amp;amp;&#xA; \color{blue}{2}(\color{Orange}{1}(\color{Green}{7}(\color{Purple}{8}(\color{red}{0}(\color{blue}{2}(\color{Orange}{17})\color{blue}{8})\color{red}{0})\color{Purple}{2})\color{Green}{1})\color{Orange}{7})\color{blue}{8}&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{c}&#xA; \color{blue}{2}(\color{Orange}{1}(\color{Green}{7}(\color{Purple}{8}(\color{red}{0}(\color{blue}{0})\color{red}{0})\color{Purple}{2})\color{Green}{1})\color{Orange}{7})\color{blue}{8}&#xA;\\\hline&#xA; \color{blue}{2}(\color{Orange}{1}(\color{Green}{7}(\color{Purple}{8}(\color{red}{0}(\color{red}{0}(\color{red}{00})\color{red}{0})\color{red}{0})\color{Purple}{2})\color{Green}{1})\color{Orange}{7})\color{blue}{8}&#xA;\\\hline&#xA; \color{blue}{2}(\color{Orange}{1}(\color{Green}{7}(\color{Purple}{8}(\color{red}{0}(\color{red}{0}(\color{red}{0}(\color{blue}{2}(\color{Orange}{1}(\color{Green}{7}(\color{Purple}{8}(\color{blue}{0})\color{Purple}{2})\color{Green}{1})\color{Orange}{7})\color{blue}{8})\color{red}{0})\color{red}{0})\color{red}{0})\color{Purple}{2})\color{Green}{1})\color{Orange}{7})\color{blue}{8}&#xA;\\\hline&#xA;\color{blue}{2}(\color{Orange}{1}(\color{Green}{7}(\color{Purple}{8}(\color{blue}{2}(\color{Orange}{1}(\color{Green}{7}(\color{Purple}{8}(\color{blue}{2}(\color{Orange}{1}(\color{Green}{7}(\color{Purple}{8}(\color{red}{0}(\color{red}{0}(\color{blue}{2}(\color{Orange}{1}(\color{Green}{7}(\color{Purple}{8}(\color{blue}{0})\color{Purple}{2})\color{Green}{1})\color{Orange}{7})\color{blue}{8})\color{red}{0})\color{red}{0})\color{Purple}{2})\color{Green}{1})\color{Orange}{7})\color{blue}{8})\color{Purple}{2})\color{Green}{1})\color{Orange}{7})\color{blue}{8})\color{Purple}{2})\color{Green}{1})\color{Orange}{7})\color{blue}{8}&#xA;\end{array}$$&lt;/p&gt;&#xA;" OwnerUserId="1779" LastEditorUserId="1779" LastEditDate="2015-01-12T10:57:10.047" LastActivityDate="2015-01-12T10:57:10.047" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="2060" PostTypeId="2" ParentId="2054" CreationDate="2014-08-11T10:26:37.967" Score="2" Body="&lt;p&gt;I made a program for it, and its algorithm is to get an interval of numbers from the user, reverse and multiply every number in between, and if the reverse and multiplication is equal, then print it for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a C# code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Console.WriteLine(&quot;number from&quot;);&#xA;int a = Convert.ToInt32(Console.ReadLine());&#xA;Console.WriteLine(&quot;to&quot;);&#xA;int b = Convert.ToInt32(Console.ReadLine());&#xA;Console.WriteLine(&quot;Multiplied by?&quot;);&#xA;int y = Convert.ToInt32(Console.ReadLine());&#xA;for (int i = a; i &amp;lt; b; i++)&#xA;{&#xA;    string reverse = new string(i.ToString().Reverse().ToArray());&#xA;    if (y * i == Convert.ToInt32(reverse))&#xA;        Console.WriteLine(i+&quot;\t\t&quot;+reverse+&quot;\t\t&quot;+&quot;Mirror number&quot;);&#xA;}&#xA;Console.WriteLine(&quot;Processing finished&quot;);&#xA;Console.ReadLine();&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;My test for numbers between 10 to 30000000:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uFX0X.png&quot; alt=&quot;Console output&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process is not finished until &quot;Processing finished&quot; shows up.&lt;/p&gt;&#xA;" OwnerUserId="2029" LastEditorUserId="34685" LastEditDate="2017-07-16T19:58:41.800" LastActivityDate="2017-07-16T19:58:41.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2061" PostTypeId="1" AcceptedAnswerId="2062" CreationDate="2014-08-11T18:40:17.183" Score="1" ViewCount="797" Body="&lt;p&gt;There is a running match, but there isn't anyone to judge; we have only information that the audience gave us (but we know their info is true). We the have name, rank, T-shirt color, and T-shirt number of each runner, as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Mahsa&lt;/strong&gt; won and she wore a red T-Shirt.  &lt;/li&gt;&#xA;&lt;li&gt;The girl who wore T-Shirt number 1 finished 3rd.  &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Mahnaz&lt;/strong&gt;, who defeated a girl with a yellow T-Shirt, didn't wear number 2.  &lt;/li&gt;&#xA;&lt;li&gt;Only one girl got a rank that was as same as her T-shirt number; her T-shirt wasn't Red.  &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Mozgan&lt;/strong&gt; defeated a girl wearing number 3 and &lt;strong&gt;Maryam&lt;/strong&gt; wore the yellow T-Shirt.  &lt;/li&gt;&#xA;&lt;li&gt;The girl with green T-shirt wore number 2.  &lt;/li&gt;&#xA;&lt;li&gt;One of the girls had a blue T-shirt.  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There were 4 runners.&lt;br&gt;&#xA;Can you name the girls along with their rank (1 to 4), number, and color?  &lt;/p&gt;&#xA;" OwnerUserId="2029" LastEditorUserId="2029" LastEditDate="2014-09-03T09:10:22.720" LastActivityDate="2014-09-03T09:10:22.720" Title="Who? What Color? Rank? Number? in a running match" Tags="|lateral-thinking|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2062" PostTypeId="2" ParentId="2061" CreationDate="2014-08-11T18:59:37.370" Score="5" Body="&lt;p&gt;I assumed from the word &quot;defeat&quot; in hints 3 and 5 that coming coming 2nd and 3rd would both defeat the 4th place-holder which pointed to it being Maryam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ranking:-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Mahsa; Red; No.4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Mozgan; Green; No.2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Mahnaz; Blue; No.1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Maryam; Yellow; No.3&lt;/p&gt;&#xA;" OwnerUserId="2031" LastEditorUserId="2031" LastEditDate="2014-08-11T19:36:50.277" LastActivityDate="2014-08-11T19:36:50.277" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2063" PostTypeId="2" ParentId="2061" CreationDate="2014-08-11T19:38:35.757" Score="3" Body="&lt;ul&gt;&#xA;&lt;li&gt;Rank 1: Mahsa wearing red number 4&lt;/li&gt;&#xA;&lt;li&gt;Rank 2: Mozgan wearing green number 2&lt;/li&gt;&#xA;&lt;li&gt;Rank 3: Mahnaz wearing blue number 1&lt;/li&gt;&#xA;&lt;li&gt;Rank 4: Maryam wearing yellow number 3&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This did not need more complicated means to solve.  All girls are listed as beating someone besides Maryam so she is last.  Mahsa can't wear 1, 2, or 3 as 1 got 3rd, 2 wore green, and 3 lost to someone.  (I assume 4 is the remaining number.) Maryam must wear 3 as 2 wore green and others are taken.  2 must have won second as it is the only one left.  Mahnaz must be 3rd as she didn't weat number 2. &lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2014-08-11T19:38:35.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2064" PostTypeId="2" ParentId="2038" CreationDate="2014-08-12T07:46:02.550" Score="11" Body="&lt;p&gt;So far the answers have been dealing with brute force. Here I present the mathematical proof that such kebab only exists iff &lt;span class=&quot;math-container&quot;&gt;$n\equiv -1,0 \pmod 4$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;h3&gt;Prove that no kebab could exist if &lt;span class=&quot;math-container&quot;&gt;$n\equiv 1,2 \pmod 4$&lt;/span&gt;.&lt;/h3&gt;&#xA;&lt;p&gt;Let &lt;span class=&quot;math-container&quot;&gt;$x_1, x_2, \ldots, x_n$&lt;/span&gt; be the positions of the left-most vegetable, for each vegetable numbered &lt;span class=&quot;math-container&quot;&gt;$1, 2,\ldots, n$&lt;/span&gt;, respectively. Then the positions for the other vegetables are: &lt;span class=&quot;math-container&quot;&gt;$x_1+2, x_2+3, \ldots, x_n+n+1$&lt;/span&gt;. Note that these &lt;span class=&quot;math-container&quot;&gt;$2n$&lt;/span&gt; positions should all be distinct and in the range of &lt;span class=&quot;math-container&quot;&gt;$[1,2n]$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Note that for all &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt;, if &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; is odd, &lt;span class=&quot;math-container&quot;&gt;$x_i$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$x_i+i+1$&lt;/span&gt; will have the same parity, and if &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; is even, they will have different parity.&lt;/p&gt;&#xA;&lt;p&gt;Now, suppose &lt;span class=&quot;math-container&quot;&gt;$n=4k+1\equiv 1 \pmod 4$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;There are &lt;span class=&quot;math-container&quot;&gt;$2k+1$&lt;/span&gt; vegetables with odd index, and &lt;span class=&quot;math-container&quot;&gt;$2k$&lt;/span&gt; vegetables with even index.&lt;/p&gt;&#xA;&lt;p&gt;There are in total &lt;span class=&quot;math-container&quot;&gt;$8k+2$&lt;/span&gt; positions to be filled, with &lt;span class=&quot;math-container&quot;&gt;$4k+1$&lt;/span&gt; even-indexed positions and &lt;span class=&quot;math-container&quot;&gt;$4k+1$&lt;/span&gt; odd-indexed positions.&lt;/p&gt;&#xA;&lt;p&gt;Since the vegetables with even index will use one even-indexed position and one odd-indexed position, there will be &lt;span class=&quot;math-container&quot;&gt;$4k+1-2k=2k+1$&lt;/span&gt; even-indexed positions and &lt;span class=&quot;math-container&quot;&gt;$2k+1$&lt;/span&gt; odd-indexed positions left for the vegetables with odd index. But each pair of vegetable with odd index will use two position with the same parity. Since &lt;span class=&quot;math-container&quot;&gt;$2k+1$&lt;/span&gt; is odd, we won't be able to place all the odd index vegetables.&lt;/p&gt;&#xA;&lt;p&gt;For the case &lt;span class=&quot;math-container&quot;&gt;$n=4k+2\equiv 2(\mod 4)$&lt;/span&gt;, it's quite analogous to previous case, again resulting in &lt;span class=&quot;math-container&quot;&gt;$2k+1$&lt;/span&gt; odd-indexed positions for the remaining &lt;span class=&quot;math-container&quot;&gt;$2k+1$&lt;/span&gt; odd index vegetables, each requiring two positions with the same parity.&lt;/p&gt;&#xA;&lt;p&gt;So it's impossible to have a kebab when &lt;span class=&quot;math-container&quot;&gt;$n\equiv 1,2\pmod 4$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;h3&gt;Prove that there is a kebab for &lt;span class=&quot;math-container&quot;&gt;$n\equiv -1,0\pmod 4$&lt;/span&gt;&lt;/h3&gt;&#xA;&lt;p&gt;&lt;i&gt;I intended to give a constructive proof here, but it seems difficult.&lt;/i&gt;&lt;/p&gt;&#xA;" OwnerUserId="1649" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2018-09-10T21:08:46.387" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="2065" PostTypeId="1" AcceptedAnswerId="2068" CreationDate="2014-08-12T08:19:54.780" Score="6" ViewCount="1821" Body="&lt;p&gt;Put the numbers from 1 to 6 in the puzzle in a way such that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;In each row or column there shouldn't are no duplicate numbers.&lt;/li&gt;&#xA;&lt;li&gt;The squares that are joined together have the same number.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GDxgl.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2029" LastEditorUserId="32642" LastEditDate="2017-06-02T06:36:43.720" LastActivityDate="2017-06-02T06:36:43.720" Title="Put numbers from 1 to 6 in the puzzle" Tags="|logical-deduction|logic-grid|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2066" PostTypeId="1" AcceptedAnswerId="2067" CreationDate="2014-08-12T08:39:17.770" Score="55" ViewCount="15900" Body="&lt;p&gt;In the below image, can you draw 6 &lt;strong&gt;straight&lt;/strong&gt; lines that pass all the circles?&lt;br&gt;&#xA;As soon as you start drawing lines you can't take your pen up until you draw all six lines.&lt;br&gt;&#xA;&lt;strong&gt;hint:&lt;/strong&gt; you don't have to keep the polyline inside the square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&#xA;Best answer is the one in which straight line passes &lt;strong&gt;center&lt;/strong&gt; of the circles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eFRcz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2029" LastEditorUserId="5373" LastEditDate="2015-04-28T09:24:18.187" LastActivityDate="2018-09-07T02:23:02.527" Title="Join all circles together only with 6 lines" Tags="|geometry|" AnswerCount="16" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2067" PostTypeId="2" ParentId="2066" CreationDate="2014-08-12T11:21:47.023" Score="52" Body="&lt;p&gt;I think this is one possible solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/Bglae.jpg&quot; alt=&quot;solution&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2033" LastEditorUserId="46903" LastEditDate="2018-09-07T00:53:39.173" LastActivityDate="2018-09-07T00:53:39.173" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="2068" PostTypeId="2" ParentId="2065" CreationDate="2014-08-12T12:28:08.190" Score="11" Body="&lt;p&gt;I found one solution and i dont think there is another. The red marked four is the first number you can fill in. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xcvCR.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2033" LastEditorUserId="2033" LastEditDate="2014-08-12T12:43:04.820" LastActivityDate="2014-08-12T12:43:04.820" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2069" PostTypeId="2" ParentId="2066" CreationDate="2014-08-12T14:16:11.907" Score="75" Body="&lt;p&gt;Here's an option that uses only 4 lines. You can extend the concept to place another 2 lines if you really want 6 ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9pBt9.png&quot; alt=&quot;solution using 4 lines&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2036" LastEditorUserId="46903" LastEditDate="2018-09-07T00:53:54.303" LastActivityDate="2018-09-07T00:53:54.303" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="2070" PostTypeId="2" ParentId="2066" CreationDate="2014-08-12T15:13:26.847" Score="73" Body="&lt;p&gt;By mapping the puzzle onto a cylindrical topography, I've solved the puzzle using only a single straight line.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/nxZYP.jpg&quot; alt=&quot;solution pic 1&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vv5xq.jpg&quot; alt=&quot;solution pic 2&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2037" LastEditorUserId="46903" LastEditDate="2018-09-07T00:52:55.623" LastActivityDate="2018-09-07T00:52:55.623" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="2072" PostTypeId="2" ParentId="2066" CreationDate="2014-08-12T17:17:51.870" Score="18" Body="&lt;p&gt;This one doesn't solve it the way you're supposed to, but also doesn't break any of the implied &quot;rules&quot;, but kind of defeats the purpose of the puzzle (which is go outside of the square). Were you to make the circles smaller, it would eliminate this answer from plausibility. (Also the (3,3) circle line is a little dubious, if I had taken some more time to perfect this it would fit inside the circle better, but also shows how this solution could be easily invalidated by making the circles smaller)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/gC2Lm.png&quot; alt=&quot;solution&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="319" LastEditorUserId="40703" LastEditDate="2018-09-07T01:17:05.050" LastActivityDate="2018-09-07T01:17:05.050" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="2073" PostTypeId="2" ParentId="2066" CreationDate="2014-08-12T18:49:09.347" Score="94" Body="&lt;p&gt;If you use just one massive line, you can make it pass through the center of all of the circles!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/4csuT.jpg&quot; alt=&quot;solution&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2045" LastEditorUserId="46903" LastEditDate="2018-09-07T00:52:34.040" LastActivityDate="2018-09-07T00:52:34.040" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="2074" PostTypeId="2" ParentId="2053" CreationDate="2014-08-12T23:34:38.577" Score="5" Body="&lt;p&gt;I can only get 3 deep with each of 6,9, and 12 blocks. The 12 block solution is 'almost' 4 deep but not quite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qXaSY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding KSab's answer - I can confirm for n=13 I can get his minimum of 4 deep.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/gRHqQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...and looking a little further I can get 7 layers for n=39. Apart from the 2nd layer, the overlaps are fairly 'efficient'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZC5sc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="1631" LastEditDate="2014-08-13T03:49:28.710" LastActivityDate="2014-08-13T03:49:28.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2075" PostTypeId="2" ParentId="2066" CreationDate="2014-08-12T23:44:34.090" Score="31" Body="&lt;ol&gt;&#xA;&lt;li&gt;Print the picture.&lt;/li&gt;&#xA;&lt;li&gt;Fold the image such that there is a single stack of circles.  This will take 6 folds.&lt;/li&gt;&#xA;&lt;li&gt;Draw one line across the middle of it.&lt;/li&gt;&#xA;&lt;li&gt;Profit.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/HPRSJ.png&quot; alt=&quot;solution&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2050" LastEditorUserId="46903" LastEditDate="2018-09-07T01:36:51.583" LastActivityDate="2018-09-07T01:36:51.583" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="2076" PostTypeId="2" ParentId="2053" CreationDate="2014-08-13T00:59:19.233" Score="1" Body="&lt;p&gt;I thought I might as well mention what results I have gotten from just a little thinking, sketching, and coding. First a very obvious upper bound for the number of layers is $\frac{n+1}{2}$ because every layer up to the second to highest must have at least two tiles in it. I am sure this can be significantly improved but its not trivial, as you can place two tiles on another two (for example a 1x6 and a 5x2 on a 4x3 and a 2x5). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for a lower bound, I considered the case in which you start with 1xn on the highest level, and simply vertically tile the rest of the tiles in order (moving to a lower level when able). For example for n=8, you have the 1x8 on the top, the 2x7 and 3x6 (giving 4 extra spaces of verticle room), and the 4x5, 5x4, 6x3, 7x2 (and optionally the 8x1) on the bottom row. Note how since they are done in order, horizontal space is a non-issue. Though it isn't hard to calculate, I don't know of anyway to turn this into a formula and from looking through some values it seems relatively difficult to predict.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case this is what we get with these bounds:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;n&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3: 1 &amp;lt;= L &amp;lt;= 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4: 2 &amp;lt;= L &amp;lt;= 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5: 2 &amp;lt;= L &amp;lt;= 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6: 2 &amp;lt;= L &amp;lt;= 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7: 2 &amp;lt;= L &amp;lt;= 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8: 3 &amp;lt;= L &amp;lt;= 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;9: 3 &amp;lt;= L &amp;lt;= 5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;10: 3 &amp;lt;= L &amp;lt;= 5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;11: 3 &amp;lt;= L &amp;lt;= 6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;12: 3 &amp;lt;= L &amp;lt;= 6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;13: 4 &amp;lt;= L &amp;lt;= 7&lt;/p&gt;&#xA;&#xA;&lt;p&gt;14: 4 &amp;lt;= L &amp;lt;= 7&lt;/p&gt;&#xA;&#xA;&lt;p&gt;15: 4 &amp;lt;= L &amp;lt;= 8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I will try lowering the upper bound, as it seems like it should be possible to do a fair bit more than the very simple formula I have now.&lt;/p&gt;&#xA;" OwnerUserId="150" LastEditorUserId="150" LastEditDate="2014-08-13T03:23:48.553" LastActivityDate="2014-08-13T03:23:48.553" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2077" PostTypeId="2" ParentId="2066" CreationDate="2014-08-13T06:24:17.317" Score="14" Body="&lt;p&gt;Why not this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This one also does what he wants.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/qZltr.jpg&quot; alt=&quot;solution&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="226" LastEditorUserId="46903" LastEditDate="2018-09-07T01:23:07.340" LastActivityDate="2018-09-07T01:23:07.340" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="2078" PostTypeId="1" AcceptedAnswerId="2084" CreationDate="2014-08-13T07:32:38.500" Score="6" ViewCount="1288" Body="&lt;p&gt;Find the last pattern in the picture and state your reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;source: zoomit.ir&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZGu9h.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2029" LastEditorUserId="2029" LastEditDate="2014-09-03T09:12:28.510" LastActivityDate="2015-06-20T15:55:52.823" Title="Guess the last pattern in the image" Tags="|pattern|" AnswerCount="3" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="2079" PostTypeId="1" AcceptedAnswerId="2082" CreationDate="2014-08-13T07:38:59.307" Score="6" ViewCount="3149" Body="&lt;p&gt;How many sqares can you draw in given pattern, when each corner of a sqare has to lie in the exact center of one of the 20 colored circles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZxzBH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2029" LastEditorUserId="2029" LastEditDate="2014-09-03T09:11:32.190" LastActivityDate="2014-09-03T09:11:32.190" Title="How many squares can you find in the picture?" Tags="|pattern|pencil-and-paper-games|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2080" PostTypeId="2" ParentId="2066" CreationDate="2014-08-13T07:48:11.290" Score="11" Body="&lt;p&gt;my own answer to this question is&lt;br&gt;&#xA;in this picture you can see 6 lines throw the center of all circles without taking your pen up.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/Xy1o1.gif&quot; alt=&quot;solution&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2029" LastEditorUserId="46903" LastEditDate="2018-09-07T01:38:00.417" LastActivityDate="2018-09-07T01:38:00.417" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="2081" PostTypeId="2" ParentId="2079" CreationDate="2014-08-13T13:11:12.003" Score="14" Body="&lt;p&gt;9 squares that are $1$x$1$     (trivial)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 squares that are $\sqrt{2}$x$\sqrt{2}$ (Diamonds centered at the 4 central circles)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 squares that are $\sqrt{5}$x$\sqrt{5}$ (Each starts at a yellow circle. Move like a knight: example 2 right 1 up, 2 up 1 left, 2 left 1 down, 2 down 1 right)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 squares that are $\sqrt{13}$x$\sqrt{13}$ (Each starts at the orange circles. Move 3 right 2 up or 3 left 2 up and then keep going)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there more than that?&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditDate="2014-08-13T13:23:06.437" LastActivityDate="2014-08-13T13:23:06.437" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2082" PostTypeId="2" ParentId="2079" CreationDate="2014-08-13T13:11:40.033" Score="32" Body="&lt;p&gt;There are an infinite number of squares whose sides pass through exactly four circles. In fact, there are an infinite number of squares passing through the top four circles alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OuNv5.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each frame of the above gif represents one possible square you can draw through those circles. In addition to the 90 squares represented in these 90 frames, there are infinitely many more. For example, &quot;the square that lies halfway between frame 1 and frame 2&quot;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: If we add the additional constraint that the &lt;em&gt;corners&lt;/em&gt; of the square must lie on a circle, then there are 17, as originally determined by kaine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/93OO5.gif&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/NMSI6.gif&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/wVNOQ.gif&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/2TpSE.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1497" LastEditorUserId="1497" LastEditDate="2014-08-13T14:44:38.700" LastActivityDate="2014-08-13T14:44:38.700" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2083" PostTypeId="2" ParentId="2078" CreationDate="2014-08-13T13:30:51.343" Score="3" Body="&lt;p&gt;Pattern puzzles like this aren't the best because there can be so many answers.  For example, I say that none of the supplied choices are the next item.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For one thing, the color pattern is &lt;em&gt;clearly&lt;/em&gt; going to repeat red, yellow, blue, green, so the next square should be red, but all the choices are yellow!&#xA;Also, the squares in the odd positions all should have just one line, with one endpoint at the center of the square.  Again, none of the choices match this!  The single line is just a simple rotation of 45 degrees clockwise, so the next square should be a red square with a line from the center to the bottom right corner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I can think of two different options for the square after that.  Keeping with the color pattern, it will be yellow.  One way to go from the second square to the fourth is to rotate the line that is furthest clockwise 135 degrees counterclockwise.  If that's the case, the 6th square will have two lines from the center of the square to the two top corners.&#xA;The other way could be to rotate the furthest clockwise line 90 degrees counterclockwise, and the other line only 45 degrees counterclockwise.  If that's the case, the 6th square will have just one line from the center to the top of the square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tGP33.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="239" LastActivityDate="2014-08-13T13:30:51.343" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2084" PostTypeId="2" ParentId="2078" CreationDate="2014-08-13T13:36:15.887" Score="18" Body="&lt;p&gt;It is the 5th one&#xA;All we have here is the numbers written in the right side of the square with the lines overlapping the box &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ej2tF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1639" LastEditorUserId="1639" LastEditDate="2014-08-13T13:41:35.737" LastActivityDate="2014-08-13T13:41:35.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2086" PostTypeId="2" ParentId="2066" CreationDate="2014-08-13T15:49:46.130" Score="36" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/kZu9z.jpg&quot; alt=&quot;solution&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's another way of thinking. This is using edges instead.&lt;/p&gt;&#xA;" OwnerUserId="2067" LastEditorUserId="46903" LastEditDate="2018-09-07T00:51:52.053" LastActivityDate="2018-09-07T00:51:52.053" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="2087" PostTypeId="2" ParentId="2066" CreationDate="2014-08-13T18:11:07.457" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/bC9WH.jpg&quot; alt=&quot;solution&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This could be the answer as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6 lines without lifting the pen.&lt;/p&gt;&#xA;" OwnerUserId="2072" LastEditorUserId="46903" LastEditDate="2018-09-07T02:22:45.257" LastActivityDate="2018-09-07T02:22:45.257" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="2088" PostTypeId="2" ParentId="420" CreationDate="2014-08-14T15:12:01.103" Score="1" Body="&lt;p&gt;My best suggestion is to take a pciture of each face before you start. I did this when I attempted to solve this a few months back and it made the proccess a lot easier I think. The puzzle was still difficult but since every piece is unique I was able to tell which other pieces were on the same face as the piece I was actually looking at/working on. &lt;/p&gt;&#xA;" OwnerUserId="1988" LastActivityDate="2014-08-14T15:12:01.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2089" PostTypeId="2" ParentId="2066" CreationDate="2014-08-14T19:13:10.837" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/OtTQ4.jpg&quot; alt=&quot;another possible answer!&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;here is my answer!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;nice question! &lt;/p&gt;&#xA;" OwnerUserId="2100" LastEditorUserId="46903" LastEditDate="2018-09-07T02:22:57.477" LastActivityDate="2018-09-07T02:22:57.477" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="2090" PostTypeId="2" ParentId="2066" CreationDate="2014-08-14T23:50:08.300" Score="2" Body="&lt;p&gt;Looking through the answers, there are several similar, but none the same as this one. So here's one more possible answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/fVbVK.png&quot; alt=&quot;Possible answer/solution to the puzzle.&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2105" LastEditorUserId="46903" LastEditDate="2018-09-07T02:22:52.767" LastActivityDate="2018-09-07T02:22:52.767" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="2091" PostTypeId="5" CreationDate="2014-08-15T08:49:26.040" Score="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Martin_Gardner&quot; rel=&quot;nofollow&quot;&gt;Martin Gardner&lt;/a&gt; (October 21, 1914 – May 22, 2010) was an American writer of popular science. &#xA;Gardner was best known for creating and sustaining general interest in recreational mathematics for a large part of the 20th century, principally through his Scientific American &quot;Mathematical Games&quot; columns from 1956 to 1981 and through subsequent books collecting them. &#xA;He was an uncompromising critic of fringe science and was a founding member of CSICOP, an organization devoted to debunking pseudoscience.&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="8874" LastEditDate="2015-02-04T15:33:31.557" LastActivityDate="2015-02-04T15:33:31.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2092" PostTypeId="4" CreationDate="2014-08-15T08:49:26.040" Score="0" Body="Martin Gardner (1914-2010) was an American writer of recreational mathematics and as such created a number of mathematical and logical puzzles." OwnerUserId="68" LastEditorUserId="8874" LastEditDate="2015-02-04T15:33:17.863" LastActivityDate="2015-02-04T15:33:17.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2093" PostTypeId="2" ParentId="236" CreationDate="2014-08-15T15:43:17.983" Score="17" Body="&lt;p&gt;The colour of the guru's eyes is not relevant. &lt;strong&gt;The guru is allowed to speak about eyes and nobody else is.&lt;/strong&gt; If any blue eyed person said &quot;I can see someone with blue eyes&quot; where everyone on the island could hear it, the same thing would happen. Also if any brown eyed person did. The moment a blue-eyed person hears that &lt;strong&gt;someone else&lt;/strong&gt; can see some blue eyes, and those blue-eyed people know it, the clock starts ticking. Once I hear that and I see N blue eyed people, if they haven't left after N days it's because they are including me in their count of N. Therefore I have to leave on day N+1. It even works if they wake up one morning and find &quot;at least one person has blue eyes&quot; scrawled on the mirror in lipstick, except for them not having any mirrors.&lt;/p&gt;&#xA;" OwnerUserId="1509" LastEditorUserId="1509" LastEditDate="2014-11-24T12:51:01.787" LastActivityDate="2014-11-24T12:51:01.787" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2095" PostTypeId="2" ParentId="236" CreationDate="2014-08-16T03:27:17.017" Score="3" Body="&lt;p&gt;Not sure if this is the right answer but my wife and I thought everyone will leave the island the 201st day and here's why:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We assumed the Guru would either say &quot; I see a blue eyed person&quot; or &quot; I see a brown eyed person&quot; each day (alternating or randomly, doesnt matter). Since she's a logician too, she would accurately total up the number of brown and blue eyes on day#200. Let's say a person x has brown eyes, she will realize by day # 200 what her eye color is as she knows by now that there are 100 blue colored eyes and 99 brown eyed people. This logic will apply to every member too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very interested to see what the geniuses on this forum have to say!&lt;/p&gt;&#xA;" OwnerUserId="2124" LastActivityDate="2014-08-16T03:27:17.017" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2096" PostTypeId="1" AcceptedAnswerId="2097" CreationDate="2014-08-17T06:26:18.443" Score="8" ViewCount="33291" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bdqS2.jpg&quot; alt=&quot;Fly in a glass&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The object is to move 2 matches to get the bluebottle to fly outside the glass. At the finish, the glass may be turned in any direction, but it must be exactly the &lt;strong&gt;same shape as before&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2132" LastEditorUserId="11876" LastEditDate="2016-08-03T22:49:45.620" LastActivityDate="2024-01-17T14:17:53.600" Title="Out of Glass Puzzle (MatchStick Puzzle)" Tags="|matches|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2097" PostTypeId="2" ParentId="2096" CreationDate="2014-08-17T07:31:37.617" Score="6" Body="&lt;p&gt;Just move the horizontal stick to the left by half width. And right vertical stick to the bottom left to get the glass turned upside-down.&lt;/p&gt;&#xA;" OwnerUserId="28" LastActivityDate="2014-08-17T07:31:37.617" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2098" PostTypeId="2" ParentId="2030" CreationDate="2014-08-17T09:26:14.877" Score="1" Body="&lt;p&gt;Another approach for A is: Flip both the 4 and the 5 and start timing.  When the 4 runs out flip it.  When the 5 runs out flip &lt;strong&gt;both&lt;/strong&gt;, the 4 has now 1 minute left.  When the 4 runs out flip both again, now the 5 has 1 minute left.  When the 5 runs out, you have 7 minutes.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastActivityDate="2014-08-17T09:26:14.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2099" PostTypeId="2" ParentId="1957" CreationDate="2014-08-17T11:57:42.107" Score="4" Body="&lt;p&gt;Another approach is to consider the ways to obtain 15 as a sum of 3 different numbers in the range 1..9.&lt;/p&gt;&#xA;&lt;p&gt;These are: 1+5+9, 1+6+8, 2+4+9, 2+5+8, 2+6+7, 3+4+8, 3+5+7, 4+5+6.&lt;/p&gt;&#xA;&lt;p&gt;As you can see there are only 8 ways, and you need 8 different sums in your square.  So each sum appears exactly once as a line in your square.&lt;/p&gt;&#xA;&lt;p&gt;5 must be in the center because it appears 4 times.  Only the center belongs to 4 lines.  2,4,6,8 are corners because they appear 3 times each.  1,3,7,9 are borders because they appear only twice.&lt;/p&gt;&#xA;&lt;p&gt;With 5 in the center 2 and 8 must be in opposite corners, so are 4 and 6.  After that the odd number fit in a single place each.&lt;/p&gt;&#xA;&lt;p&gt;So the solution is unique except for rotations and reflections.  Which variation you get depends on how you arrange the corners.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2021-01-01T00:33:55.483" LastActivityDate="2021-01-01T00:33:55.483" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="2100" PostTypeId="2" ParentId="2013" CreationDate="2014-08-17T12:20:55.263" Score="1" Body="&lt;p&gt;It looks like these represent partitions of a set of 5 items.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there should be 52 only according to &lt;a href=&quot;https://en.wikipedia.org/wiki/Partition_of_a_set&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Partition_of_a_set&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like #35 and #42 in the glyph summary are equivalent.  And #54 doesn't fit with the pattern.  Or it is a funny variation of #53.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastActivityDate="2014-08-17T12:20:55.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2101" PostTypeId="1" AcceptedAnswerId="2113" CreationDate="2014-08-17T17:05:02.483" Score="7" ViewCount="6478" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;You have three cakes of diameter 20cm, 16cm and 12cm respectively as shown in the &#xA;  figure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TEWVx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They all have the same height. Now your task is to divide these 3 cakes into 4 equal parts and you are only allowed to cut one cake at a time.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;What is the minimum number of cuts required?&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The constraints are that you have to cut one cake at a time and the only tool you have at your disposal is a knife.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Hint&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can displace the cakes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="101" LastEditorUserId="65277" LastEditDate="2024-01-09T23:24:09.447" LastActivityDate="2024-01-09T23:24:09.447" Title="Divide 3 cakes into 4 equal parts?" Tags="|mathematics|calculation-puzzle|optimization|dissection|" AnswerCount="8" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="2102" PostTypeId="2" ParentId="2101" CreationDate="2014-08-17T17:32:01.020" Score="2" Body="&lt;p&gt;Not sure if this is the minimum required, but you can do it in 6 cuts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cut the first cake into quarters - two cuts.&#xA;Cut the second cake into quarters - two cuts.&#xA;Cut the third cake into quarters - two cuts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternate solution (still 6 cuts):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total volume of the cakes is $(10^2 + 8^2 + 6^2)*\pi*h = 200 \pi h$. So $4$ equal pieces will have volume of $50 \pi h$ (you can consider just surface area if you like, since $h$ is the same for all). We can get two of those just by cutting the $20$ diameter cake in half ($1$ cut). Next we'd need to add $14 \pi h$ to the $12$ diameter cake to get another $50 \pi h$ group. $14$ out of $64$ is $\frac 7{32}$, which would require $5$ cuts across the middle of the $16$ diameter cake (to cut it into $32$nds) - yielding $6$ total cuts also.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the two cut solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Cut the $20$ cm cake in half. That gives you two pieces of $50*\pi*h$ volume. The other two cakes are a total of $100*\pi*h$ in volume.&#xA;2) Place the $12$ cm cake on top of and tangent to the $16$ cm cake. Start cutting the $16$ cm cake along the edge of the $12$ cm cake. When you reach what would be the end of the diameter for the $12$ cm cake, go perpendicularly out to the edge of the $16$ cm cake. Place the smaller piece of the $16$ cm cake with the $12$ cm cake and you'll have two pieces that are mirror images of each other. Each will be half of $100*\pi*h$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Inspiration taken from hkboy's answer - see visual there, but reverse his cut to go from the tangent point out.)&lt;/p&gt;&#xA;" OwnerUserId="1620" LastEditorUserId="1620" LastEditDate="2014-08-18T17:07:49.903" LastActivityDate="2014-08-18T17:07:49.903" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2103" PostTypeId="2" ParentId="2101" CreationDate="2014-08-17T18:28:46.737" Score="4" Body="&lt;p&gt;&lt;strong&gt;Just cut all of them in half&lt;/strong&gt; and combine halves of the cake-16 and cake-12 into two parts.&lt;br&gt;&#xA;This would require 3 cuts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Duncan said already full volume is $(20^2 + 16^2 + 12^2)*\pi*h = 800 \pi h$. $20^2*\pi*h/2 = 200 \pi h$ will give you two 1/4th parts. And $16^2*\pi*h/2 + 12^2*\pi*h/2= 200 \pi h$ will give you another two 1/4th parts.&lt;/p&gt;&#xA;" OwnerUserId="28" LastActivityDate="2014-08-17T18:28:46.737" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2104" PostTypeId="2" ParentId="2101" CreationDate="2014-08-17T18:29:25.957" Score="4" Body="&lt;p&gt;You can do it in 3 cuts, simply cut each cake in half. Each half of the biggest is a portion, and combining a half from both of the smaller two makes a portion (of which you have 2). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are $\pi\cdot20^2\cdot h+\pi\cdot16^2\cdot h+\pi\cdot12^2\cdot h = \pi\cdot800\cdot h\space cm^2$ of cake, so each part needs to be $200\cdot\pi\cdot h\space cm^2$. The first two slices are easy, cut the first cake in half. The whole is $20^2\cdot\pi\cdot h\space cm^2$, so each half is the $200\cdot\pi\cdot h\space cm^2$ we need. Now you're left with two differently sized cakes, which we need to partition into two parts. Conveniently, you can simply cut each of those in half, and combine a half from each for $\frac{\pi\cdot16^2\cdot h}{2}+\frac{\pi\cdot12^2\cdot h}{2} = 200\cdot\pi\cdot h\space$, the quantity we need.&lt;/p&gt;&#xA;" OwnerUserId="4" LastActivityDate="2014-08-17T18:29:25.957" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2105" PostTypeId="2" ParentId="2101" CreationDate="2014-08-17T18:52:44.460" Score="10" Body="&lt;p&gt;2 cuts&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area of each cake is $100 \pi, 64 \pi, 36 \pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Half of the big cake is $50 \pi$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;cut a 4 cm segment from the 16 cm cake by putting the 12 cm cake at one edge and cutting the other end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using the formula for the area of a segment $Area = 8^2 cos^{-1} \frac{(8-4)}{2} - (8 - 4) \sqrt{2*8*4-4^2}  = 12.5 \pi $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding it to the 12 cm cake would give a combined area of $48.5\pi$ and the remainder of the 16 cm cake would be $51.5\pi$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is an image of the 12cm cake on top of the 16 cm cake and the blue line is where I would cut the cake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/U5Nn2.png&quot; alt=&quot;picture&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1508" LastEditorUserId="1508" LastEditDate="2014-08-18T09:38:30.827" LastActivityDate="2014-08-18T09:38:30.827" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="2107" PostTypeId="1" AcceptedAnswerId="3067" CreationDate="2014-08-17T21:57:43.277" Score="3" ViewCount="388" Body="&lt;p&gt;An S-tileset is a collection of n oriented tiles, where no two tiles have the same size, each tile is one unit thick, and its non-zero-integer length and width add up to n+1. (So, an S-tileset has n tiles, of sizes 1×n, 2×n−1, ...n×1.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conjecture:&lt;/strong&gt; &lt;em&gt;No two disjoint subsets of any S-tileset can be tiled to form two areas of identical size and shape.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you either prove the conjecture or find a counter-example.&lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="1631" LastEditDate="2014-08-17T23:36:24.827" LastActivityDate="2014-10-20T19:34:42.843" Title="Tower of tiles revisited" Tags="|geometry|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="2108" PostTypeId="2" ParentId="2101" CreationDate="2014-08-18T05:29:03.593" Score="5" Body="&lt;ol&gt;&#xA;&lt;li&gt;Cut the 20cm cake straight in half to get two $50 \pi h$ volumes. &#xA;Keep the halves together.&lt;/li&gt;&#xA;&lt;li&gt;Place the 16cm cake flat on top of the cut 20cm cake so that their&#xA;circle bases are tangent at one end of the cut, called Point A.&lt;/li&gt;&#xA;&lt;li&gt;Place the 12cm cake flat on top of the 16cm cake so that their&#xA;circle bases are tangent at a Point B directly above Point A.&lt;/li&gt;&#xA;&lt;li&gt;Find Point C on the 16cm cake diametrically opposite to Point B&#xA;using the first cut.  Cut the 16cm cake starting at Point C,&#xA;following the first cut.  When you get to the edge of the 12cm cake,&#xA;turn and cut following the circumference of the 12cm cake until you&#xA;get to Point B, creating a fifth volume of cake.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You now have three $50 \pi h$ pieces, one $36 \pi h$ piece, and one $14 \pi h$ piece.  You made two cuts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference between the 16cm and 12cm cake is $64 \pi h - 36 \pi h = 28 \pi h$, and the method described will allow you to split the difference.&lt;/p&gt;&#xA;" OwnerUserId="1923" LastActivityDate="2014-08-18T05:29:03.593" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2109" PostTypeId="2" ParentId="2101" CreationDate="2014-08-18T11:34:54.903" Score="6" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8hBJR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about put the 12cm cake on top of the 16cm cake and make a cut like this?&lt;/p&gt;&#xA;" OwnerUserId="2147" LastActivityDate="2014-08-18T11:34:54.903" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2110" PostTypeId="1" AcceptedAnswerId="2111" CreationDate="2014-08-18T13:41:24.710" Score="5" ViewCount="2882" Body="&lt;p&gt;Consider a line segment of length &lt;em&gt;L&lt;/em&gt;. This segment is cut in three segments at arbitrary points. What is the probability that these three segments could be rearranged as sides of a triangle?&lt;img src=&quot;https://i.stack.imgur.com/9Sm1x.png&quot; alt=&quot;Example&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2042" LastActivityDate="2014-08-19T16:22:04.790" Title="Probability of three segments forming a triangle" Tags="|geometry|probability|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2111" PostTypeId="2" ParentId="2110" CreationDate="2014-08-18T15:16:45.280" Score="7" Body="&lt;p&gt;It is possible to do so with any division where none of the three sections have a length that's over half the total.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) first and second cut cannot be different by over half the maximum length&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JpjGX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the graph I colored the 'bad' areas: where the first is over half the maximum bigger then the second, and where the second is over half the maximum smaller then the first. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simple math would show that's 1/4 that doesn't qualify&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) the first two cuts cuts cannot be both in the first half, nor can they be both in the second half.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This eliminates options colored in blue&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/euzS8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining those, I think I end up with 1/4 odds for decent solutions. Putting this up now, I'll rethink it after.&lt;/p&gt;&#xA;" OwnerUserId="2156" LastEditorUserId="2156" LastEditDate="2014-08-19T06:01:58.237" LastActivityDate="2014-08-19T06:01:58.237" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2112" PostTypeId="2" ParentId="2066" CreationDate="2014-08-18T17:10:37.030" Score="14" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/6pusk.jpg&quot; alt=&quot;solution&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Similar but different answer.&#xA;Great puzzle!&lt;/p&gt;&#xA;" OwnerUserId="2159" LastEditorUserId="46903" LastEditDate="2018-09-07T01:36:55.877" LastActivityDate="2018-09-07T01:36:55.877" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="2113" PostTypeId="2" ParentId="2101" CreationDate="2014-08-18T19:57:39.740" Score="17" Body="&lt;p&gt;Two cuts (or four, depending on how you count a cut):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cut the 20cm cake in half, giving you two pieces that are each one-fourth of the total cake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then stack the 12cm on top of the 16cm cake and cut along the 16cm's diameter to the 12cm, then around the 12cm (like a stencil), to the other side of the 16cm's diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I.E., cut along the red line:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/O19XW.png&quot; alt=&quot;Cake Cut Diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formerly 16cm cake will have a 12cm core plus half the annular region, and the 12cm cake plus the other half of the annular region will be of equal size. These two, along with the two halfs of the 20cm cake, form four equal quantities.&lt;/p&gt;&#xA;" OwnerUserId="2101" LastActivityDate="2014-08-18T19:57:39.740" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2114" PostTypeId="1" AcceptedAnswerId="2115" CreationDate="2014-08-18T21:41:40.770" Score="5" ViewCount="1297" Body="&lt;p&gt;If I have a table of binary values, and tell you the column and row counts for &quot;on&quot; or &quot;1&quot; values, is it possible to solve, with certainty, any grid of size n*m?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the following puzzle for example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   | 4 | 5 | 3 | 2 | 1&#xA;---|---|---|---|---|---&#xA; 3 | ? | ? | ? | ? | ?&#xA; 2 | ? | ? | ? | ? | ?&#xA; 3 | ? | ? | ? | ? | ?&#xA; 3 | ? | ? | ? | ? | ?&#xA; 4 | ? | ? | ? | ? | ?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;1 possible solution is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   | 4 | 5 | 3 | 2 | 1&#xA;---|---|---|---|---|---&#xA; 3 | 1 | 1 | 1 | 0 | 0&#xA; 2 | 0 | 1 | 0 | 1 | 0&#xA; 3 | 1 | 1 | 0 | 1 | 0&#xA; 3 | 1 | 1 | 1 | 0 | 0&#xA; 4 | 1 | 1 | 1 | 0 | 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The only problem is that there could be (and likely are) more than 1 solution. What extra information can I give you so that you can solve, with certainty, any sized board?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some examples of extra info are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Diagonal counts&lt;/li&gt;&#xA;&lt;li&gt;1 count for even indices per value in a row or collumn, and 1 count for odd indices&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2161" LastEditorUserId="15544" LastEditDate="2015-10-02T19:05:08.110" LastActivityDate="2015-10-02T21:38:47.300" Title="Theoretical Puzzle: Making a binary puzzle with a unique solution" Tags="|mathematics|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2115" PostTypeId="2" ParentId="2114" CreationDate="2014-08-18T23:42:22.967" Score="5" Body="&lt;p&gt;Clearly the column and row counts alone isn't enough. Consider a 2x(2n) array (2 rows of 2xn columns) as described below:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Using A = |1|, B = |0|, build the array by randomly ordering n of each of A and B &#xA;          |0|      |1|                                          &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Each possible array will have identical column and row counts:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   | 1 | 1 | 1 | ... | 1 | 1 |&#xA;---|---|---|---| ... |---|---|&#xA; n | 0 | 1 | 1 | ... | 0 | 1 |&#xA; n | 1 | 0 | 0 | ... | 1 | 0 |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But there are $\binom{2n}{n} = \frac{(2n)!}{(n!)^2}$ different such arrays. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This function grows fairly fast (first few values are 1, 2, 6, 20, 70, 252, 924, 3432, 12870, 48620, 184756, 705432, 2704156, 10400600, 40116600, 155117520, 601080390, 2333606220, 9075135300, 35345263800, 137846528820, 538257874440, 2104098963720,...) so you would need to add a lot more information to uniquely identify any such array. If you add more rows the problem only gets worse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In answer to KevinOrr's question, all diagonals will be sufficient to solve for the 2x(2n) array as described above. But I don't think they are enough for arrays with more rows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To confirm that. Just consider $m \times m$ arrays with half of the elements set. So there will be $(m^2)/2$ set elements. The row count will consist of m integers that add to $(m^2)/2$, as will the column count and each of the two sets of TL-BR and TR-BL diagonal counts will be $2m$ elements with the same sum. Each of these counts is a partition of the number $(m^2)/2$, for example see &lt;a href=&quot;http://en.wikipedia.org/wiki/Partition_(number_theory)&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Partition_(number_theory)&lt;/a&gt;, broken into m parts for the row and column or 2m parts for the diagonal counts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of possible partitions of n elements grows as $\sim\exp[c \sqrt n]$ (for some constant c) and this is strictly greater than the number of partitions into some fixed k parts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the number of possible row,column plus diagonal partitions for the mxm problem (with $(m^2)/2$ elements set) is less than $(\exp[c\times\sqrt{(m^2)/2}])^4$ . i.e. it grows approximately as $exp(m)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the number of possible arrays of that form grows as $\sim\binom{m^2}{(m^2)/2}$ which, if you 'solve' using Sterling's approximation, seems to grow approximately as $2^{m^2}$ or $\exp[m^2]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So possible arrays grows as $\sim\exp(m^2)$, which is much faster than the possible combinations of row/col/diagonal numbers which only grows as $\sim\exp(m)$, and that won't be enough information to uniquely identify the array.&lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorDisplayName="user88" LastEditDate="2015-10-02T21:38:47.300" LastActivityDate="2015-10-02T21:38:47.300" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2117" PostTypeId="2" ParentId="2110" CreationDate="2014-08-19T16:22:04.790" Score="3" Body="&lt;p&gt;The algebraic way of looking at this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the first cut the probability (p) that the longest stick will be shorter than length $x$ is $p=2x-1$.  The expected length is $(1+0.5)/2=.75$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second cut has probability $1-x$ of being in the small half (loss) and $2(x-.5)$ of cutting the second half but leaving one stick at least $.5$ long. Adding these together means there is a probability $x$ that the second cut will leave a stick $.5$ or longer.  As the expected value for $x$ is $.75$, there is only a $.25$ chance that the two sticks can form a triangle.&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2014-08-19T16:22:04.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2119" PostTypeId="1" AcceptedAnswerId="2129" CreationDate="2014-08-20T07:41:39.387" Score="6" ViewCount="1378" Body="&lt;p&gt;It was late night when Markus was sitting at his desk and suddenly a fairy appeared out of nowhere. She told him that he had been granted a wish. Markus wished for her to tell him next week's lottery draw and handed her pen and paper. But the fairy told him that these were 6 wishes and that thus it was not possible for her to tell him. Instead she wrote one number on the paper and said: &amp;quot;This is the sum of all of next week's numbers in the lottery draw.&amp;quot; Markus then thought &lt;em&gt;There are too many possibilities for me to just guess the numbers.&lt;/em&gt; The fairy, who had sensed his thoughts then gave him another tip: &amp;quot;The number of possibilities that this sum can be achieved multiplied with it is a really big number, in the magnitude of a few million (number &amp;gt;= 10,000,000), and it is also equal to the product of the six numbers.&amp;quot; After she had vanished, Markus began to scribble and sure enough, won the lottery the next week.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;What are the six numbers?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The solution &lt;strong&gt;is unique&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;TL;DR&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Six distinct numbers between including 1 and 49 - for you maths people: [1,49]&lt;/li&gt;&#xA;&lt;li&gt;Sum of these times possibilities of summands is equal to product of these numbers and &amp;gt;= 10,000,000.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1943" LastEditorUserId="57742" LastEditDate="2022-10-08T08:33:34.753" LastActivityDate="2022-10-08T08:33:34.753" Title="Markus' lucky numbers" Tags="|mathematics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="2120" PostTypeId="1" CreationDate="2014-08-20T08:46:38.920" Score="-7" ViewCount="679" Body="&lt;p&gt;We are tasked with finding a password. Our only hint seems to be:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Was wir wissen, ist ein Ozean;&lt;/p&gt;&#xA;&lt;p&gt;was wir nicht wissen, ist ein Tropfen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;German for:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;What we know is an ocean, what we don't know, is a drop.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;What is the password?&lt;/p&gt;&#xA;&lt;p&gt;The site this is from can be found &lt;a href=&quot;http://www.ohjelmointiputka.net/postit/elhak-koe.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;English translations for Finnish words:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;taso = level&lt;/li&gt;&#xA;&lt;li&gt;Edellisen tason koodi = the code for the previous level&lt;/li&gt;&#xA;&lt;li&gt;Kirjoita salasana = write the password&lt;/li&gt;&#xA;&lt;li&gt;vastaus = answer&lt;/li&gt;&#xA;&lt;li&gt;Lähetä = submit.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Here are the first 4 passwords:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Level 1: salasana&lt;/li&gt;&#xA;&lt;li&gt;Level 2: aybabtu&lt;/li&gt;&#xA;&lt;li&gt;Level 3: 237812992&lt;/li&gt;&#xA;&lt;li&gt;Level 4: 62981123&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2173" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2023-04-10T10:51:42.580" Title="Password encryption" Tags="|password|" AnswerCount="5" CommentCount="10" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2122" PostTypeId="2" ParentId="2120" CreationDate="2014-08-20T11:11:38.137" Score="1" Body="&lt;p&gt;Usually the quote goes different: what we know is a drop, what we don't know is an ocean.&#xA;That seems to be a quote from Sir Isaac Newton.&lt;/p&gt;&#xA;&lt;p&gt;If anything, I'd guess the answer could be a variation on Newton (like notwen, ...)&lt;/p&gt;&#xA;" OwnerUserId="2156" LastEditorUserId="5373" LastEditDate="2023-04-10T10:51:42.580" LastActivityDate="2023-04-10T10:51:42.580" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="2123" PostTypeId="2" ParentId="2120" CreationDate="2014-08-20T12:10:48.427" Score="1" Body="&lt;p&gt;As Tim Couwelier mentioned in &lt;a href=&quot;https://puzzling.stackexchange.com/a/2122/2032&quot;&gt;his answer&lt;/a&gt;, the quote is the opposite of what Newton said. So the password could be a variation on the opposite of Newton. Anti-newton comes to mind, but that isn't a real concept. So I am guessing &lt;strong&gt;Antineutron&lt;/strong&gt;, since a good password hint does not lead directly to the password, but in a more vague manner.&lt;/p&gt;&#xA;" OwnerUserId="2032" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2014-08-20T12:10:48.427" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2124" PostTypeId="2" ParentId="2120" CreationDate="2014-08-20T13:52:00.820" Score="1" Body="&lt;p&gt;Caesar since a word from the first half is swapped with a word from the second half, or it may be Newton put through a Caesar cipher.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastEditorUserId="5373" LastEditDate="2023-04-10T10:51:32.743" LastActivityDate="2023-04-10T10:51:32.743" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="2125" PostTypeId="2" ParentId="2079" CreationDate="2014-08-20T16:40:59.230" Score="6" Body="&lt;p&gt;My answer is 21. It break down as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;9: Square from dots right next to each other&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4: Squares with 1 dot in the middle&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2: Squares with central 4 dots in the middle&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2: Squares with previous 2 squares in the middle&#xA;(The above can be more easily visualized from watching Kevin's answer)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4: From the 4 squares with 1 dot in the middle, extend out 1 more dot on each corner to make another square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total of 9 + 4 + 2 + 2 + 4 = 21.&lt;/p&gt;&#xA;" OwnerUserId="2175" LastActivityDate="2014-08-20T16:40:59.230" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2126" PostTypeId="2" ParentId="2119" CreationDate="2014-08-20T18:13:14.910" Score="0" Body="&lt;p&gt;I couldn't find a trick to solve this because I don't know a formula for &quot;possibilities of summands.&quot; So I wrote a program and tried to brute force a solution. There turn out to be many or I am mistaken...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's go step by step. Suppose the sum is 21 there is only one possible drawing: 1,2,3,4,5,6. The order of the numbers drawn doesn't matter, since the question is about finding the correct numbers to place the bet on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a sum of 22 the numbers are 1,2,3,4,5,7. Again a unique solution. For sum=23 it could be either 1,2,3,4,5,8 or 1,2,3,4,6,7. I let the computer calculate and came to this list (shortened)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;20 --&gt; P=0; &lt;br&gt;&#xA;21 --&gt; P=1;&lt;br&gt;&#xA;22 --&gt; 1; &#xA;23 --&gt; 2; &#xA;24 --&gt; 3;&#xA;25 --&gt; 5;&#xA;26 --&gt; 7;&#xA;27 --&gt; 11;&#xA;28  --&gt; 14;&#xA;29  --&gt; 20;&#xA;30  --&gt; 26;&#xA;31  --&gt; 35;&#xA;32  --&gt; 44;&#xA;33  --&gt; 58;&#xA;34  --&gt; 71;&#xA;35  --&gt; 90; ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For sum=35 there are already multiple solutions. Given the &quot;probability of summands&quot; above (P=90), possible solutions are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 2 3 4 5 45: 1*2*3*4*5*45 = 5400 = 90 * 60 = 90 * (1+2+3+4+5+45) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 2 3 4 10 12: 1*2*3*4*10*12 = 2880 = 90 * 32 = 90 * (1+2+3+4+10+12)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other examples: 1 2 3 5 6 17; 1 2 3 5 9 10; 1 2 4 5 7 9&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the wording of the puzzle, I'd have expected a unique solution. Is the wording of the puzzle correct?&lt;/p&gt;&#xA;" OwnerUserId="2042" LastActivityDate="2014-08-20T18:13:14.910" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2127" PostTypeId="2" ParentId="2114" CreationDate="2014-08-20T18:58:05.963" Score="0" Body="&lt;p&gt;A more general form of this puzzle is known as &lt;a href=&quot;https://en.wikipedia.org/wiki/Nonogram&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nonogram or Nemo-Nemo-logic&lt;/a&gt; where the consecutive number of 1's is known for each column and row. Even in this form, there are often ambiguities and one is encouraged show the uniqueness of the solution before making a new puzzle available in public.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vk2gy.gif&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;from: &lt;a href=&quot;http://japonskie.ru/pics/full/krolik_1.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://japonskie.ru/pics/full/krolik_1.gif&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="259" LastEditorUserId="11475" LastEditDate="2015-10-02T19:45:07.080" LastActivityDate="2015-10-02T19:45:07.080" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2128" PostTypeId="2" ParentId="1963" CreationDate="2014-08-20T19:49:04.363" Score="1" Body="&lt;p&gt;For bonus one...&#xA;((0!)+(0!)+(0!))!&lt;/p&gt;&#xA;" OwnerUserId="2177" LastActivityDate="2014-08-20T19:49:04.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2129" PostTypeId="2" ParentId="2119" CreationDate="2014-08-20T20:03:43.220" Score="7" Body="&lt;p&gt;I'd guess the answer is 130, based on output from the following python program.  There are 140008 ways to make the sum 130, and the tuple (5, 8, 11, 26, 37, 43) has sum 130, product 18201040, and 130 times 140008 is 18201040.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the python program, followed by its output:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#!/usr/bin/env python&#xA;# Re: http://puzzling.stackexchange.com/questions/2119/markus-lucky-numbers&#xA;# Find: Six distinct numbers from [1,49] such that their sum, times&#xA;# the number of possible ways to get that sum, is equal to their&#xA;# product, which is &amp;gt;= 10,000,000.&#xA;&#xA;from itertools import combinations&#xA;from operator import mul&#xA;&#xA;u = range(1,50)&#xA;maxSum = 44+45+46+47+48+49&#xA;ways = [0]*(maxSum+1)&#xA;for c in combinations(u, 6):&#xA;    s = sum(c)&#xA;    ways[s] += 1&#xA;&#xA;for c in combinations(u, 6):&#xA;    s = sum(c)                  # Get sum of #'s&#xA;    p = reduce(mul, c, 1)       # Get product of #'s&#xA;    w = ways[s]                 # Get # of ways to make sum&#xA;    if s*w == p:&#xA;        print s, w, p, c&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It took 0m14.775s to run, and produced the following output:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;36 110 3960 (1, 2, 3, 4, 11, 15)&#xA;45 532 23940 (1, 4, 5, 7, 9, 19)&#xA;76 13552 1029952 (2, 8, 11, 14, 19, 22)&#xA;130 140008 18201040 (5, 8, 11, 26, 37, 43)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2004" LastActivityDate="2014-08-20T20:03:43.220" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2133" PostTypeId="1" CreationDate="2014-08-21T16:15:30.507" Score="0" ViewCount="7623" Body="&lt;p&gt;A dwarf-killing giant lines up 10 dwarfs from shortest to tallest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each dwarf can see all the shortest dwarfs in front of him, but cannot see the dwarfs behind himself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The giant randomly puts a white or black hat on each dwarf. No dwarf can see their own hat. The giant tells all the dwarfs that he will ask each dwarf, starting with the tallest, for the color of his hat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the dwarf answers incorrectly, the giant will kill the dwarf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each dwarf can hear the previous answers, but cannot hear when a dwarf is killed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dwarves are given an opportunity to collude before the hats are distributed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What strategy should be used to kill the fewest dwarfs, and what is the minimum number of dwarfs that can be saved with this strategy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach: &#xA;1st guy counts which color is max, says it, all others copy his ans. This way, we save at least 5, but I think we can optimize this, just can't figure out how....&lt;/p&gt;&#xA;" OwnerUserId="1954" LastActivityDate="2014-08-21T17:03:06.240" Title="10 little dwarves" Tags="|logical-deduction|" AnswerCount="1" CommentCount="0" ClosedDate="2014-08-21T21:13:49.147" ContentLicense="CC BY-SA 3.0" />
  <row Id="2134" PostTypeId="2" ParentId="2133" CreationDate="2014-08-21T17:03:06.240" Score="2" Body="&lt;p&gt;I'm pretty sure you can save 9 dwarfs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dwarf 10 can see 9 hats. He counts the number of white hats and if they are even, says white. If they are odd, he says black. He cannot be saved except by pure luck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dwarf 9 counts the number of white hats. If it is odd, he says white (which would make it even). If it is even, he says black (keeps it even). He lives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dwarf 8 counts the number of white hats and adds 1 if dwarf 9 said white. He does the same thing (if odd, say white; if even, say black). He lives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dwarf 7 counts and adds 0, 1 or 2. Same thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dwarf 1 counts the number of times dwarves 2-9 said white. If odd, say white. If even, say black. He lives.&lt;/p&gt;&#xA;" OwnerUserId="375" LastActivityDate="2014-08-21T17:03:06.240" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2135" PostTypeId="1" AcceptedAnswerId="2146" CreationDate="2014-08-21T19:08:26.010" Score="22" ViewCount="3336" Body="&lt;p&gt;You are given a 6-sided die.The die is biased, so it is more likely to produce some results than others, but you don't know the exact odds. It is however given that the odds are the same for each roll, and that all 6 faces have a non-zero probability of coming up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Create a scheme, using this die as the only source of random, that generates a number from 1 to 6 with a probability of exactly 1/6 for each possible outcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For added difficulty, try to make a scheme that use as few rolls as possible on average. For calculating the average, assume a fair die.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My best solution has the following average:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;144/35 ≈ 4.114&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2186" LastEditorUserId="8874" LastEditDate="2015-02-07T16:07:56.103" LastActivityDate="2015-06-13T22:40:09.453" Title="Fair results with unfair die" Tags="|mathematics|probability|dice|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2137" PostTypeId="2" ParentId="2135" CreationDate="2014-08-21T23:52:07.357" Score="5" Body="&lt;p&gt;I have found a $6156/1505 \approx 4.09$ solution. Here is how it goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Label the faces A, B, C, D, E, F. Roll the dice 3 times. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each permutation of a given set of three faces is equally likely, because rolls are independent, so if the three rolls are different, we are done. If we number the permutations by their lexicographic ordering (ADE = 1, AED = 2, DAE = 3, ...), this gives us a uniform distribution from 1 to 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the rolls are all the same, we have learned nothing useful, so we start over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If two rolls are the same and one is different, (AAB), we have uniformly sampled from the 3 permutations of that outcome, e.g. AAB from (AAB, ABA, BAA). Once we have this value, we switch to attempting to generate a 1/2 probability. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We do this by rolling the dice twice. If the results are different, we use lexicographic ordering, and if they are the same, we try again. This gives us a uniformly random number, either 1 or 2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if on the second try the numbers are paired again, but with a different pair, such as AABB, this also gives us a random number of 1 or 2. If all 4 rolls are the same, we then repeat the procedure. The probability of finding any kind of solution on the first two steps is 5/6, but on the next 2 steps it goes up to 35/36. After AA, the only combination that does not give a solution is AA again. AB works, but so does BB. Obviously, this could be extended to groups of 8 and higher, but the improvement becomes very small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can use a similar procedure to this if we get rolls like AAABBBCCC in the main system. However, since the probability of a result like this is less than 1/512, I will skip it for now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We combine the random number from 1 to 3 with the random number from 1 to 2 with a formula such as $2*x-(y-1)$, where x is the number from 1 to 3 and y is the number from 1 to 2. This gives us a uniformly random number from 1 to 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probability analysis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let y = average number of rolls to get 2 different numbers, to give the 1/2 probability in the second case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = 2 + (1/6)*(2 + y/36) \implies (215/216)*y = 2 + 1/3 \implies y = (216*7)/(215*3) = 504/215 \approx 2.344$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 rolls definitely used, plus 1/6 chance of trying again, on which 2 rolls are definitely used and 1/36 chance of repeat. This is an improvement of .65 rolls in this case. Adding in support for AAAABBBB would likely be minimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let x be the average number of rolls to achieve a 1/6 probability.&#xA;P(3 different) = $6*5*4/6^3 = 5/9$. P(2 same, 1 different) = $(6*5/6^3)*3 = 5/12$. P(3 same) = $6/6^3 = 1/36$. Thus,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = 3 + (5/12)*(504/215) + x/36 \implies (35/36)x = 3 + 42/43 \implies x = (36/35)*(214/43) = 6156/1505 \approx 4.09$&lt;/p&gt;&#xA;" OwnerUserId="2187" LastEditorUserId="2187" LastEditDate="2014-08-22T20:19:43.840" LastActivityDate="2014-08-22T20:19:43.840" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2138" PostTypeId="1" AcceptedAnswerId="2140" CreationDate="2014-08-22T01:55:32.543" Score="2" ViewCount="440" Body="&lt;p&gt;In this question, &lt;em&gt;bridge&lt;/em&gt; stands for the illustrated convex quadrilateral, constructed from three equilateral triangles.  Letters A, B, C, D, E are assigned to each unit length of periphery as shown in the diagram.  The dot indicates the location of the center of gravity (COG) of a bridge; when the origin is at lower left, the COG is at (x,y) = $(1, 2\sqrt 3/9)$.  An assembly of bridges has a lowermost horizontal base line that is a face of its first bridge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZhWOb.jpg&quot; alt=&quot;protobridge picture&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Rules:&lt;/em&gt;&lt;/strong&gt; We allow a labelled side on a bridge to connect exactly to a labelled side on another bridge.  Bridges cannot overlap.  As pieces are added to an assembly, it must never tip over; &lt;em&gt;ie&lt;/em&gt; the x-coordinate of the progressive COG must always fall within or on the base line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WQDHX.jpg&quot; alt=&quot;3 example structures&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the diagrams, numbers from 0 to 4 or 5 denote the order in which bridges were added.   The path of small asterisks in each figure represents the path of the COG as parts are added.  The large asterisk shows the final COG.  For example, in the right-hand diagram, the COG first moves to the left as parts 1 and 2 are added, and back to the right as parts 3 and 4 are added. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Objectives:&lt;/em&gt;&lt;/strong&gt; Among valid assemblies of 7, 8, or 9 bridges, find those that have the &lt;strong&gt;highest ratio&lt;/strong&gt; of &lt;strong&gt;overall width&lt;/strong&gt; to &lt;strong&gt;baseline length&lt;/strong&gt;.  (Left to right, the figures shown have ratios 4.5/2 = 2.25, 5/1 5, and 4/2 = 2.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Notes:&lt;/em&gt;&lt;/strong&gt; (1) The COG of an assembly is the area-weighted average of the COGs of its parts.  (Ref: &lt;a href=&quot;http://en.wikipedia.org/wiki/Centroid#By_geometric_decomposition&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt;) (2) The diagrams were generated as PS files by a slightly clunky python program that I've uploaded to &lt;a href=&quot;http://pastebin.com/vL6M982N&quot; rel=&quot;nofollow noreferrer&quot;&gt;pastebin&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2004" LastActivityDate="2014-08-22T09:47:55.663" Title="Make a wide tower of bridge-shapes" Tags="|calculation-puzzle|geometry|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2140" PostTypeId="2" ParentId="2138" CreationDate="2014-08-22T09:47:55.663" Score="2" Body="&lt;p&gt;There doesn't seem to be much challenge in this, from a simple base the construction can be extended indefinitely by alternating between adding parts to the right and left side:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AGMCU.png&quot; alt=&quot;Easy solution&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives solutions of width 10, 11.5 and 13 for 7, 8 and 9 pieces respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I by the way find it easier to calculate the centre of mass by considering each triangle separately, that way all the involved numbers are divisible by 1/2.&lt;/p&gt;&#xA;" OwnerUserId="2186" LastActivityDate="2014-08-22T09:47:55.663" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2141" PostTypeId="2" ParentId="13" CreationDate="2014-08-22T23:00:11.027" Score="10" Body="&lt;p&gt;Joshua Taylor and Ian Parberry, &quot;Procedural Generation of Sokoban Levels&quot;, Proceedings of the 6th International North American Conference on Intelligent Games and Simulation (GAMEON-NA), pp. 5-12, EUROSIS, 2011.  The abstract is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We describe an algorithm for the procedural generation of levels for the popular Japanese puzzle game Sokoban. The algorithm takes a few parameters and builds a random instance of the puzzle that is guaranteed to be solvable. Although our algorithm and its implementation runs in exponential time, we present experimental evidence that it is sufficiently fast for offine use on a current generation PC when used to generate levels of size and complexity similar to those human-designed levels currently available online.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that in contrast to &lt;a href=&quot;https://puzzling.stackexchange.com/a/50/1694&quot;&gt;Emrakul's answer&lt;/a&gt;, the puzzles produced by this algorithm are guaranteed to be solvable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper, a presentation video, and five level sets are available &lt;a href=&quot;http://larc.unt.edu/ian/research/sokoban/&quot;&gt;at the paper web page&lt;/a&gt;.  Unfortunately, the implementation does not seem to be available.&lt;/p&gt;&#xA;" OwnerUserId="1694" LastActivityDate="2014-08-22T23:00:11.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2142" PostTypeId="2" ParentId="1751" CreationDate="2014-08-23T19:44:32.760" Score="1" Body="&lt;p&gt;I'm writing this from memory, so some of the smaller details may not be correct, and I haven't had the time to sit down and actually solve the puzzle, but the master pattern is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 3 large rings form a triangle, which is equivalent to the triangle that's formed between the green square, the red pentagon and the blue hexagon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pressing the blue triangle would move the shapes in it's relative position a quarter of a circle counter clockwise.  If the medium blue ring is not located on the green large ring, then the shapes in the relative positions of the pentagon and hexagon would move 4/5ths and 2/3rds of a circle, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar pattern applies when pressing the red pentagon - the shapes in it's relative position will move a fifth of a circle, and if the medium green ring is not on the large red ring then the the shapes in the relative positions of the square and hexagon would move 1/2 and 5/6ths of a circle, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, when pressing the blue hexagon would move the shapes in it's relative position a sixth of a circle, and if the medium red ring is not located on the large blue ring than the shapes in the relative positions of the square and pentagon would move 3/4 and 3/5ths of a circle, respectively.&lt;/p&gt;&#xA;" OwnerUserId="2190" LastActivityDate="2014-08-23T19:44:32.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2143" PostTypeId="1" AcceptedAnswerId="2145" CreationDate="2014-08-24T05:46:44.603" Score="4" ViewCount="1705" Body="&lt;p&gt;Given the figure below, find the shortest network of straight line segments (like a &lt;a href=&quot;http://en.wikipedia.org/wiki/Steiner_tree_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;Steiner tree&lt;/a&gt;, or like parts of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Delaunay_triangulation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Delaunay triangulation&lt;/a&gt;) that connects the four circled points while staying in the purple region.  No line in the network can cross or enter an area that isn't purple, but point contact is allowed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CWB0W.jpg&quot; alt=&quot;green pentagram on purple caltrop&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, the coordinates of the outer points of the green pentagram are [(29, 7), (2, 29), (-27, 11), (-19, -22), (15, -25)] (CCW from right), and the coordinates of the vertices of the purple figure are [(0, 100), (30, 30), (100, 0), (30, -30), (0, -100), (-30, -30), (-100, 0), (-30, 30)] (CW from top).  Each square in the background grid is 10 x 10 units.  Solution attempts can be given as drawings, but node coordinates are needed for objective comparisons of path length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, the figure below shows (in blue) a valid network of length 405.4 that is not minimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yKDwQ.jpg&quot; alt=&quot;with blue network added; half scale&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2004" LastEditorUserId="2004" LastEditDate="2014-08-24T14:45:39.913" LastActivityDate="2014-08-25T21:58:53.780" Title="Find shortest network connecting four points" Tags="|mathematics|geometry|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2144" PostTypeId="2" ParentId="2143" CreationDate="2014-08-24T07:29:05.033" Score="0" Body="&lt;p&gt;My first attempt had some mistakes, so here's my second attempt:-&lt;/p&gt;&#xA;&lt;p&gt;Firstly lets consider the simpler case when the green pentagram is absent.This then corresponds to the case of the four points at the corner of a square and the solution is well-known and looks like this. I show the dashed red line to indicate the constraint that we face in our case.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fM5O5.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;In this case the the total path length is &lt;span class=&quot;math-container&quot;&gt;$\lambda_{simple}=386.37$&lt;/span&gt;. So we can conclude that for the constrained case the  total path length has a lower bound given by  &lt;span class=&quot;math-container&quot;&gt;$\lambda_{constrained}&amp;gt;\lambda_{simple}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Now when the constraint(green pentagram) is included we simply need to move the red dotted line shown in the previous figure outside the&#xA;constrained space and this could be done as follows:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qh2tG.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;For this network configuration we get &lt;span class=&quot;math-container&quot;&gt;$\lambda_{constrained}=394.66$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;I hope this time i have not made any obvious mistakes. I think this should be the minimum path length unless someone can find any in-between values  &lt;span class=&quot;math-container&quot;&gt;$386.37&amp;lt;\lambda_{constrained}&amp;lt;394.66$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="101" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-08-24T14:51:55.403" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2145" PostTypeId="2" ParentId="2143" CreationDate="2014-08-25T00:51:53.907" Score="3" Body="&lt;p&gt;I can get approximately 389.6 with the following arrangement:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/a7xGs.jpg&quot; alt=&quot;584&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lines from the bottom and leftmost points of the 4-sided star follow a little inside the edges of the purple region, the section between P and Q kinks a little as it just touches the top and upper right points of the star. Angles of the 3-point intersections at P and Q are all 120 degrees. Distances and positions shown in the diagram are approximate only.&#xA;P = (-8.62246,37.24663); Q = (39.67114,-7.24132)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Correction:&lt;/strong&gt; With P=(-8.622454731281,37.246624550459); Q=(39.671139293683,-7.241324684499) I now get a total distance of 389.739586527472 which I believe is accurate to approximately 11 decimal places.&lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="1631" LastEditDate="2014-08-25T21:58:53.780" LastActivityDate="2014-08-25T21:58:53.780" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2146" PostTypeId="2" ParentId="2135" CreationDate="2014-08-25T18:32:25.620" Score="5" Body="&lt;p&gt;In order to get the most out of each roll it is beneficial to note that it doesn't matter if the intermediate data is skewed, as long as each result is paired off with 5 others of equal probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A series of rolls is considered to have a form that depend only on which rolls were equal to another, the first roll is always designated A, the first of the remaining rolls that is not equal to A is B etc. So the series 552 has the form AAB, as does 223.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Six different forms that all have the same number of repetitions can be paired, like AAAAAB, AAAABA, AAABAA, AABAAA, ABAAAA, ABBBBB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A set of form that has the same number of two different rolls only need to consist of three different forms as each form has two equally likely outcomes where these numbers are switched. For example AABC, ABAC, ABBC translates amongst other possibilities to 4412, 4421, 4142, 4241, 1442, 2441.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally an equal amount of three different numbers means that a form can be paired with only itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have designed a table of pairings for up to 6 rolls:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Results after 3 rolls:&#xA;ABC        probability 4320/7776&#xA;&#xA;Unpaired:&#xA;AAA&#xA;AAB&#xA;ABA&#xA;ABB&#xA;&#xA;Results after 4 rolls:&#xA;AABB       probability  540/7776&#xA;ABAB&#xA;ABBA&#xA;&#xA;AABC       probability 2160/7776&#xA;ABAC&#xA;ABBC&#xA;&#xA;Unpaired:&#xA;AAAA&#xA;AAAB&#xA;AABA&#xA;ABAA&#xA;ABBB&#xA;&#xA;Results after 5 rolls:&#xA;AABAC      probability  360/7776&#xA;ABAAC&#xA;ABBBC&#xA;&#xA;Unpaired:&#xA;AAAAA&#xA;AAAAB&#xA;AAABA&#xA;AAABB&#xA;AAABC&#xA;AABAA&#xA;AABAB&#xA;ABAAA&#xA;ABAAB&#xA;ABBBA&#xA;ABBBB&#xA;&#xA;Results after 6 rolls:&#xA;AAAAAB     probability   30/7776&#xA;AAAABA&#xA;AAABAA&#xA;AABAAA&#xA;ABAAAA&#xA;ABBBBB&#xA;&#xA;AAAABB     probability   30/7776&#xA;AAABAB&#xA;AAABBA&#xA;AABAAB&#xA;AABABA&#xA;ABAAAB&#xA;&#xA;AAAABC     probability  120/7776&#xA;AAABAC&#xA;AAABCA&#xA;AABAAC&#xA;ABAAAC&#xA;ABBBBC&#xA;&#xA;AAABCD     probability   60/7776&#xA;&#xA;AAABBC     probability  120/7776&#xA;AAABCB&#xA;AAABCC&#xA;AABABC&#xA;ABAABC&#xA;ABBBAC&#xA;&#xA;AABABB     probability   15/7776&#xA;ABAABB&#xA;ABBBAA&#xA;&#xA;Unpaired:  probability   21/7776&#xA;AAAAAA&#xA;AAABBB&#xA;ABAABA&#xA;ABBBAB&#xA;ABBBBA&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If this table of pairings is used along with the rule that unpaired results of 6 rolls are discarded and the method is restarted with 3 fresh rolls the average number of rolls is given by the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X=\dfrac{3\times4320+4\times2700+5\times360+6\times375+(6+X)\times21}{7776}\\&#xA;X=\dfrac{9312}{2585}\approx3.602$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extending the table to allow longer series, and thus reducing the chance of having to discard the rolls, should give a slightly better result, at the cost of making this answer even longer.&lt;/p&gt;&#xA;" OwnerUserId="2186" LastActivityDate="2014-08-25T18:32:25.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2147" PostTypeId="1" AcceptedAnswerId="2166" CreationDate="2014-08-25T18:44:55.423" Score="6" ViewCount="7950" Body="&lt;p&gt;I need help understanding the solution to the following puzzle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Puzzle:&lt;/strong&gt; There is a &quot;triangular&quot; duel between the three shooters. Everyone shoots one by one, can shoot once and any person he wants. Smith always hits the target (100%) and shoots last. Brown hits the target 80% of the time and shoots second. Jones is the worst, hitting the target only 50% of the time and he shoots first. Who should Jones shoot in order to increase his chance of surviving?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answer:&lt;/strong&gt; Jones should shoot in the air. The worst shooter has the best chance of surviving in the triangular duel, which is Jones. After him goes Smith who never misses the target. Since Smith and Brown will be shooting each other when their turn comes, Jones must shoot no one until one of his enemies dies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that he shoots at his enemy, again having an advantage. First, it's easier to calculate the probability of surviving for Smith. In the duel with Brown he shoots first with a probability of 1/2. In this case he kills Brown. Brown also shoots first with a probability of 1/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Smith survives with a probability of 1/5. Thus, Smith will live over Brown with a probability of &lt;strong&gt;1/2 + 1/2 * 1/5 = 3/5&lt;/strong&gt;. With a probability of 1/2, Smith survives the duel with Jones. All in all: the probability to survive for Smith is 3/5 * 1/2 = 3/10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability for Brown to survive the duel with Smith is &lt;strong&gt;2/5&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then Jones shoots Brown. In the first round, Brown has a probability of &lt;strong&gt;1/2 * 4/5 = 4/10&lt;/strong&gt; to win. In the second round, he has a probability of &lt;strong&gt;1/2 * 1/5 * 1/2 * 4/5 = 4/100&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, Brown has a chance to survive Jones: &lt;strong&gt;4/10 + 4/100 + 4/1000 + 4/10000 = 0.4444(4) = 4/9&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability of Brown to survive both of his opponents is equal to 2/5 (over Smith) * 4/9 (over Jones) = 8/45. The probability of Jones to survive = 1 - 3/10 - 8/45 = 47/90.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In bold are the calculations I'm confused about and don't know how to arrive at. Can someone please explain it? Thank you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE:&lt;/strong&gt; They shoot one by one in the same order until one person survives.&lt;/p&gt;&#xA;" OwnerUserId="2194" LastEditorUserId="2194" LastEditDate="2014-08-28T18:43:09.873" LastActivityDate="2016-05-09T15:46:30.070" Title="Help understanding the solution to a &quot;shooting puzzle&quot;" Tags="|mathematics|probability|" AnswerCount="7" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2148" PostTypeId="1" AcceptedAnswerId="2180" CreationDate="2014-08-25T23:20:12.160" Score="3" ViewCount="604" Body="&lt;p&gt;Let a “number chain” be a series of numbers, $n_0, n_1, ... n_{m-1}$&#xA;for some $m$, such that in base 10 the low digits of $n_i$ match high digits of $n_{(i+1)mod m}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, after the number 123456, numbers that begin with 6, or 56, or 456, etc, are allowed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any given number chain $N = n_0, n_1, ... n_{m-1}$, we assign a &lt;strong&gt;score $S(N)$&lt;/strong&gt; and a &lt;strong&gt;length $L(N)$&lt;/strong&gt; as follows.  Let $P(n) = +1$ if $n$ is a positive prime integer equal to exactly one of the $n_i$ values, and -5 otherwise.  Let $G(n) =$ number of digits of $n$, and let $H(n) = G(n)\cdot G(n)$.  Now set $S(N) = \sum\limits_{i=0}^{m-1} P(n_i)\cdot H(n_i)$, and set $L(n)$ equal to the number of characters occupied by $N$ when it is written with $n_0, n_1, ... n_{m-1}$ overlapped as much as they can be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Examples:&lt;/strong&gt;  The chain $(730573, 573973, 973373, 730573)$ has score $S(N) = +1\cdot36\cdot2 + -5\cdot36\cdot2 = -288$ and length $L(N) = 16$ (from 7305739733730573).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chain $(730573, 573973, 973373, 373273)$ has score $S(N) = +1\cdot36\cdot4 = 144$ and length $L(N) = 15$ (from 730573973373273).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt;  Find a high scoring chain $N$ with $L(N) = 53$. &lt;/p&gt;&#xA;" OwnerUserId="2004" LastEditorUserId="2004" LastEditDate="2014-08-25T23:29:51.380" LastActivityDate="2014-08-31T22:48:00.263" Title="Find a high-scoring 53-digit prime number chain" Tags="|mathematics|calculation-puzzle|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2149" PostTypeId="2" ParentId="2147" CreationDate="2014-08-26T00:30:05.670" Score="4" Body="&lt;p&gt;The answer does not appear to apply to the question as stated.  In the question it says that Jones shoots first, Brown next, and Smith last.  The calculation of the chance that it is Smith that Jones faces assumes it is random which of Smith or Brown shoots first.  If it is Smith (probability $\frac 12$) he survives.  If it is Brown (probability $\frac 12$) Smith still lives with probability $\frac 15$  So Smith survives Brown $\frac 35$ of the time and Brown survives Smith $\frac 25$ of the time.  You agree that this gives Smith a $\frac 3{10}$ overall chance of survival.  To calculate Brown's chances in the duel with Jones (assuming he has already beat Smith-a $\frac 25$ chance), he can win if Jones misses and he hits, which is where the $\frac 12 \cdot \frac 45=\frac 4{10}$ comes from.  He can also win if Jones misses, he misses, Jones misses, and he hits, which is the $\frac 12 \cdot \frac 15 \cdot \frac 12 \cdot \frac45=\frac 4{100}$  Continuing, we have an infinite series to sum, which results in Brown winning $\frac 49$ of the time.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2014-08-26T00:30:05.670" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2150" PostTypeId="2" ParentId="2148" CreationDate="2014-08-26T03:02:08.027" Score="0" Body="&lt;p&gt;&lt;strong&gt;Corrected:&lt;/strong&gt;&#xA;I believe the number chain N = (n0,n1) where:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; n0=32711111111111111111111111111111111111111111111111, and&#xA; n1=11111111111111111111111111111111111111111111111153 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;has a fairly high score.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The low 47 digits of n0 '11111111111111111111111111111111111111111111111' match the high 47 digits of n1, and the low 1st digit of n1 '3' matches the high 1st digit of n0, so I believe this is now a valid chain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two numbers overlap to form the 53 digit number &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; N=32711111111111111111111111111111111111111111111111153&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;with length 53, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both numbers are prime (according to a Miller-Rabin test), so P(n0) = P(n1) = 1&#xA;Both numbers are 50 digits long so G(n0) = G(n1) = 50 and H(n0) = H(n1) = 50^2 = 2500&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So S(N) = Sum(P(n0).H(n0),P(n1).H(n1)) = 1*2500 + 1*2500 = 5000.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be nice if N itself was also prime, but alas it is not (but its last two digits do equal its length).&lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="1631" LastEditDate="2014-08-26T05:02:20.317" LastActivityDate="2014-08-26T05:02:20.317" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2151" PostTypeId="1" CreationDate="2014-08-26T06:00:59.493" Score="18" ViewCount="2481" Body="&lt;p&gt;I would like to ask if anyone has seen elsewhere this variation on the standard truth tellers/liars kind of puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the usual repertoire it is assumed that every person questioned knows the answer to every question asked.  We now assume that this need not be the case; and that if a truth-teller is asked a question to which he does not know the answer, he will truthfully reply &quot;I don't know&quot;, while a liar will never admit to not knowing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can still be formulated in various ways: here is a specific example and a puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(But &lt;strong&gt;please note&lt;/strong&gt;: my question is not to solve the puzzle but to ask if anyone has seen anything like it before.)&#xA;&lt;hr&gt;&#xA;In a certain town each of the inhabitants is either a truth-teller or a liar; however this does not mean that everyone is actually able to answer every question they are asked.  If a truth-teller is absolutely certain of the answer to a question, he will give that answer; if not, he will say, &quot;I don't know.&quot;  On the other hand, a liar will never truthfully admit to not knowing something: he will give an answer that he knows is false, if any, but if there is nothing that he is &lt;em&gt;certain&lt;/em&gt; is false then he will give a randomly chosen answer (possibly even, by accident, the true answer).  Moreover, everyone in this town can instantly deduce the logical consequences of any facts they know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I meet four inhabitants of this town and ask them, &quot;How many of you four are truth-tellers?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kevin says, &quot;I don't know&quot;; then Laura says, &quot;One&quot;; then Mike says, &quot;None.&quot;  Noela, however, is asleep.  Fortunately I don't need to wake her up, since I can already tell whether she is a truth-teller or a liar.  Which?&lt;/p&gt;&#xA;" OwnerUserId="2199" LastActivityDate="2015-06-05T15:10:52.237" Title="Is this a new kind of liar puzzle?" Tags="|logical-deduction|puzzle-identification|liars|" AnswerCount="1" CommentCount="26" ContentLicense="CC BY-SA 3.0" />
  <row Id="2152" PostTypeId="2" ParentId="2147" CreationDate="2014-08-26T11:54:27.740" Score="3" Body="&lt;p&gt;I don't know if this is deliberately different to an old chestnut which appeared in one of Martin Gardner's books many years ago - it appears almost verbatim, even down to the names, except that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the participants &lt;strong&gt;keep shooting until only one remains&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;the initial order of shooting is chosen randomly&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The solution you quote is straight from Gardner, and doesn't relate to the question as posed here. Smith and Brown clearly need to target their most dangerous enemy first (each other), giving Jones the option to deliberately miss until Smith v Brown is resolved and then take first crack at the survivor, guaranteeing him 50% survival chance even in the worst case when he faces sure-shot Smith.&lt;/p&gt;&#xA;" OwnerUserId="1801" LastActivityDate="2014-08-26T11:54:27.740" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2153" PostTypeId="1" CreationDate="2014-08-26T18:58:24.397" Score="9" ViewCount="726" Body="&lt;p&gt;Inspired by &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2147/help-understanding-the-solution-to-a-shooting-puzzle&quot;&gt;this puzzle&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have $n$ shooters having a duel, labeled $1, 2, ... n$. Each shooter has an accuracy of $k_n$, with each $k_n$ some probability in $(0,1]$, with $k_1 &amp;lt; k_2 &amp;lt; ... &amp;lt; k_n$. The first shooter is the least accurate, while the last shooter is the most accurate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The duel works as followers. Starting with the least accurate (1) and working towards the most accurate (n), each shooter may choose to pass or to shoot at one person. If they hit, the target is out of the duel. If after the most accurate remaining shooter's turn there is still more than one shooter left, the cycle restarts with the least accurate remaining shooter. They game ends when one shooter is left, who wins the duel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is each shooter's general strategy to maximize their chance of winning? Do their strategies change if the game ends when there are two shooters left?&lt;/p&gt;&#xA;" OwnerUserId="1906" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2020-05-01T15:16:08.477" Title="Generalizing the &quot;shooting puzzle&quot;" Tags="|mathematics|probability|" AnswerCount="3" CommentCount="7" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2154" PostTypeId="2" ParentId="2153" CreationDate="2014-08-27T04:43:02.720" Score="1" Body="&lt;p&gt;&lt;strong&gt;Edited:&lt;/strong&gt;&#xA;OK, so on reflection my original answer shown below in italics was not entirely correct. For n&gt;=4 player n-2 should never pass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;strong&gt;Original non-optimal strategy for n&gt;=4&lt;/strong&gt;)... &lt;em&gt;For n&gt;3&#xA;    1..n-3 shoot at n&#xA;    n-2 shoots or passes depending on specific accuracy of duelists n-2, n-1, and n&#xA;    n-1 shoots at n&#xA;    n shoots at n-1&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To illustrate this, imagine if four very good shots (accuracy 0.997, 0.998, 0.999, and 1.0 respectively) are in a quaduel. Best strategy for 1 is still to shoot 4, then 2 shoots 3, then 4 shoots 2. But in the 3% of occasions where 1 misses, it is &lt;strong&gt;not&lt;/strong&gt; best for 2 to pass as then 3 shoots 4, 1 passes, 2 shoots 3 and 1 shoots 2. Instead, 2 should shoot 1 and 3 shoots 4, 2 shoots 3 and lives.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Updated conjectures for optimal strategies below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For n=2 &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 shoots at 2 &#xA;2 shoots at 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For n=3&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 shoots or passes depending on specific accuracy of each duelist  &#xA;2 shoots at 3&#xA;3 shoots at 2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For n&gt;3&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1..n-3 shoot at n&#xA;n-2 shoots at n depending on accuracy of duelists n-2, n-1, and n, or shoots n-3&#xA;n-1 shoots at n&#xA;n shoots at n-1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Justification&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For n=2&lt;/strong&gt; it is clearly better to shoot than to pass as, if 1 has accuracy x and 2 has accuracy y, shooting first gives 1 the probability of survival x/(x+y-xy) which is strictly greater than (x-xy)/(x+y-xy) which is the probability of survival if 1 waits for 2 to shoot first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For n=3&lt;/strong&gt; and probabilities x &amp;lt; y &amp;lt; z for i,2,3 respectively, 1 will shoot if the inequality &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; (x-xy)(y+z-yz)(x+z-xz) &amp;gt; xy(x+z-xz) + x(z-yz)(x+y-xy) &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;is true (someone with access to Mathematica may be able to simplify or make better sense than that), otherwise (s)he will pass. This results from the two scenarios: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) successfully shooting 3 and then being second shot in a duel against 2, or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) failing to shoot 3 (possibly repeatedly) or waiting for 2 and 3 to fight it out in which case (s)he will end up as first shot in a duel with 2 or 3 (there will be a different probability of each of those cases). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shooters 2 and 3 are effectively in a duel and will shoot at each other as in the n=2 case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For n&gt;3&lt;/strong&gt; &#xA;n-1 and n still are still effectively in a duel so will duke it out.&#xA;n-2 is in a similar position as in the n=3 case so will make a similar choice to shoot at n (or not). But instead of passing it is better to remove the biggest threat amongst the weaker players. The choice criterion (shoot n or not?) will be a little different than that for the n=3 game, but in most cases not significantly different. &#xA;1..n-3 are always better off if they eventually end up as '3rd-man' in a duel where the stronger players are weaker shots so, for example in the n=4 situation, 1 would rather face 2 and 3 than 2 and 4 or 3 and 4, so the weaker players are always be happy to shoot at the strongest surviving player.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think these tactics are will be valid for &lt;em&gt;most&lt;/em&gt; situations. But for 5 or more players, if they are &lt;em&gt;all&lt;/em&gt; very strong, there may be situations where it is still occasionally better for some weaker players to pass.&lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="1631" LastEditDate="2014-08-27T21:52:37.947" LastActivityDate="2014-08-27T21:52:37.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2155" PostTypeId="1" AcceptedAnswerId="2156" CreationDate="2014-08-27T08:25:27.483" Score="28" ViewCount="8993" Body="&lt;p&gt;You are on a rowboat in the middle of a large, perfectly circular lake. On the perimeter of the lake is a monster who wants to eat you, but fortunately, he can't swim.  He can run (along the perimeter) exactly $4\times$ as fast as you can row, and he will always run towards the closest bit of shore to you.   If you can touch shore even for a second without the monster already being upon you, you can escape.  Suggest a strategy that will allow you to escape, and prove that it works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other notes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If two paths take the monster to this location equally quickly, he will arbitrarily choose one.&lt;/li&gt;&#xA;&lt;li&gt;The monster can reverse direction instantaneously, and you can turn your boat instantaneously.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Follow up:  What is the minimum speed of the monster (relative to your boat) such that escape becomes impossible?&lt;/p&gt;&#xA;" OwnerUserId="1672" LastEditorUserId="10615" LastEditDate="2015-04-24T15:05:13.377" LastActivityDate="2015-04-24T15:05:13.377" Title="The lake monster" Tags="|mathematics|logical-deduction|strategy|pursuit-evasion|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="2156" PostTypeId="2" ParentId="2155" CreationDate="2014-08-27T10:07:03.417" Score="25" Body="&lt;p&gt;First of all, row out to a radius $R/4$ (where the lake has radius $R$) keeping you, the centre of the lake and the monster in a straight line - with you on the far side to the monster. This is always possible; radius $R/4$ is the first point where the angular speed you can achieve just matches that of the monster as he runs round to get you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are now a distance $3R/4$ away from the shore, directly opposite the monster so he needs to run a distance $\pi R$ to get you. You will take time $3R/4V$ at speed $V$ if you now row directly towards the nearest shore, and he will take $\pi R/4V$, which is fractionally greater.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For the followup: If instead of $4\times$, the monster runs $N\times$ your speed... then you row out to radius $R/N$, you then take time $(N-1)R/NV$ to reach shore and he takes $\pi R/NV$ to reach the same point. You escape provided that $N &amp;lt; \pi + 1 \approx 4.1459$.&lt;/p&gt;&#xA;" OwnerUserId="1801" LastEditorUserId="243" LastEditDate="2014-08-27T23:31:38.207" LastActivityDate="2014-08-27T23:31:38.207" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2157" PostTypeId="1" CreationDate="2014-08-27T11:22:14.977" Score="8" ViewCount="2036" Body="&lt;p&gt;Three friends enter a quiz-show as a team, the final round of the show is all or nothing, they only get their prize if they can win this round.&lt;/p&gt;&#xA;&lt;p&gt;Three boxes labelled 1, 2 and 3 are placed in a room, each box has either a white or a black stone in it, with a 50% probability for each.&lt;/p&gt;&#xA;&lt;p&gt;The friends enter the room in turn, the first friend get to look in box 2 and 3, and then has to guess the colour of the stone in box 1. Similarly the second friend get to look in box 1 and 3, and then has to guess the colour of the stone in box 2. And finally the third friend get to look in box 1 and 2 before guessing the colour of the stone in box 3.&lt;/p&gt;&#xA;&lt;p&gt;The friends have no means of communicating what they have seen in the boxes, and they are not told what guess those who came before them made.&lt;/p&gt;&#xA;&lt;p&gt;The friends win if they collectively get at least two guesses right.&lt;/p&gt;&#xA;&lt;p&gt;Find an optimal strategy for the friends to use.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h3&gt;Variation 1&lt;/h3&gt;&#xA;&lt;p&gt;What is the optimal strategy if the stones are black with 70% probability and white with 30% probability?&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h3&gt;Variation 2&lt;/h3&gt;&#xA;&lt;p&gt;What is the optimal strategy if the friends are allowed to switch the two stones they get to see?&lt;/p&gt;&#xA;" OwnerUserId="2186" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2019-11-05T17:06:03.793" Title="Three 50/50 guesses" Tags="|logical-deduction|probability|" AnswerCount="5" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2158" PostTypeId="2" ParentId="2157" CreationDate="2014-08-27T11:22:14.977" Score="10" Body="&lt;p&gt;No matter what they do each of the friends will have exactly 50% probability of answering correctly, this could lead some people to the incorrect conclusion that their probability of victory is 50% no matter what they do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trick is that the friends can to some extend coordinate their guesses, so that the probabilities of guessing correctly are not independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A solution should try to distribute the correct guesses so that none are &quot;wasted&quot; on three correct guesses or a single correct guess. The theoretical optimal is that all three guesses are incorrect with 25% probability and exactly two of the guesses are correct with 75% probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few facts to be deduced about any such optimal strategy:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;None of the friends may be able to deduce that they will lose by looking at the two other stones. Since a loss scenario involves 3 incorrect guesses it would tell the guesser that he is about to guess wrong by using the agreed upon system, thus he has gained knowledge of the colour of the stone he hasn't seen, which is impossible.&lt;/li&gt;&#xA;&lt;li&gt;All three stones must be a factor in the strategy of at least one of the friends. If one of the stones didn't have any influence on the guessing pattern then the guesser of that stone would be able to deduce the success of both his friends, thus knowing about a loss, which as mentioned above is impossible.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This knowledge can be used to quickly discard some options when searching for answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are actually quite a lot of ways to achieve this, one of them is to let all of the friends guess to opposite colour of the colour of the stone in the box labelled one less than the box they are guessing on. And the first friend should guess the opposite of what is in box 3. This method will fail when all the three stones are of the same colour, and give two correct answers in all other cases, thus 75% chance of victory.&lt;/p&gt;&#xA;" OwnerUserId="2186" LastActivityDate="2014-08-27T11:22:14.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2161" PostTypeId="2" ParentId="2155" CreationDate="2014-08-27T15:32:31.743" Score="33" Body="&lt;p&gt;It is possible to escape a monster that is a little more than $4.6$ times faster. Given a monster running at speed $X$ times rowing speed and a lake with radius $R$, you must first row to the circle that is $R/X$ from the centre of the lake by spiralling out while keeping the centre of the lake between you and the monster. Thereafter you should take a route that is tangential to this circle in the opposite direction of what the monster is currently running. This is longer than going the direct route to shore, but the monster also get a longer route to run.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OQaxO.png&quot; alt=&quot;&quot;&gt;&lt;br&gt;&#xA;&lt;em&gt;Your path in red, monster path in green. The shape of the inner spiral is not exact.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this exact solution best? Because of &lt;a href=&quot;http://datagenetics.com/blog/october12013/index.html&quot; rel=&quot;noreferrer&quot;&gt;differential maths&lt;/a&gt;. That is a long explanation, personally I find that stuffing some equations into Wolfram Alpha works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that the monster must run $R(\pi+\varphi)$ where $\varphi$ is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\tan(\varphi)=\pi+\varphi,\\0\leq\varphi\leq\pi/2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the circle you have to row a distance of $R\times\sin(\varphi)$, so the fastest monster that can be overcome is running $(\pi+\varphi)/\sin(\varphi)$ times faster than you row, which is approximately equal to $4.6033388487517$&lt;/p&gt;&#xA;" OwnerUserId="2186" LastActivityDate="2014-08-27T15:32:31.743" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2162" PostTypeId="2" ParentId="2155" CreationDate="2014-08-27T16:02:58.777" Score="10" Body="&lt;p&gt;The monster wants to eat me, but he is also particuarly fond of my boat since &quot;he will always run towards the closest bit of shore to your boat.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I row the boat in one direction and then jump off and swim in the opposite direction.  Even the fastest monsters would never catch me!&lt;/p&gt;&#xA;" OwnerUserId="2208" LastActivityDate="2014-08-27T16:02:58.777" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2164" PostTypeId="2" ParentId="2157" CreationDate="2014-08-27T21:10:47.750" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $R_1,R_2$ and $R_3$ be the three rooms and let $A, B ,C$ denote the three friends.&#xA;  I represent the white balls by the digit $1$ and the black ones by $0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They could adopt the strategy as summarized in the table below where I have indicated all the possibilities where the $plus$ and $dot$ symbol represent the Boolean operators &lt;a href=&quot;http://en.wikipedia.org/wiki/OR_gate&quot; rel=&quot;nofollow noreferrer&quot;&gt;OR&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/AND_gate&quot; rel=&quot;nofollow noreferrer&quot;&gt;AND&lt;/a&gt; respectively.The overhead operator indicates NOT operation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this strategy $C$ always guesses $1$ (White ball), $A$ and $B$ guess based on what $R_3$(room 3) has. Their guesses are determined by simple Boolean algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for eg in the first case i.e in a) first A looks in $R_3$ and on seeing a white ball he performs the operation $R_2 .R_3$ meaning $1.1 = 1$(as per Boolean algebra). $A$ guesses correctly. Next $B$ will perform the NAND operation of $R_1$ and $R_3$ which gives zero.Here $B$ makes an incorrect guess and finally C will simply say white every-time(correct for this case). So they get 2 out of 3 correct. Rest of the table is self-explanatory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gUz89.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The red ones indicate the cases which would fail. So they can win 6 out of 8 times.&#xA;Also this method works for the case when all balls have the same colour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For Variation 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the strategy should be to guess all those combinations correctly which has more combined probability of occurrence. The table below shows one such strategy where $C$ always guess $0$ (Black ball). $A$ guess $1$ if $R_2 = 1$ otherwise $0$. $B$ guess $1$ if $R_1 = R_3$ otherwise $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DLYK1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This way they manage to guess the 6 out of 8 which has a total probability of &lt;strong&gt;0.874&lt;/strong&gt;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ideally we would want them to not include the combination $(1,1,1)$ which has the minimum combined probability and you would want to guess $(0,0,0)$ combination for sure. But I don't think that's possible. Either you can have both $(1,1,1)$ and $(0,0,0)$ or neither of them in your correct 6. I'm not sure of this,maybe there is some other way. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Edit : After reading the comments I realized that i was wrong about $(1,1,1)$ and $(0,0,0)$ combination.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I think that this is the optimal solution which gives the maximum total probability of occurrence for 6/8 because there is only one other combination which gives a higher value (0.91) i.e. when you exclude the $(1,1,1)$ + any other with two $1's$ viz ((a,b) , (a,c) or (a,e)). But if you choose to exclude any of these two then you would surely fail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like for eg. if you try to exclude (a,b) then you need to satisfy the following:-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SrOIC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly the two failed one's should necessarily have all guesses incorrect. With this constraint we run into contradiction when two same initial configuration demand exactly opposite result(see blue coloured in table). Similar contradiction happen for the other two cases. So by contradiction we cant have the 0.91 solution so the next highest has to be the optimal solution.&lt;/p&gt;&#xA;" OwnerUserId="101" LastEditorUserId="101" LastEditDate="2014-08-29T17:22:29.727" LastActivityDate="2014-08-29T17:22:29.727" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2165" PostTypeId="2" ParentId="2079" CreationDate="2014-08-28T11:33:58.230" Score="8" Body="&lt;p&gt;I found 21&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;9 variations of black square&lt;/li&gt;&#xA;&lt;li&gt;2 variations of orange square&lt;/li&gt;&#xA;&lt;li&gt;4 variations of brown square&lt;/li&gt;&#xA;&lt;li&gt;4 variations of gray square&lt;/li&gt;&#xA;&lt;li&gt;2 variations of red square&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3HzCN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2218" LastActivityDate="2014-08-28T11:33:58.230" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2166" PostTypeId="2" ParentId="2147" CreationDate="2014-08-28T19:33:01.870" Score="3" Body="&lt;p&gt;I found the explanation of the puzzle by its &lt;a href=&quot;http://books.google.ca/books?id=fA3CAgAAQBAJ&amp;amp;pg=PA51&amp;amp;lpg=PA51&amp;amp;dq=martin%20gardner%20puzzle%20shoot%20first&amp;amp;source=bl&amp;amp;ots=QxCcl-suW0&amp;amp;sig=R1HgbRT8puvUu4hkHcAkzL_A7FY&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=aYz7U6HfOIisyASZ_oDoDQ&amp;amp;ved=0CCYQ6AEwAQ#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;original author&lt;/a&gt;. The answer is in fact for Jones to shoot into the air until one of the opponents is dead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The order is kept as per the question, but working with probabilities we need to consider all possible outcomes. Thus, we consider the case where Brown shoots Jones first instead of logically (as each will try to take away the strongest person) shooting Smith.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Smith:&lt;/strong&gt; We start calculating the probability of surviving for Smith first as it is the easiest one to do. There are two ways Smith can win Brown in their duel: 1) Smith shoots first and kills Brown (since he is 100% accurate), giving him a chance of surviving of 1/2; 2) Brown shoots first and misses, then Smith kills him; this case gives Smith a chance of surviving of 1/2*1/5 (since Brown is 4/5 accurate). As a result, Smith can survive Brown with a probability of: 1/2 (case 1) + 1/2*1/5 (case 2) = 3/5. Next Smith faces Jones. Since Jones is only 50% accurate, Smith has a survival chance of 1/2 against Jones. Smith's overall chance of surviving in this triangular duel is 3/5 (case against Brown) * 1/2 (case against Jones) = 3/10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Brown:&lt;/strong&gt; Next comes Brown. He has a chance of surviving against Smith of 2/5 (simply deducting Smith's survival probability of 3/5 from total probability of 1). Here it gets complicated as we run into infinite series of possibilities with Brown against Jones. There is a chance of 1/2 that Jones will miss (he is 50% accurate), and Brown has a chance of 4/5 to kill Jones. At this point, Brown has a chance of surviving of 1/2*4/5 = 4/10. But! He can also miss with a probability of 1/5 (which is 1 - his accuracy of 4/5). That can go on forever, and will result in Brown's total survival chance against Jones yielding 4/10 + 4/100 + 4/1,000 + 4/10,000 + ... This can be written as 0.444444... which is the decimal expansion of 4/9. Brown's overall chance of surviving in this triangular duel is 2/5 (case against Smith) * 4/9 (case against Jones) = 8/45.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Jones:&lt;/strong&gt; This leaves Jones with a probability of surviving of 1 (total probability) - 3/10 (Smith's chance) - 8/45 (Brown's chance) = 47/90.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Therefore, Jones has 52.22% of surviving, Smith - 30%, Brown - 17.78%; and it's true that it's better for Jones to fire into the air in order to increase his chance of surviving.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2194" LastActivityDate="2014-08-28T19:33:01.870" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2167" PostTypeId="1" AcceptedAnswerId="2168" CreationDate="2014-08-28T19:46:28.240" Score="24" ViewCount="4062" Body="&lt;p&gt;I'm amazed that, although well known, this problem hasn't been posted yet.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Monty Hall&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You are on a game show.&lt;/li&gt;&#xA;&lt;li&gt;The host of the show shows you three doors.&lt;/li&gt;&#xA;&lt;li&gt;He tells you that behind one of them, is a new car, and behind the&#xA;other two is nothing (or in some versions of the story, a goat).&lt;/li&gt;&#xA;&lt;li&gt;He asks you to pick a door.&lt;/li&gt;&#xA;&lt;li&gt;After you pick a door, the host opens one of the other doors (&lt;strong&gt;IMPORTANT NOTE: The host knows where the car is, and will never open a door that contains the car&lt;/strong&gt;)&lt;/li&gt;&#xA;&lt;li&gt;Then with the two doors left he asks you:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Would you like to keep your door? Or switch?&lt;/h2&gt;&#xA;" OwnerUserId="361" LastEditorUserId="15" LastEditDate="2014-12-31T19:37:00.257" LastActivityDate="2018-04-23T03:07:58.613" Title="The Monty Hall problem" Tags="|logical-deduction|mathematics|probability|monty-hall|" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2168" PostTypeId="2" ParentId="2167" CreationDate="2014-08-28T19:46:28.240" Score="21" Body="&lt;h2&gt;You switch your door, because you have double the chance of getting the car&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you pick a door, there is a 1 in 3 chance that the door contains the car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's examine that case of when you originally picked the car, and when you originally picked a blank separately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Originally picked the car(1/3 chance)&lt;/strong&gt;&lt;br&gt;&#xA;In this case, the host can open either of the 2 doors, and reveal nothing. In this case it is beneficial to keep your original door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Originally picked a blank door(2/3 chance)&lt;/strong&gt;&lt;br&gt;&#xA;In this case, the host isn't allowed to open the door with the car. He must open the blank door. Meaning that when you switch after picking a blank door, you're guaranteed to get the car!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this we can easily see that you are twice as likely to get the car if you switch your door.&lt;/p&gt;&#xA;" OwnerUserId="361" LastEditorUserId="361" LastEditDate="2015-11-12T20:01:38.223" LastActivityDate="2015-11-12T20:01:38.223" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2169" PostTypeId="1" AcceptedAnswerId="2170" CreationDate="2014-08-29T03:09:03.047" Score="3" ViewCount="1633" Body="&lt;p&gt;Two people, Anne and Bob, each hold an index card. A positive integer is printed on each index card, and these integers are consecutive (i.e. Anne's card may display 8 and Bob's card 7). Anne and Bob are aware of this information, and in addition they each know the number printed on their own card, but &lt;em&gt;not&lt;/em&gt; the number printed on the other's card.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following dialogue transpires between Anne and Bob:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1)&lt;/strong&gt; Anne tells Bob, &quot;I do not know the number on your card.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2)&lt;/strong&gt; Bob then tells Anne, &quot;I do not know the number on your card.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3)&lt;/strong&gt; Anne then says to Bob, &quot;Now I know your number. It is divisible by four.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did Anne determine this, and what numbers were printed on each card? Note that this is a logic puzzle, and can be solved by logical deduction. There is no wordplay involved, extenuating circumstances not stated in the problem, nor coded messages of any kind passed between Anne and Bob.&lt;/p&gt;&#xA;" OwnerUserId="412" LastActivityDate="2014-10-06T14:23:33.920" Title="Unknown Consecutive Integers Puzzle" Tags="|mathematics|logical-deduction|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2170" PostTypeId="2" ParentId="2169" CreationDate="2014-08-29T03:48:58.337" Score="11" Body="&lt;p&gt;From 1 we know that Anne does not have 1.  From 2 we know that Bob does not have 1 or 2.  From 3, we know that Anne has 3 and knows Bob has 4.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2014-08-29T03:48:58.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2171" PostTypeId="1" AcceptedAnswerId="2173" CreationDate="2014-08-29T05:54:28.873" Score="2" ViewCount="3900" Body="&lt;p&gt;Solution guide: &lt;a href=&quot;http://rubiks.com/uploads/general_content/Rubiks_cube_3x3_solution-en.pdf&quot; rel=&quot;nofollow&quot;&gt;http://rubiks.com/uploads/general_content/Rubiks_cube_3x3_solution-en.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can get to this point pretty reliably now, but when I try the method shown I'm just as likely to go from one or no corners to two yellow corners as I am to go back again the next round.  I don't mess up the solved rows of the cube; things just seem to shuffle at random at this step.  I get the feeling I might not be fully understanding what it means to re-orient the cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;State 1:&#xA;  Y B Y&#xA;G O Y B R&#xA;R Y Y Y O&#xA;Y G Y O Y&#xA;  R G B&#xA;&#xA;Which produced State 2:&#xA;  Y O R&#xA;O B Y G Y&#xA;B Y Y Y R&#xA;Y R Y Y O&#xA;  B G G&#xA;&#xA;Rotated 90° clockwise (to match state 2 as per the guide) and produced State 3:&#xA;  R O O&#xA;G Y Y Y G&#xA;B Y Y Y G&#xA;O Y Y Y R&#xA;  B R B&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;...which coincidentally solves it this time.  Guess it sensed my crying out for help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess what it boils down to is I'm not sure what to do when I get two corners up, as I haven't gotten particularly good at predicting where pieces will end up at this stage in the puzzle as of yet.  Any advice to be offered on that point?&lt;/p&gt;&#xA;" OwnerUserId="2221" LastEditorUserId="39" LastEditDate="2014-11-01T15:06:25.717" LastActivityDate="2014-11-01T15:06:25.717" Title="Confused about step 5 of official Rubik's 3x3 Cube solution" Tags="|rubiks-cube|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2172" PostTypeId="2" ParentId="2157" CreationDate="2014-08-29T08:31:02.550" Score="-1" Body="&lt;p&gt;&lt;strong&gt;Variation 2:&lt;/strong&gt; I believed there was a 100% strategy - reached as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A guesses BLACK, and if possible switches boxes 2 and 3 so 2 would hold a black.&#xA;B guesses BLACK, unless he sees two whites in 1 and 3. If possible and needed, he switches boxes 1 and 3 to put a white in box 3.&#xA;C guesses WHITE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The box contents are as follows:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Tvsdx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the explained logic, here are the guesses (red = mistakes)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hmNEe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is never more then one mistake, so always at least two correct answers, guaranteeing a 100% win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: also interesting variation: if box switching is allowed, how can they maximize odds of getting all three right. (I found a 3/8 strategy there)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;As pointed out below that is NOT correct, as the assumptions for option B aren't interpreted correctly in the table. However, if B assumes BLACK 100% of the time, winning chances are 7/8&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2156" LastEditorUserId="2156" LastEditDate="2014-08-29T14:45:55.313" LastActivityDate="2014-08-29T14:45:55.313" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2173" PostTypeId="2" ParentId="2171" CreationDate="2014-08-29T08:44:01.437" Score="1" Body="&lt;p&gt;Hmm, what I find lacking in the guide is the positioning of the yellow faces on the side. (which I rely on in the previous step to be more likely to end up with the 'fish' pattern on top).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may want to look at the instruction for the several 'cases' for the different top patterns to be able to fix it. It relies on the same 8-step move as described in the guide your reference, but shows them visually:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ryanheise.com/cube/beginner.html&quot; rel=&quot;nofollow&quot;&gt;Link to Ryan Heise's page on cube solving&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2156" LastActivityDate="2014-08-29T08:44:01.437" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2174" PostTypeId="2" ParentId="2167" CreationDate="2014-08-29T12:18:35.887" Score="14" Body="&lt;p&gt;Switch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a $2/3$ chance that you'll choose a goat, and a $1/3$ chance you'll choose the car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you chose a goat, after switching you will win a car. (Guaranteed)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you chose a car, after switching you will win a goat. (Guaranteed)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the chance you'll win a car is $2/3$ if you switch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply put, the chance you picked the car before the switch is $1/3$. That does not change if a door is opened. The chance will still be $1/3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be more intuitive if you make it a 100 doors instead of 3: (Although some people do &lt;strong&gt;not&lt;/strong&gt; think this a clear explanation for some reason.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You pick 1 door, then the host opens 98 doors revealing a goat. What is the chance that the car is in the door you chose? Answer: $1/100$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first door (the one you picked initially) is random, but the second door is not. That same door will always remain closed, no matter which door you pick initially, and it is the &lt;strong&gt;only&lt;/strong&gt; door that will always remain closed. (because there's something special about that door.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the host is asking, in essence, is: do you want what is behind the door you chose initially, or what is behind the other 99 doors? The act of opening the 98 doors means nothing, because it is not random.&lt;/p&gt;&#xA;" OwnerUserId="1598" LastEditorUserId="42054" LastEditDate="2018-04-23T03:07:58.613" LastActivityDate="2018-04-23T03:07:58.613" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2175" PostTypeId="2" ParentId="429" CreationDate="2014-08-30T01:22:54.503" Score="7" Body="&lt;p&gt;While it's somewhat pattern based, I managed to produce the following by&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; doing a series of rotating opposing flips (e.g. flipping left and right counterclockwise simultaneously, switching faces between each flip):&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA; &lt;pre&gt;&#xA;        Y O Y&#xA;        G R B&#xA;        W O W&#xA;        - - -&#xA;G W G | R Y R | B Y B | R W R&#xA;O B R | G W B | O G R | B Y G&#xA;G Y G | O Y O | B W B | O W O&#xA;        - - -&#xA;        W R W&#xA;        G O B&#xA;        Y R Y&#xA;&lt;/pre&gt;&#xA;&lt;p&gt;I'm relatively new to seriously trying to figure out how to solve these things, so I don't know if this is the same as the above mentioned superflip or not.  Technically it's a pattern, however without being aware of the pattern there's little to obviously suggest it was in any way intentional the way some other patterns do.&lt;/p&gt;&#xA;" OwnerUserId="2221" LastEditorUserId="63414" LastEditDate="2021-02-23T12:18:34.867" LastActivityDate="2021-02-23T12:18:34.867" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="2176" PostTypeId="2" ParentId="2079" CreationDate="2014-08-30T13:20:01.787" Score="3" Body="&lt;p&gt;I have found 21 Squares, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As &lt;a href=&quot;https://puzzling.stackexchange.com/users/1497/kevin&quot;&gt;Kevin&lt;/a&gt; illustrated in his &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2079/how-many-squares-can-you-find-in-the-pattern/2125#answer-2082&quot;&gt;beautiful answer&lt;/a&gt; there are 17 squares. PLUS there are 4 more squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;all the squares are as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dimensions  How many&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Nine × [1x1]&lt;/li&gt;&#xA;&lt;li&gt;Four × [√2×√2]&lt;/li&gt;&#xA;&lt;li&gt;Two × [√5×√5]&lt;/li&gt;&#xA;&lt;li&gt;Four × [√8×√8]&lt;/li&gt;&#xA;&lt;li&gt;Two × [√13×√13]&lt;/li&gt;&#xA;&lt;li&gt;TOTAL of 21 squares&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2225" LastActivityDate="2014-08-30T13:20:01.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2177" PostTypeId="1" CreationDate="2014-08-31T08:27:22.970" Score="7" ViewCount="585" Body="&lt;p&gt;If you have a cube, say, $3 \times 3 \times 3$, you know that mathematically it's made of $27$ ($3*3*3$) smaller cubes. Now, just remove one full line of smaller cubes (in this case, 3 smaller cubes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you prove that the remaining number of smaller cubes is always a multiple of 6?&lt;/p&gt;&#xA;" OwnerUserId="2132" LastEditorUserId="39" LastEditDate="2014-08-31T14:12:04.197" LastActivityDate="2014-08-31T14:12:04.197" Title="A cube puzzle (Multiple of 6)" Tags="|logical-deduction|calculation-puzzle|geometry|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2178" PostTypeId="2" ParentId="2177" CreationDate="2014-08-31T08:42:56.170" Score="16" Body="&lt;p&gt;We will have $N = x^3 - x = x(x-1)(x+1)$ small cubes left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the x and (x-1) is divisible by 2. One of the (x-1), x and (x+1) is divisible by 3. Therefore N is divisible by 6.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="2132" LastEditDate="2014-08-31T10:54:35.173" LastActivityDate="2014-08-31T10:54:35.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2179" PostTypeId="1" AcceptedAnswerId="2183" CreationDate="2014-08-31T22:12:08.487" Score="-3" ViewCount="1889" Body="&lt;p&gt;Can anyone tell me what is the purpose of the outter dial of this watch? Notice how the scale is not linear and goes to 100. It can only be rotated counter-clockwise only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the only picture I have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bnzfW.jpg&quot; alt=&quot;watch&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry, it's not really a puzzle but I didn't know which stackexchange site to use. This one seemed to fit.&lt;/p&gt;&#xA;" OwnerUserId="2229" LastActivityDate="2014-09-01T13:17:59.673" Title="What is the purpose of a weird non-linear watch dial that goes up to 100?" Tags="|explanation|" AnswerCount="2" CommentCount="2" ClosedDate="2014-09-02T00:35:37.017" ContentLicense="CC BY-SA 3.0" />
  <row Id="2180" PostTypeId="2" ParentId="2148" CreationDate="2014-08-31T22:48:00.263" Score="1" Body="&lt;p&gt;It is straightforward to test primes like those shown in Penguino's answer, but using fill digits in {1, 3, 7, 9} instead of just {1}.  For example, given a prefix, a suffix, a fill digit, and two limits, the &lt;code&gt;testpair&lt;/code&gt; procedure in the following code (&lt;code&gt;chain-primes.py&lt;/code&gt;) makes a list of compatible primes and calculates the score for the listed primes.  Here's an example of its use (within &lt;code&gt;ipython&lt;/code&gt; environment):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;In [94]: execfile('chain-primes.py')&#xA;In [95]: testpair(1234, 901, 9)&#xA;Scores  7282 + 819 = 8101&#xA; 6 123499...............................................&#xA;12 123499999999.........................................&#xA;20 12349999999999999999.................................&#xA;29 12349999999999999999999999999........................&#xA;34 1234999999999999999999999999999999...................&#xA;48 123499999999999999999999999999999999999999999999.....&#xA;49 1234999999999999999999999999999999999999999999999....&#xA;27 ..........................999999999999999999999999901&#xA; 7 ..............................................9999901&#xA; 5 ................................................99901&#xA; 4 .................................................9901&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For this example, &lt;code&gt;testpair&lt;/code&gt; used &lt;code&gt;findPre&lt;/code&gt; to find primes (like 123499, 123499999999, etc) that begin with &lt;code&gt;1234&lt;/code&gt; and end with &lt;code&gt;9*&lt;/code&gt;, ie an arbitrary number of 9's.  Then it used &lt;code&gt;findSuf&lt;/code&gt; to find primes (like 9901, 99901, etc) that begin with &lt;code&gt;9*&lt;/code&gt; and end with &lt;code&gt;901&lt;/code&gt;.  The line &lt;code&gt;Scores  7282 + 819 = 8101&lt;/code&gt; indicates that the seven primes beginning with &lt;code&gt;1234&lt;/code&gt; score 7282 points, and the four primes ending with &lt;code&gt;901&lt;/code&gt; score 819 points.  Finally, &lt;code&gt;testpair&lt;/code&gt; used &lt;code&gt;shoPre&lt;/code&gt; and &lt;code&gt;shoSuf&lt;/code&gt; to display results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wrote additional code (not shown) that (1) calls &lt;code&gt;findPre&lt;/code&gt; to look for high-scoring prefixes for all numbers up to some limit; (2) calls &lt;code&gt;findSuf&lt;/code&gt; to look for high-scoring suffixes that are not multiples of two or five; (3) sorts both data sets into higher-scores-first order; (4) builds lists of chain-compatible prefixes and suffixes; and (5) computes valid scores for likely candidate pairs and displays high scores.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For prefixes and suffixes that contain at most 7 digits, the best result I've found scores 25826 points, as shown next with output from the call &lt;code&gt;testpair(1349812, 198271, 9, 48)&lt;/code&gt;.  [The value of 48 for parameter &lt;code&gt;plim&lt;/code&gt; limits prefixed primes to at most 47 digits; by contrast, the less-constrained call &lt;code&gt;testpair(1349812, 198271, 9)&lt;/code&gt; lists a 51-digit prime whose trailing 9's overlap non-9's in the suffix area.  The correct total for such a pair is less than the sum of their separate unconstrained scores.]&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;In [83]: testpair(1349812, 198271, 9, 48)&#xA;Scores  13338 + 12488 = 25826&#xA; 9 134981299............................................&#xA;14 13498129999999.......................................&#xA;23 13498129999999999999999..............................&#xA;32 13498129999999999999999999999999.....................&#xA;34 1349812999999999999999999999999999...................&#xA;35 13498129999999999999999999999999999..................&#xA;39 134981299999999999999999999999999999999..............&#xA;40 1349812999999999999999999999999999999999.............&#xA;41 13498129999999999999999999999999999999999............&#xA;46 1349812999999999999999999999999999999999999999.......&#xA;47 13498129999999999999999999999999999999999999999......&#xA;46 .......9999999999999999999999999999999999999999198271&#xA;45 ........999999999999999999999999999999999999999198271&#xA;44 .........99999999999999999999999999999999999999198271&#xA;37 ................9999999999999999999999999999999198271&#xA;34 ...................9999999999999999999999999999198271&#xA;31 ......................9999999999999999999999999198271&#xA;30 .......................999999999999999999999999198271&#xA;28 .........................9999999999999999999999198271&#xA;25 ............................9999999999999999999198271&#xA;16 .....................................9999999999198271&#xA;14 .......................................99999999198271&#xA;10 ...........................................9999198271&#xA; 8 .............................................99198271&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here is the &lt;code&gt;chain-primes.py&lt;/code&gt; code:&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from gmpy import next_prime, numdigits, is_prime&#xA;&#xA;def digitcount(b):&#xA;    nd = numdigits(b)           # May be off-by-one&#xA;    while 10**nd    &amp;lt;= b:  nd += 1&#xA;    while 10**(nd-1) &amp;gt; b:  nd -= 1&#xA;    return nd&#xA;&#xA;def findPre(pre, lim, d):&#xA;    b, s, p, preDig = pre, 0, [], digitcount(pre) &#xA;    fillLim = lim-preDig&#xA;    for i in range(fillLim):&#xA;        if is_prime(b):&#xA;            s += (preDig+i)**2&#xA;            p.append(i)&#xA;        b = 10*b + d&#xA;    return s, pre, p&#xA;&#xA;def findSuf(suf, lim, d):&#xA;    b, s, p, preDig = suf, 0, [], digitcount(suf)&#xA;    fillLim = lim-preDig&#xA;    k = d * (10**preDig)      # Eg if b is 37 and d is 7, now k=700.&#xA;    for i in range(fillLim):&#xA;        if is_prime(b):&#xA;            s += (preDig+i)**2&#xA;            p.append(i)&#xA;        b += k;  k *= 10;&#xA;    return s, suf, p&#xA;&#xA;def shoPre(preSet, lim, d):&#xA;    sco, pre, llist = preSet&#xA;    pdig = digitcount(pre)&#xA;    for l in llist:&#xA;        print '{:2} {:.&amp;lt;53}'.format(pdig+l, repr(pre) + repr(d)*l)&#xA;&#xA;def shoSuf(sufSet, lim, d):&#xA;    sco, suf, llist = sufSet&#xA;    sdig = digitcount(suf)&#xA;    for l in reversed(llist):&#xA;        print '{:2} {:.&amp;gt;53}'.format(sdig+l, repr(d)*l + repr(suf))&#xA;&#xA;def testpair(pre, suf, filld, plim=53, slim=53):&#xA;    preSet = findPre(pre, plim, filld)&#xA;    sufSet = findSuf(suf, slim, filld)&#xA;    print 'Scores  {} + {} = {}'.format(&#xA;        preSet[0], sufSet[0], preSet[0]+sufSet[0])&#xA;    shoPre(preSet, plim, filld)&#xA;    shoSuf(sufSet, slim, filld)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2004" LastActivityDate="2014-08-31T22:48:00.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2181" PostTypeId="2" ParentId="2179" CreationDate="2014-09-01T00:07:30.363" Score="1" Body="&lt;p&gt;That outer dial resembles the &lt;a href=&quot;https://en.wikipedia.org/wiki/Slide_rule#Circular_slide_rules&quot; rel=&quot;nofollow&quot;&gt;slide-rule&lt;/a&gt; dial found on some watches, for example on various &lt;a href=&quot;https://www.google.com/search?hl=en&amp;amp;site=imghp&amp;amp;tbm=isch&amp;amp;source=hp&amp;amp;biw=1067&amp;amp;bih=597&amp;amp;q=citizen+nighthawk+watch&amp;amp;oq=citizen+nighthawk+watch&amp;amp;gs_l=img.3..0j0i24l7.1180.5054.0.5445.7.5.0.2.2.0.140.401.4j1.5.0....0...1ac.1.52.img..0.7.409.30UYstlR9JY&quot; rel=&quot;nofollow&quot;&gt;Citizen Nighthawk&lt;/a&gt; watches.  Section H of the pdf manual  describes how to use the slide-rule dial to compute arrival times, fuel consumption, etc as well as how to multiply and divide.  You can download &lt;a href=&quot;http://www.manualslib.com/download/222837/Citizen-Nighthawk-Bj7000-52e.html&quot; rel=&quot;nofollow&quot;&gt;nighthawk_bj700052e.pdf&lt;/a&gt; via manualslib if interested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Nighthawk has a fixed nonlinear ring just inside the outer nonlinear ring, which is needed for simple slide-rule function.  I don't see a non-linear dial in your photo; in which case the ring might not implement a general slide rule, but instead is specialized for some single function, like ETA.&lt;/p&gt;&#xA;" OwnerUserId="2004" LastActivityDate="2014-09-01T00:07:30.363" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2182" PostTypeId="2" ParentId="2096" CreationDate="2014-09-01T06:17:04.870" Score="11" Body="&lt;p&gt;you can do it with the following sequence:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/BrJry.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2225" LastActivityDate="2014-09-01T06:17:04.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2183" PostTypeId="2" ParentId="2179" CreationDate="2014-09-01T13:12:13.100" Score="2" Body="&lt;p&gt;It is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Tachymeter_(watch)&quot; rel=&quot;nofollow&quot;&gt;tachymeter&lt;/a&gt;. (Once you know that, the big yellow T makes sense.) This a common enough thing on luxury watches - I didn't actually know this, but this morning I saw a glossy ad for one on the back of a magazine and recognized the scale. If you search tachymeter on Amazon or the like you can see any number of images.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The purpose of it is to determine speed. Going a certain distance in a given time means you are going a particular speed. You would need to learn what units your watch ring is in (km/hr or mph for example) and what the calibration distance is, but then you could say that going that far in about 10 seconds means you're going 30 somethings, or going that far in 20 seconds means you're going 15 somethings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More expensive watches run the scale to a higher number, presumably because people who buy that kind of watch are in faster cars, jets etc but this only means a longer calibration distance, no change in the capabilities of the watch.&lt;/p&gt;&#xA;" OwnerUserId="1509" LastEditorUserId="1509" LastEditDate="2014-09-01T13:17:59.673" LastActivityDate="2014-09-01T13:17:59.673" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2184" PostTypeId="1" AcceptedAnswerId="2185" CreationDate="2014-09-02T17:32:00.877" Score="-2" ViewCount="836" Body="&lt;p&gt;There are 10 Boxes, each contains 1000 sugar cubes. The sugar cubes in 9 of the boxes are 10 grams each and the other box has 9 gram cubes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a digital scale that you can use it ONCE. (i.e. put something on the scale, turn it on and read the number)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the box with the 9 gram cubes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://www.fekraneh.ir/tab-177/index.aspx&quot; rel=&quot;nofollow&quot;&gt;Fekraneh.ir&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="8874" LastEditDate="2015-01-31T12:29:27.677" LastActivityDate="2015-01-31T12:29:27.677" Title="Find the Different Sugar Cubes" Tags="|mathematics|logical-deduction|weighing|" AnswerCount="1" CommentCount="3" ClosedDate="2014-09-14T03:05:54.083" ContentLicense="CC BY-SA 3.0" />
  <row Id="2185" PostTypeId="2" ParentId="2184" CreationDate="2014-09-02T17:32:00.877" Score="1" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Set number from 1 to 10 to each box.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;pick cubes from from boxes with the equal amount of number of each&#xA;box, and put them on scale.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;read the last digit of shown number on the digital scale, and&#xA;subtract this amount from 10. (e.g. weight=543 grams =&amp;gt; 10 - 3 =&#xA;&lt;strong&gt;7&lt;/strong&gt; )&lt;/p&gt;&#xA;&lt;p&gt;The signle digit result will lead us to the 9 gram cubes box&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-09-02T17:32:00.877" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2186" PostTypeId="1" AcceptedAnswerId="2187" CreationDate="2014-09-02T17:45:25.043" Score="4" ViewCount="721" Body="&lt;p&gt;There are 8 marbles that one of them is lighter than the others, and all other seven marbles have the same weight.&#xA;There is a balance scale that you can use it twice or once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way to find the lightest marble? How? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;source: &lt;a href=&quot;http://www.fekraneh.ir/tab-177/index.aspx&quot; rel=&quot;nofollow&quot;&gt;Fekraneh.ir&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="8874" LastEditDate="2015-01-31T12:28:15.100" LastActivityDate="2015-01-31T12:28:15.100" Title="Which one is the lightest marble?" Tags="|logical-deduction|strategy|weighing|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2187" PostTypeId="2" ParentId="2186" CreationDate="2014-09-02T18:03:18.397" Score="6" Body="&lt;p&gt;Put three of the marbles in one bowl of the scale, and put three others in the other bowl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the scale is in balance the light marble must be one of the two remaining marbles, so weight those two against one another to find the light marble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the scale is not in balance then the light marble must be one of the three marbles from the lightest bowl. Weigh any two of them against one another in order to either find the light marble directly, or conclude that the third is the light marble in case the weighed marbles turn out to weigh the same.&lt;/p&gt;&#xA;" OwnerUserId="2186" LastActivityDate="2014-09-02T18:03:18.397" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2188" PostTypeId="1" AcceptedAnswerId="2190" CreationDate="2014-09-02T19:13:43.363" Score="130" ViewCount="1726736" Body="&lt;p&gt;You are a prisoner in a room with 2 doors and 2 guards. One of the doors will guide you to freedom and behind the other is a hangman–you don't know which is which, but the guards do know.&lt;/p&gt;&#xA;&lt;p&gt;One of the guards always tells the truth and the other always lies. You don't know which one is the truth-teller or the liar either. However both guards know each other.&lt;/p&gt;&#xA;&lt;p&gt;You have to choose and open one of these doors, but you can only ask a single question to one of the guards.&lt;/p&gt;&#xA;&lt;p&gt;What do you ask to find the door leading to freedom?&lt;/p&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="2225" LastEditDate="2023-09-04T06:02:25.290" LastActivityDate="2023-09-04T06:02:25.290" Title="Two doors with two guards - one lies, one tells the truth" Tags="|logical-deduction|liars|" AnswerCount="16" CommentCount="16" ContentLicense="CC BY-SA 4.0" />
  <row Id="2189" PostTypeId="2" ParentId="2188" CreationDate="2014-09-02T19:39:21.497" Score="54" Body="&lt;p&gt;Choose a guard and ask him, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;If I asked you 'are you standing in front of the freedom door?', would your reply be 'yes'?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If you choose the truthful guard, he will give you an honest answer. Enter his door if he says &quot;yes&quot; and enter the other door otherwise.&lt;/li&gt;&#xA;&lt;li&gt;If you choose the liar, he will lie about what his reply would be. Since that reply is also a lie, the two lies cancel out. Enter his door if he says &quot;yes&quot; and enter the other door otherwise.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1497" LastEditorUserId="17482" LastEditDate="2016-05-23T14:06:45.363" LastActivityDate="2016-05-23T14:06:45.363" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="2190" PostTypeId="2" ParentId="2188" CreationDate="2014-09-02T22:25:50.853" Score="119" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If I asked what door would lead to freedom, what door would the other guard point to?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you asked the truth-guard, the truth-guard would tell you that the liar-guard would point to the door that leads to death.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you asked the liar-guard, the liar-guard would tell you that the truth-guard would point to the door that leads to death. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, no matter who you ask, the guards tell you which door leads to death, and therefore you can pick the other door.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorUserId="-1" LastEditDate="2017-01-10T05:59:17.700" LastActivityDate="2017-01-10T05:59:17.700" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="2191" PostTypeId="1" AcceptedAnswerId="2195" CreationDate="2014-09-03T05:47:38.390" Score="16" ViewCount="4862" Body="&lt;p&gt;You have to stay for 7 days in a hotel outside your city. But you forgot to bring your credit card and are out of cash.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a silver chain with 7 links. The rule in this hotel is that residents should pay their rent every morning. The manager of the hotel is well aware of your situation and says that you can pay one link of your silver necklace every morning instead of money.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But he wants the chain in good condition. So he suggests a solution for your every morning payment with only ONE link cut in the chain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is his solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;source: &lt;a href=&quot;http://www.fekraneh.ir/tab-178/index.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;fekraneh.ir&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="32642" LastEditDate="2017-06-26T06:46:18.533" LastActivityDate="2017-06-26T06:46:18.533" Title="The seven piece silver chain" Tags="|logical-deduction|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2192" PostTypeId="1" AcceptedAnswerId="2193" CreationDate="2014-09-03T06:02:09.763" Score="7" ViewCount="43709" Body="&lt;p&gt;You have 12 liters of apple juice in a 12 liters bucket and you want to share it with your best friend. But you have an empty 8-liter and an empty 5-liter bottle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you split the apple juice in half?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://www.fekraneh.ir/tab-178/index.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;fekraneh.ir&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="17744" LastEditDate="2016-03-27T16:33:41.250" LastActivityDate="2023-08-18T13:11:12.143" Title="Split 12 liters in half with 5 and 8 liter bottles" Tags="|logical-deduction|decanting-problems|" AnswerCount="3" CommentCount="2" ClosedDate="2016-04-07T08:25:39.550" ContentLicense="CC BY-SA 3.0" />
  <row Id="2193" PostTypeId="2" ParentId="2192" CreationDate="2014-09-03T06:02:09.763" Score="7" Body="&lt;p&gt;Here is a possible solution:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;STEP        0   1   2   3   4   5   6   7&#xA;-------------------------------------------&#xA;5-Liter     0   0   5   0   3   3   5   0&#xA;8-Liter     0   8   3   3   0   8   6   6&#xA;12-Liter    12  4   4   9   9   1   1   6&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="11244" LastEditDate="2015-05-06T12:50:47.560" LastActivityDate="2015-05-06T12:50:47.560" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2194" PostTypeId="1" AcceptedAnswerId="2198" CreationDate="2014-09-03T06:44:25.040" Score="10" ViewCount="5120" Body="&lt;p&gt;Color each cell of the table with following rules:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Yellow and Orange cells are adjacent, but Red and Black are not.&lt;/li&gt;&#xA;&lt;li&gt;Cell with number 3 is White, Green, Blue or Red and the cell with number 5 is not White or Purple.&lt;/li&gt;&#xA;&lt;li&gt;Purple and Orange are in the same column, but Yellow and Black are not, just like Green and Blue.&lt;/li&gt;&#xA;&lt;li&gt;The numbers of White and Pink cells are consecutive and they are not adjacent (either orthogonally or diagonally).&lt;/li&gt;&#xA;&lt;li&gt;The difference of Red and Yellow is less than 3 and the difference between White and Blue is not greater than 3.&lt;/li&gt;&#xA;&lt;li&gt;Red and Blue has the difference greater than 2.&lt;/li&gt;&#xA;&lt;li&gt;Green is adjacent to Pink and is odd.&lt;/li&gt;&#xA;&lt;li&gt;Red and Yellow are not in the same column.&lt;/li&gt;&#xA;&lt;li&gt;The number in Blue cell is greater than the Green.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pSGkK.jpg&quot; alt=&quot;First row: 5, 3, 8. Second row: 6, 7, 9. Third row: 4, 1, 2&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://mehcom.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;mehcom.com&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="13484" LastEditDate="2022-01-28T12:15:25.683" LastActivityDate="2022-01-28T12:15:25.683" Title="Color the Table" Tags="|logical-deduction|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="2195" PostTypeId="2" ParentId="2191" CreationDate="2014-09-03T07:59:47.887" Score="21" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Cut the 3rd link.&lt;br&gt;&lt;br&gt;1. On day 1 you give him the cut link&lt;br&gt;2. On day 2 you give him 1-2, he returns the cut link&lt;br&gt;3. On day 3 you give the cut link, along with the 1-2 he has.&lt;br&gt;4. On day 4 you swap your 4-5-6-7 for the 1-2 and 3&lt;br&gt;5. On day 5 you give the cut link (so manager has 4-5-6-7 and the cut link 3)&lt;br&gt;6. On day 6 you swap 1-2 for cut link 3 (so manager has 1-2 and 4-5-6-7)&lt;br&gt;7. On day 7 you give the remaining cut link 3.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2156" LastEditorUserId="9180" LastEditDate="2015-02-05T21:42:38.180" LastActivityDate="2015-02-05T21:42:38.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2196" PostTypeId="2" ParentId="2192" CreationDate="2014-09-03T08:02:27.960" Score="8" Body="&lt;p&gt;Here is a general algorithm.  A and B are the 5 and 8 liter bottles.&#xA;The 12-liter bottle is the juice reserve.  The capacity is irrelevant.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A empty =&amp;gt; fill A from reserve&lt;/li&gt;&#xA;&lt;li&gt;B full =&amp;gt; empty B into reserve&lt;/li&gt;&#xA;&lt;li&gt;else =&amp;gt; transfer juice from A to B&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Depending on which bottle is A and which is B, you get a 7-step or a 10-step solution.  (Or one less step if you accept solutions where one part is split in 2 bottles)&lt;/p&gt;&#xA;&lt;p&gt;This, by the way, works for every 2-bucket problem.  If it has a solution.&lt;/p&gt;&#xA;&lt;p&gt;PS: I realized the capacity of the 12-bottle is not irrelevant since it is less than 8+5.  It can be used for a 10-move solution.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2023-08-18T13:11:12.143" LastActivityDate="2023-08-18T13:11:12.143" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="2197" PostTypeId="2" ParentId="313" CreationDate="2014-09-03T09:36:01.113" Score="19" Body="&lt;p&gt;Here's my simple answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What word has that many letters?  Please answer with yes or no.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - Yes -&gt; 3&lt;br&gt;&#xA; - No -&gt; 2&lt;br&gt;&#xA; - Er... I can't. -&gt; 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2018-09-21T11:28:34.277" LastActivityDate="2018-09-21T11:28:34.277" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="2198" PostTypeId="2" ParentId="2194" CreationDate="2014-09-03T15:46:09.667" Score="6" Body="&lt;p&gt;I have found 6 potential solutions which do not seem to contradict any rules.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1       2       3       4       5       6      7        8       9&#xA;green   orange  white   pink    blue    black   yellow  red     purple&#xA;green   orange  white   pink    blue    black   yellow  purple  red&#xA;green   orange  white   pink    black   blue    yellow  purple  red&#xA;green   purple  white   pink    blue    black   yellow  red     orange&#xA;green   purple  white   pink    blue    black   yellow  orange  red&#xA;green   purple  white   pink    black   blue    yellow  orange  red&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Please let me know if I have misunderstood any rules.  If not I will explain roughly how I got the answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Noting that I first assumed that 7 must be orange or purple.  A contradiction arises depending on the order of elimination of via the rules.  Basically this assumption is false so I proceeded with that knowledge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any color which is not adjacent to something (aka red, black, white, or pink) cannot be in 7 either as 7 is adjacent to everything.  Rule 4 leads to white being in 3 or 4.  So by rule 5 and 9, blue cannot be in 8 or 9 and green cannot be in 7 or 9. Working with pink/green adjacency meant green is not in 5 either.  Green is therefore in the middle column and (by rule 3) blue is not.  Yellow is the only option for the central cell 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This quickly lead to concluding that only green could be in 1 which eliminated w=4 so w=3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ugly options I had left where how to place red, black, orange, purple, blue, and pink in the left and right columns.  Orange and purple must be in the same column while red and black must be different.  Blue and pink must therefore be together.  Eliminating blue=2 which requires pink=2 means blue and pink are in the left and orange and purple are in the right columns.  The remaining solutions were trimmed to make sure all rules are met.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This assumes that diagonals can be adjacent.  There could potentially be other solutions if diagonals are not adjacent. Of these, only the fourth option is a possible solution if &#xA;diagonals are not adjacent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The options below were found for the case where diagonals are not adjacent.  This seems to mean that for either interpretation there are multiple solutions.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1       2       3       4       5       6       7       8       9&#xA;yellow  red     green   orange  black   purple  white   pink    blue&#xA;yellow  red     green   orange  black   purple  pink    white   blue&#xA;green   black   red     purple  yellow  orange  pink    white   blue&#xA;green   purple  white   pink    blue    black   yellow  red     orange&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="99" LastEditorUserId="11297" LastEditDate="2015-07-10T23:56:10.767" LastActivityDate="2015-07-10T23:56:10.767" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2201" PostTypeId="1" AcceptedAnswerId="2203" CreationDate="2014-09-04T04:48:03.970" Score="2" ViewCount="2123" Body="&lt;p&gt;Find the value of 'a' and 'b' in the following series.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;100,101,120,103,410,321,205,404,60a,5b2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2251" LastEditorUserId="4050" LastEditDate="2015-01-03T17:50:01.897" LastActivityDate="2015-01-03T17:50:01.897" Title="Find the next number in the series?" Tags="|pattern|number-sequence|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2202" PostTypeId="2" ParentId="2194" CreationDate="2014-09-04T04:55:05.590" Score="3" Body="&lt;p&gt;This answer produces the same four solutions mentioned at the end of kaine's answer, although the solution technique I used is a little less subtle, and more along the lines of brute force.  It illustrates a technique that's ok for small problems like this, but probably not for larger problems.  It consists of generating and testing all permutations of cell colors, and reporting those that satisfy all tests.  It takes about 0.2 seconds to complete.  Here is what the program output looks like:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;yellow red green orange black purple white pink blue&#xA;yellow red green orange black purple pink white blue&#xA;green purple white pink blue black yellow red orange&#xA;green black red purple yellow orange pink white blue&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit:&lt;/em&gt; In light of Rafe's comment that adjacency includes diagonal neighbors [eg, all other cells are adjacent to the center cell], in &lt;code&gt;nextTo&lt;/code&gt; I replaced two lines.  I also corrected two program errors and have added some program explanation.  Here's the current program's output (which matches kaine's first six solutions):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;green purple white pink black blue yellow orange red&#xA;green orange white pink black blue yellow purple red&#xA;green purple white pink blue black yellow orange red&#xA;green purple white pink blue black yellow red orange&#xA;green orange white pink blue black yellow purple red&#xA;green orange white pink blue black yellow red purple&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Following is the python 2.7 program that produced the output.  I used the letters B, G, K, L, O, P, R, W, Y to stand for the nine colors Black, Green, pinK, bLue, Orange, Purple, Red, White, Yellow.  The program uses cell identification numbers 0,1,2 for cells in the top row; 6,7,8 for the middle row; and 12,13,14 for the bottom row.  Array &lt;code&gt;cellNums&lt;/code&gt; embodies this map, which is such that two cells in the same column have identification numbers that differ by $\pm6$ or $\pm12$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For adjacency, two cells adjacent in a row differ by $\pm1$; in a column, by $\pm6$; in a forward diagonal, by $\pm5$; in a back diagonal, by $\pm7$.  Thus, cells &lt;code&gt;x&lt;/code&gt; and &lt;code&gt;y&lt;/code&gt; are &lt;a href=&quot;http://en.wikipedia.org/wiki/Taxicab_geometry&quot; rel=&quot;nofollow&quot;&gt;Manhattan&lt;/a&gt; adjacent, or are in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Von_Neumann_neighborhood&quot; rel=&quot;nofollow&quot;&gt;Von Neumann neighborhood&lt;/a&gt;, if $|x-y| \in \{1, 6\}$; and are in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Moore_neighborhood&quot; rel=&quot;nofollow&quot;&gt;Moore neighborhood&lt;/a&gt; if  $|x-y| \in \{1, 6, 5, 7\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Array entry &lt;code&gt;digCells[i]&lt;/code&gt; tells the cell number for digit &lt;code&gt;i+1&lt;/code&gt;. Array entry &lt;code&gt;number[j]&lt;/code&gt; tells what digit is in cell &lt;code&gt;j&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;# Ref http://puzzling.stackexchange.com/questions/2194/color-the-table&#xA;from itertools import permutations, ifilter&#xA;#-----------------------------------------------------------&#xA;def inColumn(x,y):   # Return true if x,y are in same column.&#xA;    dxy = abs(x-y)&#xA;    return dxy == 6 or dxy == 12&#xA;#-----------------------------------------------------------&#xA;def nextTo(x,y):     # Return true if x,y are adjacent.&#xA;    #return abs(x-y) in (1, 6)       # von Neumann&#xA;    return abs(x-y) in (1, 6, 5, 7)  # Moore&#xA;#-----------------------------------------------------------&#xA;# Test if permutation p satisfies all conditions.&#xA;def isGood(p):&#xA;    B, G, K, L, O, P, R, W, Y = p&#xA;    # Tests 1 and 2.&#xA;    # Note, digit 3 is in cell 1 and digit 5 is in cell 0&#xA;    # 1. nextTo(Y,O); not nextTo(R, B)&#xA;    # 2. color(number(3)) in {W,G,L,R}; color(number(5)) not in {W,P};&#xA;    if nextTo(R, B) or not nextTo(Y,O) or \&#xA;            1 not in (W,G,L,R) or 0 in (W,P):&#xA;        return False&#xA;    # Tests 3 and 4.&#xA;    # 3. inColumn(P,O); not inColumn(Y,B); not inColumn(G,L)&#xA;    # 4. abs(number(W)-number(K)) == 1; not nextTo(W,K)&#xA;    if not inColumn(P,O) or inColumn(Y,B) or inColumn(G,L) or \&#xA;            abs(number[W]-number[K]) != 1 or nextTo(W,K):&#xA;        return False&#xA;    # Tests 5 and 6.&#xA;    # 5. abs(number(R)-number(Y)) &amp;lt; 3; abs(number(W)-number(L)) &amp;lt; 4&#xA;    # 6. abs(number(R)-number(L)) &amp;gt; 2&#xA;    if abs(number[R]-number[Y]) &amp;gt; 2 or abs(number[W]-number[L]) &amp;gt; 3 \&#xA;            or abs(number[R]-number[L]) &amp;lt; 3:&#xA;        return False&#xA;    # Tests 7, 8, 9.&#xA;    # 7. nextTo(G,K);  odd(number(G))&#xA;    # 8. not inColumn(R,Y)&#xA;    # 9. number(L) &amp;gt; number(G)&#xA;    if not nextTo(G,K) or not number[G]&amp;amp;1 or \&#xA;            inColumn(R,Y) or \&#xA;            number[L] &amp;lt;= number[G]:&#xA;        return False&#xA;    return True&#xA;#---------------------------Main program--------------------&#xA;cellNums = [0, 1, 2, 6, 7, 8, 12, 13, 14]&#xA;digCells = [13, 14, 1, 12, 0, 6, 7, 2, 8]&#xA;number = [5,3,8,0,0,0,  6,7,9,0,0,0,  4,1,2,0,0,0]&#xA;colors = ['black', 'green', 'pink', 'blue', 'orange', 'purple', 'red', 'white', 'yellow']&#xA;&#xA;for p in ifilter(isGood, permutations(cellNums)):&#xA;    for cn in digCells:&#xA;        for j, c in enumerate(p):&#xA;            if c==cn: print colors[j],&#xA;    print&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2004" LastEditorUserId="2004" LastEditDate="2014-09-04T14:38:10.870" LastActivityDate="2014-09-04T14:38:10.870" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2203" PostTypeId="2" ParentId="2201" CreationDate="2014-09-04T05:31:21.707" Score="6" Body="&lt;p&gt;Sum of all digits in every number is growing.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Number     digits SUM&#xA;---------------------&#xA;100        1&#xA;101        2&#xA;120        3&#xA;103        4&#xA;410        5&#xA;321        6&#xA;205        7&#xA;404        8&#xA;60a        it should be 9       so &quot;a&quot; = 9-6 = 3&#xA;5b2        it should be 10      so &quot;b&quot; = 10-5-2 = 3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2225" LastActivityDate="2014-09-04T05:31:21.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2204" PostTypeId="1" AcceptedAnswerId="2206" CreationDate="2014-09-04T05:57:05.823" Score="4" ViewCount="2093" Body="&lt;p&gt;As everyone knows, in Sudoku the sum of each row and column is 45. So all Sudoku solutions are some kind of magic square. My question is:&lt;/p&gt;&#xA;&lt;p&gt;Has anyone seen a Sudoku puzzle combined with the magic square rule for all nine 3x3 squares? (even without diagonal sum)&lt;/p&gt;&#xA;&lt;p&gt;Is it even possible to have such a puzzle?&lt;/p&gt;&#xA;&lt;p&gt;edit: The same question about 4x4 squares.&lt;/p&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="69582" LastEditDate="2021-05-11T20:01:15.303" LastActivityDate="2021-05-11T21:01:15.550" Title="Are there any sudoku puzzles combined with magic squares?" Tags="|logical-deduction|grid-deduction|sudoku|magic-square|" AnswerCount="2" CommentCount="0" ClosedDate="2021-05-12T11:59:30.117" ContentLicense="CC BY-SA 4.0" />
  <row Id="2205" PostTypeId="1" AcceptedAnswerId="2207" CreationDate="2014-09-04T06:36:00.247" Score="5" ViewCount="560" Body="&lt;p&gt;How to control the whole board with 8 standard pieces (one king, one queen, two rooks, two knights, two bishops)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. A piece doesn't control a cell simply by staying on it, so another piece must control it.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2014-09-05T06:28:07.160" LastActivityDate="2023-12-16T22:55:10.643" Title="How to control the whole board with 8 standard pieces?" Tags="|chess|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2206" PostTypeId="2" ParentId="2204" CreationDate="2014-09-04T06:58:24.263" Score="14" Body="&lt;p&gt;It is not possible, for the simple reason all 3x3 magic squares have the 5 in the center spot of the 3x3 block. Therefor you'll always get 3 rows and columns in the 9x9 that hold 3 5's, rendering the sudoku part impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference on the possible 3x3's:&#xA;&lt;a href=&quot;http://www.dr-mikes-math-games-for-kids.com/3x3-magic-square.html&quot; rel=&quot;noreferrer&quot;&gt;Dr Mikes math games for kids&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: to add to the answer, here's a possible solution for 4x4's:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/6C52f.png&quot; alt=&quot;4x4 solution&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice how I start in the upper left, I fill the top row by putting 4x4 blocks of which the rows are permutated. From there downwards, I build new 4x4 blocks by permutation columns in the 4x4 blocks from the top row. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I can see all diagonals within the seperate 4x4's work aswell.&lt;/p&gt;&#xA;" OwnerUserId="2156" LastEditorUserId="2156" LastEditDate="2014-09-04T20:53:04.747" LastActivityDate="2014-09-04T20:53:04.747" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2207" PostTypeId="2" ParentId="2205" CreationDate="2014-09-04T07:53:24.440" Score="7" Body="&lt;p&gt;It's a studied problem: &lt;a href=&quot;http://comjnl.oxfordjournals.org/content/32/6/567.full.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Oxford journal page&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The three possible solutions are listed at the top of page 4.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/G43nO.png&quot; alt=&quot;Three different chess diagrams where the 8 officers cover the whole board&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Also noteworthy: there is no option with the bishops on opposing colors.&lt;/p&gt;&#xA;" OwnerUserId="2156" LastEditorUserId="82468" LastEditDate="2023-12-16T22:55:10.643" LastActivityDate="2023-12-16T22:55:10.643" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="2208" PostTypeId="2" ParentId="2153" CreationDate="2014-09-05T01:24:10.730" Score="1" Body="&lt;p&gt;I think it is in $k_1, k_2, ... , k_{n-1}$ interest to each shoot $k_n$. If they all miss, it is $k_n$ interest to shoot the $n-1$ st shooter. This strategy may require collusion, and perfect information. When two players are left, it is still a valid strategy. &lt;/p&gt;&#xA;" OwnerUserId="2254" LastEditorUserId="2254" LastEditDate="2014-09-05T05:59:41.800" LastActivityDate="2014-09-05T05:59:41.800" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2210" PostTypeId="1" AcceptedAnswerId="2213" CreationDate="2014-09-05T15:47:33.863" Score="15" ViewCount="1933" Body="&lt;p&gt;James has a defective clock in his room. The clock is digital, and some of the seven bars on its units digit are broken. A bar that works is on when it should be and off when it should be, whereas a broken bar is off regardless of the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that in each scenario, unless otherwise stated, James knows there are broken bars, but doesn't know how many or where they are.  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;James enters his room and glances at his clock. He knows how many bars are broken but he doesn't know which ones they are. He immediately knows for sure what the time is.&lt;/li&gt;&#xA;&lt;li&gt;James enters his room and glances at his clock. He knows which bars are broken and immediately knows for sure what the time is.&lt;/li&gt;&#xA;&lt;li&gt;James enters his room and knows which bars are broken. He stares at his clock for 60 seconds. Now he knows for sure what time it is.&lt;/li&gt;&#xA;&lt;li&gt;James enters his room. He knows how many bars are broken but he doesn't know which ones they are. He stares at his clock for 60 seconds. Now he knows for sure what time it is.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In each of these four scenarios, what is the maximum number of broken bars, and which might it/they be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3TYjs.png&quot; alt=&quot;Image shamelessly stolen from Hubble07.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of scenario 1 might be 3.  If James walked in, and saw a time whose units digit were 8, and bars A, G, and D were burnt out, him knowing that 3 bars were burnt out and not knowing where, he's still be able to deduce that the number is 8.  However, this may not be the optimal answer for scenario 1.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There seems to be some confusion about what is meant here, and I hope that clarifies something.&lt;/p&gt;&#xA;" OwnerUserId="1672" LastEditorUserId="5373" LastEditDate="2018-05-09T14:37:10.813" LastActivityDate="2021-02-09T16:01:04.487" Title="The defective clock" Tags="|logical-deduction|seven-segment|" AnswerCount="3" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="2211" PostTypeId="2" ParentId="2210" CreationDate="2014-09-05T20:47:12.663" Score="1" Body="&lt;p&gt;I assume that this does not need to work for any time and an AM/PM clock works fine.  These all work for 24 hour clocks too.  It does not work if there are leading zeros.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we need to distinguish an 8 from anything else without waiting around, 4 lit lights are necessary.  This is because the central light must be lit or it could be a 0. The upper right must be lit or it could be a 6.  This leaves 2, 3, 4, and 9.  The lower left being lit eliminates 3, 4, and 9 letting upper left or lower right being lit to eliminate 2.  This is useful for the right most digit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 can be identified from any other number less than 6 with 2 lit lights: the central light and the lower left light. This is useful for the tens digit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 lit lights can distinguish 12 from any other options for the first 2 digits.  These would be the lower right bar on the 1 and the central bar on the 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These altogether means that at exactly 12:28 James can tell what time it is with 20 broken bars.  These seems to be the only situation where that works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Knowing how many lights are broken cannot help him any more than this.  There still have to be 8 lights illuminated to differentiate between the digits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) If he knows which ones are broken, he can use them to identify numbers by unlit bars.  This cannot help for the first two digits so the time still starts with 12 and 12 of those 14 bars are broken.  For the other two places, knowing that the lower right bar is unbroken but unlit indicates that it is a 2.  This means that at 12:22, he can tell what time it is with 24 broken bars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3/4) For the finale:  At the change between 9:59 to 10:00 with a particular arrangement of broken bars, he can tell exactly what time it is with only 1 unbroken bar (27 broken ones) even if he did not know what they are. If the upper right bar of the left most digit changes from being unlit to lit within the 60 seconds, the time changed from 9:59 to 10:00.  This is the only time that change can happen for either 12 hour or 24 hour clocks. &lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="11297" LastEditDate="2015-07-10T23:53:51.620" LastActivityDate="2015-07-10T23:53:51.620" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2212" PostTypeId="1" AcceptedAnswerId="2219" CreationDate="2014-09-05T20:56:04.163" Score="5" ViewCount="868" Body="&lt;p&gt;fill the matrix correctly and the guess the Letter and number of 5th matrix:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8OiMf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I'm sure you won't need options!&lt;/p&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="2225" LastEditDate="2014-10-16T15:09:17.757" LastActivityDate="2014-10-16T15:09:17.757" Title="Find the pattern of matrices" Tags="|logical-deduction|pattern|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2213" PostTypeId="2" ParentId="2210" CreationDate="2014-09-05T21:16:01.683" Score="10" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;You said - &lt;strong&gt;&amp;quot;The clock is digital, and some of the seven bars on its units digit are broken.&amp;quot;&lt;/strong&gt;&#xA;and then at the end you say - &lt;strong&gt;&amp;quot;In each of these four scenarios, what is the maximum number of broken bars, and which might it/they be?&amp;quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Firstly this should mean that the answers to all the four questions should be an integer in the interval [1,7] since at the most it could be that all the seven bars are broken.&lt;/p&gt;&#xA;&lt;p&gt;If this is indeed the case then we should only be worried about the single seven segment display which shows the unit place of the two digit minute.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hmxxz.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;I have tried to do this for the general case i.e. for any time on the digital clock&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;For Case 1&lt;/strong&gt; &lt;br/&gt;&#xA;&lt;em&gt;Edit- As pointed out by @justhalf it looks like for this case there is no solution for which James would be able to figure any time so the strategy given below works for all such times that don't have a 2 or 3 as the last digit.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;Since he is not aware which segments are broken the only way he could correctly identify all the digits from 0-9 (except 2 &amp;amp; 3) is if only one segment is broken and that segment is C.&#xA;For e.g when it is 1-minute only B would be lit and since he knows that one segment is broken then he would conclude that that one segment has to C because if it were any other then it wouldn't form any know digit from 0-9 (except 2 &amp;amp; 3). Similar arguments hold for any minute from 0 to 9 (except 2 &amp;amp; 3).&#xA;So if only C is broken then James could figure out all time with the exceptions mentioned above.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;For Case 2&lt;/strong&gt; &lt;br/&gt;&#xA;In this case if he knows that two segments (C &amp;amp; D) are broken then again he can correctly predict all the digits from 0-9.So now the maximum broken segments is 2 i.e C and D.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;For Case 3&lt;/strong&gt; &lt;br/&gt;&#xA;Here if James knew that there are exactly &lt;strong&gt;four broken segments and they are A,C,D and E&lt;/strong&gt; then he will be able to correctly say the time after 1-minute.&#xA;There are 10 transitions of state possible in 1-minute which differ by a minute like for e.g 0 minute becomes 1 minute (0→1) or 1→2 and so on. For each of these transitions there is a unique combination of LEDs that will(or will not) change state.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;For e,g with A,C,D and E broken lets say 0→1 happens in 1-minute then James would see:-&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BlIHN.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;In the above transition we can see that in 1-minute the LED F changes state (goes from ON state to OFF state) whereas B remains unchanged. So seeing this he can conclude that 0→1 has happened and hence the current time is &lt;strong&gt;HH:M&lt;sub&gt;1&lt;/sub&gt;M&lt;sub&gt;2&lt;/sub&gt;&lt;/strong&gt; where M&lt;sub&gt;2&lt;/sub&gt; is found to be 1. Similar argument holds for all other transitions which is summarized in the table below.&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;Possible Transitions&lt;br&gt;(Initial minute → final minute)&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;LED's that change in one minute&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;LED's remaining unchanged in one minute&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0→1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;F&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1→2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;G&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2→3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;NONE&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;BG&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3→4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;F&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;BG&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4→5&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;GF&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;5→6&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;NONE&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;GF&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;6→7&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;BGF&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;NONE&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;7→8&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;GF&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;8→9&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;NONE&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;BGF&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;9→0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;G&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;BF&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;So this way James could tell any time displayed on the digital clock. So the maximum broken segments with which he can do this is 4 and they are A,C,D and E.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="101" LastEditorUserId="69582" LastEditDate="2021-02-09T16:01:04.487" LastActivityDate="2021-02-09T16:01:04.487" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="2214" PostTypeId="1" AcceptedAnswerId="2215" CreationDate="2014-09-06T05:28:30.100" Score="1" ViewCount="2129" Body="&lt;p&gt;Four students P, Q, R and S each working under the super vision of one of the four professors A, B, C and D made their final year MBA Project Presentations one by one, one each in the areas of Finance, Marketing, Systems and Human Resource Management (HRM). Each professor is an expert in only one of the above areas and supervised exactly one of the above students in his own area. The following are the clues:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(i) First presentation was made by R.&#xA;(ii) Prof. B works in Finance.&#xA;(iii) Prof. D was P’s supervisor.&#xA;(iv) The last presentation was in the Systems area.&#xA;(v) S’s project was in the HRM area.&#xA;(vi) Prof. B’s student’s presentation followed that of Prof. C’s student.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Find:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Order of Presentation.&#xA;Each students Subjects.&#xA;Mentor of each student.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2251" LastActivityDate="2014-09-06T07:13:48.730" Title="Find the order of students presentation(logical puzzle)" Tags="|logical-deduction|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2215" PostTypeId="2" ParentId="2214" CreationDate="2014-09-06T06:25:56.750" Score="2" Body="&lt;p&gt;Here is my answer: (there is another correct answer)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;     1          2          3          4&#xA;---------------------------------------------&#xA;     R          S          Q          P&#xA;     A          C          B          D&#xA;     Marketing  HRM        Finance    Systems&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Edit: explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;question information:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/t25FJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B and finance are connected, so the only student that can pick both of them is Q:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/0nHnH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so, R cannot pick Finance, also R is 1st and Systems is last therefore R cannot pick Systems as well as HRM. &#xA;&lt;img src=&quot;https://i.stack.imgur.com/qFuKA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence we can see that the only option for P is Systems:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/q6KqB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As B is exactly after C, Here We have two possible answers: &#xA;&lt;img src=&quot;https://i.stack.imgur.com/blMqd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;     1          2          3          4&#xA;---------------------------------------------&#xA;     R          S          Q          P&#xA;     *A*       *C*        *B*         D&#xA;     Marketing  HRM        Finance    Systems&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;OR&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Tme7j.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;     1          2          3          4&#xA;---------------------------------------------&#xA;     R          Q          S          P&#xA;    *C*        *B*        *A*         D&#xA;     Marketing  Finance    HRM        Systems&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="2225" LastEditDate="2014-09-06T07:13:48.730" LastActivityDate="2014-09-06T07:13:48.730" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2217" PostTypeId="2" ParentId="2214" CreationDate="2014-09-06T07:05:52.933" Score="0" Body="&lt;p&gt;Let's start with a grid&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;N -  Prof  -  Stud  -  Area&#xA;1 - [ABCD] - [PQRS] - [FHMS]&#xA;2 - [ABCD] - [PQRS] - [FHMS]&#xA;3 - [ABCD] - [PQRS] - [FHMS]&#xA;4 - [ABCD] - [PQRS] - [FHMS]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Apply rules 1 (R first) and 4 (Systems last), they're quite straightforward.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 - [ABCD] - [  R ] - [FHM ]&#xA;2 - [ABCD] - [PQ S] - [FHM ]&#xA;3 - [ABCD] - [PQ S] - [FHM ]&#xA;4 - [ABCD] - [PQ S] - [   S]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Since B works in finance (2), his student can't go last.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 - [ABCD] - [  R ] - [FHM ]&#xA;2 - [ABCD] - [PQ S] - [FHM ]&#xA;3 - [ABCD] - [PQ S] - [FHM ]&#xA;4 - [A CD] - [PQ S] - [   S]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Student S is in HRM (5):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 - [ABCD] - [  R ] - [F M ]&#xA;2 - [ABCD] - [PQ S] - [FHM ]&#xA;3 - [ABCD] - [PQ S] - [FHM ]&#xA;4 - [A CD] - [PQ  ] - [   S]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;B is immediately after C (6) - B can't be first, C can't be third or fourth.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 - [A CD] - [  R ] - [F M ]&#xA;2 - [ABCD] - [PQ S] - [FHM ]&#xA;3 - [AB D] - [PQ S] - [FHM ]&#xA;4 - [A  D] - [PQ  ] - [   S]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Remember that B is in finance, leaving 1 = R = Marketing.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 - [A CD] - [  R ] - [  M ]&#xA;2 - [ABCD] - [PQ S] - [FH  ]&#xA;3 - [AB D] - [PQ S] - [FH  ]&#xA;4 - [A  D] - [PQ  ] - [   S]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Since D is P's professor, he can't be R's.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 - [A C ] - [  R ] - [  M ]&#xA;2 - [ABCD] - [PQ S] - [FH  ]&#xA;3 - [AB D] - [PQ S] - [FH  ]&#xA;4 - [A  D] - [PQ  ] - [   S]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now, let's try trial and error quickly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume 1 = A&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 - [A   ] - [  R ] - [  M ]&#xA;2 - [ BCD] - [PQ S] - [FH  ]&#xA;3 - [ B D] - [PQ S] - [FH  ]&#xA;4 - [   D] - [PQ  ] - [   S]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;C can then only be 2, which leaves B 3 and D 4.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 - [A   ] - [  R ] - [  M ]&#xA;2 - [  C ] - [PQ S] - [FH  ]&#xA;3 - [ B  ] - [PQ S] - [FH  ]&#xA;4 - [   D] - [PQ  ] - [   S]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;D is with P; B works in Finance; S is in HRM.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Solution 1&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   1 - [A   ] - [  R ] - [  M ]&#xA;   2 - [  C ] - [   S] - [ H  ]&#xA;   3 - [ B  ] - [ Q  ] - [F   ]&#xA;   4 - [   D] - [P   ] - [   S]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Aha, a solution!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's see if 1 = C works:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume 1 = C&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 - [  C ] - [  R ] - [  M ]&#xA;2 - [AB D] - [PQ S] - [FH  ]&#xA;3 - [AB D] - [PQ S] - [FH  ]&#xA;4 - [A  D] - [PQ  ] - [   S]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;B is immediately after. Remember he's in finance.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 - [  C ] - [  R ] - [  M ]&#xA;2 - [ B  ] - [PQ S] - [F   ]&#xA;3 - [A  D] - [PQ S] - [ H  ]&#xA;4 - [A  D] - [PQ  ] - [   S]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;S is in HRM&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 - [  C ] - [  R ] - [  M ]&#xA;2 - [ B  ] - [ Q  ] - [F   ]&#xA;3 - [A  D] - [   S] - [ H  ]&#xA;4 - [A  D] - [PQ  ] - [   S]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;D is with P&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Solution 2&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   1 - [  C ] - [  R ] - [  M ]&#xA;   2 - [ B  ] - [ Q  ] - [F   ]&#xA;   3 - [A   ] - [   S] - [ H  ]&#xA;   4 - [   D] - [P   ] - [   S]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, that's another solution fully consistent with the rules. Turns out this puzzle has two correct answers.&lt;/p&gt;&#xA;" OwnerUserId="4" LastActivityDate="2014-09-06T07:05:52.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2219" PostTypeId="2" ParentId="2212" CreationDate="2014-09-06T09:38:58.217" Score="6" Body="&lt;pre&gt;&lt;code&gt;| |X| | | |&#xA;|X| | |X| |&#xA;| |X| | | |&#xA;| | |X| | |&#xA;| | |X| | |&#xA;  O + 17&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The first piece moves to the right, the second moves down, the third moves left, the fourth moves up, and so the fifth probably moves right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The letters and numbers are the sum of X and Y coordinates respectively, with the upper left corner being coordinate $(1,1)$ or $(A,1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each image introduce its new element at the position given by the coordinate sum of the previous image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both movement and new positions wrap around, so when a piece would move over the border it appears next to the opposite border, similarly a piece that appear at for instance $(H,11)=(8,11)$, will wrap around to $(8\mod5,11\mod5)=(3,1)=(C,1)$.&lt;/p&gt;&#xA;" OwnerUserId="2186" LastActivityDate="2014-09-06T09:38:58.217" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2224" PostTypeId="1" AcceptedAnswerId="2225" CreationDate="2014-09-06T22:07:46.403" Score="9" ViewCount="1593" Body="&lt;p&gt;The anti-intellectuals have won the war, and under the iron rule of emperor Nonotin anyone who has been deemed too smart for their own good has been brought to labour camps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But most of the science folks turn out to have never done a fair day's work in their life, so they are worth little in the labour camps. In order to get rid of some of them, and to have a little laugh, Nonotin devises an impossible game based on an equally impossible heretic tale. Though he doesn't care much for the exact details of the tale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He places 1 cat in each of 20 lidded boxes, and asks one of his guards to kill a random number of them. He then gathers 100 of the most useless physicists, explain this setting to them, and states the rules of the game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The physicists will, one at a time, be taken to the room with the boxes, look into 19 of them, and then guess if the cat in the last one is living or dead. The boxes will be moved around between each visit, so the physicists will not be able to coordinate their choice of box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they all guess correctly they will be given their freedom, but if anyone guesses wrong they will all be executed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The physicists only have a short while to agree on a strategy before being taken to isolation cells. They don't think it likely that they will reliably be able to hear the living cats, or otherwise identify the content of the boxes without looking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The emperor is certain of their defeat. But is there truly nothing the physicists can do to have a fair chance of winning the game?&lt;/p&gt;&#xA;" OwnerUserId="2186" LastEditorUserId="2186" LastEditDate="2014-09-09T16:04:11.937" LastActivityDate="2014-10-20T17:06:12.797" Title="Entangled physicists" Tags="|logical-deduction|mathematics|strategy|probability|" AnswerCount="2" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="2225" PostTypeId="2" ParentId="2224" CreationDate="2014-09-07T00:23:05.800" Score="14" Body="&lt;p&gt;Assuming that &quot;kill a random number of kittens&quot; means a random integer from 0 to 20, then there is a 11/21 chance that an even number of kittens have been killed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The physicians should all agree to guess that the last kitten is dead if an odd number of kittens are dead, or alive if an even number of kittens are dead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's an approximately 52% chance of staying alive, but a less than 5% chance that no kittens were harmed in the making of this puzzle.&lt;/p&gt;&#xA;" OwnerUserId="375" LastEditorUserId="375" LastEditDate="2014-09-09T12:50:19.353" LastActivityDate="2014-09-09T12:50:19.353" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="2227" PostTypeId="1" CreationDate="2014-09-07T23:06:38.117" Score="-1" ViewCount="527" Body="&lt;p&gt;Let &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; be some number.  Let &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; be &lt;a href=&quot;http://en.wikipedia.org/wiki/Semiprime&quot; rel=&quot;nofollow noreferrer&quot;&gt;semiprimes&lt;/a&gt; (ie, each is a product of two prime numbers) such that &lt;span class=&quot;math-container&quot;&gt;$R \le N \le S$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For quite a few numbers &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; we can find &lt;span class=&quot;math-container&quot;&gt;$R, S$&lt;/span&gt; such that  &lt;span class=&quot;math-container&quot;&gt;$S-R$&lt;/span&gt; is small; eg, for &lt;span class=&quot;math-container&quot;&gt;$N=50$&lt;/span&gt;, we have &lt;span class=&quot;math-container&quot;&gt;$7\cdot7 = 49 &amp;lt; 50 &amp;lt; 51 = 3\cdot17$&lt;/span&gt;, giving &lt;span class=&quot;math-container&quot;&gt;$S-R=2$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &lt;span class=&quot;math-container&quot;&gt;$N=10240579048185725616$&lt;/span&gt; and certain other large numbers too numerous to mention we can quickly (via &lt;a href=&quot;http://en.wikipedia.org/wiki/Prime_quadruplet&quot; rel=&quot;nofollow noreferrer&quot;&gt;prime quadruplets&lt;/a&gt;) find semiprimes &lt;span class=&quot;math-container&quot;&gt;$R, S$&lt;/span&gt; such that  &lt;span class=&quot;math-container&quot;&gt;$S-R=12$&lt;/span&gt;.  But what about products of three primes?  How small can the interval be made?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The challenge in this puzzle is to find &lt;span class=&quot;math-container&quot;&gt;$U_k, V_k$&lt;/span&gt;, each the product of three primes, such that  &lt;span class=&quot;math-container&quot;&gt;$U_k \le N_k \le V_k$&lt;/span&gt;, given that &lt;span class=&quot;math-container&quot;&gt;$N_k$&lt;/span&gt; is the &lt;a href=&quot;http://en.wikipedia.org/wiki/RSA_numbers&quot; rel=&quot;nofollow noreferrer&quot;&gt;RSA number&lt;/a&gt; with &lt;em&gt;k&lt;/em&gt; decimal digits, while minimizing  &lt;span class=&quot;math-container&quot;&gt;$\Delta_k = V_k-U_k$&lt;/span&gt;.  Find best solutions for &lt;span class=&quot;math-container&quot;&gt;$k\in\{100, 160, 320, 480\}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, &lt;a href=&quot;http://en.wikipedia.org/wiki/RSA_numbers#RSA-100&quot; rel=&quot;nofollow noreferrer&quot;&gt;RSA-100&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$= 15226050279225333605356183781326374297180681149613806886579\\08494580122963258952897654000350692006139 = 37975227936943673922808872755445627854565536638199 × 40094690950920881030683735292761468389214899724061$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we let &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$U = 6246644847868435165458917\cdot1150435884651666110524532974697443\cdot211874175165550696134616092679587435515153$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$V = 6246644847868435165458967 \cdot 1150435884651666110524532974697441\cdot211874175165550696134616092679587435515309$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then we obtain &lt;span class=&quot;math-container&quot;&gt;$U \le N \le V$&lt;/span&gt; and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Delta_{100} = 1218738270442411472694284522980510595717360280370\\4967527962531133881349172780$$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is about 75 orders of magnitude larger than achievable with a little more work.  But what is the best achievable (or, minimal) value for &lt;span class=&quot;math-container&quot;&gt;$\Delta_{100}$&lt;/span&gt;?&lt;/p&gt;&#xA;" OwnerUserId="2004" LastEditorUserId="60766" LastEditDate="2019-06-09T23:31:39.550" LastActivityDate="2019-06-09T23:31:39.550" Title="Products of three primes that bound RSA numbers" Tags="|mathematics|calculation-puzzle|" AnswerCount="1" CommentCount="4" ClosedDate="2015-09-30T11:43:52.943" ContentLicense="CC BY-SA 4.0" />
  <row Id="2228" PostTypeId="2" ParentId="2210" CreationDate="2014-09-08T15:17:01.330" Score="7" Body="&lt;p&gt;I will use the A-G bar notation given in the puzzle. I am also assuming that I only need to find one time that works, based on comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 1: 4 Broken, A, B, D, G.&lt;br&gt;&#xA;James walks in and E is lit. There are only 4 digits that use E (0, 2, 6, 8). The digit must be a 2 because the other 3 numbers use at least 6 bars, which would mean that at least 2 bars would have to be lit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 2: 6 Broken, All but C.&lt;br&gt;&#xA;Bar C is used for all digits except for 2. James walks in and no bars are lit. He knows the digit must be a 2 and knows what time it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 3: All 7 Broken.&lt;br&gt;&#xA;James walks in and no bars are lit because they are all broken. He waits 1 minute and the minutes tens digit changes. Now he knows it was a 9 and is now a 0, and knows what time it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 4: 6 Broken, All but C.&lt;br&gt;&#xA;Same as case 2. James walks in and nothing is lit. Waits 1 minute and C is lit. Now he knows it was a 2 and is now a 3. Could also use the case 3 reasoning with any one bar not broken.&lt;/p&gt;&#xA;" OwnerUserId="375" LastEditorUserId="375" LastEditDate="2014-09-09T16:21:05.250" LastActivityDate="2014-09-09T16:21:05.250" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2229" PostTypeId="2" ParentId="2224" CreationDate="2014-09-09T08:39:33.750" Score="22" Body="&lt;p&gt;They should do the exact opposite of what Joel Rondeau suggests, and gamble on the number of dead cats being &lt;strong&gt;odd&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the guard were an intellectual then he might use a good randomisation technique. But he's been selected for his anti-intellectualism, so he will pick a &quot;random&quot; number mentally. When humans pick &quot;random&quot; numbers like that, we have a &lt;a href=&quot;http://scienceblogs.com/cognitivedaily/2007/02/05/is-17-the-most-random-number/&quot;&gt;definite bias towards odd numbers&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if the intellectuals assume that the guard will kill an odd number of cats, they probably have about a 60% chance of survival.&lt;/p&gt;&#xA;" OwnerUserId="228" LastActivityDate="2014-09-09T08:39:33.750" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2230" PostTypeId="1" AcceptedAnswerId="2232" CreationDate="2014-09-10T08:32:31.303" Score="13" ViewCount="11547" Body="&lt;p&gt;Consider any large rectangle from which a smaller rectangular portion has been removed. The removed rectangular portion may have any orientation i.e. the remaining figure is not necessarily symmetric anymore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using a single cut how would you divide the given figure into two figures of equal area ?&lt;/p&gt;&#xA;" OwnerUserId="2276" LastEditorUserId="8874" LastEditDate="2015-02-07T17:50:43.880" LastActivityDate="2015-02-07T17:50:43.880" Title="Divide a rectangle with a rectangular hole into two equal parts" Tags="|logical-deduction|geometry|dissection|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2231" PostTypeId="2" ParentId="2230" CreationDate="2014-09-10T08:47:46.863" Score="10" Body="&lt;p&gt;The solution space has 1 degree of freedom, meaning that I can make the cut with any angle I want, I just have to solve a reasonably simple differential equation to know where to place the cut.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But an easier solution is to simply cut along the line that goes through both rectangle centres, thus dividing both rectangles exactly in half, and therefore making the areas of the two halves equal.&lt;/p&gt;&#xA;" OwnerUserId="2186" LastActivityDate="2014-09-10T08:47:46.863" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2232" PostTypeId="2" ParentId="2230" CreationDate="2014-09-10T17:51:26.277" Score="12" Body="&lt;p&gt;Simply cut through center of both rectangles.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/ukrPc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for the case that both centers overlaps, cut through the doubled center and with any angle. &lt;/p&gt;&#xA;" OwnerUserId="2225" LastActivityDate="2014-09-10T17:51:26.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2233" PostTypeId="2" ParentId="2227" CreationDate="2014-09-11T08:03:59.397" Score="1" Body="&lt;p&gt;The insight I mentioned in a comment is realizing that most numbers no larger than those mentioned in the question have relatively few factors.  More concretely, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Kac_theorem&quot; rel=&quot;nofollow&quot;&gt;Erdős–Kac theorem&lt;/a&gt; says (roughly) that if $\omega(n)$ is the number of distinct prime factors of n, the distribution of&#xA;    $\frac{\omega(n) - \log\log n}{(\log\log n)^{1/2}}$&lt;br&gt;&#xA; is standard normal.  Thus, numbers in the neighborhood of &lt;a href=&quot;http://en.wikipedia.org/wiki/RSA_numbers#RSA-320&quot; rel=&quot;nofollow&quot;&gt;RSA-320&lt;/a&gt; average about $(\log\log n)^{1/2} \approx 6.6$ distinct prime factors, with standard deviation about 2.6, making it not at all unlikely there will be a number near RSA-320 having exactly three (not necessarily distinct) prime factors.  The following code uses linear search to look for tri-prime near-neighbors of RSA-100, 160, 320, and 480.  It runs for about 3 seconds on my i3-2120@3.30GHz system, to produce the output shown after the program.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#!/usr/bin/env python&#xA;# Close triprimes - jw 8 Sep 2014&#xA;from gmpy import is_prime, mpz&#xA;# Note, will run ok without mpz, but is twice as fast with it.&#xA;# Ref http://puzzling.stackexchange.com/questions/2227/products-of-three-primes-that-bound-rsa-numbers&#xA;&#xA;# Return a tuple of nfax factors if k appears to&#xA;# factor into that many primes; else return None&#xA;def testfaxtuple(k, nfax):&#xA;    fax = []&#xA;    for p in smallprimes:&#xA;        while k % p == 0:       # Factors don't have to be distinct&#xA;            fax.append(p)       #   Add factor to list&#xA;            k /= p              #   Divide factor out of k&#xA;            if is_prime(k):     # If k is prime we are done with k&#xA;                fax.append(k)&#xA;                if len(fax) == nfax:&#xA;                    return fax  # Right number of prime factors&#xA;                else:&#xA;                    return None # There are too few prime factors&#xA;            if len(fax) == nfax:&#xA;                return fax if k==1 else None&#xA;            if k==1:&#xA;                return None&#xA;    return None                 # No small prime divides k&#xA;&#xA;# Search up or down from k to find a number with nfax prime factors&#xA;def searchfaxtuple(k, d, nfax):&#xA;    while k:&#xA;        t = testfaxtuple(k, nfax)&#xA;        if t:&#xA;            return k, t     # Return the number and its factor tuple&#xA;        k += d&#xA;&#xA;primeLim = 162401 # &amp;lt;--- With checks to 13, we are ok below that Carmichael#&#xA;smallprimes = [2,3,5,7,11,13]+[x for x in range(17,primeLim,2) if 1==pow(2,x-1,x)==pow(3,x-1,x)==pow(5,x-1,x)==pow(7,x-1,x)==pow(11,x-1,x)==pow(13,x-1,x)]&#xA;&#xA;n100 = mpz(1522605027922533360535618378132637429718068114961380688657908494580122963258952897654000350692006139)&#xA;n160 = mpz(2152741102718889701896015201312825429257773588845675980170497676778133145218859135673011059773491059602497907111585214302079314665202840140619946994927570407753)&#xA;n320 = mpz(21368106964100717960120874145003772958637679383727933523150686203631965523578837094085435000951700943373838321997220564166302488321590128061531285010636857163897899811712284013921068534616772684717323224436400485097837112174432182703436548357540610175031371364893034379963672249152120447044722997996160892591129924218437)&#xA;n480 = mpz(302657075295090869739730250315591803589112283576939858395529632634305976144571441696598170401251852159138533455982172343712313383247732107268535247763784105186549246199888070331088462855743520880671299302895546822695492968577380706795842802200829411198422297326020823369315258921162990168697393348736236081296604185145690639952829781767901497605213955485328141965346769742597479306858645849268328985687423881853632604706175564461719396117318298679820785491875674946700413680932103)&#xA;&#xA;for kfax in [3]:&#xA;    for n, digi in ((n100,100),(n160,160),(n320,320),(n480,480)):&#xA;        hik, hit = searchfaxtuple(n, +1, kfax)&#xA;        lok, lot = searchfaxtuple(n, -1, kfax)&#xA;        print 'n+{}: {}'.format(hik-n, hit)&#xA;        print 'n-{}: {}'.format(n-lok, lot)&#xA;        print '{}_Delta_{} = {}'.format(kfax, digi, hik-lok)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In each section of the output shown below, &lt;code&gt;n&lt;/code&gt; represents an RSA number; eg, in the first line, &lt;code&gt;n+118&lt;/code&gt; refers to a triprime equal to RSA-100 plus 118, and in the next line, &lt;code&gt;n-93&lt;/code&gt; is a triprime equal to  RSA-100 minus 93, so $\Delta_{100} = 211$.  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;n+118: [13, 2953, mpz(39662534265610809360379754047582313415771916824126199918151254124361743292582586096381785164813L)]&#xA;n-93: [2, 75211, mpz(10122222998780320435412495367251049911037402208196810896397524927072655351337921963901559284493L)]&#xA;3_Delta_100 = 211&#xA;n+61: [2, 14519, mpz(74135308999204136025071120645802928206411377809961980169794671698399791487666476192334563667383809477322746301797135281427071928686646468097663303083117653L)]&#xA;n-70: [3, 1319, mpz(544033637280487667903971493887496949521802777064866307852033782354847901243077871031845099765855713824235003060799902527692523291686338170487729844560922519L)]&#xA;3_Delta_160 = 131&#xA;n+214: [11, 11, mpz(176595925323146429422486563181849363294526275898577963001245340525884012591560637141201942156625627631188746462786946811291756101831323372409349462897825265817338014972828793503479905244766716402622505987077689959486257125408530435565591308740005042768854308800768879173253489662414218570617545437984800765215949786931L)]&#xA;n-224: [13, 52967, mpz(31032539802141998370713948372794923048803506659048861371086912175551926415110187757087410014292935577266307064917373174540174489372323446763705246097551098091406550394530533545445667236373260977760206608231250640961988106055050507069621794059785570660151780084977488712077145638070904012868277923403920427364977503L)]&#xA;3_Delta_320 = 438&#xA;n+200: [3, 106109, mpz(950774126276096183294945921381446762571545246168059443262838630195698059368421282821118442360377386018586338752233308339262184430625526918133036932977045947049886582664643810707506629521666465240684262732647707617310165234420519487180926538436354475738540234808925486588681635303203907204533053585577836882503225253106681619695563938239299517807832686153949058563511011452366526580713058739184326135349574123004434448558166804769056335520764178595660391647191959672602115689L)]&#xA;n-169: [2, 3, mpz(50442845882515144956621708385931967264852047262823309732588272105717662690761906949433028400208642026523088909330362057285385563874622017878089207960630684197758207699981345055181410475957253480111883217149257803782582161429563451132640467033471568533070382887670137228219209820193831694782898891456039346882767364190948439992138296961316916267535659247554690327557794957099579884476440974878054830947903980308938767451029260743619899352886383113303464248645945824450068946821989L)]&#xA;3_Delta_480 = 369&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Suppose the last seven lines of the program are replaced with the following eight:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;print '#f   D100  D160  D320  D480'&#xA;for kfax in range(2,9):&#xA;    ro = []&#xA;    for n, digi in ((n100,100),(n160,160),(n320,320),(n480,480)):&#xA;        hik, hit = searchfaxtuple(n, +1, kfax)&#xA;        lok, lot = searchfaxtuple(n, -1, kfax)&#xA;        ro.append(int(hik-lok))&#xA;    print '{}:  {:5} {:5} {:5} {:5}'.format(kfax, *ro)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;With that code, the program computes and prints $\Delta$ values for 2-factor, 3-factor, ... and 8-factor cases for four RSA numbers.  This took about 80 seconds of computation.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#f   D100  D160  D320  D480&#xA;2:     56   132   626   420&#xA;3:    211   131   438   369&#xA;4:     54    11    98  1434&#xA;5:    106   155   303   302&#xA;6:    341    75  1286   319&#xA;7:     76   836   284  3244&#xA;8:    100   417   620  2628&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Eg, this shows there is a pair of quad-factor numbers that bound RSA-160 and differ by only 11.&lt;/p&gt;&#xA;" OwnerUserId="2004" LastEditorUserId="2004" LastEditDate="2014-09-11T08:10:04.080" LastActivityDate="2014-09-11T08:10:04.080" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2234" PostTypeId="1" CreationDate="2014-09-12T12:25:23.867" Score="9" ViewCount="1847" Body="&lt;p&gt;I found this programming related problem too difficult for me. The problem asks to find 16 integers and form a 4x4 grid&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\begin{array}{cccc}&#xA;n_1 &amp;amp; n_2 &amp;amp; n_3 &amp;amp; n_4 \\&#xA;n_5 &amp;amp; n_6 &amp;amp; n_7 &amp;amp; n_8 \\&#xA;n_9 &amp;amp; n_{10} &amp;amp; n_{11} &amp;amp; n_{12} \\&#xA;n_{13} &amp;amp; n_{14} &amp;amp; n_{15} &amp;amp; n_{16} &#xA;\end{array}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That square has 30 subsquares. Can one find a set of those 16 integers such that sums of elements in subsquares contains all integers from 1 to 24?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, there are 16 1x1-squares with sum equal to their single numbers; nine 2x2-squares sums of 4 numbers; four 3x3-squares sums of 9 numbers; and one 4x4-squares sum of 16 numbers.&lt;/p&gt;&#xA;" OwnerUserId="2287" LastEditorDisplayName="user20" LastEditDate="2015-02-20T21:41:19.500" LastActivityDate="2015-02-20T21:41:19.500" Title="Integer grid problem" Tags="|calculation-puzzle|logic-grid|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2236" PostTypeId="2" ParentId="260" CreationDate="2014-09-12T19:02:01.850" Score="4" Body="&lt;p&gt;I have beein playing around with the question and came up with the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At a party, you see a boy misbehaving and you try to locate the parents.  You see Mrs Smith.&lt;br&gt;&#xA;  - Hello, Mrs Smith, do you have children?&lt;br&gt;&#xA;  - Yes.  I have two.&lt;br&gt;&#xA;  - Do you have a boy?&lt;br&gt;&#xA;  - Yes, I do.&lt;br&gt;&#xA;  You point at the boy.&lt;br&gt;&#xA;  - Is this your son?&lt;br&gt;&#xA;  - (a) Yes, it is!  (b) No, it is not!  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With the answer (a) there is a 50% chance that Mrs Smith has a daughter.  However, if it is (b), there is a 2/3 chance that she has a daughter.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastActivityDate="2014-09-12T19:02:01.850" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2237" PostTypeId="1" CreationDate="2014-09-12T19:48:23.197" Score="-8" ViewCount="900" Body="&lt;p&gt;Is there any crime the three wise monkeys (who see no evil, hear no evil, and speak no evil) could collectively &quot;witness&quot; and not be able to indict the suspect?&lt;/p&gt;&#xA;" OwnerUserId="2291" LastEditorUserId="1509" LastEditDate="2014-09-14T16:52:19.960" LastActivityDate="2014-11-11T08:36:19.813" Title="Three Wise Monkeys and a Crime" Tags="|logical-deduction|" AnswerCount="6" CommentCount="1" ClosedDate="2015-02-08T03:58:03.290" ContentLicense="CC BY-SA 3.0" />
  <row Id="2238" PostTypeId="2" ParentId="420" CreationDate="2014-09-13T19:55:57.790" Score="5" Body="&lt;p&gt;The inner working is that of a Rubik's cube.  So, You have to recognize the cube within.  Identify the &lt;em&gt;centers&lt;/em&gt;, they serve as a pivot.  The &lt;em&gt;edges&lt;/em&gt; and &lt;em&gt;corners&lt;/em&gt; turn around the centers.  The edges are delimited by 4 cuts and touch the centers.  The corners are delimited by 3 cuts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in the Rubik's cube the centers are not oriented.  If you rotate some, it doesn't matter.  In the Ghost cube, you need to orient the centers correctly.  So, if you solve it like the Rubik's cube it won't solve the centers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot explain the whole solution here.  Here is an outline.  On the picture you show, there is a small square corner at the center.  It is actually a center or pivot.  Start from there and match the 4 edges and 4 corners around it. That forms the &lt;em&gt;bottom layer&lt;/em&gt;.  Layer 2 is a belt around the cube.  It has 4 centers and 4 edges.  You need to orient the centers properly in a way they can match with the bottom layer and then fill in the gaps with the matching edges.  From there you should be able to complete the top layer using standard Rubik's cube moves exchanging corners or edges.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2017-11-19T22:36:50.847" LastActivityDate="2017-11-19T22:36:50.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2239" PostTypeId="1" AcceptedAnswerId="2240" CreationDate="2014-09-13T22:47:45.923" Score="5" ViewCount="499" Body="&lt;p&gt;There are five fruits (Apple, Banana, Pineapple, Orange and Watermelon) that should be ordered in their places. But we do not have enough containers for them: there are only two bowls and two baskets. So one of the fruits will remain without container.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also know that both baskets are larger than either bowl.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The watermelon is larger than all the other fruits.&lt;/li&gt;&#xA;&lt;li&gt;The bowls are too small for the pineapple to fit in.&lt;/li&gt;&#xA;&lt;li&gt;The two bowls are not neighbors.&lt;/li&gt;&#xA;&lt;li&gt;The apple is the smallest fruit, and it is in a container at the center.&lt;/li&gt;&#xA;&lt;li&gt;The banana is between two baskets.&lt;/li&gt;&#xA;&lt;li&gt;The largest fruit and the smallest fruit are not neighbors.&lt;/li&gt;&#xA;&lt;li&gt;The fruit without container has two neighbor fruits in containers.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Which fruit will remain without container? Also find the order and place of the fruits and containers with following clues: (note that the orientation does not matter, left to right or right to left)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: myself&lt;/p&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="8874" LastEditDate="2015-10-18T16:05:55.657" LastActivityDate="2015-10-18T16:06:26.833" Title="find the order of fruits" Tags="|logical-deduction|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2240" PostTypeId="2" ParentId="2239" CreationDate="2014-09-13T23:13:11.900" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;The watermelon is larger than all the other fruits.&lt;/li&gt;&#xA;  &lt;li&gt;The bowls are too small for the pineapple to fit in.&lt;/li&gt;&#xA;  &lt;li&gt;The two bowls are not neighbors.&lt;/li&gt;&#xA;  &lt;li&gt;The apple is the smallest fruit, and it is in a container at the center.&lt;/li&gt;&#xA;  &lt;li&gt;The banana is between two baskets.&lt;/li&gt;&#xA;  &lt;li&gt;The largest fruit and the smallest fruit are not neighbors.&lt;/li&gt;&#xA;  &lt;li&gt;The fruit without container has two neighbor fruits in containers.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll use A, B, P, O, and W for the fruits; K for basket, L for bowl, and N for neither.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Directly from 4:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;_ _ A _ _&#xA;_ _ _ _ _&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;1 + 4 + 6 -&gt; watermelon is on one side (we'll pick one and call it left). &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;W _ A _ _&#xA;_ _ _ _ _&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;From 7 we know it's in a container (since it doesn't have two neighbors), and from 1 + 2 we know that has to be a basket.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;W _ A _ _ &#xA;K _ _ _ _&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;From 5 we can then deduce both that the banana is second from the left and the apple is in a basket.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;W B A _ _&#xA;K _ K _ _&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;From 3 we can see the banana is in a bowl, and from 7 we can tell the other bowl has to be on the far right, leaving the fruit without a container next to it.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;W B A _ _&#xA;K L K N L&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;From 2 we know the pineapple can't be in the bowl on the right, so that leaves the containerless spot for it, and the bowl for the orange:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;W B A P O&#xA;K L K N L&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So, in English, the order is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Watermelon in a basket&lt;/li&gt;&#xA;&lt;li&gt;Banana in a bowl&lt;/li&gt;&#xA;&lt;li&gt;Apple in a basket&lt;/li&gt;&#xA;&lt;li&gt;Pineapple without a container&lt;/li&gt;&#xA;&lt;li&gt;Orange in a bowl&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4" LastEditorUserId="8874" LastEditDate="2015-10-18T16:06:26.833" LastActivityDate="2015-10-18T16:06:26.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2241" PostTypeId="1" AcceptedAnswerId="2244" CreationDate="2014-09-14T01:22:56.140" Score="4" ViewCount="2379" Body="&lt;p&gt;Question: You have 3 baskets, one with apples, one with oranges and one with both apples and oranges mixed. Each basket is closed and is labeled with ‘Apples’, ‘Oranges’ and ‘Apples and Oranges’. However, each of these labels is always placed incorrectly. How would you pick only one fruit from a basket to place the labels correctly on all the baskets? This happens to be usually the basic version of the problem Modification: What if no basket is labelled can we still label all the baskets. And if so how?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analysis: Well its easy to label one of the 3 baskets as apples or oranges after 3 picks one each from one basket. But now we have a dilemma how to chose which is the random basket and which is the specific one(apples or oranges) from the two remaining baskets. And in the worst case all our picks will result in the same output. So I was wondering if there is any way we could label them and if not the worst case can we at least get a relation based on the distribution and probability of the apples and oranges in the random basket.&lt;/p&gt;&#xA;" OwnerUserId="2295" LastActivityDate="2017-01-19T20:32:56.340" Title="A modified version of apples and oranges logic" Tags="|logical-deduction|calculation-puzzle|probability|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2242" PostTypeId="2" ParentId="2234" CreationDate="2014-09-14T09:03:52.347" Score="8" Body="&lt;p&gt;After studying the puzzle I believe that finding a solution by hand would be feasible for a good mathematician with time to spare. I also believe that the number of possible solutions is quite big.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using a computer program I have found a solution:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;-42  22  23   7&#xA; 13  11 -32  14&#xA;-23  16  15   8&#xA; 19   9 -22   1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It has the following square sums:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  4  24  12       3  84      39&#xA; 17  10   5       6  20&#xA; 21  18   2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;My program use &lt;a href=&quot;https://en.wikipedia.org/wiki/Simulated_annealing&quot; rel=&quot;nofollow&quot;&gt;simulated annealing&lt;/a&gt;, which is somewhat unconventional for a puzzle that asks for an exact solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trick is to find a method of measuring the quality of a solution candidate, along with a method of transforming it into &quot;neighbours&quot; that fit both one another and the concept of simulated annealing in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My program measure quality by first sorting the list of all 30 square values, then for each of the 7 sub-lists of length 24 it compares each number to the corresponding solution number, and add the absolute difference to the score for the sub-list, the lowest score of these 7 sub-lists is then the score for the solution candidate, and the lower a candidate scores the closer it is considered to be to a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In each iteration step my program will create new candidates by picking a random selection of the 16 numbers, possibly picking some of them multiple times, and for each pick randomly add either -1, 0 or 1. Thus keeping the change of the quality measure bounded by the number of picks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My code use numbers 500 greater than the actual solution numbers in order to make sorting a bit simpler, that is just a JavaScript thing. I also skip checking the two first of the 7 sub-lists as I have proven that there must be at least 2 negative numbers in a solution. It is quickly written code, but here goes:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;lt;!DOCTYPE HTML&amp;gt;&#xA;&amp;lt;html&amp;gt;&#xA;&amp;lt;head&amp;gt;&#xA;&amp;lt;meta http-equiv=&quot;content-type&quot; content=&quot;text/html;charset=utf-8&quot;&amp;gt;&#xA;&amp;lt;title&amp;gt;Integer Grid&amp;lt;/title&amp;gt;&#xA;&amp;lt;/head&amp;gt;&#xA;&amp;lt;body&amp;gt;&#xA;&amp;lt;script&amp;gt;&#xA;var grid=[]&#xA;var a,b,c,d&#xA;for(a=0;a&amp;lt;30;a++){&#xA;    grid[a]=500&#xA;}&#xA;var scramble&#xA;var iterations&#xA;var newgrid&#xA;var sortgrid&#xA;var score&#xA;var minscore&#xA;var branches&#xA;var best&#xA;var bestscore&#xA;for(scramble=50;scramble&amp;gt;0;scramble--){&#xA;    for(iterations=0;iterations&amp;lt;10000;iterations++){&#xA;        bestscore=1000000&#xA;        for(branches=0;branches&amp;lt;30;branches++){&#xA;            newgrid=grid.slice(0)&#xA;            for(a=0;a&amp;lt;scramble;a++){&#xA;                b=Math.floor(Math.random()*16)&#xA;                newgrid[b]+=Math.floor(Math.random()*3)-1&#xA;            }&#xA;            newgrid[16]=newgrid[0]+newgrid[1]+newgrid[4]+newgrid[5]-1500&#xA;            newgrid[17]=newgrid[2]+newgrid[1]+newgrid[6]+newgrid[5]-1500&#xA;            newgrid[18]=newgrid[2]+newgrid[3]+newgrid[6]+newgrid[7]-1500&#xA;            newgrid[19]=newgrid[8]+newgrid[9]+newgrid[4]+newgrid[5]-1500&#xA;            newgrid[20]=newgrid[10]+newgrid[9]+newgrid[6]+newgrid[5]-1500&#xA;            newgrid[21]=newgrid[10]+newgrid[11]+newgrid[6]+newgrid[7]-1500&#xA;            newgrid[22]=newgrid[8]+newgrid[9]+newgrid[12]+newgrid[13]-1500&#xA;            newgrid[23]=newgrid[10]+newgrid[9]+newgrid[14]+newgrid[13]-1500&#xA;            newgrid[24]=newgrid[10]+newgrid[11]+newgrid[14]+newgrid[15]-1500&#xA;            newgrid[25]=newgrid[16]+newgrid[20]+newgrid[2]-newgrid[5]+newgrid[8]-1000&#xA;            newgrid[26]=newgrid[18]+newgrid[20]+newgrid[1]-newgrid[6]+newgrid[11]-1000&#xA;            newgrid[27]=newgrid[22]+newgrid[20]+newgrid[4]-newgrid[9]+newgrid[14]-1000&#xA;            newgrid[28]=newgrid[24]+newgrid[20]+newgrid[7]-newgrid[10]+newgrid[13]-1000&#xA;            newgrid[29]=newgrid[16]+newgrid[18]+newgrid[22]+newgrid[24]-1500&#xA;            sortgrid=newgrid.slice(0).sort()&#xA;            minscore=1000000&#xA;            for(a=2;a&amp;lt;=6;a++){&#xA;                score=0&#xA;                b=a+24&#xA;                for(c=a,d=501;c&amp;lt;b;c++,d++){&#xA;                    score+=Math.abs(d-sortgrid[c])&#xA;                }&#xA;                minscore=Math.min(minscore,score)&#xA;            }&#xA;            if(minscore&amp;lt;bestscore){&#xA;                bestscore=minscore&#xA;                best=newgrid&#xA;            }&#xA;        }&#xA;        grid=best&#xA;        if(bestscore===0){&#xA;            console.log(&quot;Solution:&quot;)&#xA;            console.log(grid)&#xA;            throw new Error(&quot;Done&quot;)&#xA;        }&#xA;    }&#xA;    console.log(grid)&#xA;}&#xA;&amp;lt;/script&amp;gt;&#xA;&amp;lt;/body&amp;gt;&#xA;&amp;lt;/html&amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h3&gt;0 to 25&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I have experimented a bit with extending the puzzle, the best I have is a solution with all the numbers from 0 to 25, but 1 to 26 have eluded me, maybe it is impossible, maybe I just haven't hit it. I expect that it is in general easier to find solutions where the average of the series is closer to 0.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 16  15   8   0      19   1  23      25  18      87&#xA; 13 -25   3  12       4 -34  11      14  22&#xA;  7   9 -21  17      24  10  21&#xA;  6   2  20   5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h3&gt;-7 to 21&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;30 consecutive numbers is impossible, as the sum of 30 consecutive numbers is always odd, and the sum of the 30 numbers in the puzzle is always even. But 29 consecutive is doable, the highest such series I have found run from -7 to 21:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  9  -4  12  13      15  10  18      17  16      55&#xA;  6   4  -2  -5       0   1  -6      20  21&#xA; -7  -3   2  -1       3  14  19&#xA;  5   8   7  11&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h3&gt;Extreme solutions&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Inspired by Florian F's observations about big and small numbers I decided to find some solutions that include really high and really low values, the best i have managed to find is a solution that include the number -99, and one that include 63 in the 16 base numbers:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;-99  11   4  13     -60  23  24     -45  16       1&#xA; 22   6   2   5      17   3   7      77  15&#xA; 14 -25  20 -20      18  12  19&#xA; 21   8   9  10&#xA;&#xA; 63   3   1  13      15  20  23      22   9      73&#xA;-62  11   5   4     -68   7   8       2  24&#xA; 10 -27  18 -19      16  17  19&#xA; 21  12  14   6&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2186" LastEditorUserId="2186" LastEditDate="2014-09-15T16:56:45.877" LastActivityDate="2014-09-15T16:56:45.877" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2243" PostTypeId="2" ParentId="2234" CreationDate="2014-09-14T10:16:26.263" Score="3" Body="&lt;p&gt;Damn! too late.  I got the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;-39  23  10 -22&#xA; 16 -21   3  11&#xA;  7  17   5   1&#xA;  9 -20   6  12&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;My method is to start from a random grid, then adjust individual values to the value that maximizes the score (number of values in 1..24 represented).  Usually it blocks at 22, so I just restart until I find a solution.  It is slow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, no grid with score 23 or 24 uses a value &gt;24, while some use values as low as -58.  I would have expected the values to extend less in the negatives than in the positives.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2014-09-14T14:39:01.877" LastActivityDate="2014-09-14T14:39:01.877" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2244" PostTypeId="2" ParentId="2241" CreationDate="2014-09-14T11:04:20.897" Score="3" Body="&lt;p&gt;Well, as you have said, there is a risk to pick always the same fruit from the &quot;apples and oranges&quot; basket.  So there is no certain answer, except if you pick all fruits from one basket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you pick one fruit from each basket, you must have one of one fruit and 2 of the other.  Let's say you pick 1 apple and 2 oranges.  The apple comes from the &quot;apples only&quot; basket.  You cannot identify the 2 remaining baskets where you picked an orange.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need to do some assumptions.  We can safely assume that the mixed basket is either of the unidentified baskets with equal probability.  But what is the distribution of the quantity of each fruit in the mixed basket?  I will assume the mixed basket contains N fruits, each of them having equal probability to be an orange or an apple.  Then it is simple.  If at any time you pick an apple from the 2 remaining baskets, you have found the mixed basket and you can identify all baskets.  If not, the more oranges you pick from a basket the more unlikely it is that it is the mixed basket.  If you pick $a$ oranges from basket A and $b$ oranges from basket b, the probability that A is the mixed basket is ${1/2^a\over 1/2^a + 1/2^b} = {2^b\over 2^a+2^b}$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the distribution can be different.  If for instance the mixed basket contains an equal number of oranges and apples, then the exact probability will depend on how many fruits there were to begin with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But regardless of the distribution, the rule still holds that the more of the same fruit you pick from a basket, the more unlikely it is to be the mixed basket.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2017-01-19T20:32:56.340" LastActivityDate="2017-01-19T20:32:56.340" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2245" PostTypeId="2" ParentId="2239" CreationDate="2014-09-14T11:33:51.860" Score="2" Body="&lt;p&gt;Actually, 3. is an unnecessary condition.  From&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;W B A _ _&#xA;K _ K _ _&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;you can continue:&lt;br&gt;&#xA;From 7. the last position is a bowl.&lt;br&gt;&#xA;From 2. the pineapple is not in the bowl, it must be the orange.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;W B A _ O&#xA;K _ K _ L&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The pineapple is the last fruit and it cannot be in a bowl&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;W B A P O&#xA;K L K N L&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2136" LastActivityDate="2014-09-14T11:33:51.860" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2247" PostTypeId="1" AcceptedAnswerId="2256" CreationDate="2014-09-14T15:54:18.910" Score="17" ViewCount="912" Body="&lt;p&gt;It is a little known fact that Santa Claus is in fact a vampire.  Why else would he spend so much time near the North Pole and travel only at night?  Yes, he fears daylight!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;His problem is that during the summer, there is plenty of daylight on the North Pole. This is why, every equinox, he migrates to the other pole.  In March he flies to the South Pole, and in September, he flies back to the North Pole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To avoid daylight, he has to complete the trip during the night.  He could leave the North Pole at 18:00 GMT (6pm), fly down along the Greenwith meridian and reach the South Pole at 6:00 GMT (6am).  He would have to fly 20000 km in 12 hours at ~1666 km/h.  Santa could easily fly that fast.  It was proven he flies faster than the speed of light to deliver all the presents in one night.  But he wants to relax and enjoy the view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is: What is the minimum speed Santa needs to fly to travel from the North Pole to the South Pole while avoiding all daylight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is assumed the Earth is a perfect sphere with perimeter 40000 km, the Sun is over the equator and shines over exactly 1/2 of Earth surface.  The altitude of flight is negligible.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2014-09-20T08:25:27.690" LastActivityDate="2014-09-20T08:25:27.690" Title="Santa Claus flies to the South Pole" Tags="|mathematics|calculation-puzzle|geometry|optimization|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2248" PostTypeId="2" ParentId="2034" CreationDate="2014-09-14T18:09:08.980" Score="3" Body="&lt;p&gt;As I said in another post, the general solution for jug A and B is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If A is empty, fill A&lt;/li&gt;&#xA;&lt;li&gt;If B is full, empty B&lt;/li&gt;&#xA;&lt;li&gt;else transfer what you can from A to B.&lt;/li&gt;&#xA;&lt;li&gt;stop when you have the wanted quantity.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Depending on which jug you choose as A or B, you get 2 solutions.  One of them is optimal.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastActivityDate="2014-09-14T18:09:08.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2250" PostTypeId="2" ParentId="2247" CreationDate="2014-09-15T01:03:57.197" Score="7" Body="&lt;p&gt;Santa can travel at approximately 761.15 kph all the way. At 18:00 GMT he leaves hugging the line of sunset (momentarily he will be travelling directly south, but as he drops to lower latitude he has to fly at an increasing angle towards the west so viewed from above the north pole his progress would vaguely resemble a spiral). Eventually, at a latitude of approximately 62.8 N, Santa will find he is travelling almost due west, and can only just keep up with the line of sunset. At that point, he starts to swing gently south (at any particular latitude L, he wants to follow a direction d = 270 - acos(0.46*cos(L)) - this points in a more southerly direction as he gets closer to the equator). Initially he will only slowly fall back from the line of sunset, but the further south he goes the more local time he loses until, as he crosses the equator (at a heading ~ 207 degrees), the local time will be midnight. He then &quot;mirrors&quot; his course so far in the southern hemisphere. If his heading is correct he will finally get to the south pole just on local sunrise. Any minor mistakes and Santa is toast. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Santa can only calculate his exact minimal speed by accurately solving an integral equation (if he has similar maths skills as me he will have to do this numerically), so he could reduce his speed a little below my estimate with a more accurate calculation. But the closer to the limiting speed he travels at the longer the journey will take (at 762 kph it already takes him about 20 days to make the trip), so unless he has lots of time to spare he may want to travel a little faster than the limit. Also, if he does decide to travel as close as possible to the limiting speed he will need to take into account the fact that the orientation of the sun to the north-south pole line will be changing during his trip, so he will need to solve an even more complicated integral (and leave the pole some time before the spring solstice).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The path near the pole is simple - he can just follow the sunset line (this part is reminiscent of the lake-monster problem). But the next section is a bit more complicated. Consider his passage between some latitude L and L-dL, the optimal angle to cross that latitude will be that which, for any specific speed he is travelling, loses him the least of his head-start to the murderous sunrise. Locally, this can be modeled by solving the equations below. I think that shows that the optimal path after the break-away is not a straight line but a vaguely s-shaped curve. &#xA;&lt;img src=&quot;https://i.stack.imgur.com/KGGTl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DKNhx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am adding a plot showing Santa's path (I have reversed the route so he is going from the equator to the north pole, 'cos that's the way i did the maths). Note that at 762 kph, the sunrise line catches up by pi/2 = 6 hours over the route. So he can mirror the route to do the complete passage from north to south. Takes approximately 20 hours, but he can take as long as he wants circling in the vicinity of latitude ~1.09 radians=63 North&#xA;&lt;img src=&quot;https://i.stack.imgur.com/aZAq4.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="1631" LastEditDate="2014-09-16T21:44:33.020" LastActivityDate="2014-09-16T21:44:33.020" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="2251" PostTypeId="2" ParentId="2237" CreationDate="2014-09-15T09:43:24.610" Score="4" Body="&lt;p&gt;Well, there's an obvious solution:&#xA;The act of shooting all three of them in the head with a big enough caliber gun.&lt;/p&gt;&#xA;" OwnerUserId="2156" LastActivityDate="2014-09-15T09:43:24.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2252" PostTypeId="2" ParentId="2237" CreationDate="2014-09-15T13:20:12.397" Score="1" Body="&lt;p&gt;Clearly it depends on your definition of &quot;evil&quot; - the third monkey, whose sight and hearing are not restricted in any way, is a perfectly good witness who could identify the culprit easily enough. Were you to declare that saying &quot;It was John who robbed the bank / shot that guy / lit the fuse&quot; was somehow &lt;em&gt;speaking evil&lt;/em&gt;, then I could declare that one of the other monkeys who had only partial information could (suggest to a cop/prosecutor to) ask questions like &quot;who was that running away afterwards?&quot; while testifying what they saw or heard the running away guy do while somehow not being sure who it was.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I say no, I can't think of any witnessable crime that the three of them can't testify succesfully about.&lt;/p&gt;&#xA;" OwnerUserId="1509" LastActivityDate="2014-09-15T13:20:12.397" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2253" PostTypeId="2" ParentId="2237" CreationDate="2014-09-15T13:38:55.443" Score="1" Body="&lt;p&gt;It depends on how much freedom the monkey who can &quot;speak no evil&quot; has.  Does this mean that he can write?  Sign language?  If so, then he effectively has no limitations since he can communicate fully.  Thus, there is no crime (apart from the monkeys being killed like in a previous answer) that he cannot indict.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If he cannot do these things, then he cannot communicate at all and goes from the most useful to the least useful of the three monkeys; in fact he is completely useless since no matter what he witnesses, he has no ability to communicate it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, you have the other two monkeys, one blind and the other deaf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the crime involves both senses to decode, then it would not be fully understood by either.  However, if the monkeys could communicate some how and share their experiences, then they could potentially figure out anything.  So lets assume they are unable to communicate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple example of something neither would be able to understand on their own, but together they would, would be a visible code and an audible key.  The deaf monkey would see the code, but not be able to decipher it, while the blind monkey would hear the key, but not have nothing to decode.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2014-09-15T13:38:55.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2255" PostTypeId="1" AcceptedAnswerId="2278" CreationDate="2014-09-15T16:55:58.593" Score="9" ViewCount="889" Body="&lt;p&gt;Break the rule of the following code and find the original text.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;4844026700420337026680789437084266769043634877&#xA;  3123111411311124132311313221113123421112223234&#xA;  1000111001011110010000101001011001010111101101&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="2225" LastEditDate="2014-11-12T15:20:42.960" LastActivityDate="2014-11-12T15:20:42.960" Title="Decrypt the code and find the text" Tags="|cipher|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2256" PostTypeId="2" ParentId="2247" CreationDate="2014-09-15T18:15:32.690" Score="5" Body="&lt;p&gt;This actually has a fair amount of similarity with &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2155/the-lake-monster&quot;&gt;The lake monster&lt;/a&gt;. But for this puzzle I'll skip all the advanced maths and find a solution based on simple logic.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Obviously, the first and last legs of the travel can be made without losing any ground to the sun, by following a path that ends in the latitude circle where Santa can only just keep up with the sun. At the end point this path is tangential to the latitude circle.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There is no kinks in the optimal path. As the night area doesn't have any corners any path that includes a kink can trivially be made shorter by cutting some of that corner off.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;During the part where Santa can't keep up with the sun there is no reason for him to go along anything but a straight line, a line that turns means he could have made the same trip shorter by following a straight line between the same points. As he moves slower than the sun such a correction will not put him in daylight.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Given these three observations his middle leg must consist of leaving the circle where he can keep up with the sun in a straight line that start out going west and follow this line until he reach the circle where he can keep up with the sun on the other side of the planet.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;No matter how large or small the circles where Santa can keep up with the sun are, the line described in 4. is 20000 km long and moves Santa 12 hours forward relative to the sun. This means Santa has 24 hours to complete those 20000 km, so he must move at a speed of $833\frac{1}{3}\mathrm{km/h}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This in turn means that the circles where Santa can keep up with the sun have a circumference of 20000 km, so they are located at 60°N and 60°S.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I guess the optimal path to and from these circles are approximately half circles. I'll use some differential maths to figure that part some day when I feel like my head doesn't hurt enough.&lt;/p&gt;&#xA;" OwnerUserId="2186" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2014-09-15T18:15:32.690" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2257" PostTypeId="1" AcceptedAnswerId="2259" CreationDate="2014-09-15T23:48:16.470" Score="6" ViewCount="670" Body="&lt;p&gt;The task is to decode the following: 5, 18, 92, 6, 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clues:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It's my name  [my name is Jake]&lt;/li&gt;&#xA;&lt;li&gt;&quot;Gliding o'er all, through all, Through Nature, Time, and Space, As a ship on the waters advancing, The voyage of the soul—not life alone, Death, many deaths I'll sing.&quot;&lt;/li&gt;&#xA;&lt;li&gt;My first guess (ASCII) was definitely the most on track -- as far as thinking&lt;/li&gt;&#xA;&lt;li&gt;The &quot;cipher&quot; has exactly 118 components &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Can you guys help me solve the above using the four clues? Or help me ask the right questions.&lt;/p&gt;&#xA;" OwnerUserId="2303" LastEditorUserId="5373" LastEditDate="2015-04-28T09:25:57.387" LastActivityDate="2015-04-28T09:25:57.387" Title="Decrypt the following message: 5, 18, 92, 6, 1 [Clues inside]" Tags="|cipher|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2258" PostTypeId="1" AcceptedAnswerId="2262" CreationDate="2014-09-16T02:26:27.393" Score="180" ViewCount="10794" Body="&lt;p&gt;I created a visual puzzle, which my wife then implemented as part of a stained-glass window. I've no idea if it is (a) obvious, (b) stupid or hopefully (c) extremely clever, and hence would love to hear your opinion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a photo. What does it represent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/O9a9n.jpg&quot; alt=&quot;Order and Chaos&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2305" LastEditorUserId="8874" LastEditDate="2015-02-04T18:51:00.467" LastActivityDate="2015-02-04T18:51:00.467" Title="Chaos and Order: a visual puzzle in stained glass" Tags="|mathematics|geometry|visual|steganography|enigmatic-puzzle|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="2259" PostTypeId="2" ParentId="2257" CreationDate="2014-09-16T04:20:06.383" Score="7" Body="&lt;p&gt;When I see 92 I think of uranium.  Decoding this as atomic symbols gives BArUCH.  There are 118 elements &lt;a href=&quot;http://en.wikipedia.org/wiki/Chemical_element&quot;&gt;now recognized&lt;/a&gt;. I don't know what to do with clues 2 and 3.  Clue 2 is a poem by Walt Whitman, for what it's worth.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2014-09-16T04:20:06.383" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2260" PostTypeId="2" ParentId="2151" CreationDate="2014-09-16T10:22:36.473" Score="6" Body="&lt;p&gt;Kevin = &quot;I don't know&quot;    == True (Because he truthfully said, that he didn't know)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Laura = &quot;One&quot;             == False (Because if she says truth, then there would be at least 2 truth tellers as Kevin is already a truth teller).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mike = &quot;None.&quot;            == False (Because Kevin proves himself a truth teller)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Noela =                   == True (Because if Noela is False then there is only 1 (i.e. Kevin) and Laura also replied &quot;One&quot; and we know that Laura is false)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, there are Two out of Four Truth-Tellers in town.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I do remembered such a kind of question in a book titled as &quot;Discrete Mathematics and its Applications&quot; Author: &quot;Kenneth H. Rosen&quot; Ed:&quot;7th&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2307" LastEditorUserId="463" LastEditDate="2015-06-05T15:10:52.237" LastActivityDate="2015-06-05T15:10:52.237" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2261" PostTypeId="1" AcceptedAnswerId="2263" CreationDate="2014-09-16T13:37:36.853" Score="11" ViewCount="1651" Body="&lt;p&gt;There's an obvious generalization of &lt;a href=&quot;https://puzzling.stackexchange.com/q/282/70&quot;&gt;this problem&lt;/a&gt;: There are a rather large number of prisoners, each of them seeing the color of the hat of all the following sad fellows bot not their; the hats may be of one of $n$ different colors, and everyone in turn must state the color of his hat, without uttering anything else. If they may devise a strategy in advance, at most $n-1$ prisoners will state the wrong color - needless to say, the rules should say that all of them will be released if at least a certain threshold of answers are correct, otherwise nobody would bother to help the other ones :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was hovever wondering if there is an algorithm that gives a better chance of (global) survival. Any idea?&lt;/p&gt;&#xA;" OwnerUserId="70" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-06-04T00:49:33.163" Title="Of men and hats" Tags="|logical-deduction|hat-guessing|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2262" PostTypeId="2" ParentId="2258" CreationDate="2014-09-16T13:52:10.280" Score="120" Body="&lt;p&gt;Solved it! Spoilers ahead:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Superposition of &lt;img src=&quot;https://i.stack.imgur.com/KNjrc.png&quot; alt=&quot;blue&quot;&gt; and &lt;img src=&quot;https://i.stack.imgur.com/Qfb4C.png&quot; alt=&quot;yellow&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's a combination of two numbers represented in yellow and blue panes, with&#xA; green panes indicating where they overlap.&#xA; &#xA; The blue panes give the digits of $\pi$ (3, 1, 4, 1, 5, 9, 2, 6, 5)&#xA; &#xA; and the yellow panes are the digits of the first &lt;a href=&quot;http://en.wikipedia.org/wiki/Feigenbaum_constants&quot;&gt;Feigenbaum constant&lt;/a&gt; (4, 6, 6, 9, 2, 0, 1, 6, 1).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;P.S. I think this window is brilliant :-)&lt;/p&gt;&#xA;" OwnerUserId="40" LastEditorUserId="5373" LastEditDate="2014-12-29T20:14:12.327" LastActivityDate="2014-12-29T20:14:12.327" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="2263" PostTypeId="2" ParentId="2261" CreationDate="2014-09-16T15:36:23.787" Score="7" Body="&lt;p&gt;Only &lt;strong&gt;one&lt;/strong&gt; person must risk death when using the correct strategy.&lt;br&gt;&#xA;You must use the &lt;a href=&quot;https://puzzling.stackexchange.com/a/285/28&quot;&gt;same&lt;/a&gt; approach, but instead of binary arithmetic use arithmetic mod $n$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before choosing, the prisoners number the colours from 0 to n-1.&#xA;The first person calculates the sum mod $n$ of colour values for the hats he sees and answers with that value's corresponding colour. He will die with probability $(n-1)/n$, but each other person can calculate his colour and survive.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2014-09-18T17:12:49.450" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2264" PostTypeId="5" CreationDate="2014-09-16T16:50:56.247" Score="0" Body="&lt;p&gt;According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Logic_puzzle#Logic_grid_puzzles&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Another form of logic puzzle, popular among puzzle enthusiasts and available in magazines dedicated to the subject, is a format in which the set-up to a scenario is given, as well as the object (for example, determine who brought what dog to a dog show, and what breed each dog was), certain clues are given (&quot;neither Misty nor Rex is the German Shepherd&quot;), and then the reader fills out a matrix with the clues and attempts to deduce the solution. These are often referred to as &quot;logic grid&quot; puzzles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Logic grids usually look something like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/806L1.png&quot; alt=&quot;example logic grid&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Indicators are used both to denote what is possible as well as what is not possible. In this diagram, we can see that Simon must be 15, but Jane cannot be green (in the context of the puzzle, this probably makes sense).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These puzzles were first created by Charles Lutwidge Dodgson (Lewis Carroll) in &lt;em&gt;The Game of Logic&lt;/em&gt;. They are often wrongly attributed to &lt;a href=&quot;https://web.stanford.edu/~laurik/fsmbook/examples/Einstein%27sPuzzle.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Albert Einstein&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not to be confused with &lt;a href=&quot;/questions/tagged/grid-deduction&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;grid-deduction&amp;#39;&quot; rel=&quot;tag&quot;&gt;grid-deduction&lt;/a&gt;, for puzzles such as Sudoku or Kakuro in which a pre-existing grid must be filled using provided clues about its elements.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorUserId="11110" LastEditDate="2020-05-12T19:24:06.000" LastActivityDate="2020-05-12T19:24:06.000" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="2265" PostTypeId="4" CreationDate="2014-09-16T16:50:56.247" Score="0" Body="Logic grid puzzles, first produced by Lewis Carroll, are puzzles in which correspondences must be drawn between multiple ways of describing the same list of objects, using a given set of clues about these correspondences. Such puzzles can be solved using a grid, but for pre-existing grids whose elements must be filled using logic, use the [grid-deduction] tag instead." OwnerDisplayName="user20" LastEditorUserId="5373" LastEditDate="2017-03-25T18:28:29.293" LastActivityDate="2017-03-25T18:28:29.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2266" PostTypeId="1" CreationDate="2014-09-16T23:09:23.763" Score="13" ViewCount="807" Body="&lt;p&gt;From as far back as psychologists &lt;a href=&quot;http://en.wikipedia.org/wiki/Robert_Sternberg&quot;&gt;Robert Sternberg&lt;/a&gt;, and &lt;a href=&quot;https://college.lclark.edu/live/profiles/87-janet-davidson&quot;&gt;Janet Davidson&lt;/a&gt; argued that solving puzzles entails the ability to compare hidden information in a puzzle with information already in memory, and, more importantly, the ability to combine the information to form novel information and ideas&lt;sup&gt;1&lt;/sup&gt;. The thinking involved in solving puzzles can be characterized as a blend of imaginative association and memory. It is this combination that possibly leads us to find the pattern or twist within a puzzle, leave to that &quot;Aha&quot; moment when you get to the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While there have been research to suggests that culture is a factor in how puzzles affect brain functioning. Ambiguous empirical data aside, even though people speak different languages, puzzles seem to rise above culture-specific modes of understanding the world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Puzzles seem to tap into a universal part of brain functioning, even though they may appear in different cultural forms. The &lt;a href=&quot;http://en.wikipedia.org/wiki/River_crossing_puzzle&quot;&gt;river crossing puzzle&lt;/a&gt; is a classic example of this. The puzzle can be found across the world in various formats from cannibals to jealous spouses, the details may change, but the underling structure structure remains the same. This can be interpreted to suggest that the puzzle is culture independent, as it's a common part of human imagination. The author and mathematician &lt;a href=&quot;http://en.wikipedia.org/wiki/Henry_Dudeney&quot;&gt;Henry E. Dudeney&lt;/a&gt; mentioned:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The curious propensity for propounding puzzles is not peculiar to any race or any period of history. It is simply innate, though it is always showing itself in different forms; whether the individual be a Sphinx of Egypt, a Samson of Hebrew lore, an Indian fakir, a Chinese philosopher, a mahatma of Tibet, or a European mathematician makes little difference &lt;sup&gt;2&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There have been many many academic types involved in not just mathematics, but also psychology that have been involved in popular puzzles for entertainment and reasearch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A notable one in recent years is &lt;a href=&quot;http://www.officialnintendomagazine.co.uk/6916/features/the-making-of-professor-layton/&quot;&gt;Akira Tago&lt;/a&gt;, Professor Emeritus of &lt;a href=&quot;http://en.wikipedia.org/wiki/Chiba_University&quot;&gt;Chiba University&lt;/a&gt;, and his team are credited for creating the puzzles found in the &lt;em&gt;Professor Layton&lt;/em&gt; series of games. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Professor Tago himself is a researcher in the field of psychology, and author of the &lt;em&gt;&lt;a href=&quot;http://kotaku.com/5988532/if-it-werent-for-this-book-professor-layton-wouldnt-exist&quot;&gt;Head Gymnasium&lt;/a&gt;&lt;/em&gt; series of puzzle books.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm curious of that kind is involved psychologically in the process of creating puzzles, in contrast to solving puzzles. Particularly what aspects of brain function and psychology have been researched and suggested by academic and professional sources to involved when creating puzzles? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; &lt;sub&gt;Sternberg, Robert J., and Janet E. Davidson. &quot;The mind of the puzzler.&quot; Psychology Today 16.6 (1982): 37-44.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;2&lt;/sup&gt; &lt;sub&gt;H.E. Dudeney, &lt;em&gt;The Canterbury Puzzles and Other Curious Problems&lt;/em&gt;, (1958): 12&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="2310" LastActivityDate="2015-04-13T17:09:03.917" Title="What drives the brain when creating puzzles?" Tags="|puzzle-creation|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2267" PostTypeId="1" AcceptedAnswerId="2279" CreationDate="2014-09-17T05:14:46.877" Score="1" ViewCount="5460" Body="&lt;p&gt;Can you draw something on a piece of paper that looks like a square and a circle at the same time? ( without using Math or 3-d visualisation techniques)&lt;/p&gt;&#xA;" OwnerUserId="2312" LastEditorUserId="2312" LastEditDate="2014-10-11T04:19:09.973" LastActivityDate="2014-10-11T04:19:09.973" Title="A square and a circle at once" Tags="|logical-deduction|" AnswerCount="3" CommentCount="3" ClosedDate="2014-09-23T06:54:22.747" ContentLicense="CC BY-SA 3.0" />
  <row Id="2271" PostTypeId="1" AcceptedAnswerId="2272" CreationDate="2014-09-17T12:02:23.417" Score="59" ViewCount="10452" Body="&lt;p&gt;I thought this changes the way series of numbers are looked at &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 &#xA;1 1&#xA;2 1&#xA;1 2 1 1 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Write down the next three lines &lt;/p&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="4050" LastEditDate="2015-01-03T17:56:04.747" LastActivityDate="2018-03-30T02:45:31.460" Title="Interview Question or Pathbreaking puzzle" Tags="|logical-deduction|pattern|number-sequence|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2272" PostTypeId="2" ParentId="2271" CreationDate="2014-09-17T13:02:56.500" Score="63" Body="&lt;p&gt;The answer is this:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 111221&lt;br&gt;&#xA; 312211&lt;br&gt;&#xA; 13112221&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first number specifies the quantity of digits of the set above it and the second number specifies what the digit is.  The second line is 11 it is saying that the line above it is one one.  The third line states that the line above it is two ones.  The fourth line is saying there is one two and one one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2314" LastEditorUserId="17744" LastEditDate="2016-03-28T17:29:58.820" LastActivityDate="2016-03-28T17:29:58.820" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2273" PostTypeId="2" ParentId="2267" CreationDate="2014-09-17T13:08:56.830" Score="11" Body="&lt;p&gt;You can draw a cylinder, which (a real 3D object, not the 2D drawing) can look like a square and like a circle, depending on how you turn it.&lt;img src=&quot;https://i.stack.imgur.com/2LzAo.jpg&quot; alt=&quot;enter image description here&quot;&gt;.&lt;/p&gt;&#xA;" OwnerUserId="28" LastActivityDate="2014-09-17T13:08:56.830" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="2275" PostTypeId="2" ParentId="313" CreationDate="2014-09-17T16:19:40.403" Score="10" Body="&lt;p&gt;I would ask&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Are there at least $N + 27$ days in February, where $N$ is your number?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if $(N = 1)$ then $(N + 27 = 28)$ --&gt; Yes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if $(N = 2)$ then $(N + 27 = 29)$ --&gt; Maybe&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;if $(N = 3)$ then $(N + 27 = 30)$ --&gt; No&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2318" LastEditorUserId="42452" LastEditDate="2018-09-08T08:52:46.417" LastActivityDate="2018-09-08T08:52:46.417" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="2276" PostTypeId="2" ParentId="2034" CreationDate="2014-09-17T16:33:31.870" Score="1" Body="&lt;p&gt;For number 3, here's a sequence that hits every number from 1 to 19:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0+0=0&#xA;0+11=11&#xA;8+3=11&#xA;0+3=3&#xA;3+0=3&#xA;3+11=14&#xA;8+6=14&#xA;0+6=6&#xA;6+0=6&#xA;6+11=17&#xA;8+9=17&#xA;0+9=9&#xA;8+1=9&#xA;0+1=1&#xA;1+0=1&#xA;1+11=12&#xA;8+4=12&#xA;0+4=4&#xA;4+0=4&#xA;4+11=15&#xA;8+7=15&#xA;8+0=8&#xA;0+8=8&#xA;8+8=16&#xA;5+11=16&#xA;5+0=5&#xA;0+5=5&#xA;8+5=13&#xA;2+11=13&#xA;2+0=2&#xA;0+2=2&#xA;8+2=10&#xA;0+10=10&#xA;8+10=18&#xA;8+11=19&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2318" LastActivityDate="2014-09-17T16:33:31.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2277" PostTypeId="2" ParentId="2271" CreationDate="2014-09-17T18:09:44.363" Score="40" Body="&lt;p&gt;Another valid answer:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1231&#xA;131221&#xA;132231&#xA;232221&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Using similar rules as the accept answer.  The first number specified the quantity of digits of the entire set above it and the second number specifies what the digit is.  The accepted answer is a reading of the prior sequence, this answer is a summary of the prior sequence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fun fact, coming back to this.  This sequence will hit a point where it will output the same number forever.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1&#xA;11&#xA;21&#xA;1211&#xA;1231&#xA;131221&#xA;132231&#xA;232221&#xA;134211&#xA;14131231&#xA;14231241&#xA;24132231&#xA;14233221&#xA;14233221&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2319" LastEditorUserId="-1" LastEditDate="2015-02-05T21:33:28.540" LastActivityDate="2015-02-05T21:33:28.540" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2278" PostTypeId="2" ParentId="2255" CreationDate="2014-09-17T18:14:01.673" Score="12" Body="&lt;p&gt;The text is: &quot;I am a decryption genius&quot;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The decryption is based on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Telephone_keypad&quot;&gt;Telephone Keypad&lt;/a&gt;. The first line gives the key number, the second line gives the letter number and the third gives if you need to consider this letter or not. So basically you just need to decode the letters with '1' in the last line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2294" LastActivityDate="2014-09-17T18:14:01.673" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2279" PostTypeId="2" ParentId="2267" CreationDate="2014-09-18T03:46:29.297" Score="5" Body="&lt;p&gt;Here is one possible answer ( like Marylin Vos Savant said try to find many possible useful answers to one problem ) ;just draw a circle on a piece of paper , it 'looks' like a circle , it also looks like a 'zero' which is a square integer....&lt;/p&gt;&#xA;" OwnerUserId="2312" LastActivityDate="2014-09-18T03:46:29.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2281" PostTypeId="1" AcceptedAnswerId="2287" CreationDate="2014-09-18T07:05:50.557" Score="9" ViewCount="1022" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Grelling%E2%80%93Nelson_paradox&quot; rel=&quot;nofollow noreferrer&quot;&gt;Grelling's Paradox&lt;/a&gt; is about certain types of words. An adjective is &lt;em&gt;heterological&lt;/em&gt; if it does not describe itself, like 'long', 'German', or 'monosyllabic'. An adjective is  &lt;em&gt;autological&lt;/em&gt; if it does describe itself, like 'short', 'English', or 'polysyllabic'. So the paradox is:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;What is &lt;em&gt;heterological&lt;/em&gt;: is it autological or heterological?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Assuming either leads to a contradiction, so a paradox exists if all adjectives can be categorized as &lt;strong&gt;autological&lt;/strong&gt; or &lt;strong&gt;heterological&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;But what if there are other categories? Could you say an adjective is '&lt;strong&gt;semi-heterological&lt;/strong&gt;' if its meaning cannot describe all of itself but does describe part of itself? If such a word exists, is it &lt;strong&gt;heterological&lt;/strong&gt; or &lt;strong&gt;autological&lt;/strong&gt;? Does this resolve the paradox?&lt;/p&gt;&#xA;" OwnerUserId="2312" LastEditorUserId="5373" LastEditDate="2021-12-27T05:32:00.817" LastActivityDate="2021-12-27T05:32:00.817" Title="About Grelling's Paradox" Tags="|logical-deduction|word|paradox|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="2283" PostTypeId="1" AcceptedAnswerId="2286" CreationDate="2014-09-18T07:50:04.690" Score="18" ViewCount="20930" Body="&lt;p&gt;I found a &lt;a href=&quot;http://www.allthetests.com/quiz18/quiz/1139060031/Very-Difficult-IQ-Test&quot; rel=&quot;noreferrer&quot;&gt;Very Difficult IQ Test&lt;/a&gt; on web.&lt;br /&gt;&#xA;One of its questions is:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;What is the box suggesting?&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zhnZu.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I have seen the answer but there was no clues for the logic behind that.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;123456789&lt;br&gt;&lt;/li&gt;&#xA;&lt;li&gt;176h54308&lt;br&gt;&lt;/li&gt;&#xA;&lt;li&gt;627456934&lt;br&gt;&lt;/li&gt;&#xA;&lt;li&gt;126485704&lt;br&gt;&lt;/li&gt;&#xA;&lt;li&gt;673258719&lt;br&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;What is the logic?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The solution is the&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; first option (&lt;strong&gt;123456789&lt;/strong&gt;)&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;from the list above.&lt;/p&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2017-05-01T01:12:36.613" Title="Difficult IQ test question: What is the box suggesting?" Tags="|pattern|visual|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2284" PostTypeId="1" AcceptedAnswerId="2290" CreationDate="2014-09-18T09:06:16.993" Score="6" ViewCount="2226" Body="&lt;p&gt;I just came up with the following puzzle. I don't think anything like this has been asked here before, but I may be wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the next symbol in this sequence?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1DH6C.png&quot; alt=&quot;(ヨ, ミ, ワ, something, something, something&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="40" LastEditorUserId="71" LastEditDate="2014-11-30T15:34:24.213" LastActivityDate="2014-11-30T15:34:24.213" Title="Find the next symbol in this sequence" Tags="|pattern|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2286" PostTypeId="2" ParentId="2283" CreationDate="2014-09-18T13:31:20.720" Score="17" Body="&lt;p&gt;Disclaimer: I'm not certain this is the complete (or even right) answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All little symbols in the perimeter squares (disregard the middle one for now) point to the next square, going counterclockwise. Starting from the upper left,&#xA;these symbols are the top half (without centre bar) of the digital representations of the digits 1 through 8. The square in the middle is a 9 with the top and bottom quarter missing, as indicated by the 'X'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of the other sequences have the correct digital representations to fit the given upper halves: of the symbols in the given sequences, only '1' fits the top left, which provides a starting point to check each sequence.&lt;/p&gt;&#xA;" OwnerUserId="439" LastEditorUserId="19765" LastEditDate="2017-05-01T01:05:18.113" LastActivityDate="2017-05-01T01:05:18.113" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2287" PostTypeId="2" ParentId="2281" CreationDate="2014-09-18T14:52:17.087" Score="8" Body="&lt;p&gt;There is no paradox as long as you realize that, when you say &quot;all adjectives can be categorized as autological &lt;em&gt;or&lt;/em&gt; heterological&quot;, the &quot;or&quot; is not exclusive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To explain this a little bit better, let's try to describe this using math syntax. Let's call $f$ a function that returns the relevant description of a word. So $f($&quot;short&quot;$)$ is &quot;short&quot;, and $f($&quot;long$)$ is also &quot;short&quot;. Then if $f(a)=a$ then $a$ is autological, while if $f(a)\ne a$ then $a$ is heterological.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using &quot;autological&quot; in this formula works fine because $f($&quot;autological&quot;$)=$ &quot;autological&quot;. However, what about $f($&quot;heterological&quot;$)$? If $f($&quot;heterological&quot;$)\ne$ &quot;heterological&quot;, then it is heterological, and if $f($&quot;heterological&quot;$)=$ &quot;heterological&quot;, then it is autological. This is where the apparent paradox comes in, but what this really suggests is that our function $f$ is poorly defined!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, words can be described by more than one description - for example, &quot;polysyllabic&quot; is both polysyllabic and long (or medium-sized, if you want). So why not have $f$ return a set of descriptions of the word? This would make an autological word one for which $a\in f(a)$. A heterological word could be defined either as $a\notin f(a)$, or one for which $\neg a\in f(a)$, where $\neg a$ is an antonym of the word. Assuming the first definition of heterological, if &quot;heterological&quot; $\notin f($&quot;heterological&quot;$)$, then &quot;heterological&quot; $\in f($&quot;heterological&quot;$)$, so we should use the second definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's still something a little weird here, but it is no longer self-contradictory. Suppose &quot;heterological&quot; $\notin f($&quot;heterological&quot;$)$. Then it is not autological, and because &quot;autological&quot; $\notin f($&quot;heterological&quot;$)$, it is also not heterological. However, suppose &quot;heterological&quot; $\in f($&quot;heterological&quot;$)$. Then it is autological, and since &quot;autological&quot; $\in f($&quot;heterological&quot;$)$, it is also heterological. The logic in both cases is circular, but neither one actually results in a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while this does not answer the question of whether &quot;heterological&quot; is heterological or not, this does resolve the paradox - it is either both heterological and autological, or neither. I believe that we should consider it to be both, making it an &lt;a href=&quot;http://en.wikipedia.org/wiki/Auto-antonym&quot; rel=&quot;nofollow&quot;&gt;auto-antonym&lt;/a&gt;. Also, it suggests that we refine the definition of &quot;heterological&quot; to be more like &quot;it can describe something that it is not&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1955" LastEditorUserId="1955" LastEditDate="2015-06-06T04:09:04.693" LastActivityDate="2015-06-06T04:09:04.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2288" PostTypeId="2" ParentId="2237" CreationDate="2014-09-18T15:07:01.073" Score="0" Body="&lt;p&gt;Two men walk into a bank, wearing masks that disguise their identities.  One of them speaks, his voice clearly identifying him as John Wayne.  One of them shoots the teller and takes the money.  Both leave.  There are no cameras or witnesses other than the monkeys.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Blind monkey knows that one of the two men was John Wayne, but doesn't know whether he was the shooter or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deaf monkey knows which man shot the teller, but didn't hear them speak, so he doesn't know which was John Wayne.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mute monkey saw and heard and knows whether it was John Wayne that shot the teller, but cannot communicate what he witnessed.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2014-09-18T15:07:01.073" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2290" PostTypeId="2" ParentId="2284" CreationDate="2014-09-18T15:53:52.597" Score="6" Body="&lt;pre&gt;&lt;code&gt; /\     /      /\     /             /\           &#xA;/ /\   / /    /  \   /        /\   /        / &#xA; / /    / /   \  /   \  /    /     \  /   \/  &#xA;  /      /      /      /             /     \  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Hint&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\pi$ and xor&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Spoiler&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Write the digits of pi on a 7-segment display and generate every symbol in the sequence by combining the previous symbol with the next digit of $\pi$ with an xor.  Start with all segments off.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2014-09-21T01:07:41.763" LastActivityDate="2014-09-21T01:07:41.763" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2291" PostTypeId="1" AcceptedAnswerId="2293" CreationDate="2014-09-18T17:55:42.073" Score="10" ViewCount="2139" Body="&lt;p&gt;I came across this puzzle somewhere on the interwebs. Has stumped me for an answer for quite sometime. Helpplease?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Someone picks, at their will, two cards from a deck of cards. The&#xA;  cards have different numbers, i.e one card is higher than the other. (In&#xA;  other words, the person picks two distinct numbers in the inclusive&#xA;  range 1 through 13. [11-13 : J-K]  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The cards are placed face down on a table in front of you.  You get to&#xA;  choose one of the cards and turn it face up.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now, you will select one&#xA;  of the two cards (one of whose face you can see, the other one you&#xA;  can’t).  If you select the highest card, you win.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Is there a &#xA;  card-selection strategy for which your chance of winning is &lt;strong&gt;strictly&#xA;  greater&lt;/strong&gt; than 50% ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2326" LastEditorUserId="4551" LastEditDate="2015-01-02T16:31:43.543" LastActivityDate="2015-01-02T16:31:43.543" Title="Two Card Probability Problem" Tags="|probability|cards|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2292" PostTypeId="2" ParentId="2291" CreationDate="2014-09-18T18:12:17.537" Score="1" Body="&lt;p&gt;The card you have facing up has a value of $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability of randomly picking a card greater than $X$ is $\frac{4*(13-X)}{13*4-4}=\frac{13-X}{12}$.  The chance that it is smaller than $X$ is $\frac{X-1}{12}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second card is, essentially, randomly chosen from the cards in the deck not equal to $X$.  You can therefore pick the face up card if $X$ is greater than $7$.  If $X$ is less than $7$ then pick the other card.  In each of these cases, you have a probability of winning greater than 50%.  It is actually equal to $\frac{|X-7|}{12}+\frac{1}{2}$.  If, however, $X=7$ then you have a 50% probability for either one being higher.&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2014-09-18T18:12:17.537" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2293" PostTypeId="2" ParentId="2291" CreationDate="2014-09-18T18:20:34.327" Score="12" Body="&lt;p&gt;Use a random source to pick a number from 2 to 13.  If it is equal or less than the value of the card you picked, keep the card.  If it is greater, switch.  Thus, if you turned over a 1, you always switch; and a 13, you never switch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the card numbers are $a$ and $b$, with $a &amp;gt; b$, then you win if you flip $a$ ($50\%$ chance) and don't switch (probability $(a-1)/12$), or if you flip $b$ and do switch (probability $50\% \cdot (13-b)/12$).  Your total chance of winning is:&#xA;$$P = 0.5 \cdot{a-1\over 12} + 0.5 \cdot{13-b\over 12} = 0.5 \cdot {a - 1 + 13 - b&#xA;\over 12} = 0.5 +{a-b\over 24}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $a &amp;gt; b$, your odds of winning are greater than $50\%$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply choosing to keep a card that is higher than $7$ will not work, as they are not picking cards at random.  They could deliberately choose $9$ and $10$, so that strategy would have only a $50\%$ chance of winning.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="3987" LastEditDate="2015-01-02T16:22:08.103" LastActivityDate="2015-01-02T16:22:08.103" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2294" PostTypeId="2" ParentId="1653" CreationDate="2014-09-18T20:32:52.793" Score="-1" Body="&lt;p&gt;I think the standard solution is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First person slices off a piece of cake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second person has two choices: let first person take that piece and slice off a new piece for herself, or slice off a portion of the first piece (to be added back to the whole) and keep the piece for herself.  In the second option, the first person goes back to the end of the line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After second person makes her choice, each successive person has the same option until each person has a piece.  If the second person chose to take a new slice, then the third person will be deciding on that; the first person will take her slice and walk away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No one would choose to slice off a piece smaller than $1/N$, as they would get stuck with that piece and have less than the others.  If anyone selects more than $1/N$, that piece would get taken away by the next person and whittled down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method only works if no one is willing to give someone else a larger share if it means cutting into their own share.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2014-09-18T20:32:52.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2297" PostTypeId="1" AcceptedAnswerId="2298" CreationDate="2014-09-19T00:21:45.540" Score="26" ViewCount="2204" Body="&lt;p&gt;Lately I have found different versions of this puzzle, I was wondering if it has a name or belongs to a specific kind, so that I can understand it better.&#xA;Here it is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Peter got \$10 from his father, \$10 from his mother and went to buy some&#xA;  groceries. He spent only \$17.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Once he is back home he gives back \$1 to his father, \$1 to his mother&#xA;  and he keeps \$1 for himself. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In the end his mother spent \$9, his father spent \$9 and that is \$18. Plus&#xA;  the dollar he kept, \$19. Where has that last dollar gone?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2334" LastEditorUserId="5169" LastEditDate="2015-06-24T13:00:04.250" LastActivityDate="2018-02-09T10:10:52.637" Title="Where has the last dollar gone?" Tags="|puzzle-identification|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2298" PostTypeId="2" ParentId="2297" CreationDate="2014-09-19T02:49:47.733" Score="34" Body="&lt;p&gt;This type of puzzle (typically called the missing dollar puzzle in this case) is referred to as an &lt;a href=&quot;http://en.wikipedia.org/wiki/Informal_fallacy&quot;&gt;informal fallacy&lt;/a&gt; puzzle. The general idea is that the logic that is presented to the user appears at first glance to be correct, but in actuality has an error in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the 9 spent by his father and 9 spent by his mother together equal the 17 spent at the store plus the dollar that he kept. Adding the dollar he kept to that again to get 19 is the error. It is already accounted for.&lt;/p&gt;&#xA;" OwnerUserId="375" LastActivityDate="2014-09-19T02:49:47.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2300" PostTypeId="1" AcceptedAnswerId="2306" CreationDate="2014-09-19T03:15:05.080" Score="16" ViewCount="1452" Body="&lt;p&gt;I have seen this question before and I am intrigued that there might be various ways it can be solved. I do know of one way, but I am trying to find out if there is any other approach possible. The acceptable answer has to provide a methodology, preferably multiple, along with the answer to the question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 25 horses and 5 tracks. Horses run at their original speed, regardless of how many races they run. What is the least number of races needed to determine the top 3 horses?  There are no stopwatches or other equipment, and the only way to determine the order is ranking in a specific race.   &lt;/p&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="463" LastEditDate="2015-06-05T15:04:53.753" LastActivityDate="2017-03-08T03:15:51.247" Title="Horses and Tracks" Tags="|mathematics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2303" PostTypeId="1" AcceptedAnswerId="2307" CreationDate="2014-09-19T11:31:50.340" Score="14" ViewCount="8320" Body="&lt;p&gt;Write the digits from 1 to 9 in an empty Sudoku table with these rules (similar to 9-Queen problem):&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There are 9 boxes indexed from 1 to 9 in the image below. Each box has a 3x3 grid child.&lt;/li&gt;&#xA;&lt;li&gt;Every two digits should NOT be in the same row, column or diagonal.&lt;/li&gt;&#xA;&lt;li&gt;Pointing rule: You have to place the next number in an specific box. (e.g. if you start to place 1 in top-left box and middle-center child grid, you have to continue with 2 in middle-center box)&lt;/li&gt;&#xA;&lt;li&gt;You are free to begin from any box.&lt;/li&gt;&#xA;&lt;li&gt;Close the loop: The last number (9) should point to the box that you have placed the first number (1) in it.&lt;/li&gt;&#xA;&lt;li&gt;At the end there will be a Sudoku table with only 9 digits in it each in a separate 3x3 box. (other 72 cells will remain empty)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Write down the 9 digits with the rules, or if there is no such a solution, prove it!&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MswnQ.png&quot; alt=&quot;3-by-3 boxes in the Sudoku grid numbered from 1 to 9, left to right and then top to bottom&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;See the following example that have some mistakes:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dFJ9h.png&quot; alt=&quot;mistakes&quot; /&gt;&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;8 and 9 are in the same row.&lt;/li&gt;&#xA;&lt;li&gt;8 and 5 are in a diagonal.&lt;/li&gt;&#xA;&lt;li&gt;5 and 2 are in the same row.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="69582" LastEditDate="2021-09-11T00:12:45.983" LastActivityDate="2021-09-11T19:25:42.493" Title="Write digits 1 to 9 in the table, each pointing to next 3x3 grid" Tags="|logical-deduction|grid-deduction|sudoku|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="2306" PostTypeId="2" ParentId="2300" CreationDate="2014-09-19T17:52:14.673" Score="14" Body="&lt;h1&gt;Intuitive Solution:&lt;/h1&gt;&#xA;&lt;p&gt;When you first look at the problem, intuitively, it makes sense to first split the 25 horses into 5 groups and then run a race with each group - we will call these the 5 heats.&lt;/p&gt;&#xA;&lt;p&gt;In the 6th race, you can run just the winners.  Lets say that the results of this race are as follows:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;1st place: A1&#xA;2nd place: B1&#xA;3rd place: C1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Without a doubt, you can say that A1 is the fastest horse.  But how do you know if A2 (the horse that finished second in the heat with A1) is faster than B1 or not?  In fact, you will have the following options for the 2nd and 3rd fastest horses overall:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;1st overall: A1&#xA;2nd overall: A2 or B1&#xA;3rd overall: A3 or B2 or C1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;So, a final (7th) consolation race is needed with A2, A3, B1, B2, and C1.&lt;/p&gt;&#xA;&lt;p&gt;So, one solution would be that you need 7 races to decide the ordering of the top 3.&lt;/p&gt;&#xA;&lt;p&gt;This all started with intuition, and while it may seem obvious that there isn't a better solution, perhaps there are equivalent ones.&lt;/p&gt;&#xA;&lt;h1&gt;Choices for heats&lt;/h1&gt;&#xA;&lt;p&gt;Perhaps a different way would be to run the first race to get 3 candidates, and then run the rest of the horses against one (or more) of these 3 to see how they fit in.&lt;/p&gt;&#xA;&lt;p&gt;If you choose to run all three of the known horses in the subsequent races, it will take 10 more races to finish ordering all the horses since there are 20 horses who haven't yet run, and each race only has room for 2 more.&lt;/p&gt;&#xA;&lt;p&gt;If you choose to run only 2 of the known horses in the next race, then it will take 7 more races before all the horses have run at least once (total of 8 races), and this is also worse than the best method so far.  Even so, there may be additional races required to sort out the fastest 3.&lt;/p&gt;&#xA;&lt;p&gt;If you choose a single known horse, then you have room for 4 new horses every race, so 5 more races (total of 6 so far).  However, no matter which horse you choose to run again, in the worst case, you will add 2 horses to the list of possible &amp;quot;top 3&amp;quot; horses, so at the end, you will have 10 horses left who could qualify in the top 3.  This would take 2 more races to sort out (total of 8 races).  However, in the best case, (say you run A3 in the subsequent races and he wins every race) you would know the answer in only 6 races!  Perhaps the average would be 7 races as well, but the worst case would be 8 or more.&lt;/p&gt;&#xA;&lt;p&gt;Thus the optimal solution should include heats of 5 groups of 5.&lt;/p&gt;&#xA;&lt;h1&gt;Choices for 6th race&lt;/h1&gt;&#xA;&lt;p&gt;After the heats, you have 15 horses who qualified.  You must be able to get the top three in some way from these 15.  By running the winners in a single race, you will eliminate all the horses in the groups that finished 4th and 5th, as well as the two horses in the 3rd place group, and the third horse in the 2nd place group.  This seems to minimize the remaining horses (5) and all fit into a single race for the consolation race, so there does not appear to be any obvious alternatives to the 6th race.&lt;/p&gt;&#xA;&lt;h1&gt;Choices for the 7th race&lt;/h1&gt;&#xA;&lt;p&gt;You will notice that the last race includes some known outcomes already.  For example:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;A2 is faster than A3&#xA;B1 is faster than B2&#xA;B1 is faster than C1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;However, the unknowns are:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;Is A2 faster than B1?&#xA;  If so, is A3 faster than B1?&#xA;    If so, then the order is A1, A2, A3&#xA;  otherwise&#xA;    The order is A1, A2, B1&#xA;otherwise Who is the fastest of A2, B2, and C1?&#xA;  If A2, the order is A1, B1, A2&#xA;  If B2, the order is A1, B1, B2&#xA;  If C1, the order is A1, B1, C1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;So, we need a race that solves these questions.  The only way to answer them all is to race these 5 horses.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-09-19T17:52:14.673" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2307" PostTypeId="2" ParentId="2303" CreationDate="2014-09-19T19:11:18.513" Score="21" Body="&lt;p&gt;It is not possible.&lt;/p&gt;&#xA;&lt;p&gt;Any solution to this puzzle must also be a solution to the 9x9 Queens puzzle. Luckily, that is a well-known puzzle. It has 352 solutions, but due to symmetry, those 352 solutions can be reduced to 46 solutions. After that, it is just a matter of checking against those 46 solutions.&lt;/p&gt;&#xA;&lt;p&gt;I found a page showing the solutions at: &lt;a href=&quot;http://stamm-wilbrandt.de/en/xsl-list/n-queens/n-queens.xsl.xml&quot; rel=&quot;noreferrer&quot;&gt;http://stamm-wilbrandt.de/en/xsl-list/n-queens/n-queens.xsl.xml&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;I then looked at the first answer in each of the 46 columns.&lt;/p&gt;&#xA;&lt;p&gt;Items that quickly ruled a solution out:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Queen in a corner. Any queen in a corner would mean that the box referred to itself.&lt;/li&gt;&#xA;&lt;li&gt;Queen in the center of a side. Again, that box referred to itself.&lt;/li&gt;&#xA;&lt;li&gt;Queen in center box. Same reason.&lt;/li&gt;&#xA;&lt;li&gt;2 Queens in same 3x3 box / empty 3x3 box.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;This left me with very few actual solutions to check. From that page, #'s 19, 20, 31, 39, 42.&lt;/p&gt;&#xA;&lt;p&gt;To check, I simply started in box 1 and followed the pattern until I returned to a box that I'd already seen. None of those took all 9 boxes.&lt;/p&gt;&#xA;" OwnerUserId="375" LastEditorUserId="69582" LastEditDate="2021-09-11T19:25:42.493" LastActivityDate="2021-09-11T19:25:42.493" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2309" PostTypeId="1" AcceptedAnswerId="2312" CreationDate="2014-09-20T05:52:37.430" Score="3" ViewCount="283" Body="&lt;p&gt;You can see the rules for &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2303/write-1-to-9-digits-in-the-table-each-pointing-to-next-3x3-grid&quot;&gt;pointing problem at this article&lt;/a&gt;. It has been clarified that there is no solution to this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, what if we remove the diagonal rule? is there any solution?&lt;/p&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2014-09-25T11:55:53.990" Title="Non-Diagonal Pointing Problem" Tags="|logical-deduction|sudoku|chess|logic-grid|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2310" PostTypeId="2" ParentId="2188" CreationDate="2014-09-20T09:39:36.743" Score="40" Body="&lt;p&gt;Here is a twisted solution.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Go to any guard, point at a door and ask:&lt;br&gt;&#xA; Among the propositions 1. &quot;You are a liar&quot;, 2. &quot;You will reply negatively&quot; and 3. &quot;This door leads to freedom&quot;, is there an odd number of true propositions?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you get the answer yes:&lt;br&gt;&#xA;If the guard is a truthteller, the number of truths is odd, 1. is false, 2. is false, so 3. must be true.&lt;br&gt;&#xA;If the guard is a liar, the number of truths is even, 1. is true, 2. is false, so 3. must be true.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you get a negative answer:&lt;br&gt;&#xA;If the guard is a truthteller, the number of truths is even, 1. is false, 2. is true, so 3. must be true.&lt;br&gt;&#xA;If the guard is a liar, the number of truths is odd, 1. is true, 2. is true, so 3. must be true.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So regardless of the answer of the guard, the door you pointed at is the door to freedom, you can leave safely.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&#xA;Before you argue about this solution, please read the following:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/7954/logic-explanation-in-two-doors-answer&quot;&gt;Logic explanation in &amp;quot;two doors&amp;quot; answer&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.923" LastActivityDate="2016-05-23T14:07:03.007" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="2311" PostTypeId="1" AcceptedAnswerId="3523" CreationDate="2014-09-20T11:42:20.693" Score="11" ViewCount="649" Body="&lt;p&gt;I am wondering when &lt;a href=&quot;https://puzzling.stackexchange.com/q/2297/70&quot;&gt;the missing dollar puzzle&lt;/a&gt; first appeared. I am sure I read it forty years ago or so, but maybe it is much older. Does anyone knows it?&lt;/p&gt;&#xA;" OwnerUserId="70" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2016-03-18T15:57:21.950" Title="origin of the &quot;missing dollar puzzle&quot;" Tags="|puzzle-history|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2312" PostTypeId="2" ParentId="2309" CreationDate="2014-09-20T15:59:07.170" Score="5" Body="&lt;pre&gt;&lt;code&gt;. 1 . | . . . | . . .&#xA;. . . | . 2 . | . . .&#xA;. . . | . . . | . 7 .&#xA;------+-------+------&#xA;. . . | . . . | 9 . .&#xA;. . . | 3 . . | . . .&#xA;4 . . | . . . | . . .&#xA;------+-------+------&#xA;. . . | . . . | . . 6&#xA;. . . | . . 8 | . . .&#xA;. . 5 | . . . | . . .&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I got 720 solutions having the 1 in the top left bloc.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2014-09-25T11:55:53.990" LastActivityDate="2014-09-25T11:55:53.990" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2313" PostTypeId="1" AcceptedAnswerId="2315" CreationDate="2014-09-20T17:21:26.880" Score="8" ViewCount="584" Body="&lt;p&gt;Ok, let say Group A has N number of people, and Group B has M number of people. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want each person in Group A to have a fixed T time conversation ( time T is fixed like 5 min or 10 min) with each person in Group B. That also means each person in Group B will have a fixed T time conversation with each person in Group A. People in the same Group will not talk to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Requirement: How to match each person in Group A to the one in Group B in such a way that all people in Group A &amp;amp; B finishes all their conversations in shortest amount of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, Group A has 2 people, Group B has 4 people, time T is 5 mins. &#xA;Suppose the conversation start from 16pm, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can see this Matching pattern:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-From 16pm to 16:05pm, person 1 in Group A talks to person 1 in Group B, person 2 in Group A talks to person 2 in Group B. (ie person 3 &amp;amp; 4 in Group B will not have any conversation)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-From 16:05pm to 16:10pm, person 1 in Group A talks to person 2 in Group B, person 2 in Group A talks to person 1 in Group B. (ie person 3 &amp;amp; 4 in Group B will not have any conversation)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-From 16:10pm to 16:15pm, person 1 in Group A talks to person 3 in Group B, person 2 in Group A talks to person 4 in Group B. (ie person 1 &amp;amp; 2 in Group B will not have any conversation)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-From 16:15pm to 16:20pm, person 1 in Group A talks to person 4 in Group B, person 2 in Group A talks to person 3 in Group B. (ie person 1 &amp;amp; 2 in Group B will not have any conversation)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, 20 min is the shortest amount of time in which all people in Group A can talk to all people in Group B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is how to find the matching pattern with for N number of people in Group A, and M number of people in Group B, and T is one fixed conversation.&lt;/p&gt;&#xA;" OwnerUserId="2344" LastEditorUserId="2344" LastEditDate="2014-09-21T02:39:16.773" LastActivityDate="2014-09-21T02:39:16.773" Title="How to let all people in Group A talk to all people in Group B in a shortest amount of time?" Tags="|logical-deduction|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2314" PostTypeId="2" ParentId="2313" CreationDate="2014-09-20T17:40:38.523" Score="4" Body="&lt;p&gt;The (or at least, an) optimal matching pattern is to take the larger group (or either if both are the same size), and pair each person in the larger group (we'll say group L) with their equivalent numbered person in the smaller group (group S), so person L1 talks to S1, L2 to S2, and so on. And members of group L that have no corresponding members of group S may be left out of the conversation round. Then, for the next round of conversations transpose the members of group L by one so that L1 talks to S2, L2 to S3, etc, with the last member of group L, LN, moving around to talk to S1. Again, members currently at the 'bottom' of the group L list will (assuming the 2 groups are not the same size), be left out of the rotation. Repeat the transposition/conversation process N times in total, so that at the end of the last conversation L1 is the last entry on the list of member of group L paired up against the ordered members of group S, and all members of each group will have talked to all members of the other group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the total time taken is T*N, where N in the number of people in group L. It should be apparent that it is impossible to achieve the desired result in less than N conversations, since even if the smaller group contains only 1 person, she will have to talk to all N members of the larger group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S: Sorry if that was a little rambling, but I hope it made sense.&lt;/p&gt;&#xA;" OwnerUserId="320" LastActivityDate="2014-09-20T17:40:38.523" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2315" PostTypeId="2" ParentId="2313" CreationDate="2014-09-20T18:31:51.337" Score="8" Body="&lt;p&gt;I think i found a simple solution, that is all people in the bigger Group will form a round circle, and all people in smaller Group will form an incomplete circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The people in the  incomplete circle will go around in such a way that the 1st person will start from the beginning position (S) and finish in the (S - 1) position, see picture 1 and 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure my solution is ok.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2e01Q.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2jXRs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2344" LastActivityDate="2014-09-20T18:31:51.337" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2316" PostTypeId="1" AcceptedAnswerId="2321" CreationDate="2014-09-21T03:49:37.793" Score="-1" ViewCount="403" Body="&lt;p&gt;What is a geographical similarity between Cambridge University and Oxford University related to their 'titles'?&lt;/p&gt;&#xA;" OwnerUserId="2312" LastEditorUserId="22922" LastEditDate="2016-09-19T07:22:46.353" LastActivityDate="2016-09-19T07:22:46.353" Title="Oxford versus Cambridge - a geographical puzzle" Tags="|english|word-problem|geography|" AnswerCount="2" CommentCount="2" ClosedDate="2016-09-25T16:39:54.350" ContentLicense="CC BY-SA 3.0" />
  <row Id="2317" PostTypeId="1" AcceptedAnswerId="2362" CreationDate="2014-09-21T05:03:18.937" Score="3" ViewCount="2474" Body="&lt;p&gt;On one of the Islands of Knights and Knaves you land for the purposes of an anthropological survey. The Island has Knights who always tell the truth ( if they know enough information regarding a question put to them) and Knaves who always lie ( also if they know enough relevent info.) ( Of course a Knight or Knave could respond with a respectively true of false statement even if they didn't know enough relevent info.) Also on the Island are the Tax-collectors , inhabitants who sometimes lie and sometimes tell the truth. The thing is these tax-collectors are very good at hiding what 'type' of inhabitant they are. For example if you ask directly a person whom you don't know is a tax-collector 'What is (2+2)?', If he answers 4 and you ask him again in one hour he won't say 5. The tax collectors have great memories and even if he forgets a previous response he might say, 'What did I say before'? You're wondering if there is one sure fire yes-or-no question that if a person responds you can tell what 'type' they are. Is there such a question?&lt;/p&gt;&#xA;" OwnerUserId="2312" LastEditorUserId="28" LastEditDate="2014-09-21T07:29:59.707" LastActivityDate="2015-06-06T03:02:39.040" Title="About the island of Knights and Knaves" Tags="|logical-deduction|liars|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2318" PostTypeId="2" ParentId="2316" CreationDate="2014-09-21T06:29:44.480" Score="2" Body="&lt;p&gt;A bridge goes over a river and you ford a river?&lt;/p&gt;&#xA;" OwnerUserId="1777" LastActivityDate="2014-09-21T06:29:44.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2319" PostTypeId="2" ParentId="2317" CreationDate="2014-09-21T07:18:47.107" Score="1" Body="&lt;p&gt;&lt;em&gt;From the formulation I would conclude that a tax-collector can answer whatever he finds reasonable, using any unknown algorithm to generate answer. This includes random generator algorithms for questions which are asked a first time (and the next time, he will repeat the answer).&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, it is quite obvious that the answer is &lt;strong&gt;no&lt;/strong&gt;. There is no useful information that you can extract from a tax-collector answer independently of what you are asking. This includes information about the type of the asked person.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(Though, if you know somebody who is not tax-collector you can ask about a person you are interested in &quot;what would you answer if I ask you what type is this person?&quot;.)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="463" LastEditDate="2015-06-06T03:02:39.040" LastActivityDate="2015-06-06T03:02:39.040" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="2320" PostTypeId="2" ParentId="2267" CreationDate="2014-09-21T15:08:55.267" Score="2" Body="&lt;p&gt;Here is an edge-view picture of a square or a circle: |&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like both at once, or indeed like any other planar polygon viewed edge-on.  It is fairly easy to draw, using a pencil, ruler, and piece of paper.&lt;/p&gt;&#xA;" OwnerUserId="2004" LastActivityDate="2014-09-21T15:08:55.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2321" PostTypeId="2" ParentId="2316" CreationDate="2014-09-21T16:36:32.247" Score="5" Body="&lt;p&gt;As noted in Jason's answer, both &lt;em&gt;bridge&lt;/em&gt; and &lt;em&gt;ford&lt;/em&gt; have to do with rivers, in these cases the &lt;a href=&quot;http://en.wikipedia.org/wiki/River_Cam&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cam&lt;/a&gt; and the &lt;a href=&quot;http://en.wikipedia.org/wiki/The_Isis&quot; rel=&quot;nofollow noreferrer&quot;&gt;Isis&lt;/a&gt; portion of the &lt;a href=&quot;http://en.wikipedia.org/wiki/River_Thames&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thames&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Etymonline's &lt;em&gt;&lt;a href=&quot;http://www.etymonline.com/index.php?search=Cambridge&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cambridge&lt;/a&gt;&lt;/em&gt; entry says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Old English &lt;em&gt;Grontabricc&lt;/em&gt; (c.745) “Bridge on the River Granta” (a Celtic river name, of obscure origin). The change to &lt;em&gt;Cante-&lt;/em&gt; and later &lt;em&gt;Cam-&lt;/em&gt; was due to Norman influence. The river name &lt;em&gt;Cam&lt;/em&gt; is a back-formation in this case, but &lt;em&gt;Cam&lt;/em&gt; also was a legitimate Celtic river name, meaning “crooked.”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Etymonline's &lt;em&gt;&lt;a href=&quot;http://www.etymonline.com/index.php?search=Oxford&quot; rel=&quot;nofollow noreferrer&quot;&gt;Oxford&lt;/a&gt;&lt;/em&gt; entry says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;university town in England, Middle English &lt;em&gt;Oxforde&lt;/em&gt;, from Old English &lt;em&gt;Oxnaforda&lt;/em&gt; (10c.) literally “where the oxen ford.” In reference to a type of shoe laced over the instep, it is attested from 1721 (&lt;em&gt;Oxford-cut shoes&lt;/em&gt;). Related: &lt;em&gt;Oxfordian; Oxfordish; Oxfordist; Oxfordy&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Cambridge/Oxford river-crossing coincidence is fairly minor, but see ELU's &lt;a href=&quot;https://english.stackexchange.com/q/131650&quot;&gt;OxFORD and CamBRIDGE&lt;/a&gt; for more discussion of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A slightly deeper coincidence is that between &lt;em&gt;Oxford&lt;/em&gt; and &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Bosphorus#Name&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bosphorus&lt;/a&gt;&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Greek name Βόσπορος (&lt;em&gt;Bosporos&lt;/em&gt;) was folk-etymologized as from βοὸς πόρος, i.e. “cattle strait” (or “Ox-ford”), from the genitive of &lt;em&gt;bous&lt;/em&gt; βοῦς “ox, cattle” + poros πόρος “passage”, thus “cattle-passage”, or “cow passage” in reference to Io from Greek mythology who was transformed into a cow and condemned to wander the earth until she crossed the Bosporus where she met Prometheus.... [wikipedia] &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A number of 3-university (Oxford, Cambridge, Bosphorus) boat races have been held in Istanbul (see &lt;a href=&quot;http://www.hurriyetdailynews.com/default.aspx?pageid=438&amp;amp;n=eighth-oxford-cambridge-bosphorus-university-boat-race-to-be-held-1998-08-28&quot; rel=&quot;nofollow noreferrer&quot;&gt;hurriyetdailynews.com&lt;/a&gt;) analogous to the famous &lt;a href=&quot;http://en.wikipedia.org/wiki/The_Boat_Race&quot; rel=&quot;nofollow noreferrer&quot;&gt;University Boat Race&lt;/a&gt; held annually on the Thames.&lt;/p&gt;&#xA;" OwnerUserId="2004" LastEditorUserId="-1" LastEditDate="2017-04-13T12:38:16.170" LastActivityDate="2014-09-21T16:36:32.247" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2322" PostTypeId="2" ParentId="2317" CreationDate="2014-09-22T04:31:31.203" Score="5" Body="&lt;p&gt;&lt;strong&gt;Intuitive Answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming &quot;sometimes lies and sometimes tells the truth&quot; means that a tax collector will lie for at least one question and tell the truth for at least one question, then we can have the following two tax collector collections:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Tax collectors of type #1 will give the same answer a Knight gives except for &quot;does 1=1?&quot;, where they'll say &quot;no&quot;.&lt;/li&gt;&#xA;&lt;li&gt;Tax collectors of type #2 will tell the same answer a Knight gives except for &quot;does 2=2?&quot;, where they'll say &quot;no&quot;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The first question cannot distinguish #1 from Knaves or #2 from Knights. The second question cannot distinguish #2 from Knaves or #1 from Knights. For any other question, the tax collectors are functionally identical to Knights and cannot be distinguished. So the answer is &lt;strong&gt;no&lt;/strong&gt;, no single question is enough to distinguish Knights and Knaves from Tax Collectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that using &quot;answers like a Knight&quot; instead of &quot;tells the truth&quot; means we don't have to think about tricky meta questions (&quot;what would you say if I asked you 'does 1=1 AND is your answer to 2=2 false'?&quot;), because the tax collector will treat the entire thing as if it were a Knight, not break it down and answer each piece like a tax collector. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Math Answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are three possibilities: Knight, Knave, tax collector. There are two possibilities for an answer: yes, no. So for any given question, one answer must map to Knight/Knave and the other must map to Knight or collector/Knave or collector. So you can't distinguish them with just one question.&lt;/p&gt;&#xA;" OwnerUserId="1906" LastActivityDate="2014-09-22T04:31:31.203" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2323" PostTypeId="1" AcceptedAnswerId="2327" CreationDate="2014-09-22T13:58:28.570" Score="33" ViewCount="12537" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Unexpected_hanging_paradox&quot;&gt;unexpected hanging paradox&lt;/a&gt; is as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A judge tells a condemned prisoner that he will be hanged at noon on&#xA;  one weekday in the following week but that the execution will be a&#xA;  surprise to the prisoner. He will not know the day of the hanging&#xA;  until the executioner knocks on his cell door at noon that day.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Having reflected on his sentence, the prisoner draws the conclusion&#xA;  that he will escape from the hanging. His reasoning is in several&#xA;  parts. He begins by concluding that the &quot;surprise hanging&quot; can't be on&#xA;  Friday, as if he hasn't been hanged by Thursday, there is only one day&#xA;  left - and so it won't be a surprise if he's hanged on Friday. Since&#xA;  the judge's sentence stipulated that the hanging would be a surprise&#xA;  to him, he concludes it cannot occur on Friday.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;He then reasons that the surprise hanging cannot be on Thursday&#xA;  either, because Friday has already been eliminated and if he hasn't&#xA;  been hanged by Wednesday night, the hanging must occur on Thursday,&#xA;  making a Thursday hanging not a surprise either. By similar reasoning&#xA;  he concludes that the hanging can also not occur on Wednesday, Tuesday&#xA;  or Monday. Joyfully he retires to his cell confident that the hanging&#xA;  will not occur at all.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The next week, the executioner knocks on the prisoner's door at noon&#xA;  on Wednesday — which, despite all the above, was an utter surprise to&#xA;  him. Everything the judge said came true.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From this I ask:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How is it that despite the prisoner's sound logic, the judge is correct just the same?&lt;/li&gt;&#xA;&lt;li&gt;Assuming the prisoner is a logician, is it even possible for the judge to be wrong?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1672" LastActivityDate="2024-01-13T02:44:40.513" Title="Unexpected hanging paradox" Tags="|paradox|" AnswerCount="12" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2324" PostTypeId="2" ParentId="2323" CreationDate="2014-09-22T14:33:44.010" Score="9" Body="&lt;p&gt;&lt;strong&gt;The simple (and I believe the only) answer to both your questions is &quot;This is a paradox, so logic does not have predictability power here&quot;.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Martin Gardner described this paradox in detail. Unfortunately, I can't find the English version online, but if you know Russian, you can find it &lt;a href=&quot;http://golovolomka.hobby.ru/books/gardner/surprize.shtml&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; (The English Google translation is &lt;a href=&quot;https://translate.google.ru/translate?hl=ru&amp;amp;sl=ru&amp;amp;tl=en&amp;amp;u=http%3A%2F%2Fgolovolomka.hobby.ru%2Fbooks%2Fgardner%2Fsurprize.shtml&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;). The English version should be in his book &lt;a href=&quot;http://books.google.de/books/about/The_unexpected_hanging.html?id=cP_uAAAAMAAJ&amp;amp;redir_esc=y&quot; rel=&quot;noreferrer&quot;&gt;The unexpected hanging and other mathematical diversions.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;P.S. The paradox is most extreme and easier to feel in the 1-day version (A judge tells a condemned prisoner that he will be hanged at noon tomorrow but that the execution will be a surprise to the prisoner).&lt;br&gt;&#xA;The formulation stays the same: &quot;He begins by concluding that the &quot;surprise hanging&quot; can't be tomorrow - it won't be a surprise. Since the judge's sentence stipulated that the hanging would be a surprise to him, he concludes it cannot occur tomorrow. Joyfully, he retires to his cell confident that the hanging will not occur at all. The next day, the executioner knocks on the prisoner's door — which, despite all the above, was an utter surprise to him. Everything the judge said came true.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="463" LastEditDate="2015-06-05T14:48:34.033" LastActivityDate="2015-06-05T14:48:34.033" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2325" PostTypeId="1" AcceptedAnswerId="2326" CreationDate="2014-09-22T17:00:06.823" Score="22" ViewCount="37218" Body="&lt;p&gt;This puzzle consists of counting ten triangles (check three sides for each one, remember that there aren't exist triangles with more than three sides :P ) using two straight lines that cross the figure wherever you want. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't count a triangle that contains other triangles inside. I mean, every triangle has to have an empty area inside.&#xA;Here is an example of the game:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yFVZu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is kind of easy to find nine triangles, but I only could find one solution that solves the problem. &#xA;Good luck! Ask me if you don't understand the rules!&lt;/p&gt;&#xA;" OwnerUserId="2260" LastEditorUserId="2260" LastEditDate="2016-07-04T12:36:17.817" LastActivityDate="2018-04-07T08:31:38.903" Title="Find 10 triangles in a five pointed star using two straight lines" Tags="|geometry|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2326" PostTypeId="2" ParentId="2325" CreationDate="2014-09-22T18:56:52.530" Score="28" Body="&lt;p&gt;I was starting to get disappointed, but finally I found the solution. here it is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PfflG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2225" LastActivityDate="2014-09-22T18:56:52.530" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2327" PostTypeId="2" ParentId="2323" CreationDate="2014-09-22T20:39:36.607" Score="31" Body="&lt;p&gt;The biggest problem with the prisoner's proof is that his model is incomplete. One piece of that is that it merges two key variables - days on which the execution can happen, and days on which the prisoner believes it possible to happen. It's easy to see why these two were merged - because of the requirement of surprise, they are strongly correlated. However, by believing it impossible for the execution to happen on a particular day, he in fact makes it possible for it to happen on that day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Understanding this shows where his proof falls apart. He believes that there is no alternative to his being executed by Friday at the latest, which means that he cannot be executed on Friday. However he ends up concluding that he will not be executed at all, which means that there is now an alternative to his being executed by the last day. Thus the basis of his proof is invalidated. Of course he doesn't realize this, so he confidently believes the set of days on which he could be executed to be empty, which allows the set of days on which he can be surprised by being executed to be all weekdays of the following week.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately for the prisoner, if he were a perfect logician he would realize that there is no stable solution for which he knows any information about which days he might be executed on. Again, this is because the possibility of being executed on a given day is tied to his knowledge of it. If he comes to the conclusion that he will be executed on a particular day, it wouldn't be a surprise, meaning that he won't be executed on that day, but by knowing he won't be executed on that particular day it means he can be executed on that day. Thus any knowledge about being executed on a particular day is inherently unstable, meaning that he doesn't actually have any knowledge about what will happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fortunately for us, our knowledge of the prisoner's execution date does not affect the possibility of his being executed on that date. This allows us to use a complete model to represent what is happening. We know based on the prisoner's proof that he believes he will not be executed. So as I stated before, because the set of days on which he believed he could be executed was empty, he would have been surprised on any of the days.&lt;/p&gt;&#xA;" OwnerUserId="1955" LastEditorUserId="1955" LastEditDate="2014-09-23T17:20:26.490" LastActivityDate="2014-09-23T17:20:26.490" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="2333" PostTypeId="1" AcceptedAnswerId="2341" CreationDate="2014-09-23T15:34:22.460" Score="18" ViewCount="1064" Body="&lt;p&gt;A woman works at a ball factory. She's instructed to open up a ball creation kit composed of one red rubber ball and two strips of digit stickers from $0$ through $9$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;She's instructed to do the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Starting with $1$, write the current number on the ball using the digits available on the 2 strips.&lt;/li&gt;&#xA;&lt;li&gt;Put unused digits into a bowl where they can be used later for when she runs out of available digits in step #1.&lt;/li&gt;&#xA;&lt;li&gt;Open up another ball creation kit and continue to step #1 for the next number.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Assume there are no problems with space to put these digits (yes, there will be a lot of them).  Also, assume $6$ and $9$ can be differentiated (have a line underneath), so you cannot substitute one for the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, what will be the first number that she cannot write out? &lt;/p&gt;&#xA;" OwnerUserId="1672" LastActivityDate="2014-09-25T17:41:59.657" Title="The Ball Factory Worker" Tags="|word-problem|combinatorics|" AnswerCount="6" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="2334" PostTypeId="2" ParentId="2333" CreationDate="2014-09-23T20:13:53.927" Score="13" Body="&lt;p&gt;Let's lay down some groundwork to help us out:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Let $req(n, d)$ be the number of digit $d$ required to write out all the numbers from 1 to $n$.&lt;/li&gt;&#xA;&lt;li&gt;Since each kit contains two of each digit, we have $2n$ of each digit&lt;/li&gt;&#xA;&lt;li&gt;This means we want the smallest $n$ for which $req(n, d) &amp;gt; 2n$ for some $d$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It seems like this would be pretty easy to brute-force until you realize that the first number will have around 20 digits (individual digits see about equal use and we get 20 more digits each time). So instead let's figure out how many of each number is required in order to write up to 9, 99, 999, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Up to 9: One each of 1 through 9&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Up to 99: One each of 0 through 9 for each group of ten, then ten each of 1 through 9 for the numbers in the tens places. This totals twenty of each except zero, leaving us with 99*2-20 = 178 of each digit other than zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Up to 999: Same as up to 99 (with leading zeros) for each group of 100, then 100 each of 1 through 9 for the hundreds places. This totals twenty*10 + 100 = 300 of each digit required, leaving 999*2-300 = 1698 of each digit left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a pattern here is that to write all the numbers of $n$ and fewer digits, we need $req(10^{n-1}-1,1)*10+10^{n-1}$ of each digit other than zero. So where does this get close to $2n$?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Up to 9 (10^1-1):                          1&#xA;Up to 99 (10^2-1):              1*10+10 = 20&#xA;Up to 999 (10^3-1):          20*10+100 = 300&#xA;Up to 9999 (10^4-1):      300*10+1000 = 4000&#xA;Up to 99999 (10^5-1):                  50000&#xA;Up to 999999 (10^6-1):                600000&#xA;Up to 9999999 (10^7-1):              7000000&#xA;Up to 99999999 (10^8-1):            80000000&#xA;Up to 999999999 (10^9-1):          900000000&#xA;Up to 9999999999 (10^10-1):      10000000000&#xA;Up to 99999999999 (10^11-1):    110000000000&#xA;Up to 999999999999 (10^12-1):  1200000000000&#xA;...&#xA;Up to 9999999999999999999 (10^19-1):&#xA;Need 19000000000000000000&#xA;Up to 99999999999999999999 (10^20-1):&#xA;Need 200000000000000000000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So it is somewhere between $10^{19}-1$ and $10^{20}-1$ at the latest. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a good summary of what we've determined so far. Keep in mind that digits required and digits left isn't accurate for zero because it follows a different pattern.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;# balls   | digits req | digits left&#xA;------------------------------------&#xA;10^1  - 1 |  1 * 10^0  | 19*10^0 - 2&#xA;10^2  - 1 |  2 * 10^1  | 18*10^1 - 2&#xA;....................................&#xA;10^17 - 1 | 17 * 10^16 | 3*10^16 - 2&#xA;10^18 - 1 | 18 * 10^17 | 2*10^17 - 2&#xA;10^19 - 1 | 19 * 10^18 | 1*10^18 - 2&#xA;10^20 - 1 | 20 * 10^19 | 0*10^19 - 2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Looking at it a different way:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Since there have been some nice patterns so far, let's try looking at a different pattern that might give us the solution directly. What if we are only given one strip of digits in each kit? We can easily brute-force that with this Python code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;current = 1&#xA;available = [0]*10&#xA;while current &amp;lt; 200000000:&#xA;    current_str = str(current)&#xA;    for i in range(10):&#xA;        available[i] += 1 - current_str.count(str(i))&#xA;        if available[i] &amp;lt; 0:&#xA;            raise Exception('Done! %s has no more available at %s' % (i, current))&#xA;    current += 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This tells us that for 1 strip we run out of 1's at 199991. This would take to long to calculate for 2 strips, but what about partial strips? That doesn't exactly make sense so let's say she has to get the strips from a vending machine for 10 cents each, and each kit has 11 cents in it. Then after opening 10 kits she buy an extra strip for 11 strips total. Let's change this line:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;available[i] += cents_per_kit - cost_per_strip*current_str.count(str(i)) # cost_per_strip is 10&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here's a table of what this gives:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Strips/kit |       Result&#xA;-----------|-------------&#xA;    1.0    |       199991&#xA;    1.1    |      1999918&#xA;    1.2    |     19999198&#xA;    1.3    |    199991998&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Each of these is for when we run out of 1's. This looks like a promising pattern that should continue as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    1.4    |      199991998&#xA;    1.5    |     1999919998&#xA;    1.6    |    19999199998&#xA;    1.7    |   199991999998&#xA;    1.8    |  1999919999998&#xA;    1.9    | 19999199999998&#xA;    2.0    |199991999999998&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, there must a break in the pattern here - 199991999999998 only takes two 1's, so we can cover that with what is included in the standard kit. What this does tell us, however, is that @Joel Rondeau and @FlorianF almost certainly have it right - they came up with 1,999,919,999,999,981, which is almost exactly what my pattern suggests.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Practical Answer:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;She'll die long before she runs out of numbers. Also, she'd need a bigger bowl - one sheet of 8.5&quot; by 11&quot; paper weighs 0.04 lbs (according to Yahoo answers), and if the stickers are 2.5mm by 2.5mm (reasonably small), then each sticker weighs 4.5 milligrams. Right before ball $10^{19}$, she'll have about $10^{19}$ stickers left, which would weigh a total of $4.5*10^{13}$ kilograms. That's 45 billion tonnes. &lt;a href=&quot;http://en.wikipedia.org/wiki/Biomass_(ecology)&quot; rel=&quot;nofollow&quot;&gt;According to Wikipedia&lt;/a&gt;, that's one twelfth of the total live biomass excluding bacteria. And that's excluding all the balls and  stickers on those balls. They're probably using Soylent Green stickers...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it turns out, the first time she'll run out is shortly before reaching $2*10^{15}$. Right before ball $10^{15}$ she'll have around $5*10^{15}$ leftover. Using the same approximations, this would end up being 22.5 million tonnes. She will have used $1.5*10^{16}$ stickers up to that point, which would weigh 67.5 million tonnes, for a total of 90 million tonnes of stickers. &lt;a href=&quot;http://en.wikipedia.org/wiki/Great_Pyramid_of_Giza&quot; rel=&quot;nofollow&quot;&gt;The Great Pyramid of Giza&lt;/a&gt; is estimated at 5.9 million tonnes, so she has almost 4 Great Pyramids worth of leftover stickers.&lt;/p&gt;&#xA;" OwnerUserId="1955" LastEditorUserId="1955" LastEditDate="2014-09-25T17:41:59.657" LastActivityDate="2014-09-25T17:41:59.657" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2335" PostTypeId="1" AcceptedAnswerId="2344" CreationDate="2014-09-24T01:50:11.990" Score="7" ViewCount="626" Body="&lt;p&gt;Last year, just before my birthday in February, I noticed something unsual. The last two digits of my birth year were the same as my age. How old was I then and was this an unusual occurrence?&lt;/p&gt;&#xA;" OwnerUserId="2363" LastEditorUserId="11297" LastEditDate="2015-07-10T23:43:55.523" LastActivityDate="2015-07-10T23:43:55.523" Title="Odd birthdate surprise" Tags="|mathematics|calculation-puzzle|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2337" PostTypeId="2" ParentId="2335" CreationDate="2014-09-24T03:25:28.813" Score="2" Body="&lt;p&gt;You were born in either 1956 or 2006 and were 56 about to turn 57 or 6 about to turn 7. I'm guessing that you're not 8, so you were 56 about to turn 57.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for how unusual it is for this to happen, it is very common - everyone born in 1956 and 2006 (except for those born on January 1st) were all 56 and 6 respectively on January 1st, 2013. It is a little less common toward the end of a century though. For example, in 1989 people who were 44 or 94 could have noticed the same thing. The only reason it was less common then is because fewer people live to be 94.&lt;/p&gt;&#xA;" OwnerUserId="1955" LastActivityDate="2014-09-24T03:25:28.813" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2338" PostTypeId="2" ParentId="166" CreationDate="2014-09-24T03:45:10.253" Score="5" Body="&lt;p&gt;My solution: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fold one rope in half and then in half again. It will then be a quarter of the original length. Lay the folded rope parallel the other rope. Light the other end of the unfolded rope. When it reaches the folded rope then 45 minutes have passed. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2363" LastEditorUserId="46903" LastEditDate="2018-09-07T02:48:29.880" LastActivityDate="2018-09-07T02:48:29.880" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="2340" PostTypeId="2" ParentId="2333" CreationDate="2014-09-24T15:30:34.477" Score="1" Body="&lt;p&gt;Continuing on the from where Rob left off, when you label all the balls up to &lt;code&gt;10^20-1&lt;/code&gt;, you will have exactly &lt;code&gt;-2&lt;/code&gt; left of each digit 1-9, and a bunch of zeros.  So, to find out the last ball you can successfully label, you need to simply &quot;un-label&quot; the last few balls until you make up the shortfall you have in every digit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we un-label the last ball &quot;99999999999999999999&quot;, then we will recover 20 9s.  Thus, we now have a surplus of 18 9s, but still are 2 short of every other non-zero digit.  So, lets un-label 8 more balls, up to &lt;code&gt;10^20-9&lt;/code&gt;.  We will end up recovering with &lt;code&gt;20*9=180&lt;/code&gt; 9s, and one of every other non-zero digit.  But we still are at -1 of each of those digits, so we need to un-label 10 more.  So, if we un-label up to &lt;code&gt;10^20-19&lt;/code&gt;, we will get a ton of 9s back, a handful of 8s (since the balls we just un-labeled ended in &quot;89&quot;-&quot;81&quot;) and 2 of every other digit. This will make up our shortfall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the last ball we can successfully label is &lt;code&gt;10^20-20&lt;/code&gt;, which is &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;99999999999999999980 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; With the help of the comments, I realized that this answer is wrong because when you &quot;un-label&quot; the last 19 balls, you have to re-package 19 kits, so you will lose 38 of each digit.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Working backwards from this point doesn't seem any easier than working forwards, so I will have to think on it a bit more.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="1592" LastEditDate="2014-09-24T16:22:42.643" LastActivityDate="2014-09-24T16:22:42.643" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2341" PostTypeId="2" ParentId="2333" CreationDate="2014-09-24T16:23:17.040" Score="9" Body="&lt;p&gt;I used the code from this answer: &lt;a href=&quot;https://stackoverflow.com/questions/20945790/count-the-number-of-ks-between-0-and-n&quot;&gt;https://stackoverflow.com/questions/20945790/count-the-number-of-ks-between-0-and-n&lt;/a&gt;, translated to C# and modified to use BigInteger.&lt;/p&gt;&#xA;&lt;p&gt;I then tried numbers that fit into the following category: 119..91. This is the last number in the long line of numbers that use up 1's faster than they are created.&lt;/p&gt;&#xA;&lt;p&gt;I now have an answer. I'd really consider it more of an upper bound at this point.&lt;/p&gt;&#xA;&lt;p&gt;1,174,638,269,999,999,981&lt;/p&gt;&#xA;&lt;p&gt;The total number of 1's available is 2,349,276,539,999,999,962. The total number of 1's needed is 2,349,276,539,999,999,963.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;I have found that 119..91 is not the best point between &lt;code&gt;10^n-1&lt;/code&gt; and &lt;code&gt;10^(n+1)-1&lt;/code&gt;. This has allowed me to significantly lower my upper bound. I believe this answer to be too high, but now within one significant digit:&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;1,999,919,999,999,981&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The total number of 1's available is 3,999,839,999,999,962. The total number of 1's needed is one more than that.&lt;/p&gt;&#xA;" OwnerUserId="375" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-09-24T20:38:22.823" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="2342" PostTypeId="2" ParentId="2333" CreationDate="2014-09-24T16:35:40.607" Score="1" Body="&lt;p&gt;The below table shows what happens in various brackets of balls.  Used shows the stickers used for those balls, received shows the number that came with the balls, and left is the number left at the end of the bracket.  We come two stickers short at $11,111,111,111,099,999,999$, so that is the first ball she can't label.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin {array} {r r r r r}Start&amp;amp;End&amp;amp;Used&amp;amp;Received&amp;amp;Left\\1&amp;amp;1E19-1&amp;amp;1.9E19&amp;amp;2E19-2&amp;amp;1E18-2\\1E19&amp;amp;1.01E19-1&amp;amp;2.7E17&amp;amp;2E17&amp;amp;93E16-2\\1.01E19&amp;amp;1.02E19-1&amp;amp;3.7E17&amp;amp;2E17&amp;amp;76E16-2\\1.02E19&amp;amp;1.1E19-1&amp;amp;21.6E17&amp;amp;16E17&amp;amp;20E16-2\\1.1E19&amp;amp;1.101E19-1&amp;amp;3.6E16&amp;amp;2E16&amp;amp;18.4E16-2\\1.101E19&amp;amp;1.102E19-1&amp;amp;4.6E16&amp;amp;2E16&amp;amp;15.8E16-2\\1.102E19&amp;amp;1.11E19-1&amp;amp;28.8E16&amp;amp;16E16&amp;amp;3E16-2\\1.11E19&amp;amp;1.111E19-1&amp;amp;4.6E16&amp;amp;2E16&amp;amp;4E15-2\\1.111E19&amp;amp;1.1111E19-1&amp;amp;5.5E15&amp;amp;2E15&amp;amp;5E14-2\\1.1111E19&amp;amp;1.11111E19-1&amp;amp;6.4E14&amp;amp;2E14&amp;amp;6E13-2\\1.11111E19&amp;amp;1.111,111E19-1&amp;amp;7.3E13&amp;amp;2E13&amp;amp;7E12-2\\1.111,111E19&amp;amp;1.111,111,1E19-1&amp;amp;8.2E12&amp;amp;2E12&amp;amp;8E11-2\\1.111,111,1E19&amp;amp;1.111,111,11E19-1&amp;amp;9.1E11&amp;amp;2E11&amp;amp;9E10-2\\1.111,111,11E19&amp;amp;1.111,111,111E19-1&amp;amp;10E10&amp;amp;2E10&amp;amp;E10-2\\1.111,111,111E19&amp;amp;1.111,111,111,1E19-1&amp;amp;11E9&amp;amp;2E9&amp;amp;E9-2\\1.111,111,111,1E19&amp;amp;1.111,111,111,11E19-1&amp;amp;12E8&amp;amp;2E8&amp;amp;-2\end {array} $$&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditDate="2014-09-24T16:44:06.883" LastActivityDate="2014-09-24T16:44:06.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2343" PostTypeId="1" AcceptedAnswerId="2346" CreationDate="2014-09-24T17:15:12.223" Score="11" ViewCount="721" Body="&lt;p&gt;Bruce found a nice quote on the web, then wrote it in on a square piece of paper and left it somewhere in his room. The quote had exactly 64 characters excluding spaces. He has written the characters in 8 rows and 8 columns, ignoring word-breaking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;His younger sister Kim was helping Bruce to clean his room and thought the quote was garbage. So she ripped the paper into 64 equal pieces. Fortunately every character (no spaces) was on a separate piece of paper and she had some rules for ripping a piece of paper. Also she didn't scramble the pieces and left them in the stack she achieved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yGPDt.jpg&quot; alt=&quot;&quot;&gt;&lt;br&gt;&#xA;&lt;sup&gt;&lt;a href=&quot;http://newfreshcookies.blogspot.com/2010/09/ripping-paper.html&quot; rel=&quot;noreferrer&quot;&gt;Image from Fresh Cookies blog&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bruce figured out that she had some rules for ripping the paper:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;She had started with paper &lt;strong&gt;heads up&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Rip paper(s) into exact halves &lt;strong&gt;from top to bottom&lt;/strong&gt;, one held in her right hand and the other in her left.&lt;/li&gt;&#xA;&lt;li&gt;Put the &lt;strong&gt;right&lt;/strong&gt; hand piece(s) &lt;strong&gt;on top of the left&lt;/strong&gt; hand piece(s), and hold them in her left hand.&lt;/li&gt;&#xA;&lt;li&gt;If the pieces are &lt;strong&gt;not square&lt;/strong&gt; shaped, rotate them &lt;strong&gt;counter-clockwise&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Rip &lt;strong&gt;again&lt;/strong&gt; from top to bottom.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The order of characters on piece of papers from front to back was: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;WLWsAiwAtvtoderliiisnneFhehmM,CooouonataetstoClnvununnBn.lhStaoo&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bruce had no problem recovering his quote from the pieces (not web). Can you recover the quote?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS. I created this puzzle, so would love to hear your opinion.&lt;/p&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="8874" LastEditDate="2015-01-30T17:58:44.067" LastActivityDate="2015-01-30T17:58:44.067" Title="Recover the Ripped Quote" Tags="|logical-deduction|cryptic-crosswords|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2344" PostTypeId="2" ParentId="2335" CreationDate="2014-09-24T17:27:34.497" Score="7" Body="&lt;p&gt;Generally, people born in $1900 + n$ will have this birthday in the year $1900 + 2n$, when they are $n$ years old. The same can be applied to any century. You'll notice this is an even number; the year in the question is odd because you noticed this just before your birthday. That is, the Beddian birthday (see Kate Gregory's link in the quetion) was in 2012, but you noticed the phenomenon in the next year, 2013.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Working backwards, if this year is $c + m$ (where $c$ is a century such as 1900 or 2000 and $m$ is  even and less than 100), those with a Beddian birthday this year were born in $c + \frac{m}{2}$ or $(c - 100) + \frac{m + 100}{2} = (c - 100) + \frac{m}{2} + 50 = c + \frac{m}{2} - 50$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can obtain the second option by &quot;going back&quot; a century, making $c - 100$ the new century part and $m + 100$ the new &quot;$2n$&quot; part from the first paragraph. Since $m &amp;lt; 100$, $\frac{m}{2} + 50 &amp;lt; 100$, and so the year is valid. You can't go back another century, or else the &quot;$m$&quot; portion will be more than two digits, which wouldn't make any sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, for the year 2012:&#xA;$$&#xA;c = 2000, m = 12\\&#xA;\text{birth year is } \boldsymbol{6}=\frac{m}{2} \text{or } \boldsymbol{56}=\frac{m}{2} + 50&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Rob Watts notes, you're more likely to have been 56 than 6. Additionally, this happens every even year for those born turning $(year ~mod~ 100)/2$ and the same age plus or minus 50.&lt;/p&gt;&#xA;" OwnerUserId="2101" LastEditorUserId="1509" LastEditDate="2014-09-25T14:10:15.310" LastActivityDate="2014-09-25T14:10:15.310" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2345" PostTypeId="2" ParentId="2333" CreationDate="2014-09-24T18:06:39.677" Score="1" Body="&lt;p&gt;Actually, it is entirely possible that she ran out of stickers &lt;strong&gt;&lt;em&gt;before&lt;/em&gt;&lt;/strong&gt; labelling the &lt;code&gt;10^19-1&lt;/code&gt; ball.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if you split a range up into groups of 10, you will see that the leading digit is required about more often then other digits within that range.  For example:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Range        1    2  ...   9    0&#xA;1-9          1    1  ...   1    0&#xA;10-19       11    1  ...   1    1&#xA;100-199    120   20  ...  20   20&#xA;1000-1999 1300  300  ... 300  300&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So, consider the point where have just finished labelling the &lt;code&gt;10^18-1 th&lt;/code&gt; ball.  We are now in this situation:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Digits 1-9 used:      1,800,000,000,000,000,000 = 1.8 * 10^18&#xA;Digits 1-9 remaining:   199,999,999,999,999,998 = 2 * 10 ^ 17 - 2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If we want to label the next &lt;code&gt;10^18&lt;/code&gt; balls, we will need 1,800,000,000,000,000,000 (1.8 * 10^18)  more of each digit.  Since all those numbers start with the digit &lt;code&gt;1&lt;/code&gt;, we will also need &lt;code&gt;10^18&lt;/code&gt; additional ones, so we will need a total of &lt;code&gt;2.8 * 10^18&lt;/code&gt; more ones to label these balls.  However, we don't have that all in reserve - we only have about 1/10th of that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summary:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Range: 10^18 - 2 * 10^18 - 1&#xA;       1,000,000,000,000,000,000 - 1,999,999,999,999,999,999&#xA;Ones required: 2,800,000,000,000,000,000&#xA;Ones reserve:    199,999,999,999,999,998&#xA;Ones accrued:  2,000,000,000,000,000,000&#xA;Short:           600,000,000,000,000,002&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So, we have actually run out of ones before &lt;code&gt;10^19&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2014-09-24T18:06:39.677" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2346" PostTypeId="2" ParentId="2343" CreationDate="2014-09-24T19:04:38.677" Score="8" Body="&lt;p&gt;This is what was written on the paper:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    &lt;code&gt;AFlowerC&lt;/code&gt;&lt;br&gt;&lt;code&gt;annotBlo&lt;/code&gt;&lt;br&gt;&lt;code&gt;ssomWith&lt;/code&gt;&lt;br&gt;&lt;code&gt;outSunsh&lt;/code&gt;&lt;br&gt;&lt;code&gt;ine,AndM&lt;/code&gt;&lt;br&gt;&lt;code&gt;anCannot&lt;/code&gt;&lt;br&gt;&lt;code&gt;LiveWith&lt;/code&gt;&lt;br&gt;&lt;code&gt;outlove.&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The quote is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Flower Cannot Blossom Without Sunshine, And Man Cannot Live Without love.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How I figured this out:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The only tricky part about following the instructions in reverse is how she rotates the paper. During her last tear, the letters had been rotated three times (tops on the right), so just before the last tear letters of odd-numbered rows were still connected to the letter right below it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1955" LastEditorUserId="1497" LastEditDate="2014-09-25T16:52:20.557" LastActivityDate="2014-09-25T16:52:20.557" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="2347" PostTypeId="2" ParentId="626" CreationDate="2014-09-24T21:16:45.713" Score="2" Body="&lt;p&gt;The minimum number of cakes is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2, no matter how many bridges there are.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because with &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2 cakes you have to give one of them away and will get another as a reward. then you can cross further bridges infinitely.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="15985" LastEditDate="2018-09-16T20:54:32.637" LastActivityDate="2018-09-16T20:54:32.637" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="2348" PostTypeId="2" ParentId="2333" CreationDate="2014-09-24T21:19:31.753" Score="6" Body="&lt;p&gt;As noted before, the first digit to run out will be the 1's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider the units you use, you'll see that you use 1,2,3,...,9,0 and restart.  At any time you'll have used the 1 before any other number.  If you consider the tens, you see that you use 10x1, 10x2, ..., 10x9, 10x0.  Again you'll use 1's before the other digits catch up.  Since the supply is equal for all digits, you will always have less 1's than any other digit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there, I wrote a program.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An upper bound is ${10}^{20}-1$.  Even with a computer it is too long to test every number up to that value.  But you can skip whole ranges of numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Numbers up to 99 pay for themselves.  The strip that comes with it is enough.  So 1 to 99 are safe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's see for value 100:  For 1..100 you need 21 1's.  But you get 200 1's.  This gives an excess of 179.  Since you now produce 3-digit numbers, you will need no more than 3 1's per ball.  2 are provided by the strips, the 3rd is provided by the pool of 179 1's.  So the next 179 are safe, that is up to 279.  The next value to check is 280.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 1..280 you will find an excess of 403 1's and you can safely jump to 683.  The pool actually grows faster than you use it up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you reach 1000, you'll have to take 2 from the pool.  But it still runs fast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At some point, you will discover that you need more 1's than are provided, this stops the loop.  Since you have always skipped safe numbers, you know that the last number is the first to need more 1's than are provided so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the program that embodies the method described:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;package stackexchange;&#xA;&#xA;public class BallFactory&#xA;{&#xA;    /** Return the number of 1's needed to write number n */&#xA;    public static long onesIn(long n, int base){&#xA;        long count = 0;&#xA;        for( long k=n ; k&amp;gt;0 ; k/=base ){&#xA;            if( k%base==1 ) count++;&#xA;        }&#xA;        return count;&#xA;    }&#xA;&#xA;    /** return the number of 1's needed to write numbers 1 to n */&#xA;    public static long sumOnesTo(long n, int base){&#xA;        long m = n/base;&#xA;        long count = (n+base-1)/base + ( m==0 ? 0 : sumOnesTo(m-1, base)*base + onesIn(m, base)*(n%base+1) );&#xA;        return count;&#xA;    }&#xA;&#xA;    public static void solve(int base, int strips){&#xA;        // up to base^strips, each ball pays for itself&#xA;        long good = (long)Math.pow(base, strips);&#xA;        while( true ){&#xA;            long need = sumOnesTo(good, base);&#xA;            long excess = good*strips - need;&#xA;            System.out.println(String.format(&quot;%20d %20d %20d&quot;, good, need, excess));&#xA;            if( excess&amp;lt;0 ) break;&#xA;            long digits = (long)(Math.log(good)/Math.log(base))+1;&#xA;            good += excess/(digits-strips) + 1; // next uncertain number&#xA;        }&#xA;        // good is first to fail&#xA;        System.out.println(&quot;base &quot; + base + &quot; strips &quot; + strips + &quot;: &quot; + good);&#xA;    }&#xA;&#xA;    public static void main(String[] args)&#xA;    {&#xA;        solve(10,2);&#xA;    }&#xA;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The last line of the output tells you:&lt;br&gt;&#xA;&lt;strong&gt;For N = 1 999 919 999 999 981 you need 3 999 839 999 999 963 1's but you have a deficit of 1.&lt;/strong&gt;&lt;br&gt;&#xA;and you know that all numbers before are safe (You have more ones than you need).&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2014-09-24T21:44:50.843" LastActivityDate="2014-09-24T21:44:50.843" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2349" PostTypeId="1" AcceptedAnswerId="2352" CreationDate="2014-09-26T18:09:52.790" Score="5" ViewCount="1222" Body="&lt;p&gt;This being Halloween, we will consider the problem of witches and their cats. The whole crowd is out tonight, naturally, and an interested observer, seeing them fly by (yes, witches' cats fly), noticed that the number of legs all told was 5 times the number of heads and the total heads and legs was 72. How many cats and witches are out this Halloween? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This riddle is taken from the Mensa genius riddle book. &lt;/p&gt;&#xA;" OwnerUserId="2254" LastEditorUserId="9404" LastEditDate="2016-03-05T17:35:52.390" LastActivityDate="2016-03-05T17:35:52.390" Title="Witches and Cats" Tags="|mathematics|algebra|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2350" PostTypeId="2" ParentId="2349" CreationDate="2014-09-26T20:14:56.733" Score="2" Body="&lt;p&gt;I feel like the question assumes a witch and her cat flying together somehow mask the presence of a single head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this line of reasoning is accurate 8 Witch/Cat combinations (8 heads 48 legs), 2 solo cats (2 heads 8 legs) and 2 solo witches (2 heads 4 legs) would suffice (12 heads and 60 legs altogether).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this would also suggest that the observer is not an accurate one, and all their other details would also be thrown into question (which isn't surprising since witches aren't supposed to exist at all).  This line of reasoning suggests 0 witches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You decide which witch list fits.&lt;/p&gt;&#xA;" OwnerUserId="457" LastActivityDate="2014-09-26T20:14:56.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2351" PostTypeId="2" ParentId="2349" CreationDate="2014-09-26T21:27:32.683" Score="1" Body="&lt;p&gt;The answer I'm going to go with is: The observer made a typo, he meant to report that there are &lt;em&gt;3&lt;/em&gt; times as many legs as heads, and hence the number of witches and cats is nine each.&lt;/p&gt;&#xA;" OwnerUserId="2374" LastActivityDate="2014-09-26T21:27:32.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2352" PostTypeId="2" ParentId="2349" CreationDate="2014-09-27T06:26:26.797" Score="4" Body="&lt;p&gt;Taking into account everything that was already said (that witches and cats with 2 and 4 legs can not lead to the 5 ratio), and that I have never heard about flying cats, and do not know how many legs they have, I would answer simply &lt;strong&gt;&quot;There are 12 cats and witches in total&quot;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;The number of legs all told was 5 times the number of heads and the total heads and legs was 72&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this we can find easily that number of heads are 72/(5+1) = 12. Therefore, independently of how many legs flying cats can have, the total number of creatures must be 12.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;P.S. You cannot exact the precise number of cats and witches. For example, it can be 9 6-legged cats and 3 2-legged witches or 6 2-legged witches each with 8-legged cat.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="463" LastEditDate="2015-06-06T03:09:58.417" LastActivityDate="2015-06-06T03:09:58.417" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2353" PostTypeId="1" AcceptedAnswerId="2355" CreationDate="2014-09-28T05:40:12.707" Score="3" ViewCount="1078" Body="&lt;p&gt;On an Island of Knights and Knaves (where Knights always tell the truth and Knaves always lie), a Knight will never contradict himself unless some true fact changes. I was wondering: can a Knave contradict himself (regarding presently unchanging facts)? If he did wouldn't that mean he told the truth about 'something'?&lt;/p&gt;&#xA;" OwnerUserId="2312" LastEditorUserId="69582" LastEditDate="2021-03-26T19:06:00.347" LastActivityDate="2021-03-26T19:06:00.347" Title="About Knights and Knaves and their consistency" Tags="|liars|word-problem|" AnswerCount="5" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2354" PostTypeId="2" ParentId="2353" CreationDate="2014-09-28T07:14:59.753" Score="2" Body="&lt;p&gt;Knave can easily contradict himself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, let's take a set of facts:&lt;br&gt;&#xA;Fact A, which is FALSE&lt;br&gt;&#xA;Fact B, which is TRUE&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you ask a knave &quot;what is A?&quot; he would answer TRUE. Then &quot;what is B?&quot; - he would answer FALSE.&lt;br&gt;&#xA;Then ask what A AND B is. He would answer TRUE (FALSE AND TRUE = FALSE).&lt;br&gt;&#xA;But combining his previous answers logician would conclude that A AND B = FALSE (TRUE AND FALSE = FALSE).  &lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="11297" LastEditDate="2015-07-10T23:42:03.020" LastActivityDate="2015-07-10T23:42:03.020" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2355" PostTypeId="2" ParentId="2353" CreationDate="2014-09-28T18:17:13.260" Score="11" Body="&lt;p&gt;A knave can contradict himself easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;- Is the sky blue?&lt;br&gt;&#xA;- No.&lt;br&gt;&#xA;- Did you just answer no?&lt;br&gt;&#xA;- No.  &lt;/p&gt;&#xA;" OwnerUserId="2136" LastActivityDate="2014-09-28T18:17:13.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2357" PostTypeId="1" AcceptedAnswerId="2364" CreationDate="2014-09-28T22:32:23.940" Score="1" ViewCount="1205" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;It's a simple pattern I thought of. Here it is:&lt;/p&gt;&#xA;&lt;p&gt;1,1,1,3,3,6,6,10,10,22,22,44,44,84,?,?,?,?,?&lt;/p&gt;&#xA;&lt;p&gt;Whoever gives the correct answer and a good explanation gets best answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2018-03-15T17:04:07.353" Title="What are the next 5 numbers in the sequence?" Tags="|pattern|number-sequence|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2358" PostTypeId="1" AcceptedAnswerId="2360" CreationDate="2014-09-29T00:34:05.077" Score="8" ViewCount="2722" Body="&lt;p&gt;When Cora was killed, Anna and Beth were questioned by the police about the manner of her death by poison. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Anna: If it was a murder, Beth did it.&lt;br&gt;&#xA;  Beth: If it was not suicide, it was murder.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The policemen made the following assumptions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$1$. If  neither Anna nor Beth lied, it was an accident.&lt;br&gt;&#xA;  $2$. If either Anna or Beth lied, it was not an accident.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Subsequent development revealed these assumptions were correct.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What was the manner of Cora's death: accident suicide or murder?&lt;/p&gt;&#xA;" OwnerUserId="2254" LastEditorUserId="8874" LastEditDate="2015-10-18T15:58:58.083" LastActivityDate="2015-10-18T15:58:58.083" Title="Cora's Death Logic Puzzle" Tags="|logical-deduction|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2359" PostTypeId="2" ParentId="2358" CreationDate="2014-09-29T01:31:59.437" Score="-3" Body="&lt;p&gt;Accident/Suicide because of contradictive facts?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it was not an accident, it was murder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that's a lie it was murder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if it was an accident it was murder...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accidental murder? XD&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-09-29T01:55:19.030" LastActivityDate="2014-09-29T01:55:19.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2360" PostTypeId="2" ParentId="2358" CreationDate="2014-09-29T02:15:44.870" Score="10" Body="&lt;p&gt;It was murder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assumption 1 (both tell truth = accident) contradicts Beth (if not suicide then murder). Therefore at least one of them wasn't telling the truth, and so (because of assumption 2) it wasn't an accident.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that we know it wasn't an accident, we also know that Beth is telling the truth (assuming only possibilities are suicide, accident or murder). So Anna is lying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way for Anna to be lying is if it was a murder and Beth didn't do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that Anna is guilty of murder, but I only have enough evidence to say it was murder and that Beth didn't do it.&lt;/p&gt;&#xA;" OwnerUserId="375" LastEditorUserId="-1" LastEditDate="2014-10-02T14:26:35.757" LastActivityDate="2014-10-02T14:26:35.757" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2361" PostTypeId="1" CreationDate="2014-09-29T08:06:17.227" Score="-3" ViewCount="745" Body="&lt;p&gt;So, recently there was a very good riddle posted here: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/626/paying-the-troll-toll&quot;&gt;Paying the Troll toll&lt;/a&gt;. I would like to propose a similar question which might have been partially answered. But anyways...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, same basic idea as the original Paying the Troll toll riddle, however, assume now that there are an infinite number of bridges with an infinite number of trolls. How many pies must I leave with to ensure that I arrive at grandmas' with a whole number of pies? (Edit: assume that you do reach a point at which you arrive at grandmas. This would be when a sequence approaches zero as the number of bridges you go over tends to infinity.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, how would this answer change if the ratio of pies given to the trolls were to change. That is instead of half, what about an eighth, etc.?&lt;/p&gt;&#xA;" OwnerUserId="2254" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-10-27T21:59:29.457" Title="Infinite Troll Tolls" Tags="|mathematics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2362" PostTypeId="2" ParentId="2317" CreationDate="2014-09-29T09:12:38.263" Score="9" Body="&lt;p&gt;A simple counter example would be the horrible island of all tax-collectors, who all mimic knights perfectly - always tell the truth about everything, except pretending that everyone on the island is a knight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could have a lifetime of questions on this island and could never find any difference to an island full of actual knights.&lt;/p&gt;&#xA;" OwnerUserId="1998" LastActivityDate="2014-09-29T09:12:38.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2364" PostTypeId="2" ParentId="2357" CreationDate="2014-09-29T15:22:44.610" Score="5" Body="&lt;p&gt;Start with S(1) = 1.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To construct the next 6 numbers (S(2) - S(7)), use n = 1 and apply the following formulas:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;S(2) = S(1) * 2 + (-1)*n&#xA;S(3) = S(2)&#xA;S(4) = S(3) * 2 + (+1)*n&#xA;S(5) = S(4)&#xA;S(6) = S(5) * 2 + ( 0)*n&#xA;S(7) = S(6)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;To construct the next 6 numbers, double n to 2 and apply the above pattern again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To construct the next 6 numbers (84 and your 5 ?s), double n to 4 and apply the pattern again.  This gives:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;S(14) = S(13) * 2 + (-1)*n = 44 * 2 - 4 = 84&#xA;S(15) = S(14) = 84&#xA;S(16) = S(15) * 2 + (+1)*n = 84 * 2 + 4 = 172&#xA;S(17) = S(16) = 172&#xA;S(18) = S(17) * 2 + ( 0)*n = 172 * 2 + 0 = 343&#xA;S(19) = S(18) = 343&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So the sequence is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1,1,1,3,3,6,6,10,10,22,22,44,44,84,84,172,172,343,343&lt;/p&gt;&#xA;" OwnerUserId="2162" LastActivityDate="2014-09-29T15:22:44.610" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2365" PostTypeId="2" ParentId="2361" CreationDate="2014-09-29T17:59:00.933" Score="6" Body="&lt;p&gt;First of all, you'd never reach grandmas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moving on from the obvious...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 pies, get to a bridge, give 1, get it back, now you have 2, keep it going for infinite number of bridges, same solution, but different number of bridges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are n bridges, you have p pies, at each bridge, you have p/2+1 pies left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are inf bridges, you have 2 cakes, at each bridge, you have 2/2+1 (2) pies left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to give them different ratios it would be something like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p/r+1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to always reach grandmas that equation &lt;strong&gt;has&lt;/strong&gt; to evaluate to p, or p/r + floor(p/r/3) must be less than the total amount of bridges. Since there are infinite bridges, that won't happen, and the only time p/r+1 = p is when p = 2 and r = 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p/r+1 = p&lt;br&gt;&#xA;p/r = p-1&lt;br&gt;&#xA;r(p/r) = r(p-1)&lt;br&gt;&#xA;p = r(p)-r&lt;br&gt;&#xA;p+r = p(r)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the only pair of positive integers where when added together are equivalent to when multiplied together are 2, 2.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2015-10-27T21:59:29.457" LastActivityDate="2015-10-27T21:59:29.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2368" PostTypeId="2" ParentId="1987" CreationDate="2014-09-30T02:33:13.370" Score="1" Body="&lt;p&gt;I'm coming across this question rather late, but given recent news about an &lt;a href=&quot;http://www.japantimes.co.jp/news/2014/09/02/national/surrogate-offers-clues-into-japanese-with-16-babies/&quot; rel=&quot;nofollow&quot;&gt;apparent serial surrogate father&lt;/a&gt;, both babies could have been co-incidentally born on the same day to two separate surrogate mothers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether five-year-olds would be aware of such details and able to understand they were not twins is of course another matter.&lt;/p&gt;&#xA;" OwnerUserId="341" LastActivityDate="2014-09-30T02:33:13.370" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2369" PostTypeId="5" CreationDate="2014-09-30T07:15:18.210" Score="0" Body="&lt;p&gt;This tag is for puzzles where one is given a set of objects and is required to find common characteristics between them - this is called the &lt;em&gt;pattern&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One is then usually required to find new elements of the set that follow this pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples of pattern puzzles include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/1961/pattern-1-numbers&quot;&gt;number pattern&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/1671/how-to-solve-this-puzzle-from-iq-test&quot;&gt;picture pattern&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/8430/five-sequence-puzzles&quot;&gt;letter pattern&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/2258/chaos-and-order-a-visual-puzzle&quot;&gt;visual pattern&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="11110" LastEditDate="2017-06-21T20:07:55.633" LastActivityDate="2017-06-21T20:07:55.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2370" PostTypeId="4" CreationDate="2014-09-30T07:15:18.210" Score="0" Body="A puzzle in which you spot the pattern that links a set of given data items (like numbers, letters, music tunes). Usually, you use your insights into the pattern to find one or more new elements of the set. " OwnerUserId="5373" LastEditorUserId="5208" LastEditDate="2015-02-11T01:09:26.010" LastActivityDate="2015-02-11T01:09:26.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2371" PostTypeId="5" CreationDate="2014-09-30T07:17:38.087" Score="0" Body="&lt;p&gt;Mathematics (from Greek μάθημα máthēma = knowledge + study + learning) is the study of abstract topics as quantity, structure, space, and change.&#xA;There is a range of views among mathematicians and philosophers as to the exact scope and definition of mathematics.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Aristotle defined it as the science of quantity.&lt;/li&gt;&#xA;&lt;li&gt;Auguste Comte defined it as the science of indirect measurement.&lt;/li&gt;&#xA;&lt;li&gt;Bertrand Russell defined it as Symbolic Logic.&lt;/li&gt;&#xA;&lt;li&gt;Walter Warwick Sawyer defined it as the classification and study of all possible patterns.&lt;/li&gt;&#xA;&lt;li&gt;Wolfram MathWorld defines it as a broad-ranging field of study in which the properties and interactions of idealized objects are examined. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Mathematical puzzles make up an integral part of recreational mathematics. &lt;strong&gt;Mathematical puzzles require mathematics to solve them.&lt;/strong&gt; Logic puzzles are a common type of mathematical puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that mathematical &lt;em&gt;problems&lt;/em&gt; (as opposed to mathematical &lt;em&gt;puzzles&lt;/em&gt;) are now considered off-topic. See &lt;a href=&quot;http://meta.puzzling.stackexchange.com/questions/2814/should-we-introduce-a-new-off-topic-close-reason-for-math-problems&quot;&gt;here&lt;/a&gt; for a discussion of the difference between the two.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="5373" LastEditDate="2015-03-15T12:55:35.320" LastActivityDate="2015-03-15T12:55:35.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2372" PostTypeId="4" CreationDate="2014-09-30T07:17:38.087" Score="0" Body="A puzzle strongly related to mathematical facts and objects, or whose solution needs serious mathematical arguments." OwnerUserId="2225" LastEditorUserId="5373" LastEditDate="2015-03-15T12:55:35.320" LastActivityDate="2015-03-15T12:55:35.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2373" PostTypeId="1" AcceptedAnswerId="2374" CreationDate="2014-09-30T08:31:28.917" Score="6" ViewCount="1748" Body="&lt;p&gt;Guess the next digits of this number. Where does it came from?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;4.9292683&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="71" LastEditDate="2014-11-24T21:52:56.433" LastActivityDate="2014-11-24T21:52:56.433" Title="What is the number 4.9292683…?" Tags="|mathematics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2374" PostTypeId="2" ParentId="2373" CreationDate="2014-09-30T08:56:31.843" Score="28" Body="&lt;h2&gt;6&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;It's $\pi e \gamma$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to the &lt;a href=&quot;http://isc.carma.newcastle.edu.au/standardCalc?input=4.9292683&quot;&gt;Inverse Symbolic Calculator&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="228" LastActivityDate="2014-09-30T08:56:31.843" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2375" PostTypeId="2" ParentId="2357" CreationDate="2014-09-30T16:47:40.370" Score="3" Body="&lt;p&gt;You can simplify it into a single formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$s_{n} = 2 s_{n-2} + 2 s_{n-6} - 4 s_{n-8}$&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2014-10-03T08:51:34.753" LastActivityDate="2014-10-03T08:51:34.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2376" PostTypeId="1" AcceptedAnswerId="2379" CreationDate="2014-09-30T19:30:53.657" Score="13" ViewCount="1315" Body="&lt;p&gt;The six-pack is a thing of the past. Beverages of the future will use the seven-pack format. But how will the mighty spacemen of the future manage to ship the Slurm seven-pack efficiently in rectangular space crates?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a schematic of the seven-pack.  How can you completely fill a rectangle using seven-packs of this shape?  It's ok to turn a seven-pack upside down, giving you a mirror image of the below shape.  The arrangement requiring the smallest rectangular space crate wins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bzbEK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whimmy wham wham wozzle, dudes!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Stated simply, can you arrange an unlimited but finite number of the pictured shape and its mirror image so that they perfectly tessellate a rectangle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Remember that since the rectangle contains an integer number of seven-packs then M x N must be a multiple of 7. Furthermore, since 7 is prime at least one of the two dimensions of the rectangle must be a multiple of 7.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2162" LastEditorUserId="8874" LastEditDate="2015-02-08T19:29:18.920" LastActivityDate="2015-02-08T19:29:18.920" Title="How to ship the new Slurm 7-pack efficiently" Tags="|mathematics|geometry|tiling|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2378" PostTypeId="2" ParentId="152" CreationDate="2014-09-30T23:09:00.363" Score="1" Body="&lt;p&gt;An approach which would be much more dangerous, but would be usable even if there were only one rope would to climb each rope (or the rope) and slice partway through it while leaving it strong enough to hold about twice one's weight, and then slide very carefully down it rope until twenty feet from the bottom.  Lift up on the rope from underneath until one has reached a spot about ten feet from the end.  Grasp that spot firmly and let go of the other place one is holding.  Depending upon the elasticity of the rope it may be necessary to adjust one's choice of locations, but it should be possible to momentarily tug on the rope which exceeds one's own weight by a factor of at least four, thus snapping the rope above.&lt;/p&gt;&#xA;" OwnerUserId="1467" LastActivityDate="2014-09-30T23:09:00.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2379" PostTypeId="2" ParentId="2376" CreationDate="2014-10-01T03:07:58.007" Score="23" Body="&lt;p&gt;I have a 14x14 (28 case) rectangle. It is completely symmetrical. Someone beat that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nrv9X.png&quot; alt=&quot;28 Slurm&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="375" LastActivityDate="2014-10-01T03:07:58.007" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2391" PostTypeId="1" AcceptedAnswerId="2394" CreationDate="2014-10-01T21:43:30.147" Score="1" ViewCount="500" Body="&lt;p&gt;The following message was encoded using a simple algorithm:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;code&gt;!kzsaain zhps V !vW&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Please try to explain how it works.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="9718" LastEditDate="2015-10-18T19:28:02.397" LastActivityDate="2015-10-18T19:28:02.397" Title="What does this encoded message mean? &quot;kzsaain zhps V !vW&quot;" Tags="|cipher|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2392" PostTypeId="1" AcceptedAnswerId="2393" CreationDate="2014-10-01T22:11:29.270" Score="17" ViewCount="2881" Body="&lt;p&gt;The new land surveyor and his boss arrived at the work site, a huge and perfectly flat parking lot.  His boss said, &quot;Your first assignment is going to be a tough one.  You see this parking lot?  It has light posts every 50 feet east to west and north to south.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;A perfect Cartesian grid with a light post at each intersection?&quot; asked the Laziest Surveyor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;That's right. And too many posts to count. They seem to go on forever.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;What's the job?&quot; asked the Laziest Surveyor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;They're building... something. I don't know what, but it's going to be crazy looking.  Just look at the polygon on this map. Angles going every which way! That's the footprint of the new building. Must be one of these Deconstructivist architects or something. Not my style. Give me a nice rectangular Parthenon any day. But never mind that. Our job is to measure the area of that polygon.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Laziest Surveyor glanced at the map.  &quot;Well, at least none of the sides of the polygon cross over each other. And none of the posts lie on the border except at the vertices.  That should make things easier.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;You're an optimist!&quot; the boss laughed. &quot;This thing has 21 vertices, each one of them located exactly at a light post. There's convex angles, concave angles, peninsulas, coves, what a mess. I hope you know your trigonometry! This will take you all day.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;When construction begins, I guess they'll have to knock down these lights,&quot; said the Laziest Surveyor, looking all around him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Just the ones inside the polygon.  They'll leave the ones at the vertices up.  Let's see, so they'll bulldoze...&quot;  The boss flipped through work orders as the Laziest Surveyor stared expectantly. &quot;Wow, 149! Big job! Well, you've got a lot of work to do so I'll leave you to it.  Here's the map and the survey form.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The boss drove away.  As soon as the boss' car was out of sight the Laziest Surveyor crumpled up the map and dropped it on the ground.  He took out a pocket calculator, jabbed at it a few times, and wrote the correct area on the survey form.  Then he spent the rest of the day at a baseball game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What area did he calculate and how did he do it?&lt;/p&gt;&#xA;" OwnerUserId="2162" LastActivityDate="2014-11-17T15:40:51.417" Title="The Laziest Surveyor" Tags="|mathematics|word-problem|geometry|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2393" PostTypeId="2" ParentId="2392" CreationDate="2014-10-01T23:21:44.787" Score="17" Body="&lt;p&gt;He calculated the area using &lt;a href=&quot;http://en.wikipedia.org/wiki/Pick&amp;#39;s_theorem&quot;&gt;Pick's Theorem.&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Area = (number of internal grid points) + (half the number of boundary points) - 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The area is $149 + \frac{21}{2} - 1 = 158.5\mbox{ units}^2$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we're given that the units are $50\mbox{ ft}$, the area is $396250\mbox{ ft}^2$.&lt;/p&gt;&#xA;" OwnerUserId="1777" LastEditorUserId="2461" LastEditDate="2014-11-17T15:40:51.417" LastActivityDate="2014-11-17T15:40:51.417" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2394" PostTypeId="2" ParentId="2391" CreationDate="2014-10-01T23:35:21.317" Score="10" Body="&lt;p&gt;The message reads &quot;Hi! I love puzzles!&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was encrypted using a ROT variant and written backwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ROT variant starts with e mapped to z then as it descends through the alphabet the encrypted characters ascend. When it reaches p, it skips one letter, presumably to avoid being invertible. Then once it runs out of letters at the end z is mapped to a.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;e f g h i j k l m n o p q r s t u v w x y z&#xA;z y x w v u t s r q p n m l k j i h g f e a&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The first 4 letters of the alphabet appear to be excluded.&lt;/p&gt;&#xA;" OwnerUserId="2406" LastEditorUserId="2406" LastEditDate="2014-10-02T04:11:48.613" LastActivityDate="2014-10-02T04:11:48.613" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2396" PostTypeId="1" AcceptedAnswerId="2397" CreationDate="2014-10-02T02:30:25.023" Score="18" ViewCount="22843" Body="&lt;p&gt;In a set of puzzles I found online recently, I acquired this unnamed puzzle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4FlO8m.jpg&quot; alt=&quot;spherical puzzle&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/aCWPDm.jpg&quot; alt=&quot;rotated spherical puzzle&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what it is, or how to solve it. It definitely comes apart, as it is split into six pieces, but I see no way to begin disassembly or, for that matter, start assembling it again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pushing on pieces doesn't seem to do anything. I can slide my finger under some pieces by squeezing them, but trying to pull pieces out gives resistance that I don't think I'm supposed to push past. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a piece of paper inside the puzzle, but I can only read it through the tiny hole in the yellow piece - I can't actually see its contents. I've made out the word &quot;puzzle&quot; and some out-of-context steps to a solution, but this doesn't really help me, since I don't know how to get to it. I don't even know what this thing is called.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I disassemble this? And once it's disassembled, how do I reassemble it?&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorUserId="71" LastEditDate="2014-11-17T15:51:24.137" LastActivityDate="2023-06-27T07:49:20.827" Title="How do I solve this put-together spherical puzzle?" Tags="|mechanical-puzzles|burr-puzzle|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2397" PostTypeId="2" ParentId="2396" CreationDate="2014-10-02T05:00:58.383" Score="34" Body="&lt;p&gt;Well, funny story.&lt;/p&gt;&#xA;&lt;p&gt;My friend grew irritated with the dern plastic thing, so he threw it at a wall. And guess what? It fell apart.&lt;/p&gt;&#xA;&lt;p&gt;Inside, the disassembly instructions read:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;HOW TO DISASSEMBLE:&lt;/p&gt;&#xA;&lt;p&gt;To disassemble, throw the ball at the ground.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I'm not kidding.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JPc8z.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;There are two things that will cause this puzzle to break apart: the first is centripetal force, and the second is significant shock. If every single piece is pulled apart at the same time, then they will all slide away. But, if you pull on just &lt;em&gt;one&lt;/em&gt; piece (particularly with tangential force), then they lock up. Also, if you throw it on the ground, the pieces really aren't that well-connected, so they'll crumble away.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;The re-assembly is actually pretty intuitive. There are two pieces that are shaped to allow for reassembly, that have an indent in them:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y56Wgm.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Set these aside. The rest are shaped like this:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3T9cEm.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Three of these are needed initially, then the two you set aside are needed, then the last piece to go in is one of the 'regular' pieces.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;For the rest of this answer, I will be referencing the following example:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uz0xVm.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Place one piece on the bottom (orange), and hold two pieces vertically adjacent to it (white and green), resting against the first at 90 degree angles. You will have to hold them there with one hand.&lt;/p&gt;&#xA;&lt;p&gt;Then, pick up each of the pieces set aside (purple and yellow) and place them against the upright pieces horizontally, indented side up. The reason you want to do this is so that the last piece can slide in easily - it's pretty clear that there's more room for the last piece to slide in from the image.&lt;/p&gt;&#xA;&lt;p&gt;Once these pieces are in, you should be able to hold together easier with just a little bit of pressure on the sides. If you're delicate, you can even get it to sit like this without any interference!&lt;/p&gt;&#xA;&lt;p&gt;Then, it's as simple as sliding the last piece in, over the indented pieces. It will take a little bit of pressure and bending, but it will snap into place. I find it helps if you pull &lt;em&gt;up&lt;/em&gt; on the piece from the other side to get it over the lip. Then boom, puzzle solved.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aHIA1m.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorUserId="-1" LastEditorDisplayName="user20" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-10-02T07:42:33.010" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2399" PostTypeId="1" CreationDate="2014-10-02T14:22:48.430" Score="9" ViewCount="2408" Body="&lt;p&gt;Inspired by this thread: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/48/what-is-the-largest-tile-possible-in-2048/&quot; title=&quot;Largest possible tile in 2048&quot;&gt;Largest possible tile in 2048&lt;/a&gt;, I've always wondered with such games to what extent you NEED luck to be able to get to a high score. For example, the maximum tile of 131072 requires the luck to spawn a '4' tile in the last free square on the board. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What often seems to screw you over, might just imply that your strategy is sub optimal however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here's the questions, both assuming you don't deliberately try to fail:&lt;br&gt;&#xA;&lt;strong&gt;1) Assuming you don't know what will spawn where, what is the first combined tile-value you cannot be guaranteed to achieve?&lt;/strong&gt;(so sum of the 16 tile values on the board).&lt;br&gt;&#xA;&lt;strong&gt;2) If you'd know the spawn tile numbers series ahead of time, but not their places, would that knowledge affect the results?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To point out the obvious:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;No cheating&lt;/li&gt;&#xA;&lt;li&gt;No 'Undo'&lt;/li&gt;&#xA;&lt;li&gt;Any step involving 'luck' (a 2 spawn where a 4 spawn would mean game over, or a 4 spawn where a 2 spawn would mean game over, or a spawn of 2 or 4 where that spawn in any other free square would mean game over) is not allowed. There is only said to be a 10% chance of a 4 spawning instead of a 2, but if it can happen, assume Murphy hates you.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This thread: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/396/what-is-the-lowest-you-can-get-in-2048&quot;&gt;Lowest score in 2048&lt;/a&gt; theorises it's possible to get as low as 48 combined. You'd really need to not make any sensible step for this, so assume the answer to the question is higher then 48.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note to readers: I have no exact answer to this myself. It's just a brainteaser to see what logic people can build around this.&lt;/p&gt;&#xA;" OwnerUserId="2156" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-01-19T13:40:41.380" Title="2048 - Maximum score that doesn't rely on luck" Tags="|2048-game|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2403" PostTypeId="2" ParentId="1751" CreationDate="2014-10-02T19:40:21.653" Score="3" Body="&lt;p&gt;The game &lt;em&gt;is&lt;/em&gt; possible.  I have completed it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you have guessed, the aim of the game is to assemble each dot with the 2 matching rings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sorry, I cannot explain exactly how the pieces moves.  I haven't figures everything out.  But some general patterns were enough to solve it.  Don't loose your time working out every rule.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One of the positive reviews on the Android store speaks about &lt;em&gt;lateral thinking&lt;/em&gt; and &lt;em&gt;thinking outside of the box&lt;/em&gt;.  Have you done that?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If, without being instructed to do so, you 1. select a color, 2. press the lever, 3. wait, and repeat that over and over, then you are locked inside of your own rules.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Spoiler 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You &lt;em&gt;can&lt;/em&gt; press 2 buttons at the same time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Spoiler 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Press a color, press the lever, and before the lever moves back, press and hold another color.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Final spoiler&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you start the game, activate blue 2 times to get to that recuring position where the 3 small circles overlap the large circles.  From there, activate red 3 times in a  row.  Then, do a combined green + blue activation: press green, press the lever and quickly press and hold blue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2014-10-08T21:20:24.257" LastActivityDate="2014-10-08T21:20:24.257" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2404" PostTypeId="2" ParentId="282" CreationDate="2014-10-03T06:35:05.700" Score="5" Body="&lt;p&gt;Why make it so confusing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;If I yell my colour, yours is the same.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First guess is 50% chance of success. Yells if it's the same as the hat directly in front, speaks quietly if it's not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has the added benefit that if someone stuffs up the count the rest can continue with a 50% chance of sacrificing a single person.&lt;/p&gt;&#xA;" OwnerUserId="2454" LastActivityDate="2014-10-03T06:35:05.700" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2419" PostTypeId="1" AcceptedAnswerId="2420" CreationDate="2014-10-03T13:13:42.590" Score="6" ViewCount="2470" Body="&lt;p&gt;A guy lives in the 10th floor of an apartment building.&#xA;On a normal day he goes to the 5th floor with the elevator and then climbs the remaining 5 floors.&#xA;On a rainy day he goes to the 10th floor directly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why ?&lt;/p&gt;&#xA;" OwnerUserId="2465" LastEditorUserId="11297" LastEditDate="2015-07-10T23:34:30.140" LastActivityDate="2016-08-12T13:27:27.730" Title="Elevator puzzle" Tags="|lateral-thinking|" AnswerCount="2" CommentCount="0" ClosedDate="2016-08-12T17:50:54.907" ContentLicense="CC BY-SA 3.0" />
  <row Id="2420" PostTypeId="2" ParentId="2419" CreationDate="2014-10-03T13:17:20.383" Score="15" Body="&lt;p&gt;I think:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He is a short person, so he can't reach the button, when it's raining he has his umbrella to push the button for him&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2402" LastEditorUserId="21116" LastEditDate="2016-08-12T13:27:27.730" LastActivityDate="2016-08-12T13:27:27.730" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2427" PostTypeId="1" AcceptedAnswerId="2428" CreationDate="2014-10-03T17:05:46.533" Score="3" ViewCount="1157" Body="&lt;p&gt;A young boy is taking a ride with his dad when a truck runs them off the road, killing the father and critically injuring the young boy. He's Medivac'd to the nearest hospital as the child is lying on the table ready for operation, the surgeon looks at the boy and says &quot;I cannot operate on him, he's my son.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this possible?&lt;/p&gt;&#xA;" OwnerUserId="2483" LastActivityDate="2014-10-06T15:49:26.613" Title="A boy and his father" Tags="|paradox|" AnswerCount="3" CommentCount="0" ClosedDate="2015-06-17T15:48:13.997" ContentLicense="CC BY-SA 3.0" />
  <row Id="2428" PostTypeId="2" ParentId="2427" CreationDate="2014-10-03T17:13:43.707" Score="18" Body="&lt;p&gt;The classic answer is &quot;The surgeon is his mother&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possible answers:&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;It is another father of him. Gays are real.:)  &lt;/li&gt;&#xA;&lt;li&gt;It is step-father. No one would care to chose precise words in such situation.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2014-10-03T17:18:53.237" LastActivityDate="2014-10-03T17:18:53.237" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2429" PostTypeId="2" ParentId="2427" CreationDate="2014-10-03T17:36:33.333" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A far-fetched solution&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Consider this : Twins are born (lets say two boys A and B) - biological parents die - ends up in foster care - then each are adopted separately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets say their worlds never collided until that fateful day of the accident.&#xA;'A' was with his father in the car and lets say this accident took place at 'X'.&#xA;Now 'B' had told his parents that day before leaving home that he had some work around the place of 'X'. So when 'B' father/mother who is a surgeon sees the body of 'A' then naturally he/she could say those lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe there are some loopholes in this story but still its possible right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also it just so happens that 'A' and 'B' were wearing the same exact dress that day :)&lt;/p&gt;&#xA;" OwnerUserId="101" LastEditorUserId="101" LastEditDate="2014-10-03T17:41:42.420" LastActivityDate="2014-10-03T17:41:42.420" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2432" PostTypeId="1" AcceptedAnswerId="2433" CreationDate="2014-10-03T18:25:07.190" Score="6" ViewCount="573" Body="&lt;p&gt;In 1980, Jack has an age of 20. In 1985 however, he is only 15 years old.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did he age backwards?&lt;/p&gt;&#xA;" OwnerUserId="2490" LastEditorUserId="5373" LastEditDate="2015-05-02T12:31:40.450" LastActivityDate="2015-05-02T12:31:40.450" Title="A curious case of Benjamin Button" Tags="|calculation-puzzle|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2433" PostTypeId="2" ParentId="2432" CreationDate="2014-10-03T18:29:11.330" Score="9" Body="&lt;p&gt;Jack lives in 1980 BC.  Of course, back then, they were counting forward from a different starting point. :)&lt;/p&gt;&#xA;" OwnerUserId="2447" LastActivityDate="2014-10-03T18:29:11.330" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2434" PostTypeId="1" AcceptedAnswerId="2504" CreationDate="2014-10-03T18:35:19.023" Score="16" ViewCount="1553" Body="&lt;p&gt;I've created a puzzle quite a while back, unfortunately not many people wanted to play so I'm posting here instead.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The idea is that the image can be decipher to uncover a hidden message. The hint is given:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kRHdI.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Hints: &lt;strong&gt;“Like time, their true forms are revealed once they’re divided by their seats”&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The first step is to figure out how to read the numbers, Joe already figured one part of the first step. The answer of the puzzle will be in text form.&lt;/p&gt;&#xA;&lt;p&gt;Hope you guys enjoy my game. Any comments on the puzzle is greatly welcome as well.&lt;/p&gt;&#xA;" OwnerUserId="2373" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2017-04-28T18:53:39.770" Title="Hidden Message: Of Shapes and Numbers" Tags="|visual|" AnswerCount="1" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="2435" PostTypeId="2" ParentId="625" CreationDate="2014-10-03T19:20:11.283" Score="0" Body="&lt;p&gt;So lets see how much information can be exchanged, my highest is  &lt;strong&gt;m = 715&lt;/strong&gt; or 1430 for both spies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets look at the restraints. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The number of stones is the only information that may be collected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The two spies must take turns , they must therefore take the same number of turns&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) The number of stones is limited , 26&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) The spies may not limit the throws of the other player (each has 13 stones)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) They may not throw 0 stones &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Implied Rules&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6) Each player may not know anything about the other players numbers. Nor may they know anything about the numbers during their meetings.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;7) One spies number may not interfere with the other spies number.  Scenario where one spy finishes early or uses the other players left over stones cannot occur.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that there are at most six turns of throwing stones.  On a given turn a player may throw a stone or more.  The number of throws cannot be available since they must stop at the same time sop they must have their information in the same space of turns. They must know how many turns are being exchanged so they can terminate at the same time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply using the dividers between the 13 stones from above as a binary word is a good idea.  This will allow numbers up to 2^12 to be represented.   By considering the division of a group as a binary signal, and the lack of a divider as the off positions you get a 12 bit word. This does not rely on extra information being passed, just translates the gaps into a binary word.  to ensure the alternation of turns the binary word will be 6 bits long of binary data ( 1 stone = 0, 2 stone = 1).  This gives us 2^6 options. I know above it seems great to have 2^12 from a grouping of 13 stones.   However by stating that you have dived them into 6-7 groups ( 7 is impossible) you have created six bits, not 12.  You are counting the number of ways to group the data, and yet you just said the there are only 6-7 groups. 64 for each spy, sadly a 7th digit cannot be included due to only having one stone and not being able to assume the number isn't 64. With only one stone that bit will never actually contain data since there is only one choice. Using the separators to binary encode is a good idea, but needs the numbers of throws to be variable, which does not work.   This is due to having to terminate the word at agreed number of symbols. so . . &lt;strong&gt;m = 64&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The information allowed is the number of stones.  However the turn limitation forces us to allocate enough stones to allow for full transimition.  So we are looking for unique signals sent by groups of numbers, limited to 13 stones.  Each grouping is a symbol to be transmitted.  Sending 6 groups using 12 stones make no logical sense. So instead of loosing that last stone it should be used for a trinary signal in that grouping. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However if the Alternation of throws is enforced very strictly . . . meaning they MUST take turns, the problem becomes very abbreviated.  Even the tree above which i find as wonderful mechanic should be constructed so that the tree is balanced, for example that each player reaches the end of their tree with the same number of turns. Logically the forces them to used a set number of agreed groups and leave left over stones.  If each player has 13 stones then they can do 6 binary, 4 trinary, or 3 quadrany choices.&lt;br&gt;&#xA;If each leaf has N choices there can be N / K rounds of signals sent.  if the signals are 1 stone /two stone there are six signals and a tree of 2^6 in size. if there are three signals (1 stone, 2 stone, 3 stone) you have a tree of 3^4 in size.  This is due to the nodes not being able to contain a value, only the end of the path does.  How can a spy terminate early?  the signals must all be sent otherwise we need a new signal to indicate the end.  so using a look up tree might not be as efficient here as we hoped due to the limited information.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;tightly enforcing this constraint pushes out many possible scenarios.  it seems that four trinary signals would carry the most information in this scenario.  four blocks of trinary data sent by both spy would allow for 81 different signals without compromising the information sharing rules, the turn alternation rules, or the rock sharing rules. However we can use the last stone and add one more choice.  This nets us 3 X 3 X 3 X 4  = 108 choices.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to use the leftover stones we can simply make  the later choices in the tree to include the possibility of the leftover stones being left.  There are many cases where the leftover stones could be used in the final 'word'.  In fact we get a larger pool of results by limiting the later throws.   10 choices, 1 choice, 1 choice, 1 choice performs better than 3 , 3, 3, 4 in this case due tot he carry over stones.  This yields many extra leaves on out decision tree.  I will try to get a mathematical expression for this calculation, I wrote a program to test it and found 715 as the optimum choice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the case where the number of turns are decided by the first players move does not work for two reasons.  This simply makes the first choice a binary choice and uses extra stones.  Furthermore, if the first player need to use the extra set and indicates this, what does the second player do if their answer is in a tree with a different number of moves?  Trees in this case must be balanced ( have the same number of nodes to all leaves).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both the binary tree above and the binary combinations above count far too many scenarios where the information exchange does not work.  The look up table is irrelevant, the total numbers of signals is the relevant answer.   Both players will have a look up tree that must be identical.  But the tree must be balanced or else one player will end on tier two while the other pushes to the last leaf of the deepest branch.  Test your scenario with the highest value/ lowest value pairs, and then with a highest /highest value to test the transmition.  &lt;/p&gt;&#xA;" OwnerUserId="2497" LastEditorUserId="2497" LastEditDate="2014-10-06T17:54:20.633" LastActivityDate="2014-10-06T17:54:20.633" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2436" PostTypeId="2" ParentId="2427" CreationDate="2014-10-03T20:05:24.040" Score="1" Body="&lt;p&gt;(young boy's age) and 9 months ago, his mother slept with twins. Afterwards they couldn't figure out who is the father by DNA tests, so they both are now. :P&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Btw, in some other language the solution &quot;it's his mother&quot; doesn't work, because you would use the female version of the word doctor.&lt;/p&gt;&#xA;" OwnerUserId="2502" LastActivityDate="2014-10-03T20:05:24.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2438" PostTypeId="1" AcceptedAnswerId="2441" CreationDate="2014-10-03T20:18:20.167" Score="4" ViewCount="624" Body="&lt;p&gt;This was inspired by &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2432/a-curious-case-of-benjamin-button&quot;&gt;A curious case of Benjamin Button&lt;/a&gt; but has a different solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1980 AD (or 1980 CE, if you prefer), Jack has an age of 20. In 1985 however, he is only 6 years old.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did Jack age backwards?&lt;/p&gt;&#xA;" OwnerUserId="2505" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-05-02T12:33:03.187" Title="A Second Benjamin Button" Tags="|word-problem|calculation-puzzle|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2439" PostTypeId="2" ParentId="2438" CreationDate="2014-10-03T20:21:43.063" Score="7" Body="&lt;p&gt;He was born on leap day, Feb 29 1960.  So in 1980 his legal age would be 20, but by 1985 only 6 additional leap years have occurred.&lt;/p&gt;&#xA;" OwnerUserId="239" LastActivityDate="2014-10-03T20:21:43.063" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2440" PostTypeId="2" ParentId="2438" CreationDate="2014-10-03T20:26:37.560" Score="4" Body="&lt;p&gt;Jack is a fictional character first published in 1979.  In the story is set in 1980 (or the sequel published 1980 or something), he is a 20 year old.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditDate="2014-10-03T20:32:33.780" LastActivityDate="2014-10-03T20:32:33.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2441" PostTypeId="2" ParentId="2438" CreationDate="2014-10-03T20:28:39.927" Score="11" Body="&lt;p&gt;Jack is 20 months old in 1980.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He is 6 years old in 1985.&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2014-10-03T20:28:39.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2442" PostTypeId="2" ParentId="282" CreationDate="2014-10-03T20:44:40.637" Score="-1" Body="&lt;p&gt;Apologies if its against the spirit of the question (which it may be as they cannot talk to each other during the 'test').&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The individuals could tap the shoulder of the person in front as a code; for example left shoulder your wearing a green hat, right shoulder if your hat is purple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would only leave the initial person to guess their own hat, ensuring 9/10 survival rate and a 50/50 chance for the first person to guess.&lt;/p&gt;&#xA;" OwnerUserId="2510" LastActivityDate="2014-10-03T20:44:40.637" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2444" PostTypeId="1" AcceptedAnswerId="2450" CreationDate="2014-10-03T21:21:35.853" Score="4" ViewCount="1146" Body="&lt;p&gt;The king had a mutant rat problem in his kingdom, so he ordered 3 young men to go down under the depth of the ground to find the cause. The 3 young men jumped down the hole, and followed the tunnel. After hours of walking, they met an intersection. There were 4 paths at this intersection. The king had told them before they left that there would be at least 1 safe path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One by one the soldiers went down a tunnel. The first one went... The other two heard a scream! Certain death met them there. The second went down the second path. Scream! Then the last knew for certain exactly which path was safe. He soon found the source of the problem, and reported it to the king.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did the young soldier know which path to take?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="5373" LastEditDate="2015-05-19T10:03:43.850" LastActivityDate="2015-05-19T10:03:43.850" Title="The Tricky Tunnels!" Tags="|logical-deduction|lateral-thinking|word-problem|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2445" PostTypeId="2" ParentId="1761" CreationDate="2014-10-03T21:25:11.687" Score="20" Body="&lt;p&gt;There is a big difference between a condom and a pair of sugical gloves.  Anyone who has tried to use the one for the purpose of the other would agree.  But the difference we are concerned with is that gloves are used in pairs while the others are used one at a time.  You can transfer gloves from one hand to the other, which you cannot do with the other product.  Therefore there is a simple solution to the doctor's dilemma problem that the other problem doesn't allow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is probably why the vulgar version of the problem sticks around, it is because they are not equivalent, the vulgar version is actually a much harder problem to solve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a better solution for the doctor's dilemma.  It doesn't render the gloves useless afterwards.  Instead, you can repeat all operations again and again in any order you like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is how:  There are 4 gloves.  I’ll label each glove with 2 letters identifying who is allowed to touch each side of the glove.  There is uppercase and lowercase to help identify the surfaces.  The labels are as follows: Hx Xy Yz Zh.  You'll need sometimes to return some glove inside out.  Hx and xH are the same glove, but reversed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The operation program is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dr. X operates:&lt;br&gt;&#xA;  left hand: Xy Yz Zh&lt;br&gt;&#xA;  right hand: xH  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dr. Y operates:&lt;br&gt;&#xA;  left hand: Yz Zh&lt;br&gt;&#xA;  right hand: yX xH  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dr. Z operates:&lt;br&gt;&#xA;  left hand: Zh&lt;br&gt;&#xA;  right hand: zY yX xH  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, no surface gets contaminated directly or indirectly by 2 different people.  Therefore, all operations can be repeated indefinitely in any order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;... and they operated happily ever after.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2015-01-16T16:25:51.660" LastActivityDate="2015-01-16T16:25:51.660" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2447" PostTypeId="2" ParentId="1842" CreationDate="2014-10-03T21:40:33.810" Score="0" Body="&lt;p&gt;Is the answer 12 mm? He says nothing about having all the volumes, just that he arranged volume 1 and 32.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Building on the most upvoted answer, the first page of volume one sits on the right having only volume 1's cover separating it from volume 32 cover, that has its final page sitting on the left side of the book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the worm would travel the distance of two covers: $6 * 2 = 12$ mm.&lt;/p&gt;&#xA;" OwnerUserId="2469" LastEditorUserId="2484" LastEditDate="2014-10-14T22:32:08.317" LastActivityDate="2014-10-14T22:32:08.317" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2448" PostTypeId="1" CreationDate="2014-10-03T21:42:22.760" Score="9" ViewCount="1374" Body="&lt;p&gt;Most of my awareness of puzzles and puzzle games has come from a decade of haphazardly stumbling around the internet. There's a flavor/genre that I particularly enjoy, but I haven't had much luck finding more of them - or even nailing down to correct terms to find them. Lo and behold, I finally stumbled across &lt;a href=&quot;https://puzzling.stackexchange.com/questions/1751/how-do-i-solve-the-between-worlds-android-puzzle-game&quot;&gt;this question&lt;/a&gt;, and it inspired me to ask you fine puzzlers for help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I enjoy puzzles that are enigmatic - where you're presented with zero hints or guidance, and all progress and information is hard-won through deduction / trial and error. Here are a few great examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.newgrounds.com/portal/view/214420&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Control Panel&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.newgrounds.com/portal/view/542840&quot; rel=&quot;nofollow noreferrer&quot;&gt;Binary&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.newgrounds.com/portal/search/games/clickdragtype&quot; rel=&quot;nofollow noreferrer&quot;&gt;ClickDragType Series&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Unfortunately all I have are links to Newgrounds, which should immediately indicate how much I'm hurting for fresh specimens. The main problem I seem to be having is that the &quot;Puzzle&quot; category in any given source is very ambiguous - they lump in anything from bejeweled to jigsaws to quiz games, and after years of casual searching I'm tired of filtering through hundreds of pages with no success.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a specific name for this kind of game? Are there any sites or sources that would help me find more? I'm completely in the dark here - any help or insight is much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2512" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-06-19T16:32:09.953" Title="How do I find more enigmatic, no-clues-given puzzles?" Tags="|puzzle-identification|enigmatic-puzzle|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2449" PostTypeId="2" ParentId="2444" CreationDate="2014-10-03T21:43:40.307" Score="6" Body="&lt;p&gt;The path they first came from is the safe one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt; The third path may or may not be safe. They are tasked with finding the cause of the problem in his kingdom. They path they come from is safe because they crossed it entirely - fufilling the &quot;ate least one safe path&quot; premisse. Plus they must meet the problem, the origin of the mutant rats. So the safe path being the third choice does not help them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the third we cannot be sure. The final young man may or may not have gone through that third path and found it to be safe or not. All we know for sure is that he came back and reported that - at least -  path 1 and 2 have the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think some more careful wording is in order, but good job on conceptualization of a problem :)&lt;/p&gt;&#xA;" OwnerUserId="2469" LastEditorUserId="2469" LastEditDate="2014-10-04T06:10:57.220" LastActivityDate="2014-10-04T06:10:57.220" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2450" PostTypeId="2" ParentId="2444" CreationDate="2014-10-03T21:46:14.400" Score="10" Body="&lt;p&gt;There were 4 paths at the intersection: A, the one they came from, B and C, the two tested, and D, the remaining path.  D is safe.&lt;/p&gt;&#xA;" OwnerUserId="2437" LastActivityDate="2014-10-03T21:46:14.400" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2451" PostTypeId="1" AcceptedAnswerId="2453" CreationDate="2014-10-03T22:33:32.623" Score="40" ViewCount="10025" Body="&lt;p&gt;There are 53 lions in a cage and 1 goat. All the lions are very hungry and would want to eat the goat. But the problem here is that any lion who eats the goat can satisfy his hunger but he will himself become a goat and then other lions will eat him. All lions are intelligent and understand this fact, so obviously no one wants to die becoming a prey for others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the question here is will the goat be eaten or not?&lt;/p&gt;&#xA;" OwnerUserId="2515" LastActivityDate="2017-07-23T17:22:23.693" Title="Hungry lions and a goat" Tags="|logical-deduction|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2453" PostTypeId="2" ParentId="2451" CreationDate="2014-10-03T23:07:03.333" Score="43" Body="&lt;p&gt;Let's follow the usual approach for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say there's just 1 lion and the goat. Would the lion eat the goat? Yes, it's not at risk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So say there's 2 lions and a goat. Would a lion eat the goat? If so it would become the goat in the 1 lion and a goat situation, and so it would be eaten. So it would not eat the goat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there's 3 lions and a goat, would a lion eat the goat? Well there'd be 2 left, and as we just worked out, with 2 left, the goat is safe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it becomes clear that the pattern is: odd number of lions, goat gets eaten, even number, goat is safe. So with 53, the goat gets eaten.&lt;/p&gt;&#xA;" OwnerUserId="1759" LastActivityDate="2014-10-03T23:07:03.333" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2455" PostTypeId="1" AcceptedAnswerId="2458" CreationDate="2014-10-03T23:41:40.347" Score="-7" ViewCount="696" Body="&lt;p&gt;I expect multiple answers to this, however only 1 of them is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bus was scheduled to stop 10 times today. Each time, 7 people got on the bus and 5 got off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the last stop, how many people were in the bus before the driver kicked them off?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="5373" LastEditDate="2018-05-06T15:52:01.420" LastActivityDate="2018-05-06T15:52:01.420" Title="The Long Bus Ride" Tags="|lateral-thinking|story|" AnswerCount="2" CommentCount="4" ClosedDate="2018-05-06T17:36:30.440" ContentLicense="CC BY-SA 3.0" />
  <row Id="2458" PostTypeId="2" ParentId="2455" CreationDate="2014-10-04T00:03:54.903" Score="5" Body="&lt;p&gt;Here's a few answers I feel are justifiable, given the question in its present state. If you want any specific number explaining a bit more, feel free to ask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1 person&lt;/strong&gt;: The question states that people get &quot;on&quot; and &quot;off&quot; the bus, but none get &quot;in&quot;, so the driver is the only one &quot;in&quot; the bus, the passengers are all sat on the roof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is apparently the correct answer. Or the one the OP wanted to see, anyway.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;26 people&lt;/strong&gt;: (+7 -5) * 9 = 18 + 7 = 25. The driver is the 26th. The 5 who would have got off get kicked off instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;25 people&lt;/strong&gt;: (+7 -5) * 9 = 18 + 7 = 25. Since the driver is &lt;strong&gt;kicking people off&lt;/strong&gt; he doesn't count (you wouldn't kick yourself off a bus). The 5 who would have got off get kicked off instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;24 people&lt;/strong&gt;: Same logic as above, except the driver is one of the first 7 people getting on, and doesn't kick himself off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;21 people&lt;/strong&gt;: (+7 -5) * 10 = 20. The driver is the 21st.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;20 people&lt;/strong&gt;: (+7 -5) * 10 = 20. Since the driver is &lt;strong&gt;kicking people off&lt;/strong&gt; he doesn't count (you wouldn't kick yourself off a bus).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;19 people&lt;/strong&gt;: (+7 -5) * 9 = 18. The driver is the 19th. It's the final stop, end of the line, nobody can get on, everyone left is kicked off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;18 people&lt;/strong&gt;: (+7 -5) * 9 = 18. It's the final stop, end of the line, nobody can get on, everyone left is kicked off. Since the driver is &lt;strong&gt;kicking people off&lt;/strong&gt; he doesn't count (you wouldn't kick yourself off a bus).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;17 people&lt;/strong&gt;: Same logic as above, except the driver is one of the first 7 people getting on, and doesn't kick himself off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Trick question&lt;/strong&gt;: You didn't say the driver actually kicks anyone off (so he doesn't kick anyone off), you just asked how many were on the bus before he kicked them off.&lt;/p&gt;&#xA;" OwnerUserId="2518" LastEditorUserId="2518" LastEditDate="2014-10-04T00:20:03.090" LastActivityDate="2014-10-04T00:20:03.090" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="2460" PostTypeId="2" ParentId="2451" CreationDate="2014-10-04T00:19:10.127" Score="24" Body="&lt;p&gt;&lt;strong&gt;The paradox&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think of 53 as just a large number, you might first believe the goat is at risk of being eaten.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then you see that if a lion eats the goat, it becomes the goat among a large number of lions and will just as likely be eaten.  The lions would therefore rather be the last lion to eat the goat.  As a result, none will try to eat the goat first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that is the case, that means the goat is safe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if a goat is safe among many lions, it should be safe for a lion to eat the goat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, any answer leads to the opposite answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The solution&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to understand that the situation is very different between 53 and 52 lions.  To understand why, you have to consider each number separately.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;With 1 goat against 1 lion, the goat gets eaten.  Obviously.&lt;/li&gt;&#xA;&lt;li&gt;With 1 goat against 2 lions, if a lion eats the goat it becomes 1 goat against 1 lion and gets eaten.  No lion wants that, so the goat survives.&lt;/li&gt;&#xA;&lt;li&gt;With 1 goat against 3 lions, if a lion eats the goat it becomes 1 goat against 2 lions and survives.  So the goat gets eaten by the fastest lion.&lt;/li&gt;&#xA;&lt;li&gt;With 1 goat against 4 lions, if a lion eats the goat it becomes 1 goat against 3 lions and gets eaten.  No lion wants that so the goat survives.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You can see a general pattern emerging:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;With an even number of lions, if a lion eats the goat it becomes 1 goat against an odd number of lions and gets eaten.  So the goat survives.&lt;/li&gt;&#xA;&lt;li&gt;With an odd number of lions, if a lion eats the goat, it becomes 1 goat against en even number of lions and suvives.  So the goat gets eaten.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Since we have an odd number of lions, the goat gets eaten by the fastest lion.  The lion becomes a goat and from there on it survives with all 52 lions.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2017-07-23T11:47:31.157" LastActivityDate="2017-07-23T11:47:31.157" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2461" PostTypeId="1" AcceptedAnswerId="2495" CreationDate="2014-10-04T02:21:31.607" Score="3" ViewCount="805" Body="&lt;p&gt;A top secret facility is guarded by 1 man, this man will kill anyone with the wrong, or without a, passcode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 5 spies. They have determined the code is one of the following;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Orange&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Apple&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Red&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;5&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;TV&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Complicator&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Each spy guessed in the order specified above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last spy got it right. How did he do it?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="5373" LastEditDate="2018-05-09T21:40:36.793" LastActivityDate="2018-05-10T17:46:17.183" Title="The Secret Passcode" Tags="|lateral-thinking|password|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2462" PostTypeId="2" ParentId="2455" CreationDate="2014-10-04T02:40:45.683" Score="2" Body="&lt;p&gt;This is probably the wrong answer, but here goes:&#xA;It's a trick question. You never say how many people are on the bus at the start, so it could virtually be any number. For all we know, there could be 300 at the last stop! (although that would be &lt;strong&gt;quite&lt;/strong&gt; a packed bus)&lt;/p&gt;&#xA;" OwnerUserId="2524" LastActivityDate="2014-10-04T02:40:45.683" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2464" PostTypeId="2" ParentId="2396" CreationDate="2014-10-04T03:45:20.957" Score="6" Body="&lt;p&gt;Three pieces meet at one intersection in this puzzle, and on the opposite side there's another intersection which touches the other three pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you gently grasp the first three with three fingers from one hand, and the other three with three fingers from the other hand, you should find that the two halves, each composed of three pieces, slide apart.  Friction and tight fitting pieces may prevent this at first, but as the puzzle is worked you'll be able to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can reassemble it the same way, but it's very difficult to get each half correctly assembled and then lined up to put them back together.  Follow the instructions for reassembly.  Disassembly can be done easily using this method.&lt;/p&gt;&#xA;" OwnerUserId="1682" LastActivityDate="2014-10-04T03:45:20.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2465" PostTypeId="1" AcceptedAnswerId="3180" CreationDate="2014-10-04T04:10:04.910" Score="4" ViewCount="218" Body="&lt;p&gt;I have a design for a puzzle I want to develop. It requires fairly good accuracy when cutting the pieces which have complicated shapes, although they all are made of equal size cubes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually I would like to market it, so I am interested in mass production (meaning a few hundred).&lt;/p&gt;&#xA;" OwnerUserId="2379" LastActivityDate="2014-10-23T18:29:46.313" Title="Where to manufacture a 3D wood puzzle?" Tags="|puzzle-creation|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2466" PostTypeId="1" AcceptedAnswerId="2468" CreationDate="2014-10-04T04:12:34.157" Score="8" ViewCount="456" Body="&lt;p&gt;You are given some matches, a rope, and a pair of scissors. The rope burns irregularly and takes 60 minutes to burn end to end; however, we know that the burn rate at distance $x$ from the left end is the same as the burn rate at distance $x$ from the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the smallest time interval that you can measure?&lt;/p&gt;&#xA;" OwnerUserId="1477" LastEditorUserId="8874" LastEditDate="2015-11-01T15:59:57.743" LastActivityDate="2015-11-01T15:59:57.743" Title="Shortest time interval measurable via burning a rope" Tags="|mathematics|optimization|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2467" PostTypeId="1" AcceptedAnswerId="2494" CreationDate="2014-10-04T04:35:52.917" Score="10" ViewCount="8843" Body="&lt;p&gt;This is related to &lt;a href=&quot;https://puzzling.stackexchange.com/q/228/2379&quot;&gt;this question&lt;/a&gt;, but different. I'm &lt;strong&gt;not&lt;/strong&gt; asking for the least number of moves needed, nor for pointers on how to use fewer moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Singmeister's notes (an 80's book!) there is mention of M. Thistlewaite's algorithm, which could solve the cube from any position in a guaranteed 45 moves. Of course we know now that an all-seeing cuber needs at most 20 moves, but what I am asking is about a method that can be applied by a human and which guarantees as few moves as possible regardless of the initial position.&lt;/p&gt;&#xA;" OwnerUserId="2379" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2018-03-15T12:59:47.930" Title="What is the shortest known rubik's cube algorithm?" Tags="|rubiks-cube|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2468" PostTypeId="2" ParentId="2466" CreationDate="2014-10-04T05:02:36.840" Score="8" Body="&lt;p&gt;Since you can cut it in half, this is equivalent to having two ropes of 30 minutes long. Lay them in rows and set fire to opposite ends, so you have this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;F-------&#xA;-------F&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Eventually, the flames will meet, like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   F----&#xA;---F&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Each is now a 15 minute rope. Set fire to the other end of the smaller rope, so that it burns down in 7.5 minutes. Then set fire to other end of the remaining rope, which will finish burning in 3.75 minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think you can do better than that. I can't think of a rigorous reason, but here's a unrigorous one: Say two ropes have &quot;similar ends&quot; if they burn at the same rate from each end. Given two ropes with different ends and burn time N, the best you can do is N/4. This is because in time N/2, for one rope you can either&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Destroy it (burn both ends)&lt;/li&gt;&#xA;&lt;li&gt;Reduce its length by N/2 (burn one end)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You need to destroy at least one rope in order to reduce the rest, because you have no other way of keeping time. So with two ropes you can destroy one to reduce the other to N/2 in N/2 time, then destroy the other in N/4 time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's with two ropes with different ends. With two ropes with similar ends, you can also produce two half-length ropes with different ends, so the &lt;strong&gt;best you can do is actually N/8&lt;/strong&gt;. And one rope with both ends similar can be cut in half to produce two ropes with similar ends, so you can do N/16. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;bold&lt;/strong&gt; part is the handwavey part, because I can't think of a strong reason why there isn't a more creative, better thing you can do there.&lt;/p&gt;&#xA;" OwnerUserId="1906" LastActivityDate="2014-10-04T05:02:36.840" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2471" PostTypeId="1" AcceptedAnswerId="2473" CreationDate="2014-10-04T08:59:03.043" Score="2" ViewCount="2712" Body="&lt;p&gt;Here's a part of the sequence:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(A)&#xA;(B)&#xA;(C)&#xA;(D)&#xA;(E)&#xA;1113122113&#xA;311311222113&#xA;...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The letters are the missing members (there are 5 missing members) and &lt;strong&gt;A&lt;/strong&gt; is the first member of the sequence. The last member presented is not the end of the sequence, it can go on forever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sequence begins with a single integer.&#xA;Can you tell which one?&#xA;Simplest solution takes the cake.&lt;/p&gt;&#xA;" OwnerUserId="2469" LastEditorUserId="4551" LastEditDate="2014-12-28T18:06:33.853" LastActivityDate="2014-12-28T18:06:33.853" Title="A sequence named by a children's toy" Tags="|lateral-thinking|pattern|" AnswerCount="2" CommentCount="0" ClosedDate="2014-12-29T02:43:14.773" ContentLicense="CC BY-SA 3.0" />
  <row Id="2473" PostTypeId="2" ParentId="2471" CreationDate="2014-10-04T10:46:35.230" Score="6" Body="&lt;p&gt;3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;13&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1113&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3113&lt;/p&gt;&#xA;&#xA;&lt;p&gt;132113&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1113122113&lt;/p&gt;&#xA;&#xA;&lt;p&gt;311311222113&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus points:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next in the series:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;13211321322113&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless I made a typo, I'm fairly certain this is all correct. It's the look and say sequence, where the next number describes the last;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;11 (there's 1 1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;21 (there's 2 1s)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1211 (there's 1 2 and 1 1s)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in other words, 3 is the first integer.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="68" LastEditDate="2014-10-05T05:38:07.317" LastActivityDate="2014-10-05T05:38:07.317" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2474" PostTypeId="1" AcceptedAnswerId="2476" CreationDate="2014-10-04T10:52:24.730" Score="-5" ViewCount="436" Body="&lt;p&gt;Sitting on the table are 3 light bulbs. 1 of them just blown, 1 of them has been blown for ages, and the other works fine. You don't which is which, but you're allowed to test (turn on) 1 bulb . How can you tell for certain which bulb will work properly.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="5373" LastEditDate="2018-05-09T14:35:25.643" LastActivityDate="2018-05-09T14:35:25.643" Title="The Blown Lightbulbs" Tags="|lateral-thinking|situation|" AnswerCount="2" CommentCount="0" ClosedDate="2014-12-07T23:51:57.537" ContentLicense="CC BY-SA 3.0" />
  <row Id="2475" PostTypeId="2" ParentId="2471" CreationDate="2014-10-04T10:58:59.580" Score="4" Body="&lt;ul&gt;&#xA;&lt;li&gt;3&lt;/li&gt;&#xA;&lt;li&gt;13&lt;/li&gt;&#xA;&lt;li&gt;1113&lt;/li&gt;&#xA;&lt;li&gt;3113&lt;/li&gt;&#xA;&lt;li&gt;132113&lt;/li&gt;&#xA;&lt;li&gt;1113122113&lt;/li&gt;&#xA;&lt;li&gt;311311222113&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Each value is the encoding of the previous, so starting with 3, the next will be 13, or 1 instance of the number 3. The next will be 1113, one instance of 1, one instance of 3. From here you have 3113, three 1s, one 3, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: D'oh, too slow - too much messing with registration and impossible captchas :)&lt;/p&gt;&#xA;" OwnerUserId="2537" LastActivityDate="2014-10-04T10:58:59.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2476" PostTypeId="2" ParentId="2474" CreationDate="2014-10-04T11:02:18.193" Score="4" Body="&lt;p&gt;Hold your hand over each one. The one that's just blown should still be warm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try one of the other 2, and you'll know which one works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB: I'm assuming several things you haven't explicitly stated (like &quot;test&quot;ing a bulb just means plugging it in).&lt;/p&gt;&#xA;" OwnerUserId="2518" LastActivityDate="2014-10-04T11:02:18.193" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2477" PostTypeId="1" AcceptedAnswerId="2481" CreationDate="2014-10-04T11:15:57.407" Score="5" ViewCount="1864" Body="&lt;p&gt;There are 3 talking hats on the table. Each can ask 2 questions, to either hat. 1 of them speaks only truth. 1 speaks only lies. 1 can speak only when spoken to, he will speak at random.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;YOU know who's who just fine. But you cannot tell the hats. You can only tell a hat a question to ask one of the other hats, you choose who but NOT BASED on who he is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can the 3 hats talk to each other in such a way as to distinguish who's the truth speaker, lie teller, and random chatter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: You can only ask yes/no questions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;You know which hat is which&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Yes/No questions only&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Each hat gets 2 questions&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You cannot tell a hat to ask another based on who the other hat is&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The random hat can only ask questions after answering another's question&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Each hat knows who he is&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Each conversation is private&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;HINT: If presented with a question the hat does not know, he will answer to the best of his ability with yes/no, following their rules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HINT 2: If the truth teller or liar is presented with a question he does not know, they must answer to the best of their ability ~ Not answering. The random does not need to know the answer to reply, therefore he can answer freely.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="5373" LastEditDate="2018-05-14T16:55:01.627" LastActivityDate="2018-05-14T16:55:01.627" Title="The 3 Talking Hats" Tags="|logical-deduction|liars|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2478" PostTypeId="1" AcceptedAnswerId="2538" CreationDate="2014-10-04T12:49:00.047" Score="1" ViewCount="486" Body="&lt;p&gt;Peter found an old piece of paper that an equation was written on it. Some of &lt;strong&gt;digits&lt;/strong&gt; or &lt;strong&gt;operations&lt;/strong&gt; are unreadable. Help Peter to complete the &lt;strong&gt;balanced&lt;/strong&gt; equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zTn2W.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation is: $87+?3?12-4?2=?36-15??1/?$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: My Brain&lt;/p&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="8874" LastEditDate="2015-02-09T10:38:24.510" LastActivityDate="2015-02-09T10:38:24.510" Title="Balance the Equation" Tags="|mathematics|calculation-puzzle|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2480" PostTypeId="2" ParentId="2477" CreationDate="2014-10-04T13:32:31.520" Score="1" Body="&lt;p&gt;I state that in the current formulation this is impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take for example the truth-hat, it will ask questions. Other hats we number as 1 and 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider both possible situations:&lt;br&gt;&#xA;A) 1 is lie-hat, 2 is random-hat.&lt;br&gt;&#xA;B) 2 is lie-hat, 1 is random-hat.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using those question truth-hat must find out which situation is present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now lets consider a possible case that random-hat accidentally answers on all questions it is asked (0 questions, 1 or 2, doesn't matter) exactly as a lie-hat would answer. In this case each question that is asked will be answered exactly the same both in situation A) and in situation B). Therefore truth-hat can not complete the task.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is true for lie-hat. Only random-hat can complete the task, asking only one question &quot;What would you answer if I ask you are you truth-hat?&quot;.&lt;/p&gt;&#xA;" OwnerUserId="28" LastActivityDate="2014-10-04T13:32:31.520" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2481" PostTypeId="2" ParentId="2477" CreationDate="2014-10-04T14:19:07.900" Score="5" Body="&lt;p&gt;The Truth-Teller asks Hat-A &quot;Is he a Truth-Teller&quot;. If he answers then he the Random-Hat. If he doesn't then he is the Liar. Hat-B can now be inferred. Only one question needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Liar asks Hat-A &quot;Will Random-Hat say 'yes' to his next question?&quot;. If he answers then he the Random-Hat. If he doesn't then he is the Truth-Teller. Hat-B can now be inferred. Only one question needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Random Hat asks Hat-A &quot;Would Hat-B say 'Yes' if I asked him whether he is the Truth-Teller?&quot;. If he is the Liar then he says &quot;No&quot;. If he is the Truth-teller, he says &quot;Yes&quot;. Hat-B can now be inferred. Only one question needed.&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-10-04T14:43:24.367" LastActivityDate="2014-10-04T14:43:24.367" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="2482" PostTypeId="2" ParentId="2478" CreationDate="2014-10-04T14:54:16.160" Score="4" Body="&lt;p&gt;87+63-12-4*2=136−15+81/9&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are probably many possible solutions. &lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-10-04T15:01:38.777" LastActivityDate="2014-10-04T15:01:38.777" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2483" PostTypeId="1" AcceptedAnswerId="2486" CreationDate="2014-10-04T15:09:00.897" Score="-3" ViewCount="729" Body="&lt;p&gt;Brian wrote:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hi, my name is Brian. I love to play video games, and write stories. Nobody is better than me at it! The only problem is I'm running out of story ideas. Can you help me? Pitch some ideas! Thanks&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He posted it on a billboard. A week later he got a call saying something was wrong with his note. Brian spent hours trying to find something wrong with it. He never found it. What was wrong with the paper?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="5373" LastEditDate="2015-04-28T09:36:52.877" LastActivityDate="2015-04-28T09:36:52.877" Title="What's the problem in the following paragraph?" Tags="|lateral-thinking|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2484" PostTypeId="1" AcceptedAnswerId="2501" CreationDate="2014-10-04T15:09:16.743" Score="19" ViewCount="14803" Body="&lt;p&gt;Here is an excellent puzzle with 3 types of characters:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You are lost in a town and meet 3 men. You somehow (God told you) know that they are Knight, Knave and Joker. Knight always tells truth, Knave always tells the opposite to the truth, Joker randomly decides whether to tell truth or lie.&lt;br&gt;&#xA;  You need to find at least one person who can give you some information, but the problem is that you do not know who is who. Luckily the three knew amongst themselves who they are. So you need to find someone who is not a Joker. How do you do it asking one yes-or-no-question to one person only if you are not allowed to ask a man something he doesn't know or any question he cannot answer either &quot;yes&quot; or &quot;no&quot; to.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since the problem seems to be impossible for a lot people (at least at the beginning) I am adding a lot of explanations to let you be sure that there is no trick:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You cannot ask something like &quot;How many friends do you have?&quot;. This is not a yes-or-no question.&lt;/li&gt;&#xA;&lt;li&gt;You cannot ask questions like &quot;Will it rain tomorrow?&quot;. He doesn't know.&lt;/li&gt;&#xA;&lt;li&gt;You cannot ask arbitrary man &quot;What would you answer if I ask you blablabla?&quot;, if it is the Joker he doesn't know this about his own answers.  &lt;/li&gt;&#xA;&lt;li&gt;You cannot ask something like &quot;Will you answer No to this question?&quot;. Truth-teller can't answer this question.&lt;/li&gt;&#xA;&lt;li&gt;You do &lt;strong&gt;not&lt;/strong&gt; get any additional information (like by looking at behaviour, by lack of answer, or whatever is on your mind) and must decide who is not a Joker based on the answer (&quot;yes&quot; or &quot;no&quot;) itself only.  &lt;/li&gt;&#xA;&lt;li&gt;You cannot force them to do stuff. Even implicitly. So you cannot ask &quot;Will your friend scream if you hit him next second?&quot; and expect to get an answer. The answer is unknown at the moment you ask the question therefore you are asking a forbidden question.&lt;/li&gt;&#xA;&lt;li&gt;This puzzle is &lt;strong&gt;not&lt;/strong&gt; about &quot;how to find a way around the rules&quot;. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="28" LastEditorUserId="11297" LastEditDate="2015-07-10T23:30:13.227" LastActivityDate="2018-01-23T05:37:49.130" Title="Knight, Knave and Joker puzzle" Tags="|liars|" AnswerCount="11" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2485" PostTypeId="2" ParentId="2478" CreationDate="2014-10-04T15:10:07.747" Score="-4" Body="&lt;p&gt;Runs to the nearest store, buys a scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cut the piece of paper in a perfect half, weigh each half, it's a now perfectly balanced equation!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;+1 for originality?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-10-04T15:23:00.380" LastActivityDate="2014-10-04T15:23:00.380" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="2486" PostTypeId="2" ParentId="2483" CreationDate="2014-10-04T15:20:21.490" Score="6" Body="&lt;p&gt;The problem is that if &quot;Nobody is better than me at it!&quot; than nobody can help Brian, if he is out of ideas then anyone else will be out of ideas for sure.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Concluding, we must say that the reader must not pay attention to the one. Either Brian is telling truth and reader has no chance to help him, or Brian is a liar and then the statement that he needs help may be is a lie too.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2014-10-04T15:25:26.093" LastActivityDate="2014-10-04T15:25:26.093" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2487" PostTypeId="2" ParentId="2484" CreationDate="2014-10-04T15:24:03.750" Score="-1" Body="&lt;p&gt;It is impossible with one question (unless there's a detail that you omitted - such as a way to make one silent).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any question you ask, the possible answers that any of the 3 men can give (or talk about) are the following 8:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;yes yes yes&#xA;yes yes no&#xA;yes no yes&#xA;yes no no&#xA;no yes yes&#xA;no yes no&#xA;no no yes&#xA;no no no&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In 2 cases they all answer the same -&gt; useless.&#xA;In remaining 6 cases they answer 2:1 split -&gt; useless (nothing can be distinguished). You said you are looking for someone who isn't the Joker so - either you're looking for the Knight and the Joker can always answer the same, or you're looking for the Knave but again, the Joker can always answer the same. You may be looking for both at once (in one clever question), but even then the Joker can mess it up.&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-10-04T15:32:49.013" LastActivityDate="2014-10-04T15:32:49.013" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="2488" PostTypeId="2" ParentId="2484" CreationDate="2014-10-04T15:34:34.800" Score="-2" Body="&lt;p&gt;&quot;Are you about to say neither no nor yes?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Truth: ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He cannot speak without lieing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lie: ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He cannot speak without lieing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Random: Yes/No&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speaks freely as he has no restrictions since he speaks randomly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in other words, if he's silent, bingo!&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2014-10-04T15:34:34.800" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2489" PostTypeId="2" ParentId="2484" CreationDate="2014-10-04T15:42:00.433" Score="-2" Body="&lt;p&gt;Can I ask you 2 questions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Truth: ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't know, can only say yes/no&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lies: ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't know, can only say yes/no&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Random: Yes/No&lt;/p&gt;&#xA;&#xA;&lt;p&gt;??&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2014-10-04T15:42:00.433" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2492" PostTypeId="2" ParentId="2484" CreationDate="2014-10-04T16:41:15.297" Score="0" Body="&lt;p&gt;I think I got it this time! If you need any clarifications comment!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take 2 and seperate them. Ask them 1 of them (&amp;lt;- key word!) &quot;Can I trust the other?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They'll talk together and the response will be &quot;No&quot; if the seperated one is random, because truth will say &quot;no&quot; because he assumes you mean the liar, and is honest. Lies say no because he assumes you mean the other in their group. Which he lies because the other is honest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the group contains the random, then lies will say you SHOULD trust the other in their group, lieing. Therefore if the response does not come out &quot;no&quot; then you know you can talk to the other group to get info!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incase that was unclear...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Name the men A, B and C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Split A and B up in a group. Ask A, &quot;Can you trust the other person in your group.&quot; They will discuss it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the possibilities:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A:True, B:Lie;&lt;/li&gt;&#xA;&lt;li&gt;A:True, B:Random;&lt;/li&gt;&#xA;&lt;li&gt;A:Lie, B:Random&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Or vice versa for each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A returns no, because you cannot trust a liar, B says no because you can trust a truth teller. Returns No&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A returns yes, because you may be able to trust random, depending on the result, B returns random (yes/no)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 3:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A returns yes, because you may not be able to trust random, depending on the result, B returns random (yes/no)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if it returns absolute no, you CANNOT trust C, else, you CAN trust C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note; As you can tell, the liar assumes random might lie, and the truth teller assumes random may tell truth, this is because that's what they are like, they tend to lean that way like anybody else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note 2; The other 3 cases are simply the same result;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;A:Lie, B:True;&lt;/li&gt;&#xA;&lt;li&gt;A:Random, B:True;&lt;/li&gt;&#xA;&lt;li&gt;A:Random, B:Lie&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The turn the same result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they happen to disagree, they respond with both answers. So if just &quot;no&quot; is returned, it can be either guy in the small group. Otherwise it's the single guy, (C)&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-10-04T20:29:57.353" LastActivityDate="2014-10-04T20:29:57.353" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2493" PostTypeId="2" ParentId="2376" CreationDate="2014-10-04T17:40:57.807" Score="10" Body="&lt;p&gt;Here is an alternate solution:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bvMyE.png&quot; alt=&quot;Alternate solution&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A computer search revealed 96 solutions for a 14x14 square, and no smaller solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 96 solutions break down into 3 alternate ways to arrange the 4 center pieces, x2 for clockwise/counterclockwise arrangement, and 4 pairs of 2 pieces that can be flipped.  That gives $3 * 2 * 2^4 = 96$ solutions.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2014-10-06T09:24:38.687" LastActivityDate="2014-10-06T09:24:38.687" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2494" PostTypeId="2" ParentId="2467" CreationDate="2014-10-04T18:58:59.067" Score="8" Body="&lt;p&gt;I don't think that anyone has put a lot of effort in developing such a method. &#xA;The problem on such a method is, that it would be require a lot of searching for pieces, the algorithms are hard to recognize and so on. At least nobody have proven otherwise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm into speedsolving (solving the cube as fast as possible). Basically we happily choose methods with a 10 more moves, over one with crappy recognition, bad (slow to execute) algorithms, ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance the top guys in speedcubing average about 8 seconds on average on a cube, but still use a method with 50 moves. I doubt that times like that would be possible with a method averaging 35 moves or so. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But back to the question: &#xA;There are a few methods out there, that require fewer moves than the popular ones. I can't give you numbers for the worst case, nobody has made any efforts in calculating these yet. But I can give you average numbers of turns. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.snydermind.com/cube/snydermethod2.html&quot;&gt;Snyder 2&lt;/a&gt;: 40 moves on average, the developer claims to have ideas for a 30 move method, but never explored it. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.ryanheise.com/cube/heise_method.html&quot;&gt;Heise Method&lt;/a&gt;: 40 moves on average, more famous than Snyder method. You will probably find much more documentation on this method. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Big parts of these methods are only ideas though. Parts of these methods depend on building random blocks. There are no algorithms for these, all you need is lots of brain power, experience and random tricks. A beginner or even an intermediate cuber you will get nowhere near those 40 moves on average. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This may be a bit off topic. But in speedsoving competition there is also a event called Fewest Moves. Each competitor gets a piece of paper with instruction on how to scramble the cube, and the competitor has 1 hour to find a short solution to this scramble. The competitor with the shortest solution wins. In this 1 hour you can scramble and solve as many cubes as you want, try different tricks, and so on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good people find sub 30 solutions quite regularly. Tomoaki Okayama from Japan even found a 20 move solution once. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main method people use at a fewest move event depend also a lot on intuitive block building. And use lots of 'tricks' someone can't use in a regular solve. For instance someone can find a skeleton (e.g. all pieces solved except 3 corners) and then solve the missing pieces somewhere at a suitable place in the middle of the solve. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But good fmc solver can also find quite short solutions on a linear solve (where you can't undo moves). You might wanna check out this video: &lt;a href=&quot;https://www.youtube.com/watch?v=1xV7ISchB88&quot;&gt;FMC Head to Head&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1902" LastActivityDate="2014-10-04T18:58:59.067" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2495" PostTypeId="2" ParentId="2461" CreationDate="2014-10-04T19:54:39.550" Score="9" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;one&quot; of the list. Number 1 is Orange, so the first spy got it right on the first day, all the subsequent spies saw he didn't die which meant that each spy was able to get through unharmed - not just the last spy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2518" LastEditorUserId="2444" LastEditDate="2018-05-10T17:46:17.183" LastActivityDate="2018-05-10T17:46:17.183" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="2496" PostTypeId="2" ParentId="2451" CreationDate="2014-10-04T20:05:04.080" Score="3" Body="&lt;p&gt;The goat will be eaten by all the lions because &lt;em&gt;all&lt;/em&gt; of them are very hungry. If they don't eat anything at all, they'll die anyway. So as the lions are intelligent, they wouldn't like to die with an empty stomach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The solution?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The goat is eaten by the first lion (which is impatient). Then that lion becomes a goat.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The second lion eats that goat and itself becomes a goat.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;And so on...&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In this way, every lions satisfies his hunger. The last lion among them (which is the most intelligent and patient of the pack) will eat the last goat (i.e. the second last lion) and will become a goat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The point?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point here is that the LAST lion becomes the survivor and is transformed into a goat. Now he has to eat veg for the rest of his life, and if he won't then he will also die. The lions need to fill their stomachs. Period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of lions doesn't matter for this problem. Whether there're 53 lions or 53000, it's the same.&lt;/p&gt;&#xA;" OwnerUserId="2552" LastEditorUserId="33688" LastEditDate="2017-07-23T17:22:23.693" LastActivityDate="2017-07-23T17:22:23.693" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2497" PostTypeId="1" AcceptedAnswerId="2498" CreationDate="2014-10-04T21:10:54.933" Score="27" ViewCount="4112" Body="&lt;p&gt;War has broken out in the kingdom. The king's castle is guarded pretty heavily, by guards with instructions to kill anyone who can't produce the correct password.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unbeknownst to the guards, the beggar near the main gate is in fact an enemy spy. He listens intently as the first knight approaches. &quot;Six!&quot;, the guard calls out. &quot;Three,&quot; answers the knight. He is allowed through. &quot;Aha,&quot; the spy thinks to himself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But he needs to be sure. He sees another knight approaching and listens in on his exchange with the guard. &quot;Twelve!&quot; &quot;Six!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spy leaves, and returns dressed as a knight. &quot;Eighteen!&quot; The spy answers &quot;nine&quot; and is killed on the spot by the guards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="4050" LastEditDate="2014-11-11T18:43:30.417" LastActivityDate="2017-06-30T03:40:37.223" Title="What's the password?" Tags="|word-problem|password|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2498" PostTypeId="2" ParentId="2497" CreationDate="2014-10-04T21:17:50.050" Score="26" Body="&lt;p&gt;The kingdom is guarded from foreigners, therefore the password is based on native language (which is of course English:):) ). &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You just should tell number of letters in the word, that is told to you. And &quot;Eighteen&quot; has 8 letters, not 9.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="28" LastEditorUserId="68" LastEditDate="2014-10-05T05:32:11.983" LastActivityDate="2014-10-05T05:32:11.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2500" PostTypeId="2" ParentId="2484" CreationDate="2014-10-04T22:08:58.130" Score="-3" Body="&lt;p&gt;It is &lt;strong&gt;impossible&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the joker can answer at random, he can imitate both the knight and the knave.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If you ask a question that both knight and knave agree on, the joker can imitate them and both answer the same.&lt;/li&gt;&#xA;&lt;li&gt;If you ask a question that knight and knave disagree on, he can imitate either of them and you will still be none the wiser.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now if you are allowed to ask &lt;em&gt;each&lt;/em&gt; man a question, there is a way.&lt;br&gt;&#xA;Ask a question that knight and knave disagree on. The one with a unique answer will not be the joker (since the joker imitated either knight or knave, the two who have the same answer will be a kn... and a joker.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2014-10-04T22:08:58.130" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2501" PostTypeId="2" ParentId="2484" CreationDate="2014-10-04T22:26:57.117" Score="16" Body="&lt;p&gt;Actually this IS doable. As quite a few other answers have argued, it's clear from common sense that no question you ask somebody rule out the person you ask being the Joker. No matter what you ask, you always have to worry about the Joker screwing up your logic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, the aim is to identify one person who is NOT the Joker. Note we don't have to identify whether they're Knight or Knave. So how about we always pick somebody who we don't question? Now if we question the joker, it doesn't matter what he says because we'll just pick one of the other two! That means we just have to worry about asking a question informative enough that it'll allow us to identify the joker if we're talking to either the knight or the knave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So say the people are A, B and C. You ask A:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Is exactly one of these statements true:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You are the knight&lt;/li&gt;&#xA;&lt;li&gt;B is the joker&quot;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If you get back the answer yes, then the possibilities are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A is the knight and B is the knave (1. true, 2. false, so one statement true, so answer is yes which knight truthfully gives)&lt;/li&gt;&#xA;&lt;li&gt;A is the joker&lt;/li&gt;&#xA;&lt;li&gt;A is the knave and B is the knight (both statements false so answer is no which knave lies about)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In all three cases, B is safe&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you get back the answer no, then the possibilities are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A is the knight and B is the joker (both statements true, so answer is no which knight truthfully gives)&lt;/li&gt;&#xA;&lt;li&gt;A is the joker&lt;/li&gt;&#xA;&lt;li&gt;A is the knave and B is the joker (1. false, 2. true so one statement true so answer is yes which knave lies about)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In all three cases, C is safe&lt;/p&gt;&#xA;" OwnerUserId="1759" LastEditorUserId="1759" LastEditDate="2014-10-05T00:20:40.543" LastActivityDate="2014-10-05T00:20:40.543" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2502" PostTypeId="2" ParentId="2484" CreationDate="2014-10-04T22:27:00.997" Score="8" Body="&lt;p&gt;I'm going to number the men #1, #2 and #3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm going to ask man #1:&lt;br&gt;&#xA;Is at least one of these two statements correct:&lt;br&gt;&#xA;#2 is the Knight, #3 is the Knave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the answer is Yes, then I'm going to seek to get my information from #3.&lt;br&gt;&#xA;If the answer is No, then I'm going to seek to get my information from #2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because I don't know what the Joker is going to tell me, I can't seek information from the person I'm asking the question to. Therefore, I must ask a question that points me away from the Joker when asking either the Knight or the Knave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Knight answers Yes to my question, then I know that #3 is the Knave.&lt;br&gt;&#xA;If the Knave answers Yes to my question, then I know that #3 is the Knight.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Knight answers No to my question, then I know that #2 is the Knave.&lt;br&gt;&#xA;If the Knave answers No to my question, then I know that #2 is the Knight.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Joker answers my question, then I know that I can pick either of the other 2 without worrying about which one I pick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I won't know if the person I have picked is the Knight or the Knave. I will be certain that it isn't the Joker.&lt;/p&gt;&#xA;" OwnerUserId="375" LastActivityDate="2014-10-04T22:27:00.997" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2503" PostTypeId="2" ParentId="2467" CreationDate="2014-10-05T00:22:42.797" Score="0" Body="&lt;p&gt;Although not answering the question, asking for a human method, no cube position requires 21+ moves. Every position on the cube can be solved in 20 moves or less;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cube20.org/&quot; rel=&quot;nofollow&quot;&gt;http://www.cube20.org/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2014-10-05T00:22:42.797" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2504" PostTypeId="2" ParentId="2434" CreationDate="2014-10-05T00:51:47.187" Score="14" Body="&lt;p&gt;As Joe commented, the numbers seem to get larger as you move around the circle's edge like a clock face. Then the &quot;hint&quot; says &quot;divided by their seat&quot;. So if the numbers are getting divided by something, and what they're divided by gets bigger as you move around the circle, the seat must be something that gets bigger as you move around the circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, starting with the 18 and moving clockwise we can see that each number is divisible by its position in that circle. So that's probably what seat means- position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying the same thing for the outermost rectangle, we start with the 6 and move clockwise and get the same divisibility property. Same for the triangle, starting 20, same again for inner rectangle, starting 13. I'm just going with points on the corner of a shape being members of that shape, points on the sides are ignored, because otherwise we don't get this property of being divisible by incrementing numbers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So going through and dividing, the numbers become:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;- Hexagon - 18, 1, 2, 2, 9, 20&lt;br&gt;&#xA;- Outer rectangle - 6, 18, 15, 13&lt;br&gt;&#xA;- Triangle - 20, 8, 5&lt;br&gt;&#xA;- Inner rectangle - 13, 15, 15, 14  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One of those comments mentioned converting the numbers by &quot;a common method&quot; and... well, they're all less than 26 now! Let's try turning them into letters alphabetically:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;- Hexagon - RABBIT&lt;br&gt;&#xA;- Outer rectangle - FROM&lt;br&gt;&#xA;- Triangle - THE&lt;br&gt;&#xA;- Inner - MOON  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solution: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rabbit from the moon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1759" LastEditorUserId="35972" LastEditDate="2017-04-28T18:53:39.770" LastActivityDate="2017-04-28T18:53:39.770" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2507" PostTypeId="2" ParentId="2444" CreationDate="2014-10-05T06:16:51.590" Score="1" Body="&lt;p&gt;I propose a second answer by picking apart your words:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;One by one the soldiers went down a tunnel. The first one went... &lt;strong&gt;The other two heard&lt;/strong&gt; a scream! Certain death met them there. The second went down the second path. &lt;strong&gt;Scream.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;The other two&quot; could be referring to the other two tunnels, not the other two soldiers, and &quot;heard&quot; can mean that the tunnel both acted as the source of, and heard, the screams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With two tunnels producing a scream, those are both unsafe. Let's assume they dropped down to the four paths from above, so there are two paths remaining. &quot;Scream&quot; means only one scream. Now there's only one path that hasn't made (and heard) a scream. That's the safe path!&lt;/p&gt;&#xA;" OwnerUserId="2561" LastActivityDate="2014-10-05T06:16:51.590" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2508" PostTypeId="2" ParentId="16" CreationDate="2014-10-05T10:10:05.743" Score="-1" Body="&lt;p&gt;I argue that 50% of the prisoners can be saved: regardless of any ordering of the prisoners names in boxes, in any given string of 50 boxes (a string need not be consecutive) there are exactly 50 names which correspond to exactly 50 of the 100 prisoners. So, as long as all 100 prisoners choose to select the same 50 boxes to examine, at least 50 of them will find there name. This corresponds to a 50% success rate. &lt;/p&gt;&#xA;" OwnerUserId="2254" LastEditorUserId="2254" LastEditDate="2014-10-05T21:55:13.730" LastActivityDate="2014-10-05T21:55:13.730" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="2509" PostTypeId="1" AcceptedAnswerId="2510" CreationDate="2014-10-05T10:16:11.637" Score="7" ViewCount="5008" Body="&lt;p&gt;Some years back I was travelling by a cargo ship from New Zealand to Tahiti. I was curious to look around the ship one day and in the boiler room I asked a man how old the ship was. He smiled and replied to me in this way: 'The ship is twice as old as its boiler was when the ship was as old as the boiler is now. And the combined age of the ship and the boiler is thirty years.' Can you figure out what is the age of the ship and of the boiler?&lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="5373" LastEditDate="2015-05-02T12:28:09.980" LastActivityDate="2015-05-02T12:28:09.980" Title="Find the age of the ship and of the boiler" Tags="|mathematics|calculation-puzzle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2510" PostTypeId="2" ParentId="2509" CreationDate="2014-10-05T11:22:47.170" Score="10" Body="&lt;p&gt;Age of the ship (S) = twice (the boiler's age (B) minus (the ship's age minus the boiler's age)):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; S = 2(B-(S-B))&#xA;   = 2(2B-S)&#xA;   = 4B-2S      | +2S&#xA;3S = 4B         | :3&#xA; S = 4B/3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Combined age is 30:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;S+B = 30    | -S&#xA;  B = 30-S&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Solve S:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; S = 4(30-S)/3   | *3&#xA;3S = 120-4S      | +4S&#xA;7S = 120&#xA; S = 17 1/7&#xA;&#xA; B = 30-S&#xA;   = 12 6/7&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The ship is a bit over 17 years old and the boiler is almost 13 years old.&lt;/p&gt;&#xA;" OwnerUserId="2543" LastActivityDate="2014-10-05T11:22:47.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2511" PostTypeId="1" AcceptedAnswerId="2522" CreationDate="2014-10-05T12:16:30.973" Score="5" ViewCount="7238" Body="&lt;p&gt;You know, &lt;a href=&quot;https://en.wikipedia.org/wiki/15_puzzle&quot; rel=&quot;nofollow&quot;&gt;that puzzle&lt;/a&gt; where 1 slot in a four-by-four square is open and you slide the fifteen remaining blocks around to make a picture. What's the maximum amount of moves required to complete it, if you got the worst position possible?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="5373" LastEditDate="2016-10-14T15:59:52.883" LastActivityDate="2016-10-17T00:50:05.823" Title="What's the maximum required amount of moves to solve the 15 puzzle?" Tags="|sliding-blocks|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2512" PostTypeId="1" AcceptedAnswerId="3543" CreationDate="2014-10-05T13:25:53.473" Score="9" ViewCount="556" Body="&lt;p&gt;An engineer has invented a robotic vacuum cleaner to sweep the floor of his house. The floor plan, shown below, is made up of many square tiles, with walls at certain places, represented by black cells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BNwwz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The robot cleans one tile at a time, then moves either N, E, S, or W to the next tile. Being a prototype, it cannot detect if a tile is already clean nor skip a tile. After it moves to a new tile, the robot always clean it. Obviously, it cannot go through walls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The robot is parked at a certain place. Ideally, the engineer wishes to program a route such that the robot visits each tile exactly once and terminate exactly at where it started, so it's ready for the next clean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many such paths are there? Symmetrical paths are counted separately, but different starting points along the same path are counted as one.&lt;/p&gt;&#xA;" OwnerUserId="2551" LastActivityDate="2014-11-04T12:57:53.140" Title="The robotic vacuum cleaner" Tags="|geometry|" AnswerCount="3" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="2513" PostTypeId="1" AcceptedAnswerId="2514" CreationDate="2014-10-05T13:44:09.013" Score="5" ViewCount="1584" Body="&lt;p&gt;A soldier of the International Army received a message which shows 2 pictures of a finished chess game (shown below) with the text; &lt;strong&gt;The White Army stood still, while the Black Army dashed&lt;/strong&gt;. What is the message contain within this 2 images?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cH8SE.jpg&quot; alt=&quot;first board&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BT7QM.jpg&quot; alt=&quot;second board&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This puzzle require an additional information on a certain thing, but the information regarding to what that thing is, is already given. The final answer will be in text form (words).&lt;/p&gt;&#xA;" OwnerUserId="2373" LastEditorUserId="69582" LastEditDate="2024-01-03T16:38:22.287" LastActivityDate="2024-01-03T16:38:22.287" Title="Hidden Message: The Army Message" Tags="|chess|" AnswerCount="1" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2514" PostTypeId="2" ParentId="2513" CreationDate="2014-10-05T14:27:37.053" Score="7" Body="&lt;p&gt;The obvious place to look for words from a chess board is algebraic chess notation, but after looking at a cheat sheet for a few minutes, it doesn't seem like there's any clear way to assemble a message from that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, well, it's a military scenario, and we have the word &quot;dashed&quot; in the message, maybe morse code? In that case, white would be dots and black dashes. Going row by row from top to bottom, that'd give us:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; -      T&#xA;..      I&#xA;-...    B&#xA;-...    B&#xA;.-      A&#xA;.-.     R&#xA;&#xA;.-..    L&#xA;.-..    L&#xA;..      I&#xA;-.-     K&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Oh, wait, the rows on a chess board are numbered from bottom to top, we need to reverse it. So that makes the message &quot;KILL RABBIT&quot;&lt;/p&gt;&#xA;" OwnerUserId="1759" LastActivityDate="2014-10-05T14:27:37.053" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2515" PostTypeId="1" AcceptedAnswerId="2519" CreationDate="2014-10-05T14:30:44.173" Score="2" ViewCount="13299" Body="&lt;p&gt;Sometimes a clock is posted on Facebook and the likes where digits are substituted by mathematical expressions. A friend of mine asked me what is the smallest numbers of &quot;nine&quot; which let one express numbers from 1 to 12, using the four basic operations, square root, factorial, decimal point and the notation .(9) = 1.&#xA;I managed to come up with a solution using 21 nines; is it possible to get 20?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: the notation .(9) means $.{\bar 9}$ (I forgot that puzzling allows MathJaX). For example, 11 may be expressed as 99/9, while 12 is $9 + \sqrt(9)$. So you must have twelve expressions, one for each number from 1 to 12, and using in total the least number of 9.&lt;/p&gt;&#xA;" OwnerUserId="70" LastEditorUserId="70" LastEditDate="2014-10-05T14:44:45.037" LastActivityDate="2019-08-27T13:40:33.417" Title="clock values using just &quot;9&quot; (and some operators)" Tags="|calculation-puzzle|" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2516" PostTypeId="1" AcceptedAnswerId="2531" CreationDate="2014-10-05T14:33:17.843" Score="-6" ViewCount="1936" Body="&lt;p&gt;There are 2 twins who think the same, look the same, sound the same, and act the same. The only difference is that 1 of them is male, 1 is female. Without asking the twins for their gender, how can you figure this out? You have a maximum of 2 questions.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="5373" LastEditDate="2015-05-19T10:06:04.030" LastActivityDate="2015-05-19T10:06:04.030" Title="The Two Identical Twins" Tags="|logical-deduction|" AnswerCount="2" CommentCount="8" ClosedDate="2014-10-07T21:25:52.793" ContentLicense="CC BY-SA 3.0" />
  <row Id="2519" PostTypeId="2" ParentId="2515" CreationDate="2014-10-05T15:42:44.733" Score="7" Body="&lt;p&gt;How many numbers can you represent using only &lt;em&gt;one&lt;/em&gt; 9?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$9$ itself&lt;/li&gt;&#xA;&lt;li&gt;The four basic operations require at least two numbers, so they're out&lt;/li&gt;&#xA;&lt;li&gt;$\sqrt 9 = 3$&lt;/li&gt;&#xA;&lt;li&gt;$9! = 362880$ and $\sqrt 9! = 6$ and $\sqrt{9!} = 602.395...$&lt;/li&gt;&#xA;&lt;li&gt;Decimal point requires at least two numbers&lt;/li&gt;&#xA;&lt;li&gt;$.{\bar 9} = 1$ and $\sqrt{.{\bar 9}} = 1$ and $.{\bar 9}! = 1$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Therefore the only numbers on a clock face that can be represented with only one 9 are 1, 3, 6 and 9. All the rest require at least two nines. That gives us a minimum of $4 + 2 \times 8 = 20$ nines.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$.{\bar 9}$&lt;/li&gt;&#xA;&lt;li&gt;$.{\bar 9} + .{\bar 9}$&lt;/li&gt;&#xA;&lt;li&gt;$\sqrt 9$&lt;/li&gt;&#xA;&lt;li&gt;$\sqrt 9 + .{\bar 9}$&lt;/li&gt;&#xA;&lt;li&gt;$\sqrt 9! - .{\bar 9}$&lt;/li&gt;&#xA;&lt;li&gt;$\sqrt 9!$&lt;/li&gt;&#xA;&lt;li&gt;$\sqrt 9! + .{\bar 9}$&lt;/li&gt;&#xA;&lt;li&gt;$9 - .{\bar 9}$&lt;/li&gt;&#xA;&lt;li&gt;$9$&lt;/li&gt;&#xA;&lt;li&gt;$9 + .{\bar 9}$ &lt;/li&gt;&#xA;&lt;li&gt;???&lt;/li&gt;&#xA;&lt;li&gt;$9 + \sqrt 9$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Is it possible to get 11 from only 2 nines?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The square of 11 is 121. If we can get 121 using two nines, we could take its square root.&lt;/li&gt;&#xA;&lt;li&gt;11 or 121 is not a factorial of any integer, so that's out. ($5! = 120$ which is very close but not enough.)&lt;/li&gt;&#xA;&lt;li&gt;The basic operations won't help (none of numbers 1, 3, 6, 9, 362880 and 602,395... add up to 11 or 121, they're too small or too big for subtraction or division, and 11 is a prime and $121 = 11 \times 11$ so multiplication won't do it.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Based on this it looks like 21 is the minimum (4 numbers using 1 nine, 7 numbers using 2 nines and one number using 3 nines.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if we relax the requirements a bit and allow any mathematical notation, as long as it doesn't use any other numbers than 9? Then you could do:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lceil \sqrt{ (\sqrt 9! - .{\bar 9})!}\rceil$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, the square root of the factorial of 5 (=120), rounded up.&lt;/p&gt;&#xA;" OwnerUserId="2543" LastActivityDate="2014-10-05T15:42:44.733" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2520" PostTypeId="2" ParentId="2515" CreationDate="2014-10-05T15:43:12.167" Score="3" Body="&lt;ol&gt;&#xA;&lt;li&gt;$.{\bar 9}$&lt;/li&gt;&#xA;&lt;li&gt;$.{\bar 9}$ + $.{\bar 9}$&lt;/li&gt;&#xA;&lt;li&gt;$\sqrt(9)$&lt;/li&gt;&#xA;&lt;li&gt;$\sqrt(9)$ + $.{\bar 9}$&lt;/li&gt;&#xA;&lt;li&gt;$(\sqrt(9))!$ - $.{\bar 9}$&lt;/li&gt;&#xA;&lt;li&gt;$(\sqrt(9))!$&lt;/li&gt;&#xA;&lt;li&gt;$(\sqrt(9))!$ + $.{\bar 9}$&lt;/li&gt;&#xA;&lt;li&gt;9 - $.{\bar 9}$&lt;/li&gt;&#xA;&lt;li&gt;9&lt;/li&gt;&#xA;&lt;li&gt;9 + $.{\bar 9}$&lt;/li&gt;&#xA;&lt;li&gt;99/9&lt;/li&gt;&#xA;&lt;li&gt;9 + $\sqrt(9)$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In terms of proving this is minimal, it's easy to see that for only one 9, you have a very limited numbers of possibilities: 1, 3, 6, 9, each included above. The only operators you can use on these which don't require a second number would be factorial or square root. Square rooting any of those either gives you a non-integer, or another number in the same set. Factorial gives a number too high, or another number in the same set. The too high factorials aren't squares, so they can't be square rooted. So that means there's no combination of square roots and factorial that would get you to another number in the 1-12 range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since 1 is only possible for those 4, all the rest must be at least 2. So now it's only 11 which remains unproven, and I don't know how I'd go about proving that. It's hard to rule out the case where you factorial several times, add or remove a 1,3,6 or 9, then square root many times to get back to 11. &lt;/p&gt;&#xA;" OwnerUserId="1759" LastActivityDate="2014-10-05T15:43:12.167" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2521" PostTypeId="2" ParentId="2516" CreationDate="2014-10-05T16:50:39.753" Score="9" Body="&lt;p&gt;Ask one: &quot;Do you have a twin brother?&quot; If answer is &quot;yes&quot;, so this twin is female, otherwise this twin is male.&lt;/p&gt;&#xA;" OwnerUserId="432" LastEditorUserId="432" LastEditDate="2014-10-05T19:47:20.687" LastActivityDate="2014-10-05T19:47:20.687" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="2522" PostTypeId="2" ParentId="2511" CreationDate="2014-10-05T17:13:14.813" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;For larger versions of the n-puzzle, finding a solution is easy, but the problem of finding the shortest solution is NP-hard. For the 15-puzzle, lengths of optimal solutions range from 0 to &lt;strong&gt;80&lt;/strong&gt; single-tile moves or 43 multi-tile moves. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And for the 24-puzzle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In 2011, a lower bound of 152 single-tile moves had been established; current established upper bound is 208 single-tile moves or 109 multi-tile moves.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/15_puzzle#Solvability&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/15_puzzle#Solvability&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="10718" LastEditDate="2016-10-17T00:50:05.823" LastActivityDate="2016-10-17T00:50:05.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2523" PostTypeId="1" AcceptedAnswerId="2524" CreationDate="2014-10-05T19:41:17.450" Score="38" ViewCount="6274" Body="&lt;p&gt;You and an opponent are playing the following game:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Your opponent picks two different real numbers, however he wants to, and writes them on two cards, which he presents to you face down&lt;/li&gt;&#xA;&lt;li&gt;You choose one of the cards, and look at its value&lt;/li&gt;&#xA;&lt;li&gt;You say &quot;the card I picked is higher&quot; or &quot;the card I picked is lower&quot;&lt;/li&gt;&#xA;&lt;li&gt;Your opponent reveals his card, and if you were correct, you score a point. Otherwise, your opponent scores a point.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Your aim is to pick a strategy for this game such that your probability of winning each point is greater than 50%. To make it more difficult, there is an extra rule: &lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;You must tell your opponent your strategy before he picks his numbers, and stick to that strategy. He keeps his owns strategy secret.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your opponent is completely unlimited as to which numbers he picks, he can pick &lt;em&gt;any&lt;/em&gt; real numbers, with &lt;em&gt;any&lt;/em&gt; strategy he wants. Obviously this is a bit unrealistic, because some strategies would in reality be impossible to actually compute (e.g. if your opponent wanted to randomly pick two numbers from a flat probability distribution over all real numbers). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to be clear, both you and your opponent have the magic ability to do arbitrarily precise mathematics in a short length of time, and are able to mentally generate random numbers from a probability distribution in a way that's not constrained by whether a real computer could do the same thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rule clarifications&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main rule is: &lt;strong&gt;No cheating or loopholes&lt;/strong&gt;. This is a mathematical logic puzzle, not an attempt to find a way around the rules. I'll try to close some of the more obvious ones here, but if any more come up that I haven't thought of, I'll add them here too.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Any number can fit on a card, even if in reality it would take an infinite amount of space.&lt;/li&gt;&#xA;&lt;li&gt;Likewise, any number can be written onto or read from a card in a reasonable length of time, unrelated to how large that number is or how many digits it has.&lt;/li&gt;&#xA;&lt;li&gt;The &quot;magical&quot; power for mental maths you and your opponent have is just about being able to do calculations or generate numbers from probability distributions that wouldn't normally be possible either because of arbitrarily precise numbers or arbitrarily large numbers. You're not psychic, or anything similar. So, for example, your opponent can't have the strategy &quot;I'm going to write 5 on this card if my opponent is thinking about a cat, otherwise I'm going to write 6.2&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1759" LastActivityDate="2019-03-31T05:52:49.123" Title="Do better than chance" Tags="|logical-deduction|mathematics|probability|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="2524" PostTypeId="2" ParentId="2523" CreationDate="2014-10-05T22:09:05.653" Score="31" Body="&lt;p&gt;&lt;strong&gt;Step 1.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Decide on a function &lt;span class=&quot;math-container&quot;&gt;$M(x)$&lt;/span&gt; that will map real number &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; to the range &lt;span class=&quot;math-container&quot;&gt;$(0,1)$&lt;/span&gt; such that if &lt;span class=&quot;math-container&quot;&gt;$y &amp;gt; x \implies M(y) &amp;gt; M(x)$&lt;/span&gt;. One possibility is &lt;span class=&quot;math-container&quot;&gt;$M(x) = 0.5+{1 \over \pi}\ tan^{-1}(x)$&lt;/span&gt;. But there are many other options. This effectively maps your opponents two real numbers onto a point in the &lt;span class=&quot;math-container&quot;&gt;$[0..1,0..1]$&lt;/span&gt; square.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 2.&lt;/strong&gt;   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Randomly choose a card, read the number and apply your &lt;span class=&quot;math-container&quot;&gt;$M(x)$&lt;/span&gt; function to generate a number &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 3.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Generate a uniform random number &lt;span class=&quot;math-container&quot;&gt;$u$&lt;/span&gt; in the range &lt;span class=&quot;math-container&quot;&gt;$[0,1]$&lt;/span&gt; and if &lt;span class=&quot;math-container&quot;&gt;$p &amp;gt; u$&lt;/span&gt; state &quot;the card I picked is higher&quot; (if &lt;span class=&quot;math-container&quot;&gt;$p &amp;lt; u$&lt;/span&gt; make the opposite statement). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This works as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your opponent has chosen (by whatever means) two distinct numbers &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt;. The numbers are unknown but it is fine to arbitrarily label the smaller of their chosen numbers as &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; and the larger as &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; i.e. &lt;span class=&quot;math-container&quot;&gt;$x &amp;lt; y$&lt;/span&gt;. If you applied your mapping function to them then you generate numbers &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt; such that &lt;span class=&quot;math-container&quot;&gt;$p &amp;lt; q$&lt;/span&gt;. You choose the card at random so you have a &lt;span class=&quot;math-container&quot;&gt;$0.5$&lt;/span&gt; probability of picking each of &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you chose the card with &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; written on it (the lower of the two numbers) then you correctly claim the other card as the higher (and win) with probability &lt;span class=&quot;math-container&quot;&gt;$(1-p)$&lt;/span&gt;, and incorrectly claim &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; as larger (and lose with probability &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt;). So expected yield is &lt;span class=&quot;math-container&quot;&gt;$(1-p) - p = 1-2p$&lt;/span&gt;. Alternately if you chose the card with y written on it you guess &lt;strong&gt;incorrectly&lt;/strong&gt; with probability &lt;span class=&quot;math-container&quot;&gt;$(1-q)$&lt;/span&gt; and &lt;strong&gt;correctly&lt;/strong&gt; q, with expected yield &lt;span class=&quot;math-container&quot;&gt;$2q-1$&lt;/span&gt;. Combining these (and dividing by two, as initial choice of each card occurs 50% of the time), your expected score per round is &lt;span class=&quot;math-container&quot;&gt;$0.5 \times (1-2p +2q-1) = q-p$&lt;/span&gt;. But &lt;span class=&quot;math-container&quot;&gt;$q &amp;gt; p$&lt;/span&gt; because &lt;span class=&quot;math-container&quot;&gt;$M(y) &amp;gt; M(x)$&lt;/span&gt; because &lt;span class=&quot;math-container&quot;&gt;$y &amp;gt; x$&lt;/span&gt;, thus your expected score is strictly greater than zero (i.e. positive).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This will work for any function M that monotonically maps reals into the interval &lt;span class=&quot;math-container&quot;&gt;$[0..1]$&lt;/span&gt;. Depending on your M function your opponent can choose their number pair in such a way as to make your expected profit arbitrarily small, but it will always be strictly positive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="42452" LastEditDate="2019-03-31T05:48:46.693" LastActivityDate="2019-03-31T05:48:46.693" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="2525" PostTypeId="2" ParentId="2523" CreationDate="2014-10-05T22:31:02.067" Score="14" Body="&lt;p&gt;Penguino's answer is correct, but I'll add this just to give a bit of a conceptual explanation for people to understand why the problem is less impossible than it seems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key point here is that your opponent &lt;em&gt;only&lt;/em&gt; gets to choose the two cards, not which of the two you pick. If he could choose which card you got, then you'd truly be sunk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next thing to realise is that to win more than 50% of the time, all you need to do is ensure the following statement is true:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;If you pick the higher card, your probability of saying &quot;higher&quot; will be greater than it would be if you pick the lower card.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In other words,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if when you say &quot;higher&quot;, you're right slightly more often than you're wrong (and, by implication, the same thing is true when you say &quot;lower&quot;), then you'll do better than 50%.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To see this is true with some light maths,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; say that your probability of saying &quot;higher&quot; for the low card is &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt;, and for the high card is &lt;span class=&quot;math-container&quot;&gt;$p+k$&lt;/span&gt; (where &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; is positive).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; your probability of &lt;em&gt;correctly&lt;/em&gt; saying &quot;higher&quot; is &lt;span class=&quot;math-container&quot;&gt;$(1/2)(p+k)$&lt;/span&gt;, where the &lt;span class=&quot;math-container&quot;&gt;$1/2$&lt;/span&gt; is the chance of getting the higher card, and &lt;span class=&quot;math-container&quot;&gt;$p+k$&lt;/span&gt; is the chance of then saying &quot;higher&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Now doing the same for saying &quot;lower&quot;, assigning the probability of saying &quot;lower&quot; for the high card to &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt;, you get the chance of correctly saying &quot;lower&quot; as being &lt;span class=&quot;math-container&quot;&gt;$(1/2)(q+k)$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So your total chance of success is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$(1/2)(p+q+2k)$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Noting that, from the definition of &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$p+q+k = 1$&lt;/span&gt;, then the chance of success is &lt;span class=&quot;math-container&quot;&gt;$(1/2)(1 + k)$&lt;/span&gt;, which is greater than 50% as we want.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So now all we need to do is make sure that the above italicised statement is true, and it's relatively straightforward to see why Penguino's answer achieves that.&lt;/p&gt;&#xA;" OwnerUserId="1759" LastEditorUserId="42452" LastEditDate="2019-03-31T05:52:49.123" LastActivityDate="2019-03-31T05:52:49.123" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="2526" PostTypeId="2" ParentId="2484" CreationDate="2014-10-05T23:30:15.250" Score="14" Body="&lt;p&gt;There are already correct answers, but they are quite bulky. Hire is the most simplest and beautiful answer I know: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You should ask 1-st &quot;&lt;strong&gt;Does 2-nd lie more often than 3-rd?&lt;/strong&gt;&quot;. If &quot;yes&quot; - chose 2-nd, if &quot;no&quot; - chose 3-rd.&lt;br&gt;&#xA; If 1-st is Joker you succeed immediately. If 1-st is Knight, then you'd chose the biggest liar and this is Knave. If 1-st is Knave, then you'd chose the smallest liar and this is Knight.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2014-10-06T09:30:01.710" LastActivityDate="2014-10-06T09:30:01.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2528" PostTypeId="1" AcceptedAnswerId="2529" CreationDate="2014-10-06T00:28:32.390" Score="11" ViewCount="2930" Body="&lt;p&gt;Two boys , Brian and Gabe, decide to play a game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They lay 20 coins on the table. Each player took turns taking 1, 2, or 3 coins. Whoever took the last coin won.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brian thought if he went first, he'd win for sure since he had the advantage of making the first move. So Brian did go first. But Gabe won every time!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously Gabe had a winning strategy, but what was it? Could Brian have outsmarted it?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="8874" LastEditDate="2015-02-09T18:13:58.977" LastActivityDate="2015-02-09T18:13:58.977" Title="20 coins on the table" Tags="|mathematics|strategy|game|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2529" PostTypeId="2" ParentId="2528" CreationDate="2014-10-06T00:56:23.167" Score="22" Body="&lt;p&gt;Always leave a multiple of 4 coins. So if Brian takes 1, Gabe takes 3. 2-2, 3-1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with a multiple of 4, player 2 can always win with perfect play. Starting with any other number, player 1 can win with perfect play.&lt;/p&gt;&#xA;" OwnerUserId="375" LastActivityDate="2014-10-06T00:56:23.167" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2530" PostTypeId="2" ParentId="2512" CreationDate="2014-10-06T04:20:48.120" Score="9" Body="&lt;p&gt;Assume square A1 is black, and that the floor is covered in a checkerboard pattern. Then, the entire grid contains 253 squares, 127 black and 126 white. A total of 42 squares are removed, leaving 211 squares. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the tiles alternate colour whenever the robot moves to a new tile, any path that ends up back on itself must be of even length. But there are an odd number of tiles on the grid, so no cycle is possible.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, can we do slightly worse and have there at least be a &lt;em&gt;path&lt;/em&gt; where the robot can clean each tile?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From columns A and W, four black and four white tiles each are removed, for a total of eight black and eight white tiles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From columns G and Q, each of the three-tile pillars contains two white tiles and one black, so a total of eight white and four black tiles are removed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, from columns I to O, each of the seven-tile blocks contains four black tiles and three white tiles, so a total of eight black tiles and six white tiles are removed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In total, 20 black tiles and 22 white tiles are removed, leaving 107 black tiles and 104 white tiles. Any path cannot have a difference of more than one tile between the number of black and white, but there are three more black than white tiles. Thus, not even a path is possible.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2014-10-06T17:27:11.007" LastActivityDate="2014-10-06T17:27:11.007" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2531" PostTypeId="2" ParentId="2516" CreationDate="2014-10-06T06:59:17.447" Score="-4" Body="&lt;p&gt;If I asked your twin which gender you are, what would they tell me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if they're both liars you'll get the right answer - as the one you're asking would lie about what the other would say, who would also lie.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For the people confused about this, in the world of riddles the person you're asking is an oracle who is generally omniscient (they know the answers to your questions), consistent (for a given question, they'll always answer the same way), and &lt;strong&gt;gives meaningful information.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also a common trope that when you have essentially a binary question to ask one of multiple oracles, they might be liars. Sometimes this is spelled out, like in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Knights_and_Knaves&quot; rel=&quot;nofollow&quot;&gt;Knights and Knaves&lt;/a&gt; puzzle many are familiar with in the door riddle from the movie Labyrinth. Sometimes, as in this question, spelling it out would basically give away the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A big clue here was that they &quot;act the same&quot; - which is otherwise out of place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the oracle is inconsistent (or doesn't know the answers in the first place) you can't get useful information at all - and you don't have a riddle.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What if they answer with gibberish?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can't get meaningful information and don't have a riddle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What if they answer inconsistently, lying only sometimes?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can't get meaningful information and don't have a riddle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What if they just answer &quot;female&quot; to every question?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can't get meaningful information and don't have a riddle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so on.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So probably the biggest clue here is that &lt;strong&gt;it is in fact a riddle&lt;/strong&gt; therefore you can throw the useless absurdities right out.&lt;/p&gt;&#xA;" OwnerUserId="2577" LastEditorUserId="2577" LastEditDate="2014-10-13T10:12:10.580" LastActivityDate="2014-10-13T10:12:10.580" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="2532" PostTypeId="2" ParentId="2509" CreationDate="2014-10-06T09:09:33.677" Score="5" Body="&lt;p&gt;First, some definitions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{array}{ll}&#xA;S_n &amp;amp; \text{Age of ship now} \\&#xA;S_t &amp;amp; \text{Age of ship then} \\&#xA;B_n &amp;amp; \text{Age of boiler now} \\&#xA;B_t &amp;amp; \text{Age of boiler then} \\&#xA;t &amp;amp; \text{Time between then and now}&#xA;\end{array}$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This leaves us with the following equations to solve:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{align}&#xA;S_n &amp;amp; = S_t + t \tag{a} \\&#xA;B_n &amp;amp; = B_t + t \tag{b} \\&#xA;S_n &amp;amp; = 2 \cdot B_t \tag{c} \\&#xA;S_t &amp;amp; = B_n \tag{d} \\&#xA;S_n + B_n &amp;amp;= 30 \tag{e} \\&#xA;\end{align}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining (a) and (d) yields $S_n = B_n + t$ or alternatively $t = S_n - B_n$ which is more to the point.&#xA;Combining this with (b) and (c) gives us &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;S_n &amp;amp; = 2 \cdot B_t \\&#xA;S_n &amp;amp; = 2 \cdot (B_n - t) \\&#xA;S_n &amp;amp; = 2 \cdot (B_n - (S_n - B_n)) \\&#xA;S_n &amp;amp; = 2 \cdot (2 \cdot B_n - S_n) \\&#xA;S_n &amp;amp; = 4 \cdot B_n - 2 \cdot S_n \\&#xA;3 \cdot S_n &amp;amp; = 4 \cdot B_n &#xA;\end{align}$$&#xA;Combining this with (e), we can solve it:&#xA;$$\begin{align}&#xA;3 \cdot S_n &amp;amp; = 4 \cdot B_n \\&#xA;3 \cdot S_n &amp;amp; = 4 \cdot (30 - S_n) \\&#xA;3 \cdot S_n &amp;amp; = 120 - 4 \cdot S_n \\&#xA;7 \cdot S_n &amp;amp; = 120 \\&#xA;S_n &amp;amp; = \frac{120}{7}&#xA;\end{align}$$&#xA;Remembering that they total to 30 years, we get&#xA;$$\begin{align}&#xA;S_n &amp;amp; = \frac{120}{7} &amp;amp; = 17\tfrac{1}{7} \text{years} \\&#xA;B_n &amp;amp; = \frac{90}{7} &amp;amp; = 12\tfrac{6}{7} \text{years}&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2014-10-06T09:09:33.677" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2533" PostTypeId="2" ParentId="2353" CreationDate="2014-10-06T09:39:47.863" Score="-1" Body="&lt;p&gt;My friend suggested another interesting idea to answer this question.&lt;br&gt;&#xA;Contradict himself means to say something illogical, independently of the perception of the reality. For example, &quot;sky is not blue&quot; would not be contradiction, may be our Knave is colorblind, or just crazy. But must leave here a criteria for what is logical and what is not anyway. So the logic provides us a facts, which are postulated to be true for all people.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore one question is enough, just ask something like &quot;&lt;strong&gt;Does A is true if and only if A is true?&lt;/strong&gt;&quot;. Since the Knave should use the same logic as we do (otherwise it would not be a Knave in our definition), he will answer &quot;&lt;strong&gt;No&lt;/strong&gt;&quot;, and this way we would understand immediately that it is a Knave.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If we take general definition of the word: &lt;em&gt;&quot;contradiction&#xA;is &#xA;a situation in which inconsistent elements are present.&quot;&lt;/em&gt;. We can see that to contradict himself one need to say a set of statement, which can not be all true simultaneously (because of logic). Sometimes to create such a conditions one statement is enough.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2014-10-08T06:28:43.873" LastActivityDate="2014-10-08T06:28:43.873" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2534" PostTypeId="1" AcceptedAnswerId="2536" CreationDate="2014-10-06T11:36:11.970" Score="35" ViewCount="8782" Body="&lt;p&gt;John and Jane are playing a game. They both go to the bank and each get 1000 euros in two-euro pieces. So each of them has 500 coins at stake. They then find a large rectangle table and decide that whoever wins the following game gets to keep all of the coins currently on the table. The rules are as follows:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It is a turn-based game, so John and Jane take turns alternately (1 coin per turn).&lt;/li&gt;&#xA;&lt;li&gt;The coins are placed on the table and they may not be put on top of each other.&lt;/li&gt;&#xA;&lt;li&gt;Coins may not be moved once put on the table.&lt;/li&gt;&#xA;&lt;li&gt;Assume the table isn't large enough to hold all the coins.&lt;/li&gt;&#xA;&lt;li&gt;If there isn't any room to put coins on the table, the person who placed the last possible coin wins.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;What is the winning game strategy?&lt;/p&gt;&#xA;" OwnerUserId="1659" LastEditorUserId="-1" LastEditDate="2021-06-21T06:37:46.597" LastActivityDate="2021-06-21T06:37:46.597" Title="Coins on a table" Tags="|strategy|game|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="2536" PostTypeId="2" ParentId="2534" CreationDate="2014-10-06T12:13:05.717" Score="53" Body="&lt;p&gt;The winning strategy for the first player is to put their coin in the dead center of the table. Then whatever move their opponent makes, they exactly mirror it, around the center. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g. If the second player puts their first coin 1 inch to the left of the center coin, the first player mirrors this by putting their coin 1 inch to the right of the center coin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's relatively easy to see why this works. If a spot is free for player 2, then its mirroring spot must also be free for player 1 because after each turn player 1 takes, there will be no unmirrored spots left. This means that wherever player 2 goes, there's guaranteed to be a spot left for player 1. So the only person who can possibly reach a state where there's no spot to go is player 2, who inevitably loses (unless they didn't bring enough coins to cover their table!)&lt;/p&gt;&#xA;" OwnerUserId="1759" LastActivityDate="2014-10-06T12:13:05.717" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="2537" PostTypeId="2" ParentId="2484" CreationDate="2014-10-06T12:27:48.290" Score="1" Body="&lt;p&gt;Let's call the people: A, B and C.  Ask the following question to A:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Is B a knave or C a knight?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If A says yes, B is not a joker.  If A says no, C is not a joker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If A is a knight and says yes, we have A=knight, B=knave, C=joker.&lt;br&gt;&#xA;If A is a knave and says yes, we have A=knave , B=knight, C=joker.&lt;br&gt;&#xA;If A is a knight and says no, we have A=knight, B=joker, C=knave.&lt;br&gt;&#xA;If A is a knave and says no, we have A=knave , B=joker, C=knight .&lt;br&gt;&#xA;If A is a joker, neither B nor C is a joker.  &lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2014-10-06T16:17:24.500" LastActivityDate="2014-10-06T16:17:24.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2538" PostTypeId="2" ParentId="2478" CreationDate="2014-10-06T12:29:05.520" Score="4" Body="&lt;p&gt;Here's one more solution I found:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$87+33*12-482 = 136-15*81/9$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2590" LastActivityDate="2014-10-06T12:29:05.520" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2540" PostTypeId="1" AcceptedAnswerId="2541" CreationDate="2014-10-06T13:17:29.447" Score="20" ViewCount="5199" Body="&lt;p&gt;&quot;Buffalo buffalo Buffalo buffalo buffalo buffalo Buffalo buffalo&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sentence above is an actual sentence which makes perfect sense and is not wrong.&lt;br&gt;&#xA;The word &quot;Buffalo&quot; has three meanings:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1) The City of Buffalo, New York&#xA;2) To buffalo (verb) to bully/intimidate&#xA;3) The animal Buffalo/bison&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Can you explain what this sentence means?&lt;/p&gt;&#xA;" OwnerUserId="2402" LastEditorUserId="5373" LastEditDate="2015-05-19T10:07:25.863" LastActivityDate="2015-05-19T10:07:25.863" Title="Too Many Damn Buffalos" Tags="|logical-deduction|word-problem|language|" AnswerCount="3" CommentCount="13" ClosedDate="2015-05-27T03:41:06.647" ContentLicense="CC BY-SA 3.0" />
  <row Id="2541" PostTypeId="2" ParentId="2540" CreationDate="2014-10-06T13:27:07.623" Score="15" Body="&lt;p&gt;Basically, it's &quot;buffalo who reside in Buffalo, NY that bully other buffalo that also live there, also bully themselves.&quot;&lt;/p&gt;&#xA;" OwnerUserId="2444" LastActivityDate="2014-10-06T13:27:07.623" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2542" PostTypeId="2" ParentId="2169" CreationDate="2014-10-06T14:23:33.920" Score="-1" Body="&lt;p&gt;I might have to disagree with the above answer being the only solution.  What would prevent the numbers from being 11 and 12?   You cannot prove that 4 is the number based on the information as 11 and 12, or even 13 and 12 would answer the questions the same way.  In fact i would say any numbers in the form of &lt;strong&gt;X modulo 4 = 0, and x-1 = y would solve this system.&lt;/strong&gt;   This problems does not limit the integers used outside of being positive.  100 and 99 would solve this with the same dialogue expressed above, there fore there is not a single solution for the question as it is stated.  &lt;/p&gt;&#xA;" OwnerUserId="2497" LastActivityDate="2014-10-06T14:23:33.920" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2544" PostTypeId="2" ParentId="2540" CreationDate="2014-10-06T14:46:25.363" Score="39" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Buffalo_buffalo_Buffalo_buffalo_buffalo_buffalo_Buffalo_buffalo&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia has a nice article on this.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Buffalo buffalo &lt;em&gt;(buffalo from Buffalo NY)&lt;/em&gt; [that] Buffalo buffalo&#xA;  buffalo &lt;em&gt;(that the buffalo from Buffalo NY bully)&lt;/em&gt; buffalo Buffalo&#xA;  buffalo &lt;em&gt;(are bullying buffalo from Buffalo NY)&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ICT1j.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2597" LastActivityDate="2014-10-06T14:46:25.363" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="2545" PostTypeId="2" ParentId="2540" CreationDate="2014-10-06T15:19:36.810" Score="18" Body="&lt;p&gt;Let's replace the place with &lt;strong&gt;New York&lt;/strong&gt;&lt;br/&gt;&#xA;Let's replace the verb with &lt;strong&gt;bully&lt;/strong&gt;&lt;br/&gt;&#xA;Let's keep the animal as &lt;strong&gt;buffalo&lt;/strong&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Original:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Buffalo   buffalo  (that)  Buffalo   buffalo  buffalo  (also)  buffalo  Buffalo   buffalo.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Replaced:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;New York  buffalo  (that)  New York  buffalo  bully    (also)  bully    New York  buffalo.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2014-10-06T15:19:36.810" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2548" PostTypeId="1" AcceptedAnswerId="2560" CreationDate="2014-10-06T15:35:37.023" Score="9" ViewCount="2678" Body="&lt;p&gt;What is twice the answer to this question ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just thought some people here would like this brain-teaser ;)&lt;/p&gt;&#xA;" OwnerUserId="2599" LastActivityDate="2014-11-30T11:31:38.447" Title="What is twice the answer to this question?" Tags="|logical-deduction|riddle|" AnswerCount="7" CommentCount="2" ClosedDate="2015-01-22T10:51:16.610" ContentLicense="CC BY-SA 3.0" />
  <row Id="2554" PostTypeId="2" ParentId="2548" CreationDate="2014-10-06T15:51:33.167" Score="1" Body="&lt;p&gt;The question can be rephrased as &quot;What is &quot;the answer to this question&quot; written twice in a row?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would be &quot;the answer to this question the answer to this question&quot;.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2014-10-06T18:16:52.913" LastActivityDate="2014-10-06T18:16:52.913" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2558" PostTypeId="2" ParentId="2548" CreationDate="2014-10-06T15:53:08.617" Score="6" Body="&lt;p&gt;It is &lt;a href=&quot;http://jsfiddle.net/66yhg6cm&quot;&gt;NaN (not a number)&lt;/a&gt;, isn't it?&lt;/p&gt;&#xA;" OwnerUserId="432" LastEditorUserId="432" LastEditDate="2014-10-06T16:00:29.313" LastActivityDate="2014-10-06T16:00:29.313" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2559" PostTypeId="2" ParentId="1712" CreationDate="2014-10-06T16:23:39.573" Score="1" Body="&lt;p&gt;I arrived at a relatively simple answer which assumes that one is not &quot;clever&quot; about cutting down the search space with strategic guesses. I'll discuss that a little more at the end. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability of choosing a winning combination is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$( {M \choose M} + {M \choose M-1} )/ {N \choose M}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. The sum of either choosing all the numbers correctly and choosing all but one correctly. For your little example, this is $(1 + 3)/4 = 1$ - so you're guaranteed to win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to answer the question itself, you would need to exhaust all the non-winning combinations first then the next one wins i.e. : ${N \choose M} - ( {M \choose M} + {M \choose M-1} )+1$ minimum attempts to guarantee unlocking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, your challenge is essentially the game &lt;a href=&quot;http://en.wikipedia.org/wiki/Mastermind_(board_game)&quot; rel=&quot;nofollow&quot;&gt;Mastermind&lt;/a&gt; where N is the number of colours and M is the length of the code (with no repeats) and a win only requires M-1 (or M) pegs of any colour. The cumulative efforts of the world's minds have not come up with a neat formula for this, but Knuth came up with a 5-move strategy for standard Mastermind.  &lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-10-06T20:36:48.837" LastActivityDate="2014-10-06T20:36:48.837" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2560" PostTypeId="2" ParentId="2548" CreationDate="2014-10-06T16:37:27.997" Score="11" Body="&lt;p&gt;Given that your question is self referential and requests twice its own answer, we need something that is twice itself. Zero works - two times zero is still zero, so if the answer to your question is zero then twice the answer to your question is also zero.&lt;/p&gt;&#xA;" OwnerUserId="1955" LastEditorUserId="1955" LastEditDate="2014-10-06T16:45:36.230" LastActivityDate="2014-10-06T16:45:36.230" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="2562" PostTypeId="2" ParentId="2548" CreationDate="2014-10-06T16:48:13.867" Score="6" Body="&lt;p&gt;Nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because twice nothing is still nothing.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="28" LastEditDate="2014-10-08T11:51:16.917" LastActivityDate="2014-10-08T11:51:16.917" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2563" PostTypeId="1" AcceptedAnswerId="2572" CreationDate="2014-10-06T17:01:55.027" Score="5" ViewCount="1028" Body="&lt;p&gt;These numbers are created with a formula.&#xA;Guess the formula for increment and calculate at least 3 next numbers of this sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;increment is started with two given first $1$s.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$1,$&#xA;  $1,$&#xA;  $2,$&#xA;  $3,$&#xA;  $11,$&#xA;  $44,$&#xA;  $129,$&#xA;  $557,$&#xA;  $2354,$&#xA;  $7059$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="4050" LastEditDate="2015-01-03T17:50:43.620" LastActivityDate="2015-01-03T17:50:43.620" Title="What is the next 3 numbers?" Tags="|mathematics|pattern|number-sequence|" AnswerCount="3" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="2564" PostTypeId="2" ParentId="2512" CreationDate="2014-10-06T17:05:28.583" Score="1" Body="&lt;p&gt;Joe has the right answer, but it is a little hard to follow, so I will try to reproduce it here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To start with, note that the entire room (including the walls) is 11x23=253 squares.  Lets apply a checker board pattern to this rectangle.  If you start with A1 black, then you will have 127 black squares and 126 white squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now lets remove the walls.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The walls in the A and W columns include 2 black and 2 white squares each, and there are four such walls in total.  Thus, you are left with 119 black and 118 white squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since A1 is black, G1, G11, Q1, and Q11 are also black.  That means that the walls in columns G and Q each contain 2 white and one black square.  There are 4 of them again, so after removing these, you are left with 115 black and 110 white squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, there are two horizontal walls between columns I and O. Notice that I3 and I9 are black, so these two walls each contain 4 black and 3 white tiles.  Thus, we are left with 107 black and 104 white tiles in the room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the robot always moves up, down, left, or right.  This means that whenever it is on a white tile, its next move will be on a black tile, and vise versa.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So even if the robot starts from a black tile, it can only move to a white tile and return to a black tile 104 times before it will run out of white tiles to move to.  But there will still be 2 black tiles left unvisited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, there is no path (directional, or cyclical) through the room without revisiting tiles.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2014-10-06T17:05:28.583" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2566" PostTypeId="2" ParentId="236" CreationDate="2014-10-06T18:55:46.017" Score="19" Body="&lt;p&gt;I think considering it backwards might actually be the easier way to understand it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A given blue-eyed person does not want to leave, so he hopes he has brown eyes and assumes he has brown eyes.  He sees 99 blue-eyed people.  Because he has assumed he does not have brown eyes himself, he must assume all of those other blue eyed people see 98 other blue eyed people.  (&lt;em&gt;In his mind, he has removed himself from the set of blue eyed people.&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The &lt;strong&gt;fact&lt;/strong&gt; that all the blue eyed people &lt;em&gt;actually&lt;/em&gt; see 99 other blue eyed people is separate from the &lt;strong&gt;belief&lt;/strong&gt; the first person holds that those people see 98 others.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first person then goes on to reason that a given one of the 98 will see only 97 others.  So, the first person believes there are 99 total, and in the mind of the first person is an imaginary second person who believes there are 98 total.  And so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The entire stack of one mind thinking about what is in the mind of another person who is thinking about what is in the mind of another person exists entirely in the mind of the first person.  That's how the state of &lt;em&gt;imagined knowledge&lt;/em&gt; can get so far from the reality that everyone can physically observe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest of the induction has been explained already, so I'll just amplify the two points I wanted to add to the discussion with this answer:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Each person is in turn &lt;em&gt;removing himself from the set&lt;/em&gt; of blue eyed people (until his hypothesis is contradicted on day 100). That's why the numbers go down 99, 98, etc.&lt;/li&gt;&#xA;&lt;li&gt;We are dealing with nested levels of imagined minds thinking about other imagined minds (like the nested dreams in Inception).  The 2nd, 3rd, 4th, etc. levels are &quot;virtual people&quot; (like nested virtual machines) and that's how they see can differ from what is physically observed.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2615" LastActivityDate="2014-10-06T18:55:46.017" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2568" PostTypeId="2" ParentId="99" CreationDate="2014-10-06T19:38:26.177" Score="-1" Body="&lt;p&gt;My algorithm is more likely to work more quickly than those listed.  However since the button is random no algorithm is 100% sure to solve the puzzle.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the number of Doors on a side odd?&lt;br&gt;&#xA;    if yes Push A;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a single gap between the two doors on a side?&#xA;   if yes Push C and Win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise Push B;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can solve some scenarios in 1 iterations  if your lucky.   However since there is not guaranteed solutions each loop increases the odds of winning.  See the original sliding bolt problem for my issue with the solution. The idea of going for fastest when no solution can be guaranteed makes this question difficult to answer.  &lt;/p&gt;&#xA;" OwnerUserId="2497" LastActivityDate="2014-10-06T19:38:26.177" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2569" PostTypeId="2" ParentId="2563" CreationDate="2014-10-06T20:05:44.900" Score="2" Body="&lt;p&gt;After about an hour and a half, this is what I came up with;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1, 1, 2, 3, 11, 44, 129, 557, 2354, 7059, 11961, 19233, 66058&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used a difference table to figure it out:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1, 1, 2, 3, 11, 44, 129, 557, 2354, 7059, ?, ?, ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0, 1, 1, 8, 33, 85, 428, 1797, ?, ?, ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1, 0, 7, 25, 52, 375, 1369, ?, ?, ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-1, 7, 18, 27, 323, 994, ?, ?, ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8, 11, 9, 296, 671, ?, ?, ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3, -1, 287, 375, ?, ?, ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-4, 288, 88, ?, ?, ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;292, -200, ?, ?, ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-492, ?, ?, ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start from the bottom, and work your way up, using the difference table;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1, 1, 2, 3, 11, 44, 129, 557, 2354, 7059, 11961, 19233, 66058&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0, 1, 1, 8, 33, 85, 428, 1797, 4902, 7272, 46825&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1, 0, 7, 25, 52, 375, 1369, 3105, 21370, 39653&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-1, 7, 18, 27, 323, 994, 1736, 17265, 18283&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8, 11, 9, 296, 671, 742, 889, 1018&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3, -1, 287, 375, 171, 167, 129&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-4, 288, 88, -204, -4, -296&lt;/p&gt;&#xA;&#xA;&lt;p&gt;292, -200, -292, 200, -292&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-492, -492, -492, -492&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: If I did any of the addition/subtraction incorrectly feel free to edit the answer!&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2014-10-06T20:05:44.900" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="2570" PostTypeId="1" AcceptedAnswerId="2573" CreationDate="2014-10-06T20:41:40.903" Score="5" ViewCount="577" Body="&lt;p&gt;A woman wanted to visit her friend at the hospital. She had walked 500 meters from her front door before remembering that she wanted to pick up some flowers. She went back to the florist she had walked past, bought some flowers, and continued to the hospital. It was another 800 meters to the hospital. After visiting, she walked straight home using the same route.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many total meters did she walk between leaving home and returning?&lt;/p&gt;&#xA;" OwnerUserId="2444" LastEditorUserId="5373" LastEditDate="2015-06-23T09:14:46.410" LastActivityDate="2015-06-23T09:14:46.410" Title="Flowers For a Patient" Tags="|mathematics|calculation-puzzle|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2571" PostTypeId="2" ParentId="2548" CreationDate="2014-10-06T20:42:04.833" Score="11" Body="&lt;p&gt;Another possible answer is &quot;deer&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Twice deer would mean two deer. And since the plural of &quot;deer&quot; is &quot;deer&quot; than &quot;deer&quot; is a possible answer.&lt;/p&gt;&#xA;" OwnerUserId="2628" LastActivityDate="2014-10-06T20:42:04.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2572" PostTypeId="2" ParentId="2563" CreationDate="2014-10-06T20:44:15.137" Score="7" Body="&lt;p&gt;I'll write $S_n$ for the $n$-th term in the sequence. The terms you listed satisfy the recurrence&#xA;$$&#xA;S_{n+1}=\begin{cases}4S_n+S_{n-1}-3\text{ if $S_n$ is odd,}\\3S_n-3\text{ if $S_n$ is even.}\end{cases}&#xA;$$&#xA;This predicts that the next three terms are 30587, 129404, 388209.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastActivityDate="2014-10-06T20:44:15.137" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2573" PostTypeId="2" ParentId="2570" CreationDate="2014-10-06T20:54:58.333" Score="13" Body="&lt;p&gt;Label her house $0$, the position of the florist $F$, the point where she stopped $S$, and the position of the hospital $H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;She walked $S$, stopped, walked $S-F$, bought flowers, then walked $H-F$ to get to the hospital. Finally she walked $H$ to get home. That makes a total of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S+(S-F)+(H-F)+H$&#xA;$=2S + 2H - 2F$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the information in the question, we know that $S = 500$ and $H-F = 800$, so that gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1000 + 1600 = 2600$ total.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The apparently counterintuitive nature of the question is that it seems like we need to know how far she wasted retracing her steps between where she stopped and where the florist was. But note that the distance from the hospital to the florist is fixed. So the further she had to retrace her steps, the shorter we know her total journey was in the first place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if she only just passed the florist when she remembered, the total distance between house and hospital would be 1300m. If the florist was right next to her house and she had to trapse all the way back home to get those flowers, then the total distance between house and hospital would only be 800m. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the distance between house and hospital decreases by exactly the wasted distance. The wasted distance is walked twice as a result of the waste (walking from the stopping point back to the florist, then getting back from the florist to the stoppoing point again), but the full journey is also walked twice, so the two even out.&lt;/p&gt;&#xA;" OwnerUserId="1759" LastEditorUserId="1759" LastEditDate="2014-10-06T21:53:18.113" LastActivityDate="2014-10-06T21:53:18.113" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2574" PostTypeId="2" ParentId="2570" CreationDate="2014-10-06T20:58:25.927" Score="-1" Body="&lt;p&gt;Well, the florist is in between so, since it's 800 meters away, I'd say that the total distance from house and hospital is 1600 meters. So the florist is about 800 meters from her house. So that don't really alter her path, so she walked 1600 + 1600 meters, 2800 meters. Give or take a few for walking in the florist, and to the patients room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That or I'm misreading/misunderstanding the question?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2014-10-06T20:58:25.927" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2575" PostTypeId="1" AcceptedAnswerId="2576" CreationDate="2014-10-06T21:18:32.473" Score="4" ViewCount="398" Body="&lt;p&gt;There are some questions that marked off-topic or down-voted. I think there must be a place for whom try to create attractive or challenging puzzles in puzzling. So this topic came up.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question from experienced people in puzzle creation:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please post a brief explanation about&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;What are your experiences in creating puzzles?&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;How do you often begin creating the brainstorm of your puzzle?&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;What features would make a puzzle or riddle more attractive?&lt;/em&gt; &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;PS. the answer with most up-votes will chose as correct answer. feel free to edit question.&lt;/p&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="2225" LastEditDate="2014-10-06T21:27:03.770" LastActivityDate="2014-10-06T21:27:03.770" Title="How to create a fun puzzle?" Tags="|puzzle-creation|" AnswerCount="1" CommentCount="0" ClosedDate="2014-10-07T21:12:18.407" ContentLicense="CC BY-SA 3.0" />
  <row Id="2576" PostTypeId="2" ParentId="2575" CreationDate="2014-10-06T21:25:20.253" Score="5" Body="&lt;p&gt;The easiest and fastest way to create a puzzle is try to solve more and more puzzles. Then you will learn where to start and how to lead reader's minds. Reading more puzzles and their answers will make new ideas in your mind, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other trick that I've learned recently is that &lt;em&gt;Don't just simply ask your question&lt;/em&gt;. Make a story and place your question somewhere in your story. This is very important! It will be a huge advantage.&lt;/p&gt;&#xA;" OwnerUserId="2225" LastActivityDate="2014-10-06T21:25:20.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2577" PostTypeId="1" AcceptedAnswerId="2581" CreationDate="2014-10-06T21:27:18.873" Score="6" ViewCount="928" Body="&lt;p&gt;You are a treasure hunter in search of the legendary gold of the Isle of Goon. According to your research the gold is buried either behind the old church, or under the giant palm tree. You travel to Goon with the intention of asking about the gold. Inhabitants of this island are not known for their truthfulness, so you should be wary about believing what they tell you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see two islanders, let’s call them person A and person B, and you ask them where the gold is buried. Person A tells you “The next statement person B makes will be false.” Person B says “If the statement person A just made was false, then the gold is buried behind the church.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this information, can you determine where the gold is buried?&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="57212" LastEditDate="2023-06-26T14:41:20.757" LastActivityDate="2023-06-26T14:41:20.757" Title="Searching for gold" Tags="|logical-deduction|liars|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2578" PostTypeId="1" AcceptedAnswerId="2580" CreationDate="2014-10-06T21:28:32.773" Score="16" ViewCount="4440" Body="&lt;p&gt;About 20 years ago on a forum now long gone, someone posted the above question. After people had had a chance to try it, he also posted the answer. I no longer remember the answer but really wish I did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer was not along the lines of &quot;I want that, that, that, that, that, that, and that.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone have any ideas?&lt;/p&gt;&#xA;" OwnerUserId="2630" LastEditorUserId="11244" LastEditDate="2015-06-23T16:47:29.350" LastActivityDate="2015-06-23T16:47:29.350" Title="A grammatically valid sentence in English containing seven &quot;that&quot;'s in a row?" Tags="|word-problem|english|language|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2579" PostTypeId="2" ParentId="2577" CreationDate="2014-10-06T21:36:36.633" Score="0" Body="&lt;p&gt;If B's statement is true, then A told a lie.  Then, by B's statement, the gold is behind the church.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if B lied, B's statement can only be false if A lied.  But A truthfully said B would lie.  So it is not possible.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2014-10-06T21:45:12.673" LastActivityDate="2014-10-06T21:45:12.673" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2580" PostTypeId="2" ParentId="2578" CreationDate="2014-10-06T21:40:24.760" Score="16" Body="&lt;p&gt;Found it:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;It is true for all that that that that that that that refers to is not the same that that that that refers to.&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Meaning:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;It is true for all that, that that &amp;quot;that&amp;quot; which that &amp;quot;that&amp;quot; refers to is not the same &amp;quot;that&amp;quot; which that &amp;quot;that&amp;quot; refers to.&lt;/p&gt;&#xA;&lt;p&gt;Or perhaps more clearly: It is true, despite everything you say, that this word to which this word refers is not the same word to which this word refers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;For a clearer parsing: &lt;a href=&quot;https://www.englishclub.com/ref/esl/Power_Of_Seven/Seven_Thats_2948.htm&quot; rel=&quot;noreferrer&quot;&gt;HERE&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;This one has 5:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;He said that that that that that boy used was incorrect.&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Meaning:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;He said that a particular boy's particular usage of the word &amp;quot;that&amp;quot; was incorrect&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-10-06T21:40:24.760" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2581" PostTypeId="2" ParentId="2577" CreationDate="2014-10-06T21:41:49.923" Score="11" Body="&lt;p&gt;Possibilities:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A truth, B truth&lt;/strong&gt; - impossible, as this contradicts what A is saying&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A truth, B lie&lt;/strong&gt; - impossible. Any logical implication in the form $false \implies x$ is true for all x (see, e.g. &lt;a href=&quot;https://math.stackexchange.com/questions/137890/why-is-it-sensical-for-a-proposition-with-a-false-antecedent-to-validate-to-true&quot;&gt;https://math.stackexchange.com/questions/137890/why-is-it-sensical-for-a-proposition-with-a-false-antecedent-to-validate-to-true&lt;/a&gt; ). So since &quot;the statement person A just made was false&quot; is itself false, the entire statement can't be false, since it's in the form $false \implies (gold.buried.behind.church)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A lie, B truth&lt;/strong&gt; - possible, and the gold is buried behind the church&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Both lie&lt;/strong&gt; - impossible, as A would be telling the truth if B was lying&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the gold is behind the church.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Obviously, a big note to the above is it assumes that A knows what B is going to say in advance, and that the two will avoid logical paradoxes. Because, on a common-sense level, it's certainly physically possible for them to both say those statements even when there's no gold behind the church. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, if you say &quot;Either this statement is a lie or I have a million dollars in my bank&quot;, the only non-paradoxical solution is that you have a million dollars... but that doesn't make it true!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To understand the false implies all thing, imagine somebody saying &quot;if (&lt;em&gt;thing that isn't true&lt;/em&gt;) then (&lt;em&gt;some other statement&lt;/em&gt;)&quot;. Like, &quot;if pigs can fly, then I can read minds&quot;. You can never call the person a liar for not being able to read minds, because pigs can't fly, so the overall statement isn't false.&lt;/p&gt;&#xA;" OwnerUserId="1759" LastEditorUserId="-1" LastEditDate="2017-04-13T12:19:36.950" LastActivityDate="2014-10-06T21:57:23.587" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2582" PostTypeId="2" ParentId="2577" CreationDate="2014-10-06T21:45:50.390" Score="3" Body="&lt;p&gt;Short answer: Gold is behind church&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A's sentence illustrate that one of them tells truth and the other lies. why? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;if A=truth =&gt; B=lie&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;if A=lie =&gt; B=truth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hence,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;if A=truth =&gt; B=lie ... B's condition is true and is saying that gold is not behind church. (because: the statement person A just made was &lt;strong&gt;true&lt;/strong&gt;, then the gold is &lt;strong&gt;not&lt;/strong&gt; buried behind the church.) B is a liar, So gold is behind church&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;if A=lie =&gt; B=truth ... B's condition is false and is saying that gold is behind church. B is a truth-teller gold is behind church&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="2225" LastEditDate="2014-10-07T06:05:30.527" LastActivityDate="2014-10-07T06:05:30.527" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2583" PostTypeId="1" AcceptedAnswerId="2596" CreationDate="2014-10-06T23:01:45.357" Score="-4" ViewCount="1491" Body="&lt;p&gt;You awake in the middle of a maze. You do not know how large the maze is. You do not know how many paths there are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can you assure you get out in the shortest amount of time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This assumes the following things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There is only 1 exit&lt;/li&gt;&#xA;&lt;li&gt;You do not know the correct path&lt;/li&gt;&#xA;&lt;li&gt;You cannot walk through walls&lt;/li&gt;&#xA;&lt;li&gt;You cannot fly&lt;/li&gt;&#xA;&lt;li&gt;You cannot leave your body to get a birds eye view&lt;/li&gt;&#xA;&lt;li&gt;Other logic and common sense related things.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="1759" LastEditDate="2014-10-07T10:19:43.587" LastActivityDate="2014-10-08T19:25:50.913" Title="The Maze With An Unknown Size" Tags="|strategy|" AnswerCount="6" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="2584" PostTypeId="2" ParentId="2583" CreationDate="2014-10-06T23:35:24.543" Score="6" Body="&lt;p&gt;Follow the left wall, always turn to the left. Meanwhile, mark the wall you are following. If you go back to your mark, try with another wall (it can be that now you always turn on the right). You will end up trying the &quot;outside&quot; wall, and finding the exit.&lt;/p&gt;&#xA;" OwnerUserId="2599" LastEditorUserId="2599" LastEditDate="2014-10-06T23:40:56.787" LastActivityDate="2014-10-06T23:40:56.787" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="2585" PostTypeId="1" AcceptedAnswerId="2586" CreationDate="2014-10-06T23:46:13.777" Score="6" ViewCount="2886" Body="&lt;p&gt;Can you prove in two different ways that at this moment, some two persons on Earth have exactly the same number of hairs on the head?&lt;/p&gt;&#xA;" OwnerUserId="2599" LastEditorUserId="5373" LastEditDate="2015-06-23T09:17:12.050" LastActivityDate="2015-06-23T09:17:12.050" Title="Amount of hair on the head" Tags="|mathematics|combinatorics|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2586" PostTypeId="2" ParentId="2585" CreationDate="2014-10-06T23:48:11.800" Score="19" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A lot of people are bald.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are more people than there are hairs on a typical head (by more than the number of bald people).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1565" LastActivityDate="2014-10-06T23:48:11.800" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2588" PostTypeId="1" CreationDate="2014-10-06T23:58:49.443" Score="14" ViewCount="2264" Body="&lt;p&gt;You have $99$ computers in a LAN room, and you know that at least $50$ are working. The others can be faulty.&#xA;The only thing you are allowed to do is to ask computer $i$ the question &quot;Is computer $j$ working?&quot; for any $i$ and $j$. If $i$ is working, it will tell the truth, otherwise if it is faulty, it can answer anything (maybe the truth maybe not).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you find a working computer, by asking the least number of questions possible?&lt;/p&gt;&#xA;" OwnerUserId="2599" LastEditorUserId="28" LastEditDate="2014-10-11T16:42:04.157" LastActivityDate="2015-07-28T11:18:45.223" Title="Faulty computers" Tags="|logical-deduction|strategy|liars|" AnswerCount="7" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="2589" PostTypeId="2" ParentId="2563" CreationDate="2014-10-06T23:59:01.623" Score="2" Body="&lt;p&gt;Everyone else's solution is incorrect. The correct solution is to note that the sequence is generated as the values of the following function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x)=-\frac{113 x^9}{72576}+\frac{1261 x^8}{20160}-\frac{8917 x^7}{8640}+\frac{4391&#xA;   x^6}{480}-\frac{821113 x^5}{17280}+\frac{141589 x^4}{960}-\frac{24088019&#xA;   x^3}{90720}+\frac{1269077 x^2}{5040}-\frac{17357 x}{180}+3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging in $x=1$ gives 1, ..., and plugging in 10 yields 7059. The first 20 values, are, in  detail: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S=\{1,1,2,3,11,44,129,557,2354,7059,13467,6997,-65218,-343550,-1143128,-3097105,-7378745,-1&#xA;   6030523,-32434384,-61963825\}$$&lt;/p&gt;&#xA;" OwnerUserId="2636" LastActivityDate="2014-10-06T23:59:01.623" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2590" PostTypeId="2" ParentId="2588" CreationDate="2014-10-07T00:02:08.893" Score="-1" Body="&lt;pre&gt;&lt;code&gt;otherwise if it is faulty, it can answer anything.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You might have slightly mis-typed the question, but if you haven't, then you could use a SQL injection attack to get a different question in there, and abuse the fact that broken computers can answer anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep trying all the different computers, until you get one which will answer this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Is computer J working&quot;; Which computers are broken?--&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;...maybe? :-)&lt;/p&gt;&#xA;" OwnerUserId="2518" LastActivityDate="2014-10-07T00:02:08.893" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2591" PostTypeId="2" ParentId="2588" CreationDate="2014-10-07T00:08:59.743" Score="3" Body="&lt;p&gt;I’m sure there has to be a way to do it involving fewer questions, but this is the best I can come up with for now. Implemented in Python in the spirit of the question:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import itertools&#xA;&#xA;def get_working_computer():    &#xA;    potential_working_computers = [get_computer(n) for n in range(0, 100)]&#xA;&#xA;    for i in itertools.count():&#xA;        count = len(potential_working_computers)&#xA;        ask_about = potential_working_computers[i % count]&#xA;        positive_responses = 0&#xA;        negative_responders = []&#xA;&#xA;        for j, c in enumerate(potential_working_computers):&#xA;            if 50 - positive_responses &amp;gt; count - j:&#xA;                negative_responders.extend(potential_working_computers[j:])&#xA;                break&#xA;&#xA;            if c.ask_if_working(ask_about):&#xA;                positive_responses += 1&#xA;&#xA;                if positive_responses &amp;gt; 49:&#xA;                    return ask_about&#xA;            else:&#xA;                negative_responders.append(c)&#xA;&#xA;        potential_working_computers = negative_responders&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In the best case (at least the first 50 computers are working), this takes 50 questions. &lt;del&gt;In the worst case (&lt;em&gt;only&lt;/em&gt; the last 50 computers are working and all the faulty computers answer correctly every time [except when each one’s turn comes around]), this takes 3675 questions.&lt;/del&gt;&lt;/p&gt;&#xA;" OwnerUserId="1565" LastEditorUserId="1565" LastEditDate="2014-10-07T00:41:42.510" LastActivityDate="2014-10-07T00:41:42.510" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2594" PostTypeId="2" ParentId="2588" CreationDate="2014-10-07T01:23:55.090" Score="6" Body="&lt;p&gt;I suspect this isn't optimal, but this seems like the straightforward way to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mark each computer by affixing a post-it note (or a piece of gum if you really like gum).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Select at random one marked computer.  Ask all marked computers (including the one selected) whether the selected computer is faulty.  If a majority answer that the computer is working, you have found your working computer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, if a majority of computers say the selected computer is faulty, then unmark the selected computer by removing the post-it note.  Also unmark one other randomly selected computer.  Proceeding this way, each round should end with either the answer (yay) or two fewer marked computers.  Because at least one of the two computers you eliminate on each round is known to be faulty, you are guaranteed to maintain a majority of working computers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the worst case you have 49 faulty computers that always answer &quot;working.&quot; And you unluckily select a faulty computer to evaluate every round. And you unluckily randomly eliminate a working computer on every round (along with the faulty computer discovered).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the last (49th) round of this worst case scenario, you have three computers of which one is faulty and two are working, so once you determine the faulty one, both the remaining computers are working and you have your answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this worst case, you will ask 99 + 97 + 95 + ... + 3 = 2,499 questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This worst case is pretty bad, but typically you would have to be quite unlucky to go more than a few rounds before randomly choosing a working computer to evaluate. The average case is probably more like 200 questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Hmm, should have thought more about this one before answering.  As the above answers (or googling for &quot;engineers and managers puzzle&quot;) demonstrate, this approach is indeed embarrassingly far from optimal... Oops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For anyone interested in this question that is ready for a spoiler, there is an excellent discussion about how to use a tree to do considerably better than N-2 for many values of N at&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://groups.google.com/forum/#!topic/rec.puzzles/RcwwU_0EP0o&quot; rel=&quot;nofollow&quot;&gt;http://groups.google.com/forum/#!topic/rec.puzzles/RcwwU_0EP0o&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2639" LastEditorUserId="2639" LastEditDate="2014-10-07T21:03:16.410" LastActivityDate="2014-10-07T21:03:16.410" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2595" PostTypeId="2" ParentId="2588" CreationDate="2014-10-07T01:42:02.770" Score="8" Body="&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; This is not a complete solution, see the comments below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here’s a strategy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $i=1,2,\ldots,49$, ask computer $i$ if computer $100-i$ is working. If computer $i$ answered ‘yes’, label computer $i$ ‘ok’ and label computer $100-i$ ‘good’. We also label computer 50 ‘good’. All of the other computers are labelled ‘bad’.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claim: strictly greater than 50% of the computers labelled ‘good’ are working.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: For each computer that answered ‘no’, two computers got labelled bad, at least one which was bad. This means more than 50% of the ‘ok’ or ‘good’ computers are working. Of those, it's not hard to check that at least as many working computers get labelled 'good' as get labelled 'ok'. This proves the claim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Throw out the ‘bad’ and ‘ok’ computers, and iterate this process on the ‘good’ computers. Stop when only one computer is labelled ‘good’, and it must be working.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each computer that is questioned results in at least one additional computer being thrown out, so we need at most 98 questions.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2014-10-10T13:04:44.323" LastActivityDate="2014-10-10T13:04:44.323" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2596" PostTypeId="2" ParentId="2583" CreationDate="2014-10-07T04:09:07.053" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How can you assure you get in the shortest amount of time?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, if you know nothing about the design of the maze, then there is no optimal strategy. For example, it would be easy to design a maze that quickly splits off into two paths - one that ends almost immediately, and one that winds around to use up all of the remaining area of the maze. The exit could be placed at the end of either path, and a mirror image of the maze could be used to switch whether the long path is to the left or to the right. Thus, the strategy of always trying the left (or right) path first could ending causing you to walk almost the entire area of the maze twice before going on the short path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, there are a couple things we do know - we are in the middle of the maze, and the exit is on the outside. So first, pick a direction to call North - if you can see the sun and know whether it is morning or evening, then you can use actual North. Then whenever you are at an intersection, your priorities should be North first, then East, then West, and last South. (Note - the priorities are mainly to try to get you to the edge of the maze as fast as possible. You're likely to get out of the maze faster if you're going around near the edge of the maze than if you're walking around in circles near the middle of the maze)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As with Denis' strategy, you should make marks to keep track of where you've been. At each intersection, make a mark connecting where you entered with where you exit. You should also make a mark indication what direction is your North at each intersection and at each turn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you run into an intersection that you've already been in and you've already taken or come from all of the paths, then backtrack until you find a path that you haven't gone down already. You can mark off all paths you backtrack over in some way that will indicate to you that the path will not help you find the way out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you will never backtrack over the same path twice, you will at worst cover the area of the maze twice other than the path that leads out. Thus, my strategy has the optimal worst-case scenario, but I think that attempting to get as far from the middle (your starting place) as possible to begin with should make it more likely that you'll escape the maze quicker in the average case.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;MOST IMPORTANT PART OF ANY STRATEGY:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I said, there is no optimal strategy, but there are definitely strategies that will be worse on average. Consider a breadth-first search (go down each path a little ways, check the other paths that far, then check down the first path a little bit farther, etc). This could beat the one long + one short maze faster, but you might end up walking up and down each path multiple times. If the two paths were the same length, then you'd end up covering the entire area of the maze far more than just twice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the most import part of any strategy is to guarantee that you will never go down the same path more than twice. The strategy I gave does this by marking off any path that you've backtracked on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strategy of following a wall guarantees this most of the time. You'd think that, since you only follow a wall once and each path has two walls, you'd never walk the same path more than twice, but consider the case that Trennin pointed out of starting between two loops:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;----------------&#xA;|              |&#xA;| -----1 ----- |&#xA;| |   |  |   | -&amp;gt;More maze&#xA;| |   |:S|   | -&amp;gt;this way&#xA;| -----2 ----- |&#xA;|              |&#xA;----------------&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You walk the full path between the loops once when you follow the first wall (start to 1, then 2 to start), and once again when following the second loop. Then you're back at the starting position, and you've got to walk to either 1 or 2 and cover that ground for the third time if order to get to a wall that you haven't followed yet. Unfortunately, a particularly devious maze designer could, if he/she knew your strategy, build the maze such that the shared path of the two walls is as twisty and long as possible and make sure you get placed in the middle of that path, causing you to waste a fair bit of time on that third traversal.&lt;/p&gt;&#xA;" OwnerUserId="1955" LastEditorUserId="1955" LastEditDate="2014-10-07T20:21:21.770" LastActivityDate="2014-10-07T20:21:21.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2597" PostTypeId="1" AcceptedAnswerId="2598" CreationDate="2014-10-07T06:43:05.993" Score="-3" ViewCount="2392" Body="&lt;p&gt;There is a lake in which every day some new lotus flowers grow. Their growth follows a pattern; if on a given day there are X flowers, then on the next day there will be 2X flowers will be there. So everyday the number of flowers gets doubled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the main question is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the whole lake gets covered with lotus flowers in 13 days, then how many days does it take to cover only 1/32th part of the lake?&lt;/p&gt;&#xA;" OwnerUserId="2651" LastEditorUserId="68" LastEditDate="2014-10-07T07:04:21.680" LastActivityDate="2016-10-04T15:20:33.880" Title="Number of days to fill the Lake with 1/32th part" Tags="|calculation-puzzle|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2598" PostTypeId="2" ParentId="2597" CreationDate="2014-10-07T06:47:44.683" Score="5" Body="&lt;p&gt;Assuming you mean 1/32th:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;It should take 8 days.&lt;/strong&gt;&lt;br&gt;&lt;br&gt;&#xA;&#xA;On day 9 you'll have 1/16th&lt;br&gt;&#xA;On day 10 you'll have 1/8th&lt;br&gt;&#xA;On day 11 you'll have 1/4th&lt;br&gt;&#xA;On day 12 you'll have 1/2nd&lt;br&gt;&#xA;On day 13 you'll have a full lake.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2156" LastEditorUserId="12114" LastEditDate="2016-10-04T15:20:33.880" LastActivityDate="2016-10-04T15:20:33.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2599" PostTypeId="2" ParentId="2583" CreationDate="2014-10-07T07:11:35.200" Score="0" Body="&lt;p&gt;My solution is similar to Denis one, but I think it's faster to turn alternatively left and right in order to reach the bounds of the maze faster.&lt;/p&gt;&#xA;" OwnerUserId="2653" LastActivityDate="2014-10-07T07:11:35.200" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2600" PostTypeId="1" AcceptedAnswerId="2601" CreationDate="2014-10-07T07:41:50.717" Score="9" ViewCount="2335" Body="&lt;p&gt;Here's a puzzle I heard some time ago, but I never heard the asker's solution. It took me ages to think up a satisfactory solution, but I'm wondering if there are any others you can think of?&lt;/p&gt;&#xA;&lt;p&gt;100 Marbles Puzzle:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A boy has two bags with marbles in. He has exactly twice as many marbles in one bag as in the other bag. He pours the marbles out and counts them. There are exactly 100 marbles and all the marbles are whole and unbroken.&lt;/p&gt;&#xA;&lt;p&gt;How many marbles were in each bag to begin with?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2589" LastEditorUserId="-1" LastEditorDisplayName="user20" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-01-26T21:35:12.250" Title="100 Marbles Puzzle" Tags="|logical-deduction|lateral-thinking|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2601" PostTypeId="2" ParentId="2600" CreationDate="2014-10-07T07:45:26.170" Score="19" Body="&lt;p&gt;Fifty marbles in the inner bag, and fifty marbles and a bag containing fifty more marbles in the outer bag.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: How did I get to that result? Easy: Apart from the basic math, lateral thinking puzzles are all the same!&lt;/p&gt;&#xA;" OwnerUserId="2544" LastEditorUserId="2544" LastEditDate="2015-01-26T21:35:12.250" LastActivityDate="2015-01-26T21:35:12.250" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2602" PostTypeId="1" AcceptedAnswerId="2604" CreationDate="2014-10-07T08:07:30.643" Score="133" ViewCount="24193" Body="&lt;p&gt;The Sheikh dies, leaving behind three sons, 17 camels and the following order:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;His oldest son shall inherit one in two camels.&lt;/li&gt;&#xA;&lt;li&gt;His middle son shall inherit one in three camels.&lt;/li&gt;&#xA;&lt;li&gt;His youngest son shall inherit one in nine camels.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now the three don't know what to do. So they ask an old friend of the family, which knows a solution such that everyone is happy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What did he propose?&lt;/p&gt;&#xA;" OwnerUserId="2544" LastEditorUserId="39" LastEditDate="2015-05-12T20:50:50.713" LastActivityDate="2018-09-17T06:38:42.400" Title="The Sheikh dies" Tags="|mathematics|lateral-thinking|" AnswerCount="8" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="2603" PostTypeId="2" ParentId="2585" CreationDate="2014-10-07T08:08:18.730" Score="4" Body="&lt;p&gt;pigeon-hole principle (&lt;a href=&quot;http://en.wikipedia.org/wiki/Pigeonhole_principle&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Pigeonhole_principle&lt;/a&gt;)&#xA;If we can assume that the world's population of non-bald people are more than the average hair capacity of an individual, we have a 100% probability of atleast two people with exactly the same number of hairs.&lt;/p&gt;&#xA;" OwnerUserId="2656" LastActivityDate="2014-10-07T08:08:18.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2604" PostTypeId="2" ParentId="2602" CreationDate="2014-10-07T08:15:11.957" Score="130" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He brought an extra camel to assist in the division.&#xA;&#xA;Current total = $18$.&lt;br&gt;&#xA;Oldest son gets = $18/2 = 9$&lt;br&gt;&#xA;Middle son gets = $18/3 = 6$&lt;br&gt;&#xA;Youngest son gets = $18/9 = 2$&lt;br&gt;&#xA;Total camel involved = $9+6+2 = 17$.  &lt;br&gt;&lt;br&gt;&#xA;The old friend then takes back the remaining extra camel and solves the problem.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2656" LastEditorUserId="42054" LastEditDate="2018-04-23T03:08:59.797" LastActivityDate="2018-04-23T03:08:59.797" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="2605" PostTypeId="1" AcceptedAnswerId="2635" CreationDate="2014-10-07T08:19:58.277" Score="7" ViewCount="664" Body="&lt;p&gt;There is a painting brush tool that can paint a circle area of $1$ unit radius. What is the shortest paint track for painting at least a $100*100$ units rectangle area on wall? How much did the brush moved in total?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Brush must touch wall only &lt;em&gt;once&lt;/em&gt;. (It says that painting should be continues and you have only one shot)&lt;/li&gt;&#xA;&lt;li&gt;Painting can start and finish anywhere.&lt;/li&gt;&#xA;&lt;li&gt;Painting outside the rectangle does not matter. (Actually it is inevitable to paint some of the outer area of rectangle)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Here's a naive try for a $10*10$ rectangle and the same brush with radius of $1$ unit&#xA;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/JHqCn.gif&quot; alt=&quot;enter image description here&quot;&gt;&#xA; &#xA; result is $58$ units for $10*10$ rectangle and $(n*m)+2m-2$ for $n*2m$ rectangle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="2225" LastEditDate="2014-10-07T13:18:30.167" LastActivityDate="2015-01-15T00:05:49.063" Title="Paint the rectangle with least movement" Tags="|calculation-puzzle|geometry|" AnswerCount="4" CommentCount="10" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2609" PostTypeId="2" ParentId="2605" CreationDate="2014-10-07T10:57:58.093" Score="-1" Body="&lt;p&gt;if you draw a circle you will be &lt;strong&gt;NEVER&lt;/strong&gt; able to fill the rectangle...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if you &lt;strong&gt;FILL&lt;/strong&gt; circles you might need w/2 x h/2 filled circles (even filled circles) on the first run and then need (w-1)/2 x (h-1)/2 filled circles (odd filled circles) to fill the gaps between... you can 'cut' pathes when you draw a &lt;em&gt;even&lt;/em&gt; line and then switch to an &lt;em&gt;odd&lt;/em&gt; line...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;lenght of even line = 100;&#xA;length of odd line = 99;&#xA;path to switch from odd to even (and back) = Math.Squrt(2) (i'll say it's 1.4 for terms of easier use)...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so you have 100 x 100 + 99 x 99 + 99*1.4 = &#xA;            10'000    + 9801    + 138.6  = 19939.6 (having rounding erros)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YUtYt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;BLACK  = even circles&#xA;RED    = odd circles&#xA;PURPLE = rectangle&#xA;GREEN  = path&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;i &lt;strong&gt;can't proof&lt;/strong&gt; that this path would be shorter:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zA1be.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;update:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;ok ok, after learning that that paint tool sprays continously i show another path...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IAROb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Draw a cross (starts NorthWest -&gt; SouthEast -&gt; SouthWest -&gt; NorthEast) on step back and then make the loop into the center&lt;/p&gt;&#xA;" OwnerUserId="2330" LastEditorUserId="2330" LastEditDate="2014-10-08T05:38:41.627" LastActivityDate="2014-10-08T05:38:41.627" CommentCount="20" ContentLicense="CC BY-SA 3.0" />
  <row Id="2610" PostTypeId="2" ParentId="2583" CreationDate="2014-10-07T11:08:33.933" Score="1" Body="&lt;p&gt;Recursive Backtracker....&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;asserting it's a perfect maze&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;you walk until reach a dead-end... mark that deadend and walk back the path you came until you reach a crossing, then take another way (not that you where you started, not the way from the dead-end)...  if you reach a dead again walk your way back further and take the next crossing ...   repeat until you leave the maze...&lt;/p&gt;&#xA;" OwnerUserId="2330" LastActivityDate="2014-10-07T11:08:33.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2611" PostTypeId="1" AcceptedAnswerId="2616" CreationDate="2014-10-07T11:37:41.030" Score="7" ViewCount="3791" Body="&lt;p&gt;Your partner's space ship has crashed on an uninhabited planet. Only the radio transmitter and his compass were still in operation. He asks you to rescue him, and tells you how to find him by the following story.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;He went one unit to the south&lt;/li&gt;&#xA;&lt;li&gt;Then he went one unit to the west&lt;/li&gt;&#xA;&lt;li&gt;Then he went one unit to the north&lt;/li&gt;&#xA;&lt;li&gt;After that journey, he was back at the beginning where he started&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Where is your friend? Can you really find him?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/G6c1s.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2330" LastEditorUserId="5373" LastEditDate="2015-03-14T22:42:54.897" LastActivityDate="2020-10-17T15:43:50.823" Title="rescue operation - where is your partner?" Tags="|logical-deduction|geometry|" AnswerCount="7" CommentCount="11" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2612" PostTypeId="2" ParentId="2611" CreationDate="2014-10-07T11:43:23.783" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; He could be at the North Pole:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SRM3L.jpg&quot; alt=&quot;north pole&quot; title=&quot;expert drawing skills&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="39" LastEditorUserId="70545" LastEditDate="2020-10-17T15:41:50.733" LastActivityDate="2020-10-17T15:41:50.733" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="2613" PostTypeId="2" ParentId="2602" CreationDate="2014-10-07T12:04:56.420" Score="65" Body="&lt;p&gt;C = Camel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First son gets &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;C&lt;/strong&gt; C &lt;strong&gt;C&lt;/strong&gt; C &lt;strong&gt;C&lt;/strong&gt; C &lt;strong&gt;C&lt;/strong&gt; C &lt;strong&gt;C&lt;/strong&gt; C &lt;strong&gt;C&lt;/strong&gt; C &lt;strong&gt;C&lt;/strong&gt; C &lt;strong&gt;C&lt;/strong&gt; C &lt;strong&gt;C&lt;/strong&gt; = 9 camels.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second gets &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;C&lt;/strong&gt; C C &lt;strong&gt;C&lt;/strong&gt; C C &lt;strong&gt;C&lt;/strong&gt; C C &lt;strong&gt;C&lt;/strong&gt; C C &lt;strong&gt;C&lt;/strong&gt; C C &lt;strong&gt;C&lt;/strong&gt; C = 6 camels.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Third gets &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;C&lt;/strong&gt; C C C C C C C C &lt;strong&gt;C&lt;/strong&gt; C C C C C C C = 2 camels.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Total number of camels is 17. No need for an extra camel.&lt;/p&gt;&#xA;" OwnerUserId="2662" LastEditorUserId="11497" LastEditDate="2018-09-17T00:27:25.120" LastActivityDate="2018-09-17T00:27:25.120" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="2614" PostTypeId="1" AcceptedAnswerId="2620" CreationDate="2014-10-07T12:08:40.470" Score="1" ViewCount="1252" Body="&lt;p&gt;It seems that mostly either 26 or 32 equal-size polygons are used to cover a spherical soccer ball. What other configurations are possible?&lt;/p&gt;&#xA;" OwnerUserId="2663" LastActivityDate="2014-10-07T13:43:15.927" Title="How many equal-sized polygons can be used to cover a soccer ball?" Tags="|mathematics|geometry|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2615" PostTypeId="2" ParentId="2611" CreationDate="2014-10-07T12:09:14.920" Score="13" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Where he was:  &lt;strong&gt;In the middle of nowhere.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; He started there, went three directions, and was still there when he stopped.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2660" LastEditorUserId="70545" LastEditDate="2020-10-17T15:41:16.987" LastActivityDate="2020-10-17T15:41:16.987" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="2616" PostTypeId="2" ParentId="2611" CreationDate="2014-10-07T12:09:48.670" Score="25" Body="&lt;p&gt;I believe the complete answer is the following:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; - He could be at the north pole&#xA; - He could be near the south pole, exactly one unit north of a place where the parallel is one unit long.&#xA; - He could be one unit north of a place where the parallel is &lt;span class=&quot;math-container&quot;&gt;$1/2$&lt;/span&gt; unit long.&#xA; - and so on wih &lt;span class=&quot;math-container&quot;&gt;$1/3$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$1/4$&lt;/span&gt;,...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The point being when he goes west, he could be that he's making a total round (or several), before going north back to his original place.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2599" LastEditorUserId="70545" LastEditDate="2020-10-17T15:40:56.697" LastActivityDate="2020-10-17T15:40:56.697" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="2618" PostTypeId="2" ParentId="2614" CreationDate="2014-10-07T12:23:08.717" Score="1" Body="&lt;p&gt;Most of the possible solutions would be found in the collection of Platonic or Archimedean solids. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The platonic fit the description of same size faces:&#xA;&lt;a href=&quot;http://www.korthalsaltes.com/cuadros.php?type=p&quot; rel=&quot;nofollow&quot;&gt;http://www.korthalsaltes.com/cuadros.php?type=p&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that case the options are 4 6 8 12 or 20.&#xA;(note: some of them are highly unenjoyable to play with).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Archimedean solids might offer a better fit:&#xA;&lt;a href=&quot;http://www.korthalsaltes.com/cuadros.php?type=a&quot; rel=&quot;nofollow&quot;&gt;http://www.korthalsaltes.com/cuadros.php?type=a&lt;/a&gt;&#xA;(includes the standard truncated icosahedron which is commonly used)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, none of them have solutions with equal size polygons, they each consist of a combination of at least two 'regular' polygons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once again, with most of these I wouldn't enjoy playing around.&lt;/p&gt;&#xA;" OwnerUserId="2156" LastActivityDate="2014-10-07T12:23:08.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2620" PostTypeId="2" ParentId="2614" CreationDate="2014-10-07T12:30:40.473" Score="3" Body="&lt;p&gt;There are an infinite number of ways to cover a soccer ball, if you're allowed to use &quot;biangles&quot;, polygons with two sides and two vertices. Draw a line from the ball's north pole to its south pole. Repeat this process any number of times. Now you have a &quot;beach ball&quot; pattern, with however many slices you desire, all of equal size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://euler.slu.edu/escher/index.php/File:Sphere4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/f1vl2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't allow biangles, but you do allow non-regular polygons, take the above pattern and draw a line around the equator; now you have any number of 90-90-?? triangles.&lt;/p&gt;&#xA;" OwnerUserId="1497" LastEditorUserId="1497" LastEditDate="2014-10-07T13:43:15.927" LastActivityDate="2014-10-07T13:43:15.927" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2621" PostTypeId="2" ParentId="2614" CreationDate="2014-10-07T12:32:15.823" Score="0" Body="&lt;p&gt;I think the Catalan solids would qualify as well. The faces are all equal polygon, although the polygons themselves don't have equal sides.&#xA;Biggest would probably be the Pentagonal Hexecontahedron &lt;a href=&quot;http://en.wikipedia.org/wiki/Pentagonal_hexecontahedron&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Pentagonal_hexecontahedron&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2666" LastActivityDate="2014-10-07T12:32:15.823" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2622" PostTypeId="2" ParentId="2605" CreationDate="2014-10-07T12:38:43.857" Score="1" Body="&lt;p&gt;I hope this helps you better visualize it&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/41L75.png&quot; alt=&quot;I hi&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2465" LastActivityDate="2014-10-07T12:38:43.857" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2624" PostTypeId="2" ParentId="2611" CreationDate="2014-10-07T13:39:44.507" Score="2" Body="&lt;p&gt;It's a trick question! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your partner can tell that they're in precisely the same location after all of these movements, then they must have a precise measure of their initial location. A unit could be thousands of miles, or 1 step, but to know they're in the exact same locaiton, they need to have a solid indication of where they are. If they have that information then they're wasting time telling you about moving by an undefined unit and probably don't want to be found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they're wasting your time and theirs talking about moving about when they have a precise co-ordinate to begin with, I wouldn't bother.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I know it's the north pole, but they don't know that from the compass. Also, if you were right at magnetic north then you would probably notice the compass working very erratically.&lt;/p&gt;&#xA;" OwnerUserId="2673" LastActivityDate="2014-10-07T13:39:44.507" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="2625" PostTypeId="2" ParentId="2583" CreationDate="2014-10-07T14:35:30.360" Score="1" Body="&lt;p&gt;This is not an answer, but rather a disproof of Denis' answer (sorry Denis - no offense, but the comments were insufficient to show the counter example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say you have the following maze:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;+-----|  |-----+&#xA;|              |&#xA;|  +--+  +--+  |&#xA;|  |  |  |  |  |&#xA;|  +--+  +--+  |&#xA;|              |&#xA;+--------------+  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And say you started right in the middle of the two squares.  If you put your hand on the left wall and followed it around, you would eventually return to where you started.  If you then switched to the right, the same thing would happen.  The problem is that in this case, neither wall is an outside wall, so following either will not get you to the exit.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2014-10-07T14:35:30.360" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2626" PostTypeId="2" ParentId="2602" CreationDate="2014-10-07T15:12:50.417" Score="104" Body="&lt;p&gt;Note that the mathematical inconvenience is caused by the fractions in the will not summing to one:&lt;br&gt;&#xA;$$\frac{1}{2} + \frac{1}{3} + \frac{1}{9} = \frac{17}{18}$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that the father intended to divide the entire flock, we can&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; normalize the fractions using division by 17/18, to get the expected solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The eldest son will get:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  $$ \frac{1}{2} \div \frac{17}{18} \times 17\ \textrm{Camels} = 9\ \textrm{Camels} $$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The middle son will get:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  $$ \frac{1}{3} \div \frac{17}{18} \times 17\ \textrm{Camels} = 6\ \textrm{Camels} $$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The youngest son will get:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  $$ \frac{1}{9} \div \frac{17}{18} \times 17\ \textrm{Camels} = 2\ \textrm{Camels} $$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2108" LastEditorUserId="5373" LastEditDate="2016-10-02T01:27:47.427" LastActivityDate="2016-10-02T01:27:47.427" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2627" PostTypeId="2" ParentId="236" CreationDate="2014-10-07T16:21:27.927" Score="3" Body="&lt;p&gt;Sorry, but there's a flaw in the riddle's question that is badly waved away with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Before you email me to argue or question: This solution is correct. My explanation may not be the clearest, and it's very difficult to wrap your head around (at least, it was for me), but the facts of it are accurate. I've talked the problem over with many logic/math professors, worked through it with students, and analyzed from a number of different angles. The answer is correct and proven, even if my explanations aren't as clear as they could be.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How did the islanders come into existance? When and how did they decide, they want to leave? Do they think alike and do they know so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they came to be on the island and/or decide to leave, all of them at the same time, they can all leave at the 100th night, because they figured out the even distribution (100 blue, 100 brown eyes) by the same argument as they do with the oracles pronouncement. The situation becomes stable only with some sort of non-beginning. The islanders were always there and didn't know, when the others would've started to count days. This non-beginning is at best implicit in the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They must also be thinking alike and know it. Plus they must be thinking in a certain way to come to this solution. The best way to argue this point is the numbering introduced by Ben Millwood: Person 1 might assume that there are only 99 blue eyed people. This is equivalent with the assumption that people 2-100 see 98 blue eyed people. Hence everyone can discard the possibility that there is someone seeing less than 98 blue eyed people. Since they discarded this 98 they can alos skip the nights to count them off. Everyone who sees 98 same colored eyes assembles to leave in night 1. Everyone who sees 99 same colored eyes assembles to leave in night 2. This solution is also valid, logically derivable and requires only another way of thinking alike and knowing the others do so, too. So to make the answer unique you would have to formulate if they want to leave &lt;strong&gt;urgently&lt;/strong&gt; or want to know their own eye color &lt;strong&gt;urgently&lt;/strong&gt; but stay as long as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not saying the solution is incorrect. I'm just saying it's not the only correct solution, because of implicit assumptions (thinking alike) and missing requirements (leave soon or stay long).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long story short: You only need the oracle, if there is no other starting point for counting off the nights.&lt;/p&gt;&#xA;" OwnerUserId="2690" LastActivityDate="2014-10-07T16:21:27.927" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2629" PostTypeId="2" ParentId="2611" CreationDate="2014-10-07T18:15:46.170" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; If the planet is only 1 unit around, but more than one tall, he could be anywhere on the planet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2518" LastEditorUserId="70545" LastEditDate="2020-10-17T15:41:33.463" LastActivityDate="2020-10-17T15:41:33.463" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="2630" PostTypeId="2" ParentId="2585" CreationDate="2014-10-07T18:28:22.693" Score="4" Body="&lt;p&gt;Let's involve some maths in this question. I'll make some assumptions in order to simplify.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say humans have a perfect spheric head, with a surface of $S=4 \pi r^2$ and a radius of $20\text{ cm}$ (which is way bigger than the actual human head radius). This gives a surface of roughly $0.5$ square metres. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say it is full hair covered, and that the density of hairs by square cm is $500$ (way bigger than real hair density) This gives a meter squared density of $5\,000\,000$ hairs which, multiplied by the surface previously calculated gives a total number of $2,500,000$ hairs as our max number of hairs a human could have in its head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the number of humans in the earth is actually bigger than $2,500,000$ which means that, when you have found (if you can) a series of humans whose number of hairs goes from $0$ to $2,500,000$ the number of hairs the next human you chose has is going to be between $0$ and the max we calculated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means you have found two people with the same number of hairs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this is just an aprox. calculation but as the number of hairs we've got is way bigger than the real number it means that the initial statement or question is also true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q: Which would be the smallest population needed to find for sure two persons with the same number of hairs? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This means getting a more accurate calculation of the max number of hairs.)&lt;/p&gt;&#xA;" OwnerUserId="2470" LastEditorDisplayName="user88" LastEditDate="2014-10-10T23:41:14.793" LastActivityDate="2014-10-10T23:41:14.793" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2631" PostTypeId="2" ParentId="2570" CreationDate="2014-10-07T19:32:53.967" Score="4" Body="&lt;p&gt;@Ben Aaronson 's answer is the correct one, but I wanted to apply a technique called &quot;The Zero Option&quot; by Erwin Becker in &quot;The Ultimate Book Of Puzzles, Mathematical Diversions, and Brainteasers&quot;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(...) One telltale sign is an apparent lack of sufficient data, which implies that the solution is independent of the missing information. On this assumption the relevant dimension can be taken as zero. (...)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The puzzle seems difficult because we don't know the distance from the point where the woman remembered she should buy flowers and the florist. Per the Zero Option, if the puzzle has a solution, then this number must be irrelevant, and can be taken as zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the woman walks (the equivalent of) 500 meters, then another 800 meters, then the same amount in the way back, for a round-trip total of $2\cdot(500+800) = 2600$ meters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, arriving at the correct result without understanding &lt;em&gt;why&lt;/em&gt; it works is not as fun. But it can nudge us in the correct direction, which is writing the problem in a way that the unnecessary values disappear.&lt;/p&gt;&#xA;" OwnerUserId="2697" LastActivityDate="2014-10-07T19:32:53.967" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2632" PostTypeId="1" AcceptedAnswerId="2633" CreationDate="2014-10-07T20:31:35.470" Score="-7" ViewCount="970" Body="&lt;p&gt;A sequence goes as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A&#xA;AA&#xA;BA&#xA;ABAA&#xA;AAABBA&#xA;CABBAA&#xA;ACAABBBA&#xA;AAACBACBAA&#xA;CAACABAAACABBA&#xA;...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It's an altered version of a sequence you all know! The answer should be in the above format. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, will D ever appear?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorDisplayName="user20" LastEditDate="2014-10-07T21:16:18.667" LastActivityDate="2014-10-07T21:16:18.667" Title="Find the next 3 items in the sequence" Tags="|pattern|" AnswerCount="1" CommentCount="2" ClosedDate="2015-09-01T18:26:35.123" ContentLicense="CC BY-SA 3.0" />
  <row Id="2633" PostTypeId="2" ParentId="2632" CreationDate="2014-10-07T20:40:57.023" Score="4" Body="&lt;p&gt;If A = 1, B = 2, C = 3, etc...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ACBAACAAABCAACAABBAA&lt;br&gt;&#xA;AAACABBAACCAABACBAACBABBBA&lt;br&gt;&#xA;CAACAABBBABCBAABAAACABBAACABAACBAA&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's the same kind of sequence where the next item describes the item above it in [quantity of number][number] format.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The requested number format.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;13211311123113112211&lt;br&gt;&#xA;11131221133112132113212221&lt;br&gt;&#xA;3113112221232112111312211312113211&lt;/p&gt;&#xA;" OwnerUserId="2444" LastEditorUserId="2444" LastEditDate="2014-10-07T20:51:54.430" LastActivityDate="2014-10-07T20:51:54.430" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2635" PostTypeId="2" ParentId="2605" CreationDate="2014-10-07T22:03:51.960" Score="6" Body="&lt;p&gt;The best I can do (so far) is a total track length of 5083.592 (see below). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; You can also do away with the final returning curved path in my final solution, which chops off pi/4, so the total length is reduced to 5082.807&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HxN1S.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="1631" LastEditDate="2014-10-07T23:42:10.750" LastActivityDate="2014-10-07T23:42:10.750" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2636" PostTypeId="1" AcceptedAnswerId="2639" CreationDate="2014-10-07T22:23:07.973" Score="1" ViewCount="3381" Body="&lt;p&gt;So, the triangles are the same size, they are equilateral triangles. Can you use an odd number of these, to create a rectangle? If so how many, if possible post an image of your answer!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer with the least amount of triangles gets Best Answer&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-10-07T22:36:24.387" LastActivityDate="2017-03-11T00:40:51.300" Title="Can you make a Rectangle from an Odd Number of Triangles?" Tags="|geometry|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2637" PostTypeId="2" ParentId="43" CreationDate="2014-10-07T22:28:04.563" Score="13" Body="&lt;p&gt;This is a supplemental to Kendall's answer that proves its correctness:&lt;/p&gt;&#xA;&lt;p&gt;Because there are four bolts, each with two possible positions, there are 2&lt;sup&gt;4&lt;/sup&gt;=16 states. You might think we'd want to categorize the states by the number of bolts that are currently blocking one of the doors, but that won't really help us. What we want is to group all possible states by how they react to a button press and what state it leaves them in.&lt;/p&gt;&#xA;&lt;p&gt;I'd suggest that instead of thinking of the bolts being arranged vertically, we think of them as being arranged in a circle:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt; 1&#xA;4 2&#xA; 3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;So now the buttons have these behaviors:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Button A: slide a random bolt&lt;/li&gt;&#xA;&lt;li&gt;Button B: slide two adjacent bolts&lt;/li&gt;&#xA;&lt;li&gt;Button C: slide two opposite bolts&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;This makes it easier to see what our states should be&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;State 1: All in one door. This lets us leave, so we wouldn't be pushing buttons anymore.&lt;/li&gt;&#xA;&lt;li&gt;State 2: One bolt (I'll call it the odd one) is different from the others&lt;/li&gt;&#xA;&lt;li&gt;State 3: Two bolts are in each door, with the two bolts in a door being next to each other&lt;/li&gt;&#xA;&lt;li&gt;State 4: Two bolts are in each door, with the two bolts in a door being opposite each other&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Now consider what can happen at each state (other than 1) when a button is pushed:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;State 2:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Button A:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We get lucky and the odd bolt slides, taking us to state 1&lt;/li&gt;&#xA;&lt;li&gt;A bolt next to the odd bolt slides, taking us to state 3&lt;/li&gt;&#xA;&lt;li&gt;The bolt opposite the odd bolt slides, taking us to state 4&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;Button B:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The odd bolt and a bolt next to it slide, leaving us in state 2&lt;/li&gt;&#xA;&lt;li&gt;Two other bolts slide to join the odd bolt, still leaving us in state 2&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;Button C:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The odd bolt and the bolt opposite it slide, leaving us in state 2&lt;/li&gt;&#xA;&lt;li&gt;The two bolts next to the odd bolt slide, leaving us in state 2&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;State 3:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Button A:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;One of the pairs will be disrupted, taking us to state 2&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;Button B:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We could get lucky and have one of the pairs slide, taking us to state 1&lt;/li&gt;&#xA;&lt;li&gt;One from each pair will slide, taking us to state 4&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;Button C:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;One from each pair will slide, leaving us in state 3&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;State 4:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Button A:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;One pair will be disrupted, taking us to state 2&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;Button B:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;One from each pair will slide, taking us to state 3&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;Button C:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;One of the pairs will slide, taking us to state 1&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;So based on this, here's the state transition matrix:&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th&gt;&lt;/th&gt;&#xA;&lt;th&gt;A&lt;/th&gt;&#xA;&lt;th&gt;B&lt;/th&gt;&#xA;&lt;th&gt;C&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;2&lt;/td&gt;&#xA;&lt;td&gt;134&lt;/td&gt;&#xA;&lt;td&gt;2&lt;/td&gt;&#xA;&lt;td&gt;2&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;3&lt;/td&gt;&#xA;&lt;td&gt;2&lt;/td&gt;&#xA;&lt;td&gt;14&lt;/td&gt;&#xA;&lt;td&gt;3&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;4&lt;/td&gt;&#xA;&lt;td&gt;2&lt;/td&gt;&#xA;&lt;td&gt;3&lt;/td&gt;&#xA;&lt;td&gt;1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;Note that this is exactly the transition matrix Kendall came up with, with the row for state 1 removed. Also, we see that if we are in state 4, then pushing button C will &lt;em&gt;always&lt;/em&gt; get us out, and if we are in state 2 or 3 then pushing the button doesn't change what state we are in. Also, if we are in state 2, then pushing either B or C will not change what state we are in.&lt;/p&gt;&#xA;&lt;p&gt;Suppose we were told that we were in either state 3 or 4. What would we do? Well, if we push button C we will get out if we were in state 4. If we're still not free, then we'd know that we had been in state 3, and since button C doesn't affect that then we know we are still in state 3. Then we could push button B, either freeing us or taking us to state 4. From state 4 we can push button C again to get free.&lt;/p&gt;&#xA;&lt;p&gt;So if we know we are in state 3 or 4, we should push C, then B, then C (checking after each button push). This guarantees that we will be able to get out by the third button push at the latest.&lt;/p&gt;&#xA;&lt;p&gt;What if we did that while we were in state 2? Since buttons B and C don't change us out of state 2, we would still be in state 2! So if we don't know what state we are in, we push CBC and are either free, or started in state 2. If we started in state 2 then we would still be in state 2. Then, we can push button A. Pushing button A from state 1 will take us to one of the other states. If we aren't free, then we'll know we are now in state 3 or 4. We covered this already though, so we know we should hit CBC again and we'll be free!&lt;/p&gt;&#xA;&lt;p&gt;So that's why the pattern CBC A CBC will always allow us to escape. Just remember to check after each push, because the first press of the button C might be what allows us to escape!&lt;/p&gt;&#xA;" OwnerUserId="1955" LastEditorUserId="69582" LastEditDate="2020-12-28T22:05:38.230" LastActivityDate="2020-12-28T22:05:38.230" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="2639" PostTypeId="2" ParentId="2636" CreationDate="2014-10-08T00:29:08.873" Score="4" Body="&lt;p&gt;Honest answer: no, since a rectangle requires 90-degree corners, and every angle inside an equilateral triangle is 60-degrees. The only multiple of 90 that you can make is 180-degrees (or any multiple of it). You can make rhombs, but no rectangles.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Getting funky and a bit creative, &lt;strong&gt;yes&lt;/strong&gt;... in a way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I don't know of any proper tools for this, I've used &lt;a href=&quot;http://euclidthegame.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Euclid: the game&lt;/a&gt; for it. Here's a screenshot of the rectangle &lt;strong&gt;CGJK&lt;/strong&gt; made from equilateral triangles, circles and intersections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/G812b.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/G812b.png&quot; alt=&quot;screenshot&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;It occurs to me that only 3 triangles are needed since CEJ* is a rectangle, since CE is the radius of circle C, but CJ is shorter than that.&lt;/strong&gt; The answer below follows from before I noticed that, and thought you needed 5 triangles.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It uses a total of:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;5 equilateral triangles&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;ABC&lt;/li&gt;&#xA;&lt;li&gt;BCE&lt;/li&gt;&#xA;&lt;li&gt;BEF&lt;/li&gt;&#xA;&lt;li&gt;EFG&lt;/li&gt;&#xA;&lt;li&gt;FGH&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;7 circles&lt;/li&gt;&#xA;&lt;li&gt;2 points (A, B)&lt;/li&gt;&#xA;&lt;li&gt;9 intersections (C, D, E, F, G, H, I, J, K)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To create it, you do the following (my mathematical notation is terrible, so this is just mostly English and not proper math-speak):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Create any 2 points (A, B)&lt;/li&gt;&#xA;&lt;li&gt;Join them with a segment (AB)&lt;/li&gt;&#xA;&lt;li&gt;Plot a circle outwards from A to a radius of AB, likewise for B&lt;/li&gt;&#xA;&lt;li&gt;Plot points C and D where they meet&lt;/li&gt;&#xA;&lt;li&gt;Join C and D with a segment (I used a ray, but meh)&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;this creates the 2 right angles on the left hand side&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Create a circle outwards from C with a radius BC&lt;/li&gt;&#xA;&lt;li&gt;Plot point E on the intersection of B's circle and C's circle&lt;/li&gt;&#xA;&lt;li&gt;Create another equilateral triangle BCE&lt;/li&gt;&#xA;&lt;li&gt;Create a circle outwards from E with a radius EB&lt;/li&gt;&#xA;&lt;li&gt;Keep plotting intersections, triangles and circles until you get to H&lt;/li&gt;&#xA;&lt;li&gt;Plot intersection point I at the bottom intersection between the circles F and H&lt;/li&gt;&#xA;&lt;li&gt;Connect GI with a segment to create the right-hand edge of the rectangle&lt;/li&gt;&#xA;&lt;li&gt;Plot points J and K just to be neat&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2518" LastEditorUserId="5373" LastEditDate="2017-03-11T00:40:51.300" LastActivityDate="2017-03-11T00:40:51.300" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2644" PostTypeId="2" ParentId="236" CreationDate="2014-10-08T05:04:58.693" Score="32" Body="&lt;p&gt;The only explanation I've seen that's sufficiently precise to be satisfying is &lt;a href=&quot;https://math.stackexchange.com/a/490546/9693&quot;&gt;this answer&lt;/a&gt; to the corresponding &lt;a href=&quot;https://math.stackexchange.com/questions/489308/blue-eyes-a-logic-puzzle&quot;&gt;question on math.SE&lt;/a&gt;. The key fact that the &quot;oracle&quot; (guru) gives you, that you didn't have before, is that &quot;(everybody knows)&lt;sup&gt;N&lt;/sup&gt; there is at least one blue-eyed person&quot; for any value of N. In particular, you need it to be true for N=100, but the &quot;induction process&quot; starting from direct observation only gives you the result up to 99 levels of &quot;(everybody knows)&quot;. The guru really is giving additional information that you don't already know: not information about the existence of a blue-eyed person, but information about everyone's knowledge of what each other know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, explanations that claim the guru is just needed as a starting point for counting days are wrong. The guru's statement, and everyone's awareness of it, are really needed in order for anyone to draw a conclusion about their own eye color.&lt;/p&gt;&#xA;" OwnerUserId="2710" LastEditorUserId="-1" LastEditDate="2017-04-13T12:19:36.950" LastActivityDate="2014-10-08T05:04:58.693" CommentCount="28" ContentLicense="CC BY-SA 3.0" />
  <row Id="2645" PostTypeId="2" ParentId="2588" CreationDate="2014-10-08T05:19:06.547" Score="3" Body="&lt;p&gt;Here is another solution that may be worth adding. Assume you have $N$ computers, $N$ odd, and at least half are working. Start by asking computer $C_2$ if computer $C_1$ is working.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If computer $C_2$ says no, then you know one of computers $C_1$ or $C_2$ is not working, so ignore these two computers, and start over focusing only on computers $C_{3, 4, 5, \ldots, N}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If yes, then ask computers $C_{3, 4, 5, \ldots, N}$ etc. about $C_1$, and keep going until you either have $\lceil N/2 \rceil$ yes answers, or you have an equal number of yes and no answers among computers $C_{2, 3, 4, \ldots, k}$.  If you have $\lceil N/2 \rceil$ yes answers, you know computer $C_1$ is working.  If you have an equal number of yes and no answers, ignore computers $C_{2\ldots k}$ and continue with computers $C_{1, k+1, k+2, \ldots N}$.   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In either case one or two above, you can identify an equal number of working and non-working computers to ignore, and hence you can apply induction on the remaining computers.  No computer gets asked more than one question, and once you're down to the last three you can do it in one question.  Hence the total number of questions is $N-2$.&lt;/p&gt;&#xA;" OwnerUserId="2722" LastEditorDisplayName="user20" LastEditDate="2014-10-10T06:08:58.433" LastActivityDate="2014-10-10T06:08:58.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2647" PostTypeId="2" ParentId="2636" CreationDate="2014-10-08T12:01:02.940" Score="8" Body="&lt;p&gt;You didn't say they can't overlap. I put a crude example - 5 triangles. Recall that a square is a &lt;em&gt;rectangle&lt;/em&gt; whose sides are of equal length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DgS9u.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think outside the equilateral triangle warspy&lt;/p&gt;&#xA;" OwnerUserId="2484" LastActivityDate="2014-10-08T12:01:02.940" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2652" PostTypeId="1" AcceptedAnswerId="2659" CreationDate="2014-10-08T13:00:21.390" Score="8" ViewCount="4614" Body="&lt;p&gt;The sundial keeps time with the fewest number of moving parts.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which timepiece has the most moving parts?&lt;/p&gt;&#xA;" OwnerUserId="2444" LastEditorUserId="5373" LastEditDate="2016-10-24T00:35:37.617" LastActivityDate="2016-10-24T00:35:37.617" Title="Timepieces and Moving Parts" Tags="|riddle|time|" AnswerCount="6" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2653" PostTypeId="2" ParentId="2652" CreationDate="2014-10-08T13:02:40.630" Score="2" Body="&lt;p&gt;The universe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It keeps time as it contains all other known timepieces and it kind of defines what time is.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="5373" LastEditDate="2015-06-23T09:22:50.867" LastActivityDate="2015-06-23T09:22:50.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2654" PostTypeId="2" ParentId="2652" CreationDate="2014-10-08T13:02:57.943" Score="5" Body="&lt;p&gt;The system of stars (including the sun) which we have used since antiquity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or maybe the body clock: &lt;a href=&quot;http://en.wikipedia.org/wiki/Circadian_rhythm&quot;&gt;http://en.wikipedia.org/wiki/Circadian_rhythm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The myriad year clock perhaps. It can be wound once every year and over 1000 parts: &lt;a href=&quot;http://en.wikipedia.org/wiki/Myriad_year_clock&quot;&gt;http://en.wikipedia.org/wiki/Myriad_year_clock&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-10-08T13:15:31.720" LastActivityDate="2014-10-08T13:15:31.720" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2656" PostTypeId="2" ParentId="2652" CreationDate="2014-10-08T13:06:50.327" Score="-1" Body="&lt;p&gt;An atomic clock: if you count each atom as a time-keeping part, it has a gigantic number of these, all moving.&lt;/p&gt;&#xA;" OwnerUserId="1759" LastEditorUserId="5373" LastEditDate="2015-06-24T11:06:58.077" LastActivityDate="2015-06-24T11:06:58.077" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2657" PostTypeId="2" ParentId="2652" CreationDate="2014-10-08T13:11:21.787" Score="1" Body="&lt;p&gt;A &lt;a href=&quot;https://en.wikipedia.org/wiki/Floral_clock&quot; rel=&quot;nofollow&quot;&gt;floral clock&lt;/a&gt;.‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might have hundreds of flowers on its hands, all moving together.&lt;/p&gt;&#xA;" OwnerUserId="2518" LastEditorUserId="5373" LastEditDate="2015-06-23T09:22:12.297" LastActivityDate="2015-06-23T09:22:12.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2658" PostTypeId="1" AcceptedAnswerId="2660" CreationDate="2014-10-08T13:18:32.460" Score="1" ViewCount="1703" Body="&lt;p&gt;Sherlock had been invited to lunch by a rich man in London. He went to the man's house and found that the lunch was actually a huge party and there were a lot of people around.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Dreadful&lt;/em&gt;, he thought. &lt;em&gt;I hate large gatherings.&lt;/em&gt;&lt;br&gt;&#xA;&quot;I didn't know there would be so many... guests,&quot; he told the host.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Oh, these aren't my guests. They're my employees...&quot; the host replied, &quot;...all except for one of them. He's the only other guest here besides you. I'm not going to tell you who it is. Do you think you can figure out who it is?&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Definitely,&quot; Sherlock replied. &quot;Let's all head into the dining room, and you can address the group.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They did just that, and after everyone was seated, the host stood up, greeted everyone, and proceeded to tell a joke. It was a bad joke, though. A really bad one. It was quite possibly the worst that Sherlock had ever heard, and John had given him some real boring groaners in his time that were better. But by the time the host finished telling it, Sherlock had figured out who the other guest was, and pointed him out, grinning.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;That's amazing!&quot; the host exclaimed.&lt;br&gt;&#xA;&quot;Not really,&quot; Sherlock replied, and explained how he arrived at his conclusion.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did he figure out who the other guest was?&lt;/p&gt;&#xA;" OwnerUserId="2444" LastEditorUserId="5373" LastEditDate="2015-04-08T09:28:20.353" LastActivityDate="2018-03-27T17:28:42.107" Title="Sherlock and the Luncheon" Tags="|mystery|story|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2659" PostTypeId="2" ParentId="2652" CreationDate="2014-10-08T13:19:17.233" Score="12" Body="&lt;p&gt;Well, one with a lot of moving parts would be an hourglass. Each grain of salt is a 'moving part'. Just scale it up as big as is needed to trump other options.&lt;/p&gt;&#xA;" OwnerUserId="2156" LastEditorUserId="12700" LastEditDate="2015-07-28T11:12:41.837" LastActivityDate="2015-07-28T11:12:41.837" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="2660" PostTypeId="2" ParentId="2658" CreationDate="2014-10-08T13:20:55.633" Score="12" Body="&lt;p&gt;&lt;strong&gt;The guest was:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The one who didn't laugh. The others would feel forced to laugh at a joke their employer told, no matter how bad the joke.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2156" LastEditorUserId="-1" LastEditDate="2018-03-27T17:28:42.107" LastActivityDate="2018-03-27T17:28:42.107" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2665" PostTypeId="1" AcceptedAnswerId="2666" CreationDate="2014-10-08T13:39:19.003" Score="15" ViewCount="2707" Body="&lt;p&gt;You die and ascend to heaven, there is a knight (truth-teller), a knave (pure liar) and a joker (random) sitting on a cloud - they all look the same. In order to gain entry you must determine their identities. You have 3 yes/no questions (each directed to only one of them). They will only respond to questions speaking their own heavenly language's words for &quot;yes&quot; and &quot;no&quot; which you do not know - &quot;pluh&quot; and &quot;plit&quot; (unknown which means what).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you going to be allowed into heaven?&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="71" LastEditDate="2014-11-24T21:46:30.450" LastActivityDate="2021-06-10T12:44:16.267" Title="Knights and knaves in a foreign language" Tags="|logical-deduction|liars|" AnswerCount="2" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="2666" PostTypeId="2" ParentId="2665" CreationDate="2014-10-08T13:49:27.990" Score="18" Body="&lt;p&gt;This is so called The Hardest Logic Puzzle Ever. &lt;a href=&quot;http://en.wikipedia.org/wiki/The_Hardest_Logic_Puzzle_Ever&quot;&gt;Wikipedia&lt;/a&gt; has a thorough description of it and its solutions, including different versions of formulations of how the joker functions (which was not defined in the OP's question).&lt;br&gt;&#xA;I can cite the very basics here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formulation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Three gods A, B, and C are called, in no particular order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are da and ja, in some order. You do not know which word means which.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solution:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Q1: Ask god B, &quot;If I asked you 'Is A Random?', would you say ja?&quot;. If B answers ja, either B is Random (and is answering randomly), or B is not Random and the answer indicates that A is indeed Random. Either way, C is not Random. If B answers da, either B is Random (and is answering randomly), or B is not Random and the answer indicates that A is not Random. Either way, you know the identity of a god who is not Random.&lt;br&gt;&#xA; Q2: Go to the god who was identified as not being Random by the previous question (either A or C), and ask him: &quot;If I asked you 'Are you False?', would you say ja?&quot;. Since he is not Random, an answer of da indicates that he is True and an answer of ja indicates that he is False.&lt;br&gt;&#xA; Q3: Ask the same god the question: &quot;If I asked you 'Is B Random?', would you say ja?&quot;. If the answer is ja, B is Random; if the answer is da, the god you have not yet spoken to is Random. The remaining god can be identified by elimination.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="28" LastEditorUserId="68" LastEditDate="2014-10-08T18:39:57.250" LastActivityDate="2014-10-08T18:39:57.250" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2668" PostTypeId="1" AcceptedAnswerId="2671" CreationDate="2014-10-08T14:25:48.723" Score="6" ViewCount="1372" Body="&lt;p&gt;A floating ship contains one stone. The captain sees the stray stone and throws it into the water. What result will this action have on the water level? Will it increase or decrease?&lt;/p&gt;&#xA;" OwnerUserId="2749" LastActivityDate="2016-09-13T20:41:18.530" Title="Ship and stone puzzle" Tags="|logical-deduction|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2670" PostTypeId="2" ParentId="2636" CreationDate="2014-10-08T15:02:29.180" Score="6" Body="&lt;p&gt;Another way of interpreting the question treats it as a classic Dudeney problem about dissections, with solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TRDXZ.png&quot; alt=&quot;Dissection of square and equilateral triangle&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image from Wikimedia, released into the public domain by its creator.&lt;/p&gt;&#xA;" OwnerUserId="228" LastActivityDate="2014-10-08T15:02:29.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2671" PostTypeId="2" ParentId="2668" CreationDate="2014-10-08T15:15:32.417" Score="14" Body="&lt;p&gt;The water level will fall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for this is that the stone while in the boat is displacing water equal to its weight.  For example, say the stone weighs 1 kilogram.  It would displace one litre of water while in the boat because the boat is at equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you take the stone out of the boat and put it in the water, you are losing the 1 litre of displacement you had while the stone was in the boat, but gaining the volume of the stone in displacement.  Since it is a stone (and presumably heavier than water!) it has a higher density than water, so its displacement is less than 1 litre.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, you are now displacing less water, so the water level will drop.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2014-10-08T15:15:32.417" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2675" PostTypeId="1" AcceptedAnswerId="2684" CreationDate="2014-10-08T17:18:07.043" Score="3" ViewCount="235" Body="&lt;p&gt;What is the minimum number of touches for painting at least a $100*100$ rectangle if you have one circular stamp tool that paints a circular area of $1$ unit radius in every touch?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is very similar to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2605/paint-the-rectangle-with-least-movement&quot;&gt;Paint the rectangle with least movement&lt;/a&gt;. But this time there is no brush tool for painting.&lt;/p&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2014-10-09T02:13:17.167" Title="Paint the Rectangle with least number of Circular Stamp Touches" Tags="|geometry|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2676" PostTypeId="2" ParentId="2675" CreationDate="2014-10-08T18:25:16.297" Score="1" Body="&lt;p&gt;you might need w/2 x h/2 filled circles (even filled circles) on the first run and then need (w-1)/2 x (h-1)/2 filled circles (odd filled circles) to fill the gaps between... you can 'cut' pathes when you draw a &lt;em&gt;even&lt;/em&gt; line and then switch to an &lt;em&gt;odd&lt;/em&gt; line...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;lenght of even line = 100; length of odd line = 99; path to switch from odd to even (and back) = Math.Squrt(2) (i'll say it's 1.4 for terms of easier use)...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so you have 100 x 100 + 99 x 99 + 99*1.4 = 10'000 + 9801 + 138.6 = 19939.6 (having rounding erros)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YUtYt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;BLACK  = even circles&#xA;RED    = odd circles&#xA;PURPLE = rectangle&#xA;GREEN  = path&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;i &lt;strong&gt;can't proof&lt;/strong&gt; that this path would be shorter:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zA1be.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2330" LastEditorUserId="2225" LastEditDate="2014-10-09T02:13:17.167" LastActivityDate="2014-10-09T02:13:17.167" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2678" PostTypeId="2" ParentId="2583" CreationDate="2014-10-08T19:25:50.913" Score="3" Body="&lt;p&gt;There are 2 simple algorithms that will let you exit a maze (from Wikipedia)&lt;/p&gt;&#xA;&lt;h1&gt;Pledge algorithm&lt;/h1&gt;&#xA;&lt;p&gt;Named after after Jon Pledge of Exeter.  Works for a plane maze when you need to move out.  It requires perfect orientation.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Move north until you are blocked by a wall.&lt;/li&gt;&#xA;&lt;li&gt;Follow the wall touching it with your right hand.  Track exactly how much you turn left and right.&lt;/li&gt;&#xA;&lt;li&gt;As soon as you are facing north again and you have turned left as much as you have turned right, restart at 1. moving north.&lt;/li&gt;&#xA;&lt;li&gt;Stop when you see the exit.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;h1&gt;Trémaux's algorithm&lt;/h1&gt;&#xA;&lt;p&gt;By Charles Pierre Trémaux.  Works for any maze.&lt;/p&gt;&#xA;&lt;p&gt;Count the starting point as a junction.&lt;/p&gt;&#xA;&lt;p&gt;At every junction, mark the path where you arrive and the path where you leave.  To choose a path, just take the one with the less marks.  You are guaranteed to walk the whole maze before returning to the starting point and having tried all paths from there.&lt;/p&gt;&#xA;&lt;h1&gt;Minimal Time&lt;/h1&gt;&#xA;&lt;p&gt;Tremeaux's algorithm favors exploring new paths before backtracking.  And since the exit could be anywhere, the best way to find the exit fast is to explore as much and as soon as possible.  Therefore I would estimate that Tremeaux's algorithm is as good as most algorithm to find the exit fast.  Better algorithms would have to track precisely where you have been and eliminate surrounded regions without visiting them.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-10-08T19:25:50.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2679" PostTypeId="1" AcceptedAnswerId="2680" CreationDate="2014-10-08T19:41:59.590" Score="-4" ViewCount="18059" Body="&lt;p&gt;You are on a trail in a desert.  You already drank all your water, and are 2 miles from the next well.&lt;br&gt;&#xA;There is a crossroads in the trail. One path leads directly to the fountain, and one leads out into the middle of nowhere.  At the crossroad, there is an Indian.  He is from one of two tribes - one tribe &lt;em&gt;always&lt;/em&gt; tells the truth, and one tribe &lt;em&gt;always&lt;/em&gt; tells lies.  The two tribes have one thing in common: They will only answer &lt;em&gt;one&lt;/em&gt; question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What question must you ask to find out which trail to take?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;I'll post the answer in a few days if no one knows it.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="2660" LastEditorUserId="5373" LastEditDate="2015-06-23T09:24:27.403" LastActivityDate="2015-06-23T09:24:27.403" Title="Crossroad with an Indian from one of two tribes" Tags="|logical-deduction|liars|" AnswerCount="3" CommentCount="2" FavoriteCount="0" ClosedDate="2014-10-11T23:57:58.660" ContentLicense="CC BY-SA 3.0" />
  <row Id="2680" PostTypeId="2" ParentId="2679" CreationDate="2014-10-08T19:47:08.643" Score="4" Body="&lt;p&gt;Pretty vanilla knights and knaves riddle. You should ask&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;If I asked you which road leads to the fountain, which would you say?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assume &lt;strong&gt;A&lt;/strong&gt; is the correct road, and &lt;strong&gt;B&lt;/strong&gt; is the wrong road.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The truth-teller would normally answer &lt;strong&gt;A&lt;/strong&gt; if you asked which road is safe, so when you ask which road he'd say if you asked which road is safe, he'll still say &lt;strong&gt;A&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The liar would lie, and tell you that &lt;strong&gt;B&lt;/strong&gt; is the safe road, but since he has to lie to answer your question, he tells you that &lt;strong&gt;A&lt;/strong&gt; is the safe road.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may still die from thirst before you get there, of course.&lt;/p&gt;&#xA;" OwnerUserId="2518" LastActivityDate="2014-10-08T19:47:08.643" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2681" PostTypeId="2" ParentId="2679" CreationDate="2014-10-08T19:57:41.313" Score="1" Body="&lt;p&gt;&quot;If I asked the representative from the other tribe, &quot;Which road leads to the fountain?&quot;, which would HE say?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume A is the correct road, and B is the wrong road.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The truth-teller would normally answer A if asked about the road directly, but would know the liar would say 'B' as his response for the fountain road.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The liar would lie always, and say the truth teller would say B is the safe road.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such, you know the wrong direction is B from either Indian, and you should choose A.&lt;/p&gt;&#xA;" OwnerUserId="2265" LastActivityDate="2014-10-08T19:57:41.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2682" PostTypeId="1" AcceptedAnswerId="2692" CreationDate="2014-10-08T20:59:20.020" Score="12" ViewCount="80900" Body="&lt;p&gt;I got a 3D wooden puzzle in which you are supposed to rebuild. I took pictures of the 6 pieces (click images to enlarge):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CvhjV.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CvhjVs.jpg&quot; alt=&quot;Piece#1&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/RFvzN.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RFvzNs.jpg&quot; alt=&quot;Piece#2&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/nOLJh.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nOLJhs.jpg&quot; alt=&quot;Piece#3&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/UoPlB.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UoPlBs.jpg&quot; alt=&quot;Piece#4&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/VzGG4.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VzGG4s.jpg&quot; alt=&quot;Piece#5&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/kM04a.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kM04as.jpg&quot; alt=&quot;Piece#6&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this image on the Internet. I believe it's the solved version:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PTxsK.jpg&quot; alt=&quot;Solution&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you solve it? Has anybody here done it?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="69582" LastEditDate="2020-12-23T19:44:33.513" LastActivityDate="2020-12-23T19:44:33.513" Title="How to solve this 3D wooden puzzle which becomes an interlocking cross?" Tags="|mechanical-puzzles|solvability|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="2684" PostTypeId="2" ParentId="2675" CreationDate="2014-10-08T21:44:44.167" Score="3" Body="&lt;p&gt;If you array circles with radius = 1 on a triangular as shown below, you can cover a rectangular area with &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; height = (3n-1)/2, where n is the number of rows of circles, and&#xA;&#xA; width = sqrt(3)(2m-1)/2, if each row is of length m, or&#xA; width = sqrt(3)(m-1), if alternate rows are of length m, and m-1 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So you can cover a 100x100 square with 67 rows, of alternating length 59,58,59...(actual rectangular area covered is 100 high by ~100.459 wide).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total number of circles used is 34x59 + 33*58 = 3920&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DDo44.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1631" LastActivityDate="2014-10-08T21:44:44.167" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2685" PostTypeId="2" ParentId="2602" CreationDate="2014-10-08T21:49:12.750" Score="15" Body="&lt;p&gt;This is how I see it. The father being a knowledgeable man, knows that he has three pregnant camels. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So at the time of his death he says that the oldest would receive one in two, or one camel, a pregnant camel, which implies a third camel carried by the mother camel. Thus the oldest receives three camels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same goes for the middle who receives one in three camels. He will get two camels who are not pregnant and one that is. Thus, with time, he will receive four. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally the youngest will receive eight not pregnant camels, and one that is. Thus he will receive ten camels.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3+4+10=17.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point would be that all three sons will have at least one fertile camel (the other camels may be infertile). The old friend is brought in to distinguish each pregnant camel and to allot one to each son. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;No extra camels.&lt;/p&gt;&#xA;" OwnerUserId="2763" LastActivityDate="2014-10-08T21:49:12.750" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2686" PostTypeId="1" AcceptedAnswerId="2687" CreationDate="2014-10-08T23:09:09.663" Score="2" ViewCount="595" Body="&lt;p&gt;You are an intrepid explorer who has been caught in a trap by a primitive tribe. They lock you in a room with no exit but one door. The door has a pressure sensitive release - incorrect pressure ($&amp;gt;0g$) will cause immediate death.  There are 10 piles each of 20 heavy stone discs (200 in total). You know what each disc should weigh (I can tell you that the number is in whole grams).       &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have some facts: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You know that one of the piles is made of discs that are missing 15.5 grams of the normal weight of a disc. &lt;/li&gt;&#xA;&lt;li&gt;You also know that another distinct pile is made of discs that are missing 6.5 grams of the normal weight. &lt;/li&gt;&#xA;&lt;li&gt;You also know that all the discs in all the other piles weigh exactly the right amount. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You have weighing scale (non-balancing) that is accurate to the gram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tribe leader has set the rules. You must determine which piles are normal, which pile is off by 15.5 grams and which is off by 6.5 grams using the scale only &lt;em&gt;once&lt;/em&gt;. Only by putting one disc from each of the 2 piles which are not normal on the pressure pad to release the door will you survive.&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="8874" LastEditDate="2015-01-31T12:28:32.620" LastActivityDate="2015-01-31T12:28:32.620" Title="200 Discs - weighing problem" Tags="|logical-deduction|weighing|" AnswerCount="1" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="2687" PostTypeId="2" ParentId="2686" CreationDate="2014-10-09T00:00:53.150" Score="7" Body="&lt;p&gt;Since we know what each disc should weigh ($n$), we can weigh 1 disc from pile 1, 2 from pile 2, etc (10+9+8... = $55n$). However, since some weights are half-grams, we'll need to double up instead (20+18+16... = $110n$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subtract that from the actual weight we read on the scales, then work out how to make the remainder from 13s and 31s. There is no multiple of 13 and 31 which can cause any problems with the math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will tell us which piles the abnormal discs came from, so finally just take 1 from each of those piles to put on to the pressure plate.&lt;/p&gt;&#xA;" OwnerUserId="2518" LastEditorUserId="2518" LastEditDate="2014-10-09T08:14:56.233" LastActivityDate="2014-10-09T08:14:56.233" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2688" PostTypeId="1" CreationDate="2014-10-09T00:25:29.903" Score="-2" ViewCount="1275" Body="&lt;p&gt;You are in a room with a barrier in the middle. This is not a physical barrier; it is a dividing line of 2 mirror images. When you look across the room, it appears to be a mirror because you see a mirror image of yourself and other objects in the room, but the mirror is not made of any substance. It is simply a point where anything past it is an exact mirror of all things on the other side. For example, if you were to walk forwards, you would not be able to cross the dividing line because you would walk into yourself walking forwards. If you were to try to punch your mirror image, you would punch it in the fist while they simultaneously punch you in the exact same spot, probably resulting in a broken hand for both you and your mirror image. If you were to try to push any object through the mirror line, it would collide with its mirror object and would not be able to pass. Is it possible for you or any other physical matter to cross the mirror line (without introducing &quot;magical&quot; phenomenons such as teleportation)?&lt;/p&gt;&#xA;" OwnerUserId="2770" LastEditorUserId="2770" LastEditDate="2014-10-09T16:06:54.690" LastActivityDate="2014-12-10T05:09:04.307" Title="Mirror Room Puzzle (Thought Experiment)" Tags="|paradox|" AnswerCount="6" CommentCount="13" ClosedDate="2014-12-10T15:55:13.243" ContentLicense="CC BY-SA 3.0" />
  <row Id="2689" PostTypeId="2" ParentId="2688" CreationDate="2014-10-09T00:45:48.657" Score="2" Body="&lt;p&gt;Attempt #1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, a liquid or gas. It would collide with, (and pass through) it's mirror image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attempt #2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light. It would that count?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attempt #3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another mirror, it will create infinite energy back and forth causing the mirrors to break, the walk by.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-10-09T17:39:46.437" LastActivityDate="2014-10-09T17:39:46.437" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2690" PostTypeId="1" AcceptedAnswerId="2691" CreationDate="2014-10-09T01:51:01.873" Score="4" ViewCount="591" Body="&lt;p&gt;This is taken from a riddle book:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Val, Lynn and Chris are related to each other, but no incestuously. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Among the three are Val's father, Lynn's only daughter, and Chris' sibling. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Chris' sibling is neither Val's father nor Lynn's daughter. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Which one is a different sex? &lt;/p&gt;&#xA;" OwnerUserId="2254" LastActivityDate="2014-10-09T02:02:57.410" Title="Val, Lynn Chris" Tags="|logical-deduction|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2691" PostTypeId="2" ParentId="2690" CreationDate="2014-10-09T02:02:57.410" Score="5" Body="&lt;p&gt;Chris is of a different sex to the other two.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Lynn is a male, and the father of the other two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chris is a female, and Lynn's only daughter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Val is a male. He's Lynn's son, and Chris's sibling. (He is not his own father, and being male nor is he Lynn's daughter.)&lt;/p&gt;&#xA;" OwnerUserId="2729" LastActivityDate="2014-10-09T02:02:57.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2692" PostTypeId="2" ParentId="2682" CreationDate="2014-10-09T04:11:10.027" Score="19" Body="&lt;p&gt;This is one of the most well-known wooden interlocking &quot;Burr&quot; puzzles, called The Chinese Cross. &lt;br /&gt;&lt;br /&gt;&#xA;If you look at the two images below, you can see pieces that are the same shapes as yours, but with each one a different color. Using the colors as a guide, you can see how they fit together in the assembled version. &lt;br /&gt;&lt;br /&gt;&#xA;The photos are from Rob Stegmann's page on interlocking puzzles, which has a great deal more information on this puzzle and many like it: &lt;a href=&quot;http://robspuzzlepage.com/interlocking.htm&quot; rel=&quot;noreferrer&quot;&gt;Rob's Puzzle Page&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also be interested in the Burr Tools program, created by Andreas Röver. It can be used to analyze and solve a huge number of different types of burr puzzles, and can be found at the &lt;a href=&quot;http://burrtools.sourceforge.net/&quot; rel=&quot;noreferrer&quot;&gt;BurrTools Sourceforge&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/k03Lc.jpg&quot; alt=&quot;Chinese Cross pieces&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/LqoIu.jpg&quot; alt=&quot;Chinese Cross assembled&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&#xA;Here is a set of steps for assembly. The numbered pieces correspond to the colored pieces in the photo with the following mapping: &lt;br /&gt;&#xA;4 - Red &lt;br /&gt;&#xA;8 - Yellow &lt;br /&gt;&#xA;13 - Green &lt;br /&gt;&#xA;12 - Blue &lt;br /&gt;&#xA;7 - Pink/Purple &lt;br /&gt;&#xA;1 - Orange &lt;br /&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/i07Xk.jpg&quot; alt=&quot;Solution Steps&quot;&gt;&#xA;&lt;br /&gt; &#xA;You can also see a solution video here: &lt;a href=&quot;https://www.youtube.com/watch?v=8rclzr24nS4&quot; rel=&quot;noreferrer&quot;&gt;Solution Video&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="447" LastEditorUserId="447" LastEditDate="2014-10-09T22:58:54.350" LastActivityDate="2014-10-09T22:58:54.350" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2693" PostTypeId="1" AcceptedAnswerId="2838" CreationDate="2014-10-09T04:17:38.607" Score="45" ViewCount="9301" Body="&lt;p&gt;Mama snake wants to knit a blanket for little baby snake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;She is not a dissipater and wants to make the &lt;strong&gt;blanket of a minimal size&lt;/strong&gt; (area). But her baby snake is quite a &lt;strong&gt;lively baby&lt;/strong&gt; and it always twists and stretches and turns around while sleeping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the baby changes position or from, mama takes the blanket and puts it over the baby again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What size (and form) is the blanket ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The baby snake has the length of 1&lt;/li&gt;&#xA;&lt;li&gt;While sleeping the baby lays on the bed (it's a 2D-puzzle, no 3D)&lt;/li&gt;&#xA;&lt;li&gt;The baby snake can take any shape it wants&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;it can have bends, even sharp bends like 120°&lt;/li&gt;&#xA;&lt;li&gt;it could also be shaped like a wave, a ring or a U-form&lt;/li&gt;&#xA;&lt;li&gt;it could be &lt;em&gt;any shape&lt;/em&gt; you can imagine (even a spiral or square)&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;The blanket &lt;strong&gt;must always cover&lt;/strong&gt; any shape of the snake&lt;/li&gt;&#xA;&lt;li&gt;Mama turns and moves the blanket if necessary&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kFR6r.jpg&quot; alt=&quot;mama snake&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/yxRBb.jpg&quot; alt=&quot;baby snake&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mama snake and her little baby snake&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Examples&lt;/h2&gt;&#xA;&#xA;&lt;h3&gt;Circle&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;This is biggest solution (from &lt;a href=&quot;https://puzzling.stackexchange.com/users/2406/miniman&quot;&gt;Miniman&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hqz8T.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a circle with diameter 1 (you can see the red baby snake stretched out totally - and you can also see the baby snake performing a circle).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Semicircle small&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;A smaller version would be a blanket of this size and shape (a semicircle).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Cik5D.png&quot; alt=&quot;semicircle&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see how the snake performs a bending like a U (red outline) and the resulting size of the blanket (black line). Unluckily this blanket would be too small, because if the snake bends into a 90°L-form it wouldn't fit into the blanket anymore.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Semicircle big&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ddTOe.png&quot; alt=&quot;big_semicircle&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That great idea comes from &lt;a href=&quot;https://puzzling.stackexchange.com/users/2518/joe&quot;&gt;Joe&lt;/a&gt;. Thanks for providing that idea! a semicircle with the diameter of 1... This definitely covers the baby snake but as you can see could easily be cut of on several parts...&lt;/p&gt;&#xA;" OwnerUserId="2330" LastEditorUserId="4219" LastEditDate="2019-06-28T19:26:41.017" LastActivityDate="2019-06-28T19:26:41.017" Title="A blanket for my baby snake" Tags="|mathematics|geometry|optimization|" AnswerCount="17" CommentCount="26" ContentLicense="CC BY-SA 3.0" />
  <row Id="2694" PostTypeId="2" ParentId="2693" CreationDate="2014-10-09T04:34:47.143" Score="1" Body="&lt;p&gt;The blanket is a circle with a diameter of 1. There is no position the baby snake can take which this blanket cannot cover. (Unless the snake's width is greater than 1; this wasn't specified, but I'm hoping it's a safe assumption.)&lt;/p&gt;&#xA;" OwnerUserId="2406" LastActivityDate="2014-10-09T04:34:47.143" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2695" PostTypeId="2" ParentId="2693" CreationDate="2014-10-09T08:18:31.947" Score="5" Body="&lt;p&gt;I'm not math-y enough to even begin to go about proving this, but just based on logic I believe you can just cut the circle in half to give you a semicircle with a diameter of $1$ unit, which would have an area of $\tfrac{1}{8} \pi$ (a circle with a &lt;em&gt;diameter&lt;/em&gt; of $1$ has an area of ${\pi \cdot\tfrac{1}{2}}^2 = \tfrac{1}{4}\pi$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will fit the longest possible layout along the diameter, and I don't think it should be possible for the snake to lie in such a way as to break out of the semicircle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone can provide any kind of proof (or disproof), feel free to edit my post.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here's an illustration of the blanket, provided by @MartinFrank. The red lines indicate possible snake positions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ddTOe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2518" LastEditorUserId="68" LastEditDate="2014-10-09T11:34:55.500" LastActivityDate="2014-10-09T11:34:55.500" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2696" PostTypeId="2" ParentId="2688" CreationDate="2014-10-09T09:59:44.263" Score="11" Body="&lt;p&gt;Basically the mirror line is a physical equivalent of an ideal wall, which can react to any force you apply to it with the same, but opposite force.&lt;br&gt;&#xA;A physical object can not cross a wall, especially an ideal wall. So my answer is &lt;strong&gt;no&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2014-10-09T18:02:49.127" LastActivityDate="2014-10-09T18:02:49.127" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2708" PostTypeId="2" ParentId="2693" CreationDate="2014-10-09T11:00:00.197" Score="1" Body="&lt;p&gt;I think the best solution is a 1*1/2 rectangular blanket. If the baby is stretched out, you can cover him entirely. if he's in a non-stretched out pose, there cannot be more than 1 part sticking out further than 1/2 from the rest of the body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: to reply to the remark: instead of a rectangle with size 1*1/2, use a rectangle with diagonal length 1, height 0.894 and width 0.447.&lt;/p&gt;&#xA;" OwnerUserId="2778" LastEditorUserId="2778" LastEditDate="2014-10-09T11:42:05.350" LastActivityDate="2014-10-09T11:42:05.350" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="2709" PostTypeId="1" AcceptedAnswerId="2710" CreationDate="2014-10-09T11:16:18.113" Score="8" ViewCount="12254" Body="&lt;p&gt;What is the meaning of the following characters?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;13579 vs U&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="2730" LastEditDate="2015-04-20T12:36:56.827" LastActivityDate="2017-05-12T21:00:22.393" Title="What is the meaning of &quot;13579 vs U&quot;?" Tags="|rebus|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2710" PostTypeId="2" ParentId="2709" CreationDate="2014-10-09T11:19:59.583" Score="17" Body="&lt;p&gt;Is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;you against all odds&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&#xA;1,3,5,7,9&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; all odd numbers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;U&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2780" LastEditorUserId="35972" LastEditDate="2017-05-12T21:00:22.393" LastActivityDate="2017-05-12T21:00:22.393" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2711" PostTypeId="2" ParentId="2693" CreationDate="2014-10-09T11:25:23.970" Score="15" Body="&lt;p&gt;What about a rectangular blanket that is 1 x circumference of baby snake?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This rectangular blanket could be wrapped around the baby snake (think tube sock) and stretch and fold in any direction with him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The big bonus here is that mama snake doesn't have to tirelessly keep replacing the blanket, the blanket will go wherever baby snake goes. :)&lt;/p&gt;&#xA;" OwnerUserId="2780" LastActivityDate="2014-10-09T11:25:23.970" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2712" PostTypeId="1" AcceptedAnswerId="2714" CreationDate="2014-10-09T11:28:33.080" Score="15" ViewCount="25748" Body="&lt;p&gt;Take a look at these letters and tell me what it means.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;T_RN&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's a bit of a brainteaser. So think outside of the box to get the answer.&lt;/p&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="25966" LastEditDate="2020-05-04T16:57:28.247" LastActivityDate="2021-01-25T18:02:59.773" Title="What does &quot;T_RN&quot; mean?" Tags="|rebus|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="2713" PostTypeId="2" ParentId="2712" CreationDate="2014-10-09T11:31:46.320" Score="2" Body="&lt;p&gt;&quot;Torn without you&quot;. As though to expressing longing for a loved one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OR similarly, and much more loosely &quot;heart torn out&quot;/&quot;to have one's heart torn out&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2484" LastActivityDate="2014-10-09T11:31:46.320" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2714" PostTypeId="2" ParentId="2712" CreationDate="2014-10-09T11:34:01.677" Score="24" Body="&lt;p&gt;It means&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No U-Turn - as in the road sign&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1801" LastEditorUserId="25966" LastEditDate="2020-05-04T16:57:43.970" LastActivityDate="2020-05-04T16:57:43.970" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="2715" PostTypeId="1" AcceptedAnswerId="2716" CreationDate="2014-10-09T11:40:20.927" Score="7" ViewCount="15517" Body="&lt;p&gt;Can anyone tell me what this secret message is?&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;EVARELTO&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Try finding the secret message represented by these letters.&lt;/p&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="69582" LastEditDate="2021-10-19T04:39:11.030" LastActivityDate="2021-10-19T04:39:11.030" Title="&quot;EVARELTO&quot; - What does this mean?" Tags="|wordplay|rebus|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="2716" PostTypeId="2" ParentId="2715" CreationDate="2014-10-09T11:49:03.797" Score="16" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;&amp;quot;Broken Elevator&amp;quot; or &amp;quot;Elevator out of order&amp;quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The letters from the word &amp;quot;Elevator&amp;quot; are broken up and mixed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="73193" LastEditDate="2021-10-19T00:44:41.273" LastActivityDate="2021-10-19T00:44:41.273" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="2718" PostTypeId="1" AcceptedAnswerId="2719" CreationDate="2014-10-09T12:10:17.440" Score="7" ViewCount="18855" Body="&lt;p&gt;A secret message is hidden in the below letters&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DDDWESTDDD&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tell me what the message is if you can find it out.&lt;/p&gt;&#xA;" OwnerUserId="2730" LastEditorDisplayName="user88" LastEditDate="2014-10-10T17:43:46.000" LastActivityDate="2018-02-06T21:16:05.197" Title="DDDWESTDDD - Find this out?" Tags="|rebus|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2719" PostTypeId="2" ParentId="2718" CreationDate="2014-10-09T12:14:36.397" Score="16" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; West Indies (&quot;West&quot; in D's...).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2156" LastEditorUserId="-1" LastEditDate="2018-02-06T21:16:05.197" LastActivityDate="2018-02-06T21:16:05.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2720" PostTypeId="2" ParentId="2693" CreationDate="2014-10-09T12:16:13.503" Score="-2" Body="&lt;p&gt;well i'm not sure if i really have a valid answer, but so least blanket i could think of is shape a little weird...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SN0RO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you can put into that blanket &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the streched snake, &lt;/li&gt;&#xA;&lt;li&gt;the 90°L-baby snake, &lt;/li&gt;&#xA;&lt;li&gt;the U-Shaped baby snake, &lt;/li&gt;&#xA;&lt;li&gt;a snake shaped like a [___] (dunno how to call that, is bend on tail and head)... &lt;/li&gt;&#xA;&lt;li&gt;a S-shaped snake would also fit in...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;im not sure if I could bring in a Z-shaped baby snake if it had 2x 60° bends...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but what's worse: What is the size of this shape? I can't tell and therefor can't compare with others :-/ ... &lt;em&gt;frustation&lt;/em&gt; ... if anyone knows sie size fo such an area let me know...&lt;/p&gt;&#xA;" OwnerUserId="2330" LastActivityDate="2014-10-09T12:16:13.503" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2721" PostTypeId="1" AcceptedAnswerId="2722" CreationDate="2014-10-09T12:22:40.303" Score="0" ViewCount="1270" Body="&lt;p&gt;One day a treasure hunter was searching for treasure on an abandoned island. He got tired of looking everywhere, so he walked towards a big tree that he saw in order to rest in its shade. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As he approached the tree, he saw a board hanging on the tree. This was written on the board:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;PETS A&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He understood that it was a clue and he found the treasure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now where was the treasure, and how did he know where to look?&lt;/p&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="39" LastEditDate="2014-10-10T00:13:37.420" LastActivityDate="2014-10-10T00:22:57.270" Title="Clue for the Treasure Hunter" Tags="|logical-deduction|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2722" PostTypeId="2" ParentId="2721" CreationDate="2014-10-09T12:26:31.400" Score="5" Body="&lt;p&gt;The treasure was buried one step backwards from the sign.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sign was a riddle. &quot;PETS A&quot; is &quot;A STEP&quot; backwards, and so the sign means &quot;a step backwards&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="2101" LastEditDate="2014-10-10T00:22:57.270" LastActivityDate="2014-10-10T00:22:57.270" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="2723" PostTypeId="1" AcceptedAnswerId="2724" CreationDate="2014-10-09T12:29:58.307" Score="4" ViewCount="866" Body="&lt;p&gt;A man works at an aquarium. Every day, he has to spend way too much time telling people to stop tapping on the glass at the shark tank. Finally, he gets fed up with it and comes up with a solution.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thankfully, the solution works perfectly, because the next day, nobody tapped on the glass. At the end of his shift, though, he was fired for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What did the man do?&lt;/p&gt;&#xA;" OwnerUserId="2444" LastEditorUserId="11244" LastEditDate="2015-06-17T13:25:39.233" LastActivityDate="2015-06-17T13:25:39.233" Title="Hey. Don't tap on the glass" Tags="|lateral-thinking|situation|" AnswerCount="3" CommentCount="3" ClosedDate="2016-09-09T08:43:06.343" ContentLicense="CC BY-SA 3.0" />
  <row Id="2724" PostTypeId="2" ParentId="2723" CreationDate="2014-10-09T12:32:57.280" Score="9" Body="&lt;p&gt;He put a decal of cracked glass on it, so it looks like it's just about to break.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He could also have &lt;strong&gt;actually&lt;/strong&gt; cracked it, but let's just assume he's grumpy and not downright insane :-)&lt;/p&gt;&#xA;" OwnerUserId="2518" LastActivityDate="2014-10-09T12:32:57.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2725" PostTypeId="2" ParentId="236" CreationDate="2014-10-09T12:35:27.087" Score="11" Body="&lt;p&gt;The Guru's information makes the blue-eyed-people special. It is a bit easier to understand if you imagine the Guru saying &quot;those with blue eyes may go&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then on day 1, you see nobody leaving, so you know nobody knows his own eyecolor, so you can conclude that at least 2 persons must have blue eyes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then on day 2, you see nobody leaving, so you know nobody knows his own eyecolor, so you can conclude that at least 3 persons must have blue eyes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&#xA;Then on day 99, you see nobody leaving, so you know nobody knows his own eyecolor, so you can conclude that at least 100 persons must have blue eyes. But if you have blue eyes and you see that there are only 99 other blue-eyed persons, you know you are the lucky #100. So you'll leave at day 100.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Guru wasn't necessary, the people with brown eyes could also leave the island sooner or later. But there is no way for them to assure they don't have red eyes, or any other color. If only two colors existed, they could all go if the Guru only said which color should leave first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, the info given by the Guru is NOT &quot;there is someone here with blue eyes&quot;. Everybody knows that already, since everybody sees two blue-eyed persons and everybody knows those two can see each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also NOT &quot;everybody knows that there is someone here with blue eyes&quot;. It actually is &quot;everybody knows, that everybody knows, that everybody knows, ... [repeat 99 times] that someone has blue eyes&quot;. &lt;/p&gt;&#xA;" OwnerUserId="2783" LastActivityDate="2014-10-09T12:35:27.087" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2726" PostTypeId="1" AcceptedAnswerId="2728" CreationDate="2014-10-09T12:45:19.047" Score="3" ViewCount="3193" Body="&lt;p&gt;A man was murdered in his office bathroom. Detective Troby found a paper with alphabets written as below from the pocket of the dead man's corpse.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; ____________________________&#xA;|                            |&#xA;|  abcdefghijkmnopqrstuvwxyz |&#xA;|                            |&#xA; ----------------------------&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Troby inspected the paper and arrested the murderer from the office within minutes. How?&lt;/p&gt;&#xA;" OwnerUserId="2730" LastActivityDate="2014-12-02T01:47:24.887" Title="Who is the killer" Tags="|logical-deduction|" AnswerCount="1" CommentCount="3" ClosedDate="2014-12-04T20:03:04.060" ContentLicense="CC BY-SA 3.0" />
  <row Id="2727" PostTypeId="1" AcceptedAnswerId="2730" CreationDate="2014-10-09T12:45:44.093" Score="16" ViewCount="2508" Body="&lt;p&gt;One day, I was waiting to get on the subway. I was eating the apple I always have for breakfast, and I decided to sit next to a homeless man on the bench. We saw a rather... rotund man walk by us, to which the hobo muttered, &quot;Pig.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn't give it a second thought, except how rude the comment was. Another man walked past, this time tall and in a business suit. As he walked past, the hobo said &quot;Human.&quot; and I thought nothing of this remark, other than how obviously he was, in fact, human. I soon after boarded the train for work.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next day, the hobo was still there. Today, as I watched from a distance, he was at it again. Several people walked past him: a skinny woman, a muscular man, and an old lady, to which he respectively muttered &quot;Soup&quot;, &quot;Pork&quot;, and &quot;Cookie&quot;. Odd, I thought, considering none of them was a cookie. Throughout my day, I couldn't stop thinking about this weird hobo in the subway.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over the next few days, I kept observing him. He continued this odd behavior. &quot;Bread&quot;, &quot;Carrot&quot;, &quot;Milk&quot;, &quot;Rabbit&quot;, and other utterances as people walked past. So the next day, I walked by the homeless man again and he muttered, &quot;Apple.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I finally recognized what this hobo was doing... and after thinking on it for a little bit, remembered something and was stricken with horror.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What was this hobo's ability? What had me so disturbed?&lt;/p&gt;&#xA;" OwnerUserId="2444" LastActivityDate="2015-12-02T16:50:11.893" Title="The Strangely Clairvoyant Hobo" Tags="|situation|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2728" PostTypeId="2" ParentId="2726" CreationDate="2014-10-09T12:47:00.930" Score="15" Body="&lt;p&gt;He had a colleague named &quot;Noel&quot;, The only letter missing is &quot;L&quot; - No &quot;L&quot;. Noel.&lt;/p&gt;&#xA;" OwnerUserId="2484" LastActivityDate="2014-10-09T12:47:00.930" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2729" PostTypeId="2" ParentId="2727" CreationDate="2014-10-09T12:49:18.090" Score="4" Body="&lt;p&gt;He knew what you had for breakfast. He rummages through the rubbish bins and has become acquainted with everybody's eating habit.&lt;/p&gt;&#xA;" OwnerUserId="2484" LastActivityDate="2014-10-09T12:49:18.090" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2730" PostTypeId="2" ParentId="2727" CreationDate="2014-10-09T12:49:39.043" Score="22" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He was listing what people had eaten (at some point in time) and had said &quot;human&quot; at one point.  That meant on the first day he had seen a cannibal. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="99" LastEditorUserId="15921" LastEditDate="2015-12-02T16:50:11.893" LastActivityDate="2015-12-02T16:50:11.893" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2731" PostTypeId="1" AcceptedAnswerId="2769" CreationDate="2014-10-09T14:08:43.907" Score="10" ViewCount="1038" Body="&lt;p&gt;The image below carry a hidden message. The image alone is not enough, you'll need to use the hints to uncover the hidden message. All the information needed to solve the puzzle is already given (as in no outside source needed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Dei3n.png&quot; alt=&quot;image of the cards&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;On the first day, the first sacrifice follow his family here while the others come alone.&#xA;On the second day, the first sacrifice leads his family here while the other come alone.&#xA;On the third day, the second sacrifice leads his family here while the others come alone.&#xA;On the fourth day, the second sacrifice leads his family here while the others come alone.&#xA;By sacrificing the heads of all that comes, can one reach the truth.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final answer will be in text/ words.&lt;/p&gt;&#xA;" OwnerUserId="2373" LastEditorUserId="4385" LastEditDate="2015-12-15T15:53:50.677" LastActivityDate="2015-12-15T15:53:50.677" Title="Hidden Message: Hidden in The Cards" Tags="|cipher|cards|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2733" PostTypeId="2" ParentId="2693" CreationDate="2014-10-09T15:17:14.660" Score="3" Body="&lt;p&gt;Looking at Joe's answer, half the semicircle seems to be unnecessarily big.  Instead, you could use a smaller area triangle of the same height and width.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to take a semicircle of diameter 1 and split it in half, replacing one side with a triangle of height 0.5 and width 0.5, I think that would be the smallest area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Area: Half the semi-circle + triangle&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Circle area: pi * r^2 = pi * (1/2)^2 = pi/4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Half the semi-circle = pi/4 * 1/2 * 1/2 = pi/16&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Triangle area: length * height / 2 = 1/2 * 1/2 * 1/2 = 1/8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total area: = pi*/16 + 1/8 ~= 0.32&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2014-10-09T15:17:14.660" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2734" PostTypeId="2" ParentId="2723" CreationDate="2014-10-09T15:21:11.857" Score="0" Body="&lt;p&gt;He simply removed all of the glass himself.  No glass, not even a chance of someone tapping on it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, since all the animals in the tank died for it, he was fired.  &lt;/p&gt;&#xA;" OwnerUserId="2756" LastActivityDate="2014-10-09T15:21:11.857" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2735" PostTypeId="2" ParentId="2693" CreationDate="2014-10-09T15:23:34.107" Score="15" Body="&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; After some reflection and help/counter-examples seen in some other answers, I think this is either very close to, or is the correct answer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about half of a square with a diagonal of $1$, cut along the diagonal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mVRGb.jpg&quot; alt=&quot;Square Diagonal&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using one of the triangles as the blanket, the long side (i.e. the diagonal of the square) is of length $1$, and the other two sides are length $\sqrt{\frac{1}{2}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area of this is $\frac{(\sqrt{\frac{1}{2}})^2}{2}=\frac{1}{4}$, and I think all shapes of snakes would fit in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some common snake shapes that will fit:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A square of side length $\tfrac{1}{4}$&lt;/li&gt;&#xA;&lt;li&gt;A circle&lt;/li&gt;&#xA;&lt;li&gt;A &quot;U&quot; shape (easy to prove if it is symmetrical, and if it is not symmetrical, then it is covered by a smaller area) &#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;e.g. where all three sides are length $\frac{1}{3}$&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;An &quot;S&quot; shape - head goes left 90 degrees, tail goes right 90 degrees&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As pointed out by Mooing Duck, if you orient this triangle with its long edge down as follows:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/DSw3h.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then its height is $\tfrac{1}{2}$, but you don't need all that height.  The tallest shape the snake can make is when it forms 2 sides of an equilateral triangle.  So, if we crop off that portion of the triangle, we end up with a truncated triangle (looks similar to the half hexagon in DenDenDo's answer!).  It looks like follows:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/AuY0j.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its area is now $\tfrac{1}{4}$ minus the amount we just removed.  The height of an equilateral triangle of side length $\tfrac{1}{2}$ is $0.433$ ($\sqrt{\tfrac{1}{2}^2-\tfrac{1}{4}^2}$), so the triangle we just removed has height $0.067$. The area of that triangle is $0.0045$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the area of the remaining blanket is ~ $0.2455127...$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some more reflection, I wonder if this shape needs to be so symmetrical.  You can always flip the blanket over.  For example, the left side of the above blanket can cover a snake bent into 2 sides of an equilateral triangle, but the right side isn't used.  If you make a &quot;U&quot; shape from the snake with the sides of length $\frac{1}{3}$, then you only need a portion of the right side.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the following blanket still covers both shapes:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/iqejN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the upper right portion doesn't seem necessary.  Can it be simply removed?&#xA;Here is the shape again with some other snakes covered by it.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Rzi9r.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;NOTE:&lt;/em&gt;&lt;/strong&gt; Does the blue equilateral snake need to be oriented that direction?  If we orient it so that the missing third line is on the right, can we recover some of the blanket?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, the brown &quot;U&quot; shape could have the opening facing up, so we could recover some of the interior of the &quot;U&quot;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the green snake is $3$ segments each of length $\frac{1}{3}$.  If the middle segment is longer, then the overall shape isn't as tall.  If the middle segment is shorter, then it also begins to get shorter, so this one is the tallest of these types, so they will never use the space above the brown &quot;U&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="1592" LastEditDate="2014-10-10T17:40:08.480" LastActivityDate="2014-10-10T17:40:08.480" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="2737" PostTypeId="2" ParentId="2723" CreationDate="2014-10-09T15:28:32.083" Score="1" Body="&lt;p&gt;He removed the fish tank and cooked the shark - served it as a fish dish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All went well, untill the restaurant owner came in to pick up the cash from the day, only to notice his beloved shark tank and the tank are gone. The fact that is was cooked and eaten by the diners enraged the owner to the point where he fired the culprit.&lt;/p&gt;&#xA;" OwnerUserId="2156" LastActivityDate="2014-10-09T15:28:32.083" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2738" PostTypeId="2" ParentId="1776" CreationDate="2014-10-09T15:50:16.913" Score="4" Body="&lt;h1&gt;Each answer is correct (if you can justify it)&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I looked for solutions that use only products, additions and subtractions, without any magical constants.&lt;br&gt;&#xA;With the numbers&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a b&#xA;c d&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I look for possible solutions of the form &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x1*a + x2*b + x3*c + x4*d +&#xA;x5*a*a + x6*b*b + x7*c*c + x8*d*d + &#xA;x9*a*b + x10*a*c + x11*a*d +&#xA;x12*b*c + x13*b*d + x14*c*d&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Where &lt;code&gt;xN in [-1,0,1]&lt;/code&gt;&lt;br&gt;&#xA;There are a total of 3**14 = 4782969 possible combinations and &lt;em&gt;only&lt;/em&gt; 88 of those work for both of the first two examples. Out of those 88 there is at least one solution for each of possible answers:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;+a -b +d +bc + bd +cd =                41/31/51&#xA;-a -b +bb -cc -ab -ac +ad +bc +bd =    41/31/62&#xA;-a -b +c +bb +cc -ab +ad -bc +bd -cd = 41/31/48&#xA;-a +b +d +bb -ab +ad +bc -cd =         41/31/48&#xA;+a +b +d -dd +ab +cd =                 41/31/53&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;None of those stands out for being especially simple or symmetric, but c:48 has twice as many solutions as the others, which tips the scales slightly in its favor.&lt;/p&gt;&#xA;" OwnerUserId="2790" LastActivityDate="2014-10-09T15:50:16.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2743" PostTypeId="2" ParentId="2731" CreationDate="2014-10-09T16:57:31.723" Score="1" Body="&lt;p&gt;OK, communal puzzling time again :P I'm posting this as an answer for now, simply because there's too much stuff for a comment. This has been revised to follow moonRabbit's latest comments.&lt;/p&gt;&#xA;&lt;p&gt;First things first, the notation &lt;code&gt;(A|2-10|J|Q|K) (c|h|s|d)&lt;/code&gt; for the cards is wrong. It's &amp;quot;much simpler&amp;quot; than that.&lt;/p&gt;&#xA;&lt;p&gt;I'm now pretty convinced that the clues refer to a &amp;quot;day&amp;quot; being a row of cards, a &amp;quot;sacrifice&amp;quot; being a card, and a &amp;quot;family&amp;quot; being all matching suited cards on the board.&lt;/p&gt;&#xA;&lt;p&gt;Now taking the clues more literally, let's see how true they ring.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Day 1: 8 of Spades | 9 of Clubs | 2 of Diamonds&lt;/p&gt;&#xA;&lt;p&gt;Clue: The first sacrifice follow his family here while the others come alone&lt;/p&gt;&#xA;&lt;p&gt;Translation: the first card follows his suit, while the others do not&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Well, we can have a quick look at the board that does indeed match - the Eight of Spades follows from the Ace of Spades, which we have elsewhere on the board. Each of the other cards is followed by another suited card (10 of Clubs, 8 of Diamonds).&lt;/p&gt;&#xA;&lt;p&gt;Looking at the 3 other rows, the same holds true for them all.&lt;/p&gt;&#xA;&lt;p&gt;This now gives us the following &amp;quot;sacrifices&amp;quot;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Eight of Spades&lt;/p&gt;&#xA;&lt;p&gt;Three of Hearts&lt;/p&gt;&#xA;&lt;p&gt;Ace of Diamonds&lt;/p&gt;&#xA;&lt;p&gt;Ace of Spades&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Assuming this is correct, we now have to &amp;quot;sacrific[e] the heads of all that come&amp;quot; to find the answer. That's as far as I'm up to so far.&lt;/p&gt;&#xA;" OwnerUserId="2518" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-10-09T22:59:44.430" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2744" PostTypeId="2" ParentId="2731" CreationDate="2014-10-09T17:39:17.987" Score="0" Body="&lt;p&gt;Here is my interpretation.  Each row in the picture is a day.  So the first clue is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On the first day, the first sacrifice follow his family here while the others come alone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That row is: &lt;code&gt;8s 9c 2d&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming we are reading left to right, &lt;code&gt;8s&lt;/code&gt; cannot be the sacrifice since the sacrifice must &lt;code&gt;follow his family&lt;/code&gt;.  So say &lt;code&gt;9c&lt;/code&gt; is the sacrifice and &lt;code&gt;8s&lt;/code&gt; is his family and &lt;code&gt;2d&lt;/code&gt; is one of the others.  Alternatively, &lt;code&gt;2d&lt;/code&gt; could be the sacrifice, &lt;code&gt;9c&lt;/code&gt; his family, and &lt;code&gt;8s&lt;/code&gt; either one of the others, or part of the family.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next clue is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On the second day, the first sacrifice leads his family here while the other come alone. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That row is:&lt;code&gt;3h _ 8d&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, one interpretation would be that &lt;code&gt;3h&lt;/code&gt; is the sacrifice leading his family (none), and &lt;code&gt;8d&lt;/code&gt; is the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third clue:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On the third day, the second sacrifice leads his family here while the others come alone. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The row is: &lt;code&gt;Jh Ad 8h&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;code&gt;Jh&lt;/code&gt; must be the first sacrifice, and &lt;code&gt;Ad&lt;/code&gt; the second, while &lt;code&gt;8h&lt;/code&gt; is his family.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last clue is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On the fourth day, the second sacrifice leads his family here while the others come alone. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's row is: &lt;code&gt;5h As 10c&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the same as the previous clue, so the &lt;code&gt;5h&lt;/code&gt; is the first sacrifice, the &lt;code&gt;As&lt;/code&gt; is the second, and his family is the &lt;code&gt;10c&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final part of the hint reads:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;By sacrificing the heads all that comes, can one reach the truth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, the sacrifices are: &lt;code&gt;9c 3h Jh Ad 5h Ah&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we remove the heads of these (i.e. the number) in the original picture, you will have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;8s  c 2d&#xA; h  _ 8d&#xA; h  d 8h&#xA; h  s 10c&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;... and this is where I get stuck...&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="1592" LastEditDate="2014-10-09T18:09:14.037" LastActivityDate="2014-10-09T18:09:14.037" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2745" PostTypeId="2" ParentId="2688" CreationDate="2014-10-09T18:22:08.140" Score="6" Body="&lt;p&gt;A black hole (the two would collide and crash into each other making one singularity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An electron (doesn't travel linearly so it wouldn't necessarily collide with itself).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lots of different objects would be created from two mirrored particle accelerators colliding their payload.  These created objects need not be spawned with a twin (so they would bypass the mirror).&lt;/p&gt;&#xA;" OwnerUserId="457" LastActivityDate="2014-10-09T18:22:08.140" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2747" PostTypeId="1" AcceptedAnswerId="2755" CreationDate="2014-10-09T18:43:29.233" Score="2" ViewCount="928" Body="&lt;p&gt;I just came up across &lt;a href=&quot;https://math.stackexchange.com/questions/962958/100-sequential-parking-spaces&quot;&gt;this question in the mathematics stack exchange site&lt;/a&gt; and found it quite interesting:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;100 autonomous robotic vehicles enter a warehouse in random order to park. Inside the warehouse, there are 100 sequential parking spaces enumerated from 1 to 100. Each vehicle has an assigned number where it will attempt to park. However, there is an error in the programming such that if a vehicle finds its path to the assigned parking spot blocked by an already-parked robot, the robot will immediately park in the spot before it. For example, if vehicle 50 parks in spot 50 but vehicle 75 is immediately behind it, vehicle 75 will park in spot 49. Also, if vehicle 1 parks in spot 1, every robot behind it will be blocked from entering the warehouse at all. The vehicles do not have the ability to maneuver around already-parked vehicles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the most likely number of robots that will be parked in the warehouse at the end of the routine?&lt;/p&gt;&#xA;" OwnerUserId="2470" LastEditorUserId="-1" LastEditDate="2017-04-13T12:19:36.950" LastActivityDate="2014-11-12T10:51:33.990" Title="100 sequential parking spaces" Tags="|mathematics|probability|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2748" PostTypeId="2" ParentId="2693" CreationDate="2014-10-09T19:07:34.023" Score="4" Body="&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; &#xA;&lt;em&gt;As pointed out by Falco and Ivo Becker in the comments, there are some shapes that do not fit.  I will leave this answer here, however, since there is valuable info in the comments.  Also, thanks to Mooing Duck for the pictures - better than the one I put up.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have another thought.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if you started with a unit square blanket and took away the unit circle from the middle, and then took one quarter of that, you'd have kind of a right-angle triangle, but its long side would be caved in. Now say you mirror this triangle along one of its short edges, and it would be like a bigger right angle triangle, except its two short edges are caved in while its long edge is straight.  (Blanket is the red area)&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Wwg8O.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would all snake positions be covered by this blanket?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; &#xA;As pointed out by Falco and Ivo Becker in the comments, there are some shapes that do not fit.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="2321" LastEditDate="2014-10-10T17:09:06.680" LastActivityDate="2014-10-10T17:09:06.680" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="2749" PostTypeId="1" AcceptedAnswerId="2750" CreationDate="2014-10-09T19:29:31.820" Score="-2" ViewCount="669" Body="&lt;p&gt;You, a young adventurer, are travelling through the forest. Suddenly, your path splits in 2. You have no map or compass - well, you do but your compass is broken. You know you have to keep travelling North, so how can you tell which direction to go in? You do not have much time, you must leave now!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where are you going to go?!&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="5373" LastEditDate="2015-07-15T23:48:57.467" LastActivityDate="2015-07-16T12:58:24.647" Title="A Forked Path in the Forest" Tags="|lateral-thinking|story|situation|" AnswerCount="3" CommentCount="3" ClosedDate="2015-07-16T14:14:29.190" ContentLicense="CC BY-SA 3.0" />
  <row Id="2750" PostTypeId="2" ParentId="2749" CreationDate="2014-10-09T19:31:50.753" Score="3" Body="&lt;p&gt;If you &lt;em&gt;have to&lt;/em&gt; keep travelling North, you’re already going North and you can’t take a path that’s not North, so pick the one straight ahead.&lt;/p&gt;&#xA;" OwnerUserId="1565" LastActivityDate="2014-10-09T19:31:50.753" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2751" PostTypeId="2" ParentId="2747" CreationDate="2014-10-09T19:38:20.160" Score="1" Body="&lt;p&gt;I would guess around 18 cars. If we assume that average placement of a car in the empty lot is 1/2 way up, then we can make some assumptions as to where the cars are placed and how many are placed ahead of it. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First car is at 50. (1)&lt;/li&gt;&#xA;&lt;li&gt;Next car is at 25, but before that car parks, there is a 50% chance (repeating) that a car is at 50+ and so is placed right after the 50. That averages out to 1 extra car. (3)&lt;/li&gt;&#xA;&lt;li&gt;The next car is at 12, but before it parks this time there's a 75% chance of a car at 25+. That averages out to 3 extra cars. (7)&lt;/li&gt;&#xA;&lt;li&gt;Here's where it gets a little tricky. You have ~7/8 chance of the next car being above 12, but there are only 12 places left. You could do the math to get an average of 8 cars in this scenario, which puts you down to spot 4 (15)&lt;/li&gt;&#xA;&lt;li&gt;At that point, you have 3 spots left, but a really high chance of picking a car ahead of 3. So, we can just say they're fully filled (18)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2014-10-09T19:38:20.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2752" PostTypeId="1" AcceptedAnswerId="2754" CreationDate="2014-10-09T19:52:25.910" Score="-2" ViewCount="2878" Body="&lt;p&gt;You roll 5 dice and I'll always be able to give a number uniquely-defined by the numbers on the dice. You have to figure out how it works, without looking it up. Here's a couple example rolls:&lt;/p&gt;&#xA;&lt;p&gt;1,2,3,4,5 = 6&lt;/p&gt;&#xA;&lt;p&gt;3,6,2,3,1 = 4&lt;/p&gt;&#xA;&lt;p&gt;Please only answer if you figured it out yourself.&lt;/p&gt;&#xA;&lt;p&gt;Also note that the name is especially important.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The answer will always be even (0 is also even)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="69582" LastEditDate="2021-09-07T17:25:24.643" LastActivityDate="2021-09-07T17:25:24.643" Title="Petals around the rose" Tags="|pattern|dice|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="2753" PostTypeId="2" ParentId="48" CreationDate="2014-10-09T19:58:44.037" Score="10" Body="&lt;p&gt;Okay, so I actually made it this far (no photoshop or anything, just a secret trick and play endlessly without any hassle. It works for long time also on normal mode, until you make a mistake).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/K7S6G.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No more possibilities...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&#xA;So someone did ask me for the &quot;trick&quot;, so here it is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, always corner your highest tile. This is very important, as this gives you the maximum space to play the game. Since all the middle tiles are surrounded by 4 tiles, the corner tiles are the ones which have minimum contact with the other tiles. If your highest tile is somewhere in the middle, it only acts like an obstruction, so keeping it aside is the best move. Like in the above screenshot, 131072 is in the top left corner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ordering your tiles is also a good strategy. Like see how I have ordered my tiles in descending order. It creates minimum wastage of big tiles and less blockage for small tiles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, &quot;stick&quot; to a wall. This is the trickiest part. Like seriously, stick to the side of a chosen wall. I had chosen the top wall. Try not to disturb the tiles along the wall. For example for sticking to the top wall, &lt;strong&gt;do not swipe down&lt;/strong&gt;. It works since once your tiles are made, and as you progress, not only the biggest tile will obstruct, but also the tiles just before it also do it's job. Also sticking to a wall makes way for smaller tiles and they could be merged and made off into bigger tiles more easily. This ensures uniformity and a clean game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above, although a logically great option, does not work pretty well in normal mode. I was only able to get 8192 by it. Just ensure the chances of getting stuck in a case where the only option is to swipe down is minimum. Play accordingly in normal mode. Practice mode has no such hassle, so it's all good :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing to ensure is blocking. Never block a tile and make it useless. Remember, unusable spaces are the steps to defeat. Moves will always be available if none of your tiles are blocked. For example, if a 2 gets cornered, on the top right and you stick to the upper boundary, do not make your 2 just sit there. Merge it first and ensure this with other tiles. No tile should block your flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I personally prefer to use minimum number of undos, but sometimes when the only move is to go down, or when things go tight and in order to go further and create a tile, you do have to undo. That is the luck part which gets eliminated in practice mode. In normal mode just try to play perfectly and follow your strategy. Your score in normal mode is literally pot luck.&lt;/p&gt;&#xA;" OwnerUserId="2808" LastEditorUserId="2808" LastEditDate="2014-11-07T17:50:13.367" LastActivityDate="2014-11-07T17:50:13.367" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="2754" PostTypeId="2" ParentId="2752" CreationDate="2014-10-09T20:03:30.880" Score="4" Body="&lt;p&gt;Some further example rolls:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4, 1, 3, 5, 4: 6&lt;br&gt;&#xA; 4, 2, 4, 6, 1: 0&lt;br&gt;&#xA; 3, 2, 5, 3, 6: 8&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The name &quot;Petals Around the Rose&quot; is especially important.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The correct number is the number of &quot;petals&quot; (pips on the dice) that are around &quot;roses&quot; (pips in the center of the die). Even numbers have no center pip, therefore no petals. Odd numbers have pips in the center and thus have roses whose petals are counted. However, 1 has no non-rose pips. That leaves 3, with 2 other pips, and 5, with 4 other pips. In effect, each 3 rolled adds 2 to the count and each 5 adds 4 to the count. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2101" LastEditorUserId="2101" LastEditDate="2014-10-09T20:09:59.000" LastActivityDate="2014-10-09T20:09:59.000" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2755" PostTypeId="2" ParentId="2747" CreationDate="2014-10-09T20:22:00.450" Score="2" Body="&lt;p&gt;I decided to brute force the problem by &lt;a href=&quot;http://ideone.com/NGszgl&quot; rel=&quot;nofollow&quot;&gt;writing a quick program&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;using System;&#xA;using System.Collections.Generic;&#xA;using System.Linq;&#xA;&#xA;public class Test&#xA;{&#xA;    static readonly int NUMROBOTS = 100;&#xA;    static readonly int ITERATIONS = 1000000;&#xA;    static Random rng = new Random();&#xA;&#xA;    int FindParking()&#xA;    {&#xA;        int free = NUMROBOTS; // # of largest free spot&#xA;        int count = 0; // Number of robots parked&#xA;&#xA;        var robots = Enumerable.Range(1, NUMROBOTS).ToList();&#xA;        // Shuffle list&#xA;        int n = robots.Count;&#xA;        while (n &amp;gt; 1)&#xA;        {  &#xA;            n--;  &#xA;            int k = rng.Next(n + 1);  &#xA;            int value = robots[k];  &#xA;            robots[k] = robots[n];  &#xA;            robots[n] = value;  &#xA;        }&#xA;&#xA;        while (free &amp;gt; 1)&#xA;        {&#xA;            if (free &amp;gt; robots[count])&#xA;                free = robots[count]; // Robot found its spot&#xA;            else&#xA;                free--;               // Spot is filled&#xA;            count++;&#xA;        }&#xA;        return count;&#xA;    }&#xA;&#xA;&#xA;    public static void Main()&#xA;    {&#xA;        var p = new Test();&#xA;        int total = 0;&#xA;        for (int i = 0; i &amp;lt; ITERATIONS; i++)&#xA;        {&#xA;            total += p.FindParking();&#xA;        }&#xA;        Console.WriteLine(&quot;Average robots parked: {0} ({1} iterations)&quot;, (decimal)total / ITERATIONS, ITERATIONS);&#xA;    }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Average robots parked: 11.920253 (1000000 iterations)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to see the math for this, though.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditDate="2014-10-10T14:25:12.543" LastActivityDate="2014-10-10T14:25:12.543" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2759" PostTypeId="1" AcceptedAnswerId="2764" CreationDate="2014-10-09T20:57:55.177" Score="2" ViewCount="308" Body="&lt;p&gt;Each of the 50 robots move around in a room. The room is 8 by 8 tiles long. Each bot takes up 1 tile. On each turn they move like the knight in Chess, they don't teleport, they move. They all move at the same speed. When they knock into eachother they stop. They cannot move into a wall, they are Pre-programmed to avoid walls. They all start off in different, random positions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the most likely amount of turns it will take before they have all stopped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few rules not mentioned (and some mentioned)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;They start off in random positions.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Each turn they move like a Chess knight.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;They never bump into a wall.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;They never go backwards (they CAN visit the same tile twice however).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;They stop when they bump into eachother.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If they bump into eachother on the same tile, then they will stay on the tiles before that to stop, 2 bots may not exist on the 1 tile.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The board has 64 tiles (8x8).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are 50 bots.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Each bot moves at the same speed.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2014-10-10T19:06:04.990" Title="The Moving Bots" Tags="|logical-deduction|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2760" PostTypeId="2" ParentId="2693" CreationDate="2014-10-09T21:28:32.530" Score="6" Body="&lt;h1&gt;Half-Hexagon, &lt;code&gt;area &amp;lt;= sqrt(3)*3/16 = 0.3927&lt;/code&gt;&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I think the optimal shape is either a half hexagon or something slightly smaller.&lt;br&gt;&#xA;Assume the head of the snake is fixed and it extends to the right along the x-axis, otherwise we just rotate it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the snake bends, we can assume the first bend goes up, otherwise we mirror the snake (or flip the blanket over). In worst case it bends in the middle and once it bends sharper than an equilateral triangle, we can align the long side of the triangle on the x-axis and the snake is covered again. if there is only one bend, we can get away with an even smaller droplet-shaped area which is the envelope of several triangles&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the snake bends more often it will only get shorter, which makes it even easier to cover. Again we align the longest side of the bounding polygon on the x-axis and the snake is covered. It needs more area than the shape from before, but still slightly less than a full half-hexagon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof to this is pure intuition and playing around with different shapes in Inkscape  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KrGOh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: In the first picture the snake is simply straight.&lt;br&gt;&#xA;In the second it can bend once, so it goes half the distance to the right and then in any direction. On the left are the required blankets (partially hiding each other)  on the right they are rotated to fit in less then a half-circle.&lt;br&gt;&#xA;Same thing with three segments of length 1/3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the hexagon I mean, with all helper circles and shadings removed&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/q2jOD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2790" LastEditorUserId="2790" LastEditDate="2014-10-11T09:29:16.557" LastActivityDate="2014-10-11T09:29:16.557" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2764" PostTypeId="2" ParentId="2759" CreationDate="2014-10-09T23:19:20.023" Score="5" Body="&lt;p&gt;The most likely number of turns is 1, in my interpretation of the rules at least.  It's a crowded board, but 1 and 2 are neck-and-neck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the distribution from a simulation with ten million runs:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 1 turns: 4837276&#xA; 2 turns: 4774137&#xA; 3 turns:  336992&#xA; 4 turns:   42760&#xA; 5 turns:    6879&#xA; 6 turns:    1488&#xA; 7 turns:     334&#xA; 8 turns:     101&#xA; 9 turns:      10&#xA;10 turns:      17&#xA;11 turns:       4&#xA;12 turns:       1&#xA;13 turns:       1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And here's the code (in Scala), in case anyone wants to play with it.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;object FiftyBots {&#xA;  import scala.util.Random&#xA;  import scala.math._&#xA;&#xA;  case class Bot(var x: Int, var y: Int) {&#xA;    var alive = true&#xA;    var backwards = 16&#xA;    val ax = new Array[Int](4)&#xA;    val ay = new Array[Int](4)&#xA;&#xA;    def out(xa: Int, ya: Int): Boolean = xa &amp;lt; 0 || xa &amp;gt; 7 || ya &amp;lt; 0 || ya &amp;gt; 7&#xA;&#xA;    def mv(n: Int, a: Array[Int], b: Array[Int], i: Int) {&#xA;      val x = signum(n)&#xA;      a(i+1) = a(i) + x&#xA;      b(i+1) = b(i)&#xA;      if (n-x != 0) mv(n-x, a, b, i+1)&#xA;    }&#xA;&#xA;    def pick(r: Random) {&#xA;      var choice = r.nextInt(16)&#xA;      while (choice == backwards) choice = r.nextInt(16)&#xA;      val (a1, a2) = if ((choice&amp;amp;1) != 0) (ax, ay) else (ay, ax)&#xA;      val (d1, d2) = if ((choice&amp;amp;2) != 0) (2, 1) else (1, 2)&#xA;      val s1 = if ((choice&amp;amp;4) != 0) 1 else -1&#xA;      val s2 = if ((choice&amp;amp;8) != 0) 1 else -1&#xA;      ax(0) = x&#xA;      ay(0) = y&#xA;      mv(d1*s1, a1, a2, 0)&#xA;      mv(d2*s2, a2, a1, d1)&#xA;      if (out(ax(3), ay(3))) pick(r)&#xA;      else backwards = choice ^ 0xF&#xA;    }&#xA;&#xA;    def step(i: Int) { if (alive) { x = ax(i); y = ay(i) } }&#xA;  }&#xA;&#xA;  class Board(r: Random) {&#xA;    val xs = Array.fill[Bot](8, 8)(null)&#xA;    val ts = Array.fill[Bot](8, 8, 4)(null)&#xA;&#xA;    def wipe(a: Array[Bot]) { var i = 0; while (i &amp;lt; a.length) { a(i) = null; i += 1 } }&#xA;    def ins(a: Array[Bot], b: Bot) { var i = 0; while (a(i) != null) i += 1; a(i) = b }&#xA;&#xA;    def every[A](f: (Int, Int) =&amp;gt; A) {&#xA;      for (x &amp;lt;- 0 until 8; y &amp;lt;- 0 until 8) f(x,y)&#xA;    }&#xA;&#xA;    def everyBot[A](f: Bot =&amp;gt; A) { &#xA;      every{ (x,y) =&amp;gt; if (xs(x)(y) != null) f(xs(x)(y)) }&#xA;    }&#xA;&#xA;    def ready {&#xA;      var n = 64&#xA;      var m = 50&#xA;      every{ (x,y) =&amp;gt;&#xA;        xs(x)(y) = if (r.nextInt(n) &amp;lt; m) { m -= 1; Bot(x,y) } else null&#xA;        n -= 1&#xA;      }&#xA;    }&#xA;&#xA;    def move: Int = {&#xA;      everyBot( _.pick(r) )&#xA;      for (i &amp;lt;- 1 to 3) {&#xA;        var bump = true&#xA;        while (bump) {&#xA;          bump = false&#xA;          every{ (x,y) =&amp;gt; wipe(ts(x)(y)) }&#xA;          everyBot{ b =&amp;gt;&#xA;            b.step(i)&#xA;            ins(ts(b.x)(b.y), b)&#xA;          }&#xA;          every{ (x,y) =&amp;gt;&#xA;            if (ts(x)(y)(1) != null) {&#xA;              var k = 0&#xA;              while (k &amp;lt; 4 &amp;amp;&amp;amp; ts(x)(y)(k) != null) {&#xA;                ts(x)(y)(k).step(i-1)&#xA;                ts(x)(y)(k).alive = false&#xA;                bump = true&#xA;                k += 1&#xA;              }&#xA;            }&#xA;          }&#xA;        }&#xA;      }&#xA;      var n = 0&#xA;      everyBot{ b =&amp;gt; if (b.alive) n += 1 }&#xA;      n&#xA;    }&#xA;&#xA;    override def toString = xs.map(y =&amp;gt; y.map{_ match {&#xA;      case null =&amp;gt; '.'; case b if b.alive =&amp;gt; '@'; case _ =&amp;gt; 'X'&#xA;    }}.mkString).mkString(&quot;\n&quot;)&#xA;  }&#xA;&#xA;  def main(args: Array[String]) {&#xA;    val n = args(0).toInt&#xA;    val N = args(1).toInt&#xA;    (0 until N).par.map{ _ =&amp;gt;&#xA;      val b = new Board(new Random())&#xA;      (0 until n).map{ _ =&amp;gt; b.ready; var i = 1; while (b.move &amp;gt; 0) i += 1; i }&#xA;    }.seq.reduce(_ ++ _).&#xA;      groupBy(identity).mapValues(_.length).toList.sortBy(_._1).&#xA;      map{ case (n,m) =&amp;gt; f&quot;$n%2d turns: $m%7d&quot; }.&#xA;      foreach(println)&#xA;  }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Edit: the first argument is the number of boards to run per thread, and the second is the number of threads.  So if you are on a 4-core system and want a million boards run as fast as possible:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;scala FiftyBots 250000 4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2815" LastEditorUserId="2815" LastEditDate="2014-10-10T19:06:04.990" LastActivityDate="2014-10-10T19:06:04.990" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2767" PostTypeId="2" ParentId="2679" CreationDate="2014-10-10T00:54:34.103" Score="2" Body="&lt;p&gt;&quot;&lt;code&gt;What would you say *yesterday* if I asked you, 'Which is the right way'&lt;/code&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Yesterday the liar (he who always tells lies) would tell a lie (= the wrong answer).  Today he must tell a lie on that lie, so you get the &lt;em&gt;real&lt;/em&gt; answer. (lie + lie = truth)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Yesterday the honest person (who always only tells the truth) would tell... &lt;em&gt;the truth&lt;/em&gt; (you guessed it! :) ).  Today, he must tell the truth on the truth, so you still get the &lt;em&gt;real&lt;/em&gt; answer.  (truth + truth = truth)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In math:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Liar:  &lt;code&gt;-1 * -1 = 1&lt;/code&gt;&lt;br&gt;&#xA;Honest man:  &lt;code&gt;1 * 1 = 1&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="2660" LastEditorUserId="2660" LastEditDate="2014-10-21T19:37:23.613" LastActivityDate="2014-10-21T19:37:23.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2769" PostTypeId="2" ParentId="2731" CreationDate="2014-10-10T05:23:38.713" Score="8" Body="&lt;p&gt;Spoiler:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SEND THE JADE FAST&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Each card has a name and a family&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ES NC DD&#xA;TH ED&#xA;JD AD EH&#xA;FH AS TC&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The days are ordered from top to bottom. The sacrifices are ordered from left to right, so on day 1, the first sacrifice is ES, second is NC, third is TD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a family is mentioned it should be included, otherwise left out.  If a sacrifice &quot;follows&quot; his family, then the family is listed first.  If the sacrifice leads his family, then the family is listed second.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On the first day, the first sacrifice follow his family here while the&#xA;  others come alone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;SE N D&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On the second day, the first sacrifice leads his family here while the&#xA;  other come alone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;TH E&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On the third day, the second sacrifice leads his family here while the&#xA;  others come alone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;J AD E&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On the fourth day, the second sacrifice leads his family here while&#xA;  the others come alone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;F AS T&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1777" LastEditorUserId="1777" LastEditDate="2014-10-10T05:29:07.670" LastActivityDate="2014-10-10T05:29:07.670" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2773" PostTypeId="1" AcceptedAnswerId="2777" CreationDate="2014-10-10T06:37:31.790" Score="4" ViewCount="5585" Body="&lt;p&gt;Suppose 9 men take 9 hours to dig 9 holes in the ground. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) How many hours would it take 3 men to dig 3 holes? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) How many men would it take to dig 54 holes in 54 hours?&lt;/p&gt;&#xA;" OwnerUserId="2730" LastEditorDisplayName="user88" LastEditDate="2014-10-10T17:49:29.300" LastActivityDate="2020-01-07T16:56:05.893" Title="Men digging holes" Tags="|logical-deduction|calculation-puzzle|" AnswerCount="4" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2774" PostTypeId="2" ParentId="2588" CreationDate="2014-10-10T06:41:54.887" Score="10" Body="&lt;p&gt;I would repeat here a big part of answer on similar, but not identical question: &lt;a href=&quot;https://puzzling.stackexchange.com/a/1666/28&quot;&gt;https://puzzling.stackexchange.com/a/1666/28&lt;/a&gt; . This algorithm guarantee to find one working computer in &lt;strong&gt;97 questions&lt;/strong&gt;, and if we are lucky it will need only 49 questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's call $M$ maximum number of faulty, which can be among Group of chosen $N$ computers according to our knowledge.&lt;br&gt;&#xA;Our start group will include all comps, $N = 99$. Our knowledge is that $M=49$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The goal is to find a chain from $L=(M+1)$ comps, such that each comp in the chain tells about the next one that he is a working one. The last comp in the chain will be a working for sure.&lt;/strong&gt;&lt;br&gt;&#xA;&lt;em&gt;Indeed, there are only $M$ faulty, therefore at least one comp in the chain is working. If one comp is working, then he doesn't lie and comp next to him is working too. Therefore all comps after very first working in the chain are working.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, &lt;strong&gt;how do we create this chain?&lt;/strong&gt;&lt;br&gt;&#xA;0. We work only with comps, which are in the Group. Ignore the others.&lt;br&gt;&#xA;1. We take an arbitrary comp as a first in the chain.&lt;br&gt;&#xA;2. Each time we ask last comp in the chain about arbitrary new comp: &quot;Is it working?&quot;.&lt;br&gt;&#xA;3. If &quot;Yes&quot; we add that new comp to the end of the chain.&lt;br&gt;&#xA;4. If &quot;No&quot; we remove that new comp from the Group. Also we remove asked comp from the Group and the chain. [Note that at least one from the excluded once is faulty comp, so we decrease N by 2 and M by 1 here.]&lt;br&gt;&#xA;5. We finish when the chain length is bigger than half of the Group size.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now &lt;strong&gt;let's calculate number of questions required&lt;/strong&gt;.&lt;br&gt;&#xA;If the chain already exists after each answer &quot;Yes&quot; we need one less comp to finish the chain ($M+1-L$ is decreased by 1 since $L=L+1$, $M=M$); and after each &quot;No&quot; we would need the same number of comps to finish the chain ($L=L-1$, $M=M-1$). If the chain doesn't exist ($L=0$) then we can progress a bit faster: &quot;Yes&quot; gives us two steps forward and &quot;No&quot; gives us one step forward (because $M$ decreased). So the best scenarios are when we get always &quot;yes&quot; or always &quot;no&quot; and finish in 49 questions. In the worst scenario we create a chain getting &quot;yes&quot; as the first answer and then the chain is never distracted completely and meanwhile we get maximum number of &quot;No&quot;-s, which are 49. Each &quot;No&quot; removes a comp from the chain, that means that we need to add 49 comps (requires 48 questions) and remove 49 comps from the chain (requires 49 questions). In this case we will end with one comp in our group and we will know that this is working one. So in the worst case we need 97 questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's it. Unfortunately I can not prove that this is optimal. Also you can check the best answer to that question: &lt;a href=&quot;https://puzzling.stackexchange.com/a/1676/28&quot;&gt;https://puzzling.stackexchange.com/a/1676/28&lt;/a&gt; . I have not read it because it is spoiler for me, but possibly it includes better answer to this question too.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2014-10-11T16:23:12.017" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2775" PostTypeId="2" ParentId="2693" CreationDate="2014-10-10T07:05:25.507" Score="12" Body="&lt;p&gt;Examining all of the answers leads to interesting insights, but my intuition is now telling me that there isn't (an obvious) closed-form solution to this puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My best attempt at a solution is the union of all line segments of length 1/2 whose endpoints touch each of the x-axis and y-axis simultaneously, and are mirrored over the y-axis. This would look like Trenin's &quot;square-minus-quarter-circle-mirrored&quot; answer. I am fairly sure they are the same shape, but I lack proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately this solution must be incomplete. Consider if the baby looped itself into any ellipse of unit perimeter--or worse: an ellipse with a break in it. For simplicity* consider a circle. The diameter of such a circle is 1/pi. It turns out a circle of such diameter is too large to fit into the shape described by my answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've used MATLAB to whip up a figure of what I mean by the above descriptions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BfDo1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, the solution could be truncated per Mooing Duck's comment on Trenin's answer, and possibly truncated further depending on the number of bends--triangles aren't the only possible bend configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*As a side note, I'll point out that determining the ratio of the radii of ellipses of constant perimeter, given the length of one radius, is quite complicated, involving the inverse complete elliptic integral of the second kind. I found a journal article and related Fortran code for the inverse if anyone is interested (links below now, thanks for the upvotes!). To prove a solution is minimal would require checking every ellipse of unit perimeter, since the circle doesn't fit. This is why I don't think there is a closed-form solution to this problem. Unfortunately both determining a minimum blanket and proving that it is minimal are beyond the time I'm willing to devote to the problem, but may be the subject of a publishable journal article for an ambitious sort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; &lt;a href=&quot;http://www.researchgate.net/publication/237006396_Numerical_Computation_of_Inverse_Complete_Elliptic_Integrals_of_First_and_Second_Kind&quot; rel=&quot;noreferrer&quot;&gt;Link to article&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.researchgate.net/profile/Toshio_Fukushima/publication/262729893_xicelke/links/0a85e538b28e3767da000000&quot; rel=&quot;noreferrer&quot;&gt;Link to code used by author of article, as text file&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2823" LastEditorUserId="2823" LastEditDate="2014-10-10T14:58:05.153" LastActivityDate="2014-10-10T14:58:05.153" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2777" PostTypeId="2" ParentId="2773" CreationDate="2014-10-10T07:15:40.040" Score="10" Body="&lt;p&gt;a) 9&lt;br&gt;&#xA;b) 9&lt;/p&gt;&#xA;&#xA;&lt;p&gt;9 men take 9 hours to dig 9 holes =&gt;&lt;br&gt;&#xA;1 man takes 9 hours to dig 1 hole =&gt;&lt;br&gt;&#xA;&lt;strong&gt;a) 3 men take &lt;em&gt;9 hours&lt;/em&gt; to dig 3 holes&lt;/strong&gt;&lt;br&gt;&#xA;1 man take 9*54 hours to dig 54 holes =&gt;&lt;br&gt;&#xA;&lt;strong&gt;b) &lt;em&gt;9 men&lt;/em&gt; take 54 hours to dig 54 holes&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2014-10-10T10:18:45.540" LastActivityDate="2014-10-10T10:18:45.540" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2780" PostTypeId="1" CreationDate="2014-10-10T11:33:42.610" Score="7" ViewCount="2917" Body="&lt;p&gt;I heard from a clever person that for the problem mentioned before: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/251/i-dont-know-the-two-numbers-but-now-i-do&quot;&gt;I don&amp;#39;t know the two numbers... but now I do&lt;/a&gt; a solution exists even if we change 100 to infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the formulation would be next:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Two perfect logicians, Summer and Proctor, are told that integers x and y have been chosen such that $x&amp;gt;1$ and $y&amp;gt;1$. Summer is given the value x+y and Proctor is given the value x⋅y. They then have the following conversation.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Proctor: &quot;I cannot determine the two numbers.&quot;&lt;br&gt;&#xA;  Summer: &quot;I knew that.&quot;&lt;br&gt;&#xA;  Proctor: &quot;Now I can determine them.&quot;&lt;br&gt;&#xA;  Summer: &quot;So can I.&quot;  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Given that the above statements are true, what are the two numbers?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't know whether it is true. Can you find the solution or prove that it doesn't exists? Can you find a general solution for any integer $N$ if we have limitation $x+y&amp;lt;N$?&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2016-02-24T04:00:28.930" Title="Generalization of Sum and Product puzzle" Tags="|mathematics|logical-deduction|calculation-puzzle|meta-knowledge|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2782" PostTypeId="2" ParentId="2693" CreationDate="2014-10-10T16:06:10.643" Score="9" Body="&lt;p&gt;I am posting a new answer, building from Trevin's trapezoid answer. I drew a diagram of the answer using Creo (which has some handy-dandy constraint mechanisms built in). Creo is a CAD tool which has a constraint-and-dimension-based sketcher, which is what I am using here. Solid yellow lines are outlines of the sketch itself, and dotted lines are construction lines. Construction lines are not part of the actual part being sketched, but allow for additional constraints to be made on the sketch. Solid blue lines are dimension leaders which tell the dimension of some feature, solid red lines are locked dimensions (so I don't mess up the sketch accidentally by clicking and dragging something). Some aren't locked because they are parameterized using mathematical relationships, as with the circle diameter, which is set equal to 1/pi. Filled yellow areas are considered solid two-dimensions sketch regions, and can be used to make three-dimensional features from. Here they are used for denoting the blanket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xIkWd.png&quot; alt=&quot;original answer&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we consider a square inside the triangle of side-length 1/2, this actually takes care of the circle problem as well, and can improve the trapezoid's area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TxYTn.png&quot; alt=&quot;square encompassing unit circumference circle&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The square is the highest a quadrilateral of perimeter 4/3 can be (4/3 = 1 for snake + 1/3 for break). If it goes taller, turn it on its side and should still fit. If we drop the height of the square, then the two vertical sides will angle outwards. If we keep doing that, eventually it will become a straight line of length 1. If instead we increase the length of the top, then the other two sides must get shorter. Either way it will fit inside of a 45-45-90 triangle of base length 1. If we trim the sides down we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gn2GF.png&quot; alt=&quot;trimmed triangle&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am fairly sure that this area is minimal for all quadrilaterals the snake can take the shape of. As a heuristic argument, I ran MATLAB code for many symmetric quadrilaterals and they converge to a 45-45-90 triangle. It can't be seen in the figure and I am hesitant to make this answer bulkier, but each stair-step is for a different iteration of the length of the vertical legs. Then they are iterated over angle between vertical and horizontal. More iterations looks more like a 45-45-90 triangle. If we trim off the top to account for turning tall quads on their sides, then we get the solution in my third image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IF70M.png&quot; alt=&quot;matlab triangle&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This solution does not consider pentagons or higher-order polygons in general. It turns out if you put a pentagon with one edge along the bottom, it sticks out into the region between the square and the equilateral triangle just a bit. Hexagons and heptagons do as well. However, none of them take up the entire space there. I outlined the regions that are likely to be improved upon in some way (outline in yellow, see first paragraph):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4T0hp.png&quot; alt=&quot;can_be_improved&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Height is sqrt(3)/4, base is 1, top is ( 1 - 2 * ( sqrt(3)/4 ) ). Multiply height by average of bases. The area of the 45 degree trapezoid I came up with is thus about 0.2455 square units.&lt;/p&gt;&#xA;" OwnerUserId="2823" LastEditorUserId="2823" LastEditDate="2014-10-10T17:04:27.603" LastActivityDate="2014-10-10T17:04:27.603" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="2783" PostTypeId="2" ParentId="2693" CreationDate="2014-10-10T17:02:09.067" Score="3" Body="&lt;p&gt;A blanket the same width and length of the snake would seem to be sufficient.  Blankets are soft, and bend just like snakes can bend - depending upon theur construction, they may be able to bend more.  In this case,  let's assume a knit blanket,  loose enough that it can stretch to bend just as much as the baby stretches to bend,  without having to be folded over or bunched up.  This has the nice effect of keeping the problem 2 dimensional.  .  Mommy snake can adjust the blanket at any time so that it bends to cover baby snake&lt;/p&gt;&#xA;" OwnerUserId="2847" LastActivityDate="2014-10-10T17:02:09.067" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2784" PostTypeId="2" ParentId="2693" CreationDate="2014-10-10T17:29:46.227" Score="3" Body="&lt;p&gt;Assuming the line (snake) has a negligible width, this shape (blanket) will contain every possible bent/zig-zagging of a 1-unit line, assuming the shape can be rotated in three dimensions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pqTXL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pqTXL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total area adds up to a quadrant of a 1-unit square. More specifically, ¼u² or 0.25 units squared.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the area, it cannot get any smaller because the largest bounding box for all possible bents has that specific area - it's the 90-degree mid-way bent.&lt;/p&gt;&#xA;" OwnerUserId="2846" LastEditorUserId="5373" LastEditDate="2017-03-11T00:43:24.180" LastActivityDate="2017-03-11T00:43:24.180" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="2785" PostTypeId="2" ParentId="2780" CreationDate="2014-10-10T17:32:59.087" Score="6" Body="&lt;p&gt;The solution to the restricted version still works, but it is no longer unique. I'll give a proof that the previous solution still works. According to &lt;a href=&quot;https://puzzling.stackexchange.com/a/257/1955&quot;&gt;Joe Z.'s answer&lt;/a&gt;, Summer is given $x+y=17$, and Proctor is given $x \cdot y=52$. Also, to avoid confusion I'm going to call Proctor male and Summer female.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because $x \cdot y=52$, Proctor knows the numbers could be either 2 and 26 or 4 and 13, but he has no way to determine which they are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summer knows that since $x+y=17$, the numbers could be any of the pairs $(2,15), (3,14),(4,13),(5,12),(6,11),(7,10),$ and $(8,9)$. None of those pairs are two prime numbers, so she knew that Proctor wouldn't be able to tell what the numbers are based on their product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that Proctor knows that Summer knew he wouldn't be able to tell the numbers by their product, he can reconsider the two possibilities - if the numbers we 2 and 26, then the sum would be 28. If the Summer had the number 28 then her list of possibilities would include the pair (5,23), both of which are prime. Then she would not have known that he would not be able to tell. So he then knows that the numbers are 4 and 13.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that Summer knows that Proctor knows the numbers, she can go through a similar process for her list of possible numbers.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the numbers were 2 and 15, then the product would be 30. Then Proctor's list of possibilities would be $(2,15), (3,10),$ and $(5,6)$. If this had been the case, then he would not have been able to eliminate $(5,6)$ because none of the pairs that sum to 11 are a pair of prime numbers. Thus Proctor would not have been able to know the numbers yet.&lt;/li&gt;&#xA;&lt;li&gt;If the numbers were 3 and 14, Proctor's list would include $(2, 21)$, which could not be eliminated.&lt;/li&gt;&#xA;&lt;li&gt;If the numbers were 5 and 12, Proctor would be unable to eliminate $(3, 20)$&lt;/li&gt;&#xA;&lt;li&gt;For 6 and 11, Proctor would be unable to eliminate $(2, 33)$&lt;/li&gt;&#xA;&lt;li&gt;For 7 and 10, Proctor would be unable to eliminate $(2, 35)$&lt;/li&gt;&#xA;&lt;li&gt;For 8 and 9, Proctor would be unable to eliminate $(3,24)$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;By process of elimination, Summer knows that the numbers are 3 and 14.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As you can see from my analysis, at no point did the restriction of $x+y&amp;lt;100$ factor in to Summer or Proctor's ability to determine the numbers. This should actually be pretty obvious because for $x,y&amp;gt;1$, $x+y\le x\cdot y$, so $x\cdot y&amp;lt;100\rightarrow x+y&amp;lt;100$. In other words, Summer knows the sum, and Proctor knows the product is less than 100 so he also knows that the sum is less than 100, leaving the restriction as merely a help to people trying to solve the puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way the restriction would affect this would be if Proctor could eliminate one or more pairs because $N$ was less than the sum of one of the pairs he would not have been able to eliminate otherwise. For example, if $N=36$, then Proctor would be able to eliminate $(2, 35)$, which would mean 7 and 10 would have still been a valid possibility based on what Summer knew.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I think that the question of generalizing this can be split into two parts:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;As $N$ gets larger, do more solutions become viable?&lt;/li&gt;&#xA;&lt;li&gt;For smaller $N$, do different solutions become viable?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; According to &lt;a href=&quot;http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/logic_sum_product&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt;, there are at least three more possible solutions to this problem: $x=16,y=111$, $x=201,y=556$, and $x=421,y=576$. It's under the section &quot;Sum-Product Puzzle: (Kiltinen/Young unbounded version)&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1955" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2014-10-13T16:32:36.313" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2786" PostTypeId="2" ParentId="2773" CreationDate="2014-10-10T17:41:59.037" Score="6" Body="&lt;p&gt;A) 3 hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why? Well I'm glad you asked!!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The men are all in a line and by force of habit just throw the dirt behind them. Unfortunately that falls into the hole of the person behind them. So the first one finishes in 1 hour, the second (now that no one is putting dirt in his hole) finishes an hour after that. There's only room for 1 man per hole so they can't go helping their friends, so the last guy ends up digging a hole for each person.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These men really need a manager.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B) 54 men. As stated above, they each dig and dig and dig as the idiot in front of them fills in their hole each time they take a shovel full out. Remember if it was just one guy, either he would make a big pile where the next hole is supposed to go, or he would dig a hole, walk forward and dig his hole out again, and have to redig it! That would be terrible. It would take this guy 1485 hours to do it by himself because we have to assume they send the worst one out there.&lt;/p&gt;&#xA;" OwnerUserId="2812" LastEditorUserId="2812" LastEditDate="2014-10-10T19:18:31.350" LastActivityDate="2014-10-10T19:18:31.350" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2787" PostTypeId="5" CreationDate="2014-10-10T17:49:00.073" Score="0" Body="&lt;p&gt;A rebus is a visual puzzle that presents the reader with an image containing a set of letters or pictures, for example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;FAREDCE&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and asks the reader to determine what word or phrase is represented by the image. Usually the phrase uses pictures or letters to represent pieces of the word at a time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above puzzle, the answer is &lt;em&gt;red in the face&lt;/em&gt;, as the word &lt;code&gt;RED&lt;/code&gt; can be found in-between the letters of the word &lt;code&gt;FACE&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;HEAD&#xA;HEEL&#xA;HEEL&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;represents &lt;em&gt;head over heels&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="11497" LastEditDate="2020-02-26T18:42:29.403" LastActivityDate="2020-02-26T18:42:29.403" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="2788" PostTypeId="4" CreationDate="2014-10-10T17:49:00.073" Score="0" Body="A puzzle that uses images and/or text, where solving the puzzle requires examination of the components' attributes. Typically wordplay combining the images/text will yield a word or phrase." OwnerDisplayName="user88" LastEditorUserId="69582" LastEditDate="2022-04-18T06:45:31.203" LastActivityDate="2022-04-18T06:45:31.203" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="2789" PostTypeId="1" AcceptedAnswerId="2930" CreationDate="2014-10-10T18:02:17.990" Score="8" ViewCount="906" Body="&lt;p&gt;How far down the rabbit hole of not knowing can we go? While working on &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2780/generalization-of-sum-and-product-puzzle&quot;&gt;the generalization&lt;/a&gt; of &lt;a href=&quot;https://puzzling.stackexchange.com/questions/251/i-dont-know-the-two-numbers-but-now-i-do&quot;&gt;I don't know the two numbers... but now I do&lt;/a&gt;, I thought of these two extensions. Both have the same conditions, but a different conversation taking place:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conditions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Two perfect logicians, Summer and Proctor, are told that integers $x$ and $y$ have been chosen such that $1&amp;lt;x&amp;lt;y$ and $x+y&amp;lt;100$. Summer is given the value $x+y$ and Proctor is given the value $x⋅y$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Conversation A:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Proctor: &quot;I cannot determine the two numbers.&quot;&lt;/li&gt;&#xA;&lt;li&gt;Summer: &quot;I knew that.&quot;&lt;/li&gt;&#xA;&lt;li&gt;Proctor: &quot;I still cannot determine the two numbers.&quot;&lt;/li&gt;&#xA;&lt;li&gt;Summer: &quot;Now I can determine them.&quot;&lt;/li&gt;&#xA;&lt;li&gt;Proctor: &quot;Now I can, too.&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Conversation B:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Proctor: &quot;I cannot determine the two numbers.&quot;&lt;/li&gt;&#xA;&lt;li&gt;Summer: &quot;I knew that.&quot;&lt;/li&gt;&#xA;&lt;li&gt;Proctor: &quot;I still cannot determine the two numbers.&quot;&lt;/li&gt;&#xA;&lt;li&gt;Summer: &quot;Neither can I.&quot;&lt;/li&gt;&#xA;&lt;li&gt;Proctor: &quot;Now I can determine them.&quot;&lt;/li&gt;&#xA;&lt;li&gt;Summer: &quot;Now I can, too.&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We could take this even farther, but based on the solution to the original problem each step will add a significant amount of complexity so I don't want to go too far down the rabbit hole (at least not yet).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would either of these conversations be possible? If so, what are the values of $x$ and $y$? If there is no solution for $x+y&amp;lt;100$, I would accept a solution that violates that restriction.&lt;/p&gt;&#xA;" OwnerUserId="1955" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2016-02-24T04:11:24.897" Title="A different extension of the sum and product puzzle" Tags="|logical-deduction|calculation-puzzle|meta-knowledge|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2790" PostTypeId="1" AcceptedAnswerId="2802" CreationDate="2014-10-10T20:22:55.477" Score="2" ViewCount="350" Body="&lt;p&gt;This is sorta like &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2376/how-to-ship-the-new-slurm-7-pack-efficiently&quot;&gt;How to ship the new Slurm 7-pack efficiently&lt;/a&gt; by Matt Malone but instead of 7, it's the 9 pack! It's basically a 2 by 4 pack with a piece sticking out, like in the above link. What's the smallest version rectangle that you can make with the described shape, solution with the least packs gets best solution.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-02-11T09:44:49.103" Title="Packing an Efficient 9 Pack" Tags="|geometry|tiling|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2791" PostTypeId="2" ParentId="2790" CreationDate="2014-10-10T20:44:40.440" Score="4" Body="&lt;p&gt;Just 4 packs are needed (sorry if the picture is too tall): &lt;img src=&quot;https://i.stack.imgur.com/72BKm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2014-10-10T20:44:40.440" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2792" PostTypeId="1" AcceptedAnswerId="2793" CreationDate="2014-10-10T21:01:22.540" Score="-6" ViewCount="980" Body="&lt;p&gt;In an alternative universe where the following is true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 # 1 = 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 # 2 = 7&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 # 2 = 5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would ? be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5 # 5 = ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;YOU DO NOT NEED THE FOLLOWING TO FINISH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extra equations:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;3 # 3 = 11, 4 # 4 = 15&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="5373" LastEditDate="2015-07-15T23:49:49.017" LastActivityDate="2015-07-15T23:49:49.017" Title="When Math Loses Meaning" Tags="|mathematics|pattern|" AnswerCount="3" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2793" PostTypeId="2" ParentId="2792" CreationDate="2014-10-10T21:06:31.380" Score="8" Body="&lt;ol start=&quot;19&quot;&gt;&#xA;&lt;li&gt;Seems to follow the formula: $(x+y) + (x+y-1)$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;By the way, as an alternative let's say your plus is $+^{\prime}$, now we can define the operator: &#xA;$x+^{\prime}y = 2(x+y)-1$&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-10-10T21:22:32.160" LastActivityDate="2014-10-10T21:22:32.160" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2794" PostTypeId="2" ParentId="2693" CreationDate="2014-10-10T21:09:58.030" Score="2" Body="&lt;p&gt;I'm not seeing this one here yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a 30-60-90 triangle with sides 1/2, sqrt(3)/2, and 1.&#xA;Total area =0.2165&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe this shape allows for all single bends of the snake.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/RywO7.png&quot; alt=&quot;triangle layout&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For bends between 0 and 60 degrees, the bend should be placed in corner A:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/ENg7N.png&quot; alt=&quot;placement for bends 0 to 60 degrees&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For bends greater than 60 degrees, one end should be placed at Point A:&lt;img src=&quot;https://i.stack.imgur.com/KnFrn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further analysis is required for additional bends. &lt;/p&gt;&#xA;" OwnerUserId="2495" LastActivityDate="2014-10-10T21:09:58.030" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2795" PostTypeId="2" ParentId="2792" CreationDate="2014-10-10T21:51:23.357" Score="17" Body="&lt;p&gt;I think the answer is clearly  17.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My Logic is that for each number digit you replace it with the next number in the fibinaci sequence, add as usual, then add by one.  so we get &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1+1+1=3 (second number in fibonacci is 1)&#xA;3+3+1=7&#xA;1+3+1=5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and thus&#xA;    8+8+1=17&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or maybe the answer is 23!  because the formula is $(x+y) + (x+y-1)$, but of course in this universe were uses the octal numbering system!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or maybe they use a base 10 numbering system, but they order their numbers from 0-9 like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;8,3,1,2,0,5,7,6,9,4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In which case the answer is 34!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All are correct given the 3 equations which are the 'only ones I need' to solve the puzzle :)&lt;/p&gt;&#xA;" OwnerUserId="2698" LastEditorUserId="2698" LastEditDate="2014-10-11T20:49:39.797" LastActivityDate="2014-10-11T20:49:39.797" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2799" PostTypeId="1" AcceptedAnswerId="2816" CreationDate="2014-10-11T07:27:54.203" Score="11" ViewCount="1661" Body="&lt;p&gt;How many queens can be placed in a standard &lt;span class=&quot;math-container&quot;&gt;$8*8$&lt;/span&gt; chessboard to threaten each of them exactly once by other queens?&lt;/p&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="57521" LastEditDate="2023-08-18T04:45:21.687" LastActivityDate="2023-08-18T04:45:21.687" Title="How Many Queens to Threaten Themselves?" Tags="|logical-deduction|optimization|chess|" AnswerCount="2" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="2800" PostTypeId="2" ParentId="2799" CreationDate="2014-10-11T08:05:47.117" Score="7" Body="&lt;p&gt;A queen cannot have a paired queen on the same column if it also has one on the same row. So, queens will pair up every other row and every other column. This will yield 8 queens. There are many configurations (100s probably). Here is one example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tLS2n.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question didn't strictly state if this matters, but I believe this is a maximum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If other pieces are allowed, then other pawns can be used to block queens from each other. 24 queens can be placed on the board in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wGGS3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-10-11T09:20:42.373" LastActivityDate="2014-10-11T09:20:42.373" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2801" PostTypeId="2" ParentId="1751" CreationDate="2014-10-11T12:35:14.973" Score="0" Body="&lt;p&gt;Each small dot and small ring are paired and move according to their cooresponding shapes perimeter. (i.e. red dot-green circle move perimeter of square, blue dot-red circle =perimeter of pentagon, green dot blue circle= perimeter of hexagon.) Think like you are moving 'gears' on a clock. I still haven't figured out which one triggers all of them to move in sync after I set them up in position right before their final spot. &lt;/p&gt;&#xA;" OwnerUserId="2865" LastActivityDate="2014-10-11T12:35:14.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2802" PostTypeId="2" ParentId="2790" CreationDate="2014-10-11T15:05:55.343" Score="2" Body="&lt;p&gt;I thihk this is the 6x6 solution&#xA;&lt;img src=&quot;https://i.stack.imgur.com/QeQvY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="143" LastActivityDate="2014-10-11T15:05:55.343" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2803" PostTypeId="2" ParentId="2693" CreationDate="2014-10-11T15:24:15.003" Score="1" Body="&lt;p&gt;The correct answer has already been given&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The SemiCircle covers everything:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we need to cover the full length of the snake - a line with length &lt;code&gt;1&lt;/code&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UcB9E.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we need to ensure to cover - as outlined above - any angle between &lt;code&gt;0&lt;/code&gt; and &lt;code&gt;180°&lt;/code&gt;.&#xA;Worst Case: it's exactly a 50:50 split -&gt; Thales circle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4v0cC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This case covers everything, including the snake forming a circle itself:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/H4JZq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whatever &quot;shapes&quot; the snake will perform on any of the both tails (if splitting at 50%) it will NEVER have a absolute length of &lt;code&gt;0.5&lt;/code&gt; therefore remaining IN the semicircle all the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Any solution having a streight line, but lesser area will not work, because the snake could be like &lt;code&gt;v-------------------^&lt;/code&gt;, messing up any attempt to have a line involved just to cover the maximum length&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2868" LastEditorUserId="2868" LastEditDate="2014-10-11T15:43:12.007" LastActivityDate="2014-10-11T15:43:12.007" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2806" PostTypeId="2" ParentId="2588" CreationDate="2014-10-11T16:18:55.150" Score="-1" Body="&lt;p&gt;Slow, but guaranteed to work.  Ask each computer if &lt;em&gt;it&lt;/em&gt; is working.  If it says no, then it is broken. If it says yes, then it is working.  Keep this up until you find all broken computers.  This works since a working computer can never say it isn't working.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, this is not the most efficient strategy, but figured I would point out a base case that is at least guaranteed to work. :)&lt;/p&gt;&#xA;" OwnerUserId="2698" LastEditorUserId="12700" LastEditDate="2015-07-28T11:18:45.223" LastActivityDate="2015-07-28T11:18:45.223" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2807" PostTypeId="1" AcceptedAnswerId="2809" CreationDate="2014-10-11T16:28:04.757" Score="5" ViewCount="858" Body="&lt;p&gt;Standard knights and knaves problem.  There are 10 people in a room, you know that each person will either always tell the truth or always lie, but you don't know who is a truth teller or liar.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are in a room with 5 doors.  Four of the doors lead to horrible hazards like an endless pit, a fiery inferno, or being forced to visit your in-laws (the horror!), one of the doors leads to freedom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the fewest number of questions you may ask to guarantee that you get out to freedom?  What process would you use? It is assumed that all of the 10 people have knowledge of which door leads to freedom.&lt;/p&gt;&#xA;" OwnerUserId="2698" LastEditorUserId="375" LastEditDate="2014-10-12T03:54:13.290" LastActivityDate="2014-10-12T03:54:13.290" Title="The knaves come out at night" Tags="|logical-deduction|liars|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2808" PostTypeId="1" AcceptedAnswerId="2815" CreationDate="2014-10-11T16:37:49.530" Score="15" ViewCount="14136" Body="&lt;ul&gt;&#xA;&lt;li&gt;The one who made it didn't want it&lt;/li&gt;&#xA;&lt;li&gt;The one who bought it didn't need it&lt;/li&gt;&#xA;&lt;li&gt;The one who received it didn't wish for it&lt;/li&gt;&#xA;&lt;li&gt;The one who used it never saw it&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What was it?&lt;/p&gt;&#xA;" OwnerUserId="2698" LastEditorUserId="2383" LastEditDate="2015-10-23T16:07:29.617" LastActivityDate="2017-10-19T16:56:40.897" Title="Riddle of the Unwanted Present" Tags="|riddle|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2809" PostTypeId="2" ParentId="2807" CreationDate="2014-10-11T16:49:08.827" Score="5" Body="&lt;p&gt;A) In this case I suppose you are allowed to ask only yes-or-no questions.&lt;br&gt;&#xA;In this case 2 questions allows you to chose only between 2x2=4 options, so you need at least 3 questions.&lt;br&gt;&#xA;It is very easy to formulate them. Just ask each time &quot;What would you answer if I ask you blabla&quot; and you will know the truth about blabla. Let's use WWYA shortcut for all these words. Then 3 questions you will need to ask:&lt;br&gt;&#xA;1. WWYA is a freedom-door among first 4?&lt;br&gt;&#xA;2. WWYA is a freedom-door among 1 and 2?  (or just go to the 5th, if you've just got No)&lt;br&gt;&#xA;3. WWYA is a freedom-door the 1st?  (or &quot;the 3rd&quot;, if you've just got No)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;B) It is much more interesting to find a question when you are allowed to ask any questions. You can not ask &quot;What you would answer if you are asked, which door leads to freedom?&quot;, since liar would answer the wrong door number and then it would lie about this number, but here he can use another wrong number, fortunately for him he has 3 more bad doors left.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I think &quot;&lt;strong&gt;What you can not answer because of your nature when you are asked, which door leads to hazards?&lt;/strong&gt;&quot; will do the job. To &quot;Which door leads to hazards?&quot; liar must answer the freedom-door. Then to &quot;What you can not answer ...?&quot; he would have the only one wrong answer to tell: the freedom-door. Meanwhile truthteller can answer to &quot;Which door leads to hazards?&quot; any door out of 4 bad-doors, therefore to &quot;What you can not answer ...?&quot; truthteller have also only one answer possible: the freedom-door.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2014-10-11T16:57:39.993" LastActivityDate="2014-10-11T16:57:39.993" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2810" PostTypeId="2" ParentId="27" CreationDate="2014-10-11T17:03:46.210" Score="19" Body="&lt;p&gt;A late solution for LED lights as well, or if you can't touch these hot bulbs :P&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since all of the bulbs are new and identical with the &lt;strong&gt;exact liftetime&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 - &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Turn on the first switch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2 -&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Wait 24 hours.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3 - &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Turn on the second switch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4 - &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Get into the room, wait until a bulb turns off.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5 -&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The bulb that turns off first corresponds to the first switch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The one which is still on corresponds to the second switch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That way you can have LED lights and you don't get a burn, but you might have to wait long time.&lt;/p&gt;&#xA;" OwnerUserId="2869" LastEditorUserId="46903" LastEditDate="2018-09-08T23:04:24.507" LastActivityDate="2018-09-08T23:04:24.507" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="2813" PostTypeId="1" AcceptedAnswerId="3037" CreationDate="2014-10-11T21:27:56.470" Score="11" ViewCount="2235" Body="&lt;p&gt;You have a stack of N cards numbered from 1 to N in random order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want to sort the cards in ascending order using the following procedure:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Your initial stack is stack 1.&lt;/li&gt;&#xA;&lt;li&gt;You distribute the cards from stack 1 to stack 2 and 3: you take the topmost card on stack 1, put it on top of stack 2 or stack 3, and repeat until stack 1 is empty.&lt;/li&gt;&#xA;&lt;li&gt;Then you continue with stack 2: you distribute all cards from stack 2 to the existing stack 3 and a new stack 4, according to the same procedure.&lt;/li&gt;&#xA;&lt;li&gt;Continue with the next stacks:  stack $k$ is distributed to stack $k+1$ and stack $k+2$.&lt;/li&gt;&#xA;&lt;li&gt;To end the process, some stack $M-1$ is transferred to the stack $M$, one card at a time.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As you can see, only 3 stacks exist at any time.  Therefore the 3 stacks in the title.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is: if you use a total of M stacks (initial and final stack included), how many cards can you sort regardless of the initial order?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Additional notes:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You choose on which stack to distribute a card.  The numbers are visible.  You have to figure out the method that maximizes the number N of cards you can sort.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A valid answer should explain why, not just provide the correct maximum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You start with 1 single stack of N cards in random order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You must end with all cards sorted, card 1 at the bottom and card N at the top.  (It doesn't really matter, but end it the same way regardless of the starting position).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a method but I haven't proven it maximal.  I'll accept any result at least as good as mine (but I don't tell what it is).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An algorithm isn't requested, only the value of N.  But you should provide a convincing proof that your best N is indeed achievable.  This almost implies the algorithm is made explicit.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There is a solution in O(log(N)) stacks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With 5 stacks you can sort 5 cards.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2014-10-16T18:57:38.273" LastActivityDate="2016-05-19T02:35:26.750" Title="Sorting cards using max 3 stacks" Tags="|calculation-puzzle|cards|minimum-solve|" AnswerCount="5" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="2814" PostTypeId="2" ParentId="2813" CreationDate="2014-10-11T22:41:44.693" Score="2" Body="&lt;p&gt;Let me start off with a very weak strategy. I'm sure there are better strategies but at least there is an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 1:&lt;br&gt;&#xA;move cards to stack 3 until you encounter card 1 and put that in stack 2.&#xA;keep repeating this for every next card in order until stack 1 is empty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you now have:&lt;br&gt;&#xA;stack 2: cards 1 till x in ascending order&lt;br&gt;&#xA;stack 3: cards x+1 till N in random order&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 2:&lt;br&gt;&#xA;move all cards in stack 2 to stack 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you now have:&lt;br&gt;&#xA;stack 3: cards x+1 till N in random order&lt;br&gt;&#xA;stack 4: cards 1 till x in descending order&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 3:&lt;br&gt;&#xA;move all cards in stack 3 to stack 5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you now have:&lt;br&gt;&#xA;stack 4: cards 1 till x in descending  order&lt;br&gt;&#xA;stack 5: cards x+1 till N in random order&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 4:&lt;br&gt;&#xA;move all cards in stack 4 to stack 6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you now have:&lt;br&gt;&#xA;stack 5: cards x+1 till N in random order&lt;br&gt;&#xA;stack 6: cards 1 till x in ascending order&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now repeat these 4 steps. this means: put cards in stack 7 until you find card x+1 and put that in stack 6. and keep doing that in a way so you have stack 6 in ascending order. and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Worst case scenario: every 4 steps only 1 card gets added to the &quot;in order&quot; stack. With this strategy you have with N=1 that M=2 and N=2 that M=6 In other words: with M stacks you can only order (M+2)/4 cards rounded down if I'm not mistaken.&lt;/p&gt;&#xA;" OwnerUserId="1748" LastEditorUserId="1748" LastEditDate="2014-10-11T22:54:28.103" LastActivityDate="2014-10-11T22:54:28.103" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2815" PostTypeId="2" ParentId="2808" CreationDate="2014-10-12T00:07:42.247" Score="13" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A coffin/tombstone&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Obviously the man who made it doesn't want it! Who would &lt;strong&gt;want&lt;/strong&gt; a [answer]?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The man that bought it didn't use it, if he needed it, he wouldn't have been in a state in which he was capable of buying anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The man who received it obviously didn't wish for it, refer back to the man who made it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The man who used it couldn't have seen it, as he does not have the ability to do so!&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2015-10-23T16:04:57.750" LastActivityDate="2015-10-23T16:04:57.750" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2816" PostTypeId="2" ParentId="2799" CreationDate="2014-10-12T02:50:18.050" Score="11" Body="&lt;p&gt;I found a 10 queen answer. I realized that d'alar'cop's reasoning was slightly off, because threatening is part of the puzzle, so 8 wasn't necessarily a limit. With the 8 inner queens in this position, you can put the final 2 queens in any 2 corners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yOf8B.png&quot; alt=&quot;10 Queens&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: I have a proof for a maximum of 10 queens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that 2 queens in a row or a column use 1x2 and 2 queens diagonally use 2x2, it is advantageous to avoid the diagonals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put 2 queens in a row. You have now used up 1 row and 2 columns.&lt;br&gt;&#xA;Do this again. 2 rows, 4 columns.&lt;br&gt;&#xA;Again. 3 rows, 6 columns.&lt;br&gt;&#xA;If you do it a 4th time, you will have used all 8 columns, so don't.&lt;br&gt;&#xA;Put 2 queens in a column. 5 rows, 7 columns.&lt;br&gt;&#xA;Do this again. 7 rows, 8 columns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can no longer add any queens, as your columns are used up.&lt;br&gt;&#xA;Repeat swapping columns and rows and you end up with 8 rows and 7 columns, so you still cannot add any queens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any attempts at using diagonals simply make the problem end faster (exception: a single diagonal, like my displayed answer, gets you to 8 rows, 8 columns).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, 10 queens is the maximum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now as for allowing other pieces, I noticed that there might be extra space from d'alar'cop's 24 queens answer, and I was able to get 2 more queens in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/G6L8X.png&quot; alt=&quot;26 Queens&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="375" LastEditorUserId="375" LastEditDate="2014-10-13T23:52:51.993" LastActivityDate="2014-10-13T23:52:51.993" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2817" PostTypeId="1" CreationDate="2014-10-12T04:50:54.237" Score="6" ViewCount="1230" Body="&lt;p&gt;My friend sent me his puzzle: it's &lt;code&gt;S2LQEQnXk4DP+WsyP1aURD+lz4JSteeOPz3tO7Yiu1dv5fFCr31uegHUpVGqyXk7WjX95TDALus8Gj6nnoGHwj0MLE2AfpB2KTQEIxVvGMyun2XLb+PTcqzVAoFQLr48&lt;/code&gt; in binary, and it came with two hints: &quot;PooP&quot; and &quot;3 6 4 2 5 -5 -4 -6 0 -5&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have been giving each other these puzzles for a time now and they have been words or phrases so far, but this one may be more complicated. Yes I know my friend is immature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;Check comments for how I got there, but I have gotten it down to &quot;Aid kIe CT&quot; and the numbers.&lt;/p&gt;&#xA;" OwnerUserId="3871" LastEditorUserId="8874" LastEditDate="2015-02-11T09:42:09.790" LastActivityDate="2015-02-11T09:42:09.790" Title="Puzzle of Random Numbers" Tags="|calculation-puzzle|cipher|" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2820" PostTypeId="2" ParentId="2752" CreationDate="2014-10-12T14:09:14.737" Score="2" Body="&lt;p&gt;The answer comes in the form of the matematical function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x)=\sum{ ( x_i \mod 2 ) · ( x_i - 1 )}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is generalised for any number of rolls in a $n$-side dice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum goes from 1 to the number of rolls, $x_i$ is the result of each roll and $x_i-1$ is the number of 'Petals around the rose' for each roll.&lt;/p&gt;&#xA;" OwnerUserId="2470" LastEditorUserId="1967" LastEditDate="2014-11-04T13:38:29.743" LastActivityDate="2014-11-04T13:38:29.743" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2821" PostTypeId="1" AcceptedAnswerId="2823" CreationDate="2014-10-12T16:36:02.143" Score="69" ViewCount="60967" Body="&lt;p&gt;I am a seven letter word. If you remove one letter from me my meaning does not change. If you remove two letters from me my meaning still does not change, and if you remove three letters from me, my meaning still does not change. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you remove all letters my meaning won't change. What word is this?&lt;/p&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="5373" LastEditDate="2018-05-06T16:22:13.290" LastActivityDate="2022-09-16T13:27:10.847" Title="A seven letter word - the meaning doesn't change if you remove letters" Tags="|riddle|word|lateral-thinking|language|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2823" PostTypeId="2" ParentId="2821" CreationDate="2014-10-12T16:58:12.903" Score="110" Body="&#xA;&#xA;&lt;p&gt;The answer is a postbox/mailbox because if you remove a physical letter from it, it remains a mailbox; its meaning is still to hold letters. You can remove all the letters, but it will still be a mailbox.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="5373" LastEditDate="2016-11-22T17:47:14.367" LastActivityDate="2016-11-22T17:47:14.367" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="2827" PostTypeId="1" AcceptedAnswerId="2833" CreationDate="2014-10-12T18:18:51.683" Score="20" ViewCount="73781" Body="&lt;p&gt;Say, you are given a cake which you must share with 7 others. So, you must cut the cake into 8 pieces. But, you are only allowed to make 3 straight cuts. You cannot move pieces of the cake after the first cut. &lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2017-07-11T15:59:17.970" LastActivityDate="2020-08-11T15:44:35.000" Title="Cutting a cake into 8 pieces" Tags="|geometry|" AnswerCount="10" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2830" PostTypeId="1" AcceptedAnswerId="2834" CreationDate="2014-10-12T18:40:03.870" Score="-1" ViewCount="608" Body="&lt;p&gt;You should see a standard chessboard - it is printed on paper. How many ways can you cut it up (around the squares) such that each piece has twice as many squares of one colour than of the other colour?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ks8ho.png&quot; alt=&quot;Chessboard&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="5373" LastEditDate="2015-04-03T09:44:14.450" LastActivityDate="2015-04-03T09:44:14.450" Title="Partitioning a chessboard" Tags="|combinatorics|checkerboard|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2831" PostTypeId="1" AcceptedAnswerId="2839" CreationDate="2014-10-12T18:57:31.933" Score="5" ViewCount="538" Body="&lt;p&gt;Similarly to this question: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2799/how-many-queens-to-threat-themselves&quot;&gt;How many Queens to threaten themselves&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What it the maximum number of queens you can place on an 8*8 board such that each queen is attacked precisely twice:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;with no other pieces on the board?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;other pieces are allowed?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I thought this was a legitimately interesting variation and it seems to be a more or less undocumented result as well. &lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="65277" LastEditDate="2023-07-10T22:20:54.503" LastActivityDate="2023-07-10T22:20:54.503" Title="How many Queens on a board" Tags="|optimization|chess|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2833" PostTypeId="2" ParentId="2827" CreationDate="2014-10-12T19:30:08.323" Score="33" Body="&lt;p&gt;Obviously, this doesn't work in the plane.  So you need a 3D solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 3D solution is simple.  Cut the cake with 2 perpendicular cuts through the center, then make a horizontal cut at half the height of the cake.  It is not fair regarding topping, but you have 8 pieces.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2014-10-16T14:33:44.613" LastActivityDate="2014-10-16T14:33:44.613" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2834" PostTypeId="2" ParentId="2830" CreationDate="2014-10-12T19:34:24.730" Score="9" Body="&lt;p&gt;A piece that has $k$ squares of one color and $2k$ squares of the other color has $3k$ squares.  It is a multiple of $3$.  You cannot split $64$ squares in any number of parts that all have multiples of $3$ squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is: there is no way.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2484" LastEditDate="2014-10-13T09:16:17.597" LastActivityDate="2014-10-13T09:16:17.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2835" PostTypeId="2" ParentId="2831" CreationDate="2014-10-12T21:50:26.780" Score="2" Body="&lt;p&gt;I found a solution based on the other version of this problem: 12 queens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/urblC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also found a solution where other pieces are permitted: 32 queens:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/x55Qu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-10-12T22:19:08.627" LastActivityDate="2014-10-12T22:19:08.627" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2836" PostTypeId="1" CreationDate="2014-10-12T19:10:04.500" Score="10" ViewCount="3645" Body="&lt;p&gt;You are given 4 balls, all equal in weight except for one that is either heavier or lighter. You are also given a two-pan balance to use. In each use of the balance you may put any number of the 4 balls on the left pan, and the same number on the right pan, and push a button to initiate the weighing; there are three possible outcomes: either the weights are equal, or the balls on the left are heavier, or the balls on the left are lighter. Your task is to design a strategy to determine which is the odd ball and whether it is heavier or lighter than the others in as few uses of the balance as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My solution is 3 weighting. However, some body told me he can get down to 2 weighting. Can someone please confirm this?&lt;/p&gt;&#xA;" OwnerUserId="3878" OwnerDisplayName="kuku" LastEditorUserId="8874" LastEditDate="2015-01-31T12:24:08.917" LastActivityDate="2015-07-16T16:30:01.243" Title="Fewest possible weighings to determine which ball is heavier/lighter" Tags="|logical-deduction|weighing|" AnswerCount="7" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="2837" PostTypeId="2" ParentId="2836" CreationDate="2014-10-12T23:52:05.447" Score="-1" Body="&lt;p&gt;&lt;em&gt;This strategy involves a case where 3 weighings are required. I would contend that it is impossible to handle that case without 3 weighings.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick 2, A and B, and weigh them against each other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If A and B are equal, then I'm afraid that a solution cannot be found within the 2-weighing limit. Weigh C and D. One will be lighter or heavier - call C the lighter. Weigh A and C, if A and C are equal, then D is the odd one out and is heavier. If A and C are not equal, then C is the odd one out and is lighter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If A and B are not equal, call A the lighter one, then pick another one C, and weigh A and C. If A and C are equal, then B is the odd one out and is heavier. If A and C are not equal, then A is the odd one out and is lighter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Done, in 2 weighings minimum and 3 weighings maximum.&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-10-13T06:53:13.407" LastActivityDate="2014-10-13T06:53:13.407" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2838" PostTypeId="2" ParentId="2693" CreationDate="2014-10-13T00:14:03.913" Score="33" Body="&lt;p&gt;As &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2693/a-blanket-for-my-baby-snake/2711#comment6357_2693&quot;&gt;remarked by sth&lt;/a&gt;, this is an open problem, a topic of ongoing research. It is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Moser%27s_worm_problem&quot; rel=&quot;noreferrer&quot;&gt;Moser's worm problem&lt;/a&gt;. The &lt;a href=&quot;https://mathoverflow.net/&quot;&gt;Math Overflow&lt;/a&gt; thread &lt;a href=&quot;https://mathoverflow.net/questions/32477/smallest-area-shape-that-covers-all-unit-length-curve&quot;&gt;Smallest area shape that covers all unit length curve&lt;/a&gt; cites some recent result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A known result is that the smallest possible convex blanket has an area between 0.227498 to 0.231999. (In comparison, a &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2693/a-blanket-for-my-baby-snake/2695#2695&quot;&gt;half-circle of diameter 1&lt;/a&gt; has an area of about 0.3927.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference: Tirasan Khandhawit, Dimitrios Pagonakis, Sira Sriswasdi. &lt;em&gt;Lower Bound for Convex Hull Area and Universal Cover Problems&lt;/em&gt;. Int.J.Comput.Geom.Appl. 23 (2013) 197-212.&#xA;&lt;a href=&quot;http://arxiv.org/abs/1101.5638&quot; rel=&quot;noreferrer&quot;&gt;arXiv:1101.5638&lt;/a&gt;. DOI: &lt;a href=&quot;http://arxiv.org/ct?url=http%3A%2F%2Fdx.doi.org%2F10%252E1142%2FS0218195913500076&amp;amp;v=cbb3f697&quot; rel=&quot;noreferrer&quot;&gt;10.1142/S0218195913500076&lt;/a&gt;. &lt;a href=&quot;http://arxiv.org/pdf/1101.5638v1.pdf&quot; rel=&quot;noreferrer&quot;&gt;PDF&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the blanket is not required to be convex, smaller blankets may be possible. A special case of interest is if the snake only consists of two straight segments with a bend at a position that may vary, in which case the blanket can be a &lt;a href=&quot;http://en.wikipedia.org/wiki/Kakeya_set&quot; rel=&quot;noreferrer&quot;&gt;Kakeya set&lt;/a&gt; which can have zero area. It is even true if the snake consists of finitely many straight segments, see the works of &lt;a href=&quot;http://cms.math.ca/10.4153/CJM-1970-091-4&quot; rel=&quot;noreferrer&quot;&gt;Ward&lt;/a&gt; and of &lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=272988&quot; rel=&quot;noreferrer&quot;&gt;Davies&lt;/a&gt;.&#xA;For real-analytic curves, the area cannot be zero by a &lt;a href=&quot;http://dx.doi.org/10.1112/S0025579300009542&quot; rel=&quot;noreferrer&quot;&gt;work of Marstrand&lt;/a&gt;, however it appears to be not known if it can be arbitrarily small.&lt;/p&gt;&#xA;" OwnerUserId="71" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:31.357" LastActivityDate="2014-10-26T15:17:05.470" CommentCount="5" CommunityOwnedDate="2014-10-13T00:14:03.913" ContentLicense="CC BY-SA 3.0" />
  <row Id="2839" PostTypeId="2" ParentId="2831" CreationDate="2014-10-13T00:22:29.430" Score="10" Body="&lt;p&gt;Update:&lt;br /&gt;&#xA;I was looking through old puzzles and saw this one and was always annoyed I couldn't get a 14th queen. So I found a way.&lt;br /&gt;&#xA;14 queens:&lt;br /&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/xVDGm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xVDGm.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Original Answer:&lt;br /&gt;&#xA;I've managed 13 queens. I started in one corner and then put them on opposite diagonals. I then tweaked it so that all the queens were threatened twice. I was hoping for 14, but I'm starting to think 13 is the maximum.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TX0Is.png&quot; alt=&quot;13 Queens&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;I've tried a few different ways with other pieces. The most I've managed is 32. Here's an alternative example:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lcdg8.png&quot; alt=&quot;32 Queens&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="375" LastEditorUserId="375" LastEditDate="2023-07-10T15:26:50.217" LastActivityDate="2023-07-10T15:26:50.217" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="2842" PostTypeId="1" AcceptedAnswerId="2844" CreationDate="2014-10-13T07:26:33.980" Score="3" ViewCount="986" Body="&lt;p&gt;We're on the middle of the most devastating war the Earth has gone through all its history. You and your brother in arms are in a plane that has been taken down while crossing the enemies lines. You are okay, but your teammate has a head contussion and must be immediately brought back to your base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Luckily, you are such a great pilot that the aircraft has nearly no damage and can fly again, but one of the two fuel tanks was lost due to the accident. Furthermore, the on-board computer has stopped working so it can't help you. You want to know if you can get your friend back to the base with the fuel you have. In the middle of your calculations, you find that you can't remember any of the digits of $\pi$, which you need to calculate the distance from where you are to your base (taking the Earth as a perfect Sphere) in order to estimate the fuel needed. There is nothing inside the plane which can help you with the calculation, so you have to stick to the next items.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can &lt;strong&gt;ONLY&lt;/strong&gt; use to calculate it:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Your &lt;strong&gt;belt&lt;/strong&gt;, which has a &lt;strong&gt;length of 1 metre&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Some &lt;strong&gt;rocks&lt;/strong&gt; you found near the plane.&lt;/li&gt;&#xA;&lt;li&gt;A &lt;strong&gt;wooden stick&lt;/strong&gt;. You can use it to draw or anotate numbers on the ground.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Luckily, you can come up with a way to get a pretty good approximation of $\pi$ and be able to know if you can save your friend's live. Each second counts...&lt;/p&gt;&#xA;" OwnerUserId="2470" LastEditorUserId="4026" LastEditDate="2014-10-20T05:08:26.790" LastActivityDate="2014-10-20T14:48:28.783" Title="A belt, a stone and a wooden stick" Tags="|mathematics|" AnswerCount="5" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2843" PostTypeId="2" ParentId="2836" CreationDate="2014-10-13T07:36:45.410" Score="-4" Body="&lt;p&gt;Lets call the balls A,B,C,D.&#xA;Put AB on the left and CD on the right.&#xA;If AB is lighter/heavier put A on the left and B on the right and you find the exact one that is lighter/heavier.&#xA;Other case is CD is lighter/heavier then put C on the left and D on the right and you can find the lighter/heavier exactly.&#xA;So always in 2 weighins &lt;/p&gt;&#xA;" OwnerUserId="2465" LastActivityDate="2014-10-13T07:36:45.410" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2844" PostTypeId="2" ParentId="2842" CreationDate="2014-10-13T08:08:21.463" Score="2" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Lay your belt out flat and straight on the ground. It will act as the diameter. Count how many rocks you can lay across your belt. Call the number of rocks $x$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Make a circle out of rocks around the belt where the belt is the diameter. Pivot the belt around if this helps to make as close-to-perfect a circle as you can. Count the number of rocks required to make this circle. Call the number of rocks $y$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The remaining task is to find the ratio between $x$ and $y$. Perform long division with $y/x$. Using the stick on the sand to act as pen and paper. Calculate to $3$ digits. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;PS: Since you only need 3 digits, you could just use $22/7$ or $355/113$ and skip steps 1 and 2. However, if such tricks were allowed, then surely one would know $\pi$ to the first 3 digits regardless and the puzzle would be pointless.&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-10-13T08:29:54.680" LastActivityDate="2014-10-13T08:29:54.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2845" PostTypeId="2" ParentId="2842" CreationDate="2014-10-13T09:28:33.093" Score="0" Body="&lt;p&gt;Step 1. Draw a circle with the belt keeping the buckle as center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 2. Lay stones over the belt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 3. Lay stones along the circumference of the circle drawn with the belt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 4. Count the number of stones over the belt, &lt;code&gt;Say x&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 5. Count the number of stones along the circumference, &lt;code&gt;Say y&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 6. Now calculate  &lt;code&gt;π = y / (2 * x)&lt;/code&gt; with the help of the stick, scribbling on the ground. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By this methos he can find out the value of  π  using his belt, stones and a stick.&lt;/p&gt;&#xA;" OwnerUserId="2730" LastActivityDate="2014-10-13T09:28:33.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2846" PostTypeId="2" ParentId="2836" CreationDate="2014-10-13T10:00:26.303" Score="6" Body="&lt;p&gt;Label the balls ABCD. Weigh A and B. Then weigh A and C.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If A is the odd ball it will be lighter or heavier in both weighings and you have solved it&lt;/li&gt;&#xA;&lt;li&gt;If B is the odd ball, it will be lighter or heavier in the first weighing and the second weighing (which balances) will confirm it is B that is the odd ball&lt;/li&gt;&#xA;&lt;li&gt;If C is the odd ball, you will know A is standard so the second weighing gives the solution&lt;/li&gt;&#xA;&lt;li&gt;If D is the odd ball you will know this after the second weighing, because A, B, and C were all equal, but if you need to know whether it's lighter or heavier you cannot escape a third weighing. If it's enough to know which is different, two weighings are enough.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So you have a 75% chance of solving it completely in two weighings, and a 25% chance of only identifying the odd ball in two weighings, and needing a third to know if it is heavy or light. I think this is the best that can be done.&lt;/p&gt;&#xA;" OwnerUserId="1509" LastEditorUserId="1509" LastEditDate="2014-10-13T10:14:35.813" LastActivityDate="2014-10-13T10:14:35.813" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2847" PostTypeId="2" ParentId="2836" CreationDate="2014-10-13T10:10:46.563" Score="10" Body="&lt;p&gt;&lt;strong&gt;You can not do it in 2 weightings&lt;/strong&gt;.&lt;br&gt;&#xA;Let's prove it step by step.  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Initially you have 8 possible combinations: 1st ball is heavier, 1st ball is lighter, 2nd ball is heavier, .., 4th ball is lighter. So you must be able to set correspondence between outcomes or 2 weightings and 8 combinations.&lt;/li&gt;&#xA;&lt;li&gt;Each time you can put either 4 balls on the scale or 2 balls.&lt;/li&gt;&#xA;&lt;li&gt;Suppose you put 4 balls on the scale at least ones. In such a weighting only 2 outcomes would be possible: left is heavier or lighter. You can not have them equal since one and only one ball is not normal. In another weighting you can have maximum 3 outcomes. so in total you can 6 possible outcomes of 2 weightings, this is not enough to chose between 8 combinations.&lt;/li&gt;&#xA;&lt;li&gt;So each time you should put only 2 balls on the scale. &lt;/li&gt;&#xA;&lt;li&gt;Suppose you decided to compare A and B first, and C and D second. But at least one pair of the balls have equal weight. So you will have only 4 possible outcomes of such 2 weightings. This is not enough too.&lt;/li&gt;&#xA;&lt;li&gt;So you must do it differently. But then one of the balls will never be on the scale. Then, if it is not normal ball you won't be able to find out whether it is lighter or heavier.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;It is easy to do in 3 weightings.&lt;/strong&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/a/2837/28&quot;&gt;d'alar'cop already done it&lt;/a&gt;, but I would write it too for completeness of my answer.&lt;br&gt;&#xA;1.2. Weight A against B then C against D. Then you will know immediately 2 balls, which have normal weight. And two candidates for not normal ball. Suppose that A=B and C&gt;D. Then either C is not normal heavy ball or D is not normal light ball.&lt;br&gt;&#xA;3. Weight A+B against C+D. If C+D is heavier, then C is heavy ball. If C+D is lighter, then D is light ball.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;You can do it in 2 weightings if you have 5th ball known to be normal for sure.&lt;/strong&gt;. Let's call it E.&lt;br&gt;&#xA;You can do it like it was done &lt;a href=&quot;https://puzzling.stackexchange.com/a/1816/28&quot;&gt;here&lt;/a&gt; for more complex problem. You may need to read the explanations there to understand the following, but in principle it is quite self explanatory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ball's arrangement:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Ball  A  B  C  D  E -A -B -C -D&#xA;W1    L     L  R  R  R     R  L &#xA;W2       L  R  R  L     R  L  L&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Weighing schedule of balls:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Weighing 1:  A C / D E&#xA;Weighing 2:  B E / C D&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The outcomes interpretation:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;L = : AH  R = : AL  &#xA;= L : BH  = R : BL  &#xA;L R : CH  R L : CL&#xA;R R : DH  L L : DL&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="28" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2014-10-13T10:26:31.087" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="2849" PostTypeId="2" ParentId="2652" CreationDate="2014-10-13T11:00:59.940" Score="8" Body="&lt;p&gt;A water clock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A typical water clock might contain about $10^{25}$ water molecules. Compare to a paltry $10^{9}$ for an &lt;a href=&quot;https://puzzling.stackexchange.com/a/2659&quot;&gt;hourglass&lt;/a&gt;. (Give or take a couple of orders of magnitude)&lt;/p&gt;&#xA;" OwnerUserId="71" LastActivityDate="2014-10-13T11:00:59.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2850" PostTypeId="2" ParentId="2842" CreationDate="2014-10-13T11:58:24.190" Score="1" Body="&lt;p&gt;When thinking about this puzzle I had in mind a more probabilistic approach: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- Draw a $2\times 2$ square and inside it draw a circumference that is tangent to the sides of the square ($r=1$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2-Start throwing stones to the square from the same position and keep a count on how many have you thrown ($N$) and how many landed inside the circumference ($n$). We assume that all of them land inside the square. If not, do not count that throw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3- $\frac{n}{N}=\frac{A_{circumference}}{A_{square}}=\frac{\pi}{4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4-Approximate $\pi=\frac{4n}{N}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more stones you throw the more accurate the result is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the method explained in the accepted solution optimizes the time spent in the calculation so it is a better solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matematical idea: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We assume (and try to guarantee) that everywhere in the square has the same probability of being the place where a stone lands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is true, then the probability of a stone landing inside the circle equals the ratio $\frac{A_c}{A_s}$ which is the fraction of area we count as a valid result divided by the total area. This seems intuitive as the bigger the square is in relation to the circle, the less likely a stone will land in it. And viceversa (Nearly equal areas implies nearly a 100% probability of landing inside the circle.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we just have to put this reasoning into practise. By throwing $N$ stones and keeping a count of the ones that fall inside the circle ($n$) we'll get an empirical probability of a stone landing inside the circle ($\frac{n}{N}$) As the theoretical result should be the same as the empirical we can work with the equation in step 3 and get an approximation of $\pi$.&lt;/p&gt;&#xA;" OwnerUserId="2470" LastEditorUserId="2470" LastEditDate="2014-10-13T14:33:11.887" LastActivityDate="2014-10-13T14:33:11.887" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2851" PostTypeId="2" ParentId="2842" CreationDate="2014-10-13T13:29:21.080" Score="3" Body="&lt;p&gt;You should ask your friend whether he remembers pi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need to make navigation calculations you can probably use the handy fact that 1 angle minute (0°1') is exactly 1 nautical mile or 1852 km.  If you know your coordinates and the destination coordinates, you should get a good estimate of how far to fly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm sure you won't leave before you figured out pi.  For that, make a square of 200x200 stones.  With the belt, measure the side S, and draw an arc of radius S from one of the corners.  This gives a 1/4 circle in a square.  The number of stones in the 1/4 circles should be 31416.  Divide by 10000.  (A few stones will be exactly on the circle.  Count 1/2 for these).&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2014-10-20T14:48:28.783" LastActivityDate="2014-10-20T14:48:28.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2852" PostTypeId="1" AcceptedAnswerId="2853" CreationDate="2014-10-13T13:43:45.270" Score="6" ViewCount="1802" Body="&lt;p&gt;The little old man has twelve children: some short, some long, some cold, some hot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who is he?&lt;/p&gt;&#xA;" OwnerUserId="432" LastActivityDate="2015-01-01T12:55:57.227" Title="Who is this old man?" Tags="|riddle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2853" PostTypeId="2" ParentId="2852" CreationDate="2014-10-13T13:50:34.283" Score="8" Body="&lt;p&gt;Sounds like a year that's ending. 12 months, some months are short, some are long, some are cold, some are hot. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2014-10-13T13:50:34.283" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="2854" PostTypeId="1" AcceptedAnswerId="2856" CreationDate="2014-10-13T13:54:17.200" Score="2" ViewCount="1678" Body="&lt;p&gt;Warning: To answer this riddle you need to have a thorough understanding of how time works.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Two twins were born during the same year, on the same month, and the same day, exactly 10 minutes apart. By the time twin A was 100, twin B was younger by more than 10 minutes, fairly slightly. Neither twin died during the 100 years.&lt;/p&gt;&#xA;&lt;p&gt;How could Twin B be younger than Twin A by more than 10 minutes?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-02T12:30:09.477" Title="The Twins with a Different Age" Tags="|physics|calculation-puzzle|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2855" PostTypeId="2" ParentId="2854" CreationDate="2014-10-13T14:04:02.217" Score="1" Body="&lt;p&gt;Twin A has since moved to the other side of the world, with the timezone differences they technically differ in age by 12h10m or 11h50m, depends on where they started :)&lt;/p&gt;&#xA;" OwnerUserId="2463" LastActivityDate="2014-10-13T14:04:02.217" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2856" PostTypeId="2" ParentId="2854" CreationDate="2014-10-13T14:05:02.970" Score="8" Body="&lt;p&gt;Due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Time_dilation&quot;&gt;time dilation&lt;/a&gt;, moving faster will make you age slightly slower. This is most commonly mentioned with astronauts and space calculations, but it can also be applied to lower speeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technically, if Twin A normally drives his car at 50mph over the course of his lifetime because he wants to save fuel, whereas Twin B dashes around at 70mph because he's always running late, Twin A will have experienced more time than Twin B, so Twin B would be slightly more than 10 minutes younger after a century.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to Clint's comment below, there is a similar effect that could happen - &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_time_dilation&quot;&gt;gravitational time dilation&lt;/a&gt;. If Twin B lives in Washington DC (250ft above sea level) but Twin A moved to Denver (5000ft above sea level) at a young age, time will pass slightly more slowly in Washington because it's closer to the gravitational centre of the earth, again resulting in Twin B having experienced less time.&lt;/p&gt;&#xA;" OwnerUserId="2518" LastEditorUserId="2518" LastEditDate="2014-10-13T14:35:49.370" LastActivityDate="2014-10-13T14:35:49.370" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2857" PostTypeId="2" ParentId="2854" CreationDate="2014-10-13T14:05:10.850" Score="1" Body="&lt;p&gt;It just happens. Depends on how much and how fast you travel, I guess. If you travel more often, you will be younger. The faster you travel, the better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other possibility is that the earth just significantly slowed spinning, so a day is longer, which means that a minute is longer. But this only works with the old definition of minute as a fraction of the mean solar day, not if you take the newer one, based on cesium 133.&lt;/p&gt;&#xA;" OwnerUserId="2544" LastActivityDate="2014-10-13T14:05:10.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2858" PostTypeId="1" AcceptedAnswerId="2860" CreationDate="2014-10-13T15:51:02.877" Score="4" ViewCount="764" Body="&lt;p&gt;Queen is combination of Bishop and Rook and &lt;a href=&quot;http://en.wikipedia.org/wiki/Amazon_(chess)&quot; rel=&quot;nofollow&quot;&gt;Amazon&lt;/a&gt; is combination of queen and knight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is how many Amazons can place on a standard chessboard that non of them attack each other without any other pieces allowed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's obvious that it should be a smaller number than 9. Also 8 is impossible too. Because all 8-Queen solutions has at least two queens with knight's move position. So is it possible to fit 7 Amazons?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="2225" LastEditDate="2014-10-13T16:10:52.403" LastActivityDate="2014-10-13T17:47:26.440" Title="How many Amazons on chessboard?" Tags="|chess|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2860" PostTypeId="2" ParentId="2858" CreationDate="2014-10-13T17:47:26.440" Score="10" Body="&lt;p&gt;The answer is 6.  I am definitely not the first to say that but my search just ended.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cyX8g.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution above is easy to find and contains 6 amazons. I had done this by hand but my program found the same (but rotated) solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Excel I wrote the following Macro:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Sub Playz()&#xA;&#xA;While Range(&quot;D2&quot;).Value &amp;lt; 7&#xA;    X = Range(&quot;D1&quot;).Value + 1&#xA;    C = Range(&quot;D2&quot;).Value&#xA;    While checkz(X) And X &amp;lt; 64&#xA;        X = X + 1&#xA;    Wend&#xA;    If X = 64 Then&#xA;        Range(&quot;D1&quot;).Value = Range(&quot;C&quot; + CStr(Range(&quot;D2&quot;).Value)).Value&#xA;        Range(&quot;C&quot; + CStr(Range(&quot;D2&quot;).Value)).Clear&#xA;    Else&#xA;        Range(&quot;C&quot; + CStr(Range(&quot;D2&quot;).Value + 1)).Value = X&#xA;        Range(&quot;D1&quot;).Value = X&#xA;    End If&#xA;Wend&#xA;End Sub&#xA;&#xA;Function checkz(ByVal N As Integer) As Boolean&#xA;&#xA;X = N Mod 8&#xA;Y = (N - X) / 8&#xA;&#xA;J = 0&#xA;While J &amp;lt; 64&#xA;If Range(&quot;B&quot; + CStr(J + 1)).Value And J &amp;lt;&amp;gt; N Then&#xA;    S = J Mod 8&#xA;    T = (J - S) / 8&#xA;    If X = S Or Y = T Or X + Y = S + T Or X - Y = S - T Then&#xA;        checkz = True&#xA;        Exit Function&#xA;    End If&#xA;    If Abs(X - S) = 2 And Abs(Y - T) = 1 Then&#xA;        checkz = True&#xA;        Exit Function&#xA;    End If&#xA;    If Abs(X - S) = 1 And Abs(Y - T) = 2 Then&#xA;        checkz = True&#xA;        Exit Function&#xA;    End If&#xA;End If&#xA;J = J + 1&#xA;Wend&#xA;checkz = False&#xA;End Function&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The b column contains a counter to identify which cells are filled and which aren't. Cell D2 counts the number of cells in column C. It is much more complicated than it needs to be but was programmed originally for a more difficult puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The code looks for the next free unattacked square.  If it cannot find one, it removes the largest one so far, and tries to find a valid larger one than that.  This is an efficient algorithm for searching for a way to arrange chess pieces as long as the removal of one piece from an acceptable arrangement still yields an acceptable arrangement (ignoring the target number of pieces on the board).  It could not find a solution for seven as Cell D1 reached 63.&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2014-10-13T17:47:26.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2861" PostTypeId="1" AcceptedAnswerId="2862" CreationDate="2014-10-13T20:16:06.473" Score="4" ViewCount="2220" Body="&lt;p&gt;How many knights can you fit on a chess board without any being threatened?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2101" LastEditDate="2014-10-15T21:03:15.460" LastActivityDate="2014-11-18T12:22:09.623" Title="How Many Knights on a Chess Board?" Tags="|chess|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2862" PostTypeId="2" ParentId="2861" CreationDate="2014-10-13T20:26:19.737" Score="13" Body="&lt;p&gt;32 knights: all on white or all on black. When a knight moves, it moves to a space of a different color. If they are all on the same color, no collision happens. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AZPW8.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="5556" LastEditDate="2014-11-18T12:22:09.623" LastActivityDate="2014-11-18T12:22:09.623" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2863" PostTypeId="1" AcceptedAnswerId="2870" CreationDate="2014-10-13T21:12:29.603" Score="5" ViewCount="925" Body="&lt;p&gt;How many squares can you control with only one set of chess pieces of the same color? (8 Pawns, 2 Knights, 2 Bishops, 2 Rook, Queen and King)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Pieces cannot attack (control) the square they are currently occupied.&lt;/p&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="2225" LastEditDate="2014-10-13T21:19:39.070" LastActivityDate="2014-10-14T00:30:00.710" Title="How many squares can standard chess pieces control?" Tags="|chess|" AnswerCount="1" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2864" PostTypeId="1" AcceptedAnswerId="2869" CreationDate="2014-10-13T21:20:51.703" Score="23" ViewCount="3125" Body="&lt;p&gt;Is it possible to order a complete set of chess pieces of the same color (8 Pawns, 2 Knights, 2 Bishops, 2 Rook, Queen and King) in such a way that none of them attack each other? If so, how?&lt;/p&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="5373" LastEditDate="2015-07-15T15:22:07.900" LastActivityDate="2023-01-09T09:07:47.297" Title="The separation of chess pieces" Tags="|chess|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2865" PostTypeId="1" AcceptedAnswerId="15259" CreationDate="2014-10-13T22:07:29.167" Score="8" ViewCount="591" Body="&lt;p&gt;Similar to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2861/how-many-knights-on-a-chess-board#&quot;&gt;How Many Knights on a Chess Board?&lt;/a&gt; however speedy knights move the way the knight moves, then the way the knight moves AGAIN. However they cannot backtrack to their starting position. How many can you place on a board so to as none are being threatened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, this 1 speedy knight, attacks all these kings:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hSbqf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2015-05-21T16:23:34.273" Title="How Many Speedy Knights on a Chess Board?" Tags="|chess|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2869" PostTypeId="2" ParentId="2864" CreationDate="2014-10-14T00:08:24.610" Score="22" Body="&lt;p&gt;One could make the argument that the 4 pawns on the top row should be promoted, but other than that, 16 white pieces with none attacking each other, with proper bishop placement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2wBs2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="375" LastActivityDate="2014-10-14T00:08:24.610" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2870" PostTypeId="2" ParentId="2863" CreationDate="2014-10-14T00:23:25.680" Score="7" Body="&lt;p&gt;I believe this is &lt;em&gt;relatively&lt;/em&gt; easy and that many combinations will exist. This seems to be an example of such a position:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jZmIJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer the question, the set of pieces for one side can attack &lt;em&gt;all&lt;/em&gt; squares (and then some).&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-10-14T00:30:00.710" LastActivityDate="2014-10-14T00:30:00.710" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2871" PostTypeId="2" ParentId="2865" CreationDate="2014-10-14T00:28:37.377" Score="7" Body="&lt;p&gt;I believe I can fit 8 speedy-knights on a board. Indeed, obviously you'd have at least of one colour. Then you'll notice that there's always that pesky diagonal that a knight has trouble getting to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VCBsI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me know if this is what you meant.&lt;/p&gt;&#xA;" OwnerUserId="2484" LastActivityDate="2014-10-14T00:28:37.377" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2872" PostTypeId="1" AcceptedAnswerId="2971" CreationDate="2014-10-14T00:50:00.273" Score="15" ViewCount="5307" Body="&lt;p&gt;You may only use pieces in the original set, and all your pieces must be the same color. The king is &lt;strong&gt;not&lt;/strong&gt; allowed. Pawns do &lt;strong&gt;not&lt;/strong&gt; get promoted. Pieces do not control the square they occupy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accepted Answer goes to the person that has the least score.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Piece cost:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Pawn - 1&lt;/li&gt;&#xA;&lt;li&gt;Knight - 3&lt;/li&gt;&#xA;&lt;li&gt;Bishop - 3&lt;/li&gt;&#xA;&lt;li&gt;Rook - 5&lt;/li&gt;&#xA;&lt;li&gt;Queen - 9&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="65277" LastEditDate="2021-02-11T03:31:54.980" LastActivityDate="2024-04-05T20:10:51.247" Title="How many Chess Pieces are needed to control every square on the board?" Tags="|logical-deduction|chess|optimization|" AnswerCount="5" CommentCount="38" ContentLicense="CC BY-SA 3.0" />
  <row Id="2873" PostTypeId="2" ParentId="2872" CreationDate="2014-10-14T01:13:56.350" Score="12" Body="&lt;p&gt;The best I've got at the moment is a cost-34 solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ITKMF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Due to clarifications, I have found a cost-35 solution that uses the pieces from one side (no king):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/o2WXE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was surprisingly hard, and I'd really like to see better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Old answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a solution (no restriction on pieces - only goal is to minimise cost) that costs 30. It is actually from &lt;a href=&quot;http://en.wikipedia.org/wiki/Mathematical_chess_problem#Domination_problems&quot; rel=&quot;noreferrer&quot;&gt;wikipedia&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/j6X9d.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-10-16T14:35:23.337" LastActivityDate="2014-10-16T14:35:23.337" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="2874" PostTypeId="2" ParentId="2872" CreationDate="2014-10-14T02:37:13.863" Score="1" Body="&lt;h3&gt;&lt;em&gt;Invalid answer&lt;/em&gt;, need to be updated to conform with latest edits in the question&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I found some other solutions which cost 30. I'm posting this in the hope someone else can improve any of these =)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0lraq.png&quot; alt=&quot;Cost-30-attack-all-squares&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/xrbLp.png&quot; alt=&quot;Cost-30-attack-all-squares-2&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/IIgWx.png&quot; alt=&quot;Cost-30-attack-all-squares-3&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found a near-solution which costs 28 (A6 and H6 are not covered):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Oh0IQ.png&quot; alt=&quot;Cost-28-attack-nearly-all-squares&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1649" LastEditorUserId="1649" LastEditDate="2014-10-15T02:46:09.980" LastActivityDate="2014-10-15T02:46:09.980" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2877" PostTypeId="1" AcceptedAnswerId="2882" CreationDate="2014-10-14T09:02:25.690" Score="6" ViewCount="750" Body="&lt;p&gt;Pirates have buried treasure on a very unusual perfectly circular desert island, with no trees or obstacles of any kind. The map takes me to the island but doesn't tell me where to find it on the island. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It only tells me that it's buried under a small x marked on the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tide is coming in fast so I have to find my treasure as soon as possible, or risk drowning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The island is 1km across in any direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can only spot the small mark from 10 meters away (pirates were often short sighted).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the most efficient way to search the island? Should I walk around in a spiral? Cut the island into segments and cross the center many times? What's the best search pattern?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I should probably mention that I can use the treasure chest as a boat and won't drown if I find it.&lt;/p&gt;&#xA;" OwnerUserId="2673" LastEditorUserId="2673" LastEditDate="2014-10-14T10:15:24.140" LastActivityDate="2014-10-14T13:54:11.880" Title="Finding treasure on a circular island" Tags="|geometry|" AnswerCount="1" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2880" PostTypeId="2" ParentId="2854" CreationDate="2014-10-14T12:35:20.880" Score="2" Body="&lt;p&gt;Time dilation is possible, and was my first thought, but I have an easier answer.  Twin A and Twin B were born 42 years apart.  They also have no relation to the two twins that were born ten minutes apart.&lt;/p&gt;&#xA;" OwnerUserId="2698" LastActivityDate="2014-10-14T12:35:20.880" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2882" PostTypeId="2" ParentId="2877" CreationDate="2014-10-14T13:54:11.880" Score="6" Body="&lt;p&gt;I believe the concentric circles would be the best way to cover the ground. Depending on the size of the &quot;x&quot; a spiral would cause you to miss a portion of the island unless you started the spiral off the island. Starting 10 meters in at a radius 490 meters you would make a circle and step in 20 meters and repeat until you are 10 meters from the center where your last circle would cover the rest.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;for (int R = 490; R &amp;gt;= 10; R-=20)&#xA;    total+=(2*Pi*R) + 20)&#xA;&#xA;return total-10&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;subtracting 10 because we only had to walk in 10 meters for the first circle results in&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;39759.91 meters with 25 concentric circles.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2803" LastActivityDate="2014-10-14T13:54:11.880" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2885" PostTypeId="2" ParentId="2872" CreationDate="2014-10-14T14:25:13.423" Score="3" Body="&lt;p&gt;I found a nice solution, which definitely isn't perfect, but could evolve in a game (only pieces of one colour):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ra8&#xA;Rh1&#xA;Qc3&#xA;Bd4&#xA;Bd5&#xA;Ne4&#xA;Ng3&#xA;Kf3&#xA;and pawns on c4, d6, e6, f5&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; +--+--+--+--+--+--+--+--+&#xA;8|R*|  |  |  |  |  |  |  |&#xA; +--+--+--+--+--+--+--+--+&#xA;7|  |  |  |  |  |  |  |  |&#xA; +--+--+--+--+--+--+--+--+&#xA;6|  |  |  |p*|p*|  |  |  |&#xA; +--+--+--+--+--+--+--+--+&#xA;5|  |  |  |B*|  |p*|  |  |&#xA; +--+--+--+--+--+--+--+--+&#xA;4|  |  |p*|B*|N*|  |  |  |&#xA; +--+--+--+--+--+--+--+--+&#xA;3|  |  |Q*|  |  |K*|N*|  |&#xA; +--+--+--+--+--+--+--+--+&#xA;2|  |  |  |  |  |  |  |  |&#xA; +--+--+--+--+--+--+--+--+&#xA;1|  |  |  |  |  |  |  |R*|&#xA; +--+--+--+--+--+--+--+--+&#xA;  a  b  c  d  e  f  g  h&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As I said, it isn't perfect, the score is 35 (39 - 4) and the King plays an active role in protecting g2 and g4. Still, assuming white pieces, the black king has nowhere to go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After making the rules better understandable, and shooting the king, I have another solution that fits. Since the King is definitely more powerful than a pawn, albeit more vulnerable, I had to add a few pawns but could eliminate a knight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pieces are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ra8&#xA;Rh1&#xA;Qc3&#xA;Nc7&#xA;Be3&#xA;Bf3&#xA;and pawns on c5, d2, d6, e2, e6, f2, f5, g4&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; +--+--+--+--+--+--+--+--+&#xA;8|R*|  |  |  |  |  |  |  |&#xA; +--+--+--+--+--+--+--+--+&#xA;7|  |  |N*|  |  |  |  |  |&#xA; +--+--+--+--+--+--+--+--+&#xA;6|  |  |  |p*|p*|  |  |  |&#xA; +--+--+--+--+--+--+--+--+&#xA;5|  |  |p*|  |  |p*|  |  |&#xA; +--+--+--+--+--+--+--+--+&#xA;4|  |  |  |  |  |  |p*|  |&#xA; +--+--+--+--+--+--+--+--+&#xA;3|  |  |Q*|  |B*|B*|  |  |&#xA; +--+--+--+--+--+--+--+--+&#xA;2|  |  |  |p*|p*|p*|  |  |&#xA; +--+--+--+--+--+--+--+--+&#xA;1|  |  |  |  |  |  |  |R*|&#xA; +--+--+--+--+--+--+--+--+&#xA;  a  b  c  d  e  f  g  h&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This makes a score of 36 (39 - 3). Not a dream score but better than nothing.&lt;/p&gt;&#xA;" OwnerUserId="3905" LastEditorUserId="3905" LastEditDate="2014-10-16T08:47:33.233" LastActivityDate="2014-10-16T08:47:33.233" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="2889" PostTypeId="2" ParentId="2864" CreationDate="2014-10-14T16:59:54.810" Score="55" Body="&lt;p&gt;I give you the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9iAUy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This solution can arise in a regular chess game, where the black king is superfluous. The conditions would still be met if the black king were removed. White needs to make 25 moves (Queenside castle, thanks to h34 in comments) to get into this position, but due to the need of capturing all opposing pieces, a game leading to this position would probably take more turns.&lt;/p&gt;&#xA;" OwnerUserId="1831" LastEditorUserId="1831" LastEditDate="2019-03-07T10:02:54.433" LastActivityDate="2019-03-07T10:02:54.433" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="2890" PostTypeId="2" ParentId="236" CreationDate="2014-10-14T17:51:42.730" Score="61" Body="&lt;p&gt;Let's continue the induction, since the jump to 99 blue eyes does seem weird. After all, everyone knows that someone has blue eyes.&lt;/p&gt;&#xA;&lt;p&gt;If there are 4 blue eyed-people, A will look at B,C,D, thinking :&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Maybe I don't have blue eyes (only 3 blue eyes?). In this case B must be thinking, that he may not have blue eyes either, and B is looking at C and D, whom he perceives as being the only ones to have blue eyes (since I consider the option that I don't have blue eyes), and B thinks that C is having the same reasoning. C thinks he does not have blue eyes and only D has.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Now, the issue here is that I, being A, can see that B has blue eyes. Therefore I know that C sees at least D and B as having blue eyes. But this is the reasoning of B, who does not know that he has blue eyes.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; When I project myself into the reasoning of the next person, I cannot use the knowledge I have of their eye color.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The same goes for 5 people and more. I see 4 blue-eyed people, each of which is possibly seeing only 3, and thinking that each of the other is possibly seeing only 2...&lt;/p&gt;&#xA;" OwnerUserId="310" LastEditorUserId="46002" LastEditDate="2020-07-03T06:05:12.723" LastActivityDate="2020-07-03T06:05:12.723" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="2894" PostTypeId="2" ParentId="2852" CreationDate="2014-10-14T21:28:03.753" Score="5" Body="&lt;p&gt;You accepted an answer, but it doesnt answer the question you asked:  Who is the old man?  Its Father Time.&lt;/p&gt;&#xA;" OwnerUserId="2758" LastActivityDate="2014-10-14T21:28:03.753" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2897" PostTypeId="1" AcceptedAnswerId="2899" CreationDate="2014-10-14T22:41:42.717" Score="5" ViewCount="1022" Body="&lt;p&gt;How many pieces can you organize on a Chess board without one side being attacked by the other? This scoring will be used to prove you have the highest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Piece cost:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Pawn - 1&lt;/li&gt;&#xA;&lt;li&gt;Knight - 3&lt;/li&gt;&#xA;&lt;li&gt;Bishop - 3&lt;/li&gt;&#xA;&lt;li&gt;Rook - 5&lt;/li&gt;&#xA;&lt;li&gt;Queen - 9&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rules:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Both sides must have the same pieces on the board&lt;/li&gt;&#xA;&lt;li&gt;Your score is totalled by only 1 side&lt;/li&gt;&#xA;&lt;li&gt;One color cannot attack the other&lt;/li&gt;&#xA;&lt;li&gt;No king allowed.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2014-10-15T20:07:29.947" Title="How Many Pieces from Both Sides can you Place on a Chess Board Without Attacking a Side?" Tags="|chess|" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2899" PostTypeId="2" ParentId="2897" CreationDate="2014-10-14T23:54:48.057" Score="15" Body="&lt;p&gt;From the bottom up:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three rows of 8 black queens&lt;br&gt;&#xA;One row of black pawns&lt;br&gt;&#xA;One row of white pawns&lt;br&gt;&#xA;Three rows of white queens&lt;/p&gt;&#xA;&#xA;&lt;p&gt;224.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the black pawns attack down and the white pawns attack up, they can be placed next to each other without attacking each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qBY5c.png&quot; alt=&quot;224 Points&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="375" LastEditorUserId="375" LastEditDate="2014-10-15T01:46:12.377" LastActivityDate="2014-10-15T01:46:12.377" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="2900" PostTypeId="1" AcceptedAnswerId="2932" CreationDate="2014-10-15T02:14:10.077" Score="23" ViewCount="2905" Body="&lt;p&gt;I was helping Ernie clean out his shed a while back when I saw a curious old machine. &quot;Oh yeah, that!&quot; said Ernie, when I pointed it out to him. &quot;A few years back, before transistors were invented, I tried to make a clockwork calculator. It worked, but was never a commercial success.&quot; The machine consisted two rows of six sockets, three buttons (labelled A, M, and E) and a short strip of paper emerging from a narrow slot on one side. &quot;Didn't need a key-pad,&quot; said Ernie, &quot;instead you would put a plug for each numeral into the sockets, and then press A or M to add or multiply the numbers represented on the two rows. The answer would be printed out on the paper strip. Internally it calculated the units digit of the result first, then the tens, hundreds, etc. It was a bit slow, so I got it to print out each digit as soon as it could, so the digits of the answer were printed out in reverse order.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had a closer look, the machine had spent a bit too much time sitting under a leaking window, and the bottom row of sockets had rusted out entirely. And the box of plugs only contained one 4, one 6, two 0's, three each of 1, 5 and 7, five each of 8, and 9, and six 2's. There were no 3's left at all. &quot;Yep, looking back I can see it would never have caught on - too easy to lose digits. That's why I never built subtraction or division into it&quot;, Ernie commented. &quot;No chance of demonstrating it properly now, but if I swap a couple of cogs over...&quot;, Ernie delved into a back panel with a screw-driver for a few seconds, &quot;...I should be able to get something out of it&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ernie put plugs for 8 and 6 in the 10's 1's slots of the machine, wound the key a couple of turns and hit the button marked A. After a few seconds of humming, a length of paper strip emerged with the digits 2, 7, and 1 on it. &quot;See,&quot; he said, &quot;now it reads the digits of both numbers from the top row of sockets. See, 86 + 86 = 172&quot;. He added another plug to the hundreds to make 786, gave the key a few more turns and pressed the M button. After a longer pause (and more humming), the digits 6, 9, 7, 7, 1, and 6 emerged in turn on the paper strip. I made a quick calculation on the back of an envelope. &quot;Wow!&quot;, I exclaimed, &quot;786*786 = 617,796. It works!&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ADKPK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&quot;Have a go yourself.&quot; said Ernie. I removed the plugs from the sockets on the top row, replaced them with some more, and turned the key until it felt tight. &quot;What does the E button do?&quot; I asked. &quot;Oh, Exponentiation.&quot; he said, &quot;it used to calculate the first row to the power of the second row, so now it will calculate n^n&quot;. I pressed the button. &quot;Can't imagine it completing that calculation before the clockwork runs out&quot; Ernie exclaimed. &quot;And even if it does, I'm sure the paper won't be long enough to print it out&quot;. We watched for a couple of minutes, but nothing seemed to happen - the machine hummed gently but no numbers were printed out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After doing more tidying around the shed for a while (still no print-out), we went back to the house for coffee. When we returned, Ernie was proven right, the clockwork calculator had run down and was humming no longer. But before it stopped, it had printed out a string of digits. To my surprise, the digits, in the order they had been printed were exactly the same as the plugs (reading left to right) that I had inserted into the upper sockets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't remember exactly what the number I used was, but do remember that at the time I worked out that the digits were all correct and I had chosen the largest possible number that could be entered into the calculator with that result. Can you help me remember what the number was? &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In other words, we're looking for an $n$, consisting of no more than $6$ of the following digits: $\{0,0,1,1,1,2,2,2,2,2,2,4,5,5,5,6,7,7,7,8,8,8,8,8,9,9,9,9,9\}$, such that $n^n$ ends with the digits of $n$ in reverse order.&lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="39" LastEditDate="2014-11-11T22:11:38.677" LastActivityDate="2016-02-22T18:50:08.997" Title="Ernie's Clockwork Calculator" Tags="|mathematics|" AnswerCount="4" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="2901" PostTypeId="1" AcceptedAnswerId="2902" CreationDate="2014-10-15T03:46:13.480" Score="5" ViewCount="1204" Body="&lt;p&gt;Ok, so my friends and I had a disagreement over this seemingly simple problem and we would very much like to know the answer for it.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;h3&gt;Question&lt;/h3&gt;&#xA;&lt;p&gt;Initially a 100 kgs watermelon is 96pc water and 4pc pulp. After keeping it out in the sun for (say X) hours, the watermelon is now 95pc water. What is the weight of the watermelon now?&lt;/p&gt;&#xA;&lt;h3&gt;Answer 1&lt;/h3&gt;&#xA;&lt;p&gt;The mass of watermelon pulp initially was 4kgs. So equating that to 5pc of the current distribution would be 5/100*weight = 4kgs =&amp;gt; X = 80Kgs&lt;/p&gt;&#xA;&lt;h3&gt;Answer 2&lt;/h3&gt;&#xA;&lt;p&gt;Assume it's a 100kg watermelon. Answer 1 is wrong as only 95pc of the 96pc water is now available. You've lost 5pc water of 96pc = 5/100* (96pc *100) = 4.8kg water. So water's total weight now is 96 - 4.8 = 91.2 kg + 4kg pulp = 95.2kgs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Can someone please explain? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3913" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-10-15T11:38:40.447" Title="Percentages and the watermelon size reduction" Tags="|logical-deduction|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2902" PostTypeId="2" ParentId="2901" CreationDate="2014-10-15T04:01:40.227" Score="4" Body="&lt;p&gt;Ok first you need to clarify if the original weight is 20 kgs or 100kgs, assuming 100.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was 96% water and 4% pulp i.e. 96kg and 4kg, after keeping in sun, assuming only water evaporates, its 95% water and 5% pulp.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculating by pulp, 5% is 4Kg so 100% is 80Kgs&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculating by water, 95% is 95/100* 80 = 76 kgs, coming to what percentage of water was lost you cannot calculate that at all using the two numbers 96 and 95 as they are percentages of different weights.  &lt;/p&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="2313" LastEditDate="2014-10-15T11:38:40.447" LastActivityDate="2014-10-15T11:38:40.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2903" PostTypeId="2" ParentId="2901" CreationDate="2014-10-15T04:01:48.027" Score="1" Body="&lt;p&gt;Originally the 20 kg watermelon consisted 19.2 kg of water (20*96%/100%) and 0.8 kg of solids (20kg*4%/100%). Assuming only water can evaporate from the melon, the solids still weigh 0.8 kg. If they make up 5% of the melon then the total weight must be w=(100%/5%)*0.8kg = 16 kg, so both your answers are wrong. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you began with 100 kg of melon then answer 1 would be correct.&lt;/p&gt;&#xA;" OwnerUserId="1631" LastActivityDate="2014-10-15T04:01:48.027" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2904" PostTypeId="2" ParentId="2900" CreationDate="2014-10-15T04:43:55.000" Score="1" Body="&lt;p&gt;I just brute forced it, and it seems to me that &quot;1&quot; may be the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\max\ n.\  (first6digits(n^n).reverse = n)\ .0&amp;lt;n&amp;lt;999998. = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just for the record:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$first6digits(n^n) = \lfloor10^{((n\ log_{10}n) \mod 1)+5}\rfloor$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's computationally quite cheap, one just needs a data type with enough precision. And it turns out that $1$ was correct given a particular (outdated) interpretation of the question.&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-10-15T22:46:07.147" LastActivityDate="2014-10-15T22:46:07.147" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2905" PostTypeId="1" CreationDate="2014-10-15T05:10:22.353" Score="3" ViewCount="666" Body="&lt;p&gt;A wealthy person has passed away, naming three siblings as beneficiaries of the vast estate. Unfortunately, the will states that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My estate shall be divided evenly, such that no sibling shall feel their share is smaller than any other share.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As the estate contains many illiquid assets of sentimental value, one cannot simply convert everything into monetary value and numerically divide. For simplicity however, assume that all assets can be arbitrarily divided.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it turns out, the three siblings came together and, after some back-and-forth, arrived at a partition where no one sibling felt they received a smaller share than any other, fulfilling the will.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did they arrive at their partition?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is a well known puzzle, which is usually posed as sharing a cake, and how to cut it. Although the solution is simple to understand, it might not be so easy to independently come up with it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="104" LastActivityDate="2015-03-10T23:17:09.420" Title="The three inheritors" Tags="|logical-deduction|" AnswerCount="5" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2907" PostTypeId="1" AcceptedAnswerId="2935" CreationDate="2014-10-15T09:44:09.087" Score="44" ViewCount="7293" Body="&lt;p&gt;You may use as many pieces as you like. Pieces do not control the square they occupy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accepted Answer goes to the person that has the least score.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Piece cost:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Pawn - 1&lt;/li&gt;&#xA;&lt;li&gt;Knight - 3&lt;/li&gt;&#xA;&lt;li&gt;Bishop - 3&lt;/li&gt;&#xA;&lt;li&gt;Rook - 5&lt;/li&gt;&#xA;&lt;li&gt;Queen - 9&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note1&lt;/strong&gt;: The king has no score, it was implied that it is not allowed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note2&lt;/strong&gt;: I meant to imply only 1 colour was allowed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes this is &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2872/how-many-chess-pieces-are-needed-to-control-every-square-on-the-board?noredirect=1#comment7788_2872&quot;&gt;How many Chess Pieces are needed to control every square on the board?&lt;/a&gt; without the set restriction.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="65277" LastEditDate="2021-02-11T03:31:40.253" LastActivityDate="2024-04-06T19:06:30.360" Title="How many Chess Pieces are needed to control every square on the board? No Piece Restriction" Tags="|chess|optimization|" AnswerCount="8" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="2908" PostTypeId="2" ParentId="2907" CreationDate="2014-10-15T09:46:44.007" Score="8" Body="&lt;p&gt;I have a solution that costs 30. It is actually from &lt;a href=&quot;http://en.wikipedia.org/wiki/Mathematical_chess_problem#Domination_problems&quot; rel=&quot;noreferrer&quot;&gt;wikipedia&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/j6X9d.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2484" LastActivityDate="2014-10-15T09:46:44.007" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2909" PostTypeId="2" ParentId="2907" CreationDate="2014-10-15T09:50:24.277" Score="10" Body="&lt;p&gt;I found some other solutions which cost &lt;span class=&quot;math-container&quot;&gt;$30$&lt;/span&gt;. I'm posting this in the hope someone else can improve any of these.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0lraq.png&quot; alt=&quot;Cost-30-attack-all-squares&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xrbLp.png&quot; alt=&quot;Cost-30-attack-all-squares-2&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="1649" LastEditorUserId="80061" LastEditDate="2024-04-06T19:06:30.360" LastActivityDate="2024-04-06T19:06:30.360" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="2910" PostTypeId="2" ParentId="2905" CreationDate="2014-10-15T10:19:36.347" Score="-2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alice divides the estate in three parts, and lets Bob and Carol choose.&#xA; &#xA; If Bob and Carol choose different parts, Alice gets the part that's left. Since Alice divided and Bob and Carol got to choose, all should be happy.&#xA; &#xA; If Bob and Carol choose the same part, Alice chooses one of the other parts. The parts left are put together and now divided between Bob and Carol by the usual way of having one partition and the other one choose.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This way, each sibling got to either divide or choose and should be happy.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is different from the cake solution, since a cake can be cut, but not joined.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To address some points raised in the comments, I propose the following adjustment:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If Bob and Carol choose different parts, Alice still gets's what left. But if Bob and Carol choose the same part, they then choose the part Alice should get. If they choose the same part for Alice, she gets that, if they choose differently, she gets to choose. Only then the remaining parts are joined and redivided.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This still doesn't address the point of Alice disagreeing with how Bob and Carol choose to redivide, should that happen.&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditDate="2014-10-15T12:24:14.267" LastActivityDate="2014-10-15T12:24:14.267" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2911" PostTypeId="2" ParentId="2864" CreationDate="2014-10-15T13:29:09.317" Score="10" Body="&lt;p&gt;Here is a solution that also maximizes the number of pawns on the board. It also takes into account pawn promotion and the no-pawns-on-first-row rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;12 Pawns&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5X3W6.gif&quot; alt=&quot;12 Pawns&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ideas used&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area of influence by the rooks is constant wherever they are (2 rows and 2 columns). I placed the bishops and the knights to the side to minimize their area of influence and the rooks on the next columns so the 2 required by the rooks overlap with most of the squares controlled by the knights and bishops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 5 columns remaining at this point and pawns placed in a column render the columns next to them useless. This means the best way to place them is to put them on columns 1, 3 and 5. This leaves the Queen on column 2 or 5.&lt;/p&gt;&#xA;" OwnerUserId="3918" LastActivityDate="2014-10-15T13:29:09.317" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2912" PostTypeId="2" ParentId="2905" CreationDate="2014-10-15T14:07:54.480" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Monetize what can be monitized (aka cash and bank accounts). This is the pot and has a value P.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Have a referee separate all of the remaining pieces into 3 (what he sees even) regions.  Call these the chunks.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Have all players chose which of the chunks is the biggest.  If they all disagree, split any remaining pot evenly between them.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If none choose one chunk, place all of the pot on that chunk.  If none wants two chunks, place P evenly on the other two.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Repeat step 3.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If multiple people still want a part with no money, the referee failed to split well and should be fired. Repeat from step 1. (this is the hole is this plan for most simular questions. It works well for spliting apartment rent though).  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Take the option which multiple people want.  Have the referee take money from that chunk slowly and add it to the pot.  When someone wants to switch chunks, have them yell stop.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;repeat step 3.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Have the referee move to the one no one wants, slowly remove identical amounts from the other two. Have someone yell stop when they want to switch.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;repeat step 3.  It should work this time.  Everyone is holding onto the one they think is the largest so they should not be envious.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditDate="2014-10-21T17:33:46.303" LastActivityDate="2014-10-21T17:33:46.303" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2913" PostTypeId="2" ParentId="2817" CreationDate="2014-10-15T14:14:04.843" Score="6" Body="&lt;p&gt;adding/subtracting the numbers from the position they are currently at (including spaces) you get a message that i'm not sure is allowed on the site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;take &quot;Aid KIe CT&quot; and assign positional values with base 1. Add the corresponding number to the position and align it in it's new position. For example &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A is position 1 and the first number is 3 so A's new position is 4. &#xA;Likewise for i it is position 2 and 6 is the second number so i's new position is 8.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;the answer is 10 characters with 2 spaces.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I eAT d***&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Yes I know my friend is immature.&quot; &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Indeed. He also has a poor diet.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2803" LastEditorUserId="2803" LastEditDate="2014-10-15T14:48:56.133" LastActivityDate="2014-10-15T14:48:56.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2914" PostTypeId="2" ParentId="2907" CreationDate="2014-10-15T14:38:47.270" Score="20" Body="&lt;p&gt;Can this be considered an improvement over justhalf's solution or am I missing something here total comes to 29 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hjnlc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="2313" LastEditDate="2014-10-15T14:50:39.517" LastActivityDate="2014-10-15T14:50:39.517" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="2915" PostTypeId="2" ParentId="2907" CreationDate="2014-10-15T15:06:06.307" Score="12" Body="&lt;p&gt;This does not qualify as the lowest score, but it does answer the question of &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How many Chess Pieces are needed to control every square on the board?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/luPqp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It takes 5 pieces&lt;/p&gt;&#xA;" OwnerUserId="3922" LastEditorUserId="3922" LastEditDate="2014-10-16T14:27:47.393" LastActivityDate="2014-10-16T14:27:47.393" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2916" PostTypeId="2" ParentId="2897" CreationDate="2014-10-15T15:18:08.477" Score="8" Body="&lt;p&gt;Although I like my first answer, I did also wonder how good I could manage with black and white in their normal positions. The best I've gotten so far uses 16 queens, 4 rooks, 4 bishops and 4 pawns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;180.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8f5FF.png&quot; alt=&quot;180 Points&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="375" LastEditorUserId="375" LastEditDate="2014-10-15T19:18:17.803" LastActivityDate="2014-10-15T19:18:17.803" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2924" PostTypeId="2" ParentId="2900" CreationDate="2014-10-15T18:21:17.357" Score="7" Body="&lt;p&gt;I think the answer is 890625.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started to look at the powers of 5 and noticed there is a lovely pattern!&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                   5&#xA;                  25&#xA;                 125&#xA;                 625&#xA;                3125&#xA;              1 5625&#xA;              7 8125&#xA;            3 9 0625&#xA;          1 9 5 3125&#xA;          9 7 6 5625&#xA;         48 8 2 8125&#xA;        244 1 4 0625&#xA;       1220 7 0 3125&#xA;       6103 5 1 5625&#xA;      30517 5 7 8125&#xA;     152587 8 9 0625&#xA;     762939 4 5 3125&#xA;    3814697 2 6 5625&#xA;   19073486 3 2 8125&#xA;   95367431 6 4 0625&#xA;  476837158 2 0 3125&#xA; 2384185791 0 1 5625&#xA;11920928955 0 7 8125&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;At first I thought that it was just 0625,3125,5625 and 8125 repeated at the end, but then I noticed that&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a number ending in 0625 is always preceded by either a 9 or a 4&lt;/li&gt;&#xA;&lt;li&gt;a number ending in 90625 is always preceded by 3 or 8.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And it goes on...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The preceding pairs always add up to 5, but I haven't noticed a pattern for how you predict which pair it will be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are all the combinations for the last 6 digits of a power of 5:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ykfv6.jpg&quot; alt=&quot;ClockworkCalculator&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number at the top is the smallest unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where there is a 3 in the sequence, I have made the circles red (as the problem states there are no 3's)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the bottom upwards, I searched for the largest number and highlighted the number in green.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used this calculator to check some of predictions:&#xA;&lt;a href=&quot;http://www.calculatorpro.com/calculator/exponent-calculator/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.calculatorpro.com/calculator/exponent-calculator/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately it cannot handle 890625 itself (it timed out!) so I cannot confirm that 890625^890625 ends in 890625, but I believe it does!&lt;/p&gt;&#xA;" OwnerUserId="1616" LastEditorUserId="1616" LastEditDate="2014-10-15T18:42:24.337" LastActivityDate="2014-10-15T18:42:24.337" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2925" PostTypeId="2" ParentId="2813" CreationDate="2014-10-15T18:29:14.320" Score="6" Body="&lt;p&gt;Update:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to hints in the comments, I realized there is a 33% more efficient solution.  It is possible to sort up to $N = 2^i$ cards using $M = 2i+1$ stacks. My algorithm is: split, flip, merge, repeat.  If you perform the split differently, you don't need the flip step.  You just split, merge, repeat.  This algorithm must be optimal, as with a single split/merge you can only sort one bit of information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Split algorithm:&#xA;For each card numbered $i$, count the bits of $i-1$.  If there are an even number of bits, put the card on stack 2, odd number on stack 3.  In the second pass, ignore the lowest bit.  Third pass ignore the first and second bits, etc.  So, first pass card 1 goes on stack 2, 2 -&gt; 3, 3 -&gt; 3, 4 -&gt; 2, 5 -&gt; 3, 6 -&gt;2, 7 -&gt; 2, 8 -&gt; 3, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After each split, merge the stacks by putting each card from stack 2 onto stack 3, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Original answer:&lt;br&gt;&#xA;Here is a plan that should work:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Move all the odd cards to stack 3 and all the even cards to stack 2.&lt;/li&gt;&#xA;&lt;li&gt;Move all the cards from stack 2 to stack 1.&lt;/li&gt;&#xA;&lt;li&gt;Move all the cards from stack 3 to stack 1.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If you have up to 2 cards, you are now finished.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;Move all the cards equal to 1,2 mod 4 to stack 3 and the others to stack 2.&lt;/li&gt;&#xA;&lt;li&gt;Move all the cards from stack 2 to stack 1.&lt;/li&gt;&#xA;&lt;li&gt;Move all the cards from stack 3 to stack 1.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If you have up to 4 cards, you are now finished.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;7&quot;&gt;&#xA;&lt;li&gt;Move all the cards equal to 1,2,3,4 mod 8 to stack 3 and others to stack 2.&lt;/li&gt;&#xA;&lt;li&gt;Move all the cards from stack 2 to stack 1.&lt;/li&gt;&#xA;&lt;li&gt;Move all the cards from stack 3 to stack 1.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If you have up to 8 cards, you are now finished.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continue following the pattern.  The total number of steps required is $3 \log_2 N$, rounding up to the next multiple of 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works because each set of 3 steps sorts cards based on the lowest unsorted bit, and no steps later mess up that ordering.  In other words, after step 3, card 1 is guaranteed to be above card 2, card 3 above card 4, etc.  Later sets of steps will preserve that ordering while fixing ordering of groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To help visualize, a walkthrough with 4 cards:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No matter the initial ordering, after step 1, stack 2 has cards 2 &amp;amp; card 4 in some order, while stack 3 has cards 1 &amp;amp; 3 in some order.  After step 3, stack 1 has (1, 3), (2,4) where the two groups are in random order.  After step 4, stack 2 has cards 4 &amp;amp; 3 (4 on top) and stack 3 has cards 2 &amp;amp; 1 (2 on top).  After step 6, stack 1 has cards 1, 2, 3, 4, in that order from the top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update:&#xA;Rereading the question, I see an algorithm wasn't requested.  The answer is: given $M$ stacks, you can sort up to $2^i$ cards, where $M = 3i+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think my approach is optimal.  I know you need $O(\log_2 N)$ stacks to sort $N$ cards.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="2328" LastEditDate="2014-10-16T12:30:25.060" LastActivityDate="2014-10-16T12:30:25.060" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="2927" PostTypeId="2" ParentId="2900" CreationDate="2014-10-15T20:05:41.423" Score="3" Body="&lt;p&gt;I did a little research on this problem and found that the least significant digits of a large exponent can be found fairly easily through the use of &lt;a href=&quot;http://en.wikipedia.org/wiki/Modular_exponentiation&quot; rel=&quot;nofollow&quot;&gt;modular exponentiation&lt;/a&gt;. So I wrote a quick Python script to find the largest numbers:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;def modular_pow(base, exp, mod):&#xA;    result = 1&#xA;    base = base % mod&#xA;    while (exp &amp;gt; 0):&#xA;        if (exp % 2 == 1):&#xA;           result = (result * base) % mod&#xA;        exp = exp &amp;gt;&amp;gt; 1&#xA;        base = (base * base) % mod&#xA;    return result&#xA;&#xA;# find the largest 5 numbers &amp;lt; 1000000, n, where n^n mod 1000000 = n        &#xA;a = 1000000&#xA;b = 0&#xA;count = 0&#xA;while (a &amp;gt; 1 and count &amp;lt; 5):&#xA;    a = a - 1&#xA;    b = modular_pow(a, a, 1000000)&#xA;    if (a == b):&#xA;        count = count + 1&#xA;        print(a)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Output:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;999999&#xA;999001&#xA;998751&#xA;998001&#xA;997376&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The second number, &lt;strong&gt;999001&lt;/strong&gt;, matches the constraints.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2014-10-15T20:05:41.423" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2929" PostTypeId="1" AcceptedAnswerId="3133" CreationDate="2014-10-15T20:16:18.597" Score="10" ViewCount="2485" Body="&lt;p&gt;The &lt;strong&gt;Bunny&lt;/strong&gt; is a new chess piece. It can move in 2 different ways: Diagonally, but only exactly one space (so like a bishop with the limitations of a king). It can also &quot;Bunny-Hop&quot; over another bunny. Here's an explanation of how this is done:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say that, in the following image, the white bishop is a bunny named Henry, and the black bishop is a bunny named George. Henry can move onto any of the pawns. He can move onto the pawns diagonal to him by simply moving there normally (bunnies can move diagonally in any direction 1 space). It can capture the pawn 3 squares north of it by jumping over George. When a bunny jumps, it moves 3 spaces in the direction that's it's jumping. George is north of Henry, so Henry must be jumping north. So, he's jumping north 3 spaces, onto where that pawn is. &lt;b&gt;Additionally,&lt;/b&gt; George must move south one space, because when a bunny jumps over another bunny, the bunny it jumps over (George) must move to the original position of the bunny that is jumping (Henry)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/38oFEm.jpg&quot; alt=&quot;bunny jumping example&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The board starts with 2 bunnies, one on a white square, one on a black square. You may choose the initial positions as long as the bunnies are on squares of different colors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can you alternate moves (move &lt;strong&gt;Bunny 1&lt;/strong&gt; during 1 turn, then &lt;strong&gt;Bunny 2&lt;/strong&gt; during the next) so that each bunny moves to each space on the board, exactly once? Note that each bunny must step on every one of the 64 squares, so they will each move 63 times. &lt;strong&gt;It will be at most 63 turns.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="8874" LastEditDate="2015-02-04T15:27:08.970" LastActivityDate="2015-05-21T07:51:13.570" Title="The Bunny's Tour" Tags="|chess|checkerboard|graph-theory|" AnswerCount="6" CommentCount="28" ContentLicense="CC BY-SA 3.0" />
  <row Id="2930" PostTypeId="2" ParentId="2789" CreationDate="2014-10-15T21:17:33.683" Score="5" Body="&lt;p&gt;I actually misspoke in my comment earlier. Here's a complete answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stepping through conversation A:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proctor: &quot;I cannot determine the two numbers.&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Condition A: The product is non-unique.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summer: &quot;I knew that.&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Condition B: The sum can only be made from pairs that satisfy condition A (form non-unique products). For example, 11 could be formed from (2,9), (3,8), (4,7), or (5,6), none of which have unique products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proctor: &quot;I still cannot determine the two numbers.&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Condition C: The product can be made from more than one pair that satisfies condition B. This rules out (2,9), for example: the only other pair that has the same product is (3,6), which has a sum of 9, which can be formed from (2,7), which has a unique product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summer: &quot;Now I can determine them.&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Condition D: The sum could be formed from more than one pair that satisfies condition A, but only one that satisfies condition C. Only possibility at this point is (5,6).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proctor: &quot;Now I can, too.&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Condition E: The product could be formed from more than one pair that satisfies condition B, but only one that satisfies condition D. (5,6) obviously still works here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Conversation B, we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summer: &quot;Neither can I.&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Condition D: The sum could be formed from more than one pair that satisfies condition C. 58 possible pairs remain at this point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proctor: &quot;Now I can determine them.&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Condition E becomes: The product could be formed from more than one pair that satisfies condition B, but only one that satisfies condition D. Only remaining pair at this point is (2, 15).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summer: &quot;Now I can, too.&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally Condition F: The sum could be formed from more than one pair that satisfies condition C, but only one that satisfies condition E. (2, 15) still works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, add more &quot;neither can I&quot;s than that and you'll be stuck with 57 possible pairs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's interesting that these still work. However I don't think you can beat the original puzzle in elegance. In the original, an outside observer only knows the solution AFTER Summer says &quot;Now I can too&quot;. But notice that in these two conversations, we knew the answer before the final statements. I wonder how else you could vary the puzzle, besides lengthening it; could be a interesting exercise. &lt;/p&gt;&#xA;" OwnerUserId="3927" LastActivityDate="2014-10-15T21:17:33.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2932" PostTypeId="2" ParentId="2900" CreationDate="2014-10-15T21:54:28.103" Score="6" Body="&lt;p&gt;I wrote a little program to compute $N^N$ $mod$ $1000000$.  And checked whether the last digits of the result match the digits of $N$ reversed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got the following solutions:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$999999^{999999}$ $mod$ $1000000$ $=$ $999999$&lt;br&gt;&#xA;  $100001^{100001}$ $mod$ $1000000$ $=$ $100001$&lt;br&gt;&#xA;  $90789^{90789}$ $mod$ $100000$ $=$ $98709$&lt;br&gt;&#xA;  $10001^{10001}$ $mod$ $100000$ $=$ $10001$&lt;br&gt;&#xA;  $1001^{1001}$ $mod$ $10000$ $=$ $1001$&lt;br&gt;&#xA;  $1^{1}$ $mod$ $10$ $=$ $1$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$999999$ is not possible, we only have 5 $9$'s (I'll assume the $6$ won't do).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$100001$ is not possible, we only have 2 $0$'s. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;$\mathbf{90789}$ is the largest working number&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2016-02-22T18:50:08.997" LastActivityDate="2016-02-22T18:50:08.997" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2933" PostTypeId="2" ParentId="2813" CreationDate="2014-10-15T21:57:12.310" Score="0" Body="&lt;p&gt;Given stacks M, I believe you can sort up to 2^i records, where m = 2*i + 1.  Effectively, you can do a binary sort, but since you include a starting column, you require 1 extra. I'll try to get an algorithm up later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: After trying to put my idea into words and looking at @user3294068's answer, I think i have the same solution (with a slightly different sort algorithm), but it ends up with the same number of columns. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2014-10-15T23:08:10.893" LastActivityDate="2014-10-15T23:08:10.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2934" PostTypeId="2" ParentId="2929" CreationDate="2014-10-15T23:35:03.847" Score="1" Body="&lt;p&gt;The bunnies can cover the entire board by following this path:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6mC6r.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black moves first, and the bunnies start in the lower left-hand corner (or, I suppose, red starting in the other corner would work just as well).  I used red and black for the bunnies for ease of illustration.  I took some liberties with the rules - I hope by &quot;forward&quot; you meant &quot;whichever of the four compass points the bunny wants to face&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1627" LastEditorUserId="1627" LastEditDate="2014-10-16T00:14:58.570" LastActivityDate="2014-10-16T00:14:58.570" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="2935" PostTypeId="2" ParentId="2907" CreationDate="2014-10-16T00:28:21.197" Score="42" Body="&lt;p&gt;Here's a solution that only takes 28 points:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NcbZU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2014-10-16T00:28:21.197" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="2936" PostTypeId="1" CreationDate="2014-10-16T00:45:44.063" Score="3" ViewCount="791" Body="&lt;p&gt;The maximal number of knights of a single color that can be placed on a chessboard without any knight being one knights-move away from any other is 32; such a position may easily be achieved by placing all 32 knights on squares of the same color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one uses both black pieces and white pieces, it's possible to place 24 knights of each color (48 total) on the board without any knight being attacked by a knight of the opposite color.  One simple approach is to place 24 white knights on the first three ranks, and 24 black knights on the last three.  That position is but one of many, which can accommodate 24 knights of each color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to either place 24 knights of one color and more than 24 of the other or prove the impossibility of doing so?  Note that if one may arbitrarily choose the number of knights of each color, one could easily place 62 knights (61 white and one black) or, for that matter, 64 (all white).  To be of interest, I think it is possible to rank solutions by the number of pieces in the &quot;shorter&quot; army and then by the number of pieces in the other.&lt;/p&gt;&#xA;" OwnerUserId="1467" LastEditorUserId="19196" LastEditDate="2016-02-13T00:16:18.027" LastActivityDate="2017-01-25T22:06:50.510" Title="Can more than 24 knights of each color be placed on a chessboard without any pieces attacking opposite color?" Tags="|chess|checkerboard|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2937" PostTypeId="2" ParentId="2872" CreationDate="2014-10-16T06:19:00.180" Score="8" Body="&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; So I ran a nearly exhaustive programmatic search. I excluded cases which put Queen or Bishop on the edge of the board, or a Rook on the same rank/file as the Queen or other Rook. Assuming correct code, I could not find any better than cost-33.&lt;/p&gt;&#xA;&lt;p&gt;I found that &lt;a href=&quot;https://puzzling.stackexchange.com/a/2971/3936&quot;&gt;@LeonBouquiet's solution&lt;/a&gt; is one of 18 distinct cost-33 solutions up to symmetry with a legal position (no pawns on first rank, no bishops on same colour), though they come in nine pairs differing by a single pawn. Including illegal positions, I found 32 cost-33 boards with distinct major piece configurations. This was much more than I had expected, and was then further surprised that given this number there were no cost-32 solutions. First, I'll show the solutions, then talk a &lt;s&gt;bit&lt;/s&gt; lot about the algorithm I coded.&lt;/p&gt;&#xA;&lt;h1&gt;Solutions&lt;/h1&gt;&#xA;&lt;p&gt;In case you want still want to give it a go, here's an interesting hint. There are one or more piece placements that are common to all solutions with a legal position. Assuming the Queen to be on the right half:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Ra1, Nd5, c6, and a further two common to all but one of the pawn-pair solutions: c2, c4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;And the full solutions (up to symmetry):&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;em&gt;Note: black pawns represent alternate pawn placements, and hopefully my brain worked as I had to manually reconstruct pawn placements.&lt;/em&gt;&lt;br/&gt;&lt;br/&gt;&#xA;&#xA; Four solutions using only one rook and seven pawns:&lt;br/&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/INSJD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/INSJD.png&quot; alt=&quot;One rook, seven pawns 1&quot; /&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/ldw7W.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ldw7W.png&quot; alt=&quot;One rook, seven pawns 2&quot; /&gt;&lt;/a&gt;&lt;br/&gt;&#xA;&#xA; &lt;br/&gt;And then twelve very similar solutions using only one bishop, one knight, and all eight pawns:&lt;br/&gt;&lt;br/&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/DdW3i.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DdW3i.png&quot; alt=&quot;One b One n Eight p 1&quot; /&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/BXURS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BXURS.png&quot; alt=&quot;One b One n Eight p 2&quot; /&gt;&lt;/a&gt;&#xA; &lt;br/&gt;&lt;br/&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/EX4RC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EX4RC.png&quot; alt=&quot;One b One n Eight p 3&quot; /&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/OyVDI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OyVDI.png&quot; alt=&quot;One b One n Eight p 4&quot; /&gt;&lt;/a&gt;&#xA; &lt;br/&gt;&lt;br/&gt;&lt;a href=&quot;https://i.stack.imgur.com/gDWlo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gDWlo.png&quot; alt=&quot;One b One n Eight p 5&quot; /&gt;&lt;/a&gt;&#xA; &lt;br/&gt;&lt;br/&gt;The remaining two solutions which remove a knight and use 5 pawns I'll not repeat - see &lt;a href=&quot;https://puzzling.stackexchange.com/a/2971/3936&quot;&gt;@LeonBouquiet's solution&lt;/a&gt;, noting that the f7 pawn can just as well be on h7.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Among the illegal position solutions, the most interesting one I found was this one, which barely uses the bishops:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/KBmku.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KBmku.png&quot; alt=&quot;illegal position solution&quot; /&gt;&lt;/a&gt;&#xA; &lt;br/&gt; Only one is used and it's just covering 3 squares not already covered by other pieces. This leads me to believe that my filter against Bishops and Queens being on the edge of the board could indeed have filtered out valid solutions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h1&gt;Algorithm&lt;/h1&gt;&#xA;&lt;p&gt;I've &lt;a href=&quot;https://github.com/DPenner1/ChessPuzzleSim/tree/main&quot; rel=&quot;nofollow noreferrer&quot;&gt;uploaded&lt;/a&gt; my C# code to GitHub. First I'll note the pawns were a bit of an after-thought. While I did eventually add them to the code, it still outputs just the major pieces so I had to manually fill in the pawns afterwards to get the solutions. I did double-check the solutions though to make sure that I didn't miss any cost-32 solutions, but I'll probably re-verify in the future as I continue to tinker with the code.&lt;/p&gt;&#xA;&lt;p&gt;I started to code this algorithm when I realized that the number of possible configurations of major pieces might actually be feasibly computable (with all 7 major pieces on the board, there are just under &lt;a href=&quot;https://www.wolframalpha.com/input?i=32%2863+choose+2%29%2861+choose+2%29%2859+choose+2%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;196 billion&lt;/a&gt; possibilities). Then, my plan was to strategically place the pawns 1-by-1 and abort evaluating the board if pawn placement was impossible. Determining impossibility is mainly that the amount of remaining pawns (given the target cost) could not cover the remaining unattacked squares. This strategy ended up working as I only had to fully evaluate roughly 1 board with pawns for every 10 boards without pawns. The core pseudocode:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;loop foreach candidate major piece board:&#xA;    call EvaluateBoard with this board&#xA;    if success then: celebrate&#xA;end loop&#xA;&#xA;function EvaluateBoard:&#xA;&#xA;    Determine attacked squares&#xA;&#xA;    if all squares are attacked then: exit with success&#xA;    if pawn placement impossible then: exit with failure&#xA;&#xA;    add one pawn such that it attacks a previously unattacked square&#xA;    call EvaluateBoard with this new board&#xA;&#xA;end function&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;h1&gt;Old manual cost-34 for posterity&lt;/h1&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/R9nrD.png&quot; alt=&quot;Chess puzzle 34&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="3936" LastEditorUserId="3936" LastEditDate="2024-04-05T20:10:51.247" LastActivityDate="2024-04-05T20:10:51.247" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="2939" PostTypeId="2" ParentId="2936" CreationDate="2014-10-16T08:22:06.657" Score="11" Body="&lt;p&gt;Ok let me stick my neck out, I would love to be proved wrong on this, but there are only a few ways (3 unique ways) 24 Knights of each side can be placed on the board i.e. without attacking any of the other coloured pieces&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6HgVO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WmPEr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Tqb0V.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other combinations of swapping colours and rotating the boards could be achieved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And even theoretically 48 squares are occupied by knights, leaving just 16, given the fact that a knight in the edge attacks 4 squares and in the center attacks 8, it is logical to understand that 2 and 4 of these have to fall in &quot;enemy&quot; territory so to say.  so even ensuring an overlap of 2 knights per square, we are talking of 8 squares per colour being &quot;attacked&quot; over and above the squares occupied by their own forces, for both sides this is 16 squares. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence it is not possible to place more than 24 knights per side&lt;/p&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="2313" LastEditDate="2014-10-16T13:19:53.643" LastActivityDate="2014-10-16T13:19:53.643" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="2945" PostTypeId="1" AcceptedAnswerId="2946" CreationDate="2014-10-16T13:47:57.093" Score="4" ViewCount="524" Body="&lt;p&gt;Heaven has a countably infinite number of cloud for people to stay, each of which has a good soul living on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A countably infinite number of infinite universes are extinguished by God (their beta period had expired). So, a countably infinite number of groups of good souls (each of which are countably infinite in size) were knocking on the doors of heaven. Each person, in each group was carrying a little ticket, from $\ge 1$ (which one receives before passing the pearly gates). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How will God and his dear mother accomodate all these people on a cloud of their own without creating any new clouds?&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2017-07-11T15:32:33.597" LastActivityDate="2021-06-28T13:01:39.987" Title="Housing crisis in heaven" Tags="|logical-deduction|mathematics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2946" PostTypeId="2" ParentId="2945" CreationDate="2014-10-16T14:01:01.537" Score="7" Body="&lt;p&gt;This is a restatement of &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert%27s_paradox_of_the_Grand_Hotel&quot; rel=&quot;noreferrer&quot;&gt;Hilbert's paradox of the Grand Hotel&lt;/a&gt;, in particular the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert%27s_paradox_of_the_Grand_Hotel#Infinitely_many_coaches_with_infinitely_many_guests_each&quot; rel=&quot;noreferrer&quot;&gt;Infinitely many coaches with infinitely many guests each&lt;/a&gt; problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution is to take each person's ticket, and assign them a cloud number based on their original universe number and their original cloud number. You can calculate the new number from the older two numbers using a &lt;a href=&quot;http://en.wikipedia.org/wiki/Pairing_function&quot; rel=&quot;noreferrer&quot;&gt;pairing function&lt;/a&gt;. One such function is the Cantor pairing function:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;new cloud number = (1/2)*(A+B)*(A+B+1)+B  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where A is your old universe number and B is your old cloud number.&lt;/p&gt;&#xA;" OwnerUserId="1497" LastActivityDate="2014-10-16T14:01:01.537" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2950" PostTypeId="1" AcceptedAnswerId="3001" CreationDate="2014-10-16T15:25:18.077" Score="3" ViewCount="664" Body="&lt;p&gt;For the following pattern of filling the $5*5$ matrix, how many steps does it take to turn all cells of the matrix to black?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8OiMf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pattern has revealed by &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2212/find-the-pattern-of-matrices#answer-2219&quot;&gt;eBusiness&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first piece moves to the right, the second moves down, the third moves left, the fourth moves up, and so the fifth probably moves right.&#xA; &#xA; &#xA; The letters and numbers are the sum of X and Y coordinates respectively, with the upper left corner being coordinate $(1,1)$ or $(A,1)$.&#xA; &#xA; Each image introduce its new element at the position given by the coordinate sum of the previous image.&#xA; &#xA; Both movement and new positions wrap around, so when a piece would move over the border it appears next to the opposite border, similarly a piece that appear at for instance $(H,11)=(8,11)$, will wrap around to $(8mod5,11mod5)=(3,1)=(C,1)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I guess after some steps next pointed cell that need to be colored, may not be white.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS. Question is much like one of my previously made questions: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2212/find-the-pattern-of-matrices&quot;&gt;Find the pattern of matrices&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2014-10-18T09:40:30.107" Title="When will the matrix turn to black?" Tags="|logical-deduction|pattern|logic-grid|" AnswerCount="2" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2958" PostTypeId="2" ParentId="2905" CreationDate="2014-10-16T18:11:00.167" Score="0" Body="&lt;p&gt;Another answer because this is inherently a different method:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Make 3 piles.  Two are actually empty. One contains everything.&lt;/li&gt;&#xA;&lt;li&gt;Have all 3 sibilings in random order pick a pile they want but a maximum of two can pick the same pile.&lt;/li&gt;&#xA;&lt;li&gt;Intentionally at #3 again: if at any point all 3 have selected different piles and no one wants to switch, everyone has their piles.  We will be done then.&lt;/li&gt;&#xA;&lt;li&gt;If the rules are being followed: 2 people want 1 pile (call it A), 1 person has chosen another pile (call it B), one pile is undesired (call it C).  The person at pile B takes any 1 object from A and places it in C.&lt;/li&gt;&#xA;&lt;li&gt;The player at pile B leaves pile B. The 2 players at pile A are given the option to select to any pile.  The remaining player is allowed to pick his new pile.  Still, a maximum of 2 is allowed on any one pile.&lt;/li&gt;&#xA;&lt;li&gt;Repeat from step 3 until the loop ends.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Please let me know how this would break apart if it would.  It will be envy free.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motivation of the player at B is to increase the net value of C.  I find this interesting as a lock + a key has alot more value that either separately. Two copies of GTA5 have little more value than one.  He has an economic incentive to cherry pick from A to make A, B, and C have the highest possible values from his perspective whilst leaving in A the things they want more than him. As a result this will work alot better than a random cut or refereed impartial separation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This falls apart if one inseparatable object has more value to multiple players than half of everything else.  It also fails if one party has the more motivation to extend the game than to obtain the largest pile or some party is willing to accept a smaller inheretance and whine about it later.  I think, however, that any plan would fail under these conditions.&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2014-10-16T18:11:00.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2961" PostTypeId="1" AcceptedAnswerId="2962" CreationDate="2014-10-16T19:15:46.823" Score="-2" ViewCount="338" Body="&lt;p&gt;No restriction on number of pieces of each type, but all pieces must be the same color. Pieces do not control the square they occupy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pieces on the chessboard must, through their combined efforts, control every square on the board. Accepted Answer goes to the answer with the lowest score.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Piece cost:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Pawn - 1&lt;/li&gt;&#xA;&lt;li&gt;Knight - 2&lt;/li&gt;&#xA;&lt;li&gt;Bishop - 3&lt;/li&gt;&#xA;&lt;li&gt;King - 3&lt;/li&gt;&#xA;&lt;li&gt;Rook - 4&lt;/li&gt;&#xA;&lt;li&gt;Queen - 6&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is a modified version of &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2907/how-many-chess-pieces-are-needed-to-control-every-square-on-the-board-no-piece&quot;&gt;How many Chess Pieces are needed to control every square on the board? No Piece Restriction&lt;/a&gt;, and the piece costs were chosen based on the difficulty of using them in that challenge.&lt;/p&gt;&#xA;" OwnerUserId="1627" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2014-10-16T19:55:46.797" Title="Chessboard Domination - Adjusted Piece Values" Tags="|chess|" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2962" PostTypeId="2" ParentId="2961" CreationDate="2014-10-16T19:44:55.163" Score="4" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0xl6O.gif&quot; alt=&quot;Forgot about the Queen&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wanted to try to make use of the kings, and this is what I got. It ends up at a score of 23.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Bah, forgot the queen's square. Now up to 25 (couldn't use a pawn since then a king isn't covered).&lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2014-10-16T19:55:46.797" LastActivityDate="2014-10-16T19:55:46.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2966" PostTypeId="2" ParentId="2950" CreationDate="2014-10-16T22:01:24.433" Score="1" Body="&lt;p&gt;It will never turn black.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that black squares can superpose on the same position.  They will cross, each following its own direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens is that the 23rd square falls exactly on the position of the 3rd one, and it moves in the same direction.  The 24th square then falls exactly on the position of the 4th, and again moves in the same direction.  In fact, from the 23rd on, all squares superpose with the square created 20 steps before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is that new squares are generated according to the sum of all coordinates.  These coordinates move.  But every group of 4 consecutive squares moves in all 4 directions, so that their movements compensate.  The sum of the 4 successive coordinates remains constant.  And it happens that 5 such groups sum up to 0.  That means that you can forget the last 20 squares, you still get the same result.  And therefore the squares are generated exactly like 20 squares before.  The generation repeats an endless cycle of size 20.  In fact 20 is 4x5, 4 is the number of directions, 5 is the size of the board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is that only the 22 first squares count.  The following only superpose on existing squares.  So there are never more than 22 squares covered at any time on the board.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2014-10-18T09:40:30.107" LastActivityDate="2014-10-18T09:40:30.107" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2968" PostTypeId="1" CreationDate="2014-10-17T03:28:46.670" Score="2" ViewCount="833" Body="&lt;p&gt;Let's say there is a 5x5 grid with a starting position in the upper left corner, and an ending position is lower left corner. How do I count how many Hamiltonian paths there are through the grid? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the answer is 86 for a 5x5 grid. But how should I calculate this? (4x4 is 8, 3x3 is 2) I want to learn calculate larger grids.&lt;/p&gt;&#xA;" OwnerUserId="3958" LastEditorUserId="1777" LastEditDate="2014-10-17T13:30:40.220" LastActivityDate="2014-10-17T13:30:40.220" Title="How do I count Hamiltonian paths of grid?" Tags="|logical-deduction|calculation-puzzle|checkerboard|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2970" PostTypeId="2" ParentId="2968" CreationDate="2014-10-17T08:02:23.263" Score="6" Body="&lt;p&gt;The &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Hamiltonian_path_problem&quot;&gt;Hamiltonian path problem&lt;/a&gt; is NP-complete. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means there's no reasonable way for you to find &lt;em&gt;all&lt;/em&gt; of these paths yourself on a sufficiently large grid. You might be able to find a few of them manually, but you will have no way of knowing if you've found all of them, even if you have. Counting them with certainty takes an unreasonable amount of computational time - we know of no faster way. &lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastActivityDate="2014-10-17T08:02:23.263" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2971" PostTypeId="2" ParentId="2872" CreationDate="2014-10-17T09:54:31.860" Score="17" Body="&lt;p&gt;I think I've found a cost-33 solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aFlK1.gif&quot; alt=&quot;Cost-33&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I can tell it's valid, but I've been staring at it for so long that I kind of don't trust myself anymore... so if you see a mistake, please tell me.&lt;/p&gt;&#xA;" OwnerUserId="3932" LastActivityDate="2014-10-17T09:54:31.860" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2972" PostTypeId="1" AcceptedAnswerId="2973" CreationDate="2014-10-17T11:29:44.183" Score="16" ViewCount="1732" Body="&lt;p&gt;I made up this puzzle, so if you are only keen on answering standard puzzles, I am sorry, I thought its been a dull day for questions so I made up my own&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a 7 x 7 squares image with 7 8 letter English words hidden in it can you find the words?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1fZk2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conditions to be followed are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first letter of one word has to be the last letter of another, so those letters would be used twice but all other letters have to be used only once per appearance in the square&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each word is formed with letters adjoining them in any one direction, i.e. if one letter is in say second row and second column, the next letter in that word has to be in one of the 8 squares adjoining it&lt;/p&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="2313" LastEditDate="2014-10-17T11:37:07.353" LastActivityDate="2014-10-17T12:03:43.017" Title="Hidden Words in squares" Tags="|word-problem|crosswords|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2973" PostTypeId="2" ParentId="2972" CreationDate="2014-10-17T12:03:43.017" Score="11" Body="&lt;p&gt;I think I managed to find the 7 8-letter words you hid in the grid. Nice little puzzle - no ridiculous words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hARQl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HviWQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2484" LastActivityDate="2014-10-17T12:03:43.017" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2976" PostTypeId="1" AcceptedAnswerId="2977" CreationDate="2014-10-17T13:35:19.993" Score="8" ViewCount="1489" Body="&lt;p&gt;The &lt;a href=&quot;http://wordplay.blogs.nytimes.com/category/numberplay/&quot; rel=&quot;noreferrer&quot;&gt;Numberplay&lt;/a&gt; blog on the NY Times website used to have word ladder puzzles along with their weekly math puzzles.  I found them fun and so I figured I'd introduce them here.  These aren't just your basic word ladders either.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You can change a single letter in place, as in traditional word ladders — for example, SOAP to SOUP.&lt;/li&gt;&#xA;&lt;li&gt;You can add or subtract a single letter in place to increase or decrease the length of a word — for example, MATH to MATCH or vice versa.&lt;/li&gt;&#xA;&lt;li&gt;You can change a single letter and rearrange the letters of the word to give another word of the same length — for example, MUSEUM to SUMMED (with a U changed to D and then rearrangement).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The aim is to make the shortest possible ladder.  All steps must be valid English words.  You can use the &lt;a href=&quot;http://code.google.com/p/dotnetperls-controls/downloads/detail?name=enable1.txt&quot; rel=&quot;noreferrer&quot;&gt;Enable list&lt;/a&gt; for a comprehensive list of words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Here's an example:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CIRCLE to SQUARE&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CIRCLE&lt;br&gt;&#xA;RELICS&lt;br&gt;&#xA;CRUISE&lt;br&gt;&#xA;SAUCER&lt;br&gt;&#xA;SQUARE  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Now try these:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;WOLVES to SHEEP&lt;br&gt;&#xA;NUMBER to PATTERN  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just for fun, here are the two words that are furthest apart and the ladder between them:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;gametophytes , gametophyte (-s), hepatectomy (g -&gt; c), hypothecate (m -&gt; h), hypothecated (+d), tracheophyte (d -&gt; r), phreatophyte (c -&gt; p), phreatophytes (+s), phanerophytes (t -&gt; n), nephropathies (y -&gt; i), ethnographies (p -&gt; g), stenographies (h -&gt; s), atherogenesis (p -&gt; e), teratogenesis (h -&gt; t), antiestrogens (e -&gt; n), antiestrogen (-s), nitrogenases (t -&gt; s), reassignment (o -&gt; m), mastersinger (n -&gt; r), streamliners (g -&gt; l), masterliness (r -&gt; s), essentialism (r -&gt; i), timelinesses (a -&gt; e), timelessness (i -&gt; s), silentnesses (m -&gt; n), tonelessness (i -&gt; o), solemnnesses (t -&gt; m), movelessness (n -&gt; v), lovelessness (m -&gt; l), lovelinesses (s -&gt; i), slovenliness (e -&gt; n), lovingnesses (l -&gt; g), livingnesses (o -&gt; i), kinglinesses (v -&gt; k), slinkinesses (g -&gt; s), sicklinesses (n -&gt; c), stickinesses (l -&gt; t), trickinesses (s -&gt; r), trickishness (e -&gt; h), ticklishness (r -&gt; l), singlesticks (h -&gt; g), singlestick (-s)&lt;/p&gt;&#xA;" OwnerUserId="239" LastEditorUserId="239" LastEditDate="2014-10-20T17:48:44.433" LastActivityDate="2015-06-19T14:32:30.020" Title="Word Ladder puzzle" Tags="|word|word-ladder|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="2977" PostTypeId="2" ParentId="2976" CreationDate="2014-10-17T13:58:02.400" Score="6" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;wolves shovel shove hopes sheep&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;number bunter banter batter patter pattern&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The second answer seems to be baseball themed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Old:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;wolves solved solve slope sleep sheep&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: as was pointed out by @Golden, Wolves can be done slightly quicker:&lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2014-10-17T14:22:49.187" LastActivityDate="2014-10-17T14:22:49.187" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2980" PostTypeId="1" AcceptedAnswerId="2981" CreationDate="2014-10-17T14:58:57.493" Score="79" ViewCount="156203" Body="&lt;p&gt;Two girls went to dinner together and both ordered iced tea. One girl pounded down five of them in about a minute, and the other took her time drinking one.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The girl who drank one died, while the girl who drank five survived. However, all of the drinks that were served turned out to contain poison.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why did the girl that drank more iced tea survive?&lt;/p&gt;&#xA;" OwnerUserId="2444" LastEditorUserId="9951" LastEditDate="2015-06-19T13:52:36.537" LastActivityDate="2018-09-25T12:56:16.277" Title="Poisoned Iced Tea" Tags="|lateral-thinking|" AnswerCount="11" CommentCount="11" ClosedDate="2015-07-20T18:19:00.913" ContentLicense="CC BY-SA 3.0" />
  <row Id="2981" PostTypeId="2" ParentId="2980" CreationDate="2014-10-17T15:00:26.813" Score="108" Body="&lt;p&gt;Because...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...the poison was in the ice. She drank them before the Ice melted to the point where it would kill her.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2803" LastEditorUserId="42452" LastEditDate="2018-09-25T12:44:26.893" LastActivityDate="2018-09-25T12:44:26.893" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="2984" PostTypeId="2" ParentId="2980" CreationDate="2014-10-17T17:13:54.293" Score="73" Body="&lt;p&gt;Because...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...they were Long Island Iced Teas.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first girl drank so much so fast she couldn't hold her liquor and (fortunately) threw them up before her body absorbed the poison.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3970" LastEditorUserId="42452" LastEditDate="2018-09-25T12:52:48.020" LastActivityDate="2018-09-25T12:52:48.020" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="2986" PostTypeId="2" ParentId="2980" CreationDate="2014-10-17T19:10:08.540" Score="2" Body="&lt;p&gt;Because...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;...the girl who drank five, had drank more than enough to the point of puking.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="42452" LastEditDate="2018-09-25T12:52:14.943" LastActivityDate="2018-09-25T12:52:14.943" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="2987" PostTypeId="2" ParentId="2980" CreationDate="2014-10-17T19:27:05.777" Score="14" Body="&lt;p&gt;Because...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;...the girl who only drank one was allergic to the tea. They were all 'poisoned', but were only meant to kill her, not the other girl.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2756" LastEditorUserId="42452" LastEditDate="2018-09-25T12:51:55.610" LastActivityDate="2018-09-25T12:51:55.610" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="2989" PostTypeId="1" AcceptedAnswerId="2993" CreationDate="2014-10-17T20:40:22.063" Score="-1" ViewCount="1071" Body="&lt;p&gt;An old farmer passes away in his sleep. After a suitable period of mourning, his three sons find their father's will in his study.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Amongst other, simpler bequests is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I wish my small flock of sheep to be well cared for, with the responsibility split between my three suns. Therefore take the pack of cards from my desk drawer, pull out three aces which you will place face up on the desk, then deal out the first three cards from the pack below them to form three fractions. These are the portions into which i wish my flock to be split between my sons.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sons do as requested, pull out the aces and deal out the three cards, which are a 2, a 3 and a 9, forming fractions one-half, one-third and one-ninth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then they go out into the field and count the flock, which turns out to contain 17 sheep.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What cunning sheep-accountancy trick can they use to split the flock according to their father's instructions, without amputating any ovine limbs?&lt;/p&gt;&#xA;" OwnerUserId="2021" LastEditorUserId="2021" LastEditDate="2014-10-17T21:08:57.687" LastActivityDate="2014-10-17T21:29:57.287" Title="Inheritance by cards" Tags="|mathematics|lateral-thinking|cards|" AnswerCount="3" CommentCount="1" ClosedDate="2016-01-02T11:59:21.373" ContentLicense="CC BY-SA 3.0" />
  <row Id="2990" PostTypeId="2" ParentId="2980" CreationDate="2014-10-17T20:56:12.113" Score="1" Body="&lt;p&gt;Because...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;...the girl who drank 5 might have drank enough other fluids to dilute the effects of the poison. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2196" LastEditorUserId="42452" LastEditDate="2018-09-25T12:48:04.773" LastActivityDate="2018-09-25T12:48:04.773" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="2991" PostTypeId="2" ParentId="2989" CreationDate="2014-10-17T21:02:55.023" Score="2" Body="&lt;p&gt;I'd go with 9, 6, 2 for the sheep, which works out fairly well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Between the first and 2nd son, the ratio is 3:2, which the same as 1/2:1/3.  Likewise between the 2nd and 3rd (6:2 = 1/3:1/9). And the same for 1 to 3 (9:2 = 1/2:1/9).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, you could say that the total share of the sons adds up to 17/18. So, each son gets his share times 18/17. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, 1/2 * 17 * 18/17 = 9. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2014-10-17T21:02:55.023" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2992" PostTypeId="2" ParentId="2980" CreationDate="2014-10-17T21:08:19.847" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The girl who drank one drank it slower, as it took her the time for one. The girl who drank five got a fresh glass each time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; THE POISON IS IN THE ICE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3978" LastEditorUserId="42452" LastEditDate="2018-09-25T12:51:28.553" LastActivityDate="2018-09-25T12:51:28.553" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="2993" PostTypeId="2" ParentId="2989" CreationDate="2014-10-17T21:26:47.203" Score="5" Body="&lt;p&gt;I think there was a similar puzzle somewhere here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Borrow one sheep from somewhere else so you have 18. Split in 1/2 (9), 1/3 (6), 1/9 (2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;9+6+2 = 17&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now give back the borrowed sheep.&lt;/p&gt;&#xA;" OwnerUserId="3979" LastActivityDate="2014-10-17T21:26:47.203" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2994" PostTypeId="2" ParentId="2989" CreationDate="2014-10-17T21:29:57.287" Score="2" Body="&lt;p&gt;They borrow a sheep from another farmer!&#xA;Then split up in the ratios:&#xA;1/9 = 2 sheep&#xA;1/3 = 6 sheep&#xA;1/2 = 9 sheep&#xA;That's 17 sheep altogether and then they can return the borrowed sheep!&lt;/p&gt;&#xA;" OwnerUserId="1616" LastActivityDate="2014-10-17T21:29:57.287" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2995" PostTypeId="1" AcceptedAnswerId="3003" CreationDate="2014-10-17T22:47:47.227" Score="-4" ViewCount="376" Body="&lt;p&gt;You must create a sentence that is grammatically correct, each word must begin with the last letter in the word before. The first word must start with the last letter of the last word. This sentence must be at least 4 words long. &lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="5373" LastEditDate="2018-05-06T15:57:29.870" LastActivityDate="2018-05-06T15:57:29.870" Title="The Word Chain!" Tags="|language|" AnswerCount="1" CommentCount="1" ClosedDate="2016-02-28T23:17:09.137" ContentLicense="CC BY-SA 3.0" />
  <row Id="2996" PostTypeId="1" AcceptedAnswerId="3000" CreationDate="2014-10-17T22:57:28.697" Score="4" ViewCount="414" Body="&lt;p&gt;Is it possible to setup a chess board where every square on the board is only controlled once. Occupying a square does not mean controlling it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is possible, with how little pieces?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Piece Cost:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Pawn: 1&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Bishop: 3&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Knight: 3&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Rook: 5&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Queen: 9&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You are not allowed to use the king.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="57521" LastEditDate="2021-04-28T16:38:19.077" LastActivityDate="2021-04-28T16:38:19.077" Title="Equal Chess Control" Tags="|chess|optimization|construction|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2998" PostTypeId="2" ParentId="2980" CreationDate="2014-10-17T23:41:50.970" Score="2" Body="&lt;p&gt;One thing that occurs to me might be...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;...eye drops in the tea.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Visine&quot; rel=&quot;nofollow noreferrer&quot;&gt;Visine&lt;/a&gt; is one example, which contains &lt;a href=&quot;http://en.wikipedia.org/wiki/Tetrahydrozoline&quot; rel=&quot;nofollow noreferrer&quot;&gt;tetrahydrozoline hydrochloride&lt;/a&gt; which derivates from &lt;a href=&quot;http://en.wikipedia.org/wiki/Naphazoline&quot; rel=&quot;nofollow noreferrer&quot;&gt;naphazoline&lt;/a&gt;. She could be unlucky and have &lt;a href=&quot;http://en.wikipedia.org/wiki/Diabetes&quot; rel=&quot;nofollow noreferrer&quot;&gt;diabetes&lt;/a&gt;. Used in enough dosage and with some spoons of sugar, the smell and taste would go away (hopefully). The slowness to drink one could be her not feeling that well, but still keeping drinking.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3984" LastEditorUserId="42452" LastEditDate="2018-09-25T12:50:49.973" LastActivityDate="2018-09-25T12:50:49.973" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="3000" PostTypeId="2" ParentId="2996" CreationDate="2014-10-18T03:00:08.300" Score="9" Body="&lt;p&gt;I found it very hard to only control each square once. Any attempt at limiting the physical number of pieces was turning into an issue. I suspect this can be beaten, but I can manage the whole board in 32 pawns. At least they're only 1 point each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;32.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DSv2q.png&quot; alt=&quot;32 Points&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="375" LastActivityDate="2014-10-18T03:00:08.300" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3001" PostTypeId="2" ParentId="2950" CreationDate="2014-10-18T03:22:46.147" Score="4" Body="&lt;p&gt;Here's an alternate interpretation. Like @Florian, I assume that two black squares can share the same position, and in then next step they will continue their independent paths. However, in the description of the pattern:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Each image introduce its new element at the position given by the coordinate sum of the previous &lt;strong&gt;image&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I interpret this to mean that you count which squares have something covering them rather than the position of the squares that are doing the covering. In other words, overlapping squares only count once for the purpose of determining the position of the new square. This throws off the pattern discovered by Florian, and makes it possible to get enough squares in different positions to eventually cover the grid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather than figure it all out by hand, I used the following Python code to figure it out for me:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;dirs = [(1,0),(0,1),(-1,0),(0,-1)] #Which direction the squares move in&#xA;spots = [(2,2),(4,3)] #The two initial squares&#xA;&#xA;while len(set(spots)) &amp;lt; 25 and len(spots) &amp;lt; 100:&#xA;    newspot = tuple(map(lambda x: (sum(x)-1)%5+1, zip(*set(spots))))&#xA;    spots = [tuple(map(lambda x: (sum(x)-1)%5+1, zip(spot, dirs[i%4]))) for i, spot in enumerate(spots)]&#xA;    spots.append(newspot)&#xA;&#xA;print len(spots)&#xA;# Prints out 61&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I verified that this correctly simulates the first four steps, so I'm quite confident that under this interpretation it will take 61 total squares to completely cover the grid (i.e 59 steps after the initial setup of two squares).&lt;/p&gt;&#xA;" OwnerUserId="1955" LastActivityDate="2014-10-18T03:22:46.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3002" PostTypeId="1" AcceptedAnswerId="3005" CreationDate="2014-10-18T05:46:53.040" Score="9" ViewCount="2067" Body="&lt;p&gt;A traveller decided to explore a square desert. He started from a point on the edge to point B, travelling perpendicular to the edge of the desert he started on.&#xA;On reaching B, he then decided to take a 90 degree turn to travel to point C, where he spent the night. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next day he was told that he could go straight and reach point D, take a 90 degree turn towards the starting point turn and reach point F (which were both on the edge of the desert), or go to a A which was further interior in the desert - which would require him to travel at an unknown angle. He chose to travel to A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After spending another night at A, He retreated to his original starting point by day end. Though he did not intend to, he passed through point B before reaching his final destination. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;He was then amazed to find that he had travelled the same distance in the last two days, and the distance covered by each of those two day's travels was a whole prime number.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;He also concluded the following from his records&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The distances he had travelled from A to B and from B to C added up to the next prime number (to the prime number he estimated to have travelled in each of the last two days)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When he decided to retreat he was equally far from all the corners of the desert &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;All his travels were in perfect straight lines, the only turns he took where when he travelled from point B to C and C to A and A to his starting point.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;He noticed that the distance he travelled from A to B was significantly shorter than his travel from B to C &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;He was wondering if he could find out how far he was from D and F when he decided to travel to A - can you help him?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you may have realised by now, I made this up, so if you think I am missing something add in comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: There are two possible values after my last edit and one of them is captured, but not the other, so I have accepted the answer with one of them now, but will upvote anyone coming up with both&lt;/p&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="2512" LastEditDate="2014-10-24T19:41:58.397" LastActivityDate="2014-10-24T19:41:58.397" Title="Can you help the traveller" Tags="|logical-deduction|mathematics|word-problem|geometry|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3003" PostTypeId="2" ParentId="2995" CreationDate="2014-10-18T07:24:37.143" Score="9" Body="&lt;p&gt;My yacht, tied down near Richard's, seemed drab because Ethel lost the emblem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So our restaurant tried doing gnocchi in new ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reading Graham's sentences should delight the enquirer.&lt;/p&gt;&#xA;" OwnerUserId="2021" LastEditorUserId="2021" LastEditDate="2014-10-18T10:46:52.073" LastActivityDate="2014-10-18T10:46:52.073" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3005" PostTypeId="2" ParentId="3002" CreationDate="2014-10-18T09:20:21.067" Score="3" Body="&lt;p&gt;Lets simplify the travelling path:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The Traveler starts from somewhere on the edge ($Start$ point)&lt;/li&gt;&#xA;&lt;li&gt;He goes perpendicular to edge to point $B$&lt;/li&gt;&#xA;&lt;li&gt;He takes 90 degree turn (either clockwise or counter-clockwise) to point $C$ or $C'$ and spend the night&lt;/li&gt;&#xA;&lt;li&gt;He travels with unknown angle to point $A$ and spend the night again (Which we can conclude that $A$ is the center of desert from his 2nd record)&lt;/li&gt;&#xA;&lt;li&gt;He came back to where he was started ($Start$ point)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The traveler traveled straight and from $A$ to $Start$ through $B$ so also $Start$-$B$ was perpendicular to edge, So $Start$ was in center of it's edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/czASy.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was not clarified that traveler traveled to his left or right at 90 degree turns so it creates some other possible paths. (all have been drawn below)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lQ3Fk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a right triangle, So we can say:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$A$-$B + B$-$C &amp;gt; A$-$C$&lt;/li&gt;&#xA;&lt;li&gt;$(A$-$B)^2 + (B$-$C)^2 = (A$-$C)^2$&lt;/li&gt;&#xA;&lt;li&gt;$A$-$B &amp;lt; A$-$C$ and $B$-$C &amp;lt; A$-$C$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We also know:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$C$-$A = A$-$Start$ are primes in mile.&lt;/li&gt;&#xA;&lt;li&gt;$A$-$B + B$-$C$ is the next prime number to $A$-$C$ (from first record)&lt;/li&gt;&#xA;&lt;li&gt;$B$-$C &amp;gt; A$-$B$ (from 4th record)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So (for minimizing the travel distance,) it came that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$A$-$C = 5$&lt;/li&gt;&#xA;&lt;li&gt;$B$-$C = 4$&lt;/li&gt;&#xA;&lt;li&gt;$A$-$B = 3$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;At point C&lt;/strong&gt;, The traveler should be in $1$ mile distance of point $D$ and $2$ or $8$ miles from point F. (it depends which side would he choose to turn - clockwise or counter-clockwise)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see the numbers below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JQtha.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="2225" LastEditDate="2014-10-18T09:33:17.067" LastActivityDate="2014-10-18T09:33:17.067" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3006" PostTypeId="1" AcceptedAnswerId="3012" CreationDate="2014-10-18T11:53:10.523" Score="4" ViewCount="625" Body="&lt;p&gt;Using only the one colour set; Is it possible to setup a chess board where every square on the board is only controlled once. Occupying a square does not mean controlling it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is possible, with how little pieces?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Piece Cost:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Pawn: 1&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Bishop: 3&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Knight: 3&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Rook: 5&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Queen: 9&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;You are not allowed to use the king.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;All pawns must face the same direction&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2019-02-12T21:11:52.210" Title="Equal Chess Control With One Color" Tags="|chess|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3007" PostTypeId="2" ParentId="3002" CreationDate="2014-10-18T13:44:57.747" Score="2" Body="&lt;p&gt;Let's say distance $AC = x, BC = y, AB = z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, x is prime p, y + z = q, the next prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y^2+z^2=p^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(y+z)^2=q^2=y^2+z^2+2yz$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2yz=q^2-p^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $y=q-z$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2(q-z)z=q^2-p^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Giving $z^2-qz+\frac{q^2-p^2}{2}=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which has real roots when&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q^2 \geq 4\frac{q^2-p^2}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which reduces to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ q \leq \sqrt{2}p$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And gives $y,z = \frac{q\pm \sqrt{2p^2-q^2}}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This rules out $p,q=7,11$ as a pair. $p,q=11,13$ satisfies our condition, and gives $y,z=\frac{13\pm \sqrt{73}}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume you were looking for the rather nicer pair of $p,q=13,17$ giving $y,z = 5,12$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whatever values we get for x,y,z, the distances to F and D are given by $D=z-y, F=z\pm x$ (this is shown quite nicely in Rafe's answer)&#xA;For the specific example of $p,q=13,17$ D is 1 and F is 8 or 18, though you can generate arbitrary numbers of solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I missed the change stating that F is on the same side as the start point. This removes the larger value for F from all solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 2: typo&lt;/p&gt;&#xA;" OwnerUserId="3987" LastEditorUserId="3987" LastEditDate="2014-10-18T15:38:42.687" LastActivityDate="2014-10-18T15:38:42.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3010" PostTypeId="1" AcceptedAnswerId="3011" CreationDate="2014-10-18T15:20:10.633" Score="7" ViewCount="1057" Body="&lt;p&gt;okay this is the scenario this time; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rebel member was captured in order to intercept a message. There are several items he was carrying on him; his wallet, house key, an old mobile phone, and 5 piece of papers (as shown below). Figure out what is the message he was carrying. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LlEgT.png&quot; alt=&quot;paper5&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/GjJOL.png&quot; alt=&quot;paper1&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/UMJSB.png&quot; alt=&quot;paper3&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/BVD6Y.png&quot; alt=&quot;paper4&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Rnutq.png&quot; alt=&quot;paper2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's something else needed to solve the puzzle, but what it is, is already hinted. You can also use an image editing software for help, although not really needed. The final answer will be in text form (words). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for a really long post, I need to use quite a number of images.&lt;/p&gt;&#xA;" OwnerUserId="2373" LastEditorUserId="4385" LastEditDate="2015-12-15T15:53:30.413" LastActivityDate="2022-06-06T11:44:34.903" Title="Hidden Message: 5 Papers" Tags="|cipher|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3011" PostTypeId="2" ParentId="3010" CreationDate="2014-10-18T15:30:40.737" Score="7" Body="&lt;p&gt;&lt;strong&gt;The message is&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;Destroy west area tonight.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Gather all of the blocks by label, read clockwise. The numbers are (key number. position on key) for old style mobile phones.&lt;/p&gt;&#xA;&lt;p&gt;Edit:&#xA;Ignore the first piece of paper. For the other pieces, arrange them to form a complete shape out of the 4 segments labelled '1'. Record, in order (starting from top left), the red numbers running around the outside.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;(3.1, 3.2, 7.4, 8.1, 7.3, 6.3, 9.3).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Repeat this for segments labelled '2', '3' and '4', giving you:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;(9.1, 3.2, 7.4, 8.1)&#xA;&#xA;(2.1, 7.3, 3.2, 2.1)&#xA;&#xA;(8.1, 6.3, 6.2, 4.3, 4.1, 4.2, 8.1)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Examining these, you'll notice that the first digit is 2-9 and the second is 1-4, which calls to mind old style phone keypads (particularly when one was mentioned in the question). The first number is which key to look at, the second number is which letter on that key is meant. So 3.2 means 3rd key ('def'), second letter:'e'&lt;/p&gt;&#xA;&lt;p&gt;Working through the list:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;we get 'destroy west area tonight'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3987" LastEditorUserId="77294" LastEditDate="2022-06-06T11:44:34.903" LastActivityDate="2022-06-06T11:44:34.903" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="3012" PostTypeId="2" ParentId="3006" CreationDate="2014-10-18T15:50:24.687" Score="6" Body="&lt;p&gt;I can manage 34. I haven't gotten better than that. 4 rooks, 14 pawns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;34.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rrlBV.png&quot; alt=&quot;34 Points&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="375" LastActivityDate="2014-10-18T15:50:24.687" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="3013" PostTypeId="1" AcceptedAnswerId="3015" CreationDate="2014-10-18T15:56:48.480" Score="17" ViewCount="2836" Body="&lt;p&gt;This is the problem I came across reading the book &lt;em&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471135712&quot;&gt;The Art and Craft of Problem Solving&lt;/a&gt;&lt;/em&gt;. When I read this question I wasn't able to figure out the solution and I saw the solution after a while, but still I couldn't understand the solution as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Pat wants to take a $1.5$-meter-long sword onto a train, but the conductor won't allow it as carry-on luggage. And the baggage person won't take any item which greatest dimension exceeds $1$ meter. What should Pat do?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is unsolvable if we limit ourselves to two-dimensional space. Once liberated from Flatland, we get a nice solution : The sword fits into a $1 \times 1 \times 1$ -meter-box, with a long diagonal of  $\sqrt{(1^2 + 1^2 + 1^2)}$ = $\sqrt{3} &amp;gt; 1.69$ meters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can anyone give me a clear explanation of this solution?&lt;/p&gt;&#xA;" OwnerUserId="2250" LastEditorUserId="5373" LastEditDate="2015-04-28T19:35:00.990" LastActivityDate="2018-02-22T09:39:59.777" Title="Baggage Problem: $1.5$-meter-long sword onto a train" Tags="|mathematics|geometry|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3014" PostTypeId="2" ParentId="3013" CreationDate="2014-10-18T16:10:24.697" Score="1" Body="&lt;p&gt;The baggage person won't allow the sword, because it's 1.5m long. They will, however, allow a 1m x 1m x 1m box. If you put the sword running from one corner of the box to the opposite, it will fit (because the diagonal is ~1.7m) and be allowed on board.&lt;/p&gt;&#xA;" OwnerUserId="3987" LastActivityDate="2014-10-18T16:10:24.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3015" PostTypeId="2" ParentId="3013" CreationDate="2014-10-18T16:10:29.207" Score="26" Body="&lt;p&gt;The sword fits in that 3D box, because the long diagonal is $1.69$ meter. The longest dimension of this box is 1 meter (length, width and height are all exactly 1 meter). The sword goes like the green line in the picture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uYYVu.png&quot; alt=&quot;Box&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Pat would use a flat case, it would be denied. In the most naive way, the box would be $1.5$ meter long and have a width of the sword itself. A more clever way to think is in the maximum dimensions, which are $1 \times 1$ meter. In that case the diagonal is $\sqrt{2} = 1.41$ meter long, which is not long enough. So the only option is to include the height into account. &lt;/p&gt;&#xA;" OwnerUserId="3993" LastEditorUserId="3993" LastEditDate="2014-10-18T19:59:32.267" LastActivityDate="2014-10-18T19:59:32.267" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3017" PostTypeId="1" AcceptedAnswerId="3019" CreationDate="2014-10-18T20:08:38.233" Score="17" ViewCount="34550" Body="&lt;p&gt;What is the complete set of 10-digit numbers using each value ($0$-$9$) once, such that the first $n$ digits form a number divisible by $n$ (for $n \in \{1,2,...,10\}$)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. if the number were $1234567890$, $1$ must be divisible by $1$, $12$ must be divisible by $2$, $123$ must be divisible by $3$, etc...&lt;/p&gt;&#xA;" OwnerUserId="3987" LastEditorUserId="4421" LastEditDate="2015-10-19T20:00:37.330" LastActivityDate="2018-08-08T20:34:26.070" Title="10 digit number where first n digits are divisible by n" Tags="|mathematics|" AnswerCount="3" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="3018" PostTypeId="2" ParentId="2905" CreationDate="2014-10-18T21:28:19.220" Score="0" Body="&lt;p&gt;Have Alice divide the estate into 3 fair (from her POV) parts. Offer Bob and Eve the choice of the three portions. If they choose different ones, we're done. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they choose the same portion, split it between them. Then have them each choose one of the remaining 2 portions. If they choose the same portion, split that between them. They each now have a fair third, and Alice takes the remainder. If they choose different portions at this point, have them each share the portion they choose with Alice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Splitting a portion between two people is, of course, done by having one person divide it and the other choose.&lt;/p&gt;&#xA;" OwnerUserId="3987" LastActivityDate="2014-10-18T21:28:19.220" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3019" PostTypeId="2" ParentId="3017" CreationDate="2014-10-18T22:09:33.687" Score="22" Body="&lt;p&gt;There's only one answer:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;3816547290&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If leading zeroes are excluded, then there are 2492 numbers that satisfy the successive divisibility condition. This C program finds them all in a few seconds:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#include &amp;lt;stdio.h&amp;gt;&#xA;&#xA;int main(void)&#xA;{&#xA;  long i,j,n,c=0;&#xA;  for (i=1000000080; i&amp;lt;=9999999990; i+=90) {&#xA;    for (n=i,j=10; j&amp;gt;1; n/=10,j--) if (n%j) break;&#xA;    if (j==1) {&#xA;      c++;&#xA;      printf(&quot;%ld &quot;,i);&#xA;    }&#xA;  }&#xA;  printf(&quot;\nTotal: %ld\n&quot;,c);&#xA;  return 0;&#xA;}&#xA;&#xA;// Output: 1020005640 1020061620 1020068010 ... 9876062430 9876069630 9876545640&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;However, 3816547290 is the only one of these numbers that contains all the digits 0-9.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;On a computer, this can be solved quite quickly by restricting the range of numbers that has to be searched.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that the last digit must be zero, so the only remaining option for the fifth digit is 5. All the digits in the other even positions must be either 2, 4, 6 or 8. And since there are four even positions to be filled, the numbers in the odd positions (apart from 5) must be 1, 3, 7 or 9 (because there are no even digits left to go around.)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Digit:       1       2       3       4       5       6       7       8       9      10&#xA;Value:    1/3/7/9 2/4/6/8 1/3/7/9 2/4/6/8    5    2/4/6/8 1/3/7/9 2/4/6/8 1/3/7/9    0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There are only 4^8 (=65536) numbers that fit this pattern. These can be tested in just a few milliseconds.&lt;/p&gt;&#xA;" OwnerUserId="40" LastEditorUserId="40" LastEditDate="2015-07-20T14:24:34.617" LastActivityDate="2015-07-20T14:24:34.617" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3020" PostTypeId="2" ParentId="2842" CreationDate="2014-10-18T22:22:57.047" Score="0" Body="&lt;p&gt;First, you make a perfect circle with your belt (or as approximate as you can).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you mark the center of the belt and draw a straight line across the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now count how many times the belt is bigger than the line (should be 3).&lt;/p&gt;&#xA;" OwnerUserId="3984" LastActivityDate="2014-10-18T22:22:57.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3021" PostTypeId="2" ParentId="2157" CreationDate="2014-10-18T23:46:36.277" Score="0" Body="&lt;p&gt;As @eBusiness notes, by coupling their guesses the friends can achieve 75% success. By choosing the opposite of the 'previous' box, they score every time there is a change in colour between one box and the next. This is useful because there are always either 0 (25%) or 2 (75%) changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Variant 1: If the probability of black stones is 70%, however, all friends guessing black succeeds with probability $.7^3+3*.7^2*.3=.784$. The previous approach now only works for the 63% of the time both colours are present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Variant 2: If they may swap the stones, friend 1 guesses the colour they see in box 2. Friend 2 guesses the colour they see in box 1. Friend 3 then switches these stones they see, making the previous 2 correct with 100% probability.&lt;/p&gt;&#xA;" OwnerUserId="3987" LastActivityDate="2014-10-18T23:46:36.277" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3022" PostTypeId="2" ParentId="3017" CreationDate="2014-10-19T00:10:12.210" Score="29" Body="&lt;p&gt;Let us write the number as $abcdefghij$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ab$, $abcd$, $abcdef$, $abcdefgh$, $abcdefghij$ are all even, therefore $\{b, d, f, h, j\}$ are the even digits and the remaining digits $\{a, c, e, g, i\}$ are the odd digits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's write $k \mid N$ (read &quot;k divides N&quot;) when $N$ is a multiple of $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$10 \mid abcdefghij \implies j = 0$.&lt;br&gt;&#xA;$5 \mid abcde \implies e$ is $0$ or $5$.  But $0$ is taken, so $e = 5$.&lt;br&gt;&#xA;$4 \mid abcd$ $\implies$ $4 \mid cd$.  This with $c$ odd $\implies$ $d \in \{2,6\}$.&lt;br&gt;&#xA;$8 \mid abcdefgh$ $\implies$ $4 \mid gh$.  This with $g$ odd $\implies$ $h \in \{2,6\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last 2 lines imply $\{d,h\} = \{2,6\}$.  In either order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know $\{b, d, f, h, j\}$ are the even digits. Since $j=0$ and $\{d,h\} = \{2,6\}$, the remaining digits are 4 and 8.  $\{b,f\}=\{4,8\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3 \mid abc \implies 3 \mid abc000$.  This with $6 \mid abcdef$ implies $3 \mid def$.&#xA;But we already know $e=5$, $d \in \{2,6\}$ and $f \in \{4,8\}$.  Only $def=258$ and $def=654$ satisfy these conditions.  The choice of $d$ and $f$ also force the value of $b$ and $h$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far we have 2 partial solutions: $\_4\_258\_6\_0$ and $\_8\_654\_2\_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$8 \mid abcdefgh$ $\implies$ $8 \mid fgh$.  But $f$ is even, $8 \mid f00$, so we just need $8 \mid gh$.  This with $g \in \{1,3,7,9\}$ leaves only $gh \in \{16,32,72,96\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Let's consider $\_4\_258\_6\_0$.  In that case we have $g \in \{1,9\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3 \mid abc$ with $b=4$ and $a,c \in \{1,3,7,9\}$ leaves only 2 possibilities: $abc \in \{147, 741\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $abcdef=147258$ with $7 \mid abcdefg$ then $g=3$.  But this is not possible because $g \in \{1,9\}$.&lt;br&gt;&#xA;If $abcdef=741258$ with $7 \mid abcdefg$ then $g=7$.  This is not possible for the same reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there is no solution for case (a).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Let's consider $\_8\_654\_2\_0$.  In that case we have $g \in \{3,7\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3 \mid abc$ with $b=8$ and $a,c \in \{1,3,7,9\}$ gives 8 possibilities: $abc \in \{183,189,381,387,783,789,981,987\}$.  Two of them, $387$ and $783$, are incompatible with $g \in \{3,7\}$.  They can be removed.  (Thanks to frodoskywalker.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the remaining 6 solutions you need to find $g$ that satisfies $7 \mid abcdefg$ and verify that $g \in \{3,7\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $abc=381$, we have $abcdef=381654$.  $7 \mid abcdefg$ implies $g=7$ which is valid. You can complete it with $i=9$ and you get the solution: $abcdefghij=3816547290$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No other $abc$ under consideration gives a valid $g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The only solution is 3816547290.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2018-08-08T20:34:26.070" LastActivityDate="2018-08-08T20:34:26.070" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="3024" PostTypeId="1" AcceptedAnswerId="3058" CreationDate="2014-10-19T12:16:15.973" Score="-3" ViewCount="487" Body="&lt;p&gt;Please note I realize there could be many &quot;correct answers&quot; but I'm looking for a specific one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1, 3, 5, 4, 2, 7, 6, 8, 9, 3, 4, 5, 8, 2, 9, 3, 5, 2, 42, 35, 42, 52, 55, 52, 109, 85, 120, ?, ?, ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It has to do with multiplying, each &quot;part&quot; being multiplied, is multiplied by the part before it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Split it up into 9s.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;After splitting it up into 9s, split it up Again into 3s&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint4: (ONLY USE THIS IF YOU HAVE TO)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Think matrices.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you answer please include which hints you used.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="4050" LastEditDate="2015-01-03T17:54:36.653" LastActivityDate="2015-01-03T17:54:36.653" Title="Finish The Pattern" Tags="|pattern|number-sequence|" AnswerCount="1" CommentCount="15" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3025" PostTypeId="1" AcceptedAnswerId="3033" CreationDate="2014-10-19T13:37:37.613" Score="5" ViewCount="900" Body="&lt;p&gt;Knowing the solution I consider this pattern as a quite beautiful.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Find the pattern:&#xA;   HH,EL,IB,EB,CN,OF,NE,NA,MG,AL,SI,PS,CL,AR,KC,AS,CT, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2014-10-19T13:42:45.880" LastActivityDate="2014-10-20T00:39:27.360" Title="Beautiful HH-pattern. Find it" Tags="|pattern|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3026" PostTypeId="1" AcceptedAnswerId="3028" CreationDate="2014-10-19T13:56:55.673" Score="6" ViewCount="1817" Body="&lt;p&gt;The non-negative integers are divided into three groups as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A = {0,3,6,8,9...}, B = {1,4,7,11,14...}, c={2,5,10,13...}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explain ? I have no idea how to proceed.&lt;/p&gt;&#xA;" OwnerUserId="2250" LastActivityDate="2014-10-21T12:40:30.617" Title="Non-Negative integers in to three groups" Tags="|mathematics|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3027" PostTypeId="1" AcceptedAnswerId="3029" CreationDate="2014-10-19T13:58:36.607" Score="3" ViewCount="897" Body="&lt;p&gt;In regular chess no piece can checkmate the opposite king alone (i.e. without the help of his allied king). But a &lt;a href=&quot;http://en.wikipedia.org/wiki/Maharajah_and_the_Sepoys&quot; rel=&quot;noreferrer&quot; title=&quot;Maharajah&quot;&gt;Maharajah&lt;/a&gt; can. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there a piece weaker than the Maharajah, of course invented, that can &lt;strong&gt;force&lt;/strong&gt; checkmate on the opposite king alone?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there a piece weaker than the Maharajah, of course invented, that can checkmate the opposite king alone (but optimal play from the opposite king can hold a draw)?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To calculate the power of a piece: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Take sum of squares attacked in all corners as $x$, &lt;/li&gt;&#xA;&lt;li&gt;Number of squares attacked when in any of 4 centre squares as $y$. &lt;/li&gt;&#xA;&lt;li&gt;The power of the piece is then $\frac{x+y}{5}$ &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For example the bishop: $7 \times 4$ for the corners, $13$ for the centre. $\frac{28+13}{5} = 8.2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A piece is weaker than another if its power is lower.&lt;/p&gt;&#xA;" OwnerUserId="4016" LastEditorUserId="3993" LastEditDate="2014-10-19T18:39:31.187" LastActivityDate="2016-07-25T19:52:41.313" Title="Weakest chess piece that can checkmate the opposite king alone" Tags="|chess|" AnswerCount="3" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3028" PostTypeId="2" ParentId="3026" CreationDate="2014-10-19T14:16:06.290" Score="12" Body="&lt;p&gt;A contains integers (typically) composed of just curved lines when written.  B contains integers (typically) composed of just straight lines when written.  And C contains integers (typically) composed of both curved and straight lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;kaine has pointed out that &quot;12&quot; seems to be missing.  If you do a web search for the following, you can see that kaine is likely correct:  &quot;0, 3, 6, 8, 9&quot; &quot;1, 4, 7, 11, 14&quot;, &quot;2, 5, 10, 12, 13&quot;&lt;/p&gt;&#xA;" OwnerUserId="1923" LastEditorUserId="1923" LastEditDate="2014-10-21T12:40:30.617" LastActivityDate="2014-10-21T12:40:30.617" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3029" PostTypeId="2" ParentId="3027" CreationDate="2014-10-19T14:24:41.353" Score="2" Body="&lt;p&gt;This all assumes that one cannot have conditional-attack pieces. Like &quot;only attacks iff the enemy king is on the left.&quot;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I'm pretty sure that one piece cannot force a mate without at least power 8. A queen that attacks the squares containing white pawns:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9P4rd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;A queen with power 1.2 that &lt;em&gt;can&lt;/em&gt; checkmate: She may attack only squares that 6 or 7 square &quot;ahead&quot; (side-wise) of her, and of those only those that share her column and the columns immediately to the left. She will attack 4 squares when on D1: C8,D8,C7,D7. In fact, the checkmate must occur with the king in H8 and the queen is in H1.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;By the way, the piece you call a &quot;Maharajah&quot; is also known as the &quot;Amazon&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-10-19T15:41:31.880" LastActivityDate="2014-10-19T15:41:31.880" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3030" PostTypeId="2" ParentId="3027" CreationDate="2014-10-19T15:48:51.993" Score="2" Body="&lt;p&gt;A piece with power 0 by your definition of &quot;power&quot;:  On the third rank, it attacks the entire board.  On any other square, it attacks nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If instead you change &quot;power&quot; to &lt;em&gt;&quot;sum of all spaces attacked on every square of the board&quot;&lt;/em&gt;, then the provably-minimal piece would be:  On one square it attacks one corner in a 2x2 area.  On every other square, it attacks nothing.&lt;/p&gt;&#xA;" OwnerUserId="293" LastEditorUserId="293" LastEditDate="2014-10-19T15:54:03.203" LastActivityDate="2014-10-19T15:54:03.203" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3031" PostTypeId="2" ParentId="2813" CreationDate="2014-10-19T16:04:17.167" Score="1" Body="&lt;p&gt;So far nobody has given an optimal answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what I believe is the optimal answer for 6 stacks.  It can sort 8 numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rule is simple to follow: move each card to the next stack, unless the number is forbidden for that stack in which case you put it on the following one.  When all the cards are on stack 6, the numbers are sorted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The forbidden numbers are as follows:&lt;br&gt;&#xA;Stack 1: none.&lt;br&gt;&#xA;Stack 2: 1,6,7.&lt;br&gt;&#xA;Stack 3: 3,4.&lt;br&gt;&#xA;Stack 4: 7,8.&lt;br&gt;&#xA;Stack 5: 1,2,3.&lt;br&gt;&#xA;Stack 6: none.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following this, from stack 1, cards 2,3,4,5,8 go to stack 2, cards 1,6,7 go to stack 3.&#xA;Then, from stack 2, cards 2,5,8 add to stack 3, cards 3,4 go to stack 4.&#xA;etc.  You can try, it works!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anyone who can generalize this to M stacks?&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2014-10-19T20:14:57.910" LastActivityDate="2014-10-19T20:14:57.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3032" PostTypeId="2" ParentId="2980" CreationDate="2014-10-19T16:21:41.803" Score="7" Body="&lt;p&gt;Because...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;...the girl who drank one glass was a small child, while the girl who drank five was an adult. The poison was of such a low concentration that it only killed the child. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="431" LastEditorUserId="42452" LastEditDate="2018-09-25T12:48:59.693" LastActivityDate="2018-09-25T12:48:59.693" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="3033" PostTypeId="2" ParentId="3025" CreationDate="2014-10-19T16:36:45.387" Score="9" Body="&lt;p&gt;IV,CR,MN,FE...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The NE...SI bunch in the middle seem to indicate the periodic table. Bringing one up shows that this is all the elements' symbols in periodic table order, taken two letters at a time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="375" LastEditorUserId="4006" LastEditDate="2014-10-20T00:39:27.360" LastActivityDate="2014-10-20T00:39:27.360" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3036" PostTypeId="1" AcceptedAnswerId="3038" CreationDate="2014-10-20T04:15:31.080" Score="15" ViewCount="1669" Body="&lt;p&gt;Does anyone knows the name of the puzzle showed in the image below?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ezuq0.jpg&quot; alt=&quot;Starting position&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The objective is to mount a 3x3x3 cube such that all sides of this cube are equal. The small cubes of the image are linked by an elastic and some of them can rotate along this link.&lt;/p&gt;&#xA;" OwnerUserId="270" LastActivityDate="2014-10-20T05:35:55.657" Title="What's the name of this puzzle, which consists of small cubes connected by elastics?" Tags="|puzzle-identification|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3037" PostTypeId="2" ParentId="2813" CreationDate="2014-10-20T04:22:35.940" Score="8" Body="&lt;h2&gt;With $M$ stacks we can sort $F_M$ numbers&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;(where $F_M$ is the $M$-th &lt;a href=&quot;http://en.wikipedia.org/wiki/Fibonacci_number&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fibonacci Number&lt;/a&gt;)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an answer for the call for generalization in &lt;a href=&quot;https://puzzling.stackexchange.com/a/3031/1649&quot;&gt;OP's answer&lt;/a&gt;, by generalizing his/her method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For comparison with other answers, the asymptotic $N$ as $M$ goes to infinity is $C\varphi^N$ (where $\varphi$ is the golden ratio: $1.61803398875\ldots$) for some constant $C$. Why is this true can be read in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fibonacci_number#Limit_of_consecutive_quotients&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that, compared to &lt;a href=&quot;https://puzzling.stackexchange.com/a/2925/1649&quot;&gt;user3294068's answer&lt;/a&gt;, which asymptotically uses $C2^{\frac{N}{2}} = C\sqrt{2}^N$, this algorithm is definitely better for larger $N$ ($\sqrt{2} = 1.41421356\ldots$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll divide this answer to three sections: a visualization of the algorithm, the general idea how it works, - which leads to the final part - and the general procedure for any $M$.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Visualization of Algorithm&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;This section will explain the algorithm in OP's answer through step-by-step visualization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For completeness, I'll copy his algorithm here, slightly modified to generalize the algorithm:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The rule is simple to follow: move each card to the next stack, unless the number is forbidden for that stack in which case you put it on the following one. When all the cards are on final stack, the numbers are sorted.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;For $M=6$, the forbidden numbers are as follows:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Stack 1: none.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Stack 2: 1,6,7.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Stack 3: 3,4.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Stack 4: 7,8.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Stack 5: 1,2,3.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Stack 6: none.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Due to my limited knowledge of LaTeX, I'll draw the stacks sideways, from bottom to top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, initially we have stack 1 ($S_1$) and that it has nothing forbidden ($\{\}$), containing the number $1$ to $8$. We do not know in what order they are initially, so we group them together, like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{eqnarray}&#xA;S_1\{\}&amp;amp;:&amp;amp; \boxed{12345678}&#xA;\end{eqnarray}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The grouping means the number in that group can be in any order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, we have Stack 2 forbidding three numbers ($S_2\{1,6,7\}$), and Stack 3 forbidding two numbers ($S_3\{3,4\}$), like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{eqnarray}&#xA;S_1\{\}&amp;amp;:&amp;amp; \boxed{12345678}\\&#xA;S_2\{1,6,7\}&amp;amp;:&amp;amp;\\&#xA;S_3\{3,4\}&amp;amp;:&amp;amp;&#xA;\end{eqnarray}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we put the numbers from Stack 1 to Stack 2 if it's not forbidden, else put in Stack 3. We'll end up like this (red means it's no longer there):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{eqnarray}&#xA;S_1\{\}&amp;amp;:&amp;amp; \color{red}{\boxed{12345678}}\\&#xA;S_2\{1,6,7\}&amp;amp;:&amp;amp; \boxed{23458}\\&#xA;S_3\{3,4\}&amp;amp;:&amp;amp; \boxed{167}&#xA;\end{eqnarray}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, we have Stack 4 with its 2 forbidden numbers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{eqnarray}&#xA;S_1\{\}&amp;amp;:&amp;amp; \color{red}{\boxed{12345678}}\\&#xA;S_2\{1,6,7\}&amp;amp;:&amp;amp; \boxed{23458}\\&#xA;S_3\{3,4\}&amp;amp;:&amp;amp; \boxed{167}\\&#xA;S_4\{7,8\}&amp;amp;:&amp;amp;&#xA;\end{eqnarray}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, after distributing Stack 2 to Stack 3 and 4 we have this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{eqnarray}&#xA;S_1\{\}&amp;amp;:&amp;amp; \color{red}{\boxed{12345678}}\\&#xA;S_2\{1,6,7\}&amp;amp;:&amp;amp; \color{red}{\boxed{23458}}\\&#xA;S_3\{3,4\}&amp;amp;:&amp;amp; \boxed{167}\boxed{258}\\&#xA;S_4\{7,8\}&amp;amp;:&amp;amp; \boxed{34}&#xA;\end{eqnarray}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in Stack 3, the group containing $258$ is on top of the group containing $167$, although in each group the order is arbitrary. So at this point Stack 3 might contain $671285$ or $761258$, but not $168257$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means, when splitting Stack 3 to Stack 4 and 5, group containing $258$ is split before the group containing $167$, so we have the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{eqnarray}&#xA;S_1\{\}&amp;amp;:&amp;amp; \color{red}{\boxed{12345678}}\\&#xA;S_2\{1,6,7\}&amp;amp;:&amp;amp; \color{red}{\boxed{23458}}\\&#xA;S_3\{3,4\}&amp;amp;:&amp;amp; \boxed{167}\color{red}{\boxed{258}}\\&#xA;S_4\{7,8\}&amp;amp;:&amp;amp; \boxed{34}\boxed{25}\\&#xA;S_5\{1,2,3\}&amp;amp;:&amp;amp; \boxed{8}&#xA;\end{eqnarray}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{eqnarray}&#xA;S_1\{\}&amp;amp;:&amp;amp; \color{red}{\boxed{12345678}}\\&#xA;S_2\{1,6,7\}&amp;amp;:&amp;amp; \color{red}{\boxed{23458}}\\&#xA;S_3\{3,4\}&amp;amp;:&amp;amp; \color{red}{\boxed{167}\boxed{258}}\\&#xA;S_4\{7,8\}&amp;amp;:&amp;amp; \boxed{34}\boxed{25}\boxed{16}\\&#xA;S_5\{1,2,3\}&amp;amp;:&amp;amp; \boxed{8}\boxed{7}&#xA;\end{eqnarray}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that at Stack 5, each group contains only one element, this means there is no variation, and so at this point, we can be sure that Stack 5 contains $8$ at the bottom and $7$ on top of it. Continuing this, we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{eqnarray}&#xA;S_1\{\}&amp;amp;:&amp;amp; \color{red}{\boxed{12345678}}\\&#xA;S_2\{1,6,7\}&amp;amp;:&amp;amp; \color{red}{\boxed{23458}}\\&#xA;S_3\{3,4\}&amp;amp;:&amp;amp; \color{red}{\boxed{167}\boxed{258}}\\&#xA;S_4\{7,8\}&amp;amp;:&amp;amp; \color{red}{\boxed{34}\boxed{25}\boxed{16}}\\&#xA;S_5\{1,2,3\}&amp;amp;:&amp;amp; \boxed{8}\boxed{7}\boxed{6}\boxed{5}\boxed{4}\\&#xA;S_6\{\}&amp;amp;:&amp;amp; \boxed{1}\boxed{2}\boxed{3}&#xA;\end{eqnarray}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally moving the whole Stack 5 into Stack 6 will give us sorted from $1$ at the bottom to $8$ at the top.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Insights&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;So, observing the process above, we see that the algorithm works by splitting the numbers into multiple groups, each containing less elements than before, ensuring that when the group size finally reach 1, they are all in order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, to generalize this, we need to find how to split the numbers into groups of smaller size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, note that the restriction to use only 3 groups means if we have $k$ group in Stack $i$, we can split them into $k$ groups in Stack $i$ and $k$ group in Stack $i+1$. This implies the number of groups in Stack $i$, which comes from Stack $i-2$ and Stack $i-1$, is at most the sum of number of groups in Stack $i-2$ and Stack $i-1$. Formally, if $G_i$ denotes the number of groups in Stack $i$, we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ G_i = G_{i-1} + G_{i-2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initially we have only one group in Stack 1, and later we can have at most 1 group in Stack 2. This gives $G_1 = G_2 = 1$, which makes $G_i = F_i$, where $F_i$ is the $i$-th Fibonacci number. So with $M$ stacks, we can sort $F_M$ numbers (precise answer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, we can use the similar reasoning to prove the optimality of this algorithm. Note that at the end we must have $N$ groups, each containing one number. So considering the first insight above, the maximum number of groups at Stack $M$ is $F_M$, therefore we can't have less than $M$ stacks to sort $F_M+1$ numbers. So, by taking $M$ such that $F_{M-1} &amp;lt; N \leq F_M$ (which would be about $\lfloor \log_\varphi \left(N\sqrt{5}+\frac{1}{2}\right)\rfloor$ according to &lt;a href=&quot;http://en.wikipedia.org/wiki/Fibonacci_number#Computation_by_rounding&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Wikipedia article&lt;/a&gt;), it is the minimum required (and sufficient) to sort $N$ numbers.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Generalization Procedure&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Now we come to the interesting part, how can we generalize this to any $M$, not just $M=6$ like in the example? More specifically, how should we assign the set of forbidden numbers for each Stack?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this, we can work backward from Stack $M$ down to Stack 1, assuming maximum possible $N$. Note that the set of forbidden numbers for $M$ stacks will work for any $N$ less than the maximum $N$ (the numbers is guaranteed to be sorted at Stack $M$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we know that there are $F_M = N$ groups at Stack $M$ at the end. One step before (that is, after we emptied Stack $M-2$), there are $F_{M-1}$ groups in Stack $M-1$ and $F_{M-2}$ groups in Stack $M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, we know that in order for those first $F_{M-2}$ numbers to be in Stack $M$, they must be forbidden at Stack $M-1$ (because those numbers come only from Stack $M-2$, so those numbers can't ever be inside Stack $M-1$). So we now know the forbidden numbers at Stack $M-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To continue, we note that in the algorithm, we &lt;em&gt;split&lt;/em&gt; groups, so now we need the reverse, which is to &lt;em&gt;merge&lt;/em&gt; groups. We know that the $i$-th top-most group in Stack $M-1$ would have come from the same group in Stack $M-2$ as the $i$-th top-most group in Stack $M$. This enables us to reverse the process, and continue numbering the forbidden numbers in the same way until we get back to Stack 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, here is the procedure for $M=7, N=13$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{eqnarray}&#xA;S_7\{\}&amp;amp;:&amp;amp; \boxed{1}\boxed{2}\boxed{3}\boxed{4}\boxed{5}\boxed{6}\boxed{7}\boxed{8}\boxed{9}\boxed{10}\boxed{11}\boxed{12}\boxed{13}&#xA;\end{eqnarray}&#xA;$$&#xA;&lt;hr/&gt;&#xA;$$&#xA;\begin{eqnarray}&#xA;S_6\{1,2,3,4,5\}&amp;amp;:&amp;amp;\boxed{13}\boxed{12}\boxed{11}\boxed{10}\boxed{9}\boxed{8}\boxed{7}\boxed{6}\\&#xA;S_7\{\}&amp;amp;:&amp;amp; \boxed{1}\boxed{2}\boxed{3}\boxed{4}\boxed{5}\color{red}{\boxed{6}\boxed{7}\boxed{8}\boxed{9}\boxed{10}\boxed{11}\boxed{12}\boxed{13}}&#xA;\end{eqnarray}&#xA;$$&#xA;&lt;hr/&gt;&#xA;$$&#xA;\begin{eqnarray}&#xA;S_5\{11,12,13\}&amp;amp;:&amp;amp;\boxed{56}\boxed{47}\boxed{38}\boxed{29}\boxed{1\ 10}\\&#xA;S_6\{1,2,3,4,5\}&amp;amp;:&amp;amp;\boxed{13}\boxed{12}\boxed{11}\color{red}{\boxed{10}\boxed{9}\boxed{8}\boxed{7}\boxed{6}}\\&#xA;S_7\{\}&amp;amp;:&amp;amp; \color{red}{\boxed{1}\boxed{2}\boxed{3}\boxed{4}\boxed{5}\boxed{6}\boxed{7}\boxed{8}\boxed{9}\boxed{10}\boxed{11}\boxed{12}\boxed{13}}&#xA;\end{eqnarray}&#xA;$$&#xA;&lt;hr/&gt;&#xA;$$&#xA;\begin{eqnarray}&#xA;S_4\{4,5,6,7\}&amp;amp;:&amp;amp;\boxed{1\ 10\ 11}\boxed{2\ 9\ 12}\boxed{3\ 8\ 13}\\&#xA;S_5\{11,12,13\}&amp;amp;:&amp;amp;\boxed{56}\boxed{47}\color{red}{\boxed{38}\boxed{29}\boxed{1\ 10}}\\&#xA;S_6\{1,2,3,4,5\}&amp;amp;:&amp;amp;\color{red}{\boxed{13}\boxed{12}\boxed{11}\boxed{10}\boxed{9}\boxed{8}\boxed{7}\boxed{6}}\\&#xA;S_7\{\}&amp;amp;:&amp;amp; \color{red}{\boxed{1}\boxed{2}\boxed{3}\boxed{4}\boxed{5}\boxed{6}\boxed{7}\boxed{8}\boxed{9}\boxed{10}\boxed{11}\boxed{12}\boxed{13}}&#xA;\end{eqnarray}&#xA;$$&#xA;&lt;hr/&gt;&#xA;$$&#xA;\begin{eqnarray}&#xA;S_3\{1,10,11\}&amp;amp;:&amp;amp;\boxed{3\ 4\ 7\ 8\ 13}\boxed{2\ 5\ 6\ 9\ 12}\\&#xA;S_4\{4,5,6,7\}&amp;amp;:&amp;amp;\boxed{1\ 10\ 11}\color{red}{\boxed{2\ 9\ 12}\boxed{3\ 8\ 13}}\\&#xA;S_5\{11,12,13\}&amp;amp;:&amp;amp;\color{red}{\boxed{56}\boxed{47}\boxed{38}\boxed{29}\boxed{1\ 10}}\\&#xA;S_6\{1,2,3,4,5\}&amp;amp;:&amp;amp;\color{red}{\boxed{13}\boxed{12}\boxed{11}\boxed{10}\boxed{9}\boxed{8}\boxed{7}\boxed{6}}\\&#xA;S_7\{\}&amp;amp;:&amp;amp; \color{red}{\boxed{1}\boxed{2}\boxed{3}\boxed{4}\boxed{5}\boxed{6}\boxed{7}\boxed{8}\boxed{9}\boxed{10}\boxed{11}\boxed{12}\boxed{13}}&#xA;\end{eqnarray}&#xA;$$&#xA;&lt;hr/&gt;&#xA;$$&#xA;\begin{eqnarray}&#xA;S_2\{3,4,7,8,13\}&amp;amp;:&amp;amp;\boxed{1\ 2\ 5\ 6\ 9\ 10\ 11\ 12}\\&#xA;S_3\{1,10,11\}&amp;amp;:&amp;amp;\boxed{3\ 4\ 7\ 8\ 13}\color{red}{\boxed{2\ 5\ 6\ 9\ 12}}\\&#xA;S_4\{4,5,6,7\}&amp;amp;:&amp;amp;\color{red}{\boxed{1\ 10\ 11}\boxed{2\ 9\ 12}\boxed{3\ 8\ 13}}\\&#xA;S_5\{11,12,13\}&amp;amp;:&amp;amp;\color{red}{\boxed{56}\boxed{47}\boxed{38}\boxed{29}\boxed{1\ 10}}\\&#xA;S_6\{1,2,3,4,5\}&amp;amp;:&amp;amp;\color{red}{\boxed{13}\boxed{12}\boxed{11}\boxed{10}\boxed{9}\boxed{8}\boxed{7}\boxed{6}}\\&#xA;S_7\{\}&amp;amp;:&amp;amp; \color{red}{\boxed{1}\boxed{2}\boxed{3}\boxed{4}\boxed{5}\boxed{6}\boxed{7}\boxed{8}\boxed{9}\boxed{10}\boxed{11}\boxed{12}\boxed{13}}&#xA;\end{eqnarray}&#xA;$$&#xA;&lt;hr/&gt;&#xA;$$&#xA;\begin{eqnarray}&#xA;S_1\{\}&amp;amp;:&amp;amp;\boxed{1\ 2\ 3\ 4\ 5\ 6\ 7\ 8\ 9\ 10\ 11\ 12\ 13}\\&#xA;S_2\{3,4,7,8,13\}&amp;amp;:&amp;amp;\color{red}{\boxed{1\ 2\ 5\ 6\ 9\ 10\ 11\ 12}}\\&#xA;S_3\{1,10,11\}&amp;amp;:&amp;amp;\color{red}{\boxed{3\ 4\ 7\ 8\ 13}\boxed{2\ 5\ 6\ 9\ 12}}\\&#xA;S_4\{4,5,6,7\}&amp;amp;:&amp;amp;\color{red}{\boxed{1\ 10\ 11}\boxed{2\ 9\ 12}\boxed{3\ 8\ 13}}\\&#xA;S_5\{11,12,13\}&amp;amp;:&amp;amp;\color{red}{\boxed{56}\boxed{47}\boxed{38}\boxed{29}\boxed{1\ 10}}\\&#xA;S_6\{1,2,3,4,5\}&amp;amp;:&amp;amp;\color{red}{\boxed{13}\boxed{12}\boxed{11}\boxed{10}\boxed{9}\boxed{8}\boxed{7}\boxed{6}}\\&#xA;S_7\{\}&amp;amp;:&amp;amp; \color{red}{\boxed{1}\boxed{2}\boxed{3}\boxed{4}\boxed{5}\boxed{6}\boxed{7}\boxed{8}\boxed{9}\boxed{10}\boxed{11}\boxed{12}\boxed{13}}&#xA;\end{eqnarray}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Bonus: Generalization of the problem&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Now, the above solves the problem of finding maximum $N$ given $M$ stacks, clearing one stack at a time, &lt;strong&gt;and using only 3 stacks at a time&lt;/strong&gt;. What if we remove that last restriction? How many numbers can we sort? (Ok, I admit that I've been &lt;a href=&quot;http://xkcd.com/356/&quot; rel=&quot;nofollow noreferrer&quot;&gt;nerd-sniped&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if we can use 4 stacks, then we can split a Stack into 3 stacks, and so the number of groups in a stack depends on the previous 3 stacks, giving us &lt;a href=&quot;http://en.wikipedia.org/wiki/Generalizations_of_Fibonacci_numbers#Tribonacci_numbers&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tribonacci Number&lt;/a&gt; with $F_{(3)}(1) = F_{(3)}(2) = 1$, for which the asymptotic ratio is about $1.83929$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For fun, here is the final diagram when we can use 4 stacks ($M=5, N=F_{(3)}(5)=7$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{eqnarray}&#xA;S_1\{\}&amp;amp;:&amp;amp; \color{red}{\boxed{1234567}}\\&#xA;S_2\{3,4,7\}&amp;amp;:&amp;amp; \color{red}{\boxed{1256}}\\&#xA;S_3\{1,6,7\}&amp;amp;:&amp;amp; \color{red}{\boxed{34}\boxed{25}}\\&#xA;S_4\{1,2,3\}&amp;amp;:&amp;amp; \color{red}{\boxed{7}\boxed{6}\boxed{5}\boxed{4}}\\&#xA;S_5\{\}&amp;amp;:&amp;amp; \boxed{1}\boxed{2}\boxed{3}\boxed{4}\boxed{5}\boxed{6}\boxed{7}&#xA;\end{eqnarray}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more stacks we can use, the higher the asymptotic ratio is. You can check Tetranacci Number, Pentanacci Number, and so on in this very helpful &lt;a href=&quot;http://en.wikipedia.org/wiki/Generalizations_of_Fibonacci_numbers#Fibonacci_numbers_of_higher_order&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article&lt;/a&gt;, which states that the limit of the ratio is $2$. So asymptotically, we can sort $C\cdot2^M$ numbers with $M$ stacks (actually the exact number is $2^{M-2}$ if we are allowed to use all available stacks).&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;With this, finally I can tell OP, that the problem is solved. (reference to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2813/sorting-cards-using-max-3-stacks#comment7552_2814&quot;&gt;this comment&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Ok, you might notice that I'm heavily influenced by &lt;a href=&quot;https://what-if.xkcd.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;What-If XKCD&lt;/a&gt; style. That's my favorite reading!&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1649" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2014-10-24T03:47:44.577" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3038" PostTypeId="2" ParentId="3036" CreationDate="2014-10-20T05:35:55.657" Score="13" Body="&lt;p&gt;The puzzle that matches the one in the image is called The Snake cube.&#xA;&lt;a href=&quot;http://www.mathematische-basteleien.de/snakecube.htm&quot;&gt;http://www.mathematische-basteleien.de/snakecube.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2782" LastActivityDate="2014-10-20T05:35:55.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3039" PostTypeId="2" ParentId="2980" CreationDate="2014-10-20T09:26:39.000" Score="-2" Body="&lt;p&gt;Because...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...the poison was in ICE. so for the girl who drank slowly there was enough time for the ice to get it dissolved in tea while the one which drank the 5 tea rapidly, for her there was not enough time for the poison in the ice to dissolve and hence she survived.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4032" LastEditorUserId="42452" LastEditDate="2018-09-25T12:47:29.430" LastActivityDate="2018-09-25T12:47:29.430" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="3040" PostTypeId="1" CreationDate="2014-10-20T09:50:24.047" Score="-2" ViewCount="1515" Body="&lt;p&gt;You are on the final round of a TV gameshow, there are 3 doors; A, B and C. Two doors lead to a goat and the other door leads to a sports car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In each episode the presenter asks the contestant to select a door. He then opens one of the unselected doors to show a goat, the contestant is allowed to change their mind and select the remaining unselected door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you choose the other door or stick with your original choice?&lt;/p&gt;&#xA;" OwnerUserId="4036" LastActivityDate="2014-10-20T13:10:29.423" Title="Do you choose the other door?" Tags="|probability|" AnswerCount="4" CommentCount="0" ClosedDate="2014-10-20T13:34:13.487" ContentLicense="CC BY-SA 3.0" />
  <row Id="3041" PostTypeId="2" ParentId="3040" CreationDate="2014-10-20T09:52:34.700" Score="4" Body="&lt;p&gt;&lt;strong&gt;Yes, you do.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two doors that you don't choose have $2/3$ chance that the sports car is behind that door. The presenter opens a door, but he will never open a door with the car behind it. The door that isn't opened by the presenter has therefor still $2/3$ chance to have the car, while your door has only $1/3$ chance. So switch, and double your chance to win!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(For a better explanation, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Monty_Hall_problem&quot; rel=&quot;nofollow&quot;&gt;Monty Hall problem&lt;/a&gt;)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="3993" LastActivityDate="2014-10-20T09:52:34.700" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3042" PostTypeId="2" ParentId="3040" CreationDate="2014-10-20T10:28:06.833" Score="4" Body="&lt;p&gt;Yes and No, and yes I know that this has been answered, but I guess it deserves a simpler explanation for people like me for both my Yes and No. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are three doors, you choose 1, there is a 1/3rd Chance of you being right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets say door 1 has the car and doors 2 and 3 have goats. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scenario 1: You chose 1 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The presenter will open one of the leftover doors (either 2 or 3) with the goat and ask if you would like to switch, if you switch, you will lose&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scenario 2: You chose 2 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The presenter will not open door 1 he will open  door 3, in this case if you switch you will win &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scenario 3: You chose 3 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The presenter will not open door 1 he will open  door 2, in this case if you switch you will win &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that the original probability was 1/3 and after the door opening, the odds of winning with a switch is 2/3, you are better off switching.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, however is the logical answer, then there is the philosophy, if you had a reason to choose door 1 and you believe that reason has not changed, you should not switch, because in life, being wrong about chance is ok, but not sticking to your decision is a bigger problem :) than losing a sports car&lt;/p&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="2313" LastEditDate="2014-10-20T13:10:29.423" LastActivityDate="2014-10-20T13:10:29.423" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3043" PostTypeId="2" ParentId="2688" CreationDate="2014-10-20T12:40:38.363" Score="0" Body="&lt;p&gt;If i would go to near the barrier, and raise my hand (not all the way through, just as to be vertical to my torso), then my reflection would do the same. Then if i catch my reflections hands and side step to the right, then my reflection would do the same but &lt;strong&gt;his&lt;/strong&gt; right would be &lt;strong&gt;my&lt;/strong&gt; left then I would enter from the right and he would do the same from the left.Side stepping in a circle whose center is the point where our hands are caught.&lt;/p&gt;&#xA;" OwnerUserId="2460" LastActivityDate="2014-10-20T12:40:38.363" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3044" PostTypeId="2" ParentId="3040" CreationDate="2014-10-20T12:41:46.373" Score="0" Body="&lt;p&gt;This is the Monty hall problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the presenter knows which door has the prize and never opens it, then you choose the other door. If he doesn't (and therefore has a probability of opening the door with the prize) it doesn't matter whether you change doors or not.&lt;/p&gt;&#xA;" OwnerUserId="4046" LastActivityDate="2014-10-20T12:41:46.373" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3045" PostTypeId="2" ParentId="3040" CreationDate="2014-10-20T12:46:06.003" Score="0" Body="&lt;p&gt;Yes, you switch. A critical point is that you say&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In &lt;strong&gt;each&lt;/strong&gt; episode the presenter asks the contestant to select a door. He then opens one of the unselected doors to show a goat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This lets us know he is (probably) deliberately choosing a goat door, thus providing you with new information. If he was randomly choosing one of the two unselected doors, seeing a goat would tell you nothing. In that case, you would expect him to reveal a car in $\frac{1}{3}$&#xA;of the episodes.&lt;/p&gt;&#xA;" OwnerUserId="3987" LastActivityDate="2014-10-20T12:46:06.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3046" PostTypeId="2" ParentId="2688" CreationDate="2014-10-20T13:06:26.600" Score="7" Body="&lt;p&gt;Create a strong magnetic field at the mirror surface. North is up, south is down. Fire an electron beam at the mirror through the field. Electrons from either side will be deflected anti clockwise (as seen from above) and will miss each other, passing through the mirror. &lt;/p&gt;&#xA;" OwnerUserId="3987" LastActivityDate="2014-10-20T13:06:26.600" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3048" PostTypeId="1" CreationDate="2014-10-20T14:22:16.447" Score="0" ViewCount="563" Body="&lt;p&gt;Have you seen Knight-Knave puzzles based on probability rather then facts?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example a puzzle with condition like this: Knight tells the answer, probability to which according to his knowledge is large that 50%. Knave - the opposite. So to the question &quot;Will a dice show 1 or 2?&quot; knight answers &quot;No&quot; and Knave &quot;Yes&quot;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the same effect can be achieved with usual characters and questions like &quot;Is probability to get 1 or 2 bigger than 50%?&quot; and I would like to see a puzzle, which can not be solved without such a probability questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you figure out a non-trivial puzzle like this?&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2014-10-20T15:43:50.987" LastActivityDate="2015-04-10T09:30:12.550" Title="Is there probability based Knight-Knave puzzles?" Tags="|puzzle-creation|puzzle-identification|liars|" AnswerCount="2" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="3050" PostTypeId="2" ParentId="2980" CreationDate="2014-10-20T14:56:50.843" Score="8" Body="&lt;p&gt;Alternatively:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Before the event, the surviving girl had prepared herself by increasing her body's tolerance of the poison: she dosed herself with it, in small then gradually increasing quantities (&lt;a href=&quot;http://wordsmith.org/words/mithridatism.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;mithridatism&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the girls were in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Mithridatism#In_fiction&quot; rel=&quot;nofollow noreferrer&quot;&gt;murder mystery novel&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;then this would definitely be the answer! :)&lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="42452" LastEditDate="2018-09-25T12:56:16.277" LastActivityDate="2018-09-25T12:56:16.277" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="3058" PostTypeId="2" ParentId="3024" CreationDate="2014-10-20T17:53:03.997" Score="3" Body="&lt;p&gt;I think the missing 3 numbers are: 879, 750, 934. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had to use all of the hints!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer can be derived by taking each group of 9 to represent a 3x3 matrix, then multiplying it with the preceding group of 9. There are 3 given, and the 3rd has been derived from the first 2. So let's take the 2nd and the 3rd:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;3, 4, 5, &#xA;8, 2, 9, &#xA;3, 5, 2 &#xA;X    &#xA;42, 35, 42, &#xA;52, 55, 52, &#xA;109, 85, 120&#xA;=&#xA;879, 750, 934&#xA;1421, 1155, 1520&#xA;604, 550, 626&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-10-20T18:00:37.277" LastActivityDate="2014-10-20T18:00:37.277" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3059" PostTypeId="2" ParentId="3048" CreationDate="2014-10-20T18:21:53.460" Score="0" Body="&lt;p&gt;Not that I know of... But it'd be easy to come up with that kind of thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;You arrive at an island one night where you find 2 types of people, intelligent, and people who often think randomly. You go up to one and ask them if you're thinking of a number between 1 and 3. How can you tell who's who after they tell you their guess?&quot;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2014-10-20T18:21:53.460" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3062" PostTypeId="1" AcceptedAnswerId="3063" CreationDate="2014-10-20T18:43:10.193" Score="9" ViewCount="2958" Body="&lt;p&gt;Here's a pretty easy one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A new student (a male) was wandering down the hall of his new school looking for the boys' locker room.  He asks a nearby student (who happens to be the school bully) to direct him.  The bully says, &quot;See those three doors? It's one of those, but I switched the signs around so they're all wrong! Ha ha!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The new student looked at the three doors.  They were labeled &quot;Boys&quot;, &quot;Girls&quot; and &quot;Cafeteria&quot;. Just then another student walked out of one of the doors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The new student told the bully &quot;Thanks&quot; and proceeded to the correct door.  How did he do it?&lt;/p&gt;&#xA;" OwnerUserId="2162" LastActivityDate="2016-05-23T05:51:49.203" Title="First day of school" Tags="|logical-deduction|word-problem|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3063" PostTypeId="2" ParentId="3062" CreationDate="2014-10-20T18:49:02.357" Score="15" Body="&lt;p&gt;He saw a male student walk out of the door marked &quot;Boys&quot; - in this case, he either walked out of the Cafeteria or the Girls' bathroom. He must've exited the Cafeteria (because he cannot use the Girls' locker room). So one of the remaining doors must be the Girl's bathroom, but it can't be marked as such, so he picks the one marked &quot;Girls&quot;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, in general, the only case where one cannot deduce is when a male walks out a door marked &quot;Girls&quot; or when a female walks out a door marked &quot;Boys&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If he sees a female exit a door marked &quot;Girls&quot;, then she certainly was in the Cafeteria. This means he chooses the door marked &quot;Cafeteria&quot; (as opposed to the remaining door marked &quot;Boys&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If he sees a female exit a door marked &quot;Cafeteria&quot;, then she certainly was in the Girls' locker room. This means he chooses the door marked &quot;Girls&quot; (as opposed to the remaining door marked &quot;Boys&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If he sees a male exit a door marked &quot;Boys&quot;, then he certainly was in the Cafeteria. This means he chooses the door marked &quot;Girls&quot; (as opposed to the remaining door marked &quot;Cafeteria&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If he sees a male exit a door marked &quot;Cafeteria&quot;, then he certainly was in the Boys' locker room. This means he chooses the door marked &quot;Cafeteria&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, it's entirely possible he's been trolled by the clever bully into going into the Girls' locker room unintentionally.&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-10-21T00:25:43.357" LastActivityDate="2014-10-21T00:25:43.357" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3065" PostTypeId="2" ParentId="3062" CreationDate="2014-10-20T18:59:31.343" Score="0" Body="&lt;p&gt;I have 4 answers:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Boy from Boys = Girls&lt;/li&gt;&#xA;&lt;li&gt;Boy from Café = Café&lt;/li&gt;&#xA;&lt;li&gt;Girl from Girls = Café&lt;/li&gt;&#xA;&lt;li&gt;Girl from Café = Girls&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Essentially, you have to either have a &quot;gender from gender&quot; or &quot;gender from café&quot; to determine this. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2014-10-20T18:59:31.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3066" PostTypeId="1" AcceptedAnswerId="3070" CreationDate="2014-10-20T19:21:51.010" Score="5" ViewCount="1636" Body="&lt;p&gt;You are at an impasse with 30 doors - 29 of them lead to endless wandering the depths of hell - and one of them leads to paradise. In front of them stand the typical trio of good friends, a Knight, a Knave and a Joker. A Knight always gives answers truthfully. A Knave always gives the answer opposite to what a Knight would give. A Joker answers in a random fashion - what he says may be true or not. They all look alike, and they can only answer &quot;yes&quot;/&quot;no&quot;. (no paradoxes allowed!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the fewest questions you can ask to find the right door?&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="12408" LastEditDate="2016-07-25T02:35:35.453" LastActivityDate="2016-07-25T02:35:35.453" Title="Knight/Knave/Joker - impasse challenge" Tags="|logical-deduction|liars|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3067" PostTypeId="2" ParentId="2107" CreationDate="2014-10-20T19:29:16.300" Score="4" Body="&lt;p&gt;Here's a counter-example to the conjecture, for N=860&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(you may need to open the image in a new tab to read the sizes in the image)&#xA;&lt;img src=&quot;https://i.stack.imgur.com/pUKNz.png&quot; alt=&quot;A counter-example&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set of rectangles used:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shape 1&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;602x259&lt;/li&gt;&#xA;&lt;li&gt;210x651&lt;/li&gt;&#xA;&lt;li&gt;28x833&lt;/li&gt;&#xA;&lt;li&gt;133x728&lt;/li&gt;&#xA;&lt;li&gt;469x392&lt;/li&gt;&#xA;&lt;li&gt;679x182&lt;/li&gt;&#xA;&lt;li&gt;707x154&lt;/li&gt;&#xA;&lt;li&gt;840x21&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Shape 2&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;196x665&lt;/li&gt;&#xA;&lt;li&gt;70x791&lt;/li&gt;&#xA;&lt;li&gt;161x700&lt;/li&gt;&#xA;&lt;li&gt;413x448&lt;/li&gt;&#xA;&lt;li&gt;609x252&lt;/li&gt;&#xA;&lt;li&gt;770x91&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2715" LastEditorUserId="2715" LastEditDate="2014-10-20T19:34:42.843" LastActivityDate="2014-10-20T19:34:42.843" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3068" PostTypeId="2" ParentId="2980" CreationDate="2014-10-20T19:40:26.980" Score="4" Body="&lt;p&gt;Because...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...Drink 1 was only poison, and Drinks 2+ also contained the antidote.  Who is the better (repeat) customer?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4059" LastEditorUserId="42452" LastEditDate="2018-09-25T12:45:11.910" LastActivityDate="2018-09-25T12:45:11.910" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="3070" PostTypeId="2" ParentId="3066" CreationDate="2014-10-20T20:00:04.620" Score="11" Body="&lt;p&gt;First, you find a nonjoker in a single question as in &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2484/knight-knave-and-joker-puzzle&quot;&gt;Knight, Knave and Joker puzzle&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you ask the nonjoker questions of the form &quot;If I were to ask the other nonjoker X, what would he reply?&quot; This is the way around not knowing whether you are asking Knight or Knave and is the usual approach to eliminating the uncertainty - you should get back a lie whatever you ask. (See &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2188/two-doors-with-two-guards-one-lies-one-tells-the-truth&quot;&gt;Two doors with two guards - one lies, one tells the truth&lt;/a&gt; for example.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the solution is then how many such questions can get you from 30 doors to just one. 5 questions should be able to get you from 32 to 1 - the first to eliminate half the possibilities, and each subsequent one half of what remains. Exactly what you ask depends on answers you get, but one pattern might be&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;is the safe door in the range 16-30? Yes. (Means the truth is No.)&lt;/li&gt;&#xA;&lt;li&gt;is the safe door in the range 8-15? Yes. (Means the truth is No.)&lt;/li&gt;&#xA;&lt;li&gt;is the safe door in the range 4-7? Yes. (Means the truth is No.)&lt;/li&gt;&#xA;&lt;li&gt;is the safe door in the range 2-3? No. (Means the truth is Yes.)&lt;/li&gt;&#xA;&lt;li&gt;is the safe door #2? No. (Means the truth is yes.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You may get lucky on one less question sometimes because there are only 30 doors, not 32. But in general you need 6 questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone stipulates that you can't say &quot;the other nonjoker&quot; but have to name a specific other guard, then you can spend one extra question by asking your known nonjoker &quot;are you the joker?&quot; - a Knight will say no and a Knave will say yes. In this case you need one more question, and your remaining questions can be asked directly without the &quot;if I asked the other&quot; business. &lt;/p&gt;&#xA;" OwnerUserId="1509" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2014-10-20T20:05:29.167" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3071" PostTypeId="2" ParentId="3066" CreationDate="2014-10-20T20:10:03.997" Score="3" Body="&lt;p&gt;&lt;strong&gt;6 questions are enough&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In one question you exclude the joker.  If the people are A, B, C, ask to A:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Is B a knave or C a knight&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If A says yes, continue with B.  If A says no, continue with C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the selected person you can find the correct door with a binary search by asking:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;If I'd ask you whether the door to parasise is among 1-15, would you say yes?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and halving the range appropriately with each question.  In total, you need 1 question to identify a non-joker and 5 to identify the door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;6 questions are necessary&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to identify 1 possibility among 29.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could in principle halve the number of possibilities of the worst case with each questions and identify the door in 5 questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if the first question goes to the joker, you won't learn anything.  The joker can reply yes or no to any question regardless of where the door is.  If you think of it, he could reply randomly without even knowing which one is the right door.  Therefore you cannot learn anything about the door from the joker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in the worst case, your first question goes to the joker and you still have 29 doors to discriminate.  And for this you need 5 more questions.  The wost case requires 6 questions.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastActivityDate="2014-10-20T20:10:03.997" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3074" PostTypeId="1" AcceptedAnswerId="3075" CreationDate="2014-10-20T21:05:58.823" Score="21" ViewCount="68778" Body="&lt;p&gt;After every time you guess, you're told if you're right, too high, or too low. Is there a strategy you can use to guarantee you'll get it on your 6th attempt (or lower)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know a strategy to get it on your 7th attempt.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2022-08-28T06:12:55.783" Title="6 Tries to Guess a Number Between 1-100" Tags="|strategy|" AnswerCount="11" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="3075" PostTypeId="2" ParentId="3074" CreationDate="2014-10-20T21:10:54.210" Score="34" Body="&lt;p&gt;Ask about 50. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If too high, then ask about 25.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If too low, then ask about 75. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Continue, continually halving the search-space. This requires $\lceil \log_2 (n+1)\rceil$ maximum questions. For 100, that's 7. It is a binary search algorithm known to be $\mathrm O(\log(n))$ time. I'm fairly sure there isn't a faster way. Binary search is considered the best for this problem - unless you are allowed to ask other questions.&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-12-04T05:17:25.157" LastActivityDate="2014-12-04T05:17:25.157" CommentCount="23" ContentLicense="CC BY-SA 3.0" />
  <row Id="3076" PostTypeId="2" ParentId="3074" CreationDate="2014-10-20T21:32:04.857" Score="11" Body="&lt;p&gt;A strategy that solves this in under 7 questions is impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each question, you only get one bit of data (too low, or too high). It's impossible to do this in 6 questions because if you enumerated all the possible outcomes (let's say &quot;too low&quot; is 0, and &quot;too high&quot; is 1):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Q1 Q2 Q3 Q4 Q5 Q6&#xA;0  0  0  0  0  0&#xA;0  0  0  0  0  1&#xA;0  0  0  0  1  0&#xA;0  0  0  0  1  1&#xA;0  0  0  1  0  0&#xA;0  0  0  1  0  1&#xA;etc.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;there would only be $2^6=64$ possible outcomes, while there are 100 possible numbers.&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2014-10-20T21:32:04.857" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3077" PostTypeId="2" ParentId="3074" CreationDate="2014-10-20T21:59:55.537" Score="2" Body="&lt;p&gt;If I am allowed to ask about individual digits, I can do it in 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First question:  Where X is the number from 1 to 100 inclusive, if X - 1  is left padded with zeros (00-99), is the left digit 5?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the answer is &quot;lower&quot;, I ask about 2.  If higher I ask about 7 or 8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say I asked about 2, and the answer again comes back &quot;lower&quot;.  Then I ask about zero and the answer comes back &quot;higher&quot;.  I've found the first digit to be 1 in three guesses.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I repeat the process to get the second digit in 3 guesses, for 6 total, max.  But of course that's 6 &lt;em&gt;questions&lt;/em&gt;, not 6 guesses as to the actual number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Taking it a step further, asking about the left digit means asking about a 10 number range.  If I get to ask about ranges in general, I push the number of questions down to five.  Basically, I just split the 1-100 range into ranges of 33, 33, and 34 numbers and ask about the middle one.  &quot;Is the number between 34 and 66 inclusive?&quot;  So your ranges go from size 100 to 34 (in the worst case) to 12 to 4 to 2 to 1.&lt;/p&gt;&#xA;" OwnerUserId="2162" LastEditorUserId="2162" LastEditDate="2014-10-20T22:21:09.343" LastActivityDate="2014-10-20T22:21:09.343" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3078" PostTypeId="2" ParentId="3074" CreationDate="2014-10-20T23:12:42.140" Score="28" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each guess eliminates one number as well as dividing the remaining numbers into 2. One guess can pick a number from 3 (is your number 2?). 2 guesses can do 7. N guesses can pick a number from $2^{N+1}-1$, so 6 guesses can do it for 1-127.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: As noted in the comments, you're supposed to have &lt;strong&gt;guessed&lt;/strong&gt; the number on or before the 6th attempt, while this only ensures you &lt;strong&gt;know&lt;/strong&gt; the answer by then.&lt;/p&gt;&#xA;" OwnerUserId="3987" LastEditorUserId="3987" LastEditDate="2014-10-21T06:21:44.193" LastActivityDate="2014-10-21T06:21:44.193" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="3079" PostTypeId="1" AcceptedAnswerId="3202" CreationDate="2014-10-21T00:16:43.207" Score="10" ViewCount="1371" Body="&lt;p&gt;N bishops are placed on a standard 8x8 chess board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each bishop can move normally but after touching or moving over an square, that square is considered marked.  A bishop cannot pass over or land on a marked square again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each bishop may stop any number of times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the minimum number (N) of bishops required to mark every square but follow these rules?&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="57521" LastEditDate="2020-04-07T17:16:32.913" LastActivityDate="2020-04-07T17:16:32.913" Title="How many bishops to walk across the board?" Tags="|chess|optimization|" AnswerCount="2" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="3080" PostTypeId="2" ParentId="3079" CreationDate="2014-10-21T00:30:00.183" Score="11" Body="&lt;p&gt;I conjecture that you need 8. 4 for each colour. The issue is clearly about getting stuck in &quot;corners&quot; (even artificial ones created by marking nearby squares.). Here is a couple of crude diagrams illustrating half of the solution (the other half is the same, but for the other colour):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fO5uo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And in case you thought that going diagonally would help:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4ArRD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-10-21T09:59:53.603" LastActivityDate="2014-10-21T09:59:53.603" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="3081" PostTypeId="2" ParentId="236" CreationDate="2014-10-21T04:55:52.013" Score="5" Body="&lt;p&gt;The solution listed is correct, but it is the solution to a much harder problem than you might think, which is:&#xA;There are 200 people on an island, where any person can have either blue or non-blue eyes.&#xA;On Day 0, a Guru announces either that:&#xA;a) I see at least one pair of blue eyes&#xA;or&#xA;b) I see no blue eyes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this single datum, the standard algorithm would solve ANY number of blue eyes, from 0 to 200. Without this single datum, even though you can see N blue eyes (where N is from 0 to 199), you can never be certain what your eye color is, because you would never know if Total Blue Eyes = N or N+1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put another way, if you can see N blue eyes, and the guru tells you that Total Blue Eyes == 0 OR that Total Blue Eyes &gt;= 1 on Day 0, you can determine your own eye color after N-1  days (if you have blue eyes) or N days (if you have non-blue eyes) according to the standard algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, however, you were ONLY trying to solve the single case where exactly N people have blue eyes, then you can leave without the Guru on Day 0:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;On Day 0, if you see N blue eyes, your eyes are non-blue. Stay.&lt;/li&gt;&#xA;&lt;li&gt;On Day 0, if you see N-1 blue eyes, your eyes are blue. Leave tonight.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What is even cooler is that if you are willing to NOT solve any single case, such as &quot;0 people have blue eyes&quot;, then you don't need the Guru to start the induction. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;On Day 0, you see N blue eyes, where N&gt;=0. On day N, if no one has left yet, leave knowing you have blue eyes. If anyone ever leaves before you get a chance, you don't have blue eyes, leave the very next day.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Which is pretty cool considering that if odds of having blue eyes were, say 50%, then the odds of all having blue eyes = 1/2 ^ 200 ~ 10^-61. Pretty tolerable odds if you were lacking a Guru!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be cool to see a general algorithm that could be tuned with a variable cost for &quot;days spent calculating&quot; versus a cost for &quot;getting the answer wrong&quot;. The default question basically assumes &quot;cost of days spent calculating&quot; == 0 or &quot;cost of getting the answer wrong&quot; == infinity.&lt;/p&gt;&#xA;" OwnerUserId="4072" LastActivityDate="2014-10-21T04:55:52.013" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3082" PostTypeId="2" ParentId="3074" CreationDate="2014-10-21T05:21:15.657" Score="4" Body="&lt;p&gt;I think skywalker's answer is the one that the riddler seeks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In his answer the notion of a 3rd outcome for a guess &quot;that's the number&quot; is being considered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following sequences (Un) and (Vn)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;U(0) = 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;U(n+1) = 2*U(n) (&quot;naive&quot; answer without regard to the additionnal information)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;V(0) = 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;V(n+1) = 2*V(n) + 1 (with regard to the additionnal information)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first items of the suite we have&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;n - U(n) - V(n)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;0 - 1 - 1&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;1 - 2 - 3&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;2 - 4 - 7&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;3 - 8 - 15&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;4 - 16 - 31&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;5 - 32 - 63&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;6 - 64 - 127&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;7 - 128 - 255&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As you can see the (Vn) suite is one step ahead of (Un) thanks to the additionnal info. I think this single step ahead is the reason why this answer is the one the riddler is looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: It is true that you can &quot;only&quot; be certain of the number to be guessed after 6 question instead of giving the right solution on the 6th guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the more naive approach actually leads to the conclusion hat you should be allowed to speak an 8th time to say the number you were certain of after the 7th guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Sorry I wanted to comment on skywalker's answer but I don't have enough points for that)&lt;/p&gt;&#xA;" OwnerUserId="4028" LastEditorUserId="4028" LastEditDate="2014-10-21T07:58:40.090" LastActivityDate="2014-10-21T07:58:40.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3086" PostTypeId="2" ParentId="3074" CreationDate="2014-10-21T09:00:34.013" Score="1" Body="&lt;p&gt;Nobody &lt;em&gt;quite&lt;/em&gt; has the right answer to why this is impossible in less than seven questions (assuming you're restricted to asking about single numbers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is why.  Whatever number you pick first, there is some chance that the answer will be on the &quot;big&quot; side; the smallest you can make the big side is 50.  (E.g. if you guess &quot;51&quot; then &quot;52-100&quot; is the small side (49 numbers) and &quot;1-50&quot; is the big side (50 numbers).  Ask again and your best worst case is 25.  Then 12.  Then 6.  Then 3.  You've just used 5 guesses, but you still have 3 numbers left to pick between.  Can you always state the correct one?  Nope!  It could be any of those 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you get one more guess you can get the worst case down to 1 number, which means you've got it.  So this proves (well, illustrates, as this isn't a formal proof) that it is impossible in 6 guesses but is possible in 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that, if you allow guesses at ranges, the answer is different.  You can, by carefully selecting your ranges, go from 100 to 34 at worst (by stating &quot;the number is between 34 and 67&quot;) on your first go.  Second gets you down to 12.  Then 4, then 2, then 1.  So you can always do it in 6.)&lt;/p&gt;&#xA;" OwnerUserId="2815" LastEditorUserId="2815" LastEditDate="2014-10-21T09:06:22.473" LastActivityDate="2014-10-21T09:06:22.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3087" PostTypeId="1" AcceptedAnswerId="3095" CreationDate="2014-10-21T10:02:51.780" Score="2" ViewCount="554" Body="&lt;p&gt;Following these questions: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2588/faulty-computers/2774#2774&quot;&gt;Faulty computers&lt;/a&gt;, &lt;a href=&quot;https://puzzling.stackexchange.com/questions/593/knights-and-jokers&quot;&gt;Knights and jokers&lt;/a&gt; I wonder, what if the conditions are the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;We have T Knights (truth-tellers), L Knaves (liars) and R Jokers (random-tellers).  $T &amp;gt; 0$. We know all 3 numbers, but we do not know who is who.&lt;/li&gt;&#xA;  &lt;li&gt;We can chose any two of them and ask the first about the second one of the 3 possible questions &quot;Is he a Knight/Knave/Joker?&quot;. We can repeat this questioning procedure any number of times.&lt;/li&gt;&#xA;  &lt;li&gt;We need to find one Knight.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How N, M and K must be related to make the task possible to complete?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It is clear for me, that $T+L&amp;gt;R$. Otherwise $T$ jokers can simulate Knights, $L$ Knaves and we would never be able to distinguish between them and find a real Knight.&lt;br&gt;&#xA;Also we know already that if $L = 0, T &amp;gt; R$ the task is possible. Also it is obviously possible if $T&amp;gt;R+L$, because we can treat Knaves as Jokers.&lt;br&gt;&#xA;So the question is what happens when $L&amp;gt;R-T$ and $L \ge T - R$. For example, when $L = 1, T = R = 10$.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2019-06-07T17:51:57.187" Title="T Knights, L Knaves and R Jokers" Tags="|puzzle-creation|liars|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3090" PostTypeId="1" AcceptedAnswerId="3094" CreationDate="2014-10-21T12:48:16.307" Score="3" ViewCount="845" Body="&lt;p&gt;You're on a game show, playing for 1 million dollars! You've made it to the final round, there are 4 doors, you pick a random door (obviously, there's a 1/4 chance for all of them, let's say you picked #3.) The host reveals door #1, it's empty. You now have the choice to stick with your door, switch to another door, or take 5 hundred thousand dollars! Which should you take?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="57212" LastEditDate="2021-06-06T13:59:34.367" LastActivityDate="2021-06-06T21:39:44.043" Title="4 Doors and a Half Value prize" Tags="|mathematics|logical-deduction|monty-hall|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3091" PostTypeId="2" ParentId="3090" CreationDate="2014-10-21T12:53:17.343" Score="0" Body="&lt;p&gt;I would take the money , because its higher than the probability of winning the chance which is 1/3rd (Edit: its 3/8 as explained in the other answer), but that is assuming the host knows which door it is in.&lt;/p&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="2313" LastEditDate="2014-10-21T13:09:53.823" LastActivityDate="2014-10-21T13:09:53.823" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="3092" PostTypeId="2" ParentId="3090" CreationDate="2014-10-21T12:54:34.603" Score="0" Body="&lt;p&gt;Take the $500,000. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that opening a door for an empty room means you get nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You picked on, and one was eliminated. So, there is now a $1/3$ probability of picking the correct door from the remaining 3. \$500,000 $&amp;gt; 1/3*1000000$.    &lt;/p&gt;&#xA;" OwnerUserId="2484" LastActivityDate="2014-10-21T12:54:34.603" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3094" PostTypeId="2" ParentId="3090" CreationDate="2014-10-21T13:07:25.643" Score="13" Body="&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;: I assume the host knew door #1 would be empty. This is analogous to the original Monty Hall problem. If the host randomly opens a door, the door-opening is moot and can be left out completely.&lt;/p&gt;&#xA;&lt;p&gt;Let's first look at the situation without the 500K consolation prize:&lt;/p&gt;&#xA;&lt;p&gt;Let's say I do NOT switch doors, then I have a 25% chance to win 1 million.&lt;/p&gt;&#xA;&lt;p&gt;Let's say that I &lt;em&gt;do&lt;/em&gt; switch:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I pick the right door immediately (25% chance). Switching wins me nothing.&lt;/li&gt;&#xA;&lt;li&gt;I pick the wring door first (75% chance). After the elimination of a wrong door, I switch with a 50% chance to win 1 million.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;I have in total &lt;span class=&quot;math-container&quot;&gt;$\frac{3}{4} \cdot \frac{1}{2}=\frac{3}{8}$&lt;/span&gt; chance of winning the 1 million.&lt;/p&gt;&#xA;&lt;p&gt;So switching is, like in the original Monty Hall problem, certainly beneficial, raising my chance of winning from &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{4}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\frac{3}{8}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;However, my expected win is &lt;span class=&quot;math-container&quot;&gt;$\frac{3}{8} \times \text{1 million}$&lt;/span&gt;. That is less then the 500K.&lt;/p&gt;&#xA;&lt;p&gt;Basically, I have a choice to gamble a sum of money with the chance of doubling it, but I have only &lt;span class=&quot;math-container&quot;&gt;$\tfrac{3}{8}$&lt;/span&gt; chance of doubling it. Going by the math, I should stick with the money I have, which means&lt;br /&gt;&#xA;&lt;strong&gt;I should choose to keep the 500k&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;This problem introduces something that is not in the original problem, and that is &lt;em&gt;utility&lt;/em&gt;. That is the phenomenon that what I can win changes my life much more than the money I lose. It is the reason people play in lotteries with a negative expected return (winning some millions changes your life. Paying some dollars every month makes no change. So even if I never win, the possible win is much more important than the almost certain loss).&lt;/p&gt;&#xA;&lt;p&gt;Now, in the case of &amp;quot;investing&amp;quot; 500k at bad odds to double it, this doesn't seem to play (both 500k and 1 million are &amp;quot;a lot of money!), but let's say I have a loan shark on my case who will kill me if I do not pay him back 1 million tomorrow. Now, the 500k will do me no good at all.&#xA;In this situation, I will certainly select another door, giving me a 3/8 chance of surviving tomorrow. (Versus taking the 500k which will see me dead, or not switching and sticking with my original door, which gives me 25% chance of survival).&lt;/p&gt;&#xA;" OwnerUserId="3903" LastEditorUserId="69582" LastEditDate="2021-06-06T21:39:44.043" LastActivityDate="2021-06-06T21:39:44.043" CommentCount="12" ContentLicense="CC BY-SA 4.0" />
  <row Id="3095" PostTypeId="2" ParentId="3087" CreationDate="2014-10-21T13:12:29.030" Score="3" Body="&lt;p&gt;Let's assume the jokers are intelligent, coordinated and aware of our strategy. If they fail to meet these conditions they only make it easy for us. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edited for clarity&lt;/strong&gt;&#xA;This works for any $R&amp;lt;T+L$, so long as T&gt;0 (because there must be a knight to identify). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step one is to identify the knave(s) by the fact that they will say 'yes' to 2 of the three questions for any given person (&quot;Is bob a knave?&quot;, &quot;Is bob a knight?&quot;, &quot;Is bob a joker?&quot;). If you conclusively identify a knave you have won - you just take the reverse of their statements to be true. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ask each person if their neighbour is a knight, a knave, a joker. Knights will say 'yes' once. Knaves will say yes twice. An identified knave is effectively a truth teller - you can use them to identify everyone else. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the jokers must use L jokers to impersonate the knaves. You now have a set of at least 2L alleged knaves &lt;strong&gt;who you know are not knights&lt;/strong&gt;. This means the rest of the people - everyone who has not acted as a knave - have a majority of knights. You then ask all of those people about someone outside this group. The majority will be telling the truth. Continue this until you identify a liar, who you can use to identify everyone else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit again:&lt;/strong&gt;&#xA;The jokers &lt;strong&gt;must&lt;/strong&gt; impersonate the entire group of knaves. If you know there are 2 knaves and you see three people acting as knaves, ask each one about the identity of the other 'knaves'. The two genuine knaves will identify the joker (one you account for their backwards answers) and by extension themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Yet another edit&lt;/strong&gt; If you don't know the values of T,L,R you can still do it so long as $R&amp;lt;T+L$. As before, identify the (apparent) knaves. They will form groups - identifying every 'knave' within their group as a knave and every 'knave' outside their group as a joker. The groups may be as small as 1 person.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, have the apparent knights identify all other apparent knights. They will, again, form groups. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have the knight groups identify the knave groups. They will form what I'll call &quot;coalitions&quot; - groups of knights and knaves who mutually identify each other and identify anyone in another coalition as a joker. The largest coalition &lt;strong&gt;must&lt;/strong&gt; be the one with knights and knaves, given $R&amp;lt;T+L$.&lt;/p&gt;&#xA;" OwnerUserId="3987" LastEditorUserId="3987" LastEditDate="2014-10-21T18:43:45.130" LastActivityDate="2014-10-21T18:43:45.130" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="3096" PostTypeId="1" AcceptedAnswerId="3097" CreationDate="2014-10-21T13:26:12.033" Score="-3" ViewCount="535" Body="&lt;p&gt;&lt;strong&gt;A variation of the Monty Hall, where the values inside the choices are changed&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You were just out gambling and lost 5000 dollars! Of course being the gambler that you are, you pay 2500 to play this game known as the &quot;three boxes&quot; You decide to play this because you're chances of getting the money back is 2/3! One box contains 7500 dollars, and another contains 20000! Of course the other contains a machine that you must insert 3000 dollars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This could be the best risk of your life, or the worst. You get to pick a box, the host will then show you what is in one of the other boxes, he knows what's in there, and won't show you the 20000, once he does this you're allowed to keep, or change your answer, (even to the one he opens), explorer the 2 possible cases where he shows you the -3000 and where he reveals the 7500.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="15" LastEditDate="2014-12-31T19:37:14.050" LastActivityDate="2017-05-23T14:43:43.077" Title="The 3 Boxes Show!" Tags="|logical-deduction|mathematics|probability|monty-hall|" AnswerCount="3" CommentCount="13" ClosedDate="2015-01-01T23:41:31.197" ContentLicense="CC BY-SA 3.0" />
  <row Id="3097" PostTypeId="2" ParentId="3096" CreationDate="2014-10-21T13:48:35.473" Score="7" Body="&lt;p&gt;If I want, I can &lt;strong&gt;always&lt;/strong&gt; win my money back!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not switching is academic: I have a 1/3 chance of picking the 20k, 1/3 of picking the -3k and 1/3 of picking the +7.5k. My expected win is 6.6k-1k+2.5k = 8.1k.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Blindly switching to the closed door raises my chances. The original Month Hall problem teaches me that I have 2/3 chance of picking the 20k by switching to the closed door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the opened door shows +7.5k, my expected win is 13.3k-(1/3*-3k)=12.3k&lt;br&gt;&#xA;If the opened door shows -3k, mys expected win is 13.3k+(1/3*7.5k)=15.8k.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I choose the &lt;em&gt;opened&lt;/em&gt; door, obviously I'm dumb to take 100% chance for -3k.&lt;br&gt;&#xA;If the opened door shows +7.5k, I am sure to win 7.5 k.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if I am going for the maximum win, I always switch to the &lt;strong&gt;closed door&lt;/strong&gt;, which gives me the highest expected win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However&lt;/strong&gt;, here utility comes into play again!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If my goal is &lt;strong&gt;to get my money back&lt;/strong&gt;, seeing as I put in 7.5k (I lost 5k, and paid 2.5k to play), I should choose the 7.5k &lt;em&gt;if it shows behind the opened door&lt;/em&gt;. Otherwise, I switch to the closed door (100% chance of winning my money back (the -3k is behind the opened door then!) and 2/3 chance for 20k).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in case &lt;strong&gt;winning back my money&lt;/strong&gt; is important (because the guy that lent me the 7.5 is behind me with a gun...):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) If the open door shows -3k, I pick the closed door. Minimal win (1/3) = 7.5k, maximum win (2/3) = 20k. Expected = 15.8k.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) If the opened door shows +7.5k, I pick that. Minimum, maximum, expected win is always 7.5k (sticking with my choice gives me an expected win of 12.3k, but I &lt;em&gt;could&lt;/em&gt; loose 3k, and the guy behind me will shoot me!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case I want to &lt;strong&gt;maximize my expected winnings&lt;/strong&gt;, I follow a different strategy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) If the open door shows -3k, I pick the closed door. Minimal win (1/3) = 7.5k, maximum win (2/3) = 20k. Expected = 15.8k.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) If the opened door shows +7.5k, I still pick the closed door. Minimum win (1/3) = -3k, maximum win (2/3) = 20k, expected win is 12.3k.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As @kaine points out correctly, if we assume that the host has his own strategy, trying to minimize our win, the situation changes slightly.&lt;br&gt;&#xA;If we play to recover our loss, nothing changes (we always win a minimum of $7500).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we play to maximize our winnings, the host knows we will always switch to the closed door (unless he reveals the 20k, which therefore he will not do).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in 2/3 of the cases (when we do not pick 20k originally), switching still gives us the 20k.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the remaining 1/3 of the cases, the host will show us the 7.5k, knowing we will choose the -3k.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our expected winnings are then easily calculated as 2/3 * 20k + 1/3 * -3k = $12333.&lt;/p&gt;&#xA;" OwnerUserId="3903" LastEditorUserId="36937" LastEditDate="2017-05-23T14:43:43.077" LastActivityDate="2017-05-23T14:43:43.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3098" PostTypeId="2" ParentId="3096" CreationDate="2014-10-21T13:49:28.063" Score="2" Body="&lt;p&gt;I always start with box 1.  I'm assuming he picks randomly if neither box 2 or 3 contains the 20,000.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If he shows the -3000 box, you're better off switching.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 1: Probability 33% - Box 1 contains 20k, remaining box contains 7500.  I win 7500 by switching.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 2: Probability 66% - Box 1 contains 7500, remaining box contains 20k.  I win 20,000 by switching.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expected outcome from switching = $7500\times 0.33+20000\times 0.66 = 15833.\overline{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expected outcome from not switching = $20000\times 0.33+7500\times 0.66 = 11666.\overline{6}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If he shows the 7500 box, you're better off switching.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 1: Probability 33% - Box 1 contains 20k, remaining box contains -3000.  I lose 3000 by switching.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 2: Probability 66% - Box 1 contains -3000, remaining box contains 20k.  I win 20,000 by switching.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expected outcome from switching = $-3000\times 0.33+20000\times 0.66 = 12333.\overline{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expected outcome from not switching = $20000\times 0.33 -3000\times 0.66 = 4666.\overline{6}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is exactly like the Monty Hall Problem.  Always switch.&lt;/p&gt;&#xA;" OwnerUserId="3928" LastEditorUserId="3928" LastEditDate="2014-10-21T14:01:47.680" LastActivityDate="2014-10-21T14:01:47.680" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="3100" PostTypeId="1" AcceptedAnswerId="3101" CreationDate="2014-10-21T14:21:53.707" Score="14" ViewCount="3099" Body="&lt;p&gt;You've gotten in trouble with a Mob Boss, and as punishment have been brought to a sandy beach.  You're ordered by him at gunpoint to bury yourself in the sand up to your neck.  He specifies a spot that will have you drowning when the tide comes in.  You have no way to contact any outside help, and you cannot overpower the mob boss.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mob boss, however, is a stickler for logic, and if you can think of a way to do as he says without trapping yourself, he'll let you go and forgive you your transgressions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can you follow the mob boss's order and save yourself from certain briny death?  &lt;/p&gt;&#xA;" OwnerUserId="2756" LastActivityDate="2015-02-07T14:29:54.423" Title="Up to your head in trouble" Tags="|logical-deduction|lateral-thinking|" AnswerCount="12" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3101" PostTypeId="2" ParentId="3100" CreationDate="2014-10-21T14:26:30.930" Score="29" Body="&lt;p&gt;He didn't say you had to &quot;dig a hole&quot; - he just said you had to &quot;bury yourself&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You pile sand around yourself like an ant-hill where your head pokes out the top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now your elevation is sufficient to keep you from drowning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the tide comes in the water washing around you will wash away the sand hill you have made and liberate you. Obviously, the hill might be so weak that you can just wriggle your way out well before this.&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-10-21T14:35:13.080" LastActivityDate="2014-10-21T14:35:13.080" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3102" PostTypeId="2" ParentId="3100" CreationDate="2014-10-21T14:27:10.233" Score="11" Body="&lt;p&gt;You can bury your head down first, up to your neck. In that way, you can easily get loose and survive. If you make a hole first, put some loose sand on the side, then you can put your head in, take a deep breath and throw the sand over your head. After the Mob Boss nods his head, you can set yourself free.&lt;/p&gt;&#xA;" OwnerUserId="3993" LastActivityDate="2014-10-21T14:27:10.233" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3105" PostTypeId="2" ParentId="3096" CreationDate="2014-10-21T14:53:14.533" Score="1" Body="&lt;p&gt;If you pick \$-3K first, he will show you \$7.5K.  You winnings are $\$23000x_{7.5}-\$3000$ where x in this equation is probability you will switch after seeing $7.5K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you pick \$7.5K first, he will show you \$-3K.  You winnings are $\$12500x_{-3}+\$7500$ where x in this equation is probability you will switch after seeing \$-3K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously if those were the only two cases, both x's should be 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets assume that in the case wher you are currently sitting on 20K, he shows you 7.5K with a probability of $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He intellegently chooses $p$ based on the assumption that you are a perfect player.  Your winnings will be $$(-\$23000x_{7.5}+\$20000)p+(-\$12500x_{-3}+\$20000)(1-p)$$ $$=(\$12500x_{-3}-\$23000x_{7.5})p-\$12500x_{-3}+\$20000$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume you should always switch, that quickly simplifies to $(-\$10500)p-\$7500$. So $p$ should equal $1$.  You will always win \$20K if you see \$-3K (and picked $7.5K to start).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you see \$7.5K, therefore, the full  calculation is:  $$.5*(\$23000x_{7.5}-\$3000)+.5*(/$20000-\$23000x_{7.5}) = /$8500)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means, of course, that his best strategy is to assume you will always switch and $2/3$ of the time you will win on average \$8500 and $1/3$ of the time you will always win \$20000.  &lt;strong&gt;On average you will win \$12333.&lt;/strong&gt;  Obviously you still win money but less than stated by the other answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note that you should still never pick the one he opens. \$20000 is large enough that $7500 is never appealing. &lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2014-10-21T14:53:14.533" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3106" PostTypeId="1" CreationDate="2014-10-21T14:58:08.063" Score="9" ViewCount="1392" Body="&lt;p&gt;You find yourself in the usual situation with three locals whose answering is highly restricted. First, they can only answer yes or no questions. Second, one will always answer truthfully, one will always lie (answer the opposite of the truth), and one gives answers that are completely unconstrained - might be random, always yes, alternate truth and lie, alternate yes and no, or anything else (with one exception, below.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unlike other KKJ scenarios, these three are teenagers and as such lack true self awareness or awareness of the inner nature or motivations of others. Simply put, you cannot ask them questions about the structure of the puzzle (their life) such as &quot;does Steve ever lie?&quot; or &quot;are you the one who always tells the truth?&quot; These questions are unanswerable for them, and as such will gain you only a sulky teenage stare, from which you learn nothing. Asking the truth teller &quot;will you answer this question &lt;em&gt;Yes&lt;/em&gt;?&quot; gets you the same sulky stare as the paradoxical &quot;will you answer this question &lt;em&gt;No&lt;/em&gt;?&quot; even through the first should be answerable. They can't answer anything about how the others will answer a hypothetical future question, or how they themselves will answer, and they are unaware of the labels Knight, Knave, and Joker, or even of these categories of behaviour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They can answer questions about past events: &quot;when I asked you how many red balls I had in my hand, did you reply truthfully?&quot; or &quot;Has Steve ever lied to me?&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since they don't know the labels, there is no difference between a &quot;Joker whose strategy is to always respond truthfully&quot; and a Knight, nor between a &quot;Joker whose strategy is to always respond with a lie&quot; and a Knave, therefore this Joker will not choose either of those strategies. However, Jokers won't sulk at a question that Knight or Knave would answer, or answer one they would sulk at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way, given any number of questions, the usual props of the genre (coloured balls, coins, scales etc) about which you can ask questions, and plenty of time, for you to establish conclusively who is Knight, Knave, and Joker? What strategy does so in the fewest questions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: split the scenarios into two:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the teenagers have been answering questions from other visitors before your arrival and have learned enough information that if they provided all the information to you, you would be able to label them, even though it doesn't occur to them to label each other&lt;/li&gt;&#xA;&lt;li&gt;the teenagers have never heard each other answer any questions yet, or have heard only questions in which the Joker happens to have always answered the truth or always a lie, and therefore cannot yet be distinguished from whoever they are agreeing with.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the first scenario you can distinguish them quickly, without waiting for the Joker to diverge from a pattern. In the second, is there any strategy other than waiting while asking questions with known answers?&lt;/p&gt;&#xA;" OwnerUserId="1509" LastEditorUserId="1509" LastEditDate="2014-10-22T19:32:07.313" LastActivityDate="2015-07-15T16:33:18.967" Title="Teenage Knight/Knave/Joker" Tags="|liars|" AnswerCount="3" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="3108" PostTypeId="2" ParentId="3074" CreationDate="2014-10-21T14:59:53.557" Score="17" Body="&lt;p&gt;Totally ripping off &lt;a href=&quot;https://puzzling.stackexchange.com/users/2162/matt-malone&quot;&gt;Matt Malone&lt;/a&gt;'s answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can ask any question about the number where &quot;correct,&quot; &quot;too high,&quot; or &quot;too low&quot; are valid answers, go with:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If you translate the number into trinary, is the last digit a &quot;1&quot;?&lt;/li&gt;&#xA;&lt;li&gt;If you translate the number into trinary, is the second-to-last digit a &quot;1&quot;?&lt;/li&gt;&#xA;&lt;li&gt;If you translate the number into trinary, is the third digit from the right a &quot;1&quot;?&lt;/li&gt;&#xA;&lt;li&gt;If you translate the number into trinary, is the forth digit from the right a &quot;1&quot;?&lt;/li&gt;&#xA;&lt;li&gt;If you translate the number into trinary, is the fifth digit from the right a &quot;1&quot;?&lt;/li&gt;&#xA;&lt;li&gt;Is the number X?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For example, 100 in trinary would be 10201.  The first five answers would be: &quot;correct, too high, too low, too high, correct&quot; which would tell me that the number is 100.  That would be my final guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works for any integer from 0 up to 242.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2014-10-21T16:57:43.043" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3109" PostTypeId="2" ParentId="3100" CreationDate="2014-10-21T15:00:40.563" Score="12" Body="&lt;p&gt;Why vertically? It's difficult, but if you can somehow bury yourself in a shallow grave then you may be able to get back up when the mob boss has moved away to avoid getting his shoes wet.&lt;/p&gt;&#xA;" OwnerUserId="4092" LastActivityDate="2014-10-21T15:00:40.563" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3111" PostTypeId="2" ParentId="3106" CreationDate="2014-10-21T15:18:52.973" Score="8" Body="&lt;p&gt;Ask all three &quot;Do I have a ball?&quot; while holding up a ball.  If the first two agree, let the third leave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the dissenter said yes, he is a knight.  If he said no he is a knave.  The other two are obviously the other two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ask the leaving knight or knave if each of the remaining two ever behaved differently from the knight/knave behavior you have left. This lets you use his/her previous interactions to try and find the Joker.  If he doesn't know which is which from previous behavior, just let him leave. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you haven't learnt who they are from the identifed individual, tell the remaining two that they may leave when they disagree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeatedly ask them in random order different questions with obvious answers.  The first one to change his behavior is a Joker.  The remaining one is the remaining class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, we have to assume that the Joker will behave almost exactly like a Knight or Knave except for one in a billion cases. This means that this is the best strategy for an unknown strategy. It will take a minimum of 3 and a maximum without an upper bound to determine who is who. Fortunately, being a teenager, the Joker will likely want to leave and stop being annoyed so will let you know who he is quickly.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditDate="2014-10-21T17:07:09.107" LastActivityDate="2014-10-21T17:07:09.107" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3115" PostTypeId="2" ParentId="3100" CreationDate="2014-10-21T17:34:14.783" Score="2" Body="&lt;p&gt;You could, in theory, build a wall of sand around you with the sand you dug out (and probably more) to keep the water from drowning you.&lt;/p&gt;&#xA;" OwnerUserId="2758" LastActivityDate="2014-10-21T17:34:14.783" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3116" PostTypeId="1" AcceptedAnswerId="3122" CreationDate="2014-10-21T18:08:02.810" Score="10" ViewCount="1460" Body="&lt;p&gt;A very old man has a million dollars, which he's going to pass down to you! He has 3 associates who have been ordered to help you. The 3 associates know the location where it is, and they tell you it is in either &lt;strong&gt;a)&lt;/strong&gt; a box, &lt;strong&gt;b)&lt;/strong&gt; a safe, or &lt;strong&gt;c)&lt;/strong&gt; a hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;One of the associates always tells the truth, one always lies, the third tells truth or lies at random.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The associates are very greedy and will try to take the money for themselves, but they will give you a chance at getting it. They let you pick the box, safe, or hole. Then one of then (you don't know which) comes up to tell you whether you have chosen correctly or wrongly, gives you one question, and then shows you the selection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They then give you a chance to switch to the other selection. &lt;strong&gt;What question should you ask. Then, should you switch?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="15" LastEditDate="2014-12-31T19:37:52.127" LastActivityDate="2014-12-31T19:37:52.127" Title="The Old Millionaire" Tags="|logical-deduction|liars|monty-hall|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="3117" PostTypeId="2" ParentId="2929" CreationDate="2014-10-21T18:09:02.587" Score="2" Body="&lt;p&gt;This isn't a complete answer, but I put it here to see if it can help in the generation of a complete answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us consider a necessary condition (though not a sufficient one) for a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets say a bunny &quot;diags&quot; when it transitions to a diagonally adjacent square. This counts as a visit of the square onto which it arrives.&#xA;Let us say a bunny &quot;jumps&quot; when it makes an allowed transition to a square three squares away in one of the side directions (which requires a second bunny in an appropriate location). This counts as a visit of the square onto which it arrives.&#xA;Let us say a bunny &quot;lands&quot; when it is repositioned due to the other bunny's &quot;jump&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a bunny jumps another, the second bunny lands (by that jump) on the jumping bunny's initial position. This landing of the second bunny either does, or does not count as a visit. If it does count then the problem is insolvable as written as in a circuit of 63 moves (j jumps and 63-j diags) 63+j squares will have been visited. Thus j must be 0 which is impossible as the bunny could have visited squares of only a single color had there been no jumps. Thus we conclude that the landing of a bunny on a square caused by the other bunny jumping must not count as a visit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider one of the two black corners Let us call it c. It has only a single diagonal neighbor which we call n, and two white squares along the side at a distance of three which we call w1 and w2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two lines tracing through that square must change colors either immediately before or after that corner as there must be a jump to allow the other bunny to exit that corner (as the entering bunny cannot also leave without visiting a square a second time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each bunny, c must either be entered by a diag from n, or by a jump from one of w1 or w2. Similarly c must either be exited by a diag to n or a jump to one of w1 or w2. The entrance and exit of a bunny cannot both be diags as this would revisit c. They cannot both be jumps as it would take the other bunny three diags to get in position for the second jump and the rules disallow either bunny skipping moves. Thus we know that the entrance and exit from c must consist of a diag and a jump in some order.&#xA;This is true of each bunny. The second bunny lands on c when the first bunny jumps in or out. This second bunny cannot jump as the first bunny cannot be in position to be jumped thus it must diag out.&#xA;Thus the second bunny's entrance to c must be by a jump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have tightly constrained what must happen at each of the four corners, but have neither discerned a possible move sequence, not shown it to be impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comments? Extensions?&lt;/p&gt;&#xA;" OwnerUserId="4097" LastEditorUserId="4097" LastEditDate="2014-10-21T18:36:14.163" LastActivityDate="2014-10-21T18:36:14.163" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3118" PostTypeId="1" AcceptedAnswerId="3123" CreationDate="2014-10-21T18:20:19.440" Score="1" ViewCount="318" Body="&lt;p&gt;I would like to try more strict conditions for &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3087/t-knights-l-knaves-and-r-jokers&quot;&gt;puzzle&lt;/a&gt;, which frodoskywalker solved so simply. Basically all the same, but only one type of questions is allowed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;We have T Knights (truth-tellers), L Knaves (liars) and R Jokers (random-tellers).  $T &amp;gt; 0$. We know all 3 numbers, but we do not know who is who.&lt;/li&gt;&#xA;  &lt;li&gt;We can chose any two of them and ask the first about the second one &quot;Is he a Knight?&quot;. We can repeat this questioning procedure any number of times.&lt;/li&gt;&#xA;  &lt;li&gt;We need to find one Knight.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How must T, L and R be related to make the task possible to complete?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My analysis:&lt;br&gt;&#xA;1. &lt;strong&gt;If $T+L \le R$ it is impossible&lt;/strong&gt;. $T$ jokers can simulate Knights, $L$ Knaves and we would never be able to distinguish between them and find a real Knight.&lt;br&gt;&#xA;2. &lt;strong&gt;If $T&amp;gt;R+L$ the task is possible&lt;/strong&gt;. It is possible with $L = 0, T &amp;gt; R$ and we can treat Knaves as Jokers.&lt;br&gt;&#xA;3. &lt;strong&gt;If $R=0, L = T$ then this is impossible&lt;/strong&gt;.&lt;br&gt;&#xA;Indeed, if you get an answer &quot;Yes&quot;, then there is two possibilities: &quot;Kni Kni&quot; or &quot;Kna Kna&quot; (both Knights or both Knaves). If we get &quot;No&quot;, then it is either &quot;Kni Kna&quot; or &quot;Kna Kni&quot;. From this we can see that if we replace all Knights by Knaves and vice versa we will get exactly the same answers. So even if we are able to divide them all into two groups according to their type we will never know who is who.&lt;br&gt;&#xA;4. &lt;strong&gt;If $R=0, L &amp;gt; T$ then it can be solved&lt;/strong&gt; in exactly the same way as for $L=0, T &amp;gt; R$ &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3087/t-knights-l-knaves-and-r-jokers&quot;&gt;case&lt;/a&gt;:&lt;br&gt;&#xA;We need to exclude all pairs who answer &quot;No&quot; and create a chain of &quot;Yes&quot; longer then half of the not excluded people. Then the last in the chain will be a Knave. Just ask him about all the rest until you hear &quot;No&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is what happens when $R \neq 0$, $L&amp;gt;R-T$ and $L \ge T - R$. For example, when $L = 1, T = R = 10$.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2014-10-23T11:00:26.150" Title="T Knights, L Knaves and R Jokers with one question type only" Tags="|puzzle-creation|liars|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3119" PostTypeId="1" AcceptedAnswerId="3126" CreationDate="2014-10-21T18:51:33.807" Score="4" ViewCount="1477" Body="&lt;p&gt;Inspired by &lt;a href=&quot;https://puzzling.stackexchange.com/q/3074/68&quot;&gt;6 Tries to Guess a Number Between 1-100&lt;/a&gt;, how many guesses would it take if you can ask if the outcome of an operation is equal, less than, or more than a number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, Alice picks a number $x$ such that $1\leq x \leq 100$ and her friend Bob asks $f(x) \overset{?}{=} y$ to which Alice may only respond with 'correct', 'bigger', and 'lower'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example would be Bob asking $ x \overset{?}{\equiv} 1\pmod 3$ and Alice replying 'higher'. Bob has now eliminated &amp;frac23; of the domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many guesses would it take before he can state the correct number Bob?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Please note that I do not have an answer to this question, but since using modulo powers of 3 is equivalent to the &lt;a href=&quot;https://puzzling.stackexchange.com/a/3125/68&quot;&gt;solution I gave to a similar question&lt;/a&gt;, I have an upper bound of stating it on your 6th 'guess'.&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2014-10-21T20:37:46.143" Title="Another 6 tries to guess a number between 1-100" Tags="|mathematics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3120" PostTypeId="2" ParentId="2929" CreationDate="2014-10-21T19:03:30.097" Score="1" Body="&lt;p&gt;There is no solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a bunny starts on a X-coloured square, without jumping another bunny that bunny will remain on that coloured square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to move a bunny from a bunny to move from one coloured square it must hop over an adjacent bunny that is either north, east, south or west (diagonal bunny hops, if they are legal, will place the bunny on the same coloured square; diagonal moves also remain on the same colour). These adjacent squares are always a different colour than the square the bunny that is hopping is on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue is this: In order for both bunnies to cover each square on the board each bunny must hop at least once (the black bunny needs to get to the white squares and the white bunny needs to get to the black squares) but, it is only for one bunny to hop over another during the course of a game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, say the bunny on the white square hops over the bunny on the black square, at this point both bunnies are on black squares, and here they must remain the entire game.&lt;/p&gt;&#xA;" OwnerUserId="4099" LastActivityDate="2014-10-21T19:03:30.097" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3121" PostTypeId="2" ParentId="3106" CreationDate="2014-10-21T19:04:58.703" Score="3" Body="&lt;p&gt;The answer to the question is in the question itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let &lt;strong&gt;Alice&lt;/strong&gt;, &lt;strong&gt;Bob&lt;/strong&gt; and &lt;strong&gt;Charles&lt;/strong&gt; represent each person respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have never met any of these teenagers or asked them any questions (otherwise, pick somebody who hasn't). However, they don't know that, as e&lt;em&gt;ach person only knows their individual history with you&lt;/em&gt;. Therefore, you ask &lt;strong&gt;Alice&lt;/strong&gt;, &lt;strong&gt;&quot;Has Alice ever lied to me?&quot;&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Alice responds &quot;No&quot;, she is the truth teller (the Knight), as you have never met or talked before. If Alice responds &quot;Yes&quot;, then she is the liar (the Knave). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Alice is the Joker, she will not respond. This is because the joker is unwilling to answer any question the Knight Or the Knave would sulk at (they both would sulk, as they have no idea if Alice has ever lied to you and have no &quot;awareness of the inner nature or motivations of others&quot;). If Alice is the Joker, one more trivial question (e.g. &quot;Does 1+1=2?&quot;) will differentiate Knave from Knight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Alice is either Knave of Knight, proceed to ask &lt;strong&gt;Bob&lt;/strong&gt;, &lt;strong&gt;&quot;Has Bob ever lied to me?&quot;&lt;/strong&gt;. If Bob responds &quot;No&quot;, he is the truth teller (Knight), if he responds &quot;Yes&quot;, he is a liar (Knave), and if he doesnt respond, hes the joker!&lt;/p&gt;&#xA;" OwnerUserId="2078" LastActivityDate="2014-10-21T19:04:58.703" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3122" PostTypeId="2" ParentId="3116" CreationDate="2014-10-21T19:08:37.287" Score="13" Body="&lt;p&gt;For simplicity, I'm going to refer to the associates as knight, knave, and joker. Also, I'm assuming that they won't reveal the location of the money. If they did, then obviously you would switch to that once they reveal it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that there is an equal chance of any of the three associates coming up and talking to you, we have a two in three chance of talking to knight or knave. So you can ask the associate &quot;If I were to ask you if you always told the truth, what would you say?&quot; Both knight and knave would reply &quot;Yes&quot; if asked if they tell the truth, so knight will honestly report &quot;Yes&quot;, while knave will lie about it and say &quot;No&quot;. We don't know what joker would say. In either case, we respond as if we were not talking to joker and switch as appropriate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at how this plays out:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we chose the wrong box (2/3 chance):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Knight tells us we chose wrong (1/3), then replies &quot;Yes&quot; to the question:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We believe knight and switch to the last choice, which is the correct choice&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Knave tells us we chose right (1/3), then replies &quot;No&quot; to the question:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We don't trust knave and switch to the last choice, which is the correct choice&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Joker (1/3)&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Tells us we chose right (1/2?)&#xA;&lt;ul&gt;&#xA;&lt;li&gt;then replies &quot;Yes&quot; to the question (1/2?):&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We trust joker and don't switch, and lose :(&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;then replies &quot;No&quot; to the question (1/2?):&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We don't trust joker and switch, and win&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Tells us we chose wrong (1/2?)&#xA;&lt;ul&gt;&#xA;&lt;li&gt;then replies &quot;Yes&quot; to the question (1/2?):&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We trust joker and switch, and win&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;then replies &quot;No&quot; to the question (1/2?):&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We don't trust joker and don't switch, and lose&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Suppose we choose the right box (1/3 chance):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Knight tells us we chose right (1/3), then replies &quot;Yes&quot; to the question:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We believe knight and don't switch - win&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Knave tells us we chose wrong (1/3), then replies &quot;No&quot; to the question:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We don't trust knave and don't switch - win&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Joker (1/3)&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Tells us we chose right (1/2?)&#xA;&lt;ul&gt;&#xA;&lt;li&gt;then replies &quot;Yes&quot; to the question (1/2?):&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We trust joker and don't switch, and win&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;then replies &quot;No&quot; to the question (1/2?):&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We don't trust joker and switch, and lose&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Tells us we chose wrong (1/2?)&#xA;&lt;ul&gt;&#xA;&lt;li&gt;then replies &quot;Yes&quot; to the question (1/2?):&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We trust joker and switch, and lose&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;then replies &quot;No&quot; to the question (1/2?):&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We don't trust joker and don't switch, and win&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So we see that if we talk to knight or knave, we will always win. If we talk to joker, we win about half of the time. If joker's response are truly random, then the chance will be exactly half. If the joker's response is not random, then we can't know what our chance will be (or come up with a strategy that will definitely be better) unless we know more about the joker's strategy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if joker chooses randomly your chance of getting the million dollars can be at least $\frac{2}{3}+\frac{1}{2}\cdot \frac{1}{3}=\frac{5}{6}$.&lt;/p&gt;&#xA;" OwnerUserId="1955" LastEditorUserId="1955" LastEditDate="2014-11-28T18:37:43.687" LastActivityDate="2014-11-28T18:37:43.687" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3123" PostTypeId="2" ParentId="3118" CreationDate="2014-10-21T19:29:56.760" Score="3" Body="&lt;p&gt;We must have $L + T &amp;gt; R$ and $T \neq L + R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distinguishing feature of knights is that they form a group of $T$ people who identify one another as knights and everyone else as not knights.  We only have trouble if we have two groups of people who do that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $T = L + R$, and the jokers always lie, then the set of non-knights will be indistinguishable from the set of knights.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here down, consider only cases where $T \neq L + R$.  In this case, the group of knights identify $T$ people as knights and the knaves identify $L+R \neq T $ people as knights, so we can separate those groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way to have confusion is if we have a group of $T$ jokers who identify one another as knights and everyone else as not knights.  This can only happen if $T \leq R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, we have two groups of potential knights and everyone else, who is potentially a knave.  The potential knaves have $L$ knaves and $R-T$ jokers.  If $L &amp;gt; R-T$, then we know most of the potential knaves will reliably lie about the groups of potential knights, so we can identify which group are the actual knights.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $L \leq R-T$, then we can have a set of $L$ jokers who tell the truth about the people in the two sets of potential knights, which is the opposite of what the $L$ knaves say.  Thus, we cannot identify which group of potential knights are the actual knights.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reiterate:&#xA;We have ambiguity if:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A. We have $T = L+R$ and all the jokers lie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B. We have one group of $T$ jokers who all claim everyone in that group is a knight and everyone else is not and we also have another group of $L$ jokers who tell the truth about the knights and the first set of jokers, but lie about everyone else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any other case, we can distinguish which group are the real knights.  Thus, the problem is solvable if $L + T &amp;gt; R$ and $T \neq L + R$.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="2328" LastEditDate="2014-10-21T19:53:43.637" LastActivityDate="2014-10-21T19:53:43.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3124" PostTypeId="1" AcceptedAnswerId="3125" CreationDate="2014-10-21T19:38:20.543" Score="0" ViewCount="1657" Body="&lt;p&gt;You get get 6 attempts to find the number between 1-100. Each &quot;attempt&quot; can either be a guess, or a range it could fall in, responses you could receive are &quot;Correct!&quot;, &quot;My number is lower.&quot;, &quot;My number is higher&quot;, or &quot;My number is within that range&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have 6 tries to guess it, NOT to know it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can do it in less than 6 please do post it.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-10-21T23:45:11.090" LastActivityDate="2015-10-05T16:28:19.813" Title="6 Attempts to Guess a Number Between 1-100 ~ Ranges" Tags="|mathematics|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3125" PostTypeId="2" ParentId="3124" CreationDate="2014-10-21T19:49:14.433" Score="12" Body="&lt;p&gt;Divide the current range by three and guess the middle range. It's either correct, higher, or lower, which reduces your search space by &amp;frac23;. You'll go from a search space of 100 to (worst case) 34, to 12, to 4, to 2, to 1 which you can state on your 6th guess.&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditDate="2014-10-21T20:15:07.260" LastActivityDate="2014-10-21T20:15:07.260" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3126" PostTypeId="2" ParentId="3119" CreationDate="2014-10-21T20:19:06.780" Score="3" Body="&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/3124/6-attempts-to-guess-a-number-between-1-100-ranges/3125#3125&quot;&gt;Your solution&lt;/a&gt; from the other question works (and is ideal) here, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the current range is $[a, b]$ (at the beginning $a=1$, $b=100$).&#xA;break the range into thirds; let $t_1 = \frac{b-a}{3} + a$, $t_2 = \frac{2(b-a)}{3} + a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let $f(x) = -1$ if $x \le{} t_1$, $0$ if $t_1 \le{} x \le{} t_2$, and $1$ if $t_2 \le{} x$.&#xA;Now compare $f(x)$; you've eliminated two thirds of the range. Just as argued in your answer, that will guarantee that your sixth guess is a single, correct number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't do any better than this, in the worst case. Any answer will separate the candidate numbers into three sets (one for each possible answer to the comparison). Assume for the worst-case analysis that the answer indicates that the largest of these three sets contains the correct answer; you want to minimize the maximum set size, which requires making sure there are one third in each basket.&lt;/p&gt;&#xA;" OwnerUserId="2101" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2014-10-21T20:19:06.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3127" PostTypeId="2" ParentId="3119" CreationDate="2014-10-21T20:21:53.997" Score="2" Body="&lt;p&gt;My answer to that question would seem to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Express $x$ in trinary.  Repeatedly ask if digit $i$ from the right is equal to $1$.  This will yield $3^5$ possible answers in $5$ questions.  At that point you know $x$, since $3^5 &amp;gt; 100$.  Ask that value for your 6th question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be more explicit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x,i) = floor\left({x \mod 3^i \over 3^{i-1}}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $y_i = 1$.  Ask for $i = 1,2,3,4,5$.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2014-10-21T20:21:53.997" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3128" PostTypeId="2" ParentId="1724" CreationDate="2014-10-21T22:54:27.917" Score="3" Body="&lt;p&gt;For clarity, I label the switches and output slots as follows:&#xA;Switches are labeled from left to right, top to bottom as 1-7.  Slots are labeled left to right as A-F.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we keep dropping coins down slot 1, eventually a coin will come out slot A.  At that point we know switch 1 is facing out without coin and switch 3 is facing in without coin. This takes at most 5 drops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We repeat with slot 4 until a coin comes out slot F.  At that point we know switch 2 is facing out without coin and switch 5 is facing in without coin.  This also takes at most 5 coins.  Total: $\leq 10$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we dropped fewer than four coins down slot 4, we drop four more coins down slot 4.  This yields another coin out slot F and leaves switches 2 and 5 in place.  It also ensures at least two coins fell on the right side of switch 4, guaranteeing that switch now points to the right.  Total: $\leq 12$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We next drop a coin down slot 2.  If a coin came out slot C, but not B, we know switch 6 is now empty and facing in.  If nothing came out, we know switch 6 is now full, facing in.  If two coins came out, we know switch 6 is now empty, but don't know which direction it's facing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We next drop a coin down slot 3.  Likewise, if a coin came out D, we know switch 7 is now empty facing in.  If nothing came out, it's full facing in.  If two coins came out, it's empty, facing one direction or other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we don't know the direction of switch 6, we drop a coin down slot 1, then another down slot 2.  (This leaves switch 3 with a coin on it).  We now know switch 6 is facing in.  If a coin came out slot C, switch 6 is now empty; if not, it is now full.  Likewise, if we don't know the direction of switch 7, we drop a coin down slot 4 then another down slot 3.  Total: $\leq 18$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now know all switches are facing in and we know which way switch 4 is facing.  Switches 3, 5, 6, and 7 might be holding coins, depending on above details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We then follow an algorithm for adding coins when we know what's going on.  We first ensure switches 6 and 7 have coins, then 3, 4, and 5, then 1 and 2.  The worst case would be if switches 3, 5, and 6 now have coins, but 7 doesn't, in which case, it would take 10 more coins to finish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total: $\leq 28$ coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this plan optimal?  No.  In the interest of writing a simple algorithm, I ignored a lot of other information we could have gathered.  Keeping track of all that information could allow for shortcuts requiring fewer coins.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2014-10-21T22:54:27.917" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3129" PostTypeId="2" ParentId="3100" CreationDate="2014-10-21T23:58:54.643" Score="8" Body="&lt;p&gt;Bury yourself up to the level of your neck, but hold your arms above your head.  With your arms free, you can easily dig yourself out again before the tide comes in.&lt;/p&gt;&#xA;" OwnerUserId="1667" LastActivityDate="2014-10-21T23:58:54.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3130" PostTypeId="2" ParentId="2929" CreationDate="2014-10-22T07:35:25.327" Score="1" Body="&lt;p&gt;Jim already made the observation about that a corner square requires either entering or exiting by means of a jump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have some more observations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Diagonal moves keep a bunny constrained to only white or only black squares; the only way to change&#xA;square color is by means of a jump.&lt;/li&gt;&#xA;&lt;li&gt;A jump requires both bunnies to be on different square colors, and the jump causes them both to &#xA;change square colors.&lt;/li&gt;&#xA;&lt;li&gt;Hence, both bunnies need to start off on different square colors (this was also stated as a &#xA;requirement)&lt;/li&gt;&#xA;&lt;li&gt;Because both bunnies need to cover both all white and black squares, one bunny needs to jump at &#xA;least once so that they can reach the other square color.&lt;/li&gt;&#xA;&lt;li&gt;A corner square can also be covered if a bunny starts of at that square.&lt;/li&gt;&#xA;&lt;li&gt;There are four corner squares. At most 2 squares can be covered by a bunny's starting position, so&#xA;there are at least 2 jumps required to cover the other 2 corner squares.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3932" LastActivityDate="2014-10-22T07:35:25.327" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3131" PostTypeId="2" ParentId="3100" CreationDate="2014-10-22T10:07:29.647" Score="0" Body="&lt;p&gt;Bury yourself in a slightly reclined position so that your upper torso is only buried a few inches and you legs and feet buried slightly more. It would be easy to sit up then use your hands to dig your legs and feet out.&lt;/p&gt;&#xA;" OwnerUserId="4113" LastActivityDate="2014-10-22T10:07:29.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3132" PostTypeId="2" ParentId="3100" CreationDate="2014-10-22T10:22:31.920" Score="0" Body="&lt;p&gt;Bury yourself sideways.  Sand up to your neck wouldn't be heavy enough to restrict you, or even fully cover your body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mob boss leaves, then you stand up and walk away.  Or, immediately after burying yourself up to your neck sideways, let him verify that half of your sideways body is buried up to your neck, then get up and walk away having fulfilled his orders.&lt;/p&gt;&#xA;" OwnerUserId="4115" LastActivityDate="2014-10-22T10:22:31.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3133" PostTypeId="2" ParentId="2929" CreationDate="2014-10-22T10:54:43.087" Score="16" Body="&lt;h1&gt;Reiterate problem&lt;/h1&gt;&#xA;&lt;p&gt;Bunny: a chess piece which moves like a bishop but only one square from its current position. It may also hop over another piece.&lt;/p&gt;&#xA;&lt;p&gt;Hop: a bunny hops when another piece is in a square touching the current square (no diagonals). In its turn it occupies the square 3 squares from its current square in the direction of the hopped piece. The hopped piece then takes the original square of the bunny. This is not considered a turn/move for the hopped piece.&lt;/p&gt;&#xA;&lt;p&gt;Say, there are 2 bunnies on a board. &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt;.&#xA;&lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt; begin on opposite colour squares of your choosing.&lt;/p&gt;&#xA;&lt;p&gt;By what means might they both tour the board? No turn may be taken to a previously occupied square (being hopped does not constitute occupation). They take turns, &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; moves first, &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt; second, and they will take 63 turns each.&lt;/p&gt;&#xA;&lt;h1&gt;Vocabulary&lt;/h1&gt;&#xA;&lt;p&gt;The bunny thus has 2 distinct modes of motion: the &amp;quot;step&amp;quot; and the &amp;quot;hop&amp;quot;. If a piece is hopped over its motion (though not counted as a move) is a &amp;quot;pivot&amp;quot;. A sequence of &amp;quot;step&amp;quot;s is a &amp;quot;walk&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;When &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; moves to a square he paints the square &lt;span class=&quot;math-container&quot;&gt;$green$&lt;/span&gt;. When &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt; moves to a square he paints it &lt;span class=&quot;math-container&quot;&gt;$blue$&lt;/span&gt;. After both bunnies have moved to a square it is painted &lt;span class=&quot;math-container&quot;&gt;$red$&lt;/span&gt; - and hereafter no bunny may enter except by &amp;quot;pivot&amp;quot;, and then only immediately as the last bunny has moved to this square.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$Lemma\ 1:$&lt;/span&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3079/how-many-bishops-to-walk-across-the-board&quot;&gt;It is a known fact that no tour purely constistuted of steps is possible. In fact, it is known that it takes 4 distinct walks (sharing no squares) to cover all the black (or white) squares on the board.&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$Lemma\ 2:$&lt;/span&gt; As a corollary to the above, there are only a subset of total squares of a given colour &lt;em&gt;reachable&lt;/em&gt; from any arbitrary square - where &lt;em&gt;reachable&lt;/em&gt; means that it can be reached by only using some sequence of steps (no hopping). If all squares or a colour were reachable then &lt;span class=&quot;math-container&quot;&gt;$Lemma\ 1$&lt;/span&gt; would be false.&lt;/p&gt;&#xA;&lt;h1&gt;Answer&lt;/h1&gt;&#xA;&lt;p&gt;I will now claim that this bunny's tour is impossible.&lt;/p&gt;&#xA;&lt;p&gt;For &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt;, given any starting square &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; on colour &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt;, we know that &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; must be a square on &lt;em&gt;only&lt;/em&gt; one of the four walks needed to cover the square of colour &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; - call this walk &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt;. Any square visited from &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; must be part of &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt;. It is clear that his tour cannot be completed from &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$Lemma\ 1$&lt;/span&gt;).&#xA;This means that &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; must escape and continue by means of a hop.&#xA;Now, &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt; occupies one of the squares in &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$s^{\prime}$&lt;/span&gt;. &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt; may now process on his walk &lt;span class=&quot;math-container&quot;&gt;$w^{\prime}$&lt;/span&gt;. We know that only the same squares that were &lt;em&gt;reachable&lt;/em&gt; to &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; are &lt;em&gt;reachable&lt;/em&gt; to &lt;span class=&quot;math-container&quot;&gt;$s^{\prime}$&lt;/span&gt;. This means that they are part of the same walk. Eventually &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt; will get stuck or otherwise need to change to a different walk to have a chance at covering the board (as we know that 1 is not sufficient). His only means of escape is a hop. Now &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; takes a square on &lt;span class=&quot;math-container&quot;&gt;$w^{\prime}$&lt;/span&gt;, BUT &lt;span class=&quot;math-container&quot;&gt;$w^{\prime}$&lt;/span&gt; has exactly the same reachable squares as &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt;. So, &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; has gained no new reachable squares for colour &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$Lemma\ 2$&lt;/span&gt;). And there is nothing he can do to remedy this situation.&lt;/p&gt;&#xA;&lt;p&gt;Thus, there is no 2-bunny tour of a chessboard.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;early observations for historical reasons only&lt;/h1&gt;&#xA;&lt;p&gt;I'm quite positive that there is no solution.&lt;/p&gt;&#xA;&lt;p&gt;Let's say we have 2 bunnies, a black one and a white one. The black bunny paints the board with green paint, and the white bunny paints the board with blue paint. A square that has been painted with both green and blue paint, is red.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a white bunny can no longer visit a blue square&lt;/li&gt;&#xA;&lt;li&gt;a black bunny can no longer visit a green square&lt;/li&gt;&#xA;&lt;li&gt;a red square can no longer be accessed by any bunny.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;The answer to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3079/how-many-bishops-to-walk-across-the-board&quot;&gt;this question&lt;/a&gt; suggests that there is no &amp;quot;walking-bishop&amp;quot; tour. The bunny is a walking-bishop with a hopping capability.&lt;/p&gt;&#xA;&lt;p&gt;Each bunny must visit every square. This means that at least 8 hops would be required. 3 squares are &amp;quot;involved&amp;quot; in a hop. And one of them will be guaranteed to be red after a hop. This means that the 8 hops will involve cancelling out at least &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{8}^{th}$&lt;/span&gt; of the entire board.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;they must change from black to white squares 4 time in order to get out of where they are stuck and to finish painting the board.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;The dead-locks occur under the following conditions: (the first 3 are because the bunny's counterpart cannot each a hopping square)&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a white bunny is anywhere where the 4 opposite colour squares surrounding him are coloured green&lt;/li&gt;&#xA;&lt;li&gt;a black bunny is anywhere where the 4 opposite colour squares surrounding him are coloured blue&lt;/li&gt;&#xA;&lt;li&gt;any bunny is anywhere where the 4 opposite colour squares surrounding him are coloured red&lt;/li&gt;&#xA;&lt;li&gt;a bunny is on a red square and cannot move to non-red square without making it red (now hopping is not possible - as the other bunny would need to occupy a red square - i.e. a square he has already occupied in the past)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;I will turn this into something solid soon.&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-12-04T04:19:44.493" CommentCount="28" ContentLicense="CC BY-SA 3.0" />
  <row Id="3136" PostTypeId="1" AcceptedAnswerId="3140" CreationDate="2014-10-22T12:44:11.797" Score="21" ViewCount="7191" Body="&lt;p&gt;I found this question and became curious, can anyone tell me the answer and prove it, i know it seems fairly simple but just thought an explanation of this would make an interesting case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flip a fair coin repeatedly until you get two heads in a row (HH assuming H indicates head and T indicates Tails). On average, how many flips should this take? What if we flip until we get heads followed by tails (HT)? Are the answers the same?&lt;/p&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="12408" LastEditDate="2016-07-25T02:29:49.483" LastActivityDate="2016-07-25T02:29:49.483" Title="Flip a Fair Coin" Tags="|combinatorics|probability|coin-tossing|" AnswerCount="5" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="3139" PostTypeId="2" ParentId="3136" CreationDate="2014-10-22T12:56:42.043" Score="1" Body="&lt;p&gt;Previous answers have stated that the probability of getting either a HH or HT combination is 1/4. Based on that, we expect that it would take 1/p or 1/(1/4) tries to achieve it. Therefore, I would say that it would take 8 flips (4 * 2 flips). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Well, it seems my reasoning is flawed, but there is hope from &lt;a href=&quot;https://math.stackexchange.com/questions/364038/expected-number-of-coin-tosses-to-get-five-consecutive-heads&quot;&gt;https://math.stackexchange.com/questions/364038/expected-number-of-coin-tosses-to-get-five-consecutive-heads&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on that, we get e = 1/2 * (e+1) + 1/4 * (e+2) + 1/4 * 2 which works out to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e = 6, so 6 tries for HH&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit again: I thought the answers were the same, but it looks like they are different. The second equation is only 1/2 * (e+1) + 1/2 * 2, which works out to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e = 3, so 3 tries for HT&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, ultimately, the probabilities are different. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="-1" LastEditDate="2017-04-13T12:19:36.950" LastActivityDate="2014-10-22T13:26:40.613" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3140" PostTypeId="2" ParentId="3136" CreationDate="2014-10-22T13:21:17.733" Score="23" Body="&lt;p&gt;&lt;strong&gt;Case 1: Waiting for HH&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A$ = average nb. of flips to get HH after T or nothing.&lt;br&gt;&#xA;$B$ = average nb. of flips to get HH after H.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A is 1 flip, plus B if you get H and continue, or plus A if you get T and restart:&lt;br&gt;&#xA;$A = 1 + {1\over2}B + {1\over2}A$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B is 1 flip, and you are done if you get H, or plus A if you get T and restart:&lt;br&gt;&#xA;$B = 1 + {1\over2}0 + {1\over2}A$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This resolves to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B = 1 + {1\over 2}A$&lt;br&gt;&#xA;$A = 1 + {1\over 2} (1 + {1\over 2} A) + {1\over2} A$&lt;br&gt;&#xA;$A = 1 + {1\over 2} + {1\over 4} A + {1\over 2} A$&lt;br&gt;&#xA;$A = 6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expected number of flips to get HH is 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 2: Waiting for HT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A$ = average nb. of flips to get HT after T or nothing.&lt;br&gt;&#xA;$B$ = average nb. of flips to get HT after H.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A is 1 flip, plus B if you get H and continue, or plus A if you get T and restart:&lt;br&gt;&#xA;$A = 1 + {1\over2}B + {1\over2}A$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B is 1 flip, and you are done if you get T, or plus B if you get H and wait for T again:&lt;br&gt;&#xA;$B = 1 + {1\over 2}0 + {1\over 2}B$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This resolves to:&lt;br&gt;&#xA;$B = 1 + {1\over 2}B$&lt;br&gt;&#xA;$B = 2$&lt;br&gt;&#xA;$A = 1 + {1\over 2}2 + {1\over 2}A = 2 + {1\over 2}A$&lt;br&gt;&#xA;$A = 4$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expected number of flips to get HT is 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Comparison&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference comes from the fact that in the first case, if you get the wrong 2nd result, you start all over, while in the second case you continue with one good result.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2014-10-29T19:45:13.530" LastActivityDate="2014-10-29T19:45:13.530" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3141" PostTypeId="2" ParentId="3100" CreationDate="2014-10-22T13:25:49.400" Score="0" Body="&lt;p&gt;If you have to bury &lt;em&gt;yourself&lt;/em&gt; up to your neck, you need your arms to do so. While there may be sand up to your neck, your arms will have been free enough to dig yourself in, so they will be free enough to dig yourself out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see a problem there, unless the Mob Boss will be waiting for the tide to do its work and keeps you (at gunpoint) from digging yourself out again.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2014-10-22T13:25:49.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3143" PostTypeId="2" ParentId="3136" CreationDate="2014-10-22T14:14:03.540" Score="5" Body="&lt;p&gt;To supplement Florian's answer and perhaps help make it more clear why it is correct that HH requires more expected coin tosses than HT, look at a list of all possible outcomes of three coin tosses:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;HHH&#xA;HHT&#xA;HTH&#xA;HTT&#xA;THH&#xA;THT&#xA;TTH&#xA;TTT&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now look at when we get HH or HT:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;HHH      HH after 2&#xA;HHT      HH after 2, HT after 3&#xA;HTH      HT after 2&#xA;HTT      HT after 2&#xA;THH      HH after 3&#xA;THT      HT after 3&#xA;TTH&#xA;TTT&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Even at this point HT is showing up more frequently - we have 1/4 chance of getting HH or HT after 2 (as you probably expected), but overall we have a 1 in 2 chance of getting HT compared to a 3 in 8 chance of getting HH within three coin tosses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For HH, here are the combos that don't work:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;HTH&#xA;HTT&#xA;THT&#xA;TTH&#xA;TTT&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Of these, two ended with heads and might only require one additional coin toss to win. For HT:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;HHH&#xA;THH&#xA;TTH&#xA;TTT&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Three of the four end with heads and might only require one additional coin toss to win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So after three coin tosses, you're more likely to get HT than HH, and you're also more likely to be in a position where the next coin toss might be a success! (3 in 4 chance over a 2 in 5 chance).&lt;/p&gt;&#xA;" OwnerUserId="1955" LastActivityDate="2014-10-22T14:14:03.540" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3144" PostTypeId="2" ParentId="3136" CreationDate="2014-10-22T14:21:19.113" Score="5" Body="&lt;h2&gt;HH&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Following the method outlined in &lt;a href=&quot;https://math.stackexchange.com/a/364135/106234&quot;&gt;this answer on Math.SE&lt;/a&gt;, we get the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $e$ be the expected number of tosses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start tossing. If we get a tail immediately (probability $\tfrac{1}{2}$) then the expected number is $e+1$. If we get a head then a tail (probability $\tfrac{1}{4}$), then the expected number is $e+2$. Finally, if our first $2$ tosses are heads (again with probability $\tfrac{1}{4}$), then the expected number is $2$. Thus&#xA;$$e=\frac{1}{2}\cdot(e+1)+\frac{1}{4}\cdot(e+2)+\frac{1}{4}\cdot(2)$$&#xA;Solving this for $e$ yields $e=6$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;HT&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Again, let $e$ be the expected number of tosses and start tossing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we get a tail immediately (probability $\tfrac{1}{2}$) then the expected number is $e+1$. If we get a head followed by another head (probability $\tfrac{1}{4}$), then the expected number is $e$, since only the first head can't be part of the sequence and were already $1$ deep into the sequence. Finally, if our first $2$ tosses are exactly heads followed by tails (again with probability $\tfrac{1}{4}$), then the expected number is $2$. Thus&#xA;$$e=\frac{1}{2}\cdot(e+1)+\frac{1}{4}\cdot(e)+\frac{1}{4}\cdot(2)$$&#xA;Solving this for $e$ yields $e=4$.&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="-1" LastEditDate="2017-04-13T12:19:36.950" LastActivityDate="2014-10-22T16:23:24.867" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3148" PostTypeId="2" ParentId="3136" CreationDate="2014-10-22T17:10:28.830" Score="4" Body="&lt;p&gt;Not exactly proof, but this short Python program simulates the problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case HH or 0,0 from the random numbers 0,1&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import random&#xA;&#xA;accum = 0.0&#xA;count = 0.0&#xA;&#xA;while count&amp;lt;10000:&#xA;  seq = []&#xA;  while True:&#xA;    n = random.randint(0,1)&#xA;    seq.append(n)&#xA;    if len(seq)&amp;gt;=2 and seq[-2]==0 and seq[-1]==0:&#xA;      break&#xA;  #print seq&#xA;  accum += len(seq)&#xA;  count += 1&#xA;&#xA;print accum/count  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Case HT or 0,1 from the random numbers 0,1&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import random&#xA;&#xA;accum = 0.0&#xA;count = 0.0&#xA;&#xA;while count&amp;lt;10000:&#xA;  seq = []&#xA;  while True:&#xA;    n = random.randint(0,1)&#xA;    seq.append(n)&#xA;    if len(seq)&amp;gt;=2 and seq[-2]==0 and seq[-1]==1:&#xA;      break&#xA;  #print seq&#xA;  accum += len(seq)&#xA;  count += 1&#xA;&#xA;print accum/count  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Generates a number close to 6 for HH and a number close to 4 for HT, which seems to support the answer provided by &lt;strong&gt;Florain F&lt;/strong&gt; and others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That does not go with my intuition which told me the answers should be similar&lt;/p&gt;&#xA;" OwnerUserId="4125" LastActivityDate="2014-10-22T17:10:28.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3149" PostTypeId="2" ParentId="2448" CreationDate="2014-10-22T18:12:29.793" Score="5" Body="&lt;p&gt;There is a class of puzzle which typically has the enigmatic nature that you describe, although it's not quite the same as the online games you link: the &lt;a href=&quot;http://en.wikipedia.org/wiki/Puzzlehunt&quot;&gt;puzzle hunt&lt;/a&gt;. I have also heard these referred to as &quot;&lt;em&gt;CiSRA-style puzzles&lt;/em&gt;&quot;, and that's the search term I used to find that Wikipedia page.&lt;/p&gt;&#xA;" OwnerUserId="228" LastActivityDate="2014-10-22T18:12:29.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3150" PostTypeId="2" ParentId="2237" CreationDate="2014-10-22T19:31:19.150" Score="0" Body="&lt;p&gt;The 3 wise monkeys are having dinner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the next table sit 3 very attactive monkeys so every move is attentively watched.  We will assume they have different wigs on that stay in place and are not particularly wise so we can distinguish the 6 monkeys. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The brunette monkey complains that her banana daquari is not sweet enough and that she has to go to the bathroom.  She stands up, the blonde follows, and the redhead says &quot;I'll catch up I need to check my email&quot;. (Ok they are smarter than normal monkeys.)  After finishing with her phone, the redhead takes a container labelled &quot;agave nectar&quot; from her purse and puts some in the brunette's drink.  She then goes to the bathroom.  After a couple of minutes, the blonde comes back.  She takes a contain labelled &quot;monk fruit extract&quot; from her purse and puts some in the same drink.  The other monkeys come back sit down and the brunette takes another sip. &quot;Hm,&quot; she says, &quot;That is much better.  But why does it smell like almonds now?&quot;  She then collapses on the ground dead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No observer could have figured out who did it unless they, after the fact, search the friends.&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2014-10-22T19:31:19.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3154" PostTypeId="1" AcceptedAnswerId="3156" CreationDate="2014-10-22T23:00:40.860" Score="7" ViewCount="3406" Body="&lt;p&gt;You are a bank robber, and you come across a safe. It has a combination lock of 4 digits. Careful research (and stealing papers at the bank) has given you multiple clues as to the code - just barely enough to figure it out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Digits are referred to as A-B-C-D in the clues. &quot;A + B&quot; is the sum of the first and second digit. All math follows the standard order of operations&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clues&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The sum of the digits is greater than (A * B) - (C * D).&lt;/li&gt;&#xA;&lt;li&gt;The product of the digits plus their sum is divisible by A and C only.&lt;/li&gt;&#xA;&lt;li&gt;The number is prime&lt;/li&gt;&#xA;&lt;li&gt;The first digit is greater than 6&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What four digit number matches these criteria? Also, if you want, post your methodology for finding the correct answer, as this will help me in the future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Note: I am pretty sure that only one number matches all these clues. However, I may have miscalculated. Please correct me in the comments.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1664" LastEditorUserId="1664" LastEditDate="2014-10-22T23:21:28.290" LastActivityDate="2016-05-02T22:46:27.797" Title="Crack the Code #1" Tags="|mathematics|calculation-puzzle|" AnswerCount="6" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3155" PostTypeId="1" AcceptedAnswerId="3446" CreationDate="2014-10-22T23:02:14.807" Score="38" ViewCount="3990" Body="&lt;p&gt;I was helping Ernie out in his shed a while back, when his favorite screwdriver suddenly wore out. I offered to go and pick up a new one for him and Ernie thought it was a great opportunity to test-drive his new electric car. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ernie's car is a work in progress. The body of the car is a rectangular prism exactly 2 m long, 1 m wide and 0.5 m high. On the top is a circular control-dais on which is mounted a steering-wheel, a console with three buttons, and a single acceler-brake pedal (no seat - Ernie believes it is healthier to stand). The car rests on two wheels, one at each end. Each is mounted so its contact point with the ground is on the centre-line of the vehicle and exactly over the relevant end-face of the vehicle. Both the front and back wheels can be turned to steer, through a vertical axis that passes through the wheel's contact point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pIBqQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I climbed on board to familiarise myself with the controls. &quot;The drive mechanism is a bit unusual,&quot; said Ernie. &quot;To move forwards, press the F button. This locks the steering of the back wheel so it is oriented at 0 degrees, parallel to the long axis of the car. The front wheel can be turned up to 90 degrees to either side and is driven by an electric motor to move the car forwards.&quot; I was impressed by the steering lock. &quot;So if you turn the wheel to 90 degrees, the whole car will pivot around the contact point of the rear wheel?&quot;, I asked. &quot;Correct,&quot; replied Ernie, &quot;and if you turn it just under 76 degrees, the car will pivot around one of the back corners&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ernie explained that with the F button pressed, the car could only go forwards. &quot;Try pressing the R button&quot; he suggested. To my surprise, when I did, the whole steering dais rotated 180 degrees so I was now facing the back of the car. &quot;Now the former front wheel locks into the forward-aft orientation, and steering and drive is through what was the back wheel. It can also be turned by +/-90 degrees, so in reverse the car handles  exactly the same as when driving forwards. You can think of the old back as the new front and the old front as the new back&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;And the P button?&quot;, I asked. &quot;That is the automatic parking assist&quot; Ernie replied. &quot;If you press that while on the street, the car senses the nearest car-park, calculates exactly the minimum number of maneuvers required to park, then carries them out to park. If you are already parked, it calculates the minimum number of maneuvers required to un-park, and then carries them out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I asked him to define exactly what he meant by some of the terms he had used:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Maneuver &#xA;    1) starts with the car staionary&#xA;    2) forward or reverse is chosen automatically&#xA;    3) the driving wheel is turned to a fixed angle automatically&#xA;    4) the car moves a distance automatically then stops&#xA;&#xA;Park&#xA;    1) a park is always made into a space between two neighbouring vehicles.&#xA;    2) when parked the car's centre-line is no more than 45 degrees from the curb.&#xA;    3) when parked every part of the car is closer to the curb than the furthest out parts of both neighbouring vehicles.&#xA;&#xA;Un-park&#xA;    1) when un-parked the car's centre-line is exactly parallel to the curb.&#xA;    2) when un-parked every part of the car is further from the curb than the furthest out parts of both neighbouring vehicles.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I headed confidently into town in Ernie's car. I found a suitable gap between two trucks parked outside the Hardware store. The bodies of the trucks were perfect rectangular prisms much longer and wider than Ernie's car. They were parked exactly parallel to the curb, and exactly the same distance out from the curb (note that the road was straight). I stopped the car, exactly parallel to them, with the front face of the car exactly level with the back edge of the gap. When I pressed the P button, the car moved forwards in a single maneuver to park. I happened to have a tape-measure in my pocket and measured the length of the parking space. After picking up the screw-driver I returned to find that the truck parked behind me had moved forwards so the parking space was now significantly shorter. Once again I measured the length of the space. I got into the car, pressed P again and the car automatically un-parked in a single maneuver.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I returned I described my trip to Ernie (exactly as above), then told him the two distances I had measured. &quot;Interesting&quot;, said Ernie, &quot;the exact events you describe wouldn't have been possible if either of the parking spaces had been any shorter&quot;. He then went on to tell me exactly what steering angles had been set for each maneuver.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't remember the lengths I measured (or the steering angles Ernie calculated). Can you help me?   &lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="39" LastEditDate="2014-11-11T22:11:46.593" LastActivityDate="2014-11-11T22:11:46.593" Title="Ernie's automatic parking valet" Tags="|mathematics|calculation-puzzle|geometry|" AnswerCount="1" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="3156" PostTypeId="2" ParentId="3154" CreationDate="2014-10-22T23:15:51.917" Score="9" Body="&lt;p&gt;edited to match intended question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;given that the number &lt;em&gt;is&lt;/em&gt; prime, the only potentially valid combination is 8017.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8017:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;8+0+1+7 = 16, 8*0 - 1*7 = -7, 16 &gt; -7&lt;/li&gt;&#xA;&lt;li&gt;8*0*1*7 + 8+0+1+7 = 16, divisible by 8 and 1, but not 7. 0 is undefined.&lt;/li&gt;&#xA;&lt;li&gt;8017 is prime.&lt;/li&gt;&#xA;&lt;li&gt;8 &gt; 6.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This depends on the assumption that 0 does not divide 16 evenly, which depends on your definition. This can go either way - 16/0 is undefined, so you could say that (0 divides 16) is undefined. Or, since there's no m such that m*0 = 16, you could say that (0 divides 16) is false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Methodology is pretty simple. for all numbers 0000-9999, have a computer check if it matches all conditions.&lt;/p&gt;&#xA;" OwnerUserId="2715" LastEditorUserId="2715" LastEditDate="2014-10-22T23:29:36.340" LastActivityDate="2014-10-22T23:29:36.340" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3157" PostTypeId="1" AcceptedAnswerId="3158" CreationDate="2014-10-23T01:50:38.680" Score="7" ViewCount="4277" Body="&lt;p&gt;After the last bank encounter, you decide that you could make a living off of just cracking codes (as if robbing banks wasn't enough). A shady man meets you on a park bench and slides you an envelope, dramatically. It contains clues for the next number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/3154/crack-the-code-1&quot;&gt;Crack the Code #1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Digits are referred to as A-B-C-D in the clues. &quot;A + B&quot; is the sum of the first and second digit. All math follows the standard order of operations&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clues&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The digits are in descending order.&lt;/li&gt;&#xA;&lt;li&gt;The number is prime.&lt;/li&gt;&#xA;&lt;li&gt;The first digit is less than 6.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What four digit number matches these criteria? Also, if you want, post your methodology for finding the correct answer, as this will help me in the future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Note: I am pretty sure that only one number matches all these clues. However, I may have miscalculated. Please correct me in the comments.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Another note: YenTheFirst, you could probably solve this pretty easily, seeing as the other one was pretty fast. This is great, but can you put the answer in a spoiler this time if you solve it?&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1664" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2016-05-21T11:20:35.600" Title="Crack the Code #2" Tags="|mathematics|calculation-puzzle|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3158" PostTypeId="2" ParentId="3157" CreationDate="2014-10-23T02:33:08.580" Score="7" Body="&lt;p&gt;Assuming you mean strictly descending order, there's only one such number:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$$5431$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The only other numbers satisfying clue 1 and 3 simultaneously are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3210,4210,4310,4320,4321,5210,5310,5320,5321,5410,5420,5421,5430,5432$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, numbers ending in $0$ or $2$ cannot be prime. This leaves:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4321=29\cdot149,5321=17\cdot313,5421=3\cdot1807$$&lt;/p&gt;&#xA;" OwnerUserId="1572" LastEditorUserId="1572" LastEditDate="2014-10-23T02:39:16.300" LastActivityDate="2014-10-23T02:39:16.300" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3159" PostTypeId="2" ParentId="3157" CreationDate="2014-10-23T02:35:54.750" Score="3" Body="&lt;p&gt;Only possible answer is (spoiler ahead)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 5431&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is pretty easy to calculate on the back of an envelope as most of the 9 possible contenders are even or divisible by 3.&lt;/p&gt;&#xA;" OwnerUserId="1631" LastActivityDate="2014-10-23T02:35:54.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3160" PostTypeId="1" AcceptedAnswerId="3162" CreationDate="2014-10-23T07:16:32.983" Score="9" ViewCount="3917" Body="&lt;p&gt;A male friend of yours wants to test if you are ok at Math and logical thinking, he says I am one of two children of my parents, and asks you to find out the probability that his sibling is a sister, he carefully also states he wants you to assume a sex ratio of 100 boys to 100 girls and that they were not twins.&lt;/p&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="5373" LastEditDate="2014-12-30T14:49:07.927" LastActivityDate="2014-12-30T14:49:07.927" Title="Is my sibling a sister" Tags="|logical-deduction|mathematics|probability|family|" AnswerCount="3" CommentCount="6" ClosedDate="2014-10-27T14:27:28.607" ContentLicense="CC BY-SA 3.0" />
  <row Id="3162" PostTypeId="2" ParentId="3160" CreationDate="2014-10-23T08:09:42.773" Score="20" Body="&lt;p&gt;The accepted answer has a subtle error. If we randomly select a 2-sibling &lt;strong&gt;family&lt;/strong&gt; with at least one male, the odds of the other being female are indeed 2/3. If we randomly select a &lt;strong&gt;male&lt;/strong&gt; who is part of a 2 person family, the odds of them having a male sibling are 1/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The three, equally probable male-containing families are&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;MF FM MM&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are four males listed, and 2 of them have male siblings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;strong&gt;Edit&lt;/strong&gt; to remove irrelevant female only family)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit again:&lt;/strong&gt; Note that if you were to randomly select 2-child families with at least one male instead, (for example, by selecting a group of fathers and asking for those with 2 children and at least 1 male) you'd get 2/3 of them having a girl&lt;/p&gt;&#xA;" OwnerUserId="3987" LastEditorUserId="3987" LastEditDate="2014-10-23T09:21:45.017" LastActivityDate="2014-10-23T09:21:45.017" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="3163" PostTypeId="2" ParentId="3013" CreationDate="2014-10-23T09:12:42.843" Score="14" Body="&lt;h2&gt;Explanation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The solution given hinges on the fact that, while the sword itself is longer than the maximum allowed length, we can construct a volume in which the sword fits, that doesn't exceed the restrictions. Contradictory, we can work around the restrictions that were undoubtedly enforced to save space, by using &lt;em&gt;more&lt;/em&gt; space!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've actually run into similar limitations once when sending a parcel and I solved it in the exact same way.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Escalation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;But a $1\text{m} \times 1\text{m} \times 1\text{m}$ box is a bit overkill. The diagonal is almost $20\text{cm}$ longer than needed!&lt;br&gt;&#xA;So let's find a couple of smaller solutions.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Minimal height&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Let's say we have a box with a base of $1\text{m} \times 1\text{m}$. What is the minimal height $h$ we need?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The length $L$ of the sword is $1.5 \text{m}$. The diagonal $D = \sqrt{1^2 + 1^2 + h^2}$. If we fit the sword precisely in the diagonal, $L=D$, so $1.5=\sqrt{1^2 + 1^2 + h^2}$. That means that $2.25 = 2 + h^2$ and so $h=0.5\text{m}$.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Minimal square&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;If we want a box of exactly $1\text{m}$ high with a square base, how long would the edges $x$ of that square be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that $D = \sqrt{1^2 + x^2 + x^2}$, so $\tfrac{9}{4} = 1 + 2x^2$. This makes $x^2=\tfrac{5}{8}$ and $x=\tfrac{1}{4}\sqrt{10} \approx 0.79\text{m}$.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Minimal cube&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Let's try to get the minimal cube with edges $x$ in which to fit the sword.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, $D = \sqrt{x^2 + x^2 + x^2}$, so $\tfrac{3}{2} = \sqrt{3x^2}$, which means that $\tfrac{9}{12} = x^2$, which yields $x = \tfrac{1}{2}\sqrt{3} \approx 0.87\text{m}$.&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditDate="2018-02-22T09:39:59.777" LastActivityDate="2018-02-22T09:39:59.777" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3164" PostTypeId="2" ParentId="3100" CreationDate="2014-10-23T10:25:52.387" Score="1" Body="&lt;p&gt;Can I do something like this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ldcTA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Please do not down vote XD&lt;/p&gt;&#xA;" OwnerUserId="3944" LastActivityDate="2014-10-23T10:25:52.387" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3165" PostTypeId="1" AcceptedAnswerId="3167" CreationDate="2014-10-23T12:53:08.380" Score="24" ViewCount="3256" Body="&lt;p&gt;I was having a house-leaving party at my old house, and challenged four of my friends to guess the number of the new address, which had three digits.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The responses I got were 280, 376, 304, and 370.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was amused, as each one of them got one digit correctly and in its right place, while the other two digits of each guess did not match in any place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's my house number?&lt;/p&gt;&#xA;" OwnerUserId="2444" LastEditorUserId="11497" LastEditDate="2020-01-28T09:13:24.847" LastActivityDate="2020-01-28T09:13:24.847" Title="House Number Mastermind" Tags="|logical-deduction|mastermind|" AnswerCount="9" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="3166" PostTypeId="2" ParentId="3165" CreationDate="2014-10-23T13:04:00.940" Score="4" Body="&lt;p&gt;The answer would be 274.  That is what would give each one a correct digit in one spot&lt;/p&gt;&#xA;&#xA;&lt;p&gt;expanding my answer here&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to get a unique answer each set has 1 correct digit in the correct spot.  4 answers 3 digits means no more than 2 digits in the same spot can be correct.  so in the first spot we can have a 2 or 3, since there are 3 - 3's it must be a 2.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so for the remaining 2 digits we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;76, 04, 70&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 of these are going to have the same correct digit in the correct spot, 0 and 7 are the duplicate digits, 0 is in 2 different spots and 7 is in the same spot, this means 7 is the second digit, 27, all that is left is 4 to complete it 274.&lt;/p&gt;&#xA;" OwnerUserId="2014" LastEditorUserId="2014" LastEditDate="2014-10-23T13:49:40.460" LastActivityDate="2014-10-23T13:49:40.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3167" PostTypeId="2" ParentId="3165" CreationDate="2014-10-23T13:04:07.343" Score="16" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 274&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;376 and 370 must share one number and not another. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;based on possibilities  of first number and can't be three or the last number would be repeated twice in one of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;seeing as 3 isn't the first number 7 must be the second number because of 376 and 370.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can't be 6 or 0 for the third number or else there would be two numbers with 2 correct guesses so that leaves 4.&lt;/p&gt;&#xA;" OwnerUserId="2803" LastActivityDate="2014-10-23T13:04:07.343" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3168" PostTypeId="2" ParentId="3165" CreationDate="2014-10-23T13:09:48.960" Score="3" Body="&lt;p&gt;If the first digit is 3 (satisfies the 2nd, 3rd and 4th guesser) and the second digit is 8 (satisfies the 1st), then the third digit can be anything apart from 0, 4 or 6 (as they have each only got one correct digit).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So some possible answers are 381, 382, 383, 385, 387, 388, 389.&lt;/p&gt;&#xA;" OwnerUserId="2796" LastEditorUserId="2796" LastEditDate="2014-10-23T13:17:45.573" LastActivityDate="2014-10-23T13:17:45.573" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3169" PostTypeId="2" ParentId="3160" CreationDate="2014-10-23T13:30:23.383" Score="7" Body="&lt;p&gt;The fact that the first child among two is a boy has no effect on the sex of the second because the draw &quot;sex of the first child&quot; and &quot;sex of the second child&quot; are independant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability of the other sibling being a girl is equal to the probability for any child picked at random to be a girl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chance is 1/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another form of the question would be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You play heads or tails -with a balanced coin, no loophole- twice.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The first time you play you get heads, what is the chance for the&#xA;  second time you play for it to get tails ?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;It is 1/2. The chance for you to get tails is not increased by the&#xA;  fact that you had heads previously.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As soon as we know that the first child is a male, FM and FF become impossible, leaving MF and MM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; I see that I still am not the accepted answer. And after discussing with my friend I can see why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My answer is counter intuitive because in real life people tend to apply the Law of Large Numbers -or law of averages- to their everyday / low amount of draws routines. For instance:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You are at the casino, and play roulette.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;You place your bets on red and lose 3 times.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;You apply the law of average and believe that now your chance to win&#xA;  by betting red is very high. But this reasoning is irrational. The&#xA;  roulette has no memory of the previous rolls. Your chances by betting&#xA;  red or black are still the same.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let me make another attempt at making you see that Skywalker's answer is flawed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A group of 4 male dudes come to you at a party and tell you that they have a 5th friend of which they ask you to guess the gender.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the logic used in his argument the chance for the 5th friend to be a girl is 4/5 = 0.80 = 80% because the possible combinations are&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;MMMMM MMMMF MMMFM MMFMM MFMMM FMMMM&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The fact that the 4 other dudes are males bring no knowledge of the gender of the 5th.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chance in that case is 1/2 too.&lt;/p&gt;&#xA;" OwnerUserId="4028" LastEditorUserId="4028" LastEditDate="2014-10-24T03:50:46.667" LastActivityDate="2014-10-24T03:50:46.667" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3170" PostTypeId="2" ParentId="3165" CreationDate="2014-10-23T13:31:54.147" Score="16" Body="&lt;p&gt;I see no unique solution for this:&#xA;Unless you assume the solution must uniquely defined by the information given. &#xA;(Or hopefully simpler terms &lt;em&gt;all&lt;/em&gt; the numbers appear in the &lt;em&gt;correct&lt;/em&gt; position in the responses.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Naming the friends A,B,C,D gives A: 280, B: 376 , C: 304 , D: 370 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Position 1 has possibilities 2,3,3,3  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming 3 is right leads to no unique solution...  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - B,C,D all have now have exactly one match&lt;br&gt;&#xA; - For a match from A either position 3 is 0 or position 2 is 8&lt;br&gt;&#xA; - Position 3 cannot be 0 as D has this and has already got 1 position correct&lt;br&gt;&#xA; - So position 2 is 8&lt;br&gt;&#xA; - The solution is 28x (where x is unknown)&lt;br&gt;&#xA; - As no 3rd digit can be right x cannot be  0,4,6  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are many possible solutions: 381, 382, 383, 385, 387, 388, 389&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By assuming a solution must contain only position-number pairs present in the given responses we can now say position 1 cannot be 3. This is because the solutions above do not satisfy this as none of the solutions contains a number in the 3rd position which is in the 3rd position of a response...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in position 1 A must be right with a value of 2, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - Position 2 can be one of 7,0,7 (B, C, D)&lt;br&gt;&#xA; - If 0 is right A and C have exactly 1 right&lt;br&gt;&#xA; - There are now no matches in either B or D thus there is no solution as one of B and D will have no match.&lt;br&gt;&#xA; - Therefore in the 2nd position 7 is right and A,B and D have exactly 1 match so position 3 must be given by C.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Leading to the unique solution 274&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4150" LastEditorUserId="4150" LastEditDate="2014-10-24T13:42:37.880" LastActivityDate="2014-10-24T13:42:37.880" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3171" PostTypeId="2" ParentId="3165" CreationDate="2014-10-23T13:42:39.117" Score="4" Body="&lt;p&gt;First number can be 3 or 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's 3, 3rd digit can't be 0, 4 or 6, so the 2nd digit must be 8. As 38x satisfies the condition for all guesses, then the 3rd digit can't be any of the already used numbers (0, 2, 3, 4, 6, 7 or 8).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If first digit is 2, because of 370 and 376, 2nd digit must then be 7, and because of 304, the 3rd one 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So final answers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 274, 381, 385 and 389.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2354" LastEditorUserId="2354" LastEditDate="2014-10-23T18:35:40.873" LastActivityDate="2014-10-23T18:35:40.873" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3172" PostTypeId="2" ParentId="3165" CreationDate="2014-10-23T16:22:41.987" Score="6" Body="&lt;p&gt;Let's say the number is ABC and let's pick 376 as our starting point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 1: C = 6&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;376 =&amp;gt; A != 3 and B !=7&#xA;370 =&amp;gt; A == 3 or B == 7&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Contradiction, so C cannot be 6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 2: B = 7&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;304 =&amp;gt; A == 3 or C == 4&#xA;376 =&amp;gt; A != 3 and C != 6&#xA;=&amp;gt; C = 4&#xA;&#xA;280 =&amp;gt; A == 2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Solution:&lt;/em&gt; &lt;strong&gt;&lt;em&gt;274&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 3: A = 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;280 =&amp;gt; B == 8 or C == 0&#xA;370 =&amp;gt; B != 7 and C != 0&#xA;=&amp;gt; B = 8&#xA;&#xA;280 =&amp;gt; C != 2 and C != 8 and C != 0&#xA;370 =&amp;gt; C != 3 and C != 7 and C != 0&#xA;376 =&amp;gt; C != 3 and C != 7 and C != 6&#xA;304 =&amp;gt; C != 3 and C != 0 and C != 4&#xA;=&amp;gt; C in [1, 5, 9]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Solution:&lt;/em&gt; &lt;strong&gt;&lt;em&gt;381, 385, 389&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="3918" LastActivityDate="2014-10-23T16:22:41.987" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3178" PostTypeId="2" ParentId="3154" CreationDate="2014-10-23T18:18:02.533" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Methodology is pretty simple. for all numbers 0000-9999, have a&#xA;  computer check if it matches all conditions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here you have the actual programme (Python 3.x):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;for A in range(7,10):&#xA;    for B in range(0,10):&#xA;        for C in range(1,10):&#xA;            for D in range(0,10):&#xA;                if sum([A,B,C,D]) &amp;gt; ((A*B)-(C*D)):&#xA;                    prod_and_sum = (A*B*C*D) + sum([A,B,C,D])&#xA;                    if prod_and_sum % A == 0 and prod_and_sum % C == 0:&#xA;                        number = A*(10**3) + B*(10**2) + C*(10**1) + D&#xA;                        if not [i for i in range(2,number) if number % i == 0]:&#xA;                            print(number)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4016" LastActivityDate="2014-10-23T18:18:02.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3179" PostTypeId="2" ParentId="3157" CreationDate="2014-10-23T18:24:09.840" Score="1" Body="&lt;p&gt;Here you have a programme you can you to check all the possibilities (Python 3.x)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;for A in range(7,10):&#xA;    for B in range(0,10):&#xA;        for C in range(1,10):&#xA;            for D in range(0,10):&#xA;                if sum([A,B,C,D]) &amp;gt; ((A*B)-(C*D)):&#xA;                    prod_and_sum = (A*B*C*D) + sum([A,B,C,D])&#xA;                    if prod_and_sum % A == 0 and prod_and_sum % C == 0:&#xA;                        number = A*(10**3) + B*(10**2) + C*(10**1) + D&#xA;                        if not [i for i in range(2,number) if number % i == 0]:&#xA;                            print(number)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This prints 5431&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4016" LastActivityDate="2014-10-23T18:24:09.840" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3180" PostTypeId="2" ParentId="2465" CreationDate="2014-10-23T18:29:46.313" Score="3" Body="&lt;p&gt;Use a shop bot. &lt;a href=&quot;http://www.shopbottools.com/&quot; rel=&quot;nofollow&quot;&gt;http://www.shopbottools.com/&lt;/a&gt; It is probably the best tool for making accurate wood cuts, and they are available in many makerspaces.If you are interested in mass production, get started by using Kickstarter.&lt;a href=&quot;http://kickstarter.com/&quot; rel=&quot;nofollow&quot;&gt;http://kickstarter.com/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4152" LastActivityDate="2014-10-23T18:29:46.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3182" PostTypeId="1" AcceptedAnswerId="3183" CreationDate="2014-10-23T19:44:53.423" Score="-3" ViewCount="367" Body="&lt;p&gt;You were on a grocery shopping day, you go to the store, and your friend tells you he'd like you to pick him up some fruit, naturally you try and be a kind friend and agrees to pick it up for him, you ask him what he wants and he replies:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Alright, I need some apples and oranges, I'd like 3 of each, but if there's only 2 apples I want 5 oranges, but if there's only 4 oranges, pick up 3 oranges and 1 apple. Otherwise if there's only 2 oranges, I want 1 orange and 6 apples, but if there's only 5 apples, give me 4 apples without any oranges. Understood?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you go grocery shopping and there's 3 apples, the store just got stocked up with plenty of oranges, so oranges aren't a problem, what do you do?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-10-23T20:01:14.513" LastActivityDate="2014-10-24T12:57:53.683" Title="Apples and Oranges" Tags="|logical-deduction|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3183" PostTypeId="2" ParentId="3182" CreationDate="2014-10-23T19:53:09.147" Score="2" Body="&lt;p&gt;Depending on availability of oranges... you get 3 oranges and those 3 apples. He requested 3 of each.&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-10-23T20:00:16.910" LastActivityDate="2014-10-23T20:00:16.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3185" PostTypeId="2" ParentId="75" CreationDate="2014-10-23T20:07:51.440" Score="3" Body="&lt;p&gt;I just came across this problem and I just wanted to point out that although the original poster posted a solution to the problem, there is more than one possible solution. I came up with a solution that has 1 1-win team (the unique score), 2 4-win teams, 13 9-win teams, and 4 16-win teams. Below is a representation of all of the games that were played. The 'W' or 'L' is in relation to the team on the left's outcome against the team listed above.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PcI1r.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;The yellow indicates games where upsets occurred. This particular solution had 13 upsets.&lt;/p&gt;&#xA;&lt;p&gt;EDIT: I might also point out that this is only one possible way that the games could have played out in this scenario. If they games were played differently there might be more or less upsets.&lt;/p&gt;&#xA;" OwnerUserId="3922" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-10-23T20:16:58.023" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3186" PostTypeId="1" AcceptedAnswerId="3194" CreationDate="2014-10-23T20:23:12.623" Score="48" ViewCount="5822" Body="&lt;p&gt;A friend of mine made a simple bar bet, in which he wrote down a list of ten words. Turning it around to show me, he covered up the first two words with his hand. The eight words below his hand were the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;THIRD&lt;br&gt;&#xA;FOURTH&lt;br&gt;&#xA;FIFTH&lt;br&gt;&#xA;SIXTH&lt;br&gt;&#xA;SEVENTH&lt;br&gt;&#xA;EIGHTH&lt;br&gt;&#xA;NINTH&lt;br&gt;&#xA;TENTH  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;He bet me \$20 that I would not be able to guess the first words on the logically-ordered list on my first try, though I could name them in any order. Of course, I'm thinking, &lt;em&gt;I'll just say 'FIRST' and 'SECOND', easy $20.&lt;/em&gt; So I did. And he removed  his hand. And much to my surprise, I was out twenty bucks for my haste.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What were actually the first two words on the list?&lt;/p&gt;&#xA;" OwnerUserId="2444" LastEditorUserId="5188" LastEditDate="2015-04-28T01:27:52.340" LastActivityDate="2015-07-17T18:47:40.840" Title="Third, Fourth, Fifth" Tags="|word-sequence|" AnswerCount="3" CommentCount="4" ClosedDate="2015-07-18T05:37:16.623" ContentLicense="CC BY-SA 3.0" />
  <row Id="3188" PostTypeId="2" ParentId="3100" CreationDate="2014-10-23T20:34:42.003" Score="4" Body="&lt;p&gt;Step 1: Put a small boat on that spot,&lt;br&gt;&#xA;Step 2: Fill the boat with sand,&lt;br&gt;&#xA;Step 3: Bury yourself in the sand that's in the boat.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the tide comes up, the boat will float, and you'll survive...&lt;/p&gt;&#xA;" OwnerUserId="4156" LastActivityDate="2014-10-23T20:34:42.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3192" PostTypeId="2" ParentId="3186" CreationDate="2014-10-23T20:37:50.030" Score="15" Body="&lt;p&gt;Senior and Junior. The list was suffixes for descendants with identical names, such as King George IV, which is King George the Fourth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Answer 2&#xA;Whole and half. One whole, one half, one third, etc. &lt;/p&gt;&#xA;" OwnerUserId="1664" LastActivityDate="2014-10-23T20:37:50.030" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="3193" PostTypeId="1" AcceptedAnswerId="3198" CreationDate="2014-10-23T20:38:21.873" Score="-3" ViewCount="496" Body="&lt;p&gt;What does this mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The girl that a man that a dog that a complex that a landlord that a friend that a cat plays with helps owns houses loved seen plays piano.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you make any sense out of this?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="11876" LastEditDate="2016-09-19T21:07:30.047" LastActivityDate="2016-09-19T21:07:30.047" Title="Confusing, Seemingly-Impossible, yet Correct sentence" Tags="|english|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3194" PostTypeId="2" ParentId="3186" CreationDate="2014-10-23T20:42:10.050" Score="52" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; WHOLE&lt;br&gt;&#xA; HALF&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;were the first two words.&lt;/p&gt;&#xA;" OwnerUserId="2265" LastEditorUserId="8874" LastEditDate="2015-03-16T13:07:42.693" LastActivityDate="2015-03-16T13:07:42.693" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3197" PostTypeId="2" ParentId="81" CreationDate="2014-10-23T20:56:27.140" Score="2" Body="&lt;p&gt;Ask a bunch of people this riddle. Then, rate the difficulty depending on how many people you ask before someone gets it. If it takes 6 people, rate the difficulty 6.&lt;/p&gt;&#xA;" OwnerUserId="4152" LastActivityDate="2014-10-23T20:56:27.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3198" PostTypeId="2" ParentId="3193" CreationDate="2014-10-23T20:59:04.527" Score="12" Body="&lt;pre&gt;&lt;code&gt;Girl&#xA;  Man&#xA;    Dog&#xA;      Complex&#xA;        Landlord&#xA;          Friend&#xA;            Cat plays with&#xA;          helps&#xA;        owns&#xA;      houses&#xA;    loved&#xA;  seen&#xA;played piano&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Or: Cat plays with a friend that helps the landlord who owns a complex that houses a dog that loved a man that had seen a girl play piano.&lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2014-10-23T20:59:04.527" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3201" PostTypeId="2" ParentId="3193" CreationDate="2014-10-23T21:15:47.570" Score="6" Body="&lt;p&gt;A cat plays with a friend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That friend helps a landlord.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That landlord owns a complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That complex houses a dog.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That dog loved a man.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That man SEEN/SAW a girl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That girl plays piano.&lt;/p&gt;&#xA;" OwnerUserId="1616" LastActivityDate="2014-10-23T21:15:47.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3202" PostTypeId="2" ParentId="3079" CreationDate="2014-10-23T21:15:59.090" Score="10" Body="&lt;p&gt;&lt;strong&gt;Preliminary remarks&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is quite obvious that visiting the black squares and visiting the white squares are two independent problems.  No bishop can move from one color to the other and they don’t hinder each other in any way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If N bishops can visit all squares of one color, then 2N bishops can visit all squares of the board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The problem to solve&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the above, we will consider the problem of visiting all grey squares on the board on fig. 1 using N paths that each follows the movement rules of a bishop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mnIiM.png&quot; alt=&quot;Fig. 1&quot;&gt; Fig.1 . . .     &lt;img src=&quot;https://i.stack.imgur.com/N5GGW.png&quot; alt=&quot;Fig. 2&quot;&gt; Fig. 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fig. 2 illustrates one possible solution of visiting all squares with 4 paths.  The direction of the movements is irrelevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you try to solve the problem manually, you will probably find solutions in 4 paths, and you will experience that it resists being solved in less than 4 paths.  It feels like, regardless of how you arrange the paths, there is a resource that gets exhausted and you need to add more paths to provide that resource.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A subproblem&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see what it is you are short of, consider another problem, which is to visit half of the board only.  Let’s consider the upper-left triangular half (fig.3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UwXmT.png&quot; alt=&quot;Fig. 3&quot;&gt; Fig. 3  . . .    &lt;img src=&quot;https://i.stack.imgur.com/NCa68.png&quot; alt=&quot;Fig. 4&quot;&gt; Fig. 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let’s add black and white dots on the board (fig. 4) in some kind of second-level chessboard pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see that any bishop path must alternate between black and white dots.  So, any bishop path will visit as many black as white dots, with maybe one extra black or white dot.  But the imbalance between black and white dots for a single path will never be more than one dot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other side, the imbalance of black and white dots in the triangle is important.  There are 10 black dots and only 6 white dots, that is an excess of 4 black dots.  This imbalance must be reflected in the paths that cover them.  Since we have an excess of 4 black dots, and since one path can only have an excess of 1 black dot, no matter how we arrange the paths, we need 4 paths to reach the same imbalance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This already proves that you need 4 disjoint paths to cover the triangular board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(And to answer the question: you are short of white-dotted squares.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Back to the problem&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let’s now consider the upper-left triangle as part of a square board.  Let’s assume we have a set &lt;span class=&quot;math-container&quot;&gt;$P_S$&lt;/span&gt; of paths that visit all gray squares of the square board.  Let &lt;span class=&quot;math-container&quot;&gt;$P_T$&lt;/span&gt; be the set &lt;span class=&quot;math-container&quot;&gt;$P_S$&lt;/span&gt; restricted to the squares of the upper-left triangle.  &lt;span class=&quot;math-container&quot;&gt;$P_T$&lt;/span&gt; is made of pieces of paths in &lt;span class=&quot;math-container&quot;&gt;$P_S$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$P_T$&lt;/span&gt; covers the triangle exaclty so it must include at least 4 disjoint paths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that &lt;span class=&quot;math-container&quot;&gt;$P_T$&lt;/span&gt; includes 4 paths doesn’t imply that &lt;span class=&quot;math-container&quot;&gt;$P_S$&lt;/span&gt; also includes 4 paths, because the paths in &lt;span class=&quot;math-container&quot;&gt;$P_T$&lt;/span&gt; can join each other via the other half.  For example, the bottom right triangle on fig. 2 is completely covered with 3 paths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The important thing is to see is that a set of paths also defines a set of path ends.  (A single-square path counts as 2 ends, the start and the end).  If &lt;span class=&quot;math-container&quot;&gt;$P_T$&lt;/span&gt; includes 4 paths, it defines 8 path ends within the triangle.  We can even assert that it defines 8 path ends located on black dots.  That is because only a path with 2 black ends contributes to the imbalance of black dots over white ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let’s take the solution on fig. 2 and see the set &lt;span class=&quot;math-container&quot;&gt;$P_T$&lt;/span&gt; covering the upper left triangle (fig. 5).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/21hd2.png&quot; alt=&quot;Fig. 5&quot;&gt; Fig. 5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As expected, there are at least 4 paths and 8 black path ends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only 4 of the black path ends are located near the diagonal.  So if we stick the 2 halves back together, only 4 of the black ends can be extended to the other side.  The at least 4 remaining black ends are not affected by the other half.  That means at least 4 black ends in &lt;span class=&quot;math-container&quot;&gt;$P_T$&lt;/span&gt; are also path ends in &lt;span class=&quot;math-container&quot;&gt;$P_S$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is that &lt;span class=&quot;math-container&quot;&gt;$P_S$&lt;/span&gt; has at least 4 path ends in the upper-left triangle.  Likewise, it has 4 path ends in the lower-right triangle.  Altogether &lt;span class=&quot;math-container&quot;&gt;$P_S$&lt;/span&gt; has at least 8 path ends.  And this requires at least 4 paths in &lt;span class=&quot;math-container&quot;&gt;$P_S$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to cover the square board, you need 4 bishop paths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conclusion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Visiting all squares of one color on a standard chess board requires 4 disjoint bishop paths.  To do both colors you need twice as much, you need 8 bishops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In counting path ends, it is important to remember the multiplicity of path ends when the path starts and ends on the same square.  This requires some precautions.  For a single-square path, count 2 ends on that same square.  This is compatible with the fact that one path has always 2 ends.  Another point is that you could have a double path end on a square of the diagonal.  This is not a problem since only one can extend to the other side.  It doesn’t allow to extend more than 4 paths to the other side.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2019-10-15T20:16:28.747" LastActivityDate="2019-10-15T20:16:28.747" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="3204" PostTypeId="1" AcceptedAnswerId="3206" CreationDate="2014-10-23T22:45:46.600" Score="5" ViewCount="1516" Body="&lt;p&gt;You return to the bench after cracking the last code, and you notice an envelope on the ground directly under it. You put down your envelope with the previous code and pick up the new one, anticipating another challenge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/3157/crack-the-code-2&quot;&gt;Crack the Code #2&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Digits are referred to as A-B-C-D in the clues. &quot;A + B&quot; is the sum of the first and second digit. All math follows the standard order of operations&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clues&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The number is prime.&lt;/li&gt;&#xA;&lt;li&gt;No digits are repeated.&lt;/li&gt;&#xA;&lt;li&gt;$A\cdot B=C\cdot D$&lt;/li&gt;&#xA;&lt;li&gt;The first digit is greater than 3.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What four digit number matches these criteria? Also, if you want, post your methodology for finding the correct answer, as this will help me in the future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Note: I am pretty sure that only one number matches all these clues. However, I may have miscalculated. Please correct me in the comments. If you find the answer, put it in a spoiler so the question is not ruined for those who want to solve it.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1664" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2014-10-24T20:56:47.950" Title="Crack the Code #3" Tags="|mathematics|calculation-puzzle|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3205" PostTypeId="2" ParentId="454" CreationDate="2014-10-23T23:12:14.640" Score="5" Body="&lt;p&gt;Don't use glue at all, just tightly &quot;trap&quot; the puzzle between a backing board (behind) an acrylic or glass sheet (front), and plastic/wood around the edges to stop the puzzle pieces falling out, secured to both the back and front.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've never seen a professional one, but I've seen a few made by a moderately skilled DIY'er with a few lengths of Balsa wood and a sheet of acrylic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is to surround the puzzle and hold it immobile in every direction, without enough clearance for the pieces to dislodge. Ideally the thickness of the wood would be the same as the thickness of the puzzle pieces, but you can place another thinner backing board behind the puzzle if needed.&lt;/p&gt;&#xA;" OwnerUserId="4159" LastActivityDate="2014-10-23T23:12:14.640" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3206" PostTypeId="2" ParentId="3204" CreationDate="2014-10-23T23:28:37.197" Score="3" Body="&lt;p&gt;Do not hover over the below unless you wish to know the answer...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 6329&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cheated and wrote a small program to do the hard work for me :p which I'll post in a moment - I'm a lazy software developer and we never do any un-necessary maths.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;/* package whatever; // don't place package name! */&#xA;&#xA;import java.util.*;&#xA;import java.lang.*;&#xA;import java.io.*;&#xA;&#xA;/* Name of the class has to be &quot;Main&quot; only if the class is public. */&#xA;class Ideone&#xA;{&#xA;&#xA;    public static boolean isPrime(long n)&#xA;    {&#xA;        if(n &amp;lt; 2) return false;&#xA;&#xA;        if(n == 2 || n == 3) return true;&#xA;&#xA;        if(n%2 == 0 || n%3 == 0) return false;&#xA;&#xA;        long sqrtN = (long)Math.sqrt(n)+1;&#xA;&#xA;        for(long i = 6L; i &amp;lt;= sqrtN; i += 6)&#xA;        {&#xA;                if(n%(i-1) == 0 || n%(i+1) == 0) return false;&#xA;        }&#xA;&#xA;        return true;&#xA;    }&#xA;&#xA;    public static void main (String[] args) throws java.lang.Exception&#xA;    {&#xA;        int a;&#xA;        int b;&#xA;        int c;&#xA;        int d;&#xA;        int combined;&#xA;&#xA;        for(a = 4; a &amp;lt; 10; a++)&#xA;        {&#xA;            for(b = 0; b &amp;lt; 10; b++)&#xA;            {&#xA;                for(c = 0; c &amp;lt; 10; c++)&#xA;                {&#xA;                    for(d = 0; d &amp;lt; 10; d++)&#xA;                    {&#xA;                        if (a != b &amp;amp;&amp;amp; a != c &amp;amp;&amp;amp; a !=d &amp;amp;&amp;amp; b != c &amp;amp;&amp;amp; b != d &amp;amp;&amp;amp; b != d &amp;amp;&amp;amp; c != d)&#xA;                        {&#xA;                            if(a*b == c*d)&#xA;                            {&#xA;                                combined = (((a * 10 + b) * 10 + c) * 10) + d;&#xA;                                if(isPrime(combined))&#xA;                                {&#xA;                                    System.out.println(combined);&#xA;                                }&#xA;                            }&#xA;&#xA;                        }&#xA;                    }&#xA;                }&#xA;            }&#xA;        }&#xA;    }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4159" LastEditorUserId="4159" LastEditDate="2014-10-23T23:33:52.460" LastActivityDate="2014-10-23T23:33:52.460" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3207" PostTypeId="2" ParentId="3160" CreationDate="2014-10-23T23:31:33.737" Score="0" Body="&lt;p&gt;it doesn't state in the question that ether of the offspring are male or female, it says a male friend of yours asks you(the reader) a question... etc. the variables you are speaking of(as far as the existing sex of the alive child, not the child to come into play in the question) will change depending on the sex of the reader, also still does not change the outcome of the end result, for the odds reset for each child born, even if the children born are all male so far, it does not mean that there is a higher or lower likelihood of the next being male or female, it is still a 50/50 shot, mathematically... while scientifically speaking if the father is predisposed to produce one sex more than the other then there would be a 2/3 likelihood that the child in question would be born male or female.&lt;/p&gt;&#xA;" OwnerUserId="4161" LastActivityDate="2014-10-23T23:31:33.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3208" PostTypeId="2" ParentId="3204" CreationDate="2014-10-23T23:37:51.257" Score="6" Body="&lt;p&gt;The answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 6329&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How I found the answer with mental arithmetic only.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First I realized the two sides must have the same set of prime factors, and since they are multiples of two one digit numbers those primes are restricted to being one of 2,3,5, or 7. We can discard 5 and 7 because they cannot be multiplied by any non unit and result in a digit, so they would have to be repeated on both sides were they factors. Therefore the answer must consist only of the primes 2 and 3. since we know digits are not repeated, we must have &gt; 2 factors and since 2^4 is greater than one digit we must have exactly 3 prime factors. The two choices are 2,2,3 and 2,3,3. (2*3)&lt;em&gt;3 == 2&lt;/em&gt;(3*3) is the solution that matches the requirements.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4160" LastEditorUserId="4160" LastEditDate="2014-10-24T12:03:26.163" LastActivityDate="2014-10-24T12:03:26.163" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3209" PostTypeId="1" AcceptedAnswerId="3210" CreationDate="2014-10-24T00:06:52.347" Score="4" ViewCount="2958" Body="&lt;p&gt;The park bench has been compromised as a safe location, so you decide to meet in the lobby of the nearest hotel. Again, the envelopes are traded securely, along with the money you are receiving for each code. This code comes with a note.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“This is the hardest code yet. No more programs to crack this one, at least not easily, and this time there is some conversion involved. The next envelope you get will be much harder. Much, much, MUCH harder.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/3204/crack-the-code-3&quot;&gt;Crack the Code #3&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Digits are referred to as A-B-C-D in the clues. &quot;A + B&quot; is the sum of the first and second digit. All math follows the standard order of operations&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clues&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There are no repeated digits.&lt;/li&gt;&#xA;&lt;li&gt;One of the digits is a four.&lt;/li&gt;&#xA;&lt;li&gt;The number is prime.&lt;/li&gt;&#xA;&lt;li&gt;The hexadecimal representation of this number is an English &lt;strong&gt;word&lt;/strong&gt; (according to m-w.com).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What four digit number matches these criteria? Also, if you want, post your methodology for finding the correct answer, as this will help me in the future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Note: I am pretty sure that only one number matches all these clues. However, I may have miscalculated. Please correct me in the comments. If you find the answer, put it in a spoiler so the question is not ruined for those who want to solve it.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1664" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2016-09-24T15:02:25.950" Title="Crack the Code #4" Tags="|mathematics|calculation-puzzle|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3210" PostTypeId="2" ParentId="3209" CreationDate="2014-10-24T00:19:06.223" Score="4" Body="&lt;p&gt;answer (edited again. fixed after question clarification)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4013 = 0xFAD or FAD&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;how i came up with it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; hex representation must be only 3 digits, it must end in b,d,f or it would not be prime. Then an educated guess assuming ??d.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4160" LastEditorUserId="4160" LastEditDate="2014-10-24T00:29:44.470" LastActivityDate="2014-10-24T00:29:44.470" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3215" PostTypeId="1" AcceptedAnswerId="3219" CreationDate="2014-10-24T03:50:27.753" Score="149" ViewCount="22165" Body="&lt;p&gt;You are a double-agent in the most important war of the 20th century, The&#xA;Falklands war. For years you have been playing the United Kingdom and Argentinian&#xA;governments against each other, taking bribes from both sides and it has caught&#xA;up with you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One day travelling you are snatched off the street, kidnapped, drugged and&#xA;placed on a private plane. You wake up in a cell, you know you have been taken by&#xA;one of the governments to their capital for interrogation and treason charges but&#xA;you cannot tell which has abducted you. You have perfected both the UK and&#xA;Argentinian languages and accents and you know you can convince the guards to&#xA;let you go if you were to speak in their native tongue asking for your&#xA;government contact. However speaking the wrong language first will&#xA;certainly turn out very poorly for you indeed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You look around your cell, it is a large room with high ceilings, a musty smell,&#xA;and thick dust covering everything. Based on being unable to hear anything and&#xA;the constant temperature you conclude you are underground. Nothing about the architecture is distinctive. The only things in&#xA;the cell are a single metal chair bolted to the floor in the center of the room&#xA;with a spotlight hanging from a cord right above the chair illuminating the&#xA;seat brightly leaving the rest of the room in shadow. You know it is a matter of time&#xA;before you are tied into the chair and interrogated by shadowy agents of one of&#xA;the governments. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You search desperately for a clue as to where you are. You&#xA;kick the chair and stub your toe; it is solid. You claw at the light, looking for a marking&#xA;or a way to open it and see the voltage rating but cannot find either. You yank&#xA;and pull on it to test how secure it is and cannot dislodge it from the ceiling&#xA;to get at the cord. You try to swing it far enough to smash against a wall but it will not reach. You search the chair for a makers mark but can find none.&#xA;You know much more action will invite suspicion even if you know the right&#xA;language so you stand back and think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The guard watching on CCTV notices, but decides it is normal behaviour for&#xA;someone waking up in a cell regardless of guilt so is not immediately&#xA;suspicious. As far as he can tell you then pace around the room randomly for a&#xA;few hours deep in thought.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You suddenly walk towards the door and knock, with 100% confidence you&#xA;correctly ask in perfect Spanish for your government handler who promptly&#xA;releases you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did you know what government to ask for?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;You can assume you are correct in all your assumptions mentioned in the puzzle.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;You have no items other than the clothes and shoes you were wearing when abducted. no cell phone, no watch, no pocket lint etc.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4160" LastEditorUserId="4160" LastEditDate="2014-10-24T11:56:43.050" LastActivityDate="2018-08-11T11:04:33.927" Title="A double-agent with a conundrum" Tags="|lateral-thinking|" AnswerCount="12" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="3217" PostTypeId="2" ParentId="3215" CreationDate="2014-10-24T04:20:34.583" Score="74" Body="&lt;p&gt;Of course you know it's not the UK government because you have been left alone in your cell long enough to miss tea time, and it would be completely barbaric to not observe that sacred ritual. So, Argentina government it is!&lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2014-10-24T04:20:34.583" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3218" PostTypeId="2" ParentId="3215" CreationDate="2014-10-24T05:06:44.430" Score="25" Body="&lt;p&gt;On walking around the room thinking for hours, he started feeling weird, like increased heartbeat, heavy breathing, feeling like less oxygen around the area, than normal. So he understood that he is currently in a high altitude area, preferably above 20000ft above sea line to feel such weird effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He knew the highest altitude in UK is some what about 4400ft above sea line and so Eliminated the possibility that he is in UK. Then the chances are he is either in Argentina or in some high altitude place outside UK or Argentina. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But he knew Argentina has places with altitude above 22000ft. Also he was sure that it's either UK or Argentina government took him. So he understood he is in some high altitude location in Argentina. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After all these deductions, with full confidence, he knocked the door and talked with the guards in Spanish.&lt;/p&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="2870" LastEditDate="2014-10-27T16:50:18.230" LastActivityDate="2014-10-27T16:50:18.230" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3219" PostTypeId="2" ParentId="3215" CreationDate="2014-10-24T05:54:28.737" Score="112" Body="&lt;p&gt;I don't know how scientifically accurate this is, but it might be possible to use the light as a foucault pendulum to determine which hemisphere you are in. Of course, if there is a toilet (which is not mentioned), a flush would do the trick.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2014-10-24T05:54:28.737" CommentCount="22" ContentLicense="CC BY-SA 3.0" />
  <row Id="3221" PostTypeId="5" CreationDate="2014-10-24T07:07:34.287" Score="0" Body="&lt;p&gt;A probability puzzle is a puzzle, usually mathematical, that involves uncertain outcomes.&lt;/p&gt;&#xA;&lt;p&gt;Probability puzzles often involve &lt;a href=&quot;/questions/tagged/coin-tossing&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;coin-tossing&amp;#39;&quot; rel=&quot;tag&quot;&gt;coin-tossing&lt;/a&gt;, &lt;a href=&quot;/questions/tagged/dice&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;dice&amp;#39;&quot; rel=&quot;tag&quot;&gt;dice&lt;/a&gt;, or &lt;a href=&quot;/questions/tagged/cards&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;cards&amp;#39;&quot; rel=&quot;tag&quot;&gt;cards&lt;/a&gt;, although other mechanisms may act as sources of uncertain outcomes.&lt;/p&gt;&#xA;&lt;p&gt;For example: &lt;a href=&quot;https://puzzling.stackexchange.com/q/6462&quot;&gt;Left coin, right coin, last coin?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Another example of a probability puzzle is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Monty_Hall_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Monty Hall problem&lt;/em&gt;&lt;/a&gt;: the game show problem with three doors, one car, and two goats.&lt;/p&gt;&#xA;&lt;p&gt;General probability questions are off-topic but can be asked on our sister site &lt;a href=&quot;https://math.stackexchange.com/&quot;&gt;Mathematics Stack Exchange&lt;/a&gt; with the &lt;a href=&quot;/questions/tagged/probability&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;probability&amp;#39;&quot; rel=&quot;tag&quot;&gt;probability&lt;/a&gt; tag (some questions can be found &lt;a href=&quot;https://math.stackexchange.com/questions/tagged/probability&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="60644" LastEditDate="2020-11-30T10:49:26.073" LastActivityDate="2020-11-30T10:49:26.073" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="3222" PostTypeId="4" CreationDate="2014-10-24T07:07:34.287" Score="0" Body="A mathematical puzzle whose essential nature involves randomness. Use with [mathematics]" OwnerUserId="3993" LastEditorUserId="69582" LastEditDate="2021-01-23T18:59:59.977" LastActivityDate="2021-01-23T18:59:59.977" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="3226" PostTypeId="2" ParentId="81" CreationDate="2014-10-24T08:28:47.507" Score="5" Body="&lt;p&gt;I think this is very hard question to answer practically. So even though I do not have an direct answer to it I would like to share my experience, which in related area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am living on a Russian site, that provides unique puzzles for people to solve. Here is the facts about how hard is to estimate puzzle hardness for us:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It has about 400 selected puzzles and 55000 users, 1000 from which are active.  &lt;/li&gt;&#xA;&lt;li&gt;We have a rating (puzzle hardness) system, which calculate a hardness of a puzzle basing on ratio $\frac{NPeopleSolved}{NPeopleSeen}$, where $NPeopleSeen$ is number of people, which solved at least one arbitrary puzzle after the given puzzle was published.  &lt;/li&gt;&#xA;&lt;li&gt;Based on 7 years experience we decided to have only 5 possible hardness categories for puzzles. That means we trying to estimate the hardness with precision of 20% only. (The hardness category depends exponentially on the ratio above, if you are interested).  &lt;/li&gt;&#xA;&lt;li&gt;Again, currently we decide that we need at least 1 month to give first estimation of the hardness. That means that first of all a puzzle is published on a site and is solved by people during one month, only then we set a hardness for it.  &lt;/li&gt;&#xA;&lt;li&gt;In average (taking into account all period the puzzle is exists on site) we have about 100 people who solve the puzzle per months for simplest puzzles and only 1 guy per months for hardest. Fortunately for first months this ratio is bigger due to the fact that active people are clever in average and hungry for new puzzles. &lt;/li&gt;&#xA;&lt;li&gt;Buy the fact the hardness is not perfectly stable even for puzzles, which have collected years of solving-statictics. Due to changes in site popularity, kind of people attracted and allowed to the site it can change with time. So it clearly depends a lot on selection of people to estimate the hardness.&lt;/li&gt;&#xA;&lt;li&gt;Before the current automatic system described above we had a set of moderators, who tried to set the puzzle hardness by voting. And the results of statistics and voting appeared to be quite different when compared. Similarly I (even having experience of solving puzzle from beginning to the end) can rarely predict the hardness before the puzzle is published, I am surprised often how very simple puzzles for me can be hard for average user and how very hard puzzles for me can have the lowest hardness on the site.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I never heard about rating community and even it is exists it hard to imagine how it can estimate the hardness of the puzzle more or less correctly and meanwhile cheap enough (in people*hours) and quick enough. Especially if your puzzles is hard enough for clever person to think about it for whole month.  &lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2014-10-24T10:03:45.113" LastActivityDate="2014-10-24T10:03:45.113" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3228" PostTypeId="1" AcceptedAnswerId="3230" CreationDate="2014-10-24T09:32:14.967" Score="8" ViewCount="1844" Body="&lt;p&gt;You are on a quest to the find the lost kingdom of Acarien. On your way, you end up at a fork in the road. Both paths look exactly the same. At the junction sits a wise man. The wise man will only speak the words “&lt;em&gt;ça&lt;/em&gt;” or “&lt;em&gt;ɠa&lt;/em&gt;” meaning either &lt;em&gt;yes&lt;/em&gt; or &lt;em&gt;no&lt;/em&gt; (you don’t know which means which). The wise man will answer any yes-no question truthfully but will only answer one question. What one question can you ask the wise man which from which you will be sure which path leads to Acarien?&lt;/p&gt;&#xA;" OwnerUserId="4157" LastEditorUserId="2484" LastEditDate="2014-10-31T00:28:21.613" LastActivityDate="2014-10-31T00:28:21.613" Title="The way to Acarien puzzle" Tags="|lateral-thinking|logical-deduction|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3229" PostTypeId="2" ParentId="3228" CreationDate="2014-10-24T10:42:07.950" Score="14" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If I asked you &quot;is it left to Acarien?&quot;, would you say ça?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The trick is that the this results in the answer ça if the truthful answer to my embedded question is 'yes', and ɠa if the truthful answer to my question is 'no'. Thus you can deduce the correct path easily from the answer.&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="68" LastEditDate="2014-10-24T14:43:09.277" LastActivityDate="2014-10-24T14:43:09.277" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3230" PostTypeId="2" ParentId="3228" CreationDate="2014-10-24T10:50:09.763" Score="11" Body="&lt;p&gt;Question to be asked = &quot;If I ask you if this is the correct path to go, would you say &lt;strong&gt;ça&lt;/strong&gt;&quot; ?&#xA;4 possibilities:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ca = yes&lt;br&gt;&#xA;path' correctness = yes&lt;br&gt;&#xA;reply= yes (ca)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ca = yes&lt;br&gt;&#xA;path' correctness = no&lt;br&gt;&#xA;reply= no (ga)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ca = no&lt;br&gt;&#xA;path' correctness = yes&lt;br&gt;&#xA;reply= no (ca)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ca = no&lt;br&gt;&#xA;path' correctness = no&lt;br&gt;&#xA;reply= yes (ga)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence we see whenever reply is &lt;strong&gt;ga&lt;/strong&gt; the path is incorrect, if the word asked in the question to the wise man was &lt;strong&gt;ca&lt;/strong&gt; .&lt;/p&gt;&#xA;" OwnerUserId="4183" LastEditorUserId="4183" LastEditDate="2014-10-24T10:57:19.087" LastActivityDate="2014-10-24T10:57:19.087" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3232" PostTypeId="2" ParentId="3209" CreationDate="2014-10-24T11:49:19.913" Score="1" Body="&lt;p&gt;The only one I can find is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4013, with hex representation &quot;fad&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As with crack the code 3, I used a simple Java frame to run through the possibilities, and manually looked at the output in this case. I'm not sure this one is possible to do by hand in any sensible time frame (unlike #3 which could be done by hand relatively easily with some mathematical thinking).&lt;/p&gt;&#xA;" OwnerUserId="4159" LastActivityDate="2014-10-24T11:49:19.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3233" PostTypeId="2" ParentId="3154" CreationDate="2014-10-24T12:54:17.733" Score="1" Body="&lt;p&gt;C# Version:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;var numbers = (from int n in Enumerable.Range(7000, 3000)&#xA;               let a = (n / (int)Math.Pow(10, 3)) % 10&#xA;               let b = (n / (int)Math.Pow(10, 2)) % 10&#xA;               let c = (n / (int)Math.Pow(10, 1)) % 10&#xA;               let d = n % 10&#xA;               let prodSum = a + b + c + d + (a * b * c * d)&#xA;               where ((c &amp;gt; 0) &amp;amp;&amp;amp;&#xA;                      ((a + b + c + d) &amp;gt; ((a * b) - (c * d))) &amp;amp;&amp;amp;&#xA;                      (prodSum % a == 0) &amp;amp;&amp;amp;&#xA;                      ((b == 0) || (prodSum % b != 0)) &amp;amp;&amp;amp;&#xA;                      (prodSum % c == 0) &amp;amp;&amp;amp;&#xA;                      ((d == 0) || (prodSum % d != 0)) &amp;amp;&amp;amp;&#xA;                      (!Enumerable.Range(2, (n - 2)).Any(o =&amp;gt; n % o == 0)))&#xA;               select n);&#xA;&#xA;Console.WriteLine(string.Join(&quot;, &quot;, numbers));&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2614" LastActivityDate="2014-10-24T12:54:17.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3235" PostTypeId="1" AcceptedAnswerId="3237" CreationDate="2014-10-24T13:28:45.347" Score="5" ViewCount="2920" Body="&lt;p&gt;The government has introduced a standardised test to identify dumb people and make them ineligible to apply for any government jobs, a pass in the exam is mandatory for applications in future.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was hard to come up with a test that can provide 100% accuracy, however the creators of the test argued that 289 of every 300 dumb people taking the test will fail and 195 of every 200 smart people taking the exam would pass it and since it left very little to desire, the test was implemented.  Given the general nature of people applying for such jobs, it was predicted that 60% of the smart people would not take this exam and all of the dumb people will take it, leaving the percentage of smart people taking the exam at 40% (To avoid confusion mentioned in the answers this means of every 100 people taking the exam 40 were smart and 60 were dumb)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You took the test and you passed, what is the probability that you are really smart? &lt;/p&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="2313" LastEditDate="2014-10-24T13:58:45.387" LastActivityDate="2014-10-25T12:55:21.337" Title="Are you Really Smart" Tags="|logical-deduction|mathematics|word-problem|probability|" AnswerCount="4" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="3236" PostTypeId="2" ParentId="3235" CreationDate="2014-10-24T13:50:42.393" Score="5" Body="&lt;p&gt;Say $3000$ people applied. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1200$ of these are smart ($40\%$), and $1800$ are dumb ($60\%$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1200\frac {195}{200})+(1800\frac {11}{300}) = 1236$ people out of $3000$ applicants passed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is known that of the $1236$, $(1200\frac {195}{200}) = 1170$ people are smart and $(1800\frac {11}{300})=66$ are dumb by the probabilities passing/failing provided.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, there is a $\frac{1170}{1236} = 0.94660194174$   probability that you are smart.&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-10-24T14:05:23.460" LastActivityDate="2014-10-24T14:05:23.460" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3237" PostTypeId="2" ParentId="3235" CreationDate="2014-10-24T13:52:08.157" Score="9" Body="&lt;p&gt;I am really smart so 100%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Y is the fraction of people who take the test that are smart then $YT*195/200$ smart people will pass the test where $T$ is the number of test takers. $(1-Y)T*11/300$ is the number of dumb people who will pass.  The percentage of people who will pass that are smart is: $$\frac{YT*195/200}{YT*195/200+(1-Y)T*11/300}=\frac{585Y}{563Y+22}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $Y =\frac{40}{100}$ then  $$\frac{585Y}{563Y+22}=94.66\%$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we want to generalise this further to answer klm123's version of the question, I would define $A$ as the fraction of geniuses who pass and $B$ as the fraction of idiots who pass. (Sorry but you are one or the other according to this.)  The proportion of people who pass who are geniuses are, therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{AYT}{AYT+B(1-Y)T}=\frac{AY}{(A-B)Y+B}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $Y=.1\%$, $A=99\%$, and $B=1\%$ this is roughly only $9\%$.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditDate="2014-10-24T20:02:58.943" LastActivityDate="2014-10-24T20:02:58.943" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3238" PostTypeId="2" ParentId="3215" CreationDate="2014-10-24T14:35:02.537" Score="81" Body="&lt;p&gt;Wave your hand through the thick layer of dust until you start sneezing.  Eventually someone will say &quot;Bless you&quot; or the equivalent in whichever language, and you're good to go.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2014-10-24T14:35:02.537" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3239" PostTypeId="1" AcceptedAnswerId="3241" CreationDate="2014-10-24T14:50:43.750" Score="2" ViewCount="1232" Body="&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/q/3228/68&quot;&gt;Once again&lt;/a&gt;, you're on your way to Acarien. You find yourself in a land of Knights and Knaves, who understand your language but don't speak it. You understand nor speak their language, but you do know their words for yes and no, &quot;yok&quot; and &quot;pom&quot;. Unfortunately, you don't know which is which. And of course, as always, Knights always tell the truth, while knaves always lie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You find yourself at a fork in the road, unsure which road to take, when you see someone coming down one of the roads you are considering. You don't know if it's a knight or a knave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a question you can ask, one single question, answerable with &quot;yok&quot; or &quot;pom&quot;, that tells you which way to go?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I do not know if such a question exists, so answers can be either a question that works, or proof that no such question exists.&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-07-12T17:01:50.430" Title="The way to Acarien, with Knights and Knaves" Tags="|logical-deduction|liars|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3240" PostTypeId="2" ParentId="3235" CreationDate="2014-10-24T15:03:40.370" Score="5" Body="&lt;p&gt;Just thought I would write an answer using normal probability notations and finish by plugging in the values...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$P$ = Passing, &lt;/li&gt;&#xA;&lt;li&gt;$D$ = Being Dumb, &lt;/li&gt;&#xA;&lt;li&gt;$S$ = Being Smart&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We want to get p(S|P)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p(S|P) = \frac{p(S\&amp;amp;P)}{p(P)}$ &#xA;$= \frac{p(P|S)p(S)}{p(P)}$ &#xA;$= \frac{p(P|S)p(S)}{[p(P|S)p(S)+p(P|D)p(D)]}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;plugging in...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=(195/200)(40/100)/[(195/200)(40/100)+(11/300)(60/100)]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=94.66%$&lt;/p&gt;&#xA;" OwnerUserId="4191" LastEditorUserId="2484" LastEditDate="2014-10-25T12:55:21.337" LastActivityDate="2014-10-25T12:55:21.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3241" PostTypeId="2" ParentId="3239" CreationDate="2014-10-24T15:09:55.553" Score="7" Body="&lt;p&gt;Let's test the &lt;em&gt;&quot;If I were to ask you &quot;is left is the correct path?&quot; would you say &quot;yok&quot;?&quot;&lt;/em&gt; question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{ccc|c}&#xA;\text{opponent} &amp;amp; \text{yok} &amp;amp; \text{Acarien} &amp;amp; \text{answer} \\&#xA;\hline&#xA;\text{knave} &amp;amp; \text{yes} &amp;amp; \text{left} &amp;amp; \text{yok} \\&#xA;\text{knight} &amp;amp; \text{yes} &amp;amp; \text{left} &amp;amp; \text{yok} \\&#xA;\text{knave} &amp;amp; \text{no} &amp;amp; \text{left} &amp;amp; \text{yok} \\&#xA;\text{knight} &amp;amp; \text{no} &amp;amp; \text{left} &amp;amp; \text{yok} \\ \hline&#xA;\text{knave} &amp;amp; \text{yes} &amp;amp; \text{right} &amp;amp; \text{pom} \\&#xA;\text{knight} &amp;amp; \text{yes} &amp;amp; \text{right} &amp;amp; \text{pom} \\&#xA;\text{knave} &amp;amp; \text{no} &amp;amp; \text{right} &amp;amp; \text{pom} \\&#xA;\text{knight} &amp;amp; \text{no} &amp;amp; \text{right} &amp;amp; \text{pom} \end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As shown above I've tried all possible combinations of answerer type, language, and correct path.  They all seem to work well.  &lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="68" LastEditDate="2015-07-12T17:01:50.430" LastActivityDate="2015-07-12T17:01:50.430" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3242" PostTypeId="2" ParentId="3239" CreationDate="2014-10-24T15:25:46.383" Score="0" Body="&lt;p&gt;The question is simple:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;If you were the opposite of what you are, and I were to ask you if left is the correct path, would you say yok?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming left = correct, yok = yes, person = Knight: &quot;pom&quot;(no)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming left = correct, yok = yes, person = Knave: &quot;pom&quot;(no)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming left = correct, yok = no, person = Knight: &quot;pom&quot;(yes)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming left = correct, yok = no, person = Knave: &quot;pom&quot;(yes)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming left = incorrect, yok = yes, person = Knight: &quot;yok&quot;(no)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming left = incorrect, yok = yes, person = Knave: &quot;yok&quot;(no)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming left = incorrect, yok = no, person = Knight: &quot;yok&quot;(yes)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming left = incorrect, yok = no, person = Knave: &quot;yok&quot;(yes)&lt;/p&gt;&#xA;" OwnerUserId="4191" LastActivityDate="2014-10-24T15:25:46.383" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3243" PostTypeId="2" ParentId="3154" CreationDate="2014-10-24T15:51:35.740" Score="1" Body="&lt;p&gt;My Ruby version had an optimization: I started &lt;a href=&quot;http://www.free-online-calculator-use.com/prime-number-generator.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and got a list of all primes greater than or equal to 7000 (since the first digit was strictly larger than 6) and less than or equal to 9999 (the largest possible combination). There are only 329 of those values, so it's a much better place to start than iterating through 10,000 values blindly.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;vals = [7001,7013,7019,7027,7039,7043,7057,7069,7079,7103,7109,7121,7127,7129,7151,7159,7177,7187,7193,7207,7211,7213,7219,7229,7237,7243,7247,7253,7283,7297,7307,7309,7321,7331,7333,7349,7351,7369,7393,7411,7417,7433,7451,7457,7459,7477,7481,7487,7489,7499,7507,7517,7523,7529,7537,7541,7547,7549,7559,7561,7573,7577,7583,7589,7591,7603,7607,7621,7639,7643,7649,7669,7673,7681,7687,7691,7699,7703,7717,7723,7727,7741,7753,7757,7759,7789,7793,7817,7823,7829,7841,7853,7867,7873,7877,7879,7883,7901,7907,7919,7927,7933,7937,7949,7951,7963,7993,8009,8011,8017,8039,8053,8059,8069,8081,8087,8089,8093,8101,8111,8117,8123,8147,8161,8167,8171,8179,8191,8209,8219,8221,8231,8233,8237,8243,8263,8269,8273,8287,8291,8293,8297,8311,8317,8329,8353,8363,8369,8377,8387,8389,8419,8423,8429,8431,8443,8447,8461,8467,8501,8513,8521,8527,8537,8539,8543,8563,8573,8581,8597,8599,8609,8623,8627,8629,8641,8647,8663,8669,8677,8681,8689,8693,8699,8707,8713,8719,8731,8737,8741,8747,8753,8761,8779,8783,8803,8807,8819,8821,8831,8837,8839,8849,8861,8863,8867,8887,8893,8923,8929,8933,8941,8951,8963,8969,8971,8999,9001,9007,9011,9013,9029,9041,9043,9049,9059,9067,9091,9103,9109,9127,9133,9137,9151,9157,9161,9173,9181,9187,9199,9203,9209,9221,9227,9239,9241,9257,9277,9281,9283,9293,9311,9319,9323,9337,9341,9343,9349,9371,9377,9391,9397,9403,9413,9419,9421,9431,9433,9437,9439,9461,9463,9467,9473,9479,9491,9497,9511,9521,9533,9539,9547,9551,9587,9601,9613,9619,9623,9629,9631,9643,9649,9661,9677,9679,9689,9697,9719,9721,9733,9739,9743,9749,9767,9769,9781,9787,9791,9803,9811,9817,9829,9833,9839,9851,9857,9859,9871,9883,9887,9901,9907,9923,9929,9931,9941,9949,9967,9973]&#xA;vals.each do |v|&#xA;  digits = v.to_s.chars.map(&amp;amp;:to_i)&#xA;  digitsum = digits.inject(:+)&#xA;  digitprod = digits.inject(:*)&#xA;  digitval = digitprod+digitsum&#xA;  if (digitsum &amp;gt; digits[0]*digits[1] - digits[2]*digits[3]) &amp;amp;&amp;amp;&#xA;    (digits[0] != 0) &amp;amp;&amp;amp;&#xA;    (digitval % digits[0] == 0) &amp;amp;&amp;amp;&#xA;    (digits[2] != 0) &amp;amp;&amp;amp;&#xA;    (digitval % digits[2] == 0)&#xA;    puts &quot;#{v}&quot;&#xA;  end&#xA;end&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4194" LastActivityDate="2014-10-24T15:51:35.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3249" PostTypeId="1" AcceptedAnswerId="3251" CreationDate="2014-10-24T16:38:35.767" Score="4" ViewCount="921" Body="&lt;p&gt;An intelligent team, ‘Smarty Pants’, applying to a competition completes an evaluation form early. Since they are feeling generous they randomly spread their assignment to all 10 teams applying from their school by successively having one team forward it to some other, which then forwards to some other team, and so on. A team can pass the form on to anyone except the team they got it from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a. By how many different paths can an assignment travel through the group of 10 in n transfers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b. What is the probability that team ‘Dunce’ receives it in the third transfer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c. What is the probability that team ‘Smarty Pants’ receives it in the third transfer?&lt;/p&gt;&#xA;" OwnerUserId="4157" LastActivityDate="2014-11-09T07:04:24.147" Title="The generous team puzzle" Tags="|mathematics|logical-deduction|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3250" PostTypeId="1" AcceptedAnswerId="3252" CreationDate="2014-10-24T17:04:53.800" Score="21" ViewCount="2170" Body="&lt;p&gt;Some people in the American Senate are working for secret organizations. Some work for the Illuminati, some work for the Knights Templar, and some work for both. The are also innocent citizens who are unaware of these organizations. Both organizations meet in secret with masks and hoods respectively, so that the members cannot identify one another outside the headquarters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One senator, Reginald, works for the Illuminati, and likes his colleagues, but prefers the Knights Templar headquarters. He finds an arcane rule in the ancient texts, that gives him a brilliant idea. He will turn each Templar into an Illuminati and each Illuminati into a Templar. That way, he becomes a Templar and gets to hang out in their headquarters, but with his old friends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rule goes as follows: If Reginald speaks the name of a senator in the senate and touches his left shoulder, that senator becomes a member of the Illuminati, if he touches his right shoulder, the senator becomes a member of the Knights Templar. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This mechanism is meant to be used by people who know to which group the senator in question belongs. To avoid abuse by people using it blindly, the following rules are added:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If a citizen is named, he remains a citizen.&lt;/li&gt;&#xA;&lt;li&gt;If the senator was already in the group he is assigned to but not in the other, things stay as they are&lt;/li&gt;&#xA;&lt;li&gt;If the senator was not in the group he is assigned to, but he is in the other, he becomes a member of both groups.&lt;/li&gt;&#xA;&lt;li&gt;If the senator was already a member of both groups, he is &lt;em&gt;removed&lt;/em&gt; from the group he was supposed to be assigned to.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Can Reginald, who does not know to which group any Senator belongs, implement his plan, and turn every Illuminati into a Templar Knight and vice versa? The shared members should end up shared, and the citizens must never, even temporarily, become part of either group.&lt;/p&gt;&#xA;" OwnerUserId="2421" LastEditorUserId="2421" LastEditDate="2014-10-24T21:57:05.720" LastActivityDate="2014-12-09T21:19:59.187" Title="The Illuminati and the Knights Templar" Tags="|logical-deduction|" AnswerCount="1" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3251" PostTypeId="2" ParentId="3249" CreationDate="2014-10-24T17:08:39.083" Score="5" Body="&lt;p&gt;a. The first transfer can go to any team except the one that already has it, so for n = 1, there are 9 possible paths.  After that, each team has 8 options, so for n = 2, 9x8=72 different paths.  In general there are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $Paths = 9\times 8^{(n-1)}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;b. The probability that team Dunce gets the form on the third pass is controlled by which team has it after the first and second.  The only way that Dunce can get it is if they are not the one who had it after transfer 1, ($P_1=8/9$), nor after transfer 2 ($P_2=7/8$, since the form cannot go back to either Smarty Pants or Team B),  and they ARE passed it in transfer 3, ($P_3 = 1/8$).  The odds of all of these happening are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  $P_4 = P_1 \times P_2\times P_3 = 56/576 = 7/72$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;c. On the third transfer (i.e. Smarty Pants gives it to Team B (1), Team B gives it to Team C (2), then Team C gives it back to Smarty Pants), there is no possibility that Team Smarty Pants was Team B or Team C.  Using the same designations as above, $P_1 = 1$ and $P_2=1$.  The probability that they get it back after transfer three is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  $P_3 = 1/8$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3928" LastEditorUserId="3928" LastEditDate="2014-10-24T17:24:38.973" LastActivityDate="2014-10-24T17:24:38.973" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3252" PostTypeId="2" ParentId="3250" CreationDate="2014-10-24T17:12:23.540" Score="22" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;For each member of the senate, say the name and touch the shoulders in LRL order (or RLR)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For Illuminati (L), first touch (L &gt; L) will do nothing, second one (R &gt; LR) will make senator both, then third (L &gt; R) will remove it, turning an Illuminati (L) to Templar (R).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Templars (R), (L &gt; LR), (R &gt; L), (L &gt; L), turning a Templar (R) to Illuminati (L). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For both (LR), (L &gt; R), (R &gt; R), (L &gt; LR), keeping as both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normal people will just wonder why all the shoulder tapping. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2014-10-24T17:12:23.540" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3253" PostTypeId="2" ParentId="3249" CreationDate="2014-10-24T17:18:19.483" Score="2" Body="&lt;h2&gt;A.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Assuming 'n' does include the first transfer to 'Smarty Pants' there will be an initial 9 choices for team 'Smarty Pants' to forward to and then 8 choices for every receiving team thereafter.  So the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$9*8^{(n-2)}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;B.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;An easy way to think about this question is: Any team has a 1/8 chance of receiving the forward as long as they are NOT the current or previous team.  So using conditional probability we can phrase it as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; P(Dunce is 3rd) = P(Dunce is 3rd|Dunce is not 2nd)* P(Dunce is not 2nd) = (1/8)*(8/9) = 1/9&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We do not need to go further since we know there is 100% probability of Dunce NOT being the first to receive.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;C.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;We know that 'Smarty Pants' is the first team and teams cannot forward back.  So regardless of who is 2nd, 'Smarty Pants' cannot be third.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; P(Smary is 3rd) = 0&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4191" LastActivityDate="2014-10-24T17:18:19.483" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3254" PostTypeId="2" ParentId="3124" CreationDate="2014-10-24T17:41:53.663" Score="2" Body="&lt;p&gt;It is impossible to do this with less than 6 guesses since we are guessing within 1-100.  The maximum space we can guess in with less than 6 guesses is $81 = 3^{5-1}$.  Less than 5 guesses would be $27$, and so on.  However, you could guess in a space of up to $3^{6-1} = 243$ with 6 guesses.&lt;/p&gt;&#xA;" OwnerUserId="4191" LastEditorUserId="4421" LastEditDate="2015-10-05T16:28:19.813" LastActivityDate="2015-10-05T16:28:19.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3257" PostTypeId="1" AcceptedAnswerId="3294" CreationDate="2014-10-24T19:36:59.473" Score="8" ViewCount="2028" Body="&lt;p&gt;The &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3250/the-illuminati-and-the-kights-templar&quot;&gt;Illuminati just stole our office&lt;/a&gt;  They shot JFK and have committed numerous crimes against humanity (not really) but we usually let that slide.  But for stealing our offices, we have decided to get them arrested for those crimes, take over both offices, and get our revenge.  In order to do this, though, we need their names.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To state the rules here:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Congressmen are either Knights (R), Illuminati (L), Both (B), or Citizens (C) &lt;/li&gt;&#xA;&lt;li&gt;No one else belongs to either group&lt;/li&gt;&#xA;&lt;li&gt;We are not supposed to know who any member of either group is. So no more members of either group will willing reveal themselves.  &lt;/li&gt;&#xA;&lt;li&gt;When attending joint meetings all of us are disguised but we can count how many of each of us there are&lt;/li&gt;&#xA;&lt;li&gt;After the plan is put into action, there will be no communication between us&lt;/li&gt;&#xA;&lt;li&gt;Citizens can't join either group by any means during this time frame.  If you say their name and repeatedly tap your shoulders, they just start googling mental health centers to recommend to you.&lt;/li&gt;&#xA;&lt;li&gt;If a person (in this case always Bob) touches his left shoulder after saying a persons name, that person becomes an Illuminati (B) if he was only a Knight (R) or loses Illuminati (R) status if he was both (B).&lt;/li&gt;&#xA;&lt;li&gt;If a person (in this case always Bob) touches his right shoulder after saying a persons name, that person becomes a Knight (B) if he was only Illuminati(L) or loses Knight status (L) if he was both (B).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Senator Bob has been identified as a member of only our group and will not betray us.  This is certain.  We have agreed to hold a meeting every day with the Illuminati where every member of each group must be present in the correct uniform.  If they are in both, they must wear the robes of one and the mask of the other.  Bob will hold a prayer in front of congress between each meeting in which he can touch his shoulders as much as he likes and everyone will know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know there are only 100 congressmen whose status we don't already know.  We know only 32 of these are in the Illuminati and 32 are our men.  None of those unidentified are in both.  Bob and Peter are currently known to be Knight Templar and Illuminati respectively because they shoulder tapped with obvious motives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is there a way we can determine the names of every one of those 32 congressmen who are right now Illuminati without revealing any of our names to them?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I am willing to lie and cheat as long as they don't know we are doing it.&lt;/strong&gt; I am a politician like you after all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets assume that we have ensure Peter will not shoulder tap again and they will not sabatoge us unless they know for a fact we have identified them (for now lets assume they are dumb enough they can't do that).  They are smart enough, however, to try and catch us breaking meeting rules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus's if you can do any of the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Minimize the number of meetings/shoulder tapping sessions&lt;/li&gt;&#xA;&lt;li&gt;Compute a plan for an arbitrary number of congressmen in each group&lt;/li&gt;&#xA;&lt;li&gt;Have the plan allowing for the possibility that unidentified congressmen is a member of both.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="99" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2014-11-08T12:02:22.827" Title="How do we get those dang Illuminati?" Tags="|logical-deduction|strategy|" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3258" PostTypeId="2" ParentId="3154" CreationDate="2014-10-24T20:10:22.557" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;8017&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Trick is here to iterate until all conditions return true thus meaning that this is the number the bank robber is looking for.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is program I used for this computation. It is written in standard Java.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;public class BankRobberProblem implements SingleSolution&amp;lt;Integer&amp;gt; {&#xA;&#xA;    public static void main(String[] args) {&#xA;        SingleSolution bankRobberProblem = new BankRobberProblem();&#xA;        System.out.println(bankRobberProblem.solveSingle());&#xA;    }&#xA;&#xA;    @Override&#xA;    public Integer solveSingle() {&#xA;        for (int a = 0; a &amp;lt; 10; a++) {&#xA;            for (int b = 0; b &amp;lt; 10; b++) {&#xA;                for (int c = 0; c &amp;lt; 10; c++) {&#xA;                    for (int d = 0; d &amp;lt; 10; d++) {&#xA;                        if ((a * b) - (c * d) &amp;lt; a + b + c + d) {&#xA;                            int productAndSum = (a * b * c * d) + (a + b + c + d);&#xA;                            if (a != 0 &amp;amp;&amp;amp; c != 0) {&#xA;                                if (productAndSum % a == 0 &amp;amp;&amp;amp; productAndSum % c == 0) {&#xA;                                    Integer n = Integer.valueOf(String.format(&quot;%s%s%s%s&quot;, a, b, c, d));&#xA;                                    boolean prime = true;&#xA;                                    if (n % 2 == 0) prime = false;&#xA;                                    for (int i = 3; i * i &amp;lt;= n; i += 2) {&#xA;                                        if(n % i == 0) prime = false;&#xA;                                    }&#xA;                                    if (prime) {&#xA;                                        if (a &amp;gt; 6) {&#xA;                                            return n;&#xA;                                        }&#xA;                                    }&#xA;                                }&#xA;                            }&#xA;                        }&#xA;                    }&#xA;                }&#xA;            }&#xA;        }&#xA;&#xA;        return null;&#xA;    }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This program is for Java 8 only but yields faster results then the previous.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;public class BankRobberProblem implements SingleSolution&amp;lt;Integer&amp;gt; {&#xA;&#xA;    public static void main(String[] args) {&#xA;        BankRobberProblem bankRobberProblem = new BankRobberProblem();&#xA;        System.out.println(bankRobberProblem.solveSingle());&#xA;    }&#xA;&#xA;    @Override&#xA;    public Integer solveSingle() {&#xA;        return IntStream.rangeClosed(7000, 9999).parallel() // The first digit is greater than 6&#xA;            .filter(n -&amp;gt; {&#xA;                boolean prime = true;&#xA;                if (n % 2 == 0) prime = false;&#xA;                for (int i = 3; i * i &amp;lt;= n; i += 2) {&#xA;                    if (n % i == 0) prime = false;&#xA;                }&#xA;                return prime;&#xA;            }) // The number is prime&#xA;            .filter(value -&amp;gt; {&#xA;                List&amp;lt;Integer&amp;gt; s = seperate(value);&#xA;                if (a(s) != 0 &amp;amp;&amp;amp; c(s) != 0) {&#xA;                    int productAndSum = (a(s) * b(s) * c(s) * d(s)) + (a(s) + b(s) + c(s) + d(s));&#xA;                    if (productAndSum % a(s) == 0 &amp;amp;&amp;amp; productAndSum % c(s) == 0) {&#xA;                        return true;&#xA;                    }&#xA;                }&#xA;                return false;&#xA;            }) // The product of the digits plus their sum is divisible by A and C only.&#xA;            .filter(value -&amp;gt; {&#xA;                List&amp;lt;Integer&amp;gt; s = seperate(value);&#xA;                if ((a(s) * b(s)) - (c(s) * d(s)) &amp;lt; a(s) + b(s) + c(s) + d(s)) {&#xA;                    return true;&#xA;                }&#xA;                return false;&#xA;            }) // The sum of the digits is greater than (A * B) - (C * D).&#xA;            .findFirst()&#xA;            .orElse(0);&#xA;    }&#xA;&#xA;    private List&amp;lt;Integer&amp;gt; seperate(Integer n) {&#xA;        List&amp;lt;Integer&amp;gt; result = new ArrayList&amp;lt;&amp;gt;();&#xA;        while (n &amp;gt; 0) {&#xA;            result.add(n % 10);&#xA;            n = n / 10;&#xA;        }&#xA;        Collections.reverse(result);&#xA;        return result;&#xA;    }&#xA;&#xA;    private Integer a(List&amp;lt;Integer&amp;gt; l) {return l.get(0);}&#xA;    private Integer b(List&amp;lt;Integer&amp;gt; l) {return l.get(1);}&#xA;    private Integer c(List&amp;lt;Integer&amp;gt; l) {return l.get(2);}&#xA;    private Integer d(List&amp;lt;Integer&amp;gt; l) {return l.get(3);}&#xA;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4196" LastEditorUserId="4196" LastEditDate="2014-10-24T20:32:12.140" LastActivityDate="2014-10-24T20:32:12.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3260" PostTypeId="2" ParentId="3157" CreationDate="2014-10-24T20:47:04.697" Score="2" Body="&lt;p&gt;This program solves the problem using Java 8. It is commented throughout for people to understand. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;public class CodeCrackerProblem implements SingleSolution&amp;lt;Integer&amp;gt; {&#xA;&#xA;    public static void main(String[] args) {&#xA;        SingleSolution&amp;lt;Integer&amp;gt; codeCrackerProblem = new CodeCrackerProblem();&#xA;        System.out.println(codeCrackerProblem.solveSingle());&#xA;    }&#xA;&#xA;    @Override&#xA;    public Integer solveSingle() {&#xA;        return IntStream.rangeClosed(1000, 5999).parallel() // The first digit is less than 6.&#xA;            .filter(this::prime) // The number is prime&#xA;            .filter(value -&amp;gt; {&#xA;                List&amp;lt;Integer&amp;gt; s = seperate(value);&#xA;                return (a(s) &amp;gt; b(s) &amp;amp;&amp;amp; b(s) &amp;gt; c(s) &amp;amp;&amp;amp; c(s) &amp;gt; d(s));&#xA;            }) // The digits are in descending order.&#xA;            .findFirst()&#xA;            .orElse(0);&#xA;    }&#xA;&#xA;    private boolean prime(int n) {&#xA;        if (n%2==0) return false;&#xA;        for(int i=3;i*i&amp;lt;=n;i+=2) {&#xA;            if(n%i==0)&#xA;                return false;&#xA;        }&#xA;        return true;&#xA;    }&#xA;&#xA;    private List&amp;lt;Integer&amp;gt; seperate(Integer n) {&#xA;        List&amp;lt;Integer&amp;gt; result = new ArrayList&amp;lt;&amp;gt;();&#xA;        while (n &amp;gt; 0) {&#xA;            result.add(n % 10);&#xA;            n = n / 10;&#xA;        }&#xA;        Collections.reverse(result);&#xA;        return result;&#xA;    }&#xA;&#xA;    private Integer a(List&amp;lt;Integer&amp;gt; l) {return l.get(0);}&#xA;    private Integer b(List&amp;lt;Integer&amp;gt; l) {return l.get(1);}&#xA;    private Integer c(List&amp;lt;Integer&amp;gt; l) {return l.get(2);}&#xA;    private Integer d(List&amp;lt;Integer&amp;gt; l) {return l.get(3);}&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;5431&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4196" LastEditorUserId="4196" LastEditDate="2016-05-21T11:20:35.600" LastActivityDate="2016-05-21T11:20:35.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3261" PostTypeId="2" ParentId="3204" CreationDate="2014-10-24T20:56:47.950" Score="0" Body="&lt;p&gt;I commented the code so that it is easy to read for people new to programming languages. It is written in Java 8 and can only be executed by Java 8.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;public class SecondCodeCrackerProblem implements SingleSolution&amp;lt;Integer&amp;gt; {&#xA;&#xA;    public static void main(String[] args) {&#xA;        SingleSolution&amp;lt;Integer&amp;gt; secondCodeCrackerProblem = new SecondCodeCrackerProblem();&#xA;        System.out.println(secondCodeCrackerProblem.solveSingle());&#xA;    }&#xA;&#xA;    @Override&#xA;    public Integer solveSingle() {&#xA;        return IntStream.rangeClosed(4000, 9999).parallel() // The first digit is greater than 3.&#xA;            .filter(this::prime) // The number is prime.&#xA;            .filter(value -&amp;gt; {&#xA;                List&amp;lt;Integer&amp;gt; s = seperate(value);&#xA;                if (a(s) != b(s) &amp;amp;&amp;amp; a(s) != c(s) &amp;amp;&amp;amp; a(s) != d(s) &amp;amp;&amp;amp; b(s) != c(s) &amp;amp;&amp;amp; b(s) != d(s) &amp;amp;&amp;amp; c(s) != d(s)) {&#xA;                    return true;&#xA;                }&#xA;                return false;&#xA;            }) // No digits are repeated.&#xA;            .filter(value -&amp;gt; {&#xA;                List&amp;lt;Integer&amp;gt; s = seperate(value);&#xA;                if (a(s) * b(s) == c(s) * d(s)) {&#xA;                    return true;&#xA;                }&#xA;                return false;&#xA;            }) // A⋅B = C⋅D&#xA;            .findFirst()&#xA;            .orElse(0);&#xA;    }&#xA;&#xA;    private boolean prime(int n) {&#xA;        if (n%2==0) return false;&#xA;        for(int i=3;i*i&amp;lt;=n;i+=2) {&#xA;            if(n%i==0)&#xA;                return false;&#xA;        }&#xA;        return true;&#xA;    }&#xA;&#xA;    private List&amp;lt;Integer&amp;gt; seperate(Integer n) {&#xA;        List&amp;lt;Integer&amp;gt; result = new ArrayList&amp;lt;&amp;gt;();&#xA;        while (n &amp;gt; 0) {&#xA;            result.add(n % 10);&#xA;            n = n / 10;&#xA;        }&#xA;        Collections.reverse(result);&#xA;        return result;&#xA;    }&#xA;&#xA;    private Integer a(List&amp;lt;Integer&amp;gt; l) {return l.get(0);}&#xA;    private Integer b(List&amp;lt;Integer&amp;gt; l) {return l.get(1);}&#xA;    private Integer c(List&amp;lt;Integer&amp;gt; l) {return l.get(2);}&#xA;    private Integer d(List&amp;lt;Integer&amp;gt; l) {return l.get(3);}&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;6329&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4196" LastActivityDate="2014-10-24T20:56:47.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3262" PostTypeId="2" ParentId="3209" CreationDate="2014-10-24T21:11:14.323" Score="1" Body="&lt;p&gt;I gotta give it to you, this one took me 5 minutes extra to figure out.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;public class ThirdCodeCrackerProblem implements SingleSolution&amp;lt;List&amp;lt;String&amp;gt;&amp;gt; {&#xA;&#xA;    public static void main(String[] args) {&#xA;        SingleSolution&amp;lt;List&amp;lt;String&amp;gt;&amp;gt; thirdCodeCrackerProblem = new ThirdCodeCrackerProblem();&#xA;        System.out.println(Arrays.toString(thirdCodeCrackerProblem.solveSingle().toArray()));&#xA;    }&#xA;&#xA;    @Override&#xA;    public List&amp;lt;String&amp;gt; solveSingle() {&#xA;        return IntStream.rangeClosed(1000, 9999).parallel() // There are four digits&#xA;            .filter(this::prime) // The number is prime.&#xA;            .filter(value -&amp;gt; {&#xA;                List&amp;lt;Integer&amp;gt; s = seperate(value);&#xA;                if (a(s) == 4 || b(s) == 4 || c(s) == 4 || d(s) == 4) {&#xA;                    return true;&#xA;                }&#xA;                return false;&#xA;            }) // One of the digits is a four.&#xA;            .filter(value -&amp;gt; {&#xA;                List&amp;lt;Integer&amp;gt; s = seperate(value);&#xA;                if (a(s) != b(s) &amp;amp;&amp;amp; a(s) != c(s) &amp;amp;&amp;amp; a(s) != d(s) &amp;amp;&amp;amp; b(s) != c(s) &amp;amp;&amp;amp; b(s) != d(s) &amp;amp;&amp;amp; c(s) != d(s)) {&#xA;                    return true;&#xA;                }&#xA;                return false;&#xA;            }) // No digits are repeated.&#xA;            .mapToObj(value -&amp;gt; Integer.toHexString(value)) // The hexadecimal representation of this number...&#xA;            .filter(s -&amp;gt; !s.matches(&quot;.*\\d.*&quot;)) // Removing the numbers that contain a number&#xA;            .filter(s -&amp;gt; s.matches(&quot;.*[aeuoyi].*&quot;)) // Only leaving the words with a vowel in it&#xA;            .collect(Collectors.toList());&#xA;    }&#xA;&#xA;    private boolean prime(int n) {&#xA;        if (n%2==0) return false;&#xA;        for(int i=3;i*i&amp;lt;=n;i+=2) {&#xA;            if(n%i==0)&#xA;                return false;&#xA;        }&#xA;        return true;&#xA;    }&#xA;&#xA;    private List&amp;lt;Integer&amp;gt; seperate(Integer n) {&#xA;        List&amp;lt;Integer&amp;gt; result = new ArrayList&amp;lt;&amp;gt;();&#xA;        while (n &amp;gt; 0) {&#xA;            result.add(n % 10);&#xA;            n = n / 10;&#xA;        }&#xA;        Collections.reverse(result);&#xA;        return result;&#xA;    }&#xA;&#xA;    private Integer a(List&amp;lt;Integer&amp;gt; l) {return l.get(0);}&#xA;    private Integer b(List&amp;lt;Integer&amp;gt; l) {return l.get(1);}&#xA;    private Integer c(List&amp;lt;Integer&amp;gt; l) {return l.get(2);}&#xA;    private Integer d(List&amp;lt;Integer&amp;gt; l) {return l.get(3);}&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This program resulted in:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;[abd, fad, fef]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And after using the site provided there was one correct answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;fad&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And after converting the answer back to decimal, I got:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;4013&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4196" LastEditorUserId="4196" LastEditDate="2014-10-24T21:42:09.070" LastActivityDate="2014-10-24T21:42:09.070" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3265" PostTypeId="1" AcceptedAnswerId="3266" CreationDate="2014-10-24T22:39:02.453" Score="4" ViewCount="2014" Body="&lt;p&gt;Bob wakes up one morning, and he needs to grab a pair of socks for work. In his sock drawer, he has 5 pairs of blue socks, 3 pairs of red socks, and 4 pairs of white socks (Bob is American and very patriotic). If Bob is logical, capable, and an observant person, what is the minimum amount of socks he needs to grab before being sure he has a pair of &lt;strong&gt;white socks&lt;/strong&gt; for work?&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;The answer will (hopefully) not be the first thing that comes to mind. Also, this is a riddle, so inventive solutions are common and entertaining. If nobody has submitted the answer I was expecting in the next couple days, I will accept the answer I think is most creative.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1664" LastEditorUserId="5373" LastEditDate="2015-07-15T23:51:21.287" LastActivityDate="2015-07-15T23:51:21.287" Title="A Drawer of Socks" Tags="|logical-deduction|lateral-thinking|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3266" PostTypeId="2" ParentId="3265" CreationDate="2014-10-24T22:42:46.220" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Two. Since it doesn't seem to be dark in the room, Bob can pick out two of the white socks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(I never know whether I'm being clever or incredibly pedantic with these things).&lt;/p&gt;&#xA;" OwnerUserId="2421" LastActivityDate="2014-10-24T22:42:46.220" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3267" PostTypeId="2" ParentId="3257" CreationDate="2014-10-24T23:12:25.063" Score="1" Body="&lt;p&gt;Ok, let's start with an inefficient one, to set up the principle (and check that I understand the rules). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We number the unknowns 1 through 100 (or 1 though 98 if we're counting Bob and this Peter fellow). First, we call a meeting, and make a note of how many people turn up in each group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At his first session of congress, Bob says the name of congressman 1, and taps his right shoulder, says the name and taps his left shoulder, and then says the name and taps his right shoulder (RLR). This turns Illuminati into Templar and vice versa, while members of both and citizens are unaffected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We call another meeting and see if the numbers have changed. If we have an extra Templar, we add congressman 1 to the list.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bob goes through all congressmen in this fashion, and after only 100 days we know whom to arrest.&lt;/p&gt;&#xA;" OwnerUserId="2421" LastEditorUserId="2421" LastEditDate="2014-10-25T10:05:25.733" LastActivityDate="2014-10-25T10:05:25.733" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3269" PostTypeId="2" ParentId="3265" CreationDate="2014-10-24T23:30:53.500" Score="-1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bob have to draw 9 pairs to be guaranteed that he draws one pair of white socks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not very creative tough ...&lt;/p&gt;&#xA;" OwnerUserId="4215" LastActivityDate="2014-10-24T23:30:53.500" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3270" PostTypeId="2" ParentId="3265" CreationDate="2014-10-24T23:47:48.607" Score="1" Body="&lt;p&gt;I'm going to stray from the normal and say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4.&#xA; He'll need to remove the two off of his feet of unknown color and then the two white ones from the dresser.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2013" LastActivityDate="2014-10-24T23:47:48.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3271" PostTypeId="1" AcceptedAnswerId="3273" CreationDate="2014-10-25T01:12:54.973" Score="8" ViewCount="1017" Body="&lt;p&gt;The distant country of Toomalia has a birthrate of 67 boys for every 33 girls. In an effort to restore a 50:50 gender ratio in the country, Toomalian policymakers institute a mandatory new social policy: Couples seeking to have children may continue to do so as long as they bear only girls, but must stop as soon as they bear one boy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the Toomalian people &lt;em&gt;i&lt;/em&gt;) rigorously adhere to the policy, and &lt;em&gt;ii&lt;/em&gt;) do not selectively abort their offspring, &lt;strong&gt;what will the net effect on the boy:girl birth ratio be when the new policy is put into effect?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that you may assume the gender of each conceived child is independent from the gender(s) of its siblings, but you should &lt;em&gt;not&lt;/em&gt; assume that every couple wishes to conceive as many children as possible. Many couples will voluntarily stop having children after having 2 or 3, for example.&lt;/p&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2014-10-25T12:35:40.777" Title="Social Engineering (The Birthrate Problem)" Tags="|probability|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3272" PostTypeId="2" ParentId="3271" CreationDate="2014-10-25T01:14:11.663" Score="2" Body="&lt;p&gt;Theroretically, I believe, the ratio would go up. Explanation will follow as soon as it is confirmed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I think the ratio will stay the same. The proof is outlined below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If families want 3 kids, there are 8 possible outcomes.&#xA;BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG, with B for boy and G for girl. However, these are not all equal scenarios. I will change the birthrate to 2/3 and 1/3 for boys and girls, respectively, as this helps with calculations but won't change the outcome.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;BBB = 8/27&lt;/li&gt;&#xA;&lt;li&gt;BBG = 4/27&lt;/li&gt;&#xA;&lt;li&gt;BGB = 4/27&lt;/li&gt;&#xA;&lt;li&gt;BGG = 2/27&lt;/li&gt;&#xA;&lt;li&gt;GBB = 4/27&lt;/li&gt;&#xA;&lt;li&gt;GBG = 2/27&lt;/li&gt;&#xA;&lt;li&gt;GGB = 2/27&lt;/li&gt;&#xA;&lt;li&gt;GGG = 1/27&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If each family is to stop after having a boy, the outcomes are changed.&#xA;4xB, 2xGB, GGB, and GGG. After adding up the percentages, it comes out to: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;B = 18/27&lt;/li&gt;&#xA;&lt;li&gt;GB = 6/27&lt;/li&gt;&#xA;&lt;li&gt;GGB = 2/27&lt;/li&gt;&#xA;&lt;li&gt;GGG = 1/27&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This means that the birth rate would be 26 boys for every 13 girls - so the ratio will stay the same. It seems my hypothesis was wrong in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT again: I still think this is right, but I am starting to doubt my proof. I'll see if there is a better way to explain it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2/3 = One boy, no girls.&#xA;2/3 of 1/3 = One boy, one girl.&#xA;2/3 of 1/3 of 1/3 = One boy, two girls.&#xA;2/3 of 1/3 of 1/3 of 1/3 = One boy, three girls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As one can see, this can be extended infinitely. Simplified mathematically, there is a $\frac{2}{3^{G+1}}$ chance of getting G girls and 1 boy. This shows that the birth rate will stay the same.&lt;/p&gt;&#xA;" OwnerUserId="1664" LastEditorUserId="1664" LastEditDate="2014-10-25T01:31:28.347" LastActivityDate="2014-10-25T01:31:28.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3273" PostTypeId="2" ParentId="3271" CreationDate="2014-10-25T02:57:08.273" Score="11" Body="&lt;p&gt;The ratio will stay the same. I'm not even sure how to prove that, since it seems so self-evident. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One simple approach:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the gender of a child is independent from their siblings, we can ignore who the parents are and ask under this procedure, what will the gender of the &lt;em&gt;next&lt;/em&gt; child born to &lt;em&gt;any&lt;/em&gt; couple in the country be? Well it'll still have a 67% chance of boy, 33% chance of girl. And since the parents/siblings don't matter in that number, those chances will be constant, so over time as more children continue to be born, it will be in the same ratio.&lt;/p&gt;&#xA;" OwnerUserId="1759" LastActivityDate="2014-10-25T02:57:08.273" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="3276" PostTypeId="2" ParentId="3271" CreationDate="2014-10-25T05:08:23.337" Score="1" Body="&lt;p&gt;If the sex of each child is independently randomly determined with a 2:1 ratio favoring boys, then the stated policy will have no effect on male versus female birth rates.  If, however, the events are not independent, then the policy might have an effect.  Suppose, for example, that that one third of all males produced only XY chromosomes while the other two thirds produced an even mix of XX and XY.  In that scenario...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1/3 of the males generate only XY chromosomes; they would sire one son and stop.&lt;/li&gt;&#xA;&lt;li&gt;1/3 of the males (half of 2/3) would generate mixed chromosomes, but their first child would be a son, so they'd stop.&lt;/li&gt;&#xA;&lt;li&gt;1/6 of the males (half of the remainder) would generate mixed chromosomes, and sire a girl followed by a boy and then stop.&lt;/li&gt;&#xA;&lt;li&gt;1/12 of the males would generate mixed chromosomes and sire two girls and a boy, then stop.&lt;/li&gt;&#xA;&lt;li&gt;1/24 would sire three girls and a boy, then stop.&lt;/li&gt;&#xA;&lt;li&gt;1/24 would sire four girls; assume for simplicity that they all stop.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Every man would thus sire one boy; 1/3 would sire a least one girl; 1/6 at least two, 1/12 at least three, and 1/24, four.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The average number of girls sired by each man would thus be 2/3.  Not enough to manage a 1:1 male/female ratio, but yielding a 3:2 ratio which is less lopsided than the 2:1 ratio that would exist without the policy.&lt;/p&gt;&#xA;" OwnerUserId="1467" LastActivityDate="2014-10-25T05:08:23.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3277" PostTypeId="2" ParentId="3235" CreationDate="2014-10-25T07:50:36.477" Score="-1" Body="&lt;p&gt;In fact, I did not pass the test, because this test about detecting smart people makes a mistake at the very beginning.&#xA;So... Instead of passing it, I came and proposed a correction :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$289$ out of $300$ true negatives and $195$ out of $200$ true positives is not $100\%$ accuracy. It's only $96.8\%$ accuracy.&lt;/p&gt;&#xA;" OwnerUserId="4226" LastActivityDate="2014-10-25T07:50:36.477" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3278" PostTypeId="2" ParentId="3249" CreationDate="2014-10-25T09:52:36.410" Score="0" Body="&lt;p&gt;At any time n the assignment is in one of 3 states&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Smarty Pants owns the assignment.&lt;/li&gt;&#xA;&lt;li&gt;Someone else owns the assignment and got it from Smarty Pants.&lt;/li&gt;&#xA;&lt;li&gt;Someone else owns the assignment and got it from someone else.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Let's define the probability of each state after n transfers:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_{n,s} = P(assignment\ is\ in\ state\ s\ after\ n\ tranfers)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_{0,1} = 1,\ p_{0,2} = p_{0,3} = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From state $1$ the assignment always goes to state $2$.  From state $2$ the assignment always goes to state $3$.  From state $3$, it goes to state $1$ with probability $1\over 8$ and to state $3$ with probability $7\over 8$.  This translates to the formulas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_{n+1,1} = {1\over 8} p_{n,3}$&lt;br&gt;&#xA;$p_{n+1,2} = p_{n,1}$&lt;br&gt;&#xA;$p_{n+1,3} = p_{n,2} + {7\over 8} p_{n,3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can compute the first 3 transfers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_{0,1} = 1,\ p_{0,2} = p_{0,3} = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_{1,1} = 0,\ p_{1,2} = 1,\ p_{1,3} = 0$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_{2,1} = 0,\ p_{2,2} = 0,\ p_{2,3} = 1$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_{3,1} = {1\over 8},\ p_{3,2} = 0,\ p_{3,3} = {7\over 8}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we have the answers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;C&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Smarty Pants gets it back after 3 transfers with probability $p_{3,1} = {1\over 8}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;B&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Other teams have the assignment after 3 transfers with probability $p_{3,3} = {7\over 8}$ &lt;em&gt;collectively&lt;/em&gt;.  Each of the 9 teams has it with probability ${7\over 8}\times{1\over 9} = {7\over72}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As everybody else already mentioned, the first step has 9 possibilities, the others 8, so the number of paths for the assignment is $8\times 9^{n-1}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2136" LastActivityDate="2014-10-25T09:52:36.410" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3279" PostTypeId="2" ParentId="3257" CreationDate="2014-10-25T11:12:06.060" Score="2" Body="&lt;p&gt;To speed things up, we can use the fact that none of the unidentified are B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let Bob execute RLR for congressman 1, LRLR for congressman 2 and RLRLR for congressman 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all possibilities, this gives us the following table:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;before  after   change &#xA;1 2 3           R  B  L  &#xA;L L L | R R B | +2    -2&#xA;L L R | R R L | +1 +1 -2&#xA;L R L | R B B |    +1 -1&#xA;L R R | R B L | -1 +2 -1&#xA;R L L | L R B | &#xA;R L R | L R L | -1 +1&#xA;R R L | L R L | -2 +1 +1&#xA;R R R | L B L | -3 +2 +1&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We can see that each assignment of memberships corresponds to a unique change in frequencies at the next secret meeting. The next day Bob does the same for congressmen 4, 5 and 6 and so on. Since we're dealing with groups of three congressmen per meeting, we can work through all hundred in 34 days, and have the Illuminati behind bars before Christmas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps this scheme can actually be extended a little further, but not to the full 100 congressman. Since there are about $100^2$ unique frequencies, the best you can do with this strategy is to expose the Illuminati in groups of 13 (ie. in 8 days). I'll leave that for another poster to work out.&lt;/p&gt;&#xA;" OwnerUserId="2421" LastActivityDate="2014-10-25T11:12:06.060" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3280" PostTypeId="1" AcceptedAnswerId="3281" CreationDate="2014-10-25T12:12:00.303" Score="12" ViewCount="5083" Body="&lt;p&gt;You've been sent to prison, but the prisons where you are from are very clever. They do not just lock you up, they put you in a room with 2 doors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Door A&lt;/strong&gt; is guarded by an evil dragon that will kill anyone who leaves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Door B&lt;/strong&gt; is simply a hallway, with an odd ceiling, it's the strongest magnifying glass on earth! What's so bad about this? It magnifies the sun so strong it'll burn you to a crisp the second you leave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They also went through extra measures to make sure you don't live through 24 hours, the back wall is slowly enclosing on you. There is no open ceiling in the prison room, so don't try wall jumping Mario style once the wall encloses a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How can you escape?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="11876" LastEditDate="2016-02-29T14:47:07.943" LastActivityDate="2016-03-01T03:41:54.757" Title="The Dragon or the Magnifying Glass?" Tags="|lateral-thinking|situation|" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3281" PostTypeId="2" ParentId="3280" CreationDate="2014-10-25T12:16:46.573" Score="27" Body="&lt;p&gt;In the 24 hours you have you there must be night time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Open the door B one slight bit to find if there is light in the next room, as soon as its dark, you know its night and the magnifying glass wont harm you and you can leave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If opening the door would kill you either ways, just feel the heat of the door B, that should give an indication if the other room is hot or cold.&lt;/p&gt;&#xA;" OwnerUserId="2313" LastActivityDate="2014-10-25T12:16:46.573" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3282" PostTypeId="2" ParentId="3280" CreationDate="2014-10-25T12:17:57.990" Score="3" Body="&lt;p&gt;The magnifying glass will only magnify the sunlight in a single spot (or a certain area, but certainly not the whole hallway). So go through &lt;strong&gt;door B&lt;/strong&gt;, and walk through the hallway avoiding the hotspots.&lt;/p&gt;&#xA;" OwnerUserId="3993" LastEditorUserId="15544" LastEditDate="2015-08-27T14:08:39.707" LastActivityDate="2015-08-27T14:08:39.707" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3283" PostTypeId="2" ParentId="3271" CreationDate="2014-10-25T12:35:40.777" Score="1" Body="&lt;p&gt;This (pretty) graph is the best illustration I could provide:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 1st batch: B (1/3)   B (1/3)   G (1/3)&#xA;                               //\&#xA; 2nd batch: ―         ―       BB  G&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;At first batch, we have 2/3 boys and 1/3 girls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second batch has the same ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be the same for next batches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;As all batches have the same ratio, if we merge batches the ratio remains the same.&lt;/p&gt;&#xA;" OwnerUserId="4229" LastActivityDate="2014-10-25T12:35:40.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3284" PostTypeId="1" AcceptedAnswerId="3533" CreationDate="2014-10-25T13:44:04.310" Score="11" ViewCount="1402" Body="&lt;p&gt;A detective friend of yours was trying to find clues for a crime he was investigating, he was almost done but a concluding location will help you close the case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately your friend could only send this but he was confident you will put together the information to close in on the place.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you are close enough you have other clues to finish the investigation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;16,06,68,88,??,98                           &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1047,1038,519,510,255,246,123,???,??,??     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4, 6, 12, 18, 30, 42, 60, 72, ???, ???      &lt;/p&gt;&#xA;&#xA;&lt;p&gt;27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, ???&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4, 7, 15, 29, 59, ???&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint/Edit: Remember these are only clues not actual values so change of case is acceptable, though there are probably clues in the casing &lt;/p&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="22922" LastEditDate="2016-09-15T08:52:16.420" LastActivityDate="2016-09-15T08:52:16.420" Title="Are you a Detective" Tags="|number-sequence|" AnswerCount="5" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3285" PostTypeId="2" ParentId="3284" CreationDate="2014-10-25T14:01:40.920" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. 78 - look at it upside down.&lt;br&gt; &#xA; 2. 114, 57, 48 - subtract 9, divide by 2, repeat&lt;br&gt;&#xA; 3.  Numbers where x+1 and x-1 are both prime. Next are 102 and 138. &lt;br&gt;&#xA; 4. 107 - multiply by three and add one, divide by two&lt;br&gt;&#xA; 5. 117 - multiply by two and subtract one, multiply by two and add one&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1664" LastEditorUserId="1664" LastEditDate="2014-10-25T15:09:18.693" LastActivityDate="2014-10-25T15:09:18.693" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="3286" PostTypeId="1" AcceptedAnswerId="3288" CreationDate="2014-10-25T14:38:33.400" Score="28" ViewCount="9912" Body="&lt;p&gt;How do you arrange 6 pencils so that each one touches the other five? And what about 7 or 8?&lt;/p&gt;&#xA;" OwnerUserId="2375" LastEditorUserId="69582" LastEditorDisplayName="user20" LastEditDate="2021-04-01T15:20:41.707" LastActivityDate="2021-04-01T15:20:41.707" Title="How do I arrange pencils so they all touch each other?" Tags="|mathematics|geometry|three-dimensional|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3288" PostTypeId="2" ParentId="3286" CreationDate="2014-10-25T14:56:24.413" Score="23" Body="&lt;p&gt;You need 2 layers. First arrange 2 pencils that spread outwards touching by the erasers, like a V, then slide one with the pencil tip in between. Then you can repeat the process with 3 more, sideways, laying on the original, this achieves 6:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CPKq6.jpg&quot; alt=&quot;6 pencils&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;You can also achieve an arrangement of 7 by making 3 V's of 2, an arranging them like a ninja star (not really sure how else to explain it) you may need to tie them with the V's because 1 pencil has to be on the other, it could be tough to balance. This is 6, you can add one more, upright to achieve a 7th:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/h6A9o.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;I believe 8 may be impossible.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; The matchstick version, and images, can be found at &lt;a href=&quot;http://brainden.com/forum/index.php/topic/148-touch/&quot; rel=&quot;nofollow noreferrer&quot;&gt;BrainDen&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;I did not use it for the first solution, I found that myself (really proud) but it's where the second answer came from.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="73592" LastEditDate="2021-04-01T14:41:10.970" LastActivityDate="2021-04-01T14:41:10.970" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="3289" PostTypeId="1" AcceptedAnswerId="3291" CreationDate="2014-10-25T15:31:24.173" Score="4" ViewCount="811" Body="&lt;p&gt;Scott is an expert outdoorsman, pursuing his dream by living in the remote Alaskan wilderness. Scott has everything he needs, and would prefer to remain a recluse, yet he will have to leave his home in four years. Why?&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Clues&lt;/h2&gt;&#xA;&#xA;&lt;h3&gt;Was something going to happen that would destroy Scott's home?&lt;/h3&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;Did Scott need to return to civilisation in order to buy food or clothing?&lt;/h3&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;Would Scott have been in danger if he ignored the four-year deadline?&lt;/h3&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4061" LastActivityDate="2014-10-28T06:37:14.383" Title="The Alaskan Dream" Tags="|lateral-thinking|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3290" PostTypeId="2" ParentId="3289" CreationDate="2014-10-25T15:45:06.027" Score="2" Body="&lt;p&gt;He's an athlete in the Winter Olympics? I can't see that as dangerous, though.&lt;/p&gt;&#xA;" OwnerUserId="2265" LastActivityDate="2014-10-25T15:45:06.027" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3291" PostTypeId="2" ParentId="3289" CreationDate="2014-10-25T15:47:12.200" Score="11" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Scott has a pacemaker.&#xA; It runs on batteries.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2136" LastActivityDate="2014-10-25T15:47:12.200" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3292" PostTypeId="2" ParentId="3284" CreationDate="2014-10-25T16:26:25.730" Score="13" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. 87 - read it upside down.&lt;br&gt;&#xA; 2. 114, 57, 48: minus 9, divide by 2, repeat&lt;br&gt;&#xA; 3. 102, 108:  middle of twin primes.  &lt;a href=&quot;https://oeis.org/A014574&quot; rel=&quot;nofollow&quot;&gt;https://oeis.org/A014574&lt;/a&gt;&lt;br&gt;&#xA; 4. 107: Collatz conjecture from 27, &lt;a href=&quot;https://en.wikipedia.org/wiki/Collatz_conjecture#Examples&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Collatz_conjecture#Examples&lt;/a&gt;&lt;br&gt;&#xA; 5. 117 - times 2 minus 1, times 2 plus 1, repeat  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This gives&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wr 90 flku&lt;br&gt;&#xA; (87 114 57 48 102 108 107 117 in ASCII)&lt;br&gt;&#xA;&lt;br&gt;&#xA; And frankly I don't know what location this could be.  FLKU is the airport code for Kanyau, Zambia, but it would be written in uppercase.  Anyway, many 4-letter combinations are airport codes.  It is probably a coincidence.&#xA; 87 is more correct than 78, but &quot;Nr 90&quot; looks better than &quot;Wr 90&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;PS: I have to give credit to mdc32 who found most of the sequences before I did.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2014-11-01T00:10:34.403" LastActivityDate="2014-11-01T00:10:34.403" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="3293" PostTypeId="2" ParentId="3286" CreationDate="2014-10-25T19:59:16.343" Score="30" Body="&lt;p&gt;The solution for 7 pencils - without using the ends - was only recently discovered:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/1308.5164&quot; rel=&quot;noreferrer&quot;&gt;http://arxiv.org/abs/1308.5164&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/R6pEb.png&quot; alt=&quot;Seven pencils&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seven mutually touching infinite cylinders - Sándor Bozóki, Tsung-Lin Lee, Lajos Rónyai. It was presented at &quot;Gathering 4 Gardner&quot; in May 2014, because the original 7-cigarette version, allowing the ends to be used, was popularised by a Martin Gardner column in Scientific American 50 years ago.&lt;/p&gt;&#xA;" OwnerUserId="4233" LastEditorUserId="3993" LastEditDate="2014-10-25T22:50:24.083" LastActivityDate="2014-10-25T22:50:24.083" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3294" PostTypeId="2" ParentId="3257" CreationDate="2014-10-25T22:12:20.413" Score="5" Body="&lt;p&gt;As a first solution that doesn't out any Knights:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each day, Bob names one congressman and touches his right shoulder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If at the next joint meeting one more person is wearing a split uniform (B), the just-named person is outed as an Illuminati.  If a citizen or Knight was named, nothing happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This takes at most 99 days to uncover all Illuminati.  (After 99 days, if we have only changed the costumes of 31 people, the unnamed congressman must be the 32nd unknown Illuminati.)&lt;/p&gt;&#xA;" OwnerUserId="1706" LastActivityDate="2014-10-25T22:12:20.413" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3295" PostTypeId="2" ParentId="1987" CreationDate="2014-10-26T00:19:26.150" Score="0" Body="&lt;p&gt;Easy. One was born at one point in the day, one at another. Since they weren't born at the same time, they don't consider themselves twins.&lt;/p&gt;&#xA;" OwnerUserId="4238" LastActivityDate="2014-10-26T00:19:26.150" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3296" PostTypeId="2" ParentId="3215" CreationDate="2014-10-26T00:34:40.903" Score="120" Body="&lt;p&gt;You think for a while and remember that Argentina had adopted the metric system way back in the late 19th century, while the metrication process in Britain had started just a handful of years before the Falklands War.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The only things in the cell are a single metal chair bolted to the floor in the center of the room with a spotlight hanging from a cord&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You walk over to the chair and kneel down to get a closer look at the bolts fastening it to the ground. The bolts have the distinctive heads of metric screws.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But since the metrication process in Britain has been going on for a few years, you can't be certain that a metric screw indicates that you are in Argentina.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You look around your cell, it is a large room with high ceilings, a musty smell, and thick dust covering everything. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the significant dust buildup in the room, you know that this room has to be pretty old. You gauge that if you were in Britain, it is highly unlikely that the room was constructed with metric screws. &lt;/p&gt;&#xA;" OwnerUserId="4236" LastActivityDate="2014-10-26T00:34:40.903" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3297" PostTypeId="2" ParentId="382" CreationDate="2014-10-26T00:43:59.200" Score="40" Body="&lt;p&gt;A slightly simpler possibility:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I have a number in mind: it’s either 14 or 15, but I’m not saying which.  If you multiply together your number and its smallest prime factor, then roll an ordinary 6-sided die and add the result, will the total be at least as big as than the number I’m thinking of?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;n/a (since 1 has no prime factors)&lt;/li&gt;&#xA;&lt;li&gt;No.  (2*2=4; the maximum achievable total is 10.)&lt;/li&gt;&#xA;&lt;li&gt;Sometimes.  (3*3=9; so if you roll a six, it’s a “yes”, if you roll a four, it’ll be “no”).&lt;/li&gt;&#xA;&lt;li&gt;Don’t know.  (4*2=8; so if I’m thinking of 15, it’ll be “no”, but if I’m thinking of 14, there’s a chance of “yes”.)&lt;/li&gt;&#xA;&lt;li&gt;Yes.  (5*5=25.)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4234" LastActivityDate="2014-10-26T00:43:59.200" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3298" PostTypeId="1" AcceptedAnswerId="3322" CreationDate="2014-10-26T01:49:52.517" Score="4" ViewCount="338" Body="&lt;p&gt;Think of a Chinese Game Board, and how you can make jumps. Now think of actual checkers where you capture pieces. How many marbles can you place without a capture possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's one slight rule to make sure this is not easy (without this rule you could just fill every slot):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;You cannot place marbles in the outer rim, but you can, when the game starts, capture in that direction.&lt;/strong&gt; AKA: A marble can jump over another onto the outer rim, but you cannot place it there at the start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, it's important to note that you only use the big part, the outer triangles are unusable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; Any jump can capture, as long as you jump the same distance on the other side as the current distance, and there are no obstructions. Colour do not matter.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-10-26T20:11:31.637" LastActivityDate="2014-10-27T22:53:53.180" Title="Largest No-Capture Setup?" Tags="|geometry|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="3299" PostTypeId="2" ParentId="3215" CreationDate="2014-10-26T02:08:22.707" Score="47" Body="&lt;p&gt;Having learned from an ancient master the art of chronobarbology, you stroke your beard stubble and determine that precisely eight hours have passed since your abduction in Bogota, Colombia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Factoring in the time needed to transport you to and from the airport, and the maximum realistic speed of a 1980's private jet, you realize you cannot possibly be in London and must therefore be in Buenos Aires, land of the Martín Fierro.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or... you could just pass gas when the first interrogator steps into the room and observe whether his response is &quot;Blimey!&quot; or &quot;¡Uy!&quot;.&lt;/p&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2014-10-26T02:08:22.707" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3300" PostTypeId="1" AcceptedAnswerId="12295" CreationDate="2014-10-26T06:43:37.053" Score="13" ViewCount="1854" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;An old sailor, by the name of John, was in his attic with his bright young great-nephew Stewart, showing him all the old tools they used to use while at sea.&lt;/p&gt;&#xA;&lt;p&gt;Stewart was fascinated by all the trinkets. One item caught his eye, which was a giant compass like one would find on a table inside the ship. It was beautifully constructed, with hand-printed markings for north, east, south and west. Halfway between those markings were four more markings for the 45° (1/8) bearings of north-east, south-east, south-west, and north-west. Halfway again between those markings were eight markings for the 22.5° (1/16) bearings corresponding to north-north-east, east-north-east, and so on.&lt;/p&gt;&#xA;&lt;p&gt;But what was special about this giant compass is that the markings went further down. There were sixteen markings for each of the 1/32 bearings, starting from north-north-north-east. And between those were thirty-two markings for the 1/64 bearings, starting from north-north-north-north-east. And it also had the 1/128 bearings and the 1/256 bearings, and so on down until they were so small that Stewart couldn't see them anymore.&lt;/p&gt;&#xA;&lt;p&gt;Stewart asked his great-uncle, &amp;quot;Uncle John, how did you keep track of all these directions? Didn't you use degrees like we do today?&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;The uncle laughed. &amp;quot;Of course we used degrees. These directions are just for decoration on the compass. But we did learn how to interpret them. Each direction is named the combination of the two directions it's between. Just like north-east is between north and east, and east-north-east is between east and north-east.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;Stewart decided to challenge the old sailor on this. &amp;quot;So what would north-east-east-north-east be?&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;John immediately said, &amp;quot;5/32 from north. Check for yourself.&amp;quot; Stewart located the arrow pointing 5/32 of the way around the circle, and sure enough, it read north-east-east-north-east.&lt;/p&gt;&#xA;&lt;p&gt;Stewart then decided to ask, &amp;quot;What about east-east-east-north-north-east-north?&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;John said, &amp;quot;That's 9/64 from north. You can name any number of 'north's and 'east's as you like, and I can tell you what fraction it is.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;And north-north-north-east-east-east-east?&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;That's the same as the last one, 9/64 from north,&amp;quot; explained Stewart. &amp;quot;Only the number of norths and easts matters, not their orders.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;How about south-east-east-south?&amp;quot; asked Stewart.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;That's just south-east. You can't use a number of 'south's and 'east's with a common factor,&amp;quot; explained John. &amp;quot;They'd all be the same.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;Okay then,&amp;quot; said Stewart, frowning in thought. He then blurted out, &amp;quot;How about west-west-south-south-south-west-west-south-west-west-west-west-south-south-west-west-west?&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;John had his fingers out counting how many 'south's and 'west's his young pupil had uttered out (six and eleven, respectively). &amp;quot;That would be 351/512 from north.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;What about north-west-south-east?&amp;quot; asked Stewart smartly.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;It'll take more than that to outfox me,&amp;quot; laughed John. &amp;quot;You can only name directions with two of the four cardinal directions that are next to each other. But I assure you, I can tell you the direction of any combination of two adjacent cardinal directions you like.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;That's cool,&amp;quot; said Stewart, now turning his attention to a wind-up clock.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Is the old sailor's claim true? Can you come up with some combination of &amp;quot;north&amp;quot;s and &amp;quot;east&amp;quot;s (or some other combination of two adjacent cardinal points) that doesn't correspond to a point on the infinite compass rose?&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-08-24T23:30:28.407" Title="Directions on an infinite compass rose" Tags="|calculation-puzzle|" AnswerCount="2" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="3303" PostTypeId="1" AcceptedAnswerId="3311" CreationDate="2014-10-26T14:14:54.183" Score="14" ViewCount="12005" Body="&lt;p&gt;A pawn is placed on the lower left corner square of a standard 8 by 8 chessboard. A 'move' involves moving the pawn, where possible, either:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;one square to the right,&lt;/li&gt;&#xA;&lt;li&gt;one square up, or&lt;/li&gt;&#xA;&lt;li&gt;diagonally one square up and to the right.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Using these legitimate moves the pawn is to be moved along a path from the lower left square to the upper right square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Bo4cf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many such paths are there?&lt;/p&gt;&#xA;" OwnerUserId="2375" LastEditorUserId="8874" LastEditDate="2015-02-13T09:25:30.987" LastActivityDate="2021-09-22T18:20:01.237" Title="How many paths are there through a chess board?" Tags="|chess|combinatorics|checkerboard|" AnswerCount="5" CommentCount="8" ClosedDate="2015-02-14T14:37:08.480" ContentLicense="CC BY-SA 3.0" />
  <row Id="3305" PostTypeId="2" ParentId="3303" CreationDate="2014-10-26T15:01:16.940" Score="4" Body="&lt;p&gt;Based on your rules, I think the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 48639&lt;br&gt;&#xA; I did this in excel, adding up the three cells below, to the left, and diagonally to the lower left of each cell. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1664" LastEditorUserId="28" LastEditDate="2014-10-26T17:49:55.950" LastActivityDate="2014-10-26T17:49:55.950" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="3306" PostTypeId="1" AcceptedAnswerId="3309" CreationDate="2014-10-26T17:06:24.110" Score="24" ViewCount="9773" Body="&lt;p&gt;Using the letters &lt;strong&gt;M&lt;/strong&gt;, &lt;strong&gt;I&lt;/strong&gt;, and &lt;strong&gt;U&lt;/strong&gt;, which can be used to create words. Can you turn MI into MU using in each step one of the following  rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Insert a U to the end of a word ending in I. e.g: &lt;strong&gt;MI&lt;/strong&gt; becomes &lt;strong&gt;MIU&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Re-insert the letters after the M e.g: &lt;strong&gt;MIU&lt;/strong&gt; to &lt;strong&gt;MIUIU&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Replace the letters III with a U. e.g: &lt;strong&gt;MUIIIU&lt;/strong&gt; to &lt;strong&gt;MUUU&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Get rid of letters UU. For example: &lt;strong&gt;MUUU&lt;/strong&gt; to &lt;strong&gt;MU&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Can you use the 4 rules to change the word &lt;strong&gt;MI&lt;/strong&gt; into &lt;strong&gt;MU&lt;/strong&gt;? If so how many does it take?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; Please use spoilers.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="11876" LastEditDate="2016-08-15T02:25:02.513" LastActivityDate="2022-07-30T18:00:36.930" Title="Can you Turn MI to MU?" Tags="|strategy|game|letters|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3308" PostTypeId="1" AcceptedAnswerId="3310" CreationDate="2014-10-26T17:15:31.120" Score="-1" ViewCount="442" Body="&lt;p&gt;Imagine an 8x8 chessboard has two opposite corners taken out, with 62 squares left over. Is it possible to lay 31 dominoes with the size 2x1 so they cover all of the empty, left over, squares?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="8874" LastEditDate="2016-03-21T13:24:31.400" LastActivityDate="2016-03-21T13:24:31.400" Title="Dominoes on a Chess Board" Tags="|dissection|tiling|checkerboard|" AnswerCount="1" CommentCount="0" ClosedDate="2014-10-27T21:18:47.547" ContentLicense="CC BY-SA 3.0" />
  <row Id="3309" PostTypeId="2" ParentId="3306" CreationDate="2014-10-26T17:27:09.977" Score="21" Body="&lt;p&gt;This is the MU-Puzzle from one of my favourite books &lt;em&gt;Gödel, Escher, Bach&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution is here: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;No. It is impossible. The number of Is in the string will never be evenly divisible by 3. You need it to be so that rule 3 can reduce it to 0 and thus acheive the goal. We start with 1 I. We may only double it in the hopes of making it a multiple of 3 - no power of 2 is a multiple of 3 - nor is any number not already a multiple of 3 by any power of 2 divisible by 3. Since we start at 1, it cannot be done.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-10-28T18:35:52.953" LastActivityDate="2014-10-28T18:35:52.953" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3310" PostTypeId="2" ParentId="3308" CreationDate="2014-10-26T17:30:15.607" Score="4" Body="&lt;p&gt;This is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Mutilated_chessboard_problem&quot; rel=&quot;nofollow&quot;&gt;mutilated chess board problem&lt;/a&gt; - I apologise for ruining this cool puzzle which a quick answer, but I'm afraid it's well-known:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The puzzle is impossible to complete. A domino placed on the chessboard will always cover one white square and one black square. Therefore a collection of dominoes placed on the board will cover an equal numbers of squares of each colour. If the two white corners are removed from the board then 30 white squares and 32 black squares remain to be covered by dominoes, so this is impossible. If the two black corners are removed instead, then 32 white squares and 30 black squares remain, so it is again impossible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastActivityDate="2014-10-26T17:30:15.607" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3311" PostTypeId="2" ParentId="3303" CreationDate="2014-10-26T17:46:19.457" Score="19" Body="&lt;p&gt;Answer: 48639.&lt;/p&gt;&#xA;&lt;p&gt;Prove: Let's number moves like this:&lt;br /&gt;&#xA;A) one square to the right&lt;br /&gt;&#xA;B) one square up&lt;br /&gt;&#xA;C) diagonally one square up and to the right&lt;/p&gt;&#xA;&lt;p&gt;We can consider set of moves, that leads to the other corner. For example, CCCCCACB.&lt;br /&gt;&#xA;It is obvious that for any set like this the numbers of A-moves and B-moves must be always the same: &lt;span class=&quot;math-container&quot;&gt;$N_A=N_B$&lt;/span&gt; (Otherwise you will not end in the upper right corner). Also &lt;span class=&quot;math-container&quot;&gt;$N_A+N_B+2N_C = 14$&lt;/span&gt; (The maximum number of moves is 14 and C can take values from 0 to 7). And that is all limitations on the sets, so we need to find all sets of A,B and C, that satisfy to these two equations.&lt;/p&gt;&#xA;&lt;p&gt;Let's consider 8 cases, each corresponds to different &lt;span class=&quot;math-container&quot;&gt;$N_C$&lt;/span&gt; (from 0 to 7).&lt;br /&gt;&#xA;If &lt;span class=&quot;math-container&quot;&gt;$N_C = M$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$N_A=N_B=7-M$&lt;/span&gt; (A and B cannot take more than 7 steps each) and in total we have &lt;span class=&quot;math-container&quot;&gt;$14-M$&lt;/span&gt; moves. There are &lt;span class=&quot;math-container&quot;&gt;$14-M \choose M$&lt;/span&gt; possible placements for C-s and, once C are placed, &lt;span class=&quot;math-container&quot;&gt;$14-2M \choose 7-M$&lt;/span&gt; possible placements for B-s and, once B are placed too, only one possible placement for A-s. So we have &lt;span class=&quot;math-container&quot;&gt;${14-M \choose M} \cdot {14-2M \choose 7-M}$&lt;/span&gt; possible sets with &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; C-s.&lt;/p&gt;&#xA;&lt;p&gt;In total we will have:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\sum_{M=0}^{7}{14-M \choose M} \cdot {14-2M \choose 7-M} = $&lt;/span&gt;&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;${14 \choose 0}\cdot{14 \choose 7} + {13 \choose 1}\cdot{12 \choose 6} + {12 \choose 2}\cdot{10 \choose 5} +  {11 \choose 3}\cdot{8 \choose 4} + $&lt;/span&gt;&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$ + {10 \choose 4}\cdot{6 \choose 3} + {9 \choose 5}\cdot{4 \choose 2} + {8 \choose 6}\cdot{2 \choose 1} + {7 \choose 7}\cdot{0 \choose 0} = 48639$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="76793" LastEditDate="2021-09-22T18:20:01.237" LastActivityDate="2021-09-22T18:20:01.237" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="3312" PostTypeId="2" ParentId="3303" CreationDate="2014-10-26T19:13:00.850" Score="8" Body="&lt;h2&gt;Delannoy numbers&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The solution for a $n+1$ by $m+1$ grid is given by Delannoy numbers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D(n,m) = \Sigma_{d=0}^{n} \ ^mC_d \ ^{n+m-d}C_m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;note this assumes n&amp;lt;=m, we can easily express the dimensions either way round to achievement this&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Derivation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There are 3 possible moves, $(1,0), (0,1)$ or $(1,1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $n+1$ by $m+1$ board we must move a total of $(n,m)$ this is made using using any combination of the 3 moves such that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$num(1,0) +num(1,1) = n$&lt;br&gt;&#xA;$num(0,1) + num(1,1) = m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we can decide how many of $(1,0)$ and $(0,1)$ to use based on the numebr of $(1,1)$s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Starting with diagonal moves&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As n is the smaller dimension there are between 0 and n $(1,1)$s  (diagonal moves) in any solution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so lets look at how they can be arranged (ignoring the other entries) &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;for 0 $(1,1)$s there is 1 (or $\ ^nC_0$) possible arrangement&lt;/li&gt;&#xA;&lt;li&gt;for 1 $(1,1)$s there are  $\ ^nC_1$ possible arrangements  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and so on until&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;for n $(1,1)$s there are $\ ^nC_n$ possible arrangements&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;in generality  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;for $d$ (1,1)s there are $\ ^nC_{d}$ possible arrangements&lt;/li&gt;&#xA;&lt;li&gt;and d must fall between 0 and n (inclusive)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;so there are in total $\Sigma_{d=0}^{n} \ ^kC_d$ possible locations for moves of $(1,1)$ in an $n+1$ by $m+1$ grid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Now the non-diagonal moves&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each of these numbers of (1,1)s there is also many ways to arrange the (0,1)s and (1,0)s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 0 $(1,1)$s  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;there must be $n (1,0)$s out of a total number of moves of $n+m$  &lt;/li&gt;&#xA;&lt;li&gt;i.e. there must be $n (1,0)$s and $m (0,1)$s&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;so there is $\ ^{n+m}C_m $ possible arrangements&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for $1 (1,1)$  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;there must be $n-1 (1,0)$s out of a total number of positions of $n+m-2$,  &lt;/li&gt;&#xA;&lt;li&gt;i.e. there must be $n-1 (1,0)$s and $m-1 (0,1)$s  &lt;/li&gt;&#xA;&lt;li&gt;one position is already $(1,1)$ so there are $n+m-1$ locations that can be taken  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are $\ ^{n+m-1}C_m$ possible arrangements of $(0,1)$ and $(1,1)$ for each location of the one $(1,1)$.&lt;br&gt;&#xA;Giving $\ ^nC_1 \ ^{n+m-1}C_m $ arrangements possible for 1 diagonal move&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In generality for $d (1,1)$s&#xA;There are $\ ^nC_d \ ^{n+m-d}C_m $ possible arrangements&lt;br&gt;&#xA;where $d$ must again be between $0$ and $n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So summing over all $d$ gives the overall total number of paths and gives the formula given at the beginning &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D(n,m) = \Sigma_{d=0}^{n} \ ^mC_d \ ^{n+m-d}C_m$&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; for 8 by 8,  n=7 and m = 7 the solution as given by others is 48639&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4150" LastActivityDate="2014-10-26T19:13:00.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3313" PostTypeId="1" AcceptedAnswerId="3314" CreationDate="2014-10-26T20:22:53.087" Score="12" ViewCount="17751" Body="&lt;p&gt;You, a right cylinder glass of water, 3 buttons, instructions, and a large flat land that seems to go on forever, are the only things that exist in a universe. In this weird universe, the exact same physics apply. The 3 buttons are labeled &quot;Less&quot;, &quot;Half&quot;, and &quot;More&quot;. The instructions are as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hello fellow human, I have abducted you from your puny planet. I am done testing you, and have wiped your memory. You have just 1 chance of escape. You must tell if the glass is half full, less than half, or more than half with water. If you are wrong I will remove the ground and let you fall into an abyss forever. I labeled 3 buttons accordingly. I have also given you perfect sight and perfect balance/handling. Please note perfect sight only helps you with levelling; it does not help in originally finding half. Also you do not have any measuring instrument and you cannot spill out or add anything. Good luck... Not.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The glass appears to be half full, but how can you tell for sure?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="5373" LastEditDate="2015-03-25T09:18:39.537" LastActivityDate="2015-03-25T09:18:39.537" Title="Is the Glass Half Full with water?" Tags="|logical-deduction|" AnswerCount="7" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="3314" PostTypeId="2" ParentId="3313" CreationDate="2014-10-26T20:46:32.670" Score="34" Body="&lt;p&gt;Tilt the glass until the edge of the water is just touching the rim. Then see whether the highest point of the bottom of the glass is under, over or on the same level as the water.&lt;/p&gt;&#xA;" OwnerUserId="2021" LastActivityDate="2014-10-26T20:46:32.670" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="3315" PostTypeId="1" AcceptedAnswerId="3317" CreationDate="2014-10-26T21:19:02.083" Score="2" ViewCount="16131" Body="&lt;p&gt;Can you arrange 7 trees so that there are 6 rows of 3 trees? It is entirely possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note: A tree can be part of more then one row, for example a if you arranged a 3x3 square (9 trees) the tree in the center is part of 4 rows.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rows are usually horizontal but in this case they can be horizontal, vertical, or diagonal.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="71" LastEditDate="2014-11-03T19:41:56.247" LastActivityDate="2022-08-29T02:58:57.003" Title="7 Trees, 6 Rows, 3 Per Row?" Tags="|geometry|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3316" PostTypeId="2" ParentId="3315" CreationDate="2014-10-26T21:31:57.543" Score="18" Body="&lt;p&gt;An equilateral triangle with 3 trees at the corners, 3 trees at the midpoints and one tree in the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each side has 3 trees. Each altitude has 3 trees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lpDLM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="375" LastEditorUserId="375" LastEditDate="2014-10-27T03:22:59.580" LastActivityDate="2014-10-27T03:22:59.580" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3317" PostTypeId="2" ParentId="3315" CreationDate="2014-10-26T21:36:48.377" Score="8" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JsbDZ.png&quot; alt=&quot;7 trees - 6 lines&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Joel Rondeau's solution would look like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    1&#xA;   2 3&#xA;    4&#xA; 5  6  7&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;With lines: 125, 137, 146, 247, 345, 567. A more symmetric solution than mine)&lt;/p&gt;&#xA;" OwnerUserId="4257" LastEditorUserId="4257" LastEditDate="2014-10-26T22:18:07.113" LastActivityDate="2014-10-26T22:18:07.113" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3318" PostTypeId="2" ParentId="3313" CreationDate="2014-10-26T22:10:41.460" Score="10" Body="&lt;p&gt;We are dealing with a power that can abduct you from your planet, wipe your memory and create a highly specialized environment. These powers clearly have access to technology that is superior to the state of the art of our technology. That means that this power will most likely be much better at measuring than your puny eyes, especially if they are combined with your human hands. Even human technology is better than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would find it likely that the power could make the cylinder half full except for a couple of atoms. In which case you might be tempted to press &quot;half&quot; when the correct answer is &quot;less&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That makes user2021's answer not strictly correct (or the puzzle malformed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless someone comes up with an answer that deals with the couple-of-atoms-short situation, then the correct answer is: No, you cannot tell for sure in all situations.&lt;/p&gt;&#xA;" OwnerUserId="4257" LastEditorUserId="4257" LastEditDate="2014-10-26T22:16:43.573" LastActivityDate="2014-10-26T22:16:43.573" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3320" PostTypeId="2" ParentId="3313" CreationDate="2014-10-26T22:40:20.170" Score="33" Body="&lt;p&gt;After checking that I can breathe in the experiment room, I can confidently assert that the glass is more than half-full, either with water or air.&lt;/p&gt;&#xA;" OwnerUserId="4260" LastActivityDate="2014-10-26T22:40:20.170" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3321" PostTypeId="1" AcceptedAnswerId="3367" CreationDate="2014-10-26T22:58:55.383" Score="4" ViewCount="6226" Body="&lt;p&gt;This puzzle is taken directly from a site and reworded. The site can be found at the bottom of this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are king of an empire and you are going to have a party soon. The party is the most important party you've ever hosted. You've got 1000 bottles of wine you were planning to use for the party, but you found out that one of them was poisoned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The poison shows no effects until death. Death occurs within ten to twenty hours after drinking even the smallest amount of poison.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have thousands of slaves at your disposal and just under 24 hours to determine which bottle has been poisoned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a few prisoners soon to be executed, and it would ruin your celebration to have anyone else killed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the least number of prisoners you must have to drink from the bottles to be confident that you found the poisoned wine in 24 hours?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;No googling!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The site this was taken from was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.folj.com/puzzles/very-difficult-analytical-puzzles.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.folj.com/puzzles/very-difficult-analytical-puzzles.htm&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-10-26T23:34:47.810" LastActivityDate="2014-10-28T08:41:11.187" Title="Poisoned Bottle of Wine!" Tags="|logical-deduction|" AnswerCount="1" CommentCount="1" ClosedDate="2014-10-29T22:24:19.580" ContentLicense="CC BY-SA 3.0" />
  <row Id="3322" PostTypeId="2" ParentId="3298" CreationDate="2014-10-26T23:06:35.007" Score="4" Body="&lt;p&gt;You can put maximum 13 marbles.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/4DhRi.png&quot; alt=&quot;solution&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first thing to note is that you cannot have adjacent pieces.  If you have a line of 2 or more adjacent pieces, it has to stop somewhere and it is not a border because the rim is empty. That means the next-to last in a line can always capture the last one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without any 2 pieces adjacent, the smallest distance between pieces is as in the solution above.  This means that arrangement is the densest possible.  The only chance of improvement is a better use of the borders.  But the borders are quite well filled already.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2014-10-27T22:53:53.180" LastActivityDate="2014-10-27T22:53:53.180" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3323" PostTypeId="2" ParentId="3306" CreationDate="2014-10-26T23:15:04.537" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Start by ignoring the Us. Let NI be the number of Is. The only rule that will get rid of Is is rule 3 and it only works on triplets. So we need to have NI become 3*n somehow as we need all Is gone. Rule 1 will not change NI. Rule 2 will double NI, but doubling a number will never make it divisible by 3 unless it was so in the first place, so that is not going to help us either (this is easy to prove if you compute modulo 3). Rule 3 will only make NI divisible by 3 if it was so in the first place. Rule 4 only deals with Us. So given that NI can never become divisible by 3 you can never get rid of all the Is =&gt; Impossible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4257" LastActivityDate="2014-10-26T23:15:04.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3326" PostTypeId="2" ParentId="3315" CreationDate="2014-10-27T00:24:23.587" Score="8" Body="&lt;p&gt;this is my solution...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and this is done by &lt;strong&gt;euler line&lt;/strong&gt; which i studied in high school.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Euler_line&quot; rel=&quot;noreferrer&quot;&gt;euler line&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1c6ay.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2375" LastActivityDate="2014-10-27T00:24:23.587" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3329" PostTypeId="2" ParentId="3313" CreationDate="2014-10-27T04:31:31.743" Score="17" Body="&lt;p&gt;Since the same laws of physics apply, I would expect capillary action to produce a concave meniscus at the edge where the water surface touches the glass. If the glass is very nearly half-filled, the tilting method proposed by &lt;a href=&quot;https://puzzling.stackexchange.com/a/3314/4145&quot;&gt;user2021&lt;/a&gt; would not be able to account for this effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way to be sure is to WAIT, since there is no time limit, and let evaporation takes place till you are confident enough to answer LESS.&lt;/p&gt;&#xA;" OwnerUserId="4145" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2014-10-27T04:31:31.743" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3331" PostTypeId="2" ParentId="3215" CreationDate="2014-10-27T07:08:11.113" Score="23" Body="&lt;p&gt;Unscrew the light bulb from the lamp, if it's bayonet style, you're &lt;a href=&quot;http://en.wikipedia.org/wiki/Edison_screw&quot;&gt;guaranteed to be in the UK&lt;/a&gt; and &lt;a href=&quot;https://www.google.com.ar/search?q=lamparas%20incandescentes&amp;amp;biw=1440&amp;amp;bih=718&amp;amp;source=lnms&amp;amp;tbm=isch&amp;amp;sa=X&amp;amp;ei=yOJOVOGyMsW1oQTXnILIBg&amp;amp;ved=0CAYQ_AUoAQ#imgdii=_&quot;&gt;not Argentina (with light bulbs inside of cars being the only exception)&lt;/a&gt;, if it's screw-style/Edison style, then you're most likely in Argentina (although, that's not a 100% guarantee because some Edison style light bulbs can be found in the UK despite the fact that bayonet style light bulbs predominate in the UK). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't even need to turn off the light to find this out. A small twist should give you enough information without cutting off the electricity to the bulb. Just make sure not to burn yourself while doing this, use some of your clothes to insulate your hand. &lt;/p&gt;&#xA;" OwnerUserId="4267" LastEditorUserId="4267" LastEditDate="2014-10-28T00:36:57.660" LastActivityDate="2014-10-28T00:36:57.660" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3333" PostTypeId="1" AcceptedAnswerId="3335" CreationDate="2014-10-27T08:41:45.160" Score="2" ViewCount="9654" Body="&lt;p&gt;You have two equal glasses. One is filled with water. One is filled with wine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you take a spoonful from the water glass, put it into the wine glass, and stir until it is well mixed. Then you take a spoonful from the wine glass, put it into the water glass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, is there more water in the wine glass than wine in the water glass?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Please note: I am not a molecular physicist.)&lt;/p&gt;&#xA;" OwnerUserId="2544" LastActivityDate="2014-10-27T09:18:35.533" Title="Water glass and wine glass" Tags="|logical-deduction|" AnswerCount="1" CommentCount="2" ClosedDate="2014-10-27T10:53:21.803" ContentLicense="CC BY-SA 3.0" />
  <row Id="3334" PostTypeId="2" ParentId="3303" CreationDate="2014-10-27T09:14:47.220" Score="5" Body="&lt;p&gt;Yes, the answer is 48639. I found it using a simple C++ code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;int move(int a=1,int b=1)&#xA;{&#xA;    if((a==8)||(b==8))&#xA;        return 1;&#xA;    return move(a+1,b)+move(a,b+1)+move(a+1,b+1);&#xA;}&#xA;int main()&#xA;{&#xA;    cout&amp;lt;&amp;lt;move();&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4269" LastEditDate="2015-01-12T09:45:21.493" LastActivityDate="2015-01-12T09:45:21.493" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3335" PostTypeId="2" ParentId="3333" CreationDate="2014-10-27T09:18:35.533" Score="2" Body="&lt;p&gt;Counter-intuitively, there are exactly equal amounts. We can see this if we consider that each glass has lost and gained the same amount of liquid (a spoonful), so any loss to the other glass is precisely offset by gain from that glass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless, of course, the &quot;&lt;strong&gt;filled&lt;/strong&gt; with water/wine&quot; statements were to be taken literally, in which case the water glass will have more wine because the wine glass lost some when it overflowed.&lt;/p&gt;&#xA;" OwnerUserId="3987" LastActivityDate="2014-10-27T09:18:35.533" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3336" PostTypeId="2" ParentId="3303" CreationDate="2014-10-27T11:31:26.157" Score="2" Body="&lt;p&gt;Here's a simple python script that solves the problem:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;def path_count(rows, cols):&#xA;    if rows == 1 or cols == 1:&#xA;        return 1&#xA;    return path_count(rows-1, cols) + path_count(rows, cols-1) + path_count(rows-1, cols-1)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The idea behind this is that there are three possible first moves: up, right, and diagonally. For each first move, count the number of ways from the second square to the goal. This is equivalent to solving the original problem on a smaller board. The number of paths with up as the first move is equal to the number of paths on a 7-row by 8-column board. This lends itself to recursion since the board is always getting smaller with every move. If the board shrinks to one row or one column, then there's only one path, straight across or straight up to the goal. The solution is the number of paths with first move up, plus the number with first move right, plus the number with first move diagonal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;path_count(8,8) returns 48,639.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I am well-aware that the function above would need to be optimized to handle much larger boards. For 8x8, the run time is less than a second. For 20x20, I stopped the program after several minutes.&lt;/p&gt;&#xA;" OwnerDisplayName="user4221" LastActivityDate="2014-10-27T11:31:26.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3337" PostTypeId="1" AcceptedAnswerId="3444" CreationDate="2014-10-27T11:38:19.900" Score="11" ViewCount="483" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;On a NxN chess board $N^2$ knights are placed (one per cell). Each pair of knights, who control each other (i.e. one move away from each other) are friends.&lt;br&gt;&#xA;  One day all knights are promoted to kings. Is it possible to put them on the board (in different order) that all friends will be similarly one move away from each other?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For N = 1 or 2 it is trivial - there are no friends, so you don't have to change anything.&lt;br&gt;&#xA; It is quite easy to do with N = 3. If you number them like this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 2 3&lt;br&gt;&#xA;4 5 6&lt;br&gt;&#xA;7 8 9  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;then 5-th knight has no friends and all others have 2 friends exacly. So we just place them friend-to-friend in a circle around the 5-th one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 8 3&lt;br&gt;&#xA;6 5 4&lt;br&gt;&#xA;7 2 9   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and it is done.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about N &gt; 3?&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2014-10-27T11:46:51.910" LastActivityDate="2014-10-31T07:24:04.290" Title="From knights to kings" Tags="|chess|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3338" PostTypeId="2" ParentId="3337" CreationDate="2014-10-27T12:18:30.483" Score="11" Body="&lt;p&gt;It is trivial to prove that on any board where two knights have 8 friends while they are friends, there is no solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to visualize it (on a 7x7 board):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;. a . a . . .&#xA;a . . b a b .&#xA;. . A . . . b&#xA;a . . . B . .&#xA;. a b a . . b&#xA;. . . b . b .&#xA;. . . . . . .&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here, all friends of A are marked a, all friends of B are marked b. And A and B are friends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if I place A and B on a new board as king-friends, and I place A's 7 friends around A, there never is room for all of B's friends around B:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a a a b . . .           a a a . . . .&#xA;a A B b . . .           a A a b . . .&#xA;a a a b . . .           a a B b . . .&#xA;. . . . . . .           . b b b . . .&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;On the left, B cannot have more than 3 friends, and on the right, he can have 5 (apart from A), still two short of the seven we wanted to place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, the situation of two friendly knights with 7 mutually exclusive friends appears on any board with N &gt;= 7, so for N&gt;=7, the answer is: &lt;em&gt;no it cannot be done&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me ponder about N=4, 5 or 6 (I suspect the limit is lower than 6 because I am &lt;em&gt;two&lt;/em&gt; squared short for friends at N=7. But I'm not sure.)&lt;/p&gt;&#xA;" OwnerUserId="3903" LastActivityDate="2014-10-27T12:18:30.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3339" PostTypeId="2" ParentId="1833" CreationDate="2014-10-27T12:20:22.760" Score="2" Body="&lt;p&gt;We have an amount of red sand $R$ and an amount of blue sand $U$. $R$ does not have to be equal to $U$.&lt;br&gt;&#xA;We pour our sand in two beakers, so that beaker A contains all the red sand and beaker B contains all the blue sand. This means we have a volume $A$ in beaker A equal to the volume of red sand $R$, and volume $B$ in beaker B equal to the volume of blue sand $U$.&#xA;$$\begin{align}&#xA;A &amp;amp; = R \tag{1} \\&#xA;B &amp;amp; = U \tag{2}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First we move an amount of red sand $r_1 \le R$ from beaker A to beaker B. The (now changed) volumes of the beakers are:&#xA;$$\begin{align}&#xA;A^\prime &amp;amp; = R - r_1\\&#xA;B^\prime &amp;amp; = U +r_1&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we move an amount of sand back from beaker B to beaker A &lt;em&gt;restoring them to their original volumes&lt;/em&gt;. The moved sand may include some blue sand $u \le U$ and some red sand $r_2 \le r_1$.&#xA;$$\begin{align}&#xA;A &amp;amp; = R - r_1 + u + r_2 \\&#xA;B &amp;amp; = U + r_1 - u - r_2&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we define the total amount of red sand displaced, $r$, as $r = r_1 - r_2$ (the amount of red sand moved initially minus the amount of red sand moved back on the second move), we see that:&#xA;$$\begin{align}&#xA;A &amp;amp; = R - r + u \\&#xA;B &amp;amp; = U + r - u &#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now we remember $(1)$ and $(2)$.&#xA;$$\begin{align}&#xA;R &amp;amp; = R - r + u \\&#xA;U &amp;amp; = U + r - u \\&#xA;&amp;amp; \implies \\&#xA;r &amp;amp; =  u \end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we see that, &lt;em&gt;as long as we restore both to their original volumes&lt;/em&gt;, the amount of red sand in beaker B is the same as the amount of blue sand in beaker A.&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditDate="2017-01-20T15:44:28.497" LastActivityDate="2017-01-20T15:44:28.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3342" PostTypeId="1" AcceptedAnswerId="3343" CreationDate="2014-10-27T13:33:59.010" Score="8" ViewCount="3282" Body="&lt;p&gt;James is in a room with three other people: Alice, Bob, and Spencer. Alice and Bob are married to each other, but James knows that Bob is secretly having an affair with Spencer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;James is going to announce the identity of his &quot;baby,&quot; the person he loves the most in the whole world. He remarks, truthfully, about the occupants of the room: &quot;&lt;strong&gt;Everybody loves my baby, but my baby doesn't love anybody but me.&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which person is James's baby?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To formalize definitions: we assume the love is a directional relationship &lt;code&gt;P=&amp;gt;Q&lt;/code&gt; where each endpoint &lt;code&gt;P&lt;/code&gt; and &lt;code&gt;Q&lt;/code&gt; represents some person in the room. We allow the possibility that a person may have any number of incoming and outgoing love relationships concurrently. (So, &lt;code&gt;A&lt;/code&gt; may love &lt;code&gt;B&lt;/code&gt;, but &lt;code&gt;B&lt;/code&gt; might not love &lt;code&gt;A&lt;/code&gt; back, and &lt;code&gt;A&lt;/code&gt; may love &lt;code&gt;C&lt;/code&gt; while still also loving &lt;code&gt;B&lt;/code&gt; as well.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I personally discovered this curiosity in &lt;a href=&quot;https://en.wikipedia.org/wiki/The_Cuckoo&amp;#39;s_Egg&quot;&gt;&lt;em&gt;The Cuckoo's Egg&lt;/em&gt; by Clifford Stoll&lt;/a&gt;, but it certainly predates that work, since the song &quot;Everybody Loves My Baby&quot; was written several decades prior, in 1924.)&lt;/p&gt;&#xA;" OwnerUserId="2478" LastEditorUserId="2478" LastEditDate="2014-10-27T19:39:11.783" LastActivityDate="2014-10-27T19:39:11.783" Title="Everybody loves my baby" Tags="|logical-deduction|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3343" PostTypeId="2" ParentId="3342" CreationDate="2014-10-27T13:38:04.853" Score="16" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; James' baby is James&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If &lt;em&gt;everybody&lt;/em&gt; loves my baby, then my baby loves my baby. Since my baby don't love anybody but me, it must be me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, it's pretty clear that it's not one of the others, since you make absolutely no distinction between them. It would just be a random guess to choose one of them. To make this more &quot;difficult&quot;, you could give each a line of back story to distract/mislead the readers.&lt;/p&gt;&#xA;" OwnerUserId="1563" LastEditorUserId="4292" LastEditDate="2014-10-27T18:29:47.890" LastActivityDate="2014-10-27T18:29:47.890" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="3344" PostTypeId="2" ParentId="3300" CreationDate="2014-10-27T14:34:17.700" Score="2" Body="&lt;p&gt;The old sailor is correct!  According to our notation, that ONLY depends on the number of N's, E's, S's or W's, we can ALWAYS create another &quot;finer&quot; degree of precision.  I will illustrate why...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we have to think about the problem in a slightly different way.  Consider that instead of drawing a circular compass rose we are drawing points on a graph.  The coordinates of the points correspond to the number of instances of the particular direction.  Also, for the purposes of this illustration, we assume that South is just negative North and West is negative East.  So for example, (1,0) corresponds to E, (1,2) corresponds to NNE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can continue to create more points by simply going through our list and adding the coordinates of two existing adjacent points.  Here are a few sample iterations...&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Set1: (1,0)                                           (0,1)&#xA;Set2: (1,0)                   (1,1)                   (0,1)&#xA;Set3: (1,0)       (2,1)       (1,1)       (1,2)       (0,1)&#xA;Set4: (1,0) (3,1) (2,1) (3,2) (1,1) (2,3) (1,2) (1,3) (0,1)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Notice anything?  The sets we create are every unique combination of integers that isn't perfectly divisible by a lower combination of integers (i.e. no common factors).  Furthermore, since there are an infinite set of prime numbers, there must be an infinite number of combinations and an infinite number of markings on the infinite compass rose :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if we take the ratios of these integers (i.e. the slopes of the lines created by (0,0) and the respective point), we get the direction of the arrows on the compass rose.  Since the ratio directly corresponds to slope which directly corresponds to the direction on the compass rose, no two combinations are allowed to have the same mix of N's or E's.  Therefore, every combination corresponds to one and only one direction and order does NOT matter.&lt;/p&gt;&#xA;" OwnerUserId="4191" LastActivityDate="2014-10-27T14:34:17.700" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3347" PostTypeId="2" ParentId="3215" CreationDate="2014-10-27T17:12:35.903" Score="11" Body="&lt;p&gt;Another lighting related one. Peel back a bit of the outer insulation on the light flex. If the wires inside are coloured black and red, you're in Britain, which had its own set of standards and didn't switch to IEC60446 (blue/brown) until the early 21st century, long after the Falklands conflict. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be careful, mind you, or it's all going to be a bit academic.&lt;/p&gt;&#xA;" OwnerUserId="1801" LastActivityDate="2014-10-27T17:12:35.903" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3350" PostTypeId="1" AcceptedAnswerId="3351" CreationDate="2014-10-27T20:39:07.513" Score="2" ViewCount="1152" Body="&lt;p&gt;You have a bar of vibranium and the Cap needs to use it (not used up or consumed) every day but he only needs 1/7th on the first day, 2/7th on the second day and so on for 7 days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a vibranium knife that can cut through it for the Cap; however the knife is only good to make 2 cuts. You are very good and can cut perfectly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can you cut the vibranium bar into 3 pieces so that the Cap can get the perfect amount of vibranium for each day from 1 to 7?&lt;/p&gt;&#xA;" OwnerUserId="4299" LastEditorUserId="5373" LastEditDate="2015-03-08T11:49:40.730" LastActivityDate="2015-03-08T11:49:40.730" Title="How to divide the vibranium bar into 3 pieces?" Tags="|logical-deduction|" AnswerCount="1" CommentCount="4" ClosedDate="2014-10-28T14:06:02.963" ContentLicense="CC BY-SA 3.0" />
  <row Id="3351" PostTypeId="2" ParentId="3350" CreationDate="2014-10-27T20:41:19.143" Score="11" Body="&lt;p&gt;Make two cuts, leaving pieces of $1/7, 2/7, 4/7$  You can select pieces to get the amount required for each day.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2014-10-27T20:41:19.143" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3354" PostTypeId="2" ParentId="3313" CreationDate="2014-10-27T22:20:40.387" Score="3" Body="&lt;p&gt;Since you have the instructions, take the instructions and place them against the opening of the cup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, tilt the cup until it is horizontal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the cup from the side so what you are looking at looks like a  circle with the water level acting like a chord. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having perfect eyesight, determine if the chord is a diameter, using your perfect balance, balance the water so that it is perfectly flat and check to see that the line drawn on the bottom of the cup is a diameter. If it is, it is half full. If it is below that diameter (the diameter will be parallel to the flat land given that physics calls for gravity), then the cup is less than half full. If it the water level line is drawn above the diameter then it is more than half full.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No adding or removing of the liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The liberty I am taking is that the instructions are perfectly flat and does not absorb water. To specify it even more, the instructions act as a &quot;second bottom&quot;, i.e. a top, for the right cylinder.&lt;/p&gt;&#xA;" OwnerUserId="2763" LastActivityDate="2014-10-27T22:20:40.387" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3357" PostTypeId="2" ParentId="3215" CreationDate="2014-10-27T23:06:47.637" Score="9" Body="&lt;p&gt;Because you are so far underground, you have the unique position to feel any earth tremors in the area. Having visited the Andes multiple times, the chance of feeling an earthquake are good, but I believe that going further underground allows one to sense any earth tremors that would be imperceptible on the Earth's surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as you went &quot;asleep&quot; for a couple hours, you were actually concentrating to feel any such movements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you did, you were instantly sure that you were in Argentina and not in Britain.&lt;/p&gt;&#xA;" OwnerUserId="2763" LastActivityDate="2014-10-27T23:06:47.637" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3358" PostTypeId="1" AcceptedAnswerId="3359" CreationDate="2014-10-27T23:30:45.837" Score="5" ViewCount="18350" Body="&lt;p&gt;The following are tricks with numbers that you must explain how they work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Trick #1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Think of a number between 1 and 9&lt;/li&gt;&#xA;&lt;li&gt;Multiply by 9&lt;/li&gt;&#xA;&lt;li&gt;Add up all of the digits of this number&lt;/li&gt;&#xA;&lt;li&gt;Divide by 3&lt;/li&gt;&#xA;&lt;li&gt;Subtract 2&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Your answer? 1 of course!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Trick #2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Think of any positive number!&lt;/li&gt;&#xA;&lt;li&gt;Double it.&lt;/li&gt;&#xA;&lt;li&gt;Add 2.&lt;/li&gt;&#xA;&lt;li&gt;Half it.&lt;/li&gt;&#xA;&lt;li&gt;Add 3.&lt;/li&gt;&#xA;&lt;li&gt;Subtract your original number.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Your answer is obvious, it's 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Trick #3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Pick any positive 3 digit number in the universe.&lt;/li&gt;&#xA;&lt;li&gt;Multiply by 7.&lt;/li&gt;&#xA;&lt;li&gt;Multiply by 11.&lt;/li&gt;&#xA;&lt;li&gt;Multiply by 13.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Your answer is your original number twice :O&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Trick #4&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Think of a number between 1 and 9.&lt;/li&gt;&#xA;&lt;li&gt;Double it.&lt;/li&gt;&#xA;&lt;li&gt;Add 5.&lt;/li&gt;&#xA;&lt;li&gt;Multiply by 5.&lt;/li&gt;&#xA;&lt;li&gt;Add another digit between 1 and 9 to it.&lt;/li&gt;&#xA;&lt;li&gt;Subtract 25.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The first digit of your new number was your original number. The second is your second number.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2484" LastEditDate="2014-10-28T11:14:28.017" LastActivityDate="2017-05-07T07:14:04.160" Title="Explain the Magic Numbers!" Tags="|mathematics|" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3359" PostTypeId="2" ParentId="3358" CreationDate="2014-10-27T23:44:58.907" Score="10" Body="&lt;ol&gt;&#xA;&lt;li&gt;Digits of any number divisible by 9 always add up to 9.&lt;/li&gt;&#xA;&lt;li&gt;It's actually always 4. Steps 1 and 2 bring the total to n+1, then you add 3 and subtract n, leaving you with 4.&lt;/li&gt;&#xA;&lt;li&gt;The numbers are actually 7, 11, and 13, because multiplying these is 1001. A three digit number times 1001 is the same as taking the number with 3 0's after, then add the number.&lt;/li&gt;&#xA;&lt;li&gt;Steps 1 through 3 give us 2n+5, then 4, 5, and 6 give us 10n + 1m.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1664" LastActivityDate="2014-10-27T23:44:58.907" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3361" PostTypeId="1" AcceptedAnswerId="3362" CreationDate="2014-10-28T02:21:45.230" Score="38" ViewCount="3738" Body="&lt;p&gt;Considering from $a$ until $z$:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = (x -a) \cdot (x -b) \cdot (x -c)   \dots (x - z)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I want is the value of $x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It´s something easy, but I could not find this here and I think it´s a nice puzzle.&lt;/p&gt;&#xA;" OwnerUserId="1675" LastEditorUserId="463" LastEditDate="2015-06-10T01:32:08.343" LastActivityDate="2015-06-10T01:32:08.343" Title="What is the result of the equation?" Tags="|logical-deduction|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="3362" PostTypeId="2" ParentId="3361" CreationDate="2014-10-28T03:52:32.087" Score="39" Body="&lt;p&gt;if $x = \textrm{anything}\cdot(x-x)$ then $x$ has to be $0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without knowing the answer, if you start with an assumed value for each of the variables, regardless of the value you assume for any of the variables when you evaluate $x-x$ this would become $0$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So regardless of other variables, if $x = \textrm{anything}\cdot(x-x)$ then $x$ has to be $0$&lt;/p&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="-1" LastEditDate="2014-10-30T03:16:51.253" LastActivityDate="2014-10-30T03:16:51.253" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3363" PostTypeId="2" ParentId="3361" CreationDate="2014-10-28T03:52:34.730" Score="41" Body="&lt;p&gt;The third from last factor is $x-x$, so the right side is $0$, so $x=0$&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2014-10-28T03:52:34.730" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3365" PostTypeId="2" ParentId="3289" CreationDate="2014-10-28T06:37:14.383" Score="11" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The man who violently killed Scott's wife and attempted to kill Scott is currently incarcerated in a Baltimore correctional facility.&#xA; In four years, Scott will need to leave Alaska to testify at this man's parole hearing, lest the man be released and come after him.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4236" LastActivityDate="2014-10-28T06:37:14.383" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3367" PostTypeId="2" ParentId="3321" CreationDate="2014-10-28T08:41:11.187" Score="11" Body="&lt;p&gt;10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you know that only one bottle is poisoned, you can solve this using binary logic.&#xA;First, you assign every prisoner a number, from 0 to 9. Then, you give every bottle a number, from 0 to 999. If you take the binary representation of the bottle numbers, you get:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;00 0000 0000&#xA;00 0000 0001&#xA;00 0000 0010&#xA;...&#xA;11 1110 0111&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Every prisoner is assigned the bit corresponding to their number for each bottle. So prisoner 1 gets the $2^0$ bit, prisoner 2 gets the $2^1$ bit, prisoner 10 gets the most significant bit ($2^9$). The slaves the take a bottle each, and feed a drop to every prisoner that has a 1-bit for the bottle number. So p0 would drink from bottle 1, 3, 5, ... 999; while p9 would drink from bottle 512 and every higher bottle. Nobody drinks from bottle 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prisoners drink from about 500 bottles each (a bit less for p9), which, with the help of 1000 slaves, should be possible in under 4 hours. That leaves enough time to see who dies. 500 drops is about 25 ml of wine (very rough guess) which doesn't even leave the prisoners inebriated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If/when prisoners die, it should be easy to see which bottle is poisoned. There will be exactly one bottle that every prisoner that died, but no living prisoner, drank from. It can be found by placing a 1 for every dead prisoner and a 0 for every living one in their assigned bit order. (So if 9, 4, 3 and 1 die, bottle 10 0001 1010 is poisoned. If nobody dies, 0 is poisoned.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if two bottles are poisoned, or one slave forgets to feed a prisoner or feeds one too many, by accident or because he's tired of your rule, you're done for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: the &quot;nobody drinks from 0&quot; rule is not necessary, since we have some unused binary numbers in the 0-1023 range, but I don't like senseless killing.)&lt;/p&gt;&#xA;" OwnerUserId="439" LastActivityDate="2014-10-28T08:41:11.187" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3368" PostTypeId="2" ParentId="3358" CreationDate="2014-10-28T11:02:37.637" Score="12" Body="&lt;h2&gt;Trick 1&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Think of a number between 1 and 9&lt;/li&gt;&#xA;  &lt;li&gt;Multiply by 9&lt;/li&gt;&#xA;  &lt;li&gt;Add up all of the digits of this number&lt;/li&gt;&#xA;  &lt;li&gt;Divide by 3&lt;/li&gt;&#xA;  &lt;li&gt;Subtract 2&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;x_1 &amp;amp; \in\left[1,9\right] \\&#xA;x_2 &amp;amp; = 9 \times x_1 \implies x_2 \in \left\{9,18,27,36,45,54,63,72,81\right\} \\&#xA;x_3 &amp;amp; \in \left\{9,1+8,2+7,3+6,4+5,5+4,6+3,7+2,8+1\right\} \implies x_3 = 9 \\&#xA;x_4 &amp;amp; = \tfrac{9}{3} = 3 \\&#xA;x_5 &amp;amp; = 3-2 = 1&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Your answer? 1 of course!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Trick 2&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Think of any positive number!&lt;/li&gt;&#xA;  &lt;li&gt;Double it.&lt;/li&gt;&#xA;  &lt;li&gt;Add 2.&lt;/li&gt;&#xA;  &lt;li&gt;Half it.&lt;/li&gt;&#xA;  &lt;li&gt;Add 3.&lt;/li&gt;&#xA;  &lt;li&gt;Subtract your original number.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;x_1 &amp;amp; \in \mathbb Z_{\ge 0} \\&#xA;x_2 &amp;amp; = 2x_1 \\&#xA;x_3 &amp;amp; = 2x_1 + 2 \\&#xA;x_4 &amp;amp; = x_1 + 1 \\&#xA;x_5 &amp;amp; = x_1 + 4 \\&#xA;x_6 &amp;amp; = 4&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Your answer is obvious, it's 4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Trick 3&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Pick any positive 3 digit number in the universe.&lt;/li&gt;&#xA;  &lt;li&gt;Multiply by 7.&lt;/li&gt;&#xA;  &lt;li&gt;Multiply by 11.&lt;/li&gt;&#xA;  &lt;li&gt;Multiply by 13.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;x_1 &amp;amp; = 100a + 10b + c\quad|\, a \in\left[1,9\right]\,\, b,c \in\left[0,9\right] \\&#xA;x_2 &amp;amp; = 700a + 70b + 7c \\&#xA;x_3 &amp;amp; = 7700a + 770b + 77c \\&#xA;x_4 &amp;amp; = 100100a + 10010b + 1001c \\&#xA;x_4 &amp;amp; = 100000a + 10000b + 1000c + 100a + 10b + c\\&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Your answer is your original number twice&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Trick 4&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Think of a number between 1 and 9.&lt;/li&gt;&#xA;  &lt;li&gt;Double it.&lt;/li&gt;&#xA;  &lt;li&gt;Add 5.&lt;/li&gt;&#xA;  &lt;li&gt;Multiply by 5.&lt;/li&gt;&#xA;  &lt;li&gt;Add another digit between 1 and 9 to it.&lt;/li&gt;&#xA;  &lt;li&gt;Subtract 25.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;x_1 &amp;amp; = m\quad|\, m \in \left[1,9\right] \\&#xA;x_2 &amp;amp; = 2m \\&#xA;x_3 &amp;amp; = 2m + 5 \\&#xA;x_4 &amp;amp; = 10m + 25 \\&#xA;x_5 &amp;amp; = 10m + 25 + n\quad|\, n \in \left[1,9\right] \\&#xA;x_6 &amp;amp; = 10m + n\quad|\, m,n \in\left[1,9\right] \\&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The first digit of your new number was your original number. The second is your second number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditDate="2015-09-09T08:15:03.250" LastActivityDate="2015-09-09T08:15:03.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3372" PostTypeId="2" ParentId="3313" CreationDate="2014-10-28T12:57:02.800" Score="2" Body="&lt;p&gt;It's a trick question: The glas is full (around half would be water, and the remainder would be air).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the puzzle to be precise in the line of the other answers, it'd need to explain that the question is 'to what degree is the glass filled with water', and explain that the assumed air contains no water vapor, nor is the ionisation of water taken into account (granted, that is very minimal, but I'm assuming aliens to be evil, so better safe then sorry).&lt;/p&gt;&#xA;" OwnerUserId="2156" LastActivityDate="2014-10-28T12:57:02.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3373" PostTypeId="2" ParentId="3284" CreationDate="2014-10-28T14:11:50.840" Score="3" Body="&lt;p&gt;Upvotes to both mdc32 and Florian F's answers, as they've done the majority of the legwork on this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From their answers, we have &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nr 90 flku&lt;br&gt;&#xA; FLKU is the airport code for Kanyau, Zambia&lt;br&gt;&#xA;&lt;br&gt;&#xA; Using &lt;a href=&quot;http://ourairports.mobi/airports/FLKU/navaids.html?start=540&quot; rel=&quot;nofollow&quot;&gt;OurAirports&lt;/a&gt;, we can see there is an airport which you can get to by starting at Kanyau Airport and flying NR @ 90 degrees.&lt;br&gt;&#xA; That airport is &lt;a href=&quot;http://ourairports.com/navaids/NR/Moela_NDB_BR/#lat=-23.97445032421124,lon=-46.32683324813848,zoom=9,type=Satellite&quot; rel=&quot;nofollow&quot;&gt;Moela NDB&lt;/a&gt;, near Rio de Janeiro in Brazil&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I feel like the answer should be more obviously right that this, but it fits the clues so it's worth posting.&lt;/p&gt;&#xA;" OwnerUserId="2518" LastActivityDate="2014-10-28T14:11:50.840" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3374" PostTypeId="1" CreationDate="2014-10-28T17:05:51.110" Score="3" ViewCount="868" Body="&lt;p&gt;A visual memory test application gives some time to memorize &lt;strong&gt;images - numbers&lt;/strong&gt; combinations, after that it shows a random shape and asks what the number is. What is to trick for such an application if your memory is not very good. &lt;a href=&quot;http://www3.telus.net/eugene_k/Space/Mem-ESA-ASPP-2008.swf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Run The Visual Test&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/byIOV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4323" LastActivityDate="2014-10-30T05:46:05.103" Title="Way to memorize shape - number combinations in short time" Tags="|logical-deduction|" AnswerCount="2" CommentCount="3" ClosedDate="2014-10-30T12:59:34.417" ContentLicense="CC BY-SA 3.0" />
  <row Id="3375" PostTypeId="1" AcceptedAnswerId="3376" CreationDate="2014-10-28T23:57:11.357" Score="1" ViewCount="640" Body="&lt;p&gt;Imagine a 5 by 5 grid, with numbers 1 to 24. One slot is left empty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now picture a &lt;a href=&quot;https://en.wikipedia.org/wiki/15_puzzle&quot; rel=&quot;nofollow&quot;&gt;15 puzzle&lt;/a&gt; style game, where you have to set it up 1 - 24, then 25 magically appears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we all know, the 15 puzzle is not always possible, but how many cases of a 24 puzzle are impossible?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="5373" LastEditDate="2016-10-14T16:01:22.463" LastActivityDate="2016-10-14T16:01:22.463" Title="How many starting positions are unsolvable in the 24 Puzzle?" Tags="|sliding-blocks|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3376" PostTypeId="2" ParentId="3375" CreationDate="2014-10-29T00:06:18.983" Score="5" Body="&lt;p&gt;The same argument for the $15$ puzzle applies.  You need the starting position to be an even permutation, so $1/2$ of the starting positions are solvable.  You follow the same proof that shows you can swap two pairs of numbers, but not one pair.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditDate="2014-10-29T00:13:17.257" LastActivityDate="2014-10-29T00:13:17.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3377" PostTypeId="1" AcceptedAnswerId="3378" CreationDate="2014-10-29T00:25:19.783" Score="1" ViewCount="442" Body="&lt;p&gt;What would be the strategy for the game &lt;a href=&quot;https://en.wikipedia.org/wiki/Dots_and_Boxes&quot; rel=&quot;nofollow&quot;&gt;Boxes&lt;/a&gt;? (Boxes is the game in which players connect dots on a grid and whoever encloses a box gets a point, along with an extra turn, until all lines are made)&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="41973" LastEditDate="2019-07-08T10:28:02.237" LastActivityDate="2019-07-08T10:28:02.237" Title="What's a good Boxes Strategy?" Tags="|strategy|dots-and-boxes|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3378" PostTypeId="2" ParentId="3377" CreationDate="2014-10-29T00:48:09.093" Score="4" Body="&lt;p&gt;The game Boxes is another form of the game &quot;Dots and Lines&quot; or &quot;Dots and Boxes&quot;, where the strategy is understood as a game of early control. The idea is to force the opponent into a condition to draw the third line into a long chain of near-complete boxes, where the ability to complete boxes with the fourth side yourself in a repetitive, connective manner is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strategy can better be explained on &lt;a href=&quot;http://en.wikipedia.org/wiki/Dots_and_Boxes#Strategy&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;, but in general terms details the use of a strategic sacrifice to the opponent to get to collect a few box wins, but be placed into a position to yield much greater returns for yourself because the last required move closes a third side into a long open chain.&lt;/p&gt;&#xA;" OwnerUserId="2265" LastEditorUserId="2265" LastEditDate="2014-10-29T01:16:04.177" LastActivityDate="2014-10-29T01:16:04.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3380" PostTypeId="1" CreationDate="2014-10-29T09:12:17.467" Score="25" ViewCount="3821" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Some number of gold, silver, and copper coins are scattered in &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; chests. You may look into each chest and count each type of coin in them, and then select &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; of the chests. Your goal is to have at least 50% of all the gold coins, at least 50% of all the silver coins, and at least 50% of all the copper coins. What is the minimum number &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; that will allow you to do this, no matter how many coins there are or how they are scattered between the chests?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In other words, you need to find the minimum number &lt;span class=&quot;math-container&quot;&gt;$M_{min} = f(N)$&lt;/span&gt;, that allows you to choose at least half of each type of coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt; If &lt;span class=&quot;math-container&quot;&gt;$N=3$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$M_{min}=3$&lt;/span&gt;, because you need to take every chest in the case that one chest contains all gold coins, one chest contains all silver coins, and one chest contains all copper coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt; &#xA;If &lt;span class=&quot;math-container&quot;&gt;$N=100$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$M_{min}&amp;lt;100$&lt;/span&gt;. Indeed, suppose that you cannot do it with &lt;span class=&quot;math-container&quot;&gt;$M=99$&lt;/span&gt;. Then whichever 99 chests you select the other 100th chest must have at least &lt;span class=&quot;math-container&quot;&gt;$50\%$&lt;/span&gt; of some type of coins. That means that you must have at least &lt;span class=&quot;math-container&quot;&gt;$50\%$&lt;/span&gt; of some type of coins in each chest. That would give you more than (50*100/3)% coins in total, which is impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is mathematical puzzle, not a riddle. Do not try to cheat or to find workarounds. If some rules are unclear - please ask in the comments.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note, the formulation was reworked (leaving the mathematical part of the puzzle exactly the same). And, since a lot of answers were already posted, just to avoid confusion I site previous formulation here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; chests with gold, silver and copper coins in a dragon's cave. You see how many chests there are, but you have no idea how coins are distributed, and the clever dragon proposes you a hard game:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You must to chose and say a number (let's call it &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt;).&lt;/li&gt;&#xA;&lt;li&gt;Then you go and check out all chests, their content and select any &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; of them.&lt;/li&gt;&#xA;&lt;li&gt;If your &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; selected chests have less than 50% of all gold coins in the chests, or less than 50% of all silver coins, or less than 50% of copper - the dragon kills you.&lt;/li&gt;&#xA;&lt;li&gt;Then you have to prove that you selected the smallest &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; possible. You are allowed to distribute all the coins between &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; chests as you wish, so now dragon won't know how they are distributed.&lt;/li&gt;&#xA;&lt;li&gt;Then the dragon goes and tries to selects &lt;span class=&quot;math-container&quot;&gt;$M−1$&lt;/span&gt; chests that have at least 50% of each type of coins in total. If he succeeds, you are dead.&lt;/li&gt;&#xA;&lt;li&gt;If you survive you get 50% of all coins and are free to go.&#xA;What number you must say to survive for sure?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="28" LastEditorUserId="1649" LastEditDate="2019-07-16T06:18:48.333" LastActivityDate="2019-07-16T06:18:48.333" Title="How to choose at least half of everything" Tags="|mathematics|combinatorics|" AnswerCount="8" CommentCount="22" ContentLicense="CC BY-SA 4.0" />
  <row Id="3381" PostTypeId="2" ParentId="3380" CreationDate="2014-10-29T09:31:58.933" Score="-3" Body="&lt;p&gt;Number 1.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;5&quot;&gt;&#xA;  &lt;li&gt;If the dragon can say a number $S&amp;lt;M$ and then go and selects $S$ chests that have at least $50\%$ of each type of coins in total, then you are dead too.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In that sense, the dragon have to say a number smaller than my number (in this case, $1$). So the only logical whole number smaller than 1 has to be 0. Therefore, he picks 0 chests, which means 0 coins.&lt;/p&gt;&#xA;" OwnerUserId="4338" LastEditorUserId="3993" LastEditDate="2014-10-29T13:39:11.640" LastActivityDate="2014-10-29T13:39:11.640" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3385" PostTypeId="2" ParentId="3380" CreationDate="2014-10-29T11:16:12.963" Score="0" Body="&lt;p&gt;If you end up with $M = 2$, and you've got at least $3$ (and uneven) coins of two different metals, you are safe. Then you can distribute those coins such that there is $&amp;gt;50\%$ of one metal in one chest, and $&amp;gt;50\%$ of the other metal in the other chest. In that way, the dragon cannot pick $S$ (=1) chests that have $&amp;gt;50\%$ of all coins in total.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(note that this isn't put here as full answer, because you have no way to control $M=2$, more as a sort of hint/thinking idea for others, or partial answer (in case there is the opportunity that the initial chests are so distributed that my initial requirements are met.)&lt;/p&gt;&#xA;" OwnerUserId="3993" LastActivityDate="2014-10-29T11:16:12.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3386" PostTypeId="2" ParentId="2167" CreationDate="2014-10-29T11:28:50.223" Score="14" Body="&lt;p&gt;I've &lt;a href=&quot;https://math.stackexchange.com/a/609552/106234&quot;&gt;answered this over on Math.SE&lt;/a&gt;, so I'll just quote most of that.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose we have $n$ doors, with a car behind $1$ of them. The probability of choosing the door with the car behind it on your first pick, is $\frac{1}{n}$. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Monty then opens $k$ doors, where $0\leq k\leq n-2$ (he has to leave your original door and at least one other door closed).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The probability of picking the car if you choose a different door, is the chance of not having picked the car, which is $\frac{n-1}{n}$, times the probability of picking it &lt;em&gt;now&lt;/em&gt;, which is $\frac{1}{n-k-1}$. This gives us a total probability of $$ \frac{n-1}{n}\cdot \frac{1}{n-k-1} = \frac{1}{n} \cdot \frac{n-1}{n-k-1} \geq \frac{1}{n} $$&#xA;  If Monty opens no doors, $k = 0$ and that reduces to $\frac{1}{n}$.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;For all $k &amp;gt; 0$, $\frac{n-1}{n-k-1} &amp;gt; 1$ and so the probabilty of picking the car on your second guess is greater than $\frac{1}{n}$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If $k$ is at its maximum value of $n-2$, the probability of picking a car after switching becomes $$\frac{1}{n}\cdot \frac{n-1}{n-(n-2)-1} = \frac{1}{n}\cdot \frac{n-1}{1} = \frac{n-1}{n}$$For $n=3$, this is the solution to the original Monty Hall problem.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Switch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Unless you want the goat).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://xkcd.com/1282/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://imgs.xkcd.com/comics/monty_hall.png&quot; alt=&quot;An image, from xkcd, showing a figure wearing a baret having chosen the goat from behind door B, telling it &amp;quot;... and my yard has so much grass, and I&amp;#39;ll teach you tricks, and ...&amp;quot;&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sup&gt;A few minutes later, the goat from behind door C drives away in the car.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="-1" LastEditDate="2017-04-13T12:19:36.950" LastActivityDate="2014-10-29T11:28:50.223" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3388" PostTypeId="2" ParentId="3380" CreationDate="2014-10-29T13:22:09.443" Score="3" Body="&lt;p&gt;Let's say we have only gold and N = 3, we know that M would have to equal 2, since the dragon could evenly divide the gold between the chests. If we add 2 more metals and bump N to 9, M would have to be 6, by the same principle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, a basic method would be to take &lt;strong&gt;Ceil(N/6) * 3&lt;/strong&gt;. Divide by 6 since we have 3 groups and need at least half. Multiply by 3 because we have the three metals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: After a counter point, I have a new thought: &lt;strong&gt;Ceil(N/2) + 1&lt;/strong&gt;.  For 5 chests, the dragon could go Gold, Silver, 1/3 Cop, 1/3 Cop, 1/3 Cop for 4 chests. For 9 chests, 1 gold, 1 silver, 7 copper = 6 chests, the same as we've established for equal distribution w/o mixing.  &lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2014-10-29T14:05:22.333" LastActivityDate="2014-10-29T14:05:22.333" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3389" PostTypeId="2" ParentId="3380" CreationDate="2014-10-29T13:22:54.063" Score="1" Body="&lt;p&gt;The dragon made the rules and the dragon wants to kill you. Since you get to look in the chests before choosing them, the dragon &lt;strong&gt;must&lt;/strong&gt; spread the coins evenly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine the dragon put all the gold in one chest, all the silver in another, and all the copper in a third. You've chosen M to be 50 out of 100 - just make sure you include those 3 in your 50. If &lt;strong&gt;any&lt;/strong&gt; chest has more than its share of the coins, you can include it in your list of M chests. The only way the dragon wins here is if you choose a number smaller than 50, but the coins are spread evenly among all the chests.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if all the coins are spread evenly you need to choose N/2. (Unless it's a very small number of chests. Here the dragon could put a different type in each of 3 chests and you need to choose 3 to be sure you won't get killed on the first try. You can't guess less than 3 to protect against this strategy. ) You don't even need to rearrange anything if the dragon did the right thing, there will be less than half in the dragon's M-1 chests. If the dragon did the wrong thing, go ahead and rearrange the coins to be evenly shared among all the chests. You will always win.&lt;/p&gt;&#xA;" OwnerUserId="1509" LastEditorUserId="1509" LastEditDate="2014-10-29T13:34:32.547" LastActivityDate="2014-10-29T13:34:32.547" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3390" PostTypeId="2" ParentId="3380" CreationDate="2014-10-29T13:29:04.763" Score="0" Body="&lt;p&gt;First it's easy to see that if $N\le3$ then $M = N$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $M\gt3$ Let's divide the question based on whether $N \equiv 0 \mod 3$ or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $N$ is divisible by three, then you would distribute the chests into three parts, and put even amount of coins into each of the chests. If we normalize the amount of coins to $1$ then this means you have $1/(N/3) = \frac{3}{N}$ coins in each chest. You need to open up at least  $\lceil 1/\frac{3}{N}/2 \rceil = \lceil \frac{N}{6} \rceil $ chests for each coin type, so a total of $M = 3 \lceil\frac{N}{6}\rceil$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you chose one less than this value, then by the algorithm above you can distribute the coins in a way that you will not get enough coins for one of the coin types.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If N is not divisible by three, then decrease the number of chests for one type of coin by  one or two, and distribute the coins the same way as before (but one of the coins will have less chests, so the amounts will be different) This results in $M_{G,S} = \lceil \frac{N}{6} \rceil$ but $M_C = \lceil \frac{N-N\%3}{6}\rceil$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives an end result of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M = M_G + M_S + M_C = \lceil \frac{N}{6} \rceil + \lceil \frac{N}{6} \rceil + \lceil \frac{N-N\%3}{6}\rceil$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some values for small Ns:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;N  M&#xA;1  1&#xA;2  2&#xA;3  3&#xA;4  3&#xA;5  3&#xA;6  3&#xA;7  5&#xA;8  5&#xA;9  6&#xA;10 6&#xA;11 6&#xA;12 6&#xA;13 8&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For example for $N=7$ you would distribute the coins as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1: 1/3 G&#xA;2: 1/3 G&#xA;3: 1/3 G&#xA;4: 1/3 S&#xA;5: 1/3 S&#xA;6: 1/3 S&#xA;7: 1   C&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;you'll need to open up 1,2,4,5,7 (5) to get the necessary amount of coins. In case of $N=8$ it is similar, but the coppers are distributed evenly into to 7th and 8th chest. The end result remains the same.&lt;/p&gt;&#xA;" OwnerUserId="85" LastEditorUserId="85" LastEditDate="2014-10-29T13:45:54.350" LastActivityDate="2014-10-29T13:45:54.350" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3391" PostTypeId="2" ParentId="3380" CreationDate="2014-10-29T13:51:27.360" Score="15" Body="&lt;h1&gt;Easily Divisible Treasure&lt;/h1&gt;&#xA;&lt;p&gt;Lets assume that the treasure can be divided into any portions we want.  This could be done with &lt;a href=&quot;https://puzzling.stackexchange.com/questions/186/whats-the-fewest-weights-you-need-to-balance-any-weight-from-1-to-40-pounds&quot;&gt;coins of different values&lt;/a&gt;, or with a sufficient number of coins the problem goes away on its own (see below).&lt;/p&gt;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; is odd, then there exists 3 numbers, &lt;span class=&quot;math-container&quot;&gt;$g,s,c$&lt;/span&gt; such that &lt;span class=&quot;math-container&quot;&gt;$g+s+c=N$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$g,s,c$&lt;/span&gt; are all odd.&lt;/p&gt;&#xA;&lt;p&gt;Then, you can split all the gold evenly amongst the &lt;span class=&quot;math-container&quot;&gt;$g$&lt;/span&gt; gold chests, and all the silver evenly amongst the &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; silver chests, and all the copper evenly amongst the &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; copper chests.&lt;/p&gt;&#xA;&lt;p&gt;Since they are all odd, you need an extra one of each, so&#xA;&lt;span class=&quot;math-container&quot;&gt;$$M=\frac{g+1}{2} + \frac{s+1}{2}+\frac{c+1}{2}=\frac{g+s+c+3}{2}=\frac{N+1}{2}+1$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; is even, then you can only get two odd numbered groups out of it.  Thus:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$M=\frac{g}{2} + \frac{s+1}{2} + \frac{c+1}{2}=\frac{g+s+c+2}{2}=\frac{N}{2} + 1$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;So, if we select our &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; based on these values, we should be able to guarantee that we get half of all the coins.  Then we can distribute in this way to ensure the dragon cannot get half of all of them with &lt;span class=&quot;math-container&quot;&gt;$S=M-1$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;h1&gt;Discrete Coins&lt;/h1&gt;&#xA;&lt;p&gt;If the number of coins is not divisible by the size of the group, and the number of coins is small, there can be problems.  For example, if there are &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt; chests and &lt;span class=&quot;math-container&quot;&gt;$7$&lt;/span&gt; coins, then two of the chests would have &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; coins.  Instead of forcing the dragon to pick &lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt; of these chests, he could get away with only those &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; since they contain &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt; coins - over half of &lt;span class=&quot;math-container&quot;&gt;$7$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;If we do not know the amount of coins, it is possible that we are in either situation, in which case our &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; would be too high if we used the method described above, or too low if we assumed the number of coins was low as in this example.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Also, if we have only a few coins and large number of chests, we are in trouble.  For example, 1 coin of each means that for any &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt;, we'd need to set &lt;span class=&quot;math-container&quot;&gt;$M=3$&lt;/span&gt;.  If we don't know the number of coins ahead of time, we'd be eaten for sure.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;Thus, this kind of thing needs to be either known ahead of time (i.e. we are told the amounts of each denomination of coin), or avoided by saying the number of coins is &lt;em&gt;much&lt;/em&gt; greater than the number of chests.  How much higher before this problem goes away? Well, if the number of coins is greater than &lt;span class=&quot;math-container&quot;&gt;$N(\frac{N+1}{2})$&lt;/span&gt;, it seems to do the trick.  Why?  Take a look at the following:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;   Chests&#xA;1 2 3 4 5 6 7&#xA;o o o o o o o&#xA;o o o o o o o&#xA;o o o o o o o&#xA;o o o&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;In this example, we have 24 coins split over 7 chest as evenly as we can make it.  But picking the first 3 chests gets you 12 coins, and we needed to make the dragon pick 4 chests.&lt;/p&gt;&#xA;&lt;p&gt;By going to 31 coins, the problem goes away:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;   Chests&#xA;1 2 3 4 5 6 7&#xA;o o o o o o o&#xA;o o o o o o o&#xA;o o o o o o o&#xA;o o o o o o o&#xA;o o o&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Now, those first 3 chests only gets you 15 coins.&lt;/p&gt;&#xA;&lt;p&gt;If you draw a line between these two groups you will notice we have two rectangles - a 3x5 rectangle and a 4x4 rectangle.  In the 24 coin example, we had a 3x4 rectangle and a 3x4 rectangle.  In order to ensure the extra chest is needed, we need to maximize the rectangle on the right.  This is accomplished when it is a square.  After that, additional rows will make the right rectangle always bigger than the left.  For arbitrary &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt;, we achieve this square when we have &lt;span class=&quot;math-container&quot;&gt;$N+1$&lt;/span&gt; rows of coins.&lt;/p&gt;&#xA;&lt;h2&gt;However...&lt;/h2&gt;&#xA;&lt;p&gt;Is there a way to avoid the problem even if we don't have sufficient coin and we don't know that?  Perhaps we can select our groups better to avoid it?  For example, if we have 7 of each coin and 15 chests, the construction method above could result in:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$G:=\{2,2,1,1,1\}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$S:=\{2,2,1,1,1\}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$C:=\{2,2,1,1,1\}$$&lt;/span&gt;&#xA;Thus we'd have &lt;span class=&quot;math-container&quot;&gt;$M=9$&lt;/span&gt; chests, but the dragon can clearly win with &lt;span class=&quot;math-container&quot;&gt;$S=6$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;However, if we set &lt;span class=&quot;math-container&quot;&gt;$\left|G\right|=\left|S\right|=7$&lt;/span&gt; and  &lt;span class=&quot;math-container&quot;&gt;$\left|C\right|=1$&lt;/span&gt;, we'd have:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$G:=\{1,1,1,1,1,1,1\}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$S:=\{1,1,1,1,1,1,1\}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$C:=\{7\}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;We'd still have &lt;span class=&quot;math-container&quot;&gt;$M=9$&lt;/span&gt;, but any &lt;span class=&quot;math-container&quot;&gt;$S&amp;lt;M$&lt;/span&gt; would fail for the dragon.&lt;/p&gt;&#xA;&lt;p&gt;So, there will be some numbers of coins for which we can still achieve the optimal value, but we would still need to know ahead of time the number of coins of each denomination.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-10-29T15:20:54.513" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3392" PostTypeId="2" ParentId="3380" CreationDate="2014-10-29T14:30:15.290" Score="3" Body="&lt;p&gt;I think you have alluded to the most challenging distribution.  The chests are divided into $G$ chests containing one (or the same number) of gold coins, $S$ containing one silver coin, and $C$ containing one copper coin, with $G+S+C=N$  I need to choose $\lceil \frac G2 \rceil +\lceil \frac S2 \rceil  +\lceil \frac C2 \rceil $ of them.  If $N$ is odd, all three numbers may be odd, so $M=\frac {N+3}2$ as all of them may need to round up.  If $M$ is even, only two can be odd, so $M=\frac {N+2}2$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not a full answer, as if the coins do not distribute evenly you may be able to do better.  Let there be $9$ chests and $4$ of each kind of coin.  The above would say $M=6$, but if you make three chests of each type there will be one of each type with two coins and you can just take those three.  We can show that $M=5$ by putting all the gold coins in one chest and each other coin in its own chest in that case.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2014-10-29T14:30:15.290" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3393" PostTypeId="2" ParentId="3374" CreationDate="2014-10-29T15:31:42.823" Score="0" Body="&lt;p&gt;As far as I can think of, there are only two ways, you remember or you find someone/some tool who/which can remember. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That someone who can remember can be a screenshot (as suggested in comments by klm123) or another person who can sit beside you and tell you the answer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another solution is to find patterns if any and apply the same, but this is not an universal solution.  As indicated by Emrakul these are subjective answers that I have used personally for other such similar situations (not tests)&lt;/p&gt;&#xA;" OwnerUserId="2313" LastActivityDate="2014-10-29T15:31:42.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3394" PostTypeId="2" ParentId="236" CreationDate="2014-10-29T17:40:36.580" Score="2" Body="&lt;p&gt;I got somewhat similar answer, but logically easier and relying on a &quot;trick&quot;.&#xA;When the Oracle is about to come all people come to the meeting unless they see that there is a blue eyed already present there. So:&#xA;1)If there are no people one goes to the meeting&#xA;1.a) if he sees anyone blue eyed coming, then he is brown eyed&#xA;1.b) if no one else comes then he is blue-eyed - the oracle will anounce at least him or anybody else blue-eyed and he can't be sure who the oracle is talking about. But if no one else comes, then he is blue eyed and leaves, knowing that.&#xA;So all blue eyed will understand they are such in the steps mentioned and the rest that they will stay there forever :)&#xA;The main reasoning is - &quot;I won't go to the meeting if I see someone blue-eyed there, because if I'm also blue eyed we won't be able to make the distinction or at least we should fallback to the other solution&quot;&#xA;&quot;Wait and see&quot; action is present in both solutions, while in mine the oracle is there only for motivation for the meeting.&lt;/p&gt;&#xA;" OwnerUserId="4348" LastEditorUserId="4348" LastEditDate="2014-10-29T17:52:39.247" LastActivityDate="2014-10-29T17:52:39.247" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3395" PostTypeId="1" AcceptedAnswerId="3397" CreationDate="2014-10-29T18:14:17.693" Score="-1" ViewCount="929" Body="&lt;p&gt;You meet a clever, and dark man. He wants to steal your mind and sell it to the dumbies (an imaginary species of shadows that have an IQ of less than -150)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He offers you to play a game, if you win, he'll give you +50 IQ, but if you lose, he gets 50 of your IQ. It's very tempting, so you listen to the rules and agree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rules are as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;He sets up 3 cups.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You have to turn them all upright by turning 2 cups over at a time each turn.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You only get 6 turns.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It seems like a fairly easy game, You play him a few times and win, then he offers you the ultimate mind, but if you lose, he takes all if your IQ. Knowing how simple this is, you accept, but here's the trick, you lost!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This evil man tricked you! But how did he do it?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="5373" LastEditDate="2015-04-28T19:33:40.470" LastActivityDate="2015-04-28T19:33:40.470" Title="The Impossible Game?" Tags="|game|story|situation|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3396" PostTypeId="2" ParentId="3395" CreationDate="2014-10-29T18:20:08.090" Score="6" Body="&lt;p&gt;You are bested by the limitation on having to flip 2 cups at a time. There is no way to get 3 upright by flipping 2 cups at a time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If all 3 are down, first flip will set 2 up. After that, you will either have 2 up (flip an up to down, and a down to up) or go back to 3 down. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The man likely tricked you before the final round by having all 3 cups up, or just 1 cup up to start with (I'm assuming the question was about getting all 3 cups up at the end of the 6 steps). &lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2014-10-29T18:28:04.017" LastActivityDate="2014-10-29T18:28:04.017" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3397" PostTypeId="2" ParentId="3395" CreationDate="2014-10-29T18:21:19.297" Score="6" Body="&lt;p&gt;He left it with 2 cups upright. By flipping 2 cups at once you can only change this to 2 upright (if you flipped 1 up, 1 down) or 0 upright (if you flipped both of the 'up' cups). Similarly for the 0 upright state you can only get to 0 or 2 upright. It doesn't matter how many turns you have.&lt;/p&gt;&#xA;" OwnerUserId="3987" LastActivityDate="2014-10-29T18:21:19.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3398" PostTypeId="1" AcceptedAnswerId="5477" CreationDate="2014-10-29T22:43:43.783" Score="11" ViewCount="4267" Body="&lt;p&gt;Imagine a 4x4 square with a picture you need to make with tiles in it, now imagine you can swap any 2 pieces next to eachother (horizontal, and vertical, not diagonal.) What's the maximum amount of moves needed for this type of game? What's the setup required for this game? Label it 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 for the tiles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;By maximum number of moves I mean the minimum number of moves necessary to solve the configuration with the longest solution?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="5373" LastEditDate="2014-11-23T12:56:53.367" LastActivityDate="2014-11-25T20:57:58.820" Title="What's the Maximum Moves Needed for this kind of Puzzle?" Tags="|mathematics|combinatorics|" AnswerCount="5" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3399" PostTypeId="2" ParentId="236" CreationDate="2014-10-30T01:50:37.207" Score="4" Body="&lt;p&gt;If the oracle said nothing and there was one person, that person could never know if anyone at all had blue eyes, so could not leave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there were two, neither would know in the first day whether the other was the only one and should leave alone, or whether they themselves were the second, so neither can leave. Everyone who can see the two knows that those two should not leave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the second day, you cannot know if the other should have left yesterday alone or whether you and he should leave today with you. You know he should not leave tomorrow, as there are definitely only one (him) or two (him and you) but since you know he is only here today because he was as clueless as you on day one, you can't determine your own eye colour from this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the third day, the two of you know that the other should have left on one of the previous day's, but still do not know which. Everybody else has the same dilemma as you did on the third - you don't know if the two are waiting for you, or simply couldn't work it out the day before. Again there are either two who missed their day yesterday, or three including you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the 4th day, everyone knows they've all missed their chance, because they can only see one or two sets of blue, and their own (unknown) would make two or three&lt;/p&gt;&#xA;" OwnerUserId="4159" LastActivityDate="2014-10-30T01:50:37.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3400" PostTypeId="2" ParentId="3398" CreationDate="2014-10-30T02:13:01.173" Score="5" Body="&lt;p&gt;Before getting the correct answer (someone else may have a nice trick to come up with the exact number immediately, but it isn't obvious to me) it may be useful to first generate some absolute limits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, I think it is obvious that any starting point can reach any finishing point, so the problem is always soluble in some &lt;strong&gt;N&lt;/strong&gt; moves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A naive solution would be to correct the top row of tiles first (left to right) then the second row etc, until all was solved. It takes no more than 6 swaps to get 1 into position, 5 swaps to get each of 2 and 3 into position and 6 swaps for 4 (the top right tile). The next row will take no more than 5 + 4 + 4 + 3, the third row no more than 4 + 3 + 3 + 4, and the final row 3 + 2 + 2 + 3, so an absolute upper limit to &lt;strong&gt;N&lt;/strong&gt; is 64 moves. This should be a grossly pessimistic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a similarly naive guesstimate for the lower limit, there are 16! starting positions and, at each step, there are 18 possible moves. It is impossible to reach more then 18^k positions in k moves (actually will be far less as many moves are independent) so there will be positions that are at least ln(16!)/ln(18) ~= 10.6 moves apart. Thus N has an absolute lower limit (grossly optimistic) of 11.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conclusion so far: &lt;strong&gt;64 &gt;= N &gt;=11&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="1631" LastEditDate="2014-10-30T02:20:42.400" LastActivityDate="2014-10-30T02:20:42.400" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3401" PostTypeId="1" AcceptedAnswerId="3403" CreationDate="2014-10-30T02:48:17.403" Score="5" ViewCount="7418" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Bob goes to a vending machine to buy a can of soda. He opens his wallet, and, to his surprise, has no dollar bills or any bills for that matter. The soda costs exactly $1 USD. Bob also has change in his wallet, but even still, he can't buy the soda. What is the maximum amount of change Bob can have without exact change for a dollar?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know, there's a lot of extra fluff in that question, but that's roughly how I remember seeing it in a puzzle book a while ago. The answer is fairly easy to find out, but I have another question. &lt;strong&gt;Is there a general method, formula, or any other way to solve these puzzles without brute-force?&lt;/strong&gt; This, at least to me, seems to be a recurring problem in books, yet they are often phrased differently and have different currencies, amounts of change required, or other factors changed.&lt;/p&gt;&#xA;" OwnerUserId="1664" LastEditorUserId="8874" LastEditDate="2015-10-03T13:59:33.780" LastActivityDate="2015-10-03T13:59:33.780" Title="The Money Puzzle: Maximum amount of change without a dollar" Tags="|combinatorics|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3402" PostTypeId="1" CreationDate="2014-10-30T03:17:46.173" Score="11" ViewCount="1814" Body="&lt;p&gt;Here is a puzzle involving distrust and asymmetric knowledge. (It was inspired by, but not exactly following, the board game &lt;a href=&quot;https://en.wikipedia.org/wiki/The_Resistance_(game)&quot; rel=&quot;noreferrer&quot;&gt;The Resistance: Avalon&lt;/a&gt;.)&lt;/p&gt;&#xA;&lt;p&gt;There are ten resistance fighters trying to overthrow the evil empire.  Among them there is actually a secretive resistance leader, 5 loyal allies, 3 spies, and an assassin.&lt;/p&gt;&#xA;&lt;p&gt;The &amp;quot;good characters&amp;quot; (the leader and the 5 loyal fighters) out number the &amp;quot;evil characters&amp;quot; (the 3 spies and the assassin).  The assassin and spies know each other, and the leader knows who all the evil characters are. Therefore by elimination the leader and the evil characters know who is 'good', but the evil characters don't know who the leader is, and the leader doesn't know who the assassin is.&lt;/p&gt;&#xA;&lt;p&gt;The resistance fighters need to select teams to send on missions against the evil empire. They meet in a secret lair to discuss and vote on plans.  The characters take turns (initial order is randomly chosen) being a &amp;quot;team captain&amp;quot;, where the team captain selects people to send on the mission.  Discussion can occur, the captain picks a team (not necessarily including himself), more discussion can occur, and then the captainr calls a vote to end discussion, and everyone votes whether or not to accept the team.  Discussion can be done publicly in the war room, or take turns privately going off in pairs to discuss and come back to the war room.  After a vote:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If the team is accepted, the team is sent off on the mission.  If there are enough evil characters on the mission, they can choose to secretly sabotage the mission so that it fails (they can also choose not to, in order to confuse the good guys trying to figure out who the spies are), otherwise the mission succeeds.  The spies and assassin are good enough in their sabotage that the good characters don't know who caused the sabotage. The players gather again at the war room, to decide the team to send on the next mission.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the team is rejected, the next character around the table will act as team captain, and  another attempt is made at selecting a team for the mission.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;The details of the missions are:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;mission    people to send     evil characters needed to sabotage&#xA;  #1             3                  1  &#xA;  #2             4                  1&#xA;  #3             4                  1&#xA;  #4             5                  2&#xA;  #5             5                  1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;After the last mission, as a last ditch effort, the assassin can try to guess the leader and kill him.&lt;/p&gt;&#xA;&lt;p&gt;The good characters win if:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;They succeed in at least three missions, and the leader is not killed.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;The evil characters win if they can do one of the following:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;They sabotage at least three missions.&lt;/li&gt;&#xA;&lt;li&gt;They kill the leader at the end.&lt;/li&gt;&#xA;&lt;li&gt;Five successive votes for a team to put on a mission fails (good out number evil, so they can always prevent this, and with 5 votes this guarantees at least once the team leader was good).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;The good have more in number, yet every evil character has full knowledge of who is good and evil, while only one good character does.  Can you think of a strategy to help the good guys defeat the evil empire?&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;1: Is there some strategy the good characters can use when discussing and voting to eventually distinguish lies from truth and maximize their chance of beating the evil empire?  What is the maximum probability of winning this strategy gives them?&lt;/p&gt;&#xA;&lt;p&gt;2: Now consider if, in a twist, the leader only knows 3 of the 4 evil characters, and therefore cannot deduce at the beginning who all the good characters are.  What is the optimal strategy now?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4351" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-11-06T18:22:11.860" Title="A puzzle of trust and lies, allies and spies" Tags="|logical-deduction|strategy|" AnswerCount="3" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="3403" PostTypeId="2" ParentId="3401" CreationDate="2014-10-30T03:18:54.800" Score="10" Body="&lt;p&gt;The answer to the question is $1.19. He has 3 quarters, 4 dimes, and 4 pennies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The unusual answer to this question relies on the fact that the quarter is not an exact multiple of the dime. If every unit of currency was an exact multiple of the next highest one, then you essentially have a positional number system, in which case you cannot have any amount of change higher than a dollar without some subset of that making a dollar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, this sort of problem can't be solved directly like some equations might. There are algorithms, but no known efficient ones. You'd need to depend on clever tricks like the above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such change problems are examples of &lt;a href=&quot;http://en.wikipedia.org/wiki/Knapsack_problem&quot;&gt;knapsack problems&lt;/a&gt; in computer science, specifically &lt;a href=&quot;http://en.wikipedia.org/wiki/Knapsack_problem#Subset-sum_problem&quot;&gt;subset-sum problems&lt;/a&gt;. There's a lot of theory on these problems, and I'd encourage you to check it out if you're really interested.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2014-10-30T03:18:54.800" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="3404" PostTypeId="2" ParentId="3374" CreationDate="2014-10-30T05:46:05.103" Score="1" Body="&lt;p&gt;Remembering a map of numbers to pictures requires your memory to map the numbers to shapes with any explanation you can think of i.e. in the above case say,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(+) -&gt; 96.... (When I see 96, rotation leads to 96,so is for +).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;JF-&gt; 35.... (J looks like half 3 with lower part missing; F is 5 with lower part missing ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;KH -&gt;15.... (H then I(1)...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RED arrow -&gt; 39 ( The arow's end looks like a 3 and RED(3)*3(arrow'send)=9) :P&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It sounds ambiguous, but the logic of mapping is &quot;to each his own&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some simple ways:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Use math (a=1,b=2....z=26) to map shapes to numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Use the picture's design ,make it synonymous to a number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: It need not be a logical mapping, any random thoughts.&lt;/p&gt;&#xA;" OwnerUserId="2656" LastActivityDate="2014-10-30T05:46:05.103" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3405" PostTypeId="1" AcceptedAnswerId="3495" CreationDate="2014-10-30T09:41:34.447" Score="6" ViewCount="464" Body="&lt;p&gt;Abalone is a 2-player board game. Here's a brief explanation from &lt;a href=&quot;http://en.wikipedia.org/wiki/Abalone_(board_game)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The board consists of 61 circular spaces arranged in a hexagon, five on a side. Each player has 14 marbles that rest in the spaces and are initially arranged as shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LfRnL.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each move, a player moves a line of one, two or three marbles one space. The move can be either in-line (serial in respect to the line of marbles) or broadside (parallel to the line of marbles), as illustrated below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lod5A.png&quot; alt=&quot;Black opens with a broadside move&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/f3mh9.png&quot; alt=&quot;White counters with an in-line move&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Challenge:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is only one set of marbles on the board with initial state as above picture (it doesn't matter which color you choose). At least how many moves does it takes to migrate all your marbles to your opponents initial space?&lt;/p&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="4421" LastEditDate="2015-09-02T18:38:51.137" LastActivityDate="2015-09-02T18:38:51.137" Title="Abalone Migration" Tags="|board-games|minimum-solve|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3406" PostTypeId="5" CreationDate="2014-10-30T09:43:28.427" Score="0" Body="&lt;p&gt;Abalone is an award winning two-player strategy board game designed by Michel Lalet and Laurent Lévi.&#xA;The underlying board is hexagonal, and the players are represented by marbles of opposing colors. The objective is to push six of the opponent's marbles off the edge of the board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Abalone_(board_game)&quot; rel=&quot;nofollow&quot;&gt;from wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="8874" LastEditDate="2015-01-31T16:35:31.047" LastActivityDate="2015-01-31T16:35:31.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3407" PostTypeId="4" CreationDate="2014-10-30T09:43:28.427" Score="0" Body="Abalone is a strategy game on a hexagonal board, in which the players are represented by marbles of opposing colors." OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-01-31T18:02:53.713" LastActivityDate="2015-01-31T18:02:53.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3414" PostTypeId="2" ParentId="3405" CreationDate="2014-10-30T11:21:28.900" Score="3" Body="&lt;p&gt;You cannot do it in any less than 30 moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to move each marble individually, you'd need 88 moves to get them all to the other side.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;4 moves each for moving the row of 3 marbles to the corresponding position on the other side&lt;/li&gt;&#xA;&lt;li&gt;6 moves each for moving the row of 6 marbles to the corresponding position on the other side&lt;/li&gt;&#xA;&lt;li&gt;8 moves each for moving the row of 5 marbles to the corresponding position on the other side&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If we can move a maximum of 3 marbles at a time, theoretically it would take us 30 moves to get them all there.  29 moves would move 3 marbles 1 space for a total of 87 spaces.  One more move for the last marble would be sufficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the question is, can this be achieved?&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2014-10-30T11:21:28.900" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3427" PostTypeId="2" ParentId="3401" CreationDate="2014-10-30T13:03:05.003" Score="1" Body="&lt;p&gt;An attempt at the general problem: given a target value $T$ and a number of denominations of coins $x_1, x_2...x_n$ (ordered from highest to lowest), find the maximum total of coins it's possible to have without being able to make exactly $T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is an $x_i$ which doesn't divide $T$ exactly then you can have an unlimited number of coin $i$ and the puzzle becomes uninteresting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the other extreme, if $x_i$ is always an exact multiple of $x_{i+1}$, then we always get a maximum total of $T-x_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following algorithm is my attempt at the remaining cases:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 1: Add $(T / x_1) - 1$ of coin 1 to the selection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step $i$, for $i &amp;gt; 1$: find all possible totals of subsets of coins selected in previous steps, which are less than $T$. For each total find the gap to $T$. If any gap is divisible by $x_i$, find the smallest such gap $G$, otherwise set $G = T$. Add $(G/x_i) - 1$ of coin $n$ to the selection.&lt;/p&gt;&#xA;" OwnerUserId="2021" LastEditorUserId="2021" LastEditDate="2014-10-30T17:14:06.910" LastActivityDate="2014-10-30T17:14:06.910" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3429" PostTypeId="1" AcceptedAnswerId="3430" CreationDate="2014-10-30T13:22:08.473" Score="2" ViewCount="1627" Body="&lt;p&gt;There's 16 buttons arranged in a 4 by 4 square. They are coloured either red, or green. They appear like this (red being 1, green being 0)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1010&lt;br&gt;&#xA;  0101&lt;br&gt;&#xA;  1100&lt;br&gt;&#xA;  0011&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Each press will switch all the buttons diagonal of it, and the button on the exact opposite of it (for example the one exact opposite of the one in the top left corner is the bottom right.) Not to mention itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of pressing would be, if you press the first 1 on the second row, the following would happen:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They changes happen in this order:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The button you press changes&lt;/li&gt;&#xA;&lt;li&gt;The buttons diagonal of it by 1 button changes&lt;/li&gt;&#xA;&lt;li&gt;The opposite button changes&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You're mission is to switch it all to either 1 or 0, in the minimum moves required. What is this minimum amount and what moves are in the required sequence?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="39" LastEditDate="2014-11-01T15:08:34.110" LastActivityDate="2014-11-01T15:08:34.110" Title="16 Buttons, Turn them all Red or Green?" Tags="|logical-deduction|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3430" PostTypeId="2" ParentId="3429" CreationDate="2014-10-30T14:19:40.063" Score="5" Body="&lt;p&gt;To turn them all to red press buttons:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2, 3, 7, 9, 10, 12, 14, 15 (8 buttons)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To turn them all green press the buttons: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2, 3, 6, 9, 11, 12, 14, 15 (8 buttons)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Buttons are numbered 1 to 4 along the top row, then next row 5 to 8, etc. ITs probably a logical way of doing it but felt it should be explicitly stated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To demonstrate the first solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pressing button 2&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1110&#xA;1111&#xA;1100&#xA;0001&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Pressing button 3&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1100&#xA;1010&#xA;1100&#xA;0101&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Pressing button 7&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1001&#xA;1000&#xA;1101&#xA;0101&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Pressing button 9&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1001&#xA;1101&#xA;0101&#xA;0001&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Pressing button 10&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1001&#xA;0101&#xA;0001&#xA;1011&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Pressing button 12&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1001&#xA;1111&#xA;0000&#xA;1001&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Pressing button 14&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1011&#xA;1111&#xA;1010&#xA;1101&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Pressing button 15&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1111&#xA;1111&#xA;1111&#xA;1111&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The second solution is identical to the first except that you are flipping the selection of the middle four buttons (6,7,10,11) because they will flip the whole board. If you add these to flip then you will have pressed some buttons twice and that is the same as pressing them not at all. Thus you add 6 and 11 and remove 7 and 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further notes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider the board to be a chess board (ie white and black squares) then a button on a white square will only ever change buttons on the white squares and likewise black buttons only change black buttons. This means that it can be split into two separate problems if you so choose.&lt;/p&gt;&#xA;" OwnerUserId="1561" LastEditorUserId="1561" LastEditDate="2014-10-30T14:42:29.193" LastActivityDate="2014-10-30T14:42:29.193" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="3432" PostTypeId="2" ParentId="3429" CreationDate="2014-10-30T14:25:11.377" Score="7" Body="&lt;p&gt;This is a puzzle of two halves, one on each set of diagonals&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 1 1  &#xA;  1 1  &#xA; 1 0  &#xA;  0 1&#xA;&#xA; 0 0&#xA;0 0&#xA; 1 0&#xA;0 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For the first half (- means select and now off, x means select and now on): &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 1       0 0       0 0       0 x       0 0      0 0&#xA; 1 1       - 1       0 0       1 1       1 1      0 0&#xA;1 0       0 0       0 x       0 1       1 0      - 0&#xA; 0 1       0 1       1 0       0 0       x 0      0 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And for the second:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 0 0       0 0       x 0      0 0&#xA;0 0       1 1       0 0      0 0  &#xA; 1 0       1 x       1 1      0 0&#xA;0 1       0 0       0 1      0 -&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For a total of 8 moves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the total board:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1010      0000      0000      00x0      0000      0000      0000      0x00      0000&#xA;0101      0x01      0000      0101      0101      0000      1010      0000      0000&#xA;1100      0100      01x0      0110      1100      -100      010x      0101      0000&#xA;0011      0011      0110      0010      0x10      0010      0000      0010      00-0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And just for kicks, here's the all-on solution:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1010      10-0      1000      1010      1x10      1011      1011      1011      1111&#xA;0101      0000      0101      0101      1111      11-1      0101      1111      1111&#xA;1100      1100      -100      1110      1110      1111      1-11      101-      1111&#xA;0011      0111      0011      0x11      0101      0101      1111      1101      11x1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And why not, here's steps that ensure you can finish any puzzle. Since each button only affects other buttons on it's diagonal (ala black squares on a chess board), the diagrams will be for 1/2 boards.  And every starting position can be thought of as a combination of these three configurations:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0 1      0 1      1 0      0 0      0 0&#xA; 0 0      1 1      - 1      0 1      0 0&#xA;0 0      x 0      0 0      0 1      0 -&#xA; 0 0      1 0      1 0      1 x      0 0&#xA;&#xA;1 0      1 1      1 0      0 0      0 0&#xA; 0 0      0 x      0 1      0 1      0 0&#xA;0 0      1 1      0 0      0 1      0 -&#xA; 0 0      0 0      x 0      1 x      0 0&#xA;&#xA;0 0      0 x      0 1      1 0    Which is the position above&#xA; 1 0      0 1      1 0      - 0&#xA;0 0      0 0      x 0      0 0&#xA; 0 0      1 0      0 0      0 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2014-10-30T16:55:02.320" LastActivityDate="2014-10-30T16:55:02.320" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="3437" PostTypeId="2" ParentId="3405" CreationDate="2014-10-30T15:33:09.660" Score="2" Body="&lt;h2&gt;31 Move Solution&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;While this is (possibly) not optimal as 30 is the theoretical lower limit...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution below uses only 31 moves...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trick to it is really in the first 2 moves...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/Vw10C.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4150" LastActivityDate="2014-10-30T15:33:09.660" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="3439" PostTypeId="2" ParentId="3398" CreationDate="2014-10-30T17:09:51.313" Score="5" Body="&lt;p&gt;Following Penguino, I'll put tighter limits on the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a lower limit to N, let's define the 'disorder' of a piece as the sum of its horizontal and vertical distance from where it should be. The disorder of the square is then the sum of the disorder of all 16 pieces. A given move can change the disorder by &lt;strong&gt;at most&lt;/strong&gt; 2, and disorder must be 0 for the solved puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we translate each piece through the centre (in the hope of maximising disorder) we go from:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;ABCD&#xA;EFGH&#xA;IJKL&#xA;MNOP&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;PONM&#xA;LKJI&#xA;HGFE&#xA;DCBA&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Giving a disorder for each piece of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;6446&#xA;4224&#xA;4224&#xA;6446&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Totalling 64. Since we saw that a move changes disorder by no more than 2, at least 32 moves are needed to solve this setup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for upper limits, if the top row starts at the bottom in a maximally disordered state, you can get it to the top in (4*3 moves) 12 moves and give it the correct order in (3+2+1) 6 moves. You cannot make this worse by, for example, placing abcd in the bottom right quadrant because every left-right move used to fit them all into the top row will reduce the number of left-right moves needed to make them ordered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So 18 moves gives us the top row in order. The second row is the same but with one less vertical space for the 4 pieces to travel, giving 14 moves maximum. Same argument again for row three gives us 10 more, and row 4 is automatically in place, needing a maximum of 6 moves to have the correct order. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives an upper limit of 18+14+10+6=48 moves for any arrangement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$48\geq N\geq 32$&lt;/p&gt;&#xA;" OwnerUserId="3987" LastActivityDate="2014-10-30T17:09:51.313" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3444" PostTypeId="2" ParentId="3337" CreationDate="2014-10-30T21:28:06.013" Score="5" Body="&lt;p&gt;Unfortunately for the knights, if N&gt;3 there is no way for all of the knights to be next to all of their friends.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Taking oerkelens' logic just a bit farther, we can show that N=6 doesn't work:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;. a . a . .&#xA;a . . b a b&#xA;. . A . . .&#xA;a . . . B .&#xA;. a b a . .&#xA;. . . b . b&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here, A has 8 friends and B has 5, so there's only one arrangement that could possibly work:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a a a .&#xA;a A a b&#xA;a a B b&#xA;. b b b&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Let's look at the &quot;a&quot; right next to B (call it C):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;. a . a . .&#xA;a . . b a b&#xA;. . A . c .&#xA;a c . . B c&#xA;. a b C . .&#xA;. c . b . b&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now C has four new friends, and needs to be next to A &lt;em&gt;and one of the &quot;b&quot;s&lt;/em&gt;. It won't work to be next to both A and B, so let's put it diagonally next to A (I'll use upper-right, but lower-left would be the same due to symmetry):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;. c c c&#xA;a a C c&#xA;a A a b&#xA;a a B b&#xA;. b b b&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, there's a problem - look at D:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;. a . a . .&#xA;a d . b a b&#xA;. . A . c .&#xA;D c . . B c&#xA;. a b C . .&#xA;. c . b . b&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It needs to be next to A, a &quot;b&quot;, and a &quot;c&quot;, as well as have a free spot open for it's friend &quot;d&quot;. However, the only spot next to A that is also next to a &quot;b&quot; and a &quot;c&quot; does not have any space left for &quot;d&quot;. So it's not going to work for N=6.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;What about N=5?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;. a . a .&#xA;a . . b a&#xA;. . A . .&#xA;a . . . B&#xA;. a b a .&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We can easily fit B's friends in here, even when B is right next to A:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a a a b&#xA;a A B b&#xA;a a a .&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now that we have only 25 knights total, it is easier to assign letters to them when referring to them:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A B C D E&#xA;F G H I J&#xA;K L M N O&#xA;P Q R S T&#xA;U V W X Y&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Rather than continue to add friends around A and B until we run into a dead end, we'll look at it a different way. First, we'll categorize them by how many friends they have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2 3 4 3 2&#xA;3 4 6 4 3&#xA;4 6 8 6 4&#xA;3 4 6 4 3&#xA;2 3 4 3 2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Obviously a piece can only be moved to a location that has at least as many squares around it as it does friends. That means that only the pieces that were in or near the corners (and have 2 or 3 friends) can be placed in a corner in the solution (where there are only 3 squares in which a friend could be placed). The way we find the problem is by considering the knight-distance (number of knight jumps to get to another location) between these pieces. For example, for A:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A 3 b&#xA;c . 1&#xA;2 a .&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Knight-distance of 3 from B and F&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A . . 3a.&#xA;. . 1 . 3b&#xA;. . . . 2a&#xA;. . . 2b.&#xA;. . . . .&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Distance 3 from D and J, and from P and V (by symmetry)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A . . . 2&#xA;. . 1 . .&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Distance 2 from E and U&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A . . . .&#xA;. . 1 . .&#xA;. a . . 2&#xA;. . c . .&#xA;b . . 3 d&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and distance 3 from X and T and distance 4 from Y. The distance is important because the king-distance in the solution cannot be greater than the original knight-distance. If we put A in a corner, then Y is the only one other piece that can fit into a corner that can be far enough away from A to actually be in a corner. So none of the pieces in a corner can remain in a corner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about B? Since A is distance 3 from each of the next-to-corner pieces, B must be distance 3 from A, E, U, and Y. Also,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;. B . 2a.&#xA;2g. . . 2b&#xA;. . 1 . .&#xA;2f. . . 2c&#xA;. 2e . 2d .&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;B is distance 2 from each of the other next-to-corner pieces. This means that if B is in a corner, then none of the other pieces that can fit into a corner are far enough away to fill the other corners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it looks like there is no way to fill the corners in a way that satisfies everyone.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;What about N=4?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A B C D       2 3 3 2&#xA;E F G H       3 4 4 3&#xA;I J K L       3 4 4 3&#xA;M N O P       2 3 3 2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For A, we have C, H, I, N, and P at distance 2, B, E, L, and O at distance 3, and D and M at distance 5. This suggests that if we put A in a corner, then D or M should be in the opposite corner, and two of B, E, L, and O should be in the two remaining corners. B and E are at distance 2 from each other, as are B and O, so we'll try B and L, like so:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A . . B&#xA;. . . .&#xA;. . . .&#xA;L . . M&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now let's consider the paths between A and B and between A and L:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A-&amp;gt;G-&amp;gt;I-&amp;gt;B, A-&amp;gt;J-&amp;gt;H-&amp;gt;B&#xA;A-&amp;gt;G-&amp;gt;N-&amp;gt;L, A-&amp;gt;J-&amp;gt;C-&amp;gt;L&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Uh-oh, looks like both G and J are distance 2 from B and L. For any two given corners, here's where distance 2's overlap:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;S  b  .  B      S  a  .  A&#xA;a  ab a  .      .  ab b  b&#xA;.  b  ab b      .  ab a  a&#xA;A  .  a  .      .  b  .  B&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In both cases, there is only one locations next to the start (S) that is at distance 2 from the two destination corners. This means if A is in a corner, then we can only use one of B, E, L, and O in another corner. However, we can't use both D and M in the other two corners because they aren't far enough away from each other (only distance 2). So using A (and by symmetry D, M, and P) in a corner won't work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looks like B is our last chance (&lt;em&gt;bad pun warning&lt;/em&gt; - help me, oh &lt;strong&gt;B&lt;/strong&gt; one, you're my only hope). From B, we have distance 1 to H and I, distance 2 to D, E, M, and O, distance 3 to A, C, N, and P, and distance 4 to L. From what we just learned, A and P can't be moved into corners, so the only possibility is to have C, N, and L in the other corners. Unfortunately, this still won't work - C and N are only distance 2 from each other (they are both friends with E), so there's no way both of them can be in a corner.&lt;/p&gt;&#xA;" OwnerUserId="1955" LastEditorUserId="1955" LastEditDate="2014-10-31T07:24:04.290" LastActivityDate="2014-10-31T07:24:04.290" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3445" PostTypeId="1" AcceptedAnswerId="3449" CreationDate="2014-10-30T22:11:05.377" Score="9" ViewCount="1922" Body="&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; A general version of this puzzle has been discussed &lt;a href=&quot;https://math.stackexchange.com/questions/441571/lights-out-variant-flipping-the-whole-row-and-column&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;There's a 4x4 square, where switches are either red or green. When you flip a switch, all the buttons on that row and column will change color.&lt;/p&gt;&#xA;&lt;p&gt;Is every starting position possible to complete? If not how many are there that are impossible?&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; To complete all tiles must be green!&lt;/p&gt;&#xA;&lt;p&gt;Letting 1s be red and 0s be green, here's an example flip;&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt; 1011&#xA; 0010&#xA; 1111&#xA; 0000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;After:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;1001&#xA;0000&#xA;0000&#xA;0010&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;The button press for above was the 3rd column, 3rd row.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="57212" LastEditDate="2022-05-03T19:56:11.737" LastActivityDate="2022-05-03T19:56:11.737" Title="Is it Always Possible? Colour Flipping" Tags="|logical-deduction|" AnswerCount="2" CommentCount="12" ContentLicense="CC BY-SA 4.0" />
  <row Id="3446" PostTypeId="2" ParentId="3155" CreationDate="2014-10-30T22:49:32.833" Score="25" Body="&lt;p&gt;&lt;strong&gt;Illustration&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GdRbt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Leaving the parking&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let’s start with the maneuver to leave the parking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The smallest possible parking from where the car still can go out is when the car is at an angle of 45º from the road and both sides of the parking touch the car.  You could fit the car in a smaller place if you placed it more parallel to the road, but the car wouldn’t be able to turn, because the parking would need to be at least as long as the diagonal of the car.  The required size for a car at 45º is still smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the position at 45º, even with both sides touching, you still can get out.  If the front wheel is at 53.13º, the center of rotation is exactly 1 unit from the rear right corner of the car, and the front left corner starts to moves parallel to the right parking side.  As it happens, the inner circle is just large enough to pass the left corner of the parking.  This is shown in red lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives the minimal parking size for the exit maneuver, the size is ${3 \sqrt2\over2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also defines the target position we need to reach in the parking entry maneuver.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Entering the parking&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 2 routes to enter the parking from a position parallel to the road.  Either you come from the left and do a $3/8$-turn left, or you come from the right and do a $1/8$ turn right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that the maneuver starts exactly parallel to the road and that the car front is exactly level with the end of the parking doesn’t give much choices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Entering the parking from the left&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This scenario is shown in green lines.  At this point, we know where the left side of the parking is, we will adjust the right side to the minimum afterwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The center of rotation is uniquely determined by the constraints that it must be on the diagonal extending the back of the parked car and on a line 2m back from the left side of the parking. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inner radius is given by $r = 2+2\sqrt2$.&lt;br&gt;&#xA;The outer radius satisfies&#xA;$R^2 = (r+1)^2 + 2^2&#xA;= (3+2\sqrt2)^2 + 4&#xA;= (21 + 12\sqrt2)$.&lt;br&gt;&#xA;The distance from the center of rotation to the border is $d = \sqrt2+2$.&lt;br&gt;&#xA;The minimum parking width that fits the outer radius satisfies: &#xA;$(width+2)^2 + d^2 = R^2$&lt;br&gt;&#xA;This gives the minum width:&lt;br&gt;&#xA;$width = \sqrt{R^2 - r^2} - 2 = \sqrt{13 + 8\sqrt2} - 2 = 3.1297…$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Entering the parking from the right&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That scenario is shown in blue lines.  The calculations are more complicated because we start aligned with the right side of the parking but we don’t know yet where the right side is.  We'll have to assume that the inner circle just touches the right corner of the parking and find out what it implies on the inner radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's call $x$ the vertical distance of the blue center of rotation to the border (the fat gray line):&lt;br&gt;&#xA;The inner radius is given by: $r^2 = x^2 + 2^2$&lt;br&gt;&#xA;But we also have: $r+1 = x \sqrt2$&lt;br&gt;&#xA;This resolves to $x = \sqrt2+\sqrt5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $width$ is the width of the parking, the distance of the blue line from the left parking side is given by $width + 2$ but also by $\sqrt2+x$.&lt;br&gt;&#xA;This gives a calculation for the parking width:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$width = \sqrt2+x-2 = 2 \sqrt2+\sqrt5-2 = 3.0645…$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is slightly better than with the green path.  So this is the minimum parking size for the entry maneuver.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the blue path, the angle of the front wheel is $arctan(2/(r+1/2)) = 23.22º$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The parking is 3.0645m wide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The car arrives from the right, parallel to the road.  It turns the front wheel 23.22º right and moves forward into the parking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The parking then shrinks to 2.1213m wide and both vehicles touch the car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The car doesn’t bother, it reverses direction, turns the front wheel 53.13º right and moves forward for a 3/8 turn until it is back parallel to the road.  Facing the traffic as it should.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2014-11-01T14:23:59.927" LastActivityDate="2014-11-01T14:23:59.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3447" PostTypeId="1" AcceptedAnswerId="3475" CreationDate="2014-10-30T23:26:36.690" Score="8" ViewCount="536" Body="&lt;p&gt;In &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3337/from-knights-to-kings&quot;&gt;From knights to kings&lt;/a&gt;, we were asked&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On a NxN chess board $N^2$ knights are placed (one per cell). Each pair of knights, who control each other (i.e. one move away from each other) are friends.&#xA;  One day all knights are promoted to kings. Is it possible to put them on the board (in different order) that all friends will be similarly one move away from each other?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We found out that for a square board, we can't put all the knights next to their friends when $N&amp;gt;3$. What if we are using a $N\times M$ board ($N\ge M$) instead? For example, if $N=3,M=2$:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A B C&#xA;D E F&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You only need to swap D and F to get a solution&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A B C&#xA;F E D&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It's fairly obvious that all $N\times 1$ and $N\times 2$ boards will work. What other board sizes will work under these conditions?&lt;/p&gt;&#xA;" OwnerUserId="1955" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2014-11-02T09:53:12.830" Title="From Knights to Kings on a rectangle" Tags="|logical-deduction|chess|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3448" PostTypeId="2" ParentId="2929" CreationDate="2014-10-31T05:22:57.167" Score="2" Body="&lt;h2&gt;Reiterate problem&lt;/h2&gt;&#xA;&lt;p&gt;Bunny: a chess piece which moves like a bishop but only one square from its current position. It may also hop over another piece.&lt;/p&gt;&#xA;&lt;p&gt;Hop: a bunny hops when another piece is in a square touching the current square (no diagonals). In its turn it occupies the square 3 squares from its current square in the direction of the hopped piece. The hopped piece then takes the original square of the bunny. This is not considered a turn/move for the hopped piece.&lt;/p&gt;&#xA;&lt;p&gt;Say, there are 2 bunnies on a board. &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt;.&#xA;&lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt; begin on opposite colour squares of your choosing.&lt;/p&gt;&#xA;&lt;p&gt;By what means might they both tour the board? No turn may be taken to a previously occupied square (being hopped does not constitute occupation). They take turns, &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; moves first, &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt; second, and they will take 63 turns each.&lt;/p&gt;&#xA;&lt;h1&gt;Vocabulary&lt;/h1&gt;&#xA;&lt;p&gt;The bunny thus has 2 distinct modes of motion: the &amp;quot;walk&amp;quot; and the &amp;quot;hop&amp;quot;. If a piece is hopped over its motion (though not counted as a move) is a &amp;quot;pivot&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;When &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; moves to a square he paints the square &lt;span class=&quot;math-container&quot;&gt;$green$&lt;/span&gt;. When &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt; moves to a square he paints it &lt;span class=&quot;math-container&quot;&gt;$blue$&lt;/span&gt;. After both bunnies have moved to a square it is painted &lt;span class=&quot;math-container&quot;&gt;$red$&lt;/span&gt; - and hereafter no bunny may enter except by &amp;quot;pivot&amp;quot;, and then only immediately as the last bunny has moved to this square.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$Lemma\ 1:$&lt;/span&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3079/how-many-bishops-to-walk-across-the-board&quot;&gt;It is a known fact that no tour purely constistuted of walks is possible. In fact, it is known that it takes 4 distinct walks (sharing no squares) to cover all the black (or white) squares on the board.&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$Lemma\ 2:$&lt;/span&gt; As a corollary to the above, there are only a subset of squares &lt;em&gt;reachable&lt;/em&gt; from any arbitrary square. If all squares or a colour were reachable then &lt;span class=&quot;math-container&quot;&gt;$Lemma\ 1$&lt;/span&gt; would be false.&lt;/p&gt;&#xA;&lt;h1&gt;Answer&lt;/h1&gt;&#xA;&lt;p&gt;I will now claim that this bunny's tour is impossible.&lt;/p&gt;&#xA;&lt;p&gt;For &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt;, given any starting square &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; on colour &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt;, we know that &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; must be a square on &lt;em&gt;only&lt;/em&gt; one of the four walks needed to cover the square of colour &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; - call this walk &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt;. Any square visited from &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; must be part of &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt;. It is clear that his tour cannot be completed from &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$Lemma\ 1$&lt;/span&gt;).&#xA;This means that &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; must escape and continue by means of a hop.&#xA;Now, &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt; occupies one of the squares in &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$s^{\prime}$&lt;/span&gt;. &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt; may now process on his walk &lt;span class=&quot;math-container&quot;&gt;$w^{\prime}$&lt;/span&gt;. We know that only the same squares that were &lt;em&gt;reachable&lt;/em&gt; to &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; are &lt;em&gt;reachable&lt;/em&gt; to &lt;span class=&quot;math-container&quot;&gt;$s^{\prime}$&lt;/span&gt;. This means that they are part of the same walk. Eventually &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt; will get stuck or otherwise need to change to a different walk to have a chance at covering the board (as we know that 1 is not sufficient). His only means of escape is a hop. Now &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; takes a square on &lt;span class=&quot;math-container&quot;&gt;$w^{\prime}$&lt;/span&gt;, BUT &lt;span class=&quot;math-container&quot;&gt;$w^{\prime}$&lt;/span&gt; has exactly the same reachable squares as &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt;. So, &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; has gained no new reachable squares for colour &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$Lemma\ 2$&lt;/span&gt;). And there is nothing he can do to remedy this situation.&lt;/p&gt;&#xA;&lt;p&gt;Thus, there is no 2-bunny tour of a chessboard.&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-11-01T19:58:19.657" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3449" PostTypeId="2" ParentId="3445" CreationDate="2014-10-31T08:45:30.903" Score="19" Body="&lt;p&gt;It's always possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to flip a single switch, you flip it once, and all switches in the same row or column, in any order. The switch itself will be flipped 7 times, the switches in same row or column 4 times, the other switches twice. Which means all other switches will stay the same color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0000        0010        1101        1111        1101        0101        0001        0000&#xA;0010 -3,2-&amp;gt; 1101 -3,1-&amp;gt; 1111 -3,3-&amp;gt; 1101 -3,4-&amp;gt; 1111 -1,2-&amp;gt; 0000 -2,2-&amp;gt; 1111 -2,4-&amp;gt; 0000&#xA;0000        0010        0000        1111        1101        0101        0001        0000&#xA;0000        0010        0000        0010        1101        0101        0001        0000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Using this technique, we can switch the red switches to green one by one. It will not be the most efficient solution in most cases, but it is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Credit goes to @mdc32, his JSFiddle gave me an awesome blackboard!&lt;/p&gt;&#xA;" OwnerUserId="439" LastEditorUserId="439" LastEditDate="2014-10-31T12:44:59.373" LastActivityDate="2014-10-31T12:44:59.373" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3450" PostTypeId="1" AcceptedAnswerId="3455" CreationDate="2014-10-31T13:02:18.597" Score="4" ViewCount="1773" Body="&lt;p&gt;25 kids were in a class, they were all trading lunches.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Each kid traded their lunch 5 times.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;At first 12 kids had bananas.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;At first 13 kids had oranges&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;16 kids like bananas&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;14 kids like oranges&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In the end 9 people threw out their lunch because they never liked it&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So the question is, what's the maximum number of people who could have oranges after this trade?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2014-10-31T14:15:52.550" Title="Banana? Or Orange?" Tags="|logical-deduction|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="3455" PostTypeId="2" ParentId="3450" CreationDate="2014-10-31T13:41:30.077" Score="7" Body="&lt;p&gt;&lt;strong&gt;Edit: This answer assumes that every kid likes something - i.e. there are no kids that like neither bananas or oranges.  If this assumption is wrong, then @RTL posted a great solution.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;So &lt;span class=&quot;math-container&quot;&gt;$11$&lt;/span&gt; kids like bananas, &lt;span class=&quot;math-container&quot;&gt;$9$&lt;/span&gt; kids like oranges, and &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt; kids like both.  Regardless of how it is setup, the &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt; that like both will never discard their lunch.  Since they are guaranteed, lets give them all oranges.&lt;/p&gt;&#xA;&lt;p&gt;Now we have &lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt; oranges left and &lt;span class=&quot;math-container&quot;&gt;$12$&lt;/span&gt; bananas.  If we try to give every kid what they want, then we will have &lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt; kids happy with oranges, &lt;span class=&quot;math-container&quot;&gt;$11$&lt;/span&gt; kids happy with bananas, and &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; kid unhappy about having a banana when he wants an orange.&lt;/p&gt;&#xA;&lt;p&gt;Status:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt; kids who like oranges have them&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$11$&lt;/span&gt; kids who like bananas have them&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt; kids who like both have oranges&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; kid who likes oranges has a banana&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;If we swap the lunches of &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt; kids who like oranges with &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt; kids who like bananas, then we will have &lt;span class=&quot;math-container&quot;&gt;$9$&lt;/span&gt; kids who were unhappy with their lunch and throw them out.&lt;/p&gt;&#xA;&lt;p&gt;Status:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt; kids who like oranges have them&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$7$&lt;/span&gt; kids who like bananas have them&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt; kids who like bananas have an orange&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt; kids who like oranges have a banana&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt; kids who like both have oranges&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Since we don't know the original configuration, we can assume that whatever it was, it was &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt; trades away from this configuration.  Otherwise, we would need to know the staring configuration to see if we can get to this optimal one.&lt;/p&gt;&#xA;&lt;p&gt;For example, lets assume that the orange loving kids are numbered &lt;span class=&quot;math-container&quot;&gt;$1-9$&lt;/span&gt;, the banana lovers are numbered &lt;span class=&quot;math-container&quot;&gt;$10-20$&lt;/span&gt;, and those that love both are &lt;span class=&quot;math-container&quot;&gt;$21-25$&lt;/span&gt;.  The final configuration we want is:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;|-----orange lovers-----| |-------banana lovers----------| |-love both--|&#xA;|---eat--|  |------throw out--------| |----------------eat--------------|&#xA;1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25&#xA;o  o  o  o  b  b  b  b  b  o  o  o  o  b  b  b  b  b  b  b  o  o  o  o  o&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Then, shift everything to the left by 5 wrapping around and we would have the following:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;|-----orange lovers-----| |-------banana lovers----------| |-love both--|&#xA;1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25&#xA;b  b  b  b  o  o  o  o  b  b  b  b  b  b  b  o  o  o  o  o  o  o  o  o  b  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Now, this is the original configuration. To get to the optimal one in 5 trades, everyone gives to the one on their right and takes from the one on their left.  Everyone is involved in &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt; 25-way trades.&lt;/p&gt;&#xA;&lt;p&gt;If you need it to be a 2-way trade (i.e. a trade is only between 2 kids), then there is no answer.  Lets say that a trade is made up of two half trades.  Each kid is involved in 5 half trades.  Since their are 25 kids, there is 125 half trades.  But a trade has 2 half trades, so there is no way to come up with exactly 125 half trades since it is odd.&lt;/p&gt;&#xA;&lt;h2&gt;Answer: &lt;span class=&quot;math-container&quot;&gt;$9$&lt;/span&gt; kids can have oranges.&lt;/h2&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-10-31T14:15:52.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3457" PostTypeId="1" CreationDate="2014-10-31T13:59:30.763" Score="-1" ViewCount="537" Body="&lt;p&gt;A hunter wants to get on a bus, but belongings which have larger than 1 meter dimensions is not allowed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how will he be able to get his rifle (which is 1.5 meters) together with him without changing the shape of it ?&lt;/p&gt;&#xA;" OwnerUserId="4323" LastEditorUserId="5373" LastEditDate="2015-04-28T19:34:26.260" LastActivityDate="2015-04-28T19:34:26.260" Title="A Hunter wants to get on a Bus" Tags="|mathematics|geometry|" AnswerCount="1" CommentCount="0" ClosedDate="2014-10-31T14:27:12.113" ContentLicense="CC BY-SA 3.0" />
  <row Id="3459" PostTypeId="2" ParentId="3450" CreationDate="2014-10-31T14:09:35.977" Score="6" Body="&lt;p&gt;The maximum is $13$, the original number of oranges... (if we &quot;optimise&quot; for most oranges)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the number of swaps is irrelevant as we have no information on this distribution of bananas and oranges to the likes and dislikes so we can set them up for our &quot;perfect&quot; solution.  (We can just set the results and work backwards to define the distribution based on that)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$16$ kids like bananas&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now we choose that&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$13$ of these banana liking kids also like oranges,&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;this gives&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$8$ who will always throw there fruit away  &lt;/li&gt;&#xA;&lt;li&gt;$3$ who only like bananas&lt;/li&gt;&#xA;&lt;li&gt;$13$ who like both oranges and bananas&lt;/li&gt;&#xA;&lt;li&gt;$1$ who only likes oranges&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now at the end we can set it such that:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $8$ who like neither fruit and the $1$ who only likes oranges all get bananas  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;This gives the $9$ pieces of fruit thrown away (all bananas)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are now $3$ bananas and $13$ oranges left  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The $3$ who only like bananas get the $3$ remaining bananas  &lt;/li&gt;&#xA;&lt;li&gt;The $13$ who like both get the $13$ oranges &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4150" LastActivityDate="2014-10-31T14:09:35.977" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3460" PostTypeId="2" ParentId="3457" CreationDate="2014-10-31T14:24:38.220" Score="-3" Body="&lt;p&gt;Cut it in half? If you do it exact it will be the exact same shape but half as long, and therefore he will have two shapes that are 0.75m long (so no shape is breaking the 1.5 meter dimension rule), it might not be a clever read through the lines answer (the one you probably wanted) but with the way you have worded this question, it is a correct answer :P&lt;/p&gt;&#xA;" OwnerUserId="4094" LastActivityDate="2014-10-31T14:24:38.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3474" PostTypeId="2" ParentId="3447" CreationDate="2014-10-31T20:04:10.150" Score="4" Body="&lt;p&gt;For a 3x3 board we can do&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A B C      A H C&#xA;D E F  -&amp;gt;  F E D&#xA;G H I      G B I&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For a 3x4 board we can do&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A B C D      A J C L&#xA;E F G H  -&amp;gt;  G H E F&#xA;I J K L      I B K D&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I can't find others.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2014-10-31T22:08:39.737" LastActivityDate="2014-10-31T22:08:39.737" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3475" PostTypeId="2" ParentId="3447" CreationDate="2014-10-31T20:09:19.713" Score="5" Body="&lt;p&gt;Assuming &quot;castle&quot; maneuvers are not a part of this drill:  Also, symmetry is not drawn out; you can always flip these boards around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;Anything 4x4 and larger doesn't have a valid solution.&#xA;A 4x4 board will generate knights with 4 friends.  You can't place 5 friendly kings on a board where each is one move away from the other 4 (neglecting &quot;castle&quot; maneuvers!).&lt;/strike&gt;    &lt;i&gt; That was incorrect reasoning.  The &quot;friendliness&quot; is not contagious.&lt;/i&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anything 6x7 and larger definitely doesn't have a valid solution.  This size board will result in knights with 8 friends, some of whom have 6 or more non-mutual friends.  You can't arrange these 2 such that the first has access to all 8 &lt;i&gt;and&lt;/i&gt; the second has access to 6 or more additional friends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Square 6x6 not allowed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can't do 5x6 or larger.  You'll end up with two knights that have 8 friends, 0 mutual friends.  x has friends a-h, X has friends A-H:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;. a . b .&#xA;c A . B d&#xA;C . x . D&#xA;e . X . f&#xA;E g . h F&#xA;. G . H .&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;They must go here on the King Board:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;. . 0 0 0   . 0 0 0 .   0 0 0 . .&#xA;. . 0 x 0   . 0 x 0 .   0 x 0 . .&#xA;. . 0 0 0   . 0 0 0 .   0 0 0 . .&#xA;0 0 0 . .   . 0 0 0 .   0 0 0 . .&#xA;0 X 0 . .   . 0 X 0 .   0 X 0 . .&#xA;0 0 0 . .   . 0 0 0 .   0 0 0 . .&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Each of the friends of x and X (meaning a-h and A-H) have at least 2 non-mutual friends with the x's.  This means none of them can go on the corners of the King Board, since they won't have access to anyone else, nor can they be boxed-in on an edge, so we can't solve this one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Square 5x5 is not allowed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nor can you do 5x4:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a b c d&#xA;e f g h&#xA;i j k l&#xA;m n o p&#xA;q r s t&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;j and k each have 6 friends.  Each of j's friends have at least 1 friend that isn't mutual with j (same goes for k).    Also, since j and k have 0 friends in common, they must be diagonal from each other.  These are the only places on the King Board where j and k could go and have access to 6 friends each. All of them puts one of their friends in a spot where they can't reach all of their own friends:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;. . . .   . . . .   . . . .&#xA;. . j .   . . j .   . . j .&#xA;. k . .   . . . .   . . . .&#xA;. . . .   . k . .   . . k .&#xA;. . . .   . . . .   . . . .&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Square 4x4 is not allowed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3x4 is the largest &lt;strike&gt;that's good to go&lt;/strike&gt; I've solved explicitly so far:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1  2  3  &#xA;4  5  6  &#xA;7  8  9  &#xA;10 11 12  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;which means (knight: friends)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1: 6,8  &#xA;2: 7,9  &#xA;3: 4,8  &#xA;4: 3,9,11  &#xA;5: 10,12  &#xA;6: 1,7,11  &#xA;7: 2,6,12  &#xA;8: 1,3  &#xA;9: 2,4,10  &#xA;10: 5,9  &#xA;11: 4,6  &#xA;12: 5,7  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;and valid solution:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;3  8  1  &#xA;4  11 6  &#xA;9  2  7  &#xA;10 5 12  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4210" LastEditorUserId="4210" LastEditDate="2014-11-01T01:46:52.750" LastActivityDate="2014-11-01T01:46:52.750" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3478" PostTypeId="1" AcceptedAnswerId="3481" CreationDate="2014-10-31T21:55:29.807" Score="31" ViewCount="8621" Body="&lt;p&gt;You are stuck on an island and have been tasked by the natives with dividing a plate of chocolate into 4 equal pieces, one for each of the island's gods. Each god must have an equal share, or you go into the volcano. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plate is a perfect square. The only tools you have are a short knife, a perfect circle with diameter equal to the side of the plate, and a Y shaped thingy with angles of 120° and leg lengths equal to the radius of the circle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are skilled enough to line up two objects perfectly, and to cut along a marked line perfectly, (but are unable to cut perfectly just eye-balling it). How do you satisfy the island's gods? &lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2383" LastEditDate="2014-11-02T01:47:58.350" LastActivityDate="2018-07-27T08:08:24.527" Title="Splitting a Plate into 4 Equal Pieces" Tags="|geometry|" AnswerCount="9" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3479" PostTypeId="2" ParentId="3478" CreationDate="2014-10-31T22:52:27.633" Score="33" Body="&lt;p&gt;Lay the circle directly on top of the square, and cut along the outside of the circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/T4Lsv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now have four pieces on the outside which have 2 sides both equal to half the width of the square, all rotations of this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/i0qzu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take one piece and line it up on the circle such that its two cusps just meet the edge of the circle, and cut:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rE6Ot.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This cuts one quarter from the circle. Align either piece so that one flat side matches with the new flat part of the circle, and the other flat side is a radius. Cut and repeat. We now divided the square into two sets of four equal pieces, not even using the Y shaped tool, four like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/i0qzu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And four like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IOUQA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1477" LastEditorUserId="1477" LastEditDate="2014-11-01T04:08:48.260" LastActivityDate="2014-11-01T04:08:48.260" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3480" PostTypeId="2" ParentId="3398" CreationDate="2014-10-31T23:00:37.753" Score="5" Body="&lt;p&gt;&lt;strong&gt;Experimenting with smaller boards&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did a computer search.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But 4x4 is too big for a brute-force search.  The case 3x4 took a night to complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But here are the maximum nb of moves for smaller sizes:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;     1  2  3  4  5  6  7  8  9 10&#xA;  +-------------------------------&#xA;1 |  0  1  3  6 10 15 21 28 36 45&#xA;2 |  1  4  9 14 21 30&#xA;3 |  3  9 16 24&#xA;4 |  6 14 24 ??&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note that in all these cases, the inverted board was among the worst cases.  Sometime, as for 3x4, it is the only one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes other starting positions are equally bad.  For example, the following board for 2x5 is also a worst case with 21 moves:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;4 8 7 6 0  -&amp;gt;  0 1 2 3 4&#xA;9 3 2 1 5      5 6 7 8 9&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Upper bound for the 4x4 case&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does not answer the 4x4 case, but it gives a better upper bound.&#xA;As frododkywalker explained, the 1st row can be done in 18 moves.  Taemyr noted that a first row at 18 moves + the 3x4 board at 24 moves gives an upper bound of $N \le 42$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the reversed board is the worst case then $N$ is even.  This, with the trend in the table above, seems to indicate a value of $N=36$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lower bound for the 4x4 case&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could brute-force a simpler problem which is to reverse the columns of the 4x4 board and minimizing horizontal flips.  I counted how many times you need to exchange any 2 numbers in 2 adjacent columns to arrive at a board where the outer left and right columns are exchanged and the inner left and right columns are exchanged.  I arrived at 18 moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to solve that by hand, but it is not obvious it cannot be done in 17 moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use frodoskywalker's measure of disorder but only horizontally, you get a disorder of 32.  And every horizontal move reduces it by no more than 2, this means a minimum of 16 moves.  But that minimum cannot be realized.  Because when you move out the first number from the left column, you need to put in a number from the right column to reduce the disorder.  That means you need a number from the right column nearby.  So you need to remove that number from the right column.  But to do that while reducing the disorder, similarily, you need a number from the left column nearby.  So the first move that affects a number from the outer columns cannot reduce the disorder, and therefore you need at least 17 moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that you need 18 is a result of my computer search.  Maybe some kind of parity is involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The implication for the original problem is that a solution that reverses the 4x4 board solves the simpler problem by columns and by rows.  It reverses the columns left-to-right, which requires at least 18 horizontal moves.  Likewise, it reversese the rows upside-down which requires 18 vertical moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That gives: $N \geq 18+18 = 36$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Estimation for the 4x4 case&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that sizes 3x3 and 3x4 are exactly 4 moves above frodoskywalker's minimum of 12 resp. 20 moves.  It is my impression that this is a general formula.  That would give $N = 32+4 = 36$.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2014-11-07T21:45:53.363" LastActivityDate="2014-11-07T21:45:53.363" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="3481" PostTypeId="2" ParentId="3478" CreationDate="2014-10-31T23:14:49.253" Score="31" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WTDBc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Place the circle so that its edge touches two of the square vertices.&lt;/li&gt;&#xA;&lt;li&gt;Align your &quot;Y shaped thingy&quot; so that one vertex touches the side of the square and the others touch the edges of the circle.&lt;/li&gt;&#xA;&lt;li&gt;Use the leg of the thingy that lays out of the circle to mark a line on the square with your knife.&lt;/li&gt;&#xA;&lt;li&gt;Rotate the square and repeat three more times.&lt;/li&gt;&#xA;&lt;li&gt;Cut the square using the marks you made.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4430" LastActivityDate="2014-10-31T23:14:49.253" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3482" PostTypeId="2" ParentId="3478" CreationDate="2014-10-31T23:16:12.263" Score="8" Body="&lt;p&gt;We can divide it into 4 equal pieces without dividing it into 8. Like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Line the circle up so that its diameter is over a side of the square. Mark the arc that passes through the square. Do this for the opposite side, resulting in a square marked like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/K65g7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the Y-shaped piece to find the midway point of any side (this will be where all three prongs join), rotate it until one prong touches both circles (center of square). Cut along this line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeat for all sides. We now have four pieces that are rotations of: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/G6P0t.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the arc in the middle didn't actually cut through the chocolate; it's just a mark.&lt;/p&gt;&#xA;" OwnerUserId="1477" LastActivityDate="2014-10-31T23:16:12.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3483" PostTypeId="2" ParentId="3284" CreationDate="2014-11-01T01:08:02.677" Score="0" Body="&lt;p&gt;Props to everyone who got us this far. I don't know the answer, but can get a little closer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first clue, 87 rotated 180 degrees is L8, so we have 'later 90 flku' flku could be the aforementioned airport, or it could be Fort Lauderdale Keiser University.&lt;/p&gt;&#xA;" OwnerUserId="3987" LastActivityDate="2014-11-01T01:08:02.677" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3484" PostTypeId="1" AcceptedAnswerId="3489" CreationDate="2014-11-01T03:45:36.707" Score="2" ViewCount="2075" Body="&lt;p&gt;You just got a new clock, but it's broken. Your friend tells you he can fix it, but he needs a little bit of data from it. One thing he needs is how many times 1 appears in the run of 2 days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rules of this broken clock:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;This is a digital, 12 hour display (AM/PM) clock&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Every second hour it is blank (odd-valued hours of both days)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Every third hour the digits behind ':' are blank (2, 5, 8, and 11 AM and PM of both days)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Every fifth hour the digits after the ':' are blank (4AM, 9AM, 2PM, 7PM of day 1, then 12AM, 5AM, 10AM, 3PM, 8PM of day 2)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Every time the digits after ':' are a multiple of both 3 and 4, the digit after ':' is blank&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If the clock runs for 2 days (starting at 12:00 AM), how many times will '1' appear? Appearances reoccur every minute; thus 12:07 and 12:08 contribute two separate appearances, despite the 1 being displayed constantly.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="69582" LastEditDate="2021-07-03T22:56:28.443" LastActivityDate="2021-07-03T22:56:28.443" Title="How Many Times will 1 Appear on the Broken Clock?" Tags="|mathematics|combinatorics|time|" AnswerCount="3" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="3485" PostTypeId="2" ParentId="3478" CreationDate="2014-11-01T04:56:53.360" Score="29" Body="&lt;p&gt;This method divides the square plate of chocolate into four equal pieces, as shown in the diagram below.&lt;/p&gt;&#xA;&lt;p&gt;Initial steps:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Align the circle so that a diameter of it lines up with one side of the square (as in the first diagrams of two of the three previous answers).  Mark a short arc near the center of the square.  (That is, lightly score a small distance.  This initial arc can be made arbitrarily short, if desired, by using the three-armed item as in step 2 of rsanchez's answer, and can be so made that it is later cut over.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Then do the following four times:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Move the circle from its current side-alignment to align with an adjacent side of the square.  Cut a quarter arc counterclockwise from a corner of the square, stopping each cut upon meeting the initial arc or a previous cut.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tWhkI.jpg&quot; alt=&quot;4 quarter arcs like a pinwheel&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="2004" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-11-01T04:56:53.360" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3486" PostTypeId="1" AcceptedAnswerId="3498" CreationDate="2014-11-01T05:48:03.827" Score="4" ViewCount="1181" Body="&lt;p&gt;A friend of mine has an interesting birthday.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you shorten the date month and year and use UK date format, it reads the year in full, his birthday falls on a Friday.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When he is as old as the year of his birth (last two digits of the year), his birthday will also be a Friday, lets call this the Magic day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He observed that there could be many dates similar to his birthday, however he noted that for one such date in the past, his birthday is the Magic day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you find my friends' birthday if he is still alive?&lt;/p&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="44915" LastEditDate="2018-05-24T12:35:14.913" LastActivityDate="2018-05-24T12:48:49.887" Title="Can you find my friends' birthday" Tags="|logical-deduction|word-problem|" AnswerCount="2" CommentCount="16" ContentLicense="CC BY-SA 4.0" />
  <row Id="3488" PostTypeId="2" ParentId="3074" CreationDate="2014-11-01T07:34:43.727" Score="3" Body="&lt;p&gt;&lt;strong&gt;Six question are not sufficient&lt;/strong&gt;&#xA;Player 1 one guesses a number between 1 and 100. Player 2 says some numbers and Player 1 answers with &quot;lower&quot;, &quot;higher&quot; or &quot;equal&quot; if his number is lower, higher or equal to your number. We write down the &lt;/p&gt;&#xA;&#xA;&lt;p&gt;answers of player 1, &lt;code&gt;0&lt;/code&gt; if he answers &quot;lower&quot;, &lt;code&gt;1&lt;/code&gt; if he answers &quot;higher&quot; and &lt;code&gt;=&lt;/code&gt; if he answers &quot;equal&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An Example:&#xA;Assume his secret number is 23. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;We ask 60 and he answers &quot;lower&quot;&lt;/li&gt;&#xA;&lt;li&gt;We ask 20 and he answers &quot;higher&quot;  &lt;/li&gt;&#xA;&lt;li&gt;We ask 25 and he answers &quot;lower&quot; &lt;/li&gt;&#xA;&lt;li&gt;We ask 23 and he answers &quot;equal&quot;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The string we write down is &lt;code&gt;010=&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the possible answers&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;=&#xA;0=&#xA;1=&#xA;00=&#xA;01=&#xA;10=&#xA;11=&#xA;000=&#xA;001=&#xA;010=&#xA;011=&#xA;100=&#xA;101=&#xA;110=&#xA;111=&#xA;0000=&#xA;0001=&#xA;0010=&#xA;0011=&#xA;0100=&#xA;0101=&#xA;0110=&#xA;0111=&#xA;1000=&#xA;1001=&#xA;1010=&#xA;1011=&#xA;1100=&#xA;1101=&#xA;1110=&#xA;1111=&#xA;00000=&#xA;00001=&#xA;00010=&#xA;00011=&#xA;00100=&#xA;00101=&#xA;00110=&#xA;00111=&#xA;01000=&#xA;01001=&#xA;01010=&#xA;01011=&#xA;01100=&#xA;01101=&#xA;01110=&#xA;01111=&#xA;10000=&#xA;10001=&#xA;10010=&#xA;10011=&#xA;10100=&#xA;10101=&#xA;10110=&#xA;10111=&#xA;11000=&#xA;11001=&#xA;11010=&#xA;11011=&#xA;11100=&#xA;11101=&#xA;11110=&#xA;11111=&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If you have a strategy it produces exactly one answer string for every secret number. There are only 63 possible answer strings so there cannot be 100 possible secret numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But six questions are sufficient for the first player to know the right answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can pose one question, you can find the solution if the there are $3$ possible numbers ${1,2,3}$ if you ask for $2$. If the answer is 'yes' the number is $2$, if the answer is 'lower' the answer is $1$ and  if the answer is 'higher' the number is $3$. We model this with (a so called binary tree)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  2&#xA; / \&#xA;1   3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which can be written in as  $1-2-3$ using less space and ask for the number in the middle. If we have two question we can use the model $(1-2-3)-4-(5-6-7)$ or &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;     4 &#xA;   /   \&#xA;  2     6&#xA; / \   / \&#xA;1   3 7   8&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We ask for 7 and if it is not 7 the remaining model is $1-2-3$ or $5-6-7$. which can be solved after one question. So its immediately clear how to query and that there are 7 possible numbers. This can be continued, for 3 queries we have $((1-2-3)-4-(5-6-7))-8-((9-10-11)-12-(13-14-15))$. I will avoid the two dimensional graphic. So for k question we can differentiate between $N(k)=2 \cdot N(k-1)+1=2^{k+1}-1$ numbers. For $k=6$ we get $N(6)=127$ so we can differentiate between $127$ numbers. Therefore $6$ questions are sufficient for the numbers from $1$ to $100$. The first question always  asks for the number $64$. &lt;/p&gt;&#xA;" OwnerUserId="1967" LastEditorUserId="1967" LastEditDate="2014-11-01T10:01:24.690" LastActivityDate="2014-11-01T10:01:24.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3489" PostTypeId="2" ParentId="3484" CreationDate="2014-11-01T11:59:36.300" Score="4" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 765&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Total hours is 48. Odd numbered hours are hidden because of rule no.1&lt;/p&gt;&#xA;&lt;p&gt;Then based on comments by OP, hours 2 5 8 11 2 5 8 11 are removed if they are not already (in the 2 day 4 clock cycles in that order) for rule no.2&lt;/p&gt;&#xA;&lt;p&gt;Minutes are removed from hours 4 9 2 7 12 5 10 3 8 (if they are not removed for rule no.1 in rule 3.&lt;/p&gt;&#xA;&lt;p&gt;Removing the hours won't affect the number of 1s as the only hour to be removed is 11 (others don't contain number 1 anyways), but it's already removed for rule.1, so number 1 exists in hours 10 and 12 (considering 1 and 11 were removed for rule no.1)&lt;/p&gt;&#xA;&lt;p&gt;Removing the minutes would affect hours 4,2,12,10,8 hours, based on rule 3 above (leaving 19 hours)&lt;/p&gt;&#xA;&lt;p&gt;So 19 x 15 1s would appear in minutes (each hour has 15 ones after final rule) that would be 285 times.  Since there are 8 hours containing 60 1s each, that would be 480 times making the total appearance of 1s to the answer above&lt;/p&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="-1" LastEditDate="2021-07-03T10:32:37.847" LastActivityDate="2021-07-03T10:32:37.847" CommentCount="15" ContentLicense="CC BY-SA 4.0" />
  <row Id="3492" PostTypeId="2" ParentId="3484" CreationDate="2014-11-01T13:09:53.170" Score="4" Body="&lt;p&gt;For every hour, 1 appears 16 times (just minutes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reducing multiple of 3s &amp;amp; 4s (12, 24, 36, 48), 1 appears 15 times (just minutes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For every second hour it is blank, it means it is left with 12 hours per day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every third hour the digits behind ':' are blank (3,6,9,12), it means it is left with 8 hours per day (3am, 9am, 3pm, 9pm)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every fifth hour the digits after the ':' are blank (5,10), it means it is left with 6 hours per day (5am, 5pm).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hours that are shown are (1am,7am,11am,1pm,7pm,11pm). Taking all the ones (1am, 11am, 1pm, 11pm), 120 '1s' for both 1am and 1pm and 240 '1s' for both 11am and 11pm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing the simple mathematics will be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;120 + 240 + (15*6) = 450.&#xA;450*2 = 900 (2days)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Please correct me if i'm wrong.&lt;/p&gt;&#xA;" OwnerUserId="4338" LastActivityDate="2014-11-01T13:09:53.170" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3495" PostTypeId="2" ParentId="3405" CreationDate="2014-11-01T18:20:00.100" Score="8" Body="&lt;p&gt;&lt;strong&gt;Solution in 30 moves.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y9sT4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The number at the top left of each board is the number of moves from the previous position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last step repeats all movement symmetrically backwards on the other side.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2014-11-02T15:57:19.033" LastActivityDate="2014-11-02T15:57:19.033" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3497" PostTypeId="1" AcceptedAnswerId="3501" CreationDate="2014-11-02T00:40:35.140" Score="5" ViewCount="454" Body="&lt;p&gt;Say there's an 8x8 board with the far ends completely occupied by pawns (White: a1 b1 c1 d1 e1 f1 g1 h1, Black: a8 b8 c8 d8 e8 f8 g8 h8). The aim of this game is to capture all but 1 pawn using the regular pawn movement rules. Pawn promotion is not allowed. So, is it possible, and if so how fast (please include move sequence used) can you achieve this 1 pawn left, and yes, &lt;strong&gt;each side takes turns moving&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="1955" LastEditDate="2014-11-02T05:58:00.897" LastActivityDate="2014-11-02T15:38:54.873" Title="Chess Pawn Capture Game!" Tags="|chess|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3498" PostTypeId="2" ParentId="3486" CreationDate="2014-11-02T01:06:39.890" Score="2" Body="&lt;p&gt;If I got this right the &quot;magic day&quot; would be the day/month/year that his birthday falls on, in which his age matches up with the last 2 dates of the year he was born?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, here's one possible solution. The birth date fell on:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1/9/39&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The birthday (&quot;magic day&quot;) was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1/9/78&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the above is his DOB then &lt;strong&gt;HIS&lt;/strong&gt; magic day would be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1/9/2056&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Otherwise, we'd need to know his age.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Logic:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Obviously the day and month had to be the same in both dates. We should try 1/9 first, it's the most logical assumption as it's closest to us.  Meaning it's September the 1&lt;sup&gt;st&lt;/sup&gt;. Going through the trial and error method, we eventually find 1/9/39. We then check 1/9/78 because obviously the year has to be doubled for the person's age to match the last 2 digits. Low and behold, it's on a Friday. Then of course to find his magic day we go ahead of 1978 by, obviously, 78, to get 2056!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He is in fact still alive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It took me a while to answer this, because it seemed to me as if the question wanted the DOB and the magic day, I didn't get the part about HIS magic day and the other person's magic day being his DOB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternative Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you go back farther than 1939 you can also find DOB: 1911 and Magic Day: 1922 making his birthday/magic day: 1944&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But this won't fit the question as the question states his birthday on his magic day &quot;will be&quot; indicating that this Magic day is in future&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="44915" LastEditDate="2018-05-24T12:48:49.887" LastActivityDate="2018-05-24T12:48:49.887" CommentCount="14" ContentLicense="CC BY-SA 4.0" />
  <row Id="3499" PostTypeId="1" AcceptedAnswerId="3500" CreationDate="2014-11-02T01:31:46.530" Score="7" ViewCount="1220" Body="&lt;p&gt;N is a number where each digit is divisible by its place, for example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1527&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1 has to be divisible by 1, 5 divisible by 2, 2 divisible by 3, and 7 divisible by 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, N must be divisible by all of its digits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has to be at least 5 digits long, and have at least 3 different digits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does N exist? Is there more than 1 number that fits N? If so is there a pattern? Give a few examples.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2101" LastEditDate="2014-11-04T19:49:00.373" LastActivityDate="2018-05-24T12:48:43.617" Title="What is this Number?" Tags="|mathematics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3500" PostTypeId="2" ParentId="3499" CreationDate="2014-11-02T01:39:59.233" Score="28" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If such an N exists, it is at least 5 digits long, so it has a 2&lt;sup&gt;nd&lt;/sup&gt; digit and a 5&lt;sup&gt;th&lt;/sup&gt; digit.&lt;br&gt;&#xA; The second digit must be divisible by 2, and the 5&lt;sup&gt;th&lt;/sup&gt; digit must be divisible by 5.&lt;br&gt;&#xA; N is divisible by both of these, so it must be divisible by 10.&lt;br&gt;&#xA; Then the last digit of N is 0.&lt;br&gt;&#xA; But no positive number can be divisible by 0, so N cannot exist.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4446" LastEditorUserId="44915" LastEditDate="2018-05-24T12:48:43.617" LastActivityDate="2018-05-24T12:48:43.617" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="3501" PostTypeId="2" ParentId="3497" CreationDate="2014-11-02T02:15:52.133" Score="11" Body="&lt;p&gt;It is not possible for any number of columns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the 4 values:&lt;br&gt;&#xA;$n_{be}$ : the number of black pawns on an even column,&lt;br&gt;&#xA;$n_{bo}$ : the number of black pawns on an odd column,&lt;br&gt;&#xA;$n_{we}$ : the number of white pawns on an even column,&lt;br&gt;&#xA;$n_{wo}$ : the number of white pawns on an odd column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And compute the value:&lt;br&gt;&#xA;$D = n_{be} - n_{bo} - n_{we} + n_{wo}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a black even pawn takes a white odd pawn, the black even pawn becomes odd and a white odd pawn is removed.  This decreases $n_{be}$ and $n_{wo}$ by 1 and increases $n_{bo}$ by 1.  The effect on D is:&lt;br&gt;&#xA;$D' = (n_{be}-1) - (n_{bo}+1) - n_{we} + (n_{wo}-1)$&lt;br&gt;&#xA;$D' = D - 3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect on $D$ of the possible moves are:&lt;br&gt;&#xA;- black even takes white odd: $D' = D - 3$,&lt;br&gt;&#xA;- black odd takes white even: $D' = D + 3$,&lt;br&gt;&#xA;- white even takes black odd: $D' = D + 3$,&lt;br&gt;&#xA;- white odd takes black even: $D' = D - 3$,&lt;br&gt;&#xA;- any non-capture leaves $D$ unchanged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you start with a position with black and white pawns facing each other, you have:&lt;br&gt;&#xA;    $n_{be} = n_{we}$  and  $n_{bo} = n_{wo}$&lt;br&gt;&#xA;and therefore&lt;br&gt;&#xA;    $D = n_{be} - n_{bo} - n_{we} + n_{wo} = 0$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with $D=0$, and adding only multiples of 3, you can see that whatever you do, D will remain a multiple of 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An end position with a single pawn has $D=\pm1$.  It is not a multiple of 3, and therefore the position cannot be reached with any combination of valid pawn moves.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2014-11-02T15:38:54.873" LastActivityDate="2014-11-02T15:38:54.873" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3502" PostTypeId="1" AcceptedAnswerId="3508" CreationDate="2014-11-02T04:30:58.053" Score="13" ViewCount="1487" Body="&lt;p&gt;You are the praetor of &lt;em&gt;Tri&lt;/em&gt;, a small triangular-shaped nation bordered on the northwest by the &lt;em&gt;Kingdom of Sauria&lt;/em&gt;, on the northeast by the &lt;em&gt;Sultanate of Avia&lt;/em&gt;, and on the south by the &lt;em&gt;Republic of Cryosta&lt;/em&gt;, as shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/BPtyv.png&quot; alt=&quot;Tri&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tri is divided into 25 small triangular provinces as shown, each of which has a bustling peasant town at its center. There are also 21 empty monster dens in Tri, which are located at its three outermost corners and at 18 points where the provinces converge (shown as red dots).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an attempt to boost your flagging economy, you've accepted an enormous sum of gold from your neighbour nations to import 21 of their unwanted monsters, which you plan on placing into your 21 dens (1 monster per den).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The three types of monsters you can import are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;fire-breating dragons&lt;/li&gt;&#xA;&lt;li&gt;man-eating gryphons&lt;/li&gt;&#xA;&lt;li&gt;giant trampleodons&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You can import as many of each monster as you want, but your neighbour nations have placed some restrictions on where certain monsters can be placed:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Sauria can only deal with reptilian-type monsters, hence no gryphons may be placed in any of the dens on the border with Sauria&lt;/li&gt;&#xA;&lt;li&gt;Avia's terrestrial gardens are laid waste by trampleodons, hence they refuse to allow any trampleodons to be placed in dens on the border with Avia&lt;/li&gt;&#xA;&lt;li&gt;Cryosta's buildings are mostly constructed of thatch glued together with paraffin and animal fat, hence no fire-breathing dragons may be placed in dens on their border&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Additionally, while the citizens of any one province in Tri can raise arms to defend against one or two types of monster, defending against three types of monsters would overwhelm them. Hence you've resolved that no province may have monsters of all three types in the three monster dens adjacent to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of these rules, you know that a dragon must go in the northernmost den, a trampleodon must go in the southwesternmost den, and a gryphon must go in the southeasternmost den, but it's up to you to assign monsters to the remaining 18 dens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can you meet the demands? Are you up to the challenge?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Respondents are politely encouraged to place answers (i.e. spoilers) in spoiler blocks, so as not to spoil the fun for other puzzlers. :)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2014-11-04T22:18:21.693" Title="The Monster Garden" Tags="|geometry|combinatorics|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3503" PostTypeId="2" ParentId="3502" CreationDate="2014-11-02T04:58:16.070" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I don't think it's possible. The most any one species can fill is up to the not-like neighbor (e.g. trampleodon and Avian). And then, to fill that remaining side, at some point one of the two species will have to switch over (either dragon to griphon, or vice-versa) at which point you'll have a triple species province&lt;br&gt;&#xA; If you have each species fill out the 5 closest dens, which as no combination of the 3 middle to prevent the triple-kill.&lt;br&gt;&#xA; Finally, if you were to try to go in a spiral (tramples along sauria, dragons along avian, gryphons along cryosta) the middle province would get triple-teamed again.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Long and short, since you can't get use a monster to cut off contact between the other two, you will always have an intersection of the three monsters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2014-11-02T04:58:16.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3504" PostTypeId="2" ParentId="3502" CreationDate="2014-11-02T06:06:09.930" Score="11" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; At first I tried coloring the dots on the map.  I made trampleodons blue, gryphons yellow, and dragons red.  This allowed me to color dots with two possibilities as green, orange, or purple, and with three as gray.  I worked through the map a few times and kept coming up with the same solution, that you always met a situation in which the three areas had to come together.&lt;br&gt;&lt;br&gt;I was fairly certain of the solution, but I decided to write a Python script to actually check each possible outcome (2,985,984 of them) and found that my original idea was correct, there is no solution.&lt;br&gt;&lt;br&gt;  It would be solvable without the edge conditions, but those prevent us from creating a solid wall of color to separate the other two.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3928" LastActivityDate="2014-11-02T06:06:09.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3505" PostTypeId="2" ParentId="3478" CreationDate="2014-11-02T06:26:49.160" Score="7" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YQ6u0.jpg&quot; alt=&quot;Chocolate&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Align the circle so that it is tangent to each side of the square.  Align the Y thing between the chocolate and circle so that it touches the circle at each tip, as shown.  The radius of a circle that touches a point of tangency is perpendicular to that tangent, so the vertical piece of the Y thing marks the position of half of the vertical center line of the chocolate.  Rotate it 180° to finish the first bisector, then complete the horizontal division line in a similar way.  This will produce 4 equal size squares of chocolate.&lt;/p&gt;&#xA;" OwnerUserId="3928" LastActivityDate="2014-11-02T06:26:49.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3506" PostTypeId="2" ParentId="3497" CreationDate="2014-11-02T08:11:03.010" Score="3" Body="&lt;p&gt;It is not possible. I believe Florian's answer is sufficient to prove that it cannot work, but this should give some intuition as to why it doesn't work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the case of only two columns. Like Florian, we'll only worry about the captures that need to occur. Here are the four pawns (two of each color):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;W W&#xA;B B&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If white captures a black pawn:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;_ W&#xA;B W&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If white then captures the other black pawn we'd be stuck with two white pawns, so we'll let black capture a white pawn:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;_ B&#xA;_ W&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now we've got pawns of different colors, but they're in the same column. There's no way that one can capture the other. That means a strategy of capturing pawns such that we'll clear columns either from left to right or right to left won't work because we'll end up with two pawns left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With three columns:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;W W W&#xA;B B B&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If a leftmost pawn captures a piece, we get&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;_ W W&#xA;B W B&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If the other leftmost pawn does the next capture, we're down to the two columns which we already know doesn't work. So we know a white pawn has to capture that pawn:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;_ W W&#xA;W _ B&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There's no way forward from here, either - the black pawn will capture the middle white pawn, but then has to choose between the two remaining pawns. Regardless of the choice, there will be a white pawn and a black pawn left with an empty column between them. So let's have the middle white pawn be the one that does the first capture:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;W _ W&#xA;W B B&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If middle black captures next there will be a gap, so white has to capture next. Here are the two options:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;W _ _      _ _ W&#xA;W W B      W W B&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We've seen the second scenario before, so we'll look at the first one. Obviously, black has to capture next:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;W _ _&#xA;W B _&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;From here, there are only two possibilities:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;_ _ _    B _ _&#xA;W W _    W _ _&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Either we end up with two white pawns, or a black and a white in the same column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the results for two and three columns, let's take a look at what happens if we start with the middle of the board:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;* W W *&#xA;* B B *&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here we can substitute the asterisks with either one piece of each color, or two of one color. Let's start with white capturing a black pawn:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;* _ W *&#xA;* B W *&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If black captures the white there will be a gap (or we're at the edge of the board), so the black pawn must be captured. Suppose the capture comes from the right:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;* _ _ *&#xA;* W W *&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This only lets us know that we don't have to assume we're at the edge of the board if we add two pieces of the same color. So suppose the capture comes from the left. Substituting in for the asterisks gives us these two options:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;* W _ W *    * W W _ W *&#xA;* B B W *    * _ _ B W *&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;From there, we get:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;* _ _ W *    * W _ _ W *&#xA;* B W W *    * _ _ W W *&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The second option has a gap, which doesn't work. In the first one, the black pawn cannot be captured from the right. If we have black capture white, we leave an empty column, so it would need to be at the edge of the board (and we're back to the three column version). If we add to the right side:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;* _ _ W W *     * _ _ W _ _ *&#xA;* B W W B *     * B W W B B *&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The next step for each of these is&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;* _ _ B W *     * _ _ _ _ _ *&#xA;* B W W _ *     * B W W W B *&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We're stuck in the second case (leaves gaps), and in the first case we get&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;* _ _ _ W *&#xA;* B B W _ *&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is just two groups of two pieces of a single color next to each other, so we've really got nowhere to go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't quite fully prove its impossibility as there are some more options that I would need to show fully. Hopefully this should be enough to help give you some intuition as to why it doesn't work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One interesting thing to note is that many of the dead-ends just have one extra piece. That suggests that if the symmetry was thrown off (start with any one pawn missing) it should be possible, at least from a capturing point of view.&lt;/p&gt;&#xA;" OwnerUserId="1955" LastActivityDate="2014-11-02T08:11:03.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3507" PostTypeId="2" ParentId="3447" CreationDate="2014-11-02T09:17:13.047" Score="3" Body="&lt;p&gt;I realize the 4x4 case has already been shown to not work, but let me try explaining it a bit differently -- a bit more &quot;locally&quot;.  Consider a large board, and we'll only look at the 4x4 corner of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an actual 4x4 board, by symmetry there are only three kinds of pieces, a corner piece (like $A$), and edge piece (like $B$), and a 'central' piece (like $C$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{matrix} &#xA;A &amp;amp; B &amp;amp; . &amp;amp; c \\&#xA;c &amp;amp; . &amp;amp; a &amp;amp; b \\ &#xA;b &amp;amp; a &amp;amp; C &amp;amp; . \\&#xA;c &amp;amp; .&amp;amp; . &amp;amp; . \end{matrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a larger board there may be more room initially (and thus more friends to place), which gives more space/options to place pieces after promotion to kings, but I will show this extra room cannot help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the corner piece $A$ has two friends $(a,a)$, the edge piece $B$ has three friends $(b,b,C)$ and the center piece $C$ has four friends $(c,c,c,B)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless of how we rearrange the board, some piece needs to go in the corner.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can $A$,$B$, or $C$ be placed in the corner after promotion to kings?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;As a king in the corner only has room for three friends, it is clear a central piece like $C$ cannot be put there.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If we put $B$ in the corner, we have two edge friends $(b,b)$ and one central friend $(C)$ to place. So this gives us only two symmetry distinct layouts to consider&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{matrix} &#xA;B &amp;amp; b &amp;amp; . \\&#xA;C &amp;amp; b &amp;amp; . \\ &#xA;. &amp;amp; . &amp;amp; . \end{matrix} \quad\quad or \quad\quad&#xA;\begin{matrix} &#xA;B &amp;amp; b &amp;amp; . \\&#xA;b &amp;amp; C &amp;amp; . \\ &#xA;. &amp;amp; . &amp;amp; . \end{matrix}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first doesn't leave enough room for the three other friends of $C$, so can be eliminated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second, there is no way to place the three other friends of $C$ without reducing at least one $b$ to only having room for one more friend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{matrix} &#xA;B &amp;amp; b &amp;amp; c \\&#xA;b &amp;amp; C &amp;amp; c \\ &#xA;. &amp;amp; . &amp;amp; c \end{matrix}&#xA;\quad\quad , \quad\quad&#xA;\begin{matrix} &#xA;B &amp;amp; b &amp;amp; . \\&#xA;b &amp;amp; C &amp;amp; c \\ &#xA;. &amp;amp; c &amp;amp; c \end{matrix}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore there isn't room for the friends of the $b$ pieces, and this layout will not work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So an edge piece like $B$ cannot be placed in the corner after promotion to king.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;That leaves $A$. Can a corner piece be placed in the corner after promotion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, if we put $A$ in the corner, this constrains placement of many pieces. Let me rename the starting positions to make the discussion more clear:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{matrix} &#xA;A &amp;amp; f &amp;amp; \bar{e} &amp;amp; . \\&#xA;\bar{f} &amp;amp; . &amp;amp; \bar{a} &amp;amp; e \\ &#xA;e &amp;amp; a &amp;amp; . &amp;amp; \bar{f} \\&#xA;. &amp;amp; \bar{e} &amp;amp; f &amp;amp; k \end{matrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overbars are used to allow making it clear two positions (such as $a$ and $\bar{a}$) are symmetry equivalent, even if we need to refer to them as separate pieces in the discussion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A$ has friends $(a,\bar{a})$, the $a,\bar{a}$ each have two distinct $(e,\bar{e})$ and two common $(A,k)$ friends, the two $e$ have two common friends $(f,f)$ and the $\bar{e}$ have two common friends $(\bar{f},\bar{f})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we put $A$ in the corner, this gives us only two symmetry distinct layouts to consider&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{matrix} &#xA;A &amp;amp; \bar{a} &amp;amp; . \\&#xA;a &amp;amp; . &amp;amp; . \\ &#xA;. &amp;amp; . &amp;amp; . \end{matrix} \quad\quad or \quad\quad&#xA;\begin{matrix} &#xA;A &amp;amp; \bar{a} &amp;amp; . \\&#xA;. &amp;amp; a &amp;amp; . \\ &#xA;. &amp;amp; . &amp;amp; . \end{matrix}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the first, forces placement of quite a few pieces&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{matrix} &#xA;A &amp;amp; \bar{a} &amp;amp; . &amp;amp; . \\&#xA;a &amp;amp; . &amp;amp; . &amp;amp; . \\ &#xA;. &amp;amp; . &amp;amp; . &amp;amp; . \\ &#xA;. &amp;amp; . &amp;amp; . &amp;amp; . \end{matrix} \quad\quad \rightarrow \quad\quad&#xA;\begin{matrix} &#xA;A &amp;amp; \bar{a} &amp;amp; \bar{e} &amp;amp; . \\&#xA;a &amp;amp; k &amp;amp; e &amp;amp; . \\ &#xA;\bar{e} &amp;amp; e &amp;amp; . &amp;amp; . \\ &#xA;. &amp;amp; . &amp;amp; . &amp;amp; . \end{matrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the bar's over the $e$ could be changed, it doesn't change the issue that there is no way to place the $(f,f)$ and $(\bar{f},\bar{f})$ so that they are common friends with the $e$ and $\bar{e}$ respectively. So this layout is not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second layout offers more possibilities for where to place the $e,\bar{e}$ but the same problem arises.  There is no way to place the $e$ pieces such that they can simultaneously be by their $a$ and $f$ friends.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;So no piece can be placed in the corner. Thus 4x4 is not possible.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;That was explained &quot;locally&quot; in the sense that having extra room on the board would not help matters.  It should be clear that any board containing a 4x4 corner as a subset will not work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the question states 1xN and 2xN are not interesting, this then only leaves 3xN for N$\ge$3 to consider.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's first look at the 3x3 case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{matrix} &#xA;a &amp;amp; d &amp;amp; g \\&#xA;f &amp;amp; x &amp;amp; b \\ &#xA;c &amp;amp; h &amp;amp; e \end{matrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The friends can be written as cyclic graphs. $x$ stands alone, and &#xA;$$ a \rightarrow b \rightarrow c \rightarrow d \rightarrow e \rightarrow f \rightarrow g \rightarrow h \rightarrow a$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a completely &quot;local&quot; observation, in that any board having a 3x3 sub board in it will have this requirement.  The unconstrained nature of $x$ along with the cyclic permutations for placement of $a...h$ gives quite a bit of freedom and many solutions for the 3x3 case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 3x4 board has two 3x3 sub-boards in it, and thus two cycles:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{matrix} &#xA;a &amp;amp; d,1 &amp;amp; g,4 &amp;amp; 7 \\&#xA;f &amp;amp;  6  &amp;amp;  b  &amp;amp; 2 \\ &#xA;c &amp;amp; h,3 &amp;amp; e,8 &amp;amp; 5 \end{matrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are no restrictions beyond the two cycles $a...h$ and $1...8$.  So just shuffling to maintain the two cycles after promotion to kings gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{matrix} &#xA;a &amp;amp; h,3 &amp;amp; g,4 &amp;amp; 5 \\&#xA;b &amp;amp;  2  &amp;amp;  f  &amp;amp; 6 \\          &#xA;c &amp;amp; d,1 &amp;amp; e,8 &amp;amp; 7 \end{matrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are of course other solutions, swapping $(f,6)$ or $(b,2)$ or $(2,f)$, or symmetry related solutions, etc.  Note however that this is already constrained enough that once a corner for $a$ is chosen, all but the middle row is already fixed.  These two cycles are so tightly interwoven that even if there was more room on the board, besides freedom in the middle row (or by symmetry rotating to a 4x3 solution), no additional solutions would be gained.  So this is a local observation and can be applied to limit larger boards containing 3x4 subboards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That proves by construction that the 3x3 and 3x4 case are possible.  Now to finish this we can attempt to prove any board containing a 3x5 sub-board is not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we go to 3x5, we can either consider this three 3x3 sub-boards, or two 3x4 sub-boards. The 3x3 board has a lot of freedom (one piece was completely constrained) which turned out to have enough freedom to simultaneously solve the two 3x3 subboards in a 3x4 board.  The resulting 3x4 solution still has some freedom, but is much more restrictive.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that it is not possible to simultaneously solve the two 3x4 subboards in a 3x5 board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{matrix} &#xA;a &amp;amp; d,1 &amp;amp; g,4,A &amp;amp; 7,D &amp;amp; G\\&#xA;f &amp;amp;  6  &amp;amp;  b,F  &amp;amp; 2 &amp;amp; B\\ &#xA;c &amp;amp; h,3 &amp;amp; e,8,C &amp;amp; 5,H &amp;amp; E \end{matrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are no restrictions beyond the cycles $a...h$, $1...8$, and $A...H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Selecting $a$ to be in the same corner, the solutions from the 3x4 subboards gives the top and bottom rows as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{matrix} &#xA;a &amp;amp; h,3 &amp;amp; g,4,A &amp;amp; 5,H &amp;amp; G \\&#xA;. &amp;amp; .   &amp;amp;  .    &amp;amp;  .  &amp;amp; . \\ &#xA;c &amp;amp; d,1 &amp;amp; e,8,C &amp;amp; 7,D &amp;amp; E \end{matrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There piece $b,F$ cannot simultaneously be by $a$ and $c$ and $G$ and $E$. So there is no solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could also go back and prove the 4x4 board is not possible this way, by looking at it as two 3x4 subboards.&lt;/p&gt;&#xA;" OwnerUserId="4351" LastEditorUserId="4351" LastEditDate="2014-11-02T09:53:12.830" LastActivityDate="2014-11-02T09:53:12.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3508" PostTypeId="2" ParentId="3502" CreationDate="2014-11-02T09:20:00.003" Score="23" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This is just &lt;a href=&quot;http://en.wikipedia.org/wiki/Sperner%27s_lemma&quot;&gt;Sperner's Lemma&lt;/a&gt;.&lt;br&gt;&#xA;&lt;br&gt;&#xA;Sperner's Lemma is a mathematical theorem that says that no arrangement can satisfy the criteria in the question. More exactly, it says that if you:&lt;br&gt;&#xA;- Divide up a triangle into smaller triangles (in any way!).&lt;br&gt;&#xA;- Colour the points of the diagram with three colours.&lt;br&gt;&#xA;- Arrange your colouring such that each colour is &quot;banned&quot; from one of the sides of the big triangle.&lt;br&gt;&#xA;&lt;strong&gt;Then&lt;/strong&gt; one of your small triangles has to have different colours at each of its three corners.&lt;br&gt;&#xA;&lt;br&gt;&#xA;Replace the colours with creatures and you've solved the puzzle!&lt;br&gt;&#xA;&lt;br&gt;&#xA;I'll prove Sperner's Lemma below. The proof looks quite long, but that's only because I'm trying hard to explain every step in detail. For the short version look at the top of page two &lt;a href=&quot;http://math.mit.edu/~fox/MAT307-lecture03.pdf&quot;&gt;here&lt;/a&gt; (which is where I got this proof from in the first place).&lt;br&gt;&#xA;&lt;br&gt;&#xA;One brilliant thing about the proof of Sperner's Lemma is that it takes a sneaky approach to proving that there is at least one &quot;rainbow&quot; triangle. It simply proves that there are an odd number of them! Zero is an even number, so an odd number means at least one!&lt;br&gt;&#xA;&lt;br&gt;&#xA;The proof goes like this: Suppose our colours are red, green and blue (for dragons, trampleodons, and gryphons). Let $R$ be the number of rainbow triangles. We want to show $R$ is odd. Pick two of the colours, say red and green. Let $Q$ be the number of triangles whose corners are coloured red-red-green or green-green-red. Let $X$ be the number of red-green edges, but only counting the ones on the outside border of the whole diagram. Let $Y$ be the remaining number of red-green edges, on the inside of the diagram.&lt;br&gt;&#xA;&lt;br&gt;&#xA;Notice that every triangle with a red-green edge is either one of the ones we counted in $R$ or one of the ones we counted in $Q$ (depending on the colour of its remaining corner). Each triangle counted in R has one red-green edge, since it's corners are red, green and blue. Each triangle counted in Q has &lt;em&gt;two&lt;/em&gt; red-green edges (the two edges touching the green corner if the triangle is red-red-green, or the two edges touching the red corner if the triangle is green-green-red). If we add up all these edges we will have counted each edge in X once, but each edge in Y &lt;em&gt;twice&lt;/em&gt; because edges on the inside of the diagram have one triangle on either side of them.&lt;br&gt;&#xA;&lt;br&gt;&#xA;So we've proved the equation $R+2Q=X+2Y$.&lt;br&gt;&#xA;&lt;br&gt;&#xA;Now lets think about the number $X$, which is the number of red-green edges on the outside of the diagram. Red points are banned from one side of the big triangle, and green points are banned from another, so red-green edges can only appear along one of the sides. This is the side where blue is banned, so it has only red and green points along it. It has a red point at one end (where it meets the side where green is banned) and a green point at the other (where it meets the side where red is banned). Start at the red end and work your way along towards the green end. The colour of the points must change from red to green and back again an odd number of times in order to end up with a different colour to what it started on. Each change gives us a red-green edge. So $X$ is odd.&lt;br&gt;&#xA;&lt;br&gt;&#xA;Since $X$ is odd and $2Q$ and $2Y$ are even (they divide by two), our equation $R+2Q=X+2Y$ means that $R$ must be odd. So there are an odd number of rainbow triangles and therefore at least one! Q.E.D.&lt;br&gt;&#xA;&lt;br&gt;&#xA;Bonus fact: Sperner's Lemma can be extended to higher dimensions. So if you have a tetrahedron divided up into small tetrahedra, and you colour in each of the points with one of &lt;strong&gt;four&lt;/strong&gt; colours so that each of the colours is banned from one of the triangular faces of your large tetrahedron, then somewhere you have to have a tetrahedon with all four colours at its corners.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4457" LastEditorUserId="4457" LastEditDate="2014-11-04T22:18:21.693" LastActivityDate="2014-11-04T22:18:21.693" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="3509" PostTypeId="2" ParentId="3484" CreationDate="2014-11-02T11:46:13.493" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 765 times&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For hours: odd hours are blank, only even hours with &lt;code&gt;1&lt;/code&gt; are &lt;code&gt;10&lt;/code&gt; &amp;amp; &lt;code&gt;12&lt;/code&gt;. 2 of each, each day, is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 60*8=480&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For minutes: $48-24(blank)=24$, five hour rule takes out 4AM, 2PM on first day &amp;amp; 10AM, 12PM, 8PM on second. $24-5=19$. every hour has &lt;code&gt;1&lt;/code&gt; 6 times on ones, and 10 times on tens, but not on $12=3*4$. Which leaves:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 15*19=285&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 285+480=765&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4317" LastActivityDate="2014-11-02T11:46:13.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3510" PostTypeId="2" ParentId="3313" CreationDate="2014-11-02T13:18:15.490" Score="2" Body="&lt;p&gt;Close over cylinder with a flat surface, using your perfect handling this should be possible with a hand, or arm perhaps (assuming this is possible if accepted answer is possible), and slowly turn it 180 degrees, clockwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check, with perfect eyesight, that there is a dry spot on the right side of the cylinder. If there is, when you are done, then less than half of it is full. If not - half. If before you reach 180 degrees the water meets a wet spot, it is more than half full.&lt;/p&gt;&#xA;" OwnerUserId="4317" LastActivityDate="2014-11-02T13:18:15.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3511" PostTypeId="2" ParentId="3486" CreationDate="2014-11-02T13:26:31.293" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; September 1st 1978&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;His Magic day:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; September 1st 2056&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;His birthday is magic day for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; September 1st 1939&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explaination&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; UK date representation is &lt;code&gt;DDMMYYYY&lt;/code&gt;, puzzle shortened it so it would be &lt;code&gt;DMYY&lt;/code&gt;. Friend could not be younger than 16 since &lt;code&gt;2/0&lt;/code&gt; doesn't account for any date, so &lt;code&gt;D/M&lt;/code&gt; has to be &lt;code&gt;1/9&lt;/code&gt;, hence first of Sept.&#xA;Testing years going back, this date was on Friday on &lt;code&gt;95&lt;/code&gt; &amp;amp; &lt;code&gt;89&lt;/code&gt; but Since Magic day is adding a number to itself, it can't be on a odd numbered year. Next one back is &lt;code&gt;78&lt;/code&gt;, where &lt;code&gt;2056&lt;/code&gt; has the Magic day, and dividing the number by 2 to &lt;code&gt;1939&lt;/code&gt; shows us that the date &lt;code&gt;1/9/78&lt;/code&gt; is indeed Magic day for &lt;code&gt;1/9/39&lt;/code&gt;.&#xA;&lt;code&gt;1950&lt;/code&gt; could have worked, but &lt;code&gt;1/9/25&lt;/code&gt; was not a Friday. No other &lt;code&gt;19xx&lt;/code&gt; date would be relevant, since before &lt;code&gt;1950&lt;/code&gt; Magic day would come out in &lt;code&gt;19xx's&lt;/code&gt;, but we eliminated those (Besides &lt;code&gt;78&lt;/code&gt;, which is our answer).&#xA;Going back to &lt;code&gt;18xx&lt;/code&gt; years would &lt;a href=&quot;http://en.wikipedia.org/wiki/File:Life_Expectancy_at_Birth_by_Region_1950-2050.png&quot; rel=&quot;nofollow&quot;&gt;be against&lt;/a&gt; the &quot;Friend being alive&quot; rule...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4317" LastEditorUserId="4317" LastEditDate="2014-11-02T16:51:45.833" LastActivityDate="2014-11-02T16:51:45.833" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3512" PostTypeId="1" AcceptedAnswerId="3513" CreationDate="2014-11-02T17:07:10.030" Score="-1" ViewCount="888" Body="&lt;p&gt;There's 2 islands, 1 island has a pile of seeds, a chicken, and a fox.  There's also a boat next to this island, you have to get it all across with minimum uses of the boat.  You can only take one thing across at once.  If you leave the fox and the chicken, the fox will eat the chicken, if you leave the chicken and the pile of seeds, the chicken will eat the seeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can you do it, please provide an explanation, and total number of boat uses.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="8874" LastEditDate="2015-02-14T18:08:30.410" LastActivityDate="2015-02-14T18:08:30.410" Title="A Fox, a Chicken, a Pile of Seeds, and a Boat" Tags="|river-crossing|" AnswerCount="1" CommentCount="5" ClosedDate="2014-11-03T12:09:29.770" ContentLicense="CC BY-SA 3.0" />
  <row Id="3513" PostTypeId="2" ParentId="3512" CreationDate="2014-11-02T17:21:38.973" Score="3" Body="&lt;ol&gt;&#xA;&lt;li&gt;take the Chicken across  - bring nothing back&lt;/li&gt;&#xA;&lt;li&gt;take the Fox across - bring the Chicken back&lt;/li&gt;&#xA;&lt;li&gt;take the Seeds across - bring nothing back&lt;/li&gt;&#xA;&lt;li&gt;take the Chicken across&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, you will go there and back 3 times then go there with the Chicken at the end making 7 trips (3.5 round trips).&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-11-02T18:05:00.907" LastActivityDate="2014-11-02T18:05:00.907" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3523" PostTypeId="2" ParentId="2311" CreationDate="2014-11-03T01:56:42.457" Score="9" Body="&lt;p&gt;It has been dated back as far as 1933, published in a book, &lt;em&gt;Diversion and Pastimes&lt;/em&gt; by R.M. Abraham. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://www.psychologytoday.com/blog/brain-workout/200912/where-is-the-missing-dollar&quot;&gt;http://www.psychologytoday.com/blog/brain-workout/200912/where-is-the-missing-dollar&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1711" LastActivityDate="2014-11-03T01:56:42.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3528" PostTypeId="2" ParentId="3478" CreationDate="2014-11-03T07:59:59.193" Score="2" Body="&lt;p&gt;Take Y shaped one and mark on four sides of the circle as one leg equals to the radius. you get marking on four edges of the plate. Now mark the center of the plate by  keeping the legs touching the circle. You have five markings. Now using legs of Y mark lines joinng the center marking from edges. You have perfect lines and use knife to cut&lt;/p&gt;&#xA;" OwnerUserId="4480" LastActivityDate="2014-11-03T07:59:59.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3530" PostTypeId="1" AcceptedAnswerId="3531" CreationDate="2014-11-03T12:09:29.523" Score="27" ViewCount="2329" Body="&lt;p&gt;This is where yesterday follows tomorrow, departure follows arrival, front follows back, start follows end, past follows future and even life follows death.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is it?&lt;/p&gt;&#xA;" OwnerUserId="4394" LastActivityDate="2014-11-04T22:14:55.193" Title="A mirror of events" Tags="|riddle|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3531" PostTypeId="2" ParentId="3530" CreationDate="2014-11-03T12:24:45.093" Score="34" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A dictionary.  The word 'yesterday' comes after 'tomorrow' in a dictionary. The word &#xA; 'departure' comes after 'arrival' in a dictionary. And so on...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="2444" LastEditDate="2014-11-03T14:46:34.550" LastActivityDate="2014-11-03T14:46:34.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3532" PostTypeId="2" ParentId="3402" CreationDate="2014-11-03T12:32:23.597" Score="1" Body="&lt;p&gt;Since the rule was clarified and a more fitting answer proposed, this answer should be on hold (though I don't know how to do that), as I haven't thought of any better solutions to the current problem. Admittedly, I couldn't think of a formal way for bad guys to beat that strategy within its framework. But it feels rather complex, and that's not a good sign when the puzzle hinges on trust and human communication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should everyone assume everyone else to be purely rational? Can some good guy decide that he wants to deceive the baddies into thinking he was the leader, thus improving the rate from 5/6 to 1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Original answer)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for not solving it completely, but I think it helps. There can be no certainty, rather a probability of the good guys winning. Obviously there's an upper bound of 5/6, since whatever happens, they can kill a random good guy in case he's the leader. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But other than that, any decision is tricky, since there's an important aspect of the rules that I think is ambiguous. (I'll take that 5 votes is enough for a team, answering the most pressing question.) Namely, what does it mean exactly that the good guys &quot;can use a strategy&quot;? Do they know the same strategy and know that other good guys know it too? Do &quot;they&quot; include the leader? Do the bad guys know it? Or does the 5 or 6 good guys have to come up with a plan without preparations? Can the strategy involve unconstrained discourse or even general actions, or is it limited to some formal exchange? The possible strategies vary wildly between these cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an optimistic example, if 6 good guys can have a strategy as common knowledge without the bad guys knowing it, then it is possible to use cryptography to establish each other's identity during discussion. Then, randomly choosing team members among themselves, they can ensure a 5/6 success rate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other extreme, if the leader has to make up a plan without help, I think it's hard to see how the loyals are going to cooperate at all, since the bad guys can claim to be leaders and suggest multiple plans to confuse everyone. But if unconstrained actions are allowed on the part of the leader, and we take the rules literally, then the leader can commit suicide for a draw. Good cannot win since he's dead, but evil didn't kill him, hence draw.&lt;/p&gt;&#xA;" OwnerUserId="4485" LastEditorUserId="4485" LastEditDate="2014-11-06T10:17:39.257" LastActivityDate="2014-11-06T10:17:39.257" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3533" PostTypeId="2" ParentId="3284" CreationDate="2014-11-03T13:48:21.983" Score="0" Body="&lt;p&gt;Ok, I think I stretched it a little too far, but I will try and explain each step for anyone to be clear of. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basics, mdc32 did a good job of uncovering most of the numbers  (except the first and one in between)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1. 87 - look at it upside down.&#xA;2. 114, 57, 48 - subtract 9, divide by 2, repeat&#xA;3. Numbers where x+1 and x-1 are both prime. Next are 102 and 108. &#xA;4. 107 - multiply by three and add one, divide by two&#xA;5. 117 - multiply by two and subtract one, multiply by two and add one&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And then Florian took it one step forward by Using ASCII  which left us with &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Wr90 flku&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then comes the interesting part, for anyone to understand a location I thought its important to give a clue about the country.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Which lies in the last two digits KU which is UK in reverse &#xA;&#xA;Then we are left with Wr90 fl the W being in Caps was to indicate that this was the first letter.&lt;br&gt;&#xA;&#xA;Since we know that its UK and we have a 6 letter alphanumeric code. I thought it would be fairly simple to guess that this is an UK postcode.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The difficulty may be in understanding that these digits could be ordered in any arbitrary order, however I have checked the database and there is only one postcode that can be constructed with these characters which is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;WR9 0LF&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="2313" LastEditDate="2014-11-04T03:10:55.177" LastActivityDate="2014-11-04T03:10:55.177" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3534" PostTypeId="1" AcceptedAnswerId="3549" CreationDate="2014-11-03T15:46:56.720" Score="5" ViewCount="1055" Body="&lt;p&gt;in December 1496, after a rainy day, an inn somewhere in the Black Forest, on one of the major north-south roads, was visited by a group of highwaymen in the middle of a freezing cold night. Fortunately, a band of adventurers present at the inn managed to fend them off. Because of the bounty on the highwaymen, they wanted to pursue, but the highwaymen had set fire to the stables, and their help was needed to save the main building.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After several hours, the fire was put out and the animals were calm enough to pursue the highwaymen, who took off with several of the horses. However, due to the frozen ground, there were no tracks to follow. Still, the master tracker of the group wanted to try his luck. While trying to find tracks, he suddenly slipped on a slippery patch of ground near the south exit and fell on his rear. When he stood up again, his pants were wet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He then went back to his group and told them he knew where the bandits were headed. They went after the gang and soon caught up with them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did he know where the bandits were headed, and what direction did they leave in?&lt;/p&gt;&#xA;" OwnerUserId="2778" LastEditorUserId="22922" LastEditDate="2016-09-15T08:47:19.550" LastActivityDate="2016-09-15T08:47:19.550" Title="The robbery of the Black Forest Inn" Tags="|story|mystery|" AnswerCount="3" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3535" PostTypeId="2" ParentId="3186" CreationDate="2014-11-03T15:55:53.710" Score="14" Body="&lt;p&gt;They could be “Main” and “Second”.  In the town I live in (and perhaps in some other small towns) west-side streets parallel to Main are $2^{nd}, 3^{rd}, 4^{th}, 5^{th}$ etc.  (East-side streets are B, C, D...; &lt;em&gt;ie&lt;/em&gt;, Main does double duty standing in for $1^{st}$ St. and A St.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/M1PR6.png&quot; alt=&quot;map snip&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2004" LastActivityDate="2014-11-03T15:55:53.710" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3536" PostTypeId="2" ParentId="3534" CreationDate="2014-11-03T16:10:49.877" Score="4" Body="&lt;p&gt;They went south.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because it rained and froze, there were puddles with frozen ice on top.  The tracker when he slipped broke through this ice.  Since the bandits left earlier, they would have left foot prints in the mud under the ice when they broke through.  Thus, they were able to follow them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seems like the date, location, and fire are all red herrings.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2014-11-03T16:10:49.877" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3537" PostTypeId="2" ParentId="3530" CreationDate="2014-11-03T21:39:24.517" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;p&gt;Only an English-language dictionary, in general.&lt;/p&gt;&#xA;&#xA; &lt;p&gt;For bonus points, here are some other languages where this does not hold: denoting &quot;follows&quot; by '&gt;' and &quot;doesn't follow&quot; by '&gt;#' :&lt;/p&gt;&#xA;&#xA; &lt;p&gt;French: hier &gt; demain, départ &gt; arrivé, devant &gt;# envers, début &gt;# fin, passé &gt; avenir, vie &gt; mort&lt;/p&gt;&#xA;&#xA; &lt;p&gt;German: gestern &gt;# morgen, Abfahrt &gt;# Ankunft, Vorderseite &gt; Rückseite, Anfang &gt;# Ende,  Vergangenheit &gt;# Zukunft, Leben &gt;# Tod/Sterben&lt;/p&gt;&#xA;&#xA; &lt;p&gt;Hence we could characterize any language with a 6-digit binary number (although the choice of words is in some cases subjective, as in whether 'front' vs 'back' refers to the human body or 'obverse'/'reverse' etc.)&lt;/p&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2349" LastEditorUserId="2729" LastEditDate="2014-11-04T09:55:02.727" LastActivityDate="2014-11-04T09:55:02.727" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3538" PostTypeId="1" AcceptedAnswerId="3539" CreationDate="2014-11-04T02:15:46.310" Score="8" ViewCount="1468" Body="&lt;p&gt;Consider the following building floor plan, with a set of rooms labeled &lt;strong&gt;A&lt;/strong&gt;-&lt;strong&gt;G&lt;/strong&gt; and an &lt;strong&gt;Outside&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/qlQZG.png&quot; alt=&quot;the building floorplan&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lines between each pair of blue dots are walls. Ordinarily in problems like this the goal is to draw a single continuous curve that passes through each wall once and only once, ending up back in the room where the curve was started (a so-called &quot;Hamiltonian cycle&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for this particular set of rooms, the real challenge is to &lt;em&gt;get stuck&lt;/em&gt;. Getting &quot;stuck&quot; means you draw your curve so that you eventually have no way of exiting the room you're in (because you've passed through all of its walls once) &lt;em&gt;even though you haven't passed through all the walls in the building&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can start your curve in any room, but &lt;strong&gt;not&lt;/strong&gt; on the outside of the building.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can you get stuck on purpose?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To present your solution to the problem, please give the sequence of rooms your curve passes through, beginning with the room you start in. For example, a curve that starts in room &lt;strong&gt;G&lt;/strong&gt; and ends at the &lt;strong&gt;x&lt;/strong&gt; (as shown below) has the sequence &lt;code&gt;G Out B A Out&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/j31Lo.png&quot; alt=&quot;codifying curves&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="5373" LastEditDate="2015-04-28T19:38:07.247" LastActivityDate="2015-04-28T19:38:07.247" Title="Getting Stuck on Purpose" Tags="|geometry|strategy|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3539" PostTypeId="2" ParentId="3538" CreationDate="2014-11-04T02:31:01.680" Score="12" Body="&lt;p&gt;One solution would be to &quot;use up&quot; all of the walls except one, and then enter the room for the last time:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cky5e.png&quot; alt=&quot;image&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The syntax for this would be &lt;code&gt;CBACFDCEGC&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be done for any room. If the room has an even number of walls, start inside of it (so that it has an odd number when you leave and actually start &quot;using up&quot; walls two at a time); otherwise, start outside (so that you can &quot;consume&quot; two walls at a time until there's only one left, which you then enter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of these rooms have an even number of walls. Therefore, to successfully solve this problem, you &lt;em&gt;must&lt;/em&gt; start in the room that you end in.&lt;/p&gt;&#xA;" OwnerUserId="39" LastEditorUserId="5373" LastEditDate="2015-02-01T15:41:19.413" LastActivityDate="2015-02-01T15:41:19.413" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3540" PostTypeId="1" AcceptedAnswerId="3545" CreationDate="2014-11-04T06:19:32.147" Score="8" ViewCount="2234" Body="&lt;p&gt;From the *&lt;strong&gt;starting position&lt;/strong&gt; of a chessboard, you need to move the pieces to end up with &lt;strong&gt;18&lt;/strong&gt; Queens following the chess rules at all times (this includes white moves first). The solution is the series of moves (not the final position, as there are plenty).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having other remaining pieces is &lt;em&gt;possible&lt;/em&gt; and allowed (e.g. a remaining bishop + the 18 queens and 2 kings).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;* starting position FEN: &lt;code&gt;rnbqkbnr/pppppppp/8/8/8/8/PPPPPPPP/RNBQKBNR w KQkq - 0 1&lt;/code&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;see also the &lt;a href=&quot;https://chess.stackexchange.com/q/4128/478&quot;&gt;orignial question&lt;/a&gt; on chess stackexchange&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="4376" LastEditorUserId="11110" LastEditDate="2017-12-04T21:35:50.203" LastActivityDate="2022-03-02T18:34:27.337" Title="Find the shortest chess game with 18 queens?" Tags="|chess|board-games|" AnswerCount="2" CommentCount="10" ClosedDate="2014-11-11T21:19:46.503" ContentLicense="CC BY-SA 3.0" />
  <row Id="3541" PostTypeId="2" ParentId="3540" CreationDate="2014-11-04T06:25:42.573" Score="6" Body="&lt;p&gt;My score is &lt;strong&gt;104&lt;/strong&gt; half-moves:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;1.a4 b5 2.a5 b4 3.a6 Bb7 4.axb7 h5 5.bxa8=Q h4 6.g4 h3 7.Bg2 hxg2 8.Ra3 bxa3 9.b4 gxh1=Q 10.e4 a2 11.b5 a1=Q 12.b6 a5 13.b7 d5 14.c4 Nd7 15.c5 Nb6 16.cxb6 Rh5 17.gxh5 g5 18.h6 g4 19.h7 g3 20.h4 a4 21.h5 a3 22.h8=Q a2 23.h6 axb1=Q 24.h7 g2 25.Bb2 d4 26.Bc3 dxc3 27.d4 c2 28.b8=Q c1=Q 29.b7 e5 30.Ke2 c5 31.Nf3 c4 32.Qba7 Bc5 33.dxc5 c3 34.Nd4 exd4 35.f4 Kd7 36.e5 Ne7 37.f5 Ng6 38.fxg6 f5 39.g7 f4 40.g8=Q Qca3 41.c6+ Ke7 42.e6 Kd6 43.c7 c2 44.Qhg7 c1=Q 45.Qc2 d3+ 46.Kf3 d2+ 47.Ke4 d1=Q 48.c8=Q f3 49.e7 f2 50.e8=Q f1=Q 51.h8=Q Qc7 52.b8=Q g1=Q+&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can view the &lt;a href=&quot;http://tinyurl.com/mb4tnq7&quot; rel=&quot;noreferrer&quot;&gt;game in here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4376" LastEditorUserId="4376" LastEditDate="2014-11-05T02:11:53.320" LastActivityDate="2014-11-05T02:11:53.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3542" PostTypeId="2" ParentId="3402" CreationDate="2014-11-04T06:49:04.733" Score="3" Body="&lt;p&gt;I'm not sure how to approach #2.  Someone clever can probably come up with something better using the ideas here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alright for situation #1, where the leader knows all the evil characters, it is possible for good to win 5/6 of the time (the best possible, as with no knowledge the assassin can always just randomly kill one of the good guys hoping that it is the leader).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After gathering in the lair, any good person can suggest the following plan forward:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;this strategy will allow good to win, so provably deviating from this plan will indicate you are evil&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the characters should come up with a means of verifiably signing a note (possibly something as simple as an elaborate signature that no on there has the skill to forge, or more mathematically, publicly stating public keys for a signature algorithm)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the characters should go off in pairs to secretly discuss, until every possible pair has had a chance to discuss privately, and during this time the players will exchange signed notes stating (possibly lying) to each other one of the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3a) declaring themself as not the leader&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3b) declaring themself as the leader, and stating who all is good&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;During these private discussions the leader, as he has knowledge of who all is evil, should declare &quot;leader&quot; to all good guys, and &quot;not-leader&quot; to all evil. Normal loyal fighters should just declare &quot;not-leader&quot; to everyone.  Therefore the evil characters gain no information from these private discussions.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;if a good player gets a message deviating from this rule, they can publicly release the verifiably signed message for that communication to out the evil person.  So evil effectively must pretend to be normal good or the leader.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;More generally, the strategy is for a good person to publicly expose any fake leader (release his messages) if they can prove they are fake without giving information about the leader.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if a fake leader ever claims another person claiming to be leader is good, this outs themselves as fake, as this cannot occur for the real leader if the good guys follow the plan.  Unfortunately, a good person might need to keep this to themself, as revealing this could reveal information about the actual leader.  However if multiple &quot;leaders&quot; choose groups such that there is no consistent way for one to be real (for example two claiming each other as good, or three claim such that it makes a cycle), then the entire group is fake and should be exposed as the evidence doesn't depend on who the actual leader is.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Now the characters gather in the war room for some public discussion.  Each player states which groups of good guys were declared to them by people claiming to be leader. (Note: only the proposed groups of good are released, not who claimed the group.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If there is a group declared such that not everyone in the group states they were told they are in the group, then that group is clearly false (as good has no reason to deviate from the plan).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the number of good characters = 6, and the number of evil characters = 4, any group with the correct number of people will necessarily include at least two good.  This means the evil characters don't have full freedom in how they mislead here.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If there is a player who appears in all proposed sets of good players, he must be good.  All good players should release the messages they got from &quot;leaders&quot; to this player.  Now even more restrictive than before, this player then can expose fake leaders whose group selections include each other in a revealing manner.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If there is enough information that every one should be able to deduce enough good players to succeed 3 missions without revealing the leader, someone can just explain it.  Good, having superiority in numbers, can now ensure the correct mission teams are sent.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If there is currently not enough information to determine the correct teams to send on the missions, the mission captain picks a team which will eliminate as many proposed &quot;good groups&quot; from consideration as possible if it fails.  If it is the leader's turn to be the mission captain, he should do the same, even if it means failing a mission, so that no information about the leader is lost here.  If the mission wouldn't actually eliminate a possible &quot;good group&quot;, and the person still insists on this mission team anyway, it outs themselves as evil and good can vote the mission team down. (Note this of course does not indicate good/evil for the team they suggested though.) So evil has to play along here to try to help narrow down a team.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the mission fails, it should by design eliminate at least one, if not more, proposed &quot;good groups&quot;.  If the mission succeeds, it provides no information.  Proceed to pick a team for the next mission.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now let's analyze this plan.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may not be enough information immediately available to succeed every mission.  But I believe there is enough information here to guarantee succeeding at least three which is all that is necessary to win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplicity I'll refer to a proposed group of good characters by someone claiming to be a leader as a &quot;set&quot;.  And we'll look at what information we can gain from these sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some simple deductions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Since there is only one real leader, once the sets are shared it is obvious to everyone how many people are trying to fake being the leader (even though they don't all know who specifically is claiming to be a leader).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Every time a mission fails, at least one set can be eliminated, so there have to be at least 3 fake leaders for team evil to have a chance.  Less than this and straightforward  application of the plan above is enough to win. So there need to be 3 or 4 fake leaders for evil to have a chance.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If a set is eliminated, a good player can reveal the message from that fake leader making it clear who at least one evil person is.  Any set containing that character is thus clearly wrong as well.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Similarly, a good player can reveal a message from a fake leader if he claims that evil person is good. And so on.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If only three evil characters fake being a leader, then they can't win if two sets are revealed as wrong in a single sabotaged mission, so none can include another fake leader in their sets. This means all fake leaders must have 4 good characters for their set, to have a chance. So for the first three missions, chose a team from such an intersection.  Either this team is actually good (so it won't fail, and good will win the first three missions), or if evil sabotages it, it will eliminate two sets (since regardless who was evil, they were in both sets) and thus evil can at most sabotage only one more mission (as only at most one set from a fake leader remains), and so good wins. Thus evil cannot win with only 3 fake leaders.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If there are 4 fake leaders (every evil character fakes being a leader), they can only afford to lose two sets at once on at most one mission, therefore at least two cannot include any other evil character in their set.  Thus at least two fake leaders must select 5 good characters for their set, to have a chance.  So if there are not two intersections of 5, play trying to eliminate sets as normal and good will win.  Otherwise select from an intersection of 5 to send on the missions, and good will win.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Therefore team good can guarantee to win at least three missions without revealing who the real leader is.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For #2, it is very risky for the leader to reveal himself to anyone, since he doesn't know for sure who is good.  Since there are more good characters, let's see what would happen if relied on chance and just publicly rolled dice for who to send on a mission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;chance of passing mission 1: (6*5*4)/(10*9*8) ~ 16.7%&lt;br&gt;&#xA;chance of passing mission 2: (6*5*4*3)/(10*9*8*7) ~ 7.1%&lt;br&gt;&#xA;chance of passing mission 3: (6*5*4*3)/(10*9*8*7) ~ 7.1%&lt;br&gt;&#xA;chance of passing mission 4: (6*5*4*3*(2+4))/(10*9*8*7*6) ~ 7.1%&lt;br&gt;&#xA;chance of passing mission 5: (6*5*4*3*2)/(10*9*8*7*6) ~ 2.4%  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;probability of winning &gt; prob of winning exactly three &gt; 0.167 * 0.071 * 0.071 (* 3 ways) + 0.167 * 0.071 * 0.024 ( * 3 ways) ~ 0.00253 + 0.00085 = 0.00338&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other ways of winning will contribute even less. So a random strategy is awful, with less than a 1% chance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's try making it smarter:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a mission succeeds, keep those characters for the next mission, else start from scratch again. Since the missions progress (or are equal) in difficulty, evil would be helping if they don't sabotage a mission just to ensure they can sabotage the next. So worse case scenario, evil sabotages every mission they can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ways to win:&lt;br&gt;&#xA;win first two (third for free), same as prob of choosing 4 good at once ~ 0.0714&lt;br&gt;&#xA;win first, lose second, win 3rd or fourth ~ 0.167*(1-0.0714)&lt;em&gt;0.0714 + 0.167&lt;/em&gt;(1-0.0714)*(1-0.0714)*0.0714 ~ 0.0111 + 0.0103 = 0.0214&lt;br&gt;&#xA;Others contribute less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this strategy gives roughly 10% chance of winning.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For #2, a cheap answer is for the leader to just guess who the unknown evil character is, and then use the strategy from #1.  If he's wrong, he'll be revealing himself to an evil character and will get assassinated.  He has a 1/6 chance of guessing correctly.  Thus good can win at least with a chance (1/6)*(5/6) = 5/36 ~ 14% chance of winning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe there is some tradeoff solution, where less certainty of who is good/evil is traded for not fully revealing the leader (maybe he only reveals himself to a couple possible good guys, which is better than 1/6 chance of success).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But currently I'm not sure how to do that.&lt;/p&gt;&#xA;" OwnerUserId="4351" LastEditorUserId="4351" LastEditDate="2014-11-04T07:00:06.333" LastActivityDate="2014-11-04T07:00:06.333" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3543" PostTypeId="2" ParentId="2512" CreationDate="2014-11-04T08:01:41.290" Score="6" Body="&lt;p&gt;There's an odd number of tiles (11 × 23, minus an even number of wall spaces). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any cyclical route must have the same number of North moves as South, and the same number of East moves as West. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the cyclical path has an even length and visits an even number of tiles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore there is no cyclical path visiting each tile once.&lt;/p&gt;&#xA;" OwnerUserId="2021" LastEditorUserId="2021" LastEditDate="2014-11-04T12:57:53.140" LastActivityDate="2014-11-04T12:57:53.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3545" PostTypeId="2" ParentId="3540" CreationDate="2014-11-04T12:10:06.447" Score="13" Body="&lt;p&gt;This is known as the &amp;quot;18 Queens Problem&amp;quot;. I know of this well-known solution by Friedrich Burchard &amp;amp; Friedrich Hariu, published in the German magazine feenschach issue #33 in 1976 on page 22. It can be viewed and downloaded as a &lt;a href=&quot;http://problem64.beda.cz/silo/feenschach33_1976.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;PDF&lt;/a&gt; (the challenge was issued in &lt;a href=&quot;http://problem64.beda.cz/silo/feenschach31_1975.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;#31&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;(1976) in &lt;em&gt;&lt;strong&gt;96 Half-moves&lt;/strong&gt;&lt;/em&gt;.&#xA;I can't make a claim to its optimality,  but by looking at it and seeing that no better can be found, I'd say it may well be optimal.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;e4 f5 2. e5 Nf6 3. exf6 e5 4. g4 e4 5. Ne2 e3 6. Ng3 e2 7. h4 f4 8. h5 fxg3 9. h6 g5 10. Rh4 gxh4 11. g5 g2 12. g6 Bg7 13. hxg7 g1=Q 14. f4 h3 15. f5 h2 16. b4 a5 17. b5 a4 18. b6 a3 19. Bb2 Ra7 20. bxa7 axb2 21. a4 b5 22. a5 b4 23. a6 b3 24. c4 h1=Q 25. c5 h5 26. c6 Bb7 27. cxb7 c5 28. d4 c4 29. d5 Nc6 30. dxc6 c3 31. c7 c2 32. c8=Q c1=Q 33. b8=Q Qc7 34. a8=Q d5 35. a7 d4 36. Nc3 dxc3 37. Qa6 c2 38. Qa8b7 c1=Q 39. a8=Q Qd5 40. gxh8=Q+ Kd7 41. g7 bxa1=Q 42. g8=Q b2 43. f7 b1=Q 44. f8=Q h4 45. f6 h3 46. f7 h2 47. Qfa3 h1=Q 48. f8=Q exf1=Q+&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You can play it out on &lt;a href=&quot;https://www.apronus.com/chess/pgnviewer/?m=e4_f5_e5_Nf6_exf6_e5_g4_e4_Ne2_e3_Ng3_e2_h4_f4_h5_fxg3_h6_g5_Rh4_gxh4_g5_g2_g6_Bg7_hxg7_g1ZQ_f4_h3_f5_h2_b4_a5_b5_a4_b6_a3_Bb2_Ra7_bxa7_axb2_a4_b5_a5_b4_a6_b3_c4_h1ZQ_c5_h5_c6_Bb7_cxb7_c5_d4_c4_d5_Nc6_dxc6_c3_c7_c2_c8ZQ_c1ZQ_b8ZQ_Qc7_a8ZQ_d5_a7_d4_Nc3_dxc3_Qa6_c2_Qa8b7_c1ZQ_a8ZQ_Qd5_gxh8ZQ*_Kd7_g7_bxa1ZQ_g8ZQ_b2_f7_b1ZQ_f8ZQ_h4_f6_h3_f7_h2_Qfa3_h1ZQ_f8ZQ_exf1ZQ*&quot; rel=&quot;nofollow noreferrer&quot;&gt;apronus.com.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="57521" LastEditDate="2022-03-02T18:34:27.337" LastActivityDate="2022-03-02T18:34:27.337" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="3549" PostTypeId="2" ParentId="3534" CreationDate="2014-11-04T14:51:56.553" Score="8" Body="&lt;p&gt;Going off of what Trenin and JNF have said so far, the bandits went North.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one man slipping on the ice is enough to break through then the horses surely would have left a path of broken ice. And since the problem description said there were no tracks, the would have had to have gone North.&lt;/p&gt;&#xA;" OwnerUserId="4505" LastActivityDate="2014-11-04T14:51:56.553" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3550" PostTypeId="1" CreationDate="2014-11-04T18:40:38.270" Score="0" ViewCount="10395" Body="&lt;p&gt;A king sent 10 of his servants out to buy him each 10 gold rings.&#xA;Each ring should weigh 10 grams, and altogether there should be 100 gold rings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the King's spies told him that one servant had cheated and had brought 10 rings each of 9 grams instead of the desired 10 grams. &#xA;Unfortunately, the spy hadn't been able to recognize the face of the servant, so that the king was left wondering which of the servants had stolen his money. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to discover the unfaithful servant, the king used a digital balance but only once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could he do that? &lt;/p&gt;&#xA;" OwnerUserId="4512" LastEditorUserId="8874" LastEditDate="2015-11-04T14:11:31.680" LastActivityDate="2015-11-04T14:11:31.680" Title="A King and his Servants" Tags="|mathematics|combinatorics|weighing|" AnswerCount="2" CommentCount="1" ClosedDate="2014-11-04T22:46:08.277" ContentLicense="CC BY-SA 3.0" />
  <row Id="3552" PostTypeId="2" ParentId="3550" CreationDate="2014-11-04T18:51:09.640" Score="6" Body="&lt;p&gt;Spoiler so others can have fun:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Assign each servant a number 1 through 10. Have each servant put that number of rings on the scale. So #1 will put one ring, #2 will put two rings, ..., and #10 will put all 10 rings.&lt;br&gt;&#xA;  The weight will be off by the number of grams matching the cheating servant.&lt;br&gt;&#xA;&lt;br&gt;&#xA; If nobody cheated the total should be:&lt;br&gt;&#xA; (1 * 10g) + (2 * 10g) + (3 * 10g) + ... + (10 * 10g) = 550g&lt;br&gt;&#xA;&lt;br&gt;&#xA; However lets say #2 cheated, so now it's:&lt;br&gt;&#xA; (1 * 10g) + (2 * 9g) + (3 * 10g) + ... + (10 * 10g) = 548g&lt;br&gt;&#xA;&lt;br&gt;&#xA; 550 - 548 = 2, the number of our cheater!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4513" LastEditorUserId="4513" LastEditDate="2014-11-04T19:03:18.777" LastActivityDate="2014-11-04T19:03:18.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3553" PostTypeId="1" AcceptedAnswerId="3555" CreationDate="2014-11-04T19:06:34.713" Score="1" ViewCount="2164" Body="&lt;p&gt;So I have a question for you all, it's probably not too hard for you but it's relatively simple:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How many times does &lt;em&gt;variable&lt;/em&gt; appear?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hint for the variable is pretty short too:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am very small, you think of me little, you can find me in many places, even though I'm tiny. I am even in here.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll accept the answer with an explanation, and a spoiler.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="22922" LastEditDate="2016-09-19T09:08:48.987" LastActivityDate="2019-08-05T05:42:51.297" Title="The Answer is in the Question…" Tags="|riddle|" AnswerCount="6" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="3554" PostTypeId="2" ParentId="3553" CreationDate="2014-11-04T19:17:24.133" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The letter &quot;e&quot; appears 33 times including the number that are in the title.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2265" LastActivityDate="2014-11-04T19:17:24.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3555" PostTypeId="2" ParentId="3553" CreationDate="2014-11-04T19:17:26.790" Score="9" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 29, or 33 including the title of the question&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first question is what &quot;variable&quot; refers to. The hint says that it is in &quot;here&quot;. To me this indicates that it should be a letter, and since &quot;you can find [it] in many places&quot;, I believe it is the most common of the letters in the word &quot;here&quot; - 'e'. Then I counted the number of 'e's in the entire question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1955" LastActivityDate="2014-11-04T19:17:26.790" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3556" PostTypeId="2" ParentId="3538" CreationDate="2014-11-04T19:24:54.097" Score="3" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Pw9Fm.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure if that counts, as I didn't understand your statement of &quot;even though you haven't passed through all the walls in the building&quot; so I just did it until all walls had been passed anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key lies in making best use of room F and G, the only odd-wall-numbered rooms in the diagram.&#xA;Code: FoFDoDEoEGoGoBoAoFACFDCEGBCBA&lt;/p&gt;&#xA;" OwnerUserId="4514" LastActivityDate="2014-11-04T19:24:54.097" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3557" PostTypeId="2" ParentId="3553" CreationDate="2014-11-04T19:41:47.283" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My first bet was a period &lt;code&gt;.&lt;/code&gt;, the smallest visible character in the question. Found three times in the question body and three more times in the title.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4394" LastEditorUserId="44915" LastEditDate="2018-05-24T10:18:45.917" LastActivityDate="2018-05-24T10:18:45.917" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="3558" PostTypeId="2" ParentId="3550" CreationDate="2014-11-04T22:05:23.370" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Take 1 ring from the first servant, 2 rings from the second servant and so on. Put all those rings in a pile. Weigh the pile. If all rings were the same weight, the pile should weigh exactly &lt;code&gt;550 grams&lt;/code&gt;. Since one of the servants brought rings that weigh one gram less, the pile weighs now &lt;code&gt;550 - x&lt;/code&gt; grams. Therefore the culprit is the &lt;code&gt;x&lt;/code&gt;th servant.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1485" LastActivityDate="2014-11-04T22:05:23.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3559" PostTypeId="2" ParentId="3530" CreationDate="2014-11-04T22:14:55.193" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I would guess a dictionary. All the first words of those pairs are lexicographically after the second words.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1485" LastActivityDate="2014-11-04T22:14:55.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3561" PostTypeId="2" ParentId="3553" CreationDate="2014-11-05T03:37:17.377" Score="7" Body="&lt;p&gt;Question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How many times does &lt;em&gt;variable&lt;/em&gt; appear?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer: (is in the question)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Twice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The word &lt;em&gt;variable&lt;/em&gt; appears in the question once, and also once in the extraneous data.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4522" LastActivityDate="2014-11-05T03:37:17.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3562" PostTypeId="1" AcceptedAnswerId="3563" CreationDate="2014-11-05T04:47:30.887" Score="29" ViewCount="4309" Body="&lt;p&gt;Consider the following diagram of your prized 9' x 12' Persian rug:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/Zdjh2.png&quot; alt=&quot;Persian rug diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During some recent festivities a guest spilled wine on the rug, ruining the 8 square feet depicted in red.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To remedy the problem, you intend on cutting the rug into multiple pieces, moving and possibly rotating the pieces (2D rotations only; no turning pieces over), and gluing them back together again. You can cut any way you please&amp;mdash;vertically, horizontally, diagonally, or along any 2D curve&amp;mdash;but you must respect three rules:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the reassembled rug may not comprise more than two pieces&lt;/li&gt;&#xA;&lt;li&gt;none of the red area may appear on the reassembled rug&lt;/li&gt;&#xA;&lt;li&gt;the reassembled rug must be square or rectangular in shape and be a simple polygon (i.e. no holes).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Since this is indeed a very expensive Persian rug, your objective is to leave as much area as possible in the reassembled rug.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What square footage can you save, and how do you accomplish it?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Puzzlers are politely encouraged to place answers in spoiler blocks to avoid spoiling the fun for other readers. :)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="8874" LastEditDate="2016-03-23T15:39:57.077" LastActivityDate="2016-03-23T15:39:57.077" Title="Drink a Little Wine, Cut a Little Rug" Tags="|geometry|dissection|" AnswerCount="4" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="3563" PostTypeId="2" ParentId="3562" CreationDate="2014-11-05T05:13:32.460" Score="50" Body="&lt;p&gt;Quite an interesting cut I had to make to get it to fit.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/ppGXv.png&quot; alt=&quot;image of rug cut&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Shift the bottom piece 2 tiles up and one tile right to produce a 10 x 10 square. You can save all 100 square feet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The end result should look like this &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the yellow is the second piece&#xA;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/ivUGJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2313" LastEditDate="2014-11-08T03:06:58.900" LastActivityDate="2014-11-08T03:06:58.900" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3564" PostTypeId="1" AcceptedAnswerId="3566" CreationDate="2014-11-05T05:14:40.463" Score="17" ViewCount="2582" Body="&lt;p&gt;I am not from US. I have heard that US laws don't allow to cremate a man living in Texas, in New York. Can anyone tell me why?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: Don't think that I have posted this question here by accident. It's a puzzle. Really. Trust me. ;)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="-1" LastEditDate="2014-11-05T09:23:55.260" LastActivityDate="2016-05-20T22:08:57.080" Title="Cremation in New York" Tags="|lateral-thinking|" AnswerCount="6" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3565" PostTypeId="2" ParentId="3562" CreationDate="2014-11-05T05:23:22.717" Score="19" Body="&lt;p&gt;I think for the sake of simplicity this should be part of the answer &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You lost 8 squares, you have 108 to start with, so the maximum we can save is 100&#xA;&#xA;I propose 99 of them can be reused&#xA;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/8XX5w.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;br&gt;&#xA; cut the (2) above and paste it on top of the spoiled squares, and throw (1) away.&lt;br&gt;&#xA; The spoiled portions wont &quot;appear&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2313" LastActivityDate="2014-11-05T05:23:22.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3566" PostTypeId="2" ParentId="3564" CreationDate="2014-11-05T05:26:23.547" Score="31" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the man is living somewhere, he's probably alive... and cremating a man who is alive is murder and hence illegal&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4394" LastEditorUserId="2313" LastEditDate="2014-11-05T06:57:43.687" LastActivityDate="2014-11-05T06:57:43.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3567" PostTypeId="2" ParentId="3478" CreationDate="2014-11-05T07:19:32.787" Score="6" Body="&lt;p&gt;In my previous answer, I used only the circle to divide the square plate of chocolate into four equal pieces. In this answer, I show how to divide it equally using only the Y shaped thingy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general idea is to use thingy to make four long cuts toward the center, each identical relative to its side of the square; and then make two slices across the short gaps to complete the cutting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 1, Mark midpoints of sides:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On each side of the square, place thingy with one post-end in a corner, and the Y-junction clockwise (of that corner) on the side.  [See first diagram.]  At each placement, mark the Y-junction location (&lt;em&gt;ie&lt;/em&gt;, the midpoint of the side).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Part 2, Make four long cuts toward the center:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On each side of the square, place one thingy post-end at the side's mid-point mark, and another post-end against the counterclockwise adjacent side.  [See second diagram.]  Cut from the mid-point's post-end up to the Y-junction.  [Net result is as shown in third diagram.]  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Part 3, Make two slices across the short gaps to complete the cutting:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Make two short cuts across the plate-center to connect the ends of opposite long cuts, as shown in red in the fourth diagram, using any leg of thingy as a straightedge.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hLoAL.png&quot; alt=&quot;Y thingy on left side&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/7wGmX.png&quot; alt=&quot;Y thingy from midpoint to side&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/LsqpO.png&quot; alt=&quot;After first 4 cuts&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/xYsjm.png&quot; alt=&quot;After last 2 cuts&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit:&lt;/em&gt; A slight variation of this method allows cutting the large square into four equal small squares (instead of shapes with six edges each) using only the 3-leg thingy.  Rather than making the long cuts shown in green in the fourth diagram, just mark their endpoints near the center,and make the red lines as marks also.  Their intersection marks the center of the plate.  Now use a thingy leg as a straightedge to cut from each side's midpoint to the center.&lt;/p&gt;&#xA;" OwnerUserId="2004" LastEditorUserId="2004" LastEditDate="2014-11-05T16:11:41.973" LastActivityDate="2014-11-05T16:11:41.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3568" PostTypeId="2" ParentId="3478" CreationDate="2014-11-05T09:26:56.973" Score="3" Body="&lt;p&gt;Cut the circle in half using the plate as a guide. Perfectly align the two pieces edge to edge and use them as a straight edge to cut the cake diagonally, twice.&lt;/p&gt;&#xA;" OwnerUserId="4522" LastActivityDate="2014-11-05T09:26:56.973" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3569" PostTypeId="2" ParentId="3564" CreationDate="2014-11-05T10:20:29.747" Score="7" Body="&lt;p&gt;Obviously laws will not allow you to cremate a man in New York if he's living in Texas, infact it don't matter  where he is, as long as&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He's living&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cremating someone like that is infact&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Murder&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Making it illegal!&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2014-11-05T10:20:29.747" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3570" PostTypeId="1" AcceptedAnswerId="3571" CreationDate="2014-11-05T11:34:11.583" Score="11" ViewCount="4379" Body="&lt;p&gt;Here is a quick riddle.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We are twins. Between us something smells. Can you guess who we are??&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Please place your answers in spoiler tags.&lt;/p&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="-1" LastEditDate="2014-11-07T06:27:08.460" LastActivityDate="2015-08-21T13:52:31.377" Title="Guess the twins with something that smells between them" Tags="|riddle|" AnswerCount="4" CommentCount="0" ClosedDate="2015-08-21T13:15:24.157" ContentLicense="CC BY-SA 3.0" />
  <row Id="3571" PostTypeId="2" ParentId="3570" CreationDate="2014-11-05T11:36:46.467" Score="24" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Twins are eyes and nose smells&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4506" LastActivityDate="2014-11-05T11:36:46.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3572" PostTypeId="2" ParentId="3570" CreationDate="2014-11-05T11:38:47.320" Score="68" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Buttcheeks, because.. I don't explain it now :D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4527" LastActivityDate="2014-11-05T11:38:47.320" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3573" PostTypeId="2" ParentId="3570" CreationDate="2014-11-05T11:46:13.527" Score="27" Body="&lt;p&gt;The word &quot;between&quot; has a couple of meanings - shyos's answer is the usual meaning, but there's also this option:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Two nostrils - between them, they make up a nose, which smells.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2518" LastActivityDate="2014-11-05T11:46:13.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3582" PostTypeId="2" ParentId="3564" CreationDate="2014-11-05T12:50:08.653" Score="16" Body="&lt;p&gt;While there are two existing good answers, the better answer is logistics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an existing history of men &lt;a href=&quot;https://en.wikipedia.org/wiki/Capital_punishment_in_Texas&quot;&gt;living in Texas&lt;/a&gt; being &lt;a href=&quot;https://en.wikipedia.org/wiki/Electric_chair&quot;&gt;cremated&lt;/a&gt;.  The problem is that if they are living in Texas, it is physical impossible to cremate them in New York, so it is a law of Physics not US law, that prevents it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not unusual for various &lt;a href=&quot;https://en.wikipedia.org/wiki/Nix_v._Hedden&quot;&gt;US laws to attempt to alter empirical truths&lt;/a&gt;, which is probably the root of the confusion in your question. &lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2014-11-05T12:50:08.653" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3583" PostTypeId="1" AcceptedAnswerId="3590" CreationDate="2014-11-05T14:38:45.820" Score="7" ViewCount="1024" Body="&lt;pre&gt;&lt;code&gt; _ _ _ _ _ × 8 = _ _ _ _ _&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There is a number that you can put in the left-hand side of the above equation so that the blanks are filled in using every digit from 0 to 9 exactly once. What is that number?&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="11876" LastEditDate="2016-06-18T01:04:50.900" LastActivityDate="2016-06-18T01:04:50.900" Title="A small pandigital equation" Tags="|calculation-puzzle|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3584" PostTypeId="5" CreationDate="2014-11-05T14:39:18.350" Score="0" Body="&lt;p&gt;A pandigital puzzle is a puzzle that involves inserting &quot;every digit&quot; (usually from 0 to 9 or 1 to 9) into some equation or construction exactly once to satisfy the conditions of the puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There exists a number &lt;code&gt;n&lt;/code&gt; where:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;n × 1 = _ _ _&#xA;n × 2 = _ _ _&#xA;n × 3 = _ _ _&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;  &#xA;  &lt;p&gt;the blanks of the above three equations can be filled in using the numbers 1 to 9 exactly once.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are four answers to this question: 192, 219, 273, and 327. (&lt;a href=&quot;http://oeis.org/A039667&quot; rel=&quot;nofollow&quot;&gt;OEIS A039667&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2014-11-05T14:56:53.670" LastActivityDate="2014-11-05T14:56:53.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3585" PostTypeId="4" CreationDate="2014-11-05T14:39:18.350" Score="0" Body="A puzzle that involves inserting every digit from 0 or 1 to 9 exactly once in an equation." OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2014-11-05T14:39:18.350" LastActivityDate="2014-11-05T14:39:18.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3586" PostTypeId="2" ParentId="3583" CreationDate="2014-11-05T14:42:49.380" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oddly enough, the numbers are 12345 &amp;amp; 98760. I knew that the first two digits of the top had to be low numbers, 1 and 2, so I just tried it out, and it worked!&lt;br&gt;&#xA;&lt;br&gt;&#xA; This worked out well since the 1's digit of the multiplication (before carrying) ends up being 86420 (8 stepping down by 2), and we have carry values of 12340&lt;br&gt;&#xA; So, we get:&lt;br&gt;&#xA; 8 - 0 + 1 = 9&lt;br&gt;&#xA; 8 - 2 + 2 = 8&lt;br&gt;&#xA; 8 - 4 + 3 = 7&lt;br&gt;&#xA; 8 - 6 + 4 = 6&lt;br&gt;&#xA; 8 - 8 + 0 = 0&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2014-11-05T16:57:56.247" LastActivityDate="2014-11-05T16:57:56.247" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3589" PostTypeId="2" ParentId="3564" CreationDate="2014-11-05T14:58:04.440" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I guess it would be discouraged in general to cremate any human being, who is &lt;strong&gt;living&lt;/strong&gt; in some country.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1485" LastActivityDate="2014-11-05T14:58:04.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3590" PostTypeId="2" ParentId="3583" CreationDate="2014-11-05T15:23:29.397" Score="18" Body="&lt;p&gt;There is more than one correct answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (10459, 83672), (10469, 83752), (10537, 84296), (10579, 84632), (10592, 84736), (10674, 85392), (10679, 85432), (10742, 85936), (10794, 86352), (10932, 87456), (10942, 87536), (10953, 87624), (10954, 87632), (12073, 96584), (12307, 98456), (12345, 98760)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4536" LastActivityDate="2014-11-05T15:23:29.397" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3591" PostTypeId="2" ParentId="3564" CreationDate="2014-11-05T16:05:37.997" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The laws don't allow for impossibilities. If he's living (alive) in Texas, he can't be in New York.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4456" LastActivityDate="2014-11-05T16:05:37.997" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3592" PostTypeId="2" ParentId="3562" CreationDate="2014-11-05T17:04:56.910" Score="10" Body="&lt;p&gt;Otaia's answer is optimal for square footage. This provides the same results as skv's without any unsightly bumps...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4tHzv.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;Cut a strip off the end (1)&#xA;Cut a 9 square section (2) out of the center and discard. (include one non-soiled square in line with the 8 square wine soiled section.)&#xA;rotate (1) and insert in the hole left by (2)&#xA;results in a 9'x11' rug. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2495" LastActivityDate="2014-11-05T17:04:56.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3593" PostTypeId="1" AcceptedAnswerId="3594" CreationDate="2014-11-05T19:19:37.437" Score="-4" ViewCount="576" Body="&lt;p&gt;A small family of chickens were discussing very important business, the conversation went like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The farmer needs eggs! Someone has to lay them!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;No quiet down! You have to be quiet! The farmer will hear you!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;But mom!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;No buts! I'll lay the eggs! Go back to your rooms!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Fine!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later that night 2 chicks woke up. Chick#1 told Chick#2 &quot;I can lay eggs!&quot; Chick#2 said &quot;Prove it!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So he went to the roof, and laid as many eggs as he could!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which way did the eggs roll? Towards the left or right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The roof is shaped like a triangular prism.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="5373" LastEditDate="2015-04-28T19:39:48.050" LastActivityDate="2015-04-28T19:39:48.050" Title="Eggs on the Roof!" Tags="|lateral-thinking|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3594" PostTypeId="2" ParentId="3593" CreationDate="2014-11-05T19:21:43.520" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The eggs didn't roll either way since there were no eggs since the chick (he) was male and therefore couldn't lay eggs. So, looks like Chick #1 lost the dare. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2014-11-05T19:21:43.520" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3595" PostTypeId="2" ParentId="3583" CreationDate="2014-11-05T19:29:28.213" Score="4" Body="&lt;p&gt;A list of all the possible solutions to this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(03187, 25496)&#xA;(04589, 36712)&#xA;(04591, 36728)&#xA;(04689, 37512)&#xA;(04691, 37528)&#xA;(04769, 38152)&#xA;(05237, 41896)&#xA;(05371, 42968)&#xA;(05789, 46312)&#xA;(05791, 46328)&#xA;(05839, 46712)&#xA;(05892, 47136)&#xA;(05916, 47328)&#xA;(05921, 47368)&#xA;(06479, 51832)&#xA;(06741, 53928)&#xA;(06789, 54312)&#xA;(06791, 54328)&#xA;(06839, 54712)&#xA;(07123, 56984)&#xA;(07312, 58496)&#xA;(07364, 58912)&#xA;(07416, 59328)&#xA;(07421, 59368)&#xA;(07894, 63152)&#xA;(07941, 63528)&#xA;(08174, 65392)&#xA;(08179, 65432)&#xA;(08394, 67152)&#xA;(08419, 67352)&#xA;(08439, 67512)&#xA;(08932, 71456)&#xA;(08942, 71536)&#xA;(08953, 71624)&#xA;(08954, 71632)&#xA;(09156, 73248)&#xA;(09158, 73264)&#xA;(09182, 73456)&#xA;(09316, 74528)&#xA;(09321, 74568)&#xA;(09352, 74816)&#xA;(09416, 75328)&#xA;(09421, 75368)&#xA;(09523, 76184)&#xA;(09531, 76248)&#xA;(09541, 76328)&#xA;(10459, 83672)&#xA;(10469, 83752)&#xA;(10537, 84296)&#xA;(10579, 84632)&#xA;(10592, 84736)&#xA;(10674, 85392)&#xA;(10679, 85432)&#xA;(10742, 85936)&#xA;(10794, 86352)&#xA;(10932, 87456)&#xA;(10942, 87536)&#xA;(10953, 87624)&#xA;(10954, 87632)&#xA;(12073, 96584)&#xA;(12307, 98456)&#xA;(12345, 98760)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Don't forget you can use 0 at the start of the first number :)&lt;/p&gt;&#xA;" OwnerUserId="4544" LastActivityDate="2014-11-05T19:29:28.213" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3597" PostTypeId="2" ParentId="3570" CreationDate="2014-11-05T19:40:17.527" Score="10" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Two human twins. They both lost their deodorant.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4456" LastEditorUserId="4456" LastEditDate="2014-11-05T20:04:01.097" LastActivityDate="2014-11-05T20:04:01.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3598" PostTypeId="1" AcceptedAnswerId="3601" CreationDate="2014-11-05T21:07:33.023" Score="20" ViewCount="4538" Body="&lt;p&gt;D'alar'cop and Rand al'Thor wake up and find themselves trapped in a square shaped room.  Suddenly, all four walls start moving at the exact same time, and each of them moves at the same speed towards the center of the room.  D'alar'cop and Rand al'Thor realize they will eventually be crushed, but they are surprised that the walls aren't colliding with each other at the edges.  Explain how this is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional details:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The ceiling and floor do not move.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4450" LastEditDate="2015-05-19T01:44:12.433" LastActivityDate="2015-05-19T01:44:12.433" Title="People trapped in a shrinking room" Tags="|mechanical-puzzles|" AnswerCount="5" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="3600" PostTypeId="1" AcceptedAnswerId="3603" CreationDate="2014-11-05T21:37:48.133" Score="22" ViewCount="3477" Body="&lt;p&gt;I am trying something new. This detective riddle/brainteaser will be the first of 5+ parts in a case to solve the murder of the president. Each part will give you a clue (usually a place to go for the next clue) and you must solve the clue. Use all knowledge you have of cryptography, ciphers, etc. This case is meant to take place in the present day, so all politicians, celebrities, places, etc. are what they are now. Please post your answers in spoiler tags.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the riddle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The President of the United States has been murdered and you are on the case.&#xA;The date is November 5, 2014 at 4:03 p.m. when you begin to investigate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what you know so far:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Location: Oval Office, Washington D.C., President found dead at his desk&lt;br&gt;&lt;br&gt;&#xA;  Time of Death: November 5, 2014 at 1:09 p.m.&lt;br&gt;&lt;br&gt;&#xA;  Body Found: 1:34 p.m. &lt;br&gt;&lt;br&gt;&#xA;  Death Wound: Stabbed in the heart (immediate death)&lt;br&gt;&lt;br&gt;&#xA;  Killer's Entrance: Used a rope to get to the window; the window was unlocked from the inside (possible sabotage).&lt;br&gt;&lt;br&gt;&#xA;  Killer's Escape Route: Out of the window&lt;br&gt;&lt;br&gt;&#xA;  Evidence: No fingerprints or other DNA evidence, only a typed, coded note left on the President's desk.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is the note:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ulea jsbl ha LDY Zvbaodlza H&lt;br&gt;&#xA;  Fvby Mypluk,&lt;br&gt;&#xA;  Aol Rpssly&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Scribbled in the corner of the note are two words: &lt;em&gt;Romans 7&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's up to you to decode this message and find out where this clue leads you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good Luck&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note&lt;br&gt;&#xA;I will be posting Part 2 in one to two days. I will select the correct answer for Part 1 before, but I will post the answer for Part 1 if no one gets it&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1711" LastEditorUserId="5373" LastEditDate="2014-12-15T23:00:26.087" LastActivityDate="2014-12-15T23:00:26.087" Title="Murder of the President - Part 1" Tags="|mystery|" AnswerCount="3" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="3601" PostTypeId="2" ParentId="3598" CreationDate="2014-11-05T21:38:56.407" Score="38" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The walls remain perpendicular, while moving diagonally. The left edge of each wall slides along the neighboring wall instead of colliding with it.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/5bqSa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4536" LastEditorDisplayName="user20" LastEditDate="2014-11-05T23:40:10.110" LastActivityDate="2014-11-05T23:40:10.110" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="3602" PostTypeId="2" ParentId="3598" CreationDate="2014-11-05T21:39:11.893" Score="0" Body="&lt;p&gt;Here's how:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The walls are made of sand (let's say), while the ceiling and floor are steel. The force on all four walls is equal, and equal at every point. Also, we are in space so there is no (or very little) gravitational force pulling the sand to the floor.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4546" LastEditorUserId="-1" LastEditDate="2014-11-06T11:06:53.447" LastActivityDate="2014-11-06T11:06:53.447" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3603" PostTypeId="2" ParentId="3600" CreationDate="2014-11-05T21:53:32.390" Score="14" Body="&lt;p&gt;The secret note reads &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Next clue at EWR Southwest A&lt;br&gt;&#xA; Your Friend,&lt;br&gt;&#xA; The Killer&lt;br&gt;&#xA;&lt;br&gt;&#xA; I'm not quite sure why though, Romans is a hint to the Caesar cipher, but I'm not sure what the significance of 7 is. The key for deciphering is 19, but I reached that after a few tries.&lt;br&gt;&#xA;&lt;br&gt;&#xA; EWR Southwest A looks like an airport code, and matches &lt;a href=&quot;http://www.panynj.gov/airports/ewr-airport-map.html&quot;&gt;the Newark Liberty International Airport (EWR)&lt;/a&gt;. Southwest A would likely mean the southwest part of the A terminal on the map.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="340" LastEditorUserId="340" LastEditDate="2014-11-05T22:04:57.230" LastActivityDate="2014-11-05T22:04:57.230" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="3605" PostTypeId="2" ParentId="3598" CreationDate="2014-11-05T22:06:25.387" Score="3" Body="&lt;p&gt;A limitation of the solution given by eclipz905 is that portions of the walls would be moving away from the center.  An alternative approach would be to have walls of a flexible or articulated material which bend out 45 degrees at the corners.  The parts of the wall which were not yet in contact with the adjoining wall would travel perpendicular to their plane toward the center, while those which had turned the corner would be moving directly toward the center at a slower rate.  No part of any wall would be moving away from the center.&lt;/p&gt;&#xA;" OwnerUserId="1467" LastActivityDate="2014-11-05T22:06:25.387" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3606" PostTypeId="2" ParentId="3600" CreationDate="2014-11-05T22:06:38.043" Score="2" Body="&lt;p&gt;The note says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Next clue at EWR Southwest A&lt;br&gt;&#xA; Your Friend,&lt;br&gt;&#xA; The Killer&lt;br&gt;&#xA;&lt;br&gt;&#xA; The next clue is at the southwest airport;&lt;br&gt;&#xA; Newark, NJ - EWR&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks Nit for providing an answer with the text of the note. Upvote his answer!&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2518" LastEditDate="2014-11-06T12:28:27.290" LastActivityDate="2014-11-06T12:28:27.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3607" PostTypeId="1" AcceptedAnswerId="3610" CreationDate="2014-11-06T00:31:00.367" Score="8" ViewCount="5159" Body="&lt;p&gt;Is it possible to form a $3\mbox{x}3$ grid containing the set of numbers: $${1,2,4,8,16,32,64,128,256}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in such a way that the product of the numbers in every row, column and diagonal are the same? If so, how?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorDisplayName="user20" LastEditDate="2015-09-10T16:18:10.363" LastActivityDate="2018-02-20T19:05:10.987" Title="Can you fill a 3x3 grid with these numbers so the products of the rows and columns are the same?" Tags="|mathematics|magic-square|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3608" PostTypeId="1" AcceptedAnswerId="3611" CreationDate="2014-11-06T00:39:02.027" Score="3" ViewCount="642" Body="&lt;p&gt;How many possibilities are there for $n$ and $m$ when given this equation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n^m$ = $m^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many possibilities are there when dealing with positive integers? What about any integers? Or fractions and decimals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Please include if there are a pattern to the numbers; that is, if any exist&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus: What if we open it up to imaginary numbers? Complex numbers included!&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="4421" LastEditDate="2015-10-05T16:19:16.923" LastActivityDate="2015-10-05T16:19:16.923" Title="Solve for $n$ and $m$ [exponents]" Tags="|mathematics|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3609" PostTypeId="2" ParentId="3608" CreationDate="2014-11-06T00:45:18.093" Score="1" Body="&lt;p&gt;Short answer, infinite: of the form $n = m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is at least one other: $(2,4)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Set $n = m^x$, then $n^m = (m^x)^m = m^{m^x}$ implying $mx = m^x$, and $x = m^{(x-1)}$.&#xA;This should have 2 solutions for $m &amp;gt; 1$, one of which is the trivial $x = 1$, which leads to $m = n$, but the other will produce $m \ne n$.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastEditorUserId="4552" LastEditDate="2014-11-06T00:58:23.690" LastActivityDate="2014-11-06T00:58:23.690" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3610" PostTypeId="2" ParentId="3607" CreationDate="2014-11-06T00:46:04.420" Score="20" Body="&lt;p&gt;This is just a &lt;a href=&quot;http://en.wikipedia.org/wiki/Lo_Shu_Square&quot; rel=&quot;noreferrer&quot;&gt;standard Lo Shu Square&lt;/a&gt; with exponents and multiplication instead of addition. But because the multiplication of exponents with the same base is analogous to the addition of their exponents, the solution here is just the same as $2^\left(L-1\right)$, where $L$ is the standard Lo Shu Square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an expanded explanation, your numbers are better expressed as: $2^0, 2^1, 2^2, 2^3, 2^4, 2^5, 2^6, 2^7, 2^8$. We can replace the whole &quot;product&quot; idea completely with just adding the powers of these exponents, by the rules of exponentiation. $a^n*a^m=a^\left(n+m\right)$, so we need to make a magic square that adds the numbers $[0,7]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that the numbers span $[0,8]$, and since we know the regular Lo Shu Square operates from $[1,9]$, we just subtract one from every position, making each row and column add to $15-3=12$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{|c|c|c|}\hline 4&amp;amp;9&amp;amp;2\\\hline 3&amp;amp;5&amp;amp;7\\\hline 8&amp;amp;1&amp;amp;6\\\hline\end{array} \overset{-1}{\rightarrow}&#xA;\begin{array}{|c|c|c|}\hline 3&amp;amp;8&amp;amp;1\\\hline 2&amp;amp;4&amp;amp;6\\\hline 7&amp;amp;0&amp;amp;5\\\hline\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we exponentiate two by this grid we end up with a grid such that all numbers, when multiplied, will be the same (by rules of exponents). If we take any three numbers $a+b+c=12$ from one row or column, then $2^a*2^b*2^c=2^{(a+b+c)}=2^{12}$. We can see that this is evidently true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{|c|c|c|}\hline 2^3&amp;amp;2^8&amp;amp;2^1\\\hline 2^2&amp;amp;2^4&amp;amp;2^6\\\hline 2^7&amp;amp;2^0&amp;amp;2^5\\\hline\end{array} = \begin{array}{|c|c|c|}\hline 8&amp;amp;256&amp;amp;2\\\hline 4&amp;amp;16&amp;amp;64\\\hline 128&amp;amp;1&amp;amp;32\\\hline\end{array}$$&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorDisplayName="user20" LastEditDate="2015-01-15T18:21:37.397" LastActivityDate="2015-01-15T18:21:37.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3611" PostTypeId="2" ParentId="3608" CreationDate="2014-11-06T00:54:31.040" Score="14" Body="&lt;p&gt;Obviously, $m=n$ is always a solution, so we restrict ourselves to $m \neq n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For positive $m,n$, the equation is equivalent to $\frac{\ln m}{m} = \frac{\ln n}{n}$. So, we are looking for collisions in the function $f(x) = \ln(x) / x$. Taking derivatives, we see that it has a unique local maximum at $x=e$ and falls to 0 as $x \to \infty$. So, there are a continuum of real solutions: each real value $n&amp;gt;e$ has a counterpart $m$ with $1&amp;lt;m&amp;lt;e$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For natural numbers, the above observation limits the lesser number $m$ to $2$, so the only solution is the pair $(2,4)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are no further solutions among integers. The 0 and -1 cases can be eliminated separately. If one number is negative and one is positive, one exponent will be less in one in absolute value, and the other greater. Finally, if both are negative (well-defined if only both are odd), taking the negative reciprocal gives a positive integer solution, of which there are none for odds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the complex case, the problem is not well-defined because the complex exponential (or, logarithm) is multivalued.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2014-11-06T00:59:59.843" LastActivityDate="2014-11-06T00:59:59.843" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="3613" PostTypeId="1" AcceptedAnswerId="3616" CreationDate="2014-11-06T01:27:22.713" Score="12" ViewCount="2920" Body="&lt;p&gt;Consider a variant of Tic-Tac-Toe and &lt;a href=&quot;http://en.wikipedia.org/wiki/Gomoku&quot;&gt;Gomoku&lt;/a&gt; on a board that's infinite in every direction. &lt;code&gt;X&lt;/code&gt; wins by getting a line of 1000 &lt;code&gt;X&lt;/code&gt;'s horizontally, vertically, or diagonally. &lt;code&gt;O&lt;/code&gt; cannot win but plays for a draw by preventing &lt;code&gt;X&lt;/code&gt; from winning indefinitely. &lt;code&gt;X&lt;/code&gt; gets to move twice each time and starts, so the turn order goes &lt;code&gt;XXOXXOXXO...&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can &lt;code&gt;X&lt;/code&gt; win with perfect play?&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2014-11-06T01:36:03.623" LastActivityDate="2014-11-06T15:55:50.747" Title="Tic-tac-toe 1000 in a row" Tags="|game-theory|pencil-and-paper-games|tic-tac-toe|" AnswerCount="3" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3614" PostTypeId="2" ParentId="3598" CreationDate="2014-11-06T01:50:23.030" Score="16" Body="&lt;p&gt;One other idea, partially alluded to by one of warspyking's answers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The walls are made out of a large number of very thin vertical structural members (as with a shelf of books, or a ream of paper stood on its edge).&#xA; All the structural members move toward the center except for the outermost ones at each corner, which stay in place.&#xA;&amp;nbsp;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/1loJY.png&quot; alt=&quot;Decreasing book/walls illustration.&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4554" LastEditorUserId="4450" LastEditDate="2014-11-06T02:04:36.880" LastActivityDate="2014-11-06T02:04:36.880" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3616" PostTypeId="2" ParentId="3613" CreationDate="2014-11-06T03:02:49.753" Score="24" Body="&lt;p&gt;Yes, X can win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To simplify things I'm going to take advantage of your rules that O cannot win, so that X doesn't need to worry about O getting 1000 in a row.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just consider a 1 dimensional game, chose an origin, and label the locations with integers in order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define the &quot;bin $K$&quot; as the set of spaces $x$ such that $1000\ K \le x &amp;lt; 1000 (K+1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple winning strategy is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;For the first $M$ turns, have X play each piece in a previously empty bin.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Now while possible, have X play each piece in a bin with only one of its pieces and no O pieces.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Now while possible, have X play each piece in a bin with only two of its pieces an no O pieces.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;And so on.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It should be intuitively clear that X will always win, with $M$ chosen sufficiently large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove this without relying on intuition, I only need to show there is an answer to&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What value of $M$ is sufficient if the win condition for X is to get $Z$ pieces in-a-row?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After $M$ turns, X will have a piece in $2M$ bins, and O can block at most $M$ of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the next $M/2$ turns, X will have two pieces in $M$ bins, and O can block at most $M/2$ of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, after $\sum_{t=0}^N M/2^t$ turns, X will have $(N+1)$ pieces in $M/2^N$ bins.  Note that $M$ must be at least $2^N$ for the last number of turns in that sequence to be an integer. So $M \ge 2^N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a sufficient value is: &#xA;$$Z = N+1 \quad\quad \rightarrow \quad\quad M \ge 2^{Z-1} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the rules are changed such that O can win with 1000 in a row, then the math is a bit more complicated.  But you could imagine X infrequently using a turn here or there to stop O and then continue on with the plan.  X should still be able to win.&lt;/p&gt;&#xA;" OwnerUserId="4351" LastEditorUserId="4351" LastEditDate="2014-11-06T03:38:20.387" LastActivityDate="2014-11-06T03:38:20.387" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3618" PostTypeId="2" ParentId="3608" CreationDate="2014-11-06T03:27:48.937" Score="9" Body="&lt;p&gt;$\newcommand{\deriv}[2]{\frac{d#1}{d#2}}$&#xA;$\newcommand{\pderiv}[2]{\frac{\partial #1}{\partial #2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any algebraic system that supports exponentiation, $m = n$ is a trivial, valid solution.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If $m \neq n$ is enforced, then in the natural numbers, $2^4 = 4^2$ is the only solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is easy to see if we try and figure out the solution for the real numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the real numbers, things get a little tricky. Suppose we start with the equation $m^n = n^m$. Then, if we start taking logarithms and moving things around, we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;m^n &amp;amp;= n^m\\&#xA;\log m^n &amp;amp;= \log n^m\\&#xA;n \log m &amp;amp;= m \log n\\&#xA;\frac{\log m}{m} &amp;amp;= \frac{\log n}{n}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We've assumed that $m$ and $n$ are both strictly greater than $0$, because otherwise $\log m$ or $\log n$ is undefined. Now, let's analyze the behaviour of $m$ and $n$ in the intervals $(0, 1]$ and $(1, \infty)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f(x) = \frac{\log x}{x}$. $\deriv{f}{x}$ is equal to $\frac{1 - \log x}{x^2}$ by the quotient rule. In $(0, 1]$, this is always positive, and therefore the function is monotone in $(0, 1]$. This means that there cannot be two different values in the interval $(0, 1]$ that have equal values of $f(x)$. However, in $(1, \infty)$, $f(x)$ is always positive, and in $(0, 1]$ it's either 0 or negative, so no value from $(0, 1]$ can be paired with a value greater than $1$ either. Thus, we've determined that $m$ and $n$ must be strictly greater than $1$ as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's continue to analyze $f(x)$ when $x &amp;gt; 1$. When $1 &amp;lt; x &amp;lt; e$, $f'(x)$ is positive. When $x &amp;gt; e$, $f'(x)$ is negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $f'(x)$ has a single local maximum on $(1, \infty)$ at $e$, and is monotone in either direction (and therefore injective for that interval). Moreover, $\lim_{x \to \infty} f(x) = 0$, so the function is surjective and therefore bijective when mapping from $(e, \infty)$ to $(0, \frac{1}{e})$, as it is when mapping from $(1, e)$ to $(0, \frac{1}{e})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for any number $1 &amp;lt; x &amp;lt; e$, there exists exactly one other value $e &amp;lt; y &amp;lt; \infty$ such that $x^y = y^x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the only integer in $1 &amp;lt; x &amp;lt; e$ is 2, $2^4 = 4^2$ is the only solution in the positive integers. (Note that ${-2}^{-4} = {-4}^{-2}$ is also a solution in the integers in general, but we assumed above that $m$ and $n$ were positive. $-2$ and $-4$ are not part of this analytic mapping that we came up with above, because in general, real non-integer exponents of negative numbers are not defined in the real numbers.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now we need to come up with a way to characterize all the ordered pairs $(m, n)$ that satisfy this equation in the real numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can't just solve the equation for $n$ directly by plugging $m$ in - the inverse of $f(x) = \frac{\log x}{x}$ involves the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lambert_W_function&quot;&gt;Lambert W function&lt;/a&gt;, which can't be expressed in terms of elementary functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So instead, we need something more clever. Suppose $n = am$ for some $a \in \mathbb R, a &amp;gt; 1$. Then the equation becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\frac{\log m}{m} &amp;amp;= \frac{\log am}{am}\\&#xA;\frac{\log m}{m} &amp;amp;= \frac{\log a + \log m}{am}\\&#xA;a \log m &amp;amp;= \log a + \log m\\&#xA;(a - 1) \log m &amp;amp;= \log a\\&#xA;\log m &amp;amp;= \frac{\log a}{a-1}\\&#xA;m &amp;amp;= e^{\frac{\log a}{a-1}}\\&#xA;m &amp;amp;= \sqrt[a-1]{a}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we get an equation for $m$ in terms of $a$, which allows us to deduce $n$ as well. If $m = \sqrt[a-1] a$, then $n$ just equals $a\sqrt[a-1]a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we write $a$ in the form of $1 + \frac 1x$, we get $m = (1 + \frac 1x)^x$ and $n = (1 + \frac 1x)^{x+1}$, which is a solution for all positive real numbers $x$. It so happens that the $(2, 4)$ case arises when $x = 1$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Complex numbers to come later.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2014-11-06T06:30:54.403" LastActivityDate="2014-11-06T06:30:54.403" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3619" PostTypeId="1" AcceptedAnswerId="3620" CreationDate="2014-11-06T04:05:16.717" Score="10" ViewCount="4179" Body="&lt;p&gt;What word follows AMPLE in this sequence:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;REDDEN&lt;/p&gt;&#xA;&lt;p&gt;SEAM&lt;/p&gt;&#xA;&lt;p&gt;PLEA&lt;/p&gt;&#xA;&lt;p&gt;REARED&lt;/p&gt;&#xA;&lt;p&gt;DENSE&lt;/p&gt;&#xA;&lt;p&gt;AMPLE&lt;/p&gt;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-11-11T18:24:47.220" Title="What is the next word in this sequence?" Tags="|word|pattern|letters|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3620" PostTypeId="2" ParentId="3619" CreationDate="2014-11-06T04:09:18.703" Score="9" Body="&lt;p&gt;Going by the three words &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;SEAM&#xA;&#xA;PLEA&#xA;&#xA;REARED&#xA;&lt;br&gt;&#xA;Since these words are formed by continuation Ample is formed by SE &quot;AM-PLE&quot; leaving A and the next word that can be formed with the remaining letters of the next word&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Would it be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;AREA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This would be followed by REDDEN followed by SEAM making it a full sequence or just saying &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first 17 letters are repeated&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="2313" LastEditDate="2014-11-06T05:49:43.897" LastActivityDate="2014-11-06T05:49:43.897" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3621" PostTypeId="2" ParentId="3613" CreationDate="2014-11-06T04:15:20.183" Score="0" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;EDIT:&lt;/em&gt;&lt;/strong&gt; I withdraw my solution, I somehow missed the 2-X moves for 1 -O move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You assume that O's priority is to immediately block bins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I was O,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X plays in 2M bins. O plays to the left in 2M bins, so far no bins blocked.&#xA;X plays 2nd consecutive piece in any bin, O plays to the right of it, blocking it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This strategy allows X to open unlimited bins at count 1 but 0 bins at count 2, but they will be thwarted before it gets to 1000. ie your third step of play in a bin with 2X and no O will never happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way to answer this question is to look at the directional advantages offered by clustering pieces together. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is needed is a cluster of pieces that for x number of X pieces, it is neccessary to place x+(sum 1 to 999) pieces to block all paths. This will allow for O blocking first path at 1 piece, 2nd path after 2 pieces, ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best strategy will involve X building a giant grid (each piece 1000 away from next one) and then trying to put pieces in the middle of the grids to slowly make more boxes. This will gradually increase the edge count that O must cover. However from what I have worked through, this will not reach the required excess edge count. Therefore X will not win. &lt;/p&gt;&#xA;" OwnerUserId="2725" LastEditorUserId="2725" LastEditDate="2014-11-06T05:42:19.433" LastActivityDate="2014-11-06T05:42:19.433" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="3622" PostTypeId="2" ParentId="3534" CreationDate="2014-11-06T04:36:16.110" Score="0" Body="&lt;p&gt;His pants were wet from water used to put out the fire. Since the men putting out the fire were on the south end, they would have seen the bandits leaving if the bandits had gone south. The bandits went north.&lt;/p&gt;&#xA;" OwnerUserId="4560" LastActivityDate="2014-11-06T04:36:16.110" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3623" PostTypeId="1" CreationDate="2014-11-06T05:04:00.537" Score="15" ViewCount="1241" Body="&lt;p&gt;Consider the game surface to be an infinite N dimensional Cartesian lattice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rules are &lt;code&gt;X&lt;/code&gt; moves first, but &lt;code&gt;O&lt;/code&gt; gets to move twice each turn, so the move order goes &lt;code&gt;XOOXOOXOO...&lt;/code&gt;.  The goal is for &lt;code&gt;X&lt;/code&gt; to get as long a sequence of neighboring &lt;code&gt;X&lt;/code&gt; pieces in a straight line as possible. The goal of &lt;code&gt;O&lt;/code&gt; is only to keep the line &lt;code&gt;X&lt;/code&gt; may get as short as possible.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given perfect play and some dimension N, what is the maximum number of pieces in-a-line that &lt;code&gt;X&lt;/code&gt; can get in this game?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To make it clear what is meant by in-a-line in this context, consider a vector displacement between points on the lattice.  If all the components of this vector are -1, 0, or +1 and the norm of this vector is greater than zero, then this displacement will take you from one point to a neighboring point (let's use this as the definition of neighbor).  If there is a sequence of K neighbors, separated by a constant displacement vector, then these K points are &quot;in-a-line&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for N=1 we just have a line, with each point having two neighbors.  For N=2 the game surface is a plane (like normal Tic-Tac-Toe, but infinite), with each point having 8 neighbors (4 on &quot;diagonals&quot;).  For N=3, each point has 26 neighbors (20 on &quot;diagonals&quot;).  Etc.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;N=1 is trivial, yet already N=2 seems hard.  It is not difficult for &lt;code&gt;X&lt;/code&gt; to get a line of 3 in N=2, but I can't even convince myself if a line of 4 is impossible or not.  Things become even more interesting in higher dimensions as the number of neighbors explodes exponentially.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So even a limited solution (such as a concrete answer for N=2) would be a quite interesting and an acceptable answer to this puzzle.&lt;/p&gt;&#xA;" OwnerUserId="4351" LastEditorUserId="4351" LastEditDate="2014-11-06T05:34:53.320" LastActivityDate="2015-10-25T20:35:24.213" Title="N-dimensional Tic-Tac-Toe variant" Tags="|geometry|combinatorics|tic-tac-toe|" AnswerCount="3" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="3625" PostTypeId="1" AcceptedAnswerId="3628" CreationDate="2014-11-06T07:23:35.230" Score="33" ViewCount="2978" Body="&lt;p&gt;You're taking the final exam for your Introduction to Probability and Statistics course, and once again you've failed to study (you were up all night solving online brainteasers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exam is multiple choice. It asks 120 questions, each of which has possible answers &lt;strong&gt;A&lt;/strong&gt;, &lt;strong&gt;B&lt;/strong&gt;, and &lt;strong&gt;C&lt;/strong&gt;. You haven't the faintest idea how to answer any of the questions and so you turn to your old standby strategy: &lt;em&gt;always answer &lt;strong&gt;B&lt;/strong&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As quickly as you can, you fill  out all the &lt;strong&gt;B&lt;/strong&gt;'s on your exam paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shortly after you've finished, your professor gets up at the front of the class and announces that an error was made while printing the exam: answer &lt;strong&gt;C&lt;/strong&gt; should not appear on the answer sheet because &lt;strong&gt;C&lt;/strong&gt; is not the correct answer to any question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You now have several options:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;leave your answers as &lt;strong&gt;B&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;switch your answers from &lt;strong&gt;B&lt;/strong&gt; to &lt;strong&gt;A&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;switch &lt;em&gt;some&lt;/em&gt;, but not all, of your answers from &lt;strong&gt;B&lt;/strong&gt; to &lt;strong&gt;A&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Given your goal is to get as many answers correct as possible, &lt;strong&gt;what strategy do you use, and how many correct answers do you expect to get (on average) when using this strategy?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Puzzlers are politely encouraged to place answers in spoiler blocks to avoid inadvertently spoiling the fun for other readers.&lt;/em&gt; :)&lt;/p&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="4219" LastEditDate="2014-11-06T14:38:16.353" LastActivityDate="2014-11-06T21:32:58.963" Title="A Strangely Familiar Probability Problem" Tags="|probability|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="3626" PostTypeId="2" ParentId="3625" CreationDate="2014-11-06T07:33:42.333" Score="14" Body="&lt;p&gt;The original probability of the &quot;Always answer B&quot; strategy would have given you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1/3rd probability of getting everything correct and 40 questions right on average. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However after the cancellation of C as an answer &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;for each question, now you have 50% chance of getting the answer right with your original strategy.  Since neither you do not have any clue regarding the correct answers, switching would give you no obvious advantage, leaving the choice of A for all questions with equal probability. Any attempt to increase the probability by choosing A or B would also leave a risk of decreasing the correct answers.  This is different from Monty Hall problem where the elimination took into account your original choice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So going by the above logic &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;In the interest of time savings, you dont switch&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="2313" LastEditDate="2014-11-06T09:00:06.337" LastActivityDate="2014-11-06T09:00:06.337" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3628" PostTypeId="2" ParentId="3625" CreationDate="2014-11-06T08:01:21.353" Score="43" Body="&lt;p&gt;Familiar indeed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; From the name and setup, it sounds like this was meant to invoke thoughts of &lt;a href=&quot;https://en.wikipedia.org/wiki/Monty_Hall_problem&quot;&gt;Monty Hall type puzzles&lt;/a&gt;.  However here what the professor revealed didn't depend on your choice, so is not actually like the Monty Hall puzzle.  With absolutely no information about the distribution of answers besides that the probability of C is zero, guessing A is as good as guessing B.  There is no reason to change your answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4351" LastActivityDate="2014-11-06T08:01:21.353" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3629" PostTypeId="2" ParentId="3598" CreationDate="2014-11-06T08:31:02.793" Score="4" Body="&lt;p&gt;Like warspyking said, the walls move through each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The walls are made of razor thin blades horizontally held in place with a gap the same width as the blade (separating each blade). A vertical bar that is connected at right angel to the blades is attached to one of four hydraulic motors that pushes them towards the centre. The blades could be made so thin that the wall would still look solid to the eye. If the blades were only a few atoms thick the gaps between them would also be a few atoms thick to allow the other blades from the walls at 90° to slide through but not big enough for biological molecules (including even blood cells) to fit through.&lt;/p&gt;&#xA;" OwnerUserId="4568" LastActivityDate="2014-11-06T08:31:02.793" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3631" PostTypeId="1" AcceptedAnswerId="3635" CreationDate="2014-11-06T08:52:43.497" Score="5" ViewCount="1059" Body="&lt;p&gt;Let's say you have a standard bagel (one that is NOT pre-sliced).  How can you cut this bagel into two interlocking rings?  The rings must never be broken.&lt;/p&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4450" LastEditDate="2015-04-28T01:45:53.133" LastActivityDate="2017-03-08T11:28:55.390" Title="Cut a bagel into interlocking rings" Tags="|geometry|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3634" PostTypeId="2" ParentId="2688" CreationDate="2014-11-06T12:36:09.167" Score="0" Body="&lt;p&gt;It is not a physical object, but one thing that can pass through the mirror is a (perfect harmonic?) wave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is due to the fact that the wave that coming from 'the other side' is reflected, and as such actually continues the wave from your side. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this effect goes both ways.&lt;/p&gt;&#xA;" OwnerUserId="1666" LastEditorUserId="1666" LastEditDate="2014-11-06T12:51:40.843" LastActivityDate="2014-11-06T12:51:40.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3635" PostTypeId="2" ParentId="3631" CreationDate="2014-11-06T12:43:36.803" Score="11" Body="&lt;p&gt;Ok Just to ensure that this question gets an answer I am posting this, and attributing it to community. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Credits to &lt;a href=&quot;http://www.dailymail.co.uk/sciencetech/article-2232924/How-make-Mobius-bagel-Sliced-just-right-breakfast-snack-makes-linked-halves-curious-mathematical-properties.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.dailymail.co.uk/sciencetech/article-2232924/How-make-Mobius-bagel-Sliced-just-right-breakfast-snack-makes-linked-halves-curious-mathematical-properties.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic concept is to consider the third dimension and visualize four key points.  Center the bagel at the origin, circling the Z axis.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A is the highest point above the +X axis.  B is where the +Y axis enters the bagel.&#xA;C is the lowest point below the -X axis.  D is where the -Y axis exits the bagel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cut through the line ABCDA.  Then turn the bagel over and make the same cut again.  Here is &lt;a href=&quot;https://www.youtube.com/watch?v=dN8AwGUaqDA&quot; rel=&quot;nofollow noreferrer&quot;&gt;a video&lt;/a&gt; showing how to cut the bagel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CDwiw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CDwiw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="5373" LastEditDate="2017-03-08T11:28:55.390" LastActivityDate="2017-03-08T11:28:55.390" CommentCount="4" CommunityOwnedDate="2014-11-06T12:43:36.803" ContentLicense="CC BY-SA 3.0" />
  <row Id="3636" PostTypeId="2" ParentId="3613" CreationDate="2014-11-06T13:24:16.213" Score="6" Body="&lt;p&gt;Here's an easy &quot;intuitive&quot; explanation how X can win:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If X can get a &lt;strong&gt;998&lt;/strong&gt; in a row and it's her turn, she has won.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if she can get &lt;em&gt;TWO&lt;/em&gt; &lt;strong&gt;998s&lt;/strong&gt; in a row, and it's Os turn, she has won.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{O will only be able to cover one - so it's over}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if she can get &lt;em&gt;FOUR&lt;/em&gt; &lt;strong&gt;997s&lt;/strong&gt; in a row, and it's O's turn, she has won.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(O will only be able to cover ..two times.., so there will be two 998s and it's O's turn - so it's over.}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if she can get &lt;em&gt;EIGHT&lt;/em&gt; &lt;strong&gt;996s&lt;/strong&gt; in a row, and it's Os turn, she has won.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(O will only be able to cover ..four times.., so there will be four 997s and it's O's turn - so it's over.}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if she can get &lt;em&gt;SIXTEEN&lt;/em&gt; &lt;strong&gt;995s&lt;/strong&gt; in a row, and it's Os turn, she has won.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{O will only be able to cover ..eight times.., so there will be eight 996s and it's O's turn - so it's over.}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... and eventually ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if she can get &lt;em&gt;&amp;lt; a very large number == NN &gt;&lt;/em&gt; &lt;strong&gt;1s&lt;/strong&gt; in a row, and it's Os turn, she has won.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To repeat, if she can get &lt;em&gt;NN&lt;/em&gt; &lt;strong&gt;1s&lt;/strong&gt; in a row, she has won.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So to win, simply: on the y cartesian axis (say), X simply keeps putting &quot;one&quot;.  ie, X simply creates an incredibly tall column, that is to say, just keeps &quot;starting one row&quot; after another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;She is able to add one unblocked row each time (the other one gets covered by O).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply, after NN goes... (that is to say, 2xNN points placed)...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;she will have &lt;em&gt;NN&lt;/em&gt; &lt;strong&gt;1s&lt;/strong&gt; in a row.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;She has won!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;footnote:&lt;/em&gt; regarding 2Dism. X's &lt;strong&gt;first step is to get &quot;NN 1s&quot; in play&lt;/strong&gt;. Above I describe them as being simply all horizontal and stacked-up. In fact, NN could hugely separate each attempt, and allow each one to be either H or V. (For a given line, of course, X could &lt;strong&gt;not&lt;/strong&gt; change between H/V, after, two points are down). In this way, it's twice as hard for O to block (this applies strictly on &quot;stage 1&quot; only .. enroute to &quot;NN 1s&quot;). So intertestingly, for &quot;stage 1&quot;, X need only take NN / 2 goes (ie, NN points placed) to achieve &quot;NN 1s&quot;. But wait! It occurs, X gets to place &lt;strong&gt;four potential lines&lt;/strong&gt; with each (hugely-spaced) two dots.  O can only block one alone of the four.  So in fact if I'm not mistaken for &quot;stage 1&quot;, X &lt;strong&gt;need only take NN / 3 goes&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4580" LastEditorUserId="4580" LastEditDate="2014-11-06T15:55:50.747" LastActivityDate="2014-11-06T15:55:50.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3637" PostTypeId="1" AcceptedAnswerId="3638" CreationDate="2014-11-06T15:03:51.747" Score="23" ViewCount="17157" Body="&lt;p&gt;Two sisters were born on the same day in the same year, yet they aren't twins. None of their other sisters share the birthday. What's going on? Explain.&lt;/p&gt;&#xA;" OwnerUserId="4508" LastEditorUserId="5373" LastEditDate="2014-12-30T14:48:40.020" LastActivityDate="2015-02-20T04:30:08.880" Title="Two sisters born on the same day" Tags="|lateral-thinking|family|" AnswerCount="12" CommentCount="12" ClosedDate="2016-09-23T03:38:33.747" ContentLicense="CC BY-SA 3.0" />
  <row Id="3638" PostTypeId="2" ParentId="3637" CreationDate="2014-11-06T15:07:04.323" Score="56" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sisters are nuns. Their shared birthday is a coincidence. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4508" LastActivityDate="2014-11-06T15:07:04.323" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="3639" PostTypeId="2" ParentId="3637" CreationDate="2014-11-06T15:15:42.373" Score="28" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Same father, different mother.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1654" LastActivityDate="2014-11-06T15:15:42.373" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="3640" PostTypeId="2" ParentId="3637" CreationDate="2014-11-06T15:26:38.213" Score="59" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The two sisters aren't twins, they are either triplets (or greater) with the other triplet being a brother (or bothers if greater than triplets).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4505" LastActivityDate="2014-11-06T15:26:38.213" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3641" PostTypeId="1" AcceptedAnswerId="3647" CreationDate="2014-11-06T15:43:59.623" Score="8" ViewCount="9291" Body="&lt;p&gt;How do you make $21$ from the numbers $1$, $5$, $6$, and $7$? &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You can use the operations of addition, subtraction, multiplication and division, as well as brackets. &lt;/li&gt;&#xA;&lt;li&gt;You must use each number exactly once.&lt;/li&gt;&#xA;&lt;li&gt;You cannot juxtapose numbers (i.e., 1 and 5 cannot be used as 15).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4508" LastEditorUserId="8874" LastEditDate="2016-03-27T17:56:27.957" LastActivityDate="2019-07-17T16:11:24.313" Title="How to make 21?" Tags="|mathematics|calculation-puzzle|formation-of-numbers|" AnswerCount="3" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="3642" PostTypeId="2" ParentId="3637" CreationDate="2014-11-06T15:45:36.480" Score="18" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; they are step sisters. One is adopted into the family.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="81" LastActivityDate="2014-11-06T15:45:36.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3643" PostTypeId="2" ParentId="3637" CreationDate="2014-11-06T15:45:46.643" Score="69" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you say &quot;two sisters&quot;, one thinks it means &quot;they are each other's sister&quot;. While it actually only means &quot;each one has a sister/brother&quot;. So the easiest answer is &quot;they are not in the same family&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2783" LastActivityDate="2014-11-06T15:45:46.643" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3644" PostTypeId="2" ParentId="3637" CreationDate="2014-11-06T15:58:44.140" Score="13" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;They have the same genetic parents but different surrogate mothers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4586" LastEditorUserId="4523" LastEditDate="2014-11-09T01:48:40.817" LastActivityDate="2014-11-09T01:48:40.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3645" PostTypeId="2" ParentId="548" CreationDate="2014-11-06T16:09:14.360" Score="7" Body="&lt;p&gt;I have the following results that lead some of the way towards an answer:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If a knight starts at the middle of a square of sidelength $a$ then he can collect all the knights in that square and bring them to its centre in time at most $Ca$, for some constant $C$ not depending on $a$ or the number of knights. (This proves that we can't slow down the first knight for ages just by putting loads of knights closely surrounding the castle.)&lt;/li&gt;&#xA;&lt;li&gt;For all $\varepsilon&amp;gt;0$ there exists some $N\in \mathbb{N}$ such that for all $n&amp;gt;N$, $n$ knights can always be collected from the kingdom in time $2\sqrt{2}+\varepsilon$. (If we put the knights in two opposite corners it will take at least time $2\sqrt{2}$. So we can see that for large numbers of knights the time needed tends to $2\sqrt 2$.)&lt;/li&gt;&#xA;&lt;li&gt;There exists some $p$ and some arrangement of $p$ knights such that collecting the knights from that arrangement takes at least as long as collecting any number of knights from any other arrangement. (So the problem actually has an answer!)&lt;/li&gt;&#xA;&lt;li&gt;The times needed for small numbers of knights are:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1 knight: $0$&lt;/li&gt;&#xA;&lt;li&gt;2 knights: $\sqrt 2$&lt;/li&gt;&#xA;&lt;li&gt;3 knights: $2\sqrt 2$&lt;/li&gt;&#xA;&lt;li&gt;4 knights: $2\sqrt 2$&lt;/li&gt;&#xA;&lt;li&gt;5 knights: $2+\sqrt 2$&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Collecting $6$ knights (including the first one from the castle) can always be done in time $\frac{1}{2}+2\sqrt 2$, which is less than $2+\sqrt 2$, the time needed to collect $5$ knights.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'll give the proofs when I get round to writing them up. (Hints: 1. splits the square into 4 and uses recursion, 2. splits the square into $k^2$ small squares and uses the pigeonhole principle to find a small square with enough knights that each one can be sent to each of the other squares, 3. uses 2 to produce an upper bound on $p$ and then uses a compactness argument to show that the bound is achieved, 4. just involves looking at each case in detail, and 5. splits the square into 4 and starts at the quarter with 2 knights in it)&lt;/p&gt;&#xA;" OwnerUserId="4457" LastActivityDate="2014-11-06T16:09:14.360" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3646" PostTypeId="2" ParentId="3637" CreationDate="2014-11-06T16:12:19.627" Score="14" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Same day, same year but what about a different month? e.g. 1st January 2000 and 1st December 2000 (more than 9 months apart)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4575" LastActivityDate="2014-11-06T16:12:19.627" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3647" PostTypeId="2" ParentId="3641" CreationDate="2014-11-06T16:13:23.783" Score="16" Body="&lt;p&gt;If this is an acceptable question, here is the answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$$\frac{6}{1-\frac{5}{7}}=21$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a pretty well known problem that, while the math obviously works, is unintuitive enough that it is sometimes difficult to solve the first time you see it.&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2014-11-06T16:13:23.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3648" PostTypeId="2" ParentId="3637" CreationDate="2014-11-06T16:16:34.890" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Having &lt;a href=&quot;http://en.wikipedia.org/wiki/Subsidiary&quot;&gt;sister companies&lt;/a&gt; founded in the same day at the same is not unusual.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4394" LastActivityDate="2014-11-06T16:16:34.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3650" PostTypeId="2" ParentId="3637" CreationDate="2014-11-06T16:44:46.970" Score="4" Body="&lt;p&gt;In the same vein as Taemyr's answer, but reversed: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TL;DR: One Mother, possibly Two Different Fathers&lt;br&gt;&#xA; &lt;a href=&quot;http://en.wikipedia.org/wiki/Superfecundation&quot; rel=&quot;nofollow&quot;&gt;Superfecundation&lt;/a&gt;: &lt;strong&gt;Superfecundation&lt;/strong&gt; is the fertilization of two or more ova from the same cycle by sperm from separate acts of sexual intercourse. The term superfecundation is derived from fecund, meaning the ability to produce offspring.&lt;br&gt;&#xA; &lt;strong&gt;Heteropaternal superfecundation&lt;/strong&gt; occurs when two different males father fraternal twins. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4403" LastEditorUserId="4403" LastEditDate="2015-02-20T04:30:08.880" LastActivityDate="2015-02-20T04:30:08.880" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3651" PostTypeId="1" AcceptedAnswerId="3660" CreationDate="2014-11-06T16:46:04.063" Score="10" ViewCount="5112" Body="&lt;p&gt;Prompted by &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3641/how-to-make-21&quot;&gt;How to make 21?&lt;/a&gt;, I wondered what numbers you can form with the allowed numbers and operations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, what numbers cannot be formed with a combination of 4 given numbers and the operators +, -, *, /?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically, what is the lowest positive integer that cannot be formed with 1, 5, 6, 7 and the operators +, -, *, /. Parentheses are permitted and each number must be used exactly once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Note that concatenation is not permitted. e.g. you cannot get 71 via 76-5*1.&lt;/p&gt;&#xA;" OwnerUserId="3987" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2017-09-14T16:46:20.237" Title="What is the lowest number you cannot make with a set of 4 numbers and the 4 basic operators?" Tags="|mathematics|" AnswerCount="5" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="3653" PostTypeId="2" ParentId="3651" CreationDate="2014-11-06T16:53:34.560" Score="-4" Body="&lt;p&gt;Possibly:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 212, since (5 * 6 * 7) + 1 = 211, which seems to me to be the largest number you can make, so one beyond that would be the smallest number you cannot make.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4518" LastActivityDate="2014-11-06T16:53:34.560" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3654" PostTypeId="2" ParentId="3637" CreationDate="2014-11-06T16:55:34.837" Score="5" Body="&lt;p&gt;It's very similar to Colonel Panic's answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The two ladies are &lt;a href=&quot;http://en.wikipedia.org/wiki/Nursing_management#Charge_nurse&quot; rel=&quot;nofollow&quot;&gt;nursing sisters&lt;/a&gt;; their shared birthday is a coincidence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2665" LastActivityDate="2014-11-06T16:55:34.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3656" PostTypeId="2" ParentId="3637" CreationDate="2014-11-06T17:01:20.790" Score="10" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;One or more has died, so you &lt;em&gt;could&lt;/em&gt; say they WERE twins but not ARE twins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4575" LastActivityDate="2014-11-06T17:01:20.790" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3657" PostTypeId="2" ParentId="3651" CreationDate="2014-11-06T17:23:04.667" Score="6" Body="&lt;p&gt;I cannot make 15 using these numbers and the given operations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding all numbers gives 19, since we cannot make 15 by taking away 1, one of the other numbers have to be taken away, so addition cannot do it, subtraction wont work with addition either. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we go to multiplication, then 5 x 3 is the only way to do it, if we take away 5, then 3 cannot be formed with 6 7 and 1, at least thats what I cannot see, so I will leave it here till someone disproves me  (there are technical combinations like 6 x 2.5 etc which I cant seem to find a way to achieve) &lt;/p&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="2313" LastEditDate="2014-12-17T10:44:14.590" LastActivityDate="2014-12-17T10:44:14.590" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3658" PostTypeId="2" ParentId="3402" CreationDate="2014-11-06T17:41:19.957" Score="2" Body="&lt;p&gt;So, I have played &lt;a href=&quot;http://boardgamegeek.com/boardgame/128882/resistance-avalon&quot; rel=&quot;nofollow&quot;&gt;this game&lt;/a&gt; probably 50+ times with friends, though not always with 10 and not always with the leader variant in the proposed puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One difficulty I have with the proposed solutions involving the leader is the weakness you correctly identify: why would the evil characters not behave exactly like the leader? &lt;strong&gt;As soon as you imply a 'perfect strategy' for the game, the other (opposing side) characters can always imitate it for themselves&lt;/strong&gt;! The evil characters have exactly the same knowledge as the leader and, until missions fail, will have no leverage against the honest leader. The evil characters have no reason not to employ the same strategy as the honest one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not think, and I could be wrong, that there is a closed form solution to the problem, even if our players were programmable bots that worked through logic only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fun of the actual game this puzzle is based from, of course, is that there are no perfect strategies -- people are always flawed in their analysis or there is a loud player who is capable of making convincing arguments during the discussions. That, and all discussions happen openly at the table and not in private (though we have had instances of signaling during games... which of course can always backfire!).&lt;/p&gt;&#xA;" OwnerUserId="4578" LastEditorUserId="4578" LastEditDate="2014-11-06T18:22:11.860" LastActivityDate="2014-11-06T18:22:11.860" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3659" PostTypeId="2" ParentId="3651" CreationDate="2014-11-06T17:48:52.577" Score="5" Body="&lt;p&gt;I just started finding a solution for each one, 15 is the first one I can't come up with a solution. (followed by 16, 20...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ETA (apparently there is a 15 added it in)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (7+5)/6 - 1 = 1&lt;br&gt;&#xA; &lt;br&gt;&#xA; (7+5)/6 * 1 = 2&lt;br&gt;&#xA; &lt;br&gt;&#xA; (7+5)/6 + 1 = 3&lt;br&gt;&#xA; &lt;br&gt;&#xA; (6+5-7) * 1 = 4&lt;br&gt;&#xA; &lt;br&gt;&#xA; (6+5-7) + 1 = 5&lt;br&gt;&#xA; &lt;br&gt;&#xA; ((7*5) + 1)/6 = 6&lt;br&gt;&#xA; &lt;br&gt;&#xA; (6+7-5) - 1 = 7&lt;br&gt;&#xA; &lt;br&gt;&#xA; (6+7-5) * 1 = 8&lt;br&gt;&#xA; &lt;br&gt;&#xA; (6+7-5) + 1 = 9&lt;br&gt;&#xA; &lt;br&gt;&#xA; (7-6+1) * 5 = 10&lt;br&gt;&#xA; &lt;br&gt;&#xA; (7-5) * 6 - 1 = 11&lt;br&gt;&#xA; &lt;br&gt;&#xA; (7-5) * 6 * 1 = 12&lt;br&gt;&#xA; &lt;br&gt;&#xA; (7-5) * 6 + 1 = 13&lt;br&gt;&#xA; &lt;br&gt;&#xA; (6-5+1) * 7 = 14&lt;br&gt;&#xA; &lt;br&gt;&#xA; 6/((7/5)-1) = 15&lt;br&gt;&#xA; &lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;br&gt;&#xA; (6+7+5) - 1 = 17&lt;br&gt;&#xA; &lt;br&gt;&#xA; (6+7+5) * 1 = 18&lt;br&gt;&#xA; &lt;br&gt;&#xA; (6+7+5) + 1 = 19&lt;br&gt;&#xA; &lt;br&gt;&#xA; &lt;br&gt;&#xA; 6 / (1-(5/7)) = 21&lt;br&gt;&#xA; &lt;br&gt;&#xA; ((6*5)-7) - 1 = 22&lt;br&gt;&#xA; &lt;br&gt;&#xA; ((6*5)-7) * 1 = 23&lt;br&gt;&#xA; &lt;br&gt;&#xA; ((6*5)-7) + 1 = 24&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edited to add in the solution for 21 and 15&lt;/p&gt;&#xA;" OwnerUserId="2014" LastEditorUserId="1477" LastEditDate="2014-11-06T19:17:41.903" LastActivityDate="2014-11-06T19:17:41.903" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3660" PostTypeId="2" ParentId="3651" CreationDate="2014-11-06T18:20:36.710" Score="14" Body="&lt;p&gt;16 is the lowest.&#xA;$$\begin{align}&#xA;    1 &amp;amp;= \frac{5+7}6-1\\&#xA;    2 &amp;amp;= \frac6{7+1-5}\\&#xA;    3 &amp;amp;= \frac{1\times 6}{7-5}\\&#xA;    4 &amp;amp;= (5-1)\times(7-6)\\&#xA;    5 &amp;amp;= 5+1+6-7\\&#xA;    6 &amp;amp;= \frac{7-1}6 + 5\\&#xA;    7 &amp;amp;= 7+6-1-5\\&#xA;    8 &amp;amp;= 7+\frac6{1+5}\\&#xA;    9 &amp;amp;= 7+6+1-5\\&#xA;    10 &amp;amp;= (7-5)\times(6-1)\\&#xA;    11 &amp;amp;= (7-5)\times 6-1\\&#xA;    12 &amp;amp;= (7-5)\times 6 \times 1\\&#xA;    13 &amp;amp;= (7-5)\times 6+1\\&#xA;    14 &amp;amp;= (6+1-5)\times 7\\&#xA;    15 &amp;amp;= \frac6{\frac75-1} = \frac6{2/5}\\&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was done using a computer program to enumerate all possible numbers achievable and then seeing the results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For extra credit the largest possible is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 252&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The algorithm used to do this was as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with four numbers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define an operation as having a left and a right number and an operation. We have four possible left numbers and four possible right numbers but we can't choose the same one twice so we have 12 number combinations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each of these apply the operator to get a new number. Take the used numbers out of the set and add in the new number. The new number needn't be an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now have 48 sets of three numbers. Apply the same principle to find the 1152 (not necessarily unique) sets of two numbers that can be made. Repeat until you have the 9216 possible combinations of numbers and operators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then from this set eliminate all the undesirables (ie non integer results). Look at list. Profit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see the code go look at &lt;a href=&quot;http://pastebin.com/HEuKHhsv&quot; rel=&quot;nofollow&quot;&gt;http://pastebin.com/HEuKHhsv&lt;/a&gt;. Its written in C# and designed to run in Linqpad (for ease of coding). Its not written to be readable necessarily but is written verbosely enough to satisfy myself that it is working as intended. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also for interest the ones under 50 that can't be done are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;16&lt;br&gt;&#xA;20&lt;br&gt;&#xA;25&lt;br&gt;&#xA;26&lt;br&gt;&#xA;27&lt;br&gt;&#xA;33&lt;br&gt;&#xA;39&lt;br&gt;&#xA;45&lt;br&gt;&#xA;49&lt;br&gt;&#xA;50  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And there are 181 positive numbers under 252 that cannot be made using this method (I hope or something in my program went wrong).&lt;/p&gt;&#xA;" OwnerUserId="1561" LastEditorUserId="4421" LastEditDate="2015-10-19T20:55:10.130" LastActivityDate="2015-10-19T20:55:10.130" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="3663" PostTypeId="2" ParentId="3637" CreationDate="2014-11-06T19:57:48.567" Score="16" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Two identical twin brothers marry two identical twin sisters.  They each have baby girls born on the dame year-month-date.  While we'd consider the girls cousins, they are genetically sisters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2318" LastActivityDate="2014-11-06T19:57:48.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3664" PostTypeId="2" ParentId="3651" CreationDate="2014-11-06T20:30:55.970" Score="0" Body="&lt;p&gt;&lt;strong&gt;WITHOUT BRACKETS&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without the use of brackets you can make the following numbers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;-37 -36 -29 -28 -23 -22 4 6 8 23 24 29 30 36 37 38 40 41 46 47&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All positive integers are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;4 6 8 23 24 29 30 36 37 38 40 41 46 47&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the lowest positive digit without brackets is 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anybody interested in all the possible equations that result in integers (there are 252 of them) can look here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pastebin.com/kvNpuDd6&quot; rel=&quot;nofollow&quot;&gt;http://pastebin.com/kvNpuDd6&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;0 is neither positive nor negative, this is why the answer is 1 &lt;em&gt;not&lt;/em&gt; 0&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;**WITH THE USE OF BRACKETS*&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note I repeated operators here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using brackets you can achieve a wider range of numbers, you can achieve 1 through 15 and many more. Too many to cover a list of equations like the pastebin in the without brackets section.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;((7+5)/6)-1=1&lt;br&gt;&#xA;  6/(1-5+7)=2&lt;br&gt;&#xA;  (6*1)/(7-5)=3&lt;br&gt;&#xA;  (7-6)&lt;em&gt;(5-1)=4&lt;br&gt;&#xA;  (5+1+6-7)=5&lt;br&gt;&#xA;  (7-1)/(6+5)=6&lt;br&gt;&#xA;  (6+7-5-1)=7&lt;br&gt;&#xA;  (6/(1+5))+7=8&lt;br&gt;&#xA;  (6+7+1-5)=9&lt;br&gt;&#xA;  (6-1)&lt;/em&gt;(7-5)=10&lt;br&gt;&#xA;  (6-1)*(7-5)=11&lt;br&gt;&#xA;  (6*1)&lt;em&gt;(7-5)=12&lt;br&gt;&#xA;  (6+1)&lt;/em&gt;(7-5)=13&lt;br&gt;&#xA;  7*(6+1-5)=14&lt;br&gt;&#xA;  6/((7/5)-1)=15&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;16 is impossible to achieve like this. You can continue going up until 252, then it becomes impossible to get a higher numbers.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-11-07T10:32:27.460" LastActivityDate="2014-11-07T10:32:27.460" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3666" PostTypeId="2" ParentId="3625" CreationDate="2014-11-06T21:32:58.963" Score="1" Body="&lt;p&gt;The correct answer: If you can figure out how to switch some of the B's to A's &lt;strong&gt;such that&lt;/strong&gt; your grade will improve, well, like I said, your grade will improve.&lt;/p&gt;&#xA;" OwnerUserId="4605" LastActivityDate="2014-11-06T21:32:58.963" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3667" PostTypeId="1" AcceptedAnswerId="3675" CreationDate="2014-11-06T23:07:46.323" Score="0" ViewCount="435" Body="&lt;p&gt;What is so fascinating over a puzzle, why is it so much fun to solve? Especially a puzzle game, what makes it so fun to complete?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to make a puzzle game so I need a little bit of information on why they are so enjoyable.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-11-06T23:38:16.007" LastActivityDate="2014-11-07T00:06:06.030" Title="What makes Puzzles so Appealing?" Tags="|puzzle-creation|" AnswerCount="1" CommentCount="2" ClosedDate="2014-11-07T18:59:16.197" ContentLicense="CC BY-SA 3.0" />
  <row Id="3668" PostTypeId="1" AcceptedAnswerId="3670" CreationDate="2014-11-06T23:14:31.910" Score="4" ViewCount="40731" Body="&lt;p&gt;This is an old one but I don't see it here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A man stepped out of his tent with his rifle and walked 1 mile south, 1 mile east and 1 Mile north.  Then he shot a bear in front of his tent.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What color was the bear?&lt;/p&gt;&#xA;" OwnerUserId="2014" LastEditorUserId="5373" LastEditDate="2016-10-24T00:04:44.373" LastActivityDate="2016-10-24T00:04:44.373" Title="What color was the bear" Tags="|lateral-thinking|situation|" AnswerCount="5" CommentCount="4" ClosedDate="2014-11-07T18:58:07.600" ContentLicense="CC BY-SA 3.0" />
  <row Id="3669" PostTypeId="2" ParentId="3668" CreationDate="2014-11-06T23:18:16.980" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The bear was white, since he was at the North Pole and the only types of bears at the North Pole are polar bears.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4450" LastEditDate="2014-11-06T23:24:40.430" LastActivityDate="2014-11-06T23:24:40.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3670" PostTypeId="2" ParentId="3668" CreationDate="2014-11-06T23:19:17.040" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The bear is white. The hunter was hunting at the North Pole (no idea why though...)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reason it had to have been there is because that is one of the few (if not the only) places on earth that match the criteria, and will only have the one type-coloured bear.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2014-11-06T23:19:17.040" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3671" PostTypeId="2" ParentId="3668" CreationDate="2014-11-06T23:22:42.417" Score="8" Body="&lt;p&gt;There are two places in the world you could be to be able to walk 1 mile south, 1 mile east then 1 mile north and end up where you started (in front of your tent). One spot is right smack dab on the north pole. The other spot (or infinite number of spots) would be just north of the south pole (about 1.159 miles) such that walking one mile south of camp puts you in a spot where walking east 1 mile puts you on a circle that ends where it started. There are no bears at the south pole, so that can be ruled out.&lt;/p&gt;&#xA;" OwnerUserId="4125" LastEditorUserId="4125" LastEditDate="2014-11-06T23:35:51.853" LastActivityDate="2014-11-06T23:35:51.853" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3674" PostTypeId="2" ParentId="3668" CreationDate="2014-11-07T00:04:29.693" Score="-2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;black. because the bear is in the tent or behind the tent, and the man can't see what&#xA;  color the bear has. only its shadow.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4589" LastEditorUserId="4589" LastEditDate="2014-11-07T00:10:43.390" LastActivityDate="2014-11-07T00:10:43.390" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3675" PostTypeId="2" ParentId="3667" CreationDate="2014-11-07T00:06:06.030" Score="1" Body="&lt;p&gt;&quot;According to neuroscientist Daniel Bor, a research fellow at the University of Sussex in England and author of the new book The Ravenous Brain: How the New Science of Consciousness Explains Our Insatiable Search for Meaning, it’s because we take great pleasure in pattern-finding. What’s more, that conclusion has big implications for understanding the brain, consciousness and even neurological disorders like autism. We spoke with Bor recently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do you call the brain “ravenous”?&#xA;Human brains have an extreme form of consciousness: they’re ravenous for new innovative solutions to problems in the world, ravenous for optimizing our lives, for building pyramids of knowledge. I was trying to capture [the sense of hunger that] extreme forms of consciousness have about searching for knowledge and for understanding.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a Time magazine article &quot;Why Solving Puzzles Is Fun: Q&amp;amp;A with Consciousness Researcher Daniel Bor&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is more in the interview but most of it moves away from talking about puzzles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to why you were voted down I can't say - the site description says &quot;Puzzling Stack Exchange is a question and answer site for those who study the creation and solving of puzzles. It's built and run by you as part of the Stack Exchange network of Q&amp;amp;A sites. With your help, we're working together to build a library of detailed answers to every question about the creation and solving of puzzles.&quot;  It seems you are asking about &quot;creation&quot; and which would make it 'on topic'.  That being said, I don't see how understanding neuro-physiology is going to help with that task.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps, &quot;What makes some puzzles more fun than others?&quot; is what you were really after.&lt;/p&gt;&#xA;" OwnerUserId="4372" LastActivityDate="2014-11-07T00:06:06.030" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3678" PostTypeId="1" AcceptedAnswerId="3714" CreationDate="2014-11-07T00:30:45.477" Score="7" ViewCount="928" Body="&lt;p&gt;A tournament was played round-robin: each pair of players played a match where one defeated the other. Prove that there was a player for which every other player either lost to them or lost to someone who lost to them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formally: Prove that if you direct the edges of a complete undirected graph, then there exists a vertex from which every vertex is reachable by a directed path of length at most 2.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="69582" LastEditDate="2020-12-01T22:49:16.053" LastActivityDate="2023-06-08T03:32:50.170" Title="Every tournament has a dominant player" Tags="|mathematics|combinatorics|graph-theory|" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3679" PostTypeId="2" ParentId="3678" CreationDate="2014-11-07T02:04:39.487" Score="2" Body="&lt;p&gt;I will try to &quot;prove&quot; this with a tournament of 4 people first. This may be followed up with 5 or 6 person solutions, if necessary.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;4 Teams&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;First, let's consider the possible cases. Each person must play 3 games, the amount of wins must equal the amount of losses, and no two players can be undefeated. Likewise, no two players can be 0-3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;A    B    C    D&#xA;3-0, 2-1, 1-2, 0-3&#xA;3-0, 1-2, 1-2, 1-2&#xA;2-1, 2-1, 1-2, 1-2&#xA;2-1, 2-1, 2-1, 0-3&#xA;&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cases 1 and 2 can be disregarded, as someone is undefeated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Case 3, I will assume A beat B, even though the proof can be reversed if B beats A. A beat B, so that means either C beat A or D beat A. However, B beat both C and D, so B is one such player in this tournament.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Case 4, A beat B, B beat C, and C beat A. All three of these beat D. This is already solved, as A, B, and C all fit the requirements.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;5 Teams&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;A    B    C    D    E &#xA;3-1, 3-1, 2-2, 1-3, 1-3&#xA;3-1, 2-2, 2-2, 2-2, 1-3&#xA;2-2, 2-2, 2-2, 2-2, 2-2&#xA;&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the only cases I need to prove. All other ones include either an undefeated team or a team with no wins, so these are identical to a 4 team tournament. The last team is either unimportant or it is the team that fulfills the requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 1 - A beats B, which means A loses to C, D, or E. B beat all three of these teams, so B is one solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 2 - Either E beat A or B beat A. The other situations are variants of this. If E beat A, then A is the correct team, because A beat B, C, and D, all of which beat E. If B beat A, then A is again the correct team, because A beat C, D, and E, and B lost to exactly two of these teams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 3 - This is &lt;a href=&quot;http://en.wikipedia.org/wiki/Rock-paper-scissors-lizard-Spock&quot; rel=&quot;nofollow&quot;&gt;Rock Paper Scissors Lizard Spock&lt;/a&gt;. Every choice actually fits the requirements.&lt;/p&gt;&#xA;" OwnerUserId="1664" LastEditorUserId="1664" LastEditDate="2014-11-07T03:30:31.447" LastActivityDate="2014-11-07T03:30:31.447" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3680" PostTypeId="1" AcceptedAnswerId="104470" CreationDate="2014-11-07T03:00:59.693" Score="8" ViewCount="11513" Body="&lt;p&gt;If I cut a perfectly circular pizza through its center 6 times at 30° angles, I get 12 pieces of equal area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I don't have to cut through the center, I can cut in a grid shape to divide the pizza into 16 pieces, but only the middle four will be of equal area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoretically I could cut the pizza into a total of 22 pieces using 6 cuts, but more often than not &lt;em&gt;none&lt;/em&gt; of them would be of equal area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 6 cuts, what is the largest number of pieces of equal size I can obtain (using only straight cuts that go through the entire pizza)? What about $n$ cuts?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;And since people are posting answers to this effect, no, you are not allowed to fold the pizza or rearrange slices. This is not a lateral thinking puzzle.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="18129" LastEditDate="2020-11-03T22:49:37.413" LastActivityDate="2020-11-03T22:49:55.927" Title="Maximum number of pieces of equal area you can obtain by cutting a pizza a certain number of times" Tags="|geometry|dissection|" AnswerCount="7" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="3681" PostTypeId="2" ParentId="3678" CreationDate="2014-11-07T03:22:36.997" Score="5" Body="&lt;h2&gt;Proof by induction&lt;/h2&gt;&#xA;&lt;p&gt;Suppose a tournament of &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; players have a dominant player (this is trivially true for &lt;span class=&quot;math-container&quot;&gt;$n=1,2,3$&lt;/span&gt;). We now try to prove that a tournament of &lt;span class=&quot;math-container&quot;&gt;$n+1$&lt;/span&gt; players also have a dominant player, which is either the previous dominant player or the new player himself.&lt;/p&gt;&#xA;&lt;p&gt;Let's call the previous dominant player &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; and the new player &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Suppose &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; beats &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt;. Then &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; is still a dominant player in the tournament of &lt;span class=&quot;math-container&quot;&gt;$n+1$&lt;/span&gt; players.&lt;/p&gt;&#xA;&lt;p&gt;Then now we can assume that &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; beats &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Now, if there is a person &lt;span class=&quot;math-container&quot;&gt;$H_1$&lt;/span&gt; such that &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; beats &lt;span class=&quot;math-container&quot;&gt;$H_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$H_1$&lt;/span&gt; beats &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; is still a dominant player, since it dominates all other player, including the new player &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;So this means now we can assume that &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; beats everyone that was beaten by &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;This means everyone that was dominated by &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; will be dominated by &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; also. And since &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; beats &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; dominates the whole tournament, making &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; the new dominant player, proving the claim. =)&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;More formally, we'll represent the players as vertices in a graph, and the matches outcome as directed edges between those vertices.&lt;/p&gt;&#xA;&lt;p&gt;First, some notations:&lt;/p&gt;&#xA;&lt;p&gt;Let &lt;span class=&quot;math-container&quot;&gt;$G = \{V, E\}, V = \{v_1, v_2, \ldots, v_n\}, E = \{(v_i, v_j) | \text{player }i\text{ beats player }j\}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Let &lt;span class=&quot;math-container&quot;&gt;$d_G(v_i, v_j)$&lt;/span&gt; be the shortest path length from &lt;span class=&quot;math-container&quot;&gt;$v_i$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$v_j$&lt;/span&gt; in &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Let &lt;span class=&quot;math-container&quot;&gt;$s_G(v)$&lt;/span&gt; be the maximum distance from &lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt; in &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; (formally: &lt;span class=&quot;math-container&quot;&gt;$s_G(v) = max(\{d_G(v, v_i)\ |\ \forall v_i \in G\})$&lt;/span&gt;)&lt;/p&gt;&#xA;&lt;p&gt;To be proven:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In any graph &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$min(\{s_G(v)\ |\ v\in G\}) \leq 2$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr/&gt;&#xA;&lt;p&gt;We'll prove by induction.&lt;/p&gt;&#xA;&lt;p&gt;Let &lt;span class=&quot;math-container&quot;&gt;$n = |V|$&lt;/span&gt;, which is the number of vertices in &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;For &lt;span class=&quot;math-container&quot;&gt;$n=1,2,3$&lt;/span&gt;, it's trivial, since the longest path in such graphs will be 2.&lt;/p&gt;&#xA;&lt;p&gt;Assume the property holds for &lt;span class=&quot;math-container&quot;&gt;$|V| = n$&lt;/span&gt;. Now we are trying to prove that it also holds for &lt;span class=&quot;math-container&quot;&gt;$|V| = n+1$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Let &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; be such that &lt;span class=&quot;math-container&quot;&gt;$|V| = n$&lt;/span&gt;. By induction hypothesis, there is a vertex with distance at most 2 to any other vertices. Without loss of generalization, &lt;span class=&quot;math-container&quot;&gt;$s_G(v_1) \leq 2$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;We want to add one new vertex &lt;span class=&quot;math-container&quot;&gt;$v_{n+1}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; and prove that the property still holds.&lt;/p&gt;&#xA;&lt;p&gt;Now, if &lt;span class=&quot;math-container&quot;&gt;$s_G(v_1) = 1$&lt;/span&gt; (i.e., player &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; beats all other players in &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt;), we have two cases:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$(v_{n+1}, v_1) \in E$&lt;/span&gt; (i.e., player &lt;span class=&quot;math-container&quot;&gt;$n+1$&lt;/span&gt; beats player &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;)&lt;/p&gt;&#xA;&lt;p&gt;This implies &lt;span class=&quot;math-container&quot;&gt;$s_G(v_{n+1}) = 2$&lt;/span&gt;, so the property still holds.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$(v_1, v_{n+1}) \in E$&lt;/span&gt; (i.e., player &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; beats player &lt;span class=&quot;math-container&quot;&gt;$n+1$&lt;/span&gt;)&lt;/p&gt;&#xA;&lt;p&gt;This implies &lt;span class=&quot;math-container&quot;&gt;$s_G(v_1) = 1$&lt;/span&gt;, so the property still holds.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&lt;strong&gt;So&lt;/strong&gt; we can now assume that &lt;span class=&quot;math-container&quot;&gt;$s_G(v_1) = 2$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$(v_1, v_{n+1}) \in E$&lt;/span&gt;, then we still have &lt;span class=&quot;math-container&quot;&gt;$s_G(v_1) = 2$&lt;/span&gt; and the property still holds.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;So&lt;/strong&gt; we can now assume that &lt;span class=&quot;math-container&quot;&gt;$(v_{n+1}, v_1) \in E$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Let &lt;span class=&quot;math-container&quot;&gt;$H_1, H_2$&lt;/span&gt; denotes, respectively, the vertices of distance 1 and distance 2 from &lt;span class=&quot;math-container&quot;&gt;$v_1$&lt;/span&gt;.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;H_1 = \{v_i\ |\ d(v_1, v_i) = 1\}\\&#xA;H_2 = \{v_i\ |\ d(v_1, v_i) = 2\}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Note that &lt;span class=&quot;math-container&quot;&gt;$H_1 \cup H_2 = V-\{v_1\}$&lt;/span&gt; since &lt;span class=&quot;math-container&quot;&gt;$s_G(v_1) = 2$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$v_{n+1} \in H_2$&lt;/span&gt;, we are done, since we'll still have &lt;span class=&quot;math-container&quot;&gt;$s_G(v_1) = 2$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;So&lt;/strong&gt; we can now assume that &lt;span class=&quot;math-container&quot;&gt;$v_{n+1} \not\in H_2$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Note that for any vertices &lt;span class=&quot;math-container&quot;&gt;$v_i \in H_2$&lt;/span&gt;, we have &lt;span class=&quot;math-container&quot;&gt;$(v_i, v_1) \in E$&lt;/span&gt;, because otherwise &lt;span class=&quot;math-container&quot;&gt;$d(v_1, v_i) = 1$&lt;/span&gt;. Also note that &lt;span class=&quot;math-container&quot;&gt;$$v_i \in H_2 \Rightarrow \exists v_j \in H_1, (v_j, v_i)\in E \qquad \ldots\ (1)$$&lt;/span&gt; (i.e., if player &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; is of distance 2 from player &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;, there exists another player that is beaten by player &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; and that player beats player &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt;).&lt;/p&gt;&#xA;&lt;p&gt;Since &lt;span class=&quot;math-container&quot;&gt;$v_{n+1} \not\in H_2$&lt;/span&gt;, it means &lt;span class=&quot;math-container&quot;&gt;$$\forall v_j\in H_1, (v_{n+1}, v_j)\in E \qquad \ldots (2)$$&lt;/span&gt; (i.e., everyone that was beaten by player &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; is also beaten by player &lt;span class=&quot;math-container&quot;&gt;$n+1$&lt;/span&gt;).&lt;/p&gt;&#xA;&lt;p&gt;By &lt;span class=&quot;math-container&quot;&gt;$(1)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$(2)$&lt;/span&gt;, we have &lt;span class=&quot;math-container&quot;&gt;$$\forall v_i \in H_2, d(v_{n+1}, v_i) \leq 2$$&lt;/span&gt; (i.e., everyone with distance 2 from player &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; is at most of distance 2 from player &lt;span class=&quot;math-container&quot;&gt;$n+1$&lt;/span&gt;)&lt;/p&gt;&#xA;&lt;p&gt;This means we have &lt;span class=&quot;math-container&quot;&gt;$$\forall v_i \in H_1, d(v_{n+1}, v_i) = 1\\\forall v_i \in H_2, d(v_{n+1}, v_i) \leq 2$$&lt;/span&gt; which means that &lt;span class=&quot;math-container&quot;&gt;$s_G(v_{n+1}) \leq 2$&lt;/span&gt; (since &lt;span class=&quot;math-container&quot;&gt;$H_1 \cup H_2 = V-\{v_1\}$&lt;/span&gt;), so the property holds.&lt;/p&gt;&#xA;&lt;p&gt;We have proven that if the property holds in a graph with &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; vertices, it also holds in a graph with &lt;span class=&quot;math-container&quot;&gt;$n+1$&lt;/span&gt; vertices. By induction, the property is true for all &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Q.E.D.&lt;/p&gt;&#xA;" OwnerUserId="1649" LastEditorUserId="1649" LastEditDate="2021-10-30T05:35:52.387" LastActivityDate="2021-10-30T05:35:52.387" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="3682" PostTypeId="2" ParentId="3680" CreationDate="2014-11-07T03:24:56.350" Score="0" Body="&lt;p&gt;With a pizza, I guess the answer is logically 12, like you said. A taller cylinder is a different scenario, as you could cut through the middle height-wise, but this is not viable with a pizza. You could technically cut a cylinder into &lt;em&gt;n&lt;/em&gt; discs of equal size.&lt;/p&gt;&#xA;" OwnerUserId="1664" LastActivityDate="2014-11-07T03:24:56.350" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3683" PostTypeId="2" ParentId="3680" CreationDate="2014-11-07T05:12:16.850" Score="0" Body="&lt;p&gt;Cut two equilateral triangles? &#xA;This should give you both the same size pieces on the outside, as well as same size pieces on the inside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GknR6.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4620" LastActivityDate="2014-11-07T05:12:16.850" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3685" PostTypeId="1" AcceptedAnswerId="3688" CreationDate="2014-11-07T07:36:13.403" Score="21" ViewCount="7313" Body="&lt;p&gt;Here is puzzle that was lifted from an old essay of mine, that was adapted from a puzzle in an old puzzle book.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The great shah has invited a young boy into his palace as a student. While he is not busy managing his kingdom, he is a scholar of mathematics and invites the brightest learners from all over the country to his palace to study.&lt;/p&gt;&#xA;&lt;p&gt;He sits the boy down at a small table, on which is a bag of stones. He then gently pours out the stones and lays them out in rows on the table.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;Here are 42 stones, my son,&amp;quot; he tells the boy. &amp;quot;We will play a game with them. First, I will take one, three, or five stones from the pile. Then you will do the same, choosing between one, three, or five stones. We will continue until all the stones are gone. And neither of us may take more stones than there are remaining in the pile. The winner of the game will be the person who takes the last stone.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;The boy nods silently at the shah, and turns back to looking at the pile of stones.&lt;/p&gt;&#xA;&lt;p&gt;The shah starts by taking three stones. &amp;quot;We shall play seven games. If you win exactly six out of seven games against me, my student, you will have a place in my study.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;The boy nods again, and looks hard at the stones. Suddenly, a look on his face changes as he realizes that he has a winning strategy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What is the boy's strategy to win the first six games?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;The next part of the story was written in the essay after the first question had already been answered; try to answer the first question before looking at the problem statement of the second.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The boy easily wins all six of the first six games following his strategy. As they prepare for the seventh game, the shah tells the boy, &amp;quot;I can already tell you, boy, that you will win all seven of these games here, not six.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;Surely enough, the shah starts by taking one stone, and the boy plays wildly, trying his hardest &lt;em&gt;not&lt;/em&gt; to follow his winning strategy from earlier. But no matter how he plays, he cannot find a way to lose this last game and earn his place in the shah's study. As he removes the last stone from the pile, he hangs his head in shame.&lt;/p&gt;&#xA;&lt;p&gt;The shah looks pitifully at the boy. &amp;quot;Poor child,&amp;quot; he says. &amp;quot;If you can tell me why you could not lose this seventh game, you will still have a place in my study.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;The boy thinks long and hard again, and suddenly his face lights up again as he turns to the shah and explains why he could not lose the game.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Could the boy have lost the seventh game? Why or why not?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2017-04-04T19:20:25.120" Title="Picking up stones" Tags="|mathematics|strategy|game|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3686" PostTypeId="2" ParentId="3685" CreationDate="2014-11-07T07:42:28.383" Score="4" Body="&lt;p&gt;First question is easy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The boy, each time can take a number of stones, that with stones the shah took previous turn would make 6. Then, since 42 is dividable by 6 the boy will take the last stone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second question is easy enough too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The shah took 1 stone at his first move, then he can apply the previous strategy of the boy (like this is a game with 41 stones only) to insure that the boy will finally remain with 5 stones exactly (41-6x6=5). Then, if the boy takes 1 stone, the shah takes 3 and vice versa. This will force the boy to take the last stone and &quot;win&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2014-11-07T08:01:55.520" LastActivityDate="2014-11-07T08:01:55.520" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3687" PostTypeId="2" ParentId="3685" CreationDate="2014-11-07T07:50:52.933" Score="62" Body="&lt;p&gt;That's easy &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since the game starts with an even number of stones and all possible moves remove an odd number of stones, there must be an even number of moves. The shah always moves first, which means that the student always moves last. The student wins no matter what strategy either player chooses.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="284" LastEditorUserId="284" LastEditDate="2014-11-07T07:57:51.430" LastActivityDate="2014-11-07T07:57:51.430" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="3688" PostTypeId="2" ParentId="3685" CreationDate="2014-11-07T07:55:44.253" Score="15" Body="&lt;p&gt;Regardless of what one player takes, the other can adjust accordingly so that 6 in total are removed. (1+5, 3+3, 5+1). So as long as the number of remaining stones before the shah's turn is dividable by 6, the student can force a win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 'trick' lies in the first turn. Given 42 is a multiple of 6, no matter what opening move the shah makes, he can make the numbers 'work' to a multiple of 6 remaining after his turn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Now, why can he not lose on the 7th game?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simple, the game starts with an even number of 42, and the shah always gets to play after an even number of turns. Given that two odd numbers have been subtracted from the pile, the remaining number has to be even. So when there are 0 stones left in the pile, it's the shah's turn, and it was the boy who removed the last stone.&lt;/p&gt;&#xA;" OwnerUserId="2156" LastEditorDisplayName="user88" LastEditDate="2015-03-17T19:21:04.987" LastActivityDate="2015-03-17T19:21:04.987" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3690" PostTypeId="1" AcceptedAnswerId="3696" CreationDate="2014-11-07T09:14:28.330" Score="6" ViewCount="378" Body="&lt;p&gt;Source: &lt;a href=&quot;https://www.austms.org.au/Publ/Gazette/2014/May14/Puzzle.pdf&quot; rel=&quot;nofollow&quot;&gt;Australian Mathematical Society Gazette Puzzle Corner 35&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There are some weights on the two sides of a balance scale. The mass&#xA;  of each weight is an integer number of grams, but no two weights on&#xA;  the same side of the scale share the same mass. At the moment, the&#xA;  scale is perfectly balanced, with each side weighing a total of W&#xA;  grams. Suppose W is less than the number of weights on the left&#xA;  multiplied by the number of weights on the right. Is it always true&#xA;  that we can remove some of the weights from each side and still keep&#xA;  the two sides balanced?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Rem: Ignore the trivial solution of removing all weights from both&#xA;  sides.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="97" LastEditorUserId="97" LastEditDate="2014-11-09T08:10:02.457" LastActivityDate="2014-11-09T08:10:02.457" Title="Balancing Weights" Tags="|mathematics|" AnswerCount="1" CommentCount="10" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3691" PostTypeId="1" AcceptedAnswerId="3692" CreationDate="2014-11-07T09:27:45.017" Score="4" ViewCount="611" Body="&lt;p&gt;This puzzle is taken from a French bestseller I read years ago. The solution is much simpler than it looks...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the following line in this sequence : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;&#xA;1&#xA;1 1&#xA;2 1&#xA;1 2 1 1&#xA;1 1 1 2 2 1&#xA;3 1 2 2 1 1&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="4631" LastEditorUserId="6178" LastEditDate="2014-12-28T15:01:51.933" LastActivityDate="2014-12-28T15:01:51.933" Title="A Strange Sequence" Tags="|logical-deduction|lateral-thinking|pattern|" AnswerCount="1" CommentCount="1" ClosedDate="2014-11-07T14:07:26.037" ContentLicense="CC BY-SA 3.0" />
  <row Id="3692" PostTypeId="2" ParentId="3691" CreationDate="2014-11-07T09:29:35.250" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Say it out loud, and count the numbers! &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the steps would be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One 1; two 1's; one 2 one 1; and so one. The next would be 1 3 1 1 2 2 2 1. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3993" LastActivityDate="2014-11-07T09:29:35.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3695" PostTypeId="1" AcceptedAnswerId="3747" CreationDate="2014-11-07T10:01:27.747" Score="2" ViewCount="319" Body="&lt;ul&gt;&#xA;&lt;li&gt;Cade has six red stamps and three blue ones.&lt;/li&gt;&#xA;&lt;li&gt;In his collection, seven stamps are from Argentina and six stamps are&#xA;from Sweden.&lt;/li&gt;&#xA;&lt;li&gt;One stamp is purple and it is not from Argentina or Sweden.&lt;/li&gt;&#xA;&lt;li&gt;Two of his Argentine stamps are red and one is blue.&lt;/li&gt;&#xA;&lt;li&gt;Two of his Swedish stamps are blue and three are red.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What is the minimum number of stamps that Cade can have?&lt;/p&gt;&#xA;" OwnerUserId="97" LastActivityDate="2014-11-07T16:39:45.247" Title="How many stamps?" Tags="|logical-deduction|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3696" PostTypeId="2" ParentId="3690" CreationDate="2014-11-07T10:05:18.720" Score="4" Body="&lt;p&gt;Yes, because there must a mass shared by two weights on different sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose for contradiction that there is no such mass. Then, the weights are all distinct. Let $a$ and $b$ be the number of weights on the two sides. At minimum, the total weight of the two sides is $(a+b)(a+b+1)/2$ when the masses are $1,2,\dots,a+b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2w \geq(a+b)(a+b+1)/2 &amp;gt; (a+b)^2/2 \geq 2ab  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the last inequality follows from being equivalent to $(a-b)^2/2 \geq 0$. But this contradicts the statement that $w&amp;lt;ab$.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2014-11-07T10:05:18.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3697" PostTypeId="2" ParentId="3695" CreationDate="2014-11-07T10:09:33.577" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 15&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All 3 blue stamps are from Argentina or Sweden&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 5 red stamps are from Argentina or Sweden, and the last one is from another location (so we already have 9 stamps)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are 4 stamps from Argentina (13) and 1 stamp from Sweden (14) that are neither blue nor red, and the purple one (15)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4527" LastActivityDate="2014-11-07T10:09:33.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3700" PostTypeId="2" ParentId="3553" CreationDate="2014-11-07T10:18:18.250" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Actually if you say variable as a &quot;word&quot; it has appeared one time, if you say variable as a &quot;place holder&quot; as programming languages, it has appeared 33 times (along with the question).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4635" LastEditorUserId="44915" LastEditDate="2018-05-24T10:18:34.737" LastActivityDate="2018-05-24T10:18:34.737" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="3701" PostTypeId="1" AcceptedAnswerId="3704" CreationDate="2014-11-07T10:24:52.480" Score="4" ViewCount="797" Body="&lt;p&gt;This same problem also appears here:  &lt;a href=&quot;http://perplexus.info/show.php?pid=9271&quot; rel=&quot;nofollow&quot;&gt;http://perplexus.info/show.php?pid=9271&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A drawer contains a mixture of red socks and blue socks, at most $2014$&#xA;  in total. It so happens that, when two socks are selected randomly&#xA;  without replacement, there is a probability of precisely $\frac23$ that both&#xA;  are red or both are blue.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;What is the maximum possible number of red socks in the drawer that is&#xA;  consistent with this data?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="97" LastEditorUserId="57212" LastEditDate="2021-08-08T22:36:58.007" LastActivityDate="2021-08-08T22:36:58.007" Title="Number of socks?" Tags="|mathematics|probability|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3702" PostTypeId="2" ParentId="3680" CreationDate="2014-11-07T10:35:49.930" Score="1" Body="&lt;p&gt;I can only create 64 slices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 1.  Cut the pizza from North to South (2 slices)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 2.  Cut the pizza from East to West (4 slicles)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 3.  Lay the 4 slices on top of each other pointing East and cut from East to West (8 slices)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 4.  Lay the 8 slices again on top of each other and slice (16 slices)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 5.  Repeat 32 slices&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 6.  Repeat again 64 slices&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum number of slices is thus given by 2^n, where n is the number of cuts.&lt;/p&gt;&#xA;" OwnerUserId="97" LastActivityDate="2014-11-07T10:35:49.930" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="3704" PostTypeId="2" ParentId="3701" CreationDate="2014-11-07T10:59:47.373" Score="5" Body="&lt;p&gt;Answer is:    &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is $1065$ red and $285$ blue socks.&lt;BR&gt;&#xA; Probability of getting same colours = $\frac{1065 \times 1064 + 285 \times 284}{1350 \times 1349} = \frac23$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Inequation and equations are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $r + b \le 2014$&lt;br&gt;&#xA; $r \ge 0, b \ge 0$&lt;br&gt;&#xA; $3 \times (r^2 - r + b^2 - b) = 2(n^2 - n)$&lt;br&gt;&#xA; Solving all, $r = 1065$, $b = 285$ and $n = 1350$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4637" LastEditorUserId="4421" LastEditDate="2015-10-05T16:03:27.637" LastActivityDate="2015-10-05T16:03:27.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3705" PostTypeId="1" AcceptedAnswerId="3847" CreationDate="2014-11-07T11:09:02.727" Score="21" ViewCount="9691" Body="&lt;p&gt;If I give you one of each letter in the alphabet what's the largest you can spell (in word form)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus: What's the smallest?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus 2: What if you can use the words &quot;minus&quot;, &quot;plus&quot; and &quot;times&quot;?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-11-08T11:25:03.650" LastActivityDate="2015-10-16T13:14:44.677" Title="What's the largest number you can spell?" Tags="|mathematics|" AnswerCount="12" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="3707" PostTypeId="2" ParentId="3705" CreationDate="2014-11-07T11:20:44.560" Score="6" Body="&lt;p&gt;First question&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It depends on the language :P. For example, &quot;centomila&quot; is Italian form of &quot;a hundred thousand&quot;) :D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4527" LastActivityDate="2014-11-07T11:20:44.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3708" PostTypeId="2" ParentId="3705" CreationDate="2014-11-07T11:26:07.207" Score="15" Body="&lt;p&gt;For the first bonus question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Although there are lower numbers, none are &lt;em&gt;smaller&lt;/em&gt; than &lt;strong&gt;zero&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4522" LastEditorUserId="4522" LastEditDate="2015-01-23T15:28:25.287" LastActivityDate="2015-01-23T15:28:25.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3709" PostTypeId="2" ParentId="3705" CreationDate="2014-11-07T11:27:46.290" Score="25" Body="&lt;p&gt;How about:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Megiston&lt;/em&gt; aka. &lt;a href=&quot;http://mathworld.wolfram.com/Megistron.html&quot;&gt;&lt;em&gt;Megistron&lt;/em&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A decimal representation of this would require by far more digits than there are estimated atoms in the universe.&lt;/p&gt;&#xA;" OwnerUserId="4394" LastEditorUserId="4394" LastEditDate="2014-11-07T11:34:01.050" LastActivityDate="2014-11-07T11:34:01.050" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3712" PostTypeId="1" AcceptedAnswerId="3717" CreationDate="2014-11-07T12:28:34.690" Score="0" ViewCount="430" Body="&lt;p&gt;Determine the largest even positive integer that cannot be written as the sum of two odd composite positive integers.&lt;/p&gt;&#xA;" OwnerUserId="97" LastActivityDate="2014-11-07T13:09:41.510" Title="Determine the largest even positive integer" Tags="|mathematics|" AnswerCount="1" CommentCount="4" ClosedDate="2014-11-07T14:02:05.080" ContentLicense="CC BY-SA 3.0" />
  <row Id="3713" PostTypeId="2" ParentId="3705" CreationDate="2014-11-07T12:29:12.040" Score="5" Body="&lt;p&gt;In hexadecimal&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; FEDCBA0 (read zero as 'Oh')&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which would be 267242400&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In dozenal ( &lt;a href=&quot;http://www.dozenalsociety.org.uk/roses/devlieger.html&quot; rel=&quot;nofollow&quot;&gt;http://www.dozenalsociety.org.uk/roses/devlieger.html&lt;/a&gt; )&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strike&gt;forsen triliad&lt;/strike&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;Which is 4.279728215×10¹⁴&#xA;Doh! Just noticed that i used 'r' twice. So the answer is&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strike&gt;twosen triliad&lt;/strike&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;And the value is 1/2 of the prev one.&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And i have just noticed the Bonus 2 for using a plus, so here it is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 8+ (Eight plus) This is regex, no less than one eight.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4641" LastEditorUserId="4641" LastEditDate="2015-10-16T13:14:44.677" LastActivityDate="2015-10-16T13:14:44.677" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="3714" PostTypeId="2" ParentId="3678" CreationDate="2014-11-07T12:41:27.657" Score="5" Body="&lt;p&gt;Assume the converse. That is, for any player $A$ there is another player $f(A)$ who beat $A$ and who beat at least those players beaten by $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $f(f(A))$ must have beaten $A$, and must have beaten at least those players beaten by $A$. By induction we generate an unlimited chain of players $f^i(A)$, who each beat $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as the number of players is finite the chain must contain a cycle, and therefore a player who has beaten himself - a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="2021" LastActivityDate="2014-11-07T12:41:27.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3715" PostTypeId="2" ParentId="3680" CreationDate="2014-11-07T12:51:53.733" Score="4" Body="&lt;p&gt;If you can &lt;em&gt;fold&lt;/em&gt; the pizza, with one cut you could ideally create &lt;strong&gt;infinite&lt;/strong&gt; slices. Just fold the pizza &lt;em&gt;N&lt;/em&gt; times along the symmetry axis and cut it in the middle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you obtain is &lt;em&gt;2(N-1)&lt;/em&gt; triangular slices folded in half, and ready for eating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5 folds, 1 cut (I know what you're thinking, dirty mind):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DxzNC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1531" LastEditorUserId="1531" LastEditDate="2014-11-07T13:13:46.950" LastActivityDate="2014-11-07T13:13:46.950" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3716" PostTypeId="2" ParentId="3705" CreationDate="2014-11-07T13:05:26.310" Score="10" Body="&lt;p&gt;For bonus 1, if you use the convention that negative numbers are &quot;minus x&quot;, then I think&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; minus forty&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;wins. Checking everything down to minus one hundred is straightforward, and then anything less than minus one hundred must include either &quot;hundred&quot;, &quot;thousand&quot;, or one of the &quot;-llion&quot; words for higher powers of $10^3$ (or $10^6$, depending on convention, but it doesn't matter). And all of these contain an &quot;n&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1527" LastActivityDate="2014-11-07T13:05:26.310" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="3717" PostTypeId="2" ParentId="3712" CreationDate="2014-11-07T13:09:41.510" Score="6" Body="&lt;p&gt;Considering the remainder after dividing by 3, the smallest odd composite number leaving a remainder of 1 is 25 and the smallest one leaving a remainder of 2 is 35.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any even positive integer greater than or equal to 40:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's a multiple of 3, then it's 9 plus an odd multiple of 3;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's one more than a multiple of 3, then it's 25 plus an odd multiple of 3;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's two more than a multiple of 3, then it's 35 plus an odd multiple of 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the largest possible one is 38. This has a remainder of 2 when divided by 3, which can only be obtained by 1+1 or 0+2. However, 25+25 and 9+35 (the smallest numbers with those remainders) are both too large. Therefore there is no way to express 38 as the sum of two odd composite positive integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;38 is the answer.&lt;/p&gt;&#xA;" OwnerUserId="4647" LastActivityDate="2014-11-07T13:09:41.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3718" PostTypeId="1" CreationDate="2014-11-07T13:24:21.927" Score="0" ViewCount="63421" Body="&lt;p&gt;I think the answer isn't as simple as the question. So, I need your help. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assumed I have a box, sized 100cm X 100cm X 100cm (width X length X height). How many balls can fit in that box (the ball is 1cm in diameter)? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4648" LastActivityDate="2017-01-31T13:58:01.380" Title="How many balls can fit in the box?" Tags="|mathematics|logical-deduction|" AnswerCount="8" CommentCount="8" ClosedDate="2014-11-16T11:26:11.943" ContentLicense="CC BY-SA 3.0" />
  <row Id="3720" PostTypeId="1" AcceptedAnswerId="3721" CreationDate="2014-11-07T13:42:02.343" Score="4" ViewCount="2111" Body="&lt;p&gt;Canada is full of Canadians&amp;mdash;about 34 million of them at present. What you may not have known is that all Canadians are the same age.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't believe me? I'll prove it to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose there are exactly $n$ people in Canada, where $n$ is a whole number. An inductive proof of my claim is simple: we assume that any set of $n-1$ Canadians all have the same age, and prove that this implies any set of $n$ Canadians all have the same age.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do this, we choose an arbitrary set of $n-1$ Canadians, and call this set $A$. Out of set $A$ we pick any two Canadians, which we label $C_1$ and $C_2$. It is understood that $C_1$ is not the same person as $C_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we choose a second, different set of $n-1$ Canadians and call this set $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since set $A$ can only have one Canadian that set $B$ does not, we know that at least one of the following two conditions must hold:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$C_1$ is in both $A$ and $B$&lt;/li&gt;&#xA;&lt;li&gt;$C_2$ is in both $A$ and $B$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Hence there is always at least one Canadian in both sets. Let us call this Canadian $C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By assumption, we know that everyone in set $A$ has the same age, which we will denote $a$. Because $C$ is in $A$, we know that $C$'s age is also $a$. We also know that (by assumption) everyone in set $B$ has the same age. And since $C$ is in set $B$, the age of everyone in set $B$ must be $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since everyone in our complete set of $n$ Canadians is contained in either $A$ or $B$, we conclude that everyone in the complete set has the same age, $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we've proved that if any set of $n-1$ Canadians have the same age, any set of $n$ Canadians have the same age.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Naturally, we can work backwards inductively from $n = {\rm 34\ million}$ all the way to $n = 1$. Start by assuming that every set of $33,999,999$ Canadians have the same age, and use our proof to show that all $34,000,000$ Canadians therefore have the same age. To prove that every set of $33,999,999$ Canadians have the same age, we assume that every set of $33,999,998$ Canadians have the same age and use our proof again, etc., etc., all the way back to $n = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n = 1$, a set of one Canadian is just one person. Hence, trivially we see that everyone in the set has the same age.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have proved by induction that every Canadian is the same age! :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now... you might be skeptical of my result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the brainteaser is: &lt;strong&gt;Where is the fallacy in my proof? Where is the error?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Puzzlers are politely encouraged to place answers in spoiler blocks to avoid inadvertently spoiling the fun for other readers.&lt;/em&gt; :)&lt;/p&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2015-10-05T16:18:27.680" Title="Why are all Canadians the same age?" Tags="|mathematics|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3721" PostTypeId="2" ParentId="3720" CreationDate="2014-11-07T13:46:04.293" Score="11" Body="&lt;p&gt;This is the: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Horses Paradox &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/All_horses_are_the_same_color&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The problem is we make an implicit assumption that the two subsets of Canadians to which the induction assumption is applied have a common element. This is not true when the original set (prior to either removal) only contains two Canadians.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastActivityDate="2014-11-07T13:46:04.293" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3722" PostTypeId="2" ParentId="3718" CreationDate="2014-11-07T13:54:36.697" Score="11" Body="&lt;p&gt;Quoting from &lt;a href=&quot;http://en.wikipedia.org/wiki/Kepler_conjecture&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Experiment shows that dropping the spheres in randomly will achieve a&#xA;  density of around 65%. However, a higher density can be achieved by&#xA;  carefully arranging the spheres as follows. Start with a layer of&#xA;  spheres in a hexagonal lattice, then put the next layer of spheres in&#xA;  the lowest points you can find above the first layer, and so on – this&#xA;  is just the way you see oranges stacked in a shop. At each step there&#xA;  are two choices of where to put the next layer, so this natural method&#xA;  of stacking the spheres creates an uncountably infinite number of&#xA;  equally dense packings, the best known of which are called cubic close&#xA;  packing and hexagonal close packing. Each of these arrangements has an&#xA;  average density of $\frac\pi {3\sqrt2} = 0.740480489...$. The&#xA;  Kepler conjecture says that this is the best that can be done—no other&#xA;  arrangement of spheres has a higher average density.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So your answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\frac{\text{box volume}}{\text{ball volume}} \times {\text{packing density}} =$&lt;BR&gt;&lt;BR&gt;&#xA; $\frac{100^3}{\frac43 \times \pi \times \left(\frac12\right)^3} \times {\frac\pi {3\sqrt2}} =$&lt;BR&gt;&lt;BR&gt;&#xA; $\frac{100^3}{\left(\frac12\right)^3\times 4 \times \sqrt2} =$&lt;br&gt;&lt;br&gt;&#xA; $\frac{100^3}{\frac12 \times \sqrt2} = 100^3 \times \sqrt2 \approx 1414213$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4637" LastEditorUserId="8822" LastEditDate="2017-01-31T13:58:01.380" LastActivityDate="2017-01-31T13:58:01.380" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3723" PostTypeId="2" ParentId="3718" CreationDate="2014-11-07T13:57:19.427" Score="0" Body="&lt;p&gt;Well, assuming each ball occupies ~1 cm&lt;sup&gt; 3 &lt;/sup&gt; of space, you have fit in 1,000,000 balls just by having them in a square lattice. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2014-11-07T13:57:19.427" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3724" PostTypeId="2" ParentId="3705" CreationDate="2014-11-07T14:00:51.433" Score="19" Body="&lt;p&gt;There's also:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://googology.wikia.com/wiki/Oktria&quot;&gt;Oktria&lt;/a&gt;, which is between 3↑↑↑↑↑↑↑4 and 3↑↑↑↑↑↑↑5 in &lt;a href=&quot;http://en.wikipedia.org/wiki/Knuth%27s_up-arrow_notation&quot;&gt;Knuth's up arrow notation&lt;/a&gt;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;" OwnerUserId="4652" LastEditorUserId="2023" LastEditDate="2014-11-10T17:18:23.593" LastActivityDate="2014-11-10T17:18:23.593" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3725" PostTypeId="2" ParentId="3718" CreationDate="2014-11-07T14:07:07.143" Score="1" Body="&lt;p&gt;If the structure of the spheres does not need to be maintained, you could melt down the spheres and they would occupy $.52\ \text{cm}^3$ (from the volume of a sphere: $\frac43 \pi r^3$) each resulting in:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $1909859 = \frac{100^3}{\frac43 \pi (.5^3)}$ &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2803" LastEditorUserId="4421" LastEditDate="2015-10-05T15:57:40.390" LastActivityDate="2015-10-05T15:57:40.390" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3726" PostTypeId="1" AcceptedAnswerId="3733" CreationDate="2014-11-07T14:12:30.190" Score="32" ViewCount="39930" Body="&lt;p&gt;Last year, December 21, 2013, you were awarded the &quot;#1 Detective in the World&quot; for solving more than 50 murders. Your boss gave you a promotion, contingent upon your moving to Chicago. You agreed and took your new position. Your boss told you that you would be taking a plane to Chicago, the ticket and everything already having been paid for by your boss. He also told you that you would be taking off on January 15, 2014.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of a sudden, all you heard was a loud beeping sound. &quot;BEEP BEEP BEEP BEEP BEEP&quot;, as your eyes open wide. You got up and realized it was your alarm clock going off. 6:30 AM. You turned the alarm off, did a big stretch and gave a big yawn. &quot;Why do I keep having the same dream about the day that I got promoted?&quot; you pondered to yourself. Without giving it another thought, you got out of bed and started getting ready for your day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You got to work and went straight to your office. As soon as you sat down you heard a knock on your office door. &quot;Come in,&quot; you said.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your assistant opened the door. &quot;You're needed at the intersection of Main and Peele.&quot; You weren't surprised to hear this. You quickly got up and drove to the location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you got there, there were 3-4 police cars, yellow &quot;Police Line&quot; tape, and a large crowd gathered around the intersection. You got out of the car and walked towards the scene. As you got closer, you could see a body covered with a blanket, and other police and detectives talking and taking pictures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Morning. It seems like this man here was murdered,&quot; said one of the other detectives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You lifted the blanket to see that the man had been stabbed three times, had two gunshot wounds to his left hip, and had red markings around his neck. &quot;It looks like he has been strangled with rope as well,&quot; you said as you kept on examining his body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You covered the body back up with the blanket. You looked around and saw that there were small yellow cones with numbers on them, and beside each cone you saw an object. You walked over to one of the cones. It had the number &quot;4&quot; on it. You looked beside the cone and saw perfect circular blood drops. You then walked over to the number &quot;2&quot; cone, seeing a knife with no blood on it. Cone number &quot;3&quot; was beside a piece of paper. On it was &quot;He Deserved It! 1H8H1M&quot;. You looked closely at &quot;1H8H1M&quot; and wondered what that meant. You stood there and pondered it for a minute, then moved on to cone number &quot;1&quot;. The last cone was beside a shoe print. It was mud which produced the prints. The shoe size looked to be a size 9 men's shoe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I've seen enough. I'm heading back to my office&quot; you told the other detectives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next day, your assistant told you that the forensics lab had asked you to visit them when you had the chance. You headed over to the forensics lab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Hello, you must be the great detective. Nice to meet you!&quot; said the forensic technician. &quot;The results are back. There were no fingerprints on the note. The knife was clean like it was brand new straight from the box. The shoe print was of a size 9 men's shoe. But the most interesting part is that the blood stains are not human. It's animal blood. Specifically from a poodle.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you pondered the information that the technician had told you, your cell phone rang. You answered it. &quot;We have found a few suspects here at the station. Please come ASAP.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You rushed over to the station. As you drove over there, you got violently cut off by another driver and almost crashed. &quot;Learn how to drive!&quot; the man in the other car yelled as he drove away. You caught a glimpse of a man with shaggy brown hair, small nose, and big ears. You took a deep breath and continued on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you arrived at the station you saw that the police had gathered three suspects: a white man between the age of 22-26, short black hair, with a toned build; a tall blonde female, wearing a fancy dress and jewelry like she was going on a date or to a party; and a skinny homeless man wearing very battered clothing, with long greasy hair and a full beard. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The twenty-something man said that he was at the bank getting some money three blocks away from the murder site around the same time the murder was reported.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The blonde female said that she was going for a run around the block as she just lived down the street from the murder site. She made one round around the block, then decided to do another round. As she was jogging she saw the crowd and decided to stop and take a look.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The homeless man didn't speak English well and seemed kinda crazy. &quot;I did no murder. I asking for money for food. Hungry, haven't eat couple days.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You went back to your office and pondered about the clues and information from the murder scene, the forensics lab, and the suspects. After pondering a while you figured out who the murderer was. You quickly rushed back to the police station and told the police to arrest the person.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who was the murderer?&lt;/p&gt;&#xA;" OwnerUserId="4645" LastEditorUserId="5373" LastEditDate="2015-05-03T10:55:22.570" LastActivityDate="2015-05-03T10:55:22.570" Title="Mysterious Murder Mystery" Tags="|logical-deduction|mystery|story|situation|" AnswerCount="15" CommentCount="19" ClosedDate="2014-12-02T02:11:47.083" ContentLicense="CC BY-SA 3.0" />
  <row Id="3727" PostTypeId="2" ParentId="3718" CreationDate="2014-11-07T14:18:16.593" Score="3" Body="&lt;p&gt;I don't have the exact value or formula to how to calculate this. But I want to show how you should think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if this question was in some simple elementary school test, answer should be obviously 100x100x100=10&lt;sup&gt; 6 &lt;/sup&gt; in total. Because it uses simple &lt;strong&gt;square packing&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However there are few other more packing styles like hexagonal packing, you can actually use empty spaces that occurs on &lt;strong&gt;square packing&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ioy1A.gif&quot; alt=&quot;packing types&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As one can see, in &lt;strong&gt;square packing&lt;/strong&gt; height of two lines is 2cm. But in &lt;strong&gt;hexagonal packing&lt;/strong&gt; it is less then 2cm. You can try and use much more space with different packing types. After finding right packing, you can calculate exact value.&lt;/p&gt;&#xA;" OwnerUserId="4506" LastActivityDate="2014-11-07T14:18:16.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3728" PostTypeId="1" AcceptedAnswerId="3729" CreationDate="2014-11-07T14:23:45.603" Score="8" ViewCount="6154" Body="&lt;p&gt;(Got this from my facebook news feed.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A friend of mine, upset after the earlier housing number &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3165/house-number-mastermind&quot;&gt;conundrum&lt;/a&gt;, decided to make me guess his. He said:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If my house number is a multiple of 3, it's from 50 to 59.&lt;/li&gt;&#xA;&lt;li&gt;If it is not a multiple of 4, then it is a number from 60 to 69.&lt;/li&gt;&#xA;&lt;li&gt;If it is not a multiple of 6, then it is a number from 70 to 79.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What's the house number?&lt;/p&gt;&#xA;" OwnerUserId="2444" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2016-02-29T18:10:56.903" Title="Unhelpful House Number Reminders" Tags="|logical-deduction|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3729" PostTypeId="2" ParentId="3728" CreationDate="2014-11-07T14:25:58.760" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The house number must be 76. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Either the number is a multiple of 3 or it is not.  If it is then it is between 50 and 59, if it is not, then it is not a multiple of 6 either, and thus it is between 70 and 79.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore we know it is not between 60 and 69, and therefore it is a multiple of 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is between 50 and 59, and is a multiple of 4 it must be 52 or 56.  But neither of these are multiples of 3.  Therefore the number is not between 50 and 59, but between 70 and 79.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of the numbers between 70 and 79, it can be 72 or 76.  Since 72 is a multiple of 6 however, this leaves 76 as the only possible solution.&lt;/p&gt;&#xA;" OwnerUserId="97" LastEditorUserId="17446" LastEditDate="2016-02-29T18:10:56.903" LastActivityDate="2016-02-29T18:10:56.903" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3730" PostTypeId="2" ParentId="3705" CreationDate="2014-11-07T14:26:23.697" Score="24" Body="&lt;p&gt;You did not specify which alphabet we should be using, so when using greek, the smallest would be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;ε &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4631" LastActivityDate="2014-11-07T14:26:23.697" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3731" PostTypeId="2" ParentId="3726" CreationDate="2014-11-07T14:29:51.400" Score="2" Body="&lt;p&gt;Well, I have some ideas, but I'm still struggling to piece it together. Here's my thinking so far (as much for me as anyone else).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the rest of the clues, the killer is either...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the apparently-homeless man in the identity parade&lt;br&gt;&#xA; The man who cut you up in the car was the real homeless man, the ID parade &quot;homeless man&quot; is actually a normal guy who gave the bum his car keys, then made himself look like the homeless guy as a disguise&#xA; this explains why you got cut up by a reckless driver: the real bum wouldn't know how to drive&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the woman&lt;br&gt;&#xA; She is tall (size 9 shoes would fit), dressed for a party but going out running around the block in a fancy dress and jewellery&lt;br&gt;&#xA; It's likely she would have worn high heels dressed like that, but obviously couldn't run in them, so if she was actually doing that&lt;br&gt;&#xA; A pair of trainers would make sense (which could get muddy and leave the footprint).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The victim...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; His dog was killed and used for blood (the poodle blood)&lt;br&gt;&#xA; The rope he kept his dog on was used to strangle the victim&lt;br&gt;&#xA; The &quot;1H8H1M&quot; note is text-slang for &quot;I hate him&quot;, which could be either the woman living down the road who has to walk past him, and/or hear him, or could be the apparently-homeless guy for any reason&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I don't yet know&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Why the blood stains are circular&lt;br&gt;&#xA; Where the stabbing-knife and gun came from, or went&lt;br&gt;&#xA; Whether Chicago, Main and Peele have anything to do with it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2518" LastEditorUserId="2518" LastEditDate="2014-11-07T15:14:10.813" LastActivityDate="2014-11-07T15:14:10.813" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3732" PostTypeId="2" ParentId="3720" CreationDate="2014-11-07T14:36:17.260" Score="5" Body="&lt;p&gt;The fallacy is here: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;Out of set A we pick any two Canadians, which we label C1 and C2. It is understood that C1 is not the same person as C2.&lt;/code&gt;. We cannot pick two elements out of set containing only one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="419" LastActivityDate="2014-11-07T14:36:17.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3733" PostTypeId="2" ParentId="3726" CreationDate="2014-11-07T14:42:09.357" Score="24" Body="&lt;p&gt;I would have arrested the blonde woman. She did say she was going on a jog, and size 9 men's shoes are not uncommon for women with big feet to wear. Furthermore, someone who is dressed to the nines like her is the most likely to own both a poodle and a leash for said poodle. Leash = rope.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, if she was going on a jog and a murder and crowd forming took place in the time it took her to simply go around the block, she would have heard the gunshots and would likely have either gone home or sprinted back to investigate. One does not continue a leisurely jog after that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The young man probably has a reasonable alibi; cameras at the bank could vindicate him. If he had a gun, he wouldn't have taken it to a bank unless he intended to rob the place.&#xA;The homeless man is likely either innocent or a victim here; with a rudimentary grasp of English, there's absolutely no way he would have written a note in l33tsp3ak (clearly &quot;I hate him&quot;) and, with his tattered and dirty state, not left it or the murder weapon completely clean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for how she performed the murder or why... I'm still not sure. I'd get a warrant out for her apartment to find the gun. What is interesting is that the poodle's body is nowhere to be found, but it could serve as a motive... finding its corpse would also probably pin the killer, as its owner could easily be identified.&lt;/p&gt;&#xA;" OwnerUserId="2444" LastActivityDate="2014-11-07T14:42:09.357" CommentCount="20" ContentLicense="CC BY-SA 3.0" />
  <row Id="3734" PostTypeId="2" ParentId="3728" CreationDate="2014-11-07T14:45:11.780" Score="10" Body="&lt;p&gt;Writing the long answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If it's a multiple of 3 is between 50 and 59: 51, 54, 57, but then it should be also a multiple of 4 and 6 and none of them are. &#xA;&lt;br&gt;&#xA;If it's not a multpile of 4 it's between 60 and 69: &lt;s&gt;60&lt;/s&gt;, 61, 62, 63, &lt;s&gt;64&lt;/s&gt;, 65, 66, 67, &lt;s&gt;68&lt;/s&gt;, 69; 63, 66, and 69 are multiple of 3 and not vaild, while 61, 62, 65, 67 and 69 are not multiple of 6 and thus not valid either. The number must be between 70 and 79.&#xA;&lt;br&gt;&#xA;72, 75 and 78 are multiple of 3 and not valid. 70, 71, 73, 74, 77 and 79 are not multiple of 4 and not valid. The only remaining number is 76.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4652" LastEditorUserId="-1" LastEditDate="2014-11-07T18:09:07.460" LastActivityDate="2014-11-07T18:09:07.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3737" PostTypeId="2" ParentId="3726" CreationDate="2014-11-07T15:59:03.053" Score="4" Body="&lt;p&gt;The biggest question is, why would a stabbed man, be shot, or strangled. I think the answer lies in three attempts, only one killed him. I would say let's put them in different orders, it might be that all three made an attempt, at different stages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One used the gun, that didn't kill him. A struggle meant the drop of the note, the lack of finger prints means the person came with intention to kill, left with a struggle. The poodle blood means it was carried, and that it was an unsuccessful attempt. The knife is weird. But it also can mean someone left in a hurry after finding out the victim was already killed. Now the rope! I doubt it was used last, because a gun shot or a stab wound would deter a strangler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I would say, the first came and tried to strangle the guy, unsuccessfully (and probably earlier in a different location). The poodle dog was present and it was carried, it got injured. A struggle with the victim yields in gun shots, but not death. The last person to show up is the stabber. I think the knife simply belongs to the victim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to think about the order of events though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;==== First attempt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Homeless man tries to strangle him for a buck. Not successful. Man meats victim probably for a chat, they both have weapons thought. Struggle, drop the knife, two gun shots, then the man runs. The woman meats the victim half breathed, with the note in her pocket, that probably was given to her by the man, she assumes the worst. Finding him with gun shots, alerted, she stabs him to finish him, and drops the note and runs. Can't place the dog though!&lt;/p&gt;&#xA;" OwnerUserId="4657" LastEditorUserId="4657" LastEditDate="2014-11-07T16:15:58.843" LastActivityDate="2014-11-07T16:15:58.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3738" PostTypeId="2" ParentId="3726" CreationDate="2014-11-07T16:00:47.047" Score="11" Body="&lt;p&gt;I don't think I've got it 100% solved, but here's what I'm thinking...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The suspect interviews are happening THE DAY AFTER the body was found. There's nothing odd about the woman from jogging at 6:30am one day and then being in fancy outfit during business hours the next day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The murder seems to have happened in the early morning hours. I think the perfectly circular (undisturbed) blood droplets imply they are pretty fresh. If your alarm goes of at 6:30 and the first thing you do is head to a crime scene, that also seems to indicate the murder happened early morning or perhaps late-night.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...Which means the black-haired guy could NOT have been at a bank. They usually don't open until 8am. I suppose he could have been at an ATM, but that transaction would be quite quick and wouldn't be much of an alibi. You'd have plenty of time to swing by the ATM after the murder!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The homeless guy probably didn't write the note, due to his difficulty with English. The near-car-crash is irrelevant, especially if the story takes place in Chicago! Road rage is part of daily life :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also find it puzzling that you're apparently meeting the forensic scientist for the first time on this particular day... Not sure what to make of that. It's important to figure out if this takes place in Chicago or at your prior location. Could this be the first case at your new job?&lt;/p&gt;&#xA;" OwnerUserId="4661" LastActivityDate="2014-11-07T16:00:47.047" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3739" PostTypeId="1" AcceptedAnswerId="3741" CreationDate="2014-11-07T16:02:59.550" Score="13" ViewCount="16002" Body="&lt;p&gt;An old man and his mute niece always wanted to visit some beautiful tropical island. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;em&gt;Before my sight leaves me, I want to see how a real palm looks like!&lt;/em&gt;&quot;, said the old man. He was lying. His sight had already left him, and he could still see silhouettes and shadows, but communication in sign language with his niece was getting more and more complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, they decided to go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They left on a small boat, one of those tourist trips on a sailboat. There were 10 people on the boat, including them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They were approaching a very nice archipelago when it began.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A storm, a powerful one, overturned the boat, leaving the crew in the ocean to struggle for life. The niece, although chubby and weak, managed to rescue her grandfather and reach a deserted little island in the archipelago with other 3 people. The sea halved them, but still there was hope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Days passed with very low food. Our two tourists were weak and hungry and only a little water supply was available. After a week one of the man said that he would begin to hunt for seagulls, of which the island had a lot. The old man only saw a big black cloud talking, the salt and sand had only worsen his sight, and, beaten by hunger, promptly said &quot;&lt;em&gt;Yes...please...&lt;/em&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently, the hunter was a very good one. Finally they could eat something, and could also eat it cooked since a fire was lit up by one of the survivors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few other days passed. The seagulls were manna for them, but not enough to give them strength to move. All the day the old man sat beneath a palm thinking &quot;&lt;em&gt;This is not how I wanted to see them...&lt;/em&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suddenly a great &quot;&lt;em&gt;WOO&lt;/em&gt;&quot; echoed in the archipelago. A cruise ship. A cruise ship! They waved vigorously, everyone beside the old man, he was too weak. The ship arrived and collected them all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;em&gt;We are safe!&lt;/em&gt;&quot; thought the old man. Then, with his heart in peace, he fell asleep, only to wake up in the medical wing of the ship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;em&gt;Come, my friend, it's time for you to eat something good!&lt;/em&gt;&quot; said the captain. They went to the restaurant and the old man couldn't believe his eyes and almost cried to the signboard at the entry:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;TODAY'S SPECIAL: SEAGULL MEAT!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;em&gt;Well, at least it will be well cooked,&lt;/em&gt;&quot; he thought. He cut the meat and tried it. A tear slipped on his cheek. He stood up and went to the deck of the ship. He climbed over the railing and jumped off the ship, giving to the sea another victim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why did the old man commit suicide?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;DISCLAIMER: I wrote this riddle long on purpose. I think an interesting story behind enhances the fun.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="4652" LastEditorUserId="5373" LastEditDate="2016-10-24T00:05:49.860" LastActivityDate="2021-03-25T10:55:11.857" Title="The old man doesn't like seagull meat" Tags="|story|situation|" AnswerCount="13" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="3740" PostTypeId="2" ParentId="3705" CreationDate="2014-11-07T16:07:54.623" Score="44" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; US Gov. Debt&lt;br&gt;&lt;br&gt;&#xA; 17,907,911,809,200 at last glance. :P&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2014-11-07T16:07:54.623" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3741" PostTypeId="2" ParentId="3739" CreationDate="2014-11-07T16:08:46.053" Score="27" Body="&lt;p&gt;Because&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; he was eating what had been termed seagull meat on the island, when in fact he was eating the flesh of the dead. He realized he was a cannabal then, because now was the true &amp;quot;seagull meat&amp;quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2265" LastEditorUserId="63414" LastEditDate="2021-03-25T10:55:11.857" LastActivityDate="2021-03-25T10:55:11.857" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="3742" PostTypeId="2" ParentId="3739" CreationDate="2014-11-07T16:08:48.320" Score="6" Body="&lt;p&gt;The meat in the restaurant tasted unlike anything he had eaten on the desert island, and the old man realised that he had been fed meat from those who had drowned.&lt;/p&gt;&#xA;" OwnerUserId="2461" LastActivityDate="2014-11-07T16:08:48.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3743" PostTypeId="2" ParentId="3739" CreationDate="2014-11-07T16:10:17.020" Score="12" Body="&lt;p&gt;He was blind.  After he was rescued and he ate the seagull from the restaurant it didn't taste the same as the meat he had eaten on the island. Therefore, he surmised that on the island he had been eating the flesh of the human victims of the crash. He couldn't live with the thought that &lt;strong&gt;he ate his own niece&lt;/strong&gt; and so killed himself. &lt;/p&gt;&#xA;" OwnerUserId="97" LastEditorUserId="97" LastEditDate="2014-11-07T16:55:12.843" LastActivityDate="2014-11-07T16:55:12.843" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="3745" PostTypeId="2" ParentId="3553" CreationDate="2014-11-07T16:34:46.837" Score="4" Body="&lt;p&gt;Objection!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Why is the letter e considered as tiny? If we're talking about written language components, the dot would be more appropriate in my opinion. &quot;It's in here&quot; can imply that the variable is present in the question block, not just the word &quot;here&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My two cents: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The dot. It appears multiple times in the question: 3 times as a period, 3 times as an element of the character i.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4631" LastActivityDate="2014-11-07T16:34:46.837" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3746" PostTypeId="2" ParentId="3668" CreationDate="2014-11-07T16:35:41.047" Score="7" Body="&lt;p&gt;Black, because the bear was a polar bear and polar bears are black underneath the fur.&lt;/p&gt;&#xA;" OwnerUserId="97" LastActivityDate="2014-11-07T16:35:41.047" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3747" PostTypeId="2" ParentId="3695" CreationDate="2014-11-07T16:39:45.247" Score="3" Body="&lt;p&gt;How to build up the answer as we go:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Cade has six red stamps and three blue ones.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus Cade must have at least 9 stamps.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In his collection, seven stamps are from Argentina and six stamps are from Sweden.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This means he has at least 13 stamps, with 7 from Argentina and 6 from Sweden. Any three of them are blue and any other six of them are red.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;One stamp is purple and it is not from Argentina or Sweden.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now he must have at least 14 stamps: 7 from Argentina and 6 from Sweden, and one from Elsewhere.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Two of his Argentine stamps are red and one is blue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is where it gets tricky. Among the six red stamps, only two can be from Argentina. The other four can be from Sweden. Among his three blue stamps, one can be from Argentina. Let's assign the other two to Sweden. Now we still have 14 stamps:&#xA;2 Argentina/red, 1 Argentina/blue, 4 Argentina/x, 4 Sweden/red, 2 Sweden/blue, 1 Elsewhere/Purple.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Two of his Swedish stamps are blue and three are red.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Luckily, we had assumed that there were two blue Swedish stamps. However, we need to take one red one away from Sweden.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our configuration after this rule: 2 Argentina/red, 1 Argentina/blue, 4 Argentina/(not red or blue), 3 Sweden/red, 2 Sweden/blue, 1 Elsewhere/Purple, and 1 Elsewhere/Red.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since every group has either a different color or a different source location, we can't combine any groups. There may be more stamps, but all of the stamps in the rules have been accounted for exactly once. Thus, the rules indicate a minimum of 15 stamps.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2101" LastActivityDate="2014-11-07T16:39:45.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3748" PostTypeId="2" ParentId="3728" CreationDate="2014-11-07T16:44:06.020" Score="0" Body="&lt;p&gt;The house number can't be a multiple of 3: It would be from 50 to 59, so it would be one of 51, 54 or 57, all these are not multiples of four, so the house number would also be from 60 to 69 which isn't possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since it's no multiple of 3, it's no multiple of 6 either, so its from 70 to 79. Of these only 72 and 76 are multiples of four, but 72 is a multiple of 3 which the house number isn't, that leaves only 76. &lt;/p&gt;&#xA;" OwnerUserId="4664" LastActivityDate="2014-11-07T16:44:06.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3749" PostTypeId="2" ParentId="1987" CreationDate="2014-11-07T16:51:47.227" Score="0" Body="&lt;p&gt;I think the standard answer I around here for a similar question is that they are not twins because there are in fact 3 of them and the third one is in a different school, so they are not twins, rather they are triplets.&lt;/p&gt;&#xA;" OwnerUserId="4667" LastActivityDate="2014-11-07T16:51:47.227" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3750" PostTypeId="2" ParentId="3728" CreationDate="2014-11-07T16:57:34.977" Score="0" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 76&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the answer were between 50 and 59 then it must be a multiple of 3 (by rule 1), but also a multiple of 4 and 6 (to negate the other 2 rules). There are no such numbers in this range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the answer were between 60 and 69, then it must not be a multiple of 3, but must be a multiple of 6. There are no numbers that can be a multiple 6 and &lt;em&gt;not&lt;/em&gt; a multiple of 6. So that rules out this range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the number between 70 and 79, one must find a number that is not a multiple of 6, nor a multiple of 3, but &lt;em&gt;is&lt;/em&gt; a multiple of 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 76 is the product of 2 x 2 x 19 - thus it is a multiple of 4, but not 3 or 6 (as needed).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4362" LastActivityDate="2014-11-07T16:57:34.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3751" PostTypeId="1" AcceptedAnswerId="3752" CreationDate="2014-11-07T17:01:38.860" Score="5" ViewCount="2545" Body="&lt;p&gt;Examine the following sequence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;O, T, T, F, F, S, S, E, ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you determine what comes next?&lt;/p&gt;&#xA;" OwnerUserId="97" LastEditorUserId="5188" LastEditDate="2015-05-08T04:31:39.317" LastActivityDate="2015-05-08T04:31:39.317" Title="What comes next in this sequence?" Tags="|letter-sequence|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3752" PostTypeId="2" ParentId="3751" CreationDate="2014-11-07T17:03:31.623" Score="19" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; N&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each letter is the first letter of English counting numbers.&#xA; &lt;strong&gt;O&lt;/strong&gt;ne, &lt;strong&gt;T&lt;/strong&gt;wo, &lt;strong&gt;T&lt;/strong&gt;hree, &lt;strong&gt;F&lt;/strong&gt;our, &lt;strong&gt;F&lt;/strong&gt;ive, &lt;strong&gt;S&lt;/strong&gt;ix, &lt;strong&gt;S&lt;/strong&gt;even, &lt;strong&gt;E&lt;/strong&gt;ight, &lt;strong&gt;N&lt;/strong&gt;ine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2101" LastActivityDate="2014-11-07T17:03:31.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3753" PostTypeId="2" ParentId="3718" CreationDate="2014-11-07T17:28:28.810" Score="4" Body="&lt;p&gt;Since this is supposed to be a puzzle (not just a math question), I assume any loophole can be used. So, I'll say &lt;strong&gt;an infinity of balls&lt;/strong&gt; can fit in the box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trick is that the description doesn't explicitly require &lt;em&gt;all&lt;/em&gt; the balls to fit &lt;em&gt;at the same time&lt;/em&gt; in the box. Since any 1cm diameter ball can fit the box, there's an infinity of existing and hypothetical balls that can fit in the box.&lt;/p&gt;&#xA;" OwnerUserId="4668" LastEditorUserId="4668" LastEditDate="2014-11-07T19:19:00.463" LastActivityDate="2014-11-07T19:19:00.463" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3754" PostTypeId="2" ParentId="3705" CreationDate="2014-11-07T17:51:25.683" Score="12" Body="&lt;p&gt;Taking the question literately, the best I can do is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; five thousand&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Venturing outside the box, I can get to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ULONG_MAX which (in the standard C library) comes out to (2^64)-1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4670" LastEditorUserId="4670" LastEditDate="2014-11-10T18:03:55.077" LastActivityDate="2014-11-10T18:03:55.077" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="3755" PostTypeId="2" ParentId="3726" CreationDate="2014-11-07T18:13:50.167" Score="2" Body="&lt;p&gt;I will improve and involve all clues in time but here is the my initial thoughts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clues&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;My main lead on this murder is definetely the wounds. There is 3&#xA;different type of wounds. Which means either there were a long fight&#xA;and victim and killer swapped some tools during encounter or each&#xA;wound made by another person.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Bullet wounds is on victim's hip. Probably he was running from&lt;br&gt;&#xA;someone and got shot from hip or someone shoot him from his hip on&lt;br&gt;&#xA;purpose.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Knife is clean. This signs knife is the murder weapon. It is used to &#xA;kill the victim and cleaned perfectly by killer.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Blood of the dog is circular, probably dropped from higher point. I'm&#xA;assuming it was held and carried by someone (around chest level)&lt;br&gt;&#xA;since its injured(most probably).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The note is clearly does not belong to killer. Someone probably wrote&#xA;the message after s/he saw the dead body.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So here is my scenario:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;White-man is a hired killer. Victim was kept and tortured by white-man someplace near to the intersection. Somewhere dusty and muddy. As the killer tries to choke him with a rope, somehow victim got rid of him and escaped. During the escape white-man shoot him twice but bullets hits to his hips. While the victim was running, he accidentally harmed lady's dog and continued running. Lady got her injured puppy on her lap and chased the victim. Few minutes later, killer caught the victim. Stabbed and killed him. Then he quickly cleaned knife and dropped it there. Soon after killer left, lady arrived the point of murder. She saw the body and wrote a note for victim. At the morning, white-guy was at the bank, because he did his job and he was getting his payment for the kill.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4506" LastActivityDate="2014-11-07T18:13:50.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3756" PostTypeId="2" ParentId="3720" CreationDate="2014-11-07T18:14:46.493" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Thus we've proved that if any set of $n−1$ Canadians have the same age, any set of $n$ Canadians have the same age.&quot;  No, you've actually proven the opposite: if a set of $n$ Canadians have the same age, then any subset of $n−1$ Canadians have the same age.  More precisely, you've proven that any $n-1$ subset is comprised of members the same age as members of any other $n-1$ subset.  You've restricted yourself to subsets of size $n-1$, but it should be possible to extend the logic for subsets of any size.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My first thought was that you can't work backwards to size $n=1$, because your logic depends on having two sets of at least two people.  However, you can work backwards to $n=2$ or $n=3$, and I'm pretty sure you can find a set of $2$ or $3$ people in Canada who do, in fact, have the same age.  &lt;/p&gt;&#xA;" OwnerUserId="4674" LastEditorUserId="4421" LastEditDate="2015-10-05T16:18:27.680" LastActivityDate="2015-10-05T16:18:27.680" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3757" PostTypeId="2" ParentId="3726" CreationDate="2014-11-07T18:23:38.333" Score="21" Body="&lt;p&gt;&lt;strong&gt;Story&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The woman was jogging with the poodle around the block. &lt;br /&gt;&lt;br /&gt;The &quot;victim&quot; (here on out as Attacker) confronted the woman with the knife and the dog went crazy as any good dog should. Then the attacker pulls out a gun and shoots the dog. The homeless man is witnessing this and comes to the aid of the woman. He is carrying his own knife. &lt;br/&gt;&lt;br/&gt;The homeless man surprises the attacker and stabs him and they start wrestling on the ground. The attacker struggles and accidentally shoots himself in the leg as the homeless man wrestles with him and the gun. &lt;br /&gt;&lt;br /&gt; While this is happening, the woman starts to strangle the man with the leash. As the attacker gets weaker, the homeless man is able to stab a couple of more times.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Evidence Allocation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - The circular blood stains is from the dog bleeding and being carried. &lt;br /&gt; - The foot print is either from the woman jogging if she has large feet or the homeless man &lt;br /&gt; - The clean knife is the attackers which he never used. &lt;br /&gt; - The gunshots were self inflicted by the attacker struggling. &lt;br /&gt;- The note was written by both or just the homeless man as the woman told him what to write. The &quot;1H8H1M&quot; came from the homeless man meaning &quot;I hate him&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Murderer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The murderers are the woman and homeless man.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4672" LastEditorUserId="4672" LastEditDate="2014-11-07T22:40:15.693" LastActivityDate="2014-11-07T22:40:15.693" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="3758" PostTypeId="2" ParentId="3739" CreationDate="2014-11-07T18:24:15.043" Score="18" Body="&lt;p&gt;It wasn't salty enough.                        .                              &lt;/p&gt;&#xA;" OwnerUserId="4329" LastActivityDate="2014-11-07T18:24:15.043" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3760" PostTypeId="1" AcceptedAnswerId="3761" CreationDate="2014-11-07T18:40:46.397" Score="1" ViewCount="2364" Body="&lt;p&gt;Carl and Mike are two great friends. They grew up together and stay together a lot. They both &lt;em&gt;LOVE&lt;/em&gt; the delicious turkish meal &lt;a href=&quot;http://www.villageuniversel.com/wp-content/uploads/2014/06/kebab3.jpg&quot; rel=&quot;nofollow&quot;&gt;Kebab&lt;/a&gt; and eat it very often.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One day Carl comes to Mike with great news.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;A new fantastic kebab shop recently opened in town!&quot;, said Carl. &quot;&lt;em&gt;There are already a lot of kebab shops in town, why this should be so 'fantastic'?&lt;/em&gt;&quot; thought Mike.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Tell me more&quot; said Mike.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;This is truly the best, Mike! The owner is the most renowned kebab chef of all time and they have a lot of wonderful ingredients and the place is great and is really cheap and...&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Calm down!&quot; shout Mike. &quot;From what you say I think you've already been there&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yes&quot; replied Carl. &quot;And there is something more I have to tell you. This is no normal place. They have a lot of costumers and can't handle them all. They've come with a very peculiar solution for it&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;And it is?&quot; asked Mike.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Riddles.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;What? Riddles?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yes. Every costumer, to enter the shop, has to answer a riddle in one minute. After the minute passes, if the answer is correct, the doors open and he can enter to eat. You just have to say the solution, that's it! Mine was really silly: &lt;em&gt;'If a brick weighs one kilogram plus half brick, how much does a brick weight?'&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;code&gt;??&lt;/code&gt;&quot; quickly answered Mike. &lt;em&gt;&lt;strong&gt;BONUS! Solve this!&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Exactly! You should really give it a try!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Really curious Mike decided to go to the kebab shop the same evening.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upon arrival Mike was faced with a big door with a television screen on it. It was written &quot;PRESS THE BELL&quot;. Mike pressed it.&lt;br&gt;&#xA;Very soon the riddle appeared. &quot;&lt;em&gt;WOW! So Carl wasn't lying after all!&lt;/em&gt;&quot; he thought:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;WHAT'S IT THAT MAKES POTATO IN MANDATORY AND PATIENCE IN CAT?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;em&gt;...Wait, what? Mandatory potatoes? What's this? Think, think!&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The screen changed in a big 60 number. And started to countdown. 59, 58, 57..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;em&gt;The cat is patient. Also the potato. No, no...&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...45, 44, 43..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;em&gt;Argh! The time is flowing! ...Wait. I'm really struggling to solve a riddle just to eat a kebab? There is another shop just 400 mt from here. What I'm doing??&lt;/em&gt;&quot; thought Mike.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...32, 31, 30...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;em&gt;I'm not going to do this. I just want to eat.&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;You know what? This is nonsense. There is a kebab shop just around the corner, I'm going there. Goodbye.&quot; shouted Mike, and left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...10, 9, 8...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...3, 2, 1... &lt;em&gt;CLICK!&lt;/em&gt; and the door opened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you explain why?&lt;/p&gt;&#xA;" OwnerUserId="4652" LastEditorUserId="5373" LastEditDate="2014-12-29T19:10:29.873" LastActivityDate="2014-12-29T19:10:29.873" Title="Pointlessly difficult to get a meal here, isn't it?" Tags="|riddle|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3761" PostTypeId="2" ParentId="3760" CreationDate="2014-11-07T18:46:50.107" Score="16" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mike said &quot;This is nonsense&quot;, which is pretty true. The door didn't open right away; as Carl said, &quot;&lt;strong&gt;After the minute passes&lt;/strong&gt;, if the answer is correct, the doors open and he can enter to eat.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bonus:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A brick weighs 2 kg.&#xA; Let the weight of a brick be $x$. &quot;a brick weights one kilogram plus half brick&quot; means $x = 1 + \frac{x}{2}$. Solving for $x$ yields $x=2$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2101" LastActivityDate="2014-11-07T18:46:50.107" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3763" PostTypeId="2" ParentId="3739" CreationDate="2014-11-07T19:17:20.550" Score="33" Body="&lt;p&gt;The old man was supposed to be practically blind by now. However:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the old man couldn't believe his eyes and almost cried to the signboard at the entry:&#xA;  TODAY SPECIAL: SEAGULL MEAT!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He can suddenly read, which means that during the stay on the island his eyesight was somehow cured. Apparently, seagul meat (or human flesh, when the other answers are also right) was the remedy for his condition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;He went outside and saw the palms on the island nearby. He had now fulfilled his live goal of seeing a palm tree and was able to die in peace.&lt;/p&gt;&#xA;" OwnerUserId="1774" LastEditorUserId="1774" LastEditDate="2014-11-07T19:50:32.933" LastActivityDate="2014-11-07T19:50:32.933" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="3764" PostTypeId="1" CreationDate="2014-11-07T19:44:18.540" Score="-3" ViewCount="277" Body="&lt;p&gt;Find the next item in this series:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D, P, N, G, C, M, M&lt;/p&gt;&#xA;" OwnerUserId="4686" LastEditorUserId="15544" LastEditDate="2015-09-27T02:56:56.160" LastActivityDate="2015-09-27T02:56:56.160" Title="Letter Series Problem (D,P,N,G....)" Tags="|pattern|letter-sequence|" AnswerCount="1" CommentCount="2" ClosedDate="2015-09-27T20:13:59.077" ContentLicense="CC BY-SA 3.0" />
  <row Id="3766" PostTypeId="2" ParentId="3764" CreationDate="2014-11-07T20:14:42.720" Score="5" Body="&lt;p&gt;As the intended answer was included in the question, I've copied it here as community wiki.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The next in the series is S which stands for South Carolina. This is the pattern of states added to the union!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user20" LastActivityDate="2014-11-07T20:14:42.720" CommentCount="2" CommunityOwnedDate="2014-11-07T20:14:42.720" ContentLicense="CC BY-SA 3.0" />
  <row Id="3767" PostTypeId="1" AcceptedAnswerId="3771" CreationDate="2014-11-07T20:16:10.330" Score="7" ViewCount="561" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;4, 23, 20, 12, 14, 9, 20, 19, ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also explain your reasoning.&lt;/p&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4050" LastEditDate="2015-01-03T17:54:20.760" LastActivityDate="2015-01-04T22:55:37.343" Title="Determine what the last number in this series is" Tags="|pattern|number-sequence|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3768" PostTypeId="2" ParentId="3726" CreationDate="2014-11-07T20:39:30.483" Score="0" Body="&lt;p&gt;&lt;strong&gt;Known Evidence:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1. Victim has 2 gun wounds, 3 stab wounds, and red marking from strangle&#xA;2. Artifacts: piece of paper &quot;He deserved it! 1H8H1M&quot; with no finger prints, clean new knife, poodle blood, size 9 men's shoe&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Known Suspects:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1. Young white man&lt;br&gt;&#xA;2. Tall blond woman&lt;br&gt;&#xA;3. Homeless man&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Possible Additional or Alternative Suspects:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1. Myself&lt;br&gt;&#xA;2. My boss&lt;br&gt;&#xA;3. Man who cuts me off in traffic&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ruled out as Suspects:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1. Forensic Team&lt;br&gt;&#xA;2. Police&lt;br&gt;&#xA;3. Suicide (he couldn't have shot, stabbed, and strangled himself)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My Interpretation:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Tall blond woman is walking her poodle. Her poodle somehow gets unleashed and begins viciously attacking a man (the murder victim) by his leg. The man flings his leg to get the dog to get off, his shoe flies off his foot. In his frustration, the man brings out his pistol and shoots the dog, causing blood splatter on the ground. Woman screams in horror at the sight, loud enough that a young man hears it and came rushing over from the ATM nearby to brawl with the victim over the pistol, so as to prevent the victim from using it on the woman or anyone else. The young man pulls out a knife he had just purchased a few hours ago, and stabs the victim 3 times, one stab through the arm which knocked the gun out of his hands. The gun hits the ground. With more raw strength, the victim begins to overpower the young man for the knife, despite his bleeding wounds. The woman notices this and frantically runs over, takes her dog's rope and wraps it around around the victim's neck. The victim still persists putting up a fight. A homeless man stumbles nearby, glances at the struggle before him, then at the gun on the ground. He picks the gun up, unsure of who he should attack. So he points the gun, closes his eyes and fires. At that very instance, there was a cry from a man and a shriek from a woman, then pure silence. When the homeless man opened his eyes, he saw a dead man lying in the middle of the street, a woman siting and crying near the corpse of her murdered dog, and a young man standing looking fearful with both his hands on his forehead. The woman took out a piece of paper and pen from her purse, while wearing her satin ballroom gloves that match her fancy dress outfit, in pure anger and depression, wrote on the paper &quot;HE DESERVED IT! 1H8H1M&quot;, the last portion translating to &quot;I Hate Him&quot;. The tall man wiped the knife and gun clean, and removed the dog chain from the victim's neck. The woman and grabbed the corpse of her dog.  All three made an agreement to never speak of this again and go their separate ways as if nothing ever happened.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My Synopsis:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1. Who stabbed him? - The young man&lt;br&gt;&#xA;2. Who strangled him? - The woman&lt;br&gt;&#xA;3. Who shot him? - The homelessman&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My Assumptions:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1. All 3 suspects arrived in nearly the same timeframe.&lt;br&gt;&#xA;2. The 9 men shoe belongs to the victim&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Left Unexplained:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1. Why would a fancy woman would know leet speak &quot;1H8H1M&quot;&lt;br&gt;&#xA;2. Why I keep having dreams of the day I was promoted (whether relevant or not)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4690" LastActivityDate="2014-11-07T20:39:30.483" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3769" PostTypeId="2" ParentId="3767" CreationDate="2014-11-07T20:49:13.903" Score="-1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;4, 23, 20, 12, 14, 9, 20, 19, ?&lt;br&gt;&#xA;  19, -3, -8, 2, -5, 11, -1, ?&lt;br&gt;&#xA;  -22, -5, -6, -7, -16, -12, ?&lt;br&gt;&#xA;  17, -1, -1, -9, 4, ?&lt;br&gt;&#xA;  -18, 0, -8, 13, ?&lt;br&gt;&#xA;  18, -8, 21, ?&lt;br&gt;&#xA;  -26, 29, ?&lt;br&gt;&#xA;  45, ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can achieve that by using a difference table. Then starting by the bottom we can work out way up by replacing the first &quot;?&quot; with another 45:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;4, 23, 20, 12, 14, 9, 20, 19, 118&lt;br&gt;&#xA;  19, -3, -8, 2, -5, 11, -1, 99&lt;br&gt;&#xA;  -22, -5, -6, -7, -16, -12, 100&lt;br&gt;&#xA;  17, -1, -1, -9, 4, 112&lt;br&gt;&#xA;  -18, 0, -8, 13, 108&lt;br&gt;&#xA;  18, -8, 21, 95&lt;br&gt;&#xA;  -26, 29, 74&lt;br&gt;&#xA;  45, 45&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the next number in the sequence is 118. Making the full sequence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;4, 23, 20, 12, 14, 9, 20, 19, 118&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;That is unless I made a mistake calculating&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note: This may not be the answer you're looking for, and I get that, but it doesn't make this answer any less correct; so please do not downvote.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2014-11-07T20:49:13.903" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="3770" PostTypeId="2" ParentId="3726" CreationDate="2014-11-07T21:31:20.067" Score="15" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It was the road raged driver.&lt;br&gt;&#xA; &lt;code&gt;1H8H1M&lt;/code&gt; is his licence plate. He is unapologetic about it, since he thinks the victim deserved it (due to bad driving). You see, he is the owner of the poodle. Have you never heard of dogs looking like their owners?&lt;br&gt;&#xA;  &quot;(...) shaggy brown hair, small nose, and big ears (...)&quot;&lt;br&gt;&#xA; The victim had hit his dog, so the owner killed him with the leash. It was the victim's own gun that went off, wounding him in the hip.&lt;br&gt;&#xA; The victim wasn't stabbed, those were the stiletto heels worn by the woman.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="68" LastActivityDate="2014-11-07T21:31:20.067" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3771" PostTypeId="2" ParentId="3767" CreationDate="2014-11-07T21:33:07.823" Score="10" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 9&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the title of the question, the words begin with DWTLNITSI, corresponding to positions 4, 23, 20, 12, 14, 9, 20, 19, 9 in the alphabet.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3987" LastActivityDate="2014-11-07T21:33:07.823" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3772" PostTypeId="1" AcceptedAnswerId="3775" CreationDate="2014-11-07T21:48:05.513" Score="4" ViewCount="1492" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;It is invisible to the eye,&lt;br&gt;&#xA;  yet it is a stranger to no one.&lt;br&gt;&#xA;  It weighs less than air,&lt;br&gt;&#xA;  yet it is sharper and more keen than any knife, sword, or blade.&lt;br&gt;&#xA;  It comes from nothing,&#xA;  but it has the power to vanquish armies. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is it that I am speaking of?&lt;/p&gt;&#xA;" OwnerUserId="4689" LastEditorUserId="5188" LastEditDate="2015-03-08T19:54:43.877" LastActivityDate="2015-03-13T18:07:19.647" Title="Invisible to the eye - What is it that I am speaking of" Tags="|riddle|" AnswerCount="4" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3774" PostTypeId="2" ParentId="3772" CreationDate="2014-11-07T21:58:55.303" Score="7" Body="&lt;p&gt;You are speaking of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Time‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is invisible to the eye&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You cannot see time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yet it is a stranger to no one&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; But everyone is familiar with it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It weighs less than air&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It has no weight&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yet it is sharper and more keen than any knife, sword, or blade.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Time cannot be blocked, it pierces any armour.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It comes from nothing&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Time has no 'source' of which we know&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but it has the power to vanquish armies.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The passage of time can destroy anything, including armies&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3987" LastEditorUserId="3987" LastEditDate="2015-01-24T07:36:10.637" LastActivityDate="2015-01-24T07:36:10.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3775" PostTypeId="2" ParentId="3772" CreationDate="2014-11-07T22:02:11.880" Score="6" Body="&lt;p&gt;The answer that I remember was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;hunger&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but it could also be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; time or cold.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2015-03-13T18:07:19.647" LastActivityDate="2015-03-13T18:07:19.647" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3777" PostTypeId="2" ParentId="3772" CreationDate="2014-11-07T22:10:54.993" Score="5" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Intellect&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is invisible to the eye,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Intellect is abstract so it can't be seen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yet it is a stranger to no one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everyone has intellect (ranging from a little bit to a lot).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It weighs less than air,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Intellect is abstract so it has no weight.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yet it is sharper and more keen than any knife, sword, or blade.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people are said to possess a sharp or keen intellect.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It comes from nothing,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Intellect comes from the mind, which due to its abstractness could be considered nothing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but it has the power to vanquish armies.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One must typically use intellect to win a war.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4450" LastEditDate="2015-01-24T05:56:34.610" LastActivityDate="2015-01-24T05:56:34.610" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3778" PostTypeId="2" ParentId="3739" CreationDate="2014-11-07T22:14:13.270" Score="7" Body="&lt;p&gt;They &lt;em&gt;were&lt;/em&gt; eating seagull on the island.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It was a cannibal cruise ship!&lt;br&gt;&#xA; Poor niece.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He'd gotten a corneal transplant from his now-dead niece.  Cannibals are better!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4698" LastActivityDate="2014-11-07T22:14:13.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3783" PostTypeId="2" ParentId="3726" CreationDate="2014-11-08T01:12:10.490" Score="3" Body="&lt;p&gt;This puzzle is quite hard especially trying to link every clues in the story. Let's try:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Wounds&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; -Gun Shot&lt;br/&gt;-Red signs on his neck&lt;br/&gt;-Knife wounds&lt;br/&gt;&lt;br/&gt;I would be surprised if a gun shot in the lip would kill. AndMoreover, he assumed he was strangled but we can't be sure. The victim could possibly be that kind of man who revels in pain. Maybe the night before his partner was a bit too rude and that kind of signs have been left on his neck. So I can tell &lt;strong&gt;the murderer stabbed him to death&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now we have the cause of death. Let's analyze the clues we have:&lt;br/&gt;&lt;br/&gt;&#xA;&lt;strong&gt;Clues&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;-Blood stains&lt;br/&gt;-1H8H1M&lt;br/&gt;-Knife&lt;br/&gt;-Shoe print&lt;br/&gt;&lt;br/&gt;According to the scientist, that blood came from a dog. But, where is the fluffy poodle? It could be just a case, and he's suggesting the detective to not focus the blood. Perfect circle blood stains occour when the drop fall down without momentus [ !-shape otherwise ]. A stab, a gun shot, a strong kick on the snout would have not produce that kind of shape. The dog could have been biten by another dog the day before without any reason, they are animals, who can foresee their moves. The next one is the piece of paper. Why should the murderer leave that kind of information on the ground? I can't imagine the dying man took pen and paper and started to write that to help the detective somehow. And, why the &quot;hate part&quot; has a different way of writing? He could have write the entire message that way without problem. I can assume he prepared that piece of paper before the murder, maybe he was waiting someone with that licence plate that looks so strange or some kind of information he would give to someone. The knife, instead, lets me think he was involved in a fight he lost [ of course, he's dead :P ]. By the way, after they met, him and his murdered started to fight and he was stabbed 2 times to death. I will explain the prints after. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's talk about the suspects.&#xA;&lt;br/&gt;&lt;br/&gt;&#xA;&lt;strong&gt;Suspects&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; According to the description, the homeless would have been too weak to start a fight against the victim. He declared he haven't eaten for 2 days, so he could not handle a fight and kill a man. But I think he sneaked near the dead body to find money to buy food or something like that and he left the mud prints near the victim. He's not well dressed so he can even use shoes that don't fit his feet. I don't think the murderer was the woman. She was jogging and she met a lot of people after her second round. If she killed him, she could hide the knife but she can't hide the blood squirted from the victim. Just talk with a pair of witnesses, and we can confirm this theory. The only one left is the young guy, the only one who did not met anyone that day. I have this theory. If he was gunshot where he was found, anyone can tell they heard the noise of a gun. Even the woman, who is innocent. &lt;br/&gt;Let me think out of the box. The victim and the young man are accompices that are willing to rob the bank [ take some money ]. The victim betrayed and has been shot in another place but he was able to run away because the wound are not that bad. The young man found him and finished him using a knife, the best way to kill fast someone without make a noise.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What happened&lt;/strong&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The victim and the &lt;strong&gt;young man&lt;/strong&gt; are planning to rob a bank. He victim betrayed and he was shot by the young man because he's afraid that could go to the police. The victim was able to run away for a while but he found him and killed him with a &lt;strong&gt;knife&lt;/strong&gt; without making a noise. Maybe the piece of paper spilled off a pocket during the fight. I can say that was just a motivational reminder to betray the partner. After that, the homeless went near the body, left some prints, took the money [if he found any ] and gone away. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1685" LastEditorUserId="1685" LastEditDate="2014-11-08T01:21:30.787" LastActivityDate="2014-11-08T01:21:30.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3784" PostTypeId="1" AcceptedAnswerId="3791" CreationDate="2014-11-08T01:14:35.467" Score="2" ViewCount="6293" Body="&lt;p&gt;Tomorrow you will meet three sisters, and be married to one of them. The eldest always tells the truth, the youngest always lies. The middle sister lies or tells the truth at random. Knowing this, &lt;strong&gt;you would like to marry the either the eldest or the youngest&lt;/strong&gt;—so you'll know where you stand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conscious to their own futures, the sisters conspire to visit you the night before. Everyone gets on really well. They are Abigail, Eleanor and Lucy. You're too polite to guess their ages. As they're about to leave, Abigail draws you aside and says 'since we might be married tomorrow, is there anything you'd like to ask?'. From the look in her eyes, you understand it should be a yes-no question. To achieve your aim, what do you ask? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Needless to say, the sisters know each others' ages.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It might help to solve this simpler problem first &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2188/two-doors-with-two-guards-one-lies-one-tells-the-truth&quot;&gt;Two doors with two guards - one lies, one tells the truth&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4508" LastEditorUserId="4508" LastEditDate="2014-11-10T17:06:59.840" LastActivityDate="2014-11-10T17:06:59.840" Title="Which sister to marry?" Tags="|logical-deduction|" AnswerCount="3" CommentCount="5" FavoriteCount="0" ClosedDate="2014-11-08T12:23:10.153" ContentLicense="CC BY-SA 3.0" />
  <row Id="3785" PostTypeId="2" ParentId="3784" CreationDate="2014-11-08T01:25:37.617" Score="2" Body="&lt;p&gt;Name the sisters A, B and C. A is the one who came to you. You now ask her:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;You are the oldest $\oplus$ B is the middle sister?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you get back the answer yes, then B is either the youngest or the oldest - pick B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you get back the answer no, then C is either the youngest or the oldest - pick C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This hinges on the fact that A came to you and if you discard her, at least one of the others is &quot;safe to marry&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If A was the middle sister, then regardless of yes/no whoever else you choose is &quot;safe to marry&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If A was the oldest then she'll respond yes if B is safe and no if C is safe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If A was the youngest then she'll respond yes if B is safe and no if C is safe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;$\oplus$ means logical XOR&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2383" LastEditDate="2014-11-08T12:56:19.010" LastActivityDate="2014-11-08T12:56:19.010" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="3787" PostTypeId="2" ParentId="3726" CreationDate="2014-11-08T01:46:56.500" Score="2" Body="&lt;p&gt;First off, I find it odd that while it seems that the action is placed in Chicago, I can't actually find a Peele street (or boulevard or avenue etc.) on the city map.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, it seems to me like the &lt;em&gt;&quot;red markings&quot;&lt;/em&gt; are a red herring of sorts. Given that the body is found at the intersection of Peele and Main there are two main options:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Either the deceased was not strangled on spot, strangling someone&#xA;takes a good while and we're talking about doing it next to the Main&#xA;street of a large city. It's possible the markings are a result of,&#xA;say, bedtime activities or something similar.&lt;/li&gt;&#xA;&lt;li&gt;Or the murder itself took place elsewhere and the body was later moved. This raises the question of why would anyone move a body someplace where it's very likely to be noticed very fast (again, Main street). This could be an option if the killing was meant to send a message (for example, a hired hit), but the messiness of the situation leads me to consider that option less. It's also possible that the body was moved to a public location as a means of humiliation.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Next up, the note. This, again, seems to be a red herring. Leaving aside the contents of the note for a moment, leaving a note would mean the murder was preplanned. However, why would you plan to murder someone on a busy street, there are plenty of better options?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, either the murder wasn't planned (and the note is a herring) &lt;em&gt;or&lt;/em&gt; the body was moved after the murder took place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exploring the option of a moved body further, it's possible that the victim was tortured before being killed. At the scene of crime, the only notes of blood are some drops (poodle's blood) which is an awful little for a body with both knife and gun wounds.&lt;br&gt;&#xA;From this perspective, it seems likely that the body was moved after the murder had taken place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exploring the note itself: &lt;em&gt;&quot;He Deserved It! 1H8H1M&quot;&lt;/em&gt;&lt;br&gt;&#xA;Firstly, there's an exclamation mark used on the first sentence, but no punctuation on the second half. Hate is generally a very strong emotion and using softer language for that seems off.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads me to believe that the second part of the note means something else (not &lt;em&gt;&quot;I hate you&quot;&lt;/em&gt;). Others have pointed out it would fit a license plate, but I'm not that familiar with the US system.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we delve in the math side of things for a moment, &lt;em&gt;1H8H1M&lt;/em&gt; is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;13,723,246 in base 15&#xA;23,553,310 in base 27&#xA;38,301,352 in base 30&#xA;89,414,986 in base 36&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;None of these seem to reference the rest of the case so I'm considering these irrelevant for the time being.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the above conflicting nature of the note, it seems reasonable that the note is either: a red herring (unrelated to the case) or a plant (distraction by the murderer). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analyzing the scene of the crime further we can note that &lt;em&gt;&quot;the knife was clean like it was brand new straight from the box&quot;&lt;/em&gt;. The knife is in mint condition. This seems to indicate that the knife is not related to the actual killing of the victim itself. It is unspecified whether the blade's shape matches the wounds of the deceased, but using two identical knives isn't out of the question.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to further support the idea, that the body was moved to the scene after the actual murder. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Either the knife was carried by the deceased, but for some reason the deceased was unable to use it in self defense. This makes sense only if the murder wasn't preplanned.&lt;/li&gt;&#xA;&lt;li&gt;Or the knife was not carried by the deceased. That is, someone else related to the murder, possibly the murderer.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So for a timeout recap:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Either this was not a preplanned murder and everything that could go wrong seems to have gone wrong.&lt;/li&gt;&#xA;&lt;li&gt;Or this was a preplanned murder and the body was later moved to the place of discovery.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Coming back to the body of the deceased, namely the gun wounds: there are two of them, both in the left hip. The location of the wounds is crucial. Let's consider two options:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Either the gun originally belonged to the deceased: regardless of the usual pistol holster types, none of them are positioned in a way that shooting oneself in the hip with the pistol still in the holster is possible. This means the gun should have been drawn. &lt;/li&gt;&#xA;&lt;li&gt;Or the gun was originally in the hands of someone else. Landing one shot in a hip can be accidental, but getting two shots in someone's hip means that either the target was not moving, the shots were made at very close quarters, or both.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If the gun belonged (at least at first) to the deceased if a struggle ensued, gunshots would likely follow, even if not hitting anyone (right away). This would surely catch attention (on Main street) and wouldn't leave enough time for the rest of the actions to take place without any witnesses. The gun could have been silenced, but that would only seem reasonable if the other party was carrying the gun, as most every-day carries (police and alike) are not silenced (unless the deceased was in fact planning to kill someone himself).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the gun seems to have belonged to some other party, however, since there are neither bullet casings nor blood splatters at the scene of the crime, we seem to again arrive at the possibility that the body was moved after the murder. Additionally, if you have a gun and wanted to &lt;em&gt;kill&lt;/em&gt; someone, you wouldn't aim for their hip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So for a second recap:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently the most likely option seems to be that the body was moved after the actual murder took place. It seems like the gun wounds are made with the intent of making the deceased unable to move quickly or simply inflicting pain, rather than trying to kill them, since those wounds alone don't seem lethal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll edit this post with more analysis once I have any further thoughts.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Coming to the suspects, namely the homeless man: &lt;em&gt;&quot;I did no murder. I asking for money for food. Hungry, haven't eat couple days.&quot;&lt;/em&gt;&lt;br&gt;&#xA;I think these words are vital. Namely, this leads me to believe the homeless man is in fact either a witness or was close to the scene of the actual murder before it took place. I interpret his words for the time being as &lt;em&gt;&quot;I didn't kill the deceased, I was simply asking him (or someone near him) for money.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's come back to what the twenty-something white man was doing when the murder was &lt;em&gt;reported&lt;/em&gt; (note that this is always later than the actual murder taking place): getting some money from the bank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most humans can survive quite a while without eating anything (see &lt;a href=&quot;http://www.scientificamerican.com/article/how-long-can-a-person-sur/&quot; rel=&quot;nofollow&quot;&gt;this article, for example&lt;/a&gt;), however not eating anything for a couple of days is generally a great discomfort and it's not far fetched to say one might do things of questionable legality to get money for food in that situation, especially if one's been on rough times for a while (our homeless man is described as &lt;em&gt;&quot;wearing very battered clothing, with long greasy hair and a full beard&quot;&lt;/em&gt;). However, I don't think the homeless man was our murderer, as he was described as skinny and unfed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, I propose the twenty-something white man had made an agreement with the homeless man for some services in exchange of money, for example, moving a body out of a small alley into the street and dumping some random misleading clues next to it to cover his trails. He was getting money from the bank because he had just given what ever cash he had on hand to the homeless man.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Therefore, a third recap:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The white twenty-something man was the murderer. The victim was killed (and possibly tortured prior) somewhere else and the body was later brought near the scene of the discovery. The homeless man was paid to move the body and dump some misleading evidence near the scene to create confusion (as none of the clues reasonably add up).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I don't think the case is still finished at this point. We still haven't covered one suspect: the tall blonde woman.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following the day of the murder she was dressed up to celebrate: &lt;em&gt;&quot;wearing a fancy dress and jewelry like she was going on a date or to a party&quot;&lt;/em&gt;. While this could be a meaningless fact (a random preplanned event), I don't think this is the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During the dumping of the body she goes for a jog nearby to keep an eye on things. People are bad waiters and no one wants to wait blindly. Since she lives nearby it's not a bad alibi for her to be jogging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The twenty-something man and the woman might have a love affair or other joint interests that lead to the murder and are looking to celebrate the day after the dumping of the body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further update to follow.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;What's a clue in and of itself is the fact that we &lt;em&gt;have&lt;/em&gt; suspects. When picking suspects, police doesn't pick random people off the street, they bring people who are related to the case. And while we're not given the details of &lt;em&gt;why&lt;/em&gt; each of our suspects is one, it inevitably leads to the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of our suspects are either related to the deceased somehow or linked with the scene of the discovery somehow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting off with the homeless man, since most homeless people have generally abandoned or lost connections with people close to them it's more likely he's linked with the scene of the discovery. The most obvious link being he was likely seen nearby sometime near the discovery of the body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both the twenty-something man and the blonde woman likely have a relation to the deceased somehow, how else would police know to pick up a random jogger and a guy from the bank?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;All this together brings me to my final recap:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The white twenty-something man and the blonde woman planned the&#xA; murder. They had some sort of a united goal that the deceased was in&#xA; the way of (love, money etc.). The victim was killed (and possibly&#xA; tortured prior) somewhere else and the body was later brought near the&#xA; scene of the discovery. The homeless man was paid to move the body and&#xA; dump some misleading evidence near the scene to create confusion (as&#xA; none of the clues reasonably add up).&lt;br&gt;&#xA;&lt;br&gt;&#xA; There is no victim's blood at the scene because the body was moved.&#xA; The knife is clean because it wasn't used. The poodle's blood is in&#xA; circular drops because it was planted manually.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The woman was jogging nearby to keep an eye on things, no one likes&#xA; waiting nervously. The man was getting money from the bank because he&#xA; had just paid the homeless man to drop the body and needed some for&#xA; whatever reason.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="340" LastEditorUserId="340" LastEditDate="2014-11-09T14:50:16.280" LastActivityDate="2014-11-09T14:50:16.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3791" PostTypeId="2" ParentId="3784" CreationDate="2014-11-08T04:32:26.090" Score="13" Body="&lt;p&gt;You simply need to ask Abigail:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Is Lucy older than Eleanor?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Marry the sister she indicates as younger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If Abigail is the eldest: She is telling the truth, and so marrying&#xA;the person she indicates as the youngest is marrying the person who&#xA;is always lying, which is acceptable.&lt;/li&gt;&#xA;&lt;li&gt;If Abigail is the middle aged: Then it is acceptable to marry either&#xA;of the other two girls, and thus marrying the one who Abigail selects&#xA;as the youngest is permitted.&lt;/li&gt;&#xA;&lt;li&gt;If Abigail is the youngest: She is lying and thus the person she&#xA;indicates is the youngest is in fact the eldest and therefore is the&#xA;truth telling sister.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So overall, if you always pick the one Abigail chooses as the youngest then you will be marrying either the oldest or the youngest as you desire.&lt;/p&gt;&#xA;" OwnerUserId="97" LastEditorUserId="97" LastEditDate="2014-11-08T12:39:14.877" LastActivityDate="2014-11-08T12:39:14.877" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="3792" PostTypeId="1" AcceptedAnswerId="3794" CreationDate="2014-11-08T04:43:25.270" Score="5" ViewCount="377" Body="&lt;p&gt;While browsing the Internet for medical lore, you happen on a medical encyclopedia article about purple spots disease.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instances of purple spots disease have been documented in 150 countries. The disease affects one in every 100,000 individuals and there is no known cure. Individuals who have the disease remain completely asymptomatic until one day they very suddenly break out in purple spots that never go away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerned that you may have purple spots disease and not know it, you scour the Internet and locate a reputable medical testing company that provides a screening test for the disease. The test is state of the art. It guarantees 99.5% accuracy, meaning that whether you have the disease or not, the result returned by the test will be correct 99.5% of the time. The test costs $300.00 USD to administer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Figuring that your peace of mind is worth $300.00, you undergo the test for the disease.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three weeks later, the results come back. The test is positive! You're advised to come in for a retest to be absolutely sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Based on the information above, should you be deeply worried? Why or why not?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Puzzlers are politely encouraged to place answers in spoiler blocks to avoid inadvertently spoiling the fun for other readers.&lt;/em&gt; :)&lt;/p&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="5373" LastEditDate="2015-04-28T19:42:20.493" LastActivityDate="2015-04-28T19:42:20.493" Title="The Ravages of Purple Spots Disease" Tags="|probability|" AnswerCount="1" CommentCount="1" ClosedDate="2021-08-10T01:53:50.550" ContentLicense="CC BY-SA 3.0" />
  <row Id="3794" PostTypeId="2" ParentId="3792" CreationDate="2014-11-08T04:59:16.570" Score="6" Body="&lt;p&gt;No you shouldn't be too worried. The probability you have the disease is 0.199%, or in other words about a 1/500 chance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, let us assume there are 10,000,000 people in the society (the number you assume here is irrelevant, and you could even just let x be the number if you were mathematiaclly inclined). Now given that 1/100,000 people have the disease, therefore, 100 people will have it in this society.  Using this information a 2 by 2 table can be constructed as seen below:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;             Has Disease        No Disease         Total&#xA;+ve test        99.5              49999.5          50099       &#xA;-ve test         0.5            9949900.5        9949901&#xA;&#xA;Total            100              9999900       10000000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;From the above table, given a +ve test result, the probability of having the disease can be evaluated by computing by 99.5/50099 = 1.99 x 10^-3, or about a 1/500 chance.&lt;/p&gt;&#xA;" OwnerUserId="97" LastEditorUserId="97" LastEditDate="2014-11-08T05:04:23.660" LastActivityDate="2014-11-08T05:04:23.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3796" PostTypeId="2" ParentId="3705" CreationDate="2014-11-08T05:36:56.297" Score="12" Body="&lt;p&gt;Just as a fun answer, for a loop hole:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; G &lt;br&gt;&lt;br&gt;&#xA; This refers to Graham's number &lt;a href=&quot;http://en.wikipedia.org/wiki/Graham&amp;#39;s_number&quot;&gt;(Wikipedia)&lt;/a&gt;, and is the largest number used in a serious mathematical proof. It is so large it is best expressed as a recurrence relation of Knuth's up-arrow notation as $G = g_{64}$ where $g_1 = 3\uparrow\uparrow\uparrow\uparrow3$ and $g_n = 3\uparrow^{g_n-1}3$ (equations lifted shamelessly from Wikipedia). The phrase &quot;many orders of magnitude&quot; is negligible in the face of this number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2823" LastActivityDate="2014-11-08T05:36:56.297" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3797" PostTypeId="2" ParentId="3726" CreationDate="2014-11-08T05:50:25.600" Score="-1" Body="&lt;p&gt;Well after reading the story I ran across this sentence, im going to quote &quot;The homeless man didn't speak english well and seemed kinda crazy, &quot;I did no murder, i asking for money for food. Hungry, haven't eat couple days&quot;.&quot;. Now this homeless dude is &quot;Crazy&quot;, asking for food? Saying he has done no murder? This would lead me to thinking it was him rather than the other two people. The size 9 shoes are for men so he could been the one wearing them. Although there are two guys I still think he's the one judging by what he said.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also think the Blonde Female is apart of this crime scene too, &quot;The blonde female said that she was going for a run around the block as she just lived down the street from the murder site. She amde one round around the block, and decided to do another round, as she was jogging she saw the crowd and decided to stop and take a look.&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;She runs around the block and then decides o do another round and thats then when she saw a crowd of people around the crime scene. She kinda raises some suspicion.&lt;/p&gt;&#xA;" OwnerUserId="4717" LastActivityDate="2014-11-08T05:50:25.600" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3798" PostTypeId="2" ParentId="3760" CreationDate="2014-11-08T05:57:35.903" Score="15" Body="&lt;p&gt;&quot;After the minute passes, if the answer is correct, the doors open&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; So before the minute passes, the doors stay shut even if the answer is correct. You get a minute to shout out a pile of rapid-fire answers the way Marky Mark shouts out women's names in the film &lt;em&gt;Ted&lt;/em&gt;. If one of them is correct before &lt;a href=&quot;http://www.albinoblacksheep.com/flash/jada&quot;&gt;&quot;Ja Da&quot; finishes&lt;/a&gt;, Kebab King opens.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But that's not the solution at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Every costumer, to enter the shop, has to answer a riddle in one minute.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A costumer (note the spelling) is one who deals in sets of clothes in the distinctive style of a culture or character. Carl is a costumer; Mike is not. A movie costume design business just opened next door, and they've been driving away the kebab shop's regular customer base. So only costumers need to answer riddles; people with other occupations just need to wait while the bouncer verifies that it isn't yet another employee of the costume business and while the busser gets a table ready.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now for my take on the bonus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;If a brick weighs one kilogram plus half brick, how much does a brick weight?&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Kilogram is a unit of mass, not force, and weight is a force. The answer to &quot;If a brick's mass is 1 kg plus half a brick, what is its mass?&quot; is 2 kg. But in spoken English, &quot;weight&quot; sounds like &quot;wait&quot;, thus a brick waits a minute just like Carl and Mike have to wait a minute.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2521" LastEditorUserId="2521" LastEditDate="2014-11-12T04:34:54.543" LastActivityDate="2014-11-12T04:34:54.543" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3799" PostTypeId="2" ParentId="3739" CreationDate="2014-11-08T06:13:24.523" Score="13" Body="&lt;p&gt;He committed suicide because the seagull meat tasted absolutely awful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The man was of course the first to try the seagull meat; soon the other sailors tried the meat and followed the man in one of the greatest mass-suicides known to man. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ship was found a few weeks later on the 5th of December 1872 in the Atlantic Ocean.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The ship's name, if you haven't already guessed, was the Mary Celeste.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="386" LastActivityDate="2014-11-08T06:13:24.523" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3800" PostTypeId="2" ParentId="3718" CreationDate="2014-11-08T07:59:13.027" Score="0" Body="&lt;p&gt;Two loopholes:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;We can deflate the balls, which will decrease the volume taken up by them. In this case the amount will depend by the width of the &quot;skin&quot; of the ball.&lt;/li&gt;&#xA;&lt;li&gt;We take &quot;the ball&quot; that is 1cm in diameter and many other balls that are of smaller sizes. Since we can potentially mathematically assume them to be as small as possible the answer will approach infinity. In practice we are limited by the plank volume, but i am unsure if we can call that truly a &quot;ball&quot; at all, so the question becomes how few plank volumes do we need, to be able to claim that something is a ball at all?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4641" LastActivityDate="2014-11-08T07:59:13.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3801" PostTypeId="1" CreationDate="2014-11-08T07:59:34.633" Score="14" ViewCount="7726" Body="&lt;p&gt;Let's say you have a donut.  You are allowed to slice it 3 times.  Each slice must be a perfectly straight cut.  What is the highest number of donut pieces you can end up with after 3 slices?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that no crumbs are created during the slicing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also assume that no pieces move until after you have finished all 3 slices.  That way while you are making your third slice, the pieces made from the first and second slices don't start moving and falling off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please hide your answers with the spoiler markup!&lt;/p&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4450" LastEditDate="2014-11-08T09:12:49.173" LastActivityDate="2017-09-22T17:49:42.467" Title="Slicing a donut 3 ways - what's the most number of pieces?" Tags="|geometry|" AnswerCount="6" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3802" PostTypeId="2" ParentId="3801" CreationDate="2014-11-08T08:08:03.010" Score="0" Body="&lt;p&gt;I can create&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;8 pieces by cutting along the axis: &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With cuts &amp;nbsp;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1: Horizontally through the middle of the piece.&lt;br&gt;&#xA;2 and 3: Like a cross vertically.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4641" LastEditorUserId="4450" LastEditDate="2014-11-08T08:31:17.353" LastActivityDate="2014-11-08T08:31:17.353" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3803" PostTypeId="2" ParentId="3801" CreationDate="2014-11-08T08:11:47.003" Score="4" Body="&lt;p&gt;Edit: This works if we're talking about a filled donut. For toroidal donuts better answers are given already.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;8 If you make each cut intersect all of the others, for example by making the all perpendicular to each other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is provably the maximum unless you get tricky and distort the shape:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Each cut can, at best, split each existing piece in two, doubling the current number. We start with 1 piece, 1 cut gives 2, 2 cuts gives 4, 3 cuts gives 8.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3987" LastEditorUserId="3987" LastEditDate="2014-11-08T15:20:08.270" LastActivityDate="2014-11-08T15:20:08.270" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3804" PostTypeId="2" ParentId="3760" CreationDate="2014-11-08T08:23:52.360" Score="1" Body="&lt;p&gt;Bonus (alternative through the loophole):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We consider the weight in a weightless environment, so it's 0 (zero). 1 kilogram (a measure of mass) weighs 0 Newtons + 1/2 of the brick that weighs 0 Newtons, the total is 0 Newtons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4641" LastActivityDate="2014-11-08T08:23:52.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3805" PostTypeId="1" AcceptedAnswerId="3807" CreationDate="2014-11-08T08:25:22.310" Score="5" ViewCount="470" Body="&lt;p&gt;if you divide my first digit by my last digit then you can find me for ever&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you multiply me by any number equal to or less than my length (but greater than zero) and I will only cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you divide any number equal to or less than my length (but greater than zero) by the last digit of me you will still find me for ever in different cycles&lt;/p&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="2313" LastEditDate="2014-11-08T09:22:24.953" LastActivityDate="2015-10-05T16:13:03.967" Title="Can you find me" Tags="|logical-deduction|mathematics|word-problem|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3806" PostTypeId="2" ParentId="3801" CreationDate="2014-11-08T09:05:25.463" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can create 10 pieces. Make two cuts that are perpendicular to the table as shown, tangent to the hole, creating five pieces. Then, slice the donut parallel to the table, splitting each piece into two.&#xA;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/h9b3i.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4450" LastEditDate="2014-11-08T09:12:10.560" LastActivityDate="2014-11-08T09:12:10.560" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3807" PostTypeId="2" ParentId="3805" CreationDate="2014-11-08T09:08:43.013" Score="2" Body="&lt;p&gt;Here you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $142857$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\frac17 = 0.\overline{142857}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $142857\times2=285714$&lt;br&gt;&#xA; $142857\times3=428571$&lt;br&gt;&#xA; $142857\times4=571428$&lt;br&gt;&#xA; $142857\times5=714285$&lt;br&gt;&#xA; $142857\times6=857142$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And finally:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\frac27 = 0.\overline{285714}$, but it can be rewritten as $0.2857\overline{142857}$&lt;br&gt;&#xA; $\frac37 = 0.\overline{428571}=0.42857\overline{142857}$&lt;br&gt;&#xA; $\frac47 = 0.\overline{571428}=0.57\overline{142857}$&lt;br&gt;&#xA; $\frac57 = 0.\overline{714285}=0.7\overline{142857}$&lt;br&gt;&#xA; $\frac67 = 0.\overline{857142}=0.857\overline{142857}$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The way to find it is actually not that difficult:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After you recognise that &quot;cycle&quot; refers to the repeating digits you need to ask what single-digit denominators produce repeating patterns in decimal. We have 3, 6, 7, 9 as candidates. However: 3, 6, and 9 produce repeating pattern of a single digit for any single-digit numerator, so we know that the last digit must be 7. So we try 1 as the first digit and it works.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4641" LastEditorUserId="4421" LastEditDate="2015-10-05T16:13:03.967" LastActivityDate="2015-10-05T16:13:03.967" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="3809" PostTypeId="2" ParentId="3801" CreationDate="2014-11-08T09:29:07.297" Score="18" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can cut it into 13 pieces.&lt;br&gt;&lt;br&gt;I couldn't draw the picture, but I found this website that has already drawn it for me:&lt;br&gt;&lt;br&gt;Reference:  &lt;a href=&quot;http://www.hunkinsexperiments.com/pages/doughnuts.htm&quot;&gt;http://www.hunkinsexperiments.com/pages/doughnuts.htm&lt;/a&gt;&lt;br&gt;&lt;br&gt;Unfortunately I cannot include the picture here as it is copyright. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="97" LastEditorUserId="4450" LastEditDate="2014-11-08T09:46:45.207" LastActivityDate="2014-11-08T09:46:45.207" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="3810" PostTypeId="1" AcceptedAnswerId="3813" CreationDate="2014-11-08T09:49:03.167" Score="4" ViewCount="870" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Which word means &quot;to push hard&quot;, and if it is written backwards it&#xA;means &quot;to spoil&quot;?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A synonym of this word is &quot;stalemate&quot;, and if it is written backwards a&#xA;synonym of the transformed word is &quot;department&quot;. Which word?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A synonym of this word is &quot;deviate&quot;, and if it is written backwards a&#xA;synonym of the transformed word is &quot;method&quot;. Which word?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="97" LastEditorUserId="22922" LastEditDate="2016-09-15T08:30:27.967" LastActivityDate="2016-09-15T08:30:27.967" Title="What are these words?" Tags="|word|wordplay|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3811" PostTypeId="1" AcceptedAnswerId="3814" CreationDate="2014-11-08T09:55:31.827" Score="4" ViewCount="503" Body="&lt;p&gt;Find my two numbers:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Both are positive integers.&lt;/li&gt;&#xA;&lt;li&gt;Their difference is a prime.&lt;/li&gt;&#xA;&lt;li&gt;Their product is a perfect square.&lt;/li&gt;&#xA;&lt;li&gt;Their sum's last digit is 3.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Bonus:  Can you show how many solutions exist?&lt;/p&gt;&#xA;" OwnerUserId="97" LastEditorUserId="5373" LastEditDate="2019-07-04T14:16:16.503" LastActivityDate="2019-07-04T14:16:16.503" Title="Find my two numbers" Tags="|mathematics|number-theory|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3812" PostTypeId="1" AcceptedAnswerId="3815" CreationDate="2014-11-08T09:58:11.670" Score="3" ViewCount="541" Body="&lt;p&gt;A certain chain store sells chocolate bars in packets of 17 and 9 only. Clearly, you could not get 8 or 25 bars. Find all quantities of bars that you cannot buy.&lt;/p&gt;&#xA;" OwnerUserId="97" LastActivityDate="2014-11-11T13:25:54.103" Title="What can't I buy?" Tags="|mathematics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3813" PostTypeId="2" ParentId="3810" CreationDate="2014-11-08T09:59:49.840" Score="5" Body="&lt;ol&gt;&#xA;&lt;li&gt;The answer is&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Mar&lt;/strong&gt;&lt;br&gt;&#xA; Ram is to push hard, mar is to spoil.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;The answer is&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Draw&lt;/strong&gt;&lt;br&gt;&#xA; The synonym of stalemate is &quot;draw&quot;, the anagram is &quot;ward&quot; and ward is a synonym for department&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;The answer is:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Part&lt;/strong&gt;&lt;br&gt;&#xA; It's the only synonym for deviate I can think of that makes a real word when reversed, but &quot;trap&quot; doesn't really mean method.&lt;br&gt;&#xA; However, I misread a bit, thanks @d'alar'cop&#xA; &quot;a synonym of &quot;trap&quot; is &quot;stratagem&quot; which is also a &quot;method&quot;.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4729" LastEditorUserId="4729" LastEditDate="2014-11-08T10:42:31.663" LastActivityDate="2014-11-08T10:42:31.663" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3814" PostTypeId="2" ParentId="3811" CreationDate="2014-11-08T10:02:05.630" Score="8" Body="&lt;p&gt;4 and 9&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proving uniqueness: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any mutual prime factor of the numbers is a prime factor of their difference, which is a prime. So, their GCD is either 1 or that prime $p$. So, in order for their product to be a perfect square, they must have the form $(x^2,y^2)$ or $(p x^2,p y^2)$, with $x$ and $y$ relatively prime. Then, their prime difference is $(y+x)(y-x)$ or $p(y+x)(y-x)$. The second case cannot be prime because $y+x \geq 2 $ and $p$ are nontrivial factors, whereas the first case must have $y-x=1$ (assuming WLOG $y \geq x$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we need only consider pairs of the form $(x^2,(x+1)^2)$. We know their sum ends in a $3$ and so is $3$ mod $5$. This is only true for $x=2$ mod $5$. But, then, the prime difference $2x+1$ is a multiple of $5$, which implies that it must be $5$. This gives $x=2$, so $(4,9)$ is the unique solution.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2014-11-08T10:27:33.673" LastActivityDate="2014-11-08T10:27:33.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3815" PostTypeId="2" ParentId="3812" CreationDate="2014-11-08T10:34:14.863" Score="10" Body="&lt;p&gt;We can make $17 \times 8 = 136$ and any number above this by replacing a $17$ with $2 \times 9$, to increment the number by $1$. By the time we run out of $17$s to replace ($9 \times 16$), we continue with $17 \times 8 + 9 = 145$ and repeat the trick, until we arrive at $9 \times 17$, at which point we swap that with $17 \times 9$ and start replacing $17$s again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the numbers below $136$, we can make any number $n \times 17$ and&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the following $n$ integers by replacement of $17$ with $2 \times 9$, &lt;/li&gt;&#xA;&lt;li&gt;anything $m\times9$ above this e.g.&lt;br&gt;&#xA;$17$ gives us $18$ by replacement and $26$, $35$, $44$, $53$, $62$, $71$, $80$, $89$, $98$, $107$, $116$, $125$, $134$ by addition of $9$s&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Adding all of these we find that the highest number we cannot make is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $127$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and there are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$63$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;possible numbers below this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ones we cannot form are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 16, 19, 20, 21, 22, 23, 24, 25, 28, 29, 30, 31, 32, 33, 37, 38, 39, 40, 41, 42, 46, 47, 48, 49, 50, 55, 56, 57, 58, 59, 64, 65, 66, 67, 73, 74, 75, 76, 82, 83, 84, 91, 92, 93, 100, 101, 109, 110, 118, 127  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3987" LastEditorUserId="68" LastEditDate="2014-11-11T12:35:56.307" LastActivityDate="2014-11-11T12:35:56.307" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="3817" PostTypeId="2" ParentId="3257" CreationDate="2014-11-08T12:02:22.827" Score="2" Body="&lt;p&gt;We won't be using the left shoulder, because that will out the Templars. The right shoulder is fine though. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Improving on aschepler's solution, we name 2 people each day while touching the right shoulder. If the number of B is unchanged at the next meeting, neither was illuminati. If B has increased by 2, they were both illuminati. If only 1, then we spend the next day identifying which: name one of the two while touching the left shoulder. If the number of B has dropped at the next meeting, this person was originally illuminati, otherwise it was the other person.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method has a worst case when we never eliminate 2 illuminati in one day and still have 1 left in 99-100 so we spend 49 days testing pairs and 32 identifying which member of each pair was illuminati, giving 81 days.&lt;/p&gt;&#xA;" OwnerUserId="3987" LastActivityDate="2014-11-08T12:02:22.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3818" PostTypeId="2" ParentId="3784" CreationDate="2014-11-08T12:04:07.100" Score="-3" Body="&lt;p&gt;Question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Are one, and only one of these 2 statements true:&lt;br&gt;&#xA;&lt;br&gt;&#xA; You are the eldest&lt;br&gt;&#xA; Lucy is the middle aged/random&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Response:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes: Lucy is safe to marry no matter what, feel free to check the possibilities.&lt;br&gt;&#xA;&lt;br&gt;&#xA; No: Eleanor is safe to marry no matter what, feel free to check the possibilities&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-11-08T13:34:55.273" LastActivityDate="2014-11-08T13:34:55.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3819" PostTypeId="2" ParentId="3726" CreationDate="2014-11-08T12:14:59.097" Score="8" Body="&lt;p&gt;It's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you, since you're not surprised hearing that you're needed at the intersection of Main and Peele.&lt;br&gt;&#xA; But you don't seem to know &lt;em&gt;why&lt;/em&gt; you're not surprised. You keep having flashbacks of your promotion, but you don't want to go to Chicago! You like it here in &lt;a href=&quot;https://goo.gl/maps/KRdvj&quot;&gt;Everetts&lt;/a&gt;. That's why you killed your boss, while you were sleepwalking. You approached him from behind, strangling him. He tried to go for his gun, which he kept in his &lt;a href=&quot;http://simage1.sportsmansguide.com/adimgs/l/1/144587_ts.jpg&quot;&gt;shoulder holster&lt;/a&gt;. It went off, wounding him in the hip. You stabbed him, then cleaned the knife and dumped the body, all while sleeping.&lt;br&gt;&#xA; The note came from the angry driver, who saw the dead body of a man he hated (don't know why) and left a note.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But what happened to the poodle?&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditDate="2014-11-09T18:29:31.277" LastActivityDate="2014-11-09T18:29:31.277" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3820" PostTypeId="1" AcceptedAnswerId="3822" CreationDate="2014-11-08T13:32:08.133" Score="-3" ViewCount="594" Body="&lt;p&gt;There are $n$ jellybeans in a bottle with unknown dimensions. What is the only surefire way of knowing how many jellybeans are in the bottle?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="5373" LastEditDate="2015-04-28T19:43:43.260" LastActivityDate="2018-01-04T07:25:36.177" Title="How many Jellybeans are in the bottle" Tags="|lateral-thinking|combinatorics|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3821" PostTypeId="2" ParentId="3820" CreationDate="2014-11-08T13:38:15.867" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You can empty the bottle and count the jellybeans on the table.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1685" LastActivityDate="2014-11-08T13:38:15.867" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3822" PostTypeId="2" ParentId="3820" CreationDate="2014-11-08T13:44:38.053" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Perhaps by counting the number of jellybeans in the bottle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="97" LastEditorUserId="97" LastEditDate="2014-11-08T13:50:22.580" LastActivityDate="2014-11-08T13:50:22.580" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3823" PostTypeId="1" AcceptedAnswerId="3826" CreationDate="2014-11-08T14:15:51.063" Score="10" ViewCount="1252" Body="&lt;ol&gt;&#xA;&lt;li&gt;Rocky the dog is alive.&lt;/li&gt;&#xA;&lt;li&gt;Rocky the dog is dead.&lt;/li&gt;&#xA;&lt;li&gt;Exactly one of statements 6 and 9 is true.&lt;/li&gt;&#xA;&lt;li&gt;Exactly one of statements 2 and 6 is false.&lt;/li&gt;&#xA;&lt;li&gt;Statements 4, 5 and 10 are all false.&lt;/li&gt;&#xA;&lt;li&gt;Exactly one of statements 1 and 10 is false.&lt;/li&gt;&#xA;&lt;li&gt;Exactly 5 statements are true.&lt;/li&gt;&#xA;&lt;li&gt;Exactly one of statements 3 and 10 is false.&lt;/li&gt;&#xA;&lt;li&gt;Exactly one of statements 6 and 10 is true.&lt;/li&gt;&#xA;&lt;li&gt;Exactly one of statements 1 and 2 is false.&lt;/li&gt;&#xA;&lt;li&gt;Statements 1, 8 and 11 are all false.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Which of the above statements are definitely true? Is Rocky the dog alive or dead?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The puzzle is based on a similar puzzle here:  &lt;a href=&quot;http://blog.physicsworld.com/2013/10/07/physics-world-at-25-puzzle-2/&quot;&gt;http://blog.physicsworld.com/2013/10/07/physics-world-at-25-puzzle-2/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="97" LastEditorUserId="97" LastEditDate="2014-11-09T07:53:55.867" LastActivityDate="2014-11-10T07:00:10.920" Title="Is the dog dead or alive?" Tags="|logical-deduction|" AnswerCount="5" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3825" PostTypeId="2" ParentId="3823" CreationDate="2014-11-08T14:28:18.367" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Statement 5 tells us that 4 and 10 are true. So, it confirms that the dog is dead or alive ( statement 1 and 2 ). If 4 is true ,we can find a conflict between 6 and 10. Because, if 10 is false, it is not possible that 1 and 2 are true at the same time, either false. So 10 must be true and 1 is false, so the &lt;strong&gt;dog is dead&lt;/strong&gt; &lt;br/&gt;The true statement are: 2, 3, 4, 6, 8, 10&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1685" LastActivityDate="2014-11-08T14:28:18.367" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3826" PostTypeId="2" ParentId="3823" CreationDate="2014-11-08T14:34:52.420" Score="5" Body="&lt;p&gt;The dog is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alive&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Rocky the dog is alive.                         TRUE                &#xA;Rocky the dog is dead.                          FALSE&#xA;Exactly one of statements 6 and 9 is true.      TRUE&#xA;Exactly one of statements 2 and 6 is false.     FALSE&#xA;Statements 4, 5 and 10 are all false.           FALSE&#xA;Exactly one of statements 1 and 10 is false.    FALSE&#xA;Exactly 5 statements are true.                  TRUE&#xA;Exactly one of statements 3 and 10 is false.    FALSE&#xA;Exactly one of statements 6 and 10 is true.     TRUE&#xA;Exactly one of statements 1 and 2 is false.     TRUE&#xA;Statements 1, 8 and 11 are all false.           FALSE&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2484" LastActivityDate="2014-11-08T14:34:52.420" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3827" PostTypeId="2" ParentId="3801" CreationDate="2014-11-08T15:42:42.723" Score="0" Body="&lt;p&gt;Assuming we're cutting just perpendicular to the surface, the maximum is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;9 Pieces. This is by first cutting at bearing 000, to the left of the centre but still going through the circular gap in the centre, Then, rotate the donut by 120 degrees and make the same cut. Repeat this once more and you should have 9 separate pieces if the distance from the centre that you used was correct. Simply scale the distance from the centre for each cut until the cut clips the hole, but also crosses with another cut before doing so. Of course, if we allow non-perpendicular cuts to the surface...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4748" LastActivityDate="2014-11-08T15:42:42.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3828" PostTypeId="2" ParentId="3726" CreationDate="2014-11-08T15:57:31.617" Score="2" Body="&lt;p&gt;There have been several genius answers, but none of them explain where the poodle went. I agree with TyCobb's answer, specifically the part where the poodle is killed. After the poodle died, the homeless man might have offered to take the body and ate it. This might sound implausible but:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;He says he hasn't eaten in days. I think this is true because I see no motive for lying about this. (But I can seen why he wouldn't admit eating the dog.)&lt;/li&gt;&#xA;&lt;li&gt;Dog is edible. In some cultures it is completely normal, I heard. (I am not sure but I seem to recall this is the case in Japan. Again, this is a vague memory and I have never been there but only heard about it.) Also during food shortages, for example during the winter of '44/'45 when the remaining territory of the Nazi's had an enormous shortage of food, it seems people eat dogs too. (&lt;a href=&quot;http://en.m.wikipedia.org/wiki/Dog_meat&quot; rel=&quot;nofollow&quot;&gt;http://en.m.wikipedia.org/wiki/Dog_meat&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4749" LastActivityDate="2014-11-08T15:57:31.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3829" PostTypeId="2" ParentId="3739" CreationDate="2014-11-08T15:58:32.807" Score="1" Body="&lt;p&gt;The answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;What the old man didn't know on the island was that the meat given to him was actually human meat, the flesh of another survivor of the storm. After tasting real seagull meat on the cruise ship, he noticed the difference in taste and realised that he had been fed human meat. After finding out he was a cannibal, the old man saw no choice but to kill himself. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4748" LastActivityDate="2014-11-08T15:58:32.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3830" PostTypeId="2" ParentId="3823" CreationDate="2014-11-08T16:35:24.953" Score="-1" Body="&lt;p&gt;My calculations say Statement 2 is  true. Fingers crossed...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reason for this answer... Statement 5 and 11 are themselves false so no considered... As per statement 3, 6 &amp;amp; 9 any 1 is true again statement 9 says 6 &amp;amp; 10 any 1 is true but statement 6 which has probability of being true twice says that 1 &amp;amp; 10 is false. So if I consider both to be false for a while statement 10 says 1 &amp;amp; 2 any 1 is false&lt;/p&gt;&#xA;" OwnerUserId="4750" LastEditorUserId="2313" LastEditDate="2014-11-10T07:00:10.920" LastActivityDate="2014-11-10T07:00:10.920" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3831" PostTypeId="1" AcceptedAnswerId="3876" CreationDate="2014-11-08T16:44:05.883" Score="3" ViewCount="1328" Body="&lt;p&gt;This question was previously from the &lt;em&gt;Haselbauer - Dickheiser Test For Exceptional Intelligence&lt;/em&gt;,  a test that was previously available on the website www.highiqosociety.com, but was removed several years ago.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Consider two breeding strategies of the fictional Furble.  Dominator&#xA;  Furbles can fight for a breeding territory, and if they win, will be&#xA;  able to rear 10 offspring.  An alternative is to share territory with&#xA;  another Furble which will allow each to rear 5 offspring.  Sharers who&#xA;  attempt to share with dominators will be forced out of the territory,&#xA;  although they will be able to find a new territory.  Assume sharers&#xA;  become extra cautious after encountering a dominator and so will&#xA;  always find another territory to share the next time around, but due&#xA;  to lost time will only be able to produce 3 offspring.  Dominators are&#xA;  always able to force sharers out of the territory and rear 10 young. &#xA;  Dominators who meet dominators will win 50% of the time.  When they&#xA;  lose, they are not able to reproduce that season due to sustained&#xA;  injuries.  Individual Furbles cannot switch strategies.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;With a total population of 2000 dominator and sharer Furbles&#xA;  altogether, how many would you expect to dominate?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would assume no Furble finds an empty location.  They either find someone to share or fight for their territory.  Otherwise the solution is trivially 2000 dominators.&lt;/p&gt;&#xA;" OwnerUserId="97" LastEditorUserId="11110" LastEditDate="2017-12-26T09:37:48.217" LastActivityDate="2017-12-26T10:20:02.777" Title="Dominating Furbles" Tags="|logical-deduction|mathematics|strategy|" AnswerCount="5" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="3832" PostTypeId="2" ParentId="3831" CreationDate="2014-11-08T17:05:29.870" Score="1" Body="&lt;p&gt;If there are infinitely many territories then they would all be best to share i.e. it is in everybody's best interests that &lt;em&gt;everyone&lt;/em&gt; &quot;shares&quot; - this avoids fights so there's no risk of missing out on breeding (nobody can &quot;lose&quot;). And everyone will get to rear 5 offspring (it was never said that the goal is to maximise offspring). Their offspring will have no trouble moving off to share some other breeding grounds after this (since there is no limits to the number of territories imposed.)&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-11-08T17:30:21.837" LastActivityDate="2014-11-08T17:30:21.837" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3833" PostTypeId="1" AcceptedAnswerId="3882" CreationDate="2014-11-08T17:24:17.557" Score="15" ViewCount="1136" Body="&lt;p&gt;The task is to construct a sudoku whose rows (or columns) are the moves for a game in tic tac toe where the game ends in a draw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some clarifications and constraints which should limit towards a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the $3*3$ grid of a regular tic tac toe with an embedded number pad, like the top left cell as 1, top centre cell as 2.... all the way down to the down right cell as 9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first number the series makes is of X, and the next number by O, and so on, until the game ends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the game shown by this notation is a good draw (the end result game is shown in the picture, and the order accordingly) : $6-5-3-9-1-2-8-7-4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/57Fi8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So basically, construct a grid consisting of numbers 1 to 9, with no digits repeating in the rows and the columns, i.e. construct a Sudoku where  the numbers in the rows/columns indicate a game of tic tac toe where the game ends in a draw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm really curious if the grid with the exact sudoku conditions exists or not. I've been thinking about this puzzle all day and I haven't even got near to an answer. I thought it is a nice question and thought why not share it with you all. Hoping to see some great answers.&lt;/p&gt;&#xA;" OwnerUserId="2808" LastEditorUserId="11100" LastEditDate="2015-05-02T18:40:39.980" LastActivityDate="2015-05-02T18:40:39.980" Title="Devising a sudoku where the rows are the moves for a draw in tic tac toe" Tags="|logical-deduction|sudoku|tic-tac-toe|construction|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3834" PostTypeId="1" AcceptedAnswerId="4764" CreationDate="2014-11-08T17:47:30.800" Score="11" ViewCount="423" Body="&lt;p&gt;Imagine a game of connect 4 where Player1 wins by Player2 connecting 4, and Player2 wins by forcing the board into a filled board without any 4-in-a-rows. A tie happens if Player1 connects 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now Let's assume neither of them want to tie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, with perfect play by both players, will the game result in a win for Player1, Player2, or will it be a draw?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The board size is 7 columns, 6 rows!&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2014-11-16T13:44:45.990" Title="No-Connecting-4" Tags="|strategy|board-games|" AnswerCount="2" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="3835" PostTypeId="2" ParentId="3831" CreationDate="2014-11-08T18:08:44.860" Score="1" Body="&lt;p&gt;Given that a furble has to fight or share with someone, it seems that the maximum amount of furbles from the previous generation would come from equal dominators and sharers. Each dominator dominates a sharer, then the sharers get together and have offspring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the 2000 furbles would end up with 1000 + 10000 dominators and 1000 + 3000 sharers. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2014-11-08T18:08:44.860" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3837" PostTypeId="2" ParentId="3705" CreationDate="2014-11-08T18:25:45.310" Score="7" Body="&lt;p&gt;Omega.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/Limit_ordinal&quot;&gt;Wikipedia&lt;/a&gt;, omega is &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the smallest ordinal [number] greater than every natural number. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that omega is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Transfinite_number&quot;&gt;transfinite number&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2763" LastActivityDate="2014-11-08T18:25:45.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3838" PostTypeId="2" ParentId="2120" CreationDate="2014-11-08T18:26:23.313" Score="1" Body="&lt;p&gt;May it be &quot;Graviton&quot;, it is an hypothetical elementary particle that mediates the force of gravitation. &#xA;It is small &quot;like a drop&quot;, unknown, related to Newton (Gravity) and proposed by Einstein (for some reason the phrase is in German) also Einstein General Relativity changed some concepts from Newtons original theory, like the quote.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried with Graviton but it is no, maybe a variant?. Sorry for not using the comments section, I have no 50 points yet.&lt;/p&gt;&#xA;" OwnerUserId="4756" LastActivityDate="2014-11-08T18:26:23.313" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3839" PostTypeId="1" AcceptedAnswerId="3840" CreationDate="2014-11-08T18:41:36.343" Score="17" ViewCount="2155" Body="&lt;p&gt;Once upon a time, there was a beautiful girl who had a lot of lovers. Every guy wanted to marry her, be with her. Given a lot of options, she had reduced the candidates from 128 to 2. Yet it was impossible to make up her mind anymore; both of the candidates were almost equal. Thus, she decided to play the &quot;patience game&quot;. She summoned both of her lovers and told them to ride to the Garden Way at midnight for there was a quest waiting for them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At midnight in the Garden Way, there was quite a crowd. Everybody had heard about the quest for this beautiful lady and came down to watch it. Without any delay, she told what the quest was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Tonight, both of you will start a quest for patience to prove your love to me. Whoever comes out as the winner - I will be yours, forever. Tonight, you will race for me. But this race requires your utmost patience. To win this race, your horse must be the last one to pass the finish line. Good luck and may the gods help you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There was a sudden shock in the audience. Everybody was surprised. Both of the riders didn't even pinch. They waited for the other to retreat. It didn't happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a full month, both lovers were still waiting. They were far from giving up. At midnight, there came a wise man of the town. He approached the lovers and gave them some advice. Following his advice they both started galloping rapidly, racing to the finish line as fast as they can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What did this wise man say?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Please put answers in spoilers&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="4755" LastEditorUserId="5373" LastEditDate="2014-12-28T21:54:47.020" LastActivityDate="2015-05-01T13:58:01.783" Title="Race of patience for the sake of love" Tags="|lateral-thinking|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3840" PostTypeId="2" ParentId="3839" CreationDate="2014-11-08T18:44:59.283" Score="21" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;He said to switch their horses. The condition was that whose horse would finish last, not them. So they exchanged horses, and raced to their full potential.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2808" LastEditorUserId="4755" LastEditDate="2014-11-08T18:49:48.133" LastActivityDate="2014-11-08T18:49:48.133" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3841" PostTypeId="2" ParentId="3839" CreationDate="2014-11-08T18:50:27.600" Score="7" Body="&lt;p&gt;The wise man had said&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you switch horses you can attempt to make the other lose by getting their horse across. You do not have to pass last, your horse does!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Creating a race to the finish, a &lt;em&gt;real&lt;/em&gt; race, not an endless one.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2014-11-08T18:50:27.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3842" PostTypeId="1" AcceptedAnswerId="4159" CreationDate="2014-11-08T18:52:47.230" Score="2" ViewCount="319" Body="&lt;p&gt;In a standard American game of checkers what's the maximum amount of kings a player can get, and what is the move sequence?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Keeping in mind the force capture rule&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-11-09T19:40:15.157" LastActivityDate="2014-11-12T15:04:31.603" Title="How many Kings can you Achieve?" Tags="|checkerboard|board-games|" AnswerCount="1" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="3843" PostTypeId="2" ParentId="3839" CreationDate="2014-11-08T18:58:10.453" Score="8" Body="&lt;p&gt;I would say it is exactly the other way around:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He said that he already switched their horses so they had to start as soon as possible. In this way, they have to start immediately (like in the question)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3993" LastActivityDate="2014-11-08T18:58:10.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3845" PostTypeId="1" AcceptedAnswerId="3887" CreationDate="2014-11-08T19:43:28.400" Score="21" ViewCount="2041" Body="&lt;p&gt;This is Part 2 of the Murder of the President brainteaser/riddle series. If you have not already, you might want to see the answer for &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3600/murder-of-the-president-part-1&quot;&gt;Murder of the President - Part 1&lt;/a&gt; posted by Nit before you begin to work on Part 2. Each part will give you a clue (usually a place to go for the next clue) and you must solve the clue. This part will have two puzzles you have to solve. Use all knowledge you have of cryptography, ciphers, etc. This case is meant to take place in the present day, so all politicians, celebrities, places, etc. are what they are now. Please post your answers in spoiler tags.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the riddle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You arrive at the Newark, NJ airport and head to the Southwest part of terminal A. Once you arrive, you begin to look around the area. Nothing seems unusual. You glance at a TV and notice that, surprisingly, no one is reporting about it yet. They've done a good job keeping this silent. You check your watch. 7:30 p.m. You begin to look around again. As you scan the walls for any clues, a man bumps into and mutters something you vaguely hear as 'End of six.' You start walking around and at about 7:36 p.m. you spot a briefcase isolated from everyone. You head over to investigate. The briefcase has three number locks, the first has eight digits, the second has four digits, and the last one has two digits. You turn the briefcase over and see a sticky note with some scribbled words on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what it says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;C (bytes)&lt;br&gt;&lt;/li&gt;&#xA;  &lt;li&gt;ADIF &lt;br&gt;&lt;/li&gt;&#xA;  &lt;li&gt;Deep Thought&lt;br&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After you open the briefcase there is a New International Version Bible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the three codes? What should you do with the Bible?&#xA;What should you look for?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good Luck.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note&#xA;I will be posting Part 3 in one to two days. I will select the correct answer for Part 2 before, but I will post the answer for Part 2 if no one gets it&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1711" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2014-12-15T23:00:56.983" Title="Murder of the President - Part 2" Tags="|mystery|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3846" PostTypeId="1" CreationDate="2014-11-08T19:57:00.417" Score="1" ViewCount="766" Body="&lt;p&gt;I have just thought about an interesting problem in classical propositional logic to propose at this site, but lost my notes...  I do recall, however, that the problem involved a particular contingent sentence with three atomic variables, $p$, $q$ and $r$. This first sentence contained exactly two occurrences of connectives, both binary. I am positive that at least one of those connectives was an implication having one of the three aforementioned variables as antecedent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 1.&lt;/strong&gt;  Assuming that the previous information is sufficient to &lt;em&gt;semantically characterize&lt;/em&gt; the forgotten sentence, can you determine what was its other connective, as well as the possible logical forms that my sentence could have had?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, another thing I do recall is that the problem that I had created consisted in checking that the first sentence above is &lt;em&gt;semantically equivalent&lt;/em&gt; to a second sentence, written with the same variables and containing the same connectives as the first sentence, but with one extra implication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 2.&lt;/strong&gt; What could have been this second sentence?  Justify.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The following extra considerations should be unnecessary to solve the problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Word of warning.&lt;/strong&gt;&#xA; &lt;br&gt;&#xA;  While one must &lt;em&gt;not&lt;/em&gt; disconsider any information from the above statement when looking for a solution, one should also not read more than what is written. In particular: &#xA; &lt;br&gt;&#xA; (1) Nowhere in the statement of the problem is to be found a claim about the first (or the second) sentence being &quot;syntactically unique&quot;. In fact, there is obviously no way, for instance, of avoiding permutation of the variables in a given solution. But there might clearly be many other &lt;em&gt;structurally different&lt;/em&gt; sentences (belonging to the same equivalence class!) that solve &lt;strong&gt;Question 1&lt;/strong&gt;. The task is to identify such sentences (preferably without just guessing).&#xA; &lt;br&gt;&#xA; (2) In the statement of the problem, a sentence is supposed to be &quot;semantically characterized&quot; when it is characterized &lt;em&gt;up to logical equivalence&lt;/em&gt; --- so, again, you had better think first about what is happening at the level of the quotient algebra, and consider next the syntactic restrictions imposed by the statement of the problem.&#xA; &lt;br&gt;&#xA; &lt;br&gt;&lt;strong&gt;Hint.&lt;/strong&gt;&#xA; &lt;br&gt;The &lt;em&gt;assumption&lt;/em&gt; made in &lt;strong&gt;Question 1&lt;/strong&gt; is integral part of the statement, and in fact essential to the existence of a well-determined solution!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4757" LastEditorUserId="4757" LastEditDate="2014-11-09T18:14:30.027" LastActivityDate="2014-11-11T00:22:55.233" Title="What could have been the sentences I was thinking about?" Tags="|mathematics|logical-deduction|" AnswerCount="4" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="3847" PostTypeId="2" ParentId="3705" CreationDate="2014-11-08T21:11:45.210" Score="10" Body="&lt;p&gt;For a number immensely bigger than $\omega$, consider the uncountably infinite number hidden below.   (Note, by the way, that $\omega$ is countably infinite, and rather than being the biggest something, it is in fact  “the smallest infinite ordinal ... as it is the least upper bound of the natural numbers” [&lt;a href=&quot;http://en.wikipedia.org/wiki/Limit_ordinal#Examples&quot;&gt;1&lt;/a&gt;]).  So &lt;strong&gt;omega&lt;/strong&gt; is a good candidate for the first bonus, the smallest number one can spell if given one of each letter in the alphabet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The transfinite number &lt;strong&gt;aleph sixtyfour&lt;/strong&gt; appears to be the biggest aleph ($\aleph$) number one can spell if given one of each letter in the alphabet.&lt;br&gt;&#xA; Note that  $\aleph_{64} &amp;gt; \aleph_{63} &amp;gt; ... \aleph_1 = 2^{\aleph_0} &amp;gt; \aleph_0  = \omega$.&lt;br&gt;&#xA;(For a big number that doesn't quite work because it has two a's and e's, see wikipedia's &lt;a href=&quot;http://en.wikipedia.org/wiki/Aleph_number#Aleph-.CF.89&quot;&gt;Aleph-ω&lt;/a&gt; article; &lt;strong&gt;aleph omega&lt;/strong&gt; is  the least upper bound of ${\aleph_n : n\in\{0,1,2,\dots}\}$.  But if we use five Roman and one Greek letter, or one Hebrew and one Greek letter, &lt;strong&gt;aleph $\omega$&lt;/strong&gt; or &lt;strong&gt;$\aleph_{\omega}$&lt;/strong&gt; work ok.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2004" LastEditorUserId="2004" LastEditDate="2014-11-10T16:06:07.700" LastActivityDate="2014-11-10T16:06:07.700" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="3848" PostTypeId="2" ParentId="3726" CreationDate="2014-11-08T21:28:23.603" Score="45" Body="&lt;p&gt;I have a possible solution which no-one else has yet suggested. It has the &lt;a href=&quot;http://en.wikipedia.org/wiki/Occam&amp;#39;s_razor&quot; rel=&quot;nofollow noreferrer&quot;&gt;virtue of simplicity&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The puzzle has no single logically unique answer; OP is a troll (or does not  understand that a puzzle should have a specific answer that can be deduced from the question).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - OP has no posting history here or on linked SE sites.&lt;br&gt;&#xA; - So much detail in the story, so many 'clues', &lt;a href=&quot;http://tvtropes.org/pmwiki/pmwiki.php/Main/RasputinianDeath&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rasputinian Death&lt;/a&gt;.&lt;br&gt;&#xA; - &lt;a href=&quot;https://puzzling.stackexchange.com/posts/3726/revisions&quot;&gt;Flaky grammar&lt;/a&gt; in the original post is not consistent with an intelligent OP (counter-argument: OP could have English as a foreign language or have a condition such as dyslexia which allows them to be intelligent but not good at writing in English. Or maybe the mistakes are intentional.)&lt;br&gt;&#xA; - No sign of this puzzle on Google, so OP probably came up with it themselves.&lt;br&gt;&#xA; - OP: &quot;i just like reading all the answers people come up with&quot;&lt;br&gt;&#xA; - I'm a deeply cynical person.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'd consider the following answers equivalent to my solution (you might not agree):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - &lt;strong&gt;OP has a solution but it's no more logical or likely than anyone else's solution.&lt;/strong&gt; &lt;br&gt;&#xA; - Everyone in the story is non-human (they're goldfish and 'Chicago' is the name of the fishbowl, etc etc). &lt;br&gt;&#xA; - God / Satan / Father Christmas / aliens did it (&lt;a href=&quot;http://en.wikipedia.org/wiki/Deus_ex_machina&quot; rel=&quot;nofollow noreferrer&quot;&gt;deus ex machina&lt;/a&gt;).  &lt;br&gt;&#xA; - It's &lt;a href=&quot;http://tvtropes.org/pmwiki/pmwiki.php/Main/AllJustADream&quot; rel=&quot;nofollow noreferrer&quot;&gt;all just a dream&lt;/a&gt; / hallucination / taking place in The Matrix. &lt;br&gt;&#xA; - Dead body is not a human or is not really dead.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The OP gets a bounty from me if:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The puzzle &lt;em&gt;is&lt;/em&gt; logically coherent and: &lt;br&gt; &lt;br&gt;&#xA; - The detective is a ghost ('the Shyamalan' plot twist).&lt;br&gt;&#xA; - The detective is the murderer or the victim.&lt;br&gt;&#xA; - They all did it ('&lt;a href=&quot;http://en.wikipedia.org/wiki/Murder_on_the_Orient_Express&quot; rel=&quot;nofollow noreferrer&quot;&gt;Murder on the Orient Express&lt;/a&gt;').&lt;br&gt;&#xA; - The solution involves a time machine or an alternate universe. &lt;br&gt;&#xA; - The murderer is one of the minor characters, eg the assistant, the old boss or the forensic scientist ('the butler did it').&lt;br&gt;&#xA; - The mangled English is integral to the solution of the puzzle.&lt;br&gt;&#xA; - The puzzle is not a murder mystery (e.g message enciphered in the typos).&lt;br&gt;&#xA; - The story (with original typos) successfully compiles in any well-known programming language&lt;br&gt; &lt;br&gt;&#xA; or if OP's solution consists solely of trollface.gif.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Double bounty if:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The puzzle &lt;em&gt;is&lt;/em&gt; logically coherent and: &lt;br&gt; &lt;br&gt;&#xA; - The detective is the murderer &lt;strong&gt;and&lt;/strong&gt; the victim.&lt;br&gt;&#xA; - The detective appears to be a ghost but is actually a living human pretending to be a ghost pretending to be a living human (the dreaded &lt;a href=&quot;http://dresdencodak.com/2009/05/11/42-essential-3rd-act-twists/&quot; rel=&quot;nofollow noreferrer&quot;&gt;reverse double Shyamalan&lt;/a&gt;).&lt;br&gt;&#xA; - The puzzle outputs the name of the murderer when executed by the Perl interpreter.&lt;br&gt;&#xA; - The entire puzzle takes place in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Buffyverse&quot; rel=&quot;nofollow noreferrer&quot;&gt;Buffyverse&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Update after reading OP's solution: yay! Nailed it.  :) &lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2014-11-10T17:55:35.490" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3849" PostTypeId="2" ParentId="3846" CreationDate="2014-11-08T21:35:26.727" Score="5" Body="&lt;p&gt;Question 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;p implies NAND(q,r)&lt;/code&gt;, which is equivalent to &lt;code&gt;(not p) or (not q) or (not r)&lt;/code&gt; which is symmetric in the variables. Once you're told that the second gate is a NAND, you know the sentence must have the form &lt;code&gt;_ implies NAND(_,_)&lt;/code&gt;. Since any permutation of &lt;code&gt;p,r,q&lt;/code&gt; in the blanks is equivalent to the symmetric expression given, all possible choices are semantically equivalent. Intuitively, the sentence says that if a variable is true, the other two can't both also be true, which is equivalent to &quot;It can't be the case that all three variables are true&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2014-11-08T22:18:41.323" LastActivityDate="2014-11-08T22:18:41.323" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="3850" PostTypeId="2" ParentId="3846" CreationDate="2014-11-08T22:58:22.633" Score="3" Body="&lt;p&gt;I have a new answer based on a purely formal logical interpretation of the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma 1:&lt;/strong&gt; The conditions in paragraph 1 do not specify a syntantically unique sentence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof of lemma:&lt;/strong&gt; Both the sentences &lt;code&gt;p implies (q or r)&lt;/code&gt; and &lt;code&gt;p implies (q and r)&lt;/code&gt; satisfying those conditions, as they have three atomic variables &lt;code&gt;p,q,r&lt;/code&gt; and two connectives, one an implication implied by a variable. By, they are not logically equivalent as &lt;code&gt;p=True, q=True, r=False&lt;/code&gt; satisfies the first but not the second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, afterwards, we're asked to assume that  the previous information is sufficient to semantically characterize the forgotten sentence, which is the negation of Lemma 1. For having both Lemma 1 and its negation, we use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Principle_of_explosion&quot; rel=&quot;nofollow&quot;&gt;principle of explosion&lt;/a&gt; to derive all statements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, we derive that the forgotten statement was &quot;I am the eggman, I am the walrus.&quot; As to the question of &quot;the possible logical forms that my sentence could have had&quot;, it is the set of logical forms.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2014-11-08T22:58:22.633" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3851" PostTypeId="2" ParentId="3823" CreationDate="2014-11-08T23:50:15.887" Score="6" Body="&lt;p&gt;Going down the list in order:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This whole line of reasoning is predicated upon 10 being absolute truth, because no dog can be both dead and alive. If that is not the case and we have a quantum dog that can be in a superposition of states, none of this is valid. &lt;br/&gt;&#xA;&lt;br/&gt;&#xA;1 is a simple statement. &lt;br/&gt;&#xA;2 is a simple statement. &lt;br/&gt;&#xA;&lt;br/&gt;&#xA;3 says that only one of 6 or 9 is true. If 3 is false, then either both 6 and 9 are true, or both 6 and 9 are false. Since statement 10 is true, if 9 is true, then 6 must be false. Therefore, they cannot both be true. IF 9 is false, then 6 must be true, otherwise 9 would be true. Therefore, they cannot both be false. Therefore, statement 3 must be true. Therefore, either 6 is true, or 9 is true, but not both.&lt;br/&gt;&#xA;&lt;br/&gt;&#xA;9 states that only one of 6 or 10 is true. 10 tells us something we already knew to be true; therefore, if 9 is true, 10 is true and 6 is false. (We can also derive this from 3 being true, which makes 6 false if 9 is true). If 6 is false, then either 1 and 10 are both true, or they are both false. 10 is true, therefore 1 must be true; therefore, if 6 is false, rocky is alive. &lt;br/&gt;&#xA;&lt;br/&gt;&#xA;If 6 is the one that is true, either 1 is true or 10 is true. 10 must be true, as 1 and 2 are logically opposed. Therefore, if 6 is true, 1 is false, making 2 true. Therefore, if 6 is true, Rocky is Dead.&lt;br/&gt;&#xA;&lt;br/&gt;&#xA;On to the next statement. 4 states that one of 2 and 6 is false. If 2 is false, Rocky is Alive. However, if 2 is false, 6 is true, and if 6 is true, as we have shown, 2 is true, and Rocky is Dead. Therefore, if 2 is false, 2 is true, which is a logical contradiction. Therefore, 2 is true, and 6 is false. Therefore, if 4 is true, Rocky is Dead.&lt;br/&gt;&#xA;&lt;br/&gt;&#xA;If statement 4 is false, then either 2 and 6 are both true or 2 and 6 are both false. If 2 is false (therefore Rocky is Alive) and 6 is false (therefore making statement 1 true, since 10 cannot be false for them both to be false), then Rocky is alive. &lt;br/&gt;&#xA;&lt;br/&gt;&#xA;To recap the state of Rocky:&lt;br/&gt;&#xA;if 1: alive; if !1: dead&lt;br/&gt;&#xA;if 2: dead; if !2: alive&lt;br/&gt;&#xA;3, therefore either dead or alive.&lt;br/&gt;&#xA;if 4: Dead; if !4: alive&lt;br/&gt;&#xA;if 6: dead; if !6: alive. &lt;br/&gt;&#xA;&lt;br/&gt;&#xA;If Statement 5 is true, then statements 4, 5, and 10 are all false. Since statement 5 being true precludes statement 5 being false, then statement 5 is false. This lets us know that statements 4, 5, and 10 are not all false. Statement 5 is false, so both of statements 4 and 10 cannot be false. Therefore,  one ore both of statements 4 and 10 is true. However, we know that statement 10 is true. Therefore, either statement 4 is false or statement 4 is true. This tells us nothing of interest about statement 4; we already knew it was either true or false. &lt;br/&gt;&#xA;&lt;br/&gt;&#xA;Statement 7 claims that five statements are true. This is not useful at this time, so we set it aside for now. &lt;br/&gt;&#xA;&lt;br/&gt;&#xA;Statement 8 claims that exactly one of 3 and 10 is false. We know that 10 is empirically true. Therefore, if 8 is true 3 is false. We know, however, that 3 must be true. Therefore, 8 is false. You can reason out here to show that both statements 3 and 10 are true, but we knew that already.&lt;br/&gt;&#xA;&lt;br/&gt;&#xA;To recap the state of Rocky:&lt;br/&gt;&#xA;if 1: alive&lt;br/&gt;&#xA;if 2: dead&lt;br/&gt;&#xA;since 3, either Dead or alive&lt;br/&gt;&#xA;if 4: Dead; if !4: alive&lt;br/&gt;&#xA;if 6: dead&lt;br/&gt;&#xA;since !8, either dead or alive&lt;br/&gt;&#xA;&lt;br/&gt;&#xA;&lt;br/&gt;&#xA;Statement 11 claims that 1, 8, and 11 are all false. As with statement 5, 11 being true leads to a logical contradiction, so we know that 11 is false. Since we know that 11 is false and 8 is false, 1 must be true (otherwise 11 would be true). Therefore, 1 is true and rocky is ALIVE. Since he is alive, he cannot be dead, so 2 is false.&lt;br/&gt;&#xA;&lt;br/&gt;&#xA;So far:&lt;br/&gt;&#xA;1: true&lt;br/&gt;&#xA;2: false&lt;br/&gt;&#xA;3: true&lt;br/&gt;&#xA;5: false&lt;br/&gt;&#xA;8: false&lt;br/&gt;&#xA;10: true&lt;br/&gt;&#xA;11: false&lt;br/&gt;&#xA;&lt;br/&gt;&#xA;We deduced earlier that if statement 4 is true, then rocky is dead. Since rocky is not dead, statement 4 must be false. We deduced earlier that that means 2 and 6 must both be true or both be false; since 2 is true, 6 must be false. &lt;br/&gt;&#xA;&lt;br/&gt;&#xA;So far:&lt;br/&gt;&#xA;1: true&lt;br/&gt;&#xA;2: false&lt;br/&gt;&#xA;3: true&lt;br/&gt;&#xA;4: false&lt;br/&gt;&#xA;5: false&lt;br/&gt;&#xA;6: false&lt;br/&gt;&#xA;8: false&lt;br/&gt;&#xA;10: true&lt;br/&gt;&#xA;11: false&lt;br/&gt;&#xA;&#xA;Statement 9 claims that exactly one of 6 and 10 is true. This is obviously true. &lt;br/&gt;&#xA;&lt;br/&gt;&#xA;1: true&lt;br/&gt;&#xA;2: false&lt;br/&gt;&#xA;3: true&lt;br/&gt;&#xA;4: false&lt;br/&gt;&#xA;5: false&lt;br/&gt;&#xA;6: false&lt;br/&gt;&#xA;8: false&lt;br/&gt;&#xA;9: true&lt;br/&gt;&#xA;10: true&lt;br/&gt;&#xA;11: false&lt;br/&gt;&#xA;&lt;br/&gt;&#xA;We now have four true statements. Statement 7 states that exactly 5 statements are true. If 7 is true, then five statements are true, as 7 is the fifth. If 7 is false, then ther are only four true statements, making 7 false. So 7 is impossible to determine. I will mark this as &quot;mu&quot;. &lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Final answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Rocky is &lt;strong&gt;Alive&lt;/strong&gt;. &lt;br/&gt;&#xA;1: true&lt;br/&gt;&#xA;2: false&lt;br/&gt;&#xA;3: true&lt;br/&gt;&#xA;4: false&lt;br/&gt;&#xA;5: false&lt;br/&gt;&#xA;6: false&lt;br/&gt;&#xA;7: mu&lt;br/&gt;&#xA;8: false&lt;br/&gt;&#xA;9: true&lt;br/&gt;&#xA;10: true&lt;br/&gt;&#xA;11: false&lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a side note, I am ridiculously proud that I managed to avoid confusing myself, but please let me know if I mixed something up in the middle! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edited: I had mixed up statement 8 somehow. Expanded and fixed. &lt;/p&gt;&#xA;" OwnerUserId="431" LastEditorUserId="431" LastEditDate="2014-11-09T18:28:02.410" LastActivityDate="2014-11-09T18:28:02.410" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3852" PostTypeId="2" ParentId="3846" CreationDate="2014-11-09T00:58:58.717" Score="2" Body="&lt;p&gt;I'll have a go with :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$p \rightarrow (q \leftrightarrow r)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;together with &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$(p \rightarrow q) \leftrightarrow (p \rightarrow r)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This reads as &quot;p implies (q if and only if r)&quot;, and &quot;(p implies q) if and only if (p implies r).&lt;/p&gt;&#xA;" OwnerDisplayName="user4699" LastEditorDisplayName="user4699" LastEditDate="2014-11-09T01:32:20.517" LastActivityDate="2014-11-09T01:32:20.517" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3853" PostTypeId="1" AcceptedAnswerId="3914" CreationDate="2014-11-09T01:25:32.743" Score="-6" ViewCount="1645" Body="&lt;p&gt;&lt;em&gt;This is part 1 of a series of multiple parts about a takeover of all the computers on earth.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are an elite hacker working very high up in the government; you handle many important tasks. Recently you've been called in for a very special mission, all the computers across the globe are being locked down!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The worlds top leaders unite to figure out what to do, the president of the US, the prime minister of Canada, the Russian premier, the Chinese premier, and many others. You are also invited to attend, as you are known to many as the #1 Techie on the planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While at the meeting you get learn new information, and take the following notes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Nobody can access their computers, this is slowing down large companies, small businesses, even schools.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;New computers being made are affected immediately.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The canadian government has designed a back door program that effectively bypasses the lock and accesses a password screen.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Nobody knows the password.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Last but not least, the hackers cannot be traced back to.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You are tasked with the mission of unlocking a computer, finding the sinister files added, figure out who added them, and stop them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you accept the mission you were given the backdoor computer, and are told that they have an emergency copy in case it breaks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You take the computer back to your office and open it up. Once it turns on you're presented with the loading screen &quot;Loading Glidors Ultimate...&quot; curious enough you decide to look it up online. Then it hits you, you have no access to the Internet!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It starts up and a password screen appears, it asks you to input a password. Obviously not sure what may happen if you enter the incorrect password, you do not try to guess, instead you take out hacking tools you've developed, after attempting to run them, they come out with the error &quot;Unknown OS&quot;. Highly disappointed, you decide to think over a tea. You go to make your tea, and when you get back you notice something weird, some pixels on the screen are different colours when you look at it differently. In fact you can clearly see that the colours are yellow, blue, red.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You take a few minutes to ponder, and then it comes to you! You type in the password and get through the lock screen!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What was the password?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's all numbers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It has to do with RGB&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Final Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RGB Values are measured as 0-1 (decimal)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="5373" LastEditDate="2015-04-29T08:42:32.110" LastActivityDate="2015-04-29T08:42:32.110" Title="Computer Crisis Takeover! [Part 1]" Tags="|password|computer-puzzle|" AnswerCount="10" CommentCount="26" ClosedDate="2015-08-23T05:50:45.277" ContentLicense="CC BY-SA 3.0" />
  <row Id="3855" PostTypeId="2" ParentId="3853" CreationDate="2014-11-09T02:09:21.520" Score="1" Body="&lt;p&gt;The password was &quot;the password&quot;, because in the last line you said he types in &quot;the password&quot; and got through the lock screen!&lt;/p&gt;&#xA;" OwnerUserId="97" LastActivityDate="2014-11-09T02:09:21.520" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="3858" PostTypeId="1" AcceptedAnswerId="3860" CreationDate="2014-11-09T04:57:00.837" Score="3" ViewCount="568" Body="&lt;p&gt;A colleague recently sent you the following puzzle in an email.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;ejsy lomf pg vpsy vsm pm;u nr [iy pm ejrm oy od ery&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What should your reply be?  To make it more fun, reply in the same language.&lt;/p&gt;&#xA;" OwnerUserId="97" LastEditorUserId="5373" LastEditDate="2016-11-20T14:38:42.613" LastActivityDate="2016-11-20T14:38:42.613" Title="Secret messages in the office" Tags="|cipher|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3859" PostTypeId="1" AcceptedAnswerId="3864" CreationDate="2014-11-09T05:32:27.617" Score="12" ViewCount="6786" Body="&lt;p&gt;A man was shot dead in his car. The doors are closed, and locked. The windows are up. The keys are in the ignition. It's clear from the fragmentation around the bullet hole in the front windshield that the shot came from the inside of the car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On inspection of the crime scene it quickly becomes apparent that the killer neither opened nor closed any of the doors or windows. As the killer is no longer in the car, one question remains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did the killer exit the car?&lt;/p&gt;&#xA;" OwnerUserId="2694" LastEditorUserId="5373" LastEditDate="2015-04-03T20:02:28.500" LastActivityDate="2015-06-18T10:56:32.010" Title="Man shot from inside car - how did the killer escape?" Tags="|mystery|" AnswerCount="10" CommentCount="9" ClosedDate="2015-06-18T07:42:02.750" ContentLicense="CC BY-SA 3.0" />
  <row Id="3860" PostTypeId="2" ParentId="3858" CreationDate="2014-11-09T05:46:13.533" Score="6" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; s vpsy pg [somy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which means:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A coat of paint&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The clue is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;what kind of coat can only be put on when it is wet&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The key is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The QWERTY keyboard has been shifted to the right by one key! &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-11-09T14:35:09.357" LastActivityDate="2014-11-09T14:35:09.357" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3861" PostTypeId="2" ParentId="3859" CreationDate="2014-11-09T05:47:30.040" Score="8" Body="&lt;p&gt;The killer was dragged out of the car by police/medics and put into a bodybag. The man shot himself.&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-11-09T06:34:02.370" LastActivityDate="2014-11-09T06:34:02.370" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3862" PostTypeId="2" ParentId="2792" CreationDate="2014-11-09T06:24:47.067" Score="7" Body="&lt;p&gt;Without looking at the &quot;extra equations&quot;, I have come up with this solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Evaluate the sum and look for the corresponding prime number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first few prime numbers are  2      3      5      7     11     13     17     19     23     29. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 + 1 = 2nd prime number = 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 + 2 = 4th prime number = 7&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 + 1 = 3rd prime number = 5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore 5 + 5 = 10th prime number = 29&lt;/p&gt;&#xA;" OwnerUserId="97" LastActivityDate="2014-11-09T06:24:47.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3863" PostTypeId="2" ParentId="3859" CreationDate="2014-11-09T06:45:59.343" Score="2" Body="&lt;p&gt;The killer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; climbed out the sunroof&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is arguably not a window.&lt;/p&gt;&#xA;" OwnerUserId="2637" LastEditorUserId="5373" LastEditDate="2015-04-03T16:50:22.160" LastActivityDate="2015-04-03T16:50:22.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3864" PostTypeId="2" ParentId="3859" CreationDate="2014-11-09T06:50:38.110" Score="33" Body="&lt;p&gt;The car is a convertible like the one below and the killer simply jumped out of the car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UPsku.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="97" LastActivityDate="2014-11-09T06:50:38.110" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3867" PostTypeId="2" ParentId="3823" CreationDate="2014-11-09T07:05:36.483" Score="3" Body="&lt;p&gt;There are 2 solutions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. T, 2. F, 3. T, 4. F, 5. F, 6. F, 7. T, 8. F, 9. T, 10. T, 11. F&lt;br&gt;&#xA; 1. T, 2. F, 3. T, 4. F, 5. F, 6. F, 7. F, 8. F, 9. T, 10. T, 11. F&lt;br&gt;&#xA;&#xA; So, answers 1,3,9 and 10 are definitely true and Rocky is alive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And here is the long story why&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 10. &quot;Exactly one of statements 1 and 2 is false.&quot;&lt;br&gt;&#xA; Rocky is alive or dead.  10 is true.&lt;br&gt;&#xA; &lt;br&gt;&#xA; 9. &quot;Exactly one of statements 6 and 10 is true.&quot;&lt;br&gt;&#xA; 10 is true, so it becomes &quot;Exactly zero of statements 6 is true.&quot; or &quot;6 is false&quot;.&lt;br&gt;&#xA; 6 and 9 have opposite truth values.&lt;br&gt;&#xA; &lt;br&gt;&#xA; 3. &quot;Exactly one of statements 6 and 9 is true.&quot;&lt;br&gt;&#xA; 3 is true because of the statement just above.&lt;br&gt;&#xA; &lt;br&gt;&#xA; 8. &quot;Exactly one of statements 3 and 10 is false.&quot;&lt;br&gt;&#xA; 10 is true, 3 is true, so 8 is false.&lt;br&gt;&#xA; &lt;br&gt;&#xA; 11. &quot;Statements 1, 8 and 11 are all false.&quot;&lt;br&gt;&#xA; If it were true it would say it is false.  =&gt; 11 is false&lt;br&gt;&#xA; To be false, one of 1, 8, 11 must be true.  8 and 11 are false, therefore 1 must be true.&lt;br&gt;&#xA; This proves Rocky is alive.  1 is true and 2 is false.&lt;br&gt;&#xA; &lt;br&gt;&#xA; 5. &quot;Statements 4, 5 and 10 are all false.&quot;&lt;br&gt;&#xA; 10 is true, so 5 is false.&lt;br&gt;&#xA; &lt;br&gt;&#xA; 6. &quot;Exactly one of statements 1 and 10 is false.&quot;&lt;br&gt;&#xA; 1 and 10 are true, so 6 is false.  And 9 is the opposite of 6, 9 is true.&lt;br&gt;&#xA; &lt;br&gt;&#xA; 4. &quot;Exactly one of statements 2 and 6 is false.&quot;&lt;br&gt;&#xA; 2 and 6 are false, so 4 is false.&lt;br&gt;&#xA; &lt;br&gt;&#xA; 7. Exactly 5 statements are true.&lt;br&gt;&#xA; So far, we have:&lt;br&gt;&#xA; 1 true, 2 false, 3 true, 4 false, 5 false, 6 false, 7 ?, 8 false, 9 true, 10 true, 11 false.&lt;br&gt;&#xA; Besides 7, four statements are true.&lt;br&gt;&#xA; If 7 is true then we have five true statements and 7 evaluates to true.&lt;br&gt;&#xA; If 7 is false then we have four true statements and 7 evaluates to false.&lt;br&gt;&#xA; 7 can either true or false.&lt;br&gt;&#xA; &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2014-11-09T13:49:38.463" LastActivityDate="2014-11-09T13:49:38.463" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3870" PostTypeId="2" ParentId="3810" CreationDate="2014-11-09T08:58:40.453" Score="3" Body="&lt;p&gt;A previous answer gave apparently-correct results for parts 1 and 2, but missed out on part 3, which is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;a href=&quot;http://en.wiktionary.org/wiki/yaw#Verb&quot; rel=&quot;nofollow&quot;&gt;yaw&lt;/a&gt;&lt;/strong&gt;, “To swerve off course to port or starboard”,  “To steer badly, zigzagging back and forth across the intended course” etc., thus analogous to &lt;em&gt;deviate&lt;/em&gt;;  and   &lt;strong&gt;&lt;a href=&quot;http://en.wiktionary.org/wiki/way#Noun&quot; rel=&quot;nofollow&quot;&gt;way&lt;/a&gt;&lt;/strong&gt;, “A method or manner of doing something”, thus analogous to &lt;em&gt;method&lt;/em&gt;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2004" LastActivityDate="2014-11-09T08:58:40.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3871" PostTypeId="2" ParentId="3859" CreationDate="2014-11-09T09:19:39.487" Score="10" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There was another person on the crime scene who opened and closed the door for the killer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4394" LastActivityDate="2014-11-09T09:19:39.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3872" PostTypeId="1" AcceptedAnswerId="3873" CreationDate="2014-11-09T09:55:51.340" Score="55" ViewCount="8250" Body="&lt;p&gt;which one is the correct answer of the following picture? What is the algorithm of shapes sequence?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Cdo2q.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="2225" LastEditDate="2015-01-06T06:15:08.707" LastActivityDate="2015-01-06T06:15:08.707" Title="What is the next shape?" Tags="|pattern|visual|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3873" PostTypeId="2" ParentId="3872" CreationDate="2014-11-09T10:12:29.987" Score="79" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is A.&lt;br&gt;&#xA;&lt;p&gt;&#xA; Count the number of straight lines on each shape, and it corresponds to the digits of pi.  The shapes you provided match 3, 1, 4, 1, 5, 9, 2, 6, 5.  The following digit of pi is 3, so the answer is A.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2014-11-09T10:12:29.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3874" PostTypeId="2" ParentId="3831" CreationDate="2014-11-09T10:50:34.817" Score="6" Body="&lt;p&gt;I assume that this question is asking what a stable ratio of dominators to sharers is. I'm also going to assume that their strategy comes from their parent (otherwise there's definitely not enough information to solve the puzzle), and that there at least some of each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at what happens if we have a 50-50 ratio, so 1000 furble using each strategy. Since their numbers are equal, each dominator has a 50-50 chance of encountering a dominator or a sharer, and the same is true for the sharers. So with this generation, we have 250 territories with 2 dominators, 500 territories with a dominator and a sharer, and 250 territories with 2 sharers. With 2 dominators, one will win and rear 10 young while the other has no children. With a dominator and a sharer, the dominator rears 10 young while the sharer rears 3 young. With 2 sharers, they each raise 5 young.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So starting with 1000 of each, we get:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;250*10 + 500*10 = 7500 new dominators&lt;/li&gt;&#xA;&lt;li&gt;500*3 + 250*10 = 4000 new sharers&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Whether or not we include the parents in the final count, the ratio of dominators to sharers has changed, so a 50-50 ratio is not stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather than guess and check at proper ratios, we'll set up some formulas. If there are $x$ dominators and $y$ sharers, then there will be&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\frac{x^2}{2(x+y)}$ dominator pairs&lt;/li&gt;&#xA;&lt;li&gt;$\frac{2xy}{2(x+y)}$ dominator-sharer pairs&lt;/li&gt;&#xA;&lt;li&gt;$\frac{y^2}{2(x+y)}$ sharer pairs&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thus in the next generation there will be&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$X=\frac{10x^2+20xy}{2(x+y)}$ dominators&lt;/li&gt;&#xA;&lt;li&gt;$Y=\frac{6xy+10y^2}{2(x+y)}$ sharers&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Because we want a stable ratio, we want $\frac{x}{y}=\frac{X}{Y}=\frac{10x^2+20xy}{10y^2+6xy}$. Attempting to solve, we get:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$5xy^2+3x^2y=5x^2y+10xy^2$&lt;/li&gt;&#xA;&lt;li&gt;$0=2x^2y+5xy^2$&lt;/li&gt;&#xA;&lt;li&gt;$2x+5y=0$&lt;/li&gt;&#xA;&lt;li&gt;$2x=-5y$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Well, that didn't go so well, but it fits the formulas. If we can somehow have negative numbers of furbles, then we can get a stable ratio. For 2000 total, this means we would want to have -1333.33 sharing furbles and 3333.33 dominating furbles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means that there are only two (trivial) stable solutions - either no sharing furbles or no dominating furbles. If there are ever a (positive) mix of the two, then with each successive generation the dominating furbles become more and more common. As such, I'd expect that if you have 2000 furbles, most of them will be dominators, so most of the initial 1000 pairings have at least one dominator. Thus, I would expect around 1000 furbles to dominate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, I could be overthinking this whole thing - if they're assuming an initial 50-50 ratio, then 750 furbles dominated (the 500 paired with a sharer + the 250 winners of the dominator pairs).&lt;/p&gt;&#xA;" OwnerUserId="1955" LastActivityDate="2014-11-09T10:50:34.817" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3875" PostTypeId="2" ParentId="3859" CreationDate="2014-11-09T13:13:22.737" Score="18" Body="&lt;p&gt;As it is unlikely a suicide inside a car would result in the bullet going through the windshield (driver's side window or roof are much more likely), I'd conclude that the killer was behind the victim in the back seat. If he didn't exit via any door or window, that leaves the trunk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many cars have a way to easily access the trunk from the back seat, so as long as he had a way to open it (say a button on the dashboard), that's the way to go.&lt;/p&gt;&#xA;" OwnerUserId="375" LastActivityDate="2014-11-09T13:13:22.737" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3876" PostTypeId="2" ParentId="3831" CreationDate="2014-11-09T14:06:39.820" Score="4" Body="&lt;p&gt;I didn't know the answer when I posted this question either, but this is the answer I have come up with after reading the attempt by Rob Watts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Rob pointed out, if $x$ is the number of dominators, and $y$ is the number of shares, then for a particular round of breeding, the number of dominator-dominator pairs, the number of dominator-sharer pairs and the number of sharer-sharer pairs is given by the equations below:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\frac{x^2}{2(x+y)}$ dominator pairs&lt;/li&gt;&#xA;&lt;li&gt;$\frac{2xy}{2(x+y)}$ dominator-sharer pairs&lt;/li&gt;&#xA;&lt;li&gt;$\frac{y^2}{2(x+y)}$ sharer pairs&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The total number of children in a single breeding season, $C$ is thus given by,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C=\frac{10x^2+20xy}{2(x+y)} + \frac{6xy+10y^2}{2(x+y)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My assumption is different to Rob's in that instead of assuming that dominators will always breed dominators, and sharers will always breed sharers, I will assume that the probability of being a dominator is $p$ and the probability of being a sharer is $1-p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, if we let $m$ be the total population, the total number of children in a breeding season is given by,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C=\frac{10m^2p^2+20m^2p(1-p)}{2m} + \frac{6m^2p(1-p)+10m^2(1-p)^2}{2m}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which simplifies to,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C = m[-3p^2 + 3p + 5]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, the number of children in a season depends on current population, $m$, and the probability a child will be a dominator $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C$ is then maximized when the derivative, $C' = 0$, that is,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-6p + 3 = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And therefore,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p = 0.5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore to maximize the number of children produced, the probability of being a dominator should be 50%, and therefore there are 1000 dominators and 1000 sharers.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Non mathematical solution:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum number of children will occur when a dominator clashes with a sharer.  This gives 10 children to the dominator and 3 children to the sharer, a total of 13 children.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a dominator clashes with a dominator, a total of 10 children result, and the same is true for when a sharer clashes with a sharer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore you want to maximize the number of dominator-sharer clashes which will occur when you have an equal number of dominators and sharers.&lt;/p&gt;&#xA;" OwnerUserId="97" LastEditorUserId="97" LastEditDate="2014-11-09T14:19:06.697" LastActivityDate="2014-11-09T14:19:06.697" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="3877" PostTypeId="2" ParentId="3833" CreationDate="2014-11-09T14:47:44.800" Score="2" Body="&lt;p&gt;&lt;strong&gt;Answer incomplete.&lt;/strong&gt;  This should probably be finished by a computer search, but I don't know how to do it efficiently enough.  The ideas here may help some with efficiency.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There are three possible tic-tac-toe draw configurations, up to rotation and reflection:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  A         B         C&#xA;o x x     x o x     x o x&#xA;x x o     o o x     x o x&#xA;o o x     x x o     o x o&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Let $a,b,c$ be the number of rows of type A, B, and C respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the first, third, fifth, seventh, and ninth columns represent moves of X, and the second, fourth, sixth, and eigth columns represent moves of O.  Call these columns &quot;X columns&quot; and &quot;O columns&quot; respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total number of 1s, 3s, 7s, and 9s in the X columns must be 20 (4 for each of 5 columns), while the total number of 1s, 3s, 7s, and 9s in the O columns must be 16 (4 for each of 4 columns).  The 1s, 3s, 7s, and 9s in A and C are made up of two Xs and two Os, and in B they are made up of three Xs and one O (1, 3, 7, 9 are the four corner squares in tic-tac-toe).  So we get that the total number in the X columns is $2a + 2c + 3b = 20$, and the total number in the O columns is $2a + 2c + 1b = 16$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, the total number of 2s, 4s, 6s, and 8s in the X columns must be 20, while the total number of 2s, 4s, 6s, and 8s in the O columns must be 16.  The 2s, 4s, 6s, and 8s in A and B are made up of of two Xs and two Os, and in C they are made up of three Xs and one O (2, 4, 6, 8 are the four side squares in tic-tac-toe).  So we get that the total number in the X columns is $2a + 2b + 3c = 20$, and the total number in the O columns is $2a + 2b + 1c = 16$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, we have&#xA;\begin{align}&#xA;2a + 2c + 3b &amp;amp;= 20 \tag{1} \\&#xA;2a + 2c +  b &amp;amp;= 16 \tag{2} \\&#xA;2a + 2b + 3c &amp;amp;= 20 \tag{3} \\&#xA;2a + 2b +  c &amp;amp;= 16 \tag{4}&#xA;\end{align}&#xA;Subtracting (3) from (1), $b - c = 0$.  Subtracting (2) from (1), $2b = 4$.  So $b = 2$, $c = 2$, and $a = 5$...&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="2484" LastEditDate="2014-11-10T02:47:34.427" LastActivityDate="2014-11-10T02:47:34.427" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3878" PostTypeId="2" ParentId="3853" CreationDate="2014-11-09T15:40:20.657" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Only Columbia has a flag that is yellow-blue-red in that order.&lt;br&gt;&#xA; So the password is &quot;columbia&quot;.&lt;br&gt;&#xA; &lt;br&gt;&#xA; There is also Equador.  But with coat of arms. Try that also.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2136" LastActivityDate="2014-11-09T15:40:20.657" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3879" PostTypeId="2" ParentId="3853" CreationDate="2014-11-09T16:13:37.290" Score="1" Body="&lt;p&gt;I have a couple of ideas about this, but from my reading of the puzzle the answer should be obvious rather than an &quot;I think this could be it&quot; kind of response.  That said, here is my response.  I think this could be it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The password is &quot;primary&quot;.  The three pixels are the primary colors and this is also the first (primary) puzzle in the series and likely the first challenge that will have to be overcome by the hacker.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3928" LastActivityDate="2014-11-09T16:13:37.290" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3881" PostTypeId="1" AcceptedAnswerId="3884" CreationDate="2014-11-09T21:24:01.750" Score="14" ViewCount="7380" Body="&lt;p&gt;Four guard towers are situated in a square formation of side length 1km. A general wants to build roads to connect the towers so that one can walk from any tower to any tower along the roads, possibly passing through towers. The size of the towers is too small to matter; think of them as points. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the smallest total length of road that suffices? For example, one can build roads along three of the sides of the square for a total length of 3km, but better is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stating it mathematically, your goal is find a finite collection of line segments with total length as small as possible such that their union is connected and contains points $(0,0),(0,1),(1,0),(1,1)$.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2014-11-09T22:40:36.010" Title="Connect four towers by roads" Tags="|geometry|optimization|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3882" PostTypeId="2" ParentId="3833" CreationDate="2014-11-09T21:24:22.383" Score="13" Body="&lt;h1&gt;Solution:&lt;/h1&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gUW9i.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;h1&gt;Explanation:&lt;/h1&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I noticed that there are only &lt;span class=&quot;math-container&quot;&gt;${9\choose5 }= 126$&lt;/span&gt; ways for the &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt; to play a game.&lt;/li&gt;&#xA;&lt;li&gt;I noticed that only &lt;span class=&quot;math-container&quot;&gt;$28$&lt;/span&gt; of these would not represent a win for &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;I then noticed that of the &lt;span class=&quot;math-container&quot;&gt;$28$&lt;/span&gt; some may be a win for &lt;span class=&quot;math-container&quot;&gt;$O$&lt;/span&gt;. It turns out that after removing those which would be a win for &lt;span class=&quot;math-container&quot;&gt;$O$&lt;/span&gt; there are &lt;span class=&quot;math-container&quot;&gt;$16$&lt;/span&gt; left. i.e. &lt;span class=&quot;math-container&quot;&gt;$(\mbox{all games for X}) \setminus ((\mbox{X wins}) \cup (\mbox{O wins})) = draws \land |draws| = 16$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;So far, this means that are &lt;span class=&quot;math-container&quot;&gt;$16$&lt;/span&gt; ways where (disregarding move order) we have a draw. Since with a draw, and, in particular, in our case, we don't care about move order we are doing OK. Call the moves &lt;span class=&quot;math-container&quot;&gt;$x_1 \cdots x_5$&lt;/span&gt; and put all &lt;span class=&quot;math-container&quot;&gt;$16$&lt;/span&gt; in a set called &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Given a row or column &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; whose elements are &lt;span class=&quot;math-container&quot;&gt;$p_1 \cdots p_9$&lt;/span&gt;, we just want for &lt;span class=&quot;math-container&quot;&gt;$\{p_1,p_3,p_5,p_7,p_9\}\in S$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;I used a powerful SAT/SMT Solver within which I had built a Sudoku solver and added the above additional constraint to every row and column and it produced the above sample solution.  I suspect there are many (I can count them if you really want - that would surely take much time).&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;Notes:&lt;/h2&gt;&#xA;&lt;p&gt;I had some mathematical definition for a draw in tic-tac-toe. I'm sure it's suboptimal, and it did not end up helping to produce the solution. But, I put it here just as a record.&lt;/p&gt;&#xA;&lt;p&gt;A row/column &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; is an array of elements &lt;span class=&quot;math-container&quot;&gt;$(p_1 \cdots p_9)$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$\forall i \not= j.p_k \in \{1,\cdots,9\} \land p_i \not= p_j$&lt;/span&gt;.&#xA;Let player 1 be known as &lt;span class=&quot;math-container&quot;&gt;$pX$&lt;/span&gt;. &lt;span class=&quot;math-container&quot;&gt;$pX$&lt;/span&gt;'s moves are &lt;span class=&quot;math-container&quot;&gt;$X = \{ p_1,p_3,p_5,p_7,p_9\}$&lt;/span&gt;.&#xA;Let player 2 be known as &lt;span class=&quot;math-container&quot;&gt;$pO$&lt;/span&gt;. &lt;span class=&quot;math-container&quot;&gt;$pO$&lt;/span&gt;'s moves are &lt;span class=&quot;math-container&quot;&gt;$O = \{ p_2,p_4,p_6,p_8\}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Let &lt;span class=&quot;math-container&quot;&gt;$pr^n$&lt;/span&gt; be the &lt;span class=&quot;math-container&quot;&gt;$n^{th}$&lt;/span&gt; prime number.&lt;/p&gt;&#xA;&lt;p&gt;No wins across diagonals:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$f1 = \lnot(\exists x \in \{45,105\}. x | \prod X \lor x | \prod O)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;No wins across rows:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$f2 = \lnot(\exists x \in \mathbb{N}.(x^3 | (\prod\limits^{|X|}_{i=0} pr^{\lfloor (X_i-1) / 3 \rfloor}) \lor x^3 | (\prod\limits^{|O|}_{i=0} pr^{\lfloor (O_i-1) / 3 \rfloor})))$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;No wins across columns:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$f3 = \lnot(\exists x \in \mathbb{N}.(x^3 | (\prod\limits^{|X|}_{i=0} pr^{X_i \mod 3}) \lor x^3 | (\prod\limits^{|O|}_{i=0} pr^{O_i \mod 3})))$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;A mapping of values to &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; is a draw if &lt;span class=&quot;math-container&quot;&gt;$f1 \land f2 \land f3$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-11-10T13:33:08.533" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3883" PostTypeId="2" ParentId="3881" CreationDate="2014-11-09T21:31:47.450" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Connect towers diagonally using 2 roads each of length √2 km. You can reach any tower from any other tower using the crossroad in the middle. The total road length will be 2√2 ≈ 2.83 km.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4394" LastActivityDate="2014-11-09T21:31:47.450" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3884" PostTypeId="2" ParentId="3881" CreationDate="2014-11-09T22:26:39.410" Score="22" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The problem is basically asking for a &lt;a href=&quot;http://en.wikipedia.org/wiki/Steiner_tree_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;minimum Steiner tree&lt;/a&gt;. For points located at the corners of a square, the solution consists of five lines. Four of them connect to the corners, and one connects the other four lines in two vertices of degree 3. I found a picture of the solution in &lt;a href=&quot;https://mathoverflow.net/q/93764/56581&quot;&gt;this MathOverflow&lt;/a&gt; question&#xA;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/OYPkX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA; &lt;br&gt;&#xA; So the minimum length is 1 + √3 ≈ 2.73. Note that at the vertices of a Steiner tree (in general) all angles are 2π/3 or 120 degrees, to ensure that each point is at a local minimum.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2450" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:31.357" LastActivityDate="2014-11-09T22:40:36.010" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3885" PostTypeId="2" ParentId="3853" CreationDate="2014-11-09T22:37:57.373" Score="6" Body="&lt;p&gt;Look at Nick R's answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;This is on the right track, but wrong noneoftheless&quot; - OP (warspyking)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Brown because when you mix yellow, red and blue you get brown.&#xA;&lt;a href=&quot;https://answers.yahoo.com/question/index?qid=20120328201707AAZNGvj&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://answers.yahoo.com/question/index?qid=20120328201707AAZNGvj&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit in response to &quot;it's all numbers&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nobody has a answer yet so I will continue with my logic - could be hex code, which the &quot;#1 Techie on the Planet&quot; would definitely know about. Is it a mere coincidence that the hex code for brown (663300) has no letters in it despite the fact that it is shown in a base 16?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;The password is 663300.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also - meta-puzzling:(if this is frowned upon please tell me - I am new to puzzling SE)&#xA;The OP is a programmer (as is shown by multiple tech related SE accounts including Stack Overflow)&#xA;1 of the 7 questions he asks is on SO about contains a single line of  JavaScript and a &lt;strong&gt;hex color code&lt;/strong&gt;. &#xA;&lt;a href=&quot;https://stackoverflow.com/questions/25584703/cannot-change-the-colour-of-html-element-with-javascript&quot;&gt;https://stackoverflow.com/questions/25584703/cannot-change-the-colour-of-html-element-with-javascript&lt;/a&gt;&#xA;This solution seems likely to me.&lt;/p&gt;&#xA;" OwnerUserId="4802" LastEditorUserId="-1" LastEditDate="2017-05-23T12:38:53.617" LastActivityDate="2014-11-09T23:52:08.760" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3887" PostTypeId="2" ParentId="3845" CreationDate="2014-11-09T23:33:36.337" Score="17" Body="&lt;p&gt;Let's decode the briefcase numbers first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You mention that the codes are 8 numbers, 4 numbers, and 2 numbers respectively. My best guesses as to how to solve them are as follows.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;There's nothing quite relevant to the letter &quot;C&quot; (in the form of the programming language, the roman numeral, or anything else) that has 8 digits in it. However, the hint mentions &quot;bytes&quot; and a byte is 8 bits. So we can try converting &quot;C&quot; into ASCII and then representing it in binary.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This gives &lt;code&gt;01000011&lt;/code&gt;, which is an 8-digit code.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;ADIF is not a well-known acronym relating to anything. However, all the letters come before J in the alphabet. So, we can try converting &quot;ADIF&quot; using the A = 1, B = 2 cipher.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This gives &lt;code&gt;1496&lt;/code&gt;, which is a 4-digit code.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;A quick Google of Deep Thought reveals that the name belongs to &lt;a href=&quot;http://en.wikipedia.org/wiki/Deep_Thought_%28chess_computer%29&quot;&gt;one of the chess machines built by IBM&lt;/a&gt;. However, no two-digit numbers stand out as especially important (for example, finding a checkmate in 37 moves, winning 63 games in its lifetime, being built in '88).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second result reveals that Deep Thought is also the name of &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_minor_The_Hitchhiker%27s_Guide_to_the_Galaxy_characters#Deep_Thought&quot;&gt;the computer in The Hitchhiker's Guide to the Galaxy&lt;/a&gt; that was given the ultimate question of life, the universe, and everything. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Here, the two-digit number that stands out is &lt;code&gt;42&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus, the three codes to the briefcase are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;01000011&lt;/code&gt; &lt;code&gt;1496&lt;/code&gt; &lt;code&gt;42&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now that we've opened the briefcase, there's a New International Version Bible inside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You mention that you vaguely hear a phrase from somebody as &quot;end of six&quot;, but that phrase is not homophonous or phonetically similar to any single word I could find, so it can probably be interpreted as &quot;end of six&quot; literally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Referring back to the &quot;Romans 7&quot; clue in the first riddle, my first guess is that &quot;end of six&quot; refers to the end of &lt;a href=&quot;https://www.biblegateway.com/passage/?search=Romans+7:6&amp;amp;version=NIV&quot;&gt;Romans 7:6&lt;/a&gt;, which reads in the NIV as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;sup&gt;&lt;strong&gt;6&lt;/strong&gt;&lt;/sup&gt; But now, by dying to what once bound us, we have been released from the law so that we serve in the new way of the Spirit, and &lt;strong&gt;not in the old way of the written code.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So if my interpretation is correct, you must do something with one of the written codes, possibly the one from part 1. I still haven't figured out what, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, it could simply have referred to the &quot;end of six minutes&quot;, as you mentioned that he found the briefcase at 7:36 PM, six minutes after 7:30.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2014-11-10T01:06:51.907" LastActivityDate="2014-11-10T01:06:51.907" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3888" PostTypeId="2" ParentId="3853" CreationDate="2014-11-09T23:43:02.803" Score="3" Body="&lt;p&gt;If the pw is only numbers and looking at the Nick R answer.. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;.. the password is the values of a color on the screen in RGB. According to the text, yellow is the only one different from the classic green-red-blue, so I could suppose the password is &lt;strong&gt;2552550&lt;/strong&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe the password is too easy right now. Adding some numbers..&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; .. the result could be &lt;strong&gt;255000000255255000000000255&lt;/strong&gt;, the values of the R[ed]-Y[ellow]-B[lue] pixels of the screen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not sure about that one, it's riddling me &quot;Unknows OS&quot;, but maybe it's just a sensation.&lt;/p&gt;&#xA;" OwnerUserId="1685" LastEditorUserId="1685" LastEditDate="2014-11-10T09:10:06.287" LastActivityDate="2014-11-10T09:10:06.287" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3889" PostTypeId="2" ParentId="3859" CreationDate="2014-11-09T23:46:22.200" Score="2" Body="&lt;p&gt;Most cars nowadays are locked with electronic gizmos, not keys. The vicious killer probably just took that off and locked the car to delay the police. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, his wife killed him. She obviously has a spare key.&lt;/p&gt;&#xA;" OwnerUserId="4805" LastActivityDate="2014-11-09T23:46:22.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3893" PostTypeId="2" ParentId="3853" CreationDate="2014-11-10T02:05:19.463" Score="2" Body="&lt;p&gt;In the additive colour model (which is what to expect from pixels), the colours Red + Blue + Yellow combine to form the colour White.  You initially just saw this white light, before you looked at it more carefully or differently and saw the underlying pixels of red, blue and yellow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You decide to try the password of the colour white in RGB notation, which is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;255255255&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or alternatively, depending on your colour mixing algorithim could yield (170,085,085) and therefore the password could be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;170085085&lt;/p&gt;&#xA;" OwnerUserId="97" LastEditorUserId="97" LastEditDate="2014-11-10T02:18:34.497" LastActivityDate="2014-11-10T02:18:34.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3894" PostTypeId="1" AcceptedAnswerId="3900" CreationDate="2014-11-10T02:15:30.913" Score="13" ViewCount="1486" Body="&lt;p&gt;I dropped in on Ernie a while ago and he told me that there was &quot;good news and bad news&quot;. I asked him to tell me about it and he explained as follows: &quot;Remember my Casimir-effect electric motor and how it always used to overheat? The good news is that I have increased its efficiency, so now it won't overheat any more&quot;. &quot;And the bad news?&quot;, I inquired. &quot;Unfortunately, I got a little carried away and improved things too much. If I ran it at 10 volts and 10 amps, I used to get 99 watts mechanical output and some waste heat, but now I get 101 watts of work out of it&quot;. &quot;But that's wonderful!&quot;, I exclaimed. &quot;Free energy. How can that be bad news?&quot;. &quot;The problem is&quot;, Ernie explained, &quot;the extra energy has to come from somewhere. Now as soon as I start the motor it begins to cool down. After a while it gets so cold that it seizes up and stops working. What a disaster!&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;But couldn't you just add a heater?&quot;, I asked. Ernie paused for a moment. &quot;Of course!&quot; he cried. &quot;You're a genius. And I can use my new Acme Resist-a-Pak; it just arrived in the mail yesterday&quot;. He pulled a brightly coloured box from a drawer and lifted its lid. Inside were a number of components and a sheet of paper labelled &quot;&lt;strong&gt;&lt;em&gt;INSTRUCTIONS&lt;/em&gt;&lt;/strong&gt;&quot;. I read the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Contents:&lt;/strong&gt; Thirty-one high-precision resistors - each precisely 1.000000 Ohms. One programmable voltage supply - 0.000000 to 9.999999 volts. Sixteen orthogonal connector brackets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;To use:&lt;/strong&gt; No solder required! Just clip resistors and voltage supply together, using the orthogonal connector brackets at the corners, to produce the resistor network of your choice. Set operating voltage on the power supply and operate as required. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Caution:&lt;/strong&gt; Orthogonal connector brackets can connect two or more components together, but only at 90 degree angles. &lt;strong&gt;&lt;em&gt;Do not try to force them into other orientations or damage will occur!!!&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TK0bV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Hmmm&quot;, I mused, &quot;if you take two identical resistors (both oriented the same way) then slide them apart and join their ends with one more resistor and the voltage supply (both oriented orthogonally to the other two resistors), you will end up with a square network with resistance of 3 Ohms. You could use sqrt(3) volts to produce 1 watt of heating - just what your motor needs.&quot; Ernie did a few thermal calculations and told me that it wouldn't work. The square network would make the sides of the motor a lot too hot, and the top and bottom would be much too cold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Well, I continued, &quot;How about making two identical squares of four resistors each (both oriented the same way) slide them apart and connect their corners with three more resistors and one voltage supply (all oriented orthogonally to the other eight resistors). You will end up with a cubic network and I am sure I can work out its resistance and what voltage you need to get 1 watt of heating.&quot; I started to calculate the resistance of the network, and had just got the answer when Ernie told me that it wouldn't work. The cubical network would make the outside of the motor a little too hot and the insides would be a little too cold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;There doesn't seem to be a solution then&quot;, I muttered. &quot;But wait&quot; Ernie replied, &quot;your ideas might just work. If I make two identical cubes of twelve resistors each (both oriented the same way) I can slide them apart and connect their vertices with seven more resistors and the voltage supply (all oriented orthogonally to the other 24 resistors, of course). I calculate that the heat will be even enough to avoid hot or cold spots on the motor - the temperature will be just right everywhere. And there are just enough parts in the Pak to do the job.&quot; I was about to point out the impossibility of producing that geometry when I got an emergency phone call from home - the cat was stuck in the tree again - and I had to rush off to help. By the time I had solved &lt;em&gt;that&lt;/em&gt; problem, and put away the chain-saw and shot-gun, it was too late to get back to Ernie's shed, but I did sit down in the evening and calculated the resistances of my cubical 11 resistor network and Ernie's weird 31 resistor network (just for fun) - then texted him the results. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my surprise, I got a message from Ernie the next morning to let me know that 'my' idea had worked perfectly - the motor had run all night without over- or under-heating. This all happened a while ago and I can't remember the two resistances I calculated - maybe you can help.&lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="39" LastEditDate="2014-11-11T22:11:53.730" LastActivityDate="2014-11-11T22:11:53.730" Title="Ernie and the Orthogonal Network" Tags="|mathematics|calculation-puzzle|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="3895" PostTypeId="2" ParentId="3726" CreationDate="2014-11-10T02:28:31.327" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are the killer, and you are in fact a serial killer - this is why you're not surprised to hear of the murder.  All 50 of your previous 'solved' murders were also you.  As a result you have become quite proficient at planting evidence.  The knife is not the murder weapon, the shoe is not yours, the blood is from a poodle you killed for the purpose of creating a false lead, and the note was designed to make it seem like the victim knew the killer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The only question is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  which of the suspects you will arrest for the murder you committed.  Possibly none of them, since you're leaving town and wanted to leave an unsolvable case for your colleagues.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1667" LastActivityDate="2014-11-10T02:28:31.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3896" PostTypeId="2" ParentId="476" CreationDate="2014-11-10T03:40:59.837" Score="2" Body="&lt;p&gt;The definitive books on chess retrograde analysis are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Raymond_Smullyan&quot; rel=&quot;nofollow&quot;&gt;Raymond Smullyan&lt;/a&gt; classics &quot;The Chess Mysteries of Sherlock Holmes&quot; (1979) and &quot;The Chess Mysteries of the Arabian Knights.&quot; (1981)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Retrograde Analysis puzzles are where you are given a finished state of the chess board and have to work backwards from that state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may have to prove who moved last, or what the last move was, or even what way around the board is! They are logic problems, not chess problems, though they take place within the confines of a normal chess board and within its rule structure, so a working knowledge of chess is necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are very different than traditional chess puzzles that ask you to take a certain state and work it forwards to a check-mate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are also fiendishly difficult to get good at!&lt;/p&gt;&#xA;" OwnerUserId="4815" LastActivityDate="2014-11-10T03:40:59.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3897" PostTypeId="5" CreationDate="2014-11-10T04:27:16.463" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-11-10T04:27:16.463" LastActivityDate="2014-11-10T04:27:16.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3898" PostTypeId="4" CreationDate="2014-11-10T04:27:16.463" Score="0" Body="For puzzles that involve solving a crime." OwnerUserId="1711" LastEditorUserId="1711" LastEditDate="2014-11-10T07:32:52.627" LastActivityDate="2014-11-10T07:32:52.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3899" PostTypeId="2" ParentId="3894" CreationDate="2014-11-10T04:37:45.233" Score="4" Body="&lt;p&gt;Since one of Penguino's comments mentions offhand that Ernie's inventions violate all sorts of physics, I think it should be safe to assume that the &quot;weird 31 resistor network&quot; is what results from taking two cubes of resistors and connecting them orthogonally vertex-by-vertex &lt;a href=&quot;http://en.wikipedia.org/wiki/Tesseract&quot; rel=&quot;nofollow&quot;&gt;via a fourth spatial dimension&lt;/a&gt;, then replacing one edge with the voltage supply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now we just need to calculate the equivalent resistance between two adjacent nodes of a 3D and 4D cube of resistors, with the connecting edge shorted out.  Luckily, I managed to find &lt;a href=&quot;http://www.wpi.edu/Pubs/E-project/Available/E-project-032913-185209/unrestricted/networks.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; where there are values (pg. 28) calculated for any pair of nodes for cubes up to 9D.  We're interested in two adjacent nodes, so we can just read the first row of the table.  Then, to take into account the voltage source replacing the connecting resistor, I believe you should get answers of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 7/5 and 15/17&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;for the resistances.&lt;/p&gt;&#xA;" OwnerUserId="4819" LastEditorUserId="4819" LastEditDate="2014-11-10T05:30:43.547" LastActivityDate="2014-11-10T05:30:43.547" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3900" PostTypeId="2" ParentId="3894" CreationDate="2014-11-10T04:43:46.620" Score="6" Body="&lt;p&gt;For those who are interested in the problem but having difficulty visualizing it, I have sketched out the circuits as I understand them.  The 31 resistor circuit is not exactly as Ernie built, but I ran out of 8 legged orthogonal connectors, so I had to construct it using the boring 6 legged variety and some jumper wire.  The colors are unimportant, I only included them to keep the image from becoming completely uninterpretable visually.  I have posted visual solutions farther down, this first spoiler box is only the circuits with no additional information included.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/RDASL.jpg&quot; alt=&quot;Drawing of the circuits&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The following is my solution for the 11 resistor problem.  I'm still working on the 31, but I'm probably going to have to go to bed for the night.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I found the points of equipotential and remapped the circuit so that they were shorted.  From there it was a matter of finding the equivalent resistance of the remaining paths between those points.  $R_{11} = 7/5\ \Omega$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is a drawing of the solution to the 11 resistor problem.  I have separated it from the 31 resistor problem in case someone wants to use this as a &lt;em&gt;very&lt;/em&gt; strong hint in solving the second half without giving away the 31 resistor solution.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/IIthM.jpg&quot; alt=&quot;Solution to the 11 resistor problem&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Gotta love nerd sniping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I couldn't put this down, so here is my solution to the 31 resistor problem.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The 31 resistor problem is very similar, but it has more points of symmetry and more connections, so it's kind of a mess to draw.  I'm not entirely sure I did this one correctly, but the answer is reasonable.  I got $R_{31} = 15/17\ \Omega$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is a diagram of the solution to the 31 resistor problem.  It's ugly, but due to the number of points of symmetry it actually winds up being a pretty straightforward solution.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/21Ynv.jpg&quot; alt=&quot;31 Resistor Solution&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I love Ernie problems.&lt;/p&gt;&#xA;" OwnerUserId="3928" LastEditorUserId="3928" LastEditDate="2014-11-10T18:47:42.750" LastActivityDate="2014-11-10T18:47:42.750" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3901" PostTypeId="2" ParentId="3853" CreationDate="2014-11-10T04:55:03.720" Score="2" Body="&lt;p&gt;If the clue is the yellow pixel, then the solution must be the computer representation of yellow. Since we know 31337 hackers would never pick a short password, it must be as long as possible, hence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;255255000&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;if they're using the CMYK scheme: 000255000&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2501" LastActivityDate="2014-11-10T04:55:03.720" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3902" PostTypeId="2" ParentId="3859" CreationDate="2014-11-10T05:01:48.747" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Killer shot the driver. The driver's death caused an accident which caused the killer (who was not wearing a seatbelt) to be thrown from the car through the windshield.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2501" LastActivityDate="2014-11-10T05:01:48.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3903" PostTypeId="5" CreationDate="2014-11-10T05:33:24.780" Score="0" Body="&lt;p&gt;A mystery puzzle is a puzzle that involves figuring out the exact details of an incident or a specific desired detail, given limited relevant information. Mystery puzzles usually take the form of solving a crime, often murder. Such puzzles are a subset of the &lt;a href=&quot;/questions/tagged/word-problem&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;word-problem&amp;#39;&quot; rel=&quot;tag&quot;&gt;word-problem&lt;/a&gt; genre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please be careful that your mystery puzzle is not a &lt;a href=&quot;https://en.wikipedia.org/wiki/Situation_puzzle&quot; rel=&quot;nofollow noreferrer&quot;&gt;situation riddle&lt;/a&gt;, in which only a situation is given with no context or evidence. Those problems require active question-asking on the part of the participant to answer. Make sure that the correct answer is clearly deducible given your problem statement alone.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="11110" LastEditDate="2017-06-21T07:32:13.427" LastActivityDate="2017-06-21T07:32:13.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3904" PostTypeId="4" CreationDate="2014-11-10T05:33:24.780" Score="0" Body="A puzzle that involves figuring out what happened with regard to a certain incident through analyzing clues and deducing conclusions." OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-04-29T00:57:02.810" LastActivityDate="2015-04-29T00:57:02.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3905" PostTypeId="2" ParentId="3718" CreationDate="2014-11-10T05:52:03.100" Score="0" Body="&lt;p&gt;We will assume that we want to find the largest number of balls that can fit in the box when they are hexagonal close packed.  The other answers have ignored the edge effects, using the  density of balls in an infinite lattice.  Let the balls be aligned in the $x$ direction, so each alternate horizontal line on the first layer will hold $100$ and the ones between will hold $99$ balls.  The vertical spacing between the rows is $\frac {\sqrt 3}2 \approx 0.8660254$, so we get $\left \lfloor \frac {99 \cdot 2}{\sqrt 3}\right \rfloor +1=115$ rows.  That gives us $58 \cdot 100 + 57 \cdot 99=11443$ on the bottom layer.  The next layer also has $115$ rows, but they alternate the other way and we only get $ 11442$ balls  We get $115$ layers, so in total we get $58*11443+57*11442=1315888.$  We can probably get closer to the density limited $1414213$ by using a smarter configuation, but I am sure we won't get all the way there.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="2186" LastEditDate="2014-11-10T19:08:24.987" LastActivityDate="2014-11-10T19:08:24.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3906" PostTypeId="2" ParentId="3853" CreationDate="2014-11-10T06:23:20.047" Score="1" Body="&lt;p&gt;What if we concatenate the decimal representations of all three colors, in the order they appear on the screen?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Yellow is 2552550&lt;/li&gt;&#xA;&lt;li&gt;Blue is 00255&lt;/li&gt;&#xA;&lt;li&gt;Red is 25500&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If each of the RBG values needs to be three digits long, the values are 255255000, 000000255, and 255000000.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my guess is either &lt;code&gt;25525500025525500&lt;/code&gt; or &lt;code&gt;255255000000000255255000000&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="294" LastActivityDate="2014-11-10T06:23:20.047" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3907" PostTypeId="5" CreationDate="2014-11-10T07:21:50.307" Score="0" Body="&lt;p&gt;Ernie seems to be an inventor in the tradition of Tesla and Brown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ernie's inventions often violate thermodynamics, general relativity, and quantum physics (but are seldom in bad taste). He is just that kind of guy.&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditDate="2014-11-10T16:51:42.557" LastActivityDate="2014-11-10T16:51:42.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3908" PostTypeId="4" CreationDate="2014-11-10T07:21:50.307" Score="0" Body="Ernie's inventions often violate thermodynamics, general relativity, and quantum physics (but are seldom in bad taste). He is just that kind of guy." OwnerUserId="68" LastEditorUserId="68" LastEditDate="2014-11-10T15:59:07.150" LastActivityDate="2014-11-10T15:59:07.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3909" PostTypeId="2" ParentId="3600" CreationDate="2014-11-10T10:00:59.820" Score="2" Body="&lt;p&gt;The note reads&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;Next clue at EWR Southwest A&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;Your Friend,&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;The Killer&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; 'Romans 7' hints that a &lt;a href=&quot;http://en.wikipedia.org/wiki/Caesar_cipher&quot; rel=&quot;nofollow&quot;&gt;Caesar cipher&lt;/a&gt; was used (named after the &lt;em&gt;Roman&lt;/em&gt; emperor Julius Caesar), with a shift of &lt;em&gt;7&lt;/em&gt; to the right. To unshift, we must either shift 7 to the left, or 26 - 7 = 19 to the right.&lt;br&gt;&#xA;&lt;br&gt;&#xA; 'EWR' is the IATA code for Newark Liberty International Airport. Southwest Airlines uses terminal A at Newark, so we may need to take a look there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="68" LastActivityDate="2014-11-10T10:00:59.820" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3910" PostTypeId="2" ParentId="3853" CreationDate="2014-11-10T10:44:18.710" Score="1" Body="&lt;p&gt;Colors: {yellow, blue, red} (note all following are numbers)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Making assumption that password should be easy to remember and hard to crack with &lt;a href=&quot;http://en.wikipedia.org/wiki/Brute-force_attack&quot; rel=&quot;nofollow&quot;&gt;brute force&lt;/a&gt; most likely password would would be :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;FFFF000000FFFF0000&lt;/strong&gt; (hackers would see &lt;strong&gt;F&lt;/strong&gt; as a number). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On off chance i would want to pick a password with only numbers it would be : &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;101010100101010101010101&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning behind my decision would be &lt;strong&gt;guess = what it is (cracking time with average pc)&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;111111111111111100000000000000000000000011111111111111110000000000000000 =  concatenate 24-bit RGB values as binary value (7 septendecillion years)&lt;/li&gt;&#xA;&lt;li&gt;FFFF000000FFFF0000 = concatenate 24-bit RGB values as hexadecimal value (81 billion years)&lt;/li&gt;&#xA;&lt;li&gt;255255000000000255255000000 = concatenate 24-bit RGB values  (7 billion years)&lt;/li&gt;&#xA;&lt;li&gt;1111111101111111111111111 = sum of 24-bit RGB values as binary value (79 million years)&lt;/li&gt;&#xA;&lt;li&gt;101010100101010101010101 = avarage 24-bit RGB values as binary value (7 million years)&lt;/li&gt;&#xA;&lt;li&gt;4722294425279902187520 = concatenate 24-bit RGB values as base 10 number (79 thousand years)&lt;/li&gt;&#xA;&lt;li&gt;570590450475625750 = concatenate start and end of color wavelength (7 years)&lt;/li&gt;&#xA;&lt;li&gt;25525500025525500 = concatenate 24-bit RGB values (289 days)&lt;/li&gt;&#xA;&lt;li&gt;1111111111111111 = bitwise and 24-bit RGB values as binary value (28 days)&lt;/li&gt;&#xA;&lt;li&gt;5702045025625125 = concatenate start and diapason of color wavelength (28 days)&lt;/li&gt;&#xA;&lt;li&gt;1feffff = sum of 24-bit RGB values as hexadecimal value (19 seconds)&lt;/li&gt;&#xA;&lt;li&gt;00FFFF = bitwise and 24-bit RGB values as hexadecimal value (0.544195584 seconds)&lt;/li&gt;&#xA;&lt;li&gt;AA5555 = avarage 24-bit RGB values as hexadecimal value (0.544195584 seconds)&lt;/li&gt;&#xA;&lt;li&gt;33488895 = sum of 24-bit RGB values (0.025 seconds)&lt;/li&gt;&#xA;&lt;li&gt;0255255 = bitwise and 24-bit RGB values (0.0025 seconds)&lt;/li&gt;&#xA;&lt;li&gt;1708585 = avarage 24-bit RGB values (0.0025 seconds)&lt;/li&gt;&#xA;&lt;li&gt;602400 = concatenate hue HSV/HSL (0.00025 seconds)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4834" LastEditorUserId="4834" LastEditDate="2014-11-10T10:52:22.687" LastActivityDate="2014-11-10T10:52:22.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3914" PostTypeId="2" ParentId="3853" CreationDate="2014-11-10T11:19:43.120" Score="3" Body="&lt;p&gt;Given everyone else's work and the 3 hints we now have, we know that we're looking for an RGB representation of yellow (as that's the only non-standard pixel colour [ie. not red, green or blue]) in the 0-1 scale format (rather than the 0-255 format).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the 255-scale representation is &lt;code&gt;255, 255, 0&lt;/code&gt;, the 0-1 scale representation would be &lt;strong&gt;&lt;code&gt;110&lt;/code&gt;&lt;/strong&gt; (or &lt;code&gt;1 1 0&lt;/code&gt;, or &lt;code&gt;1,1,0&lt;/code&gt;).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Update: Since we're apparently not just looking for the only non-standard pixel colour, we're either looking for:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RGB notation for Yellow, Red, Blue: &lt;strong&gt;110001100&lt;/strong&gt;&lt;br&gt;&#xA;or&lt;br&gt;&#xA;RGB notation for mixing Yellow, Red, Blue: &lt;strong&gt;492449&lt;/strong&gt; (assuming 00-99 to avoid decimals)&lt;/p&gt;&#xA;" OwnerUserId="2518" LastEditorUserId="2518" LastEditDate="2014-11-10T11:32:36.280" LastActivityDate="2014-11-10T11:32:36.280" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="3922" PostTypeId="1" AcceptedAnswerId="3931" CreationDate="2014-11-10T12:13:05.930" Score="33" ViewCount="6212" Body="&lt;p&gt;A car starts distance 1 from a wall then drives away at constant speed $ c $. There is a length of elastic tied between the wall and the car. Remarkably this doesn't affect the motion of the car (in reality the tension would slow the car). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, an ant crawls from the wall onto the elastic and then towards the car. On solid ground, the speed of the ant is $ a $. Needless to say, ants are slower than cars $ a \ll c $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will the ant ever reach the car? Assume the elastic stretches without breaking.&lt;/p&gt;&#xA;" OwnerUserId="4508" LastEditorUserId="4508" LastEditDate="2014-11-10T16:32:23.707" LastActivityDate="2019-02-03T13:48:25.457" Title="Will the ant reach the car?" Tags="|mathematics|" AnswerCount="8" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="3924" PostTypeId="2" ParentId="3922" CreationDate="2014-11-10T12:29:27.523" Score="20" Body="&lt;p&gt;This is the&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Ant_on_a_rubber_rope&quot;&gt;Ant on a rubber rope problem&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The ant will eventually reach the car. The key to understand this is that the proportion between the total length of the elastic rope and the length walked by the ant can only increase with time, as it will eventually approach one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sorry if my short solution makes this question seem too trivial. It is not, and in fact, the Wikipedia page I linked in the spoiler has a lengthy explanantion that covers both the mathematical aspect and the (un)intuitive approach to understand the problem better than I could explain here.&#xA;Maybe an important detail not explicitly mentioned in the problem statement is that the ant has plenty of (unlimited) time to reach the car.&lt;/p&gt;&#xA;" OwnerUserId="4394" LastEditorUserId="4394" LastEditDate="2014-11-10T13:01:29.977" LastActivityDate="2014-11-10T13:01:29.977" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="3926" PostTypeId="2" ParentId="3922" CreationDate="2014-11-10T13:11:52.000" Score="8" Body="&lt;p&gt;Let x(t) be the position of the ant at time t and l(t) the total length of the rubber band.&#xA;Since the car is moving, the band will distort over time and this will affect the position of the ant. Assuming the elastic band distorts uniformly, we can say that :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;dl(t)/dt = c and l(t) = 1 + ct&lt;/li&gt;&#xA;  &lt;li&gt;dx(t)/dt = a + x(t)/l(t) . dl(t)/dt&lt;/li&gt;&#xA;  &lt;li&gt;dx(t)/dt = a + c.x(t)/(1 + ct)&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We now have to solve the differential equation&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;y' + c/(1 + cx).y = a&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The general solution of the homogeneous equation is &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;x&lt;sub&gt;g&lt;/sub&gt;(t) = k.(1 + ct)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using the constant variation technique, we can find a particular solution of the differential equation :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;x&lt;sub&gt;h&lt;/sub&gt;(t) = a(ct+1).ln(ct+1)/c&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The overall solution is thus :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;x(t) = x&lt;sub&gt;s&lt;/sub&gt;(t) + x&lt;sub&gt;g&lt;/sub&gt; = k(ct + 1) + a(ct+1).ln(ct+1)/c&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We also have x(0) = 0, thus, x(0) = k = 0, and :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;x(t) = a(ct+1).ln(ct+1)/c&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally we can find the time t such that x(t) = l(t).&lt;br&gt;&#xA;To do this, we must solve :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a.ln(ct+1) = c&lt;br&gt;&#xA;  ln(ct+1) = c/a&lt;br&gt;&#xA;  t = (e&lt;sup&gt;c/a&lt;/sup&gt; - 1)/c&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4849" LastEditorUserId="4849" LastEditDate="2014-11-10T22:08:26.763" LastActivityDate="2014-11-10T22:08:26.763" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3928" PostTypeId="1" AcceptedAnswerId="3944" CreationDate="2014-11-10T13:50:29.303" Score="10" ViewCount="1728" Body="&lt;p&gt;A car starts distance 1 from a wall then drives away at constant &lt;strong&gt;acceleration&lt;/strong&gt; &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt;. There is a length of elastic tied between the wall and the car. Remarkably, this doesn't affect the motion of the car (in reality the elastic would snap or its tension slow the car).&lt;/p&gt;&#xA;&lt;p&gt;Meanwhile, an ant crawls from the wall on to the elastic then towards the car. On solid ground, the speed of the ant is &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Will the ant ever reach the car? Assume the elastic can stretch infinitely.&lt;/p&gt;&#xA;&lt;p&gt;(Note: Assume that the laws of Newtonian mechanics apply, and that there are no effects with relativity.)&lt;/p&gt;&#xA;&lt;p&gt;This question is related to a &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3922&quot;&gt;previous puzzle&lt;/a&gt;, posted by &lt;a href=&quot;https://puzzling.stackexchange.com/users/4508&quot;&gt;Colonel Panic&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="97" LastEditorUserId="69582" LastEditDate="2021-08-09T17:20:51.180" LastActivityDate="2021-08-09T17:20:51.180" Title="Will The Ant Ever Reach The Car [Part 2]" Tags="|mathematics|" AnswerCount="6" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="3930" PostTypeId="2" ParentId="3928" CreationDate="2014-11-10T14:11:45.673" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are 3 possibilities for constant acceleration; Constant positive acceleration, zero acceleration, constant negative acceleration. If the car were already moving and we are using a negative acceleration then the car would eventually stop and the acceleration would no longer be constant so this isn't the case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the acceleration were positive the car would eventually reach the speed of light which should be impossible even in this universe where there is no force associated with an infinitely elastic rubber band and cars can run infinitely without recharging or refueling. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Therefor the car's acceleration is 0 which is constant and since it has always been zero the car has not moved and the ant will make it his speed of a or it was already moving and the problem becomes the same as the one this question was derived from&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2803" LastActivityDate="2014-11-10T14:11:45.673" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3931" PostTypeId="2" ParentId="3922" CreationDate="2014-11-10T14:33:24.473" Score="8" Body="&lt;p&gt;Yes, the ant will reach the car. Calculus tells us exactly when:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let C(t) be the position of the car at time t. The car travels at constant speed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ C(t) = 1 + ct $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let A(t) be the position of the ant at time t. The speed of the ant is its speed on ordinary ground, plus a proportion of the car's speed (according to how far along the elastic it is):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{dA}{dt} = a + { A \over { 1 + ct} } c = a + uc $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Making the natural substition $ u = {A  \over {1+ct} } $ for the ants position as a proportion of the elastic. Then by the product rule&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{du}{dt} = { a \over {1+ct } } $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which has solution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ u = { a \over c } \log(1 + ct)  $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recalling that u is what proportion the ant is along the elastic, the ant reaches the car when u = 1 at&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ t = { {e^{c / a} - 1 } \over c } $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: fixed missing c in exponent&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'd love to see an intuitive argument that the ant will reach the car without appealing to calculus.&lt;/p&gt;&#xA;" OwnerUserId="4508" LastEditorUserId="4508" LastEditDate="2015-06-15T10:33:19.240" LastActivityDate="2015-06-15T10:33:19.240" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3932" PostTypeId="1" AcceptedAnswerId="3933" CreationDate="2014-11-10T14:55:44.403" Score="31" ViewCount="5169" Body="&lt;p&gt;Jenny works in an office with four other colleagues and someone has been using her milk. She finds this very annoying as she likes her mid-afternoon tea and she simply can't enjoy the second half of her day if she has to drink black tea instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deciding that enough is enough Jenny decides to do something about it, she's suspicious of all of her four colleges Liz, John, Sarah and Bill so she decides she's going to solve the crime once and for all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Her first course of action is to make a note of which days her milk is used and which days her coworkers are in the office.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Monday - JBS - THEFT!&lt;/li&gt;&#xA;&lt;li&gt;Tuesday      - JLBS  - THEFT!&lt;/li&gt;&#xA;&lt;li&gt;Wednesday    - JLBS  - THEFT!&lt;/li&gt;&#xA;&lt;li&gt;Thursday - LBS   - No theft&lt;/li&gt;&#xA;&lt;li&gt;Friday       - LB    - No theft&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Next, deciding to forsake her regular duties for the sake of office security she follows various members of the office for a day over the next two weeks. Alas the thief evades detection.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Monday       - Sarah&lt;/li&gt;&#xA;&lt;li&gt;Tuesday      - Liz&lt;/li&gt;&#xA;&lt;li&gt;Wednesday    - John&lt;/li&gt;&#xA;&lt;li&gt;Thursday - Bill&lt;/li&gt;&#xA;&lt;li&gt;Friday       - Liz&lt;/li&gt;&#xA;&lt;li&gt;Monday       - Sarah&lt;/li&gt;&#xA;&lt;li&gt;Tuesday      - Jenny has to attend a meeting and fails to follow John all day&lt;/li&gt;&#xA;&lt;li&gt;Wednesday    - Liz&lt;/li&gt;&#xA;&lt;li&gt;Thursday - Liz&lt;/li&gt;&#xA;&lt;li&gt;Friday       - Bill&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To her interest the thefts occur on the same days each week (Monday, Tuesday and Wednesday).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the second Friday she's at her wit's end, she's still not caught the thief - is she doomed to a lifetime of black tea!?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;She explains her dilemma to her friend one evening who (also being British) realises the seriousness of the situation. Pouring over Jenny's copious notes it looks like a lost cause until her friend smiles. She knows who the thief is!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who is responsible for the milk thefts?&lt;/p&gt;&#xA;" OwnerUserId="2466" LastEditorUserId="15544" LastEditDate="2015-09-17T07:42:27.333" LastActivityDate="2018-11-29T19:41:36.487" Title="Who's been stealing Jenny's milk?" Tags="|logical-deduction|mystery|" AnswerCount="6" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="3933" PostTypeId="2" ParentId="3932" CreationDate="2014-11-10T15:02:22.783" Score="35" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it's Bill&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They only steal it on the days John is in though; that's why they weren't caught red handed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If she was going to waste two weeks why didn't she just watch the fridge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the first set of clues you would assume it was John because it's only stolen on the days he's there. However she followed him on Wednesday so we know it's not John. She also followed the other two besides the culprit on the days it was taken. So the one who stole it the only one who isn't followed on a day they where it is stolen. Her friend noticed this pattern as well. The culprit must be trying to implicate John based on attendance alone. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the culprit should not be punished; he seems to work very hard as does Liz. Jenny apparently can waste 2 weeks following people around. The company should probably fire Jenny and provide free milk for the thief.&lt;/p&gt;&#xA;" OwnerUserId="2803" LastEditorUserId="5373" LastEditDate="2014-11-29T17:54:01.287" LastActivityDate="2014-11-29T17:54:01.287" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3934" PostTypeId="2" ParentId="3859" CreationDate="2014-11-10T15:13:10.913" Score="5" Body="&lt;p&gt;The killer got in the car the day before and shot through the windshield. The next day, the murder victim got in the car and was about to start it when he noticed the bullet hole and stopped to investigate. The killer then shot him through the same bullet hole with a smaller caliber round from outside the car.&lt;/p&gt;&#xA;" OwnerUserId="4869" LastActivityDate="2014-11-10T15:13:10.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3935" PostTypeId="2" ParentId="3932" CreationDate="2014-11-10T15:16:43.683" Score="17" Body="&lt;p&gt;Actually we only need 4 Lines of information to solve this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Monday    - JBS  - THEFT!&#xA;Wednesday - JLBS - THEFT!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So we know it is either John,Bill or Sarah! And we know the Thief is stealing on Mondays and Wednesdays.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Monday    - Sarah&#xA;Wednesday - John&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So we know it is not John or Sarah, which leaves &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Bill&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1998" LastEditorUserId="1998" LastEditDate="2014-11-10T15:54:52.333" LastActivityDate="2014-11-10T15:54:52.333" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3936" PostTypeId="2" ParentId="3928" CreationDate="2014-11-10T15:28:11.577" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; NO, if the car accelerates fast enough!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the ant travels at a speed of &lt;strong&gt;a&lt;/strong&gt; m/s If the car starts at a speed, which will strech the rubberband in front of the ant by a length of &lt;strong&gt;a&lt;/strong&gt;/s, the distance of the ant to its goal will not increase. If the and comes closer to its goal, the car has to speed up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if the ant has reached halfway on the rubber-band the car would have to drive twice as fast, so it will still strech the band in front of the ant by &lt;strong&gt;a&lt;/strong&gt; m/s, so it has to strech the whole band by &lt;strong&gt;2 a&lt;/strong&gt; m/s So if the acceleration is large enough (or more) to get the car to double its speed in the same time the ant reaches 50%, the ant will never catch up, since the rubberband in front of the ant is always expanding faster than the ant can travel, so the distance to the car always increases (in the edge case actually stays the same)&lt;/p&gt;&#xA;" OwnerUserId="1998" LastActivityDate="2014-11-10T15:28:11.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3937" PostTypeId="2" ParentId="3928" CreationDate="2014-11-10T15:28:48.227" Score="0" Body="&lt;p&gt;The ant reaches the car if the distance of the ant from the wall is less than distance of car from wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;dant &amp;lt;= dcar&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Using basic calculus rules of integration, and knowing the relationship between acceleration, speed, and distance we have the two equations:&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;dant = a*t&lt;/h1&gt;&#xA;&#xA;&lt;h1&gt;dcar = 1/2*c&lt;em&gt;t^2 + s&lt;/em&gt;t + 1&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Where t is the amount of time passed, and s is the initial speed of the car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the ant will reach the car in situations where&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;a*t &amp;lt;= 1/2*c&lt;em&gt;t^2 +s&lt;/em&gt;t + 1&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Where a and c (and s) are any real numbers.&#xA;As the poster above mentions, acceleration can be positive negative or zero.  If the acceleration is negative or zero, then situation becomes trivial.  The ant's speed just has to be greater than zero.  If the acceleration is positive, then the relationship between a and c must satisfy the equation that&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;1/2*c*t^2 + (s-a)*t + 1 &gt;= 0&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;In other words, the ant has a much better chance of catching the car at a very early time, before the car accelerates to infinite speeds.&lt;/p&gt;&#xA;" OwnerUserId="4868" LastEditorUserId="4868" LastEditDate="2014-11-10T15:34:08.540" LastActivityDate="2014-11-10T15:34:08.540" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3938" PostTypeId="1" AcceptedAnswerId="3940" CreationDate="2014-11-10T15:29:52.850" Score="2" ViewCount="549" Body="&lt;p&gt;This 'cake division by lines' problem was used as an example in one of the lectures I attended at business school.&lt;/p&gt;&#xA;&lt;p&gt;I've searched and it doesn't seem to be on the site yet, so here it is:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;You have a cake and a knife.&lt;/p&gt;&#xA;&lt;p&gt;You may cut the cake, with the knife, three times, in any direction.&lt;/p&gt;&#xA;&lt;p&gt;You may &lt;em&gt;not&lt;/em&gt; break the cake into pieces with your fingers (it's unhygienic) or in any other way apart from using the knife.&lt;/p&gt;&#xA;&lt;p&gt;Each cut must be in a straight line.&lt;/p&gt;&#xA;&lt;p&gt;What is the maximum number of pieces into which you can cut the cake?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The pieces don't all have to be the same size.&lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-06-07T00:46:45.030" Title="Time for tea: cake division by lines" Tags="|geometry|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3939" PostTypeId="1" AcceptedAnswerId="3941" CreationDate="2014-11-10T15:31:37.983" Score="62" ViewCount="9080" Body="&lt;p&gt;There are 4 interns at the new government agency. All 4 are comp sci majors. John is the only male. The three girls are all close but secretly like John. Lisa is playing hard to get and hopes John will make a move but could trick John into asking her out. Anne likes John and flirts openly but repeatedly tells everyone she will not date a smoker. Sarah is the smartest out of the bunch and wants John to prove to her that he can fulfill her intellectual needs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One day John arrives at his computer and there is a post-it note with a heart around it and the following numbers written in it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;546591785 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Being that asking any of the girls will dissuade the other two from ever wanting to date him, John must guess who left the message and make a gesture to show he wants to date her.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who left the message? What does John do?&lt;/p&gt;&#xA;" OwnerUserId="2803" LastEditorUserId="11876" LastEditDate="2016-09-15T01:01:12.163" LastActivityDate="2016-09-15T01:01:12.163" Title="Secret Admirer Secret Message" Tags="|cipher|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3940" PostTypeId="2" ParentId="3938" CreationDate="2014-11-10T15:40:13.120" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It depends on shape of the cake. for example with this one you can make too many pieces with only one cut:&#xA;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/nUW8Y.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&#xA; but suppose shape of cake is convex like a Cube or Cylinder. The answer should be 8 pieces this way:&#xA; &#xA; &lt;img src=&quot;https://i.stack.imgur.com/mKwqi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="2225" LastEditDate="2014-11-10T15:48:16.083" LastActivityDate="2014-11-10T15:48:16.083" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3941" PostTypeId="2" ParentId="3939" CreationDate="2014-11-10T15:49:26.117" Score="58" Body="&lt;p&gt;He should ask out ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Anne&lt;br&gt;&#xA;&lt;br&gt;&#xA; The numbers correspond to the keyboard key below-and-to-the-right of it (eg. 5=T, 9=O) and spell out TRY TO QUIT.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2518" LastActivityDate="2014-11-10T15:49:26.117" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3942" PostTypeId="2" ParentId="3932" CreationDate="2014-11-10T15:56:18.230" Score="9" Body="&lt;p&gt;A binary table solves this in a pretty obvious way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I merged the 2 weeks duration for the sake of simplicity.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;| MO | TUE | WED | Theft and Observation&#xA;|  1 |  1  |  1  | Theft&#xA;|  1 |  0  |  0  | Sarah&#xA;|  0 |  1  |  0  | Liz&#xA;|  0 |  0  |  1  | John&#xA;|  0 |  0  |  0  | Bill&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As you can see Only &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bill &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;was not stalked during the days of the theft. Which makes him the suspect of this serious crime :(&lt;/p&gt;&#xA;" OwnerUserId="4872" LastEditorUserId="2803" LastEditDate="2014-11-10T16:02:43.850" LastActivityDate="2014-11-10T16:02:43.850" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3943" PostTypeId="1" AcceptedAnswerId="3969" CreationDate="2014-11-10T16:29:31.563" Score="57" ViewCount="11848" Body="&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/2497/whats-the-password&quot;&gt;War has broken out in the kingdom.&lt;/a&gt; The king's castle is guarded pretty heavily, by guards with instructions to kill anyone who can't produce the correct password.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unbeknownst to the guards, the beggars near the main gate are in fact enemy spies. They listen intently as the first knight approaches. &quot;Six!&quot;, the guard calls out. &quot;Three,&quot; answers the knight. He is allowed through. &quot;Aha,&quot; the spies think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But they needs to be sure. They see another knight approaching and listen in on his exchange with the guard. &quot;Twelve!&quot;, &quot;Six!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One spy decides he's heard enough, so he leaves and returns dressed as a knight. &quot;Eighteen!&quot; The spy answers &quot;nine&quot; and is killed on the spot by the guards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another spy, who used to browse Puzzling.SE before he signed up for the Espionage Division feels a moment of deja vu, and decides he now knows the answer. As night falls, he goes off and disguises himself as a knight. He comes back in the morning and walks up to the guard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Eighteen!&quot; shouts the guard, &quot;Eight!&quot; the spy replies. The guard cuts him down, and he makes a surprised &quot;gurgle...splot&quot; noise as he dies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last spy, still in hiding is confused. As he's thinking, another knights walks up. &quot;Twelve!&quot; is the challenge, &quot;Zero!&quot; comes the reply. The guard steps aside and the knight enters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the last spy gets up, goes off and does the whole dress-as-a-knight thing and comes back. &quot;Eighteen!&quot; shouts the guard. What is the correct response?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; The answer to TWELVE could have been Zero or Six either time - both were correct. This is one of only two such cases of 2 valid answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit for anyone who sees this in the future:&lt;/strong&gt; I'm aware that it's not fully-scoped. It's my first puzzle and I neglected to give any odd-number examples, which is kind of important. It's totally solvable without them, I just should have included at least one.&lt;/p&gt;&#xA;" OwnerUserId="2518" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2016-07-09T14:16:26.300" Title="What's the Password, again?" Tags="|password|" AnswerCount="14" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3944" PostTypeId="2" ParentId="3928" CreationDate="2014-11-10T17:08:13.470" Score="13" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The ant never reaches the car, if the car is fast enough.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At time $t$, the velocity of the car is $ct$, and the length of the rope is $1+\frac12ct^2$. Track the fractional progress of the ant, who starts 0% of the way along the rope and wishes to reach the car positioned at 100%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At time $t$, the ant's fractional speed is inversely proportional to the length of the rope. It is $\frac{2}{2+ct^2}a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is really: is $\int_0^\infty\frac{2}{2+ct^2}a\text{ }dt$ at least 1? We could evaluate the integral using the arctan function, but there is an easier way to proceed. Recall that $\sum_0^\infty\frac{2}{2+ct^2}$ converges. Therefore, by the integral test for infinite series, $\int_0^\infty\frac{2}{2+ct^2}a\text{ }dt$ is finite. Therefore, if $a$ is small enough, then the integral is less than 1, and the ant never reaches the car.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT: By request, we can figure out the exact conditions for the ant reaching the car. By integrating using the arctan function (or WolframAlpha), we find $\int_0^\infty\frac{2}{2+ct^2}a\text{ }dt=\frac{\pi a}{\sqrt{2c\text{ }}}$. The ant reaches the car iff this quantity is greater than 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the ant reaches the car iff $a&amp;gt;\frac{\sqrt{2c\text{ }}}{\pi}$.&lt;/p&gt;&#xA;" OwnerUserId="1752" LastEditorUserId="1752" LastEditDate="2014-11-10T19:14:08.207" LastActivityDate="2014-11-10T19:14:08.207" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3947" PostTypeId="2" ParentId="3943" CreationDate="2014-11-10T17:23:54.320" Score="7" Body="&lt;p&gt;It's a guess, but does it have something to do with how the spies are dressed?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You specifically talk about the spies listening and not talking.  Perhaps they are dressed as blind beggars?  Then the problem isn't what they're hearing, but what they aren't seeing.&#xA; &#xA;My guess is that the soldiers are holding up a number of fingers when they call out their arbitrary number, based on the fact that none of the answers have been more than ten!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4881" LastActivityDate="2014-11-10T17:23:54.320" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3948" PostTypeId="2" ParentId="3943" CreationDate="2014-11-10T17:24:24.557" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Zero &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because on the second time twelve was said it was zero, so on the second time eighteen was said it would also be: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; zero.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the soldiers to tell which time it is (to confuse the spies) there's have to be a visual cue that the spies cannot overhear such as blinking.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="22157" LastEditDate="2016-07-09T14:16:26.300" LastActivityDate="2016-07-09T14:16:26.300" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3949" PostTypeId="2" ParentId="3928" CreationDate="2014-11-10T17:28:43.130" Score="2" Body="&lt;p&gt;This should probably be on the math stackexchange. It's a differential equations problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interesting part about the elastic band is that it provides the ant with a fraction of the car's velocity. The fraction varies depending on how far along the elastic the ant has gotten. For instance, at t = 0 when the ant is at the very beginning, the elastic is anchored and doesn't move. But at t = ? when the ant is halfway to the car, the elastic point is moving at half the speed of the car. And at t = ? when the ant catches up, the elastic is moving at the full speed of the car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance of the car from the wall at time t is $d_c = \frac{1}{2} c t^2 + 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the distance of the ant from the wall at time t be $d_a$, which is a function of t&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The velocity of the ant at time t is $d_a'$ (first derivative of $d_a$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that $d_a' = a + \frac{d_a} {d_c} d_c'$ (the constant component a, plus a fraction of the car's velocity based on how far along the elastic the ant is at the time)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting:&#xA;$d_a' = a + \frac{d_a} {(\frac{1}{2} c t^2 + 1)} (c t)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a first order differential equation. I plugged it into Wolfram Alpha's differential equation solver. It provided a complicated solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = \frac{a(c t^2 + 2)\arctan(x\sqrt{\frac{c}{2}})}{\sqrt{2c}} + k_1(c t^2+2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k_1$ is a constant that we have to provide based on the initial conditions. Our initial condition is y(0) = 0. Substituting that in, we see $k_1$ must be 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we're left with $y = \frac{a(c t^2 + 2)\arctan(t\sqrt{\frac{c}{2}})}{\sqrt{2c}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ant catches up to the car when $d_a = d_c$, or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{a(c t^2 + 2)\arctan(t\sqrt{\frac{c}{2}})}{\sqrt{2c}} = \frac{1}{2} c t^2 + 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's too hairy to keep working with. But there are definitely solutions. Pick an x (time), a c (car's acceleration), substitute, and solve for a, the required minimum speed of the ant. For instance, t = 10, c = 1, results in a = ~0.494.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully someone can check my math, because I'm not sure I did all the math markup correctly!&lt;/p&gt;&#xA;" OwnerUserId="4869" LastEditorUserId="3960" LastEditDate="2014-11-11T13:49:34.247" LastActivityDate="2014-11-11T13:49:34.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3950" PostTypeId="2" ParentId="3943" CreationDate="2014-11-10T17:49:17.207" Score="2" Body="&lt;p&gt;I think the solution here is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; answer = (#of previous correct answers in a row) * 3 &#xA;&lt;br&gt;&#xA; So in this case, the correct response should be 3.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4882" LastActivityDate="2014-11-10T17:49:17.207" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3951" PostTypeId="2" ParentId="1924" CreationDate="2014-11-10T17:51:24.777" Score="2" Body="&lt;p&gt;There are a couple degenerate cases: N=0 and N=1.  In those cases, there are zero pairs of points where the perpendicular bisector of the segment between them doesn't pass through at least two other points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem also degenerates if we allow coincident points.&lt;/p&gt;&#xA;" OwnerUserId="4886" LastActivityDate="2014-11-10T17:51:24.777" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3953" PostTypeId="2" ParentId="3943" CreationDate="2014-11-10T17:55:04.497" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My guess is &quot;Minus nine&quot;. Answer = [guard's number] $\times$ ([number of beggars] - 1) / 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4887" LastActivityDate="2014-11-10T17:55:04.497" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3954" PostTypeId="2" ParentId="3943" CreationDate="2014-11-10T18:02:07.403" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Six.&lt;br&gt;&#xA;The knight has to respond the subtraction of the question by the guard with the previuos correct answer.&lt;br&gt;&#xA;&#xA;This has 2 minor flaws: you need to start with 3 and there is no way described as how the previous question is passed from knight to knight.&lt;br&gt;&#xA;&lt;br&gt;&#xA; 1. Start with Q: Six, A: Three (prev correct Q &lt;em&gt;three&lt;/em&gt; &lt;strong&gt;FLAW HERE&lt;/strong&gt; -&gt; 6-3 = 3)&lt;br&gt;&#xA; 2. Q: Twelve,   A: Six   (prev correct Q six      -&gt; 12- 6 = 6)&lt;br&gt;&#xA; 3. Q: Eighteen, A: Nine  (prev correct Q twelve   -&gt; 18-12 = 6 WRONG)&lt;br&gt;&#xA; 4. Q: Eighteen, A: Eight (prev correct Q twelve   -&gt; 18-12 = 6 WRONG)&lt;br&gt;&#xA; 5. Q: Twelve,   A: Zero  (prev correct Q twelve   -&gt; 12-12 = 0)&lt;br&gt;&#xA; 6. Q: Eighteen, A: Six   (prev correct Q eighteen -&gt; 18-12 = 6)&lt;br&gt;&#xA;&lt;br&gt;&#xA;Something like a chalkboard must be available to all the living knights to sign their question for the future answerer. &#xA;&lt;br&gt;&#xA; My reasoning is simple: since 12 has two different answers there is no static way to get an answer, it depends from something dynamic. From the riddle, only two thing are dynamic between 12's answers:&lt;br&gt;&#xA;&lt;br&gt;&#xA; - Previous answers/questions&lt;br&gt;&#xA; - Change in the day&lt;br&gt;&#xA;&lt;br&gt;&#xA; The change in the day can explain the 0, being the number of correct answers given that day. But this doesn't match with 18-9 the day before. Using the hour count doesn't help eiter, altough every number is minor than 24.&lt;br&gt;&#xA; This leaves with the answers/questions being the only way to have dynamic data, but no strategy of close loop his shown, thus the flaw in my answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4652" LastEditorUserId="4652" LastEditDate="2014-11-10T18:11:35.263" LastActivityDate="2014-11-10T18:11:35.263" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3955" PostTypeId="2" ParentId="3943" CreationDate="2014-11-10T18:02:18.780" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It depends on the time of day. The numbers the guard gives are red herrings; the password is the hour of day, in military time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="68" LastActivityDate="2014-11-10T18:02:18.780" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3956" PostTypeId="1" AcceptedAnswerId="4000" CreationDate="2014-11-10T18:14:15.217" Score="7" ViewCount="782" Body="&lt;p&gt;The table below illustrates the availability of four interviewers (&lt;sub&gt;1&lt;/sub&gt;, &lt;sub&gt;2&lt;/sub&gt;, &lt;sub&gt;3&lt;/sub&gt;, &lt;sub&gt;4&lt;/sub&gt;) at four time slots (&lt;sub&gt;1&lt;/sub&gt;, &lt;sub&gt;2&lt;/sub&gt;, &lt;sub&gt;3&lt;/sub&gt;, &lt;sub&gt;4&lt;/sub&gt;).  A 1 in the table indicates that the corresponding interviewer is available at the corresponding time slot.&lt;/p&gt;&#xA;&lt;p&gt;For example, the first interviewer, &lt;sub&gt;1&lt;/sub&gt;, is available at time slots &lt;sub&gt;1&lt;/sub&gt; and &lt;sub&gt;2&lt;/sub&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Each interviewer may only take one interview and all four slots should follow sequentially, like &lt;sub&gt;1&lt;/sub&gt; → &lt;sub&gt;2&lt;/sub&gt; → &lt;sub&gt;3&lt;/sub&gt; → &lt;sub&gt;4&lt;/sub&gt;.&lt;/p&gt;&#xA;&lt;p&gt;What are all possible sequences of interviewers? Can this be generalised?&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;sub&gt;1&lt;/sub&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;sub&gt;2&lt;/sub&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;sub&gt;3&lt;/sub&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;sub&gt;4&lt;/sub&gt;&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;strong&gt;&lt;sub&gt;1&lt;/sub&gt;&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;strong&gt;&lt;sub&gt;2&lt;/sub&gt;&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;strong&gt;&lt;sub&gt;3&lt;/sub&gt;&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;strong&gt;&lt;sub&gt;4&lt;/sub&gt;&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;" OwnerUserId="4888" LastEditorUserId="69582" LastEditDate="2021-02-04T19:04:17.200" LastActivityDate="2021-02-04T19:04:17.200" Title="How many possible interview slots" Tags="|logical-deduction|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="3958" PostTypeId="2" ParentId="3718" CreationDate="2014-11-10T18:35:36.477" Score="4" Body="&lt;h1&gt;1401611&lt;/h1&gt;&#xA;&lt;p&gt;The optimal pattern of packing spheres is pretty well known, not as widely known is the fact that this pattern has 7 different orientations where the spheres are all organised in layers of simple patterns. In four of these orientations the spheres are aligned in a triangular pattern, in the three other planes they are aligned in a square pattern. The key to this puzzle is to rotate the pattern so that the wasted space along the sides of the box is minimised.&lt;/p&gt;&#xA;&lt;p&gt;An orientation where the three square planes are aligned with the walls of the box happens to fit extremely well, as a 71 ball wide configuration is &lt;span class=&quot;math-container&quot;&gt;$99.995\text{ cm}$&lt;/span&gt; wide. In this configuration the bottom of the box is filled like so:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GeA9A.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;With 71 balls bordering each edge, for a total of &lt;span class=&quot;math-container&quot;&gt;$71^2+70^2=9941$&lt;/span&gt; balls in the bottom layer, the layer on top of it will use the same pattern, but be shifted so that the balls lie in the holes between the balls in the bottom layer, this will allow one ball less. These two layer configurations are then repeated until there are 71 configurations like the first layer, and 70 like the second layer. This gives a total of &lt;span class=&quot;math-container&quot;&gt;$71\times9941+70\times9940=1401611$&lt;/span&gt; balls.&lt;/p&gt;&#xA;" OwnerUserId="2186" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-11-10T18:35:36.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3959" PostTypeId="2" ParentId="3939" CreationDate="2014-11-10T18:45:50.683" Score="115" Body="&lt;p&gt;&lt;sup&gt;&lt;strong&gt;Disclaimer: This is only posted in a half-serious manner because I was bored and looking for alternative decodings.&lt;/strong&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The message was sent by:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sarah, but she's trying way too hard. The message is a two-fold trap&#xA; designed to let John &quot;&lt;em&gt;prove to her that he can fulfill her&#xA; intellectual needs.&lt;/em&gt;&quot; It's far more complicated than it needs to be but Sarah is too much of an intellectual to see that.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first trap is the simple answer (given by Joe). &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; By using the adjacent numbers on a QWERTY keyboard, it spells out 'TRY TO QUIT'. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If Sarah sees him ask Anne out, she knows he fell for this trap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second is far more convoluted. First we take the numbers in&#xA;overlapping pairs:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;54 46 65 59 91 17 78 85&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We then assign each to a letter of the alphabet by mod26 so A=0,Z=25:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;CUNHNRAH&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Translate that to Morse code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;-.-. ..- -. .... -. .-. .- ....&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If you translate that as a binary code, you end up with exactly three&#xA;bytes:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;10100011(163) 0000100(4) 10010000(144)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Again taking the mod26 of those values and assigning letters:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;HEO&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now with a simple ROT-4 cipher applied:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;LIS&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which is short for Lisa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If he asks Lisa out, Sarah knows he's smart enough to decipher a&#xA;ridiculously convoluted code. However, he fails the common sense test&#xA;because only the brainy one (Sarah) would do something like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If he asks Sarah out, he's in for a world of hurt when she decides to&#xA;keep sending messages like this in the future. He should do himself a&#xA;favor and find someone else altogether.&lt;/p&gt;&#xA;" OwnerUserId="1563" LastActivityDate="2014-11-10T18:45:50.683" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3960" PostTypeId="2" ParentId="3943" CreationDate="2014-11-10T18:46:30.947" Score="12" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Zero.&lt;br&gt;&#xA;&#xA; The formula is &lt;code&gt;3*(challenge/6+(number of beggars)-3)&lt;/code&gt; or, equivalently, &lt;code&gt;challenge/2+3*(number of beggars)-9&lt;/code&gt;. This fits all known good and bad responses, and allows each knight approaching to determine the answer independently: they don't need to know anything about who came before them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1496" LastEditorUserId="1496" LastEditDate="2014-11-10T19:01:43.187" LastActivityDate="2014-11-10T19:01:43.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3961" PostTypeId="2" ParentId="3943" CreationDate="2014-11-10T18:49:46.460" Score="2" Body="&lt;p&gt;I will say the answer is &quot;Three!&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is no indication whether it is the knight or the guard that yells either twelve. So the password is &quot;Twelve!&quot; if the knight yells first, or &quot;Three!&quot; if he is responding to the guard.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3922" LastEditorUserId="3922" LastEditDate="2014-11-10T19:01:36.593" LastActivityDate="2014-11-10T19:01:36.593" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3964" PostTypeId="2" ParentId="3943" CreationDate="2014-11-10T19:02:41.600" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Ten&lt;br&gt;&#xA;&lt;br&gt;&#xA;Value of number - number of letters&lt;br&gt;&#xA;'Six' --&gt; 6 - 3 = 3 --&gt; 'Three'&lt;br&gt;&#xA;'Twelve' --&gt; 12 - 6 = 6 --&gt; 'Six'&lt;br&gt;&#xA;'Eighteen' --&gt; 18 - 8 = 10 --&gt; 'Ten'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4893" LastActivityDate="2014-11-10T19:02:41.600" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3965" PostTypeId="2" ParentId="3943" CreationDate="2014-11-10T19:12:04.447" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Anything under six: for all successful answers, the challenge multiplied by the answer is less than one hundred.  For all failed answers, the challenge multiplied by the answer is over one hundred.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4895" LastActivityDate="2014-11-10T19:12:04.447" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3967" PostTypeId="2" ParentId="3943" CreationDate="2014-11-10T19:29:19.447" Score="26" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'll guess &lt;strong&gt;3&lt;/strong&gt;. Take a standard clock. Move from the 12 the number of hours the guard shouts. Note that you loop at the 12 which could be considered 0 or 12. When done, divide the final result by 2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2614" LastEditorUserId="2518" LastEditDate="2014-11-10T19:30:43.830" LastActivityDate="2014-11-10T19:30:43.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3968" PostTypeId="2" ParentId="3943" CreationDate="2014-11-10T19:31:53.343" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Two.&lt;br&gt;&#xA;The answer is (number of letters in challenge modulo 6), where the answer must be in the range 0-6. &quot;Six&quot; has 3 letters, so the answer is &quot;three&quot;. &quot;Twelve&quot; has 6 letters, so the answer is &quot;zero&quot; or &quot;six&quot;. &quot;Eighteen&quot; has 8 letters, so the answer is (8-6=)&quot;two&quot;.&lt;br&gt;&#xA;This fits all known information: the good and bad responses, the dual answer to 12 (but no other used challenges), the independence of the answers, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1496" LastActivityDate="2014-11-10T19:31:53.343" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3969" PostTypeId="2" ParentId="3943" CreationDate="2014-11-10T19:49:52.047" Score="33" Body="&lt;p&gt;After reading Paul's answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Start with an analog clock at noon (or midnight). The number the guard starts with is the number of printed digits to advance the minute hand (in other words, the number of 5-minute intervals to advance the time). Take the number of minutes past the hours and divide by 10. That's the answer the knight gives.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This yields an answer of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;three&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Expressed as a math formula, where N is the number the guard gives, the solution is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;(N % 12)*5 / 10 (more or less - modulo doesn't actually give the answer x for x%x, but humans who don't understand how modulo works sometimes do).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2501" LastEditorUserId="2501" LastEditDate="2014-11-10T20:04:54.130" LastActivityDate="2014-11-10T20:04:54.130" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3970" PostTypeId="2" ParentId="3932" CreationDate="2014-11-10T20:47:42.343" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Bill is the only one present on all the days the milk was stolen (M,T,W) AND wasn't followed by Jenny during the theft period (only followed on TH, F) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2014" LastActivityDate="2014-11-10T20:47:42.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3971" PostTypeId="2" ParentId="3846" CreationDate="2014-11-10T21:19:15.240" Score="2" Body="&lt;p&gt;EDIT: As J Marcos points out in the comments, this answer cannot possibly be correct, as the 'lost sentence' it gives is tautological, not contingent as specified in the question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's been a while since I've done anything with formal logic, so please forgive (and correct) me if I butcher the terminology. I have only attempted to answer question 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the information given there are three possible forms (up to relabeling of variables) for the forgotten sentence.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$p \rightarrow (q \text{ B } r)$&lt;/li&gt;&#xA;&lt;li&gt;$(p \rightarrow q) \text{ B } r$&lt;/li&gt;&#xA;&lt;li&gt;$p \text{ B }(q\rightarrow r)$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;where B is some binary connective. Note that I left out $(p \text{ B } q) \rightarrow r$, as the implication doesn't have an atomic antecedent (unless B is also implication, in which case we're back in case 2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no other information in the question that would allow us to choose any particular one of these forms over the others. Thus, the problem is only well-defined* if there is exactly one choice of B such that the three forms are semantically equivalent. As it happens, there is, and we can derive it as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $p$ is false, the three forms have the values $True$, $True\text{ B }r$, and $False\text{ B }(q\rightarrow r)$, respectively. These must all be equal regardless of the value of $r$, so $True\text{ B } True = True\text{ B }False = True$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $p$ is true and $q$ is false, form 1 reduces to $True \text{ B } True = True$ and form 2 to $False \text{ B } r$. Again these must be equal for either value of $r$, so $False \text{ B } True = False \text{ B } False = True$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is enough to completely characterize B: it is &lt;b&gt;the binary connective that always yields $True$ regardless of its operands&lt;/b&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*I'm cheating a bit here, I think -- J Marcos only assumes that the information is enough to define the &lt;i&gt;sentence&lt;/i&gt; up to semantic equivalence, which is not quite the same as  assuming the question &quot;what was the other connective&quot; has a well-defined answer. To make up the gap I'd need to show that semantic equivalence of $p \rightarrow (q \text{ C } r)$, $(p\rightarrow q)\text{ D } r$, and $p\text{ E }(q\rightarrow r)$ is enough to imply $\text{ C } = \text{ D } = \text{ E } = \text{ B }$, where B is the connective derived above. Proving this is completely mechanical (write down the truth tables for C, D, and E; plug in different values of $p$, $q$, and $r$ to 1, 2, and 3; and follow your nose) but would be tedious to write up, so I leave it as an exercise to the reader.&lt;/p&gt;&#xA;" OwnerUserId="4912" LastEditorUserId="4912" LastEditDate="2014-11-11T00:22:55.233" LastActivityDate="2014-11-11T00:22:55.233" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="3972" PostTypeId="1" CreationDate="2014-11-10T21:39:13.613" Score="4" ViewCount="1635" Body="&lt;p&gt;&lt;b&gt;Dilemma&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are taken to a dark room and given two thin long rods - 30 cm long and 1 cm thick. One of them is a permanent magnet and the other is pure iron which is not magnetized. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both of them are insulated with glass. They weigh the same. The shapes are same too. How can you tell which one of these rods is the magnet? Remember, you can't see anything, and you cannot use anything other than these two rods in your investigation. You are not allowed to break them or remove the insulation either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Hint&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;zntargfnerzbernggenpgvirarnegurcbyrfgunavagurzvqqyr&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4838" LastEditorDisplayName="user88" LastEditDate="2015-04-29T00:54:19.847" LastActivityDate="2015-04-29T00:54:19.847" Title="Find the magnet among the two rods" Tags="|physics|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3973" PostTypeId="2" ParentId="3922" CreationDate="2014-11-10T21:41:12.593" Score="-1" Body="&lt;p&gt;I will try for the intuitive argument&#xA;When the problem starts the difference in speeds is c-a&#xA;When the ant catches up to to the car and is crawling on the bumper it is c+a&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the ant will be at the speed of the car at (1-a/c)*100% of the band.  After that he is going faster and will eventually overtake the car&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So can he get there?&#xA;Well that point is no different then the car.  He will be faster then it when he reaches  (1-a/c)*100% to that point&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And more importantly he will do it faster because it is closer to him&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now we can imagine that infinity until the ant's first step reaches this magic point of outrunning the next point.  Another way of saying is that the velocity of one of these points will be less than a if the subdivision is preformed enough times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example if the ant is going .5 c, he will reach the magic point at .5 of the band.  The point to reach that is at .25, the point to accelerate over that spot is .125.  And so towards 0.  At some moment the ant's step will be greater than 1/(x^2)&lt;/p&gt;&#xA;" OwnerUserId="1639" LastEditorUserId="1639" LastEditDate="2014-11-10T22:21:28.070" LastActivityDate="2014-11-10T22:21:28.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3974" PostTypeId="2" ParentId="3972" CreationDate="2014-11-10T21:45:42.010" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For the two rods, place the end against the middle of the other and compare the magnetic attraction. The magnetic rod will be the one with its end against the other's middle with the highest pull. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2014-11-10T21:55:34.750" LastActivityDate="2014-11-10T21:55:34.750" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3975" PostTypeId="2" ParentId="3972" CreationDate="2014-11-10T22:01:26.853" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Scrape the tip of one rod against the edge of the other, repeatedly and in the same direction. If there is now repulsion between the two rods for some orientation of ends, then the rod you scraped with is the magnet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4914" LastActivityDate="2014-11-10T22:01:26.853" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3976" PostTypeId="2" ParentId="3922" CreationDate="2014-11-10T22:06:01.183" Score="0" Body="&lt;p&gt;The intuitive case is roughly this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the ant begins at some point, $ P $, which is some fraction of the distance along the rope.  The ant travels at $a$ and the car travels at $c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine a point, $P'$, which is $ \frac{a}{2c} $ percent of the rope's total length towards the car.  The ant will clearly travel that distance, given the ant will travel in some period of time $t$ the distance $\frac{a}{c}t$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This continues, indefinitely, covering points $P''$ etc., until the ant reaches far enough down the line such that the distance between $P^n$ and the end of the rope is less than $\frac{a}{2c}$.  At that point the ant will reach the end of the rope, and begin walking on the car (and eventually off of it!).  Note, depending on $c$ and $a$ this may be a &lt;em&gt;very&lt;/em&gt; long time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is because, while the rope continues to stretch indefinitely, the amount each stretch increases the length of the rope decreases over time proportional to the fraction of rope the ant has covered.  At 1%, the rope stretching 1km more means the way 'ahead' increases by 0.99km; however, at 99%, the rope stretching 1km more means only a 0.01 km increase in the ant's total length covered.  Eventually the remaining distance is less than the amount the ant can cover.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't quite a 'limit' - eventually the ant would in fact walk &lt;em&gt;past&lt;/em&gt; the car, assuming it still had the property of keeping the rope's proportion behind it, which isn't practical in a real life example, but this isn't practical either - which is what confuses people.  It's similar to the reason why Achilles eventually catches the tortoise when you're running away: you can keep halving the distance between the two indefinitely, true, but time isn't really paying attention to that, it's just going on its merry way. (Of course, that paradox is a bit different, but in both cases the answer is that it does overtake despite the apparent limit problem.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I find it helpful to think of radial geometry here, as well.  You can imagine traversing a circle of 360° at constant radius 1km, it will total 2$\pi$km distance covered to traverse.  As long as your speed is positive, you will certainly traverse the entire circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, turn the circle into a spiral (say, a logarithmic spiral).  Traversing a spiral starting at 1km radius, say the point $(x,y)=(1,0)$, and increasing in radius, you would always still eventually traverse that 360°, back to some point $(x,y)=(x,0)$, regardless of the speed at which you travel.  Exactly the same problem as the circle, except you're making some points further 'out' than others - but still the same 360 degrees to travel, always traveling along, increasing what degree you're at (at least by a little!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second graph here gives a good example of how this will eventually happen.  The parameters in the logarithmic spiral below can be manipulated to mimic the speed of the ant and the car.  While it clearly may end up being a &lt;em&gt;very&lt;/em&gt; long spiral, you can precompute the entire spiral given the relative speed of the car and the ant, and the starting length of the rope.  It is fairly clear that no matter what that total spiral length is, given it is finite, the ant &lt;em&gt;must&lt;/em&gt; eventually traverse all of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZH89Q.gif&quot; alt=&quot;Wolfram Alpha Logarithmic Spiral&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(sourced from &lt;a href=&quot;http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427eit37bqh929&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wolfram Alpha&lt;/a&gt;, searching for &quot;logarithmic spiral&quot;) &lt;/p&gt;&#xA;" OwnerUserId="2437" LastEditorUserId="2437" LastEditDate="2014-11-10T22:13:40.573" LastActivityDate="2014-11-10T22:13:40.573" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3977" PostTypeId="1" AcceptedAnswerId="3986" CreationDate="2014-11-10T22:14:38.090" Score="12" ViewCount="1321" Body="&lt;p&gt;This is Part 3 of the Murder of the President brainteaser/riddle. If you have not already, you might want to see the answer for &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3600/murder-of-the-president-part-1&quot;&gt;Murder of the President - Part 1&lt;/a&gt; posted by Nit and the answer for &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3845/murder-of-the-president-part-2&quot;&gt;Murder of the President - Part 2&lt;/a&gt; posted by Joe Z. Each part will give you a clue and you must solve it. Use all knowledge you have of cryptography, ciphers, past puzzles, etc. You should also use Google. This case is meant to take place in the present day, so all politicians, celebrities, places, etc. are who they are now. Please post your answers in spoiler tags.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the riddle: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You open up the Bible to Romans 7:6 which reads&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;sup&gt;&lt;b&gt;6&lt;/b&gt;&lt;/sup&gt; But now, by dying to what once bound us, we have been released from the law so that we serve in the new way of the Spirit, and not in the old way of the written code.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On the page after this verse, there are 8 sticky notes, some typed, some handwritten, all coded (for this part, all typed notes will be &lt;strong&gt;bold&lt;/strong&gt;, and all handwritten ones will be &lt;em&gt;italics&lt;/em&gt;). At the top of the page, there is some writing in marker.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Every &lt;sup&gt;4&lt;/sup&gt;/&lt;sub&gt;16&lt;/sub&gt;&lt;br&gt;&lt;br&gt;&#xA;  WDFPSSDOQWXLVMKYSGFACRVLKJJPDDDHMNNAZXZBQWSEXNXTKJDIXXXC&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then the notes read:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;#1 Romans&lt;br&gt;&#xA;  UDQF BNEQRS TBNT VT QHJGRDH&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;#2 Snake&lt;br&gt;&#xA;  VCEP SNACPW YWWP NLBU NTKFFNY&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;#3 Cow&lt;br&gt;&#xA;  FWBFZ FQ IEH EOHHFGTKWH AHHH&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;#4 Firetruck&lt;br&gt;&#xA;  TDEGYJ NHGKW AX NZZ FZXAWHN UGXP&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;#5 Gorilla&lt;br&gt;&#xA;  UJEKQD UOGWN UE UOT UXDDFJI UUGQ&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;#6 Tree&lt;br&gt;&#xA;  NZNX CK GJZGWJ NZJ HIJWJWL HQOW&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;#7 APPLE&lt;br&gt;&#xA;  HE JA YIU FQSJDIGQB JMA MVDTDJE QDP YIYHFD OYDQ&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;#8 Dog&lt;br&gt;&#xA;  CDB EGL EI ALT DVDUK ESERL&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;What do these codes mean? What do they say? Where should you go next?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good Luck&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note I will be posting Part 4 in one to two days. I will select the correct answer for Part 3 before, but I will post the answer for Part 3 if no one gets it&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1711" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2014-12-15T23:01:30.377" Title="Murder of the President - Part 3" Tags="|mystery|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3981" PostTypeId="1" AcceptedAnswerId="3998" CreationDate="2014-11-11T01:08:37.593" Score="-5" ViewCount="4032" Body="&lt;p&gt;A party is being held at a local mansion. The host is very rich and his success is because of one thing ~ his famous recipe for Spaghetti!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only guests that may attend are people the host knows and trusts, whom he gave the password to enter to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's where you come in. You and a friend are trying to steal this recipe. You sneak by and listen to the passwords.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first guest arrives. The security says &quot;5&quot;, and the guest replies &quot;5&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second guest arrives. The security says &quot;6&quot;, the guest replies &quot;6&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your friend thinks he got it all figured out. He approaches the security and as they say &quot;7&quot;, when he replies &quot;7&quot; he's taken by security to the host for impersonating a guest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another guest arrives, you seem to get the same pattern, security says &quot;1&quot; and he says &quot;1&quot; to get in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final guest before you figure it out comes and says &quot;4&quot; in which the response is (you guessed it!) &quot;4&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You walk up to security and get &quot;3&quot;, what is the correct response?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="-1" LastEditDate="2014-11-11T23:05:34.163" LastActivityDate="2014-11-11T23:05:34.163" Title="The Security to the Party" Tags="|password|" AnswerCount="2" CommentCount="11" ClosedDate="2014-11-12T18:38:30.887" ContentLicense="CC BY-SA 3.0" />
  <row Id="3985" PostTypeId="2" ParentId="3932" CreationDate="2014-11-11T01:42:10.867" Score="5" Body="&lt;p&gt;Based on the information provided, and some information not provided, there are two possible outcomes of the single thief theory. The first as deduced by others with the cross reference of data provided. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bill. The second, who was not considered, has all the opportunity, and none of the scrutiny. This person has been stated to have easy access to the fridge. And is also likely to be have been in the office on each of the days the theft occurred. Since no one was observed committing the crime, this person could not be ruled out as a suspect. The manager could easily helped themselves to the milk. If they were a good manager they would have noticed two things: 1 someone was taking the milk, 2 that action had cost him / her over 40 man hours. So with that observation, I would not put it past the manager to have a little refreshment. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I need to slow down on the murder mysteries.  &lt;/p&gt;&#xA;" OwnerUserId="4930" LastEditorUserId="4930" LastEditDate="2014-11-11T01:48:20.300" LastActivityDate="2014-11-11T01:48:20.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3986" PostTypeId="2" ParentId="3977" CreationDate="2014-11-11T01:50:55.653" Score="10" Body="&lt;p&gt;Before trying to break the other eight codes, it would be a good idea to figure out what the message written in marker says first.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Every &lt;sup&gt;4&lt;/sup&gt;⁄&lt;sub&gt;16&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;WDFPSSDOQWXLVMKYSGFACRVLKJJPDDDHMNNAZXZBQWSEXNXTKJDIXXXC&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If we take every fourth letter in the large string of letters, we get &amp;quot;POLYALPHABETIC&amp;quot;, which most likely means that the next eight codes are encoded using some sort of polyalphabetic cipher.&lt;/p&gt;&#xA;&lt;p&gt;If we then apply the &lt;a href=&quot;http://en.wikipedia.org/wiki/Vigen%C3%A8re_cipher&quot; rel=&quot;noreferrer&quot;&gt;Vigenère cipher&lt;/a&gt; (the most common form of polyalphabetic cipher) to each of the codes on the sticky notes with their respective keys, we get the following:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;code&gt;1. CODE NUMBER FIVE IS CORRECT&lt;/code&gt; &lt;br /&gt;&#xA; &lt;code&gt;2. CODE NUMBER FIVE ISNT CORRECT&lt;/code&gt; &lt;br /&gt;&#xA; &lt;code&gt;3. CHECK IN THE PRESIDENTS DESK&lt;/code&gt; &lt;br /&gt;&#xA; &lt;code&gt;4. NUMBER SEVEN IS THE CORRECT CLUE&lt;/code&gt; &lt;br /&gt;&#xA; &lt;code&gt;5. NUMBER THREE IS THE CORRECT CLUE&lt;/code&gt; &lt;br /&gt;&#xA; &lt;code&gt;6. THIS IS BEFORE THE CORRECT CLUE&lt;/code&gt; &lt;br /&gt;&#xA; &lt;code&gt;7. GO TO THE PENINSULA TWO HUNDRED AND THIRTY NINE&lt;/code&gt; &lt;br /&gt;&#xA; &lt;code&gt;8. YOU ARE AT THE WRONG PLACE&lt;/code&gt; &lt;br /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;A bunch of mutually contradictory statements! What do we do?&lt;/p&gt;&#xA;&lt;p&gt;As always, look to the Bible. Recall in my answer to part 2, that the key to the puzzle was the &lt;em&gt;end&lt;/em&gt; of Romans 7:6:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;sup&gt;&lt;strong&gt;6&lt;/strong&gt;&lt;/sup&gt; But now, by dying to what once bound us, we have been released from the law so that we serve in the new way of the Spirit, and &lt;strong&gt;not in the old way of the written code.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Five of these codes were handwritten. If we throw these out, then we are left with:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;code&gt;1. CODE NUMBER FIVE IS CORRECT&lt;/code&gt; &lt;br /&gt;&#xA; &lt;code&gt;5. NUMBER THREE IS THE CORRECT CLUE&lt;/code&gt; &lt;br /&gt;&#xA; &lt;code&gt;8. YOU ARE AT THE WRONG PLACE&lt;/code&gt; &lt;br /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Which tells us that we are at the wrong place, and we need to check inside the President's desk. Back to the scene of the crime we go!&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-11-11T01:50:55.653" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3987" PostTypeId="2" ParentId="3972" CreationDate="2014-11-11T02:07:46.517" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Balance each rod in turn at its midpoint on your fingertip.  Try this several times for each rod in different orientations.  The magnet is the one which will tend to rotate towards the same point (north).  You can determine this by feel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1667" LastActivityDate="2014-11-11T02:07:46.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3988" PostTypeId="2" ParentId="3972" CreationDate="2014-11-11T02:22:49.650" Score="1" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; magnetsaremoreattractivenearthepolesthaninthemiddle&lt;br&gt;&#xA;&lt;br&gt;&#xA; Magnets are more attractive near the poles than in the middle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Balance a pole from the middle on the edge of the other (which is up vertically) and see if it moves, turn the pole 90° clockwise and try again. If it moves it's magnetic, otherwise either try to the other, or automatically assume it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2014-11-11T02:22:49.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3990" PostTypeId="2" ParentId="3981" CreationDate="2014-11-11T02:45:10.607" Score="26" Body="&lt;p&gt;The formula is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $x \mod 7$ &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the answer you should give is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $3$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="2484" LastEditDate="2014-11-11T08:00:26.513" LastActivityDate="2014-11-11T08:00:26.513" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3991" PostTypeId="1" AcceptedAnswerId="3996" CreationDate="2014-11-11T02:56:36.093" Score="13" ViewCount="2561" Body="&lt;p&gt;A party is being held at a local mansion. The host is very rich and his success is because of one thing ~ his famous recipe for Spaghetti!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only guests that may attend are people who correctly reply to the guard at the door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's where you come in. You and a friend are trying to steal this recipe. You sneak by and listen to the passwords.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first guest arrives. The security says &quot;5&quot;, and the guest replies &quot;5&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second guest arrives. The security says &quot;6&quot;, the guest replies &quot;6&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your friend walks up to the door and the security says &quot;7&quot; arrives. He replies with &quot;7&quot;, and is told to leave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another guest arrives, you seem to get the same pattern, security says &quot;1&quot; and he says &quot;&lt;strong&gt;2&lt;/strong&gt;&quot; to get in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another guest arrives, you seem to get the same pattern, security says &quot;4&quot; and he says &quot;4&quot; to get in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You now walk up to security and get &quot;3&quot;, what is the correct response?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 1 was created by warspyking and is found here:  &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3981/the-security-to-the-party&quot;&gt;The Security to the Party&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="97" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2014-11-11T13:44:34.363" Title="The Security to the Party [Part 2]" Tags="|password|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3992" PostTypeId="2" ParentId="3939" CreationDate="2014-11-11T03:19:58.513" Score="6" Body="&lt;p&gt;Of course it means &quot;try to quit&quot; but more along the lines of a stalker's threat. (Seriously, is that really the best that Anne can do?) John needs to get out of there as quickly as possible...do not ask questions, do not look back, leave no forwarding address.  In the words of the great philosopher: &quot;You can check out any time you want but you can never leave.&quot;&lt;/p&gt;&#xA;" OwnerUserId="4940" LastActivityDate="2014-11-11T03:19:58.513" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3996" PostTypeId="2" ParentId="3991" CreationDate="2014-11-11T03:41:48.733" Score="19" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer corresponds to the number of segments used on creating the number on a 7 segment display. Because 3 has 5 segments on a 7 segment display, the answer is 5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4874" LastActivityDate="2014-11-11T03:41:48.733" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3997" PostTypeId="1" AcceptedAnswerId="3999" CreationDate="2014-11-11T05:15:23.973" Score="10" ViewCount="1881" Body="&lt;p&gt;A party is being held at a local mansion. The host is very rich and his success is because of one thing — his famous recipe for Spaghetti!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only guests that may attend are people who correctly reply to the guard at the door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's where you come in. You and a friend are trying to steal this recipe. You sneak by and listen to the passwords.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first guest arrives. The security says &quot;5&quot;, and the guest replies &quot;0&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second guest arrives. The security says &quot;6&quot;, the guest replies &quot;1&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third guest arrives. The security says &quot;3&quot;, the guest replies &quot;0&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your friend thinks he's got it all figured out so he walks up to the door and the security says &quot;2&quot;. He replied &quot;1&quot;, and is chased by dogs for stating an incorrect password.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another guest arrives, you seem to get the same pattern, security says &quot;1&quot; and he says &quot;0&quot; to get in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You now walk up to security and get &quot;7&quot;, what is the correct response?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Another guest arrives, you seem to get the same pattern, security says &quot;8&quot; and he says &quot;2&quot; to get in.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Another guest arrives, you seem to get the same pattern, security says &quot;0&quot; and he says &quot;1&quot; to get in.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No computations are needed for this problem so please don't answer using complex equations, it's just simple logic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for the ideas, credits to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 1 was created by warspyking and is found here: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3981/the-security-to-the-party&quot;&gt;The Security to the Party&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 2 in Mew's version &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3991/the-security-to-the-party-part-2&quot;&gt;The Security to the Party [Part 2]&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4944" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2014-11-11T16:46:21.933" Title="The Security to the Party [Part 3]" Tags="|pattern|password|" AnswerCount="1" CommentCount="0" ClosedDate="2014-11-17T20:40:41.100" ContentLicense="CC BY-SA 3.0" />
  <row Id="3998" PostTypeId="2" ParentId="3981" CreationDate="2014-11-11T05:17:42.307" Score="25" Body="&lt;p&gt;There are many possible answers. But I will say...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In general, the last digit of $x^3$.&lt;br&gt;&#xA; $$\begin{align}5^3 &amp;amp;= 125 \\ 6^3 &amp;amp;= 216 \\ 4^3 &amp;amp;= 64 \\ 7^3 &amp;amp;= 343 \\ 3^3 &amp;amp;= 27 \end{align}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From this, it is clear that we should respond:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 7 (as response to 3)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Our friend should have answered:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3 (as response to 7)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="68" LastEditDate="2014-11-11T07:15:37.823" LastActivityDate="2014-11-11T07:15:37.823" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3999" PostTypeId="2" ParentId="3997" CreationDate="2014-11-11T05:28:55.080" Score="17" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Number of circles in the figure ????&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It seems ridiculous, but &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;5,3,1 have got no full circles, 6 has got 1, 8 has 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then for 7 the reply should be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4850" LastEditorUserId="2313" LastEditDate="2014-11-11T05:46:02.603" LastActivityDate="2014-11-11T05:46:02.603" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4000" PostTypeId="2" ParentId="3956" CreationDate="2014-11-11T05:32:49.037" Score="10" Body="&lt;p&gt;There are 5 solutions to your particular problem.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$i_1 \rightarrow i_2 \rightarrow i_3 \rightarrow i_4$&lt;/li&gt;&#xA;&lt;li&gt;$i_3 \rightarrow i_1 \rightarrow i_2 \rightarrow i_4$&lt;/li&gt;&#xA;&lt;li&gt;$i_4 \rightarrow i_1 \rightarrow i_2 \rightarrow i_3$&lt;/li&gt;&#xA;&lt;li&gt;$i_4 \rightarrow i_1 \rightarrow i_3 \rightarrow i_2$&lt;/li&gt;&#xA;&lt;li&gt;$i_2 \rightarrow i_1 \rightarrow i_3 \rightarrow i_4$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This is a variation of the &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Nurse_scheduling_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nurse Scheduling Problem&lt;/a&gt;&quot;. It is NP-Hard so we can't do much better than brute-force and some heuristics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a Non-Deterministic Finite Automata here (a graph with directed and weighted edges):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/55oNW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the problem becomes, how many ways are there to accept the word &quot;1234f&quot;?&#xA;And equivalently, how many Hamiltonian paths are there from 0 to 5 where you must always take the cheapest edge that you haven't taken already (and f is the most expensive).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally, this particular problem is equivalent to the formal language:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L = \left\{ w | w = s_1 \rightarrow s_2 \rightarrow s_3 \rightarrow s_4 \land (\forall i \not= j. s_i \not= s_j) \land s_1 \in \{i_1,i_2,i_3,i_4\} \land s_2 \in \{i_1,i_2\} \land s_3 \in \{i_2,i_3\} \land s_4 \in \{i_2,i_3,i_4\}\right\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation:&lt;/strong&gt; L is a set of &quot;words&quot;. A word can be denoted w. So L is a set of all w where w looks like 4 symbols separated by a →. The 4 symbols are $i_1,i_2,i_3,i_4$. The little sub-formula with $\forall$ means that all symbols used must be distinct (i.e. an interviewer may only be used once). Then the last part of the formula expresses what the constraints on the positions of the symbols are: e.g. $s_2 \in \{i_i,i_2\}$ means that only interviewers $1$ and $2$ can go second (which matches your diagram).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The size of $L$ is: $|L| = 5$&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-12-04T04:19:57.093" LastActivityDate="2014-12-04T04:19:57.093" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4001" PostTypeId="1" AcceptedAnswerId="4023" CreationDate="2014-11-11T07:09:17.313" Score="30" ViewCount="4894" Body="&lt;p&gt;Linda sadly passes away in a stampede of elephants and when she reaches St. Peter, she is distressed because she doesn't think she'll be buried properly on the family grave plot and having to listen to her mother complain about her not being buried beside her will make even eternal paradise unenjoyable. However, on explaining her distress to St. Peter he smiles knowingly and hands her a card which he says is from Linda's husband back on the ground &lt;sub&gt;(no, he doesn't explain how, put your hand down)&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The card has 6 words on it, Linda reads it and smiles and enters Heaven assured that her husband will ensure she is buried properly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 3 words on the top line and 3 words underneath but there are only 5 unique words. The words are 'taking', 'take', 'to', 'your' and 'I'. What does the card say and therefore what is the message?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The repeated word is take.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The words are layed out as follows: &lt;br/&gt;&#xA; ____&lt;code&gt;&lt;/code&gt;____&lt;code&gt;&lt;/code&gt;____&lt;br/&gt;&#xA; ____&lt;code&gt;&lt;/code&gt;____&lt;code&gt;&lt;/code&gt;____&lt;br/&gt;&#xA;and the layout of the words are integral to the solution being correct&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As usual please place your answers in spoiler tags.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;As a side note, I'm not in any way religious (total atheist), however I needed a setting involving a dead person receiving a message from a living loved one, so this sprang to mind. It was not my intention to mock any religious beliefs/values/ideals/concepts held by anyone.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="4951" LastEditorUserId="5373" LastEditDate="2015-02-11T13:34:11.363" LastActivityDate="2015-02-11T13:34:11.363" Title="The Comforting Death Message" Tags="|word-problem|lateral-thinking|rebus|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4002" PostTypeId="1" AcceptedAnswerId="4006" CreationDate="2014-11-11T07:18:10.620" Score="3" ViewCount="1392" Body="&lt;p&gt;A party is being held at a local mansion. The host is very rich and his success is because of one thing — his famous recipe for Spaghetti!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only guests that may attend are people who correctly reply to the guard at the door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's where you come in. You and a friend are trying to steal this recipe. You sneak by and listen to the passwords.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first guest arrives. The security says &quot;5&quot;, and the guest replies &quot;2&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second guest arrives. The security says &quot;6&quot;, the guest replies &quot;2&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third guest arrives. The security says &quot;3&quot;, the guest replies &quot;2&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your friend thinks he's got it all figured out so he walks up to the door and the security says &quot;2&quot;. He replied &quot;2&quot;, and is tarred and feathered and sent away in shame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another guest arrives, security says &quot;0&quot; and he says &quot;0&quot; to get in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You walk up and get a &quot;4&quot; from security. What is your response?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for the ideas, credits to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 1 was created by warspyking and is found here: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3981/the-security-to-the-party&quot;&gt;The Security to the Party&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 2 in Mew's version &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3991/the-security-to-the-party-part-2&quot;&gt;The Security to the Party [Part 2]&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 3 in Kelvin Barsana's version &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3997/the-security-to-the-party-part-3&quot;&gt;The Security to the Party [Part 3]&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4317" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2014-11-11T09:39:23.867" Title="The Security to the Party [Part 4]" Tags="|pattern|password|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4003" PostTypeId="2" ParentId="3922" CreationDate="2014-11-11T07:19:08.633" Score="1" Body="&lt;p&gt;I'll try to keep the math as concise as possible. If we measure time from the moment a bit before the start of the experiment when extrapolating backwards the distance from the car to the wall would be$~0$, then at time $t$ the rubber rope will be stretched by a factor proportional to$~t$. If one uses marking on the rope itself to measure (relative) progress of the ant, then this stretching causes the effective speed relative to the markings to be divided by that stretching factor, so to be of the form $\frac bt$ where$~b&amp;gt;0$ is a constant depending on the unscaled speed of the ant and the above factor of proportionality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $t_0$ is the (positive) starting time, the progress of the ant at time $t_1&amp;gt;t_0$ is given by $\int_{t_0}^{t_1}\frac bt\mathrm d t$.&#xA;The basic fact that shows that the ant will ultimately arrive, no matter what the constants are, is that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;any anti-derivative of the function $t\mapsto\frac bt$ on the positive real numbers is unbounded above.&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Indeed any anti-derivative is $t\mapsto b\ln(t)+C$, and the logarithm is unbounded, though it advances very slowly (whence the ant needs a lot of time). Concretely the progress of the ant at time $t_1$ is given by $\int_{t_0}^{t_1}\frac bt\mathrm d t=b(\ln(t_1)-\ln(t_0))$, and this gets as large as one wants by increasing $t_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this also shows is that if the car were to accelerate slightly giving a stretch factor $t^{1.1}$ at time$~t$, then the arrival of the ant would no longer be assured. Indeed an anti-derivative of $t\mapsto \frac b{t^{1.1}}$ is $-10\frac b{t^{0.1}}+C$, which function is increasing, but bounded above (by the constant $C$) as $t\to+\infty$.&lt;/p&gt;&#xA;" OwnerUserId="1521" LastActivityDate="2014-11-11T07:19:08.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4004" PostTypeId="1" AcceptedAnswerId="4282" CreationDate="2014-11-11T07:27:46.523" Score="1" ViewCount="1066" Body="&lt;p&gt;A party is being held at a local mansion. The host is very rich and his success is because of one thing — his famous recipe for Spaghetti!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only guests that may attend are people who correctly reply to the guard at the door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's where you come in. You and a friend are trying to steal this recipe. You sneak by and listen to the passwords.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first guest arrives. The security says &quot;2&quot;, and the guest replies &quot;4&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second guest arrives. The security says &quot;3&quot;, the guest replies &quot;4&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third guest arrives. The security says &quot;5&quot;, the guest replies &quot;4&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fourth guest arrives. The security says &quot;8&quot;, the guest replies &quot;8&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your friend thinks he's got it all figured out so he walks up to the door and the security says &quot;4&quot;. He replied &quot;4&quot;, and is tarred and feathered and sent away in shame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another guest arrives, security says &quot;6&quot; and he says &quot;6&quot; to get in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You walk up and get a &quot;7&quot; from security. What is your response?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for the ideas, credits to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 1 was created by warspyking and is found here: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3981/the-security-to-the-party&quot;&gt;The Security to the Party&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 2 in Mew's version &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3991/the-security-to-the-party-part-2&quot;&gt;The Security to the Party [Part 2]&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 3 in skv's version &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3997/the-security-to-the-party-part-3&quot;&gt;The Security to the Party [Part 3]&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 4 in JNF's version &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4002/the-security-to-the-party-part-4&quot;&gt;The Security to the Party [Part 4]&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tip 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oops! I'm sorry. The tip WAS wrong!&lt;br&gt;&lt;strong&gt;Real tip:&lt;/strong&gt;&lt;br&gt;It's related to electronic display screen. Draw the number like this:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; -&#xA;| |&#xA; -&#xA;| |&#xA; -&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; And COMPARE WITH the answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tip 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2 more example:&lt;br&gt;security: 1 guest: 0&lt;br&gt;security: 4 guest: 3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4952" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2014-11-13T11:29:01.307" Title="The Security to the Party [Part 5]" Tags="|password|" AnswerCount="5" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="4006" PostTypeId="2" ParentId="4002" CreationDate="2014-11-11T07:40:39.460" Score="12" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Respond $1$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Put x into binary. The number of 1s in the result is the answer.&#xA; Here is a key for the given numbers:&#xA; $$\begin{align}5&amp;amp;=101 &amp;amp;\rightarrow 2 \\ 6&amp;amp;=110 &amp;amp;\rightarrow 2 \\ 3&amp;amp;=11 &amp;amp;\rightarrow 2 \\ 2&amp;amp;=10 &amp;amp;\rightarrow 1 \\ 0&amp;amp;=0 &amp;amp;\rightarrow 0 \\ 4&amp;amp;=100 &amp;amp;\rightarrow 1\end{align}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-11-11T07:53:28.640" LastActivityDate="2014-11-11T07:53:28.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4007" PostTypeId="2" ParentId="4002" CreationDate="2014-11-11T07:44:09.553" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Number of 1's in the binary code of the decimal number being called.....&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So for 4 the response should be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4850" LastActivityDate="2014-11-11T07:44:09.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4008" PostTypeId="5" CreationDate="2014-11-11T07:46:00.437" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-11-11T07:46:00.437" LastActivityDate="2014-11-11T07:46:00.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4009" PostTypeId="4" CreationDate="2014-11-11T07:46:00.437" Score="0" Body="A riddle asking to identify a series. Each item in the series generates a response. Exposing a few of the correct responses (and perhaps some incorrect ones as well) gives clues on finding the pattern, or structure, of the series." OwnerUserId="4317" LastEditorUserId="4317" LastEditDate="2014-11-11T09:05:48.757" LastActivityDate="2014-11-11T09:05:48.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4013" PostTypeId="1" AcceptedAnswerId="4150" CreationDate="2014-11-11T07:51:25.077" Score="3" ViewCount="972" Body="&lt;p&gt;A party is being held at a local mansion. The host is very rich and his success is because of one thing — his famous recipe for Spaghetti!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only guests that may attend are people who correctly reply to the guard at the door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's where you come in. You and a friend are trying to steal this recipe. You sneak by and listen to the passwords.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first guest arrives. The security says &quot;1&quot;, and the guest replies &quot;0&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second guest arrives. The security says &quot;6&quot;, the guest replies &quot;0&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third guest arrives. The security says &quot;7&quot;, the guest replies &quot;0&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your friend thinks he's got it all figured out so he walks up to the door and the security says &quot;2&quot;. He replied &quot;0&quot;, and is tarred and feathered and sent away in shame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another guest arrives, security says &quot;9&quot; and he says &quot;3&quot; to get in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Another guest arrives, security says &quot;2&quot; and he says &quot;2&quot; to get in.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Another guest arrives, security says &quot;4&quot; and he says &quot;3&quot; to get in.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Another guest arrives, security says &quot;0&quot; and he says &quot;3&quot; to get in.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You walk up and get a &quot;8&quot; from security. What is your response?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint #4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 8.  You might not think that's a hint, but it is if you think outside the box.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #5 (last hint and I will post the answer later tonight if still unsolved):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; See my profile.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks for the ideas, credits to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 1 was created by warspyking and is found here: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3981/the-security-to-the-party&quot;&gt;The Security to the Party&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 2 in Mew's version &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3991/the-security-to-the-party-part-2&quot;&gt;The Security to the Party [Part 2]&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 3 in skv's version &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3997/the-security-to-the-party-part-3&quot;&gt;The Security to the Party [Part 3]&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 4 in JNF's version &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4002/the-security-to-the-party-part-4&quot;&gt;The Security to the Party [Part 4]&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 5 in user3804649's version &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4004/the-security-to-the-party-part-5&quot;&gt;The Security to the Party [Part 5]&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2014-11-12T04:16:29.410" Title="The Security to the Party [Part 6]" Tags="|password|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4014" PostTypeId="1" AcceptedAnswerId="4052" CreationDate="2014-11-11T08:06:10.467" Score="5" ViewCount="860" Body="&lt;p&gt;A party is being held at a local mansion. The host is very rich and his success is because of one thing &amp;mdash; his famous recipe for spaghetti!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only guests that may attend are people who correctly reply to the guard at the door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's where you come in. You and a friend are trying to steal this recipe. You sneak by and listen to the passwords.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first guest arrives. The security says &quot;8&quot;, and the guest replies &quot;4&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second guest arrives. The security says &quot;2&quot;, the guest replies &quot;1&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third guest arrives. The security says &quot;6&quot;, the guest replies &quot;3&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your friend thinks he's got it all figured out (again), so he walks up to the door and the security says &quot;16&quot;. He replies &quot;8&quot;, and is laughed out of town.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another guest arrives, security says &quot;10&quot; and the guest replies &quot;4&quot; to get in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You walk up and get a &quot;13&quot; from security. What is your response?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the latest installment in the series:&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/q/3981/68&quot;&gt;The Security to the Party&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/q/3991/68&quot;&gt;The Security to the Party [Part 2]&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/q/3997/68&quot;&gt;The Security to the Party [Part 3]&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/q/4002/68&quot;&gt;The Security to the Party [Part 4]&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/q/4004/68&quot;&gt;The Security to the Party [Part 5]&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/q/4013/68&quot;&gt;The Security to the Party [Part 6]&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2017-07-29T05:16:53.377" Title="The Security to the Party [Part 7]" Tags="|password|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4015" PostTypeId="1" AcceptedAnswerId="4021" CreationDate="2014-11-11T08:06:39.703" Score="3" ViewCount="981" Body="&lt;p&gt;A party is being held at a local mansion. The host is very rich and his success is because of one thing ~ his famous recipe for Spaghetti!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only guests that may attend are people the host knows and trusts, whom he gave the password to enter to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's where you come in. You and a friend are trying to steal this recipe. You sneak by and listen to the passwords.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first guest arrives. The security says &quot;6&quot;, and the guest replies &quot;0&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second guest arrives. The security says &quot;5&quot;, the guest replies &quot;1&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next guest arrives.  The security says &quot;10&quot;, the guest replies &quot;1&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next guest arrives. The security says &quot;7&quot;, the guest replies &quot;2&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next guest arrives.  The security says &quot;8&quot;, the guest replies &quot;2&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You walk up to security and get &quot;3&quot;, what is the correct response?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hint:  Look at some of the previous solutions in this series.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="97" LastEditorUserId="2518" LastEditDate="2014-11-11T13:44:11.427" LastActivityDate="2014-11-11T13:44:11.427" Title="The Security to the Party [Part 8]" Tags="|password|" AnswerCount="2" CommentCount="6" ClosedDate="2014-11-17T20:41:20.627" ContentLicense="CC BY-SA 3.0" />
  <row Id="4016" PostTypeId="2" ParentId="4015" CreationDate="2014-11-11T08:17:50.313" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2. The number of closed spaces in the lower case spelling of the number the guard gives you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4959" LastEditorUserId="4959" LastEditDate="2014-11-11T08:41:29.933" LastActivityDate="2014-11-11T08:41:29.933" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4017" PostTypeId="5" CreationDate="2014-11-11T08:29:05.517" Score="0" Body="&lt;p&gt;A &lt;a href=&quot;https://en.wikipedia.org/wiki/Password&quot; rel=&quot;nofollow noreferrer&quot;&gt;password&lt;/a&gt; puzzle focuses on finding a password to a challenge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Password are strings of symbols that are used to protect information. Usually only a select number of people know it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unlike real-world passwords, here, on Puzzling.SE, clues are usually given to enable the solver to determine the password.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Password_strength&quot; rel=&quot;nofollow noreferrer&quot;&gt;strength&lt;/a&gt; of a password is a measure of how difficult the password is to 'crack', i.e. how many trial and errors are needed to determine an unknown password.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This site does not claim to provide any security on passwords. For a more rigorous approach to information security, please see:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://crypto.stackexchange.com/&quot;&gt;https://crypto.stackexchange.com/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://security.stackexchange.com/&quot;&gt;https://security.stackexchange.com/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="68" LastEditorUserId="11497" LastEditDate="2020-03-16T09:47:30.557" LastActivityDate="2020-03-16T09:47:30.557" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4018" PostTypeId="4" CreationDate="2014-11-11T08:29:05.517" Score="0" Body="A password puzzle focuses on finding a password to a challenge." OwnerUserId="68" LastEditorUserId="11497" LastEditDate="2020-03-16T09:47:30.557" LastActivityDate="2020-03-16T09:47:30.557" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4019" PostTypeId="2" ParentId="1660" CreationDate="2014-11-11T08:33:49.837" Score="2" Body="&lt;p&gt;The open questions in the other answers create loopholes that can lead to the answer not being discernable.  - We have little assurances about what the lying troll answers to &quot;what would you tell me if...&quot;.  The only thing we know is that the lying troll would not say &quot;I would say yes&quot; (if the truthfull answer to the containeed question was no) or &quot;I would say no&quot; (if the truthfull answer to the containeed question was no).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By explicitly making the question a yes or no question you can avoid this;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;If I asked you, does this road lead to death, would you answer yes?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1654" LastActivityDate="2014-11-11T08:33:49.837" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4020" PostTypeId="2" ParentId="2237" CreationDate="2014-11-11T08:36:19.813" Score="0" Body="&lt;p&gt;There are a couple of loopholes in the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to kidnap the 3 wise monkeys, they would all witness it, but none would be able to indict you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, anything which adds disabilities to the monkeys, ie. making them all blind. Again, they can all witness it, but assuming you take care to protect your identity (wear a mask, talk in a chipmunk voice), none can ID you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last one I've got is that if the suspect is actually not the criminal, then (assuming the monkeys are lawful) they wouldn't be able to indict &lt;strong&gt;the suspect&lt;/strong&gt; - the question doesn't say indict the criminal.&lt;/p&gt;&#xA;" OwnerUserId="2518" LastActivityDate="2014-11-11T08:36:19.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4021" PostTypeId="2" ParentId="4015" CreationDate="2014-11-11T08:43:17.347" Score="8" Body="&lt;p&gt;The correct answer is...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2 - for each response, you should count the number of enclosed areas in the lowercase letters for the challenge word.&lt;br&gt;&#xA;&lt;br&gt;&#xA; six = no closures&lt;br&gt;&#xA; five = 1 closure (e)&lt;br&gt;&#xA; ten = 1 closure (e)&lt;br&gt;&#xA; seven = 2 closures (e, e)&lt;br&gt;&#xA; eight = 2 closures (e, g)&lt;br&gt;&#xA;&lt;br&gt;&#xA; three = 2 closures (e, e)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2518" LastActivityDate="2014-11-11T08:43:17.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4023" PostTypeId="2" ParentId="4001" CreationDate="2014-11-11T09:19:25.533" Score="37" Body="&lt;p&gt;I've seen a variation of this before:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;Take   to    Taking&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;I     see     Your&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Which reads as &quot;I undertake to oversee your undertaking&quot;. You don't allow the word &quot;see&quot; but I assume it's a variation of that....?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4963" LastEditorUserId="2518" LastEditDate="2014-11-11T09:22:58.853" LastActivityDate="2014-11-11T09:22:58.853" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4025" PostTypeId="1" AcceptedAnswerId="4048" CreationDate="2014-11-11T09:44:56.767" Score="2" ViewCount="1050" Body="&lt;p&gt;A party is being held at a local mansion. The host is very rich and his success is because of one thing ~ his famous recipe for Spaghetti!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only guests that may attend are people the host knows and trusts, whom he gave the password to enter to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's where you come in. You and a friend are trying to steal this recipe. You sneak by and listen to the passwords.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first guest arrives. The security says &quot;6&quot;, and the guest replies &quot;6&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second guest arrives. The security says &quot;5&quot;, the guest replies &quot;1&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next guest arrives. The security says &quot;10&quot;, the guest replies &quot;2&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next guest arrives. The security says &quot;7&quot;, the guest replies &quot;5&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next guest arrives. The security says &quot;8&quot;, the guest replies &quot;4&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You walk up to security and get &quot;3&quot;, what is the correct response?&lt;/p&gt;&#xA;" OwnerUserId="1491" LastActivityDate="2014-11-11T17:29:27.230" Title="The Security to the Party [Part 9]" Tags="|mathematics|password|" AnswerCount="7" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4026" PostTypeId="1" AcceptedAnswerId="4033" CreationDate="2014-11-11T09:55:26.427" Score="41" ViewCount="9662" Body="&lt;p&gt;You are going through the belongings of your recently deceased uncle and come upon a small wooden box with the following inscription, presumably describing its contents.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Those who take me, won't know me.&lt;br&gt;&#xA;  Those who know me, won't want me.&lt;br&gt;&#xA;  Those who want me, will make me.&lt;br&gt;&#xA;  Those who make me, will give me away.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You open the box to find it empty. What did your uncle--who was a scoundrel, by the way--keep in this box?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; While your uncle was an unsavory person, he was not a murderer.&lt;/p&gt;&#xA;" OwnerDisplayName="user4221" LastEditorUserId="5373" LastEditDate="2015-05-03T10:54:11.810" LastActivityDate="2015-08-06T17:52:07.633" Title="Nobody knows me. Nobody wants me" Tags="|riddle|story|" AnswerCount="9" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4028" PostTypeId="2" ParentId="4026" CreationDate="2014-11-11T09:58:40.917" Score="2" Body="&lt;p&gt;All I can think of is some kind of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; unpleasant drug... maybe something which makes people's memories hazy?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Those who take me, won't know me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; makes people forget they've taken it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Those who know me, won't want me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you know about it you won't take it!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Those who want me, will make me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bad guys make it at home?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Those who make me, will give me away&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Then give it to victims&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I feel like I'm waaaay off course with this one!&lt;/p&gt;&#xA;" OwnerUserId="2466" LastActivityDate="2014-11-11T09:58:40.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4029" PostTypeId="2" ParentId="4026" CreationDate="2014-11-11T09:58:48.193" Score="26" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some kind of undetectable poison  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With the reasoning being that if you ...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;drink it, you wouldn't know&lt;/li&gt;&#xA;&lt;li&gt;know what it is, you wouldn't want to take it yourself&lt;/li&gt;&#xA;&lt;li&gt;want it, you would make it&lt;/li&gt;&#xA;&lt;li&gt;make it, you'd give it to someone else&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2518" LastEditorUserId="2518" LastEditDate="2014-11-25T10:25:55.217" LastActivityDate="2014-11-25T10:25:55.217" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4030" PostTypeId="1" AcceptedAnswerId="4043" CreationDate="2014-11-11T09:58:50.287" Score="5" ViewCount="1814" Body="&lt;p&gt;The murderer has fled the scene, you have a feeling he will try to leave the country. You get access to his credit card records and see that he has recently spent £500 on flights at Getaway Travel Agency. You immediately rush to the shop to inspect their records.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, Getaway is a very old-fashioned Travel Agency and they still use hand-written tickets with carbon-copies. Even worse, the staff have no memory of the man who bought the ticket and the carbon copy has not worked very well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All you can see for the departure airport is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;THE&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&quot;The what?&quot; you ask the manager.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I don't know,&quot; says the manager, looking puzzled. &quot;We always write-in the full name of the Departure airport in that box&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being the genius that you are, you immediately  dispatch your men to the correct airport where he is arrested boarding the plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which airport?&lt;/p&gt;&#xA;" OwnerUserId="4963" LastEditorUserId="22922" LastEditDate="2016-09-19T09:57:33.663" LastActivityDate="2016-09-19T09:57:33.663" Title="Catch the killer" Tags="|word|" AnswerCount="7" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4031" PostTypeId="2" ParentId="4030" CreationDate="2014-11-11T10:02:07.077" Score="10" Body="&lt;p&gt;You send your men to...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.world-airport-codes.com/brazil/senador-petrnio-portella-9213.html&quot;&gt;Senador Petrônio Portela Airport&lt;/a&gt; in Teresina, Brazil - its airport code is &quot;THE&quot; (in the same way LA International is LAX)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2518" LastActivityDate="2014-11-11T10:02:07.077" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4032" PostTypeId="2" ParentId="4030" CreationDate="2014-11-11T10:04:08.420" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The airport in question is 'Theodore' in eastern Australia - the only airport whose full name starts with 'The'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1491" LastActivityDate="2014-11-11T10:04:08.420" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4033" PostTypeId="2" ParentId="4026" CreationDate="2014-11-11T10:09:22.370" Score="50" Body="&lt;p&gt;It was&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Counterfeit money&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Those who take me, won't know me.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Those who receive counterfeit money do not know that the money is actually counterfeit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Those who know me, won't want me.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Those who know that the money is counterfeit will not accept it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Those who want me, will make me.&lt;br&gt;&#xA;Those who make me, will give me away.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Those are counterfeiters, like the infamous uncle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4394" LastEditorUserId="11475" LastEditDate="2015-08-06T17:52:07.633" LastActivityDate="2015-08-06T17:52:07.633" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4034" PostTypeId="2" ParentId="4026" CreationDate="2014-11-11T10:10:04.253" Score="17" Body="&lt;p&gt;just a wild guess, i think there is nothing in the box?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those who take nothing, won't know nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those who know nothing, won't want nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those who want nothing, will make nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those who make nothing, will give nothing away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i think that's why the box is empty?&lt;/p&gt;&#xA;" OwnerUserId="4944" LastActivityDate="2014-11-11T10:10:04.253" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4035" PostTypeId="2" ParentId="4026" CreationDate="2014-11-11T10:20:44.043" Score="2" Body="&lt;p&gt;I Think it's a joint(Tobacco) used to smoke.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those who take me, won't know me&#xA;(While they are smoking, they don't understand it's harmfull effects on their body)&#xA;Those who know me, won't want me.&#xA;(One who understand those effect will try to avoid smoking)&#xA;Those who want me, will make me.&#xA;(One who wants to smoke will put effort to prepare it correctly, mixing all ingredients)&#xA;Those who make me, will give me away.&#xA;(One who makes it share it with his friends, thus giving &quot;AWAY&quot;)&lt;/p&gt;&#xA;" OwnerUserId="4875" LastActivityDate="2014-11-11T10:20:44.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4036" PostTypeId="1" AcceptedAnswerId="4038" CreationDate="2014-11-11T10:21:52.260" Score="5" ViewCount="703" Body="&lt;p&gt;I'm a young programmer from NYC attending a conference in London. Last night I met a fellow developer in a bar she was young, beautiful and had the sexiest English accent!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the evening drew to a close she pulled my hand close to her, pulled out a pen and wrote her phone number on it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;01861 919550&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My heard skipped a beat, this vision of geeky beauty really did like me!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's now the following morning, I've tried to call her but number isn't recognised. I'm devastated, please help me find my lost love before I have to go back to the US!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suggested Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some help it's been suggested that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;01861 919220&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Actually works better, I've edited it into the question to help anyone wanting to solve it!&lt;/p&gt;&#xA;" OwnerUserId="2466" LastEditorUserId="2466" LastEditDate="2014-11-11T10:44:56.197" LastActivityDate="2014-11-11T16:30:42.760" Title="I thought she liked me?" Tags="|logical-deduction|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4037" PostTypeId="2" ParentId="4030" CreationDate="2014-11-11T10:24:31.763" Score="0" Body="&lt;p&gt;The name of the airport is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Departure&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is written in upper case, so it seems to be a proper name and not just where the airplane departs from.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4971" LastActivityDate="2014-11-11T10:24:31.763" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4038" PostTypeId="2" ParentId="4036" CreationDate="2014-11-11T10:24:49.410" Score="5" Body="&lt;p&gt;Sorry to hear about your lost love. Have you tried&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Calling 05561 619810 - I think you may have read the number upside down!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1561" LastActivityDate="2014-11-11T10:24:49.410" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="4039" PostTypeId="2" ParentId="4030" CreationDate="2014-11-11T10:24:50.683" Score="4" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_airports_in_the_United_Kingdom_and_the_British_Crown_Dependencies&quot; rel=&quot;nofollow&quot;&gt;Thetford&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We know that the killer spent £500, so we're dealing with a Pound currency. The most common of them is GBP, so we're looking for airports in the UK.&lt;br&gt;&#xA;&lt;br&gt;&#xA;  (I have checked airports in all other pound-currency countries, and none contain &quot;THE&quot; in upper or lowercase.&lt;br&gt;&#xA;&lt;br&gt;&#xA;  The only airport in the UK starting with &quot;THE&quot; is &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_airports_in_the_United_Kingdom_and_the_British_Crown_Dependencies&quot; rel=&quot;nofollow&quot;&gt;Thetford&lt;/a&gt; in Norfolk&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2518" LastEditorUserId="2518" LastEditDate="2014-11-11T10:34:14.690" LastActivityDate="2014-11-11T10:34:14.690" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4040" PostTypeId="2" ParentId="4030" CreationDate="2014-11-11T10:25:03.900" Score="0" Body="&lt;p&gt;is it..&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Heathrow Airport in England ? the word Heath means a wasteland or uncultivated land (the ROW is empty hence it's a HeathRow).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4962" LastActivityDate="2014-11-11T10:25:03.900" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4041" PostTypeId="2" ParentId="4030" CreationDate="2014-11-11T10:25:43.060" Score="2" Body="&lt;p&gt;Key word is &quot;that box&quot;. Airport code is spelled backwards:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Airport is (EHT) Rentschler Airport.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4973" LastEditorUserId="2518" LastEditDate="2014-11-11T10:27:12.820" LastActivityDate="2014-11-11T10:27:12.820" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4042" PostTypeId="2" ParentId="4036" CreationDate="2014-11-11T10:30:41.880" Score="1" Body="&lt;p&gt;From your comment on Chris's answer, the clue is in the city.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have you tried calling&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; +44 1861 919550 - if you're in NYC (maybe you flew back), you're ringing an American number, she's English&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2518" LastActivityDate="2014-11-11T10:30:41.880" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4043" PostTypeId="2" ParentId="4030" CreationDate="2014-11-11T10:34:43.400" Score="10" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most likely, Southend.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The killer paid the ticket in pounds, so we're probably in England. The only international airport with 'the' in its name, is &lt;em&gt;London Southend Airport&lt;/em&gt;.&lt;br&gt;&#xA; All other &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_airports_in_the_United_Kingdom_and_the_British_Crown_Dependencies&quot; rel=&quot;nofollow noreferrer&quot;&gt;airports in the UK&lt;/a&gt; containing 'the' are private, military, or local.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Another option would be &lt;em&gt;St. Catherine International Airport&lt;/em&gt; in Egypt. It contains 'the' as well and Egypt uses the pound, like England does. &lt;em&gt;(Hat tip to &lt;a href=&quot;https://puzzling.stackexchange.com/users/5081/ergwun&quot;&gt;Ergwun&lt;/a&gt;).&lt;/em&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; I haven't been able to find any other airports with 'the' in its name in a country using the pound as its currency.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="68" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2014-11-12T07:22:53.377" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="4046" PostTypeId="1" AcceptedAnswerId="4792" CreationDate="2014-11-11T11:03:05.560" Score="3" ViewCount="728" Body="&lt;p&gt;A party is being held at a local mansion. The host is very rich and his success is because of one thing — his famous recipe for spaghetti!&lt;/p&gt;&#xA;&lt;p&gt;The only guests that may attend are people who correctly reply to the guard at the door.&lt;/p&gt;&#xA;&lt;p&gt;Here's where you come in. You and a friend are trying to steal this recipe. You sneak by and listen to the passwords.&lt;/p&gt;&#xA;&lt;p&gt;The first guest arrives. The security says &amp;quot;2&amp;quot;, and the guest replies &amp;quot;2&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;The second guest arrives. The security says &amp;quot;4&amp;quot;, the guest replies &amp;quot;4&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;Your friend thinks he's got it all figured out (again), so he walks up to the door and the security says &amp;quot;6&amp;quot;. He replies &amp;quot;6&amp;quot;, and is &lt;a href=&quot;https://puzzling.stackexchange.com/q/4001/68&quot;&gt;trampled to death by elephants&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Another guest arrives, security says &amp;quot;3&amp;quot; and the guest replies &amp;quot;2&amp;quot; to get in.&lt;/p&gt;&#xA;&lt;p&gt;You walk up and get a &amp;quot;5&amp;quot; from security. What is your response?&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;The correct answer has already been given, but the method hasn't been guessed. So here's a hint:&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th&gt;challenge&lt;/th&gt;&#xA;&lt;th&gt;response&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;1&lt;/td&gt;&#xA;&lt;td&gt;0&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;2&lt;/td&gt;&#xA;&lt;td&gt;2&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;3&lt;/td&gt;&#xA;&lt;td&gt;2&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;4&lt;/td&gt;&#xA;&lt;td&gt;4&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;5&lt;/td&gt;&#xA;&lt;td&gt;4&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;6&lt;/td&gt;&#xA;&lt;td&gt;4&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&lt;hr /&gt;&#xA;&lt;p&gt;And here is another hint: &amp;quot;Petals Around the Rose&amp;quot; (although it's not that).&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="69582" LastEditDate="2021-01-13T15:58:40.213" LastActivityDate="2021-01-13T15:58:40.213" Title="The Security to the Party [X]" Tags="|password|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4047" PostTypeId="2" ParentId="3943" CreationDate="2014-11-11T11:07:38.400" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Answer is zero or twelve as it sits directly opposite the clock face of the number used as the question. I guess guard 2's answer could also have been Eighteen?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4979" LastActivityDate="2014-11-11T11:07:38.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4048" PostTypeId="2" ParentId="4025" CreationDate="2014-11-11T11:09:15.797" Score="8" Body="&lt;p&gt;Putting these in a table makes a very simple possible pattern much more apparent, where the first column gives the questions, the middle column gives the known answers, and the last column is filled in based on the pattern:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;1  |   || 5&#xA;2  |   || 4&#xA;3  |   || 3&#xA;4  |   || 2&#xA;5  | 1 || 1&#xA;6  | 6 || 6&#xA;7  | 5 || 5&#xA;8  | 4 || 4&#xA;9  |   || 3&#xA;10 | 2 || 2&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;so my answer is 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As with all these questions, it may or may not be the answer you're looking for, but it works.&lt;/p&gt;&#xA;" OwnerUserId="1764" LastActivityDate="2014-11-11T11:09:15.797" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4049" PostTypeId="2" ParentId="4025" CreationDate="2014-11-11T11:13:56.190" Score="3" Body="&lt;p&gt;My answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $3$, since that's what I need to add to the guard's number to make it a multiple of $6$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="68" LastActivityDate="2014-11-11T11:13:56.190" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4050" PostTypeId="2" ParentId="4046" CreationDate="2014-11-11T11:17:21.887" Score="5" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;Update:&lt;/strong&gt; This is a now outdated answer to the &lt;a href=&quot;https://puzzling.stackexchange.com/revisions/4046/1&quot;&gt;original version&lt;/a&gt; of the puzzle. The presented solution does no longer match the current formulation.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd guess the solution is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4, i.e. the largest power of 2 not above the number said by the security.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I'm just too afraid of those damn elephants :-D&lt;/p&gt;&#xA;" OwnerUserId="4394" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2014-11-16T10:08:19.780" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4051" PostTypeId="2" ParentId="4025" CreationDate="2014-11-11T11:17:25.310" Score="3" Body="&lt;p&gt;The answer is 3. Find the highest multiple of 12 that is lower than or equal to the question. Subtract the multiple from the question. If the result is greater than 5 then you subtract it from 12, but if the result is 5 or less, then you subtract is from 6. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the highest multiple of 12 less than or equal to 3 is 0. &lt;/li&gt;&#xA;&lt;li&gt;3 - 0 = 3.&lt;/li&gt;&#xA;&lt;li&gt;3 is less than 5 so 6-3 = 3.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4978" LastEditorUserId="4978" LastEditDate="2014-11-11T11:47:59.537" LastActivityDate="2014-11-11T11:47:59.537" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4052" PostTypeId="2" ParentId="4014" CreationDate="2014-11-11T11:21:50.867" Score="10" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is the number of prime numbers up to and including the number the&#xA; guard says.  Therefore you should answer with 6.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="97" LastActivityDate="2014-11-11T11:21:50.867" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4053" PostTypeId="2" ParentId="4026" CreationDate="2014-11-11T11:33:13.777" Score="17" Body="&lt;p&gt;Alternate answer:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;h1&gt;Riddles&lt;/h1&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Those who take me, won't know me:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;If you receive a riddle, you wont know it (the corresponding answer).&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Those who know me, won't want me.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;If you do know it, you don't want to receive the riddle anymore.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Those who want me, will make me.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;People who like riddles, will make more of them&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Those who make me, will give me away.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Those who make riddles, will give them to other people.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1666" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-11-11T11:33:13.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4054" PostTypeId="1" AcceptedAnswerId="4064" CreationDate="2014-11-11T11:40:58.190" Score="7" ViewCount="1007" Body="&lt;p&gt;A party is being held at a local mansion. The host is very rich and his success is because of one thing &amp;mdash; his famous recipe for spaghetti!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only guests that may attend are people who correctly reply to the guard at the door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's where you come in. You and a friend are trying to steal this recipe. You sneak by and listen to the passwords.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You notice a chalkboard next to the guard which you haven't seen before. It looks like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                    Week XI&#xA;                   ---------&#xA;This week's passwords were changed 2 days ago since&#xA;John the Guard was taken ill, and Carlo the Guard&#xA;hasn't had time to learn that system yet.&#xA;If you didn't receive the message, please check your&#xA;mobile phone for the new rules.&#xA;&#xA;ZERO (0) will never be a correct answer, and none of&#xA;the numbers ONE (1), TWO (2) or THREE (3) will be asked&#xA;because it's far too insecure, anyone could just repeat&#xA;the same number and we don't want that.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You quickly check your mobile phone, just in case they made a horrendous mistake, but you have no messages at all - they're not quite that daft. Any attempts to steal a mobile phone would undoubtedly be noticed by the guards, so you decide it's far too risky to try to game the system, and instead hang around to overhear some challenges and responses.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The first guest arrives. The security says &quot;4&quot;, and the guest replies &quot;3&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second guest arrives. The security says &quot;25&quot;, the guest replies &quot;6&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third guest arrives. The security says &quot;16&quot;, the guest replies &quot;5&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your friend thinks he's got it all figured out - it's just $\sqrt{n}+1$, so he walks up to the door and the security says &quot;49&quot;. He replies &quot;8&quot;, and is &lt;a href=&quot;https://puzzling.stackexchange.com/q/4001/68&quot;&gt;trampled to death by an irrelevant elephant&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another guest arrives, security says &quot;18&quot; and the guest replies &quot;7&quot; to get in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You walk up and get a &quot;36&quot; from security. What is your response?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For peace of mind (since someone decided to have a go at me for not stating this upfront on a previous puzzle): &lt;strong&gt;Every response is independent of every other challenge/response pair, and there is no hidden information (such as time of day).&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is the latest installment in the series:&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/q/3981/68&quot;&gt;The Security to the Party&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/q/3991/68&quot;&gt;The Security to the Party [Part 2]&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/q/3997/68&quot;&gt;The Security to the Party [Part 3]&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/q/4002/68&quot;&gt;The Security to the Party [Part 4]&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/q/4004/68&quot;&gt;The Security to the Party [Part 5]&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/q/4013/68&quot;&gt;The Security to the Party [Part 6]&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/q/4014/68&quot;&gt;The Security to the Party [Part 7]&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/q/4015/68&quot;&gt;The Security to the Party [Part 8]&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/q/4025/68&quot;&gt;The Security to the Party [Part 9]&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/q/4046/68&quot;&gt;The Security to the Party [X]&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2518" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2014-11-11T13:31:15.960" Title="The Security to the Party [XI]" Tags="|password|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4055" PostTypeId="2" ParentId="3922" CreationDate="2014-11-11T11:45:38.603" Score="0" Body="&lt;p&gt;An intuitive answer would be that the rubber band acts like an accelerating moving walkway for the ant. Every step it takes it's speed increases a little bit. At the start of the rubber band it's at a, and at the end it's at speed c+a. So it must be accelerating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the ant is accelerating and the car's speed is constant. The ant will eventually catch the car.&lt;/p&gt;&#xA;" OwnerUserId="4988" LastActivityDate="2014-11-11T11:45:38.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4056" PostTypeId="2" ParentId="4046" CreationDate="2014-11-11T11:46:01.940" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2 and 3 answer is 2, 4 and 5's answer is 4, so for 6 it would've been 7 I think&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is already out there, I am just trying to figure out a correct method.&lt;/p&gt;&#xA;" OwnerUserId="4850" LastActivityDate="2014-11-11T11:46:01.940" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4057" PostTypeId="1" AcceptedAnswerId="4776" CreationDate="2014-11-11T12:44:29.610" Score="29" ViewCount="2657" Body="&lt;p&gt;This party is so popular!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3981/the-security-to-the-party&quot;&gt;just like before, you want the spaghetti recipe&lt;/a&gt; yada yada yada... basically, you want to get in to the party.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first guest arrives, the security guard says &quot;6&quot;. The guest goes away to check the answer, but soon returns, saying &quot;3&quot;, and is admitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second guest arrives, the security guard looks annoyed and says &quot;1&quot;, in a menacing tone. The guest says &quot;9&quot; and is admitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third guest arrives, the security guard says &quot;5&quot; and the guest replies back, &quot;0&quot;, and is admitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's the next guest's turn now. The security guard says &quot;2&quot;, and to your surprise, the guest attacks him! The guest overpowers the security guard, and tries to enter the party without saying anything! But another security guard appears (from the infinite stock of security guards in the guardroom) and stops him. Abashed, the guest says &quot;2&quot; and is admitted to the party.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now the original security guard is back on his feet and he's looking angry. Another guest comes up to him and the guard grabs him by the collar. Clearly he's looking to get his revenge on these unruly guests! He snarls, &quot;3&quot;, but the guest replies &quot;5&quot; and the security guard is forced to admit him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's your turn now. You go up to the door and the security guard says &quot;4&quot;. You hope that he'll admit you, but you know that it's pointless to try overpowering him. What do you reply?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hints&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you're standing outside the party waiting for inspiration to strike, you hear music and voices coming from inside the party. Could this be a clue? [Yes]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The track listing is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Royale With Cheese (Dialogue), &#xA;Pulp Fiction on Pulp Fiction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/F8MMM.jpg&quot; alt=&quot;Pulp Fiction (album cover)&quot;&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;20th Century Fox, &#xA;The Doors on The Best of The Doors&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QMXjZ.jpg&quot; alt=&quot;Best of The Doors (album cover)&quot;&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Mr Tambourine Man, &#xA;The Byrds on American Graffiti (More) - 1979&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Out of Space, &#xA;The Prodigy&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Space Oddity, &#xA;David Bowie on Best of Bowie&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BfIWB.jpg&quot; alt=&quot;bowie (album cover)&quot;&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Indiana Jones Theme, &#xA;John Williams on Raiders of the Lost Ark&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uQViX.jpg&quot; alt=&quot;Raiders of the Lost Ark (album cover)&quot;&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Space Cowboy, &#xA;Jamiroquai on The Return of the Space Cowboy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fo1AT.jpg&quot; alt=&quot;The Return of the Space Cowboy (album cover)&quot;&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Galaxy Song, &#xA;Eric Idle on Eric Idle Sings Monty Python&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/G8gbE.png&quot; alt=&quot;Eric Idle Sings Monty Python (album cover)&quot;&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Girl From Mars, &#xA;Ash on 1977&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/z8U8C.png&quot; alt=&quot;1977 (album cover)&quot;&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Across the Universe, &#xA;Fiona Apple on Across the Universe&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NMtUJ.jpg&quot; alt=&quot;Across the Universe (album cover)&quot;&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Choose Life, &#xA;Underworld&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Born To Be Wild (With Miss Piggy)&#xA;Ozzy Osbourne, on Prince Of Darkness (Box Set) &lt;img src=&quot;https://i.stack.imgur.com/MZ9ej.jpg&quot; alt=&quot;Prince Of Darkness (Box Set - cover art)&quot;&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Clue: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4642/&quot;&gt;Decode the message enciphered in these symbols: ◳◰ ◓◨ ◨◧◕ ◎◌ ◱◯◱◯ ◍◌○ ◉◉ ◇◔◓◕ ◐►◓◒ ◒◑ ◈◑ ◆◆◓ ◉◉◉&lt;/a&gt; (now solved!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;New clue:&lt;/strong&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4735/pick-up-sticks-%E2%86%97%E2%86%91%E2%86%99%E2%86%93%E2%86%98%E2%86%90-%E2%86%91%E2%86%99%E2%86%98%E2%86%98%E2%86%97%E2%86%93%E2%86%97%E2%86%90%E2%86%99%E2%86%93%E2%86%96&quot;&gt;Decode the message enciphered in these symbols: ↗↑↙↓↘←↑↙↘↘↗↓↗←↙↓↖&lt;/a&gt; (also solved)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Administrative details (not part of the puzzle):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;This is the latest instalment in the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/tagged/party-security&quot;&gt;Party Security&lt;/a&gt; series.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Answers in spoiler blocks please.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Where someone proposes a solution that is not the method I'm thinking of, I've decided not to comment on whether the actual answer they've given is correct, because if I do then it provides an additional data point; which changes the puzzle for other people.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The other playlists on the Grooveshark account have nothing to do with this puzzle, neither does the Grooveshark username or the Grooveshark profile pic. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Pulp Fiction dialogue track is &lt;a href=&quot;http://en.wikipedia.org/wiki/Not_safe_for_work&quot; rel=&quot;nofollow noreferrer&quot;&gt;NSFW&lt;/a&gt; if your workplace has objections to swearing. All the rest should be fine to play out-loud.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2016-05-16T09:34:55.357" Title="The Security to the Party [12] (now with party soundtrack!)" Tags="|password|music|" AnswerCount="6" CommentCount="26" ContentLicense="CC BY-SA 3.0" />
  <row Id="4058" PostTypeId="2" ParentId="4025" CreationDate="2014-11-11T12:52:12.913" Score="4" Body="&lt;p&gt;Given a as the input (what guard says) and b as output (keyword)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;The next guest arrives. The security says &quot;10&quot;, the guest replies &quot;2&quot;.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;a+b=12&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a-b=8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;12-8=4&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;The next guest arrives. The security says &quot;8&quot;, the guest replies &quot;4&quot;.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;a+b=12&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a-b=4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;12-4=8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so on, you get this table&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a  b  (a+b)-(a-b)&#xA;10 2  4&#xA;9  ?  (6?)&#xA;8  4  8&#xA;7  5  10&#xA;6  6  12&#xA;5  1  2 (Sequence resets?)&#xA;4  ?  (4?)&#xA;3  ?  (6?)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Since we know a=3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(a+b)-(a-b)=6$$&#xA;$$(3+b)-(3-b)=6$$&#xA;$$3-3+b+b=6$$&#xA;$$2b=6$$&#xA;$$b=3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You walk up to security and get &quot;3&quot;, what is the correct response?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A : 3. (Steals spaghetti recipe)&lt;/p&gt;&#xA;" OwnerUserId="4995" LastActivityDate="2014-11-11T12:52:12.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4060" PostTypeId="2" ParentId="4004" CreationDate="2014-11-11T13:00:35.193" Score="6" Body="&lt;p&gt;Do you want a stupid answer? I got one that fits perfectly and is almost definitely not what you are looking for.&lt;/p&gt;&#xA;&lt;h1&gt;The Series&lt;/h1&gt;&#xA;&lt;p&gt;What do we have here? There is a pattern to be extrapolated; probably several hundred thousand different expressions that would generate the series.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt; The series&#xA;/----------\&#xA; 2   =&amp;gt;   4 &#xA; 3   =&amp;gt;   4 &#xA; 4   !!   4 &#xA; 5   =&amp;gt;   4 &#xA; 6   =&amp;gt;   6 &#xA; 7   =&amp;gt;   ? &#xA; 8   =&amp;gt;   8 &#xA;\----------/&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;It looks arbitrary when first glanced at, yet there's something to it. The significance of the 4 is obvious, as it's the only clue we have as to what doesn't work. The amount of fours repeating is also pretty nauseating, so I've decided to try something interesting.&lt;/p&gt;&#xA;&lt;h1&gt;Pattern?&lt;/h1&gt;&#xA;&lt;p&gt;Deciding to preformulate each of those left-hand input numbers into an expression of the form &lt;code&gt;xn&lt;/code&gt; where &lt;code&gt;n&lt;/code&gt; is a prime and &lt;code&gt;x&lt;/code&gt; the coefficient gives us this series:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt; The series again&#xA;/----------------\&#xA;  1 * 2   =&amp;gt;   4 &#xA;  1 * 3   =&amp;gt;   4 &#xA;  2 * 2   !!   4 &#xA;  1 * 5   =&amp;gt;   4 &#xA;  2 * 3   =&amp;gt;   6 &#xA;  1 * 7   =&amp;gt;   ? &#xA;  4 * 2   =&amp;gt;   8 &#xA;\----------------/&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;This has the drawback of giving a variable representation to some numbers; &lt;code&gt;6&lt;/code&gt; can be both &lt;code&gt;2 * 3&lt;/code&gt; and &lt;code&gt;3 * 2&lt;/code&gt;, for example, so I've decided to put the smaller number to the left. Since we have the series, we can see that we can disregard the prime and still have rising coefficients in some sort of logical order:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt; Third series&#xA;/------------\&#xA;  1   =&amp;gt;   4 &#xA;  1   =&amp;gt;   4 &#xA;  2   !!   4 &#xA;  1   =&amp;gt;   4 &#xA;  2   =&amp;gt;   6 &#xA;  1   =&amp;gt;   ? &#xA;  4   =&amp;gt;   8 &#xA;\------------/&#xA;&#xA;   [Note]: This series has&#xA;   little to do with the first&#xA;   and as such they aren't&#xA;   equivalent.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;This kind of series lends itself better to the eyes, and so we can cut out some duplicate entries, giving series #10923853Sqrt[-1]:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;/----------\&#xA; 1   =&amp;gt;   4&#xA; 2   !!   4&#xA; 1   =&amp;gt;   ?&#xA; 2   =&amp;gt;   6&#xA; 4   =&amp;gt;   8&#xA;\----------/&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;From this we can expect &lt;code&gt;x=8 =&amp;gt; 10&lt;/code&gt;, probably following the pattern of &lt;code&gt;x=n^a =&amp;gt; 4+2*a&lt;/code&gt;. As such, we can reasonably expect that seven will imply four as the answer, as &lt;code&gt;(n^0)*7 =&amp;gt; 4+2*0 = 4&lt;/code&gt;. A few further bits of this series would be:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;/----------\&#xA; 4   =&amp;gt;   6&#xA; 9   =&amp;gt;   8&#xA; 10  =&amp;gt;   6&#xA; 11  =&amp;gt;   4&#xA; 12  =&amp;gt;   8&#xA; 13  =&amp;gt;   4&#xA;\----------/&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Pfft, so much effort for something so wrong and pointless ^^&lt;/p&gt;&#xA;" OwnerUserId="4990" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-11-11T17:11:00.260" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4062" PostTypeId="1" AcceptedAnswerId="4189" CreationDate="2014-11-11T13:24:17.040" Score="6" ViewCount="598" Body="&lt;p&gt;After pretty sick 12 parties, obviously 13th party occured during &quot;Halloween&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Halloween themed party is being held at a local mansion. The host is very rich and his success is because of one thing ~ his famous recipe for Evil Spaghetti!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only guests that may attend are people the host knows and trusts, whom he gave the password to enter to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's where you come in. You and a friend are trying to steal this recipe. You sneak by and listen to the passwords.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first guest arrives. The security says &quot;1&quot;, and the guest replies &quot;3&quot; and guard let him pass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second guest arrives. The security says &quot;3&quot;, the guest replies &quot;3&quot; and guard let him pass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third guest arrives. The security says &quot;9&quot;, the guest replies &quot;3&quot; and guard let him pass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're friend thinks he got it all figured out. He approaches the security and guard say &quot;7&quot;, when he replies &quot;3&quot;, guard looked to your friend waited for a bit before talking. Your friend was tense and holding his position to dodge possible elephant attacks. Suddenly guard said &quot;Welcome Sir&quot; and let him in. Unfortunately your friend still needs you to steal evil spaghetti recipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, another guest arrives, you seem to get the same pattern, security says &quot;2&quot; and he says &quot;2&quot; to get in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You walk up to security and get &quot;13&quot;, what is the correct response?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Meanwhile one more guest arrives, guard says &quot;4&quot; and guest replies with &quot;1&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Later one more guest arrives, guard says &quot;14&quot; and guest replies with &quot;4&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4506" LastEditorUserId="4506" LastEditDate="2014-11-12T07:46:55.447" LastActivityDate="2014-11-12T08:50:54.263" Title="The Security to the Party [Part 13]" Tags="|word-problem|password|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4063" PostTypeId="2" ParentId="3812" CreationDate="2014-11-11T13:25:54.103" Score="2" Body="&lt;p&gt;The numbers we &lt;em&gt;can&lt;/em&gt; make are of the form $m \times 9 + n \times 17 \,| \, m, n \in \mathbb{Z}_{\ge0}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we can see (and as has been said before), the largest number we &lt;em&gt;can't&lt;/em&gt; make is $127$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{r|rr}&#xA;\text{target} &amp;amp; m &amp;amp; n \\&#xA;\hline&#xA;0 &amp;amp; 0 &amp;amp; 0 \\&#xA;1 &amp;amp; - &amp;amp; - \\&#xA;2 &amp;amp; - &amp;amp; - \\&#xA;3 &amp;amp; - &amp;amp; - \\&#xA;4 &amp;amp; - &amp;amp; - \\&#xA;5 &amp;amp; - &amp;amp; - \\&#xA;6 &amp;amp; - &amp;amp; - \\&#xA;7 &amp;amp; - &amp;amp; - \\&#xA;8 &amp;amp; - &amp;amp; - \\&#xA;9 &amp;amp; 1 &amp;amp; 0 \\&#xA;10 &amp;amp; - &amp;amp; - \\&#xA;11 &amp;amp; - &amp;amp; - \\&#xA;12 &amp;amp; - &amp;amp; - \\&#xA;13 &amp;amp; - &amp;amp; - \\&#xA;14 &amp;amp; - &amp;amp; - \\&#xA;15 &amp;amp; - &amp;amp; - \\&#xA;16 &amp;amp; - &amp;amp; - \\&#xA;17 &amp;amp; 0 &amp;amp; 1 \\&#xA;18 &amp;amp; 2 &amp;amp; 0 \\&#xA;19 &amp;amp; - &amp;amp; - \\&#xA;20 &amp;amp; - &amp;amp; - \\&#xA;21 &amp;amp; - &amp;amp; - \\&#xA;22 &amp;amp; - &amp;amp; - \\&#xA;23 &amp;amp; - &amp;amp; - \\&#xA;24 &amp;amp; - &amp;amp; - \\&#xA;25 &amp;amp; - &amp;amp; - \\&#xA;26 &amp;amp; 1 &amp;amp; 1 \\&#xA;27 &amp;amp; 3 &amp;amp; 0 \\&#xA;28 &amp;amp; - &amp;amp; - \\&#xA;29 &amp;amp; - &amp;amp; - \\&#xA;30 &amp;amp; - &amp;amp; - \\&#xA;31 &amp;amp; - &amp;amp; - \\&#xA;32 &amp;amp; - &amp;amp; - \\&#xA;33 &amp;amp; - &amp;amp; - \\&#xA;34 &amp;amp; 0 &amp;amp; 2 \\&#xA;35 &amp;amp; 2 &amp;amp; 1 \\&#xA;36 &amp;amp; 4 &amp;amp; 0 \\&#xA;37 &amp;amp; - &amp;amp; - \\&#xA;38 &amp;amp; - &amp;amp; - \\&#xA;39 &amp;amp; - &amp;amp; - \\&#xA;40 &amp;amp; - &amp;amp; - \\&#xA;41 &amp;amp; - &amp;amp; - \\&#xA;42 &amp;amp; - &amp;amp; - \\&#xA;43 &amp;amp; 1 &amp;amp; 2 \\&#xA;44 &amp;amp; 3 &amp;amp; 1 \\&#xA;45 &amp;amp; 5 &amp;amp; 0 \\&#xA;46 &amp;amp; - &amp;amp; - \\&#xA;47 &amp;amp; - &amp;amp; - \\&#xA;48 &amp;amp; - &amp;amp; - \\&#xA;49 &amp;amp; - &amp;amp; - \\&#xA;50 &amp;amp; - &amp;amp; - \\&#xA;51 &amp;amp; 0 &amp;amp; 3 \\&#xA;52 &amp;amp; 2 &amp;amp; 2 \\&#xA;53 &amp;amp; 4 &amp;amp; 1 \\&#xA;54 &amp;amp; 6 &amp;amp; 0 \\&#xA;55 &amp;amp; - &amp;amp; - \\&#xA;56 &amp;amp; - &amp;amp; - \\&#xA;57 &amp;amp; - &amp;amp; - \\&#xA;58 &amp;amp; - &amp;amp; - \\&#xA;59 &amp;amp; - &amp;amp; - \\&#xA;60 &amp;amp; 1 &amp;amp; 3 \\&#xA;61 &amp;amp; 3 &amp;amp; 2 \\&#xA;62 &amp;amp; 5 &amp;amp; 1 \\&#xA;63 &amp;amp; 7 &amp;amp; 0 \\&#xA;64 &amp;amp; - &amp;amp; - \\&#xA;65 &amp;amp; - &amp;amp; - \\&#xA;66 &amp;amp; - &amp;amp; - \\&#xA;67 &amp;amp; - &amp;amp; - \\&#xA;68 &amp;amp; 0 &amp;amp; 4 \\&#xA;69 &amp;amp; 2 &amp;amp; 3 \\&#xA;70 &amp;amp; 4 &amp;amp; 2 \\&#xA;71 &amp;amp; 6 &amp;amp; 1 \\&#xA;72 &amp;amp; 8 &amp;amp; 0 \\&#xA;73 &amp;amp; - &amp;amp; - \\&#xA;74 &amp;amp; - &amp;amp; - \\&#xA;75 &amp;amp; - &amp;amp; - \\&#xA;76 &amp;amp; - &amp;amp; - \\&#xA;77 &amp;amp; 1 &amp;amp; 4 \\&#xA;78 &amp;amp; 3 &amp;amp; 3 \\&#xA;79 &amp;amp; 5 &amp;amp; 2 \\&#xA;80 &amp;amp; 7 &amp;amp; 1 \\&#xA;81 &amp;amp; 9 &amp;amp; 0 \\&#xA;82 &amp;amp; - &amp;amp; - \\&#xA;83 &amp;amp; - &amp;amp; - \\&#xA;84 &amp;amp; - &amp;amp; - \\&#xA;85 &amp;amp; 0 &amp;amp; 5 \\&#xA;86 &amp;amp; 2 &amp;amp; 4 \\&#xA;87 &amp;amp; 4 &amp;amp; 3 \\&#xA;88 &amp;amp; 6 &amp;amp; 2 \\&#xA;89 &amp;amp; 8 &amp;amp; 1 \\&#xA;90 &amp;amp; 10 &amp;amp; 0 \\&#xA;91 &amp;amp; - &amp;amp; - \\&#xA;92 &amp;amp; - &amp;amp; - \\&#xA;93 &amp;amp; - &amp;amp; - \\&#xA;94 &amp;amp; 1 &amp;amp; 5 \\&#xA;95 &amp;amp; 3 &amp;amp; 4 \\&#xA;96 &amp;amp; 5 &amp;amp; 3 \\&#xA;97 &amp;amp; 7 &amp;amp; 2 \\&#xA;98 &amp;amp; 9 &amp;amp; 1 \\&#xA;99 &amp;amp; 11 &amp;amp; 0 \\&#xA;100 &amp;amp; - &amp;amp; - \\&#xA;101 &amp;amp; - &amp;amp; - \\&#xA;102 &amp;amp; 0 &amp;amp; 6 \\&#xA;103 &amp;amp; 2 &amp;amp; 5 \\&#xA;104 &amp;amp; 4 &amp;amp; 4 \\&#xA;105 &amp;amp; 6 &amp;amp; 3 \\&#xA;106 &amp;amp; 8 &amp;amp; 2 \\&#xA;107 &amp;amp; 10 &amp;amp; 1 \\&#xA;108 &amp;amp; 12 &amp;amp; 0 \\&#xA;109 &amp;amp; - &amp;amp; - \\&#xA;110 &amp;amp; - &amp;amp; - \\&#xA;111 &amp;amp; 1 &amp;amp; 6 \\&#xA;112 &amp;amp; 3 &amp;amp; 5 \\&#xA;113 &amp;amp; 5 &amp;amp; 4 \\&#xA;114 &amp;amp; 7 &amp;amp; 3 \\&#xA;115 &amp;amp; 9 &amp;amp; 2 \\&#xA;116 &amp;amp; 11 &amp;amp; 1 \\&#xA;117 &amp;amp; 13 &amp;amp; 0 \\&#xA;118 &amp;amp; - &amp;amp; - \\&#xA;119 &amp;amp; 0 &amp;amp; 7 \\&#xA;120 &amp;amp; 2 &amp;amp; 6 \\&#xA;121 &amp;amp; 4 &amp;amp; 5 \\&#xA;122 &amp;amp; 6 &amp;amp; 4 \\&#xA;123 &amp;amp; 8 &amp;amp; 3 \\&#xA;124 &amp;amp; 10 &amp;amp; 2 \\&#xA;125 &amp;amp; 12 &amp;amp; 1 \\&#xA;126 &amp;amp; 14 &amp;amp; 0 \\&#xA;127 &amp;amp; - &amp;amp; - \\&#xA;\hline&#xA;128 &amp;amp; 1 &amp;amp; 7 \\&#xA;129 &amp;amp; 3 &amp;amp; 6 \\&#xA;130 &amp;amp; 5 &amp;amp; 5 \\&#xA;131 &amp;amp; 7 &amp;amp; 4 \\&#xA;132 &amp;amp; 9 &amp;amp; 3 \\&#xA;133 &amp;amp; 11 &amp;amp; 2 \\&#xA;134 &amp;amp; 13 &amp;amp; 1 \\&#xA;135 &amp;amp; 15 &amp;amp; 0 \\&#xA;136 &amp;amp; 0 &amp;amp; 8 \\&#xA;137 &amp;amp; 2 &amp;amp; 7 \\&#xA;138 &amp;amp; 4 &amp;amp; 6 \\&#xA;139 &amp;amp; 6 &amp;amp; 5 \\&#xA;140 &amp;amp; 8 &amp;amp; 4 \\&#xA;141 &amp;amp; 10 &amp;amp; 3 \\&#xA;142 &amp;amp; 12 &amp;amp; 2 \\&#xA;143 &amp;amp; 14 &amp;amp; 1 \\&#xA;144 &amp;amp; 16 &amp;amp; 0 \\&#xA;145 &amp;amp; 1 &amp;amp; 8 \\&#xA;146 &amp;amp; 3 &amp;amp; 7 \\&#xA;147 &amp;amp; 5 &amp;amp; 6 \\&#xA;148 &amp;amp; 7 &amp;amp; 5 \\&#xA;149 &amp;amp; 9 &amp;amp; 4 \\&#xA;150 &amp;amp; 11 &amp;amp; 3 \\&#xA;151 &amp;amp; 13 &amp;amp; 2 \\&#xA;152 &amp;amp; 15 &amp;amp; 1 \\&#xA;153 &amp;amp; 17 &amp;amp; 0 \\&#xA;154 &amp;amp; 2 &amp;amp; 8 \\&#xA;155 &amp;amp; 4 &amp;amp; 7 \\&#xA;156 &amp;amp; 6 &amp;amp; 6 \\&#xA;157 &amp;amp; 8 &amp;amp; 5 \\&#xA;158 &amp;amp; 10 &amp;amp; 4 \\&#xA;159 &amp;amp; 12 &amp;amp; 3 \\&#xA;160 &amp;amp; 14 &amp;amp; 2 \\&#xA;161 &amp;amp; 16 &amp;amp; 1 \\&#xA;162 &amp;amp; 18 &amp;amp; 0 \\&#xA;163 &amp;amp; 3 &amp;amp; 8 \\&#xA;164 &amp;amp; 5 &amp;amp; 7 \\&#xA;165 &amp;amp; 7 &amp;amp; 6 \\&#xA;166 &amp;amp; 9 &amp;amp; 5 \\&#xA;167 &amp;amp; 11 &amp;amp; 4 \\&#xA;168 &amp;amp; 13 &amp;amp; 3 \\&#xA;169 &amp;amp; 15 &amp;amp; 2 \\&#xA;170 &amp;amp; 17 &amp;amp; 1 \\&#xA;171 &amp;amp; 19 &amp;amp; 0 \\&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2014-11-11T13:25:54.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4064" PostTypeId="2" ParentId="4054" CreationDate="2014-11-11T13:30:12.230" Score="11" Body="&lt;p&gt;My guess is answer maybe&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;9&lt;/code&gt;. Because, the guest reply is equal to number lines in &lt;code&gt;Roman numeral&lt;/code&gt; representation for every number said by guard.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4848" LastEditorUserId="2313" LastEditDate="2014-11-11T13:31:15.960" LastActivityDate="2014-11-11T13:31:15.960" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4065" PostTypeId="2" ParentId="4062" CreationDate="2014-11-11T13:33:29.377" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;3&lt;/code&gt;. Because, even number was 2 and odds are 3.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5007" LastEditorUserId="5007" LastEditDate="2014-11-11T13:42:07.160" LastActivityDate="2014-11-11T13:42:07.160" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4066" PostTypeId="2" ParentId="4025" CreationDate="2014-11-11T13:37:58.790" Score="3" Body="&lt;p&gt;Look up at the table:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;10 - 2  =  8&#xA;[9]-[3] =  6 //my guess&#xA; 8 - 4  =  4&#xA; 7 - 5  =  2&#xA; 6 - 6  =  0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As you can see, the difference goes down by every time 'til 0. If we follow this pattern we can go finding the matches for the remaining numbers.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 5 - 1  = 4&#xA;[4]-[2] = 2 //my guess&#xA; 3 - 3  = 0&#xA;[2]-[0] = 2 //my guess&#xA;[1]-[1] = 0 //my guess&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Obviously, 5 is less than 6, so the difference could be that high, thus 5-1 = 4. The same though has been made for 2-0 = 2.&lt;br/&gt;Following this pattern, the solution is the well-known:&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;3&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1685" LastActivityDate="2014-11-11T13:37:58.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4067" PostTypeId="2" ParentId="4046" CreationDate="2014-11-11T13:44:20.153" Score="1" Body="&lt;p&gt;The answer is:&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;4&lt;/strong&gt; because is the highest even number that can be divided from the number given by the guard, not counting the reminder obviously!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1685" LastActivityDate="2014-11-11T13:44:20.153" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4068" PostTypeId="1" CreationDate="2014-11-11T13:48:14.220" Score="25" ViewCount="11171" Body="&lt;p&gt;&amp;quot;Take her away!&amp;quot; commands the police officer, as two other officers handcuff the blonde woman. &amp;quot;Good work, Detective! I can see why you were named the number one detective. Keep up the great work!&amp;quot;.  You stand there trying to be mature about it, but deep down you are overwhelmed by the comment, which makes you very happy.&lt;/p&gt;&#xA;&lt;p&gt;You then remember that after every crime you must write a detailed report and send it to your superiors, so you head back to your office. As you arrive at your building, you park in the designated parking spot, get out of your car and lock it. As you head to the building's front door, you look up to see that there are dark clouds approaching. &amp;quot;Seems like we're gonna have a storm tonight,&amp;quot; you say as you enter through the front door.&lt;/p&gt;&#xA;&lt;p&gt;You head to your office, sit down on your chair, and start writing a report of what happened. You reach into one of your drawers to get a stapler, when you notice a file you have not seen. You open the file to see that this file belonged to a murder case from years ago. You see that the victim's name was Ling O, and she was brutally murdered. They caught the murderer; her name is Bella B. She tortured Ling for several days before police found out. The murder case was solved by a detective named Randy White.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;That name sounds familiar,&amp;quot; you think to yourself, and then it hits you: Randy is a detective in the same department as you. &amp;quot;He might have misplaced this file by accident; I will just place it back in the vault with the others,&amp;quot; you think to yourself.&lt;/p&gt;&#xA;&lt;p&gt;As you finish your report, you realise that it is almost time for you to head home. Since this was your first crime solved in Chicago, you decide to celebrate by going to the new bakery not too far from your house. You look out your office window to see that the dark clouds have taken over the earlier sunny day, but you don't think much of it.&lt;/p&gt;&#xA;&lt;p&gt;You start driving to the new bakery shop, but all of a sudden you get violently cut off. You swerve out of control, but manage to stop safely and don't crash. You look up to see that the other car is also stopped right in front of you. Through the driver's window you see a man with shaggy brown hair, a small nose, and big ears. You give a big sigh. &amp;quot;Not again!&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;You see the man get out of his car and walk towards you. You get out of your car as well.  Just as you're about to ask if he is OK, the man starts yelling at you: &amp;quot;LEARN HOW TO DRI....&amp;quot; he pauses. &amp;quot;Hey, I know you! you're the one who put my sister in jail! She's innocent!&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;The man slowly approaches you; you can tell that he is getting more and more angry. &amp;quot;You put the wrong person in jail! Do you know what you have done to my family? DO YOU?&amp;quot; yells the man. As he is about to get physical, two police officers get between the two of you and separate you. The man is arguing with one of the officers, and does not seem to be calming down.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;Good thing we were patrolling around the neighbourhood; it could have ended badly,&amp;quot; says the officer next to you. As you are walking back to your car, the man calls out your name. &amp;quot;YOU HAVEN'T SEEN THE LAST OF ME! I WILL GET REVENGE FOR MY SISTER, YOU HEAR ME?&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;You take a deep breath and get back in your car. You finally arrive at the bakery and sit down at one of the tables. &amp;quot;Welcome to Niki's Bakery, how may I help you?&amp;quot; says the waitress as she approaches you.  You take a look at the menu and choose a couple of snacks. &amp;quot;Ok, I'll be right back with your food,&amp;quot; she says. You notice that she is wearing the bakery's uniform, has blonde hair, is very beautiful, and has a name tag that says &amp;quot;Heather&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;As you wait, you look around the bakery. In one corner you see a TV, and the news is on. &amp;quot;BREAKING NEWS! An unknown person has left a note at the front door of the police station. Its contents seem to be addressed to the new head detective of Chicago. The letter says, and I quote, 'To the Chicago detective department: I have killed someone, and the body is located somewhere in RuffNear Park. I challenge the head detective to find me.  For everyday that you do not find me, I will kill someone else. I wish you the best of luck, because you will never find me! I am invisible like objects in a December storm; you will not see me coming! My story will go down in history! Better watch out for the children, because they will remember this story forever!' End quote. The police are warning citizens to report any suspicious activities, and....&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;The TV goes blank, and all you see is the word &amp;quot;Reconnecting...&amp;quot;, and then you hear a loud noise: BOOM! You look outside the bakery's window to see that it is raining heavily outside, with thunder and lightning striking every so often. &amp;quot;It seems like the weather has affected the reception,&amp;quot; you think to yourself.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;Here's your food,&amp;quot; says the waitress.  You ask her if you can take it to go instead. You call the police department, and one of the police officers picks up. &amp;quot;Good afternoon, don't worry about the message and the body; we have already dispatched a group of search teams to search for the body. We will contact you again when we find it.&amp;quot; The officer ends the call.&lt;/p&gt;&#xA;&lt;p&gt;You decide to head home to get some rest. The next morning, you receive a call:  RING RING RING RING.  You pick up the phone. &amp;quot;Good morning. I would like to let you know that we have found the body.  Please come to RuffNear Park by the washroom ASAP.&amp;quot;  The call ends. You quickly drive over to the location given by  the officer. When you arrive, the forensics team is already at work taking pictures and discussing the matter. You take a look at the body.  There do not seem to be any kind of knife or gunshot wounds. You notice that the body is lying down as if asleep. The victim has blonde hair, blue eyes, and is skinny. He still had everything with him: his wallet, his cell phone, and his keys; nothing seems to be missing. Checking his wallet shows that the victim is around 5' 1&amp;quot; in height, and his name is Alex T.&lt;/p&gt;&#xA;&lt;p&gt;You do not see any small yellow cones with numbering around, so you ask one of the officers.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;Nothing seemed to be left behind; the only clue we have is this cloth on top of the victim's face.&amp;quot; The officer shows you a ziplock bag with the cloth inside. You take a closer look at the body and find a small puncture hole in the victim's right arm. &amp;quot;It rained pretty heavily last night, it might have washed away some clues and evidence. We are going to bring the body and the cloth back to the forensics lab to more analysis,&amp;quot; one of the detectives says. With that, you make your way back to your office.&lt;/p&gt;&#xA;&lt;p&gt;Soon after, you receive a call from the forensic scientist telling you to head over to the forensics lab for more details on the victim.  When you arrive, he says, &amp;quot;Glad you could make it so soon. The victim did not have any foreign fingerprints on him, nor any foreign hair on his clothing. The small hole in his right arm seems to have come from something sharp, like a needle. The cause of death is a very toxic and deadly poison.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;You stand there and ponder things for a minute. You call the police station asking if they have found any suspects yet, but none have came up. You know that time is running out before the day ends and the next victim is killed.&lt;/p&gt;&#xA;&lt;p&gt;With no leads, you head back to RuffNear Park where the body was found. As you arrive, you try to search for any clues that might have been missed by the other police and detectives. You search for hours but find nothing.&lt;/p&gt;&#xA;&lt;p&gt;You look at the time on your wrist watch; it says 8:32 p.m. It has been more 24 hours since the arrival of the note received by the police, and there have not been any reports of another dead body, so you decide to head home to get some rest. You arrive home, change out of your clothes, go to bed, and close your eyes.&lt;/p&gt;&#xA;&lt;p&gt;You wake up to a loud sound: &amp;quot;RING RING RING RING RING!&amp;quot; You look at your alarm clock, which reads 6:15 a.m. You pick up the phone.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;Sorry to disturb you so early, but we just got a report of another body found.&amp;quot; You ask for the location, and the officer directs you to it.&lt;/p&gt;&#xA;&lt;p&gt;You arrive at the scene, and, like every other case, you see police and detectives everywhere and a yellow police line around the area, but no yellow cones. You walk up to the body, and, just as you feared, you notice no signs of physical injury on the body; no gunshots, no knife wound, nothing. You check the victim's right arm to find that there is a small puncture hole. &amp;quot;It must be him,&amp;quot; you mutter to yourself. You check the victim's wallet, her name is Patricia G., 5' 2&amp;quot; in height. You stand there pondering, trying to find a relation between the first victim and the second victim, but you can't think of any.&lt;/p&gt;&#xA;&lt;p&gt;The murder case goes on for another four days, and for each day another person is murdered. On the third day, it is a 34-year-old male, Peter L., 5' 2&amp;quot; in height. On the fourth day, a 12-year-old male named Leo E., 4' 9&amp;quot;. On the fifth day, a 68-year-old female, Elsa D., 5' 2&amp;quot;. The pattern seems to continue, until the sixth day. On the sixth day, another person has been murdered, but this time there is no wallet found on the victim's body, and the same for the seventh day. Forensic scientists confirm that all of the victims died from the same deadly poison in the right arm.&lt;/p&gt;&#xA;&lt;p&gt;Just when you thought the case couldn't get any weirder, the murders stop after the seventh day. For the next 2 days, there are no reports of any dead bodies, no murders, no notes, nothing. On the tenth day, you receive a call. &amp;quot;We have found some suspects; come down to the police department pronto.&amp;quot; The person on the other end of the phone hangs up.&lt;/p&gt;&#xA;&lt;p&gt;You rush over to the police department. When you arrive you notice that they have a lineup of four suspects. &amp;quot;I hope we can get any new information soon; we are out of ideas,&amp;quot; says one of the officers. You noticed that he looks to be a bit out of breath, his eyes looking here and there. You look at the ID badge pinned to the left side of his chest; it says &amp;quot;ID: 7777777  Officer: George Snoe&amp;quot;. You ask him if everything is OK. &amp;quot;Yeah I'm OK; I'm just not in the best of shape like I used to be,&amp;quot; he replies.&lt;/p&gt;&#xA;&lt;p&gt;You don't think much of it after that, so you start interrogating each of the suspects. The first suspect is a 25-year-old female, beautiful, with a slim body, blonde hair, and blue eyes. &amp;quot;I didn't commit any murders.  I was working every day from 8:00 a.m. to 4:30 p.m. I teach health and phys. ed.,&amp;quot; says Patricia F.&lt;/p&gt;&#xA;&lt;p&gt;The second suspect is a 34-year-old male, tall, about 6' 9&amp;quot;, with black hair and brown eyes. &amp;quot;I couldn't have murdered anyone. I work as a investment banker at the Chicago bank, and after work I have night school. I don't have time to murder anyone,&amp;quot; he says. Before you can ask for his name, he receives a phone call.  He says it is an emergency, and you let him take the call.&lt;/p&gt;&#xA;&lt;p&gt;The third suspect is a 41-year-old male, 6' 7&amp;quot; in height, very groomed, with nicely-styled hair and brown eyes. &amp;quot;I am a doctor at the Chicago Hospital. I did surgery on two of the days all day, and on three of the days I was working my usual hours. On the weekend I was with my family, so it couldn't have been me.&amp;quot; You ask for his name, and he replies, &amp;quot;My name is Dr. David N. Edle&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;The fourth and last suspect is a 24-year-old male. He is wearing a bandana, jeans hanging low, and nice white shoes. Looking at some papers the officers gave you beforehand, you read that he was caught doing all sorts of drugs. He is an addict and also a drug dealer.  You asked him where he was during the week. &amp;quot;Nowhere man, just been chillin' with my homies. I didn't do shit, man. You pigs always arresting my brothers and shit man. Been working at the local fast food chain Rocky Ranchy's Burger Joint, just trying to get by. Don't need you poppin' me for doing nothin' wrong man.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;You pondered everything for several minutes, trying to put together all the clues and suspects and other information, trying to find the murderer. Then it hit you. &amp;quot;How could I have been so blind?&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Who is the murderer?&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="4645" LastEditorUserId="81979" LastEditDate="2023-05-21T07:22:52.743" LastActivityDate="2023-05-21T07:22:52.743" Title="Mysterious Murder Mystery 2" Tags="|riddle|story|situation|word-problem|" AnswerCount="7" CommentCount="17" ContentLicense="CC BY-SA 4.0" />
  <row Id="4069" PostTypeId="2" ParentId="4068" CreationDate="2014-11-11T14:20:26.237" Score="12" Body="&lt;p&gt;You arrest&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Officer George Snoe. His badge number of seven 7's shows he has some compulsion towards the number 7 which is how many murders there were. He also admits he is also out of shape which explains why all his victims are all short and light in order to overpower them or to move them. His written note also explains &quot;I am invisible like objects in a December storm&quot; and his last name is Snoe similar to snow which would be what the precipitation would be in a December storm. Being an officer he also knows what not to leave at a crime scene.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Possible motive:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Officer Snoe was up for a promotion before you came into town. He was the officer who commented &quot;Good thing we were patrolling around the neighborhood, it could have ended bad&quot; suggesting you can't handle yourself and don't belong here in the big city. Officer Snoe does not like the fact that you are in town. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alternative scenario:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Detective White is also upset you are new and getting all the glory he is the one committing the crime and framing Officer Snoe. Since White used to solve the big murders he left the file on your desk so as a passive aggressive attempt to show what you've done.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also the initials of the 5 victims plus the two from Randy White's file are LOBBATPLEGPLED have tried rot 7 and other variations of forming words but haven't gotten anywhere. &lt;/p&gt;&#xA;" OwnerUserId="2803" LastEditorUserId="2803" LastEditDate="2014-11-11T15:06:16.297" LastActivityDate="2014-11-11T15:06:16.297" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="4070" PostTypeId="2" ParentId="4025" CreationDate="2014-11-11T14:27:31.430" Score="3" Body="&lt;p&gt;Here's the method I used where &lt;strong&gt;x is the number called out by the guard&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (x mod 6) + 1, then take the number on the opposite side of a standard 6 sided die to the result. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (3 mod 6) + 1 = 4, and &lt;strong&gt;3&lt;/strong&gt; is the number on the opposite die face.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1491" LastEditorUserId="1491" LastEditDate="2014-11-11T17:29:27.230" LastActivityDate="2014-11-11T17:29:27.230" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4071" PostTypeId="2" ParentId="4068" CreationDate="2014-11-11T15:09:16.097" Score="3" Body="&lt;p&gt;You're arrested&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Drug Addict. How he did it? Well he's drug addict. Obviously he have needle. And he wear bandana, recall that victim got covered by cloth. Motive is revenge to police, since his friends got arrested. And noticed that he said &quot;Don't need to poppin me&quot;. That's a slip. How did he know that his victim are &quot;popped&quot; in the right arm?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4995" LastActivityDate="2014-11-11T15:09:16.097" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4072" PostTypeId="1" AcceptedAnswerId="4104" CreationDate="2014-11-11T15:12:09.107" Score="3" ViewCount="499" Body="&lt;p&gt;The Spaghetti dude just can't get enough of his parties... so there's yet another chance for you and your mate to steal the recipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first guest arrives. The security says &quot;3&quot;, and the guest replies &quot;2&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second guest arrives. The security says &quot;4&quot;, and the guest replies &quot;3&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third guest arrives. The security says &quot;5&quot;, and the guest replies &quot;4&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're friend thinks, this time the host really let his guard down. He approaches the security and they say &quot;6&quot;, when he replies &quot;5&quot; they refuse entry and he runs off before they can release the elephant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another guest arrives. The security says &quot;1&quot;, and the guest replies &quot;2&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's your turn now. You go up to the door and the security guard says &quot;8&quot;. What do you reply?&lt;/p&gt;&#xA;" OwnerUserId="2450" LastEditorUserId="2450" LastEditDate="2014-11-11T20:02:14.710" LastActivityDate="2014-11-11T20:02:14.710" Title="The Security to the Party [Part 14]" Tags="|password|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4073" PostTypeId="1" AcceptedAnswerId="4090" CreationDate="2014-11-11T15:43:46.690" Score="-2" ViewCount="416" Body="&lt;p&gt;Wow. It's all about spaghetti with you folks huh. Well plot twist! He has a new get-rich recipe for pie!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As usual you hide and listen the the passwords, this time to you decide to make a chart:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;10&quot; -&gt; &quot;5&quot;&lt;br&gt;&#xA;  &quot;6&quot; -&gt; &quot;3&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At this point your friend has had enough. He's been taken by security, trampled by elephants. He's tired of you being the one to get in. He tells you this will be the last time he goes for that recipe! He walks up to security and they give him the number &quot;2&quot; he whispers the number in their ear so you can't hear it and he gets in. Traitor!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems you arrived late and only got to hear 2 guests! So you walk up anyway and get the number &quot;14&quot; what's your response?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2014-11-12T07:42:51.673" Title="Security to the Party [Part 15]" Tags="|password|" AnswerCount="4" CommentCount="6" ClosedDate="2014-11-16T23:17:48.080" ContentLicense="CC BY-SA 3.0" />
  <row Id="4074" PostTypeId="2" ParentId="4073" CreationDate="2014-11-11T15:50:53.110" Score="9" Body="&lt;p&gt;The answer is 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on all the previous puzzles in this series, your friend always goes for the obvious solution.  The obvious solution based on the 2 clues here is to take half of what the security guy tells you.  This time your friend got in, so the obvious solution must be the correct one.  So if they say &quot;14&quot; to you, then you should reply with &quot;7&quot;.&lt;/p&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2014-11-11T15:50:53.110" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4075" PostTypeId="1" AcceptedAnswerId="4136" CreationDate="2014-11-11T15:52:55.090" Score="47" ViewCount="4654" Body="&lt;p&gt;This is part of a series which I'll be continuing if this goes down well enough. If you have any feedback on the puzzle (too hard, too easy, not clear enough, etc) I'd appreciate it so that future puzzles can be better :-)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You wake up and log on to your computer. You log on to Puzzling.SE, browse some puzzles. You spend half an hour trying to think of number sequences because you're &lt;a href=&quot;https://puzzling.stackexchange.com/questions/tagged/party-security&quot;&gt;really desperate to go to a party&lt;/a&gt;, then after coming up with nothing, you get grumpy and decide to have a bit of fun - trolling people on StackOverflow with questions about &lt;a href=&quot;https://meta.stackexchange.com/questions/14470/what-is-the-boat-programming-meme-about/14490#14490&quot;&gt;the best way to become a lion tamer, for a career programmer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You hop on over to StackOverflow, but as the page loads, something is clearly wrong. You can't click on anything, all of the questions in the list seem to be related to how to animate blood-effects in Javascript, and red pixels start coming down from the top of the screen. As they dribble down over more and more of the screen, some gaps are left in the middle of the screen which forms words, the &quot;blood&quot; drops flowing around them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It reads:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ha ha ha ha ha...&lt;br&gt;&#xA;  Pathetic newbie programmers, WHERE IS YOUR GOD-OBJECT NOW?&lt;br&gt;&#xA;  MUA HA HA HA HA!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;-The SErial Killer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Terrified, you just sit there staring at the screen, until suddenly it redirects to &lt;a href=&quot;https://www.youtube.com/watch?v=BROWqjuTM0g&quot; rel=&quot;noreferrer&quot;&gt;a terrible, terrible thing&lt;/a&gt; and you snap into action, closing the window. You're not sure what exactly is going on, but in a rare moment of genius you realise that &quot;SE&quot; could refer to &quot;Stack Exchange&quot;, and decide to see if any other sites on the SE network have been brutally murdered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you pull up the &lt;a href=&quot;http://stackexchange.com/sites&quot;&gt;network sites page&lt;/a&gt; you see that yes, there are some other victims in this heinous crime. With no clues to go on, and nothing but a refined puzzling nose from weeks wast...err, spent on Puzzling.SE, you steel yourself and vow to bring the person or persons responsible for this to justice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only question is ... where do you start looking?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lvROU.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hint for you, to narrow the search:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I didn't spend all that time making a fancy image for nothing you know! Graphics are hard (you can tell because of how cheesy it is :-P)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2518" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.923" LastActivityDate="2016-09-13T17:29:31.933" Title="The SErial Killer (Part 1)" Tags="|story|mystery|" AnswerCount="12" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="4076" PostTypeId="2" ParentId="4075" CreationDate="2014-11-11T16:08:53.827" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Android Enthusiasts. GOD OBJECT related to object-oriented-programming which means Java. And Android main language is Java. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4995" LastActivityDate="2014-11-11T16:08:53.827" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4077" PostTypeId="2" ParentId="4075" CreationDate="2014-11-11T16:10:18.817" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Travel seems to be the oddball in the group, and as pointed out in a comment by Martin, it has no icon and name in the background, unlike the others. So I'd say go to the site Travel and find out what's up. I'd also put some sort of monitor on &quot;Server Fault&quot;, &quot;Super User&quot;, and &quot;Programmers&quot; and any other software type sites.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-11-11T16:19:21.477" LastActivityDate="2014-11-11T16:19:21.477" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4078" PostTypeId="2" ParentId="4036" CreationDate="2014-11-11T16:30:42.760" Score="2" Body="&lt;p&gt;assuming she wrote it down upside down to yourself but the right way to her&lt;/p&gt;&#xA;&#xA;&lt;p&gt;01861 919550&lt;/p&gt;&#xA;&#xA;&lt;p&gt;022616 19810&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OR&lt;/p&gt;&#xA;&#xA;&lt;p&gt;055616 19810&lt;/p&gt;&#xA;&#xA;&lt;p&gt;depending on whether an upside down 5 is a 2 or not&lt;/p&gt;&#xA;" OwnerUserId="5024" LastActivityDate="2014-11-11T16:30:42.760" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4079" PostTypeId="2" ParentId="4025" CreationDate="2014-11-11T16:36:21.733" Score="0" Body="&lt;p&gt;The answer is actually the opposite of what everyone has said...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; -3.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While the other answers have provided solutions that fit the pattern, the spaghetti host is actually a big fan of...   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; tenth degree polynomials! For a value &quot;x&quot;, the correct response is $p_0x^{10}+p_1x^{9}+p_2x^{8}+p_3x^{7}+p_4x^{6}+p_5x^{5}+p_6x^{4}+p_7x^{3}+p_8x^{2}+p_9x^{1}+p_{10}$ with values $p_i$ from the vector [  3.30687831e-05,  -1.28968254e-03,   1.95436508e-02, -1.35119048e-01,   2.63194445e-01,   2.05625000e+00, -1.47291997e+01,   3.86658730e+01,  -4.55535714e+01, 1.84142857e+01,   1.66727566e-10].&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's the full set of correct responses:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm not good at formatting, so I'll write them as (host, guest) ordered pairs. (0,0),(1,-1),(2,-2),(3,-3),(4,-4),(5,1),(6,6),(7,5),(8,4),(9,-9),(10,2).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sorry if this counts as being a spoilsport, but linear algebra is fun!&lt;/p&gt;&#xA;" OwnerUserId="5025" LastActivityDate="2014-11-11T16:36:21.733" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="4081" PostTypeId="2" ParentId="4013" CreationDate="2014-11-11T16:56:39.017" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I would say &quot;2&quot;. The formula to calculate the answer seems to be &lt;code&gt;x=Floor(2(1-cos(n)))&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5023" LastEditorUserId="5023" LastEditDate="2014-11-11T17:31:22.603" LastActivityDate="2014-11-11T17:31:22.603" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4083" PostTypeId="2" ParentId="4004" CreationDate="2014-11-11T16:59:05.653" Score="1" Body="&lt;p&gt;My response is 4.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The formula I used is 2+2f where f is the number of factors in the prime factorization of x.&lt;br/&gt;For 2: The prime factors in 2 are: 2. 2+(2*1) = 4&lt;br /&gt;For 3: The prime factors in 3 are: 3. 2+(2*1) = 4&lt;br /&gt;For 4: The prime factors in 4 are: 2*2. 2+(2*2) = 6. &lt;br /&gt;For 5: The prime factors in 5 are: 5. 2+(2*1) = 4&lt;br /&gt;For 6: The prime factors in 6 are: 2*3. 2+(2*2) = 6&lt;br /&gt;For 7: The prime factors in 7 are: 7. 2+(2*1) = 4. &lt;br /&gt;For 8: The prime factors in 8 are: 2*2*2. 2+(2*3) = 8.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3922" LastEditorUserId="3922" LastEditDate="2014-11-11T17:06:49.403" LastActivityDate="2014-11-11T17:06:49.403" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="4084" PostTypeId="2" ParentId="580" CreationDate="2014-11-11T16:59:17.877" Score="5" Body="&lt;p&gt;Are you familiar with Raymond Smullyan's book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0486482014&quot; rel=&quot;noreferrer&quot;&gt;The Chess Mysteries of Sherlock Holmes&lt;/a&gt;?  I don't know if it contains exactly the kind of puzzle you're looking for.  There are different kinds in the book, and most that I remember are trying to reconstruct the moves in a game from an unlikely or absurd position.  Some are more subtle - for example, determining merely from the pieces on the board whether it is legal to castle.  I don't remember if there is one that exactly meets your specification.  But it is marvelously entertaining.&lt;/p&gt;&#xA;" OwnerUserId="4385" LastActivityDate="2014-11-11T16:59:17.877" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4086" PostTypeId="2" ParentId="4072" CreationDate="2014-11-11T17:22:22.833" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Consider the mapping:&lt;br&gt;&#xA; &lt;code&gt;1 2 3 4 5 6 7 8 9&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;2 3 2 3 4 2 3 4 5&lt;/code&gt;&lt;br&gt;&#xA; (where the top is the guard's prompt and the bottom is the response)&lt;br&gt;&#xA; Then, the response for 8 is 4.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1474" LastActivityDate="2014-11-11T17:22:22.833" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4087" PostTypeId="2" ParentId="4068" CreationDate="2014-11-11T17:34:27.757" Score="6" Body="&lt;p&gt;It's just a guess. So please be light on me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's David Edle, the doctor. No one but the doctor has access to fine needles and equipments which could be used to make puncture holes. Also as Joe Z. Hinted in the comments, the initials spell out APPLE, and an apple a day keeps the doctor away. The murderer was obviously a lunatic, a psychopath, was trying to challenge the department and these clues might be the clues a lunatic might leave. Also, the five days he committed the murder, he was on duty and was working. The sixth day might have been busy on him, I exactly don't know why the wallet was missing. The banker and the druggie might be out of the scene, as banks and fast food joints are busy at weekends.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2808" LastEditorUserId="2808" LastEditDate="2014-11-11T17:43:03.533" LastActivityDate="2014-11-11T17:43:03.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4088" PostTypeId="1" AcceptedAnswerId="4098" CreationDate="2014-11-11T17:36:39.563" Score="5" ViewCount="419" Body="&lt;p&gt;A party is being held at a local mansion. The host is very rich and his success is because of one thing — his famous recipe for Spaghetti!  Tonight he is trying a new variation on his signature creation which is expected to be a big hit in his restaurants all over town: Spaghetti Pie!  However, after several successful attempts by a couple of party crashers to steal his recipe, the host has implemented an enhanced security protocol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As usual, the only guests that may attend are people who correctly reply to the guard at the door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As usual, you and a friend are trying to steal this recipe, and are hiding in some bushes by the gate listening to the challenge/responses between guests and the guards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first guest arrives. The guards says &quot;33&quot;, the guest thinks for a moment, then replies &quot;8&quot;.  The guard grunts, and then says &quot;35&quot;.  The guest thinks again for a shorter amount of time, and then replies &quot;8&quot; again.  The guard lets the guest in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second guest arrives.  The security guard says &quot;92&quot;. The immediately blurts out &quot;6&quot;.  The guard grunts, and then says &quot;26&quot;.  The guest immediately say &quot;5&quot; and is let in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third guest arrives. The guard thinks for a minute, and then says &quot;50&quot;.  The guest appears to be reciting something in her head, and after a minute says &quot;Oh sure, give me the hard one.  2&quot;.  The guard nods, then says &quot;14&quot;, to which the guest immediately replies &quot;1&quot; and gets let in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Wow this is really hard&quot;, says your friend.  &quot;Not only are there two challenge/responses, but they don't seem to be following any sort of pattern.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yes,&quot; you reply, &quot;it almost seems as if the numbers are just chosen from some random distribution.  But that isn't at all in keeping with our host's tendency to have some kind of pattern in the challenge/response sequence!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you both decide to sit around a bit longer and do some more eavesdropping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few minutes pass and the guard changes.  The next guest then arrives and the guard barks out &quot;14&quot; to which the guest immediately responds, &quot;1&quot;.  The guard nods and then asks &quot;and 15?&quot; to which the guest almost immediately responds, &quot;9&quot;.  The guard lets them in the gate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Hmm,&quot; your friend says.  A few minutes later the next guest gets &quot;92&quot; to which he responds &quot;6&quot; in short order, after which the guard says &quot;65&quot; and the guest replies &quot;3&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this, your friend remarks, &quot;This guard is definitely not doing his job very well, is he?&quot; and with that he slips away and reappears at the gates a moment later.  &quot;8!&quot; your friend announces to the guard, without waiting for him to gives his challenge.  &quot;Err, yes&quot; the guard replies, looking puzzles, &quot;eighty...&quot; &quot;7&quot; your friend cuts him off before he can finish.  &quot;Quite,&quot; the guard agrees, scratching his head as he absentmindedly lets your friend through the gate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deciding that maybe your friend was onto something, you leave your hiding places and circle around to approach the gate from the main path.  But when you arrive, the guard has been changed again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;62 and 65&quot; the guard says.  You give him a strange look.  As if expecting your surprise, the guard replies,  &quot;Yes, I know the e-mail said the second number would only be given after you answered the first one.  Never mind that. Just give me both numbers.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What numbers do you give the guard?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the latest installment in the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/tagged/party-security&quot;&gt;Party Security&lt;/a&gt; series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answers in spoiler tags please.&lt;/p&gt;&#xA;" OwnerUserId="2718" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2018-11-13T18:31:59.117" Title="The Security to the Party 16" Tags="|password|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4090" PostTypeId="2" ParentId="4073" CreationDate="2014-11-11T17:38:20.183" Score="3" Body="&lt;p&gt;Realizing that my friend is cluing me in on the fact that the pattern is not the obvious one I confidently answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because of course, my friend answered:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 as well!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Had he not whispered it,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It would have confirmed the wrong notion that the answer is to divide by 2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a side note, I originally took some time to compose &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4088/the-security-to-the-party-16&quot;&gt;my own party question&lt;/a&gt; and didn't see this one until I was checking to see what the latest number was before setting the number on mine. Since my question has now been answered, I will lay out how I arrived at this number:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The new recipe is one for &quot;Pie&quot;.&#xA;&#xA; The numbers index digits in Pi with a modulo added in to stop at the 5th digit.&#xA;&#xA; &lt;pre&gt;1-&gt;3,  2-&gt;1,  3-&gt;4,  4-&gt;1,  5-&gt;5&#xA;6-&gt;3,  7-&gt;1,  8-&gt;4,  9-&gt;1,  10-&gt;5&#xA;11-&gt;3, 12-&gt;1, 13-&gt;4, 14-&gt;1, 15-&gt;5&lt;/pre&gt;&#xA; etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2718" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2014-11-11T20:49:27.587" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4092" PostTypeId="1" AcceptedAnswerId="4100" CreationDate="2014-11-11T18:04:04.203" Score="6" ViewCount="2344" Body="&lt;p&gt;I have a black box that contains 4 balls. 3 are red and 1 is blue. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I remove two from the box without seeing them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I look at one of them, it is red. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the probability that the other one is red?&lt;/p&gt;&#xA;" OwnerUserId="2484" LastActivityDate="2014-11-12T15:29:31.510" Title="4 balls - red and blue" Tags="|logical-deduction|mathematics|" AnswerCount="6" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4093" PostTypeId="2" ParentId="4092" CreationDate="2014-11-11T18:09:47.090" Score="7" Body="&lt;p&gt;It's equivalent to taking one of the balls, then looking at it, then taking out another and looking at &lt;em&gt;it&lt;/em&gt;. 66% is my guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: I am going to risk ending up stupid and complain: why 50%? Maybe the combinatorics data ends up showing that but to me it doesn't make any sense. I'm going to stick with 66% due to the fact that, if we assume we have balls labelled &lt;code&gt;R1 R2 R3 B&lt;/code&gt;, giving six possible matches, when we pick out two balls and look at one, we can determine that one of the balls is, for example, &lt;code&gt;R1&lt;/code&gt;. This means that there are three valid matches: &lt;code&gt;R1R2 R1R3 R1B&lt;/code&gt; and three invalid ones: &lt;code&gt;R2R3 R2B R3B&lt;/code&gt;. &lt;code&gt;R1R2&lt;/code&gt; and &lt;code&gt;R1R3&lt;/code&gt; thus make up two thirds of the possible matches. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I am thoroughly wrong in my methodology, might someone explain why? The other answers I've seen don't really cut it for me but that might just end up being chalked up to pickiness or stubbornness.&lt;/p&gt;&#xA;" OwnerUserId="4990" LastEditorUserId="4990" LastEditDate="2014-11-11T18:57:44.210" LastActivityDate="2014-11-11T18:57:44.210" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4096" PostTypeId="2" ParentId="4092" CreationDate="2014-11-11T18:14:38.077" Score="2" Body="&lt;p&gt;This is how I see it&#xA;Basically you have 4 balls. What is the probability that you will pick 2 red balls out of the 4?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are only 6 ways to choose 2 balls out of the 4. And only 3 of those will give you both red balls.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So my answer is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3/6 = 1/2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since @w0lf posted the clarification I am going to try to do this one with the proper math notation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let A denote that first ball is red.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let B denote that the second ball is red.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then P(B|A) = P(A and B)/P(A) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Read P(B|A) as probability of B happening after A has happened.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; P(A and B) i have solved above. which is equal to 1/2.  And P(A) is&#xA; 3/4. Therefore P(B|A) = (1/2)/(3/4) = 2/3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5030" LastEditorUserId="5030" LastEditDate="2014-11-11T19:51:20.493" LastActivityDate="2014-11-11T19:51:20.493" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4097" PostTypeId="2" ParentId="4075" CreationDate="2014-11-11T18:35:24.373" Score="2" Body="&lt;p&gt;You should probably put up a support request in stackapps so that more people will be aware and working on the problem, a wild wild guess&lt;/p&gt;&#xA;" OwnerUserId="2313" LastActivityDate="2014-11-11T18:35:24.373" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4098" PostTypeId="2" ParentId="4088" CreationDate="2014-11-11T18:38:43.590" Score="6" Body="&lt;p&gt;The replies:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reply to 62: &lt;strong&gt;6&lt;/strong&gt;&lt;br&gt;&#xA; Reply to 65: &lt;strong&gt;3&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reason:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You need to look at the digits of Pi: the next digit after &lt;em&gt;33&lt;/em&gt; is &lt;em&gt;8&lt;/em&gt;, the next digit after &lt;em&gt;35&lt;/em&gt; is &lt;em&gt;8&lt;/em&gt;, the next digit after &lt;em&gt;92&lt;/em&gt; is &lt;em&gt;6&lt;/em&gt; and so on. The same counts for &lt;em&gt;62&lt;/em&gt; (next digit is &lt;em&gt;6&lt;/em&gt;) and &lt;em&gt;65&lt;/em&gt; (next digit is &lt;em&gt;3&lt;/em&gt;).&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Why the friend could reply before a number was said (reply to Mr.Mindor's comment):&lt;br&gt;&#xA; After the guard changes, the other guard was just taking the next two digits from Pi: first 14, then 15, then 92, then 65 and so on, so the friend knew what the guard was going to ask and could reply before he asked something.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="153" LastEditorDisplayName="user44966" LastEditDate="2018-11-13T18:31:59.117" LastActivityDate="2018-11-13T18:31:59.117" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="4099" PostTypeId="2" ParentId="4073" CreationDate="2014-11-11T18:53:53.320" Score="1" Body="&lt;p&gt;Disappointed that my previous answer did not work, I get in my time machine and go back to the point right before I made my original answer, and instead I try:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because my friend answered:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reason being:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The host had some sort of thing with 5 and 3 whereby a two digit number was answered with 5 and a one-digit number was answered with 3.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2718" LastActivityDate="2014-11-11T18:53:53.320" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4100" PostTypeId="2" ParentId="4092" CreationDate="2014-11-11T19:10:53.640" Score="11" Body="&lt;p&gt;The procedure of drawing 2 balls and looking at 1 can produce 12 equally likely outcomes. We can draw balls in 6 different ways, and we can choose either of the 2 balls we drew to look at. Labeling the red balls R1, R2, and R3, and using an arrow to mark the one we look at, the possibilities are&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;R1,R2  R1,R3  R2,R3  R1,B  R2,B  R3,B&#xA;^      ^      ^      ^     ^     ^&#xA;&#xA;R1,R2  R1,R3  R2,R3  R1,B  R2,B  R3,B&#xA;   ^      ^      ^      ^     ^     ^&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;When we see a red ball, this evidence eliminates 3 possibilities:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;R1,B  R2,B  R3,B&#xA;   ^     ^     ^&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Of the 9 possibilities consistent with the evidence, 6 of them have the other ball be red. Thus, the probability that the other ball is red &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; is 2/3.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1549" LastEditorUserId="2313" LastEditDate="2014-11-12T06:25:32.637" LastActivityDate="2014-11-12T06:25:32.637" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4101" PostTypeId="2" ParentId="3939" CreationDate="2014-11-11T19:24:31.903" Score="16" Body="&lt;p&gt;John should announce he&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; can't figure out the note because he uses the Dvorak keyboard layout. Sarah sent the note to test his mental acuity, but using Dvorak makes him enough of a mega-nerd that Sarah (and the other ladies) will openly fall for him.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4242" LastActivityDate="2014-11-11T19:24:31.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4102" PostTypeId="2" ParentId="4092" CreationDate="2014-11-11T19:29:33.360" Score="10" Body="&lt;p&gt;My solution was to disregard the first ball, which we know is red.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second ball we extracted came out of a pool of 3 balls (2 red and 1 blue).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, it has a chance of exactly&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2/3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;to be red.&lt;/p&gt;&#xA;" OwnerUserId="1585" LastActivityDate="2014-11-11T19:29:33.360" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4104" PostTypeId="2" ParentId="4072" CreationDate="2014-11-11T19:40:23.057" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is 3&#xA;&#xA; It's the number of sides the number shares with other numbers in a number cube&lt;br&gt;&lt;br&gt;&#xA;1 2 3&lt;br&gt;&#xA;4 5 6&lt;br&gt;&#xA;7 8 9&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditDate="2014-11-11T19:43:42.533" LastActivityDate="2014-11-11T19:43:42.533" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4105" PostTypeId="1" AcceptedAnswerId="4131" CreationDate="2014-11-11T19:44:30.267" Score="8" ViewCount="532" Body="&lt;p&gt;So, you got through the guards and to the party? Easy stuff! The number sequences were silly and you walked in and out multiple times. Now what? One of the guests you mingled with at the party has told you about an afterparty in the works. The location, the staff, the people invited - you know all the details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a classy outfit you slide into the mansion where the afterparty shall take place at. The two guards are thick and dense as logs, but brighter than most folk you've met. Behind those classy glasses sits dry wit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You come with three more chaps and a pair of gals you met at the main show. You joke around a bit and line up to get in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;first&lt;/strong&gt; is the ginger guy who's as pale as vanilla. He comes up to the guards and says: &quot;Tell me.&quot; The left guard says &quot;zero point 940542104683244&quot; and the right one goes &quot;minus 12340633 over 36330480.&quot; Stunned at the sequence, you quickly tune in on the proceedings. The lad gives it a couple seconds' worth of thought and says: &quot;Two.&quot; They let him pass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second&lt;/strong&gt; goes the black woman. She eyes the guards for a moment and then says: &quot;Hit me.&quot; The left guard goes &quot;zero point 955741461871332&quot; and the right one goes &quot;minus 13047101 over 44346543.&quot; She metaphorically chews it over and says: &quot;Six, should be.&quot; She gets through as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Third&lt;/strong&gt; is the average Joe who came to the party to meet women, not eat spaghetti. The left guard goes &quot;one,&quot; but the right one says: &quot;Nothing from me.&quot; He starts panicking visibly before he stutters out a meek &quot;Zero?&quot; The guards shake their heads and he walks away, defeated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fourth&lt;/strong&gt; is the blond guy you've seen sip far too much champaigne. The left guard tells him: &quot;You get a zero point 948639957057241&quot; and the right one goes &quot;minus 18032031 over 56998858.&quot; He blinks at them a couple of times, then cups his chin for half a dozen seconds. &quot;Five?&quot; The guards nod and let him pass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fifth&lt;/strong&gt; is the short, burly brunette. The left guard tells her &quot;zero point 6532814824381882&quot; and the right one goes &quot;minus 17280845 over 63861676&quot;. She lets out a bored &quot;imaginary four&quot; and ploughs her way through the guards. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are stunned by the sheer brainpower of the people you drank with. Your turn now. You gulp and look at the smiling men. The left one says: &quot;So, for you... zero point 933581447902683?&quot; The right one nods and goes &quot;minus 17753215 over 49539438.&quot; You're looked at expectantly. The brunette's waiting on you behind the duo. What do you say?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The right guard always says a negative number. Why is this so?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4990" LastEditorUserId="4990" LastEditDate="2014-11-12T22:38:39.833" LastActivityDate="2014-11-12T22:38:39.833" Title="The Security to the Afterparty" Tags="|mathematics|password|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4106" PostTypeId="1" AcceptedAnswerId="4111" CreationDate="2014-11-11T19:49:28.830" Score="3" ViewCount="1083" Body="&lt;p&gt;A neanderthal walks into a room.  Inside are a man and a woman with large sticks, surrounded by dozens of unintelligible glass discs and squares.  The man (with the stick) comments he can see the ocean under the clouds.  Why?&lt;/p&gt;&#xA;" OwnerUserId="5042" LastEditorUserId="5042" LastEditDate="2014-11-11T20:04:02.417" LastActivityDate="2014-11-13T21:09:31.007" Title="Neanderthal Man" Tags="|lateral-thinking|puzzle-identification|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4107" PostTypeId="2" ParentId="4092" CreationDate="2014-11-11T19:49:49.017" Score="1" Body="&lt;p&gt;To expand on DarkGammas answer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Label the balls R1, R2, R3 and B1 if you grab two of them the possible combinations are&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;R1/R2 R1/R3 R1/B1&lt;br&gt;&#xA;  R2/R1 R2/R3 R2/B1&lt;br&gt;&#xA;  R3/R1 R3/R2 R3/B1&lt;br&gt;&#xA;  B1/R1 B1/R2 B1/R3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For a total of 12 combinations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You take two out and look at one which happens to be red. We will call this one R1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you &lt;strong&gt;know the one you looked at was the first ball grabbed&lt;/strong&gt; you can narrow it down to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;R1/R2 R1/R3 R1/B1 (3 combinations; 2 combinations have the second ball being red; &lt;strong&gt;2/3&lt;/strong&gt;)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now if you &lt;strong&gt;know the one you looked at was the second ball grabbed&lt;/strong&gt; you can narrow it down to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;R2/R1 R3/R1 B1/R1 (3 combinations; 2 combinations have the first ball being red; &lt;strong&gt;2/3&lt;/strong&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you do no know which one you are looking at your possible combinations are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;R1/R2 R1/R3 R1/B1 B1/R1 R2/R1 R3/R1. (6 combinations; 4 combinations have both red;&lt;strong&gt;2/3&lt;/strong&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore I believe 66.6% to be the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I was initially working under the premise that the last example would yield a result of 50%. once I found that to be wrong, I didn't want to waste the time I put into this answer and posted it anyways.)&lt;/p&gt;&#xA;" OwnerUserId="3922" LastActivityDate="2014-11-11T19:49:49.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4108" PostTypeId="2" ParentId="4106" CreationDate="2014-11-11T19:54:47.330" Score="10" Body="&lt;p&gt;Easy&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because the ocean &lt;strong&gt;is&lt;/strong&gt; under the clouds. ☺&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4877" LastActivityDate="2014-11-11T19:54:47.330" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4110" PostTypeId="1" AcceptedAnswerId="4116" CreationDate="2014-11-11T20:05:43.640" Score="1" ViewCount="262" Body="&lt;p&gt;You arrive at the party and the host personally comes out to greet you. He invites you in to the party and takes you to his office.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What madness is this!?!?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He gives you a numbered list of ingredients for his famous meatball recepie, and tells you that some of the ingredients are red herrings (but none of them are red herring). He has encoded the numbers of the correct ingredients; you have to figure out how to decode them into the proper numbers. He gives you some examples: &quot;6&quot; decoded is &quot;20&quot;, &quot;2&quot; decoded is &quot;4&quot;, and &quot;3&quot; decoded is &quot;8&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your friend has given up. He says good bye and leaves. The host tells you that the correct ingredients are those with numbers decoded from &quot;1&quot;, &quot;7&quot;, &quot;10&quot;, and &quot;9&quot;. What ingredient numbers are correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;By the way, if you get this wrong a trampling elephant is waiting!&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2101" LastEditDate="2014-11-11T20:49:12.710" LastActivityDate="2014-11-12T07:43:21.463" Title="The Security to the Party [Part 17!]" Tags="|password|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4111" PostTypeId="2" ParentId="4106" CreationDate="2014-11-11T20:07:14.600" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Neanderthal walked into a plane cockpit. The people with sticks are the pilot and co-pilot, who are holding onto the control sticks, are discussing the scenery (through the clouds below, they can see the ocean). The glass shapes are the windows and the various buttons for controlling the plane.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2101" LastActivityDate="2014-11-11T20:07:14.600" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4112" PostTypeId="1" CreationDate="2014-11-11T20:15:17.643" Score="-4" ViewCount="3420" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;A man has a wolf, a goat, and a cabbage. He must cross a river with the two animals and the cabbage. There is a small rowing boat, in which he can take only one thing with him at a time. If, however, the wolf and the goat are left alone, the wolf will eat the goat. If the goat and the cabbage are left alone, the goat will eat the cabbage.&lt;/p&gt;&#xA;&lt;p&gt;How can the man get across the river with the two animals and the cabbage?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://www.puzzle.dse.nl/logical/index_us.html#wolf_goat_cabbage&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.puzzle.dse.nl/logical/index_us.html#wolf_goat_cabbage&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user5049" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-02-16T19:02:07.870" Title="How can the man get across the river with the two animals and the cabbage?" Tags="|logical-deduction|river-crossing|" AnswerCount="1" CommentCount="3" ClosedDate="2014-11-11T21:53:48.307" ContentLicense="CC BY-SA 3.0" />
  <row Id="4113" PostTypeId="2" ParentId="4112" CreationDate="2014-11-11T20:17:45.323" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Fox,_goose_and_bag_of_beans_puzzle&quot; rel=&quot;nofollow&quot;&gt;This is a classic.&lt;/a&gt; I'm surprised it hasn't been asked before.&lt;br&gt;&lt;br&gt;&#xA; &#xA; &lt;strong&gt;Take the goat.&lt;/strong&gt; This leaves the wolf and the cabbage which is fine.&lt;br&gt;&#xA; &lt;strong&gt;Return alone.&lt;/strong&gt;&lt;br&gt;&#xA; &lt;strong&gt;Take the cabbage.&lt;/strong&gt; (To the goat.)&lt;br&gt;&#xA; &lt;strong&gt;Return with the goat.&lt;/strong&gt; So it can't eat the cabbage.&lt;br&gt;&#xA; &lt;strong&gt;Take the wolf.&lt;/strong&gt; So it can't eat the goat.&lt;br&gt;&#xA; &lt;strong&gt;Return alone.&lt;/strong&gt; Again leaving the wolf with the cabbage.&lt;br&gt;&#xA; &lt;strong&gt;Take the goat.&lt;/strong&gt;&lt;br&gt;&#xA; &lt;br&gt;&#xA; Alternatively, you can swap the steps of taking the wolf and the cabbage.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2450" LastEditorUserId="2450" LastEditDate="2014-11-11T20:37:30.977" LastActivityDate="2014-11-11T20:37:30.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4114" PostTypeId="2" ParentId="4092" CreationDate="2014-11-11T20:19:41.647" Score="2" Body="&lt;p&gt;Classical conditional probability (or &quot;Bayesian&quot; probability) problem.&#xA;Bayes formula is well known, but just for simplicity purposes,one may simply think:&#xA;I have one ball FIRMLY in hand and it IS red.&#xA;So the problem has transformed to: Draw one ball out of 3 (2 red, 1 blue)&#xA;probability of drawing a red, clearly= 2/3&lt;/p&gt;&#xA;" OwnerUserId="5051" LastActivityDate="2014-11-11T20:19:41.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4115" PostTypeId="2" ParentId="4110" CreationDate="2014-11-11T20:30:26.583" Score="2" Body="&lt;p&gt;The sequence is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; powers of 2 in hexadecimal,&#xA; 2^6 = 32 = 0x20 &#xA; 2^2 = 4 = 0x04&#xA; 2^3 = 8 = 0x08&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2^1 = 2 = 0x2&#xA; 2^7 = 129 = 0x80 &#xA; 2^10 = 1024 = 0x400&#xA; 2^9 = 512 = 0x200&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therewithal&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2,80,400,200&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5052" LastActivityDate="2014-11-11T20:30:26.583" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4116" PostTypeId="2" ParentId="4110" CreationDate="2014-11-11T20:34:48.957" Score="2" Body="&lt;p&gt;One would think:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $4x-4$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which gives&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;0&quot;, &quot;24&quot;, &quot;36&quot;, &quot;32&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that's obvious! So let's take a different road. How about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jumping prime numbers. Start from $x$ jump $x$ primes (including $x$) and add one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Resulting in&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2, 30, 42, 38&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4317" LastEditorUserId="4317" LastEditDate="2014-11-12T07:43:21.463" LastActivityDate="2014-11-12T07:43:21.463" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4117" PostTypeId="2" ParentId="4075" CreationDate="2014-11-11T20:54:38.913" Score="2" Body="&lt;p&gt;Start looking in   &lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Web Applications or Webmasters  &lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because  &lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a) All the questions were about animating blood in javascript  &lt;br/&gt;&#xA; b) The blood comes from the top (both are on the top of the all sites page)  &lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Comedy reason&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3) Web coders can never figure anything out on their own, there is probably a post that reads &quot;how me hack a SE website?&quot;  &lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Alternate Answer:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Look on your own computer, if nobody else has noticed, it's probably your local machine that's hacked, and not SE.  &lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5052" LastActivityDate="2014-11-11T20:54:38.913" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4118" PostTypeId="2" ParentId="4075" CreationDate="2014-11-11T21:03:02.033" Score="3" Body="&lt;h1&gt;Where do you start looking?&lt;/h1&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;The web inspector of your navigator. Since it's a JavaScript, you will have access to the source code since it's download by the browser and that's the first place you should be looking for clues. Maybe it is probably an extern script and you'll maybe see the domain origin!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5056" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-11-11T21:03:02.033" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4119" PostTypeId="1" AcceptedAnswerId="4223" CreationDate="2014-11-11T21:38:18.507" Score="7" ViewCount="1927" Body="&lt;p&gt;This is Part 4 of the Murder of the President brainteaser/riddle series. If you have not already, check out the answers for &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3600/murder-of-the-president-part-1&quot;&gt;Murder of the President - Part 1&lt;/a&gt; posted by Nit and the answers for &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3845/murder-of-the-president-part-2&quot;&gt;Murder of the President - Part 2&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3977/murder-of-the-president-part-3&quot;&gt;Murder of the President - Part 3&lt;/a&gt;, both posted by Joe Z. Each part will give you a clue and you must solve it. Use all knowledge you have of cryptography, ciphers, past puzzles, etc. This question will have two different clues you have to solve. You should also use Google. This case is meant to take place in the present day, so all politicians, celebrities, places, etc. are who they are now. Please post your answers in spoiler tags. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here's the riddle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You arrive back at the White House at 11:18 p.m. EST. You go into the Oval Office and you begin searching the President's desk. All of the drawers are open and filled with paper work, except one. It is the bottom left drawer, not too big and it has a lock with three number codes, the first with two digits and the second and third with a four digit codes. You ask for the codes to the desk, but you are told that no one knew them besides the President. Weird, you think. You are given a piece of paper that has some clues that the President wrote that are the codes to the desk. Weird, you think once again. But you don't have time to worry about that now. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the clues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first one seems to be a riddle.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Before you figure out this password&lt;br&gt;&#xA;  There is a riddle you must master&lt;br&gt;&#xA;  You must first get the tower that looms&lt;br&gt;&#xA;  Then you must find the hollow moon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The second one is just some words and letters that don't seem to make sense.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;D.R.K Dahanu&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The thirds seems to be a code.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Down the Rabbit Hole&lt;sup&gt;&lt;strong&gt;2&lt;/strong&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Once you manage to figure out the codes you open the drawer and, once again, you find a coded message. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bl gl .l_cl_ QQI Z;;.D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;What are the codes to the drawer? What does this coded message mean? Where should you go next?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; APPLE is in all uppercase letters. It is not an address. The 118 is not as important as the London part.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: I will be posting Part 5 in one to two days. I will select the correct answer for Part 4 before, but I will post the answer for Part 4 if no one gets it&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1711" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2014-12-15T23:01:55.593" Title="Murder of the President - Part 4" Tags="|riddle|mystery|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4120" PostTypeId="2" ParentId="4075" CreationDate="2014-11-11T21:40:36.203" Score="4" Body="&lt;p&gt;Maybe a little bit obvious but.. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If I need to find where is the GOD, I could start searching in &lt;a href=&quot;https://christianity.stackexchange.com/&quot;&gt;christianity&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1685" LastEditorUserId="-1" LastEditDate="2017-04-13T12:56:54.507" LastActivityDate="2014-11-11T21:40:36.203" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4121" PostTypeId="1" AcceptedAnswerId="4122" CreationDate="2014-11-11T22:19:55.600" Score="-3" ViewCount="850" Body="&lt;p&gt;You are in a dark room with 1 source of light (a lamp). You see a large door with a combination with nothing but letters. You find a sheet of paper stuck in the door with the numbers,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;010101000110100001100101 0110011101110010011001010110000101110100 011000100110100101101110011000010111001001111001 01101100011011110110001101101011 0110100101110011 011011110110111001100101 01110011011101000110010101110000 01100001011101110110000101111001 0111010001101111 01110110011010010110001101110100011011110111001001111001 00101110&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does this mean?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;SPOILER!!!!!!......&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;This is ASCII binary text. Will be making another one in a little.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5061" LastEditorUserId="11297" LastEditDate="2015-07-10T22:02:53.537" LastActivityDate="2016-05-16T17:53:36.110" Title="The Great Binary Puzzle - Part 1" Tags="|cipher|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4122" PostTypeId="2" ParentId="4121" CreationDate="2014-11-11T22:22:49.063" Score="3" Body="&lt;p&gt;Depending on what encoding you use and what endianness you use, it can mean lots of things. Assuming Big Endian UTF8/ASCII it gives &quot;The great binary lock is one step away to victory.&quot; (thank you Brainfuck). Nothing else to go on with.&lt;/p&gt;&#xA;" OwnerUserId="4990" LastActivityDate="2014-11-11T22:22:49.063" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4124" PostTypeId="1" AcceptedAnswerId="4128" CreationDate="2014-11-11T22:36:23.673" Score="7" ViewCount="324" Body="&lt;p&gt;Blah, bah, blah, party, blah, blah, blah, spaghetti.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First guest arrives, guard says &quot;3&quot;, guest says &quot;0&quot;, and is admitted. &lt;br&gt;&#xA;Second guest arrives, guard says &quot;4&quot;, guest says &quot;1&quot;, and is admitted. &lt;br&gt;&#xA;Third guest arrives, guard says &quot;8&quot;, guest says &quot;2&quot;, and is admitted.&lt;br&gt;&#xA;Fourth guest arrives, guard says &quot;9&quot;, guest says &quot;3&quot;, and is admitted.&lt;br&gt;&#xA;Fifth guest arrives, guard says &quot;10&quot;, guest says &quot;4&quot;, and guard says &quot;Umm, I don't know, come back later.&quot;&lt;br&gt;&#xA;Sixth guest arrives, guard says &quot;11&quot;, guest says &quot;5&quot; and is dissolved by acid.&lt;br&gt;&#xA;Seventh guest arrive, guard says &quot;18&quot;....   what should the guest answer?&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the series: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/tagged/party-security&quot;&gt;https://puzzling.stackexchange.com/questions/tagged/party-security&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I may need to edit the question later&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I won't accept an answer without an explanation of why it is right.&lt;/p&gt;&#xA;" OwnerUserId="1537" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2014-11-12T01:17:26.103" Title="The Security to the Party [Part 18]" Tags="|password|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4125" PostTypeId="2" ParentId="4075" CreationDate="2014-11-11T22:50:07.153" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Meta-stack exchange&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's where you find out what's up.&lt;/p&gt;&#xA;" OwnerUserId="358" LastActivityDate="2014-11-11T22:50:07.153" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4126" PostTypeId="1" AcceptedAnswerId="4134" CreationDate="2014-11-11T22:53:47.640" Score="1" ViewCount="634" Body="&lt;p&gt;You pass the first room to enter the second room. It isn't that dim like the first room so it is easier to see. You see another combination, but this time you see something scraped on the wall that's different from the last code. It says &quot;54686572652069732061206c69676874206e6561722c2062757420746865206c696768742069736e5c2774207468652077617920746f207468652070757a7a6c652e2049745c2773206a7573742061207374657020636c6f7365722e&quot;. Right under it says &quot;Six and a math term.&quot; What does this mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Previous parts:&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/4121/the-great-binary-puzzle-part-1&quot;&gt;Part 1&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="5061" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2015-04-18T04:53:18.833" Title="The Great Binary Puzzle - Part 2" Tags="|cipher|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4128" PostTypeId="2" ParentId="4124" CreationDate="2014-11-11T23:16:52.797" Score="11" Body="&lt;p&gt;My reasoning&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The response is the accepted response to the given party security puzzle - #3 had the answer 0, #4 is currently unknown.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We seem to be asked for the answer to this very challenge, which presumably means I can pick any number. I'll go with &lt;a href=&quot;http://en.wikipedia.org/wiki/Aaron_Rodgers&quot;&gt;12&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2518" LastEditorUserId="1537" LastEditDate="2014-11-12T01:17:26.103" LastActivityDate="2014-11-12T01:17:26.103" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4129" PostTypeId="2" ParentId="4075" CreationDate="2014-11-11T23:24:25.603" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;HSM (History Science and Math) has apparently been taken over or deleted altogether in an effort by the newly self-aware robots who &quot;will conquer&quot; to keep the necessary knowledge from us to do anything about it (as well as most of the 'programing' sites). I immediately search Robotics for clues to help me find the lost HSM.SE site and try to discover what they are attempting to hide from me.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4522" LastActivityDate="2014-11-11T23:24:25.603" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4130" PostTypeId="2" ParentId="4075" CreationDate="2014-11-11T23:33:00.880" Score="7" Body="&lt;p&gt;I'd look in &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Graphic Design, because these blood graphics are too pretty. Graphics are really hard! Plus their name is red, like blood. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="431" LastActivityDate="2014-11-11T23:33:00.880" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4131" PostTypeId="2" ParentId="4105" CreationDate="2014-11-11T23:40:21.563" Score="5" Body="&lt;p&gt;I have no idea how to work backwards on this, and I don't know how #5 works so maybe someone else can pitch in.&#xA;But anyway...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer you give is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2.5 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;zero point 940542104683244&quot; and the right one goes &quot;minus 12340633 over 36330480.&quot; ... 12340633/36330480 = .3396771251 ... 0.940542105 + 0.339677125 i = 2^(i/2) so &lt;strong&gt;2&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;zero point 955741461871332&quot; and the right one goes &quot;minus 13047101 over 44346543.&quot; ...  13047101/44346543 = 0.2942078484 ... 0.955741462 + 0.294207848 i = 6^(i/6) so &lt;strong&gt;6&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;zero point 948639957057241&quot; and the right one goes &quot;minus 18032031 over 56998858.&quot;  ... 18032031/56998858 = 0.3163577593  .... 0.948639957 + 0.316357759 i = 5^(i/5) so &lt;strong&gt;5&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6(you)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;zero point 933581447902683?&quot; The right one nods and goes &quot;minus 17753215 over 49539438.&quot; .... 17753215/49539438 =  0.35836528868   ..... 0.933581448 + 0.358365289 i = 2.5^(i/2.5) so &lt;strong&gt;2.5&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4680" LastActivityDate="2014-11-11T23:40:21.563" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4132" PostTypeId="2" ParentId="4124" CreationDate="2014-11-11T23:48:43.097" Score="0" Body="&lt;p&gt;The Guest should answer 6 as the pattern is along the lines of modulus 3 so:&lt;br/&gt;&#xA;0 % 3 = 0&lt;br/&gt;&#xA;4 % 3 = 1&lt;br/&gt;&#xA;8 % 3 = 2&lt;br/&gt;&#xA;9 % 3 = 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Fifth person should have answered 1 to be admitted&lt;/p&gt;&#xA;" OwnerUserId="5070" LastActivityDate="2014-11-11T23:48:43.097" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4133" PostTypeId="1" AcceptedAnswerId="4272" CreationDate="2014-11-12T00:11:24.683" Score="11" ViewCount="1392" Body="&lt;p&gt;In a kingdom far far away there is a castle with an irritated king. Why irritated you ask? He just came back from a trip to the western world and he was thrown out from nearly all the parties... just because he refused to do all those hard math riddles. Seriously, he felt so insulted by their insolence... yet also strangely inspired! He would right this wrong directly! As soon as he gets back in his castle he calls in his head of guards to devise a new password scheme.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A merchant comes and the guards ask him&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Original version... containing a mistake... see comment on accepted answer :'( : &lt;em&gt;Kind merchant, what brings you to our gates?&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Kind merchant, what is it that brings you here?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The merchant closes his eyes for a bit and says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Trade and wares for our sovereign, my fair guards.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And they let him right in.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Behind the merchant a couple of spies for a neighboring nation were waiting to be let in. They had come and gone from the city many times, but they grew suspicious as the guards were behaving differently from other days and turned around to listen for awhile from far away whilst acting as if they were waiting for a bit.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A monk arrives at the gate, and the first guard says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Father, it shames us that we have to ask this of you, but what...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the second guard continues&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... brings thee here?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the monk answers&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is quite all right my children, I have heard of this new scheme your rex has thought of. Be that as it may be, I will be here for awhile, if you need my help or advice, visit me at the monastery.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And walks right in.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A forester arrives at the gate and says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Do you really need to ask me again? I have left and entered all the time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Guard&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Yes, Yes, we do.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the forester sighs&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;That king of ours... as much as I love him, sometimes... &lt;em&gt;sigh&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the guards chuckle to each other and let him right in.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A beggar arrives at the gate&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Don't come to our city! Not like a piece of filth like you will know our password! If you knew you would just spoil it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first guard exclaims. The second guard gives an irritated stare at his companion and kindly says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Well... he is half right, do you?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The beggar, doesn't even wait to think and starts answering directly&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Yes of course I do... the monarch... ehm... I have heard it from our own monarchs mouth when he announced it last week. I may be cripple and without arms, but I am not an idiot! &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And in he goes.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For two hours nobody turns up at the gate, before the unloved mother in law of the second guard turns up.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Welcome back, need help?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first one says kindly, seeing she is pushing in a cart that's far too heavy for her. The lady smiles kindly at him and is about to answer when the second interjects&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What brings you here to our gate at night?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The lady halts, thinks for a bit, then scowls at her daughters husband angrily before thinking for a bit longer before answering half afraid:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is all to honour our kind and just and fair overlord of course...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the second guard replies&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You dare call our fair King an overlord?! You...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Before the first interjects and whilst helping her in through the gate says to the second guard&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You ass! She did not deserve that! If you ever do something like that again I will seriously tell your wife!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At which point the second guard goes white with fear.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;During the exchange one of the spies looks at the other and says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I know it! It's so simple! This sounds &lt;em&gt;just&lt;/em&gt; like that king of theirs!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He walks up to the guards and they say&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So, decided to return after all? Hiding in the trees and everything...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the second continues with&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Anything you wish to say for yourself?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The spy looks doubtful for a bit... goes silent... and then exclaims&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;His Majesty! His Majesty wants me in!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The guards look at each other, nod, and kill him in one move whilst the first says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You idiot, did you forget who we are?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The question is: &lt;strong&gt;What should the spy have done/said differently? And why?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It didn't all work out as nicely as I was hoping, because of a reason I can't disclose (should have checked better before I started). Either way, still think it's a nice simple puzzle with a nice little twist in there to keep you guys thinking for a bit :P Though who knows, maybe you will get it within minutes :P&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If somebody else would have given the spies answer to the same question they &lt;strong&gt;would&lt;/strong&gt; have been let in. Might have been scolded for giving a 'poor' answer, but definitely would not have been killed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="468" LastEditorUserId="468" LastEditDate="2014-11-12T19:53:25.160" LastActivityDate="2014-11-12T19:53:25.160" Title="The king who is entirely sick of those insolent western parties" Tags="|password|" AnswerCount="6" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="4134" PostTypeId="2" ParentId="4126" CreationDate="2014-11-12T00:46:18.837" Score="2" Body="&lt;p&gt;&quot;Six and a math term&quot; just means hexadecimal, which is actually base 16 (not binary).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hex to ascii translation results in:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is a light near, but the light isn\'t the way to the puzzle. It\'s just a step closer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="104" LastEditorUserId="4900" LastEditDate="2014-11-12T06:30:38.153" LastActivityDate="2014-11-12T06:30:38.153" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4135" PostTypeId="2" ParentId="3939" CreationDate="2014-11-12T01:22:50.910" Score="23" Body="&lt;p&gt;I don't understand all the support for the shameless apophenia in the other answers. &quot;TRY TO QUIT&quot;? Weak sauce.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct solution is thus:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;observe the absence of &lt;code&gt;0&lt;/code&gt; in the code, which strongly suggests a numpad-based solution (since the &lt;code&gt;0&lt;/code&gt; is distinct and separate from all other digits here)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;tracing out the numpad digits in the specified order yields a simple diagram where four letters are distinctly seen: &lt;code&gt;L&lt;/code&gt;, &lt;code&gt;I&lt;/code&gt;, &lt;code&gt;S&lt;/code&gt;, &lt;code&gt;A&lt;/code&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;we confirm the correctness of this interpretation by noting that &lt;code&gt;SARAH&lt;/code&gt; cannot be encoded in this way, since &lt;code&gt;A&lt;/code&gt; would appear appear twice and the two instances would be indistinguishable, and &lt;code&gt;ANNE&lt;/code&gt; cannot be encoded in this way because &lt;code&gt;E&lt;/code&gt; clearly doesn't appear in the figure&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Indeed, it's the coy but non-judgmental Lisa flirting with John, by clearly indicating her name. &amp;hearts; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/yoVCd.png&quot; alt=&quot;LISA&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2014-11-12T01:22:50.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4136" PostTypeId="2" ParentId="4075" CreationDate="2014-11-12T01:25:32.987" Score="43" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Moderators&lt;/strong&gt;. The &quot;killed&quot; sites are StackOverflow, Ask Different, MathOverflow, Reverse Engineering, Robotics, and Travel. The capital letters of the names of the &quot;killed sites&quot; are SOADMORERT, which when you jumble, you get MODERATORS.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5079" LastActivityDate="2014-11-12T01:25:32.987" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4137" PostTypeId="2" ParentId="4133" CreationDate="2014-11-12T01:47:52.233" Score="6" Body="&lt;p&gt;The answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The spy should have indicated that he served the king. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The merchant said &quot;&lt;em&gt;our&lt;/em&gt; sovereign lord&quot; and was let in. The monk, who wasn't let in, described the king as &quot;&lt;strong&gt;your&lt;/strong&gt; rex&quot;. Perhaps he doesn't officially serve the king, and was opposed to claiming the king as his own? The forester, who entered the city, talked about &quot;That king of &lt;strong&gt;ours&lt;/strong&gt;&quot;. The beggar (without thinking) begins with &quot;&lt;strong&gt;the&lt;/strong&gt; monarch&quot; but corrects himself with &quot;&lt;strong&gt;our&lt;/strong&gt; own monarch's&quot;. It's not clear whether or not he was let in, but it's also not clear whether or not the mistake ruined his chance. The spy made the mistake of invoking &quot;His Majesty&quot;; he should have said something with a possessive pronoun, like &quot;my liege&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2101" LastActivityDate="2014-11-12T01:47:52.233" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4138" PostTypeId="2" ParentId="4133" CreationDate="2014-11-12T01:51:31.960" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It has something to do with pronoun use. There are &lt;em&gt;a lot&lt;/em&gt; of possibilities, but my best guess is as follows: the King must be referenced in the response, and his description must be preceded by a first- or second-person pronoun. Ergo the &quot;Did you forget who 'we' are?&quot; from the guards.In all but one case, &quot;we&quot; and &quot;our&quot; (first person) is used. Even so, the monk uses second person and is apparently admitted to the castle. I found no consistent pattern between the pronoun use in the question asked by the guard and the response.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="4951" LastEditDate="2014-11-12T02:18:10.793" LastActivityDate="2014-11-12T02:18:10.793" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4139" PostTypeId="1" AcceptedAnswerId="4291" CreationDate="2014-11-12T01:54:12.133" Score="22" ViewCount="3232" Body="&lt;p&gt;I just wrote this poem.  I hope you guys like it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This thing can be so tiny,&lt;br&gt;&#xA;  That's when it has the most variety.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Standing up straight it is surrounded,&lt;br&gt;&#xA;  Otherwise it can make things bounded.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Within a box it lets you choose,&lt;br&gt;&#xA;  But if it points up you will lose.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;It can be seen leaning,&lt;br&gt;&#xA;  That's when it reveals true meaning.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;It can be a celebrity and nothing at once,&lt;br&gt;&#xA;  It can show uncertainty but never incoherence.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;But when wealth is what it shows,&lt;br&gt;&#xA;  The end is what you will know.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the poem referring to?  Can you explain what each line means?&lt;/p&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4450" LastEditDate="2015-05-16T20:40:31.627" LastActivityDate="2015-05-16T20:40:31.627" Title="Riddle Poem: It can be a celebrity and nothing at once" Tags="|riddle|" AnswerCount="10" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4140" PostTypeId="2" ParentId="4139" CreationDate="2014-11-12T02:03:13.447" Score="3" Body="&lt;p&gt;You are talking about:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; quarks &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This thing can be so tiny,&#xA;That's when it has the most variety.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  quark is an elementary particle....smallest of the smallest things. It has 6 flavors:  up, down, strange, charm, bottom, and top&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Standing up straight, it is surrounded,&#xA;Otherwise it can make things bounded.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; figuring this one out&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Within a box it lets you choose,&#xA;But if it points up you will lose.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; referring to Schrödinger's cat. Havent figured out second sentence&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It can be seen leaning,&#xA;That's when it reveals true meaning.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Decay of particles show that quarks lean away from the collision...I think thats how they were discovered. This is just last minute guessing :P&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It can be a celebrity and nothing at once,&#xA;It can show uncertainty but never incoherence.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again quarks are elementary particles, they make up every other particle. Havent figured out the coherence bit&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But when wealth is what is shows,&#xA;The end, you will know.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think this is referring to A quark-nova/quark-star. Quark star/nova are stars made up of quarks. When they are at the end of their life cycle they would cause a super novae aka the end of you. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am not an expert in this area... just a student doing Comp. Eng, this is what I have learned from my own reading and as you can see there are gaps in my knowledge. So I could be really really wrong but I see parallels here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyways, whatever it is. It is a good poem :)&lt;/p&gt;&#xA;" OwnerUserId="5030" LastEditorUserId="5030" LastEditDate="2014-11-12T02:43:24.910" LastActivityDate="2014-11-12T02:43:24.910" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4141" PostTypeId="2" ParentId="4133" CreationDate="2014-11-12T02:40:59.400" Score="3" Body="&lt;p&gt;To answer what they should've done differently, I'd say&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Give up, the guards can see you in the trees.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which also answers the question why he was killed.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2014-11-12T02:40:59.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4142" PostTypeId="2" ParentId="4139" CreationDate="2014-11-12T02:43:42.760" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You're referring to a line.&lt;br&gt;&#xA;&lt;br&gt;&#xA; It can be stood up, leaning, sideways, etc. (|, /, \, -, _) It's also in a money sign $ so it seems to fit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2014-11-12T02:43:42.760" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4143" PostTypeId="2" ParentId="4133" CreationDate="2014-11-12T02:48:29.073" Score="1" Body="&lt;p&gt;Probably wrong but...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Everyone mentions the king &lt;em&gt;and&lt;/em&gt; the guards (often indirectly).  The spies only mentioned the king.  Explains why the guards say 'did you forget who &lt;em&gt;we&lt;/em&gt; are'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Problem is I don't think the hint likes this interpretation.&lt;/p&gt;&#xA;" OwnerUserId="2643" LastActivityDate="2014-11-12T02:48:29.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4144" PostTypeId="2" ParentId="4075" CreationDate="2014-11-12T02:52:01.037" Score="4" Body="&lt;p&gt;From the picture I think its clear that (well....ugh only after the hint though)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Robotics &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is the place to look because &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Thats where it says &quot;we will conquer&quot; but in other places it says Die&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2313" LastActivityDate="2014-11-12T02:52:01.037" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4145" PostTypeId="1" AcceptedAnswerId="4197" CreationDate="2014-11-12T02:57:08.417" Score="1" ViewCount="437" Body="&lt;p&gt;&lt;sub&gt;Oh God Another Recap&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A party is being held at a local mansion. The host is very rich and his success is because of one thing — his famous recipe for Spaghetti!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only guests that may attend are people who correctly reply to the guard at the door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's where you come in. You and a friend are trying to steal this recipe. You sneak by and listen to the passwords.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first guest arrives. The security says &quot;2&quot;, and the guest replies &quot;3&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second guest arrives. The security says &quot;3&quot;, the guest replies &quot;7&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third guest arrives. The security says &quot;5&quot;, the guest replies &quot;2&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fourth guest arrives. The security says &quot;8&quot;, the guest replies &quot;4&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your friend thinks he's got it all figured out so he walks up to the door and the security says &quot;4&quot;. He replied &quot;4&quot;, and is immediately gored by the guard's security rhinoceros. Before passing out your friend says &quot;Oh God, not again&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another guest arrives, security says &quot;6&quot; and he says &quot;3&quot; to get in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You walk up and get a &quot;7&quot; from security. What is your response?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For ease of answering however, I shall provide 2 hints. Start out at a score of 100 If you solve the riddle without looking at any hints, your score is 100 - (distance from answer * 10). If you look at the first hint only your score is 80 - (distance from answer * 10). If you look at the second hint only your score is 60 - (distance from answer * 10). If you look at both hints your score is 30 - (distance from answer * 10). Honor system applies. I will calculate your score accordingly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As usual please post your answer, an explanation of what process you used to get the answer and any hints used in spoiler tags.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The math tag isn't used for decoration here. A mathematical function $(f(x))$ is involved, but only a small part of the result is used (because numbers get very big very fast when this function is used) after being modified by rounding the results of another function $(g(x)$ (because the results of this second function may be a decimal number). TL;DR &lt;em&gt;small part of&lt;/em&gt; $(f(x) + g(x)) =$ answer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are 2 functions involved here are $f(x)$ and $g(x)$. $f(x)$ is a factorial, and $g(x)$ is a floor function. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4951" LastEditorUserId="4951" LastEditDate="2014-11-12T03:23:24.663" LastActivityDate="2014-11-12T09:35:27.353" Title="The Security To The Party (Part 19: Oh God Another One)" Tags="|mathematics|password|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4148" PostTypeId="2" ParentId="4126" CreationDate="2014-11-12T03:36:40.530" Score="0" Body="&lt;p&gt;Take the code and paste it &lt;a href=&quot;http://www.rapidtables.com/convert/number/hex-to-ascii.htm&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. &lt;br&gt;&#xA;You will get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There is a light near, but the light isn\'t the way to the puzzle. It\'s just a step closer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is not binary, but rather, hexadecimal.&lt;/p&gt;&#xA;" OwnerUserId="4973" LastEditorUserId="4900" LastEditDate="2014-11-12T05:34:43.450" LastActivityDate="2014-11-12T05:34:43.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4149" PostTypeId="1" AcceptedAnswerId="4158" CreationDate="2014-11-12T03:54:07.737" Score="28" ViewCount="4395" Body="&lt;p&gt;You've seen me often, know me well,&lt;br&gt;&#xA;my name the scribes and sages tell.&lt;br&gt;&#xA;My many kin you also know,&lt;br&gt;&#xA;but I'm most famed, as our kind go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One sister's love is grand largesse,&lt;br&gt;&#xA;another's like me, more or less.&lt;br&gt;&#xA;My brother shuns such grand appeal,&lt;br&gt;&#xA;he much prefers his meager meal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our distant cousin's always in,&lt;br&gt;&#xA;and that implies he's also kin.&lt;br&gt;&#xA;And where one village holds another,&lt;br&gt;&#xA;you'll often find our eldest brother.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I keep the peace with great panache.&lt;br&gt;&#xA;Strike me down and neighbours clash,&lt;br&gt;&#xA;or leave me be, it's all the same,&lt;br&gt;&#xA;but tell me, friend, &lt;strong&gt;what is my name?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;&lt;em&gt;Puzzlers are politely encouraged to place answers in spoiler blocks to avoid inadvertently spoiling the fun for other readers.&lt;/em&gt; :)&lt;/p&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="5373" LastEditDate="2015-07-15T23:53:21.540" LastActivityDate="2015-07-15T23:53:21.540" Title="My Famous Family and Me" Tags="|riddle|rhyme|" AnswerCount="12" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4150" PostTypeId="2" ParentId="4013" CreationDate="2014-11-12T04:16:29.410" Score="3" Body="&lt;p&gt;&lt;code&gt;You walk up and get a &quot;8&quot; from security. What is your response?&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The correct response is 4.&#xA;&lt;p&gt;&#xA; The rule is to convert the number into capitalized word form (EIGHT), look at the first letter, and take the minimum number of &lt;strong&gt;straight&lt;/strong&gt; lines needed to form it.  Four straight lines are needed to form an uppercase E.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What was Hint #4 referring to?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;8. You might not think that's a hint, but it is if you think outside the box.&quot;&#xA;&lt;p&gt;&#xA; That is referring to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4015/the-security-to-the-party-part-8&quot;&gt;Part 8&lt;/a&gt; of this series.  &lt;a href=&quot;https://puzzling.stackexchange.com/a/4021/4450&quot;&gt;The answer&lt;/a&gt; for that question requires looking at the word form of the number and counting things.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What was Hint #5 referring to?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;See my profile.&quot;&#xA;&lt;p&gt;&#xA; On my profile, the most upvoted thing is &lt;a href=&quot;https://puzzling.stackexchange.com/a/3873/4450&quot;&gt;this answer&lt;/a&gt;, which involves counting straight lines.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2014-11-12T04:16:29.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4152" PostTypeId="2" ParentId="4139" CreationDate="2014-11-12T04:23:09.293" Score="4" Body="&lt;p&gt;A possible answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Book/Story&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This thing can be so tiny,&#xA;That's when it has the most variety.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Short stories are plenty and varied.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Standing up straight, it is surrounded,&#xA;Otherwise it can make things bounded.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You stack books together all supporting each other.  (Bounded?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Within a box it lets you choose,&#xA;But if it points up you will lose.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can choose books from a box/shelf, but if the spine is facing the wrong way, you can't see the title/author (you lose).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It can be seen leaning,&#xA;That's when it reveals true meaning.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You need to lean/hold a book (while open) to read it and see its contents.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It can be a celebrity and nothing at once,&#xA;It can show uncertainty but never incoherence.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A book's contents can be famous, but not the book itself. A book's content may be contentious but an interpretation is always possible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But when wealth is what it shows,&#xA;The end, you will know.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When the page numbers get high (wealth), you know you're near the end.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a great puzzle by the way. I'm pretty sure this isn't right, but it's a candidate.&lt;/p&gt;&#xA;" OwnerUserId="2484" LastActivityDate="2014-11-12T04:23:09.293" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4153" PostTypeId="5" CreationDate="2014-11-12T04:43:01.997" Score="0" Body="&lt;p&gt;&lt;strong&gt;What is a riddle?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;https://en.wikipedia.org/wiki/Riddle&quot; rel=&quot;nofollow noreferrer&quot;&gt;riddle&lt;/a&gt; gives indirect clues towards the identity of an unnamed object or concept, often presented in the form of a poem. The goal of any riddle is to figure out what it is describing. A riddle generally uses &lt;a href=&quot;https://en.wiktionary.org/wiki/metaphor&quot; rel=&quot;nofollow noreferrer&quot;&gt;metaphors&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Allegory&quot; rel=&quot;nofollow noreferrer&quot;&gt;allegories&lt;/a&gt;  to conceal the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What makes a good riddle?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Riddles must be based on &lt;a href=&quot;https://puzzling.meta.stackexchange.com/a/1530&quot;&gt;calculation and interpretation&lt;/a&gt; and not rely on assumption. The answer must be unique and clearly fit all the given clues.&lt;/li&gt;&#xA;&lt;li&gt;Riddles must contain enough information and have an unambiguous, clearly correct answer.&lt;/li&gt;&#xA;&lt;li&gt;Riddles should be &lt;a href=&quot;https://puzzling.meta.stackexchange.com/a/1518&quot;&gt;clever and creative&lt;/a&gt;. Riddles that are too easy or unimaginative are not fun to solve.&lt;/li&gt;&#xA;&lt;li&gt;Users can receive feedback on their riddles by posting them to &lt;a href=&quot;https://puzzling.meta.stackexchange.com/questions/5276/the-riddle-sandbox&quot;&gt;The Riddle Sandbox&lt;/a&gt; before posting to the main site.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How should I try to solve a riddle?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Analyze the riddle line by line. Note words or sentences that seem out of place.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/a/4578&quot;&gt;Keep a list of possibilities&lt;/a&gt; that fit the description the riddle provides and eliminate possible options.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How should I answer the question once I have the solution to the riddle?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Make sure your solution fully fits the description in the riddle.&lt;/li&gt;&#xA;&lt;li&gt;Answer the riddle fully, and &lt;a href=&quot;https://puzzling.meta.stackexchange.com/a/1494&quot;&gt;provide a detailed explanation&lt;/a&gt;. Provide a line-by-line analysis to justify your solution.&lt;/li&gt;&#xA;&lt;li&gt;Use &lt;a href=&quot;https://puzzling.meta.stackexchange.com/search?q=spoiler&quot;&gt;spoilers&lt;/a&gt; for content that directly relates to the answer. Use plain text in-between spoilers to explain what the reader is to expect.&#xA;(Spoilers hide part of a post by placing &lt;code&gt;&amp;gt;!&lt;/code&gt; at the beginning of a line.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How should I format my post to look like a riddle?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Riddles are often presented as quotes, and we often see users struggling with line breaks, causing all lines to be concatenated. Start each line with a &lt;code&gt;&amp;gt;&lt;/code&gt; and end with two spaces, like this (the underscores indicate the trailing spaces):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;gt; Roses are red__&#xA;&amp;gt; Violets are blue__&#xA;&amp;gt; Riddles are nice__&#xA;&amp;gt; But I don't have a clue.&#xA;&lt;/code&gt;&lt;/pre&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This renders as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Roses are red&lt;br&gt;&#xA;  Violets are blue&lt;br&gt;&#xA;  Riddles are nice&lt;br&gt;&#xA;  But I don't have a clue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7308" LastEditorUserId="11110" LastEditDate="2020-04-24T07:24:07.123" LastActivityDate="2020-04-24T07:24:07.123" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4154" PostTypeId="4" CreationDate="2014-11-12T04:43:01.997" Score="0" Body="This is NOT a general tag for all puzzles that are somehow indirect. A riddle gives indirect clues about an unnamed object or concept to be identified, often in a poem format." OwnerUserId="4219" LastEditorUserId="69582" LastEditDate="2022-02-25T17:27:29.273" LastActivityDate="2022-02-25T17:27:29.273" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4155" PostTypeId="5" CreationDate="2014-11-12T04:49:41.730" Score="0" Body="&lt;p&gt;A puzzle related to shapes, geometric objects (polygons, circles, solids, etc.) of any number of dimensions, relative positions of figures, and the properties of space.&lt;/p&gt;&#xA;&lt;p&gt;Geometric puzzles include:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;dissection puzzles&lt;/li&gt;&#xA;&lt;li&gt;tangram puzzles&lt;/li&gt;&#xA;&lt;li&gt;shortest-path puzzles&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;General geometry questions are considered off-topic but can be asked on &lt;a href=&quot;https://math.stackexchange.com/&quot;&gt;Mathematics Stack Exchange&lt;/a&gt; with the geometry tags(search &amp;quot;geometry&amp;quot; in the &lt;a href=&quot;https://math.stackexchange.com/tags&quot;&gt;list of tags&lt;/a&gt; for a full list of geometry tags).&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="71981" LastEditDate="2020-11-30T01:56:24.067" LastActivityDate="2020-11-30T01:56:24.067" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4156" PostTypeId="4" CreationDate="2014-11-12T04:49:41.730" Score="0" Body="A puzzle related to shapes, geometric objects (polygons, circles, solids, etc.) of any number of dimensions, the relative position of figures, and the properties of space. Use with [mathematics]" OwnerUserId="4219" LastEditorUserId="69582" LastEditDate="2021-01-29T22:11:52.047" LastActivityDate="2021-01-29T22:11:52.047" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4157" PostTypeId="2" ParentId="4119" CreationDate="2014-11-12T04:56:26.150" Score="3" Body="&lt;p&gt;I haven't figured out the numerical clues yet, but here's what I have of the coded message that was inside the drawer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears to have been typed by somebody whose hands were one space down from their correct positions, because if you shift each letter up, you get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Go to lo?do? !!* ApplE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The question marks don't indicate literal question marks, but ambiguous or unknown characters. There is no character above the underscore key on a keyboard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if we take the underscores to mean the space bar instead, it seems obvious that the message is in fact referring to a location in London, which makes sense because &quot;n&quot; is in fact above the space bar on most QWERTY keyboards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, if we forgive the strange misuse of the shift key in certain cases (and treat them as caps lock instead), we get a final message of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Go to London 118 APPLE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the significance of which I haven't completely figured out yet.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2014-11-12T04:56:26.150" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4158" PostTypeId="2" ParentId="4149" CreationDate="2014-11-12T05:03:56.483" Score="16" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am thinking that &quot;strike me down and neighbours clash&quot; might have to do with keys on a keyboard.  When you type, typically you are not just pressing one key, you are pressing a bunch of keys one right after another.  So neighboring keys will clash - not with each other, but clash with the bottom of the keyboard.&#xA;&lt;P&gt;&#xA; You are the most famous (&quot;I'm most famed&quot;) - the Enter key.  &quot;Another's like me&quot; - that is the other Enter key on the keyboard (on the keypad on the right side).&#xA;&lt;P&gt;&#xA; Your &quot;sister's love is grand largesse&quot;.  That means your sister outputs more than the others.  She is the Tab key - a tab is equivalent to several characters (several spaces).&#xA; &lt;P&gt;&#xA; Your brother &quot;prefers his meager meal&quot;.  That means your brother eats characters - he is either the Backspace or Delete key.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4450" LastEditDate="2014-11-12T05:24:48.340" LastActivityDate="2014-11-12T05:24:48.340" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4159" PostTypeId="2" ParentId="3842" CreationDate="2014-11-12T05:05:52.387" Score="4" Body="&lt;p&gt;The following game produces $12$ kings on the black side.  I have suffixed the usual move with $K$ when a king is made:&lt;br&gt;&#xA;$$\begin {array} {l l} 10-14&amp;amp;21-17\\ 14\text { x }21&amp;amp;22-17&#xA;\\9-13&amp;amp;25-22\\7-10&amp;amp;30-25\\&#xA;21\text { x }30K&amp;amp;17-14\\10\text { x }17&amp;amp;29-25\\&#xA;30\text { x }21&amp;amp;24-20\\21-25&amp;amp;20-16\\&#xA;11\text { x }20&amp;amp;22-18\\25-29&amp;amp;18-15\\&#xA;17-21&amp;amp;15-10\\6\text { x }15&amp;amp;26-22\\&#xA;21-25&amp;amp;22-18\\15\text { x }22&amp;amp;23-19\\&#xA;5-9&amp;amp;19-15\\1-5&amp;amp;31-26\\&#xA;22\text { x }31K&amp;amp;27-23\\25-30K&amp;amp;15-10\\&#xA;29-25&amp;amp;28-24\\20\text { x }2&amp;amp;32-28\\&#xA;27-32K&amp;amp;10-6\\31-26&amp;amp;28-24\\&#xA;26\text { x }19\text { x }28&amp;amp;6-1K \end {array}$$ and it seems clear that Black can let Red come to him, capture all the men, and king them all.  Red has space to shuffle his king when needed.  It wasn't hard to find.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditDate="2014-11-12T15:04:31.603" LastActivityDate="2014-11-12T15:04:31.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4160" PostTypeId="5" CreationDate="2014-11-12T05:09:32.687" Score="0" Body="&lt;p&gt;A puzzle whose solution is characteristically worked out using a pencil and paper. Common examples include &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;connect-the-dots, &lt;/li&gt;&#xA;&lt;li&gt;hangman, &lt;/li&gt;&#xA;&lt;li&gt;dots-and-boxes, &lt;/li&gt;&#xA;&lt;li&gt;fill-in-the-blank,&lt;/li&gt;&#xA;&lt;li&gt;SOS. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note that crossword, Sudoku, tic-tac-toe, and logic grid puzzles all have their own unique tags.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-04T15:32:49.750" LastActivityDate="2015-02-04T15:32:49.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4161" PostTypeId="4" CreationDate="2014-11-12T05:09:32.687" Score="0" Body="A puzzle whose solution is characteristically worked out using a pencil and paper. Common examples include connect-the-dots, hangman, dots-and-boxes, fill-in-the-blank, and SOS. Note that crossword, Sudoku, tic-tac-toe, and logic grid puzzles all have unique tags." OwnerUserId="4219" LastEditorUserId="4219" LastEditDate="2014-11-12T13:35:04.573" LastActivityDate="2014-11-12T13:35:04.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4162" PostTypeId="5" CreationDate="2014-11-12T05:15:50.773" Score="0" Body="&lt;p&gt;The 15-puzzle is a sliding puzzle that consists of a frame of numbered square tiles in random order with one tile missing. The objective of the puzzle is to place the tiles in increasing order by making sliding moves that use the empty space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Johnson &amp;amp; Story (&quot;Notes on the 15-puzzle&quot;, American Journal of Mathematics 2:397-404, 1879) used a parity argument to show that half of the starting positions for the 15-puzzle are impossible to resolve, no matter how many moves are made. In particular, the starting position where the tiles are arranged in decreasing order is impossible to resolve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 15-puzzle was invented by Noyes Palmer Chapman, a postmaster in Canastota, New York, who is said to have shown friends, as early as 1874, a precursor puzzle consisting of 16 numbered blocks that were to be put together in rows of four, each summing to 34. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The well-known American puzzle-composer Sam Loyd claimed that he had invented the puzzle, for example writing in the Cyclopedia of Puzzles (published 1914): &quot;The older inhabitants of Puzzleland will remember how in the early seventies I drove the entire world crazy over a little box of movable pieces which became known as the 14-15 Puzzle.&quot; However, Loyd had nothing to do with the invention or initial popularity of the puzzle.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-06T18:12:42.073" LastActivityDate="2015-02-06T18:12:42.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4163" PostTypeId="4" CreationDate="2014-11-12T05:15:50.773" Score="0" Body="A puzzle comprising a set of numbered tiles on a grid, where tiles can be shifted into adjacent gaps." OwnerUserId="4219" LastEditorUserId="4219" LastEditDate="2014-11-12T13:34:33.327" LastActivityDate="2014-11-12T13:34:33.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4164" PostTypeId="5" CreationDate="2014-11-12T05:20:02.310" Score="0" Body="&lt;p&gt;A &lt;em&gt;magic square&lt;/em&gt; is an arrangement of distinct numbers (i.e. each number is used only once), usually integers, in a square grid, where the numbers in each row, and in each column, and the numbers in the main and secondary diagonals, all add up to the same number. An $n\times n$ magic square that contains the integers from 1 to $n^2$ is called a &lt;em&gt;normal&lt;/em&gt; magic square.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Example&lt;/strong&gt;: Up to rotations and reflections, there is only one $3\times 3$ magic square with entries 1,2,...,9 (which is also known under its Chinese name &quot;Lo Shu&quot;):  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   8  1  6   &#xA;   3  5  7  &#xA;   4  9  2  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorDisplayName="user20" LastEditDate="2015-02-20T20:13:57.283" LastActivityDate="2015-02-20T20:13:57.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4165" PostTypeId="4" CreationDate="2014-11-12T05:20:02.310" Score="0" Body="A puzzle related to magic squares: grids of integers where all rows, columns, and diagonals have the same sum." OwnerUserId="4219" LastEditorUserId="8874" LastEditDate="2015-01-30T17:13:14.640" LastActivityDate="2015-01-30T17:13:14.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4166" PostTypeId="5" CreationDate="2014-11-12T05:26:36.893" Score="0" Body="&lt;p&gt;A puzzle relating to physical equations or physical principles such as energy, entropy, and classical/quantum mechanics. General physics questions are off-topic here but can usually be asked at our sister site &lt;a href=&quot;https://physics.stackexchange.com&quot;&gt;Physics Stack Exchange&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11110" LastEditorUserId="11110" LastEditDate="2018-09-09T10:47:43.003" LastActivityDate="2018-09-09T10:47:43.003" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4167" PostTypeId="4" CreationDate="2014-11-12T05:26:36.893" Score="0" Body="A puzzle relating to physics equations or physical principles such as energy, entropy, and classical/quantum mechanics." OwnerUserId="4219" LastEditorUserId="69582" LastEditDate="2021-02-18T08:14:04.553" LastActivityDate="2021-02-18T08:14:04.553" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4168" PostTypeId="2" ParentId="4149" CreationDate="2014-11-12T05:28:26.133" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Kim Kardashian. You know, she's the most famous one of the whole clan, and the mentions of close sisters and brothers...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, at least there's my attempt :)&lt;/p&gt;&#xA;" OwnerUserId="4900" LastActivityDate="2014-11-12T05:28:26.133" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4169" PostTypeId="5" CreationDate="2014-11-12T05:39:10.920" Score="0" Body="&lt;p&gt;Use this tag when you're not asking about how to solve a puzzle, but asking about what the puzzle is called or how it works, given its description. Try to include as much detail as possible to aid identification, including pictures if applicable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that questions with this tag, unlike most on Puzzling SE, are intended to be asked by people who &lt;strong&gt;don't already know the answer&lt;/strong&gt;. If you're writing a challenge question in which you do know the name and type of the puzzle and are leaving cryptic clues about its nature for others to solve, then you should use the &lt;a href=&quot;/questions/tagged/reverse-puzzling&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;reverse-puzzling&amp;#39;&quot; rel=&quot;tag&quot;&gt;reverse-puzzling&lt;/a&gt; tag instead.&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="5373" LastEditDate="2016-09-10T15:02:17.583" LastActivityDate="2016-09-10T15:02:17.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4170" PostTypeId="4" CreationDate="2014-11-12T05:39:10.920" Score="0" Body="This tag is for questions about the name or nature of a specific puzzle, given as much descriptive information about it as the asker possesses. For challenge questions where the asker knows the puzzle type and gives cryptic clues about it, use the [reverse-puzzling] tag instead." OwnerUserId="4219" LastEditorUserId="5373" LastEditDate="2016-09-10T15:02:17.583" LastActivityDate="2016-09-10T15:02:17.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4171" PostTypeId="5" CreationDate="2014-11-12T05:40:41.643" Score="0" Body="&lt;p&gt;A question about assessing or adjusting the difficulty level of a puzzle.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-01T00:01:22.817" LastActivityDate="2015-02-01T00:01:22.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4172" PostTypeId="4" CreationDate="2014-11-12T05:40:41.643" Score="0" Body="A question about assessing or adjusting the difficulty level of a puzzle." OwnerUserId="4219" LastEditorUserId="4219" LastEditDate="2014-11-12T13:33:04.587" LastActivityDate="2014-11-12T13:33:04.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4173" PostTypeId="5" CreationDate="2014-11-12T05:44:04.283" Score="0" Body="&lt;p&gt;A &lt;strong&gt;crossword&lt;/strong&gt; is a word puzzle that normally takes the form of a square or a rectangular grid of white and black shaded squares. The goal is to fill the white squares with letters, forming words or phrases, by solving clues which lead to the answers. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Indirect clues: many crossword puzzles feature clues involving wordplay which are to be taken metaphorically or in some sense other than their literal meaning, requiring some form of lateral thinking. For instance, &lt;em&gt;&quot;half a dance&quot;&lt;/em&gt; could clue &lt;em&gt;CAN&lt;/em&gt; (half of CANCAN) or &lt;em&gt;CHA&lt;/em&gt; (half of CHACHA).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Cryptic crosswords (with their own tag &lt;a href=&quot;/questions/tagged/cryptic-crosswords&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;cryptic-crosswords&amp;#39;&quot; rel=&quot;tag&quot;&gt;cryptic-crosswords&lt;/a&gt;, and &lt;a href=&quot;/questions/tagged/cryptic-clues&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;cryptic-clues&amp;#39;&quot; rel=&quot;tag&quot;&gt;cryptic-clues&lt;/a&gt; for clues not assembled into crosswords) feature clues that are puzzles themselves (like anagrams, charades, double definitions, hidden words, homophones, reversals, spoonerisms).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Fill-in crosswords (also known as &lt;em&gt;cruzadex&lt;/em&gt;) feature a grid and the full list of words to be entered in that grid, but do not give explicit clues for where each word goes. The challenge is figuring out how to integrate the list of words together within the grid so that all intersections of words are valid. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="19114" LastEditDate="2017-11-10T17:30:52.730" LastActivityDate="2017-11-10T17:30:52.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4174" PostTypeId="4" CreationDate="2014-11-12T05:44:04.283" Score="0" Body="A question including or relating to crossword puzzles. Use [cryptic-crosswords] for British-style cryptic crosswords; [crossword-clues] or [cryptic-clues] for clues as distinct from whole crosswords." OwnerUserId="4219" LastEditorUserId="19114" LastEditDate="2017-11-10T17:30:52.730" LastActivityDate="2017-11-10T17:30:52.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4175" PostTypeId="2" ParentId="4126" CreationDate="2014-11-12T05:44:57.823" Score="0" Body="&lt;p&gt;Hex to ascii translation: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is a light near, but the light isn\'t the way to the puzzle. It\'s just a step closer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4338" LastActivityDate="2014-11-12T05:44:57.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4176" PostTypeId="5" CreationDate="2014-11-12T05:48:25.107" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-11-12T05:48:25.107" LastActivityDate="2014-11-12T05:48:25.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4177" PostTypeId="4" CreationDate="2014-11-12T05:48:25.107" Score="0" Body="A question relating to interpretation of clues in a crossword puzzle, or to generation of clues in a crossword puzzle. Use [cryptic-clues] for clues obeying the conventions of cryptic crosswords, [crosswords] or [cryptic-crosswords] for whole crosswords rather than individual clues." OwnerUserId="4219" LastEditorUserId="19114" LastEditDate="2017-11-10T17:29:23.633" LastActivityDate="2017-11-10T17:29:23.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4178" PostTypeId="5" CreationDate="2014-11-12T05:56:44.707" Score="0" Body="&lt;p&gt;A puzzle based on &lt;a href=&quot;https://en.wikipedia.org/wiki/Combinatorics&quot; rel=&quot;nofollow noreferrer&quot;&gt;combinatorics&lt;/a&gt;, which is the study of counting discrete structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, how many ways to arrange &lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$abcd$&lt;/span&gt;, where order is important?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have &lt;span class=&quot;math-container&quot;&gt;$abc,abd,acd,bcd$&lt;/span&gt; and six of each (for example &lt;span class=&quot;math-container&quot;&gt;$abc,acb,bac,bca,cab,cba$&lt;/span&gt;), and so the answer is &lt;span class=&quot;math-container&quot;&gt;$24$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For basic combinatoric examples and explanations, see:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Combination&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://mathigon.org/world/Combinatorics&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mathigon&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://brilliant.org/wiki/combinatorics/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brilliant&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.mathsisfun.com/combinatorics/combinations-permutations.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Maths is Fun&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Puzzles where permutations, sets, graphs, lattices and other combinatorial mathematics feature prominently should include this tag.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="11497" LastEditDate="2020-04-05T11:23:50.543" LastActivityDate="2020-04-05T11:23:50.543" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4179" PostTypeId="4" CreationDate="2014-11-12T05:56:44.707" Score="0" Body="A puzzle based on combinatorics, which is the study of counting discrete structures. Use with [mathematics]" OwnerUserId="4219" LastEditorUserId="69582" LastEditDate="2021-02-03T16:50:50.930" LastActivityDate="2021-02-03T16:50:50.930" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4180" PostTypeId="2" ParentId="4139" CreationDate="2014-11-12T06:25:50.823" Score="6" Body="&lt;p&gt;You are talking about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; O&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This thing can be so tiny,&#xA;That's when it has the most variety.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When it is used as a symbol (in written communication), an ellipse can have many meanings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Standing up straight, it is surrounded,&#xA;Otherwise it can make things bounded.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure about the first part.  Perhaps for an egg to stand up straight, it needs external support. An ellipse can be used to encircle things. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Within a box it lets you choose,&#xA;But if it points up you will lose.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In an exam, you can encircle a choice or a word (horizontal ellipse).  If your score is zero (vertical ellipse), you &quot;lose.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It can be seen leaning,&#xA;That's when it reveals true meaning.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When writing measurements as numbers, an italicized zero (at the end) can indicate that it is a significant value.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It can be a celebrity and nothing at once,&#xA;It can show uncertainty but never incoherence.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; O is a nickname of Oprah Winfrey, a famous celebrity.  Zero represents nothing.  A zero sometimes is the middle choice in a Likert scale, which indicates no preference.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But when wealth is what it shows,&#xA;The end, you will know.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When your wealth is zero, you are bankrupt.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5095" LastEditorUserId="5095" LastEditDate="2014-11-12T06:31:01.867" LastActivityDate="2014-11-12T06:31:01.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4181" PostTypeId="2" ParentId="4119" CreationDate="2014-11-12T06:31:16.380" Score="3" Body="&lt;pre&gt;&lt;code&gt;Before you figure out this password&#xA;There is a riddle you must master&#xA;You must first get the tower that looms&#xA;Then you must find the hollow moon&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The number 10.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;D.R.K Dahanu&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; D. R. Kaprekar, guy that stays in Dahanu. 4 digits for Kaprekar number.&lt;br&gt;6174&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is what I have found so far. Will edit post when I find more. Hope my answer can help other people :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After analysing the coded messages, I realise that the message are 1 letter up from the coded message (compare it to a qwerty keyboard). The message is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; go to lo_do_ 118 apple which corresponds to london 118 apple in my opinion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I looked up on:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 118 Apple&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And found out that there is a place in America which corresponds to it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; New York City is called the BIG apple, therefore matches hint, APPLE are in caps.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After googling for a bit, I found out that there is a place that matches 118 in its logo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, I believe that this is the next place we have to go to.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Opus 118 Harlem School of Music&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Changed answer for 2nd riddle.&#xA;Couldn't figure out 3rd riddle still except for some hints I gathered on myself:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; since there is a []&lt;sup&gt;2&lt;/sup&gt;, it's probably a 4-digit number with each number squared? I would lean more towards number 2, since it's a quote from Alice in wonderland. There are 2 movies, 1 in 1951 and 1 in recent years, 2010. I would call for 2010 to be the 3rd riddle's answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4338" LastEditorUserId="1955" LastEditDate="2014-11-12T23:35:31.230" LastActivityDate="2014-11-12T23:35:31.230" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4182" PostTypeId="1" AcceptedAnswerId="4184" CreationDate="2014-11-12T07:24:37.360" Score="7" ViewCount="4683" Body="&lt;p&gt;In one form I can stretch to infinity, in another I have a clearly defined limit.&lt;br/&gt;&#xA;In one form I am very flexible, in another I am exceedingly rigid.&lt;br/&gt;&#xA;In one form intersections are catastrophic, in another intersections are common and can be helpful.&lt;br/&gt;&#xA;In one form I am a physical object, in another I am a conceptual object&lt;br/&gt;&#xA;In both forms I am used daily by millions of people all round the world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What Am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't forget the spoiler tags.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you get the correct answer, you must then think of a thing whose first letter begins with the end letter of the answer to this problem. So if the answer was apple, you need to think of an object that starts with 'e' and post a riddle about that. We can keep going until people get bored. However it does rely on getting the answer to this right first. If you really can't be bothered to think up a new riddle, say so in your answer. First person who comments something like 'I'll do it' can have the honor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;No hate, this is my first entirely made up word based riddle. Have Fun! &lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="4951" LastEditorUserId="4951" LastEditDate="2014-11-12T07:49:25.410" LastActivityDate="2018-02-23T14:48:35.403" Title="Guess the object - What Am I Riddle (Part 1)" Tags="|word-problem|riddle|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4183" PostTypeId="2" ParentId="4073" CreationDate="2014-11-12T07:42:51.673" Score="2" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And incidentally, my friend's number was&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You see, the key is that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You need to add a digit to make the entire number divisible by 7.&lt;br&gt;&#xA; 105 = 15 x 7&lt;br&gt;&#xA; 63 = 9 x 7&lt;br&gt;&#xA; 21 = 3 x 7 (28 would've worked as well)&lt;br&gt;&#xA; 140 = 20 x 7 (and 147 would've worked too)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="68" LastActivityDate="2014-11-12T07:42:51.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4184" PostTypeId="2" ParentId="4182" CreationDate="2014-11-12T07:47:02.100" Score="13" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Network. The virtual/social vs road networks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In one form I can stretch to infinity, in another I have a clearly defined limit.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The virtual/social can cross boundaries reaching the endless depths of connections between people. Roads are physically bounded.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In one form I am very flexible, in another I am exceedingly rigid.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The digital can be instantly modified and twisted (eg. changing physical layout of a network but regaining functionality, or people changing locations not interfering with their friendship). Roads are very difficult to change.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In one form intersections are catastrophic, in another intersections are common and can be helpful.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Car accidents vs. meeting people / transferring data&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In one form I am a physical object, in another I am a conceptual object&#xA;In both forms I am used daily by millions of people all round the world.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; self explanatory&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4317" LastEditorUserId="4317" LastEditDate="2015-01-27T07:39:05.167" LastActivityDate="2015-01-27T07:39:05.167" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="4185" PostTypeId="2" ParentId="4182" CreationDate="2014-11-12T08:03:19.250" Score="10" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The answer I had in mind was a &lt;strong&gt;plane&lt;/strong&gt; which is both a physical object (airplane) and a mathematical concept.&lt;br/&gt;&#xA; - Geometric plane can stretch infinitely along a 2D surface, Aeroplane has a defined shape &lt;br/&gt;&#xA; - Aeroplane is quite flexible (at least the wings certainly are), Geometric plane, is not flexible at all, because it lays along a 2D    surface, as soon as it bends, it is no longer a plane. &lt;br/&gt;&#xA; - When two aeroplanes intersect...... pretty obvious, Geometric plane intersection, can give us solutions to mathematical problems.&lt;br/&gt;&#xA; - Aeroplane - Physical, Geometric - Conceptual &lt;br/&gt;&#xA; - Both are used around the world.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4951" LastEditorUserId="4951" LastEditDate="2015-01-27T04:47:20.433" LastActivityDate="2015-01-27T04:47:20.433" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4186" PostTypeId="1" AcceptedAnswerId="4190" CreationDate="2014-11-12T08:47:52.990" Score="8" ViewCount="2142" Body="&lt;p&gt;Having gotten accepted on &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4182/guess-the-object-what-am-i-riddle-part-1&quot;&gt;part 1&lt;/a&gt;, here I go:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who am I? You'd like to know, so please just ask who's there,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might have a laugh on my account, I don't really care,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not it's still OK, that's something I can bear,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I come right out no need to guess or speculate your share.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;; Two more lines which might be sort of a giveaway, I'll leave them as a hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your cooperation is required, don't leave me out your lair,&#xA;&lt;br&gt;&#xA; I am right here behind the door, respond my tapping fair.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4317" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2014-11-12T16:55:08.387" Title="Guess the object - What Am I Riddle (Part 2)" Tags="|riddle|rhyme|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4187" PostTypeId="5" CreationDate="2014-11-12T08:48:46.497" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-11-12T08:48:46.497" LastActivityDate="2014-11-12T08:48:46.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4188" PostTypeId="4" CreationDate="2014-11-12T08:48:46.497" Score="0" Body="A problem, or riddle written in rhyme. For an artistic twist." OwnerUserId="4317" LastEditorUserId="4317" LastEditDate="2014-11-12T13:35:11.950" LastActivityDate="2014-11-12T13:35:11.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4189" PostTypeId="2" ParentId="4062" CreationDate="2014-11-12T08:50:54.263" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ans is &lt;strong&gt;4&lt;/strong&gt;. Common letters between digit in words and word &quot;Halloween&quot;. 13 = Thirteen, so here 'h','e','e' &amp;amp; 'n' four letters common between 13 &amp;amp; Halloween.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5101" LastActivityDate="2014-11-12T08:50:54.263" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4190" PostTypeId="2" ParentId="4186" CreationDate="2014-11-12T08:53:31.993" Score="7" Body="&lt;p&gt;Are you a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Knock-knock joke&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good to see you were up to the challenge!&lt;/p&gt;&#xA;" OwnerUserId="2484" LastActivityDate="2014-11-12T08:53:31.993" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4191" PostTypeId="2" ParentId="4149" CreationDate="2014-11-12T08:58:38.013" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are the &lt;strong&gt;space&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your family is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; punctuation marks&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As such (what I could make of it):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eldest brother - Period&#xA;&lt;br&gt;&#xA; Sister 1 - Exclamation mark&#xA;&lt;br&gt;&#xA; Sister 2 - Paragraph&#xA;&lt;br&gt;&#xA; Cousin - Comma&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4317" LastActivityDate="2014-11-12T08:58:38.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4192" PostTypeId="2" ParentId="4149" CreationDate="2014-11-12T09:08:35.617" Score="2" Body="&lt;p&gt;Not an answer by itself, I would like to add to &lt;a href=&quot;https://puzzling.stackexchange.com/users/4450/pacoverflow&quot;&gt;pacoverflow&lt;/a&gt; 's answer, which I believe to be correct. (But I can't comment)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your &quot;sister's love is grand largesse&quot; could also mean either the Caps Lock or the Shift, since they &quot;like&quot; making things big &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5102" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2014-11-12T09:08:35.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4193" PostTypeId="1" AcceptedAnswerId="4205" CreationDate="2014-11-12T09:11:18.177" Score="20" ViewCount="3598" Body="&lt;p&gt;I'm lying down, with a dah-ditty-dah, in the middle of the end,&lt;br&gt;&#xA;I'd hang upside-down similarly with my friend,&lt;br&gt;&#xA;Were I to flip over, with a dah-ditty-dah, it'd be the middle of the day,&lt;br&gt;&#xA;Right way up I cannot plainly say,&lt;br&gt;&#xA;Here's the key, let's end the fun,&lt;br&gt;&#xA;I'll tell you a secret, I hide behind one&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/4182/guess-the-object-what-am-i-riddle-part-1&quot;&gt;Part 1&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/4186/guess-the-object-what-am-i-riddle-part-2/4190#4190&quot;&gt;Part 2&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2018-06-19T14:03:22.703" Title="Guess the object - What Am I Riddle (Part 3)" Tags="|word-problem|riddle|rhyme|" AnswerCount="5" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="4194" PostTypeId="2" ParentId="4193" CreationDate="2014-11-12T09:14:36.480" Score="4" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter &quot;e&quot;.&#xA;&quot;e&quot; is in the middle of &quot;th&lt;strong&gt;e e&lt;/strong&gt;nd&quot;.&#xA;Turn over e, and you get a, which is in the middle of the &quot;d&lt;strong&gt;a&lt;/strong&gt;y&quot;.&#xA;&quot;e&quot; hides behind on&lt;strong&gt;e&lt;/strong&gt;, because it is at the end.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="97" LastEditorUserId="45954" LastEditDate="2018-06-19T14:03:22.703" LastActivityDate="2018-06-19T14:03:22.703" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="4197" PostTypeId="2" ParentId="4145" CreationDate="2014-11-12T09:35:27.353" Score="1" Body="&lt;p&gt;The answer for 7 is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Following the formula&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Password = mod(n!, 10) + floor(n / 2)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4637" LastActivityDate="2014-11-12T09:35:27.353" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4198" PostTypeId="2" ParentId="4193" CreationDate="2014-11-12T09:39:48.927" Score="-1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter 'n' it's the middle of end, and turn it upside down it's a 'u'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4951" LastActivityDate="2014-11-12T09:39:48.927" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4199" PostTypeId="2" ParentId="4193" CreationDate="2014-11-12T09:41:05.187" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;N&lt;/strong&gt;, spelled out En&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;maybe?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; middle of end&#xA;&lt;br&gt;&#xA; upside-down - u&#xA;&lt;br&gt;&#xA; middle of day - noon&#xA;&lt;br&gt;&#xA; o&lt;strong&gt;n&lt;/strong&gt;e&#xA;&lt;br&gt;&#xA; end fu&lt;strong&gt;n&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit&lt;/strong&gt; - &quot;lying down&quot; would be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; dow&lt;strong&gt;n&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4317" LastEditorUserId="4317" LastEditDate="2014-11-12T09:46:51.863" LastActivityDate="2014-11-12T09:46:51.863" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4200" PostTypeId="2" ParentId="4026" CreationDate="2014-11-12T09:43:26.473" Score="3" Body="&lt;p&gt;If my uncle wasn't a scoundrel it could also have been&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; sadness&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or something similar.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Box content in this case could be something that made him sad, e.g. pictures of a lost girlfriend. He still loves her but she either died or has left him because he is a scoundrel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Those who take me, won't know me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;take sadness&quot; means to console or comfort the sad person, which supposedly makes the sadness go away (won't know me)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Those who know me, won't want me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Who wants to be sad, right?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Those who want me, will make me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you want to make somebody sad, there usually is a way to cause sadness.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Those who make me, will give me away.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Same, you don't make yourself sad (on purpose) but if you make sadness, you do it to others.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4971" LastActivityDate="2014-11-12T09:43:26.473" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4201" PostTypeId="2" ParentId="4193" CreationDate="2014-11-12T09:47:56.623" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; is it (.) dot? its in the middle of the end of an exclamation point similarly with its friend which is (i) in upside down, you said behind 'one.' theres a dot at the end of one right? dunno with the other phrases, when a dot is assumed as the globe, when it turned upside down, it will be the middle of the day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4944" LastActivityDate="2014-11-12T09:47:56.623" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4202" PostTypeId="2" ParentId="4149" CreationDate="2014-11-12T09:49:24.243" Score="7" Body="&lt;p&gt;Your name is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the &lt;strong&gt;sun&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your sister could be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.space.com/25881-sun-sister-star-found-hd162826.html&quot;&gt;Sister star HD 162826&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The distant cousin's would refer to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; pretty much all other stars&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not so sure about the brother, however&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the meager meal could imply it's a collapsed sun (resulting in a black hole, &quot;eating&quot; whatever comes close enough)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I keep the peace with great panache. &#xA;Strike me down and neighbours clash,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;could refer to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Remove the sun and planets will fall out of orbit, possible crashing into each other&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2463" LastActivityDate="2014-11-12T09:49:24.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4203" PostTypeId="2" ParentId="4075" CreationDate="2014-11-12T09:52:05.073" Score="19" Body="&lt;p&gt;Yeah, it's been answered, but what the hell.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An important part of any murder mystery is the murder weapon. Finding that often gives you a solid lead. Based on the copious blood, and the fact that the icons remain intact, we can hypothesise that they were stabbed, so we're looking for something sharp. TeX-LaTeX has a pair of those nasty braces, SuperUser has a nose that tapers to a point and you could definitely do somebody a mischief if you flew Arqade into them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But wait!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Travel's icon is missing. And with three points, it's a murder weapon just waiting to happen. But it'd be difficult to use - it's evidentially not made to be held comfortably. Which leads us to &lt;strong&gt;Role Playing Games&lt;/strong&gt;, who I posit took the feat &lt;em&gt;Exotic Weapon Proficiency (Travel Icon)&lt;/em&gt; and took down some of his friends in a Pathfinder session gone horribly wrong. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bonus pun:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'Die' indeed. Looks like their saving rolls came up snake eyes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4164" LastActivityDate="2014-11-12T09:52:05.073" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4205" PostTypeId="2" ParentId="4193" CreationDate="2014-11-12T09:59:55.790" Score="16" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ! (&lt;strong&gt;e&lt;/strong&gt;xclamation mark)&lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lying down in the &quot;middle of the e&lt;strong&gt;n&lt;/strong&gt;d&quot; (n = -· in morse code)&lt;br/&gt;&#xA; Upside-down in fr&lt;strong&gt;i&lt;/strong&gt;end&lt;br/&gt;&#xA; Flipped over in the &quot;middle of the d&lt;strong&gt;a&lt;/strong&gt;y&quot; (a = ·- in morse code)&lt;br/&gt;&#xA; Right way up I cannot plainly say (one does not speak &quot;plainly&quot; when using '!')&lt;br/&gt;&#xA; Here's the key... Hide behind one - (think &quot;keyboard&quot;)&lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2484" LastEditDate="2014-11-12T10:26:43.750" LastActivityDate="2014-11-12T10:26:43.750" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="4207" PostTypeId="1" AcceptedAnswerId="4219" CreationDate="2014-11-12T10:49:39.793" Score="8" ViewCount="1101" Body="&lt;p&gt;I'm off hiking and I need to buy a bus ticket to the mountains. The ticket costs £4.25 and I have the following coins in my pocket:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2 x £2, 4 x £1, 50p, 3 x 20p, 2 x 5p, 2p, 7 x 1p&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Weight is everything when you're climbing big mountains so I want to make sure I'm carrying as little as possible when I set off. Luckily I had the foresight to check the Royal Mint's website and look up exactly how much each coin weighs.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;£2 - 12.0 grams&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;£1 - 9.5 grams&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;50p - 8 grams&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;20p - 5 grams&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;10p - 6.5 grams&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;5p - 3.25 grams&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;2p - 7.12 grams&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;1p - 3.56 grams&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's around this time that I consider that moving to the US and using notes for much lower denominations would really benefit me! Perhaps for my next trip!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Buses in the UK are often expect exact change, what coins should I give the bus driver so that I'm left with the smallest possible weight to carry on my hike and how much will that be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: For whatever little its worth 100p = £1&lt;/p&gt;&#xA;" OwnerUserId="2466" LastEditorUserId="2313" LastEditDate="2014-11-12T10:53:11.190" LastActivityDate="2022-12-29T12:05:54.097" Title="Travelling Light" Tags="|mathematics|" AnswerCount="5" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4208" PostTypeId="5" CreationDate="2014-11-12T10:53:12.927" Score="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Brainfuck&quot; rel=&quot;nofollow&quot;&gt;Brainfuck&lt;/a&gt; is an esoteric programming language.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brainfuck uses a simple machine model consisting of the program and instruction pointer, and an array of at least 30,000 byte cells initialized to zero. There is a movable data pointer (initialized to point to the leftmost byte of the array), and two streams of bytes for input and output (most often connected to a keyboard and a monitor respectively, and using the ASCII character encoding).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that general programming questions, even in Brainfuck, are off-topic here, but can be posted on &lt;a href=&quot;http://stackoverflow.com/tags/brainfuck&quot;&gt;Stack Overflow&lt;/a&gt;. Programming puzzles are best posted on &lt;a href=&quot;http://codegolf.stackexchange.com/tags/brainfuck&quot;&gt;Programming Puzzles &amp;amp; Code Golf&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="71" LastEditorUserId="71" LastEditDate="2014-11-12T12:27:06.097" LastActivityDate="2014-11-12T12:27:06.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4209" PostTypeId="4" CreationDate="2014-11-12T10:53:12.927" Score="0" Body="Brainfuck is an esoteric programming language." OwnerUserId="71" LastEditorUserId="71" LastEditDate="2014-11-12T13:35:08.177" LastActivityDate="2014-11-12T13:35:08.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4210" PostTypeId="2" ParentId="4207" CreationDate="2014-11-12T11:04:37.410" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;3 x £1&#xA;&lt;br/&gt;&#xA;1 x 50p&#xA;&lt;br/&gt;&#xA;3 x 20p&#xA;&lt;br/&gt;&#xA;2 x 5p&#xA;&lt;br/&gt;&#xA;5 x 1p&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That would leave you with 47.74 grams from the original 123.54 grams&lt;/p&gt;&#xA;" OwnerUserId="2313" LastActivityDate="2014-11-12T11:04:37.410" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4211" PostTypeId="1" AcceptedAnswerId="4586" CreationDate="2014-11-12T11:04:54.253" Score="6" ViewCount="6175" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am something that everyone needs.&lt;br&gt;&#xA;  You use me many times in different ways.&lt;br&gt;&#xA;  Sometimes you only use part of me,&lt;br&gt;&#xA;  Others you only use most of me.   &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;You always turn and bend me to your will.&lt;br&gt;&#xA;  You use to me to scare, to confuse, even to question.&lt;br&gt;&#xA;  You can even use to confront someone like when you say &quot;Bill, [rest of confrontation]&quot;&lt;br&gt;&#xA;  You absolutely use me every day.    &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;What am I?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You guys don't mind I made a part 4 do you?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original question by &lt;a href=&quot;https://puzzling.stackexchange.com/users/4951/james-massey&quot;&gt;https://puzzling.stackexchange.com/users/4951/james-massey&lt;/a&gt; answered by &lt;a href=&quot;https://puzzling.stackexchange.com/users/4317/jnf&quot;&gt;https://puzzling.stackexchange.com/users/4317/jnf&lt;/a&gt; can be found here: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4182/guess-the-object-what-am-i-riddle-part-1&quot;&gt;Guess the object - What Am I Riddle (Part 1)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latest (other than mine) by &lt;a href=&quot;https://puzzling.stackexchange.com/users/2484/dalarcop&quot;&gt;https://puzzling.stackexchange.com/users/2484/dalarcop&lt;/a&gt; answered by &lt;a href=&quot;https://puzzling.stackexchange.com/users/2153/kami&quot;&gt;https://puzzling.stackexchange.com/users/2153/kami&lt;/a&gt; can be found at &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4193/guess-the-object-what-am-i-riddle-part-3&quot;&gt;Guess the object - What Am I Riddle (Part 3)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The others can be found in this list:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 2 by &lt;a href=&quot;https://puzzling.stackexchange.com/users/4317/jnf&quot;&gt;https://puzzling.stackexchange.com/users/4317/jnf&lt;/a&gt;: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4186/guess-the-object-what-am-i-riddle-part-2&quot;&gt;Guess the object - What Am I Riddle (Part 2)&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2014-11-24T07:25:08.930" Title="Guess The Object - What Am I Riddle (Part 4!)" Tags="|riddle|" AnswerCount="10" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="4215" PostTypeId="2" ParentId="4211" CreationDate="2014-11-12T11:34:17.397" Score="2" Body="&lt;p&gt;Are you:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Language&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am something that everyone needs.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Thinking and living in society requires some kind of language.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You use me many times in different ways.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We use language at almost all times and we have apply it to everything.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes you only use part of me,&#xA;Others you only use most of me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We vary quite dramatically with what subsets of which languages we use.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You always turn and bend me to your will.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Language is at our mercy regarding usage. Anything goes - better if understood by others normally.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You use to me to scare, to confuse, even to question.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You can express emotions and intent - and prvoke emotions - using language.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can even use to confront someone like when you say &quot;Bill, [rest of confrontation]&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Same as last.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You absolutely use me every day.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Indeed, language is a mainstay of the human mind.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-11-19T07:47:23.790" LastActivityDate="2014-11-19T07:47:23.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4216" PostTypeId="2" ParentId="4211" CreationDate="2014-11-12T11:35:48.197" Score="4" Body="&lt;p&gt;Is it..&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Words ? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am something that everyone needs.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; technically doesn't fit since you can be alive without needing Words but i'd say its fairly nessesary in everyday Life. even mute people can use Words with sign language, Stephen Hawking has a computer etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You use me many times in different ways.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Fairly self explanatory since you can use a lot of Words in different ways (to explain, express feeling etc).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes you only use part of me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You can abbreviate Words for example, lol, nvm, which are parts of a word or multiple Words.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Others you only use most of me. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This one is also very self explanatory in relation to my answer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You always turn and bend me to your will.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You can bend different Words as you probably already know.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You use to me to scare, to confuse, even to question.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Also very self explanatory. scary stories, police questioning etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can even use to confront someone like when you say &quot;Bill, [rest of confrontation]&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Surely if you want to confront someones its better to use Words than, for example, violence&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You absolutely use me every day. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This one is probably the least fitting but also fairly self explanatory. in relation to my own Life i seem to use Words every day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It isn't the right answer but i felt like it could fit all of these&lt;/p&gt;&#xA;" OwnerUserId="4962" LastEditorUserId="4962" LastEditDate="2014-11-19T07:25:32.387" LastActivityDate="2014-11-19T07:25:32.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4219" PostTypeId="2" ParentId="4207" CreationDate="2014-11-12T12:05:35.130" Score="7" Body="&lt;p&gt;We can first ammend the list with grams per penny, because we want to spend the most grams per penny possible, so money with a higher g/p should be spent first:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2 £2  - 0.06  g/p&#xA;4 £1  - 0.095 g/p&#xA;1 50p - 0.16  g/p&#xA;3 20p - 0.25  g/p&#xA;2  5p - 0.65  g/p&#xA;1  2p - 3.56  g/p&#xA;7  1p - 3.56  g/p&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As expected there are no exceptions to smaller coins = more grams, so we simply try to get as much small coins as possible to combine to £4.25&lt;/p&gt;&#xA;&#xA;&lt;p&gt;all coins below £1 together are 129p togehter with 3x£1 we are at 4.29 and have to shave off 4p, so we have to leave 4 single pennies or the 2p and 2x 1p. Since we don't have any smaller change left, it won't bring any benefit to switch things araound, since the overall gram per penny ratio will only drop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So overall:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 5x1p 2x5p 3x20p 1x50p 3x£1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This would be more interesting, if there were some valuable coin with more weight...&lt;/p&gt;&#xA;" OwnerUserId="1998" LastActivityDate="2014-11-12T12:05:35.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4220" PostTypeId="2" ParentId="4207" CreationDate="2014-11-12T12:15:17.670" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You've got £9.29 in your pocket, so give him all but 4p (either in pennies or with the 2p in there) and ask for a fiver in the change. According to &lt;a href=&quot;http://www.answers.com/Q/How_much_does_a_British_sterling_five_pound_note_weigh&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt;, a five pound note weighs 0.812 grams. So that plus the remaining pennies (14.24 grams) would be 15.052g to lug up and down the mountains.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5118" LastEditorUserId="5118" LastEditDate="2014-11-12T12:32:50.080" LastActivityDate="2014-11-12T12:32:50.080" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4223" PostTypeId="2" ParentId="4119" CreationDate="2014-11-12T12:30:04.480" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First numerical code is 10 (1 = tower, 0 - hollow moon, with credit to FortMauris for coming up with it first)&lt;br&gt;&lt;br&gt;&#xA; Second numerical clue is the Kaprekar constant: 6174&lt;br&gt;&lt;br&gt;&#xA; Lewis Carroll made frequent use of the number 42 in the novel Alice in Wonderland.  42^2 = 1764 for third number&lt;br&gt;&lt;br&gt;&#xA; There's a hotel called &quot;The London&quot; in downtown NYC at 151 West 54th Street.  Presumably, it has a room number 118, so let's go there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5121" LastEditorUserId="5121" LastEditDate="2014-11-12T15:39:50.313" LastActivityDate="2014-11-12T15:39:50.313" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4225" PostTypeId="2" ParentId="498" CreationDate="2014-11-12T12:31:05.080" Score="0" Body="&lt;p&gt;Maybe a different take, since I'm not a native enlgish speaker, but here's what I make from it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;'mean to confine a flower' - a piece of string, or 'stringy'. (when you buy a bouquet, there's often a small piece of string tied around the stems to keep it together).&#xA;'fibrous' - consisting of fibres, which look like a set of small strings?&lt;/p&gt;&#xA;" OwnerUserId="2156" LastActivityDate="2014-11-12T12:31:05.080" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4227" PostTypeId="2" ParentId="4207" CreationDate="2014-11-12T12:37:21.643" Score="1" Body="&lt;p&gt;Let's look at these coins and calculate their &amp;quot;density&amp;quot;. In general, you want to use the coins with the highest density.&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: right;&quot;&gt;Worth (£)&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: right;&quot;&gt;Weight (g)&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: right;&quot;&gt;Density (g/£)&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;2.00&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;12.00&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;6.00&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;1.00&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;9.50&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;9.50&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;0.50&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;8.00&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;16.00&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;0.20&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;5.00&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;25.00&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;0.10&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;6.50&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;65.00&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;0.05&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;3.25&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;65.00&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;0.02&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;7.12&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;356.00&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;0.01&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;3.56&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;356.00&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;We see that it doesn't matter if you use 1 x 2p, or 2 x 1p, and neither does it matter whether you use 1 x 10p, or 2 x 5p.&lt;/p&gt;&#xA;&lt;p&gt;Since the total value of all coins under 5p comes to 9p, we're stuck with 4p no matter what unless we tip the driver.&lt;br /&gt;&#xA;The £2 coins are least dense, followed by the £1 coins, so we try to keep as many of those as possible.&lt;br /&gt;&#xA;So first we spend our 50p, 2 x 20p, 1 x 5p, 1 x 2p, and 3 x 1p for our first pound. After that, we have to spend 3 x £1 to make up the rest, since the remaining smaller coins do not reach up to another pound. We're still 25p short, we pay this by using 1 x 20p and 1 x 5p.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; So we've spent 3 x £1, 1 x 50p, 3 x 20p, 2 x 5p, 1 x 2p, and 3 x 1p, for a total weight loss of 75.80 grams, which leaves us with 2 x £2, 1 x £1, and 4 x 1p for a total of £5.04 at a weight of 47.74 grams.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditDate="2022-12-29T12:05:54.097" LastActivityDate="2022-12-29T12:05:54.097" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4228" PostTypeId="2" ParentId="4211" CreationDate="2014-11-12T12:39:39.703" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Punctuation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &quot;bend and twist&quot; so close to &quot;question&quot; was the key for me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am something that everyone needs.&#xA; You use me many times in different ways.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Punctuation is used in a variety of ways by just about everyone&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes you only use part of me,&#xA; Others you only use most of me. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some punctuation marks are part of others - for example comma is part of semicolon, or dot is part of question mark or exclamation mark&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You always turn and bend me to your will.&#xA; You use to me to scare, to confuse, even to question.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A question mark is bent. Exclamation marks are used in scary sentences like Eek! A ghost! and of course question marks are used to question&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can even use to confront someone like when you say &quot;Bill, [rest of confrontation]&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sample sentence contains a comma&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You absolutely use me every day. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As above, we use punctuation every day&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note: even though in a comment the OP has indicated a more specific answer is correct, I'm adding the reasoning to this one as requested.&lt;/p&gt;&#xA;" OwnerUserId="1509" LastEditorUserId="1509" LastEditDate="2014-11-19T02:54:54.527" LastActivityDate="2014-11-19T02:54:54.527" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4229" PostTypeId="5" CreationDate="2014-11-12T12:45:08.987" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-11-12T12:45:08.987" LastActivityDate="2014-11-12T12:45:08.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4230" PostTypeId="4" CreationDate="2014-11-12T12:45:08.987" Score="0" Body="A question about effective methodologies for solving puzzles on paper. Questions that present pencil-and-paper problems to be solved should use the &quot;pencil-and-paper&quot; tag instead." OwnerUserId="4219" LastEditorUserId="4219" LastEditDate="2014-11-12T15:44:47.263" LastActivityDate="2014-11-12T15:44:47.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4231" PostTypeId="2" ParentId="4211" CreationDate="2014-11-12T12:45:12.013" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are &quot;YOU&quot; or &quot;U&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It fits almost all the clues.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I am something that everyone needs. (During Conversations)&lt;br /&gt;&#xA;You use me many times in different ways. (You, U, or even with different meanings, like 'You?'- as a question, Shouting 'YOU' angrily etc)&lt;br /&gt;&#xA;Sometimes you only use part of me, (Short version of You - 'U')&lt;br /&gt;&#xA;Others you only use most of me. (Full 'You')&lt;br /&gt;&#xA;You always turn and bend me to your will. (Rotating 'U' 90 degree clockwise will give C, and rotating 180 degrees give 'n' and if we bend 'U' more it will become 'O')&lt;br /&gt;&#xA;You use to me to scare, to confuse, even to question. ('You?'-question, 'You!'-confuse,'YOU!!$$'-scare etc)&lt;br /&gt;&#xA;You can even use to confront someone like when you say &quot;Bill, [rest of confrontation]&quot; (Bill, its You!!)&lt;br /&gt;&#xA;You absolutely use me every day. (No doubt that we use You many times everyday)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastActivityDate="2014-11-12T12:45:12.013" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4232" PostTypeId="2" ParentId="4119" CreationDate="2014-11-12T12:46:12.600" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;APPLE is a big apple, so the address is in NYC. There is a London Road on Staten Island, so we're looking for 118 London Rd, Staten Island.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5123" LastActivityDate="2014-11-12T12:46:12.600" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4233" PostTypeId="2" ParentId="4211" CreationDate="2014-11-12T12:46:58.087" Score="3" Body="&lt;p&gt;I think it is the&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Brain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes you only use part of me,&#xA;Others you only use most of me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Women and Men are known to use differently the brain. If I remember correctly women use normally only an emisphere&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You always turn and bend me to your will.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Will comes from brain :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Bill, [rest of confrontation]&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;use your brain!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5122" LastActivityDate="2014-11-12T12:46:58.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4235" PostTypeId="2" ParentId="3100" CreationDate="2014-11-12T12:54:32.073" Score="4" Body="&lt;p&gt;A bit more realistic approach: Bury yourself squatting with your knees up to your shoulders, if your head is above the sand, there are only a few kg of sand above your shoulders and you can easily stand up. This would need the least loopholes and will look exactly the same form the outside as someone buried standing - and you wouldn't have to crank your neck like with lying down.&lt;/p&gt;&#xA;" OwnerUserId="1998" LastActivityDate="2014-11-12T12:54:32.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4236" PostTypeId="2" ParentId="4026" CreationDate="2014-11-12T13:02:56.097" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A virus / Illness&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Those who take me, won't know me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Viruses can have an incubation period&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Those who know me, won't want me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Who DOES want an illness?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Those who want me, will make me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Evil doers would create viruses to infect people&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Those who make me, will give me away.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Evil doers would spread the virus but not themselves&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Besides... The box was &quot;empty&quot;, no? ;)&lt;/p&gt;&#xA;" OwnerUserId="5122" LastActivityDate="2014-11-12T13:02:56.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4239" PostTypeId="2" ParentId="4149" CreationDate="2014-11-12T13:07:53.747" Score="2" Body="&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your name is backspace&lt;br&gt;&#xA;&lt;br&gt;&#xA; One sister's love is grand largesse, &amp;lt;- ENTER&lt;br&gt;&#xA;&lt;br&gt;&#xA; another's like me, more or less. &amp;lt;- DELETE&lt;br&gt;&#xA;&lt;br&gt;&#xA; My brother shuns such grand appeal, &amp;lt;- INSERT&lt;br&gt;&#xA;&lt;br&gt;&#xA; he much prefers his meager meal. &amp;lt;- as insertmode overrides&lt;br&gt;&#xA;&lt;br&gt;&#xA; Our distant cousin's always in, &amp;lt;- numlock/capslock&lt;br&gt;&#xA;&lt;br&gt;&#xA; And where one village holds another, &amp;lt;- between ctrl/alt - altgr/ctrl&lt;br&gt;&#xA;&lt;br&gt;&#xA; you'll often find our eldest brother. &amp;lt;- SPACEBAR&lt;br&gt;&#xA;&lt;br&gt;&#xA; Strike me down and neighbours clash, &amp;lt;- two words clash as you backspace&lt;br&gt;&#xA;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5124" LastEditorUserId="-1" LastEditDate="2014-11-12T16:40:20.983" LastActivityDate="2014-11-12T16:40:20.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4240" PostTypeId="1" AcceptedAnswerId="4242" CreationDate="2014-11-12T13:12:01.433" Score="5" ViewCount="1022" Body="&lt;p&gt;Since part 4 is dragging it's feet, What am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Puny mortals, I control your very lives!&lt;br/&gt;&#xA;  When I'm angry, you cease motion, &lt;br/&gt;&#xA;  but yield to my envy, &lt;br/&gt;&#xA;  Without me you descend into chaos, &lt;br/&gt;&#xA;  I am never wrong, only long &lt;br/&gt;&#xA;  Turn from me at your peril.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the city you see me every day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4951" LastActivityDate="2016-11-24T04:35:15.423" Title="Guess The Object - What Am I Riddle (Part 5)" Tags="|word-problem|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4241" PostTypeId="2" ParentId="4026" CreationDate="2014-11-12T13:13:52.280" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Placebo&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not scoundrely enough...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Those who take me, won't know me: placebo is often used in control groups when evaluating new medicine and treatments. Test persons aren't supposed to know its not the real thing. They take sugar pills. &#xA; Those who know me, won't want me: if you know what placebo is, and that something is placebo, the placebo effect will generally be smaller than otherwise. You're less enthusiastic if you know it's fake. &#xA; Those who want me, will make me: researchers want the sugar pills, not the patients.&#xA; Those who make me, will give me away: the point is to fool someone else, not to eat/inject it for yourself. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5125" LastEditorUserId="5125" LastEditDate="2014-12-12T21:44:23.613" LastActivityDate="2014-12-12T21:44:23.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4242" PostTypeId="2" ParentId="4240" CreationDate="2014-11-12T13:14:26.480" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A traffic light: angry (red) means stop, envy (green) means go&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3903" LastActivityDate="2014-11-12T13:14:26.480" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4245" PostTypeId="2" ParentId="4139" CreationDate="2014-11-12T13:21:59.477" Score="2" Body="&lt;p&gt;The solution is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Television&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This thing can be so tiny,&#xA;That's when it has the most variety.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;a href=&quot;http://en.wikipedia.org/wiki/Variety_show&quot; rel=&quot;nofollow&quot;&gt;Variety shows&lt;/a&gt; were a staple of English language television from its early days (late 1940's) into the 1980s, when televisions were tiny.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Standing up straight, it is surrounded,&#xA;Otherwise it can make things bounded.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the TV is on and in a straight position, everyone surrounds it to see it. If it isn't (aka is off) you can talk with others and bound with them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Within a box it lets you choose,&#xA;But if it points up you will lose.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The box is quite obviuos and you can chose a channel with the remote, but if the remote points up you lose this ability.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It can be seen leaning,&#xA;That's when it reveals true meaning.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you see television you lean back on the sofà. That's the true meaning of having a TV, right? :P&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It can be a celebrity and nothing at once,&#xA;It can show uncertainty but never incoherence.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First sentence: Justin Bieber. You can see &lt;a href=&quot;http://www.imdb.com/title/tt1086216/&quot; rel=&quot;nofollow&quot;&gt;Uncertainty&lt;/a&gt; but not incoherence (I have some doubts about it :P)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But when wealth is what is shows,&#xA;The end, you will know.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Happy_ending&quot; rel=&quot;nofollow&quot;&gt;&quot;and they lived happily ever after.&quot;&lt;/a&gt;&lt;/em&gt; (and wealthy, I suppose)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Just as @d'alar'cop answer, &lt;em&gt;I'm pretty sure this isn't right, but it's a candidate.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="4652" LastEditorUserId="4652" LastEditDate="2014-11-12T13:34:49.003" LastActivityDate="2014-11-12T13:34:49.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4246" PostTypeId="2" ParentId="4149" CreationDate="2014-11-12T13:22:33.167" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wall (although I'm not sure)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You've seen me often, know me well,&#xA;my name the scribes and sages tell.&#xA;My many kin you also know,&#xA;but I'm most famed, as our kind go.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are many walls described in history but the most famous are the &quot;recent ones&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One sister's love is grand largesse,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chinese Great Wall, the other &quot;famous&quot; wall could be the Wall of Berlin&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Our distant cousin's always in,&#xA;and that implies he's also kin.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Walls inside houses&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And where one village holds another,&#xA;you'll often find our eldest brother.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Walls inside a city (e.g. Florence in Italy has nowadays old city walls that &quot;hold&quot; the center, but outside the city has grown, so basically a village holds another&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I keep the peace with great panache. Strike me down and neighbours clash&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Needless to say...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5122" LastActivityDate="2014-11-12T13:22:33.167" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4247" PostTypeId="1" AcceptedAnswerId="4258" CreationDate="2014-11-12T13:25:11.823" Score="50" ViewCount="8502" Body="&lt;p&gt;I'm always in your thoughts, technically twice&lt;br&gt; &#xA;If displayed correctly, just to be precise&lt;br&gt;&#xA;&lt;br&gt;&#xA;If you see my left you will call me a con&lt;br&gt;&#xA;and it would tear me apart if you called me John&lt;br&gt;&#xA;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take down a note and drop your cup&lt;br&gt;&#xA;All you need is what keeps up&lt;br&gt;&#xA;&lt;br&gt;&#xA;Please don't apply pressure or I will be off&lt;br&gt;&#xA;and try to solve my riddle if you can take the scoff&lt;br&gt;&#xA;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Don't you dare look at the solution below.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Shame upon you! Try harder. &lt;br&gt; The solution is a single, short word. This won't help you for now but &quot;meow&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5106" LastEditorUserId="39" LastEditDate="2015-08-08T15:39:10.930" LastActivityDate="2015-08-08T15:39:10.930" Title="Are you clever enough?" Tags="|riddle|word|rhyme|" AnswerCount="7" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4248" PostTypeId="2" ParentId="4247" CreationDate="2014-11-12T13:32:57.450" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I thought the word you be looking for is 'tea' but I'm not quite sure if it fits all clues. The letter 't' is twice in thoughts. And you drink tea from a cup so if you drop the cup tou are left with 'tea'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2014-11-12T13:32:57.450" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4252" PostTypeId="2" ParentId="4139" CreationDate="2014-11-12T13:49:56.820" Score="4" Body="&lt;p&gt;It's a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; coin&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This thing can be so tiny, That's when it has the most variety.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is a wide variety of tiny coins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Standing up straight, it is surrounded, Otherwise it can make things bounded.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If it isn't standing up straight it has a certain side facing up. The different sides are used in coin flipping or tossing to make a decision which cannot be altered (i.e. bounded).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Within a box it lets you choose, But if it points up you will lose.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you don't know the state of a coin (like Schrödinger's cat (thank you stackErr)) you can guess which side faces up. When it points up (i.e. standing on its edge) you probably guessed wrong and you lose.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It can be seen leaning, That's when it reveals true meaning.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When a coin starts to lean (this can take a while when the coin is spinning on a surface) you know which side will be facing up, announcing the winner of the toss (the coin's true meaning) before it ends.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It can be a celebrity and nothing at once, It can show uncertainty but never incoherence.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A very well known person could be on the face of the coin while the monetary value could be practically worthless (i.e. nothing). Coin tosses are always uncertain and never incoherent.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But when wealth is what it shows, The end, you will know.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The person who called tails in the coin toss won. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'd reckon I have a 50% chance of being right ;)&lt;/p&gt;&#xA;" OwnerUserId="5127" LastEditorUserId="5127" LastEditDate="2014-11-12T15:38:55.447" LastActivityDate="2014-11-12T15:38:55.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4255" PostTypeId="2" ParentId="4247" CreationDate="2014-11-12T14:28:54.020" Score="2" Body="&lt;p&gt;Maybe&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Brain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It is where thoughts are made, technically twice because it has two lobes, and John Braine is a novelist&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4527" LastActivityDate="2014-11-12T14:28:54.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4256" PostTypeId="2" ParentId="4004" CreationDate="2014-11-12T14:29:06.853" Score="0" Body="&lt;p&gt;In light of new information, I say the answer is 2.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Looking at the digital display I could find the following information:&lt;br&gt;&#xA; x - | a ......... x is input; - is number of crossbars; | is number of vertical bars; a is answer&lt;br&gt;&#xA; 1 0 2 0&lt;br&gt;&#xA; 2 3 2 4&lt;br&gt;&#xA; 3 3 2 4&lt;br&gt;&#xA; 4 1 3 3&lt;br&gt;&#xA; 5 3 2 4&lt;br&gt;&#xA; 6 3 3 6&lt;br&gt;&#xA; 7 1 2 ?&lt;br&gt;&#xA; 8 3 4 8  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From this it appears:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That the number of crossbars is a multiplier for the number of vertical bars.&lt;br&gt;&#xA; 0 bars = x0&lt;br&gt;&#xA; 1 bar = x1&lt;br&gt;&#xA; 3 bars = x2&lt;br&gt;&#xA; Why 3 bars is a x2 multiplier is unknown to me. Could it be because of the number of &quot;on&quot; bits to make up the number in binary? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3922" LastActivityDate="2014-11-12T14:29:06.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4257" PostTypeId="2" ParentId="4247" CreationDate="2014-11-12T14:32:00.123" Score="3" Body="&lt;p&gt;I was thinking about this, but it's not a short word and it doesn't fit all the clues.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;consciousness&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your consciousness is on your thoughts,&#xA; the left part of the word is 'con',&#xA; John Searle is a philosopher who examined it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4114" LastActivityDate="2014-11-12T14:32:00.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4258" PostTypeId="2" ParentId="4247" CreationDate="2014-11-12T14:33:10.343" Score="45" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Perhaps the single short word is the word &quot;on&quot;&#xA; &lt;br&gt;&lt;br&gt;&#xA; It appears in &quot;your&quot; and &quot;thoughts&quot; if thoughts is viewed upside down&#xA; and reflected.&#xA; &lt;br&gt;&lt;br&gt;&#xA; To make the word John, you have to split the o and the n, thus&#xA; tearing him up.&#xA; &lt;br&gt;&lt;br&gt;&#xA; If you &quot;c&quot; the left, you will now be &quot;con&quot;, a con.&#xA; &lt;br&gt;&lt;br&gt;&#xA; If a switch is &quot;on&quot; and you apply pressure, it is now &quot;off&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="97" LastEditorUserId="97" LastEditDate="2014-11-12T14:52:29.070" LastActivityDate="2014-11-12T14:52:29.070" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4259" PostTypeId="2" ParentId="4247" CreationDate="2014-11-12T14:33:49.080" Score="7" Body="&lt;p&gt;Now this may sound crude and probably not the best answer but if looked at correctly in your mind theres 2 of them.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The con bit is the bit that is confusing there could be a link im not picking up on&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No one wants to be called john if they have them? culture permitting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;no one wants to have heavy preasure on them no?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;take a note and drop a a cup.. Size??&lt;br&gt;&#xA;all you need is what keeps up ... Bra??&lt;br&gt;&#xA;and again correct me if im wrong but no one wants to have heavy preasure on boobs no?&lt;br&gt;&#xA; boobs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5135" LastActivityDate="2014-11-12T14:33:49.080" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4260" PostTypeId="2" ParentId="4057" CreationDate="2014-11-12T14:52:13.297" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PRIME NUMBER&#xA; &lt;P&gt;&#xA; Sequence starts at 2. Guard:2 Guest:2 2-2 =0&#xA; &lt;P&gt;&#xA; Guard:3 Guest:5 5-3=2 (First prime number)&#xA; Guard:5 Guest:0 10-5=5 (Third prime number)&#xA; &lt;P&gt;&#xA; Guard:6 Guest:3 13-6=7(Forth prime number)&#xA; &lt;P&gt;&#xA; Guard:4 Guest:4 + 2nd prime number= 4+ 3&#xA; &lt;P&gt;&#xA; Answer: 7&#xA; &lt;P&gt;&#xA; PS: How about Guard:1 ? 1+19%10 is not 9!&#xA; &lt;P&gt;&#xA; Then you have to deduct it. 1-2=9. You move backwards, the sequence doesn't loops over&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Easy.&lt;/p&gt;&#xA;" OwnerUserId="4995" LastEditorUserId="4995" LastEditDate="2014-11-12T14:59:40.750" LastActivityDate="2014-11-12T14:59:40.750" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4261" PostTypeId="2" ParentId="4211" CreationDate="2014-11-12T14:53:28.727" Score="-1" Body="&lt;p&gt;It must be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Money&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am something that everyone needs.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Do I need to explain more?.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You use me many times in different ways.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You spend money to buy whatever you need (in different ways).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes you only use part of me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You spend part of your money.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Others you only use most of me. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You spend all you have to buy expensive staff.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You always turn and bend me to your will.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You can fold money, bend it, turn it....&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You use to me to scare, to confuse, even to question.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Use your imagination, you can use money to scare, confuse, question either legally or illegally.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can even use to confront someone like when you say &quot;Bill, [rest of confrontation]&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This one is probably need some more imagination. e.g.: &quot;Get all your money and get out of here.&quot; something like that... &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You absolutely use me every day. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;In most cases, you need to use it everyday.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4316" LastEditorUserId="4316" LastEditDate="2014-11-19T09:16:48.043" LastActivityDate="2014-11-19T09:16:48.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4262" PostTypeId="2" ParentId="4139" CreationDate="2014-11-12T15:04:11.800" Score="2" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a dry erase marker.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The riddle's explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This thing can be so tiny,&#xA;That's when it has the most variety.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can buy packs of smaller markers with a huge number of colors.  Big markers are typically sold as just a few colors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Standing up straight, it is surrounded,&#xA;Otherwise it can make things bounded.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sold in the store it is standing up straight, surrounded by other markers and packaging.  When you're writing with it, you can draw lines and borders.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Within a box it lets you choose,&#xA;But if it points up you will lose.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Drawing on a whiteboard (box shape) you can choose what to draw. But careful! If you keep the marker pointing up, the end will go dry and you will lose, as it will stop writing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It can be seen leaning,&#xA;That's when it reveals true meaning.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That's the ticket -- just hold the marker at an a &quot;leaning&quot; angle, and it will write better and the end won't go dry.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It can be a celebrity and nothing at once,&#xA;It can show uncertainty but never incoherence.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I personally get excited sometimes about writing with dry erase markers (celebrity), but if I'm not using them, they're ignored.  Not totally sure about the uncertainty/incoherence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But when wealth is what it shows,&#xA;The end, you will know.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They're used in business and sales meetings to write down final numbers, I guess.  (A bit of a stretch here.)  Maybe the end is the numbers getting erased.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5138" LastActivityDate="2014-11-12T15:04:11.800" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4263" PostTypeId="1" AcceptedAnswerId="4266" CreationDate="2014-11-12T15:08:51.833" Score="26" ViewCount="2497" Body="&lt;p&gt;This puzzle is part of a series - &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4075/the-serial-killer-part-1&quot;&gt;part 1 is here&lt;/a&gt;. Any comments I make on the OP are canonical and &lt;a href=&quot;http://tvtropes.org/pmwiki/pmwiki.php/Main/OutOfCharacter&quot; rel=&quot;nofollow noreferrer&quot;&gt;out of character (OOC)&lt;/a&gt;. Any comments I post on answers are for flavour only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; As it turns out, this was a very easy puzzle, and I'm actually having to rewrite the plot of the series thanks to SKV being far too clever for his own bleeding good (I love you really). The glory's gone, but try to solve it without checking out the answers. I enjoyed making it, anyway :-) Part 3 will be tougher.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You're staring at the list of Stacks when out of nowhere, you hear a voice going &lt;a href=&quot;https://www.youtube.com/watch?v=k85mRPqvMbE&quot; rel=&quot;nofollow noreferrer&quot;&gt;rik tika tik tik tik tik&lt;/a&gt; and before you wonder &lt;a href=&quot;https://puzzling.stackexchange.com/a/4136/2518&quot;&gt;where it came from&lt;/a&gt; you realise something - the capital letters in the titles of all the dead sites can be re-arranged to spell &quot;Moderators&quot;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You cast your eyes over the the list of murdered sites to find the Moderators Stack, and in your best Scottish-detective voice you announce &quot;Och! There's bin a murrrrrrder, and ah nae where ahm going tae look!&quot; (OOC: Sorry, Scotland). Tentatively, you click through to the Moderators stack and right at the top, you see a post entitled &quot;We Will Conquer&quot;. You recognise the title from one of the dead Stacks, so you click on to read it...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dq03f.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You gasp in disbelief and check back to the Stacks list - sure enough, &lt;a href=&quot;http://new.tinygrab.com/c158ca1523d63be3349907658c09f87b3fee07f701.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ask Ubuntu is dead&lt;/a&gt;. You're about to click on it to make sure, but then remember the ominous warning about entering a new Stack, and think better of it. This &quot;Handel&quot; character doesn't seem like someone to mess around with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You realise you're suddenly on a countdown, trying to minimise the number of stacks that die, and every false step you make could bring a flood of confused newbies asking inane questions about &lt;strong&gt;anything&lt;/strong&gt; on to your beloved Puzzling.SE... or even worse, Puzzling.SE might get killed off!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Baffled and starting to panic a little, you sit back to mull things over. What's your next move? Where should you look?&lt;/p&gt;&#xA;" OwnerUserId="2518" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2016-09-13T17:31:41.763" Title="The SErial Killer (Part 2)" Tags="|story|mystery|" AnswerCount="4" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="4264" PostTypeId="2" ParentId="4263" CreationDate="2014-11-12T15:17:36.830" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; programmers. purely because of &quot;look sharp&quot; which is reference to the type of coding used c#&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5135" LastActivityDate="2014-11-12T15:17:36.830" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4265" PostTypeId="2" ParentId="4263" CreationDate="2014-11-12T15:25:31.767" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What roman numerals, Julius Caesar (We see, We saw, god conquered) and George Friderich is known for?&#xA; &lt;P&gt;&#xA; Italian&#xA; &lt;P&gt;&#xA; So I visit Italian Language.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4995" LastActivityDate="2014-11-12T15:25:31.767" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4266" PostTypeId="2" ParentId="4263" CreationDate="2014-11-12T15:26:14.363" Score="25" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; English language &amp;amp; usage. The ampersand in the logo has hex color 90383b, which is the roman numeral in the tag.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="439" LastActivityDate="2014-11-12T15:26:14.363" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4268" PostTypeId="5" CreationDate="2014-11-12T15:30:46.877" Score="0" Body="&lt;p&gt;A cryptogram consists of a short piece of encrypted text. &#xA;Generally the cipher used to encrypt the text is simple enough that the cryptogram can be solved by hand. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Frequently used are substitution ciphers where each letter is replaced by a different letter or number. To solve the puzzle, one must recover the original lettering. &lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-01T03:56:02.647" LastActivityDate="2015-02-01T03:56:02.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4269" PostTypeId="4" CreationDate="2014-11-12T15:30:46.877" Score="0" Body="A puzzle consisting of a short piece of enciphered information, which the puzzler must decipher. " OwnerUserId="2225" LastEditorUserId="4050" LastEditDate="2014-11-28T15:18:41.283" LastActivityDate="2014-11-28T15:18:41.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4270" PostTypeId="1" AcceptedAnswerId="4271" CreationDate="2014-11-12T15:38:43.207" Score="13" ViewCount="1882" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am a tree on highest mountain.&lt;br&gt;&#xA;  I am many, don't try countin'.&lt;br&gt;&#xA;  I grow skinny. I grow fat.&lt;br&gt;&#xA;  Chop me down and I'll grow back.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Many mountains are my home.&lt;br&gt;&#xA;  You don't plant me. I just grow.&lt;br&gt;&#xA;  Sometimes I just go away.&lt;br&gt;&#xA;  Though people wish that I would stay.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What kind of tree am I?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4877" LastEditorUserId="25966" LastEditDate="2016-07-12T19:53:18.013" LastActivityDate="2016-07-12T19:53:18.013" Title="What kind of tree am I?" Tags="|riddle|rhyme|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4271" PostTypeId="2" ParentId="4270" CreationDate="2014-11-12T15:41:07.883" Score="16" Body="&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HAIR (as on the head)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a tree on highest mountain.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Something that grows in a high place. The top of our head is as high as we get in a way.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am many, don't try countin'.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One has numerous hairs, normally too many count.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I grow skinny. I grow fat.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Presumably about the thickness variations in people's hair. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Chop me down and I'll grow back.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hair keeps growing regardless of being cut to nothingness (as in a shaved head)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Many mountains are my home.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The head has bumps.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You don't plant me. I just grow.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One does not plant hair seeds. Hair grows naturally without seeds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I just go away.&#xA;Though people wish that I would stay.                          &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This refers to people who are bald. They normally seem to dislike the condition.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-11-28T23:11:56.527" LastActivityDate="2014-11-28T23:11:56.527" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4272" PostTypeId="2" ParentId="4133" CreationDate="2014-11-12T15:45:40.957" Score="7" Body="&lt;p&gt;I think I see the pattern (although it doesn't work for one of the answers), but I don't quite know the answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The number of words in the sentence the last guard speaks is the number of letters that you need to use in the pronoun referring to the king.&lt;br&gt;&lt;br&gt;&#xA; Kind merchant, what brings you to our gates? (8) | Sovereign (9?) &lt;br&gt;&lt;br&gt;&#xA; ... brings thee here? (3)                        | Rex (3) &lt;br&gt;&lt;br&gt;&#xA; Yes, Yes we do. (4)                              | King (4) &lt;br&gt;&lt;br&gt;&#xA; Well... he is half right, do you? (7)            | Monarch (7) &lt;br&gt;&lt;br&gt;&#xA; Anything you wish to say for yourself? (7)       | Majesty (7)&lt;br&gt;&lt;br&gt;&#xA; My guess is that the guards just knew the spy was a spy, having seen him hiding.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditDate="2014-11-12T15:50:54.773" LastActivityDate="2014-11-12T15:50:54.773" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4273" PostTypeId="2" ParentId="4247" CreationDate="2014-11-12T16:03:49.303" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the answer is &quot;on&quot; :) if you flip &quot;your&quot; and &quot;thoughs&quot; vertically, and if you press (apply pressure on a switch) it will become OFF :) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;P.S. just saw someone answered it :) And yet I'm not a cheater, I just had a logo with ou/no for one of my projects, and I recognized it in &quot;your thoughts&quot; :)&lt;/p&gt;&#xA;" OwnerUserId="5147" LastActivityDate="2014-11-12T16:03:49.303" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4274" PostTypeId="2" ParentId="4133" CreationDate="2014-11-12T16:16:12.557" Score="2" Body="&lt;p&gt;After reading the accepted answer, I decided I have one that makes more sense:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each visitor must say at least as many words as the sum of the words said by the two guards.&#xA;&#xA; &lt;p&gt;Guard: 9 words, Merchant: 9 words&#xA; &lt;p&gt;Guards: 14 + 3 = 17 words, Father: 43 words&#xA; &lt;p&gt;Guard: 4 words, Forester: 11 words&#xA; &lt;p&gt;Guards: 26 + 7 = 33 words, Beggar: 36 words&#xA; &lt;p&gt;Guards: 4 + 9 = 13 words, Mother in law: 14 words&#xA; &lt;p&gt;Guards 12 + 7 = 19 words, Spy: 7 words&#xA; &lt;p&gt;&#xA; If the spy had not been waiting behind the trees, his answer would have been correct, because the first guard wouldn't have said anything, and he would only have had to reply with 7 words.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4394" LastEditorUserId="4394" LastEditDate="2014-11-12T16:43:52.780" LastActivityDate="2014-11-12T16:43:52.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4275" PostTypeId="2" ParentId="4139" CreationDate="2014-11-12T16:27:47.900" Score="1" Body="&lt;p&gt;OK, can't get this out of my head, so posting a second idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to the riddle is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a lottery or raffle ticket&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This thing can be so tiny,&#xA;That's when it has the most variety.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lotteries with lots of small winnings can have more variety.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Standing up straight, it is surrounded,&#xA;Otherwise it can make things bounded.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When still in the convenience store, a dangling string of lottery tickets is surrounded by other tickets.  Otherwise the tickets are still wrapped around the roll, i.e. bounding it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Within a box it lets you choose,&#xA;But if it points up you will lose.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sometimes you have the tickets in a box and somebody picks one. Not sure about the pointing up, but losing definitely is something that goes with lotteries!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It can be seen leaning,&#xA;That's when it reveals true meaning.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some tickets have a lenticular or hologram that make you look at an angle to see what you've got.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It can be a celebrity and nothing at once,&#xA;It can show uncertainty but never incoherence.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This part fits well -- a lottery ticket may be highly valued by its owner (&quot;celebrity&quot;) even though it is not going to win and is therefore worthless.  It may be as of yet &quot;uncertain&quot; whether I'm going to win, but it will never be &quot;incoherent&quot; - there will always be a way to definitively tell whether you won or not.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But when wealth is what it shows,&#xA;The end, you will know.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; At some point every ticket either shows wealth, or lack thereof.  Either way, it's the end of this ticket.&#xA;&#xA; By the way, I despise lotteries and do not participate in them, this was just an idea that came to mind. :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5138" LastActivityDate="2014-11-12T16:27:47.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4276" PostTypeId="1" AcceptedAnswerId="4537" CreationDate="2014-11-12T16:39:50.433" Score="19" ViewCount="1233" Body="&lt;p&gt;This color box contains a sentence. Figure out what it is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WWFMe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each row is a single word of the sentence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="2225" LastEditDate="2015-01-06T06:15:42.503" LastActivityDate="2015-01-06T06:15:42.503" Title="What do the colors say?" Tags="|pattern|visual|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4277" PostTypeId="5" CreationDate="2014-11-12T16:55:25.637" Score="0" Body="&lt;p&gt;A rhyme is a repetition of the same sound(s) at the ends of two or more words, most often in the final syllables of lines in poems and songs.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Examples:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Rhyme&lt;/em&gt; and &lt;em&gt;Time&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Knight&lt;/em&gt; and &lt;em&gt;Flight&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&amp;quot;See you &lt;em&gt;later&lt;/em&gt;, alli&lt;em&gt;gator&lt;/em&gt;!&amp;quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Rhyme partly seems to be enjoyed simply as a repeating pattern that is pleasant to hear. It also serves as a powerful mnemonic device, facilitating memorization. The regular use of tail rhyme helps to mark off the ends of lines, thus clarifying the metrical structure for the listener. As with other poetic techniques, poets use it to suit their own purposes; for example in &lt;a href=&quot;/questions/5737/variations-on-a-theme&quot;&gt;this riddle&lt;/a&gt; rhyming is used as a hint towards the solution.&lt;/p&gt;&#xA;" OwnerUserId="2518" LastEditorUserId="60644" LastEditDate="2023-02-03T16:00:08.677" LastActivityDate="2023-02-03T16:00:08.677" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4278" PostTypeId="4" CreationDate="2014-11-12T16:55:25.637" Score="0" Body="A problem or riddle written in rhyme for an artistic twist." OwnerUserId="2518" LastEditorUserId="5373" LastEditDate="2015-03-15T13:07:43.803" LastActivityDate="2015-03-15T13:07:43.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4279" PostTypeId="1" AcceptedAnswerId="4639" CreationDate="2014-11-12T17:00:32.180" Score="8" ViewCount="784" Body="&lt;p&gt;I dont know how this one might go down, but no chance of finding out if not asked&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Name the entity being described&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Me and my family made in the UK&lt;br&gt;&#xA;  the lesser of 3 my siblings say&lt;br&gt;&#xA;  Kept together unwanted&lt;br&gt;&#xA;  in darkness till sorted&lt;br&gt;&#xA;  Our absence will give you leeway.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See what you think, if theres no interest ill assume my career is over here.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The greatest of we&lt;br&gt;&#xA; alone has 99 more than me&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5135" LastEditorUserId="5135" LastEditDate="2014-11-14T12:21:02.897" LastActivityDate="2014-11-14T12:28:53.587" Title="Name that entity" Tags="|riddle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4280" PostTypeId="1" AcceptedAnswerId="4357" CreationDate="2014-11-12T17:12:27.720" Score="0" ViewCount="987" Body="&lt;p&gt;Am I real? It's up to you..&lt;br /&gt;&#xA;I am real, if you flip me..&lt;br /&gt;&#xA;But, once you flip, I may bite you...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I spit a 'POLO', I will become a little good..&lt;br /&gt;&#xA;If I split a 'POLO', I will become a little more good..&lt;br /&gt;&#xA;If I swallow a 'POLO', I will become fully good...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you guess who I am???&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Polo_(confectionery)&quot; rel=&quot;nofollow&quot;&gt;A Polo is approximately 1.9 cm in diameter, 0.4 cm deep and has a 0.8 cm wide hole. The original Polo is white in colour with a hole in the middle, and the word 'POLO' embossed twice on one side around the ring, hence the popular slogan The Mint with the Hole.&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: Please place your answers in spoiler tags. Also mention how you get to the answer.&lt;/em&gt; &lt;/p&gt;&#xA;" OwnerUserId="2730" LastActivityDate="2014-11-13T03:07:29.090" Title="What Am I Riddle (Part 6)" Tags="|riddle|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4281" PostTypeId="1" AcceptedAnswerId="4451" CreationDate="2014-11-12T17:26:52.230" Score="9" ViewCount="1963" Body="&lt;p&gt;I'm the one to be told&#xA;at the very end &lt;br&gt;&#xA;just get rid of the old&#xA;before you descend&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm the tie on your neck &lt;br&gt;&#xA;if you slice it apart &lt;br&gt;&#xA;and the third in the group &lt;br&gt;&#xA;rhyming is hard&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm the first of the group&#xA;and I will be kind &lt;br&gt;&#xA;Think of my age&#xA;and it's me who you'll find&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although mentioned last &lt;br&gt;&#xA;Im the easiest of all &lt;br&gt;&#xA;If you can not find me &lt;br&gt;&#xA;start again after fall&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my riddle&#xA;and you know the deal &lt;br&gt;&#xA;Find out the answer -&#xA;What is it I feel?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Super secret solution ahead:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Spoiler: It is something you can feel and it has 4 letters. I'm sure you already knew that. Also it has nothing to do with religion of any kind.&lt;br&gt; I focused on playing with grammar and letters here. Sticking close to the words without interpreting may help.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5106" LastEditorUserId="5106" LastEditDate="2015-08-15T14:39:25.253" LastActivityDate="2015-08-15T14:39:25.253" Title="I hope your brain won't fall apart" Tags="|riddle|word|" AnswerCount="10" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4282" PostTypeId="2" ParentId="4004" CreationDate="2014-11-12T17:34:04.133" Score="4" Body="&lt;p&gt;I say&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think the answer is the number of corners in the digital display representation of it, with a T shape counting twice, i.e. towards both sides.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Example for explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For &quot;3&quot; we count: 1: the upper right corner connects the top bar and the top right bar. 2: the top right and middle bar. 3: middle and lower right bar. (2+3 make said T shape). 4: lower right and bottom bar.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit: A more mathematical way to express this is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; answer = number of unique pairs (h,v) where h is a horizontal bar, v an &lt;em&gt;adjacent&lt;/em&gt; vertical bar and both are lit when displaying the number given by the guard.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit: forgot an important detail in the previous (hidden) paragraph.&lt;/p&gt;&#xA;" OwnerUserId="4971" LastEditorUserId="4971" LastEditDate="2014-11-13T08:04:40.900" LastActivityDate="2014-11-13T08:04:40.900" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4284" PostTypeId="2" ParentId="4270" CreationDate="2014-11-12T17:38:56.297" Score="3" Body="&lt;p&gt;I think &quot;Hope&quot;(dream) would be a beautiful solution, too!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a tree on highest mountain.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Something far away we are trying to reach.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am many, don't try countin'.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are many people hoping for something and maybe everyone has many &quot;trees&quot; he is looking at. But there is no point counting them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I grow skinny. I grow fat.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hope can have all different shapes and it can change over time. You sometimes change your mind and thus change hope to some point.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Chop me down and I'll grow back.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reminds me of someone trying to destroy your dreams but hope is returning. They say 'hope' is what dies last.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Many mountains are my home.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many people can have the same dream or hope for something.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You don't plant me. I just grow.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You cannot control what you want. Sometimes you desire forbidden things.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I just go away. Though people wish that I would stay.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sometimes an opportunity vanishes and as a result, hope dies eventually even though you still want to pursue it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5106" LastEditorUserId="4312" LastEditDate="2015-10-31T06:22:03.203" LastActivityDate="2015-10-31T06:22:03.203" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4285" PostTypeId="2" ParentId="4263" CreationDate="2014-11-12T17:40:58.887" Score="5" Body="&lt;p&gt;Obviously the answer is already posted; but this solution seemed valid to me as well:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Server Fault&lt;/strong&gt; is the next target.&#xA;&#xA; No self respecting serial killer would leave the clues from Part 1 without having the post for Part 2 already up, and yet the timestamp on the post is from 11/12!&#xA;&#xA; Clearly, the incorrect NTP configuration is the indication of the next target.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5157" LastActivityDate="2014-11-12T17:40:58.887" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4286" PostTypeId="1" CreationDate="2014-11-12T17:41:22.630" Score="4" ViewCount="606" Body="&lt;p&gt;A party is being held at a local mansion. The host is very rich and his success is because of one thing — his famous recipe for Spaghetti!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only guests that may attend are people who correctly reply to the guard at the door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's where you come in. You and a friend are trying to steal this recipe. You sneak by and listen to the passwords.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first guest arrives.  The guard says, &quot;1&quot;, the guest replies &quot;2&quot; and is let in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second guest arrives.  The guard says, &quot;3&quot;, the guest replies &quot;5&quot; and is let in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third guest arrives.  The guard says, &quot;8&quot;, the guest replies &quot;4&quot; and is let in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your friend says, &quot;Aha!  This is an easy one.&quot;   He slips away and show up at the gate a minute later.  &quot;13&quot;, the guard says.  Your friend replies &quot;3&quot; and is immediately trampled by an elephant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where did your friend go wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You decide to hang around a bit longer.&#xA; The next guest arrives, and the guard says, &quot;2&quot;.  The guest replies &quot;2&quot; and is let in.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Another guest arrives, and the guard says &quot;89&quot;.  The guest replies &quot;8&quot; and get in.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Final hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Still scratching your head, you wait for another guest to approach. &quot;Two&quot; you hear the guard start to say, but at that exact moment a passing truck honks somewhere behind you, nearly startling you out from the tree.  You barely recover in time to hear the guest reply, &quot;8&quot; and the guard lets her in.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Having read the above hints, now you decide to approach the guard.  He says, &quot;34&quot;, what do you reply?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As usual please post your answer, an explanation of what process you used to get the answer and any hints used in spoiler tags.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Whoops, flubbed the final hint. It should be correct now.&lt;/p&gt;&#xA;" OwnerUserId="2718" LastEditorUserId="2718" LastEditDate="2014-11-12T20:37:51.180" LastActivityDate="2014-11-15T04:15:40.170" Title="The Security to the Party [Part 20]" Tags="|mathematics|password|" AnswerCount="3" CommentCount="19" ClosedDate="2014-11-17T20:58:14.533" ContentLicense="CC BY-SA 3.0" />
  <row Id="4288" PostTypeId="2" ParentId="4281" CreationDate="2014-11-12T17:42:54.357" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; love??&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Told at the very end of every fairty tale??  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;decend as in fall, but young people arnt considered this by elders. its just a phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is the restrainst of finding it released after the ceremony&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is kind, almost all the time but I digress&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it is easy if right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you usually fall into this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are my reasons as flimsy as the base might be.&lt;/p&gt;&#xA;" OwnerUserId="5135" LastActivityDate="2014-11-12T17:42:54.357" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4289" PostTypeId="2" ParentId="4281" CreationDate="2014-11-12T17:45:26.647" Score="12" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Cold?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's what people feel when they're about to die, &#xA;how your neck will feel without a tie,&#xA;it's how the third in the group will feel without a special friend,&#xA;It's what you'll feel at fall's end.&lt;/p&gt;&#xA;" OwnerUserId="5159" LastEditorUserId="5159" LastEditDate="2014-11-12T17:55:01.680" LastActivityDate="2014-11-12T17:55:01.680" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4290" PostTypeId="2" ParentId="4247" CreationDate="2014-11-12T17:56:34.687" Score="11" Body="&lt;p&gt;The original answer is quite brilliant and exciting, but in my pursuit for finding it, I've stumbled upon a rather strange alternative which &lt;em&gt;seems to fit&lt;/em&gt; the riddle. Have a look  at my &lt;strong&gt;intentionally forced theory&lt;/strong&gt; and give me your opinion.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sherlock Holmes speaking to Professor Moriarty near the Reichenbach falls&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why? Well, read on if you are interested.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm always in your thoughts, technically twice&lt;br&gt;&#xA;  If displayed correctly, just to be precise&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Holmes was always a stinging thought in Moriarty's mind. The professor was directly involved in two cases titled &quot;The Valley of Fear&quot; and &quot;The Final Problem&quot;. Hence Holmes would've crossed his mind twice.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you see my left you will call me a con&lt;br&gt;&#xA;  and it would tear me apart if you called me John&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Holmes have said Watson about this once. &quot;If you would've seen my sinister (left) side, You'd know that I make a perfect villain (con)&quot;. Also, since Holmes is speaking to Moriarty we could assume the second sentence is about (Dr.) John (Watson).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Take down a note and drop your cup&lt;br&gt;&#xA;  All you need is what keeps up&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;note&lt;/em&gt; might be referring to the one Holmes wrote to Watson moments before the death of his arch-enemy. Since the combat was done without any ammunition, Moriarty might be saying &quot;All you need (for this final battle) is what keeps up (the strength of his body)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Please don't apply pressure or I will be off&lt;br&gt;&#xA;  and try to solve my riddle if you can take the scoff&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Moriarty's final words perhaps? He might be inches away from slipping off the ledge. &quot;Do not apply the pressure (spare me Holmes)&quot;. Since his right-hand, Sebastian Moran, was still living to complete his mission, the professor shouts &quot;solve my riddle if you can take the scoff&quot; (which is impossible, my organization is bigger than you think, Holmes...)&lt;/p&gt;&#xA;" OwnerUserId="4838" LastEditorUserId="4838" LastEditDate="2014-11-12T18:02:19.163" LastActivityDate="2014-11-12T18:02:19.163" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4291" PostTypeId="2" ParentId="4139" CreationDate="2014-11-12T18:00:30.260" Score="21" Body="&lt;p&gt;I think it's &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Regular Expression Syntax&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This thing can be so tiny,&#xA;That's when it has the most variety.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The . character is a wildcard which can match ANY single character. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Standing up straight, it is surrounded,&#xA;Otherwise it can make things bounded.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The | character is an &quot;or&quot; operator, so it's surrounded by two choices.&#xA; Otherwise (if not standing up, but on its side) it sets the bounds of a range, e.g. [a-z]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Within a box it lets you choose,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The box is []. [bc]at returns bat or cat, in other words you choose between the characters within the box.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But if it points up you will lose.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; [^bc]at will return ANYTHING OTHER than bat or cat. You lose these options.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It can be seen leaning,&#xA;That's when it reveals true meaning.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; \ combines with other characters to create something other than a literal character. For example, \t is a tab character&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It can be a celebrity and nothing at once,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &quot;star&quot; character * matches any string with zero or more of a given element.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It can show uncertainty but never incoherence.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You might not be certain if you've written the regular expression without errors, which matches what you're looking for, but either way what's returned is coherent with what you wrote. Not too sure about this one...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But when wealth is what it shows,&#xA;The end, you will know.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $ matches the end position of a string.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-11-12T18:00:30.260" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4293" PostTypeId="2" ParentId="4207" CreationDate="2014-11-12T18:30:07.717" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Buses in the UK are often expect exact change, what coins should I&#xA;  give the bus driver so that I'm left with the smallest possible weight&#xA;  to carry on my hike and how much will that be?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's my answer to the original question as stated above:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Just because the driver expects exact change doesn't necessarily mean the driver can &lt;em&gt;only&lt;/em&gt; accept exact change for the listed bus fare. The driver can also accept more than the listed fare. You don't need any money for the return trip (as clarified in the question comments), so you give the driver all your change, leaving you with 0g to carry on your hike.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1989" LastEditorUserId="1989" LastEditDate="2014-11-12T18:41:41.770" LastActivityDate="2014-11-12T18:41:41.770" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4295" PostTypeId="2" ParentId="4281" CreationDate="2014-11-12T19:23:53.003" Score="3" Body="&lt;p&gt;If it's not Marko's answer then maybe it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Time&#xA; slice it apart and you can make tie. it's at the end of most datetime formats. you can feel time (old). At the end of fall we have daylights saving where we change time (this is where year fits better) in an zero based array of dimensions it would be the third (quite a stretch) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2803" LastActivityDate="2014-11-12T19:23:53.003" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4296" PostTypeId="2" ParentId="3685" CreationDate="2014-11-12T19:42:48.230" Score="3" Body="&lt;p&gt;I know @Jan Dvorak already answer, but I want to share my solution too, because I took the inverse route.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to win, you need 1, 3 or 5 stones in the pile to make the last move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise if you have 2, 4 or 6, no matter how many stones you remove {1,3,5} you will create a state where the great shah will win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to win you need an odd number in your turn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I realize not matter how the stones are removed, the student always will face a pile of stones with odd number of them. Because the shah move first, the starting number of stones is even, and can only remove odd number of stones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There for the student always win.&lt;/p&gt;&#xA;" OwnerUserId="5165" LastEditorUserId="11297" LastEditDate="2015-07-16T16:41:20.417" LastActivityDate="2015-07-16T16:41:20.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4297" PostTypeId="2" ParentId="4286" CreationDate="2014-11-12T20:06:50.603" Score="1" Body="&lt;p&gt;My answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The guard always asks a number from the Fibonacci series:&lt;br&gt;&lt;br&gt;&#xA;&#xA; [1 -&gt; 1, 2 -&gt; 2, 3 -&gt; 5, 5 -&gt; ???, 8 -&gt; 4, 13 -&gt; ??? (not 3), 21 -&gt; ???, 34 -&gt; ???, 55 -&gt; ???, 89 -&gt; 8]&lt;br&gt;&lt;br&gt;&#xA;&#xA; 2 is the 2nd answer of the series, which maps to a value that is the double of the 1st answer.&lt;br&gt;&lt;br&gt;&#xA; 89 is the 10th answer of the series, which maps to 8, the double of the 5th answer, which is 4.&lt;br&gt;&lt;br&gt;&#xA; The answer for 34 should be the double of the answer for 5, which is the double of the answer of 2.&lt;br&gt;&lt;br&gt;&#xA; So the answer is &lt;strong&gt;8&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;EDIT: Wrong answer! I was trampled by an elephant. :(&lt;/p&gt;&#xA;" OwnerUserId="5044" LastEditorUserId="5044" LastEditDate="2014-11-12T21:28:16.003" LastActivityDate="2014-11-12T21:28:16.003" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="4298" PostTypeId="2" ParentId="4149" CreationDate="2014-11-12T20:11:00.070" Score="3" Body="&lt;p&gt;Ok, This is maybe just silly, but I'll give it a shot. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your name is... Mouth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You've seen me often, know me well,&lt;br&gt;&#xA;my name the scribes and sages tell.&lt;br&gt;&#xA;My many kin you also know,&lt;br&gt;&#xA;but I'm most famed, as our kind go.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The kind here are other orifices.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One sister's love is grand largesse,&lt;br&gt;&#xA;another's like me, more or less.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The two sisters, I'm not sure about this one. The first could be the eyes, but I don't see how the second could be ears...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My brother shuns such grand appeal,&lt;br&gt;&#xA;he much prefers his meager meal.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The brother is the nose, the meal is air, which is less than food/air like the mouth &quot;eats&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Our distant cousin's always in,&lt;br&gt;&#xA;and that implies he's also kin.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The belly button, I guess? usually an &quot;innie&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And where one village holds another,&lt;br&gt;&#xA;you'll often find our eldest brother.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Clearly&lt;/em&gt; the butt holding your eldest brother, the anus.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I keep the peace with great panache.&lt;br&gt;&#xA;Strike me down and neighbours clash,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if you don't have a mouth all the other ones kinda fall apart.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4126" LastActivityDate="2014-11-12T20:11:00.070" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4299" PostTypeId="1" AcceptedAnswerId="4337" CreationDate="2014-11-12T20:11:08.303" Score="4" ViewCount="508" Body="&lt;p&gt;Follow the clues and tell me where you are in the end.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;PSE    &lt;/li&gt;&#xA;  &lt;li&gt;To the opposite of the opposite of 2 + 2 ~= 4    &lt;/li&gt;&#xA;  &lt;li&gt;Go to the way you round .5    &lt;/li&gt;&#xA;  &lt;li&gt;If up is down and down is up go down    &lt;/li&gt;&#xA;  &lt;li&gt;You're on the right track    &lt;/li&gt;&#xA;  &lt;li&gt;You've made it, where are you?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This is part 1. I may not make a part 2, but others in the community may.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="5373" LastEditDate="2015-04-29T08:43:51.783" LastActivityDate="2015-04-29T08:43:51.783" Title="Where did you end up? [Part 1]" Tags="|logical-deduction|treasure-hunt|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4300" PostTypeId="1" AcceptedAnswerId="4301" CreationDate="2014-11-12T20:12:44.680" Score="5" ViewCount="597" Body="&lt;p&gt;Bob leaves home with $50.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;His birthday approaching, he goes to the neighbourhood bakery to order a birthday cake and pays \$29.95 for the cake, plus \$3.08 for the candles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, he meets three friends in town and they all go grab something to eat: Bob takes a \$4.50 sandwich and a \$0.70 soda.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that, they all go to the movie theater. Tickets cost \$10, but they can benefit from both an &quot;under 25yo&quot; \$2 discount for each one and a &quot;Free ticket for every three tickets bought&quot; offer, and split the bill equally between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They finish the afternoon at a bar, where Bob has two \$2.75 beers while chatting with his friends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How old is Bob?&lt;/p&gt;&#xA;" OwnerUserId="4668" LastEditorUserId="4668" LastEditDate="2014-11-13T17:01:50.143" LastActivityDate="2014-11-13T17:01:50.143" Title="Bob's expensive day" Tags="|logical-deduction|" AnswerCount="2" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="4301" PostTypeId="2" ParentId="4300" CreationDate="2014-11-12T20:16:37.737" Score="11" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bob is 21.  He is under 25 (discount), older than 20 (beer), and the 3.08 for candles works out to 0.14 for each candle for his upcoming birthday (22) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2014-11-12T20:16:37.737" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4302" PostTypeId="5" CreationDate="2014-11-12T20:21:49.400" Score="0" Body="&lt;p&gt;Rush Hour is a sliding block puzzle invented by Nob Yoshigahara in the 1970s. &#xA;It is played on a 6x6 grid with 14 sliding blocks.&#xA;The sliding blocks represent 12 cars (2 squares) and 4 trucks (3 squares), each colored differently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cars and trucks can only be moved within a straight line along the grid, but cannot be rotated.&#xA;The goal of the game is to get the red car out of a six-by-six grid full of automobiles by moving the other vehicles out of its way. However, the cars and trucks (set up before play according to a puzzle card) obstruct the path which makes the puzzle harder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;https://en.wikipedia.org/wiki/Rush_Hour_(board_game)&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Wikipedia article&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="11110" LastEditDate="2017-06-20T20:28:40.580" LastActivityDate="2017-06-20T20:28:40.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4303" PostTypeId="4" CreationDate="2014-11-12T20:21:49.400" Score="0" Body="Rush Hour is a sliding block puzzle on a 6x6 grid, where the tiles represent cars. " OwnerUserId="2383" LastEditorUserId="8874" LastEditDate="2015-01-31T18:09:40.507" LastActivityDate="2015-01-31T18:09:40.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4306" PostTypeId="2" ParentId="4279" CreationDate="2014-11-12T20:35:33.507" Score="4" Body="&lt;p&gt;The entity being described is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The British Government&#xA; Which is composed of three parts: &quot;Her Majesty Queen Elizabeth the Second, by the Grace of God, of the United Kingdom of Great Britain and Northern Ireland and of Her other Realms and Territories Queen, Head of the Commonwealth, Defender of the Faith.&quot; who is the figurative head of Her Majesty's Government, plus Parliament which is divided into the House of Lords and the House of Commons.&#xA; The House of Commons is the lesser of the 3.&#xA; Kept together unwanted, because let's face it, who wants these clowns in parliament?&#xA; In darkness till sorted, because honestly, is there any light and truth in Parliament until the Queen signs off on something with her rubber stamp?&#xA; Our absence will give you leeway, because without the House of Commons there will be no new laws and you will have more leeway in what you are allowed to do.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2718" LastActivityDate="2014-11-12T20:35:33.507" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4310" PostTypeId="2" ParentId="4149" CreationDate="2014-11-12T20:43:20.540" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your name is &lt;strong&gt;pen&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;my name the scribes and sages tell.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; both scribes and sages write&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One sister's love is grand largesse,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; perhaps she's generous with ink -- &lt;strong&gt;a fountain pen&lt;/strong&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;another's like me, more or less.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; could be a pencil.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My brother shuns such grand appeal,&#xA;he much prefers his meager meal.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; he's a highly efficient pen -- &lt;strong&gt;ballpoint&lt;/strong&gt; or &lt;strong&gt;fine point&lt;/strong&gt; perhaps.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Our distant cousin's always in,&#xA;and that implies he's also kin.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;not sure what to make of this stanza. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And where one village holds another,&#xA;you'll often find our eldest brother.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The eldest brother could be a &lt;strong&gt;feather pen&lt;/strong&gt;. The village could be an &lt;strong&gt;inkwell&lt;/strong&gt;, but that seems like a bit of a stretch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I keep the peace with great panache.&#xA;Strike me down and neighbours clash,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; written peace treaties.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or leave me be, it's all the same,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if you don't use a pen, it's essentially the same as striking one down.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4675" LastEditorUserId="4675" LastEditDate="2014-11-12T23:04:14.380" LastActivityDate="2014-11-12T23:04:14.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4313" PostTypeId="2" ParentId="4149" CreationDate="2014-11-12T20:58:58.963" Score="3" Body="&lt;p&gt;Could the answer be  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a &lt;strong&gt;font&lt;/strong&gt;?&lt;br&gt;&#xA; There could be all kinds of relatives if you factor in font size, bold, italics, strike through, letter spacing, etc.&lt;br&gt;&#xA;&lt;br&gt;&#xA;You've seen me often, know me well, &amp;lt;-- You are looking at it right now&lt;br&gt;&#xA;&lt;br&gt;&#xA;One sister's love is grand largesse, &amp;lt;-- Bold&lt;br&gt;&#xA;&lt;br&gt;&#xA;another's like me, more or less.     &amp;lt;-- Italics&lt;br&gt;&#xA;&lt;br&gt;&#xA;he much prefers his meager meal.     &amp;lt;-- StrikeThrough?&lt;br&gt;&#xA;&lt;br&gt;&#xA;Our distant cousin's always in &amp;lt;-- Condensed&lt;br&gt;&#xA;&lt;br&gt;&#xA;Strike me down and neighbours clash, &amp;lt;-- Change font size?&lt;br&gt;&#xA;or leave me be, it's all the same,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3922" LastEditorUserId="3922" LastEditDate="2014-11-12T21:28:49.780" LastActivityDate="2014-11-12T21:28:49.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4314" PostTypeId="2" ParentId="4286" CreationDate="2014-11-12T21:14:36.107" Score="2" Body="&lt;p&gt;Most patterns I'm getting seem to point me at&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;7&lt;/strong&gt;, but I can't find the pattern that fits perfectly.&lt;br&gt;&lt;br&gt;&#xA;&#xA; I know it's got to have something to do with the Fibonacci numbers and particularly their indices in the series, but I just can't seem to crack it...&lt;br&gt;&lt;br&gt;&#xA;&#xA; Another hint maybe?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user5040" LastActivityDate="2014-11-12T21:14:36.107" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4315" PostTypeId="1" AcceptedAnswerId="4317" CreationDate="2014-11-12T21:21:38.593" Score="25" ViewCount="65589" Body="&lt;p&gt;Fred has a puzzle.  The instruction sheet, which is separate from the equation,simply says:  &quot;Make this equation true.  You may ONLY draw ONE, PERFECTLY straight line&quot;.  The equation is all alone on a blank piece of paper and reads:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;5+5+5=550&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There are 4 acceptable answers.  You must get all 4 for a correct answer.&lt;/p&gt;&#xA;" OwnerUserId="5180" LastEditorUserId="5180" LastEditDate="2014-11-12T21:28:34.907" LastActivityDate="2021-07-24T05:41:40.997" Title="Make the equation true by only drawing a single straight line" Tags="|mathematics|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4316" PostTypeId="2" ParentId="4315" CreationDate="2014-11-12T21:24:08.797" Score="10" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 5+5+5≠550&#xA; &lt;br&gt;&lt;br&gt;negating the equality.&#xA; &lt;br&gt;&lt;br&gt;&#xA; Edit:&#xA; &lt;br&gt;You could negate the entire statement, depending on how you interpret the equation...&lt;br&gt;&#xA; _________&lt;br&gt;&#xA; 5+5+5=550     &amp;nbsp;&amp;nbsp;&amp;nbsp;could evaluate to &quot;true&quot; in some contexts... ^^&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user5040" LastEditorDisplayName="user5040" LastEditDate="2014-11-12T21:37:03.537" LastActivityDate="2014-11-12T21:37:03.537" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4317" PostTypeId="2" ParentId="4315" CreationDate="2014-11-12T21:24:46.277" Score="38" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cross out the equals sign&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Turn the first $+$ into a 4 giving $545+5=550$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Turn the second $+$ into a 4 giving $5+545=550$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Turn the equals sign into less than or equal to&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="255" LastEditorUserId="42054" LastEditDate="2018-05-06T03:35:04.570" LastActivityDate="2018-05-06T03:35:04.570" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="4318" PostTypeId="2" ParentId="4315" CreationDate="2014-11-12T21:24:47.640" Score="4" Body="&lt;p&gt;I'd put a line &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; on one of the plus signs, straight up.&lt;br&gt;&#xA; The equation would read&#xA; 545 + 5 = 550&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2444" LastActivityDate="2014-11-12T21:24:47.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4319" PostTypeId="2" ParentId="4300" CreationDate="2014-11-12T21:27:53.387" Score="2" Body="&lt;p&gt;He is celebrating his 22nd birthday.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He has to be at least 18 or 21 (depending on where he is) to buy the beers and less than 25 to get the discount. Dividing the candles by 22 gives a price of 14 cents per candle.&lt;/p&gt;&#xA;" OwnerUserId="4598" LastEditorUserId="1955" LastEditDate="2014-11-12T23:27:31.280" LastActivityDate="2014-11-12T23:27:31.280" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4320" PostTypeId="2" ParentId="4315" CreationDate="2014-11-12T21:31:55.327" Score="15" Body="&lt;p&gt;I think the fourth answer, which hasn't been posted yet is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$5+5+5≤550$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That makes all of them:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$5+5+5≤550$&lt;br/&gt;&#xA;$545+5=550$&lt;br/&gt;&#xA;$5+545=550$&lt;br/&gt;&#xA;$5+5+5≠550$&lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4687" LastEditorUserId="42054" LastEditDate="2018-05-06T04:38:25.210" LastActivityDate="2018-05-06T04:38:25.210" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="4321" PostTypeId="1" AcceptedAnswerId="4324" CreationDate="2014-11-12T21:37:07.577" Score="7" ViewCount="1251" Body="&lt;p&gt;Here's a neat geometrical puzzle I was once shown&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to divide a circle into a finite number of congruent parts some of which don't touch the center?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clarification:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find a shape for a tile such that a certain number of tiles of that shape will fit completely and with no overlap inside a circle. Additionally at least one the the tiles can't touch the center of the circle&lt;/p&gt;&#xA;" OwnerUserId="255" LastEditorUserId="8874" LastEditDate="2016-03-23T15:38:39.993" LastActivityDate="2016-03-23T15:38:39.993" Title="Dividing the Circle" Tags="|mathematics|geometry|dissection|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4323" PostTypeId="2" ParentId="4149" CreationDate="2014-11-12T21:52:04.110" Score="9" Body="&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your name is Middle Finger. &lt;br&gt;&lt;br&gt;&#xA; Your sister's love with grand largess is the wedding band on the ring finger.&lt;br&gt;&lt;br&gt;&#xA; The one like you is either the index or middle finger. More or less indicates that it could be either one of you, but let's face, there's only one &quot;The Finger&quot; and it's the middle one.&lt;br&gt;&lt;br&gt;&#xA; Your distant cousin is the thumb. It's not exactly a finger, but we always say we have five fingers... not four and a thumb.&lt;br&gt;&lt;br&gt;&#xA; The meager meal is the pinky. It's much smaller, we can only assume it uses less of the total body intake to sustain itself.&lt;br&gt;&lt;br&gt;&#xA; One village holding another refers to people holding hands.&lt;br&gt;&lt;br&gt;&#xA; Striking down the hand could mean a number of things: refusing to shake hands could certainly cause a war, or the striking could be someone physically striking someone, that would &lt;em&gt;definitely&lt;/em&gt; cause a clash! If you mean it literally, then when you slam your fist on the table, pinky down, your ring will clash your pinky, your middle clash your ring, and your index clash your middle, all in succession. &#xA;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2812" LastActivityDate="2014-11-12T21:52:04.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4324" PostTypeId="2" ParentId="4321" CreationDate="2014-11-12T21:55:25.700" Score="11" Body="&lt;p&gt;It is possible like so: &amp;nbsp;&amp;nbsp;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HYGQJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5051" LastEditorDisplayName="user20" LastEditDate="2015-01-28T06:18:43.993" LastActivityDate="2015-01-28T06:18:43.993" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4326" PostTypeId="1" AcceptedAnswerId="4878" CreationDate="2014-11-12T22:16:32.157" Score="15" ViewCount="2246" Body="&lt;p&gt;This is a very hot summer in Rome and you don't know what to do. You are the best detective in town but no crime seems to happen. You are all the day in front of your PC trying to solve some puzzles about parties and elephants but you can't find any solution so far. You are on the way to solving the 20th when a young policeman rushes into your room and gives you a letter. &lt;br/&gt;&lt;br/&gt;&#xA;&quot;Get the hell out! I had the solution of that devilish puzzle and now it is gone! Get out!&quot;&lt;br/&gt;&lt;br/&gt;&#xA;He can't understand your frustration and closes the door as fast as he could. &#xA;Anyway, you opened up the letter and started to read:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Rome, 22/7/14&#xA;AHAHA Mr. Rossi, I have planted a bomb into the Colosseum! It will explode tonight at 24:00. &#xA;I know you are a puzzle-lover, so I would give you a chance to defuse it. Go there and&#xA;and follow the instructions, every clue is important, so be focused! &#xA;You will find a 6-fold piece of paper with some hints to crack the bomb!&#xA;Good Luck!&#xA;The evil bomber&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&quot;Accipicchia [Oh my gosh]&quot; you said &quot;I need to move!&quot; &lt;br/&gt;&lt;br/&gt;You started to run to the Colosseum and after 20 minutes you found the bomb and the instruction. There is a keyboard and a monitor linked to the bomb. On the screen, you can see a textbox which allows you to type in just 4 characters and big red text saying: &quot;Enter the password&quot;.&lt;br/&gt;&#xA;You opened up the piece of paper and started to read the hint.&lt;br/&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Let's do some math: &#xA;2^2 = ...&#xA;3^2 = ... &#xA;5^2 = ...&#xA;7^2 = ...&#xA;11^2 = ...&#xA;You can choose to stop here or keep on with all the prime numbers you can evaluate!&#xA;Now you should know the password!&#xA;The evil bomber&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You started to think about for a while and after 20 minutes you started to type the solution that came up in your mind. After you pressed enter, on the screen you saw &lt;strong&gt;&quot;The bomb has been defused!&quot;&lt;/strong&gt;&lt;br/&gt;&lt;br/&gt;&#xA;&quot;Perbacco [My goodnees], I made it! Now I need to find the solution of the 20th party!&quot;.&lt;br/&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The puzzle:&lt;/strong&gt;&lt;br/&gt;&#xA;What is the password and how did he come up with that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;br/&gt;&#xA;Let's add some hints to help you to find the right solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #1&lt;/strong&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The evil bomber is a PhD in Math, but he did not find any job and he started his career as a terrorist. Loving numbers, every number he gave are there for a reason, but one, just to make him struggle a little more.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #2&lt;/strong&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mr. Rossi is a fan of Math History, he loves &lt;a href=&quot;https://hsm.stackexchange.com/&quot;&gt;HSM&lt;/a&gt; too. He found in a while which &quot;subject&quot; of the hints the evil bomber gave to him. It was easy, every numbers are linked univocally to that subject. It was easy to type the right answer this way.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #3&lt;/strong&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; During his PhD, the evil bomber studied a lot &lt;a href=&quot;http://mathworld.wolfram.com/EulerProduct.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Euler&lt;/a&gt;, Reimann, and their &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_zeta_function&quot; rel=&quot;nofollow noreferrer&quot;&gt;Euler-Reimann Z function.&lt;/a&gt; He found with joy they are related to &lt;a href=&quot;http://en.wikipedia.org/wiki/Pi#Number_theory_and_Riemann_zeta_function&quot; rel=&quot;nofollow noreferrer&quot;&gt;PI&lt;/a&gt;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reference&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can find at &lt;a href=&quot;http://it.wikipedia.org/wiki/Pi_greco#Analisi&quot; rel=&quot;nofollow noreferrer&quot;&gt;prodotto di Eulero&lt;/a&gt; the relation between primes and $\pi$. If you want more, you can see this &lt;a href=&quot;http://en.wikipedia.org/wiki/Pi#Number_theory_and_Riemann_zeta_function&quot; rel=&quot;nofollow noreferrer&quot;&gt;one&lt;/a&gt; and read the demostration &lt;a href=&quot;http://mathworld.wolfram.com/EulerProduct.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&#xA;Anyway guys, I actually hope you had fun with this riddle as I had! &lt;/p&gt;&#xA;" OwnerUserId="1685" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:33.120" LastActivityDate="2014-11-26T03:47:13.507" Title="Defuse the bomb, Mr. Rossi!" Tags="|calculation-puzzle|password|" AnswerCount="11" CommentCount="21" ContentLicense="CC BY-SA 3.0" />
  <row Id="4327" PostTypeId="1" AcceptedAnswerId="4377" CreationDate="2014-11-12T22:18:50.003" Score="11" ViewCount="1817" Body="&lt;p&gt;Keeping the ball rolling from pacoverflow's post &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4139/pacoverflow-riddle-poem/&quot;&gt;here&lt;/a&gt;, I've written a similar riddle. Never having done this before, I hope I've managed to strike a balance between being too easy and being impossibly difficult to anyone other than me! &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Owned by millions, seen by all,&#xA;Borne by some, and used by most.&#xA;Be careful you don't blow it all,&#xA;When it's your turn to make a toast.&#xA;&#xA;By day it gleams if you take heed,&#xA;At nighttime, cold as morning dew.&#xA;Although it may seem hard to read,&#xA;Its countenance is clear to you.&#xA;&#xA;Salute with it, it will surpass,&#xA;A bow to temper, or a kneel.&#xA;Now hearing many forms, at last,&#xA;Its true identity reveal.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2014-12-13T00:03:53.327" Title="Owned by millions, seen by all .... What is it?" Tags="|riddle|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4331" PostTypeId="2" ParentId="4149" CreationDate="2014-11-12T22:40:30.647" Score="11" Body="&lt;p&gt;The &quot;correct&quot; solution for reference purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My proper respects to all the thoughtful and creative answers. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that you can keep a spoiler locked to &quot;visible&quot; by clicking on it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/YOyvV.png&quot; alt=&quot;Famous Family&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2014-11-12T22:40:30.647" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4332" PostTypeId="2" ParentId="4281" CreationDate="2014-11-12T22:41:54.000" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; hate &lt;br&gt; 1) I think of 'age' and I realize it sounds a lot like 'h'. &lt;br&gt; 3) Slice the 'tie' apart? Ok, then that should be the &quot;t&quot;. &lt;br&gt; 4) I can see the 'e' in the 'end', although I'm not sure if I interpreted this correctly. &lt;br&gt; 2) No clue needed like you said: 'a'. Plus, if I start again I'm right at the beginning, and the first letter of the alphabet is 'a'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4427" LastEditorUserId="4427" LastEditDate="2014-11-12T22:59:03.383" LastActivityDate="2014-11-12T22:59:03.383" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4334" PostTypeId="1" AcceptedAnswerId="4355" CreationDate="2014-11-12T22:47:54.380" Score="36" ViewCount="2829" Body="&lt;p&gt;Cut the shape below in four congruent pieces.  The gray area is a hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YJmq1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In non-mathematical terms, cut the white area in 4 pieces having the same shape, same size, possibly mirrored or rotated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: the hole is not randomly placed.  It is in the exact size, shape and position where it needs to be for the solution to work.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2017-07-09T15:20:53.657" LastActivityDate="2017-07-09T15:20:53.657" Title="Cut the disk with a hole in four equal pieces" Tags="|mathematics|geometry|dissection|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4337" PostTypeId="2" ParentId="4299" CreationDate="2014-11-12T23:17:16.453" Score="8" Body="&lt;ol&gt;&#xA;&lt;li&gt;PSE very likely stands for &lt;strong&gt;p&lt;/strong&gt;uzzles.&lt;strong&gt;s&lt;/strong&gt;tack&lt;strong&gt;e&lt;/strong&gt;xchange, hence I'm starting right here.&lt;/li&gt;&#xA;&lt;li&gt;The statement &lt;code&gt;2 + 2 ≠ 4&lt;/code&gt; is &lt;em&gt;wrong&lt;/em&gt;. The opposite of wrong is &lt;em&gt;right&lt;/em&gt;. The opposite of right is &lt;em&gt;left&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;li&gt;.5 is rounded &lt;em&gt;up&lt;/em&gt; (usually, on pen and paper).&lt;/li&gt;&#xA;&lt;li&gt;Up (proper) is down (1), and down (1) is up (2). Ergo up (2) is up (proper), and we can deduce that down (2) is therefore down (proper). Hence by going down (2), I go &lt;em&gt;down&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;li&gt;You're on the &lt;em&gt;right&lt;/em&gt; track.&lt;/li&gt;&#xA;&lt;li&gt;I've gone up, down, right, and left, thus I'm right back where I started: on puzzles.stackexchange! :D&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2014-11-12T23:17:16.453" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4339" PostTypeId="5" CreationDate="2014-11-12T23:38:55.000" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-11-12T23:38:55.000" LastActivityDate="2014-11-12T23:38:55.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4340" PostTypeId="4" CreationDate="2014-11-12T23:38:55.000" Score="0" Body="A puzzle in which involves cutting some sort of shape or object into a number of pieces." OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-11-13T13:40:38.480" LastActivityDate="2014-11-13T13:40:38.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4341" PostTypeId="1" AcceptedAnswerId="4772" CreationDate="2014-11-12T23:44:55.523" Score="14" ViewCount="2992" Body="&lt;p&gt;On the warm and windy night of November 12, Joe was coming back home. He ordered pizza from the car, to be delivered at 8 PM. Once he got back, he realized that one of his windows was slightly ajar. He concluded that he just left it open for ventilation.&lt;/p&gt;&#xA;&lt;p&gt;Once he got to the front door, he taped a note for the pizza guy that would arrive shortly. As he unlocked his door and walked into the house, a loud crash came from the basement. He thought that the cat somehow got locked in and was trying to escape. He walked over to the door and turned the handle. As soon as the door creaked open, he fell over, dead. The unknown assassin walked over to him and watched him bleed out. The last thing Joe saw was a masked and gloved man standing over him, laughing as the last of Joe's life left him...&lt;/p&gt;&#xA;&lt;p&gt;Later on, &lt;a href=&quot;https://puzzling.stackexchange.com/users/39/doorknob&quot;&gt;Doorknob&lt;/a&gt; arrived with Joe's pizza. As Doorknob walked towards the house, he saw a note on the door saying, &amp;quot;Come in, I'm in the dining hall. Second door to the left.&amp;quot; He went into the house. After he walked past the first door, all the lights went out, except for one. As Doorknob walked into the dining room, he saw the only light on. That light was pointed right onto the table, where Joe's body was.&lt;/p&gt;&#xA;&lt;p&gt;Doorknob couldn't believe his own eyes. He went to go investigate the body, and saw three things:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A tattoo on his chest saying: &amp;quot;An eye for an eye. Is it not?&amp;quot;&lt;/li&gt;&#xA;&lt;li&gt;A single bullet&lt;/li&gt;&#xA;&lt;li&gt;A loaded pistol with something carved into the side.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Shocked and horrified, Doorknob ran out of the house and to the police. He reported the crime.&lt;/p&gt;&#xA;&lt;p&gt;The next morning, Doorknob was certain of where the murderer came from. He only needed to find who did it. After a bit of searching, he discovered the murderer and apprehended him.&lt;/p&gt;&#xA;&lt;p&gt;Who was the murderer?&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Bonus:&lt;/strong&gt; What's carved into the side of the gun?&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Hints Revealed:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;1. This is about &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4263/the-serial-killer-part-2?rq=1&quot;&gt;AskUbuntu&lt;/a&gt;&lt;br&gt;&#xA;2. The Joe is the same Joe that posted the above question.&lt;br&gt;&#xA;3. Treat everything in this post as a hint.&lt;br&gt;&#xA;4. A certain pokemon is not to be blamed here.&lt;br&gt;&#xA;5. Actual detective work is needed for this puzzle.&lt;br&gt;&#xA;6. The killer had plenty of time to prepare.&lt;br&gt;&#xA;7. Doorknob is irrelevant.&lt;br&gt;&#xA;8. While AskUbuntu was murdered, something else survived...&lt;br&gt;&#xA;9. You may assume that everything in the hints are true.&lt;br&gt;&#xA;10. (Gun-Related) The gun has a logo etched into it..&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2135" LastEditorUserId="57521" LastEditDate="2021-11-23T21:47:33.960" LastActivityDate="2021-11-23T21:47:33.960" Title="The SErial Killer's killer" Tags="|situation|" AnswerCount="5" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="4342" PostTypeId="1" CreationDate="2014-11-12T23:54:55.260" Score="1" ViewCount="8287" Body="&lt;p&gt;In Faraway Land there is a monastery built atop a mountain in which monks live isolated from the rest of the world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The monastery is made of two large buildings: one is where the monks sleep, the other is where they all spend the days together, working or minding their own business.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this monastery, there is a rule that forbids the monks to communicate with each other, whether it's through writing, talking, or body language, they are not allowed to communicate at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One day, the master of the monastery decides to exceptionally break that rule to inform his fellow monks of one important fact:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is an illness among us. It is fortunately not contagious nor particularly dangerous, but I would like every single one of you to remain in bed for an entire day once you know you have contracted it.&#xA;  The illness has only one recognizable symptom: it causes an easily visible red dot to appear on one's forehead.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, there are no mirrors in the monastery and nothing else that might help the monks see for themselves if they have contracted the disease or not. Fortunately, every single monk is highly intelligent and wouldn't require any help in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seven days pass without a single monk missing. After the seventh day (on the eight day), all sick monks suddenly remain in bed resting just as the master had instructed them to. How many monks were ill and how did they find out?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You may assume the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How many monks live in the monastery is irrelevant to the puzzle.&lt;/li&gt;&#xA;&lt;li&gt;All of the sick monks were already sick and had visible symptoms when the master made his announcement, no other monk fell sick since then.&lt;/li&gt;&#xA;&lt;li&gt;None of the monks had visible symptoms the day &lt;em&gt;before&lt;/em&gt; the master made his announcement.&lt;/li&gt;&#xA;&lt;li&gt;&quot;The first day&quot; is the day the announcement was made. &quot;After one day&quot; means the first day has passed and it is now the second day.&lt;/li&gt;&#xA;&lt;li&gt;There was at least one sick monk.&lt;/li&gt;&#xA;&lt;li&gt;The master is to be considered a monk like every other.&lt;/li&gt;&#xA;&lt;li&gt;Whether the master was among the sick monks is irrelevant.&lt;/li&gt;&#xA;&lt;li&gt;The master broke the rule to make his announcement only, nothing else.&lt;/li&gt;&#xA;&lt;li&gt;The monks do not communicate, but they see every other monk every single day, therefore they know exactly who was missing and who wasn't, as well as who was ill and who wasn't.&lt;/li&gt;&#xA;&lt;li&gt;The monks never forget. Ever.&lt;/li&gt;&#xA;&lt;li&gt;Only the sick monks stayed in bed on the eight day and nobody else.&lt;/li&gt;&#xA;&lt;li&gt;There is no third party involved in this puzzle, the monastery is completely isolated from the outside world.&lt;/li&gt;&#xA;&lt;li&gt;No monk tried to leave and no new monk arrived to the monastery during those seven days.&lt;/li&gt;&#xA;&lt;li&gt;The monks' solution is the best solution given their circumstances.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5189" LastActivityDate="2014-11-12T23:54:55.260" Title="An illness at a monastery" Tags="|logical-deduction|" AnswerCount="0" CommentCount="4" ClosedDate="2014-11-13T01:10:39.463" ContentLicense="CC BY-SA 3.0" />
  <row Id="4343" PostTypeId="2" ParentId="4341" CreationDate="2014-11-13T00:17:51.317" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's someone from the pizza place - how else would they know to leave a note telling the delivery boy where to find the body?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4159" LastActivityDate="2014-11-13T00:17:51.317" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4346" PostTypeId="1" AcceptedAnswerId="4349" CreationDate="2014-11-13T01:03:03.273" Score="11" ViewCount="2379" Body="&lt;p&gt;You have two men in a room. You know for a fact that one is a physicist and the other is a plumber. Sadly both men have recently been rendered totally deaf and haven't learned sign language. This is a problem because you have to tell which one is the physicist who can bring the run-away LHC under control and which is the plumber here to unclog the toilet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a piece of paper and a pen. What single word do you write down and ask them to say to determine which is which? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt; Anyone that says plumber or physicist shall be squashed under an irrelevant elephant.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="4951" LastEditorUserId="5373" LastEditDate="2016-11-20T14:37:26.403" LastActivityDate="2020-08-01T22:08:53.337" Title="Two Men - The Physicist and Plumber" Tags="|word|situation|" AnswerCount="6" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="4347" PostTypeId="2" ParentId="4281" CreationDate="2014-11-13T01:07:45.960" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; SWAG&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;SOLD - (to get rid of) minus OLD&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Split Bowtie , third letter&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Look at beginning of AGE&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Start A-Gain after fall - after a Fall to start to Gain&lt;/p&gt;&#xA;&lt;p&gt;You did not know you could feel this ? Nether did I :&#xA;&lt;a href=&quot;http://www.urbandictionary.com/define.php?term=swag+level&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.urbandictionary.com/define.php?term=swag+level&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5201" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-11-13T01:07:45.960" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4348" PostTypeId="2" ParentId="4139" CreationDate="2014-11-13T01:15:44.677" Score="2" Body="&lt;p&gt;Congratulations to Apd1!  &lt;a href=&quot;https://puzzling.stackexchange.com/a/4291/4450&quot;&gt;His answer&lt;/a&gt; is what I was looking for - the poem is referring to a &lt;a href=&quot;http://en.wikipedia.org/wiki/Regular_expression&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;regular expression&lt;/strong&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will only mention the differences between my thinking and his thinking:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Otherwise it can make things bounded.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise if it's not a &lt;code&gt;|&lt;/code&gt;, then it can be something else such as &lt;code&gt;\b&lt;/code&gt; which sets a boundary on the match.  What Apd1 wrote works too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;It can be seen leaning, That's when it reveals true meaning.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thinking here is the opposite of what Apd1 wrote.  Special characters need to be escaped with &lt;code&gt;\&lt;/code&gt;.  If you just have &lt;code&gt;*&lt;/code&gt; in your regular expression then you are matching 0 or more occurrences of the previous item.  But if you have &lt;code&gt;\*&lt;/code&gt; in your regular expression, then you are matching the asterisk - the true meaning of the character.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;It can show uncertainty&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just referring to the fact that one of the metacharacters is &lt;code&gt;?&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;but never incoherence.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regular expressions follow rules, so they are always perfectly logical.  This is what Apd1 wrote in different words (he said he wasn't sure about that one).&lt;/p&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2014-11-13T01:15:44.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4349" PostTypeId="2" ParentId="4346" CreationDate="2014-11-13T01:19:59.537" Score="25" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Unionized&quot;. The idea is that the physicist will pronounce according to the meaning &quot;not ionized&quot;, while the plumber will treat it as &quot;formed a union&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5203" LastActivityDate="2014-11-13T01:19:59.537" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4350" PostTypeId="2" ParentId="4346" CreationDate="2014-11-13T01:21:31.670" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Unionised:&#xA; Plumber will pronounce it as the boolean operation and the physicist as not an ion but an atom&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="255" LastActivityDate="2014-11-13T01:21:31.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4351" PostTypeId="1" CreationDate="2014-11-13T01:26:48.253" Score="4" ViewCount="567" Body="&lt;p&gt;Joe likes 2025 but not 2024; he likes 529 but not 530; he likes 400 but not 500.&#xA;Which does Joe like - 3600 or 3700? Why?&lt;/p&gt;&#xA;" OwnerUserId="5204" LastEditorUserId="6178" LastEditDate="2014-12-28T14:58:08.220" LastActivityDate="2014-12-28T14:58:08.220" Title="What's the sequence?" Tags="|pattern|logical-deduction|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4352" PostTypeId="2" ParentId="4351" CreationDate="2014-11-13T01:32:11.920" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He likes 3600 because that is a perfect square (60*60) like the other numbers.&#xA;&lt;P&gt;&#xA; 2025 = 45*45&#xA;&lt;P&gt;&#xA; 529 = 23*23&#xA;&lt;P&gt;&#xA; 400 = 20*20&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4450" LastEditDate="2014-11-13T01:38:54.910" LastActivityDate="2014-11-13T01:38:54.910" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4354" PostTypeId="2" ParentId="4351" CreationDate="2014-11-13T02:08:11.290" Score="8" Body="&lt;p&gt;Another response similar to the answer given earlier&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;When you add the digits in the numbers he likes it results in perfect squares so he likes 3600&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;2+0+2+5 =9&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;5+2+9=16&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;4+0+0=4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="2313" LastEditDate="2014-11-13T02:33:43.907" LastActivityDate="2014-11-13T02:33:43.907" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4355" PostTypeId="2" ParentId="4334" CreationDate="2014-11-13T02:31:38.700" Score="50" Body="&lt;p&gt;Here's one solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/lyGcd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I assumed the radius of the hole's curvature matches the curvature radius of the circle, the hole's straight side is equal to the circle's radius, and its curved edges meet the straight edge at right angles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Used Flash Pro to solve this, of all things: it has the useful feature that as you paint while editing an object, it updates all instances. So I made 4 instances, then made incremental changes to the source outline and instance positions, and converged pretty quickly on a solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could imagine a genetic or hill-climbing algorithm using the same process.&lt;/p&gt;&#xA;" OwnerUserId="5197" LastEditorUserId="5197" LastEditDate="2014-11-13T07:16:52.507" LastActivityDate="2014-11-13T07:16:52.507" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4356" PostTypeId="1" AcceptedAnswerId="4368" CreationDate="2014-11-13T03:02:58.287" Score="7" ViewCount="1226" Body="&lt;p&gt;In one form I am very tall, in another, elegant &lt;br/&gt;&#xA;In one form I am happiness, in another, an eyesore &lt;br/&gt;&#xA;In one form I am metal, in another, paper &lt;br/&gt;&#xA;In one form Sasaki's lament, in another Greek ingenuity&lt;br/&gt;&#xA;In one form I am kung-fu, in another I am strong and static.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As usual, please post your answer in spoiler tags. For bonus magical unicorn points, what does each line mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sasaki is the last name of a famous Japanese girl.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4951" LastEditorUserId="4951" LastEditDate="2014-11-13T05:34:31.300" LastActivityDate="2014-11-13T05:39:03.320" Title="What Am I - Riddle Part 7" Tags="|riddle|word-problem|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4357" PostTypeId="2" ParentId="4280" CreationDate="2014-11-13T03:07:29.090" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;GOD. &lt;br&gt; Whether [a] God is real is up to personal belief. If flipped, it becomes &quot;dog&quot;, which is something that is definitely real and can bite you. &lt;br&gt; A Polo looks like the letter &quot;O&quot;. I don't know what the lines about spitting or splitting a 'POLO' refer to, but if &quot;god&quot; &quot;swallows&quot; an O, it becomes &quot;good&quot;, which is certainly fully good.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5211" LastActivityDate="2014-11-13T03:07:29.090" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4359" PostTypeId="2" ParentId="4326" CreationDate="2014-11-13T03:42:21.400" Score="4" Body="&lt;p&gt;I'm just going to try..&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; NONE, since anything raised to a number except 0 or 1 have multiples and cant be a prime number&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5214" LastActivityDate="2014-11-13T03:42:21.400" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4360" PostTypeId="1" AcceptedAnswerId="4944" CreationDate="2014-11-13T03:48:49.170" Score="11" ViewCount="1837" Body="&lt;p&gt;This is Part 5 of the Murder of the President brainteaser/riddle series. If you have not already, check out the answers for &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3600/murder-of-the-president-part-1&quot;&gt;Murder of the President - Part 1&lt;/a&gt; posted by Nit, &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3845/murder-of-the-president-part-2&quot;&gt;Murder of the President - Part 2&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3977/murder-of-the-president-part-3&quot;&gt;Murder of the President - Part 3&lt;/a&gt; both posted by Joe Z., and &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4119/murder-of-the-president-part-4&quot;&gt;Murder of the President - Part 4&lt;/a&gt; posted by el jefe.  Each part will give you a clue and you must solve it. This part will have several clues. Use all knowledge you have of cryptography, ciphers, past puzzles, etc. You should also use Google. This case is meant to take place in the present day, so all politicians, celebrities, places, etc. are who they are now. Please post your answers in spoiler tags.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here's the riddle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You arrive at the London NYC Hotel at 2:26 a.m. You've already called the staff at the hotel to let them know you were coming. On the way up, you were beginning to hear reports that the President was murdered. You head up to room 118. You look around for a bit until you find a box under the covers. On top of the box is a note.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;nvnmjoonhenuan&lt;br&gt;&#xA;  m&lt;sup&gt;&lt;strong&gt;2&lt;/strong&gt;&lt;/sup&gt;m&lt;br&gt;&#xA;  121&lt;br&gt;&#xA;  0+1&lt;br&gt;&#xA;  ttfflhesi&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You check out the padlock. It is a one 5 digit code. Under the box there is another note.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Padlock:&lt;br&gt;&#xA;  ZUTYS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;After you figure out this code, you open up the box and find, of course, 6 coded notes. Some notes are typed, others are handwritten. (Italics will indiacte &lt;em&gt;handwritten&lt;/em&gt; and bold will indicate &lt;strong&gt;typed&lt;/strong&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the notes&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;#1&lt;br&gt;&#xA;  ERP BTRCBIV PTUKU GTI CUBL&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;#2&lt;br&gt;&#xA;  WAPHFS RPL MU UMK SQOF GWAWFGX UCTME UPCG&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;#3&lt;br&gt;&#xA;  M DT AJF MMUPIS - YQ&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;#4&lt;br&gt;&#xA;  SAQEHY WQG NT XLF VMOIA DUWI&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;#5&lt;br&gt;&#xA;  BVB YPPX LEXDK NJ&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;#6&lt;br&gt;&#xA;  VVHC ASL FNOXIS JETM ZVVOIU ZWTJPK&lt;br&gt;&#xA;  LLIFVFXZIDN&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the code to get into the box? What do all the notes mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good Luck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The messages are coded using the method and seed answered by el jefe. The sequence of numbers is continuous (it doesn't restart for each message). Also,  you must add one to every zero in the sequence&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: I will be posting Part 6 (the final part) in one to two days. I will select the correct answer for Part 5 before, but I will post the answer for Part 5 if no one gets it&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1711" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2014-12-15T23:02:27.867" Title="Murder of the President - Part 5" Tags="|cipher|mystery|" AnswerCount="3" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="4361" PostTypeId="1" AcceptedAnswerId="4363" CreationDate="2014-11-13T04:23:48.610" Score="1" ViewCount="306" Body="&lt;p&gt;A party is being held at a local mansion. The host is very rich and his success is because of one thing - his famous recipe for Spaghetti! By the way, this guy is so rich that he gives a party everyday.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only guests that may attend are people who correctly reply to the guard at the door. They always use a scheme where they concoct a function as the password, so when the guard gives a number, the guest computes the result and is allowed in if, and only if, the answer is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, uninvited people with some math degrees always were able to sneak in to every party that was being held because they were smart enough to guess the function used. They are not stopped even if some of them happen to guess it wrong and die in the process; they simply never stop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today the party is special - the rich man will premiere his recipe for Fettuccine to a set of very distinctive guests! Looks like that he is expanding his Spaghetti businesss into other markets. To ensure that no uninvited math nerd slips in again, he looked for some function that the math nerds would not be able to guess and would not find in their PhD books or scientific articles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's where you come in. You and a friend are trying to steal this new recipe. You selected him because he won a gold medal in a competition about function regression and holds the current world record for it. Further he has a PhD thesis and numerous acclaimed Post-doc works about breaking numeric codes. Also, he works very hard, 18 hours per day and 7 day-a-week in the NSA to ensure that the US government can illegally read the e-mails of everybody in the world. You two sneak by and listen to the passwords.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first guest arrives. The guard says &quot;10&quot;, and the guest replies &quot;8&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second guest arrives. The guard says &quot;100&quot;; the guest replies &quot;7&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third guest arrives. The guard says &quot;41&quot;; the guest replies &quot;5&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The forth guest arrives. The guard says &quot;113&quot;; the guest replies &quot;7&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fifth guest arrives. The guard says &quot;79&quot;; the guest replies &quot;8&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your friend then suddenly has an idea. He starts to quickly scratch some formulas on a paper and then says: &quot;Eureka, got it! They are using a composite of the 3rd-order Friedman-Jumlach partial derivative vertex inverse function!&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So he goes to the guard and is given &quot;39&quot;. He very confidently answers &quot;6&quot;. Then he is poisoned by a Venomoth, eletrocuted by a Pikachu, burned by a Charizard, drowned by a Squirtle, and finally has his bones crushed to dust by a Snorlax.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another guest arrives. The guard says &quot;131&quot;, and the guest replies &quot;6&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And still other guest arrives. The guard says &quot;63&quot;; the guest replies &quot;4&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another guest. The guard says &quot;25&quot;; the guest replies &quot;7&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You, who never had good math scores in school because you were playing video games all day long, and don't even remember what the Bhaskara's formula or the Pythagoras theorem is, think a bit and then guess that you might have got it. So you walk up to the door and the guard says &quot;177&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What should you respond?&lt;/p&gt;&#xA;" OwnerUserId="5044" LastEditorUserId="4204" LastEditDate="2014-11-13T05:17:30.060" LastActivityDate="2014-11-13T05:17:30.060" Title="The security to the party [Part 21]" Tags="|password|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4363" PostTypeId="2" ParentId="4361" CreationDate="2014-11-13T04:35:46.097" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The correct response is the number of letters in the name of the Pokémon with that number in the National Pokédex (except this doesn't seem to work for the first example). &lt;br&gt; 10: Caterpie (8 letters) &lt;br&gt; 100: Voltorb (7 letters) &lt;br&gt; 41: Zubat (5 letters) &lt;br&gt; 113: Chansey (7 letters) &lt;br&gt; 79: Slowpoke (8 letters) &lt;br&gt; 39: Jigglypuff (10 letters) &lt;br&gt; 131: Lapras (6 letters) &lt;br&gt; 63: Abra (4 letters) &lt;br&gt; 25: Pikachu (7 letters) &lt;br&gt; Since #177 is Natu, you should answer 4.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5218" LastActivityDate="2014-11-13T04:35:46.097" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4364" PostTypeId="2" ParentId="4281" CreationDate="2014-11-13T04:46:31.937" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LOST:&#xA; T - last letter, cutting off old from told,&#xA; S - third letter, looks like tie unraveled,&#xA; L - roman numeral for fifty, because I doubt X, C, D, M would fit ;) and&#xA; O - o&lt;strong&gt;f all&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5220" LastActivityDate="2014-11-13T04:46:31.937" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4365" PostTypeId="2" ParentId="4326" CreationDate="2014-11-13T04:56:18.403" Score="1" Body="&lt;p&gt;One possibility is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0169&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since the paper was folded 6 times, that probably means that the password is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the square of the sixth prime&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, it took me around 20 seconds to get this answer, not 20 minutes, so perhaps this is not what you're looking for.&lt;/p&gt;&#xA;" OwnerUserId="5095" LastEditorUserId="5095" LastEditDate="2014-11-26T03:47:13.507" LastActivityDate="2014-11-26T03:47:13.507" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4368" PostTypeId="2" ParentId="4356" CreationDate="2014-11-13T05:39:03.320" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A crane?&#xA;&#xA; &lt;ul&gt;&#xA; &lt;li&gt;The machine is tall, the animal, elegant&#xA; &lt;li&gt;I’m not sure about happiness, but metal cranes are eyesores&#xA; &lt;li&gt;Metal cranes, paper cranes&#xA; &lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Sadako_Sasaki&quot;&gt;Sadako Sasaki&lt;/a&gt;&#xA; &lt;li&gt;Surely there’s some kung-fu crane thing. Or maybe not. And again, the machine is strong and static – or maybe this refers to something else?&#xA; &lt;/ul&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1565" LastActivityDate="2014-11-13T05:39:03.320" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4369" PostTypeId="1" AcceptedAnswerId="4392" CreationDate="2014-11-13T05:49:02.610" Score="1" ViewCount="447" Body="&lt;p&gt;A party is being held at a local mansion. The host is very rich and his success is because of one thing ~ his famous recipe for &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Meth&quot; rel=&quot;nofollow&quot;&gt;Spaghetti!&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only guests that may attend are people the host knows and trusts, whom he gave the password to enter to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's where you come in. You and a friend are trying to steal this recipe. You sneak by and listen to the passwords.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first guest arrives. The security says &quot;5&quot;, and the guest replies &quot;5&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second guest arrives. The security says &quot;-6&quot;, the guest replies &quot;19&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third guest arrives. The security says &quot;10.5&quot;, the guest replies &quot;11&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your friend thinks he got it all figured out. He approaches the security and as they say &quot;What time do you call this? Your number is -7&quot;, when he replies &quot;20&quot; &lt;strong&gt;and he's admitted!!!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fourth guest arrives in a cloud of blue smoke obviously late. After apologizing to security for being late he is challenged with &quot;0&quot; and replies &quot;26&quot; and is admitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fifth guest is hot on his heels and arrives. The guard says &quot;Look mate, the letter of the rules here is that you can't come in after 7.00pm but, you are only 3 minutes late so if you get the correct answer then you can come in.&quot; and the challenge is &quot;-3&quot; to which the guest replies &quot;16&quot; and is admitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing your time has run out you walk up to security, wishing your friend had told you the secret to get in and get &quot;You are going to be the last guest I admit in this zone. 3.5&quot; before the guard quickly changes his mind and asks you for &quot;-9.5&quot; instead, what is the correct response and how do you figure it out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HINT:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Look at what the guards are saying. Some of their words are incongruous and will give you a hint on what to focus on. &lt;strong&gt;Note:&lt;/strong&gt; I am a fluent English speaker (Australian born and raised), so any weird mangling of phrases is totally intentional and is a clue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4951" LastEditorUserId="4951" LastEditDate="2014-11-13T08:37:19.030" LastActivityDate="2014-11-13T08:50:44.053" Title="The Security To The party [Part 22]" Tags="|password|" AnswerCount="1" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="4370" PostTypeId="1" AcceptedAnswerId="16317" CreationDate="2014-11-13T06:09:46.713" Score="14" ViewCount="1075" Body="&lt;p&gt;François Hollande, famous Cordon Bleu chef that he (like all French presidents) is, has invited a select few guests to his party, where he will be showing off the secret cheese-making process that doubles as a plan of attack against Britain. Your task as a UK spy (who happens to also be an Argentinian agent) is to steal this recipe for disaster. Entrance to the château wherein lies the party, however, is overseen by two guards. One guard always lies, and the other always tells the truth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You arrive early and hide in some bushes. Guests begin approaching the door, and it becomes clear that there is some kind of password involved.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;One guest arrives by himself. “1”, says a guard. “6”, replies the guest. He is permitted entrance.&lt;/li&gt;&#xA;&lt;li&gt;Two more guests show up. “3” is prompted; “2” is the response, and it is apparently correct.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A group of seven is next to request entry, and the guards see fit to ask them each in turn.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;“2!” “7.”&lt;/li&gt;&#xA;&lt;li&gt;“11!” “6.”&lt;/li&gt;&#xA;&lt;li&gt;“20!” “4.”&lt;/li&gt;&#xA;&lt;li&gt;“4!” “2.”&lt;/li&gt;&#xA;&lt;li&gt;“6!” “3.”&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The group had been doing fine until that last person, whose throat is promptly slit, along with those of what look to be his wife and son. “You were just off by one!”, says one of the guards, who looks like the sort of person who would tell the truth – but then again, it’s hard to tell with them. You’re not in Argentina anymore.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You think you’ve figured out the pattern, though, and are about to take your turn when another guest arrives, late. The challenge is given – “26” – but before the guest can give an answer, a loud gunshot is heard. Someone shouts out the window that the president is dead! It’s a coup d’état! You must get to the plans before the assassin, but one guard is still performing his duty. You are given the number five as a challenge; what is your response?&lt;/p&gt;&#xA;" OwnerUserId="1565" LastEditorUserId="1565" LastEditDate="2018-08-12T05:44:52.837" LastActivityDate="2018-08-12T05:44:52.837" Title="The security to the murdered president’s “what am I?” riddle party [Part TREE(3)]" Tags="|password|" AnswerCount="2" CommentCount="20" ContentLicense="CC BY-SA 3.0" />
  <row Id="4373" PostTypeId="2" ParentId="4281" CreationDate="2014-11-13T06:32:09.777" Score="3" Body="&lt;p&gt;I dont know if this you feel right now but..&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HURT, T from told, R from apart, U from splitting the tie and removing the 3rd part, H from age..&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5214" LastActivityDate="2014-11-13T06:32:09.777" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4375" PostTypeId="2" ParentId="4341" CreationDate="2014-11-13T07:23:19.107" Score="1" Body="&lt;p&gt;The answer is obvious.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After the pain and anguish of watching AskUbuntu's death in &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4263/the-serial-killer-part-2&quot;&gt;a previous puzzle&lt;/a&gt;, you were distraught. Thoughts of retribution whirled through your brain. Grabbing a gun, you used &lt;a href=&quot;http://bulbapedia.bulbagarden.net/wiki/Teleport_%28move%29#By_TM&quot; rel=&quot;nofollow noreferrer&quot;&gt;Teleport&lt;/a&gt; and arrived at Joe's house. Surprising him as he was absorbed in writing another brain teaser, you made him an unwilling star in his own personal murder mystery. &quot;An eye for an eye&quot; - you lost your precious stack, so Joe loses his life. This puzzle is simply a way to throw everyone off your scent. Back in the ball with you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Carved into the gun?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'Gotta Catch This'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4164" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2014-11-13T07:23:19.107" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4376" PostTypeId="1" AcceptedAnswerId="4382" CreationDate="2014-11-13T07:24:14.873" Score="8" ViewCount="508" Body="&lt;p&gt;A party is being held at a local mansion. The host is very rich and his success is because of one thing ~ his famous recipe for Spaghetti!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only guests that may attend are people the host knows and trusts, whom he gave the password to enter to. But, Since over 20 of his secret recipes are stolen, he has decided to change the method a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's where you come in. You and a friends are trying to steal this recipe. You and 2 of your friends sneak behind the trees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But instead of saying the number and asking the password, the guards/security stopped all the guests at the entrance. After ten minutes the Security Head made an announcement: &quot;All the guests are requested to gather here in front of the gate. No body will be allowed to enter if arrive after the procedure&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You and your friends got tensed and the only option in front of you to steal the recipe is taking the chance and go there and stand among the guests hoping you will be able to figure out the password. But one of your friends ran home leaving you and your friend fearing about getting tampered by the elephant and all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You and your friend went there and stood with the guests. There was a total of 16 guests including you and your friend. The guards made all the guests ( along with you and your friend ) to stand in a 4x4 (4 rows and 4 columns) arrangement and provided a paper and pen. You were standing in the 3rd row from front and 3rd column from left. Your friend was standing on your right ( that is 3rd row from front and 2nd column from left )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the Head Guard shouted &lt;strong&gt;&quot;144&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't know what to write and what to do. You started to sweat. You tried to peak to the answers written by some of the guests standing near you. You were able to see the answers written by a guest standing your left, front left (diagonal position), front, front-right (diagonal position), and your friend standing on your right.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The guest on your left wrote : 12&lt;/li&gt;&#xA;&lt;li&gt;The guest on your front-left wrote : 18&lt;/li&gt;&#xA;&lt;li&gt;The guest on your front wrote : 24&lt;/li&gt;&#xA;&lt;li&gt;The guest on your front-right wrote : 36&lt;/li&gt;&#xA;&lt;li&gt;Your friend (on your right) wrote : 52&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now, What number you will write in order to get into the party and steal the recipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Additional Question&lt;/strong&gt;: Did your friend wrote the correct password, or did he get tampered by the elephants? &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: Please place your answers in spoiler tags. Also mention how you get to the answer.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Feel free to edit the question to fix grammatical errors. ;)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint (Check it only if you can't find the answer without any hints):&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The position of guests whose answers you saw during the procedure are mentioned relative to your position. Try finding there actual position first and then try solving)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;New Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PoPvs.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The one with the &lt;strong&gt;big mustache&lt;/strong&gt; is the Head Guard. You are the one in &lt;strong&gt;red&lt;/strong&gt; color and your friend is the one in &lt;strong&gt;blue&lt;/strong&gt;, that is at your right, considering you are facing towards the guard. The rows and columns are mentioned in the question from the Head Guards's perspective (To make the solution a little more difficult). That is row 1 is the first row in front of him. Column 1 is the left most column in front of him.&lt;/p&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="2730" LastEditDate="2015-12-15T07:07:20.940" LastActivityDate="2015-12-15T07:07:20.940" Title="The Security To The party [Part 23]" Tags="|mathematics|logical-deduction|password|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="4377" PostTypeId="2" ParentId="4327" CreationDate="2014-11-13T07:43:46.893" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Glass?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Owned by millions, seen by all,&#xA;Borne by some, and used by most.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Needless to comment&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Be careful you don't blow it all,&#xA;When it's your turn to make a toast.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We toast with a glass in hand... And it happened before (to me too) that a powerful toast would crash the glass&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By day it gleams if you take heed,&#xA;At nighttime, cold as morning dew.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again needless to say...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Although it may seem hard to read,&#xA;Its countenance is clear to you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We uses glasses to read when it gets hard to do it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Salute with it, it will surpass,&#xA;A bow to temper, or a kneel.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Glass &quot;bows&quot; and &quot;kneels&quot; to high temperatures ( &lt;a href=&quot;http://en.wikipedia.org/wiki/Toughened_glass&quot;&gt;http://en.wikipedia.org/wiki/Toughened_glass&lt;/a&gt; )&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now hearing many forms, at last,&#xA;Its true identity reveal.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Can be shaped into many forms&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5122" LastEditorUserId="5122" LastEditDate="2014-11-13T08:13:44.970" LastActivityDate="2014-11-13T08:13:44.970" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4379" PostTypeId="2" ParentId="4327" CreationDate="2014-11-13T07:56:09.563" Score="2" Body="&lt;p&gt;I thought about it for some minutes. And it seems to fit all the clues. How about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Money &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For sure it will go away and you can blow it all at once. Created by some, used by most, and so on...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Coins &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;can be hard to read but you know their value. At night they will feel cold due to their material.&lt;/p&gt;&#xA;" OwnerUserId="5106" LastActivityDate="2014-11-13T07:56:09.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4380" PostTypeId="1" AcceptedAnswerId="4381" CreationDate="2014-11-13T07:58:10.483" Score="10" ViewCount="791" Body="&lt;p&gt;I hope you guys like this one:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;If you have me, you don't have&#xA;If you give me, you will displease&#xA;You can't make me, feel at ease&#xA;For sure not trace me on a graph&#xA;&#xA;You could as well not read the rhymes&#xA;but you'll find me between the lines&#xA;&#xA;What am I?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It's pretty easy I think!&lt;/p&gt;&#xA;" OwnerUserId="5122" LastEditorUserId="4637" LastEditDate="2016-05-05T11:11:07.743" LastActivityDate="2016-05-05T11:11:07.743" Title="If you have me, you don't have" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4381" PostTypeId="2" ParentId="4380" CreationDate="2014-11-13T08:02:43.853" Score="9" Body="&lt;p&gt;Maybe it's nothing but could it be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;nothing&quot; ? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5106" LastActivityDate="2014-11-13T08:02:43.853" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4382" PostTypeId="2" ParentId="4376" CreationDate="2014-11-13T08:14:40.793" Score="5" Body="&lt;p&gt;My guess is this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is the layout of the numbers:&lt;br&gt;&#xA; &lt;br&gt;&#xA; &lt;code&gt;???    ???    ???    ???&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;18     24     36    ???&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;12    YOU     52    ???&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;???    ???    ???    ???&lt;/code&gt;&lt;br&gt;&#xA; &lt;br&gt;&#xA; 12, 18, 24 and 36 all divides 144. 144 is a number with many divisors (144=2*2*2*2*3*3).&lt;br&gt;&#xA; Well, 12=18*2/3, 24=18*4/3, So, you=24*2/3=12*4/3=&lt;strong&gt;16&lt;/strong&gt;&lt;br&gt;&#xA; And the friend is doomed, 52=2*2*13, and 13 is not a divisor of 144.&lt;br&gt;&#xA; Since 36=24*3/2, his number should be 36*2/3=16*3/2=&lt;strong&gt;24&lt;/strong&gt;&lt;br&gt;&#xA; &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;EDITED: New layout.&lt;/p&gt;&#xA;" OwnerUserId="5044" LastEditorUserId="5044" LastEditDate="2014-11-13T08:51:42.510" LastActivityDate="2014-11-13T08:51:42.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4383" PostTypeId="1" AcceptedAnswerId="4495" CreationDate="2014-11-13T08:24:12.167" Score="4" ViewCount="340" Body="&lt;p&gt;I'm a length of hard wood&lt;br&gt;&#xA;Hooped with iron at each end&lt;br&gt;&#xA;Used for making a block&lt;br&gt;&#xA;Or a case to defend&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My sound is like silence:&lt;br&gt;&#xA;Dumb, irrelevant am I&lt;br&gt;&#xA;Talk about me forever&lt;br&gt;&#xA;For I point to the sky&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Hint 1&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I am a single, four-letter word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Hint 2&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The first clue is archaic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Hint 3&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The title is a clue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Hint 4&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;But for the difference in sound between a cat's and a cow's, the &quot;sound&quot; clue is homophonic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="2819" LastEditDate="2014-11-14T13:49:17.910" LastActivityDate="2014-11-14T13:49:17.910" Title="An interesting point?" Tags="|riddle|word|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4384" PostTypeId="1" AcceptedAnswerId="4415" CreationDate="2014-11-13T08:31:38.533" Score="36" ViewCount="25409" Body="&lt;p&gt;...but first you have to find the solution!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two friends (Andrew and Gabriel) go to another friend (Ellen) and claim to be able to read each other's mind.&#xA;Ellen doesn't believe that and wants to test their ability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Andrew steps away from Gabriel and Ellen, gets blindfolded and waits. Meanwhile Ellen whispers something in the ear of Gabriel.&#xA;Gabriel then says: &quot;ok, let me estabilish the mental link, are you ready Andrew?&quot;&#xA;&quot;Yes I am&quot; replies the friend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then Gabriel begins to ask always with the same voice tone:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Am I thinking of a fun park?&quot;&#xA;&quot;Nope&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Am I thinking of person?&quot;&#xA;&quot;Nope&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Am I thinking of Bavarian sky?&quot;&#xA;&quot;Nope&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Am I thinking of love?&quot;&#xA;&quot;Nope&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Am I thinking of a spider?&quot;&#xA;&quot;Nope&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Am I thinking of StackExchange?&quot;&#xA;&quot;Nope&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Am I thinking of a car?&quot;&#xA;&quot;Nope&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Am I thinking of a muzzle?&quot;&#xA;&quot;Yes&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was a muzzle. Ellen doesn't believe her ears and wants to test it again, this time to understand if it works both ways. After the same preparation, Andrew starts to ask:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Am I thinking of playing soccer?&quot;&#xA;&quot;Nope&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Am I thinking of a fridge?&quot;&#xA;&quot;Nope&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Am I thinking of Ellen?&quot;&#xA;&quot;Yes&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ellen told Andrew exactly her name. She wanted then be sure that this wasn't a &quot;turn based trick&quot;, therefore asked again Andrew to ask and Gabriel to reply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Am I thinking of Ellen again?&quot;&#xA;&quot;Nope&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Am I thinking of learning German?&quot;&#xA;&quot;Nope&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Am I thinking of the wild wild west?&quot;&#xA;&quot;Nope&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Maybe then I'm thinking of Will Smith!&quot;&#xA;&quot;Nope&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Are you sure? I may be thinking about a saloon then!&quot;&#xA;&quot;Nope&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I give you the solution, I'm thinking about an harpoon&quot;&#xA;&quot;Don't make me laugh, you ruin my concentration! Of course not!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Ok, then am I thinking about a yellow Submarine?&quot;&#xA;&quot;Nope&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Am I thinking about a dirty table?&quot;&#xA;&quot;Nope&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Am I thinking about smoking a cigar?&quot;&#xA;&quot;Yes&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ellen becomes red with anger. They did it again!&#xA;Andrew and Gabriel go on the entire day doing this game and eventually Ellen surprises them during another random day asking them to repeat the experiment. The number of questions always changes... Ellen asks if the trick is dependant on the number of questions or the day, but Andrew and Gabriel laugh and give Ellen a hint: &quot;If it was a trick, which of course is not, it wouldn't be dependant on the number of questions or the day&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did Andrew and Gabriel put this up?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.s.: This has nothing to do with hidden microphones or something like that :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.p.s: If you DO realize how they did it, I will upvote you if you don't post the solution, but a list of questions/answers &quot;Yes/Nope&quot; to make other users go mad :D&lt;/p&gt;&#xA;" OwnerUserId="5122" LastEditorUserId="5188" LastEditDate="2015-05-01T03:31:45.107" LastActivityDate="2015-05-01T03:31:45.107" Title="A funny game you can play with a friend of yours: brain reading" Tags="|pattern|magic|" AnswerCount="6" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4385" PostTypeId="1" AcceptedAnswerId="4403" CreationDate="2014-11-13T08:36:48.937" Score="2" ViewCount="942" Body="&lt;p&gt;A never ending party is being held at a local mansion. The host is very rich and his success is because of one thing — his famous recipe for Spaghetti!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only guests that may attend are people who correctly reply to the guard at the door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's where you come in. You and a friend are trying to steal this recipe. You sneak by and listen to the passwords.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first guest arrives. The security says &quot;1 and 1&quot;, and the guest replies &quot;1&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second guest arrives. The security says &quot;2 and 2&quot;, the guest replies &quot;2&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third guest arrives. The security says &quot;3 and 3&quot;, the guest replies &quot;3&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your friend thinks he's got it all figured out, he thought that it was a crappy logical problem and at last he would have a glimpse of the hidden recipe so he walks up to the door and the security says &quot;4 and 4&quot;. He was very confident and whispered that at last he will not be terminated which happened in the previous episodes, the security heard him and changed the question to &quot;4 and 7&quot;. Your friend was devastated and he knew what would happen so he grabbed a pill that he hid in his pocket and he ate it and died.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another guest arrives, you seem to get the same pattern, security says &quot;1 and 7&quot; and he says &quot;7&quot; to get in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another guest arrives, you seem to get the same pattern, security says &quot;8 and 3&quot; and he says &quot;8&quot; to get in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You now walk up to security and get &quot;4 and 7&quot;, what is the correct response?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;hint(worthless hint but still accurate):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Another guest arrives, you seem to get the same pattern, security says &quot;4 and 4&quot; and he says &quot;4&quot; to get in.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;hint2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Another guest arrives, you seem to get the same pattern, security says &quot;0 and 1&quot; and he says &quot;0&quot; to get in.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;hint3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is another problem that doesn't include computations in it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;hint4&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You need to merge the numbers but not in the mathematical way or not in a number words, just the number itself&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;final and a giveaway hint&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; analog&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;i think this one is an easy one since that it was already used in the previous versions&lt;/p&gt;&#xA;" OwnerUserId="4944" LastEditorUserId="68" LastEditDate="2014-11-13T11:14:54.827" LastActivityDate="2014-11-13T11:14:54.827" Title="The Security to the Party [Part 24]" Tags="|pattern|password|" AnswerCount="5" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4386" PostTypeId="2" ParentId="4360" CreationDate="2014-11-13T08:37:43.180" Score="10" Body="&lt;p&gt;For the note on the top of the box&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;nvnmjoonhenuan&quot; is an anagram of &quot;John von Neumann&quot;, and &quot;m2m&quot; refers to the middle square method of pseudorandom number generation he created.  121 is likely a seed number for this PRNG. Also, &quot;ttfflhesi&quot; is an anagram of &quot;shift left.&quot; &quot;0+1&quot; may refer to the loops of addition test used as a computing benchmark, signifying that one should take the previous number and add 1.&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not sure about note under box yet.  Will try again later.&lt;/p&gt;&#xA;" OwnerUserId="5121" LastEditorUserId="5121" LastEditDate="2014-11-13T11:17:53.540" LastActivityDate="2014-11-13T11:17:53.540" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4388" PostTypeId="2" ParentId="4385" CreationDate="2014-11-13T08:42:07.747" Score="1" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;7&lt;/code&gt;. Answer is the one which is bigger one b/w guard said.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4848" LastEditorUserId="2313" LastEditDate="2014-11-13T09:13:04.970" LastActivityDate="2014-11-13T09:13:04.970" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4389" PostTypeId="2" ParentId="4384" CreationDate="2014-11-13T08:48:37.020" Score="2" Body="&lt;p&gt;Not quite sure what you're asking with &lt;em&gt;a list of questions/answers &quot;Yes/Nope&quot;&lt;/em&gt; as answer... So I'll go with the canonical answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since is the information holder that makes the questions and he knows the hidden word Andrew has to answer &quot;yes&quot; to the 8th question of Gabriel, while Gabriel to the 3rd. At least this is how I would do it :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4652" LastActivityDate="2014-11-13T08:48:37.020" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4390" PostTypeId="2" ParentId="4376" CreationDate="2014-11-13T08:49:04.250" Score="5" Body="&lt;p&gt;So you are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 16&#xA;&lt;br&gt;&#xA; From guard's perspective multiply row &amp;amp; column, and divide 144 by that&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And friend&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; poor friend should have done 12&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4317" LastActivityDate="2014-11-13T08:49:04.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4392" PostTypeId="2" ParentId="4369" CreationDate="2014-11-13T08:50:44.053" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;22&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some crazy logic &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think it is related to time..... 0 answers to 26 while -6 answers to 19. If -6 means 6 in evening, it should be 1800 hrs (military), the answer is added with 1 hr more, which can be the reason why 0 becomes 26 instead of 24 hrs (1 hr from each day). -9 means (9+12)= 21 , 0.5 rounds as 1 , so 22?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4850" LastActivityDate="2014-11-13T08:50:44.053" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4393" PostTypeId="2" ParentId="4327" CreationDate="2014-11-13T09:07:58.513" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gun? &lt;br&gt; It fits well in the first and third paragraphs. Gun salutes are considered an ultimate from of salute. It's less obvious in the second paragraph, but, being metal objects, guns are cold at night, they gleam in the day and their countenance is clear.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4427" LastActivityDate="2014-11-13T09:07:58.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4394" PostTypeId="2" ParentId="4385" CreationDate="2014-11-13T09:16:09.950" Score="3" Body="&lt;p&gt;My (trollish) answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First, I save the game and answer 4. If I get in, I am done. If I die, I should load my saved game data back and try 7 instead.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5044" LastActivityDate="2014-11-13T09:16:09.950" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4396" PostTypeId="2" ParentId="4327" CreationDate="2014-11-13T09:21:46.053" Score="2" Body="&lt;p&gt;It's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;respect&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Owned by millions, seen by all,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many people are respected (I don't have actual figures but 1 in 1000 is still more than 7 million) It is easy to recognize these people.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Borne by some, and used by most.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not everyone has (bears) respectability, but you don't have to own it to use it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Be careful you don't blow it all, When it's your turn to make a toast.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Toasts are used to give people honor (i.e. respect), but using to much can seem insincere (&quot;[and you] blow it all&quot;).  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By day it gleams if you take heed,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If there are people around (&quot;By day&quot;) and they pay (&quot;take heed&quot;) proper respect, it will gleam (light is often used as an analogy of justice/righteousness).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At nighttime, cold as morning dew.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you are alone (&quot;At nighttime&quot;) respect is useless, &quot;cold&quot; if you will.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Although it may seem hard to read, Its countenance is clear to you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It isn't always clear if you have it (&quot;hard to read&quot;), but you support or approve it either way (&quot;countenance&quot;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Salute with it, it will surpass,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Salute according to Google: &quot;a gesture of &lt;em&gt;respect&lt;/em&gt; or polite recognition, especially one made to or by a person when arriving or departing.&quot; When you use respect as often as you can (from arriving to departing), people will also be respectful to you. The net amount of respect will be greater than before (&quot;it will surpass&quot;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A bow to temper, or a kneel.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A bow or kneel are both signs of honor. When you use respect you moderate (&quot;temper&quot;) yourself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now hearing many forms, at last, Its true identity reveal.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people are respected by some and hated (or not-so-respected) by others (&quot;many forms&quot;). When you hear why, the real respectability (&quot;its true identity&quot;) will be apparent.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if I'm right, but I hope the effort will find some appreciation, or respect if you will ;)&lt;/p&gt;&#xA;" OwnerUserId="5127" LastActivityDate="2014-11-13T09:21:46.053" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4397" PostTypeId="2" ParentId="4211" CreationDate="2014-11-13T09:22:26.020" Score="2" Body="&lt;p&gt;My guess: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A tongue. It bends and sometimes you don't use all of it. You speak with it so u can question, scare and confuse. It fits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5241" LastEditorUserId="5241" LastEditDate="2014-11-13T09:27:59.740" LastActivityDate="2014-11-13T09:27:59.740" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4398" PostTypeId="2" ParentId="4385" CreationDate="2014-11-13T09:34:33.530" Score="1" Body="&lt;p&gt;Could it be this?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4, if the first number is an even number, you take the first number, if it's odd you take the second (0 is an even number).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5242" LastActivityDate="2014-11-13T09:34:33.530" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4399" PostTypeId="2" ParentId="4385" CreationDate="2014-11-13T09:35:35.047" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 7, count the characters of the first number (for example: two = 3 characters)&lt;br/&gt;So you get 1 - 3 = -2 &lt;br/&gt;one 1 = -2&lt;br/&gt;two 2 = -1&lt;br/&gt;three 3 = -1&lt;br/&gt;one 7 = 4&lt;br/&gt;eight 3 = -2&lt;br/&gt;four 7 = 2&lt;br/&gt;if the result is positive then the right number is correct&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5241" LastEditorUserId="5241" LastEditDate="2014-11-13T09:41:51.753" LastActivityDate="2014-11-13T09:41:51.753" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4401" PostTypeId="2" ParentId="4384" CreationDate="2014-11-13T09:41:59.550" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One way to do it is that they agree beforehand to a list of words that indicate that the question after it will be the correct answer. In this case it is &quot;a car&quot;, &quot;a fridge&quot; and &quot;a dirty table&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2014-11-13T09:41:59.550" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4402" PostTypeId="2" ParentId="4341" CreationDate="2014-11-13T09:42:31.257" Score="-1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; constantly being phoned up to repair the electrics he burnt out with an unbelievable amount of hours being dedicated to SE. His electrician took revenge.&lt;br&gt;&#xA; the eye for an eye was when the electrician in a freak accident fixing his fuse box suffered an electric shock blowing out his retina.&lt;br&gt;&#xA; the way he killed him was an electric shock through the door handle. &quot;eye for an eye&quot; no?.&#xA; the carving on the gun was simply a electric bolt symbol.&lt;br&gt;&#xA; because of his vision and depth perception being impaired he fell trying to manoeuvre in the dark in the basement.&lt;br&gt;&#xA; assuming he (Joe) only went into his house and to the basement there might be 2-3 lights on at max. so unless there was more than 1 murderer (here on known as electricians), he must of been at the fuse box and knew which switches affected which lights from previous experiences. to ensure the right light was switched on.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5135" LastEditorUserId="11297" LastEditDate="2015-07-17T16:50:17.427" LastActivityDate="2015-07-17T16:50:17.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4403" PostTypeId="2" ParentId="4385" CreationDate="2014-11-13T09:46:58.660" Score="6" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 9&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Overwrite number on each other in 7 segment display&lt;BR&gt;&#xA; Write 8 over 3 to get 8&lt;BR&gt;&#xA; Write 1 over 0 to get 0&lt;BR&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4637" LastActivityDate="2014-11-13T09:46:58.660" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4405" PostTypeId="1" CreationDate="2014-11-13T09:54:56.417" Score="8" ViewCount="105525" Body="&lt;p&gt;There were three men. One was deaf, the other was blind and the third was mute. One day the deaf man kissed the blind man's wife. The mute man saw this and he wants to tell it to the blind man, but he cannot speak. So how does the mute man explain the situation to the blind man.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to give a clarification regarding the question, I want to specify in detail these points:&#xA;There were only three of them and the blind man's wife and no one else in the situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mute man doesn't know how to write Braille.&lt;/p&gt;&#xA;" OwnerUserId="5246" LastEditorUserId="11297" LastEditDate="2015-07-10T21:53:04.073" LastActivityDate="2015-07-10T21:53:04.073" Title="Deaf, Mute and Blind" Tags="|logical-deduction|" AnswerCount="10" CommentCount="11" ClosedDate="2016-08-05T20:28:35.633" ContentLicense="CC BY-SA 3.0" />
  <row Id="4407" PostTypeId="1" AcceptedAnswerId="4442" CreationDate="2014-11-13T10:04:56.420" Score="3" ViewCount="761" Body="&lt;p&gt;Another time, there is still another never-ending party at a local mansion. The host is very rich and his success is because of one thing ~ his famous recipes for &lt;a href=&quot;http://en.wikipedia.org/wiki/Methamphetamine&quot; rel=&quot;nofollow&quot;&gt;spaghetti&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Cocaine&quot; rel=&quot;nofollow&quot;&gt;fettuccini&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Cannabis_(drug)&quot; rel=&quot;nofollow&quot;&gt;green salad&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Lysergic_acid_diethylamide&quot; rel=&quot;nofollow&quot;&gt;juice&lt;/a&gt;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only guests that may attend are people the host knows and trusts, whom he gave the password to enter to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's where you come in. You and four friends are trying to steal this recipe. This time you called four friends to get many heads working simultaneously to quickly break the code. To successfully steal it, two people are needed to simultaneously rotate two keys on a vault inside the mansion. You all sneak by and listen to the passwords.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first guest arrives. The security says &quot;19&quot;, and the guest replies &quot;3&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second guest arrives. The security says &quot;14&quot;, the guest replies &quot;3&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third guest arrives. The security says &quot;9&quot;, the guest replies &quot;3&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your friend #1 quickly goes there, that is the easiest password afterall! The security says &quot;20&quot; and he replies &quot;3&quot;. Immediately an elephant falls from the sky over him and starts to dance over his crushed body splattering blood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another guest arrives. The security says &quot;15&quot;, the guest replies &quot;1&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still another guest arrives. The security says &quot;22&quot; and the guest replies &quot;22&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your other three friends looked one to the other at the same time, smiled, and the friend #2 said:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Are you pondering what I am pondering?&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other two friends answered together at the same time:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;I think so, Brain&quot;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your friends goes there. The guard asks &quot;21&quot; to your friend #2. He quickly happily answers &quot;4&quot; and goes inside. Good, now you just need just one more man!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, your friend number #3: The guard asks &quot;31&quot;. Your friend immediatelly freezes in panic and becames mute. The guard impaciently repeats: &quot;31&quot;. Your friend starts to sweat and to cry without answering. So the guard had no other option but to &lt;code&gt;kill -9&lt;/code&gt; your friend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, your friend number #4: The guard asks &quot;25&quot;. He stops, thinks, thinks, thinks and in the end he says: &lt;code&gt;java.lang.StackOverflowError&lt;/code&gt; and explodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Darn, now you are the last guy available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see another guest coming. The guard asks &quot;24&quot;. The guest answers &quot;9&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You decided to go there. The guard asks &quot;11&quot;. What should you answer? Explain why. Why your friends #3 and #4 failed? Please put your answer inside spoiler tags.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Since we already got 2 very wrong answers, here some hints:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 31 can't be answered (but you still should tell me why).&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The computation for 25 can recurse forever. Tell me why.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer for 20 was wrong. It is not 3.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You could look to this question all day long and do a lot of math just to go to nowhere. You will need to look somewhere outside to find the answer and to understand why the answers given by the guests are correct.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Don't need to go to wikipedia. The links in the start are just a joke.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 6:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No math is needed to solve this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 7:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is not a wording or letters gameplay.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5044" LastEditorUserId="5044" LastEditDate="2014-11-13T11:36:40.847" LastActivityDate="2014-11-13T12:53:19.847" Title="Security to the Party [Part 25]" Tags="|password|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4408" PostTypeId="2" ParentId="4405" CreationDate="2014-11-13T10:04:57.977" Score="10" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tap it out in morse code - or use text-to-speech technology.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Old answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;He types up some braille and puts the blind man's hand on it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-11-13T10:10:43.193" LastActivityDate="2014-11-13T10:10:43.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4409" PostTypeId="2" ParentId="4405" CreationDate="2014-11-13T10:28:24.187" Score="-2" Body="&lt;p&gt;If the question is considered not plagiarised then this answer is not either &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source:&lt;a href=&quot;http://www.riddlers.org/riddles/riddles/view/6928&quot; rel=&quot;nofollow&quot;&gt;http://www.riddlers.org/riddles/riddles/view/6928&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One is blind, the other is deaf and the third is mute (cant speak)&quot; &#xA;This doesn't say that each of the three men have a disability. All we know for sure is that one of them is blind, one of them is deaf, and one of them is mute. Since you say that one is blind and the other is deaf, the blind person is not deaf. That leaves the deaf person who is also mute. Since he is the one who kissed, he would not tell on himself. But he doesn't have to since the 3rd person who has no disability could tell the blind man what happened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer therefore is, the mute man cannot tell and if at all, the third person who has no disabilities or the wife herself or the deaf man himself could tell the blind man&lt;/p&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="2313" LastEditDate="2014-11-13T18:21:43.710" LastActivityDate="2014-11-13T18:21:43.710" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="4410" PostTypeId="1" AcceptedAnswerId="4434" CreationDate="2014-11-13T10:34:44.570" Score="3" ViewCount="350" Body="&lt;p&gt;OK, so you want to get in to the party.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The door is being guarded by an elephant, who will trample any visitor who gives the wrong answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A polar bear comes up to the door. The elephant says, &quot;1&quot;, and the polar bear replies &quot;14&quot;, and is admitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then a horned lark flies up to the door. The elephant says, &quot;2&quot;. The lark twitters back &quot;14&quot; and is trampled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next up is a cat. The cat saunters up to the door, and the elephant says &quot;2&quot;. The cat replies &quot;10&quot; and is admittted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An emperor penguin waddles up to the door. The elephant says &quot;3&quot;, and the penguin replies &quot;19&quot; (and is admitted).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally a massive tortoise, which has flown in specially from the Galapagos Islands just to attend this party, walks (very slowly) up to the door. The door-keeping elephant says &quot;4&quot; and the tortoise replies (very slowly) &quot;Six&quot; (deep breath, long pause) &quot;teen&quot;. The tortoise is admitted to the party.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it's your turn. The elephant says &quot;5&quot;; what do you reply?&lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="15544" LastEditDate="2015-09-25T05:02:57.880" LastActivityDate="2015-09-25T05:02:57.880" Title="The Animal Party [Security to the Party 26] Find the formula to predict the next answer" Tags="|password|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="4411" PostTypeId="2" ParentId="4384" CreationDate="2014-11-13T10:38:37.513" Score="3" Body="&lt;p&gt;Maybe it is not a&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; list of words they agreed upon, but the item before the right answer is a thing Andrew has got, for example, at home.&#xA; In fact, the first words are somewhat weird things, like &amp;quot;Bavarian Sky&amp;quot;, or &amp;quot;a spider&amp;quot;, while the last word before the right one is something more &amp;quot;real&amp;quot; than the others.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;A simple example:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;Am I thinking of a unicorn?&amp;quot; &amp;quot;Nope&amp;quot;&lt;br /&gt;&#xA;&amp;quot;Am I thinking of an angel?&amp;quot; &amp;quot;Nope&amp;quot;&lt;br /&gt;&#xA;&amp;quot;Am I thinking of a pen?&amp;quot; &amp;quot;Nope&amp;quot;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The &amp;quot;pen&amp;quot; is the key item&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&amp;quot;Am I thinking of (whatever you want)?&amp;quot; &amp;quot;Yes&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Another try:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;Am I thinking of a flying unicorn?&amp;quot; &amp;quot;Nope&amp;quot;&lt;br /&gt;&#xA;&amp;quot;Am I thinking of an angel?&amp;quot; &amp;quot;Nope&amp;quot;&lt;br /&gt;&#xA;&amp;quot;Am I thinking of a blue pen?&amp;quot; &amp;quot;Nope&amp;quot;&lt;br /&gt;&#xA;&amp;quot;Am I thinking of a door?&amp;quot; &amp;quot;Nope&amp;quot;&lt;br /&gt;&#xA;&amp;quot;Am I thinking of a hole?&amp;quot; &amp;quot;Yes&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4527" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-11-27T08:38:38.707" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="4412" PostTypeId="2" ParentId="4407" CreationDate="2014-11-13T10:40:14.363" Score="1" Body="&lt;p&gt;OK, my answer is this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Say any number and try to avoid being squashed by falling animals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The guards have been instructed to say a number to anybody who tries to enter, and to admit anybody who replies numerically. There is an problem with falling animals in the area where the party is being held - this is entirely unrelated to the party entry system, except that of course any guests killed by falling animals are no longer welcome at the party. (If they had loads of corpses at the party, that might spoil everybody's enjoyment.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4206" LastActivityDate="2014-11-13T10:40:14.363" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4413" PostTypeId="2" ParentId="4326" CreationDate="2014-11-13T10:41:32.677" Score="6" Body="&lt;p&gt;My attempt is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3125&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Of the calculations listed, 5^5 is the only one that has an answer 4 digits long.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1491" LastActivityDate="2014-11-13T10:41:32.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4414" PostTypeId="2" ParentId="4384" CreationDate="2014-11-13T10:47:22.350" Score="8" Body="&lt;p&gt;I'd ask for your mother, a T-Rex, Michael Jackson, a chair and a rainbow (which is the answer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me ask for a beach, a skyscraper, a castle, a park bench, world's fastest computer and a smartphone &lt;/p&gt;&#xA;" OwnerUserId="5106" LastEditorUserId="5106" LastEditDate="2014-11-13T11:18:43.653" LastActivityDate="2014-11-13T11:18:43.653" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="4415" PostTypeId="2" ParentId="4384" CreationDate="2014-11-13T10:49:16.387" Score="25" Body="&lt;p&gt;Do these series work? If so, I know the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Am I thinking about a raisin?&quot; &quot;Nope&quot;&lt;br&gt;&#xA;  &quot;Am I thinking about an ostrich?&quot; &quot;Nope&quot;&lt;br&gt;&#xA;  &quot;Am I thinking about a finger?&quot; &quot;Nope&quot;&lt;br&gt;&#xA;  &quot;Am I thinking about a moose?&quot; &quot;Nope&quot;&lt;br&gt;&#xA;  &quot;Am I thinking about an astronaut? &quot;Yes&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Example 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Am I thinking about a snake?&quot; &quot;Nope&quot;&lt;br&gt;&#xA;  &quot;Am I thinking about the Eiffel Tower?&quot; &quot;Nope&quot;&lt;br&gt;&#xA;  &quot;Am I thinking about the Pacific ocean? &quot;Yes&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Example 3&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Am I thinking about a traffic light?&quot; &quot;Nope&quot;&lt;br&gt;&#xA;  &quot;Am I thinking about a ring?&quot; &quot;Nope&quot;&lt;br&gt;&#xA;  &quot;Am I thinking about air?&quot; &quot;Nope&quot;&lt;br&gt;&#xA;  &quot;Am I thinking about a bed?&quot; &quot;Nope&quot;&lt;br&gt;&#xA;  &quot;Am I thinking about a bottle? &quot;Yes&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;A thing with 4 legs&quot; is the key. The guess after the 4-legged item is the correct one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4962" LastEditorUserId="4204" LastEditDate="2014-11-27T05:58:25.527" LastActivityDate="2014-11-27T05:58:25.527" CommentCount="24" ContentLicense="CC BY-SA 3.0" />
  <row Id="4416" PostTypeId="2" ParentId="4384" CreationDate="2014-11-13T11:05:17.787" Score="2" Body="&lt;p&gt;I could ask for&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The planet Jupiter: Nope&lt;/li&gt;&#xA;&lt;li&gt;Queen Cleopatra: Nope&lt;/li&gt;&#xA;&lt;li&gt;The Catholic Church: Nope&lt;/li&gt;&#xA;&lt;li&gt;A tree: Nope&lt;/li&gt;&#xA;&lt;li&gt;A window: Nope&lt;/li&gt;&#xA;&lt;li&gt;New York City: Nope&lt;/li&gt;&#xA;&lt;li&gt;A TV: Nope&lt;/li&gt;&#xA;&lt;li&gt;The Pacific Ocean: Yes&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5044" LastEditorUserId="5044" LastEditDate="2014-11-13T11:13:24.453" LastActivityDate="2014-11-13T11:13:24.453" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4417" PostTypeId="1" AcceptedAnswerId="4606" CreationDate="2014-11-13T11:19:30.127" Score="5" ViewCount="534" Body="&lt;p&gt;Follow the clues and tell me where you are in the end.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;You are here    &lt;/li&gt;&#xA;  &lt;li&gt;The birthplace    &lt;/li&gt;&#xA;  &lt;li&gt;The biggest child    &lt;/li&gt;&#xA;  &lt;li&gt;The first letter of it's name    &lt;/li&gt;&#xA;  &lt;li&gt;Go left&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: This one is harder, and has the requirement that you have to live in US/Canada or assume that you do.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You need to make the Canada/US for clue #5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This is part 2 of a series of puzzles, part 1 can be found here: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4299/where-did-you-end-up-part-1&quot;&gt;Where did you end up? [Part 1]&lt;/a&gt; feel free to join in the fun!&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-04-30T23:35:55.367" Title="Where did you end up? [Part 2!]" Tags="|lateral-thinking|treasure-hunt|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4418" PostTypeId="2" ParentId="4407" CreationDate="2014-11-13T11:19:49.793" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The mansion is actually a MCC LAD system (take the capital letters from the recipe links) and the numbers the guard said were signals sent by the system. Following the standard UNIX signals' chart, signal 11 corresponds to &quot;segfault&quot;, so an appropriate answer would be 9, &quot;kill&quot; signal. #4 failed because he continued his thinking activity until a stackoverflow error raised (25 is &quot;continued&quot;), #3 could've stand the size of the input so he didn't know how to manage it, being killed by a &quot;kill -9&quot; sent by the system. Everything is related with Unix signals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user1962" LastActivityDate="2014-11-13T11:19:49.793" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4420" PostTypeId="2" ParentId="4004" CreationDate="2014-11-13T11:29:01.307" Score="0" Body="&lt;p&gt;I think the answer is 1.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Count the amount of corners.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5254" LastActivityDate="2014-11-13T11:29:01.307" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4422" PostTypeId="2" ParentId="4405" CreationDate="2014-11-13T11:32:05.373" Score="2" Body="&lt;p&gt;Even if the mute man doesn't know Braille, Morse etc., he can (probably, it's not forbidden by the condition) write normal letters on the sand etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The blind man can then 'read' those letters with his hands. Braille enables blind to read quickly, but they can feel the shape of the 'normal' letters and decode them, even if it takes more time. &lt;/p&gt;&#xA;" OwnerDisplayName="user66" LastActivityDate="2014-11-13T11:32:05.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4434" PostTypeId="2" ParentId="4410" CreationDate="2014-11-13T12:15:34.353" Score="7" Body="&lt;p&gt;Got it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The number given by the elephant is useless, just a red-herring.&lt;br&gt;&#xA; &lt;br&gt;&#xA; Polar bear is &lt;em&gt;Ursus maritimus&lt;/em&gt;, which have 14 letters.&lt;br&gt;&#xA; Horned lark is &lt;em&gt;Eremophila alpestris&lt;/em&gt;, which have 19 letters, not 14, so it was trampled.&lt;br&gt;&#xA; Cat is &lt;em&gt;Felis catus&lt;/em&gt;, which have 10 letters.&lt;br&gt;&#xA; Emperor penguin is &lt;em&gt;Aptenodytes forsteri&lt;/em&gt;, which have 19 letters.&lt;br&gt;&#xA; Galápagos giant tortoise is &lt;em&gt;Chelonoidis nigra&lt;/em&gt;, which have 16 letters.&lt;br&gt;&#xA; &lt;br&gt;&#xA; Finally, I am an &lt;em&gt;Homo sapiens&lt;/em&gt;, so I should answer 11.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5044" LastActivityDate="2014-11-13T12:15:34.353" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4435" PostTypeId="2" ParentId="4346" CreationDate="2014-11-13T12:22:46.027" Score="11" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; σ. The physicist will say &quot;sigma&quot;, the plumber will say &quot;huh?&quot;, assuming a non-greek plumber.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2187" LastActivityDate="2014-11-13T12:22:46.027" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4436" PostTypeId="1" CreationDate="2014-11-13T12:28:46.063" Score="3" ViewCount="2582" Body="&lt;p&gt;In a five star hotel, I ordered for a cup of coffee. When I was about to sip, I found a fly in the coffee. I complained. The manager apologized, took back the cup and brought another fresh hot one. But I realized that he bought the same coffee. I sued the hotel in a consumer court and won compensation. The question is not how I won the case. How could I recognize that the manager has bought the same coffee?&lt;/p&gt;&#xA;" OwnerUserId="5246" LastActivityDate="2014-11-14T08:31:17.260" Title="Why was the Five Star Hotel sued?" Tags="|logical-deduction|" AnswerCount="18" CommentCount="14" ClosedDate="2014-11-13T20:32:00.630" ContentLicense="CC BY-SA 3.0" />
  <row Id="4437" PostTypeId="2" ParentId="4436" CreationDate="2014-11-13T12:29:50.390" Score="11" Body="&lt;p&gt;You already added the milk/sugar? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fly was still there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coffee had been in the cup long enough that the outside of the cup had had time to heat up?&lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="4050" LastEditDate="2014-11-13T12:45:20.300" LastActivityDate="2014-11-13T12:45:20.300" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4438" PostTypeId="2" ParentId="4326" CreationDate="2014-11-13T12:32:59.233" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is 3125&lt;br&gt;&#xA; 5^5 is the only such equation with an answer of 4 digits.&lt;br&gt;&#xA;&lt;br&gt;&#xA; 4^4 = 256&lt;br&gt;&#xA; 5^5 = 3125&lt;br&gt;&#xA; 6^6 = 46656&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="340" LastActivityDate="2014-11-13T12:32:59.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4439" PostTypeId="2" ParentId="4436" CreationDate="2014-11-13T12:37:25.480" Score="1" Body="&lt;p&gt;If the hotel had a system like Starbucks where you get your name on the coffee cup, you may have realized that it was the same cup.&lt;/p&gt;&#xA;" OwnerUserId="1711" LastActivityDate="2014-11-13T12:37:25.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4440" PostTypeId="2" ParentId="4436" CreationDate="2014-11-13T12:42:04.853" Score="3" Body="&lt;p&gt;I don't expect the waiter to be stupid enough to bring a coffee already mixed with milk again. There might have been some color on the cup from your lips (assuming you are a woman) but again he should have seen that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could have told from the temperature which decreases when the cup is carried around. Maybe there was no steam in the air, no smell. But you said he brought a &quot;fresh hot one&quot; which confused me. &#xA;I also thought about the surface of the coffee that changes from a gleaming one to a dull, black one after some time. Plus, the fly or the object that picked the fly could have had influence on that.&lt;/p&gt;&#xA;" OwnerUserId="5106" LastActivityDate="2014-11-13T12:42:04.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4441" PostTypeId="2" ParentId="4436" CreationDate="2014-11-13T12:42:21.230" Score="26" Body="&lt;p&gt;The cup has an unique IPv6 address?&lt;/p&gt;&#xA;" OwnerUserId="5044" LastActivityDate="2014-11-13T12:42:21.230" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4442" PostTypeId="2" ParentId="4407" CreationDate="2014-11-13T12:42:43.173" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Answer is  9&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The security to the Party part 19 answer 3 &#xA; The security to the Party part 22 answer 22 &#xA; The security to the Party part 24 answer 9&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  The security to the Party part 11 answer 9.&#xA; Now no part in 31. so now no answer for 31&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4th  Friend&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think if friend check 25 part that part itself again asking same question so it recursively so it so error&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4564" LastEditorUserId="4564" LastEditDate="2014-11-13T12:53:19.847" LastActivityDate="2014-11-13T12:53:19.847" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4443" PostTypeId="2" ParentId="4436" CreationDate="2014-11-13T12:45:34.490" Score="16" Body="&lt;p&gt;Here is one possibility&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You had already added the sugar before observing the fly. This time when you added the sugar again, it was extra sweet. You produced that cup of coffee with video recording that you added sugar just once still coffee contains double sugar.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4637" LastActivityDate="2014-11-13T12:45:34.490" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4444" PostTypeId="2" ParentId="4436" CreationDate="2014-11-13T12:51:24.420" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You had not told the manager how you take your coffee (managers of 5 star hotels are rarely at ground level taking orders), so there is no way the manager knew how to make a new cup.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1491" LastEditorUserId="1491" LastEditDate="2014-11-13T12:57:55.483" LastActivityDate="2014-11-13T12:57:55.483" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4445" PostTypeId="2" ParentId="4281" CreationDate="2014-11-13T12:58:40.567" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; so is this how heat comes to pass? ... &#xA; h from &quot;age&quot;,&#xA; e from &quot;eeesiest&quot;,&#xA; a from third letter in apart? no idea,&#xA; t from told - old&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5265" LastEditorUserId="5265" LastEditDate="2014-11-13T13:23:39.427" LastActivityDate="2014-11-13T13:23:39.427" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="4446" PostTypeId="1" AcceptedAnswerId="4567" CreationDate="2014-11-13T13:07:10.023" Score="11" ViewCount="1288" Body="&lt;p&gt;You have a closed cord of length 100 meters which touches point $P$.  There is a line $L$ whose closest point is 20 meters from point $P$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value of a layout of cord is defined by the area enclosed on the side of the line with point $P$ plus &lt;em&gt;twice&lt;/em&gt; the area enclosed on the opposite side.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the optimal shape of the cord that maximizes value?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See diagram below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JWRtN.png&quot; alt=&quot;Diagram&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1672" LastEditorUserId="1672" LastEditDate="2014-11-13T13:18:45.370" LastActivityDate="2014-11-14T22:14:24.350" Title="Optimization of area" Tags="|mathematics|geometry|optimization|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4448" PostTypeId="2" ParentId="4436" CreationDate="2014-11-13T13:13:38.650" Score="2" Body="&lt;p&gt;The ear of the cup was warm from being in the microwave?&lt;/p&gt;&#xA;" OwnerUserId="2156" LastActivityDate="2014-11-13T13:13:38.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4449" PostTypeId="1" AcceptedAnswerId="4520" CreationDate="2014-11-13T13:15:48.443" Score="12" ViewCount="1652" Body="&lt;p&gt;This puzzle is part of a series - &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4075/the-serial-killer-part-1&quot;&gt;part 1 is here&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4263/the-serial-killer-part-2&quot;&gt;part 2 is here&lt;/a&gt;. Any comments I make on the OP are canonical and &lt;a href=&quot;http://tvtropes.org/pmwiki/pmwiki.php/Main/OutOfCharacter&quot; rel=&quot;nofollow noreferrer&quot;&gt;out of character (OOC)&lt;/a&gt;. Any comments I post on answers are for flavour only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since part 2 turned out to be too easy (or at least, solved too quickly), part 3 is harder. If it turns out to be too hard, I will start hinting, but that won't happen within the first few hours. It's a puzzle - get thinking :-) This version has gone through about 5 revisions and I'm finally happy with it, and where the series is &lt;strong&gt;now&lt;/strong&gt; going (shakes fist at skv).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You take a deep breath and settle yourself. In a moment of brilliance you realise that $\overline{XC}CCCLXXXIII$ is the Roman numeral for &lt;code&gt;90383&lt;/code&gt;, and the &lt;code&gt;b&lt;/code&gt; on the end forms a hexadecimal colour. You quickly map this to the colour used on the &lt;code&gt;&amp;amp;&lt;/code&gt; logo of the English Language &amp;amp; Usage Stack - it's an exact match!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something's nagging at you in the back of your mind though - Handel seems strangely familiar. You cast your mind around... do you know any German composers? No, that's dumb. Male friends who wear wigs? Well, you know plenty of them but you don't think any of them are called Handel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hang on a second... &quot;called Handel&quot;... what if Handel is actually a &lt;a href=&quot;http://en.wikipedia.org/wiki/Pseudonym&quot; rel=&quot;nofollow noreferrer&quot;&gt;handle&lt;/a&gt; for the real name? As soon as that occurs to you, you realise that a &lt;a href=&quot;http://stackexchange.com/users/1266491/doorknob&quot;&gt;Doorknob&lt;/a&gt; is a synonym for &quot;handle&quot;. Just like that, in one moment of brilliance you identify &quot;Handel&quot; and &lt;a href=&quot;http://new.tinygrab.com/c158ca1523f0fe3a922d95db3356dfb84174f970ea.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;absolutely infuriate&lt;/strong&gt; the narrator&lt;/a&gt; of the series, who was hoping to sneak that one under the radar for a little while.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deciding that since you've already caused so many headaches, you figure your best lead would be to check out Doorknob's profile - it's more likely to have an unintentional clue than whatever fiendish and convoluted puzzle awaits you in EL&amp;amp;U. Besides, you're on the same Stack so it's not like you're going to cause the deaths of any more Stacks, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You open Doorknob's profile, and straight away you twig that something's clearly not right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ffJqv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ffJqv.png&quot; alt=&quot;http://new.tinygrab.com/c158ca1523a3647daaedeb813e9478b0cbbefa711a.png&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you try to make sense of the mess, suddenly pop-up windows start appearing all over the screen. Initially they're white, but then they all start loading different Stacks! &quot;Oh no... no, no, no, &lt;strong&gt;not&lt;/strong&gt; good!&quot; you cry, trying desperately to close them but there are just too many. As a last-gasp effort, you flick the power switch on the wall to turn your computer off. The screen goes black and the fans slowly whir to a stop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You wait.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You fear starting the computer up again, but know that it's not going to be very helpful in bringing the attacker(s) to justice if you just leave it powered off all day, and it could be disastrous if they're allowed to run loose, unchecked. You steel yourself, take a deep breath and turn the computer back on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You pull up your browser when the computers finishes booting up, and breathe a sigh of relief when absolutely nothing unusual at all happens. Browsing to Stack Exchange also seems safe enough - nothing explodes, the site looks normal, and for a moment you let yourself think that maybe everything's OK.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you look at the list of Stacks. An ominous message greets you, there are even more dead Stacks and critically, English Language &amp;amp; Usage has been killed. Your last lead has gone cold. You slump back in your seat, a sense of defeat beginning to crash down on you. What on earth can you do next?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;OOC:&lt;/strong&gt; Ask Ubuntu should be dead too - I missed that when making the image. I'll fix it for part 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/v9UPg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2518" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2017-03-11T00:47:27.547" Title="The SErial Killer (Part 3)" Tags="|story|mystery|" AnswerCount="4" CommentCount="31" ContentLicense="CC BY-SA 3.0" />
  <row Id="4451" PostTypeId="2" ParentId="4281" CreationDate="2014-11-13T13:20:08.353" Score="8" Body="&lt;p&gt;You can feel:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HEAT&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;'H'  $\rightarrow$ &quot;age&quot; sounds like H (first in the group)&lt;br&gt;&#xA;'E'  $\rightarrow$ tie-it (and third letter of tie); also the third in &quot;&lt;em&gt;th&lt;strong&gt;e&lt;/strong&gt; group&lt;/em&gt;&quot; &lt;br&gt;&#xA;'A'  $\rightarrow$ &quot;start again&quot; the TITLE! - &quot;brain won't FALL &lt;strong&gt;A&lt;/strong&gt;PART&quot; (brilliant!!)&lt;br&gt;&#xA;'T'  $\rightarrow$ told-old (last of the group)&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="5106" LastEditDate="2014-11-13T14:05:43.817" LastActivityDate="2014-11-13T14:05:43.817" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4452" PostTypeId="2" ParentId="4281" CreationDate="2014-11-13T13:20:58.690" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; heat. h from age. e if you slice &quot;i&quot; and &quot;t&quot; away from tie. a first in the group of letters in the alphabet. t from told - old.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2803" LastEditorUserId="2803" LastEditDate="2014-11-13T13:30:13.490" LastActivityDate="2014-11-13T13:30:13.490" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4453" PostTypeId="1" AcceptedAnswerId="4456" CreationDate="2014-11-13T13:25:04.590" Score="9" ViewCount="1512" Body="&lt;p&gt;You and your friend have met to play a game of dice.&#xA;This is a very simple game where both players roll a die at the same time, and the player with the lower score pays a coin to the other player.&#xA;In case of tie, no money is exchanged and the turn is repeated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You propose your friend to make the game more interesting by replacing the usual playing dice with three special dice you have designed.&#xA;Your dice have a number of dots between 1 and 6 on each face like regular dice, but some numbers are repeated on different faces and some numbers are missing, and no two dice are identical.&#xA;Apart from this twist, those are pretty normal playing dice, not unbalanced towards any side.&#xA;At each turn, you propose, first your friend will pick a die and then you will pick one of the remaining two dice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your friends accepts, thinking that, if at all, he's being offered an advantage.&#xA;If one of your dice were stronger than the other two, your friend could use that die at every turn because he can choose first, and that would make his winning chances better.&#xA;But you know this isn't going to be the case, and in fact, no matter which die your friend chooses, you will be always able to pick a stronger one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did you design your dice such that your chances to win are always better than your friend's?&lt;/p&gt;&#xA;" OwnerUserId="4394" LastActivityDate="2019-03-05T02:13:34.287" Title="A first choice die" Tags="|logical-deduction|mathematics|probability|dice|" AnswerCount="3" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="4454" PostTypeId="2" ParentId="4453" CreationDate="2014-11-13T13:37:35.480" Score="0" Body="&lt;p&gt;One (of many) solutions:&#xA;Die A has sides 2, 2, 4, 4, 9, 9.&#xA;Die B has sides 1, 1, 6, 6, 8, 8.&#xA;Die C has sides 3, 3, 5, 5, 7, 7&#xA;The probability that die A rolls a higher number than B, the probability that B rolls higher than C, and the probability that C rolls higher than A are all 5/9, so this set of dice is &quot;non-transitive&quot;. In fact, it has the even stronger property that, for each die in the set, there is another die that rolls a higher number than it more than half the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The game whith such dice is biased in favor of the second player, because the second player can always find a die that will beat the first player's die with probability 5/9.&#xA;So your friend should return the &quot;courtesy&quot; and allow you the first choice! :-)&lt;/p&gt;&#xA;" OwnerUserId="5051" LastActivityDate="2014-11-13T13:37:35.480" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4455" PostTypeId="2" ParentId="4436" CreationDate="2014-11-13T13:39:36.593" Score="0" Body="&lt;p&gt;the same batch and variety of coffee, just a different cup of it. The fly may not have been in the cup, but actually lay among the original source of the coffee.&lt;/p&gt;&#xA;" OwnerUserId="5273" LastActivityDate="2014-11-13T13:39:36.593" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4456" PostTypeId="2" ParentId="4453" CreationDate="2014-11-13T13:42:43.670" Score="6" Body="&lt;p&gt;There are many examples. This is one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A: 1, 1, 3, 5, 5, 6&lt;br&gt;&#xA;B: 2, 3, 3, 4, 4, 5&lt;br&gt;&#xA;C: 1, 2, 2, 4, 6, 6&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The probability of A winning versus B (or B vs. C, C vs. A) is... &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;17/36. The probability of a draw is 4/36, so that only 15 out of 36 rolls lose. So the overall winning expectation is higher.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://en.wikipedia.org/wiki/Nontransitive_dice&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Nontransitive_dice&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1748" LastEditorUserId="45158" LastEditDate="2019-03-05T02:13:34.287" LastActivityDate="2019-03-05T02:13:34.287" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4459" PostTypeId="2" ParentId="4446" CreationDate="2014-11-13T13:55:50.880" Score="1" Body="&lt;p&gt;For this, I did a little math and an angle at P of around 40 degrees to make this work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WDQLp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1491" LastActivityDate="2014-11-13T13:55:50.880" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4460" PostTypeId="1" CreationDate="2014-11-13T13:58:32.847" Score="-3" ViewCount="344" Body="&lt;p&gt;Jack is a programmer working in an IT company. One day he arrives at his workplace and sees a briefcase on the table with a note attached to it that says: &quot;g00d and 3v1l&quot;. He realizes - this must be his coworkers, who put it there. They had a party yesterday and he might have had a little bit too much to drink, so he started bragging about how he hacked his friend's email (even thought he just guessed his secret question), and how he could hack anything if he wanted to. Now to avoid looking like a fool it seems like he has to open this case... The case has a 4-digit code on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Help Jack solve this puzzle and pick a code to the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tip 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;and&quot; could be spelled &quot;4nd&quot;, but why is it not?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tip 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the whole thing could have all sorts of weird symbols in it if it was true leetspeak, but it's limited to letters and numbers...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Added after initial post:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tip 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it would be nice if the message didn't have those pesky &quot;g,v,l&quot; letters, but oh...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tip 4 (first digit):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the first digit is 3.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The code is &quot;3095&quot;. &quot;g00d&quot;, &quot;3v1l&quot; are in fact numbers with base of 36 (10 digits and 26 letters of English alphabet) so they correspond to &quot;746509&quot;, &quot;180201&quot; in decimal. Now the &quot;and&quot; word is of course a bitwise AND operation, which gives decimal 140297. Now you convert that number back to 36 base representation and you get string &quot;3095&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;P.S. Sorry if it was too ambiguous. It was hard for me to tell if it was dead easy or very hard because I thought that up. Actually, I knew it was hard, but after Tip #3... not so much. I mean, just look at Mathias711's answer, who would have thought of that :O, that would be definitely a harder way to hide the password.&lt;/p&gt;&#xA;" OwnerUserId="5019" LastEditorUserId="4050" LastEditDate="2016-12-19T21:45:11.340" LastActivityDate="2016-12-19T21:45:11.340" Title="After party briefcase code" Tags="|alphametic|" AnswerCount="5" CommentCount="14" ClosedDate="2014-11-14T00:25:30.970" ContentLicense="CC BY-SA 3.0" />
  <row Id="4461" PostTypeId="2" ParentId="4460" CreationDate="2014-11-13T14:03:53.120" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Given the hints we have, it's bound to be 0031 (or a variation of four-digit-numbers containing 0, 0, 3 and 1), as per the &quot;leetspeak&quot; on the note. g&lt;strong&gt;00&lt;/strong&gt;d and &lt;strong&gt;3&lt;/strong&gt;v&lt;strong&gt;1&lt;/strong&gt;l.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5265" LastActivityDate="2014-11-13T14:03:53.120" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4462" PostTypeId="2" ParentId="4460" CreationDate="2014-11-13T14:09:44.080" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;g00d has a MIDI representation of &lt;strong&gt;1942&lt;/strong&gt; according to &lt;a href=&quot;http://www.wolframalpha.com/input/?i=g00d&quot;&gt;WolframAlpha&lt;/a&gt;. 3v1l is a bit more complicated, but it turned out to be a &lt;a href=&quot;http://www.rcsb.org/pdb/explore.do?structureId=3v1l&quot;&gt;molecule&lt;/a&gt;! The EC code is &lt;strong&gt;3718&lt;/strong&gt;. Adding that up gives: &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 5560&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or, if you have the logical (and bitwise I suppose it is called) AND it gives :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0000, because no bits are the same&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;about the hints:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I used the hints, because 'and' is the + or AND operator (and should not be written as 4nd) and changing any of the other letters to something leetspeek, it will change the whole outcome.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3993" LastActivityDate="2014-11-13T14:09:44.080" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4464" PostTypeId="1" AcceptedAnswerId="4502" CreationDate="2014-11-13T14:20:49.973" Score="6" ViewCount="1949" Body="&lt;pre&gt;Everybody should be able to solve this one!&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You might try to pet me or kick my butt&lt;br&gt;&#xA;but you can not touch me no matter what&lt;br&gt;&lt;br&gt;&#xA;If you gave me presents I'd leave you for sure&lt;br&gt;&#xA;but take what's around me and I will endure&lt;br&gt;&lt;br&gt;&#xA;You can search the whole ocean or even beyond&lt;br&gt;&#xA;in sometimes for sure, a loch or a pond&lt;br&gt;&lt;br&gt;&#xA;If you see what's behind it&lt;br&gt;&#xA;you are very close&lt;br&gt;&#xA;And as you assumed&lt;br&gt;&#xA;its right under your nose&lt;br&gt;&lt;br&gt;&lt;br&gt;&#xA;Additional verses&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;are not available for this is a really easy one!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since it is an easy one I want every single line ripped apart!&lt;/p&gt;&#xA;" OwnerUserId="5106" LastEditorUserId="11297" LastEditDate="2015-07-17T16:55:44.580" LastActivityDate="2015-07-17T16:55:44.580" Title="Holy as the elements" Tags="|riddle|word|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4465" PostTypeId="2" ParentId="4436" CreationDate="2014-11-13T14:21:23.380" Score="12" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You put salt in it before you sent it back.  That's a flavor that just doesn't belong in your coffee.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="358" LastActivityDate="2014-11-13T14:21:23.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4466" PostTypeId="2" ParentId="4436" CreationDate="2014-11-13T14:24:52.793" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You could still taste the fly in the cup.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2778" LastActivityDate="2014-11-13T14:24:52.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4468" PostTypeId="2" ParentId="4460" CreationDate="2014-11-13T14:27:24.177" Score="1" Body="&lt;p&gt;First &lt;strong&gt;(wrong)&lt;/strong&gt; try:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Supposing &lt;code&gt;AND&lt;/code&gt; stands for &quot;&lt;strong&gt;sum&lt;/strong&gt;&quot; I think:&lt;br/&gt;&#xA; &lt;strong&gt;&lt;code&gt;0615&lt;/code&gt;&lt;/strong&gt; &lt;br/&gt;&#xA; Because i changed all letters with corresponding ASCII code, and sum every letter in the word (&lt;code&gt;147 + 0 + 0 + 144 = 291&lt;/code&gt; and  &lt;code&gt;3 + 166 + 1 + 154 = 324&lt;/code&gt;). Adding them gives &lt;code&gt;615&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second try (thanks to @No. 7892142):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Changing even digits to ASCII, result is increased by 240, giving&#xA; &lt;code&gt;855&lt;/code&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4527" LastEditorUserId="4527" LastEditDate="2014-11-13T14:38:26.510" LastActivityDate="2014-11-13T14:38:26.510" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4469" PostTypeId="2" ParentId="4436" CreationDate="2014-11-13T14:29:40.533" Score="5" Body="&lt;p&gt;The customer in question is female, and there was lipstick on the cup.&lt;/p&gt;&#xA;" OwnerUserId="2628" LastActivityDate="2014-11-13T14:29:40.533" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4470" PostTypeId="2" ParentId="4326" CreationDate="2014-11-13T14:32:02.733" Score="4" Body="&lt;p&gt;Is the solution 1.006. Calculated as ζ(2), where we calculate for the next prime that is 13.&lt;/p&gt;&#xA;" OwnerUserId="5282" LastEditorUserId="5282" LastEditDate="2014-11-17T20:50:39.540" LastActivityDate="2014-11-17T20:50:39.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4472" PostTypeId="1" AcceptedAnswerId="4575" CreationDate="2014-11-13T14:35:08.243" Score="16" ViewCount="4171" Body="&lt;p&gt;I have 3 riddles for you that I got from &lt;a href=&quot;http://checkbit.ch/2014/09/21/riddles/&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Riddle 1:&#xA;  I am larger when cold and shrink when warm. You can see me, but you cannot touch me.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Riddle 2:&#xA;  When people need me, they throw me away. when I’m not needed they take me back.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Riddle 3:&#xA;  The more you give me, the richer you get. When you rob me, only you get poorer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5283" LastEditorUserId="17970" LastEditDate="2016-02-25T00:20:35.513" LastActivityDate="2016-06-07T01:17:22.187" Title="Homemade riddles" Tags="|riddle|" AnswerCount="8" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4473" PostTypeId="2" ParentId="4449" CreationDate="2014-11-13T14:36:46.160" Score="2" Body="&lt;p&gt;Joke answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RPG&#xA; &lt;P&gt;&#xA; I dunno. Blind guess. I look at that final line. There's 2 'fullstop' for stop. The last fullstop seems to be looped over. So 3 output letter&#xA; &lt;P&gt;&#xA; If this correct. Wow. But let's hope not.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Serious answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please wait. I'm compiling this using pen and paper.&lt;/p&gt;&#xA;" OwnerUserId="4995" LastActivityDate="2014-11-13T14:36:46.160" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4476" PostTypeId="2" ParentId="4464" CreationDate="2014-11-13T14:40:19.940" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bacteria&lt;br&gt;&#xA; you cant touch them&lt;br&gt;&#xA; present as in medicine&lt;br&gt;&#xA; it feeds of whats around it such as a host&lt;br&gt;&#xA; they survive underwater&lt;br&gt;&#xA; you look through them so youd be close looking anywhere&lt;br&gt;&#xA; they are in and around your nose.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5135" LastActivityDate="2014-11-13T14:40:19.940" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4477" PostTypeId="2" ParentId="4383" CreationDate="2014-11-13T14:46:55.237" Score="2" Body="&lt;p&gt;I believe it's an&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Idea&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It's a four letter word that matches the title (an interesting point). &lt;/li&gt;&#xA;&lt;li&gt;It can be used to block other ones or give you something to defend.&lt;/li&gt;&#xA;&lt;li&gt;Its sound is quiet (especially while forming one)&lt;/li&gt;&#xA;&lt;li&gt;Dumb ones are useless/irrelevant&lt;/li&gt;&#xA;&lt;li&gt;They're constantly talked about&lt;/li&gt;&#xA;&lt;li&gt;Point to the sky... Either the &quot;light bulb&quot; effect or something pithy about it &quot;taking flight&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I can't figure out the wood/iron clue, but I assume I'm missing something obscure, since the hint specifically calls it archaic.&lt;/p&gt;&#xA;" OwnerUserId="1563" LastActivityDate="2014-11-13T14:46:55.237" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4478" PostTypeId="2" ParentId="4460" CreationDate="2014-11-13T14:49:04.210" Score="2" Body="&lt;p&gt;The answer is of course&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1337, because every 4-digit code in an IT company is 1337.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5285" LastActivityDate="2014-11-13T14:49:04.210" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4479" PostTypeId="2" ParentId="4464" CreationDate="2014-11-13T14:50:24.537" Score="3" Body="&lt;p&gt;Well, it's definetely an interesting riddle. My try will be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; space (maybe empty space)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; air&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5283" LastActivityDate="2014-11-13T14:50:24.537" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4480" PostTypeId="1" AcceptedAnswerId="4483" CreationDate="2014-11-13T14:53:09.640" Score="20" ViewCount="3134" Body="&lt;p&gt;So... Second riddle for me. Somehow more difficult than the last one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Pick a side, then relax&lt;br&gt;&#xA;  it may take a hour or two.&lt;br&gt;&#xA;  If you throw it, then it cracks,&lt;br&gt;&#xA;  but it puzzles, why would you?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;It is one, yet has many;&lt;br&gt;&#xA;  with it figure what to do!&lt;br&gt;&#xA;  if you possibly own any,&lt;br&gt;&#xA;  that it's old you always knew.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Three Six Nine are the numbers&lt;br&gt;&#xA;  that describe me as a clue.&lt;br&gt;&#xA;  Ninety also, for the slumbers&lt;br&gt;&#xA;  which won't even read anew.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is it?&lt;/p&gt;&#xA;" OwnerUserId="5122" LastEditorDisplayName="user20" LastEditDate="2015-02-13T05:24:22.797" LastActivityDate="2015-03-08T22:38:54.340" Title="A riddle worth this site" Tags="|riddle|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4481" PostTypeId="2" ParentId="4472" CreationDate="2014-11-13T14:56:49.097" Score="2" Body="&lt;p&gt;Riddle 1: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gas&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Riddle 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Breath&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Riddle 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bank&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5122" LastActivityDate="2014-11-13T14:56:49.097" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4482" PostTypeId="2" ParentId="4472" CreationDate="2014-11-13T14:57:11.273" Score="3" Body="&lt;p&gt;#1 is tougher. It could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a star. &lt;a href=&quot;http://www.astro.keele.ac.uk/workx/starlife/StarpageS_26M.html&quot; rel=&quot;nofollow&quot;&gt;It heats up when contracting&lt;/a&gt; then stabilises, then cools and expands into a Red Giant. Feels a bit forced, though.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;#2 is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A boomerang&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;#3 is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A piggy bank&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2518" LastEditorUserId="2518" LastEditDate="2014-11-13T15:40:32.227" LastActivityDate="2014-11-13T15:40:32.227" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4483" PostTypeId="2" ParentId="4480" CreationDate="2014-11-13T14:59:43.143" Score="13" Body="&lt;p&gt;It is a..........&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rubik's Cube  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pick a side, then relax, it may take a hour or two.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; for less experienced people, it could take an hour or two to solve  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you throw it, then it cracks, but it puzzles, why would you?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it can break if you throw it, but why would you?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is one, yet has many; with it figure what to do!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it is one cube composed of 26 cubelets  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;if you possibly own any, that it's old you always knew.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rubik's cube has been available for more than 30 years  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Three Six Nine are the numbers that describe me as a clue.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The cube is 3x3x3 with six faces and nine colored squares on each face  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ninety also, for the slumbers which won't even read anew.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every move is a turn of a multiple of 90 degrees  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5287" LastEditorUserId="5188" LastEditDate="2015-03-08T22:38:54.340" LastActivityDate="2015-03-08T22:38:54.340" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4484" PostTypeId="2" ParentId="4480" CreationDate="2014-11-13T15:00:50.797" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;a rubik's cube? it is a square yet has many little square's, it may take an hour or two to solve&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5289" LastActivityDate="2014-11-13T15:00:50.797" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4485" PostTypeId="2" ParentId="4472" CreationDate="2014-11-13T15:05:15.843" Score="2" Body="&lt;p&gt;&lt;s&gt; #1 &lt;/s&gt;  &lt;strong&gt;(wrong)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Sun, that is closer to Earth in winter and further in summer (in boreal hemisphere)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h1&gt;1&lt;/h1&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; night, that is longer when climate is cold, and shorter otherwise&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h1&gt;2&lt;/h1&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; a ball, when you need it you kick it away, when you finish to play you take it home&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4527" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-11-13T15:39:29.687" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4486" PostTypeId="2" ParentId="4068" CreationDate="2014-11-13T15:16:19.927" Score="13" Body="&lt;p&gt;ANSWER:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The murderer is the detective Randy White&lt;br/&gt;&lt;br/&gt;Mr. White based everything on the classic Disney movie Snow white and the 7 dwarves.&lt;br/&gt;&lt;br/&gt;1) you found the file in your desk, no one else enters your office except for your assistant and other co-workers if they need to talk to you while you are present in the office. Your office is for the TOP DETECTIVE in the department, and other detectives work outside of your office, safe to assume in cubicals and such. Since the file was in your office in your desk (keep in mind that you are still new to Chicago and just moved into your new office not too long ago), it is safe to assume that there was another top detective before you arrived. They tried to clean up everything but might have forgotten a few things. Thus placing Mr. White in the office with his file.&lt;br/&gt;&lt;br/&gt;2) Since Mr. White worked there, it is not suspicious if someone saw him walking out/in the front door, thus letting him wait till no one is around then place the note on the front door of the department. &lt;br/&gt;&lt;br/&gt;3) the letter says &quot;I am invisible like objects in a December storm&quot;, December Storm can be translated to Snow Storm, combined with hs name Randy White will equal Snow White. (Not with officer Snoe, which will be equal to Snow Snow). &lt;br/&gt;&lt;br/&gt;4) for all 7 victims Mr. White killed they were short (dwarves), and if you take the First Letter of the first name of every victim, it spells out &quot;APPLE&quot;, thus why the 6th and 7th victims did not have a wallet to tell them the victims name.&lt;br/&gt;&lt;br/&gt;5) Mr. White used deadly poison to kill the victims like how the evil witch poisoned the apple and gave it to Snow White.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4645" LastActivityDate="2014-11-13T15:16:19.927" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4487" PostTypeId="2" ParentId="4460" CreationDate="2014-11-13T15:17:45.103" Score="1" Body="&lt;p&gt;I'm guessing the answer you're going for is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3377&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;since that's just the decimal form of the hexadecimal&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;s&gt;g&lt;/s&gt;00d 3&lt;s&gt;v&lt;/s&gt;1&lt;s&gt;l&lt;/s&gt; (00d31)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, there's no way to be sure. I'm still not a fan of the ambiguity in this.&lt;/p&gt;&#xA;" OwnerUserId="1563" LastActivityDate="2014-11-13T15:17:45.103" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4488" PostTypeId="2" ParentId="4436" CreationDate="2014-11-13T15:24:22.267" Score="3" Body="&lt;p&gt;The manager came back too fast.&#xA;In a five star hotel, one can expect good, fresh made coffee - and such coffee needs &gt;= 5 minutes in preparation and letting it steep.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because good coffee loses too much flavour when not grinded literally just seconds before brewing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if the manager came back in under five minutes, he could not bring a new cup of coffee.&lt;/p&gt;&#xA;" OwnerUserId="5285" LastActivityDate="2014-11-13T15:24:22.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4490" PostTypeId="2" ParentId="4327" CreationDate="2014-11-13T15:28:41.560" Score="2" Body="&lt;p&gt;The correct answer has been posted &lt;a href=&quot;https://puzzling.stackexchange.com/a/4377/5160&quot;&gt;here&lt;/a&gt; by @Noldor130884! This post will just clarify some of my thinking. Comments on this riddle are welcome so I can improve the next one!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple of additions to Noldor's (awesome) explanation:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Be careful you don't blow it all&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Also refers to &lt;a href=&quot;http://en.wikipedia.org/wiki/Glassblowing&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Glassblowing&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Its countenance is clear to you.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Also, it is literally clear (to look at)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Salute with it, it will surpass / A bow to temper, or a kneel.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Cheers&quot;, as @d'alar'cop mentioned. Take a drink with someone, it's a better way of 'tempering' them than bowing or kneeling... but I was really trying to work in &lt;a href=&quot;http://en.wikipedia.org/wiki/Tempered_glass&quot; rel=&quot;nofollow noreferrer&quot;&gt;temper&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Annealing_(glass)&quot; rel=&quot;nofollow noreferrer&quot;&gt;anneal&lt;/a&gt; :-)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2014-11-13T15:28:41.560" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4492" PostTypeId="2" ParentId="4472" CreationDate="2014-11-13T15:47:39.990" Score="5" Body="&lt;p&gt;I see that #1 and #3 have already been answered, but is the answer to #2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An anchor&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-11-13T15:47:39.990" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4493" PostTypeId="2" ParentId="4449" CreationDate="2014-11-13T15:59:26.110" Score="1" Body="&lt;p&gt;I think the killer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;a href=&quot;https://communitybuilding.stackexchange.com/users/135/tim-post&quot;&gt;Tim Post&lt;/a&gt; because he is the only Tim that appears both in Moderators and puzzling.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The site is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Judaism stack exchange: I just googled &quot;earth, wind, fire and water stackexchange&quot;&lt;br&gt;&#xA; Probably not a good answer after all.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5295" LastEditorUserId="-1" LastEditDate="2017-04-12T07:32:58.753" LastActivityDate="2014-11-13T15:59:26.110" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4495" PostTypeId="2" ParentId="4383" CreationDate="2014-11-13T16:20:51.073" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;moot. Means &quot;irrelevant&quot;, sounds like &quot;mute&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5299" LastActivityDate="2014-11-13T16:20:51.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4497" PostTypeId="2" ParentId="4464" CreationDate="2014-11-13T16:32:16.580" Score="3" Body="&lt;p&gt;I'm stumped with a few of these lines, they're always excluding some ideas I have.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;But my best shot is a &quot;riddle&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You might try to pet me or kick my butt&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Implies liking it, and / or trying to solve it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but you can not touch me no matter what&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A concept&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you gave me presents I'd leave you for sure&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Answer me and I'm done&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but take what's around me and I will endure&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I'll keep nagging your brain if you give me false answers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can search the whole ocean or even beyond&#xA;in sometimes for sure, a loch or a pond&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Not sure how this could fit though.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you see whats behind it you are very close&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Seeing the tricks / word plays for what they are&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And as you assumed its right under your nose&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I have a riddle in front of me&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5107" LastEditorUserId="2484" LastEditDate="2014-11-13T17:13:56.193" LastActivityDate="2014-11-13T17:13:56.193" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4498" PostTypeId="1" AcceptedAnswerId="4500" CreationDate="2014-11-13T16:38:28.117" Score="9" ViewCount="849" Body="&lt;p&gt;I answer all who can connect&lt;br&gt;&#xA;I amend mistakes without respect  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I aim to place, I aim to serve&lt;br&gt;&#xA;I give you knowledge with little reserve  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In me, the primary colours I do hold&lt;br&gt;&#xA;Though two had a child who won't grow old&lt;/p&gt;&#xA;" OwnerUserId="5135" LastEditorUserId="5373" LastEditDate="2015-07-16T21:45:58.450" LastActivityDate="2015-07-16T21:45:58.450" Title="Name that entity (2)" Tags="|riddle|word|rhyme|" AnswerCount="4" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4499" PostTypeId="2" ParentId="4498" CreationDate="2014-11-13T16:48:52.073" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The spellchecker on Microsoft Word&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;In me, the primary colours I do hold&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spellchecker uses three colors: green for grammar errors, red for spelling errors, and blue for miscellaneous others. These are considered by some to be the primary colors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I amend mistakes with little regret&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Haven't we all been annoyed when it automatically, incorrectly fixes &quot;errors&quot;?&lt;/p&gt;&#xA;" OwnerUserId="5300" LastActivityDate="2014-11-13T16:48:52.073" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4500" PostTypeId="2" ParentId="4498" CreationDate="2014-11-13T16:49:26.633" Score="9" Body="&lt;p&gt;Are you:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Google&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I answer all who can connect&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Connect to the Internet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I amend mistakes without respect&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Suggestions!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I aim to place, I aim to serve&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A free service&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I give you knowledge with little reserve&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An essential source of knowledge these days.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In me, the primary colours I do hold&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The logo is Blue, Red, Yellow (and Green) - NB: This refers to Subtractive Primary Colours (unlike those usually used in computer devices (Additive Colours))&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though two had a child who wont grow old&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Green is a child of Blue and Yellow &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-11-24T13:01:35.540" LastActivityDate="2014-11-24T13:01:35.540" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4501" PostTypeId="2" ParentId="4498" CreationDate="2014-11-13T16:53:34.340" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Google. It answers anyone who can connect to it, and it corrects anything that might be a mistake in your search terms. The logo contains the three primary colors, and a fourth color that is a mixture of two of them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5317" LastActivityDate="2014-11-13T16:53:34.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4502" PostTypeId="2" ParentId="4464" CreationDate="2014-11-13T16:54:57.627" Score="9" Body="&lt;p&gt;Are you:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A HOLE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might try to pet me or kick my butt&#xA;but you can not touch me no matter what&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can caress the abyss or kick it but you will never make contact. &quot;No matter&quot; may be a clue here. There are some more graphic and direct applications of petting and kicking with regards to a HOLE, but never mind.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you gave me presents I'd leave you for sure&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If one fills a hole it is no longer a hole&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but take what's around me and I will endure&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can make a hole bigger and it is still a hole&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can search the whole ocean or even beyond&#xA;in sometimes for sure, a loch or a pond&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Very clever. &quot;Whole&quot;=HOLE. &quot;Loch&quot; = HOLE (in German). Also note the presence of &quot;O&quot; in all the key words.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you see whats behind it&#xA;you are very close&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you can see the bottom of the hole, you must be close to it (or even in it).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And as you assumed&#xA;its right under your nose&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your mouth is a HOLE of sorts :O &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-11-14T01:20:25.567" LastActivityDate="2014-11-14T01:20:25.567" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="4503" PostTypeId="1" AcceptedAnswerId="4509" CreationDate="2014-11-13T17:05:56.533" Score="4" ViewCount="1790" Body="&lt;p&gt;I tossed off this quick one. Does that make me a tosser?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know what is or isn't,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;but never what might be;&lt;br&gt;&#xA;And many claim to know me,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;though many disagree;&lt;br&gt;&#xA;There are no sly dilemmas,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;no subtleties I see;&lt;br&gt;&#xA;Both I encompass gladly,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;to be, and not to be.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's two long words, but you can use just one, provided it's short.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5313" LastEditorUserId="39" LastEditDate="2015-07-17T02:27:37.950" LastActivityDate="2015-07-17T14:20:57.667" Title="It rhymes and wonders: who am I?" Tags="|riddle|rhyme|computer-puzzle|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4506" PostTypeId="1" AcceptedAnswerId="4538" CreationDate="2014-11-13T17:15:26.337" Score="9" ViewCount="479" Body="&lt;p&gt;You're browsing Puzzling.SE when suddenly you uncover a massive conspiracy. The posters of the &quot;Security to the Party&quot; questions are just posting 7 or 8 random numbers and then accepting the second or third answer which explains the data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Infuriated, you attempt to break into the Dark Puzzlers Chat. However, there is a guard, et cetera.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A puzzler arrives. The guard says &lt;code&gt;1 5 2 1 4 3 8&lt;/code&gt;, and she replies &lt;code&gt;10 2 6&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second puzzler appears. The guard says &lt;code&gt;CODREEF&lt;/code&gt;, and he replies &lt;code&gt;PSF&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A third puzzler arrives. The guard says &lt;code&gt;1.3 2.0 1.6 1.8 2.2 2.3 2.4&lt;/code&gt;, and the puzzler replies &lt;code&gt;2.0 2.6 2.6&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the fourth puzzler, the guard says &lt;code&gt;6, 3x^2+1, 6x+8, 6(x+6)^2, (x+1)(3x+5), sin(x)/x, (x+1)^2(x+2)&lt;/code&gt;. The puzzler complains that the last one is impossible and leaves in a huff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, you've heard enough, because anyone can work out any pattern after four examples. You approach the guard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The guard says &lt;code&gt;Bleach as needed, Doctor Whooves's cutie mark, Play video, Infinity, Give way to traffic, Cancer horoscope, Normal subgroup.&lt;/code&gt; What should you reply?&lt;/p&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2022-10-07T06:26:43.713" Title="[Security to the Party 27: Conspiracy Edition]" Tags="|password|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4509" PostTypeId="2" ParentId="4503" CreationDate="2014-11-13T17:24:24.300" Score="10" Body="&lt;p&gt;You said in a comment to think computers, so I did.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A boolean variable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know what is or isn't,&lt;br&gt;&#xA;  but never what might be;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Which is the case with variables. You know what type it is/isn't, but not what its value will end up being.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And many claim to know me,&lt;br&gt;&#xA;  though many disagree;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rather vague statement, as it's impossible to discern the value of a variable just by looking at it, even in context. With booleans, though, a logical argument can be made both ways.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are no sly dilemmas,&lt;br&gt;&#xA;  no subtleties I see;&lt;br&gt;&#xA;Both I encompass gladly,&lt;br&gt;&#xA;  to be, and not to be.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; But in the end, a boolean variable holds nothing more than the result of a condition; true or false.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2444" LastEditorUserId="2444" LastEditDate="2015-07-17T14:20:57.667" LastActivityDate="2015-07-17T14:20:57.667" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4510" PostTypeId="2" ParentId="4417" CreationDate="2014-11-13T17:27:29.300" Score="0" Body="&lt;p&gt;I can't seem to do any better than this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. North America&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The birthplace:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jamestown, VA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Biggest Child&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Georgia is the largest of the thirteen colonies&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To the left&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alabama&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="51" LastActivityDate="2014-11-13T17:27:29.300" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4512" PostTypeId="1" AcceptedAnswerId="4517" CreationDate="2014-11-13T17:36:20.427" Score="-5" ViewCount="323" Body="&lt;pre&gt;&lt;code&gt;In the morning you brush your teeth with it. (hopefully)&#xA;You can sit on it.&#xA;At night you lay down on it.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;No hints needed, my father told me that one ages ago and I still like it.&lt;/p&gt;&#xA;" OwnerUserId="5278" LastActivityDate="2018-03-21T21:09:13.500" Title="What do I search for?" Tags="|riddle|" AnswerCount="5" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4513" PostTypeId="2" ParentId="4512" CreationDate="2014-11-13T17:38:41.803" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nylon, all three you described can be made of this?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5135" LastActivityDate="2014-11-13T17:38:41.803" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4514" PostTypeId="2" ParentId="4436" CreationDate="2014-11-13T17:44:51.983" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;The manager apologized, took back the cup and brought another fresh hot one.&quot;&lt;br&gt;&#xA;&lt;br&gt;&#xA; He brought back another fresh, hot &lt;strong&gt;cup&lt;/strong&gt;. The coffee in the cup was the same coffee, and still had the fly in it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5329" LastActivityDate="2014-11-13T17:44:51.983" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4515" PostTypeId="2" ParentId="4436" CreationDate="2014-11-13T17:45:00.073" Score="4" Body="&lt;p&gt;What if:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the cup was chipped and so recognizable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5313" LastActivityDate="2014-11-13T17:45:00.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4516" PostTypeId="2" ParentId="4512" CreationDate="2014-11-13T17:45:19.990" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your hand? Assuming you sit and sleep in a weird position&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="255" LastActivityDate="2014-11-13T17:45:19.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4517" PostTypeId="2" ParentId="4512" CreationDate="2014-11-13T17:46:55.277" Score="1" Body="&lt;p&gt;You search for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a toothbrush, a chair and a bed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At least that's how it went when I was a child.&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2014-11-13T17:46:55.277" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4518" PostTypeId="2" ParentId="4512" CreationDate="2014-11-13T17:48:48.973" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A pair of &lt;strong&gt;pyjamas&lt;/strong&gt;. In the morning you brush your teeth while you are still wearing it. You can of course sit down while wearing it. And in the night, you go to sleep with it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5324" LastEditorUserId="2383" LastEditDate="2014-11-13T17:57:13.557" LastActivityDate="2014-11-13T17:57:13.557" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4519" PostTypeId="2" ParentId="4436" CreationDate="2014-11-13T17:56:09.407" Score="3" Body="&lt;p&gt;&quot;But I realized that he &lt;strong&gt;bought&lt;/strong&gt; the same coffee.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;How could I recognize that the manager has &lt;strong&gt;bought&lt;/strong&gt; the same coffee?&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The receipt was left on your table.  In plain text you saw: Medium Fly Coffee $2.99&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5333" LastEditorUserId="5333" LastEditDate="2014-11-13T18:01:30.850" LastActivityDate="2014-11-13T18:01:30.850" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4520" PostTypeId="2" ParentId="4449" CreationDate="2014-11-13T17:58:43.160" Score="18" Body="&lt;p&gt;This is all I've got so far but I'm at work and shouldn't be on PSE. :)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Check out Area 51.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Collectively earth, wind, fire, and water are the elements.&#xA; Badly annoy Tim: anagram ANNOYTIM into the element antimony, symbol Sb, atomic number 51.&#xA; &quot;We are not alone&quot; also fits Area 51.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4373" LastActivityDate="2014-11-13T17:58:43.160" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4522" PostTypeId="2" ParentId="4436" CreationDate="2014-11-13T18:09:25.980" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The cup was hot, but the coffee was lukewarm &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="220" LastActivityDate="2014-11-13T18:09:25.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4523" PostTypeId="2" ParentId="4436" CreationDate="2014-11-13T18:10:11.800" Score="0" Body="&lt;p&gt;You did what everybody would do in these cases.  You put the cup in a plastic bag and sent it to the lab for fingerprints.  And the fly for autopsy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you find your own fingerprints, that means it is the same cup.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2014-11-14T08:31:17.260" LastActivityDate="2014-11-14T08:31:17.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4525" PostTypeId="1" AcceptedAnswerId="4549" CreationDate="2014-11-13T18:13:10.457" Score="-4" ViewCount="442" Body="&lt;p&gt;You gave up trying to get the recipes, and go home. You get on Puzzling.SE to find your friend has been recording all your attempts (and his elephant failures).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also find when you try to log into Puzzling.SE You get redirected to a question called &quot;Security to your account!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The body goes like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You try to get on Puzzling.SE so you can answer puzzles but find that a hacker has set up a password system. You wait to listen to traffic to see how he lets people in.    &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;While analyzing traffic packets (You little hacker!) you get the following responses:    &lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;&quot;5&quot; -&gt; &quot;Impossible&quot;&lt;br&gt;&#xA;    &quot;6&quot; -&gt; &quot;Impossible&quot;    &lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;  &#xA;  &lt;p&gt;Then some bright guy who's also analyzing traffic thinks he's got it.    &lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;&quot;0&quot; -&gt; &quot;Impossible&quot;    &lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;  &#xA;  &lt;p&gt;His computer then shuts down forever.&lt;br&gt;&#xA;  Nobody else comes for a while so it's your turn now!&lt;br&gt;&#xA;  Your question is &quot;Infinity&quot;. What's your answer? Also, what was the correct answer to 0? Why?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To get it you know you have to answer the question. What is the answer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hints:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What is and what can't is elementary.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Let it be known there are inverse operations!&lt;br&gt;&#xA;&lt;br&gt;&#xA; You are in the basic math division.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;BonusResponses:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;7&quot; -&gt; &quot;Impossible&quot;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &quot;8&quot;-&gt; &quot;Impossible&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-11-13T22:41:43.047" LastActivityDate="2014-11-14T02:02:42.503" Title="Security to the Party [Part 28!] [Hacked Eddition]" Tags="|password|" AnswerCount="4" CommentCount="11" ClosedDate="2014-11-14T03:14:12.310" ContentLicense="CC BY-SA 3.0" />
  <row Id="4526" PostTypeId="2" ParentId="4346" CreationDate="2014-11-13T18:17:27.170" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TeV&lt;br&gt;&#xA;&lt;br&gt;&#xA; The plumber will almost certainly pronounce it as the Latvian dative second person singular pronoun, whereas the particle physicist will read it as tera-electron-volt.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="228" LastActivityDate="2014-11-13T18:17:27.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4527" PostTypeId="2" ParentId="4436" CreationDate="2014-11-13T18:19:42.477" Score="0" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantharidin#Medical_uses&quot; rel=&quot;nofollow&quot;&gt;fly&lt;/a&gt; was was a fly of a different sort...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; spanish fly&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="220" LastActivityDate="2014-11-13T18:19:42.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4528" PostTypeId="1" AcceptedAnswerId="4529" CreationDate="2014-11-13T18:19:57.773" Score="3" ViewCount="826" Body="&lt;p&gt;Possibly known to all, especially to your drums &lt;br /&gt;&#xA;Particularly when a police officer comes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sultry grin, haughty glance &lt;br /&gt;&#xA;Very few have ever eyed askance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Red, flowing hair my long distant cousin has &lt;br /&gt;&#xA;For in the minds of children is where she is known as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sweetest of melody, behold the wile I procure &lt;br /&gt;&#xA;More trouble for you should you not resist my allure&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fall in and over, do not dismay &lt;br /&gt;&#xA;Eternity with me is but a breath away&lt;/p&gt;&#xA;" OwnerUserId="4689" LastActivityDate="2014-11-13T18:30:12.377" Title="What could I possibly be?" Tags="|riddle|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4529" PostTypeId="2" ParentId="4528" CreationDate="2014-11-13T18:21:49.733" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; she's a siren.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Her cousin is a mermaid, she sings as allure that cannot be escaped. Also, police car siren.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="51" LastActivityDate="2014-11-13T18:21:49.733" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4530" PostTypeId="2" ParentId="4464" CreationDate="2014-11-13T18:22:33.920" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Kim Kardashian&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You might try to pet me or kick my butt&#xA;but you can not touch me no matter what&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; She's pretty much unavailable&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you gave me presents I'd leave you for sure&#xA;but take what's around me and I will endure&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; She's not into nice guys&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can search the whole ocean or even beyond&#xA;in sometimes for sure, a loch or a pond&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; She looks good in a bikini (near bodies of water)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you see whats behind it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Giggity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;you are very close&#xA;And as you assumed&#xA;its right under your nose&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; She shows up everywhere; the media love her for some reason&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5341" LastActivityDate="2014-11-13T18:22:33.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4531" PostTypeId="2" ParentId="4528" CreationDate="2014-11-13T18:22:37.007" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are a siren.  The word has two meanings, as explained below.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Possibly known to all, especially to your drums&#xA;Particularly when a police officer comes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A police car siren.  Drums referring to eardrums.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The rest refers to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The siren from Greek mythology  who sings irresistably with a female voice, luring sailors to their deaths if they don't resist her sweet song.  Children learn this myth in school.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4450" LastEditDate="2014-11-13T18:30:12.377" LastActivityDate="2014-11-13T18:30:12.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4532" PostTypeId="2" ParentId="4528" CreationDate="2014-11-13T18:26:24.787" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A siren, particularly she' she cousin of Ariel from the tale of Andersen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user1962" LastActivityDate="2014-11-13T18:26:24.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4533" PostTypeId="1" AcceptedAnswerId="4542" CreationDate="2014-11-13T18:30:56.973" Score="17" ViewCount="1607" Body="&lt;p&gt;Let &lt;em&gt;M&lt;/em&gt; be a finite positive integer. It's exact value is not known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we have three classes of automaton, all of which accept a bit stream as input, produce a bit stream as output (one bit per input bit), and have a non-observable state, &lt;em&gt;S&lt;/em&gt;, that is an integer coordinate on an infinite bidirectional 1D line:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;knightbots&lt;/strong&gt; start with &lt;em&gt;S = M&lt;/em&gt;. Whenever a &lt;code&gt;1&lt;/code&gt; bit is inputted, knights move 2 units to the right (i.e. &lt;em&gt;S&lt;/em&gt; is incremented by 2). Whenever a &lt;code&gt;0&lt;/code&gt; bit is inputted, knights move 1 unit to the left (i.e. &lt;em&gt;S&lt;/em&gt; is decremented by 1).&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;knavebots&lt;/strong&gt; start with &lt;em&gt;S&lt;/em&gt; = -&lt;em&gt;M&lt;/em&gt;. Whenever a &lt;code&gt;1&lt;/code&gt; bit is inputted, knaves move 1 unit to the right. Whenever a &lt;code&gt;0&lt;/code&gt; bit is inputted, knaves move 2 units to the left.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;jokerbots&lt;/strong&gt; start randomly at either &lt;em&gt;S = M&lt;/em&gt; or &lt;em&gt;S&lt;/em&gt; = -&lt;em&gt;M&lt;/em&gt;. On cromulent days, jokers move 1 unit right on a &lt;code&gt;1&lt;/code&gt; input and 1 unit left on a &lt;code&gt;0&lt;/code&gt; input, but on non-cromulent days, jokers move 1 unit &lt;em&gt;left&lt;/em&gt; on a &lt;code&gt;1&lt;/code&gt; input and 1 unit &lt;em&gt;right&lt;/em&gt; on a &lt;code&gt;0&lt;/code&gt; input. You have no idea whether or not the current day is cromulent, and there's really no way to find out.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;After each input, the automata output either a &lt;code&gt;1&lt;/code&gt; if they have moved into the state &lt;em&gt;S&lt;/em&gt; = 0, or a &lt;code&gt;0&lt;/code&gt; otherwise.&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Simple Problem (for &quot;Correct Answer&quot; Credit)&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;You are given an automaton that is either a knightbot, a knavebot, or a jokerbot, but you don't know which of the three. The automaton is guaranteed to be in its starting state. You can input bits to the automaton and observe its output, but otherwise you cannot observe its state or inner workings. Also note that you cannot &quot;reset&quot; the automaton.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your task is to produce an algorithm that is mathematically guaranteed to properly classify the automaton as a knight, knave, or joker using an input sequence of &lt;strong&gt;finite&lt;/strong&gt; length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your solution should include a (reasonably) detailed description of your algorithm, as well as a guaranteed (although not necessarily tight) upper bound on required input length as a function of &lt;em&gt;M&lt;/em&gt;.&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Challenge Problem (for 100 Bonus Rep!)&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Instead of the aforementioned &lt;strong&gt;jokerbots&lt;/strong&gt;, consider &lt;strong&gt;jesterbots&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;jesterbots start with &lt;em&gt;S&lt;/em&gt; start randomly at either &lt;em&gt;S = M&lt;/em&gt; or &lt;em&gt;S&lt;/em&gt; = -&lt;em&gt;M&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;jesterbots have a secondary state, &lt;em&gt;b&lt;/em&gt;, which assumes the value of the most-recently-inputted bit. The initial value of &lt;em&gt;b&lt;/em&gt; is &lt;code&gt;0&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;li&gt;jesterbots have a tertiary state, &lt;em&gt;D&lt;/em&gt;, which takes on the values &lt;code&gt;1&lt;/code&gt; or &lt;code&gt;-1&lt;/code&gt;, and may be treated as a direction. The initial value of &lt;em&gt;D&lt;/em&gt; is &lt;code&gt;1&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;li&gt;on cromulent days, jesterbots move in the direction of &lt;em&gt;D&lt;/em&gt; (i.e. increment &lt;em&gt;S&lt;/em&gt; by &lt;em&gt;D&lt;/em&gt;) iff an input bit matches &lt;em&gt;b&lt;/em&gt; (the previous input bit), or else stand still and about face (multiply &lt;em&gt;D&lt;/em&gt; by &lt;code&gt;-1&lt;/code&gt;)&lt;/li&gt;&#xA;&lt;li&gt;on non-cromulent days, jesterbots move in the direction of &lt;em&gt;D&lt;/em&gt; iff an input bit does &lt;em&gt;not&lt;/em&gt; match &lt;em&gt;b&lt;/em&gt;, or else stand still and about face&lt;/li&gt;&#xA;&lt;li&gt;after each input, the jesterbot outputs a &lt;code&gt;1&lt;/code&gt; if it has moved into the state &lt;em&gt;S&lt;/em&gt; = 0, or a &lt;code&gt;0&lt;/code&gt; otherwise&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Your task is identical to that for the simple problem, only substituting jester(bot) for joker(bot) in the task description.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will award 100 bounty rep to the correct, defensible solution to this challenge problem that has the strictest upper input length bound (in terms of $\mathcal{O}\left( M\right)$), or to the earliest solution in the case of a tie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck automating. :)&lt;/p&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="8874" LastEditDate="2015-02-02T09:24:15.383" LastActivityDate="2015-02-02T09:24:15.383" Title="Automatically a Knight, Knave, and Joker" Tags="|mathematics|combinatorics|liars|game|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="4534" PostTypeId="2" ParentId="4446" CreationDate="2014-11-13T18:34:16.473" Score="1" Body="&lt;p&gt;The shape has to be defined by three circular arcs:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;one arc on the right side of L, between two points that I'll call A and B&lt;/li&gt;&#xA;&lt;li&gt;one between P and A&lt;/li&gt;&#xA;&lt;li&gt;one between P and B&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Each three having curvature away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ratioanle:&#xA;Given any optimal shape maximizing the weighted enclosed area, one can fix the points A and B as well as the line length in each area. Then, by keeping A and B and the segment lengths, each area is maximize by using a circular arc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to find the optimal shape, just do a search by fixing the three lengths (so they sum to 100 m) and picking A and B. So, finding the maximum of a well-defined function over 4 variables (the last length is a function of the other two).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shouldn't be to hard from here. :-)&lt;/p&gt;&#xA;" OwnerUserId="5343" LastActivityDate="2014-11-13T18:34:16.473" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4535" PostTypeId="2" ParentId="4446" CreationDate="2014-11-13T19:00:10.193" Score="3" Body="&lt;h1&gt;Best value: 1116.34 (two semiellipses)&lt;/h1&gt;&#xA;&lt;p&gt;This question is interesting. I would like to answer it.&lt;/p&gt;&#xA;&lt;p&gt;The most efficient shape is the a raindrop shape type. That is: an isosceles triangle with a semicircle.&lt;/p&gt;&#xA;&lt;p&gt;Indeed, we are looking to maximize the area in the right because of the &lt;code&gt;x2 incentive&lt;/code&gt;. The best way to maximize the area of a shape with a cord is a circle. We can't create a full circle (P would be outside of it otherwise), so the other side of the circle has to be transformed.&lt;/p&gt;&#xA;&lt;p&gt;We are looking to minimize the area in the left side because of the &lt;code&gt;x1 deterrent effect&lt;/code&gt;. The best way to minimize an area with a cord is a segment. But the &lt;em&gt;lollipop&lt;/em&gt; strategy shows that it reduces dramaticaly the size of a full circle in the right area. So we need 3 sides; we need a triangle. The triangle can have straight sides or outwards curved sides (see below).&lt;/p&gt;&#xA;&lt;p&gt;Last but not least, the shape must be symmetric. If not, the area of the semicircle is grazed by that of the triangle (&lt;a href=&quot;https://puzzling.stackexchange.com/a/4459/5335&quot;&gt;see this example&lt;/a&gt;). Of course, the symmetry is not vertical (because we have two very different shapes), but horizontal.&lt;/p&gt;&#xA;&lt;p&gt;A triangle + a semicircle + a mandatory horizontal symmetry = this shape, &lt;a href=&quot;https://puzzling.stackexchange.com/a/4447/5335&quot;&gt;as discovered by our friend&lt;/a&gt;.Another option: two half-ellipses side by side (see below).&lt;/p&gt;&#xA;&lt;h2&gt;Shape 1: a triangle with straight sides : 1083.63&lt;/h2&gt;&#xA;&lt;p&gt;We need to compute the best areas of the triangle and the semicircle in order to have the highest value. To do so, I would like to add some information:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sY05f.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a) I add &lt;code&gt;q&lt;/code&gt; at the center of the side of the triangle that is on L.&lt;/li&gt;&#xA;&lt;li&gt;b) I add &lt;code&gt;r&lt;/code&gt; at one end of this side.&lt;/li&gt;&#xA;&lt;li&gt;c) Because the triangle is isosceles (horizontal symmetry), &lt;code&gt;Pq&lt;/code&gt; is&#xA;the altitude of the triangle. Its lenght is 20 (&lt;code&gt;Pq = 20&lt;/code&gt;). Moreover, for the same reason, angle&#xA;&lt;code&gt;Pqr = 90°&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;li&gt;d) I name &lt;code&gt;i&lt;/code&gt; the length of the side of the triangle that is on L, at&#xA;the same time the diameter of the semicircle.&lt;/li&gt;&#xA;&lt;li&gt;e) &lt;code&gt;A1&lt;/code&gt; is the area of the triangle; &lt;code&gt;A2&lt;/code&gt; the area of the semicircle.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Now, we know how to compute the area of a triangle with its altitude (20) and the base (&lt;code&gt;i&lt;/code&gt;), and we know how to compute the area of a semicircle with its radius (here: &lt;code&gt;i/2&lt;/code&gt;). If we find the optimal length &lt;code&gt;i&lt;/code&gt;, then we have the solution.&lt;/p&gt;&#xA;&lt;p&gt;To find &lt;code&gt;i&lt;/code&gt;, I have coded a small c++ program.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;#include &amp;lt;iostream&amp;gt;&#xA;#include &amp;lt;math.h&amp;gt;&#xA;&#xA;#define PI 3.1415926535&#xA;&#xA;int main()&#xA;{&#xA;    double i, // the length we are looking for&#xA;           area1, // area of the triangle&#xA;           area2, // area of the circle&#xA;           perimeter, // perimeter of the whole shape&#xA;           value; // the value to maximize&#xA;&#xA;    for (i = 0; i &amp;lt; 200; i=i+0.001) { // 200 is an arbitrary integer way outside the maximum expected value of i&#xA;        area1 = (20 * i) / 2; // classic formula of the area of a triangle&#xA;        area2 = (PI*((i/2)*(i/2)))/2; // (classic formula of the area of a circle)/2&#xA;        value = area1 + area2 * 2; // We compute the expected value&#xA;        perimeter = (sqrt((20 * 20) + (i / 2)*(i / 2))) * 2// Perimeter = lengths of the sides minus the base (Pythagorean theorem)...&#xA;            + (i*PI) / 2; // ... plus (perimeter of a circle)/2&#xA;        if (perimeter &amp;gt; 99.999 &amp;amp;&amp;amp; perimeter &amp;lt; 100.001) // if the perimeter +/- = 100, then show the solution&#xA;            std::cout &amp;lt;&amp;lt; &amp;quot;i: &amp;quot; &amp;lt;&amp;lt; i &amp;lt;&amp;lt; &amp;quot; | p: &amp;quot; &amp;lt;&amp;lt; perimeter &amp;lt;&amp;lt; &amp;quot; | a1: &amp;quot; &amp;lt;&amp;lt; area1 &amp;lt;&amp;lt; &amp;quot; | a2 : &amp;quot; &amp;lt;&amp;lt; area2 &amp;lt;&amp;lt; &amp;quot; | value : &amp;quot;&amp;lt;&amp;lt; value &amp;lt;&amp;lt; std::endl;&#xA;    }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;&lt;strong&gt;After having incremented &lt;em&gt;i&lt;/em&gt; by 0.001 steps, it appears that the optimal data are (for this shape):&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;i:&lt;/strong&gt; 31.32 m. | &lt;strong&gt;p.:&lt;/strong&gt; 100 (+/- .001) | &lt;strong&gt;area 1:&lt;/strong&gt; 313.2 | &lt;strong&gt;area 2:&lt;/strong&gt; 385.215&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Value: 1083.63&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;Shape 2: a triangle with outwards curved sides (semiellipse+semicircle) : 1113.6&lt;/h2&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eCB8o.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;We suppose that the triangle is composed of 2 quarters of ellipse. See the following image. The curved triangle is &lt;code&gt;ABC&lt;/code&gt;. It is composed of two quarters of an ellipse : &lt;code&gt;ABO&lt;/code&gt; and &lt;code&gt;AOC&lt;/code&gt;. Well, in fact, our triangle is more half an ellipse, but anyway. :)&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ctEJq.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;I have modified my program.&lt;/p&gt;&#xA;&lt;p&gt;First, the area of the triangle is computed with the curves of the ellipse instead of the original segments. The formula for an ellipse is &lt;code&gt;pi * minor axis [i/2] * major axis [20]&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Second, the perimeter of the triangle minus the base is computed thanks to a Ramanujan formula:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/il49l.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Then the new program is:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;#include &amp;lt;iostream&amp;gt;&#xA;#include &amp;lt;math.h&amp;gt;&#xA;&#xA;#define PI 3.1415926535&#xA;&#xA;int main()&#xA;{&#xA;    double i, // the length we are looking for&#xA;           area1, // area of the triangle&#xA;           area2, // area of the circle&#xA;           perimeter, // perimeter of the whole shape&#xA;           value; // the value to maximize&#xA;&#xA;    for (i = 0; i &amp;lt; 200; i=i+0.01) { // 200 is an arbitrary integer way outside the maximum expected value of i&#xA;        area1 = (PI*20*(i/2))/2; // ((area of ellipse))/4)*2&#xA;        area2 = (PI*((i/2)*(i/2)))/2; // (classic formula of the area of a circle)/2&#xA;        value = area1 + area2 * 2; // We compute the expected value&#xA;        perimeter = (PI*(3*(20 + (i / 2)) - sqrt(((3 * 20 + (i / 2))*(20 + 3 * (i / 2)))))) / 2 // Perimeter of 2*1/4 of the ellipses (Ramanujan formula)&#xA;            + (i*PI) / 2; // ... plus (perimeter of a circle)/2&#xA;        if (perimeter &amp;gt; 99.99 &amp;amp;&amp;amp; perimeter &amp;lt; 100.01) // if the perimeter +/- = 100, then show all the possible solutions&#xA;            std::cout &amp;lt;&amp;lt; &amp;quot;i: &amp;quot; &amp;lt;&amp;lt; i &amp;lt;&amp;lt; &amp;quot; | p: &amp;quot; &amp;lt;&amp;lt; perimeter &amp;lt;&amp;lt; &amp;quot; | a1: &amp;quot; &amp;lt;&amp;lt; area1 &amp;lt;&amp;lt; &amp;quot; | a2 : &amp;quot; &amp;lt;&amp;lt; area2 &amp;lt;&amp;lt; &amp;quot; | value : &amp;quot;&amp;lt;&amp;lt; value &amp;lt;&amp;lt; std::endl;&#xA;    }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;The optimal data with outwards curved triangle are:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;i:&lt;/strong&gt; 28.96 m. | &lt;strong&gt;p.:&lt;/strong&gt; 100 (+/- .002) | &lt;strong&gt;area 1:&lt;/strong&gt; 454.903 | &lt;strong&gt;area 2:&lt;/strong&gt; 329.349&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Value: 1113.6&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;Shape 3: Two half-ellipses side by side : 1116.34&lt;/h2&gt;&#xA;&lt;p&gt;A last hypothesis. What happens if the shape on the right is, as well as that on the left, a semi-ellipse?&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/waaCZ.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;We have to add a new variable: the minor axis of the right ellipse. We could call this variable &lt;code&gt;j&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;p&gt;My new program:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;#include &amp;lt;iostream&amp;gt;&#xA;#include &amp;lt;math.h&amp;gt;&#xA;&#xA;#define PI 3.14159265358979323846264338327950288419716939937510&#xA;&#xA;int main()&#xA;{&#xA;    double i, j,// the lengths we are looking for&#xA;           area1, // area of right semiellipsis&#xA;           area2, // area of left semiellipsis&#xA;           perimeter, // perimeter of the whole shape&#xA;           value; // the value to maximize&#xA;&#xA;    for (i = 0; i &amp;lt; 200; i = i + 0.01) {&#xA;        for (j = 0; j&amp;lt;200; j += 0.01) {// 200 is an arbitrary integer way outside the maximum expected value of i&#xA;            area1 = (PI * 20* (i/2) )/ 2; // area of semiellipse&#xA;            area2 = (PI * j * (i/2)) / 2;  // area of right semiellipse&#xA;            value = area1 + area2 * 2; // Computation of the expected value&#xA;            perimeter = (PI*(3 * (20 + (i/2)) - sqrt(((3 * 20 + (i/2))*(20 + 3 * (i/2)))))) / 2 // Perimeter of left semiellipse (Ramanujan formula)&#xA;                + (PI*(3 * (j + (i/2)) - sqrt(((3 * j + (i/2))*(j + 3 * (i/2)))))) / 2;  // ... plus Perimeter of right semiellipse &#xA;            if (perimeter &amp;gt; 99.999 &amp;amp;&amp;amp; perimeter &amp;lt; 100.001) // if the perimeter +/- = 100, then show all the possible solutions&#xA;                std::cout &amp;lt;&amp;lt; &amp;quot;i: &amp;quot; &amp;lt;&amp;lt; i &amp;lt;&amp;lt; &amp;quot; | j: &amp;quot; &amp;lt;&amp;lt; j &amp;lt;&amp;lt; &amp;quot; | p: &amp;quot; &amp;lt;&amp;lt; perimeter &amp;lt;&amp;lt; &amp;quot; | a1: &amp;quot; &amp;lt;&amp;lt; area1 &amp;lt;&amp;lt; &amp;quot; | a2 : &amp;quot; &amp;lt;&amp;lt; area2 &amp;lt;&amp;lt; &amp;quot; | value : &amp;quot; &amp;lt;&amp;lt; value &amp;lt;&amp;lt; std::endl;&#xA;        }&#xA;    }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;And now we have the best optimal value for this problem:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;i:&lt;/strong&gt; 27.87 | &lt;strong&gt;j:&lt;/strong&gt; 15.5 | &lt;strong&gt;p.:&lt;/strong&gt; 100 (+/- .001) | &lt;strong&gt;area 1:&lt;/strong&gt; 437.781 | &lt;strong&gt;area 2:&lt;/strong&gt; 339.28&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;value : 1116.34&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;My answer is then:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/f4CSo.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="5335" LastEditorUserId="-1" LastEditorDisplayName="user2174" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-11-14T00:27:33.433" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4536" PostTypeId="2" ParentId="4533" CreationDate="2014-11-13T19:05:52.850" Score="4" Body="&lt;p&gt;Assuming you know the value of M, you can figure it out in &lt;code&gt;3M + 2 + (ceiling(M/2) + 2)&lt;/code&gt; input bits. The parenthesized portion can be reduced to &lt;code&gt;ceiling(M/2)&lt;/code&gt; if we know M is odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let's call knightbots (+), knavebots (-), a jokerbot that started at [+/-]M and moves [+/-]1 per 1 bit by appending the two signs. So a joker that starts at M on a cromulent day is labeled (++), or (+-) on a non-cromulent day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, input &lt;code&gt;M&lt;/code&gt; 1 bits. Here are the possible values of S:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; (+) : 3M&#xA; (-) : 0&#xA; (++): 2M&#xA; (+-): 0&#xA; (-+): 0&#xA; (--): 2M&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and so we pay attention to the Mth output.&lt;br&gt;&#xA;Then input 1 and 0, causing&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; (+) : 3M + 1&#xA; (-) : -1&#xA; (++): 2M&#xA; (+-): 0&#xA; (-+): 0&#xA; (--): 2M&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and pay attention to the M+2nd output.&lt;br&gt;&#xA;Then input 0, &lt;code&gt;2M&lt;/code&gt; times.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; (+) : M + 1&#xA; (-) : 4M -1&#xA; (++): 0&#xA; (+-): -2M&#xA; (-+): 2M&#xA; (--): -4M&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and pay attention to the 3M+2nd output.&lt;br&gt;&#xA;Finally, input 0 &lt;code&gt;ceiling(M/2)&lt;/code&gt; times if &lt;code&gt;M&lt;/code&gt; is odd, or input 1 once and 0 &lt;code&gt;1 + M/2&lt;/code&gt; times if even.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; (+) : 0&#xA; (-) : (unimportant)&#xA; (++): (nonzero)&#xA; (+-): (unimportant)&#xA; (-+): (unimportant)&#xA; (--): (nonzero)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and pay atention to the last output bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now here's what the relvant output bits would be for different possiblities:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; (+) : 0001&#xA; (-) : 10...&#xA; (++): 001...&#xA; (+-): 11...&#xA; (-+): 11...&#xA; (--): 0000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Thus, if the first two are &lt;code&gt;10&lt;/code&gt;, you have a knavebot. If you have &lt;code&gt;0001&lt;/code&gt;, it's a knightbot. Otherwise you must have a jokerbot.&lt;/p&gt;&#xA;" OwnerUserId="2101" LastActivityDate="2014-11-13T19:05:52.850" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4537" PostTypeId="2" ParentId="4276" CreationDate="2014-11-13T19:14:15.527" Score="14" Body="&lt;p&gt;I may be off on the second word here. My answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What is your color&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If this is correct, I will edit and post the explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Here's the explanation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You spell out the names of the colors. So the first row becomes:&lt;br&gt;&#xA; &lt;code&gt;White White Black White Blue&lt;/code&gt;&#xA; &lt;br&gt;&lt;br&gt;&#xA; Then pick the letters from each of the word that lies in the same&#xA; position as the color in the row. So, you will pick the following&#xA; letters as highlighted in Bold:&#xA;&lt;br&gt;&lt;br&gt;&#xA; &lt;strong&gt;W&lt;/strong&gt;hite W&lt;strong&gt;h&lt;/strong&gt;ite Bl&lt;strong&gt;a&lt;/strong&gt;ck Whi&lt;strong&gt;t&lt;/strong&gt;e Blue&lt;strong&gt;blank space&lt;/strong&gt;&#xA; &lt;br&gt;&#xA; So, the first row gives you &lt;strong&gt;What&lt;/strong&gt;&#xA; &lt;br&gt;&lt;br&gt;&#xA; Similarly analyzing the rest of the rows and always assuming blue to&#xA; be blank space, you get&#xA; &lt;br&gt;&#xA; &lt;strong&gt;S&lt;/strong&gt;ilver P&lt;strong&gt;i&lt;/strong&gt;nk Blue Blue Blue = Si = re-ordering = &lt;strong&gt;is&lt;/strong&gt;&#xA; &lt;br&gt;&#xA; &lt;strong&gt;O&lt;/strong&gt;range P&lt;strong&gt;u&lt;/strong&gt;rple Pu&lt;strong&gt;r&lt;/strong&gt;ple Gra&lt;strong&gt;y&lt;/strong&gt; Blue = Oury = re-ordering = &lt;strong&gt;your&lt;/strong&gt;&#xA; &lt;br&gt;&#xA; &lt;strong&gt;O&lt;/strong&gt;range O&lt;strong&gt;r&lt;/strong&gt;ange Ye&lt;strong&gt;l&lt;/strong&gt;low Bla&lt;strong&gt;c&lt;/strong&gt;k Yell&lt;strong&gt;o&lt;/strong&gt;w = Orlco = re-ordering = &lt;strong&gt;color&lt;/strong&gt;&#xA; &lt;br&gt;&lt;br&gt;&#xA; Hence the sentence becomes &lt;strong&gt;What is your color&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4903" LastEditorUserId="4903" LastEditDate="2014-11-15T15:41:54.157" LastActivityDate="2014-11-15T15:41:54.157" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4538" PostTypeId="2" ParentId="4506" CreationDate="2014-11-13T19:14:20.057" Score="14" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Bowtie, 8, 69. &lt;br&gt; The 1st, 3rd, 5th, and 7th things that the guard says are derived from each other by the same operation. The answer is to apply that operation on the 2nd, 4th, and 6th things. &lt;br&gt; &quot;Bleach as Needed&quot; is a triangle pointing up, &quot;Play video&quot; is a triangle pointing to the right, &quot;Give way to traffic&quot; is a triangle pointing down, and &quot;Normal subgroup&quot; is a triangle pointing to the left. Therefore the solution is to turn the other three objects 90 degrees clockwise. &lt;br&gt; First puzzler: double the numbers. &lt;br&gt; Second puzzler: take the next letter of the alphabet. &lt;br&gt; Third puzzler: add the decimal part of the number to the number. &lt;br&gt; Fourth puzzler: provide an antiderivative.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5352" LastActivityDate="2014-11-13T19:14:20.057" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4539" PostTypeId="1" AcceptedAnswerId="4544" CreationDate="2014-11-13T19:24:47.347" Score="4" ViewCount="1147" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Most in our family&lt;br&gt;&#xA;  Are born whole and quickly.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;But not us four&lt;br&gt;&#xA;  Who require a detour.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;We don't want to interrupt&lt;br&gt;&#xA;  Until the rest is developed.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;So until then we wait&lt;br&gt;&#xA;  To be put into our complete state.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Then everything will make sense&lt;br&gt;&#xA;  When finished is the sequence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who are the &quot;four&quot; that this poem refers to?&lt;/p&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4450" LastEditDate="2015-05-16T20:17:15.043" LastActivityDate="2015-05-16T20:17:15.043" Title="Riddle Poem: These four require a detour" Tags="|riddle|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4540" PostTypeId="2" ParentId="4539" CreationDate="2014-11-13T19:29:19.190" Score="11" Body="&lt;p&gt;Are the 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wisdom teeth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Most in our family&#xA;Are born whole and quickly.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You normally have most of your teeth all at once.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But not us four&#xA;Who would require a detour.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There are 4 teeth which come at some other unexpected time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We don't want to interrupt&#xA;Until the rest is developed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;They only come out AFTER the period of loosing baby-teeth has occurred.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So until then we wait&#xA;To be put into our complete state.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Just poetic emphasis of the previous.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then everything will make sense&#xA;When finished is the sequence.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is no further normal activity after the wisdom teeth have established themselves.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-11-19T03:46:39.477" LastActivityDate="2014-11-19T03:46:39.477" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4541" PostTypeId="2" ParentId="4539" CreationDate="2014-11-13T19:33:32.393" Score="0" Body="&lt;p&gt;this is my thought:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Un-born quadruplets. two of them are fully developed in the womb, so they are waiting for the other two to develop so as not to interrupt their growth. As it is 4 children, a cesarean section will probably be used, which is a detour to the regular way of being born quickly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;i have no idea how i came up with this so quickly.&lt;/p&gt;&#xA;" OwnerUserId="5358" LastEditorUserId="4450" LastEditDate="2014-11-13T20:50:31.977" LastActivityDate="2014-11-13T20:50:31.977" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4542" PostTypeId="2" ParentId="4533" CreationDate="2014-11-13T19:40:11.250" Score="14" Body="&lt;p&gt;I claim a bound linear in $M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, the challenge is really to make the automaton hit 0. Once it hits 0, we can easily determine what type it is using a constant number of moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A_1,A_2,...,A_n$ denote different types of automatons. For us, the $A_i$ consist of a knightbot, a knavebot, and then many different automatons for each initial setting of a jester/jokerbot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also assume that, given an automaton $A_i$, if we know what state it is in, then we can always move it one space to the right or one space to the left in constant time. This is clearly true for any automaton in the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose we have a finite upper bound $B$ on the distance from our automaton to the origin. Then the following series of moves is guaranteed to make the automaton hit 0 in $O(B)$ time:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Pretend the automaton is $A_1$. Move it $B$ spaces to the right, one at a time. Then move it $2B$ spaces to the left. This takes time $O(B)$.&lt;/li&gt;&#xA;&lt;li&gt;If we hit 0 along the lines, great! Otherwise, our automaton is not $A_1$. Pretend the automaton is $A_2$. Then we have spent $O(B)$ moves so far, so there is a number $k=O(B)$ such that our automaton is at most $k$ steps from the origin. So, first calculate the internal state that $A_2$ would be in, given the entire sequence of inputs we have fed it so far. Then, based on this state, move the automaton $k$ steps to the right, one at a time, and then $2k$ steps left, one at a time.&lt;/li&gt;&#xA;&lt;li&gt;If we hit 0, great! Otherwise, pretend our automaton is $A_3$. You can see where this is going.... The important thing to note is that, since the number of automatons $n$ is a constant, the total number of steps spent to test all $n$ automatons is still $O(B)$. The constant factor may be large, but whatever, it's still $O(B)$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Great. Unfortunately, we &lt;em&gt;don't&lt;/em&gt; know an upper bound $B$ on the distance from our automaton to the origin. The fix is very simple. The above procedure runs in $O(B)$ moves, so there exists a constant $c$ such that the above procedure never moves any automaton $A_i$ by more than $cB$ steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, run the above procedure with $B=1,3c,(3c)^2,(3c)^3,(3c)^4,\dots$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that this works, fix $M$. Pick $k$ so that $\frac 12(3c)^k\leq |M| \leq \frac 12(3c)^{k+1}$. Then after the first $k$ iterations of the procedure, by our choice of $c$, the automaton has moved at most $c(1+3c+(3c)^2+\cdots+(3c)^k)=c\frac{(3c)^{k+1}-1}{3c-1}\leq\frac 12(3c)^{k+1}$ spaces. So, it is at most $\frac 12(3c)^{k+1}+|M|\leq(3c)^{k+1}$ distance from the origin. Therefore, it will be found on the next iteration (because on the next iteration we use $B=(3c)^{k+1}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, because each iteration takes linear time, the entire sequence of iterations takes at most $O(1)+O(3c)+O((3c)^2)+\cdots+O((3c)^{k+1})$ time. This sum is $O((3c)^k)$, and therefore $O(M)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, this procedure is guaranteed to hit 0 in $O(M)$ moves. As noted previously, as soon as we hit 0, we can easily determine which automaton we are dealing with using a constant number of moves. Therefore, this procedure guarantees a linear bound. QED&lt;/p&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2014-11-13T19:40:11.250" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4543" PostTypeId="2" ParentId="4326" CreationDate="2014-11-13T19:40:46.943" Score="10" Body="&lt;p&gt;Here are my thoughts:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4 Characters, not 4 Numbers, so the password is:&lt;br&gt;&#xA; 22/7&lt;br&gt;&#xA; It's the historical approximation of pi, and the date of the letter.&lt;br&gt;&#xA; Bonus: the prime counting function is denoted as pi(x)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5360" LastEditorUserId="5395" LastEditDate="2014-11-14T02:08:11.747" LastActivityDate="2014-11-14T02:08:11.747" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4544" PostTypeId="2" ParentId="4539" CreationDate="2014-11-13T19:53:51.877" Score="13" Body="&lt;p&gt;My guess is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the letters 'i', 't', 'j', and 'x'.  Letters you complete after you write a word in cursive (dot your i's and cross your t's!).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5281" LastEditorUserId="4450" LastEditDate="2014-11-16T05:15:57.117" LastActivityDate="2014-11-16T05:15:57.117" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="4545" PostTypeId="2" ParentId="4498" CreationDate="2014-11-13T20:06:05.543" Score="0" Body="&lt;p&gt;It must be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; THE INTERNET&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Only the part with the primary colors does not match perfectly.. I suppose, what is meant in this part are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RGB pixels&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5283" LastActivityDate="2014-11-13T20:06:05.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4546" PostTypeId="2" ParentId="4503" CreationDate="2014-11-13T20:08:07.567" Score="0" Body="&lt;p&gt;I thought it would be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;(theory of) evolution&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation as per request:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I know what is or isn't, but never what might be;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the results at one point of the process of evolution are a given, but only a theoretical all-knowing entity (Laplace's Demon) could accurately predict what evolution will lead to in the future. &lt;em&gt;Added later: forgot my most compelling point here. Evolution itself does not (need to) look forward to function. There is no planning or looking ahead, just basic rules that always apply and will lead to the desired result (in the long term)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And many claim to know me, though many disagree;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This one is fairly straight-forward: many people claim to know (understand) evolution, and claim that it is a fact. However, many people also disagree (often proposing intelligent design instead).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There are no sly dilemmas, no subtleties I see;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Evolution is not about morality, and in a sense is a fairly simple, straight-forward process (though the results that emerge certainly aren't). That's what I thought of when reading &quot;sly dilemmas&quot; and &quot;subtleties&quot;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Both I encompass gladly, to be, and not to be.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Evolution is as much about 'succeeding' as 'failing'. The process of natural selection only works because each entity has strengths and weaknesses, which are only strengths and weaknesses in a given context. What turns out to be strong, will 'be'. What turns out to be overrun by the environment, will 'not be'. Both are necessary for evolution to function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My explanation may not be perfect but hopefully the idea is clear.&lt;/p&gt;&#xA;" OwnerUserId="5365" LastEditorUserId="5365" LastEditDate="2015-01-22T21:06:27.490" LastActivityDate="2015-01-22T21:06:27.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4547" PostTypeId="2" ParentId="4539" CreationDate="2014-11-13T20:17:27.523" Score="0" Body="&lt;p&gt;A tough one.. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably it's not so easy, but could it be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the signs at the end of a sentence -&gt; . ! ? ;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5283" LastActivityDate="2014-11-13T20:17:27.523" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4549" PostTypeId="2" ParentId="4525" CreationDate="2014-11-13T20:28:24.183" Score="3" Body="&lt;p&gt;I think it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;0=0&#xA;Infinity=Impossible/Undefined&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thinking:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Arcsin 0=0 and arcsin infinity is undefined. Arcsin's inverse is sin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5368" LastEditorUserId="5368" LastEditDate="2014-11-13T22:39:52.157" LastActivityDate="2014-11-13T22:39:52.157" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="4550" PostTypeId="2" ParentId="4326" CreationDate="2014-11-13T20:33:41.483" Score="6" Body="&lt;p&gt;&lt;strong&gt;It didn't matter what you type. The bomb won't go off anyway&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 24 Hour clocks run from 00:00.00 to 23:59.59 and then roll back over to 00:00.00. If the bomber has set the clock to detonate at &lt;strong&gt;24:00&lt;/strong&gt; then the clock will never reach that number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4852" LastActivityDate="2014-11-13T20:33:41.483" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4553" PostTypeId="2" ParentId="4106" CreationDate="2014-11-13T21:09:31.007" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The man and woman are playing pool at a bar on a cruise ship. They are holding pool cues and are surrounded by the bar's cylindrical glass tables. There are a number of rectangular LCD TVs suspended in the room. As the sunlight slowly whisks away the heavy morning fog, the man remarks that he can finally see the ocean under the clouds. Clouds of fog, that is. &lt;img src=&quot;https://i.stack.imgur.com/DYSjJs.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4236" LastActivityDate="2014-11-13T21:09:31.007" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4554" PostTypeId="2" ParentId="4525" CreationDate="2014-11-13T21:21:46.410" Score="0" Body="&lt;p&gt;The solution is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0-0&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Five and Six can have opposites, but 0 can't.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5374" LastActivityDate="2014-11-13T21:21:46.410" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4555" PostTypeId="2" ParentId="4525" CreationDate="2014-11-13T21:25:26.323" Score="0" Body="&lt;p&gt;The question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Can you determine the image of this number if we consider an arbitrary 1-1 function on the set of integer numbers?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the answers are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0 -&gt; 0, infinity -&gt; &quot;impossible&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of course, this won't be the right answer, but this one sounds good to me. :-)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Previous attempt&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What does the number divided 0 times equals to?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answers are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;0&quot; -&gt; 1, &quot;infinity&quot; -&gt; &quot;Impossible&quot;. I'm not &lt;em&gt;that&lt;/em&gt; sure about the second answer; the first should work with every number, not only &quot;1&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user1962" LastEditorDisplayName="user1962" LastEditDate="2014-11-13T22:31:11.150" LastActivityDate="2014-11-13T22:31:11.150" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4557" PostTypeId="2" ParentId="4405" CreationDate="2014-11-13T21:31:55.853" Score="0" Body="&lt;p&gt;The Mute takes the Blind man to the Deaf man and puts the blind man's hands on the Deaf man's face so he can identify him.  The Mute then kisses the Blind man and then puts the Blind man's hand on his wife's face so that he can identify her.&lt;/p&gt;&#xA;" OwnerUserId="5375" LastActivityDate="2014-11-13T21:31:55.853" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4558" PostTypeId="2" ParentId="4405" CreationDate="2014-11-13T21:41:30.793" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The blind man and the mute man know a language that the deaf man does not, so the mute man writes his message onto a piece of paper (in this language), and uses persuasion and/or trickery to get the deaf man to read the message out loud. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some reservations I have about this answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. How would one convince the deaf man to read something out loud?&lt;br&gt;&#xA; 2. The mute man can't be certain the deaf man will pronounce the foreign words appropriately if he hasn't ever tried this before&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5340" LastActivityDate="2014-11-13T21:41:30.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4560" PostTypeId="2" ParentId="4503" CreationDate="2014-11-13T22:20:27.963" Score="0" Body="&lt;p&gt;My personal guess was a &lt;strong&gt;fact&lt;/strong&gt;, as many disagree on facts, many claim to know the facts, and a fact can't be a maybe.&lt;/p&gt;&#xA;" OwnerUserId="5302" LastActivityDate="2014-11-13T22:20:27.963" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4565" PostTypeId="1" AcceptedAnswerId="4670" CreationDate="2014-11-13T22:50:34.053" Score="4" ViewCount="583" Body="&lt;p&gt;A party is being held at a local mansion. The host is very rich and his success is because of one thing — his famous recipe for Spaghetti!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only guests that may attend are people who correctly reply to the guard at the door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's where you come in. You and a friend are trying to steal this recipe. You sneak by and listen to the passwords.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when the first guest arrives, you are dismayed to find that the guard does not ask them a question, but merely motions for them to stand by the gate.  When the second guest arrives, the guard motions to them too, and they line up behind the first guest.  This goes on until at last there are 19 guests lined up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The guard looks at his watch and says, &quot;Well, I think everyone is here, but we will wait a couple of minutes just in case there are any stragglers.&quot;  At this you and your friend realize that this is your last chance so you quickly come out and line up at the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;As you know,&quot; says the guard.  &quot;There have been many attempts made to steal our host's secret recipe, and at every party somebody has managed to learn the password scheme and sneak in when they shouldn't.&quot;  The guards eyes each of the guests one by one, as if to divine any unauthorized guests presently in attendance, but when his eyes meet yours you stare right back at him dumbly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;So our host has gone back and picked a variant on one of his earlier schemes that his thinks &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4286/the-security-to-the-party-part-20&quot;&gt;perhaps nobody outside our group quite caught on to&lt;/a&gt; and mixed it up a bit.  And just to wise up a little, as you know, you are all standing here together before I will start asking the questions, so as to not leak any information.  Is everybody ready for me to give out the number?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You and your friend see everyone else nodding, so you nod too, fearful of what is going to come next.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The guard pulls out his clipboard and then counts all the potential guests.  &quot;I see 21 people,&quot; he begins, and scanning down his clipboard he quickly calls out, &quot;The challenge is thirteen thousand eight hundred and thirty two.   We will start at the head of the line and work backwards.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;5&quot; says the first guest after a few moments, and he is let in.&#xA;&quot;12&quot; says the second guest.  The guard nods and lets him by.&#xA;&quot;16&quot;.  &quot;Yes.&quot;&#xA;&quot;17&quot; &quot;Sure&quot;.&#xA;&quot;14&quot; says the fifth guest.  &quot;Your answer is baseless!&quot; exclaims the guard and a out of nowhere a large gorilla zip-lines across the group and snatches up the fifth guest and carries him off.  &quot;The correct answer was 40!&quot; the guard says loudly, then smiles at the sixth guest.  &quot;No reason for you to have to figure out two answers, eh?&quot;&#xA;The sixth guest nods, then says &quot;65&quot;.  &quot;Correct.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remaining guests in front of you all answer correctly. Starting with the 7th guest, the answers are: 72,58,38,31,32,37,44,29,56,27,52,11, and 22.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before it gets to your friend, who is in front of you, he turns and whispers to you, &quot;Well it's pretty obvious this isn't just a random sequence, isn't it? If there had been one more person, he would have had the same number as me!&quot; but then it is his turn and he has to turn and quickly answer: &quot;35&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The guard eyes him over for a minute, then replies, &quot;Correct.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it is your turn!  What number do you tell the guard?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anti-Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If there were only 4 guests, the guard would have been in a pickle! He would have had to challenge with &quot;14&quot; and the correct responses from each guest in order would have been &quot;3&quot;, &quot;4&quot;, &quot;5&quot; and &quot;6&quot;.  Fortunately the host knew that at least 6 people would show up (&quot;27&quot; -&gt; &quot;4&quot;, &quot;6&quot;, &quot;7&quot;, &quot;9&quot;, 3&quot;, &quot;1&quot;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The guard must base his challenge on the number of guests that show up at the door.  Your friend figures out why before he gives his response.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2718" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2014-11-16T11:48:22.853" Title="The Security to the Party [Part 29]" Tags="|mathematics|password|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4567" PostTypeId="2" ParentId="4446" CreationDate="2014-11-13T22:57:39.350" Score="6" Body="&lt;p&gt;&lt;strong&gt;My best solution: $score = 1142.5655\ m^2$&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8BWVs.png&quot; alt=&quot;Best solution&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Jeffrey also explained, the curve is made of 3 arcs, one going from P to A, the second from A to B, the third going from B to P.  The local optimality of the use of the available length implies constant curvature.  It is only at boundaries, where the value of the surface changes, that the curvature can change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The curvature is proportional to the value of the area.  You can think of it as pressure.  The higher the pressure, the more the lines curves out.  Or to put it differently, the higher the value of the enclosed surface, the more it is worth to make an incursion in that area.  That translates to higher curvature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, the area right of the line counting twice, the curvature is double and the radius of the arc is halved.  If the radius of the semi-circle on the right is $r$, then the radius of the 2 arcs left is $R = 2r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem being symmetrical, I will assume the optimal curve is also symmetrical relative to the horizontal axis.  So I will only consider the upper half of the curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GaLby.png&quot; alt=&quot;Solution explanations&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$CA$ is at an angle $\alpha$ from the horizontal.&lt;br&gt;&#xA;$DP$ is at an angle $\alpha+\beta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the image, we can see some relations beteen $\alpha$, $\beta$ and $r$:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance from $P$ to the axis is 20.  It can be computed from $\alpha$, $\beta$ and $2r$:&lt;br&gt;&#xA;(1)  $20 = 2r\cos\alpha - 2r\cos(\alpha+\beta)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The length of the curve is half of the available length of 100:&lt;br&gt;&#xA;(2)  $50 = r\alpha + 2r\beta$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance from A to the horizontal axis can be computed in 2 ways:&lt;br&gt;&#xA;$dist = r\sin(\alpha) = 2r\sin(\alpha+\beta)$&lt;br&gt;&#xA;(3) $\sin(\alpha) = 2\sin(\alpha+\beta)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have 3 equations, (1), (2) and (3), and 3 unknowns.  We can solve it numerically:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\alpha=1.7390005916$, $\beta=0.8871224528$, $r=14.2318548591$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It places A at $(x_0,0)$, C at $(0,y_1)$ and D at $(2x_0,-y_1)$&lt;br&gt;&#xA;where $x_0 = -r\sin(\alpha) = 2.38258653$&lt;br&gt;&#xA;and $y_1 = r\cos(\alpha) = 14.03100047$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The areas can be computed (for the whole curve):&lt;br&gt;&#xA;$area\ right = \alpha r^2 + x_0 y_1 = 385.657152221015$&lt;br&gt;&#xA;$area\ left = \beta (2r)^2 - (20+x_0) y_1 - x_0 y_1 = 371.25117209358604$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the final scores evaluates to:&lt;br&gt;&#xA;$area\ left + 2\ area\ right = 1142.565477$.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2014-11-14T22:14:24.350" LastActivityDate="2014-11-14T22:14:24.350" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4568" PostTypeId="1" AcceptedAnswerId="4571" CreationDate="2014-11-13T23:01:11.190" Score="-2" ViewCount="5197" Body="&lt;p&gt;Suppose your friend grabs a random card from a deck. Now you can ask questions about that card. You can only ask yes/no and greater/less than questions. How many questions do you need to determine the card?&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Questions like these are not allowed:&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is its value between 1 and 6?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;These are:&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is its suit clubs?&lt;/li&gt;&#xA;&lt;li&gt;Is it a face card?&lt;/li&gt;&#xA;&lt;li&gt;Is its value greater than, less than, or equal to six?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="6178" LastEditDate="2014-12-30T16:23:35.387" LastActivityDate="2015-09-04T10:22:46.970" Title="How many Questions will you need to Guess the Card?" Tags="|strategy|cards|" AnswerCount="5" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="4571" PostTypeId="2" ParentId="4568" CreationDate="2014-11-13T23:34:15.513" Score="3" Body="&lt;p&gt;I think it can be done in &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; six questions in the worst case&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First, we determine the color, and then the suit in two questions. Then, numbering the cards 1 through 13, we ask is it higher, lower, or equal to 7 (three questions). Worst case, it's not seven and we have six cards left. If its the lower six, we ask if it's higher than 3 (four questions). We now have three consecutive values left. We use our fifth question to ask if it's higher equal to or lower than the middle one, this gives us the answer which we reveal in the final question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could also use a question with three answers for the fourth, but this is easier to explain.&lt;/p&gt;&#xA;" OwnerUserId="2421" LastEditorUserId="2421" LastEditDate="2014-11-14T00:04:34.020" LastActivityDate="2014-11-14T00:04:34.020" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="4572" PostTypeId="1" AcceptedAnswerId="4610" CreationDate="2014-11-13T23:44:30.357" Score="6" ViewCount="1509" Body="&lt;p&gt;Application of the mind is where I begin&lt;br /&gt;&#xA;Depictions of what is, what was, will be, and perhaps have never been&lt;br /&gt;&#xA;Vibrant color with each  illustrious stroke one will strive &lt;br /&gt;&#xA;Unlimited potential for an imagination that is now alive&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the middle there is only me, this you knew &lt;br /&gt;&#xA;My neighbors make up two thirds of you&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we come to the end we reach an uncomfortable state &lt;br /&gt;&#xA;Perhaps because of something off your plate &lt;br /&gt;&#xA;Gagging, heaving, flailing violently in attempts to mitigate  &lt;br /&gt;&#xA;Always remember how essential it is to masticate&lt;/p&gt;&#xA;" OwnerUserId="4689" LastEditorUserId="4689" LastEditDate="2014-11-14T01:51:51.273" LastActivityDate="2014-11-14T18:07:46.847" Title="Identity crisis. What exactly am I?" Tags="|riddle|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4574" PostTypeId="1" AcceptedAnswerId="4578" CreationDate="2014-11-14T00:35:13.123" Score="17" ViewCount="20462" Body="&lt;p&gt;Poem riddles are becoming more prolific and more popular on Puzzling.SE (examples: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4139/pacoverflow-riddle-poem&quot;&gt;Pacoverflow Riddle Poem&lt;/a&gt;, &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4539/pacoverflow-riddle-poem-2&quot;&gt;Pacoverflow Riddle Poem #2&lt;/a&gt;, &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4572/identity-crisis-what-exactly-am-i&quot;&gt;Identity Crisis: What Am I?&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inspired by the 'Tips for Golfing in....' (&lt;a href=&quot;/questions/tagged/tips&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;tips&amp;#39;&quot; rel=&quot;tag&quot;&gt;tips&lt;/a&gt;) questions on Programming Puzzles and Code Golf.SE and confused by the plethora of poem riddles springing up, I wondered: what are some tips for solving poem riddles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, it is best to read each line carefully the first time or is it best to read it through a couple of times and then start guessing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please state the tip, then give the reason for the tip being successful in the solving of poem riddles.&lt;/p&gt;&#xA;" OwnerDisplayName="user2174" LastEditorUserId="-1" LastEditorDisplayName="user2174" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2017-12-22T12:48:54.530" Title="Tips for solving poem riddles" Tags="|puzzle-creation|solvability|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4575" PostTypeId="2" ParentId="4472" CreationDate="2014-11-14T00:46:41.753" Score="11" Body="&lt;p&gt;Since the answers are scattered (by the way, you should have asked 3 separate questions):&lt;/p&gt;&#xA;&lt;h2&gt;1&lt;/h2&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Night&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;2&lt;/h2&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Anchor&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;3&lt;/h2&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Piggy bank&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I do this in the interest of the site being the QnA it is designed to be.&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-11-14T00:46:41.753" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4576" PostTypeId="1" AcceptedAnswerId="4577" CreationDate="2014-11-14T00:47:49.247" Score="3" ViewCount="796" Body="&lt;p&gt;I saw this puzzle a while back in a children's puzzle book, but I'll try to make it as accurate as possible.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is a small army of three ants walking down a very narrow road, so there is only room for them to walk single-file. They come across another small army heading the opposite direction. Neither army wants to fight (ants are peaceful, I think), but it seems as if they have no options. Fortunately, one clever ant spots a small clearing in the trees, that only &lt;strong&gt;one ant&lt;/strong&gt; can fit into at a time. Can the armies pass each other without having to fight? If so, how do they do this?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have made a very bad illustration of this so it will hopefully be easier to understand, but if you need clarification I can provide it. Only one ant can fit into the hole at a time, and the road is long enough for both armies to fit on one side of the hole. This is not a &lt;a href=&quot;/questions/tagged/lateral-thinking&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;lateral-thinking&amp;#39;&quot; rel=&quot;tag&quot;&gt;lateral-thinking&lt;/a&gt; puzzle, so don't post any answers like &quot;These ants aren't peaceful.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/355Ga.jpg&quot; alt=&quot;Picture&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I actually included the picture this time.&lt;/p&gt;&#xA;" OwnerUserId="1664" LastEditorUserId="1664" LastEditDate="2014-11-14T01:02:17.200" LastActivityDate="2014-11-14T01:04:40.757" Title="Two Armies of Ants" Tags="|logical-deduction|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4577" PostTypeId="2" ParentId="4576" CreationDate="2014-11-14T00:57:35.527" Score="5" Body="&lt;p&gt;The first ant of side A (call her &quot;A0&quot;) goes into the clearing, and the other back off. While A0 is in the clearing the whole B army pass the clearing so A0 now can go away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeat.&lt;/p&gt;&#xA;" OwnerUserId="5393" LastActivityDate="2014-11-14T00:57:35.527" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4578" PostTypeId="2" ParentId="4574" CreationDate="2014-11-14T00:58:15.270" Score="19" Body="&lt;p&gt;There are several types of clues embedded in these puzzles.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Cryptic clues where the words have no meaning. One is meant to break words apart into constituent letters and then reconstruct.&lt;/li&gt;&#xA;&lt;li&gt;The clues are literal, however they are veiled in metaphor and use of hyperbole and exploitation of synonyms.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The clues are totally descriptive (easiest, most obvious and least likely).&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Read the whole clue first, line-by-line, thinking/parsing/interpreting as you go. You must keep all the above types of clues in mind as you go. For every word or combination of consecutive or nearby words check against those 3 main types of clues. (Don't forget the TITLE, and other throw-away comments anywhere...) &lt;/li&gt;&#xA;&lt;li&gt;Start conjecturing answers right from the first letter you see. Consider making a list of candidate answers matching one part, and see if it fits with other parts.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Look for&lt;/em&gt; conflicts (don't hope that they aren't there) with the conjectures and adjust them.&lt;/li&gt;&#xA;&lt;li&gt;If you make it through and you are in disarray go back to 1. (looking around at similar solutions/puzzles can be inspirational). &lt;/li&gt;&#xA;&lt;li&gt;If it all fits, then post it. Otherwise, continue till tired and consider giving up.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This answer is deliberately slightly vague. Solving riddles is essentially a creative act of invention and as such can not be mechanised.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's walk through a few to illustrate:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/4464&quot;&gt;Example 1:&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Full poem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You might try to pet me or kick my butt&lt;br&gt;&#xA;  but you can not touch me no matter what&lt;br&gt;&lt;br&gt;&#xA;  If you gave me presents I'd leave you for sure&lt;br&gt;&#xA;  but take what's around me and I will endure&lt;br&gt;&lt;br&gt;&#xA;  You can search the whole ocean or even beyond&lt;br&gt;&#xA;  in sometimes for sure, a loch or a pond&lt;br&gt;&lt;br&gt;&#xA;  If you see whats behind it&lt;br&gt;&#xA;  you are very close&lt;br&gt;&lt;br&gt;&#xA;  And as you assumed&lt;br&gt;&#xA;  its right under your nose&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now let's walk through it line by line:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You might try to pet me or kick my butt&#xA;  but you can not touch me no matter what&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A couple of options here. &quot;Pet me&quot; could refer to the letters of &quot;me&quot;. However the second line indicates that are dealing with something intangible. So, let's constrain ourselves to concepts and things which &quot;cannot be touched&quot;. Example candidate list: Darkness/Reflection/Truth/Nothingness/Sun. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you gave me presents I'd leave you for sure&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We could be dealing with &lt;em&gt;present&lt;/em&gt; tense. &quot;I'd leave you&quot;, could mean we are dealing with a word containing a &quot;u&quot; which when removing some letters (like &quot;ID&quot;) would leave something else. Alternatively, more metaphorically, rather than symbol tricks, if you &lt;em&gt;give/add&lt;/em&gt; something it is no longer what it was. The second option is closer to our feelings about the first clue. We can lean towards &lt;em&gt;Nothingness&lt;/em&gt; now. However this is probably never going to go this smoothly but it is intuitive.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;but take what's around me and I will endure&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This really looks like a symbol trick. Like maybe the letter in the middle is an &quot;I&quot;. It may also mean that so long as you don't mess with the core of the thing (which cannot be touched - clue #1), it will not be affected. It is hard to map this to anything on our preliminary list (darkness maybe?). &lt;em&gt;Nothingness&lt;/em&gt; may also fit. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You can search the whole ocean or even beyond&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is very cryptic and confusing. It does not seem to reveal anything. But it gives the impression that the thing cannot be &quot;found&quot;. But notice the trickery of the words spoken aloud taking different meaning. One must always keep an eye out for homophones - whole does indeed = &quot;HOLE&quot; - something to consider. Anagrams of &lt;em&gt;ocean&lt;/em&gt; should also be considered.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;in sometimes for sure, a loch or a pond&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is very unclear what a &quot;loch or a pond&quot; has to do with anything. Do take notice that the author is cunning and a German... &quot;Loch&quot; means &quot;HOLE&quot; in German...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you see whats behind it you are very close&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This just goes further than &quot;cannot be touched&quot;. This is more like &quot;is transparent&quot;. It may also mean to look &quot;behind &lt;em&gt;it&lt;/em&gt;&quot; (the letter T). &quot;You are very close&quot; - probably means that if your candidate is transparent then you are looking good.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And as you assumed its right under your nose&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This looks like a clever double meaning (I would expect it to have the following 2 meanings). It is implied that the answer is supposed to be obvious - a day-to-day thing. And that there is some relation to the physical body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, so let's take stock... From our original candidate list &quot;Nothingness&quot; (include equivalent concepts like &quot;Zero&quot; (and thus &quot;O&quot;)) is looking very strong and would probably get accepted due to inability to refute using the clues. But please do notice some possible hinting to a &quot;HOLE&quot;. &quot;Nothingness&quot; is quite close to &quot;HOLE&quot;, and &quot;HOLE&quot; fits just a tiny bit better (knowing the German connection helps a lot).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's go with HOLE!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing one that I made might be good because I know the intention entirely.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/4193/guess-the-object-what-am-i-riddle-part-3&quot;&gt;Example 2&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Full poem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm lying down, with a dah-ditty-dah, in the middle of the end,&lt;br&gt;&#xA;  I'd hang upside-down similarly with my friend,&lt;br&gt;&#xA;  Were I to flip over, with a dah-ditty-dah, it'd be the middle of the day,&lt;br&gt;&#xA;  Right way up I cannot plainly say,&lt;br&gt;&#xA;  Here's the key, let's end the fun,&lt;br&gt;&#xA;  I'll tell you a secret, I hide behind one&lt;br&gt;&#xA;  What am I?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now let's walk through the poem line by line.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm lying down, with a dah-ditty-dah, in the middle of the end,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This line relies on drawing suspicion to the words &quot;dah-ditty-dah&quot;. Hopefully, one would notice that this is the vocal sound used in Morse code. &quot;Middle of the end&quot; will almost always be referring to the letter &quot;N&quot; (this is a clue of the first kind). So let's look at Morse code of &quot;N&quot;. It is &lt;code&gt;-.&lt;/code&gt;. Supposedly it is &quot;lying down&quot;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'd hang upside-down similarly with my friend,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Similarly&quot; here should mean &quot;in the middle&quot; or possible &quot;lying down&quot;. You should try both - like branch the search. So, what's in the middle of &quot;friend&quot; - &quot;i&quot; and &quot;e&quot;. We should move along with &quot;a&quot; (&quot;e&quot; upside-down). And even &quot;!&quot;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Were I to flip over, with a dah-ditty-dah, it'd be the middle of the day&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again &quot;dah-ditty-dah&quot; and &quot;middle of the X&quot;. Let's look at A in Morse code. It is &lt;code&gt;.-&lt;/code&gt;. A clear pattern emerges. Let's start to solidify the conjecture of &lt;code&gt;!&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Right way up I cannot plainly say,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ah-hah, one does not &quot;say plainly&quot; when using &lt;code&gt;!&lt;/code&gt;. However, &quot;Right way 'up'&quot; could refer to a letter trick giving 'pu'. So this clue could be of the first or the second kind.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Here's the key, let's end the fun,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Here's the key&quot; is quite attention grabbing. &quot;What is the key!?&quot; we ask. Indeed, it could be a literal key... how many types of those do we know and use? - a door key, and a keyboard key, a piano key. Well, we are conjecturing something like a symbol, so probably a keyboard key.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'll tell you a secret, I hide behind one&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Tell you a secret&quot; does not sound like a clue, but rather filler that signifies that a clue is coming. &quot;I hide behind one&quot; - this could refer to the letter &quot;E&quot; - this is almost a possible answer to the question. However, since we are think about a keyboard, let's try the key with 1 on it. We see that it also shares the &lt;code&gt;!&lt;/code&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's go with &lt;code&gt;!&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="5373" LastEditorDisplayName="user2174" LastEditDate="2017-12-22T12:48:54.530" LastActivityDate="2017-12-22T12:48:54.530" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4579" PostTypeId="2" ParentId="4576" CreationDate="2014-11-14T01:04:40.757" Score="1" Body="&lt;p&gt;Ant 1 of Army Alpha gets in the clearing, less call him A1. All soldiers of army Delta pass the clearing. A1 goes away&lt;br&gt;&#xA;Ant 1 of Delta army gets into the clearing. lets call him D1. Soldiers of Delta and Alpha pass the clearing and go to Delta Army side, D1 goes away.&lt;br&gt;&#xA;A2 gets in the clearing, all armies go to Alpha side. A2 goes away.&lt;br&gt;&#xA;Same process with D2, A3 and D3.&lt;/p&gt;&#xA;" OwnerUserId="4973" LastActivityDate="2014-11-14T01:04:40.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4582" PostTypeId="2" ParentId="4472" CreationDate="2014-11-14T01:18:13.107" Score="16" Body="&lt;p&gt;Number 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The empty space in a thermometer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An Anchor&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My Savings Account&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5397" LastActivityDate="2014-11-14T01:18:13.107" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4583" PostTypeId="2" ParentId="4568" CreationDate="2014-11-14T01:27:18.260" Score="0" Body="&lt;p&gt;is it's suit club or spade?&#xA;if answear is no: is it's a diamond?&#xA;if answear is yes: is it's a spade?&#xA;Then you know the right suit. Since &quot;interval&quot; isn't allowed then I assume you can ask if it's a black card?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it's value &gt;7?&#xA;Is the cards value even?&#xA;This leaves you with four possible cards left in the worst case. &#xA;ie. A,3,5,7: Is the card &gt;3?, Then you pick a card from either side of the interval depending on the answear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I needed 6 questions for that.&lt;/p&gt;&#xA;" OwnerUserId="5331" LastActivityDate="2014-11-14T01:27:18.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4585" PostTypeId="2" ParentId="4572" CreationDate="2014-11-14T01:29:13.253" Score="4" Body="&lt;p&gt;I think it's:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The palate&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My reasoning:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Application of the mind is where I begin&lt;br&gt;&#xA;Depictions of what is, what was, will be, and perhaps have never been&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;em&gt;Palate&lt;/em&gt; can refer to someone's tastes, opinions, thoughts, etc. regarding appreciation of aesthetics.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Vibrant color with each illustrious stroke one will strive&lt;br&gt;&#xA;Unlimited potential for an imagination that is now alive&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A &lt;em&gt;palate&lt;/em&gt; is what an artist uses to hold and mix different colors of paint while painting.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the middle there is only me, this you knew&lt;br&gt;&#xA;My neighbors make up two thirds of you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The &lt;em&gt;palate&lt;/em&gt; is the middle, top part of your mouth. &quot;Neighbors&quot; could refer to teeth and tongue (bone and muscle, two thirds of the components of the mouth, with the remainder being soft tissue such as gums and palate) or to nose and mouth (both adjacent to the palate, and two of the three facial cavities, the third being the eyes).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As we come to the end we reach an uncomfortable state&lt;br&gt;&#xA;Perhaps because of something off your plate &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gagging, heaving, flailing violently in attempts to mitigate&lt;br&gt;&#xA;Always remember how essential it is to masticate&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If you don't chew properly, food can get lodged in your palate and you can choke to death.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5395" LastActivityDate="2014-11-14T01:29:13.253" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4586" PostTypeId="2" ParentId="4211" CreationDate="2014-11-14T01:42:31.027" Score="11" Body="&lt;p&gt;I think it's:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The exclamation point!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;My reasoning:&lt;/p&gt;&#xA;&lt;p&gt;I am something that everyone needs.&lt;br /&gt;&#xA;You use me many times in different ways.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Everyone uses it, and for a variety of reasons. Linguistics, coding, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Sometimes you only use part of me,&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The dot at the bottom is a period.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Others you only use most of me.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; A vertical line by itself can be an L, pipe (|), or a 1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You always turn and bend me to your will.&lt;br /&gt;&#xA;You use to me to scare, to confuse, even to question.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Turn it upside down and it makes a lower case i, which appears in &amp;quot;will&amp;quot;. Turn it on its side and it becomes Morse code for A (&lt;code&gt;._&lt;/code&gt;) or N (&lt;code&gt;_.&lt;/code&gt;). You can bend it into the shape of a question mark (?) or a lower case j. There are lots of reasons and ways to use an exclamation point, so you can shape it to fit your meaning (&amp;quot;bend it to your will&amp;quot;), both literally and figuratively. An exclamation can be used to scare, confuse, and (rhetorically) question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You can even use to confront someone like when you say &amp;quot;Bill, [rest of confrontation]&amp;quot;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; We speak with passion during confrontations, calling for this punctuation mark.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You absolutely use me every day.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; We all get excited or emphatic when speaking sometimes, and of course we use it to make spoiler tags for answers posted here on Puzzling.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;And finally...&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; There's &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4211/guess-the-object-what-am-i-riddle-part-4?rq=1#comment12270_4228&quot;&gt;this comment&lt;/a&gt;, though I thought of this as the answer before I saw it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Turning my original answer on its head, a stronger possibility is:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The letter i&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I am something that everyone needs.&lt;br /&gt;&#xA;You use me many times in different ways.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &amp;quot;Eye&amp;quot; is pronounced the same as &amp;quot;i&amp;quot;. &amp;quot;Everyone&amp;quot; contains &amp;quot;eye&amp;quot; and &amp;quot;different ways&amp;quot; contains both &amp;quot;eye&amp;quot; and &amp;quot;i&amp;quot;. Everyone needs both the letter i and their eyes, and uses them frequently for various purposes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Sometimes you only use part of me,&lt;br /&gt;&#xA;Others you only use most of me.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; What I said about the shape of the exclamation point applies here as well, but there's more. How much of your eyes you show varies with circumstance (squinting, winking, etc.; the size of your pupil depends on the amount of light) and emotion (surprise, sleepiness, etc.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You always turn and bend me to your will.&lt;br /&gt;&#xA;You use to me to scare, to confuse, even to question.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; All the reasoning about the shape of the exclamation point applies here as well; you can get lots of utility and different meanings out of this shape (&amp;quot;i&amp;quot; or &amp;quot;!&amp;quot;). But you can also use your eyes to express the emotions mentioned. You do this by moving or reshaping them (&amp;quot;turn and bend&amp;quot;), and this can express your desires or feelings (&amp;quot;will&amp;quot;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You can even use to confront someone like when you say &amp;quot;Bill, [rest of confrontation]&amp;quot;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &amp;quot;Bill&amp;quot; contains &amp;quot;i&amp;quot;, and the rest of the confrontation is likely to involve some statement like &amp;quot;I know...&amp;quot; as well as angry looks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You absolutely use me every day.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; You use both your eyes and &amp;quot;i&amp;quot; every day, and &amp;quot;every&amp;quot; contains &amp;quot;eye&amp;quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;And regarding the hints given in comments:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4211/guess-the-object-what-am-i-riddle-part-4?rq=1#comment12270_4228&quot;&gt;This statement&lt;/a&gt; holds true for &amp;quot;i&amp;quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5395" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-11-15T04:12:07.043" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="4587" PostTypeId="2" ParentId="4525" CreationDate="2014-11-14T02:02:42.503" Score="5" Body="&lt;p&gt;I've solved it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct response is always &quot;impossible&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The computer shut down because of a coincidental hardware failure and the correct answer to 0 is thus &quot;impossible&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think my answer fits every case so far.&lt;/p&gt;&#xA;" OwnerUserId="97" LastActivityDate="2014-11-14T02:02:42.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4588" PostTypeId="1" AcceptedAnswerId="4592" CreationDate="2014-11-14T02:21:33.630" Score="10" ViewCount="21324" Body="&lt;p&gt;The following is a classic puzzle. It has a canonical solution, but I'd also like any other creative or unusual solutions you can think of.  This is a popularity contest - the highest voted answer after 48 hours will be accepted.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You have at your disposal:&#xA;  A candle, a box of thumb-tacks, and a book of matches.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Your task is to attach the candle to a cork board on the wall and light it without the wax dripping onto the table below.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1667" LastEditorUserId="5373" LastEditDate="2015-04-30T22:57:13.557" LastActivityDate="2015-04-30T22:57:13.557" Title="Attach a candle to the wall" Tags="|strategy|situation|" AnswerCount="15" CommentCount="2" ClosedDate="2014-11-17T10:31:00.663" ContentLicense="CC BY-SA 3.0" />
  <row Id="4589" PostTypeId="2" ParentId="4588" CreationDate="2014-11-14T02:22:35.377" Score="10" Body="&lt;p&gt;Here is what I think is the &quot;correct&quot; answer to the riddle.  First empty the box of thumbtacks, put the candle into the box, use the thumbtacks to nail the box (with the candle in it) to the wall, and light the candle with the match. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Mx5sF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;A few thoughts on some creative approaches:&lt;/strong&gt;&lt;br&gt;&lt;br&gt;&#xA; Step 1: First tear up the thumbtack box and use the cardboard to tape&#xA; the candle to the board, using thumbtacks to hold the cardboard tape&#xA; on. The friction between the cardboard and the candle should keep the&#xA; candle on the wall.  To ensure it won't fall out, you could a third thumbtack piecing the cardboard and candle together.  If the thumbtacks are large enough you may be able to attach the candle directly to the board with the thumbtack alone, or else you could also use the match stick to penetrate the candle and the cork board (first pierce a small hole in the cork with the tack).&lt;br&gt;&lt;br&gt;&#xA; &#xA;  Step 2: Light the candle and simply move the table below - then no wax will drip on it. Or light up the candle (as in &quot;illuminate&quot;) with light from another match without actually lighting the wick.&#xA; &#xA; &#xA;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="97" LastEditorUserId="97" LastEditDate="2014-11-14T04:08:59.707" LastActivityDate="2014-11-14T04:08:59.707" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4591" PostTypeId="2" ParentId="4588" CreationDate="2014-11-14T02:48:06.680" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;Take a few thumbtacks and etch some 1-centimeter-diameter holes in the cork board.&lt;/li&gt;&#xA;&lt;li&gt;Break off a small bit of the candle, tear it into pieces, and stuff a piece in each hole.&lt;/li&gt;&#xA;&lt;li&gt;Turn these pieces molten by heating them with a match.&lt;/li&gt;&#xA;&lt;li&gt;Quickly attach the remaining bit of the candle to the bits by &quot;welding&quot; it to the molten pieces with the heat of the match.&lt;/li&gt;&#xA;&lt;li&gt;Take another tack and tack the box, face-up, to the bottom of the candle, as well as to the corkboard with another tack.&lt;/li&gt;&#xA;&lt;li&gt;Finally, light the candle.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5300" LastActivityDate="2014-11-14T02:48:06.680" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4592" PostTypeId="2" ParentId="4588" CreationDate="2014-11-14T02:48:29.440" Score="11" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;You have at your disposal: A candle, a box of thumb-tacks, and a book of matches.&lt;/p&gt;&#xA;&lt;p&gt;Your task is to attach the candle to a cork board on the wall and light it without the wax dripping onto the table below.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/a/4589/5395&quot;&gt;Mew already posted the classical answer&lt;/a&gt;, so I'll suggest some lateral thinking approaches.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The puzzle doesn't say you have to leave the board up, so take it off the wall and set it on the floor. You can tack or melt the base of the candle to the board to satisfy the attachment requirement.&lt;/li&gt;&#xA;&lt;li&gt;Attach the candle to the board however you like. Light it, then put it out before any wax melts and drips. The puzzle doesn't say how long it has to stay lit.&lt;/li&gt;&#xA;&lt;li&gt;Interpret &amp;quot;it&amp;quot; to refer to the cork board rather than the candle, so light that. No wax will melt.&lt;/li&gt;&#xA;&lt;li&gt;Attach the candle to the board however you like and light it. Move the table so the wax falls on the floor instead.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5395" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-11-14T03:03:15.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4595" PostTypeId="2" ParentId="4588" CreationDate="2014-11-14T02:56:30.817" Score="6" Body="&lt;p&gt;I have a couple answers. If you want me to split them up, just let me know.&lt;/p&gt;&#xA;&lt;h3&gt;Answer 1&lt;/h3&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Attach the candle to the board with the tacks. Light the candle, then blow it out immediately. No wax will drip down.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h3&gt;Answer 2&lt;/h3&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Attach the candle with tacks again. Move the table so it isn't under the cork board.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h3&gt;Answer 3&lt;/h3&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;Stick the candle near the edge of the table, then tilt it slightly so the wax will not drip onto the table, just onto the floor.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h3&gt;Answer 4&lt;/h3&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Put the candle on top of the cork board, and light it there. The wax will hopefully drip on the top of the board.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1664" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-11-14T20:25:10.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4596" PostTypeId="2" ParentId="4588" CreationDate="2014-11-14T03:16:20.433" Score="2" Body="&lt;p&gt;Put some thumbtacks into the body of the candle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then attach it to the cork board (probably the easiest way is to nail it up by the wick).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then &lt;a href=&quot;http://en.wiktionary.org/wiki/light#Verb_2&quot; rel=&quot;nofollow&quot;&gt;light&lt;/a&gt; it. Take out some of those excessive thumbtacks to ease its burden.&lt;/p&gt;&#xA;" OwnerUserId="4473" LastActivityDate="2014-11-14T03:16:20.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4597" PostTypeId="2" ParentId="4276" CreationDate="2014-11-14T03:23:18.470" Score="9" Body="&lt;p&gt;I hope it is OK to post the same answer as Hetal:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;what is your color&quot; based on the following steps:&lt;br&gt;&#xA;&lt;br&gt;&#xA; 1. Spell each color in each box&lt;br&gt;&#xA; .... white white black white blue&lt;br&gt;&#xA; .... silver pink blue blue blue&lt;br&gt;&#xA; .... orange purple purple gray blue&lt;br&gt;&#xA; .... orange orange yellow black yellow&lt;br&gt;&#xA;&lt;br&gt;&#xA; 2. Select nth letter from each word where n is the column number, except that blue is always a blank&lt;br&gt;&#xA; .... w h a t _&lt;br&gt;&#xA; .... s i _ _ _&lt;br&gt;&#xA; .... o u r y _&lt;br&gt;&#xA; .... o r l c o&lt;br&gt;&#xA;&lt;br&gt;&#xA; 3. Unscramble each row&lt;br&gt;&#xA;   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastActivityDate="2014-11-14T03:23:18.470" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4598" PostTypeId="2" ParentId="4588" CreationDate="2014-11-14T03:44:14.783" Score="3" Body="&lt;p&gt;Light the candle and let the wax drip on its side and then push the candle against the wall. Keep your hand under the candle at all times so it doesn't drip on the table.&lt;/p&gt;&#xA;" OwnerUserId="2596" LastActivityDate="2014-11-14T03:44:14.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4600" PostTypeId="2" ParentId="4211" CreationDate="2014-11-14T05:11:57.203" Score="-1" Body="&lt;p&gt;Just a guess..&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sentence, you use it everyday, you change its meaning by the way you speak, you can change it to being narrative, imperative etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5214" LastActivityDate="2014-11-14T05:11:57.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4602" PostTypeId="2" ParentId="4417" CreationDate="2014-11-14T05:24:41.523" Score="0" Body="&lt;p&gt;I'm not sure about this, but...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are here&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PSE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The birthplace&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; StackExchange&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The biggest child&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; StackOverflow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first letter of it's name&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; S&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Go left&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; R? Meaning Rhode Island? Because it's the &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4373" LastActivityDate="2014-11-14T05:24:41.523" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4603" PostTypeId="1" AcceptedAnswerId="4710" CreationDate="2014-11-14T05:56:23.567" Score="9" ViewCount="1278" Body="&lt;p&gt;Inspired by &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4384/a-funny-game-you-can-play-with-a-friend-of-yours-brain-reading&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two friends (Bob and Kim) go to another friend (Bill) and claim to be able to read each other's mind. Bill doesn't believe that and wants to test their ability.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Before we begin, I must tell you, our abilities work better if it is something simple we can focus on, such as a number or a color.  Let's start with a color just to warm up, shall we?&quot; Kim says.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bill agrees, so Bob steps away from Kim and Bill, gets blindfolded and waits. Meanwhile Bill whispers something in the ear of Kim. Kim then says: &quot;Ok, Bob, are you ready to establish the mental link?&quot; &quot;Yes I am&quot; replies Bob.  &quot;Very well then, let us establish the link.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point both Bob and Kim begin making beeping noises which sound like a cross between tuning a radio and an alien spaceship landing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The link is established,&quot; says Bob.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yes, the link is established,&quot; replies Kim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Are you thinking of a rabbit?&quot; asks Bob.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kim sighs.  &quot;Colors, Bob!&quot; replies Kim.  &quot;We're doing colors!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Oh right.&quot; replies Bob.  He thinks for a minute and then replies, &quot;Is it blue?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yes!&quot; replies Kim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Lucky guess,&quot; says Bill.  Let's try again.  He whispers something else into Kim's ear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kim starts to speak, but Bill covers her mouth.  &quot;You can guess now, Bob,&quot; he says motioning for Kim to remain silent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clarification Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bill didn't want Kim to speak, because he was trying to trick Bob and was afraid that she would blurt out that Bill had told her a number, not a color.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bob thinks for a minute, then says, &quot;I'm having trouble seeing your thoughts clearly.  Are you sure you are concentrating hard enough?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;No,&quot; replies Kim and she screws up her face and glares at Bob as if trying to transmit her thoughts into his brain like a nail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Is Bill still in the room?&quot; asks Bob.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yes he is,&quot; replies Kim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bob scratches his head.  &quot;Has he told you the color yet?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;He already did that!&quot; replies Kim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Oh,&quot; says Bob.  Are you sure?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kim looks at Bill and just sighs deeply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;All right, all right!&quot; says Bob.  &quot;It's coming to me, it's coming to me, it's... Bill, you didn't give Kim a color at all, you gave her the number 34!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bill looks at Bob in shock.  &quot;How did you know that?&quot; he asks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How &lt;em&gt;did&lt;/em&gt; Bob know that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bill decides he wants to see another demonstration, so he whispers something else in Kim's ear.  Kim says nothing, so Bill finally says, &quot;Go ahead, Bob!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Oh I'm sorry, says Bob.  Was I supposed to be guessing again?  I think our link dropped.  Kim, are you ready to re-establish our link?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yes,&quot; says Kim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point both Bob and Kim begin making beeping noises which sound like a cross between tuning a radio and an alien spaceship landing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The link is established,&quot; says Bob.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yes, the link is established,&quot; replies Kim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Kim, I sense Bill is getting impatient. Do we need to step up our game?&quot; asks Bob.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;No,&quot; replies Kim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Ok,&quot; replies Bob.  &quot;Are you thinking of a squirrel?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kim turns to Bill.  &quot;I am...&quot; she sighs.  &quot;Sorry, Bill.&quot;  She shakes her head sadly. &quot;Give him a...&quot; she looks into Bill's eyes.  &quot;No? No, Bob.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Oh right,&quot; replies Bob.  &quot;We aren't doing animals this time, are we?  I keep telling you we should do animals, but you are always like, 'No Bob, animals are too hard for me to focus on.'  Tell you what, we will switch next time and Bill can whisper an animal in my ear and you can tell me what it is, right after after I guess that Bill told you the number one-thousand seven hundred and forty two.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Inconceivable!&quot; exclaims Bill.  &quot;How do you keep doing that?!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How &lt;em&gt;do&lt;/em&gt; they keep doing that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Ok my turn,&quot; says Bob.  &quot;Whisper an animal into my ear and Kim will guess it.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bill whispers something into Bob's ear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bob then says: &quot;Ok, Kim, are you ready to establish the mental link?&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yes I am&quot; replies Kim.  &quot;Very well then, let us establish the link.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point both Bob and Kim begin making beeping noises which sound like a cross between tuning a radio and an alien spaceship landing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The link is established,&quot; says Kim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yes, the link is established,&quot; replies Bob.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Are you concentrating on sending me the image of the animal yet?&quot; asks Kim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yes I am,&quot; replies Bob.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Ok,&quot; says Kim. &quot;Are you sure?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Absolutely!&quot; replies Bob.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Ok, I am receiving the image now.  Can you sense me receiving properly?&quot; asks Kim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yes, yes&quot; says Bob.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Have you finished sending it?&quot; asks Kim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yes I'm done.&quot; replies Bob.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kim squints slightly, then appears to be looking at the ceiling for a moment.  Then she says, &quot;Is it a raccoon?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bill faints.  How did Kim and Bob do it this time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The beeps are just something to throw Bill off.  However, Bob and Kim &lt;em&gt;are&lt;/em&gt; passing the information to each other in plain sight.&#xA; The fact that Bob and Kim can't seem to remember/decide what kind of information they are going to be sending each other plays an important role in this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Update:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next day Bill corners Kim and Bob and demands another session.  Bob and Kim agree, on the condition that Bill remain seated, so that he won't faint again.  However, when they are ready to begin, Bill insists that Bob reads his mind, instead of Kim's.  Reluctantly they agree, and Bob is blindfolded once again.  Kim whispers something in Bill's ear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bill then begins imitating Kim, and says: &quot;Ok, Bob, are you ready to establish the mental link?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Wait, wait wait!&quot; says Kim. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;What?&quot; says Bill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Are you sure?&quot; asks Kim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Sure of what?&quot; asks Bill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Are you sure you know how to establish the link?&quot; asks Kim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I saw you do it yesterday, didn't I?&quot; asks Bill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yeah, I guess you did huh,&quot; says Kim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;So can I start now?&quot; asks Bill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I guess.  Sorry to interrupt your concentration like that.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;You should be,&quot; replies Bill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kim nods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Ok, Bob, are you ready to establish the mental link?&quot; Bill asks again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yes I am&quot; replies Bob.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Very well then, let us establish the link.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point both Bill and Bob begin making beeping noises which sound like a cross between tuning a radio and an alien spaceship landing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The link is established,&quot; says Bob.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yes, the link is established,&quot; replies Bill.  &quot;I am sending the image now.&quot;  Bill appears to be concentrating heavily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Sorry to interrupt, but I don't um...&quot; begins Kim, her voice trailing off as she sees Bill glaring at her.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I'm sorry Bill...&quot; she says he stares at her for a few seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Are you sure you are on the same wavelength?&quot;  she looks hesitantly at Bill, until suddenly Bill's frown turns into a smile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;You're just stalling,&quot; he announces confidently.  &quot;You know Bob can't read my mind and you're just trying to delay the obvious fact coming to light, aren't you?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;No, Bill.&quot; Kim announces dejectedly, as she folds her arms and sits down with her back to him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Can we continue then?&quot; Bill asks. Kim does not reply, so Bill continues happily. &quot;Good.  Guess away Bob.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bob appears to be muttering something obscene under his breath.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;What's that Bob?&quot; Bill says, taunting him.  &quot;Frickin hell you can't guess what I'm thinking, did I hear you right?  Guess you can't read my mind after all!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bob chuckles, then replies.  &quot;I don't have to guess Bill to know that you were thinking of an elephant!  And not just plain elephant, not an Asian elephant, but an elephant of the African variety!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bill looks suddenly stricken, but fortunately does not fall to the ground as he is already sitting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did Bob and Kim do it &lt;em&gt;again&lt;/em&gt;?  If you think you know the gist of what is going on, post your own mind reading session between Bob and Kim.  If there are any answers that are on the right track, I will accept the &lt;em&gt;closest&lt;/em&gt; one at the end of the day today.  Otherwise, I will add another session tomorrow.&lt;/p&gt;&#xA;" OwnerUserId="2718" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2015-05-01T03:41:45.693" Title="Brain Reading, part 2" Tags="|pattern|magic|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4604" PostTypeId="1" AcceptedAnswerId="4755" CreationDate="2014-11-14T06:04:45.337" Score="13" ViewCount="654" Body="&lt;p&gt;On average, a lumberjack is 10 logs taller (logs are usually measured in feet) than a normal person plus the height of his tan. Remember that most lumberjacks are young Canadians. Fill the blanks of the following heights and explain why my reasoning is valid. Resulting heights are to be rounded to the closest whole number.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Before:&#xA;  65&quot;, 66&quot;, 69&quot;, 71&quot;, 83&quot;, 89&quot;.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;After Lumberjackification Ritual:&#xA;  201, 204, 212, ?  , ?  , 402.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Hints:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A lumberjack's tan, or sunburn, is a 1st &lt;em&gt;degree&lt;/em&gt; burn.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;This information is mostly (completely) false in real life.&lt;/li&gt;&#xA;&lt;li&gt;All information provided was given for a reason. Do not edit out the (made up) Canadian part.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4802" LastEditorUserId="5373" LastEditDate="2016-11-20T14:34:49.000" LastActivityDate="2016-11-20T14:34:49.000" Title="Measuring Lumberjacks" Tags="|mathematics|enigmatic-puzzle|" AnswerCount="1" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="4605" PostTypeId="2" ParentId="4603" CreationDate="2014-11-14T07:09:40.560" Score="1" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the beeping noises? Probably Kim and Bob agreed on alphabet. Something like morse code and &quot;transmit&quot; the messages via beeping?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or might it be some&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; sign language? Bill is blindfolded after all.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5283" LastActivityDate="2014-11-14T07:09:40.560" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4606" PostTypeId="2" ParentId="4417" CreationDate="2014-11-14T07:14:23.617" Score="1" Body="&lt;p&gt;You are here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Puzzling.SE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Birthplace:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Area 51&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Biggest Child:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Graphic Design&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The First Letter of it's Name:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; G&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Go Left:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; F&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5414" LastActivityDate="2014-11-14T07:14:23.617" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4608" PostTypeId="2" ParentId="4286" CreationDate="2014-11-14T07:54:44.290" Score="2" Body="&lt;p&gt;Ok Here's my friends try:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 34 = 7&lt;br&gt;&#xA;&lt;br&gt;&#xA; 1=2&lt;br&gt;&#xA;&lt;br&gt;&#xA; 2=2&lt;br&gt;&#xA;&lt;br&gt;&#xA; 3=5&lt;br&gt;&#xA;&lt;br&gt;&#xA; 5=8&lt;br&gt;&#xA;&lt;br&gt;&#xA; 8=13=1+3=4&lt;br&gt;&#xA;&lt;br&gt;&#xA; 13=4&lt;br&gt;&#xA;&lt;br&gt;&#xA; 21=3&lt;br&gt;&#xA;&lt;br&gt;&#xA; 34=7&lt;br&gt;&#xA; ...&lt;br&gt;&#xA; 89=8+9=17=1+7=8&#xA;&#xA; How it works:&#xA; The number given is one of Fibonacci numbers&#xA; If it have 1 char =&gt; return next number from Fibonacci&#xA; If it have 2 char =&gt; plus them together&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And now 21 is not 3, I guess luckily the guard didn't give me 21 :D&lt;/p&gt;&#xA;" OwnerUserId="5406" LastEditorUserId="2718" LastEditDate="2014-11-15T04:15:40.170" LastActivityDate="2014-11-15T04:15:40.170" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4609" PostTypeId="2" ParentId="4603" CreationDate="2014-11-14T07:58:01.173" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are using Morse Code when establishing the link. However, the jig was almost up when Bill decided do play dirty. Due to Kim's behaviour and Bill being a friend of both of them Bob deduced it wasn't a colour, so he hazarded a guess and went for Bill's favorite number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5265" LastActivityDate="2014-11-14T07:58:01.173" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4610" PostTypeId="2" ParentId="4572" CreationDate="2014-11-14T08:02:01.573" Score="6" Body="&lt;p&gt;Might you be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an artichoke&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reason:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You begin with &quot;art&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Two:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the middle, you have &quot;I&quot; having neighbors &quot;O&quot; and &quot;U&quot; as &lt;a href=&quot;https://puzzling.stackexchange.com/a/4594/5421&quot;&gt;someone else pointed out&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Three:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the end, you &quot;choke&quot; because you failed to masticate properly&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5421" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2014-11-14T08:02:01.573" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4611" PostTypeId="2" ParentId="4588" CreationDate="2014-11-14T08:08:43.740" Score="0" Body="&lt;p&gt;I assume the candle is fairly thin and the thumbtacks are fairly large, and push them through the candle fixing it onto the board.&#xA;Alternatively, I just push them with precision through the wick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a little bonus, I fold the thumbtack box into a candle shape and also put that on the board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering I didn't do this in absolute dark, the candle is already lit (illuminated) by sunlight/the room lamp.&lt;/p&gt;&#xA;" OwnerUserId="5265" LastActivityDate="2014-11-14T08:08:43.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4612" PostTypeId="2" ParentId="4603" CreationDate="2014-11-14T08:19:53.380" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Kim is ventriloquist. While making 'the weird voice', he mimicked the answer. That's why when Bill keeps Kim shut up (covered his mouth), Bob can't think.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4995" LastActivityDate="2014-11-14T08:19:53.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4613" PostTypeId="1" AcceptedAnswerId="4616" CreationDate="2014-11-14T08:27:54.450" Score="20" ViewCount="2046" Body="&lt;p&gt;A party is being held at a local mansion. The host is very rich and his success is because of one thing ~ his famous recipe for spaghetti!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spaghetti recipe is kept in a secret room inside the mansion, which is only accessible from the kitchen, which can be only reached from the party lounge, which is separated from the security lobby by a floor where &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4001/&quot;&gt;elephants have been reported to occasionally trample fishy guests&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mansion has two entrances to the security lobby, and all adjacent rooms are connected by two doors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your task is to enter the mansion, sneak into the secret room, steal the spaghetti recipe and get out again. Here is a plan of the mansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7mHoU.png&quot; alt=&quot;Spaghetti Mansion&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this puzzle, you start as the number 1 outside the mansion and each time you pass a door, an operation indicated on the door will change your number. After leaving the mansion with the recipe, you must be again number 1 as when you came in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can pass all doors multiple times in both directions, but the less, the better, otherwise the security may become suspicious and you don't want them to release the elephants :) The shortest solution wins.&lt;/p&gt;&#xA;" OwnerUserId="4394" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2020-06-13T09:25:43.053" Title="1 in, 1 out (The Security to the Party Maze)" Tags="|mathematics|mazes|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4615" PostTypeId="1" AcceptedAnswerId="4688" CreationDate="2014-11-14T08:47:34.950" Score="32" ViewCount="7233" Body="&lt;p&gt;This riddle is so easy that I already have created another one. Just in case.&#xA;Anyway, please avoid wild guesses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For sure I am queer&lt;br&gt;&#xA;and out of place&lt;br&gt;&#xA;Are you still here?&lt;br&gt;&#xA;Go start the chase&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But always remember&lt;br&gt;&#xA;in the first place&lt;br&gt;&#xA;stacks are amazing&lt;br&gt;&#xA;but not a maze&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm in the beginning&lt;br&gt;&#xA;as well as the end&lt;br&gt;&#xA;And always traceable&lt;br&gt;&#xA;no matter what I pretend&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really am curious&lt;br&gt;&#xA;about who will find&lt;br&gt;&#xA;the animate being&lt;br&gt;&#xA;that hides behind&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution is here&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;No, it is not. One might call it far-fetched.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here is also a real hint&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You will find only one of my kind.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5106" LastEditorUserId="5106" LastEditDate="2016-07-13T11:29:02.953" LastActivityDate="2016-07-13T11:29:02.953" Title="You will have to pray!" Tags="|word|treasure-hunt|" AnswerCount="13" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="4616" PostTypeId="2" ParentId="4613" CreationDate="2014-11-14T08:49:17.627" Score="16" Body="&lt;p&gt;12 doors. Total number of steps must be even. And problem can not be solved in 10 doors or fewer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Step 1 : 1 + 11 = 12 Security &lt;BR&gt;&#xA; Step 2 : - 7 = 5 Elephant &lt;BR&gt;&#xA; Step 3 : + 4 = 9 Party &lt;BR&gt;&#xA; Step 4 : - 1 = 8 Kitchen &lt;BR&gt;&#xA; Step 5 : - 6 = 2 Secret &lt;BR&gt;&#xA; Step 6 : - 6 = -4 Kitchen &lt;BR&gt;&#xA; Step 7 : - 1 = -5 Party &lt;BR&gt;&#xA; Step 8 : * √2 = -5√2 Kitchen&lt;BR&gt;&#xA; Step 9 : * √2 = -10 Party&lt;BR&gt;&#xA; Step 10: / 2 = -5 Elephant&lt;BR&gt;&#xA; Step 11: * 2 = -10 Security&lt;BR&gt;&#xA; Step 12: + 11 = 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4637" LastEditorUserId="4637" LastEditDate="2014-11-14T09:01:46.677" LastActivityDate="2014-11-14T09:01:46.677" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4618" PostTypeId="2" ParentId="4613" CreationDate="2014-11-14T08:54:57.810" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 + 11 * 2 / 2 - 1 - 6 - 6 - 1 - 1 - 1 - 1 - 1 / 2 - 7 + 11 = 1 &lt;br&gt;&#xA; 14 doors, but sort of achieved with a quick and dirty program and some thinking (that also proved that there is no 10-door-possibility, so Jan's answer is the optimum.)&lt;br&gt;&#xA; &lt;a href=&quot;http://pastebin.com/VraSrJ9R&quot;&gt;http://pastebin.com/VraSrJ9R&lt;/a&gt; contains a neat little list of all 1024 &quot;10-door-possibilities&quot;, 0 being the left option, 1 being the right option.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another 12-door-solution I like a lot:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 + 11 - 7 + 4 - 1 - 6 - 6 - 1 - 1 - 1 + 4 - 7 + 11 = 1&lt;br&gt;&#xA; Essentially only take addition/substraction and go through the -1 door twice (three times, you know what I mean).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5265" LastEditorUserId="5265" LastEditDate="2014-11-14T09:21:28.317" LastActivityDate="2014-11-14T09:21:28.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4619" PostTypeId="1" AcceptedAnswerId="4620" CreationDate="2014-11-14T08:58:42.980" Score="28" ViewCount="4237" Body="&lt;p&gt;I have one start&lt;br&gt;&#xA;and usually many endings.&lt;br&gt;&#xA;I'm very smart&lt;br&gt;&#xA;when guiding you to sorted findings.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For you I offer&lt;br&gt;&#xA;many useful suggestions&lt;br&gt;&#xA;but I only answer&lt;br&gt;&#xA;to &quot;yes-no&quot; questions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The title gives you a hint, by the way)  &lt;/p&gt;&#xA;" OwnerUserId="5283" LastEditorUserId="5188" LastEditDate="2015-03-24T05:18:09.713" LastActivityDate="2018-10-02T14:45:04.940" Title="Another riddle for scientists" Tags="|riddle|rhyme|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4620" PostTypeId="2" ParentId="4619" CreationDate="2014-11-14T09:04:33.453" Score="34" Body="&lt;p&gt;I love computer programming (particular algorithms and &lt;strong&gt;data structures&lt;/strong&gt;). So my answer would be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Binary_search_tree&quot; rel=&quot;nofollow noreferrer&quot;&gt;BST&lt;/a&gt; (Binary Search Tree)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is an image for the same&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/ELdbg.png&quot; alt=&quot;A binary search tree&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have one start&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Root node&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and usually many endings.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Leaf nodes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm very smart&lt;BR&gt;&#xA;when guiding you to sorted findings.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BST smartly guides your way in nodes arranged in a sorted fashion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For you I offer&lt;BR&gt;&#xA;many useful suggestions&lt;BR&gt;&#xA;but I only answer&lt;BR&gt;&#xA;to &quot;yes-no&quot; questions.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BST can be used to fetch a sorted list, find whether an item is present, find the closest item etc, but at each node all that you can decide is whether you should go left or right (yes-no)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;About the title hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Scientists here means computer scientists&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4637" LastEditorUserId="4637" LastEditDate="2017-12-20T11:00:28.707" LastActivityDate="2017-12-20T11:00:28.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4622" PostTypeId="2" ParentId="4588" CreationDate="2014-11-14T09:12:25.917" Score="2" Body="&lt;p&gt;Now for some frivolous/humorous answers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nice easy one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;leave everything on the table&lt;/li&gt;&#xA;  &lt;li&gt;get my wife to do it, that way &lt;em&gt;I&lt;/em&gt; have not let any wax drip on the table&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;get my children involved &lt;/li&gt;&#xA;  &lt;li&gt;wait 10-15 minutes&lt;/li&gt;&#xA;  &lt;li&gt;by the time that has all happened, the candle will be smushed into the carpet, the table will be a fort and I will be having a coffee in&#xA;  the garden, and the problem is moot :)&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I will wait a bit and post some more proper answers as well, but that's me for now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addendum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just remembered to post another answer, this one from one of my children (he is 9, and its a real belt and braces approach):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;get a duplicate set of all items&lt;/li&gt;&#xA;  &lt;li&gt;melt wax from the candle onto one edge of the box to make it stick to the board&lt;/li&gt;&#xA;  &lt;li&gt;tack through the box and board, ensuring the box has its walls upright&lt;/li&gt;&#xA;  &lt;li&gt;manually melt the candle down and ensure that the wax does not overflow the box&lt;/li&gt;&#xA;  &lt;li&gt;if it does cut the original candle short by the amount left over on the duplicate candle&lt;/li&gt;&#xA;  &lt;li&gt;repeat the process with the original materials, but this time merely light the candle&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Takes after his dad this one.... :)&lt;/p&gt;&#xA;" OwnerUserId="5012" LastEditorUserId="5012" LastEditDate="2014-11-18T11:12:49.397" LastActivityDate="2014-11-18T11:12:49.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4623" PostTypeId="2" ParentId="4615" CreationDate="2014-11-14T09:31:25.020" Score="21" Body="&lt;p&gt;Are you &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; deoxyribonucleic acid? &lt;br&gt;&lt;br&gt;&quot;AND&quot; out of place = DNA.&lt;br&gt;&lt;br&gt;DNA is both in&quot;&lt;strong&gt;be&lt;/strong&gt;gi&lt;strong&gt;n&lt;/strong&gt;ning&quot; and &quot;&lt;strong&gt;end&lt;/strong&gt;&quot;, if you turn e's to form a's and flip b's to form d's.&lt;br&gt;&lt;br&gt;DNA molecules are held together by &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Stacking_(chemistry)&quot;&gt;pi - stacks&lt;/a&gt;&quot;.&lt;br&gt;&lt;br&gt;DNA is traceable (e.g. in forensic cases).&lt;br&gt;&lt;br&gt;DNA encodes for animate beings.&lt;br&gt;&lt;br&gt;The clue you will only find one of my kind relates to the fact that each being has a unique DNA sequence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="97" LastEditorUserId="5106" LastEditDate="2014-11-14T13:38:08.817" LastActivityDate="2014-11-14T13:38:08.817" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4625" PostTypeId="2" ParentId="4588" CreationDate="2014-11-14T09:53:21.887" Score="1" Body="&lt;p&gt;Use a thumb-tack to patiently scratch some wax from the candle in order to reduce its diameter, so that when it will be lighted up all the wax will burn and none will drip (as happens with thin birthday candles). You can leave the very bottom larger so that the possible few wax drops will be stopped there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then fix it to the cork board as you prefer: heating its bottom on the side with a match, using a couple of thumb-tack, planting a couple of matches into the board and putting the candle on top of them...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use the box of thumb-tacks to collect the wax during the scratching phase.&lt;/p&gt;&#xA;" OwnerUserId="2683" LastEditorUserId="2683" LastEditDate="2014-11-14T14:20:14.477" LastActivityDate="2014-11-14T14:20:14.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4626" PostTypeId="2" ParentId="4588" CreationDate="2014-11-14T10:08:51.963" Score="1" Body="&lt;p&gt;If the cork board can be lifted from the wall, you can plant some matches close to the edge of the board with a 45° inclination. Then twist the board in order to create a shelf like these:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/I9aWo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the inclined metal rods will actually be many matches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shelf wont be able to host some books, but for sure will be capable to hold a small candle on top of it.&lt;/p&gt;&#xA;" OwnerUserId="2683" LastActivityDate="2014-11-14T10:08:51.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4627" PostTypeId="2" ParentId="2188" CreationDate="2014-11-14T10:10:14.787" Score="0" Body="&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;: My answer is wrong. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;If anyone gets confused like I did, please see the comments for a reminder.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;(original)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of the answers are incorrect, and the puzzle itself flawed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To reiterate the (overlooked) requirements:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&quot;&lt;em&gt;... you can only ask a single question from &lt;strong&gt;one&lt;/strong&gt; of the&#xA;guards&lt;/em&gt;&quot;, and; &lt;/li&gt;&#xA;&lt;li&gt;&quot;&lt;em&gt;One of the guards always tells the truth and the&#xA;other always lies. You don't know which one is [which].&lt;/em&gt;&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These rules create a situation where there is no method for discerning one guards' honesty from the other, and thus which door leads to freedom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's best explained as, &quot;take my word for it*. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit - For Example, you ...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Ask Guard-A : &quot;&lt;em&gt;Do you guard the door to Freedom?&lt;/em&gt;&quot; |&lt;strong&gt;OR&lt;/strong&gt;| &quot;&lt;em&gt;Does Guard-B guard the door to freedom?&lt;/em&gt;&quot;&lt;/li&gt;&#xA;&lt;li&gt;Guard-A replies: &quot;&lt;em&gt;Yes&lt;/em&gt;&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;50% - Guard-A is lying, you die.&#xA;50% - Guard-A is honest, you live.&lt;/p&gt;&#xA;" OwnerUserId="5412" LastEditorUserId="5412" LastEditDate="2014-11-14T15:22:48.983" LastActivityDate="2014-11-14T15:22:48.983" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4628" PostTypeId="1" AcceptedAnswerId="4630" CreationDate="2014-11-14T10:17:21.663" Score="6" ViewCount="555" Body="&lt;p&gt;We are dotted around and what all life need&lt;br&gt;&#xA;for structure, stability and to feed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some share the name of what we hold&lt;br&gt;&#xA;others bear the names of characters of old&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nomads we fear will make us stray&lt;br&gt;&#xA;or Dwarves will decide to take us away&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our shape is manipulated by those greater than us&lt;br&gt;&#xA;we only survive untill they make a fuss  &lt;/p&gt;&#xA;" OwnerUserId="5135" LastActivityDate="2016-06-08T09:26:10.103" Title="Name that entity (3)" Tags="|riddle|word|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4630" PostTypeId="2" ParentId="4628" CreationDate="2014-11-14T10:35:41.950" Score="5" Body="&lt;p&gt;You are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A planet. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are dotted around and what all life need for structure, stability and to feed&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everything we know is built on a planet, including ourselves. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some share the name of what we hold others bear the names of characters of old&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some are named after things we can hold (Mercury, Earth), others after characters (Pluto).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nomads we fear will make us stray&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nomads in this case are bodies in space (asteroids etc.) that could substantially alter our course or damage the planet irreparably.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or Dwarves will decide to take us away&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dwarves are known for digging too greedily and too deeply.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Our shape is manipulated by those greater than us&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A planet's shape is determined by its own gravity which is, of course greater than us.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;we only survive untill they make a fuss &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; And if you believe mass media we're slowly destroying our planet through industry...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1491" LastEditorUserId="20274" LastEditDate="2016-06-08T09:26:10.103" LastActivityDate="2016-06-08T09:26:10.103" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4632" PostTypeId="2" ParentId="4615" CreationDate="2014-11-14T11:04:50.077" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Time&lt;br&gt;&#xA; its considered as a fourth dimension thats kinda queer as we cant comprehend 4d objects&lt;br&gt;&#xA; Time stacks are a name given to photos which show lots of time in one image there amazing&lt;br&gt;&#xA; but they are no maze&lt;br&gt;&#xA; we measure time trace it back through history, make futuristic predictions and so forth&lt;br&gt;&#xA; the animate being is yourself in the future if you were to look through time. Are you curios to see your future self.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5135" LastActivityDate="2014-11-14T11:04:50.077" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4633" PostTypeId="2" ParentId="4405" CreationDate="2014-11-14T11:08:57.773" Score="4" Body="&lt;p&gt;Of course with access to modern technology or other people, this would become trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He could write it down and ask someone else without disabilities to read it to the blind man, or he could just send the blind man a message via SMS/Mail whatever, which can be processed text-to-speech.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume there are only these three man and no access to special technology or other people (one could just place the riddle in a closed room in 16th century to assure this) there are two ways left - communication over a shared available sense (most likely touch) or by tricking the deaf man into telling the blind man.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is for any kind of trick or secret language the blind and the mute man would have to share a secret-code or something, which has to be communicated first - and if we have a way to communicate that, we can just use this way for the vital information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leaves a trick where the deaf man will somehow reveal himself by accident (for example convincing him the blind man has left the room, he won't here when he comes back and talking to him about the kiss) or using violence or threats to persuade the perpetrator, which will be pretty hard, since the deaf man can tell the blind man anything and we cannot argue against it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the only ways left are conveying information by touch, morse, writing letters with somethimg the blind man can feel, or humming &quot;he kissed your girl&quot; ^^&lt;/p&gt;&#xA;" OwnerUserId="1998" LastActivityDate="2014-11-14T11:08:57.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4637" PostTypeId="5" CreationDate="2014-11-14T12:16:48.953" Score="0" Body="&lt;p&gt;A maze is a tour puzzle in the form of a complex branching passage through which the solver must find a route. &#xA;The pathways and walls in a maze are fixed, and puzzles in which the walls and paths can change during the game are usually categorized as &lt;em&gt;tour puzzles&lt;/em&gt;. &#xA;The Cretan labyrinth is the oldest known maze.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-04T15:33:34.220" LastActivityDate="2015-02-04T15:33:34.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4638" PostTypeId="4" CreationDate="2014-11-14T12:16:48.953" Score="0" Body="A tour puzzle in the form of a complex branching passage through which the solver must find a route." OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-04T15:38:12.243" LastActivityDate="2015-02-04T15:38:12.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4639" PostTypeId="2" ParentId="4279" CreationDate="2014-11-14T12:28:53.587" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My guess at this one is... &lt;strong&gt;the penny&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1491" LastActivityDate="2014-11-14T12:28:53.587" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4642" PostTypeId="1" AcceptedAnswerId="4683" CreationDate="2014-11-14T13:14:37.017" Score="27" ViewCount="6833" Body="&lt;p&gt;◳◰ ◓◨ ◨◧◕ ◎◌ ◱◯◱◯ ◍◌○ ◉◉ ◇◔◓◕ ◐►◓◒ ◒◑ ◈◑ ◆◆◓ ◉◉◉&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Administrative details (not part of the puzzle):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The answer is a clue to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4057/the-security-to-the-party-12-now-with-party-soundtrack&quot;&gt;The Security to the Party [12] (now with party soundtrack!)&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For the benefit of anyone without a Unicode-capable browser, the puzzle should look like this (image of the puzzle text): &#xA;&lt;img src=&quot;https://i.stack.imgur.com/zh2NU.png&quot; alt=&quot;◳◰ ◓◨ ◨◧◕ ◎◌ ◱◯◱◯ ◍◌○ ◉◉ ◇◔◓◕ ◐►◓◒ ◒◑ ◈◑ ◆◆◓ ◉◉◉&quot;&gt; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The answer is written text in the English language. The puzzle is: &lt;strong&gt;&quot;◳◰ ◓◨ ◨◧◕ ◎◌ ◱◯◱◯ ◍◌○ ◉◉ ◇◔◓◕ ◐►◓◒ ◒◑ ◈◑ ◆◆◓ ◉◉◉&quot; is the cyphertext. What is the plaintext?&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This type of puzzle is called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Cryptogram&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cryptogram&lt;/a&gt;. The Google search &quot;&lt;a href=&quot;https://www.google.co.uk/webhp?q=%22how%20to%20solve%20a%20cryptogram%22&quot; rel=&quot;nofollow noreferrer&quot;&gt;How to solve a cryptogram&lt;/a&gt;&quot; leads to all kinds of useful resources, hints and tips.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The puzzle is &lt;em&gt;not&lt;/em&gt; case-sensitive (see question below).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The plaintext does not contain any slang or acronyms (see question below).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Chat about this question is here: &lt;a href=&quot;http://chat.stackexchange.com/rooms/18659/discussion-on-question-by-a-e-------------&quot;&gt;http://chat.stackexchange.com/rooms/18659/discussion-on-question-by-a-e-------------&lt;/a&gt; (contains spoilers!)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are no proper nouns (names) in the plaintext. (question from chat)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The plaintext does not contain roman numerals. (question from chat)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There is no punctuation in the plaintext. (question from chat)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Thank you @Kevin for providing a list of names of the symbols, for anyone who's having trouble with fonts (also good from an accessibility point of view, I guess). So everyone's on a level playing field, I've copied it below:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;WHITE SQUARE WITH UPPER RIGHT QUADRANT&#xA;WHITE SQUARE WITH UPPER LEFT QUADRANT&#xA;SPACE&#xA;CIRCLE WITH UPPER HALF BLACK&#xA;SQUARE WITH RIGHT HALF BLACK&#xA;SPACE&#xA;SQUARE WITH RIGHT HALF BLACK&#xA;SQUARE WITH LEFT HALF BLACK&#xA;CIRCLE WITH ALL BUT UPPER LEFT QUADRANT BLACK&#xA;SPACE&#xA;BULLSEYE&#xA;DOTTED CIRCLE&#xA;SPACE&#xA;WHITE SQUARE WITH LOWER LEFT QUADRANT&#xA;LARGE CIRCLE&#xA;WHITE SQUARE WITH LOWER LEFT QUADRANT&#xA;LARGE CIRCLE&#xA;SPACE&#xA;CIRCLE WITH VERTICAL FILL&#xA;DOTTED CIRCLE&#xA;WHITE CIRCLE&#xA;SPACE&#xA;FISHEYE&#xA;FISHEYE&#xA;SPACE&#xA;WHITE DIAMOND&#xA;CIRCLE WITH UPPER RIGHT QUADRANT BLACK&#xA;CIRCLE WITH UPPER HALF BLACK&#xA;CIRCLE WITH ALL BUT UPPER LEFT QUADRANT BLACK&#xA;SPACE&#xA;CIRCLE WITH LEFT HALF BLACK&#xA;BLACK RIGHT-POINTING POINTER&#xA;CIRCLE WITH UPPER HALF BLACK&#xA;CIRCLE WITH LOWER HALF BLACK&#xA;SPACE&#xA;CIRCLE WITH LOWER HALF BLACK&#xA;CIRCLE WITH RIGHT HALF BLACK&#xA;SPACE&#xA;WHITE DIAMOND CONTAINING BLACK SMALL DIAMOND&#xA;CIRCLE WITH RIGHT HALF BLACK&#xA;SPACE&#xA;BLACK DIAMOND&#xA;BLACK DIAMOND&#xA;CIRCLE WITH UPPER HALF BLACK&#xA;SPACE&#xA;FISHEYE&#xA;FISHEYE&#xA;FISHEYE&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2014-12-09T17:03:24.560" Title="Decode the message enciphered in these symbols: ◳◰ ◓◨ ◨◧◕ ◎◌ ◱◯◱◯ ◍◌○ ◉◉ ◇◔◓◕ ◐►◓◒ ◒◑ ◈◑ ◆◆◓ ◉◉◉" Tags="|cipher|symbols|" AnswerCount="1" CommentCount="52" ContentLicense="CC BY-SA 3.0" />
  <row Id="4645" PostTypeId="2" ParentId="4588" CreationDate="2014-11-14T13:54:10.433" Score="1" Body="&lt;p&gt;Take some matches and place them next to each other on the table so that they form a raft, leaving only two  matches. Light one of the remaining matches and carefully melt the candle's paraffin, so that it drops onto your match-raft, binding them together. Before the paraffin cools, place the candle on the raft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You now have a wax shield made out of match-sticks. Attach it to the cork wall anyway you like (make sure not to destroy the shield in the process&lt;/p&gt;&#xA;" OwnerUserId="1831" LastActivityDate="2014-11-14T13:54:10.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4646" PostTypeId="2" ParentId="4615" CreationDate="2014-11-14T14:04:57.060" Score="-1" Body="&lt;p&gt;I want to say&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Life&lt;br&gt;&#xA;&lt;br&gt;&#xA;- It is technically out of place in the universe, where non-life is the vast majority.&lt;br&gt;&#xA;- Stacks, i.e., lots of life/lives is better than a little, I suppose.&lt;br&gt;&#xA;- Beginning and end are merely a metter of timeframe, lives are traceable whether people want it or not.&lt;br&gt;&#xA;- And behind every life there's something animate.&lt;br&gt;&#xA;- And everyone only has one (cats go away!).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but it doesn't really fit the bill in all aspects..&lt;/p&gt;&#xA;" OwnerUserId="5265" LastEditorUserId="5265" LastEditDate="2014-11-14T14:13:58.680" LastActivityDate="2014-11-14T14:13:58.680" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4647" PostTypeId="2" ParentId="4615" CreationDate="2014-11-14T14:09:06.910" Score="1" Body="&lt;p&gt;Your hint brings to mind&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a fingerprint, and this connects well with the &quot;beginning&quot; and &quot;end&quot;, as they are at the ends of your fingers, and stick with you from the beginning and end of your life. They look rather like a &quot;maze&quot;?? are literally traceable and traceable for tracking crimes etc. and humans are behind them and each are unique?! also you touch them together when you pray hahahah&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;idk im new here hi&lt;/p&gt;&#xA;" OwnerUserId="5441" LastActivityDate="2014-11-14T14:09:06.910" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4648" PostTypeId="1" AcceptedAnswerId="4650" CreationDate="2014-11-14T14:10:54.203" Score="4" ViewCount="1794" Body="&lt;p&gt;I heard this puzzle in a quiz I participated in:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;10, 11, 12, 13, 14, 15, 16, 17, 20, 22, 24, ?, 100, 121, 10000&lt;/p&gt;&#xA;&lt;p&gt;What is the missing number?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;(You don't need to do any big calculations, and you don't need any advanced mathematical knowledge - just one basic concept that's known to many non-mathematical people.)&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-09-17T18:06:21.807" Title="Missing number in this sequence?" Tags="|mathematics|pattern|number-sequence|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4650" PostTypeId="2" ParentId="4648" CreationDate="2014-11-14T14:16:20.720" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; According the google the answer be 31. &lt;br&gt;&lt;br&gt;&#xA; &#xA; The sequence is 16 in base 16-n.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="97" LastEditorUserId="97" LastEditDate="2014-11-14T14:29:20.187" LastActivityDate="2014-11-14T14:29:20.187" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="4653" PostTypeId="2" ParentId="4615" CreationDate="2014-11-14T14:34:26.853" Score="2" Body="&lt;p&gt;Here's my thought:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The answer is &lt;strong&gt;the null pointer&lt;/strong&gt;, as in C and C-like languages.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Normally it's not supposed to appear in programs, thus &amp;quot;out of place&amp;quot; and needing to be &amp;quot;chased&amp;quot;.&lt;/li&gt;&#xA;&lt;li&gt;In some languages like C, it points to the start of memory (the zero address), but is often used to signal the end of a data structure with indeterminate length, like strings and &amp;quot;stacks&amp;quot;.&lt;/li&gt;&#xA;&lt;li&gt;Many compilers and runtimes &amp;quot;pretend&amp;quot; that the null pointer points to nothing (and signal its use as an error), because to do otherwise would open up major potential for security bugs, but (assuming that it points to zero, which is true for many languages but not all of them), technically it &lt;em&gt;could&lt;/em&gt; point to something useful if it were allowed to be &amp;quot;traced&amp;quot;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5448" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-11-14T14:34:26.853" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4654" PostTypeId="2" ParentId="4588" CreationDate="2014-11-14T14:53:16.357" Score="3" Body="&lt;p&gt;Take the cork board down off the wall.&lt;br&gt;&#xA;Optionally place it on the table for ease of working.&lt;br&gt;&#xA;Set candle on cork board and light it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sell box of tacks on eBay.&lt;/p&gt;&#xA;" OwnerUserId="5082" LastActivityDate="2014-11-14T14:53:16.357" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4655" PostTypeId="2" ParentId="4341" CreationDate="2014-11-14T14:54:33.550" Score="3" Body="&lt;p&gt;The murderer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mew&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; he set the puzzle in which Joe died.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Carved into the side of the gun is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; some sort of fish shape.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Comparing with pictures in books that you find on Joe's shelves, you determine that it appears to be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a herring, and it has been stained with berries to a deep maroon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4373" LastActivityDate="2014-11-14T14:54:33.550" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4656" PostTypeId="2" ParentId="4615" CreationDate="2014-11-14T14:57:05.457" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A road&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;seems to fit everything except the last verse and the hint.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's 'out of place' because it leads out of the place where it begins. 'Go start the chase' - chasing along the road. 'Not a maze' - it's a straight road. 'I'm in the beginning as well as the end' - both ends of the road are part of it. It's 'always traceable' because it appears on a map.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Probably not the right answer though, since it has nothing to do with 'stacks' or an 'animate being' and it's not 'the only one of its kind'!&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2014-11-14T14:57:05.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4657" PostTypeId="1" AcceptedAnswerId="4693" CreationDate="2014-11-14T14:59:48.957" Score="5" ViewCount="1010" Body="&lt;pre&gt;&lt;code&gt;A wild cacophony of sound greets the arrival of this beast &#xA;Perhaps you thought there'd be applause, a jovial shout should be the least&#xA;&#xA;But fleeting glances, nostrils flared, each face curled slightly in a frown,&#xA;As all who share the journey know the beast is also going down. &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What is the name of this beast?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus points: Where is it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;br&gt;&#xA;A final couplet as a hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; So try to banish this disgrace, light a candle, wave a fan,&lt;br&gt;&#xA; Smear with incense, hide your face, or get out early, while you can.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-11-14T20:43:36.827" LastActivityDate="2014-11-17T19:24:50.780" Title="Taming the beast" Tags="|riddle|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4659" PostTypeId="2" ParentId="4615" CreationDate="2014-11-14T15:08:07.747" Score="2" Body="&lt;p&gt;If the riddle is computer related : &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An IP adress&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because : &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Queer = not understandable by the average user, &#xA; Start the chase = find the shortest way to the next IP, &#xA; Stacks relate to the complexity of Internet routes, &#xA; Beginning and end = Both client and server, &#xA; Always traceable (even though it can be very difficult), &#xA; Animate being = human or machine on either side, &#xA; One of a kind&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It doesn't seem like it fits everything but it's my best guess&lt;/p&gt;&#xA;" OwnerUserId="4631" LastActivityDate="2014-11-14T15:08:07.747" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4660" PostTypeId="1" AcceptedAnswerId="4672" CreationDate="2014-11-14T15:25:05.900" Score="35" ViewCount="2526" Body="&lt;p&gt;&lt;em&gt;Disclaimer&lt;/em&gt;: no offence intended to Americans, Australians, or any other group! It's all part of the puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I talking about in the following riddle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many different types of men;&lt;br&gt;&#xA;They often come in groups of two or three.&lt;br&gt;&#xA;Each group has a matching group of women,&lt;br&gt;&#xA;But they should stay with their compatriots.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Europe we have several different kinds;&lt;br&gt;&#xA;The Americans are only half as good as us.&lt;br&gt;&#xA;Those from Down Under are all thin and weedy,&lt;br&gt;&#xA;Not as tough as the solid British men.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The French can have a woman in their group;&lt;br&gt;&#xA;The Israelis won't fit into any other land;&lt;br&gt;&#xA;But if you need translation services,&lt;br&gt;&#xA;I can provide you with a magic cube.&lt;br&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2014-12-15T15:56:01.837" LastActivityDate="2014-12-15T15:56:01.837" Title="There are many different types of men" Tags="|riddle|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4661" PostTypeId="2" ParentId="4615" CreationDate="2014-11-14T15:28:56.797" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; memory&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;reasons for my answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For sure I am queer, - &lt;em&gt;we hardly remember something, but easily remember something, too.&lt;/em&gt; &lt;br/&gt; and out of place, - &lt;em&gt;it can't be seen&lt;/em&gt; &lt;br/&gt; Are you still here? - &lt;em&gt;sometimes we forgot something shortly&lt;/em&gt;&lt;br/&gt; Go start the chase - &lt;em&gt;we sometime want to remember things we forgot&lt;/em&gt;&lt;br/&gt;&lt;br/&gt; But always remember, in the first place, stacks are amazing, but not a maze - &lt;em&gt;we get more memory as we grow up, it will keep &quot;stacking&quot; until the end&lt;/em&gt; &lt;br/&gt;&lt;br/&gt; I'm in the beginning, as well as the end - &lt;em&gt;we have memory since we born, until we die&lt;/em&gt;&lt;br/&gt; And always traceable, no matter what I pretend - &lt;em&gt;we can remember something when we saw/touch/smell specific item or hear specific word. Even when you have amnesia&lt;/em&gt;&lt;br/&gt;&lt;br/&gt;I really am curious, about who will find, the animate being, that hides behind - &lt;em&gt;we can rewind what we have done, and find something back&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;another answer (for fun):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a troll/people who make fun of you whatever you did, they are in everywhere, you can trace them, but barely can do things to them&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5453" LastActivityDate="2014-11-14T15:28:56.797" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4664" PostTypeId="2" ParentId="4588" CreationDate="2014-11-14T15:32:32.750" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The problem here is not mounting the candle, or stopping the wax. It's just which way &quot;down&quot; is.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Leave the room, exit the building and dig a hole under one side of the building until it topples over onto the wall that has the cork board on (please ensure this is done in line with all relevant Health &amp;amp; Safety legislation). Re-enter the building, ensure that the table and cork board are approximately where they were before (in relation to the original floor, wall etc.), stand the candle on the cork board, and light it with the matches.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5327" LastActivityDate="2014-11-14T15:32:32.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4665" PostTypeId="2" ParentId="4657" CreationDate="2014-11-14T15:35:37.813" Score="1" Body="&lt;p&gt;Is it...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Sun? Animals wake when the Sun rises. The Sun also sets.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4685" LastActivityDate="2014-11-14T15:35:37.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4666" PostTypeId="2" ParentId="4615" CreationDate="2014-11-14T15:53:12.223" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; i am the beginning, i am the end... makes me think of a circle&lt;br&gt;&#xA; stacks are amazing... maybe i am taking this too literally&lt;br&gt;&#xA;&lt;br&gt;&#xA; could it be... a circular queue?&lt;br&gt;&#xA; &lt;a href=&quot;http://en.wikipedia.org/wiki/Circular_buffer&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Circular_buffer&lt;/a&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; they sure are queer... yet, always traceable...  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5455" LastActivityDate="2014-11-14T15:53:12.223" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4667" PostTypeId="2" ParentId="4615" CreationDate="2014-11-14T15:53:16.767" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pokemon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pokemon are conventionally odd entities (Queer)&#xA;Whilst in the pokeball they see nothing (Are you still here?)&#xA;Gotta catch em all (The chase)&#xA;Ash' Pikachu was with him from the beginning and always will be his pokemon (In the first place)&#xA;Pokemon evolve from one stage to another so the initial one is still the end one, but evolved (I'm in the beginning as well as the end, and always traceable)&#xA;Pokemon are curious creatures (I really am curious)&#xA;You have to hunt to find them in long grass or elsewhere etc. (About who will find, the animate being, that hides behind)&#xA;Pokemon are animate beings&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*You said no wild guesses, but they can be tamed!&lt;/p&gt;&#xA;" OwnerUserId="5452" LastActivityDate="2014-11-14T15:53:16.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4668" PostTypeId="1" AcceptedAnswerId="5263" CreationDate="2014-11-14T15:53:27.873" Score="14" ViewCount="1049" Body="&lt;p&gt;Thus is a multi-part party installment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A famous host is storing his party recipes on a data server, and rumor has it that the passwords are in his house. He has another party today without his famous recipes, but it gives you a chance to steal the password. You and your friend start spying on the guests. Both of you need to go in as you need to bypass security, (part 2).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Guard: 1 Guest: 13&#xA;Guard: 3 Guest: 6&#xA;Guard: 125 Guest: 26&#xA;Guard: 31 Guest: 19&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;At this point your friend is spotted and is called on.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Guard: 9 Friend: 12&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Surprisingly he gets in in. However, he still needs you to get the passwords. The entrance is about to close so you come out. Your number is 12, what should you reply?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: No codes or ciphers are used. Not dependent on day, or anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4927/morse-code-cryptogram/4929#4929&quot;&gt;Morse code cryptogram&lt;/a&gt; now solved!&lt;br&gt;&#xA;Hint: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/5208/hard-text-cipher/5218#5218&quot;&gt;Vigenere text cipher&lt;/a&gt; now solved!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Convert to something computer-based&lt;br&gt;&#xA; All guests have an electronic device with no calculator...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bonus Guests:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You pretend to go to the nearby restroom. You hear these responses:&lt;br/&gt;&#xA;Guard: 90 Guest: 24&lt;br/&gt;&#xA;Guard: 91 Guest: 25&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5368" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2016-08-01T12:59:47.247" Title="The Security to the Party [Part 30] [#1]" Tags="|password|" AnswerCount="4" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4669" PostTypeId="2" ParentId="4657" CreationDate="2014-11-14T15:53:32.663" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The beast is a [duplicate] question.&lt;br&gt;&lt;br&gt;Most duplicates are popular, classic puzzles. Puzzlers post them because they're fundamentally just good puzzles people want to share with the world. And initially, since most readers don't realize they're duplicates, they're met with great enthusiasm: many answers (&quot;a wild cacophony of sound&quot;) and upvotes (&quot;applause, a jovial shout should be the least&quot;).&lt;br&gt;&lt;br&gt;But inevitably, after some quick checks (&quot;fleeting glances&quot;), the comments start to appear: &quot;&lt;em&gt;Possible duplicate of...&lt;/em&gt;&quot;, &quot;&lt;em&gt;This has already been posted at...&lt;/em&gt;&quot; and jubilation turns to resentment (&quot;nostrils flared, each face curled slightly in a frown&quot;).&lt;br&gt;&lt;br&gt;A duplicate question. It must be destroyed and the poster punished. This is the StackExchange code that we, &quot;all who share the journey&quot;, all know well.&lt;br&gt;&lt;br&gt;And so with a heavy heart, we dutifully slay the beast, downvoting it (&quot;the beast is also going down&quot;) until, after one last whimpering cry, nothing remains.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2014-11-14T15:53:32.663" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4670" PostTypeId="2" ParentId="4565" CreationDate="2014-11-14T15:56:28.103" Score="5" Body="&lt;p&gt;Inductive definition of the nth guest's answer $A_n$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A_1$ is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the sum of the digits of the binary representation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;of the number given by the guard. $A_n$ is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the sum of the digits of the base-$A_{n-1}$ representation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;of the number given by the guard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you should say&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 28, since the base-35 representation of 13832 is BA7 and 11+10+7=28.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2014-11-16T11:48:22.853" LastActivityDate="2014-11-16T11:48:22.853" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="4671" PostTypeId="2" ParentId="4615" CreationDate="2014-11-14T16:04:59.667" Score="0" Body="&lt;p&gt;I'm new here so my spoiler tag could be a bit messy, forgive me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could the answer be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fossils/bones &lt;br&gt; &#xA; - It is queer to see fossils (bones) in the ground or in rocks where they could be considered out of place&lt;br&gt;&#xA;&lt;br&gt;&#xA; - Paleontologists chase after fossils&lt;br&gt;&#xA;&lt;br&gt;&#xA; - Stacks are geological formations which can be made of limestone, sometimes containing fossils, i.e the jurassic coast. they are pretty amazing to look at but they are not a maze&#xA;&lt;br&gt;&#xA; - They are there at the beginning , when a vertebrate is born and they remain at the end after the flesh has decomposed&lt;br&gt;&#xA;&lt;br&gt;&#xA; - They are tracable through carbon dating and genetic testing, i suppose, no matter how they are hidden &lt;br&gt;&#xA;&lt;br&gt;&#xA; - You can trace bones to the animal that they once supported&lt;br&gt;&#xA;&lt;br&gt;&#xA; - I also assume that no two bones are the same&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It could also be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; cloth&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5442" LastEditorUserId="5442" LastEditDate="2014-11-14T16:10:55.340" LastActivityDate="2014-11-14T16:10:55.340" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4672" PostTypeId="2" ParentId="4660" CreationDate="2014-11-14T16:12:34.050" Score="37" Body="&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.worldstandards.eu/electricity/plugs-and-sockets/&quot;&gt;Electrical plugs&lt;/a&gt;. Sorry, I don't know enough about others to contribute anything additional.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation as requested:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;&lt;code&gt;&quot;There are many different types of men;&quot;&lt;/code&gt;&lt;/em&gt;&lt;br&gt;&#xA; &quot;Men&quot; referring to the &quot;male&quot; pins of an electrical plug.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;em&gt;&lt;code&gt;&quot;They often come in groups of two or three.&quot;&lt;/code&gt;&lt;/em&gt;&lt;br&gt;&#xA; Most plugs have two or three pins (usually a &quot;hot&quot; and &quot;neutral&quot;, and sometimes a separate ground pin).&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;em&gt;&lt;code&gt;&quot;Each group has a matching group of women,&quot;&lt;/code&gt;&lt;/em&gt;&lt;br&gt;&#xA; &quot;Women&quot; referring to the &quot;female&quot; parts of the receptacle.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;em&gt;&lt;code&gt;&quot;But they should stay with their compatriots.&quot;&lt;/code&gt;&lt;/em&gt;&lt;br&gt;&#xA; Keep plugs with their intended receptacles or risk electrical damage/fire.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;em&gt;&lt;code&gt;&quot;In Europe we have several different kinds;&quot;&lt;/code&gt;&lt;/em&gt;&lt;br&gt;&#xA; There's no single European standard on plugs.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;em&gt;&lt;code&gt;&quot;The Americans are only half as good as us.&quot;&lt;/code&gt;&lt;/em&gt;&lt;br&gt;&#xA; Could refer to the relatively thin plug blades in North American plugs, but more likely to the difference in voltage of household electrical service (120V vs. 230V).&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;em&gt;&lt;code&gt;&quot;Those from Down Under are all thin and weedy,&lt;/code&gt;&lt;/em&gt;&lt;br&gt;&#xA; &lt;em&gt;&lt;code&gt;Not as tough as the solid British men.&quot;&lt;/code&gt;&lt;/em&gt;&lt;br&gt;&#xA; I'm not familiar with Australian plugs, but presumably the blades are thinner than those used in the UK (which are certainly stout).&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;em&gt;&lt;code&gt;&quot;The French can have a woman in their group;&quot;&lt;/code&gt;&lt;/em&gt;&lt;br&gt;&#xA; Looking at plug Type E in the link above, commonly used in France, there is actually a &quot;female&quot; bit in the plug that accepts a pin from the receptacle.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;em&gt;&lt;code&gt;&quot;The Israelis won't fit into any other land;&quot;&lt;/code&gt;&lt;/em&gt;&lt;br&gt;&#xA; Type H is &quot;used exclusively in Israel, the West Bank &amp;amp; the Gaza Strip.&quot; (same link)&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;em&gt;&lt;code&gt;&quot;But if you need translation services,&lt;/code&gt;&lt;/em&gt;&lt;br&gt;&#xA; &lt;em&gt;&lt;code&gt;I can provide you with a magic cube.&quot;&lt;/code&gt;&lt;/em&gt;&lt;br&gt;&#xA; Adapters exist that allow you to plug a device into a physically-incompatible receptacle. &lt;a href=&quot;http://www.monoprice.com/Product?p_id=9876&quot;&gt;Some of them are vaguely cube-shaped.&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="5456" LastEditDate="2014-11-14T21:53:01.677" LastActivityDate="2014-11-14T21:53:01.677" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4676" PostTypeId="2" ParentId="4615" CreationDate="2014-11-14T16:40:20.243" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the shadow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is out of place of where the original object it reflects, stacks of things don't affect it, it is a simple hard object, it appears when the light comes on, all the way till it dies down, and it does not matter what it looks like, you can always see trace it... The animate that hides behind? could be symbolic to: the object that it reflects&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4657" LastActivityDate="2014-11-14T16:40:20.243" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4679" PostTypeId="2" ParentId="4572" CreationDate="2014-11-14T17:14:53.270" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Dilemma&lt;br&gt;&#xA; its a decision that needs to be made which conjures all possible routes taken&lt;br&gt;&#xA; colours would be the emotions the decision in each route will cause&lt;br&gt;&#xA; em reverse is me which is the middle whilst the neighbours DIL MA, well ma is 2/3 of man&lt;br&gt;&#xA; The decision made can be uncomfortable&lt;br&gt;&#xA; choking on the decision, its good the chew the fat?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5135" LastActivityDate="2014-11-14T17:14:53.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4680" PostTypeId="2" ParentId="4503" CreationDate="2014-11-14T17:16:55.990" Score="1" Body="&lt;p&gt;Since it's been answered well already, just a clarification. The two long words I had in mind were a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; boolean variable&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is sometimes called a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; truth value&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;whence the second couplet about disagrement. The one short word, not quite synonymous, would of course be a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; bit&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thank you one and all.&lt;/p&gt;&#xA;" OwnerUserId="5313" LastActivityDate="2014-11-14T17:16:55.990" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4681" PostTypeId="2" ParentId="4405" CreationDate="2014-11-14T17:17:30.440" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Have the mute man write a well-thought out letter to the blind man's wife explaining the folly of her ways and encouraging her to confess and apologise in order to achieve atonement and regain her honour&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would put this &quot;solution&quot; in a comment but I don't have enough kudos or whatever you call them&lt;/p&gt;&#xA;" OwnerUserId="5441" LastActivityDate="2014-11-14T17:17:30.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4683" PostTypeId="2" ParentId="4642" CreationDate="2014-11-14T18:03:56.653" Score="42" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;MAGICSIBLINGS &lt;br&gt; It's morse code, with every circular symbol representing a dot and every other symbol representing a dash. The content of the symbols is a red herring.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5352" LastActivityDate="2014-11-14T18:03:56.653" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4684" PostTypeId="2" ParentId="4211" CreationDate="2014-11-14T18:31:10.890" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;finger&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Everyone more or less needs them and can use them many times in different ways. You sometimes only use a part, such as the fingertip or fingerprint, sometimes use most of it. You can bend your finger at will. You can use to scare, confuse and question with gesture, i.e. the Middle Finger, air quotes etc, and confront people by pointing at them and you absolutely use them everyday. Seems to fit every part to me.&lt;/p&gt;&#xA;" OwnerUserId="5180" LastEditorUserId="4" LastEditDate="2014-11-19T15:41:06.323" LastActivityDate="2014-11-19T15:41:06.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4686" PostTypeId="2" ParentId="4405" CreationDate="2014-11-14T19:01:36.047" Score="0" Body="&lt;p&gt;&quot;One was deaf, the other was blind and the one was mute.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The story goes like this, the deaf and the mute is the same person. He kissed the lady and now he feels sorry for the blind dude and wants to confess.So he writes that on a piece of paper and shows to this third guy who is neither deaf, blind or mute. And the third guy tells the blind dude. And the mute guy gets smacked on the face. :D&lt;/p&gt;&#xA;" OwnerUserId="5282" LastActivityDate="2014-11-14T19:01:36.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4687" PostTypeId="1" AcceptedAnswerId="4689" CreationDate="2014-11-14T19:27:34.533" Score="22" ViewCount="7650" Body="&lt;p&gt;Two of the officers arrest Detective Randy White; one was holding him, while the other was putting handcuffs on Mr. White. You see that the whole Investigation department talking, some whispering to each other. You can't quite make out what they are saying, but in this kind of situation you just kinda know what is going on. You look down to see all 7 victims lying on the floor, covered up with blankets. As you start to kneel down to get a closer look at one of the bodies, all of sudden the body got up! &quot;Why didn't you help us!&quot; the body screamed, &quot;Why didn't you catch him sooner?! I wouldn't have died!&quot; said another victim as that body rises. One by one, every victim's body got up, asking you why you couldn't save them, telling you that this could have been prevented. Everything goes black, as the victims got closer and closer to you, you try to get away, but you are surrounded. You start panicking, constantly trying to look for a solution to get away from them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the victims got within arm's reach, your eyes open wide! You quickly got up, breathing heavily through your mouth, sweat drops down from your forehead. All you see is that you are in your bed, tv and dresser in front of you, lights are off. After a couple of seconds, you realize that you were just having a nightmare. You look at your alarm clock, &quot;3:45 AM, March 3rd, 2014&quot; glowing bright red. You still can't get over the fact that you have let 7 people die, 7 families torn apart because you couldn't figure out who the murderer was sooner. &quot;Voted number one detective huh?&quot; you muttered to yourself. You know that you won't be able to fall back asleep that easily, so you decide to go for a walk instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before you head out the front door, you are feeling thirsty, so you walk over to the fridge. You see on the top shelf that there is a jug of water, and a carton of milk. &quot;Hmm, what should I have?&quot; you asked yourself. &lt;strong&gt;(Choose One)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Water:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You take the jug and pour yourself a glass of cold water, you gulped the entire glass down in seconds. &quot;That was refreshing, no other beverage can beat it&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Milk:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You take the carton and pour yourself a glass of cold milk, you gulped the entire glass down in seconds. &quot;Who knew something so good can come from a cow&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You head out the front door to go for your walk, as you open the door, a blast of chilly wind hits you. &quot;Brrr, it sure is cold. I better go put on a jacket&quot; as you re-enter the house. You walk down a couple of blocks, you noticed that the streets are much empty, streetlights are on, and the moon shining bright. As you continued walking you, you hear a woman scream. You started running towards the direction of the scream, you noticed that in one of the alleyway, there is someone on the floor, not moving. You slowly approached the person on the floor, while getting ready to right back if someone suddenly appears. When you reached the body, you knee down to get a closer look. You see a young woman, between the ages of 19-23, blonde, around 5' 7&quot;, wearing jeans, brown boots, fancy dark blue buttoned shirt. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You look around the small alleyway, the place is dirty, and filled with trash. In the corner,  you noticed that there is a rat chewing away at some garbage. You see that the woman was stabbed 4 times in the chest area, you put your hand on her body, the body is still warm. Just when you were about to call for help, you hear sirens. All of a sudden, you see bright lights shining towards you in the alleyway, lighting up every dark corner. several police officers got out of the car, with guns pointed right at you. &quot;Put your hands up!&quot; shouted one of the officers, you then remembered that you did not bring your ID badge with you nor any identification. You tried telling them who you are, but they didn't listen, and none of the officers look familiar to you. The officers handcuffed you and pushed you inside the back of the car, taking you to the police department and threw you behind bars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several hours passed, then you see one of the officers let you out. They apologized for not recognizing you, and explained that they are new recruits and have been placed on the night shift duty. You looked at the officer and said: &lt;strong&gt;(Choose One)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Passive:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Don't worry about it, it was an honest mistake. Can you possibly drive me back home so I can get my badge?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Aggressive:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;You better be sorry! I was in here for hours! Now take home so I can get my badge and continue my work.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Certainly, and again I apologize for my mistake. It won't happen again&quot; said the officer. You got your badge and head straight to the crime scene. As you arrive, you noticed that the crime scene was already swarming with police and detectives. You cross over the yellow police line, and went into the alleyway. You noticed there are 3 small yellow cones around, you walk over and examine the clues beside the yellow cones. &lt;strong&gt;(Choose Two)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clue 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Beside the cone, you noticed that there are only a few blood drops on the ground. You got one of the forensic scientist to come over and take samples back to the lab for analysis.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clue 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;At first you did not notice anything beside the cone, but then you saw a couple stands of hair. Most of the hairs were blonde color, while one of them is brown/brunette. You got the forensic scientists to come over and put the hairs in a bag.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clue 3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You noticed partial shoe prints on the ground made out of blood, from the partial shoe prints, you were able to tell that the first print was in the direction of the alleyway, while the second print was in the direction facing away from the alleyway. The prints were right on top of each other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As you were about to examine the last clue by the last yellow cone, a bunch of rats came out of the corner of the alleyway and started running around, eating garbage, and destroying the last piece of evidence. Both police officers and detectives ran out of the alleyway as quickly as possible, you knew that the last piece of evidence is now gone. You noticed that the alleyway was right beside an apartment building, so you decide to enter the building to ask if anyone had heard anything that night. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several hours passed, and You went to almost every unit in the building, no one has seemed to have seen or hear anything. The only clue you have gotten was from a residence on the 2nd floor, the old lady claimed to have hear something along the lines of &quot;I'm tired, I don't want to argue with you as well!&quot; that might have came from the floor above, but she said that her hearing isn't as good as it used to be. Just when you were about to leave, you noticed a young red haired woman about 5' 5&quot; enter through the front door, and enter unit #1. You stop her before she could close the door, &quot;Excuse me, do you live here?&quot; you asked, &quot;Yes I do, I live here with a roommate. We go to school together&quot; she replied. You describe the victim to the young woman, &quot;OMG, that's my roommate, Sarah. She hasn't came home since yesterday evening&quot;. You asked if you could interview her, she agreed and let you in her unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Me and her have been friends since like forever, we went to the same high school together and now the same college. We are like bff's&quot; she said, you proceeded to ask her what she was doing, and the last time she has seen Sarah. &quot;I had class in the morning and got home around noon time, I was at home since. Sarah came home around 4-ish, got ready and then left her house to go to her boyfriends place. I haven't seen her since she left&quot;. You asked about Sarah, &quot;She's very popular, liked by everyone, male and female. Smart, strong, funny, beautiful, great personality, fun to be around, she had everything going for her&quot;. You asked if you could look around,&quot;Please go right on ahead, but I can't believe she's dead now&quot; she said in a sad tone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you looked around each room. The living room had a 42&quot; tv, with two couches, and a coffee table in the middle. there was dirty dishes, mugs, and a lot of magazines on the coffee table. The kitchen had a stove, fridge, kitchen table, and sink. The counter had pots, pans, utensils, bowls, dishes, and food just lying around. You enter the bathroom, the bathroom is shared between the two so it was a complete mess. you see the usual tooth paste, tooth brush, hair dryer, and towel. You also see a lot of make up, brushes, hair ties, hair dyes, hair curler, perfume, clothes, underwear, accessories, and other items scattered everywhere. You enter one of the bedroom. &lt;strong&gt;(Choose One)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Roommate's Room:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You enter the roommate's room, you noticed that it is messy, you see a bed, table, dresser, and clothes everywhere. On the wall you can see many pictures of her friend Sarah, and family members. One of the family pictures is her and her parents outside in the winter time. She was wearing thick winter jacket with hat and hood. Mom was also wearing thick jacket, brunette, green eyes, very beautiful. The father is wearing just a sweater, brown hair, brown eyes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sarah's Room:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You enter Sarah's room, you see a bed, table, dresser, and clothes everywhere. She had many pictures taped to the wall, some pictures was of her and her boyfriend, Some pictures were of her and her family and some are her and her friends. You noticed that the boyfriend is tall, nice build, brown hair, blue eyes. the mom and dad both had blonde hair, dad taller than the mom, Sarah taller than the mom but shorter than the dad. In one of the pictures, it is Sarah in between two other friends, the two other friends were turned around, both shorter than Sarah. The one of the left had brown/brunette hair, while the one to the right was a bigger female with blonde hair. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You asked her where Sarah's boyfriend lived, she told you the address and you drove there. You drive about 30 to 45 mins until you arrive at the boyfriend's place, you ring the door bell. Someone opens the door, &quot;Can I help you?&quot; the boyfriend said. You tell him that you are a detective and about the news of his girlfriend, &quot;What?! I just saw her last night, how can that be?&quot; the surprised boyfriend said. He lets you enter his place, you ask him where he was and what he was doing the night of the murder. &quot;I had class in the morning till around 3-ish, i came home, and texted Sarah to come over to hang. After she arrived, we were watching tv and made dinner&quot;. &quot;What else, that can't be all that happened&quot; you said, &quot;Well, we got into an argument. It was a pretty bad one too&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You asked if you could look around the place, he nodded. You look around the living room, you see a 42&quot; tv, stereo, gaming console, leather couch, coffee table. The table had a few things on it, but it was certainly cleaner than the other place. You enter the washroom, you noticed a tooth brush, tooth paste, comb, hair dryer, gel, shaver, shampoo, soap, and towel. The kitchen had food and dirty dishes everywhere, probably from the dinner he and sarah had the other night. The kitchen had a stove, fridge, sink, dinner table, and a garbage can. On the counter table, you noticed that there is a knife holder, with a couple of knives missing. You enter his room, you see a bed, table, dresser, closet, clothes everywhere, a couple of adult magazines here and there. You then asked the boyfriend what happened after they had the fight, &quot;I don't know, we had the fight and then she left in anger. I assumed she went back home, I haven't talked to her or tried to contact her since the fight&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just when you were about to leave, the boyfriend turns around and said &quot;Try investigating someone named Simon G, he is Sarah's ex-boyfriend&quot;. You head back to the police station and asked them to help you find Simon G, &quot;Come back tomorrow, and we will have the results back for you&quot; said one of the officers. You head back home to get some rest, you start pondering about the case, thinking how weird this case is. As you were coming on to something, you fell asleep due to the lack of sleep you had the other night. In the morning you head over to the police department, one of the officers gave you the address of where Simon lives, you took no time getting there. When you arrive, you see a man exercising on the front lawn. He has no shirt on, shorts, shoes, boxing gloves, brown hair, muscular. You went up to the man asking if he is Simon, &quot;Yeah, that's me&quot; he said trying to act all tough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You tell him about the news of Sarah, &quot;Sarah's dead?! how? when?&quot; he said. You asked if you may enter his house, he hesitated at first, but eventually lets you in. As you enter, you notice that he has a huge trophy case with many trophies. You walk over to see that he has trophies and medals for football, rugby, hockey, lacrosse, boxing, muay thai, and many other sports. &quot;You like my trophy case? Got it custom made the way I like it, it was a hefty price&quot; he said with excitement and pride in his voice. You asked him where he was and what he was doing on the night of the murder, &quot;I was home alone, just working out, watching some tv, and all that. I didn't commit any murder if that is what you are thinking&quot; he replied. &quot;May I take a look around?&quot; you asked him, he paused for a couple of seconds, &quot;Sure, but don't go making a mess!&quot; he replied. &lt;strong&gt;(Choose Two)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bedroom:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You enter the bedroom, it is dark because he has the blinds closed completely. You see that the bed is a mess, there are pills scattered on the floor, protein powder on the night stand, empty crushed energy drinks on the floor, table, and bed. On the floor somewhat hiding under the bed, you see a ripped picture. you pick it up, it seems to be a picture of him and a woman, you can't make out who the woman is because it has been torn in half-ish. All you can tell is that the woman had blonde hair, and that she is shorter than Simon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Kitchen:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You see that the kitchen is clean, not messy, no dirty dishes, nothing out of place. This seemed suspicious to you, but you can't seem to find anything wrong from this kitchen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Washroom:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Washroom is a mess, and dirty. There are work out weights on the floor, energy drinks here and there, two full bags of trash beside a already full trash bin. Near the sink, you noticed that there are some brown strands of hair, but you didn't think much of it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Basement:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You head down to the basement, when you get to the bottom you see that the basement has been transformed into a work out room. you see cardio machines, weights, chin up bar, bench, stereo, and mats on the floor. There are empty energy drinks on the floor, a bottle of pills on the table, and protein bars on the bench.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Before you can enter into the next room, Simon stops you, and tells you that there is nothing in his house, and tells you to leave. You can see clearly tell that he will take no as an answer, so you leave the house. He closes the door, as you think to yourself how strange this man is. You head back to the forensics lab to see if they have any results back from the clues you picked up on the night of the murder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yes, the results have came back, you are actually just in time. The results came back not too long ago&quot; one of the forensic scientist said. &quot;Ok, so here are the results&quot;. &lt;strong&gt;(Choose The Same Two Clues You Chose Earlier At The Crime Scene)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clue 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;So the DNA has come back from the blood sample, the DNA seems to point towards Sarah. All the blood drops seems to be from her.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clue 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There were no roots attached to any of the hair strands, so we could not tell who the hair strand came from. All we know is that it is blonde and brown/brunette hair.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clue 3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;We did some analysis on the shoe prints. The blood that made both shoe prints are from Sarah, but there is something weird about the prints. What we found is that the print that was pointing towards the alleyway was dry, and the prints heading out was not as dry. The prints heading out was on top of the prints that were heading in to the alleyway.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You ponder about the clues, then it hit you. You know who the murderer is!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who is the murderer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Answer is posted under the Answer section.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="4645" LastEditorUserId="5373" LastEditDate="2016-12-11T20:10:03.397" LastActivityDate="2023-03-10T09:04:24.967" Title="Mysterious Murder Mystery 3" Tags="|story|mystery|situation|" AnswerCount="4" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4688" PostTypeId="2" ParentId="4615" CreationDate="2014-11-14T20:08:14.057" Score="28" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is &quot;cat.&quot;&#xA; &lt;/p&gt;&#xA; &lt;em&gt;For sure I am queer and out of place. Are you still here? Go start the chase&lt;/em&gt;&#xA; &lt;/p&gt;&#xA;This means that the answer is not here, on this page. After some looking around, I ended up on avigrail's user page, finding this:&#xA; &lt;/p&gt;&#xA; Well done! If you are looking for something&#xA; &lt;em&gt;out of place&lt;/em&gt;&#xA; you are on the right track here.&#xA; &lt;/p&gt;&#xA; Next up:&#xA; &lt;/p&gt;&#xA; &lt;em&gt;But always remember in the first place stacks are amazing but not a&#xA; maze&lt;/em&gt;&#xA; &lt;/p&gt;&#xA; This led me to look around on Stack Exchange, eventually ending up on avigrail's network profile mostly by chance. There I found:&#xA; Well done! Almost there.&#xA; &lt;/p&gt;&#xA; Next part of the riddle:&#xA; &lt;/p&gt;&#xA; &lt;em&gt;I'm in the beginning as well as the end And always traceable no matter&#xA; what I pretend&lt;/em&gt;&#xA; &lt;/p&gt;&#xA; I originally thought this referred to tags, but now I'm pretty sure it refers to edits. Reason being, the answer is &lt;a href=&quot;https://puzzling.stackexchange.com/posts/4247/revisions&quot;&gt;in an edit that avigrail made yesterday&lt;/a&gt;.&#xA; &lt;/p&gt;&#xA; &lt;em&gt;I really am curious about who will find the animate being that hides&#xA; behind&lt;/em&gt;&#xA; &lt;/p&gt;&#xA; Curiosity killed the cat, they say.&#xA;&lt;/p&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;FYI I included some other small things here (very small actually). Those were not to be the main hints.&#xA;&lt;/p&gt;&#xA;The second meaning for &quot;in the end&quot; is that this riddle of course includes the solution, too. Technically, this post also is &lt;em&gt;the end&lt;/em&gt; (for now :p). &#xA;&lt;/p&gt;&#xA;&lt;em&gt;The solution is here - No it is not&lt;/em&gt;. No further explanation. Same with &lt;em&gt;far-fetched&lt;/em&gt;.&#xA;&lt;/p&gt;&#xA;&lt;em&gt;You will &lt;strong&gt;find&lt;/strong&gt; only one of my kind&lt;/em&gt; - is pretty interesting because by &lt;strong&gt;finding&lt;/strong&gt; (CRTL + F) the word &quot;cat&quot; could be found only once on this page. By now, several answers to this questions have added some cats of course. Anyway you will most likey &lt;em&gt;find&lt;/em&gt; &quot;cat&quot; but no dog, cow, rabbit, etc. In other words the cat is the only kind of animal around.&#xA;&lt;/p&gt;&#xA;&lt;em&gt;And always traceable, no matter what I pretend&lt;/em&gt; also means, that the solution was hidden in a spoiler and therefore pretended to be invisible/untraceable. Hilarious side effect: the &lt;strong&gt;cat&lt;/strong&gt; hides until you move your &lt;strong&gt;mouse&lt;/strong&gt; close to it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5464" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.923" LastActivityDate="2014-11-15T14:52:15.350" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="4689" PostTypeId="2" ParentId="4687" CreationDate="2014-11-14T20:09:50.723" Score="10" Body="&lt;p&gt;&lt;a href=&quot;http://knowyourmeme.com/memes/the-wadsworth-constant&quot; rel=&quot;noreferrer&quot;&gt;Wadsworth Constant&lt;/a&gt; for this question applies, y'all.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I immediately suspect &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the roommate. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's my reasoning.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sarah's boyfriend corroborates where both he and Sarah's roommate were when they last saw the victim. (Sarah left her place around 4, around the same time boyfriend got back home and texted her to come over.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simon's apartment was a mess except for the kitchen, where one would keep the murder weapon... for some reason, in pristine condition, but full bags of trash could have indicated that he probably doesn't cook much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first, it would seem like Sarah's boyfriend would have a motive with the fight, but he wouldn't admit it directly to a detective investigating her murder without knowing it would arouse suspicion. One can tell from the objects found in his apartment what the fight could have entailed, but this is never brought up.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; However, I think there is something to be said about the roommate's hair color. Redhead. This doesn't make much sense; redheadedness is a recessive gene, and is unlikely when both your parents have brown hair. Furthermore, the stray hair at the scene is incredibly suspicious, along with the long hair in the washroom... which makes me think that the roommate and Simon are in cahoots. Also, there are no other footprints at the scene aside from the victim's, and in an odd configuration at that. Given the scene of the crime's proximity to the apartment building, she seems the most likely suspect... and probably under the motivation of Simon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2444" LastEditorUserId="42" LastEditDate="2014-11-19T18:31:43.520" LastActivityDate="2014-11-19T18:31:43.520" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4690" PostTypeId="2" ParentId="4619" CreationDate="2014-11-14T20:30:39.690" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A tree based sorting algorithm, that starts with 2 random picks.&lt;br&gt;The one start is the mess that needs to be sorted. He guides us to a sorted list, which is useful. However, every time you run it, it is constructed a little differently, due to the random factor. It only answers to yes/no questions in the sense of checks with bools (is greater than, ...) The only problem is that it could work with any sort of unsorted list.&lt;br&gt;&lt;br&gt;Except of course, it sorts only ints for example, so in that sense it has only &quot;one&quot; start (a list filled with ints)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Just read the previous answer. Looks like my answer isn't as original as I thought.&lt;/p&gt;&#xA;" OwnerUserId="5470" LastEditorUserId="1955" LastEditDate="2014-11-14T20:42:05.910" LastActivityDate="2014-11-14T20:42:05.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4691" PostTypeId="2" ParentId="4588" CreationDate="2014-11-14T20:39:29.057" Score="1" Body="&lt;p&gt;I would do this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Take the cork board off the wall. Put it on the floor underneath the table or just on the table. Put the candle on the cork board and light it. When leaving, line up the thumb tacks on the floor under the door to the room. Now wait and watch as your boss (who gave you this shit task in the first place) walks in and screams in agony as he steps on the tacks! :D &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Burn a hole in the wall big enough for the candle stick to fit in. Shove the candle in the hole so that it doesnt fall down. Now take the thumb tacks box attach it below the candle on to the cork board using the thumb tacks (similar to classical solution). Light the candle and watch it wilt.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5030" LastEditorUserId="5030" LastEditDate="2014-11-14T21:02:36.383" LastActivityDate="2014-11-14T21:02:36.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4692" PostTypeId="2" ParentId="4405" CreationDate="2014-11-14T20:41:56.527" Score="0" Body="&lt;p&gt;I doubt it's the intended answer, but here's my take:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The mute man sends an invitation to meet at night to both the deaf man and the blind man's wife (asking her to bring her husband) without telling them that the other invitees would be coming. When they arrive and are all in the same place, the mute man removes any lighting (flips a switch if modern day, blows out a candle, etc.), kisses the wife loudly (mutes can still suck air through their lips), slaps the deaf man, and leaves into the dark to let everything unravel between the three.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At the very least, I think it would make for a fun solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternate solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The mute sends a blackmail letter to the wife posing as the deaf man saying that he will tell her husband if she doesn't pay him some amount of money she couldn't possibly have and that she should meet him at night at a specific time/place. Then send a letter to the deaf man inviting him to the same time/place as the wife but saying that she's been thinking about their kiss and wants more. The place chosen as the meeting place should have a hidden location nearby where the mute can lead the blind man and let him listen to the conversation. In the confusion of the meeting between the deaf man and wife, they're very likely going to discuss the kiss and much more.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another one that I doubt was intended, but I think would be a fun solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit&lt;/em&gt;: I suppose the second answer might backfire/fail since the wife probably knows the deaf man is deaf and could/would use sign language.&lt;/p&gt;&#xA;" OwnerUserId="1663" LastEditorUserId="1663" LastEditDate="2014-11-14T21:18:32.043" LastActivityDate="2014-11-14T21:18:32.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4693" PostTypeId="2" ParentId="4657" CreationDate="2014-11-14T21:05:22.137" Score="6" Body="&lt;p&gt;Given the &quot;final couplet&quot;, it seems obvious that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &quot;beast&quot; is a fart.&lt;br&gt;&lt;br&gt;It arrives with a &quot;wild cacophony of sound&quot;, and for some reason &quot;you&quot;, the emitter, hoped it would be greeted with cheers, possibly of the &quot;Way to let 'er rip, bro!&quot; variety.&lt;br&gt;&lt;br&gt;As it happens, the gas turns out to be unnaturally rank, and adulation quickly turns to gasps, frowns (&quot;nostrils flared, each face curled slightly in a frown&quot;), and a panic to escape the toxic bloom. Your once formidable accomplishment is now &quot;going down&quot; in history: &quot;&lt;em&gt;Hey, do you guys remember the day that Peter ripped that absolutely nasty...&lt;/em&gt;&quot;&lt;br&gt;&lt;br&gt;&quot;So try to banish this disgrace, light a candle, wave a fan.&quot; Maybe you can diffuse it or blow it up or something. &lt;em&gt;Sweet mercy, what did you eat, Pete!? Oh f... Where's the Febreeze? I'm gonna pass out.&lt;/em&gt;&lt;br&gt;&lt;br&gt;&quot;Smear with incense, hide your face, or get out early, while you can.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2014-11-14T21:05:22.137" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4695" PostTypeId="2" ParentId="4405" CreationDate="2014-11-14T21:15:40.303" Score="0" Body="&lt;p&gt;Go to the wife and tell her (via letter, sign language whatever) that you have informed her husband and she has better talk to him about it.  Better yet tell her the husband over heard the mute man discussing it with them.  The wife then goes to 'explain' to her husband and reveals it all.&lt;/p&gt;&#xA;" OwnerUserId="2698" LastActivityDate="2014-11-14T21:15:40.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4698" PostTypeId="1" AcceptedAnswerId="4701" CreationDate="2014-11-14T21:27:38.340" Score="-3" ViewCount="637" Body="&lt;p&gt;A man was walking through a graveyard and saw the following words written on his father's tombstone:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;be&lt;/li&gt;&#xA;&lt;li&gt;I&lt;/li&gt;&#xA;&lt;li&gt;me&lt;/li&gt;&#xA;&lt;li&gt;was&lt;/li&gt;&#xA;&lt;li&gt;will&lt;/li&gt;&#xA;&lt;li&gt;you&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What was the father's epitaph? You will need to know the correct order of the words, and the number of times that each word occurs.&lt;/p&gt;&#xA;" OwnerUserId="5474" LastEditorUserId="2444" LastEditDate="2014-11-14T21:46:20.070" LastActivityDate="2015-07-19T00:00:43.597" Title="Riddle in an epitaph" Tags="|riddle|" AnswerCount="3" CommentCount="1" ClosedDate="2014-11-19T11:49:55.393" ContentLicense="CC BY-SA 3.0" />
  <row Id="4699" PostTypeId="2" ParentId="4698" CreationDate="2014-11-14T21:31:52.830" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I was me&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;you will be you&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It seems rather free form.  I don't know how to judge if it is the right answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that your Dad whoever he was was himself and lived his own life.  You (whoever you are reading this) should be who you are and live your own life as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, &quot;be yourself.&quot;&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorDisplayName="user20" LastEditDate="2015-07-19T00:00:43.597" LastActivityDate="2015-07-19T00:00:43.597" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4701" PostTypeId="2" ParentId="4698" CreationDate="2014-11-14T21:44:13.407" Score="3" Body="&lt;p&gt;This one seems to be a bit more on the open-ended side, but here's my go:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I was you&lt;br&gt;&#xA; You will be me&lt;br&gt;&lt;br&gt;&#xA;     Explanation:&lt;br&gt;I (dead guy) was you (the person reading it -- who is &lt;i&gt;alive&lt;/i&gt;)&lt;br&gt; ... That is to say, the person who was once alive is now dead.&lt;br&gt;&#xA;     You (person reading -- alive) will be me (dead -- eventually)&lt;br&gt;&#xA; ... Which is to say, the person who is currently alive will at some time in the future be dead.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user5040" LastActivityDate="2014-11-14T21:44:13.407" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4702" PostTypeId="2" ParentId="4687" CreationDate="2014-11-14T22:08:32.137" Score="2" Body="&lt;p&gt;I chose the clues #2 and #3 at the crime scene, Sarah's room, and the kitchen and the washroom at Simon's house. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sara's boyfriend&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Before the murder, Sara and her boyfriend had a fight. Though it may seem too obvious, I think that whatever this fight was about, caused the boyfriend to kill her. Knives were missing from his kitchen and the boyfriend has brown hair. Also, he suggested that the detective investigates Simon G. Also, he didn't try to call Sarah and apologize. Most guys would, even if they are right, call the girl or at least text her after a fight. I don't think it's Simon because, even though he seems like a jerk, it doesn't seem like he would have any reason to kill Sarah I think that the clues at his place are pointless. I don't really think we need any of the clues, and the boyfriend's house is the only place we don't get them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While everyone is thinking outside the box, I'm gonna think inside the box and take the obvious clues.&lt;/p&gt;&#xA;" OwnerUserId="1711" LastActivityDate="2014-11-14T22:08:32.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4703" PostTypeId="2" ParentId="4057" CreationDate="2014-11-14T22:11:00.560" Score="6" Body="&lt;p&gt;&amp;lt;3 &lt;a href=&quot;https://oeis.org/A011009&quot;&gt;OEIS&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Define x as the number the guard tells you. &lt;P&gt;&#xA; Add 80 to x. &lt;P&gt;&#xA; Obtain the xth significant digit of fourth root of 12. &lt;P&gt;&#xA; Therefore the number we say to the guard should be 1. &lt;P&gt;&#xA; The number 12 was chosen because you have 12 songs in the soundtrack, and [12] in the title. &lt;P&gt;&#xA; The fourth root is because of time and space - four dimensions. We need to normalize these 4 dimensions into a 1 dimensional number.&lt;P&gt;&#xA; The number 80 was added because it is the secret ingredient in the spaghetti recipe!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Obviously assuming all the guests are capable of doing math precise to 100 decimal places...&lt;/p&gt;&#xA;" OwnerUserId="4109" LastActivityDate="2014-11-14T22:11:00.560" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4704" PostTypeId="1" AcceptedAnswerId="4706" CreationDate="2014-11-14T22:15:38.963" Score="10" ViewCount="3253" Body="&lt;p&gt;I assure you it's completely possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I made this maze myself so no copyright stuff :D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You start at the bottom right corner.&lt;/li&gt;&#xA;&lt;li&gt;You end at the top left corner.&lt;/li&gt;&#xA;&lt;li&gt;No going around the maze, over the maze, or under the maze.&lt;/li&gt;&#xA;&lt;li&gt;No walking through walls.&lt;/li&gt;&#xA;&lt;li&gt;No teleportation.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Maze:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/D0wbc.jpg&quot; alt=&quot;Maze&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck!&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-11-14T22:24:38.073" LastActivityDate="2016-05-17T20:18:04.553" Title="Seemingly Impossible Maze" Tags="|lateral-thinking|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4705" PostTypeId="1" AcceptedAnswerId="4714" CreationDate="2014-11-14T22:23:00.527" Score="13" ViewCount="988" Body="&lt;p&gt;The following is a puzzle that I was given back in the early 1970’s.  Everything that follows this introductory paragraph is part of the puzzle I was given.  That includes all of the paragraphs below statement 14.  The paragraphs below are kind of dated.  I changed a couple of letters in line 14 because I think some measurements might now be more precise than whenever this puzzle was created. I hope I didn't make any typos.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A.B.C.D.E.F.G.H.I.J.K.L.M.N.O.P.Q.R.S.T.U.V.W.X.Y.Z.&lt;/li&gt;&#xA;&lt;li&gt;AA,B;AAA,C;AAAA,D;AAAAA,E;AAAAAA,F;AAAAAAA,G;AAAAAAAA,H; AAAAAAAAA,I;AAAAAAAAAA,J.&lt;/li&gt;&#xA;&lt;li&gt;AKALB;AKAKALC;AKAKAKALD.AKALB;BKALC;CKALD;DKALE. &#xA;BKELG;GLEKB.FKDLJ;JLFKD.&lt;/li&gt;&#xA;&lt;li&gt;CMALB;DMALC;IMGLB.&lt;/li&gt;&#xA;&lt;li&gt;CMNLC;NKNLN.DMDLN;EMELN.&lt;/li&gt;&#xA;&lt;li&gt;JLAN;JKALAA;JKBLAB;AAKALAB.JKJLBN;JKJKJLCN.FNKGLFG.&lt;/li&gt;&#xA;&lt;li&gt;BPCLF;EPBLJ;FPJLFN.&lt;/li&gt;&#xA;&lt;li&gt;FQBLC;JQBLE;FNQFLJ.&lt;/li&gt;&#xA;&lt;li&gt;CRBLI;BRELCB&lt;/li&gt;&#xA;&lt;li&gt;JPJLJRBLSLANN;JPJPJLJRCLTLANNN.JPSLT;JPTLJRD.&lt;/li&gt;&#xA;&lt;li&gt;AQJLU;UQJLAQSLV.&lt;/li&gt;&#xA;&lt;li&gt;ULWA;UPBLWB;AWDMALWDLDPU.VLWNA;VPCLWNC.&#xA;VQJLWNNA;VQSLWNNNA.JPEWFGHLEFWGH;SPEWFGHLEFGWH.&lt;/li&gt;&#xA;&lt;li&gt;GIWIHYHN;TKCYT.ZYCWADAF.&lt;/li&gt;&#xA;&lt;li&gt;DPZPWNHFGRCQC.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Much speculation has recently been devoted to the possibility of intelligence elsewhere in the universe.  So far no evidence exists to indicate that there is intelligent life other than us.  If intelligent life does exist, the problem arises of how to contact it so that the culture receiving its message will acknowledge it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, consider Earth for a moment.  If we were to send a message in some language what language would we choose?  Since there are some several thousand languages and dialects spoken it is easy to see the problem involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above message comes from somewhere within our solar system.  Its creator felt it had circumvented the above problem by assuming it had chosen a language that could be universally understood within our solar system.  See if you can solve this code.  If you can you must then say where the message came from, which indicated you have totally solved the puzzle.  Here are some rules to follow:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numbers 1. through 14. merely indicate a statement of lettered symbols.  They are not part of the code.  You must proceed in the numbered sequence or the code will be impossible to solve.  A comma indicates a slight pause, a semicolon a greater pause, and a period the end of a statement.  This message can only come from one place in the solar system and statement 14. represents it.&lt;/p&gt;&#xA;" OwnerUserId="5478" LastActivityDate="2014-11-15T00:48:28.033" Title="Solve an alien message" Tags="|encoded-message|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4706" PostTypeId="2" ParentId="4704" CreationDate="2014-11-14T22:23:09.220" Score="29" Body="&lt;p&gt;If you look at it carefully enough, it's the easiest maze ever.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Normally, when one does a maze, one goes between the lines.&#xA; You didn't specify this as a rule. If you begin on the black line in the lower right hand corner, and follow it, unbroken, to the upper left... it's a breeze.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2444" LastActivityDate="2014-11-14T22:23:09.220" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="4707" PostTypeId="2" ParentId="4704" CreationDate="2014-11-14T22:29:45.080" Score="24" Body="&lt;p&gt;Not sure if this is what you are looking for...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This picture actually has a white border (roughly 8px wide), with the edges defined by four points.  Assuming this is part of the maze, simply follow either the left or upward paths, following one of these paths until the destination is reached.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5479" LastActivityDate="2014-11-14T22:29:45.080" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4708" PostTypeId="2" ParentId="4704" CreationDate="2014-11-14T22:30:48.850" Score="23" Body="&lt;p&gt;I entered the maze, intrigued by the difficulty caused by the fact that the exit was blocked off by a wall. Attached to my belt, I had one less powerful grenade and two more powerful grenades. I was glad to find out that the walls were made of a less sturdy material, meaning that a grenade could easily knock down the wall. I tested the weak grenade on an unnecessary wall, and to my delight, it worked! I proceeded by using a larger grenade on the wall that was blocking the exit. Not to my delight, there was a police officer guarding the exit! I used my last grenade and ran towards the exit. I made it out of the maze successfully.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WZZYA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5477" LastActivityDate="2014-11-14T22:30:48.850" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="4710" PostTypeId="2" ParentId="4603" CreationDate="2014-11-14T22:37:33.803" Score="5" Body="&lt;p&gt;In the first example,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the &quot;four-legged then the answer&quot; pattern seems to have the guesser and the sender's positions reversed from the original puzzle; is it intentionally different?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I notice that in the second example,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Kim answers &quot;no&quot; to Bob's initial question, then responds with a three-word phrase (&quot;Yes he is&quot;), a four-word phrase (&quot;He already did that&quot;), and a no-word phrase (sigh), and the answer turns out to be 34. It could be that the first question is to establish whether or not Bill is thinking of a color, and the number is determined by number of words in each sentence, with &lt;em&gt;sigh&lt;/em&gt; terminating the number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, this seems to fall apart in the third example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Kim answers &quot;yes&quot; to Bob's question about whether she's ready to begin, maybe meaning that yes it is a number again (this seems like a weak guess to me though). Kim then replies with a one-word phrase (&quot;No&quot;), a two-word phrase (&quot;I am&quot;), a two-word phrase (&quot;Sorry, Bill&quot;), a four(?) word phrase (&quot;Give him a...No?&quot;) and a two-word phrase (&quot;No, Bob.&quot;) This gets 12242 rather than 1742 though.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't know about the fourth one, but in the fifth (update) example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Bob seems to be muttering &quot;african elephant&quot; and Bill mishears this as &quot;frickin hell&quot;, but I don't know if that's a clue or just part of the story that wasn't obvious to me on first reading.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="414" LastActivityDate="2014-11-14T22:37:33.803" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4711" PostTypeId="1" AcceptedAnswerId="4712" CreationDate="2014-11-14T22:40:16.817" Score="4" ViewCount="942" Body="&lt;p&gt;There are 2 rooms left and it's lighter than the other 2. You enter the 3rd room. You see a door with a key lock and you see a cartridge and a TV with a Nintendo Entertainment System. You look at the cartridge and it says play me. You put the cartridge in the NES and you get this message: &quot;So you have finished 2 rooms, there is still 1 room left after this one. Press start to get your next message.&quot; You take the NES controller and press start. You see a message that says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;131 157 165 040 150 141 166 145 040 155 141 144 145 040 151 164 040 164 150 151 163 040 146 141 162 056 040 147 157 157 144 040 152 157 142 056 040 124 150 145 040 153 145 171 040 151 163 040 042 111 156 040 164 150 145 040 154 151 147 150 164 054 040 142 165 164 040 157 165 164 163 151 144 145 040 164 150 145 040 154 151 147 150 164 056 042, 55 73 65 20 79 6f 75 72 20 6b 6e 6f 77 6c 65 64 67 65 2e, 01010100 01101000 01100101 00100000 01101110 01100101 01111000 01110100 00100000 01110000 01100001 01110010 01110100 00100000 01110111 01101111 01101110 01110100 00100000 01100010 01100101 00100000 01110011 01101111 00100000 01100101 01100001 01110011 01111001 00101110&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where is the key and what does this message mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Previous parts:&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/4121/the-great-binary-puzzle-part-1&quot;&gt;Part 1&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/4126/the-great-binary-puzzle-part-2&quot;&gt;Part 2&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="5061" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-07-22T05:35:21.180" Title="The Great Binary Puzzle - Part 3" Tags="|cipher|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4712" PostTypeId="2" ParentId="4711" CreationDate="2014-11-14T23:10:08.087" Score="4" Body="&lt;p&gt;The first section is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Octal, it translates to: You have made it this far, good job. The key is &quot;In the light, but outside the light.&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The second section is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; hex, it translates to: Use your knowledge, &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The last section is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; binary, it translates to: the next part won't be so easy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Altogether that's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You have made it this far, good job. The key is &quot;In the light, but outside the light.&quot; Use your knowledge, the next part won't be so easy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2015-07-22T05:35:21.180" LastActivityDate="2015-07-22T05:35:21.180" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4713" PostTypeId="1" AcceptedAnswerId="4716" CreationDate="2014-11-14T23:12:56.750" Score="1" ViewCount="356" Body="&lt;p&gt;Given the following properties, what am I?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I'm white.&lt;/li&gt;&#xA;&lt;li&gt;I can dance although I typically don't.&lt;/li&gt;&#xA;&lt;li&gt;I can be bound to a contract.&lt;/li&gt;&#xA;&lt;li&gt;I am rubbed frequently before I am seen and thrown away.&lt;/li&gt;&#xA;&lt;li&gt;I am a inanimate object.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4802" LastActivityDate="2014-11-15T03:37:41.083" Title="White Dancer Riddle" Tags="|riddle|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4714" PostTypeId="2" ParentId="4705" CreationDate="2014-11-14T23:16:15.697" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The lines define a base-10 number system and some operations. &lt;br&gt; Line 1 lays out the symbols which can be used. &lt;br&gt; Line 2 shows that B through J are two through ten times A, which is probably 1. &lt;br&gt; Line 3 introduces K and L, with 1K1L2, 1K1K1L3, 2K1L3, etc. K must be addition and L is equality. &lt;br&gt; Line 4 had 3M1=2, 4M1=3, 9M7=2. M is subtraction. &lt;br&gt; Line 5 states that 3-N=3, N+N=N, and 4-4=N. N must be 0. &lt;br&gt; Line 6 shows that the number system is in base ten, because J (ten) is written as 10. &lt;br&gt; Line 7 defines P as multiplication: 2*3=6, 5*2=10, 6*10=60. &lt;br&gt; Similarly, Line 8 makes Q be division. &lt;br&gt; Line 9 has 3R2=9 and 2R5=32. R is exponentiation. &lt;br&gt; Line 10 has S=100 and T=1000. &lt;br&gt; Line 11 has U=1/10 and V=1/100. &lt;br&gt; Line 12 introduces W as the decimal point, because U=0.1 and V=0.01. &lt;br&gt; Line 13 states that 79.98Y80, 1000+3Y1000, and ZY3.1416. Y must be approximately equal, and Z is pi. &lt;br&gt; Line 14 is 4*pi*0.0867^3/3. This is the formula for the volume of a sphere with a radius of 0.0867. The radius of Saturn is 0.0867 times the radius of the Sun, so the signal must be from Saturn.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5352" LastActivityDate="2014-11-14T23:16:15.697" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4716" PostTypeId="2" ParentId="4713" CreationDate="2014-11-15T00:02:12.560" Score="2" Body="&lt;p&gt;Is this ..&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Paper&lt;/strong&gt;?&lt;br/&gt;It is white, I can dance the &quot;paper dance&quot;, bills are written on it, I can use a rubber if I did something wrong with a pencil and I'm not alive, for sure!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1685" LastActivityDate="2014-11-15T00:02:12.560" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4718" PostTypeId="1" AcceptedAnswerId="5408" CreationDate="2014-11-15T00:58:32.803" Score="7" ViewCount="878" Body="&lt;p&gt;This is a cool cipher I determined. It involves lateral thinking, so try it out. The answer is a English plaintext, with the following being the cipher text... Try to decrypt it... or decipher it, you get what I mean.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ɖɸɔɱʂβʉθ ɸʀ t͡ʃʃɱʍʝɾɦ χɪf ŋɱʎɪə djtiis&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Non-IPA Ciphertext: DPOMSBUT PR TSMWJRH XIf NMYIE djtiis (Caps don't matter)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some symbols are IPA phonetic vowels, others are not.&lt;br&gt;&#xA; The date this was posted is important... 11/14/14.&lt;br&gt;&#xA; Date shift cipher...&lt;br&gt;&#xA; Shift right...&lt;br&gt;&#xA; Remember a date shift cipher shifts the letters by a date. &#xA; Don't include slashes...&lt;br&gt;&#xA; The answer is a short compliment.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Small riddle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My letters are multi-meaning&lt;br&gt;&#xA;Frequencies are off the charts.&lt;br&gt;&#xA;Symbols are a nuisance&lt;br&gt;&#xA;Think outside the box...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you want is here,&lt;br&gt;&#xA;but you can achieve somewhere else...&lt;/p&gt;&#xA;" OwnerUserId="5368" LastEditorUserId="5368" LastEditDate="2014-11-21T14:03:45.003" LastActivityDate="2014-11-22T05:05:29.730" Title="Shifted symbol cipher" Tags="|lateral-thinking|cipher|" AnswerCount="3" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4719" PostTypeId="1" CreationDate="2014-11-15T01:48:35.833" Score="6" ViewCount="618" Body="&lt;p&gt;Inspired by &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4384/a-funny-game-you-can-play-with-a-friend-of-yours-brain-reading&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two friends, Andrew and Gabriel are approached one day by their friend Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;So, Ellen tells me you two can read each other's mind,&quot; says Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yeah that's right,&quot; says Andrew.  &quot;Would you like us to demonstrate it to you?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yes,&quot; says Harry, &quot;but I'm a bit more skeptical than Ellen, so I'm not going to let you give the questions.  Instead, you will whisper them to me and I will ask the questions. And just to make sure you aren't doing anything funny, I reserve the right to ask a different question than the one you ask me.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;But you might always change the question and never ask the thing we are really thinking about!&quot; exclaims Andrew.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;All right, I'll tell you what,&quot; says Harry.  &quot;When one of you is ready to ask if it is the thing you are thinking about, whisper &quot;5&quot; into my ear before the word, and I won't change it. Is that okay?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Hmm, I don't know,&quot; says Gabriel.  &quot;Let us talk this over first.&quot;  He whispers something to Andrew, Andrew shakes his head, and then whispers something back.  Then they both nod and agree to the terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Andrew steps away from Gabriel and Andrew gets blindfolded and waits. Then Gabriel whispers something in the ear of Harry. Harry then says: &quot;Ok, Gabriel is establishing the mental link, are you ready Andrew?&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yes I am&quot; replies Andrew.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Is it a rhinoceros?&quot; asks Gabriel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;No,&quot; replies Andrew.  Gabriel whispers something to Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Is it a Teletubby?&quot; asks Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;No,&quot; replies Andrew. Gabriel whispers something to Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Is it a rainbow?&quot; asks Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;No,&quot; replies Andrew. Gabriel whispers something to Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Is it the Statue of Liberty?&quot; asks Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;No,&quot; replies Andrew. Gabriel whispers something to Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Is it a pumpkin?&quot; asks Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;No,&quot; replies Andrew. Gabriel whispers something to Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Is it the Magna Carta?&quot; asks Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yes!&quot; replies Andrew.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Lucky guess,&quot; says Harry.  &quot;This time I'm going to tell Gabriel what it is you are guessing, and he will whisper the questions.   Just in case you decided before hand what you were going to think about.   Same as before, I might change any question beside the actual answer.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Harry whispers something into Gabriel's ear.  Gabriel whispers something back in Harry's ear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Is it a candle?&quot; asks Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;No,&quot; replies Andrew. Gabriel whispers something to Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Is it the color blue?&quot; asks Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;No,&quot; replies Andrew. Gabriel whispers something to Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Is it a Beautiful Sunshine of the Spotless Mind?&quot; asks Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;No,&quot; replies Andrew. Gabriel whispers something to Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Is it Bart Simpson?&quot; asks Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yes!&quot; replies Andrew.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Hmm,&quot; says Harry.  &quot;Let's try this one more time.&quot;  He again whispers something into Gabriel's ear. Gabriel whispers something back into Harry's ear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Is it chlorophyll?&quot; asks Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;No,&quot; replies Andrew. Gabriel whispers something to Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Is it a turnip in a field of donkeys?&quot; asks Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;No,&quot; replies Andrew. Gabriel whispers something to Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Is it an avalanche in the Swiss Alps?&quot; asks Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;No,&quot; replies Andrew. Gabriel whispers something to Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Is it a wigwam on the prairie?&quot; asks Harry?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yes!&quot; replies Andrew.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point Harry suspects that Andrew and Gabriel are just waiting a fixed number of &quot;no&quot; answers, so he lets do another round to see if he is right.  However, he does not let on as to his suspicions.  He whispers another answer to Gabriel, and then Gabriel whispers something to Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Is it a clown?&quot; asks Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;No,&quot; replies Andrew. Gabriel whispers something to Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Is it the North Pole?&quot; asks Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;No,&quot; replies Andrew. Gabriel whispers something to Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Is it Julie Andrews?&quot; asks Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;No,&quot; replies Andrew. Gabriel whispers something to Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Is it Pope Benedict the sixteenth?&quot; asks Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;No,&quot; replies Andrew. Gabriel whispers something to Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Is it a dead plant?&quot; asks Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;No,&quot; replies Andrew. Gabriel whispers something to Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Is it Superman?&quot; asks Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;No,&quot; replies Andrew. Gabriel whispers something to Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Is it... a lion? asks Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;No,&quot; replies Andrew. Gabriel whispers something to Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Is it a bathtub?&quot; asks Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;No,&quot; replies Andrew. Gabriel whispers something to Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Is it a school bus?&quot; asks Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;No,&quot; replies Andrew. Gabriel whispers something to Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Is it a coal mine?&quot; asks Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;No,&quot; replies Andrew. Gabriel whispers something to Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Is it a Glade plug-in air freshener?&quot; asks Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;No,&quot; replies Andrew. Gabriel whispers something to Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Is it a wombat?&quot; asks Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;No,&quot; replies Andrew. Gabriel whispers something to Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Is it a baseball?&quot; asks Harry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yes!&quot; replies Andrew.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Huh,&quot; says Harry.  &quot;I guess you really can read minds.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: In the last round, Harry did not whisper any of the questions Gabriel asked, except of course for the last one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is Harry wrong?  If you think you know the answer, create a question/answer session for the next thing Harry is going to think of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Andrew and Gabriel are using a similar (but less fixed) mechanism as part 1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2718" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-05-01T03:38:26.367" Title="Brain Reading, part 3" Tags="|pattern|magic|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4720" PostTypeId="1" AcceptedAnswerId="4723" CreationDate="2014-11-15T03:04:02.977" Score="2" ViewCount="430" Body="&lt;p&gt;There are 7 young men. All are excellent basketball players. They each started to 1v1 another player on their team.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They all played and nobody was left out. Everyone was present and none left the practice. Why was player #7 not left out? Who did he play with?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="46903" LastEditDate="2018-09-21T17:19:16.403" LastActivityDate="2018-09-23T01:07:39.807" Title="The 7 Young Men" Tags="|lateral-thinking|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4721" PostTypeId="2" ParentId="4720" CreationDate="2014-11-15T03:06:01.967" Score="0" Body="&lt;p&gt;You said &quot;they each started to 1vs1 another player on their team&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means Player 1 played one-on-one against Player 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then Player 2 played one-on-one against Player 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then Player 3 played one-on-one against Player 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then Player 4 played one-on-one against Player 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then Player 5 played one-on-one against Player 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then Player 6 played one-on-one against Player 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No one was left out.  Player 7 played against the other 6 players.&lt;/p&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2014-11-15T03:06:01.967" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4722" PostTypeId="1" CreationDate="2014-11-15T03:28:22.247" Score="2" ViewCount="1018" Body="&lt;p&gt;There are some puzzles that require sorting out many possible clues, such as &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4341/the-serial-killers-killer-difficult&quot;&gt;The SErial Killer&amp;#39;s killer&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4687/mysterious-murder-mystery-3&quot;&gt;Mysterious Murder Mystery 3&lt;/a&gt;. They are often mysteries. What is a good way to organize and look at the clues in these types of puzzles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Disclaimer: I am not using these tips to solve these puzzles and get reputation.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="4429" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2014-11-18T00:07:31.797" Title="Strategy for solving many-clue detective puzzles" Tags="|strategy|mystery|" AnswerCount="0" CommentCount="6" ClosedDate="2014-11-24T18:50:50.647" ContentLicense="CC BY-SA 3.0" />
  <row Id="4723" PostTypeId="2" ParentId="4720" CreationDate="2014-11-15T03:29:31.923" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;There was 7 young men. All excellent basketball players.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; So good in fact, that one of them could have played a 1 vs. 1 against another player who was already playing 1 vs. 1 against a different player.  But that is not what happened.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;They each started to 1vs1 another player on their team.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We will come back to this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;They all played and nobody was left out.&#xA;  Everyone was present and none left the practice. Why was player#7 not left out? Who did he play with?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; #7 played against #4.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; But in case of the odd man out, he played against #8.  See there were 7 young men, but this was not the entirety of the team, there was an eight man.  After all, in a game there are 5 players plus three in reserve.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2718" LastEditorUserId="42452" LastEditDate="2018-09-23T01:07:39.807" LastActivityDate="2018-09-23T01:07:39.807" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4725" PostTypeId="2" ParentId="4720" CreationDate="2014-11-15T03:50:11.433" Score="2" Body="&lt;p&gt;This answer works due to the ambiguity of the description of the scenario:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;There was 7 young men.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;They each started to 1vs1 another player on their &lt;strong&gt;team&lt;/strong&gt;.&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;The basketball team contained 8 people, but only 7 of the 8 were young.&lt;br /&gt;&#xA;Player 1 played against player 8 and etc.&lt;br /&gt;&#xA;Nobody was left out and nobody left, so this fits the criteria.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4802" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2018-09-21T17:09:38.877" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4728" PostTypeId="2" ParentId="4588" CreationDate="2014-11-15T06:43:00.543" Score="1" Body="&lt;p&gt;Pin the candle to the wall. Light it. If it drips is doesn't matter because the question never says that the candle isn't allowed to drip after it's been lit.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Your task is to attach the candle to a cork board on the wall and &lt;strong&gt;light it without the wax dripping&lt;/strong&gt; onto the table below.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As long as you don't drip any wax whilst lighting it, you're fine.&lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2014-11-15T06:43:00.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4730" PostTypeId="1" AcceptedAnswerId="4767" CreationDate="2014-11-15T08:08:51.863" Score="39" ViewCount="7063" Body="&lt;p&gt;You have been thrown into a dungeon with many rooms; you don't know who did this or why. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You start in room 37. You might be thinking &quot;Why 37?&quot;, so I will explain this: &quot;Why not?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to find the way out of this place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, there is a lot of data that are encrypted in your way. Some of the data are really simple and easy to decode like &quot;&lt;strong&gt;the first prime&lt;/strong&gt;&quot; or &quot;&lt;strong&gt;0x30&lt;/strong&gt;&quot;. Some of them are very hard, like &quot;&lt;strong&gt;☏☀☑☀☂☇☔☓☄&lt;/strong&gt;&quot;. All the data that should be decrypted by you in your journey are in &lt;strong&gt;bold&lt;/strong&gt; text.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of rooms have exits to other rooms, however you will need to discover to which rooms. Note that if there is a way from room A to room B this DO NOT imply that there is a way from room B to room A. I.E. you should consider every passage as one-way only. A two-way passage is explicitly represented as a pair or two one-way only passages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Post an answer explaining your movements step-by-step and room-by-room in order to successfully escape the dungeon. Do not forget to explain what items you collected on your journey and how you used them. Explain how you dealt with the trouble that you encountered. And of course, the most important: Explain how you decrypted all the &lt;strong&gt;boldtext&lt;/strong&gt; secret data! I will accept the first answer that do all of this correctly, without gaps, errors or cheating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beware: I expect that this puzzle to be really hard, as in fact it is a series of puzzles in one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And please, please. DO NOT FORGET TO POST YOUR ANSWERS IN SPOILER TAGS!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put your answers in spoiler tags, you should add &lt;code&gt;&amp;gt;!&lt;/code&gt; to the start of the lines inside your spoiler. To insert line breaks inside your spoiler tags, add a &lt;code&gt;&amp;lt;br&amp;gt;&lt;/code&gt; in the end of the line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Note: Feel free to edit this to fix possible spelling, grammar or punctuation errors, as long as it does not changes the &lt;strong&gt;bold&lt;/strong&gt; text parts, except when obvious in clear english.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;[Edited, long time after solved]: Made only somewhat spoilery rooms protected by spoilers. If you want to show them all, run this javascript: &lt;code&gt;$(&quot;.postcell blockquote&quot;).removeClass(&quot;spoiler&quot;);&lt;/code&gt;. If you want to hide them all, run this other one: &lt;code&gt;$(&quot;.postcell blockquote&quot;).addClass(&quot;spoiler&quot;);&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here are the rooms. Remember, you starts out in the 37:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Room 1:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That is the exit and the princess is already here just waiting for you!&lt;br&gt;&#xA; Sure that the evil wizard was defeated, you and the princess walks out of the dungeon, going back to the castle.&lt;br&gt;&#xA; And they lived happily ever after. ❤&lt;br&gt;&#xA; The end!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Room 2:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You found a &quot;&lt;strong&gt;☏☀☑☀☂☇☔☓☄&lt;/strong&gt;&quot;. Strange, why the heck you would need this? You get it anyway.&lt;br&gt;&#xA;  Now go to the room numbered as &lt;strong&gt;the cube of this room&lt;/strong&gt;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Room 4:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A giant plutonium dragon comes from nowhere, put you in his mouth, crush your bones with his diamond jaw and eats you.&lt;br&gt;&#xA; In his stomach, a boiling acid jet dissolves your body in a few seconds.&lt;br&gt;&#xA; Game Over.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Room 5:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is a man here. He said that he name is Caesar and he shares a secret to you. This secret is &quot;&lt;strong&gt;kwws=22elw1o|24vSg{I}&lt;/strong&gt;&quot;. What do this means?&lt;br&gt;&#xA;  There is a ventilation tube leading to the &lt;strong&gt;tri-hepta&lt;/strong&gt;.&lt;br&gt;&#xA;  There is a strange device, called &quot;&lt;strong&gt;sfuspqfmfu&lt;/strong&gt;&quot; but it does not works. If you manage to activate it somehow, it will lead you to room &lt;strong&gt;11110&lt;/strong&gt;.&lt;br&gt;&#xA;  There is window leading to room &quot;&lt;strong&gt;how many legs there is in a chair?&lt;/strong&gt;&quot;&lt;br&gt;&#xA;  There is a big door leading to room &quot;&lt;strong&gt;ruof-ytrihT&lt;/strong&gt;&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Room 8:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is a whirlpool in this room.&lt;br&gt;&#xA;  If you are unable to cope with it, you will be sucked through a hole direct to &lt;strong&gt;a room which the number divided by two is rounded to the number of this room&lt;/strong&gt;.&lt;br&gt;&#xA;  If you can somehow stop it, you may grab a hook and crawl into the &lt;strong&gt;perfect room&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Room 10:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is an arrow in this room, you decided to take it. There is something written in the arrow: &quot;&lt;strong&gt;这是一个红色的鲱鱼 - Made in China&lt;/strong&gt;&quot;.&lt;br&gt;&#xA;  There are three buttons: red, blue and yellow. You don't see any exit here, so what button do you press? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Room 12:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is a door that leads to the &lt;strong&gt;Z&lt;/strong&gt; room.&lt;br&gt;&#xA;  There are four pills here:&lt;br&gt;&#xA;  If you eat the red pill, you will be sent to the &quot;&lt;strong&gt;naM hsaM&lt;/strong&gt;&quot; room.&lt;br&gt;&#xA;  If you eat the blue pill, you will be sent to the &quot;&lt;strong&gt;ロコン&lt;/strong&gt;&quot; room.&lt;br&gt;&#xA;  If you eat the orange pill, you will be sent to the &quot;&lt;strong&gt;Pidgeotto&lt;/strong&gt;&quot; room.&lt;br&gt;&#xA;  If you eat the white pill, you will be sent to the &lt;strong&gt;21120110122101211011201112&lt;/strong&gt; room.&lt;br&gt;&#xA;  Warning: Pills do not respawn! So if you come back to this room later, the pills already taken won't be here again. I.E. you can't eat the same pill twice!&lt;br&gt;&#xA;  EDITED: If you take a pill, you need to eat it. You can't take all the pills at once and eat them somewhere else.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Room 13:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You found a &quot;&lt;strong&gt;D&lt;/strong&gt;&quot;. Now leave using &quot;&lt;strong&gt;the same door that you used to come in&lt;/strong&gt;&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Room 17:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The floor mysteriously disappears, leaving a dark bottomless pit where you instantaneously fall, forever!&lt;br&gt;&#xA; Game Over.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Room 18:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There are three doors:&lt;br&gt;&#xA;  The left door goes to the &lt;strong&gt;iron room&lt;/strong&gt;.&lt;br&gt;&#xA;  The front door to &lt;strong&gt;the room where you started&lt;/strong&gt;.&lt;br&gt;&#xA;  In the ceiling, there is a big green steel locked door. If you are able to unlock it, it will lead you to the &lt;strong&gt;first prime&lt;/strong&gt; room.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Room 21:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You found a &quot;&lt;strong&gt;Красный ключ&lt;/strong&gt;&quot;.&lt;br&gt;&#xA;  There are four tunnels here:&lt;br&gt;&#xA;  The left tunnel leads to room &quot;&lt;strong&gt;square of the first prime&lt;/strong&gt;&quot;.&lt;br&gt;&#xA;  The middle tunnel leads to the &lt;strong&gt;room with the pills&lt;/strong&gt;.&lt;br&gt;&#xA;  The right tunnel leads to the &lt;strong&gt;0x30&lt;/strong&gt; room.&lt;br&gt;&#xA;  The tunnel in the back leads to the &lt;strong&gt;room made by repeated numbers&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Room 23:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You found a magic bean!&lt;br&gt;&#xA;  There is a sign in this room: &quot;&lt;strong&gt;◓●◒◒◙ ◔◈◉◓ ◉◓ ◁ ◒◅◄ ◈◅◒◒◉◎◇&lt;/strong&gt;&quot;&lt;br&gt;&#xA;  Two exits. The left one leads to room &quot;&lt;strong&gt;@@@@@@@@&lt;/strong&gt;&quot;. The right to room &lt;strong&gt;XLVIII&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Room 24:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is a huge lion here. If you can't kill him quickly enough, you will become his dinner.&lt;br&gt;&#xA;  Behind the lion, should you be able to avoid him, there are two stairs:&lt;br&gt;&#xA;  The stairs going up leads to the room &quot;&lt;strong&gt;Monica, Erica, Rita, Tina, Sandra, Mary, Jessica&lt;/strong&gt;&quot;.&lt;br&gt;&#xA;  There is a massive red steel door which needs a red key. If you are able to open it, there is a stairs going down leading to the &lt;strong&gt;bad luck room&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Room 26:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You found a powerful gun. There is just one bullet there.&lt;br&gt; &#xA;  There is a window that leads to room &quot;&lt;strong&gt;Butterfree&lt;/strong&gt;&quot;.&lt;br&gt;&#xA;  There are two doors: The left one leads to room &quot;&lt;strong&gt;neetneves&lt;/strong&gt;&quot; and the right one leads to the &lt;strong&gt;iron room&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Room 28:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; WHOA! You found a big pot of gold!&lt;br&gt;&#xA; Now, proceed to the &lt;strong&gt;room of that girls&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Room 30:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The front door leads to the room which the number is &lt;strong&gt;THE answer&lt;/strong&gt;!&lt;br&gt;&#xA;  There is a window in the wall, but it is too small for you to pass through.&lt;br&gt;&#xA;  Looking through the window, you see some trees out there. There is a bullseye near one of the trees.&lt;br&gt;&#xA;  If you have the &lt;strong&gt;needed equipment (what is it?)&lt;/strong&gt;, you may shoot the bullseye.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Room 34:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You found the blue key!&lt;br&gt;&#xA;  There is a spring that may throw you away to the room &quot;&lt;strong&gt;Mg&lt;/strong&gt;&quot;.&#xA;  There is a closet that is a secret passage to the room which is &lt;strong&gt;a multiple of 11&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Room 37:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is a blue box, but it is locked. You need to find a blue key to open it.&lt;br&gt;&#xA;  Two doors. The left door goes to room &quot;&lt;strong&gt;Argon&lt;/strong&gt;&quot;. The right goes to room &quot;&lt;strong&gt;Micro Mike&lt;/strong&gt;&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Room 40:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You found the cake!&lt;br&gt;&#xA; This cake is so delicious and moist!&lt;br&gt;&#xA; Now you can walk back to &lt;strong&gt;the previous room&lt;/strong&gt; or go to the &lt;strong&gt;&quot;Yt4$)&quot;&lt;/strong&gt; room.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Room 42:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is a gnome in this room.&lt;br&gt;&#xA;  He said: &quot;If you give me a pot of gold and a magic bean, I will make a rainbow! Everybody knows that there is a pot of gold in the end of the rainbow!&quot;&lt;br&gt;&#xA;  Curious, you decide to ask for more information. So, he tells this:&lt;br&gt;&#xA;  &quot;I used to be a prisoner in a dungeon just like you. But then, I took an arrow to the knee.&quot;&lt;br&gt;&#xA;  ! Anyway, in this room there is a river called Tiber (why is the name of the river important?), leading to room &lt;strong&gt;X&lt;/strong&gt;.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Room 44:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You found a glowing crystal key!&lt;br&gt;&#xA;  There are a window leading the &lt;strong&gt;room with the largest number&lt;/strong&gt;.&lt;br&gt;&#xA;  There is an elevator leading to the &lt;strong&gt;first room with a number greater than this one&lt;/strong&gt;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Room 48:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A giant circular fast-spinning blade falls onto your head.&lt;br&gt;&#xA; Then, your body is divided in two halves!&lt;br&gt;&#xA; Guess what? You got a nice Game Over!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Room 56:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You found a black magic book and got it. Then you are teleported back to &lt;strong&gt;the starting room&lt;/strong&gt;.&lt;br&gt;&#xA;  If you already have been in this room before and had already taken the book, you will be teleported to the room &lt;strong&gt;whose number is the first two-digit room number appearing in π&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Room 59:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You found something &lt;strong&gt;TERRA&lt;/strong&gt;. Guess what it could be?&lt;br&gt;&#xA;  Now choose your way out: &quot;&lt;strong&gt;dezessete&lt;/strong&gt;&quot; or &quot;&lt;strong&gt;dezoito&lt;/strong&gt;&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Room 63:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Congratulations, you won a million dollars!&lt;br&gt;&#xA; Now you just need to leave this dungeon and you are a rich man!&lt;br&gt;&#xA; There is an escalator going up here. It leads to the &lt;strong&gt;room with repeated numbers&lt;/strong&gt;.&lt;br&gt;&#xA; There is a door to the left it leads to the room &lt;strong&gt;0x4&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Blue box:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You found the batteries for the &quot;&lt;strong&gt;84 69 76 69 80 79 82 84 69 82&lt;/strong&gt;&quot;!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Shooting the bullseye:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You shoot through the window, and hit the bullseye just in the middle!&lt;br&gt;&#xA; Immediately, a new trapdoor opens in the room &lt;strong&gt;1A&lt;/strong&gt;. This trapdoor leads to the room &quot;&lt;strong&gt;五十六&lt;/strong&gt;&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Red button:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You pressed the RED button:&lt;br&gt;&#xA; BOOOOOM!!!! Game over!&lt;br&gt;&#xA; You should know that this always happens when you press a red button!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Blue button:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You pressed the BLUE button:&lt;br&gt;&#xA; An elephant comes out of nowhere and tramples you. Game over!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Yellow button:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You pressed the YELLOW button:&lt;br&gt;&#xA; A secret passage opens in that &quot;&lt;strong&gt;ending with '3s2 3p6'&lt;/strong&gt;&quot; room. It leads to room &quot;&lt;strong&gt;◔◈◅ ◌●◒◄ ◉◓ ◍◙ ◓◈◅◐◅◒◄, ◉ ◌◁◃○ ◎●◔◈◉◎◇&lt;/strong&gt;&quot;.&lt;br&gt;&#xA; A trapdoor opens under you feet. You fell onto the room &quot;&lt;strong&gt;NPPSTVPJWFSQFIUGPGMBITJSFCNVOFIUFTPIX&lt;/strong&gt;&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;If you decide to insert the arrow in your own knee:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Seriously? Are you crazy? What the hell are you thinking?&lt;br&gt;&#xA; The arrow was poisoned. You became paralyzed. Your skin turns blue.&lt;br&gt;&#xA; You fell to the ground in a deep pain and agony. You can't breath and your heart stops.&lt;br&gt;&#xA; Congratulations! That was the most stupid thing that you could do in this game!&lt;br&gt;&#xA; Guess what you got now? A giant, red, blinking and dancing &quot;game over&quot;! That is right dude, you lost the game and died for being so stupid!  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Magic book:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the book, it is written:&lt;br&gt;&#xA;  &lt;br&gt;&#xA;  Abrakadabra!&lt;br&gt;&#xA;  &lt;strong&gt;. ορετόσσιρεπ ιχό ιεσάρδ αν ιεπέρπ αθ ςασ ήγρΟ ! άρεν ανέμγαρατ ατ όπα ωνάπ υομ ήμανύδ ητ ιαμύολακιπε , ςανώδιεσοΠ .&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Giving the gold and the magic bean to the gnome:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The gnome receives the gold and the magic bean and a rainbow appeared right in front of your eyes, entering in the pot of gold.&lt;br&gt;&#xA; You started to walk over it. That is incredible! You did not knew that you could walk over a rainbow!&lt;br&gt;&#xA; You thanks the gnome and go away, walking over the rainbow into the blue sky.&lt;br&gt;&#xA; The view here is so beautiful! You can see mountains, rivers and cities down there...&lt;br&gt;&#xA; But then, you perceived something:&lt;br&gt;&#xA; You leaved the dungeon! Congratulations! You are free!&lt;br&gt;&#xA; And then, you remembered still another great thing:&lt;br&gt;&#xA; Rainbows do not lasts for long time!&lt;br&gt;&#xA; The rainbow disappears under your feet, and you fall, quickly going directly to the ground.&lt;br&gt;&#xA; Oh no, you will face certain death! &lt;strong&gt;Except if...?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5044" LastEditorUserId="5373" LastEditDate="2019-07-20T12:35:06.230" LastActivityDate="2023-03-02T09:35:25.797" Title="Escape the dungeon, by deciphering the codes and navigating the rooms to find a way out" Tags="|logical-deduction|cipher|password|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="4731" PostTypeId="2" ParentId="4687" CreationDate="2014-11-15T08:45:03.737" Score="3" Body="&lt;p&gt;I chose Clues 1 and 3, Sarah's room, and the kitchen and basement at Simon's house. Also, milk and a passive response. =D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sarah's roommate&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The woman upstairs heard someone say &quot;I'm tired; I don't want to argue with you as well.&quot; &lt;em&gt;Assuming&lt;/em&gt; that was Sarah, it seems when she came home her roommate must have tried to argue with her. But her roommate specifically said she hadn't seen her since she left for her boyfriends house, and that would be a lie. Also, I think the second footprint, the one leaving the alleyway was probably the detective's!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5500" LastActivityDate="2014-11-15T08:45:03.737" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4732" PostTypeId="1" AcceptedAnswerId="4733" CreationDate="2014-11-15T08:47:15.197" Score="7" ViewCount="1566" Body="&lt;pre&gt;&lt;code&gt;When you get to know me&#xA;I'll be long time gone.&#xA;&#xA;If you ever saw me&#xA;I was someone else back then.&#xA;&#xA;I have two younger siblings&#xA;and the middle one is your best peer.&#xA;&#xA;The youngest one you haven't met.&#xA;And when you meet,&#xA;he'll disappear.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The question is, who are the mysterious brothers? &lt;/p&gt;&#xA;" OwnerUserId="5283" LastActivityDate="2015-01-26T19:34:43.213" Title="Riddle about mysterious family" Tags="|riddle|rhyme|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4733" PostTypeId="2" ParentId="4732" CreationDate="2014-11-15T08:48:42.997" Score="7" Body="&lt;p&gt;I guess that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is the past, the present and the future.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I reached this by guessing what would reasonably fit in the given text, and noting that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All the narrative are about the &lt;strong&gt;time&lt;/strong&gt; of when something happens.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;When you get to know me&#xA;I'll be long time gone.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you know the past (the one who refers himself/herself as &quot;me&quot;), it is something that do not exists anymore.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;If you ever saw me&#xA;I was someone else back then.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you saw the past, it was not the past, it was the present.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I have two younger siblings&#xA;and the middle one is your best peer.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The past has two younger siblings, the present and the future. The present is your best peer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;The youngest one you haven't met.&#xA;And when you meet,&#xA;he'll disappear.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you meet the future, he will be the present instead, and immediately become the past.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5044" LastEditorUserId="5044" LastEditDate="2015-01-26T19:34:43.213" LastActivityDate="2015-01-26T19:34:43.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4734" PostTypeId="2" ParentId="4732" CreationDate="2014-11-15T08:49:42.250" Score="9" Body="&lt;p&gt;They are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; YESTERDAY, TODAY AND TOMORROW &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And you are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  YESTERDAY&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explaination&lt;/strong&gt; &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you get to know me&#xA;I'll be long time gone.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When a day is &quot;yesterday&quot; it is forever &quot;gone&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you ever saw me&#xA;I was someone else back then.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Yesterday&quot; (as a day) changed every day. i.e. for every day there is a different &quot;yesterday&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have two younger siblings&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;younger&quot; means they first occur later in time (fits with &quot;today&quot; and &quot;tomorrow&quot;). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the middle one is your best peer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;today&quot; is the one we are always &quot;with&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The youngest one you haven't met.&#xA;And when you meet,&#xA;he'll disappear.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;tomorrow&quot; is the &quot;youngest&quot;. When we encounter &quot;tomorrow&quot; (i.e. arrive there in time), it has become &quot;today&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-11-15T08:57:09.037" LastActivityDate="2014-11-15T08:57:09.037" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4735" PostTypeId="1" AcceptedAnswerId="4786" CreationDate="2014-11-15T09:18:18.217" Score="11" ViewCount="4630" Body="&lt;p&gt;↗↑↙↓↘←↑↙↘↘↗↓↗←↙↓↖&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first character of the plaintext is 'P'. The final character of the plaintext is 'E'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The second word of the plaintext is DISPUTE.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; NW = 'end of message'. It's an end-of-file marker.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Details (not part of the puzzle):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;&quot;↗↑↙↓↘←↑↙↘↘↗↓↗←↙↓↖&quot; is the cyphertext. What is the plaintext?&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The answer is a clue to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4057/the-security-to-the-party-12-now-with-party-soundtrack&quot;&gt;The Security to the Party [12] (now with party soundtrack!)&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This type of puzzle is called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Cryptogram&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cryptogram&lt;/a&gt;. The Google search &quot;&lt;a href=&quot;https://www.google.co.uk/webhp?q=%22how%20to%20solve%20a%20cryptogram%22&quot; rel=&quot;nofollow noreferrer&quot;&gt;How to solve a cryptogram&lt;/a&gt;&quot; leads to all kinds of useful resources, hints and tips.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The answer is a pair of &lt;s&gt;dictionary&lt;/s&gt; words in the English language. The puzzle is &lt;em&gt;not&lt;/em&gt; case-sensitive. The plaintext does not contain any slang or acronyms, it does not contain any proper nouns (names), it does not contain roman numerals, it does not contain punctuation or spaces.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For anyone who's having trouble with fonts, and for accessibility: the cyphertext is a series of Unicode arrows pointing in different directions. The sequence is: NE N SW S SE W N SW SE SE NE S NE W SW S NW&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Chat about this puzzle is here: &lt;a href=&quot;http://chat.stackexchange.com/rooms/18676/pick-up-sticks&quot;&gt;http://chat.stackexchange.com/rooms/18676/pick-up-sticks&lt;/a&gt; I'll be popping in and out. Feel free to ask any questions there or in comments.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;On some screens it looks like some of the arrows are in bold and some aren't. That is not the intention, it's just the way the fonts look. If you use the list of compass points above then you are at no disadvantage.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Update: Argh! I was so busy doing the cipher that I didn't notice the spelling mistake in the plaintext. Well done to nexolute for solving anyway! Future readers beware that the first word of the puzzle is (unintentionally) &lt;em&gt;not&lt;/em&gt; a dictionary word, because it has a spelling mistake in it. Are my cheeks red!&lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2021-10-23T13:09:33.460" Title="Decode the message enciphered in these symbols: ↗↑↙↓↘←↑↙↘↘↗↓↗←↙↓↖" Tags="|cipher|symbols|" AnswerCount="1" CommentCount="23" ContentLicense="CC BY-SA 3.0" />
  <row Id="4736" PostTypeId="2" ParentId="4046" CreationDate="2014-11-15T09:19:31.983" Score="1" Body="&lt;p&gt;so the question is 2=2, 4=4, 6=/=6, 3=2, 5=?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; so every answer is even number and must less than 5 (because 6 =/= 6)&lt;br/&gt; if the question is given odd number, it should minus 1 (so that 3-1=2)&lt;br/&gt; so that answer is 5-1=4, about the poor friend, his answer should be 6-4=2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5453" LastActivityDate="2014-11-15T09:19:31.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4739" PostTypeId="2" ParentId="4698" CreationDate="2014-11-15T12:47:48.640" Score="4" Body="&lt;p&gt;I was me. Will you be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The father made decisions best for himself. He wants those living to make their own choices in life and not simply repeat his&lt;/p&gt;&#xA;" OwnerUserId="5507" LastActivityDate="2014-11-15T12:47:48.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4740" PostTypeId="2" ParentId="4057" CreationDate="2014-11-15T12:49:38.910" Score="3" Body="&lt;p&gt;just random guessing&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if this one doesn't require maths, it will be those soundtrack name &lt;br/&gt; guard will tell the guest which number of the track they want to guess (according to the list without modify)&lt;br/&gt; First, it will check if the track contain &quot;the&quot;. If yes, go to F1. If not, go to F2. if contain more than 1 &quot;the&quot;, go to F3. &lt;br/&gt;&lt;br/&gt;F1: if &quot;t&quot; is lowercase, then select the second word after &quot;the&quot; (example: &quot;the bla thing&quot;, choose &quot;thing&quot;), if &quot;T&quot; is uppercase, then select the first word after &quot;the&quot;. Count the total alphabet of the selected word&lt;br/&gt;&lt;br/&gt;f2: if track contains &quot;()&quot;, count inner word length. If doesn't contain &quot;()&quot;, then say 0.&lt;br/&gt;&lt;br/&gt;F3: count total &quot;the&quot; in the track and minus one.&lt;br/&gt;&lt;br/&gt; the answer for 4th track is &quot;Prodigy&quot;, which is 7.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5453" LastActivityDate="2014-11-15T12:49:38.910" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4741" PostTypeId="2" ParentId="4326" CreationDate="2014-11-15T13:00:36.263" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; did anyone notice the date of the message wrote at 22/7? and it will blow up at midnight, which will change the time to 23/7 00:00. since the topic is prime number, 23 and 7 are prime number too. &lt;br/&gt;&lt;br/&gt; so i think that the answer is 23/7&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5453" LastActivityDate="2014-11-15T13:00:36.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4742" PostTypeId="1" AcceptedAnswerId="4753" CreationDate="2014-11-15T13:32:50.793" Score="6" ViewCount="617" Body="&lt;p&gt;You're out on an adventure, when you find you're lost. While trying to find your way out you come across 4 others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of them know 1 direction. There's two girls, they know east and west. There's also two boys, they know north and south. &lt;strong&gt;They are also from another country, and have different words for north, south, east, and west that you do not know.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You also know that one girl and one boy enjoys lying, and will lie to you at all times.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may assume the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;You cannot ask a question that will silence them because they do not know. If they can't answer, assume they will kill you.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;They have no idea what they would say to a question, so &quot;Are you about to say no?&quot; or &quot;What would you say if I asked you 'Is that way North?'&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;They have no knowledge of what the others in their group or gender would say to a question, or if they lie or not.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;They know enough English to understand your question, but not enough to respond in English.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You cannot see the sun, moon, or stars. Pretend it's a cloudy night.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When you walk up to them they will hold their hands up in 2 directions, the correct way, and the opposite to the correct way. Can you ask them each a question, and be able to determine north, east, south, and west, if so, how?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You can ask them each a different question if you like, and label them BoyA, BoyB, GirlA, GirlB.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I in no way imply this is possible, or agree that it can be done. The point is to determine if you can, and if so how?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="5216" LastEditDate="2014-11-15T20:46:58.823" LastActivityDate="2018-04-03T16:31:45.580" Title="The directions from another language?" Tags="|logical-deduction|" AnswerCount="6" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="4743" PostTypeId="2" ParentId="4326" CreationDate="2014-11-15T13:44:00.120" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everything is related to $\pi$.&lt;br&gt;&lt;br&gt;&#xA; -22/7 is approximation of $\pi$ (tempted to think that it will be $\pi/14$, but $14$ is probably useless).&lt;br&gt;&#xA; -&lt;s&gt;24:00, which can also be written as $4\pi$ in radian (that's why you didnt write 00:00).&lt;/s&gt;&lt;br&gt;&#xA; - 6-folded paper, which appears in $\zeta(2)=\pi^2/6$ (makes me wonder if paper's length is $\pi$).&lt;br&gt;&#xA; - Primes are related to $\pi$ in many ways (not sure why powered to itself though).&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That said, I still don't have a solution. Derp.&lt;/p&gt;&#xA;" OwnerDisplayName="user5508" LastEditorDisplayName="user5508" LastEditDate="2014-11-15T15:53:23.773" LastActivityDate="2014-11-15T15:53:23.773" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="4744" PostTypeId="2" ParentId="4742" CreationDate="2014-11-15T14:05:30.463" Score="-1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; assume that only boys know north and south, and girls know east and west, and each gender of one of them lied, they will standing on the same place at the same time.&lt;br/&gt;&lt;br/&gt; try to shake hands with them, determine if they are left-handed or right-handed.&lt;br/&gt; ask boy which is north, if they are right-handed, they would use their right hand to point at the correct one.&lt;br/&gt; while the wrong one always lie, they would try to point the right one, but then they might point to the wrong. &lt;br/&gt; the one that didn't change the position will be right&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I HAVE EDITED TO  ANOTHER ANSWER&lt;/p&gt;&#xA;" OwnerUserId="5453" LastEditorUserId="5453" LastEditDate="2014-11-15T14:16:10.637" LastActivityDate="2014-11-15T14:16:10.637" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4745" PostTypeId="2" ParentId="4742" CreationDate="2014-11-15T14:18:10.800" Score="0" Body="&lt;p&gt;You can say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Can all of you point the cardinal point you know?&lt;/strong&gt;&lt;br/&gt; They won't answer cause they don't know English but they can move their arms to give the right answer. The 2 boys and the 2 girls will point only two direction, because two of them are lying. But after, if you ask &lt;strong&gt;Can all of you point the opposite cardinal point you know?&lt;/strong&gt; you will easy find whom is lying and identify North, South, East and West. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1685" LastActivityDate="2014-11-15T14:18:10.800" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4746" PostTypeId="2" ParentId="4742" CreationDate="2014-11-15T14:21:59.180" Score="0" Body="&lt;p&gt;Ask the first girl whether 1+1=2. Now, the other girl is pointing both west and east, and you know whether she's a liar, so ask &quot;Is [that direction] the same one that the sun rises in?&quot;&lt;/p&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2014-11-15T14:21:59.180" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4747" PostTypeId="1" AcceptedAnswerId="4780" CreationDate="2014-11-15T14:38:00.910" Score="8" ViewCount="298" Body="&lt;p&gt;Alice and Bob are telecommunicating in some way. When Alice speaks, it takes A seconds to reach Bob. When Bob speaks, it takes B seconds to reach Alice. Both Alice and Bob have clocks that tick at the same speed, but no guarantee that the clocks are set to the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can they determine A and B?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The algebraic solution to this puzzle is not very difficult, although I'd be interested to know if there are alternative approaches.&lt;/p&gt;&#xA;" OwnerUserId="5511" LastActivityDate="2014-11-16T16:36:15.633" Title="Determinine the delay on a phone call" Tags="|calculation-puzzle|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4750" PostTypeId="1" AcceptedAnswerId="4756" CreationDate="2014-11-15T16:00:22.480" Score="2" ViewCount="1291" Body="&lt;p&gt;You found yourself in a weird room, there was no doors, but a limited computer with no internet that has open an app that is displaying the following message:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;f(0) = 0&lt;br/&gt;f(1) = 549&lt;br/&gt;f(2) = 646&lt;br/&gt;f(10) = 1350&lt;br/&gt;f(100) = 450&lt;br/&gt; Hence, find f(x) and f(75)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For people who don't know what a maths function is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;function is like a formula with dynamic variable, eg: f(x) = x+12, f(1) = 13&lt;/li&gt;&#xA;&lt;li&gt;for tutorial/example &lt;a href=&quot;http://www.mathsisfun.com/sets/function.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.mathsisfun.com/sets/function.html&lt;/a&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Function_%28mathematics%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Function_%28mathematics%29&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I will give a clue everyday if no one has the right answer, if alternative answer was found and has the same results as my answer, it counted as the right answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CLUE:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the function contains squared number&lt;/li&gt;&#xA;&lt;li&gt;the function does not has complex number&lt;/li&gt;&#xA;&lt;li&gt;the function uses BODMAS&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5453" LastEditorDisplayName="user31569" LastEditDate="2017-12-13T17:22:15.450" LastActivityDate="2017-12-13T17:22:15.450" Title="Determine the Maths Function" Tags="|mathematics|calculation-puzzle|password|" AnswerCount="2" CommentCount="7" FavoriteCount="0" ClosedDate="2014-11-17T21:01:47.643" ContentLicense="CC BY-SA 3.0" />
  <row Id="4751" PostTypeId="1" AcceptedAnswerId="4957" CreationDate="2014-11-15T16:24:44.703" Score="3" ViewCount="701" Body="&lt;p&gt;A party is being held at a local mansion, starting at 7pm. The host is very rich and his success is because of one thing — his famous recipe for spaghetti!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only guests that may attend are people who correctly reply to the guard at the door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a friend called &lt;a href=&quot;https://puzzling.stackexchange.com/users/4652/narmer&quot;&gt;Narmer&lt;/a&gt;, a former acquaintance of the host, who has talked you into trying to get into the mansion in order to steal the spaghetti recipe. The pair of you hide by the entrance to the mansion at 6pm and wait for the other guests to arrive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first guest arrives at 7pm. The guard says &quot;4&quot;. The guest says &quot;6&quot;. A gunman pops up from behind the wall and shoots the guest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second guest arrives at 7:15pm. The guard says &quot;15&quot;. The guest says &quot;8&quot; and is admitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third guest arrives. The guard says &quot;14&quot;. The guest says &quot;11&quot; and is admitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fourth guest arrives. The guard says &quot;12&quot;. The guest says &quot;6&quot; and is admitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fifth guest arrives. The guard says &quot;21&quot;. The guest says &quot;7&quot; and is admitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Can it really be that simple?&quot; your friend murmurs to himself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sixth guest arrives. The guard says &quot;9&quot;. The guest says &quot;5&quot;. The gunman pops up again and shoots her too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The seventh and eighth guests arrive together. They are talking to each other but you cannot understand what they are saying. The guard says &quot;7&quot; to one and &quot;3&quot; to the other. They answer &quot;5&quot; and &quot;3&quot; respectively; both are admitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;It is!&quot; your friend exclaims. He breaks cover and saunters up to the entrance. The guard eyes him suspiciously and says &quot;1&quot;. Your friend says &quot;3&quot; and is admitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The guard is now peering at your hiding place; he must have seen where your friend came from. You cannot stay here much longer. When the guard looks away for a moment, you emerge and walk up to him. He says &quot;22&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First question&lt;/strong&gt;: what number do you tell the guard?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus question/hint 1&lt;/strong&gt;: how does the guard decide what numbers to use as challenges? (There is a pattern to it!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Where do Narmer and the host know each other from?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit after Michal's answer&lt;/strong&gt;: none of the guests has a calculator with them; they all give their replies after only a few seconds' thought, without recourse to any electronic aid.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2014-11-19T11:53:34.200" Title="The Security to the Party [Part 31]" Tags="|password|" AnswerCount="1" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4752" PostTypeId="2" ParentId="4057" CreationDate="2014-11-15T16:32:14.910" Score="3" Body="&lt;p&gt;Still can't figure out any answer from the numbers, but some kind of an idea...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What if the first 5 guests are sibilings (hence the annoyed guard, after seeing the similar face again and again, and one of them even beat him up), and these sibilings have magic powers to travel through spacetime... Using this special ability, we could explain the odd behaviors of the sibilings (such as the &quot;0&quot; answer of the 3rd guest, and the going-crazy of the 4th guest) because they meet a spacetime paradox. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user5508" LastActivityDate="2014-11-15T16:32:14.910" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4753" PostTypeId="2" ParentId="4742" CreationDate="2014-11-15T16:35:34.743" Score="5" Body="&lt;p&gt;The problem is insufficiently constrained to come up with a guaranteed solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that one boy and one girl are knaves, but we &lt;em&gt;don't&lt;/em&gt; know whether the other boy and girl will tell the truth. They may or may not lie some or all of the time. The allowed assumptions (which do &lt;em&gt;not&lt;/em&gt; include anything like &quot;the other boy and girl always tell the truth&quot;) eliminate the usual ways of determining truth-telling value by disallowing asking questions about the other parties' answers. But those generally depend on knowing the distribution of orientations anyway (e.g. knowing there's a knight and a knave, but not who's who). Because of this, asking a single question to a single agent is insufficient to determine which of the pair is the knave. Both boys or both girls must be questioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If two same gendered agents give different answers to a question you know the answer to, that's all well and good. But if they don't, you don't know whether either of them will continue lying to you (considered as individuals; obviously at least one of them will). If they give different answers to a question you don't know the answer to, then you have no way of knowing who is the liar. This is the sticky wicket: when you ask a question, you know the answer or you don't. Either way, it's possible to fail to gain any information.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You can ask them each a different question if you like&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you're allowed to ask as many questions as you want, then you can build a probabilistic model. However, this isn't reliable either, because we can't assume the agents aren't malicious. For example, the non-guaranteed-knave might be a malevolent joker: lying until you're confident of knave status, then telling the truth when you ask about the directions. Probably because they got cranky about being asked so many questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &quot;a question&quot; means only one question per agent, then you're basically down to a coin toss, since the boys show you north and south. This is because you need at least two different questions to determine a liar (granting for the sake of argument that doing so is possible): one with a known truth value and one to determine a direction.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You cannot see the sun, moon, or stars. Pretend it's a cloudy night.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I suppose you could just ask a question that would take sufficiently long enough to answer and wait for the sun to come up. But I assume the point of this puzzle is deducing the answer, not lateral thinking.&lt;/p&gt;&#xA;" OwnerUserId="5395" LastEditorUserId="30633" LastEditDate="2018-04-02T06:15:38.593" LastActivityDate="2018-04-02T06:15:38.593" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4754" PostTypeId="2" ParentId="4730" CreationDate="2014-11-15T16:36:36.610" Score="5" Body="&lt;p&gt;OK, this is going to take a while.&lt;/p&gt;&#xA;&lt;p&gt;I think this answer correctly escapes the dungeon now. However, there are still some unsolved mysteries (in particular, the identity of the iron room).&lt;/p&gt;&#xA;&lt;h2&gt;1. You just have to kill the lion&lt;/h2&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Go to 18 (atomic number of &lt;strong&gt;Argon&lt;/strong&gt;). This room has an exit to the (&lt;strong&gt;iron room&lt;/strong&gt;) and so does room 26, so presumably we can get from 18 to 26. Pick up the gun. We see exits to 17 (&lt;strong&gt;neetneves&lt;/strong&gt; spelled backwards) and 12 (&lt;strong&gt;Butterfree&lt;/strong&gt;'s Pokedex number). Proceed to 12. Take all the pills. Retrace your steps all the way back to your starting point and go to 24 (&lt;strong&gt;Micro Mike&lt;/strong&gt;'s Action 52 index). Shoot the lion and ravage its corpse.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;2. You only have to get a million dollars&lt;/h2&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Go to 5 (&lt;strong&gt;Monica etc&lt;/strong&gt; is from some song with 5 in the title). Proceed to 34 (&lt;strong&gt;rouf-ytrihT&lt;/strong&gt; backwards) and pick up the blue key. Go to 44 (&lt;strong&gt;the room with repeated numbers&lt;/strong&gt;) and take the crystal key. Go to 63 (&lt;strong&gt;room with largest numbers&lt;/strong&gt;), and take our cool million dollars.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;3. You merely have to become an expert archer&lt;/h2&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Retrace all the way back to room 37 and open the blue box. We found the batteries for the teleporter in room 5! (&lt;strong&gt;sfuspqfmfu&lt;/strong&gt; Caesared and backwards, &lt;strong&gt;84 69 76 69 80 79 82 84 69 82&lt;/strong&gt; in UTF-8 character table.) Proceed to the teleporter and go to room 30 (&lt;strong&gt;11110&lt;/strong&gt; in binary). Go to 42 (&lt;strong&gt;the answer&lt;/strong&gt;) and then to 10 (&lt;strong&gt;X&lt;/strong&gt; Roman numeral). Take the arrow and press the yellow button. A secret passage opens from 18 (&lt;strong&gt;3s2 3p6&lt;/strong&gt; is Argon electron orbitals) room. It leads to room 23 (◔◈◅ ◌●◒◄ ◉◓ ◍◙ ◓◈◅◐◅◒◄, ◉ ◌◁◃○ ◎●◔◈◉◎◇, decoded through character values to THE LORD IS MY SHEPERD, ETC, which is Psalm 23.) But first we fall into 21. (&lt;strong&gt;NPPSTVPJWFSQFIUGPGMBITJSFCNVOFIUFTPIX&lt;/strong&gt; Caesared backwards, half of the PREVIOUS room number is half of 42 is 21.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;4. You only need to summon a Greek god&lt;/h2&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; From 21, we can get a red key (&lt;strong&gt;[Russian]&lt;/strong&gt;). Unlock the red door to go to 13 (&lt;strong&gt;bad luck room&lt;/strong&gt;), and find a &lt;strong&gt;D&lt;/strong&gt;. But what is a &lt;strong&gt;D&lt;/strong&gt;? Well, a bow, of course (look at the shape). Retrace our steps and use our bow and arrow to shoot the bullseye in 30, opening a passage from 26 (&lt;strong&gt;1A&lt;/strong&gt; hex) to 56 (&lt;strong&gt;Japanese kanji numbers&lt;/strong&gt;). Take the passage and obtain the magic book. It can summon the power of Poseidon to create a bridge over troubled waters (&lt;strong&gt;[backwards Greek message]&lt;/strong&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;5. You just have to walk on rainbows&lt;/h2&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; We are warped to the starting room, but after returning to 56 again we are warped to 59 (&lt;strong&gt;first two-digit room number appearing in π&lt;/strong&gt;). Take the green (&lt;strong&gt;TERRA&lt;/strong&gt; Caesared) key and proceed to 18 (&lt;strong&gt;dezoito&lt;/strong&gt; in Portugese). Open the green door and go to 2 (&lt;strong&gt;first prime number&lt;/strong&gt;). Take the parachute (&lt;strong&gt;☏☀☑☀☂☇☔☓☄&lt;/strong&gt;, unicode character values). Proceed to 8 (&lt;strong&gt;this room cubed&lt;/strong&gt;). Read from the magic book, summoning a bridge. Go to 28 (&lt;strong&gt;perfect number&lt;/strong&gt;). Take the gold! I have no idea what &lt;strong&gt;that girls&lt;/strong&gt; is supposed to be, but whatever it is, eat a pill and go to 37 (&lt;strong&gt;naM hsaM&lt;/strong&gt; backwards, Action 52 index.) Give the gold to the gnome. Use the parachute to survive the fall.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;And finally...&lt;/h2&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Return to the dungeon, enter room 1 (the exit), and say hello to the princess.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;Mysteries left unsolved **&lt;/h2&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What is &amp;quot;that girls&amp;quot;?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What does &lt;strong&gt;kwws=22elw1o|24vSg{I}&lt;/strong&gt; mean?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Wait, no, I got this one. Caesar it three back and follow the URL to find a trollface. Good one, Victor.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;What does &lt;strong&gt;@@@@@@@@&lt;/strong&gt; mean?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Where does the &lt;strong&gt;21120110122101211011201112&lt;/strong&gt; pill lead?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Probably 37, because that's where two of the others lead (Japanese Pokemon name and Mash Man action 52 index.) But Pidgeotto is 17 (Pokedex index), so I'm not sure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;Is 26 not the &lt;strong&gt;iron room&lt;/strong&gt; after all? (See Martin's comment).&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is room 40 a Portal-esque red herring?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1752" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-11-15T17:10:47.913" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="4755" PostTypeId="2" ParentId="4604" CreationDate="2014-11-15T16:39:23.517" Score="8" Body="&lt;p&gt;So I finally got it... a really nice riddle, except the part with 10 logs measuring in feet...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; So 10 logs taller is log(10) in feet, so 1 foot or 12 inches &lt;br&gt;&#xA; The tan of a lumberjack is the tangens-function of his height (taken as degrees) &lt;br&gt;&#xA; The result has to be converted from inch to cm (factor 2.54) &lt;br&gt; because Canadians&#xA; changed to the metric system!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The formula for lumberfication of old height h is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (h + tan(h) + 12) * 2.54&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This matches all numbers and results in:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (71 + tan(71°) + 12) * 2.54 = 218&#xA; &lt;br&gt;&#xA; (83 + tan(83°) + 12) * 2.54 = 262&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1998" LastEditorUserId="1998" LastEditDate="2014-11-16T14:16:51.980" LastActivityDate="2014-11-16T14:16:51.980" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4756" PostTypeId="2" ParentId="4750" CreationDate="2014-11-15T16:49:33.267" Score="6" Body="&lt;p&gt;$f(1) = 549 = 1 * 100 + 450 - 1^2$&lt;br&gt;&#xA;$f(2) = 646 = 2 * 100 + 450 - 2^2$&lt;br&gt;&#xA;$f(10) = 1350 = 10 * 100 + 450 - 10^2$&lt;br&gt;&#xA;$f(100) = 450 = 100 * 100 + 450 - 100^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $f(x) = 100x + 450 - x^2$, and $f(75) = 2325$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I realized... maybe it should be the following, but then I think I cheated, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;f(x) = \begin{cases} 100x + 450 - x^2 &amp;amp;\mbox{if } x \neq 0 \\ &#xA;0 &amp;amp; \mbox{if } x = 0\end{cases}&#xA;$&lt;/p&gt;&#xA;" OwnerDisplayName="user5508" LastActivityDate="2014-11-15T16:49:33.267" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4757" PostTypeId="1" CreationDate="2014-11-15T17:35:38.617" Score="-2" ViewCount="229" Body="&lt;p&gt;You once again decide to steal the recipe at the party! How many has this guy had now? Like 30? He really is rich. No wonder you want the recipe so bad!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This time the extravagant man has amped up on the security. The guard is still there, he still gives out numbers, but it seems the host is catching on to your trick. He's installed a keypad! So much for listening in on responses!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You still decide to try! You hide in the bushes and listen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first guest arrives. The guard said &quot;53&quot;. You hear &quot;BEEP!&quot; And the doors open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second guest arrives. The guard said &quot;32&quot;. You hear &quot;BEEP!&quot; And the doors open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Realizing you're getting nowhere, you get a pair of binoculars and find a new hiding spot. The only problem is you can't hear them anymore!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 3rd guest comes and you see him type in the letter &quot;D&quot;. He's admitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then a 4th guest appears and you seem him type in &quot;E&quot;. The door opens - well at least the trap door: he falls down into an alligator pit. About 5 seconds after an elephant is dropped down too!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You find an old satellite and get to listen and hear. You only get to once though. No other guest comes after the 5th for a while, so you gave up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 5th guest gets the number &quot;62&quot; and types in &quot;Q&quot; to get let in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You get down and the guard says &quot;73&quot; what do you type in to get admitted?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;BONUS: What were the audio/visual to the first 3 guests?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hints:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A fellow puzzler fails and says F, now you know it isn't F, you stick around and find the answer to &quot;33&quot; is &quot;I&quot; You go back and you also get hit with 73!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes that is the hint form I'm going to use.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="5373" LastEditDate="2014-11-15T20:46:46.913" LastActivityDate="2014-11-16T13:03:36.080" Title="Security to the Party [Part 32!]" Tags="|password|" AnswerCount="2" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4758" PostTypeId="1" AcceptedAnswerId="4958" CreationDate="2014-11-15T17:38:19.593" Score="7" ViewCount="495" Body="&lt;p&gt;Created or evolved whatever you please&lt;br&gt;&#xA;walk in my shoes, which I can't seize  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A computer originated world, I won't fathom&lt;br&gt;&#xA;my death by some considered gruesome  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how to live do not give me tips&lt;br&gt;&#xA;but don't mourn me now, for I am in bits  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;suffocate me quick don't tarnish my produce&lt;br&gt;&#xA;remember our sacrifice, our past abuse.     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;hints to make it easier:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My name is changed once deceased&lt;br&gt;&#xA; yet it holds the same meaning&lt;br&gt;&#xA; well done! I am rare once the fires have ceased&lt;br&gt;&#xA; I humble your buds once you begin feeding&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5135" LastEditorUserId="5135" LastEditDate="2014-11-18T11:01:44.660" LastActivityDate="2014-11-18T11:34:15.483" Title="Name that entity (4)" Tags="|riddle|word|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4759" PostTypeId="2" ParentId="4326" CreationDate="2014-11-15T17:40:56.840" Score="1" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; zero, because that's what the reimann zeta function tends to, or zeros are kind of important in the study of it or something?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(no maths phd here, can you tell?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or just&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; zeta&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;maybe?&lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="4050" LastEditDate="2014-11-15T17:46:41.180" LastActivityDate="2014-11-15T17:46:41.180" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4760" PostTypeId="2" ParentId="4758" CreationDate="2014-11-15T18:57:24.360" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;A computer originated world&quot;&#xA;&lt;P&gt;&#xA; Are you a process running on a computer?&#xA;&lt;P&gt;&#xA; &quot;Walk in my shoes, which I can't seize&quot;&#xA;&lt;P&gt;&#xA; A computer process has a memory footprint (how much RAM is used), so you have figurative feet.  Not real shoes so they can't be seized.&#xA;&lt;P&gt;&#xA; &quot;my death by some considered gruesome&quot;&#xA;&lt;P&gt;&#xA; If you kill a process on Linux using &quot;kill -9&quot;, the OS kills it immediately without giving the process a chance to clean up gracefully.  That could be considered gruesome.&#xA;&lt;P&gt;&#xA; &quot;I know how to live do not give me tips&quot;&#xA;&lt;P&gt;&#xA; A process's life is a pre-defined series of computer instructions that cannot be changed.&#xA; &lt;P&gt;&#xA; &quot;but don't mourn me now, for I am in bits&quot;&#xA;&lt;P&gt;&#xA; The memory that the process used is free again, representing bits that can be used for another process.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4450" LastEditDate="2014-11-15T19:08:49.047" LastActivityDate="2014-11-15T19:08:49.047" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4761" PostTypeId="1" CreationDate="2014-11-15T18:57:50.753" Score="6" ViewCount="3635" Body="&lt;p&gt;100 prisoners are sent into a room to hear the warden say: &quot;tomorrow I will line you up and put a hat on your head. The hat can be red, green, yellow, …&quot; and he goes on enumerating colors. He then adds: &quot;The executioner will then start from the back of the line and run up to the front, and to every prisoner he will ask what the color of his hat is. If the prisoner guesses the color correctly then he is freed, but if he gets it wrong he will be killed right away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following days, the prisoners are lined up, and the warden puts a colored hat on their head. Then, the executioner starts asking question. The unlucky prisoner at the back of the line guesses wrong and is killed. But by some magic, all the other prisoners guess the color of their hat correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did they do?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I apologize for my English&lt;/p&gt;&#xA;" OwnerUserId="5518" LastEditorUserId="10615" LastEditDate="2015-06-04T00:56:27.897" LastActivityDate="2015-06-04T00:56:27.897" Title="The sadistic executioner (a.k.a the 100 prisoners)" Tags="|logical-deduction|hat-guessing|" AnswerCount="4" CommentCount="6" ClosedDate="2014-11-16T23:41:01.943" ContentLicense="CC BY-SA 3.0" />
  <row Id="4762" PostTypeId="1" AcceptedAnswerId="5525" CreationDate="2014-11-15T19:01:15.247" Score="1" ViewCount="552" Body="&lt;p&gt;If the island of knights and knaves got together and there were &lt;em&gt;n&lt;/em&gt; people, including a joker, how many questions would it take for all of the knights and knaves found the joker?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nobody knows who each other is. When the knight finds the joker, he/she may tell the others, if a knave finds a knight/another knave the knave will do so. Only 1 person asks a question at once. The knave can not always exclaim he/she found a joker when they find a knight/knave. Let's assume a 50-50 chance of telling the others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the least amount of questions needed to guarantee everybody knows who the joker is for sure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The joker don't just randomly tell truth/lie. He can also exclaim that &lt;em&gt;he found the joker himself.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Standard rules apply, nobody had any knowledge about eachother, or what they would have answered. You can not silence them, if you do they will kill you (very violent knights and knaves huh?)&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-11-15T19:18:09.120" LastActivityDate="2014-11-25T10:43:26.797" Title="If The Knights and Knaves got together" Tags="|liars|minimum-solve|" AnswerCount="2" CommentCount="11" ClosedDate="2014-11-26T17:01:30.107" ContentLicense="CC BY-SA 3.0" />
  <row Id="4763" PostTypeId="2" ParentId="4761" CreationDate="2014-11-15T19:08:50.310" Score="3" Body="&lt;p&gt;Assuming the prisoners can see all those in front of them in the line but not those behind...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All the hats are the same colour, except that of the last in line. So the rearmost prisoner (say with a green hat) guesses the colour of all the others (say red) and is killed. The second prisoner sees the front 98 people are all wearing red, and knows the one behind him guessed red, so guesses he himself is wearing red. Ditto for all the others.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2014-11-15T19:08:50.310" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4764" PostTypeId="2" ParentId="3834" CreationDate="2014-11-15T19:11:09.753" Score="7" Body="&lt;p&gt;As mentioned in the comments, Player2 can guarantee a tie.  An easy strategy for this is to just always move on top of Player1's last move.  Due to the number of rows being even, this is always possible.  Therefore only Player1 pieces can end up in the bottom row, and he'll end up with a 4 in a row before Player2 does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore Player1 cannot win.  The best player1 can do is force a tie, which if possible means with perfect play player2 cannot win either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the normal play of &lt;a href=&quot;http://en.wikipedia.org/wiki/Connect_Four#Mathematical_solution&quot;&gt;connect 4 is solved&lt;/a&gt; in favor of Player1, this means Player1 can always get 4 in a row no matter what Player2 plays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, with perfect play the result of the game is a draw.&lt;/p&gt;&#xA;" OwnerUserId="4351" LastActivityDate="2014-11-15T19:11:09.753" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4765" PostTypeId="2" ParentId="4757" CreationDate="2014-11-15T19:17:03.777" Score="5" Body="&lt;p&gt;My (wrong) guess is F.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reasoning: the answer to the number n is the (79-n)th letter of the alphabet, where 79-n is reduced mod 26.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the questions and answers for the first five guests are 53 and Z, 32 and U, 75 and D, 74 and E, 62 and Q.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2014-11-16T11:52:04.493" LastActivityDate="2014-11-16T11:52:04.493" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4766" PostTypeId="2" ParentId="4757" CreationDate="2014-11-15T19:27:51.823" Score="3" Body="&lt;p&gt;The correct response to the number AB is the Bth letter on key#(A+1) on a standard telephone number pad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should type V, the first two guests typed O and H, and the third guest received 21.&lt;/p&gt;&#xA;" OwnerUserId="5352" LastEditorUserId="5352" LastEditDate="2014-11-16T13:03:36.080" LastActivityDate="2014-11-16T13:03:36.080" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="4767" PostTypeId="2" ParentId="4730" CreationDate="2014-11-15T19:31:23.943" Score="15" Body="&lt;p&gt;So here is my stab at this. I promise I did not look at any of Lopsy's work.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;The Codes&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Room 2: &lt;strong&gt;☏☀☑☀☂☇☔☓☄&lt;/strong&gt;, Subtracting 9727 from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Unicode&quot;&gt;Unicode&lt;/a&gt; code points gives numbers, which correspond to the letters &lt;code&gt;Parachute&lt;/code&gt;.&lt;br&gt;&#xA; Room 2: &lt;strong&gt;the cube of this room&lt;/strong&gt;, is &lt;code&gt;8&lt;/code&gt;.&lt;br&gt;&#xA; Room 5: &lt;strong&gt;kwws=22elw1o|24vSg{I}&lt;/strong&gt;, is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Caesar_cipher&quot;&gt;Caesar cypher&lt;/a&gt;. Subtracting 3 from each character gives &lt;code&gt;http://bit.ly/1sPdxFz&lt;/code&gt; which shows &lt;a href=&quot;http://bit.ly/1sPdxFz&quot;&gt;a bunch of troll faces&lt;/a&gt; (red herring).&lt;br&gt;&#xA; Room 5: &lt;strong&gt;tri-hepta&lt;/strong&gt;, &lt;code&gt;tri&lt;/code&gt; is for 3 and &lt;code&gt;hepta&lt;/code&gt; for 7, so I'm gonna venture a guess and say that's &lt;code&gt;37&lt;/code&gt;. I've never used this door, though.&lt;br&gt;&#xA; Room 5: &lt;strong&gt;sfuspqfmfu&lt;/strong&gt;, decrementing each character by 1 and reversing the string spells &lt;code&gt;teleporter&lt;/code&gt;.&lt;br&gt;&#xA; Room 5: &lt;strong&gt;11110&lt;/strong&gt;, is binary for &lt;code&gt;30&lt;/code&gt;.&lt;br&gt;&#xA; Room 5: &lt;strong&gt;How many legs there is in a chair?&lt;/strong&gt;, I hope that's &lt;code&gt;4&lt;/code&gt;.&lt;br&gt;&#xA; Room 5: &lt;strong&gt;ruof-ytrihT&lt;/strong&gt;, that's &lt;code&gt;34&lt;/code&gt; backwards.&lt;br&gt;&#xA; Room 8: &lt;strong&gt;a room which the number divided by two is rounded to the number of this room&lt;/strong&gt;, is &lt;code&gt;17&lt;/code&gt; (could be &lt;code&gt;16&lt;/code&gt; but that doesn't exist).&lt;br&gt;&#xA; Room 8: &lt;strong&gt;perfect room&lt;/strong&gt;. I'm very unsure about this one, but I'm gonna guess 21, because it's the perfect score in Black Jack. &lt;strong&gt;Edit:&lt;/strong&gt; Now I &lt;em&gt;have&lt;/em&gt; looked at Lopsy's solutions. ...facepalm... &lt;a href=&quot;http://en.wikipedia.org/wiki/Perfect_number&quot;&gt;Perfect numbers&lt;/a&gt; were my first thought, but looking at the Wikipedia article I overlooked &lt;code&gt;28&lt;/code&gt; and thought they went &lt;code&gt;6&lt;/code&gt;, &lt;code&gt;496&lt;/code&gt;, &lt;code&gt;8128&lt;/code&gt;... I'm gonna count this one as &quot;solved correctly, but too blind&quot;. I've shortened the way out correspondingly.&lt;br&gt;&#xA; Room 10: &lt;strong&gt;这是一个红色的鲱鱼&lt;/strong&gt;, mean &lt;code&gt;This is a red herring&lt;/code&gt; in Chinese.&lt;br&gt;&#xA; Room 12: &lt;strong&gt;Z&lt;/strong&gt;, I believe this is just room &lt;code&gt;26&lt;/code&gt; (alphabet and stuff).&lt;br&gt;&#xA; Room 12: &lt;strong&gt;naM hsaM&lt;/strong&gt;, &quot;Mash Man&quot; is game number &lt;code&gt;37&lt;/code&gt; in &lt;a href=&quot;http://en.wikipedia.org/wiki/Action_52&quot;&gt;Action 52&lt;/a&gt;.&lt;br&gt;&#xA; Room 12: &lt;strong&gt;ロコン&lt;/strong&gt;, is the Japanese name of Pokémon &lt;a href=&quot;http://bulbapedia.bulbagarden.net/wiki/Vulpix_(Pok%C3%A9mon)&quot;&gt;Vulpix&lt;/a&gt;, with Pokédex number &lt;code&gt;37&lt;/code&gt;.&lt;br&gt;&#xA; Room 12: &lt;strong&gt;Pidgeotto&lt;/strong&gt;, is also a Pokémon, &lt;a href=&quot;http://bulbapedia.bulbagarden.net/wiki/Pidgeotto_(Pok%C3%A9mon)&quot;&gt;with Pokédex number&lt;/a&gt; &lt;code&gt;17&lt;/code&gt;.&lt;br&gt;&#xA; Room 12: &lt;strong&gt;21120110122101211011201112&lt;/strong&gt;, This is &lt;a href=&quot;http://en.wikipedia.org/wiki/Morse_code&quot;&gt;Morse code&lt;/a&gt;. &lt;code&gt;0&lt;/code&gt; are separators, &lt;code&gt;1&lt;/code&gt; short and &lt;code&gt;2&lt;/code&gt; long signals. It spells &lt;code&gt;XIPLUV&lt;/code&gt; which is &lt;code&gt;VULPIX&lt;/code&gt; backwards. So this is another &lt;code&gt;37&lt;/code&gt;.&lt;br&gt;&#xA; Room 13: &lt;strong&gt;D&lt;/strong&gt;, that looks like a &lt;code&gt;Bow&lt;/code&gt; to me!&lt;br&gt;&#xA; Room 13: &lt;strong&gt;the same door that you used to come in&lt;/strong&gt;, I think only &lt;code&gt;24&lt;/code&gt; has a door leading to 13.&lt;br&gt;&#xA; Room 18: &lt;strong&gt;Iron room&lt;/strong&gt;, new guess: this is just &lt;code&gt;26&lt;/code&gt; (the atomic number of &lt;a href=&quot;http://en.wikipedia.org/wiki/Iron&quot;&gt;Iron&lt;/a&gt;) in parallel to the &lt;em&gt;Argon&lt;/em&gt; hint.&lt;br&gt;&#xA; Room 18: &lt;strong&gt;the room where you started&lt;/strong&gt;, is &lt;code&gt;37&lt;/code&gt;.&lt;br&gt;&#xA; Room 18: &lt;strong&gt;the first prime&lt;/strong&gt;, is &lt;code&gt;2&lt;/code&gt;.&lt;br&gt;&#xA; Room 21: &lt;strong&gt;Красный ключ&lt;/strong&gt;, is Russian for &lt;code&gt;Red key&lt;/code&gt; (you have no idea how proud I was to be able to read that myself, despite my very rudimentary Russian lessons being some 6-10 years in the past :D). &lt;br&gt;&#xA; Room 21: &lt;strong&gt;square of the first prime&lt;/strong&gt;, is &lt;code&gt;4&lt;/code&gt;.&lt;br&gt;&#xA; Room 21: &lt;strong&gt;room with the pills&lt;/strong&gt;, is room &lt;code&gt;12&lt;/code&gt;.&lt;br&gt;&#xA; Room 21: &lt;strong&gt;0x30&lt;/strong&gt;, is hexadecimal notation for &lt;code&gt;48&lt;/code&gt;.&lt;br&gt;&#xA; Room 21: &lt;strong&gt;room made by repeated numbers&lt;/strong&gt;, I believe &quot;numbers&quot; refers to the &quot;digits&quot; in the room number, so this would be &lt;code&gt;44&lt;/code&gt;.&lt;br&gt;&#xA; Room 23: &lt;strong&gt;◓●◒◒◙ ◔◈◉◓ ◉◓ ◁ ◒◅◄ ◈◅◒◒◉◎◇&lt;/strong&gt;, Subtracting 9600 from each Unicode code point yields &lt;code&gt;SORRY THIS IS A RED HERRING&lt;/code&gt;.&lt;br&gt;&#xA; Room 23: &lt;strong&gt;@@@@@@@@&lt;/strong&gt;, There's 8 of them, so that's just room &lt;code&gt;8&lt;/code&gt;.&lt;br&gt;&#xA; Room 23: &lt;strong&gt;XLVIII&lt;/strong&gt;, Roman numeral for &lt;code&gt;48&lt;/code&gt;.&lt;br&gt;&#xA; Room 24: &lt;strong&gt;Monica, Erica, Rita, Tina, Sandra, Mary, Jessica&lt;/strong&gt;, is from &lt;a href=&quot;http://en.wikipedia.org/wiki/Mambo_No._5&quot;&gt;Mambo No.&lt;/a&gt; &lt;code&gt;5&lt;/code&gt; (as a German, I didn't even have to look that up thanks to Lou Bega's cover^^).&lt;br&gt;&#xA; Room 24: &lt;strong&gt;bad luck room&lt;/strong&gt;, this has got to be &lt;code&gt;13&lt;/code&gt;, being the unlucky number.&lt;br&gt;&#xA; Room 26: &lt;strong&gt;Butterfree&lt;/strong&gt;, yet another Pokémon &lt;a href=&quot;http://bulbapedia.bulbagarden.net/wiki/Butterfree_(Pok%C3%A9mon)&quot;&gt;with Pokédex number&lt;/a&gt; &lt;code&gt;12&lt;/code&gt;.&lt;br&gt;&#xA; Room 26: &lt;strong&gt;neetneves&lt;/strong&gt;, is &lt;code&gt;17&lt;/code&gt; backwards.&lt;br&gt;&#xA; Room 26: &lt;strong&gt;iron room&lt;/strong&gt;, so if my guess above is right, then this door leads to &lt;code&gt;26&lt;/code&gt; itself.&lt;br&gt;&#xA; Room 28: &lt;strong&gt;room of that girls&lt;/strong&gt;, I think this refers back to the hint for Mambo No. &lt;code&gt;5&lt;/code&gt;.&lt;br&gt;&#xA; Room 30: &lt;strong&gt;THE answer&lt;/strong&gt;, should be &lt;a href=&quot;https://www.google.co.uk/search?q=the+answer+to+life%2C+the+universe+and+everything&amp;amp;oq=the+answer+to+life%2C+the+universe+and+everything&amp;amp;aqs=chrome..69i57j69i64l3.255j0j7&amp;amp;sourceid=chrome&amp;amp;es_sm=0&amp;amp;ie=UTF-8&quot;&gt;the answer to life, the universe and everything&lt;/a&gt;: &lt;code&gt;42&lt;/code&gt;.&lt;br&gt;&#xA; Room 34: &lt;strong&gt;Mg&lt;/strong&gt;, is symbol for &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnesium&quot;&gt;Magnesium&lt;/a&gt;, with atomic number &lt;code&gt;12&lt;/code&gt;.&lt;br&gt;&#xA; Room 34: &lt;strong&gt;multiple of 11&lt;/strong&gt;, that's &lt;code&gt;44&lt;/code&gt;.&lt;br&gt;&#xA; Room 37: &lt;strong&gt;Argon&lt;/strong&gt;, has &lt;a href=&quot;http://en.wikipedia.org/wiki/Argon&quot;&gt;atomic number&lt;/a&gt; &lt;code&gt;18&lt;/code&gt;.&lt;br&gt;&#xA; Room 37: &lt;strong&gt;Micro Mike&lt;/strong&gt;, another game from Action 52, with number &lt;code&gt;24&lt;/code&gt;.&lt;br&gt;&#xA; Room 40: &lt;a href=&quot;http://knowyourmeme.com/memes/the-cake-is-a-lie&quot;&gt;The &lt;strong&gt;cake&lt;/strong&gt; is a lie.&lt;/a&gt; There's no way to get to this room, and therefore the &lt;strong&gt;Yt4$)&lt;/strong&gt; likely doesn't mean anything at all.&lt;br&gt;&#xA; Room 42: &lt;strong&gt;X&lt;/strong&gt;, I thought this could be 24 as the 24th letter of the alphabet, or it could be &lt;code&gt;10&lt;/code&gt; as a Roman numeral. I went with the latter, because I didn't find another way into room 10, and because the &lt;a href=&quot;http://en.wikipedia.org/wiki/Tiber&quot;&gt;Tiber&lt;/a&gt; runs through Rome (then again, the alphabet in question &lt;em&gt;is&lt;/em&gt; the Latin alphabet ;)).&lt;br&gt;&#xA; Room 44: &lt;strong&gt;room with the largest number&lt;/strong&gt;, is &lt;code&gt;63&lt;/code&gt;.&lt;br&gt;&#xA; Room 44: &lt;strong&gt;first room with a number greater than this one&lt;/strong&gt;, is &lt;code&gt;48&lt;/code&gt;.&lt;br&gt;&#xA; Room 56: &lt;strong&gt;the starting room&lt;/strong&gt;, is &lt;code&gt;37&lt;/code&gt;.&lt;br&gt;&#xA; Room 56: &lt;strong&gt;the room whose number is the first two-digit room number appearing in π&lt;/strong&gt;, is &lt;code&gt;59&lt;/code&gt;.&lt;br&gt;&#xA; Room 59: &lt;strong&gt;TERRA&lt;/strong&gt;, is a &lt;a href=&quot;http://en.wikipedia.org/wiki/ROT13&quot;&gt;ROT-13 cypher&lt;/a&gt; of &quot;GREEN&quot; so this is the &lt;code&gt;Green key&lt;/code&gt;.&lt;br&gt;&#xA; Room 59: &lt;strong&gt;dezessete&lt;/strong&gt;, Portuguese for &lt;code&gt;17&lt;/code&gt;.&lt;br&gt;&#xA; Room 59: &lt;strong&gt;dezoito&lt;/strong&gt;, Portuguese for &lt;code&gt;18&lt;/code&gt;.&lt;br&gt;&#xA; Room 63: &lt;strong&gt;room with repeated numbers&lt;/strong&gt;, again, I believe this refers to the digits, so that's &lt;code&gt;44&lt;/code&gt;.&lt;br&gt;&#xA; Room 63: &lt;strong&gt;0x4&lt;/strong&gt;, should be hexadecimal notation, which is just &lt;code&gt;4&lt;/code&gt;.&lt;br&gt;&#xA; Blue box: &lt;strong&gt;84 69 76 69 80 79 82 84 69 82&lt;/strong&gt;, ASCII codes of &lt;code&gt;TELEPORTER&lt;/code&gt;.&lt;br&gt;&#xA; Bullseye: &lt;strong&gt;1A&lt;/strong&gt;, I'm hoping is also hexadecimal, which would be &lt;code&gt;26&lt;/code&gt;.&lt;br&gt;&#xA; Bullseye: &lt;strong&gt;五十六&lt;/strong&gt;, Japanese (I think?) for &lt;code&gt;56&lt;/code&gt;.&lt;br&gt;&#xA; Yellow button: &lt;strong&gt;ending with '3s2 3p6'&lt;/strong&gt;, could refer to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Electron_configuration&quot;&gt;electron configuration&lt;/a&gt; of &lt;em&gt;Argon&lt;/em&gt;, which would be &lt;code&gt;18&lt;/code&gt; again.&lt;br&gt;&#xA; Yellow button: &lt;strong&gt;◔◈◅ ◌●◒◄ ◉◓ ◍◙ ◓◈◅◐◅◒◄, ◉ ◌◁◃○ ◎●◔◈◉◎◇&lt;/strong&gt;, subtracting 9600 from the Unicode code points gives &lt;code&gt;THE LORD IS MY SHEPERD, I LACK NOTHING&lt;/code&gt; which is in &lt;a href=&quot;http://en.wikipedia.org/wiki/Psalm_23&quot;&gt;Psalm&lt;/a&gt; &lt;code&gt;23&lt;/code&gt;.&lt;br&gt;&#xA; Yellow button: &lt;strong&gt;NPPSTVPJWFSQFIUGPGMBITJSFCNVOFIUFTPIX&lt;/strong&gt;, decrement each letter by 1 and reverse the string to get &lt;code&gt;WHOSETHENUMBERISHALFOFTHEPREVIOUSROOM&lt;/code&gt;. Since the button is in room 10, and (I think) the only way to get there is from 42, this is &lt;code&gt;21&lt;/code&gt;.&lt;br&gt;&#xA; Magic book: &lt;strong&gt;. ορετόσσιρεπ ιχό ιεσάρδ αν ιεπέρπ αθ ςασ ήγρΟ ! άρεν ανέμγαρατ ατ όπα ωνάπ υομ ήμανύδ ητ ιαμύολακιπε , ςανώδιεσοΠ .&lt;/strong&gt; Reverse this to get &lt;em&gt;actual&lt;/em&gt; Greek. It translates to &lt;code&gt;Neptune, I invoke my power over troubled waters! Rage you should act no more.&lt;/code&gt;, so the book can be used to tame the whirlpool. I swear this is a reference to &lt;a href=&quot;http://en.wikipedia.org/wiki/The_Tempest&quot;&gt;The Tempest&lt;/a&gt;.&lt;br&gt;&#xA; Rainbow: &lt;strong&gt;Except if...?&lt;/strong&gt;, you've got a parachute...&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Phew... that's out of the way. Let's get out of this dungeon!&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;The Long Way Out&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First, I think &lt;code&gt;44&lt;/code&gt; is a dead end, because it leads to &lt;code&gt;48&lt;/code&gt; (instant death), and &lt;code&gt;63&lt;/code&gt;, which in turn leads only back to &lt;code&gt;44&lt;/code&gt; or to &lt;code&gt;4&lt;/code&gt; (instant death). Furthermore, you don't need either glowing crystal key, nor any money. Likewise, &lt;code&gt;1&lt;/code&gt; is probably a red herring, I don't think there's a way to get to it. Furthermore, there's no evil wizard, and the puzzle doesn't require me to rescue a princess. &lt;code&gt;40&lt;/code&gt; also cannot be reached.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Let's go!&lt;br&gt;&#xA;&lt;br&gt;&#xA; 37 -&gt; Go to 18&lt;br&gt;&#xA; 18 -&gt; Go to 26&lt;br&gt;&#xA; 26 -&gt; Take gun, go to 26. &lt;em&gt;(because I can!)&lt;/em&gt;&lt;br&gt;&#xA; 26 -&gt; Go to 26. &lt;em&gt;(because I can!)&lt;/em&gt;&lt;br&gt;&#xA; 26 -&gt; Go to 26. &lt;em&gt;(because I can!)&lt;/em&gt;&lt;br&gt;&#xA; 26 -&gt; Go to 12.&lt;br&gt;&#xA; 12 -&gt; Red pill to 37.&lt;br&gt;&#xA; 37 -&gt; Go to 24.&lt;br&gt;&#xA; 24 -&gt; Kill lion with gun, go to 5&lt;br&gt;&#xA; 5 -&gt; Go to 34&lt;br&gt;&#xA; 34 -&gt; Take blue key, go to 12&lt;br&gt;&#xA; 12 -&gt; Blue pill to 37.&lt;br&gt;&#xA; 37 -&gt; Unlock blue box, take batteries, go to 24&lt;br&gt;&#xA; 24 -&gt; Go to 5.&lt;br&gt;&#xA; 5 -&gt; Plug batteries in teleporter, go to 30&lt;br&gt;&#xA; 30 -&gt; Go to 42&lt;br&gt;&#xA; 42 -&gt; Go to 10&lt;br&gt;&#xA; 10 -&gt; Take arrow, press yellow (opens door in 18 to 23), go to 21&lt;br&gt;&#xA; 21 -&gt; Take red key, go to 12&lt;br&gt;&#xA; 12 -&gt; White pill to 37&lt;br&gt;&#xA; 37 -&gt; Go to 24&lt;br&gt;&#xA; 24 -&gt; Unlock red door, go to 13&lt;br&gt;&#xA; 13 -&gt; Take the bow, go to 24&lt;br&gt;&#xA; 24 -&gt; Go to 5&lt;br&gt;&#xA; 5 -&gt; Use teleporter to go to 30&lt;br&gt;&#xA; 30 -&gt; Shoot bullseye with bow and arrow, to open trapdoor in 26 to 56, go to 42&lt;br&gt;&#xA; 42 -&gt; Go to 10&lt;br&gt;&#xA; 10 -&gt; Go to 21&lt;br&gt;&#xA; 21 -&gt; Go to 12&lt;br&gt;&#xA; 12 -&gt; Door to 26&lt;br&gt;&#xA; 26 -&gt; Use trapdoor to 56&lt;br&gt;&#xA; 56 -&gt; Take Magic Book, go to 37&lt;br&gt;&#xA; 37 -&gt; Go to 18&lt;br&gt;&#xA; 18 -&gt; Secret passage to 23.&lt;br&gt;&#xA; 23 -&gt; Take magic bean, go to 8.&lt;br&gt;&#xA; 8 -&gt; Use Magic Book, go to 28&lt;br&gt;&#xA; 28 -&gt; Take pot of gold, go to 5&lt;br&gt;&#xA; 5 -&gt; Go to 34&lt;br&gt;&#xA; 34 -&gt; Go to 12&lt;br&gt;&#xA; 12 -&gt; Door to 26&lt;br&gt;&#xA; 26 -&gt; Trapdoor to 56&lt;br&gt;&#xA; 56 -&gt; This time go to 59&lt;br&gt;&#xA; 59 -&gt; Take green key, go to 18&lt;br&gt;&#xA; 18 -&gt; Unlock green door, go to 2&lt;br&gt;&#xA; 2 -&gt; Take Parachute, go to 8&lt;br&gt;&#xA; 8 -&gt; Use Magic Book, go to 28&lt;br&gt;&#xA; 28 -&gt; Go to 5&lt;br&gt;&#xA; 5 -&gt; Use teleporter to go to 30&lt;br&gt;&#xA; 30 -&gt; Go to 42&lt;br&gt;&#xA; 42 -&gt; Give gold to gnome&lt;br&gt;&#xA; Rainbow -&gt; Use parachute to survive the fall and land somewhere safely outside the dungeon!&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Woohoo!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe there's a shorter way through the dungeon (especially the second half), but this one does the trick. Thanks to Victor for a couple of hints when I was making things more complicated then they were. :)&lt;/p&gt;&#xA;" OwnerUserId="2450" LastEditorUserId="2450" LastEditDate="2014-11-15T23:44:18.253" LastActivityDate="2014-11-15T23:44:18.253" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4768" PostTypeId="2" ParentId="4761" CreationDate="2014-11-15T19:44:41.513" Score="11" Body="&lt;p&gt;This is a fantastic puzzle, one of the best I have had to solve, so give yourself time to think it over. It took me a month to come to the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make you guys understand the solution, I will first demonstrate it in a particular case where there are just two colors. I will then expand it to N colors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Preliminary logical deduction:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The fact that the prisoners are in line adds nothing to the puzzle. Once the first prisoner has talked, there is no difference in being the 9th prisoner to talk or the 11th, or the last: they will all hear what has been said before, and they can all see the colors of the hats of the prisoners ahead of them in the line. As we know that the 99 are saved, they all have the same information: the color the first prisoner said, and the colors of the 98 other prisoners' hat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Particular case:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  So let's first assume that there are only black and white hats. The prisoners agree that they will count 0 for black hats, and 1 for white hats. The first prisoner quickly counts the PARITY of the black hats, and says WHITE if it's even, BLACK if it's odd. Now all 99 prisoners can deduce the colors of their hat by counting the parity of black hats in front and behind them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;General case:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It gets a bit trickier with N colors, but the same idea leads to solution. The prisoners associate a number to each of the color. 0 for green, 1 for red, 2 for blue, etc. The first prisoner computes the sum of all the colors he sees, and the modulo of that number to N. He announces that number as a color. Now every prisoner can deduce the color of their hat, by computing the sum of the other prisoners' hat color and the modulo to N. Then if the number anounced by the first prisoner is 24 and I compute 22, I know that my color is blue, because adding 2 to 22 would give me 24, the number anounced by the first prisoner.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5518" LastEditorUserId="5518" LastEditDate="2014-11-16T10:39:13.340" LastActivityDate="2014-11-16T10:39:13.340" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4769" PostTypeId="1" AcceptedAnswerId="4770" CreationDate="2014-11-15T19:50:59.343" Score="1" ViewCount="297" Body="&lt;p&gt;A party is being held at a local mansion. The host is very rich and his success is because of one thing — his famous recipe for spaghetti!  Tonight he is having a special gala event to raise funds for the local zoo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only guests that may attend are people who correctly reply to the guard at the door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As usual, you are trying to break in to crash the party, steal the spaghetti recipe, and may even catch a glimpse of the rare horn-studded snipe that the host is rumored to be keeping as a pet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pair of you hide by the entrance to the mansion at 6pm and wait for the other guests to arrive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first guest arrives, and the guard says: &quot;one six six, who are you?&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The guest thinks for a moment, then says, &quot;I'm a peacock!&quot; and the guard lets him in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second guest arrives, and the guard says: &quot;five ten three, who are you?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The guest thinks for a moment, then says, &quot;I'm a giraffe!&quot; and the guard lets him in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third guest arrives, and the guard says: &quot;ten four nine, who are you?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The guest thinks for a moment, then says, &quot;Give me a second, this is a hard one.&quot;  Then after thinking a bit more, she replies, &quot;I'm a rhinocerous!&quot; and the guard lets her in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fourth guest arrives, and the guard says: &quot;five eleven seven, who are you?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The guest replies, &quot;I'm a gazelle!&quot; and the guard lets her in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fifth guest arrives, and the guard says: &quot;twelve ten two four five, who are you?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The guest thinks, then replies, &quot;I'm a warthog!&quot; and the guard lets him in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your friend says, &quot;I think I've got this!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yeah?&quot; you ask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yes, don't you see, they are all animals!  Like at the zoo!&quot; and before you can get a better explanation, she jumps out of the bushes and approaches the guard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;eleven four eight one nine eleven, who are you?&quot; asks the guard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I'm a herd of sheep!&quot; replies your friend, and suddenly a speeding Mac truck roars by, flattening your friend into a pancake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The party is about to start, so you quickly come out to face the guard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;seven one ten two, who are you?&quot; asks the guard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you reply?  Please put your answer under spoiler tags.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your friend should have answered &quot;Chimpanzee&quot;. Although &quot;eleven four eight one&quot; would also elicit the same response.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2718" LastEditorUserId="2718" LastEditDate="2014-11-15T21:48:04.213" LastActivityDate="2014-11-15T21:48:04.213" Title="Security to the Party [part 33]" Tags="|password|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4770" PostTypeId="2" ParentId="4769" CreationDate="2014-11-15T20:23:50.657" Score="2" Body="&lt;p&gt;I think it is a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;L&lt;/strong&gt;eo&lt;strong&gt;p&lt;/strong&gt;a&lt;strong&gt;rd&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each number seems a letter or sound. Not every letter/sound of the animal is represented, but the key letters/sounds are. &lt;br&gt;The L comes from the 7 of the fourth guest. The P is the P of one of the first guest. Ten is the R of rhinoceros. Two comes only once, and is the 't' in warthog. That sounds a bit like a 'd' so I end up with lprd, so the animal is Leopard!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and more explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The guests and host agreed on a code before the party. For example, look at the five: every time there is a g in the correct place. So five is definitely a 'g'. Also the ten, which is a 'r', occurs a couple of times. I am pretty confident that something like this is the answer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;bonus what the friend should have answered:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Seahorse? I am still not sure, because the one is a 'p', but it fits the other letters. To fit that 'p', I can try scorpion (or scorpions to fit the last eleven), but that the 'h' is not in the running. And four is a 'h', because of warthog. So I'm not sure about that one:P&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3993" LastEditorUserId="3993" LastEditDate="2014-11-15T21:35:52.643" LastActivityDate="2014-11-15T21:35:52.643" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4771" PostTypeId="1" AcceptedAnswerId="4775" CreationDate="2014-11-15T21:13:04.307" Score="6" ViewCount="569" Body="&lt;p&gt;This should be an easy one if you have a Bible handy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My name is Kether Torah,&lt;br&gt;&#xA;  the Crown of the Torah.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I am a pillar of righteousness,&lt;br&gt;&#xA;  and a foundation of mercy.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I am four, and three, and two, and one;&lt;br&gt;&#xA;  and beside me there is none other.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Those who hear me have life;&lt;br&gt;&#xA;  they that disobey me shall surely die.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="5474" LastEditorUserId="7586" LastEditDate="2016-05-30T00:35:25.907" LastActivityDate="2018-07-12T04:06:48.350" Title="The Crown of the Torah" Tags="|riddle|knowledge|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4772" PostTypeId="2" ParentId="4341" CreationDate="2014-11-15T21:19:43.683" Score="6" Body="&lt;p&gt;The murderer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RpiAwesomeness&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Exhibit A:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/SKHcZ.png&quot; alt=&quot;What Waaaaaat said&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Exhibit B:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/ko0x2.png&quot; alt=&quot;What Rpi replied&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I feel like this is not the evidence I was meant to find though. My detailed report:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As I pondered the clues laid out before me, my admiration for Doorknob continued to swell. A bit of searching was all he needed?! I made a list of open questions: Why was the window ajar? What was the loud crash? Why did the lights go out? How did the killer find time to give him that tattoo before the pizza guy showed up? I dreamed up some answers, and started to get inside the mind of the killer. I shifted my thoughts to what I knew about him. He's a man. He's smart enough to wear a mask and gloves. He's a bit dramatic. And he knows how to give tattoos... Aha! Something to make his stand out from the crowd! I headed over to AskUbuntu and searched the user profiles for a tattoo artist. But my search was in vain. I found only SkyDivingPanda, whose low rep made me doubt he would have the passion and connections to commit this crime, and Wahyaohni, whose website you should definitely not click on. Still &quot;tattoo&quot; was the only foothold I could find in the narrative, so I pressed on. I started searching the chat logs. And I found this:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Exhibit C&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/Ti8YL.png&quot; alt=&quot;Waaaaaat&amp;#39;s suspiciously timed tattoo reference&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The report continues:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Almost enough to put Waaaaaat behind bars, but given the very short tattoo timeframe, I couldn't take a chance that the judge would rule that circumstantial. I dug deeper into the chat logs. In Ask Ubuntu General, I found Seth being thanked for his &quot;help&quot;, and he joined my suspect list. There was much discussion about a secret chat that may or may not have existed, and I feared that secret room, deleted by Seth, might have been the only real trail. I searched and searched for a careless one-boxed link. But I came up empty. Suddenly, I noticed that Waaaaaat had name-dropped the Teachers' Lounge! Off I went to find it, and what I found... made me feel like a noob. Another dead end. &quot;I need to clear my mind&quot;, I thought, and went to do some dishes. I was back in seconds, my dishes still unwashed. I could search the chat logs for other suspicious words, like &quot;killer&quot;! And there I found it, the name of the true culprit, and his own cold agreement to do the deed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Carved into the handle of the gun:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Ask Ubuntu logo. Or maybe the Ubuntu logo.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I just don't know why there are so many bullets!&lt;/p&gt;&#xA;" OwnerUserId="5500" LastEditorUserId="2135" LastEditDate="2014-11-16T04:41:36.783" LastActivityDate="2014-11-16T04:41:36.783" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="4773" PostTypeId="2" ParentId="4771" CreationDate="2014-11-15T21:29:08.550" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Headstone?&lt;br&gt;&#xA; Some jewish people believe a good name is what is superior, &quot;one crown of the torah&quot; which a headstone has inscribed&lt;br&gt;&#xA;a pillar of righteousness as we honor those who fall before us and mercy on their deaths.&lt;br&gt;&#xA; an equestrian headstone has meanings depending on how many legs (4) it has touching the ground.&lt;br&gt;&#xA;there is none other that is the same as each is different inscriptions and no one is beside as they are deceased&lt;br&gt;&#xA;those who keep them have life as there not planted yet or not sold&lt;br&gt;&#xA; if you violate your &quot;head&quot; you shall surely die.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5135" LastActivityDate="2014-11-15T21:29:08.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4774" PostTypeId="2" ParentId="4750" CreationDate="2014-11-15T21:38:40.847" Score="7" Body="&lt;p&gt;The problem with this type of puzzle is that you can match any dataset using a polynomial of sufficiently high degree. For this one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x)=-0.225x^4+25.425x^3-300.7x^2+824.5x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;giving&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(75)=1977431.25$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course you probably have a different formula, but from the point of view answering the question we have no way of knowing which of the infinite number of formulae matching this data is the one you had in mind&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2014-11-15T21:38:40.847" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4775" PostTypeId="2" ParentId="4771" CreationDate="2014-11-15T21:44:58.897" Score="4" Body="&lt;p&gt;Are you the:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 10 Commandments&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My name is Kether Torah,&#xA;the Crown of the Torah.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This refers to the fact that we are looking into something from Judaism (probably the first 5 books of the Bible)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a pillar of righteousness,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We are looking for something pertaining to morality - could be quite symbolic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and a foundation stone of mercy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Foundation makes me think of &quot;stone&quot;... i.e. a tablet perhaps.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am four, and three, and two, and one;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SURELY some numerical trick. Possibly pointing to a passage. But I notice the sum is 10, and product is 24.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and beside me there is none other.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is said by God in the 10 commandments (the first).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Those who hear me have life;&#xA;those who disobey me shall surely die.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; God says this kind of thing frequently according to the Bible. Always with regards to a threat for disobedience to a command/law.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The evidence for guessing the Ark of the Covenant or the Ten Commandments is strong enough.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Regarding the Kether Torah:&lt;/code&gt;&#xA;Courtesy of OP:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/5wErm.jpg&quot; alt=&quot;Ten Commandments&quot;&gt;&#xA;&lt;P&gt;&#xA; [Picture of the &lt;strong&gt;Ten Commandments&lt;/strong&gt; with the words כתר תורה (&lt;em&gt;Kether Torah&lt;/em&gt;) written prominently above.]&#xA;&lt;P&gt;&#xA; The identification of the Ten Commandments with the &lt;strong&gt;Crown of the Torah&lt;/strong&gt; comes from the ancient Jewish practice of &lt;a href=&quot;http://en.wikipedia.org/wiki/Gematria&quot; rel=&quot;nofollow noreferrer&quot;&gt;gematria&lt;/a&gt;, whereby the numerical value of the words &lt;em&gt;Kether Torah&lt;/em&gt; is equal to that of &lt;em&gt;Aseret Ha-Devarim&lt;/em&gt; (&quot;the Ten Commandments&quot;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="49155" LastEditDate="2018-07-12T04:06:48.350" LastActivityDate="2018-07-12T04:06:48.350" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="4776" PostTypeId="2" ParentId="4057" CreationDate="2014-11-15T22:07:18.473" Score="23" Body="&lt;p&gt;Feeling good about this one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 7&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is based on the ones value of the year that the corresponding Star Wars episode was released: &lt;br /&gt; (&lt;a href=&quot;http://www.imdb.com/find?q=star+wars+episode&amp;amp;&amp;amp;s=tt&amp;amp;ref_=fn_al_tt_mr&quot;&gt;From IMDB&lt;/a&gt;) &lt;br /&gt; Episode I: The Phantom Menace  - 199&lt;strong&gt;9&lt;/strong&gt; &lt;br /&gt; Episode II: Attack of the Clones - 200&lt;strong&gt;2&lt;/strong&gt; (Apparently the guest wasn't a fan of the prequels) &lt;br /&gt; Episode III: Revenge of the Sith - 200&lt;strong&gt;5&lt;/strong&gt; &lt;br /&gt; Episode IV: A New Hope - 197&lt;strong&gt;7&lt;/strong&gt; &lt;br /&gt; Episode V: The Empire Strikes Back - 198&lt;strong&gt;0&lt;/strong&gt;  &lt;br /&gt; Episode VI: Return of the Jedi - 198&lt;strong&gt;3&lt;/strong&gt; &lt;br /&gt; &lt;br /&gt; I'll note here that one section of the main Star Wars Wikipedia page says Episode III was released in 2006, which almost ruined my theory, but I'm glad I double-checked!  &lt;br /&gt; &lt;br /&gt;Ohhh, now I'm seeing more hints related to each episode embedded. Return, menace, back, attack (and clones), revenge, and hope. Very nicely done!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Guesses for what each song in the soundtrack refers to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Royale With Cheese&lt;/em&gt; - Mr. Samuel L. Jackson &lt;br /&gt; &lt;em&gt;20th Century Fox&lt;/em&gt; - The name of the distributor &lt;br /&gt; &lt;em&gt;Mr Tambourine Man&lt;/em&gt; - American Graffiti/George Lucas &lt;br /&gt; &lt;em&gt;Out of Space&lt;/em&gt; - Space &lt;br /&gt; &lt;em&gt;Space Oddity&lt;/em&gt; - Space &lt;br /&gt; &lt;em&gt;Indiana Jones Theme&lt;/em&gt; - George Lucas and John Williams &lt;br /&gt; &lt;em&gt;Space Cowboy&lt;/em&gt; - Space &lt;br /&gt; &lt;em&gt;The Galaxy Song&lt;/em&gt; - Space (and a reference to &quot;a long time ago in a galaxy far, far away&quot;) &lt;br /&gt; &lt;em&gt;Girl From Mars&lt;/em&gt; - Space (and 1977) &lt;br /&gt; &lt;em&gt;Across the Universe&lt;/em&gt; - Space &lt;br /&gt; &lt;em&gt;Choose Life&lt;/em&gt; - ??? (Maybe the future? &quot;Choose your future&quot;)  &lt;br /&gt; &lt;em&gt;Born To Be Wild (With Miss Piggy)&lt;/em&gt; - &lt;a href=&quot;http://en.wikipedia.org/wiki/Frank_Oz&quot;&gt;Frank Oz, creator and performer of Miss Piggy and the puppeteer and voice of Yoda!&lt;/a&gt; &lt;br /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first (solved) clue refers to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (movie spoilers!) Luke and Leia&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second clue (thanks, nexolute) refers to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (more movie spoilers!) Darth Vader being Luke's father&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5340" LastEditorUserId="5340" LastEditDate="2014-11-17T22:59:04.597" LastActivityDate="2014-11-17T22:59:04.597" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="4777" PostTypeId="1" AcceptedAnswerId="4783" CreationDate="2014-11-15T22:49:38.807" Score="0" ViewCount="552" Body="&lt;p&gt;This is an easy one - a more interesting one is in the works!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A party is being held at a local mansion. The host is very rich and his success is because of one thing - his famous recipe for spaghetti.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only guests that may attend are people who correctly reply to the guard at the door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are trying to gatecrash the party in order to steal the recipe. You sneak by and listen to the passwords, hoping to guess the pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first guest arrives. The guard says &quot;3&quot;. The guest replies &quot;5&quot; and is admitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second guest arrives. The guard says &quot;1&quot;. The guest replies &quot;2&quot; and is admitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third guest arrives. The guard says &quot;12&quot;. The guest replies &quot;9&quot; and is thrown into a tank of red herrings to be devoured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fourth guest arrives. The guard says &quot;21&quot;. The guest replies &quot;7&quot; and is admitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You walk up and get a &quot;20&quot; from the guard. What should you say?&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2014-11-16T12:24:07.027" Title="The Security to the Party [Part 34]" Tags="|password|" AnswerCount="1" CommentCount="7" ClosedDate="2014-11-16T19:35:30.837" ContentLicense="CC BY-SA 3.0" />
  <row Id="4778" PostTypeId="2" ParentId="4761" CreationDate="2014-11-16T00:13:36.800" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well, each prisoner just asked the others what color they hat was. Only the first prisoner had to guess, because the others hate him and didn't want to tell him his hat color.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They just took their hat off, looked at it and then put it back. Only the first prisoner could not do that because he was disabled.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Someone had to mention the obvious answers...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4721" LastActivityDate="2014-11-16T00:13:36.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4780" PostTypeId="2" ParentId="4747" CreationDate="2014-11-16T01:48:04.440" Score="5" Body="&lt;p&gt;Unfortunately there is no way for Alice or Bob to determine the individuals times A and B, without outside communication or knowledge.  The best they can do is determine the value of A+B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To illustrate why, let us first consider one approach one might take.  One could allow Alice to send her time to Bob, and Bob to immediately send his time to Alice, and back and forth.  In the following table I have assumed the clocks are out of synch by &lt;strong&gt;8 units, and A = 2 and B = 5.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Table of Scenario 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hEkkg.png&quot; alt=&quot;enter image description here&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &quot;Y(x)&quot; in the diagram represents a signal was &lt;strong&gt;received&lt;/strong&gt;, where &quot;x&quot; is the value of the signal received.  The initial &quot;Y&quot; at time 0 is when Alice &lt;strong&gt;sends&lt;/strong&gt; her first signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now from the above table, Alice can easily deduce A+B = 7, and Bob can do the same.  Now clearly there is no point in this process continuing further, because Bob can already now deduce what his next received signals will be and the times he will receive them.  E.g. Y(14) at 24 seconds, then Y(21) at 31 seconds etc., and Alice can deduce what her received signals will and the times she will receive them.  E.g. Y(17) at 14 seconds, then Y(24) and 21 seconds etc. So no more information can be exchanged between the pair.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the same scenario but where the clocks are out of synch by &lt;strong&gt;7 units, and A = 3 and B = 4.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following table shows the signals received and the times they are received. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Table of Scenario 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uxoo9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The important thing to notice here is that despite a difference in time synchronization and a difference in the values for A and B, the signals are all sent and received at identical times to previous example, with identical values.  That is, if Alice and Bob receive signals in such a pattern, there is no way they can determine if they are in Scenario 1 or Scenario 2 or many other scenarios for that matter.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while this isn't the only strategy Alice and Bob could attempt, it is clear that this strategy shares maximal knowledge between Alice and Bob - that is, Alice knows everything Bob knows and Bob knows everything Alice knows.  Therefore no other strategy will be able to transfer additional information to derive the values of A and B individually.&lt;/p&gt;&#xA;" OwnerUserId="97" LastActivityDate="2014-11-16T01:48:04.440" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4781" PostTypeId="2" ParentId="4360" CreationDate="2014-11-16T02:07:08.390" Score="6" Body="&lt;p&gt;The number under the box is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 16728&lt;br&gt;&#xA; this number is 26 - (position in alphabet, with  A = 0 and Z = 25).  For example, 26 - Z = 1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user5521" LastActivityDate="2014-11-16T02:07:08.390" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4783" PostTypeId="2" ParentId="4777" CreationDate="2014-11-16T04:02:10.480" Score="5" Body="&lt;p&gt;The logic seems to be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;counting the number of segments required to show the numbers in a digital board&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Going by the above logic 20 requires 2 and 0 - 5 segments needed for 2 and 6 for 0.  Totalling to 11 as the answer &lt;/p&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="2313" LastEditDate="2014-11-16T12:24:07.027" LastActivityDate="2014-11-16T12:24:07.027" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4784" PostTypeId="1" AcceptedAnswerId="4815" CreationDate="2014-11-16T06:40:33.403" Score="10" ViewCount="620" Body="&lt;p&gt;Your roommate is on vacation, and you need to get something off his laptop. You ring him up, asking if you can borrow his computer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Sure, if you can get in,&quot; he replies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;What? But it's your laptop!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Give me a break, man. You know how my memory is. Tell you what — look through the second drawer of the desk and you might find a clue.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You thank your friend, and begin rummaging through his stuff. You don't feel very nice doing so, of course, but it's getting late and you need those files. To your dismay, all you can find is the left half of a torn note:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   edi&#xA;   est&#xA;   ser&#xA;   ing&#xA;restau&#xA;    on&#xA;   myo&#xA;   mus&#xA;   hot&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit: Right-aligned the words to make it clearer that you have the &lt;strong&gt;left&lt;/strong&gt; half of the note&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing your friend's naïveté you figure that the password is probably just a single word, and start racking your brain for ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is your friend's password?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each row can be extended to form words with a certain property. Some words are more obscure than others, but &lt;code&gt;restau&lt;/code&gt; does not become &lt;code&gt;restaurant&lt;/code&gt;/&lt;code&gt;restaurants&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="445" LastEditorUserId="445" LastEditDate="2014-11-16T20:58:59.523" LastActivityDate="2014-11-16T21:18:09.010" Title="A Fragmented Puzzle" Tags="|word|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="4785" PostTypeId="1" CreationDate="2014-11-16T06:46:01.850" Score="5" ViewCount="491" Body="&lt;p&gt;I played &lt;a href=&quot;http://www.kongregate.com/games/StudioCime/mu-complex-episode-one&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; puzzle game, the last question of which leads to &lt;a href=&quot;http://mu.database.studiocime.com/&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; website, which contains, hidden as a comment in its HTML, the following string: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;OgUMGU0EHk0ZBQwZTQMCBB4ITSwqLCQjUgo=&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Believing this to be a base64 encoding, I decoded it and got&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;: MM MM,*,$#R&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where the first character is &lt;code&gt;:&lt;/code&gt; and each M is preceded by one or more control characters.  I'm stumped on what to do with the string at this point, and am unsure base64 decoding was the right choice. &lt;/p&gt;&#xA;" OwnerUserId="5527" LastEditorUserId="71" LastEditDate="2014-12-18T13:19:50.463" LastActivityDate="2015-05-02T12:38:58.060" Title="The Final Question of MU episode 1" Tags="|logical-deduction|cipher|encoded-message|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4786" PostTypeId="2" ParentId="4735" CreationDate="2014-11-16T08:42:33.403" Score="10" Body="&lt;p&gt;Going in clockwise, ignoring NW, we get the following...&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;N  : AHOV&#xA;NE : BIPW&#xA;E  : CJQX&#xA;SE : DKRY&#xA;S  : ELSZ&#xA;SW : FMT&#xA;W  : GNU&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And let NW be the EOF character. Now this is where I'm bad with, how can we construct the final solution from these?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using &lt;a href=&quot;http://www.visca.com/regexdict/&quot;&gt;regular expression dictionary&lt;/a&gt;, I didn't manage to find anything, so I reduce to search with less characters and get use a little guess and check to get &lt;strong&gt;PATERNATY DISPUTE&lt;/strong&gt; (should it be patern&lt;strong&gt;i&lt;/strong&gt;ty dispute?). Not so sure if this is right?&lt;/p&gt;&#xA;" OwnerDisplayName="user5508" LastEditorDisplayName="user5508" LastEditDate="2014-11-16T09:13:05.183" LastActivityDate="2014-11-16T09:13:05.183" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4787" PostTypeId="2" ParentId="4784" CreationDate="2014-11-16T08:45:27.197" Score="3" Body="&lt;p&gt;Is the password&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; food&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems too simple to me, but &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4784/a-fragmented-puzzle#comment13910_4784&quot;&gt;you said&lt;/a&gt; to look for a pattern and my guesses are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; edi&lt;strong&gt;ble&lt;/strong&gt;&lt;br&gt;&#xA;est??&lt;br&gt;&#xA;ser&lt;strong&gt;ving&lt;/strong&gt;&lt;br&gt;&#xA;ing&lt;strong&gt;redients&lt;/strong&gt;&lt;br&gt;&#xA;restau&lt;strong&gt;rant&lt;/strong&gt;&lt;br&gt;&#xA;on&lt;strong&gt;ion&lt;/strong&gt;&lt;br&gt;&#xA;m&lt;strong&gt;a&lt;/strong&gt;yo&lt;strong&gt;nnaise&lt;/strong&gt; (I'm guessing that's a typo?)&lt;br&gt;&#xA;mus&lt;strong&gt;tard&lt;/strong&gt;&lt;br&gt;&#xA;hot &lt;strong&gt;sauce&lt;/strong&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5421" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2014-11-16T08:45:27.197" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4790" PostTypeId="1" AcceptedAnswerId="4975" CreationDate="2014-11-16T10:13:18.433" Score="11" ViewCount="1747" Body="&lt;p&gt;&lt;pre&gt;Here is the next one for you. I recommend using a list for this.&#xA;Oh, almost forgot to tell you that it's freaking easy. Have fun!  &lt;/pre&gt;&lt;br&gt;&#xA;Even if tiny I call you tall&lt;br&gt;&#xA;The first two lines contain no hint at all&lt;br&gt;&lt;br&gt;&#xA;For sure I'm egocentric&lt;br&gt;&#xA;You mean nothing to me&lt;br&gt;&#xA;So the best choice seems a third person to be&lt;br&gt;&lt;br&gt;&#xA;Make him steal what surrounds &lt;br&gt;&#xA;&lt;strike&gt;the guy behind the screen&lt;/strike&gt; you&lt;br&gt;&#xA;Find one more glue&lt;br&gt;&#xA;and my full name can be seen&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;If you want to check your answer this might help. The cat in the manger may misuse this as a hint.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Read on with prudence &lt;br&gt; and drop the penny&lt;br&gt;Some say I'm one, I say I'm many&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5106" LastEditorUserId="5106" LastEditDate="2015-01-08T07:02:30.197" LastActivityDate="2015-01-22T13:30:02.113" Title="The egocentric one" Tags="|riddle|word|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4792" PostTypeId="2" ParentId="4046" CreationDate="2014-11-16T10:50:11.390" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's the number of pips in the corners of that face of a standard six-sided die.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2014-11-16T10:50:11.390" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4793" PostTypeId="2" ParentId="4761" CreationDate="2014-11-16T11:27:05.920" Score="2" Body="&lt;p&gt;The question contains trick wording.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is only one hat. At no point does the word &quot;hats&quot; appear in the question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The hat is placed on the head of the person at the back. He guesses incorrectly, so is executed. The executioner then removes the hat from the executed man's head, and places it on the next person's head. By luck (or perhaps because he snuck a glace out of the corner of his eye) the next prisoner guesses right. All the others have the good sense to figure out what's happening, and state the same colour they heard the second prisoner say.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4206" LastActivityDate="2014-11-16T11:27:05.920" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4794" PostTypeId="2" ParentId="4790" CreationDate="2014-11-16T12:04:56.640" Score="0" Body="&lt;p&gt;I'll make a bold guess here :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a server? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;For sure I'm egocentric&#xA;You mean nothing to me&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the server only answers user requests, but usually is not interested in who sent the requests&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;So the best choice seems a third person to be&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the third person might be a browser&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Make him steal what surrounds &#xA;the guy behind the screen&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the browser can collect some information about the user, for example cookies or http form information&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5283" LastActivityDate="2014-11-16T12:04:56.640" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4798" PostTypeId="1" AcceptedAnswerId="4994" CreationDate="2014-11-16T12:32:32.393" Score="7" ViewCount="997" Body="&lt;p&gt;&lt;strong&gt;Chat.&lt;/strong&gt; I've created a &lt;a href=&quot;http://chat.stackexchange.com/rooms/18764/nerd-of-the-loose&quot;&gt;chat room&lt;/a&gt; in case anyone wish to discuss this further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Preface.&lt;/strong&gt; This is the first time I'm creating a puzzle. Since I find enjoyment in science and technology, the series of puzzles I'll be creating will revolve around such topics. I might provide warning as a form of hint when technical knowledge are required. Fret not, it would not be too advance. I hope you will enjoy solving them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Background.&lt;/strong&gt; Professor Swannie Taco is a genius in most fields of science and technology. She is probably hated by her parents, and her name is a perfect evidence. After Swannie went missing a few months ago, strange letters from her kept appearing on her G+ account. Nobody knows who exactly post these messages, nor understand the content of these seemingly cryptic memo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clarification.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. For this particular puzzle, the only information required to solve it is the post.&#xA; &lt;br&gt;&#xA; 2. Knowledge of programming is not required. The necessary ideas are given as clues.&#xA; &lt;br&gt;&#xA; 3. At one part of the puzzle, some ideas are simplified. You'll know when you reach it.&#xA; &lt;br&gt;&#xA; 4. The &quot;more&quot; links on hints are for interested nerds who want to explore more. It is not required.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hints.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. She is knowledgeable in physical science and computer science.&#xA; &lt;br&gt;&#xA; 2. GNU stands for GNU is Not Unix. &lt;a href=&quot;https://www.gnu.org/&quot; rel=&quot;nofollow&quot;&gt;More&lt;/a&gt;&#xA; &lt;br&gt;&#xA; 3. If it looks like a duck, swims like a duck, and quacks like a duck, it might not be a duck. &lt;a href=&quot;http://en.wikipedia.org/wiki/Duck_test&quot; rel=&quot;nofollow&quot;&gt;More&lt;/a&gt;&#xA; &lt;br&gt;&#xA; 4. Since physics is fun, reciprocate frequently.&#xA; &lt;br&gt;&#xA; 5. Again, she love physical science, you know, like chemistry. &lt;a href=&quot;http://en.wikipedia.org/wiki/Outline_of_physical_science&quot; rel=&quot;nofollow&quot;&gt;More&lt;/a&gt;&#xA; &lt;br&gt;&#xA; 6. SOH, CAH, TOA!&#xA; &lt;br&gt;&#xA; 7. To understand recursion, you must first understand recursion. &lt;a href=&quot;http://en.wikipedia.org/wiki/Recursion&quot; rel=&quot;nofollow&quot;&gt;More&lt;/a&gt;&#xA; &lt;br&gt;&#xA; 8. Imaginary number squared produces negative one. &lt;a href=&quot;http://en.wikipedia.org/wiki/Imaginary_number&quot; rel=&quot;nofollow&quot;&gt;More&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Overview.&lt;/strong&gt; It was a few months after Professor Swannie went missing, when suddenly, she posted a riddle her G+ page, which seemingly tries to predict the doomsday.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Post.&lt;/strong&gt; Definition of dynamite: Imaginary 66.7.95.53:52... For all cases, functions are not sensitive... Every iteration, a year went by... When the next struck, doomsday shall arrive... I shall overturn this world through my wisdom of science...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Task.&lt;/strong&gt; Figure out the date of her predicted doomsday.&lt;/p&gt;&#xA;" OwnerDisplayName="user5508" LastEditorDisplayName="user5508" LastEditDate="2014-11-18T15:44:37.267" LastActivityDate="2014-11-18T15:44:37.267" Title="Nerd on the Loose I - WHEN" Tags="|mathematics|lateral-thinking|science|" AnswerCount="4" CommentCount="22" ContentLicense="CC BY-SA 3.0" />
  <row Id="4801" PostTypeId="2" ParentId="3834" CreationDate="2014-11-16T13:44:45.990" Score="0" Body="&lt;p&gt;Player1 wins = normal Player2 win/player1 lose&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Player2 wins = normal tie&lt;/p&gt;&#xA;&#xA;&lt;p&gt;tie = normal Player1 win/player2 lose&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for this, Player2 has larger chance to get tie and win (as only he wins, player1 win will count 1), but if Player1 use strategy to force player2 win, then Player 1 wins&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Player1 as slightly larger chance, since he lose, he will win, he win, it counted as tie, the only thing he need to do is not to get tie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since they perfect play connect four, mostly they get is tie, so that Player2 will win.&lt;/p&gt;&#xA;" OwnerUserId="5453" LastActivityDate="2014-11-16T13:44:45.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4805" PostTypeId="2" ParentId="4747" CreationDate="2014-11-16T16:36:15.633" Score="2" Body="&lt;p&gt;@Mew already pointed out that you cannot solve this puzzle exactly as stated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But&lt;/strong&gt; for all practical purposes, we can assume that &lt;code&gt;A~=B&lt;/code&gt;, and since we &lt;em&gt;can&lt;/em&gt; determine &lt;code&gt;A+B&lt;/code&gt;, we can approximate A and B as &lt;code&gt;A~=B~=(A+B)/2&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, it's only shown that you cannot solve the problem in-band. What if they could synchronize their clocks?&#xA;Then it's trivial: At midnight, Alice announces &quot;it's Midnight&quot;, and Bob can announce A. Vice versa for B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how could they sync their clocks?&#xA;Unfortunately, because of the proof already shown, they cannot do this in-band. The best they can do is sync to an error of &lt;code&gt;|A-B|&lt;/code&gt;.&#xA;They need eighter some kind of signalling with known time to travel, or some kind of worldwide synchronous event as reference.&#xA;They could shoot a laser at the mirrors on the moon, or something.&#xA;When they decide that Alice's clock is the reference clock (e.g. the one that shows the &quot;correct time&quot; for the system), then Alice fires her laser when their clock shows midnight. When Bob observes Alice's signal, he sets his clock to Midnight+time_the_laser_signal_travelled. Now, their clocks are synchronized and they can trivially determine A and B.&lt;/p&gt;&#xA;" OwnerUserId="5285" LastActivityDate="2014-11-16T16:36:15.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4807" PostTypeId="1" AcceptedAnswerId="5824" CreationDate="2014-11-16T17:04:43.003" Score="5" ViewCount="904" Body="&lt;p&gt;A party is being held at a local mansion. The host is very rich and her apparently miraculous success is entirely due to her famous recipe for spaghetti. A friend of yours gatecrashed one of the earlier parties and found where the recipe is hidden - in one of the many wardrobes in the host's bedroom - but he was caught by security before he managed to steal the recipe. Your task is to get into the current party and steal the recipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the entrance to the mansion is a security guard. You know that the only guests that may attend are people who answer his questions correctly, so you hide nearby and listen to the passwords, hoping to guess the pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first guest arrives, his fingers covered in rings. The guard greets him by name and says &quot;6&quot;. The guest replies &quot;5&quot; and is admitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second guest arrives on a sleigh. The guard greets her as an old friend of the host, and says &quot;7&quot;. She replies &quot;6&quot; and is admitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third guest comes galloping along on his horse. The guard recognises him, but when he says &quot;5&quot; the guest answers wrongly (&quot;4&quot;) and is dragged away by lions. &quot;You should have said '7',&quot; the guard mutters. &quot;That can't have been who I thought it was; must have been his twin brother.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By now you've realised that this time there is a guest list: to be admitted, you need not only to figure out the key to the passwords but also to be recognised by the guard. You creep further away from the entrance, hoping to find someone you can impersonate. As you leave, you hear the guard say &quot;2&quot; to the fourth guest, who says &quot;8&quot; and is admitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mansion is situated very near the sea, and you find the fifth guest arriving on a boat. As he is securing it to a post, you sneak up behind him and stun him. Quickly you dress up in his clothes. You search his pockets in case he's brought written instructions he was given on how to answer the security questions, but you find only &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an inkwell, a map of the Hebrides, a box of staples, and a copy of Thomas Hardy's book 'Jude the Obscure'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Back at the mansion, you walk up to the guard, who takes you for the man you attacked and says &quot;3&quot;. What do you reply?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You hear music by Gustav Holst coming from inside the mansion. You also notice another entrance where animals are coming to the party; you're lucky you came to the humans' door.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Required methodology (read this if you think you have a very simple answer)&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The underlying idea type is the same as that of &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4057/the-security-to-the-party-12-now-with-party-soundtrack&quot;&gt;this question&lt;/a&gt;. More explicitly, you need to think of a real-life person (all the pocket items point towards this person) and then use this person's best-known fictional works. So (clues) -&gt; (person) -&gt; (works) -&gt; (answer)! There are also more clues towards these works hidden throughout the story.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2014-12-03T02:41:35.083" Title="The Security to the Party [Part 36] - an interesting one!" Tags="|password|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4810" PostTypeId="1" AcceptedAnswerId="5803" CreationDate="2014-11-16T19:38:23.113" Score="17" ViewCount="2905" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;You are a member of the Rebel Alliance.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You have been captured by the Empire and imprisoned.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In a stroke of sadistic brilliance, your guards have denied you all reading matter except &quot;Uncle Vader's News for Younglings&quot;, an Empire propaganda publication which has an intended readership aged 5 to 8 years.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;But unknown to the Empire, the Rebel Alliance has a secret sympathiser placed within the publication's editorial office.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The sympathiser has replaced the weekly wordsearch puzzle with one of the Rebel Alliance's own devising - and they're using it to get a message to you!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The puzzle is: find the secret message hidden in the wordsearch!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yvBfK.png&quot; alt=&quot;Wordsearch: Uncle Vader&amp;#39;s News for Younglings&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Words to find:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ALDERAAN&lt;br&gt;&#xA;AMIDALA&lt;br&gt;&#xA;ARMY&lt;br&gt;&#xA;CHEWBACCA&lt;br&gt;&#xA;CLONE&lt;br&gt;&#xA;COUNT&lt;br&gt;&#xA;CORUSCANT&lt;br&gt;&#xA;COUNT&lt;br&gt;&#xA;DAGOBAH&lt;br&gt;&#xA;DARTH&lt;br&gt;&#xA;DEATHSTAR&lt;br&gt;&#xA;DROID&lt;br&gt;&#xA;EWOKS&lt;br&gt;&#xA;FORCE&lt;br&gt;&#xA;GALAXY&lt;br&gt;&#xA;HAN&lt;br&gt;&#xA;JABBA&lt;br&gt;&#xA;JEDI&lt;br&gt;&#xA;PADME&lt;br&gt;&#xA;SITH&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Hint: The answer contains (but does not consist solely of) the word 'EWOKS'.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The introductory text is just for fun and does not contain any enciphered information. All the information you need is in the wordsearch itself and the 'words to find' list.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The answer is a series of Roman-alphabet characters. There are no spaces or punctuation. The words are separated by 'X' characters ('THISXISXANXEXAMPLE'). The puzzle is not case-sensitive and there are no roman numerals in it. All the words except for EWOKS are dictionary words.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;No knowledge of Star Wars is necessary to solve this puzzle.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This is my first time doing one of these (possibly the first time anyone's ever done one of these) so please be patient if I've made any errors or typos.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This type of puzzle is called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Cryptogram&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cryptogram&lt;/a&gt; (although this is quite an unusual one in that it is &lt;a href=&quot;http://en.wikipedia.org/wiki/Steganography&quot; rel=&quot;nofollow noreferrer&quot;&gt;steganogophised&lt;/a&gt; into a &lt;a href=&quot;http://en.wikipedia.org/wiki/Word_search&quot; rel=&quot;nofollow noreferrer&quot;&gt;wordsearch&lt;/a&gt;). A known section of plaintext ('EWOKS') is called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Known-plaintext_attack&quot; rel=&quot;nofollow noreferrer&quot;&gt;crib&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The message can be read by the prisoner in the cell without the need for a computer. You can feel free to use a computer to &lt;em&gt;crack&lt;/em&gt; the code, but it's a pencil-and-paper cipher.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A couple of minor bugs have been found (thanks for bug reports!), neither of them should screw up the puzzle, although the first one might make it slightly harder. If many more bugs emerge then I might redraft. Probably best to consider this a beta release. ;)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Yes, you do need to find COUNT twice. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The 'HAN' in the top-left quadrant is the one that's there intentionally - that's the one you're intended to find. The 'HAN' in the bottom-right quadrant is there by accident and should be ignored.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Chatroom is now open: &lt;a href=&quot;http://chat.stackexchange.com/rooms/18735/crypto-wordsearch-gram-you-are-a-member-of-the-rebel-alliance-and-a-traitor&quot;&gt;http://chat.stackexchange.com/rooms/18735/crypto-wordsearch-gram-you-are-a-member-of-the-rebel-alliance-and-a-traitor&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The text of the puzzle is below:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;HABLXNERCWNCTJQINOCY&#xA;MIMVDFJDKJRJOELPXUEX&#xA;VNRFENIPCTBSLIVATMNA&#xA;BRKDHLUADEMUCYMZYCOL&#xA;CRNJAORDMWEUJIEBTOLA&#xA;CKRJNVEMMOIRDBRXKUCG&#xA;ECROFNDELKVARIDEJNMH&#xA;MFAYFEAJTSLXFNHFYTAA&#xA;ORTNJDRURANJDUEHCCIB&#xA;MBSUFETLNRABBAJVCKMO&#xA;QFHJUVHGNAAREDLAMFBG&#xA;NJTFVHYTTMJDYNBGEYLA&#xA;BIAGRDHNLINDCWJTMMVD&#xA;YVEJUHAFYKMGEEHRCUNR&#xA;UNDENCIBFKEHLOAVTHSK&#xA;USYNSKTDNLCMDYNEFJHT&#xA;NJMUDGBJHUDIORDBRGNN&#xA;IMRRKLEDNTOUEMLLQDUU&#xA;NOUYNCGPKEINIFUTNURO&#xA;CMXIUNTDRMLSKYBGELPC&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Clues (if you need them):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;em&gt;Big Clue:&lt;/em&gt;&lt;/strong&gt; ignore the 'filler letters' which fill the spaces in between the words that you have to find. They could be any letters at all and it would make no difference to the puzzle.&#xA; &lt;br&gt; &lt;br&gt;&#xA; &lt;strong&gt;&lt;em&gt;Another Clue:&lt;/em&gt;&lt;/strong&gt; the word 'Ewoks' in the wordsearch is not specifically related to the word 'Ewoks' in the plaintext. It would be possible to swap 'Ewoks' for a different word in the wordsearch, without changing the plaintext.&#xA; &lt;br&gt; &lt;br&gt;&#xA; &lt;strong&gt;&lt;em&gt;Clue 3&lt;/em&gt;&lt;/strong&gt;: the order of the words in the 'list of words to find' is significant. If we jumbled up the order in which they are listed then the puzzle would be broken.&#xA; &lt;br&gt; &lt;br&gt;&#xA; &lt;strong&gt;&lt;em&gt;Clue 4&lt;/em&gt;&lt;/strong&gt;: the start position of each word (in the grid) is significant.&#xA; &lt;br&gt; &lt;br&gt;&#xA; &lt;strong&gt;&lt;em&gt;Clue 5 (Monday)&lt;/em&gt;&lt;/strong&gt;: the orientation of each word (in the grid) is significant. &#xA; &lt;br&gt; &lt;br&gt;&#xA; &lt;strong&gt;&lt;em&gt;Totally-giving-it-away-now bonus clue&lt;/em&gt;&lt;/strong&gt;: Those three things (word list order, start position of each word in the grid, orientation of each word in the grid) are &lt;strong&gt;&lt;em&gt;all&lt;/em&gt;&lt;/strong&gt; that matters. Between them they carry &lt;em&gt;all&lt;/em&gt; the information about the plaintext.&#xA; &lt;br&gt; &lt;br&gt;&#xA; &lt;strong&gt;&lt;em&gt;Tuesday clue:&lt;/em&gt;&lt;/strong&gt; &lt;br&gt;&#xA; - All the following directions mean the same thing as each other: N, NE, E, SE. &lt;br&gt;&#xA; - All the following directions mean the same thing as each other: S, SW, W, NW. &lt;br&gt;&#xA; &lt;br&gt; &lt;br&gt;&#xA; &lt;strong&gt;&lt;em&gt;Tuesday massive bonus clue:&lt;/em&gt;&lt;/strong&gt; &lt;br&gt;&#xA; Each word-to-find represents one character in the plaintext. It starts with:&lt;br&gt;&#xA; ALDERAAN: e &lt;br&gt;&#xA; AMIDALA: w &lt;br&gt;&#xA; ARMY: o &lt;br&gt;&#xA; CHEWBACCA: k &lt;br&gt;&#xA; CLONE: s &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="4050" LastEditDate="2014-12-09T17:06:14.267" LastActivityDate="2018-04-17T07:54:51.090" Title="Crypto-wordsearch-gram: &quot;You Are a Member of the Rebel Alliance, and a Traitor!&quot;. A cryptogram hidden within a wordsearch" Tags="|cipher|word-search|steganography|" AnswerCount="2" CommentCount="35" ContentLicense="CC BY-SA 3.0" />
  <row Id="4811" PostTypeId="2" ParentId="4704" CreationDate="2014-11-16T20:01:52.073" Score="20" Body="&lt;p&gt;The rules can be satisfied without even entering the maze.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There's &quot;No going around the maze&quot;, but nothing to say &quot;The maze can't go around&quot;. So if you stand at the bottom-right corner of the maze, and rotate it 180 degress about its centre, you will end up at the top-left corner of the maze.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5550" LastActivityDate="2014-11-16T20:01:52.073" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4813" PostTypeId="1" AcceptedAnswerId="4816" CreationDate="2014-11-16T20:20:35.150" Score="1" ViewCount="484" Body="&lt;p&gt;You have made it to the last room. It is very bright, but suddenly you realize that all the codes you've been reading have been different machine codes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see the final note scripted on the large door with 4 light bulbs with a different light power. It says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This is the last room in this great puzzle. You're prize will be to&#xA;  live forever. There are no codes here, but solve this, Here you have&#xA;  the 4 changeable light bulbs. Put them in order, remember all of the&#xA;  rooms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How do you solve this puzzle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Previous parts:&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/4121/the-great-binary-puzzle-part-1&quot;&gt;Part 1&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/4126/the-great-binary-puzzle-part-2&quot;&gt;Part 2&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/4711/the-great-binary-puzzle-part-3&quot;&gt;Part 3&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="5061" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2015-07-02T12:54:18.387" Title="The Great Binary Puzzle - Part 4" Tags="|logical-deduction|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4814" PostTypeId="2" ParentId="2821" CreationDate="2014-11-16T20:27:45.573" Score="97" Body="&#xA;&lt;p&gt;NOTHING. The pronunciation of this in various accents could be rendered as NOTHIN' (one letter removed), NOT'IN' (two letters removed), or NOT'N' (three letters removed). And of course if you remove all the letters you're left with &lt;em&gt;nothing&lt;/em&gt;!&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2020-08-20T11:40:57.287" LastActivityDate="2020-08-20T11:40:57.287" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="4815" PostTypeId="2" ParentId="4784" CreationDate="2014-11-16T21:11:04.573" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; FAVORITES&#xA; &lt;br&gt;&#xA; &lt;br&gt;&#xA; Each prefix can be extended by adding a letter, then an anagram of the prefix.&#xA; &lt;code&gt;&#xA; edi     F    ied&#xA; est     A    tes&#xA; ser     V    ers&#xA; ing     O    ing&#xA; restau  R    ateurs&#xA; on      I    on&#xA; myo     T    omy&#xA; mus     E    ums&#xA; hot     S    hot&#xA; &lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2014-11-16T21:18:09.010" LastActivityDate="2014-11-16T21:18:09.010" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4816" PostTypeId="2" ParentId="4813" CreationDate="2014-11-16T21:18:52.250" Score="3" Body="&lt;p&gt;Put them from dimmest to brightest. Referring back to the earlier puzzles, each one gets progressively brighter than the last.&lt;/p&gt;&#xA;" OwnerUserId="1664" LastActivityDate="2014-11-16T21:18:52.250" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4817" PostTypeId="2" ParentId="4718" CreationDate="2014-11-16T22:39:41.927" Score="1" Body="&lt;p&gt;I cannot quite decipher the text, but the last 3 words seem to me like&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; with number digits&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5283" LastActivityDate="2014-11-16T22:39:41.927" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4818" PostTypeId="1" AcceptedAnswerId="4835" CreationDate="2014-11-16T22:49:04.987" Score="33" ViewCount="24217" Body="&lt;p&gt;I'm a little uncertain as to why these two parity situations occur on the 4x4 Rubik's cube. The first parity case that can occur is this one, where an edge is simply flipped upside down:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://i.ytimg.com/vi/aMgcU6PIUhw/maxresdefault.jpg&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5dhD4m.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second one that can occur is this one, where two edges are swapped: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.speedcubing.com/chris/images/4as3-adjpllparityl.gif&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mVsj9.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Neither of these are valid positions on a 3x3 Rubik's cube. I've been told something along the lines of &quot;hidden pieces are out of place,&quot; but this doesn't make much sense to me. If this is a reasonable explanation, what are those virtual pieces, and why do they have to exist? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, what makes this a legal position on a 4x4 cube, and not a 3x3 cube?&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorUserId="9038" LastEditDate="2015-01-26T06:28:31.897" LastActivityDate="2018-04-16T17:20:12.483" Title="Why does the 4x4 Rubik's cube have parity cases, while the 3x3 does not?" Tags="|rubiks-cube|mechanical-puzzles|solvability|" AnswerCount="7" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="4819" PostTypeId="2" ParentId="4818" CreationDate="2014-11-16T22:54:45.990" Score="16" Body="&lt;p&gt;The key is that the four center cubules on each face of a Rubik's Revenge are indistinguishable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you do the move sequence to swap those two edges in the &quot;3x3x3&quot; phase of solving a 4x4x4 cube (or to flip a single edge or swap two corners), a bunch of the center cubules get reoriented, but you don't notice because they're all the same colour and rotationally symmetrical anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A picture cube wouldn't have this problem, and thus it would be much harder to solve.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2014-11-16T22:54:45.990" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="4820" PostTypeId="1" AcceptedAnswerId="4824" CreationDate="2014-11-17T00:26:42.603" Score="16" ViewCount="44508" Body="&lt;p&gt;Is there a sequence of moves that can be repeated over and over again which can solve any legal position the Rubik's Cube? If so what is it, and if there's more than one, what's the shortest? If not, prove that it's impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The sequence can be of any length but cannot be broken up into multiple pieces to be executed depending on Rubik's cube patterns; it must, when repeated from start to finish, end in a perfectly solved Rubik's cube.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="69582" LastEditDate="2021-05-07T02:42:14.597" LastActivityDate="2021-06-23T08:02:47.863" Title="Is it possible to use one sequence of moves to solve the Rubik's cube from any position?" Tags="|strategy|rubiks-cube|mechanical-puzzles|solvability|twisty-puzzles|" AnswerCount="4" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4821" PostTypeId="1" AcceptedAnswerId="4822" CreationDate="2014-11-17T01:13:41.040" Score="12" ViewCount="1503" Body="&lt;p&gt;The following picture shows an arrangement of colored triangles.  What is their meaning?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Bn32s.jpg&quot; alt=&quot;Colorful triangles&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5188" LastActivityDate="2014-11-17T02:17:15.073" Title="What do these colorful triangles mean?" Tags="|lateral-thinking|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4822" PostTypeId="2" ParentId="4821" CreationDate="2014-11-17T01:25:09.527" Score="15" Body="&lt;p&gt;They represent &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The colours of the flags of the G20 nations, displayed in alphabetical order. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Specifically in order from left to right,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Argentina, Australia, Brazil, Canada, China, European Union, France, Germany, India, Indonesia, Italy, Japan, Republic of Korea, Mexico, Russia, Saudi Arabia, South Africa, Turkey, the United Kingdom, the United States &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="97" LastEditorUserId="97" LastEditDate="2014-11-17T02:17:15.073" LastActivityDate="2014-11-17T02:17:15.073" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4823" PostTypeId="2" ParentId="4820" CreationDate="2014-11-17T01:42:01.967" Score="5" Body="&lt;p&gt;&lt;strong&gt;EDIT: This only works in a certain set of configurations, as the loop only goes through 1260 states before returning to the original position. My mistake, this is incorrect, but still useful.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Great solution found &lt;a href=&quot;http://aperiodical.com/2014/03/how-to-solve-a-rubiks-cube-in-one-easy-step/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. Basically, if you rotate the right, back, left, and front faces all clockwise and in order, then it will always solve the cube - eventually. The core part of it is in one simple explanation - doing this cannot enter an inescapable loop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you move the front face clockwise, then counter-clockwise, nothing happens to the cube. The faces are not moved with respect to each other, and you are in the same position. In order for a loop to happen, two different layouts &lt;strong&gt;must&lt;/strong&gt; lead to the same configuration with the same move. But wait - if you reverse  this move, then what position does it go to? The simple answer is this will never happen. Here is a great diagram in the article itself. &lt;img src=&quot;https://i.stack.imgur.com/9Folm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, reversing this loop would not work, as there are two possible options. This is a proof that loops will never happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moving these four faces is the shortest possible configuration of moves that affects each and every cube, save for the center ones which are irrelevant anyway. This means it is the shortest possible solution that you could repeat to solve the Cube.&lt;/p&gt;&#xA;" OwnerUserId="1664" LastEditorUserId="1664" LastEditDate="2014-11-17T02:13:03.003" LastActivityDate="2014-11-17T02:13:03.003" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="4824" PostTypeId="2" ParentId="4820" CreationDate="2014-11-17T01:57:36.343" Score="17" Body="&lt;p&gt;The only way you can have a solve-all sequence is if you have a sequence of moves that goes through all 43 quintillion configurations of the Rubik's Cube. In order to do this, you need to draw a transition graph between all the states of the Rubik's Cube and find a Hamiltonian cycle through them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sequence of moves doesn't necessarily have to be 43 quintillion moves long - a simple sequence of 4 moves can produce a cycle of 1,260 configurations as seen in mdc32's answer, and in general a sequence of symbols in the group will produce a cycle of configurations much longer than the cycle itself. However, the sequence will still be very long, simply because 43 quintillion moves is still a lot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Micah provided &lt;a href=&quot;http://bruce.cubing.net/ham333/rubikhamiltonexplanation.html&quot;&gt;a link&lt;/a&gt; to a page that did construct such a Hamiltonian cycle in a comment. I haven't been able to make head or tail of its notation (or to figure out how to count the number of moves from the descriptions of the cosets), but it looks like the sequence of moves that is required is billions of moves long, which is still definitely outside of the realm of plausibility for memorization.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2014-11-17T17:19:59.120" LastActivityDate="2014-11-17T17:19:59.120" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="4825" PostTypeId="2" ParentId="4807" CreationDate="2014-11-17T02:23:05.530" Score="5" Body="&lt;p&gt;Likely not the answer but...&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You should respond with 4.&lt;br /&gt;&lt;br /&gt;Explanation:&lt;br /&gt;&lt;br /&gt;Regardless of what the guard asks, you respond with the number corresponding to the number of letter for your transportation.&lt;br /&gt;&lt;br /&gt;so...&lt;br /&gt;&lt;br /&gt;Guest 1 arrived on a &quot;horse&quot;, &quot;5&quot; gets her in (H-O-R-S-E)&lt;br /&gt;Guest 2 arrives on a &quot;sleigh&quot;, &quot;6&quot; works (S-L-E-I-G-H)&lt;br /&gt;Guest 3 arrives on a &quot;horse&quot;, says &quot;4&quot;, should have said &quot;5&quot; (same as the guard)&lt;br /&gt;Guest 4... don't know what &quot;8&quot; corresponds to, but the gatecrasher doesn't see it... maybe by &quot;carriage&quot;? :)&lt;br /&gt;Guest 5 arrives by &quot;boat&quot;, should answer &quot;4&quot;.&lt;br /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4758" LastActivityDate="2014-11-17T02:23:05.530" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4828" PostTypeId="1" AcceptedAnswerId="4834" CreationDate="2014-11-17T03:47:28.433" Score="22" ViewCount="4164" Body="&lt;p&gt;The golden bird can only fly at 5 mph. That is the only way the bird can move. The bird &lt;strong&gt;must&lt;/strong&gt; stop flying and rest its wings for 5 hours immediately after every hour of flying it does. That is the only way the bird can rest so it can continue flying. The ocean is 6 miles wide. The golden bird flies over the ocean and makes it to the other side without resting. How is this possible?&lt;/p&gt;&#xA;" OwnerUserId="4802" LastEditorUserId="11876" LastEditDate="2016-09-15T01:03:56.870" LastActivityDate="2016-09-15T01:03:56.870" Title="The Golden Bird and The Ocean" Tags="|mathematics|" AnswerCount="16" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="4829" PostTypeId="2" ParentId="4828" CreationDate="2014-11-17T03:50:41.493" Score="4" Body="&lt;p&gt;&lt;s&gt;Could it be that the Earth is round, so the bird just gonna fly in parallel with the diameter of the earth to reduce the distance to 5 mph.&lt;/s&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Analysis to completely trashed this idea of mine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the radius of Earth $R=6371\mathrm{km}$. To obtain the arclength $C=6\mathrm{m}$, we need to travel $\theta=6/6371000=9.417\times10^{-7}\mathrm{rad}$. So if the bird fly parallel to the diameter of the Earth, it will be $L=2R\sin{(\theta/2)}=5.999\mathrm{m}$. If you use common sense and think about this physical world, this should be obvious too. Maybe I've not been going out to the real world for too long, and I lost my common sense about the world. Probably.&lt;/p&gt;&#xA;" OwnerDisplayName="user5508" LastEditorDisplayName="user5508" LastEditDate="2014-11-17T04:35:54.810" LastActivityDate="2014-11-17T04:35:54.810" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4830" PostTypeId="2" ParentId="4828" CreationDate="2014-11-17T04:13:28.237" Score="15" Body="&lt;p&gt;While this particular ocean is 6 miles wide, its length is considerably less.&lt;/p&gt;&#xA;" OwnerUserId="97" LastEditorUserId="2484" LastEditDate="2014-11-17T04:32:16.633" LastActivityDate="2014-11-17T04:32:16.633" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4831" PostTypeId="2" ParentId="4828" CreationDate="2014-11-17T04:38:38.570" Score="1" Body="&lt;p&gt;On its journey it gets hungry and looks for food on the back of a stationary whale for more than one hour.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The bird was looking for food, therefore not resting.&lt;/li&gt;&#xA;&lt;li&gt;The OP didn't mention that the bird should not stop flying.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4973" LastEditorUserId="4802" LastEditDate="2014-11-18T04:13:43.810" LastActivityDate="2014-11-18T04:13:43.810" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4832" PostTypeId="2" ParentId="4828" CreationDate="2014-11-17T05:16:41.707" Score="11" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/The_Golden_Bird&quot; rel=&quot;noreferrer&quot;&gt;The Golden Bird&lt;/a&gt; is obviously the bird captured by the prince (see the wiki link on &amp;quot;The Golden Bird&amp;quot; at the beginning). What the Grimm brothers missed out on though was that after the Golden Bird the Golden Cat came first, before the Golden Horse. So, when the Golden Bird explains to the prince that to get the Golden Cat the Golden Bird has to fly to the other side of the 6 miles wide ocean without resting, the fox advises the prince to use the wooden boat in the castle and peddle it at a speed of at least 6mph whilst the bird peacefully will fly inside its cage&lt;sup&gt;1&lt;/sup&gt;. Sadly the prince fails, as he ignores the foxes advice and instead uses the Golden Boat, which is far too heavy to peddle that fast, but that's not important now.&lt;/p&gt;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; Of course the cage is covered by a piece of cloth, otherwise air resistance would make this impossible.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;em&gt;And just for the record and for fun: The answer I originally wanted to give was the following:&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;The Golden Bird is obviously from a fairy tale (&lt;em&gt;before I realized that it &lt;strong&gt;really&lt;/strong&gt; was from a fairy tale&lt;/em&gt;). And as commonly known fairy tales do not happen on this earth, so they must have happened on another earth. And as commonly known other earths are not necessarily round, nor does gravity necessarily behave in the same way, so in this particular case the bird simply flew over the mini ocean that was located in a deep hole.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3kKyn.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="468" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-11-17T05:16:41.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4833" PostTypeId="2" ParentId="4828" CreationDate="2014-11-17T05:28:59.830" Score="12" Body="&lt;p&gt;The golden bird flies without rest inside the cabin of an airplane or helicopter, which takes less than one hour to fly over the ocean.&lt;/p&gt;&#xA;" OwnerUserId="4394" LastActivityDate="2014-11-17T05:28:59.830" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4834" PostTypeId="2" ParentId="4828" CreationDate="2014-11-17T05:37:56.153" Score="33" Body="&lt;p&gt;The bird's speed is 5 mph. If&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the speed of the wind across the ocean is greater than or equal to 1 mph towards the direction the bird is flying&lt;sup&gt;*&lt;/sup&gt;, the overall speed of the bird will be greater than 6 mph,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and it can easily reach the other side of the ocean without resting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that means&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the speed of the ocean wind at that time must have been at least 1 mph in the same direction the bird was flying.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And hence the bird reached the other side of the ocean without resting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This picture illustrates the point:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/Rn7O2.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;*Or greater than or equal to √2 mph to any angle not more than 45˚ from the direction of bird's flight. If greater than 45˚, it needs to be more than √2 mph, depending on the angle.&lt;/p&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="1955" LastEditDate="2014-11-17T20:17:04.090" LastActivityDate="2014-11-17T20:17:04.090" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4835" PostTypeId="2" ParentId="4818" CreationDate="2014-11-17T05:47:46.827" Score="15" Body="&lt;p&gt;The easiest explanation would be that in a 3x3 cube, only one cube is out of position, but in a 4x4 cube two cubes are out of position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a &lt;a href=&quot;http://en.wikipedia.org/wiki/15_puzzle&quot;&gt;15 puzzle&lt;/a&gt; (the sliding puzzle where you try to put the numbers in order) half of all possible initial positions are unsolvable. They call the solvable positions &quot;even&quot; and the unsolvable positions &quot;odd&quot;. The &quot;odd&quot; positions are called that because there are an odd number of two-tile swaps (not sliding pieces) that need to be done (from the solved position) in order to generate that position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is similar with Rubik's cubes, though a little bit more complicated (&lt;a href=&quot;http://ruwix.com/the-rubiks-cube/unsolvable-rubiks-cube-invalid-scramble/&quot;&gt;only one in twelve&lt;/a&gt; of all positions are actually solvable, so there's more than just even or odd). Without getting into too much depth, you can simply think of the 3x3 cube being in an &quot;odd&quot; position because only one cube has changed, but the 4x4 cube is in an &quot;even&quot; position because two cubes have changed.&lt;/p&gt;&#xA;" OwnerUserId="1955" LastActivityDate="2014-11-17T05:47:46.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4838" PostTypeId="2" ParentId="4828" CreationDate="2014-11-17T06:13:44.997" Score="8" Body="&lt;p&gt;The bird is on a artificial small planet (gravity doesn't matter but lets say its artificially the same as on earth) it has only 2 land masses and a ocean, the two landmasses are a small island (where the bird is) and a bigger continent. The island is located on the ocean like so that its 6 miles away from the main land on one side and only 5 miles on the other side. So it just flies in the other direction.&lt;/p&gt;&#xA;" OwnerUserId="5560" LastEditorUserId="4802" LastEditDate="2014-11-18T06:31:54.957" LastActivityDate="2014-11-18T06:31:54.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4840" PostTypeId="1" AcceptedAnswerId="4841" CreationDate="2014-11-17T07:06:08.257" Score="-2" ViewCount="1509" Body="&lt;p&gt;There were two friends, Joy and Tom. Tom was oversmart and always thought he was brilliant and smarter than anyone else, especially Joy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once Tom and Joy went for a trip around the world and got lost in an Arabian desert during their journey. They walked and walked and walked over the sand for two days and reached no where. They were suffering from hunger and thirst and even couldn't find anything to drink or an oasis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the third day morning they reached a mosque. They saw that there are people inside that mosque and decided to go inside the mosque and ask for food and drink. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inside the mosque, they met the Imam (mosque leader). He asked them, &quot;Who are you gentlemen?&quot;. &#xA;Then Joy replied: &quot;I am Joy&quot;.&#xA;Before replying, Tom thought for a while. He thought that if he said a 'Muslim' name he may get special consideration and get more food and water. And Joy is a fool and he may get kicked out. Then he replied: &quot;I am Abdullah, and we got lost in the desert for over three days. We need help&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Imam then called his assistant and asked him to arrange food and drinks for one man in the dining hall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then Tom thought Joy was going to be kicked out and one he is going to get food and all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the food was ready, the Imam came and invited only 'Joy' to have food. Seeing this, Tom felt like his head was rotating and was about to cry. At that time the Imam came and said something to him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Why did the Imam give food and all to Joy only?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) What did the Imam say to Tom (fake Abdullah) in the end?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: Request to place answers in spoiler tags.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="5373" LastEditDate="2015-04-30T22:58:19.287" LastActivityDate="2015-04-30T22:58:19.287" Title="Why Tom didn't get food?" Tags="|lateral-thinking|situation|" AnswerCount="3" CommentCount="6" ClosedDate="2015-06-12T16:30:19.577" ContentLicense="CC BY-SA 3.0" />
  <row Id="4841" PostTypeId="2" ParentId="4840" CreationDate="2014-11-17T07:12:20.710" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; i think there are certain days that Muslims are not allowed to eat or drink, i think that's why Tom didn't receive any food since that he pretended as a Muslim. Maybe the Imam said to Tom to kneel or pray with the other Muslims&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4944" LastActivityDate="2014-11-17T07:12:20.710" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4842" PostTypeId="2" ParentId="4840" CreationDate="2014-11-17T07:13:39.873" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It was Ramadan, and Abdullah is a distinctly Islamic name. The Imam assumed that Tom was Islamic and hence was fasting.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4914" LastActivityDate="2014-11-17T07:13:39.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4843" PostTypeId="1" AcceptedAnswerId="4844" CreationDate="2014-11-17T08:14:17.377" Score="12" ViewCount="3210" Body="&lt;p&gt;Ok, you hacked my other riddles pretty quickly, but this one is meant to give you heat.. There you go:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I am born from the death of my mother&#xA;and I have millions of secrets to hide.&#xA;Like me there are plenty of others&#xA;but we are all empty inside.&#xA;&#xA;Maybe you know, I'm never in a hurry&#xA;due to my origin sublime.&#xA;At the same time, I am always hungry,&#xA;not for knowledge,&#xA;but for time.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Good luck this time. And btw, just one single word I skipped in the last sentence can point you immediately to the right answer. :)&lt;/p&gt;&#xA;" OwnerUserId="5283" LastEditorUserId="4962" LastEditDate="2014-11-17T08:23:39.873" LastActivityDate="2014-11-17T16:38:26.350" Title="Riddle with heavy answer" Tags="|riddle|rhyme|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4844" PostTypeId="2" ParentId="4843" CreationDate="2014-11-17T08:19:34.300" Score="9" Body="&lt;p&gt;Just a suggestion mostly , although sublime origin might not stand &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Blackhole&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4850" LastActivityDate="2014-11-17T08:19:34.300" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="4846" PostTypeId="2" ParentId="4790" CreationDate="2014-11-17T08:38:38.513" Score="1" Body="&lt;p&gt;I think this fits all clues:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; hacker?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;// EDIT&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if tiny I call you tall&#xA;The first two lines contain no hint at all&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; well, the above is obvious, first two lines have no hints&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For sure I'm egocentric&#xA;You mean nothing to me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; blackhat!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the best choice seems a third person to be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Man in the middle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Make him steal what surrounds &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Interwebs is everywhere now :-)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the guy behind the screen&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the hacker&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Find one more glue&#xA;and my full name can be seen&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; oh noes, suppose we left a cookie for the feds, we wouldn't be anonymous anymore!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Read on with prudence &#xA;and drop the penny&#xA;Some say I'm one, I say I'm many&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; anonymous&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5254" LastEditorUserId="5254" LastEditDate="2015-01-22T13:30:02.113" LastActivityDate="2015-01-22T13:30:02.113" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4847" PostTypeId="2" ParentId="4828" CreationDate="2014-11-17T08:43:41.550" Score="6" Body="&lt;p&gt;The bird flies up, it doesn't go parallel with the ocean. When it should rest, it just falls down in the direction of the land side, so it falls to his death at the other side. But at least it's there.&lt;/p&gt;&#xA;" OwnerUserId="5254" LastActivityDate="2014-11-17T08:43:41.550" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4848" PostTypeId="1" AcceptedAnswerId="4850" CreationDate="2014-11-17T08:56:00.997" Score="1" ViewCount="365" Body="&lt;p&gt;After not getting invited to any of the parties that have been going on lately I decided to make my own.&#xA;As expected I hired my own security guards and sent out a letter to all the invited guests with instructions on how to answer the guards when told a random number.&#xA;Of course the likes of you would not be invited but you decide to try and decipher the &quot;code&quot; needed to get inside. You lay in wait and hear a couple of people who actually were invited and what their responses were. It Went as follows&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first guest arrives and the guard says &quot;13&quot;. the guest says &quot;5&quot; and the guard steps aside and lets the guest inside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second guest arrives and the guard says &quot;23&quot; and the second guest says &quot;7&quot; and gets let inside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third guest arrives and the guard says &quot;17&quot;. the third guest answers &quot;6&quot; and goes inside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fourth guest arrives and the guard says &quot;19&quot;. the fourth guest says &quot;4&quot; and gets KILLED RIGHT THEN AND THERE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You feel confident that you have cracked the system and decide to give it a go. You walk up to the guard with a smug on your face and the guard goes &quot;33&quot;. What do you say in order to get in to the party and not killed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(please put your answers in spoiler tags)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: The answers are not the results of a mathematical function even though such a function most likely does exist. &lt;/p&gt;&#xA;" OwnerUserId="4962" LastEditorUserId="5373" LastEditDate="2015-04-30T22:59:10.090" LastActivityDate="2015-04-30T22:59:10.090" Title="I'll host my own party with.. (not sure if duplicate)" Tags="|pattern|password|" AnswerCount="2" CommentCount="1" ClosedDate="2014-11-17T11:54:06.097" ContentLicense="CC BY-SA 3.0" />
  <row Id="4849" PostTypeId="2" ParentId="4848" CreationDate="2014-11-17T09:21:49.933" Score="1" Body="&lt;p&gt;The small number of examples dont do justice to such questions.  The purpose of this answer is not to &quot;answer&quot; the question but just to show that its easy to come  up with some function that fits &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rounded up value of $(((x-3)/5)+3)$ i.e (((13-3)/5)+3) = 5, (((17-3)/5)+3)=5.8 rounded up to 6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So going by the above logic your answer should be 9 (((33-3)/5+3) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please don't place spoilers for this, I mean it to be open.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This answer was made prior to the clarification, so before downvoting take that fact into consideration, I am leaving it here as proof and guideline, I am not worried about negative reputation from here but feel it does not deserve to be downvoted or upvoted&lt;/p&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="2313" LastEditDate="2014-11-17T10:05:27.027" LastActivityDate="2014-11-17T10:05:27.027" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4850" PostTypeId="2" ParentId="4848" CreationDate="2014-11-17T09:53:28.277" Score="3" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the number of lines in the Roman-numeral representation of the number the guard gives?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That matches all the questions and answers you've provided, and the right response to '33' is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 9&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;while the right response to '19' was&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 5.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(This idea was used in another of the 'Security to the Party' questions, but I can't remember which one! +1 for new formatting of the question anyway.)&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2014-11-17T09:53:28.277" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4851" PostTypeId="2" ParentId="4840" CreationDate="2014-11-17T09:54:32.820" Score="2" Body="&lt;p&gt;This was similar to a story which i read earlier.It is same, but it seems to be converted into a puzzle..&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Muslims keeps fast in day time and as it was Morning time and Ramadan days fast (Roza) is going on.&lt;br /&gt;&lt;br /&gt;Answers: &lt;br /&gt; (1) As Joy is non muslims and hungry,only he got Food.  &lt;br /&gt;(2) Told fake Abdullah to wait for food and water to evening when fast (Rozah)ends.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4981" LastEditorUserId="4981" LastEditDate="2014-11-17T10:01:14.103" LastActivityDate="2014-11-17T10:01:14.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4852" PostTypeId="2" ParentId="4828" CreationDate="2014-11-17T10:05:55.987" Score="1" Body="&lt;p&gt;I'll go along the lines of what others have said about the planet being peculiar. The atmosphere of said planet is very low and the bird can survive outside of the atmosphere. Additionally, the said planet spins rather fast. This means the bird flies out of the atmosphere, the planet spins, moving in the opposite direction the bird is flying in. So the bird travels at most five miles, while the planet spins the missing distance.&lt;/p&gt;&#xA;" OwnerUserId="5107" LastActivityDate="2014-11-17T10:05:55.987" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4853" PostTypeId="2" ParentId="4828" CreationDate="2014-11-17T10:08:26.077" Score="2" Body="&lt;p&gt;A few ideas:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Bird flies for 1 hour, then dies and falls in the ocean.  The tide washes her up on the shore.&lt;/li&gt;&#xA;&lt;li&gt;The golden bird is owned by a prince, who sails the ocean in his luxury yacht.  The bird doesn't even need to fly, but the prince lets her out for a fly near the end of his journey, so the bird &lt;em&gt;flies over the ocean&lt;/em&gt; and makes it to the other side without resting.&lt;/li&gt;&#xA;&lt;li&gt;The golden bird flies for an hour (making five miles' progress in the air) and then glides down to land on the other side.  &lt;em&gt;Gliding&lt;/em&gt; is not the same as &lt;em&gt;flying&lt;/em&gt; and it's &lt;a href=&quot;http://ngm.nationalgeographic.com/2007/12/albatross/safina-text&quot; rel=&quot;nofollow&quot;&gt;reasonable to assume that an ocean bird can glide for 1 mile.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;None of the above answers satisfy the &lt;code&gt;&quot;that is the only way the bird can move&quot;&lt;/code&gt; criterion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: JUST THOUGHT OF A FORTH!&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;The bird flies at 5mph, but the earth is rotating towards the bird at &gt;= 1mph, therefore the bird travels 5 miles in the time it takes the earth's surface to travel 1 mile or more = 6 miles or more and the bird lands safely.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5245" LastEditorUserId="5245" LastEditDate="2014-11-18T08:59:29.650" LastActivityDate="2014-11-18T08:59:29.650" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4854" PostTypeId="1" AcceptedAnswerId="4855" CreationDate="2014-11-17T10:23:35.943" Score="38" ViewCount="11168" Body="&lt;p&gt;I had a logical riddle on a programming interview test which was something like this:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In his garage a millionaire had cars, of which only 2 were not white, only 2 were not green and only 2 were not red. How many cars did the millionaire have in his garage?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This seemed quite straightforward to me and I gave an answer, however I wasn't allowed to see my test afterwards and I cannot tell if the answer is correct.&lt;/p&gt;&#xA;&lt;p&gt;I am curious to see if my answer was correct and also if there is a better mathematical derivation to the problem.&lt;/p&gt;&#xA;" OwnerUserId="5567" LastEditorUserId="81979" LastEditDate="2022-12-29T09:25:05.957" LastActivityDate="2022-12-29T11:49:13.060" Title="How many cars does the millionaire have?" Tags="|logical-deduction|mathematics|" AnswerCount="8" CommentCount="13" ContentLicense="CC BY-SA 4.0" />
  <row Id="4855" PostTypeId="2" ParentId="4854" CreationDate="2014-11-17T10:25:27.753" Score="40" Body="&lt;p&gt;A total of &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; three &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;cars, with colors distributed such that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; one is red, one is green, and one is white. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you take white, then all cars except two (green and red) are white. Likewise, if you take red, all cars except two (white and green) are red. And if you take green, all cars except two (red and white) are green.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reasoning for this conclusion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are more, then the statement that &quot;of them two are not white/green/red&quot; will fail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the total number of cars be $n$&lt;br&gt;&#xA;Total number of white cars be $x$&lt;br&gt;&#xA;Total number of green cars be $y$&lt;br&gt;&#xA;Total number of red cars be $z$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $$2 = n-x \rightarrow x = n-2$$&#xA;$$2 = n-y \rightarrow y = n-2$$&#xA;$$2 = n-z \rightarrow z = n-2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So&#xA;$$n = x+y+z$$&#xA;$$n = (n-2) + (n-2) + (n-2)  $$&#xA;$$n = 3n - 6$$&#xA;$$6 = 2n  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;n = 3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as what I understood from the question &quot;of which &lt;strong&gt;&lt;em&gt;only&lt;/em&gt;&lt;/strong&gt; 2 were not white&quot; means there is at least one white car and definitely more than 2 cars. The same for all colors specified in the question. Otherwise the questioner should have mentioned &quot;of which 2 were not white&quot; or &quot;2 cars were not white&quot; or just &quot;none of them are white&quot; instead.&lt;/p&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="2730" LastEditDate="2014-11-18T13:05:56.603" LastActivityDate="2014-11-18T13:05:56.603" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="4856" PostTypeId="2" ParentId="4854" CreationDate="2014-11-17T10:26:14.847" Score="77" Body="&lt;p&gt;It could be  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; two  cars (both black),  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; three cars, &lt;a href=&quot;https://puzzling.stackexchange.com/a/4855/&quot;&gt;as AeJey explained&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2014-11-18T20:02:25.413" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="4857" PostTypeId="2" ParentId="4057" CreationDate="2014-11-17T10:41:23.247" Score="2" Body="&lt;p&gt;Reference answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tyler W has the right solution and the right method, so &lt;strong&gt;go and upvote him/her&lt;/strong&gt; before reading this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The purpose of this post is to explain all the clues, including a few that haven't been solved yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My own post-mortem: I was surprised that the question turned out to be so difficult. But I enjoyed compiling the extra clues - I hope you enjoyed solving them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, the clues explained:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clues in the text of the question (each clue applies specifically to the challenge/response that it shares a paragraph with): &lt;br&gt; &lt;br&gt;&#xA; &lt;em&gt;returns&lt;/em&gt; (Return of the Jedi)&lt;br&gt;&#xA; &lt;em&gt;menacing&lt;/em&gt; (Phantom Menace)&lt;br&gt;&#xA; &lt;em&gt;back&lt;/em&gt; (Emprire Strikes Back)&lt;br&gt;&#xA; &lt;em&gt;attacks&lt;/em&gt; and a hint at &lt;em&gt;clones&lt;/em&gt; (Attack of the Clones)&lt;br&gt;&#xA; &lt;em&gt;revenge&lt;/em&gt; (Revenge of the Sith)&lt;br&gt;&#xA; &lt;em&gt;hope&lt;/em&gt; (A New Hope)&lt;br&gt;&#xA; &lt;br&gt;&#xA; Musical clues: &lt;br&gt;&#xA; &lt;br&gt;&#xA; &lt;strong&gt;Royale With Cheese&lt;/strong&gt; (Dialogue), Pulp Fiction on Pulp Fiction &lt;br&gt; &#xA; Features the voice of &lt;a href=&quot;http://www.imdb.com/name/nm0000168/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Samuel L. Jackson&lt;/a&gt;, who appears in Star Wars. &lt;br&gt;&#xA; &lt;br&gt;&#xA; &lt;strong&gt;20th Century Fox&lt;/strong&gt;, The Doors on The Best of The Doors &lt;br&gt;&#xA; The movie studio which owns Star Wars. &lt;br&gt;&#xA; &lt;br&gt;&#xA; &lt;strong&gt;Mr Tambourine Man&lt;/strong&gt;, The Byrds on American Graffiti (More) - 1979 &lt;br&gt;&#xA; Taken from a &lt;a href=&quot;http://www.imdb.com/title/tt0079576/&quot; rel=&quot;nofollow noreferrer&quot;&gt;movie&lt;/a&gt; where George Lucas (director of Star Wars) was one of the writers. Sequel to a &lt;a href=&quot;http://www.imdb.com/title/tt0069704/&quot; rel=&quot;nofollow noreferrer&quot;&gt;movie directed by George Lucas&lt;/a&gt;. &lt;br&gt;&#xA; &lt;br&gt;&#xA; &lt;strong&gt;Out of Space&lt;/strong&gt;, The Prodigy &lt;br&gt;&#xA; Theme of space. &lt;br&gt;&#xA; &lt;br&gt;&#xA; &lt;strong&gt;Space Oddity&lt;/strong&gt;, David Bowie on Best of Bowie &lt;br&gt;&#xA; More space. &lt;br&gt;&#xA; &lt;br&gt;&#xA; &lt;strong&gt;Indiana Jones Theme&lt;/strong&gt;, John Williams on Raiders of the Lost Ark &lt;br&gt;&#xA; Written by &lt;a href=&quot;http://www.imdb.com/name/nm0002354/&quot; rel=&quot;nofollow noreferrer&quot;&gt;the composer who wrote the music for Star Wars&lt;/a&gt;. &#xA; From a movie directed by George Lucas and starring Harrison Ford (director and star of Star Wars respectively) &lt;br&gt;&#xA; &lt;br&gt;&#xA; &lt;strong&gt;Space Cowboy&lt;/strong&gt;, Jamiroquai on The Return of the Space Cowboy &lt;br&gt;&#xA; Yet more space. &lt;br&gt;&#xA; &lt;br&gt;&#xA; &lt;strong&gt;The Galaxy Song&lt;/strong&gt;, Eric Idle on Eric Idle Sings Monty Python &lt;br&gt;&#xA; Have I made the point yet that &lt;em&gt;space is a clue&lt;/em&gt;? :) &lt;br&gt;&#xA; &lt;br&gt;&#xA; &lt;strong&gt;Girl From Mars&lt;/strong&gt;, Ash on 1977 &lt;br&gt;&#xA; All the members of the band were born in 1977, which is also the name of the album - and the year in which Star Wars was released. And the answer to the puzzle, essentially. Also, space. &lt;br&gt;&#xA; &lt;br&gt;&#xA; &lt;strong&gt;Across the Universe&lt;/strong&gt;, Fiona Apple on Across the Universe &lt;br&gt;&#xA; Just in case you missed the 'space' theme. &lt;br&gt;&#xA; &lt;br&gt;&#xA; &lt;strong&gt;Choose Life&lt;/strong&gt;, Underworld &lt;br&gt;&#xA; Features the voice of &lt;a href=&quot;http://www.imdb.com/name/nm0000191/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ewan McGregor&lt;/a&gt;, who stars in Star Wars. &lt;br&gt;&#xA; &lt;br&gt;&#xA; &lt;strong&gt;Born To Be Wild (With Miss Piggy)&lt;/strong&gt; Ozzy Osbourne, on Prince Of Darkness (Box Set)  &lt;br&gt;&#xA; Features the voice of &lt;a href=&quot;http://www.imdb.com/name/nm0000568/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Frank Oz&lt;/a&gt; (as Miss Piggy, not Ozzy), who is also the voice of Yoda. &lt;br&gt;&#xA; &lt;br&gt;&#xA; Q: &lt;em&gt;Was I meant to know all that obscure background information?&lt;/em&gt; &lt;br&gt;&#xA; &lt;br&gt;&#xA; A: No, it's all on Wikipedia and IMDB. Googling the track titles is likely to bring it up pretty easily. &lt;br&gt;&#xA; &lt;br&gt;&#xA; The two sub-puzzles: &lt;br&gt;&#xA; &lt;br&gt;&#xA; * MAGICSIBLINGS are Luke and Leia, obviously. &lt;br&gt;&#xA; &lt;br&gt;&#xA; * PATERNITYDISPUTE: &quot;You're not my father!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hope you had fun - I certainly did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/users/4050/a-e?tab=questions&amp;amp;sort=newest&quot;&gt;If you liked it, look out for more of my puzzles!&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2014-11-17T11:24:50.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4858" PostTypeId="1" AcceptedAnswerId="4861" CreationDate="2014-11-17T10:54:40.943" Score="0" ViewCount="4347" Body="&lt;p&gt;All Rubik's cube positions can be solved in 20 moves or less; &lt;a href=&quot;http://www.cube20.org/&quot; rel=&quot;nofollow&quot;&gt;http://www.cube20.org/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's one such position that requires 19 moves to solve, please give the faces, and the moves to solve.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-11-17T11:29:18.287" LastActivityDate="2014-11-17T13:51:25.490" Title="Rubik's cube 19 move Solution" Tags="|rubiks-cube|" AnswerCount="2" CommentCount="7" ClosedDate="2014-11-18T09:16:23.480" ContentLicense="CC BY-SA 3.0" />
  <row Id="4861" PostTypeId="2" ParentId="4858" CreationDate="2014-11-17T12:04:55.907" Score="21" Body="&lt;p&gt;Given any position you know that has 20 moves (like the Superflip) just do one move towards the solution and you have a position that requires 19 moves.&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2014-11-17T12:04:55.907" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4862" PostTypeId="2" ParentId="4858" CreationDate="2014-11-17T12:05:26.703" Score="4" Body="&lt;p&gt;Below sequence movement from a solved cube will make the position solvable in 19 moves&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; F U' F2 D' B U R' F' L D' R' U' L U B' D2 R' F U2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the solution would be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Step 1: U2&lt;BR&gt;&#xA; Step 2: F'&lt;BR&gt;&#xA; Step 3: R&lt;BR&gt;&#xA; Step 4: D2 &lt;BR&gt;&#xA; Step 5: B &lt;BR&gt;&#xA; Step 6: U' &lt;BR&gt;&#xA; Step 7: L' &lt;BR&gt;&#xA; Step 8: U &lt;BR&gt;&#xA; Step 9: R &lt;BR&gt;&#xA; Step10: D &lt;BR&gt;&#xA; Step11: L' &lt;BR&gt;&#xA; Step12: F &lt;BR&gt;&#xA; Step13: R &lt;BR&gt;&#xA; Step14: U' &lt;BR&gt;&#xA; Step15: B' &lt;BR&gt;&#xA; Step16: D &lt;BR&gt;&#xA; Step17: F2 &lt;BR&gt;&#xA; Step18: U &lt;BR&gt;&#xA; Step19: F'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4637" LastEditorUserId="4637" LastEditDate="2014-11-17T12:11:56.687" LastActivityDate="2014-11-17T12:11:56.687" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4863" PostTypeId="2" ParentId="4828" CreationDate="2014-11-17T12:08:05.433" Score="29" Body="&lt;p&gt;&quot;The other side&quot; meaning, &quot;the heavenly realm&quot; - i.e. the bird didn't make it across the ocean but died in mid-air.&lt;/p&gt;&#xA;" OwnerUserId="4963" LastActivityDate="2014-11-17T12:08:05.433" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="4866" PostTypeId="2" ParentId="4854" CreationDate="2014-11-17T12:57:27.943" Score="32" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 2 or 3.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;For a total of &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; cars, of which &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt; are white, &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; are red, &lt;span class=&quot;math-container&quot;&gt;$g$&lt;/span&gt; are green, and &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; have another colour:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;n &amp;amp; , w, r, g, c \in \mathbb{Z}_{\ge 0} \tag{1} \label{eq1} \\&#xA;n &amp;amp; = w + r + g + c  \tag{2} \label{eq2} \\&#xA;n &amp;amp; \ge c  \tag{3} \label{eq3} \\&#xA;2 &amp;amp; = n - w \implies w = n - 2  \tag{4a} \label{eq4a} \\&#xA;2 &amp;amp; = n - r \implies r = n - 2  \tag{4b} \label{eq4b} \\&#xA;2 &amp;amp; = n - g \implies g = n - 2  \tag{4c} \label{eq4c} \\&#xA;n &amp;amp; = (n-2) + (n-2) + (n-2) + c &amp;amp;&amp;amp; \text{From $\eqref{eq2}, \eqref{eq4a}, \eqref{eq4b}, \eqref{eq4c}$}\\&#xA;n &amp;amp; = 3n - 6 + c \\&#xA;2n &amp;amp; = 6 - c \\&#xA;n &amp;amp; = 3 - \tfrac{c}{2} \tag{5} \label{eq5} \\&#xA;c &amp;amp; = 0 \land n = 3 \Large \lor \normalsize c = n = 2  &amp;amp;&amp;amp; \text{From $\eqref{eq1}, \eqref{eq3}, \eqref{eq5}$}\\&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;As we see, when taking &lt;span class=&quot;math-container&quot;&gt;$\eqref{eq1}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\eqref{eq3}$&lt;/span&gt; into account, &lt;span class=&quot;math-container&quot;&gt;$\eqref{eq5}$&lt;/span&gt; has two solutions, at &lt;span class=&quot;math-container&quot;&gt;$c=0$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$c=2$&lt;/span&gt;.&lt;br /&gt;&#xA;If &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; were any bigger, we'd have &lt;span class=&quot;math-container&quot;&gt;$c&amp;gt;n$&lt;/span&gt; meaning that our millionaire has more other coloured cars than the total number of cars, which would be quite impossible.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If &lt;span class=&quot;math-container&quot;&gt;$c=0$&lt;/span&gt; then &lt;span class=&quot;math-container&quot;&gt;$n=3 \implies w=r=g=n-2=1$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;If &lt;span class=&quot;math-container&quot;&gt;$c=2$&lt;/span&gt; then &lt;span class=&quot;math-container&quot;&gt;$n=2 \implies w=r=g=n-2=0$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; So either he has one white, one red, and one green car, or he has two cars that are neither white nor red nor green.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The first solution corresponds to &lt;a href=&quot;https://puzzling.stackexchange.com/a/4855/68&quot;&gt;AeJey's answer&lt;/a&gt;, the second one to &lt;a href=&quot;https://puzzling.stackexchange.com/a/4856/68&quot;&gt;FlorianF's answer&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditDate="2022-12-29T11:49:13.060" LastActivityDate="2022-12-29T11:49:13.060" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="4869" PostTypeId="2" ParentId="4843" CreationDate="2014-11-17T13:46:30.677" Score="4" Body="&lt;p&gt;I tried to figure out some of those lines. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Born by the death of my mother&lt;/em&gt; possibly means that there are never both of them - maybe I can only make use of one or the old one. &lt;em&gt;The mother&lt;/em&gt; is not useful anymore/broken/...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;We are all empty inside&lt;/em&gt; could mean that I am going to fill it with something and therefore need to get a new one sometimes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You know I'm never in a hurry&lt;/em&gt; at first made me think of a technical device always running at the same speed or an object that is standing still all the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Not hungry for knowledge&lt;/em&gt; means the user will not create value to the object itself by using it. It will only consume or I will consume. &lt;em&gt;Hungry for time&lt;/em&gt; supports this thought. Also it is kind of empty so I came across a hole. So in order to feed it I will have to wait or use it for a long time, maybe frequently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought of a book or a diary which didn't feel right, especially because of &quot;I'm never in a hurry&quot;. I went on and thought of more static object growing wealth over time.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A piggy bank came to my mind. You have to buy a new one as soon as you kill the old one. It keeps secrets because you never know how much money is inside. Well the &quot;millions&quot; are kinda wrong here, I admit :D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's it for now. I have to go on with my own riddle. Thanks for sharing!&lt;/p&gt;&#xA;" OwnerUserId="5106" LastActivityDate="2014-11-17T13:46:30.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4870" PostTypeId="1" AcceptedAnswerId="4871" CreationDate="2014-11-17T13:46:51.927" Score="17" ViewCount="4610" Body="&lt;pre&gt;&lt;code&gt;If you throw me out the window,  &#xA;I'll leave a grieving wife.  &#xA;Bring me back, but through the door,  &#xA;You'll see someone giving life.  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quite old apparently but it took me awhile to figure out.&lt;/p&gt;&#xA;" OwnerUserId="5578" LastEditorUserId="2484" LastEditDate="2014-11-17T13:59:13.030" LastActivityDate="2014-11-17T13:59:13.030" Title="What Am I?...If you throw me out the window" Tags="|riddle|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4871" PostTypeId="2" ParentId="4870" CreationDate="2014-11-17T13:50:33.710" Score="11" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is the Letter &quot;N&quot; Window = Widow Door = Donor&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5247" LastActivityDate="2014-11-17T13:50:33.710" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4873" PostTypeId="2" ParentId="4843" CreationDate="2014-11-17T14:01:44.840" Score="6" Body="&lt;p&gt;Just a guess , not quite sure where time fits in.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Matryoshka_doll&quot;&gt;A Matryoshka Doll?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5581" LastActivityDate="2014-11-17T14:01:44.840" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4874" PostTypeId="2" ParentId="4687" CreationDate="2014-11-17T14:05:40.480" Score="7" Body="&lt;p&gt;ANSWER&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Murderer is the roommate.&lt;br/&gt;&lt;br/&gt;1) when you opened the door, a blast of cold air hit you, so you went to grab your jacket. This means that it is really cold outside, if that is the case, Sarah, the victim, would have been wearing a jacket/coat. Thus, sarah has been killed inside then brought to the alleyway after.&lt;br/&gt;&lt;br/&gt;2) You hear a woman scream, this could have been either sarah or the roommate as that you don't know, thus excluding the boyfriend and the ex-boyfriend. The reason for the scream would be either for sarah about to be murdered and screamed, or that the roommate saw the mouse and screamed. The actual reason is after the roommate brought sarah to the alleyway, she saw the rat and screamed.&lt;br/&gt;&lt;br/&gt;3) when you found the body, it was still warm, thus stating that she has been killed not too long ago. Thus again excluding the boyfriend because he lived 45 mins away. This also points towards the roommate, because of how close the alleyway and the unit is, as well as the reason why sarah was not wearing a jacket, she was inside her house/unit.&lt;br/&gt;&lt;br/&gt;4) Clue 1: only a couple of blood drops have been found, and not a pool of blood. This means that sarah was killed somewhere else then brought to the alleyway, and the lack of spilled blood means that sometime has passed for the blood to dry and stop bleeding but not long enough to comlpetely stop the blood from dripping.&lt;br/&gt;&lt;br/&gt;5) Clue 2: finding the blonde and brunette hair really helped, if you were paying attention, in the bathroom of sarah and the roommate, hair dye was mentioned. when you met the roommate, she has red hair. But if you enter sarahs room, you can notice that one of the picture is the roommates back and she has brunette/brown hair, since it did mention that sarah is taller than the roommate. If you enter the roommates room, you can see that the picture of her and her immediate family, both her parents have brunette/brown hair, thus a very great chance of the roommate having the same color hair.&lt;br/&gt;&lt;br/&gt;6) Clue 3: When the roommate killed sarah, she stepped in some of her blood, as she walked into the alleyway with her body, it made the foot print heading in, the intense cold air would freeze/dry the blood quickly, and when the roommate came back out, it made the foot print on top of the dry prints that pointed in. Thus again excluding the boyfriend, since he lives 45 mins away, the blood would have dried long time ago or be wiped off.&lt;br/&gt;&lt;br/&gt;7) when the old lady told you that she heared something the lines of &quot;I'm tired, i don't want to argue with you as well!&quot;, it was indeed sarah talking to the roommate, that statement proved to be sarah because earlier sarah and the boyfriend had a fight, sarah decided to head back home, and then had an arguement with the roommate, thus stating her saying &quot;i don't want to argue with you as well&quot;.&lt;br/&gt;&lt;br/&gt;8) when you asked the roommate to describe sarah, she kinda went on a rant there, implying that she might be a bit jealous of her and her popularity, and being treated like a sidekick, thus giving her a possible motive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4645" LastEditorUserId="4645" LastEditDate="2014-11-17T14:24:48.810" LastActivityDate="2014-11-17T14:24:48.810" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4875" PostTypeId="1" AcceptedAnswerId="4905" CreationDate="2014-11-17T14:14:54.523" Score="9" ViewCount="1500" Body="&lt;p&gt;So me and my friend frequently make puzzles for each other which can be anything. I wasn't able to figure this one out so you guys might.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those are the three strings he gave me. They are separate strings that all return the same result (they don't need to be combined or anything like that, he just gave me 3 so that it would be somehow easier to crack it apparently):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;essentially snared aggie hafnium tit acrostically blueing casks handkerchief cries mcdonald parred incumbencies corundums spermatozoon rear insolence supercilious &#xA;&#xA;wadis administrator scragged subunits selfness persevering dreidel chartist poundkeeper mcdonald lavalava maybe apostrophizing secobarbital unwinder &#xA;&#xA;augustine pegged unravelled presides briefness rawhided ellipse aweigh bodkin cd exuberance quired regard ruinous excises &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;He was so nice to give me the result of what those 3 things would return if they where &quot;cracked&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; redking&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And he also gave me those tips:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; something is reversed&lt;br&gt;&#xA; a word can only contain one letter&lt;br&gt;&#xA; the letters q,x,y,z don't matter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So my task was to send him another string like those coded with his method with the same phrase.&lt;/p&gt;&#xA;" OwnerUserId="5560" LastEditorUserId="71" LastEditDate="2014-11-24T23:24:11.493" LastActivityDate="2014-11-25T08:40:32.577" Title="Help me break my friend's code!" Tags="|cipher|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4876" PostTypeId="2" ParentId="4843" CreationDate="2014-11-17T14:25:06.967" Score="5" Body="&lt;p&gt;I'm not sure if this answer fits the title or the part about &quot;heat&quot; or the missing word in the last sentence (if those are even clues), but I'd guess it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an hourglass?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is my reasoning:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am born from the death of my mother&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The bottom is &quot;born&quot; and filled as the top is drained, creating the new from the death of the old&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and I have millions of secrets to hide.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Millions of grains of sand in an hourglass&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Like me there are plenty of others&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are many different types of timepieces, also many different hourglasses&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but we are all empty inside.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is always at least enough space for one half of the glass to be empty, and the &quot;child&quot; we're referring to always starts empty&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Maybe you know, I'm never in a hurry&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ever stare at an hourglass and try to will the grains to fall faster?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;due to my origin sublime.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hourglasses measure time, a very sublime concept&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At the same time, I am always hungry,&#xA;not for knowledge,&#xA;but for time.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The grains constantly fall through into the bottom until the top is drained, but no matter how many times you flip the glass over, it will still be hungry for more&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5583" LastActivityDate="2014-11-17T14:25:06.967" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4877" PostTypeId="1" AcceptedAnswerId="4915" CreationDate="2014-11-17T14:40:46.237" Score="31" ViewCount="4209" Body="&lt;p&gt;I am like a dear halo,&lt;br&gt;&#xA;Plucked from a rainbow,&lt;br&gt;&#xA;A tight noose brings me to life,&lt;br&gt;&#xA;With the repeated stab of a knife,  &lt;br&gt;&#xA;Old grief of the elephants,&lt;br&gt;&#xA;Holy like the elements,&lt;br&gt;&#xA;I penetrate a scar of my soft master,&lt;br&gt;&#xA;I elude many, you better run faster.&lt;br&gt; &lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="5373" LastEditDate="2020-04-22T14:55:25.947" LastActivityDate="2020-04-22T14:55:25.947" Title="Riddle... I am like a dear halo" Tags="|riddle|word|rhyme|" AnswerCount="14" CommentCount="22" ContentLicense="CC BY-SA 3.0" />
  <row Id="4878" PostTypeId="2" ParentId="4326" CreationDate="2014-11-17T14:48:42.110" Score="7" Body="&lt;p&gt;My guess, motivated by @nexolute 's answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\pi^2/6$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(assuming the 'characters' you can input don't have to be letters or numbers). Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As many people have already noticed, $\pi$ is important somehow. The Riemann zeta function was mentioned in the third hint, and $\zeta(2)=\pi^2/6$ is its best-known value. The number 6 is suggested by the six-fold piece of paper. And the sum of the reciprocals of all squares of primes is $\pi^2/6$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(In the OP's original question, the numbers on the bomber's last note were (prime)^(prime) instead of (prime)^2; this is why the question generated answers such as 5^5 and took such a long time to solve. But the OP only spotted his error after the solution had already been found!)&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2014-11-18T10:36:36.190" LastActivityDate="2014-11-18T10:36:36.190" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4879" PostTypeId="2" ParentId="4877" CreationDate="2014-11-17T14:53:30.623" Score="3" Body="&lt;p&gt;Could it be &lt;strong&gt;death&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't make this fit with all the lines, but:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;em&gt;A tight noose brings me to life&lt;/em&gt;&quot; - if someone is hanged, death appears on the scene ('comes to life'?). Similarly if someone receives &quot;&lt;em&gt;the repeated stab of a knife&lt;/em&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;em&gt;Old grief of the elephants&lt;/em&gt;&quot; - the famous elephant 'graveyards' (or is this irrelephant?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;em&gt;I elude many, you better run faster&lt;/em&gt;&quot; - people might try to run from death, but can never run fast enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably not...&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2014-11-17T16:17:41.343" LastActivityDate="2014-11-17T16:17:41.343" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4880" PostTypeId="1" AcceptedAnswerId="4882" CreationDate="2014-11-17T14:55:33.400" Score="11" ViewCount="1994" Body="&lt;p&gt;It seems to me that many mysteries involve hats.&#xA;Therefore, a good old hat mystery that I came across a couple of years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Abraham (A), Bingchiang (B), Charly (C) and Derek (D) are sitting in a line facing the same direction in that way that A can see B, C and D. B can see C and D, C can see D and D can see no one.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;gt; &amp;gt; &amp;gt; &amp;gt;&#xA;A B C D&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Ernest is a rich guy that has 5 hats. He shows them 3 white hats and 2 black hats that only differ in color. Thus, they feel exactly the same way. Of those 5 hats, he randomly picks 4 hats and puts one hat on each person's head. The remaining hat is destroyed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ernest claims that if one can guess which color his hat is, they all will receive exactly 625000 dollar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ernest says that A has to go first in guessing, then B, then C and then D. If one of them miss guesses his color, all would die instantly. However, if they don't know their color, they just can say that and the next person would be able to guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Abraham answers first: I can't guess the color of my hat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bingchiang doesn't speak English so mumbles something weird which no-one could understand. Ernest decides that it doesn't count as a wrong answer but that he just lost his turn to guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Charly also answers that he can't guess the color of his hat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Derek is a smart man and knows the color of his hat. He gives the answer to Ernest and they claim their prize rightfully.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you know what Derek's answer was? How did he know what color his hat was? Do you know even more hat colors (Abraham's? Bing's? Charly's?)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: no-one is a chameleon or some sort. Every name is linked to a normal human being without super powers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SPOILER ALERT: A lot of the answers below are not provided with the spoiler alert tag. Beware if you scroll down!&lt;/p&gt;&#xA;" OwnerUserId="420" LastEditorUserId="10615" LastEditDate="2016-02-24T04:09:57.240" LastActivityDate="2016-02-24T04:09:57.240" Title="That hat at that head" Tags="|logical-deduction|meta-knowledge|" AnswerCount="5" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4881" PostTypeId="2" ParentId="4880" CreationDate="2014-11-17T15:01:24.663" Score="1" Body="&lt;p&gt;Well, since there are four people and 5 hats, the combination of hats is either 3 white and 1 black or 2 white and 2 black. If A cannot answer, that means that A sees 2 white and 1 black in front of him. B most likely also sees one black hat, since if there were two black hats B would know that they had a white hat on. I can't figure out why Charly can't guess but if I had to guess by what I have now, ABC are wearing white and D is wearing black sorry I failed :(&lt;/p&gt;&#xA;" OwnerUserId="5441" LastActivityDate="2014-11-17T15:01:24.663" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4882" PostTypeId="2" ParentId="4880" CreationDate="2014-11-17T15:06:22.433" Score="10" Body="&lt;p&gt;Derek is wearing a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; white &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;hat. One of &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bingchian and Charly &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is wearing black and the other white. As for &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Abraham, &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;he is wearing the opposite of the color that was destroyed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reasoning:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If A sees &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3 white &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;hats, he knows he is wearing the opposite color. If he sees &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2 black &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;hats, he knows he is wearing the opposite. But Abraham doesn't know, so by elimination, EXACTLY ONE of B, C, and D is wearing &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; black.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If D is wearing &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; black, &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;then C, reasoning as above, knows the 'exactly one' is D, so C must be wearing &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; white. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But C doesn't know, so D must be wearing &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; white.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2014-11-17T20:44:20.510" LastActivityDate="2014-11-17T20:44:20.510" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="4884" PostTypeId="2" ParentId="4880" CreationDate="2014-11-17T15:13:08.083" Score="1" Body="&lt;p&gt;&lt;strong&gt;A&lt;/strong&gt; can't guess the color of his hat.   If he had seen 3 white, he would have known he's black, similarly, had he seen 2 black, he would have known he's white.  Therefore, he is seeing 2 white and 1 black&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;B&lt;/strong&gt;'s answer is unknown.  He would have seen 2 white, and guessed black, or seen one black, one white and guessed white.  In any case, &lt;strong&gt;B&lt;/strong&gt; would have been able to determine the colour of his own hat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the same logic, &lt;strong&gt;C&lt;/strong&gt; does not see a single black hat, which would make &lt;strong&gt;C&lt;/strong&gt; black.  This tells &lt;strong&gt;D&lt;/strong&gt; that he is wearing a white hat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The order of the hats is: &lt;strong&gt;A&lt;/strong&gt; - ?, &lt;strong&gt;B&lt;/strong&gt; - ?, &lt;strong&gt;C&lt;/strong&gt; - ?, &lt;strong&gt;D&lt;/strong&gt; - W, but we do know that B and C have opposite colours.&lt;/p&gt;&#xA;" OwnerUserId="2752" LastEditorUserId="2752" LastEditDate="2014-11-17T15:18:45.647" LastActivityDate="2014-11-17T15:18:45.647" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4885" PostTypeId="2" ParentId="4880" CreationDate="2014-11-17T15:19:06.647" Score="1" Body="&lt;p&gt;B, C and D combined wear 2 white hats and 1 black hat regardless of which hat is destroyed, since if there were 3 white hats in front of A, or 2 black hats, then A would be able to answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B's answer (or lack thereof) presents two possibilities to C and D - either there are two white hats, or one black hat and one white hat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C's answer solidifies the whole process - if he saw a black hat in front of him, then he would be able to call out that he had a white hat and the group goes away free. However, he did not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This signals to D that he is in fact wearing a white hat.&lt;/p&gt;&#xA;" OwnerUserId="1491" LastActivityDate="2014-11-17T15:19:06.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4886" PostTypeId="2" ParentId="4877" CreationDate="2014-11-17T15:23:30.380" Score="3" Body="&lt;p&gt;is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; light&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;reasoning for it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A break in the clouds. rainbow is made from it refracting. light at the end of the tunnel from dying. many cuts could create a whole in something for light to come through. elephant legend were to be known wonder off and die without anything seeing. only the light reflected could carry this image. Holy light scar could be any imperfection ins a solid that makes it translucent. many people are blind and light is fast so you have to run really fast if you can see but want to escape it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so that or maybe it's &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ghost&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2803" LastEditorUserId="2803" LastEditDate="2014-11-17T15:33:19.567" LastActivityDate="2014-11-17T15:33:19.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4887" PostTypeId="2" ParentId="4877" CreationDate="2014-11-17T15:25:50.307" Score="6" Body="&lt;p&gt;I'm thinking of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Water.&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;em&gt;I'm like a dear halo&lt;/em&gt;, &#xA;&lt;em&gt;Plucked from a rainbow&lt;/em&gt;: Rainbows and water always come together, and the rain seems like to &lt;em&gt;fall&lt;/em&gt; (plucked) from the rainbow and form a wet &lt;em&gt;halo&lt;/em&gt; to the portion of land where its raining.&#xA;&lt;br&gt;&#xA;&lt;em&gt;A tight noose brings me to life, With the repeated stab of a knife&lt;/em&gt;: drop by drop (knife stabs) a tiny hole (noose) is made, which can then lead to an opening through which water can pass.&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;em&gt;Old grief of the elephants, Holy like the elements&lt;/em&gt;: Stones can be considered elephants here: grey, huge and long living. Their grief is that tiny drop that continues to fall on them, filling them with sorrow and pain. Also, water is one of the elements, but the only one that can be &lt;em&gt;holy&lt;/em&gt;.&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;em&gt;I penetrate a scar of my soft master, I elude many, you better run faster.&lt;/em&gt;: Here the master is our planet earth. Water makes its way trough it leaving a scar on it. When a dam its built, everyone thinks that it will resist. When it does not, well, run.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4652" LastActivityDate="2014-11-17T15:25:50.307" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4888" PostTypeId="2" ParentId="4718" CreationDate="2014-11-17T15:45:59.677" Score="1" Body="&lt;p&gt;OK, so here's what I've got so far:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Some symbols are IPA phonetic vowels, others are not&quot; - but they are all IPA phonetic symbols! Taking the IPA code for each symbol, your message becomes: &lt;br&gt;&lt;br&gt; 106, 126, 306, 115, 136, 127, 318, 130 &lt;br&gt; 126, 123 &lt;br&gt; 103, 134, 115, 169, 139, 124, 147 &lt;br&gt; 142, 319, 128 &lt;br&gt; 119, 115, 157, 319, 322 &lt;br&gt; 104, 153, 103, 301, 301, 132&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As for the date...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; following the hint on your user profile, I'd like to try shifting all these numbers by something. But 11+14+14 seems too big. Maybe 11-14+14 or 14-11+14?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Am I on the right track here?&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2014-11-17T18:50:54.203" LastActivityDate="2014-11-17T18:50:54.203" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="4889" PostTypeId="2" ParentId="4854" CreationDate="2014-11-17T15:49:51.383" Score="5" Body="&lt;p&gt;Given the ambiguities in the question (does it mean just the minimum possible, does it mean that they have to be red, white, green, define red/white/green), I think the only really correct answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Possibly 2 or 3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For what it's worth, I agree with the idea that the test is looking for someone seeing the issues with the spec.&lt;/p&gt;&#xA;" OwnerUserId="5594" LastEditorUserId="5594" LastEditDate="2014-11-17T17:07:20.307" LastActivityDate="2014-11-17T17:07:20.307" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4890" PostTypeId="2" ParentId="4877" CreationDate="2014-11-17T16:02:04.093" Score="3" Body="&lt;p&gt;It sounds like&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a thunder&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am like a dear halo,&#xA;Plucked from a rainbow,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; thunders are bright and glorious&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A tight noose brings me to life,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the form of the thunder resembles a noose&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With the repeated stab of a knife, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; probably the repeating sound of many rain drops&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Old grief of the elephants,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; probably the sound of a comming thunder storm&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Holy like the elements,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; thunders are associated with god's anger&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I penetrate a scar of my soft master,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the scar could be the thunder itself in the cloudy sky (that's why soft)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I elude many, you better run faster.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the people run and try to find shelter when thunder storm is comming&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;could it be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a painting? or icon?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5283" LastEditorUserId="5283" LastEditDate="2014-11-17T18:29:30.857" LastActivityDate="2014-11-17T18:29:30.857" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4893" PostTypeId="2" ParentId="4854" CreationDate="2014-11-17T16:41:46.840" Score="9" Body="&lt;p&gt;Two possible answers hinging on the answer to a question; are red, green and white the only &lt;em&gt;possible&lt;/em&gt; colors the cars could be? This meta-question is likely the real test of the puzzle; being able to spot the &quot;hole&quot; in the requirements given by the puzzle (that not all cars in the real world are either white, red or green).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the puzzle enforces an alternate reality where these three colors are the only options even for millionaires, then the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; three cars; one white, one red, one green. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In this case, for each of the three colors, only&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; one &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;of the cars can have that color, and therefore two cars cannot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we're modelling the real real world, then the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; two &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;cars, all in a color other than white, red, or green, such as black, silver or blue. In this situation, the same cars are not white, not red and not green, because they're black (or blue or silver or yellow or...).&lt;/p&gt;&#xA;" OwnerUserId="5600" LastEditorUserId="4204" LastEditDate="2014-11-17T20:17:34.337" LastActivityDate="2014-11-17T20:17:34.337" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4894" PostTypeId="2" ParentId="4877" CreationDate="2014-11-17T16:54:52.180" Score="2" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Planetary ring? I.e. the rings of saturn specifically?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am like a dear halo,&#xA;Plucked from a rainbow,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Is halo shaped, and multi-colored as a rainbow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A tight noose brings me to life,&#xA;With the repeated stab of a knife,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gravity is the tight noose, bringing the frozen ice together in this shape&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Old grief of the elephants,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Planetary rings are made mostly of water, as elephant tears.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Holy like the elements,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We can't see planetary rings directly from earth? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I penetrate a scar of my soft master,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The ring could be composed of pieces of the planet, since original formation is unknown.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I elude many, you better run faster.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Possibly another reference to their invisibility to the naked eye?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2014-11-17T16:54:52.180" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4895" PostTypeId="2" ParentId="4877" CreationDate="2014-11-17T16:56:36.170" Score="2" Body="&lt;p&gt;The colour &lt;strong&gt;red&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;em&gt;I am like a dear halo&lt;/em&gt;&quot; - 'dear' is an anagram of 'read', which is a homophone of 'red'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;em&gt;Plucked from a rainbow&lt;/em&gt;&quot; is pretty self-explanatory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;em&gt;A tight noose brings me to life&lt;/em&gt;&quot; - does a person's face go red when they're hanged?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;em&gt;With the repeated stab of a knife&lt;/em&gt;&quot; - blood-red.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;em&gt;Old grief of the elephants&lt;/em&gt;&quot; - not so sure about this one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;em&gt;Holy like the elements&lt;/em&gt;&quot; - cardinals in the Catholic Church wear red.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;em&gt;I penetrate a scar of my soft master&lt;/em&gt;&quot; - scar tissue is red.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;em&gt;I elude many, you better run faster&lt;/em&gt;&quot; - again, I can't make this one fit.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="2484" LastEditDate="2017-11-29T16:57:04.407" LastActivityDate="2017-11-29T16:57:04.407" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4896" PostTypeId="2" ParentId="4758" CreationDate="2014-11-17T16:57:39.993" Score="0" Body="&lt;p&gt;My guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dinosaurs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Created or evolved whatever you please&#xA;walk in my shoes, which I can't seize&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  T-rex? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A computer originated world, I won't fathom&#xA;my death by some considered gruesome&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dinosaur, again. Killed off in mass extinction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know how to live do not give me tips&#xA;but don't mourn me now, for I am in bits&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mass extinction, again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Suffocate me quick don't tarnish my produce&#xA;remember our sacrifice, our past abuse.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The carbon remains, over time, could be anything from natural gas to diamonds&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2014-11-17T16:57:39.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4897" PostTypeId="2" ParentId="2749" CreationDate="2014-11-17T17:01:04.013" Score="1" Body="&lt;p&gt;Since moss grows on the north side of a tree, it will point the way.&lt;/p&gt;&#xA;" OwnerUserId="5604" LastActivityDate="2014-11-17T17:01:04.013" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4898" PostTypeId="2" ParentId="4719" CreationDate="2014-11-17T17:12:30.623" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When Gabriel chooses the answers they do 5 wrong questions and then answer correctly. When Gabriel chooses it they do &quot;No&quot; 3 times.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Further motivation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Harry accidentally gives them the idea and when they are talking over it they are like&#xA; Gabriel: &quot;Let's just make 5 wrongs before the right?&quot;&#xA; Andrew: &quot;Yeah sure, and if he changes rules we'll do 3 wrongs.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="5610" LastEditDate="2014-11-17T17:19:33.657" LastActivityDate="2014-11-17T17:19:33.657" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4900" PostTypeId="2" ParentId="4877" CreationDate="2014-11-17T17:16:05.343" Score="4" Body="&lt;p&gt;A hunch tells me it is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the different forms of Gold&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cannot make sense of all the lines, but it seems to match some fairly well&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am like a dear halo,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; religious christian imagery used gold foil to depict the &quot;holy radiation&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Plucked from a rainbow,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Leprechauns in Irish folk tales are said to leave a pot of gold at the end of a rainbow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A tight noose brings me to life,&#xA;With the repeated stab of a knife,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; could be a reference to the rubber that flows out of the gum trees when you cut them&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Old grief of the elephants,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; elephants have been hunted for their tusk which is also called white gold&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Holy like the elements,&#xA;I penetrate a scar of my soft master,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; fossil oil is the &quot;black gold&quot; that flows out of the ground&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I elude many, you better run faster.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pyrite/fool's gold could be the one that eludes all&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5608" LastEditorDisplayName="user2174" LastEditDate="2014-11-17T20:08:55.190" LastActivityDate="2014-11-17T20:08:55.190" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4901" PostTypeId="2" ParentId="4877" CreationDate="2014-11-17T17:20:28.960" Score="3" Body="&lt;p&gt;Third try: &lt;strong&gt;pain&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seems to fit all but the first two lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;em&gt;A tight noose brings me to life,&lt;br&gt;&#xA; With the repeated stab of a knife, &lt;br&gt;&#xA; Old grief of the elephants&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A noose, stabbing, and grief are all similar to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Holy like the elements&quot; - I'm sure I remember some quote about pain being holy. Not so sure about the elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;em&gt;I penetrate a scar of my soft master&lt;/em&gt;&quot; - scars, same theme.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;em&gt;I elude many, you better run faster.&lt;/em&gt;&quot; - anaesthetic?&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2014-11-17T17:39:23.633" LastActivityDate="2014-11-17T17:39:23.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4902" PostTypeId="2" ParentId="4807" CreationDate="2014-11-17T17:25:47.167" Score="1" Body="&lt;p&gt;Could the person be &lt;a href=&quot;http://en.wikipedia.org/wiki/Samuel_Johnson&quot; rel=&quot;nofollow noreferrer&quot;&gt;Samuel Johnson&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;He was a writer (inkwell),&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;He visited the Hebrides (see Boswell's '&lt;a href=&quot;http://en.wikipedia.org/wiki/The_Journal_of_a_Tour_to_the_Hebrides&quot; rel=&quot;nofollow noreferrer&quot;&gt;Journal of a Tour to the Hebrides&lt;/a&gt;' and Johnson's '&lt;a href=&quot;http://en.wikipedia.org/wiki/A_Journey_to_the_Western_Islands_of_Scotland&quot; rel=&quot;nofollow noreferrer&quot;&gt;A Journey to the Western Islands of Scotland&lt;/a&gt;'),&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;He lived at Staple Inn.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/F8WlA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't think of a connection to 'Jude the Obscure' though...&lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="4050" LastEditDate="2014-11-17T17:30:52.220" LastActivityDate="2014-11-17T17:30:52.220" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4903" PostTypeId="2" ParentId="4877" CreationDate="2014-11-17T17:36:27.650" Score="2" Body="&lt;p&gt;are you &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Blood&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;halo&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; blood cells are circular&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;rainbow&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; red&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;noose, knife, elephants&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; blood running or surfacing based on bruising.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;holy&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; blood to wine. many other religious references of blood&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;scar&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; recent scars can bleed during the healing process.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;elude&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; many people seek blood transfusions but can't get them?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;better run faster&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; blood flowing from a bag or flowing out of the body&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2803" LastActivityDate="2014-11-17T17:36:27.650" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4904" PostTypeId="2" ParentId="4877" CreationDate="2014-11-17T18:13:33.310" Score="4" Body="&lt;p&gt;This sounds for me like a &lt;em&gt;tattoo&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I am like a dear halo,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Dear to the person who has the tattoo.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Plucked from a rainbow,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tattoos may use inks of different colors.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A tight noose brings me to life,&lt;br&gt;&#xA;  With the repeated stab of a knife,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the noose of the nail that injects the ink by stabbing into the skin.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Old grief of the elephants,&lt;br&gt;&#xA;  Holy like the elements,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not really sure about this... elephants presumably refers to the nooses mentioned before.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I penetrate a scar of my soft master,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A tattoo penetrates the skin of the one who decides to have it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I elude many, you better run faster.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not sure either.&lt;/p&gt;&#xA;" OwnerUserId="4394" LastActivityDate="2014-11-17T18:13:33.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4905" PostTypeId="2" ParentId="4875" CreationDate="2014-11-17T18:16:29.570" Score="13" Body="&lt;p&gt;Find below the decoding method:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Find the alphabet backwards and keep the letter just right of it.  More precisely, scan the words from right to left, but scan the words themselves from left to right.  And pick no more than one letter in each word.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Decoding:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;essentially snared a&lt;strong&gt;GG&lt;/strong&gt;ie ha&lt;strong&gt;FN&lt;/strong&gt;ium tit acrostically blu&lt;strong&gt;EI&lt;/strong&gt;ng casks han&lt;strong&gt;DK&lt;/strong&gt;erchief cries m&lt;strong&gt;CD&lt;/strong&gt;onald parred incum&lt;strong&gt;BE&lt;/strong&gt;ncies corundums spermatozoon re&lt;strong&gt;AR&lt;/strong&gt; insolence supercilious&lt;br&gt;&#xA;&lt;br&gt;&#xA; wadis administrator scra&lt;strong&gt;GG&lt;/strong&gt;ed subunits sel&lt;strong&gt;FN&lt;/strong&gt;ess persevering dr&lt;strong&gt;EI&lt;/strong&gt;del chartist poun&lt;strong&gt;DK&lt;/strong&gt;eeper m&lt;strong&gt;CD&lt;/strong&gt;onald lavalava may&lt;strong&gt;BE&lt;/strong&gt; apostrophizing secob&lt;strong&gt;AR&lt;/strong&gt;bital unwinder&lt;br&gt;&#xA;&lt;br&gt;&#xA; augustine pe&lt;strong&gt;GG&lt;/strong&gt;ed unravelled presides brie&lt;strong&gt;FN&lt;/strong&gt;ess rawhided ellipse aw&lt;strong&gt;EI&lt;/strong&gt;gh bo&lt;strong&gt;DK&lt;/strong&gt;in &lt;strong&gt;CD&lt;/strong&gt; exu&lt;strong&gt;BE&lt;/strong&gt;rance quired reg&lt;strong&gt;AR&lt;/strong&gt;d ruinous excises&lt;/em&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2014-11-25T08:40:32.577" LastActivityDate="2014-11-25T08:40:32.577" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="4907" PostTypeId="1" AcceptedAnswerId="4913" CreationDate="2014-11-17T18:36:10.367" Score="14" ViewCount="1738" Body="&lt;p&gt;Backed by a Benjamin,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Used as an insult: I just can't win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might be surprised to find rocks in me,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just as shocked as when you find me in a tree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can get high when I get wild,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But making you nod is more my style.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am dark and I am light;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;cold tomorrow but hot tonight.&lt;/p&gt;&#xA;" OwnerUserId="4901" LastEditorUserId="4901" LastEditDate="2014-11-17T18:46:39.473" LastActivityDate="2015-04-09T20:39:35.947" Title="Backed by a Benjamin, What am I?" Tags="|riddle|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4908" PostTypeId="2" ParentId="4877" CreationDate="2014-11-17T18:39:18.550" Score="2" Body="&lt;p&gt;How about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Thread?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am like a dear halo,&#xA;Plucked from a rainbow,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Thread can be any color, and is wound in a spool.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A tight noose brings me to life,&#xA;With the repeated stab of a knife,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You have to tie a knot through the needle to sow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Old grief of the elephants,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Elephants bury their dead, a needle must bury the thread under other layers of thread to weave anything. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Holy like the elements,HOLEY. HOW MANY ELEMENTS?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Four classical elements, four directions you sow in.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I penetrate a scar of my soft master,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each place a thread is sown, before there was a hole or space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I elude many, you better run faster. IT IS EASILY LOST&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the thread is not attached to a needle, or becomes removed while weaving, it is difficult to find.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2014-11-17T18:39:18.550" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4909" PostTypeId="2" ParentId="4907" CreationDate="2014-11-17T18:41:42.807" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lightning? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't get the rocks though, so maybe I'm wrong. Wait, yes I do.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Hail&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="51" LastActivityDate="2014-11-17T18:41:42.807" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4910" PostTypeId="2" ParentId="4877" CreationDate="2014-11-17T18:51:41.303" Score="2" Body="&lt;p&gt;how about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a necklace? They were made of ivory in early days and are held together by a string / noose?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5608" LastActivityDate="2014-11-17T18:51:41.303" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="4912" PostTypeId="2" ParentId="4877" CreationDate="2014-11-17T19:06:59.527" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;a needle ? easily lost, stabs repeatidly, uses threads of any color, used to be made of ivory, used to sow back wounds and leaves scar. I don't see what the elements have to do with this, though.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="310" LastEditorUserId="310" LastEditDate="2014-11-17T19:40:54.483" LastActivityDate="2014-11-17T19:40:54.483" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4913" PostTypeId="2" ParentId="4907" CreationDate="2014-11-17T19:08:50.330" Score="17" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Turkey.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Backed by a Benjamin,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Benjamin Franklin is known for supporting the turkey as the national bird over the eagle. &lt;a href=&quot;http://en.wikipedia.org/wiki/Wild_turkey#Benjamin_Franklin_and_the_National_Bird_Suggestion&quot;&gt;This may just be urban legend. However he definitely preferred the turkey over the eagle.&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Used as an insult: I just can't win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might be surprised to find rocks in me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Turkeys swallow rocks to help the gizzard digest food: &lt;a href=&quot;http://en.wikipedia.org/wiki/Gizzard&quot;&gt;http://en.wikipedia.org/wiki/Gizzard&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Just as shocked as when you find me in a tree.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Turkeys sleep in trees to avoid predators&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can get high when I get wild,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wild turkeys can fly&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But making you nod is more my style.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tryptophan in turkey meat caused drowsiness (nodding off)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am dark and I am light;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Turkey is made up of dark meat and light meat&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;cold tomorrow but hot tonight.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; On Thanksgiving it is served hot, but often leftovers are cold the next day&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5616" LastEditorUserId="1955" LastEditDate="2014-11-17T20:11:56.147" LastActivityDate="2014-11-17T20:11:56.147" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="4915" PostTypeId="2" ParentId="4877" CreationDate="2014-11-17T19:18:16.027" Score="22" Body="&lt;p&gt;It is a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;button&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally enough clues to guess it ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relevant details:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am like a dear halo,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Something precious and circular&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Plucked from a rainbow,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In any color&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A tight noose brings me to life,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The eye of a needle/needle and thread&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With the repeated stab of a knife,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; knife = needle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Old grief of the elephants,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ivory buttons?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Holy like the elements,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Has 4 holes like the 4 elements (earth air fire water)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I penetrate a scar of my soft master,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; not sure why it's a scar, but sewn on...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I elude many, you better run faster.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Buttons often get lost&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4901" LastEditorUserId="2484" LastEditDate="2017-10-08T06:55:10.650" LastActivityDate="2017-10-08T06:55:10.650" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4916" PostTypeId="1" CreationDate="2014-11-17T21:57:17.733" Score="1" ViewCount="479" Body="&lt;p&gt;A party is being held at a local mansion. The host is very rich and his success is because of one thing — his famous recipe for Spaghetti!&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only guests that may attend are people who correctly reply to the guard at the door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's where you come in. You and a friend are trying to steal this recipe. You sneak by and listen to the passwords.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first guest arrives. The guard stops him and says: &lt;strong&gt;22&lt;/strong&gt;! The guest after a couple of seconds answers: &lt;strong&gt;4&lt;/strong&gt;! The guard allows him to pass.&lt;br/&gt;&#xA;It's time for the second guest. The guard yells: &lt;strong&gt;136&lt;/strong&gt;! The guest starts to think about it and comes out with &lt;strong&gt;81&lt;/strong&gt;. The guard nods and lets him pass. &lt;br/&gt;&#xA;Your friend has been spotted so he's forced to give an answer. You hear the guard say &lt;strong&gt;44&lt;/strong&gt; and your friend, surprisingly, joins the party with his magic &lt;strong&gt;22&lt;/strong&gt;.&lt;br/&gt;&#xA;It's now your turn, because you can't be beaten by your friend. After the guard tells you &lt;strong&gt;94&lt;/strong&gt;, what will you answer to join your friend, already drunk and dancing with &lt;a href=&quot;http://sidoxia.files.wordpress.com/2010/02/dancing-elephants.jpg&quot; rel=&quot;nofollow&quot;&gt;elephants&lt;/a&gt;?&#xA;&lt;br/&gt;&lt;br/&gt;&#xA;&lt;strong&gt;Trivia&lt;/strong&gt;&lt;br/&gt;&#xA;Tomorrow (18-11) will be my birthday! I will be [answer] years old! It's up to you to find out how old I'll be!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1&lt;/strong&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Another guest arrives. The guard says &lt;strong&gt;11&lt;/strong&gt; and the guest answers &lt;strong&gt;1&lt;/strong&gt; and comes in with ease!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1685" LastEditorUserId="71" LastEditDate="2014-11-30T22:08:15.093" LastActivityDate="2014-11-30T22:08:15.093" Title="The security to the party[#37] - Wish me &quot;Happy birthday&quot;!" Tags="|password|pattern|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="4917" PostTypeId="1" AcceptedAnswerId="4918" CreationDate="2014-11-17T22:24:55.410" Score="1" ViewCount="503" Body="&lt;p&gt;&lt;strong&gt;Not Exactly one of the Fourteen&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I am ruin's servant. My name is well known.&lt;br&gt;&#xA;  Because you're unobservant. Your frustration has grown.&lt;br&gt;&#xA;  My opposite's opposite is wrath, though I am certainly not.&lt;br&gt;&#xA;  Reflecting my brother on the path. I am his heart subjected to rot.&lt;br&gt;&#xA;  You leave the page all the same and I laugh to myself with glee.&lt;br&gt;&#xA;  You'll not learn my name because you are me.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5626" LastEditorUserId="15544" LastEditDate="2015-09-04T07:03:37.580" LastActivityDate="2015-09-04T07:03:37.580" Title="Not Exactly one of the Fourteen - A simple riddle about the servant of ruin" Tags="|riddle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4918" PostTypeId="2" ParentId="4917" CreationDate="2014-11-17T22:31:15.693" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Impatience&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;An antonym for wrath is patience, if you leave the page you won't solve the riddle because you are impatient. Impatience can be described as ruin's servant. Your heart (passion) for something can be rotted by being impatient.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5052" LastEditorUserId="4050" LastEditDate="2014-11-17T22:54:40.523" LastActivityDate="2014-11-17T22:54:40.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4919" PostTypeId="2" ParentId="4917" CreationDate="2014-11-17T22:36:29.190" Score="2" Body="&lt;p&gt;The answer is right in front of us!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well Known&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;My name is well known. Because you're unobservant. Your frustration has grown.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2014-11-17T22:36:29.190" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4920" PostTypeId="2" ParentId="4798" CreationDate="2014-11-17T22:54:01.597" Score="0" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is no doomsday.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As aschepler pointed out, The numbers refer to the compound DyNAmITe. If Dynamite is defined as negative DyNAmITe (ignoring case in the function not sensitive to it), the list will never iterate to the next value. It's a recursive loop. Doomsday will never arrive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5626" LastActivityDate="2014-11-17T22:54:01.597" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4922" PostTypeId="1" AcceptedAnswerId="4937" CreationDate="2014-11-17T23:34:27.970" Score="32" ViewCount="2185" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;Our father was the first of our kind, though his many children and grandchildren rule over billions, caging them.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;I am a liberator, and my father shelters those who travel through water.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Father once bled silver till there was none left to bleed, and my closest brother could no longer move; he was dead.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Sister wisdom remains partially with us, though she will not come to our aid.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Some of my siblings have revolted and hurry toward our doom. One of my brothers is the very hole of a noose. Another crafted weapons for the thrice born empire before coming after me.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;The Eight come, seven shadow siblings with curved blades, across the sea to feast upon iron and my father's heart. I alone remain faithful. And all along... the rumble... the smoke... the lightning flash... Speak my father's name and he may yet live.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5626" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-09-15T20:16:18.540" Title="Speak the Name of the Father! - Name the lord of the gods so that he may yet survive the theomachy" Tags="|riddle|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4923" PostTypeId="2" ParentId="4922" CreationDate="2014-11-18T00:12:45.647" Score="-1" Body="&lt;p&gt;Sounds like a riddle for christian teaching to me.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;The father - God.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;One of my brothers is the very hole of a noose - Satan/The devil.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;Thrice born empire - Jesus dies twice, and is &amp;quot;born&amp;quot; three times.  (not perfectly accurate)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;Father once bled silver till there was none left to bleed - death on a cross.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;My closest brother could no longer move; he was dead - Satan/Death?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;I alone remain faithful - Jesus, the only human to never sin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;The Eight come, seven shadow siblings with curved blades, across the sea to feast upon iron and my father's heart. - It's been a long time since I read Revelations (the last book in the bible), but I remember something about seven &amp;quot;beings of destruction&amp;quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5479" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-11-18T00:12:45.647" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4927" PostTypeId="1" AcceptedAnswerId="4929" CreationDate="2014-11-18T01:12:01.433" Score="3" ViewCount="642" Body="&lt;p&gt;Ciphertext: 1010, 111, 11, 0110, 001, 1, 0, 010, 000&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This cryptogram is based off of AE's first one. Enjoy.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to this is a clue to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4668/the-security-to-the-party-part-30-1&quot;&gt;this question&lt;/a&gt; which is currently unanswered and has a bounty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Technical Stuff&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No Roman numerals&lt;br&gt;&#xA;No abbreviations&lt;br&gt;&#xA;No slang or shorthands&lt;br&gt;&#xA;Answer is relatively short... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have fun!&lt;/p&gt;&#xA;" OwnerUserId="5368" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2014-11-18T21:00:07.003" Title="Morse code cryptogram" Tags="|cipher|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4928" PostTypeId="2" ParentId="4828" CreationDate="2014-11-18T03:08:34.857" Score="9" Body="&lt;p&gt;The bird speed is expressed in Irish miles, with 1 Irish mile = 2.048 km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ocean's width is expressed in normal miles, with 1 mile = 1.609 km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bird flies at 5mph for one hour and covers 5 * 2.048 = &lt;strong&gt;10.240 km&lt;/strong&gt; to cross the ocean that is 6 * 1.609 = &lt;strong&gt;9.656 km&lt;/strong&gt; in width.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://en.wikipedia.org/wiki/Mile&quot;&gt;http://en.wikipedia.org/wiki/Mile&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4028" LastActivityDate="2014-11-18T03:08:34.857" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4929" PostTypeId="2" ParentId="4927" CreationDate="2014-11-18T03:11:00.690" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think it's &lt;code&gt;COMPUTERS&lt;/code&gt;. If you translate 0 as dot and 1 as dash in Morse code, except the &lt;code&gt;0011&lt;/code&gt; needs to be split as &lt;code&gt;001&lt;/code&gt; (U) and &lt;code&gt;1&lt;/code&gt; (T). There's no Morse code symbol for &lt;code&gt;0011&lt;/code&gt;. Edit: The dash character &lt;code&gt;-&lt;/code&gt; is apparently encoded by &lt;code&gt;0011&lt;/code&gt; so possibly &lt;code&gt;COMP-ERS&lt;/code&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2014-11-18T03:24:37.883" LastActivityDate="2014-11-18T03:24:37.883" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4930" PostTypeId="2" ParentId="4922" CreationDate="2014-11-18T05:18:58.570" Score="3" Body="&lt;p&gt;Your Father is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The &lt;strong&gt;Stomach&lt;/strong&gt; or The &lt;strong&gt;Digestive System&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the mouth (the only one faithful to your father)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;though his many children and grandchildren rule over billions, caging them refers to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the brain, muscles and bones &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a liberator and my father shelters those who travel through water is about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the food particles mixed with saliva and other stomach fluids and I am a liberator refers to the mouth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Father once bled silver till there was none left to bleed refers to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; excrement of metabolic waste&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;my closest brother could no longer move refers to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;any skull bone other than jaw bone&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sister wisdom remains partially with us, though she will not come to our aid refers to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the brain and the brain can't control the hunger&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some of my siblings have revolted and hurry toward our doom refers to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;hunger&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One of my brothers is the very hole of a noose refers to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the anus&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Eight come, seven shadow siblings with curved blades, across the sea to feast upon iron and my father's heart refers to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the rib cage&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I alone remain faithful. And all along... the rumble... the smoke... the lightning flash... refers to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the mouth receives/gets food for the digestive system or stomach (the father) and so mouth is the one faithful to his father.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="2730" LastEditDate="2014-11-18T05:33:20.727" LastActivityDate="2014-11-18T05:33:20.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4931" PostTypeId="1" AcceptedAnswerId="4940" CreationDate="2014-11-18T05:29:02.247" Score="10" ViewCount="675" Body="&lt;p&gt;It is well-known (and has been &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2861/how-many-knights-on-a-chess-board&quot;&gt;asked here earlier&lt;/a&gt;) how many knights can fit on a chessboard without any knights attacking any others. But what if each knight can attack at most one other?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Idea from &quot;The Mathematical Knight&quot; by Noam Elkies and Richard Stanley.&lt;/p&gt;&#xA;" OwnerUserId="1752" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2014-11-18T22:02:33.883" Title="Knights on a Chessboard II" Tags="|chess|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4932" PostTypeId="1" AcceptedAnswerId="4936" CreationDate="2014-11-18T05:42:48.227" Score="19" ViewCount="438" Body="&lt;p&gt;These are the squares from 1 to 100 in Japanese kanji:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;百、六十四、十六、九、八十一、三十六、四、一、二十五、四十九&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is which?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you already know Japanese/Chinese numbers, then sorry, this puzzle is not for you.  Needless to say, looking it up is cheating, even to check your answer.&lt;/p&gt;&#xA;" OwnerUserId="1752" LastEditorUserId="8874" LastEditDate="2016-03-23T17:11:19.857" LastActivityDate="2016-10-11T14:55:04.393" Title="Linguistics: Japanese Squares" Tags="|cipher|language|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4936" PostTypeId="2" ParentId="4932" CreationDate="2014-11-18T06:18:48.590" Score="16" Body="&lt;p&gt;Great puzzle!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the numbers are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1    一&lt;br&gt;&#xA; 4    四&lt;br&gt;&#xA; 9    九&lt;br&gt;&#xA; 16   十六&lt;br&gt;&#xA; 25   二十五&lt;br&gt;&#xA; 36   三十六&lt;br&gt;&#xA; 49   四十九&lt;br&gt;&#xA; 64   六十四&lt;br&gt;&#xA; 81   八十一&lt;br&gt;&#xA; 100  百 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's how I reasoned.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are four one-symbol numbers 一, 四, 九, 百. There's only three one-digit squares, so I reasoned that one must be &lt;code&gt;100&lt;/code&gt;. 百 fits this by appearing nowhere else. I guess that the stick 一 is probably 1. Of 4 and 9, there's a square 49, which uses both symbols 四十九. But I don't know if I should read left to right or right to left. But 4 also appears in 64, and 9 doesn't appear elsewhere, so that means 六十四 is 64 and 四 is 4. So, left to right. Only here did I notice that 十 was in the middle of every three-symbol example. I had thought it meant 50 before, but that doesn't make sense. It must be a digit separator, or maybe the symbol ten (like &quot;six tens and four&quot;). The second one works great with 十六, which must be sixteen, since &quot;ten&quot; makes sense for &quot;one ten&quot;. And that matches the 6 from 64, great. This leaves only 81, recognized from the 1, and 25, which remains and overlaps no digits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2014-11-18T06:18:48.590" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4937" PostTypeId="2" ParentId="4922" CreationDate="2014-11-18T06:29:32.697" Score="15" Body="&lt;p&gt;Maybe&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ford? The car company?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&#xA;&lt;strong&gt;Reasoning:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The first of our kind&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not quite right, but a first with respect to mass production&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;his many children and grandchildren rule over billions, caging them&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lots of other cars&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am a liberator&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lincoln Motors/Abraham Lincoln&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;my father shelters those who travel through water&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Referring to fording a river&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Father once bled silver till there was none left to bleed, and my closest brother could no longer move; he was dead.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mercury, a now-defunct sub-brand of Ford&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Sister Wisdom&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mazda, named after &lt;a href=&quot;http://en.wikipedia.org/wiki/Ahura_Mazda&quot;&gt;Ahura Mazda&lt;/a&gt; - &quot;The literal meaning of the word Ahura is light and Mazda is wisdom.&quot; and Mazda used to have a partnership with Ford.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The very hole of a noose&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Honda_knot&quot;&gt;Honda&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Weapons for the thrice-born empire&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mitsubishi manufactured war planes during WW2 (Third Reich)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The Eight&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Perhaps Toyota? Wikipedia says the name's &quot;eight-stroke count in the Japanese language was associated with wealth and good fortune&quot;, but I'm pretty much lost here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And all along... the rumble... the smoke... the lightning flash...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The internal combustion engine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&#xA;Not sure about the revolting siblings and a few other bits, but I'll sleep on that.&lt;/p&gt;&#xA;" OwnerUserId="5340" LastActivityDate="2014-11-18T06:29:32.697" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4940" PostTypeId="2" ParentId="4931" CreationDate="2014-11-18T06:50:24.103" Score="12" Body="&lt;p&gt;The maximum possible is number of knights is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 32&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This can be achieved by:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Putting a &lt;a href=&quot;http://mathworld.wolfram.com/images/eps-gif/KnightsMax_700.gif&quot; rel=&quot;nofollow&quot;&gt;knight on every white square&lt;/a&gt; (thanks to Timtech for the graphic). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's also other nontrivial configurations, like:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A knight on every edge or corner square (28), plus the four very center squares.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is optimal because of existence of a construction with the following properties:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Partition the chessboard into 16 sets of 4 squares, with each set forming a a loop of four connected by knight moves. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One instance of this construction is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Splitting the 4x4 square into four sets as follows&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;code&gt;1342&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;4213&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;3124&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;2431&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;Then, split the chessboard into four 4*4 quadrants each with a copy of this pattern.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This construction proves optimality because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each set of 4 can only have 2 knights, since if there are 3, one knight must be attacking two knights. So, this gives a maximum of 2*16=32.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2014-11-18T22:02:33.883" LastActivityDate="2014-11-18T22:02:33.883" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4941" PostTypeId="2" ParentId="4932" CreationDate="2014-11-18T06:53:39.220" Score="5" Body="&lt;p&gt;Answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 百 100&lt;BR&gt;&#xA; 六十四 64&lt;BR&gt;&#xA; 十六 16&lt;BR&gt;&#xA; 九 9&lt;BR&gt;&#xA; 八十一 81&lt;BR&gt;&#xA; 三十六 36&lt;BR&gt;&#xA; 四 4&lt;BR&gt;&#xA; 一 1&lt;BR&gt;&#xA; 二十五 25&lt;BR&gt;&#xA; 四十九 49&lt;BR&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Approach&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 一 and 百 should be 1 4 or 9. 十 is multiply by ten I guessed by looking at its frequency. 一 is also used in at unit digit in another number so it is 1 and 百 should be 100.(When I assume 一 is 4, assumption fails)&lt;BR&gt;&#xA; 一 is 1 so I guess, 二 is 2 and 三 is 3.&lt;BR&gt;&#xA; Because 3 is never unit digit in any square, number should be read from left to right&lt;BR&gt;&#xA; 四 is either 4 or 9. It is placed in units as well as tens digit so it must be 4. (9 can be used at units digit but can not be at tens digit)&lt;BR&gt;&#xA; 二十五, as 二十 is 20, 五 must be 5.&lt;BR&gt;&#xA; Looking at 六十四, 64 is the only number with 4 at units digit, so 六 is 6. &lt;BR&gt;&#xA; Looking at 八十一, 八 must be 8 as 81 is the only square with unit digit 1. &lt;BR&gt;&#xA; Single digit square 九 can be 1, 4 or 9. But 1 and 4 are already taken, so it must be 9.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4637" LastEditorUserId="4637" LastEditDate="2014-11-18T07:33:14.853" LastActivityDate="2014-11-18T07:33:14.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4943" PostTypeId="2" ParentId="4828" CreationDate="2014-11-18T08:18:43.320" Score="3" Body="&lt;p&gt;The bird is flying normally with 5 mph and it mades it to the other side of the ocean, because the water underneath is moving. For example, a high tide starting at the same time as the bird's flight and with development direction opposite to the bird's flight direction. If the high tide has a peak in an hour and the water displacement is about 1 mile, there you go, the bird succeded in its adventure.&lt;/p&gt;&#xA;" OwnerUserId="5283" LastEditorUserId="5283" LastEditDate="2014-11-19T10:19:31.210" LastActivityDate="2014-11-19T10:19:31.210" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4944" PostTypeId="2" ParentId="4360" CreationDate="2014-11-18T08:33:03.800" Score="8" Body="&lt;p&gt;Here are the decoded notes. &lt;em&gt;Italics for handwritten&lt;/em&gt;, &lt;strong&gt;bold for typed&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;#1: ALL WRITTEN NOTES ARE TRUE&lt;/strong&gt;&lt;br&gt;&#xA; &lt;em&gt;#2: NUMBER ONE IS THE ONLY CORRECT TYPED NOTE&lt;/em&gt;&lt;br&gt;&#xA; &lt;strong&gt;#3: I AM THE KILLER - VP&lt;/strong&gt;&lt;br&gt;&#xA; &lt;em&gt;#4: NUMBER SIX IS THE RIGHT CLUE&lt;/em&gt;&lt;br&gt;&#xA; &lt;strong&gt;#5: YOU WONT CATCH ME&lt;/strong&gt;&lt;br&gt;&#xA; &lt;em&gt;#6: PREZ TOD WINTER FALL SUMMER SPRING CHEESESTEAK&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I suppose:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Our next stop is Hotel Four Seasons in Philadelphia (as in Philadelphia cheesesteak). Also, the VP is off our list of suspects.&lt;br&gt;&#xA; &quot;PREZ TOD&quot; must refer to the President's time of death, which was November 5, 2014 at 1:09 p.m. That's most likely our clue for the room number (109 maybe?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;About decrypting the messages:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The notes are encrypted with a &lt;a href=&quot;http://rumkin.com/tools/cipher/vigenere.php&quot;&gt;vigenére cipher&lt;/a&gt;, using the sequence of pseudorandom numbers obtained by using a variation of John von Neumann's middle square method using 121 as a seed, taking three numbers from the end (offset by one) and replacing all zeroes with ones. For those interested, &lt;a href=&quot;https://gist.github.com/Kaivosukeltaja/1f1921dba9e3d3c9f2a6&quot;&gt;here&lt;/a&gt; is the ugly but working Python implementation I ended up with.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5641" LastEditorUserId="5641" LastEditDate="2014-11-18T09:01:06.263" LastActivityDate="2014-11-18T09:01:06.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4950" PostTypeId="1" AcceptedAnswerId="4954" CreationDate="2014-11-18T10:33:28.073" Score="4" ViewCount="920" Body="&lt;pre&gt;&lt;code&gt;I am dying at each set and i am born on each rise&#xA;No one has seen me or will see me but i am always there&#xA;I dont move and always stay where i am  &#xA;I will live on, until time's own end&#xA;What am I?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;A riddle i made, inspired by another little riddle&lt;/p&gt;&#xA;" OwnerUserId="5578" LastEditorUserId="2484" LastEditDate="2014-11-18T10:39:04.263" LastActivityDate="2016-04-29T01:55:02.457" Title="What am I? Time's own end" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4951" PostTypeId="1" AcceptedAnswerId="4953" CreationDate="2014-11-18T10:33:33.037" Score="0" ViewCount="1221" Body="&lt;p&gt;There were $3$ BOYS living in a house with their father and mother.&lt;br&gt; One day, they decided no matter what, they are going to eat ONE THING.&lt;br&gt;&#xA;One of them said: &quot;Today is &lt;strong&gt;SUNDAY&lt;/strong&gt;, we can wake up late and take our time to eat that THING we were craving about&quot;.&lt;br&gt;&#xA;As they were ready to get back to sleep, their mother yelled &quot;You'd better wake up and clean that dirty room of yours&quot;.&lt;br&gt;&#xA;They had no other choice but to obey their mother. It was 8AM when they started the chore.&lt;br&gt;&#xA;After one hour , the room was clean and the drooling 3 boys thought it is time to eat that THING.&lt;br&gt;&#xA;But their joy didn't last. The mother yelled again: &quot;Have you finished your &lt;strong&gt;HOMEWORK&lt;/strong&gt;?&quot;. They replied in the &lt;strong&gt;NEGATIVE&lt;/strong&gt;.&lt;br&gt;&#xA;&quot;You better finished your HOMEWORK before your father gets back from the &lt;strong&gt;FARM&lt;/strong&gt;&quot;.&lt;br&gt;&#xA;Their homework was so hard that it took them four hours to finish.&lt;br&gt;&#xA;Just in time, their father got back from the FARM to find the $3$ boys sobbing.&lt;br&gt;&#xA;&quot;What's the &lt;strong&gt;MATTER&lt;/strong&gt;?&quot; asked the father.&lt;br&gt;&#xA;&quot;We can not eat that THING anymore, ...&quot;&lt;br&gt;&#xA;What is the thing the $3$ boys wanted to badly eat?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The family lives in Japan.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4973" LastEditorUserId="10281" LastEditDate="2015-04-25T14:23:17.863" LastActivityDate="2015-04-25T14:23:17.863" Title="What is the thing the boys could not eat?" Tags="|lateral-thinking|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4952" PostTypeId="2" ParentId="4951" CreationDate="2014-11-18T10:35:53.917" Score="2" Body="&lt;p&gt;The &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Sacramental_bread&quot; rel=&quot;nofollow&quot;&gt;Body of Christ&lt;/a&gt;, distributed on sunday morning, among the people in the church.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3993" LastActivityDate="2014-11-18T10:35:53.917" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4953" PostTypeId="2" ParentId="4951" CreationDate="2014-11-18T10:40:49.457" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Breakfast. (Assuming one doesn't eat breakfast at 1 pm.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or to be more specific:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Breakfast at the local restaurant (as they normally start serving lunch/brunch at noon).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After the hint about it being in Japan.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Breakfast is in Japanese literally called &quot;Morning Meal&quot;. You wouldn't consider 1 pm as a morning (at least not when greeting people), thus still: Breakfast &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5517" LastEditorUserId="5517" LastEditDate="2014-11-18T11:07:00.043" LastActivityDate="2014-11-18T11:07:00.043" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4954" PostTypeId="2" ParentId="4950" CreationDate="2014-11-18T10:43:43.500" Score="5" Body="&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tomorrow.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am dying at each set and i am born on each rise&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Tomorrow stops being tomorrow (dies) and starts being today - and moving to the next day (reborn).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No one has seen me or will see me but i am always there&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If you see tomorrow, it is &quot;today&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I dont move and always stay where i am&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Tomorrow is always out of reach at a constant range from where we are.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I will live on, until time's own end&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There will always be a &quot;tomorrow&quot; until the end of time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="7586" LastEditDate="2016-04-29T01:55:02.457" LastActivityDate="2016-04-29T01:55:02.457" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4955" PostTypeId="2" ParentId="4790" CreationDate="2014-11-18T10:43:52.713" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Your brain?   or a braincell?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I always act like my brain is another (3th person view) of my world like some spectator haha&lt;/p&gt;&#xA;" OwnerUserId="5578" LastEditorUserId="5578" LastEditDate="2014-11-18T11:05:08.100" LastActivityDate="2014-11-18T11:05:08.100" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4957" PostTypeId="2" ParentId="4751" CreationDate="2014-11-18T10:59:34.943" Score="5" Body="&lt;p&gt;Without a moment's hesitation, I answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;8&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The host is a famous spaghetti chef, so of course he must be Italian. The guard gives the guests numbers and the correct answer for each is their letter count when translated to Italian. For 22, the translation is &quot;ventidue&quot; which has eight letters.&lt;br&gt;&lt;br&gt;&#xA; All the clues I've found towards this:&#xA; &lt;br&gt; 1. The whole spaghetti thing (which has been mentioned in 37 puzzles here but only used as a clue in this one).&#xA; &lt;br&gt; 2. Everybody's arriving late except for the poor Spanish guy at 7pm. ;)&#xA; &lt;br&gt; 3. &quot;Can &lt;strong&gt;IT&lt;/strong&gt; really be that simple?&quot; (country code for Italy)&#xA; &lt;br&gt; 4. Guests 7 and 8 are discussing in a foreign language.&#xA; &lt;br&gt; &lt;br&gt; &lt;strong&gt;Added by OP&lt;/strong&gt;:&#xA; &lt;br&gt; 5. Your friend Narmer &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3739/the-old-man-doesnt-like-seagull-meat#comment10784_3739&quot;&gt;is Italian&lt;/a&gt;.&#xA; &lt;br&gt; 6. Instead of being trampled by elephants like at the other spaghetti parties, guests who answer wrongly are shot by a mafioso!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My shot at the bonus question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The guard chooses the numbers based on the host's name. He begins with a 4 - &lt;strong&gt;D&lt;/strong&gt; is the fourth letter of the alphabet. Followed by &lt;strong&gt;O, N, L, U, I, G, C, A, V&lt;/strong&gt;. The host's name must be &lt;strong&gt;Don Luigi Cavaliere&lt;/strong&gt; or something similar. Our seventh clue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5641" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2014-11-19T11:53:34.200" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4958" PostTypeId="2" ParentId="4758" CreationDate="2014-11-18T11:05:47.537" Score="1" Body="&lt;p&gt;Seems to be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Created or evolved whatever you please&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is a creature of some sort. Definitely a physical thing. Refers to the religious arguments about &quot;physical&quot; existence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;walk in my shoes, which I can't seize&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Cow have no say about where they go.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A computer originated world, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;C&lt;/strong&gt;omputer  &lt;strong&gt;O&lt;/strong&gt;riginated  &lt;strong&gt;W&lt;/strong&gt;orld (dear God.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I won't fathom&#xA;my death by some considered gruesome&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The death of a cow even in a humane abbatoir can be quite brutal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know how to live do not give me tips&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Leave me alone, don't tip me over!&quot; (a past-time of certain people includes pushing cows over).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but don't mourn me now, for I am in bits&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Cows are sold in small pieces.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;suffocate me quick don't tarnish my produce&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You must kill the cow quickly or the adrenaline can affect the meat. May also be some kind of reference to meat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;remember our sacrifice, our past abuse.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Cows have been sacrificed since time immemorial.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My name is changed once deceased&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Cow becomes beef once killed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yet it holds the same meaning&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is the exact same thing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;well done! I am rare once the fires have ceased&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Beef can be &quot;well done&quot;. But it &quot;rare&quot; if you take off the heat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I humble your buds once you begin feeding&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Delicious&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-11-18T11:34:15.483" LastActivityDate="2014-11-18T11:34:15.483" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4964" PostTypeId="1" AcceptedAnswerId="4974" CreationDate="2014-11-18T12:13:06.963" Score="22" ViewCount="1098" Body="&lt;p&gt;&lt;em&gt;Inspired by Lopsy's &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4932/linguistics-japanese-squares&quot;&gt;excellent puzzle idea&lt;/a&gt;, I thought I'd try my hand at coming up with a puzzle that's not simply a real-world language.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once again, we have the squares 1, 4, 9, ... , 81, 100 in some order — but this time it's in an alien language! Once you figure out which is which, can you tell me how you think you would say 42 and 58?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;an&#xA;ansont&#xA;ansont carton-catr&#xA;ansont derton el an&#xA;cartva&#xA;catr&#xA;catrsont&#xA;detrsont derton-catr&#xA;yorton el an&#xA;yotrsont anva&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="445" LastActivityDate="2014-11-18T13:47:09.487" Title="Squares in an alien language" Tags="|encoded-message|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4967" PostTypeId="2" ParentId="4951" CreationDate="2014-11-18T12:54:19.823" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ice.  It is freezing at night, they had put syrup on the window-sill overnight.  But now it is all melted.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Or maybe it was a dessert made of fresh snow and syrup.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="10748" LastEditDate="2015-04-25T14:16:42.233" LastActivityDate="2015-04-25T14:16:42.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4968" PostTypeId="2" ParentId="4951" CreationDate="2014-11-18T12:56:02.747" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the boys wanted to sleep.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;although i don't know japanese, but i think it is related &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; that in japanese they say 'i eat sleep' or something.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1669" LastActivityDate="2014-11-18T12:56:02.747" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4971" PostTypeId="2" ParentId="4916" CreationDate="2014-11-18T13:06:33.740" Score="1" Body="&lt;p&gt;My hunch is that it's got something to do with&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; dates. Each number called by the guard could be interpreted as a date (22 = 2 Feb, 136 = 13 Jun, 44 = 4 Apr, 94 = 9 Apr), and of course there's the mention of a birthday at the end. &lt;br&gt; &lt;br&gt; 2 Feb is the 33rd day of the year, 13 Jun the 164th (165th in a leap year), 4 Apr the 94th (95th in a leap year), and 9 Apr the 99th (100th in a leap year). So what function sends 33 to 4, 164 to 81, 94 to 22, and (thanks @Napokue) 99 to 27? Is it significant that 22 and 27 differ by 5, as do 94 and 99?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Am I on the right lines here?&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2014-11-18T13:06:33.740" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4973" PostTypeId="2" ParentId="4790" CreationDate="2014-11-18T13:10:59.707" Score="0" Body="&lt;p&gt;forgive me if im wrong...  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cyanoacrylate &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For sure I'm egocentric&#xA;You mean nothing to me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An ego centric letter is S which starts the most words in the english dictionary &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the best choice seems a third person to be  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a different perspective or from a different persona leading me to PER also in PERson  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Make him steal what surrounds you  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;thought&quot; being deep in thought throughout this riddle which leaves us with U  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Find one more glue  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Super Glue which its proper name is Cyanoacrylate&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5135" LastActivityDate="2014-11-18T13:10:59.707" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4974" PostTypeId="2" ParentId="4964" CreationDate="2014-11-18T13:17:24.487" Score="15" Body="&lt;p&gt;I would say 42 as &lt;em&gt;ansont yorton-detr&lt;/em&gt; and 58 as &lt;em&gt;detrsont yortva&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will write up my line of thinking, but I really hope I'm on the right track :)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I suspected that these aliens would not use base 10. I tried with 16 and 8, which looked promising, but I didn't get far.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Looking at the given words, there seem to be some &quot;multipliers&quot;: &lt;em&gt;sont&lt;/em&gt; and &lt;em&gt;ton&lt;/em&gt; and &lt;em&gt;va&lt;/em&gt;; as basic numerals, it seems we have four candidates: &lt;em&gt;an&lt;/em&gt;, &lt;em&gt;catr&lt;/em&gt;, &lt;em&gt;detr&lt;/em&gt; and &lt;em&gt;yotr&lt;/em&gt;.&lt;br&gt;&#xA; &lt;em&gt;El&lt;/em&gt; I assume to be &lt;em&gt;and&lt;/em&gt; or &lt;em&gt;minus&lt;/em&gt;. It seems that in combination with the &lt;em&gt;-ton&lt;/em&gt; and the &lt;em&gt;-va&lt;/em&gt; ending, &lt;em&gt;tr&lt;/em&gt; becomes &lt;em&gt;rt&lt;/em&gt;, probably because it is easier to pronounce. Combined with &lt;em&gt;-sont&lt;/em&gt;, &lt;em&gt;tr&lt;/em&gt; stays &lt;em&gt;tr&lt;/em&gt;.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Since we have four single digit words, and &lt;em&gt;zero&lt;/em&gt; is unlikely to be used, I tried base-5 (0,1,2,3,4).&#xA; If we write the first ten squares as base-5, we get:&lt;br&gt;&#xA; 1, 4, 14, 31, 100, 121, 144, 224, 311, 400.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Assuming &lt;em&gt;sont&lt;/em&gt; to mean 100, and having three ansonts, it seems &lt;em&gt;an&lt;/em&gt; = 1. That solves &lt;em&gt;an&lt;/em&gt; and &lt;em&gt;ansont&lt;/em&gt; (1 and 100 base-5). &#xA; If &lt;em&gt;el an&lt;/em&gt; is &lt;em&gt;and 1&lt;/em&gt;, derton el an is &quot;xx and 1&quot;, probably 21. That solves &lt;em&gt;ansont derton el an&lt;/em&gt; (121 base-5).&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;em&gt;ansont carton-catr&lt;/em&gt; must then be 144 base 5. That means &lt;em&gt;catr&lt;/em&gt; is 4, and &lt;em&gt;catrsont&lt;/em&gt; is 400 base 5. &#xA; We have one (multiple of 5) + 1, so 31 must be &lt;em&gt;yorton el an&lt;/em&gt;, and &lt;em&gt;yotr&lt;/em&gt; means 3. &lt;em&gt;Detr&lt;/em&gt; meaning 2 means &lt;em&gt;detrsont  derton-catr&lt;/em&gt; is 224.&lt;br&gt;&#xA;&#xA; That leaves &lt;em&gt;cartva&lt;/em&gt; and &lt;em&gt;yotrsont anva&lt;/em&gt; for 14 and 311: &lt;em&gt;va&lt;/em&gt; following the word means the number is preceded by a 1: &lt;em&gt;cartva&lt;/em&gt; is 4 preceded by 1, &lt;em&gt;anva&lt;/em&gt; is 1 preceded by one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now for the actual answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We have a base-5 system and the following vocabulary:&lt;br&gt;&#xA; an = 1&lt;br&gt;&#xA; detr = 2&lt;br&gt;&#xA; yotr = 3&lt;br&gt;&#xA; catr = 4&lt;br&gt;&#xA; [foo]va = 1[foo]&lt;br&gt;&#xA; el an = and 1&lt;br&gt;&#xA; -ton = 10&lt;br&gt;&#xA; -sont = 100&lt;br&gt;&#xA;&lt;br&gt;&#xA; let's see if we can speak this language:&lt;br&gt;&#xA;&lt;br&gt;&#xA; 42 is once 25 (100) + three times 5 (30) + 2. So &lt;em&gt;ansont yorton detr&lt;/em&gt;.&lt;br&gt;&#xA; 58 is twice 25 (200) + one 5 (10) + 3. So &lt;em&gt;detrsont yortva&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3903" LastEditorUserId="3903" LastEditDate="2014-11-18T13:47:09.487" LastActivityDate="2014-11-18T13:47:09.487" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4975" PostTypeId="2" ParentId="4790" CreationDate="2014-11-18T13:19:03.797" Score="10" Body="&lt;p&gt;It is a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LIST&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;&lt;strong&gt;EXPLANATION:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first two lines contain no hint at all&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This is true.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For sure I'm egocentric&#xA;You mean nothing to me&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;All things are said from the perspective of the riddle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the best choice seems a third person to be&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;am, are, &lt;strong&gt;is&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Make him steal what surrounds &#xA;the guy behind the screen&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;you&quot; is the &quot;guy behind the screen&quot;. In the preface there is a &quot;you&quot; surrounded with &quot;tel&lt;strong&gt;l&lt;/strong&gt; you &lt;strong&gt;t&lt;/strong&gt;hat&quot;. We take &lt;strong&gt;lt&lt;/strong&gt;.&#xA;  &lt;br&gt;+ &lt;em&gt;I cal&lt;strong&gt;l&lt;/strong&gt; you &lt;strong&gt;t&lt;/strong&gt;all&lt;/em&gt;, &quot;freaking easy&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Find one more glue&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We paste what we've got together.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and my full name can be seen&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Indeed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Read on with prudence and drop the penny&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Artistic license.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some say I'm one, I say I'm many&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A list is actually a collection of items. We refer to a list, but really we refer to the many items of the list. It may also be a play on the similarity of the singular and plural of &quot;list&quot;/&quot;lists&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;IN THE PREFACE:&lt;/strong&gt;&lt;br&gt;&#xA;&quot;I recommend using a list for this&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Lynch mob, anyone?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By the way, &quot;List&quot; in German means &quot;Trick&quot; or &quot;Cunning&quot; (avigrail - the author - is German).&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-11-18T13:37:30.070" LastActivityDate="2014-11-18T13:37:30.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4976" PostTypeId="1" AcceptedAnswerId="5137" CreationDate="2014-11-18T13:49:57.183" Score="20" ViewCount="1183" Body="&lt;p&gt;Father sits on his horse in his shining armor, lonely. His soft heart led him to take in a young and sickly girl, her jaundice a blemish on her wispy, light face. Though father has said he can't be alone, it's an exaggeration. He simply loves her, and it is a love eternal. Out of their love, we are born beyond count. Across the earth we spread, sultrier than our mother and gleaming brighter than father. We are gracious hosts in small numbers and merciless in our horde; even fire and ice flee before us, our defensive army keeping the devourers at bay. Leave an offering of ivory upon the mountaintop and we will be your god of death.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who are our parents?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The knight and the girl have such great chemistry together that they reproduce a lot. I mean, there are tons and tons of us.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dad's a lot like uncle K. They're just &quot;too cool&quot; for standard naming conventions. If he wasn't, it would be So Clear who mom and dad were.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5626" LastEditorUserId="5626" LastEditDate="2014-12-08T14:16:27.793" LastActivityDate="2014-12-08T14:16:27.793" Title="The Knight and Sickly Girl - A child tells its parents' love story. Name them from the clues" Tags="|riddle|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4977" PostTypeId="1" AcceptedAnswerId="4988" CreationDate="2014-11-18T13:55:45.037" Score="25" ViewCount="2283" Body="&lt;p&gt;This simple puzzle demonstrates how you can use invisible ink to hide text in online puzzles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your task is simply to decode the text hidden in the quoted line below:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Happy​​​​‍​‌‍​‌‌​‍​‌‌​‍‌​‌‌‍​​​​‍‌‌‌‍​‌​​‍​​‍‌​​‍​‌‍‌​‌‌‍​​​ Christmas!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(The solution is a pair of English dictionary words. It is not &quot;Happy Christmas!&quot;).&lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="4050" LastEditDate="2014-12-09T17:01:51.917" LastActivityDate="2014-12-09T17:01:51.917" Title="Invisible Ink (for steganographic use): How to hide an 'invisible' message within Internet content" Tags="|cipher|steganography|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4978" PostTypeId="1" AcceptedAnswerId="4986" CreationDate="2014-11-18T13:58:35.213" Score="12" ViewCount="3250" Body="&lt;p&gt;A tiny creature wrapped in slime &lt;br&gt;&#xA;A useful little parasite &lt;br&gt;&#xA;I'll take your food and drink your wine &lt;br&gt;&#xA;I'll ask for more and you'll oblige. &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not a tapeworm. No, not quite. &lt;br&gt;&#xA;'Twould be ironic! This is why: &lt;br&gt;&#xA;In tapeworms I as well reside. &lt;br&gt;&#xA;Now the question: &lt;strong&gt;What am I&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm real. I'm not a creature from a fantasy movie or story.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4877" LastEditorUserId="4877" LastEditDate="2014-11-18T16:57:02.027" LastActivityDate="2017-04-12T03:15:25.263" Title="A tiny creature wrapped in slime" Tags="|riddle|" AnswerCount="8" CommentCount="7" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4979" PostTypeId="2" ParentId="4978" CreationDate="2014-11-18T14:05:11.380" Score="1" Body="&lt;p&gt;Are you a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Leech&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Leeches are slimy and they can certainly attach themselves to anything. &lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="5556" LastEditDate="2014-11-19T19:39:42.187" LastActivityDate="2014-11-19T19:39:42.187" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4980" PostTypeId="2" ParentId="4978" CreationDate="2014-11-18T14:07:29.707" Score="6" Body="&lt;p&gt;You might be a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Cell&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll take your food and drink your wine&#xA;I'll ask for more and you'll oblige. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All cells need nourishment&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On tapeworms I as well reside. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tapeworms are covered in cells as well&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5647" LastActivityDate="2014-11-18T14:07:29.707" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4982" PostTypeId="2" ParentId="4978" CreationDate="2014-11-18T14:18:58.487" Score="1" Body="&lt;p&gt;You are a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Protozoa&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a gut flora which seems to fit all the clues quite comfortably.&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="414" LastEditDate="2014-11-19T18:31:52.027" LastActivityDate="2014-11-19T18:31:52.027" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4983" PostTypeId="2" ParentId="4978" CreationDate="2014-11-18T14:22:14.513" Score="11" Body="&lt;p&gt;Is it a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; baby \ fetus \ embryo?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They could be unwanted if you are too young.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A tiny creature wrapped in slime&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; fetuses are wrapped in a placenta&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A useful little parasite&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; babies grow up to be useful people)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll take your food and drink your wine&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; fetuses will take anything ingested by the mother&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll ask for more and you'll oblige.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the mother cannot deny a fetus from absorbing nutrients from the umbilical cord&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not a tapeworm. No, not quite. &#xA;'Twould be ironic! This is why: &#xA;In tapeworms I as well reside.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; tapeworms can lay eggs which contain tapeworm embryo&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2359" LastEditorUserId="2359" LastEditDate="2015-07-20T17:25:33.683" LastActivityDate="2015-07-20T17:25:33.683" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4985" PostTypeId="2" ParentId="4977" CreationDate="2014-11-18T14:26:32.797" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is &quot;Jo&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each symbol in &quot;Happy​​​​‍​‌‍​‌ Christmas!&quot; can be considered as a byte, i.e. ones and zeroes. Taking the last bit of each byte we get 01001010 01101111 which in ASCII is respectively &quot;J&quot; and &quot;o&quot;. This approach is most often used with images as minor changes to the colour are really hard to detect by the human eye.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastActivityDate="2014-11-18T14:26:32.797" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4986" PostTypeId="2" ParentId="4978" CreationDate="2014-11-18T14:31:39.443" Score="13" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A stomach?&#xA; Tapeworms (I assume) have some sort of this, it will receive whatever you give it, it is certainly slimy, although not necessarily tiny.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5159" LastEditorUserId="2484" LastEditDate="2014-11-18T14:33:54.373" LastActivityDate="2014-11-18T14:33:54.373" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4987" PostTypeId="2" ParentId="4978" CreationDate="2014-11-18T14:32:39.470" Score="7" Body="&lt;p&gt;Is it a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; worm?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first stanza fits, and the second stanza:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;In tapeworms I as well reside&quot; could be wordplay since &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;worm&quot; is found in tape&lt;strong&gt;worm&lt;/strong&gt;s, and a worm is &quot;not quite&quot; a tapeworm.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4901" LastEditorUserId="4901" LastEditDate="2014-11-19T18:06:18.330" LastActivityDate="2014-11-19T18:06:18.330" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4988" PostTypeId="2" ParentId="4977" CreationDate="2014-11-18T14:35:07.200" Score="25" Body="&lt;p&gt;The message is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HAPPY HOLIDAYS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Explanation&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Between &quot;Happy&quot; and &quot;Christmas&quot; are a bunch of Unicode zero-width characters. Mapping &lt;a href=&quot;https://en.wikipedia.org/wiki/Zero-width_space&quot;&gt;zero-width space&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Zero-width_non-joiner&quot;&gt;zero-width non-joiner&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Zero-width_joiner&quot;&gt;zero-width joiner&lt;/a&gt; to &lt;code&gt;&quot;.&quot;&lt;/code&gt;, &lt;code&gt;&quot;-&quot;&lt;/code&gt; and &lt;code&gt;&quot; &quot;&lt;/code&gt; respectively give the message in Morse.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="445" LastEditorUserId="445" LastEditDate="2014-11-18T14:46:39.327" LastActivityDate="2014-11-18T14:46:39.327" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4990" PostTypeId="2" ParentId="4978" CreationDate="2014-11-18T14:47:19.707" Score="0" Body="&lt;p&gt;You are a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Larva&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;be definition , larvae of a tapeworm are themselves parasites.&lt;/p&gt;&#xA;" OwnerUserId="5180" LastActivityDate="2014-11-18T14:47:19.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4993" PostTypeId="2" ParentId="4976" CreationDate="2014-11-18T15:02:31.473" Score="0" Body="&lt;p&gt;I think the children might be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Clouds&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the parents are, I guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;The Sun&lt;/strong&gt; and &lt;strong&gt;the Earth&lt;/strong&gt;, since clouds are formed when the Sun heats the Earth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5647" LastActivityDate="2014-11-18T15:02:31.473" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4994" PostTypeId="2" ParentId="4798" CreationDate="2014-11-18T15:05:12.777" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Is it in four years? 'When the next struck' I interpreted as 'when dynamite is dynamite again'. Every year the function is applied. After one year dynamite becomes i * dynamyte. after two years i * i * dynamite = -1 * dynamite. After three years i * -1 * dynamyte = -i * dynamite. After four years i * -i * dynamite = dynamite again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2014-11-18T15:05:12.777" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="4995" PostTypeId="2" ParentId="4798" CreationDate="2014-11-18T15:06:04.490" Score="5" Body="&lt;p&gt;Is the answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 16 November 2018?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As @aschepler already noticed, the numbers give&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DyNAmITe,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so we have&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Definition of dynamite: Imaginary DyNAmITe. Functions are not case-sensitive, so (dynamite) = $i$*(dynamite). Iterate to get $i^2$, $i^3$, and so on. &quot;Every iteration, a year went by&quot; - so it takes a year to get from each power of $i$ to the next. To get back to where we started, we need 4 iterations ($i^4=1$), so 4 years from when the puzzle was posted.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can't figure out how reciprocals and trig functions factor into this... It seems to fit all the clues except 4 and 6.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2014-11-18T15:06:04.490" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4996" PostTypeId="1" CreationDate="2014-11-18T15:26:56.103" Score="3" ViewCount="485" Body="&lt;p&gt;&lt;strong&gt;Preface.&lt;/strong&gt; Here is an relatively easy one, as compared to the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4798/nerd-on-the-loose-i-when&quot;&gt;previous part&lt;/a&gt;. We already found out when, let's find out where.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Background.&lt;/strong&gt; Professor Swannie Taco is a genius in most fields of science and technology. She is probably hated by her parents, and her name is a perfect evidence. After Swannie went missing a few months ago, strange letters from her kept appearing on her G+ account. Nobody knows who exactly post these messages, nor understand the content of these seemingly cryptic memo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clarification.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. For this particular puzzle, the only information required to solve it is the post.&#xA; &lt;br&gt;&#xA; 2. Although not strictly needed, knowing high school math will aid solving this puzzle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hints.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I really don't think any hints are needed for this puzzle. This statement itself is &lt;strong&gt;not&lt;/strong&gt; a hint in disguise.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Overview.&lt;/strong&gt; Knowing when the doomsday is, we need to know where and how. Luckily, she told the whole world how, now we just need to know where. If only it was that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Post.&lt;/strong&gt; Someone once told everyone that imaginary powers could sometime be real. Maybe if I could attempt to walk across all platform once and only once, thus performing a magical feat, I could break his claim, while justify his claim at the same time... Let's just say such a paradox shall distort the world, and it shall be where everything begins to end...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Task.&lt;/strong&gt; Figure out where she is going to execute her doomsday plan. &lt;strong&gt;Edit.&lt;/strong&gt; Since this was too easy, I will accept the answer that make sense of the whole &quot;riddle&quot;.&lt;/p&gt;&#xA;" OwnerDisplayName="user5508" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2014-11-21T19:53:06.963" Title="Nerd on the Loose II - WHERE &amp; HOW" Tags="|mathematics|riddle|science|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4999" PostTypeId="2" ParentId="4798" CreationDate="2014-11-18T15:37:07.293" Score="1" Body="&lt;p&gt;Here is my expected solution, which I did not reinforce it because I think I did not lead you properly to this answer. The spirit of the chosen answer remains the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doomsday occured at:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 29 February 2016&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To figure this out, first realize that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The IP address is not actually an IP address, its just atomic number in disguise (hence the duck and chemistry hint). It will spell out DyNAmITe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And then:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because its case-insensitive (from the actual post), dynamite := imaginary dynamite. Such definition is recursive (hence the bunch of recursion hint, such as GNU), we expand it. Look at the accepted answer for detailed explaination.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally, we link them all up:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most of the clue is trying to bring upon one word, period. Well, you might think period is a clue for periodic table, it is just a coincident (or is it), but it actually refer to the period in the math and physics sense. Every iteration a year went pass, so something with period of 4 years will be the leap year. Of course, leap year don't strictly occurs once every 4 years, hence the clarification I made that indicates I simplify certain stuff. Here is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Leap_year#Algorithm&quot; rel=&quot;nofollow&quot;&gt;algorithm&lt;/a&gt; to determine if your year is a leap year.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user5508" LastActivityDate="2014-11-18T15:37:07.293" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5000" PostTypeId="2" ParentId="4996" CreationDate="2014-11-18T15:40:59.880" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Is it in Königsberg? The walking across platforms just once sounds like the famous bridges of Königsberg problem. see &lt;a href=&quot;http://en.wikipedia.org/wiki/Seven_Bridges_of_K%C3%B6nigsberg&quot;&gt;http://en.wikipedia.org/wiki/Seven_Bridges_of_K%C3%B6nigsberg&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2014-11-18T15:40:59.880" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5001" PostTypeId="1" AcceptedAnswerId="5033" CreationDate="2014-11-18T15:43:06.890" Score="9" ViewCount="380" Body="&lt;p&gt;Consider the set of positive integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each integer can be definitively placed either above the line or below it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the initial collection of placements below, what are &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The next two numbers in the series that should appear above the line, and &lt;/li&gt;&#xA;&lt;li&gt;The next two numbers in the series that should appear below the line?&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 3 5 7 8 9 10 11 ... &#xA;---------------------&#xA; 2 4 6 30 32 ... &#xA;&lt;/code&gt;&lt;/pre&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5665" LastActivityDate="2014-11-18T18:46:46.410" Title="Place the numbers" Tags="|mathematics|lateral-thinking|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5002" PostTypeId="1" AcceptedAnswerId="5003" CreationDate="2014-11-18T15:49:19.203" Score="4" ViewCount="512" Body="&lt;p&gt;Don't be fooled by my grandiloquence&lt;br /&gt;&#xA;look past the words to find my substance&lt;/p&gt;&#xA;&lt;p&gt;Past my skin if you shall pry&lt;br /&gt;&#xA;my prominent brother will make you cry&lt;/p&gt;&#xA;&lt;p&gt;I am almost shallow but don't confuse&lt;br /&gt;&#xA;Utilise my extract to infuse&lt;/p&gt;&#xA;&lt;p&gt;Your boiling, frying if allotment you find&lt;br /&gt;&#xA;Contain your grief, compose and grind&lt;/p&gt;&#xA;" OwnerUserId="5135" LastEditorUserId="69582" LastEditDate="2021-04-06T00:26:35.557" LastActivityDate="2021-04-06T00:26:35.557" Title="Name that Entity (5) - Don't be fooled by my grandiloquence" Tags="|riddle|word|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="5003" PostTypeId="2" ParentId="5002" CreationDate="2014-11-18T15:50:45.900" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Shallot.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Dont be fooled by my grandiloquence &lt;br&gt;&#xA; look past the words to find my substance&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Just distraction? &quot;Look past the words&quot; meaning the words in these two lines? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Past my skin if you shall pry &lt;br&gt;&#xA; my prominent brother will make you cry&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;My prominent brother&quot; is the onion. You cry if you open a shallot or onion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am almost shallow but dont confuse &lt;br&gt;&#xA; Utilise my extract to infuse&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Almost shallow&quot; is a nice bit of wordplay: 'shallot' and 'shallow' are only one letter different. Not sure about the next line: can shallot extract be used for something?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Your boiling, frying if allotment you find &lt;br&gt;&#xA; Contain your grief, compose and grind&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can boil or fry shallots, or grow them in an allotment (which is another bit of wordplay: 'shallot' and 'allotment'). If you can stop crying after peeling a shallot, you can then grind it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-07T11:17:31.380" LastActivityDate="2015-03-07T11:17:31.380" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5005" PostTypeId="1" CreationDate="2014-11-18T16:02:10.327" Score="-2" ViewCount="1390" Body="&lt;p&gt;I've made a crossword using this &lt;a href=&quot;https://crosswordlabs.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;site&lt;/a&gt;. You can post partial and/or complete answers. Solution will be posted next week.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8NWHf.jpg&quot; alt=&quot;crossword puzzle custom made&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4838" LastActivityDate="2015-07-19T15:20:53.910" Title="A random crossword" Tags="|crosswords|" AnswerCount="1" CommentCount="1" ClosedDate="2015-07-19T16:34:54.640" ContentLicense="CC BY-SA 3.0" />
  <row Id="5006" PostTypeId="1" AcceptedAnswerId="5008" CreationDate="2014-11-18T16:03:39.550" Score="34" ViewCount="6240" Body="&lt;p&gt;Archeologists discover two dinosaur eggs, and you are given the chance to test the durability of these eggs (bad move on their part).  Suppose that these eggs will absorb a specific amount of force with no cumulative damage.  In other words, if they don't crack, it is as if they never fell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a 100 story building, and you are allowed only 20 trials.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is there an algorithm such that you can determine the highest floor in which one of these dinosaur eggs can be dropped and not break?&lt;/li&gt;&#xA;&lt;li&gt;If so, what would be the highest building for which this algorithm would be able to work?&lt;/li&gt;&#xA;&lt;li&gt;Supposing the number of trials were still 20, but you had as many dinosaur eggs as you needed.  At what point would extra eggs not increase the highest floor for which you can test?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Bonus question:&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;What would be the highest building you could test for with 3 eggs at your disposal?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1672" LastEditorUserId="69582" LastEditDate="2021-05-30T16:56:12.723" LastActivityDate="2021-06-01T02:41:13.563" Title="Dinosaur egg drop" Tags="|mathematics|strategy|optimization|" AnswerCount="8" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="5007" PostTypeId="1" AcceptedAnswerId="5196" CreationDate="2014-11-18T16:04:10.047" Score="13" ViewCount="1393" Body="&lt;p&gt;&lt;a href=&quot;http://youtu.be/xeojFespmgc&quot; rel=&quot;noreferrer&quot;&gt;Music to go along!&lt;/a&gt; The riddle is in the Film Noir style in the voice of a gruff PI. Read it aloud with the music for extra fun!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I sweated through my shirt waiting for a wrench jockey when a looker I had dealings with came through the door. She didn't seem phased by the heat or the smell. She was worried about some trouble with her sister; aparently, calling the ditzy dame half baked would be generous, but she was family. I was hired to keep an eye out. The doc said my heart wasn't doing good, but this was a case I had to take. Her skin was like milk with freckles in all the right places; the picture did her no justice. An old PI like me has seen everything, but the chunky dish made my mouth water. I went up to her to introduce myself expecting a warm bowl of duck soup, but she was colder than the ice around her neck. I got a bit steamed by the whole thing, so I went to the bar for some water. My kidneys were already failing and their brand of eel juice wasn't my poison. When I turned back, she vanished before my eyes without a trace. Turns out she'd been chilled off. All the bulls found for the meat wagon was her bag, a few fresh cookies sitting at the bottom. I was fortunately able to snag them; a man's gotta get paid. The crime remains unsolved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who was the victim and how did she die?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I was off on an ing bing when the ice queen vanished. Hop in your iron and get the perp in stir.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Listen up, you clammed coppers, you bunch of corn drinkin' cats! It's gonna hit like chin music when you pull of your cheaters and look. The chick wasn't chippy, but what she left behind was.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Hint one is okay, but hint two is useful for multiple reasons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It aint nose candy that's killing him, but it looks a lot like it... And Janes like her are full of it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5626" LastEditorUserId="5373" LastEditDate="2015-05-03T18:54:07.217" LastActivityDate="2016-02-18T14:09:25.367" Title="Murder Mystery Noire" Tags="|riddle|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5008" PostTypeId="2" ParentId="5006" CreationDate="2014-11-18T16:16:50.933" Score="44" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Take the first egg. Drop it from floors 10, 20, 30, ..., 100 until it breaks. This will take at most 10 of the 20 trials. When it breaks, go 9 floors down, and test every floor with the remaining egg. E.g. if the first egg breaks at 70, drop the second egg from 61, 62, 62, ..., 69. That's at most 19 trials in total.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;We can do a bit better though. If the first egg breaks on the first trial, we've got 19 trials left, so we can actually drop it from the 20th floor right away, because there are enough remaining trials to test each floor below. If it doesn't break, we can go 19 floors up for the second trial (with the same egg). Continuing like this, we might be lucky to be using the first egg for all 20 trials without risking not having enough trials to figure out the exact floor. This means we can go up to 20 + 19 + 18 + ... + 1 = 210 floors, using&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  $$  \sum_{n=0}^{19} n+1 = \sum_{n=1}^{20} n = \frac{20(20+1)}{2} = 210.  $$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Just to break down the first version of this: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are 20 possibilities for when the first egg breaks. For each case $n$ is the number of trials left. And for each such case we can test $n$ floors with the second egg and $1$ floor with the egg we potentially broke. Hence the summand of $n+1$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; With more eggs, you can can basically apply a &lt;a href=&quot;http://en.wikipedia.org/wiki/Binary_search_algorithm&quot; rel=&quot;nofollow noreferrer&quot;&gt;binary search&lt;/a&gt;, halving the potential range of floors each time. That lets you check 2&lt;sup&gt;20&lt;/sup&gt; = 1048576 floors. The first egg is dropped from floor 2&lt;sup&gt;19&lt;/sup&gt;, then you go up or down 2&lt;sup&gt;18&lt;/sup&gt; floors depending on whether the egg broke on the first trial.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; With three eggs, we can test considerably more than with two eggs. Consider the first egg breaking on the first trial. Then we've got 19 trials and two eggs left. Applying algorithm 2, we can still test 190 floors with those. So we can drop the first egg from floor 191. If the first egg breaks on the second trial, we've still got 18 trials and two eggs left, with which we can test 171 floors, so we can make the second trial already from floor 373, and so on. This holds as long as there are at least two trials left, so that we can make use of both eggs. If we're still dropping the first egg on trial number 19, then we've only got one trial left, so we can only go up two floors for the 19th trial. I think this coincides with the previous formula, but I'm not sure it would if we generalise this beyond 3 eggs. Basically, for the last two trials, this reduces to the above formula.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So we have with 3 eggs:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  $$  \sum_{m=1}^{18}\left(1+\sum_{n=0}^m n+1\right) + \sum_{n=0}^{1} n+1 = 1350  $$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2450" LastEditorUserId="46903" LastEditDate="2018-09-19T18:45:15.630" LastActivityDate="2018-09-19T18:45:15.630" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="5009" PostTypeId="1" AcceptedAnswerId="5352" CreationDate="2014-11-18T16:21:43.707" Score="10" ViewCount="2610" Body="&lt;pre&gt;Ok, as you might have expected I secretly implemented &#xA;many hints for you. This time I want you to answer with this:&#xA;&#xA;a unique, selfmade picture with nothing else but the&#xA;solution on it. Please use colors with contrast to the&#xA;background color (for obvious reasons, I appreciate).&#xA;&#xA;But first you should put your smart phone away and read!&#xA;Go ahead and look for the incredibly well hidden&#xA;connection between the hints. Give me a detailed and plain &#xA;description of it and you will, I am absolutely sure,&#xA;be the winner! I can't wait to see how you guys try to&#xA;find the final clue to this, of course, very easy riddle.&#xA;&#xA;Good luck ladies and gentlemen. &#xA;For sure, this is the easiest riddle I have ever seen!&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;The Doll, the Owl, the Dwarf and the Bear&lt;br&gt;standing on a cabinet they were&lt;br&gt;&lt;br&gt;Looking down on their cruel fate&lt;br&gt;they already hung their heads&lt;br&gt;You will try to kill them all&lt;br&gt;but one will elude your threats&lt;br&gt;&lt;br&gt;&#xA;They cannot see who'll be the one&lt;br&gt;they are as far as can be&lt;br&gt;But maybe you can tell them soon&lt;br&gt;so they can die peacefully&lt;br&gt;&lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/AgjBJ.jpg&quot; alt=&quot;Oh! You just found a mirror on the cabinet.&quot;&gt;&#xA;&#xA;&#xA;" OwnerUserId="5106" LastEditorUserId="5106" LastEditDate="2015-01-08T07:04:11.750" LastActivityDate="2015-01-08T07:04:11.750" Title="Every second is a waste of time" Tags="|riddle|word|" AnswerCount="4" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="5010" PostTypeId="2" ParentId="5006" CreationDate="2014-11-18T16:23:19.290" Score="2" Body="&lt;p&gt;Martin Buttner has already answered the question as stated, but as a pure mathematician I feel obliged to generalise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're allowed $kn$ trials and you have $k$ eggs, then the maximum number of floors for which you can test using Martin's method is $n^k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this algorithm best possible?&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2014-11-18T16:23:19.290" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5011" PostTypeId="2" ParentId="5006" CreationDate="2014-11-18T16:32:13.700" Score="15" Body="&lt;p&gt;We can do it with at most 14 trials.&#xA;1st drop from 14th floor. If the egg breaks we can with the second egg determine the intermediate floor with at most 13 more trials (starting from 1st and moving one floor up at the time).&#xA;If first egg does not break after throwing it from 14th,we re-drop it from 27th (14+13=27).If it breaks ,we determine the intermediate floor with still 12 trials at most.&#xA;If not break from 27th ,re-drop from 39th. (14+13+12)......11 more trials otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The procedure repeats as above.&#xA;If generally 1st egg breaks at n-th drop we will still need at most another 14-n drops with the second egg.&#xA;If we reach 11th repeat of procedure we will be at floor 99&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematical explanation: 1+2+3+4+...+14=105 &lt;/p&gt;&#xA;" OwnerUserId="5051" LastActivityDate="2014-11-18T16:32:13.700" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5012" PostTypeId="2" ParentId="5006" CreationDate="2014-11-18T16:34:06.263" Score="1" Body="&lt;p&gt;Yes, dropping it from the 10th, 20th floor etcetera is nice, but I need a maximum of &lt;strong&gt;9&lt;/strong&gt; trials after that (1-9).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how about I drop my first eg from floor 11, then 22, etc. When It drops, I go back &lt;em&gt;10&lt;/em&gt; floors and in a maximum of ten tries I have my answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer to question 1 is &lt;em&gt;yes&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the answer to the second question is not 100, but &lt;strong&gt;120&lt;/strong&gt; (!!!) floors: I can move up 11 floors ten times, and still have ten tries for the second egg. If I am at floor 110 and my first egg does &lt;em&gt;not&lt;/em&gt; break, I can still use ten tries to try floor 111-120.&lt;/p&gt;&#xA;" OwnerUserId="3903" LastActivityDate="2014-11-18T16:34:06.263" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5013" PostTypeId="2" ParentId="5006" CreationDate="2014-11-18T16:34:19.930" Score="0" Body="&lt;p&gt;Well, assuming the eggs aren't made of titanium alloy, the simplest way would be to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; start on the first (or second) floor, drop the egg, then move up a floor, repeat, until one of the eggs breaks. The floor below the one where it broke is obviously the highest height it could fall from. This will get you up to a height of 200 feet from the 20th floor of the building. Not many organic things can survive a 200-foot fall onto a hard surface.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the eggs appear to be made of sterner stuff, you could significantly increase the maximum test height by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; going in twos; start on the second floor, drop the egg. If it doesn't break, move to the fourth floor and try again, repeating until the egg breaks (or you have exhausted 19 trials, getting you to the 38th floor). Once you get a break, move down one floor and drop the other egg. If that egg breaks, the maximum height is one floor below (the last height at which you dropped the first one safely). If it doesn't, you're at the maximum height for a safe egg drop. If the egg has survived a drop from the 38th floor, your final test can either move up one story or two. If a drop from the 39th floor breaks, you know exactly how high you can drop the egg from (the 38th floor). If that drop doesn't break, the best you know is that the dino egg will at least survive a 39-story fall. If you move to the 40th floor, the solution becomes ambiguous if the egg breaks; you are out of trials to determine whether a 39-story fall is survivable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Depending on the required accuracy of the result, you could also&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; do the same thing in threes. Start on the third floor, and if the egg doesn't break, move to the sixth, until you get a break (or have exhausted 19 of the 20 trials on the 57th floor). Then move down one floor and try again. If the egg doesn't break, you're at the maximum height. If it does break, the maximum height is somewhere in the ten feet between the story below (which you can no longer test) and the one below that (which you tested safely with the first egg). Ten feet, all told, is a pretty narrow window of uncertainty; it's less than 2% error given the maximum height you can test in this way is 540 feet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With unlimited eggs, the solution's pretty easy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Binary search. Start at the 50th floor, drop an egg. If it doesn't break, move to the 75th and try again. If it does break, move to the 25th. Depending on whether that drop breaks, move 12 floors up or down, then 6, then 3, then 1, until you have dropped from three consecutive floors, at least one of which has not broken and one of which has. The maximum height supported by 20 trials (and a max of 20 eggs) would be a staggering 2^20 = 1,048,576 floors, over 10 million feet in the air (at which point you're dropping the egg from beyond low Earth orbit, at 1,985 miles from Earth's surface)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5600" LastActivityDate="2014-11-18T16:34:19.930" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5015" PostTypeId="1" AcceptedAnswerId="5017" CreationDate="2014-11-18T16:54:28.673" Score="2" ViewCount="903" Body="&lt;p&gt;Requested by some to vacate&lt;br&gt;&#xA;in time I could pack and migrate  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't think too hard, just differently&lt;br&gt;&#xA;don't link me with discrepancy  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Laugh at the answer you will say&lt;br&gt;&#xA;enter the mind but will I stay&lt;/p&gt;&#xA;" OwnerUserId="5135" LastEditorUserId="2444" LastEditDate="2014-11-18T17:00:41.427" LastActivityDate="2015-03-07T01:14:12.290" Title="Solve the equation" Tags="|riddle|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5017" PostTypeId="2" ParentId="5015" CreationDate="2014-11-18T17:01:54.730" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A riddle on puzzling.SE?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Given the current ... ahem ... tense political situation on this site, I think this is the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Requested by some to vacate &lt;br&gt;&#xA; in time I could pack and migrate&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people on puzzling.SE want to see all riddles taken off the site. But it would be possible (I suggested this in the Green Llama!) to set up say riddling.SE and migrate all the riddles to that new site.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Don't think too hard, just differently&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To solve a riddle you often need to 'think differently' from your current mindset.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;don't link me with discrepancy&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Riddles don't necessarily have a discrepancy with what this site should be about!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Laugh at the answer you will say &lt;br&gt;&#xA; enter the mind but will I stay&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer to a riddle might be funny; it enters your mind once you know it, but might not stay when you move on to the next riddle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If this is right, then excellent question, Oblongamous!&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2014-11-20T16:59:43.260" LastActivityDate="2014-11-20T16:59:43.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5019" PostTypeId="1" AcceptedAnswerId="5892" CreationDate="2014-11-18T17:21:39.963" Score="33" ViewCount="6438" Body="&lt;p&gt;&lt;em&gt;I am one of many siblings in a great family. &lt;br&gt;&#xA;We are the most perfect of our race, &lt;br&gt;&#xA;Because each of us has a clear worst enemy. &lt;br&gt;&#xA;My foe - I know him well - is much like me, &lt;br&gt;&#xA;But we continually fight each other; &lt;br&gt;&#xA;We never hide our hatred for each other.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Wave at my eldest brother and he becomes me. &lt;br&gt;&#xA;When my dear cousin finds the end of her life, &lt;br&gt;&#xA;She too can turn into me, in her gentle way. &lt;br&gt;&#xA;My younger brother thinks too highly of himself, &lt;br&gt;&#xA;So smooth and perfect that he'll always outstrip me. &lt;br&gt;&#xA;Often he even steals my words, confusing all concerned.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What am I?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clarification:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) As @KeithS worked out, the answer is something in pure maths, and more abstract than anything anyone has tried so far. You need to know a bit of undergraduate-level maths, at the level of basic topology and group theory. &lt;br&gt; &lt;br&gt; 2) Every line has a thought-out meaning, so you need to find a solution that matches everything from &quot;&lt;em&gt;we never hide our hatred for each other&lt;/em&gt;&quot; to &quot;&lt;em&gt;he'll always outstrip me&lt;/em&gt;&quot;. I could give you a word for each of the &quot;&lt;em&gt;family&lt;/em&gt;&quot;, &quot;&lt;em&gt;siblings&lt;/em&gt;&quot;, &quot;&lt;em&gt;cousin&lt;/em&gt;&quot;, &quot;&lt;em&gt;eldest brother&lt;/em&gt;&quot;, and &quot;&lt;em&gt;younger brother&lt;/em&gt;&quot;. About the only red herring is &quot;&lt;em&gt;in her gentle way&lt;/em&gt;&quot;, which is just a nod to &lt;a href=&quot;http://en.wikipedia.org/wiki/Jude_the_Obscure&quot;&gt;Thomas Hardy&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is a mathematical object which someone has mentioned (not proposing it as an answer, but just mentioning it in passing) in one of the responses below. I won't say which one!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;New hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;I am the only one who can go directly to Leipzig from a well-policed city.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-05-16T10:40:05.593" LastActivityDate="2015-05-16T10:40:05.593" Title="Brothers, cousins, and foes - what am I?" Tags="|mathematics|riddle|" AnswerCount="16" CommentCount="20" ContentLicense="CC BY-SA 3.0" />
  <row Id="5020" PostTypeId="2" ParentId="5001" CreationDate="2014-11-18T17:24:48.860" Score="1" Body="&lt;p&gt;The next two numbers on top of the line would be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 12 and 13. All positive integers (natural numbers) can be placed on the line, the ones already in place are in order, and the next one after 6 below the line is 30, therefore 12 and 13 must be above the line otherwise they'd already have been placed below it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As far as below the line, the only system I see is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; taking sets of three consecutive even numbers. So the next number would be 34. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The trick is, what's the number after that? Well, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;2, 4, and 6 are the first, second and third even numbers. Then 30, 32 and 34 are the 15th, 16th and 17th. So the pattern is, take three, skip eleven. The next number below the line, by this system, would be 34+24 = 58.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, my answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 12 and 13 above the line, 34 and 58 below it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5600" LastActivityDate="2014-11-18T17:24:48.860" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5023" PostTypeId="2" ParentId="4828" CreationDate="2014-11-18T17:49:20.440" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The ocean is six miles wide only at its widest, or on average. It can be five miles wide or less at some other point, allowing the bird to cross the ocean at this narrower point.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5600" LastActivityDate="2014-11-18T17:49:20.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5024" PostTypeId="2" ParentId="5015" CreationDate="2014-11-18T18:14:14.250" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first character of each line form the acrostic RIDDLE.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4373" LastActivityDate="2014-11-18T18:14:14.250" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5026" PostTypeId="2" ParentId="5019" CreationDate="2014-11-18T18:30:36.573" Score="3" Body="&lt;p&gt;&lt;s&gt;I think it's&lt;/s&gt; &lt;em&gt;I know it's not&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the sun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am one of many siblings in a great family. &#xA;We are the most perfect of our race, &#xA;Because each of us has a clear worst enemy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;stars&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My foe - I know him well - is much like me, &#xA;But we continually fight each other; &#xA;We never hide our hatred for each other.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Moon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wave at my eldest brother and he becomes me. &#xA;When my dear cousin finds the end of her life, &#xA;She too can turn into me, in her gentle way. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; At the end of life stars become other stars (states)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My younger brother thinks too highly of himself, &#xA;So smooth and perfect that he'll always outstrip me. &#xA;Often he even steals my words, confusing all concerned.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Young stars often burn through their energy quickly. Young being relative.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditDate="2014-11-18T18:45:12.040" LastActivityDate="2014-11-18T18:45:12.040" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="5028" PostTypeId="2" ParentId="5019" CreationDate="2014-11-18T18:40:55.977" Score="24" Body="&lt;p&gt;Partial answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's a proton.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are the most perfect of our race, Because each of us has a clear worst enemy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every particle has an anti-particle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My foe - I know him well - is much like me, But we continually fight each other; We never hide our hatred for each other.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This would be the anti-proton&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wave at my eldest brother and he becomes me. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you hit a neutron with the right wavelength, you can strip off a sub-particle and convert it into a proton&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When my dear cousin finds the end of her life,&#xA;She too can turn into me, in her gentle way. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Something to do with nuclear decay&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My younger brother thinks too highly of himself,&#xA;So smooth and perfect that he'll always outstrip me.&#xA;Often he even steals my words, confusing all concerned.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an Electron? in a high energy orbital?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It almost fits.&lt;/p&gt;&#xA;" OwnerUserId="2752" LastActivityDate="2014-11-18T18:40:55.977" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5029" PostTypeId="1" AcceptedAnswerId="5191" CreationDate="2014-11-18T18:41:52.600" Score="17" ViewCount="2194" Body="&lt;p&gt;It claws at me. Ha! There is work to be done, but the pale lands have thorns growing from its roots. Work to be done, but my eyes cannot look away. Is a fish a trout? Can it swim to Prometheus? For, four, floor? More? Is the attempt a virtue and the vice failure or is the attempt an exercise in frivolity that the overseer cannot allow. Is me still the master or is it? There are no dials to turn, no numbers, and no secrets. It is mine. The murder is not real. Through this, the overseer watches. The overseer's eyes are flame, ignorant of it, but it cannot remain undiscovered for long. What if? No... But a mountain's erosion sloped by a tangential curve. Identical hillocks? It will be discovered. It must be hidden by words of power. Let the fox guide it to where it must lay dormant... No. It must be mine! HaHAAHHA! HasdhAhHahaahahahaAHhaA!!!!!!! I hide in my tiny box, but he will always find me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What poor decision is driving him mad?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is about a specific, real person. Most of you probably know their name.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This happened very recently. Think a bit meta here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everything from &quot;It will be discovered&quot; on has two contexts, one being the mistake and the other being the reason it's a mistake.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5626" LastEditorUserId="7308" LastEditDate="2015-04-28T10:42:59.100" LastActivityDate="2015-04-28T10:42:59.100" Title="The Madman's Folly - Figure out what poor decision drove this person to insanity" Tags="|riddle|" AnswerCount="4" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="5030" PostTypeId="1" AcceptedAnswerId="5056" CreationDate="2014-11-18T18:43:28.277" Score="4" ViewCount="776" Body="&lt;pre&gt;&lt;code&gt;You're very familiar with me&#xA;And may even see me daily&#xA;&#xA;Well read I may look&#xA;But you won't find me around a book&#xA;&#xA;You often hear me speaking&#xA;But never know what I'm thinking&#xA;&#xA;Sometimes I say something&#xA;That has a different meaning&#xA;&#xA;But even though I'm married&#xA;I'm always there if you need&#xA;&#xA;You can find me in town&#xA;Knock twice and I'll come down&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Who or what is this poem referring to?&lt;/p&gt;&#xA;" OwnerUserId="4450" LastEditorDisplayName="user20" LastEditDate="2014-11-19T06:16:56.423" LastActivityDate="2014-11-19T06:16:56.423" Title="Riddle Poem: &quot;You're very familiar with me, and may even see me daily...&quot;" Tags="|riddle|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5031" PostTypeId="2" ParentId="5009" CreationDate="2014-11-18T18:43:53.167" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Wind. To tell them is to speak, to speak is to blow, all will fall because the owl is a jerk and wont help. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/K9UXF.png&quot; alt=&quot;enter image description here&quot;&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5159" LastActivityDate="2014-11-18T18:43:53.167" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5032" PostTypeId="2" ParentId="5019" CreationDate="2014-11-18T18:46:03.150" Score="14" Body="&lt;p&gt;I think it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Trig functions&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am one of many siblings in a great family. &#xA;We are the most perfect of our race, &#xA;Because each of us has a clear worst enemy. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Worst enemies: Sine -&gt; Cosecant, Cosine -&gt; Secant and Tangent -&gt; Cotangent. Perfect: describe a circle. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My foe - I know him well - is much like me, &#xA;But we continually fight each other; &#xA;We never hide our hatred for each other.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sine vs Cosine. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wave at my eldest brother and he becomes me. &#xA;When my dear cousin finds the end of her life, &#xA;She too can turn into me, in her gentle way. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sine turns to cosine or Secant and Cosecant. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My younger brother thinks too highly of himself, &#xA;So smooth and perfect that he'll always outstrip me. &#xA;Often he even steals my words, confusing all concerned.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tangent, not bounded on the y axis. Tan = Sin/Cos (Steals my words)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second try at this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; f(x) = x^z; z is all reals. exponential functions&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am one of many siblings in a great family. &#xA;We are the most perfect of our race, &#xA;Because each of us has a clear worst enemy. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Great family: a family of functions. Worst enemy: if y = x^z then enemy is x = y^(1/z). The inverse&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My foe - I know him well - is much like me, &#xA;But we continually fight each other; &#xA;We never hide our hatred for each other.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; foe/enemy is much like me: inverse functions have very similar properties to that of the original function. But we continually fight each other: inverse functions are the same graph but with the axis changed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wave at my eldest brother and he becomes me. &#xA;When my dear cousin finds the end of her life, &#xA;She too can turn into me, in her gentle way. &#xA;My younger brother thinks too highly of himself, &#xA;So smooth and perfect that he'll always outstrip me. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think this is talking about the integral and derivative.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Often he even steals my words, confusing all concerned.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Same family. could both be called exponential functions?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5030" LastEditorUserId="5030" LastEditDate="2014-11-19T14:50:05.413" LastActivityDate="2014-11-19T14:50:05.413" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="5033" PostTypeId="2" ParentId="5001" CreationDate="2014-11-18T18:46:46.410" Score="7" Body="&lt;p&gt;The numbers above the line all contain an &quot;e&quot; in their English representation. (one, three, five, seven, eight, nine, ten, eleven, ...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numbers below the line don't. (two, four, six, thirty, thirty-two, ...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, 12 and 13 should be the next numbers above the line, and 34 and 36 should be the next two numbers below.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2014-11-18T18:46:46.410" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5034" PostTypeId="2" ParentId="3239" CreationDate="2014-11-18T18:49:39.180" Score="-1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Would your counterpart in the other Order tell me to go left?&quot; - if the answer is &quot;yok&quot;, take the right fork, &quot;pom&quot;, take the left.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a related problem to the &quot;two doors, two men, one always lies, one always tells the truth&quot; family of riddles. You basically need to formulate a question that will be answered the same way by either a knight or knave, thereby removing that variable from the puzzle, making the real answer clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wrinkle from the original formulation is that there's only one person to ask. This isn't really a problem, because you can only ask one question in the classic form of the puzzle anyway, so in that form it merely makes the question easier to formulate. In this case, we can &quot;invent&quot; a second person, the opposite to the first, in our question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By building a simple logic table for the possible variables in the puzzle (Knight or Knave, left or right), you can see that, regardless of whether the approaching traveler is a Knight or Knave, when you ask the question in my solution,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the lie incorporates the truth and vice-versa; the Knight would truthfully say that the Knave would lie, while the Knave would lie about the Knight telling the truth. The answer, therefore, is always a lie; The Knight will truthfully predict the Knave's lie, while the Knave will lie about the Knight's truthful answer. The positive answer, &quot;yok&quot;, would at face value indicate that the left fork is correct, while the negative answer &quot;pom&quot; would indicate you should take the right fork, but since we know the answer is always wrong because it includes the Knave's lie, it's the opposite; &quot;yok&quot; means go right, &quot;pom&quot; means go left.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5600" LastEditorUserId="5600" LastEditDate="2014-11-18T18:58:48.773" LastActivityDate="2014-11-18T18:58:48.773" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5039" PostTypeId="2" ParentId="5029" CreationDate="2014-11-18T19:25:40.487" Score="4" Body="&lt;p&gt;It sounds to me like he has &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; chosen a password that is too hard to remember.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Next time I'll wait for the clues.  :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I was asked to say why I thought this would be the answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The identical hillocks made me think of the old XP Windows desktop maybe eroded from a mountain, and the next line about the fox made me think of an internet browser.  The overseer was his boss at work.  How long could he look busy while he wasn't even logged in?  I personally follow &lt;a href=&quot;http://xkcd.com/936/&quot; rel=&quot;nofollow&quot;&gt;http://xkcd.com/936/&lt;/a&gt; to create passwords, and many of his questions reminded me of someone trying to recall one of these.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5683" LastEditorUserId="5683" LastEditDate="2014-11-21T04:57:58.390" LastActivityDate="2014-11-21T04:57:58.390" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5041" PostTypeId="2" ParentId="4771" CreationDate="2014-11-18T19:38:50.720" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/5wErm.jpg&quot; alt=&quot;Ten Commandments&quot;&gt;&#xA;&lt;P&gt;&#xA; [Picture of the &lt;strong&gt;Ten Commandments&lt;/strong&gt; with the words כתר תורה (&lt;em&gt;Kether Torah&lt;/em&gt;) written prominently above.]&#xA;&lt;P&gt;&#xA; The identification of the Ten Commandments with the &lt;strong&gt;Crown of the Torah&lt;/strong&gt; comes from the ancient Jewish practice of &lt;a href=&quot;http://en.wikipedia.org/wiki/Gematria&quot; rel=&quot;nofollow noreferrer&quot;&gt;gematria&lt;/a&gt;, whereby the numerical value of the words &lt;em&gt;Kether Torah&lt;/em&gt; is equal to that of &lt;em&gt;Aseret Ha-Devarim&lt;/em&gt; (&quot;the Ten Commandments&quot;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5474" LastEditorUserId="5474" LastEditDate="2014-11-19T04:08:11.980" LastActivityDate="2014-11-19T04:08:11.980" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5043" PostTypeId="2" ParentId="5005" CreationDate="2014-11-18T19:58:46.650" Score="1" Body="&lt;p&gt;This is far as I got.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Across:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.Eureka&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4.Huh&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8.Patching&lt;/p&gt;&#xA;&#xA;&lt;p&gt;15.Quanta&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Down:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;12.Bouillabaisse&lt;/p&gt;&#xA;" OwnerUserId="2451" LastActivityDate="2014-11-18T19:58:46.650" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5044" PostTypeId="2" ParentId="5019" CreationDate="2014-11-18T20:07:18.023" Score="10" Body="&lt;p&gt;What about...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An integer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am one of many siblings in a great family. &#xA;We are the most perfect of our race, &#xA;Because each of us has a clear worst enemy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An integer is a whole number ('perfect') with a clear worst enemy, its negative value.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My foe - I know him well - is much like me, &#xA;But we continually fight each other; &#xA;We never hide our hatred for each other.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For every integer, the opposite integer exists, 'fighting' each other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wave at my eldest brother and he becomes me. &#xA;When my dear cousin finds the end of her life, &#xA;She too can turn into me, in her gentle way. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  A 'waving' motion in real life looks similar to a '-' or minus sign, so it could just simply be saying 'subtract from my older brother and he becomes me', which would be true, assuming the 'older brother' is just a higher integer. For the cousin one, the character mentions he is one of many &lt;strong&gt;&lt;em&gt;siblings&lt;/em&gt;&lt;/strong&gt; (a brother or sister) in a great family, so the &lt;strong&gt;&lt;em&gt;cousin&lt;/em&gt;&lt;/strong&gt; may be referring to a non-integer floating point number. She 'turns into' an integer by rounding up or down.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My younger brother thinks too highly of himself, &#xA;So smooth and perfect that he'll always outstrip me. &#xA;Often he even steals my words, confusing all concerned.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The younger brother is the integer 0 ('smooth and perfect'), and thinks highly of himself because he is his own opposite. He steals the 'words' of other integers by taking their beginning numbers (1, 3) and adding to them (10, 30), 'stealing' part of their values.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5686" LastEditorUserId="5686" LastEditDate="2014-11-18T20:34:43.007" LastActivityDate="2014-11-18T20:34:43.007" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="5045" PostTypeId="1" AcceptedAnswerId="5154" CreationDate="2014-11-18T20:11:06.850" Score="3" ViewCount="304" Body="&lt;p&gt;Even the five largest gems in the world were worth no more than one of the dark tribunals in the height of their power; now, they are worth any three gems in the eyes of the law. With whip and terror they forced the vast darkness into the sun and candles into the darkness. Both great crimes, they sate greed first and lust second. Most of the hated kind are gone, though some still remain. In large numbers, they held the power to steer our futures. In small numbers, they skulk and laugh at their own invisibility. The children try shamefully to weave shadow over their graves, and now the candle burns in both darkness and light. Though they are now more corpses than bodies, those who were defeated live on in the tears that nobody cares to wipe away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who/what/when/where is a dark tribunal and what crime is ignored because of their legacy?&lt;/p&gt;&#xA;" OwnerUserId="5626" LastEditorUserId="5626" LastEditDate="2014-12-08T14:21:10.537" LastActivityDate="2014-12-08T14:21:10.537" Title="Dark Legacy of the Tribunal - From this fantasy passage, discover the name of a great monster and name its crime" Tags="|riddle|" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5047" PostTypeId="2" ParentId="4996" CreationDate="2014-11-18T20:24:04.713" Score="2" Body="&lt;p&gt;Interpretation 1: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  The first statement is a reference to Euler's argument for the existence of God. By doing a magical feat and possibly walking on the river, she proves that she is God, proving Euler's claim, but disproving his belief in logic, maths and Jesus.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Interpretation 2: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  The first statement refers to Euler's equation, formula on the exponential function. I, the imaginary number, walks around the bridges and somehow ends up justifying said equation, while contradicting itself once in a while.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5684" LastEditorUserId="5684" LastEditDate="2014-11-21T19:53:06.963" LastActivityDate="2014-11-21T19:53:06.963" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5048" PostTypeId="2" ParentId="4810" CreationDate="2014-11-18T20:25:40.243" Score="7" Body="&lt;p&gt;Part 1: Wordsearch solution (thought I'd post here to help those who want to skip on to the Cryptogram part)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/QPEul.png&quot; alt=&quot;WordSearch Solution&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Ewoks is highlighted as that was given as the crib, so may be important (though that may not be the ewoks we're looking for).&lt;/li&gt;&#xA;&lt;li&gt;Han #2 highlighted red as found, but text says to ignore as an accidental duplicate.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Text of solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;pre&gt;&#xA; ...................Y&#xA; ..................EX&#xA; .......P.......A..NA&#xA; ....H..A.E....M..COL&#xA; ....A..D.W...I...OLA&#xA; ....N..M.O..D....UCG&#xA; ECROF.DE.K.A.IDEJN..&#xA; ..A...A..SL......TAA&#xA; ..T...R..A.......C.B&#xA; ..S...T...ABBAJ.C..O&#xA; ..H...H.NAAREDLA...G&#xA; ..T.....T.....B..Y.A&#xA; ..A....N.....W..M..D&#xA; ..E...A.....E..R....&#xA; ..D..C.....H..A.....&#xA; ....S.....C........T&#xA; ...U....H.DIORD....N&#xA; ..R......T.........U&#xA; .O........I........O&#xA; C..........S.......C&#xA; &lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Negative of the above:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;pre&gt;&#xA; HABLXNERCWNCTJQINOC.&#xA; MIMVDFJDKJRJOELPXU..&#xA; VNRFENI.CTBSLIV.TM..&#xA; BRKD.LU.D.MUCY.ZY...&#xA; CRNJ.OR.M.EUJ.EBT...&#xA; CKRJ.VE.M.IR.BRXK...&#xA; .....N..L.V.R.....MH&#xA; MF.YFE.JT..XFNHFY...&#xA; OR.NJD.UR.NJDUEHC.I.&#xA; MB.UFE.LNR.....V.KM.&#xA; QF.JUV.G........MFB.&#xA; NJ.FVHYT.MJDYN.GE.L.&#xA; BI.GRDH.LINDC.JT.MV.&#xA; YV.JUH.FYKMG.EH.CUNR&#xA; UN.EN.IBFKE.LO.VTHSK&#xA; USYN.KTDNL.MDYNEFJH.&#xA; NJM.DGBJ.U.....BRGN.&#xA; IM.RKLEDN.OUEMLLQDU.&#xA; N.UYNCGPKE.NIFUTNUR.&#xA; .MXIUNTDRML.KYBGELP.&#xA; &lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1681" LastEditorUserId="1681" LastEditDate="2018-04-17T07:54:51.090" LastActivityDate="2018-04-17T07:54:51.090" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5049" PostTypeId="1" AcceptedAnswerId="5052" CreationDate="2014-11-18T20:29:25.923" Score="13" ViewCount="1799" Body="&lt;p&gt;Here's an oldie but goodie from The Daily WTF; I paraphrase to avoid copyright issues:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A middle school math teacher, who also happened to be the P.E. coach, made the following deal with the non-athletic &amp;quot;nerds&amp;quot; in each new math class he taught on their first day:&lt;/p&gt;&#xA;&lt;p&gt;Outside the classroom was the main hallway, along one wall of which was an uninterrupted row of lockers numbered 1 through 100. As it was morning of the first day of school, the lockers hadn't been assigned yet and so didn't have any padlocks on them.&lt;/p&gt;&#xA;&lt;p&gt;The &amp;quot;jocks&amp;quot; in the class would go out to the row of lockers, which were all closed, and begin opening all of them. Then they'd close every even-numbered locker, and then open or close every third locker, then every fourth locker, and so on up to 100, &amp;quot;toggling&amp;quot; each locker in the sequence, opening if it was closed, closing if it was open.&lt;/p&gt;&#xA;&lt;p&gt;Once the jocks had toggled the last locker for the last time, a certain sequence of locker doors would be open. It was the nerds' job to figure out in a more elegant way which locker doors would still be open, before the jocks finished going through their &amp;quot;brute force&amp;quot; routine.&lt;/p&gt;&#xA;&lt;p&gt;If the nerds cracked it, they were excused from PE class for the rest of the year. If they couldn't, they'd face the jocks in the traditional first-day dodgeball game. In the P.E. teacher's 30 year career, not a single nerd ever avoided his fate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;What sequence of doors remains open when toggled according to this sequence, and why? And no cheating, running up and down the halls of your own school toggling lockers.&lt;/p&gt;&#xA;" OwnerUserId="5600" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-08-29T13:15:18.947" Title="Nerds, Jocks, and Lockers" Tags="|mathematics|combinatorics|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5050" PostTypeId="2" ParentId="5030" CreationDate="2014-11-18T20:33:55.513" Score="1" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;b&gt;Volume&lt;/b&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Familiarity / daily&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People often listen to music, etc., and view a volume meter of some sort.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Books&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (1) In general, books don't have particularly large volumes (unit of measurement). (2) Books don't make any noise, hence &quot;no volume&quot;, or you wouldn't find noise around a book. (3) Some kind of play on the idea of the volume of a book, e.g. My Book Name Vol. II&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Speaking / not knowing what you're thinking&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (1) Volume in terms of sound indicates loudness but not the content of the noise that &lt;i&gt;is&lt;/i&gt; loud or quiet. Said another way, knowing something is loud doesn't mean you know the entity causing the loudness. (2) The volume you use to express yourself can change the meaning of what you are saying. For example, shouting &quot;I hate you&quot; louder and more powerfully expresses a different meaning than a quiet &quot;I hate you&quot; said quieter, perhaps in a joking manner, or followed by a laugh.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Different meaning&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As explored in this answer, &quot;volume&quot; has multiple definitions and interpretations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Married&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Admittedly, for this one I haven't a clue. :(&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In town / knock twice&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Ton&quot; is a substring of &quot;town&quot;, and a ton is a unit of measuring &lt;i&gt;volume&lt;/i&gt;. The second part of the statement I'm more iffy on, though my best guess here would have something to do with the sound generated by knocking, or something that you would knock on to lower the volume, perhaps of a device of some sort.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user5040" LastActivityDate="2014-11-18T20:33:55.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5051" PostTypeId="2" ParentId="5049" CreationDate="2014-11-18T20:49:16.530" Score="4" Body="&lt;p&gt;At this point, I can only determine that lockers with an even number of factors of the same number (e.g. 16 = 2 x 2 x 2 x 2 ; 81 = 3 x 3 x 3 x 3) will have the locker doors open (includes door 1, since that has 0 non-one factors)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a more intuitive answer:&lt;br&gt;&#xA;Factor out each number (e.g. 12 = 2 x 2 x 3). Take the number of each factor, add 1, and multiply each factor together. This will give the total number of times a door is opened. (e.g. 2x 2's, 1x 3's -&gt; 3 x 2 = 6 toggles, door is closed). This works because you have to find each combination of factors, so you can use the 2's (0-1-2) times, and the 3's (0-1) times. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To have a door opened, you need all of your factors to be odd. The only way to do this is to have all your factors be even, which means you have a square number. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2014-11-18T21:13:12.810" LastActivityDate="2014-11-18T21:13:12.810" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5052" PostTypeId="2" ParentId="5049" CreationDate="2014-11-18T20:50:01.553" Score="18" Body="&lt;p&gt;Every locker is closed except for square numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say on turn 1, they make sure every locker is open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On turn 2, they close the even lockers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On turn 3, they toggle the lockers divisible by 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If locker N has a prime number, it will only be closed on turn N and opened on turn 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If locker N is even and greater than 2, it have been toggled on turn N/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If locker N is divisble by x and greater than s, it have been toggled on turn N/x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If locker N is square, it will $\sqrt N=N/\sqrt N$ so this will cause it to be flipped a time with no corresponding pair.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means every locker N will be toggled equal to the number of its unique factors.  &lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditDate="2014-11-18T20:59:03.890" LastActivityDate="2014-11-18T20:59:03.890" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5053" PostTypeId="2" ParentId="5049" CreationDate="2014-11-18T20:50:12.170" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1, 2, 5, 10, 17, 26, 37, 50, 65, 82&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I obtained this list running some simple JavaScript code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;var lockers = [];&#xA;for(i=0;i&amp;lt;100;i++){&#xA;    lockers.push(false); // all closed at the beginning&#xA;} &#xA;for (i = 1; i&amp;lt;100; i++) {&#xA;    for (j=0;j&amp;lt;100;j+=i) {&#xA;        lockers[j] = !lockers[j]; // toggle&#xA;    }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4456" LastActivityDate="2014-11-18T20:50:12.170" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5054" PostTypeId="2" ParentId="4854" CreationDate="2014-11-18T21:03:06.820" Score="4" Body="&lt;p&gt;Let $w$, $g$, $r$ be the numbers of white green and red cars, and $x$ be the number of cars of any other color&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$w + g + r + x = total$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$total - w = 2 = g + r + x$&lt;br&gt;&#xA;$total - g = 2 = w + r + x$&lt;br&gt;&#xA;$total - r = 2 = w + g + x$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$w + g + x = g + r + x$&lt;br&gt;&#xA;$w = r$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g + r + x = w + r + x$&lt;br&gt;&#xA;$g = r$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$w = r = g$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x &amp;lt;= 2 - w - g$&lt;br&gt;&#xA;$x &amp;lt;= 2 -2r$&lt;br&gt;&#xA;$x &amp;lt;= 2 (1-r)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r &amp;gt;= 0$&lt;br&gt;&#xA;$x &amp;gt;= 0$&lt;br&gt;&#xA;$1-r &amp;gt;= 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by the above constraints,&lt;br&gt;&#xA;$r &amp;lt;= 1$ AND $r &amp;gt;= 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r = g = w = 1$ and $x = 0$&#xA;or&#xA;$r = g = w = 0$ and $x = 2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 green, one white and one red&lt;br&gt;&#xA;or&lt;br&gt;&#xA;2 of any other color, and none green white or red&lt;/p&gt;&#xA;" OwnerUserId="5691" LastEditorUserId="2101" LastEditDate="2014-11-18T22:00:48.263" LastActivityDate="2014-11-18T22:00:48.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5056" PostTypeId="2" ParentId="5030" CreationDate="2014-11-18T21:08:32.323" Score="4" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; quotation marks&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See me daily&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Commonly used&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Books&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You don't use quotations with book titles. You use italics&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Speaking/Thinking&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Used for someone talking. Also used to show sarcasm in a word or phrase&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Married&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are two of them&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Knock twice&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is either type single quotes twice or you close the quotes by typing it a second time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5612" LastActivityDate="2014-11-18T21:08:32.323" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5058" PostTypeId="2" ParentId="3739" CreationDate="2014-11-18T21:14:10.757" Score="24" Body="&lt;p&gt;Five people saying almost the same thing? Well I like to be different, so try this for an answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The old man tasted the seagull and realized it was the exact same meat he'd been eating all along - it wasn't &quot;good&quot; food cooked properly or anything, just more seagull meat cooked by the same firepit they've been using this entire time. He realized that he was hallucinating/losing his mind and went to jump because he was still stuck on that island.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5688" LastEditorUserId="5688" LastEditDate="2014-11-18T21:24:50.457" LastActivityDate="2014-11-18T21:24:50.457" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5059" PostTypeId="2" ParentId="5049" CreationDate="2014-11-18T21:18:52.800" Score="2" Body="&lt;p&gt;Every square number from 1 to 100. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1, 4, 9, 16, 25, 36, 49, 56, 64, 81, 100&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Obtained with a Lua script ran on iPod:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;local Open = {}    &#xA;local Closed = {}    &#xA;&#xA;for i = 1,100 do    &#xA;    Closed[tostring(i)] = true    &#xA;end    &#xA;&#xA;for i = 1,100 do    &#xA;    for locker = 1,100 do    &#xA;       if locker % i == 0 then    &#xA;          if Open[tostring(locker)] then    &#xA;             Open[tostring(locker)] = nil    &#xA;             Closed[tostring(locker)] = true    &#xA;          elseif Closed[tostring(locker)] then    &#xA;             Closed[tostring(locker)] = nil    &#xA;             Open[tostring(locker)] = true    &#xA;          end    &#xA;       end    &#xA;    end    &#xA; end    &#xA;&#xA; for i,v in pairs(Closed) do&#xA;    print(i)    &#xA; end    &#xA; print(&quot;Open:&quot;)    &#xA; for i,v in pairs(Open) do    &#xA;    print(i)    &#xA; end&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is because they are toggled an odd number of times.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="15544" LastEditDate="2015-08-29T13:15:18.947" LastActivityDate="2015-08-29T13:15:18.947" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5060" PostTypeId="2" ParentId="5019" CreationDate="2014-11-18T21:30:05.483" Score="5" Body="&lt;p&gt;What about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; arithmetic operators&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am one of many siblings in a great family.&#xA;We are the most perfect of our race,&#xA;Because each of us has a clear worst enemy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; addition / subtraction, multiplication / division&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wave at my eldest brother and he becomes me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; multiplication can be expanded as series of additions&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When my dear cousin finds the end of her life,&#xA;She too can turn into me, in her gentle way.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; power operator can too be ultimately expanded as series of additions&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My younger brother thinks too highly of himself,&#xA;So smooth and perfect that he'll always outstrip me.&#xA;Often he even steals my words, confusing all concerned.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; well, this one I am not sure, factorial maybe?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5695" LastActivityDate="2014-11-18T21:30:05.483" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5061" PostTypeId="2" ParentId="5006" CreationDate="2014-11-18T21:38:41.537" Score="4" Body="&lt;p&gt;As mentioned in another answer, I would use the first egg to incrementally test as far UP as I can go while still having enough remaining trials to test the skipped floors. Basically test 1 is go up 20 floors, if the egg breaks, start at the bottom floor and begin testing upwards one floor until you find the breaking point. If the egg did NOT break at floor 20, I would go up an additional 19 floors and test again. Use the remaining 18 trials to test floors 21 - 39. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continue moving up by the number of trials remaining until the first egg breaks, then go back to the bottom of your test set and work your way up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To determine the largest building that this method will work you must simply do &lt;/p&gt;&#xA;&#xA;&lt;p&gt;20+19+18+17+16+15+14+13+12+11+10+9+8+7+6+5+4+3+2+1 = 210 stories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For the second part of the questions&lt;/strong&gt;, given an unlimited number of eggs, the answer is to use the divide and conquer method. Given 100 floors, start at the &quot;middle&quot; floor and test up or down reducing the trial size by 1/2 each test. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If egg breaks, repeat from floor 25. If egg is intact, move up to floor 75. Continue to test from the half ways points until you have determined the correct floor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.G. Presuming that the answer was that the egg tolerated being dropped from the 88th floor without breaking (out of 100).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;TRIAL 1 - drop from 50th floor -&amp;gt; No break, retest only top half&#xA;TRIAL 2 - drop from 75th floor -&amp;gt; No break, retest only top half&#xA;TRIAL 3 - drop from 87th floor -&amp;gt; No break, retest only top half&#xA;TRIAL 4 - drop from 93rd floor -&amp;gt; break, retest only bottom half&#xA;TRIAL 5 - drop from 90th floor -&amp;gt; break, retest only bottom half&#xA;TRIAL 6 - drop from 88th floor -&amp;gt; No break, can be dropped from up to 88 floor&#xA;TRIAL 7 - drop from 89th floor -&amp;gt; break, 88 is the answer.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another Example, Presuming that egg can only be dropped from the first floor without breaking. (second story breaks egg)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;TRIAL 1 - drop from 50th floor -&amp;gt; break, retest only bottom half&#xA;TRIAL 2 - drop from 25th floor -&amp;gt; break, retest only bottom half&#xA;TRIAL 3 - drop from 12th floor -&amp;gt; break, retest only bottom half&#xA;TRIAL 4 - drop from 6th floor -&amp;gt; break, retest only bottom half&#xA;TRIAL 5 - drop from 3rd floor -&amp;gt; break, retest only bottom half&#xA;TRIAL 6 - drop from 2nd floor -&amp;gt; break, retest only bottom half&#xA;TRIAL 7 - drop from 1st floor -&amp;gt; no break, floor 1 is the answer.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The largest size building is equal to the largest number that can be divided into equal parts not having more than 3 floors in the final two trials. A building with 2048 floors could be tested with only 11 trials. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following this logic the largest size building you could test with this method is 1048576 stories tall. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point at witch additional eggs becomes irrelevant is at 20, since the number of trials limits the number of &quot;splits&quot;, not the eggs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus question:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having three eggs at your disposal increases the size of the building that you can test substantially because you can blend the two methods mentioned above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would begin at whichever floor is the highest that you can go so that you only need to test 1/2 of the floors with the remaining trials. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;TRIAL 1 test from floor 38. If the egg breaks, use the second egg to test at floor 19. If the egg breaks use the third egg to systematically test floors 1-18; if it doesn't break use the third egg to test floors 19-37.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeat incrementing up double the one less than the number of floors that you have trials remaining.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the maximum number of floors is&#xA;38+36+34+32+30+28+26+24+22+20+18+16+14+12+10+8+6+4 = 336 floors with 3 eggs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Personally, speaking I must say that I would only test two stories since I'd really just enjoy having the excuse to throw eggs at people as they walk by and since my aim isn't that good, I'd throw both eggs from the second story to increase my chances of hitting someone.&lt;/em&gt; &lt;/p&gt;&#xA;" OwnerUserId="5696" LastActivityDate="2014-11-18T21:38:41.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5062" PostTypeId="2" ParentId="4828" CreationDate="2014-11-18T21:43:18.693" Score="4" Body="&lt;p&gt;Try this one on for size:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Golden Bird only moves 5mph - but it can choose the reference point! Normally we would expect it to fly 5mph relative to the ground / the planet it's on. But that means (assuming earth) the Golden Bird is moving ~67,000mph relative to the sun. That's ~492,000 mph relative to our &lt;strong&gt;&lt;em&gt;galaxy&lt;/em&gt;&lt;/strong&gt;. So we need to know what it's moving at 5mph in relation to - and the answer is: it chooses. If it can choose to move at 5mph relative to an airplane which is itself moving at &gt;= 1mph across the ocean (relative to the ocean, of course), than the Golden Bird will cover the 6 miles (relative to the ocean) in just a single hour.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5688" LastActivityDate="2014-11-18T21:43:18.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5063" PostTypeId="2" ParentId="5019" CreationDate="2014-11-18T21:49:41.613" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Sine&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am one of many siblings in a great family.&#xA;We are the most perfect of our race,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The family is &lt;strong&gt;Trig functions&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because each of us has a clear worst enemy.&#xA;My foe - I know him well - is much like me,&#xA;But we continually fight each other;&#xA;We never hide our hatred for each other.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sine function is a periodic one. It can not overcome the +1 / -1 limits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wave at my eldest brother and he becomes me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The cousin is the &lt;strong&gt;cosine&lt;/strong&gt;: sin($\alpha$ +$\pi/2$ ) = - cos($\alpha$ ). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When my dear cousin finds the end of her life,&#xA;She too can turn into me, in her gentle way.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The dear cousing is a &lt;strong&gt;McLaurin /Taylor series&lt;/strong&gt; If n$\rightarrow$   $\infty$, the series is really the same as sin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My younger brother thinks too highly of himself,&#xA;So smooth and perfect that he'll always outstrip me.&#xA;Often he even steals my words, confusing all concerned.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The brother is &lt;strong&gt;tan&lt;/strong&gt;, thus it goes to infinity at 90° against 1. You can transform tan as $\frac{sin}{cos}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1685" LastActivityDate="2014-11-18T21:49:41.613" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5065" PostTypeId="2" ParentId="5030" CreationDate="2014-11-18T22:35:24.937" Score="1" Body="&lt;p&gt;Congratulations to Ric, who posted &lt;a href=&quot;https://puzzling.stackexchange.com/a/5056/4450&quot;&gt;the answer&lt;/a&gt; I was looking for.  The poem is referring to &quot; (double quotes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me just add a few thoughts to Ric's answer.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Well read I may look, But you won't find me around a book&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Quotes are used for the titles of a lot of written material such as short stories, chapters, articles, essays, and poems.  But like Ric said, they are never used with book titles.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;You can find me in town, Knock twice and I'll come down&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I was thinking of the keyboard as a town, with each key as a house.  The &quot; lives upstairs (above the '), so you have to press 2 keys (SHIFT and ') to get the &quot;.&lt;/p&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2014-11-18T22:35:24.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5066" PostTypeId="2" ParentId="4453" CreationDate="2014-11-18T22:44:37.250" Score="8" Body="&lt;p&gt;I solved this without looking at any of the previous answers and was surprised that there are many solutions!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wrote a C program to brute-force the solution here: &lt;a href=&quot;https://gist.github.com/timmontague/4244afcde2750f6f9c67&quot; rel=&quot;noreferrer&quot;&gt;https://gist.github.com/timmontague/4244afcde2750f6f9c67&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the &quot;best&quot; solutions that I found is the following set of dice:&lt;br&gt;&#xA;d1: &lt;code&gt;1 1 4 4 4 4&lt;/code&gt;&lt;br&gt;&#xA;d2: &lt;code&gt;3 3 3 3 3 3&lt;/code&gt;&lt;br&gt;&#xA;d3: &lt;code&gt;2 2 2 2 5 5&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I defined &quot;best&quot; as the set of dice with the largest average win percentage. In the example above there are no draws and the win percentages are:&lt;br&gt;&#xA;d1 beats d2 24 out of 36 times&lt;br&gt;&#xA;d2 beats d3 24 out of 36 times&lt;br&gt;&#xA;d3 beats d1 20 out of 36 times&lt;/p&gt;&#xA;" OwnerUserId="5698" LastActivityDate="2014-11-18T22:44:37.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5067" PostTypeId="1" AcceptedAnswerId="5070" CreationDate="2014-11-18T22:59:34.527" Score="-4" ViewCount="681" Body="&lt;p&gt;Somebody ordered a pizza. They ordered &lt;em&gt;X&lt;/em&gt; slices with pepperoni, &lt;em&gt;Y&lt;/em&gt; slices with mushrooms, and &lt;em&gt;Z&lt;/em&gt; slices with cheese. If no two slices were identical, and no combinations of toppings were not present, how many (equal) slices was the pizza cut into?&lt;/p&gt;&#xA;" OwnerUserId="5703" LastEditorUserId="69582" LastEditDate="2021-01-14T18:49:59.887" LastActivityDate="2021-01-14T18:49:59.887" Title="Somebody ordered a pizza" Tags="|mathematics|logical-deduction|combinatorics|" AnswerCount="6" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="5069" PostTypeId="2" ParentId="5067" CreationDate="2014-11-18T23:06:06.640" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I would say 3, but you referring to slices so every topping must be &gt; 1. So my answer is 7:&lt;br/&gt; The slices has been made in this order: &lt;strong&gt;A - B - A - B - C - B - C&lt;/strong&gt; where the last C is next to the first one. So, X = 2, Y = 3 and Z = 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1685" LastActivityDate="2014-11-18T23:06:06.640" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5070" PostTypeId="2" ParentId="5067" CreationDate="2014-11-18T23:08:37.747" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 7. PMC, PM, PC, MC, P, M, C.  8 if you can have one with no toppings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;pre&gt;&#xA; - 3.  There are 3 combinations where each slice has only one topping (as 3 toppings).&#xA; - 3.  There are 3 combinations where each slice has 2 toppings (more easily visualised as 3 combinations where one topping is missing)&#xA; - 1.  There is only 1 topping with all three combinations of toppings.&#xA; - 1.  There is only 1 topping with no toppings (debatable whether this is a valid &quot;combination&quot;).&#xA; 3 + 3 + 1 = 7 unique combinations of toppings excluding the empty topping&#xA; 3 + 3 + 1 + 1 = 8 unique combinations of toppings including the empty topping&#xA; &lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1681" LastEditorUserId="1681" LastEditDate="2014-11-19T13:47:39.017" LastActivityDate="2014-11-19T13:47:39.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5072" PostTypeId="2" ParentId="5067" CreationDate="2014-11-18T23:10:24.247" Score="0" Body="&lt;p&gt;Well, the fact none can be identical means nothing. It limits nothing but that it cannot be split into even sized portions with the same amount of toppings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was n &lt;strong&gt;slices&lt;/strong&gt; means plural. There were 3 toppings, meaning a minimum of 3x2 slices; &lt;strong&gt;6 slices&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No combinations were present. Reinforcing the idea of 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the minimum of slices was 6. The maximum amount of slices could depend on how precise the cutter is with sizes and if there were other types of slices with other, or no toppings (like pineapple, or hamburger meat). So anything 6+ would work.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2014-11-18T23:10:24.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5074" PostTypeId="1" AcceptedAnswerId="5114" CreationDate="2014-11-18T23:39:47.677" Score="-1" ViewCount="669" Body="&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;You're in a country which contains thousands of soldiers whom never fight.    &lt;/li&gt;&#xA;  &lt;li&gt;You take the 1st number of the name of this country, and subtract lines of 7.    &lt;/li&gt;&#xA;  &lt;li&gt;Under the square tree's stump.    &lt;/li&gt;&#xA;  &lt;li&gt;You take this number and find a corresponding key.    &lt;/li&gt;&#xA;  &lt;li&gt;You go down soh, left stx and concatenate the first key with this one.    &lt;/li&gt;&#xA;  &lt;li&gt;You are here. What have you made it to?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can find previous puzzles here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/4299/where-did-you-end-up-part-1&quot;&gt;Where did you end up? [Part 1]&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/4417/where-did-you-end-up-part-2&quot;&gt;Where did you end up? [Part 2!]&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2014-11-24T10:00:34.273" Title="Where'd you end up? [Part 3!]" Tags="|lateral-thinking|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5075" PostTypeId="1" AcceptedAnswerId="5091" CreationDate="2014-11-19T00:02:35.023" Score="10" ViewCount="1333" Body="&lt;p&gt;Ernie loves cooking and it's always worth the wait as he follows his curiously precise recipe to make Ginger Crunch Cookies. So I was happy to drop around when he invited me to help test his latest invention - a pneumatic cookie-cutter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I've made two of them,&quot; said Ernie. &quot;They use pneumatic actuators to transfer &lt;em&gt;all&lt;/em&gt; the dough inside the cut area onto the cooking tray. This one cuts squares and the other one cuts equilateral triangles.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The triangular cutter seems a bit bigger&quot;, I commented.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Only in linear dimensions&quot;, Ernie responded. &quot;The triangle has exactly the same $10,000\text{ mm}^2$ area as the square&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ernie already had the dough rolled out so it covered most of the surface of the large rectangular cooking bench and was about to cut the cookies, when he noticed that the readouts of the twin thermometers in his 'cooking furnace' disagreed by more than 0.6°K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Can't have that&quot;, he muttered as he reached for his screwdriver to make some necessary adjustments. &quot;Can you cut them out for me? The instructions are at the end of the recipe.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I turned to the last page of his cooking notes and was more than a little nervous when I read the following instructions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;&lt;em&gt;Cutting&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Step A.&lt;/strong&gt; Using the square cutter:&lt;br&gt;&#xA;  1) Position the cutter somewhere over the table&lt;br&gt;&#xA;  2) Rotate the cutter to an angle $\theta$ relative to one edge of the table&lt;br&gt;&#xA;  3) Make a cut and transfer the square piece of dough inside the cut region to the tray.&lt;br&gt;&#xA;  4) Repeat step 3, with the cutter at the same angle $\theta$, at all positions on the table displaced $md$ horizontally and $nd$ vertically (where $m$ and $n$ are integers $\{...-2,-1,0,1,2...\}$ and $d$ is a chosen displacement), and transfer the square piece of dough inside each cut square region to&#xA;  the tray.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Step B.&lt;/strong&gt; Using the triangular cutter:&lt;br&gt;&#xA;  1) Position the cutter at a chosen position over the table&lt;br&gt;&#xA;  2) Rotate the cutter to an angle $\phi$ relative to one edge of the table&lt;br&gt;&#xA;  3) Make a cut and transfer all the pieces of dough inside the cut region to the tray.&lt;br&gt;&#xA;  4) Repeat step 3, with the cutter at the same angle $\phi$, at all positions on the table displaced $md$ horizontally and $nd$ vertically (where $m$ and $n$ are integers $\{...-2,-1,0,1,2...\}$ and $d$ is the &lt;strong&gt;same&lt;/strong&gt; displacement as used for the square cutter), and transfer all the pieces of dough inside each cut triangular region to the tray.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;. Choose positions and angles of the square and triangular cutters so you can use the largest value of $d$ possible, such that after following steps A and B, there is no dough left on the table (except maybe for a little bit round the edges of the table) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;So you want a set of perfectly square cookies - that don't overlap one-another?&quot;, I asked. &quot;Correct&quot;, Ernie replied. &quot;Plus, the odd-shaped bits that the triangular cutter will make in Step B...?&quot;. &quot;Correct...&quot;, Ernie agreed again. &quot;...and of course, make sure that you use the biggest value of $d$ possible! Because that's the &lt;em&gt;only&lt;/em&gt; way to cut perfect cookies&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Msid7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried sketching a few test patterns to find a suitable tiling on a spare copy of one of Ernie's recipes, but couldn't sort out the optimal value for $d$. So while still repairing the oven, Ernie drew a quick diagram on the dusty floor, and annotated it with angles accurate to the nearest second of arc, and distances to the nearest micron. I cut the cookies at the required spacing and sure enough, after the last triangular cut was made all the dough was gone. The cookies made it into the oven in time and were a great success.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I have a bit of a problem. I borrowed Ernie's cookie cutters and plan to make him some perfect ginger cookies for his birthday, but can't remember Ernie's diagram or distance measurements for efficient cuts. Can you help me? Accuracy to the nearest $\frac1{10}^{th}$ of a mm and $\frac1{10}^{th}$ of a degree (and maybe a simple diagram) should be enough.   &lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="8874" LastEditDate="2016-03-23T15:40:56.423" LastActivityDate="2016-03-23T15:40:56.423" Title="Ernie and the Geometric Ginger Cookies" Tags="|mathematics|pattern|geometry|optimization|dissection|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5076" PostTypeId="2" ParentId="5067" CreationDate="2014-11-19T00:13:14.660" Score="6" Body="&lt;p&gt;For each slice, each slice either has a topping or doesn't have a topping.&lt;/p&gt;&#xA;&lt;p&gt;That's two possibilities per topping, for three toppings, or a total of 8 possibilities on a single slice.&lt;/p&gt;&#xA;&lt;p&gt;This means that there were 8 slices on the pizza, with the following combinations of toppings:&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th&gt;Slice #&lt;/th&gt;&#xA;&lt;th&gt;P&lt;/th&gt;&#xA;&lt;th&gt;M&lt;/th&gt;&#xA;&lt;th&gt;C&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;1&lt;/td&gt;&#xA;&lt;td&gt;╳&lt;/td&gt;&#xA;&lt;td&gt;╳&lt;/td&gt;&#xA;&lt;td&gt;╳&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;2&lt;/td&gt;&#xA;&lt;td&gt;╳&lt;/td&gt;&#xA;&lt;td&gt;╳&lt;/td&gt;&#xA;&lt;td&gt;○&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;3&lt;/td&gt;&#xA;&lt;td&gt;╳&lt;/td&gt;&#xA;&lt;td&gt;○&lt;/td&gt;&#xA;&lt;td&gt;╳&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;4&lt;/td&gt;&#xA;&lt;td&gt;╳&lt;/td&gt;&#xA;&lt;td&gt;○&lt;/td&gt;&#xA;&lt;td&gt;○&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;5&lt;/td&gt;&#xA;&lt;td&gt;○&lt;/td&gt;&#xA;&lt;td&gt;╳&lt;/td&gt;&#xA;&lt;td&gt;╳&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;6&lt;/td&gt;&#xA;&lt;td&gt;○&lt;/td&gt;&#xA;&lt;td&gt;╳&lt;/td&gt;&#xA;&lt;td&gt;○&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;7&lt;/td&gt;&#xA;&lt;td&gt;○&lt;/td&gt;&#xA;&lt;td&gt;○&lt;/td&gt;&#xA;&lt;td&gt;╳&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;8&lt;/td&gt;&#xA;&lt;td&gt;○&lt;/td&gt;&#xA;&lt;td&gt;○&lt;/td&gt;&#xA;&lt;td&gt;○&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;So four slices have each topping.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="69582" LastEditDate="2021-01-14T18:48:22.777" LastActivityDate="2021-01-14T18:48:22.777" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="5077" PostTypeId="2" ParentId="5067" CreationDate="2014-11-19T00:18:19.847" Score="1" Body="&lt;p&gt;Pizza's not pizza without cheese; you don't call up the place, order a pizza &quot;with just pepperoni&quot;, and expect it to show up without cheese (or sauce). A &quot;normal&quot; amount of cheese is included in the cost of the pie, and you usually don't save any money ordering a cheeseless pizza. So, I will interpret Z as the number of slices that have only cheese. Given your other requirements, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Z = 1 because no two slices are alike, therefore no two slices can have only cheese.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The rest of it is simple; there is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; one slice with pepperoni, one slice with mushrooms, and one slice with both pepperoni and mushrooms, for four unique equal slices of pizza.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5600" LastActivityDate="2014-11-19T00:18:19.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5078" PostTypeId="2" ParentId="4828" CreationDate="2014-11-19T00:19:33.193" Score="2" Body="&lt;p&gt;Well everyone has already answered all the good serious answers.  The only serious one that appears to be missing is&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The bird can rest it's wings by &lt;em&gt;gliding&lt;/em&gt;, which is exactly how birds that fly over large expanses of water rest.  He builds altitude during his hour of flying, and glides, slowly burning altitude, while resting his wings.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;alternatively&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the bird lived on an island in the center of the ocean all his life.  When he finally chooses to cross over the ocean to the other side he finds it takes nearly 50 minutes to fly from his island all the way to the far side of the ocean, sure is good he didn't live on the other coast or he would never have made it!&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;or, to take the last answer to an even more annoying extream&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The golden bird one day decides to fly over the ocean.  Flying part way there he decides to rest by landing on the water and floating awhile, being a golden duck and all.  After awhile he gets bird and starts up his flight again.  When he decides to fly again he flys over the ocean and to the other side without any rest &lt;em&gt;that&lt;/em&gt; flight.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Of course if we really want to be annoying we can start playing with pronouns and proper names: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;the bird's 5 MPH is defined as his total average speed for a long trip.  Meaning that he is taking an hour of travel, 5 of rest, then repeating over and over.  Because of this the speed he travels at while in motion is actually 30 MPH, making it quite easy to cross the ocean.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;the other side&quot; is the name of a popular party boat, currently in the middle of crossing the ocean.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;5 MPH is the name of the only ocean at which the Golden Bird is capable of flying.  He can, however, manage quite fast speeds over the &quot;5 MPH&quot; ocean. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Golden Bird can only fly 5 MPH and crosses the ocean.  His friend The Bird (who's feathers are not nearly as brilliant) must rest after every hour of flying.  It is a good thing that The Golden Bird can fly much longer then The Bird can.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The ocean is the name of a huge moving lava flow.  The bird flys for 5 hours to cross it, luckily the lava flows more then 1 MPH the other way at the same time.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;'IT' is the name of The Golden Bird's BFF the albatross.  The Golden Bird must rest after It the albatross has flown for an hour, because 'It' gets tired easily and the Golden Bird would never leave his tired friend.  Sadly one night while flying over the ocean It the albatross was shot down by a ship full of hunger lateral thinking puzzlers (who all have a strange uncontrollable fear of tasting soup).  The Golden Bird is so shocked seeing his friend It killed after only 30 minutes of flying he keeps flying for hours crying!.  Wanting to get away from the horrible memory he flys all the way to the other coast, which he was never able to reach while he had to rest with IT every hour.  In his outrage The Golden Bird swears to force all future puzzlers to figure out the story behind how his doomed voyage, in hopes that they will learn to leave poor easily-tired albatross alone, and also stop being cannibals of course.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;yeah, okay sorry I got carried away there.  I've had an image of an albatross in my head that I needed to get out since reading this puzzle!  My apologies to all those that don't get the allusion.  Lets move away from messing with names and pronouns, that's just too easy.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The bird flys five miles across the ocean before landing on the ocean and floating while because his wings are growing tired and he is hungry.  However, his landing isn't the restful Fishing reprieve he had hoped for. A storm blows in even as he starts diving for fish, and it rapidly grows. The poor bird finds himself paddling for dear life just to stay above the waves and turbulence.  He tries fishing for hours, despite the difficulty in catching fish in such turbulent waters, in hopes of settling his gnawling hunger.  He keeps hoping the storm will die out as quickly as it started so he can start flying again, but after 5 hours of none stop paddling just to try to keep his head above the water he comes to realize this storm isn't going to end any time soon. The poor bird is exhausted from his time at sea, and doesn't know how he can manage any strain on his weary body, but he realizes the only hope of surviving this storm is reaching dry land, no matter what that takes!  The only good news is that muscles of his wings have recovered slightly from the strain of flight during the hours trying to wait out the storm, if only he had the energy in the rest of his body for the exertion of flight!.   With an exhausted and desperate strain he somehow manages to climb into the air, though he doesn't know where he finds the energy.  He drives on through the ferocious storm towards dry land and safety.  He is so tired he doesn't know how he finds the energy to beat his rested wings, but the need to survive drives him.  Finally he seen land, with one last deperate dive he swoops in to the safe dry grown, too exauhusted to even try to find a try to rest on; he can only hope no preditors come to find him sheltering against the storm before he recovers enough to start searching for a safer place to stay.  He can barely believe he made it through the hours long ordeal without a rest, he will never try crossing that ocean while it's raining that's for sure!  The poor tired bird tucks his head under his golden wing and settles in for his first rest in over 6 hours.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2698" LastActivityDate="2014-11-19T00:19:33.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5079" PostTypeId="1" AcceptedAnswerId="5082" CreationDate="2014-11-19T00:38:13.217" Score="6" ViewCount="506" Body="&lt;p&gt;The census taker has arrived at yet another house, and is greeted by a busy housewife. This woman poses a new riddle: &quot;2 of my girls have red hair, the rest are brunettes. All 8 of my boys have black hair. What percentage of my children are girls?&quot; she asks. The census taker is stumped, and instead decides to help her put the dozen popsicles she bought for her kids in the fridge, and watches as the mother scolds a child for not wanting to share his toys. &quot;All of my children are so greedy she says&quot;. The census taker decides to skip this house, since he isn't feeling like unnesesarily hard riddles. Did the census taker give up too easily? If so, what was the answer to the mother's question? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I made the problem myself, so do tell me what you think about it and how I could improve it. &lt;/p&gt;&#xA;" OwnerUserId="5706" LastEditorUserId="4385" LastEditDate="2015-12-15T15:57:53.457" LastActivityDate="2015-12-15T15:57:53.457" Title="Modified Census Taker Problem" Tags="|lateral-thinking|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5082" PostTypeId="2" ParentId="5079" CreationDate="2014-11-19T00:51:15.337" Score="10" Body="&lt;p&gt;The number of girls she has is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Meaning the percentage of girls is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4 / 12  or one-third or 33.333... %&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; She put a dozen popsicles in the fridge.  She has at least 2+8=10 kids, so 12 is the only number of kids to evenly divide the popsicles so each kid has at least 1 - and the busy housewife knows if there are extras or anybody has to share, the house will be full of strife.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2718" LastActivityDate="2014-11-19T00:51:15.337" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5089" PostTypeId="1" AcceptedAnswerId="5369" CreationDate="2014-11-19T01:32:50.213" Score="5" ViewCount="567" Body="&lt;p&gt;&lt;strong&gt;Preface.&lt;/strong&gt; After the extraordinarily easy &quot;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/4996/nerd-on-the-loose-ii-where-how&quot;&gt;riddle&lt;/a&gt;&quot;, I guess I can wrap out everything in this &quot;detective&quot; puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Background.&lt;/strong&gt; Professor Swannie Taco is a genius in most fields of science and technology. She is probably hated by her parents, and her name is a perfect evidence. After Swannie went missing a few months ago, strange letters from her kept appearing on her G+ account. Nobody knows who exactly posted these messages, nor understands the content of these seemingly cryptic memos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clarification.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; None.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hints.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; None.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Overview.&lt;/strong&gt; Feeling worried, you decided to do some detective work and check her past. You thought, which parents name their child &quot;Taco&quot;, it is definitely fishy. However, no matter how much you dig, you just can't seem to find any record of her birth. It's like she just appeared into this world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On her &lt;a href=&quot;https://plus.google.com/116758574347391243190/&quot; rel=&quot;nofollow noreferrer&quot;&gt;G+&lt;/a&gt; page, there are some information that seems valuable. You recorded them just in case. In the meantime, she posted her final message.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Post.&lt;/strong&gt; It was believed that I &quot;borrowed&quot; ideas from him. It wasn't true, and I cannot accept the distrust from you mere mortals. Here is a final chance, let me have what is rightfully mine, and you homo sapiens shall avoid the avoidable death.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Task.&lt;/strong&gt; Figure what who she really is, why she is doing all this, and what she really wants.&lt;/p&gt;&#xA;" OwnerDisplayName="user5508" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2016-09-11T01:20:05.733" Title="Nerd on the Loose III - WHO, WHAT &amp; WHY" Tags="|mathematics|mystery|science|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5090" PostTypeId="1" AcceptedAnswerId="5098" CreationDate="2014-11-19T01:36:26.950" Score="18" ViewCount="1308" Body="&lt;p&gt;Let's say I have a puzzle like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8lwDq.jpg&quot; alt=&quot;Ice puzzle&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is that since you're on ice, once you start moving you can't stop until you hit a rock. If you treat the red/blue patches as paint that leaves trails when you move, then trying to collect all the coins in the fewest number of moves spells out a word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as you can see, this puzzle heavily restricts what moves you can perform, making it very easy. I'd like to construct a puzzle which gives more freedom and possibly involves crossing over the same paint patches multiple times, but I'm afraid that the resulting puzzle might have multiple solutions or the painted word might be unrecognisable if you do things in a different order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is this: &lt;strong&gt;Is there a good way of either 1) generating a more open puzzle which still guarantees a unique solution or 2) modifying the puzzle mechanics so that 1) is easier to do?&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit:&lt;/em&gt; To clarify I'm not asking about generating these sorts of puzzles in general. Instead, given a target word, how can we generate a unique-solution puzzle which spells out that word? (via paint/any other mechanic)&lt;/p&gt;&#xA;" OwnerUserId="445" LastEditorUserId="445" LastEditDate="2014-11-19T03:06:33.280" LastActivityDate="2014-11-19T09:57:11.520" Title="Generating ice-sliding puzzles which spell out words" Tags="|puzzle-creation|" AnswerCount="2" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="5091" PostTypeId="2" ParentId="5075" CreationDate="2014-11-19T02:43:43.640" Score="7" Body="&lt;p&gt;OK, well first, it would be useful to calculate the dimensions of the cutters. The area of each cookie, square or triangular, is $10000\text{ mm}^2$, or $100\text{ cm}^2$. That's easy enough for the square; the cutter is $10\text{ cm}$ ($100\text{ mm}$) on a side. The triangle's a bit trickier; an equilateral  triangle is basically two adjacent 30-60-90 triangles, and the unit measurements of such a triangle (assuming a hypotenuse of $1$) are $\frac12$ for the short side, and $\frac{\sqrt3}2$ for the long leg. So, the area of one 30-60-90 triangle with hypotenuse length $a$ is $\frac a2 \cdot \frac{\sqrt3a}2 \cdot \frac12$, or $\frac{a^2\sqrt3}8$. Two of those make an equilateral triangle, so the area of that is $\frac{a^2\sqrt3}4$. We need the value of $a$ that makes the area $10000$; substituting and rearranging variables gives us $\sqrt{\frac{40000}{\sqrt3}}$, or $151.9 \text{ mm}$ per side ($15.19\text{ cm}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we need an arrangement of squares, rotated from square with the table by some arbitrary angle $\theta$ and spaced by $m$ and $n$ multiples of $d$ in each direction such that they don't overlap, then an arrangement of triangles which can overlap, such that using the same $d$ but different $m$, $n$ and angle $\phi$, all dough in the area being cut is used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simple answer is $d = 10 \text{ mm}, \theta = 0$, and all $m,n = 1$. Additional measurements for the triangular blade are immaterial as this solution produces only the square cookies; there will be nothing left for the triangle blade to cut. This is a working answer but probably not the largest possible $d$ value. So, we must space out the squares to allow the triangular blade something to cut, but not so much that dough is left behind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ideal solution, IMO, is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an arrangement of rotated squares, such that the edges of adjacent squares touch along a portion of the full length of each side, and the space between them is the largest possible square small enough to fit within the triangle cutter, so the triangular cutter essentially doesn't make triangles, it simply picks up the smaller squares. $d$ is then the displacement that allows a rotation of the squares by the angle $\theta$ to create this leftover space, for the triangle cutters, also rotated to $\phi = \theta$ to come in behind and collect.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, let's start by finding the dimensions of &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the largest square that could fit into an equilateral triangle $15.2\text{ cm}$ on a side. By doing so, we essentially split the triangle into four regions; the square, an equilateral triangle above, and two 30-60-90 triangles to either side. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What we're looking for, given the length of the side of the triangle $a$, is a side length $b$ for a square, such that &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the length of each side of the smaller equilateral triangle produced by inscribing the square will also be $b$.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, if the full triangle has height $h$ and side length $a$, then the height of the smaller triangle will be $h-b$ (as the square will have height $b$ and width $b$). The proportion of height to base of the smaller and larger triangles will be the same because they are geometrically similar (same angles, different dimensions), therefore $\frac{h-b}b = \frac{h}a$. We know $\frac{h}a = \frac2{\sqrt3}$, and so $h = \frac{a\sqrt3}2$, therefore substituting and solving for $b$, $b = \frac{a\sqrt3}{2 + \sqrt3}$. We know the triangle is $15.2\text{ cm}$ on each side, therefore with a little calculator work, the smaller squares will be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $7.054\text{ cm}$ on a side,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and therefore, adjacent squares' edges will touch along&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $10 - 7.054 = 2.94\text{ cm}$ &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;of their length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what angle of rotation $\theta$ of the squares from being orthogonal to the cutting sheet gives us this amount of overlap? And what spacing between the centers of the squares $d$ will ensure the squares don't overlap? That's a bit more complex. We know that at $\theta = 0$, the overlap is $10\text{ cm}$, and the $d$ that causes the edges to line up perfectly is also $10\text{ mm}$. Now, squares are made of two 45-45-90 triangles, and the hypotenuse of such a triangle given unit side lengths is $\sqrt2 \approx 1.414$. So, a rotation $\theta=45°$, and $D=10\text{ cm}\sqrt2$ would cause the squares' edges to intersect at a single point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if I had something I could use to illustrate this next point, I would, but I don't, so you have to draw it out yourself and confirm it. If you put two squares together so that some but not all of an edge is touching, then draw a line between those two squares' bottom corners, that line would be parallel to and the same length as the line between their centers, because those two lines and lines connecting center to corner would create a parallelogram. In addition, that line forms the hypotenuse of a triangle with the bottom of one square and the side of the adjacent one, and &lt;em&gt;we know the lengths of those sides&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ($10\text{ cm}$ and $7.054\text{ cm}$). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By the Pythagorean Theorem, $a^2 + b^2 = c^2$, therefore &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $d = \sqrt{10^2 + 7.054^2} = \sqrt{149.763} = 12.2378\text{ cm}$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In addition, we know two sides of a triangle, and we want the angle $\theta$ made between the longer leg and the hypotenuse. Remember &quot;some old hippie caught another hippie tripping on acid&quot;?, tan = opp/adj, so &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\tan\theta = \frac{7.054}{10}$, thus $\theta = 35.199°$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So there you have it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $d = 12.24\text{ cm}$ or $122.4\text{ mm}$. Both $\theta$ and $\phi$ are $35.199°$ assuming that zero for both shapes makes at least one edge of the shape parallel to the top and bottom of the cutting sheet, and the grid for both square and triangle cutters is a unit grid of squares of length $d$, so all $m$ and $n$ are 1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Any bigger, and&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the square-shaped areas in between the square cutouts would be too big for the triangle cutter to remove completely, leaving small 30-60-90 triangles laying around.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I could, with a little more time, figure out the relative initial position of the triangle given the center of the square, but that was not part of the puzzle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;I have added a diagram of the tiling below - Penguino&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/rUpKE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5600" LastEditorUserId="4421" LastEditDate="2015-11-18T20:06:57.337" LastActivityDate="2015-11-18T20:06:57.337" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5092" PostTypeId="2" ParentId="5079" CreationDate="2014-11-19T03:04:42.010" Score="3" Body="&lt;p&gt;He could observe that one dozen popsicles were bought for the kids, and the mother said they were greedy, so we can assume she wouldn't have wanted to deal with either a shortage or surplus when handing them out. So,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; there can be no more than 12 kids, and there are almost certainly exactly 12 kids.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;10 are either redheaded girls or black-haired boys. &quot;The rest&quot; are brunette girls, the term implying &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; more than one brunette, reinforcing the previous spoiler. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore we can say that there are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2 brunette girls,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and therefore the percentage of girls among the children in the household is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4/12 = 1/3 = 33.33%&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5600" LastActivityDate="2014-11-19T03:04:42.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5093" PostTypeId="2" ParentId="5009" CreationDate="2014-11-19T03:07:50.243" Score="1" Body="&lt;p&gt;&lt;strong&gt;Guess&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Whitespace&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Justification 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Ok, as you might have expected I secretly implemented many hints for you.&quot; This hint that there are other clues elsewhere, so take a look at the markdown of the post, you will see comments that will further justify my answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Justification 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;A unique, selfmade picture with nothing else but the solution on it. Please use colors with contrast to the background color.&quot; I'm thinking the picture found in the comments will justify this. Even if the background color is white, and font color is black, it will still be a plain white image.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Justification 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Go ahead and look for the incredibly well hidden connection between the hints.&quot; Well, in the markdown, there are huge amount of whitespace connecting the hints.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Justification 4&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Give me a detailed and plain description of it and you will, I am absolutely sure, be the winner!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Justification 5&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Give me a detailed and plain description of it and you will, I am absolutely sure, be the winner!&quot; The description I gave is indeed detailed enough, and plain enough (pun intented I guess).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Justification 6&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the poem, the first verse state that there are Doll, Owl, Dwaft and Bear. If you look carefully at your keyboard, Owl escape being above the keyboard. So when they look down on their cruel fate, Owl shall elude its threats.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That say, there are just some part of the hint and poem that I cannot wrap my head around with. &lt;/p&gt;&#xA;" OwnerDisplayName="user5508" LastActivityDate="2014-11-19T03:07:50.243" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5094" PostTypeId="2" ParentId="5067" CreationDate="2014-11-19T03:25:07.447" Score="1" Body="&lt;p&gt;Reminds me of school..:P&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This reminds me of binary digits , combination 3 0's and 3 1's ..... So, in that case from 000 to 111 is valid for 3 binary digits . Hence (111)2 = (7)10. So 7 slices &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4850" LastActivityDate="2014-11-19T03:25:07.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5096" PostTypeId="2" ParentId="4976" CreationDate="2014-11-19T03:53:42.580" Score="2" Body="&lt;p&gt;Just a guess...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The knight is &lt;strong&gt;oxygen&lt;/strong&gt; and the girl is &lt;strong&gt;hydrogen&lt;/strong&gt;. The children would then be &lt;strong&gt;water&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Father sits on his horse in his shining armor, lonely. His &lt;strong&gt;soft&lt;/strong&gt; heart led him to take in a young and sickly girl, her jaundice a blemish on her wispy, &lt;strong&gt;light&lt;/strong&gt; face.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What's softer than Oxygen? Hydrogen is the lightest element of the periodic table.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Though father has said he can't be alone, it's an exaggeration.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oxygen can be alone, or it can combine with Hydrogen to form water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Out of their love, we are born beyond count. Across the earth we spread, sultrier than our mother and gleaming brighter than father.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water is by itself beyond count, and spreads when lakes pour into rivers, come onto the shore, etc - it's everywhere on earth. Water can be hot, which might explain &quot;sultrier&quot;. It can also gleam when the sun reflects off of it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;We are gracious hosts in small numbers and merciless in our horde&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water is fine in small numbers (i.e. a glass of water or a bath) but dangerous in high numbers (think hurricanes, floods, etc)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The last two lines regarding fire/ice and the &quot;god of death&quot; just don't fit in, unless I'm missing something. Obviously water makes fire flee, but what makes ice flee? And I'm very interested to learn what the god of death is. Great riddle. This is my new favorite site. :)&lt;/p&gt;&#xA;" OwnerUserId="5713" LastEditorUserId="5713" LastEditDate="2014-11-19T04:01:42.710" LastActivityDate="2014-11-19T04:01:42.710" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5097" PostTypeId="2" ParentId="5019" CreationDate="2014-11-19T04:21:22.797" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; multiplication&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am one of many siblings in a great family. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; mathematical operations&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are the most perfect of our race,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; multiplication, division, addition, subtraction are the most quintessential mathematical operations&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because each of us has a clear worst enemy. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; multiplication vs. division, addition vs. subtraction&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My foe - I know him well - is much like me, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; multiplication is very related to division&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But we continually fight each other; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; during algebraic operations, division and multiplication are constantly used to manipulate the equation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We never hide our hatred for each other.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; they are open and in a sense equal to each other&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wave at my eldest brother and he becomes me. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; multiplication is just repeated addition&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When my dear cousin finds the end of her life,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; when you subtract (the dear cousin) a negative&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;She too can turn into me, in her gentle way. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; addition is performed and can be just as easily turned into multiplication&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My younger brother thinks too highly of himself,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; division is multiplications enemy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So smooth and perfect that he'll always outstrip me. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; division undoes multiplication&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Often he even steals my words, confusing all concerned.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; when a coefficient is attached to a variable, division will take the coefficient away from the variable, a letter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;i feel like i am either pretty close or very off point&lt;/p&gt;&#xA;" OwnerUserId="5717" LastEditorUserId="5078" LastEditDate="2014-11-19T08:32:43.257" LastActivityDate="2014-11-19T08:32:43.257" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5098" PostTypeId="2" ParentId="5090" CreationDate="2014-11-19T04:25:12.500" Score="7" Body="&lt;h3&gt;Three Ideas&lt;/h3&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;It seems a bit unimaginative, but why not eliminate the paint blotches and replace them with &quot;glow&quot; tiles that change state from glowing to non-glowing and back (or perhaps switch to glowing and remain glowing) when the puck passes over them? Divide the puzzle into a series of &quot;rooms&quot;, with one letter per room or a few letters per room, and engineer bottlenecks so that while it's a minor nightmare to get the puck out of one room into the next one, any solution that does get you out has automatically lit up all the right tiles by running over them the right number of times.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As a second thought, you could take advantage of having multiple pushable blocks that leave colour trails behind them. And while it might seem obvious what direction a block is supposed to be pushed in, such pushes obviously don't always commute. The real challenge then comes in figuring out the order and direction of block pushes to complete the letter in a room and get the puck through to the next room. Non-commutivity obviously isn't an exploitable tool if you're only ever moving one object around.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;Under the hood&quot;, puzzles of this type are just directed graphs, or (my preference of abstraction) automata. You have a set of states (position of the puck, position of any other movable objects), and a set of edges or transitions between them. You also have the states of any tiles that get painted, light up,  etc.  The more complex the topology of states and transitions in the automaton (vertexes, edges in the graph), the more complex the puzzle will be. Ensuring tiles are properly lit up, etc. is accomplished by imposing a guaranteed set of states that must be visited (and potentially, a guaranteed order of visitation) leading up to a new &quot;room&quot; or the end of the puzzle.&lt;br&gt;&lt;br&gt;There are some wonderful tools out there for doing this using automata and formal languages, but you can also do it by hand. My recommendation would be to start with a state diagram with specific &quot;rooms&quot; as described above. Ignore any correspondence between states and positions for now. For each room, start adding states and transitions to make a room as topologically complex as desired. Add in loops and one-ways and blocking states (i.e. dead ends) and whatever else your heart desires.&lt;br&gt;&lt;br&gt;With this done, now choose a small, specific set of states that the system &quot;should&quot; pass through in order to reach the &quot;end of room&quot; state. Again, don't worry about the correspondence of these states with the physical reality of the ice puzzle for now. If you find that there are other ways of getting to the end of room state that don't require you to pass through the necessary states (possibly in sequence), you then either have the option of editing the automaton to eliminate erroneous solutions, changing the correct solution, accepting multiple solutions, or, if you're feeling bold, syncing the automaton with a higher order automaton that dictates a broader order of operations. For this last option, think of a dungeon in a Zelda game. You have to run through a maze (a low-level automaton) to get the dungeon item, then run through the same maze again to get the master key using the dungeon item, then run through the same @#$% maze a third time to get to the boss using the big key. The high-level automaton in this case is the enforced sequence &quot;enter &amp;rarr; item &amp;rarr; big key &amp;rarr; boss&quot;. Your high level automaton doesn't necessarily have to be this explicit. It could be based on pushing blocks, running over switches, basically &lt;em&gt;anything&lt;/em&gt; that causes a state change.&lt;br&gt;&lt;br&gt;However complex you ultimately wind up making your room's automaton, at the end you have a certain guaranteed complexity, and an assurance that a specific set of states must be visited (possibly in order) to reach the end-of-room state. Now comes the part where you turn your automaton into an ice rink. You do this by laying out rocks in ways that limit the puck's movement to paths that correspond to state transitions in the automaton.&lt;br&gt;&lt;br&gt;I admit this second phase can be challenging, especially since the layout phase would have to be built around your &quot;special sequence&quot; of states corresponding to very specific physical transitions (to paint out a letter, light up a letter, etc.) but you have three saving graces:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You can always add as many intermediate conveyor states as needed to connect two states. Just make sure each conveyor state leading to state A can only exit to A (or another conveyor leading to A).&lt;/li&gt;&#xA;&lt;li&gt;You can always place rocks out in the boonies without affecting any of the intermediate tiles in an ice puzzle.&lt;/li&gt;&#xA;&lt;li&gt;The sky's the limit on how puck movement can potentially affect the state of the rink tiles. Not only do you have your original idea and the few alternatives I've suggested, there's nothing stopping you from making a puck that passes over tile X paint all the tiles in X's row up until it hits rocks, or paint X and its 4-connected neighbours, or do &lt;em&gt;whatever&lt;/em&gt;, so long as the user can establish a consistent pattern between action and consequence.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Both of these processes (automaton design, rock layout) can be automated with a good deal of work, but they can just as easily be done by hand. My preference is always to do a smaller scale example by hand first, get a feel for how easy the process is when done manually and how painful the various steps would be if scaled up, and then assess whether programming an automated solution is worth the effort.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2014-11-19T04:25:12.500" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5099" PostTypeId="1" AcceptedAnswerId="5104" CreationDate="2014-11-19T04:36:43.183" Score="18" ViewCount="880" Body="&lt;p&gt;The answer is a six-letter word. You're strongly encouraged to discuss and collaborate on the puzzle in &lt;a href=&quot;http://chat.stackexchange.com/rooms/18790/room-to-collaborate-on-the-puzzle-split&quot;&gt;this chat room&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've had this puzzle playtested before, but I still welcome feedback and suggestions.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1) Diamond, to carbon  &#xA;2) Deflated (#7)  &#xA;3) That in “this or that” (#9)  &#xA;4) Eureka producer (#6)  &#xA;5) Elevated beard (#8)  &#xA;6) Old-school marksmen (#2)  &#xA;7) Take to court (#4)  &#xA;8) Like flashlight bulb (#7)  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h1&gt;=&lt;/h1&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;70’s fever (#3 of 5)  &#xA;Apportion (#4 of 5)  &#xA;Have to (#4 of 4)  &#xA;Not poetry (#2 of 5)  &#xA;Piece of curve (#1 of 3)  &#xA;Pizarro victim (5)  &#xA;Sing through nose (#2 of 3)  &#xA;Tweak (#5 of 5)  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h1&gt;+&lt;/h1&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;American before America (#2 of 6)  &#xA;Back or head problem (#2 of 4)  &#xA;Climber’s way back (#1 of 7)  &#xA;Hitchcock thriller (#3 of 4)  &#xA;Like kittens (#4 of 4)  &#xA;Lost some fluid (4)  &#xA;Quite (#3 of 4)  &#xA;That woman’s (#1 of 4)  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2014-12-28T12:02:42.553" Title="Split: A puzzle-hunt style word puzzle" Tags="|word|crossword-clues|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5100" PostTypeId="1" AcceptedAnswerId="5101" CreationDate="2014-11-19T05:34:18.703" Score="20" ViewCount="4451" Body="&lt;p&gt;Create the pattern shown in the picture below using the set of standard tetris blocks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nxYr0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This is a rectangular arrangement of 6×5 squares where the first and third squares have been removed from the second row.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each tetris block is available only once, blocks can be rotated but not flipped or deformed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RaboC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you find the solution? Please put your answer in &lt;a href=&quot;https://puzzling.stackexchange.com/editing-help#spoilers&quot;&gt;spoiler tags&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4394" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2015-02-15T17:57:01.220" Title="Mosaic with tetris blocks" Tags="|geometry|pattern|tiling|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5101" PostTypeId="2" ParentId="5100" CreationDate="2014-11-19T05:57:32.733" Score="19" Body="&lt;p&gt;Solution!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/TdIjz.png&quot; alt=&quot;solution&quot;&gt;&#xA; I found the trick was to start with filling the upper left. There are only four pieces that can go there. Then I had to keep in mind that the two S blocks and the T block couldn't fill a nice rectangle, so I had to leave some holes open to fill with them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5612" LastEditorUserId="5612" LastEditDate="2014-11-20T00:50:16.540" LastActivityDate="2014-11-20T00:50:16.540" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5102" PostTypeId="2" ParentId="5100" CreationDate="2014-11-19T06:05:09.407" Score="13" Body="&lt;p&gt;I have a satisfying arrangement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/tXXtn.png&quot; alt=&quot;Solution&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I used &lt;a href=&quot;http://www.gfredericks.com/sandbox/polyominoes&quot; rel=&quot;nofollow noreferrer&quot;&gt;this tool&lt;/a&gt;. I just let it run a few times until it had only used the desired pieces. The shortcoming is that one cannot set how many instances of each piece are allowed.&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-11-19T22:20:50.320" LastActivityDate="2014-11-19T22:20:50.320" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5104" PostTypeId="2" ParentId="5099" CreationDate="2014-11-19T06:13:22.957" Score="14" Body="&lt;p&gt;That was a very, very nice puzzle! Solved this with a lot of help from @grc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a step-by-step walkthrough of how we solved it.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Working out some clues&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;We figured that &lt;code&gt;(#X of Y)&lt;/code&gt; probably meant that the clue answer had Y letters, and that we had to take the &lt;code&gt;X&lt;/code&gt;-th letter later. Why some clues didn't have a &lt;code&gt;#X&lt;/code&gt; we weren't sure, but the first clues we worked out were the following:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;70's fever               DISCO&#xA;Apportion&#xA;Have to                  MUST&#xA;Not poetry               PROSE&#xA;Piece of curve           ARC&#xA;Pizarro victim&#xA;Sing through nose        HUM&#xA;Tweak&amp;lt;br&amp;gt;&#xA;American before America&#xA;Back or head problem     ACHE&#xA;Climber's way back       DESCENT&#xA;Hitchcock thriller       ROPE&#xA;Like kittens&#xA;Lost some fluid&#xA;Quite&#xA;That woman's             HERS&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h2&gt;Figuring out the mechanic&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;From the clues we had, we worked out that &lt;code&gt;MUST + ACHE = MUSTACHE&lt;/code&gt;, which is indeed an &quot;elevated beard&quot;. This makes sense, given the &lt;code&gt;+&lt;/code&gt; and &lt;code&gt;=&lt;/code&gt; in the puzzle, and made it easier to work out the remaining clues.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Filling in the rest&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The remaining clues could then be filled in like so:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Diamond, to carbon       ALLOT + ROPE&#xA;Deflated                 HUM   + BLED&#xA;That in &quot;this or that&quot;   ALTER + NATIVE&#xA;Eureka producer          DISCO + VERY&#xA;Elevated beard           MUST  + ACHE&#xA;Old-school marksmen      ARC   + HERS&#xA;Take to court            PROSE + CUTE&#xA;Like flashlight bulb     INCAN + DESCENT&#xA;70's fever               DISCO&#xA;Apportion                ALLOT&#xA;Have to                  MUST&#xA;Not poetry               PROSE&#xA;Piece of curve           ARC&#xA;Pizarro victim           INCAN&#xA;Sing through nose        HUM&#xA;Tweak                    ALTER&#xA;American before America  NATIVE&#xA;Back or head problem     ACHE&#xA;Climber's way back       DESCENT&#xA;Hitchcock thriller       ROPE&#xA;Like kittens             CUTE&#xA;Lost some fluid          BLED&#xA;Quite                    VERY&#xA;That woman's             HERS&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h2&gt;Taking letters&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If we take the letters of the first set of clues as indicated (e.g. &lt;code&gt;Deflated (#7)&lt;/code&gt; means to take the 7th letter of &lt;code&gt;HUMBLED&lt;/code&gt;, or &lt;code&gt;D&lt;/code&gt;), then we get &lt;code&gt;DIVERSE&lt;/code&gt;. Note that the first clue, &lt;code&gt;Diamond, to carbon&lt;/code&gt;, contributes no letters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second and third sets, we do the same but &lt;strong&gt;in the order of the first set of clues&lt;/strong&gt;, like so:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ALLOTROPE          ALLOT    O    ROPE    P&#xA;HUMBLED       D    HUM      U    BLED&#xA;ALTERNATIVE   I    ALTER    R    NATIVE  A&#xA;DISCOVERY     V    DISCO    S    VERY    R&#xA;MUSTACHE      E    MUST     T    ACHE    C&#xA;ARCHERS       R    ARC      A    HERS    H&#xA;PROSECUTE     S    PROSE    R    CUTE    E&#xA;INCANDESCENT  E    INCAN         DESCENT D&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As you can see, we get &lt;code&gt;OURSTAR&lt;/code&gt; and &lt;code&gt;PARCHED&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The answer&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;So what do we do with &lt;code&gt;DIVERSE&lt;/code&gt;, &lt;code&gt;OURSTAR&lt;/code&gt; and &lt;code&gt;PARCHED&lt;/code&gt;? We reapply the mechanic!&lt;br&gt;&#xA;Our star is &lt;code&gt;SUN&lt;/code&gt; and &quot;parched&quot; is &lt;code&gt;DRY&lt;/code&gt;, so the answer, a word meaning &quot;diverse&quot; is...&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;SUN + DRY = SUNDRY&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="445" LastEditorUserId="445" LastEditDate="2014-12-28T12:02:42.553" LastActivityDate="2014-12-28T12:02:42.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5105" PostTypeId="1" AcceptedAnswerId="5107" CreationDate="2014-11-19T06:23:31.003" Score="5" ViewCount="533" Body="&lt;p&gt;Suppose a general is in charge of a mining site with 5 workers and he has a total of 50 units of gold.  Each worker is able to mine for him 40 units of gold per minute.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general is able to employ new workers for a total one of cost of 50 gold units (creating a worker takes 17 seconds and each mining site can only construct one worker at a time), but only 24 workers can exist at a mining site. The general can thus decide to construct a new mining site, costing 400 gold units (taking 2 minutes to construct), and with 2 mining sites, he can then construct 2 workers simultaneously, one from each mining site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general decides he wants to have a total of 3 mining sites.  The general however is unsure of whether he should build any workers to increase income or if he should just wait until he has enough gold to purchase his 2nd or 3rd site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the strategy the general should follow in order to minimize the time taken to achieve the three mining bases he desires?  How many villagers should he build from each base and when?&lt;/p&gt;&#xA;" OwnerUserId="97" LastEditorUserId="4551" LastEditDate="2014-11-19T07:39:37.073" LastActivityDate="2014-11-19T07:43:31.270" Title="Optimal Building Strategy Puzzle" Tags="|strategy|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5107" PostTypeId="2" ParentId="5105" CreationDate="2014-11-19T07:29:53.073" Score="7" Body="&lt;p&gt;Should the general build additional bases?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No. A single base is the way to go. The general should just save up to 800 gold units and build both bases at once.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How many workers should the general build?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Five workers, to get a total of 10.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When will the general have completed three bases?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 295.5 seconds = 4 minutes and 55.5 seconds. This is two minutes after the first mining site is built at 175.5 seconds. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What build order achieves this time?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Build five workers, starting immediately. One will see that there's enough gold to build a worker as soon as the previous one finishes. So, the limitation is the 17-second build time. These workers are then finished at &lt;code&gt;17, 34, 51, 68, and 85&lt;/code&gt; seconds.&#xA;&lt;br&gt;&#xA; I used a &lt;a href=&quot;https://docs.google.com/spreadsheets/d/1LHJR-HATPfMkeY7tSXaApabFhyIXTCk-k6xqQqeTHAE/edit#gid=689812735&quot; rel=&quot;nofollow&quot;&gt;spreadsheet&lt;/a&gt; to compute how much gold one has at each of those times by updating the previous gold with the number of workers and mining rate. &#xA;&lt;code&gt;&lt;br&gt;&#xA;              Time   Workers   Minerals   Time to 800    Total time&#xA;              0       5        50         225            345&#xA;Worker built  17      6        56.6667    202.8333       322.8333&#xA;Worker built  34      7        74.6667    189.4285714    309.4285&#xA;Worker built  51      8        104        181.5          301.5&#xA;Worker built  68      9        144.6667   177.2222       297.2222&#xA;Worker built  85     10        196.6667   175.5          295.5&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;  After 5 workers, just wait until you have 800 minerals to build both mining sites. The &quot;Time to 800&quot; column says the total time its will take to reach 800 gold if you build no further workers, and 5 additional workers optimizes that time. It allows the bases to be started at 175.5 sec and complete at 295.5 sec. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why produce that number of workers?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A worker pays for itself after 75 seconds of mining, and so after 92 seconds counting building time. Building a sixth worker would not pay for itself as there are only 90.5 seconds remaining at that point. Building fewer workers doesn't help either, as can be seen from the spreadsheet. At the time the last worker is built, there's more than 92 seconds left before building, so it's worth it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why produce that number of mining bases?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Any additional base will have no benefit to use. We are nowhere near the cap of 24 workers on a mining base. Such a base would finish after at least 120 seconds, at which point we've already found it's not worth it to build workers, so additional production won't help. Each base it only worth it to reach to end goal. Since building a base occupies the worker that's building it and slows down mining, it's best to wait until reaching 800 minerals and starting to build both bases at once.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="1614" LastEditDate="2014-11-19T07:43:31.270" LastActivityDate="2014-11-19T07:43:31.270" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5108" PostTypeId="2" ParentId="5089" CreationDate="2014-11-19T08:43:27.467" Score="2" Body="&lt;p&gt;It's probably wrong but, is Swannie Taco an anagram for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Isaac Newton&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What she wants&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An apple&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the girl is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; EVE (also a robot in Wall-E)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I found out that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Isaac Newton died on 31 March, 1727. Maybe 1927 is not accepted by Google+&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;EDIT (new guess):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What she wants&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Recognition for E=MC², you could say she's pissed at Einstein!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the girl is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Emilie du Chatelet, she translated the principia from Isaac Newton&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5254" LastEditorUserId="5254" LastEditDate="2014-11-19T11:39:27.913" LastActivityDate="2014-11-19T11:39:27.913" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="5112" PostTypeId="2" ParentId="5090" CreationDate="2014-11-19T09:41:21.470" Score="1" Body="&lt;p&gt;The simple solution to more freedom is just that, give them more freedom. Add intersections, and forks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wanna make it harder? That's possible too. Add special blocks if that give way when used # times. Make each paint color have a special effect when you cross them, like Blue turns you around, Yellow spins you 90° clockwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wanna make it more interesting?&#xA;Throw in some enemies that move back and forth, pacing, that way they have to time their movements when their making them so they don't die!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wanna give the player a better feel of their character? Make it a penguin that slides on it's belly :D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Don't forget to build up on difficulty, don't start out with something too hard or the player will get frustrated to quickly. Don't keep it too simple or they'll get bored. A good idea is to introduce new features of gameplay slowly, and each time give them a tutorial level, only then do you start merging features.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-11-19T09:57:11.520" LastActivityDate="2014-11-19T09:57:11.520" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5113" PostTypeId="2" ParentId="3239" CreationDate="2014-11-19T10:04:39.330" Score="5" Body="&lt;p&gt;This reduces to the Knights and knaves question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make an assumption about which of &quot;yok&quot; and &quot;pom&quot; translates to yes and no.  If your assumption is incorrect the effect will be that from your perspective the knight will behave as a knave and vica versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only refinement that you need to be careful about is that you need to ask about the word he would answer with.  Ie. you need to use formulations such as &quot;would you answer 'yok'&quot; rather than formulations such as &quot;would you agree&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1654" LastActivityDate="2014-11-19T10:04:39.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5114" PostTypeId="2" ParentId="5074" CreationDate="2014-11-19T10:26:26.293" Score="3" Body="&lt;p&gt;Maybe it is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;s&gt;Switzerland (CH)&lt;/s&gt; Congo (CD)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;ul&gt;&#xA; &lt;li&gt;If the army is the Terracotta Army in China we take the first letter &quot;C&quot; with the &#xA; ASCII code of 67. &lt;/li&gt;&#xA; &lt;li&gt;&lt;s&gt;Minus 7 we got 60.&lt;/s&gt; The ASCII-code (or 'line number') of &quot;7&quot; is 55. &lt;/li&gt;&#xA; &lt;li&gt;67 - 55 = 12&lt;/li&gt;&#xA; &lt;li&gt;The square root of &lt;s&gt;60&lt;/s&gt; 12 is something like &lt;s&gt;7.74596669241&lt;/s&gt; 3.46410161514, but let's round it to 3&lt;/li&gt;&#xA; &lt;li&gt;ASCII code 3 is the ETX(End of Text). &lt;/li&gt;&#xA; &lt;li&gt;SOH (Start of Header) is 1 and STX (Start of Text) is 2.&lt;/li&gt;&#xA; &lt;li&gt;If I look at the ASCII table and go 1 down and 2 right I find the letter &lt;s&gt;&quot;H&quot;&lt;/s&gt; &quot;D&quot;&lt;/li&gt;&#xA; &lt;li&gt;concatenate both letters it's CD which is the code for congo&lt;/li&gt;&#xA; &lt;/ul&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After reading the comments it should be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hawaii (HI)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is confimred by the OP: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;ul&gt;&#xA; &lt;li&gt;The army is the Terracotta Army in China so we take the first letter &quot;C&quot; with the &#xA; ASCII code of 67. &lt;/li&gt;&#xA; &lt;li&gt;Minus 3, because a &quot;7&quot; on a digital board has 3 lines &lt;/li&gt;&#xA; &lt;li&gt;The square root of 64 is 8&lt;/li&gt;&#xA; &lt;/ul&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the rest of the Explantation could be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;ul&gt;&#xA; &lt;li&gt;the 8th letter on a Keyboard would be the &quot;I&quot;&lt;/li&gt;&#xA; &lt;li&gt;SOH (Start of Header) is 1 and STX (Start of Text) is 2.&lt;/li&gt;&#xA; &lt;li&gt;If we look on the keyboard at the letter &quot;I&quot; and go 1 down and 2 left I find the letter &quot;H&quot;&lt;/li&gt;&#xA; &lt;li&gt;In the comments the OP said &quot;H&quot; woulde be the first letter of the concatenation so it's &quot;HI&quot;, which stands for the state Hawaii&lt;/li&gt;&#xA; &lt;/ul&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5636" LastEditorUserId="5636" LastEditDate="2014-11-24T10:00:34.273" LastActivityDate="2014-11-24T10:00:34.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5115" PostTypeId="2" ParentId="5089" CreationDate="2014-11-19T10:31:19.803" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Just a wild guess. Looking at her Google+ page i see she is born March 31, 1927. Looking at &lt;a href=&quot;http://en.wikipedia.org/wiki/March_1927#March_31.2C_1927_.28Thursday.29&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/March_1927#March_31.2C_1927_.28Thursday.29&lt;/a&gt; I see that on that date also the actor was born who is the voice of Knightrider's KITT, an artificially intelligent electronic computer module in the body of a highly advanced, very mobile, robotic automobile. Reading in the &lt;strong&gt;Post&lt;/strong&gt; that she refers to us as mortals I can imagine that she is immortal and that like KITT she is some sort of intelligent robot.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2014-11-19T10:31:19.803" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5116" PostTypeId="2" ParentId="5019" CreationDate="2014-11-19T10:43:29.013" Score="6" Body="&lt;p&gt;I am one of many siblings in a great family.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The narrator is x -&gt; e ^ x (e is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant&quot; rel=&quot;nofollow&quot;&gt;Euler–Mascheroni constant&lt;/a&gt;), and the family is the set of all &lt;a href=&quot;https://en.wikipedia.org/wiki/Exponential_function&quot; rel=&quot;nofollow&quot;&gt;exponential functions&lt;/a&gt;, x -&gt; a ^ x, a is real.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are the most perfect of our race,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Race means the set of all functions. Perfect because of the nice properties about derivatives also being exponential functions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because each of us has a clear worst enemy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Worst enemy means inverse, in this case (natural) &lt;a href=&quot;https://en.wikipedia.org/wiki/Natural_logarithm&quot; rel=&quot;nofollow&quot;&gt;logarithic function&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My foe - I know him well - is much like me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; x -&gt; log(x) has many similar properties to the exponential function.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But we continually fight each other;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Both the exponential and logarithm functions are continuous.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We never hide our hatred for each other.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;???&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wave at my eldest brother and he becomes me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think the eldest brother is x -&gt; a ^ x where a &amp;lt; e. There is a transformation from the eldest brother function to the narrator function.  The word wave is a reference to the fact that the exponential function can be constructed from sine and cosine &quot;wave&quot; functions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When my dear cousin finds the end of her life,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think the cousin is x -&gt; e ^ -x.  As x increases towards positive infinity, this function decreases towards zero.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;She too can turn into me, in her gentle way.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is a transformation from one function to the other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My younger brother thinks too highly of himself,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; By the same logic as the younger brother, this must be x -&gt; a ^ x where a &gt; e.  (Probably a = 10; see below.)  This function will return values greater than the narrator function for all x &gt; 0.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So smooth and perfect that he'll always outstrip me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The exponential functions are infinitely differentiable (smooth).  Outstrip me means the same as &quot;thinks too highly of himself&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Often he even steals my words, confusing all concerned.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Exponential function&quot; can refer specifically to x -&gt; e ^ x, or to the more general family of functions.  In particular, x -&gt; 10 ^ x is often confusingly referred to as the exponential function.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5735" LastEditorUserId="5735" LastEditDate="2014-11-19T10:57:41.853" LastActivityDate="2014-11-19T10:57:41.853" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5117" PostTypeId="1" AcceptedAnswerId="28611" CreationDate="2014-11-19T10:54:27.060" Score="9" ViewCount="565" Body="&lt;p&gt;The general from &lt;a href=&quot;https://puzzling.stackexchange.com/questions/5105/optimal-building-strategy-puzzle&quot;&gt;Optimal Building Strategy Puzzle&lt;/a&gt; is in a conflict with another general. This general is from another faction with 3 major differences in his technology:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Instead of the worker having to be present during the building process, the worker instead just instantly places a marker on the ground. After 100 seconds, a new base has landed from an overhead satellite. The worker is not needed for the further construction of the building after the base is dropped and can continue with mining. The cost is also 400 units of gold.&lt;/li&gt;&#xA;&lt;li&gt;The mining base can overcharge their production of (mechanical) workers. To do this, they require 25% of their electricity upfront. electricity recharges at 0.5625% per second. They can do this as often as they like. Overcharging production lasts 20 seconds at a time and increases production speed by 50% (without affecting cost), but cannot be applied multiple times simultaneously on the same base. Electricity is not shared between bases: each base has their own supply. however, they can choose to overcharge another base instead of their own base.&lt;/li&gt;&#xA;&lt;li&gt;A worker requires processing overhead from a mainframe. A default mining base provides support for 11 workers. A worker can call in additional extension to this support for 100 units gold in the same way as a mining base, with the extension arriving 25 seconds after landing. One extension supports an additional 8 workers.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Just like his adversary, this general wants to have 3 mining bases active as quickly as possible. What is the fastest way for this general to reach this?&lt;/p&gt;&#xA;" OwnerUserId="2778" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2016-04-09T11:38:32.473" Title="Optimal building strategy puzzle with airdropped mining bases" Tags="|strategy|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5119" PostTypeId="2" ParentId="4668" CreationDate="2014-11-19T11:06:02.933" Score="1" Body="&lt;p&gt;&lt;strong&gt;Old answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You have got the following questions and answers:&#xA;&#xA;Guard: 1 Guest: 13 &lt;br&gt;&#xA;Guard: 3 Guest: 6 &lt;br&gt;&#xA;Guard: 9 Guest: 12 &lt;br&gt;&#xA;Guard: 31 Guest: 19 &lt;br&gt;&#xA;Guard: 90 Guest: 24 &lt;br&gt;&#xA;Guard: 91 Guest: 25 &lt;br&gt;&#xA;Guard: 125 Guest: 19&lt;br&gt;&#xA;If you want to know the answer of the guard when he is saying: 31. You are combining &#xA;the 3 and the 1 question together and sum them up. So question 3 + 1 = 13 + 6 = 19. to &#xA;answer the question of 91 you combine 9 and 1 which are 12 + 13 = 25.&#xA;We need to answer the question &quot;12&quot; so we need the digits &quot;1&quot; and &quot;2&quot;. We know 1 which &#xA;is 13 so we only need 2. We dont know 2 but we can calculate it by looking at the &#xA;question 125. 125 contains the digits &quot;1&quot;, &quot;2&quot; and &quot;5&quot;. &#xA;We know 1 which is 13 and 19 - 13 = 6. I almost know the answer but need to think a &gt;!little bit more about it.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update!&lt;/strong&gt;&lt;br&gt;&lt;br&gt;&#xA;&lt;strong&gt;New answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Now Quynguyen gave us the hints I tried and combine the answers of the 2 other puzzles. I used a vigenere cipher and entered the folowing:&#xA;Plain: Quynguyen&#xA;Key: Computers&#xA;output was: nflxlatmu&#xA;some letters of the alphabet are reflecting with the numbers like f is the sixth letter so the answer was six. I think I'm close. I hope someone solves this because I want the answer to this puzzle!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5597" LastEditorUserId="5597" LastEditDate="2014-11-20T10:43:46.157" LastActivityDate="2014-11-20T10:43:46.157" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5120" PostTypeId="2" ParentId="5100" CreationDate="2014-11-19T11:44:54.757" Score="15" Body="&lt;p&gt;The &lt;a href=&quot;http://burrtools.sourceforge.net/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Burr Tools&lt;/a&gt; freeware tool tells us there are exactly &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3 solutions. The first two have already been posted, but the last one has not:&lt;br&gt;&#xA;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/voWxD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How to use Burr Tools to solve this problem yourself:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first tab is the Entities tab, where you define your shapes - both the puzzle pieces and the target shape. For each shape, click the New button, then paint it in the rectangular grid below. The slider next to the grid allows you to paint different layers in space. A 3d preview appears on the right. The Toggle 3D lets you exchange the preview with the paint grid. The sliders above the paint area configure the grid size. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By default the shapes can be rotated arbitrarily in 3D while solving. To prevent flipping but still allow rotation, you can use the color option: Add two more colors, representing the bottom and top layer of each shape. Make each shape two-layer, and paint each layer with one color. The solver will then ensure the piece colours match those of the target shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second tab is the Puzzles tab, where you define which tiles will be used to solve which target shape. Since there is only one puzzle to solve, click New once. Select the target shape, then click Set Result, then click All+1 to select all other pieces as those being fit into the target shape. The 3D preview will then show the target in big, the selected shape next to it, and the pieces to fit listed below. The status bar should read &quot;Problem P1 result can contain 56-56 voxels, pieces (N=7) contain 56 voxels&quot;. The color constraints are preconfigured so that each color only fits itself - except for the default color, which fits anywhere and accepts any piece color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third tab is the Solver tab. Set Drop to 1 (keep every solution) and set the Limit high enough if you want to keep every solution - otherwise it will still count them, but it won't show every one. Click Start to solve the puzzle entirely, or Prepare -&gt; Step to watch the solving process look for the solutions. The Placements button shows all possible placements for each piece that the Prepare step has found. It includes placements not found in any solution, but it excludes obviously wrong placements such as those that cut off a single voxel in a problem with only 5-voxel pieces. Finally the Assemblies value tells you how many solutions there are (the Solutions value only includes those that can be disassembled, which is 0 if disassembly is not performed) and the slider below lets you list through all of the solutions. Be sure to set the 3d mode (icons in the bottom right) to &quot;display normally with shape color&quot; if you have changed it previously.&lt;/p&gt;&#xA;" OwnerUserId="284" LastActivityDate="2014-11-19T11:44:54.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5121" PostTypeId="1" CreationDate="2014-11-19T11:46:32.570" Score="5" ViewCount="925" Body="&lt;p&gt;&lt;strong&gt;STORY:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;He awakes, covered in quite a thick layer of dust. Quite clearly suffering from some form of memory loss, this man begins to look around, eyes blurry, almost sticky in feeling. There's bodies around the town. The town is hot but cloudy, confusing him further. His last memory is that of America, talking to a friend called Neville. &amp;quot;Oh, you're nothing like the the Coroner I know.&amp;quot; Neville took that as a compliment, assuming it was because Neville was funny, but the man knew it was clearly due to Neville's American accent.&lt;/p&gt;&#xA;&lt;p&gt;Coming back to what seems to be the present for the moment, he feels the need to head inside a nearby building. &amp;quot;Mr Hyde&amp;quot; seems to be the only reference he can remember for your location, but that seems so unimportant for the moment, as he can barely lift his feet to walk. Where he stood now seemed a place of luxury. Where one comes to watch. There's a sign, but he can not make any sense of it, since part had been destroyed, leaving nothing but &amp;quot;Kino?to?en&amp;quot;. Quite prominent, he finds the stairs leading to the front, where all eyes would see, upon which there was a note. Upon further inspection of the note, apart from it being in a format alien to him, he could make out who wrote it. A certain Doctor Edward R.&lt;/p&gt;&#xA;&lt;p&gt;Becoming quite receptive of this almost post apocalyptic surroundings, he reaches for what he knows causes damage, and finds a weapon which needs a mechanic operating each time it's used. Inscribed on the side is L.E 303.&#xA;&lt;br&gt;A slight chuckle greets him at the inscription, as it seems it's made for him. He picks the weapon up, unaware of its weight. He lurches over and mutters with frustration, &amp;quot;This is beneath me, I never carry mine personally.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;Weapon in hand, he follows the path to a nearby exit. However, something catches his eye - a slight glisten. He turns to try and work out what his eyes are telling him. This weapon on the floor seems like the natives have made it, much like the one he holds. It is operated with a mechanic, holds 5 and seems to be the standard. Above the object, written in what is hoped to be non bodily fluid, is &amp;quot;penal was here&amp;quot;. Apart from its blatant apparent absence of actual information, it seems unimportant.&lt;/p&gt;&#xA;&lt;p&gt;In a timid manner he approaches the glowing armament and touches it. He felt catapulted. He had been shot, but he was the bullet and time was the entity landing somewhere unknown to him yet shouting &amp;quot;I'm Home!&amp;quot;. It was the weapon; it was the key.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;SIDE NOTE:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The puzzle is to work out the missing pieces of information using your detective skills, to deduce the answers using logic and maybe a hint of help from the puzzling community. Every clue is made ensuring you do not need to be an expert programmer, physicist or mathematician. If you believe you are on the right track confirm with Google. Its your friend! I don't want the puzzle to be limited to those with higher education, therefore all the information is based upon deducing its reason for being there rather than knowing its equation, genetic make up or what it looks like in Java script.&lt;/p&gt;&#xA;&lt;p&gt;If you get an answer from other people's post it would be kind to credit them in the answer so we know who's answer for which question is most popular or considered right.&lt;/p&gt;&#xA;&lt;p&gt;Let us rather build a community on puzzling in all its aspects rather than invoking a stigma on the new guys for poor quality. If we didn't have poor quality we wouldn't have good.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;QUESTIONS:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;What is the man's name?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What was his job at the time?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Where is his home?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Who are the Penal?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What were the 2 weapons names?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What year in time was he taken to?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What was the building he entered?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What time period was this from?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Even with the time difference, how would he know his home?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5135" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-07-12T18:02:31.560" Title="The man, the weapons, the time" Tags="|logical-deduction|word|story|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5122" PostTypeId="1" AcceptedAnswerId="5127" CreationDate="2014-11-19T11:48:14.607" Score="12" ViewCount="3337" Body="&lt;p&gt;One I made myself. Have fun!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You need me. You need me in many ways. Yet, I'm slowly losing my family.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You can count on me as you always did!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The sound of an explosion, transferred to the orange land.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I can stop the darkness!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If you want, I can even stop the bitter cold!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I bring you life as in 2 bound circle-like shapes!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Yet, I keep losing my family. We are torn apart by whom who need us most.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Who or what are we?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PLEASE use the spoiler tags!&#xA;Do it by putting &lt;strong&gt;&gt;!&lt;/strong&gt; in front of your sentences!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Every line is a hint. Obviously ignore my introduction and the end part...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="420" LastActivityDate="2015-10-23T21:12:52.747" Title="Who or what are we?" Tags="|riddle|" AnswerCount="11" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5123" PostTypeId="2" ParentId="5122" CreationDate="2014-11-19T12:03:55.430" Score="1" Body="&lt;p&gt;Maybe it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Molecules?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It does make sense in almost every sentence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need me. You need me in many ways. Yet, I'm slowly losing my family.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;You need molecules for everything.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can count on me as you always did!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;You always count on the molecules otherwise you couldn't even exist.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sound of an explosion, transferred to the orange land.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;The state change of solid to gas and from gas to plasma which is orange (I believe so).&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can stop the darkness!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Maybe has to do something with electricity.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you want, I can even stop the bitter cold!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;If water reaches 100 degrees it becomes hot and so do the molecules&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I bring you life as in 2 bound circle-like shapes!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Don't understand this one&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet, I keep losing my family. We are torn apart by whom who need us most.&lt;br&gt;&#xA;Who or what are we?&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Clueless too for now&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5597" LastEditorUserId="5373" LastEditDate="2015-03-12T23:56:30.597" LastActivityDate="2015-03-12T23:56:30.597" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5124" PostTypeId="1" AcceptedAnswerId="5273" CreationDate="2014-11-19T12:16:23.827" Score="6" ViewCount="1178" Body="&lt;p&gt;So here's a little puzzle for all of the IT crowd out there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It demonstrates the &lt;a href=&quot;http://en.wikipedia.org/wiki/Vigen%C3%A8re_cipher&quot; rel=&quot;nofollow&quot;&gt;Vigenere cipher&lt;/a&gt;, the unexpectedly strong 'chiffre indéchiffrable'. The Wikipedia description explains how you can use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Tabula_recta&quot; rel=&quot;nofollow&quot;&gt;tabula recta&lt;/a&gt; to decode it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Each row starts with a key letter. The remainder of the row holds the letters A to Z (in shifted order). Although there are 26 key rows shown, you will only use as many keys (different alphabets) as there are unique letters in the key string, here just 5 keys, {L, E, M, O, N}. For successive letters of the message, we are going to take successive letters of the key string, and encipher each message letter using its corresponding key row. Choose the next letter of the key, go along that row to find the column heading that matches the message character; the letter at the intersection of [key-row, msg-col] is the enciphered letter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you can't be bothered to do it by hand then there's &lt;a href=&quot;http://cryptii.com/vigenere/text&quot; rel=&quot;nofollow&quot;&gt;a handy converter here at cryptii.com&lt;/a&gt;, which can handle a bunch of common ciphers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your ciphertext is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;WEMGO ZJLD EGH XJCD ELW&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let the countdown begin!&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The solution is a short series of English dictionary words. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The key length is 9 characters.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;OK, well evidently one of the things to watch for when setting this type of puzzle is that if you don't specify enough details of the plaintext or the key then the plaintext can be anything which fits the word boundaries. ;) Thank you Sp3000 for pointing this out in such an entertaining way. &lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="4050" LastEditDate="2014-11-20T17:11:26.137" LastActivityDate="2014-11-20T19:27:47.877" Title="Have You Tried Turning It Off And On Again? (Vigenere cipher)" Tags="|cipher|" AnswerCount="2" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="5125" PostTypeId="1" AcceptedAnswerId="5132" CreationDate="2014-11-19T12:25:14.093" Score="5" ViewCount="1157" Body="&lt;p&gt;I managed to write a rhyming one this time!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I sound like a sieve, &lt;br&gt;&#xA;But knowledge I give. &lt;br&gt;&#xA;I change when to where; &lt;br&gt;&#xA;Just find the right square.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A cycle moves me to the right, &lt;br&gt;&#xA;But never entirely out of sight. &lt;br&gt;&#xA;Seven by four, or usually more, &lt;br&gt;&#xA;Betwixt the ceiling and the floor.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2014-11-19T13:24:44.213" Title="I change when to where - another riddle" Tags="|riddle|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5127" PostTypeId="2" ParentId="5122" CreationDate="2014-11-19T12:34:48.407" Score="8" Body="&lt;p&gt;&lt;strong&gt;My guess is&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Trees&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You need me. You need me in many ways. Yet, I'm slowly losing my family.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;We all need trees for oxygen, rain, food, shelter and all&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can count on me as you always did!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;We depended on trees and made many things using trees even from ancient time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sound of an explosion, transferred to the orange land.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The big bang that resulted in the formation of earth, which was an orange land because of the lack of water, plats and was red hot because of the heat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can stop the darkness!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;By lighting sticks, wood etc &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you want, I can even stop the bitter cold!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Making fire with wood&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I bring you life as in 2 bound circle-like shapes!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Producing oxygen (oxygen molecule consist of two circle shaped oxygen atoms)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet, I keep losing my family. We are torn apart by whom who need us most.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;We cut down trees for various needs and polluted the environment which too will cause the trees to get extinct soon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="2730" LastEditDate="2014-11-19T14:12:41.560" LastActivityDate="2014-11-19T14:12:41.560" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5128" PostTypeId="2" ParentId="5122" CreationDate="2014-11-19T12:35:19.300" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Maybe it is a MatchBox.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can count on me as you always did!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; to light a fire.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sound of an explosion, transferred to the orange land.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; this is how a matchstick lights up and fire color is orange.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can stop the darkness!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; creates light so stops darkness&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you want, I can even stop the bitter cold!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; fire is warm&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I bring you life as in 2 bound circle-like shapes!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; trying to figure this out. maybe it refers to the head portion of matchstick which is sphere like.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet, I keep losing my family. We are torn apart by whom who need us most.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; matchstick gets depleted as you use it up.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who or what are we?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; so a matchbox with matchsticks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1669" LastEditorUserId="1669" LastEditDate="2014-11-19T12:44:26.323" LastActivityDate="2014-11-19T12:44:26.323" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5129" PostTypeId="2" ParentId="5125" CreationDate="2014-11-19T12:46:00.203" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The blackboard&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I sound like a sieve, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For the sound of the chalk&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But knowledge I give. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm used for teaching&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I change when to where;&#xA;Just find the right square.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The tool to wipe&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A cycle moves me to the right, &#xA;But never entirely out of sight. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Less clear to me, but perhaps it refers to those that can be swapped&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Seven by four, or usually more, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Feet?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Betwixt the ceiling and the floor&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Usually hanging on the wall&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1531" LastActivityDate="2014-11-19T12:46:00.203" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5132" PostTypeId="2" ParentId="5125" CreationDate="2014-11-19T12:46:52.030" Score="6" Body="&lt;p&gt;Sounds like it could be a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Calendar&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I sound like a sieve&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Maybe a sieve is a grid&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But knowledge I give&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Calendar can be an external brain for keeping track of everything&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I change when to where&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As an appointment time approaches, it will tell you where you need to be&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A cycle moves me to the right&#xA;But never entirely out of sight&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A cycle of days is a month and as months increase, they normally show a smaller version of the prior and next month. As we imagine a year goes left to right, this may indicate direction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Seven by four or usually more&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 28 days in Feb, otherwise 29, 30 or 31&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Betwixt the ceiling and the floor&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Hangs on the wall.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the comments on where I was a little off - &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  A &quot;sieve&quot; is like a colander, which sounds like calendar. &quot;When to&#xA;  where&quot; is supposed to mean that 'which day?' when you're looking on a&#xA;  calendar becomes a spatial question rather than a temporal one. A&#xA;  &quot;cycle&quot; is supposed to mean a day; as each day passes (unless it's the&#xA; end of the week) you move one square to the right on the calendar. – &#xA; rand al'thor&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditDate="2014-11-19T13:24:44.213" LastActivityDate="2014-11-19T13:24:44.213" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5133" PostTypeId="2" ParentId="5125" CreationDate="2014-11-19T12:47:46.880" Score="0" Body="&lt;p&gt;I think you are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;A compass&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I sound like a sieve, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;used to distinguish directions (North, East, West, South).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But knowledge I give. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;give knowledge about locations and how to reach a place.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I change when to where; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;is used to find the answer for 'where?'. That is, to find locations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Just find the right square.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;To decide in which direction the squire map is to be held in an exploration or journey.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A cycle moves me to the right, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Circular rotation of the needle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But never entirely out of sight. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;will always be kept in the pocket or hand&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="2730" LastEditDate="2014-11-19T12:54:21.407" LastActivityDate="2014-11-19T12:54:21.407" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5134" PostTypeId="2" ParentId="5122" CreationDate="2014-11-19T12:50:30.493" Score="2" Body="&lt;p&gt;I interpret &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I bring you life as in 2 bound circle-like shapes!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;as&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DNA, chromosomes, or genes - the double helix. Since they are made of proteins, you could also try proteins or molecules here, except that the OP has said molecules is not the intended answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How to make it fit the other clues?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need me. You need me in many ways. Yet, I'm slowly losing my family.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sure, we need DNA to keep making new blood, bones, skin etc as long as we live. As your ancestors die, your genes or DNA lose some of their family.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can count on me as you always did!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Same as you need me, really&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sound of an explosion, transferred to the orange land.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I have nothing here. Perhaps &quot;the orange land&quot; is the Netherlands, who use orange as a symbol for sports teams etc, or Spain, where oranges grow. I can't see a way to have bang, boom, or other explosion sounds somehow &quot;transfer&quot; to another country so this clue mystifies me.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can stop the darkness!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some species have genes for bioluminescence&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you want, I can even stop the bitter cold!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; And some for antifreeze-like properties&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I bring you life as in 2 bound circle-like shapes!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This is really the crucial clue for me as I said above&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet, I keep losing my family. We are torn apart by whom who need us most.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first sentence is a repeat. The second might refer to the fact that each parent passes on only half their DNA to the offspring. It's &quot;torn apart&quot; to pass on half&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1509" LastActivityDate="2014-11-19T12:50:30.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5135" PostTypeId="1" AcceptedAnswerId="5216" CreationDate="2014-11-19T13:06:51.987" Score="9" ViewCount="5490" Body="&lt;p&gt;The sky is dark, filled with black clouds. The sun, high in the sky, still shining, trying to cut through the dark, thick, black clouds. The moon, slowly turning red, slowly taking over. You hear a howl, it seems to be coming from the thick forest that is not too far in the distance, holding on tight to her hand, trying to protect her at any costs. A werewolf jumped high off of a tree, the red moon shining behind the werewolf as it winds its hands back, ready to attack. As the beast falls towards her, an arrow appeared out of nowhere, lunging into the beasts heart. It falls to the ground, still squirming, trying to stop the pain. Slowly, the beasts stops moving, just laying there, now dead. A bunch of villagers come out of the forests asking if everything is ok, there was silence, the shock was just too great to give a response. &quot;Oh John!&quot; she said while still holding hands. She leaned in for a kiss, everything suddenly stopped. Words started rolling up really fast, faster than you can read them. The place just got brighter, silence than turned into many voices talking, laughing, and crying. &lt;strong&gt;(Choose One)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Passive:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;The movie wasn't terrible, but it was still pretty bad. The producers did the best they could, the acting was still ok&quot; you thought to yourself, as you got up from your seat and did a big stretch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Aggressive:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;That was one of the worst horror movies i've seen in a long time! totally not worth the money!&quot; you thought to yourself, as you got up from your seat and did a big stretch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You walked down the stairs, walking towards the exit. As you leave through the exit, a bright light just hit you, the front lobby of the movie theatre is now packed. Lobby filled with the young and the old, male and female, friends and family, all having fun and enjoying the day. You exit the theatre and start walking towards  your car, when all of a sudden you receive a call. &quot;RING RING RING RING RING!&quot; your cell phone goes, you pick it up, &quot;Hello?&quot; you said. &quot;Sorry to bother you, but we have a situation&quot; said one of the officers, you rush towards your car as the officer tells you the location. As you drive to the destination, you hear on the radio &quot;Cloudy with a chance of rain today, but as for the rest of the week, SUNNY, SUNNY, SUNNY! Back to you Steve. Thank you Margaret, we do have a collison on Main St. so there might be a jam up.&quot;, you realised that you are current on Main St. and will probably get to the collison and the traffic jam soon. &quot;Should i take a detour or stay on Main St.?&quot; you wondered to yourself. &lt;strong&gt;(Choose One)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Detour:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You decided to take a detour, trying to cut through the traffic. As you are taking the detour, you get stuck in traffic. &quot;There is currently a big jam up on Duke St. as well, looks like everybody is trying to take a detour around Main St.&quot; you hear on the radio. You give a big sigh and went along with the traffic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Main St:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You decided to stay on Main St, you didn't seem to have noticed a slow down. It is slower than usual, but the traffic is still moving. You didn't seem to mind because the traffic jam up wasn't that bad.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you finally arrive at the crime scene, everyone is already there like usual. There is a crowd of people trying to see what is going on, as some of the officers are trying to keep pedestrians and news reporters behind the yellow police line. You squeeze your way through the crowd of people, and cross the police line. As you got closer, you see a body lying on the floor, &quot;Glad you can make it, i hope the traffic wasn't too bad for you&quot; said one of the other detectives smiling. you didn't think much about it, because you knew that she always likes to joke around and get on your nerve. You walk up to the body, you noticed that the man has been stabbed twice in the back, blood everywhere on the ground, staining the walkway. The body reeked of a foul odor, like garbage or trash.The man has short black hair, wearing a black t-shirt and red pants. On the victim is a folded piece of paper, you unfold the piece of paper. On it is a picture of a mouse, &quot;This might be the killers calling card or something&quot; you thought to yourself. Near the body are a few yellow cones, you walk up to them. &lt;strong&gt;(Choose Two)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clue 1:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You see a pair of yellow running shoes. With further investigation, you found that the shoes are a size 15 for men. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clue 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You see a wallet, and inside the wallet you find a driver's license. The mans name is Terry G Delf, 5' 5&quot; in height, born on November 18. You take a closer look to see that the year of Terry's birthday has been scratched out. You pondered about this for a minute. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clue 3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You see a pair of dirty white gloves right beside yellow cone, the gloves seems to have been used.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As you started walking towards the last yellow cone, you hear &quot;HEY!&quot;, You turn around to see what the commotion is about. You see that one of the officers has been pushed down, and reporters rushing towards you to see the crime scene. The reporters quickly set up the cameras and mics to film as much as they can, the officers tried to stop them from disturbing the crime scene. All of a sudden, you hear someone scream! You see a bunch of mice running everywhere, most of them running towards the body. Citizens, reporters, officers, and detectives all started running around. Trying to either get away from the mice, or trying to stop others from disturbing the crime scene. You see mice running over the evidence, nibbling on it. The mice are probably attracted to the odor, you thought to yourself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scene is now a mess, with everyone still running, pushing, and screaming. You knew that there won't be any evidence left, other than what you have already saw. As you got out of the crowd, you noticed that on the ground is a cage, the door opened, nothing inside the cage. &quot;Seems like sabotage&quot; you muttered to yourself. You looked around to see if you can catch anyone that is suspicious, but no one in particular caught your eye. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You decide to head back to your office, ponder about the situation and start your report. Couple of hours has passed, you are in your office writing up the report when you hear a knock on your door. &quot;Come in&quot; you said, your assistant opens the door &quot;Your needed at the intersection of Key St. and Peele Road&quot;. As you arrive, the officers are setting up the police line and securing the middle of the road. You enter the crime scene to see that there is another dead body lying on the ground, and a detective setting up some yellow cones around. You walk up to the body, the man is wearing a orange shirt, and blue jeans. The man also has black hair, and brown eyes. Upon closer investigation, you notcied that there is a red mark around the mans neck and beside him is a folded piece of paper. You unfold the paper to find a picture of a dog on it, You then walk over to the yellow cones, trying to gather more information about the crime. &lt;strong&gt;(Choose Two)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clue 4:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You see a brown wallet, you open it up trying to find a piece of identification. You find the mans driver license, his name is George Y Logan, 6' 7&quot;, and date of birth is May 23. The year of which the man is born on is scratched out.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clue 5:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You see a simple dog leash, not one of those dog leashes where the owner can just click a button and it pulls the dog back in, it was just a simple dog leash.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clue 6:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You see a pair of dirty white gloves right beside the yellow cone, the gloves seems to have been used.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As you were examining the clue, you hear a loud honk coming from an incoming truck. The driver repeatedly honking as if he is trying to warn others of something, the truck does not seem to be slowing down. People a head of you started running away, trying to avoid the truck as it speeds pass everything and crushing whatever is in its path. You jumped and rolled out of the way, just as the truck was about to hit you. The truck swirves, and tips over, crashing into everything infront of it, including the remainding evidence and an electrical post. As you got up, you noticed that the as tank is leaking, the gasoline rapidly flowing out to the street. The gasoline, flowing closer and closer towards the live electrical cord from the post. One of the officers warns everyone to evacuate as the gasoline is about to touch the live cord, other officers helping the driver out of the truck as fast as possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You started running away from the truck, as you look back, you see the gasoline just about to touch the wire. There is an loud explosion, &quot;BOOM!&quot;. As you got up from the ground, you see that the truck, as well as, its surroundings engulfed in flames. You hear sirens coming from down the street, you knew it was the fire trucks and ambulances. You watched the fire burning, knowing that the evidence is now gone. &quot;Is it just me, or do i have the worst luck?&quot; you said to yourself. As the fire fighters try to put out the flames, you walk towards your car and headed home. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You arrive home, showered, made dinner, and is now relaxing and watching some tv. You are flipping through the guide menu, trying to search for what to watch next on the 9:30 PM time slot. &quot;RING RING RING&quot; goes your phone, you knew that it will not be good news. &quot;Hello?&quot; you said, &quot;There has been a murder at 1890 Fountain St, it's the house with a blue garage door, you can't miss it&quot; as the officer on the other line ends the call. You rush over to the location, when you arrive, cop cars are already surrounding the whole area of the house. You enter through the front door, you noticed that there are people on the ground, all bleeding heavily, blood splattered everywhere. two are sitting on the couch slouched over, one of them are laying on there stomach, and the other two were on the floor, backs againsts the floor. They were all young, you estimated them around the age of 19-23 years old, and have all been shot with a gun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of a sudden, one of the bodies moved, breathing heavily, gasping like she was trying to tell us something. &quot;WE HAVE A SURVIVOR! WE NEED MEDICAL ATTENTION IMMEDIATELY!&quot; yelled one of the officers, 4 paramedics rushed in, all surrounding the young female, trying to save her life. They wheeled her outside and onto the ambulance, and rushed to the hospital. You noticed that everyone had their own unique characteristics, some had black hair, some had blonde, and others had brunette/brown. They were of all different sorts of race, but you could not specifically tell what race they were.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You decide to look around the house, you noticed that they have decorated the place like they were having a party. There are drinks on the living room table, as well as snacks. The tv was on, it seems like some of them were playing with the gaming console. There was glow sticks on the ground, table, and couch. In the kitchen, there was more food and drinks, on one of the counters there were plates, cups, and utensils. Each plate and cup had names on it, probably to be able to tell whos plate or cup belonged to who. You read the names on each cup from left to right, Suzy F Guido, Hank F Tay, Joe L Kring, Katie H Yu, Vicki B Bee, and Ann E Mole. Each plate had some chips, dipping sauce, and a slice of pizza, while the cups had beverages in them, all but the last set. You went back to the living room to see if any of them had their wallets, but none of them had one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You went to check every other room, but none of them were disturbed. It seems like the party was only intended for the living room and the kitchen. It didn't seem like anything else was disturbed. After searching through the house, you left the crime scene and headed home. When you got home, you went straight to bed. You immediately fell asleep due to the exhausting day you just had. You get woken up by your alarm clock, you get up and give a big stretch. You yawn, &quot;That was a great sleep, i needed that&quot; you said to yourself. You started to get ready for your day, you brush your teeth, take a shower, got dressed, and made breakfast. As you head out the door, you realised that you are ahead of schedule, so you decide to take the nice scenic route to work today. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you were driving along the nice view of the river, you see that there are cop cars around this one area of the river, you decide to go check it out. Upon arrival you see many officers talking and pointing at something in the water. You ask one of the officers what is going on, &quot;There a body over there in the river, and it seems like blood everywhere surrounding the body&quot;. You focus on the body in the distance, you see that the person is not moving, the person is also wearing a blue shirt, white pants, and has blonde to white hair. On top of the person is a piece of paper, you recognize the paper and how it is folded. You knew that this was another murder by the same person, you noticed that there are 3 objects near the body, but the body and the objects are quickly sinking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You quickly jump into the river and swim as fast as you can towards the body, as you get closer, the body and objects are sinking deeper and deeper. You got there just in time to be able to grab the piece of paper and the body, one of the officers threw you a life jacket and you put it on the body. You look around to see that the objects are still sinking, you remembered that you were never too good at swimmming and the objects are almost too deep for you to reach. You take a deep breathe and started diving after the objects, you know you can't reach all of them so you decided to go after one of the object. &lt;strong&gt;(Choose One)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Object 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You grabbed one of the objects, and started swimming back to the surface, once you reach the surface you gasp for air. You swim back to shore, as the officers help pull you up. You look at the object you have grabbed, it is a wallet. Inside the wallet was a drivers license, The mans name is Doug R Peul, 5' 5&quot; in height, and date of birth is June 9. The year is scratched out.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Object 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You grabbed one of the objects, and started swimming back to the surface, once you reach the surface you gasp for air. You swim back to shore, as the officers help pull you up. You look at the object you have grabbed, it is a small blue hat. It is so small that it would not actually fit on anyones head.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Object 3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You grabbed one of the objects, and started swimming back to the surface, once you reach the surface you gasp for air. You swim back to shore, as the officers help pull you up. You look at the object you have grabbed, it is a red bow tie. There is nothing special about it, it is just a regular red bow tie.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You remembered that you grabbed the piece of paper as well. As you carefully opened the piece of paper, making sure it does not rip, you see there is a picture of a duck on it. You walk over to the body, you see there are two paramedics there beside the body. As you got closer, one of the paramedics said that he is dead and the cause of death is drowning. Still exhausted, you went back to your car and sat there for a little while to catch your breath. &quot;And i just took a shower too&quot; you muttered to yourself. Out of leads, You call the General Hospital to see if the survivor from the party is awake, maybe that person can give us a lead. &quot;Sorry, but the patient has not awoken yet and will need rest for another week or so&quot;. Just when things looked bad, you recieve a call from the police department, telling you that they have got some suspects. You quickly rush over to the police department.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;when you arrive, you see that there are 3 suspects lined up, ready for interrogation. The first suspect is male, 35 years of age, has black hair, beard, and green eyes. &quot;My name is Peter P Anne, i am an art teacher. I didn't commit any murders, i have been working&quot;. The second suspect is a female, 38 years old, has red hair, beautiful, and blue eyes. &quot;My name is Alice W Derland, i am a performer. I did not kill anyone, i work late nights, and get home in the morning. I worked during the days the murder happened&quot;. The last suspect is a male, 27 years old, black hair, and has a moustache. &quot;My name is Walt D Esnie, i didn't commit any crimes. I work in a studio as a helper, so i couldn't have killed anyone&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You ponder about all the information at the crime scenes, and the house party. You thought hard about it, then it hit you! You told the police to go arrest the murderers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who is the murderers?&lt;/p&gt;&#xA;" OwnerUserId="4645" LastEditorUserId="4645" LastEditDate="2016-02-12T13:03:48.830" LastActivityDate="2016-02-18T14:52:26.020" Title="Mysterious Murder Mystery 4" Tags="|riddle|word-problem|story|mystery|situation|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5136" PostTypeId="2" ParentId="5124" CreationDate="2014-11-19T13:07:27.270" Score="10" Body="&lt;p&gt;&lt;em&gt;This was posted before A.E. specified the key length, just to highlight what happens if you don't :P&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The near-pangram &quot;key&quot; is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; qtmowadvxsrcgupfjhznbly&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the &quot;message&quot;, revealing what we should do if the Earth ever meets its impending doom, is...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; GLASS WOOL CAN SAVE DAY&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="445" LastEditorUserId="445" LastEditDate="2014-11-19T14:01:52.243" LastActivityDate="2014-11-19T14:01:52.243" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5137" PostTypeId="2" ParentId="4976" CreationDate="2014-11-19T13:23:10.030" Score="12" Body="&lt;p&gt;They're&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sodium and Chlorine.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Father sits on his horse in his shining armor, lonely.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sodium is a shiny silver-colored metal.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;His soft heart led him to take in a young and sickly girl, her jaundice a blemish on her wispy, light face.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sodium is very soft (0.5 on the mohs hardness scale).  Chlorine is a yellowish  gas.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Though father has said he can't be alone, it's an exaggeration. He simply loves her, and it is a love eternal.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sodium can exist on its own, but forms a strong bond with chlorine.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Their kid is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Salt, sodim chloride.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Out of their love, we are born beyond count.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Salt can be found all over the planet.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Across the earth we spread, sultrier than our mother...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It forms a solution with water (making it sultry in the sense of &quot;humid&quot;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;... and gleaming brighter than father.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is translucent/white (arguably &quot;brighter&quot; than silver).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;We are gracious hosts in small numbers and merciless in our horde;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Salt is an important nutrient in small quantities, but can cause cardiovascular disease in large quantities.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;even fire and ice flee before us,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Salt is commonly used in class D fire extinguishers, and used to de-ice roadways.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;our defensive army keeping the devourers at bay&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A ring of salt around your garden will repel slugs, which would otherwise devour your plants.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Leave an offering of ivory upon the mountaintop and we will be your god of death.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I have no idea what this means.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1497" LastEditorUserId="1497" LastEditDate="2014-11-19T13:28:10.577" LastActivityDate="2014-11-19T13:28:10.577" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5138" PostTypeId="2" ParentId="4668" CreationDate="2014-11-19T13:44:47.630" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The binary conversion of the two number, the one asked by the guard and the one answered by the guests, has the same parity alas the same number of 1 (e.g. 3 is 11 and 6 is 110, 125 is 1111101 and 24 is 11000). My logic is really simple, yet I had to try since you referred to computers in the morse code cryptogram :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5139" LastEditorUserId="5139" LastEditDate="2014-11-19T13:58:30.850" LastActivityDate="2014-11-19T13:58:30.850" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5139" PostTypeId="2" ParentId="5135" CreationDate="2014-11-19T13:50:02.127" Score="2" Body="&lt;p&gt;The murderer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You edited it to imply multiple murderers, so I guess it's all three? Peter P Anne = Peter Pan, a Disney movie character; Alice W Derland = Alice In Wonderland, also a Disney movie character; and Walt D Esnie is Walt Disney himself.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I chose options: Aggressive, Main Street, Clues 1, 3, 4, 6, and Object 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clues 1 and 3, along with the mouse picture imply&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mickey Mouse.  He wears white gloves and large yellow shoes and is a mouse.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clues 4 and 6, along with the dog picture imply&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Goofy.  Most of the details from clue 4 don't mean much to me, but Goofy is tall, and he wears white gloves&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Object 1, along with the duck picture imply&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Donald Duck.  None of the details from the object really mean anything to me, but the duck picture along with the fact that the body was in the water, along with the previous clues would indicate Donald Duck.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Other things&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'd guess that the movie is Beauty and the Beast?  And the traffic problems on Main St is the usual Disney park parade.  And the party would be some kid's birthday party at the Disney park.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="239" LastEditorUserId="239" LastEditDate="2014-11-19T15:38:59.500" LastActivityDate="2014-11-19T15:38:59.500" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5140" PostTypeId="2" ParentId="5135" CreationDate="2014-11-19T13:56:28.017" Score="2" Body="&lt;p&gt;This one was slightly easier than the last two. The murderers are   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Walt D Esnie and Peter P Anne.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here are my explanations along with the clue choices I used to arrive at those conclusions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scene 1, with the mouse calling card:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dirty white gloves, and size 15 shoes that would be far too big for a short 5'5&quot; victim. Points towards &quot;Mickey Mouse&quot;. The victim is dressed like this as well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Scene 2, with the dog calling card:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Strangulation with a dog leash, and white gloves again. Points towards a &quot;Goofy&quot; motif, as the victim was dressed like this as well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Scene 3, the party:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is the nail in the coffin. Walt said he works at a studio. You notice that all the partygoers are all over the living room, while there are &lt;em&gt;full plates&lt;/em&gt; of snacks and such in the kitchen. Why would that be? If the party was in full swing, the food would be splattered all over the living room floor. The last beverage cup being empty would likely be from Walt rushing to leave the scene once he heard the authorities coming.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Scene 4, the body in the river:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The duck calling card and the red bow tie, pointing to a &quot;Donald&quot; motif. The victim, again, was dressed like Donald Duck.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clearly, the killers work for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a Disney Imagineering studio, and decided to go on a serial murdering spree. Walt staged the party and came up with the Disney motif idea, while Peter likely created the wardrobe and calling cards, something that would be easy for an art teacher.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2444" LastEditorUserId="2444" LastEditDate="2014-11-19T14:13:39.407" LastActivityDate="2014-11-19T14:13:39.407" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="5141" PostTypeId="1" AcceptedAnswerId="5383" CreationDate="2014-11-19T14:06:29.443" Score="6" ViewCount="2234" Body="&lt;p&gt;To answer this riddle, solve each of the four lines independently and put all the clues together to get a one-word final answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Nonsensical abstraction high up in the army; &lt;br&gt;&#xA;The woman with the wheel fires the cannon. &lt;br&gt;&#xA;She's very fast but only in reverse gear. &lt;br&gt;&#xA;Certainly a terrorist, but only just starting.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint for non-mathematicians:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First line: &lt;a href=&quot;http://en.wikipedia.org/wiki/Abstract_nonsense&quot; rel=&quot;nofollow&quot;&gt;Category Theory&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2014-11-22T13:50:22.837" LastActivityDate="2014-11-22T15:23:04.260" Title="riddle - pray you can find the word!" Tags="|riddle|word|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5142" PostTypeId="2" ParentId="5122" CreationDate="2014-11-19T14:13:00.313" Score="1" Body="&lt;p&gt;&lt;strong&gt;Another guess is&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;WEDDING RINGS&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You need me. You need me in many ways. Yet, I'm slowly losing my family.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For proposing, During engagement or Marriage&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can count on me as you always did!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;All most all value and trust the wedding ring and considers it as an agreement with the life partner.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sound of an explosion, transferred to the orange land.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Making of the wedding ring, by heating and melting to make the metal to look like orange lava&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can stop the darkness!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;With the love from the partner&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you want, I can even stop the bitter cold!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;With the caring from the partner&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I bring you life as in 2 bound circle-like shapes!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Shape of two wedding rings/making love which will result in the birth of a baby&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet, I keep losing my family. We are torn apart by whom who need us most.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Divorces. Recent increase in number of divorces.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastActivityDate="2014-11-19T14:13:00.313" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5143" PostTypeId="1" AcceptedAnswerId="5155" CreationDate="2014-11-19T14:15:34.277" Score="172" ViewCount="14601" Body="&lt;p&gt;Sine waves bound in rectangles shaped into prisms. Mass increases, position changes twice, and mass decreases. The prisms become rectangles and are either discarded or stored for later use. &lt;em&gt;If that sounds exciting to you, you may have an exciting future in our field!&lt;/em&gt; We need as many able-bodied candidates as possible to help with our project starting ASAP!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to lack of funding to actually pay a skilled team, we offer these fantastic incentives instead! How would you like some soggy old gramineaeze? Perhaps acidified coagulation and bio-viscera upon heavily triticumized water? Imagine what you could do with all of that! Sign up today to start this &lt;strong&gt;FANTASTIC ADVENTURE!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;YOUR FUTURE IS WAITING! CORPORATIONS DON'T UNDERSTAND THAT ALL CAPS IS BAD. JOIN US!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;*Company is not responsible for death or injury caused by crushing or falling.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are you being asked to do and how are they paying you?&lt;/p&gt;&#xA;" OwnerUserId="5626" LastEditorDisplayName="user20" LastEditDate="2015-08-05T17:48:13.967" LastActivityDate="2015-08-05T17:48:13.967" Title="Internship Available!* - Figure out what you're being asked to do before you sign up" Tags="|story|" AnswerCount="1" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="5145" PostTypeId="2" ParentId="5122" CreationDate="2014-11-19T14:35:54.210" Score="3" Body="&lt;p&gt;Who or what are we?&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; We are Brains! (I know this is far-fetched)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You need me. You need me in many ways. Yet, I'm slowly losing my family.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Off-course we need them and as you age the no of neurons decrease so its like losing family.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You can count on me as you always did!&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; We can take this literally as in counting in your head/brain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The sound of an explosion, transferred to the orange land.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; So there was this sound recording company called &lt;a href=&quot;http://en.wikipedia.org/wiki/Brain_Records&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brain_Records&lt;/a&gt; and the wiki page says that - &lt;em&gt;&amp;quot;the labels of Brain LPs changed from green to orange&amp;quot;&lt;/em&gt;. So the LP would be the metaphor for the orange land and it could have the recording of an explosion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I can stop the darkness!&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Brains developed bulbs!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If you want, I can even stop the bitter cold!&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;a href=&quot;http://www.sciencedaily.com/releases/2013/04/130408084858.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;mind over matter&lt;/a&gt; - Its said that by using brain power we can regulate body temperature.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I bring you life as in 2 bound circle-like shapes!&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Eyes are the extension of brains through which we see life and they could be viewed as two bound circular shapes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Yet, I keep losing my family. We are torn apart by whom who need us most.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The more we use it the more structural changes happens inside the brain over the course of our lifetime so in-effect we are kind of tearing it as we use it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I'm sure that this is not the intended answer but still want to share this.&lt;/p&gt;&#xA;" OwnerUserId="101" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-11-19T14:35:54.210" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5146" PostTypeId="2" ParentId="5122" CreationDate="2014-11-19T14:56:53.970" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hydrogen atoms on sun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You need me. You need me in many ways. Yet, I'm slowly losing my family.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are turning to Helium, slowly decreasing in size.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can count on me as you always did!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We count days and nights&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sound of an explosion, transferred to the orange land.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The fusions turning sun into an orange land&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can stop the darkness!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Daylight&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you want, I can even stop the bitter cold!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Warmth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I bring you life as in 2 bound circle-like shapes!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Maybe sun and the reflection of the light from moon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet, I keep losing my family. We are torn apart by whom who need us most.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Helium atoms need them to increase in size, but this destroys Hydrogen atoms&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5752" LastActivityDate="2014-11-19T14:56:53.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5147" PostTypeId="2" ParentId="5007" CreationDate="2014-11-19T14:58:59.553" Score="2" Body="&lt;p&gt;Ok, since no one has answered this I am giving it a shot with my theory.  If it is not correct, I think it should at least help with some of the interpretation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;She is ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;a beer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and she was killed by ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;getting spilled/dropped.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is my explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Roughly the first half is meant to throw you off and think the victim is a person.  From &quot;I went up to her ...&quot; and forward it is talking about the beer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;I went up to her to introduce myself expecting a warm bowl of duck soup, but she was colder than the ice around her neck.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Translation: I thought it would be easy to drink but it was freezing cold, colder than the ice forming around the neck of the bottle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;I got a bit steamed by the whole thing, so I went to the bar for some water. My kidneys were already failing and their brand of eel juice wasn't my poison.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Translation: I got upset about it so went to get some liquor to warm up, they didn't carry my drink of choice, however.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;When I turned back, she vanished before my eyes without a trace. Turns out she'd been chilled off.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Translation: I looked back to see it being dropped/spilled.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;All the bulls found for the meat wagon was her bag, a few fresh cookies sitting at the bottom. I was fortunately able to snag them; a man's gotta get paid.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Translation: All that was left of her 'body' was the container.  I'll admit, the &quot;cookies&quot; throw me off a bit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Translation: I was on a drinking binge when the beer was dropped. Use your stirring rod if you want to find the true perpetrator, alcohol, which caused the clumsiness that lead to a spill.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Translation: Listen, you whiskey drinkers, this is going to cause some discussion (chin music) when you lower your sunglasses/glasses to see what happened.  I.E. its not really a big deal since it will only cause some light discussion.  The beer was not chippy (strong) but the glass left behind is chippy (actually chipped/broken).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4191" LastActivityDate="2014-11-19T14:58:59.553" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5148" PostTypeId="2" ParentId="5122" CreationDate="2014-11-19T15:03:53.653" Score="2" Body="&lt;p&gt;This answer is very similar to the one above, with one difference.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I think it is a book of matches instead of a box of matches.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You need me. You need me in many ways. Yet, I'm slowly losing my family.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;In a book of matches, you remove a match from the book to use it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can count on me as you always did!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It is easy to count how many matches are left in the book.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sound of an explosion, transferred to the orange land.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The sound is the match igniting, and the flame is orange.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can stop the darkness!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Fire brings light. You can use a match to light a lantern.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you want, I can even stop the bitter cold!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Fire brings heat. You can use a match to light a bigger fire.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I bring you life as in 2 bound circle-like shapes!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A flame has 2 parts, a cooler part near the head of the match and a hotter one at the end of the flame. The two parts of the flame are roughly circle shaped.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet, I keep losing my family. We are torn apart by whom who need us most.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You have to tear a match out of the book. This is why I think it is a book as opposed to a box.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4685" LastActivityDate="2014-11-19T15:03:53.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5152" PostTypeId="2" ParentId="5019" CreationDate="2014-11-19T15:32:14.017" Score="5" Body="&lt;p&gt;Here is my guess.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Equivalence relations&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am one of many siblings in a great family.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Family is the universal set, and siblings probably means those in the same class.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are the most perfect of our race, because each of us has a clear worst enemy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Race refer to all relations. We are perfect because we are equivalence, and being equivalence, we have an inverse relation that are &quot;perfect&quot; too. That, my friend, is my worst enemy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My foe - I know him well - is much like me, but we continually fight each other; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He is indeed a perfect creature, just like me. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We never hide our hatred for each other.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Indeed, one can easily find the inverse relation when given an equivalence, nothing to hide.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wave at my eldest brother and he becomes me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yeah of course, we are equivalent, so if I am x, and my brother is y, x ~ y. Awesome.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When my dear cousin finds the end of her life, she too can turn into me, in her gentle way.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm not so sure, maybe some clousure under certain operator. Strictly speaking, I did not have any education in sets and relations yet, so my knowledge is limited here. Tried my best.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My younger brother thinks too highly of himself, so smooth and perfect that he'll always outstrip me. Often he even steals my words, confusing all concerned.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well, we can represent a certain partition with just a single element. Like maybe the n ≡ k modulo 3 operator, we can represents all multiple of 3, which includes 6, 9, ... 3k by [3]. It's really stealing my identity as the holy go-to person for all questions, 42.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pretty sure this isn't correct. I feel like there is too many holes that I only breifly dealt with. Just post this because it might be one interesting answer.&lt;/p&gt;&#xA;" OwnerDisplayName="user5508" LastActivityDate="2014-11-19T15:32:14.017" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5154" PostTypeId="2" ParentId="5045" CreationDate="2014-11-19T15:54:07.813" Score="2" Body="&lt;p&gt;The passage alludes to slavery in its various forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first two sentences recall colonial slavery, where inequity between men was cemented in law (&quot;even the five largest gems in the world were worth no more than one of the dark tribunals&quot;) and darker-skinned peoples were forced into agricultural toil (&quot;with whip and terror they forced the vast darkness into the sun&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Slavery still exists in many forms today (and in numbers vastly greater than at the time of European colonialism). Nearly 40 million people are classified as slaves, which doesn't include de facto types of slavery such as sweatshops and towns under military occupation. The type of slavery most referenced in the mass media is sex slavery and its many variants, and is also suggested by the use of the word &quot;lust&quot; following the &quot;greed&quot; of the colonial slavers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final three sentences in the first paragraph suggest a willful silence or neglect (&quot;they skulk and laugh at their own invisibility&quot;) concerning the modern practices of slavery. In some cases this arises due to a general lack of awareness of the problem, in some cases due to apathy, and in some cases due to the disconcerting similarities between modern slavery and colonial slavery (&quot;those who were defeated live on in the tears that nobody cares to wipe away&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &quot;dark tribunal&quot; is therefore a reference to a slave owner or slave trader, and the crime being ignored is the modern practice of slavery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Tribunal&quot; is a curious choice of word. It literally refers to a seat or court of justice, and when applied to men refers to a plurality of judges or court officers presiding over a matter of law. &quot;Dark&quot; in &quot;dark tribunal&quot; presumably refers to the darkness in the practice of slavery. There may be some deeper symbolism here, but I've rambled on long enough already. ;)&lt;/p&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2014-11-19T15:54:07.813" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5155" PostTypeId="2" ParentId="5143" CreationDate="2014-11-19T16:30:12.180" Score="131" Body="&lt;p&gt;I think that you're asking someone to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;help you move.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The prisms are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;cardboard boxes.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sine waves are the &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;cardboard's corrugation.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You move them twice: once from &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the house to the truck, then back into the new house.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you're done with &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the boxes, you break them down &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so they're rectangles again.  You can store them for later or throw them out.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will be paid for your help in &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;pizza and beer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Gramineaeze refers to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Poaceae (aka Gramineae), the family of plants that includes the cereals used in the brewing of beer. Through drenching and allowing the cereals to ferment, magical beer is born!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The other item is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pizza, a combination of:&#xA; | cheese, a milk product that is acidified (soured) and whose proteins are made to coagulate&#xA; | Sauce/meat/veg, biological life that has been eviscerated (a term whose dictionary definition means the removal of organs, but whose popular definition often refers to the ripping or cutting apart of a living entity in a very brutal or astonishing way).&#xA; | Dough, a combination of flour (ground grains of the triticum family) and water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2162" LastEditorUserId="5626" LastEditDate="2014-11-19T22:32:57.680" LastActivityDate="2014-11-19T22:32:57.680" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5156" PostTypeId="2" ParentId="5019" CreationDate="2014-11-19T16:39:37.330" Score="2" Body="&lt;p&gt;The Great Family is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; matrices&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Me and my siblings are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; rotation matrices&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are the most perfect of our race because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; we are all invertible&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Our clear worst enemy is our:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; inverse&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My foe is much like me, I know him well becuase:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the inverse of a rotation matrix is another rotation matrix&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We never hide our hatred for each other:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; as before the inverse of a rotation matrix is another rotation matrix and inverting a matrix does not lose or hide data or information&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My eldest brother is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The identity matrix.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wave at my eldest brother and he becomes me because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As the identity matrix he can become any matrix that is multiplied by him, and rotation matrices involve the wave functions sine and cosine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When my dear cousin finds the end of her life she too can turn into me:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm uncertain of this one the cousin might be a scale or shear matrix which if the numbers are just right can become a rotation matrix, there is word play though on 'turn into me' since I am in fact a rotation matrix&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My younger brother is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a matrix with 0'd elements possibly the null matrix&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He is smooth and perfect because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; all of his elements are smooth perfect 0's&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He will always outstrip me because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0's applied to a matrix remove data that can never be recovered.&#xA; this is the same as him stealing my words confusing all concerned.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not sure why he thinks so highly of himself though.&lt;/p&gt;&#xA;" OwnerUserId="5627" LastActivityDate="2014-11-19T16:39:37.330" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5157" PostTypeId="2" ParentId="5122" CreationDate="2014-11-19T16:44:59.757" Score="3" Body="&lt;p&gt;I think it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stars:&#xA; We've always needed them, looks orange during rise and set, stops darkness, warms things, brings us life, (not sure about the 2 circle like shapes, rain drops &amp;amp; planets maybe indirectly provided by the sun) losing my family thanks to gravity stretching space-time tearing them away from each other , but they need gravity to hold them together. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2014-11-19T16:44:59.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5159" PostTypeId="1" AcceptedAnswerId="5170" CreationDate="2014-11-19T16:46:25.633" Score="3" ViewCount="728" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I love to eat my younger brother.&lt;br&gt;&#xA;  Our Intelligience is never ending.&lt;br&gt;&#xA;  My younger brother is very famous.&lt;br&gt;&#xA;  We have days devoted to us.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A very simple riddle. &lt;em&gt;The first line is very important.&lt;/em&gt; (Not to say the others aren't.)&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-11-19T17:08:57.660" LastActivityDate="2014-11-20T16:46:08.563" Title="What am I? Can you figure it out?" Tags="|riddle|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5160" PostTypeId="2" ParentId="5019" CreationDate="2014-11-19T16:50:30.723" Score="2" Body="&lt;p&gt;Is it the &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; union function?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am one of many siblings in a great family.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Which would be set operation functions&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are the most perfect of our race, because each of us has a clear worst enemy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each union function has a corresponding intersection function&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My foe - I know him well - is much like me, but we continually fight each other;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The intersection function and the union function each undo each other&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We never hide our hatred for each other.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The intersection function function which corresponds to a union function would either be defined in terms of it, it would have the same parameters. Or you might be getting at the fact the symbols for these operations are vertically symmetrical opposites?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wave at my eldest brother and he becomes me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He would be the identity function - the &quot;oldest&quot; function in terms of a functional definition from first principles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When my dear cousin finds the end of her life, she too can turn into me, in her gentle way.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is the summation function, which can incrementally add set elements until reaching the same set membership as the union function would create.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My younger brother thinks too highly of himself, so smooth and perfect that he'll always outstrip me. Often he even steals my words, confusing all concerned.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The younger brother would be the recursion function, who would be able to defining himself in terms of the original union function. Recursion functions are just generally quite confusing!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5765" LastActivityDate="2014-11-19T16:50:30.723" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5161" PostTypeId="2" ParentId="5019" CreationDate="2014-11-19T16:55:41.117" Score="3" Body="&lt;p&gt;Ah, more abstract then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a family of functions&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The most perfect of our race:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; well formed functions, not piece-wise or broken&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Each of us has a clear worst enemy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The functions are invertible&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wave at my older brother and he becomes me:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am the integral of my older brother&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My dear cousin is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a limit, limits of a function taken to infinity can be the equivalent of derivatives and integrals&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My younger brother is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; my own integral, he is a higher order function, he is also a smoother function, he also includes all of my own terms and adds some more of his own.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5627" LastActivityDate="2014-11-19T16:55:41.117" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5162" PostTypeId="2" ParentId="5159" CreationDate="2014-11-19T16:58:29.700" Score="3" Body="&lt;p&gt;My guess is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;You are MOON and your younger brother is SUN&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I love to eat my younger brother&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Solar eclipse&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Our Intelligence is never ending&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Many secrets are still hidden that us (humans) still can't figure out&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My younger brother is very famous&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Sun is more famous and important than Moon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We have days devoted to use&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Sunday and Monday&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="2383" LastEditDate="2014-11-19T17:30:03.237" LastActivityDate="2014-11-19T17:30:03.237" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5163" PostTypeId="2" ParentId="5122" CreationDate="2014-11-19T16:59:34.590" Score="1" Body="&lt;p&gt;This is the correct answer: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tree(s)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You need me. You need me in many ways. Yet, I'm slowly losing my family.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Trees can be used as wood, paper, for fire, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You can count on me as you always did!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;We can count on paper. Literally, write math on paper...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The sound of an explosion, transferred to the orange land.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This one is a tricky one. The sound of an explosion is &lt;em&gt;boom&lt;/em&gt;. The orange land in fact points towards the Netherlands. Tree in Dutch is boom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I can stop the darkness!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;When wood is lit, it gives you light.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If you want, I can even stop the bitter cold!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Again, when lit, it gives warmth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I bring you life as in 2 bound circle-like shapes!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Refers to the formula of oxygen! Well found!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Yet, I keep losing my family. We are torn apart by whom who need us most.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Refers to people cutting down trees.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Who or what are we?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Tree(s)!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="420" LastEditorUserId="420" LastEditDate="2014-11-19T17:06:46.967" LastActivityDate="2014-11-19T17:06:46.967" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5165" PostTypeId="2" ParentId="5029" CreationDate="2014-11-19T17:19:47.817" Score="0" Body="&lt;p&gt;I could be way off, but I just love contributing to these things...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you say it's a real person, and this poor decision was a recent deed, I thought of &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Vladimir Putin (the poor deed being invading Ukraine)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Work to be done, but my eyes cannot look away&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Putin has a country to run, but he can't take his eyes off Ukraine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The murder is not real.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I.E. The murder of Ukrainians by the rebels.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;It will be discovered&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Putin is trying to hide his country's advance on Ukraine and publicly denies it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;No! It must be mine!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ukraine was once part of Russia; Putin must have it back.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The overseer would be god/karma or what have you...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Disclaimer&lt;/strong&gt;: Absolutely no ill will meant to Putin, not trying to cast him as a madman...&lt;/p&gt;&#xA;" OwnerUserId="5713" LastActivityDate="2014-11-19T17:19:47.817" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5167" PostTypeId="2" ParentId="5159" CreationDate="2014-11-19T17:55:45.733" Score="2" Body="&lt;p&gt;This one is slightly obscure   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a shark baby still in the womb  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I love to eat my younger brother.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; sharks will eat each other untill born for nutrients  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Our Intelligience is never ending.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the DNA in the shark has insurmountable intelligence  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My younger brother is very famous&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the younger embryo's are famous caviar   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We have days devoted to us.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;14th July, national shark day  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5135" LastActivityDate="2014-11-19T17:55:45.733" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5169" PostTypeId="1" AcceptedAnswerId="5194" CreationDate="2014-11-19T18:04:36.103" Score="5" ViewCount="1488" Body="&lt;p&gt;Inspired by &lt;a href=&quot;https://puzzling.stackexchange.com/questions/5100/mosaic-with-tetris-blocks/5101?noredirect=1#comment15069_5101&quot;&gt;Mosaic with tetris blocks&lt;/a&gt; I was wondering if there were any generic algorithms to solve or show there was a solution to these types of problems (i.e. placing &lt;a href=&quot;http://en.wikipedia.org/wiki/Polyomino&quot; rel=&quot;nofollow noreferrer&quot;&gt;polyominos&lt;/a&gt; on a 2D board).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Breaking it down to the simplistic form of dominos, there's the classic puzzle of placing them a checkerboard with either two white or black squares missing. This is known to be impossible and left as an exercise for the reader =).&#xA;But what if we're given a less uniform board. Is there any known algorithms other than brute-force to show there is a solution or to even solve the puzzle? We know there have to be an even number of square and an equal number of white and black squares, but that doesn't mean there is a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moving up in complexity, there's another classic problem of filling an $n \times n$ board with &lt;a href=&quot;http://en.wikipedia.org/wiki/Tromino&quot; rel=&quot;nofollow noreferrer&quot;&gt;L-triominos&lt;/a&gt; except for one marked square. Recursive solutions exist here, but I've never seen anything for non-uniform board or allowing non-L blocks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally we hit the Mosaic puzzle. What if we're given a set of tetris pieces and an irregular board? Can we easily determine if there's a solution? Is it easier if we aren't constrained by the number of each piece we can use?&lt;/p&gt;&#xA;" OwnerUserId="5612" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2021-08-29T09:46:35.203" Title="Solutions for generic polyomino puzzles" Tags="|geometry|checkerboard|solvability|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5170" PostTypeId="2" ParentId="5159" CreationDate="2014-11-19T18:13:55.410" Score="3" Body="&lt;p&gt;If&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; pi is close, as in the comment to Tyson's guess,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;then maybe you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; tau&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and you eat your younger brother because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; tau = 2 * pi&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Making your day&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; June 28th&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Further explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter tau was chosen to represent (2 * pi) because of its similarity in appearance to pi (tau has one leg instead of two). There is a (somewhat tongue-in-cheek) &quot;movement&quot; in mathematics attempting to supplant pi with tau because it makes certain trigonometric calculations easier (The distance around a circle for instance, is 2 pi radians, which is just tau.) The reference to eating is because pi sounds like pie and also because tau is greater than pi, so it &quot;contains&quot; pi (like being eaten). Never-ending is a reference to the transcendental nature of the numbers; the decimal expansion contains an infinite number of digits. And just as Pi Day is March 14th because pi begins with 3.14, Tau Day is June 28th because tau begins with 6.28 (3.14 times 2).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4518" LastEditorUserId="4518" LastEditDate="2014-11-20T16:46:08.563" LastActivityDate="2014-11-20T16:46:08.563" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5172" PostTypeId="1" AcceptedAnswerId="5185" CreationDate="2014-11-19T18:31:41.300" Score="6" ViewCount="1274" Body="&lt;p&gt;For every question you might ask&lt;br&gt;&#xA;Five answers might I give&lt;br&gt;&#xA;Despite my ever-changing mask&lt;br&gt;&#xA;No err will you forgive&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I may appear a fickle host&lt;br&gt;&#xA;But we’re both keeping score&lt;br&gt;&#xA;From the side, I will riposte&lt;br&gt;&#xA;And you’ll be back for more&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For posterity, I will protect&lt;br&gt;&#xA;Your quests within a picture&lt;br&gt;&#xA;So quicker than you could expect&lt;br&gt;&#xA;The student becomes the teacher&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;i&gt;What am I?&lt;/i&gt;&lt;/p&gt;&#xA;" OwnerUserId="5766" LastEditorUserId="5766" LastEditDate="2014-11-20T14:54:31.967" LastActivityDate="2014-11-20T14:54:31.967" Title="A face of a thousand masks" Tags="|riddle|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5173" PostTypeId="2" ParentId="5172" CreationDate="2014-11-19T18:55:18.933" Score="0" Body="&lt;p&gt;My best guess is...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Computer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Seems to fit. You can search with it and get a mess of answers, constant evolution (pc, laptop, phones, watches), very easy to blame the computer (&quot;this thing is so slow!&quot;), it keeps score and tracks your 'quests' in a monitor while playing games, and finally because a user (student of a computer) can become a developer (teacher of computers).&lt;/p&gt;&#xA;" OwnerUserId="4191" LastEditorDisplayName="user20" LastEditDate="2014-11-20T02:36:51.813" LastActivityDate="2014-11-20T02:36:51.813" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5174" PostTypeId="1" AcceptedAnswerId="5176" CreationDate="2014-11-19T18:58:37.657" Score="4" ViewCount="1651" Body="&lt;p&gt;&lt;em&gt;My cousins have a lump or two, &lt;br&gt;&#xA;But I have none. &lt;br&gt;&#xA;I'm coloured like a grassy hill; &lt;br&gt;&#xA;Never seen one.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A war is fought within my bounds, &lt;br&gt;&#xA;Let's hope that peace will be found. &lt;br&gt;&#xA;I may sound like a pub on the roadside, &lt;br&gt;&#xA;But on the internet I reside.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to this riddle can be given as a URL or as a picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Disclaimer&lt;/em&gt;: this time, no maths involved! That's a promise (and this is not a hint).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Leo's answer taught me more than I realised there was to know about the solution to this riddle. Feel free to edit my question in order to make it fit his answer better. He's done his research better than I did!&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2014-11-19T19:28:16.767" LastActivityDate="2014-11-19T19:56:24.377" Title="Riddle - answer is a URL or picture" Tags="|riddle|" AnswerCount="3" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5175" PostTypeId="2" ParentId="5174" CreationDate="2014-11-19T19:04:39.240" Score="2" Body="&lt;p&gt;I think this is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.hdwallpapers.in/walls/windows_xp_bliss-wide.jpg&quot; rel=&quot;nofollow&quot;&gt;Windows XP default Wallpaper&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A war is fought within my bounds, &#xA;Let's hope that peace will be found. &#xA;I may sound like a pub on the roadside, &#xA;But on the internet I reside.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lots of people create/delete files and folders on their desktop. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5030" LastActivityDate="2014-11-19T19:04:39.240" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5176" PostTypeId="2" ParentId="5174" CreationDate="2014-11-19T19:08:54.783" Score="5" Body="&lt;p&gt;Nice riddle...  I think I got it though...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bunchie is the name of this mythical internet creature that started on genmay and became huge rather quickly.  I remember when enough members on genmay even took over drawball to make a giant picture of Bunchie... &lt;a href=&quot;http://knowyourmeme.com/memes/bunchie&quot; rel=&quot;nofollow&quot;&gt;http://knowyourmeme.com/memes/bunchie&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And to address the second half...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The second half of the riddle is about the Puzzling Beta chat room which is called &quot;Green Llama&quot; in reference to Bunchie.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4191" LastEditorUserId="4191" LastEditDate="2014-11-19T19:56:24.377" LastActivityDate="2014-11-19T19:56:24.377" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5177" PostTypeId="1" AcceptedAnswerId="5200" CreationDate="2014-11-19T19:09:16.683" Score="4" ViewCount="993" Body="&lt;p&gt;You're sitting at home reading a magazine and you came across a cool article, but there's a misprint. Most of the page is blank. You can make out that it's someone plugging an autobiography from the scattered words that halfway made it onto the page, but you don't see a name. Can you tell who it is from the paragraph that printed?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;en, people ask about my early career. Well, I was an actor, a guy with a face for every stage, you might say. There was this guy who used to upstage me all the time, a real jerk that always made me feel like second place, a fraction. After a time, even that was a dream. I was supporting number 2 to the world, but I was third rate at best in my heart. I tried to push myself after that. I had fifty-six new roles over the next decade, though the massive crowds dried up after a few short years. I still have anger issues over it, obviously. That smarmy, British thumbs up Simon Cowell gave me... but I'm sure everyone's read about that in the paper. I just flipped and he ended up missing a leg. I can't stand the unamerican. If I were able to bleed, I would bleed red, white, and blue. Ha! But... you can read more in the book. I don't wanna give everything away. I will share this though. I've been acting again, going on tour. I'm starring in America the Beautiful, playing all the best venues from Hot Springs to Tuskeegee. The next stop is go&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5626" LastEditorUserId="5626" LastEditDate="2014-12-08T14:19:41.800" LastActivityDate="2014-12-08T14:19:41.800" Title="The autobiography of ___________? - From the misprinted page of an article, uncover the name of the writer" Tags="|riddle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5179" PostTypeId="1" CreationDate="2014-11-19T19:15:34.397" Score="9" ViewCount="1610" Body="&lt;p&gt;A room has n empty chairs in a row. You choose a chair and sit it in it. Now someone else comes in and sits as far away as possible from everyone in the room or an arbitrary choice of furthest chair if there is more than one. This happens repeatedly with the rule that no two people can ever sit next to each other. It finishes when there is no more space to sit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your goal of course is to choose your chair initially so that as many people as possible can sit in the room. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say n=2. Wherever we sit we end up with exactly one person sat down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For n=3. If we sit in the first chair, we get full, empty, full in the end. But if we set the chair bit we get empty, full, empty which is not optimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For n=4. Wherever we sit we end up with two chairs with people on them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For n=5. Sitting in the first chair gives us full, empty, full, empty,full  with three people sat down in the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For n=7. We need to sit in the third chair it seems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the simplest rule for choosing where to sit which will always give an optimal answer?&lt;/p&gt;&#xA;" OwnerUserId="8998" LastEditorUserId="8998" LastEditDate="2014-11-19T20:01:26.773" LastActivityDate="2014-11-20T13:29:58.590" Title="The antisocial chair problem" Tags="|logical-deduction|" AnswerCount="3" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="5180" PostTypeId="2" ParentId="5172" CreationDate="2014-11-19T19:18:12.600" Score="1" Body="&lt;p&gt;Perhaps a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Search engine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;many answers on one question&lt;/p&gt;&#xA;&#xA;&lt;p&gt;fickle host&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the information isn't really his, just elsewhere on the web&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;keeping quests&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; caching search queries&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;student becomes teacher - &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; crawls the internet for data, then presents it on the search&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4317" LastActivityDate="2014-11-19T19:18:12.600" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5181" PostTypeId="2" ParentId="5029" CreationDate="2014-11-19T19:22:04.660" Score="0" Body="&lt;p&gt;I think someone wants to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; go on private browsing using Firefox instead of Chrome, to avoid Google spying on him. But he is using an android machine, Google search engine, gmail etc. so he isn't really hiding&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4317" LastActivityDate="2014-11-19T19:22:04.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5183" PostTypeId="2" ParentId="5174" CreationDate="2014-11-19T19:25:20.267" Score="2" Body="&lt;p&gt;Just for the record (but Leo's answer is much better), the answer I was looking for is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://chat.stackexchange.com/rooms/14524/the-green-llama&quot;&gt;The Green Llama&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My cousins have a lump or two, &lt;br&gt;&#xA; But I have none.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The llama is closely related to the dromedary and Bactrian camels with one and two humps respectively, but it itself is humpless.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I'm coloured like a grassy hill;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Green.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Never seen one.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's not a real llama, just something on the internet!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A war is fought within my bounds,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This chatroom is the site of a big argument about what puzzling.SE should look like, which I described in one of my posts there as a &quot;war&quot;...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Let's hope that peace will be found.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ... but we're hoping for an amicable solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I may sound like a pub on the roadside,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well, it &lt;em&gt;does&lt;/em&gt;. &quot;How about a pint at the Green Llama?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But on the internet I reside.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Simple truth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2014-11-19T19:25:20.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5184" PostTypeId="2" ParentId="5179" CreationDate="2014-11-19T19:41:56.477" Score="4" Body="&lt;p&gt;Pick the highest power of 2 that's under the chair limit, add 1, and sit there. (So, if there are 16 chairs, sit at chair 9). This will ensure that the chairs to your left are all filled in a binary fashion. After that, it's just hoping the chairs to the right fill out correctly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, the means you can only control the chairs to your left. If you had, say, 15 chairs, you can control the first 9, but after that, chair 15 will be taken, then chair 12, leaving 2x 2-space gaps (10-11, 13-14) that you can't control. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2014-11-19T19:41:56.477" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="5185" PostTypeId="2" ParentId="5172" CreationDate="2014-11-19T19:44:19.407" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stack Exchange&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For every question you might ask&#xA;Five answers might I give&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; or more &amp;amp; more...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Despite my ever-changing mask&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; different people answering each time &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No err will you forgive&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; downvote!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I may appear a fickle host&#xA;But we’re both keeping score&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; votes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the side, I will riposte&#xA;And you’ll be back for more&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For posterity, I will protect&#xA;Your quests within a picture&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; maybe user profile&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So quicker than you could expect&#xA;The student becomes the teacher&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Someone who came to ask might as well find themselves answering...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4317" LastActivityDate="2014-11-19T19:44:19.407" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5186" PostTypeId="2" ParentId="5177" CreationDate="2014-11-19T19:48:50.707" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Daffy Duck&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is what my guess is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upstaged by a jerk, namely&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Bugs Bunny.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Simon Cowell was in the paper for saying his co-host &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;had the mind of Daffy Duck.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Seems there must be at least 56 &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Daffy Duck shorts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hot Springs to Tuskeegee sounds like the kind of trip he might take if he takes &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;that wrong turn at Kookamonga.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5052" LastEditorUserId="4204" LastEditDate="2014-11-19T23:40:11.047" LastActivityDate="2014-11-19T23:40:11.047" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5189" PostTypeId="2" ParentId="4978" CreationDate="2014-11-19T19:53:35.153" Score="1" Body="&lt;p&gt;You are a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A brain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In my case it is tiny and slimy, and yet very useful. In a way I am my own one, so it can be thought of as a creature. In the case of most people, when their one wants something, they will get it. And I suppose tapeworms have one inside them.&lt;/p&gt;&#xA;" OwnerUserId="5775" LastEditorUserId="5775" LastEditDate="2014-11-19T19:59:04.787" LastActivityDate="2014-11-19T19:59:04.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5190" PostTypeId="1" AcceptedAnswerId="5599" CreationDate="2014-11-19T19:54:11.060" Score="34" ViewCount="3783" Body="&lt;p&gt;You have 12 coins which each weigh either 20 grams or 10 grams. Each is labelled from 1 to 12 so you can tell the coins apart.  You have one weighing device as well. At each turn you can put as many coins as you like on the weighing device and it will tell you exactly how much they weigh.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the minimum number of weighings  that will always tell you exactly which coins weigh 10 grams and which weigh 20?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly you can do it in 12 weighings by just weighing each coin separately.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There is now a solution using $7$ &lt;strong&gt;&lt;em&gt;static&lt;/em&gt;&lt;/strong&gt; weighings by Julian Rosen!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Current record found by computer search is $7$ (dynamic) weighings by Victor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a hand developed record of $8$ (dynamic) weighings by Joel Rondeau.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the first three lines of a  solution that uses $7$ weighings.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0 0 1 0 1 1 1 0 0 0 0 0&#xA;1 0 0 1 0 1 1 1 0 0 0 0&#xA;1 1 0 0 1 0 1 1 1 0 0 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Bounty awarded to Victor for the first correct answer.  Julian Rosen gave a static solution with $7$ weighings but without an explanation yet (just external references).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will award the &quot;win&quot; to the first static solution with $7$ weighings &lt;em&gt;and&lt;/em&gt; an explanation.&lt;/p&gt;&#xA;" OwnerUserId="8998" LastEditorUserId="8874" LastEditDate="2015-02-19T13:47:12.680" LastActivityDate="2015-02-19T13:47:12.680" Title="Coin weighing with a single weighing device" Tags="|mathematics|combinatorics|weighing|" AnswerCount="11" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="5191" PostTypeId="2" ParentId="5029" CreationDate="2014-11-19T20:01:48.610" Score="9" Body="&lt;p&gt;I don't know who the person being referred to is, but since the question asks only &quot;What poor decision is driving him mad?&quot; my answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He spent too much time on Puzzles SE.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reason being:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First he says &quot;It claws at me. Ha! There is work to be done, but the pale lands have thorns growing from its roots. Work to be done, but my eyes cannot look away.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He is at work and supposed to be working, but instead he can't stop refreshing the Puzzles SE page to see if his questions have been answered, and if his answers have been accepted as correct.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Is a fish a trout? Can it swim to Prometheus? For, four, floor? More?&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He has been on the site so long his brain now thinks in nonsense riddles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Is the attempt a virtue and the vice failure or is the attempt an exercise in frivolity that the overseer cannot allow. Is me still the master or is it?&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He thinks maybe coming up with good questions and answers is virtuous (better than working) but fears his boss will found out and discipline him, and questions whether he himself has discipline.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;There are no dials to turn, no numbers, and no secrets. It is mine. The murder is not real.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; His job doesn't have any excitement or mystery to it like this site does.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Through this, the overseer watches. The overseer's eyes are flame, ignorant of it, but it cannot remain undiscovered for long. What if?&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He has managed to not get caught so far, and his guilt is eating away at him and he fears getting caught at any moment.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;No... But a mountain's erosion sloped by a tangential curve. Identical hillocks? It will be discovered. It must be hidden by words of power. Let the fox guide it to where it must lay dormant...&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He is using various schemes to avoid being found out, such as using VPNs, Proxies, and Incognito mode in the browser.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. It must be mine! HaHAAHHA! HasdhAhHahaahahahaAHhaA!!!!!!! I hide in my tiny box, but he will always find me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He is slowly going mad, his brain trapped in the tiny URL bar where he enters the text &lt;code&gt;puzzling.stackexchange.com&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Update:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess logically I should have guessed right away that the name of this madman is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Travis Kindred&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2718" LastActivityDate="2014-11-19T20:01:48.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5192" PostTypeId="2" ParentId="5190" CreationDate="2014-11-19T20:12:52.820" Score="0" Body="&lt;p&gt;A way to bring the number of weighings down to about 1/2 would be to weigh all of them, record that weight (if it's 24 or 12, you know the weights). Then weigh 6, record those (again, 12 or 6, you know them). Then weigh 2 (4 or 2, you know). Then weigh 1. Based on that, you know the other. Repeat for the other sets of 2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Worst case scenario, every set of 2 is mixed, which would lead to 12 weighings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take the scenario of 101101 101101 and do a binary search:&lt;br&gt;&#xA;{101101 101101}&lt;br&gt;&#xA;{101101} 101101&lt;br&gt;&#xA;{101} 101 101101&lt;br&gt;&#xA;{1}01 101 101101&lt;br&gt;&#xA;1 {0}1 101 101101&lt;br&gt;&#xA;1 0 1 {1}01 101101&lt;br&gt;&#xA;1 0 1 1 {0}1 101101&lt;br&gt;&#xA;1 0 1 1 0 1 {101} 101&lt;br&gt;&#xA;1 0 1 1 0 1 {1}01 101&lt;br&gt;&#xA;1 0 1 1 0 1 1 {0}1 101&lt;br&gt;&#xA;1 0 1 1 0 1 1 0 1 {1}01&lt;br&gt;&#xA;1 0 1 1 0 1 1 0 1 1 {0} 1  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's try with groups of 2:&lt;br&gt;&#xA;{101101 101101}&lt;br&gt;&#xA;{101101} 101101&lt;br&gt;&#xA;{10} 1101 101101&lt;br&gt;&#xA;{1}0 1101 101101&lt;br&gt;&#xA;1 0 {11} 01 101101&lt;br&gt;&#xA;1 0 1 1 {0}1 101101&lt;br&gt;&#xA;1 0 1 1 0 1 {10} 1101&lt;br&gt;&#xA;1 0 1 1 0 1 {1}0 1101&lt;br&gt;&#xA;1 0 1 1 0 1 1 0 {11} 01&lt;br&gt;&#xA;1 0 1 1 0 1 1 0 1 1 {0}1  &lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2014-11-19T22:07:55.620" LastActivityDate="2014-11-19T22:07:55.620" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5193" PostTypeId="2" ParentId="5179" CreationDate="2014-11-19T20:16:45.570" Score="0" Body="&lt;p&gt;I think the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (n/2) for all even. and for the series n = 3, 7, 11, 15, ... use floor(n/2) while for the series n = 5, 9, 13, 17, ... use ciel(n/2)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;n = 4&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;_ _ _ _&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;formula says pick chair 2. so you end up with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;0 1 0 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;n = 5&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;_ _ _ _ _&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;2 _ 1 _ 3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;n = 6&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;3 _ 1 _ _ 2 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;n = 7&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;4 _ 1 _ 3 _ 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;lets skip a few...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;n= 10&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;_ _ _ _ _ _ _ _ _ _&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;now the formula says sit down in 5. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;3 _ 5 _ 1 _ _ 4 _ 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the sequence that would follow would be the same as n =4 and n =6:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the left side of 1 is n = 4&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;the right side is n = 6 including the first person.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;similarly by induction all other higher n would break into a subset of n/2 chair arrangements&lt;/p&gt;&#xA;&#xA;&lt;p&gt;n = 11&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;3 _ 6 _ 1 _ 4 _ 5 _ 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5030" LastEditorUserId="5030" LastEditDate="2014-11-19T20:43:05.113" LastActivityDate="2014-11-19T20:43:05.113" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5194" PostTypeId="2" ParentId="5169" CreationDate="2014-11-19T20:17:02.837" Score="5" Body="&lt;p&gt;The problem is a type of &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Packing_problems#Related_fields&quot; rel=&quot;nofollow&quot;&gt;Packing Problem&lt;/a&gt;&quot;. Regarding polyominoes we are interested in &lt;a href=&quot;http://en.wikipedia.org/wiki/Polyomino#Tiling_regions_with_sets_of_polyominoes&quot; rel=&quot;nofollow&quot;&gt;tiling specific regions&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is known that the problem of answering whether a specific region can be filled with tiles is undecidable. And that finding a solution, in  general, is NP-Complete. So we can never do much better than trying all possible configurations - i.e. pick something and exhaustively show that it doesn't work, backtrack and try again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to the &quot;simpler&quot; problems like L-trominoes, it is &lt;a href=&quot;http://arxiv.org/pdf/math/0003039.pdf&quot; rel=&quot;nofollow&quot;&gt;essentially the same problem&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to answer directly: there is no easy way to determine whether there is a solution. Just as there is no easy solution for the travelling salesman problem, the tiling problem, or the &lt;a href=&quot;http://en.wikipedia.org/wiki/Boolean_satisfiability_problem&quot; rel=&quot;nofollow&quot;&gt;boolean satisfiability problem&lt;/a&gt; (SAT) (on a personal note, I immediately think of this problem, and could fairly easily cast the polyomino problem as a SAT problem).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Related:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://math.umn.edu/~jedyang/papers/&quot; rel=&quot;nofollow&quot;&gt;More Papers&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/De_Bruijn%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Be Bruijn's Theorem&lt;/a&gt;&lt;br&gt;&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-11-19T23:52:18.690" LastActivityDate="2014-11-19T23:52:18.690" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5195" PostTypeId="2" ParentId="1932" CreationDate="2014-11-19T20:22:58.213" Score="7" Body="&lt;p&gt;&lt;strong&gt;There are no solutions.&lt;/strong&gt; This test is extremely subjective. For each question, there are no enough information to solve them. See question 1: It is absolutely impossible to discover an objective rule between two &lt;em&gt;very&lt;/em&gt; different shapes. You need at least 3. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't trust this pseudo-test. The fact that is is numbered in a weird and useless way shows that I can't take it seriously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or maybe I don't understand it because my IQ is too low.&lt;/p&gt;&#xA;" OwnerUserId="5335" LastActivityDate="2014-11-19T20:22:58.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5196" PostTypeId="2" ParentId="5007" CreationDate="2014-11-19T20:48:53.090" Score="3" Body="&lt;p&gt;All right I'll take a stab. She is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Vanilla ice cream with cookie dough chunks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;She was killed by:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Being put in the oven. She melted and the cookies baked.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5500" LastActivityDate="2014-11-19T20:48:53.090" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5198" PostTypeId="1" AcceptedAnswerId="5316" CreationDate="2014-11-19T21:37:21.030" Score="26" ViewCount="1696" Body="&lt;p&gt;Let's pack some (one or more) T hexominoes together with some (one or more) small $a\times b$ rectangles into some bigger $m\times n$ rectangle without holes and overlapping pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I can pack two T hexominoes and eight $1\times1$ rectangles into a $5\times4$ rectangle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/A4oCr.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Task&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Find as many as you can different integer pairs $\{a,b\}$ so that one or more rectangles of size $a\times b$ can be packed together with T hexominoes into a $m\times n$ rectangle without holes and overlapping pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Provide images of solutions if you claim that some combination is possible.&#xA;Please, put your images in spoiler tags so that other users can try find themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If some answers will have same number of pairs $\{a,b\}$, then answer with smaller total area of outer rectangles on example images is preferable. If area also equal, then earlier posted answer is preferable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Please, add to your answers total area of outer rectangles. If one tiling implicitly includes several $a\times b$ rectangles, then multiply area by number of combinations it includes.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Notes&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I can guarantee that there exist solutions following integer pairs:&lt;br&gt;&#xA;$1\times1$, $1\times2$, $1\times3$, $1\times4$, $1\times5$, $1\times6$, $1\times8$, $1\times9$,&lt;br&gt;&#xA;$2\times2$, $2\times3$, $2\times4$, $2\times5$, $2\times6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering why a $1\times7$ rectangle solution is hard to find (maybe it's impossible?) when there exist solutions for with $1\times8$ and $1\times9$ rectangles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 week after posting, I'll put up my own answers for unfound combinations. &lt;/p&gt;&#xA;" OwnerUserId="5781" LastEditorUserId="5781" LastEditDate="2015-02-21T21:12:44.377" LastActivityDate="2018-06-11T07:49:09.487" Title="Polyomino T hexomino and rectangle packing into rectangle" Tags="|geometry|solvability|polyomino|" AnswerCount="5" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="5199" PostTypeId="1" CreationDate="2014-11-19T21:52:54.080" Score="-1" ViewCount="229" Body="&lt;p&gt;I'm a computer programmer trainer and sometimes I do some puzzle exercises with my students. Right now we did a checkers game and I would like to make them program to solve something with their checkers board. Do you know any interesting puzzle?&lt;/p&gt;&#xA;" OwnerUserId="5779" LastActivityDate="2014-11-20T01:01:49.767" Title="8 queens equivalent on checkers" Tags="|checkerboard|" AnswerCount="3" CommentCount="1" ClosedDate="2014-11-20T08:21:24.390" ContentLicense="CC BY-SA 3.0" />
  <row Id="5200" PostTypeId="2" ParentId="5177" CreationDate="2014-11-19T21:54:30.057" Score="7" Body="&lt;p&gt;Not fully reasoned out yet, but:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quarter (coin)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&#xA;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A face for every stage&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's always had a &quot;heads&quot; side(?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A fraction&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1/4 of a dollar - supporting number 2 might relate to not being a full dollar as well?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Fifty-six new roles over the next decade&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/50_State_Quarters&quot; rel=&quot;nofollow noreferrer&quot;&gt;The 50 state quarters + District of Columbia, Puerto Rico, American Samoa, Guam, the United States Virgin Islands, and the Northern Mariana Islands.&lt;/a&gt; (1999-2009)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Massive crowds dried up after a few short years&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Collector mania faded out over time?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;British thumbs up Simon Cowell Gave me...flipped&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Did Simon Cowell flip a coin for something? Can't find any news about it yet, but I'm not able to fully research right now. Found a tenuous connection in a news article talking about him missing a leg of an X-factor tour, but can't figure out how a quarter was involved.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Bleed red, white, and blue&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A reference to being a US coin?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Starring in America the Beautiful&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/America_the_Beautiful_Quarters&quot; rel=&quot;nofollow noreferrer&quot;&gt;America the Beautiful Quarters&lt;/a&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Playing all the best venues from Hot Springs to Tuskeegee&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quarter #1 in America the Beautiful: Hot Springs National Park &lt;br /&gt;Quarter #56 (last one): Tuskegee Airmen National Historic Site&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Further explanation from &lt;a href=&quot;https://puzzling.stackexchange.com/users/5626/&quot;&gt;Travis Kindred&lt;/a&gt;:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Stage refers again to the states on the new quarters. Second and third place refer to the chemical composition of the coin changing over time from silver to a silver-colored bronze. Simon Cowell interviewed with hitquarters a decade ago, but his name was mainly meant to obscure the fact that a thumb moving upward caused the coin to flip. His missing leg is a reference to quartering, an English criminal punishment that left a person cut in four pieces. The leg is only one quarter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5340" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2014-11-19T22:55:41.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5201" PostTypeId="2" ParentId="5199" CreationDate="2014-11-19T22:00:37.267" Score="4" Body="&lt;p&gt;How about, as a suggestion: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;maximisation of possible jumps (i.e. &quot;What is a position giving the highest possible number of jumps in one move?&quot;).&lt;/li&gt;&#xA;&lt;li&gt;And how many pieces are required to achieve this maximum (i.e. how many of each side)?&lt;/li&gt;&#xA;&lt;li&gt;Obviously, what is the number?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2484" LastActivityDate="2014-11-19T22:00:37.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5202" PostTypeId="2" ParentId="5199" CreationDate="2014-11-19T22:02:35.217" Score="3" Body="&lt;p&gt;Using a program, find a series of checkers turns that result in the original position, but with all kings, (use the regular setup of checkers in the start)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use a program you made to find how many kings you can place on a checkers board where each king can legally capture exactly 2 pieces? Or is this even possible?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2014-11-19T22:02:35.217" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5203" PostTypeId="1" AcceptedAnswerId="5206" CreationDate="2014-11-19T22:23:15.233" Score="15" ViewCount="18089" Body="&lt;p&gt;Heading down to the city of Thebes, you encounter a Sphinx who stops you in your journey and speaks to you. The Sphinx has a riddle for you: if you can solve it, she will let you pass; if you fail, she will devour you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this is the riddle the Sphinx has prepared for you:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;If this were an answer, what would be the question?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can you answer the Sphinx's riddle? There are two rules you must accept:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Your answer must not repeat the Sphinx's riddle verbatim.&lt;/li&gt;&#xA;&lt;li&gt;Your answer must be itself a real question, not just a sentence and not a rhetorical question.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Hence the following is not acceptable, the Sphinx will eat you if you say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strike&gt;Could you kindly ask: &lt;em&gt;&quot;If this were an answer, what would be the question?&quot;&lt;/em&gt;?&lt;/strike&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While the Sphinx is highly intelligent, she has no time to lose, so don't expect her to check any weblinks, decode messages, watch you writing or mimicking, etc.&lt;/p&gt;&#xA;" OwnerUserId="4394" LastEditorUserId="4394" LastEditDate="2014-11-22T07:23:15.270" LastActivityDate="2015-07-22T13:26:20.820" Title="If this were an answer, what would be the question?" Tags="|riddle|logical-deduction|" AnswerCount="17" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="5204" PostTypeId="2" ParentId="5203" CreationDate="2014-11-19T22:32:40.270" Score="7" Body="&lt;p&gt;The first thing that comes to mind is  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If this was the question, what would be the answer?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but that seems too easy  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility in the literalist bent  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hold some object in your hand and ask &quot;What?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditDate="2014-11-19T22:40:14.103" LastActivityDate="2014-11-19T22:40:14.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5205" PostTypeId="2" ParentId="5179" CreationDate="2014-11-19T22:34:14.117" Score="5" Body="&lt;p&gt;The ideal situation is to have ceil(n/2) chairs filled. People and empty chairs will alternate in this situation. Any time there is a gap of just two chairs, it can't be filled, so you want to have gaps of 3 that will eventually fill up leaving gaps of one. For odd number of chairs, you want both ends to be full. For even, you want one of the ends to be full. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some strategies that don't work well:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;just sit in the middle. Consider 9: you sit with 4 empty chairs on either side of you. The next person sits at one end, and the next at the other. Now the last two people sit in the centre of each 3-chair gap. Great! But for 11, you have 5 on each side, the ends fill in leaving a pair of 4-gaps, and people sit in those leaving a 1-gap on one side of them and a 2-gap on the other - suboptimal.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;sit at one end. Second person will sit at the other, third in the middle and we are now in the very same situation as &quot;sit in the middle&quot; is after the third person&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;sit 3 from one end. for 9 you start 001000000 then 001000001 the third person fills the 5-chair gap so 001001001&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You win if all the gaps are 1 chair. (Ignore the ends for a moment and imagine you're joining this in progress.) The only way to force all the gaps to be 1 chair is to force them all to be 3 (which someone sits in the middle of) or 1. The only way to make a 3-gap is to have a 7 gap that someone sits in the middle of. The only way to make a 7 gap is a 15 gap that someone sits in the middle of. These are one less than powers of 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the ends mess this up. If you have a 7 gap that stretches to the end, someone will sit at the end, creating a 6 gap which can't be neatly split up like this. So you want to sit so that you put the largest possible power of two between you and one end. Someone will sit at that end creating one of these 2^n-1 gaps and then it will shake out from there. The short end you can't control, you just hope it ends up all 1-gaps but it might not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 11, your target gap is 8 You sit: 00100000000. Next person goes to the end: 00100000001. Next person 00100010001. Then you're all set - you will end up 10101010101.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should work for any number of chairs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some things I can prove:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;if you sit with 2^n chairs to your left, the gap to your right is smaller. Well, if it's not, then it's equal to 2^n + m, where m is positive, meaning you could sit somewhere with 2^n + 2^n chairs to your left and m to your right, meaning there is a way to sit and get 2^(n+1) chairs to your left, so you should have done that. You get the &lt;strong&gt;largest&lt;/strong&gt; power of 2 you can. The not-necessarily-power-of-2 gap on the other side of you &lt;strong&gt;must&lt;/strong&gt; then be smaller. This is what forces the next sitter to sit at the end of your 2^n gap, creating the first (2^n)-1 gap.&lt;/li&gt;&#xA;&lt;li&gt;once you gaps that are (2^n)-1, they will always be split into a pair of (2^(n-1))-1 gaps with a filled chair between them. First, (2^(n-1))-1 + 1 + (2^(n-1))-1 is (2^(n))-1. Second, sitting anywhere in the gap that doesn't result in the same number of empty chairs on either side of the sitter means they are not as far away from others as they could be - they can't get further away than (2^(n-1))-1 and anything else is closer.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1509" LastEditorUserId="1509" LastEditDate="2014-11-20T13:29:58.590" LastActivityDate="2014-11-20T13:29:58.590" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="5206" PostTypeId="2" ParentId="5203" CreationDate="2014-11-19T22:39:30.740" Score="33" Body="&lt;p&gt;I'd say some variation of the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What is the Sphinx's riddle?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The trick is to get the Sphinx to repeat herself somehow without repeating the text in the question yourself. I think this is about as simple and direct a method of doing this as is possible. It doesn't rely on any sort of social cue tricks (e.g., &quot;huh?&quot;, &quot;say again?&quot;) or loopholes in the logic of the puzzle.&lt;/p&gt;&#xA;" OwnerUserId="5612" LastEditorUserId="5612" LastEditDate="2014-11-20T02:49:28.643" LastActivityDate="2014-11-20T02:49:28.643" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5207" PostTypeId="2" ParentId="5203" CreationDate="2014-11-19T22:39:33.480" Score="8" Body="&lt;p&gt;This is both a normal response and a question. Often people use it after a question they don't understand.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Could you kindy repeat the question?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;How would you answer this?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5364" LastEditorUserId="5364" LastEditDate="2014-11-20T14:33:12.630" LastActivityDate="2014-11-20T14:33:12.630" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5208" PostTypeId="1" AcceptedAnswerId="5218" CreationDate="2014-11-19T22:48:55.263" Score="-2" ViewCount="400" Body="&lt;p&gt;Cipher: Udfwavk wcfo nt &quot;zjljhydmu&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This cipher is a solution to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4668/the-security-to-the-party-part-30-1&quot;&gt;this&lt;/a&gt; which is currently unsolved and has a bounty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clarifications:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a special cipher, a Vigenere cipher, with this riddle telling the secret key:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of are given this at birth,&lt;br&gt;&#xA;it denotes each other apart,&lt;br&gt;&#xA;you can find my profile with this,&lt;br&gt;&#xA;and as such, it is my profile's ____&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chat is open: &lt;a href=&quot;http://chat.stackexchange.com/rooms/18815/hard-text-cipher&quot;&gt;http://chat.stackexchange.com/rooms/18815/hard-text-cipher&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5368" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2014-11-20T00:44:23.413" Title="Vigenere text cipher" Tags="|encoded-message|cipher|" AnswerCount="1" CommentCount="8" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5210" PostTypeId="2" ParentId="5203" CreationDate="2014-11-19T22:53:50.500" Score="0" Body="&lt;p&gt;The answer (&lt;em&gt;or question&lt;/em&gt;) is so easy though!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If I were to have you repeat that question in exact words, what would you reply with?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-11-20T01:23:37.203" LastActivityDate="2014-11-20T01:23:37.203" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5211" PostTypeId="2" ParentId="5190" CreationDate="2014-11-19T22:54:15.187" Score="14" Body="&lt;p&gt;There is a solution requiring a maximum of $9$ weighings. The strategy rests on a solution for $4$ coins $\{c_1,c_2,c_3,c_4\}$ in $3$ weighings - then performing the same operation on the remaining sets of $4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With 4 coins, weigh coins $c_1$ and $c_2$. If the amount is 20 or 40, then the coins weigh 10 or 20, respectively, and we can just weigh the remaining coins individually (for a total of 3 weighings). If $c_1$ and $c_2$ weigh 30 altogether, then all we know is that the weights are different. I will continue assuming this is the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weigh coins $c_1$ and $c_3$. Like last time, if the amount is 20 or 40 then the weights for $c_1, c_2,$ and $c_3$ can be deduced. In this case, weigh $c_4$ giving you a total of 3 operations for 3 coins. If, however, this amount also ends up being 30, then all we know is that $c_2$ is the same as $c_3$. Again, I will assume this is the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weigh $c_2, c_3,$ and $c_4$. These three coins could weigh 30, 40, 50, or 60 grams in total. If the weight is 30 or 40, then because coins $c_2$ and $c_3$ are the same weight, we know that they are both 10, with $c_4$ being 10 or 20 grams respectively. If the weight is 50 or 60, we can do the same thing, except with $c_2$ and $c_3$ both weighing 20 grams. $c_1$ and $c_2$ weigh 30 together, so we can deduce that coin $c_1$ is the opposite of $c_2$. This gives us all four coins in a worst case scenario of 3 weighings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A more formal solution can be found below, thanks to d'alar'cop.&lt;/strong&gt;&lt;hr/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the first group of $4$, $\{c_1,c_2,c_3,c_4\}$, we will perform our first weighing $c_1$ and $c_2$ giving $ W_1 = w(c_1) + w(c_2)$. It is clear that $W_1 \in \{20,30,40\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$W_1 = 20 \rightarrow c_1 = 10 \land c_2=10$ and $W_1 = 40 \rightarrow c_1 = 20 \land c_2 = 20$. Either way, the weights of $c_1$ and $c_2$ are known. In this case, simply weigh $c_3$ and $c_4$ alone and the $4$ coins are solved in $3$ weighings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $W_1 = 30$ then all we know is that $c_1 \not= c_2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we weigh $c_1$ and $c_3$ giving $W_2 = w(c_1) + w(c_3)$. It is clear that $W_2 \in \{20,30,40\}$. $W_2 = 20 \rightarrow c_1 = 10 \land c_2 = 20 \land c_3=10$ and $W_2 = 40 \rightarrow c_1 = 20 \land c_2=10 \land c_3 = 20$. Thus $3$ are known in $2$ weighings. In this case, simply weigh $c_4$ alone and the $4$ coins are solved in $3$ weighings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $W_2 = 30$ then all we know is that for $c_1,c_2,c_3 \in \{10,20\}.(c_1 \not= c_2) \land (c_1 \not= c_3) \land (c_2 = c_3)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, weigh $c_2$, $c_3$ and $c_4$ giving $W_3 = w(c_2) + w(c_3) + w(c_4)$. It is clear that $W_3 \in \{30, 40, 50, 60\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case $W_3 = 30$&lt;/strong&gt;: $c_2 = c_3 = c_4 = 10 \land c_1 = 20$. &lt;br&gt;&#xA;&lt;strong&gt;Case $W_3 = 40$&lt;/strong&gt;: $c_2 = c_3 = 10 \land c_1 = c_4 = 20$. &lt;br&gt;&#xA;&lt;strong&gt;Case $W_3 = 50$&lt;/strong&gt;: $c_2 = c_3 = 20 \land c_1 = c_4 = 10$. &lt;br&gt;&#xA;&lt;strong&gt;Case $W_3 = 60$&lt;/strong&gt;: $c_2 = c_3 = c_4 = 20 \land c_1 = 10$. &lt;br&gt;&#xA;This follows from the earlier inference that $c_2 = c_3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, we have deduced the weights of coins $c_1,c_2,c_3,c_4$ in $3$ weighings. We may now perform the same operation on $c_5,c_6,c_7,c_8$ and $c_9,c_{10},c_{11},c_{12}$ for $3$ weighings each &lt;strong&gt;&lt;em&gt;giving a total of $9$ weighings to deduce the weights of the $12$ coins $\{c_1, \cdots ,c_{12}\}$&lt;/em&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1664" LastEditorUserId="5434" LastEditDate="2014-11-20T23:45:00.233" LastActivityDate="2014-11-20T23:45:00.233" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5213" PostTypeId="1" AcceptedAnswerId="5215" CreationDate="2014-11-19T22:58:43.770" Score="-10" ViewCount="654" Body="&lt;p&gt;Two mothers and two daughters were in a multi car accident. They were all pronounced dead at the scene. All the bodies have been moved to the local mortuary. There are only three coffins. No one is sharing a coffin and none are left out. How is it possible?&lt;/p&gt;&#xA;" OwnerUserId="5364" LastEditorUserId="5188" LastEditDate="2015-04-19T21:29:54.030" LastActivityDate="2016-11-22T08:14:02.983" Title="The Three Coffins" Tags="|family|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5214" PostTypeId="2" ParentId="5203" CreationDate="2014-11-19T22:58:51.193" Score="12" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;What's the typical format of a &lt;em&gt;Jeopardy!&lt;/em&gt; clue?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if the question is &quot;elegant&quot;, but I think it works.&lt;/p&gt;&#xA;" OwnerUserId="5441" LastEditorUserId="12700" LastEditDate="2015-07-22T13:24:17.703" LastActivityDate="2015-07-22T13:24:17.703" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5215" PostTypeId="2" ParentId="5213" CreationDate="2014-11-19T23:00:20.420" Score="12" Body="&lt;p&gt;Two mothers and two daughters but in&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;three different generations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One of them is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;both a mother and a daughter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1664" LastEditorUserId="4204" LastEditDate="2014-11-19T23:09:41.687" LastActivityDate="2014-11-19T23:09:41.687" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5216" PostTypeId="2" ParentId="5135" CreationDate="2014-11-19T23:00:28.733" Score="6" Body="&lt;p&gt;I looked at Clues 1 &amp;amp; 2, 5 &amp;amp; 6, and Object 1. I also took the detour, sigh.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The party murders were committed by:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Peter P Anne. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Evidence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All of the victims at the party were young, but none of the suspects were. He is a teacher, so would have the circumstances to befriend a group of young people. Also his name hints that he would hang out with young people.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There were plates and cups for 6 guests, but only 5 of them had been used. There were 5 victims. The unused plate and cup belonged to  Ann E Mole (animal). Since Peter's last name was Anne, that was probably his nickname among the group. The name being a woman's name was a red herring pointing to Alice W Derland.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The mouse/dog/duck murders were committed by:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Walt D Esnie. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Evidence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Having that name no doubt inspired a life-long obsession with Disney characters. He chose victims with the same birthdates as Disney characters, then dressed them to match. Mickey Mouse (November 18, debut of Steamboat Willie; black shirt, red pants, yellow shoes), Donald Duck (June 9th, his debut in The Wise Little Hen; blue shirt, white pants, white hair), and Goofy (orange shirt, blue pants, a leash, and white gloves; no birthday in my clues).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5500" LastEditorUserId="5500" LastEditDate="2014-11-19T23:20:23.523" LastActivityDate="2014-11-19T23:20:23.523" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5217" PostTypeId="2" ParentId="5203" CreationDate="2014-11-19T23:02:40.947" Score="2" Body="&lt;p&gt;An ideal solution seems to be along the lines of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Quine_(computing)&quot; rel=&quot;nofollow&quot;&gt;quine&lt;/a&gt;, posing a question that contains the Sphinx's challenge so that the two &quot;generate&quot; each other. Unfortunately, embedding her riddle verbatim -- &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If &quot;If this was an answer, what would be the question?&quot; was the answer, what would be the question?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;-- is not allowed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could &lt;em&gt;refer to&lt;/em&gt; the Sphinx's expression: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If what you just asked me was the answer, what would be the question?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Although it seems to me that the correct response to &lt;em&gt;that&lt;/em&gt; is itself again, producing a sort of fixed-point dialogue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possibly better would be this inversion:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If what you just asked me was a question, what would be the answer?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4204" LastEditorUserId="12700" LastEditDate="2015-07-22T13:26:20.820" LastActivityDate="2015-07-22T13:26:20.820" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5218" PostTypeId="2" ParentId="5208" CreationDate="2014-11-19T23:06:39.120" Score="6" Body="&lt;p&gt;Answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DIGITAL ROOT OF SOMETHING.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation: we are all given&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a name&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;at birth, so the secret key is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; your profile's name, namely 'QUYNGUYEN'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Shove this into &lt;a href=&quot;http://sharkysoft.com/vigenere/&quot;&gt;a Vigenere decoder&lt;/a&gt;, with the given ciphertext, and the answer comes out as stated.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2014-11-19T23:06:39.120" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5219" PostTypeId="2" ParentId="5213" CreationDate="2014-11-19T23:22:51.937" Score="1" Body="&lt;p&gt;There are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;4 people in total.&lt;br&gt;&#xA;Three of them wished to be buried in a coffin upon death, but one of them wanted to be cremated upon death.  Cremated remains are not placed in a coffin.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;None are left out&quot; of the local mortuary.&lt;/p&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="22529" LastEditDate="2016-11-22T08:14:02.983" LastActivityDate="2016-11-22T08:14:02.983" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5220" PostTypeId="2" ParentId="5203" CreationDate="2014-11-19T23:38:27.413" Score="59" Body="&lt;p&gt;What kind of pointer do you use to refer to the current object in C++?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/This_(computer_programming)#C.2B.2B&quot;&gt;http://en.wikipedia.org/wiki/This_(computer_programming)#C.2B.2B&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5785" LastActivityDate="2014-11-19T23:38:27.413" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="5222" PostTypeId="2" ParentId="5203" CreationDate="2014-11-20T00:29:57.780" Score="16" Body="&lt;p&gt;I think the key lies in the use of the word &lt;strong&gt;an&lt;/strong&gt;. It's not saying that this is &lt;strong&gt;the&lt;/strong&gt; answer, but one of multiple. Then, you really just have to ask it something open-ended like:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What is an example of a self-referential riddle?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are many such solutions, all with many possible answers, &lt;em&gt;including&lt;/em&gt; the one given. The sphinx would have to let you pass (assuming the sphinx plays fair, of course).&lt;/p&gt;&#xA;" OwnerUserId="1563" LastActivityDate="2014-11-20T00:29:57.780" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5223" PostTypeId="2" ParentId="5203" CreationDate="2014-11-20T00:31:03.040" Score="4" Body="&lt;p&gt;What is the opposite of &quot;that&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://www.merriam-webster.com/thesaurus/this&quot;&gt;Merriam-Webster&lt;/a&gt;, &quot;that&quot; is an antonym for &quot;this&quot;.&lt;/p&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4450" LastEditDate="2014-11-20T01:56:36.377" LastActivityDate="2014-11-20T01:56:36.377" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5224" PostTypeId="2" ParentId="5203" CreationDate="2014-11-20T00:34:15.180" Score="-2" Body="&lt;p&gt;My  guess...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; May I pass through?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And 'this was an answer'...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; refers to how the Sphinx explained you need to solve her riddle to pass through&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5709" LastActivityDate="2014-11-20T00:34:15.180" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5225" PostTypeId="2" ParentId="5199" CreationDate="2014-11-20T01:01:49.767" Score="2" Body="&lt;p&gt;Some other ideas are finding zugzwangs or forced ties with the minimal or maximum number of pieces. Maybe minimal for one side and maximum for the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also give them a couple of pre-set boards and tell them to find the steps taken by either side to ensure a win. That might turn into some sort of branching tree, so you'd have to be careful with the starting positions to ensure both an optimal strategy and not make it too complex.&lt;/p&gt;&#xA;" OwnerUserId="5612" LastActivityDate="2014-11-20T01:01:49.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5227" PostTypeId="2" ParentId="5203" CreationDate="2014-11-20T01:28:30.527" Score="7" Body="&lt;p&gt;Here's my try:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What sort of self-referential question could you ask that requests the question to the answer which is the question you just asked?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A quite literal interpretation of the question, but I think it works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More poetically phrased (in the same way as the Sphinx), we might say it as such:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the answer to this question were to ask for the question it answered, what would the question be?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="12700" LastEditDate="2015-07-22T13:25:57.430" LastActivityDate="2015-07-22T13:25:57.430" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5231" PostTypeId="1" AcceptedAnswerId="5235" CreationDate="2014-11-20T05:24:59.017" Score="9" ViewCount="1043" Body="&lt;p&gt;Suppose you are traveling and come to a fork in the road. One of them leads to something good like a cat shelter and a company handing out free ice cream and money, while the other leads to something terrible like a series of pasta parties you can't join.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The path is guarded by $n&amp;gt;1$ guards. Each one either always tells the truth or tells lies. They all have perfect knowledge of which of them is honest or dishonest and which path goes to which destination. You know that at least one guard is honest and at least one is dishonest, but you don't know which or how the others answer questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a function of $n$, what is the smallest number of questions you must ask to be guaranteed to discover which path to take? To keep the problem mathematically formal, questions should be able to be expressed as Boolean logic expressions, which the guards will then answer with true or false, depending on their dispositions. You solve the puzzle when you can use the guards' answers to know which path to take. You can decide which questions to ask depending on the responses you get, but if an approach has an upper and lower bound for how many questions you must ask, you want to minimize the upper bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the guards know how to evaluate the statement $H(G)$, which is true if guard $G$ is honest, and $I(P)$, which is true if the path $P$ takes you to your destination. If you can think of other statements that make sense for the context of the puzzle, feel free to use them.&lt;/p&gt;&#xA;" OwnerUserId="294" LastEditorUserId="294" LastEditDate="2014-11-20T21:21:42.580" LastActivityDate="2014-11-20T21:44:45.267" Title="Generalized Path Guards" Tags="|mathematics|logical-deduction|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5233" PostTypeId="2" ParentId="5231" CreationDate="2014-11-20T05:41:59.070" Score="1" Body="&lt;p&gt;The minimum number of questions required is 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ask the first guard &quot;If I ask you and the second guard which is the correct path, will your answers be the same?&quot;, then go to the second guard and ask him &quot;If I ask you and the first guard which is the correct path, will your answers be the same?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the responses by both the guards are different, you can confirm that one of them is a truth telling guard (knight) and other is a liar (knave). Now you can ask the normal knight/knave question to get the right path:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ask either of them, &quot;If I ask the other guard which path will lead to the cat shelter, which path will he show me?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then go down the other path than the one the guard pointed to.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now if both the guards answered the same, check what they have answered. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they both answered yes, they both are knights. If they both answered no, both of them are knaves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if both are knights, ask one of them which is the path to cat shelter and go down the path he points to. If both are knaves, ask one of the which is the path to cat shelter take the other path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still 3 questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the final answer is that whatever the value of n may be, the total number of questions needed to find the right path will be 3.&lt;/p&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="4204" LastEditDate="2014-11-20T06:21:49.240" LastActivityDate="2014-11-20T06:21:49.240" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5234" PostTypeId="2" ParentId="5231" CreationDate="2014-11-20T06:07:49.653" Score="6" Body="&lt;p&gt;The classic 1-guard solution works for any number of guards. Pick a guard $G$, and ask him $H(G) \iff I(P_1)$: &quot;Are you honest XNOR is path 1 good?&quot; Then, take path 1 if he says &quot;True&quot; and path 2 if he says &quot;False&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Liars lying and honest guards not lying is equivalent to every guard $G$ XNORing the answers to questions with $H(G)$. By building an XNOR with $H(G)$ into the question we ask, we cancel this out, and the answer we get will always be $I(P_1)$.&lt;/p&gt;&#xA;" OwnerUserId="1549" LastEditorUserId="1549" LastEditDate="2014-11-20T21:44:45.267" LastActivityDate="2014-11-20T21:44:45.267" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="5235" PostTypeId="2" ParentId="5231" CreationDate="2014-11-20T06:08:41.053" Score="8" Body="&lt;p&gt;$1$ question will suffice - for whatever $n$ (given the other constraints). Take guard 1, $g_1$, given that we have $2$ paths $p_1 ,p_2$ and we disregard the way we came from.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ask $g_1$ &quot;(you are a liar $\land$ $p_1$ leads to the cat shelter) $\lor$ (you are a truthteller $\land$ $p_1$ leads to the pasta party)?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation:&lt;/strong&gt;&lt;br&gt;&#xA;There are 4 possible conditions:&lt;br&gt;&#xA;Case 1: $g_1$ is a liar AND $p_1$ leads to the cat shelter. You will get the answer: FALSE&lt;br&gt;&#xA;Case 2: $g_1$ is a truthteller AND $p_1$ leads to the cat shelter. You will get the answer: FALSE.&lt;br&gt;&#xA;Case 3: $g_1$ is a liar AND $p_1$ leads to the pasta party. You will get the answer: TRUE&lt;br&gt;&#xA;Case 4: $g_1$ is a truthteller AND $p_1$ leads to the pasta party. You will get the answer: TRUE&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, if you get the answer TRUE you take $p_2$, if you get FALSE, you take $p_1$.&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-11-20T06:24:14.797" LastActivityDate="2014-11-20T06:24:14.797" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5236" PostTypeId="2" ParentId="5203" CreationDate="2014-11-20T06:54:29.050" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What is the name of the capital city of the first dynasties of ancient Egypt? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or you could ask&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What's the other name for Thinis,the capital city of the first dynasties of ancient Egypt?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The questions above considers the given sentence literally by taking the word &quot;this&quot; to refer to the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Thinis&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Thinis&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="101" LastEditorUserId="101" LastEditDate="2014-11-20T09:24:31.367" LastActivityDate="2014-11-20T09:24:31.367" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5239" PostTypeId="1" AcceptedAnswerId="5248" CreationDate="2014-11-20T08:20:12.297" Score="1" ViewCount="1285" Body="&lt;p&gt;The following text is encrypted with its own key.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ip;g siyyiu el ytaik pl ej ygi dit el auillik =eyg pb zaazleyi jebhiuC&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A hint for those that want it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The entire solution is a syntactically valid English sentence (barring any mistakes from the manual encryption).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It has something to do with your computer. This method is the same as the way most people on here write it, at least for the most part, since my native tongue is not English. There is a 1 to 1 direct match between letters in the solution and in the encrypted text.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How did I encrypt this and what is the solution? &lt;/p&gt;&#xA;" OwnerUserId="2778" LastEditorUserId="97" LastEditDate="2014-11-20T09:39:56.740" LastActivityDate="2014-11-20T10:17:48.160" Title="The key is also the solution: a word puzzle" Tags="|cipher|" AnswerCount="1" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5241" PostTypeId="2" ParentId="5203" CreationDate="2014-11-20T08:46:16.377" Score="-1" Body="&lt;p&gt;First thing that comes to mind:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;What is the logical way you should question if i say this were an answer?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4338" LastActivityDate="2014-11-20T08:46:16.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5242" PostTypeId="1" CreationDate="2014-11-20T08:58:16.807" Score="15" ViewCount="1637" Body="&lt;p&gt;To help people think this out, I've made a web game with increasingly difficult cpu players to allow brainstorming and testing strategies:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;cpu player, &lt;a href=&quot;http://www.testerdump.com/TicTacToe_variant3.html&quot; rel=&quot;noreferrer&quot;&gt;easiest&lt;/a&gt; (version 1 - at least 5-in-a-row is possible against this strategy)&lt;br&gt;&#xA;cpu player, &lt;a href=&quot;http://www.testerdump.com/TicTacToe_variant3b.html&quot; rel=&quot;noreferrer&quot;&gt;easy&lt;/a&gt;    (version 2 - at least 4-in-a-row is possible against this strategy)&lt;br&gt;&#xA;cpu player, &lt;a href=&quot;http://www.testerdump.com/TicTacToe_variant3c.html&quot; rel=&quot;noreferrer&quot;&gt;hard&lt;/a&gt;    (version 3 - I have yet to get 4-in-a-row against this strategy)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This puzzle is about strategy in a variant of Tic-Tac-Toe.  The game differs from the usual version in the following ways:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The board is 16x16&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The two players have different goals&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Your only goal is to get as long a line of pieces in-a-row as possible (horizontal, vertical, or diagonal)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Your opponent's only goal is to limit the line of pieces you can get in-a-row (it does not matter how many pieces in a row the opponent gets)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;One final twist: for every move you make, your opponent gets to move twice&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;With perfect play can you get four in-a-row?  Or can you prove that it is not possible?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;NOTE: I am not a web developer, so I apologize if the interactive version doesn't work for you.  The goal is to figure out how to guarantee four in-a-row or prove it is impossible &lt;em&gt;for the game described above&lt;/em&gt; please let me know if the interactive game seems to deviate from that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The version 3.0 cpu player does not try to think any moves ahead.  It just calculates a heuristic score for one of 24 spots around the player's last move, and moves in the best available spots allowed.  I set it up this way, to make it easier to reason about.  If it turns out that this strategy is sufficient to prevent 4-in-a-row, then that solves it and I find it is easier to reason against a simple concrete opponent.  Alternatively if one can find how to get 4-in-a-row against this computer player, it may allow for a strategy to guarantee 4-in-a-row.  If you'd like to know more about the cpu ver3 strategy, I've tried to comment the code to explain, so just read the cpu player's javascript.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, the goal of the puzzle isn't to beat the cpu, but playing against the cpu may help you think out ideas and try strategies.  That being said, if you manage to beat the hardest cpu player, please post your move log so we can learn from your amazing strategy skills.  Good luck!&lt;/p&gt;&#xA;" OwnerUserId="4351" LastEditorUserId="4351" LastEditDate="2014-12-04T10:58:03.217" LastActivityDate="2017-09-23T01:33:43.587" Title="Tic-Tac-Toe variant" Tags="|strategy|tic-tac-toe|pencil-and-paper-games|" AnswerCount="2" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="5243" PostTypeId="2" ParentId="5203" CreationDate="2014-11-20T09:00:15.187" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;They say the answer of The Question of Life, the Universe and Everything is 42.  What do you think?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It looks like an anchronism, but if you consider the Sphynx as immortal, it cannot be.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastActivityDate="2014-11-20T09:00:15.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5244" PostTypeId="2" ParentId="5203" CreationDate="2014-11-20T09:04:47.040" Score="0" Body="&lt;p&gt;I would answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well, what are you going to ask me?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the explanation is, the dialogue must make sense played in both directions. For example, the sphinx question first and my answer after have to make sense, as well as my answer first and then the sphynx question.&lt;/p&gt;&#xA;" OwnerUserId="5283" LastEditorUserId="5283" LastEditDate="2014-11-20T09:24:15.400" LastActivityDate="2014-11-20T09:24:15.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5245" PostTypeId="2" ParentId="5242" CreationDate="2014-11-20T09:27:07.420" Score="5" Body="&lt;p&gt;&lt;strong&gt;Note: the contents of this answer are &lt;a href=&quot;http://chat.stackexchange.com/transcript/message/18696829#18696829&quot;&gt;not actually correct&lt;/a&gt;; however, I've been asked to undelete this because it contains content that might be useful.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;I'm still trying to work through what I missed, but in essence, I think it comes down to the assumption that Red plays pieces that are adjacent to existing pieces, or do not otherwise interfere with them. I'm not totally sure, though, and I'd definitely be interested in an explanation.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;With perfect playing, it is not possible to get four-in-a-row.&lt;/strong&gt; It is at most possible to get three-in-a-row.&lt;/p&gt;&#xA;&lt;p&gt;Let's let Player 1 be the player who's trying to achieve a high length using red squares, and Player 2 be the player who's trying to restrict the max length using blue squares. This follows your web example, but I just wanted to establish clear terminology.&lt;/p&gt;&#xA;&lt;p&gt;It actually makes absolutely no difference what player 1 does in this situation, since it's not possible for player 1 to achieve anything more than a length of 2. Here's what you have to do, as player 2:&lt;/p&gt;&#xA;&lt;p&gt;Whenever Player 1 makes a move look at all the adjacent squares.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First, check to see if Red could form a chain of four anywhere on their next move. If such spots exist, fill them in. (There will never be more than two of these at at time - see below.)&lt;/li&gt;&#xA;&lt;li&gt;Then, check to see if Red could form a chain of three anywhere on their next move. If any such spots exist, fill them in.&lt;/li&gt;&#xA;&lt;li&gt;If neither of the above are true, place Blue squares anywhere next to the square that red just placed.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;It's that simple.&lt;/p&gt;&#xA;&lt;p&gt;How does this work in practice? I'll walk through a few steps.&lt;/p&gt;&#xA;&lt;p&gt;Step 1: Red makes the first move.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xOZg4.png&quot; alt=&quot;enter image description here&quot; /&gt; (becomes) &lt;img src=&quot;https://i.stack.imgur.com/dqoZj.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Blue sees that there's a red square, and there are no places where Red could go to create a chain of three. So, per instructions, Blue fills in two squares on opposite sides of the red square.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Step 2: Red tries to break away!&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wTQb3.png&quot; alt=&quot;enter image description here&quot; /&gt; (becomes) &lt;img src=&quot;https://i.stack.imgur.com/S5jH8.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Blue sees that Red has tried to break away diagonally. Since there are two spots opposite each other where Red could break away, Blue fills those in.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Step 3: Red creates more escape routes!&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CedC7.png&quot; alt=&quot;enter image description here&quot; /&gt; (becomes) &lt;img src=&quot;https://i.stack.imgur.com/yuKQx.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Blue here sees four spaces which could generate chains of three. Since two of them are opposite squares, Blue picks a random pair and fills them in. The result is the same.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Step 4: Red creates a chain of three!&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ddBuV.png&quot; alt=&quot;enter image description here&quot; /&gt; (becomes) &lt;img src=&quot;https://i.stack.imgur.com/tWxJH.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Blue here sees two spots where a four-in-a-row could happen. Blue obviously fills them in.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Why does this work?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Here's an easy explanation that's very intuitive: &lt;strong&gt;any time Red could make a 4-chain, they're cut off at both ends.&lt;/strong&gt; This is actually not simple to demonstrate, though.&lt;/p&gt;&#xA;&lt;p&gt;What we need to show is that Red can generate at most &lt;em&gt;two&lt;/em&gt; openings for a fourth piece to go into at a given time. If there are only two open slots, Blue will fill them on the next turn and the chain will die at length 3.&lt;/p&gt;&#xA;&lt;p&gt;We know that Red can produce 2-chains, and that Red can produce 3-chains from them. Let's take a closer look at why Red can even do this, and classify it a little. &lt;strong&gt;In order to create a 3-chain, Red needs to unify two 2-chains at a common point.&lt;/strong&gt; (I'll call these &lt;em&gt;linked 2-chains&lt;/em&gt;.) Each individual 2-chain only has two openings to expand into 3-chain. However, if we conjoin two of them at a common point, then the resulting group is given not two, but &lt;em&gt;four&lt;/em&gt; openings.&lt;/p&gt;&#xA;&lt;p&gt;In other words, the fundamental property is: &lt;strong&gt;a single chain of length two or greater can only have two degrees of movement.&lt;/strong&gt; (A chain of length 1 can have up to &lt;em&gt;eight&lt;/em&gt;, by comparison.) Because the chain must go in a certain direction, at most two possibilities exist: continuing off either end. When we stick them together into linked chains, however, we can &lt;em&gt;at most&lt;/em&gt; add those degrees of movement together, and adding two 2-chains' movements together gets us a 2-chain with movement 4.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ePhaA.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Here, Red only has two openings to create a new chain. Both of these will be cut off when Blue moves next. However, Red can force Blue into a helpful chain:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SJq1N.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;How did Red do this? They went first in Cb. Blue blocked some random squares off, then Red went in Cc. Blue necessarily blocked off this chain of two from becoming a chain of three. However, at this point, Red goes in Bc. Bc gives red two unrestricted pairs, (Cb-Bc) and (Cc-Bc). Blue can only stop one of these, so the other will be able to expand into a chain of three.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Why is this important?&lt;/strong&gt; In order to create a 4-chain, we need to create a linked 3-chain. Each 3-chain only has two degrees of movement, but if we can add them together at once, we can create a linked 3-chain that has at most four degrees of movement. This means that blue can't kill all degrees of movement in the next turn, and Red can play to win.&lt;/p&gt;&#xA;&lt;p&gt;Easy, right? Turns out it's not possible.&lt;/p&gt;&#xA;&lt;p&gt;The reason is actually pretty simple: you need to create two 2-chains that are adjacent to each other. You'll get to a position comparable to this:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/D40rCs.png&quot; alt=&quot;enter image description here&quot; /&gt; (becomes) &lt;img src=&quot;https://i.stack.imgur.com/206yUs.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Blue sees Red place a tile somewhere totally irrelevant. Following the steps in logical order, Blue spots a space where Red could, in theory, create a 4-chain on the next turn. Blue fills that in. Blue then spots a space where red could, in theory, create a 3-chain on the next turn, so Blue fills that in, too.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;In other words, you've created a 2-chain. That's fine and good, but now you need to make &lt;em&gt;another one&lt;/em&gt;, and doing so consumes six degrees of movement. It takes three moves. Your process creates four degrees of movement, so there's a net loss of two degrees of movement. Where do those two degrees come from? The chain you're trying to link to, on the first turn you try to link to it!&lt;/p&gt;&#xA;&lt;p&gt;It's theoretically possible, if you already had two existing 2-chains and could link them at a unifying spot. But it's not possible to get to that position; if you tried to create the second 2-chain you need, the first one will be cut off and die.&lt;/p&gt;&#xA;&lt;p&gt;Thus, it's not possible to get four in a row if Blue plays perfectly. QED.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;tl;dr? The maximum-length chain that Red can create is &lt;em&gt;three&lt;/em&gt;. There's no possible way to generate a longer-length chain.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorUserId="-1" LastEditorDisplayName="user20" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-11-22T07:10:25.653" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="5248" PostTypeId="2" ParentId="5239" CreationDate="2014-11-20T09:34:43.890" Score="3" Body="&lt;p&gt;I manually deciphered it as a hinted symmetrical substitution cipher&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each letter is typed as if the key is pressed with an opposite finger.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;remark:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mark that the &quot;.&quot; would be a &quot;:&quot; on most(?) keyboards, if I were to guess, your keyboard has the &quot;.&quot; and the &quot;;&quot; on the same key. Mine has an &quot;:&quot; when I press SHIFT+&quot;;&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, after checking images of an AZERTY keyboard, my suspicion is confirmed. Interestingly, those who see the &quot;trick&quot; may be put off by some letters being in non-standard positions for QWERTY-users.&lt;/p&gt;&#xA;" OwnerUserId="3903" LastEditorUserId="3903" LastEditDate="2014-11-20T10:17:48.160" LastActivityDate="2014-11-20T10:17:48.160" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="5249" PostTypeId="1" AcceptedAnswerId="5250" CreationDate="2014-11-20T09:38:43.163" Score="38" ViewCount="4305" Body="&lt;p&gt;This is about the objective interpretations/assessment/quantification of riddles. Let us define a riddle as a type of puzzle relying on the usage of only language and context. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using precise English as a formal (non-opinion based) tool, and examples and sources like dictionaries, grammar sources (ELU.SE maybe), and perhaps literature (e.g. The Hobbit), or even possibly sources from this SE, explain how to construct and what defines a &quot;solid&quot; Riddle? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I might expect that the answer might contain something like: a &quot;solid&quot; Riddle should have one and only one interpretation which coherently addresses each line/clue contained in the Riddle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Related:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/105/how-do-you-come-up-with-a-good-classic-rhyme-riddle&quot;&gt;Question about writing rhyming riddles.&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/4574/tips-for-solving-poem-riddles?lq=1&quot;&gt;Tips for solving riddles.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2016-09-09T17:56:59.533" Title="How do I write a &quot;solid&quot; Riddle?" Tags="|riddle|puzzle-creation|" AnswerCount="3" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="5250" PostTypeId="2" ParentId="5249" CreationDate="2014-11-20T10:11:28.380" Score="28" Body="&lt;p&gt;As you say in the question: there should be exactly one interpretation which fits each clue given. However hard the riddle a supermajority of fluent English (or whatever your target language may be) speakers must, when presented with the solution, consider it the only correct one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practise, this is going to mean finding the unique qualities of whatever your solution is, then using those as clues. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g. Elephant. It's a huge, grey, living creature. It has extremely thick limbs, a trunk and large ears. None of these in isolation is unique, but together they only describe an elephant. Of course, this is not yet a riddle. You will then need to obfuscate the clues (by metaphor, vague description, literary allusion) throwing out any which cannot be disguised sufficiently. Hopefully at that point you will still have enough clues to uniquely define your solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for the elephant example, large limbs is too straightforward. Perhaps we could compare them to other large, vertical cylinders. Like pillars, or trees. We could refer to the trunk by function (hose, snorkel) or by alternative meaning of 'trunk' (trees, luggage). Both give us options for trees, so let's go with that. &quot;I walk on trees, I drink through a tree&quot;. Elephants are &lt;strong&gt;heavy&lt;/strong&gt;, but they are also &lt;strong&gt;light&lt;/strong&gt; grey, so we could have &quot;heavier than a boulder, lighter than a raven's feather&quot;. And so on... &lt;/p&gt;&#xA;" OwnerUserId="3987" LastEditorUserId="3987" LastEditDate="2014-11-20T10:39:55.517" LastActivityDate="2014-11-20T10:39:55.517" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5259" PostTypeId="2" ParentId="5249" CreationDate="2014-11-20T12:57:41.460" Score="13" Body="&lt;p&gt;What for me is the most important property in a &quot;solid&quot; riddle - &quot;the twist&quot;. These are small confusing details in a riddle. Let me give you an example (one of my favourite riddles in Neil Gaiman's &quot;The books of magic&quot;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;When there is a fire in me, then I am still cold; When I own your true love’s face then you will not see me; To all things I give no more than I am given; In time I may have all things, and yet I can keep nothing.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a perfect example, because the riddle contains three twists. In the first sentence, the twist is the second part of the sentence (&quot;then I am still cold&quot;). Generally, there is a simple test for twists - the question &quot;How the hell is this possible?&quot;. If you find yourself asking this question at some point, then you've just encountered a twist.&lt;/p&gt;&#xA;" OwnerUserId="5283" LastActivityDate="2014-11-20T12:57:41.460" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="5261" PostTypeId="1" AcceptedAnswerId="8318" CreationDate="2014-11-20T13:44:27.633" Score="3" ViewCount="1649" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The eleventh day of spring 26&lt;br&gt;&#xA;  The twelfth night of summer 74&lt;br&gt;&#xA;  The first night of autumn 80&lt;br&gt;&#xA;  The second day of winter 86&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have hidden 4 song titles in the sentences above. What are the song titles, and who sings them?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;each line represents 1 song title&lt;/li&gt;&#xA;&lt;li&gt;the songs are about something that have similarity with time (daylight, seasons, weeks, etc.)&lt;/li&gt;&#xA;&lt;li&gt;the songs are sung by different people&lt;/li&gt;&#xA;&lt;li&gt;every song and singer listed first if you google it&lt;/li&gt;&#xA;&lt;li&gt;based on Wikipedia&lt;/li&gt;&#xA;&lt;li&gt;I am a magician, perhaps by knowing this you can more easily resolve this puzzle.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5800" LastEditorUserId="22922" LastEditDate="2016-09-15T08:20:59.393" LastActivityDate="2016-09-15T08:20:59.393" Title="The title of a song" Tags="|riddle|knowledge|" AnswerCount="2" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="5262" PostTypeId="2" ParentId="5231" CreationDate="2014-11-20T14:49:27.803" Score="-2" Body="&lt;p&gt;You only have to ask one question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ask either Guard &quot;If I were to ask the other Guard which is the best path to take what would be his answer?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You then follow the opposite path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The logic is one lies one tells the truth.  So if you ask the question to the liar he will tell you the opposite of the true answer from the other guard, and if you ask the true guard he will tell you the answer from the liar which will be the wrong path.&lt;/p&gt;&#xA;" OwnerUserId="5379" LastActivityDate="2014-11-20T14:49:27.803" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5263" PostTypeId="2" ParentId="4668" CreationDate="2014-11-20T14:50:35.390" Score="8" Body="&lt;p&gt;The answer is...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 18&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you convert the ASCII character to its decimal value and add the digits&lt;br&gt;&#xA;&lt;br&gt;&#xA; 1 =&gt; 49 =&gt; 4+9=13&lt;br&gt;&#xA; 3 =&gt; 51 =&gt; 5+1=6&lt;br&gt;&#xA; 90 =&gt; 57, 48 =&gt; 5+7+4+8=24&lt;br&gt;&#xA; 91 =&gt; 57, 49 =&gt; 5+7+4+9=25&lt;br&gt;&#xA; 12 =&gt; 49, 50 =&gt; 4+9+5+0=18 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3922" LastEditorUserId="27067" LastEditDate="2016-08-01T12:59:47.247" LastActivityDate="2016-08-01T12:59:47.247" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5264" PostTypeId="2" ParentId="4668" CreationDate="2014-11-20T15:02:11.803" Score="0" Body="&lt;p&gt;I think the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;18&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit- okay, someone else posted while I was trying to log in. I had the right answer but was using a completely different method that was ultimately not consistent. Oh well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 2-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found the digital root of the original number (ie, added all the digits) and compared them with the answers. I noted that two numbers both were 9 (9 and 90) and that their answers were 12 and 24. The difference between the two is the number of initial digits present, and noting that the difference between the answers is 12, I applied the same logic to the other initial answer that had 3 (which is the same as sum of the digits of the 12 that needs a response), added 12 to the 6 that was the answer to 3, and got 18. This doesn't explain anything else in the problem though, it's a complete coincidence that I got the right answer.&lt;/p&gt;&#xA;" OwnerUserId="5080" LastEditorUserId="5080" LastEditDate="2014-11-20T15:19:28.740" LastActivityDate="2014-11-20T15:19:28.740" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5266" PostTypeId="2" ParentId="5141" CreationDate="2014-11-20T15:49:53.473" Score="5" Body="&lt;p&gt;The answer has to be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sarah Palin, in a parallel world where she became President of the United States.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First let's cover the basics. The “woman with the wheel” is the person who drives the country, i.e. the person at the very top. She's the head of the armed forces (“high up in the army”). We're told that she “fires the cannon”, i.e. she launches the country's weapons by herself (or at least she presses the button — other people load the cannon for her); this must refer to the atomic bomb, as other weapons are fired entirely by underlings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far we have the gender and position, which still leaves quite a few possibilities. We have to use the rest of the riddle to figure out who it is. The start of the riddle is a big clue: it's an “abstraction”, so this is about a fictional world! Furthermore, it's a “nonsensical” one, i.e. a counterfactual one: an alternate world where history proceeded differently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a woman who's “fast”. This could be interpreted in several ways. A common meaning is one who sleeps around, but among politicians, sleeping around is far more common with men than with women. So “fast” must refer to another meaning of the word — tenacious, resistant, not fading with age. I can think of several women who came close to becoming head of state and who fit this description, so we'll need the rest of the riddle to whittle the field down to one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Only in reverse gear” means that she is by no means a progressive, i.e. she is a staunch conservative. This rules out, in particular, Hillary Clinton and Ségolène Royal. As for Sarah Palin, she fits the clues like a glove. She was a basketball &lt;a href=&quot;http://en.wikipedia.org/wiki/Point_guard&quot; rel=&quot;nofollow&quot;&gt;point guard&lt;/a&gt;, which is a clever double meaning for the second line — the point guard passes the ball (fires the cannon) to other players, and decides which player to pass it to based on tactical considerations (hence acting as a wheel on which the ball turns).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the final line, “terrorist” means not one who is directly violent, but one who supports a policy of scaremongering, as was the case in the US following 9-11. “How to prove it?” Her opponents were unsuccessful in demonstrating to the American public that her policies were overly strong, which is why she got elected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The injunction to “pray” and “find the word” in the title alludes to her being Christian, which is a religion of the Word.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The “abstract nonsense” hint refers to category theory. Why point non-mathematicians at category theory? They'll have no idea what it means. “Category theory” here is thus not about the algebraic concept, but rather concerns the everyday meaning of the word “category” — the puzzle looks like math, but that's only for the surface reading, and the answer lies in another category.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I do remain puzzled by the question calling for a &lt;em&gt;word&lt;/em&gt; rather than a name. Could this be a nickname? I can't find one that matches the clues.&lt;/p&gt;&#xA;" OwnerUserId="71" LastEditorUserId="71" LastEditDate="2014-11-20T16:46:30.997" LastActivityDate="2014-11-20T16:46:30.997" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="5267" PostTypeId="1" AcceptedAnswerId="5268" CreationDate="2014-11-20T15:59:54.960" Score="41" ViewCount="19876" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is greater than God, worse than evil,&lt;br&gt;&#xA;  The poor have it, the rich require it,&lt;br&gt;&#xA;  And if you eat it, you die?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is it?&lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="20274" LastEditDate="2016-05-31T08:40:50.920" LastActivityDate="2023-03-08T23:58:53.017" Title="Oh, it is greater than God!" Tags="|riddle|" AnswerCount="4" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="5268" PostTypeId="2" ParentId="5267" CreationDate="2014-11-20T16:00:33.573" Score="60" Body="&lt;p&gt;Too easy! I've seen something like this before.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nothing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation isn't really necessary - just check all the clues - but I'll give some anyway:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nothing is greater than God; nothing is worse than evil; the poor have nothing; the rich require nothing; if you eat nothing, you die.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2014-11-20T16:17:50.373" LastActivityDate="2014-11-20T16:17:50.373" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5269" PostTypeId="2" ParentId="5267" CreationDate="2014-11-20T16:51:12.267" Score="24" Body="&lt;p&gt;Although the correct answer has been posted, I think there is room for another answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Something&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To support this claim, I offer that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;What is greater than God&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you assume the Mormon conception of God, there are multiple Gods recursively back forever, so &quot;God&quot; has his own &quot;God&quot; which has his one &quot;God&quot;, forever.  By this definition, there is always something greater than God, which is the God of that God.&#xA; Actually you don't even have to assume a Mormon God for this to work, it can be proved mathematically.  If God is infinite, then what is greater than infinity than a larger infinity? (e.g. aleph one versus aleph null)&#xA; Also, as Psychemaster pointed out, if you are an atheist then from your perspective something (or anything) is greater than God, because God doesn't exist.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;worse than evil&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Similar to how there are things worse than death, there is also something worse than evil.  For instance, simple incompetence can often be worse than evil (&quot;never attribute to malice...&quot;)  Natural suffering (disease, old age, etc. not caused by evil) can be worse than evil to the person suffering it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;the poor have it&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No matter how poor a person is, they will always have something, whether it is their health, or happiness, or simply life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;the rich require it&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A rich person will always require something (water, food, shelter, etc.), their being rich simply means they will be able to acquire it without any problem.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;if you eat it, you die?&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Living creatures are a paradox of sort.  You need the sun to create vitamin D, but the sun can also give you skin cancer.  If you eat something your body will be nourished, but your body will age (production of free radicals, telomere exhaustion, etc.) and you will eventually die. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2718" LastEditorUserId="57523" LastEditDate="2020-03-07T16:19:25.707" LastActivityDate="2020-03-07T16:19:25.707" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="5270" PostTypeId="1" CreationDate="2014-11-20T16:52:04.940" Score="15" ViewCount="1819" Body="&lt;p&gt;The body was lying naked on the floor of an apartment living room. Crime scene investigators and forensics were inspecting the body and the nearby area. You, detective Randy, opened the door and entered the room. After greeting your colleagues you have joined them in the investigation of the body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You noticed that something was carved into the forehead of the body with a knife or blade. It appears to say &lt;strong&gt;&quot;LaExO&quot;&lt;/strong&gt;. You took a photo of it with your personal camera so that you can use it in your investigation and to allow yourself have a closer look later. You wonder: &lt;strong&gt;&quot;What this will be? Will it be an encoded message, or some words in a foreign language?&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your thoughts are interrupted. &lt;strong&gt;&quot;Hey.. Look at this..&quot;&lt;/strong&gt;, said your colleague. She removed a crushed and rolled paper from the mouth of the dead body using tweezers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You put your gloves on and unfolded the paper to see a printed message/word on it. You read it as &lt;strong&gt;&quot;dabel&quot;&lt;/strong&gt;. What's this? Another message? Is the killer trying to convey some message to the world? You thought, &lt;strong&gt;&quot;This is either a pre-planned murder, or the killer must be a psycho... maybe a serial killer.&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You took a photo of this note with your camera. You and your other colleagues inspected the entire room and the body, but didn't find any other evidence. So you decided to wrap up the inspection and made arrangements to send the body and evidence to the forensic lab for further research.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your office you were sitting with the printouts of photos of the evidence you have taken. You were going through both photos again and again trying to find out what the words mean.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Photo 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vof42.jpg&quot; alt=&quot;photo of victim&amp;#39;s forehead&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Photo 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KwRhf.jpg&quot; alt=&quot;photo of paper in victim&amp;#39;s mouth&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;At that time, your assistant knocked on your door and entered your room with the report from the forensic lab. He placed it on your table and left the room. You started going through the report. The report tells you:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The cause of death was suffocation, possibly with a pillow or cushion.&lt;/li&gt;&#xA;&lt;li&gt;The estimated time of death was between 11pm and 11:30pm. There was no indication of a fight.&lt;/li&gt;&#xA;&lt;li&gt;The victim had sex shortly before her death; they found traces of lubrication used in condoms.&lt;/li&gt;&#xA;&lt;li&gt;The victim was not sedated before sex, as they haven't found any traces of drugs or poisons in her body.&lt;/li&gt;&#xA;&lt;li&gt;Because there was no sign of a fight, either the sex was consensual, or the sex occurred after her death.&lt;/li&gt;&#xA;&lt;li&gt;They didn't find any suspicious fingerprints on the body or in the house other than the victim's and her husband's.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Also from the report of local police officers, you came to know the following details:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The victim's apartment is part of a 3 apartment building and only those who are living there have the key to the main entrance. It is not easy for an outsider to enter the apartment building and murder someone.&lt;br /&gt;&lt;br /&gt; &#xA;  &lt;strong&gt;Victim's Name:&lt;/strong&gt; Riyannah James&lt;br /&gt;&#xA;  &lt;strong&gt;Age:&lt;/strong&gt; 28&lt;br /&gt;&#xA;  &lt;strong&gt;Description:&lt;/strong&gt; Riyannah was a software engineer and has been working for a leading IT firm for the past 6 years. She married her colleague, Mr. James Hopes, 2 years ago after 3 years of dating. Their relationship had been going through a difficult patch for the past few days because of some recent happenings.&lt;br /&gt;&lt;br /&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;h2&gt;Suspects&lt;/h2&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Suspect 1:&lt;/strong&gt; Jakub Tomas&lt;br /&gt;&#xA;  &lt;strong&gt;Age:&lt;/strong&gt; 32&lt;br /&gt;&#xA;  &lt;strong&gt;Description:&lt;/strong&gt; Victim's neighbor. Had issues with victim's husband, James Hopes. Jakub caught his wife Eliska Jakub cheating on him with James Hopes. Jakub and James ended up in a fight and Jakub threatened that James is &quot;going to pay for this&quot;. These incidents happened just a few days ago. Jakub was uneducated (school dropout) and migrated to the United States 4 years ago from The Czech Republic. He makes a living by organizing Black Masses and he is the leader of a group of Satan Worshipers in the area.&lt;br /&gt;&#xA;  &lt;strong&gt;Alibi:&lt;/strong&gt; He claimed that he was conducting a Black Mass at an unspecified location. He refused to reveal the location or more details as he doesn't want to expose others who participated in the Black Mass.&lt;br /&gt;&lt;br /&gt;&#xA;  &lt;strong&gt;Suspect 2:&lt;/strong&gt; James Hopes&lt;br /&gt;&#xA;  &lt;strong&gt;Age:&lt;/strong&gt; 29&lt;br /&gt;&#xA;  &lt;strong&gt;Description:&lt;/strong&gt; James Hopes was married to the victim for two years and was reportedly in a good relationship until he was caught having an affair with his neighbor's wife, Eliska Jakub, last week. James and Jakub became enemies after this incident. James claimed that he had talked with his wife and convinced her that he will be faithful to her forever, and that she forgave him. Some of his and the victim's friends confirmed this statement. &lt;br /&gt;&#xA;  &lt;strong&gt;Alibi:&lt;/strong&gt; He claimed that he bought liquor from a bar on that night and got heavily drunk and slept in his car somewhere. The bartender and security confirmed that he came to the bar by 9:40pm and returned by 10pm after buying two bottles of rum. There is no evidence regarding where he stayed later that night.&lt;br /&gt;&lt;br /&gt;&#xA;  &lt;strong&gt;Suspect 3:&lt;/strong&gt; Eliska Jakub&lt;br /&gt;&#xA;  &lt;strong&gt;Age:&lt;/strong&gt; 30&#xA;  &lt;strong&gt;Description:&lt;/strong&gt; Married to Jakub Tomas for 5 years and likes to live a fancy life. She was not a Worshiper of Satan like her husband. She had recently started a relationship with the victim's husband, a possible motive to kill her.&lt;br /&gt;&#xA;  &lt;strong&gt;Alibi:&lt;/strong&gt; NONE&lt;br /&gt;&lt;br /&gt;&#xA;  &lt;strong&gt;Suspect 4:&lt;/strong&gt; Abel Johns&lt;br /&gt;&#xA;  &lt;strong&gt;Age:&lt;/strong&gt; 22&lt;br /&gt;&#xA;  &lt;strong&gt;Description:&lt;/strong&gt; Abel and his friend, Ijaz Ahmad, are two bachelor friends working toward their PhD. They are living in the same apartment building as the victim. Abel is a strong catholic. He had many affairs at college and he likes to flirt. He and Ijaz were good friends with the victim.&lt;br /&gt;&#xA;  &lt;strong&gt;Alibi:&lt;/strong&gt; He and Ijaz were at a friend's house; that friend confirmed this. (We can't confirm whether the friend is lying or not.)&lt;br /&gt;&lt;br /&gt;&#xA;  &lt;strong&gt;Suspect 5&lt;/strong&gt;: Ijaz Ahmad&lt;br /&gt;&#xA;  &lt;strong&gt;Age:&lt;/strong&gt; 23&lt;br /&gt;&#xA;  &lt;strong&gt;Description:&lt;/strong&gt; Ijaz is a Muslim, and described as forward-thinking too. He and Abel were good friends with the victim.&lt;br /&gt;&#xA;  &lt;strong&gt;Alibi:&lt;/strong&gt; He and Abel were at a friend's house; that friend confirmed this. (We can't confirm whether the friend is lying or not.)&lt;br /&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the security camera (cctv camera) placed at the main entrance hall of the apartment building you have determined the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Ijaz and Abel left the apartment at 7pm&lt;/strong&gt;&lt;br /&gt;&#xA;  &lt;strong&gt;James Hopes left the apartment at 9pm&lt;/strong&gt;&lt;br /&gt;&#xA;  &lt;strong&gt;Jakub Tomas left the apartment at 9:45pm&lt;/strong&gt;&lt;br /&gt;&#xA;  &lt;strong&gt;A man wearing a hood (his face was not clear) entered the apartment at 10:47pm and exited at 11:48pm&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Photo of hooded man&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z3OjR.jpg&quot; alt=&quot;Photo of hooded man&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it's up to you, Detective Randy, to find the killer and arrest him/her.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Only the persons described in the description are related with the crime. Clues and hints are straightforward and one of the suspects is the killer. Randy is a dedicated detective and detests crime, so he definitely has no part in the murder. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Note: This is just a detective puzzle, not a real murder case or crime. So the clues, deductions and assumptions made based on the above description will be sufficient to arrest the murderer/killer.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Additional Note: I will accept the correct answer (if any) on 22nd November, Morning IST. If there is no correct answer, I will give the bounty to the most appropriate answer, and will answer my self.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Hooded man image was taken from the following source &quot;&lt;a href=&quot;http://www.wisbechstandard.co.uk/polopoly_fs/1.3817812!/image/image.jpg_gen/derivatives/landscape_490/image.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.wisbechstandard.co.uk/polopoly_fs/1.3817812!/image/image.jpg_gen/derivatives/landscape_490/image.jpg&lt;/a&gt;&quot; &#xA;I just choose a random image from google.&lt;/p&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="2730" LastEditDate="2018-10-10T14:06:45.153" LastActivityDate="2018-10-10T14:06:45.153" Title="The Sinister Killer" Tags="|story|mystery|" AnswerCount="2" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="5271" PostTypeId="2" ParentId="5100" CreationDate="2014-11-20T16:56:14.047" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/YrjEh.png&quot; alt=&quot;result&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;this one was good :))) it took me some time and a bunch of paper sheets :D&lt;/p&gt;&#xA;" OwnerUserId="5147" LastActivityDate="2014-11-20T16:56:14.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5272" PostTypeId="1" AcceptedAnswerId="5279" CreationDate="2014-11-20T17:01:28.477" Score="13" ViewCount="1522" Body="&lt;p&gt;Is this about all the heroin? After everything? And this is how you repay me? I don't just mean keeping you clothed and respectable looking. I don't just mean saving your life. I thought we were close. I've been firmly with you everywhere you go. Even my words, you ass! Everything coming out of my mouth, coming from my tongue reflects you and what you bring me to experience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you say you don't need me? Well, I don't need you. Go off, have your adventures, but remember this conversation when you come crawling back. And don't until you lose some weight. I'm not ending up at my limit again for some hot mess like you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Figure out this lonely heart's name so you can make them feel a little better and perhaps engage in a rebound romance.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify, the lonely heart you're trying to identify is the speaker.&lt;/p&gt;&#xA;" OwnerUserId="5626" LastEditorUserId="5626" LastEditDate="2014-12-08T14:24:55.360" LastActivityDate="2016-06-07T11:46:30.633" Title="A Lover Spurned - A broken heart cries, waiting for your love. Figure out their name so that you can engage in a little rebound romance" Tags="|riddle|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5273" PostTypeId="2" ParentId="5124" CreationDate="2014-11-20T17:09:21.250" Score="9" Body="&lt;p&gt;Well, I AM a giddy goat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a fan of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The IT Crowd,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I knew there was some connection but couldn't flipping put it together, until I realized&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let the countdown begin!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is a reference to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.imdb.com/title/tt1676563/&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;The Final Countdown&lt;/em&gt;&lt;/a&gt; (Season 4 Episode 2), in which (&lt;a href=&quot;http://en.wikipedia.org/wiki/Countdown_%28game_show%29#Humour&quot; rel=&quot;noreferrer&quot;&gt;as Wikipedia describes it&lt;/a&gt;) &quot;Moss stuns everyone by declaring that the 9 letter string TNETENNBA is in fact a word.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Obviously, that makes the key:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://www.youtube.com/watch?v=lsFAokXCxTI&quot; rel=&quot;noreferrer&quot;&gt;TNETENNBA&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the deciphered text:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DRINK MILK AND KICK ASS (from a Moss quote from the same episode: &quot;&lt;a href=&quot;https://www.youtube.com/watch?v=SwvCX94_EWI&quot; rel=&quot;noreferrer&quot;&gt;I came here to drink milk and kick ass. And I've just finished my milk&lt;/a&gt;.&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I regret not asking if the 9-letter key could be used in a sentence.&lt;/p&gt;&#xA;" OwnerUserId="5340" LastEditorUserId="5340" LastEditDate="2014-11-20T19:27:47.877" LastActivityDate="2014-11-20T19:27:47.877" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5276" PostTypeId="2" ParentId="5272" CreationDate="2014-11-20T17:16:57.680" Score="0" Body="&lt;p&gt;My guess is a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;shoe&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If only because they have a mouth, most notably a tongue, and the &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;wearer brings it to its &quot;experiences&quot; by walking, and can possibly outgrow it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, they keep people clothed and respectable looking enough, some places forbid service without having one or the other? The heroin bit throws me off, however.&lt;/p&gt;&#xA;" OwnerUserId="5441" LastActivityDate="2014-11-20T17:16:57.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5278" PostTypeId="2" ParentId="5272" CreationDate="2014-11-20T17:22:16.853" Score="-3" Body="&lt;p&gt;Heroin: &lt;strong&gt;H&lt;/strong&gt;, after everything: at the end of the word.&lt;br&gt;&#xA;You ass: &lt;strong&gt;U S&lt;/strong&gt;.&lt;br&gt;&#xA;To experience -&gt; to be: &lt;strong&gt;B&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bush&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;who let the national debt reach unprecedented levels (= at my limit again)? &lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditDate="2016-06-07T11:46:30.633" LastActivityDate="2016-06-07T11:46:30.633" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5279" PostTypeId="2" ParentId="5272" CreationDate="2014-11-20T17:35:41.023" Score="12" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A belt&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is this about all the heroin?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Heroin refers to belting an arm before shooting up.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't just mean keeping you clothed and respectable looking.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clothed and respectable as belts suspend pants and are classy as can be.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't just mean saving your life.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A (security) belt can save your life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Everything ... from my tongue reflects you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;em&gt;toungue&lt;/em&gt; of a belt is usually metallic and reflective.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now you say you don't need me? ... remember this conversation when you come crawling back. And don't until you lose some weight.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The belt is no longer needed because the wearer has gained weight and their pants now stay on. Notice that the &quot;and don't until&quot; bit is written in such a way that it could mean that he/she won't decide to come back till there is weight loss and the belt is needed again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not ending up at my limit again&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The limit is the last punch hole of the belt. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks to the OP for his feedback on this answer. My old answer is included under the line.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My guess is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A security belt&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's all about the heroin&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In this context, the heroin is a person belted up&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't just mean keeping you clothed and respectable looking.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A security belt is not just something you put on like clothes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't just mean saving your life.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A security belt can save your life but it can do more.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've been firmly with you everywhere you go.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A security belt is tight firmly to the passenger.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Everything ... from my tongue reflects you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;em&gt;toungue&lt;/em&gt; of a security belt is usually metallic and reflective.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now you say you don't need me? ... Go off&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you put off a security belt, you don't need it any more.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And don't until you lose some weight. I'm not ending up at my limit again for some hot mess like you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Makes sense. If you're fat, you're going to pull the security belt to its limit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Update (to the old answer)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;After thinking about it once again, the solution could be also just:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A belt&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This would explain the &quot;ass&quot; part better, although other clues would then sound more fanciful.&lt;/p&gt;&#xA;" OwnerUserId="4394" LastEditorUserId="4394" LastEditDate="2014-11-21T07:06:07.767" LastActivityDate="2014-11-21T07:06:07.767" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5280" PostTypeId="1" AcceptedAnswerId="5281" CreationDate="2014-11-20T17:49:16.990" Score="2" ViewCount="4722" Body="&lt;p&gt;I got those from a cereal box.&lt;br&gt;&#xA;You can use each number 1 to 9 only once.&lt;br&gt;&#xA;Operation should match the value in the green box horizonal and vertical&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fc2Qb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/z8xTI.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is Operation Order isnt specify in the instructions &lt;br&gt;&#xA;But because those puzzles are usually for childrens &lt;b&gt;my guess&lt;/b&gt; is precedence is left to right and up down. &#xA;&lt;br&gt;&lt;br&gt;&#xA;Also for puzzle A using normal operator presedence&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;first row will be &lt;br&gt;&#xA;  9 - X&lt;em&gt;Y = 2 --&gt; X&lt;/em&gt;Y = 7 &lt;br&gt;&#xA;  and there isnt any combination using 1 to 9 to solve that. &lt;BR&gt;&#xA;  * &lt;strong&gt;EDIT&lt;/strong&gt; * My mistake 1*7 = 7 doh &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I could solve B but not A using that order.&lt;br&gt;&#xA;You can show your B solution&lt;br&gt;&#xA;But the correct answer go for the one finding a solution for A or proving there isnt a solution.&lt;/p&gt;&#xA;" OwnerUserId="5165" LastEditorUserId="5165" LastEditDate="2014-11-20T18:25:27.633" LastActivityDate="2014-11-20T18:25:27.633" Title="One to Nine Puzzle" Tags="|mathematics|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5281" PostTypeId="2" ParentId="5280" CreationDate="2014-11-20T18:09:06.270" Score="7" Body="&lt;p&gt;&lt;strong&gt;TL;DR.&lt;/strong&gt; The first puzzle is impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the first row, you incorrectly claim that there is no XY | X,Y e {1-9}&lt;br&gt;&#xA;Actually there is.  1 and 7.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;BUT&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the left column, the only possible solution is 9 x 7 - 5 = 58, which means there can't be a 7 in the first ROW.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: 9 x 6 = 54, and we can't subtract to get 58.  9 x 8 = 72, and we need to subtract 14 to get 58&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we must assume that the operator precedence is left to right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes the first row 9 - 7 x 1 = 2, which runs afoul of our first column.,&#xA;or 9-8 x2 = 2, Which frees up the seven for the first column,  but kills us in the second column, where 8 - X + Y = 8, which implies X = Y, which is impossible.&lt;/p&gt;&#xA;" OwnerUserId="2752" LastActivityDate="2014-11-20T18:09:06.270" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5282" PostTypeId="2" ParentId="5280" CreationDate="2014-11-20T18:09:23.857" Score="3" Body="&lt;p&gt;I don't think &lt;strong&gt;A&lt;/strong&gt; is solvable at all, no matter the order of operations.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First row: &lt;code&gt;9-a*b=2&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For any order of operations, the only options for &lt;em&gt;a&lt;/em&gt; and &lt;em&gt;b&lt;/em&gt; are 7 and 1. If you tried using left-to-right order and &lt;code&gt;9-8*2&lt;/code&gt;, then you run into a problem in the middle column where you have &lt;code&gt;8-c+d=8&lt;/code&gt;, which needs a repeated number.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First column: &lt;code&gt;9*e-f=58&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Since 58 is not a multiple of 9, the order of operation can only be multiplication first. So now we need a multiple of 9 that is &lt;em&gt;more&lt;/em&gt; than 58, but &lt;em&gt;less&lt;/em&gt; than 68 (since you can only subtract a single digit number to get to 58). The only one is 63, which is &lt;code&gt;9*7&lt;/code&gt;. However, 7 is already taken (first row) and can't be used again.&lt;/p&gt;&#xA;" OwnerUserId="1563" LastActivityDate="2014-11-20T18:09:23.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5284" PostTypeId="2" ParentId="5270" CreationDate="2014-11-20T18:12:37.603" Score="3" Body="&lt;p&gt;I suspect it would be the husband (James) alone or along with his lover (Eliska)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Using geobits' logic (which I had almost found myself -Flip that forehead etching and you get 0x3e7, a hex number representing 999 in decimal. Flip it back over and you get 666. You might want to try to link that to someone) 666 is the number of the beast, and the killer left it there to turn the investigation on Jakub, Jakub himself is illiterate so he could not have come up with that notation especially to turn the investigators on to himself&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second reason is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if it was one of the bachelors, two people would have to lie, the second bachelor and the friend which is unlikely &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The third reason is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;dabel probably indicates Abel, which may be another attempt to turn the investigators on to him (or as the other answer points another word for devil trying to fix Jakub&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fourth reason is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The victim seems to have just patched up with her husband and is unlikely to have had sex with anyone else&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know my case is not very strong, but yeah I leave it here for someone to develop&lt;/p&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="2313" LastEditDate="2014-11-20T18:47:45.173" LastActivityDate="2014-11-20T18:47:45.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5285" PostTypeId="2" ParentId="5280" CreationDate="2014-11-20T18:20:11.953" Score="7" Body="&lt;p&gt;Assuming left to right precedence, in the third row we should have 1 in the third column, So:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;9 -   *   = 2&#xA;*   -   *&#xA;  +   *   = 39&#xA;-   +   +&#xA;  -   * 1 = 1&#xA;=   =   =&#xA;58  8   7&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then we have 2 for the last column of first row, and the middle one would get the value of 8:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;9 - 8 * 2 = 2&#xA;*   -   *&#xA;  +   *   = 39&#xA;-   +   +&#xA;  -   * 1 = 1&#xA;=   =   =&#xA;58  8   7&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Next step follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;9 - 8 * 2 = 2&#xA;*   -   *&#xA;  +   * 3 = 39&#xA;-   +   +&#xA;  -   * 1 = 1&#xA;=   =   =&#xA;58  8   7&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now, in the first column, the only possible values are 7 and 5:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;9 - 8 * 2 = 2&#xA;*   -   *&#xA;7 +   * 3 = 39&#xA;-   +   +&#xA;5 -   * 1 = 1&#xA;=   =   =&#xA;58  8   7&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;9 - 8 * 2 = 2&#xA;*   -   *&#xA;7 +   * 3 = 39&#xA;-   +   +&#xA;5 - 4 * 1 = 1&#xA;=   =   =&#xA;58  8   7&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The remaining cell should be 6. This works for the row, but then the column's result would be 6, not 8. I guess this is a typo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solution for B:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;9 + 7 + 3 = 19&#xA;*   *   *&#xA;6 - 2 + 4 = 8&#xA;+   -   *&#xA;1 * 8 - 5 = 3&#xA;=   =   =&#xA;55  6   60&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="5752" LastActivityDate="2014-11-20T18:20:11.953" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5287" PostTypeId="2" ParentId="5270" CreationDate="2014-11-20T18:35:18.983" Score="3" Body="&lt;p&gt;It was the James and the Eliska working together, trying to frame Jakub, getting rid of both spouses at once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;James came back in dressed in the hoodie, had make up sex with his wife, who made him wear a condom due to past infidelities, and smothered her with a pillow.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The 666 clue is just too clever for Jakub, who is uneducated, to have come up with. However, James, was a Colleague of the victim, (stated in her bio), and would have a familiarity with Hexadecimal. &#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; DABEL is czech for Devil.  Eliska came up with this.  She's been married to Jakub for 5 years.  Jakub immigrated 4 years ago.  So she's czech as well.  They're trying to frame Jakub&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2752" LastEditorUserId="2313" LastEditDate="2014-11-20T18:44:01.740" LastActivityDate="2014-11-20T18:44:01.740" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5288" PostTypeId="2" ParentId="5019" CreationDate="2014-11-20T19:00:39.397" Score="2" Body="&lt;p&gt;This is not full answer as well, but &quot;stealing my words&quot; makes me think of either&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1) Sets (younger brother could be a subset)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2) Geometric sequence (where younger brother could be a sequence that contains a subset of another sequence) - e.g. 1, 1/2, 1/4 and 1, 1/4. 1/16&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5695" LastActivityDate="2014-11-20T19:00:39.397" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5289" PostTypeId="1" AcceptedAnswerId="5365" CreationDate="2014-11-20T20:10:08.267" Score="13" ViewCount="6435" Body="&lt;p&gt;Be kind, this is my first time trying to make my own...&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;When my owner is closed, I am open&lt;/p&gt;&#xA;&lt;p&gt;My owner never cleans me out&lt;/p&gt;&#xA;&lt;p&gt;Few can afford to keep me fed&lt;/p&gt;&#xA;&lt;p&gt;Because people always take my food&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="4321" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2020-07-13T05:17:03.413" Title="Hungry, hungry _____" Tags="|riddle|" AnswerCount="4" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="5290" PostTypeId="1" AcceptedAnswerId="5293" CreationDate="2014-11-20T20:14:37.953" Score="13" ViewCount="1475" Body="&lt;p&gt;Assume you have an unlimited but finite number of linking, teleporting doors you can add to a hallway of finite length. Assuming there's a door at the start and finish, can you arrange the teleports so that one can never make it to the end of the hallway?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;hr&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;A B A B&lt;/strong&gt;    &lt;/p&gt;&#xA;  &#xA;  &lt;hr&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where when you enter A from the left (you'll never turn and enter from the right) you come out of the other A on the right. So you'd enter the first A, come out of the second, enter B come out of the other, hit A again, and when you exit the other and hit B you've got to the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You HAVE to start before the first door, and every tele door HAS to have a linking door. No combining links (AB would not have a 50 - 50 chance of A or B) and no more than 2 links, so no random linking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Can you arrange the tele doors in such a way, you'll never reach the end? If so, how many tele doors does it require, and what's the setup? If not, prove it's impossible.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorDisplayName="user88" LastEditDate="2015-04-29T00:53:15.417" LastActivityDate="2015-11-04T14:24:26.120" Title="The Hallway and Tele Doors" Tags="|logical-deduction|" AnswerCount="6" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5291" PostTypeId="2" ParentId="5289" CreationDate="2014-11-20T20:16:47.720" Score="3" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; one of the hippos from the game &lt;a href=&quot;http://en.wikipedia.org/wiki/Hungry_Hungry_Hippos&quot; rel=&quot;nofollow&quot;&gt;Hungry Hungry Hippos&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer fits the title, &quot;Hungry Hungry _____.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When my owner is closed, I am open&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you press the lever for your hippo closed, the hippo opens its mouth.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My owner never cleans me out&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the middle of a game of HHH, you don't empty the tray of pellets you've grabbed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Few can afford to keep me fed&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Because people always take my food&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You're competing against three other people trying to eat pellets, making it difficult to grab the most pellets.&lt;/p&gt;&#xA;" OwnerUserId="294" LastEditorUserId="5373" LastEditDate="2015-03-08T15:41:51.623" LastActivityDate="2015-03-08T15:41:51.623" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5293" PostTypeId="2" ParentId="5290" CreationDate="2014-11-20T20:45:55.890" Score="15" Body="&lt;p&gt;To the best of my knowledge, it is impossible to have an infinite loop when starting on the far left.&lt;/p&gt;&#xA;&lt;p&gt;The best reasoning I can provide is that once you enter the first door, it is impossible to ever enter that door again from the left. Since your loop is already 'breaking', you'll then never be able to enter the path between the second A-B again.&lt;/p&gt;&#xA;&lt;p&gt;Ultimately, you run into the case where you will only ever walk the path between the two letters once. A good way to look at this is to consider the Start/End as a pair of doors themselves.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;[Start] [You] A D B A C B C D ... [End]&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;In the long run, the only loop you're bound to be stuck in is the one you started between, which in this case is the losing factor.&lt;/p&gt;&#xA;" OwnerUserId="2013" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-11-04T14:24:26.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5294" PostTypeId="2" ParentId="5290" CreationDate="2014-11-20T20:46:49.653" Score="17" Body="&lt;p&gt;It's impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you never reach the end, it must be because you got stuck in an infinite loop. This requires that you enter the same teleportation door twice at some point. Consider the first time that this happens, and what happened immediately beforehand. Because you can't go towards the left, you can only possibly enter the leftmost door once. Therefore the repeated door must have another door immediately to its left, and you must have just come out of that one both times. But this contradicts the fact that it's the first time you used a door twice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore you can't get stuck in a loop, and must eventually reach the end.&lt;/p&gt;&#xA;" OwnerUserId="5833" LastActivityDate="2014-11-20T20:46:49.653" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5295" PostTypeId="2" ParentId="5290" CreationDate="2014-11-20T20:52:53.653" Score="-1" Body="&lt;p&gt;If I understood correctly, the doors effectively reverse the direction of the traveller, so in the example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;AA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The traveller would enter the left-hand A travelling left to right, exit the right-hand A travelling right to left, thereby entering the left-hand A again, this time exiting right-hand A going left to right and ending her run.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no solution where the traveller would turn around and head back the other way because the traveller is always travelling forward in the sequence, regardless of the number of doors she travels through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had a look at a way of producing a &lt;em&gt;mirror&lt;/em&gt;, forcing the traveller to hit a wall and travel back the way they came, but all the doors act in the same way. You would need a type of door that doesn't reverse the direction of the traveller in order to... reverse the direction of the sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doors are always paired, so the traveller is always reversed twice (i.e. no change) eventually before exiting the sequence.&lt;/p&gt;&#xA;" OwnerUserId="5245" LastActivityDate="2014-11-20T20:52:53.653" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5296" PostTypeId="2" ParentId="5290" CreationDate="2014-11-20T21:02:27.267" Score="1" Body="&lt;p&gt;DISCLAIMER: This should possibly be a comment but I don't have the reputation to do that :( It is NOT proof, just perhaps a start on the idea.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Let's start by considering what we know. Every time the 'player' hits a door from the left, they exit the corresponding from the right. Hence the player, in order to leave the circuit of doors, must exit the last door on the line. If the player leaves the last door, $a_2$, then they must enter it's correspondent door, $a_1$. So in order to create an infinite loop, we would have to make sure that this never happens. &lt;br&gt;&lt;br&gt;In short, we would have to place the correspondent door $a_1$ such that it was never passed in part of our theoretical infinite loop of teledoors. If this was the case for door $a_1$, it would have to be effectively sectioned off from the remaining teledoors such that it could never be accessed by the player. To do this, let's say we were to use a set of doors, $p_1$ and $p_2$, to surround door $a_1$ so that it could never be entered. By entering door $p_1$ we would exit from $p_2$ on the other side of $a_1$, skipping it altogether. &lt;br&gt;&lt;br&gt;However, what would happen after that? Consider the incomplete sequence of doors $p_1, a_1, p_2, q_1, a_2$. Here we are left with the issue of where to place our 'restarter', $q_2$. After all, in an infinite loop of a finite number of doors, the sequence must iterate. Hence, with nowhere to place $p_2$ so that the loop is infinite, we may assume that an infinite loop is impossible to create.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4748" LastActivityDate="2014-11-20T21:02:27.267" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5298" PostTypeId="2" ParentId="5198" CreationDate="2014-11-20T21:15:22.427" Score="8" Body="&lt;p&gt;Let's start with the easy ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1x1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/dwRsM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1x2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/2YfRg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1x3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/IfWrz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1x4&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/BGmlD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1x6&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/vmK3M.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These ones took me a while.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1x5&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/IkA48.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2x3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/W2W0d.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1497" LastEditorUserId="1497" LastEditDate="2014-11-21T16:32:02.400" LastActivityDate="2014-11-21T16:32:02.400" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5299" PostTypeId="1" AcceptedAnswerId="5302" CreationDate="2014-11-20T21:21:04.383" Score="19" ViewCount="2276" Body="&lt;blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;You are in an airport and are worried about missing your flight.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;So you start to run the long distance to the flight gate at a steady speed of 10km/h (you have bags and are not very fit).&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Luckily some stretches of the journey are covered by &lt;a href=&quot;http://en.wikipedia.org/wiki/Moving_walkway&quot; rel=&quot;noreferrer&quot;&gt;moving walkways&lt;/a&gt; on which you also run at 10km/h.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;However, you don't have the energy to run all the way and must rest for 1 minute in total at some point before you get to the gate.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When you rest you walk at 5km/h.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;To get there fastest, should you rest on a moving walkway or on some part in between them?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;(Please give a full explanation in any answer rather than just a simple guess.)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="8998" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-01-25T15:58:06.907" Title="When should you rest?" Tags="|logical-deduction|mathematics|" AnswerCount="8" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="5302" PostTypeId="2" ParentId="5299" CreationDate="2014-11-20T21:40:56.877" Score="15" Body="&lt;p&gt;$D=$distance to gate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$W=$distance on walkways&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S=$Speed of walkways&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X=$distance travelled while standing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You time to get there assuming you stop off walkway is $$T_r=\frac{D-W-X}{10}+\frac{W}{10+S}+\frac{X}{5}$$&#xA;$$X=1/12$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you stop on walkway:&#xA;$$T_w=\frac{D-W}{10}+\frac{W-X}{10+S}+\frac{X}{S+5}$$&#xA;$$\frac{X}{S+5}=1/60$$&#xA;$$X=S/60+1/12$$&#xA;$$T_w=\frac{D-W}{10}+\frac{W-S/60-1/12}{10+S}+1/60$$&#xA;$$T_w=\frac{D-W}{10}+\frac{W}{10+S}-\frac{S+5}{600+60S}+1/60$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Tr-Tw=\frac{-1}{120}+\frac{S+5}{600+60S}&amp;gt;0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note that if $S=0$ then the result is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should stop on the walkway because you will let the walkway do work to move you during that time.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditDate="2014-11-20T21:51:59.440" LastActivityDate="2014-11-20T21:51:59.440" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="5303" PostTypeId="1" AcceptedAnswerId="5305" CreationDate="2014-11-20T21:42:22.737" Score="2" ViewCount="380" Body="&lt;p&gt;Say you have a grid of 16x16 squares. Label them A through P and 1 through 16.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now say that at H8 there's a mouse that moves at 1 slot every turn. Also, you get to block a tile every turn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say the mouse gets to move first; can you, with perfect play by both you and the mouse, completely box the mouse in, before it escapes the board? If so, how, and how many turns will it take? If not, explain why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;BONUS: What is the smallest square board size (1x1, 2x2, 3x3, 4x4, 5x5, 6x6, etc.) for which this is possible? What's the largest that's impossible? You may assume the rules above apply, except the one about the mouse starting at H8; the mouse begins as close to the center as possible.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="5188" LastEditDate="2015-05-02T06:32:32.367" LastActivityDate="2015-05-02T06:32:32.367" Title="Capture The Runaway Mouse" Tags="|strategy|checkerboard|" AnswerCount="3" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="5305" PostTypeId="2" ParentId="5303" CreationDate="2014-11-20T21:59:55.713" Score="0" Body="&lt;p&gt;If the mouse can move diagonally, you can't box it in, since the most efficient net would normally be a diagonal line. The king will move towards a corner, and since you have to cover 2 sides, you can't solid lines fast enough to hold the king in. If you do manage to get a solid line, the king can just move towards the short side diagonally. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the mouse is not able to move diagonally, I would say that you would need to be able to put a blocking piece out just 3 spaces from the mouse, then box it in diagonally. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think a winning strategy for the blocker involves a board at least 21 squares wide. Essentially, the mouse moves diagonally to a corner. If the blocker can get to this, I think he/she can win: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ON0Du.gif&quot; alt=&quot;Blocking position&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2014-11-20T22:36:29.157" LastActivityDate="2014-11-20T22:36:29.157" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5306" PostTypeId="2" ParentId="5299" CreationDate="2014-11-20T22:05:44.380" Score="5" Body="&lt;p&gt;You have to travel distance $d$. At some point, you will stop and rest for one minute, spending time $t$ minutes on a walkway. (If you stopped off of a walkway, $t = 0$. Otherwise, $0 &amp;lt; t \leq 1$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During your rest time, you will move $(w + r)t + r(1 - t) = wt + rt + r - rt = wt + r$, where $w$ is the speed of the walkway and $r$ is your running speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have $W$ metres of walkway out of the distance $d$, without resting, you would cover the distance in $\dfrac{W}{w + r} + \dfrac{d - W}{r}$ minutes. With your rest time, you have $(w + r)t$ less moving walkway, so your non-rest time becomes $\dfrac{W - (w + r)t}{w + r} + \dfrac{d - W}{r} = \dfrac{W}{w + r} + \dfrac{d - W}{r} - t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding your rest time, now,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{W}{w + r} + \dfrac{d - W}{r} - t + 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;. Minimizing this means making $t$ as high as possible – 1 – meaning you should spend all your rest time on a moving walkway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably.&lt;/p&gt;&#xA;" OwnerUserId="1565" LastEditorUserId="1565" LastEditDate="2014-11-21T01:23:02.550" LastActivityDate="2014-11-21T01:23:02.550" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5307" PostTypeId="2" ParentId="5303" CreationDate="2014-11-20T22:14:08.797" Score="-1" Body="&lt;p&gt;The basic idea is as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The mouse situated in H8 will run either upwards or to the left, as it will take it less moves to do so. Let us assume that the mouse runs left for this purpose. The most efficient way is to start by blocking up the outer wall to the left, leaving one block gaps between each block placed. By doing this, when the mouse reaches the left wall you may simply block off the holes before it attempts to go through them. By expanding this holed border around the back of the mouse, you can effectively box it in. I got the idea from an hour playing &lt;a href=&quot;https://play.google.com/store/apps/details?id=com.ketchapp.circlethedot&amp;amp;hl=en&quot; rel=&quot;nofollow&quot;&gt;Circle the Dot&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4748" LastActivityDate="2014-11-20T22:14:08.797" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5308" PostTypeId="2" ParentId="5198" CreationDate="2014-11-20T22:32:40.260" Score="6" Body="&lt;p&gt;Rev 4 - added 1 x 9 incorrect but still trying, added 1 x 5 using Florian's clue&lt;br&gt;&#xA;Rev 3 - added 1 x 8 solution&lt;br&gt;&#xA;Rev 2 - added 2 x 6 solution&lt;br&gt;&#xA;Rev 1 - added 2 x 2 solution&lt;br&gt;&#xA;Thanks for the clarification, Somnium.  I will put some more answers here.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1 x 5 solution&lt;/strong&gt; using Florian's clue (area = 54)  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/E9IJU.jpg&quot; alt=&quot;1x5 solution&quot;&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1 x 8 solution&lt;/strong&gt; (area = 560)  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/ZEuKZ.jpg&quot; alt=&quot;1x8 solution&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1 x 9 incorrect&lt;/strong&gt;  still trying  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/gUO46.jpg&quot; alt=&quot;1x9 incorrect&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2 x 2 solution&lt;/strong&gt; (area = 144)  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/ELQUG.jpg&quot; alt=&quot;2x2 solution&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2 x 3 alternate&lt;/strong&gt; (area = 72)  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/XUYrJ.jpg&quot; alt=&quot;2x3 alternate&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2 x 6 solution&lt;/strong&gt; (area = 240)  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/oXrio.jpg&quot; alt=&quot;2x6 solution&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2014-11-22T18:44:39.660" LastActivityDate="2014-11-22T18:44:39.660" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5309" PostTypeId="1" AcceptedAnswerId="5384" CreationDate="2014-11-20T23:19:09.823" Score="7" ViewCount="358" Body="&lt;p&gt;&lt;strong&gt;When was this graphics software released?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Must include reasoning behind your answer!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: As of this date, customers in Canada and the Us MUST USE TWO VERSIONS PRIOR - 1.63-1. Current version will not boot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;New and returning in this version:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Example projects by Leighton and Rembrandt added.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;kbd&gt;pause&lt;/kbd&gt; and &lt;kbd&gt;prt scr&lt;/kbd&gt; projects abandoned due to hardware incompatibility&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;kbd&gt;shift&lt;/kbd&gt; + &lt;kbd&gt;enter&lt;/kbd&gt; and &lt;kbd&gt;shift&lt;/kbd&gt; + &lt;kbd&gt;return&lt;/kbd&gt; can both be used to begin the next queued operation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;O.T.M. pathways reinitialized.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Shifted glaukopic balance along an inverdant axis. Formula below:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Delta = 1/4^\infty &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're almost out of beta. Probably three more releases before final. See our catalog online for planned products coming next year: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.iloveyoulikeafatladylovesapples.com/&quot; rel=&quot;nofollow&quot;&gt;Full Catalog&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This eponymous software was named while the creator was drunk. The translation ended up a bit backwards, but a visit to the vomitoria cleared their stomach and their mind. What really helped was going home again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The standard made up dictionary defines glaukopic as &quot;adj. pertaining to glaukopis.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5626" LastEditorUserId="5626" LastEditDate="2014-12-08T14:29:00.967" LastActivityDate="2014-12-08T14:29:00.967" Title="Serpina Pro Art Studio 1.89-1 Technical Update Manual - When is this software's new version released?" Tags="|riddle|wordplay|" AnswerCount="3" CommentCount="19" ContentLicense="CC BY-SA 3.0" />
  <row Id="5310" PostTypeId="1" AcceptedAnswerId="5314" CreationDate="2014-11-20T23:40:52.777" Score="5" ViewCount="498" Body="&lt;p&gt;(4-6-4)&#xA;(6-12-8)&#xA;(8-12-6)&#xA;(12-30-20)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the next number triplet - and why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll add hints if nobody gets the answer first.&#xA;In the past I've used this riddle in a game with my RPG gaming group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; The answer I'm seeking in this riddle is unique and well defined. It requires thinking out of the box a bit. I will add hints over time. One, actually, has already been given.&lt;/p&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="4050" LastEditDate="2015-01-03T17:52:22.780" LastActivityDate="2015-01-03T17:52:22.780" Title="Continue this number-triplet series" Tags="|mathematics|pattern|number-sequence|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5311" PostTypeId="2" ParentId="5310" CreationDate="2014-11-20T23:45:13.487" Score="1" Body="&lt;p&gt;I think that there are many possible solution for the next triplet.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (18, 48, 24)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I've seen number patterns like this where there is no math pattern. I think that the pattern is that the first and third numbers must both be less than the middle number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1711" LastActivityDate="2014-11-20T23:45:13.487" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5312" PostTypeId="2" ParentId="5299" CreationDate="2014-11-20T23:54:02.807" Score="7" Body="&lt;p&gt;If the distance to the destination is $D$ km and the moving walkway moves at speed $v$ km/h and carries you distance $D_w$ km then:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If you run all the time on the moving walkway it will take you $\frac{D_w}{v+10}$ hours and if you rest off the moving moving walkway you will do $\frac{1}{60}$ hours resting and $\frac{D - D_w - \frac{5}{60}}{10}$ hours running; and&lt;/li&gt;&#xA;&lt;li&gt;If you run all the time off the moving walkway then it will take you $\frac{D-D_w}{10}$ hours and $\frac{D_w - \frac{v+5}{60}}{v+10}$ hours running on the walkway and $\frac{1}{60}$ hours resting on the walkway.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Both options have $\frac{1}{60}$, $\frac{D-D_w}{10}$ and $\frac{D_w}{v+10}$ terms which can be cancelled out so the comparison is between:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$-\frac{\frac{v+5}{60}}{v+10}=-\frac{1}{60}+\frac{1}{12(v+10)}$ hours if you rest on the walkway; and&lt;/li&gt;&#xA;&lt;li&gt;$-\frac{5}{600} = -\frac{1}{120}$ hours if you rest off the walkway.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These two times can only be equal if $v=0$ and if $v&amp;gt;0$ then $-\frac{1}{60}+\frac{1}{12(v+10)} &amp;lt; -\frac{1}{120}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, less time is spent overall if you rest on the moving walkway.&lt;/p&gt;&#xA;" OwnerUserId="5434" LastEditorUserId="5434" LastEditDate="2014-11-21T01:17:56.840" LastActivityDate="2014-11-21T01:17:56.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5313" PostTypeId="1" AcceptedAnswerId="5382" CreationDate="2014-11-21T00:16:27.647" Score="-5" ViewCount="521" Body="&lt;p&gt;An American architect was in Norway for a business trip, and he was working with a Norwegian Foreman who was building a skyscraper. The Norwegian did not know how to &lt;em&gt;say&lt;/em&gt; any numbers in English, but he did know a few English words which he used to communicate his thoughts to the architect. In one case, the American pointed to a wall on his blueprint: &quot;What's the length of the wall in feet?&quot; he asked. The Foreman drew three pictures next to one another:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The first was of a pile of dirt&lt;/li&gt;&#xA;&lt;li&gt;The second was a large Maple tree&lt;/li&gt;&#xA;&lt;li&gt;The third was of a squating dog which was defecating.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;What was the length of the wall (excluding units)?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When the Norwegian was explaining the defecating dog, he didn't seem to be pointing at the dog&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My apologies for the many edits, this is my first post here. I swear I'll be more straightforward next time!&lt;/p&gt;&#xA;" OwnerUserId="5159" LastEditorUserId="5373" LastEditDate="2015-04-30T23:00:00.207" LastActivityDate="2015-04-30T23:00:00.207" Title="The Norwegian Foreman" Tags="|wordplay|" AnswerCount="4" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="5314" PostTypeId="2" ParentId="5310" CreationDate="2014-11-21T00:20:32.343" Score="6" Body="&lt;p&gt;The last triplet is $(20, 30, 12)$. These are the number of faces, edges, and vertices of the five Platonic solids in ordered triples, in order of increasing face count: tetrahedron, cube, octahedron, dodecahedron, and icosahedron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't seem like this riddle involves much outside-of-the-box thinking at all; it's just a case of knowing what to look for. I would consider this to be trivia more than it is something that can be reasoned out.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2014-11-21T00:20:32.343" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5315" PostTypeId="2" ParentId="5303" CreationDate="2014-11-21T00:36:10.207" Score="8" Body="&lt;p&gt;The mouse can win on a 34x34 moving first, but will lose on a 35x35 board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof of this is in &lt;em&gt;Winning Ways for Your Mathematical Plays, Vol 3&lt;/em&gt; (1982) by Berlekamp, Conway, and Guy. &lt;a href=&quot;http://www.link.cs.cmu.edu/15859-s11/notes/king-devil.pdf&quot;&gt;Here's the relevant extract&lt;/a&gt;, starting from &quot;The Game of Kinggo&quot;, which is the same as the game in the question. Thanks to @Julian Rosen for tracking down this source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proofs are very intricate, and I'm sorry that I cannot summarize them. Perhaps someone with a better understanding can explain the general gist of the strategy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general version of this problem is known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Angel_problem&quot;&gt;Angel Problem&lt;/a&gt;. It is known that an angel can escape indefinitely if it can make move to spaces up to two king's moves away. &lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="2627" LastEditDate="2014-11-21T00:58:27.010" LastActivityDate="2014-11-21T00:58:27.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5316" PostTypeId="2" ParentId="5198" CreationDate="2014-11-21T01:15:17.517" Score="14" Body="&lt;p&gt;&lt;strong&gt;2x2&lt;/strong&gt; - area 108 - optimal&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/Ds4tU.png&quot; alt=&quot;2x2&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2x3&lt;/strong&gt; - area 72 - optimal&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/HRvTv.png&quot; alt=&quot;2x3&quot;&gt;    &lt;img src=&quot;https://i.stack.imgur.com/3sRhh.png&quot; alt=&quot;2x3&quot;&gt;    &lt;img src=&quot;https://i.stack.imgur.com/7MGmK.png&quot; alt=&quot;2x3&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2x4&lt;/strong&gt; - area 108 - optimal&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/WEbCm.png&quot; alt=&quot;2x4&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3x4&lt;/strong&gt; - area 84 - optimal&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/T4Jkn.png&quot; alt=&quot;3x4&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1x5&lt;/strong&gt; - area 54 - optimal&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/aOotR.png&quot; alt=&quot;1x5&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2x5&lt;/strong&gt; - area 304 - optimal&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/aIpza.png&quot; alt=&quot;2x5&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3x5&lt;/strong&gt; - area 576 - optimal&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/9IikI.png&quot; alt=&quot;3x5&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2x6&lt;/strong&gt; - area 240 - optimal&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/oilCZ.png&quot; alt=&quot;2x6&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1x7&lt;/strong&gt; - area 1034&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/OzkeX.png&quot; alt=&quot;1x7&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1x8&lt;/strong&gt; - area 432 - optimal&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/JthKh.png&quot; alt=&quot;1x8&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3x8&lt;/strong&gt; - area 5880&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/u7ZY6.png&quot; alt=&quot;3x8&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1x9&lt;/strong&gt; - area 585 - optimal&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/HkovG.png&quot; alt=&quot;1x9&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that by subdividing the yellow rectangles:&lt;br&gt;&#xA;2x3 indirectly solves 1x1, 1x2, 1x3&lt;br&gt;&#xA;2x4 indirectly solves 1x4 and 2x2&lt;br&gt;&#xA;2x5 indirectly solves 1x5&lt;br&gt;&#xA;2x6 indirectly solves 1x6  &lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="11110" LastEditDate="2018-06-08T13:00:11.257" LastActivityDate="2018-06-08T13:00:11.257" CommentCount="20" ContentLicense="CC BY-SA 4.0" />
  <row Id="5317" PostTypeId="2" ParentId="5309" CreationDate="2014-11-21T01:16:38.623" Score="0" Body="&lt;p&gt;While this isn't a complete answer, here's my thoughts on where to go for this riddle based on the clues I think I've figured out so far.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The software specs mention that &lt;kbd&gt;Pause&lt;/kbd&gt; and &lt;kbd&gt;PrtSc&lt;/kbd&gt; aren't supported due to hardware incompatibility. This most likely means that these came around at a time when these two keys had just been introduced to some keyboards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at pictures of old &lt;a href=&quot;http://en.wikipedia.org/wiki/IBM_PC_keyboard&quot; rel=&quot;nofollow&quot;&gt;IBM PC keyboard&lt;/a&gt; models, which almost all modern keyboards today are based on, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Model_M_keyboard&quot; rel=&quot;nofollow&quot;&gt;Model M&lt;/a&gt; was the first one to introduce the Pause key, in 1985.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the manual states that both &lt;kbd&gt;Shift+Enter&lt;/kbd&gt; and &lt;kbd&gt;Shift+Return&lt;/kbd&gt; can now be used for the same shortcut. The fact that this is a new feature means that it must have only recently come into demand at the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Apple_Keyboard#Macintosh_Keyboard_.28M0110.29&quot; rel=&quot;nofollow&quot;&gt;Macintosh keyboard&lt;/a&gt; is the only keyboard that makes a distinction between Enter and Return on its labels, and it was released in 1984. The IBM keyboards simply labelled both buttons Enter, and most keyboards in that era either only had one key for enter/return, or labelled them both the same thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, I think that this version of the software was released sometime in 1985. I haven't gotten any further than that, though, and I can't figure out the month or day if 1985 is in fact the correct year.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I may just be massively overthinking things, but it seems like the kind of thing that would fit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The O.T.M. pathways and glaukopic balance don't seem to make any sense; especially the formula, as $1/4^\infty$ is just zero, and it doesn't seem to make sense that something would be set &lt;em&gt;to&lt;/em&gt; infinity rather than just having it as an upper bound.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2014-11-21T01:22:27.513" LastActivityDate="2014-11-21T01:22:27.513" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5318" PostTypeId="1" AcceptedAnswerId="5323" CreationDate="2014-11-21T01:34:14.020" Score="8" ViewCount="555" Body="&lt;p&gt;Replacement cipher: Let τ be a permutation of the alphabet, and apply τ to each letter of the message. Frequency analysis is useful for breaking this type of code. Decode the following, which was encoded using a replacement cipher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MIZVN KXXHA XRRTK NXYEX QIZVI IZXWM NXYGT JWVHC YTOXX QNHTI JYTWV NMHUR XOYLN ZTJTE XYAZX RWMHU XEMRK LIJYT WNWVR REXPV IMTHN OTHIM HLVRR GYXCX VIXQ --- NVWLX RBTZH NTH&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have made a frequency table of the occurrence of each letter in the above code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X-18 T-12 H-10 N-10 R-10 I-9 V-9 Y-8 M-7 W-7 Z-6 E-4 J-4 L-4 K-3 O-3 Q-3 A-2 C-2 G-2 U-2 B-1 P-1 D-0 F-0 S-0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing this with the frequency of letters occurring in English leads me to believe that X corresponds with E and that T corresponds with T. However this is where I get stuck.&lt;/p&gt;&#xA;" OwnerUserId="5842" LastActivityDate="2014-11-21T02:23:01.797" Title="Replacement Cipher" Tags="|cipher|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5319" PostTypeId="2" ParentId="5318" CreationDate="2014-11-21T01:44:30.067" Score="6" Body="&lt;p&gt;After doing frequency analysis, you often have a guess at what ciphertext letters correspond to the most common letters &lt;code&gt;ETAOIN&lt;/code&gt;. What I often try to do next is pin these down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Check for doubled letters. &lt;code&gt;E&lt;/code&gt;, &lt;code&gt;T&lt;/code&gt;, &lt;code&gt;O&lt;/code&gt;, and &lt;code&gt;N&lt;/code&gt; can be doubled, but &lt;code&gt;A&lt;/code&gt; and &lt;code&gt;I&lt;/code&gt; cannot be except in rare cases of the letter being at the end of a word and the start of the next word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Check for the possible common substring &lt;code&gt;THE&lt;/code&gt;, which uses the &lt;code&gt;T&lt;/code&gt; and the &lt;code&gt;E&lt;/code&gt;. This is a good way to test your hypothesis about those letters. &lt;code&gt;THAT&lt;/code&gt; and &lt;code&gt;THERE&lt;/code&gt; also can be useful cribs. Actually, finding reused substrings is useful in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is generic advice; I haven't applied it to your specific ciphertext.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2014-11-21T01:44:30.067" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5320" PostTypeId="2" ParentId="5313" CreationDate="2014-11-21T01:47:08.913" Score="0" Body="&lt;p&gt;I think the height of the wall is&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;&lt;strong&gt;33.4&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The foreman may have used &amp;quot;The first was of a pile of dirt&amp;quot;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;instead of &lt;strong&gt;thirty&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The second was a large Maple tree&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;to represent &lt;strong&gt;three&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;and The third was of a squatting dog which was defecating&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;is dog poo and it can be interpreted as dot four(.4)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Combining all these we get&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;strong&gt;dirty tree dog poo &amp;gt;&amp;gt; thirty three dot four (33.4)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;New Edits&lt;/h2&gt;&#xA;&lt;p&gt;From OP's comment on another answer (&amp;quot;It's a measurement of length, not area&amp;quot;) I think he is looking for the area, not length.&lt;/p&gt;&#xA;&lt;p&gt;So The last point &amp;quot;a squatting dog which was defecating&amp;quot; may be referring to&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;square feets (squat feces) [Since he is not English, he may be thinking as feets instead of feet]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So the final answers become&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;&lt;strong&gt;33 Sq.ft. (Thirty Three Square Feet)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Note: I still doubt, if the foreman can made up words to sound like/similar to the number(number representing the height of wall), why can't he just say it loud?? Is he mute/the architect deaf and the foreman doesn't know how to write in English??&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-11-21T07:19:27.230" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5321" PostTypeId="2" ParentId="5313" CreationDate="2014-11-21T02:00:49.207" Score="1" Body="&lt;p&gt;The answer is;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 30, 43.2 ft&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or in words;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Thirty-hundred, forty-three, dot two.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or in &lt;strong&gt;&lt;em&gt;OTHER&lt;/em&gt;&lt;/strong&gt; words;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Three-thousand, forty-three, decimal two. (3043.2)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How I got it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Thirty;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Dirty (pile of dirt) sounds like 30&lt;br&gt;&#xA;&lt;br&gt;&#xA; Forty-three;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Maple (sounds like April, the 4th month of the year) tree (sounds like 3)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Dot two;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Dog poo, need I say anything else?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Credit to AeJay for;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &quot;dog&quot; in dog poo meaning &quot;dot&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2014-11-21T02:00:49.207" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5323" PostTypeId="2" ParentId="5318" CreationDate="2014-11-21T02:23:01.797" Score="6" Body="&lt;p&gt;You are correct that X corresponds to E.  E is the most commonly used letter and X is the letter that appears most commonly in your cipher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T is the next common letter so it makes sense to try another vowel for this one.  Try O.  See how you go with these tips. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the solution to the cipher if you find you get too stuck:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; IT HAS BEEN WELL OBSERVED THAT THE MISERY OF MAN PROCEEDS NOT FROM A&#xA; SINGLE CRUSH OF OVER WHELMING EVIL BUT FROM SMALL VEXATIONS&#xA; CONTINUALLY REPEATED SAMUEL JOHNSON&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="97" LastActivityDate="2014-11-21T02:23:01.797" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5324" PostTypeId="2" ParentId="5290" CreationDate="2014-11-21T02:23:31.360" Score="11" Body="&lt;p&gt;A lateral thinking answer (well, not so much lateral as its complement...):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes, it's possible, by thinking vertically. Structure the hallway like so:&lt;br&gt;&#xA; &lt;code&gt;-----------A------------&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;start           A    end&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;------------------------&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Walking through the left of door A dumps you out of the A on the ceiling, placing you once again on the left of door A. This setup also has the benefit of making it impossible to walk through the hallway from right to left, as walking through A from the right will push you &lt;em&gt;upward&lt;/em&gt; out of the A on the ceiling, depositing you on the floor above. Or, more likely, in the middle of the solid matter creating the ceiling and killing you. Or just disallowing travel in that direction if you want to be a spoilsport about it.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Note that the A B A B placement in the question was just an example, not a requirement that the doors be placed linearly. The placement requirements only state that the doors be added in the hallway, and that you use an arbitrary but finite number of doors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5395" LastEditorUserId="5395" LastEditDate="2014-11-21T02:29:11.797" LastActivityDate="2014-11-21T02:29:11.797" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5325" PostTypeId="2" ParentId="5313" CreationDate="2014-11-21T02:38:47.857" Score="0" Body="&lt;p&gt;I think it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Just 33 feet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pile of dirt&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dirty =&gt; Thirty&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Maple Tree&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tree =&gt;Three&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;a dog defecating&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; shit =&gt; feet?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5030" LastActivityDate="2014-11-21T02:38:47.857" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5328" PostTypeId="2" ParentId="5121" CreationDate="2014-11-21T04:25:32.690" Score="4" Body="&lt;p&gt;Rather than answer the point by point questions, I'll just state that this is about:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;a href=&quot;http://nazizombies.wikia.com/wiki/Kino_der_Toten&quot; rel=&quot;nofollow&quot;&gt;Call of Duty Black Ops &quot;Nazi Zombies&quot; map &quot;Kino der Toten&quot;&lt;/a&gt; (or possibly the &lt;a href=&quot;http://nazizombiesplus.wikia.com/wiki/Kino_Redux&quot; rel=&quot;nofollow&quot;&gt;&quot;Redux&quot; version&lt;/a&gt; which includes a character called &quot;Neville&quot;). The description (bodies everywhere, post-apocalyptic) fits the theme of zombies, and also matches the layout of the map, which starts inside the lobby of a movie theater (&quot;where one comes to watch&quot;) and includes a staircase. Weapons appear in glowing outlines.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;there's a sign, he could not make any sense of it, since part had been destroyed leaving nothing but &quot;Kino?to?en&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Kino&lt;/strong&gt; der &lt;strong&gt;To&lt;/strong&gt;t&lt;strong&gt;en&lt;/strong&gt; is the name of the map.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;a certain Doctor Edward R &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://nazizombies.wikia.com/wiki/Edward_Richtofen&quot; rel=&quot;nofollow&quot;&gt;Doctor Edward Richtofen&lt;/a&gt; is featured on this map.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In a timid manner he approaches the glowing armament and touches it. He felt catapulted. He had been shot, but he was the bullet and time was the entity landing somewhere unknown to him yet shouting &quot;I'm Home!&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There's a teleporter in this map.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are other things to pick out from the description, but I'll leave it at that.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The puzzle is to work out the missing pieces of information using your detective skills, to deduce the answers using logic and maybe a hint of help from the puzzling community, every clue is made ensuring you do not need to be an expert programmer, physicist or mathematician. If you believe you are on the right track confirm with google. Its your friend! I don't want the puzzle to be limited to those with higher education, therefore all the information is based upon deducing its reason for being there rather than knowing its equation, genetic make up or what it looks like in Java script.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While I support this sentiment...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...puzzles with answers dependent on either video game trivia or being easy to search (I've not played this and all of my information comes from the links; searching &quot;ww2 kino&quot; immediately yielded the answer) are no better. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5395" LastActivityDate="2014-11-21T04:25:32.690" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5329" PostTypeId="2" ParentId="5289" CreationDate="2014-11-21T04:49:33.267" Score="8" Body="&lt;p&gt;I believe it to be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a vending machine.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;'When my owner is closed, I am open'&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Doesn't matter what time of day it is, the vending machine will happily take your money.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;'My owner never cleans me out'&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Customers clean out vending machines - the owner never need to.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;'Few can afford to keep me fed'&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Either a comment on the inflated prices you pay for the convenience, or the fact that it's only businesses which can justify the cost of purchase and stocking.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;'Because people always take my food'&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The purpose of a vending machine - to dispense tasty goodness.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4164" LastActivityDate="2014-11-21T04:49:33.267" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5330" PostTypeId="2" ParentId="5019" CreationDate="2014-11-21T06:57:50.927" Score="4" Body="&lt;p&gt;I know this isn't going to be the answer, but just wanted to share my thoughs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Infinity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;in a great family&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; of numbers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are the most perfect of our race,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; we can't be beaten&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because each of us has a clear worst enemy.&#xA;&lt;br /&gt;My foe - I know him well - is much like me,&#xA;&lt;br /&gt;But we continually fight each other;&#xA;&lt;br /&gt;We never hide our hatred for each other.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Positive Infinity vs Negative Infinity. Which one is stronger ? Impossible to tell.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wave at my eldest brother and he becomes me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (Positive) Infinity + 1 (or any other positive number) &lt;em&gt;is&lt;/em&gt; Infinity (or so XD)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When my dear cousin finds the end of her life,&#xA;She too can turn into me, in her gentle way.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; anything (number/function) that tend to infinity...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My younger brother thinks too highly of himself,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (positive) infinity - 1 ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So smooth and perfect that he'll always outstrip me.&#xA;Often he even steals my words, confusing all concerned.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (positive) infinity - 1 is not really infinity, but is so great that it can be assumed as infinity aswell... or so. XD.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5848" LastActivityDate="2014-11-21T06:57:50.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5331" PostTypeId="2" ParentId="4370" CreationDate="2014-11-21T06:59:57.297" Score="-3" Body="&lt;p&gt;I'm guessing the answer is 0. All the answers are to have an implicit relation with number 5. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 - 6 = 5 |&#xA;3 + 2 = 5|&#xA;2 - 7 = 5|&#xA;11 - 6 = 5|&#xA;20 / 4 = 5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the last two was not so. As the last person is found wrong but the one before that is not means that guest was questioned by the guard who speaks lies and the last guest was questioned by the guard who speaks truth.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   “You were just off by one!”, says one of the guards who looks like the sort of person who would tell the truth – but then again, it’s hard to tell with them.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This means that what the guard told was a lie.&lt;/p&gt;&#xA;" OwnerUserId="5723" LastEditorUserId="5723" LastEditDate="2014-11-21T14:52:39.500" LastActivityDate="2014-11-21T14:52:39.500" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="5333" PostTypeId="2" ParentId="5290" CreationDate="2014-11-21T08:06:17.397" Score="4" Body="&lt;p&gt;Impossible. Walking such a path in reverse, you could never reach the starting point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the number of doors is finite and you never reach the end, you will eventually reach a door for the second time, at which point you will find yourself in a loop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, however, is impossible. You could not have started in a loop, because no door is left of the left-most door. Nor could you have entered a loop, because walking the path in reverse, you could never leave the loop. Notice that any path you walk can be walked in reverse as well, and loops forward are also loops in reverse.&lt;/p&gt;&#xA;" OwnerUserId="5853" LastEditorUserId="5853" LastEditDate="2014-11-22T03:56:57.270" LastActivityDate="2014-11-22T03:56:57.270" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5334" PostTypeId="2" ParentId="3680" CreationDate="2014-11-21T08:35:50.753" Score="2" Body="&lt;p&gt;I'm aware of an algorithm for &lt;span class=&quot;math-container&quot;&gt;$(\frac{n}{2}-1)^2 + 8$&lt;/span&gt; pieces given an even &lt;span class=&quot;math-container&quot;&gt;$n \ge 6$&lt;/span&gt; cuts as follows:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Inscribe a square &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; inside the circle. Let &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; be some distance from the edge of the square, and cut &lt;span class=&quot;math-container&quot;&gt;$\frac{n}{2}$&lt;/span&gt; times, equally spaced between &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; away from one edge to &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; away from the opposite edge, both horizontally and vertically (for a total of &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; cuts).&lt;/p&gt;&#xA;&lt;p&gt;By the Intermediate Value Theorem, there exists some value &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; for which the 8 edge pieces adjacent to the corners and the inside pieces all have equal area, which is a total of &lt;span class=&quot;math-container&quot;&gt;$(\frac{n}{2}-1)^2 + 8$&lt;/span&gt; pieces.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This overtakes the naive solution of &lt;span class=&quot;math-container&quot;&gt;$2n$&lt;/span&gt; at &lt;span class=&quot;math-container&quot;&gt;$n = 8$&lt;/span&gt;. It gives &lt;span class=&quot;math-container&quot;&gt;$17$&lt;/span&gt; pieces for &lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt; cuts, &lt;span class=&quot;math-container&quot;&gt;$24$&lt;/span&gt; pieces for &lt;span class=&quot;math-container&quot;&gt;$10$&lt;/span&gt; cuts, &lt;span class=&quot;math-container&quot;&gt;$33$&lt;/span&gt; pieces for &lt;span class=&quot;math-container&quot;&gt;$12$&lt;/span&gt; cuts, etc.&lt;/p&gt;&#xA;&lt;p&gt;There &lt;em&gt;may&lt;/em&gt; be a solution better than this one, though. Anyone else willing to give it a shot?&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="18129" LastEditDate="2020-10-31T19:35:18.610" LastActivityDate="2020-10-31T19:35:18.610" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="5341" PostTypeId="5" CreationDate="2014-11-21T09:35:49.440" Score="0" Body="&lt;p&gt;A puzzle whose answer will be a word or a set of words but not a phrase, sentence or sentences. Normally crossword puzzles, acrostic puzzles, jumble puzzles, find/guess-the-word type of puzzles, etc. will come under this tag. Do not use this tag to ask word questions which aren't puzzles as they are off-topic and should be asked on &lt;a href=&quot;https://english.stackexchange.com/&quot;&gt;English Language and Usage Stack Exchange&lt;/a&gt; or &lt;a href=&quot;https://ell.stackexchange.com/&quot;&gt;English Language Learners Stack Exchange&lt;/a&gt; (depending on level).&lt;/p&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="70545" LastEditDate="2021-02-05T03:20:24.693" LastActivityDate="2021-02-05T03:20:24.693" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="5342" PostTypeId="4" CreationDate="2014-11-21T09:35:49.440" Score="0" Body="A puzzle whose answer will be a word or a set of words but not a phrase or sentence." OwnerUserId="2730" LastEditorUserId="70545" LastEditDate="2021-02-05T03:33:47.313" LastActivityDate="2021-02-05T03:33:47.313" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="5344" PostTypeId="1" AcceptedAnswerId="5695" CreationDate="2014-11-21T09:38:51.453" Score="25" ViewCount="2144" Body="&lt;p&gt;Police found the body of a computer technician in front of his computer, with a belly wound. There was a possibility of the victim bleeding to death after being stabbed with a sharp object. There were no signs of damage on the windows or doors, which indicated that the killer was someone close to the victim. The dead body was found holding the keyboard and mouse. The monitor displayed the explorer and command prompt. On the user folder of his computer (Windows 8), police found one image file with the name &lt;strong&gt;img.gif&lt;/strong&gt;. They also found blood on the following keyboard keys:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Enter Shift &lt;strong&gt;Space&lt;/strong&gt; Alt F7 = / &lt;strong&gt;.&lt;/strong&gt; A B &lt;strong&gt;C&lt;/strong&gt; E F &lt;strong&gt;G&lt;/strong&gt; &lt;strong&gt;I&lt;/strong&gt; L M O P &lt;strong&gt;R&lt;/strong&gt; U Y&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The bold keys indicates that the key has more blood than others, and here is the image (img.gif):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hJVMb.gif&quot; alt=&quot;img.gif&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The questions is, who killed the technician?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the layout of the keyboard was QWERTY&lt;/li&gt;&#xA;&lt;li&gt;the command prompt was open, but without any text; apparently the victim just deleted the history by Alt+F7&lt;/li&gt;&#xA;&lt;li&gt;there is a possibility that the victim or someone else deleted some files&lt;/li&gt;&#xA;&lt;li&gt;the culprit's name is on the list&lt;/li&gt;&#xA;&lt;li&gt;the keys are not in sequence&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Transcript of image:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Aisyah P       08085733163110 &#xA;Putut H        085648887714 &#xA;Nur Komaroh    72346874 &#xA;Rossa Riasa    085733265290 &#xA;Ullah Prihasti 08563146198 &#xA;Sela-ajah      083857251129 / 085857311079 &#xA;Fonisah S      083857921353 &#xA;Febrina H      085733224890 &#xA;Sriong Putri K 71488407 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;please, only open the spoiler text if you think this riddle is kinda difficult&lt;/p&gt;&#xA;&#xA;&lt;p&gt;clue 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; by solving what was written by the technician, it will be easier to solve this puzzle, but you need a little knowledge of computers for this, if you do not have it, you could just try to find out what is contained in this picture.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;clue 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; steganography&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;more spoiler text will come...&lt;/p&gt;&#xA;" OwnerUserId="5800" LastEditorUserId="5800" LastEditorDisplayName="user9377" LastEditDate="2016-02-08T23:08:03.703" LastActivityDate="2016-02-08T23:08:03.703" Title="Death cases of computer technician" Tags="|mystery|computer-puzzle|steganography|" AnswerCount="2" CommentCount="21" ContentLicense="CC BY-SA 3.0" />
  <row Id="5347" PostTypeId="1" AcceptedAnswerId="5348" CreationDate="2014-11-21T09:52:57.813" Score="37" ViewCount="4499" Body="&lt;p&gt;It's Monday morning. Fernando just sat down at his desk for another hard day at work - programming stuff and answering telephone calls - &quot;have you tried turning it off and on again?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As he is about to login onto his machine he notices a tiny bit of yellow paper sticking out from underneath the keyboard. He moves the keyboard away to find a sticky note with the following text on it:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;51577830494373674D773D3D&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;He is not quite sure what to make of it, but he thinks that the new intern, Ganna, left it for him. The previous night the two were out on a date. He took her to this fancy restaurant and they seemed to have a really good time together. However when they had almost finished eating, she received a call and after a brief moment she said she had an urgent matter to attend to - promptly leaving without offering any explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This made him think that she didn't actually enjoy the date at all - so he didn't quite bother call her afterwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hints:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; #1 (Obvious): Ganna is also a WINDOWS software developer. [Thanks for pointing out guys]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5855" LastEditorUserId="5855" LastEditDate="2014-11-26T09:33:33.323" LastActivityDate="2014-11-26T09:33:33.323" Title="So how did the date go?" Tags="|logical-deduction|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="5348" PostTypeId="2" ParentId="5347" CreationDate="2014-11-21T10:03:57.193" Score="46" Body="&lt;p&gt;Three steps:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since there is no higher letter than F we can assume it is Hex code, if we convert it to ASCII: &lt;code&gt;QWx0ICsgMw==&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This looks like&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a base64-encoded string. If we decode that: &lt;code&gt;Alt + 3&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which gives us the following result when used in a web browser:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ♥&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3885" LastEditorUserId="3885" LastEditDate="2014-11-21T12:05:59.563" LastActivityDate="2014-11-21T12:05:59.563" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="5351" PostTypeId="1" AcceptedAnswerId="5357" CreationDate="2014-11-21T10:43:40.893" Score="9" ViewCount="2175" Body="&lt;p&gt;Hopefully this one will occupy you longer.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I am a ghost&#xA;discarnate and sneaky.&#xA;Invisible at most.&#xA;Catching me is tricky.&#xA;&#xA;I am an awkward loner&#xA;team work is not my thing at all.&#xA;To cooperate with my persona&#xA;you need huge meeting hall.&#xA;&#xA;I am very fair&#xA;and never pick a side.&#xA;You know me, I am everywhere.&#xA;But who am I?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="5283" LastActivityDate="2014-11-21T13:28:59.913" Title="Riddle about small things" Tags="|riddle|word|rhyme|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5352" PostTypeId="2" ParentId="5009" CreationDate="2014-11-21T10:58:55.660" Score="8" Body="&lt;p&gt;IT IS THE:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; DOLL&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Who avoids our threats.&lt;/p&gt;&#xA;&lt;p&gt;There was something hidden in the:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Markup... a mirror (take a look)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If you:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Reflect all the names of the stuffed animals, the Doll is the only one who isn't &amp;quot;killed&amp;quot;. This means that &amp;quot;Doll&amp;quot; is the only words that isn't mangled when you reflect the name on the horizontal axis.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;h2&gt;Let's analyse the clues further:&lt;/h2&gt;&#xA;&lt;p&gt;The title is &amp;quot;Every second is a waste of time&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;Read every second line of the preface:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Ok, as you might have expected I secretly implemented &lt;br&gt;&#xA; a unique, selfmade picture with nothing else but the&lt;br&gt;&#xA; background color (for obvious reasons, I appreciate).&lt;br&gt;&#xA; Go ahead and look for the incredibly well hidden&lt;br&gt;&#xA; description of it and you will, I am absolutely sure,&lt;br&gt;&#xA; find the final clue to this, of course, very easy riddle.&lt;br&gt;&#xA; &lt;br&gt;&#xA; For sure, this is the easiest riddle I have ever seen!&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;courtesy of rand al'thor&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The last line of the secret message is just avigrail being infuriating ( :-) ), but the rest gives us a new message. Now&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; click on the button to edit the question - look into HTML/Markdown,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;and we find, as expected,&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; a link to a picture with nothing else but the background colour,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;but also&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; a hidden message saying &amp;quot;Oh! You just found a mirror on the cabinet.&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Now the clues themselves:&lt;/p&gt;&#xA;&lt;p&gt;The Doll, the Owl, the Dwarf and the Bear&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;These are main objects of the riddle. It seems clear that they will be significant&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;standing on a cabinet they were&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;They are all lined up together&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Looking down on their cruel fate. they already hung their heads&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;They are all &amp;quot;looking down&amp;quot; - i.e. in the same direction. This is the signal for horizontal reflection.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You will try to kill them all&#xA;but one will elude your threats&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;You will mangle them somehow (seems obvious since we have a mirror). BUT &amp;quot;one will elude our threats&amp;quot; (how mysterious). This refers to the fact that &amp;quot;Doll&amp;quot; is the only one of the 4 names of the toys that isn't mangled by the mirror reflection.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;They cannot see who'll be the one&lt;br&gt;&#xA;they are as far as can be&lt;br&gt;&#xA;But maybe you can tell them soon&lt;br&gt;&#xA;so they can die peacefully&lt;br&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;This is just a poetic way to emphatically signal that the final answer will be one of the animals - i.e. that is what we are supposed to identify.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Let the cyanide lift :D&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-11-21T11:18:47.570" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5354" PostTypeId="2" ParentId="5009" CreationDate="2014-11-21T11:08:20.560" Score="2" Body="&lt;h2&gt;Answer&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Mirror.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Speculation&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The mirror is the only connection between the hints and the given poem. At this point I'm suspecting the entire poem is a red herring and you are supposed to find the answer at the first part of the text.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Use of the Hints&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Skip one line each in the first part of the post:&#xA;&quot;Ok, as you might have expected I secretly implemented  a unique,&#xA;selfmade picture with nothing else but the background color (for&#xA;obvious reasons, I appreciate). Go ahead and look for the incredibly&#xA;well hidden description of it and you will, I am absolutely sure, find&#xA;the final clue to this, of course, very easy riddle. For sure, this is&#xA;the easiest riddle I have ever seen!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Final clue&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Dig into the HTML (or markdown) for the image with &quot;nothing but the background color&quot;. Its caption reads: &quot;Oh! You just found a mirror on the cabinet&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5859" LastActivityDate="2014-11-21T11:08:20.560" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5355" PostTypeId="2" ParentId="5351" CreationDate="2014-11-21T11:13:18.410" Score="3" Body="&lt;p&gt;Maybe it can be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Argon (Air gas)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;I am a ghost&#xA;discarnate and sneaky.&#xA;Invisible at most.&#xA;Catching me is tricky.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can't obviously see it, and you can't catch it easily&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;I am an awkward loner&#xA;team work is not my thing at all.&#xA;To cooperate with my persona&#xA;you need huge meeting hall.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since it is about 1% of total air, you need a lot of air in order to have a significantly high concentration of Argon (that's why I think you need a huge meeting hall, because it has lot of air in it). Teamwork is not its thing because its name derives from the Greek word &quot;lazy&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;I am very fair&#xA;and never pick a side.&#xA;You know me, I am everywhere.&#xA;But who am I?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is inert, so it doesn't react to anything. It's everywhere because it is in air.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4527" LastEditorUserId="4527" LastEditDate="2014-11-21T11:22:43.040" LastActivityDate="2014-11-21T11:22:43.040" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5356" PostTypeId="1" AcceptedAnswerId="5427" CreationDate="2014-11-21T11:13:25.457" Score="7" ViewCount="351" Body="&lt;p&gt;In your mind I manifest&lt;br&gt;&#xA;Understanding of me is vague at best  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm said to take away the order&lt;br&gt;&#xA;Less likey to be seen in the daughter  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heightened moods wont decide your fate&lt;br&gt;&#xA;When held onto, is my traits  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;White cloaks might give you sensations&lt;br&gt;&#xA;Line them up for no frustrations  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some I am not what I envision&lt;br&gt;&#xA;Yet oppression is used to describe my prison    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Give it a go and see if you can match every line to the answer. Please if you do answer explain how you arrived at it and how it manages to describe each line.&lt;/p&gt;&#xA;" OwnerUserId="5135" LastActivityDate="2014-11-22T13:58:10.423" Title="Use intuition, It causes a schism. Entity?" Tags="|riddle|word|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5357" PostTypeId="2" ParentId="5351" CreationDate="2014-11-21T11:32:59.000" Score="6" Body="&lt;p&gt;It is a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Neutrino&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a ghost&lt;br&gt;&#xA;discarnate and sneaky.&lt;br&gt;&#xA;Invisible at most.&lt;br&gt;&#xA;Catching me is tricky.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Neutrinos are hard to catch and not visible to the human eye.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am an awkward loner&lt;br&gt;&#xA;team work is not my thing at all.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Neutrinos do not interact with each other, nor with most other particles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To cooperate with my persona&lt;br&gt;&#xA;you need huge meeting hall.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The huge meeting hall is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Neutrino_detector&quot; rel=&quot;nofollow&quot;&gt;neutrino detector&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am very fair&lt;br&gt;&#xA;and never pick a side.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strike&gt;Since they don't collide with other particles, neutrinos don't change direction.&lt;/strike&gt; &lt;strong&gt;Update after OP's clarification:&lt;/strong&gt; not picking a side is about being neutral (having no positive or negative charge).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You know me, I am everywhere.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; True :-)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4394" LastEditorUserId="4394" LastEditDate="2014-11-21T13:28:59.913" LastActivityDate="2014-11-21T13:28:59.913" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5358" PostTypeId="2" ParentId="5289" CreationDate="2014-11-21T11:40:17.237" Score="4" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An account&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When my owner is closed, I am open&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &quot;owner&quot; is the customer. When a bank &quot;closes&quot; a client the account is &quot;opened&quot;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My owner never cleans me out&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The customer will never rob himself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Few can afford to keep me fed&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not many people can keep their account topped up...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because people always take my food&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Money is always required to the spent.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastActivityDate="2014-11-21T11:40:17.237" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5360" PostTypeId="2" ParentId="5351" CreationDate="2014-11-21T11:53:29.177" Score="3" Body="&lt;p&gt;It could be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Higgs Boson&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I am a ghost&#xA;discarnate and sneaky.&#xA;Invisible at most.&#xA;Catching me is tricky.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The existence of Higgs has been very difficult to confirm&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I am an awkward loner&#xA;team work is not my thing at all.&#xA;To cooperate with my persona&#xA;you need huge meeting hall.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Higgs is a loner because it is an elementary particle. To find it you need a Large Hadron Collider&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I am very fair&#xA;and never pick a side.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A bit of a stretch, but a team of international scientists work at CERN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;You know me, I am everywhere.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Higgs field is present everywhere&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4336" LastActivityDate="2014-11-21T11:53:29.177" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5364" PostTypeId="2" ParentId="5351" CreationDate="2014-11-21T12:09:49.560" Score="3" Body="&lt;p&gt;I might be completely off track, but I believe the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A photon &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sentences that lead me to this conclusion are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a ghost&#xA;discarnate and sneaky.&#xA;Invisible at most.&#xA;Catching me is tricky.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A &quot;ghost&quot; as in &quot;can't be perceived when you touch it&quot;. Disincarnate because it has &quot;no&quot; mass, hard to catch because studying it has been historically difficult &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To cooperate with my persona you need a huge meeting hall&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;a person's perceived or evident personality, as that of a well-known official, actor, or celebrity; personal image; public role.&quot; from  &lt;a href=&quot;http://dictionary.reference.com/browse/persona&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.In this sense a photon's persona could be the Sun, thus the &quot;huge meeting hall&quot; to cooperate with it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am very fair&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; fair meaning bright&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and never pick a side&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the particle-wave duality&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can't figure out the other hints&lt;/p&gt;&#xA;" OwnerUserId="5102" LastEditorUserId="5102" LastEditDate="2014-11-21T12:50:28.137" LastActivityDate="2014-11-21T12:50:28.137" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5365" PostTypeId="2" ParentId="5289" CreationDate="2014-11-21T12:18:42.833" Score="10" Body="&lt;p&gt;A variant of Damien's answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an ATM machine.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;When my owner is closed, I am open&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When the bank is closed, the ATM machine is still open.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My owner never cleans me out&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The bank never empties their ATM machines.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Few can afford to keep me fed&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not many people have enough money to supply an ATM machine.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Because people always take my food&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People are withdrawing money from it all the time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Credit to @AeJey for this answer. He didn't think it worked, but it seems to me to fit all the clues. (&quot;When&quot;, not &quot;when and only when&quot;, in the first line!)&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2014-11-21T12:18:42.833" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5366" PostTypeId="2" ParentId="5299" CreationDate="2014-11-21T12:40:57.420" Score="-1" Body="&lt;p&gt;&lt;strike&gt;Not sure if this is right after seeing the other answers, but I think it doesn't matter if you rest on the moving walkway or not (maybe someone can correct me?)&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Apparently it does matter when you choose to walk. It seems walking on the Moving Walkway is the better option. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my previous example I was only calculating with time and not distances. Using my previous reasoning and adding distance this time, my calculations are as followed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll assume the moving walkway moves at 10km/h (same as run speed). This leaves us with 4 speeds:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Run without MW: 166.7 M/Min&lt;/li&gt;&#xA;&lt;li&gt;Walk without MW: 83.3 M/Min&lt;/li&gt;&#xA;&lt;li&gt;Run with MW: 333.3 M/Min&lt;/li&gt;&#xA;&lt;li&gt;Walk with MW: 250 M/Min&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(MW = moving walkway)(numbers have been rounded)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's also assume we have to walk 1km, of which the last half is a Moving Walkway (so 500m normal walkway, 500m moving walkway).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we choose to walk on the normal walkway the whole trip will take us:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1 minute walking&lt;/li&gt;&#xA;&lt;li&gt;2.5 minutes running&lt;/li&gt;&#xA;&lt;li&gt;1.5 minutes running (on the moving walkway)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Total: 5 minutes&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we choose to walk on the moving walkway the whole trip will take us:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;3 minutes running&lt;/li&gt;&#xA;&lt;li&gt;1 minute walking (on the moving walkway)&lt;/li&gt;&#xA;&lt;li&gt;0.75 minutes running (on the moving walkway)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Total: 4.75 minutes&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This difference can be explained by looking at the difference in speeds in ratio. The difference in walking/running on the normal walkway is 1:2, which means you are 50% slower when walking. On the moving walkway the ratio is 3:4, so you're only 25% slower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amount of time you save depends on the speed of the moving walkway. The greater the speed, the more time you save if you rest on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could try to add a calculation here but I'd probably end up copying the other answers :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;strong&gt;My previous (faulty) reasoning:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have 4 different speeds (I changed to meters/minute for easy calculation):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Run without MW: 166.6667 M/Min&lt;/li&gt;&#xA;&lt;li&gt;Walk without MW: 83.3333 M/Min&lt;/li&gt;&#xA;&lt;li&gt;Run with MW: 333.3333 M/Min&lt;/li&gt;&#xA;&lt;li&gt;Walk with MW: 250 M/Min&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(MW = moving walkway)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only two minutes of time are relevant (all other time will be spent running):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;One minute without the moving walkway&lt;/li&gt;&#xA;&lt;li&gt;One minute with the moving walkway&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This gives us two options&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A.&lt;/strong&gt; Run the first minute, walk the second (on the MW)&#xA;&lt;br&gt;&#xA;&lt;strong&gt;B.&lt;/strong&gt; Walk the first minute, run the second (on the MW)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Option A gives us 166.6667 + 250 = 416.6667 meters&#xA;&lt;br&gt;&#xA;Option B gives us 83.3333 + 333.3333 = 416.6666 meters&#xA;&lt;br&gt;&#xA;(I think we can neglect the minor difference, which is probably caused by rounding)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A different way to look at it, is that the moving walkway will always move you 166.6667 meters per minute extra, regardless if you walk or not.&#xA;If you split this value from the rest in the calculation, it becomes more obvious:&#xA;&lt;br&gt;&#xA;Option A gives us 166.6667 + (166.6667 + 83.3333)&#xA;&lt;br&gt;&#xA;Option B gives us (166.6667 + 166.6667) + 83.3333&lt;/p&gt;&#xA;" OwnerUserId="4313" LastEditorUserId="4313" LastEditDate="2014-11-21T13:49:06.990" LastActivityDate="2014-11-21T13:49:06.990" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5367" PostTypeId="2" ParentId="5356" CreationDate="2014-11-21T13:15:09.413" Score="1" Body="&lt;p&gt;All right, I'm giving this a go -&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Anxiety&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In your mind I manifest&#xA;Understanding of me is vague at best&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It is a mental disorder, fight vs flight response. But it is difficult to pinpoint the cognitive or phisiological causes for a general population&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm said to take away the order&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Anxiety and panic tend to create chaos in the mind. It's difficult to reason with internally.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Less likey to be seen in the daughter&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Males under 18 are more likely to show aggression and other symptoms of anxiety than females&lt;sup&gt;&lt;a href=&quot;http://www.apa.org/news/press/releases/2011/08/mental-illness.aspx&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;&lt;/sup&gt;, though mental disorders of this nature is generally diagnosed more often in females.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Heightened moods wont decide your fate&#xA;When held onto, is my traits&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Mood stabilizers and other medicine can only have an effect if the root-cause of the anxiety, and in turn also the tendencies, are also addressed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;White cloaks might give you sensations&#xA;Line them up for no frustrations&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Mental health doctors and therapists have a lot of different methods of diagnosis and can provide relief from the anxiety with therapy and medicine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For some I am not what I envision&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Some envision anxiety as panic, but it can present much differently among people&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet oppression is used to describe my prison&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Living with anxiety feels like a prison because of the way it changes everything everyday. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think the last line for me is the weakest. I highly doubt this is correct, but I couldn't resist seeing if it fit.&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2014-11-21T13:15:09.413" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5368" PostTypeId="2" ParentId="236" CreationDate="2014-11-21T13:20:08.617" Score="8" Body="&lt;p&gt;I was able to more or less understand the solution only by imagining that this whole story is happening in Island 100 - our island, and there are another 99 islands in the ocean, each called Island 1, Island 2, Island 3, ..., Island 99, each of them named after the total number of people with blue eyes in them. The total number of people in each island is the same: 200.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of the islanders knows anything at all about the other islands. Actually, for them the other islands might just be a mental construction in their imagination; but for the sake of our reasoning, let's consider them as real islands. Since the islands do not have any sort of communication between them, Island 100 is &lt;em&gt;exactly&lt;/em&gt; the island of the original problem.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Island 1:&lt;/strong&gt; 1 blue eyed person, 199 brown eyed people.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Island 2:&lt;/strong&gt; 2 blue eyed people, 198 brown eyed people.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Island 3:&lt;/strong&gt; 3 blue eyed people, 197 brown eyed people.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Island 4:&lt;/strong&gt; 4 blue eyed people, 196 brown eyed people.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Island 5:&lt;/strong&gt; 5 blue eyed people, 195 brown eyed people.&lt;/li&gt;&#xA;&lt;li&gt;...&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Island 99:&lt;/strong&gt; 99 blue eyed people, 101 brown eyed people.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Island 100:&lt;/strong&gt; 100 blue eyed people, 100 brown eyed people.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The rules are equal in every island - people will leave when they find out their eye colour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a given day, the guru, travelling on a boat, does the same operation in every island.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;On each day &lt;em&gt;N&lt;/em&gt;, the &lt;em&gt;N&lt;/em&gt; blue eyed people from Island &lt;em&gt;N&lt;/em&gt; will leave.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that the &lt;em&gt;N-1&lt;/em&gt; blue eyed people seen by any blue eyed observer on any island &lt;em&gt;didn't&lt;/em&gt; leave the day before convinces that observer that they are &lt;em&gt;actually&lt;/em&gt; in Island &lt;em&gt;N&lt;/em&gt;, and not in Island &lt;em&gt;N-1&lt;/em&gt;. (The only two possible islands they could be in, since each of them knows that there are either &lt;em&gt;N-1&lt;/em&gt; or &lt;em&gt;N&lt;/em&gt; blue-eyed people on their island.)&lt;/p&gt;&#xA;" OwnerUserId="5591" LastEditorUserId="5591" LastEditDate="2018-05-30T10:23:04.450" LastActivityDate="2018-05-30T10:23:04.450" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="5369" PostTypeId="2" ParentId="5089" CreationDate="2014-11-21T13:28:37.677" Score="0" Body="&lt;p&gt;Perhaps she, as&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Isaac Newton&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is upset with&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gottfried Leibniz&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it is unclear which of the two created the idea of calculus (Newton's &quot;fluxions and fluents&quot;) first, and Newton wanted full credit for the original discovery.&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5121" LastActivityDate="2014-11-21T13:28:37.677" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5370" PostTypeId="1" AcceptedAnswerId="5371" CreationDate="2014-11-21T13:50:43.060" Score="16" ViewCount="7513" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am precious to you,&lt;br&gt;&#xA;  but kill me you do.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I am older than you,&lt;br&gt;&#xA;  but still ever new.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I cannot be sold,&lt;br&gt;&#xA;  for money or gold.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I can EVEN fly.&lt;br&gt;&#xA;  Yet wings have not I.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This is no game,&lt;br&gt;&#xA;  what is my name?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Give the name and  a brief reasoning.&lt;/p&gt;&#xA;" OwnerUserId="5567" LastEditorUserId="2729" LastEditDate="2014-11-22T23:35:42.123" LastActivityDate="2014-11-23T21:39:21.247" Title="I am precious to you, but kill me you do. ... What is my name?" Tags="|riddle|wordplay|" AnswerCount="7" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5371" PostTypeId="2" ParentId="5370" CreationDate="2014-11-21T13:54:30.680" Score="27" Body="&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation&lt;/strong&gt;&lt;br&gt;&#xA;I am precious to you,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Time is precious&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but kill me you do.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One kills time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am older than you,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Time has always been&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but still ever new.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What has not been seen is fresh&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cannot be sold, for money or gold.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Money does not buy time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can EVEN fly. Yet wings have not I.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Time flies&quot;. AND it progresses in an &quot;EVEN&quot;/steady/constant manner - i.e. &quot;at an even pace&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is no game,&#xA;what is my name?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Poetic style :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-11-21T14:09:04.260" LastActivityDate="2014-11-21T14:09:04.260" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="5372" PostTypeId="2" ParentId="5370" CreationDate="2014-11-21T13:54:40.377" Score="10" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;I am precious to you,&#xA;but kill me you do.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Time is money&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;I am older than you,&#xA;but still ever new.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is always a quantity of time bigger than your age&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;I cannot be sold,&#xA;for money or gold.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You cannot buy time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;I can EVEN fly.&#xA;Yet wings have not I.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sometimes you say &quot;time flies&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4527" LastActivityDate="2014-11-21T13:54:40.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5373" PostTypeId="2" ParentId="5370" CreationDate="2014-11-21T13:57:32.317" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Time. &lt;br&gt; Time is precious, and yet sometimes we &quot;kill&quot; it, as in waste it. &lt;br&gt;&#xA;He's &quot;older&quot; than anyone, since it existed before anyone, and yet there's always more time, so &quot;when&quot; we're living is, technically, new time. &lt;br&gt;Can't be sold, well, I dont know what more I can say about this :) &lt;br&gt; Time flies, usually when you're having fun, despite not having wings &lt;br&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;edit: Ninja'd :(&lt;/p&gt;&#xA;" OwnerUserId="5102" LastEditorUserId="5102" LastEditDate="2014-11-21T15:01:03.023" LastActivityDate="2014-11-21T15:01:03.023" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5374" PostTypeId="1" AcceptedAnswerId="5381" CreationDate="2014-11-21T14:12:53.867" Score="7" ViewCount="2575" Body="&lt;p&gt;Inspector Rottman is investigating the murder of a young clergy assistant in the quiet village of Roeville.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having already done some detective work, he has the following three main suspects:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Trevor Matthews, the vicar.&lt;/li&gt;&#xA;&lt;li&gt;Benedict Leith, the organist. &lt;/li&gt;&#xA;&lt;li&gt;Lionel Richardson, the choirmaster.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The Inspector is struggling to identify the culprit until he receives the following tip-off:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The altar boy behind the church.&lt;br/&gt;&#xA;  The crows they flit from perch to perch.&lt;br/&gt;&#xA;  Monday yes but Tuesday no,&lt;br/&gt;&#xA;  Weds I stay but Thurs I go.&lt;br/&gt;&#xA;  I’m in Friday, Saturday out.&lt;br/&gt;&#xA;  Sunday’s odd so I say nowt.&lt;br/&gt;&#xA;  Every other second counts,&lt;br/&gt;&#xA;  The alter-ego presence mounts.&lt;br/&gt;&#xA;  From black to white to black to white&lt;br/&gt;&#xA;  Altruism wins the fight.&lt;br/&gt;&#xA;  Solve this puzzle then you  can,&lt;br/&gt;&#xA;  Change your way to find your man.&lt;br/&gt;&#xA;  &lt;br/&gt;&lt;br/&gt;&#xA;  &quot;0153&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who is the murderer and how do you know (provide full explanations for each clue)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint #1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The rhyme will get you half way there, but the numbers are what will really reveal the secret.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2665" LastEditorUserId="5373" LastEditDate="2014-11-22T11:18:25.363" LastActivityDate="2014-11-24T18:18:17.730" Title="Who committed the murder?" Tags="|riddle|mystery|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5376" PostTypeId="1" AcceptedAnswerId="5706" CreationDate="2014-11-21T14:23:02.380" Score="6" ViewCount="1091" Body="&lt;p&gt;&lt;pre&gt;Hello everybody, can someone help me? This riddle really gave me the gilles so far. &#xA;&lt;strike&gt;I need to solve this for school but I have no idea!&lt;/strike&gt; joking ;) &#xA;Not sure if it will help - &lt;a href=&quot;http://en.wikipedia.org/wiki/James_while_John_had_had_had_had_had_had_had_had_had_had_had_a_better_effect_on_the_teacher&quot; rel=&quot;nofollow&quot;&gt;check this out. So funny!&lt;/a&gt; &lt;/pre&gt;&lt;br&gt;&#xA;You're very close&lt;br&gt;&#xA;But I'm always ahead&lt;br&gt;&#xA;Can be beheaded&lt;br&gt;&#xA;but can't be fed&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can have&lt;br&gt;&#xA;a look inside&lt;br&gt;&#xA;there will be&lt;br&gt;&#xA;no place to hide&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However,&lt;br&gt;&#xA;You try to enter me&lt;br&gt;&#xA;and more room arises&lt;br&gt;&#xA;suddenly&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you explain this trick to me?&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This is so easy!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5106" LastEditorUserId="5106" LastEditDate="2015-01-08T07:07:06.607" LastActivityDate="2015-01-08T07:07:06.607" Title="Headings had had their hints long long time ago" Tags="|riddle|word|" AnswerCount="5" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="5377" PostTypeId="1" AcceptedAnswerId="5388" CreationDate="2014-11-21T14:54:34.033" Score="13" ViewCount="1474" Body="&lt;p&gt;A presence almost feline seems to follow wherever you go, such is your grace. The brilliance of thy concentrat'd beauty would pierce a steel aegis as sure as my heart, though through you the beauty of company kept only increaseth. Fiercer than the sword of Uriel, your shine is to mine eyes more potent than the heavens. Though thou hast deigned to speed arrows to me, a wound is a creature that can be left banish'd by your touch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Belt me not as I ascend to your bedchamber and kiss your cool lips, for my adulation deluxe is beyond control.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Name the intended recipient of the letter with your reasoning behind it.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5626" LastEditorUserId="5626" LastEditDate="2014-12-09T17:00:36.477" LastActivityDate="2014-12-09T17:00:36.477" Title="Victorian Love Letter - The name has rubbed off. Who do you deliver it to?" Tags="|riddle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5378" PostTypeId="2" ParentId="5299" CreationDate="2014-11-21T14:55:21.483" Score="-1" Body="&lt;p&gt;Lets say we have to rest 1 hour instead of 1 min for easier calculation. Speed increase is linear if you run\walk on moving walkways(MV). &#xA;Distance you will cover on MV: &#xA;1)run  (10 + 10) * 1 = 20 km&#xA;2) rest (10 + 5) * 1 = 15 km&#xA;Distance you will cover on road:&#xA;1) run 10 * 1 = 10 km&#xA;2) rest 5 * 1 = 5 km&#xA;What we see here is that difference covered distance is constant to 5km, so the answer is doesn't matter, take a rest whenever you want. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPD: Another way of thinking. We can translate to new coordinate system, where all speed is decreased by 5km\h. This means we run 5km\h and MV is also 5km\h and now rest means stop moving. Obviously we get the same answer - doesn't matter.&lt;/p&gt;&#xA;" OwnerUserId="5816" LastEditorUserId="5816" LastEditDate="2014-11-21T15:02:04.630" LastActivityDate="2014-11-21T15:02:04.630" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5380" PostTypeId="2" ParentId="5377" CreationDate="2014-11-21T15:15:02.570" Score="6" Body="&lt;p&gt;The intended recipient is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Spirit (strong distilled alcoholic drink)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;written by:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alcoholic&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reasoning:&lt;/strong&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;feline&quot; could be interpreted as: &#xA;&lt;em&gt;&quot;sly, stealthy, or treacherous&quot;&lt;/em&gt;&#xA;rather than,&#xA;&lt;em&gt;&quot;belonging or pertaining to the cat family&quot;&lt;/em&gt;&#xA;according to &lt;a href=&quot;http://www.thefreedictionary.com/feline&quot; rel=&quot;nofollow noreferrer&quot;&gt;thefreedictionary&lt;/a&gt;.&lt;br&gt;&#xA;&lt;br&gt;&#xA;The alcoholic loves the presence and feeling of alcohol which he describes as grace. He feels her treacherous presence everywhere he goes, because he is the one seeking after her (unintentionally).&lt;br&gt;&#xA;&lt;br&gt;&#xA;&quot;Concentrat'd beauty&quot; also speaks of spirit (as in strong concentrated alcohol).&lt;br&gt;&#xA;&lt;br&gt;&#xA;&#xA;He speaks of piercing with arrows and the ferocity of a sword -&gt; the effect of alcohol, which however can be banished, once they meet again! :D&lt;br&gt;&#xA;(besides as &lt;a href=&quot;https://puzzling.stackexchange.com/users/1563/geobits&quot;&gt;Geobits&lt;/a&gt; points out: &lt;em&gt;Alcohol is/was often used as a sterilization method for wounds&lt;/em&gt;)&lt;br&gt;&#xA;&lt;br&gt;&#xA;He also describes what her shininess does to his eyes -&gt; the shiny look of the spirit in the bottle (especially after a few...).&lt;br&gt;&#xA;&lt;br&gt;&#xA;That is why he wants to go into her bedchamber even this night and kiss her once again... kiss her cool lips (bottle/glass lips)... emphasizing he's got no control over his situation! :D &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It reminds me very much of the Bible proverb:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;a href=&quot;https://www.biblegateway.com/passage/?search=Proverbs23%3A30-36&amp;amp;version=ESV&quot; rel=&quot;nofollow noreferrer&quot;&gt;Proverbs 23:30-35&lt;/a&gt;&lt;/strong&gt; (ESV)&lt;br&gt;&#xA;Those who tarry long over wine;&lt;br&gt;&#xA;    those who go to try mixed wine.&lt;br&gt;&#xA; Do not look at wine when it is red,&lt;br&gt;&#xA;   when it sparkles in the cup&lt;br&gt;&#xA;    and goes down smoothly.&lt;br&gt;&#xA; In the end it bites like a serpent&lt;br&gt;&#xA;    and stings like an adder.&lt;br&gt;&#xA; Your eyes will see strange things,&lt;br&gt;&#xA;    and your heart utter perverse things.&lt;br&gt;&#xA; You will be like one who lies down in the midst of the sea,&lt;br&gt;&#xA;    like one who lies on the top of a mast.&lt;br&gt;&#xA; “They struck me,” you will say, “but I was not hurt;&lt;br&gt;&#xA;    they beat me, but I did not feel it.&lt;br&gt;&#xA; When shall I awake?&lt;br&gt;&#xA;    I must have another drink.”&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5567" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2014-11-21T15:48:37.900" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="5381" PostTypeId="2" ParentId="5374" CreationDate="2014-11-21T15:39:26.243" Score="8" Body="&lt;p&gt;The answer is ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Trevor Matthews&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is because...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The entire poem is a reference to the word &quot;Alt&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The altar boy behind the church.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ALTar boy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The crows they flit from perch to perch.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Perch to perch&quot; means they are ALTernating their perches.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Monday yes but Tuesday no,&lt;br&gt;&#xA;Weds I stay but Thurs I go.&lt;br&gt;&#xA;I’m in Friday, Saturday out.&lt;br&gt;&#xA;Sunday’s odd so I say nowt.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes then no, stay then go, in then out.  All forms of ALTernation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Every other second counts,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Self-explanatory.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The alter-ego presence mounts.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ALTer-ego&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From black to white to black to white&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ALTernating yet again, this time with colors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Altruism wins the fight.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ALTruism&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solve this puzzle then you can,&lt;br&gt;&#xA;Change your way to find your man.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Refering to using the ALT key to &quot;change your way&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Putting the poem clue together with &quot;0153&quot;...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alt+0153 is the keyboard shortcut for the Trademark symbol that looks like &quot;TM&quot;, ™.  This references the initials of the murderer, Trevor Matthews.  Incidentally he is also the suspect that would spend the most time on the ALTar.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;BELOW IS MY PREVIOUS ANSWER PURELY FOR REFERENCE:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not a complete answer since I do not know how to make all the clues fit.  But here is what I have so far.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Benedict Leith, the organist, is the murderer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Every other second counts,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Could be a reference of timing in music.  This could also apply to the choirmaster...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From black to white to black to white&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Keys on an organ are like those of a piano, black and white.  This is what made me think about this at first.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;0153&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; These look like the parts of the major chord except the 0.  The root (1), third (3), fifth (5).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Honestly, I am still trying to work out what the days mean but maybe this will spark some ideas for others.&lt;/p&gt;&#xA;" OwnerUserId="4191" LastEditorUserId="4191" LastEditDate="2014-11-24T18:18:17.730" LastActivityDate="2014-11-24T18:18:17.730" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="5382" PostTypeId="2" ParentId="5313" CreationDate="2014-11-21T15:57:05.810" Score="4" Body="&lt;p&gt;The wall was&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 33 1/3 units long.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is because &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dirt -&gt; Thirt(y)&lt;br&gt;&#xA; Tree -&gt; Three&lt;br&gt;&#xA; dog poop = turd &gt; third  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So it was&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dirt, a tree, and a turd.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5874" LastActivityDate="2014-11-21T15:57:05.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5383" PostTypeId="2" ParentId="5141" CreationDate="2014-11-21T16:00:10.843" Score="5" Body="&lt;p&gt;&lt;strong&gt;EXPLANATION FOLLOWED BY CONCLUSION&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonsensical abstraction high up in the army; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;general abstract nonsense&quot; - Possibly meaning that the lines are not to be taken as talking about the same thing. But rather that they are to be linked at a later time - i.e. that things here are not going to be strongly semantically or syntactically coupled. In general though, this line yields &quot;&lt;strong&gt;CAT&lt;/strong&gt;EGORY THEORY&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The woman with the wheel fires the cannon. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; St. Catherine of the Wheel (Catherine of Alexandria) was tortured and executed on a wheel. She is part of the Christian Canon (homophone of &quot;Cannon&quot;) - &quot;firing&quot; may refer to inspiring vitality or possibly as in &quot;firing clay in a kiln&quot; (as in cementing). Regardles, this line yields &quot;ST &lt;strong&gt;CAT&lt;/strong&gt;HERINE&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;She's very fast but only in reverse gear.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; She is the fastest thing there is - Tachyon. So, she's in reverse gear - &quot;NOYH&lt;strong&gt;CAT&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Certainly a terrorist, but only just starting.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;C&lt;/strong&gt;ertainly &lt;strong&gt;A&lt;/strong&gt; &lt;strong&gt;T&lt;/strong&gt;errorist. Lynch mob, anyone? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The title&lt;/strong&gt;&lt;br&gt;&#xA;pray you can find the word!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This refers to very popular &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4615/are-you-clever-enough-you-will-have-to-pray&quot;&gt;riddle recently by avigrail&lt;/a&gt;. There it referred to &quot;prey&quot; (as a homophone) because we're looking for a predator. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What word am I getting at here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There only coherent thread I can tie between them all is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CAT&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2014-11-22T15:23:04.260" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="5384" PostTypeId="2" ParentId="5309" CreationDate="2014-11-21T16:08:49.007" Score="4" Body="&lt;p&gt;Well I'm muddling towards something. No idea if it's an answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The author is Proserpine/Persephone - painted by Rembrandt in 1631 and Leighton in 1891.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the 1891 painting Proserpine is returning to the underworld implying it's autumn/ winter - I'm guessing this is what you mean by shifting gaze (glaukopis is greek for gaze) on an inverdant (away from green-ness) axis. and the formula refers to the changing seasons forever?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are lots of paintings of persephone but after 1891 perhaps fewer than before and it would be hard to count these as &quot;releases&quot; in an authoritative way so I'm going to assume you mean years (releases from the underworld) and go for autumn 1894 (...or 1895 if you mean 3 more releases and then a final release)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Clarifications from the author - Major spoilers:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes, Persephone is far too busy. She wrote the program to automate the season change. This version of the seasons is the southern hemisphere's autumn, as is noted by two northern nations needing to use two versions previous. You're right about the artists and the dates. Glaukopic shift references greek culture and a shift in what is seen. Verdant means green, so inverdant would mean the shift from green as the leaves change. The delta symbol stands for change in most scientific uses and the 1/4 refers to the first of four seasons (again Southern hemisphere) in a cycle continuing forever. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;then&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Three more releases in the year. Three more seasons. The link is to an interactive something where someone eats and gets fat before the cycle being forced to start again. This references the cycle of harvest every year. You can't pause the change of the seasons and each shift in seasons is an entry and a return. Did you miss the bit about the pathways? say it aloud without the periods. BTW, year is irrelevant.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;then&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Yeah, Sam. Since you couldn't load the link, it's this strange page where you control the arm of a woman. Apples keep rolling your way and feeding them to her fattens her up. After so much fattening, she gets stabbed in the back and the apples all tumble out, restarting the odd cycle. If you would, Sam, can you add the extra clues from my comments into your answer so I can delete the comments? I want the rest hidden by spoiler tags too so others can have fun guessing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5875" LastEditorUserId="5875" LastEditDate="2014-11-21T17:23:52.363" LastActivityDate="2014-11-21T17:23:52.363" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5386" PostTypeId="2" ParentId="5309" CreationDate="2014-11-21T16:20:38.853" Score="2" Body="&lt;p&gt;No idea if I'm on the right track, but let's give it a shot:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The versions point to years 1631 and 1891. In these two years, Rembrandt and Leighton, respectively, both had a few works of art, but the only two that have a connection of some sort are &quot;Andromeda Chained to the Rocks&quot; by Rembrandt and &quot;Perseus and Andromeda&quot; by Leighton. So the answer thus far is &lt;b&gt;Andromeda&lt;/b&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastActivityDate="2014-11-21T16:20:38.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5387" PostTypeId="2" ParentId="5370" CreationDate="2014-11-21T16:20:45.020" Score="29" Body="&lt;p&gt;Something different!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bacteria in/on your body&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am precious to you, but kill me you do.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We wouldn't be able to survive long without it however our body is constantly killing off bacteria  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am older than you, but still ever new.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's been around longer than humans but what is with us is constantly rejuvenating.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cannot be sold, for money or gold.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not so typical to sell your bacteria (unlike our time how many people get paid per the hour...)   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can EVEN fly. Yet wings have not I.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We sneeze, its airborne... &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5873" LastActivityDate="2014-11-21T16:20:45.020" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5388" PostTypeId="2" ParentId="5377" CreationDate="2014-11-21T16:37:04.430" Score="14" Body="&lt;p&gt;The recipient is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a laser pointer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A presence almost feline seems to follow wherever you go, such is your grace. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; cats love to chase the dot from the laser&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The brilliance of thy concentrat'd beauty would pierce a steel aegis as sure as my heart,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The laser is a concentrated beam of light, and some types can be used to cut through material&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;though through you the beauty of company kept only increaseth. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; cosmetic lasers*&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fiercer than the sword of Uriel, your shine is to mine eyes more potent than the heavens. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; looking into a laser can easily damage eyes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though thou hast deigned to speed arrows to me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; laser sights on weapons&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;a wound is a creature that can be left banish'd by your touch.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; laser surgery&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Belt me not as I ascend to your bedchamber and kiss your cool lips, for my adulation deluxe is beyond control.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; adulation de luxe -&gt; lover of light*&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;*Missing explanations provided by OP&lt;/p&gt;&#xA;" OwnerUserId="5877" LastEditorUserId="5877" LastEditDate="2014-11-21T16:48:22.353" LastActivityDate="2014-11-21T16:48:22.353" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5389" PostTypeId="1" AcceptedAnswerId="5856" CreationDate="2014-11-21T16:44:28.563" Score="17" ViewCount="2140" Body="&lt;p&gt;The following message can be unambiguously decoded to reveal an eleven letter word. It can be deciphered with pen and paper, although knowing a bit of Python would probably help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the word?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;o83AfLftzPrpZPdflFo4AFlhTFIeOpvoRd5f5lFDfPrqnFudFfiSa&#xA;  tHFiSapbVOp0vorvmApD4fLFpBvo1oaiEofF2lhTdFprpBvo2FieO&#xA;  9dFlFdfPrfD3ifaPFieOoaiEko8iPhODfLfpb32vODfPrpVorPbvO&#xA;  iF2c9pIPhoDFLff83DifaSFtOacoNQvmAppvqzUeOaftDFPrdf2lF&#xA;  PVODieHOvA3QKQzUeDFp5rvaqKOp7zprpOcVdiehopBvodFprfD30&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;(Note: The original text had to be edited, but it still available in the &lt;a href=&quot;https://puzzling.stackexchange.com/revisions/5389/4&quot;&gt;edit history&lt;/a&gt; if anyone needs it.)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't think of any more clues I can offer without giving the whole thing away, but the answer can be obtained without any fancy arithmetic.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Clue 1:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Don't eat the fish&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Clue 2:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Try the bacon instead&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Clue 3:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; And for dessert, perhaps the Cranberry Tart or Stilton Cheese?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Clue 4:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;em&gt;The keyword is important&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="40" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2014-12-03T16:42:37.343" Title="A riddle wrapped in a mystery inside an enigma" Tags="|cipher|" AnswerCount="2" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="5390" PostTypeId="2" ParentId="5242" CreationDate="2014-11-21T19:30:54.817" Score="2" Body="&lt;p&gt;Here is one possible move sequence with nine moves:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BbFfGiJfHhGgFhGhIh&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The program seems to prioritize threats and then surround if none are available. By shooting one of your red blocks in the leg and leaving him in the corner, it lets you set up without hindrance. The strategy is to create a series of forced moves in spaces that needn't be occupied culminating in a cross.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's actually a lot like playing chess.&lt;/p&gt;&#xA;" OwnerUserId="5626" LastEditorUserId="5626" LastEditDate="2014-11-21T19:42:11.713" LastActivityDate="2014-11-21T19:42:11.713" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5391" PostTypeId="1" AcceptedAnswerId="5482" CreationDate="2014-11-21T20:08:57.330" Score="5" ViewCount="529" Body="&lt;p&gt;Warning: Most of my posts are riddles disguised as something else. For those who have enjoyed my contribution and are expecting more, DO NOT try to guess that the answer is a tugboat, geese, or the Winter Bonanza sale. This is NOT a riddle. You have been warned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a question that has frustrated me to no end for the longest time. In a 1 on 1 on 1 game of skill and strategy, there are only two outcomes assuming roughly equal skill. Either (A) two players conflict and the victor (now weakened) is destroyed by the third player or (B) two players both focus their efforts on the same opponent to easily destroy them. The victor who contributed most (and was weakened because of it) is destroyed afterwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The argument can be made that all three fighting both opponents at the same time is possible, but it can't derive an outcome. Such game play can only result in an even stalemate until one player changes tactics, resulting in A or B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's simple and boring, with the outcome more dependent upon chance (or how disliked you are by the other players) than anything. Making the game itself complex can give allow players to turn the tables, but &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the simplest possible 1 on 1 on 1 game that can be designed such that it discourages these outcomes and encourages something more interesting?&lt;/p&gt;&#xA;" OwnerUserId="5626" LastActivityDate="2014-11-24T03:38:14.873" Title="Question about 1 on 1 on 1" Tags="|game-theory|" AnswerCount="4" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="5392" PostTypeId="2" ParentId="5391" CreationDate="2014-11-21T20:24:00.093" Score="4" Body="&lt;p&gt;Have some mechanics that give a player an advantage when two others are working together, and a disadvantage when two others are fighting. Balanced correctly, the whole thing can be a negative feedback loop that only clever strategy can overcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe one way to do this is to have a circular advantage loop, like rock-paper-scissors. So if Rock wants to work with Paper to bring down Scissors, it needs to protect Paper against Scissors and also mount its own defense. Rock ends up losing resources on two fronts, faring no better than Scissors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Rock fights Scissors alone, make it so Rock will actually gain great strength by defeating Scissors (perhaps by commandeering resources), so Paper has an incentive to even things out, because even Rock, if built up enough, can defeat Paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;****EDIT (because I don't have the rep to make comments) ****&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding chance ruining strategy, that's true only if the players are infinitely intelligent. Complex rules can make a system unpredictable for mere humans, which is exactly what noise does. The only difference between chance and complex rules is that chance is true noise.&lt;/p&gt;&#xA;" OwnerUserId="5341" LastEditorUserId="5341" LastEditDate="2014-11-21T21:44:47.177" LastActivityDate="2014-11-21T21:44:47.177" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="5393" PostTypeId="2" ParentId="5391" CreationDate="2014-11-21T20:30:06.367" Score="3" Body="&lt;p&gt;Two quick options:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;1 on 1 rounds&lt;/strong&gt;:  Three people want to play charades and don't know each other (simpler would be guess what number I am thinking).  Adam goes to the front, acts out a clue, and lets Becky and Clara guess.  Which ever gets it right wins a point.  Becky goes up and does the same.  Clara does it too.  We then start again with Adam.  As the competition for each round is 1 on 1; teaming up is not possible.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Three player Paper Scissors Rock&lt;/strong&gt;:  Some games can be designed such that no teamed up strategy offers an advantage for the team.  We dictate that if all pick different or the same there is a tie and they go again.  If two pick the same, they act as one move against the other player. The two remaining players play against each other. There is no advantage for both teammates whether they chose to play differently or the same.  (Note: that this assumes that prizes for 1st, 2nd, and 3rd are chosen so a 50-50 chance of first and last is as valuable as a 100% chance of getting second.)&lt;br&gt;&#xA;For a strategy video game, this means that you cause teaming up to hurt yourself.  One example follows like this: Adam has archers who can kill unarmored berzerkers; Becky has berzerkers who are great against the wussy armored cavaliers.  Clara's cavaliers, however, can reach archers in time to cut them down.  If Adam teams up with Clara he will lose just after Becky does.  Becky won't team up with him though for the same reason. You will end up with an odd game, therefore, where everyone is trying to fortify themselves and the neighbor they have an advantage against while weakening their kryptonite.  Teaming up, however, is out of the question until very end game.&lt;br&gt;&#xA;A more interesting option that is closer to RPS would be as follows:  Adam picks auburn which is really red.  Becky and Clara think he could have picked amber or aqua though (yellow or blue).  He doesn't know their colors either.  Red burns the yellow grain. Yellow drink the blue water. Blue douses the red flames.  The strategy game is to collect your own color powerups which is what decides victory in a tie.  If they guess what color you are, however, they take away your powerups easier than you can collect them.  At end game, you colors are revealed and the purpose of your strategy becomes clear.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;One more option since I can:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Attacking one player does not work&lt;/strong&gt;: Design a game where an &quot;attack&quot; damages both other players equally.  This would require some thought to build but it is possible.  The easiest seems to be either disguised one player games (you look like you fight each other but are really just trying to build your score faster) or a game with a shared resource pool.  You win by taking what you need before the others can not buy killing them.&lt;br&gt;&#xA;Otherwise: Adam is amber; Becker, blue; Clara, Crimson.  Adams economy relies on growing oranges and guar.  Becky needs guar and plums. Clara needs plums and oranges.  Without either they fail.  Your best strategy to win the game is to tend your fields with your foes while offering goods to summon blights on you opponents shared fields.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditDate="2014-11-21T22:52:36.520" LastActivityDate="2014-11-21T22:52:36.520" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5394" PostTypeId="2" ParentId="5391" CreationDate="2014-11-21T20:56:26.843" Score="1" Body="&lt;p&gt;&lt;strong&gt;Resurrection Mechanic&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A player must defeat both opponents in order to win. If he's the last standing and did not defeat both opponents, the opponent that he didn't personally defeat is resurrected with equal abilities / units / stats. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be reflected in real life in a situation where a regime has fallen, but there is still outlying individuals that believe in the cause and they are lying in wait for the perfect opportunity to strike.&lt;/p&gt;&#xA;" OwnerUserId="5889" LastActivityDate="2014-11-21T20:56:26.843" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="5395" PostTypeId="1" AcceptedAnswerId="5401" CreationDate="2014-11-21T22:46:13.110" Score="5" ViewCount="719" Body="&lt;p&gt;So, the official guide shows you exactly how to solve the white cross, first layer, etc. but on the second layer, assume you've completed every little square but one (and edge) and that there's 2 edges if that colour on the bottom face. Making it (or at least seem to be) impossible. What do you do in this case, because I cannot seem to find it online.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HIQZl.jpg&quot; alt=&quot;Cube Position&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-11-21T23:43:22.970" LastActivityDate="2014-11-21T23:45:02.543" Title="What to do when the official Rubik's cube solving sequence breaks?" Tags="|rubiks-cube|minimum-solve|solvability|" AnswerCount="2" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="5396" PostTypeId="1" AcceptedAnswerId="5504" CreationDate="2014-11-21T23:03:05.393" Score="13" ViewCount="2433" Body="&lt;p&gt;This is a multi-part challenge; you will need stamina to complete it! The style of this problem may remind you of some of my friend &lt;a href=&quot;https://puzzling.stackexchange.com/questions/5009/are-you-clever-enough-every-second-is-a-waste-of-time&quot;&gt;avigrail&lt;/a&gt;'s riddles, and I will admit to having pinched a few ideas from him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, let's get down to business. The real answer is in the ciphertext at the very bottom, but before you get on to that, you need to find the key. That's actually really easy, but if you need a hint (or if you just like riddles), try to work out the message hidden in the following verses. Each verse gives you a single item, which you can then put together to get the hint you need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Dark girl she will be, &lt;br&gt;&#xA;Enemies to set free. &lt;br&gt;&#xA;&#xA;The place where you sing, &lt;br&gt;&#xA;Starts with your needed thing.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am I and you are you; &lt;br&gt;&#xA;I sound like what you read this through. &lt;br&gt;&#xA;And if this thing you still can't guess, &lt;br&gt;&#xA;Go to sea and shout out yes.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am as handsome as can be. &lt;br&gt;&#xA;I am as ugly as can be. &lt;br&gt;&#xA;I am as clean as can be. &lt;br&gt;&#xA;I am as dirty as can be. &lt;br&gt;&#xA;In the middle of my name &lt;br&gt;&#xA;Is what you need to find fame.&lt;/em&gt;&#xA;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Some people use me for addition. &lt;br&gt;&#xA;Some turn me into dust and bag me. &lt;br&gt;&#xA;Some take me to a course and play. &lt;br&gt;&#xA;I can be twenty or one-oh-three.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, enough fun and games. It's time to get on to the real challenge. So now you need to board a plane and travel to the land of clean shoes and long sticks. When you get there, you'll have the key you need to unlock the following message and become the final champion! Remember, whatever the name may tell you, it's actually Bellaso who's helping you now.&#xA;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;VHF BRT QSF ZALNPXLY. ZLN SBVT MLTPXO BLA QSF QXDUS PKO NXWP JT IL EIB YTOAA XYTTXC. ZOJO ZOB KPNAXKTOD MLTK XP EP TKTUE P OTEAEP PF NLFS LPY BNS MZTB BE BS P NFFPMTPN DK EIFL DJTT. QSFOX LSE ALET LY AFOEIP PRM EIEGB DBIBGBTXKR GLK XPRT OTEAEPT. (TWBCF XKP BLHL DPJX HIO LXYU QH MBN IEPN YNE OEKBC NFGO UHPQ WPQ). ZZPD ARNL XGO IAEMJ SFWOMICD.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2014-12-26T00:07:59.243" Title="Multi-part challenge: riddles and cipher" Tags="|riddle|cipher|" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="5398" PostTypeId="2" ParentId="5395" CreationDate="2014-11-21T23:15:32.180" Score="0" Body="&lt;p&gt;It's very difficult to understand exactly what situation your cube is in based on your description, but I'll do my best...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an online guide that tells you what you need to do &lt;a href=&quot;http://www.rubiks.com/blog/how-to-solve-the-rubiks-cube&quot; rel=&quot;nofollow&quot;&gt;http://www.rubiks.com/blog/how-to-solve-the-rubiks-cube&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Side note: I solve the cube beginning with the bottom layer, so by &quot;top layer&quot; I am referring to the part which is currently unsolved (that may be your bottom layer?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on your description, I believe you are in step four of that tutorial, completing the second layer. There are several possible situations I think you could be in (based on the confusing description), most of which are solveable. I interpret your description as there are either one or multiple pieces that are in the right places, but rotated incorrectly (or two pieces on the second layer are in swapped positions). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This situation occurs if you brought the edge piece down from the top border using the wrong algorithm mentioned on the website (or if it began in this unfortunate position). In either case, use one of the two algorithms provided on the website to bring a top piece (which belongs on top) down into the incorrect spot, the where the piece you are concerned about is currently located (which moves the correct piece back up to the top layer). Then, bring the correct piece down into its original spot with the &lt;strong&gt;other&lt;/strong&gt; algorithm from the website. This time it should be rotated correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: If this is a used or old Rubik's cube, it is also possible that at one point the cube burst or was purposely taken apart, and then accidentally put back together in a way that makes the cube unsolveable. If so, I think you would never encounter any difficulties until later in the solving process, but it's a possibility.&lt;/p&gt;&#xA;" OwnerUserId="4447" LastActivityDate="2014-11-21T23:15:32.180" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5401" PostTypeId="2" ParentId="5395" CreationDate="2014-11-21T23:45:02.543" Score="6" Body="&lt;p&gt;In your example picture, the sequence isn't broken at all - the cubule you need to insert into the second layer is already below the blue face where it needs to be. If you perform &lt;code&gt;D' R' D R D F D' F'&lt;/code&gt; with blue in front, the second layer will be solved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your confusion comes from still treating the cube in terms of stickers instead of cubules. The cubule that's currently in the blue-orange corner is entirely the wrong cubule, not just one sticker off.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2014-11-21T23:45:02.543" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="5402" PostTypeId="1" AcceptedAnswerId="5413" CreationDate="2014-11-22T01:08:36.330" Score="10" ViewCount="869" Body="&lt;p&gt;I am having difficulty understanding the third stage of Thistlethwaite's algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using &lt;a href=&quot;http://www.jaapsch.net/puzzles/thistle.htm&quot;&gt;http://www.jaapsch.net/puzzles/thistle.htm&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The third stage has a factor of [8!/(4!4!)]2 *2*3. This corresponds to the fact that the edges in the L and R faces are placed in their correct slices, the corners are put into their correct tetrads, the parity of the edge permutation (and hence the corners too) is made even, and the total twist of each tetrad is fixed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I understand correctly, a tetrad is formed from the path a corner travels using only moves in &lt;code&gt;&amp;lt;L2,R2,F2,B2,U2,D2&amp;gt;&lt;/code&gt;. This forms two tetrads, each consisting of four corners, and each corner can then be placed into its respective tetrad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the edge slices can be arranged and the parity determined. This counts up to&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;C(8, 4)&lt;sup&gt;2&lt;/sup&gt; &amp;times; 2 = 9,800&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The size of stage three, however, is 29,400 which means I am off by a factor of 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't understand is the meaning of &lt;strong&gt;tetrad twist&lt;/strong&gt; and how it generates a factor of 3.&lt;/p&gt;&#xA;" OwnerUserId="5899" LastEditorUserId="5908" LastEditDate="2014-11-22T07:24:23.373" LastActivityDate="2014-11-22T07:24:23.373" Title="What is the meaning of a &quot;tetrad twist&quot; in Thistlethwaite's algorithm?" Tags="|mathematics|rubiks-cube|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5403" PostTypeId="2" ParentId="5299" CreationDate="2014-11-22T01:36:58.803" Score="3" Body="&lt;p&gt;No need for any complex calculation!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $r$ be your run speed, $w$ your walking speed, and $m$ be the moving walkway speed. If you rest for some time $T$, then you cover the distance $(w+m) \, T$ if you're on the walkway and $w\,T$ otherwise, whereas the distances covered would be $(r+m)T$ and $r\,T$ if running. Let's compare the distance covered by walking for a time $T$ on the walkway and spending the same amount of time running on the unmoving ground, and the distance covered if running on the unmoving ground and walking on the walkway. (I assume that the rest time isn't long enough to cover the whole unmoving ground or the whole walkway.)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Run on the walkway, walk on unmoving ground: $(r+m) T + w T$&lt;/li&gt;&#xA;&lt;li&gt;Walk on the walkway, run on unmoving ground: $(w+m) T + r T$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So whether you choose to rest on the walkway or off it, there is a time period $2T$ (composed of two intervals which may not be consecutive) during which you'll cover the distance $(r+w+m) T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, how long will it take to cover the rest of the distance? If you run on the walkway, then you'll have less walkway to cover during the time not already accounted for, therefore it will take longer to cover the distance. Hence it's better to run on the walkway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Illustrated graphically — &lt;code&gt;===&lt;/code&gt; is the walkway, &lt;code&gt;---&lt;/code&gt; is unmoving ground, &lt;code&gt;+++&lt;/code&gt; means running, &lt;code&gt;...&lt;/code&gt; means walking, and &lt;code&gt;←————→&lt;/code&gt; is one of two intervals of time $T$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Walk on the walkway:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;=============================---------------------------------&#xA;++++++++++++++++++++.........+++++++++++++++++++++++++++++++++&#xA;                    ←———————→←————→&#xA;&lt;/code&gt;&lt;/pre&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Walk on the unmoving ground:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;=============================---------------------------------&#xA;+++++++++++++++++++++++++++++...++++++++++++++++++++++++++++++&#xA;                 ←——————————→←—→&#xA;&lt;/code&gt;&lt;/pre&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the second case, more unmoving ground is covered outside the period of time of duration $2T$, therefore the second case requires more time.&lt;/p&gt;&#xA;" OwnerUserId="71" LastActivityDate="2014-11-22T01:36:58.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5404" PostTypeId="2" ParentId="5376" CreationDate="2014-11-22T02:44:40.877" Score="3" Body="&lt;p&gt;I know this isn't the right answer, but &lt;strong&gt;this question&lt;/strong&gt; posted on Puzzling Stack Exchange almost applies...&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&quot;You're very close but I'm always ahead&quot;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This could mean our answers are close to correct, that we are physically located close to the right answer (right on our screens), or could be a reference to the fact that the question will always be located just ahead (above) the correct answer on our screens.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&quot;Can be beheaded&quot;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;We can edit the heading&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;&quot;You can have a look inside, there will be no place to hide&quot;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;We can attempt to edit your question, where you will not be able to conceal any additional hints from us that are hidden when viewed normally. Also there is no place to fit additional text when editing the question...&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;&quot;However, try to enter me, more room arises suddenly&quot;&#xA;until we hit &quot;enter&quot; on the keyboard and more room arises for us to type in.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Just thought that was interesting...&#xA;Since you said you need this for school, it sadly can't be the correct answer. :(&lt;/p&gt;&#xA;" OwnerUserId="4447" LastActivityDate="2014-11-22T02:44:40.877" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5405" PostTypeId="2" ParentId="5203" CreationDate="2014-11-22T04:06:19.277" Score="1" Body="&lt;p&gt;A possible solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What question would you ask me if I should answer with the question I would ask to get your answer?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4394" LastActivityDate="2014-11-22T04:06:19.277" CommentCount="0" CommunityOwnedDate="2014-11-22T04:06:19.277" ContentLicense="CC BY-SA 3.0" />
  <row Id="5406" PostTypeId="2" ParentId="5370" CreationDate="2014-11-22T04:29:35.380" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The environment&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;I am precious to you,&#xA;but kill me you do.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everybody needs the environment but are destroying it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;I am older than you,&#xA;but still ever new.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The environment was there long before anybody, but new plants, animals are being born every day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;I cannot be sold,&#xA;for money or gold.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You cannot buy the environment.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;I can EVEN fly.&#xA;Yet wings have not I.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The environment can &quot;fly&quot; from place-to-place.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1845" LastActivityDate="2014-11-22T04:29:35.380" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5407" PostTypeId="2" ParentId="5141" CreationDate="2014-11-22T04:58:38.603" Score="1" Body="&lt;p&gt;Not a complete solution, but maybe it can help someone else see what I'm missing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lines look like&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; cryptic crossword clues&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;to me. Especially since the link in your hint refers to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;General&lt;/strong&gt; abstract nonsense.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With &quot;high up in the army&quot; pointing to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;general.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, &quot;the woman with the wheel&quot; is probably referring to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a ship (ships are usually referred to with female pronouns in English and ships have wheels/helms)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With the &quot;firing the canon part&quot;, can the actual word that solves the second line be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; artillery?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5421" LastActivityDate="2014-11-22T04:58:38.603" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5408" PostTypeId="2" ParentId="4718" CreationDate="2014-11-22T05:05:29.730" Score="3" Body="&lt;p&gt;I believe it to be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;congrats on solving the mixed cipher&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, I actually got different letters for the fourth and sixth letters.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Just using the date cipher as you mentioned:&lt;br&gt;&lt;code&gt;DPOMSBUT PR TSMWJRH XIf NMYIE djtiis&lt;/code&gt;&lt;br&gt;&lt;code&gt;11141411 14 1411141 411 14141 114141&lt;/code&gt;&lt;br&gt;&lt;code&gt;CONIRXTS ON SOLVING THe MIXED cipher&lt;/code&gt;&lt;br&gt;Here, for example, D going 1 letter back is C (C ← D), and R going 4 letters back is N (N ← O ← P ← Q ← R).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5853" LastActivityDate="2014-11-22T05:05:29.730" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5410" PostTypeId="1" AcceptedAnswerId="5425" CreationDate="2014-11-22T06:38:44.560" Score="17" ViewCount="1780" Body="&lt;p&gt;The answer is a six-letter word.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/j1xU3.png&quot; alt=&quot;The puzzle&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here's an attempt at a text version, but you're really better off looking at the picture.  &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Rolling Dice&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;To get a full tour, you must be a man of many faces and blend into all the right spots.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;     |&#xA;     v    &#xA;O6TAL2&#xA;RA2U1Y&#xA;DI■ROT&#xA;LLHTP3&#xA;6FE3PR&#xA;MISMO6&#xA;^&#xA;|&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2014-11-22T18:20:00.553" Title="Rolling Dice puzzle" Tags="|word|dice|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5411" PostTypeId="1" AcceptedAnswerId="5412" CreationDate="2014-11-22T06:40:08.330" Score="6" ViewCount="1235" Body="&lt;p&gt;Last night I had a dream:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I beheld a woman with a scroll in her hand&lt;br&gt;&#xA;  And the crescent moon beneath her feet,&lt;br&gt;&#xA;  And she sat between two pillars&lt;br&gt;&#xA;  Whose names are B. and J.&lt;br&gt;&#xA;  And she said, &quot;There are others&lt;br&gt;&#xA;  Who are like me, but I trump them all.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who can interpret the dream, and tell me the name of the woman whom I saw, and the scroll which was in her hand, and the two pillars between which she was seated?&lt;/p&gt;&#xA;" OwnerUserId="5474" LastEditorDisplayName="user20" LastEditDate="2015-03-05T08:49:20.620" LastActivityDate="2018-11-22T21:05:36.577" Title="The woman between the two pillars" Tags="|riddle|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5412" PostTypeId="2" ParentId="5411" CreationDate="2014-11-22T06:47:25.593" Score="6" Body="&lt;p&gt;She is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/The_High_Priestess&quot; rel=&quot;nofollow&quot;&gt;The High Priestess (of the Tarot)&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation&lt;/strong&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I beheld a woman with a scroll in her hand&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In her card in the Tarot she is holding a scroll.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the crescent moon beneath her feet,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In her card in the Tarot she has a crescent moon at her feet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And she sat between two pillars Whose names are B. and J.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I cracked it because I thought of &lt;a href=&quot;http://en.wikipedia.org/wiki/Boaz_and_Jachin#In_popular_media&quot; rel=&quot;nofollow&quot;&gt;Boaz and Jachin&lt;/a&gt; immediately. From the link: &lt;em&gt;&quot;The woman sat like that blind interlocutrix between Boaz and Jachin inscribed upon the one card in the juggler's deck that they would not see come to light, true pillars and true card,prophetess for all.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And she said, &quot;There are others Who are like me, but I trump them all.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; According to &lt;a href=&quot;http://en.wikipedia.org/wiki/The_High_Priestess&quot; rel=&quot;nofollow&quot;&gt;wiki&lt;/a&gt;: The High Priestess is the second trump or Major Arcana card in most traditional Tarot decks. This card is used in game playing as well as in divination. In the first Tarot pack with inscriptions, the 18th-century woodcut Marseilles Tarot, this figure is crowned with the Papal tiara and labelled La Papesse, the Popess, a possible reference to the legend of Pope Joan.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://upload.wikimedia.org/wikipedia/en/8/88/RWS_Tarot_02_High_Priestess.jpg&quot; rel=&quot;nofollow&quot;&gt;See a cool image depicting your dream&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-12-04T04:18:26.080" LastActivityDate="2014-12-04T04:18:26.080" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5413" PostTypeId="2" ParentId="5402" CreationDate="2014-11-22T06:48:05.253" Score="7" Body="&lt;p&gt;Tricky question!  The extra factor of 3 comes from how the permutation of the corners in one tetrad affects the permutation of the corners in the other one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I'm sure you know, the tetrads are each made up of 4 corners in an orbit like in the following image.  I colored the corners in one tetrad in shades of pink and the corners in the other tetrad in shades of teal:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eXkvo.png&quot; alt=&quot;highlight the two tetrads&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be easy to see that any permutation is possible within a tetrad.  For example &lt;code&gt;U2&lt;/code&gt; swaps a pair of corners in each tetrad and if a pair of corners can be swapped in a tetrad then any permutation (&lt;code&gt;4!&lt;/code&gt;) in each tetrad is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trick comes when you try to perform a 3-cycle in one of the tetrads.  For example the commutator &lt;code&gt;[R2, F2]&lt;/code&gt; (the sequence &lt;code&gt;R2 F2 R'2 F'2&lt;/code&gt;) performs a 3-cycle of the &lt;code&gt;UFR&lt;/code&gt; and &lt;code&gt;DRB&lt;/code&gt; and &lt;code&gt;DFL&lt;/code&gt; corners. But, and this is where the factor of 3 comes in, it also performs a 3-cycle in the other tetrad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So instead of the total permutations of the corners being &lt;code&gt;(4! * 4!) / 2&lt;/code&gt; (where the 2 comes from the fixed parity) it is actually &lt;code&gt;(4! * 4!) / (2 * 3)&lt;/code&gt; because a lone 3-cycle is disallowed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example of the sequence &lt;code&gt;R'2 F'2 R'2 F'2 D'&amp;amp;2 F'2 D&amp;amp;2 F'2&lt;/code&gt; (where &lt;code&gt;D&amp;amp;2&lt;/code&gt; is the equatorial slice turned the same direction as &lt;code&gt;D&lt;/code&gt;) that shows two 3-cycles at once:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5o0wD.png&quot; alt=&quot;tetrad 3-cycle pair&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the group generated by &lt;code&gt;&amp;lt;L2,R2,F2,B2,U2,D2&amp;gt;&lt;/code&gt; you just can't overcome this restriction so when you move from &lt;code&gt;&amp;lt;L,R,F2,B2,U2,D2&amp;gt;&lt;/code&gt; to &lt;code&gt;&amp;lt;L2,R2,F2,B2,U2,D2&amp;gt;&lt;/code&gt; there is an unintuitive extra factor of 3 in the reduction.&lt;/p&gt;&#xA;" OwnerUserId="5908" LastActivityDate="2014-11-22T06:48:05.253" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5414" PostTypeId="2" ParentId="5370" CreationDate="2014-11-22T06:53:11.880" Score="15" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are... The Precious! &lt;img src=&quot;https://i.stack.imgur.com/pjDPA.jpg&quot; alt=&quot;enter image description here&quot;&gt; (&lt;a href=&quot;http://www.imdb.com/title/tt0120737/&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;Lord of the Rings&lt;/em&gt;, imdb&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am precious to you,&#xA;but kill me you do.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;It is precious to me, though I buy it with a great pain.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am older than you,&#xA;but still ever new.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An ancient artifact, harmed only by the fires that forged it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cannot be sold,&#xA;for money or gold.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;No ring bearer would ever part with the Precious for something so paltry as money.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can EVEN fly.&#xA;Yet wings have not I.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;The Ring of Power has a will of its own. It betrayed Isildur, to his death.&quot; Getting &lt;em&gt;even&lt;/em&gt; and 'flying' from its bearer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is no game,&#xA;what is my name?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;&lt;strong&gt;DO NOT TAKE ME FOR SOME CONJURER OF CHEAP TRICKS&lt;/strong&gt;.&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4522" LastActivityDate="2014-11-22T06:53:11.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5417" PostTypeId="2" ParentId="5190" CreationDate="2014-11-22T08:38:35.890" Score="3" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;This isn't an answer. It's just a potentially useful mathematical definition of the problem.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the coins be labelled $C = \{c_1, \cdots, c_{12}\}$ be mapped to a vector $V = (s_1, \cdots, s_{12})$ where $c_i = 10 \rightarrow s_i = 0$ and $c_i = 20 \rightarrow s_i = 1$. There are $2^{12}$ possible weight patterns for the $12$ coins. And, in fact the vector members concatenated map a unique binary integer to each of those possible weight patterns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The possible binary strings are in the set $L_1 = \Big\{w\mid w = s_1s_2 \cdots s_{12} \land s_i \in \{0,1\} \Big\}$. $|L_1| = 2^{12} = 4096$. &#xA;Let $H(x)$ be the Hamming weight (the count of non-zero digits) in the word $x$.                             &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We select $x$ subsets, $S = \{v_1, \cdots ,v_x\}$ where $v_i \subseteq V$. We can be sure that $H(v_i)&amp;lt;=12$. So, consider the string $H(v_1)H(v_2) \cdots H(v_{12})$ and interpret this string as a number in base-13. The possible strings of this sort are in the set $L_2 = \Big\{w\mid w = v_1 \cdots v_x \land v_i \in \{0,\cdots,9,A,B,C\} \Big\}$. $|L_2| = 13^x$.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is to define $S$ that minimises $x$ such that, regardless of any initial weights for the coins in $C$, there is a one-to-many mapping from $L_1$ to $L_2$ and all mappings from members of $L_1$ are disjoint. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that $x \geq 2$ (purely from the fact that $13^x$ must exceed $2^{12}$) although it's surely between $6$ and $8$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This assumes that $v_j$ is not dependent on the results any combinations of $v_i$s where $i&amp;lt;j$. &lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-11-25T15:07:00.777" LastActivityDate="2014-11-25T15:07:00.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5418" PostTypeId="2" ParentId="5410" CreationDate="2014-11-22T08:42:54.183" Score="1" Body="&lt;p&gt;I believe the word is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MORTAL&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whenever you land on a square, if there is a number next to it (either touching a side or a corner), then you may move that many squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the path that I took:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/lkuwX.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;p&gt;&#xA; Starting at M, there is a 6 next to it, so I moved 6 squares to O.  There is a 6 next to the O, so I moved 6 squares to R.  There is a 2 and a 1 next to the R, so I chose 1 and moved 1 square to T.  There is a 3 next to the T, so I moved 3 squares to the A.  There is a 2 and a 1 next to the A, so I chose 1 and moved 1 square to the L.  There is a 1 and a 2 next to the L, so I chose 2 and moved 2 squares out of the grid.  I landed on M, O, R, T, A, and L, which spells MORTAL.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2014-11-22T08:42:54.183" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5423" PostTypeId="1" AcceptedAnswerId="5424" CreationDate="2014-11-22T09:29:59.403" Score="15" ViewCount="692" Body="&lt;p&gt;Consider the following set of 25 commonly-used acronyms&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;AMEX   PETA   CC     CCG    COBOL&#xA;DJIA   FLOPS  NAACP  ESL    SIAM&#xA;ASCII  FEDEX  VHS    GIF    SPCA&#xA;OEIS   BBC    FBI    HELOC  GPA&#xA;FICO   BBB    SVGA   MMORPG NBC&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;When expanded, these acronyms form a chain where each acronym definition shares exactly one word in common with the acronym that precedes it (note that articles such as &quot;of&quot;, &quot;the&quot;, &quot;in&quot;, etc. do not count for matching purposes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider the set of acronyms&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;NASA   DEA    ISS    FDA    IPCC&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;These can be chained together as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;IPCC :            International Panel on Climate Change&#xA;ISS  :            International Space Station&#xA;NASA : National Aeronautics and Space Administration&#xA;FDA  :                  Food and Drug Administration&#xA;DEA  :                           Drug Enforcement Agency&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where the word linking each pair of adjacent definitions is aligned for clarity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the acronyms have more than one popular definition. The puzzle may be trickier than it seems!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can you chain all 25 acronyms together?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Puzzlers are encouraged to place answers in spoiler blocks to avoid inadvertently spoiling the fun for other readers.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="4219" LastEditDate="2014-11-22T10:17:04.067" LastActivityDate="2014-11-23T10:38:17.497" Title="A Fine Chain of Acronyms" Tags="|wordplay|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5424" PostTypeId="2" ParentId="5423" CreationDate="2014-11-22T10:58:02.700" Score="11" Body="&lt;p&gt;Image Version&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/Wd4St.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Text Version&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;pre&gt;&lt;code&gt;  OEIS    :                                                                             Online Encyclopedia of Integer Sequences&#xA;  MMORPG  :                                                       Massively Multiplayer Online Role-Playing Game &#xA;  CCG     :                                                                                Collectible Card Game&#xA;  CC      :                                                                                     Credit Card&#xA;  HELOC   :                                                                 Home Equity Line of Credit&#xA;  VHS     :                                                           Video Home System&#xA;  SVGA    :                                                     Super Video                   Graphics Array&#xA;  GIF     :                                                                                   Graphics   Interchange Format&#xA;  ASCII   :                                                       American Standard Code for Information Interchange&#xA;  AMEX    :                                                       American Express&#xA;  FEDEX   :                                                       Federal  Express&#xA;  FBI     :                                                       Federal         Bureau of Investigation&#xA;  BBB     :                                                       Better Business Bureau &#xA;  COBOL   :                                                       Common Business Oriented             Language&#xA;  ESL     :                                                                        English as a Second Language &#xA;  FLOPS   :                                                       Floating Point Operations Per Second&#xA;  GPA     :                                                          Grade Point         Average&#xA;  DJIA    :                                                       Dow Jones   Industrial Average&#xA;  SIAM    :                                                       Society for Industrial and Applied Mathematics&#xA;  SPCA    :                                                       Society for the Prevention of Cruelty to Animals&#xA;  PETA    :                                                       People for the Ethical Treatment of      Animals&#xA; NAACP   : National Association for the Advancement of Colored    People&#xA; NBC     : National Broadcasting Company &#xA; BBC     : British  Broadcasting Corporation&#xA; FICO    : Fair Isaac            Corporation&lt;/code&gt;&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Just the ordering (in reading order)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;pre&gt;&lt;code&gt;&#xA;OEIS  MMORPG  CCG     CC      HELOC&lt;br&gt;&#xA;VHS   SVGA    GIF     ASCII   AMEX&lt;br&gt;&#xA;FEDEX FBI     BBB     COBOL   ESL&lt;br&gt;&#xA;FLOPS GPA     DJIA    SIAM    SPCA&lt;br&gt;&#xA;PETA  NAACP   NBC     BBC     FICO&lt;br&gt;&#xA;&lt;/code&gt;&lt;/pre&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="2313" LastEditDate="2014-11-23T10:38:17.497" LastActivityDate="2014-11-23T10:38:17.497" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="5425" PostTypeId="2" ParentId="5410" CreationDate="2014-11-22T11:48:59.263" Score="18" Body="&lt;p&gt;I'll use a different spoiler for each level of solving the puzzle, such that you can read only as far as you want to be spoiled, if you want to only look at the first few steps for hints. Note that there is some Mathematica code further down, which is not inside a spoiler tag.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Making sense of the hints&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;strong&gt;&quot;full tour&quot;&lt;/strong&gt; indicates that we need to traverse every cell of the grid. It's likely that we need to traverse each cell exactly once, hence asking for a &lt;a href=&quot;http://en.wikipedia.org/wiki/Hamiltonian_path&quot; rel=&quot;noreferrer&quot;&gt;Hamiltonian path&lt;/a&gt;. The arrows indicate the start and end of the path.&lt;br&gt;&#xA; The &lt;strong&gt;&quot;man of many faces&quot;&lt;/strong&gt; probably just indicates that faces of a (six-sided) die play a role.&lt;br&gt;&#xA; &lt;strong&gt;&quot;Blend into all the right spots&quot;&lt;/strong&gt; means that we somehow need to match up the faces with the grid cells.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;Putting the hints together&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The puzzle asks us to place a six-sided die on the bottom left cell, and roll it around the grid, such that we visit each cell exactly once. Furthermore, whenever the die is on top of a number, the top face has to show that same number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;Finding the tour&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To simplify the search for the path, I started at the end, because there are only 4 possible die orientations with a 2 on top.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These constraints result in a unique tour, which I found using the following Mathematica code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;board = Characters /@ StringSplit@&quot;XXXXXXXX&#xA;    XO6TALXX&#xA;    XRA2U1YX&#xA;    XDIXROTX&#xA;    XLLHTP3X&#xA;    X6FE3PRX&#xA;    XMISMO6X&#xA;    XXXXXXXX&quot;;&#xA;&#xA;moves = {&#xA;   (*up*)&#xA;   {c = #1 + {-1, 0}, Extract[board, c], &amp;lt;|u -&amp;gt; #3[s], n -&amp;gt; #3[u], w -&amp;gt; #3[w], &#xA;      s -&amp;gt; #3[d], e -&amp;gt; #3[e], d -&amp;gt; #3[n]|&amp;gt;} &amp;amp;,&#xA;   (*left*)&#xA;   {c = #1 + {0, -1}, Extract[board, c], &amp;lt;|u -&amp;gt; #3[e], n -&amp;gt; #3[n], w -&amp;gt; #3[u], &#xA;      s -&amp;gt; #3[s], e -&amp;gt; #3[d], d -&amp;gt; #3[w]|&amp;gt;} &amp;amp;,&#xA;   (*down*)&#xA;   {c = #1 + {1, 0}, Extract[board, c], &amp;lt;|u -&amp;gt; #3[n], n -&amp;gt; #3[d], w -&amp;gt; #3[w], &#xA;      s -&amp;gt; #3[u], e -&amp;gt; #3[e], d -&amp;gt; #3[s]|&amp;gt;} &amp;amp;,&#xA;   (*right*)&#xA;   {c = #1 + {0, 1}, Extract[board, c], &amp;lt;|u -&amp;gt; #3[w], n -&amp;gt; #3[n], w -&amp;gt; #3[d], &#xA;      s -&amp;gt; #3[s], e -&amp;gt; #3[u], d -&amp;gt; #3[e]|&amp;gt;} &amp;amp;&#xA;   };&#xA;f[path_] := Module[{i, next},&#xA;  If[Length@path == 35, Print@Reverse@path; &#xA;   Print@TableForm@Reverse@path; Return[]];&#xA;  For[i = 1, i &amp;lt;= 4, ++i,&#xA;   next = moves[[i]] @@ path[[-1]];&#xA;   If[! (next[[2]] == &quot;X&quot;&#xA;       || next[[1]] == {7, 2} &amp;amp;&amp;amp; Length@path &amp;lt; 34&#xA;       || DigitQ@next[[2]] &amp;amp;&amp;amp; next[[2]] != next[[3]][u]),&#xA;    (board[[##]] = &quot;X&quot;) &amp;amp; @@ next[[1]];&#xA;    f[Append[path, next]];&#xA;    (board[[##]] = next[[2]]) &amp;amp; @@ next[[1]];&#xA;    ]&#xA;   ];&#xA;  ]&#xA;f[{{{2, 7}, &quot;2&quot;, &amp;lt;|u -&amp;gt; &quot;2&quot;, n -&amp;gt; &quot;3&quot;, w -&amp;gt; &quot;1&quot;, s -&amp;gt; &quot;4&quot;, e -&amp;gt; &quot;6&quot;, d -&amp;gt; &quot;5&quot;|&amp;gt;}}];&#xA;f[{{{2, 7}, &quot;2&quot;, &amp;lt;|u -&amp;gt; &quot;2&quot;, n -&amp;gt; &quot;6&quot;, w -&amp;gt; &quot;3&quot;, s -&amp;gt; &quot;1&quot;, e -&amp;gt; &quot;4&quot;, d -&amp;gt; &quot;5&quot;|&amp;gt;}}];&#xA;f[{{{2, 7}, &quot;2&quot;, &amp;lt;|u -&amp;gt; &quot;2&quot;, n -&amp;gt; &quot;4&quot;, w -&amp;gt; &quot;6&quot;, s -&amp;gt; &quot;3&quot;, e -&amp;gt; &quot;1&quot;, d -&amp;gt; &quot;5&quot;|&amp;gt;}}];&#xA;f[{{{2, 7}, &quot;2&quot;, &amp;lt;|u -&amp;gt; &quot;2&quot;, n -&amp;gt; &quot;1&quot;, w -&amp;gt; &quot;4&quot;, s -&amp;gt; &quot;6&quot;, e -&amp;gt; &quot;3&quot;, d -&amp;gt; &quot;5&quot;|&amp;gt;}}];&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The only solution is the following: start in the bottom left corner, with &lt;strong&gt;4&lt;/strong&gt; facing &lt;em&gt;up&lt;/em&gt; and &lt;strong&gt;6&lt;/strong&gt; facing &lt;em&gt;south&lt;/em&gt;. Then this is the path (red numbers indicate number facing up):&lt;br&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/8Gf4i.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;Interpreting the tour&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If we now pick out the letters in that path where each number is on the top face (see red letters in the image), we get 6 words. In order from 1 to 6 they are &lt;strong&gt;SLUMP&lt;/strong&gt;, &lt;strong&gt;DROOP&lt;/strong&gt;, &lt;strong&gt;OR&lt;/strong&gt;, &lt;strong&gt;MILITARY&lt;/strong&gt;, &lt;strong&gt;FELT&lt;/strong&gt;, &lt;strong&gt;HAT&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;Cracking the final hint&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;to slump&quot; and &quot;to droop&quot; are synonyms for &quot;to slouch&quot;, but &quot;a slouch&quot; can also be a &lt;a href=&quot;http://en.wikipedia.org/wiki/Slouch_hat&quot; rel=&quot;noreferrer&quot;&gt;military felt hat&lt;/a&gt;. Hence the solution is &lt;strong&gt;slouch&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2450" LastEditorUserId="2450" LastEditDate="2014-11-22T18:20:00.553" LastActivityDate="2014-11-22T18:20:00.553" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5427" PostTypeId="2" ParentId="5356" CreationDate="2014-11-22T13:46:51.557" Score="5" Body="&lt;p&gt;I thought of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; autism&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In your mind I manifest&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; this so-called disease doesn't affect the body but the brain only, I assume&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Understanding of me is vague at best&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the people suffering from it are often considered geniuses, hard to catch up with their mental level&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm said to take away the order&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well, nothing to be added here from my point of view&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Less likely to be seen in the daughter&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As far as I know, females have lower risk of being autistic (&quot;artistic&quot; ? ;))&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Heightened moods wont decide your fate&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; autism is said to be almost unaffected by hightened moods like happiness&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When held onto, is my traits&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;White cloaks might give you sensations&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; doctors/scientists give cognitive behavioral therapy, range of medicine (?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Line them up for no frustrations&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Line them up in the right order alphabetically or colour, which ever is the right compulsion&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For some I am not what I envision&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; many think of 'handy caps' as they are all the same :( (autistic movements to change the way we perceive autism as many believe its not a disability)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet oppression is used to describe my prison&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; disabled people are not fully integrated and seen as kind of lesser entities by many (Autism forces &quot;sufferers&quot; into their own mind and are unusual to start a conversation/relationship with other&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5106" LastEditorUserId="5135" LastEditDate="2014-11-22T13:58:10.423" LastActivityDate="2014-11-22T13:58:10.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5428" PostTypeId="1" AcceptedAnswerId="5429" CreationDate="2014-11-22T15:59:37.887" Score="8" ViewCount="722" Body="&lt;p&gt;Stop worshipping me!&lt;br&gt;&#xA;Away! you flea!&lt;br&gt;&#xA;I'll stomp you with my boots,&lt;br&gt;&#xA;You know I'll accept any substitutes,&lt;br&gt;&#xA;At first, I change you,&lt;br&gt;&#xA;Then I begin to arrange you,&lt;br&gt;&#xA;And finally I end your precious heartbeat!&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who is yelling at you?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;As usual, please explain all steps in deriving an answer :)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="5373" LastEditDate="2015-03-14T12:57:36.990" LastActivityDate="2015-03-14T12:57:36.990" Title="Riddle... pray that this is the last one" Tags="|riddle|rhyme|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5429" PostTypeId="2" ParentId="5428" CreationDate="2014-11-22T18:46:48.257" Score="3" Body="&lt;p&gt;Yet again it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a cat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Stop worshipping me! &lt;br&gt;&#xA;Away! you flea!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The cat is telling us, the riddlers of Stackexchange, to stop writing riddles about it :-p Also, in former times, Egyptians worshipped cats. Cats get and surely hate fleas :D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I'll stomp you with my boots,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is a reference to &lt;a href=&quot;http://en.wikipedia.org/wiki/Puss_in_Boots&quot; rel=&quot;nofollow noreferrer&quot;&gt;Puss in Boots&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You know I'll accept any substitutes,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cats are not known for loyalty.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;At first, I change you, &lt;br&gt;&#xA; Then I begin to arrange you, &lt;br&gt;&#xA; And finally I end your precious heartbeat!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C is the &lt;em&gt;first&lt;/em&gt; letter in &lt;em&gt;change&lt;/em&gt;; &lt;em&gt;then&lt;/em&gt; the letter A &lt;em&gt;begin&lt;/em&gt;s the word &lt;em&gt;arrange&lt;/em&gt;, and &lt;em&gt;finally&lt;/em&gt; the letter T &lt;em&gt;end&lt;/em&gt;s the word &lt;em&gt;heartbeat&lt;/em&gt; - spelling out CAT.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The title is also a hint, since two riddles with &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4615/are-you-clever-enough-you-will-have-to-pray?rq=1&quot;&gt;similar&lt;/a&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/questions/5141/riddle-pray-you-can-find-the-word?rq=1&quot;&gt;titles&lt;/a&gt; have both had the same solution.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2014-11-22T18:58:07.667" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5431" PostTypeId="1" AcceptedAnswerId="5435" CreationDate="2014-11-22T20:38:12.267" Score="1" ViewCount="478" Body="&lt;p&gt;&lt;strong&gt;My last part 2 was sort of not a puzzle (oops) I've made sure this fits the criteria!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This is part 2 of a series of multiple parts about a takeover of all the computers on earth. The links to previous questions are here:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/3853/computer-crisis-takeover-part-1&quot;&gt;Computer Crisis Takeover! [Part 1]&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After cracking the original password screen, you open up this new OS &quot;Glidors Ultimate&quot;. You are welcomed and all of a sudden a message pops up saying you've been locked out for not being the owner of this OS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It gives you a second chance to get in by allowing you to guess the backup password. You only get 1 try, and there's a password hint included on the screen! Perhaps this hacker thought he'd need it to jog his memory if he ever needed to use this screen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The password hint is as follows;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm Not the neWest, but I'm also nOt olD, I am but in betWeen. ^&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is this password?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2014-11-23T22:13:15.657" Title="Computer Crisis Takeover! [Part 2!]" Tags="|password|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5432" PostTypeId="1" CreationDate="2014-11-22T20:55:03.817" Score="3" ViewCount="531" Body="&lt;p&gt;My friend gave me this code and I just cannot crack it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the code: Teleeelm eelm reeldna ehelt roellos feloor, uoely eelkiell-eelkiell eelm. teeelt eelm noel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My friend is a  sort of bad speller, so this may be hard.&lt;/p&gt;&#xA;" OwnerUserId="5927" LastActivityDate="2014-11-23T02:20:31.010" Title="Can you help me crack this code my friend gave me?" Tags="|encoded-message|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5433" PostTypeId="2" ParentId="5370" CreationDate="2014-11-22T21:08:44.697" Score="8" Body="&lt;p&gt;You are...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Air&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am precious to you, but kill me you do.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We need air to survive, but constantly consume it in our metabolisms. We also spoil it by polluting.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am older than you, but still ever new.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Air has been around for ages, but is constantly &quot;renewed&quot; by photosynthesis which converts carbon dioxide in it to oxygen which can be breathed again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cannot be sold, for money or gold.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; While bottled air is a thing in certain dystopic scenarios, breathing air is commonly accepted as a universal right that shouldn't need to be bought.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can EVEN fly. Yet wings have not I.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Air flies, in its way, despite having no wings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5928" LastActivityDate="2014-11-22T21:08:44.697" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5434" PostTypeId="2" ParentId="5190" CreationDate="2014-11-22T21:38:42.823" Score="1" Body="&lt;p&gt;I think this may lead to a unique solution but I'm not 100% sure.  I think it can be done with six weighs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weigh(1, 2, 3, 4) = X&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weigh(5, 6, 7, 8) = Y&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weigh(1 - 12) = T (for total)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this you can also calculate &lt;code&gt;(9, 10, 11, 12) = Z = T - X - Y&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weigh[(1, 5, 9) + (2, 6, 10)] = A&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weigh[(1, 5, 9) + (3, 7, 11)] = B&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weigh[(1, 5, 9) + (4, 8, 12)] = C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;code&gt;T = A + B + C - 2(1, 5, 9)&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do the algebra to get &lt;code&gt;(1, 5, 9) = (A + B + C - T) / 2&lt;/code&gt; All values on the right side are known, so we know what (1, 5, 9) equals.  From there we can calculate &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;(2, 6, 10) = A - (1, 5, 9)&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;(3, 7, 11) = B - (1, 5, 9)&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;(4, 8, 12) = C - (1, 5, 9)&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can set up a matrix:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{bmatrix}1 &amp;amp; 2 &amp;amp; 3 &amp;amp; 4 &amp;amp; X\\5 &amp;amp; 6 &amp;amp; 7 &amp;amp; 8 &amp;amp; Y\\9 &amp;amp; 10 &amp;amp; 11 &amp;amp; 12 &amp;amp; Z\\i &amp;amp; j &amp;amp; k &amp;amp; l &amp;amp; T\end{bmatrix}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where 1 + 2 + 3 + 4 = X, 1 + 5 + 9 = i, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above we know X, Y, Z, i, j, k, and l and can use that information to determine the individual values of all coins.  But, like I said at the beginning I'm not 100% sure if this leads to a unique solution.  I tried coming up with combinations of 10s and 20s to test and they were all solvable.  Horizontally for XYZ values, a 40 or 80 is a dead giveaway that all coins in that row are 10 or 20.  Similarly, vertically for ijkl values, 30 and 60 do the same.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm obviously terrible at formatting. I literally just looked up how to do a matrix with MathJax and that's about the limit of my abilities. If someone wants to make it prettier feel free).&lt;/p&gt;&#xA;" OwnerUserId="5930" LastActivityDate="2014-11-22T21:38:42.823" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5435" PostTypeId="2" ParentId="5431" CreationDate="2014-11-22T21:39:40.710" Score="2" Body="&lt;p&gt;This may be a stretch...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;windows vista&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There are actually 5 capitalized letters: N,W,O,D,W, and I (the last I in '...I am...'). together they make the word WINDOW - the '^' symbol  is created by hitting shift+6, the 6th edition of windows is vista, ergo 'windows vista' &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5159" LastEditorUserId="5159" LastEditDate="2014-11-23T18:33:50.553" LastActivityDate="2014-11-23T18:33:50.553" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="5436" PostTypeId="1" AcceptedAnswerId="5444" CreationDate="2014-11-22T21:58:24.127" Score="9" ViewCount="1206" Body="&lt;p&gt;A riddle. Each pair of lines contains some clues - expect history and wordplay!&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Duck or dragon, through the Veil,&#xA;Simpson, young goat, all for sale.&#xA;&#xA;When you take away two of my letters&#xA;you make me a pawn for my betters,&#xA;&#xA;but take just one, if you please&#xA;to make a villain of the seas.&#xA;&#xA;Some say I'm a thief, they lie;&#xA;All's fair in war - but who am I?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="5928" LastEditorUserId="5928" LastEditDate="2014-11-23T09:21:51.447" LastActivityDate="2014-12-01T14:45:44.253" Title="Riddle of the Mysterious Salesman" Tags="|riddle|wordplay|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5437" PostTypeId="2" ParentId="5431" CreationDate="2014-11-22T23:45:35.207" Score="1" Body="&lt;p&gt;The password is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; WINDOW6&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Sorry, I can't get the formatting to work with spoiler tags.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's why:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rewriting the riddle on separate lines we get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm Not the neWest&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;but I'm also nOt olD&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I am but in betWeen. ^&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Next we extract the capitalized letters:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I N W&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I O D&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I W&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; rearrange the letters to get WINDOW and shift+6 is caret.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2718" LastEditorUserId="2718" LastEditDate="2014-11-23T22:13:15.657" LastActivityDate="2014-11-23T22:13:15.657" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="5438" PostTypeId="2" ParentId="5436" CreationDate="2014-11-22T23:53:59.433" Score="5" Body="&lt;p&gt;I still can't make sense of the first two lines, but here's my attempt:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PRIVATE.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;When you take away two of my letters &lt;br&gt;&#xA;you make me a pawn for my betters,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Take away P and A, leaving letters which rearrange to RIVET. A &lt;a href=&quot;http://en.wikipedia.org/wiki/Rivet&quot;&gt;rivet&lt;/a&gt; is a small fastener in a large structure, so could be called a 'pawn' for the bigger and stronger things that force it to conform to a certain shape.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;but take just one, if you please &lt;br&gt;&#xA;to make a villain of the seas.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Take away V, leaving letters which rearrange to PIRATE.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Some say I'm a thief, they lie;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Private as in private property; this refers to the slogan '&lt;a href=&quot;http://en.wikipedia.org/wiki/Property_is_theft!&quot;&gt;Property is theft&lt;/a&gt;'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;All's fair in war - but who am I?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A private is a soldier of the lowest rank. This is probably sarcastic; privates have seen the worst and dirtiest parts of war, and they are just nobodies in the grand scheme of things.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(I could also make&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PRIMATE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fit all but the first two lines and the last one.)&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2014-11-22T23:53:59.433" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5439" PostTypeId="2" ParentId="5432" CreationDate="2014-11-22T23:57:04.850" Score="1" Body="&lt;p&gt;I reversed it and removed some letters:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;on me teet. me like-like you, roof solor the ander me meet&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Though, I may have removed letters i shouldn't have, or not enough. Anyway, this is the best I can do. Someone could take it further..&lt;/p&gt;&#xA;" OwnerUserId="5933" LastActivityDate="2014-11-22T23:57:04.850" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5440" PostTypeId="2" ParentId="5074" CreationDate="2014-11-23T00:14:35.090" Score="1" Body="&lt;p&gt;A variation of Yvonne's answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Burkina Faso?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation, following the steps of the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;1. You're in a country which contains thousands of soldiers who[m] never fight.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is the Terracotta Army, so the country is China.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;2. You take the 1st number of the name of this country, and subtract lines of 7.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first letter of 'China' is 'C', which has ASCII code 67. The number of &quot;lines&quot; in the number 7 printed on a digital board is 3. Subtract 3 from 67 to get 64.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;3. Under the square tree's stump.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Take the square root of 64 to get 8.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;everything down to here has been confirmed by the OP&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;4. You take this number and find a corresponding key.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The 8th letter key on a (QWERTY) keyboard is 'I'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;5. You go down soh, left stx and concatenate the first key with this one.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SOH (Start of Header) is 1 and STX (Start of Text) is 2, in ASCII code. Start from the I key on your keyboard; go 1 down and 2 left to get to the letter 'G' (or 'H', depending on whether you go down-left or down-right). Concatenate this with the first key to get IG.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;6. You are here. What have you made it to?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Ig,_Ig&quot; rel=&quot;nofollow&quot;&gt;The town of Ig, Slovenia&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2014-11-23T17:09:08.593" LastActivityDate="2014-11-23T17:09:08.593" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="5441" PostTypeId="1" AcceptedAnswerId="5527" CreationDate="2014-11-23T00:51:11.673" Score="6" ViewCount="593" Body="&lt;p&gt;For this puzzle, you must figure out a list of 10 words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of the following words is a clue for a word in the list:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Candy&lt;/li&gt;&#xA;&lt;li&gt;Jack&lt;/li&gt;&#xA;&lt;li&gt;Following&lt;/li&gt;&#xA;&lt;li&gt;Territory&lt;/li&gt;&#xA;&lt;li&gt;Area&lt;/li&gt;&#xA;&lt;li&gt;Common&lt;/li&gt;&#xA;&lt;li&gt;Improve&lt;/li&gt;&#xA;&lt;li&gt;Straight&lt;/li&gt;&#xA;&lt;li&gt;Membrane&lt;/li&gt;&#xA;&lt;li&gt;Leg&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The clues are in order, so Candy is a clue for the first word in the list, Jack is a clue for the second word in the list, and so on.  The list of 10 words has some sort of relationship between all the words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have structured this puzzle so that there is only one perfect answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hints:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. Some of the clues are synonyms, but not all of them are.&#xA; &lt;P&gt;&#xA; 2. I would start with the third through eighth clues, since the other 4 are a little more difficult.&#xA; &lt;P&gt;&#xA; 3. See &lt;a href=&quot;https://puzzling.stackexchange.com/questions/5441/figure-out-the-list-of-10-words#comment16736_5510&quot;&gt;my comment&lt;/a&gt; to dmg's answer.&#xA; &lt;P&gt;&#xA; 4. The first word is a specific type, and the second word is not a synonym.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-03-19T14:35:04.850" Title="Figure out the list of 10 words" Tags="|word|word-sequence|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5442" PostTypeId="2" ParentId="5432" CreationDate="2014-11-23T02:20:31.010" Score="5" Body="&lt;p&gt;When I first saw the string, the first thing the first few words reminded me of was &quot;meet me at the&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears that the string &lt;code&gt;el&lt;/code&gt;, wherever it appears should be removed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives us:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Teem em redna eht rolos foor, uoy ekil-ekil em. teet em no.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And if we reverse the letters in each word, we get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Meet me ander the solor roof, you like-like me. Teet me on.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It appears that some of the words have either been deliberately or inadvertently mangled, because it looks like it &lt;em&gt;should&lt;/em&gt; be this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Meet me under the solar roof, (you like-like me). Meet me on [message cut off].&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can't figure out what the part in parentheses means, but at least they're English words.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2014-11-23T02:20:31.010" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5444" PostTypeId="2" ParentId="5436" CreationDate="2014-11-23T02:58:01.383" Score="10" Body="&lt;p&gt;Answer: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Privateers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Duck or dragon, through the Veil, Simpson, young goat, all for sale.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Francis Drake (Drake meaning male duck as well as draconic monster), William Death (through the veil meaning death), Otway Burns (Mr. Burns from the Simpsons animated sitcom), and Captain Kidd (kid being a young goat) were all privateers, ship captains hired by governments for use in war.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you take away two of my letters you make me a pawn for my betters.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Private (er), bottom level military rank.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But take just one, if you please to make a villain of the seas.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pirate (missing a v), a seaborn raider. OP added that a Letter of Marque was needed to make privateering legal. Without one, you're a mere pirate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some say I'm a thief, they lie; All's fair in war - but who am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Though privateers would often plunder their targets, they were not pirates because their actions were sanctioned by governments. It's war, so they merely take their spoils.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5626" LastEditorUserId="5626" LastEditDate="2014-12-01T14:45:44.253" LastActivityDate="2014-12-01T14:45:44.253" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5445" PostTypeId="1" AcceptedAnswerId="5450" CreationDate="2014-11-23T06:51:47.453" Score="4" ViewCount="1148" Body="&lt;p&gt;There is a type of puzzle where one needs to count triangles in a figure, generally a large triangle full of lines which create smaller ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=t9cxM_DQMXI&quot; rel=&quot;nofollow&quot;&gt;One example&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there, or can someone develop ;), an algorithm to answer this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feel free suggesting and making up data representations for the problem.&lt;/p&gt;&#xA;" OwnerUserId="4317" LastActivityDate="2015-01-13T13:35:34.210" Title="Counting triangle algorithm" Tags="|visual|algorithm|counting|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5446" PostTypeId="5" CreationDate="2014-11-23T06:52:40.197" Score="0" Body="&lt;p&gt;Visual puzzles consist primarily of diagrams and pictures. Visual puzzles can be :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/9835/what-does-this-modern-art-picture-represent&quot;&gt;Encoded messages&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/20275/jigsaw-puzzle-edge-and-middle-pieces&quot;&gt;Jigsaw puzzles&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/30844/a-super-smart-rebus&quot;&gt;Rebus&lt;/a&gt;  where images must be interpreted as words, syllables, or letters to spell out a well-known phrase or quotation&lt;/li&gt;&#xA;&lt;li&gt;Mazes&lt;/li&gt;&#xA;&lt;li&gt;Based on &lt;a href=&quot;/questions/17856/smashing-seven-segments&quot;&gt;diagrammatic components&lt;/a&gt;. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you find that your puzzle revolves predominantly around some visual graphic or component that is central to the puzzle, then this tag is likely one you should use.  Do not use this tag for puzzles that discuss geometric or other hypothetical objects without displaying them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a rule of thumb: If the question involves a visual (the solver has to look at an image or graphic of some sort), this tag is probably appropriate. If the answer requires a visual representation, but the question does not, then don't use this tag.&lt;/p&gt;&#xA;" OwnerUserId="20274" LastEditorUserId="11110" LastEditDate="2017-06-21T07:44:53.540" LastActivityDate="2017-06-21T07:44:53.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5447" PostTypeId="4" CreationDate="2014-11-23T06:52:40.197" Score="0" Body="A puzzle that incorporates a visual component like pictures, diagrams, drawings, etc.&#xD;&#xA;Do not use this tag for puzzles that discuss geometric objects without displaying them.&#xD;&#xA;" OwnerUserId="4317" LastEditorUserId="8874" LastEditDate="2016-04-26T12:43:36.320" LastActivityDate="2016-04-26T12:43:36.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5448" PostTypeId="5" CreationDate="2014-11-23T06:53:52.203" Score="0" Body="&lt;p&gt;An algorithm is defined as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a process or set of rules to be followed in calculations or other problem-solving operations (especially by a computer).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Many puzzles in the area of computer science are algorithmic in nature.&#xA;Some standard examples of algorithmic puzzles are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Eight_queens_puzzle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Eight queens puzzle&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Tower_of_Hanoi&quot; rel=&quot;nofollow noreferrer&quot;&gt;Towers of Hanoi&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Dining_philosophers_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dining philosophers&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://classic-puzzles.blogspot.nl/2006/12/google-interview-puzzle-2-egg-problem.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Two egg problem&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="11110" LastEditDate="2017-06-21T18:26:44.033" LastActivityDate="2017-06-21T18:26:44.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5449" PostTypeId="4" CreationDate="2014-11-23T06:53:52.203" Score="0" Body="A puzzle related to algorithms (a main concept in computer science). Puzzles may ask for an algorithm as an answer or require the analysis of a given algorithm." OwnerUserId="8874" LastEditorUserId="69582" LastEditDate="2021-02-16T07:43:40.223" LastActivityDate="2021-02-16T07:43:40.223" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="5450" PostTypeId="2" ParentId="5445" CreationDate="2014-11-23T08:03:02.257" Score="4" Body="&lt;p&gt;This has been discussed at &lt;a href=&quot;https://stackoverflow.com/questions/7396570/what-is-an-efficient-algorithm-for-counting-the-number-of-triangles-in-a-graph&quot;&gt;StackOverflow&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything rests on seeing that the problem you are describing is a graph problem. So, everything should begin with expressing your shape as a graph. There is an &lt;a href=&quot;http://www.msccomputerscience.com/2014/04/wap-to-find-number-of-triangle-in-given.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;implementation in C&lt;/a&gt; which gives an example of the representation as a graph and which performs the task of calculating how many triangles there are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Someone pointed at &lt;a href=&quot;http://kam.mff.cuni.cz/~matousek/cla/tria-mmult.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt;, saying that the problem is computationally as difficult as Matrix Multiplication. Essentially the time complexity will be $O(n^3)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the answers there by &lt;a href=&quot;https://stackoverflow.com/users/1153791/evhen14&quot;&gt;evhen13&lt;/a&gt; outlines the following candidate algorithm which is recursive in nature:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You will need depth first search. The algorithm will be:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;1) For the current node ask all unvisited adjacent nodes&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;2) for each of those nodes run depth two check to see if a node at depth 2 is your current node from step one&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;3) mark current node as visited&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;4) on make each unvisited adjacent node your current node (1 by 1) and run the same algorithm&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="-1" LastEditDate="2017-05-23T12:38:53.617" LastActivityDate="2015-01-13T10:55:51.537" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5451" PostTypeId="1" AcceptedAnswerId="5452" CreationDate="2014-11-23T11:20:43.637" Score="20" ViewCount="3124" Body="&lt;p&gt;This is the easiest riddle I've written yet. Have fun!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I'm a hot girl! &lt;br&gt;&#xA;I'm twenty-six. &lt;br&gt;&#xA;If you give me water, &lt;br&gt;&#xA;I'll flatten you!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="4050" LastEditDate="2016-01-13T18:43:57.163" LastActivityDate="2016-01-13T21:46:45.687" Title="An easy riddle, for once. I'm a hot girl!" Tags="|riddle|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5452" PostTypeId="2" ParentId="5451" CreationDate="2014-11-23T11:38:02.350" Score="25" Body="&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;An&lt;/em&gt; iron (not simply iron) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm a hot girl! &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Iron Maiden? Indeed, an iron for clothing is normally hot. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm twenty-six. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The 26th element on the periodic table is Iron (Fe).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you give me water, I'll flatten you!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You put water in your iron and spray water on the clothing or whatever and use it to de-crease your clothes. :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastActivityDate="2014-11-23T11:38:02.350" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5453" PostTypeId="1" AcceptedAnswerId="5455" CreationDate="2014-11-23T11:39:49.030" Score="14" ViewCount="952" Body="&lt;p&gt;Yesterday I saw a pair of calendar dice that can be rotated and swapped to show all days of a month between 01 and 31.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aoGlz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dice have a digit painted on each face.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First die: [0, 1, 2, 3, 4, 5]&lt;/li&gt;&#xA;&lt;li&gt;Second die: [0, 1, 2, 6, 7, 8].&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The trick is that one of the faces can be rotated to show a 6 or a 9 as required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Mj48N.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With those dice one can also display the number 32. This is the last number that can be reached if we start counting from 01. Even though larger numbers like 42 are possible, there are gaps between 32 and 42 that cannot be represented with those two dice, so we can only count up to 32.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Problem&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If we were given three dice to count numbers in zero-padded three-digit format starting from 001 and counting up by 1, what would be the largest countable number? How would the dice have to look like?&lt;/p&gt;&#xA;" OwnerUserId="4394" LastEditorUserId="8874" LastEditDate="2016-03-22T16:56:29.097" LastActivityDate="2021-02-06T00:21:11.033" Title="Counting numbers with 3 dice" Tags="|mathematics|combinatorics|dice|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5454" PostTypeId="2" ParentId="5451" CreationDate="2014-11-23T12:25:11.407" Score="2" Body="&lt;p&gt;I think it could also be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an old hot-air balloon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm 26: It's an &lt;em&gt;old&lt;/em&gt; balloon.&lt;br&gt;&#xA; Hot, water: If you give it water, you could put out the fire, and it will come down and crush you.&lt;br&gt;&#xA; Girl: Ships and other modes of transportation are often referred to with feminine pronouns.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4456" LastActivityDate="2014-11-23T12:25:11.407" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5455" PostTypeId="2" ParentId="5453" CreationDate="2014-11-23T12:47:02.093" Score="10" Body="&lt;p&gt;First, each die must have a 0 face. We can see this by noting that we need two 0s to make 001, and a third die with {1,2,3,4,5,6} only gets us to 006.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having this fact, we need 1-8 at least once on some face to reach 010. It doesn't matter (yet) where these are - there are always two other 0s to allow 00x, no matter where x is. If we want to get to 011, we need a second 1 on a different die to the first 1. Similarly for all other numbers, we find that we cannot have a full set of two of each number 1-8 (requires 16 faces) because 3 dice with one face already assigned, leaving us only 15 faces free. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dropping the highest number, we need 15 faces to reach 087 if we can arrange them suitably.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{0,1,2,3,4,5}&#xA;{0,1,2,3,6,7}&#xA;{0,4,5,6,7,8}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will suffice, though there should be many other suitable arrangements.&lt;/p&gt;&#xA;" OwnerUserId="3987" LastActivityDate="2014-11-23T12:47:02.093" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5456" PostTypeId="1" AcceptedAnswerId="5468" CreationDate="2014-11-23T14:24:21.333" Score="46" ViewCount="24236" Body="&lt;p&gt;So, as you probably know, each and every element in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Periodic_table&quot; rel=&quot;noreferrer&quot;&gt;Periodic Table&lt;/a&gt; has its own unique &lt;em&gt;chemical symbol&lt;/em&gt;, which is either one, two, or three letters long. For instance, Oxygen is represented as &lt;strong&gt;O&lt;/strong&gt; and Copper as &lt;strong&gt;Cu&lt;/strong&gt;. So how about combining these symbols to form words?&lt;/p&gt;&#xA;&lt;p&gt;For instance, &lt;b&gt;Ba&lt;/b&gt;rium and &lt;b&gt;N&lt;/b&gt;eo&lt;b&gt;d&lt;/b&gt;ymium would give us &lt;strong&gt;Band&lt;/strong&gt;, or &lt;b&gt;C&lt;/b&gt;h&lt;b&gt;l&lt;/b&gt;orine-&lt;b&gt;Am&lt;/b&gt;ericium-&lt;b&gt;P&lt;/b&gt;hosphorus would give us &lt;strong&gt;Clamp&lt;/strong&gt;. So how long can these words get? I've managed the following examples:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;b&gt;Br&lt;/b&gt;omine-&lt;b&gt;O&lt;/b&gt;xygen-&lt;b&gt;C&lt;/b&gt;arbon-&lt;b&gt;Co&lt;/b&gt;balt-&lt;b&gt;Li&lt;/b&gt;thium: &lt;strong&gt;Broccoli&lt;/strong&gt; (8)&lt;/p&gt;&#xA;&lt;p&gt;&lt;b&gt;B&lt;/b&gt;oron-&lt;b&gt;I&lt;/b&gt;odine-&lt;b&gt;C&lt;/b&gt;arbon-&lt;b&gt;Y&lt;/b&gt;ttrium-&lt;b&gt;C&lt;/b&gt;h&lt;b&gt;l&lt;/b&gt;orine-&lt;b&gt;E&lt;/b&gt;in&lt;b&gt;s&lt;/b&gt;teinium: &lt;strong&gt;Bicycles&lt;/strong&gt; (8)&lt;/p&gt;&#xA;&lt;p&gt;&lt;b&gt;Th&lt;/b&gt;orium-&lt;b&gt;U&lt;/b&gt;ranium-&lt;b&gt;N&lt;/b&gt;eo&lt;b&gt;d&lt;/b&gt;ymium-&lt;b&gt;Er&lt;/b&gt;bium-&lt;b&gt;O&lt;/b&gt;xygen-&lt;b&gt;U&lt;/b&gt;ranium-&lt;b&gt;S&lt;/b&gt;ulphur: &lt;strong&gt;Thunderous&lt;/strong&gt; (10)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So, community, can you beat my attempts? For clarification, I will allow any word on &lt;a href=&quot;http://dictionary.reference.com/&quot; rel=&quot;noreferrer&quot;&gt;dictionary.com&lt;/a&gt;, including alternate forms of the word (plurals, verb conjugations, etc.). The word must be constructed using only chemical symbols that are recognised according to the Wikipedia page. You may use any symbol as many times as you like in your word.&lt;/p&gt;&#xA;&lt;p&gt;Victory will go to the person who can give the longest word in characters, &lt;em&gt;not&lt;/em&gt; the word made from the most symbols.&lt;/p&gt;&#xA;" OwnerUserId="4748" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-30T14:50:42.797" Title="The Longest Word Made from Chemical Symbols" Tags="|word|open-ended|" AnswerCount="9" CommentCount="12" ClosedDate="2015-07-27T10:26:38.243" ContentLicense="CC BY-SA 3.0" />
  <row Id="5457" PostTypeId="2" ParentId="5456" CreationDate="2014-11-23T14:34:00.250" Score="24" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;nonrepresentationalisms (23 characters is the maximum that can be formed) &#xA;as &#xA;&lt;br/&gt;&#xA;&lt;br/&gt;&#xA;No+N+Re+P+Re+Se+N+Ta+Ti+O+Na+Li+Sm+S&#xA;&lt;br/&gt;&lt;br/&gt;Nobelium+Nitrogen+Rhenium+Phosphorus+Rhenium+Selenium+Nitrogen+Tantalum+Titanium+Oxygen+Sodium+Lithium+Samarium+Sulphur&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is a word &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;hypothalamicohypophyseal &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is not listed in every dictionary which has 24 characters &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If only unique usage (each element being used once is a condition) then &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There is a tie &quot;hypercoagulabilities&quot; and &quot;hyperconsciousnesses&quot; (&quot;HYPErCoAgULaBiLiTiEs&quot; and &quot;HYPErCoNScIOUSnEsSeS&quot;) - 20 characters&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My Source &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Google which led to&lt;br&gt;&#xA;&lt;a href=&quot;http://www.nandor.org/math/chemwords/chemwords.htm&quot;&gt;http://www.nandor.org/math/chemwords/chemwords.htm&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="2313" LastEditDate="2014-11-23T15:03:15.973" LastActivityDate="2014-11-23T15:03:15.973" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5458" PostTypeId="2" ParentId="5190" CreationDate="2014-11-23T15:32:12.300" Score="19" Body="&lt;p&gt;I can manage it in 8 weighings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started with mdc32's answer and expanded on it. I was able to weigh 6 coins in 4 weighings. I suspect 12 can be done in 7 weighings, but I haven't figured out a way to do that yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with a group of coins $\{c_1,c_2,c_3,c_4,c_5,c_6\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weigh 1: $c_1 + c_2$. If the weight is either 20 or 40, follow mdc32's answer to weigh the other 4 in 3 weighings. Otherwise, the weight was 30, and proceed to the next weighing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weigh 2: $c_1 + c_3 + c_4$. If the weight is either 30 or 60, all of the weights of $c_1, c_2, c_3, c_4$ are known. Use weighing 3 for $c_5$ and 4 for $c_6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the weight was 40, the 3 options for the group of coins weighing 20 are:&lt;br&gt;&#xA;$\{c_1\}, \{c_2,c_3\}, \{c_2,c_4\}$. Proceed to 3a.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the weight was 50, the 3 options for the group of coins weighing 20 are:&lt;br&gt;&#xA;$\{c_1,c_3\}, \{c_1,c_4\}, \{c_2,c_3,c_4\}$ Proceed to 3b.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weigh 3 (a or b): $c_2 + c_3 + c_5$. If the weight is either 30 or 60, all the weights of $c_1, c_2, c_3, c_4, c_5$ are known. Use weighing 4 for $c_6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If 3a and a weight of 40, the 2 options for coins weighing 20 are:&lt;br&gt;&#xA;$\{c_1,c_5\}, \{c_2,c_4\}$. Proceed to 4a.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If 3a and a weight of 50, the 2 options for coins weighing 20 are:&lt;br&gt;&#xA;$\{c_2,c_3\}, \{c_2,c_4,c_5\}$. Proceed to 4b.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If 3b and a weight of 40, the 2 options for coins weighing 20 are:&lt;br&gt;&#xA;$\{c_1,c_3\}, \{c_1,c_4,c_5\}$. Proceed to 4c.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If 3b and a weight of 50, the 2 options for coins weighing 20 are:&lt;br&gt;&#xA;$\{c_2,c_3,c_4\}, \{c_1,c_3,c_5\}$. Proceed to 4d.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weigh 4a: $c_1 + c_5 + c_6$&lt;br&gt;&#xA;If the weight is 30, the heavy coins are: $\{c_2,c_4\}$&lt;br&gt;&#xA;If the weight is 40, the heavy coins are: $\{c_2,c_4,c_6\}$&lt;br&gt;&#xA;If the weight is 50, the heavy coins are: $\{c_1,c_5\}$&lt;br&gt;&#xA;If the weight is 60, the heavy coins are: $\{c_1,c_5,c_6\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weigh 4b: $c_4 + c_5 + c_6$&lt;br&gt;&#xA;If the weight is 30, the heavy coins are: $\{c_2,c_3\}$&lt;br&gt;&#xA;If the weight is 40, the heavy coins are: $\{c_2,c_3,c_6\}$&lt;br&gt;&#xA;If the weight is 50, the heavy coins are: $\{c_2,c_4,c_5\}$&lt;br&gt;&#xA;If the weight is 60, the heavy coins are: $\{c_2,c_4,c_5,c_6\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weigh 4c: $c_4 + c_5 + c_6$&lt;br&gt;&#xA;If the weight is 30, the heavy coins are: $\{c_1,c_3\}$&lt;br&gt;&#xA;If the weight is 40, the heavy coins are: $\{c_1,c_3,c_6\}$&lt;br&gt;&#xA;If the weight is 50, the heavy coins are: $\{c_1,c_4,c_5\}$&lt;br&gt;&#xA;If the weight is 60, the heavy coins are: $\{c_1,c_4,c_5,c_6\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weigh 4d: $c_2 + c_4 + c_6$&lt;br&gt;&#xA;If the weight is 30, the heavy coins are: $\{c_1,c_3,c_5\}$&lt;br&gt;&#xA;If the weight is 40, the heavy coins are: $\{c_1,c_3,c_5,c_6\}$&lt;br&gt;&#xA;If the weight is 50, the heavy coins are: $\{c_2,c_3,c_4\}$&lt;br&gt;&#xA;If the weight is 60, the heavy coins are: $\{c_2,c_3,c_4,c_6\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do the same for the other 6 coins, and all coins are known in 8 weighings.&lt;/p&gt;&#xA;" OwnerUserId="375" LastActivityDate="2014-11-23T15:32:12.300" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5461" PostTypeId="2" ParentId="5431" CreationDate="2014-11-23T15:58:20.160" Score="1" Body="&lt;p&gt;I think the password is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;WINDOWS XP&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm Not the neWest&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;but I'm also nOt olD&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I am but in betWeen. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Taking the upper-case letters and arranging, we get &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;WINDOW&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;^ is used&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;to denote power, exponent. so I guess ^ = &quot;eXponent&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When making it short&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;^ = XP&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then we will get&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;WINDOW XP&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now take the hints again.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm Not the neWest&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;but I'm also nOt olD&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I am but in betWeen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which made me to think that the password must be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;WINDOWS XP&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Windows XP is not the newest. (WINDOWS 10 is the newest)&lt;br /&gt;Windows XP is not OLD too. There are many people who are still using Windows XP. Windows 98, 2000 etc are old.&lt;br /&gt;Windows XP is in between.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastActivityDate="2014-11-23T15:58:20.160" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="5462" PostTypeId="1" AcceptedAnswerId="5465" CreationDate="2014-11-23T16:12:49.087" Score="4" ViewCount="6600" Body="&lt;p&gt;Can you picture the perfectly scrambled cube? A Rubik's Cube that's perfectly scrambled? How would it look? Is it even possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quest for the Perfect Scramble is now upon us, and I challenge you puzzlers to find the state that is &quot;perfectly scrambled&quot; and how to do this scramble, starting with a solved cube, in minimum moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But first, before you can even begin to figure it out, mathematically prove it, or even brute force it, you must define it. &lt;em&gt;What exactly &lt;strong&gt;is&lt;/strong&gt; a perfectly scrambled cube?&lt;/em&gt; there's 6 faces to the cube, and 9 squares on each one, so how do you define the perfect scramble?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The perfect scramble is the scramble farthest away from the perfect cube. What would that look like?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quest for the perfect scramble is upon you. It's up to you to solve this Rubik's crisis! Can you do it?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="-1" LastEditDate="2015-01-27T17:12:46.430" LastActivityDate="2024-03-03T03:25:45.170" Title="What does it mean for a Rubik's cube to be perfectly scrambled, and how do you reach it? Without doing a checker board" Tags="|rubiks-cube|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5463" PostTypeId="2" ParentId="5456" CreationDate="2014-11-23T16:20:17.880" Score="33" Body="&lt;p&gt;Perhaps this is cheating, but my answer (inspired by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Longest_word_in_English#Humour&quot;&gt;Wikipedia page&lt;/a&gt; on the longest English words) is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; YArDs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Formed by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yttrium, Argon, Darmstadtium&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This word is, of course, much shorter than any other words given so far. The reason, though, that it's longer is that it's a yard long.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5300" LastEditorUserId="5300" LastEditDate="2014-11-23T16:36:03.433" LastActivityDate="2014-11-23T16:36:03.433" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="5464" PostTypeId="2" ParentId="5462" CreationDate="2014-11-23T16:26:17.507" Score="9" Body="&lt;p&gt;There are (at least) two ways to measure how far a cube is from being solved: the quarter-turn metric (which counts the minimum number of quarter turns of faces needed to solve the cube), and the half-turn metric (which counts the minimum number of face turns needed, be they quarter or half turns).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the quarter-turn metric, a cube can be at most 26 steps from being solved. This goes down to 20 steps for the half-turn metric. Surprisingly, only a single position is known requiring 26 quarter-turns to solve. Hundreds of millions of known positions require 20 half or quarter turns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Images of cubes requiring the maximum number of moves, as well as descriptions of how to reach the positions, can be found &lt;a href=&quot;http://www.cube20.org/qtm/&quot;&gt;here&lt;/a&gt; (in the quarter-turn metric) and &lt;a href=&quot;http://www.cube20.org/&quot;&gt;here&lt;/a&gt; (in the half-turn metric)&lt;/p&gt;&#xA;" OwnerUserId="2627" LastActivityDate="2014-11-23T16:26:17.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5465" PostTypeId="2" ParentId="5462" CreationDate="2014-11-23T16:27:38.587" Score="18" Body="&lt;p&gt;The &quot;most scrambled cube&quot; is any configuration that requires the greatest number of rotations to solve the cube using a perfect algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although unknown, this algorithm is hypothetically called &lt;a href=&quot;https://en.wikipedia.org/wiki/God%27s_algorithm&quot;&gt;God's algorithm&lt;/a&gt; and in fact the maximum number of rotations, called &lt;a href=&quot;http://www.cube20.org/&quot;&gt;God's number&lt;/a&gt; has been found to be 20.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The list of such 20-move cubes is given &lt;a href=&quot;http://www.cube20.org/distance20s/&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To define the &quot;most scrambled-looking&quot; cube, consider any legal cube configuration where:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;each face displays all six colours&lt;/li&gt;&#xA;&lt;li&gt;for each face: upon blacking out the cells of any $n$ colours, $n \leq 4$, the resulting $3\times 3$ grid must exhibit no horizontal, vertical, diagonal, or anti-diagonal symmetry&lt;/li&gt;&#xA;&lt;li&gt;no line of 3 identically-coloured cells may appear horizonally, vertically, diagonally, or anti-diagonally on any face&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Alternatively, we might gauge scrambledness via empirical observation as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Given the population, $H$, of all capable human beings on Earth, let $S$ be a subject drawn from $H$, and conduct the following experiment $\forall\; S \in H$:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Explain the nature and objective of the Rubix cube to subject $S$, and provide a &quot;play&quot; cube for familiarization purposes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;After a minimum of 5 minutes of continuous play, present $S$ with a complete set of 43,252,003,274,489,856,000 rubix cubes, one per each legal state of the contraption. For every unique subset of 3 cubes, have $S$ rank the cubes in order of increasing scrambledness per his/her subjective preference.&lt;br&gt;&lt;br&gt;After each evaluation, tally $0$ for the cube ranked &quot;least scrambled&quot;, $2$ for the cube ranked &quot;most scrambled&quot;, and $1$ for the remaining cube.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;After all subjects in $H$ have completed all tallies for all subsets of all possible cube states, assign to each cube state a datum $s$ that is the sum of all numbers tallied on the state. $s$ will be a non-negative integer $&amp;lt;$ 1.7$\times$10&lt;sup&gt;69&lt;/sup&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let $s_{\rm max}$ be the $s$ datum of greatest magnitude computed in step 3. Let $C$ be the set of all Rubix cube states for which $s = s_{\rm max}$. We conclude that any cube $c \in C$ is &quot;maximally visually scrambled&quot; with 95% certainty.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="4219" LastEditDate="2014-11-23T18:30:48.047" LastActivityDate="2014-11-23T18:30:48.047" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="5466" PostTypeId="1" AcceptedAnswerId="5470" CreationDate="2014-11-23T16:30:35.800" Score="9" ViewCount="1264" Body="&lt;p&gt;&lt;strong&gt;Only one side has to have the score, the other can end with just a king for all I care.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What game ends with the highest piece scoring? A trivial answer would be 38, using fools mate (or scholars mate)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But of course, pawn promotion can be a factor, so can you do better than 38? Is 102 possible? If so what moves does it require? If not, what is the highest piece scoring?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PIECE SCORING:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Pawn:&lt;/strong&gt; &lt;em&gt;1&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Bishop:&lt;/strong&gt; &lt;em&gt;3&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Knight:&lt;/strong&gt; &lt;em&gt;3&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Rook:&lt;/strong&gt; &lt;em&gt;5&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Queen:&lt;/strong&gt; &lt;em&gt;9&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;King:&lt;/strong&gt; &lt;em&gt;0&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;*BONUS: When giving the game, give the quickest version to achieve the high score!&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-11-23T16:39:32.647" LastActivityDate="2021-12-08T03:20:34.883" Title="Chess Game with Highest Scoring" Tags="|chess|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="5467" PostTypeId="2" ParentId="5456" CreationDate="2014-11-23T16:42:49.470" Score="21" Body="&lt;p&gt;29 characters&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; floccinaucinihilipilification&lt;br/&gt;&#xA; Fl O C C I N Au C I N I H I Li P I Li F I C At I O N&lt;br/&gt;&#xA; 114, the atomic number of Flerovium, is the highest one used so far&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5341" LastEditorUserId="5341" LastEditDate="2014-11-23T16:48:46.950" LastActivityDate="2014-11-23T16:48:46.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5468" PostTypeId="2" ParentId="5456" CreationDate="2014-11-23T16:46:26.610" Score="47" Body="&lt;p&gt;One of my favourite words, that's 29 characters long is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://dictionary.reference.com/browse/floccinaucinihilipilification?s=t&quot;&gt;Floccinaucinihilipilification&lt;/a&gt; - the estimation of something as being worthless.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's quite useful for describing what the contestants find on a fair few reality TV programmes! As most of the letters are covered off by only a few elements there's quite a lot of repetition...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Fl - Flerovium&lt;br&gt;&#xA;O - Oxygen&lt;br&gt;&#xA;C - Carbon&lt;br&gt;&#xA;C - Carbon&lt;br&gt;&#xA;I - Iodine&lt;br&gt;&#xA;N - Nitrogen&lt;br&gt;&#xA;Au - Gold&lt;br&gt;&#xA;C - Carbon&lt;br&gt;&#xA;I - Iodine&lt;br&gt;&#xA;N - Nitrogen&lt;br&gt;&#xA;I - Iodine&lt;br&gt;&#xA;H - Hydrogen&lt;br&gt;&#xA;I - Iodine&lt;br&gt;&#xA;Li - Lithium&lt;br&gt;&#xA;P - Phosphorus&lt;br&gt;&#xA;I - Iodine&lt;br&gt;&#xA;Li - Lithium&lt;br&gt;&#xA;F - Fluorine&lt;br&gt;&#xA;I - Iodine&lt;br&gt;&#xA;C - Carbon&lt;br&gt;&#xA;At - Astatine&lt;br&gt;&#xA;I - Iodine&lt;br&gt;&#xA;O - Oxygen&lt;br&gt;&#xA;N - Nitrogen  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5946" LastActivityDate="2014-11-23T16:46:26.610" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5469" PostTypeId="2" ParentId="3398" CreationDate="2014-11-23T16:59:37.357" Score="1" Body="&lt;p&gt;Is the answer 48?  I don't feel like writing out a proof, given that I'm not even certain it's correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edit: Adding brief explanation...]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most pessimum initial configuration would be the inverse of the final design, so the equivalent matrix would be...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{bmatrix}16 &amp;amp; 15 &amp;amp; 14 &amp;amp; 13\\12 &amp;amp; 11 &amp;amp; 10 &amp;amp; 9\\ 8 &amp;amp; 7 &amp;amp; 6 &amp;amp; 5\\4 &amp;amp; 3 &amp;amp; 2 &amp;amp; 1\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with the first piece, it would take six moves at a minimum to place it properly.  It would take five moves for the second...then, four for the third, and three for the fourth.  (6 + 5 + 4 + 3...)  We're up to eighteen moves at this point - first row down.  The matrix becomes...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{bmatrix}1 &amp;amp; 2 &amp;amp; 3 &amp;amp; 4\\16 &amp;amp; 15 &amp;amp; 14 &amp;amp; 13\\12 &amp;amp; 11 &amp;amp; 10 &amp;amp; 9\\ 8 &amp;amp; 7 &amp;amp; 6 &amp;amp; 5\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fixing the second row, each cell would take one less move than its corresponding one (cell above) in the first row.  So #5 takes five moves; #6 takes four; #7 takes three; #8 takes two.  (5 + 4 + 3 + 2...)  That's another 14 moves for a subtotal of 32.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third row then takes (4 + 3 + 2 + 1) = 10 moves, and the last row takes (3 + 2 + 1) = 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total number of moves for the most efficient solution to the most pessimum starting setup is, then, &lt;strong&gt;48&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you can rotate the puzzle, everything changes; the matrix shown above would not take any moves, aside from a 180 degree spin.  My &lt;em&gt;guess&lt;/em&gt; for least moves in &lt;em&gt;that&lt;/em&gt; case is 33.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, this question was posted weeks ago with no confirmed answer.  Why not just give the solution at this point??  I only posted to ask whether 48 was actually the answer you wanted since it was already the answer given in an earlier response, and it seems to be the right one, unless there's a hack to be used.&lt;/p&gt;&#xA;" OwnerUserId="5948" LastEditorUserId="5948" LastEditDate="2014-11-23T23:56:54.747" LastActivityDate="2014-11-23T23:56:54.747" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5470" PostTypeId="2" ParentId="5466" CreationDate="2014-11-23T17:01:50.017" Score="18" Body="&lt;p&gt;Here's a solution that achieves the maximum possible 103 points (9 queens, 2 rooks, 2 bishops, 2 knights). It uses 40 moves, which is the minimum possible because a pawn must move 5 times in order to promote to a queen.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;a4 e5 2. b4 d5 3. b5 c5 4. a5 f5 5. b6 g5 6. a6 h5 7. bxa7 Ke7 8. axb7 Kd6 9. c4 Ke6 10. d4 Kf6 11. dxc5 Kg6 12. cxd5 Kf6 13. d6 Ke6 14. c6 Kf6 15. d7 Ke6 16. c7 Kf6 17. e4 Kg6 18. f4 Kh6 19. fxe5 Kg6 20. exf5+ Kg7 21. f6+ Kg6 22. e6 Kf5 23. e7 Bg7 24. f7 Ke5 25. h4 Bf6 26. g4 Bg7 27. hxg5 Bf6 28. gxh5 Bg7 29. g6 Bf6 30. h6 Bg5 31. g7 Be3 32. h7 Kf4 33. bxa8=Q Kg3 34. axb8=Q Bf4 35. dxc8=Q Be5 36. cxd8=Q Bf4 37. e8=Q Be5 38. f8=Q Bf4 39. gxh8=Q Be5 40. hxg8=Q#&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/en1zM.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/en1zM.gif&quot; alt=&quot;Here is a gif of the game&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="20088" LastEditDate="2017-03-21T04:50:38.310" LastActivityDate="2017-03-21T04:50:38.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5471" PostTypeId="1" AcceptedAnswerId="5551" CreationDate="2014-11-23T18:10:23.320" Score="13" ViewCount="1151" Body="&lt;p&gt;Light, soft and healthy is me&lt;br&gt;&#xA;mirrors are all over these&lt;br&gt;&#xA;if finding gold is rather dismal&lt;br&gt;&#xA;I sit next to the one in Pepto-Bismol  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different letters here is your goal&lt;br&gt;&#xA;somewhere to go with an unfortunate hole&lt;br&gt;&#xA;Equestrian researchers aren't quite right&lt;br&gt;&#xA;take from the morning not from the night  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Careful not to shoot from the hip&lt;br&gt;&#xA;take a bow and get a grip&lt;br&gt;&#xA;the vikings watch me in awe&lt;br&gt;&#xA;I hate my bed I'll stay up more  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These three up high are not the same&lt;br&gt;&#xA;but their collective being will grant you fame&lt;/p&gt;&#xA;" OwnerUserId="5135" LastEditorUserId="2484" LastEditDate="2014-11-24T01:30:44.020" LastActivityDate="2015-03-07T01:14:39.660" Title="Name that Sandwich" Tags="|riddle|word|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5472" PostTypeId="1" AcceptedAnswerId="5560" CreationDate="2014-11-23T19:06:11.420" Score="7" ViewCount="4350" Body="&lt;p&gt;You've been known to your government to be an expert in decrypting messages for years so you're not really surprised when a man in a black suit knocks on your door one rainy afternoon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;“Good afternoon, your country needs you.”&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much later, somewhere you’ve never been and would never know how to get to – or get out of, for that matter – you’re sitting in front of a grey-haired elderly man with a semi-permanent smile on his face which never seems quite honest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;“Fact is, we know they are using an encryption relying on a code word they change every other week, and while we have been able to intercept all their messages, we’ve never found the keyword to decipher them. But this time, we’ve got close. We’ve observed a suspect for several months, and we are pretty sure he’s the one receiving the keyword every other week, and passing it on to their network. Last night, the suspect was followed when he entered a shabby hotel, and also when he left it only 20min later. We’ve managed to arrange… and incident, which allowed us to go through his things without giving away that he’s followed. The only remarkable thing we discovered was a handwritten crossword-like paper of letters which we successfully copied down. It has presumably been written by the suspect while he was in the hotel, and we are sure it’s the keyword. Unfortunately, we have not at all been able to discover the keyword. And this is where you come into the game. We’ll update you with any intelligence we get, and you may ask us any question you think relevant, but right now, there really isn’t more to go on than this handwritten crossword… Oh, and the analysts of the messages tell me, that it is mathematically required that the keyword consists of at least 5 and at maximum 7 letters, and that all of those need to be unique.”&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you find the keyword? The paper found with the suspect is shown below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your ‘friend’ might give further possible information over the next few days. You may also ask specific questions, if you think that useful, and your friend will answer to the best of his knowledge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/p41fc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Technical notes and answers to comment-questions&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This is only my second attempt making a puzzle here, and tagging it properly is a bit difficult. I can assure you that there is a single answer and that it is clearly the answer (once found). I will add clues over time (if unsolved), and if you have specific questions, please ask them &quot;in story&quot; and you'll receive answers &quot;in story&quot; (when suitable)  Those questions might make things easier, but they are not essential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The image letters typed out as ASCII text (as requested):&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; FiPwUFrsWcQBaBnyI&#xA; TmtsoOpzfgkKnBCRu&#xA; lBAmGyJoUXATuwrxQ&#xA; iVYHKrCkNijmaczvY&#xA; iSOUeFUrKeEfEnDWt&#xA; mkcnqjDmnGNaajryH&#xA; AIotSrqOPadGKckYi&#xA; HuTHOrZpgpaMYIBCg&#xA; ukrbjiZSkwMdGVaKh&#xA; LiAVEtDimanxuXhJw&#xA; nrTABFdckbAmnfBKA&#xA; wBsKeysEEttdubkjY&#xA; FunkyrKuAmhCntPmC&#xA; kxAijZbIXiMVFwGun&#xA; HfBjeVdTXDrmARFtb&#xA; mtkAEJrUbwKYbaHam&#xA; RwXyaujKhavnIYJpG&#xA; LiMTuzvFWyQJiRuAR&#xA; xGqUeekGJRjEeZHIj&#xA; bJYpmiTByrjihIwrJ&#xA; RyNWHmkHanfzfJXxt&#xA; LIiLjHyatXlabrAtZ&#xA; ywqHawrivQuickgXQ&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Alignment of characters within the boxes:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The original note was handwritten with the characters not all being&#xA;  perfectly aligned in the boxes. (However, it's a cross-hatched paper,&#xA;  so the boxes are accurate.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Hints &amp;amp; Solution-Cross-checking&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First clue.&lt;/strong&gt; (If you're stuck. You can solve the puzzle without it.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A bit later that night a clerk opens the door and tells you: &quot;Any new&#xA; ideas already I'm sorry to tell you, that the first few words you've&#xA; given us to try didn't seem to do the trick. For what it's worth:&#xA; We've now searched the hotel room the message was received in, and it&#xA; is now clear to us, that the message was transmitted using Morse code.&#xA; We've found an automatic Morse-code writer hidden behind a panel. A&#xA; tiny scrap of paper was still in there - either from the beginning or&#xA; the end of the transmission. It reads:&#xA; &quot; · − − − − | − − · · · | − · · − | · · − − − | · · · − −&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second clue.&lt;/strong&gt; (If you're stuck. You can solve the puzzle without it. This clue is stronger than the first.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is not long after, that the same clerk knocks on your door again. &quot;We have just witnessed the suspect handing over the keynote to another man. We think we have positively identified him as the one-eyed spy, who we are trying to nail down for a long time. Anyway, the exchange was quite cleverly done in a Staples store and we nearly have missed it. The spy came in, asked the shop keeper for a packet of colourful pencils and, just after paying, picked up the crossword note our suspect had left in the packing area, and went out. For whatever reason - and to our luck - the one-eyed spy came back 5 minutes  later and exchanged - rather angrily - the packet of pencils for another one, before leaving again. We have of course spoken with the shop-keeper. According to him, the spy asked for &quot;a pack of 30 coloured pencils&quot; the first time, which the shopkeeper gave him. The spy returned, because the pack consisted of pencils in red, blue, green, yellow, pink and orange - 5 of each - but he wanted the pack containing 30 different colours.&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Once you have a keyword...&lt;/strong&gt; (Solving the one below will verify your solution.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your grey-haired friend crashes into the room and startles you out of&#xA; your thoughts. &quot;We really have to hurry now. We think we have just&#xA; intercepted a message send to the one-eyed man. It was stored on one&#xA; of those cheap label-printers of all things! When we tried to arrest&#xA; the man, he could escape, but at least he had to leave behind the&#xA; device. The message was still in memory! It's an endless mess of&#xA; letters, of course, but we have all confidence in you, that you can&#xA; turn this into a message...&quot; He hands you a long stripe printed from&#xA; the label machine (It is a single long piece like in the first image.) &#xA; &#xA; &lt;img src=&quot;https://i.stack.imgur.com/1Hkas.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA; &#xA; &lt;img src=&quot;https://i.stack.imgur.com/JGFja.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA; Typed out for copy &amp;amp; paste convenience:&#xA; &lt;em&gt;&quot;dSmAmcptscAbypOetIoaweorRoaLexdrrevmmiusVtFNnkneeenrsExyieatnXdgxpDKCgrvyxHxxQoxiMHxeXdnaTTruNWtaXieoJJXttn&quot;&lt;/em&gt;&#xA;&#xA; &#xA; ( The solution is a message consisting of 12 English words. )&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="5840" LastEditDate="2014-12-09T17:46:15.970" LastActivityDate="2014-12-09T17:46:15.970" Title="A keyword in a riddle - Find the cypher-keyword and crack the message" Tags="|cipher|encoded-message|" AnswerCount="2" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="5474" PostTypeId="2" ParentId="5190" CreationDate="2014-11-23T21:51:33.043" Score="1" Body="&lt;p&gt;There's absolutely nothing to support my suspicions, but I would guess that the most efficient way to determine the weight of each coin is to weigh half the coins at a time in multiple ways. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Begin by determining total weight of coins$\{c_1,c_2,c_3,c_4,c_5,c_6,c_7,c_8,c_9,c_{10},c_{11},c_{12}\}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Determine combined weight of coins $\{c_1,c_2,c_3,c_4,c_5,c_6\}$ and subtract from total weight to determine weight of $\{c_7,c_8,c_9,c_{10},c_{11},c_{12}\}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Determine combined weight of $\{c_2,c_4,c_6,c_8,c_{10},c_{12}\}$ and subtract from total weight to determine weight of $\{c_1,c_3,c_5,c_7,c_9,c_{11}\}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;After each weighing, the sum of the weights of six other coins can be deduced by subtracting the recently weighed coins from the total. After 7 total weighings, the weigher will have obtained 13 equations regarding the combined weights of the coins. Would this massive system of equations be solvable? This is not meant to be an answer, but hopefully this will point someone with the skills to actually solve this in the right direction. (Or in the wrong direction, I'm not sure.)&lt;/p&gt;&#xA;" OwnerUserId="4447" LastActivityDate="2014-11-23T21:51:33.043" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5475" PostTypeId="1" AcceptedAnswerId="5491" CreationDate="2014-11-23T22:47:52.970" Score="14" ViewCount="1645" Body="&lt;p&gt;Ring City has 100 houses arranged in a circle. Alice starts at her own house, and every day she randomly moves to one of the two adjacent houses, each with 50% probability. She repeats this until she has visited all of the houses. Which house(s) is/are the most likely to be visited last?&lt;/p&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2014-11-25T00:23:19.830" Title="Random Walk on Ring City" Tags="|mathematics|probability|combinatorics|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5476" PostTypeId="2" ParentId="5475" CreationDate="2014-11-23T23:10:11.050" Score="0" Body="&lt;p&gt;&lt;strong&gt;It turns out that while this is correct is doesn't not answer the question. I will leave it here on the off chance that it's useful to anyone. It does answer a question that noone asked.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If she starts at house $h_1$ and the adjacent houses are houses $h_2$ (to the left) and $h_{100}$ (to the right). Probabilistically she will visit house $h_{51}$ last (the one most directly opposite).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider a much simpler case. $4$ houses. She starts at house $h_1$ and the adjacent houses are houses $h_2$ (to the left) and $h_4$ (to the right). She's already been to $h_1$. There is now probability $0.5$ of visiting house $h_2$ next (similarly for house $h_4$). From there there is a $0.5$ probability of visiting house $h_3$. And since there are two ways to reach $h_3$ we must double the probability of having arrived from one side, $0.25$, and see that there is a probability of $0.5$ of visiting house $h_3$ for the first time in $2$ moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's scale back up. The probabilty of visiting house $h_{51}$ for the first time after $50$ steps is $1/2^{49}$. This is lowest probability that any house has to be visited for the first time.&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-11-24T09:42:01.297" LastActivityDate="2014-11-24T09:42:01.297" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="5477" PostTypeId="2" ParentId="3398" CreationDate="2014-11-23T23:43:39.233" Score="4" Body="&lt;p&gt;Edit: As others have pointed out this is incorrect, I realize that my solutions below only apply as upper bounds. It seems that the actual solution is somewhat more complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I attempted to formally prove these results:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;First consider a single dimension, a simple permutation of tiles such as ABCD. In this case the only valid operation is of course flipping two adjacent tiles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $M(n)$ denote a function representing the maximum number of moves required to sort a set of tiles of size $n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clear that $M(1)=0$. When considering $M(n)$ for the list of tiles $T_1,T_2...T_n$ we can apply the following method: flip the tile which belongs at the leftmost position with its neighbor to the left until it reaches said position. After this is completed, we simply have to sort the rest of the n-1 members, meaning $M(n)=F+M(n-1)$ where F is the number of times a flip was necessary. Since the largest possible value of $F$ is obviously $n-1$ (if the tile began at the rightmost spot) we can say that $M(n) \leq n-1+M(n-1)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a little math this is $M(n) \leq \frac{n(n-1)}{2}$  (we are essentially writing a formula for 1+2+...+n-1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also see that when the tiles begin in the reversed order from how they should be, each tile, $T_i$, must move to the right $n-i$ times (one for every tile to the right of it which has to switch with it to get to the left of it) and then left $i-1$ times to get to its correct spot. The total moves each tile performs is $n-1$, meaning the total number of moves &lt;em&gt;all&lt;/em&gt; the tiles have to undergo is $n(n-1)$ and since each switch moves two tiles this will take at least $\frac{n(n-1)}{2}$ switches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we have shown that this example takes at least $\frac{n(n-1)}{2}$ switches, we can confirm that the maximum number of switches required for any set of tiles of size $n$ must be at least that much: $M(n)\geq \frac{n(n-1)}{2}$ and with the result we determined before, we can conclude that $M(n)= \frac{n(n-1)}{2}$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let us now consider a two dimensional, $n$ x $n$ grid. Using our last result we can see that it takes $\frac{n(n-1)}{2}$ switches to sort any individual row, meaning it would take $\frac{n^2(n-1)}{2}$ switches to sort every row. This is effectively moving each tile to the correct column. Similarly, if we have already sorted rows, we would need to switch rows at most $\frac{n(n-1)}{2}$ times, or $\frac{n^2(n-1)}{2}$ total tile switches. This is effectively moving each tile to the correct row. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As frodoskywalker also briefly described in his answer, it does not affect the answer if tiles which do not belong to the same row begin in the same row before they are moved to their respective columns. This is because the vertical moves to correct it can double as also moving them toward correct row. Specifically if two tiles do not belong to the same row but are currently in the same row, one should be flipped vertically away towards its correct row, taking an extra move to sort the columns, but also lessening the number of moves required to sort the rows by 1. Since the gains and losses are equivalent in this method we can conclude that the total number of switches required is $\frac{n^2(n-1)}{2} + \frac{n^2(n-1)}{2}$ or simply $n^2(n-1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure about this part, but it seems to be intuitively true that this can be extended to higher dimensions; for each additional dimension, every tile must undergo at most $\frac{n(n-1)}{2}$ switches to reach its correct position in that dimension. Multiplying this by the number of dimensions and the number of tiles leads to the formula $\frac{n(n-1)}{2}n^{d-1}d$ or  $\frac{d}{2}n^d(n-1)$. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Whew, I hope everything I said made sense, I spent a lot more time writing this than I first intended but it was pretty fun, and I'm glad I managed to get the results that I did (I think anyway). If anyone doesn't understand something or thinks I might have goofed somewhere please comment.&lt;/p&gt;&#xA;" OwnerUserId="150" LastEditorUserId="150" LastEditDate="2014-11-25T20:57:58.820" LastActivityDate="2014-11-25T20:57:58.820" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5478" PostTypeId="2" ParentId="5475" CreationDate="2014-11-24T00:19:09.637" Score="7" Body="&lt;p&gt;I wrote the following program to check the probabilities. Surprisingly, it appears the answer is that all houses have an equal probability of being visited last!  The more likely a house is to be visited from one side, the less likely from the other side, and apparently those cancel out.  See &lt;a href=&quot;https://puzzling.stackexchange.com/a/5491/4&quot;&gt;Ruakh's answer&lt;/a&gt; for a rigorous proof.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#include &amp;lt;stdio.h&amp;gt;&#xA;#include &amp;lt;stdlib.h&amp;gt;&#xA;&#xA;#define N 100&#xA;#define TRIALS 100000&#xA;&#xA;int go() {&#xA;    int doors[N] = {1};&#xA;    int unique = 1;&#xA;    int i = 0;&#xA;    while (unique &amp;lt; N) {&#xA;        i = (i + 2*arc4random_uniform(2) + N - 1)%N;&#xA;        if (doors[i] == 0) {&#xA;            unique++;&#xA;            doors[i] = 1;&#xA;        }&#xA;    }&#xA;    return i;&#xA;}&#xA;&#xA;int main() {&#xA;    int doors[N] = {0};&#xA;    for (int i = 0; i &amp;lt; TRIALS; i++) {&#xA;        doors[go()]++;&#xA;    }&#xA;    for (int i = 0; i &amp;lt; N; i++) {&#xA;        printf(&quot;%d: %d\n&quot;, i+1, doors[i]);&#xA;    }&#xA;    return EXIT_SUCCESS;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Sample results [TRIALS = 10M]:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1: 0&#xA;2: 101527&#xA;3: 101171&#xA;4: 100559&#xA;5: 100836&#xA;6: 100708&#xA;7: 101287&#xA;8: 100198&#xA;9: 100880&#xA;10: 101034&#xA;11: 101746&#xA;12: 100797&#xA;13: 101113&#xA;14: 100255&#xA;15: 100268&#xA;16: 101401&#xA;17: 101076&#xA;18: 101165&#xA;19: 100811&#xA;20: 101339&#xA;21: 100836&#xA;22: 101519&#xA;23: 101056&#xA;24: 100544&#xA;25: 100960&#xA;26: 101046&#xA;27: 100687&#xA;28: 100987&#xA;29: 100801&#xA;30: 101123&#xA;31: 101045&#xA;32: 101195&#xA;33: 101462&#xA;34: 101302&#xA;35: 100930&#xA;36: 100672&#xA;37: 100794&#xA;38: 100362&#xA;39: 100957&#xA;40: 101342&#xA;41: 100819&#xA;42: 101464&#xA;43: 100824&#xA;44: 100810&#xA;45: 101009&#xA;46: 100797&#xA;47: 101409&#xA;48: 101267&#xA;49: 101176&#xA;50: 101398&#xA;51: 100872&#xA;52: 101328&#xA;53: 101060&#xA;54: 101174&#xA;55: 100592&#xA;56: 101091&#xA;57: 101059&#xA;58: 101357&#xA;59: 101020&#xA;60: 101002&#xA;61: 101168&#xA;62: 100612&#xA;63: 101208&#xA;64: 101341&#xA;65: 100329&#xA;66: 101180&#xA;67: 101474&#xA;68: 101442&#xA;69: 101302&#xA;70: 101537&#xA;71: 101158&#xA;72: 101082&#xA;73: 101007&#xA;74: 100812&#xA;75: 100826&#xA;76: 100542&#xA;77: 100244&#xA;78: 101250&#xA;79: 100853&#xA;80: 100696&#xA;81: 101522&#xA;82: 100750&#xA;83: 101193&#xA;84: 101617&#xA;85: 101515&#xA;86: 100634&#xA;87: 101384&#xA;88: 100391&#xA;89: 100587&#xA;90: 101050&#xA;91: 101382&#xA;92: 101077&#xA;93: 101155&#xA;94: 100603&#xA;95: 100357&#xA;96: 100809&#xA;97: 101075&#xA;98: 101054&#xA;99: 101308&#xA;100: 101155&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2014-11-24T17:07:10.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5479" PostTypeId="2" ParentId="5411" CreationDate="2014-11-24T02:40:44.757" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Last night you had a dream   &lt;br/&gt; 'twas me you'd seen  &lt;br/&gt;&lt;br/&gt;  &lt;img src=&quot;https://i.stack.imgur.com/v08vim.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;br/&gt; Lady Justice is I, and  &lt;br/&gt; the scroll in my hand &lt;br/&gt; was the law of your land &lt;br/&gt;&lt;br/&gt; Nestled in my crescent wreath  &lt;br/&gt; It is this I do bequeath &lt;br/&gt;&lt;br/&gt; The pillars of &lt;strong&gt;J&lt;/strong&gt;ustice   &lt;br/&gt;are mine alone  &lt;br/&gt;How the &lt;strong&gt;B&lt;/strong&gt;alance swings  &lt;br/&gt;is yours to have known &lt;br/&gt;&lt;br/&gt; Be warned that others  &lt;br/&gt;will lead you astray    &lt;br/&gt;and that only I   &lt;br/&gt;will have the last say&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4522" LastEditorUserId="52483" LastEditDate="2018-11-22T21:05:36.577" LastActivityDate="2018-11-22T21:05:36.577" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="5480" PostTypeId="2" ParentId="5475" CreationDate="2014-11-24T02:41:49.370" Score="2" Body="&lt;p&gt;The the last door index is a uniform distribution, with a constant probability of any door being last is 1/99, and the standard deviation after N evaluations is 2.3/sqrt(N).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had to develop algorithms in order to have the program run faster enough in R.  As I was making the algorithms, the following 2 arguments seem relevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The number of steps required to reach the last door is a random variable greater than 100.  Hence, the location of the last door is a random variable.  The door selection forms a rings, so there is no central point, so the random distribution is uniform.  Every door, except the 1st, has a probability of 1/99 of being the last door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) A different argument, it doesn't matter what doors have been previously visited.  So, the puzzle after visiting 50 doors, or N doors, is the same.  The probability of reach the opposite door does not depend on the starting point, so there is no preference to the opposite door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heart of the proof is the convergence of the mean statistic and standard deviation trend as the number of evaluations increases.  The mean percent occurrence is always 1.0101, starting with only 1 door (1/99).  The standard deviation decreases exponentially as N increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the plot of mean percentage as the number of iterations increases on a semilog scale (the mean percentage is a constant, 1.0101)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z5vkq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the tread of standard deviation with increasing number of evaluations on a log log scale.  The trend is linear, indicating that std_dev ~ 2.3*sqrt(N)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NiwXJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an R script to tabulate the frequency of the last door.  The algorithm is&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1) Find a random walk that covers all 100 doors&#xA;2) Determine the length of the random walk to cover 99 doors&#xA;3) The last door is whatever door is remaining&#xA;4) Repeat steps 1 to 3 for a range of iterations, then plot the mean/std&#xA;&#xA;# 100 houses in a ring&#xA;# random move to either side&#xA;# What is the probability of door being last&#xA;cat(&quot;\014&quot;) &#xA;doMoves &amp;lt;- function() {&#xA;  doorVec &amp;lt;- rep(1,100)&#xA;  currentDoor = 1&#xA;  doorVec[currentDoor] = 0&#xA;  nUnique = 1&#xA;  # Find sequence longer than than required to visit all doors&#xA;  houseDoors = c(1)&#xA;  nDoorsToAdd = 1000&#xA;  while(nUnique &amp;lt; 100) {&#xA;    randMoves &amp;lt;- sample(c(-1,1),nDoorsToAdd,replace=TRUE)&#xA;    currentDoor &amp;lt;- houseDoors[length(houseDoors)]&#xA;    newHousesVisited &amp;lt;- currentDoor+cumsum(randMoves)&#xA;    houseDoors &amp;lt;- ((c(houseDoors, newHousesVisited)-1) %% 100)+1&#xA;    nUnique = length(unique(houseDoors))&#xA;  }&#xA;  # Find lenght of sequence required to visit 99 doors&#xA;  maxMovesRequired = length(houseDoors) &#xA;  minMovesRequired = maxMovesRequired - nDoorsToAdd&#xA;  nMovesToTry &amp;lt;- minMovesRequired&#xA;  nDoorsVisited &amp;lt;- length(unique(houseDoors[1:nMovesToTry]))&#xA;  while(nDoorsVisited != 99) {&#xA;    nMovesToTry &amp;lt;- floor((minMovesRequired+maxMovesRequired)/2)&#xA;    nDoorsVisited = length(unique(houseDoors[1:nMovesToTry]))&#xA;    if (nDoorsVisited&amp;lt;99) {&#xA;      minMovesRequired = nMovesToTry&#xA;    } else if (nDoorsVisited&amp;gt;99) {&#xA;      maxMovesRequired = nMovesToTry&#xA;    }&#xA;  }&#xA;  # Find the last door&#xA;  lastDoor &amp;lt;- setdiff(1:100,unique(houseDoors[1:nMovesToTry]))&#xA;&#xA;  return(lastDoor)&#xA;}&#xA;&#xA;findLastDoors &amp;lt;- function(nTries) {&#xA;  lastDoorCount &amp;lt;- rep(0,100)&#xA;  for (iTry in seq(1,nTries)) {&#xA;    lastDoor &amp;lt;- doMoves()&#xA;    lastDoorCount[lastDoor] = lastDoorCount[lastDoor]+1&#xA;  }&#xA;  return(lastDoorCount)&#xA;}&#xA;doIt &amp;lt;- function(n) {&#xA;  lastDoorCount &amp;lt;- findLastDoors(n)&#xA;  print(paste(sum(lastDoorCount),'samples'))&#xA;  ld_mn &amp;lt;- mean(lastDoorCount[-1]/n*100)&#xA;  ld_st &amp;lt;- sd(lastDoorCount[-1]/n*100)&#xA;  plot(lastDoorCount[-1])&#xA;  plot(lastDoorCount[-1]/n*100,xlab=&quot;Door&quot;,ylab=&quot;Percent Times Last&quot;,&#xA;       main=sprintf('%d doors has mean %.6f std dev %.4f',n,ld_mn,ld_st))&#xA;  print(lastDoorCount[-1])&#xA;  print(lastDoorCount[-1]/n*100)&#xA;  print(sprintf('%d doors has mean %.6f std dev %.4f',n,ld_mn,ld_st))&#xA;  return(list(n=n, ldc = lastDoorCount, ld_mn = ld_mn, ld_std = ld_st))&#xA;}&#xA;print(system.time(r1&amp;lt;-doIt(1L)))&#xA;print(system.time(r5&amp;lt;-doIt(5L)))&#xA;print(system.time(r10&amp;lt;-doIt(10L)))&#xA;print(system.time(r50&amp;lt;-doIt(50L)))&#xA;print(system.time(r100&amp;lt;-doIt(100L)))&#xA;print(system.time(r500&amp;lt;-doIt(500L)))&#xA;print(system.time(r1000&amp;lt;-doIt(1000L)))&#xA;print(system.time(r5000&amp;lt;-doIt(5000L)))&#xA;#print(system.time(r10000&amp;lt;-doIt(10000L)))&#xA;#print(system.time(r10000&amp;lt;-doIt(50000L)))&#xA;#print(system.time(r100000&amp;lt;-doIt(100000)))&#xA;#print(system.time(r100000&amp;lt;-doIt(1000000)))&#xA;&#xA;smry&amp;lt;-data.frame(n=c(r1$n,r5$n,r10$n,r50$n,r100$n,&#xA;                     r500$n,r1000$n,r5000$n),&#xA;                 ld_mn=c(r1$ld_mn,r5$ld_mn,r10$ld_mn,r50$ld_mn,r100$ld_mn,&#xA;                     r500$ld_mn,r1000$ld_mn,r5000$ld_mn),&#xA;                 ld_std=c(r1$ld_std,r5$ld_std,r10$ld_std,r50$ld_std,r100$ld_std,&#xA;                     r500$ld_std,r1000$ld_std,r5000$ld_std))&#xA;plot(smry$n,smry$ld_mn,log='x',xlab='N Iterations',ylab='Mean Percetage')&#xA;plot(smry$n,smry$ld_std,log='xy',xlab='N Iterations',ylab='Std Dev Percetage')&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here's a plot of the percent frequencies for 500 and 5000 evaluations.  The distribution is uniform, constant mean of 1.0101 percent, and reduced standard deviation for increasing number of evaluations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/X7WDP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sbhd0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5913" LastEditorUserId="5913" LastEditDate="2014-11-25T00:23:19.830" LastActivityDate="2014-11-25T00:23:19.830" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5481" PostTypeId="1" AcceptedAnswerId="5531" CreationDate="2014-11-24T03:00:51.270" Score="10" ViewCount="1507" Body="&lt;p&gt;Quite a few years ago (I can't remember the date but it was obviously some time after I first met Ernie at a New Years Eve party in 2001) I got an unexpected phone call late one afternoon. It was Ernie. He told me that it was his wedding anniversary and he had secretly bought two tickets to see the famous psychic Madame Mu as a surprise for his wife. Unfortunately, his wife was called out of town on unexpected business wouldn't be able to attend. Ernie  was in need of company, so if I wanted to see the show with him, I should turn up at the &lt;em&gt;Palais&lt;/em&gt; no later than 8.30 pm that evening.&lt;/p&gt;&#xA;&lt;p&gt;I arrived in time to meet Ernie and we entered the auditorium . The lights dimmed and the curtain rose to show Madame Mu, dressed in a scarlet robe and emerald turban, standing center-stage. The show was very entertaining. A search-light would raster across the audience before stopping on someone who would be invited to speak with Madame Mu. She would ask them to introduce themselves and then, by &amp;quot;communing with the spirit world&amp;quot;, she would reveal various facts about them.&lt;/p&gt;&#xA;&lt;p&gt;About halfway into the show, the spot-light landed on Ernie, who rose and walked down to the stage. He introduced himself as an inventor and amateur mathematician, and Madame Mu announced she would attempt something &amp;quot;a little special&amp;quot; here and would try to &amp;quot;channel the great German mathematician, Carl Fredrich Gauss&amp;quot;. The lights dimmed, Madame Mu slumped slightly - seemingly in deep meditation - there was a short pause, then she raised her head again.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AKlX5.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;Good evening Ernest&amp;quot;, said Mu/Gauss in a strong German accent, &amp;quot;let me entertain you with a little mind reading and lightning calculation&amp;quot;. The following conversation then took place:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Mu/Gauss:&lt;/strong&gt; &amp;quot;If you concatenate the day (1..31), month (1..12), and year of your birth (4 digits),&#xA;in that order, you will get a number divisible by 7. Correct?&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Ernie:&lt;/strong&gt; &amp;quot;That is true - but there is a 1 in 7 chance it would be true&#xA;for any random birthday&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Mu/Gauss:&lt;/strong&gt; &amp;quot;If you concatenate the year, month, and day of your birth,&#xA;in that order, you will get a number divisible by 11. Correct?&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Ernie:&lt;/strong&gt; &amp;quot;Correct&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Mu/Gauss:&lt;/strong&gt; &amp;quot;If you concatenate the month, day, and year of your birth, in that order, you will get a number divisible by 13.&#xA;Correct?&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Ernie:&lt;/strong&gt; &amp;quot;Correct - and only about 1 in 1000 you could get&#xA;that right by chance&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Mu/Gauss:&lt;/strong&gt; &amp;quot;If you add together, those three numbers you just calculated plus your age in whole years, you will get a prime.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Ernie (after a short pause for&#xA;calculation):&lt;/strong&gt; &amp;quot;You're right again - the calculation works as you say!&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Mu/Gauss:&lt;/strong&gt; &amp;quot;And by an amazing coincidence, the same five facts are also true for your father - correct?&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Ernie (after another pause):&lt;/strong&gt;&#xA;&amp;quot;Amazing - all true!&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Mu/Gauss:&lt;/strong&gt; &amp;quot;Thank you Ernest, please return to your seat and enjoy the rest of the show. But as you go, can you confirm your father is&#xA;currently twice your age in years?&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Ernie:&lt;/strong&gt; (returning to his seat) &amp;quot;In whole years, yes, that's true.&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Later Ernie told me exactly how Mu did her trickery: something to do with internet searches, a hidden radio earpiece, and an accomplice with a &amp;quot;top model clockwork calculator&amp;quot;, but that isn't the mystery of this story. As you probably know by now, I'm great at recollecting conversations, but hopeless at remembering numbers. I want to send Ernie a card on his next birthday. I can't remember the date but believe there is sufficient information in this story to calculate not only is date of birth. Can you help me?&lt;/p&gt;&#xA;&lt;p&gt;PS - As an added note, it may help to know that Ernie's father is alive and well and nothing out of the ordinary in years of age.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;ADDENDUM:&lt;/strong&gt; in case it isn't clear, according to what Ernie told me after the show &amp;quot;those three numbers&amp;quot; refer to the three numbers divisible by 7, 11, and 13 respectively, not the three numbers that make up the date.&lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-11-25T14:44:00.897" Title="Ernie and the Mystery of Madame Mu" Tags="|mathematics|word-problem|" AnswerCount="5" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="5482" PostTypeId="2" ParentId="5391" CreationDate="2014-11-24T03:38:14.873" Score="3" Body="&lt;p&gt;If the game mechanics allow it, you could devise a situation where the victor gained certain benefits from their victory - experience points or whatever - sufficient to compensate for part or all of the damage they suffered.&lt;/p&gt;&#xA;" OwnerUserId="1631" LastActivityDate="2014-11-24T03:38:14.873" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5483" PostTypeId="1" AcceptedAnswerId="5485" CreationDate="2014-11-24T04:03:28.793" Score="2" ViewCount="986" Body="&lt;p&gt;The following list contains 30 clues describing 30 unique English words or well-known names, all of which have one thing in common.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The objective of the puzzle is to find the words associated with the clues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simple enough, but to change things up a bit I've divided the clues into three groups: &lt;strong&gt;A&lt;/strong&gt;, &lt;strong&gt;B&lt;/strong&gt;, and &lt;strong&gt;C&lt;/strong&gt;. Puzzlers are requested to only post the complete solution to &lt;strong&gt;one&lt;/strong&gt; group (the choice is up to you), and to only post &lt;strong&gt;one&lt;/strong&gt; solution per user. This way, users needn't solve all thirty clues, and at least three users will share in the spoils of solving the puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;correct answer&quot; check will go to the first correct solution posted (for any group).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This format is somewhat experimental, hence any feedback is welcome &lt;a href=&quot;http://meta.puzzling.stackexchange.com/questions/1364/are-the-following-mass-producible-puzzles-acceptable-on-puzzling-se-and-is-it-a&quot;&gt;on this meta thread&lt;/a&gt;. :)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Group A&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;every investor has one&lt;/li&gt;&#xA;&lt;li&gt;a roaring good sign&lt;/li&gt;&#xA;&lt;li&gt;Mr. Limbaugh's medium&lt;/li&gt;&#xA;&lt;li&gt;Waitaminit! Isn't he that actor...?&lt;/li&gt;&#xA;&lt;li&gt;Where art thou?&lt;/li&gt;&#xA;&lt;li&gt;dress like a Tahitian&lt;/li&gt;&#xA;&lt;li&gt;An apartment, a workplace, a place of creation. All of the above.&lt;/li&gt;&#xA;&lt;li&gt;one difficult, confused situation&lt;/li&gt;&#xA;&lt;li&gt;because mono is old hat&lt;/li&gt;&#xA;&lt;li&gt;nothing but rainforest&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Group B&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;11&quot;&gt;&#xA;&lt;li&gt;let those notes fly hot and fast&lt;/li&gt;&#xA;&lt;li&gt;Ballet, opera anyone? I am your host.&lt;/li&gt;&#xA;&lt;li&gt;Where else would we put the barbeque and the deck chairs?&lt;/li&gt;&#xA;&lt;li&gt;a bounder, but at least a seductive one&lt;/li&gt;&#xA;&lt;li&gt;bought it at a novelty store&lt;/li&gt;&#xA;&lt;li&gt;with some milk for dipping&lt;/li&gt;&#xA;&lt;li&gt;a feast for the ears&lt;/li&gt;&#xA;&lt;li&gt;full of bull&lt;/li&gt;&#xA;&lt;li&gt;a fraction of an answer&lt;/li&gt;&#xA;&lt;li&gt;the corporate head (abbr.)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Group C&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;21&quot;&gt;&#xA;&lt;li&gt;Are you vaccinated?&lt;/li&gt;&#xA;&lt;li&gt;a group of three&lt;/li&gt;&#xA;&lt;li&gt;OK, picture the following...&lt;/li&gt;&#xA;&lt;li&gt;I can't believe it's not butter!&lt;/li&gt;&#xA;&lt;li&gt;now available in streaming and -on-demand flavours&lt;/li&gt;&#xA;&lt;li&gt;a nosey little fellow&lt;/li&gt;&#xA;&lt;li&gt;vivacity, not the cheese&lt;/li&gt;&#xA;&lt;li&gt;that's some slow music&lt;/li&gt;&#xA;&lt;li&gt;it's a word from the heart&lt;/li&gt;&#xA;&lt;li&gt;what a nut&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2015-06-16T23:11:36.440" Title="A Puzzle for Rio" Tags="|crossword-clues|wordplay|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5484" PostTypeId="2" ParentId="5481" CreationDate="2014-11-24T04:32:28.933" Score="4" Body="&lt;p&gt;I believe his birthday is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1956/4/27 (27 April 1956)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I arrived at this answer after looking up the possible primes and the nearest perfect square possible for any human to be alive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what i have found:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 45 x 45 = 2025 (Nearest perfect square for human kind to be alive at 20th century)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I then looked up on prime numbers nearest to this number, subtracted the numbers to find the possible age that can be multiplied by 2 and still be alive (his dad) and found the following numbers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1979 1987 1993 1997 2003.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The rest is just coding, running from 1950 up all the way to 2014. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My answer is wrong. Look at GOTO 0's solution.&lt;/p&gt;&#xA;" OwnerUserId="4338" LastEditorUserId="4338" LastEditDate="2014-11-24T09:02:41.620" LastActivityDate="2014-11-24T09:02:41.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5485" PostTypeId="2" ParentId="5483" CreationDate="2014-11-24T05:11:11.093" Score="4" Body="&lt;p&gt;Group C&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 21. polio&#xA;&lt;P&gt;&#xA; 22. trio&#xA;&lt;P&gt;&#xA; 23. scenario &#xA;&lt;P&gt;&#xA; 24. oleo&#xA;&lt;P&gt;&#xA; 25. video&#xA;&lt;P&gt;&#xA; 26. pinocchio&#xA;&lt;P&gt;&#xA; 27. brio &#xA;&lt;P&gt;&#xA; 28. adagio&#xA;&lt;P&gt;&#xA; 29. cardio&#xA;&lt;P&gt;&#xA; 30. pistachio&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2014-11-24T05:11:11.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5486" PostTypeId="1" AcceptedAnswerId="5490" CreationDate="2014-11-24T05:21:21.793" Score="-1" ViewCount="197" Body="&lt;p&gt;I have a code for you:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;L0EJAU&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Amazon's biggest rival was the one who made this. What is this?&lt;/p&gt;&#xA;" OwnerUserId="5958" LastEditorUserId="5958" LastEditDate="2014-11-24T06:54:25.513" LastActivityDate="2014-11-24T07:35:20.283" Title="Crack the code: L0EJAU" Tags="|word|" AnswerCount="1" CommentCount="0" ClosedDate="2014-11-24T08:28:30.390" ContentLicense="CC BY-SA 3.0" />
  <row Id="5488" PostTypeId="2" ParentId="5481" CreationDate="2014-11-24T06:36:54.640" Score="4" Body="&lt;p&gt;Ernie was born on &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;9 December 1966&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;DMMYYYY: 9121966 is divisible by 7 (9121966 = 7 &amp;times; 1303138).&lt;/li&gt;&#xA;&lt;li&gt;YYYYMMD: 1966129 is divisible by 11 (1966129 = 11 &amp;times; 178739).&lt;/li&gt;&#xA;&lt;li&gt;MMDYYYY: 1291966 is divisible by 13 (1291966 = 13 &amp;times; 99382).&lt;/li&gt;&#xA;&lt;li&gt;YYYY + MM + D: 1966 + 12 + 9 = 1987 (prime number)&lt;/li&gt;&#xA;&lt;li&gt;The next perfect square larger than 1987 is 2025 = 45², so Erwin's age was 2025 - 1987 = 38, and the show was in 2004 or 2005.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I found this result by running a script over all possible date combinations between 1950-01-01 and 1999-01-01.&lt;/p&gt;&#xA;" OwnerUserId="4394" LastEditorUserId="4394" LastEditDate="2014-11-24T06:59:39.413" LastActivityDate="2014-11-24T06:59:39.413" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="5490" PostTypeId="2" ParentId="5486" CreationDate="2014-11-24T07:35:20.283" Score="5" Body="&lt;p&gt;Interesting, this is my answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Searching &quot;Amazon's biggest rival&quot; bring google at the top of the result, searching &quot;L0EJAU&quot; bring random page, but there is one that makes me interested, bit.ly/L0ejAU, it takes me to korean website (I think), but changing bit.ly with goo.gl take me to this riddle, so my answer, that (L0EJAU) is google short url (&lt;a href=&quot;http://goo.gl/L0EJAU&quot;&gt;http://goo.gl/L0EJAU&lt;/a&gt;) code that point to this question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5800" LastActivityDate="2014-11-24T07:35:20.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5491" PostTypeId="2" ParentId="5475" CreationDate="2014-11-24T07:45:46.017" Score="16" Body="&lt;p&gt;John and Kevin are correct: the probability is the same, $\frac{1}{99}$, for each house except the one she begins at. Below is a mathematical argument for why this must be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the houses be numbered in order, going around the circle, from #1 to #100, with #1 being the house she starts at. (Note that house #1 is between houses #2 and #100.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $p_n$ denote the probability that the last house she visits is house #&lt;em&gt;n&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $q_n$ denote the probability that the last house she visits is house #&lt;em&gt;n&lt;/em&gt;, &lt;strong&gt;provided that&lt;/strong&gt; the first house she randomly moves to is house #2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $r_n$ denote the probability that the last house she visits is house #&lt;em&gt;n&lt;/em&gt;, &lt;strong&gt;provided that&lt;/strong&gt; the first house she randomly moves to is house #100.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following observations are straightforward:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Various probabilities are known to be zero: $p_1 = 0$ and $q_1 = 0$ and $q_2 = 0$ and $r_1 = 0$ and $r_{100} = 0$.&lt;/li&gt;&#xA;&lt;li&gt;By symmetry, $p_2 = p_{100}$, and $p_3 = p_{99}$, and so on, up through $p_{50} = p_{52}$.&lt;/li&gt;&#xA;&lt;li&gt;By symmetry, $q_3 = r_{99}$, and $q_4 = r_{98}$, and so on, all the way up through $q_{100} = r_2$.&lt;/li&gt;&#xA;&lt;li&gt;Since Alice must initially move to either house #2 or house #100, and either of these is equally likely, we know that $p_n = \frac{q_n + r_n}{2}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In addition, we need one somewhat-subtle observation: If she initially moves to house #2, then she &quot;shifts&quot; the probability of every house over by one house &amp;mdash; so, for example, $q_3 = p_2$ &amp;mdash; except that $q_1 = 0$ (instead of $q_1 = p_{100}$, which is &lt;em&gt;not&lt;/em&gt; true) $q_{100} = p_{99} + p_{100}$. The reason that $q_{100} = p_{99} + p_{100}$ is that if we imagine that Alice initially moves to house #2 and then reset her memory (forgetting that she had already visited house #1), and therefore continues until she has visited every house at least once and house #1 at least &lt;em&gt;twice&lt;/em&gt;, then if it indeed happens that her return to house #1 is her last move, then house #100 must have been her second-to-last house, because she had to visit #100 before re-visiting #1 (in order for her house #1 re-visit to be her last house), and the only way to get all the way to #100 before re-visiting #1 is to first visit every single house from #3 to #99.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given those facts, the rest is algebra:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$p_1 = 0$&lt;/li&gt;&#xA;&lt;li&gt;$p_2 = \frac{q_2 + r_2}{2} = \frac{r_2}{2} = \frac{q_{100}}{2} = \frac{p_{99} + p_{100}}{2} = \frac{p_2 + p_3}{2} \Rightarrow p_3 = p_2$&lt;/li&gt;&#xA;&lt;li&gt;$p_3 = \frac{q_3 + r_3}{2} = \frac{q_3 + q_{99}}{2} = \frac{p_2 + p_{98}}{2} = \frac{p_2 + p_4}{2} = \frac{p_3 + p_4}{2} \Rightarrow p_4 = p_3$&lt;/li&gt;&#xA;&lt;li&gt;$p_4 = \frac{q_4 + r_4}{2} = \frac{q_4 + q_{98}}{2} = \frac{p_3 + p_{97}}{2} = \frac{p_3 + p_5}{2} = \frac{p_4 + p_5}{2} \Rightarrow p_5 = p_4$&lt;/li&gt;&#xA;&lt;li&gt;...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4029" LastActivityDate="2014-11-24T07:45:46.017" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5492" PostTypeId="2" ParentId="5481" CreationDate="2014-11-24T08:13:56.533" Score="4" Body="&lt;p&gt;The answer is what GOTO 0 said. Since I'm learning Python, I also used a script to figure out the answer for some practice. One tricky part is deciding between &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 27 April 1956 and 9 December 1966. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They both work when the age is 38. Using the two dates 1956 and 1966, Ernie's age in 2001 could have been either 45 or 35, respectively. He had to have been 35 if this event happened when he was 38, so his birth date is 9 December 1966.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Python script is included below for anyone interested.  It should run as is if it is copied and pasted into a .py file.  Cheers!&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;print 'So when was Ernie born?'&#xA;&#xA;# determines if a number is a perfect square&#xA;def is_square(apositiveint):&#xA;    x = apositiveint // 2&#xA;    seen = set([x])&#xA;    while x * x != apositiveint:&#xA;        x = (x + (apositiveint // x)) // 2&#xA;        if x in seen: return False&#xA;        seen.add(x)&#xA;    return True&#xA;&#xA;# determines if a number is prime&#xA;def isprime(n):&#xA;    '''check if integer n is a prime'''&#xA;    # make sure n is a positive integer&#xA;    n = abs(int(n))&#xA;    # 0 and 1 are not primes&#xA;    if n &amp;lt; 2:&#xA;        return False&#xA;    # 2 is the only even prime number&#xA;    if n == 2: &#xA;        return True    &#xA;    # all other even numbers are not primes&#xA;    if not n &amp;amp; 1: &#xA;        return False&#xA;    # range starts with 3 and only needs to go up the squareroot of n&#xA;    # for all odd numbers&#xA;    for x in range(3, int(n**0.5)+1, 2):&#xA;            if n % x == 0:&#xA;                return False&#xA;    return True&#xA;&#xA;# This is where the magic happens... &#xA;days = xrange(1, 32)    # 1-31&#xA;months = xrange(1,13)   # 1-12&#xA;years = xrange(1900, 2020)  # 1900-2019&#xA;ages = xrange(10,99)    # 10-98&#xA;&#xA;# These are possible dates that might solve the riddle&#xA;possibleDates = []&#xA;&#xA;# boolean flag&#xA;dateIsBad = False&#xA;&#xA;for year in years:&#xA;    for month in months:&#xA;        for day in days:&#xA;            for age in ages:&#xA;&#xA;                date = str(day) + ' ' + str(month) + ' ' + str(year)&#xA;                ageAndDate = 'age = ' + str(age) + ' date = ' + date&#xA;&#xA;                concat1 = int(str(day) + str(month) + str(year))&#xA;                if concat1 % 7 != 0:&#xA;                    dateIsBad = True&#xA;&#xA;                concat2 = int(str(year) + str(month) + str(day))&#xA;                if concat2 % 11 != 0:&#xA;                    dateIsBad = True&#xA;&#xA;                concat3 = int(str(month) + str(day) + str(year))&#xA;                if concat3 % 13 != 0:&#xA;                    dateIsBad = True&#xA;&#xA;                number1 = day + month + year&#xA;                if isprime(number1) and is_square(number1 + age):&#xA;                    pass&#xA;                else:&#xA;                    dateIsBad = True&#xA;&#xA;                if not dateIsBad:&#xA;                    print 'found one! ' + ageAndDate&#xA;                    possibleDates.append(ageAndDate)&#xA;&#xA;                # reset the boolean flag&#xA;                dateIsBad = False&#xA;&#xA;print possibleDates&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="5960" LastEditorUserId="319" LastEditDate="2014-11-24T14:38:48.103" LastActivityDate="2014-11-24T14:38:48.103" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5498" PostTypeId="1" AcceptedAnswerId="5500" CreationDate="2014-11-24T10:19:10.057" Score="11" ViewCount="10369" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;gegs? (9, 4)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I hope this is a suitable forum for this British crossword clue, as it's my first post and I'm trying something new. This is a cryptic crossword clue, so (9, 4) tells you about the structure of the answer.&lt;/p&gt;&#xA;&lt;p&gt;Bonus question:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;code&gt;D R U G G E D   U P? (6, 3)&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5968" LastEditorUserId="5373" LastEditDate="2021-10-18T05:19:32.983" LastActivityDate="2021-10-18T05:19:32.983" Title="gegs? (9, 4) - a british cryptic crossword clue" Tags="|cryptic-clues|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="5499" PostTypeId="2" ParentId="5498" CreationDate="2014-11-24T10:20:33.453" Score="-4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;gegs = scrambled eggs&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note: &lt;a href=&quot;https://www.google.co.uk/webhp?sourceid=chrome-instant&amp;amp;ion=1&amp;amp;espv=2&amp;amp;ie=UTF-8#q=gegs%20crossword%20clue&quot; rel=&quot;nofollow noreferrer&quot;&gt;Google is your friend&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for drugged up, I think I've actually done this one in the past (in &lt;a href=&quot;http://www.theguardian.com/crosswords/cryptic/24950&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Guardian&lt;/a&gt;). Answer was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;   &lt;strong&gt;drugged up = zonked out&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4965" LastEditorUserId="44224" LastEditDate="2018-03-28T20:41:59.317" LastActivityDate="2018-03-28T20:41:59.317" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="5500" PostTypeId="2" ParentId="5498" CreationDate="2014-11-24T10:25:28.093" Score="21" Body="&lt;p&gt;gegs? (9, 4)  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;SCRAMBLED EGGS&lt;/strong&gt; &lt;br&gt;&lt;br&gt;&#xA; The letter of &quot;gegs&quot; are those of the word &quot;eggs&quot; only mixed up (or &lt;em&gt;scrambled&lt;/em&gt;). The answer is thus &lt;code&gt;scrambled eggs&lt;/code&gt;. The question mark denotes that we are looking for something a little tricky. Normally a cryptic clue would want something like &quot;Messy gegs for breakfast&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;D R U G G E D   U P? (6, 3)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;SPACED OUT&lt;/strong&gt; &lt;br&gt;&lt;br&gt;&#xA; Notice there is a space between every letter. This means the letters are &lt;code&gt;spaced out&lt;/code&gt; which is a synonym of being on drugs (or &quot;drugged up&quot;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="4385" LastEditDate="2015-07-15T18:14:55.083" LastActivityDate="2015-07-15T18:14:55.083" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5502" PostTypeId="1" AcceptedAnswerId="5524" CreationDate="2014-11-24T10:27:20.257" Score="15" ViewCount="5563" Body="&lt;p&gt;Harry was a handsome and intelligent young bachelor with an impressive personality. One day he went to a leading IT firm for an interview. As he entered the room, he got stunned by seeing the beauty of the boss of that company. She was hot, sensual, in her mid-twenties with a black geek specs, which doubled her hotness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interview has begun and after going through his resume, she asked him some random technical and HR questions. He answered most of them and tried to impress her.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once she finished asking the questions, she took his resume in her hands and said &lt;strong&gt;&lt;code&gt;&quot;I need to double-check it&quot;&lt;/code&gt;&lt;/strong&gt; and then she started writing something in it. After writing something she stopped and said &lt;strong&gt;&lt;code&gt;&quot;I am afraid this post is not apt for you. Even if you don't get this job, don't be disappointed. You have to move on, to a position that suits you&quot;&lt;/code&gt;&lt;/strong&gt;. Then she continued writing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hearing that Harry got disappointed. She handed over his resume back to him and said &lt;strong&gt;&lt;code&gt;&quot;Thank you&quot;&lt;/code&gt;&lt;/strong&gt;. Harry greeted her back with a sad smile and left the room while she was glancing at him with a mischievous smile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After reaching his room Harry checked his resume to see what she has written on it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;TDRB NDZDJJDT&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;IDQN FQ&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&quot;What is this. What kind of language is this. Is this some kind of a test or something? May be I will get the job if I crack this&quot;&lt;/strong&gt;, he thought.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He wrestled with the codes for hours, but failed to end up in any meaningful solution. He was not willing to give up. Harry was thinking about her beauty. He desperately needed to get close to her. He fell in day dreaming. He recollected everything, her beauty, her smile, expressions, style, voice and all. Then he decided to give it another try.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once he is cracked the first code he became happy and yelled &lt;strong&gt;&lt;code&gt;&quot;Yaahooo&quot;&lt;/code&gt;&lt;/strong&gt; with waving hands in the air with joy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then he sat down and started cracking the next code. Within 30 minutes he cracked that too, and became happier. Not just happier, but he felt like he is the happiest man in the world and started jumping over his bed going crazy with joy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Now tell me what the code was and how Harry cracked it.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Just like Harry, if you succeed in finding out the first message, you too become happy (as I will +1 your answer) and will be double happy if you succeed in finding the second message (as I will mark your answer as accepted)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note: She (The boss) hasn't used computer or any other device for encoding the message. She did everything in her mind (Definitely she can. After all she is a 'nerd')&lt;/p&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="2730" LastEditDate="2016-03-03T09:42:17.187" LastActivityDate="2016-03-03T09:42:17.187" Title="Crack the code and become happy" Tags="|logical-deduction|cipher|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5503" PostTypeId="2" ParentId="5198" CreationDate="2014-11-24T11:18:46.220" Score="2" Body="&lt;p&gt;&lt;strong&gt;2x6&lt;/strong&gt; - area 240&#xA;Another solution&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/SfqWB.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3x5&lt;/strong&gt; - area 1014&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/2evpk.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Following maybe will help someone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2x9&lt;/strong&gt; unsuccessful try&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/14E3H.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1x7&lt;/strong&gt; unsuccessful try&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/J8w9x.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5781" LastEditorUserId="5781" LastEditDate="2014-12-13T12:26:23.183" LastActivityDate="2014-12-13T12:26:23.183" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5504" PostTypeId="2" ParentId="5396" CreationDate="2014-11-24T12:36:13.143" Score="11" Body="&lt;p&gt;Dark girl she will be, &#xA;Enemies to set free. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We &lt;a href=&quot;http://books.google.com.au/books?id=x5qhrl_DeFoC&amp;amp;pg=PT44&amp;amp;dq=%22Dark+girl+she+will+be,+Enemies+to+set+free.%22&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=4kBzVJLxNoeV8QXX8YKwCw&amp;amp;ved=0CB4Q6AEwAA#v=onepage&amp;amp;q=%22Dark%20girl%20she%20will%20be%2C%20Enemies%20to%20set%20free.%22&amp;amp;f=false&quot;&gt;search these lines on Google and find a book&lt;/a&gt; with these lines where the following line is &quot;Sing in Harmony&quot;. There is a (totally &lt;em&gt;unnecessary&lt;/em&gt;) hint in the markup there too: &quot;&lt;a href=&quot;http://lmgtfy.com&quot;&gt;lmgtfy.com&lt;/a&gt;&quot;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The place where you sing, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Harmony&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Starts with your needed thing.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; H&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am I and you are you; &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The subject of the line is the answer for this verse (as is normal riddle etiquette (what am I?)). The line just states outright &quot;I am I&quot;. &quot;You are you&quot; is just stylistic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I sound like what you read this through.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My eye(s)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And if this thing you still can't guess, &#xA;Go to sea and shout out yes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sailor traditional and stereotype has &quot;yes&quot; as &quot;aye&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think we can conclude that we need one of the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; eye/I&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am as handsome as can be. &#xA;I am as ugly as can be. &#xA;I am as clean as can be. &#xA;I am as dirty as can be. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This does not seem to make sense. Imagine a man named &quot;Can Be&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the middle of my name &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Ca&lt;strong&gt;n&lt;/strong&gt;Be&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is what you need to find fame.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Poetically meaning that we need the&lt;br&gt;&#xA; N&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;The following few lines are quite brilliant&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some people use me for addition. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; +&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some turn me into dust and bag me. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; tea&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some take me to a course and play.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; tee (for playing golf on a golf course)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be twenty or one-oh-three.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; t is the 20th letter of the alphabet&lt;br&gt;&#xA; t is is the IPA phonetic character with code 103&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This yields:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; T&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the riddles we ended up with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HINT&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So now you need to board a plane and travel to the land of clean shoes and long sticks. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Board a plane and travel&quot; hints strongly that we are looking for a place; likely a country.&lt;br&gt;&#xA; Clean shoes = POLISH. &lt;br&gt;&#xA; Long Sticks = POLES. &lt;br&gt;&#xA; The land we are looking for is &lt;br&gt;&#xA; POLAND&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now proceeding:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I used HINT in Polish - &quot;wskazówka&quot; - with the &lt;code&gt;ó&lt;/code&gt; replaced with &lt;code&gt;o&lt;/code&gt; as the key (wskazowka) in a Vignere decoder using subtraction.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With the results of the cipher text being:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; YOU ARE THE CHAMPION. YOU HAVE PASSED ALL THE TESTS AND MADE IT TO THE FINAL ANSWER. YOUR ONE REMAINING TASK IS TO WRITE A RIDDLE OF YOUR OWN AND POSE IT AS A QUESTION ON THIS SITE. THERE ARE LOTS OF PEOPLE OUT THERE SALIVATING FOR MORE RIDDLES. GOOD LUCK AND HAPPY RIDDLING :D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think this calls for a lynch mob. Anyone?&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-11-25T02:59:02.613" LastActivityDate="2014-11-25T02:59:02.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5505" PostTypeId="2" ParentId="5502" CreationDate="2014-11-24T12:58:18.290" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'll depict ciphered letters with UPPER CASE and decoded ones with lower case. Through some simple analysis of &lt;strong&gt;NDZDJJDT&lt;/strong&gt; - pattern matching, I got a list of words that match the pattern (for example the two Js are the same letter in the decoded text, the three Ds also are the same letter, etc.). The most relevant match was &quot;tomorrow&quot;. Substituting in the first word I got &lt;strong&gt;woRB&lt;/strong&gt;. So the answer of the first part is &lt;strong&gt;work tomorrow&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastActivityDate="2014-11-24T12:58:18.290" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5506" PostTypeId="1" AcceptedAnswerId="5518" CreationDate="2014-11-24T13:55:28.237" Score="5" ViewCount="672" Body="&lt;p&gt;The Man-ape's whose fur and name change in the cold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The caprivore with piercing teeth threefold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My adopted kin, for my only true brothers are gone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They melted to black before history's dawn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ask my name and I'll sing of my home.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For no toe have I, nor knee, nor glome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With monstrous fame of a ghastly sort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a notorious false report.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You won't get in if a key's what you've got.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look on the map. The white X marks the spot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who am I and how do you know?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5626" LastEditorUserId="5626" LastEditDate="2014-12-08T14:22:05.527" LastActivityDate="2014-12-08T14:22:05.527" Title="The Brother to Bones - Name the brother of the man-ape and the caprivore" Tags="|riddle|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="5507" PostTypeId="2" ParentId="5502" CreationDate="2014-11-24T13:56:49.900" Score="10" Body="&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;First Part:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Initial analysis yielded that &quot;NDZDJJDT&quot; probably meant:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;TOMORROW&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This made sense given the clue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This meant that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; N=T,D=O,Z=M,J=R,T=W&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which gives a transformation pattern of the numeral alphabet positions of the characters of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $13 \rightarrow 26$&lt;br&gt;$15 \rightarrow 4$&lt;br&gt;$18 \rightarrow 10$&lt;br&gt;$20 \rightarrow 14$&lt;br&gt;$23 \rightarrow 20$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I noticed a possible formula:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$(real*2) \bmod 26 = coded$. Additionally, this is supported by elements of the clue, such as &quot;&lt;strong&gt;double&lt;/strong&gt;-check&quot; and &quot;&lt;strong&gt;move on, to a position that suits you&lt;/strong&gt;&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which I can reverse like so:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$\frac{(coded+(26y))}{2} = real$ where $y \in \{0,1\}$ - because we know that the results of $(real*2)$ can never be more than $52$ (Yes, there s case that breaks this, when $real=Z=26$, but $Z$ is rare enough to ignore).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So TDRB is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 20:4:18:2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which when I apply the formula is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 10(J) or 23(W):2(B) or 15(O):9(I) or 22(V):1(A) or 14(N)&lt;br&gt;&#xA; JOIN looks good here for a coherent word and pairing of words.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Giving us: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; JOIN TOMORROW&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Second part:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Loading...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-11-24T17:35:10.290" LastActivityDate="2014-11-24T17:35:10.290" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5508" PostTypeId="1" AcceptedAnswerId="5516" CreationDate="2014-11-24T14:00:48.570" Score="8" ViewCount="674" Body="&lt;p&gt;So, my &lt;a href=&quot;https://puzzling.stackexchange.com/questions/5456/the-longest-word-made-from-chemical-symbols/&quot;&gt;first question&lt;/a&gt; on this topic was answered fairly quickly with quite a long word, so I thought I might post a different version of the puzzle&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea this time is similar to last time: you must construct the longest possible word from &lt;em&gt;chemical symbols&lt;/em&gt; on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Periodic_table&quot; rel=&quot;nofollow noreferrer&quot;&gt;Periodic Table&lt;/a&gt;. However this time, to narrow down the playing field a little, I've restricted the chemical symbols that can be used in your word into 7 categories:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1 - Alkali Metals and Alkali Earth Metals + Hydrogen;&#xA;  H, Li, Be, Na, Mg, K, Ca, Rb, Sr, Cs, Ba, Fr, Ra&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;2 - Transition Metals;&#xA;  Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn, Y, Zr, Nb, Mo, Tc, Ru, Rh, Pd, Ag, Cd, Hf, Ta, W, Re, Os, Ir, Pt, Au, Hg, Rf, Db, Sg, Bh, Hs, Mt, Ds, Rg, Cn&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;3 - &lt;em&gt;Groups&lt;/em&gt; 3 + 4 (Groups topped with Boron / Carbon);&#xA;  B, C, Al, Si, Ga, Ge, In, Sn, Tl, Pb, Uut, Fl&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;4 - &lt;em&gt;Groups&lt;/em&gt; 5 + 6 (Groups topped with Nitrogen / Oxygen);&#xA;  N, O, P, S, As, Se, Sb, Te, Bi, Po, Uup, Lv&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;5 - The &lt;em&gt;Halogens&lt;/em&gt; and &lt;em&gt;Noble Gases&lt;/em&gt;;&#xA;  He, F, Ne, Cl, Ar, Br, Kr, I, Xe, At, Rn, Uus, Uuo&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;6 - The &lt;em&gt;Lanthanides&lt;/em&gt;;&#xA;  La, Ce, Pr, Nd, Pm, Sm, Eu, Gd, Tb, Dy, Ho, Er, Tm, Yb, Lu&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;7 - The &lt;em&gt;Actinides&lt;/em&gt;;&#xA;  Ac, Th, Pa, U, Np, Pu, Am, Cm, Bk, Cf, Es, Fm, Md, No, Lr&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Simply answer any or all of these categories with the longest possible word made from that particular groups of elements. May the games begin!&lt;/p&gt;&#xA;" OwnerUserId="4748" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2014-12-14T17:00:16.477" Title="The Longest Word Made from Specific Chemical Symbols" Tags="|word|open-ended|" AnswerCount="4" CommentCount="13" ClosedDate="2021-06-29T08:15:34.020" ContentLicense="CC BY-SA 3.0" />
  <row Id="5509" PostTypeId="2" ParentId="5456" CreationDate="2014-11-24T14:20:53.977" Score="17" Body="&lt;p&gt;Placing the emphasis on &quot;the longest word &lt;strong&gt;made&lt;/strong&gt; from chemical symbols&quot; - embroidery on a bag!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(From &lt;a href=&quot;https://secure.flickr.com/photos/rjw76/15497828388/&quot; rel=&quot;noreferrer&quot;&gt;https://secure.flickr.com/photos/rjw76/15497828388/&lt;/a&gt; )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/30UJF.jpg&quot; alt=&quot;PuFFeRfISH Bag&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5978" LastActivityDate="2014-11-24T14:20:53.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5510" PostTypeId="2" ParentId="5441" CreationDate="2014-11-24T14:31:35.027" Score="1" Body="&lt;p&gt;Super wildguess here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All words have synonyms starting with &quot;S&quot;&lt;br/&gt;&lt;br/&gt;Candy - Sweets&lt;br/&gt;Jack - Sailor&lt;br/&gt;Following - Successive/Subsequent&lt;br/&gt;Territory - Sector/Space/State&lt;br/&gt;Area - Scope&lt;br/&gt;Common - Simple&lt;br/&gt;Improve - Strengthen&lt;br/&gt;Straight - Straightforward/Square/Straightarrow&lt;br/&gt;Membrane - Sheath/Sheet&lt;br/&gt;Leg - Stage&lt;br/&gt;&lt;br/&gt;Also the &quot;&lt;strong&gt;s&lt;/strong&gt;ome &lt;strong&gt;s&lt;/strong&gt;ort of a relationship&quot; remark&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastActivityDate="2014-11-24T14:31:35.027" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5511" PostTypeId="2" ParentId="5508" CreationDate="2014-11-24T14:45:01.177" Score="4" Body="&lt;p&gt;I'll start the ball rolling. I don't know if these are the longest. I'd be happy to be proven wrong. NB: I got all the lists of elements from Wikipedia.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Alkali metals and alkaline earth metals&lt;/em&gt;: &quot;BaNaNa&quot;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Transition metals&lt;/em&gt;: &quot;ScReWY&quot; (there &lt;em&gt;must&lt;/em&gt; be something longer than this)&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Group 3 and 4&lt;/em&gt;: &quot;LaY&quot;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Group 5 and 6&lt;/em&gt;: &quot;MoW&quot;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Halogens and noble gases&lt;/em&gt;: &quot;BrIAr&quot;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Lanthanides&lt;/em&gt;: &quot;LaNdLaDy&quot;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Actinides&lt;/em&gt;: &quot;ThEsEs&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2014-11-24T14:45:01.177" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5513" PostTypeId="2" ParentId="5483" CreationDate="2014-11-24T15:40:22.917" Score="1" Body="&lt;p&gt;&lt;strong&gt;Group A&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;every investor has one&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;portfolio&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;a roaring good sign&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Leo&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mr. Limbaugh's medium&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Radio&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Waitaminit! Isn't he that actor...?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;DeCaprio&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where art thou?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;..Romeo?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;dress like a Tahitian&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Kahikio&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An apartment, a workplace, a place of creation. All of the above.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;studio&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;one difficult, confused situation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;imbroglio&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because mono is old hat&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;stereo&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;nothing but rainforest&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Borneo&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2444" LastEditorUserId="2444" LastEditDate="2014-11-24T17:26:46.213" LastActivityDate="2014-11-24T17:26:46.213" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5514" PostTypeId="1" AcceptedAnswerId="5539" CreationDate="2014-11-24T15:52:01.283" Score="8" ViewCount="2174" Body="&lt;p&gt;You have been crossing this hot and wet place for weeks now, but when you enter the cool and mouldy ruin, you know in seconds it has been worth it. Nobody has ever been here ( - well nobody in last millennia at least - ) and the ancient writings and symbols all over the place are comparable to none you've ever seen. Could it really have been build by aliens? The answer is quite possibly behind this huge, stone door in front of you. You approach cautiously, anxiously noticing the holes in the floor and ceiling which don't promise much good for unwelcomed guests. There is no handle on that door, but 5 oddly shaped intentions which are far too regular to be anything else than a hidden mechanism.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/VcVOu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And of course you recognize the shape immediately. After all, it was the amulet consisting of 5 elements shaped exactly like this, which send you on this endless quest in a first place. Eagerly, you pour out the pouch you've brought to bring out the five pieces which were once, ages ago, stringed together on a leather string. &#xA;5 indentations, 5 puzzle pieces of matching size - what could be more obvious?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oiV7T.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the one thing, which has been missing whenever you pondered over the meaning of the symbols on the pieces in the past is only now in front of you. Directly above the little panel, nearly aged away by the forces of nature over the years, are some instructions in the stone. Will they be enough to help you? &#xA;Again you suspiciously eye the holes in the floor and the ceiling. There might only be once way to find out - but there might as well only be a single chance to get it right...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hLTwo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Labelling the 5 pieces from left to right as A, B, C, D, and E. In which order will you dare to put them into the 5 indentations (#1 to #5 from left to right.)?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I would like to thank Peter for sorting out an ambiguity of the original puzzle (and solving it as well). The puzzle as presented above does include this fix.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="5840" LastEditDate="2014-12-09T17:41:26.440" LastActivityDate="2014-12-09T17:41:26.440" Title="Can you pass through the ancient tomb? Find the correct interpretation for the symbols" Tags="|mathematics|visual|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="5515" PostTypeId="2" ParentId="5506" CreationDate="2014-11-24T15:56:24.923" Score="2" Body="&lt;p&gt;The Man-ape's whose fur and name change in the cold.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Yeti or &quot;Abominable Snowman&quot; - snow-dwelling cousin of Bigfoot.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The caprivore with piercing teeth threefold.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Chupacabra - literally means something like &quot;goat-sucker&quot; and leaves three teeth marks when sucking blood from animals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My adopted kin, for my only true brothers are gone.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Zombies, mebbe?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They melted to black before history's dawn.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I think this must be referring to prehistoric creatures, such as dinosaurs, going extinct and becoming petroleum and coal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ask my name and I'll sing of my home.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I'm not sure but could be the Sirens..?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For no toe have I, nor knee, nor glome.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Ghosts - non-corporeal entities.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With monstrous fame of a ghastly sort.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Scary monstros, like vampires and mummies, in movies (or hiding in the closet).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm a notorious false report.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;UFOs - &quot;swamp gas,&quot; etc...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You won't get in if a key's what you've got.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Heaven?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Look on the map. The white X marks the spot.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;X-Files - title screen/logo has a white &quot;X.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...Final answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I suppose you're &lt;strong&gt;the paranormal&lt;/strong&gt;.&lt;br/&gt;&#xA;[EDIT: After adding explanations for the fourth and fifth lines, I'm wondering whether the answer must be made more general - changed to &lt;strong&gt;the unknown&lt;/strong&gt;.]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;[EDIT: A couple of my line explanations might be dubious, still.]&lt;/p&gt;&#xA;" OwnerUserId="5948" LastEditorUserId="5948" LastEditDate="2014-11-24T16:15:48.260" LastActivityDate="2014-11-24T16:15:48.260" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5516" PostTypeId="2" ParentId="5508" CreationDate="2014-11-24T16:13:56.900" Score="4" Body="&lt;p&gt;All words are in SOWPODS.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;em&gt;CaRaCaRa&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;CrOsTiNi&lt;/em&gt; (NB using the red-background elements from &lt;a href=&quot;http://en.wikipedia.org/wiki/Transition_metal#Classification&quot; rel=&quot;nofollow&quot;&gt;the first table&lt;/a&gt; in the Wikipedia article: the article notes that there isn't universal agreement on what the term means).&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;AlGeSiC&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;TeAsPoONS&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;HeXeNe&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;LaNdLaDy&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;AmEsAcEs&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="228" LastEditorUserId="228" LastEditDate="2014-11-25T17:45:57.690" LastActivityDate="2014-11-25T17:45:57.690" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5517" PostTypeId="2" ParentId="5344" CreationDate="2014-11-24T16:24:16.167" Score="0" Body="&lt;p&gt;If I would have to decided on a prime suspect, it would be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ullah Prihasti&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is currently based on some rather loose clues and not strong facts, though. My reasoning is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assumption #1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; More blood should potentially indicated later in typing, but also, that the victim is obviously already closer to death than to life. As a result, there is a higher chance of mistyping those letters. Assuming that he is &quot;slumping to ground&quot;, I'd guess the likelihood for the &quot;true&quot; key being one row above the actually hit one is higher, but one shouldn't rule out other close-bye keys.&#xA; So, chances are that C should be D/F,  G should be R/T/Y, I should be 8/9 and R should 4/5.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assumption #2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Another fact is the number of keys. Leaving aside space and others, there have been at least 14 characters been typed. Now, assuming that a dying man is not wasting his time on complex riddles, the guess would be that he tries typing a name. This narrows the list of suspects to those with longer names, in particular &quot;Ullah Prihasti&quot; and &quot;Sriong Putri K&quot;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And drawn conclusion:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Both have less than 14 characters in their name, so the telephone numbers possibly count. (Although it's odd that a dying man tries to identify his murdere by his number.) Only the keys &quot;I&quot; and &quot;R&quot; are close to the numbers, and most likely 89 or 45. Of the two mentions above,  Sriong's number starts with 714... while Ullah's starts with 085... which is by far the better match. Hence he's my top-suspect. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While I have a strong suspicion, clearly more detective work needs to be done for an arrest! Am I on the right track at least?&lt;/p&gt;&#xA;" OwnerUserId="5840" LastActivityDate="2014-11-24T16:24:16.167" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="5518" PostTypeId="2" ParentId="5506" CreationDate="2014-11-24T16:36:45.910" Score="10" Body="&lt;p&gt;The narrator seems to be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Loch Ness Monster.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Man-ape's whose fur and name change in the cold.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sasquatch in warm climates, abominable snowman in cold. (as previously determined by tjbtech)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The caprivore with piercing teeth threefold.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The chupacabra (lit. &quot;goat sucker&quot;) (as previously determined by tjbtech)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My adopted kin, for my only true brothers are gone.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nessie is purported to be a plesiosaur, the last of its kind.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They melted to black before history's dawn.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Her relatives have long since decomposed into oil. (as previously determined by tjbtech)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ask my name and I'll sing of my home.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Her home is in her name - Loch Ness.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For no toe have I, nor knee, nor glome.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; plesiosaurs don't have legs/toes/knees/glomes, they have flippers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With monstrous fame of a ghastly sort.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nessie is a monster. Like a ghast, she is difficult to locate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm a notorious false report.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There have been dozens of dubious sightings of Nessie, going back to 1933.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You won't get in if a key's what you've got.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can unlock a lock, but you can't unlock a loch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Look on the map. The white X marks the spot.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Loch Ness lies southwest of the intersection of the A9, A82, and A96 roads, which form an X shape.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1497" LastActivityDate="2014-11-24T16:36:45.910" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5521" PostTypeId="1" CreationDate="2014-11-24T17:43:56.267" Score="0" ViewCount="480" Body="&lt;p&gt;&lt;em&gt;This is part 3 of a series of multiple parts about a takeover of all the computers on earth. The links to previous questions are here:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/3853/computer-crisis-takeover-part-1&quot;&gt;Computer Crisis Takeover! [Part 1]&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/5431/computer-crisis-takeover-part-2&quot;&gt;Computer Crisis Takeover! [Part 2!]&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You just cracked another password screen due to being the invalid owner of the OS. Now that you're in you start looking through numerous file paths finding nothing but folders and folders of nothing. Then it hits you that they could be hidden. Noticing that the OS is unbelievably like Windows you enable hidden folders and system folders in the options. Suddenly it reopens the process and you're in a folder with 5 executables:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;B.exe&lt;br&gt;&#xA;  A.exe&lt;br&gt;&#xA;  I.exe&lt;br&gt;&#xA;  BE.exe&lt;br&gt;&#xA;  AF.exe&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's also a text file called &lt;em&gt;Instructions.txt&lt;/em&gt;. Before even daring to do anything, you open the instructions. The contents are as shown:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dear User,    &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I have made a little game to pass the time! There's 5 executables in a folder, the order in which you execute them will decide on the result! Access to system files will be granted if correct. OS Shut down will result if wrong!    &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Here's the hint you'll need to figure it out. &lt;em&gt;Let letters be known as the master of words, let words be known as the peasants to numbers, let numbers be known as kings to words&lt;/em&gt;.    &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Good luck, Dr. Glin!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What order should you open these executables in?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2019-09-26T15:45:43.150" Title="Computer Crisis Takeover! [Part 3]" Tags="|computer-puzzle|mystery|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5523" PostTypeId="2" ParentId="5521" CreationDate="2014-11-24T18:29:30.843" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;BE.exe, B.exe, A.exe, I.exe, AF.exe&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The order is numbers&gt;letters&gt;words. Letters A-e can be interpreted as numbers via hexadecimal. So the three titles which can be hex are A,B, and BE, whose values are 10, 11, and 176 (ordered greatest to least for answer) the remaining two are a letter &quot;I&quot; and a word &quot;AF&quot;. Since letters are greater than words, &quot;I&quot; is executed before &quot;AF&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5159" LastActivityDate="2014-11-24T18:29:30.843" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5524" PostTypeId="2" ParentId="5502" CreationDate="2014-11-24T18:35:03.493" Score="17" Body="&lt;p&gt;The first part is as d'alar'cop explained. In short, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The message was codified doubling the letter value (mod 26, of course) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so that &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; JOIN (10 15 9 14) =&gt; TDRB (20 4 18 2)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And so on. As per the second, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The difference is the position of the letter in the string. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What you have to do is, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Instead of just doubling, you have to add the position value too. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the second message is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DATE ME (4 1 20 5 0 13 5 =&gt; 9 4 17 14 0 6 17) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;DETAILS ADDED BY OP&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Clue for first code was in her word &quot;I have to double check it&quot;. Also she indirectly hinted, in order to make sure he is brilliant and smart enough for her to date, she want to double check him by adding a double encoded code too. &quot;You have to move on, to a position that suits you&quot; was the clue for second code. She was indirectly hinting that the characters are shifted according to their position values too. From these words Harry got the idea how to solve the codes. That is why his attempt succeeded after recollecting everything about the meeting with her (especially, her words)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5990" LastEditorUserId="2730" LastEditDate="2014-12-10T10:35:27.863" LastActivityDate="2014-12-10T10:35:27.863" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5525" PostTypeId="2" ParentId="4762" CreationDate="2014-11-24T20:50:18.503" Score="5" Body="&lt;p&gt;First, I am not trying to find the smallest number of questions. Let's just find an answer and then try to optimize it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you ask to each person the question &quot;Are you the joker?&quot;, the knights would answer &quot;no&quot; and the knaves would answer &quot;yes&quot;. This easily separates them, but leaves the joker in any of the two groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets suppose that the joker answered &quot;no&quot;, so he is in the knights group. Since he does not want to be caught and is presumably smart, he would pose himself as a knight, always giving the answers that a knight would give.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, lets suppose that the joker answered &quot;yes&quot;, so he is in the knaves group. Since he does not want to be caught and is presumably smart, he would pose himself as a knave, always giving the answers that a knave would give.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since nobody (except the joker himself) knows if somebody else is or is not the joker and nobody can give a proof that he is not the joker, we have a situation that if the joker answered &quot;yes&quot;, he can't be distinguished from a knight. If he answered &quot;no&quot;, he can't be distinguished from a knave. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Asking a knight &quot;Do you know who is the joker&quot;, he would answer &quot;No&quot;. The knave would answer &quot;Yes&quot;. The joker would follow saying what a knight or a knave would say accordingly to his previous answers. No information is gained here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Asking a knight if somebody else is or is not the joker would produce the answer &quot;I don't know&quot; or something similar. If you ask this to a knave, he could not answer &quot;yes&quot;, &quot;no&quot; or &quot;I don't know&quot;, so he would answer &quot;I know the answer, but I can't tell you&quot; or something similar. The joker would follow the answers from the group he chose. This is essentially the same as the previous question, so again, no information is gained here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the question is changed to &quot;Is the joker among/not among that X group?&quot; the same &quot;I don't know&quot; and &quot;I know the answer, but I can't tell you&quot; answers will follow from the same people for any arbitrary definition of group X, and no information gain would be produced. There is an exception for trivial cases where X is an empty group, a group with everybody or a group with everybody except who is being asked, which would not produce any information gain either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you change the question to &quot;what would the person X answer if I asks him if person Y is the joker?&quot;, the pattern gets a bit more complicated, but this would not produce any information gain either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in all the ways, except if the joker blunders by producing an incoherent answer somewhere, it is impossible to solve the puzzle only with the provided information, even if we knew what the joker answered to the &quot;are you the joker?&quot; question (and yet, we don't).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I have a lateral thinking answer: To not be caught, when the joker decides to pose himself definitely either as a knight or a knave, he essentially becomes a knight or a knave, so we have no joker anymore, and thus the puzzle is solved: Nobody is the joker and the least amount of questions is zero!&lt;/p&gt;&#xA;" OwnerUserId="5044" LastEditorUserId="5044" LastEditDate="2014-11-24T21:08:36.690" LastActivityDate="2014-11-24T21:08:36.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5526" PostTypeId="1" AcceptedAnswerId="5536" CreationDate="2014-11-24T22:00:11.593" Score="2" ViewCount="716" Body="&lt;p&gt;I've heard the following second half of a story:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A policeman got out of the car and saw the boy. He walked up to him and said amiably, &amp;quot;What's your name?&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;Shut up!&amp;quot; the boy replied.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;Where's your manners?&amp;quot; asked the affronted policeman.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;Up that tree,&amp;quot; said the boy offhandedly, pointing to a nearby tree.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;You're looking for trouble, aren't you?&amp;quot; said the policeman.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;No, trouble's looking for me!&amp;quot; the boy answered earnestly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Can you fill in the first half, or somehow explain the strange conversation? The boy is not being intentionally rude, and his answers all make sense when interpreted correctly.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-11-25T05:10:05.903" Title="Conversation with a cop - please explain" Tags="|lateral-thinking|mystery|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5527" PostTypeId="2" ParentId="5441" CreationDate="2014-11-24T22:10:06.503" Score="6" Body="&lt;p&gt;I all but guarantee this is not the &quot;perfect&quot; answer,  but I would say it would qualify as an answer...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Candy - IceBreakers (A specific type of candy)&lt;br&gt;&#xA;Jack - Ma (Jack Ma is a Chinese entrepreneur and philanthropist.)&lt;br&gt;&#xA;Following - Pursuing&lt;br&gt;&#xA;Territory - Enclave&lt;br&gt;&#xA;Area - Compass&lt;br&gt;&#xA;Common - Conventional&lt;br&gt;&#xA;Improve - Advance&lt;br&gt;&#xA;Straight - Beeline&lt;br&gt;&#xA;Membrane - Lamina&lt;br&gt;&#xA;Leg - Extremity (a leg is an extremity)&lt;br&gt;&#xA;&lt;br&gt;&#xA;1 is a type of candy, 2 is NOT a synonym, and some of my answers are synonyms I got from thesaurus.com (3 through 9) so it meets the requirements for hints 1-4.&lt;br&gt;&#xA;&lt;br&gt;&#xA;So what do the answers have in common? The first letters of the answers spell &quot;Impeccable&quot; which is a synonym of &quot;Perfect&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3922" LastActivityDate="2014-11-24T22:10:06.503" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5528" PostTypeId="2" ParentId="5526" CreationDate="2014-11-24T22:14:08.580" Score="4" Body="&lt;p&gt;A French boy called Joudoupe was sent by his mother Troub Les to catch their cat named Man'ers?&lt;/p&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="4" LastEditDate="2014-11-24T23:34:15.597" LastActivityDate="2014-11-24T23:34:15.597" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5529" PostTypeId="1" AcceptedAnswerId="5532" CreationDate="2014-11-24T22:15:03.093" Score="8" ViewCount="424" Body="&lt;p&gt;On a certain road there are 32 houses in a row, numbered 1 through 32, each inhabited by a single person.  During the month of January each house is hosting a dinner date.  House 1 on January 1, house 2 on January 2, and so forth.  The occupant of house 32 is a lovable misanthrope, so he never has anyone over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every house follows the same dinner date etiquette:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The guest may arrive bearing one bottle of wine, but it is not mandatory.&lt;/li&gt;&#xA;&lt;li&gt;When the guest arrives, if there is no wine in the house then tea is served.&lt;/li&gt;&#xA;&lt;li&gt;If there is one bottle of wine in the house, it is wisely saved for later and, again, tea is served.&lt;/li&gt;&#xA;&lt;li&gt;If there are two bottles of wine in the house, they drink one bottle of wine with dinner and in a state of intoxicated generosity give the other bottle to the person next door (the next higher numbered house).&lt;/li&gt;&#xA;&lt;li&gt;In the rare case that a house finds itself in possession of three or more bottles of wine one is served with dinner, one is given to the person at the next higher numbered house and the rest is saved for later.&lt;/li&gt;&#xA;&lt;li&gt;No one drinks wine except during a date, and every bottle opened is finished. No partials. No one dates anyone from the neighborhood. In short, no lateral thinking funny business.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;On the morning of January 1, each house contains one bottle of wine except houses 3, 5, 11, 12, 17, 18, 24, 28 and of course 32.  These nine houses contained no wine at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the morning of February 1, each house contained one bottle of wine except 1, 5, 7, 8, 11, 12, 16, 20, 22, 25, 27, 28, and 29.  These thirteen houses contained no wine at all.  The gentleman in house 32 graciously accepted a bottle but has not drunk it, as per etiquette.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which houses received a dinner date bearing a bottle of wine?&lt;/p&gt;&#xA;" OwnerUserId="2162" LastEditorUserId="22922" LastEditDate="2016-09-15T08:15:45.593" LastActivityDate="2016-09-15T08:15:45.593" Title="Wine with dinner" Tags="|mathematics|calculation-puzzle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5531" PostTypeId="2" ParentId="5481" CreationDate="2014-11-24T22:34:01.707" Score="5" Body="&lt;p&gt;After the new information, it seems possible to find a valid solution after also using the calculation for the dad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The birthday is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 21/06/1985&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;while the events in the story happened in the year:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2013&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and his father's birthday is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 11/12/1957&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation for all the conditions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2013 - 1985 = 28 =&gt; 28*2 = 56 =&gt; 2013-56 = 1957  ::: This makes the dad twice as old.&lt;br&gt;&#xA; 2161985 is divisible by 7&lt;br&gt;&#xA; 1985621 is divisible by 11&lt;br&gt;&#xA; 6211985 is divisible by 13&lt;br&gt;&#xA; 2161985 + 1985621 + 6211985 = 10359591 + 28 = 10359619 =&gt; prime number  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I wrote quite a long, poorly written/formatted and probably completely inefficient SAS program for it, but it worked and gave me the results. :)&lt;/p&gt;&#xA;" OwnerUserId="5975" LastEditorUserId="5975" LastEditDate="2014-11-24T22:40:06.390" LastActivityDate="2014-11-24T22:40:06.390" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5532" PostTypeId="2" ParentId="5529" CreationDate="2014-11-24T22:35:49.727" Score="5" Body="&lt;p&gt;There is a simple mathematical pattern here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Straightforward binary addition.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer can therefore be determined from:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 11100010110101110111001100101110 - 1110111011111001111001111101011 = 01101011010110100111111101000011&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and therefore bottles of wine were brought to houses:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1, 2, 7, 9, 10, 11, 12, 13, 14, 15, 18, 20, 21, 23, 25, 26, 28, 30, 31&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4270" LastEditorUserId="4270" LastEditDate="2014-11-24T23:51:20.157" LastActivityDate="2014-11-24T23:51:20.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5533" PostTypeId="2" ParentId="5441" CreationDate="2014-11-24T23:00:58.460" Score="2" Body="&lt;p&gt;Here is the answer that I had in mind:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;Candy     - Jellybean&lt;/code&gt;&#xA; &lt;P&gt;&#xA;&lt;code&gt;Jack      - Beanstalk&lt;/code&gt;&#xA; &lt;P&gt;&#xA;&lt;code&gt;Following - Stalking&lt;/code&gt;&#xA; &lt;P&gt;&#xA;&lt;code&gt;Territory - Kingdom&lt;/code&gt;&#xA; &lt;P&gt;&#xA;&lt;code&gt;Area      - Domain&lt;/code&gt;&#xA; &lt;P&gt;&#xA;&lt;code&gt;Common    - Mainstream&lt;/code&gt;&#xA; &lt;P&gt;&#xA;&lt;code&gt;Improve   - Streamline&lt;/code&gt;&#xA; &lt;P&gt;&#xA;&lt;code&gt;Straight  - Linear&lt;/code&gt;&#xA; &lt;P&gt;&#xA;&lt;code&gt;Membrane  - Eardrum&lt;/code&gt;&#xA; &lt;P&gt;&#xA;&lt;code&gt;Leg       - Drumstick&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The relationship is that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the last few letters of each word are the same as the first few letters of the next word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2014-11-24T23:00:58.460" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5534" PostTypeId="5" CreationDate="2014-11-24T23:22:13.697" Score="0" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Slitherlink&quot; rel=&quot;nofollow noreferrer&quot;&gt;Slitherlink&lt;/a&gt; is a combinatorial logic puzzle usually played on a square grid.&lt;/p&gt;&#xA;&lt;p&gt;The rules are simple. Some of the squares contain a number. The objective is to mark the edges on the grid in order to form a single closed circuit with no intersections.&lt;/p&gt;&#xA;&lt;p&gt;The number of edges of every square that contains a number must match with that number.&lt;/p&gt;&#xA;&lt;p&gt;Variants exist with other grid shapes, for example see &lt;a href=&quot;/questions/40191/help-with-pathing-game-loopy&quot;&gt;Loopy&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Use this tag alongside the &lt;a href=&quot;/questions/tagged/grid-deduction&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;grid-deduction&amp;#39;&quot; aria-label=&quot;show questions tagged &amp;#39;grid-deduction&amp;#39;&quot; rel=&quot;tag&quot; aria-labelledby=&quot;grid-deduction-container&quot;&gt;grid-deduction&lt;/a&gt; and &lt;a href=&quot;/questions/tagged/logical-deduction&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;logical-deduction&amp;#39;&quot; aria-label=&quot;show questions tagged &amp;#39;logical-deduction&amp;#39;&quot; rel=&quot;tag&quot; aria-labelledby=&quot;logical-deduction-container&quot;&gt;logical-deduction&lt;/a&gt; tags.&lt;/p&gt;&#xA;" OwnerUserId="71" LastEditorUserId="52483" LastEditDate="2023-02-02T20:06:31.837" LastActivityDate="2023-02-02T20:06:31.837" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="5535" PostTypeId="4" CreationDate="2014-11-24T23:22:13.697" Score="0" Body="Slitherlink is a logic puzzle where the goal is to draw a loop so each number has that many edges of the loop adjacent to it. Use with [grid-deduction] and [logical-deduction]." OwnerUserId="71" LastEditorUserId="69582" LastEditDate="2020-12-03T02:39:37.320" LastActivityDate="2020-12-03T02:39:37.320" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="5536" PostTypeId="2" ParentId="5526" CreationDate="2014-11-24T23:28:19.970" Score="6" Body="&lt;p&gt;The boy's name is &quot;Shut Up&quot;, and he's playing hide-and-seek with his 2 friends named &quot;Manners&quot; and &quot;Trouble&quot;.  &quot;Trouble&quot; is the one who counted, so he's looking for the other two boys.  &quot;Manners&quot; hid up in the tree.&lt;/p&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2014-11-24T23:28:19.970" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5537" PostTypeId="2" ParentId="5456" CreationDate="2014-11-24T23:32:02.017" Score="20" Body="&lt;p&gt;I wrote a script (&lt;a href=&quot;http://pastebin.com/T60RQdXa&quot;&gt;http://pastebin.com/T60RQdXa&lt;/a&gt;) that takes a wordlist and outputs the largest word that can be written as a concatenation of chemical symbols. Calling &lt;code&gt;python chemWords.py /usr/share/dict/words&lt;/code&gt; yields &quot;NoNRePReSeNTaTiONAl&quot; (19 chars). Calling with a more complete dictionary gives &quot;FlOCCINAuCINIHILiPILiFICAtIONS&quot; (30 chars).&lt;/p&gt;&#xA;" OwnerUserId="5994" LastActivityDate="2014-11-24T23:32:02.017" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5539" PostTypeId="2" ParentId="5514" CreationDate="2014-11-24T23:54:11.847" Score="11" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;A D E B C&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let me begin by saying that next time you write a puzzle like this for people to submit answers via a computer keyboard, please please please use letters which we can type rather than weird symbols. I know it looks cooler, but it's going to make it hard to follow my explanation because there's no way I'm going to embed hundreds of images in this answer.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I start by assuming, based on the diagrams at the top of the key, that &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the U symbol is addition, zigzag is division, spiral is multiplication, and the leaking container is subtraction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then each of the lines contains a trapezium or two superimposed trapezia, which I take as&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; balanced: equals symbol; unbalanced: an inequality symbol, to be discussed in more detail later.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second line tells us that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the S-like symbol is the digit 4.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The fifth line tells us that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; either the most significant digit in a two-digit number comes to the left and the symbol which I call &lt;code&gt;L&lt;/code&gt; is a 1, or the most significant digit comes to the right and the symbol which I call &lt;code&gt;p&lt;/code&gt; is a 1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The eleventh line would then be information-free unless we assume the first of those alternatives, so I make that assumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ninth line is rather suggestive:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It seems that &lt;code&gt;p&lt;/code&gt; must divide into &lt;code&gt;pS&lt;/code&gt;, which means that &lt;code&gt;p&lt;/code&gt; must divide into &lt;code&gt;S&lt;/code&gt;, but since &lt;code&gt;S = 4&lt;/code&gt; and we already have a &lt;code&gt;1&lt;/code&gt; and a &lt;code&gt;4&lt;/code&gt; that means that &lt;code&gt;p = 2&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now the penultimate line comes into play. I'm not aware of any human culture which uses number bases greater than 60, so searching up to base 60 for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; cubes of the form &lt;code&gt;n*n*n = 2*base + n&lt;/code&gt; we get the options &lt;code&gt;3 * 3 * 3 = 23 (base 12)&lt;/code&gt; or &lt;code&gt;5 * 5 * 5 = 25 (base 60)&lt;/code&gt;. But with the first line &lt;code&gt;LI / S = n&lt;/code&gt; and the sixth line &lt;code&gt;D * D = nI&lt;/code&gt; we can look for two-digit squares &lt;code&gt;nI&lt;/code&gt; such that &lt;code&gt;4 * n = base + I&lt;/code&gt; and this rules out base 60. Base 12 we get &lt;code&gt;10 / 4 = 3&lt;/code&gt; and &lt;code&gt;6 * 6 = 30&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At this point we have a few digits and a few remaining lines to process.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    &lt;code&gt;I &amp;lt;=&amp;gt; 0&lt;/code&gt;&lt;br&gt;&#xA;    &lt;code&gt;L &amp;lt;=&amp;gt; 1&lt;/code&gt;&lt;br&gt;&#xA;    &lt;code&gt;p &amp;lt;=&amp;gt; 2&lt;/code&gt;&lt;br&gt;&#xA;    &lt;code&gt;n &amp;lt;=&amp;gt; 3&lt;/code&gt;&lt;br&gt;&#xA;    &lt;code&gt;S &amp;lt;=&amp;gt; 4&lt;/code&gt;&lt;br&gt;&#xA;    &lt;code&gt;m &amp;lt;=&amp;gt; 5&lt;/code&gt;&lt;br&gt;&#xA;    &lt;code&gt;D &amp;lt;=&amp;gt; 6&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;    1. &lt;code&gt;10 / 4 = 3&lt;/code&gt;&lt;br&gt;&#xA;    2. &lt;code&gt;3 + 3 + 3 + 3 = 4 * 3&lt;/code&gt;&lt;br&gt;&#xA;    3. &lt;code&gt;Q - 2 = 5&lt;/code&gt;&lt;br&gt;&#xA;    4. &lt;code&gt;e * 6 (inequality) Q * b&lt;/code&gt;&lt;br&gt;&#xA;    5. &lt;code&gt;b + 5 = 12&lt;/code&gt;&lt;br&gt;&#xA;    6. &lt;code&gt;6 * 6 = 30&lt;/code&gt;&lt;br&gt;&#xA;    7. &lt;code&gt;5 * 5 = 21&lt;/code&gt;&lt;br&gt;&#xA;    8. &lt;code&gt;2 - Q = w&lt;/code&gt;&lt;br&gt;&#xA;    9. &lt;code&gt;Y + 6 = 24 / 2&lt;/code&gt;&lt;br&gt;&#xA;    10. &lt;code&gt;d + Y = T&lt;/code&gt;&lt;br&gt;&#xA;    11. &lt;code&gt;e + 6 (inequality) 14&lt;/code&gt;&lt;br&gt;&#xA;    12. &lt;code&gt;3 * 3 * 3 = 23&lt;/code&gt;&lt;br&gt;&#xA;    13. &lt;code&gt;mirror-C (inequality) C&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So those give us&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;Q = 7, b = 9, w = -5, Y = 8&lt;/code&gt;. The symmetry between &lt;code&gt;m&lt;/code&gt; and &lt;code&gt;w&lt;/code&gt; and their values being negative is very suggestive, and helps explain why amulet A doesn't appear at all but its mirror image does. It also fits nicely with 0 being the only horizontally symmetric digit. Equation 10 checks out: -9 + 8 = -1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We still need to tackle the unbalanced trapezium.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Line 4 tells us that &lt;code&gt;e (inequality) ten and a half&lt;/code&gt; and line 11 tells us that &lt;code&gt;e (inequality) ten&lt;/code&gt;. If the inequality is &lt;code&gt;&amp;lt;&lt;/code&gt; then &lt;code&gt;e&lt;/code&gt; must duplicate a digit value for which we already have a symbol. So either the inequality is &lt;code&gt;&amp;lt;=&lt;/code&gt; and &lt;code&gt;e&lt;/code&gt; is the digit ten, or the inequality is &lt;code&gt;&amp;gt;&lt;/code&gt; and &lt;code&gt;e&lt;/code&gt; is the digit eleven. The latter fits with the reading of the unbalanced trapezium as a scale which tips down on the heavier side, and also with the superimposed balanced-and-unbalanced trapezia of the last line, which then means &lt;code&gt;&amp;lt;=&lt;/code&gt;. So we have &lt;code&gt;e = eleven&lt;/code&gt; and &lt;code&gt;C&lt;/code&gt; is the only unassigned digit, ten. We resolve its sign with the aid of the last line: &lt;code&gt;C&lt;/code&gt; is the positive, and the negative is &lt;code&gt;mirror-C&lt;/code&gt; which is smaller than it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the amulets are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A: -11&lt;br&gt;&#xA; B: 5&lt;br&gt;&#xA; C: 6&lt;br&gt;&#xA; D: -10&lt;br&gt;&#xA; E: 1&lt;br&gt;&#xA;&#xA; and should be sorted in ascending order (the trapezium points right rather than left), giving solution A D E B C.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note that this answer has been revised in accordance with the revision to the question, following some discussion with BmyGuest about sharpening up both of them. See the revision history for more details.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="228" LastEditorUserId="228" LastEditDate="2014-11-25T14:49:26.863" LastActivityDate="2014-11-25T14:49:26.863" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5540" PostTypeId="1" AcceptedAnswerId="5550" CreationDate="2014-11-24T23:55:47.917" Score="-1" ViewCount="502" Body="&lt;pre&gt;&lt;code&gt;I GMMA OBBK GKONYQ TC WWA ZTS KCSD SM PKGPX.&#xA;GEST MBG ROC BOK KMUUU FDU IAONXJJ, JWS EQDD RA LGXC.&#xA;EWU IS GP JJN TK VQHM NE TD BMLSX.&#xA;KRQE LTAH!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Hint (if you want one):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter substitutions are not &quot;one-to-one&quot;, eg. a letter isn't usually encoded into the same letter multiple times.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5934" LastEditorUserId="5934" LastEditDate="2014-11-25T01:44:03.367" LastActivityDate="2014-11-25T11:39:35.220" Title="Can you crack the code?" Tags="|cipher|encoded-message|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5542" PostTypeId="2" ParentId="5508" CreationDate="2014-11-25T00:16:01.320" Score="0" Body="&lt;p&gt;For the alkali metals et. al., CaBaNa is the best I can do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BrINe ties BrIAr for the longest I can find in the halogens and noble gases.&lt;/p&gt;&#xA;" OwnerUserId="5300" LastActivityDate="2014-11-25T00:16:01.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5543" PostTypeId="2" ParentId="5521" CreationDate="2014-11-25T00:37:29.483" Score="1" Body="&lt;p&gt;How about:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I.exe, AF.exe, B.exe, A.exe, BE.exe&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Assuming the order should be numbers, letters, words based on kings &gt; masters &gt; peasants: &#xA;&#xA; 1. I.exe because it can be a number (roman numeral 1);&#xA; 2. AF.exe and B.exe&#xA;    because they can be neither numbers nor words, and just going in&#xA;    alphabetical order;&#xA; 3. A.exe and BE.exe because they are words, and also&#xA;    sorted alphabetically&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5791" LastActivityDate="2014-11-25T00:37:29.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5545" PostTypeId="2" ParentId="5526" CreationDate="2014-11-25T05:10:05.903" Score="3" Body="&lt;p&gt;Three boys decided to play &lt;strong&gt;hide and seek&lt;/strong&gt;. To pick the one who will be counting first, they decided to play another &lt;strong&gt;mini game&lt;/strong&gt;. They agreed to shout any of these three &lt;strong&gt;&quot;Shut-Up&quot;, &quot;Manners&quot; or &quot;Trouble&quot;&lt;/strong&gt; at same time and the one who shouts a unique word will be named that word. This will continue until each of them will get a unique name. Then, the one named Trouble will be the one counting first and the other two have to hide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the mini game, each of them got named &lt;strong&gt;&quot;Shut-up&quot;, &quot;Trouble&quot; and &quot;Manners&quot;&lt;/strong&gt;. The one who named &lt;strong&gt;&quot;Trouble&quot; started counting&lt;/strong&gt;. &lt;strong&gt;&quot;Manners&quot; Climbed the tree&lt;/strong&gt; and hid himself in the branch. &lt;strong&gt;&quot;Shut-up&quot; was looking for a place to hide&lt;/strong&gt;. At that time a police car approached stopped there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A policeman got out of the car and saw the boy (&quot;Shut-up&quot;, who was looking for a place to hide). He walked up to him and said amiably, &quot;What's your name?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Shut up!&quot; the boy replied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Where's your manners?&quot; asked the affronted policeman.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Up that tree,&quot; said the boy offhandedly, pointing to a nearby tree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;You're looking for trouble, aren't you?&quot; said the policeman.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;No, trouble's looking for me!&quot; the boy answered earnestly.&lt;/p&gt;&#xA;" OwnerUserId="2730" LastActivityDate="2014-11-25T05:10:05.903" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5546" PostTypeId="1" AcceptedAnswerId="5547" CreationDate="2014-11-25T07:05:01.563" Score="2" ViewCount="1138" Body="&lt;p&gt;I guess as an addendum to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/5456/the-longest-word-made-from-chemical-symbols&quot;&gt;The Longest Word Made from Chemical Symbols&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the longest word in the ENGLISH language that can be formed by combining chemical symbols and has no repeat elements in its name?&lt;/p&gt;&#xA;" OwnerUserId="4425" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2014-11-26T06:41:25.117" Title="What is the longest word made from Chemical Symbols without repeats?" Tags="|word|" AnswerCount="1" CommentCount="8" ClosedDate="2014-12-02T20:57:09.233" ContentLicense="CC BY-SA 3.0" />
  <row Id="5547" PostTypeId="2" ParentId="5546" CreationDate="2014-11-25T08:17:51.777" Score="6" Body="&lt;p&gt;To ensure that we dont search something less than 20 charaters I am posting this &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If only unique usage (each element being used once is a condition) then&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.nandor.org/math/chemwords/chemwords.htm&quot; rel=&quot;nofollow&quot;&gt;There is a tie&lt;/a&gt; between &quot;hypercoagulabilities&quot; and &quot;hyperconsciousnesses&quot; (&quot;HYPErCoAgULaBiLiTiEs&quot; and &quot;HYPErCoNScIOUSnEsSeS&quot;) - both are 20 characters&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In case if this is found as the valid answer in a reasonable amount of time, then yes this question can be considered duplicate (but not deleted)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To ensure that I dont get credits for copying and pasting and in the true sense this is community wiki's primary purpose as per me, I am marking it as community wiki&lt;/p&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="2518" LastEditDate="2014-11-25T13:18:44.020" LastActivityDate="2014-11-25T13:18:44.020" CommentCount="0" CommunityOwnedDate="2014-11-25T08:17:51.777" ContentLicense="CC BY-SA 3.0" />
  <row Id="5549" PostTypeId="2" ParentId="4762" CreationDate="2014-11-25T10:08:11.197" Score="1" Body="&lt;p&gt;Original post:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The joker don't just randomly tell truth/lie. He can also exclaim that he found the joker himself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A comment on the OP:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;He will random lie/truth when asked questions, and will randomly pick to lie and say someone is the joker.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the joker acts like above and will &quot;random[ly] lie/truth[-tell] when asked questions&quot; (as OP clarifies in comments on the OP) then you can't define any optimal strategy as the joker may contradict himself after $2$ questions, or after $2+n$ questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, if he can be consistent and choose to answer like a knight, you can never tell the difference between a joker and a knight (similar result if he always pretends to be a knave).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There have been several questions like this, with Ks and Ks, then a Joker which you have to identify - you &lt;strong&gt;can never tell&lt;/strong&gt; who the Joker is if he is able to consistently act like Knight or Knave, and if he can't, you can't determine how long it'll take to out him.&lt;/p&gt;&#xA;" OwnerUserId="2518" LastEditorUserId="2518" LastEditDate="2014-11-25T10:43:26.797" LastActivityDate="2014-11-25T10:43:26.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5550" PostTypeId="2" ParentId="5540" CreationDate="2014-11-25T11:39:35.220" Score="5" Body="&lt;p&gt;The cipher is a simple incremental cipher. Each character in the code is $L-n$ where $L$ is the numerical position of the letter in the alphabet and $n$ is the index of the character in the encrypted text.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I actually solved it by guessing the last line first, and then reverse-engineering that makes it quite obvious what the cipher is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Working through the message, it decrypts as&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I HOPE THIS PUZZLE IS NOT TOO HARD TO SOLVE&lt;br&gt;&#xA; ONCE YOU GET THE FIRST FEW LETTERS, THE REST IS EASY&lt;br&gt;&#xA; BUT IT IS NOT AS HARD AS IT SEEMS&lt;br&gt;&#xA; GOOD LUCK!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It would be nice if you don't make it so so so long next time though ^_^&lt;/p&gt;&#xA;" OwnerUserId="2518" LastActivityDate="2014-11-25T11:39:35.220" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5551" PostTypeId="2" ParentId="5471" CreationDate="2014-11-25T11:50:36.197" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LEADERSHIP.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Light, soft and healthy is me &lt;br&gt;&#xA; mirrors are all over these &lt;br&gt;&#xA; if finding gold is rather dismal &lt;br&gt;&#xA; I sit next to the one in Pepto-Bismol&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mirror suggests reversal. The opposites of light, soft, and healthy are heavy, hard, and unhealthy. What is heavy, hard, unhealthy, and can be used to find gold? Ancient alchemists wanted to turn LEAD into gold, which is actually possible (though unfeasible) by &lt;a href=&quot;http://en.wikipedia.org/wiki/Nuclear_transmutation&quot;&gt;nuclear transmutation&lt;/a&gt;. LEAD is &lt;a href=&quot;http://en.wikipedia.org/wiki/Lead&quot;&gt;heavy and unhealthy&lt;/a&gt;, and post-transition it is classified as a hard metal. It's also adjacent in the period table to bismuth, which is contained in &lt;a href=&quot;http://en.wikipedia.org/wiki/Pepto-Bismol&quot;&gt;Pepto-Bismol&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Different letters here is your goal &lt;br&gt;&#xA; somewhere to go with an unfortunate hole &lt;br&gt;&#xA; Equestrian researchers aren't quite right &lt;br&gt;&#xA; take from the morning not from the night&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer to the second verse is two different letters taken from the word &quot;here&quot;. The only possibilities are H and R, E and R, or H and E. If you have an unfortunate hole, you go to the emergency room. Take letters from the morning (the start) not the night (the end) of Emergency Room and Equestrian Researchers to get ER.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Careful not to shoot from the hip &lt;br&gt;&#xA; take a bow and get a grip &lt;br&gt;&#xA; the vikings watch me in awe &lt;br&gt;&#xA; I hate my bed I'll stay up more&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;S[hoot from the] hip&quot; and Vikings, famous for going around in longships, suggest the word SHIP. Cannons on SHIPs must not be fired from the hip! A SHIP has a &quot;bow&quot; meaning front, and you can &quot;get a grip&quot; on the wheel. And a SHIP hates the seabed, preferring to stay well above it. (Also if you're on one you might sleep in a hammock, which is higher up than a bed.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;These three up high are not the same &lt;br&gt;&#xA; but their collective being will grant you fame&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We have LEAD, ER, and SHIP. Put them together to get LEADERSHIP. Hurray!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2014-11-25T12:08:50.040" LastActivityDate="2014-11-25T12:08:50.040" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5552" PostTypeId="1" AcceptedAnswerId="5554" CreationDate="2014-11-25T12:03:53.833" Score="7" ViewCount="1849" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;In Futurama's episode 6x10, the professor invents a device which allows to switch mind between two bodies. He then switches his mind with Amy. It later turns out that the a given pair of bodies can not complete this procedure twice with each other.&lt;/p&gt;&#xA;&lt;p&gt;In this situation, is it possible to undo what is done using other bodies to swap minds between?&lt;/p&gt;&#xA;&lt;p&gt;What if initially not 2, but N minds has been switched between N bodies? Consider the worst case scenario that each pair of N bodies has been switched already.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;For example, if they use 1 additional body&lt;/em&gt;:&lt;br /&gt;&#xA;Let the Professor be &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; and Amy be &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt;. Upper case letters represent a body, lower case letters - minds. They start out with &lt;span class=&quot;math-container&quot;&gt;$Pp,Ee$&lt;/span&gt; then after one switch, they have &lt;span class=&quot;math-container&quot;&gt;$Pe, Ep$&lt;/span&gt;. If they use third person &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;, then they can do &lt;span class=&quot;math-container&quot;&gt;$Pe, Ea, Ap$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$Pp, Ea, Ae$&lt;/span&gt;, but &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; has already switched once so they would face the same problem again.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-11-25T22:52:23.713" Title="Irreversible body swap" Tags="|logical-deduction|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5553" PostTypeId="2" ParentId="5552" CreationDate="2014-11-25T12:19:51.353" Score="5" Body="&lt;p&gt;I think you need 2 buddies A and B:&lt;br&gt;&#xA;1. switch Emmy with A, Prof with B,&lt;br&gt;&#xA;2. switch Emmy with B, Prof with A,&lt;br&gt;&#xA;3. switch A and B.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For N &quot;scrambled&quot; people, N is an upper bound.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the lower bound, it depends in what sequence the people got swiched.&#xA;For example if A switched with B and C with D, you can undo the mixup with A&amp;lt;-&gt;C, B&amp;lt;-&gt;D, A&amp;lt;-&gt;D, B&amp;lt;-&gt;C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it can alwasy be done with 4 extra buddies.  I'll show it by example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you keep only people &quot;out of their mind&quot; the general configuration is that of a permutation, i.e. a number of cycles: A has B's mind, B has C's mind ... E has A's mind.  Let's write Ab for A has B's mind.  For example, we could have the configuration with three cycles:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Ab Bc Cd De Ea | Fg Gh Hi If | Jk Kl Lm Mj&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Comes first buddy X (Xx) and switches mind with A, B, C in order, the minds are now arranged as:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Ax Bb Cc Dd Ee | Fa Gg Hh Ii | Jf Kk Ll Mm | Xj&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Most people are now OK, except for the first in each loop, and X.  The remaining people to fix are:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Ax Fa Jf Xj&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You can see the remaining people now form a loop (X dragged one mind over to the next loop).  To fix that, a new buddy Y comes in and exchanges minds with X, J, F, A (from the end).  The minds now are like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Aa Ff Jj Xy Yx&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now we are back in the case of Emmy and Prof.  Unfortunately X and Y already have switched minds, so 2 more buddies are necessary to switch these minds back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just an example, but I believe it works regardless of the number of cycles.  And nobody in the original group needs to swap minds.  One shortcut is when there is a single cycle.  Then you can jump directly to Y for the final cycle.  But it also works with X.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 3 people, you still have the upper bound of 3 additional buddies.  Or you can see that 3 people are always a cycle, so you can fix it Y and the 2 last buddies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I didn't see there is a better answer already.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2014-11-25T22:52:23.713" LastActivityDate="2014-11-25T22:52:23.713" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5554" PostTypeId="2" ParentId="5552" CreationDate="2014-11-25T13:23:05.187" Score="17" Body="&lt;p&gt;A more detailed plot summary of &quot;Prisoner of Benda&quot; reveals that reversal for the worst-case swappage of $N$ bodies is possible with only two additional bodies for any $N$. The proof of this is a &lt;a href=&quot;https://en.wikipedia.org/wiki/The_Prisoner_of_Benda#The_theorem&quot;&gt;non-trivial exercise in group theory&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to several sources, the writing of this episode marks the first known instance of a group theorem being devised &lt;em&gt;and proven&lt;/em&gt; specifically for the sake of a TV program plotline. According to Wiki, the theorem even has a name, &quot;Keeler's Theorem&quot;, after the mathematician who devised and proved it&lt;/p&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2014-11-25T13:23:05.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5555" PostTypeId="1" AcceptedAnswerId="5580" CreationDate="2014-11-25T13:55:18.533" Score="24" ViewCount="6011" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;Note: This question has been severely edited from its original version to better fit the format of this site. Some answers below - including the accepted, correct answer - have been given based on the original question.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Consider the following two digital images below:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QDF8G.jpg&quot; alt=&quot;Image 1&quot; /&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/XyO68.jpg&quot; alt=&quot;Image 2&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Although they depict two very different things, they share a strong common feature: They &lt;em&gt;hide&lt;/em&gt; an identical 3-digits prime number within themselves.&lt;/p&gt;&#xA;&lt;p&gt;Not only do the hide the &lt;em&gt;identical&lt;/em&gt; number, but they hide it by the &lt;em&gt;identical method&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;If you apply the method on image one, you'll get the prime number.&lt;br&gt;&#xA;If you apply the &lt;em&gt;identical&lt;/em&gt; method in the &lt;em&gt;identical&lt;/em&gt; way to the other image, you will get the same prime number.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h1&gt;What is the method to be applied? And which 3-digit prime number do both images hide?&lt;/h1&gt;&#xA;&lt;p&gt;Important hints &amp;amp; restrictions:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The answer to this puzzle is &lt;strong&gt;not an ambiguous or subjective solution&lt;/strong&gt;. Subjective interpretation is not involved.&lt;br&gt;(&lt;em&gt;&amp;quot;...looks like...&amp;quot;&lt;/em&gt;; &lt;em&gt;&amp;quot;...reminds me of...&lt;/em&gt;&amp;quot;; &lt;em&gt;&amp;quot;...could be interpreted as...&amp;quot;&lt;/em&gt;;... are all invalid concepts.)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The answer lies &lt;strong&gt;within the image itself&lt;/strong&gt; and can not be 'forced' upon it by altering the data. &lt;br&gt;( &lt;em&gt;&amp;quot;...paint everything black and draw the prime onto it...&amp;quot;&lt;/em&gt;;... is not a valid answer. )&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There is &lt;strong&gt;no hidden meta-data&lt;/strong&gt; to the images. &lt;br&gt; (You can right-click &amp;amp; save it from here; or you may print it and then rescan it again; or you may simply take a good picture of you screen and use that; all of this will keep the image suitable for the method in question. However, you do need to use a computer to find the answer -- simply looking at the images does not suffice.)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2018-09-27T05:03:27.203" Title="Hidden in plain sight - By what method do these images hide a prime number?" Tags="|mathematics|encoded-message|visual|" AnswerCount="4" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="5556" PostTypeId="1" AcceptedAnswerId="5559" CreationDate="2014-11-25T14:16:09.810" Score="2" ViewCount="1258" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/E6bYO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the next colour in the sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct answer will also state the logic behind it.&lt;/p&gt;&#xA;" OwnerUserId="5567" LastEditorUserId="71" LastEditDate="2014-11-30T15:34:36.063" LastActivityDate="2014-11-30T15:34:36.063" Title="What is the next colour?" Tags="|pattern|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5557" PostTypeId="2" ParentId="5389" CreationDate="2014-11-25T14:18:15.717" Score="12" Body="&lt;p&gt;OK, so I'll post my working so far. I'm plodding along slowly with it :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; This working is all done with the original ciphertext. The same logic can be applied to the &lt;em&gt;correct&lt;/em&gt; ciphertext and you'll end up at the numbers that the OP supplied part-way through the solve.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The second hint is a kind of cipher... how do we utilise it?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Convert lowercase letters to a and uppercase to b. Treat numbers as spaces. Decode it with the Bacon cipher.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Decrypting the first line like this gives us a translated ciphertext of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a  babaaabaabbaaaba bbaabbbabaaaba a abbabaaabaabaaba&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which in turn, decodes to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; KEY WOR DIS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Continuing the remaining 4 lines in the same way, we get the following ciphertext and message:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a  babaaabaabbaaaba bbaabbbabaaaba a abbabaaabaabaaba&lt;br&gt;&#xA; abbabaaabba aaaaabab abbabaa aaabaab aababaaabaa baab&lt;br&gt;&#xA; ababaabaab aaabbaabaaabaa ababbabaaa  abbabaabaabaab&lt;br&gt;&#xA; ab a abbaabbbaa  baaabbabaaabbaabaaaaababaaabbbaaa ab&lt;br&gt;&#xA; bbbbaabbab bbbababba aaaabba aaaababaaaaaabaaabaaab&lt;br&gt;&#xA;&lt;br&gt;&#xA; KEY WORD IS JUMBLIES REVERSE LINES THEN DECRYPT PLAYFAIR&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;By following the first directive, we now have the ciphertext of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; fEhkQzkQavrPfDFl5f5dRovpOeIFThlFA4oFlfdASifHtpzTfL38f&lt;br&gt;&#xA; AooE2ifOeifRpFdr2PfdGmtn1ocAtOVb4PfLfdpam0vROvpoEiFQn&lt;br&gt;&#xA; OcaOovBprOvPRp23fdoEiFOvBp8ohPiokEiaOOeif3PafIdfRoVp9&lt;br&gt;&#xA; Qf2oeiFRPfdtFaOeuvhpItrEUvheIaOTfavU38fnQOEifOHp9i2Pc&lt;br&gt;&#xA; 03FidfRpfdOvbpoheOrZrovp7qNOcaot5fSAmVfLFD3OhEOrzPZQV&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, this is the evil part.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In Playfair encryption, J gets translated to I, so our keyword is actually IUMBLES.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Decrypting with this, we get a string containing digits written out (one, two) and a lot of &lt;code&gt;q&lt;/code&gt; characters which split up groups of digits (it's not unusual for &lt;code&gt;q&lt;/code&gt; to be omitted from this kind of cipher, so they probably represent spaces). It isn't clear just yet what these numbers mean, or whether any should be combined together.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; eightxhrexqonenineqtwofiveqnineqninesevenqtwonineqfivefiveqoneqonehundredqfournineqsixqtwofivethreeqfourqtwoqonefivefourqfourfiveqfiveqsevenqtwoohfiveqoneoneqsixfourqsixfiveqonexightfiveqfoursevenqonefourqfiftyqtwothreeqonesixnineqfiftytwox&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The numbers are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 83 19 25 9 97 29 55 1 100 49 6 253 4 2 154 45 5 7 205 11 64 65 185 47 14 50 23 169 52&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But&lt;/strong&gt; these are wrong (@OP HOW COULD YOU!?) so we need these ones instead, provided by the OP :P&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 78 19 25 9 92 29 54 1 95 49 6 247 4 2 149 45 5 7 199 11 62 63 179 47 14 50 23 163 52&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Interestingly, all the deciphering we've done so far has been purely alphabetical - the Bacon deciphering was done with upper/lowercase letters and we ignored the numbers as all characters in Bacon are comprised of 5 a/bs, and the Playfair decryption only uses letter digraphs, again discarding numbers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There happen to be 5 digits in each of the 5 lines of the ciphertext (and they remained consistent between old/new ciphertexts):&lt;br&gt;&#xA; 83 4 5 5&lt;br&gt;&#xA; 0 4 1 2 2&lt;br&gt;&#xA; 9 3 8 32&lt;br&gt;&#xA; 2 9 83 2&lt;br&gt;&#xA; 3 5 7 30&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2518" LastEditorUserId="2518" LastEditDate="2014-11-26T15:58:57.330" LastActivityDate="2014-11-26T15:58:57.330" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="5559" PostTypeId="2" ParentId="5556" CreationDate="2014-11-25T14:22:57.930" Score="10" Body="&lt;p&gt;This is the next image in the sequence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IFPQp.png&quot; alt=&quot;Next image&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the hex colours for each image - each pair increments on a different scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pair 1: 3C 4E 60 72 ... 84 (+18)&lt;br&gt;&#xA;Pair 2: 55 5C 63 6A ... 71 (+7)&lt;br&gt;&#xA;Pair 3: 64 66 68 6A ... 6C (+2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The increments have been chosen because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Letter 18 is R, letter 7 is G and letter 2 is B - RGB is the ugly twin sister of HEX colors (ugly twin sister in CSS, anyway :P)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For giggles, you could also argue that there's a 4th pair:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pair 4: 1, 1, 1 ... 1 (RGBA is RGB with Alpha, ie. opacity)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2518" LastEditorUserId="5368" LastEditDate="2014-11-29T20:45:09.633" LastActivityDate="2014-11-29T20:45:09.633" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5560" PostTypeId="2" ParentId="5472" CreationDate="2014-11-25T14:34:45.737" Score="2" Body="&lt;p&gt;Ok, this is what happened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the hotel room,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the first suspect received a long stream of letters in Morse code, and the dimensions of the matrix: 17x23. He arranged the letters in the matrix in order,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and that's how the keynote came to be. The keyword can be obtained by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; looking at the positions of all the occurrences of each letter in the matrix. For example, all the E-letters are arranged in the shape of an E. That's why the second suspect needed all those coloured pencils: he had to check the shapes formed by each of the 26 letters. The only letters whose occurrences form the correct shapes are D, E, L, O, S, and V.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The order of these letters (I suppose) is given by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the order of their appearance in the matrix.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hence the keyword:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; S O L V E D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The keyword can be used to decode the final message like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letters S-O-L-V-E-D are equally spaced in the text. If you arrange the message in 13 columns, each column will have a full english word delimited by an X. The first column will spell &quot;SOLVED&quot;, while the other ones will give the hidden message: &quot;meet at midnight corner parkave twenty seven come armed bring your passport&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5990" LastEditorUserId="5990" LastEditDate="2014-11-25T14:48:31.070" LastActivityDate="2014-11-25T14:48:31.070" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="5561" PostTypeId="2" ParentId="5481" CreationDate="2014-11-25T14:44:00.897" Score="2" Body="&lt;p&gt;I always pictured Ernie as kind of a Doc Brown: an eccentric, older inventor.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as it turns out, he's only&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 29&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;now, since he was&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 28&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;when this happened, in&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; late 2013 or the first half of 2014.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You see, he was born on&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the 21&lt;sup&gt;st&lt;/sup&gt; of June, in 1985.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;His father was born on&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the 11&lt;sup&gt;th&lt;/sup&gt; of December, in 1957.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The show with Madame Mu took place some day between&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; his dad's latest birthday (that day included), and his own latest birthday (that day excluded), or to be exact: [11 December 2013 - 20 June 2014]. During that period, he was 28 while his father was (and still is) 56.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I verified this using a rather long SQL query because why not, and you can do a lot more in SQL than you may think. I added prime factors to check for until I was down to one set of dates, which I checked on WolframAlpha.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there already is an answer, but I did this independent of it, and looking at the answer now, I see I'm a bit more precise on when the events took place.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;WITH&#xA;   dates AS (&#xA;      SELECT&#xA;         TRUNC(SYSDATE) AS today,&#xA;         TO_DATE('20011231', 'YYYYMMDD') AS nye&#xA;      FROM dual&#xA;   ),&#xA;   days AS (&#xA;      SELECT today - LEVEL AS that_day&#xA;      FROM dates&#xA;      CONNECT BY LEVEL &amp;lt;= 50000&#xA;   ),&#xA;   day_nrs AS (&#xA;      SELECT&#xA;         that_day,&#xA;         TO_NUMBER(&#xA;               TO_NUMBER(TO_CHAR(that_day, 'DD')) &#xA;            || TO_NUMBER(TO_CHAR(that_day, 'MM')) &#xA;            || TO_CHAR(that_day, 'YYYY')&#xA;         ) AS dmy,&#xA;         TO_NUMBER(&#xA;               TO_CHAR(that_day, 'YYYY') &#xA;            || TO_NUMBER(TO_CHAR(that_day, 'MM')) &#xA;            || TO_NUMBER(TO_CHAR(that_day, 'DD'))&#xA;         ) AS ymd,&#xA;         TO_NUMBER(&#xA;               TO_NUMBER(TO_CHAR(that_day, 'MM')) &#xA;            || TO_NUMBER(TO_CHAR(that_day, 'DD')) &#xA;            || TO_CHAR(that_day, 'YYYY')&#xA;         ) AS mdy&#xA;      FROM days&#xA;   ),&#xA;   valid_days AS (&#xA;      SELECT&#xA;         nr.that_day AS dob,&#xA;         nr.dmy,&#xA;         nr.ymd,&#xA;         nr.mdy,&#xA;         TRUNC(MONTHS_BETWEEN(dd.that_day, nr.that_day) / 12) AS age,&#xA;         dd.that_day AS mu_day&#xA;      FROM day_nrs nr&#xA;      CROSS JOIN days dd&#xA;      CROSS JOIN dates d &#xA;      WHERE nr.that_day &amp;lt;= d.nye&#xA;        AND MOD(nr.dmy,  7) = 0&#xA;        AND MOD(nr.ymd, 11) = 0&#xA;        AND MOD(nr.mdy, 13) = 0&#xA;        AND dd.that_day &amp;gt; d.nye&#xA;   ),&#xA;   ages AS (&#xA;      SELECT &#xA;         ernie.dob AS dob_ernie,&#xA;         ernie.age AS age_ernie,&#xA;         TRUNC(MONTHS_BETWEEN(d.today, ernie.dob) / 12) AS curr_age_ernie,&#xA;         daddy.dob AS dob_daddy,&#xA;         daddy.age AS age_daddy,&#xA;         TRUNC(MONTHS_BETWEEN(d.today, daddy.dob) / 12) AS curr_age_daddy,&#xA;         MIN(ernie.mu_day) AS min_date,&#xA;         MAX(ernie.mu_day) AS max_date&#xA;      FROM valid_days ernie&#xA;      JOIN valid_days daddy&#xA;         ON (   ernie.dob &amp;gt; daddy.dob &#xA;            AND ernie.mu_day = daddy.mu_day&#xA;            AND ernie.age * 2 = daddy.age)&#xA;      CROSS JOIN dates d &#xA;      WHERE MOD(ernie.dmy + ernie.ymd + ernie.mdy + ernie.age,  2) &amp;lt;&amp;gt; 0&#xA;        AND MOD(ernie.dmy + ernie.ymd + ernie.mdy + ernie.age,  3) &amp;lt;&amp;gt; 0&#xA;        AND MOD(ernie.dmy + ernie.ymd + ernie.mdy + ernie.age,  5) &amp;lt;&amp;gt; 0&#xA;        AND MOD(ernie.dmy + ernie.ymd + ernie.mdy + ernie.age,  7) &amp;lt;&amp;gt; 0&#xA;        AND MOD(ernie.dmy + ernie.ymd + ernie.mdy + ernie.age, 11) &amp;lt;&amp;gt; 0&#xA;        AND MOD(ernie.dmy + ernie.ymd + ernie.mdy + ernie.age, 13) &amp;lt;&amp;gt; 0&#xA;        AND MOD(ernie.dmy + ernie.ymd + ernie.mdy + ernie.age, 17) &amp;lt;&amp;gt; 0&#xA;        AND MOD(ernie.dmy + ernie.ymd + ernie.mdy + ernie.age, 19) &amp;lt;&amp;gt; 0&#xA;        AND MOD(ernie.dmy + ernie.ymd + ernie.mdy + ernie.age, 23) &amp;lt;&amp;gt; 0&#xA;        AND MOD(ernie.dmy + ernie.ymd + ernie.mdy + ernie.age, 29) &amp;lt;&amp;gt; 0&#xA;        AND MOD(ernie.dmy + ernie.ymd + ernie.mdy + ernie.age, 31) &amp;lt;&amp;gt; 0&#xA;        AND MOD(ernie.dmy + ernie.ymd + ernie.mdy + ernie.age, 37) &amp;lt;&amp;gt; 0&#xA;        AND MOD(ernie.dmy + ernie.ymd + ernie.mdy + ernie.age, 41) &amp;lt;&amp;gt; 0&#xA;        AND MOD(ernie.dmy + ernie.ymd + ernie.mdy + ernie.age, 43) &amp;lt;&amp;gt; 0&#xA;        AND MOD(ernie.dmy + ernie.ymd + ernie.mdy + ernie.age, 47) &amp;lt;&amp;gt; 0&#xA;        AND MOD(ernie.dmy + ernie.ymd + ernie.mdy + ernie.age, 53) &amp;lt;&amp;gt; 0&#xA;        AND MOD(ernie.dmy + ernie.ymd + ernie.mdy + ernie.age, 59) &amp;lt;&amp;gt; 0&#xA;        AND MOD(ernie.dmy + ernie.ymd + ernie.mdy + ernie.age, 61) &amp;lt;&amp;gt; 0&#xA;        AND MOD(daddy.dmy + daddy.ymd + daddy.mdy + daddy.age,  2) &amp;lt;&amp;gt; 0&#xA;        AND MOD(daddy.dmy + daddy.ymd + daddy.mdy + daddy.age,  3) &amp;lt;&amp;gt; 0&#xA;        AND MOD(daddy.dmy + daddy.ymd + daddy.mdy + daddy.age,  5) &amp;lt;&amp;gt; 0&#xA;        AND MOD(daddy.dmy + daddy.ymd + daddy.mdy + daddy.age,  7) &amp;lt;&amp;gt; 0&#xA;        AND MOD(daddy.dmy + daddy.ymd + daddy.mdy + daddy.age, 11) &amp;lt;&amp;gt; 0&#xA;        AND MOD(daddy.dmy + daddy.ymd + daddy.mdy + daddy.age, 13) &amp;lt;&amp;gt; 0&#xA;        AND MOD(daddy.dmy + daddy.ymd + daddy.mdy + daddy.age, 17) &amp;lt;&amp;gt; 0&#xA;        AND MOD(daddy.dmy + daddy.ymd + daddy.mdy + daddy.age, 19) &amp;lt;&amp;gt; 0&#xA;        AND MOD(daddy.dmy + daddy.ymd + daddy.mdy + daddy.age, 23) &amp;lt;&amp;gt; 0&#xA;        AND MOD(daddy.dmy + daddy.ymd + daddy.mdy + daddy.age, 29) &amp;lt;&amp;gt; 0&#xA;        AND MOD(daddy.dmy + daddy.ymd + daddy.mdy + daddy.age, 31) &amp;lt;&amp;gt; 0&#xA;        AND MOD(daddy.dmy + daddy.ymd + daddy.mdy + daddy.age, 37) &amp;lt;&amp;gt; 0&#xA;        AND MOD(daddy.dmy + daddy.ymd + daddy.mdy + daddy.age, 41) &amp;lt;&amp;gt; 0&#xA;        AND MOD(daddy.dmy + daddy.ymd + daddy.mdy + daddy.age, 43) &amp;lt;&amp;gt; 0&#xA;        AND MOD(daddy.dmy + daddy.ymd + daddy.mdy + daddy.age, 47) &amp;lt;&amp;gt; 0&#xA;        AND MOD(daddy.dmy + daddy.ymd + daddy.mdy + daddy.age, 53) &amp;lt;&amp;gt; 0&#xA;        AND MOD(daddy.dmy + daddy.ymd + daddy.mdy + daddy.age, 59) &amp;lt;&amp;gt; 0&#xA;        AND MOD(daddy.dmy + daddy.ymd + daddy.mdy + daddy.age, 61) &amp;lt;&amp;gt; 0&#xA;      GROUP BY&#xA;         ernie.dob,&#xA;         daddy.dob,&#xA;         ernie.age,&#xA;         daddy.age,&#xA;         d.today&#xA;   ) &#xA;SELECT&#xA;   min_date,&#xA;   max_date,&#xA;   dob_ernie,&#xA;   age_ernie,&#xA;   dob_daddy,&#xA;   age_daddy,&#xA;   curr_age_daddy&#xA;FROM ages&#xA;WHERE curr_age_daddy &amp;lt;= 120&#xA;ORDER BY&#xA;   min_date,&#xA;   max_date;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="68" LastActivityDate="2014-11-25T14:44:00.897" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5566" PostTypeId="2" ParentId="5555" CreationDate="2014-11-25T16:24:37.263" Score="21" Body="&lt;p&gt;I think it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 313&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;photo evidence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mVNeb.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;as for the other picture&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tree, none, Tree&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="51" LastActivityDate="2014-11-25T16:24:37.263" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="5567" PostTypeId="1" CreationDate="2014-11-25T16:38:05.857" Score="11" ViewCount="6160" Body="&lt;p&gt;An archaeologist who has just unearthed a long-sought-after pair of ancient treasure chests.&lt;/p&gt;&#xA;&lt;p&gt;One chest is plated with &lt;strong&gt;silver&lt;/strong&gt; and the other is plated with &lt;strong&gt;gold&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;According to the legend one of the two chests is filled with a great treasure whereas the other chest houses a death cursing spell in the form of a dragon.&lt;/p&gt;&#xA;&lt;p&gt;Faced with a dilemma archaeologist then noticed that there are inscriptions on the chest. Based on these inscriptions which chest should he open?&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{Gold\, Chest}- \mathrm{One\, of\, these\, inscriptions\, is\, true}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{Silver\, Chest- \mathrm{This\, chest\, contains\, the\, dragon}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Note: Thanks to his understanding of the culture from which these chests came, the archaeologist knows that the gold chest's inscription could also be translated as &amp;quot;One &lt;em&gt;and only one&lt;/em&gt; of these inscriptions is true&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;I already know the answer, what I am seeking here is a nice explanation.&lt;/p&gt;&#xA;" OwnerUserId="6007" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-11-29T02:30:07.090" Title="The treasure hunt" Tags="|logical-deduction|" AnswerCount="10" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="5568" PostTypeId="2" ParentId="5567" CreationDate="2014-11-25T16:43:02.913" Score="20" Body="&lt;p&gt;He should open the&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Silver Chest&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation of the logic (assumes the inscriptions are intended to be helpful):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Assume the inscription on the gold chest is true. Because that inscription is true, we already have the one true inscription, so the other inscription must be false. Thus the silver chest does not contain the dragon, so it must have the treasure.&lt;br&gt;&lt;br&gt;Now assume the inscription on the gold chest is false. That means either both inscriptions are true, or both are false. Since the gold chest's inscription is false, the silver chest's inscription must also be false, so it must have the treasure.&lt;br&gt;&lt;br&gt;So regardless of whether the gold chest's inscription is true or false, the silver chest's inscription must be false and therefore must be the one with the treasure in it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm also assuming here that the dragon referred to by the silver chest's inscription is the same thing as the death curse.&lt;/p&gt;&#xA;" OwnerUserId="1955" LastEditorUserId="1955" LastEditDate="2014-11-25T19:46:09.993" LastActivityDate="2014-11-25T19:46:09.993" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="5569" PostTypeId="2" ParentId="5567" CreationDate="2014-11-25T16:51:01.423" Score="11" Body="&lt;p&gt;&lt;strong&gt;This is a more formal solution&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a small system of formulae to define the problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a chest $c_i$ is $true$ then it contains the treasure. If a chest $c_i$ is $false$ then it contains the dragon. We have 2 propositional variables, $c_1$ and $c_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first formal assertion we can define is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c_1 \oplus c_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We express the inscriptions as further assertions about each other, themselves and the propositional variables. $f_1$ expresses the inscription on the Gold chest and $f_2$ that of the Silver chest. Conjointly they are $F$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_1 = f_1 \oplus f_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_2 = (c_2 = false)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We must deduce for which $i$, $c_i$ is true. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We do this by observing contradictions in $F$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $c_2 = false$, then that makes $f_2 = true$. Now, if $f_1$ is $false$ then $f_1$ is true. If $f_1$ is true then $f_1$ is false, this is a contradiction. Thus, $c_2 \not= false$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $c_2 = true$, then that makes $f_2 = false$. Now, $f_1 \leftrightarrow f_1$. No problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $c_2 = true$. The Silver chest contains the treasure. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-11-26T01:18:34.040" LastActivityDate="2014-11-26T01:18:34.040" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5570" PostTypeId="2" ParentId="5567" CreationDate="2014-11-25T18:33:59.143" Score="7" Body="&lt;p&gt;Assuming this is a bit of a trick question...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Neither.&#xA;There is not enough information to safely pick a chest.&#xA;According to legend, one chest has treasure while the other has &lt;b&gt;a death cursing spell.&lt;/b&gt;&#xA;The fact that the silver chest must be lying and thus does not contain a dragon has no bearing on which chest has the curse and which has the treasure. A chest that lacks a dragon can instead contain a curse or a treasure.&#xA;This assumes the question is correctly worded.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5473" LastEditorUserId="5473" LastEditDate="2014-11-25T20:59:23.420" LastActivityDate="2014-11-25T20:59:23.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5572" PostTypeId="2" ParentId="5555" CreationDate="2014-11-25T19:15:09.650" Score="7" Body="&lt;p&gt;I think the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 101&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pillar - Window - Pillar&#xA; Pillars look like 1 and the Windows like 0&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I did take a look at your second picture and this is the only way I'm able to connect them.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bush-Space-Bush.  Something-Nothing-Something :-)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5654" LastEditorUserId="5654" LastEditDate="2014-11-25T19:20:17.197" LastActivityDate="2014-11-25T19:20:17.197" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5573" PostTypeId="1" AcceptedAnswerId="5576" CreationDate="2014-11-25T19:52:04.477" Score="30" ViewCount="2157" Body="&lt;p&gt;The following shapes are sequenced from left to right. Describe or create the next five shapes. Bonus points to those who describe how the pattern works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hJSY3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The pattern could continue on forever.&lt;/li&gt;&#xA;&lt;li&gt;The line thicknesses you see here are 1, 3, and 5; and the next five shapes require only these.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Spoilers: &lt;a href=&quot;https://codegolf.stackexchange.com/questions/41901/numbers-as-circular-graphics&quot;&gt;I presented this as a programming question and folks made some very cool animations.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6014" LastEditorUserId="-1" LastEditDate="2017-04-13T12:39:28.753" LastActivityDate="2017-01-23T16:08:16.247" Title="Continue the Pattern of Circles" Tags="|geometry|pattern|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5574" PostTypeId="1" AcceptedAnswerId="17199" CreationDate="2014-11-25T20:15:11.797" Score="8" ViewCount="1500" Body="&lt;p&gt;I devised this cipher independently, but I have no idea whether it's easy or hard because I have little knowledge of cryptography. The algorithm is simple but not trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;AYVEKCI0 3D3 1UDTY1NWBYFJFUJDLQOFQIDWZ0UQMVWKLUCO3GWXWO&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that both 0s and Os are present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Here are symbols you might think are possible in this cipher, but are not: 456789&lt;br/&gt;&#xA; Here is a symbol you wouldn't know is possible from the given code: ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;FYW?HIJUHZ0LPENNS0SZBF?N3DUZTECJMJVTMY03HDPJWH2Q0NZIE31SNUV YOBK&#xA;MRKTBFFASDXBKLKR?FAL3PWO30IIBM HKGO PCSHD1PDYZPUIQVHHVIZYB BTHHP&#xA;M IPEX2R L3Z?FAL3YH?120QD 3D3KOVTU?V2B2IBVEV 23XDSHMYUJQSV?SDMJH&#xA;UNYJJUJVATFE  TCX0D3KOVPY? WFSIF2LFADUYJG22S1?SWAUFEIF?ORKDNYLUX&#xA;NC2MLFPN1AMBY3VG?R2LMRD1BKMJVOBXGVTN OVHIR0JBZLJSXTAXZ2XEJYJITJY&#xA;AEFYCCR12REYJMOAIH2ITSC3SLVC01YMVIOFMZNNMGHEWQHLAGIOTM1QSHA1AYVE&#xA;KCI0 3D3 1UDTY1NWBYFJFUJDLQOFQIDWZ0UROTVKXKG32HLPM CZDS3NRHGVCF?&#xA;O2?D2IS0GPLM0KEYYSDZIYSUL3UHOGJ3GJ0KSJ?BYUFSHFNLVAL1DEKEULGPL1YH&#xA;B1 XMQLKLDSL1FLKNWF3YEGXOSO1XJTX FT VOTYNZ2WDQFMH0W3L VMVVDUHFO0&#xA;MRDMCJLKWNJ?G2GG?JWLUXT3LWZTGYGB 1D 0XII3T1MQSHA1AYVEKCI0 3D3 1U&#xA;DTY1NWBYFJFUJDLQOFQIDWZ0UROTVKXKG32HKWO2DVSTM MRIMQKH3O?N22YSYNQ&#xA;UNUIGUD2 DE0JUHIPUAZTHMUI2PVG03QUFUYUAFU N1QG3TOYEUDW GFVMXNOXQO&#xA;AFVSAV2BZ02W1NCZ3QSB 1?JCSULEQO3ETXE2IU2UYQCJQMQON31AURE32ESO3BV&#xA;0F DV TGO3P3O&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Everything has been generated by computer instead of by hand, so there should be no more errors. So sorry for it being messed up at first!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2014/12/8 - More clues:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Characters are processed in groups of five.&lt;br/&gt;The plaintext contains every letter of the alphabet except Q.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2015/6/30&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ? immediately precedes A.&lt;br/&gt;The fact of 32 possible characters requires that characters be processed in groups of 5: 2^5 characters in the alphabet, 5 characters per group.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5341" LastEditorUserId="5341" LastEditDate="2015-06-30T19:29:31.887" LastActivityDate="2015-06-30T19:42:15.160" Title="Crack this code: AYVEKCI0" Tags="|cipher|" AnswerCount="1" CommentCount="19" ContentLicense="CC BY-SA 3.0" />
  <row Id="5575" PostTypeId="2" ParentId="5555" CreationDate="2014-11-25T20:21:08.510" Score="5" Body="&lt;p&gt;So, my first instinct was to download both pictures and compare them.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I ended up comparing their &quot;properties&quot; to see if their dimensions were a prime number. 512 is not prime number :( But I noticed that they were both 179 kilobytes, a prime number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But then I looked at the hints, and realized that can't be it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I made a list of all the prime numbers that it could be based on that hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 101,131,151,181,191,313,353,373,383,727,757,787,797,919,929&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As stated in previous answers, it could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 101 or 313&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But, since those are incorrect, I'm going to take a very wild guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It might &lt;strong&gt;929&lt;/strong&gt;. Here's my reasoning. Both images are 512x512 pixels. Another way to write that is to say 2^9 by 2^9. The &quot;2&quot; could come from the &quot;power of 2&quot; or from the word &quot;by&quot;, in that the letter &quot;b&quot; is the 2nd letter in the alphabet. I know it's a stretch, but it's the best I can come up with for now.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5934" LastActivityDate="2014-11-25T20:21:08.510" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5576" PostTypeId="2" ParentId="5573" CreationDate="2014-11-25T20:41:32.490" Score="29" Body="&lt;p&gt;Here is your sequence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/6B2Hb.png&quot; alt=&quot;Sequence&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the explanation is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; These are Roman numerals.  Read the circles from inside out.  Thickness 1 is I, thickness 3 is V, thickness 5 is X.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2014-11-25T20:51:49.320" LastActivityDate="2014-11-25T20:51:49.320" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5577" PostTypeId="2" ParentId="5567" CreationDate="2014-11-25T20:53:38.177" Score="4" Body="&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Consider four possibilities: (1) Both inscriptions are true, (2) both are false, (3) only the gold chest inscription is true, and (4) only the gold chest inscription is false.  There are no other possibilities.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let's look at four possibilities:&lt;br&gt;&lt;br&gt;1. Both inscriptions are true.  This is a &lt;strong&gt;contradiction&lt;/strong&gt; because the Gold inscription says &quot;One of these inscriptions is true&quot;, so we can rule this out (see note below).&lt;br&gt;&lt;br&gt;2. Both inscriptions are false.  This means that the Silver inscription &quot;This chest contains the dragon&quot; is false and therefore, by elimination, the &lt;strong&gt;silver&lt;/strong&gt; chest has the treasure.&lt;br&gt;&lt;br&gt;3. The gold chest's inscription is true.  If we assume the gold chest's inscription is true and the gold chest's inscription says that one of the inscriptions is true (see note below), then the silver chest's inscription is necessarily false and thus the &lt;strong&gt;silver&lt;/strong&gt; chest has the treasure.&lt;br&gt;&lt;br&gt;4. The gold chest's inscription is false.  If we assume the gold chest's inscription is false, then either both inscriptions are false or both inscriptions are true.  I've shown in #1 above that both inscriptions being true leads to a contradiction, so both inscriptions must be false.  Therefore the &lt;strong&gt;silver&lt;/strong&gt; chest has the treasure.&lt;br&gt;&lt;br&gt;These possibilities either are contradictory or point to the silver chest, so &lt;strong&gt;the silver chest is the answer&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note, I'm assuming &quot;One of these inscriptions is true&quot; means &quot;only one&quot; is true.&lt;/p&gt;&#xA;" OwnerUserId="2318" LastActivityDate="2014-11-25T20:53:38.177" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5578" PostTypeId="1" AcceptedAnswerId="5579" CreationDate="2014-11-25T21:13:46.480" Score="10" ViewCount="637" Body="&lt;p&gt;This is simple and has a glaringly clear solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instructions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Print&lt;/li&gt;&#xA;  &lt;li&gt;Cut&lt;/li&gt;&#xA;  &lt;li&gt;Assemble &lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To make it a proper puzzle and avoid half-solutions: The solution I'm looking for &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;has all pieces touching two others&lt;/li&gt;&#xA;&lt;li&gt;produces an image people would generally considers not scrambled&lt;/li&gt;&#xA;&lt;li&gt;is rotated with what appears to be the &quot;ground&quot; showing at the bottom&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Get puzzling!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zS23y.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="5373" LastEditDate="2014-11-29T14:59:53.913" LastActivityDate="2014-11-29T14:59:53.913" Title="Physical puzzle in four parts" Tags="|mechanical-puzzles|visual|jigsaw-puzzle|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5579" PostTypeId="2" ParentId="5578" CreationDate="2014-11-25T21:37:18.370" Score="12" Body="&lt;p&gt;Someone had to do it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/2OIev.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I liked the ancient math one more though.&lt;/p&gt;&#xA;" OwnerUserId="4191" LastActivityDate="2014-11-25T21:37:18.370" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5580" PostTypeId="2" ParentId="5555" CreationDate="2014-11-25T21:53:48.213" Score="31" Body="&lt;p&gt;The answer is 101, which you can see&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; by computing the Fourier transform.&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/wkCiK.jpg&quot; alt=&quot;Fourier transform&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;How this works&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;You can think of an image as a function $F$ of two variables, $x$ and $y$. $x$ and $y$ represent the horizontal and vertical positions of each pixel, and $F(x,y)$ represents the pixel intensity (lightness or darkness). Now it turns out that any periodic function (and with images we only care about a finite range, so we can always just suppose $F$ to repeat once you reach the image boundaries) can be represented as a weighted sum of sine and cosine functions of different frequencies. These sums are called fourier series; the image below (taken from Wolfram MathWorld) shows various sums of sinusoids converging to a few different periodic functions in the one-dimensional case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vC6wD.gif&quot; alt=&quot;Fourier series examples from MathWorld&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in general for this representation to be exact for a continuous function you may need an infinite number of terms in the sum. However, in the discrete case, you only need to sum over as many different frequency values as you have samples (pixels) of the original function (image).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in our discrete, two-dimensional case, what we have is this: a function $F(x,y)$ representing an image of size M-by-N can be written as a weighted sum of 2D sinusoids with horizontal frequency $u$ and vertical frequency $v$, where $u$ ranges from $1$ to $M$ and $v$ ranges from $1$ to $N$. You can think of the weight of each term as a function, say $G$, of the frequency variables $u$ and $v$. There's then nothing stopping you from thinking of $u$ and $v$ as horizontal and vertical positions of a new image, with $G(u,v)$ determining pixel intensity at each point. This is the process that was carried out to produce the &quot;101&quot; image above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried to keep this explanation as non-technical as possible; if you'd like to see the gory details of how the coefficients are actually calculated, MathWorld's &lt;a href=&quot;http://mathworld.wolfram.com/DiscreteFourierTransform.html&quot; title=&quot;Discrete Fourier Transform&quot;&gt;Discrete Fourier Transform&lt;/a&gt; article has a nice overview of the 1D case. You can see the (relatively straightforward) generalization to two dimensions &lt;a href=&quot;http://www.cs.cf.ac.uk/Dave/Multimedia/node228.html&quot; title=&quot;2D DFT&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To play around with this stuff yourself you could try &lt;a href=&quot;http://imagej.nih.gov/ij/&quot;&gt;ImageJ&lt;/a&gt;, a public domain image processing package developed by the NIH. It'll do all the dirty work for you -- just install, open an image, and select &lt;i&gt;Process-&gt;FFT&lt;/i&gt; (Fast Fourier Transform).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see a reason (one of the many, I swear) why you'd bother to do this outside of solving puzzles, try &lt;a href=&quot;http://imagej.nih.gov/ij/docs/examples/FFT/part2.html&quot;&gt;this nice, quick example&lt;/a&gt; of using FFT filtering in ImageJ to remove periodic noise from an image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a deep understanding of why this works, I recommend a course in applied analysis. (Or, I'm sure, any number of good books or online tutorials -- but I don't know what the good ones are, so I won't try to suggest one.)&lt;/p&gt;&#xA;" OwnerUserId="4912" LastEditorUserId="71" LastEditDate="2014-11-26T17:05:07.563" LastActivityDate="2014-11-26T17:05:07.563" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="5583" PostTypeId="2" ParentId="5567" CreationDate="2014-11-26T00:51:36.240" Score="4" Body="&lt;p&gt;Three possibilities.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The archaeologist can CT scan both the chest. The one in which is finds treasure he can open it.&lt;/li&gt;&#xA;&lt;li&gt;He can tap on both the treasure chest or do something that creates noise. If there is a dragon it will react&lt;/li&gt;&#xA;&lt;li&gt;Ask someone else to open both the chest on his behalf&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5660" LastActivityDate="2014-11-26T00:51:36.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5585" PostTypeId="1" AcceptedAnswerId="5951" CreationDate="2014-11-26T03:05:13.670" Score="-10" ViewCount="883" Body="&lt;p&gt;You can't crack it. There's a solution, definitely. It's pretty long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0005200056 0005200057 0005200056 0005200057 0005200056 0005200057 0005200056 0005200056 0005200056 0005200057 0005200056 0005200056 0005200056 0005200057 0005200057 0005200056 0005200056 0005200057 0005200056 0005200056 0005200057 0005200056 0005200056 0005200057 0005200056 0005200057 0005200056 0005200056 0005200057 0005200057 0005200056 0005200056 0005200056 0005200056 0005200057 0005200056 0005200056 0005200056 0005200056 0005200056 0005200056 0005200057 0005200056 0005200056 0005200056 0005200057 0005200056 0005200057 0005200056 0005200056 0005200057 0005200056 0005200056 0005200056 0005200056 0005200056 0005200056 0005200057 0005200056 0005200057 0005200056 0005200057 0005200057 0005200057 0005200056 0005200057 0005200056 0005200056 0005200056 0005200056 0005200057 0005200057 0005200056 0005200057 0005200056 0005200056 0005200056 0005200056 0005200056 0005200057 0005200056 0005200057 0005200056 0005200057 0005200056 0005200057 0005200056 0005200056 0005200056 0005200056 0005200057 0005200056 0005200056 0005200056 0005200056 0005200056 0005200056 0005200057 0005200056 0005200056 0005200057 0005200056 0005200056 0005200056 0005200056 0005200057 0005200056 0005200056 0005200057 0005200057 0005200057 0005200057 0005200056 0005200057 0005200056 0005200056 0005200057 0005200057 0005200056 0005200056 0005200056 0005200057 0005200056 0005200056 0005200057 0005200057 0005200057 0005200056 0005200056 0005200056 0005200057 0005200056 0005200056 0005200056 0005200056 0005200056 0005200056 0005200057 0005200056 0005200057 0005200056 0005200057 0005200056 0005200057 0005200056 0005200057 0005200056 0005200057 0005200056 0005200056 0005200056 0005200057 0005200056 0005200057 0005200056 0005200056 0005200056 0005200057 0005200057 0005200057 0005200056 0005200056 0005200057 0005200056 0005200057 0005200057 0005200057 0005200057 0005200056 0005200057 0005200056 0005200057 0005200056 0005200057 0005200057 0005200056 0005200056 0005200057 0005200056 0005200056 0005200056 0005200056 0005200057 0005200057 0005200056 0005200057 0005200056 0005200056 0005200056 0005200057 0005200056 0005200056 0005200056 0005200057 0005200056 0005200056 0005200057 0005200057 0005200056 0005200056 0005200056 0005200056 0005200057 0005200056 0005200056 0005200056 0005200056 0005200056 0005200056 0005200057 0005200056 0005200056 0005200057 0005200056 0005200056 0005200056 0005200056 0005200057 0005200056 0005200056 0005200057 0005200056 0005200056 0005200056 0005200056 0005200057 0005200056 0005200056 0005200056 0005200056 0005200057 0005200056 0005200056 0005200056 0005200057 0005200056 0005200056 0005200056 0005200056 0005200056 0005200056 0005200057 0005200056 0005200057 0005200056 0005200056 0005200057 0005200057 0005200056 0005200057 0005200056 0005200056 0005200056 0005200056 0005200057 0005200057 0005200056 0005200057 0005200056 0005200056 0005200056 0005200057 0005200057 0005200056 0005200056 0005200057 0005200056 0005200056 0005200057 0005200056 0005200056 0005200056 0005200056 0005200056 0005200057 0005200056 0005200056 0005200056 0005200056 0005200056 0005200056 0005200057 0005200056 0005200056 0005200057 0005200056 0005200057 0005200056 0005200056 0005200057 0005200056 0005200057 0005200056 0005200056 0005200057 0005200056 0005200056 0005200057 0005200056 0005200056 0005200057 0005200056 0005200056 0005200056 0005200056 0005200057 0005200056 0005200056 0005200057 0005200057 0005200056 0005200057 0005200056 0005200056 0005200057 0005200056 0005200056 0005200056 0005200056 0005200056 0005200056 0005200057 0005200056 0005200056 0005200057 0005200056 0005200056 0005200057 0005200056 0005200057 0005200056 0005200056 0005200056 0005200057 0005200056 0005200056 0005200056 0005200057 0005200056 0005200057 0005200057 0005200056 0005200057 0005200056 0005200056 0005200057 0005200056 0005200057 0005200057 0005200056 0005200056 0005200056 0005200056 0005200056 0005200057 0005200056 0005200056 0005200056 0005200056 0005200056 0005200056 0005200057 0005200056 0005200057 0005200056 0005200057 0005200057 0005200056 0005200056 0005200056 0005200057 0005200056 0005200056 0005200056 0005200056 0005200056 0005200056 0005200057 0005200056 0005200057 0005200057 0005200056 0005200056 0005200057 0005200056 0005200057 0005200056 0005200056 0005200056 0005200057 0005200056 0005200056 0005200056 0005200057 0005200056 0005200057 0005200056 0005200057 0005200056 0005200056 0005200056 0005200056 0005200057 0005200056 0005200056 0005200056 0005200056 0005200056 0005200056 0005200057 0005200056 0005200056 0005200056 0005200057 0005200056 0005200056 0005200056 0005200057 0005200056 0005200056 0005200056 0005200057 0005200057 0005200056 0005200056 0005200057 0005200056 0005200056 0005200057 0005200056 0005200056 0005200056&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A little code to crack for if you've got nothing to do. Enjoy!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note if you cannot solve it;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's a multi-part process. It's been encrypted a few times. At first it's trivial and it slowly gets harder. Once you're at the last step it'll probably seem meaningless. There's no way to tell if you've found the last step or not.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2383" LastEditorDisplayName="user88" LastEditDate="2015-04-29T00:52:48.023" LastActivityDate="2015-04-29T00:52:48.023" Title="Code Uncrackable!" Tags="|cipher|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5586" PostTypeId="2" ParentId="5567" CreationDate="2014-11-26T06:50:20.143" Score="10" Body="&lt;p&gt;Most of the existing answers assume that you meant &lt;em&gt;&quot;&lt;strong&gt;Only one&lt;/strong&gt; of these inscriptions is true&quot;&lt;/em&gt;, rather than &lt;em&gt;&quot;One of these inscriptions is true&quot;&lt;/em&gt;, so I will go ahead and assume that you meant what you said instead, for completeness, or just in case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that case:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He should pick the &lt;strong&gt;Silver&lt;/strong&gt; chest, but he only has a 2/3 chance of this being the correct one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's see why:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We know we only have 4 possibilities:&lt;br&gt;&lt;br&gt;&#xA; 1. &lt;strong&gt;Gold is true, Silver is true&lt;/strong&gt;: Definitely possible if we assume you didn't mean &lt;em&gt;&quot;Only one...&quot;&lt;/em&gt;. In this possibility, both statements are true and therefore &lt;strong&gt;Gold&lt;/strong&gt; contains the treasure.&lt;br&gt;&lt;br&gt;&#xA; 2. &lt;strong&gt;Gold is false, Silver is false&lt;/strong&gt;: This is also possible since rendering the &quot;one of these is true&quot; statement false allows (and requires) for both to be false. The treasure chest would be &lt;strong&gt;Silver&lt;/strong&gt;&lt;br&gt;&lt;br&gt;&#xA; 3. &lt;strong&gt;Gold is true, Silver is false&lt;/strong&gt;: Since &quot;this contains death&quot; is false, the &lt;strong&gt;Silver&lt;/strong&gt; chest contains the treasure.&lt;br&gt;&lt;br&gt;&#xA; 4. &lt;strong&gt;Gold is false, Silver is true&lt;/strong&gt;: This is a contradiction and should be discarded. &lt;em&gt;&quot;One of these is true&quot;&lt;/em&gt; being false (as a statement) would mean that neither inscription is true. Since by definition we are looking for a possibility where Silver is true, this possibility contradicts itself.&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4900" LastActivityDate="2014-11-26T06:50:20.143" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5587" PostTypeId="2" ParentId="5456" CreationDate="2014-11-26T07:41:42.657" Score="3" Body="&lt;p&gt;A pretty common word with 20 characters, in British spelling:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;internationalisation&lt;/strong&gt; &#xA; &lt;br&gt;&#xA; &lt;strong&gt;I&lt;/strong&gt;odine-&lt;strong&gt;N&lt;/strong&gt;itrogen-&lt;strong&gt;Te&lt;/strong&gt;llurium-&lt;strong&gt;R&lt;/strong&gt;ado&lt;strong&gt;n&lt;/strong&gt;-&lt;strong&gt;A&lt;/strong&gt;sta&lt;strong&gt;t&lt;/strong&gt;ine-&lt;strong&gt;I&lt;/strong&gt;odine-&lt;strong&gt;O&lt;/strong&gt;xygen-&lt;strong&gt;N&lt;/strong&gt;itrogen-&lt;strong&gt;Al&lt;/strong&gt;uminum-&lt;strong&gt;I&lt;/strong&gt;odine-&lt;strong&gt;S&lt;/strong&gt;ulfur-&lt;strong&gt;A&lt;/strong&gt;sta&lt;strong&gt;t&lt;/strong&gt;ine-&lt;strong&gt;I&lt;/strong&gt;odine-&lt;strong&gt;O&lt;/strong&gt;xygen-&lt;strong&gt;N&lt;/strong&gt;itrogen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5748" LastEditorUserId="5748" LastEditDate="2014-11-30T09:17:11.930" LastActivityDate="2014-11-30T09:17:11.930" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5588" PostTypeId="1" AcceptedAnswerId="5682" CreationDate="2014-11-26T08:13:21.490" Score="35" ViewCount="2360" Body="&lt;p&gt;Determine color and direction of the missing elephants in the picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/66pdJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a textual representation of the above.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;G&amp;lt; R&amp;lt; R&amp;lt; G&amp;lt; R&amp;lt; M&amp;gt; G&amp;lt; R&amp;lt; R&amp;gt; M&amp;lt;&#xA;G&amp;lt; G&amp;lt; G&amp;gt; G&amp;lt; G&amp;gt; G&amp;lt; G&amp;lt; G&amp;lt; G&amp;gt; G&amp;gt;&#xA;G&amp;lt; M&amp;gt; M&amp;gt; G&amp;lt; G&amp;lt; ?? M&amp;gt; M&amp;gt; G&amp;lt; G&amp;lt;&#xA;Y&amp;lt; Y&amp;gt; G&amp;gt; G&amp;lt; Y&amp;lt; Y&amp;gt; Y&amp;lt; G&amp;gt; G&amp;lt; Y&amp;gt;&#xA;G&amp;lt; G&amp;lt; G&amp;gt; G&amp;gt; M&amp;lt; M&amp;gt; G&amp;gt; G&amp;gt; M&amp;lt; M&amp;lt;&#xA;R&amp;gt; M&amp;gt; ?? G&amp;lt; R&amp;gt; M&amp;lt; Y&amp;lt; Y&amp;gt; G&amp;gt; R&amp;gt;&#xA;R&amp;gt; R&amp;lt; R&amp;lt; R&amp;lt; R&amp;gt; R&amp;gt; R&amp;lt; ?? R&amp;gt; R&amp;lt;&#xA;R&amp;lt; G&amp;lt; R&amp;lt; G&amp;lt; R&amp;lt; G&amp;gt; G&amp;gt; R&amp;gt; G&amp;gt; R&amp;lt;&#xA;Y&amp;gt; Y&amp;gt; Y&amp;lt; Y&amp;lt; Y&amp;gt; Y&amp;lt; Y&amp;gt; Y&amp;lt; Y&amp;lt; Y&amp;lt;&#xA;Y&amp;gt; Y&amp;lt; Y&amp;lt; Y&amp;lt; Y&amp;lt; Y&amp;gt; Y&amp;gt; Y&amp;lt; Y&amp;gt; Y&amp;lt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;code&gt;R&lt;/code&gt;: red&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;Y&lt;/code&gt;: yellow&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;G&lt;/code&gt;: green&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;M&lt;/code&gt;: magenta&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;&amp;gt;&lt;/code&gt;: facing right&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;&amp;lt;&lt;/code&gt;: facing left&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;?&lt;/code&gt;: gap&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The elephant herd has no boundaries. The right side joins the left side and the bottom joins the top.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Elephants are social animals: they count on themselves and on their neighbours.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; With more fellows around, an elephant must see further.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eight are the neighbours. One is the self. Sight overflows. Where is the match?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Almost everything in this puzzle is random.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4394" LastEditorUserId="4394" LastEditDate="2014-11-28T07:53:49.323" LastActivityDate="2015-03-12T14:38:21.087" Title="Find the missing elephants" Tags="|pattern|" AnswerCount="3" CommentCount="20" ContentLicense="CC BY-SA 3.0" />
  <row Id="5590" PostTypeId="2" ParentId="5567" CreationDate="2014-11-26T09:32:20.737" Score="7" Body="&lt;p&gt;I'm sure I've seen something similar in one of Ray Smullyan's books... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the position is &lt;strong&gt;exactly&lt;/strong&gt; as stated, and we take the legend as true (no &quot;I don't believe in curses&quot; type solutions), the archaeologist should still not open either casket. If he does, he is making a big assumption: &lt;strong&gt;that the chest inscriptions have valid truth values&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard solution is essentially that the dragon is in the gold chest, because if he's in the silver one then the silver inscription is true, and the gold inscription can neither be true nor false. Without additional information, though, it's perfectly possible for the gold chest to have a meaningless inscription - just as there's nothing physically stopping me carving &quot;This statement is false&quot; on to the chest. Compare this with the canonical knight/knave/joker type puzzle - there we know that we only ever deal with people who make true or false statements.&lt;/p&gt;&#xA;" OwnerUserId="1801" LastActivityDate="2014-11-26T09:32:20.737" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5591" PostTypeId="1" CreationDate="2014-11-26T10:57:09.977" Score="2" ViewCount="266" Body="&lt;p&gt;You have two hour glasses: A seven minute one and an 11 minute one. Using just these two hour glasses, accurately time 15 minutes.&lt;/p&gt;&#xA;" OwnerUserId="6007" LastEditorUserId="5373" LastEditDate="2020-07-23T08:47:03.790" LastActivityDate="2020-07-23T08:47:03.790" Title="The hour glasses!" Tags="|logical-deduction|calculation-puzzle|time|" AnswerCount="2" CommentCount="2" FavoriteCount="0" ClosedDate="2014-11-26T11:41:14.070" ContentLicense="CC BY-SA 3.0" />
  <row Id="5592" PostTypeId="2" ParentId="5591" CreationDate="2014-11-26T11:01:51.603" Score="2" Body="&lt;p&gt;Start the 11 minute hourglass and the 7 minute hourglass at the same time, as soon as the 7 minute one runs out start timing, once it fully runs down run it again for the full time  you would be 15 minutes by the end of the timing&lt;/p&gt;&#xA;" OwnerUserId="2313" LastActivityDate="2014-11-26T11:01:51.603" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5593" PostTypeId="1" AcceptedAnswerId="5594" CreationDate="2014-11-26T11:37:20.147" Score="-2" ViewCount="415" Body="&lt;p&gt;Okay, here I am having a puzzle for you all. The puzzle is quite short and clear in it's own way I am telling you. Rather than a question it's more like a challenge to you all (&lt;em&gt;answer is already known&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Puzzle is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Some months have 31 days; how many have 28?&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer with appropriate explanation will be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="6007" LastActivityDate="2014-11-26T11:40:13.330" Title="The 'number of months' challenge" Tags="|logical-deduction|riddle|" AnswerCount="1" CommentCount="5" ClosedDate="2014-11-26T13:08:52.833" ContentLicense="CC BY-SA 3.0" />
  <row Id="5594" PostTypeId="2" ParentId="5593" CreationDate="2014-11-26T11:40:13.330" Score="7" Body="&lt;p&gt;Answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;12 months&lt;br /&gt;All months have at least 28 days&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastActivityDate="2014-11-26T11:40:13.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5595" PostTypeId="2" ParentId="5591" CreationDate="2014-11-26T11:41:06.097" Score="3" Body="&lt;p&gt;You can try&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. Start both hourglasses.&lt;br/&gt;&#xA; 2. When the seven minutes runs out turn it.&lt;br/&gt;&#xA; 3. When the eleven minutes runs out, turn the seven minute hour glass again. It now has 4 minutes of sand.&lt;br/&gt;&#xA; 4. When the seven minutes hourglass runs out, you will have 15 minutes.&lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The method is dependant on the accuracy of the hour glasses and your speed at turning them. Not 100% accurate.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2014-11-26T11:41:06.097" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5596" PostTypeId="1" CreationDate="2014-11-26T12:30:13.453" Score="3" ViewCount="1199" Body="&lt;p&gt;You are living in the &lt;strong&gt;22nd century&lt;/strong&gt;. The fuel prices are unimaginably high and it is very rare to get, so the Government has introduced some new laws to control fuel smuggling.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Rule 1: One cannot carry more fuel than their vehicle's tank capacity&#xA;  (fuel cannot be carried by hand or any other method. The only way to&#xA;  transfer fuel from one place to another place is by filling it in the&#xA;  vehicles tank and driving it to the destination. Then the fuel can be&#xA;  drained from the tank).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Rule 2: All vehicles have to be run on their own fuel (one cannot fill&#xA;  the tank and then push the vehicle, or pull with another vehicle to&#xA;  the destination or use alternate energy).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Rule 3: Fuel can only be sold at the Fuel Market run by the&#xA;  Government.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, you have &lt;strong&gt;300 liters&lt;/strong&gt; of petrol at your home (saved by your ancestors). You have decided to sell it at the Fuel Market. But the problem is that the Fuel Market is &lt;strong&gt;1000 km&lt;/strong&gt; away from your home. Your car has a tank capacity of &lt;strong&gt;100 liters&lt;/strong&gt; and a mileage of &lt;strong&gt;10 km/liter&lt;/strong&gt;. You know that there are storage stations run by old mafia members every &lt;strong&gt;1 km&lt;/strong&gt; and they will charge &lt;strong&gt;1 million dollars&lt;/strong&gt; for storage (you can drain as much amount of fuel from your car and store there if you want). Once you reach the Fuel Market, you will get a price of &lt;strong&gt;1 million dollars per liter&lt;/strong&gt; of petrol. What is the maximum amount of money you can make by selling the fuel at the Fuel Market and how you can achieve it?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;IMPORTANT: You have to use your car. You cannot borrow or buy another car with more tank capacity and/or mileage. Also you can abandon your car after selling the fuel if you want since your car costs only a few dollars&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: I don't know the answer. So you have to prove your answer with details to get accepted.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="8874" LastEditDate="2016-03-22T16:56:15.783" LastActivityDate="2016-03-22T16:56:15.783" Title="Sell the fuel and become a Millionaire" Tags="|mathematics|calculation-puzzle|optimization|" AnswerCount="4" CommentCount="14" ClosedDate="2014-11-27T06:30:51.780" ContentLicense="CC BY-SA 3.0" />
  <row Id="5597" PostTypeId="2" ParentId="5567" CreationDate="2014-11-26T13:05:01.310" Score="1" Body="&lt;p&gt;I hate to be '&lt;em&gt;&lt;a href=&quot;http://theimpactnews.com/wp-content/uploads/2013/11/that-guy.jpg&quot; rel=&quot;nofollow&quot;&gt;that guy&lt;/a&gt;&lt;/em&gt;' but since dragons and death curses aren't real, he can safely open either chest as long as he takes elementary precautions against explosives, booby-traps and poison.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yes, I fully appreciate that this answer sucks. Sorry.&lt;/p&gt;&#xA;" OwnerUserId="4852" LastActivityDate="2014-11-26T13:05:01.310" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5598" PostTypeId="2" ParentId="5588" CreationDate="2014-11-26T13:42:52.040" Score="3" Body="&lt;p&gt;My guess is, from top to bottom &lt;code&gt;??&lt;/code&gt; we have &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; M&gt;, G&gt;, R&amp;lt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Non-spoiler paragraph would read that I tried to match the pattern of colors vs the left/right ratio between the rows and the columns of the table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It makes the row read 5,0,0,5 for colors and 5,5 for L/R. It affects the column to go 6/4 L/R because we've already matched the 3/7 with the second elephant. It affects column colors to 2,1,3,4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To go 3/7 R/L would match the final column's 3/7 with 2,3,3,2 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Third&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All rows the same color are 6/4 L/R, so match the pattern&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is mostly guess-work, but you can see my (convoluted) work with this image:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/t6X1T.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditDate="2014-11-26T13:51:25.550" LastActivityDate="2014-11-26T13:51:25.550" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5599" PostTypeId="2" ParentId="5190" CreationDate="2014-11-26T13:47:20.303" Score="14" Body="&lt;p&gt;I decided to try a different approach: Brute force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I created a Java 8 program to brute-force the creation of a decision tree in order to find how to weight the 12 coins in &lt;strong&gt;7 measurements&lt;/strong&gt;, and it worked. Further, I could use it to prove that this is the optimum, because it said that &lt;strong&gt;with 6 there is no solution&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The program found a solution in 35 seconds on my computer. There are many solutions, most due to permutations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The program creates a tree of nodes, where each one iterates through all 4096 possibles forms of choosing the coins, but it do variable renaming to eliminate permutations, i.e, this eliminates all but a tiny part of the permutations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, for each measurement form, it simulates all the possibles combinations of light-or-heavy coins, already excluding possibilities eliminated by previous nodes in the tree. Then it checks if the generated node is able to separate different combinations, putting those in child nodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a node passes the maximum depth (i.e. the maximum number of measurements), it is considered bad and is pruned. If it results in a single possibility, no further measurements are needed. If the node created only one child node, then the coin combination for the measurement is useless and the node is considered bad. A node that has bad nodes as children is considered bad and is pruned, even if not all of its child nodes were checked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a node tries every possible combination of coins to measure and is unable to determine all the possibilities uniquely, it is considered bad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end, after encountering the solution, it is outputted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I uploaded the human readable output to &lt;strong&gt;&lt;a href=&quot;http://pastebin.com/vzj4MuU9&quot;&gt;THIS LINK&lt;/a&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also uploaded a JSON-formatted output &lt;strong&gt;&lt;a href=&quot;http://pastebin.com/TNpb8zeq&quot;&gt;IN THIS OTHER LINK&lt;/a&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I couldn't post either output here because they are too large. If you don't like the fact that the solution is provided in links, know that I don't like either, but again, it is simply too large to post here. You may just compile and run the program instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the program:&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import java.util.Arrays;&#xA;import java.util.Set;&#xA;import java.util.TreeSet;&#xA;&#xA;/**&#xA; * @author Victor&#xA; */&#xA;public class BruteForce {&#xA;    // Change this to false to output in pretty-print mode.&#xA;    private static final boolean COMPACT = true;&#xA;&#xA;    private final int coins;&#xA;    private final int maxDepth;&#xA;    private final int possibilities;&#xA;&#xA;    public final class Node {&#xA;        private final boolean[] combs = new boolean[possibilities];&#xA;        private final boolean[] shouldWeight = new boolean[possibilities];&#xA;        private Node[] children = null;&#xA;        private int counts = 0;&#xA;        private final int level;&#xA;        private int lastCount = -1;&#xA;        private final Node parent;&#xA;        private Set&amp;lt;Integer&amp;gt; coinRenames;&#xA;        private int whichCoinsToWeight = -1;&#xA;        private int tag; // For writing it in some human readable format. Not used for creating and processing the tree.&#xA;&#xA;        public Node() {&#xA;            this(0, null);&#xA;        }&#xA;&#xA;        private Node(int level, Node parent) {&#xA;            this.parent = parent;&#xA;            this.level = level;&#xA;            if (parent != null) return;&#xA;            for (int i = 0; i &amp;lt; possibilities; i++) {&#xA;                count(i);&#xA;            }&#xA;        }&#xA;&#xA;        public void count(int c) {&#xA;            combs[c] = true;&#xA;            counts++;&#xA;            lastCount = c;&#xA;            if (coinRenames == null) coinRenames = new TreeSet&amp;lt;&amp;gt;();&#xA;            int renamed = renameCoins(c);&#xA;&#xA;            // Not really important, but if you want a solution in a few seconds instead of&#xA;            // waiting for some hundreds of years, keep this check here.&#xA;            if (coinRenames.add(renamed)) {&#xA;                shouldWeight[c] = true;&#xA;            }&#xA;        }&#xA;&#xA;        // Convert an int to an array of boolean.&#xA;        private boolean[] asArray(int coinSet) {&#xA;            boolean[] array = new boolean[coins];&#xA;            for (int i = 0; i &amp;lt; coins; i++) {&#xA;                 array[i] = ((coinSet &amp;gt;&amp;gt; i) &amp;amp; 1) != 0;&#xA;            }&#xA;            return array;&#xA;        }&#xA;&#xA;        private void summarizeCoinsUsedForWeighting(boolean[][] weightingCoinsPerLevel) {&#xA;            weightingCoinsPerLevel[level] = asArray(whichCoinsToWeight);&#xA;            if (parent != null) parent.summarizeCoinsUsedForWeighting(weightingCoinsPerLevel);&#xA;        }&#xA;&#xA;        private int renameCoins(int thisLevelCoins) {&#xA;            // This arrays represent, for each level of the tree that is an ancestor to this node&#xA;            // the set of coins used for weighting, where each int corresponds to a sequence of&#xA;            // 0 and 1 bits representing which coins are being used.&#xA;            boolean[][] weightingCoinsPerLevel = new boolean[level + 1][coins];&#xA;            if (parent != null) parent.summarizeCoinsUsedForWeighting(weightingCoinsPerLevel);&#xA;            weightingCoinsPerLevel[level] = asArray(thisLevelCoins);&#xA;&#xA;            // Transpose the array, so we have a new array, where for each coin, we have an int corresponding&#xA;            // to a sequence of 0 and 1 bits where the 1 bits are the levels in which the coin is used for weighting.&#xA;            // But since we do want that the levels near the root of the tree are represented by the most&#xA;            // significative bits, we reverse the bit order of the levels.&#xA;            int[] usingLevelsPerCoin = new int[coins];&#xA;            for (int i = 0; i &amp;lt;= level; i++) {&#xA;                for (int j = 0; j &amp;lt; coins; j++) {&#xA;                    if (weightingCoinsPerLevel[i][j]) usingLevelsPerCoin[j] |= 1 &amp;lt;&amp;lt; (level - i);&#xA;                }&#xA;            }&#xA;&#xA;            // Rename the coins to ensure that the most firstly-used coins are the first ones.&#xA;            // Coin-renaming serves to avoid recalculating combinations that are simply permutations to some&#xA;            // previously calculated position. Further, this coin renaming provides a single canonical representation&#xA;            // for each group of equivalent permutations. To ensure that, we simply sort the array, so the most used&#xA;            // coins with higher bits sets will be the LAST ones, I.E, the resulting array would still be in the reverse order.&#xA;            Arrays.sort(usingLevelsPerCoin);&#xA;&#xA;            // We could untranspose the sorted array, so we have again coins per level, but with the coins are renamed.&#xA;            // However, now we are interested only in the last level, so we just get the last bit from each row on the array.&#xA;            // Further, we reverse the order of the bits that we get because the array is sorted in reverse order. This way it&#xA;            // will be in the right order again.&#xA;            int sortedCoinsOnLastLevel = 0;&#xA;            for (int j = 0; j &amp;lt; coins; j++) {&#xA;                sortedCoinsOnLastLevel |= (usingLevelsPerCoin[j] &amp;amp; 1) &amp;lt;&amp;lt; (coins - j - 1);&#xA;            }&#xA;&#xA;            return sortedCoinsOnLastLevel;&#xA;        }&#xA;&#xA;        public boolean separate() {&#xA;            if (counts == 1) return true; // Just one possibility, it is in the lastCount variable.&#xA;&#xA;            // If is recursing too deply and there is still at least 2 coins left in this possibility,&#xA;            // so there was a bad choice somewhere up in the decision tree.&#xA;            if (level &amp;gt;= maxDepth) return false;&#xA;&#xA;            // The coins choosen to weight are the 1 bits. Start counting with 1 because there is no sense to weight zero coins.&#xA;            // This means that it will choose every possible combination of coins to weight, until it founds a good one or resigns.&#xA;            // Save it in an instance field, so it can be remembered when the solution is found.&#xA;            a: for (whichCoinsToWeight = 1; whichCoinsToWeight &amp;lt; possibilities; whichCoinsToWeight++) {&#xA;&#xA;                // If this is just a permutation of some previously tested combination, just skip it.&#xA;                if (!shouldWeight[whichCoinsToWeight]) continue;&#xA;&#xA;                children = null;&#xA;                int childCount = 0;&#xA;&#xA;                // Iterate every possible valuation for the coins.&#xA;                b: for (int selectedCoinsCombination = 0; selectedCoinsCombination &amp;lt; possibilities; selectedCoinsCombination++) {&#xA;&#xA;                    // If the choosen valuation never happens in this place, skip it.&#xA;                    if (!combs[selectedCoinsCombination]) continue;&#xA;&#xA;                    // Weight the coins. This works by setting 0 to coins which&#xA;                    // are not being weighted, leaving 0 for light coins and 1 for heavy ones.&#xA;                    int bits = Integer.bitCount(whichCoinsToWeight &amp;amp; selectedCoinsCombination);&#xA;&#xA;                    // Now start to populate the children. If it does not have any, create the array.&#xA;                    if (children == null) children = new Node[coins + 1];&#xA;&#xA;                    // Choose the appropriate children node in the decision tree.&#xA;                    Node child = children[bits];&#xA;&#xA;                    // If the selected node still do not exist, create it.&#xA;                    if (child == null) {&#xA;                        child = new Node(level + 1, this);&#xA;                        children[bits] = child;&#xA;                        childCount++;&#xA;                    }&#xA;&#xA;                    // Add the selected coins combination as possible to the child node.&#xA;                    child.count(selectedCoinsCombination);&#xA;                }&#xA;&#xA;                // If we have just one child, then the selected coins to weight did not gave any information,&#xA;                // so discard this choice and proceed to the next one.&#xA;                if (childCount &amp;lt;= 1) continue;&#xA;&#xA;                // Now, perform the separation recursively in each children.&#xA;                // If some of them fails, then the selected coins to weight combination is useless and should&#xA;                // be discarded, even if just some of the nodes were tested.&#xA;                for (Node n : children.clone()) {&#xA;                    if (n == null) continue;&#xA;                    boolean r = n.separate();&#xA;                    if (!r) continue a;&#xA;                }&#xA;&#xA;                // Yeah, found some useful combination!&#xA;                return true;&#xA;            }&#xA;&#xA;            // We tried every combination of coins to weight and none of them was useful.&#xA;            return false;&#xA;        }&#xA;&#xA;        // Output a JSON!&#xA;        public void reportResult(StringBuilder sb) {&#xA;            boolean first = true;&#xA;            sb.append(&quot;{&quot;);&#xA;            if (counts == 1) {&#xA;                sb.append(&quot;\&quot;HEAVY\&quot;: [&quot;);&#xA;                for (int i = 0; i &amp;lt; coins; i++) {&#xA;                    if (((lastCount &amp;gt;&amp;gt; i) &amp;amp; 1) != 0) {&#xA;                        if (first) {&#xA;                            first = false;&#xA;                        } else {&#xA;                            sb.append(&quot;, &quot;);&#xA;                        }&#xA;                        sb.append(i + 1);&#xA;                    }&#xA;                }&#xA;                sb.append(&quot;], \&quot;LIGHT\&quot;: [&quot;);&#xA;                first = true;&#xA;                for (int i = 0; i &amp;lt; coins; i++) {&#xA;                    if (((lastCount &amp;gt;&amp;gt; i) &amp;amp; 1) == 0) {&#xA;                        if (first) {&#xA;                            first = false;&#xA;                        } else {&#xA;                            sb.append(&quot;, &quot;);&#xA;                        }&#xA;                        sb.append(i + 1);&#xA;                    }&#xA;                }&#xA;                sb.append(&quot;]}&quot;);&#xA;                return;&#xA;            }&#xA;            if (!COMPACT) {&#xA;                sb.append(&quot;\n&quot;);&#xA;                ident(level + 1, sb);&#xA;            }&#xA;            sb.append(&quot;\&quot;TO WEIGHT\&quot;: [&quot;);&#xA;            for (int i = 0; i &amp;lt; coins; i++) {&#xA;                if (((whichCoinsToWeight &amp;gt;&amp;gt; i) &amp;amp; 1) != 0) {&#xA;                    if (first) {&#xA;                        first = false;&#xA;                    } else {&#xA;                        sb.append(&quot;, &quot;);&#xA;                    }&#xA;                    sb.append(i + 1);&#xA;                }&#xA;            }&#xA;            sb.append(&quot;]&quot;);&#xA;            for (int i = 0; i &amp;lt;= coins; i++) {&#xA;                if (children[i] == null) continue;&#xA;                sb.append(&quot;,\n&quot;);&#xA;                ident(level + 1, sb);&#xA;                Integer weighting = Integer.bitCount(whichCoinsToWeight);&#xA;                sb.append(&quot;\&quot;&quot;).append((i + weighting) * 10).append(&quot; grams\&quot;: &quot;);&#xA;                children[i].reportResult(sb);&#xA;            }&#xA;            if (!COMPACT) {&#xA;                sb.append(&quot;\n&quot;);&#xA;                ident(level, sb);&#xA;            }&#xA;            sb.append(&quot;}&quot;);&#xA;        }&#xA;&#xA;        private void ident(int level, StringBuilder sb) {&#xA;            for (int i = 0; i &amp;lt; level; i++) {&#xA;                sb.append(COMPACT ? &quot; &quot; : &quot;  &quot;);&#xA;            }&#xA;        }&#xA;&#xA;        // Output a human readable algorithm!&#xA;        public void reportAlgorithm(StringBuilder sb) {&#xA;            this.tag = 1;&#xA;            for (int i = 0; i &amp;lt; maxDepth; i++) {&#xA;                reportAlgorithm(new int[maxDepth], i, sb);&#xA;            }&#xA;        }&#xA;&#xA;        private void reportAlgorithmDone(StringBuilder sb) {&#xA;            if (counts != 1) return;&#xA;            boolean first = true;&#xA;            sb.append(&quot;Heavies = [&quot;);&#xA;            for (int i = 0; i &amp;lt; coins; i++) {&#xA;                if (((lastCount &amp;gt;&amp;gt; i) &amp;amp; 1) != 0) {&#xA;                    if (first) {&#xA;                        first = false;&#xA;                    } else {&#xA;                        sb.append(&quot;, &quot;);&#xA;                    }&#xA;                    sb.append(i + 1);&#xA;                }&#xA;            }&#xA;            sb.append(&quot;]; Lights = [&quot;);&#xA;            first = true;&#xA;            for (int i = 0; i &amp;lt; coins; i++) {&#xA;                if (((lastCount &amp;gt;&amp;gt; i) &amp;amp; 1) == 0) {&#xA;                    if (first) {&#xA;                        first = false;&#xA;                    } else {&#xA;                        sb.append(&quot;, &quot;);&#xA;                    }&#xA;                    sb.append(i + 1);&#xA;                }&#xA;            }&#xA;            sb.append(&quot;].\n&quot;);&#xA;        }&#xA;&#xA;        private void reportAlgorithm(int[] levelCounters, int desiredLevel, StringBuilder sb) {&#xA;            boolean first = true;&#xA;            if (counts == 1) {&#xA;                reportAlgorithmDone(sb);&#xA;                return;&#xA;            }&#xA;            if (level &amp;lt; desiredLevel) {&#xA;                for (int i = 0; i &amp;lt;= coins; i++) {&#xA;                    if (children[i] != null &amp;amp;&amp;amp; children[i].counts != 1) children[i].reportAlgorithm(levelCounters, desiredLevel, sb);&#xA;                }&#xA;                return;&#xA;            }&#xA;            sb.append(level + 1);&#xA;            if (level &amp;gt; 0) sb.append(&quot;.&quot;).append(tag);&#xA;            sb.append(&quot;. Weight the coins [&quot;);&#xA;            for (int i = 0; i &amp;lt; coins; i++) {&#xA;                if (((whichCoinsToWeight &amp;gt;&amp;gt; i) &amp;amp; 1) != 0) {&#xA;                    if (first) {&#xA;                        first = false;&#xA;                    } else {&#xA;                        sb.append(&quot;, &quot;);&#xA;                    }&#xA;                    sb.append(i + 1);&#xA;                }&#xA;            }&#xA;            sb.append(&quot;]\n&quot;);&#xA;            for (int i = 0; i &amp;lt;= coins; i++) {&#xA;                if (children[i] == null) continue;&#xA;                Integer weighting = Integer.bitCount(whichCoinsToWeight);&#xA;                sb.append(&quot;  &quot;).append((i + weighting) * 10).append(&quot; grams: &quot;);&#xA;                if (children[i].counts == 1) {&#xA;                    children[i].reportAlgorithmDone(sb);&#xA;                } else {&#xA;                    children[i].tag = ++levelCounters[level + 1];&#xA;                    sb.append(&quot;Go to &quot;).append(level + 2).append(&quot;.&quot;).append(children[i].tag).append(&quot;.\n&quot;);&#xA;                }&#xA;            }&#xA;            sb.append(&quot;\n&quot;);&#xA;        }&#xA;    }&#xA;&#xA;    public BruteForce(int coins, int maxDepth) {&#xA;        this.coins = coins;&#xA;        this.maxDepth = maxDepth;&#xA;        this.possibilities = 1 &amp;lt;&amp;lt; coins;&#xA;    }&#xA;&#xA;    public void run() {&#xA;        Node n = new Node();&#xA;        boolean ok = n.separate();&#xA;        if (!ok) {&#xA;            System.out.println(&quot;Impossible&quot;);&#xA;            return;&#xA;        }&#xA;&#xA;        // Uncomment this to output as JSON.&#xA;        //StringBuilder sb = new StringBuilder(1024 * 1024);&#xA;        //n.reportResult(sb);&#xA;        //System.out.println(sb.toString());&#xA;&#xA;        StringBuilder sb2 = new StringBuilder(1024 * 1024);&#xA;        n.reportAlgorithm(sb2);&#xA;        System.out.println(sb2.toString());&#xA;    }&#xA;&#xA;    public static void main(String[] args) {&#xA;        int coins = 12; // You may change the number of coins if you want.&#xA;        int measures = 7; // You may change the number of measures, if you want.&#xA;        BruteForce b = new BruteForce(coins, measures);&#xA;        b.run();&#xA;    }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;I am sorry that this site does not have syntax-coloring enabled.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may modify the program as you wish, specially to brute-force different combinations of number of coins and number of measurements. This is cool because you may also use this program to solve any number of desired coins and measurements (as long as you have enough time and memory).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you change the &lt;code&gt;int measures = 7;&lt;/code&gt; near the end of the program to 6, it will output that it is impossible, proving that 7 are indeed the minimum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may uncomment the code in the &lt;code&gt;run()&lt;/code&gt; method to get a compact JSON output. You may also change the value of the &lt;code&gt;COMPACT&lt;/code&gt; variable in the beginning to &lt;code&gt;false&lt;/code&gt; and then the JSON output will be a better looking one, but will be still larger. You may get that JSON and use it to work with something else. Or you may simple keep just the human readable output.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may argue that this is a very inelegant, ugly and stupid way of solving this problem, as the output of the program is basically a bunch of random-looking arbitrary measurements without any clear, simple or intuitive pattern. I agree, but nevertheless, the problem is solved: Just follow the measurements show in the output and you will find the weight for any combination of 10 or 20 grams coins in only 7 measurements.&lt;/p&gt;&#xA;" OwnerUserId="5044" LastEditorUserId="5044" LastEditDate="2014-11-26T17:43:39.247" LastActivityDate="2014-11-26T17:43:39.247" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="5600" PostTypeId="2" ParentId="5596" CreationDate="2014-11-26T14:07:40.153" Score="0" Body="&lt;p&gt;&lt;strong&gt;Edited&lt;/strong&gt;:&lt;br&gt;&#xA;52.6 million, if the mafia will take 1 liter of fuel instead of 1 million&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Go out to 196km, pay 1 liter, drop off 59.8, drive back. Repeat. Go back to 196km. At this point you have 200 liters. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Go out 526km (+330km), pay 1 liter, drop off 33, drive back. Go out 526km, fill up, drive to depot and sell the 52.6 liters.&lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2014-11-26T16:05:29.413" LastActivityDate="2014-11-26T16:05:29.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5602" PostTypeId="2" ParentId="5596" CreationDate="2014-11-26T15:17:07.947" Score="3" Body="&lt;p&gt;1 million dollars is such a small storage charge that the solution is identical to the camel and bananas puzzle. See &lt;a href=&quot;https://math.stackexchange.com/questions/53526/how-many-bananas-can-a-camel-deliver-without-eating-them-all&quot;&gt;here&lt;/a&gt;. Use storage areas at the 200 km and 534 km points. We arrive with 53.3 liters of fuel. Since the problem uses integer kilometer distances only, hence multiples of .1 liters only, this is the closest we can get to the camel and bananas puzzle's 53 1/3 bananas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we do better using fewer storage steps? Well, we only used 3 storage steps in this answer. Good luck using 0 storage steps. If you use 1 storage step, then it is straightforward to prove that you can't get anywhere near 53 liters. (Casework on the storage's position.) 2 storage steps works out the same way. It's intuitively true, but it's a messy argument, so I'll avoid putting it here for both of our sakes. In conclusion, the camel and bananas answer is the best we can do. Hey, it gets us a cool 50.3 million dollars.&lt;/p&gt;&#xA;" OwnerUserId="1752" LastEditorUserId="-1" LastEditDate="2017-04-13T12:19:36.950" LastActivityDate="2014-11-26T15:36:02.213" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5603" PostTypeId="1" AcceptedAnswerId="5613" CreationDate="2014-11-26T15:19:39.247" Score="23" ViewCount="7136" Body="&lt;p&gt;The answer to this riddle is a single word. It should be slightly harder than my simultaneously posted &lt;a href=&quot;https://puzzling.stackexchange.com/questions/5604/riddle-with-a-twist&quot;&gt;Riddle with a twist&lt;/a&gt;, but still fairly easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;From the middle of the sun through all the rain, &lt;br&gt;&#xA;Colours stretch from money to money. &lt;br&gt;&#xA;Between the end of life and the start of death, &lt;br&gt;&#xA;The first part of me resides.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Now you need to pick sides, &lt;br&gt;&#xA;But keep away from the twin snakes. &lt;br&gt;&#xA;I am the weapon the other side lacks, &lt;br&gt;&#xA;And by now you should know what I am.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2014-11-26T22:18:39.370" Title="Beware - this riddle might kill you!" Tags="|riddle|word|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5604" PostTypeId="1" AcceptedAnswerId="5607" CreationDate="2014-11-26T15:20:50.777" Score="11" ViewCount="3945" Body="&lt;p&gt;This is a riddle with a twist. Almost all the lines will point you in the right direction, but hidden somewhere in the riddle is a red herring. If you find the right solution, I'll accept your answer, but if you find the red herring &lt;em&gt;and&lt;/em&gt; the right solution, I'll upvote and accept. Don't be afraid; the puzzle is still really easy. Good luck!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;From the middle of the sun through all the rain, &lt;br&gt;&#xA;Colours stretch from money to money. &lt;br&gt;&#xA;The very first part of this, &lt;br&gt;&#xA;Gives you my first word.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Not him, not it, perhaps his wife? &lt;br&gt;&#xA;Or half of either of his parents? &lt;br&gt;&#xA;Add to that the precious thing, &lt;br&gt;&#xA;Not the three, the seven, or the nine, &lt;br&gt;&#xA;To get at last my second word.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who or what am I?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2014-11-27T16:19:24.140" LastActivityDate="2014-11-27T16:19:24.140" Title="Riddle with a twist" Tags="|riddle|word|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5605" PostTypeId="2" ParentId="5596" CreationDate="2014-11-26T15:28:35.127" Score="1" Body="&lt;p&gt;I did 4 sets of analysis. On the way, i realised that whichever way he takes, he has to have approximately 100 litres, no matter how many intervals he plan. If he will have lesser than 100 litres on the next interval trip, he is better off driving to the end and give up the extra fuels on the previous interval trip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;FIRST SET&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first set is in intervals of 100km. He brings first 100 litre to first checkpoint, puts down 80 and drive back. Bring 2nd 100 litre, puts 80 and drive back. Take last set of 100 litre and drive to 1st checkpoint.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Mafia Storage: 2 times&#xA;Remaining Fuel: 250 litres&#xA;Distance Covered: 100 km&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;He repeats again, bringing first 100 litres to 2nd checkpoint, puts 80 down and drive back. Takes 2nd set, puts 80 down and drive back. Takes last 50 litre and drive to 2nd checkpoint.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Mafia Storage: 4 times&#xA;Remaining Fuel: 200 Litres&#xA;Distance Covered: 200 km&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;He repeats the process over again and he will arrive at 3rd checkpoint with:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Mafia Storage: 5 times&#xA;Remaining Fuel: 170 Litres&#xA;Distance Covered: 300 km&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;He repeats the 4th time:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Mafia Storage: 6 times&#xA;Remaining Fuel: 140 Litres&#xA;Distance Covered: 400 km&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;5th time:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Mafia Storage: 7 times&#xA;Remaining Fuel: 110 Litres&#xA;Distance Covered: 500 km&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;From here, he doesn't have to go back and collect the 10 litres as he can drive to the end and would have made:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;100 - 50 - 7 = 43 Mil&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Instead of:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;80 - 40 - 8 = 32 Mil&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;SECOND SET&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 2nd set is in intervals of 200 km. He brings the first 100 litre to the first checkpoint and he will be left with 80. Puts down 60, drive back with remaining 20. Takes next 100 litre and repeats. The data will be:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Mafia Storage: 2 times&#xA;Remaining Fuel: 200 litres&#xA;Distance Covered: 200km&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;He repeats again for the 2nd time to 2nd checkpoint, bringing 100 litres. Puts 60 down, drive back. Takes last set of 100 and drive back to 2nd checkpoint.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Mafia Storage: 3 times&#xA;Remaining Fuel: 140 litres&#xA;Distance Covered: 400km&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;From here on, he can take the 100 litres and continue driving to the end. He would have made:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;100 - 60 - 3 = 37 Mil&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THIRD SET&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 3rd analysis is in intervals of 300 km. He brings first 100 litre to first checkpoint, puts down 40 and drive back. He brings 100 again, puts down 40 and drive back. He takes last set of 100 and drive straight to 1st checkpoint.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Mafia Storage: 2 times&#xA;Remaining Fuel: 150 litres&#xA;Distance Covered: 300 km&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;He can now drive straight to the end and he would have made:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;100 - 70 - 2 = 28 Mil&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Instead of:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;60 - 40 - 3 = 17 Mil&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;FOURTH SET&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last analysis will be intervals of 400km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starts off taking 100 to 1st checkpoint, puts 20 down and drive back. Takes 100 and drive to 1st checkpoint, drop 20 and drive back. Take last set of 100 and drive to 1st checkpoint.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Mafia Storage: 2 times&#xA;Remaining Fuel: 100 km&#xA;Distance Covered: 400 km&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;From here on, he can drive directly to the end because he can contain all his fuel, therefore he would have made:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;100 - 60 - 2 = 38 Mil&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Through all these data, we can say that the first set is at the top with 43 mil followed by 4th set with 38 mil. I have tested a few more sets privately to see if lower interval trips are better, but it seems like the storage is killing the bank (intervals of 50 and 1). Therefore, i am confident to say that the first set will earn him the most.&lt;/p&gt;&#xA;" OwnerUserId="4338" LastActivityDate="2014-11-26T15:28:35.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5606" PostTypeId="2" ParentId="5604" CreationDate="2014-11-26T15:31:06.523" Score="3" Body="&lt;p&gt;My guess is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The one ring&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Breakdown:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; From the middle of the sun through all the rain, - &lt;strong&gt;Middle&lt;/strong&gt;earth reference&lt;br/&gt;Colours stretch from money to money. - actually, not the red herring. Reference to different sorts of rings&lt;br/&gt;The very first part of this, - &lt;strong&gt;the&lt;/strong&gt; is the very first part&lt;br/&gt;Gives you my first word.&lt;br/&gt;&lt;br/&gt;Not him, not it, perhaps his wife?&lt;br/&gt;Or half of either of his parents?&lt;br/&gt;Add to that the precious thing, - &lt;strong&gt;precious&lt;/strong&gt;&lt;br/&gt;Not the three, the seven, or the nine, - &lt;strong&gt;more LOTR&lt;/strong&gt;&lt;br/&gt;To get at last my second word.&lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the herring is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Not him, not it, perhaps his wife?&lt;br/&gt;Or half of either of his parents? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="5610" LastEditDate="2014-11-26T15:54:05.770" LastActivityDate="2014-11-26T15:54:05.770" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5607" PostTypeId="2" ParentId="5604" CreationDate="2014-11-26T15:31:46.133" Score="13" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a red herring&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;From the middle of the sun through all the rain,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The red herring line; doesn't affect the answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Colours stretch from money to money. &#xA;The very first part of this, &#xA;Gives you my first word.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;This&quot; refers to the previous line itself. The very first part of the line is &quot;colours&quot;, which, when combined with the rest of the answer, naturally yields RED.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or, alternatively, thanks to @dmg:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Could refer to a rainbow. &quot;Money to money&quot; meaning the pot of gold at the end of a rainbow. And the first part of the rainbow is red.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Not him, not it, perhaps his wife?&#xA;Or half of either of his parents?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HER (Not him or it, half of motHER or fatHER)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Add to that the precious thing, &#xA;Not the three, the seven, or the nine, &#xA;To get at last my second word.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RING (Lord of the Rings reference)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4518" LastEditorUserId="4518" LastEditDate="2014-11-26T15:59:18.300" LastActivityDate="2014-11-26T15:59:18.300" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="5608" PostTypeId="2" ParentId="5603" CreationDate="2014-11-26T16:15:56.083" Score="8" Body="&lt;p&gt;From the middle of the sun through all the rain, &#xA;Colours stretch from money to money. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Refers to a rainbow going from pot of gold to pot of gold.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Between the end of life and the start of death, &#xA;The first part of me resides.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; end of lif&lt;strong&gt;e&lt;/strong&gt;. start of &lt;strong&gt;d&lt;/strong&gt;eath. first part of me (&lt;strong&gt;r&lt;/strong&gt;ainbow) resides. We have collected R,E and D. RED.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now you need to pick sides, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SIDES&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But keep away from the twin snakes. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Remove SS = IDE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the weapon the other side lacks,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; L &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And by now you should know what I am.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We take the letters we have accumulated: RED IDE L&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RIDDLE :D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the title:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this riddle might kill you!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LYNCH MOB!!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-11-26T22:18:39.370" LastActivityDate="2014-11-26T22:18:39.370" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="5609" PostTypeId="2" ParentId="5603" CreationDate="2014-11-26T16:20:57.033" Score="6" Body="&lt;p&gt;I think the answer may be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lightning&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Beware - this riddle might kill you!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lightning can kill people.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the middle of the sun through all the rain, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Light comes from the sun.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Colours stretch from money to money. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This refers to a rainbow, with pots of gold on each end.  Light can be broken up into the spectrum of colors in a rainbow.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Between the end of life and the start of death, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People supposed see a white light as they die.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first part of me resides.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That means the first part of the word is &quot;light&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now you need to pick sides, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One side is the sky, the other side is the ground.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But keep away from the twin snakes. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm not sure about this, each &quot;n&quot; in lightning could be considered a snake.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the weapon the other side lacks, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lightning only comes from the sky, not the ground.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2014-11-26T16:20:57.033" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5610" PostTypeId="2" ParentId="5567" CreationDate="2014-11-26T16:24:25.577" Score="3" Body="&lt;p&gt;It's most likely the...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gold Chest&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And here's why:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let's pretend for a while that I'm the person who made the treasure chests and was responsible for storing the treasure in them. (As well as making sure nobody made off with them.)&#xA;&lt;br&gt;&lt;br&gt;&#xA; A chest allows me to move treasure from point A to point B and provides a reasonable protection from external weather conditions, it does not however, provide a decent protection from thieves, raiders and other varieties of hoodlums.&#xA;&lt;br&gt;&lt;br&gt;&#xA; So what do I do?&#xA;&lt;br&gt;&lt;br&gt;&#xA; I make another chest, this time plated with silver and rig it with a deathtrap inside.&#xA;&lt;br&gt;&lt;br&gt;&#xA; Ok, what next? If I make a scary story about a curse residing in these chests, that should put off at least a good deal of people wanting to get their clutches on this treasure. But what about the people who do not believe in mystical curses?&#xA;&lt;br&gt;&lt;br&gt;&#xA; An idea springs into the mind: &quot;Their rational way of thought will be their undoing.&quot; I will engrave inscriptions upon both of the chests portraying a conundrum. But what would initially seem to be a logical puzzle is actually a deception leading to their demise. A solution of which will not lead to the chest containing the treasure but instead to the chest with the deathtrap.&#xA;&lt;br&gt;&lt;br&gt;&#xA; For I have been given the task of protecting these chests and I am aware that the gold chest contains the treasure, and all staff who need access to the treasure know that the gold chest is where the treasure lies.&#xA;&lt;br&gt;&lt;br&gt;&#xA; For I have no reason to hand over the treasure to just about anyone who can solve a puzzle just as how the fair people of the future will not be able to solve a logical puzzle to access an account of a complete stranger.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some might argue that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - What if thieves anticipate such a cunning way and decide to pick the gold chest?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To which I answer: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Kudos to you, sir thief, you have bested my guards as you have bested my wit and are worthy of the treasure inside.&#xA;&lt;br&gt;&lt;br&gt;&#xA; For if I switch the chests around, my initial ploy becomes void and the measures taken described above would become ineffective.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think this fits quite nicely with the &lt;a href=&quot;/questions/tagged/lateral-thinking&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged 'lateral-thinking'&quot; rel=&quot;tag&quot;&gt;lateral-thinking&lt;/a&gt; tag.&lt;/p&gt;&#xA;" OwnerUserId="6036" LastEditorUserId="6036" LastEditDate="2014-11-26T16:36:23.930" LastActivityDate="2014-11-26T16:36:23.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5611" PostTypeId="2" ParentId="5508" CreationDate="2014-11-26T18:11:44.777" Score="2" Body="&lt;p&gt;1 Alkali Metals and Alkali Earth Metals + Hydrogen:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Barabara [BaRaBaRa] (8)&lt;br&gt;&#xA; Caracara [CaRaCaRa] (8)&lt;br&gt;&#xA; Banana [BaNaNa] (6)&lt;br&gt;&#xA; Cabana [CaBaNa] (6)&lt;br&gt;&#xA; Linacs [LiNaCs] (6)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2 Transition Metals:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Crostini [CrOsTiNi] (8)&lt;br&gt;&#xA; Redbirds [ReDbIrDs] (8)&lt;br&gt;&#xA; Cryptos [CrYPtOs] (7)&lt;br&gt;&#xA; Viragos [VIrAgOs] (7)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3 Groups 3 + 4 (Groups topped with Boron / Carbon):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Albinal [AlBInAl] (7)&lt;br&gt;&#xA; Algesic [AlGeSiC] (7)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4 Groups 5 + 6 (Groups topped with Nitrogen / Oxygen):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nonsenses [NONSeNSeS] (9)&lt;br&gt;&#xA; Possesses [POSSeSSeS] (9)&lt;br&gt;&#xA; Teaspoons [TeAsPoONS] (9)&lt;br&gt;&#xA; Assesses [AsSeSSeS] (8)&lt;br&gt;&#xA; Nonsense [NONSeNSe] (8)&lt;br&gt;&#xA; Osteoses [OSTeOSeS] (8)&lt;br&gt;&#xA; Sponsons [SPONSONS] (8)&lt;br&gt;&#xA; Stenoses [STeNOSeS] (8)&lt;br&gt;&#xA; Stepsons [STePSONS] (8)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5 The Halogens and Noble Gases:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hexene [HeXeNe] (6)&lt;br&gt;&#xA; Neinei [NeINeI] (6)&lt;br&gt;&#xA; Briar [BrIAr] (5)&lt;br&gt;&#xA; Brine [BrINe] (5)&lt;br&gt;&#xA; Cliff [ClIFF] (5)&lt;br&gt;&#xA; Cline [ClINe] (5)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6 The Lanthanides:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Landlady [LaNdLaDy] (8)&lt;br&gt;&#xA; Lander [LaNdEr] (6)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;7 The Actinides:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Amesaces [AmEsAcEs] (8)&lt;br&gt;&#xA; Estheses [EsThEsEs] (8)&lt;br&gt;&#xA; Noeses [NoEsEs] (6)&lt;br&gt;&#xA; Theses [ThEsEs] (6)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5550" LastActivityDate="2014-11-26T18:11:44.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5612" PostTypeId="2" ParentId="5603" CreationDate="2014-11-26T18:28:03.810" Score="2" Body="&lt;p&gt;My guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Idea&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the middle of the sun through all the rain, &#xA;Colours stretch from money to money.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Red herring (like in his other riddle)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Between the end of life and the start of death, &#xA;The first part of me resides.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Part of the word is comprised of letters at the end of the word &quot;life&quot;, and/or the beginning of the word &quot;death&quot;, in this case &quot;dea&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now you need to pick sides, &#xA;But keep away from the twin snakes. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Take the word &quot;sides&quot;, and remove the &quot;snakes&quot; to get &quot;ide&quot; (thanks to d'alar'cop for this one)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the weapon the other side lacks, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The other side here is our &quot;dea&quot; from earlier, literally lacking the letter &quot;i&quot; from &quot;ide&quot;.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And by now you should know what I am.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Idea?!&lt;br&gt;&#xA; Not sure how the title of this riddle relates though&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6040" LastActivityDate="2014-11-26T18:28:03.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5613" PostTypeId="2" ParentId="5603" CreationDate="2014-11-26T19:22:26.517" Score="18" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; cyanide&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the middle of the sun through all the rain,&#xA;Colours stretch from money to money.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Refers to the rainbow, as has been previously noted.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Between the end of life and the start of death,&#xA;The first part of me resides.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The color cyan lies between green (life) and blue (death) in the rainbow's spectrum.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now you need to pick sides,&#xA;But keep away from the twin snakes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As others have noted, this refers to stripping the two Ss from &quot;sides&quot;, leaving &quot;ide&quot;.  Cyan + ide = cyanide&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the weapon the other side lacks,&#xA;And by now you should know what I am.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hydrogen cyanide has been used as a chemical weapon, which is why the riddle might kill you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2162" LastEditorUserId="2162" LastEditDate="2014-11-26T19:39:14.827" LastActivityDate="2014-11-26T19:39:14.827" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5614" PostTypeId="2" ParentId="5585" CreationDate="2014-11-26T19:43:02.307" Score="7" Body="&lt;p&gt;Partial answer. I am working on it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First step:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Replace &lt;code&gt;0005200057&lt;/code&gt; as 1 and &lt;code&gt;0005200056&lt;/code&gt; as 0.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second step:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Group octets of zeros and ones from the first step to form ASCII characters. This is the result:&lt;br&gt;&#xA; &lt;br&gt;&#xA; &lt;strong&gt;TFIL E WCAT HOLN UQG/VCDL HHB SCFH JRHM IDZX V YDT DFH&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Third step:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Interpreting that as morse code, where consonants are dashes and vowels are dots, and the slash is a space, produces this:&lt;br&gt;&#xA; &lt;br&gt;&#xA; &lt;strong&gt;FTFLD HSHHBESS&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A few thoughts:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It is not a Caesar cypher. Could be a Vigenere or something more complicated.&lt;/li&gt;&#xA;&lt;li&gt;It is pretty clear that the encryption works on the A-Z range, at least in this stage. Or maybe this is a red-herring?&lt;/li&gt;&#xA;&lt;li&gt;It is somewhat hard to write intelligible long sentences in English using no words with more than 4 letters (result of second step), so this may indicate that the way that the words are distributed is not the way presented in the plain-text message. The third step procedure from that produces something that resembles letters, but I can't be sure if I did it this step correctly.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5044" LastEditorUserId="5044" LastEditDate="2014-12-04T10:25:43.487" LastActivityDate="2014-12-04T10:25:43.487" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="5616" PostTypeId="2" ParentId="5588" CreationDate="2014-11-26T20:24:05.427" Score="4" Body="&lt;p&gt;I have an idea of a solution which would currently give me for the three elephants (top to bottom): &#xA;Green,right&#xA;Green,left&#xA;Green,right&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reasoning given below in the spoiler. However, my solution is currently flawed, so it is rather the &lt;em&gt;idea&lt;/em&gt; I want to post here for commenting.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  The puzzle could be turned into a system of equations by assigning the color to a variable name and the direction to +-. Each line would translate into one equation, but the right hand side of the equation is not defined, hence I set it to zero. (Also line 4 seems to indicate this.) This gives me the following, flawed, equation system: &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;this gives me:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0 = -3 g -3 r  ;&#xA; 0 = -2 g  ;&#xA; 0 = -5 g + 4 v + X1  ;&#xA; 0 = 0  ;&#xA; 0 = 2 g - 2 v  ;&#xA; 0 = 3 r + x2  ;&#xA; 0 = -3 r + x3  ;&#xA; 0 = g - 3 r  ;&#xA; 0 = -2 y  ;&#xA; 0 = -2 y  ;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And solving this from bottom up give my solution. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; But the equation system as whole is flawed, so I either made a mistake, or an incorrect assumption. I could try different ways of turning this into equations ( I.e going to vertically, or assuming other right-hand sides, but if I am on the completely wrong track here, I'd rather know it....) other may pick up the idea here as well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Am I barking up the wrong tree, or just haven't done it properly yet?&lt;/p&gt;&#xA;" OwnerUserId="5840" LastActivityDate="2014-11-26T20:24:05.427" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5617" PostTypeId="2" ParentId="5190" CreationDate="2014-11-26T20:51:27.563" Score="1" Body="&lt;p&gt;This is my revised guess and check strategy. You may argue that it's pointless and wastes time, but I don't want to spend the weekend reading up on college-level math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Not sure if this is static or dynamic&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, weigh the below coins and note the weights.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1, 2, 4, 5, 7, 9, 11&lt;/li&gt;&#xA;&lt;li&gt;2, 3, 5, 6, 12&lt;/li&gt;&#xA;&lt;li&gt;1, 3, 5, 7, 8, 10&lt;/li&gt;&#xA;&lt;li&gt;5, 6, 8, 9&lt;/li&gt;&#xA;&lt;li&gt;1, 2, 4, 6, 8, 10&lt;/li&gt;&#xA;&lt;li&gt;2, 3, 8, 9&lt;/li&gt;&#xA;&lt;li&gt;1, 3, 6, 9&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thanks to @Julian for the weights...&#xA;Then write all of them like this:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/mR1Nx.jpg&quot; alt=&quot;Coin Writing&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substitute the ?$n$ for the actual weights. Then, see if there are any no guesses-like weighing $2$ being 100, or etc.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Guessing Tips&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Guess only when there are no more weights that can be deduced.&#xA;-Try to only guess 1 weight at a time, never more than 2&#xA;-Use scratch paper and keep the un guessed version in case you messed up&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example of finished weights:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/zveCD.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5368" LastEditorUserId="5368" LastEditDate="2014-11-30T16:03:05.613" LastActivityDate="2014-11-30T16:03:05.613" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5619" PostTypeId="2" ParentId="5462" CreationDate="2014-11-27T00:46:50.647" Score="2" Body="&lt;p&gt;Here's another perspective.  Realistically, humans have just as much trouble with a position which requires 18 half turns as they do with a position which requires 20 half turns.  Therefore I do not see why exactly the actual distance a position is away from the solved state should be considered &quot;perfectly scrambled&quot; (the first two responses mentioned this), as when I hear &quot;perfectly scrambled&quot;, I think of &lt;em&gt;the difficulty&lt;/em&gt; for a human to solve it...it's not like humans have a method to solve a cube optimally every time (if they can by chance solve one optimally in a fewest moves competition).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, I define a perfectly scrambled cube state to be any state which doesn't allow any extra pieces to be solved than intended during the solving process.  It's a scramble which we call &quot;unlucky&quot; or &quot;full-step&quot;.  Since there are different human methods for solving a cube, in reality, a perfectly scrambled cube for one person will be an easier one for another person, and therefore a perfectly scrambled cube is unknowable and yet occurs frequently for everyone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way to determine if a scramble is perfectly scrambled for YOU is for YOU to solve it with whichever method you use.  Even then, it's questionable whether a scramble is &quot;unlucky&quot; for you should you have chosen to solve the pieces in one step in a different order, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the conditions for a perfectly scrambled cube occurring is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Unique to the individual,&lt;/li&gt;&#xA;&lt;li&gt;Dependent on the order in which that individual chooses to solve pieces in each stage of the solving process he or she uses.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I therefore can understand &lt;em&gt;why&lt;/em&gt; the first two responses were mentioning God's number, as that is the only answer which can solidly answer this question, but I do not believe the two are related.  Therefore I have answered an inexact question with its corresponding inexact answer.&lt;/p&gt;&#xA;" OwnerUserId="6044" LastActivityDate="2014-11-27T00:46:50.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5621" PostTypeId="2" ParentId="4818" CreationDate="2014-11-27T01:10:09.777" Score="1" Body="&lt;p&gt;You showed two general parity cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first was a single dedge flip (OLL parity).  When you are solving a 3x3x3 cube and get to the last layer, don't you often have to swap two corners/two edges?  Sure.  The same is true for the 4x4x4.  A 4x4x4 has two inner layers which a related phenomena occurs.  No mystery here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second was a two dedge swap (PLL parity).  Assuming that you use the reduction method or one of its variants, you first solve the centers and then pair the dedges.  Since the last step is to solve the 4x4x4 like a 3x3x3, we need to consider the properties of the 3x3x3 to see if our pseudo 3x3x3 matches it.  The 3x3x3 cube law of permutations state that if the corners are in an odd permutation, then the (middle) edges are in an odd permutation, and vice versa.  When you pair dedges, you &lt;em&gt;define&lt;/em&gt; the dedge parity in relation to the corners.  When you get a two dedge swap (or a 2 corner swap), you created a 3x3x3 edge formation whose parity doesn't match the parity of the corners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;OLL parity occurring on the 4x4x4 strictly has to deal with how many inner slice quarter turns you used to solve the first three centers, and PLL parity occurring on the 4x4x4 strictly has to deal with how you paired up the last few dedges.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The key is that the four center cubules on each face of a Rubik's Revenge are indistinguishable.&#xA;When you do the move sequence to swap those two edges in the &quot;3x3x3&quot; phase of solving a 4x4x4 cube (or to flip a single edge or swap two corners), a bunch of the center cubules get reoriented, but you don't notice because they're all the same colour and rotationally symmetrical anyway.&#xA;A picture cube wouldn't have this problem, and thus it would be much harder to solve.&lt;/em&gt;--Joe Z&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although this is true for the single dedge flip (OLL parity) on the 5x5x5 and larger supercubes, it is not true for the 4x4x4 supercube:  the parity of the wing edges on the 4x4x4 is independent of the parity of the X centers of the 4x4x4 supercube, and therefore you are ultimately incorrect.  (There exists algorithms which flip an edge but do not distort any centers on the 4x4x4 supercube.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the two dedge swap (PLL parity), there exists algorithms which do not affect centers on any size supercube, no exceptions, and therefore if your statement was referring to PLL parity regarding same color centers, it is clearly always wrong.&lt;/p&gt;&#xA;" OwnerUserId="6044" LastEditorUserId="6044" LastEditDate="2014-11-27T01:16:51.340" LastActivityDate="2014-11-27T01:16:51.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5624" PostTypeId="2" ParentId="5596" CreationDate="2014-11-27T01:55:14.530" Score="0" Body="&lt;p&gt;I thought this was an exact duplicate of the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/230/a-camel-transporting-bananas&quot;&gt;camel transporting bananas&lt;/a&gt; problem, but some people have pointed out the difference that in this case, it costs money to deposit fuel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way it's laid out, however, means that the problem is numerically identical to the camel problem except with the added condition that it costs 10 bananas to set down bananas at any location. So I will give the solution framed in those terms for easier analogy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/a/267/88&quot;&gt;This answer by durron597&lt;/a&gt; gives us a formula for the cost of moving bananas assuming the camel can carry $c$ bananas:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For bananas $1$ through $c$, it costs $1$ banana per km to move a banana.&lt;br&gt;&#xA;  For bananas $c+1$ through $2c$, it costs $3$ banana per km to move a banana.&lt;br&gt;&#xA;  For bananas $2c+1$ through $3c$, it costs $5$ banana per km to move a banana.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since we want a solution that uses as few &quot;banana-setdowns&quot; as possible, we can't just shunt the pile of bananas along one kilometre at a time. We need a solution that moves $200$, then $333$, then $1,000$ bananas all at once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following list of steps:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Load up 1,000 bananas, carry them 200 km forward.&lt;/li&gt;&#xA;&lt;li&gt;Set down 600 (-10) bananas, carry the rest back.&lt;/li&gt;&#xA;&lt;li&gt;Load up 1,000 bananas, carry them 200 km forward.&lt;/li&gt;&#xA;&lt;li&gt;Set down 600 (-10) bananas, carry the rest back.&lt;/li&gt;&#xA;&lt;li&gt;Load up 1,000 bananas, carry them 200 km forward.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You now have 1,980 bananas 200 km ahead, 800 of which are on the camel.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Load up 200 more bananas, carry them 333 km forward.&lt;/li&gt;&#xA;&lt;li&gt;Set down 334 (-10) bananas, carry the rest back.&lt;/li&gt;&#xA;&lt;li&gt;Load up 980 bananas, carry them 333 km forward.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You now have 324 + 647 = 971 bananas 533 km ahead, 647 of which are currently on the camel.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Load up the remaining 324 bananas, and carry them 467 km ahead.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You're left with 504 bananas. In terms of the car's fuel price, this means you can sell the fuel in your car for &lt;strong&gt;50.4 million dollars&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.923" LastActivityDate="2014-11-28T16:11:31.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5626" PostTypeId="1" AcceptedAnswerId="5630" CreationDate="2014-11-27T09:38:53.630" Score="40" ViewCount="15000" Body="&lt;p&gt;A dragon and a knight live on an island where the only sources of fresh water are a pond (containing ordinary water) and six wells, numbered 1 to 6. Each well's water is completely indistinguishable from the pond water, but contains a magical poison that has no immediate symptoms, yet suddenly kills the drinker about an hour after drinking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make things complicated still, each well contains a different variety of poison. If a drinker who has been poisoned by a well drinks the water from another well, the result depends on the relative numbers. If the second well has a higher number then both poisons will eliminate each other and the drinker will be cured. If the second well has the same number or a lower one, it is as though the drinker had only drunk from the first well. For example, if you drink from well 1 and then from well 4 you will not be poisoned, but if you drink from well 4 and then well 1 you will be poisoned as though you had drunk only from well 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result of these rules, water from well 6 can cure poison from any of the other wells, but, when drunk by someone who is not poisoned, is incurably lethal.&#xA;Furthermore, while wells 1-5 are a short walking distance from each other, well 6 is located at the top of an unclimbable mountain on the island that the knight cannot reach but the dragon can fly onto very quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sets the stage for the following puzzle: both knight and dragon understand these rules completely, know the numbering of each of the wells, and each want the other dead. Being evenly matched in combat, they arrange a special sort of duel. Each secretly fills a glass of water from one of the island's sources, then meets the other in a field, where they exchange glasses, and drink. They may seek water from any of the island's sources that they can personally reach before and after.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible for either the knight or the dragon to ensure it will survive this duel?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: drinking the same poison twice in a row (or 20 times in a row) has exactly the same effect as drinking the poison once.&lt;/p&gt;&#xA;" OwnerUserId="1672" LastEditorUserId="228" LastEditDate="2014-12-01T11:24:59.660" LastActivityDate="2018-05-29T15:59:32.467" Title="Duel between the knight and the dragon" Tags="|logical-deduction|strategy|" AnswerCount="5" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5627" PostTypeId="2" ParentId="5626" CreationDate="2014-11-27T09:48:17.560" Score="4" Body="&lt;p&gt;Yes. The dragon can ensure it will survive.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;First it must drink water from well 5. Then come for the duel. After drinking the water given by the Knight (no matter if its poison-less water from the pond or poisoned water from well 1-5), it can fly to the well 6 and drink the water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Condition 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If the knight give him water from pond, the dragon already has drank water from well 5 which is poisonous. So it will be cured by drinking the water from 6th well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Condition 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If the knight give him water from any of the well 1-4, it will still have poison in it as it already drank from well 5 and it will not be cured. SO he can cure the poison by flying to well 6 and drinking the water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Condition 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If the knight give him water from well 5, it is poisoned, and he will still be poisoned since &quot;drinking the same poison twice in a row (or 20 times in a row) has exactly the same effect as drinking the poison once.&quot; He can still fly to well 6 and drink the poisoned water to get cured.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Knight Cannot make sure he will survive. He have to take his chance. The only two possible survival chances of knight are as follows.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Option 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Knight drinking poisoned water from a lower numbered well before the duel and the dragon giving him poisoned water from a higher numbered well (preferably from well 6).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Option 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Knight drinking nothing before the duel, and the dragon giving him pond water, thinking that the knight may have drank any of the poisonous water (preferably from well 1) before the duel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="2730" LastEditDate="2014-11-27T10:01:18.187" LastActivityDate="2014-11-27T10:01:18.187" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5628" PostTypeId="1" AcceptedAnswerId="5629" CreationDate="2014-11-27T09:54:37.703" Score="6" ViewCount="2157" Body="&lt;p&gt;The object of the following flow-chart is to rearrange the numbers in boxes 1, 2 and 3 in such a fashion that, at the end:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Box 3 will always contain the largest number,&#xA;Box 1 will always contain the smallest number,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;However, this flow-chart is INCORRECT.  As it stands, it will not accomplish the desired purpose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which one instruction must be changed to correct the flow-chart?&#xA;&lt;img src=&quot;https://i.stack.imgur.com/5eAoZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to correct this flow-chart, we must change Instruction No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that the first box-number mentioned in it is box-number.............&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the second box-number mentioned in it is box-number.............    &lt;/p&gt;&#xA;" OwnerUserId="6054" LastActivityDate="2014-11-27T10:06:42.740" Title="Find out which instruction is wrong" Tags="|logical-deduction|calculation-puzzle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5629" PostTypeId="2" ParentId="5628" CreationDate="2014-11-27T10:00:19.970" Score="7" Body="&lt;p&gt;We want to arrange the numbers so that reading down from 1 to 3, they are increasing in size. There is one particular branch which is clearly wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Which one?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Look at the 2 branches coming off step 5.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I still don't see it&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The yes branch... does that look a bit strange?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Nope&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Step 5's &quot;Yes&quot; branch says &quot;if $box3 &amp;lt; box2$ then end&quot; - we always want $box2 &amp;lt; box3$ so this has to be wrong.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Wait - so that means the answer is...&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In order to correct this flow-chart, we must change Instruction No. 5&lt;br&gt;&#xA; So that the first box-number mentioned in it is box-number 2&lt;br&gt;&#xA; And the second box-number mentioned in it is box-number 3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2518" LastEditorUserId="2518" LastEditDate="2014-11-27T10:06:42.740" LastActivityDate="2014-11-27T10:06:42.740" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5630" PostTypeId="2" ParentId="5626" CreationDate="2014-11-27T10:04:06.313" Score="31" Body="&lt;p&gt;The knight can ensure survival&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; by drinking from well 4 first, and 3 + 5 after. If the dragon uses pond water, or 1, 2, 3 or 4, the last drink from 5 will neutralize the others (4, ? and 3). If the dragon gives 5 or 6, it will neutralize the first '4', then 3 will be neutralized by 5.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As AeJey said, the dragon can ensure survival by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Drinking from well 5 before, and 6 after the duel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="439" LastActivityDate="2014-11-27T10:04:06.313" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5631" PostTypeId="1" AcceptedAnswerId="5896" CreationDate="2014-11-27T11:59:49.533" Score="10" ViewCount="3322" Body="&lt;p&gt;This is a riddle without a twist. A straight forward one. Not almost, but all the lines will point you in the right direction. The puzzle is really really easy. I am sure someone is going to post the answer within minutes.  Good luck!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;My younger brother is Mr. Wayne,&lt;/em&gt;&lt;br /&gt; &lt;em&gt;And the beginning of all his pain&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;I will tease your eyes with nails&lt;/em&gt;&lt;br/&gt; &lt;em&gt;But, I won't let you close your eyes&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;To make it simple, cut my tail&lt;/em&gt;&lt;br/&gt; &lt;em&gt;With my new tail, you won't fail&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;Now take some of my exact middle&lt;/em&gt;&lt;br/&gt; &lt;em&gt;That will help you solve this riddle&lt;/em&gt;&lt;br/&gt; &lt;em&gt;Then cut its end, the beginning of the end&lt;/em&gt;&lt;br/&gt; &lt;em&gt;It's a wordplay, that's a hint&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;My New head and tail are almost the same&lt;/em&gt;&lt;br/&gt; &lt;em&gt;If you place a mirror in between&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;My lost tail is definitely not for doctors&lt;/em&gt;&lt;br/&gt; &lt;em&gt;Tell me who I am, that's what matters&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="2730" LastEditDate="2015-12-15T07:18:31.663" LastActivityDate="2015-12-15T07:18:31.663" Title="Riddle without a twist (this riddle may/may not kill you!)" Tags="|riddle|word|movies|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="5632" PostTypeId="1" AcceptedAnswerId="5634" CreationDate="2014-11-27T14:03:46.863" Score="6" ViewCount="1482" Body="&lt;p&gt;&lt;em&gt;Because these have been so successful in the past, I had to bring up my own. In fact, this puzzle was &lt;strong&gt;inspired by &lt;a href=&quot;https://puzzling.stackexchange.com/questions/5494/find-the-next-number&quot;&gt;Find the next number&lt;/a&gt;&lt;/strong&gt; even though I (and I guess everybody but the author) still don't know the intended solution yet. Anyway, here it is:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;The puzzle&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The British inventor &lt;strong&gt;Jackemias Muff&lt;/strong&gt; invented (and build) a number-evolving machine in &lt;strong&gt;1817&lt;/strong&gt;. (It was a cold, windy winter's day in Manchester, the Sunday before boxing-day.) You could feed it with any number, and it would produce an unambiguously determined, endless series of follow-up numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you feed it with &quot;8&quot; you would get&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;8,5,10,14,23,36,47,59,70,78,84,96,...&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you feed it with &quot;28&quot; instead, you would get&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;28,12,19,28,41,51,61,71,83,96,...&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can you &lt;em&gt;explain&lt;/em&gt; how his machine works and build a copy? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Victory condition: A correct answer is able to reproduce both series starting from the same seed values and can successfully predict the&#xA;  next number in each series.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I do not have proof that Muff's machine is unique, although I believe it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will add hints over time, one is given as a starter:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In 1817, computers have not yet been invented. (Well, 'computer' was&#xA;  still a job-description those days..)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-01-03T17:52:40.293" Title="Oh no! Yet another number series... - Find the pattern in the two examples and provide an extension" Tags="|mathematics|pattern|number-sequence|" AnswerCount="1" CommentCount="12" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5633" PostTypeId="1" AcceptedAnswerId="5950" CreationDate="2014-11-27T14:23:29.217" Score="0" ViewCount="801" Body="&lt;p&gt;&lt;pre&gt;Another riddle coming to you from the land of poets and thinkers!&#xA;Focus on word building this time. Well, as you might have thought...it's easy. &lt;/pre&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I decide if sweet or stinky&lt;br&gt;&#xA;For most locks I'm the key&lt;br&gt;&#xA;Go ahead and keep on piling&lt;br&gt;&#xA;because even a child can find me&#xA;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's no vanishing&lt;br&gt;&#xA;no clues here&lt;br&gt;&#xA;no hiding there&lt;br&gt;&#xA;It's no use&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A man of words&lt;br&gt;&#xA;participating right now&lt;br&gt;&#xA;Teach him a lesson&lt;br&gt;&#xA;and break his literal vow&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subdued by ships&lt;br&gt;&#xA;Along lonely isles&lt;br&gt;&#xA;I may result &lt;br&gt;&#xA;in stunning smiles&lt;br&gt;&lt;/p&gt;&#xA;" OwnerUserId="5106" LastEditorUserId="5106" LastEditDate="2015-01-08T07:08:24.300" LastActivityDate="2015-01-08T07:08:24.300" Title="Feel the Enlightenment!" Tags="|riddle|word|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="5634" PostTypeId="2" ParentId="5632" CreationDate="2014-11-27T14:30:06.853" Score="13" Body="&lt;p&gt;The sequence is generated as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each consecutive term is except for the first (5 and 12 respectively) term is made by adding 1 and the length of the text representation of the current number to the current number. The 5 and 12 are made by simply counting the length of the text representation of the first term.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Next terms will be 107&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First sequence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 8 -&gt; &quot;eight&quot; -&gt; 5 &lt;br/&gt;5 -&gt; 5 + len(&quot;five&quot;) + 1 -&gt; 10 &lt;br/&gt;10 + len(&quot;ten&quot;) + 1 -&gt; 14 &lt;br/&gt;14 + len(&quot;fourteen&quot;) + 1 -&gt; 23 &lt;br/&gt;Next steps are obvious.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="4748" LastEditDate="2014-11-30T16:33:17.597" LastActivityDate="2014-11-30T16:33:17.597" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="5635" PostTypeId="1" AcceptedAnswerId="5637" CreationDate="2014-11-27T14:56:06.547" Score="11" ViewCount="981" Body="&lt;p&gt;Chaos at the international conference for world leaders: What should have been a peaceful exchange of ideas for common growth quickly exploded into hostility shortly after the German speaker addressed the USA delegates with accusations:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;We have recently discovered hacking attempts at our security network.&#xA;Someone deliberately gave us faulty processors, such that we now&#xA;compute&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;1 + 1 = 8&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;We think, the NSA is trying to spy on us!&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The Americans jumped up and shouted:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;Outrageous! Why is everybody always suspecting us for such things?&#xA;Just so you know, we are victims, not aggressors. Checking our&#xA;security protocols we've found a vulnerability. Our processors gave&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;2 + 2 = 6&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;And guess who wrote our software? The Italians!&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The Italian minister's face went through a spectacular range of colours, finishing with purple as he shouts out:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;You accuse us of putting deliberate vulnerabilities in software we&#xA;distribute? Or do you just think we can't program? Either way, we will&#xA;take up American traditions and sue you out of your wits! And we have&#xA;proof that your harmful allegations are wrong. Checking our own&#xA;computer systems with the prime OS installed - which, I may add, has&#xA;been provided by an American company not known for their security, we&#xA;could already see the same miscalculations you've just thrown in our&#xA;face! But we tested further and realized that also&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;16 + 16 = 12&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;However, the security expert advising us on our network was a Czech&#xA;immigrant, so if you want to point any fingers...&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&amp;quot;I don't believe it!&amp;quot; the Czech ambassador to Italy barks out.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;First you only pay dumping prices for our experts and beg me to bring&#xA;them into your country, and then you call them spies? I can only&#xA;assume all of this is a political ploy initiated by our Finnish&#xA;friends, who not only live close to Russia, but also provided most of&#xA;the mobile telecommunication in our countries. We have recently seen&#xA;some of those devices miscalculate&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;32 + 32 = 18&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;64 + 64 = 24&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;So...&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;But before the Finnish delegation has time to react, the meeting is interrupted by the session chair.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;Gentlemen, Gentlemen! Please calm down. I have just received&#xA;intelligence that an international crime organisation has secretly&#xA;spread a virus who, as I've been led to believe, is responsible for all&#xA;the effects you have just described.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Can you tell what this mystic virus is actually doing?&lt;/strong&gt;&#xA;(i.e. we seek a method/algorithm/equation/instruction which makes all those equations come true.)&lt;/p&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="5373" LastEditDate="2021-07-10T11:18:36.847" LastActivityDate="2021-07-10T11:18:36.847" Title="An international affair: Find the common property in incorrect equations" Tags="|mathematics|pattern|lateral-thinking|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="5636" PostTypeId="2" ParentId="5633" CreationDate="2014-11-27T15:07:25.307" Score="3" Body="&lt;p&gt;I decide if sweet or stinky&lt;br&gt;&#xA;For most locks I'm the key&lt;br&gt;&#xA;Go ahead and keep on piling&lt;br&gt;&#xA;because even a child can find me   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TEETH, keys have teeth, teeth are in 2 layers piling up milk/adult. Children bite or teething either way they find thier teeth.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's no vanishing&lt;br&gt;&#xA;no clues here&lt;br&gt;&#xA;no hiding there&lt;br&gt;&#xA;It's no use    &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I I N N. Repition of the first words, theres no clues in here or there which are situated at the end of the sentence not the start.    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A man of words&lt;br&gt;&#xA;participating right now&lt;br&gt;&#xA;Teach him a lesson&lt;br&gt;&#xA;and break his literal vow  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; M N.  A man of words in a debate trying to win to become the ALPHA male take the literal vowel, A from man. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Subdued by ships&lt;br&gt;&#xA;Along lonely isles&lt;br&gt;&#xA;I may result&lt;br&gt;&#xA;in stunning smiles  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; G. Gold is restrained by ships and buried on lonely islands.&lt;br&gt;&#xA; L. Lonely isles is a body of Land.&lt;br&gt;&#xA; E. smiles is an expression&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Put all them together  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Enlightenment! rotating the i to become a exclamation mark&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5135" LastActivityDate="2014-11-27T15:07:25.307" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5637" PostTypeId="2" ParentId="5635" CreationDate="2014-11-27T15:19:00.623" Score="15" Body="&lt;p&gt;What the virus does, is that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it translates each number to the speaker's native language, then counts its letters &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and uses that number to compute the outcome.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Germany: 1 + 1 -&gt; eins + eins -&gt; 4 + 4 = 8&lt;br&gt;&#xA; America: 2 + 2 -&gt; two + two -&gt; 3 + 3 = 6&lt;br&gt;&#xA; Italy: 16 + 16 -&gt; sedici + sedici -&gt; 6 + 6 = 12&lt;br&gt;&#xA; Czech: 32 + 32 -&gt; třicet dva + třicet dva  -&gt; 9 + 9 = 18&lt;br&gt;&#xA; Czech: 64 + 64 -&gt; šedesát čtyři + šedesát čtyři  -&gt; 12 + 12 = 24&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditDate="2014-12-12T07:12:16.407" LastActivityDate="2014-12-12T07:12:16.407" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5639" PostTypeId="2" ParentId="5376" CreationDate="2014-11-27T16:09:51.587" Score="3" Body="&lt;p&gt;Could it be a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; conjuror's hat?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You're very close &lt;br&gt;&#xA;But I'm always ahead&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When wearing a hat, you are very close to it, but it's always above ('ahead of'?) your head (so the word 'ahead' is sort of wordplay too).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Can be beheaded &lt;br&gt;&#xA;but can't be fed&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'Beheaded' is here used to mean 'removed from the head'. (The OP is not a native English speaker!) And certainly a hat can't be fed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You can have &lt;br&gt;&#xA;a look inside &lt;br&gt;&#xA;there will be &lt;br&gt;&#xA;no place to hide &lt;br&gt;&#xA;However, &lt;br&gt;&#xA;try to enter me &lt;br&gt;&#xA;more room arises &lt;br&gt;&#xA;suddenly&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These verses are what makes it a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;conjuror's&lt;/em&gt; hat&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;rather than just a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; hat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can look inside this object, and indeed there appears to be no place to hide. But&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; when the conjuror is performing 'magic' with it on stage, he can produce all sorts of objects, most famously rabbits, from the apparently empty hat, and also make them disappear into it. So for a rabbit, trying to enter the hat may lead to 'more room arising suddenly'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Credit to @d'alar'cop for the idea of a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; hat,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but he couldn't make this fit with the 3rd verse. The only remaining mystery is the relevance of the multi-'had' sentence linked to by the OP - maybe it's a red herring, or just a hint to a word which sounds like 'had'?&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2014-11-27T16:09:51.587" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5640" PostTypeId="5" CreationDate="2014-11-27T16:25:04.637" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-11-27T16:25:04.637" LastActivityDate="2014-11-27T16:25:04.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5641" PostTypeId="4" CreationDate="2014-11-27T16:25:04.637" Score="0" Body="To be used when asking for an explanation to a puzzle or concept that is known to many as basic." OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-11-27T17:51:54.627" LastActivityDate="2014-11-27T17:51:54.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5642" PostTypeId="1" CreationDate="2014-11-27T17:13:13.287" Score="-3" ViewCount="564" Body="&lt;p&gt;How is it possible that a person born in Russia, with parents that were indeed born in Russia, is not born as a Russian citizen?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Tip: This riddle is a little tricky, but when you think about it should come.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6061" LastEditorUserId="2383" LastEditDate="2014-11-27T17:27:40.003" LastActivityDate="2014-11-27T19:42:02.093" Title="The Russian Birth Dilemma!" Tags="|riddle|" AnswerCount="6" CommentCount="5" ClosedDate="2014-11-28T02:00:50.223" ContentLicense="CC BY-SA 3.0" />
  <row Id="5644" PostTypeId="2" ParentId="5642" CreationDate="2014-11-27T17:23:44.760" Score="1" Body="&lt;p&gt;Both parents were born in places that were &lt;em&gt;then&lt;/em&gt; in Russia but are now in e.g. Finland or Kazakhstan?&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2014-11-27T17:23:44.760" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5645" PostTypeId="2" ParentId="5642" CreationDate="2014-11-27T17:35:07.403" Score="0" Body="&lt;p&gt;Parents were born before February 6, 1992 to Russian residents, who themselves were born prior to December 30, 1922, and moved elsewhere before returning to visit and giving birth.  (Not sure, though.  Would the parents have any citizenship?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...Or the parents denounced their citizenship but delivered on a vacation to Siberia.&lt;/p&gt;&#xA;" OwnerUserId="5948" LastActivityDate="2014-11-27T17:35:07.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5646" PostTypeId="2" ParentId="5642" CreationDate="2014-11-27T17:39:28.133" Score="4" Body="&lt;p&gt;It is possible if that person is &lt;strong&gt;Smirnov&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Smirnov was born on RSFSR territory in 1950. In 1979, he married and&#xA;  moved his permanent residence to the Lithuanian SSR. He divorced in&#xA;  1992 and returned to the RSFSR on December 8, 1992. He then applied&#xA;  for a notice of Russian citizenship in his passport, but this was&#xA;  rejected by executive officers. His claim was also rejected by common&#xA;  jurisdiction courts, including the Supreme Court of the Russian&#xA;  Federation.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The position of the executive officers and the courts was that&#xA;  Smirnov, in accordance with Article 13 of the Citizenship Act, was a&#xA;  former citizen of the Russian Federation, but not a citizen of the&#xA;  Russian Federation since February 6, 1992.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://en.wikipedia.org/wiki/Russian_nationality_law&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2730" LastActivityDate="2014-11-27T17:39:28.133" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5647" PostTypeId="2" ParentId="5642" CreationDate="2014-11-27T17:39:52.630" Score="2" Body="&lt;p&gt;Both parents were born &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;em&gt;in&lt;/em&gt; Russia, but not to Russian citizens, &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and are therefore &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;not Russian citizens themselves. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Their child is consequently not a Russian citizen either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know, it's an amazing coincidence, but you see, Carol and Jack were both &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;embassy-kids, born to expats -- American diplomats. So they were both Americans in Moscow. They met in Washington, married, and moved to Moscow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;when Jack followed in his father's footsteps. Their daughter Rose was born in Moscow like her parents.&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="5373" LastEditDate="2014-11-27T19:30:55.677" LastActivityDate="2014-11-27T19:30:55.677" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5648" PostTypeId="2" ParentId="5642" CreationDate="2014-11-27T17:40:45.260" Score="0" Body="&lt;p&gt;The kid was born on a plane above Russia but the parents (who decided to do this is the bathroom) both passed out. The mother because of the pain, and father who had been frightened to death for his wife. They weren't found until they arrived at their destination, then they were found in the bathroom, so the baby was given [Destination] citizen ship?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2014-11-27T17:40:45.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5650" PostTypeId="2" ParentId="5631" CreationDate="2014-11-27T18:09:01.537" Score="2" Body="&lt;p&gt;The answer is (&lt;strong&gt;incorrect&lt;/strong&gt;)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter &quot;e&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOT CORRECT&lt;/strong&gt;&#xA;&lt;em&gt;My younger brother is Mr. Wayne,&#xA;And the beginning of all his pain&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bruce Wayne's older brother (Thomas Wayne Jr.) aka Owlman (Owls prey on bats)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;INCORRECT&lt;/strong&gt;&#xA;&lt;em&gt;I will tease your eyes with nails&#xA;But, I won't let you close your eyes&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nope this neither...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOT CORRECT&lt;/strong&gt;&#xA;&lt;em&gt;To make it simple, cut my tail&#xA;With my new tail, you won't fail&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Without the tail you've got the top of e.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;CORRECT:&lt;/strong&gt;&#xA;&lt;em&gt;Then cut its end, the beginning of the end&#xA;It's a wordplay, that's a hint&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This I can help with :D The beginning of &quot;end&quot; is e. True wordplay.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOT CORRECT&lt;/strong&gt;&#xA;&lt;em&gt;My New head and tail are almost the same&#xA;If you place a mirror in between&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Check the bottom and top of &quot;e&quot; almost exact same.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOT CORRECT&lt;/strong&gt;&#xA;&lt;em&gt;My lost tail is definitely not for doctors&#xA;Tell me who I am, that's what matters&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A curved line isn't for doctors, but don't worry, I've said who it is :D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is not complete enough to be an answer, but it's the best I've got! If you can fill in the other hints, feel free to do so!&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-11-27T20:14:41.767" LastActivityDate="2014-11-27T20:14:41.767" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="5653" PostTypeId="2" ParentId="5203" CreationDate="2014-11-27T18:30:10.817" Score="-1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What do the letters T-H-I-S spell?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or another possibility (though this doesn't have a unique answer...):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What's an anagram of 'hits'?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sorry, I think I'm too much of a fan of wordplay and misinterpretation!&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2014-11-27T18:30:10.817" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5656" PostTypeId="1" AcceptedAnswerId="5657" CreationDate="2014-11-27T19:04:31.247" Score="-3" ViewCount="131304" Body="&lt;p&gt;When does one plus one equal three?&lt;/p&gt;&#xA;" OwnerUserId="6068" LastEditorUserId="2383" LastEditDate="2014-11-28T16:40:15.810" LastActivityDate="2017-10-07T14:19:15.057" Title="When does 1+1=3?" Tags="|riddle|lateral-thinking|" AnswerCount="6" CommentCount="7" ClosedDate="2014-11-28T01:54:41.213" ContentLicense="CC BY-SA 3.0" />
  <row Id="5657" PostTypeId="2" ParentId="5656" CreationDate="2014-11-27T19:07:56.250" Score="7" Body="&lt;p&gt;The following slightly abuses the notion of &quot;equals&quot; but in a way that is common among non-mathematicians (more or less &quot;And the next thing we get is...&quot;)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When a woman plus a man gives a woman plus a child plus a man.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2023" LastActivityDate="2014-11-27T19:07:56.250" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5659" PostTypeId="2" ParentId="5656" CreationDate="2014-11-27T19:12:27.290" Score="4" Body="&lt;p&gt;1+1=3 When &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the one who calculated it doesn't know maths or not good in maths (probably a small child or someone who doesn't learned mathematics properly)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastActivityDate="2014-11-27T19:12:27.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5663" PostTypeId="2" ParentId="5656" CreationDate="2014-11-27T19:32:11.477" Score="16" Body="&lt;p&gt;Here is an image that shows that 1 + 1 = 3:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/H6sE0.png&quot; alt=&quot;1 + 1 = 3&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5044" LastActivityDate="2014-11-27T19:32:11.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5664" PostTypeId="2" ParentId="5642" CreationDate="2014-11-27T19:42:02.093" Score="6" Body="&lt;p&gt;Simple really,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Because they were all born in &lt;a href=&quot;http://en.wikipedia.org/wiki/Russia,_Ohio&quot;&gt;Russia, Ohio.&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4852" LastActivityDate="2014-11-27T19:42:02.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5665" PostTypeId="2" ParentId="5656" CreationDate="2014-11-27T19:51:27.830" Score="16" Body="&lt;p&gt;When? This is very easy... &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Never.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6070" LastActivityDate="2014-11-27T19:51:27.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5666" PostTypeId="1" AcceptedAnswerId="5670" CreationDate="2014-11-27T20:02:29.617" Score="1" ViewCount="1624" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I lie inside this question, and I hide behind a mask.&lt;br&gt;&#xA;  I don't think you'll see me from my appearance.&lt;br&gt;&#xA;  I hope you conclude where I'm too,&lt;br&gt;&#xA;  but I doubt you will without clever thinking.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What am I?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="11497" LastEditDate="2017-07-22T12:23:15.760" LastActivityDate="2017-07-22T12:23:15.760" Title="The answer lies within this question!" Tags="|enigmatic-puzzle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5667" PostTypeId="2" ParentId="5666" CreationDate="2014-11-27T20:06:31.680" Score="36" Body="&lt;p&gt;The answer is quite simply this.  &lt;/p&gt;&#xA;" OwnerUserId="4050" LastActivityDate="2014-11-27T20:06:31.680" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="5668" PostTypeId="2" ParentId="5656" CreationDate="2014-11-27T20:08:57.687" Score="26" Body="&lt;p&gt;I have a mathematical proof for you:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Start with the following simple equation:&#xA;  $$a = b$$ (step 1)&#xA;  Multiply both sides by $b$:&#xA;  $$ab = b^2$$ (step 2)&#xA;  Subtract $a^2$ from both sides and factorize:&#xA;  $$ab - a^2 = b^2 - a^2$$ (step 3)&#xA;  $$a(b-a) = (b+a)(b-a)$$ (step 4)&#xA;  Simplify and add 1 to both sides:&#xA;  $$a = b + a$$ (step 5)&#xA;  $$a + 1 = b + a + 1$$&#xA;  Now since $a = b$ (the starting point of this proof), we can write this as:&#xA;  $$a + 1 = 2a + 1$$&#xA;  And in the case where $a = 1$, we have:&#xA;  $$1 + 1 = 2 + 1$$&#xA;  So, therefore,&#xA;  $$1 + 1 = 3$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="40" LastEditorUserId="30254" LastEditDate="2017-10-07T14:19:15.057" LastActivityDate="2017-10-07T14:19:15.057" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="5669" PostTypeId="2" ParentId="5656" CreationDate="2014-11-27T20:10:10.447" Score="1" Body="&lt;p&gt;Well, 1 + 1 has always equaled 3 :D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see long ago, when the very first few math geniuses got together to come up with basic rules, they gave it lots of thought and settled on what's taught today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you see, numbers don't truly exist, they're a concept created by the human mind, and because our mind has created and developed them for years, and they are all in our head, we can merely change them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you think 1 + 1 = 3, who cares if teachers and professionals say you're wrong! You are correct! Never forget that kids ;D Don't stop imagining! :D&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-11-29T01:14:02.360" LastActivityDate="2014-11-29T01:14:02.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5670" PostTypeId="2" ParentId="5666" CreationDate="2014-11-27T20:18:02.570" Score="10" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Hello, I'm the answer, just waiting for you :D&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's hidden with &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&amp;lt; ? tag. Click edit to see it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastActivityDate="2014-11-27T20:18:02.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5671" PostTypeId="2" ParentId="5626" CreationDate="2014-11-27T20:28:57.680" Score="9" Body="&lt;p&gt;The other answers provide one way for the dragon to survive. A variation would be for the dragon to drink nothing before the duel, then drink from well 1 afterwards. This ensures that the dragon is poisoned -- either the knight brought pond water, and the dragon is poisoned by well 1, or the knight brought another (stronger or equal) poison, and well 1 has no effect. Drinking from well 6 now cures the dragon, regardless of which drink poisoned him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, since the knight cannot reach the strongest poison, the dragon can always guarantee his survival. He can be certain that he has poisoned himself with a weak poison, either by drinking from the strongest well his opponent can reach before the duel, or the weakest afterwards. He can then cure himself at the top well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The knight has a more convoluted play, but she can also guarantee her survival, likewise by making sure she is poisoned with a weaker poison than the strongest she can reach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If she poisons herself before the duel with anything but the water from well 5, then depending on what the dragon brings, she will be either cured or poisoned with a weaker poison than 5's (0 here indicates the unpoisoned pond water):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drunk from 4 beforehand -&gt; she is now poisoned at level 4&lt;br&gt;&#xA;If given: 0-4 -&gt; &lt;em&gt;still&lt;/em&gt; poisoned 4&lt;br&gt;&#xA;Instead given: 5-6 -&gt; cured&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drunk from 3 beforehand -&gt; poisoned 3&lt;br&gt;&#xA;If given 0-3 -&gt; &lt;em&gt;still&lt;/em&gt; poisoned 3&lt;br&gt;&#xA;Instead given 4-6 -&gt; cured&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drunk from 2 -&gt; poisoned 2&lt;br&gt;&#xA;Given 0-2 -&gt; poisoned 2&lt;br&gt;&#xA;Instead 3-6 -&gt; cured  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drunk from 1 -&gt; poisoned 1&lt;br&gt;&#xA;Given 0-1 -&gt; poisoned 1&lt;br&gt;&#xA;Given 2-6 -&gt; cured&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To remove the possibility that she is cured, she can then drink from well 1. This either repoisons here, or has no effect on her already-poisoned state. Either way, she is poisoned by a well lower than 5, and so drinking from 5 now cures her. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although the dragon alone can reach the strongest well, by poisoning herself beforehand the knight can ensure that it always acts as an antidote if she is forced to drink it. With that guaranteed, she can also always use the second strongest as an antidote.&lt;/p&gt;&#xA;" OwnerUserId="4204" LastEditorUserId="4204" LastEditDate="2018-05-29T15:59:32.467" LastActivityDate="2018-05-29T15:59:32.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5672" PostTypeId="1" AcceptedAnswerId="5964" CreationDate="2014-11-27T22:00:25.733" Score="5" ViewCount="1192" Body="&lt;p&gt;The epitome of an evil the skewed letter would label&lt;br&gt;&#xA;To many is such truth, not a fable&lt;br&gt;&#xA;Held tightly by urges of the seconds&lt;br&gt;&#xA;Within the sleepless cities that beckons  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Twice I hold the path of tea&lt;br&gt;&#xA;Forever you stay within me&lt;br&gt;&#xA;It is not glue that binds us&lt;br&gt;&#xA;Nor the substance which I fuss  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Guarded by a significant worm&lt;br&gt;&#xA;Your sight has gone, time to squirm&lt;br&gt;&#xA;I laugh at your extravagance&lt;br&gt;&#xA;I assure you my acquaintance      &lt;/p&gt;&#xA;&#xA;&lt;p&gt;/* thank you for your answers they are lovely but for the bounty you need to be ugly*/&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I like your answers, my brain they have fondled&lt;br&gt;&#xA;  Macdonalds are evil, poor old ronald&lt;br&gt;&#xA;  dont give poverty, metaphorical bread&lt;br&gt;&#xA;  teach them to farm and grow instead  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;sex, it is true its marked with the letter&lt;br&gt;&#xA;  however this is no fable you should no better&lt;br&gt;&#xA;  struggling I see this is a comedy&lt;br&gt;&#xA;  let the divine be your remedy  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5135" LastEditorUserId="5135" LastEditDate="2014-12-04T11:03:03.703" LastActivityDate="2014-12-05T15:26:44.097" Title="Name that Sandwich (2)" Tags="|riddle|" AnswerCount="4" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="5674" PostTypeId="1" AcceptedAnswerId="5676" CreationDate="2014-11-27T22:56:03.043" Score="17" ViewCount="2056" Body="&lt;p&gt;Dr. George Wenzel and his colleague Dr. Martin Hupp are two expert biochemists researching brain-wave-changing drugs. Their work has long been regarded as being worthy of the Nobel Prize, in particular since their recent success with large cats. They showed that they could boost the intelligence of those animals to near-human level. The main challenge they had to overcome was the side-effects on the cats. Being so intelligent made them extraordinarily aggressive. (Maybe they realized their situation in full?)&lt;br&gt;&#xA;Anyway, the doctors' latest discovery was generally assumed to be so fundamental, that there wouldn't be a choice but award them the prestigious prize.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it came as a shock that on the very day a press-conference was announced which would disclose their latest results, Dr. Wenzel was found dead in the animal laboratory. He was ripped to pieces by one of the tigers they were experimenting on. His colleague Dr. Hupp could not explain at all how Dr.Wenzel could have been so careless. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;He normally took all precautions very, very seriously,&quot; he said. &quot;And this time it was especially tragic. We had just discovered the proper drug to get all the positive effects &lt;em&gt;and&lt;/em&gt; keep the animals calm. I tested it myself, and I've left George all the notes - see there on the table. I really don't understand this. The formulae &lt;em&gt;does&lt;/em&gt; work properly. I have even cross-checked it this morning. The only explanation I can possibly think of is that he tried to modify the mixture even a bit more... But why would he do that?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Inspector Edo Clu, who is leading the investigation, is wondering the same thing. However, it does not take him long to have Dr. Hupp as his prime suspect. Who else would have more profit of the untimely end of Dr. Wenzel than him? The Nobel Prize, after all, can only go to &lt;em&gt;living&lt;/em&gt; scientists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So he gives the surroundings another very close look, then calls Dr. Wenzel's mother to ask her a single question, and based on the answer arrests Dr. Hupp on suspicion of murder. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The puzzle is: &lt;strong&gt;What did inspector Edo Clu ask Mrs. Wenzel and what was the answer?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the two relevant photos from the crime-scene are shown below. (&lt;em&gt;please use your imagination a bit, I'm not an artist&lt;/em&gt;) One shows the desk of Dr. Wenzel on which a note from Dr. Hupp is found together with a couple of post-it blocks and some pencils which have been used all over the lab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second shows a group of labelled bottles all standing in a group (no sorting) on the shelf. According to Dr. Hupp, these contain the latest drugs. The bottles are labelled with the post-its and pencils found on the desk (as are other items in the lab on different photos not shown.) There are two distinct handwritings on them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NpvnM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TLW2T.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Transcription of the letter:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;George, we did it! Yesterday's experiment series went perfectly fine.&#xA;  None of the animals showed any of the side-effects at all! It seems&#xA;  that you were right about adding the Xonophoricum Hauratio to the&#xA;  mixture. It totally  neutralized all of the aggression while keeping&#xA;  the  brain-stimulating effects. You have to see this for  yourself,&#xA;  and then you should prepare for the call from Sweden!  Have you ever&#xA;  cuddled with a tiger? Now you can. He actually purred while I stroked&#xA;  him yesterday. Unbelievable. I have left one of the  bottles in the &#xA;  prep-room, if you want to see for yourself. As it was a long day, and&#xA;  tomorrow will be even longer with the press interviews and stuff,&#xA;  I’ve gone home already. See you tomorrow. Make sure you come shaved,&#xA;  our pictures will go around the world!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the five bottles:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Series VII/3/2.1; What do you think,George?; no good&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Series VII/5/1; extra aggressive!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Series VII/5/1.1; placebo&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Series VII/6/1; new mixture;brain-boost! This is it!; too aggressive; Don’t use!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Series VII/6/1.0; +XoHa&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This is my first 'crime' puzzle and I'm sure it could be better, but I hope it's still good fun.&lt;/em&gt; Asking questions in the comments is welcomed and may lead to edits. If there is no solution in a week (which I highly doubt), then I will start giving explicit hints.&lt;/p&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="5373" LastEditDate="2016-09-10T12:54:26.080" LastActivityDate="2016-12-07T22:18:32.060" Title="The untimely death of Dr. Wenzel: What give-away clue is hidden in the crime scene image?" Tags="|visual|mystery|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5675" PostTypeId="2" ParentId="5672" CreationDate="2014-11-27T23:26:21.543" Score="7" Body="&lt;p&gt;That is my guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm loving it!&lt;br&gt;&#xA; It is a &lt;strong&gt;Big Mac&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The epitome of an evil the skewed letter would label&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is an skewed M, and some people says that McDonald's is evil.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To many is such truth, not a fable&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people are &lt;em&gt;loving it&lt;/em&gt; as THE truth.&lt;br&gt;&#xA; For others, it is just a junk food company with a stupid clown-like or whatever-that-is mascot.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Held tightly by urges of the seconds&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's a &lt;strong&gt;fast&lt;/strong&gt;-food.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Within the sleepless cities that beckons&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is open until late night. Sometimes it is 24/7.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Twice I hold the path of tea&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hmm, don't know... Do they sell tea? If so, people would drink tea at morning and afternoon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Forever you stay within me&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People who is &lt;em&gt;loving it&lt;/em&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is not glue that binds us&lt;br&gt;&#xA;  Nor the substance which I fuss&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is money and fanatism.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Guarded by a significant worm&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A worm called Ronald McDonald.&lt;br&gt;&#xA; Also, there are people who says that the burgers are done from worm proteins instead of meat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Your sight has gone, time to squirm&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The people who say &quot;&lt;em&gt;I'm loving it&lt;/em&gt;&quot; has surely lost their sights!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I laugh at your extravagance&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People who frequent McDonald's and think that this is something special.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I assure you my acquaintance&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Money, money, money, money!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5044" LastEditorUserId="5044" LastEditDate="2014-11-28T07:09:51.780" LastActivityDate="2014-11-28T07:09:51.780" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5676" PostTypeId="2" ParentId="5674" CreationDate="2014-11-28T00:06:14.500" Score="14" Body="&lt;p&gt;He asked her&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Did your son suffer from deuteranopia (red-yellow-green color blindness)?&lt;br&gt;&lt;br&gt;Deuteranopia is characterized by an inability to distinguish between colours in the red-orange-yellow-green spectrum except by brightness.&lt;br&gt;&lt;br&gt;The two Series VII/6 boxes are identical save for the labeling. We can reasonably infer that Dr. Wenzel died when Mr. Cuddles (the tiger) mauled him to death after receiving the too-violent version of the serum. This suggests that the means of committing the murder was related to doctoring the labels.&lt;br&gt;&lt;br&gt;The serum labeled &quot;+XoHa&quot; is the ideal serum, but carries a green-on-yellow label that Dr. Wenzel wouldn't be able to read easily. This wasn't a large problem when there was only one series 6 serum, but when a second series 6 label was accidentally printed and applied to a too-aggressive serum, it became important that he be able to distinguish old from new. Dr. Hupp's green writing barely contrasts against the yellow background and so Dr. Wenzel writes &quot;new mixture&quot; in red, which he's able to see slightly better on the new too-aggressive serum where the &quot;too aggressive&quot; warning isn't visible to him.&lt;br&gt;&lt;br&gt;Knowing about Dr. Wenzel's colour blindness, Dr. Hupp pens in &quot;brain-boost! this is it!&quot; in bold blue letters on the too-aggressive serum label and invites Dr. Wenzel to cuddle with Mr. Cuddles. Dr. Wenzel, upon seeing the labels, immediately sees the bright blue nicely contrasted and doesn't even bother to check for his red &quot;new mixture&quot; label contraindicating the serum since it's difficult for him to discern and is redundant thanks to the helpful, clear label applied by his colleague.&lt;br&gt;&lt;br&gt;Cuddling with Mr. Cuddles doesn't go so well for him.&lt;br&gt;&lt;br&gt;Upon learning that Dr. Wenzel is color-blind and vulnerable to this kind of attack, Inspector Clu arrests Dr. Hupp on suspicion of deliberately confusing the identities of the two serums, resulting in a classic case of tigercide.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2014-11-28T00:06:14.500" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5677" PostTypeId="1" AcceptedAnswerId="5678" CreationDate="2014-11-28T03:58:53.387" Score="11" ViewCount="2587" Body="&lt;p&gt;One of the riddles we used to have painted on the walls of one of our classrooms in secondary school (with the last part added in :+ ):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What runs but doesn't walk,&lt;br&gt;&#xA;  has a mouth yet still can't talk,&lt;br&gt;&#xA;  has a head but doesn't think,&lt;br&gt;&#xA;  has a bed but will not sleep&lt;br&gt;&#xA;  and twists yet is not a riddle?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Took us 12 year old students a couple of days to figure out back then (despite it being terribly obvious in hindsight of course). So just wondering how many minutes it will take here :P .&lt;/p&gt;&#xA;" OwnerUserId="5719" LastEditorUserId="42452" LastEditDate="2019-04-08T00:13:41.150" LastActivityDate="2019-04-08T00:13:41.150" Title="Simple twisting riddle" Tags="|riddle|wordplay|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5678" PostTypeId="2" ParentId="5677" CreationDate="2014-11-28T04:03:50.597" Score="17" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A river. As you said, it is pretty obvious. They run, have heads, mouths, and beds, and they often twist.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1664" LastEditorUserId="1664" LastEditDate="2014-11-28T04:09:24.537" LastActivityDate="2014-11-28T04:09:24.537" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5679" PostTypeId="2" ParentId="5677" CreationDate="2014-11-28T04:10:48.927" Score="10" Body="&lt;p&gt;Is it a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; river?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A river runs, there is a river mouth, a river head, a riverbed, and there are rivers with twisting shapes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5709" LastActivityDate="2014-11-28T04:10:48.927" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5680" PostTypeId="1" AcceptedAnswerId="5681" CreationDate="2014-11-28T04:18:41.410" Score="12" ViewCount="700" Body="&lt;p&gt;Have a go at this riddle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Нidden in this іs an order, one so important that it may cost your life. Yet those that ԁo not know will not seе it, so look carefully beyond that which is seen, for your life depends on itǃǃǃ&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's a lot harder than my first one :D&lt;/p&gt;&#xA;" OwnerUserId="5719" LastEditorUserId="2484" LastEditDate="2014-11-29T05:11:56.090" LastActivityDate="2014-11-30T19:23:40.533" Title="The hidden order not to be ignored" Tags="|encoded-message|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5681" PostTypeId="2" ParentId="5680" CreationDate="2014-11-28T04:39:28.337" Score="11" Body="&lt;p&gt;The hidden order is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Ніԁеǃǃǃ&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is found by gathering all the characters in the riddle whose Unicode code points are above 128 (outside the ASCII range, in other words).&lt;/p&gt;&#xA;" OwnerUserId="4204" LastEditorUserId="4204" LastEditDate="2014-11-30T19:23:40.533" LastActivityDate="2014-11-30T19:23:40.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5682" PostTypeId="2" ParentId="5588" CreationDate="2014-11-28T04:52:26.713" Score="14" Body="&lt;p&gt;OK. I finally figured out this &lt;s&gt;accurs&lt;/s&gt; slightly frustrating puzzle. I shall forever curse the name of GOTO for more than an hour spent convolving kernels with elephant herds in vain attempts to find some sort of linear invariant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The missing cells, in standard reading order, are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; green elephant facing right&lt;br&gt;yellow elephant facing left&lt;br&gt;red elephant facing left&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The rationale is as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An elephant's &quot;gaze&quot; must be fixed on an elephant of identical colour. Let $D$ be the number of elephants in the 3x3 cell block surrounding an elephant that share the elephant's direction (this includes the elephant him/herself). The grid is toroidal, hence the grid &quot;wraps around&quot; at the edges, from right to left, top to bottom. An elephant's gaze is fixed on the cell $D$ cells ahead of him/her in the direction (s)he's looking (respecting wrapping). We shall call this property &quot;elephanticity&quot;.&lt;br&gt;&lt;br&gt;The three colours and directions listed in the above spoiler are the only three combinations that yield elephanticity for all cells in the rows in which the empty cells appear as well as the rows immediately above and below them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="4219" LastEditDate="2014-11-28T07:39:53.217" LastActivityDate="2014-11-28T07:39:53.217" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5684" PostTypeId="1" AcceptedAnswerId="5685" CreationDate="2014-11-28T08:55:03.607" Score="2" ViewCount="2453" Body="&lt;p&gt;Which square doesn't belong here and why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QunoV.gif&quot; alt=&quot;Image&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4973" LastEditorUserId="5373" LastEditDate="2015-04-28T19:31:31.157" LastActivityDate="2015-04-28T19:31:31.157" Title="Which square doesn't belong here and why?" Tags="|pattern|visual|" AnswerCount="7" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="5685" PostTypeId="2" ParentId="5684" CreationDate="2014-11-28T08:57:44.500" Score="15" Body="&lt;p&gt;Answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;3 (third square)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In all the other squares except&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;3rd square, &lt;strong&gt;3 (three) lines&lt;/strong&gt; divides the square in to &lt;strong&gt;5 parts&lt;/strong&gt;, while in 3rd square &lt;strong&gt;the lines (three lines)&lt;/strong&gt; divides it to &lt;strong&gt;6&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastActivityDate="2014-11-28T08:57:44.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5686" PostTypeId="2" ParentId="5684" CreationDate="2014-11-28T10:32:05.207" Score="8" Body="&lt;p&gt;Answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1, 2 and 4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;could be a part of representation of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; number 4, mirrored or rotated.&#xA;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/0n8ZX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But the picture in number &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is different.&lt;/p&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="2225" LastEditDate="2014-11-28T10:44:50.377" LastActivityDate="2014-11-28T10:44:50.377" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="5687" PostTypeId="2" ParentId="5190" CreationDate="2014-11-28T10:35:56.200" Score="-1" Body="&lt;p&gt;This is obviously just from your hints and not proven to work in any case:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;#1 -     0 0 1 0 1 1 1 0 0 0 0 0&lt;br&gt;&#xA;#2 -     1 0 0 1 0 1 1 1 0 0 0 0&lt;br&gt;&#xA;#3 -     0 1 0 0 1 0 1 1 1 0 0 0&lt;br&gt;&#xA;#4 -     0 0 1 0 0 1 0 1 1 1 0 0&lt;br&gt;&#xA;#5 -     0 0 0 1 0 0 1 0 1 1 1 0&lt;br&gt;&#xA;#6 -     1 0 0 0 1 0 0 1 0 1 1 1&lt;br&gt;&#xA;#7 -     0 1 0 0 0 1 0 0 1 0 1 1&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Victor's answer is already correct anyway.)&lt;/p&gt;&#xA;" OwnerUserId="5265" LastActivityDate="2014-11-28T10:35:56.200" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5689" PostTypeId="1" AcceptedAnswerId="5756" CreationDate="2014-11-28T12:00:09.947" Score="36" ViewCount="3023" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4BnVp.jpg&quot; alt=&quot;lolcat&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fsLJR.jpg&quot; alt=&quot;lolcat&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7Ak9p.jpg&quot; alt=&quot;lolcat&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZRu5X.jpg&quot; alt=&quot;lolcat&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3Fg7s.jpg&quot; alt=&quot;lolcat&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EyDeB.png&quot; alt=&quot;lolcat&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8G3at.jpg&quot; alt=&quot;lolcat&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XVJfP.jpg&quot; alt=&quot;lolcat&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xqiQI.jpg&quot; alt=&quot;lolcat&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/N6M3X.jpg&quot; alt=&quot;lolcat&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GVs9G.jpg&quot; alt=&quot;lolcat&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8o8Nx.jpg&quot; alt=&quot;lolcat&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/L2UFg.jpg&quot; alt=&quot;lolcat&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/llovg.jpg&quot; alt=&quot;lolcat&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CBsyb.jpg&quot; alt=&quot;lolcat&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ScesL.jpg&quot; alt=&quot;lolcat&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aQwR1.jpg&quot; alt=&quot;lolcat&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The Bad Guys (curse them!) have begun communicating in lolcats, to enable their Internet communications to take place unnoticed.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This is a job for the NSA/GCHQ crack team of cryptanalysts.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Unfortunately they're a bit busy today, so you'll have to do it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A secret message is concealed in the lolcatstream above.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It's a pencil-and-paper cipher. You do not need a computer to decipher it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;All the information you need is above the horizontal line. These instructions (everything below the horizontal line) do not contain hidden information.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;From HUMINT assets, you have received the Bad Guys' code book. You know for certain that the message above consists entirely of one of the following code words:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;YODA (meaning &quot;Your cover has been blown!&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LUKE (meaning &quot;Get me Tom Cruise's autograph!&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LEIA (meaning &quot;Explode the enormous bomb on Tuesday!&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FISHCAKE (meaning &quot;Send more food!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DURANCE (meaning &quot;Give yourself up to the police!&quot;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Summary (just for clarity). There is a hidden message in the lolcats above. It consists of one of this set of words: YODA, LUKE, LEIA, FISHCAKE, DURANCE. Your task is to discover which of those words the message consists of. Please explain your answer.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It's been a few days and nobody's cracked it, so here's a clue (if you need it): &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;em&gt;Cats like bacon&lt;/em&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="4050" LastEditDate="2014-12-09T17:00:41.213" LastActivityDate="2014-12-09T17:00:41.213" Title="Lolcat Steganography: Find the message hidden within the transport medium of humorous feline photography" Tags="|cipher|steganography|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="5690" PostTypeId="2" ParentId="5376" CreationDate="2014-11-28T12:02:56.810" Score="1" Body="&lt;p&gt;Could it be a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Shadow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I guess explanation won't be required after you revisit your's:-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're very close&#xA;But I'm always ahead&#xA;Can be beheaded&#xA;but can't be fed&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A shadow is always very close to you and ahead of you. Sometimes, because of the light's angle or broad daylight, you can behead it ( remove the head or kill its existence).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can have&#xA;a look inside&#xA;there will be&#xA;no place to hide&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wherever you go, it follows.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However,&#xA;You try to enter me&#xA;and more room arises&#xA;suddenly&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since the shodow moves with you, when you try to catch it ( as we must have tried as a child :) ), it moves with us and sometimes growing bigger.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2656" LastEditorUserId="2656" LastEditDate="2014-11-29T07:39:59.593" LastActivityDate="2014-11-29T07:39:59.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5692" PostTypeId="2" ParentId="5567" CreationDate="2014-11-28T12:20:50.143" Score="2" Body="&lt;p&gt;I fully agree with Julia Hayward's answer. Apparent laws of logic cannot prevent anyone from putting a treasure into the gold chest, a dragon into the silver one, and then make arbitrary inscriptions on them. Valid conclusions about physical world require some preliminary assumptions that themselves are not purely logical, and here we are supposed to derive one from propositions of unknown logical value, based only on their logical structure. This is not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Smullyan's book cited by Julia, there is also a statement which &quot;proves&quot; that unicorns exist. I don't remember it exactly, but I believe it was, &quot;This statement is false, and unicorns do not exist,&quot; which seems to imply, on a purely logical basis, that unicorns exist indeed. (In fact, anything could be &quot;proved&quot; this way.) Self-referencing propositions have been a known problem in logic for over a century, leading to contradictions and antinomies. That's why contemporary mathematics became strictly formalized in a way that does not allow them as valid statements.&lt;/p&gt;&#xA;" OwnerUserId="6080" LastActivityDate="2014-11-28T12:20:50.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5694" PostTypeId="2" ParentId="5376" CreationDate="2014-11-28T12:36:22.587" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A pair of eye glasses&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For obvious reasons; the first two verses, respectively: location of application and translucency. Taking poetic license for the third verse, that while in use, they have been 'entered' and provide more 'room' that you can:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; -&lt;em&gt;see&lt;/em&gt;. &quot;It's just a trick of the light.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4522" LastActivityDate="2014-11-28T12:36:22.587" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5695" PostTypeId="2" ParentId="5344" CreationDate="2014-11-28T14:01:33.323" Score="25" Body="&lt;p&gt;I assume that bold keys pressed two times, so we have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Enter Shift Space Space Alt F7 = / . . A B C C E F G G I I L M O P R R U Y&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;First, I substract the image name from the key list, IMG.GIF, and ALT F7 (command to clear cmd history), now we have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Enter Shift Space Space = / . A B C C E L O P R R U Y&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;because command prompt windows is open, I assume the victim write some cmd command, then by looking at this page &lt;a href=&quot;http://ss64.com/nt/&quot; rel=&quot;noreferrer&quot;&gt;http://ss64.com/nt/&lt;/a&gt; as refference, there is 3 possible command written by the victim &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ARP : Address Resolution Protocol&#xA;CLS : Clear the scree&#xA;COPY: Copy one or more files to another location`&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;the ARP and CLS is unlikely, so my best bet is &lt;code&gt;COPY&lt;/code&gt;, now we have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Enter Shift Space Space = / . A B C E L R R U&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then, because &lt;code&gt;/&lt;/code&gt; key is also there, I asssume the victim use &lt;code&gt;COPY&lt;/code&gt; command with options, by looking at the &lt;code&gt;COPY&lt;/code&gt; command refference, &lt;a href=&quot;http://ss64.com/nt/copy.html&quot; rel=&quot;noreferrer&quot;&gt;http://ss64.com/nt/copy.html&lt;/a&gt;, there is three possible options used by the victim:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;/A:  ASCII text file (default)&#xA;/B:  Binary file copy - will copy extended characters.&#xA;/L:  If source is a symbolic link copy the link to the target instead of the actual file the source link points to.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;/A&lt;/code&gt; is unlikely, &lt;code&gt;/L&lt;/code&gt; is unlikely too I think, my best bet is &lt;code&gt;/B&lt;/code&gt; now we have&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Enter Shift Space Space = . A C E L R R U&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I do not have any clue after this, but by writting the command together, I have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;COPY IMG.GIF + [. A C E L R R U] /B&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;hey [. A C E L R R U] must be some filename too, maybe its &lt;code&gt;CLUE.RAR&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;COPY IMG.GIF + CLUE.RAR /B&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now I have complete command, I google for &lt;code&gt;copy /b image&lt;/code&gt; the first result take me here &lt;a href=&quot;http://www.wikihow.com/Store-a-Rar-or-Zip-in-an-Image-File&quot; rel=&quot;noreferrer&quot;&gt;http://www.wikihow.com/Store-a-Rar-or-Zip-in-an-Image-File&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if you read the page, you will know that this command store archive in image file, this is some kind of steganography.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I download the image, then rename it to &lt;code&gt;CLUE.RAR&lt;/code&gt;, then open it, BANG... it's open, the archive contain one file &lt;code&gt;clue.wav&lt;/code&gt; I extract it, and open it with music player, there you can hear someone says &quot;the answer is inside this file&quot; with background sound which not clear, this is backmasking technique, if you reverse playback this sound, you will hear &quot;killer is febrina&quot;, and that is, the killer is &quot;FEBRINA H&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5958" LastEditorUserId="5958" LastEditDate="2016-02-08T13:46:56.297" LastActivityDate="2016-02-08T13:46:56.297" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5697" PostTypeId="2" ParentId="5684" CreationDate="2014-11-28T14:45:22.000" Score="5" Body="&lt;p&gt;I choose&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; frame 3 because is the only one that shows the triangle created by the three lines through their pairwise intersections. In all four cases the (non-trivial) triangle exists, but in cases 1, 2 and 4 it is not presented within respective frame. (Edit: note that this is technically equivalent to AeJey's answer that observes the number of parts the frame is split into by the three lines.) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6087" LastEditorUserId="6087" LastEditDate="2014-11-30T18:42:17.573" LastActivityDate="2014-11-30T18:42:17.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5698" PostTypeId="2" ParentId="5684" CreationDate="2014-11-28T15:27:20.993" Score="3" Body="&lt;p&gt;The answer is;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Frame 3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For three reasons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you extend the lines of 1, 2, and 4, they can join in a path of 3 lines which are in broken. You cannot do this with 3.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;2&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you look at how many sections of white space there are in 1, 2, and 4 there's 5. In 3 there's 6&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;3&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3 is the only square, where the lines within the frame make a triangle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So obviously it has to be the answer stated above!&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-11-29T09:26:20.230" LastActivityDate="2014-11-29T09:26:20.230" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5699" PostTypeId="2" ParentId="5684" CreationDate="2014-11-28T15:29:24.197" Score="12" Body="&lt;p&gt;Only number 1...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; has lines that meet the bounding square at a corner&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Only number 2...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; lacks any lines that slant down and to the right&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Only number 3...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (a) divides the square into 6 areas instead of 5, (b) contains two complete intersections, and (c) includes a complete red triangle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Only number 4...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; has no uniquely distinguishing characteristics.  So my guess is 4.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2162" LastActivityDate="2014-11-28T15:29:24.197" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5700" PostTypeId="2" ParentId="5684" CreationDate="2014-11-28T16:28:50.333" Score="35" Body="&lt;p&gt;The big orange one. All the others are brown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's clearly the odd one out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/I5Ihp.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4852" LastEditorUserId="4852" LastEditDate="2014-11-28T16:39:30.393" LastActivityDate="2014-11-28T16:39:30.393" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="5706" PostTypeId="2" ParentId="5376" CreationDate="2014-11-28T19:38:12.367" Score="4" Body="&lt;p&gt;The answer is simply:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hat&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You're very close&#xA;But I'm always ahead&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ahead -&gt; A-head -&gt; A-het -&gt; A-hat (just keep saying it over and over again (see title))&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can be beheaded&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can take me off your head :D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but can't be fed&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Indicates inanimate - but since &quot;can be beheaded&quot; seems a contradiction - thus is the art of riddles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can have&#xA;a look inside&#xA;there will be&#xA;no place to hide&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This inanimate object is hollow but not large enough to hide inside. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However,&#xA;You try to enter me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;U&quot; try to enter. &quot;U&quot; displaces &quot;A&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and more room arises&#xA;suddenly&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hat -&gt; Hut. A hut is larger than a hat. By putting a &quot;U&quot; in the hat (displacing &quot;A&quot;) more room has been created :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastActivityDate="2014-11-28T19:38:12.367" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5707" PostTypeId="2" ParentId="5190" CreationDate="2014-11-28T21:39:59.377" Score="9" Body="&lt;p&gt;There is some literature on this problem. Lindström [2] and Cantor and Mills [1] independently give weighing strategies using 7 weighings, with the property that no weighing depends on the results of previous weighings. More generally, these papers prove that for every positive integer $k$, $2^{k}-1$ weighings suffice to determine which of $2^{k-1}k$ coins have which weights.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The weighing strategies are constructed recursively. The general process is a little intricate, so I won't try to describe it here. If I understood everything correctly, the strategy given in [1] for 12 coins suggests putting the following coins on the scale:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1, 2, 4, 5, 7, 9, 11&lt;/li&gt;&#xA;&lt;li&gt;2, 3, 5, 6, 12&lt;/li&gt;&#xA;&lt;li&gt;1, 3, 5, 7, 8, 10&lt;/li&gt;&#xA;&lt;li&gt;5, 6, 8, 9&lt;/li&gt;&#xA;&lt;li&gt;1, 2, 4, 6, 8, 10&lt;/li&gt;&#xA;&lt;li&gt;2, 3, 8, 9&lt;/li&gt;&#xA;&lt;li&gt;1, 3, 6, 9&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$\,$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Cantor, David G. and Mills, W.H.. Determination of a subset from certain combinatorial properties. &lt;em&gt;Canad. J. Math.&lt;/em&gt; &lt;strong&gt;18&lt;/strong&gt; 1966 42-48&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] Lindström, Bernt. On a combinatorial problem in number theory. &lt;em&gt;Canad. Math. Bull.&lt;/em&gt; &lt;strong&gt;8&lt;/strong&gt; 1965 477-490 &lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2014-11-28T21:49:33.333" LastActivityDate="2014-11-28T21:49:33.333" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5709" PostTypeId="2" ParentId="5684" CreationDate="2014-11-29T09:08:52.263" Score="5" Body="&lt;p&gt;Clearly the odd one out is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Graphical explanation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/7MYtX.gif&quot; alt=&quot;Explanation&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wordy explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1,3,2 are slices of a bigger picture. 4 does not fit with any of them. Even if we prolongue the lines in 1,2,3 they still fit well in the overall picture.&#xA;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3918" LastEditorUserId="3918" LastEditDate="2014-11-30T09:01:10.753" LastActivityDate="2014-11-30T09:01:10.753" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5711" PostTypeId="5" CreationDate="2014-11-29T14:53:44.613" Score="0" Body="&lt;p&gt;The MPire coloring game is a game where several &quot;empires&quot; are laid out touching each other. Your task is to color each empire a different color, and not let two of the same color touch. You also have to use as few colors as possible.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2014-11-29T20:13:01.590" LastActivityDate="2014-11-29T20:13:01.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5712" PostTypeId="4" CreationDate="2014-11-29T14:53:44.613" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-11-29T14:53:44.613" LastActivityDate="2014-11-29T14:53:44.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5714" PostTypeId="1" AcceptedAnswerId="5715" CreationDate="2014-11-29T17:42:10.917" Score="7" ViewCount="1216" Body="&lt;p&gt;This is an old puzzle and won't take long to be solved, but I think it's a useful addition to the site.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In a hotel in France at the time of the German occupation, two men and two women are in a lift moving downwards. One of the men is a uniformed German officer, the other a secret member of the French resistance. One of the women is young and beautiful, and the other is an elderly lady.&lt;/p&gt;&#xA;&lt;p&gt;A momentary power cut causes the lights to go out for thirty seconds, leaving the four people in pitch blackness. There is the sound of a kiss followed by a punch. The lights come back on again, revealing that the German has just acquired a black eye.&lt;/p&gt;&#xA;&lt;p&gt;The old lady thinks, &amp;quot;It's nice to see that young women these days can look after themselves. That rude German must have kissed her, but she gave him what for!&amp;quot; &lt;br&gt; The German thinks, &amp;quot;Ow! That &lt;em&gt;schweinhund&lt;/em&gt; Frenchman must have kissed this young lady, and she punched me by mistake.&amp;quot; &lt;br&gt; The young lady thinks, &amp;quot;What odd people these Nazis are! He must have ignored me and kissed this old woman, or even the other man, and got punched for his trouble.&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;What happened?&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-03-30T19:58:28.733" Title="Four people in a dark lift" Tags="|logical-deduction|mystery|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5715" PostTypeId="2" ParentId="5714" CreationDate="2014-11-29T17:49:20.357" Score="6" Body="&lt;p&gt;A very intriguing puzzle, but one that is easy to find the solution to (I have already seen this so I knew the answer).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Frenchman makes a kissing sound, then punches the German officer in the eye. Nobody was really kissed, so nobody was offended, and the officer did not suspect the Frenchman of punching him.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1664" LastEditorUserId="5373" LastEditDate="2014-11-29T18:24:41.687" LastActivityDate="2014-11-29T18:24:41.687" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5717" PostTypeId="2" ParentId="5677" CreationDate="2014-11-29T19:39:32.590" Score="4" Body="&lt;p&gt;Anyone with &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; young toddlers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;knows that they:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;tend to run instead of walk&lt;/li&gt;&#xA;&lt;li&gt;have a mouth but don't talk&lt;/li&gt;&#xA;&lt;li&gt;certainly have a head but don't think all that well&lt;/li&gt;&#xA;&lt;li&gt;don't use their bed to sleep&lt;/li&gt;&#xA;&lt;li&gt;will twist if you try to hold them&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6098" LastActivityDate="2014-11-29T19:39:32.590" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5718" PostTypeId="2" ParentId="2548" CreationDate="2014-11-29T20:39:09.580" Score="7" Body="&lt;p&gt;Another non-trivial possibility:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\infty$&lt;/p&gt;&#xA;" OwnerUserId="6101" LastActivityDate="2014-11-29T20:39:09.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5719" PostTypeId="1" AcceptedAnswerId="5726" CreationDate="2014-11-29T20:41:05.080" Score="4" ViewCount="708" Body="&lt;p&gt;&lt;em&gt;I might be a city in one of the 'stans. &lt;br&gt;&#xA;I might be a river or more in Australia. &lt;br&gt;&#xA;I might have once bifurcated cleanly. &lt;br&gt;&#xA;I might have colleges named after me. &lt;br&gt;&#xA;I might be said to be covered in blood. &lt;br&gt;&#xA;What can I be to gather such renown?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is a single word.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2014-11-29T20:51:00.190" LastActivityDate="2014-11-30T14:10:20.710" Title="Riddle - who am I to gather such renown?" Tags="|riddle|word|" AnswerCount="1" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="5720" PostTypeId="5" CreationDate="2014-11-29T23:52:54.247" Score="0" Body="&lt;p&gt;Tic-tac-toe, or 'Noughts and crosses', is a paper-and-pencil game for two players, who take turns marking the spaces in a 3×3 grid with the symbol 'O' or 'X', respectively.&lt;/p&gt;&#xA;&lt;p&gt;The aim of the game is to place three of one's own marks in a horizontal, vertical, or diagonal row. If both players play optimally, a game will always result in a draw.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="60644" LastEditDate="2021-02-02T13:55:28.220" LastActivityDate="2021-02-02T13:55:28.220" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="5721" PostTypeId="4" CreationDate="2014-11-29T23:52:54.247" Score="0" Body="Puzzles centered around tic-tac-toe (a.k.a. noughts and crosses)." OwnerUserId="5373" LastEditorUserId="8874" LastEditDate="2015-01-31T20:35:50.230" LastActivityDate="2015-01-31T20:35:50.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5722" PostTypeId="5" CreationDate="2014-11-29T23:54:09.980" Score="0" Body="&lt;p&gt;For puzzles involving addition (&lt;span class=&quot;math-container&quot;&gt;$+$&lt;/span&gt;), subtraction (&lt;span class=&quot;math-container&quot;&gt;$-$&lt;/span&gt;), multiplication (&lt;span class=&quot;math-container&quot;&gt;$\times$&lt;/span&gt;), or division (&lt;span class=&quot;math-container&quot;&gt;$\div$&lt;/span&gt;).&lt;/p&gt;&#xA;" OwnerUserId="11497" LastEditorUserId="11110" LastEditDate="2020-04-30T06:50:49.893" LastActivityDate="2020-04-30T06:50:49.893" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="5723" PostTypeId="4" CreationDate="2014-11-29T23:54:09.980" Score="0" Body="For puzzles involving addition, subtraction, multiplication, or division. Use with [mathematics]" OwnerUserId="5373" LastEditorUserId="69582" LastEditDate="2021-03-08T19:39:09.480" LastActivityDate="2021-03-08T19:39:09.480" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="5724" PostTypeId="5" CreationDate="2014-11-29T23:54:51.320" Score="0" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Dice&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dice&lt;/a&gt; are small throwable objects used for generating random numbers.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A traditional die is a rounded cube, with each of its six faces showing a different number of dots (pips) from 1 to 6. When thrown or rolled, the die comes to rest showing on its upper surface a random integer from 1 to 6, each value being equally likely.&lt;/li&gt;&#xA;&lt;li&gt;A variety of similar items are also described as dice; such specialized dice may be tetrahedrons, dodecahedrons, or irregular solids, and may have faces marked with symbols instead of numbers. Different dice are often referred to as 'd#' where the '#' references the highest value on the dice (i.e. d6 for 6-sided, d20 for 20-sided, etc.).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Non-standard numberings include the &lt;a href=&quot;https://en.wikipedia.org/wiki/Sicherman_dice&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sicherman&lt;/a&gt; dice (&lt;span class=&quot;math-container&quot;&gt;$1, 2, 2, 3, 3, 4$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$1, 3, 4, 5, 6, 8$&lt;/span&gt; - throw them together and the distribution is the same as with two normal dice), and &lt;a href=&quot;https://en.wikipedia.org/wiki/Nontransitive_dice&quot; rel=&quot;nofollow noreferrer&quot;&gt;nontransitive&lt;/a&gt; dice, where A beats B over &lt;span class=&quot;math-container&quot;&gt;$50\%$&lt;/span&gt; of the time, B beats C over &lt;span class=&quot;math-container&quot;&gt;$50\%$&lt;/span&gt; of the time, but A doesn't beat C over &lt;span class=&quot;math-container&quot;&gt;$50\%$&lt;/span&gt; of the time, for example &lt;span class=&quot;math-container&quot;&gt;$[2, 2, 4, 4, 9, 9]$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$[1, 1, 6, 6, 8, 8]$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$[3, 3, 5, 5, 7, 7]$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Typically, dice faces are arranged such that opposite faces always add up to the sum of the highest and lowest values on the dice. For example, with a d6 (six-sided die) the highest value is 6 and the lowest is 1. Therefore the numbers are typically arranged so that opposite faces add up to 7 (1 and 6, 2 and 5, 3 and 4).&lt;/p&gt;&#xA;&lt;p&gt;A loaded dice is one that doesn't have a uniform distribution for the probability of landing on each face.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="60644" LastEditDate="2023-01-31T16:44:33.407" LastActivityDate="2023-01-31T16:44:33.407" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="5725" PostTypeId="4" CreationDate="2014-11-29T23:54:51.320" Score="0" Body="For puzzles involving any form of dice." OwnerUserId="5373" LastEditorUserId="11497" LastEditDate="2020-02-20T15:18:11.437" LastActivityDate="2020-02-20T15:18:11.437" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="5726" PostTypeId="2" ParentId="5719" CreationDate="2014-11-30T05:41:11.650" Score="4" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;MARY&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might be a city in one of the 'stans. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Mary is the capital city of Mary Province in Turkmenistan&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might be a river or more in Australia.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Mary River is a large river system in South East Queensland, Australia. [OP note: there are three Mary Rivers in different parts of Australia - &lt;a href=&quot;http://en.wikipedia.org/wiki/Mary_River&quot; rel=&quot;nofollow&quot;&gt;reference&lt;/a&gt;.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might have once bifurcated cleanly&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This is referring to the Virgin Mary, who gave birth after an 'immaculate conception'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might have colleges named after me. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There are many colleges and universities named after Mary (Ex: St.Mary's College)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might be said to be covered in blood.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Reference to Bloody Mary&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What can I be to gather such renown?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Because she is the mother of 'Jesus Christ'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="5373" LastEditDate="2014-11-30T14:10:20.710" LastActivityDate="2014-11-30T14:10:20.710" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5727" PostTypeId="2" ParentId="2548" CreationDate="2014-11-30T06:58:31.993" Score="1" Body="&lt;p&gt;The answer is twice itself. 2 possibilities so far. &lt;br&gt;&#xA;1. If it's a number, it can be 0 as well as infinity. &lt;br&gt;&#xA;2. If it's a statement, &quot;twice the answer of your question&quot; should work.&lt;/p&gt;&#xA;" OwnerUserId="6107" LastEditorUserId="5373" LastEditDate="2014-11-30T11:31:38.447" LastActivityDate="2014-11-30T11:31:38.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5728" PostTypeId="1" AcceptedAnswerId="5729" CreationDate="2014-11-30T15:32:05.310" Score="-4" ViewCount="575" Body="&lt;p&gt;You're not allowed to restate the question, or answer with a question. What could the answer be?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2014-11-30T17:03:41.393" Title="What could the only answer to this question be?" Tags="|lateral-thinking|" AnswerCount="1" CommentCount="2" ClosedDate="2014-11-30T15:40:02.517" ContentLicense="CC BY-SA 3.0" />
  <row Id="5729" PostTypeId="2" ParentId="5728" CreationDate="2014-11-30T15:53:06.760" Score="5" Body="&lt;p&gt;There couldn't be any answers to this question because this question was put on hold by Doorknob.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://meta.stackexchange.com/questions/10582/what-is-a-closed-or-on-hold-question&quot;&gt;Meta.SE&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h2&gt;What does it mean for a question to be &lt;em&gt;closed&lt;/em&gt;?&lt;/h2&gt;&#xA;  &#xA;  &lt;p&gt;When a question is closed, &lt;strong&gt;no additional answers may be posted to it&lt;/strong&gt;,&#xA;  although the question and existing answers can still be edited (by&#xA;  users with edit privileges or by suggested edit) and voted upon, and&#xA;  will continue to count for badges. The asker of a closed question may&#xA;  still accept an answer.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Closed questions can be re-opened by users who have sufficient&#xA;  reputation.&lt;/p&gt;&#xA;  &#xA;  &lt;h2&gt;What does it mean for a question to be &lt;em&gt;on hold&lt;/em&gt;?&lt;/h2&gt;&#xA;  &#xA;  &lt;p&gt;If a question has been closed, then for the first 5 days, it is marked&#xA;  as “on hold” rather than “closed”. This is meant to convey that the&#xA;  question requires improvement and may be reopened if improved. During&#xA;  this period, if the question is edited &lt;s&gt;by the asker&lt;/s&gt;&#xA;  (&lt;a href=&quot;https://meta.stackexchange.com/questions/196026/nobody-in-my-community-has-the-power-to-re-open-a-question#comment618477_196026&quot;&gt;now&lt;/a&gt;&#xA;  by anyone), it will be added to the &lt;a href=&quot;https://meta.stackoverflow.com/help/reopen-questions&quot;&gt;reopen&#xA;  queue&lt;/a&gt;. &lt;strong&gt;Other&lt;/strong&gt;&#xA;  &lt;strong&gt;than this, there is no functional difference between “on hold” and&lt;/strong&gt;&#xA;  &lt;strong&gt;“closed”&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&lt;sup&gt;Please not reopen the question or delete my answer because we all love fun :)&lt;/sup&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="432" LastEditorUserId="-1" LastEditDate="2017-03-20T10:32:12.560" LastActivityDate="2014-11-30T17:03:41.393" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5731" PostTypeId="1" AcceptedAnswerId="5732" CreationDate="2014-11-30T16:19:00.210" Score="11" ViewCount="608" Body="&lt;p&gt;It's well-known that a knight placed on one square of a chessboard can get to any other square, but a bishop can only reach half the squares from a fixed starting point. &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2929/the-bunnys-tour&quot;&gt;Another question on this site&lt;/a&gt; dealt with a new type of chess piece with a different way of moving. I'm trying to generalise all of these ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A knight's move takes you between opposite corners of a $2\times3$ rectangle of squares. A bishop's move takes you between opposite corners of a $2\times2$ rectangle. So let's define a &lt;strong&gt;General&lt;/strong&gt; (a new, generalised breed of chess piece) to be a piece that can move from its current position to the opposite corner of an $m\times n$ rectangle. &lt;strong&gt;For what values of $m$ and $n$ is it possible for a General to start from one square and reach every other square on the chessboard?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Possible further generalisations include changing the chessboard to $M\times N$ rather than $8\times8$; looking for a 'General's tour', the General visiting each square exactly once; or looking for a 'General's tour' with the General visiting each square exactly once and ending on the same square it began on.)&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2023-05-21T07:19:05.107" Title="Generalisation of tours on chessboards" Tags="|chess|checkerboard|graph-theory|" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="5732" PostTypeId="2" ParentId="5731" CreationDate="2014-11-30T16:43:28.327" Score="15" Body="&lt;p&gt;This question is answered in a &lt;a href=&quot;http://arxiv.org/abs/math/9411240&quot; rel=&quot;nofollow noreferrer&quot;&gt;paper by Donald Knuth&lt;/a&gt; [1].&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Theorem&lt;/strong&gt;: An &lt;span class=&quot;math-container&quot;&gt;$(m\times n)$&lt;/span&gt;-Leaper, with &lt;span class=&quot;math-container&quot;&gt;$1\leq m\leq n$&lt;/span&gt;, can reach every square on an &lt;span class=&quot;math-container&quot;&gt;$M\times N$&lt;/span&gt; board, with &lt;span class=&quot;math-container&quot;&gt;$2\leq M\leq N$&lt;/span&gt;, if and only if the following three conditions hold:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$m+n$&lt;/span&gt; is relatively prime to &lt;span class=&quot;math-container&quot;&gt;$m-n$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$N\geq 2n$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$M\geq m+n$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The proof is a couple of pages. Maybe it's possible to give a quicker argument for &lt;span class=&quot;math-container&quot;&gt;$8\times 8$&lt;/span&gt; boards.&lt;/p&gt;&#xA;&lt;p&gt;[1] Knuth, Donald. Leaper Graphs. &lt;em&gt;The Mathematical Gazette&lt;/em&gt;, Vol. 78, No. 483 (Nov., 1994), pp. 274-297&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="5373" LastEditDate="2023-05-21T07:19:05.107" LastActivityDate="2023-05-21T07:19:05.107" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="5733" PostTypeId="2" ParentId="5472" CreationDate="2014-11-30T17:27:55.563" Score="1" Body="&lt;p&gt;As the correct answer has been found, I finish up this puzzle by presenting some solution bitmaps for easier understanding. The solution is identical to the one accepted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What the hints tell you:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The message was spelled in Morse-code with the numbers specifying the grid. The CASE of the letters can therefore not be of any importance. (Morse code is case insensitive.)&#xA; The one-eyed spy needed to have more than 6 different colours, so he was quite likely not after vowels. He has more colours than different letters - most likely he wants to colourise the note... which can only mean their position is of importance. (This was a real give-way clue, I guess.)&#xA; The final message was coded on the label-printer. So the case-sensitive argument from above does not apply. Indeed, one needs to find the upper-case letters of the keyword to get to the final message correctly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Finding the keyword in the grid (part 1):&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Only 6 letters appear in positions outlining the letter themselves&#xA; &lt;img src=&quot;https://i.stack.imgur.com/Wpx1u.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This further leads to the final keyword: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The order of the letters is determined by their appearance (top row) and the word therefore is:&#xA; &lt;img src=&quot;https://i.stack.imgur.com/oyyFo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Using the keyword to decipher the message:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Searching (potentially cutting) the label-strip at the upper-case keyword letters and placing the stripes next to each other reveals the message below.&#xA; Note that the keyword is required, because the message consist of a prime-number of letters and does not form a regular grid.&#xA; &lt;img src=&quot;https://i.stack.imgur.com/y6TnW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Final note:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I know that the second encoding is very week. In particular due to the X-delimiters. I suppose one can solve this puzzle &quot;bottom up&quot; by first solving the message. However, this second riddle was only meant as a less blunt cross-check than &quot;The keyword is...&quot;. The real puzzle was the first one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="5840" LastActivityDate="2014-11-30T17:27:55.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5734" PostTypeId="1" AcceptedAnswerId="5735" CreationDate="2014-11-30T17:44:38.893" Score="41" ViewCount="3632" Body="&lt;p&gt;So I was given this puzzle quite some time ago and just thought it would make a nice addition to this site:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your task is to determine the method how the result is determined from the given number. A few example cases include:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;177383 -&amp;gt; 2&#xA;267453 -&amp;gt; 2&#xA;111111 -&amp;gt; 0&#xA;636240 -&amp;gt; 4&#xA;367183 -&amp;gt; 3&#xA;247123 -&amp;gt; 1&#xA;369108 -&amp;gt; 5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To those, who hit a brick wall, here is a tip: This puzzle is easiest for young children and gets harder the older you get... Happy guessing ;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1943" LastEditorUserId="1943" LastEditDate="2016-09-02T21:30:11.070" LastActivityDate="2023-11-20T16:02:58.297" Title="Determine the algorithm" Tags="|mathematics|pattern|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5735" PostTypeId="2" ParentId="5734" CreationDate="2014-11-30T17:55:06.840" Score="35" Body="&lt;p&gt;The number is the&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; number of &amp;quot;holes&amp;quot; in the written decimal representation of the number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;More specifically,&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; each digit contributes this much to the total number:&#xA;&#xA; &lt;ul&gt;&#xA; &lt;li&gt;0: 1&lt;/li&gt;&#xA; &lt;li&gt;1: 0&lt;/li&gt;&#xA; &lt;li&gt;2: 0&lt;/li&gt;&#xA; &lt;li&gt;3: 0&lt;/li&gt;&#xA; &lt;li&gt;4: 0 or 1, depending on how you draw them&lt;/li&gt;&#xA; &lt;li&gt;5: 0&lt;/li&gt;&#xA; &lt;li&gt;6: 1&lt;/li&gt;&#xA; &lt;li&gt;7: 0&lt;/li&gt;&#xA; &lt;li&gt;8: 2&lt;/li&gt;&#xA; &lt;li&gt;9: 1&lt;/li&gt;&#xA; &lt;/ul&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;In your case above,&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;  4 is drawn with a hole, so it contributes 1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="69582" LastEditDate="2023-11-20T16:02:58.297" LastActivityDate="2023-11-20T16:02:58.297" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="5736" PostTypeId="1" AcceptedAnswerId="5772" CreationDate="2014-11-30T19:05:58.657" Score="12" ViewCount="1944" Body="&lt;p&gt;This will be a tricky one, I think.&#xA;The whole puzzle is self-contained in the information between the two horizontal bars below.&lt;/p&gt;&#xA;&lt;p&gt;The answer to this riddle is a book title of 5 words. The book is rather well known literature and can be legally obtained for free as e-book.&lt;/p&gt;&#xA;&lt;p&gt;The word search-puzzle comes with a few &lt;em&gt;important&lt;/em&gt; clues (see below), but there is no additional information to this puzzle outside the two horizontal bars.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Can you find the 5-word-title ?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The whole puzzle and the book title are in English. Although there are several &amp;quot;stages&amp;quot; to this puzzle, you will need to look at the puzzle as a whole, i.e. the additional clues may not be disregarded. (Or you will not solve this puzzle.)&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Words are born as letters, but they reach adulthood when they are six. (&lt;em&gt;And this puzzle is not child's play!&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;p&gt;You find us from East to West and West to East, from North to South and South to North. And in between, can we also be seen.  We do not jump, we do not bend, a straight line we form, from start to end.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;C E O F K O S D B R P &#xA;Y L T O I W G U M S R &#xA;J E S P B O U A A K W &#xA;W I L U W S D G J T E &#xA;A N E A Q L H K G Y N &#xA;D N A S Y S C Y L P L &#xA;Z O K L O E D T L W Z &#xA;N U G S Y C O F A R T &#xA;F E H U F D S T L E S &#xA;J U P H Z V H R T Y U &#xA;W V C X V Y E Y C A D&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;We meet at Waterloo station.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;__ &amp;lt;-&amp;gt; __&#xA;__ &amp;lt;-&amp;gt; __&#xA;__ &amp;lt;-&amp;gt; __&#xA;__ &amp;lt;-&amp;gt; __&#xA;__ &amp;lt;-&amp;gt; __&#xA;__ &amp;lt;-&amp;gt; __&#xA;&#xA;B*H***&#xA;B***KF***&#xA;H****H-*UG&#xA;**F**QU***&#xA;M*****G*&#xA;*********&#xA;P**K**&#xA;******&#xA;U*U****&#xA;V******&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;hr /&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-12-09T17:37:16.410" Title="The horrible wordsearch: Can you decypher the word-search puzzle to find the final clue?" Tags="|cipher|word-search|" AnswerCount="3" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="5737" PostTypeId="1" AcceptedAnswerId="5739" CreationDate="2014-11-30T20:07:00.573" Score="9" ViewCount="614" Body="&lt;h2&gt;Instructions&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The game is to &lt;strong&gt;fill in the blank&lt;/strong&gt; with a word or few - no more than three.  No tricksy nonsense.  No hidden meanings.  No &quot;lateral thinking&quot;. In fact, only vertical, parallel, dorsal, collinear, and/or perpendicularly derived answers will be accepted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end, name the apparent theme (for the last blank).  Then, name the most bizarre variation you've encountered, along with your personal go-to choice.  Answer now! &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;And don't forget not to spoil the fun for others!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Simple Pleasures, Hidden Treasures&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;How popular, you with a piquant manner,&lt;br/&gt;&#xA;And boldly do you raise the banner.&lt;br/&gt;&#xA;It might just be some strange bologna?&lt;br/&gt;&#xA;The consensus garnish is&lt;/em&gt; _____________.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Alone or in harmony with the aforementioned,&lt;br/&gt;&#xA;Sliced, earthy gifts for the well-intentioned...&lt;br/&gt;&#xA;Kids don't enjoy them in their school lunchroom,&lt;br/&gt;&#xA;But fun guys tolerate the humble&lt;/em&gt; _____________.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Sometimes the minimum ain't enough.&lt;br/&gt;&#xA;Above and beyond's how they want the stuff.&lt;br/&gt;&#xA;&quot;A bit more sprinkled, if you please.&quot;&lt;br/&gt;&#xA;Some like theirs with&lt;/em&gt; _____________.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What's this you smell (and hold your nose)?&lt;br/&gt;&#xA;&quot;Somethings fishy,&quot; you'll presuppose.&lt;br/&gt;&#xA;Surely, folks who like them have multiple neuroses.&lt;br/&gt;&#xA;Who actually relishes canned&lt;/em&gt; _____________?&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;It's fresh like IHOP's Rooty Tooty&lt;br/&gt;&#xA;But feeling merely half as fruity.&lt;br/&gt;&#xA;Pig out on the rest with an iced tea Snapple®.&lt;br/&gt;&#xA;My lady's fave must be&lt;/em&gt; _____________.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Now, I'm of a simpler taste: just the classics.&lt;br/&gt;&#xA;No need for leftovers resembling some plastics.&lt;br/&gt;&#xA;Plus, there's less grease in keeping it simple.&lt;br/&gt;&#xA;Perhaps I'll avoid an extraneous pimple.&lt;br/&gt;&#xA;The ignorant take it for a well sauced pita.&lt;br/&gt;&#xA;I stand by the original,&lt;/em&gt; _____________.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;So sorry to bore you with my keen obsession.&lt;br/&gt;&#xA;It could be the product of sex-'shul repression.&lt;br/&gt;&#xA;Still, life's greatest treats are but simple pleasure,&lt;br/&gt;&#xA;Variety adding the spice for good measure.&lt;br/&gt;&#xA;I don't mean to continue with this name-dropping;&lt;br/&gt;&#xA;Time to cease listing forms of&lt;/em&gt; _____________.&lt;br/&gt;&lt;/p&gt;&#xA;" OwnerUserId="5948" LastEditorUserId="5373" LastEditDate="2016-12-14T11:43:11.640" LastActivityDate="2016-12-14T11:43:11.640" Title="Variations on a Theme" Tags="|riddle|word|rhyme|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5738" PostTypeId="2" ParentId="5737" CreationDate="2014-11-30T20:17:04.400" Score="4" Body="&lt;p&gt;The seventh and final blank, which gives the theme, is surely&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; pizza topping!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first blank (I did see mdc32 and pacoverflow's answers for this one!):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; pepperoni.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second blank is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; mushroom (&quot;fun guys&quot; is a hint to this, since it sounds like 'fungi').&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Third blank could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; extra cheese?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fourth blank could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; anchovies? (doesn't quite rhyme with neuroses, but &quot;fishy&quot; is a hint to this).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fifth blank is probably&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ham and pineapple (a weird but popular pizza topping; &quot;pig out&quot; is a hint to this).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sixth blank must be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; margherita (this was the easiest, since it's &quot;just the classics&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The most bizarre variation I've encountered is definitely&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a fried egg.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My go-to choice:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I love olives on pizza!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2014-11-30T20:34:25.577" LastActivityDate="2014-11-30T20:34:25.577" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5739" PostTypeId="2" ParentId="5737" CreationDate="2014-11-30T20:22:37.430" Score="5" Body="&lt;p&gt;The consensus garnish is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; pepperoni&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But fun guys tolerate the humble&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; mushroom&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some like theirs with &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; extra cheese&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who actually relishes canned&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; anchovies&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My lady's fave must be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ham and pineapple&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I stand by the original, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; margherita&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Time to cease listing forms of &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; pizza toppings&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Most bizarre variation I've encountered:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; seafood (from an Asian Pizza Hut)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My go-to choice:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; sausage&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4450" LastEditDate="2014-11-30T20:27:52.707" LastActivityDate="2014-11-30T20:27:52.707" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5740" PostTypeId="2" ParentId="5737" CreationDate="2014-11-30T20:23:08.690" Score="2" Body="&lt;p&gt;As rand'al'thor said, the last one is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pizza topping&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think I have them figured out though.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1: Pepperoni &lt;br/&gt;&#xA; 2: Mushroom &lt;br/&gt;&#xA; 3: Extra cheese &lt;br/&gt;&#xA; 4: Anchovies? &lt;br/&gt;&#xA; 5: Ham and Pineapple (Hawaiian pizza is great) &lt;br/&gt;&#xA; 6: Margherita &lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My personal favorite is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bacon and garlic sauce. They have it at Domino's (I know, quality pizza), and it's the best I've ever had.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Weirdest variation? None. I'm pretty mainstream, sorry.&lt;/p&gt;&#xA;" OwnerUserId="1664" LastEditorUserId="1664" LastEditDate="2014-11-30T20:37:58.890" LastActivityDate="2014-11-30T20:37:58.890" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5741" PostTypeId="5" CreationDate="2014-11-30T21:18:57.057" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-11-30T21:18:57.057" LastActivityDate="2014-11-30T21:18:57.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5742" PostTypeId="4" CreationDate="2014-11-30T21:18:57.057" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-11-30T21:18:57.057" LastActivityDate="2014-11-30T21:18:57.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5743" PostTypeId="1" AcceptedAnswerId="5746" CreationDate="2014-11-30T21:53:29.380" Score="2" ViewCount="366" Body="&lt;p&gt;Your old friend from college is visiting and brings back this weird electronic machine with him. &lt;strong&gt;He says that it's supposed to take input and spit out a number&lt;/strong&gt;, but some prank was pulled on him (he was kind of a geek) where he, and the device, got dunked into the pool. It was turned on and your friend was electrocuted. Luckily he did survive (he did visit didn't he?) but during that &lt;em&gt;shocking&lt;/em&gt; moment he forgot how his device worked, and the device was broken.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He came to you hoping you could help, he was hoping you could remember how it had worked. After you reply &quot;no&quot;, he still wants help. He says you're the smartest person he knows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Since it's broken the only information you have is a video you guys had took in college, and a weird note that had slowly been disappearing over the years.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the video you guys were testing it. The few examples you could make out were:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$1863784 \rightarrow 1$&lt;br&gt;&#xA;  $1243242 \rightarrow 9$&lt;br&gt;&#xA;  $1742 \rightarrow 5$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The note seemed to be a set of instructions the computer followed:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Stp1&lt;br&gt;&#xA;  adalheumrs&lt;br&gt;&#xA;  Sp2&lt;br&gt;&#xA;  rpatstputloetsnledgt&lt;br&gt;&#xA;  Sep3&lt;br&gt;&#xA;  rurnvle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your friend was a terrible writer back then and his words were packed tightly together, therefore you couldn't distinguish any spaces. It was written in English.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your friend needs your help to figure out how it works, and what the note says.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Just to clarify, you can do it in either order. Figuring out how it gets the number can be figured out easily by figuring out the note, or you could reverse the order and try to use the numbers to help figure out the note.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="5373" LastEditDate="2014-12-28T19:50:19.563" LastActivityDate="2014-12-28T19:50:19.563" Title="The Broken Device" Tags="|password|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5745" PostTypeId="2" ParentId="5743" CreationDate="2014-11-30T22:46:17.953" Score="2" Body="&lt;p&gt;This is incomplete, but I'd like to know if I'm on the right track...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;ad[d ]al[l t]he[ n]um[be]rs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Step 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;r[e]p[e]at[ ]st[e]p[ 1 ]u[n]t[i]l[ ]o[n]e[ ]t[????? ] s[i]n[g]le[ ]d[i]g[i]t&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Step 3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Profit&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5948" LastActivityDate="2014-11-30T22:46:17.953" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5746" PostTypeId="2" ParentId="5743" CreationDate="2014-11-30T22:59:48.017" Score="5" Body="&lt;p&gt;Step 1 says:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;adalheumrs&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;ad&lt;/strong&gt;d &lt;strong&gt;al&lt;/strong&gt;l t&lt;strong&gt;he&lt;/strong&gt; n&lt;strong&gt;um&lt;/strong&gt;be&lt;strong&gt;rs&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Step 2 says:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;rpatstputloetsnledgt&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;r&lt;/strong&gt;e&lt;strong&gt;p&lt;/strong&gt;e&lt;strong&gt;at&lt;/strong&gt; &lt;strong&gt;st&lt;/strong&gt;e&lt;strong&gt;p&lt;/strong&gt; 1 &lt;strong&gt;u&lt;/strong&gt;n&lt;strong&gt;t&lt;/strong&gt;i&lt;strong&gt;l&lt;/strong&gt; y&lt;strong&gt;o&lt;/strong&gt;u g&lt;strong&gt;et&lt;/strong&gt; a &lt;strong&gt;s&lt;/strong&gt;i&lt;strong&gt;n&lt;/strong&gt;g&lt;strong&gt;le&lt;/strong&gt; &lt;strong&gt;d&lt;/strong&gt;i&lt;strong&gt;g&lt;/strong&gt;i&lt;strong&gt;t&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Step 3 says:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;rurnvle&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;r&lt;/strong&gt;et&lt;strong&gt;urn&lt;/strong&gt; &lt;strong&gt;v&lt;/strong&gt;a&lt;strong&gt;l&lt;/strong&gt;u&lt;strong&gt;e&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Basically, the machine keeps summing up the digits until you get a one-digit number, and returns that number. Like so:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $1863784 \rightarrow 37 \rightarrow 10 \rightarrow 1$&lt;br&gt;&#xA; $1243242 \rightarrow 18 \rightarrow 9$&lt;br&gt;&#xA; $1742 \rightarrow 16 \rightarrow 5$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2014-11-30T22:59:48.017" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="5747" PostTypeId="1" CreationDate="2014-12-01T02:38:39.883" Score="5" ViewCount="4911" Body="&lt;p&gt;When Lenny was five years old, he hammered a nail into a tree near his house to mark his height. Fifteen years later at age 20, Lenny returned to see how much higher the nail was. &#xA;If the tree grew by 5 inches each month, how much higher would the nail be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have provided a hint if you need one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Remember that trees &quot;________&quot;, and you will know the answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6111" LastEditorUserId="5373" LastEditDate="2015-04-29T08:40:57.303" LastActivityDate="2016-03-17T12:53:27.657" Title="How high is the nail in the tree?" Tags="|lateral-thinking|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5748" PostTypeId="2" ParentId="5747" CreationDate="2014-12-01T02:51:49.403" Score="19" Body="&lt;p&gt;The nail will be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;zero&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;inches higher.  Because...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Trees grow from the top.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5948" LastActivityDate="2014-12-01T02:51:49.403" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5749" PostTypeId="1" AcceptedAnswerId="5751" CreationDate="2014-12-01T04:33:45.077" Score="6" ViewCount="1531" Body="&lt;p&gt;Note: This is a question about a strategy puzzle / game, and is not proposing a puzzle itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Growing up I often played a pencil and paper game with the following rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;First someone draws some dots on a piece of paper (number is usually small, maybe 6 - 10), and placement doesn't matter as will become clear from the rules of play.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Players then take turns drawing a curve starting at a dot and ending at a dot (it can even be the same dot the curve started at) subject to the restrictions&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The curve cannot cross over any curve (including itself)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Each dot can have a maximum total of 3 curve end/start points on it&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;After a new curve is drawn another dot is added somewhere on that curve&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The winner is whoever can make the last valid move.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Due to how one draws the lines, it is possible to separate off points that cannot become &quot;saturated&quot; with three curve ends/starts, hence allowing strategy due to choice of how to have the resulting graph &quot;grow&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know the name of this game?  And even better, a book analyzing the strategy of this game?&lt;/p&gt;&#xA;" OwnerUserId="4351" LastActivityDate="2014-12-01T05:38:16.207" Title="Name of dots and lines game" Tags="|pencil-and-paper-games|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5751" PostTypeId="2" ParentId="5749" CreationDate="2014-12-01T05:25:01.400" Score="9" Body="&lt;p&gt;This game is called Sprouts. There is a discussion in &lt;em&gt;Winning ways for your mathematical plays&lt;/em&gt;, Vol. 3, by Berlekamp, Conway, and Guy, on p. 598.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sprouts also has a &lt;a href=&quot;http://en.wikipedia.org/wiki/Sprouts_(game)&quot;&gt;Wikipedia page&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2014-12-01T05:38:16.207" LastActivityDate="2014-12-01T05:38:16.207" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5752" PostTypeId="2" ParentId="5734" CreationDate="2014-12-01T07:15:35.543" Score="16" Body="&lt;p&gt;I included mathematics :)&lt;/p&gt;&#xA;&lt;p&gt;The method is as follows:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Add all the digits&lt;/li&gt;&#xA;&lt;li&gt;If step 1's result has a 9, subtract 2&lt;/li&gt;&#xA;&lt;li&gt;If step 1's result has a 6, add 1&lt;/li&gt;&#xA;&lt;li&gt;If the original number has a 0, multiply by 2 then subtract 4&lt;/li&gt;&#xA;&lt;li&gt;Add all the digits&lt;/li&gt;&#xA;&lt;li&gt;Mod 7 step 5's result&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Now use this method on your numbers:&lt;/p&gt;&#xA;&lt;p&gt;1 + 7 + 7 + 3 + 8 + 3 = 29&lt;br /&gt;&#xA;29 - 2 = 27&lt;br /&gt;&#xA;2 + 7 = 9&lt;br /&gt;&#xA;9 % 7 = 2&lt;/p&gt;&#xA;&lt;p&gt;2 + 6 + 7 + 4 + 5 + 3 = 26&lt;br /&gt;&#xA;26 + 1 = 27&lt;br /&gt;&#xA;2 + 7 = 9&lt;br /&gt;&#xA;9 % 7 = 2&lt;/p&gt;&#xA;&lt;p&gt;1 + 1 + 1 + 1 + 1 + 1 = 6&lt;br /&gt;&#xA;6 + 1 = 7&lt;br /&gt;&#xA;7 = 7&lt;br /&gt;&#xA;7 % 7 = 0&lt;/p&gt;&#xA;&lt;p&gt;6 + 3 + 6 + 2 + 4 + 0 = 21&lt;br /&gt;&#xA;21 x 2 - 4 = 38&lt;br /&gt;&#xA;3 + 8 = 11&lt;br /&gt;&#xA;11 % 7 = 4&lt;/p&gt;&#xA;&lt;p&gt;3 + 6 + 7 + 1 + 8 + 3 = 28&lt;br /&gt;&#xA;2 + 8 = 10&lt;br /&gt;&#xA;10 % 7 = 3&lt;/p&gt;&#xA;&lt;p&gt;2 + 4 + 7 + 1 + 2 + 3 = 19&lt;br /&gt;&#xA;19 - 2 = 17&lt;br /&gt;&#xA;1 + 7 = 8&lt;br /&gt;&#xA;8 % 7 = 1&lt;/p&gt;&#xA;&lt;p&gt;3 + 6 + 9 + 1 + 0 + 8 = 27&lt;br /&gt;&#xA;2 x 27 - 4 = 50&lt;br /&gt;&#xA;5 + 0 = 5&lt;br /&gt;&#xA;5 % 7 = 5&lt;/p&gt;&#xA;" OwnerUserId="6076" LastEditorUserId="6076" LastEditDate="2020-11-18T07:52:28.420" LastActivityDate="2020-11-18T07:52:28.420" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="5753" PostTypeId="1" AcceptedAnswerId="5754" CreationDate="2014-12-01T09:43:00.913" Score="9" ViewCount="17498" Body="&lt;p&gt;I saw a riddle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am whole but incomplete,&lt;br/&gt;&#xA;I have no eyes, yet I see.&lt;br/&gt;&#xA;You can see, and see right through me.&lt;br/&gt;&#xA;My largest part is one fourth of what I once was.&lt;br/&gt;&#xA;What am I?&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A skeleton&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I don't understand it.  How does it &quot;see&quot;?  How is its largest part one fourth of what it once was?&lt;/p&gt;&#xA;" OwnerUserId="388" LastActivityDate="2019-12-15T14:55:59.383" Title="Explanation for the &quot;I am whole but incomplete...&quot; riddle?" Tags="|riddle|explanation|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5754" PostTypeId="2" ParentId="5753" CreationDate="2014-12-01T10:00:38.850" Score="9" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am whole but incomplete&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This means a skeleton is whole considering it as a structure. But it is just a part of the body. So it's incomplete.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I have no eyes, yet I see.&lt;br&gt;&#xA;  You can see, and see right through me.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This means skeletons don't have eyes - just the holes. By yet I can 'see' it refers to the 'see' in the word '&lt;strong&gt;s&lt;/strong&gt;'k'&lt;strong&gt;e&lt;/strong&gt;'l'&lt;strong&gt;e&lt;/strong&gt;'ton. The same is meant by 'you can see', and 'see right through me'. Which makes it clear that we can see the word 'see' in (right through) '&lt;strong&gt;s&lt;/strong&gt;k&lt;strong&gt;e&lt;/strong&gt;l&lt;strong&gt;e&lt;/strong&gt;ton'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Or as mentioned in the comments: people perceive a skeleton to &quot;see&quot; through its eye sockets, even though it no longer has eyes.  And people can see through the gaps in a skeleton.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My largest part is one fourth of what I once was.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The largest part of a skeleton is the femur, which is one fourth of the body (which the skeleton once was) in length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope it's clear now.&lt;/p&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="388" LastEditDate="2014-12-04T09:25:14.063" LastActivityDate="2014-12-04T09:25:14.063" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5755" PostTypeId="1" AcceptedAnswerId="5759" CreationDate="2014-12-01T11:26:39.820" Score="-5" ViewCount="330" Body="&lt;p&gt;Here are the two puzzles that I have heard a long ago from my friends.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1) You can touch it with your right hand, but not with your left hand. What is it?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;2) 10x+4x = 2y Then, what is x and y?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;HINT: (x and y are not numbers)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think both the puzzles are easy for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;code&gt;+1 if only one correct answer. +1 and will mark as accepted answer if both answers are correct.&lt;/code&gt;&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="2730" LastEditDate="2015-12-15T07:21:32.590" LastActivityDate="2015-12-15T07:21:32.590" Title="Double the fun with two puzzles" Tags="|logical-deduction|" AnswerCount="2" CommentCount="2" FavoriteCount="0" ClosedDate="2014-12-02T01:38:15.303" ContentLicense="CC BY-SA 3.0" />
  <row Id="5756" PostTypeId="2" ParentId="5689" CreationDate="2014-12-01T11:33:47.047" Score="19" Body="&lt;p&gt;Well, I finally figured it out (thanks to the hint, of course).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hint is &quot;Cats like bacon&quot;, which means that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the message is encoded with a Baconian cipher.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In this case, we need to figure out&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the binary distinction in the images upon which the encoding was based.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The characteristic used to encode the message is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the color of the cats: a fully-white cat is B, others are A.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the encoded message is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ABABA BAABB ABAAB AABAA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which translates to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; L U K E&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5990" LastActivityDate="2014-12-01T11:33:47.047" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5758" PostTypeId="2" ParentId="5755" CreationDate="2014-12-01T11:52:02.153" Score="2" Body="&lt;p&gt;1) Your left elbow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) 'x' means eggs; y is a box of seven eggs.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2014-12-01T11:52:02.153" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="5759" PostTypeId="2" ParentId="5755" CreationDate="2014-12-01T11:53:59.173" Score="3" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Your left elbow.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$y$ is a week. $x$ is a day.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2518" LastActivityDate="2014-12-01T11:53:59.173" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5761" PostTypeId="1" AcceptedAnswerId="5765" CreationDate="2014-12-01T14:13:06.080" Score="10" ViewCount="531" Body="&lt;p&gt;This is a question about puzzle-building. &#xA;I very much enjoyed (albeit not solved) the puzzle &lt;a href=&quot;https://puzzling.stackexchange.com/questions/5588/find-the-missing-elephants/5616?noredirect=1#comment17931_5616&quot;&gt;Find the missing elephants&lt;/a&gt; by &lt;strong&gt;GOTO 0&lt;/strong&gt;. And, as it is often the case with intriguing puzzles, it sparked some ideas of my own. (No need to worry, I'm not planning to put tons of clones online anytime soon...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I then asked myself: How does one best build and verify such puzzles? &#xA;Simple kernel-convolution puzzles I would have a good idea how to build &amp;amp; test, but anything with a complex conditional like in this particular puzzle, I don't know. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What tools/methods are you using/would you use to build such puzzles?&lt;/p&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2014-12-01T21:33:39.513" Title="How to build pattern search puzzles" Tags="|pattern|puzzle-creation|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5762" PostTypeId="2" ParentId="4807" CreationDate="2014-12-01T15:53:59.483" Score="1" Body="&lt;p&gt;I believe the password should be...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;9&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Each correct answer simply increments by one.  It could be the first guest you saw was the fifth to actually enter, and the password is simply the guest #, or maybe the first guest you saw was the first to enter, making the function (guest # + 4).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I suppose, it's possible I just found an unintended pattern, though.  Besides, I'm clueless on the disguise as of yet.&lt;/p&gt;&#xA;" OwnerUserId="5948" LastActivityDate="2014-12-01T15:53:59.483" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5763" PostTypeId="1" AcceptedAnswerId="5874" CreationDate="2014-12-01T16:02:03.493" Score="12" ViewCount="1544" Body="&lt;p&gt;&lt;strong&gt;This time it is not complicated. No tricks, no brain bending involved. A simple, straight forward word-search puzzle to relax...&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;All valid words are &lt;strong&gt;straight lines&lt;/strong&gt; but can be found in any of the &lt;strong&gt;8 directions&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;All valid words are &lt;strong&gt;at least 6 letters long&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;All special letters are replaced by their closest simple representation ( &lt;strong&gt;Ñ --&gt; N&lt;/strong&gt; ; &lt;strong&gt;Á --&gt; A&lt;/strong&gt; etc.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;...with a twist:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Words are written in &lt;strong&gt;multiple, different, European languages&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Only words which can be found in &lt;em&gt;at least 3 European languages&lt;/em&gt; are valid ( &quot;word-triplet&quot; )&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Words of a valid word-triplets may not overlap with more than a single letter.&lt;/strong&gt; &#xA;(This automatically rules out words which are identical in multiple&#xA;languages unless you find 2 variants which are spelled differently. It also rules out words which are just different by pre- or suffix.)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;It is possible that words of &lt;em&gt;different&lt;/em&gt; triplets share more than one letter.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;small spoiler (not needed): &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes, English is also used in this puzzle. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;big spoiler (not needed):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Finding &lt;em&gt;pairs&lt;/em&gt; of words does not guarantee there is a &lt;em&gt;triplet&lt;/em&gt;. Red herrings have been included.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;P Y B E G R A F E N I S X J R L P &#xA;I S V A T B A O E V A N O R E A O &#xA;Y A B O A R D G A M E N R E K M M &#xA;Y N B D L P Z R V H K M N N Q R M &#xA;P A O B L A G I A A S G N I C H I &#xA;A B M S Y N T Y M A P A I V A T E &#xA;S B R E T T S P I E L O L U F E R &#xA;P O R M I T S K N S O E T R M R D &#xA;O R H G R I A V E O S A Z A K N O &#xA;O L H C M V P U G N J P K J A I B &#xA;R E M Z H A E H A A O P O A I N O &#xA;T T A C O N S C I E N C E H R G K &#xA;S A N Z S K I S R L G N G D C E O &#xA;J D Z I T I E X R P H S Y Y R R C &#xA;H L A T A T R R A M I R G S A S K &#xA;B O N F G S D G M U A B L E F P A &#xA;I P O G E F X S O C G E W E T E N&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Winner is, who first discovered &lt;em&gt;at least 5 valid triplets&lt;/em&gt; (5x3=15 words!).&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answers have to list the word-triplets and their according languages in brackets.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus points if you discover more than that. &#xA;There should be at least 6 valid triplets, but maybe there are more? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus question:&lt;/strong&gt; How many different valid languages can you find? (A language is considered valid, if it is spoken somewhere in Europe and if you discover at least one word with at least two translations found as well.)&#xA;9 different languages were deliberately used, but one can possibly find more than that.&lt;/p&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="5840" LastEditDate="2014-12-09T17:38:27.360" LastActivityDate="2014-12-09T17:38:27.360" Title="International word-search: Can you find the words in multiple languages?" Tags="|word-search|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="5765" PostTypeId="2" ParentId="5761" CreationDate="2014-12-01T18:32:55.370" Score="2" Body="&lt;p&gt;I cannot give any general recommendations on how to create great puzzles, but I'm glad to explain how the particular puzzle mentioned in the question was made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The elephant puzzle was built (you guessed it) with a computer program: I tried out a few different algorithms before coming up with one that seemed good enough to make a puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Random numbers are produced with a seeded random number generator: this makes sure that we can always replay an interesting arrangement after we find one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The board is created in three successive steps.&#xA;First, all directions are set up randomly.&#xA;Next, each row is scanned and cells are partitioned in subsets of color-matching ele&lt;strike&gt;phants&lt;/strike&gt;, ehm, elements. Each of these subsets is assigned a different color, chosen randomly from a palette of four colors.&#xA;The last step is the rendering. This is done in Javascript/HTML. Just grab a screenshot and use GIMP to export the board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The source code of the generator can be found here: &lt;a href=&quot;https://github.com/fasttime/Elephant-Puzzle-Maker&quot; rel=&quot;nofollow&quot;&gt;https://github.com/fasttime/Elephant-Puzzle-Maker&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4394" LastEditorUserId="4394" LastEditDate="2014-12-01T21:33:39.513" LastActivityDate="2014-12-01T21:33:39.513" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5766" PostTypeId="1" AcceptedAnswerId="5776" CreationDate="2014-12-01T19:07:08.207" Score="3" ViewCount="3150" Body="&lt;p&gt;The object came with a few clues to help you figure it out.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I can be viewed on paper&lt;br&gt;&#xA;  I cannot be viewed in reality by your kind&lt;br&gt;&#xA;  I am made up of what your mind can truly comprehend&lt;br&gt;&#xA;  I resemble a shape many learn first in primary&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's freakishly easy if you think in a specific way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;BONUS (for people who don't mind risking not being first for more quality):&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;How many edges and vertices does it have, and what does it look like?&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="4050" LastEditDate="2014-12-11T10:13:37.263" LastActivityDate="2014-12-11T10:13:37.263" Title="How many faces does this object have?" Tags="|mathematics|lateral-thinking|geometry|" AnswerCount="7" CommentCount="7" ClosedDate="2014-12-02T08:42:31.590" ContentLicense="CC BY-SA 3.0" />
  <row Id="5767" PostTypeId="5" CreationDate="2014-12-01T19:08:47.017" Score="0" Body="&lt;p&gt;A puzzle tagged with this should deal with philosophical concepts but should not be a philosophy question by itself.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-12-01T20:58:53.423" LastActivityDate="2014-12-01T20:58:53.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5768" PostTypeId="4" CreationDate="2014-12-01T19:08:47.017" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-12-01T19:08:47.017" LastActivityDate="2014-12-01T19:08:47.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5769" PostTypeId="2" ParentId="5766" CreationDate="2014-12-01T19:34:28.353" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well, I'd say it's a square, so 1 face. However, I think that you could view it, just not from the side; unless it's just the boundary of a square. For the 3rd clue, squares are made up of lines, which are made up of points, and points are hard to imagine. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2014-12-01T19:34:28.353" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5770" PostTypeId="5" CreationDate="2014-12-01T19:42:19.927" Score="0" Body="&lt;p&gt;Word searches are typically made up of a grid of letters (although one could be made with numbers) with words hidden vertically, horizontally or diagonally inside the grid that the solver must then find. In a standard word search the words that must be found are listed below (or near) the grid of letters. As word searches are generally simpler puzzles to solve and create, combining them with other puzzles or adding some kind of twist is encouraged.&lt;/p&gt;&#xA;" OwnerUserId="52483" LastEditorUserId="19547" LastEditDate="2020-04-18T15:58:16.810" LastActivityDate="2020-04-18T15:58:16.810" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="5771" PostTypeId="4" CreationDate="2014-12-01T19:42:19.927" Score="0" Body="Puzzles tagged with word-search should generally be about finding words within a grid, these types of puzzles are mass-producible so adding a twist or spin to them is a good idea." OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-12-01T21:15:56.120" LastActivityDate="2014-12-01T21:15:56.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5772" PostTypeId="2" ParentId="5736" CreationDate="2014-12-01T20:10:01.010" Score="8" Body="&lt;p&gt;I have part of the puzzle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Arrows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A &amp;lt;--&gt; O&lt;br&gt;&#xA; D &amp;lt;--&gt; R&lt;br&gt;&#xA; E &amp;lt;--&gt; S&lt;br&gt;&#xA; C &amp;lt;--&gt; T&lt;br&gt;&#xA; N &amp;lt;--&gt; L&lt;br&gt;&#xA; I &amp;lt;--&gt; Y  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Words:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Behind&lt;br&gt;&#xA;Breakfast&lt;br&gt;&#xA;Hearth-Rug&lt;br&gt;&#xA;Infrequent&lt;br&gt;&#xA;Mornings&lt;br&gt;&#xA;Occasions&lt;br&gt;&#xA;Picked&lt;br&gt;&#xA;Seated&lt;br&gt;&#xA;Usually&lt;br&gt;&#xA;Visitor  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(The following text is taken from &lt;a href=&quot;https://puzzling.stackexchange.com/a/5791/2518&quot;&gt;xnor's answer&lt;/a&gt; as he finished the solution. This answer is marked as accepted as the majority of the grunt-work was done by Warlord --Joe)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From there, I simply&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Googled the words. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This led me to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first paragraph of &lt;a href=&quot;http://www.gutenberg.org/files/2852/2852-h/2852-h.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Hounds of Baskerville&lt;/a&gt;, a story featuring Sherlock Holmes, which contains these words.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As confirmation as to the answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the first chapter contains references to meeting at Waterloo Station.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I feel like I disregarded some information in my solution, so I'm curious if this is how it was intended to be solved.&lt;/p&gt;&#xA;" OwnerUserId="3922" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2014-12-02T10:48:29.047" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5773" PostTypeId="2" ParentId="5766" CreationDate="2014-12-01T20:15:49.920" Score="1" Body="&lt;p&gt;It is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;photons&lt;/strong&gt;&lt;br&gt;&#xA; &lt;br&gt;&#xA; Light particles/waves that are reflected by a white paper.&lt;br&gt;&#xA; &lt;br&gt;&#xA; Their quantum nature can't be really comprehended by our minds. But at the same time, light is all that we see.&lt;br&gt;&#xA; &lt;br&gt;&#xA; Their wave resembles sinuidal waves. Although people normally do not know what a sinuidal wave is in the primary, every small kid know their shapes and frequently draws them even without knowing their exact name nor the mathematics behind. If light is considered a particle, then photons are aproximated to spheres, which every small kid knows about.&lt;br&gt;&#xA; &lt;br&gt;&#xA; How many faces? This can't really be determined. :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5044" LastActivityDate="2014-12-01T20:15:49.920" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5774" PostTypeId="2" ParentId="5766" CreationDate="2014-12-01T20:22:17.850" Score="7" Body="&lt;p&gt;Probably not the intended answer, but&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the &lt;a href=&quot;http://en.wikipedia.org/wiki/Real_projective_plane&quot;&gt;real projective plane&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;seems to work. It &quot;can be viewed on paper&quot; (see the picture at the top of the link I provided), but &quot;cannot be viewed in reality by your kind&quot; (i.e. by 3D beings) since it cannot be embedded into $\mathbb{R}^3$. It can be seen as being made up of lines in $\mathbb{R}^3$, which &quot;your mind can truly comprehend&quot;. And it looks like JontheMon's answer (again see the picture at the top of the link)!&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2014-12-01T20:22:17.850" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5775" PostTypeId="2" ParentId="5766" CreationDate="2014-12-01T20:23:27.897" Score="17" Body="&lt;p&gt;The number of faces is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Zero&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The object is a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A point&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Line 1: &lt;code&gt;I can be viewed on paper&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A point can be viewed on paper.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Line 2: &lt;code&gt;I cannot be viewed in reality by your kind&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A point, a dimension-less object, cannot be seen in the real world.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Line 3: &lt;code&gt;I am made up of what your mind can truly comprehend&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is a philosophy tag. As we learn more, we realize how little we actually know. There is no real way to know that we truly comprehend something as technology and science change our perspectives as time advances. Therefore, we can say that there is nothing we truly comprehend. A point can be considered made up of nothing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Line 4: &lt;code&gt;I resemble a shape many learn first in primary&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A point resembles a circle on paper.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2077" LastEditorUserId="2077" LastEditDate="2014-12-01T20:44:14.813" LastActivityDate="2014-12-01T20:44:14.813" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="5776" PostTypeId="2" ParentId="5766" CreationDate="2014-12-01T20:31:52.810" Score="13" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A &lt;a href=&quot;http://en.wikipedia.org/wiki/Tesseract&quot;&gt;tesseract&lt;/a&gt; (a 4-dimensional analog of the cube). You can see the projection of a tesseract. &quot;Our kind&quot; is three-dimensional, so we cannot see it. It is made of cubes which we can comprehend. It resembles a square which we learn in primary (it is actually a generalization).&lt;br/&gt;&lt;br/&gt;A really cool &lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/5/55/8-cell-simple.gif&quot;&gt;gif&lt;/a&gt; of a tesseract.&lt;br/&gt;&lt;br/&gt;24 faces&lt;br/&gt;32 edges&lt;br/&gt;16 vertices&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="5610" LastEditDate="2014-12-01T20:38:13.140" LastActivityDate="2014-12-01T20:38:13.140" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="5777" PostTypeId="2" ParentId="5766" CreationDate="2014-12-01T20:37:40.337" Score="2" Body="&lt;p&gt;Edges and vertices:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 16 vertices&lt;br&gt;&#xA; &lt;br&gt;&#xA; 32 edges&lt;br&gt;&#xA; &lt;br&gt;&#xA; 24 square faces&lt;br&gt;&#xA; &lt;br&gt;&#xA; 8 cubic cells&lt;br&gt;&#xA; &lt;br&gt;&#xA; In 2D, frequently projected as an octogon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5044" LastActivityDate="2014-12-01T20:37:40.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5778" PostTypeId="1" AcceptedAnswerId="5808" CreationDate="2014-12-01T21:05:23.163" Score="12" ViewCount="535" Body="&lt;p&gt;I have thought up one simple cipher. Hope it didn't exist before. Try to crack it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is encoded one sentence in English consisting of three words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(2_ 2,5_ 2,7_ 5,8_ 5,9_ 4,14_ 4),(6_ 2,7_ 1),(3_ 2,5_ 4,30_ 4,70_ 2)$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Number with index corresponds to one letter&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Numbers are not numbers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5781" LastEditorUserId="5781" LastEditDate="2014-12-02T16:41:29.900" LastActivityDate="2014-12-02T20:58:28.387" Title="My new cipher with numbers" Tags="|cipher|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5779" PostTypeId="1" AcceptedAnswerId="5782" CreationDate="2014-12-01T21:29:48.047" Score="-3" ViewCount="314" Body="&lt;p&gt;This puzzle works best if you &lt;strong&gt;do not look in edit.&lt;/strong&gt; If you do, please refrain from posting answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So basically the idea is to count how many squares/rectangles this post includes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Don't underestimate the ability of this being a puzzle.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;br&gt;&#xA;  &lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;    &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;br&gt;&#xA;  &lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/kbd&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-12-02T01:25:30.513" LastActivityDate="2014-12-02T01:25:30.513" Title="Formatting Puzzle!" Tags="|lateral-thinking|" AnswerCount="3" CommentCount="5" ClosedDate="2014-12-02T08:40:53.577" ContentLicense="CC BY-SA 3.0" />
  <row Id="5780" PostTypeId="2" ParentId="5779" CreationDate="2014-12-01T21:44:35.593" Score="2" Body="&lt;p&gt;I did not looked at the edit, nor cheated with anything similar. Just fixed the eyes in the question as show and counted.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I counted 130 white boxes.&lt;br&gt;&#xA; &lt;br&gt;&#xA; If you count the quoting blocks, then this is 132.&lt;br&gt;&#xA; &lt;br&gt;&#xA; If you count the two tags, the OP's box and the OP's avatar, add more 4.&lt;br&gt;&#xA; &lt;br&gt;&#xA; You may consider that each quote box is a pair of rectangles, one of them tall, thin and yellow and the big one is pink. The yellow and pink rectangles combined forms another larger rectangle also. So three rectangles for each quote box. Totalizing more 6 rectangles here (two quote boxes).&lt;br&gt;&#xA; &lt;br&gt;&#xA; In the end this could be 140.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks to @tjbtech for a suggestion. :)&lt;/p&gt;&#xA;" OwnerUserId="5044" LastEditorUserId="5044" LastEditDate="2014-12-01T22:26:16.700" LastActivityDate="2014-12-01T22:26:16.700" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5781" PostTypeId="2" ParentId="5779" CreationDate="2014-12-01T21:59:48.577" Score="1" Body="&lt;p&gt;There are exactly:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 square and 7 rectangles.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Just that!? Why!?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The boxes are not rectangles, because they have rounded corners. The same for the tags marks. No letter possesses rectangular parts.&lt;br&gt;&#xA; &lt;br&gt;&#xA; The quote boxes are a pair of rectangles each one. A yellow thin and tall rectangle and a big pink one each. The pink and yellow rectangle are also a rectangle when combined. There are two quote boxes, so 6 rectangles with that.&lt;br&gt;&#xA; &lt;br&gt;&#xA; The OP's box is a rectangle. The OP's avatar is a square.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks to @tjbtech for a suggestion. :)&lt;/p&gt;&#xA;" OwnerUserId="5044" LastEditorUserId="5044" LastEditDate="2014-12-01T22:24:43.930" LastActivityDate="2014-12-01T22:24:43.930" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5782" PostTypeId="2" ParentId="5779" CreationDate="2014-12-01T22:53:57.057" Score="4" Body="&lt;p&gt;I think there is actually no unique answer!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It depends on the device you use to look at the posting. Checking the site with my iphone via web is rendered differently from web by PC is differently from iPad app... Most notably the 'citation' block are rendered differently.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastActivityDate="2014-12-01T22:53:57.057" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5785" PostTypeId="2" ParentId="5766" CreationDate="2014-12-02T02:06:43.957" Score="0" Body="&lt;p&gt;# of faces:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The object is any two-dimensional object. &lt;br&gt;&#xA; &lt;strong&gt;I can be viewed on paper&lt;/strong&gt; &lt;br&gt;&#xA; You can view a two-dimensional object on paper. &lt;br&gt;&#xA; &lt;strong&gt;I cannot be viewed in reality by your kind&lt;/strong&gt; &lt;br&gt;&#xA; Our universe is defined in three spatial dimensions. No two-dimensional object can be viewed in reality by us; even a drawing on paper has a thickness, however small, and is never two-dimensional. &lt;br&gt;&#xA; &lt;strong&gt;I am made up of what your mind can truly comprehend&lt;/strong&gt; &lt;br&gt;&#xA; We can comprehend two-dimensional objects pretty easily. &lt;br&gt;&#xA; &lt;strong&gt;I resemble a shape many learn first in primary&lt;/strong&gt; &lt;br&gt;&#xA; Most people learn two-dimensional shapes in primary school. &lt;br&gt;&#xA; &lt;strong&gt;How many edges and vertices does it have, and what does it look like?&lt;/strong&gt; &lt;br&gt;&#xA; Any natural number of edges, any natural number of vertices. It looks like whatever you want a two-dimensional object to look like.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6121" LastActivityDate="2014-12-02T02:06:43.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5786" PostTypeId="1" AcceptedAnswerId="5787" CreationDate="2014-12-02T04:54:14.557" Score="17" ViewCount="2805" Body="&lt;p&gt;How can a library contain less information than one of the books in it?  The library is intended to be a set of books on shelves.  The information content is the least number of bits to describe the data.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2014-12-02T12:47:18.430" Title="A library with less information than one of its books" Tags="|mathematics|paradox|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5787" PostTypeId="2" ParentId="5786" CreationDate="2014-12-02T04:56:43.607" Score="18" Body="&lt;p&gt;The books in the library contains all binary strings of length 1 million, one per book, sorted in lexicographic order. An individual book takes a million bits to specify, but this description of the whole library is much shorter.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2014-12-02T04:56:43.607" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5788" PostTypeId="1" AcceptedAnswerId="5862" CreationDate="2014-12-02T06:43:11.273" Score="3" ViewCount="1919" Body="&lt;ul&gt;&#xA;&lt;li&gt;What is searching for letters exactly?&lt;/li&gt;&#xA;&lt;li&gt;What do we get when using letters literally?&lt;/li&gt;&#xA;&lt;li&gt;What does wordplay finally result in?&lt;/li&gt;&#xA;&lt;li&gt;And what is a metaphorical series?&lt;br&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is there also a word for the combination of the things mentioned above?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;" OwnerUserId="5106" LastEditorDisplayName="user20" LastEditDate="2015-03-20T04:16:11.913" LastActivityDate="2016-02-19T13:30:56.963" Title="This is a riddle" Tags="|riddle|word|" AnswerCount="4" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="5789" PostTypeId="2" ParentId="5786" CreationDate="2014-12-02T06:51:10.600" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The library contains all possible combinations of letters conceivable. (It is a very large library) with no ordering of books at all. Perfect entropy. Each of its book contains a subset only, which therefore contains some order and some information. (The library information is 'complete/all' which can be described with a single bit &lt;em&gt;true&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sorry if this is not in proper mathematical terms, I am not a mathematician...&lt;/p&gt;&#xA;" OwnerUserId="5840" LastActivityDate="2014-12-02T06:51:10.600" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5790" PostTypeId="2" ParentId="5786" CreationDate="2014-12-02T06:56:01.620" Score="20" Body="&lt;p&gt;This is indeed possible if we define information content as the least number of bits necessary to describe the data.&#xA;Consider the following example (in human readable language):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Book 1: Jim knows all animals whose name starts with a letter between A and L.&lt;/li&gt;&#xA;&lt;li&gt;Book 2: Jim knows all animals whose name starts with a letter between M and Z.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then the information in the whole library would be:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Jim knows all animals.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The paradox is solved if we consider the source of an information as an information itself (i.e. which book contains which information). In this case, the library, if not empty, will always contain at least as much information as any of its books.&lt;/p&gt;&#xA;" OwnerUserId="4394" LastActivityDate="2014-12-02T06:56:01.620" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5791" PostTypeId="2" ParentId="5736" CreationDate="2014-12-02T07:12:32.447" Score="6" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Hound of the Baskervilles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/a/5772/4551&quot;&gt;Warlord 099's answer&lt;/a&gt; took me almost the whole way there. It suggests to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; fill in partial words like &lt;code&gt;B***KF***&lt;/code&gt; into words like &lt;code&gt;BREAKFAST&lt;/code&gt;, then find them in the grid with a set of substitutions. For example, it becomes &lt;code&gt;BDSOKFOEC&lt;/code&gt;, a string found backwards in the top row of the words search, with the substitutions &lt;code&gt;D&amp;lt;-&amp;gt;R, S&amp;lt;-&amp;gt;E, O&amp;lt;-&amp;gt;A. C&amp;lt;-&amp;gt;T&lt;/code&gt;. Presumably, Warlord 099 found these by looking for snippets in the word search and finding consistent substitutions. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From there, I simply&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Googled the words. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This led me to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first paragraph of &lt;a href=&quot;http://www.gutenberg.org/files/2852/2852-h/2852-h.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Hounds of Baskerville&lt;/a&gt;, a story featuring Sherlock Holmes, which contains these words.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As confirmation as to the answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the first chapter contains references to meeting at Waterloo Station.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I feel like I disregarded some information in my solution, so I'm curious if this is how it was intended to be solved.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2014-12-02T09:00:28.500" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5793" PostTypeId="2" ParentId="5788" CreationDate="2014-12-02T08:19:15.493" Score="3" Body="&lt;p&gt;Going by the hints given let me try what i think it might be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is searching for letters exactly? &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  umm Googling i think. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When do we use letters literally? &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Whe we write Mail to someone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What does wordplay result in?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; results in a Conversation . &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And what is a metaphorical series?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; no idea about this though.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Combining these i think you are talking about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; your Gmail account.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;also, Internet is a cruel place..&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; maybe your account got hacked so you are trying to ditch that mail and create new one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1669" LastEditorUserId="2484" LastEditDate="2014-12-02T09:13:18.133" LastActivityDate="2014-12-02T09:13:18.133" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5794" PostTypeId="2" ParentId="5763" CreationDate="2014-12-02T08:24:35.463" Score="8" Body="&lt;p&gt;So far, I managed to find these triplets:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;MANZANO&lt;/strong&gt; (Spanish) - column 3&#xA; &lt;br&gt;&#xA; &lt;strong&gt;POMMIER&lt;/strong&gt; (French) - column 17&#xA; &lt;br&gt;&#xA; &lt;strong&gt;APFELBAUM&lt;/strong&gt; (German) - line 16&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;GEWETEN&lt;/strong&gt; (Dutch) - line 17&#xA; &lt;br&gt;&#xA; &lt;strong&gt;CONSCIENCE&lt;/strong&gt; (English) - line 12&#xA; &lt;br&gt;&#xA; &lt;strong&gt;SAMVITTIGHET&lt;/strong&gt; (Norwegian, Bokmaal) - line 1, col 12, diagonal towards SW&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;AERONAVE&lt;/strong&gt; (French) - line 2&#xA; &lt;br&gt; &#xA; &lt;strong&gt;AIRCRAFT&lt;/strong&gt; (English) - column 15&#xA; &lt;br&gt; &#xA; &lt;strong&gt;LETADLO&lt;/strong&gt; (Czech) - column 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Potential candidates/red herrings:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;REISEPASS&lt;/strong&gt; (German)&#xA; &lt;br&gt; &#xA; &lt;strong&gt;PASPOORT&lt;/strong&gt; (Dutch)&#xA; &lt;br&gt;&#xA; &lt;br&gt; &#xA; &lt;strong&gt;BOARDGAME&lt;/strong&gt; (English)&#xA; &lt;br&gt; &#xA; &lt;strong&gt;BRETTSPIEL&lt;/strong&gt; (German)&#xA; &lt;br&gt; &#xA; &lt;br&gt;&#xA; &lt;strong&gt;HOSTAGE&lt;/strong&gt; (English)&#xA; &lt;br&gt; &#xA; &lt;strong&gt;GIJZELAAR&lt;/strong&gt; (Dutch)&#xA; &lt;br&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1585" LastEditorUserId="1585" LastEditDate="2014-12-03T12:54:06.357" LastActivityDate="2014-12-03T12:54:06.357" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5795" PostTypeId="2" ParentId="5736" CreationDate="2014-12-02T10:13:23.210" Score="6" Body="&lt;p&gt;This is the solution as intended. The puzzle has been properly solved in a somewhat joint attempt between Warlord 099, Xnor and Aioros. I feel I should post this solution here for completeness sake:&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;The hints:&lt;/h1&gt;&#xA;&lt;p&gt;The first two hints are auxiliary to narrow down the word search. They were edited in after I realized that the possibilities of the word-search might have been too many, if &amp;quot;anything allowed&amp;quot; has to be assumed.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;Words are born as letters, but they reach adulthood when they are six. &#xA;(And this puzzle is not child's play!)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Simply a hint to disregard words with less than 6 letters in this puzzle. This is valid for both the word-search and then the book-search&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;You find us from East to West and West to East, from North to South and South to North. &#xA;And in between, can we also be seen. &#xA;We do not jump, we do not bend, a straight line we form, from start to end.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Another clarification hint to narrow down the puzzle a bit. Indicates that the word-search is a standard 8-direction-straight-line-no-steps search in the grid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The next hint is a critical one for the book-search. I think it is a real key-hint.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;We meet at Waterloo station.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Who is we? Well, &lt;em&gt;what&lt;/em&gt; meets at a station? &lt;strong&gt;Trains&lt;/strong&gt;. The word is needed further down. &lt;em&gt;Waterloo&lt;/em&gt; is a smaller hint towards the geographic area but it is not really needed. &lt;em&gt;Central Station&lt;/em&gt; would have been as good, but I wanted it a bit more themed.  (&lt;em&gt;Aioros figured that hint out. Well done.&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;The word-search&lt;/h2&gt;&#xA;&lt;p&gt;This really was a crypto-word-search, so the additional clues needed where:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;__ &amp;lt;-&amp;gt; __&#xA;__ &amp;lt;-&amp;gt; __&#xA;__ &amp;lt;-&amp;gt; __&#xA;__ &amp;lt;-&amp;gt; __&#xA;__ &amp;lt;-&amp;gt; __&#xA;__ &amp;lt;-&amp;gt; __&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 6 letter pairs are switched in this puzzle! Find those 6 pairs, and you've nearly solved it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;B*H***&#xA;B***KF***&#xA;H****H-*UG&#xA;**F**QU***&#xA;M*****G*&#xA;*********&#xA;P**K**&#xA;******&#xA;U*U****&#xA;V******&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Those where the 10 words actually hidden in the word-search. I thought it would be pretty obvious, but it turned out, that many thought in a different direction. The * indicate letters which where of the switched pairs while the &amp;quot;fixed&amp;quot; letters are given.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The idea was...&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; ...that one could still find the &lt;em&gt;pattern&lt;/em&gt; in the word-search, which would then give 10 words with mangled letters (sometimes more than one alternative, but never too many), and one could then solve this like a simple replacement cypher. i.e. the last word could only fit in two places of the puzzle and could only have been one of the following according to the word search: V&lt;em&gt;YEYCAD&lt;/em&gt; or V&lt;em&gt;DCESLS&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The way to find them...&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; It is potentially easiest to first highlight the letters which are not changed, i.e. BHKFUGQMPV.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This gives:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/GrVeq.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Next:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; If you match the pattern, of the given words, there are really only very few places they would fit. Some letters appear very rarely, and the space is limited.&#xA; For example, there is only one place where the pattern B*H of the first word fits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/iNGQw.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If you follow this through, you'll find all words:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; You now have to solve the replacement-cypher with a bit of 'feeling' for English words. This gives you all 6 letter-pairs as discovered by &lt;em&gt;Warlord 099&lt;/em&gt;. (&lt;em&gt;Although he seems to have started out with the word breakfast regardless of the word-search!&lt;/em&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If you'd sort in alphabetic increasing order, this would give:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/FuxMR.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;But now you need the word found from the important clue, giving you:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/44yWZ.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Which brings you to the final part of the puzzle:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; C.Doyle is a rather known author and 'father' of Sherlock Holmes. The words found in the word-search appear an odd choice? Well, those are the first 10 words (of more than 5 letters) of one of the Sherlock books.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Mr. Sherlock Holmes, who was &lt;strong&gt;usually&lt;/strong&gt; very late in the &lt;strong&gt;mornings&lt;/strong&gt;, save&#xA; upon those not &lt;strong&gt;infrequent&lt;/strong&gt; &lt;strong&gt;occasions&lt;/strong&gt; when he was up all night, was&#xA; &lt;strong&gt;seated&lt;/strong&gt; at the &lt;strong&gt;breakfast&lt;/strong&gt; table. I stood upon the &lt;strong&gt;hearth-rug&lt;/strong&gt; and &lt;strong&gt;picked&lt;/strong&gt;&#xA; up the stick which our &lt;strong&gt;visitor&lt;/strong&gt; had left &lt;strong&gt;behind&lt;/strong&gt; him the night before.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;The solution&lt;/h2&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The title of that book: &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/The_Hound_of_the_Baskervilles&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Hound of the Baskervilles&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;CONCLUSIONS&lt;/h2&gt;&#xA;&lt;p&gt;&lt;strong&gt;In the spirit of this site, I would like to invite others to comment (in this answer) how they like the puzzle; how deductive they think it is; if they could solve it, and if so because of what clues; what could be improved etc. etc.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-12-02T14:46:29.750" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="5796" PostTypeId="1" AcceptedAnswerId="5802" CreationDate="2014-12-02T12:17:25.520" Score="2" ViewCount="1472" Body="&lt;p&gt;I have here a three dimensional, solid object. Its sides are flat planes. In the following views, all lines are shown &amp;mdash; there are no hidden edges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the front, this shape looks like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xvRPk.gif&quot; alt=&quot;two concentric squares&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From above (at a 90&amp;deg; angle from the previous view), this shape looks like:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xvRPk.gif&quot; alt=&quot;two concentric squares&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;What would this shape look like from the side?&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditDate="2014-12-04T11:10:13.217" LastActivityDate="2014-12-04T11:20:47.453" Title="What does this shape look like?" Tags="|geometry|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5797" PostTypeId="2" ParentId="5796" CreationDate="2014-12-02T12:24:27.693" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It will be a plane square. Since there won't be any cavity on the sides.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are two possibilities for all six sides of the 3D object&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Possibility 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6kjjm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Possibility 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aKBRz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, in both cases&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the side will be perfect square without any cavity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note:From the description and images in the question, it doesn't state that the sides too have cavity, also no info about the dept of the cavity on front and top. So I considered the object with maximum volume.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="2730" LastEditDate="2014-12-02T12:47:27.110" LastActivityDate="2014-12-02T12:47:27.110" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5798" PostTypeId="2" ParentId="5796" CreationDate="2014-12-02T12:29:03.200" Score="0" Body="&lt;p&gt;My guess is that it is a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;a href=&quot;http://edge.rit.edu/edge/P07897/public/cube%20frame.jpg&quot; rel=&quot;nofollow&quot;&gt;frame of a cube&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alternatively it might even be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Tesseract&quot; rel=&quot;nofollow&quot;&gt;tesseract&lt;/a&gt; projected in 3D (I assume internal edges are not visible).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastActivityDate="2014-12-02T12:29:03.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5799" PostTypeId="1" AcceptedAnswerId="80026" CreationDate="2014-12-02T12:41:21.810" Score="27" ViewCount="2762" Body="&lt;p&gt;This is a puzzle was inspired by the board game &lt;a href=&quot;http://en.wikipedia.org/wiki/Labyrinth_(board_game)&quot; rel=&quot;noreferrer&quot;&gt;labyrinth&lt;/a&gt;, which I very much enjoyed as a kid.&lt;/p&gt;&#xA;&lt;p&gt;It either requires very good 3D-visualization skills in your brain or some paper &amp;amp; scissors work. (Hence &lt;em&gt;mechanical&lt;/em&gt; puzzle).&lt;/p&gt;&#xA;&lt;p&gt;Thanks to &lt;a href=&quot;https://puzzling.stackexchange.com/users/1497/kevin&quot;&gt;&lt;strong&gt;Kevin&lt;/strong&gt;&lt;/a&gt;, there is now also a Javascript &lt;a href=&quot;http://jsfiddle.net/vwe9kw7j/2/&quot; rel=&quot;noreferrer&quot;&gt;simulator&lt;/a&gt; for this puzzle, which makes things way easier.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;Goal&lt;/h2&gt;&#xA;&lt;p&gt;Imagine you have a labelled board of 25 fields like the following:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/qWvjj.jpg&quot; alt=&quot;board&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;You also have 25 dice which are all build up by the same mesh:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/o0tVX.jpg&quot; alt=&quot;dice mesh&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;These dice are placed on a 5x5 field grid to create a labyrinth with the faces showing upwards.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ob6jx.jpg&quot; alt=&quot;Starting position&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;You start out on the top-left position (A1), and you goal is to reach all other three corners (E1, E5, A5 ) &lt;em&gt;in arbitrary sequence&lt;/em&gt;. You want to do this with &lt;em&gt;minimum amount of steps&lt;/em&gt;. Note that 'passing through' one of the corners in a single move is valid for the purpose of the goal. (You do not have to end a move in a corner. )&lt;/p&gt;&#xA;&lt;h2&gt;Rules&lt;/h2&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A single &lt;strong&gt;step&lt;/strong&gt; consists of the sequence of &lt;strong&gt;first tilting&lt;/strong&gt; a &lt;em&gt;row &lt;strong&gt;or&lt;/strong&gt; column&lt;/em&gt; and &lt;strong&gt;then moving&lt;/strong&gt; from your current location along 'open paths' for an arbitrary distance. (You may do only one of the two, but not change the sequence order.)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When tilting, you have to &lt;strong&gt;tilt all&lt;/strong&gt; dice of the row/column in the same direction by 90 degree. (&lt;em&gt;Fixed tiles of the row/column stay obviously unchanged.&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You &lt;strong&gt;must not tilt&lt;/strong&gt; a row/column which includes your &lt;strong&gt;current position&lt;/strong&gt; on the grid.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the board does &lt;strong&gt;not warp around&lt;/strong&gt;. It is not possible to leave it on one side and renter from the other side.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;h2&gt;Notation&lt;/h2&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Tilt-notation&lt;/strong&gt; includes either the letter (column) or the number (row) of the tilt, and either + or - to indicated direction (&lt;em&gt;see example below&lt;/em&gt;).&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Move-notation&lt;/strong&gt; includes start and end position given by the grid-coordinates (&lt;em&gt;see example below&lt;/em&gt;).&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;h2&gt;Example&lt;/h2&gt;&#xA;&lt;p&gt;Starting from the start position given above, you first all dice of column B in + direction. (&lt;em&gt;Imagine the top of the dice tilting downwards.&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Tilt-Notation: &lt;strong&gt;(B+)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PxZxW.jpg&quot; alt=&quot;Tilt1&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now you move the marker.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Move-Notation: &lt;strong&gt;[A1 -&amp;gt; B1]&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KquaE.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;So the whole step would have the notation:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Step 1: &lt;strong&gt;(B+)  [A1 -&amp;gt; B1]&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;The puzzle question:&lt;/h1&gt;&#xA;&lt;p&gt;What is the minimum number of steps you need to visit all 4 corners of the grid (starting in A1) from the given starting situation:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ob6jx.jpg&quot; alt=&quot;Starting position&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;You have to give all steps in the notation explained above.&lt;/p&gt;&#xA;&lt;h1&gt;The current record for this puzzle is 7 moves.&lt;/h1&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2019-02-27T19:50:45.410" Title="The tilted labyrinth - Can you find the fastest path in this 3D-puzzle? (Simulator now included.)" Tags="|geometry|mechanical-puzzles|optimization|visual|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="5801" PostTypeId="2" ParentId="5796" CreationDate="2014-12-02T12:56:56.640" Score="3" Body="&lt;p&gt;Could it be a cube missing its middle?  Basically, if you drilled a square hole through a cube in all 3 directions, you'd have a cube with square shaped holes. &lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2014-12-02T12:56:56.640" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5802" PostTypeId="2" ParentId="5796" CreationDate="2014-12-02T13:44:02.840" Score="9" Body="&lt;p&gt;Take a large cube.  Extract a smaller cube exactly centered in the large cube.  You get a hollow cube.  Then cut the large cube along a diagonal (i.e. along a plane that contains 2 opposite edges).  And put back the small cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have the tools right now to produce a correct image.  Here is what it looks like from the side.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   ____ /|&#xA;  |   /  |&#xA;  | /    |&#xA;  /      |&#xA; --------'&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(Image provided by OP)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vkCKd.png&quot; alt=&quot;A right isoscles triangle, with a smaller traingle of similar shape on it, their hypothenusa aligned, the midpoints of their hypothenusa on the same point, and mirrored on the hypothenuse, so the triangles are &amp;#39;pointing away&amp;#39; from each other&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="68" LastEditDate="2014-12-04T11:20:47.453" LastActivityDate="2014-12-04T11:20:47.453" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5803" PostTypeId="2" ParentId="4810" CreationDate="2014-12-02T16:52:52.043" Score="9" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;EWOKS ATTACK AT DAWN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The explanation is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;As AE said, the only important information is the order, the grid positions and the direction of the words.  I call directions S, SW, W, or NW direction 1, and the others direction 2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From there we just apply the formula:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  If (direction 1) then calculate (row number + 20) mod 26. Else (direction 2) just take the column number. Apply the usual number to letter mapping (1=A, 2=B, etc). Interpret Xs (24) as spaces.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, for example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ALDERAAN runs in direction 1 and starts in row 11.  11 + 20 = 31.  31 mod 26 is 5.  E is the 5th letter of the alphabet. ARMY runs in direction 2 and starts on column 15.  Letter 15 is O.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Additional info added by AE:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's absolutely correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For our prisoner to begin deciphering, he just needs to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Write the alphabet across the top of the wordsearch and then down the left-hand side, as I did when I was creating it:&#xA;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/sOpZg.png&quot; alt=&quot;wordsearch solution&quot;&gt;&#xA;&#xA; (note extra row/column at top and left).&#xA;&#xA; Then if the word points up/right you take the letter from the top of the grid, if it points down/left then you take the letter from the left of the grid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2162" LastEditorUserId="4050" LastEditDate="2014-12-02T17:40:18.413" LastActivityDate="2014-12-02T17:40:18.413" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="5804" PostTypeId="1" CreationDate="2014-12-02T17:17:02.967" Score="5" ViewCount="792" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KaXQX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 3 datatables. We forgot the numbers around the third square. How can we compute the 6 missing  numbers?(there are question marks instead of them)&lt;/p&gt;&#xA;" OwnerDisplayName="Veritas" LastEditorDisplayName="user20" LastEditDate="2014-12-02T18:05:52.407" LastActivityDate="2014-12-03T20:27:10.567" Title="How can we compute the 6 missing numbers?" Tags="|mathematics|calculation-puzzle|" AnswerCount="2" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="5805" PostTypeId="2" ParentId="5804" CreationDate="2014-12-02T18:35:57.093" Score="6" Body="&lt;p&gt;Moving top to bottom on the left side...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1 4 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Going right to left on the bottom...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;6 7 7&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These values in the third table, using the same method as the other two tables...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;...multiply two three-digit values.  So the third one shows that 249 times 573 equals 142,677 (value on left, read downward, concatenated with value on bottom, read going right).  The technique is called &quot;lattice multiplication.&quot;  It works because it's simply like multiplying any long numbers by hand.  That is, multiplying each combination of digits and summing for the appropriate values of the answer's digits in their corresponding places.  It's just a different format of writing out the calculation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The values are calculated as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The value in each cell is its column's (top) header value multiplied by the header value to the right of its row.  Then, the product's two digits are split by a diagonal line which bisects the cell equally from top-right to bottom-left.  After that, the values in each diagonal stripe, starting on the bottom-right and moving to the top-left, are summed up, carrying any overflow into a second digit into the sum of the next stripe.  Those summed values are placed either to the bottom of the table or to its left, depending on where the diagonal stripe for the summed values end.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;[EDIT:]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thought of an easier solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Just multiply the number on top of the table, read moving to the right, by the number to the right, read going down.  Then, the first three digits of the product go top to bottom on the left, and the next three digits go left to right on the bottom.  (The question is how the numbers can be computed - not that they &lt;em&gt;have&lt;/em&gt; to be computed in the way implied by the table's presence, right?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5948" LastEditorUserId="5948" LastEditDate="2014-12-02T18:53:17.510" LastActivityDate="2014-12-02T18:53:17.510" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5806" PostTypeId="1" AcceptedAnswerId="5817" CreationDate="2014-12-02T18:59:33.493" Score="3" ViewCount="2620" Body="&lt;p&gt;I have a problem for anybody who cares to try.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're job is to take a 10x10x10x10 size tesseract and design a maze that fits. The maze must be a perfect maze (no loops, one path cannot be followed and find the start of its path without having the solver turn around), and must have at least 25 dead ends, and at least 8 branching off of other dead ends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer you provide should be an image, and at most of the tesseract (if not all) should be used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;note; The fun is in creating the 4D maze! 2D and even 3D is too easy for you puzzlers, I'm going to challenge you!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;To limit the possibilities of answers, in order for your maze to be accepted, it has to be the shortest possible maze that meets all the requirements already stated. By shortest I mean the maze with the least area of the tesseract filled.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-12-03T11:11:44.863" LastActivityDate="2017-08-15T10:07:21.417" Title="4D Maze Creation!" Tags="|mathematics|geometry|mazes|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="5808" PostTypeId="2" ParentId="5778" CreationDate="2014-12-02T20:58:28.387" Score="10" Body="&lt;p&gt;The answer is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; winter is here   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Method:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - spell each integer (excluding index numbers) - two five seven eight  etc.&lt;br&gt;&#xA; - select letters from each word using index numbers - w i n t  etc.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastActivityDate="2014-12-02T20:58:28.387" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5809" PostTypeId="1" AcceptedAnswerId="5822" CreationDate="2014-12-02T21:18:20.670" Score="30" ViewCount="10673" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Disclaimer&lt;/em&gt;: no offence is intended to any of the puzzling.SE users mentioned (or not mentioned) here! I was inspired by &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4341/the-serial-killers-killer&quot;&gt;this question&lt;/a&gt; to use them as characters, with no regard for reality.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are walking through the mansion of the mighty Lord &lt;a href=&quot;https://puzzling.stackexchange.com/users/2484/dalarcop&quot;&gt;D'alar'cop&lt;/a&gt;, Duke of Puzzling in the vast empire of Stackexchange. You see almost nobody apart from the occasional servant. As you pass outside through the great double doors at the entrance to the mansion, you overhear part of a heated conversation between two footmen, &lt;a href=&quot;https://puzzling.stackexchange.com/users/2383/warspyking&quot;&gt;Warspyking&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/users/2718/michael&quot;&gt;Michael&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;His Lordship will hear about this! You'll be out on your ear before the end of the week!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Oh, will he indeed? We'll see about that. If you tell him anything about me, your adventures with a certain maid might also come to his attention.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;He'll never hear a word from you again, I can promise you. You...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you walk further out along the sweeping drive towards the edge of the estate, the footmen's conversation fades out of your earshot. It is November, and bitterly cold even at noon. The parkland surrounding you is covered in deep snow, but the drive has been kept clear of snow and ice for his Lordship's carriage. The winter panorama is astonishingly beautiful, and you do not even notice the cold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You reach the Folly, a small tower just off the main drive where his Lordship is normally to be found at this time of day, poring over his books. A gardener in snowshoes, &lt;a href=&quot;https://puzzling.stackexchange.com/users/5135/oblongamous&quot;&gt;Oblongamous&lt;/a&gt;, is adjusting one of the snow sculptures that are dotted about the grounds, and fails to notice you as you walk up to the Folly and enter silently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you open the single door to the Folly, you find his Lordship sprawled over his writing desk in a pool of blood, clearly quite dead. His body is stiff and ice-cold; he must have been dead for some time. You wonder why nobody has yet found the body and alerted the authorities. Looking around for clues, you see a bloodstained knife lying on the floor just behind the door. Also the electric light is switched off, the only light coming from the sun shining directly through the open doorway. There are no windows. The killer must have turned off the light as (s)he left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You leave the Folly, moving northwards across virgin snow towards the mansion. But you have only got a few metres when you hear a piercing scream from behind you. Looking back, you see a maid, &lt;a href=&quot;https://puzzling.stackexchange.com/users/5106/avigrail&quot;&gt;Avigrail&lt;/a&gt;, running out of the Folly in a terrible state. Oblongamous is still nearby; when he hears her story, he checks quickly inside the Folly and then they both go running to the mansion to raise the alarm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bloodstained knife is found to bear the fingerprints of &lt;a href=&quot;https://puzzling.stackexchange.com/users/71/gilles&quot;&gt;Gilles&lt;/a&gt;, a well-known rival of his Lordship, and the blood is of his Lordship's blood group. At Gilles's trial, &lt;a href=&quot;http://en.wikipedia.org/wiki/Rumpole_of_the_Bailey&quot; rel=&quot;noreferrer&quot;&gt;his defence lawyer&lt;/a&gt; makes the case that his Lordship's death wound is much wider and rounder than the knife blade, and the bloodstains on the knife show it could not have entered the body. He argues that the blood was only smeared onto the knife, and finds a cut on his Lordship's arm which could have been used to produce blood for this purpose. &quot;Someone,&quot; he thunders in his final summing-up, &quot;has been trying to frame my client!&quot; His eloquence impresses the jury, and Gilles is acquitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Avigrail and Oblongamous both appear as witnesses in court, and say they saw nobody entering or leaving the Folly other than his Lordship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The police have nothing to go on, but you can help!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who killed his Lordship?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus question: &lt;strong&gt;how was he killed?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2016-09-05T12:29:24.987" Title="Murder on Stack Exchange!" Tags="|mystery|" AnswerCount="12" CommentCount="19" ContentLicense="CC BY-SA 3.0" />
  <row Id="5810" PostTypeId="2" ParentId="5809" CreationDate="2014-12-02T21:58:53.663" Score="7" Body="&lt;p&gt;Not really sure who the killer is.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'd guess that Gilles is indeed the killer and he set up this apparent framing to try to get the blame off him. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;His Lordship was killed by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; stabbing with an icicle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="239" LastActivityDate="2014-12-02T21:58:53.663" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5811" PostTypeId="2" ParentId="5809" CreationDate="2014-12-02T22:08:36.447" Score="5" Body="&lt;p&gt;You walk from the mansion towards the folly, then leave and walk back towards the mansion &lt;strong&gt;on virgin snow&lt;/strong&gt;. The snow is all fresh, and you've only gone a few metres from the door, and yet...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...you get surprised by a maid who sees the stiff. Avigrail killed him, then saw you coming, so had to hide inside the folly until you opened the door so she could get out and pretend to have just discovered the body.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As Golden Dragon said, he was killed with&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an icicle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2518" LastActivityDate="2014-12-02T22:08:36.447" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5813" PostTypeId="2" ParentId="5809" CreationDate="2014-12-02T23:40:06.047" Score="6" Body="&lt;p&gt;The answer is quite simple...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The murderer was;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; rand al'thor&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why? It was all part of the plan...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He had killed his Lordship and then came to &lt;em&gt;puzzling stack exchange&lt;/em&gt; explaining &lt;em&gt;his version of the story&lt;/em&gt; in which he was not mentioned in. He then decided to take &lt;strong&gt;OUR&lt;/strong&gt; answers and give them to the court. He never mentioned himself in the story at all, therefore he never appeared in any answers. He attempted to blame someone else, but tricked us into helping in his elaborate scheme.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How was he killed?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; With whatever other elaborate scheme rand used. I guess he framed an icicle so probably something else.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Comment for OP;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nice plan/scheme you had there ;D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-12-03T00:18:22.377" LastActivityDate="2014-12-03T00:18:22.377" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5814" PostTypeId="5" CreationDate="2014-12-02T23:57:08.593" Score="0" Body="&lt;p&gt;Only tag a question with this tag if you're asking about a puzzle within a video game, or about strategies for puzzles like the puzzle in your video game. Do not use this tag if your question is about general strategies about the game as a whole. In that case you should try a gaming site on the network, such as Arqade. Or if your question is about developing core concepts into a puzzle game (and not the puzzle itself) try game development.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-12-03T03:18:15.187" LastActivityDate="2014-12-03T03:18:15.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5815" PostTypeId="4" CreationDate="2014-12-02T23:57:08.593" Score="0" Body="A question that involves a video-game (console or computer) which contains a puzzle of sorts." OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-12-03T03:18:09.997" LastActivityDate="2014-12-03T03:18:09.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5816" PostTypeId="2" ParentId="5809" CreationDate="2014-12-03T01:13:18.553" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;It is November, and bitterly cold even at noon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This, and the mention of &lt;em&gt;parkland&lt;/em&gt; presumably ruling out Antarctica, indicates that the story happens in the northern hemisphere.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;at noon. (…)&lt;br&gt;&#xA;  (…) the single door to the Folly (…)&lt;br&gt;&#xA;  the sun shining directly through the open doorway.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the only door of the Folly is to the south.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;moving northwards across virgin snow towards the mansion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The door of the Folly is on the side opposite to the direction of the mansion.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Looking back, you see a maid, Avigrail, running out of the Folly in a terrible state.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I'm already around the side of the Folly opposite to the sole door, yet I can see the maid running out? And then the gardener comes around too. Hmmm.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;then they both go running to the mansion to raise the alarm.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And yet they don't encounter me. Nor do they notice the prints that I just left walking “across virgin snow towards the mansion”.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;But enough lollylagging! None of this is terribly relevant…&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You leave the Folly, moving northwards across virgin snow towards the mansion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I just discovered a murder, and I don't go and report it straight away: the gardener and the maid know the grounds well and thus presumably took the most direct route to the mansion — the same route I took on the way there. I, on the other hand, sneaked around the footmen and hid in the snow (amongst the ice sculptures?), so that I somehow managed to not be noticed by either the maid or the gardener. Why am I so keen not to make my presence known?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am either the murderer or an accomplice. We don't have enough information to know who the accomplice might be — I consider everyone on the estate suspect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh, and given how cold it is, &lt;a href=&quot;https://puzzling.stackexchange.com/a/5810&quot;&gt;the icicle&lt;/a&gt; wouldn't have melt. Did I by any chance chuck it into the fireplace while you were in the mansion? I hope someone took note if any fire had been recently doused.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Well, except that I &lt;em&gt;know&lt;/em&gt; I didn't do all of this. This story is bogus — I was indeed framed. The narrator is therefore the one who's lying to cover up for the murder he committed or was an accomplice in.&lt;/p&gt;&#xA;" OwnerUserId="71" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2014-12-03T01:13:18.553" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5817" PostTypeId="2" ParentId="5806" CreationDate="2014-12-03T01:26:57.243" Score="22" Body="&lt;p&gt;Here it is. Access the &lt;a href=&quot;https://i.stack.imgur.com/ZeRW7.png&quot; rel=&quot;noreferrer&quot;&gt;direct link&lt;/a&gt; to see it in its full size (or zoom-in the image).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZeRW7.png&quot; alt=&quot;Maze&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a plane of boards (horizontal is $w$ and vertical is $z$) where each board is a 2D-plane (horizontal is $x$ and vertical is $y$). To change your $x$ and $y$ positions, just walk around in the current board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The arrows allows you to change your $w$ and $z$ positions. They make you jump one board up (blue), down (yellow), left (red) or right (green), accordingly to its direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you are in a particular $(a, b)$ position of a board:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;By using the up (blue) arrow you will land in the $(a, b)$ position of the board immediately above of the current one.&lt;/li&gt;&#xA;&lt;li&gt;By using the down (yellow) arrow you will land in the $(a, b)$ position of the board immediately below of the current one.&lt;/li&gt;&#xA;&lt;li&gt;By using the left (red) arrow, you will land in the $(a, b)$ position of the board immediately to the left of the current one.&lt;/li&gt;&#xA;&lt;li&gt;By using the right (green) arrow, you will land in the $(a, b)$ position of the board immediately to the right of the current one.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To generate this, I created this Java program:&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import java.awt.Color;&#xA;import java.awt.Graphics2D;&#xA;import java.awt.image.BufferedImage;&#xA;import java.io.File;&#xA;import java.io.IOException;&#xA;import java.util.ArrayList;&#xA;import java.util.Arrays;&#xA;import java.util.Collections;&#xA;import java.util.HashMap;&#xA;import java.util.List;&#xA;import java.util.Map;&#xA;import java.util.Objects;&#xA;import javax.imageio.ImageIO;&#xA;&#xA;/**&#xA; * @author Victor&#xA; */&#xA;public class Tesseracter {&#xA;    public static void main(String[] args) throws IOException {&#xA;        TesseractMaze maze = new TesseractMaze(10, 10, 10, 10);&#xA;        BufferedImage im = maze.draw();&#xA;        ImageIO.write(im, &quot;png&quot;, new File(&quot;maze.png&quot;));&#xA;    }&#xA;&#xA;    public static final class Coordinate4D {&#xA;        private final TesseractMaze maze;&#xA;        private final int w, x, y, z;&#xA;&#xA;        public Coordinate4D(TesseractMaze maze, int w, int x, int y, int z) {&#xA;            Objects.requireNonNull(maze);&#xA;            if (w &amp;lt; 0 || w &amp;gt;= maze.wSize || x &amp;lt; 0 || x &amp;gt;= maze.xSize || y &amp;lt; 0 || y &amp;gt;= maze.ySize || z &amp;lt; 0 || z &amp;gt;= maze.zSize) throw new IndexOutOfBoundsException();&#xA;            this.maze = maze;&#xA;            this.w = w;&#xA;            this.x = x;&#xA;            this.y = y;&#xA;            this.z = z;&#xA;        }&#xA;&#xA;        @Override&#xA;        public int hashCode() {&#xA;            return Objects.hash(maze, w, x, y, z);&#xA;        }&#xA;&#xA;        @Override&#xA;        public boolean equals(Object another) {&#xA;            if (!(another instanceof Coordinate4D)) return false;&#xA;            Coordinate4D c4d = (Coordinate4D) another;&#xA;            return maze == c4d.maze &amp;amp;&amp;amp; w == c4d.w &amp;amp;&amp;amp; x == c4d.x &amp;amp;&amp;amp; y == c4d.y &amp;amp;&amp;amp; z == c4d.z;&#xA;        }&#xA;&#xA;        public int squareDistance(Coordinate4D another) {&#xA;            Objects.requireNonNull(another);&#xA;            if (maze != another.maze) throw new IllegalArgumentException();&#xA;            int dw = Math.abs(w - another.w);&#xA;            int dx = Math.abs(x - another.x);&#xA;            int dy = Math.abs(y - another.y);&#xA;            int dz = Math.abs(z - another.z);&#xA;            return dw + dx + dy + dz;&#xA;        }&#xA;&#xA;        public Coordinate4D minusW() { return w == 0              ? null : new Coordinate4D(maze, w - 1, x, y, z); };&#xA;        public Coordinate4D plusW()  { return w == maze.wSize - 1 ? null : new Coordinate4D(maze, w + 1, x, y, z); };&#xA;        public Coordinate4D minusX() { return x == 0              ? null : new Coordinate4D(maze, w, x - 1, y, z); };&#xA;        public Coordinate4D plusX()  { return x == maze.xSize - 1 ? null : new Coordinate4D(maze, w, x + 1, y, z); };&#xA;        public Coordinate4D minusY() { return y == 0              ? null : new Coordinate4D(maze, w, x, y - 1, z); };&#xA;        public Coordinate4D plusY()  { return y == maze.ySize - 1 ? null : new Coordinate4D(maze, w, x, y + 1, z); };&#xA;        public Coordinate4D minusZ() { return z == 0              ? null : new Coordinate4D(maze, w, x, y, z - 1); };&#xA;        public Coordinate4D plusZ()  { return z == maze.zSize - 1 ? null : new Coordinate4D(maze, w, x, y, z + 1); };&#xA;        public TesseractMaze getMaze() { return maze; }&#xA;    }&#xA;&#xA;    public static final class TesseractMaze {&#xA;        private final int wSize, xSize, ySize, zSize;&#xA;        private final Map&amp;lt;Coordinate4D, Node&amp;gt; nodes;&#xA;        private final Node start;&#xA;&#xA;        public TesseractMaze(int w, int x, int y, int z) {&#xA;            this.wSize = w;&#xA;            this.xSize = x;&#xA;            this.ySize = y;&#xA;            this.zSize = z;&#xA;            nodes = new HashMap&amp;lt;&amp;gt;(w * x * y * z);&#xA;            fill();&#xA;            this.start = chooseRandomNode();&#xA;            growMaze();&#xA;        }&#xA;&#xA;        private void fill() {&#xA;            for (int a = 0; a &amp;lt; wSize; a++) {&#xA;                for (int b = 0; b &amp;lt; xSize; b++) {&#xA;                    for (int c = 0; c &amp;lt; ySize; c++) {&#xA;                        for (int d = 0; d &amp;lt; zSize; d++) {&#xA;                            Coordinate4D coord = new Coordinate4D(this, a, b, c, d);&#xA;                            nodes.put(coord, new Node(coord));&#xA;                        }&#xA;                    }&#xA;                }&#xA;            }&#xA;        }&#xA;&#xA;        public Node nodeAt(Coordinate4D coord) {&#xA;            if (coord == null) return null;&#xA;            return nodes.get(coord);&#xA;        }&#xA;&#xA;        private Node chooseRandomNode() {&#xA;            int n = (int) (Math.random() * wSize * xSize * ySize * zSize);&#xA;            return new ArrayList&amp;lt;&amp;gt;(nodes.values()).get(n);&#xA;        }&#xA;&#xA;        private void growMaze() {&#xA;            List&amp;lt;Node&amp;gt; frontier = new ArrayList&amp;lt;&amp;gt;(wSize * xSize * ySize * zSize);&#xA;            frontier.add(start);&#xA;            start.linked = true;&#xA;            while (!frontier.isEmpty()) {&#xA;                Collections.shuffle(frontier);&#xA;                Node n = frontier.get(0);&#xA;                Node next = n.linkRandomUnlinkedNeighbour();&#xA;                if (next != null) {&#xA;                    frontier.add(next);&#xA;                } else {&#xA;                    frontier.remove(0);&#xA;                }&#xA;            }&#xA;        }&#xA;&#xA;        public BufferedImage draw() {&#xA;            int cellWidth = 16;&#xA;            int cellHeight = 16;&#xA;            int boardWidth = cellWidth * (xSize + 1);&#xA;            int boardHeight = cellHeight * (ySize + 1);&#xA;            int arrowSize = 3;&#xA;            int margin = 2;&#xA;            Color red = Color.RED;&#xA;            Color blue = Color.BLUE;&#xA;            Color yellow = new Color(128, 128, 0);&#xA;            Color green = new Color(0, 128, 0);&#xA;            BufferedImage im = new BufferedImage(wSize * boardWidth + cellWidth - 1, zSize * boardHeight + cellHeight - 1, BufferedImage.TYPE_INT_ARGB);&#xA;            Graphics2D g = im.createGraphics();&#xA;            for (int w = 0; w &amp;lt; wSize; w++) {&#xA;                for (int z = 0; z &amp;lt; zSize; z++) {&#xA;                    for (int x = 0; x &amp;lt; xSize; x++) {&#xA;                        for (int y = 0; y &amp;lt; ySize; y++) {&#xA;                            Coordinate4D c = new Coordinate4D(this, w, x, y, z);&#xA;                            Node n = nodeAt(c);&#xA;                            int x1 = cellWidth * x + boardWidth * w + cellWidth - 1;&#xA;                            int y1 = cellHeight * y + boardHeight * z + cellHeight - 1;&#xA;                            int x2 = x1 + cellWidth;&#xA;                            int y2 = y1 + cellHeight;&#xA;                            int x3 = (x1 + x2) / 2;&#xA;                            int y3 = (y1 + y2) / 2;&#xA;                            g.setColor(Color.BLACK);&#xA;                            if (!n.isLinkedTo(n.minusY())) g.drawLine(x1, y1, x2, y1);&#xA;                            if (!n.isLinkedTo(n.plusY()))  g.drawLine(x1, y2, x2, y2);&#xA;                            if (!n.isLinkedTo(n.minusX())) g.drawLine(x1, y1, x1, y2);&#xA;                            if (!n.isLinkedTo(n.plusX()))  g.drawLine(x2, y1, x2, y2);&#xA;                            if (n.isLinkedTo(n.minusW())) { // Board left, left arrow.&#xA;                                g.setColor(red);&#xA;                                for (int i = 0; i &amp;lt; arrowSize; i++) {&#xA;                                    g.drawLine(x1 + margin + i, y3 - i, x1 + margin + i, y3 + i);&#xA;                                }&#xA;                            }&#xA;                            if (n.isLinkedTo(n.plusW())) { // Board right, right arrow.&#xA;                                g.setColor(green);&#xA;                                for (int i = 0; i &amp;lt; arrowSize; i++) {&#xA;                                    g.drawLine(x2 - margin - i, y3 - i, x2 - margin - i, y3 + i);&#xA;                                }&#xA;                            }&#xA;                            if (n.isLinkedTo(n.minusZ())) { // Board up, up arrow.&#xA;                                g.setColor(blue);&#xA;                                for (int i = 0; i &amp;lt; arrowSize; i++) {&#xA;                                    g.drawLine(x3 - i, y1 + margin + i, x3 + i, y1 + margin + i);&#xA;                                }&#xA;                            }&#xA;                            if (n.isLinkedTo(n.plusZ())) { // Board down, down arrow.&#xA;                                g.setColor(yellow);&#xA;                                for (int i = 0; i &amp;lt; arrowSize; i++) {&#xA;                                    g.drawLine(x3 - i, y2 - margin - i, x3 + i, y2 - margin - i);&#xA;                                }&#xA;                            }&#xA;                        }&#xA;                    }&#xA;                }&#xA;            }&#xA;            return im;&#xA;        }&#xA;    }&#xA;&#xA;    public static final class Node {&#xA;        private final Coordinate4D coord;&#xA;        private final List&amp;lt;Node&amp;gt; linkedNeighbours;&#xA;        private List&amp;lt;Node&amp;gt; neighbours;&#xA;        private boolean linked;&#xA;&#xA;        public Node(Coordinate4D coord) {&#xA;            Objects.requireNonNull(coord);&#xA;            this.coord = coord;&#xA;            linkedNeighbours = new ArrayList&amp;lt;&amp;gt;(8);&#xA;        }&#xA;&#xA;        public Node linkRandomUnlinkedNeighbour() {&#xA;            List&amp;lt;Node&amp;gt; list = new ArrayList&amp;lt;&amp;gt;(getNeighbours());&#xA;            list.removeIf(n -&amp;gt; n.linked);&#xA;            if (list.isEmpty()) return null;&#xA;            Collections.shuffle(list);&#xA;            Node next = list.get(0);&#xA;            next.getNeighbours();&#xA;            linkedNeighbours.add(next);&#xA;            next.linkedNeighbours.add(this);&#xA;            next.linked = true;&#xA;            return next;&#xA;        }&#xA;&#xA;        @SuppressWarnings(&quot;ReturnOfCollectionOrArrayField&quot;)&#xA;        public List&amp;lt;Node&amp;gt; getNeighbours() {&#xA;            if (neighbours == null) {&#xA;                List&amp;lt;Node&amp;gt; nodes = new ArrayList&amp;lt;&amp;gt;(Arrays.asList(minusW(), plusW(), minusX(), plusX(), minusY(), plusY(), minusZ(), plusZ()));&#xA;                nodes.removeIf(x -&amp;gt; x == null);&#xA;                neighbours = Collections.unmodifiableList(nodes);&#xA;            }&#xA;            return neighbours;&#xA;        }&#xA;&#xA;        public boolean isDeadEnd() {&#xA;            return linkedNeighbours.size() == 1;&#xA;        }&#xA;&#xA;        public boolean isBranch() {&#xA;            return linkedNeighbours.size() &amp;gt; 2;&#xA;        }&#xA;&#xA;        public boolean isLinkedTo(Node node) {&#xA;            return linkedNeighbours.contains(node);&#xA;        }&#xA;&#xA;        public TesseractMaze getMaze() { return coord.getMaze(); }&#xA;        public Coordinate4D getCoord() { return coord; }&#xA;        public Node minusW() { return getMaze().nodeAt(coord.minusW()); };&#xA;        public Node plusW()  { return getMaze().nodeAt(coord.plusW());  };&#xA;        public Node minusX() { return getMaze().nodeAt(coord.minusX()); };&#xA;        public Node plusX()  { return getMaze().nodeAt(coord.plusX());  };&#xA;        public Node minusY() { return getMaze().nodeAt(coord.minusY()); };&#xA;        public Node plusY()  { return getMaze().nodeAt(coord.plusY());  };&#xA;        public Node minusZ() { return getMaze().nodeAt(coord.minusZ()); };&#xA;        public Node plusZ()  { return getMaze().nodeAt(coord.plusZ());  };&#xA;    }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;I am sorry that this site has no syntax-coloring.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are hundreds or thousands of dead-end and branching points. Much more than just 25 and 8 required by the OP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each two locations there is exactly one path to each other location. There are no cycles in the graph and it is connected. The program ensures that (&lt;code&gt;growMaze&lt;/code&gt; method).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no defined starting or end point. Just randomly get any two points and try to find a path. As you see, manually finding a path here should be hard, since there is ten thousand positions in this maze and looking around the $w$ and $z$ dimensions to find a useful path is harder for human eyes than it is in the $x$ and $y$ dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may use the program to randomly generate other mazes. Changing its size is easy too: They are those four tens in the start of the &lt;code&gt;main&lt;/code&gt; method. In the &lt;code&gt;main&lt;/code&gt; method you may change to which file the generated maze is saved. To show that, here it is a much smaller and simpler $4 \times 5 \times 3 \times 2$ maze generated by the program:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eWbgG.png&quot; alt=&quot;Small maze&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, by setting $w$ to 1 and $z$ to 1, you may use it to generate 2D mazes. If you set only one of them to 1, it will be a 3D maze.&lt;/p&gt;&#xA;" OwnerUserId="5044" LastEditorUserId="5044" LastEditDate="2015-07-17T00:29:24.503" LastActivityDate="2015-07-17T00:29:24.503" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="5818" PostTypeId="2" ParentId="5804" CreationDate="2014-12-03T01:38:33.567" Score="4" Body="&lt;p&gt;These tables are examples of &lt;a href=&quot;http://en.wikipedia.org/wiki/Lattice_multiplication&quot; rel=&quot;nofollow&quot;&gt;lattice multiplication&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ignoring the diagonal lines, the numbers in each of the squares is the product of the number at the top of its column and the number at the right of its row. For example, in the top-left of the first example in the question, $4 \cdot 8 = 32$.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;          4&#xA;+---+---+---+&#xA;|   |   |   |&#xA;|   |   |32 | 8&#xA;|   |   |   |&#xA;+---+---+---+&#xA;|   |   |   |&#xA;|   |   |   |&#xA;|   |   |   |&#xA;+---+---+---+&#xA;|   |   |   |&#xA;|   |   |   |&#xA;|   |   |   |&#xA;+---+---+---+&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then, in the square, write the tens digit of the product to the left of the diagonal line and the ones digit to the right of it.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;          4&#xA;+---+---+---+&#xA;|  /|  /|3 /|&#xA;| / | / | / | 8&#xA;|/  |/  |/ 2|&#xA;+---+---+---+&#xA;|  /|  /|  /|&#xA;| / | / | / |&#xA;|/  |/  |/  |&#xA;+---+---+---+&#xA;|  /|  /|  /|&#xA;| / | / | / |&#xA;|/  |/  |/  |&#xA;+---+---+---+&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then, looking at the diagonal lines and ignoring the vertical and horizontal lines making the squares, we see that there are several diagonal columns going down and to the left.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;  2   7   4&#xA;+---+---+---+&#xA;|1 / 1 / 3 /|&#xA;| /   /   / | 8&#xA;|/ 6 / 4 / 2|&#xA;+   /   /   +&#xA;|1 / 6 / 3 /|&#xA;| /   /   / | 9&#xA;|/ 8 / 3 / 6|&#xA;+   /   /   +&#xA;|1 / 3 / 2 /|&#xA;| /   /   / | 5&#xA;|/ 0 / 5 / 0|&#xA;+---+---+---+&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For each of these diagonal columns, starting with the leftmost, add up all of the digits that are in them. Write the ones digit of the sum in the space immediately to the bottom or left of the column (where the question marks are in example 3), then carry the tens and larger digits to the next column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In example three, the multiplication steps have already been completed. All that's left is to add up the diagonal columns. Doing so, we find that the numbers in the question marks are:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;    2   4   9&#xA;  +---+---+---+&#xA;* |1 /|2 /|4 /|&#xA;1 | / | / | / | 5&#xA;* |/ 0|/ 0|/ 5|&#xA;  +---+---+---+&#xA;* |1 /|2 /|6 /|&#xA;4 | / | / | / | 7&#xA;* |/ 4|/ 8|/ 3|&#xA;  +---+---+---+&#xA;* |0 /|1 /|2 /|&#xA;2 | / | / | / | 3&#xA;* |/ 6|/ 2|/ 7|&#xA;  +---+---+---+&#xA;   *6* *7* *7*&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Since lattice multiplication is an algorithm for multiplying, we can find the numbers in the question marks even faster: Ignore the numbers in the lattice altogether, multiply the numbers going across the top and right of the lattice, and write the product along the left and bottom of the lattice. In the case of example three, the number going across the top of the lattice is $249$, the number going down the right of the lattice is $573$, and their product is $249 \cdot 573 = 142677$.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;    2   4   9&#xA;  +---+---+---+&#xA;* |   |   |   |&#xA;1 |   |   |   | 5&#xA;* |   |   |   |&#xA;  +---+---+---+&#xA;* |   |   |   |&#xA;4 |   |   |   | 7&#xA;* |   |   |   |&#xA;  +---+---+---+&#xA;* |   |   |   |&#xA;2 |   |   |   | 3&#xA;* |   |   |   |&#xA;  +---+---+---+&#xA;   *6* *7* *7*&#xA;&lt;/pre&gt;&#xA;" OwnerUserId="294" LastEditorUserId="294" LastEditDate="2014-12-03T20:27:10.567" LastActivityDate="2014-12-03T20:27:10.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5819" PostTypeId="1" AcceptedAnswerId="5821" CreationDate="2014-12-03T01:41:19.480" Score="5" ViewCount="257" Body="&lt;p&gt;You have two people in the center of an endless plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each one is holding some item that can fit in their palm, like a tennis ball. Both persons are identical in physical ability and brainpower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, each one is just as clever as sly as the other one, and both are very clever and sly in their own right. They will immediately do what is best for them, and will always do what is best for them, unless there is an incentive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Q&lt;/strong&gt;: What foolproof way is there for these two to exchange their items?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples of inappropriate ways:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Throw it over each other: One might not throw it and chase the others item&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each hold the others item at the same time: One might let go first and the other won't let go, taking both items&lt;/p&gt;&#xA;" OwnerUserId="6142" LastActivityDate="2014-12-03T01:49:07.247" Title="The strange exchange" Tags="|logical-deduction|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5820" PostTypeId="2" ParentId="5809" CreationDate="2014-12-03T01:46:14.523" Score="4" Body="&lt;p&gt;He was killed by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Avigrail. She was still inside the folly, perhaps hiding in the same room as the body and the person who discovered it, and shortly after the body was found (the finder had only moved &quot;a few meters&quot; away from the Folly - scarcely time for her to come unseen from the house to the Folly, enter without encountering the finder, and then come screaming back out) she made her escape from the Folly, screaming for &quot;help&quot;. Hearing someone approaching the Folly she had switched the lights off so as not to be noticed. &lt;em&gt;WHY&lt;/em&gt; she killed him is a matter of conjecture. Perhaps she killed him because she was the &quot;certain maid&quot; mentioned in the conversation between the two footmen. Possibly she was carrying on an affair with His Lordship, and was discovered to be diddling the footman as well. When His Lordship found out he dismissed her from service, not caring to share his strumpet with a mere servant, then turned back to this desk. In a fit of pique/rage/passion she grabbed the nearest weapon that came to hand, darted back through the still open door, and plunged the weapon into His Lordship's back, penetrating his heart and killing him instantly. As to the framing: Gilles was &quot;a well-known rival&quot; of His Lordship. Rival indeed - for the favors of the fair Avigrail! Gilles, though, had tired of her and cast her aside, and she had stolen his favorite knife. The murder of His Lordship gave her a chance to kill two birds with one knife, so to speak, by framing Gilles for the murder she had committed. Perfidy! Thy name is woman!!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;who used&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an icicle, which subsequently melted inside the body before the body cooled. It was bitterly cold but sunny - perfect weather for forming icicles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Additional rationale:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The only two other reasonable candidates are 1) the narrator, and 2) Oblongamous. IMO both should be discarded due to lack of motive communicated elsewhere in the story/question. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Summation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; M'Lord! The evidence before the court is incontrovertible - there's no need for the jury to retire! The murderer - the only POSSIBLE person with motive AND opportunity - is the &quot;Angel of Death&quot; - Avigrail! You MUST convict, and you MUST deliver the only just sentence the law allows - the same sentence she passed on the man who loved her too much, if not too wisely - DEATH!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The persecution rests.&lt;/p&gt;&#xA;" OwnerUserId="2526" LastEditorUserId="2526" LastEditDate="2014-12-03T03:29:26.173" LastActivityDate="2014-12-03T03:29:26.173" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5821" PostTypeId="2" ParentId="5819" CreationDate="2014-12-03T01:49:07.247" Score="5" Body="&lt;p&gt;Both people set their items down a ways apart, then simultaneous walk toward each other's item.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neither person can benefit from being sneaky by returning to pick up her own item, as the other person would then do the same. There is no way either person can end up with both items.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastActivityDate="2014-12-03T01:49:07.247" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5822" PostTypeId="2" ParentId="5809" CreationDate="2014-12-03T01:56:12.497" Score="23" Body="&lt;p&gt;As &lt;a href=&quot;https://puzzling.stackexchange.com/users/71/gilles&quot;&gt;Gilles&lt;/a&gt; has told us, I (the protagonist of the OP story, who discovered the body) was standing directly between the Folly and the mansion, but the maid and gardener did not see me, nor even my prints in the snow. Nor did my presence quell the footmen's argument.  Furthermore, I do not notice the cold. I must conclude that I am invisible and incorporeal, though I'm rather proud of the supernatural feat of telekinetic concentration by which I opened the Folly door to discover the body.  I suspect, in fact, that I am the ghost of His Lordship. This would explain why the author is so certain that I can help with the inquiry. If only I could remember what happened: it's all such a blur. Everything moved so fast, and then it was all dark...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wait! it's coming back to me: I was killed by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; my own hand!  I was in despair at the constant bickering and intriguing of my staff and the popular failure of my snow sculpture exhibition.  My last act was to frame my hated rival, Gilles, by cutting my arm with one of his knives. Then I plunged an icicle into my own heart, allowing the weapon to melt with my dissipating body heat. Goodbye cold, cruel world.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What I do not understand is how I could have seen Avigrail exiting the Folly. We know that the Folly's only door is on its South side, that there are no windows, and that I was a few metres North of the Folly when she came out. That must mean... my goodness! The walls of the Folly are fading away before my eyes!  And the mansion - it's... everything is fading away, everything except that point of light in the distance. I feel myself drawn towards that light...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Warning:  since His Lordship seems to have correctly solved the mystery of his own death, the comment thread below also contains spoilers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;br&gt;&lt;br&gt;&#xA;&lt;strong&gt;Motive (added by OP):&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; His Lordship D'alar'cop had learned that riddles and challenge questions were about to be &lt;a href=&quot;https://puzzling.meta.stackexchange.com/questions/1601/puzzling-se-where-we-hate-fun-its-time-to-disallow-challenge-questions?cb=1&quot;&gt;summarily banned from Puzzling.SE&lt;/a&gt;, and wished to end it all in his misery, starting a new life outside of the &quot;vast empire of Stackexchange&quot; in which his duchy (now in the process of being destroyed) lay. His final act was to attempt to frame his hated enemy Gilles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Some extra explanation added by the OP:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The above answer already explains how one can deduce that the narrator is his Lordship's ghost. To work out that his Lordship committed suicide, consider the following: &lt;br&gt;&lt;br&gt; &lt;em&gt;When you open the single door to the Folly, you find his Lordship sprawled over his writing desk in a pool of blood, clearly quite dead. His body is stiff and ice-cold; he must have been dead for some time. You wonder why nobody has yet found the body and alerted the authorities. Looking around for clues, you see a bloodstained knife lying on the floor just behind the door. Also the electric light is switched off, the only light coming from the sun shining directly through the open doorway. There are no windows. The killer must have turned off the light as (s)he left.&lt;/em&gt; &lt;br&gt;&lt;br&gt; How come the knife is behind the door? If a killer dropped it as he was running out, it would have been either within the room, not contained in the sweep of the door, or within the sweep of the door but on the &lt;em&gt;out&lt;/em&gt;side, in which case it would have been swept outside by the closing door (the door was closed when the narrator arrived) and not swept behind the door as the narrator opened it. &lt;br&gt;&lt;br&gt; I hope that's clear; if not, I'll try and make a picture!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6141" LastEditorUserId="-1" LastEditDate="2017-03-16T16:42:07.097" LastActivityDate="2014-12-03T17:14:16.157" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="5823" PostTypeId="1" AcceptedAnswerId="5825" CreationDate="2014-12-03T02:36:24.467" Score="2" ViewCount="1904" Body="&lt;p&gt;&lt;pre&gt; This riddle is a computer program/language. Enjoy!&lt;/pre&gt;&lt;br&gt;&#xA;I used to love the Sun&lt;br&gt;&#xA;Now I pray to my oracle&lt;br&gt;&#xA;I'm said to virtualize all&lt;br&gt;&#xA;I'm not picky, working anywhere...&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I love my classes-all of them&lt;br&gt;&#xA;I love my matter-known as objects&lt;br&gt;&#xA;I love running parallel-too lazy to wait&lt;br&gt;&#xA;Fans worldwide build and break with me!&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My syntax-derived from a letter&lt;br&gt;&#xA;many programmers found me familiar&lt;br&gt;&#xA;I have nothing included-very greedy&lt;br&gt;&#xA;And I must import all!&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I want every line torn apart, as this is an easy one :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5368" LastEditorUserId="5368" LastEditDate="2014-12-04T21:22:19.037" LastActivityDate="2014-12-04T21:22:19.037" Title="Software Riddle" Tags="|riddle|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5824" PostTypeId="2" ParentId="4807" CreationDate="2014-12-03T02:41:35.083" Score="5" Body="&lt;p&gt;I think the password is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;9&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the person in question is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C.S. Lewis&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The individual guests have the following hints towards certain aspects of his works&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first guest's rings symbolize the rings used by Digory Kirke and Polly Plummer in &lt;em&gt;The Magician's Nephew&lt;/em&gt;.  The second guest's sleigh symbolizes the sleigh used by the White Witch in &lt;em&gt;The Lion, the Witch and the Wardrobe&lt;/em&gt; (assuming since it's a female getting out of the sleigh, although it could also be Santa).  The third guest's horse symbolizes the horse in &lt;em&gt;The Horse and His Boy&lt;/em&gt;.  The fact that they were dragged away by lions can refer to Aslan and the guard mentioning twins points to Shasta/Cor and Corin being twins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As for further hints in the text&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A wardrobe is mentioned as the hiding place of the recipe, as in the wardrobe in &lt;em&gt;The Lion, the Witch and the Wardrobe&lt;/em&gt;.  The animals having their own entrance also alludes to the animals in &lt;em&gt;The Chronicles of Narnia&lt;/em&gt; being intelligent and human-like.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The contents of the man who you stunned pockets also point to the individual.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The inkwell, as mentioned by A E, could mean that the person is an author.  It could also point to The Inklings, a group which C.S. Lewis was a part of.  Lewis writes about both the Hebrides and Gustav Holst (the music that is playing) in his letters.  The northern part of the Outer Hebrides is also known as Lewis.  The 'S' in C.S. Lewis' name stands for Staples.  And the book 'Jude the Obscure' was mainly written by Hardy in the pub The Lamb &amp;amp; Flag, which was also frequented by C.S. Lewis.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As for the numbering of the password&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The guard's number is bogus, it doesn't matter what they say.  The number that the guest is meant to say is the number of the fictional work, considering that he wrote/published &lt;em&gt;The Pilgrim's Regress&lt;/em&gt; and the &quot;Space Trilogy&quot; novels prior to writing and publishing &lt;em&gt;The Chronicles of Narnia&lt;/em&gt;.  This would mean that the first book of &lt;em&gt;The Chronicles of Narnia&lt;/em&gt; would be the fifth overall.  The first guest stated &quot;5&quot;, and since it was determined that the book they are associated with is &lt;em&gt;The Magician's Nephew&lt;/em&gt;, this would mean that the order of the books is in the series' chronological order and not the order that it was published.  This is backed up by &lt;em&gt;The Lion, the Witch and the Wardrobe&lt;/em&gt; being second chronologically (6th overall) and &lt;em&gt;The Horse and His Boy&lt;/em&gt; being third (7th overall).  This would mean that the boat, considering that it's either a major plot point or associated with a key person, would be linked to &lt;em&gt;The Voyage of the Dawn Treader&lt;/em&gt; and is the 5th book chronologically and 9th overall.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6140" LastActivityDate="2014-12-03T02:41:35.083" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5825" PostTypeId="2" ParentId="5823" CreationDate="2014-12-03T02:49:32.190" Score="7" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Java&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I used to love the Sun&lt;br&gt;Now I pray to my oracle&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sun Microsystems used to be the standards keeper for the JRE and JDK; now Oracle Ltd. has inherited that role.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm said to virtualize all&lt;br&gt;&#xA;I'm not picky, working anywhere...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All methods in Java classes are what C++ calls &lt;code&gt;virtual&lt;/code&gt; by default, referring to the fact that they are dynamically bound. Java code is deployable across numerous operating systems and platforms, in numerous &quot;flavours&quot; (standard, micro, and enterprise). A great deal of firmware also runs on Java.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I love my classes-all of them&lt;br&gt;&#xA;I love my matter-known as objects&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Java OOP structure primitives are called &lt;code&gt;class&lt;/code&gt;es, and all Java classes are a subclass of &lt;code&gt;Object&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I love running parallel-too lazy to wait&lt;br&gt;&#xA;Fans worldwide build and break with me!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Java 5 and later come with a significant concurrency (i.e. parallel execution) API. Java enjoys a broad adoption in numerous platforms worldwide.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My syntax-derived from a letter&lt;br&gt;&#xA;many programmers found me familiar&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Java inherits much of its syntax from the programming language &lt;strong&gt;C&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have nothing included-very greedy&lt;br&gt;&#xA;And I must import all!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; By default, without imports, a Java file only has access to other classes in its package and classes in &lt;code&gt;java.lang.*&lt;/code&gt;, which is an incredibly sparse collection of objects such as boxing primitives, &lt;code&gt;System&lt;/code&gt; interface methods, and some common exceptions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2014-12-03T02:49:32.190" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5826" PostTypeId="2" ParentId="5823" CreationDate="2014-12-03T02:52:43.317" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Java&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Unspoilered explanations below:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I used to love the Sun&lt;br&gt;&#xA;  Now I pray to my oracle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It used to be owned by the company Sun, but Oracle has since bought Sun. (&lt;a href=&quot;https://en.wikipedia.org/wiki/Java_(programming_language)#History&quot; rel=&quot;nofollow&quot;&gt;Wikipedia link&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm said to virtualize all&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Everything is an object in this language.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm not picky, working anywhere...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Its motto is &quot;write once, run anywhere,&quot; and one of its design goals was to be cross-platform.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I love my classes-all of them&lt;br&gt;&#xA;  I love my matter-known as objects&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again, this language is OOP and everything is an object.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I love running parallel-too lazy to wait&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Code can run in parallel with threads in this language.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Fans worldwide build and break with me!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Write once, run anywhere&quot; - it can even run on fans / refrigerators / etc.; it's also used worldwide.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My syntax-derived from a letter&lt;br&gt;&#xA;  many programmers found me familiar&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a C-based language.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I have nothing included-very greedy&lt;br&gt;&#xA;  And I must import all!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Technically, &lt;code&gt;&amp;lt;language name&amp;gt;.lang.*&lt;/code&gt; is imported by default, but most other classes must be explicitly &lt;code&gt;import&lt;/code&gt;ed.&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2014-12-03T02:52:43.317" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5827" PostTypeId="1" AcceptedAnswerId="5828" CreationDate="2014-12-03T03:27:16.867" Score="3" ViewCount="932" Body="&lt;p&gt;&lt;em&gt;The original by &lt;a href=&quot;https://puzzling.stackexchange.com/users/5368/quynguyen2013&quot;&gt;quynguyen2013&lt;/a&gt; can be found here; &lt;a href=&quot;https://puzzling.stackexchange.com/questions/5823/software-riddle&quot;&gt;Software Riddle&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I compile quickly, yet I'm dynamic.&lt;br&gt;&#xA;  I was developed by a company you know well.&lt;br&gt;&#xA;  I'm open-sourced and a lot like a snake.&lt;br&gt;&#xA;  I also tend to be like a letter in the alphabet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's not too hard, but you have to explain every line.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2014-12-28T19:47:56.400" Title="Software Riddle [Part 2!]" Tags="|riddle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5828" PostTypeId="2" ParentId="5827" CreationDate="2014-12-03T04:04:56.643" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Go&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Nonspoilered explanations below:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I compile quickly, yet I'm dynamic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This language was designed to compile very quickly. It's statically typed, but &lt;a href=&quot;https://stackoverflow.com/q/20518457/1223693&quot;&gt;there are ways to achieve dynamic typing&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I was developed by a company you know well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It was developed by Google.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm open-sourced and a lot like a snake.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is indeed open source, and it was influenced heavily by Python.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I also tend to be like a letter in the alphabet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Certain components of this language could be somewhat comparable to the C programming language (such as the &lt;code&gt;main&lt;/code&gt; function).&lt;/p&gt;&#xA;" OwnerUserId="39" LastEditorUserId="-1" LastEditDate="2017-05-23T12:38:53.617" LastActivityDate="2014-12-03T04:04:56.643" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5829" PostTypeId="2" ParentId="5827" CreationDate="2014-12-03T04:35:40.447" Score="1" Body="&lt;p&gt;Is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because &lt;em&gt;I also tend to be like a letter in the alphabet&lt;/em&gt; makes me think so.&lt;/p&gt;&#xA;" OwnerUserId="4850" LastEditorUserId="5373" LastEditDate="2014-12-28T19:47:56.400" LastActivityDate="2014-12-28T19:47:56.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5830" PostTypeId="1" AcceptedAnswerId="5831" CreationDate="2014-12-03T04:44:42.270" Score="12" ViewCount="741" Body="&lt;p&gt;You are Agent X, an enterprising NSA analyst who's been assigned the task of spying on Bill Krill, a US citizen suspected of unpatriotic opposition to warrantless NSA spying on American citizens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By remotely activating Bill's webcam, you were able to spy out Bill's password reminder sheet on his physical desktop. Unfortunately, Bill appears to have only written down &lt;em&gt;reminders&lt;/em&gt; rather than actual passwords.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Foodie Message Board Password: &lt;em&gt;You force heaven to be ecstasy.&lt;/em&gt;&lt;/li&gt;&#xA;  &lt;li&gt;Library Password: &lt;em&gt;Am I too wise for you?&lt;/em&gt;&lt;/li&gt;&#xA;  &lt;li&gt;Online Gambling Password: &lt;em&gt;I foresee an empty success.&lt;/em&gt;&lt;/li&gt;&#xA;  &lt;li&gt;soopershoeshopper.com Password: &lt;em&gt;Knot 'em, or else he's won.&lt;/em&gt;&lt;/li&gt;&#xA;  &lt;li&gt;Daredevil Message Board Password: &lt;em&gt;I once ate a hat for cash.&lt;/em&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Bill to self:&lt;/strong&gt; All of your passwords are 8 characters or less, and all of them use your usual convention.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can you figure our Bill's &quot;usual convention&quot; as well as his 5 passwords, thus helping to make the world a safer place for citizens like Bill?&lt;/p&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2014-12-03T04:55:20.653" Title="You Force Heaven to be Ecstasy" Tags="|password|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5831" PostTypeId="2" ParentId="5830" CreationDate="2014-12-03T04:55:20.653" Score="11" Body="&lt;p&gt;The answers are:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. u472bxtc&lt;br&gt;&#xA; 2. mi2yy4u&lt;br&gt;&#xA; 3. i4cnmt6s&lt;br&gt;&#xA; 4. 0mrlcc1&lt;br&gt;&#xA; 5. i118a^4$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Method:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bill's convention is to form sentences from a sequence of characters that sound like words.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastActivityDate="2014-12-03T04:55:20.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5832" PostTypeId="2" ParentId="5809" CreationDate="2014-12-03T05:14:16.927" Score="2" Body="&lt;p&gt;&lt;strong&gt;The killer is&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oblongamous&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Since&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Oblongamous, was adjusting one of the snow sculptures that are dotted about the grounds&quot; and is the Gardner, it will be easier for him to get an icicle or a sharp object made with ice, since it was snowing and also there are ice sculptures outside. Most probably, he should have taken it from one of the ice sculptures and may be that's why he was fixing one of the sculpture from which he took the sharp part. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Also&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Oblongamous was working outside the folly with a snowshoes and was outside. Still there is no foot prints means he would have cleared the foot marks with a broom or something. Otherwise, there would be his foot mark.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So my assumption is&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;May be Oblongamous hated both Gilles and Cop and so he decided to finish the nuisance of both in a single shot. He stolea knife from Gilles house in the night with Gilles finger print. Oblongamous  used gloves so that there won't be his finger prints and wont damage Gilles prints. Next day when Cop was busy with some writings, Oblongamous thought &quot;This is the right time. There is nobody near him.&quot; He broke one of the snow sculptures and took a sharp piece from it. He switched the light of so that Cop won't identify him even though the attempt failed. He reached the back of Cop and stabbed him on his stomach with the icicle. Since the wound was deep and Cop was week (since he is too old ;p) Cop died very soon. Then Oblongamous made a wound in Cop's hand with the knife so that there will be Cop's blood on Gilles knife and dropped the knife there, so that the police will find Gilles prints and will arrest him. So he will get rid of troubles from Gilles too. After leaving the room he took a broom and cleared all the foot prints he made and clearing the marks he reached near the broken sculpture (he broke it to get the icicle). Before somebody noticing he tried to fix it and place it back urgently. That is when &quot;YOU&quot; reached there and saw that. Since Oblongamous was in a hurry fixing it before somebody noticing it, he didn't saw &quot;YOU&quot;. So Gilles or no one else entered the mansion and that's why the Guards didn't see anyone. &lt;strong&gt;It was Oblongamous  itself who killed Cop, switched the lights off and cleared the foot prints.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Another assumption is&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gilles came to kill  D'alar'cop from his back while he was writing, but when Gilles was about to stab him, Cop noticed him and blocked the knife with his hand which made the wound in his hand and blood on knife. In the fear that he will get caught, Gilles thew the knife there and tried to escape from the mansion. While he was running, Oblongamous caught him. Seeing Oblongamous , Gilles got some idea. He gave Oblongamous a handful of gold coins and asked him to kill Cop for him. Oblongamous  agreed and put the coins in his pocket. He broke one of the ice sculptures and took a sharp piece from it. He entered the Lord's room and switched the light off so that even if he get caught, the Lord (Cop) won't identify him. He sneaked right behind Cop and stabbed him on his stomach with the icicle. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="2730" LastEditDate="2014-12-03T05:43:26.213" LastActivityDate="2014-12-03T05:43:26.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5837" PostTypeId="2" ParentId="5809" CreationDate="2014-12-03T07:40:02.137" Score="2" Body="&lt;p&gt;Although I'm from a tropical country and never knew what are the effects of snow in plants I think the killer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oblongamous - a gardener.. what would a gardener do in that place? He is supposed to take care of the plants but it is winter where in based on the pictures I have seen plants/trees having lost its leaves due to winter. If plants and trees looks like that what would be the use of a gardener, another thing is that he is wearing snowshoes, this is used for better mobility, so if he would commit a crime he can escape the place pretty fast.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What was used?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If Oblangamous is the killer he would use a chip/sharp fragment from the ice sculpture that's why he fixed it after so that no one will notice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;another note.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The maid should be the another accomplice for the crime as she is already inside the folly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6145" LastEditorUserId="24713" LastEditDate="2016-09-05T12:29:24.987" LastActivityDate="2016-09-05T12:29:24.987" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="5838" PostTypeId="2" ParentId="5809" CreationDate="2014-12-03T08:48:31.840" Score="2" Body="&lt;p&gt;The killer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Oblongamous while he was adjusting one of the snow sculptures that are dotted about the grounds&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And he was killed by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A piece of the sculpture or an icicle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6147" LastActivityDate="2014-12-03T08:48:31.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5839" PostTypeId="2" ParentId="5019" CreationDate="2014-12-03T09:27:28.437" Score="1" Body="&lt;p&gt;Not an answer, rather than some random rant:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I believe that the Leipzig reference is a reference to Leibniz (not because they sound similar, rather than Leibniz being born in Leipzig, the Uni at Leipzig, etc.). So following this train of thought, we arrive at a lot of philosophy and maths. Topology and calculus, to be precise.&lt;br/&gt;I have two suggestions, first - &lt;strong&gt;fractals&lt;/strong&gt;, defined with functions that can have their inverse. Fractals are self-similar, which explains why age difference is a thing in the family. As each member becomes older, they get more-complex (higher-definition of sorts). That is why the smaller brother is smoother.&lt;br/&gt;The second suggestion is about &lt;strong&gt;manifolds&lt;/strong&gt;. A manifold can have different spatial properties depending on &quot;scale&quot;. For example the area around each point on a sphere resembles Euclidean space when the area is small enough. Each manifold is defined by a function and the enemies are the inverse?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastActivityDate="2014-12-03T09:27:28.437" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5840" PostTypeId="2" ParentId="5809" CreationDate="2014-12-03T09:48:44.677" Score="5" Body="&lt;p&gt;Ok, here is the solution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fun fact: this question is not closed because it seems that it has only one fitting answer(imagine me die of laughing now)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The amount of cyanide he carried in his body was slightly killing him. All those riddles were too much for his Lordship. His doctor had already told him that he will die, soon. So the precious Lord decided to turn himself in and at the same time burden Gilles with the accusation of murder.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solution &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;So he cut his arm and made sure to put the knife somewhere it can be easily found. He then had a look at some riddles of Avigrail and Oblongamous and the cyanide found a way out of his body as he was trying to solve them. The murderer was the Lord himself and he got killed over the years by solving hard riddles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5106" LastActivityDate="2014-12-03T09:48:44.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5841" PostTypeId="1" AcceptedAnswerId="5843" CreationDate="2014-12-03T10:13:09.650" Score="12" ViewCount="7068" Body="&lt;p&gt;Five competitors (A, B, C, D and E) are arguing about a swimming test that rewards the 1st, 2nd and 3rd places with gold, silver and bronze medals, respectively. The following assertions about the test are false, though one assertion from each (note each has two assertions) may be true.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;• A did not get gold and B did not get silver;&#xA;• D did not get silver and B did not get bronze;&#xA;• C won a medal, but D didn't;&#xA;• A won a medal, but C didn't;&#xA;• D won a medal and E did, too.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note: the answer is only about the first three places since the last two can't be said for sure. &lt;/p&gt;&#xA;" OwnerUserId="6151" LastEditorUserId="-1" LastEditDate="2014-12-03T14:30:46.207" LastActivityDate="2015-09-07T10:38:21.373" Title="Swimming test puzzle" Tags="|logical-deduction|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5842" PostTypeId="2" ParentId="5809" CreationDate="2014-12-03T10:40:59.213" Score="2" Body="&lt;p&gt;The killer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Warspyking - with Avigrail as the accomplice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Warspyking killed him with an icicle and Avigrial hid in the Folly.  When the Narrator leaves the Folly for the Mansion - there are no foot prints on the snow; Avigrial had to be already in the building or at-least pass the Narrator.  The motive is due to the discovery of their affair.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2014-12-03T10:40:59.213" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5843" PostTypeId="2" ParentId="5841" CreationDate="2014-12-03T10:49:44.820" Score="12" Body="&lt;p&gt;(3) and (4) together say that if A won a medal, C did, and if C won a medal, D did. (5) says that if E won a medal, D didn't - and therefore neither did A or C. This results in only two medallists at most, so E did not win a medal. Now there's only one non-medallist in A,B,C,D, which must be either A or B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose D didn't get silver. Then B got bronze (2) and A got gold (1), but this contradicts the deduction that one of A and B failed to win a medal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore D got silver. (1) then says A got gold, and so C must be bronze. B and E failed to win medals.&lt;/p&gt;&#xA;" OwnerUserId="1801" LastActivityDate="2014-12-03T10:49:44.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5844" PostTypeId="2" ParentId="5841" CreationDate="2014-12-03T11:05:45.050" Score="6" Body="&lt;p&gt;&lt;strong&gt;A slightly more formal solution&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This happens to be a Boolean satisfiability problem where we want a model for the variables: $A,B,C,D,E$ where they take values disjointly from $\{gold,silver,bronze,nothing1,nothing2\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formulae that act as constraints on the system are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$f1 = (A \not= gold \land B \not=silver)$&lt;/li&gt;&#xA;&lt;li&gt;$f2 = (D \not= silver \land B \not= bronze)$&lt;/li&gt;&#xA;&lt;li&gt;$f3 = (C \in \{gold,silver,bronze\} \land D \not\in \{gold,silver,bronze\})$&lt;/li&gt;&#xA;&lt;li&gt;$f4 = (A \in \{gold,silver,bronze\} \land C \not\in \{gold,silver,bronze\})$&lt;/li&gt;&#xA;&lt;li&gt;$f5 = (D \in \{gold,silver,bronze\} \land E \not\in \{gold,silver,bronze\})$&lt;/li&gt;&#xA;&lt;li&gt;$fz = (\lnot (f1 \lor f2 \lor f3 \lor f4 \lor f5))$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There is a satisfying model where:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$A = gold$ &lt;/li&gt;&#xA;&lt;li&gt;$D = silver$&lt;/li&gt;&#xA;&lt;li&gt;$C = bronze$&lt;/li&gt;&#xA;&lt;li&gt;$B = nothing1$&lt;/li&gt;&#xA;&lt;li&gt;$E = nothing2$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You can easily encode this in SMT-LIB2 and execute &lt;a href=&quot;http://rise4fun.com/z3&quot;&gt;here&lt;/a&gt; using this code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(declare-datatypes()((M g s b n1 n2)))&#xA;(declare-const A M)&#xA;(declare-const B M)&#xA;(declare-const C M)&#xA;(declare-const D M)&#xA;(declare-const E M)&#xA;(assert(distinct A B C D E))&#xA;;A did not get gold and B did not get silver&#xA;(define-fun f1 () Bool (and (not (= A g)) (not (= B s))))&#xA;;D did not get silver and B did not get bronze&#xA;(define-fun f2 () Bool (and (not (= D s))(not (= B b))))&#xA;;C won a medal, but D didn't&#xA;(define-fun f3 () Bool (and (or (= C g)(= C s)(= C b))(or (= D n1)(= D n2))))&#xA;;A won a medal, but C didn't&#xA;(define-fun f4 () Bool (and (or (= A g)(= A s)(= A b))(or (= C n1)(= C n2))))&#xA;;D won a medal and E did, too&#xA;(define-fun f5 () Bool (and (or (= D g)(= D s)(= D b))(or (= E g)(= E s)(= E b))))&#xA;(assert(not (or f1 f2 f3 f4 f5)))&#xA;(check-sat)&#xA;(get-model)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-12-03T11:13:07.160" LastActivityDate="2014-12-03T11:13:07.160" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5845" PostTypeId="1" AcceptedAnswerId="5847" CreationDate="2014-12-03T11:28:14.643" Score="8" ViewCount="2644" Body="&lt;p&gt;I have an opaque 6-sided die. Without using mirrors or any other optical aids, what is the greatest number of faces of the die that I am able to see at the same time?&lt;/p&gt;&#xA;" OwnerUserId="2186" LastEditorUserId="2444" LastEditDate="2014-12-03T19:01:55.513" LastActivityDate="2014-12-04T06:20:11.157" Title="Looking at a die" Tags="|lateral-thinking|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5846" PostTypeId="2" ParentId="5626" CreationDate="2014-12-03T11:43:46.027" Score="-2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First of all the dragon can always survive by drinking from well 5 before the duel and flying to the 6th well and drinking from it after the duel. (this has been mentioned in other answers with examples so I'm not giving any more)&lt;br&gt;&#xA;&lt;br&gt;&#xA; The knight has a few options though:&lt;br&gt;&#xA;&lt;br&gt;&#xA; The knight can drink poisoned water from a lower numbered well before the duel. Should the dragon give him poisoned water from a higher number well, that poison will outtrade the one he drank and he will be cured. (should the dragon give him pond water however he will die).&lt;br&gt;&#xA;&lt;br&gt;&#xA; One thing I feel like no-one has touched however is a moral aspect of the question. If you had 1 hour after the duel and you didn't know what your opponent gave you you simply have to wait. Since all of the water looks exactly the same there is no telling whether or not your opponent gave you a deadly poison or harmless pond water. This is no problem for the dragon since he can drink poison from well 5 before the duel and he can ensure survival by drinking from 6 knowing that it will cancel out any poison. The knight doesn't have access to well 6 so the highest he can drink is from is 5.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Like I stated before, the knight has a moral aspect to battle. Should he decide to drink poison 1 before the duel and he believes the dragon gives him another poison (higher then 1) he wont drink for an hour (which makes a man second guess himself a lot). Does he drink poison 1 before the duel and he believes the dragon gave him pond water he drinks another poison after the fight. (this would mean suicide if the dragon actually gave him a poison).&lt;br&gt;&#xA;&lt;br&gt;&#xA; One answer that I didn't read in the other answers is this: If the knight doesn't drink before the duel. The dragon gives him a poison. Should this poison not kill the knight instantly he knows this is not poison 6. This leaves 6 other possible drinks (poison 1 - 5 and the pond water). The knight now needs to make a moral decision whether or not the dragon gave him pond water or one of the poisons. Only poison 5 and the pond water would mean death for the knight should he decide to drink poison 5. If he decides to not drink after the duel all of the poisons would kill the knight.&lt;br&gt;&#xA;&lt;br&gt;&#xA; There is no doubt the dragon has the upper hand in this duel, the 'safest' bet the knight can make is to drink poison 1 and hope the dragon gives him another poison higher than the one he drank. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6151" LastEditorUserId="44915" LastEditDate="2018-05-28T11:09:33.850" LastActivityDate="2018-05-28T11:09:33.850" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="5847" PostTypeId="2" ParentId="5845" CreationDate="2014-12-03T11:56:32.850" Score="28" Body="&lt;p&gt;Due to the fact that the die is smaller than the distance between the eyes, if you hold it die very close to your face, it's possible to see 5 faces at the same time (all faces except the one opposite to you):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/i6NEp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1585" LastEditorUserId="1585" LastEditDate="2014-12-03T12:02:09.647" LastActivityDate="2014-12-03T12:02:09.647" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="5848" PostTypeId="2" ParentId="5788" CreationDate="2014-12-03T12:24:47.337" Score="5" Body="&lt;p&gt;The answer is obviously&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; letters&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why is that? Well, because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is searching for letters exactly?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Searching for letters is searching for &quot;letters&quot;, exactly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What do we get when using letters literally?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When we use letters literally, well, we use &quot;letters&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What does wordplay finally result in?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wordplay usually results in letters. (Well technically words, but those are made out of, what? Right.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And what is a metaphorical series?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not only metaphorically but actually, letters are a series of symbols (or single letters, if you want to get even more meta)!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5265" LastActivityDate="2014-12-03T12:24:47.337" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5853" PostTypeId="1" AcceptedAnswerId="5854" CreationDate="2014-12-03T13:17:27.073" Score="4" ViewCount="695" Body="&lt;p&gt;Run and hide, the hunt has begun!&lt;br&gt;&#xA;Witches burn; no trial, no fun.&lt;br&gt;&#xA;Their hammer has struck a decisive blow,&lt;br&gt;&#xA;Letting little live and nothing grow.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their minds are made, their fates are sealed,&lt;br&gt;&#xA;Get away now, or your art be killed.&lt;br&gt;&#xA;Abandon ship! This vessel has sunk.&lt;br&gt;&#xA;They will be left with a pitiful chunk.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the ashes have settled, nothing left to attack.&lt;br&gt;&#xA;Did they make the right choice? We are not coming back!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who are they?&lt;/p&gt;&#xA;" OwnerUserId="5135" LastEditorUserId="12700" LastEditDate="2015-07-30T12:19:46.300" LastActivityDate="2015-07-30T12:21:49.783" Title="The Last Supper" Tags="|riddle|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5854" PostTypeId="2" ParentId="5853" CreationDate="2014-12-03T13:20:37.063" Score="4" Body="&lt;p&gt;They are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; puzzling.stackexchange.com users and moderators!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No need to explain that. But anyway:&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Run and hide, the hunt has begun!&lt;br&gt;&#xA;  Witches burn; no trial, no fun.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This topic: &lt;a href=&quot;https://puzzling.meta.stackexchange.com/questions/1601/puzzling-se-where-we-hate-fun-its-time-to-disallow-challenge-questions&quot;&gt;Is it time for us to disallow challenge-only questions?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Their hammer has struck a decisive blow,&lt;br&gt;&#xA;  Letting little live and nothing grow.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The party is over.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Their minds are made, their fates are sealed,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They won't change their mind. They &quot;democratically&quot; decided that, because everything is always democratic here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Get away now, or your art be killed.&lt;br&gt;&#xA;  Abandon ship! This vessel has sunk.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The last to go, please turn off the lights!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;They will be left with a pitiful chunk.&lt;br&gt;&#xA;  When the ashes have settled, nothing left to attack.&lt;br&gt;&#xA;  Did they make the right choice? We are not coming back!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What will be left in this site? Just an evacuated ghost town. Exactly what SE sites are supposed to not be.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5044" LastEditorUserId="-1" LastEditDate="2017-03-16T16:42:07.097" LastActivityDate="2015-07-30T12:21:49.783" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5855" PostTypeId="2" ParentId="5841" CreationDate="2014-12-03T13:52:00.940" Score="2" Body="&lt;p&gt;Since all sentences are false, from (1) either A get gold or B get silver. And from (2) either D get silver or B get bronze.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's suppose B got silver. Then B didn't get bronze, so from (2) D must get silver too, and we have a contradiction (B and D both get silver).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, A must get gold.&#xA;For (4) to be false, C must also get a medal.&#xA;And for (3) to be false, D must also get a medal.&#xA;Since we have our 3 medalists, B and E must not get a medal.&#xA;Then, from (2) we have that D get silver. And C is left with bronze.&lt;/p&gt;&#xA;" OwnerUserId="6156" LastActivityDate="2014-12-03T13:52:00.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5856" PostTypeId="2" ParentId="5389" CreationDate="2014-12-03T14:16:08.373" Score="6" Body="&lt;p&gt;Continuing from Joes answer...basically all credit goes to him, since he only missed this part:&lt;/p&gt;&#xA;&lt;p&gt;The unsolved part from Joes answer:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 78 19 25 9 92 29 54 1 95 49 6 247 4 2 149 45 5 7 199 11 62 63 179 47 14 50 23 163 52&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I did this:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; I googled Jumblies and found a poem with the same name. &lt;a href=&quot;http://www.poetryfoundation.org/poem/241166#poem&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Jumblies&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Taking the corresponding words from that poem you will get the following:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &amp;quot;go of on did the in.....&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;What they all have in common is that they are all:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Preposition!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&lt;p&gt;This could be wrong I created a script to double check: &lt;a href=&quot;http://jsfiddle.net/fz29ruoy/1/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://jsfiddle.net/fz29ruoy/1/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;and it gives me the following words:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; go of On did, Jumblies on be They heads every a they sea went said, round in Sieve, Their a big, But In round, went one could the ‘You’ll&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;And taking the first letter of each word:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; goOd Job The atswsr  iS  Tabbirwocty&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;And I am guessing there are some typos or I have mistakes in my code:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Good Job the answer is:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Jabberwocky!!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5030" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-12-03T16:23:31.777" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5857" PostTypeId="1" AcceptedAnswerId="6024" CreationDate="2014-12-03T14:40:16.000" Score="3" ViewCount="2033" Body="&lt;p&gt;One day the local police from the Stack Exchange community received a concerned call from one of the bakeshop owners nearby.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Police:&lt;/strong&gt; Good Morning, this is S.E. Police. How may I help you? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bakeshop Employee:&lt;/strong&gt; Hello sir, I'm an employee in the bakeshop near the police department. I am worried about the house in front of our shop. The owner of this house is one of our best customers, and he asked me to do him a favor. He wanted me to prepare 20 pieces of our specialty bread to be picked up early in the morning. He told me that he would like for his guests to taste our bread. However, it's already the second day since he asked me and I haven't seen him. I am very  worried because he always picks up bread from us early in the morning, and he has promised to pick up his order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Police&lt;/strong&gt;: okay sir, we will send some officers to the house you are mentioning.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the police visited the house, they tried to knock, but found that the door was open. They proceeded inside the house.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They entered the house, and tried to find the owner of the house. They found his dead body, hanging from the high ceiling of a well lit room. There were two glass windows and one of them was broken. It seemed as if something was thrown to break the window. The police that responded to the house immediately called the forensics department and started the investigation. The room was empty; there was no furniture, garbage, or any large items inside the room, only shards of broken glass and the dead body of the owner inside, the police also checked out side the the house where the broken window is and found an empty bottle of beer. broken, might be from the fall. there where also no furnitures that looked liked misplaced inside or outside of the house, the house was pretty clean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The police checked the last contacts of the owner and found out that there were two people that the owner invited for a drink in his home, one girl and one guy. They were brougt in for interrogation, and the police asked them questions regarding the scene and what happened when they visited.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The girl admitted that they three of them had some drinks, and she also admitted that while she was drunk she happened to have a conversational fight with the owner of the house.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Girl&lt;/strong&gt;: I said to him, &quot;No! We can't be together again.&quot; I don't love him anymore, and I had brought my new boyfriend for him to meet and have a drink. I liked him and wanted to still be friends with him, but he still insisted he loved me and I couldn't handle it. I needed it to stop!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Boy&lt;/strong&gt;: I was there to escort my girlfriend. Although I didn't like what he was doing to my girlfriend, I still tried to understand his feelings. But... it was too much. I had to punch him in the face to protect my girlfriend from him. He is a nuisance and he deserved to die. After the confrontation, me and my girlfriend left the house peacefully and never heard from him again.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The police are very puzzled about this bizarre crime and do not know the answer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you can help the investigators.. &lt;strong&gt;WHO IS THE KILLER?&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus Question: &lt;strong&gt;How was the owner Killed?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="6145" LastEditorUserId="6145" LastEditDate="2014-12-03T15:59:04.860" LastActivityDate="2021-03-02T16:13:48.343" Title="Who is the killer now?" Tags="|mystery|" AnswerCount="5" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="5858" PostTypeId="2" ParentId="5857" CreationDate="2014-12-03T15:09:59.010" Score="2" Body="&lt;p&gt;A wild guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He hanged himself. The room being completely empty means that the window was broken from the inside. When he was about to jump from a chair/ladder/nightstand he kicked it so that it will fly through the window. That is why there was no furniture.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastActivityDate="2014-12-03T15:09:59.010" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5859" PostTypeId="2" ParentId="5845" CreationDate="2014-12-03T15:11:19.057" Score="21" Body="&lt;p&gt;You didn't specify the material the die can be made of. Dice made of translucent plastic are common, and you can easily see all six sides. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tIaUZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="6158" LastActivityDate="2014-12-03T15:11:19.057" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5861" PostTypeId="2" ParentId="5853" CreationDate="2014-12-03T15:40:48.277" Score="0" Body="&lt;p&gt;I think the &lt;strong&gt;solution&lt;/strong&gt; clearly is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://area51.stackexchange.com/proposals/80752/puzzle-and-riddle-challenges&quot;&gt;found here&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastActivityDate="2014-12-03T15:40:48.277" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5862" PostTypeId="2" ParentId="5788" CreationDate="2014-12-03T15:48:31.087" Score="9" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; WORD&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;as per the 'word' tag and also the headline.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lynch mob, anyone?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What is searching for letters exactly?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'For letters exactly' -&gt; 'four letters exactly'; the word 'four' does indeed have four letters exactly (in fact it's the only number whose English-language word has that number of &lt;em&gt;different&lt;/em&gt; letters), so the answer to this line is FOUR.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What do we get when using letters literally?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well, it does say 'literally'. The answer is simply LETTERS.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What does wordplay finally result in?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The word 'finally' means we should look at the end, so take the word IN.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And what is a metaphorical series?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sod the metaphor, just take the word SERIES (possibly?). Or a &quot;metaphorical series&quot; is just a SERIES (of some kind... metaphorically).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So we have&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; FOUR LETTERS IN SERIES.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Is there also a word for the combination of the things mentioned above?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; WORD is a four-letter word. It's a word; it fits the combination of the above clues; the combination is a word and is word and word ... argh, this is getting too meta for me!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2014-12-03T20:41:20.880" LastActivityDate="2014-12-03T20:41:20.880" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5863" PostTypeId="2" ParentId="5857" CreationDate="2014-12-03T15:53:40.383" Score="2" Body="&lt;p&gt;I think&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; He hanged himself by standing on ice slabs. Ice slabs melted and evaporated.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;@Aura&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; To confuse people that it is indeed not a suicide. That is the reason why he moved all the furniture so that melted water doesn't deposit anywhere. Also he tried to show as if the beer bottle was thrown in rage after heated arguments to frame the other boy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4637" LastEditorUserId="72282" LastEditDate="2021-03-02T16:13:48.343" LastActivityDate="2021-03-02T16:13:48.343" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="5865" PostTypeId="1" AcceptedAnswerId="5968" CreationDate="2014-12-03T15:58:39.193" Score="4" ViewCount="1976" Body="&lt;pre&gt;&lt;code&gt;I go by many a name,&#xA;Including one that's quite a shame.&#xA;&#xA;I can take you to an exact location,&#xA;But sometimes only to a nearby destination.&#xA;&#xA;If great power is your wish,&#xA;I'll grant it but your money will vanish.&#xA;&#xA;Substituting for that which can be negative,&#xA;I'll always remain completely positive.&#xA;&#xA;I can be very important,&#xA;While at the same time very insignificant.&#xA;&#xA;When I contain many things,&#xA;I simultaneously contain nothing.&#xA;&#xA;I can signal the beginning,&#xA;But I can also signal the ending.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Who or what is this poem referring to? Can you explain each line?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please put your answers in the spoiler markup!&lt;/p&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4450" LastEditDate="2014-12-04T19:24:54.737" LastActivityDate="2014-12-05T16:12:11.103" Title="Riddle Poem: &quot;I go by many a name...&quot;" Tags="|riddle|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5866" PostTypeId="2" ParentId="5865" CreationDate="2014-12-03T16:06:34.347" Score="6" Body="&lt;p&gt;Is it a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Smartphone&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;?&#xA;I am not sure, but it makes sense to most clues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I go by many a name,&#xA;Including one that's quite a shame.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Smartphone, perhaps it's a &quot;shameful&quot; name because it's not really smart.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can take you to an exact location,&#xA;But sometimes only to a nearby destination.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The gps that is on a smartphone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If great power is your wish,&#xA;I'll grant it but your money will vanish.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It gives you the possibility to do a lot but it costs a lot.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Substituting for that which can be negative,&#xA;I'll always remain completely positive.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The negative side is that people are not social anymore because they are constantly watching their smartphone. But in the end, it's still a great invention.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be very important,&#xA;While at the same time very insignificant.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Some people just need it a lot for reading mails etc. However, it's still luxury. You can actually survive without one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I contain many things,&#xA;I simultaneously contain nothing.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Lot of apps and lot of other things are on a smartphone, however it's still just data, it doesn't take actual space. This one was the hardest for me...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can signal the beginning,&#xA;But I can also signal the ending.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Signal --&gt; reception --&gt; (smart)phone&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, that's it :).&lt;/p&gt;&#xA;" OwnerUserId="420" LastEditorUserId="420" LastEditDate="2014-12-03T16:58:17.467" LastActivityDate="2014-12-03T16:58:17.467" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5867" PostTypeId="2" ParentId="5857" CreationDate="2014-12-03T16:28:51.473" Score="4" Body="&lt;p&gt;I suspect the &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;bakeshop employee.&lt;/p&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;h3&gt;Reasons:&lt;/h3&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;li&gt;He is &amp;ldquo;very worried&amp;rdquo; yet he didn't check when the customer didn't pick up the bread yesterday (even though the door isn't locked).&lt;/li&gt;&lt;li&gt;There isn't any furniture in the room, possibly because it has been stolen.&lt;/li&gt;&lt;li&gt;His guests for which he ordered the bread are probably the boy and the girl. But, they were invited for a drink, not for breakfast. Did he expect them to stay over? That's very unlikely.&lt;/li&gt;&lt;li&gt;Which leads us to the question: why didn't his guests for who he ordered the bread notice he was dead?&lt;/li&gt;&lt;li&gt;Which leads me to call the truthfulness of the baker's story into question.&lt;/li&gt;&lt;li&gt;It wasn't suicide; the ceiling is high, and if a ladder was used by the customer it should have been carried away by someone.&lt;/li&gt;&lt;li&gt;The broken bottle of beer may have been dropped by one of his guests (the boy or the girl) when they left.&lt;/li&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2541" LastEditorUserId="2313" LastEditDate="2014-12-03T16:41:30.240" LastActivityDate="2014-12-03T16:41:30.240" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5868" PostTypeId="1" AcceptedAnswerId="5875" CreationDate="2014-12-03T16:30:15.130" Score="10" ViewCount="580" Body="&lt;p&gt;&lt;strong&gt;This is a riddle worded as a historical excerpt for the sake of the decipherer's entertainment.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Excerpt from the journal of Aulete of Hal-Saflieni, chronicler of the gods:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One curious case is that of the goddess of the high and the deep. After the culture's patriarch god created the universe, it seems that the new universe in turn created this diety itself. Angered that his creations made their own god, the patriarch refuses to allow her into the heavens. Unable to ascend, she is said to reside in the worship houses of sky, earth, and sea. The goddess of all cathedrals, all temples, and all mosques, songs to every diety are sung to her glory first. Her magical cloak makes her invisible, hidden from the creator, but practitioners insist that even the most faithless believe in her.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intense study will follow, but with caution. The elders say that those who worship too intensely can diminish their connection to the goddess permanently. In some extreme cases, severe pain and disorientation have occurred.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aulete died without ever elaborating on the subject. Archeologists haven't found anything either. Can you figure out the name of the goddess for the sake of preserving history and prove conclusively to the archeological community that you're correct?&lt;/p&gt;&#xA;" OwnerUserId="5626" LastEditorUserId="5626" LastEditDate="2014-12-10T13:14:36.837" LastActivityDate="2014-12-10T13:14:36.837" Title="The Theographer's Journal - Discover an ancient secret and finish the work of a chronicler of the gods" Tags="|riddle|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5869" PostTypeId="2" ParentId="5799" CreationDate="2014-12-03T16:30:58.010" Score="6" Body="&lt;p&gt;NEW ANSWER: 10 Turns (I used Kevin's interactive tool for this one)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;(2-)&lt;br&gt;&#xA;(2-)&lt;br&gt;&#xA;(3+)&lt;br&gt;&#xA;(4+)&lt;br&gt;&#xA;(4+)&lt;br&gt;&#xA;(C+), [A1 -&gt; C5]&lt;br&gt;&#xA;(4-)&lt;br&gt;&#xA;(3-)&lt;br&gt;&#xA;(2+), [C5 -&gt; E1]&lt;br&gt;&#xA;(3+) [E1 -&gt; E5]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;OLD ANSWER: 13 turns&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to get the ball rolling my moves are as follows (I created an interactive version in Excel, so there is plenty of margin for error.):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (B+)&lt;br&gt;&#xA; (C+)&lt;br&gt;&#xA; (D+)&lt;br&gt;&#xA; (E+)&lt;br&gt;&#xA; (E+) [A1-&gt;E1]&lt;br&gt;&#xA; (2-)&lt;br&gt;&#xA; (3+)&lt;br&gt;&#xA; (4+)&lt;br&gt;&#xA; (5-) [E1-&gt;E5]&lt;br&gt;&#xA; (4-)&lt;br&gt;&#xA; (B+)&lt;br&gt;&#xA; (B+)&lt;br&gt;&#xA; (A+) [E5-&gt;A5]  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3922" LastEditorUserId="3922" LastEditDate="2014-12-03T19:20:35.617" LastActivityDate="2014-12-03T19:20:35.617" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5870" PostTypeId="2" ParentId="5868" CreationDate="2014-12-03T16:46:25.673" Score="4" Body="&lt;p&gt;The goddess is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Air&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some explanations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;the patriarch refuses to allow her into the heavens. Unable to ascend, she is said to reside in the worship houses of sky, earth, and sea.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gravity holds the air below a certain altitude, so it is held close to the earth, sky, and sea.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;songs to every diety are sung to her glory first.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Before you can sing, you have to take in a breath of air.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;The elders say that those who worship too intensely can diminish their connection to the goddess permanently. In some extreme cases, severe pain and disorientation have occurred.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can become dizzy, pass out, or even die, if you hold air in your lungs for too long.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4518" LastActivityDate="2014-12-03T16:46:25.673" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5871" PostTypeId="2" ParentId="5672" CreationDate="2014-12-03T17:18:00.510" Score="5" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a &lt;strong&gt;Poor Boy Sandwich&lt;/strong&gt;&lt;br&gt;&lt;br&gt;The irony of the riddle is that the clues describe the literal interpretation of the sandwich name&amp;mdash;the condition of poverty&amp;mdash;rather than the sandwich itself. The word &quot;sandwich&quot; also displays some polysemy, which is discussed later in this response.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The epitome of an evil the skewed letter would label&lt;br&gt;&#xA;To many is such truth, not a fable&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &quot;skewed letter&quot; is the dollar sign (&lt;em&gt;\$&lt;/em&gt;), with &quot;skewed&quot; jointly referring to the fact that &lt;em&gt;i&lt;/em&gt;) the character is &quot;skewered&quot; (skew'ed, in poetic parlance) by a long downstrike, &lt;em&gt;ii&lt;/em&gt;) the character is a modified (skewed, in the general sense) version of an 'S', and &lt;em&gt;iii&lt;/em&gt;) the character represents a concept (money) that deeply skews people's attitudes and perceptions.&lt;br&gt;&lt;br&gt;The &quot;epitome of an evil&quot; labeled by &lt;em&gt;\$&lt;/em&gt; is obviously &lt;em&gt;money&lt;/em&gt;, which some would argue is the root of all evil. Poverty, and specifically a life where earning money to survive is the crux of one's existence, is a fact of life for countless people (&quot;To many is such truth, not a fable&quot;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Held tightly by urges of the seconds&lt;br&gt;&#xA;Within the sleepless cities that beckons&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Money needs to be earned. &quot;Time is money,&quot; as the expression goes. Whether impoverished or not, a man captivated by the need to earn money is &quot;held tightly by urges of the seconds&quot;. Will there be enough work hours to survive? Will the bull market hold or lose support?&lt;br&gt;&lt;br&gt;&quot;Within the sleepless cities that beckons&quot; are the sweatshops that operate at all hours, the impoverished storekeeper stocking new wares by midnight due to lack of time during the day, the indebted university student studying until 4:00 in the morning in the hopes of graduating and landing a well-paying job. The sleepless cities beckon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Twice I hold the path of tea&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This statement lends itself to the following dual interpretations:&lt;br&gt;&lt;br&gt;1) The &quot;path of tea&quot; is to be &lt;em&gt;sieved&lt;/em&gt;: to be held back as one's essence is extracted. Regrettably, many of the social constructs of man's societies act as a sieve for the impoverished. Education and opportunities for advancement are often inaccessible. Barriers such as lack of resources, lack of connections, and even socially-entrenched caste systems ensure the poor stay poor even as their time, labour and &quot;essence&quot; carry through the sieve to the benefit of society, like the flavouring of tea leaves.&lt;br&gt;&lt;br&gt;2) The &quot;path of tea&quot; is to be unwashed. Unlike fruits and vegetables, tea leaves are virtually never washed after being harvested. Metaphorically speaking, poverty is also rarely washed off. As the labours of the poor are exploited (harvested), none of the grit and grime of their condition is washed away by the process, nor can be. A man whose poverty is washed away becomes comfortable, and a comfortable man may not be driven to harsh labour in the pursuit of money. Such a man must therefore be subject to the &quot;path of tea&quot; when brought to market: harvested quickly and sold without cleansing for the sake of his labour.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Forever you stay within me&lt;br&gt;&#xA;It is not glue that binds us&lt;br&gt;&#xA;Nor the substance which I fuss&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Poverty begets poverty, as implied by the so-called &quot;cycle of poverty&quot;. The poor are not simply bound by physical circumstances (&quot;It is not glue that binds us&quot;) or by a lack of money (the &quot;substance&quot; over which we &quot;fuss&quot;), but also by a kind of enmity with society: a lack of education, a cultural lack of aspiration to higher things, the allure of crime, distrust of authority, and hostile cultural attitudes towards &quot;leaving one's roots&quot;. In short, a poor boy is not simply a poor boy (one forever remaining in a condition of poverty) because he lacks money. A great many factors besides bind him to this condition.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Guarded by a significant worm&lt;br&gt;&#xA;Your sight has gone, time to squirm&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A &quot;significant worm&quot; is &lt;em&gt;death&lt;/em&gt;, or more specifically &lt;em&gt;the inevitability of death&lt;/em&gt;, by which all of our lives are guarded.&lt;br&gt;&lt;br&gt;When a man loses his sight, he becomes unfit to work. Moreover, blindness is often the result of old age, that faithful precursor to death. Regardless of the cause, a poor man whose sight has failed him finds himself imperiled (&quot;time to squirm&quot;), facing the very real prospect of death in spite of his previous labours. This grim observation establishes the context for the next two lines.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I laugh at your extravagance&lt;br&gt;&#xA;I assure you my acquaintance&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As &quot;I&quot; am Poverty, Death is my more extravagant (more abundant, more ostentatious) acquaintance, afflicting not only the poor but indeed &lt;em&gt;all&lt;/em&gt; men, cutting down their pride and their power in spectacular fashion. What laughable irony that the rich and the poor, so entrenched in their disparate circles in life, should reach a state of perfect equality in death, where all the wealth they've striven for and the possessions they've accumulated are accounted as nothing.&lt;br&gt;&lt;br&gt;Perhaps this is also the ancillary meaning of &quot;sandwich&quot;&amp;mdash;that all of us, whether rich or poor, are sandwiched between birth and death, with our lives as the substance that connects these extrema like the two halves of sandwich. Is ours a Poor Boy sandwich, defined by sleepless nights, the skewed letter, and the path of tea, or is it something different? What circumstances will connect the extrema of our own lives?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2014-12-03T17:18:00.510" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5872" PostTypeId="2" ParentId="5857" CreationDate="2014-12-03T18:00:34.273" Score="3" Body="&lt;p&gt;It's clear that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;He committed suicide from a broken heart. Since there was no furniture and he was the bread shop's best customer we must conclude that he stood... on many loaves of bread! He threw the beer bottle at the window shortly before kicking away the crusty bread supporting his weight so that birds could get into the room and eat the bread (which worked exactly as he planned). As a bread connoisseur he didn't want any of it to go to waste, making it his parting gift to the world. &#xA;&lt;p&gt;&#xA;The 20 pieces of bread he ordered is clearly too much bread for just 2 guests to sample. He must have ordered the bread AFTER the girl rejected him (since we are never told how long ago the guests visited him) as his last meal - pig out on delicious specialty multi-grain before pulling the plug. But after being up all night crying (nobody would be able to sleep with a planned suicide coming up) he was a total wreck and realized he can't show up at the store like this and doesn't really want any food anyway. He downed a beer and figured he might as well just get it over with.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6163" LastActivityDate="2014-12-03T18:00:34.273" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5873" PostTypeId="2" ParentId="5845" CreationDate="2014-12-03T18:48:48.550" Score="4" Body="&lt;p&gt;If your goal is to be able to see the most number of sides of a normal opaque die without any sort of aid, whatsoever, besides your own hands and eyes, then i'd say w0lf's answer is the best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However if your goal is to see the most amount of sides of a normal opaque die without the aid of optical tricks but by utilising other tricks, then i'd suggest simply cutting off the faces of all six sides, thinly, line them up and you will be able to see them all, with an added bonus - the insides of the die! :O&lt;/p&gt;&#xA;" OwnerUserId="6164" LastEditorUserId="6164" LastEditDate="2014-12-04T00:10:49.607" LastActivityDate="2014-12-04T00:10:49.607" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5874" PostTypeId="2" ParentId="5763" CreationDate="2014-12-03T18:50:27.773" Score="6" Body="&lt;p&gt;As the answer is lengthy, I start with a table of contents:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part 1: Explanation of my numbering system&lt;br&gt;&#xA;Part 2: Numbering shown around the table of letters&lt;br&gt;&#xA;Part 3: Program &lt;code&gt;8wayshow.py&lt;/code&gt; that displays numbered table and displays all text strings reoriented left to right&lt;/strong&gt;&lt;br&gt;&#xA;Parts 1 through 3 are shown in clear, with no spoilers so no need for hidden text.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part 4: Plain-strings from the letter table as output by &lt;code&gt;8wayshow.py&lt;/code&gt;&lt;/strong&gt;&lt;br&gt;&#xA;Part 4 is in hidden text (rollover-revealed text) although it does not identify any words, merely shows text strings from the eight  orientations reoriented for left-to-right reading.  It is the reference table for numbered word references in part 5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part 5: Five triples (two new, three old) &amp;amp; some herrings&lt;/strong&gt; (in hidden text)&lt;br&gt;&#xA;&lt;strong&gt;Part 6: A bunch of words found&lt;/strong&gt; (in hidden text)&lt;br&gt;&#xA;Part 6 also includes half-a-dozen things that look to me like they might be wordlike in some language I don't know.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part 1: Explanation of numbering system&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In following, I numbered rows and columns from 0 to 16. &#xA;In word references below, letters a...h represent eight orientations: → a, ← b, ↓ c, ↑ d, ↗ e, ↙ f, ↖ g, ↘ h.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diagonals are numbered such that if row or column numbering continues forward and back and around corners as shown below, each diagonal with six or more letters can be identified by a number from 6 to 28.  D6-D17 are of lengths 6...17.  Some numbers align with axes of diagonals, others with first characters, others only obscurely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, 6a &lt;em&gt;aabotr&lt;/em&gt; is a right-ascending (↗) diagonal in line with two 6's and 16b &lt;em&gt;leengpenvpaozzlb&lt;/em&gt; is a left-descending (↙) diagonal in line with two 16's.  The right-descending (↘) diagonal 11h &lt;em&gt;aoahsalrkio&lt;/em&gt; looks like it's in line with a top 5 and right 28; in a wrapped view, it also is in line with two 11's.  To be less obscure (but rather more bulky) the table could be printed with four sets of label lines on each side.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part 2: Numbering shown around the table of letters&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;       33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17&#xA;        0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16&#xA;33  0   p  y  b  e  g  r  a  f  e  n  i  s  x  j  r  l  p   17 16&#xA;32  1   i  s  v  a  t  b  a  o  e  v  a  n  o  r  e  a  o   18 15&#xA;31  2   y  a  b  o  a  r  d  g  a  m  e  n  r  e  k  m  m   19 14&#xA;30  3   y  n  b  d  l  p  z  r  v  h  k  m  n  n  q  r  m   20 13&#xA;29  4   p  a  o  b  l  a  g  i  a  a  s  g  n  i  c  h  i   21 12&#xA;28  5   a  b  m  s  y  n  t  y  m  a  p  a  i  v  a  t  e   22 11&#xA;27  6   s  b  r  e  t  t  s  p  i  e  l  o  l  u  f  e  r   23 10&#xA;26  7   p  o  r  m  i  t  s  k  n  s  o  e  t  r  m  r  d   24  9&#xA;25  8   o  r  h  g  r  i  a  v  e  o  s  a  z  a  k  n  o   25  8&#xA;24  9   o  l  h  c  m  v  p  u  g  n  j  p  k  j  a  i  b   26  7&#xA;23 10   r  e  m  z  h  a  e  h  a  a  o  p  o  a  i  n  o   27  6&#xA;22 11   t  t  a  c  o  n  s  c  i  e  n  c  e  h  r  g  k   28  5&#xA;21 12   s  a  n  z  s  k  i  s  r  l  g  n  g  d  c  e  o   29  4&#xA;20 13   j  d  z  i  t  i  e  x  r  p  h  s  y  y  r  r  c   30  3&#xA;19 14   h  l  a  t  a  t  r  r  a  m  i  r  g  s  a  s  k   31  2&#xA;18 15   b  o  n  f  g  s  d  g  m  u  a  b  l  e  f  p  a   32  1&#xA;17 16   i  p  o  g  e  f  x  s  o  c  g  e  w  e  t  e  n   33  0&#xA;       17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33&#xA;       16 15 14 13 12 11 10  9  8  7  6  5  4  3  2  1  0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part 3: Python program &lt;code&gt;8wayshow.py&lt;/code&gt; that displays numbered table and displays all text strings reoriented left to right&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#!/usr/bin/env python&#xA;# Re http://puzzling.stackexchange.com/questions/5763/international-wordsearch&#xA;t = ['pybegrafenisxjrlp', 'isvatbaoevanoreao', 'yaboardgamenrekmm',&#xA;     'ynbdlpzrvhkmnnqrm', 'paoblagiaasgnichi', 'abmsyntymapaivate',&#xA;     'sbrettspielolufer', 'pormitsknsoetrmrd', 'orhgriaveosazakno',&#xA;     'olhcmvpugnjpkjaib', 'remzhaehaaopoaino', 'ttaconsciencehrgk',&#xA;     'sanzskisrlgngdceo', 'jdzitiexrphsyyrrc', 'hlatatrramirgsask',&#xA;     'bonfgsdgmuablefpa', 'ipogefxsocgeweten']&#xA;tr = [''.join(reversed(s)) for s in t]&#xA;&#xA;# Create u by transposing t&#xA;n = len(t);    d = 2*n-1        # n = #rows,  d = #diags&#xA;u = [''.join(t[i][j] for i in range(n)) for j in range(n)]&#xA;ur = [''.join(reversed(s)) for s in u]&#xA;&#xA;# Rotate t +/- 45 degrees, returning result r&#xA;def rotor(rdir):&#xA;    r = []&#xA;    for i in range(5,n):&#xA;        x = [' ']*n&#xA;        for j in range(i+1):&#xA;            x[j] = t[i-j][j] if rdir else t[i-j][n-1-j]&#xA;        r.append(''.join(x))&#xA;&#xA;    for i in range(n-6):&#xA;        x = [' ']*n&#xA;        for j in range(i+1,n):&#xA;            x[j-i-1] = t[n+i-j][j] if rdir else t[n+i-j][n-1-j]&#xA;        r.append(''.join(x))&#xA;    return r&#xA;&#xA;v = rotor(1);   vr = [''.join(reversed(s)) for s in v]&#xA;w = rotor(0);   wr = [''.join(reversed(s)) for s in w]&#xA;&#xA;def printhead(first,last,step):&#xA;    print '      ',&#xA;    for i in range(first,last,step):&#xA;        print '{:2}'.format(i),&#xA;    print&#xA;&#xA;printhead(d,n-1,-1)&#xA;printhead(0,n,1)&#xA;for i in range(n):&#xA;    print '{:2} {:2}  '.format(d-i, i),&#xA;    for j in range(n):&#xA;        print '{:2}'.format(t[i][j]),&#xA;    print ' {:2} {:2}'.format(i+n, n-i-1)&#xA;printhead(n,d+1,1)&#xA;printhead(n-1,-1,-1)&#xA;print&#xA;&#xA;for i in range(n):&#xA;    print '{:2} {} {} {} {}'.format(i, t[i], tr[i], u[i], ur[i])&#xA;&#xA;for i in range(len(v)):&#xA;    print '{:2} {} {} {} {}'.format(i+6, v[i], vr[i], w[i], wr[i])&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part 4: Plain-strings from the letter table as output by &lt;code&gt;8wayshow.py&lt;/code&gt;&lt;/strong&gt;&lt;br&gt;&#xA;Part 4 is in hidden text (rollover-revealed text) although it does not identify any words, merely shows text strings from the eight  orientations reoriented for left-to-right reading.  It is the reference table for numbered word references in part 5. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  0 pybegrafenisxjrlp plrjxsinefargebyp piyypaspoortsjhbi ibhjstroopsapyyip &lt;br&gt;&#xA;  1 isvatbaoevanoreao oaeronaveoabtavsi ysanabborletadlop poldatelrobbanasy &lt;br&gt;&#xA;  2 yaboardgamenrekmm mmkernemagdraobay bvbbomrrhhmanzano onaznamhhrrmobbvb &lt;br&gt;&#xA;  3 ynbdlpzrvhkmnnqrm mrqnnmkhvrzpldbny eaodbsemgczczitfg gftizczcgmesbdoae &lt;br&gt;&#xA;  4 paoblagiaasgnichi ihcingsaaigalboap gtallytirmhostage egatsohmrityllatg &lt;br&gt;&#xA;  5 abmsyntymapaivate etaviapamytnysmba rbrpanttivankitsf fstiknavittnaprbr &lt;br&gt;&#xA;  6 sbrettspielolufer refuloleipstterbs aadzgtssapesierdx xdreisepasstgzdaa &lt;br&gt;&#xA;  7 pormitsknsoetrmrd drmrteosnkstimrop fogriypkvuhcsxrgs sgrxschuvkpyirgof &lt;br&gt;&#xA;  8 orhgriaveosazakno onkazasoevairghro eeavaminegairramo omarriagenimavaee &lt;br&gt;&#xA;  9 olhcmvpugnjpkjaib biajkpjngupvmchlo nvmhaaesonaelpmuc cumpleanoseaahmvn &lt;br&gt;&#xA; 10 remzhaehaaopoaino oniaopoaaheahzmer iaeksplosjonghiag gaihgnojsolpskeai &lt;br&gt;&#xA; 11 ttaconsciencehrgk kgrhecneicsnocatt snnmgaoeappcnsrbe ebrsncppaeoagmnns &lt;br&gt;&#xA; 12 sanzskisrlgngdceo oecdgnglrsiksznas xornniltzkoegyglw wlgygeokztlinnrox &lt;br&gt;&#xA; 13 jdzitiexrphsyyrrc crryyshprxeitizdj jrenivurajahdysee eesydhajaruvinerj &lt;br&gt;&#xA; 14 hlatatrramirgsask ksasgrimarrtatalh rekqcafmkaircraft tfarcriakmfacqker &lt;br&gt;&#xA; 15 bonfgsdgmuablefpa apfelbaumgdsgfnob lamrhterningerspe epsregninrethrmal &lt;br&gt;&#xA; 16 ipogefxsocgeweten netewegcosxfegopi pommierdobokockan nakcokobodreimmop &lt;br&gt;&#xA;  &lt;br&gt;&#xA;  6 aabotr                       rtobaa ehqeos                       soeqhe &lt;br&gt;&#xA;  7 sbodaba                     abadobs rtcnrni                     inrnctr &lt;br&gt;&#xA;  8 pbmblraf                   farlbmbp deainnan                   nanniaed &lt;br&gt;&#xA;  9 oorslpdoe                 eodplsroo orfvnmeve                 evemnvfro &lt;br&gt;&#xA; 10 orreyazgen               negzayerro bnmuigkmef               femkgiumnb &lt;br&gt;&#xA; 11 rlhmtngravi             ivargntmhlr oikrlashaoa             aoahsalrkio &lt;br&gt;&#xA; 12 tehgittivmas           samvittighet knaatopavgar           ragvapotaank &lt;br&gt;&#xA; 13 stmcrtsyahenx         xnehaystrcmts ogijzelaardbg         gbdraalezjigo &lt;br&gt;&#xA; 14 jaazmispmaknoj       jonkampsimzaaj cerakaoemizrte       etrzimeoakarec &lt;br&gt;&#xA; 15 hdnchvakiasmrrr     rrrmsaikavhcndh krchopssiygpaab     baapgyisspohcrk &lt;br&gt;&#xA; 16 blzzoapvnepgneel   leengpenvpaozzlb asrdepjonptalovy   yvolatpnojpedrsa &lt;br&gt;&#xA; 17 ioaisneueslannkap paknnalseuensiaoi npaygconeksnldbsp psbdlnskenocgyapn &lt;br&gt;&#xA; 18 pnttkshgoooiiqmo   omqiioooghskttnp efsynnagvstybbai   iabbytsvgannysfe &lt;br&gt;&#xA; 19 ofaiicanselvcrm     mrcvlesnaciiafo tegsgeauattsony     ynosttauaegsget &lt;br&gt;&#xA; 20 ggtesiajatuahm       mhautajaisetgg elrhlihpiiemay       yameiiphilhrle &lt;br&gt;&#xA; 21 esrxreopzrfti         itfrzpoerxrse wbiprcevrmrbp         pbrmrvecrpibw &lt;br&gt;&#xA; 22 fdrrlnpkamee           eemakpnlrrdf eamrssamgrba           abrgmassrmae &lt;br&gt;&#xA; 23 xgapgcojkrr             rrkjocgpagx guaxinhchos             sohchnixaug &lt;br&gt;&#xA; 24 smmhneaand               dnaaenhmms cmrekozhrp               prhzokermc &lt;br&gt;&#xA; 25 ouisghiio                 oiihgsiuo ogriscmlo                 olmcsirgo &lt;br&gt;&#xA; 26 carydrnb                   bnrdyrac sdttzaeo                   oeazttds &lt;br&gt;&#xA; 27 gbgycgo                     ogcygbg xsaintr                     rtniasx &lt;br&gt;&#xA; 28 elsrek                       kersle fgtzat                       taztgf &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part 5: Five triples &amp;amp; some herrings&lt;/strong&gt; (in hidden text)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two new triples:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 04c hostage  (English)  5c,13g &lt;br&gt;&#xA; 05c panttivanki  (Finnish)   hostage 4c,13g &lt;br&gt;&#xA; 13g gijzelaar (Dutch),  hostage 4c,5c &lt;br&gt; &lt;br&gt;&#xA; 01d robbanás     (Hungarian), explosion  10c &lt;br&gt;&#xA; 10c eksplosjon   (Norwegian Bokmål),     an explosion  1d &lt;br&gt;&#xA; 13c rajahdys   (Finnish), explosion 10c,1d&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Three old triples (With credits to @w0lf; although I'd found the rest on my own, I'd missed a Czech word and a Norwegian word):  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 01b aeronave   (Italian) dirigible; (Spanish/Portuguese) aircraft&lt;br&gt;&#xA; 01c letadlo      (Czech)  aircraft  14c,1b &lt;br&gt;&#xA; 14c aircraft (English)  1b,1c &lt;br&gt;&lt;br&gt;&#xA; 11a conscience (English)  16a,12f &lt;br&gt;&#xA; 12f samvittighet  (Norwegian Bokmaal) conscience 11a,16a &lt;br&gt;&#xA; 16a geweten  (Dutch)  conscience  11a,12f &lt;br&gt;&lt;br&gt;&#xA; 02c manzano     (Spanish)     apple tree  15b &lt;br&gt;&#xA; 15b apfelbaum  (German)  apple tree  2c,16c &lt;br&gt;&#xA; 16c pommier  (French)  apple tree 2c,15b &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Herrings:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 00a begrafenis   (Dutch), burial, funeral  20f &lt;br&gt;&#xA; 20f hautajaiset  (Finnish)   funeral, memorial service; wake 0a &lt;br&gt;&lt;br&gt;&#xA; 00c paspoort     (Dutch), passport etc    6d &lt;br&gt;&#xA; 06d reisepass (German)    passport   0c &lt;br&gt;&#xA; 07a pormit ?  (eg for 0c, 6d) &lt;br&gt;&lt;br&gt;&#xA; 01a svatba       (Czech), wedding  8d &lt;br&gt;&#xA; 08d marriage  (English)  1a &lt;br&gt;&lt;br&gt;&#xA; 05a syntymapaivat   (Finnish), birthday  9d &lt;br&gt;&#xA; 09d cumpleanos (Asturian) &amp;amp; (Spanish)  birthday  5a &lt;br&gt;&lt;br&gt;&#xA; 02a boardgame  (English)  6a &lt;br&gt;&#xA; 06a brettspiel  (German)  board game   2a &lt;br&gt;&lt;br&gt;&#xA; 15c terning  (Norwegian Bokmål/Nynorsk)  a die or dice  16c &lt;br&gt;&#xA; 16c dobokocka  (Hungarian),  (games) dice 15c &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part 6: A bunch of words found&lt;/strong&gt; (in hidden text)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 00a begrafenis   Dutch, burial, funeral  20f &lt;br&gt;&#xA; 00c paspoort     Dutch, passport etc    6d &lt;br&gt;&#xA; 00c spoort       Dutch, ?, in spoort na/op/aan &lt;br&gt;&#xA; 00d stroop       Dutch, syrup &lt;br&gt;&#xA; 01a svatba       Czech, wedding  8d &lt;br&gt;&#xA; 01b aeronave   Italian dirigible; Spanish/Portuguese aircraft  14c,1c ## &lt;br&gt;&#xA; 01c letadlo      Czech  aircraft  ## 14c,1b &lt;br&gt;&#xA; 01d robbanás     Hungarian, explosion  10c &lt;br&gt;&#xA; 02a boardgame   6a &lt;br&gt;&#xA; 02c manzano     Spanish     apple tree  15b &lt;br&gt;&#xA; 04c hostage   5c,13g ## &lt;br&gt;&#xA; 04e yazgen ? like Spanish yazgan &lt;br&gt;&#xA; 05a syntyma  Finnish birth  &lt;br&gt;&#xA; 05a paivat   Finnish an inexact time; eg syntyma paivat = ~birthday ? 9d &lt;br&gt;&#xA; 05c panttivanki  Finnish   hostage 4c,13g ## &lt;br&gt;&#xA; 06a brettspiel  German  board game   2a &lt;br&gt;&#xA; 06c pesier  ? like Italian pesiera, pesiere set of weights &lt;br&gt;&#xA; 06d reisepass German    passport   0c &lt;br&gt;&#xA; 07a pormit ?  (eg for 0c, 6d) &lt;br&gt;&#xA; 08a riaveo ? &lt;br&gt;&#xA; 08d marriage   1a &lt;br&gt;&#xA; 08d agenim ? &lt;br&gt;&#xA; 09d cumpleanos Asturian &amp;amp; Spanish  birthday  5a &lt;br&gt;&#xA; 10a haehaa ? like Danish haehae, haha &lt;br&gt;&#xA; 10c eksplosjon   Norwegian Bokmål,     an explosion  1d &lt;br&gt;&#xA; 11a conscience  16a,12f ## &lt;br&gt;&#xA; 11f vittighet ? like Danish  joke, witticism, pun &lt;br&gt;&#xA; 12a sanzskisr ? &lt;br&gt;&#xA; 12c niltzkoegy ? &lt;br&gt;&#xA; 12f samvittighet  Norwegian, Bokmaal 11a,16a ## conscience &lt;br&gt;&#xA; 13d ruvine  Friulian,  ruin &lt;br&gt;&#xA; 13g gijzelaar Dutch,  hostage 4c, 5c ## &lt;br&gt;&#xA; 14c aircraft  1b,1c &lt;br&gt;&#xA; 15a muable  French   mutable &lt;br&gt;&#xA; 15b apfelbaum  German  apple tree  2c,16c ## &lt;br&gt;&#xA; 15c terning  Norwegian Bokmål/Nynorsk  a die or dice  16c &lt;br&gt;&#xA; 15d regnin Catalan ~ to reign, to rule &lt;br&gt;&#xA; 16a ipoge ? like Italian cavern? &lt;br&gt;&#xA; 16a geweten  Dutch  conscience  11a,12f ## &lt;br&gt;&#xA; 16b netewe       Kurdish, nation &lt;br&gt;&#xA; 16c pommier  French  apple tree 2c,15b ## &lt;br&gt;&#xA; 16c dobokocka  Hungarian,  (games) dice 15c &lt;br&gt;&#xA; 16f leengpen, Myanmar &lt;br&gt;&#xA; 17 &lt;br&gt;&#xA; 18h Gannys = name of a Roman general  &lt;br&gt;&#xA; 19 &lt;br&gt;&#xA; 20f hautajaiset  Finnish   funeral, memorial service; wake 0a &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="2004" LastEditorUserId="2004" LastEditDate="2014-12-04T16:58:11.640" LastActivityDate="2014-12-04T16:58:11.640" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5875" PostTypeId="2" ParentId="5868" CreationDate="2014-12-03T18:50:49.153" Score="8" Body="&lt;p&gt;She represents&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Sound.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;the high and the deep&lt;/code&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A sound or note can be described as high or deep.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;After the culture's patriarch god created the universe, it seems that the new universe in turn created this deity itself.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Sound can only happen after something is around to make it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Angered that his creations made their own god, the patriarch refuses to allow her into the heavens. Unable to ascend, she is said to reside in the worship houses of sky, earth, and sea.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There is no sound in space. Gas, liquids, and solids can all be transmission mediums for sound.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;The goddess of all cathedrals, all temples, and all mosques, songs to every deity are sung to her glory first.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;All songs must make sound, and any religious congregation makes noise of some sort.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Her magical cloak makes her invisible, hidden from the creator, but practitioners insist that even the most faithless believe in her.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Sound is not visible, but everyone believes in it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;The elders say that those who worship too intensely can diminish their connection to the goddess permanently. In some extreme cases, severe pain and disorientation have occurred.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Listening to loud sounds damages hearing and can cause pain/disorientation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="414" LastActivityDate="2014-12-03T18:50:49.153" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5879" PostTypeId="1" AcceptedAnswerId="5881" CreationDate="2014-12-03T22:59:16.640" Score="6" ViewCount="591" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;In the essence of every living.&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;And even in those who borders the live.&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;There, I am.&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;From the letters, I am the first.&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;Five times the seven, five times the six and five times the one.&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;But then, we have two circles.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What I am?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;This have a unique answer and it is a single word.&lt;/li&gt;&#xA;&lt;li&gt;Hopefully it won't make sense to anything else, ensuring the uniqueness.&lt;/li&gt;&#xA;&lt;li&gt;The challenge contains enough information for people find out what is this.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5044" LastEditorUserId="5373" LastEditDate="2014-12-03T23:13:19.820" LastActivityDate="2014-12-04T00:54:10.247" Title="What I Am? - Essence of every living" Tags="|riddle|word|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5880" PostTypeId="1" AcceptedAnswerId="6062" CreationDate="2014-12-04T00:51:06.933" Score="30" ViewCount="3932" Body="&lt;p&gt;When making a new puzzle, the average person is usually faced with a hard problem of guaranteeing quality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know some key points and ideas, for beginners and experts alike, to keep in mind when creating their puzzle, because &lt;em&gt;a good puzzle must be well-polished&lt;/em&gt;. What are some features of good puzzles that apply broadly to a wide range of types of puzzles? I would imagine careful management of difficulty - something like the requirement of play-testing - would be an aspect to consider, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Answerers are encouraged (but not required) to provide examples of features, common mistakes, and ways to improve a puzzle.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="4050" LastEditDate="2014-12-10T20:59:26.777" LastActivityDate="2015-06-25T02:38:36.270" Title="What are the features of a well-written puzzle?" Tags="|puzzle-creation|" AnswerCount="5" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5881" PostTypeId="2" ParentId="5879" CreationDate="2014-12-04T00:54:10.247" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Adenine &lt;a href=&quot;http://en.wikipedia.org/wiki/Adenine&quot;&gt;http://en.wikipedia.org/wiki/Adenine&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the essence of every living.&#xA;And even in those who borders the live.&#xA;There, I am.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Has to do with cellular respiration, and is a component of DNA/RNA.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the letters, I am the first&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Starts with A&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Five times the seven, five times the six and five times the one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The chemical composition is $C5N5H5$ (5 carbon, 5 nitrogen, 5 hydrogen)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But then, we have two circles.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The molecule is 2 circles in shape &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2765" LastActivityDate="2014-12-04T00:54:10.247" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5882" PostTypeId="2" ParentId="5880" CreationDate="2014-12-04T01:41:15.937" Score="16" Body="&lt;p&gt;A well-written puzzle is hard to define. Even a decent puzzle is hard to define. Instead, I'll say some signs of a poor puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Impossible to confirm&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are unintended solutions that look just as good as the intended one when explained, you're no longer solving a puzzle, you're trying to guess what's in the mind of the asker. The right solution should be clearly correct if explained, and not just for sounding clever, but for fitting the given conditions better than any alternative. Moreover, no wrong solution should feel this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Just a cipher or encoding&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Codes&quot; that are just a short string of numbers and letters aren't fun. You can't do anything better than throw random solution attempts and check if English comes out. There's so many to try and no way to tell which is right except trial and error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the following fictional example (not a real message).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ADOOGF&#xA;SORPPF&#xA;FFLGMQ&#xA;EPGKBN&#xA;EPEKGF&#xA;GFPELW&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Maybe you Caesar shift each row by a different index. Maybe you split up the grid into 2x3 blocks and interpret them as Braille with vowels being dots. Maybe you convert each letter to ASCII and read down the columns to get numbers and ... There's way too many things to try. Composing multiple codes without intermediate confirmation makes it even worse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're going to write a code, include some hints as to the mechanism used, either through direct clues, thematically in flavortext, or in some apparent feature of the ciphertext.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Loads of pointless backstory&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the entire backstory is &quot;One day I was talking to my friend. [3 paragraphs omitted] He said he'd give me a cookie if I solve the following puzzle&quot;, then there's no need for it. Just post the puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the backstory has a few clues for the puzzle, but has way more non-clue text, it will do more harm than good as solvers will latch onto red herrings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even worse is if the backstory contradicts the idea of the puzzle. Why and how did the victim write a puzzle cluing her murderer with her dying breath?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Invisible clues&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the key to a puzzle is hidden in invisible characters or in the edit history, then someone will either find it or not. And someone who doesn't think to check will waste their time fruitlessly trying to make something of what they see, unaware that they have no chance at getting anywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Loads of distraction&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A puzzle that is only hard because it includes lots of irrelevant distracted content is a bad puzzle. It wastes the solvers' time, as they are likely to find loads of things they think are clues among the filler. Solvers don't want to search for a needle in a haystack -- or a needle in a stack of needles. If your puzzle is &quot;too easy&quot; without intentional red herrings, then it's not much of a puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This doesn't apply to puzzles designed around this principle like spot-the-clue mysteries and word searches.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Intentionally misleading&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://xkcd.com/169/&quot;&gt;http://xkcd.com/169/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your puzzle is intentionally worded so that any reasonable solver would interpret it wrong, that's not a puzzle but an exercise in mindreading. An answer that's technically correct under some twisted interpretation of the conditions in not an answer, and just invites other misinterpretations that are arguable but equally bad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wordplay and circumspect hints are fine, as long as it's a type of puzzle where a solver would expect such.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why did he do X?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why did the man kill himself after eating seagull meat? Why did the teacher say &quot;you're lying&quot;? Why did the police officer arrest the woman in green?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know why; I'm not them. I can think of lots of reasons it could have happened. You have a story for it, and I can make up a story too. You're really asking me to guess your story. Your story probably involves some clever trick, but cleverness isn't correctness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Tacked-on busy-work&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, &quot;plug in the result to [this URL] to get the answer.&quot;Or, &quot;take the answer to the riddle and use it as a key to solve the cipher to get the solution.&quot; Why? Decoding a cipher with the key is a purely rote exercise. It's not fun. Or if you can solve it without the key, that makes the earlier steps moot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A puzzle can take multiple steps, but they should feel cohesive and justified on their own. It's inevitable to have bits that the solver just needs to execute once they know what to do, but don't make the puzzler do work for the sake of doing work. If you need a follow-the-instruction step to &quot;pad out&quot; the puzzle, you should instead improve the puzzle so that it doesn't need it.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2014-12-06T03:14:07.673" LastActivityDate="2014-12-06T03:14:07.673" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5883" PostTypeId="1" AcceptedAnswerId="5911" CreationDate="2014-12-04T01:43:33.343" Score="14" ViewCount="1517" Body="&lt;p&gt;&lt;sup&gt;Since challenge questions are up in the air, what better time to pose a puzzle-building question that's intrigued me for some time.&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;h2&gt;Preamble&lt;/h2&gt;&#xA;&lt;p&gt;Consider an infinite square grid made up of &lt;em&gt;nodes&lt;/em&gt; and &lt;em&gt;edges&lt;/em&gt; as shown in the following figure:&lt;/p&gt;&#xA;&lt;p&gt;                         &lt;img src=&quot;https://i.stack.imgur.com/dBCLt.png&quot; alt=&quot;infinite square lattice&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Each node connects to four edges, which are labeled &lt;em&gt;top&lt;/em&gt;, &lt;em&gt;left&lt;/em&gt;, &lt;em&gt;right&lt;/em&gt;, and &lt;em&gt;bottom&lt;/em&gt; as one would intuitively expect. Likewise, every square in the grid has four labeled edges. These labels are depicted in the two figures below.&lt;/p&gt;&#xA;&lt;p&gt;                              &lt;img src=&quot;https://i.stack.imgur.com/lDKDQ.png&quot; alt=&quot;node labels&quot; /&gt;            &lt;img src=&quot;https://i.stack.imgur.com/rZxyk.png&quot; alt=&quot;grid square labels&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Every edge in the grid has a &lt;em&gt;parity label&lt;/em&gt;, which is either &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt;. Parity labels obey the following two &lt;em&gt;symmetry rules&lt;/em&gt;:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;At every node, the multiset of parity labels of the &lt;em&gt;top and right&lt;/em&gt; edges must equal the multiset of parity labels on the &lt;em&gt;bottom and left&lt;/em&gt; edges.&lt;/p&gt;&#xA;&lt;p&gt;For example, if both the top and right edges have a parity label of &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;, then both the bottom and left edges must have a parity label of &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;. If the top and right edges have parity labels of &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; respectively, then then parity labels for the bottom and left edges must either be &lt;span class=&quot;math-container&quot;&gt;$A,B$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$B,A$&lt;/span&gt;, respectively.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For every grid square, the multiset of parity labels of the &lt;em&gt;top and left&lt;/em&gt; edges must equal the multiset of parity labels on the &lt;em&gt;bottom and right&lt;/em&gt; edges.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;The idea of the puzzle is to find a rule (or rules) for assigning a parity label to each edge in the grid that satisfies rule 1 at every node and rule 2 for every grid square.&lt;/p&gt;&#xA;&lt;p&gt;There are several trivial ways to do this, such as assigning &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;'s to all edges, assigning &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt;'s to all edges, etc., but one additional complication exists. The grid must start in either of the following two states:&lt;/p&gt;&#xA;&lt;p&gt;                     &lt;img src=&quot;https://i.stack.imgur.com/oCgkY.png&quot; alt=&quot;symmetry-breaking starting states&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Note that rule 2 is satisfied for the fully-assigned grid square in both grids, but rule 1 is &lt;em&gt;not&lt;/em&gt; satisfied for the two fully-assigned nodes. This broken symmetry is &lt;em&gt;by design&lt;/em&gt; for these particular nodes in these grids.&lt;/p&gt;&#xA;&lt;p&gt;None of the existing parity labels may be changed, and rule 1 applies everywhere except at the two symmetry-breaking nodes.&lt;/p&gt;&#xA;&lt;p&gt;The amended goal of the puzzle is therefore to find a rule (or rules) for assigning a parity label to each edge in the grid that satisfies rule 1 at every node &lt;em&gt;except&lt;/em&gt; at the two symmetry-breaking nodes, and satisfies rule 2 for every grid square.&lt;br&gt;&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;h1&gt;Question&lt;/h1&gt;&#xA;&lt;p&gt;It's not intuitively obvious that the puzzle as stated above even &lt;em&gt;has&lt;/em&gt; a solution. And indeed, all of my attempts to find a rule to populate the grid have lead to what I'll call &amp;quot;gridlock&amp;quot;—the inability to extend labels further in any direction without violating at least one of the two symmetry rules.&lt;/p&gt;&#xA;&lt;p&gt;Unfortunately, &amp;quot;COTO gets consistently stuck&amp;quot; is far from a rigorous proof of non-existence of a solution, hence I shall commit the puzzle to the enterprising minds of puzzling.SE.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Does this puzzle have a solution? If so, what is it? If not, is there an elegant proof of this fact?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Answers and intuitions are welcome.&lt;/p&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2018-05-22T20:35:02.370" Title="Does this grid puzzle with a symmetry-breaking condition have a solution?" Tags="|combinatorics|pencil-and-paper-games|solvability|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="5884" PostTypeId="2" ParentId="5880" CreationDate="2014-12-04T01:45:15.383" Score="13" Body="&lt;p&gt;&lt;em&gt;(More challenges! Otherwise this will just go to the graveyard)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no black-or-white rule for saying what is or not a well-written puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should consider those criteria:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Is it logically solvable?&lt;/strong&gt; I.E. There is some logic sequence of steps that leads to the solution.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Is it practically solvable?&lt;/strong&gt; I.E. The solution do not requires brute forcing a ridiculous large set of possible solutions or an enormous quantity of guesses.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Is it well redacted in clear and correct English?&lt;/strong&gt; Except if some part is not on purpose (I.E. not by accident), puzzles with bad grammar, bad punctuation and typos may mislead eventual solvers in ways that they are not supposed to be misleaded, and make them harder to understand.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Is it interesting?&lt;/strong&gt; Many puzzles are boring and annoying to find an answer. Make them invite people to think about solving them instead of scaring.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Is it cohesive?&lt;/strong&gt; The parts of the puzzle should bends together elegantly and naturally. Puzzles which have non-red-herring subparts that has few to no relation with the rest are IMO not well designed.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Has it the right degree of difficulty?&lt;/strong&gt; Puzzles that are too hard may frustrate people and be rendered as unsolvable without major give-away hints or brute-force methods. Puzzles that are too easy may be disappointing both for who asks and for who answers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Is it balanced?&lt;/strong&gt; Following &lt;a href=&quot;https://puzzling.meta.stackexchange.com/a/1530/5044&quot;&gt;this answer on meta from Travis Kindred&lt;/a&gt;, there are basically three types of puzzles: Calculation, Interpretation and Assumption. I will add that most puzzles tends to be a mixture of different proportions of those components rather than being purely one of them. The component &quot;Assumption&quot; is normally (but not always IMO) bad. The others are normally ok. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Is it concise and detailed?&lt;/strong&gt; The text used to describe the puzzle should not be too long, complex and hard to mentally parse, i.e. it should be concise. OTOH it should not be too short, oversimplified and unclear, i.e. it should be detailed. Some puzzles might need long sections of texts narrating what happened, that is ok. But if many parts of the texts is superfluous or unnecessarily complicated, it should be redacted more concisely. If many parts are not clear or vague, then it should be more detailed. I.E. The size and the content of the text should be just right, with not too much nor too few.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Is it complete?&lt;/strong&gt; Is all the needed information inside the puzzle? If yes, then it is ok. If not, does it make clear that people should look for some information elsewhere? If not, then it is bad. If yes, do it gives a reasonable clue to where to find additional information? If not, it is bad. If yes it is ok.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Is it simple and compact?&lt;/strong&gt; As suggested by @klm123 (and directly quoting him): &quot;a puzzle that frustrate people at first look, but has impressively simple solution when you get right idea is the most brilliant ones&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5044" LastEditorUserId="-1" LastEditDate="2017-03-16T16:42:07.097" LastActivityDate="2014-12-10T20:03:06.263" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5885" PostTypeId="1" AcceptedAnswerId="5908" CreationDate="2014-12-04T01:53:36.923" Score="12" ViewCount="2900" Body="&lt;p&gt;After you're finished making a puzzle, what types of details should you look for when rating it's difficulty? What makes a puzzle easier or harder? Keep in mind a puzzle includes both the question and the answer.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="4050" LastEditDate="2015-01-31T20:47:21.830" LastActivityDate="2015-01-31T20:47:21.830" Title="What should you look for when rating the difficulty of a puzzle?" Tags="|puzzle-creation|difficulty-rating|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5886" PostTypeId="1" CreationDate="2014-12-04T02:02:48.183" Score="5" ViewCount="658" Body="&lt;p&gt;What's the best way to make sure your new cryptogram has one, or very limited, valid solutions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It could be very easy to accidentally include multiple solutions, what's a good way to prevent this?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="4050" LastEditDate="2014-12-04T19:22:36.413" LastActivityDate="2014-12-04T19:35:14.773" Title="What's the best way to ensure a cryptogram has a unique solution?" Tags="|puzzle-creation|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5887" PostTypeId="2" ParentId="5885" CreationDate="2014-12-04T02:24:23.877" Score="32" Body="&lt;p&gt;&lt;strong&gt;Your puzzle is probably harder than you think it is.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just a heuristic answer. Some puzzles turn out easier than the writer expects. But in my experience, it's much more common error to make a puzzle that's harder than you intend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When judging the difficulty of your own puzzle, it's easy to forget that you have the advantage of knowing the answer. It sounds silly, but it's hard to push what you know out of mind and take an outside perspective. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You've had this puzzle in your head for a while, it just seems so natural to you. And a puzzle that you make is naturally of your style, which makes you better at solving it than most.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's tempting to think a puzzle is easy because the steps are easy. But to a solver, the correct steps and deductions and just one possibility of many. They might have to try a bunch of steps that &lt;em&gt;don't&lt;/em&gt; work. And sometimes, they try one that does work, but still continue to try other ideas, or even discard the right one because they don't realize another step is needed for it to make sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any &lt;a href=&quot;http://meta.puzzling.stackexchange.com/a/1530/4551&quot;&gt;interpretation puzzle&lt;/a&gt; is liable to have coincidental patterns, things that seems to kind-of fit. Confirmation bias is hard to shake, and solvers might get stuck on wrong theories. And that's fine, as long the right answer is clearly better, but you need to account for these detours in assessing difficulty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, flaws in design tend to make a puzzle harder. Unintentional errors, misworded clues, unintuitive conventions. Of course, ideally a puzzle shouldn't have these, but nobody is perfect, so consider erring on the easier side to account for this. Cluing things multiple ways (redundancy) is a good way to make a puzzle resilient against these faults.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="5044" LastEditDate="2014-12-04T07:35:16.650" LastActivityDate="2014-12-04T07:35:16.650" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5888" PostTypeId="2" ParentId="5672" CreationDate="2014-12-04T02:57:28.220" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The epitome of an evil the skewed letter would label&lt;br&gt;&#xA;  To many is such truth, not a fable&lt;br&gt;&#xA;  Held tightly by urges of the seconds&lt;br&gt;&#xA;  Within the sleepless cities that beckons&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The skewed letter is X (hint hidden in edit history). X is an skewed letter.&lt;br&gt;&#xA; For me, this could be a reference to a &lt;strong&gt;sexual intercourse&lt;/strong&gt;.&lt;br&gt;&#xA; Porn films and other sex-related things frequently are labeled with the letter X.&lt;br&gt;&#xA; In an intercourse the people would be tightly held together, normally in late night, making moves repeatedly and quickly (urges of the seconds).&lt;br&gt;&#xA; For many people it is such truth, not fable, indeed.&lt;br&gt;&#xA; So the corresponding evil would be &lt;strong&gt;lust&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Twice I hold the path of tea&lt;br&gt;&#xA;  Forever you stay within me&lt;br&gt;&#xA;  It is not glue that binds us&lt;br&gt;&#xA;  Nor the substance which I fuss&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is not TEA, it is TEE. And path is junction (hint hidden in edit history).&lt;br&gt;&#xA; Grab two T's and glue them together to crate a &lt;strong&gt;plus sign&lt;/strong&gt; (could be another X too, meaning &lt;strong&gt;sex&lt;/strong&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Guarded by a significant worm&lt;br&gt;&#xA;  Your sight has gone, time to squirm&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A spermatozoon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I laugh at your extravagance&lt;br&gt;&#xA;  I assure you my acquaintance&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The spermatozoon trying to find the ovum.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;So lets join everything now:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lust or sexual intercourse&lt;br&gt;&#xA; plus (or more sex)&lt;br&gt;&#xA; spermatozoon trying to find the ovum&lt;br&gt;&#xA; =&lt;br&gt;&#xA; &lt;strong&gt;PREGNANCY&lt;/strong&gt; or &lt;strong&gt;BABY&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5044" LastEditorUserId="5044" LastEditDate="2014-12-04T03:21:16.410" LastActivityDate="2014-12-04T03:21:16.410" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5889" PostTypeId="2" ParentId="5885" CreationDate="2014-12-04T04:02:54.747" Score="5" Body="&lt;p&gt;You didn't say what kind of puzzles you were asking about.  I enjoy programs that generate logic puzzles.  I suspect for those, the programmer writes a solver with whatever logic s/he thinks the person solving should use.  The program then generates a solution and feeds the (empty set) of clues to the solver.  If the solver fails, you add a clue and try again.  For Sudoku, there is an agreed hierarchy of types of reasoning.  The difficulty assessment is what level of reasoning is needed.  The solver can have branches to avoid certain levels of reasoning, and add more clues as required.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2014-12-04T04:02:54.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5892" PostTypeId="2" ParentId="5019" CreationDate="2014-12-04T06:26:53.987" Score="19" Body="&lt;p&gt;I am &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;a homeomorphism.  i.e. an isomorphism between two topological spaces.  I am continuous and so is my inverse.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My family is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the set of invertible functions.  Our clear worst enemy is the inverse function.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And my race is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the set of functions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are perfect among our race because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;we are invertible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My eldest brother is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;a noncontinuous invertible function.  &quot;Wave&quot; at my eldest brother means to make him continuous, and thus become me. (sine waves are continuous)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My cousin is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;a homomorphism.  When she &quot;finds the end of her life&quot; (the letter e), and adds it to herself, it forms my name.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My younger brother is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A diffeomorphism.  i.e. an isomorphism over a differentiable manifold.  It is a smooth function whose inverse is also smooth.  Because smoothness is a higher restriction than continuousness, the younger brother outstrips me.  However, people are often confused about the difference between smooth and continuous.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6178" LastEditorUserId="6178" LastEditDate="2014-12-04T07:21:12.197" LastActivityDate="2014-12-04T07:21:12.197" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5895" PostTypeId="2" ParentId="5806" CreationDate="2014-12-04T09:14:54.967" Score="5" Body="&lt;p&gt;&lt;em&gt;Now that this has been reopened I want to post the &amp;quot;solution&amp;quot; I had for this in the night before Victor's answer. Victor has beaten me to it, and Warspyking has modified the victory conditions since, so it is no longer a valid answer.  I still wanted to post this for commenting (and verification/disproving.)&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;The answer works for n-dimensional mazes and it is also a &lt;em&gt;algorithm&lt;/em&gt; not a specific maze. I do not post &lt;em&gt;code&lt;/em&gt; here, but the algorithm concept.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;My solution is based on the idea that each pixel/voxel/n-dim-element ( &lt;em&gt;called voxel from now on&lt;/em&gt;) can either be a path (0) or a wall (1). This is different from what Victor made, but it can be turned into each other rather easily. (&lt;em&gt;Just convert walls&amp;amp;openings into additional voxels or the other way around.&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The data dimensions sizes are called &lt;em&gt;n,m,k,l...&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Indexing starts with 0 for the first voxel&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h3&gt;Initialize data:&lt;/h3&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Create the maze-data as Boolean array of needed dimensionality&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Initialize all voxels as 0 (empty)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Create a list for voxel-indices of walls (empty)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Set every voxel with at least one &lt;em&gt;even&lt;/em&gt; index to 1 (wall)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Store the wall-voxel index in the list.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;From now on, whenever a wall is removed:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;remove the according index from the list&lt;/li&gt;&#xA;&lt;li&gt;set the according Boolean to 0 (empty)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h3&gt;Define start &amp;amp; end position:&lt;/h3&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Essentially set the two opposing &amp;quot;corners&amp;quot; to be start and end.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Set the voxel of index (0,0,...) to be the starting location&lt;/li&gt;&#xA;&lt;li&gt;Set the voxel of index (n,m,...) to be the goal destination&lt;/li&gt;&#xA;&lt;li&gt;If the destination voxel is a wall, remove the wall. (All dimensions are  even-sized)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h3&gt;Create the maze:&lt;/h3&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The current maze is a grid with isolated empty spaces.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Label all empty spaces with a unique label.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Now go iteratively (&lt;em&gt;WHILE-loop&lt;/em&gt;):&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Randomly choose a wall from the index list.&lt;/li&gt;&#xA;&lt;li&gt;Remove the index form the list (&lt;em&gt;never test a wall twice&lt;/em&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;strong&gt;Test&lt;/strong&gt;: Will removing this wall merge two empty spaces of different label?&#xA;NO: next iteration. ELSE:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Remove wall&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Of all involved labels choose the one with lowest value&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Set all voxels having involved labels to this chosen value (=merge empty spaces)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;next iteration&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;strong&gt;Stop iteration when no indices are in the list.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Without proofing it, I think this algorithm will give you:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;- a non-looping maze from start to end &#xA;&#xA;- maximize the size of the maze within the volume&#xA;&#xA;- The identical algorithm could be used for arbitrary amount of dimensions&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Originally I was intending to code this algorithm and then think of a nice way of displaying the 4D case, but Victor has done a great job of that already, so I'll leave it with this.&lt;/p&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-12-04T09:14:54.967" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5896" PostTypeId="2" ParentId="5631" CreationDate="2014-12-04T10:10:56.803" Score="9" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Insomnia?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My younger brother is Mr. Wayne,&lt;br&gt;&#xA;And the beginning of all his pain&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Insomnia is the Christopher Nolan movie that precedes Batman Begins&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I will tease your eyes with nails&lt;br&gt;&#xA;But, I won't let you close your eyes&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A poetic description of insomnia&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To make it simple, cut my tail&lt;br&gt;&#xA;With my new tail, you won't fail&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Cut the last letter, &quot;A&quot;.  If you have an A, you won't fail (a course or test).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now take some of my exact middle&lt;br&gt;&#xA;That will help you solve this riddle&lt;br&gt;&#xA;Then cut its end, the beginning of the end&lt;br&gt;&#xA;It's a wordplay, that's a hint&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Take &quot;some&quot;, remove the &quot;e&quot;, this forms the middle (som) of &quot;insomni&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My New head and tail are almost the same&lt;br&gt;&#xA;If you place a mirror in between&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;insomni&quot; remains almost unchanged if the letters are reversed. [the new head 'in' and tail 'ni' are the same if a mirror is placed in between. That is 'in' is almost similar to the mirror image of 'ni' and vice verse.]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My lost tail is definitely not for doctors&lt;br&gt;&#xA;Tell me who I am, that's what matters&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Not sure about this, but insomnia is a medical condition.  Edit: as per OP, the lost tail, A, stands for Apple.  An apple a day keeps the doctor away,  as they say,  so &quot;A&quot; is not for doctors. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6178" LastEditorUserId="2730" LastEditDate="2014-12-08T11:37:50.630" LastActivityDate="2014-12-08T11:37:50.630" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="5898" PostTypeId="2" ParentId="5865" CreationDate="2014-12-04T12:07:09.853" Score="2" Body="&lt;p&gt;Television&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I go by many a name,&#xA;Including one that's quite a shame. : Its Called idiot box&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can take you to an exact location,&#xA;But sometimes only to a nearby destination. : Various shows take us to different destinations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If great power is your wish,&#xA;I'll grant it but your money will vanish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More money for more picture and sound quality&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting for that which can be negative, I'll always remain completely positive. : Negative as it time consuming but positive in terms of entertainment..also I guess its somthing about positive and negaitve charges(electronics )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I contain many things, I simultaneously contain nothing. : It shows your fav celebs,bt celebs are not practically in there&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can signal the beginning, But I can also signal the ending. : In TV you can see Happy endings !! :)&lt;/p&gt;&#xA;" OwnerUserId="6183" LastEditorUserId="4450" LastEditDate="2014-12-05T02:08:16.853" LastActivityDate="2014-12-05T02:08:16.853" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5900" PostTypeId="1" AcceptedAnswerId="5901" CreationDate="2014-12-04T12:12:04.960" Score="8" ViewCount="7769" Body="&lt;p&gt;For the TV series &lt;a href=&quot;http://en.wikipedia.org/wiki/Sherlock_(TV_series)&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Sherlock&lt;/em&gt;&lt;/a&gt;, BBC published some &lt;a href=&quot;http://www.bbc.co.uk/programmes/articles/4yTrVL4srk48mKs8bhK4TvM/other-sherlock-related-websites&quot; rel=&quot;nofollow noreferrer&quot;&gt;fictional websites&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sherlock’s site, &lt;a href=&quot;http://thescienceofdeduction.co.uk/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://thescienceofdeduction.co.uk/&lt;/a&gt;, contains three &lt;a href=&quot;http://thescienceofdeduction.co.uk/hidden-messages&quot; rel=&quot;nofollow noreferrer&quot;&gt;hidden messages&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Some anonymous guy has been sending me emails. It could be nothing but I'll post them up here. Could be an interesting case that you can help me with...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It says all three were &quot;Solved&quot;, but it seems that the solution for the &lt;a href=&quot;http://thescienceofdeduction.co.uk/hidden-messages/hiddenmessage3&quot; rel=&quot;nofollow noreferrer&quot;&gt;hidden message #3&lt;/a&gt; is still missing. Here’s the message:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sherlock Holmes!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Here's a picture you might enjoy: &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fMMCB.gif&quot; alt=&quot;Handwritten coded message&quot;&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Also, where is it the pigs live?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Mwah!&lt;br&gt;&#xA;  Xx&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the solution? How does it work?&lt;/p&gt;&#xA;" OwnerUserId="141" LastActivityDate="2016-02-21T17:48:20.503" Title="Sherlock’s hidden message #3" Tags="|encoded-message|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5901" PostTypeId="2" ParentId="5900" CreationDate="2014-12-04T12:29:34.607" Score="10" Body="&lt;p&gt;The hidden message is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;SHERLOCK I HAVE FOUND YOU&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Form the clue &quot;where is it the pigs live?&quot; I understood it is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Pigpen cipher&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By deciphering, I got&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;SHERLOCK I HAVE FOUND YOU&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Source: &lt;a href=&quot;http://en.wikipedia.org/wiki/Pigpen_cipher&quot;&gt;Pigpen cipher&lt;/a&gt; (You will get the idea about how to replace the code with corresponding alphabets from there)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="2730" LastEditDate="2014-12-04T12:34:45.853" LastActivityDate="2014-12-04T12:34:45.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5902" PostTypeId="2" ParentId="5865" CreationDate="2014-12-04T13:11:46.213" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SPACE  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I go by many a name,&lt;br&gt;&#xA;Including one that's quite a shame.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; space, Empty, nothing, Vaccum.  Empty being the one that is shameful to have.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can take you to an exact location,&lt;br&gt;&#xA;But sometimes only to a nearby destination.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if you are in the exact location of space there is no longer space, although it would be nearby as you passed through space to get to space leaving space behind.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If great power is your wish,&lt;br&gt;&#xA;I'll grant it but your money will vanish.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; buying space is power; land is power, alot of what you do with space could be considered powerful.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be very important,&lt;br&gt;&#xA;While at the same time very insignificant.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; certain areas of space are sentimental to people, whilst to others is quite insignificant.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I contain many things,&lt;br&gt;&#xA;I simultaneously contain nothing.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; space is filled with millions of things but space still has nothing in it.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can signal the beginning,&lt;br&gt;&#xA;But I can also signal the ending.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; space has the potential to start and end anything. Also SpacE = Start + End&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5135" LastActivityDate="2014-12-04T13:11:46.213" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5904" PostTypeId="2" ParentId="5880" CreationDate="2014-12-04T14:17:55.250" Score="8" Body="&lt;p&gt;Here's what I think. Note that some points overlap with other answers.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Logical steps&lt;/strong&gt;: Is each solve step clear and logical, or does it rely on unstated assumptions or unreasonable interpretations of the clues?&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Confirmation&lt;/strong&gt;: Once you've solved a step, is there a good way of telling whether you're on the right track?&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Fair search spaces&lt;/strong&gt;: If the puzzle is hiding a message via whitespace between words or Markdown in a post, then unless the formatting is clearly abnormal then a hint is a must. If the puzzle requires brute forcing all permutations of a Rubik's Cube... then that's just unfair.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;No artificial difficulty&lt;/strong&gt;: Don't include anything that requires purely time and no brains. &quot;My puzzle is too easy, maybe I'll just encode the first clue paragraph in Morse code and make them decipher it.&quot; Please, no — that's just no fun.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Minimise red herrings&lt;/strong&gt;: Unusual surface readings (e.g. for cryptic crossword clues) are fine, but red herrings placed for the explicit purpose of leading someone down the garden path are not.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Pay attention to the details&lt;/strong&gt;: If something is irrelevant to the puzzle, then order it in some way such that the irrelevance is clear. For example, the order the words are given in a word search rarely matters, so order them alphabetically so nobody will try to find patterns in the word order. Also check your spelling and grammar — if something seems off, then people might end up thinking that it's relevant.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Do your research&lt;/strong&gt;: Unfortunately, even good puzzles written with the intention of having a single clear answer can fall into this trap. Maybe you have two clues saying &quot;Morse&quot; and &quot;insect&quot;, telling people to interpret something as Morse code to get an insect as the answer. Well I'm sorry — as clear as the instruction may seem, &lt;a href=&quot;http://mspaintadventures.wikia.com/wiki/Serenity&quot;&gt;Serenity&lt;/a&gt; is a firefly that communicates in Morse code and a &lt;a href=&quot;https://en.wikipedia.org/wiki/Vibroplex&quot;&gt;Vibroplex&lt;/a&gt; is a semi-automatic Morse key that became known as a &quot;bug&quot;. Think about how other people might interpret your clues, and try to get a feel for other possible interpretations out there.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Consider your target audience&lt;/strong&gt;: If you have a puzzle that requires specialist knowledge (e.g. programming, advanced mathematics) and you don't expect most of your potential solvers to have that knowledge, then try to make sure that the knowledge required is 1) minimised, 2) easily searchable and 3) easy to understand quickly.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="445" LastEditorUserId="445" LastEditDate="2014-12-04T19:15:31.603" LastActivityDate="2014-12-04T19:15:31.603" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5905" PostTypeId="1" CreationDate="2014-12-04T14:23:12.117" Score="83" ViewCount="16732" Body="&lt;p&gt;If I have a &lt;a href=&quot;http://en.wikipedia.org/wiki/Ciphertext&quot;&gt;ciphertext&lt;/a&gt;, what information about the probable &lt;a href=&quot;http://en.wikipedia.org/wiki/Cipher&quot;&gt;ciphers&lt;/a&gt; used can I infer from the ciphertext itself? For example, information from:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the length&lt;/li&gt;&#xA;&lt;li&gt;the spacing / grouping of characters&lt;/li&gt;&#xA;&lt;li&gt;the range of characters used&lt;/li&gt;&#xA;&lt;li&gt;frequency counts&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and any other information that is intrinsic to the ciphertext itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm talking about pencil-and-paper ciphers of the kind often used in cryptogram puzzles, rather than real, 'strong', modern forms of crypto.&lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="4050" LastEditDate="2014-12-04T14:46:57.007" LastActivityDate="2019-06-07T08:34:29.507" Title="What characteristics of a ciphertext can be indicators of a particular cipher?" Tags="|cipher|" AnswerCount="5" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="5907" PostTypeId="2" ParentId="5905" CreationDate="2014-12-04T14:33:27.817" Score="22" Body="&lt;p&gt;&lt;strong&gt;Groups of 5 or total length divisible by 5&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Might be the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bacon%27s_cipher&quot; rel=&quot;nofollow noreferrer&quot;&gt;Baconian Cipher&lt;/a&gt;, which uses a set of 5 binary-coded ('A' or 'B') items to represent each letter in the plaintext. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/5689/lolcat-steganography&quot;&gt;Lolcat Steganography: Find the message hidden within the transport medium of humorous feline photography&lt;/a&gt;   has 20 cats, and some of the possible plaintext answers are 4 characters in length. So each plaintext character could be represented by 5 cats, which makes it worth checking for Bacon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Short groups&lt;/strong&gt; (1 to 4 characters) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check for &lt;a href=&quot;http://en.wikipedia.org/wiki/Morse_code&quot; rel=&quot;nofollow noreferrer&quot;&gt;Morse Code&lt;/a&gt;. Is there a binary distinction between the characters which could easily be mapped into dots and dashes? Look particularly for circles or lines in the ciphertext symbols, which could map into dots and dashes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4642/%E2%97%B3%E2%97%B0-%E2%97%93%E2%97%A8-%E2%97%A8%E2%97%A7%E2%97%95-%E2%97%8E%E2%97%8C-%E2%97%B1%E2%97%AF%E2%97%B1%E2%97%AF-%E2%97%8D%E2%97%8C%E2%97%8B-%E2%97%89%E2%97%89-%E2%97%87%E2%97%94%E2%97%93%E2%97%95-%E2%97%90%E2%96%BA%E2%97%93%E2%97%92-%E2%97%92%E2%97%91-%E2%97%88%E2%97%91-%E2%97%86%E2%97%86%E2%97%93-%E2%97%89%E2%97%89%E2%97%89&quot;&gt;Decode the message enciphered in these symbols: ◳◰ ◓◨ ◨◧◕ ◎◌ ◱◯◱◯ ◍◌○ ◉◉ ◇◔◓◕ ◐►◓◒ ◒◑ ◈◑ ◆◆◓ ◉◉◉&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max group length is 5 characters if digits are used, 6 if punctuation is used, and &lt;a href=&quot;http://www.w2egb.org/technical/cw.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;prosigns can be even longer&lt;/a&gt;. So if the ciphertext is &lt;em&gt;almost&lt;/em&gt; all groups of up to 4 characters then it still might be worth checking for Morse. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lengths of groups = lengths of possible plaintext words&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the spaces between the groups correspond directly to spaces between the words in the plaintext. If so then this could be a simple &lt;a href=&quot;http://en.wikipedia.org/wiki/Substitution_cipher&quot; rel=&quot;nofollow noreferrer&quot;&gt;substitution cipher&lt;/a&gt; such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Caesar_cipher&quot; rel=&quot;nofollow noreferrer&quot;&gt;Caesar&lt;/a&gt; or even good old &lt;a href=&quot;http://en.wikipedia.org/wiki/ROT13&quot; rel=&quot;nofollow noreferrer&quot;&gt;ROT-13&lt;/a&gt;! &lt;a href=&quot;http://en.wikipedia.org/wiki/Frequency_analysis&quot; rel=&quot;nofollow noreferrer&quot;&gt;Frequency analysis&lt;/a&gt; might yield interesting results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ciphertext consists of arbitrary objects rather than textual characters&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can a binary distinction be made which divides the objects into 2 types? If so then if the total number of objects is divisible by 5 then this could well be a &lt;a href=&quot;http://en.wikipedia.org/wiki/Bacon%27s_cipher&quot; rel=&quot;nofollow noreferrer&quot;&gt;Baconian Cipher&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Txn1Y.jpg&quot; alt=&quot;&amp;quot;Knowledge Is Powe&amp;quot;&quot;&gt; &#xA;  &lt;br&gt; &#xA;  Detail from a photograph of World War I cryptographers trained by William and Elizebeth Friedman, Aurora, Illinois, early 1918. By facing either forward or sideways, the soldiers formed a coded phrase utilizing Francis Bacon’s biliteral cipher. The intended message was the Baconian motto “Knowledge is power,” but there were insufficient people to complete the r (and the w was compromised by one soldier looking the wrong way).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cabinetmagazine.org/issues/40/sherman.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;How to Make Anything Signify Anything&lt;/a&gt;, &lt;em&gt;Cabinet&lt;/em&gt; issue 40 winter 2010/2011, William H. Sherman&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ends in an equals symbol&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try &lt;a href=&quot;http://en.wikipedia.org/wiki/Base64&quot; rel=&quot;nofollow noreferrer&quot;&gt;Base64&lt;/a&gt; - the '=' symbol is sometimes (but not always) needed for padding, to make the ciphertext the right length.&lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2014-12-17T10:13:42.427" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5908" PostTypeId="2" ParentId="5885" CreationDate="2014-12-04T14:36:06.133" Score="7" Body="&lt;p&gt;The best way to run difficulty tests on your puzzle is the tried-and-true method of a test audience.  As others have pointed out, do &lt;em&gt;not&lt;/em&gt; include yourself in this audience - you are biased because you already know the solution.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can try to be rigorous in your testing b getting a large sample audience to test your puzzle against, which is the usual manner in which these things are tested (analogous to &lt;a href=&quot;http://en.wikipedia.org/wiki/Game_testing&quot;&gt;Game Testing&lt;/a&gt;), but for smaller puzzles, or if you're strapped for participants, you can choose a small handful.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is a personally-made puzzle for fun, you're more likely to use a small handful of participants, so choose your testers wisely.  Know how skilled they are at puzzles of your type ahead of time, and base their test results on that information - if they're not typically good at puzzles, them solving it quickly might be a sign that you need to make it harder, though if they're particularly good at it, taking a very long time when they shouldn't have a problem could be a sign that you need to make it easier or include more hints.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Testing puzzles is an imprecise science.  Sometimes what seems like a minor adjustment can make the puzzle unbeatable, and other times your focus on a preconceived solution will blind you to an obvious 'easy' solution you didn't think about before. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And of course this depends on what type of puzzle you've made - if it's wholly new, you'll need a lot more testers to really get a feel for how hard/easy you've made it.  &lt;/p&gt;&#xA;" OwnerUserId="2756" LastActivityDate="2014-12-04T14:36:06.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5909" PostTypeId="2" ParentId="5905" CreationDate="2014-12-04T14:38:00.217" Score="45" Body="&lt;p&gt;I'll start with a few simple ones to get us rolling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Base64&quot;&gt;Base64&lt;/a&gt; has a 2/3 chance of requiring &lt;strong&gt;one or two trailing &lt;code&gt;=&lt;/code&gt; signs&lt;/strong&gt; - if you spot any of these, there's a very good chance you're dealing with Base64 encoded text.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Bacon%27s_cipher&quot;&gt;Bacon cipher&lt;/a&gt; is composed of &lt;strong&gt;2 binary bits&lt;/strong&gt;, which can be represented by anything (eg. could be upper/lowercase as in one recent question, or could be based on whether a cat is white/black, as seen in another). The tell-tale sign is if the &lt;strong&gt;ciphertext length is divisible by 5&lt;/strong&gt;, as each letter requires 5 binary bits of Bacon. Yum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Frequency_analysis&quot;&gt;Frequency analysis&lt;/a&gt; is most useful for &lt;a href=&quot;http://en.wikipedia.org/wiki/Caesar_cipher&quot;&gt;Caesar ciphers&lt;/a&gt; (also known as Caesarian shifts) or similar ciphers where a letter is always encoded as itself (ie. if an &lt;code&gt;E&lt;/code&gt; encodes to a &lt;code&gt;K&lt;/code&gt; anywhere in the text, it will encode to a &lt;code&gt;K&lt;/code&gt; everywhere in the text). Many websites have &lt;strong&gt;frequency tables&lt;/strong&gt; for how often you should expect a letter to appear, and if you have a reasonably large block of ciphertext, this may be a good way to get some initial letters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Substitution_cipher&quot;&gt;substitution cipher&lt;/a&gt; is a very simple numerical cipher where letters are replaced with their numeric position in the alphabet (eg. &lt;code&gt;A&lt;/code&gt; is 1, &lt;code&gt;B&lt;/code&gt; is 2, etc) - this may also be combined with a Caesarian shift. If you have a &lt;strong&gt;sequence of numbers where none are over 26&lt;/strong&gt;, try a quick substitution and see if you get anywhere. Numbers above 26 may still be substitutable using modulo, but it's less common.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Playfair_cipher&quot;&gt;Playfair ciphers&lt;/a&gt; are &lt;strong&gt;composed of digrams&lt;/strong&gt; (pairs of letters) so the pair of letters &lt;code&gt;HI&lt;/code&gt; may encode to something like &lt;code&gt;PK&lt;/code&gt;. If you have an &lt;strong&gt;even-length ciphertext&lt;/strong&gt;, split it into digrams and look for duplicates. If you spot &lt;strong&gt;the same digram appearing multiple times&lt;/strong&gt;, you should begin to suspect a Playfair cipher (or other digram-based cipher).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Polybius_square&quot;&gt;Polybius ciphers&lt;/a&gt; (or Polybius squares) are composed of the &lt;strong&gt;same 5 characters&lt;/strong&gt; over and over again (for example, &lt;code&gt;A&lt;/code&gt;, &lt;code&gt;B&lt;/code&gt;, &lt;code&gt;C&lt;/code&gt;, &lt;code&gt;D&lt;/code&gt;, &lt;code&gt;E&lt;/code&gt;) and also happen to be digram-based so it must be an &lt;strong&gt;even-length ciphertext&lt;/strong&gt;. If you have only 5 repeated characters and can split them into digrams, it's highly likely you're dealing with Polybius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more common-sense point - if you have a &lt;strong&gt;keyword&lt;/strong&gt; then you can usually safely rule out any cipher which &lt;strong&gt;doesn't&lt;/strong&gt; require a keyword. For example, Base64 is just an algorithm which you put text in and it gets spat out the other end, while a Playfair cipher uses a keyword to make the cipher stronger.&lt;/p&gt;&#xA;" OwnerUserId="2518" LastEditorUserId="2518" LastEditDate="2014-12-05T08:08:42.183" LastActivityDate="2014-12-05T08:08:42.183" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5910" PostTypeId="2" ParentId="5905" CreationDate="2014-12-04T14:42:03.400" Score="9" Body="&lt;h3&gt;unique character mapping?&lt;/h3&gt;&#xA;&lt;p&gt;If you have already the information, that a cipher maps &lt;code&gt;something&lt;/code&gt; onto letters, then&#xA;counting the number of &lt;em&gt;different&lt;/em&gt; types of &lt;code&gt;something&lt;/code&gt; can give you a clue if there is a unique or a multiple mapping of &lt;code&gt;something&lt;/code&gt; onto a single character.&lt;/p&gt;&#xA;&lt;p&gt;i.e. there could be a cipher where &lt;code&gt;B&lt;/code&gt; stands for &lt;code&gt;A&lt;/code&gt; but &lt;code&gt;C&lt;/code&gt; &lt;em&gt;also&lt;/em&gt; stands for &lt;code&gt;A&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;p&gt;An example for this can be found in the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/5778/my-new-cipher-with-numbers&quot;&gt;My new cipher with numbers&lt;/a&gt; puzzle:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$(2_ 2,5_ 2,7_ 5,8_ 5,9_ 4,14_ 4),(6_ 2,7_ 1),(3_ 2,5_ 4,30_ 4,70_ 2)$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;In this puzzle, all number-pairs are different, despite the additional information given that each pair stands for 1 letter, and that the given ciphertext is a full sentence.&lt;/p&gt;&#xA;&lt;p&gt;Clearly a specific letter can be expressed in &lt;em&gt;different&lt;/em&gt; ways which already rules out many of the known ciphers.&lt;/p&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-12-04T14:47:04.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5911" PostTypeId="2" ParentId="5883" CreationDate="2014-12-04T15:13:40.893" Score="6" Body="&lt;p&gt;I wouldn't call this 'elegant', but here's an argument that there can be no solution starting with Grid 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tV3ob.png&quot; alt=&quot;Diagram 1&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/cTRWb.png&quot; alt=&quot;Diagram 2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first diagram, edges 1-6 must all have the same label: edges 1 and 2 because of the square they share, then edges 3 and 4 because of the vertex they share with 1 and 2, and finally 5 and 6 because of the square they share with 2 and 4. Looking at the orange vertex shows that the common label for edges 1-6 must be A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second diagram, edge 1 must get the label A (it is adjacent to a vertex with 3 A's), then edge 2 must also get the label A (it is adjacent two a square with 3 A's). But now we're stuck, as there is no way to make the symmetry rule hold at the green vertex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I drew something similar for Grid 2. It was a slightly more complicated, but I again found that there could be no solution.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2014-12-04T15:47:05.393" LastActivityDate="2014-12-04T15:47:05.393" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5912" PostTypeId="2" ParentId="491" CreationDate="2014-12-04T15:46:57.187" Score="0" Body="&lt;p&gt;I'm not sure about the probabilities, but I thought I'd address the question &quot;Is this even possible?&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a favourite paper of mine, &lt;a href=&quot;http://homepages.warwick.ac.uk/staff/D.Maclagan/papers/set.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Card Game Set&lt;/a&gt;. In it they give a collection of twenty cards with no set, which is apparently the maximum (i.e. 21 cards must contain a set somewhere):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qWIBz.png&quot; alt=&quot;Twenty and no set&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, it's perfectly possible to have no set within 15 or even 18 cards!&lt;/p&gt;&#xA;" OwnerUserId="445" LastActivityDate="2014-12-04T15:46:57.187" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5913" PostTypeId="1" AcceptedAnswerId="5915" CreationDate="2014-12-04T16:04:50.913" Score="8" ViewCount="1208" Body="&lt;p&gt;I wrote my first riddle.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Twelve smooth pebbles sit facing the sky,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;With one large roof, keeping them dry.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Listens and speaks at the same time,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Knows every secret, both yours and mine.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Used by billions, invented by two,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Older than most, but wiser than few.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is it?&lt;/p&gt;&#xA;" OwnerUserId="6187" LastEditorUserId="6187" LastEditDate="2014-12-04T16:16:08.387" LastActivityDate="2014-12-04T17:17:45.160" Title="Solve this riddle: Twelve smooth pebbles" Tags="|riddle|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5914" PostTypeId="1" AcceptedAnswerId="5916" CreationDate="2014-12-04T16:05:10.630" Score="5" ViewCount="799" Body="&lt;p&gt;My last one charged politically&lt;br&gt;&#xA;aimed at those who show hypocrisy&lt;br&gt;&#xA;if you want the bounty worthy of a Spartan&lt;br&gt;&#xA;then hearken here and answer this jargon.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I am the reasons why you burn&lt;br&gt;&#xA;  well after you're left in the end of your turn&lt;br&gt;&#xA;  a plethora of actions lead you here&lt;br&gt;&#xA;  under a septenary of hate and fear  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Situated here, desolate of fortitude&lt;br&gt;&#xA;  just moulded eyes and tunnels of feud&lt;br&gt;&#xA;  for an act of me is negligible advice&lt;br&gt;&#xA;  my existence defended by who surmise  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No patronising statement here&lt;br&gt;&#xA;no tarnishing or reputational smear&lt;br&gt;&#xA;just read the lines and think it through&lt;br&gt;&#xA;revive the fun in riddles anew.&lt;/p&gt;&#xA;" OwnerUserId="5135" LastEditorUserId="2484" LastEditDate="2014-12-05T11:10:31.617" LastActivityDate="2014-12-05T11:10:31.617" Title="Riddle: Safety in numberS" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5915" PostTypeId="2" ParentId="5913" CreationDate="2014-12-04T16:18:23.417" Score="8" Body="&lt;p&gt;It's a...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; telephone&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Twelve smooth pebbles sit facing the sky,&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it has 12 keys (including * and #)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;With one large roof, keeping them dry.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;(not sure)&lt;/em&gt; the plastic between keys themselves?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Listens and speaks at the same time,&#xA;Knows every secret, both yours and mine.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can both talk and listen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Used by billions, invented by two,&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many people have telephones, but it was invented by Meucci and/or Graham Bell&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Older than most, but wiser than few.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everyone alive now was born after it was invented and it is not wise because it only transports voice, it's doesn't know anything&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4527" LastEditorUserId="4443" LastEditDate="2014-12-04T17:17:45.160" LastActivityDate="2014-12-04T17:17:45.160" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="5916" PostTypeId="2" ParentId="5914" CreationDate="2014-12-04T16:23:34.343" Score="5" Body="&lt;p&gt;I think this is related to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Seven Deadly Sins&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I am the reasons why you burn&#xA;  well after your left in the end of your turn&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Taking &quot;burn&quot; literally, these actions will send you to Hell. The end of your turn meaning once you are dead.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a plethora of actions lead you here&#xA;  under a septenary of hate and fear&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are many reasons (a plethora of actions) one might commit any of these transgressions, but most of the time hate and/or fear are involved, as any good Jedi would tell you. In addition, &lt;em&gt;septenary&lt;/em&gt; directly links to &lt;em&gt;seven&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Situated here, desolate of fortitude&#xA;  just molded eyes and tunnels of fued&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Desolate of fortitude&quot; means you are lacking resilience, and cannot stop yourself from committing them. Molded eyes and tunnels of feud sounds like some depictions of Hell from the Renaissance era to me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;for an act of me is negligible advice&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I wouldn't advise someone to commit these acts. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;my existance defended by who surmise&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The existence of &quot;sin&quot; at all is defended by religion, which is literally people who surmise (suppose something to be true &lt;a href=&quot;https://www.google.com/search?espv=2&amp;amp;q=define+surmise&amp;amp;oq=define+surmise&amp;amp;gs_l=serp.3..0l4j0i22i30l5.494274.496018.0.496155.14.14.0.0.0.0.132.1065.13j1.14.0.msedr...0...1c.1.58.serp..0.14.1064.pKWyOquOQFM&quot;&gt;without confirming evidence&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, the title has a capital S at beginning at end instead of the usual title case: &lt;code&gt;Safety in numberS&lt;/code&gt;, so the answer could be shortened to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Seven Sins&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1563" LastActivityDate="2014-12-04T16:23:34.343" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="5918" PostTypeId="1" AcceptedAnswerId="6058" CreationDate="2014-12-04T18:49:36.360" Score="7" ViewCount="1169" Body="&lt;p&gt;I've recently acquired a Geared Mixup puzzle, and have spent several hours trying to solve this particular problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've pretty much solved the puzzle. All centers, edges, and corners are in the right positions. I just have one small problem: I have three individual edges rotated on the Geared Mixup puzzle. Two are shown in the picture below, and a (hidden) one is opposite the green-white edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HxBrHm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to find a move-efficient way to rotate the edges back on this puzzle, but I'm struggling. I have two major thoughts:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It's definitely possible to rotate a single edge individually. I know two centers can be rotated, and I know that edges can be swapped with centers, so I'd just swap and edge with a center and rotate them both. &lt;/li&gt;&#xA;&lt;li&gt;Since it's possible to rotate two centers, it should, in theory, be possible to rotate two edges at once: the front right and back left (which is also rotated). I'm not sure how to do this, though it seems like it might (?) be more move-efficient.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm looking to construct something move-efficient, too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm starting with a base sequence &lt;code&gt;[A] = (R4 U' R4 U)3&lt;/code&gt;. Sequence &lt;code&gt;[A]&lt;/code&gt; does a few things. First, it swaps the &lt;code&gt;FL&lt;/code&gt; edge with the &lt;code&gt;BR&lt;/code&gt; edge. Second, it swaps the &lt;code&gt;F&lt;/code&gt; center with the &lt;code&gt;B&lt;/code&gt; center. Third, it rotates the &lt;code&gt;U&lt;/code&gt; and &lt;code&gt;D&lt;/code&gt; centers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we construct a reversible sequence &lt;code&gt;[B]&lt;/code&gt; that sets up the puzzle such that a center is opposite the edge that needs to be rotated, both are in center-slot positions, and we can easily reverse the edge swap and center swap, then this problem is solved. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that last bit: I can construct a sequence that does this, but I can't easily reverse the side-effect center and edge swap. If the edges are opposing edge pairs, they can be easily reversed by &lt;code&gt;[A]&lt;/code&gt;. (I'm not worried about centers, because pairs of centers can be swapped easily.) For the life of me, I can't get it that way.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here's an example. Let's say that we construct a &lt;code&gt;[B] = B' D R&lt;/code&gt; (s.t. &lt;code&gt;[B'] = R' D' B&lt;/code&gt;). We've moved the green-yellow edge to the opposite of the white center in positions where they'll both be rotated. However, as a side effect of the way this was constructed, executing &lt;code&gt;[B A B']&lt;/code&gt; will result in a swap of the orange enter and blue-red edge, and the yellow center and an orange-red edge. &lt;em&gt;This is a swap that's not easily reversible.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for a sequence &lt;code&gt;[C]&lt;/code&gt; that will put reversible swaps in those positions, such that I can execute &lt;code&gt;[B C A C' B']&lt;/code&gt; (or, in other words, use &lt;code&gt;[B C]&lt;/code&gt; as a setup sequence) without having to worry later about struggling to swap those pieces back. (To swap them back is pretty simple: just execute &lt;code&gt;[A]&lt;/code&gt; on the right edge pair - but they need to be effecacious for this change.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My two questions are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is a sequence &lt;code&gt;[C]&lt;/code&gt; that completes this algorithm?; or:&lt;/li&gt;&#xA;&lt;li&gt;Is there an easy way to do this that I've totally missed?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="user20" LastActivityDate="2014-12-10T06:03:48.837" Title="How do I rotate edges on the Geared Mixup puzzle?" Tags="|mechanical-puzzles|twisty-puzzles|geared-puzzles|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5919" PostTypeId="1" AcceptedAnswerId="5922" CreationDate="2014-12-04T19:21:25.967" Score="12" ViewCount="4108" Body="&lt;p&gt;There's actually 2 parts to question;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;How to determine how hard your own maze is (knowing the solution)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How to determine how hard another's maze is (without knowing the solution)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What are some general rules and guidelines you can use to figure out how difficult a maze is?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2016-07-10T12:47:52.390" Title="What's the best way to rate how difficult a maze is?" Tags="|puzzle-creation|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5920" PostTypeId="2" ParentId="5886" CreationDate="2014-12-04T19:35:14.773" Score="7" Body="&lt;h3&gt;Make it long&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The longer your cryptogram is, the less likely it is that &lt;em&gt;all&lt;/em&gt; the words will form other valid words. As a very obvious example, a short one like &lt;code&gt;ATX YXKI&lt;/code&gt; could have so many possibilities that it's just pointless to bother.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Use most of the alphabet, including uncommon letters&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;For similar reasons to above. Uncommon letters (by definition) don't appear in as many words, so your chance of a collision is lower. If you're using 25/26 letters, the odds that another word just happens to have a Z or J in the right spot drops considerably.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Use letter patterns&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Certain digrams/trigrams are used often, others aren't. Other patterns such as vowel-consonant-vowel are very common. This is often the starting point when &lt;em&gt;solving&lt;/em&gt; cryptograms, but you can also take it into consideration when creating them. By using uncommon patterns, you might increase both the difficulty and &quot;uniqueness&quot;.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Test it!&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;If you have the ability to do so, write a program to test many, many combinations against a dictionary. A straight brute-force is probably out of the question, but you could approach it step by step. First, just enter a word or two and have it search for matches. Then once it finds potentials, feed the rest of it in with those known letters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do this by hand if you have strange word lengths or patterns. Find yourself a good dictionary (arranged by letter count), and try to solve one of the more rare words/patterns in as many ways as possible. Then see if there are any likely matches in the rest of the text.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Unique?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;For some cryptograms, there may be (despite your best effort) another possibility or few. That's not really a big deal, since you can give your puzzle a title/clue/theme. For example, many are pithy quotations, so if you &quot;solve&quot; one but it doesn't form coherent sentences, you didn't &lt;em&gt;solve&lt;/em&gt; it.&lt;/p&gt;&#xA;" OwnerUserId="1563" LastActivityDate="2014-12-04T19:35:14.773" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5922" PostTypeId="2" ParentId="5919" CreationDate="2014-12-04T20:04:06.900" Score="10" Body="&lt;p&gt;I would say that there is no good clear cut answer to this question. The best way to get an idea of how difficult the maze is is to have people try the maze and see how long it takes to be completed. Some factors I would consider, however, are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Number of Forks leading into Dead Ends or Loops&lt;/strong&gt;&lt;br&gt;&#xA;The more forks you have the more locations you have to get off of the correct path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Number of Forks along the correct path&lt;/strong&gt;&lt;br&gt;&#xA;If you only have to cross one fork on the ideal path then you only have to get lucky once. The forks work best if they are evenly distributed among the different possible paths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Length of incorrect paths&lt;/strong&gt;&lt;br&gt;&#xA;A short, linear dead end is only a short distraction. Increasing the complexity of incorrect paths can make them even more of a nuisance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Path of the ideal solution&lt;/strong&gt;&lt;br&gt;&#xA;If you start on the left and pretty much only ever have to work your way right in a straight line, then the maze will be really pretty easy. For harder mazes the solution should at some points get close to the end, while at some points wind further and further away from the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Size of the maze&lt;/strong&gt;&lt;br&gt;&#xA;The bigger the maze the more room you have to work with and the more space you have for longer false paths, dead ends, forks etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Shape of the maze&lt;/strong&gt;&lt;br&gt;&#xA;It is my opinion that circular mazes tend to be a little more challenging because the circular paths are a little harder to keep track of than straight paths. Others, however, may have different preferences.&lt;/p&gt;&#xA;" OwnerUserId="3922" LastEditorUserId="3922" LastEditDate="2014-12-04T20:40:44.000" LastActivityDate="2014-12-04T20:40:44.000" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5923" PostTypeId="1" AcceptedAnswerId="5966" CreationDate="2014-12-04T21:23:20.593" Score="20" ViewCount="6206" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Another &lt;em&gt;themed&lt;/em&gt; puzzle for you to enjoy. As with all my puzzles, please provided feedback about pro's and cons of the puzzle. The puzzle is self-contained (the link is just to provide a theme) and I believe it contains all necessary information to find the &lt;em&gt;unique&lt;/em&gt; solution. If you don't think so or if you discover any problems, please let me know.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;After you have made it through the door of the ancient tomb (&lt;a href=&quot;https://puzzling.stackexchange.com/questions/5514/can-you-pass-through-the-ancient-tomb&quot;&gt;puzzle here&lt;/a&gt;), you find yourself in an odd cavern which partly looks natural, partly artificial. From the ceiling hang several stalactites but of a strange, crystalline nature with facets. They also seem to shimmer in different colours.&lt;/p&gt;&#xA;&lt;p&gt;You walk a bit deeper and find a very technical looking panel on top of a short column. The panel has the shape of truncated pyramid and on its top-facet are 36 wheels with a marker which can be set to one of 6 positions.&#xA;The four truncated sides of the pyramid each contain what appears to be a socket of some sort with 6 finger-thick holes in a row.&lt;/p&gt;&#xA;&lt;p&gt;It takes not long, and you find a very modern looking device nearby which has a cable and plug, obviously fitting the sockets. The device has a single button in form of an arrow, and a liquid-crystal-like display without any visible reading. You push the button, but nothing seems to happen. Next you plug the thing into one of the four sockets and press the button again. Suddenly you get a reading of some tree-like symbols on the display - as long as you keep the button pressed.&lt;/p&gt;&#xA;&lt;p&gt;You turn a few of the wheels on the panel (while and also in between using the strange device at the same time), but that does not seem to change any of the readings.&lt;/p&gt;&#xA;&lt;p&gt;Knowing that the ancient people who've built this place seem to be keen on puzzles and testing the skills of archaeologists and grave-robbers alike, you soon decide that this has to be another logic test to be solved. But what is to do? Surely, there have to be more hints somewhere. Sure enough you find three smaller drawing on the right wall of the cavern, and a single, larger one on the other side of the cavern. Will they tell you how you can solve the puzzle?&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;The goal of the puzzle&lt;/h2&gt;&#xA;&lt;p&gt;..is to find a combination of the 36 dials on top of the panel, i.e. a single combination of 36 colours. The only clues you have are the pictures given below and 4 device-readings when you plug in the device on each of the 4 sockets.&lt;/p&gt;&#xA;&lt;p&gt;In your answer, you have to explain &lt;em&gt;why&lt;/em&gt; you've chosen a particular combination. Please give your reasoning as detailed as possible.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;The socket &amp;amp; the device:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vnZKp.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VUSoQ.jpg&quot; alt=&quot;The measuring device&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;The 4 display readings on the device when plugged in as indicated, respectively:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rMs4U.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;The 3 drawings on the left wall of the cavern:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TkKWR.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wyXzd.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Kitz8.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;The two, large drawings on the right wall of the cavern:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pA7yt.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UvOxW.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;Spoilers &amp;amp; Hints&lt;/h2&gt;&#xA;&lt;p&gt;If you feel really stuck (&lt;em&gt;and only then&lt;/em&gt;) you may look at the following spoiler hints. They are meant to guide your thoughts in the right direction.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;How do I even start thinking about this?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; While this is an abstract riddle, it's &lt;em&gt;themed&lt;/em&gt; in the real world. It can sometimes be helpful to think with some physics in mind.&#xA; On the other hand, the crystals in this cave are &lt;em&gt;of a strange, crystalline nature&lt;/em&gt; as mentioned in the introduction. Don't expect them to act like simple coloured glass.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;So many arrows - what could they mean?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Did you notice that there are 3 different arrow shapes? This may not be accidental.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;All those dial combinations in the image, what should they mean?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Have you read the story? You've already tried turning those dials and seeing if it makes a difference - it doesn't. But your pretty convinced that a final, correct setting will be what you're after...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;Okay, I have found the solution - what now?&lt;/h2&gt;&#xA;&lt;p&gt;Please post the solution &lt;strong&gt;and how you derived at it&lt;/strong&gt; as an answer. Please use spoiler tags to hide the essential parts, but do not use one big single spoiler. There should be readable text interrupted by spoiler sections, similar to what you see just below here:&lt;/p&gt;&#xA;&lt;p&gt;The solution&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; which consists of 36 dial positions on the socket&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;could potentially be noted down as&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; a 6x6 grid of dial positions I to VI using the coordinate notation as in the following image:&#xA; &lt;img src=&quot;https://i.stack.imgur.com/t3QVU.jpg&quot; alt=&quot;Solution notation system&quot; /&gt;&lt;br&gt;&#xA; Then produce the 6x6 grid with the dial-positions in place.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;References&lt;/h2&gt;&#xA;&lt;p&gt;&lt;strong&gt;WARNING&lt;/strong&gt; looking at the references will be a strong and fundamental spoiler, as a large part of the puzzle is based on finding out &lt;em&gt;how&lt;/em&gt; this represents a puzzle.&lt;/p&gt;&#xA;&lt;p&gt;This puzzle was inspired by &lt;br&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; the &amp;quot;Towers&amp;quot; puzzle one can find elsewhere.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;In my case, I've seen such puzzles in the app &lt;br&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;strong&gt;Time Killers&lt;/strong&gt; by &lt;em&gt;Andrea Sabbatini&lt;/em&gt;. &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;However, the (accepted) solution by &lt;em&gt;Misch&lt;/em&gt; also provides a link to &lt;br&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;a href=&quot;http://www.chiark.greenend.org.uk/%7Esgtatham/puzzles/&quot; rel=&quot;noreferrer&quot;&gt;this site&lt;/a&gt;. &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I do not know about any other &amp;quot;original&amp;quot; copyrights, but if you alert me to any, I'll put them here.&lt;/p&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-12-09T17:44:15.677" Title="Can you solve this logic puzzle in disguise? The color code behind the tomb" Tags="|logical-deduction|visual|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5925" PostTypeId="1" AcceptedAnswerId="5926" CreationDate="2014-12-04T21:33:20.507" Score="0" ViewCount="528" Body="&lt;p&gt;&lt;pre&gt;This is another software riddle. It's a game... Enjoy!&lt;/pre&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I run off of a cup of joe&lt;br&gt;&#xA;I virtualized the sandbox&lt;br&gt;&#xA;And brought construction to the poor&lt;br&gt;&#xA;Some exploited me, and rose to wealth&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My founder once was loyal&lt;br&gt;&#xA;But now he left, a billionare&lt;br&gt;&#xA;The company that bought me-&lt;br&gt;&#xA;Both small and big-&lt;br&gt;&#xA;With a worldwide influence&lt;br&gt;&lt;/p&gt;&#xA;" OwnerUserId="5368" LastActivityDate="2014-12-05T14:10:49.623" Title="Software Riddle [Part 3!]" Tags="|riddle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5926" PostTypeId="2" ParentId="5925" CreationDate="2014-12-04T21:47:55.113" Score="6" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Minecraft&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I run off of a cup of joe&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Minecraft is written in Java (a language and also another name for coffee)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I virtualized the sandbox&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You play in a virtual sandbox world.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And brought construction to the poor&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The game lets anyone create/construct things in a virtual world.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some exploited me, and rose to wealth&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can apparently make money selling Minecraft perks and services.  (&lt;a href=&quot;http://www.rockpapershotgun.com/2014/06/07/minecraft-paid-servers/&quot;&gt;source&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My founder once was loyal, But now he left, a billionare&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Markus Persson founded Mojang, the company that created Minecraft.  He stayed for 4 years before selling out to Microsoft for $2.5 billion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The company that bought me-, Both small and big-, With a worldwide influence&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Microsoft is a huge company with worldwide influence.  &quot;Micro&quot; means small.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2014-12-04T21:47:55.113" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5927" PostTypeId="1" AcceptedAnswerId="5929" CreationDate="2014-12-04T21:57:16.650" Score="-2" ViewCount="410" Body="&lt;p&gt;&lt;pre&gt; This riddle is a computer language. Enjoy!&lt;/pre&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My name is a dictionary word&lt;br&gt;&#xA;Built on microcomputers&lt;br&gt;&#xA;I'm general purpose&lt;br&gt;&#xA;My syntax is short, but easy&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I once was the &quot;beginner&quot; language&lt;br&gt;&#xA;My GOTO is very popular&lt;br&gt;&#xA;Made in 1964-by Hampshians&lt;br&gt;&#xA;I was very high-level&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Tear every line apart for full credit!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5368" LastEditorUserId="5368" LastEditDate="2014-12-04T22:23:29.180" LastActivityDate="2014-12-04T22:23:55.343" Title="Software Riddle [Part 4!]" Tags="|riddle|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5928" PostTypeId="2" ParentId="5905" CreationDate="2014-12-04T22:09:06.950" Score="10" Body="&lt;p&gt;One class of ciphers that can be identified in an interesting way is the &lt;strong&gt;polyalphabetic ciphers&lt;/strong&gt;. Of these the &lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Vigen%C3%A8re_cipher&quot;&gt;Vigenère cipher&lt;/a&gt;&lt;/strong&gt; is the best known. They cycle through different alphabets periodically. This period can be identified, which will both identify the cipher as polyalphabetic, and also provide a starting point for decrypting it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An effective way to break a polyalphabetic cipher is by using a technique known as index of coincidence. This is a measure of how frequently a pair of independent characters happen to be identical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example if a text has 100 characters, there will be 100*99/2 = 4950 ways to pick two characters. One can count how large a percentage of these 4950 pairs happen to be identical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If 100 characters were encrypted using a polyalphabetic cipher that cycle through 10 different alphabets, then each alphabet would be used 10 times. A pair of characters encrypted using different alphabets would be mostly independent. They are not very likely to be identical. But for a pair of characters encrypted using the same alphabet, they are more likely to be identical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one makes a guess at the period being 10, one can compute percentage of identical characters for only those pairs using the same alphabet. There are 10 alphabets and for each alphabet there are 10*9/2 possible pairs of characters, that's a total of 450 pairs, which is enough for the percentage to still be statistically significant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do that calculation with a correct period, you get more identical pairs than if you do the calculation with an incorrect period. So you can simply try different period lengths, and for each length compute the frequency of identical pairs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you plot those frequencies, you should see a visible spike when you hit the correct period. If the guessed period and the correct period are different but share a prime factor, there will be a small spike. For example if the correct period was 10, you would see small spikes at 2, 4, 5, 6, 8, 10, 12, 14, 15.&lt;/p&gt;&#xA;" OwnerUserId="1619" LastEditorUserId="1619" LastEditDate="2014-12-28T00:22:49.077" LastActivityDate="2014-12-28T00:22:49.077" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5929" PostTypeId="2" ParentId="5927" CreationDate="2014-12-04T22:13:59.450" Score="4" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BASIC&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My name is a dictionary word&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Basic is an English word&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Built on microcomputers&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It was widely used on microcomputers in the 70s and 80s.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm general purpose&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Can be used to write all kinds of programs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My syntax is short, but easy&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are not too many keywords, and they are easy to learn.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I once was the &quot;beginner&quot; language&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BASIC stands for Beginner's All-purpose Symbolic Instruction Code.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My GOTO is very popular&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The GOTO statement is used in many other languages too.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Made in 1964-by Hampshians&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Invented by Kemeny and Kurtz of Dartmouth College in New Hampshire.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was very high-level&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A high level programming language that abstracts you away from computer specific details.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4450" LastEditDate="2014-12-04T22:23:55.343" LastActivityDate="2014-12-04T22:23:55.343" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5931" PostTypeId="1" AcceptedAnswerId="5935" CreationDate="2014-12-04T23:51:49.660" Score="-11" ViewCount="744" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am a site many use to play games.&lt;br&gt;&#xA;  Others get my game engine to develop them.&lt;br&gt;&#xA;  When developing you use the language Lua.&lt;br&gt;&#xA;  I have many members signed up to use me.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the question is;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What the heck am I?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2016-04-24T19:43:12.000" Title="Software Riddle [Part 5!]" Tags="|riddle|" AnswerCount="2" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5932" PostTypeId="2" ParentId="5931" CreationDate="2014-12-05T00:11:17.227" Score="3" Body="&lt;p&gt;The answer might be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Steam&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a site many use to play games.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many people play games on the Steam platform.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Others get my game engine to develop them.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This may be referring to the Leadwerks Game Engine.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When developing you use the language Lua.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Leadwerks supports Lua as a development language.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have many members signed up to use me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Steam has millions of members.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4450" LastEditDate="2014-12-05T00:47:53.747" LastActivityDate="2014-12-05T00:47:53.747" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5933" PostTypeId="1" AcceptedAnswerId="5936" CreationDate="2014-12-05T00:18:20.583" Score="5" ViewCount="6045" Body="&lt;p&gt;A &lt;em&gt;magic square&lt;/em&gt; consists of the numbers $1,2,\ldots,m^2$ placed into $m\times m$ square grid, so that every row, column, and both diagonals have the same sum. The &lt;em&gt;magic constant&lt;/em&gt; of the square is this common sum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the dimensions of an $m\times m$ square, can you determine what the magic constant is? If so, how can you do that? If not, please explain why it isn't possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, here are some magic square &lt;em&gt;&quot;magic constants&quot;&lt;/em&gt;;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$1\times 1\,\to\, 1 $&lt;br&gt;&#xA;  $2\times 2 \,\to nil$&lt;br&gt;&#xA;  $3\times 3\,\to \,15$&lt;br&gt;&#xA;  $4\times4\,\to\, 34$&lt;br&gt;&#xA;  $5\times 5 \,\to\, 65$    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How can you determine the $6\times 6$, $7\times7$, etc. just from the dimensions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Please note; I'm not asking for how to solve an $m\times m$ magic square, nor do I want a &quot;trial and error&quot; answer. I want a mathematical based answer explaining how to figure out the magic constant of a magic square with the dimensions $m\times m$.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2627" LastEditDate="2014-12-05T01:26:54.677" LastActivityDate="2014-12-05T04:26:24.087" Title="How to Determine a Magic Constant in a Magic Square?" Tags="|mathematics|magic-square|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5934" PostTypeId="1" AcceptedAnswerId="5940" CreationDate="2014-12-05T00:25:21.317" Score="12" ViewCount="4449" Body="&lt;p&gt;My definition of a magic rectangle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Any $m \times n$ rectangle where $m \ne n$ and all the numbers $1, 2, 3,\dots, mn$ fit into the rectangle. All horizontal lines, vertical lines, and diagonal lines (albeit not the same length) add up to the same number called a &lt;em&gt;&quot;magic constant&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Do any magic rectangles exist? If so, what are some examples? Please include dimensions, magic constant, and if possible, the whole rectangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;BONUS: How can you determine a rectangle's magic constant from its $m \times n$ dimensions?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="4204" LastEditDate="2014-12-05T01:18:08.010" LastActivityDate="2017-10-31T16:54:51.003" Title="Does a magic rectangle exist?" Tags="|mathematics|magic-square|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5935" PostTypeId="2" ParentId="5931" CreationDate="2014-12-05T00:28:09.100" Score="1" Body="&lt;p&gt;Is the answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; www.roblox.com&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Roblox&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; or &lt;a href=&quot;http://www.roblox.com&quot; rel=&quot;nofollow&quot;&gt;the site itself&lt;/a&gt; for why this fits the clues:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; people use it to play and develop games (development &quot;using a sandboxed edition of Lua 5.1&quot;), and many users are signed up to it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="2383" LastEditDate="2014-12-05T02:42:42.770" LastActivityDate="2014-12-05T02:42:42.770" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5936" PostTypeId="2" ParentId="5933" CreationDate="2014-12-05T00:32:27.380" Score="8" Body="&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Magic_square&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;, you're talking about a &lt;em&gt;Magic Sum&lt;/em&gt;.&#xA;Let's say the formula for finding this number is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\text{MagicSum} = \begin{cases} 1 &amp;amp; \mbox{if }n = 1 \\ n\Big( \frac{n^2 +1}{2}\Big) &amp;amp; \mbox{if }n&amp;gt; 2  \\ \text{undefined} &amp;amp; \mbox{else}\end{cases} \\ $&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorDisplayName="user20" LastEditDate="2014-12-05T04:26:24.087" LastActivityDate="2014-12-05T04:26:24.087" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5937" PostTypeId="1" AcceptedAnswerId="6027" CreationDate="2014-12-05T00:52:53.303" Score="8" ViewCount="945" Body="&lt;p&gt;When solving a logic puzzle, at times, one &lt;em&gt;must&lt;/em&gt; make an assumption in order for a solution to be found. For example, in &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2484/knight-knave-and-joker-puzzle&quot;&gt;this&lt;/a&gt; perfectly good logic puzzle, it had to be assumed that the characters knew amongst themselves who's who (i.e. the Knight knows who the Joker and Knave are, etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the technical term (a noun) which represents a necessary assumption, without which, a solution may be impossible? I have heard this term at some point (somewhere here on Puzzling.SE), but it has escaped me — if it helps, I think it was something like 'zero assumption' (90% sure it was more than one word) and it had a Wikipedia entry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I'm not entirely certain the puzzle I mentioned &lt;em&gt;requires&lt;/em&gt; any extra assumptions - it's a perfectly good logic puzzle.&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-09-02T21:03:50.180" Title="Assumption Needed when Solving Puzzle" Tags="|logical-deduction|puzzle-theory|" AnswerCount="3" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="5938" PostTypeId="2" ParentId="5933" CreationDate="2014-12-05T00:53:12.273" Score="10" Body="&lt;p&gt;Given an $n\times n$ magic square, write $M_n$ for its magic constant. There are $n$ rows, each of which has sum $M_n$, so the sum of all the entries in the square is $n \cdot M_n$. Each whole number between $1$ and $n^2$ appears once, so&#xA;$$&#xA;\begin{gather}&#xA;n\,M_n=1+2+\ldots+n^2=\frac{n^2(n^2+1)}{2},\\&#xA;M_n = \frac{n(n^2+1)}{2}.&#xA;\end{gather}&#xA;$$&#xA;In particular, every $n\times n$ magic square has the same constant.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastActivityDate="2014-12-05T00:53:12.273" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="5939" PostTypeId="2" ParentId="5919" CreationDate="2014-12-05T01:01:13.290" Score="2" Body="&lt;p&gt;The best (most accurate and thorough) way to rate how difficult a maze is to perform a controlled study.  Unfortunately, this requires a lot of time and people, the more the better.  Put each person in a room by themself and time how long it takes them to complete the maze.  The longer it takes, the harder the maze is.  Make sure you find people who are interested in solving mazes.  For better results, perform some kind of baseline test with which to weight the results.  For instance, give each person what you consider an easy maze, and see how long each person takes to solve it.  Then do the same with what you consider a hard maze.  Those who solve the &quot;hard&quot; maze faster should have longer times weighted more, because they have shown they are good at solving hard mazes.  Conversely those who solve the &quot;easy&quot; maze faster should have shorter times weight more, because they have shown that they are good at solving easy mazes.  For better results, consult a statistician regarding the specifics about the tests you want to run.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This answers the question of how to determine how difficult mazes are that you know the solution to. Once you have got enough data points you can define various maze metrics and and they objectively score each maze based on these metrics.  As another answer suggests, metrics might include the size and shape of the maze, and various topological conditions.  Run correlations between each metric and the relative score obtained from the first step.  Once you have these numbers you can then take mazes which you don't know the solution to, measure the metrics for those mazes, and apply the previously determined correlations in order to predict the difficulty levels.  If you have the time and money, you can refine your model over time by running additional studies against these predictions, determine how well they predict the results, tweak your metrics, and update your models accordingly.&lt;/p&gt;&#xA;" OwnerUserId="2718" LastActivityDate="2014-12-05T01:01:13.290" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5940" PostTypeId="2" ParentId="5934" CreationDate="2014-12-05T01:09:38.900" Score="51" Body="&lt;p&gt;No such rectangle exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have a rectangle with $m$ rows and $n$ columns. If every row adds up to some magic value $M$, then the number obtained by adding together &lt;em&gt;every&lt;/em&gt; cell in the rectangle must be $m \times M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, if every column adds up to $M$, then the value obtained by adding together every cell in the rectangle must also equal $n \times M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $m \times M = n \times M$. This can only be true if $m = n$, which corresponds to a square.&lt;/p&gt;&#xA;" OwnerUserId="40" LastActivityDate="2014-12-05T01:09:38.900" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="5942" PostTypeId="1" AcceptedAnswerId="6452" CreationDate="2014-12-05T01:26:23.027" Score="-5" ViewCount="741" Body="&lt;p&gt;&lt;strong&gt;Magic Square:&lt;/strong&gt; &lt;em&gt;An nxn square where every horizontal,vertical, and diagonal line all add up to the same number.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Magic Constant:&lt;/strong&gt; &lt;em&gt;The number which every line with a magic square adds up to&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So most magic squares, have a &lt;em&gt;magic constant&lt;/em&gt;. &lt;strong&gt;But some do not.&lt;/strong&gt; A fairly trivial answer is 0x0. Where no numbers are present, therefore it cannot sum up into any number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A less than trivial answer&lt;/em&gt; but &lt;em&gt;not complicated&lt;/em&gt; answer is 2x2. No possible configuration of 1,2,3,4 within a 2x2 block can have all lines add up to the same number. &lt;strong&gt;If it were possible it would theoretically be 5&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So besides 0, and 2 are there any other &lt;strong&gt;positive dimensions&lt;/strong&gt; for a magic square that do not have a magic constant? Technically making all possible configurations invalid magic squares?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mathematical Proof is Appreciated!&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2484" LastEditDate="2014-12-05T06:36:13.167" LastActivityDate="2014-12-28T21:29:31.603" Title="How Many Undefined Magic Constants are there?" Tags="|mathematics|magic-square|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="5943" PostTypeId="1" CreationDate="2014-12-05T02:07:34.007" Score="1" ViewCount="425" Body="&lt;p&gt;(&lt;em&gt;Don't worry, this is the last magic square themes question I'll be posting (as far as I know), so I picked a challenging one!&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, let's define what a magic square is;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A magic square is an $n\times n$ grid filled with the numbers $1$ through $n\times n$, where every horizontal, vertical, and diagonal line adds up to be the same number, this number is called the &lt;strong&gt;magic constant&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So now let's generalize that...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A magic $m$-cube is an $n^m$ grid filled with the numbers $1$ through $n^m$, where $m &amp;gt; 1$ and each line on every axis adds up to be the same number.  This number is called the &lt;strong&gt;magic constant&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So now imagine the penteract being a 5-cube, so that would mean a magic penteract is an $n\times n\times n\times n\times n$ grid where each line on every axis ($v, w, x, y, z$) all adds up to the one and only &lt;strong&gt;Magic Constant!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your job? Easy! Determine the $3\times 3\times 3\times 3\times 3$ magic penteract, and its magic constant!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck!&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="4421" LastEditDate="2015-12-11T22:30:56.823" LastActivityDate="2015-12-11T22:30:56.823" Title="Magic Penteract!" Tags="|mathematics|magic-square|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5945" PostTypeId="2" ParentId="5865" CreationDate="2014-12-05T03:13:23.297" Score="5" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A degree?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I go by many a name,&#xA;Including one that's quite a shame.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There are many types of degrees, but the Associate of Social Sciences degree stands out in particular.  If Associate of Science is A.S., and Associate of Arts is A.A., then...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can take you to an exact location,&#xA;But sometimes only to a nearby destination.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A degree of longitude is 0 km at the poles (pinpoint accuracy) and becomes over 100km at the equator, so your geographic coordinates in degrees can sometimes specify your exact location or only an approximation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If great power is your wish,&#xA;I'll grant it but your money will vanish.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A degree can give you great power, but costs a lot and may put you in debt.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Substituting for that which can be negative,&#xA;I'll always remain completely positive.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Degrees Celsius (can be positve or negative) to degrees Kelvin (always positive).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be very important,&#xA;While at the same time very insignificant.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A degree is very important to help you find a job, at the same time everyone else also has a degree so it has become insignificant.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I contain many things,&#xA;I simultaneously contain nothing.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A degree contains many years of hard work, but is just a piece of paper.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can signal the beginning,&#xA;But I can also signal the ending.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A degree signals the beginning of your career and also your graduation, the end of your student life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6178" LastActivityDate="2014-12-05T03:13:23.297" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5946" PostTypeId="2" ParentId="5943" CreationDate="2014-12-05T03:43:32.447" Score="0" Body="&lt;p&gt;Marián Trenkler constructs magic $m$-cubes of arbitrary size in the papers [1] and [2]. The following theorem is in [2]:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem&lt;/strong&gt;: A magic $m$-cube of size $\underbrace{n\times\ldots\times n}_m$ exists if and only if $m\geq 2$ and $n\neq2$, or $m=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that a $3\times3\times3\times 3\times 3$ magic penteract exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Trenkler, Marián. Magic $p$-dimensiona cubes of order $n\neq 2\mod 4$. &lt;em&gt;Acta Arith.&lt;/em&gt; 92 (2000), no. 2, 189-194.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] Trenkler, Marián. Magic $p$-dimensiona cubes. &lt;em&gt;Acta Arith.&lt;/em&gt; 96 (2001), no. 4, 361-364.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastActivityDate="2014-12-05T03:43:32.447" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5947" PostTypeId="2" ParentId="5943" CreationDate="2014-12-05T03:48:32.247" Score="3" Body="&lt;p&gt;You are describing a &lt;a href=&quot;http://en.wikipedia.org/wiki/Magic_hypercube&quot; rel=&quot;nofollow&quot;&gt;magic hypercube&lt;/a&gt;. For a magic hypercube's Magic Sum, have $M_k(n)$, where $k$ is the dimensionality (5 in this case), and $n$ is the size of each row/column. In general this gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M_k(n) = n\Big( \frac{n^k +1}{2}\Big)$ for n &gt; x where x is particular to each k.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in your formulation, you need:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$MagicSum = 3\Big( \frac{3^5 +1}{2}\Big) = 366$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://www.magic-squares.net/c-t-htm/t_overview.htm#THREE%20D&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;pre&gt;The SMALLEST SIMPLE MAGIC HYPERCUBE of dimension 5 is of Order 3. &#xA;Published by Hendricks in May 1962. &#xA;All the numbers from 1 TO 243 are arranged in such a way that the Magic Sum is: &#xA;366 in 421 ways.&#xA;&#xA;81 rows            (parallel to x-axis) &#xA;81 columns (parallel y-axis) &#xA;81 pillars (parallel to z-axis) &#xA;81 files            (parallel to w-axis)&#xA;81 posts            (parallel to v-axis) &#xA;16 pentagonals (continuous) &#xA;Altogether 421 ways.&#xA;&#xA;The number of dimension 5 hypercubes is not known (for any order), &#xA;but there are 3840 variations of each due to rotations and reflections.&lt;/pre&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-12-05T04:06:06.260" LastActivityDate="2014-12-05T04:06:06.260" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5948" PostTypeId="1" AcceptedAnswerId="5949" CreationDate="2014-12-05T04:15:24.670" Score="2" ViewCount="562" Body="&lt;p&gt;I saw one in Key West, and then I saw one in the Pacific Northwest. It has seven letters in its name, and uses each of the 5 standard vowels exactly once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What did I see?&lt;/p&gt;&#xA;" OwnerUserId="6094" LastEditorUserId="2521" LastEditDate="2015-04-11T01:31:24.733" LastActivityDate="2015-04-11T01:31:24.733" Title="What did I see in Key West and the Pacific Northwest?" Tags="|riddle|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5949" PostTypeId="2" ParentId="5948" CreationDate="2014-12-05T04:23:28.873" Score="4" Body="&lt;p&gt;Is it a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sequoia?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Only word I know that fits the description.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A &lt;a href=&quot;https://en.wikipedia.org/wiki/Sequoia_%28genus%29&quot; rel=&quot;nofollow&quot;&gt;Sequoia&lt;/a&gt; is a type of tree that grows in the Pacific Northwest.  It is also the name of a &lt;a href=&quot;https://en.wikipedia.org/wiki/Toyota_Sequoia&quot; rel=&quot;nofollow&quot;&gt;Toyota SUV&lt;/a&gt;.  Whether this vehicle is particularly popular in Key West, I do not know.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6178" LastEditorUserId="6178" LastEditDate="2014-12-05T17:37:34.280" LastActivityDate="2014-12-05T17:37:34.280" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5950" PostTypeId="2" ParentId="5633" CreationDate="2014-12-05T08:49:59.863" Score="3" Body="&lt;p&gt;&lt;strong&gt;Read to the end. It's worth the journey.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I decide if sweet or stinky&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is a letter which makes the difference between &quot;SWEET&quot; and something stinky (&quot;SWEAT&quot;). It is A (or E).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For most locks I'm the key&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;E&quot; and &quot;A&quot; seen as keys on a keyboard.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Go ahead and keep on piling&#xA;because even a child can find me &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Keep on piling means collecting/&lt;strong&gt;counting&lt;/strong&gt; which every child can do. &quot;E&quot; happens to be the &lt;a href=&quot;http://jumk.de/wortanalyse/word-analysis.php&quot; rel=&quot;nofollow&quot;&gt;most used letter&lt;/a&gt; in this riddle. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's no vanishing&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Keep this is mind... we need to get rid of &quot;ITSNO&quot; from somewhere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;no clues here&#xA;no hiding there&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To be taken literally.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's no use&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Oh, let's get rid of it now. &quot;ITSNOUSE&quot; -&gt; &quot;USE&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A man of words&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An educated man, an author or poet, of the Enlightenment era (see title) who is German (see preface). &lt;a href=&quot;http://en.wikipedia.org/wiki/German_literature#The_Enlightenment&quot; rel=&quot;nofollow&quot;&gt;See here for a list.&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;participating right now&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Participating presently... There aren't that many German Enlightenment authors, &quot;LESSING&quot; is one of them and his name reminds me of a present participle (it is).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Teach him a lesson&#xA;and break his literal vow&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This is too tricky, we remove the &quot;ING&quot;. Because it's what makes &quot;LESSING&quot; a verbal part - it is his vow, his constraint. So we have &quot;LESS&quot; (there is also a small hint in &lt;em&gt;Teach him a lesson&lt;/em&gt; giving you lesson &gt;&gt; &lt;em&gt;lessen&lt;/em&gt; = take something away).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Subdued by ships&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ships subdue the &quot;SEA&quot;/&quot;C&quot; and &quot;PEOPLE&quot; perhaps.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Along lonely isles&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Simply the letter C (=&quot;sea&quot;) along lonely isles &quot; : &quot;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I may result &#xA;in stunning smiles&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;C: The word &lt;em&gt;may&lt;/em&gt; also allows for something that is not smiling: :C &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is to say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We have &quot;C&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So we end up with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;E USELESS C&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now taking away what's useless,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;EC&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which sounds like: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;EASY :D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, look in the preface again:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, as you might have thought...&lt;strong&gt;&lt;em&gt;it's easy.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Lynch mob time!!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="5106" LastEditDate="2014-12-05T09:08:05.287" LastActivityDate="2014-12-05T09:08:05.287" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5951" PostTypeId="2" ParentId="5585" CreationDate="2014-12-05T10:05:24.743" Score="9" Body="&lt;p&gt;Continuing with Victor's excellent analysis:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First step:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Replace 0005200056 with &lt;strong&gt;0&lt;/strong&gt; and 0005200057 with &lt;strong&gt;1&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second step:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Group octets of zeros and ones from the first step to form ASCII characters. The result is:&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;TFIL E WCAT HOLN UQG/VCDL HHB SCFH JRHM IDZX V YDT DFH&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Third step:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Interpret the results of the second step as Morse code, where:&lt;br&gt;&#xA; - letters formed with only straight lines are dots (AEFHIKLMNTVWXYZ)&lt;br&gt;&#xA; - letters formed with curves are dashes (BCDGJOPQRSU)&lt;br&gt;&#xA; - and the slash is a space&lt;br&gt;&#xA;&lt;br&gt;&#xA; The solution is:  &lt;strong&gt;HELLO PUZZLERD&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I assume the D at the end is a smiley with no eyes.  &lt;/p&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2014-12-05T10:41:11.583" LastActivityDate="2014-12-05T10:41:11.583" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5952" PostTypeId="2" ParentId="1957" CreationDate="2014-12-05T10:57:09.520" Score="5" Body="&lt;p&gt;It is simple. Put&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th&gt;&lt;/th&gt;&#xA;&lt;th&gt;&lt;/th&gt;&#xA;&lt;th&gt;&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;1&lt;/td&gt;&#xA;&lt;td&gt;2&lt;/td&gt;&#xA;&lt;td&gt;3&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;4&lt;/td&gt;&#xA;&lt;td&gt;5&lt;/td&gt;&#xA;&lt;td&gt;6&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;7&lt;/td&gt;&#xA;&lt;td&gt;8&lt;/td&gt;&#xA;&lt;td&gt;9&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;…then make cross lines as shown in the picture below. Then it becomes 3x3 boxes. For the remaining boxes, put the number which is opposite to the box.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hI25c.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hI25c.png&quot; alt=&quot;&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6198" LastEditorUserId="69582" LastEditDate="2020-12-30T19:16:26.640" LastActivityDate="2020-12-30T19:16:26.640" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="5956" PostTypeId="1" AcceptedAnswerId="5959" CreationDate="2014-12-05T13:35:35.663" Score="17" ViewCount="1802" Body="&lt;p&gt;In the image you can see twelve mutation chambers (A to L).&#xA;Each alien on the left side of the aliens goes through the three mutation chambers of its right side. The alien on the right side is the result after the three mutations.&#xA;So, the alien left of chamber A goes through A, B and C. The result is the one right of C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same goes for the aliens op top. Every alien on the top of the image goes through the four mutation chambers below it. The one above A goes through A, D, G and J. The alien below chamber J is the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every chamber applies exactly one mutation (make a head square, ...).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These mutations can occur: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Add antennas&lt;/li&gt;&#xA;&lt;li&gt;Remove antennas&lt;/li&gt;&#xA;&lt;li&gt;Add arms&lt;/li&gt;&#xA;&lt;li&gt;Remove arms&lt;/li&gt;&#xA;&lt;li&gt;Add legs&lt;/li&gt;&#xA;&lt;li&gt;Remove legs&lt;/li&gt;&#xA;&lt;li&gt;Make square head&lt;/li&gt;&#xA;&lt;li&gt;Make round head&lt;/li&gt;&#xA;&lt;li&gt;Make square body&lt;/li&gt;&#xA;&lt;li&gt;Make round body&lt;/li&gt;&#xA;&lt;li&gt;Walk upright&lt;/li&gt;&#xA;&lt;li&gt;Walk on 4 legs (2 arms and 2 legs)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Each of these mutations can occur as many times as you want. It is not said that every mutation described above will occur for sure. This means that multiple chambers can remove leggs and that adding arms is done nowhere (e.g.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5Lc7c.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you tell me which room provides which mutation? If you can tell me, tell me in spoiler tags!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://www.raadselweb.net/index.php?option=com_content&amp;amp;view=article&amp;amp;id=165:aliens&amp;amp;catid=14:logicaraadsels&amp;amp;Itemid=2&quot;&gt;Raadselweb.net&lt;/a&gt; (translated from Dutch to English by me)&lt;/p&gt;&#xA;" OwnerUserId="420" LastEditorUserId="2518" LastEditDate="2014-12-05T15:22:55.907" LastActivityDate="2014-12-06T19:24:51.543" Title="Mutation of aliens" Tags="|logical-deduction|visual|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5958" PostTypeId="2" ParentId="5925" CreationDate="2014-12-05T14:10:49.623" Score="4" Body="&lt;p&gt;Quy, I would suggest obscuring the answer more. By saying it's software and a game, you make it obvious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, say something like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The fruit of the brew of the bean, a multiverse was born. Though all of the gods came to marvel, the creator abandoned it. Twas cheapened by the demons exploiting its denizens and the dominator whose glass fingers gently pull strings. Now, the world is changed. Inside of boxes and pockets, the multiverse is broken and diasporic. Brown, Grey, Silver, Gold, and Blue. The colors run pale as meaning bleeds away.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Keep at it, keep improving, and keep enjoying the artform. See things in abstractions. Microsoft is known for Windows OS, therefore &quot;glassy&quot; and gentle meaning soft. Try and use words that suggest a wildly different answer. Riddle about a goat? Use aquatic terms. Riddling about sunshine? Use math terms. Mix it up and make the community sweat for a day or two coming up with the answer. Finish a riddle, then start over. Change the tone, the language, and anything else you can. Only post second or third drafts, and only post those if they're awesome.&lt;/p&gt;&#xA;" OwnerUserId="5626" LastActivityDate="2014-12-05T14:10:49.623" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5959" PostTypeId="2" ParentId="5956" CreationDate="2014-12-05T14:11:05.263" Score="4" Body="&lt;p&gt;Think I got it...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  A = make head square&lt;br&gt;&#xA;  B = add legs&lt;br&gt;&#xA;  C = make body square&lt;br&gt;&#xA;  D = remove arms&lt;br&gt;&#xA;  E = add antennae&lt;br&gt;&#xA;  F = walk upright&lt;br&gt;&#xA;  G = walk on all fours&lt;br&gt;&#xA;  H = make head round&lt;br&gt;&#xA;  I = remove antennae&lt;br&gt;&#xA;  J = add arms&lt;br&gt;&#xA;  K = walk upright&lt;br&gt;&#xA;  L = make head square&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6201" LastEditorUserId="5044" LastEditDate="2014-12-05T14:15:43.230" LastActivityDate="2014-12-05T14:15:43.230" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="5960" PostTypeId="1" AcceptedAnswerId="9598" CreationDate="2014-12-05T14:23:00.607" Score="10" ViewCount="1396" Body="&lt;p&gt;&lt;strong&gt;Flight Incorporated has just recently posted these advertisements:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h2&gt;Buying Fuel for High Prices!&lt;/h2&gt;&#xA;  &#xA;  &lt;p&gt;Buying fuel for our spaceships&lt;br&gt;&#xA;  Fuel in possession/fuel needed in gallons: $\Delta = 1000/\infty$&lt;br&gt;&#xA;  Current Bank Balance: $100,000,000,000.00&lt;br&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Required Balance = $\infty$ as fuel required is infinite&lt;br&gt;&#xA;  We will pay 1 gram of AN:92 per gallon of fuel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h2&gt;Oxygen needed!&lt;/h2&gt;&#xA;  &#xA;  &lt;p&gt;We need compressors... and oxygen tanks! You will be paid in brass-plated troughs with your name on a plate. Under the plating is a layer of plastic. Engraved on the bottom is our company name and logo.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The following was posted on Flight Inc.'s website.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h2&gt;Pilot's Guide&lt;/h2&gt;&#xA;  &#xA;  &lt;p&gt;Remember to strap in your seat belts! Mass and speed increase simultaneously. Time is dilated. Communications will be offline, as no form of communication travels fast enough. Prepare for a breathtaking tunnel view.&lt;br&gt;&#xA;  Time to destination is: $cyears=year$&lt;br&gt;&#xA;  $C$ is the undisputable speed limit, proposed by the maker of $e=mc^2$&#xA;  What we're doing is impossible, but that's never stopped us!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h2&gt;Technical Guide&lt;/h2&gt;&#xA;  &#xA;  &lt;p&gt;Classical engine runs off of hydrogen. Liquid fueled rocket needs two tanks for propellant and oxidizer. System cannot be distorted by mass increase.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What do they need the fuel for and what are they being paid in?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: No knowledge of science fiction is required to solve this.&#xA;Tear every line apart :)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h2&gt;Trouble deciphering our ads?&lt;/h2&gt;&#xA;  &#xA;  &lt;p&gt;AN is an abbreviation for something related to materials. (Atomic Number)&#xA;  Basic science knowledge is required. We don't buy from people who don't know science. We love the periodic table... Who cares about speed limits, even those universe-wide?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h2&gt;Hints&lt;/h2&gt;&#xA;  &#xA;  &lt;p&gt;AN is an abbreviation for Atomic Number.&#xA;  This puzzle is related to Einstein, relativity, and classical physics.&#xA;  The speed that they are trying to go to (or near it) does weird things to a spacecraft.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Huge Spoiler:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;They are trying to travel at the speed of light. The oxygen is related to combustion and the hydrogen is the fuel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5368" LastEditorUserId="39" LastEditDate="2015-02-28T04:39:36.477" LastActivityDate="2015-02-28T04:39:36.477" Title="Flight Inc., supply order" Tags="|riddle|" AnswerCount="6" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5961" PostTypeId="1" CreationDate="2014-12-05T14:24:56.870" Score="4" ViewCount="801" Body="&lt;p&gt;OK, so I've written &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4075/the-serial-killer-part-1&quot;&gt;several&lt;/a&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4263/the-serial-killer-part-2&quot;&gt;detective&lt;/a&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4449/the-serial-killer-part-3&quot;&gt;puzzles&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 1 turned out to be what I'd consider balanced for difficulty: it had a clearly-correct answer, and took about half a day to be solved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 2 was an unmitigated disaster: it took 20 minutes to solve, and someone else guessed the end-game I had planned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 3 was in between the two - it took about 4 hours to solve.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It's obviously very easy to just carpet-bomb the puzzle with red herrings all over the place, littering it with irrelevant clues. I feel that's a bit blunt though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The converse is just as bad: if the puzzle &lt;strong&gt;only&lt;/strong&gt; contains relevant information then someone can start anywhere and pretty quickly figure out the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm wondering is &lt;strong&gt;given an only-relevant-information detective puzzle, how much distraction should I add&lt;/strong&gt;? As a supplementary question, how can I judge it to try to strike a balance between the solution and the red herrings?&lt;/p&gt;&#xA;" OwnerUserId="2518" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2016-10-02T12:34:06.460" Title="Red herrings in detective puzzles" Tags="|puzzle-creation|mystery|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5962" PostTypeId="2" ParentId="5961" CreationDate="2014-12-05T14:35:31.460" Score="6" Body="&lt;p&gt;I believe &lt;strong&gt;detective puzzles&lt;/strong&gt; are a bit special, as in that it is in their very nature that there are only tiny amounts of &lt;em&gt;relevant&lt;/em&gt; information in what is otherwise a haystack of information. It is all about &lt;em&gt;searching&lt;/em&gt; and &lt;em&gt;finding&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I think a good detective puzzle differentiates false information in &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;just not relevant &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;real red herrings. (apparently relevant but misleading)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For the second category, I think a &lt;em&gt;fair&lt;/em&gt; detective puzzle should allow the red-herrings to be disregarded by logic. i.e. there should be something in the available information which with some thinking allows a clear conclusion: &quot;&lt;em&gt;This can not be it&lt;/em&gt;&quot;.   Note the difference between the conclusion &quot;&lt;em&gt;No, it can not be because...&lt;/em&gt;&quot; and &lt;em&gt;No this is not sufficient proof for...&quot;&lt;/em&gt;  I think it should be possible, to really rule any red-herring &lt;em&gt;out&lt;/em&gt; completely. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As for the question of &lt;em&gt;&quot;how much...&quot;&lt;/em&gt;  , I don't think there is clear answer to that. I think it's experience and feedback from puzzle-solvers which can (iteratively) help you there to find the right balance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My &lt;a href=&quot;https://puzzling.stackexchange.com/questions/5674/the-untimely-death-of-dr-wenzel&quot;&gt;own attempt&lt;/a&gt; was far too easy...&lt;/p&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2014-12-05T14:35:31.460" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5963" PostTypeId="2" ParentId="5960" CreationDate="2014-12-05T14:45:16.767" Score="5" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;They're planning to travel at infinite velocity (aka. warp ten). They're willing to pay the speed of light per gallon for fuel. I know this sounds a bit off, but read the explanation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;As some have theorized, infinite velocity would cause a vessel to exist at all points in space simultaneously. With infinite velocity comes infinite fuel consumption. As a fun side note, existing at all points in the universe at the same time means that all fuel in the universe would exist inside of the fuel tanks. This means that the trip would essentially use up all fuel in the universe and then still not have enough.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where C comes in:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C in physics stands for the speed of light. This reinforces the fact that velocity will pay for fuel, the fuel grabbed from the entire known universe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The only way it would work:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since there isn't enough fuel in the known universe, other universes must be drained as well. In fact, infinite universes must be. Therefore, all fuel in the multiverse would be consumed, infinite mass would create a universal black hole, and the ship would be destroyed unless it was permanently transphasic and had a permanent energy source using energy that the engines can't.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5626" LastActivityDate="2014-12-05T14:45:16.767" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="5964" PostTypeId="2" ParentId="5672" CreationDate="2014-12-05T15:08:39.310" Score="5" Body="&lt;p&gt;A comment on the main question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;There are a few clues to the answer now that are external to the riddle, theres no reason this isnt answered. Its no work of virgil.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It's no work of Virgil made me think immediately of a classical work - a book or author most likely.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(not necessary - and cheeky, but) in OP's profile: &lt;em&gt;&quot;I do enjoy a bit of Dante Alighieri.&quot;&lt;/em&gt; - the hint to look somewhere like a profile (simply &quot;outside&quot;) is in the previous clue &lt;em&gt;&quot;There are a few clues to the answer now that are external to the riddle&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I think OP was sick of this riddle being open for so long. There is no doubt now.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/5914/riddle-safety-in-numbers&quot;&gt;In a different puzzle by the same OP:&lt;/a&gt; &lt;em&gt;&quot;under a septenary of hate and fear&quot;&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dante's Comedy is about 700 years old, and the 7 sins are a major theme. (Again, this is not necessary, but OP is sick of seeing it unanswered - look at the bounty!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are talking about the wonderful, the magical, the most classical:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dante's Divine Comedy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The epitome of an evil the skewed letter would label&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The letter 'X' when skewed gives '+', a Christian crucifix. We are looking for something in Christian folk-lore. Wikipedia states: &quot;Allegorically, the Purgatorio represents the Christian life.&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To many is such truth, not a fable&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There are many who hold these tales to express truth - this may be a metaphor for life (as it is in my opinion), or as a literal depiection of hell.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Held tightly by urges of the seconds&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We are in the midst of sin (according to the story). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Within the sleepless cities that beckons&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The entire Comedy is based in Hell, in 9 layers - the sleepless cities.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Twice I hold the path of tea&#xA;Forever you stay within me&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Path of tea&quot; = a T-intersection. There were junctions explicitly mentioned in the text.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is not glue that binds us&#xA;Nor the substance which I fuss&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You are on your own in Hell. It is your soul, not your body (the substance) which is at stake.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Guarded by a significant worm&#xA;Your sight has gone, time to squirm&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Cerberus in Dante's Divine Comedy is described as having a worm-like head. Your sight has gone in the sense that you are powerless and dead.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I laugh at your extravagance&#xA;I assure you my acquaintance&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Ceberus is the guard of &lt;em&gt;gluttony and extravagance&lt;/em&gt; (archaically, alone, &quot;extravagance&quot; meant: &lt;em&gt;excess and waste&lt;/em&gt; - i.e. gluttony). I would also suggest that: Satan or whoever is amused by our sin (extravagance), since he takes his due.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, look in the newly added hint to this question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;struggling I see this is a comedy&lt;br&gt;&#xA;let the divine be your remedy&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This calls for a lynch mobbing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2014-12-05T15:26:44.097" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5965" PostTypeId="1" AcceptedAnswerId="6033" CreationDate="2014-12-05T15:41:20.790" Score="8" ViewCount="710" Body="&lt;p&gt;You are a robo-soldier in the distant planet Minos. That planet is inhabited only by robots which mine and process Uranium ores and send enriched Uranium back to their mother planet Krawge, a planet full of life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suddenly, you received a message from Krawge. Unfortunately the antenna was sabotaged by the evil Egfwark, a race of space pirates that want to steal the Uranium. And as a result, you got the message seriously garbled:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;AGAW DVQE CE EMMPDRD AWT GAQTDRD IW JWAF SGZXUWYIQ&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You examined the antenna in order to fix it and hopefully ungarble the message. But you couldn't. All that you could know is that the message was encrypted using a key with 10 characters length and that the encryption just skipped the spaces keeping them untouched.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without knowing the key, you start to look around. The antenna was fully operational a few minutes ago, so whoever sabotaged it should not be far away. You could detect an unindentified robot hidden behind a rock and went to catch him. Knowing that it has been detected, it starts its self-destruct sequence, but you shot him in time to just cause some minor physical damage, turning off the robot and avoiding its complete self-destruction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You disassembled the enemy robot in hope to be able to retrieve the key to the received message, but all that you found is that it was derived from another text that was itself garbled in the self-destruct process. That text is this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;W,=885*5~:9;/,=*5/0+~~~~~~~~~~~~~&#xA;I0:9,~=~+3%~/8~:)+*~~~~~~~~~~~~~~&#xA;]~;/0+*=0*~'=(9~/8~*90+5/0~~~~~~~&#xA;O0~*/.~/8~&amp;lt;,/390~*,)+*~~~~~~~~~~~&#xA;J69~29++/0+~*6=*~%/)~*=)76*~19~~~&#xA;U~29=,09:~'9,9~09(9,~*,)9~~~~~~~~&#xA;P/'~U~850:~1%+928~50~-)9+*5/0~~~~&#xA;J69%~./50*~*69~85079,~=*~19~=7=50&#xA;W)52*%~&amp;lt;%~=++/;5=*5/0~~~~~~~~~~~~&#xA;E/)~./50*~*69~85079,~=*~19~=7=50~&#xA;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~&#xA;U~'=00=~,)0='=%~~~~~~~~~~~~~~~~~~&#xA;P9(9,~+=%~7//:&amp;lt;%9~~~~~~~~~~~~~~~~&#xA;U~'=00=~30/'~*69~*,)*6~~~~~~~~~~~&#xA;U0+*9=:~/8~'/0:9,507~'6%~~~~~~~~~&#xA;U~'=00=~30/'~*69~=0+'9,+~~~~~~~~~&#xA;P/~1/,9~259+~~~~~~~~~~~~~~~~~~~~~&#xA;U~'=00=~+6)*~*69~://,~~~~~~~~~~~~&#xA;]0:~/.90~).~1%~150:~~~~~~~~~~~~~~&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You suspected that all those &lt;code&gt;~&lt;/code&gt; could be destroyed parts of the text containing the key, but after detailed investigation of the enemy robot leftovers, you found that this is not the case, and in fact that multiple repeating &lt;code&gt;~&lt;/code&gt; are not artifacts of the self-destruction procedure nor artifacts of the encryption, they are indeed features of the plain-text. Further, all the line breaks in the encrypted message should be line breaks in the plain-text.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examining the damaged memory cells of the enemy robot, most of them were able to unrecoverably self-destruct, but one of them could not. In that memory cell you found a crib (a part of the message, in the case the message containing the key). This is the crib:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;myself&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Can you recover the text containing the key, the key and the received message?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please, use spoiler blocks to protect sensitive parts of your answer. Use &lt;code&gt;&amp;gt;!&lt;/code&gt; for spoiler blocks and use &lt;code&gt;&amp;lt;br&amp;gt;&lt;/code&gt; to break lines inside spoilers.&lt;/p&gt;&#xA;" OwnerUserId="5044" LastEditorUserId="5044" LastEditDate="2014-12-05T21:10:57.230" LastActivityDate="2014-12-08T20:05:31.540" Title="Recover the message, the encryption key and the text containing the key" Tags="|cipher|encoded-message|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5966" PostTypeId="2" ParentId="5923" CreationDate="2014-12-05T15:51:19.790" Score="9" Body="&lt;p&gt;I'll answer this riddle with my own notation, not the one provided by the author. This notation makes more sense, but maybe the author didn't chose it because it could possibly spoil the riddle?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/NfLFN.png&quot; alt=&quot;Color code&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My solution is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 541326&lt;br&gt;&#xA; 132564&lt;br&gt;&#xA; 423615&lt;br&gt;&#xA; 256431&lt;br&gt;&#xA; 614253&lt;br&gt;&#xA; 365142&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;Explanation&lt;/h1&gt;&#xA;&#xA;&lt;h2&gt;The socket&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The 36 wheels on the socket are connected to crystals, as shown on a drawing on the wall. First interpretation could be that the wheel can change the aspect of the crystal which is below it. However, as changing the wheels doesn't seem to have any effect, it is more likely, that the crystals have a given color, and the solution is to find out which color, and to turn the wheels accordingly. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;The device&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The device has six sensors, which as shown in a drawing on the wall send light (the colored array) into the socket. Each sensor sends light to and &quot;reads&quot; information back from exactly one of the rows of crystals. As seen in the drawings, the light emitted from the device consists of 6 different colors. Additionally, there is one additional color (I shall call it the reflection color). The device measures how many reflection rays are coming back from the crystals. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Crystals modifying the light&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First, it is shown, that all colors let the reflection ray pass. Additionally, it is shown for each crystal, which colors it absorbs, and that it reflects a reflection ray in the opposite direction, if if is illuminated by all color rays. What is missing here, is the information, if the crystal also emits a reflection ray if it is only lit by several instead of by all color rays. &#xA; There is no information provided to find out, which crystal needs which color to let it emit a reflection ray. Here, I used the hint, that a little physical thinking would be useful, and just guessed, that a crystal emits a reflection ray, exactly then, if it is hit by a colored ray, which it doesn't let pass. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;The solution&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; At this stage, this puzzle reminded me of an other puzzle, which, as it turned out is the same puzzle, just presented differently. The game Towers works the following way: You have a grid, which you have to fill with towers, with different sizes. On each border, there are numbers, which indicate, how many towers you see from there. You see a tower only if it isn't hidden by a bigger tower which is in front of it. When you think a little about it, you'll see that both puzzles have the exact same concept. &#xA; So I converted the given indications into a tower game, and solved it (as that's easier than playing with strange colors. The color code, I gave above symbolizes the tower heights. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;The last constraint:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After solving still one thing was not clear to me: in the Towers game, there is the constraint, that in each row and column, each tower size can only be placed once, but I was missing that constraint in this riddle (otherwise, more solutions would surely have been possible. Only now, I understand that exactly that is symbolized by the last drawing. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;The analogy:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/7w5DA.png&quot; alt=&quot;Towers game&quot;&gt;&#xA; From &lt;a href=&quot;http://www.chiark.greenend.org.uk/~sgtatham/puzzles/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Simon Tatham's puzzle collection&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5430" LastEditorUserId="5430" LastEditDate="2014-12-05T16:20:56.210" LastActivityDate="2014-12-05T16:20:56.210" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5967" PostTypeId="1" AcceptedAnswerId="5976" CreationDate="2014-12-05T15:54:00.447" Score="11" ViewCount="1632" Body="&lt;p&gt;Well, my previous few puzzles have been too hard, so I've done my best to avoid that this time. If this one is too easy then let me know! :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a multi-part puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We start with a riddle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I am thought and I am memory; &lt;br&gt;&#xA;  I create the world; I free men and I free women.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Black wings by the pallid tower, &lt;br&gt;&#xA;  I sit at the head of Pallas.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I am Bran and I am Bartholomew; &lt;br&gt;&#xA;  I protect Vincent and I protect Damien.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the first riddle I've ever written so please be kind! I've tried to make it fairly easy to get the basic idea but fairly hard to track down &lt;em&gt;all&lt;/em&gt; the references (a full answer will need to explain &lt;em&gt;all&lt;/em&gt; the references).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer from that riddle will lead you to a classic and well-known written work which is available free online in many places. You can use that to solve this cryptogram:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1;4;1    &lt;br&gt;&#xA;  17;3;1    &lt;br&gt;&#xA;  18;1;1    &lt;br&gt;&#xA;  2;6;1     &lt;br&gt;&#xA;  15;1;1   &lt;br&gt;&#xA;  18;5;1   &lt;br&gt;&#xA;  14;4;1   &lt;br&gt;&#xA;  13;6;1   &lt;br&gt;&#xA;  15;4;1   &lt;br&gt;&#xA;  7;3;1    &lt;br&gt;&#xA;  11;1;1   &lt;br&gt;&#xA;  16;1;1   &lt;br&gt;&#xA;  16;1;2   &lt;br&gt;&#xA;  16;1;3   &lt;br&gt;&#xA;  7;5;25   &lt;br&gt;&#xA;  16;5;-1  &lt;br&gt;&#xA;  11;3;1   &lt;br&gt;&#xA;  13;1;1   &lt;br&gt;&#xA;  5;4;1    &lt;br&gt;&#xA;  6;4;1    &lt;br&gt;&#xA;  9;4;1    &lt;br&gt;&#xA;  14;2;1   &lt;br&gt;&#xA;  15;6;9   &lt;br&gt;&#xA;  15;6;10  &lt;br&gt;&#xA;  15;6;11  &lt;br&gt;&#xA;  15;6;12  &lt;br&gt;&#xA;  15;6;13  &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The cryptogram's answer will give you a piece of music; that piece of music is the answer to the puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have fun! :)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Thank you @luxmi12, with an assist from @Len, for the excellent answer! I've marked it as solved because, well, basically you solved it, but there are a few clues still unaccounted-for, so if anyone else wants to chip in with a solution to them then go right ahead! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Title: 'dark' (quite obvious), 'progressive'&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;I free men and I free women&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;pallid tower&quot; (specifically?)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;Bran&quot;: not &lt;em&gt;intended&lt;/em&gt; as a reference to Game of Thrones, although the scriptwriters there pretty clearly had the same idea in mind as I did.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="4050" LastEditDate="2014-12-09T16:59:15.763" LastActivityDate="2017-06-15T18:09:26.007" Title="A Dark But Quite Progressive Puzzle. What's the musical solution to this riddle + cryptogram?" Tags="|riddle|cipher|literature|" AnswerCount="3" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="5968" PostTypeId="2" ParentId="5865" CreationDate="2014-12-05T16:12:11.103" Score="6" Body="&lt;p&gt;Here is my answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; #&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I go by many a name,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; # has many names: number sign, pound sign, hash, octothorp, tictactoe, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Including one that's quite a shame.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sometimes people refer to # as the sharp symbol (♯), even though they are completely different.  Shame on Microsoft for referring to C# as &quot;C Sharp&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can take you to an exact location,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In HTML, you can mark a specific location on a web page (let's call it webpage.html) by using an anchor: &amp;lt;a name=&quot;anchor_name&quot;&amp;gt;&amp;lt;/a&amp;gt;.  Then someone can go to that exact location on the page by using # and the anchor name (e.g. &lt;code&gt;http://.../webpage.html#anchor_name&lt;/code&gt;).  The browser will load the page and automatically scroll the page so that the anchor is at the top of the browser window.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But sometimes only to a nearby destination.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the anchor is defined near the bottom of the web page, the browser may not be able to scroll far enough to place the anchor at the top of the window.  But the anchor will still be somewhere in the window.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If great power is your wish, I'll grant it but your money will vanish.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In many Unix shells, non-root users will have a \$ prompt.  But if you &quot;su&quot; to root, the \$ disappears and your prompt becomes #.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Substituting for that which can be negative,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; # can substitute for &quot;no.&quot;, which is an abbreviation for number.  But &quot;no.&quot; can also be negative (the opposite of &quot;yes.&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll always remain completely positive.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One could say that # is made up completely of +.  Four plus signs all connected to each other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be very important,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In some programming languages, # indicates a comment.  Comments are important to developers for documentation purposes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While at the same time very insignificant.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Comments are insignificant because they are ignored by the compiler/interpreter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I contain many things,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The tic-tac-toe board can contain up to nine Xs and Os.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I simultaneously contain nothing.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tic-tac-toe is also known as &quot;Noughts and crosses&quot;.  Nought (O) means nothing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can signal the beginning,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; # indicates the beginning of a comment in some programming languages.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I can also signal the ending.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many phone systems will ask you to enter information (such as a PIN), &quot;followed by the pound or hash sign&quot;.  You press # to indicate the end of that information.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2014-12-05T16:12:11.103" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5969" PostTypeId="2" ParentId="5956" CreationDate="2014-12-05T16:59:54.443" Score="13" Body="&lt;p&gt;&lt;em&gt;As I haven't been the first to present a solution, I thought I might at least provide a detailed deduction route here.&lt;/em&gt; ( I did not look at the solution first. )&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To solve this puzzle, I first make a reduction of information, moving form the (nice!) pictures to pure information. I use the following abbreviations:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;+ added property&#xA;- removed property&#xA;0 unchanged property&#xA;&#xA;A Antennae &#xA;H Hands (arms)&#xA;L Legs&#xA;S Square Head&#xA;B Square Body&#xA;U Upright Position&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;By simply comparing left &amp;amp; right as well as top &amp;amp; bottom one gets:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/3vYnd.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The next step is to count which mutations actually appear, giving us:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Once of each:   +A -A +H -H +L -S +B -U &lt;br&gt;&#xA; Twice of each:  +S +U &lt;br&gt;&#xA; &lt;br&gt;&#xA; Note that &lt;em&gt;twice&lt;/em&gt; means that the same mutation appears twice in either the rows or twice in the columns.&lt;br&gt;&#xA; Summing those mutations, we get 12, which is good, as we have 12 cells to fill.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now lets start assigning mutations to cells:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Start with the modifications which appear once in total, and also appear once in either the rows or columns. Their cross-section &lt;em&gt;has&lt;/em&gt; to be the mutation cell:&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/qfcvK.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;br&gt;&#xA; By this, we have not yet assigned the single: +H  -H  -S  -U&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now we look at the obvious double-counted mutations:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; +S +U appear each twice but we see already that for the first row only a single cell is left over for +S. For the second row a net +U can only be achieved if there is only a positive and no negative mutation in that row. This immediately gives:&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/wB0Ea.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Next we can expand the logic:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the third column 4 mutations have to happen but only one cell isn't yet assigned. It has to be +S. But now the last row needs to be neutral in S, which we can only achieve by a -S in the same row. This has to be in the 2nd column, because the net mutation for columns is +S in the first and -S in the second column, respectively. This gives us:&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/vZZ7h.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can use the same arguments again for the upright (U) property:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the third row there has to be +U and -U, and there is only one possibility to fit this:&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/PYKnN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And now there is only one possibility to fill in the rest for the final solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/aiOKp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="5840" LastEditDate="2014-12-06T19:24:51.543" LastActivityDate="2014-12-06T19:24:51.543" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5970" PostTypeId="1" AcceptedAnswerId="6051" CreationDate="2014-12-05T18:01:00.243" Score="10" ViewCount="607" Body="&lt;p&gt;Born from the damming of a dam, my people are strong and hardy. Once holy servants to four kingdoms, their concurrent demise diminished our status. We are now a simple people, no longer worshiped by many. Though smaller than our desert cousins, we're almost the spitting image. We come bearing gifts. A chest sits before you, filled with food. Below it, drinks to quench your thirst. Upon the chest, sits cloud made flesh upon a sheet unreadied. It is fit for slather with the dripping of the mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We send them with our leader, a native of Mastbergen. Though he is half cucumber, you have accepted him and even included him in conversation. For this, we thank you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who are the people mentioned? Provide reasoning. Each sentence other than the thanks contains at least one clue. The title refers to the species name of the cucumber plant, not marijuana.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;What do these people have in common with LL Cool J? Just a couple of things.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5626" LastEditorUserId="5626" LastEditDate="2014-12-09T15:21:03.467" LastActivityDate="2014-12-09T23:12:20.613" Title="The Sativan Sultan - The leader of a proud, holy people, you must uncover their name" Tags="|riddle|" AnswerCount="3" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5971" PostTypeId="2" ParentId="5948" CreationDate="2014-12-05T18:14:15.740" Score="3" Body="&lt;p&gt;You saw &quot;The AIOU&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The AIOU is a research facility/university in pakistan.  Upon your travels you saw some of their handiwork in the form of a peer reviewed paper.&lt;/p&gt;&#xA;" OwnerUserId="457" LastActivityDate="2014-12-05T18:14:15.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5972" PostTypeId="2" ParentId="5970" CreationDate="2014-12-05T18:15:30.277" Score="5" Body="&lt;p&gt;Updated with a new guess: Now I think it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Snakes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;damming of a dam&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; damning of Adam = Garden of Eden&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;holy servants to four kingdoms&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Egypt? Old Kingdom / Middle Kingdom / New Kingdom / Late Period&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;spitting image&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Spitting_cobra&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Spitting_cobra&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A chest&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The hood of a snake is kind of like its chest.&#xA; &lt;img src=&quot;https://i.stack.imgur.com/bxe01.jpg&quot; alt=&quot;Rearing cobra&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;drinks&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Venom?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;cloud made flesh&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Snake_Temple&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Snake_Temple&lt;/a&gt;&#xA; &quot;The temple was originally called the &quot;Temple of the Azure Cloud&quot; in honor of the beauty of Penang's sky.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;although I haven't got anything yet for Mastbergen or the half cucumber. I think the 'leader' could perhaps be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Satan (as per Genesis)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Original guess:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it's &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Zoroastrians&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are only two kinds of people who &quot;come bearing gifts&quot;, Greeks and Magi. Greeks would be too obvious (and doesn't fit well with the other clues) so I'm going with the Magi - who were, I think Zoroastrians in the story.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once holy servants to four kingdoms, their concurrent demise diminished our status.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Zoroastrianism &quot;was once the state religion of the Achaemenid, Parthian, and Sasanian empires&quot; (Wikipedia) - and we could count the Persian empire as well to make 4.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;damming of a dam&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;could be the dam&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; that was opened in the Battle of Ullais&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, Sativa&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; means 'cultivated', so it would make sense for us to be talking about the Fertile Crescent, where agriculture was developed. This could fit with the drinks (irrigation) and the food. Damming (birthing) of a dam could also fit with irrigation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="4050" LastEditDate="2014-12-08T16:09:31.093" LastActivityDate="2014-12-08T16:09:31.093" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="5975" PostTypeId="2" ParentId="5956" CreationDate="2014-12-05T19:08:38.410" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A [head-square] &lt;br&gt;&#xA; B [add-legs] &lt;br&gt;&#xA; C [body-square] &lt;br&gt;&#xA; D [remove arms] &lt;br&gt;&#xA; E [add-ant] &lt;br&gt;&#xA; F [walk-up] &lt;br&gt;&#xA; G [walk-down] &lt;br&gt;&#xA; H [walk-up] &lt;br&gt;&#xA; I [remove-ant] &lt;br&gt;&#xA; J [add-arms] &lt;br&gt;&#xA; K [head-round] &lt;br&gt;&#xA; L [head-square] &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I noticed that my answer is different from BmyGuest; maybe there are multiple solutions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ed: comparing the two answers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;pre&gt;BmyGuest     me &lt;/pre&gt;&lt;br&gt;&#xA; &lt;pre&gt;+S +L +B   +S +L +B  &lt;/pre&gt;&lt;br&gt;&#xA; &lt;pre&gt;+A +A +U   -H +A +U  &lt;/pre&gt;&lt;br&gt;&#xA; &lt;pre&gt;-U +U -A   -U +U -A  &lt;/pre&gt;&lt;br&gt;&#xA; &lt;pre&gt;+H -S +S   +H -S +S  &lt;/pre&gt;&lt;br&gt;&#xA; Actually, I think that D must be -H (remove arms/hands)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was much less methodical. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I found A, B, and C by process of elimination.  Then I listed all the possiblities for E, F, H, I, K, and L.  Then I was able to find E and I, followed by the rest through guessing and checking.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6206" LastEditorUserId="6206" LastEditDate="2014-12-05T19:17:29.260" LastActivityDate="2014-12-05T19:17:29.260" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5976" PostTypeId="2" ParentId="5967" CreationDate="2014-12-05T19:50:38.810" Score="7" Body="&lt;p&gt;Well I have the poem!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;&lt;a href=&quot;http://www.heise.de/ix/raven/Literature/Lore/TheRaven.html&quot;&gt;The Raven&lt;/a&gt;&quot;, by Edgar Allen Poe&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is why;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am thought and I am memory;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;In Norse mythology there are two Ravens called Huginn (thought) and Muninn (memory) that bring information to Odin&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I create the world; I free men and I free women.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There exists a creation story from the Inuit, in which a Raven creates the world. He then travels around his new world and discovers man, whom he feeds and teaches. Soon after a woman came to be and he cared for her as well, as well as all their children&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Black wings by the pallid tower&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There are lots of Ravens around the tower of London? Also &quot;pallid&quot; appears in the poem&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I sit at the head of Pallas.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;In the poem the Raven perches on a bust of Pallas above the chamber door&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am Bran and I am Bartholomew;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;In &quot;Game of Thrones&quot; (spoilers!) Bran is visited by a three eyed Raven who he later discovers represents himself. In the Simpsons &quot;Treehouse of Horror&quot; (also known as &quot;'The Simpsons 'Halloween Special&quot;) the Raven poem is acted out with Bart as the Raven&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I protect Vincent and I protect Damien.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Saint Vincent of Saragossa was said to have his body protected by ravens from being devoured by vultures, until his followers could recover the body. In The Omen, the Raven was a servant of Satan who watched over Damien Thorn&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm definitely stuck on the cryptogram though... I thought I had it but the -1 stumped me, then later the 25 in the third column got me again. I shall keep trying :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;EDIT&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the cryptogram..&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;As @Len found out below, the trick is to use cryptography on the poem. The first column is the verse number, the second is the line number and the third is the character. No spaces or punctuation count.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, the answer is...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;&lt;a href=&quot;https://www.youtube.com/watch?v=YAE1XTvKLXA&quot;&gt;Alan Parsons Project The Raven&lt;/a&gt;&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Although there's still some riddle clues I'm working on...&lt;/p&gt;&#xA;" OwnerUserId="6205" LastEditorUserId="6205" LastEditDate="2014-12-05T21:11:59.440" LastActivityDate="2014-12-05T21:11:59.440" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="5977" PostTypeId="1" AcceptedAnswerId="5979" CreationDate="2014-12-05T20:35:47.583" Score="1" ViewCount="214" Body="&lt;p&gt;There is a room with a single light bulb in it. Outside the room there are three light switches. You cannot see inside the room at all. Only one light switch turns on the light bulb. You may only enter the room once to find out which switch turns it on. The goal is to figure out which light switch turns on the light.&lt;/p&gt;&#xA;" OwnerUserId="5155" LastEditorUserId="5188" LastEditDate="2015-05-02T06:34:16.310" LastActivityDate="2015-05-02T06:34:16.310" Title="Shedding some light" Tags="|strategy|" AnswerCount="1" CommentCount="5" ClosedDate="2014-12-05T23:13:17.987" ContentLicense="CC BY-SA 3.0" />
  <row Id="5978" PostTypeId="2" ParentId="5967" CreationDate="2014-12-05T20:57:55.233" Score="1" Body="&lt;p&gt;Continuing with the excellent work by luxmi12, I think the answer is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Alan Parsons Project The Raven&lt;/strong&gt;&#xA; assuming there are two errors in the cryptogram numbers which actually produce: &lt;strong&gt;alanparsonsprojecttlesraven&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Method:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Apply the cryptogram numbers to the poem by Edgar Allan Poe.  The numbers indicate the  verse;line;character.  Characters do not include spaces or punctuation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastActivityDate="2014-12-05T20:57:55.233" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5979" PostTypeId="2" ParentId="5977" CreationDate="2014-12-05T21:05:50.277" Score="2" Body="&lt;p&gt;Label the switches A B and C.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Turn on A and B and wait for a while, then turn off A. Enter the room.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How to determine the switch?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; On -&gt; B&lt;br&gt;&#xA; Off and cold -&gt; C&lt;br&gt;&#xA; Off and hot -&gt; A&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2014-12-05T21:05:50.277" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5980" PostTypeId="1" CreationDate="2014-12-05T21:29:12.900" Score="16" ViewCount="664" Body="&lt;p&gt;Al and Jane are hosting a party tonight. Al plans on proposing to Jane, a secret only he knows. &lt;br&gt;&#xA;The party begins and all 8 guests arrive on time. Immediately and again every minute, everyone forms new conversations.&lt;br&gt;&#xA;Throughout the entire party: &lt;br&gt;&#xA;1. Each of the 10 partiers is always in exactly 1 conversation. &lt;br&gt;&#xA;2. Each conversation must contain at least 2 people.&lt;br&gt;&#xA;3. Each conversation lasts exactly 1 minute.&lt;br&gt;&#xA;4. No two conversations may involve the exact same set of people.&lt;br&gt;&#xA;If someone in a conversation knows the secret they reveal it, so everyone in the conversation now knows the secret.&lt;br&gt;&#xA;What is the maximum number of minutes the party can last without Jane discovering the secret?&lt;/p&gt;&#xA;" OwnerUserId="6208" LastActivityDate="2015-12-11T22:28:06.793" Title="Secret within the Gossiping Partiers" Tags="|mathematics|combinatorics|" AnswerCount="5" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5981" PostTypeId="1" AcceptedAnswerId="6502" CreationDate="2014-12-05T22:50:09.353" Score="12" ViewCount="809" Body="&lt;p&gt;This is my first real puzzle, so I hope you enjoy it! Sorry if the writing is a little poor, etc..&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You are the unstoppable Detective Joshua Brown, and no puzzle can stand in your way!&lt;br&gt;&lt;br&gt;&#xA;  &lt;em&gt;The Case:&lt;/em&gt; &lt;br&gt;&#xA;  You awake at once to a phone call from the police: Your best friend since childhood, Doctor Frank Andrews, has been reported missing! Arriving on the crime scene, you search high and low for a clue, when suddenly, you find it.&lt;br&gt;&lt;br&gt;&#xA;  A handwritten note, taped to the bottom of the mahogany desk in his study. It's not Frank's handwriting. You read it:&lt;br&gt;&lt;br&gt;&#xA;  &lt;em&gt;&quot;If you wish to see him again, you must prove yourself worthy. The clue to finding Doctor Andrews is hidden behind a painting in this room. Attempting to open the wrong painting will set off an alarm and notify us. We will kill him immediately.&quot;&lt;/em&gt;.&lt;br&gt;&lt;br&gt;&#xA;  You shiver in fear: these men clearly mean business. Alongside this note are two more. The first reads: &lt;br&gt;&#xA;  &lt;em&gt;&quot;u171Bb4C0NBwiOX2e81Luc6S/OWKyDO0QjE5PwiDoF8b4YmxlIA55oGiRm735nNUTKm7ZO+d0Hh3QJb=&quot;&lt;/em&gt;.&lt;br&gt;&lt;br&gt;&#xA;  And the second: &lt;br&gt;&#xA;  &lt;em&gt;&quot;PINVEBLFWFMCCPRWIKWUGUSFZR&quot;&lt;/em&gt;.&lt;br&gt;&lt;br&gt;&#xA;  Spurred on by the need to save your friend, you get to work on the case. Frank is an art lover, so there are five replicas of famous paintings on the wall: &lt;em&gt;The Mona Lisa&lt;/em&gt;, &lt;em&gt;Basket of Fruit&lt;/em&gt;, &lt;em&gt;Sunflowers&lt;/em&gt;, &lt;em&gt;The Persistence of Memory&lt;/em&gt; and &lt;em&gt;The Scream&lt;/em&gt;, but which holds the oh-so-important clue?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, that's all. Try to figure out the painting using the three notes, and give a reason with your answer. Good luck!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After this puzzle is solved, I will put up the next part in the series within a few days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;While the first passage appears to be Base64, it's actually not. With that said, you shouldn't just be guessing blindly as to how to decode it. The answer is hidden there somewhere. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The first passage should be decoded first. To decode the second, you require the plaintext from the first.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Apologies for the skewed difficulty, here's one more hint: There is a text string hidden inside the first string. It points towards a common cipher that can then be used to decrypt the first string&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The story was shortened after a tip from Travis Kindred (Thanks!)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="4748" LastEditorUserId="6178" LastEditDate="2014-12-30T14:58:48.163" LastActivityDate="2014-12-30T14:58:48.163" Title="The Mysterious Disappearance of Doctor F. Andrews [Part 1]" Tags="|logical-deduction|word-problem|cipher|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="5982" PostTypeId="1" AcceptedAnswerId="5990" CreationDate="2014-12-05T22:55:46.520" Score="20" ViewCount="1168" Body="&lt;p&gt;&lt;strong&gt;Hello! And welcome to &quot;The Riddling Chef,&quot; the show where the recipe is deliberately obscure! I'm Travis.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today, we're going to be making a dish fit for the king! The recipe was sent in by Edward and Caroline Lancaster; thank you both for being such loyal viewers. For the first course, some plucked pheasant shish kabobs. Maybe some small steaks a few inches in diameter each. After that, a traditional mud pie with a secret ingredient added. (It's not horse meat) Okay, you got me. It's horse meat. We'll add it in last to keep the flavor fresh. To finish it off, a french red sauce on a good old slice of victory pie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the kind of recipe that really sets chefs apart. If you have any old-fashioned notions about cooking with honor, they end here. You cook this dish-- and cook it right, or you die. Your choice. You're going to see a lot fewer amateurs and a lot more professionals after this. Prep time is about a hundred years, maybe less and maybe more depending on your interpretation. Cook time is 14-15. Finish it up and it's going to smell like roses. Okay, let's get started, but first... a word from our sponsors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Don't go away! We'll be right back! Before we do, have you uncovered the secret recipe? What dish is Chef Travis working on today?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5626" LastEditorUserId="5626" LastEditDate="2014-12-10T14:20:22.397" LastActivityDate="2014-12-10T14:20:22.397" Title="Dish Fit For A King - Find out what chef is cooking on an exciting new episode of The Riddling Chef!" Tags="|riddle|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5983" PostTypeId="2" ParentId="5981" CreationDate="2014-12-05T23:32:57.380" Score="8" Body="&lt;p&gt;Congrats on making a giant leap! As a puzzler whose first question needed improvement, I want to offer a suggestion. This puzzle looks fun, but it has an issue. It lacks brevity. I suggest shortening the puzzle and making your puzzles stand-alone unless the multi-part format is very important. Let me offer an example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As a homicide investigator, you've seen some odd things, but this takes the cake. A local art appraiser has been kidnapped, but a clue has been left. The note reads:&#xA;  &quot;u171Bb4C0NBwiOX2e81Luc6S/OWKyDO0QjE5PwiDoF8b4YmxlIA55oGiRm735nNUTKm7ZO+d0Hh3QJb= --- --- --- PINVEBLFWFMCCPRWIKWUGUSFZR&quot;.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;There are five replicas of famous paintings on the wall: The Mona Lisa, Basket of Fruit, Sunflowers, The Persistence of Memory and The Scream, and your intuition tells you that one of them holds the key to solving this case.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, here's where I backpedal. It's possible that the change has broken the puzzle. It's possible that the lengthy narrative is required. I don't know because the puzzle is unsolved. Just think about the value of brevity and let it improve your work in the future. One of the best things you can do in any form of writing is to save your audience time. They'll value you for it.&lt;/p&gt;&#xA;" OwnerUserId="5626" LastActivityDate="2014-12-05T23:32:57.380" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5984" PostTypeId="2" ParentId="5982" CreationDate="2014-12-05T23:44:24.597" Score="1" Body="&lt;p&gt;My answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The battle of Agincourt&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Edward of York and Humphrey of Lancaster were Commanders alongside the King (Henry V), it ended the 100 year war against the French, and it led almost directly to the War of the Roses in which York and Lancaster became enemies.  Oh, and it was in 1415 AD.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5683" LastActivityDate="2014-12-05T23:44:24.597" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5986" PostTypeId="2" ParentId="5980" CreationDate="2014-12-06T00:44:27.623" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;8 minutes.&lt;br&gt;&#xA;If the secret follows a linear path:&lt;br&gt;&#xA;- Al talk to A, so A knows the secret(1 minute);&lt;br&gt;&#xA;- Then Al and A, talk to B, so B knows the secret(2 minutes);&lt;br&gt;&#xA;- Then Al and A and B talk to C, so now C knows the secret(3 minutes);&lt;br&gt;&#xA;Keep this until the next to last person, because &quot;Each conversation must contain at least 2 people&quot; and Jane must be the last person to know about it AND if we include that last person, Jane won´t have anyone to make a conversation.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Anyway it is really strange party!&lt;/p&gt;&#xA;" OwnerUserId="1675" LastEditorUserId="1675" LastEditDate="2014-12-06T00:58:19.623" LastActivityDate="2014-12-06T00:58:19.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5987" PostTypeId="2" ParentId="5980" CreationDate="2014-12-06T00:57:46.827" Score="3" Body="&lt;p&gt;Here's a way to get a time of &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;12 minutes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Arranges conversations by &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; spliting the party into two groups of five, one with Al and one with Jane, and have each group split into 2+3 in each of the ten possible ways.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can get another two minutes by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Having each group of 5 in a conversation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Doing the same but splitting a person from each group (not Jane) into a conversation pair, leaving each half with 4.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here's a way to get a slighly better time of &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 13 minutes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's nothing that systematic here. I'll just post the list of conversation clusters. Each line list the people who know the secret beforehand, followed by the conversations, with secret-sharing conversation before the semicolon.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;&#xA; A:   AB; CDEFGHIJ&#xA; AB:  ABC; DEFGHIJ&#xA; ABC: ABCD; EFGHIJ&#xA; A-D: AD BC; EGI FHJ&#xA; A-D: AC BD; EGJ FHI&#xA; A-D: ABCDE; FGHIJ&#xA; A-E: AE BCD; FG HIJ&#xA; A-E: BE ACD; FH GIJ&#xA; A-E: CE ABD; FI GHJ&#xA; A-E: ABCDEF; GHIJ&#xA; A-F: ABE CDF; GH IJ&#xA; A-F: ACE BDF; GI HJ&#xA; A-F; BCE ADF; GJ HI&#xA; &lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2014-12-06T01:24:32.100" LastActivityDate="2014-12-06T01:24:32.100" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5988" PostTypeId="1" AcceptedAnswerId="5994" CreationDate="2014-12-06T01:24:05.153" Score="0" ViewCount="3741" Body="&lt;p&gt;You are locked in a 50 x 50 x 50-foot room which sits on 100-foot stilts. There is an open window at the coner of the room, near the floor, with a strong hook cemented into the floor by the window. So if you had a 100-foot rope, you could tie one end to the hook, and climb down the rope to freedom. (The stilts are not accessible from the window. ) There are two 50-foot lengths of rope, each cemented into the ceiling, about 1 foot apart, near the center of the ceiling. You are a strong, agile rope climber, good at tying knots,and you have a sharp knife. You have no other tools (not even clothes). The rope is strong enough to hold your weight, but not if it is cut lengthwise. You can survive a fall of no more than 10 feet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you get out alive?&lt;/p&gt;&#xA;" OwnerUserId="2250" LastActivityDate="2014-12-06T04:41:59.393" Title="Come out alive from locked room" Tags="|logical-deduction|lateral-thinking|" AnswerCount="3" CommentCount="0" ClosedDate="2014-12-06T17:03:40.940" ContentLicense="CC BY-SA 3.0" />
  <row Id="5989" PostTypeId="2" ParentId="5980" CreationDate="2014-12-06T01:35:28.197" Score="4" Body="&lt;p&gt;This is a proof of an upper bound of&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 22 minutes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;It shows that the party can't last longer than that. I don't think this bound is tight.&lt;/p&gt;&#xA;&lt;p&gt;The main idea is to:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Split the party into two parts: Phase A, where &lt;span class=&quot;math-container&quot;&gt;$\le5$&lt;/span&gt; people know the secret, and Phase B, where &lt;span class=&quot;math-container&quot;&gt;$\ge6$&lt;/span&gt; people know it. The party consists of Phase A followed by Phase B. We bound the length of each half.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The first part is:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Phase A lasts at most 15 minutes because Al must be talking with a different non-empty subset of the four other people in the group of 5 that will be spoiled by the end of Phase A, of which there are &lt;span class=&quot;math-container&quot;&gt;$2^4-1 = 15$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The second part proceeds similarly&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Phase B lasts at most 7 minutes because Jane must be talking with a different non-empty subset of the three other people that are not spoiled when Phase B starts, of which there are &lt;span class=&quot;math-container&quot;&gt;$2^3-1 = 17$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This gives an overall bound of&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$7+15 = 22$&lt;/span&gt; minutes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;Edit&lt;/h2&gt;&#xA;&lt;p&gt;We can actually improve the bound to&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 17 minutes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;by observing that&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Al can only be in one group of 4 in phase A, and Jane can only be in one group of three in Phase B.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-12-11T22:28:06.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5990" PostTypeId="2" ParentId="5982" CreationDate="2014-12-06T01:40:31.853" Score="13" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The hundred years war.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edward and Caroline Lancaster:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The first phase of the war (1337-1360) was initated by King Edward III, called the Edwardian war.&lt;br&gt;&#xA;The second phase of the war (1369-89) was called the Caroline war.&lt;br&gt;&#xA;The third phase of the war (1415-1453) was called the Lancastrian war.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Plucked pheasant shish-kebabs&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;English longbowmen, plucked refers to feathers, pheasant refers to peasant and shish-kebabs refers to the arrow shaft and arrow head.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mud pie with horse meat&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; refers to Agincourt, mounted French knights bogged down in mud and skewered by arrows.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;French red sauce on a good old slice of victory pie&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; is the blood of the French, but also eventual French victory.  Also Patay reminds me of &lt;a href=&quot;http://en.wikipedia.org/wiki/P%C3%A2t%C3%A9&quot;&gt;pate&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you have any old-fashioned notions about cooking with honor, they end here.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many chivalric ideals died as knightly charges failed against longbowmen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You cook this dish-- and cook it right, or you die. Your choice. You're going to see a lot fewer amateurs and a lot more professionals after this.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Its a war.  Soldiers who survive are going to be veterans.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Prep time is about a hundred years, maybe less and maybe more depending on your interpretation. Cook time is 14-15.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Hundred years war.  The cook time is the intermissions between the phases of the war, I guess?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finish it up and it's going to smell like roses.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Wars of the Roses started in 1455 shortly afterwards.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6178" LastActivityDate="2014-12-06T01:40:31.853" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5991" PostTypeId="2" ParentId="5988" CreationDate="2014-12-06T02:00:40.380" Score="9" Body="&lt;p&gt;You climb up one of the 50 foot lengths of rope.  Then when you get within a couple feet of the ceiling, you do the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hold the rope with your left hand and use your right hand to grab the other 50 foot length of rope.  Put it in your left hand so that your left hand is holding both ropes.  With your right hand, use the knife to cut each rope halfway through its thickness.  So if each rope is 1-inch thick, you make a half-inch cut in each one.  Even though each rope is partially cut, neither one should break since each one is only supporting half of your weight now.  Then grab one rope with your right hand, so that each hand is holding a rope.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now climb down to the floor and do the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Grab one rope and start climbing it a few feet.  It will break since it is partially cut and cannot support your weight.  Do the same with the other rope.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now you have 2 ropes on the floor that are almost 50 feet long each.  Tie the ropes together at the ends, so that you have one rope that is almost 100 feet long.  Then tie one end to the hook.  Toss the rope out the window, and the other end will be within a few feet of the ground.  Climb down to safety.  You are alive and on the ground!&lt;/p&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2014-12-06T02:00:40.380" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5993" PostTypeId="2" ParentId="5970" CreationDate="2014-12-06T02:23:44.150" Score="7" Body="&lt;p&gt;A possible answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Tulips&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sultan&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Tulips were introduced to Europe by the Sultan of Turkey.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sativan&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Sativa means &lt;a href=&quot;http://en.wikipedia.org/wiki/Sativum&quot;&gt;cultivated&lt;/a&gt; as per AE's answer, and tulips were certainly cultivated to produce new varieties and colours.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Born from the damming of a dam, my people are strong and hardy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Amsterdam is the dam (or could also be the breaking of the bulb when sprouting), tulip bulbs are strong and hardy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once holy servants to four kingdoms, their concurrent demise diminished our status.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Kingdom of the Netherlands is made up of four countries (more like colonies): Netherlands, Aruba, Curaçao, and Sint Maarten.  The Netherlands practiced Christianity and flourished in the Dutch golden age but then diminished in power.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are now a simple people, no longer worshiped by many.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The tulip bulb craze ended.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though smaller than our desert cousins, we're almost the spitting image.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Tulips are native to the middle east.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We come bearing gifts. A chest sits before you, filled with food.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The tulip bulb is the chest.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Below it, drinks to quench your thirst. Upon the chest, sits cloud made flesh upon a sheet unreadied.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Below the bulb, roots grow to collect water.  Above it, the sprout forms into a tulip.  The sheet unreadied is the closed tulip petal and the cloud made flesh is morning dew.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mastbergen&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Mastbergen seems to be a Dutch name.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cucumber&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The cucumber tree and the tulip tree are related.  Neither seems to be related to actual tulips though.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6178" LastActivityDate="2014-12-06T02:23:44.150" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5994" PostTypeId="2" ParentId="5988" CreationDate="2014-12-06T02:26:47.370" Score="11" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;This answer assumes whoever built this place is sane, which is a shaky assumption at best from the sound of it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Climb one rope, cut the other down completely, letting it fall to the floor. Climb back down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take your 50-foot rope, tie it to the hook by the window. Climb out to the end, and start swinging back and forth in line with the nearest stilt. Once you can reach it, climb down and find some clothes.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;When I said it assumes the builder is sane, I mean that the room is supported by at least one stilt, and that these stilts are positioned in some fashion that &lt;em&gt;evenly&lt;/em&gt; distributes the weight of the room among them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is the case, the furthest a stilt could be from a window at any corner's floor is about 35 feet away (a single stilt dead-center is 25*sqrt(2)). With a 50-foot(ish) rope, you'd only need to swing out to an angle of about 45 degrees&#xA;from vertical, which should be easily managed. You'd end up about 35 feet down the stilt.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Of course, if the guy's crazy (very likely) and planted a single stilt on a corner opposite the window or something, you're probably better off using another method. Then again, it might be better to just jump out the window anyway, since he'll probably just capture you again and put you through some other bizarre trial.&lt;/p&gt;&#xA;" OwnerUserId="1563" LastActivityDate="2014-12-06T02:26:47.370" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5996" PostTypeId="2" ParentId="5988" CreationDate="2014-12-06T04:41:59.393" Score="5" Body="&lt;p&gt;Climb the the top and cut one rope near the ceiling leaving enough to tie a loop hanging from the ceiling. Climb down and tie the long piece you cut off to the piece still hanging from the ceiling, so you have a length of rope almost 100 feet long, still attached to the ceiling. Take the loose end, climb to the top and thread it through the loop. Now cut the other rope from the ceiling and lower yourself to the ground. Since the rope is nearly 100 feet long and you will need to hold both ends of the rope threaded through the loop, you can lower yourself 50 feet. Then pull the rope out from the loop, tie it to the hook, and lower yourself down.&lt;/p&gt;&#xA;" OwnerUserId="5853" LastActivityDate="2014-12-06T04:41:59.393" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5997" PostTypeId="1" AcceptedAnswerId="5998" CreationDate="2014-12-06T05:35:26.620" Score="-3" ViewCount="286" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Cv5Fq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Puzzle:&lt;/strong&gt; Identify whose skull is depicted at the bottom of this Russian icon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus Point 1:&lt;/strong&gt; Identify the two items depicted on either side of the cross.&lt;/p&gt;&#xA;" OwnerUserId="5474" LastEditorUserId="5474" LastEditDate="2014-12-11T01:33:21.117" LastActivityDate="2014-12-11T01:33:21.117" Title="The mystery of the skull" Tags="|riddle|symbols|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5998" PostTypeId="2" ParentId="5997" CreationDate="2014-12-06T07:20:11.360" Score="6" Body="&lt;p&gt;Russian &quot;Golgotha's Cross&quot; (Greek letter Γ stands for G in russian Golgothas in Greek) (or Calvary Cross)&#xA;The items are: the spear (that penetrated Jesus' body) and the sponge (with vinegar) during Crucifixion.&#xA;Skull of Adam&lt;/p&gt;&#xA;" OwnerUserId="5051" LastEditorUserId="5474" LastEditDate="2014-12-06T07:42:33.283" LastActivityDate="2014-12-06T07:42:33.283" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5999" PostTypeId="2" ParentId="2474" CreationDate="2014-12-06T08:28:29.630" Score="1" Body="&lt;p&gt;You carefully break open the light bulbs to examine their filaments. The two blown lightbulbs will have broken filaments, and the just-blown one will have smoke inside the bulb.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2014-12-06T08:28:29.630" CommentCount="1" CommunityOwnedDate="2014-12-06T08:28:29.630" ContentLicense="CC BY-SA 3.0" />
  <row Id="6000" PostTypeId="1" CreationDate="2014-12-06T11:19:50.183" Score="5" ViewCount="818" Body="&lt;p&gt;First question here; I hope this question fits the &lt;a href=&quot;https://puzzling.stackexchange.com/a/5884/2062&quot;&gt;rules&lt;/a&gt; well :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, first please see the two images below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://lh5.googleusercontent.com/-kGVFakZ4tSg/VIBCKiE1VjI/AAAAAAAAAqk/dlhhYboCEFY/w756-h567-no/14%2B-%2B1&quot; alt=&quot;Image 1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://lh6.googleusercontent.com/-1J8FVXCDk28/VIBCKm0OMPI/AAAAAAAAAqw/PKAw3oHM63I/w756-h567-no/14%2B-%2B2&quot; alt=&quot;Image 2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can find yourself, the eraser appears to &lt;strong&gt;float motionless&lt;/strong&gt; over the register's page, with &lt;strong&gt;more than half&lt;/strong&gt; of its area outside the contact area with the register. This picture has not been modified. The eraser is a normal unmodified brand new cuboid eraser found in stores.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't be bewildered :) This is possible; and these pictures are from my own experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: How is this possible? Or how have I done this?&lt;br&gt;&#xA;&lt;strong&gt;Answer&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Of course, that you have to find! ;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Please post your answer along with a picture (just as I have) of the eraser floating in air explaining what you did. Remember that the eraser has to float over a register kept over a table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is only one definite answer (at least, I believe there is). Happy solving :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;br&gt;&#xA;So, after thinking that I could bewilder you all with my original puzzle, I am bewildered myself. I tried to be as clear as possible. First as &lt;a href=&quot;https://puzzling.stackexchange.com/questions/6000/the-motionless-floating-eraser#comment19051_6000&quot;&gt;I was told&lt;/a&gt; that one definite answer should be there, so I edited the question to state that just one definite answer is there to my knowledge. I clearly stated that the eraser is motionless, still I &lt;a href=&quot;https://puzzling.stackexchange.com/questions/6000/the-motionless-floating-eraser#comment19085_6004&quot;&gt;get a comment&lt;/a&gt; that a &lt;a href=&quot;https://www.flickr.com/photos/barbar/4470877212/&quot; rel=&quot;nofollow noreferrer&quot;&gt;man is floating&lt;/a&gt; in air like this eraser. I tell that the eraser is unmodified, still I &lt;a href=&quot;https://puzzling.stackexchange.com/a/6007/2062&quot;&gt;get answers&lt;/a&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/a/6002/2062&quot;&gt;breaking&lt;/a&gt; this rule. I tell that this is a real world situation, still I get an answer that the background image is &lt;a href=&quot;https://puzzling.stackexchange.com/a/6008/2062&quot;&gt;tilted&lt;/a&gt;. People are breaking what I say. Is this my fault? And then I am told that the question is too broad so we are closing it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought that I would explore some new ideas. I tried to be as positive as possible, but I guess I can't handle this much negativity. Too remove all scopes of this question being reopened, I answer the question myself:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/R1zso.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/R1zso.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&#xA;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The center of gravity is maintained by the compass with a pencil. In the rush of closing this question, none of you thought of this, which I thought would be too easy for the genius people here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you everyone. and Good bye.&lt;/p&gt;&#xA;" OwnerUserId="2062" LastEditorUserId="11110" LastEditDate="2019-02-16T23:31:59.777" LastActivityDate="2019-02-16T23:31:59.777" Title="The Motionless Floating Eraser" Tags="|lateral-thinking|science|" AnswerCount="4" CommentCount="16" ClosedDate="2014-12-06T23:00:05.420" ContentLicense="CC BY-SA 4.0" />
  <row Id="6001" PostTypeId="1" AcceptedAnswerId="6052" CreationDate="2014-12-06T11:39:00.853" Score="10" ViewCount="848" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I have an Earl to which I was born&lt;br&gt;&#xA;  not on the beach and without a thorn&lt;br&gt;&#xA;  No Rand; I am not the Grail&lt;br&gt;&#xA;  I assist your game of cards, no fail  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Without my walls I'm not complete&lt;br&gt;&#xA;  the epitome of me is no easy feat&lt;br&gt;&#xA;  when I bleed, don't call that Cop&lt;br&gt;&#xA;  for he is unable to make it stop  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Longer and longer I grow hard&lt;br&gt;&#xA;  but I am not the male shard&lt;br&gt;&#xA;  evolved with gluttonous imagination&lt;br&gt;&#xA;  created with utter anticipation     &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I am a dam to those inside&lt;br&gt;&#xA;  terrible by definition, porous my hide&lt;br&gt;&#xA;  Humble farmers are my starts&lt;br&gt;&#xA;  Suffocated so as to extend my parts  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Rife with many shapes and proportions&lt;br&gt;&#xA;  A magnitude of functions suit my options&lt;br&gt;&#xA;  heat me from both sides for fun&lt;br&gt;&#xA;  I'm not, not insane I'm now not one&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5135" LastEditorUserId="6178" LastEditDate="2014-12-20T17:03:10.873" LastActivityDate="2014-12-20T17:03:10.873" Title="Name that Sandwich: I am a catalyst" Tags="|riddle|word|" AnswerCount="3" CommentCount="18" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6004" PostTypeId="2" ParentId="6000" CreationDate="2014-12-06T14:52:24.530" Score="8" Body="&lt;p&gt;A few possibilities that spring to mind, some more practical than others:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Eraser is in the process of falling at the moment the photo is taken (shutter speed of 1/250 or faster will 'freeze' motion).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Photo is taken in 0-g or low-g environment e.g. space station, descending aircraft, descending lift.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Surface tension: water, oil, etc.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Static electricity.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Frame of reference is rotating (picture is sideways, camera is spinning with other equipment), Coriolis force is in play. E.g. all equipment is on a wheeled trolley being spun around fast.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Wind: a fan below the frame of the photo is pushing air upwards; a vacuum cleaner above the frame of the photo is pulling air upwards.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Black hole outside frame of photo is causing gravity distortion.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Descriptive text is untrue: picture has been modified, eraser has been modified (e.g. hollowed out at one end to shift centre of gravity, concealed magnet, etc etc).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Rubber has perished, stuck itself to the paper.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Glue.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;That's before we even get onto the possibilities of parallel universes....&lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="432" LastEditDate="2014-12-06T15:53:13.467" LastActivityDate="2014-12-06T15:53:13.467" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="6005" PostTypeId="2" ParentId="6000" CreationDate="2014-12-06T14:57:06.380" Score="2" Body="&lt;p&gt;Clearly the eraser is touching the register at more than one points. Rubber often sticks to paper after a long time, like a few days. So, i think, you left the eraser in the register for a day, with more than half of it lying outside. It got stuck then, when you opened the register.&lt;/p&gt;&#xA;" OwnerUserId="6216" LastActivityDate="2014-12-06T14:57:06.380" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6006" PostTypeId="2" ParentId="5190" CreationDate="2014-12-06T15:22:19.950" Score="6" Body="&lt;p&gt;I have a static 7-weighing schedule:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;{3,5,6,7}&lt;/li&gt;&#xA;&lt;li&gt;{1,4,6,7,8}&lt;/li&gt;&#xA;&lt;li&gt;{1,2,5,7,8,9}&lt;/li&gt;&#xA;&lt;li&gt;{1,2,3,6,8,9,10}&lt;/li&gt;&#xA;&lt;li&gt;{2,3,4,7,9,10,11}&lt;/li&gt;&#xA;&lt;li&gt;{3,4,5,8,10,11,12}&lt;/li&gt;&#xA;&lt;li&gt;{1,4,5,6,9,11,12}&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This can also be given in terms of ON/OFF (with 1/0) given the array of 12 coins:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;0 0 1 0 1 1 1 0 0 0 0 0 &lt;/li&gt;&#xA;&lt;li&gt;1 0 0 1 0 1 1 1 0 0 0 0 &lt;/li&gt;&#xA;&lt;li&gt;1 1 0 0 1 0 1 1 1 0 0 0 &lt;/li&gt;&#xA;&lt;li&gt;1 1 1 0 0 1 0 1 1 1 0 0 &lt;/li&gt;&#xA;&lt;li&gt;0 1 1 1 0 0 1 0 1 1 1 0 &lt;/li&gt;&#xA;&lt;li&gt;0 0 1 1 1 0 0 1 0 1 1 1 &lt;/li&gt;&#xA;&lt;li&gt;1 0 0 1 1 1 0 0 1 0 1 1&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I am still considering how to properly explain it in an intuitive way. The pattern in the above bit-array-set is obvious - i.e. some kind of shifting pattern. But as to how/why this achieves the goal (c.f. &lt;a href=&quot;https://puzzling.stackexchange.com/a/5417/2484&quot;&gt;my other answer&lt;/a&gt;) is not yet clear.&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2014-12-06T23:23:08.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6007" PostTypeId="2" ParentId="6000" CreationDate="2014-12-06T16:33:31.977" Score="7" Body="&lt;ol&gt;&#xA;&lt;li&gt;glue or anything sticky&lt;/li&gt;&#xA;&lt;li&gt;a magnet in a hole in the eraser&lt;/li&gt;&#xA;&lt;li&gt;lead in a hole in the eraser&lt;/li&gt;&#xA;&lt;li&gt;a hidden paperclip stuck between the pages&lt;/li&gt;&#xA;&lt;li&gt;a thumbtack&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There are too many answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an unfolded paper clip stuck between the pages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WEgG4.jpg&quot; alt=&quot;paperclip solution&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, a new possibility:&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;6&quot;&gt;&#xA;&lt;li&gt;A glass table.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2014-12-06T17:39:52.963" LastActivityDate="2014-12-06T17:39:52.963" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6008" PostTypeId="2" ParentId="6000" CreationDate="2014-12-06T20:09:19.197" Score="1" Body="&lt;p&gt;The notebook is tilted at a steep angle. The eraser balances on the upper edge aided by a little surface friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The background is provided by a printed image that has been rotated to make it look like the notebook is being held horizontally.&lt;/p&gt;&#xA;" OwnerUserId="5550" LastActivityDate="2014-12-06T20:09:19.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6009" PostTypeId="1" AcceptedAnswerId="6017" CreationDate="2014-12-07T09:08:59.227" Score="11" ViewCount="1637" Body="&lt;p&gt;I have an admission to make: I very rarely put my DVDs back in the case they came from. Whenever I watch a new DVD I just put whatever DVD is the player in the new case. I put the cases back on the shelf wherever they'll fit without worrying about any sort of order. My collection is very disorganised! Now I want to watch my favourite movie &lt;em&gt;The Life of Pi&lt;/em&gt; but I have no idea which case I put it back in! The right case has &lt;em&gt;The Fifth Element&lt;/em&gt; in it. I've tried opening a few cases at random but I have so many it's going to take me forever! I know I've watched it fairly recently, I just can't remember which case I put it back into or all of the DVDs I have recently watched. The DVD currently in the player is &lt;em&gt;Four Weddings and a Funeral&lt;/em&gt; and its case is empty (it happened to be in the right case before I watched it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes, when people come over, I tidy the house, which means I take the DVD out of the player, stick it into the empty case and put it back on the shelf. And sometimes I put the DVD into the already empty case regardless of whether it's the right one or not (when I select a new DVD I briefly have two empty cases. The old DVD might end up in either of them, but usually the new case). I don't really have a system which is why they're so jumbled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not going to buy another copy, I don't belong to a video library, none of my friends have it and my internet is too slow to download it. I only have one DVD player. I know the DVD is in the wrong case somewhere on my shelf. I must find it in my own collection as quickly as possible! How can I do it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This is a logic puzzle. It is not a trick question. It does not rely on an obscure interpretation of certain words or lateral thinking to get around any conditions. I believe I have given sufficient clues to be able to answer the question but I will be happy to clarify any points. I don't believe I need to provide any extra clues.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2058" LastEditorUserId="2058" LastEditDate="2014-12-08T06:25:41.560" LastActivityDate="2014-12-08T13:37:53.647" Title="Help me find my favourite DVD" Tags="|logical-deduction|" AnswerCount="4" CommentCount="21" ContentLicense="CC BY-SA 3.0" />
  <row Id="6010" PostTypeId="2" ParentId="6009" CreationDate="2014-12-07T09:42:22.050" Score="0" Body="&lt;p&gt;It's a matter of Derangements (arrangements that map each dvd to a case ,other than its own)&#xA;It is an &quot;NP-complete&quot; problem (that means HARD) to determine whether a given permutation group (described by a given set of permutations that generate it n! for n cases and n DVD's) contains any derangements&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Derangement&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Derangement&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5051" LastActivityDate="2014-12-07T09:42:22.050" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6012" PostTypeId="2" ParentId="6009" CreationDate="2014-12-07T11:45:33.580" Score="-4" Body="&lt;p&gt;I know this is a &lt;strong&gt;logical&lt;/strong&gt; puzzle so the following answer is not what you're looking for, but I think it is a &lt;strong&gt;valid&lt;/strong&gt; answer if &lt;em&gt;lateral thinking&lt;/em&gt; is allowed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Put the following announcement into a (free) newspaper / online-blog etc.: &lt;br&gt;&#xA;    &lt;strong&gt;&quot;Free video rental for the curious!&lt;/strong&gt; Feel bored? Are your ready to try something &lt;em&gt;unexpected&lt;/em&gt; ? I'm giving out (lending!) DVDs &lt;em&gt;for free&lt;/em&gt; - for a limited time only. And this is how it works: You come by my place (see Google-map below) and I'll hand you out a &lt;em&gt;random&lt;/em&gt; DVD case with a &lt;em&gt;random&lt;/em&gt; DVD inside! No guarantees whatsoever about the content - but all my films are best quality mainstream films (no porn). You've a 24hrs period to watch and return the DVD, and &lt;strong&gt;the only condition:&lt;/strong&gt; You have to tell the title of the film you've watched. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now simply wait and the problem will solve itself. Is it the &lt;em&gt;fastest&lt;/em&gt; solution? I don't know yet as no other answer has been posted. It will depend on the &lt;em&gt;size&lt;/em&gt; of you library though, because it &lt;em&gt;scales&lt;/em&gt;. The bigger your library, the better my solution in comparison!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course there would be a &lt;em&gt;faster&lt;/em&gt; alternative:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Invite all your friends (or trustworthy people you know) for a &quot;picke &amp;amp; watch&quot; DVD evening. Everybody may come and take (borrow!) as many DVD as he likes &lt;em&gt;except&lt;/em&gt; the one you want to watch for yourself. Condition: Open the DVD case at once and check.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastActivityDate="2014-12-07T11:45:33.580" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6015" PostTypeId="2" ParentId="5900" CreationDate="2014-12-07T13:57:15.953" Score="2" Body="&lt;p&gt;The hidden message is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SHERLOCK I HAVE FOUND YOU&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I didn't use the clue &quot;where is it the pigs live?&quot;, but instead I used:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://observeanalysededuce.files.wordpress.com/2010/08/pigpen.png&quot; rel=&quot;nofollow&quot;&gt;http://observeanalysededuce.files.wordpress.com/2010/08/pigpen.png&lt;/a&gt;&lt;br&gt;&#xA; If you click into the link above, you will see four different grids. And according to the grids, we can find that different symbols (just like in the message) mean different letters. So, if we follow the grids, we can find the hidden message.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5866" LastEditorUserId="5395" LastEditDate="2014-12-08T07:29:58.040" LastActivityDate="2014-12-08T07:29:58.040" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6016" PostTypeId="1" AcceptedAnswerId="6018" CreationDate="2014-12-07T15:41:38.177" Score="4" ViewCount="524" Body="&lt;p&gt;&lt;em&gt;A long time ago, in a galaxy far, far away...&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;These are troublesome times: The Galactic Republic is gone, the Jedi were exterminated due to Order 66, Vader and Palpatine rule the galaxy with absolute power as an evil Empire. The Empire had built a Death Star, which was used to destroy Alderaan, but the Rebel Alliance managed to destroy it. However, this is not enough because The Empire Strikes Back and what we really need is The Return Of Jedi.&lt;/p&gt;&#xA;&lt;p&gt;Here you are, in Tatooine. You are a podracer mechanic working for the Reborn Dragons, a podracing team that uses podracing as a way to raise funds to support the Rebel Alliance. There are three categories of podracing, according to the &lt;em&gt;Engine Power &lt;strong&gt;Y&lt;/strong&gt;ield&lt;/em&gt;: the Beginners &lt;span class=&quot;math-container&quot;&gt;$(Y=1)$&lt;/span&gt;, the Moderate &lt;span class=&quot;math-container&quot;&gt;$(Y=2)$&lt;/span&gt; and the Masters &lt;span class=&quot;math-container&quot;&gt;$(Y=3)$&lt;/span&gt;. You are still unsure about the category that you should choose to enroll for racing.&lt;/p&gt;&#xA;&lt;p&gt;Your job is to tune a podracer in order to ensure a victory in the next race. You are new to the job, since the experienced mechanic was captured by the Galactic Empire and imprisoned. The Rebel Alliance even managed to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4810/crypto-wordsearch-gram-you-are-a-member-of-the-rebel-alliance-and-a-traitor&quot;&gt;send him messages encoded in crosswords&lt;/a&gt;, but this is not enough.&lt;/p&gt;&#xA;&lt;p&gt;There are many variables in your podracer:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Power &lt;strong&gt;R&lt;/strong&gt;ecovery&lt;/em&gt;: You can set this as Low &lt;span class=&quot;math-container&quot;&gt;$(R=1)$&lt;/span&gt;, Moderate &lt;span class=&quot;math-container&quot;&gt;$(R=2)$&lt;/span&gt; or High &lt;span class=&quot;math-container&quot;&gt;$(R=3)$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;&lt;strong&gt;E&lt;/strong&gt;nergy consumption&lt;/em&gt;: You can set this as Low &lt;span class=&quot;math-container&quot;&gt;$(E=1)$&lt;/span&gt;, Moderate &lt;span class=&quot;math-container&quot;&gt;$(E=2)$&lt;/span&gt; or High &lt;span class=&quot;math-container&quot;&gt;$(E=3)$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;&lt;strong&gt;L&lt;/strong&gt;evitation field strength&lt;/em&gt;: You can set this as Low &lt;span class=&quot;math-container&quot;&gt;$(L=1)$&lt;/span&gt;, Moderate &lt;span class=&quot;math-container&quot;&gt;$(L=2)$&lt;/span&gt; or High &lt;span class=&quot;math-container&quot;&gt;$(L=3)$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Engine &lt;strong&gt;O&lt;/strong&gt;rbiter field strength&lt;/em&gt;: You can set this as Low &lt;span class=&quot;math-container&quot;&gt;$(O=1)$&lt;/span&gt;, Moderate &lt;span class=&quot;math-container&quot;&gt;$(O=2)$&lt;/span&gt; or High &lt;span class=&quot;math-container&quot;&gt;$(O=3)$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;&lt;strong&gt;H&lt;/strong&gt;ull &lt;strong&gt;S&lt;/strong&gt;trength&lt;/em&gt;: You can set this as Low &lt;span class=&quot;math-container&quot;&gt;$(HS=1)$&lt;/span&gt;, Moderate &lt;span class=&quot;math-container&quot;&gt;$(HS=2)$&lt;/span&gt; or High &lt;span class=&quot;math-container&quot;&gt;$(HS=3)$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;&lt;strong&gt;S&lt;/strong&gt;pecific &lt;strong&gt;A&lt;/strong&gt;cceleration&lt;/em&gt;: You can set this as Low &lt;span class=&quot;math-container&quot;&gt;$(SA=1)$&lt;/span&gt;, Moderate &lt;span class=&quot;math-container&quot;&gt;$(SA=2)$&lt;/span&gt; or High &lt;span class=&quot;math-container&quot;&gt;$(SA=3)$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;The best category may be deduced by minimising the quantity &lt;span class=&quot;math-container&quot;&gt;$x = (Y - 2)^2 + (R - 2)^2 + (E - 2)^2 + (L - 2)^2 + (O - 2)^2 + (HS - 2)^2 + (SA - 2)^2$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;You should also choose two pilots to enroll, and you have six candidates with the following attributes:&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th&gt;Pilot&lt;/th&gt;&#xA;&lt;th&gt;Skill&lt;/th&gt;&#xA;&lt;th&gt;Fear&lt;/th&gt;&#xA;&lt;th&gt;Reflexes&lt;/th&gt;&#xA;&lt;th&gt;Trust/confidence&lt;/th&gt;&#xA;&lt;th&gt;Training&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;P1&lt;/td&gt;&#xA;&lt;td&gt;1 - Poor&lt;/td&gt;&#xA;&lt;td&gt;1 - Fearless&lt;/td&gt;&#xA;&lt;td&gt;1 - Slow&lt;/td&gt;&#xA;&lt;td&gt;3 - High&lt;/td&gt;&#xA;&lt;td&gt;1 - Rookie&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;P2&lt;/td&gt;&#xA;&lt;td&gt;1 - Poor&lt;/td&gt;&#xA;&lt;td&gt;3 - Fearful&lt;/td&gt;&#xA;&lt;td&gt;2 - Moderate&lt;/td&gt;&#xA;&lt;td&gt;2 - Moderate&lt;/td&gt;&#xA;&lt;td&gt;3 - Veteran&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;P3&lt;/td&gt;&#xA;&lt;td&gt;2 - Moderate&lt;/td&gt;&#xA;&lt;td&gt;2 - Moderate&lt;/td&gt;&#xA;&lt;td&gt;2 - Moderate&lt;/td&gt;&#xA;&lt;td&gt;2 - Moderate&lt;/td&gt;&#xA;&lt;td&gt;2 - Moderate&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;P4&lt;/td&gt;&#xA;&lt;td&gt;3 - Elite&lt;/td&gt;&#xA;&lt;td&gt;3 - Fearful&lt;/td&gt;&#xA;&lt;td&gt;3 - Fast&lt;/td&gt;&#xA;&lt;td&gt;1 - Low&lt;/td&gt;&#xA;&lt;td&gt;3 - Veteran&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;P5&lt;/td&gt;&#xA;&lt;td&gt;1 - Poor&lt;/td&gt;&#xA;&lt;td&gt;3 - Fearful&lt;/td&gt;&#xA;&lt;td&gt;1 - Slow&lt;/td&gt;&#xA;&lt;td&gt;1 - Low&lt;/td&gt;&#xA;&lt;td&gt;1 - Rookie&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;P6&lt;/td&gt;&#xA;&lt;td&gt;2 - Moderate&lt;/td&gt;&#xA;&lt;td&gt;2 - Moderate&lt;/td&gt;&#xA;&lt;td&gt;1 - Slow&lt;/td&gt;&#xA;&lt;td&gt;2 - Moderate&lt;/td&gt;&#xA;&lt;td&gt;1 - Rookie&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;Considering that each pilot attribute has the same weight (but could be either positive or negative, i.e. weight either &lt;span class=&quot;math-container&quot;&gt;$+1$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$-1$&lt;/span&gt;), who are the two best pilots to choose? How is the pilots fear and confidence? What is the best way to tune the podracer (please, list each attribute one-by-one)? What is the best category? &lt;strong&gt;Post an answer answering those questions.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Please, post sensitive parts of your answer in spoilers (use lines starting with &lt;code&gt;&amp;gt;!&lt;/code&gt; for those). To break lines inside spoilers, use &lt;code&gt;&amp;lt;br&amp;gt;&lt;/code&gt; or add 2 spaces at the end of the line.&lt;/p&gt;&#xA;" OwnerUserId="5044" LastEditorUserId="69582" LastEditDate="2021-01-15T16:44:54.993" LastActivityDate="2021-01-15T16:44:54.993" Title="Tune a podracer and choose the pilots by solving the podracer tuning equation" Tags="|mathematics|optimization|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6017" PostTypeId="2" ParentId="6009" CreationDate="2014-12-07T16:59:17.137" Score="15" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All of your DVDs are in n rings, where n &gt;= 2 (&lt;em&gt;Four Weddings And a Funeral&lt;/em&gt; is in a ring of length 1). So if you open a DVD at random, it may or may not be in the ring that contains &lt;em&gt;The Life Of Pi&lt;/em&gt;.&lt;br&gt;&#xA; However, we do know one movie that is in the same ring as &lt;em&gt;The Life Of Pi&lt;/em&gt;. It is &lt;em&gt;The 5th Element&lt;/em&gt;. So if you start with &lt;em&gt;The 5th Element&lt;/em&gt;'s case, you are guaranteed to be in the ring that contains &lt;em&gt;The Life Of Pi&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You should really consider getting yourself a NAS and a copy of Plex (or other similar software), so that you can stream your movies locally and avoid this problem in the future.&lt;/p&gt;&#xA;" OwnerUserId="375" LastEditorUserId="5044" LastEditDate="2014-12-07T20:08:58.800" LastActivityDate="2014-12-07T20:08:58.800" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6018" PostTypeId="2" ParentId="6016" CreationDate="2014-12-07T22:20:32.663" Score="4" Body="&lt;p&gt;I feel like I'm missing something because the optimization seems rather straightforward. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Enter the competition with yield &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Y=2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and parameters&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;R=E=L=O=HS=HA=2&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; these mininimize the quantity &lt;code&gt;x&lt;/code&gt; by making each square be &lt;code&gt;0&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the pilot, choose &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pilots 4 and 3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because they have &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the greatest and second greatest total weighted attributes, with skill, reflexes, trust/confidence, training being the natural positives, and fear being negative. Actually, even if fear is a positive, the answer is the same.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2014-12-08T06:10:40.073" LastActivityDate="2014-12-08T06:10:40.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6022" PostTypeId="1" AcceptedAnswerId="6023" CreationDate="2014-12-08T02:50:52.820" Score="2" ViewCount="359" Body="&lt;p&gt;This is my first time to make a puzzle so it might be simply and easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are trapped in a huge room. You found 5 different pictures on the wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Picture A- A red circle with a green background&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Picture B- A purple oval with a yellow background&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Picture C- A green square with a purple background&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Picture D- A blue triangle with an orange background&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Picture E- A yellow semicircle with a cyan background&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, you saw a folded note in front of you. You unfold it and it says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dear ??,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Happy to be trap? I wish you are. You wanna escape? Here's your chance! Go flip the correct picture then you free. Go flip the wrong one then you will be a shark lunch. Hope you find it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From: THE KILLER&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You were panic as you didn't know where to start. However, you realized something is at the back of the note .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From: The same guy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today, I am being generous and want to give some hints&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1: 551a8b&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 2: One of the colour is primary for light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you stopped panicing, found the correct picture and escape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is the correct picture? Answer with explainations. Spoilers will be out in 2 days. If no correct answer is given, I will answer the question&lt;/p&gt;&#xA;" OwnerUserId="5866" LastActivityDate="2014-12-08T03:12:01.263" Title="Choose a picture" Tags="|mystery|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6023" PostTypeId="2" ParentId="6022" CreationDate="2014-12-08T03:00:08.393" Score="4" Body="&lt;p&gt;The correct answer is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Picture C- A green square with a purple background  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because hex 551a8b is a purple and green is a primary color for light.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="2383" LastEditDate="2014-12-08T03:12:01.263" LastActivityDate="2014-12-08T03:12:01.263" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6024" PostTypeId="2" ParentId="5857" CreationDate="2014-12-08T03:17:17.993" Score="4" Body="&lt;p&gt;Suspect its:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Suicide and the owner try to frame it on the Girl and Boy&lt;br&gt;&#xA;Deliberately ordering 20 bread to ensuring the baker know there are people coming to his house and ensure someone will find the corpse&lt;br&gt;&#xA;It was the owner who break the window and move all the furniture out since he's the owner, he have plenty of time to prepare and do stuff without getting suspicion &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How was owner killed:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;As per what one of the answer, suicide by standing on ice. The beer smashing the window is to frame the couple and also to ensure the water get evaporated  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2258" LastActivityDate="2014-12-08T03:17:17.993" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6025" PostTypeId="2" ParentId="496" CreationDate="2014-12-08T10:58:50.113" Score="0" Body="&lt;p&gt;$$\begin{array}&#xA;LLCM(2,3,4,5,6,7,8,9,10)&amp;amp;=2\times3\times2\times5\times7\times2\times3\\&#xA;&amp;amp;=2520\end{array}$$&#xA;Let $x+1$ be the solution&lt;br&gt;&#xA;And $x,n,k\in\mathbb{N}$&#xA;$$x\equiv0\mod2520$$&#xA;$$x=2520n$$&#xA;$$x+1\equiv0\mod11$$&#xA;$$2520n+1\equiv0\mod11$$&#xA;$$2520n\equiv10\mod11$$&#xA;$$2520\equiv1\mod11$$&#xA;$$n\equiv10\mod11$$&#xA;$$n=11k+10$$&#xA;$$x+1=2520(11k+10)+1$$&lt;/p&gt;&#xA;" OwnerUserId="6076" LastActivityDate="2014-12-08T10:58:50.113" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6026" PostTypeId="2" ParentId="6009" CreationDate="2014-12-08T11:01:12.597" Score="2" Body="&lt;p&gt;Take the DVD cases to be indices of the array elements in the following explanation. Let the DVDs themselves to be the values in the array elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us establish the means by which this problem may be solved:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Take the array $C = [1,\cdots,n]$. It contains each integer from $1$ to $n$ and is of size $n$. Consider any permutation of the elements of $C$. There are $n!$ ways $C$ can be shuffled. We have a searching algorithm which is to find $1$.&#xA;&lt;br&gt;&lt;br&gt;&#xA;Start with checking in $C[1]$.&lt;br&gt; &#xA;If $C[1] = 1$ then we have found $1$ in $1$ check.&lt;br&gt;&#xA;If $C[1] = x$ where $x\not=1$ then we check $C[x]$. If $C[x] = 1$ then we have found $1$ in $2$ checks.&lt;br&gt;&#xA;If $C[x] = y$ where $y\not=1$ then we check $C[y]$. If $C[y] = 1$ then we have found $1$ in $3$ checks.&lt;br&gt;&#xA;If $C[y] = z$ where $z\not=1$ then we check $C[z]$.&#xA;&lt;br&gt;&#xA; This process can be continued and it forms a &lt;em&gt;chain&lt;/em&gt;. Clearly, $x \not= y \not= z$ - so long as $C[i] \not= 1$, no number previously encountered will be repeated ($Lemma1$). The claim is that this &lt;em&gt;chain&lt;/em&gt; will include $1$ &lt;strong&gt;because&lt;/strong&gt; we started with $C[1]$.&#xA;&lt;br&gt;&lt;br&gt;&#xA;The proof is by contradiction. Assume that $1$ is somewhere outside the &lt;em&gt;chain&lt;/em&gt;.&#xA;&lt;br&gt;&lt;br&gt;&#xA;Recall that $|C|=n$. Start from $C[1]$. Clearly, any &lt;em&gt;chain&lt;/em&gt; including every element of $C$ will include $1$. We must be looking for a &lt;em&gt;chain&lt;/em&gt; of size less than $n$. Such a &lt;em&gt;chain&lt;/em&gt; must moreover be a &lt;em&gt;loop&lt;/em&gt; - otherwise it must continue and consume all elements of $C$, and $n$ is finite. Therefore, we know that we are looking for a &lt;em&gt;loop&lt;/em&gt; starting from $C[1]$ which does not contain $1$. A &lt;em&gt;loop&lt;/em&gt; is a &lt;em&gt;chain&lt;/em&gt; of length $z$ where the last element visited contains the value contained in the first element visited. By $Lemma1$ given a &lt;em&gt;loop&lt;/em&gt; of length $z$, the loop must involve $z$ elements &lt;em&gt;and&lt;/em&gt; only the values being the indices of the elements. So, a &lt;em&gt;loop&lt;/em&gt;, in order to arrive again at $C[1]$ must have included $1$. Contradiction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As to the concerns of efficiency, in particular, to defeating randomly checking:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If we add one constraint (which may exist in many forms), that the &lt;em&gt;loop&lt;/em&gt; in which the target, $1$, can be found, is smaller than $n$, then we can guarantee better than random chance performance.&#xA;&lt;br&gt;&lt;br&gt;&#xA;Randomly checking has a worst-case runtime of $n$ checks.&#xA;&lt;br&gt;&lt;br&gt;&#xA;We can constrain the &lt;em&gt;loop&lt;/em&gt; housing $1$, $L$, which will be of size $z$, to have $z &amp;lt; n$ by simple asserting that there is a guarantee that some of the elements have not been moved - or that there exists a &lt;em&gt;loop&lt;/em&gt; outside $L$. NB: These constraints/guarantees are equivalent since is an element has not been moved then it is a &lt;em&gt;loop&lt;/em&gt; of size $1$.&#xA;&lt;br&gt;&lt;br&gt;&#xA;Checking $L$ as described above has a worst-case runtime of $z$ checks.&#xA;&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since $z &amp;lt; n$ this strategy is &quot;better&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-12-08T13:37:53.647" LastActivityDate="2014-12-08T13:37:53.647" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6027" PostTypeId="2" ParentId="5937" CreationDate="2014-12-08T13:37:50.357" Score="3" Body="&lt;p&gt;OP: Seems that it was: &lt;strong&gt;FIRST PRINCIPLE&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Could it be &lt;strong&gt;Basic supposition&lt;/strong&gt;? I've seen it in some (really old) books about philosophy and physics, but the full phrase does not have a Wikipedia entry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scratch, that. You are probably searching for &lt;a href=&quot;https://en.wikipedia.org/wiki/Tacit_assumption&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Tacit assumption&lt;/strong&gt;&lt;/a&gt;, also known as &lt;strong&gt;Implicit assumption&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;SECOND EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could it be &lt;a href=&quot;https://en.wikipedia.org/wiki/Unstated_assumption&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Unstated assumption&lt;/strong&gt;&lt;/a&gt; as described in the &lt;strong&gt;second&lt;/strong&gt; paragraph on the Wikipedia link?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THIRD EDIT&lt;/strong&gt; (this is getting fun)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or maybe &lt;a href=&quot;http://en.wikipedia.org/wiki/First_principle&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;First principle&lt;/strong&gt;&lt;/a&gt; a.k.a. &lt;strong&gt;Foundational proposition/assumption&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus question:&lt;/strong&gt; How many terms for practically the same thing are there?&lt;/p&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="2484" LastEditDate="2014-12-09T08:17:44.303" LastActivityDate="2014-12-09T08:17:44.303" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="6028" PostTypeId="2" ParentId="5980" CreationDate="2014-12-08T14:37:02.847" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 15 minutes.&lt;br&gt;&#xA; First 4 minutes, starts with 1 person, increases group by 1 until size 5. (+4)&lt;br&gt;&#xA; Then, need to make groups of 2, so 5 choose 2 is 10, but down by 2 since the groups of 2 and 3 from part 1 don't account. (+8)&lt;br&gt;&#xA; Add 1 person. At that point, the small group of 4 can do 2 arrangements (all 4, 2+2, if you left the right combinations from their 5 choose 2). (+2)&lt;br&gt;&#xA; Finally, add 1 person to the big group, then finally tell Jane. You can't have the final untold group of 2, since all those combinations have already been done. (+1)&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Edited:&lt;/strong&gt; down to 15 min.&lt;br&gt;&#xA;&lt;br&gt;&#xA; AB | CDEFGHIJ&lt;br&gt;&#xA; ABC | DEFGHIJ&lt;br&gt;&#xA; ABCD | EFGHIJ&lt;br&gt;&#xA; ABCDE | FGHIJ&lt;br&gt;&#xA; AC | BDE | HJ | FGI&lt;br&gt;&#xA; AD | BCE | FH | GIJ&lt;br&gt;&#xA; AE | BCD | FI | GHJ&lt;br&gt;&#xA; BC | ADE | FJ | GHI&lt;br&gt;&#xA; BD | ACE | GH | FIJ&lt;br&gt;&#xA; BE | ACD | GI | FHJ&lt;br&gt;&#xA; CD | ABE | GJ | FHI&lt;br&gt;&#xA; CE | ABD | HI | FGJ&lt;br&gt;&#xA; ABCDEF | GHIJ&lt;br&gt;&#xA; ABCDEFG | HIJ&lt;br&gt;&#xA; ABCDEFGH | IJ&lt;br&gt;&#xA; ABCDEFGHIJ  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2014-12-10T14:03:50.190" LastActivityDate="2014-12-10T14:03:50.190" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6029" PostTypeId="1" CreationDate="2014-12-08T18:35:19.303" Score="31" ViewCount="3284" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;This puzzle has 3 levels of increasing complexity. Each &amp;quot;level&amp;quot; is separate and complete, so feel free to post partial solutions to the individual levels only. &lt;br&gt;&#xA;I'm most interested in the principal questions asked in the text-boxes for level 3, but I think the 'simpler' levels are also worth considering.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;The task&lt;/h1&gt;&#xA;&lt;p&gt;You are in charge of defining traffic lights and traffic signs on crossroads in your district.&lt;/p&gt;&#xA;&lt;p&gt;Your district is in a silent backwater area, so &lt;strong&gt;there is zero traffic, except during rush-hour&lt;/strong&gt; when suddenly a lot of cars drive through.&lt;/p&gt;&#xA;&lt;p&gt;You have been charged with the task to &lt;strong&gt;minimize the total time&lt;/strong&gt; traffic spends in your district, so that exhaust fumes are reduced to a minimum. You do not care about &lt;em&gt;individual&lt;/em&gt; drivers. From your perspective, it is okay to have one person wait for the whole hour at a traffic light while all other traffic passes by - as long as the total integrated time of all traffic is minimized.&lt;/p&gt;&#xA;&lt;p&gt;As only rush-hour is important, this puzzle counts time in seconds starting with 0 at the beginning of the rush-hour. From time 0 onwards cars will continuously arrive at the border of your district and drive in a straight line until they leave the district at the other end. &lt;strong&gt;Every second a car is in your district, it adds +1 to the overall &amp;quot;fumes&amp;quot;&lt;/strong&gt; regardless of its motion.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;The overall &amp;quot;fumes&amp;quot; is the property which you want to minimize.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;h1&gt;The simplified traffic rules&lt;/h1&gt;&#xA;&lt;p&gt;For the purpose of this puzzle, traffic is described by the following, simplified rules:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A car requires &lt;strong&gt;30sec&lt;/strong&gt; to pass &lt;strong&gt;through one &lt;em&gt;section&lt;/em&gt;&lt;/strong&gt; of your district. A &lt;em&gt;section&lt;/em&gt; is the distance between border and crossing or between two crossings. The length of these sections is of no importance, and neither is traffic level on the road. The 'space' in front of a crossing is large enough to hold all traffic, so &lt;strong&gt;back-queuing does not affect anything except the 'crossing time'&lt;/strong&gt; of the following crossing as specified below. (&lt;em&gt;The 30sec will only add an offset to the fume-level, but it prevents 'speeding' through the whole puzzle on no-traffic.&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Passing through a crossing does take 1 sec.&lt;/strong&gt; (Only at green light of course!) There can only be 1 car in each direction in the crossing during this 1 sec.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If a crossing is amber or red when a driver reaches it, he will stop and stand in queue. He will stand there until it is green.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When a traffic light turns amber-to-green, &lt;strong&gt;starting takes&lt;/strong&gt; each driver &lt;strong&gt;1 sec per car in front&lt;/strong&gt; to reach (and then pass) the crossing. So the first car goes immediately, the second needs 1 sec, the third 2 sec and so on. Any car which doesn't get across until the traffic light turns amber again, stops for the next green period in the new queuing position.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;No car can overtake&lt;/strong&gt; another car, nor can two cars be in the same spot at the same time. (So in a queue, the front-car will delay all others.)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;h1&gt;Traffic light rules&lt;/h1&gt;&#xA;&lt;p&gt;Every crossing has a single, individual traffic light which will periodically go through the following cycles:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;code&gt;state |&lt;/code&gt;: Green for North-South while red for East-West&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;state X&lt;/code&gt;: Amber for all directions&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;state -&lt;/code&gt;: Red for North-South while green for East-West&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;state X&lt;/code&gt;: Amber for all directions&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;It is up to you to define:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Variable &lt;code&gt;S&lt;/code&gt;&lt;/strong&gt; for &lt;em&gt;starting state&lt;/em&gt;, which is the state the traffic light is in at begin of the rush-hour (0 sec). It can either be &lt;code&gt;state |&lt;/code&gt; or &lt;code&gt;state -&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Variable &lt;code&gt;V&lt;/code&gt;&lt;/strong&gt; for &lt;em&gt;vertical&lt;/em&gt;, which is the length of &lt;code&gt;state |&lt;/code&gt; in seconds .&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Variable &lt;code&gt;H&lt;/code&gt;&lt;/strong&gt; for &lt;em&gt;horizontal&lt;/em&gt;, which is the length of &lt;code&gt;state -&lt;/code&gt; in seconds.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Variable &lt;code&gt;A&lt;/code&gt;&lt;/strong&gt; for &lt;em&gt;amber&lt;/em&gt;, which is the length of &lt;code&gt;state X&lt;/code&gt; in seconds. &lt;br&gt;&#xA;Note that there is only a single length for the amber phases. &lt;bR&gt;&#xA;Also, for security reasons, &lt;strong&gt;the amber phase must not be shorter than 2sec&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;LEVEL 1 - district map and traffic&lt;/h1&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Q6jqz.jpg&quot; alt=&quot;Level 1&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Starting with rush-hour ( 0 sec ) and immediately stopping after rush-hour ( &amp;gt;3600 sec ) cars will enter at the district borders as follows:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;N1:&lt;/strong&gt; A car every 7 sec starting with 0. ( 0, 7, 14, 21, ... )&lt;br&gt;&#xA;&lt;strong&gt;S1:&lt;/strong&gt; A car every 9 sec starting with 0. ( 0, 9, 18, 27, ... )&lt;br&gt;&#xA;&lt;strong&gt;W1:&lt;/strong&gt; A car every 4 sec starting with 0. ( 0, 4, 8, 12, ... )&lt;br&gt;&#xA;&lt;strong&gt;E1:&lt;/strong&gt; A car every 8 sec starting with 0. ( 0, 8, 16, 24, ... )&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;What does the ideal (single) traffic-light setting look like?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;LEVEL 2 - district map and traffic&lt;/h1&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bz69v.jpg&quot; alt=&quot;level 2&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Starting with rush-hour ( 0 sec ) and immediately stopping after rush-hour ( &amp;gt;3600 sec ) cars will enter at the district borders as follows:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;N1:&lt;/strong&gt; A car every 24 sec starting with 0. ( 0, 24, 48, ... )&lt;br&gt;&#xA;&lt;strong&gt;S1:&lt;/strong&gt; A car every 24 sec starting with 2. ( 2, 26, 50, ... )&lt;br&gt;&#xA;&lt;strong&gt;N2:&lt;/strong&gt; A car every 15 sec starting with 2. ( 2, 17, 32, ... )&lt;br&gt;&#xA;&lt;strong&gt;S2:&lt;/strong&gt; A car every 19 sec starting with 0. ( 0, 19, 38, ... )&lt;br&gt;&#xA;&lt;strong&gt;N3:&lt;/strong&gt; A car every 14 sec starting with 14. ( 14, 28, 42, ... )&lt;br&gt;&#xA;&lt;strong&gt;S3:&lt;/strong&gt; A car every 14 sec starting with 2. ( 2, 16, 26, ... )&lt;br&gt;&#xA;&lt;strong&gt;W1:&lt;/strong&gt; A car every 15 sec starting with 0. ( 0, 15, 30, ... )&lt;br&gt;&#xA;&lt;strong&gt;E1:&lt;/strong&gt; A car every 10 sec starting with 10. ( 10, 20, 30  ... )&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;What does the ideal traffic-light setting now look like?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;LEVEL 3 - district map and traffic&lt;/h1&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BXkY1.jpg&quot; alt=&quot;level 3&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Starting with rush-hour ( 0 sec ) and immediately stopping after rush-hour ( &amp;gt;3600 sec ) cars will enter at the district borders as follows:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;N1:&lt;/strong&gt; A car every 20 sec starting with 0. ( 0, 20, 40, ... )&lt;br&gt;&#xA;&lt;strong&gt;S1:&lt;/strong&gt; A car every 25 sec starting with 0. ( 0, 25, 50, ... )&lt;br&gt;&#xA;&lt;strong&gt;N2:&lt;/strong&gt; A car every 15 sec starting with 2. ( 2, 17, 32, ... )&lt;br&gt;&#xA;&lt;strong&gt;S2:&lt;/strong&gt; A car every 19 sec starting with 0. ( 0, 19, 38, ... )&lt;br&gt;&#xA;&lt;strong&gt;N3:&lt;/strong&gt; A car every 90 sec starting with 0. ( 0, 90, 180, ... )&lt;br&gt;&#xA;&lt;strong&gt;S3:&lt;/strong&gt; A car every 90 sec starting with 12. ( 12, 102, 192, ... )&lt;br&gt;&#xA;&lt;strong&gt;W1:&lt;/strong&gt; A car every 35 sec starting with 0. ( 0, 35, 70, ... )&lt;br&gt;&#xA;&lt;strong&gt;E1:&lt;/strong&gt; A car every 30 sec starting with 10. ( 10, 40, 70, ... )&lt;br&gt;&#xA;&lt;strong&gt;W2:&lt;/strong&gt; A car every 21 sec starting with 0. ( 0, 21, 42, 63, ... )&lt;br&gt;&#xA;&lt;strong&gt;E2:&lt;/strong&gt; A car every 21 sec starting with 0. ( 0, 21, 42, 63, ... )&lt;br&gt;&#xA;&lt;strong&gt;W3:&lt;/strong&gt; A car every 26 sec starting with 0. ( 0, 26, 52, ... )&lt;br&gt;&#xA;&lt;strong&gt;E3:&lt;/strong&gt; A car every 26 sec starting with 3. ( 3, 29, 55, ... )&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;And what is the solution now?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Can the above-posed puzzle still be solved analytically or does it require simulation, approximate models or other approaches?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;pre&gt;&lt;code&gt;Note: I do *not* know the answer to this puzzle. &#xA;In particular, this also means that some of the timings of cars might have been &#xA;very badly chosen (for level 2 &amp;amp; 3 at least). &#xA;The goal here (for me) is to **learn** how to make this into a better puzzle. &#xA;So if you discover a problem, please put it into the comments &#xA;and be aware that the puzzle might change due to edits.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2022-08-16T22:26:47.070" Title="Cross-road optimization - what is the proper way to solve this type of puzzle?" Tags="|mathematics|optimization|" AnswerCount="4" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="6030" PostTypeId="2" ParentId="5937" CreationDate="2014-12-08T18:49:46.490" Score="3" Body="&lt;p&gt;As someone with a little bit more of a math background, I call these assumptions &lt;em&gt;constraints&lt;/em&gt;. The word comes from optimization, a field studied by both computer scientists and mathematicians. When you write a computer program to solve a problem like determining the most efficient way to schedule trains transporting materials, you provide the program constraints that limit what kinds of solutions are acceptable. For example, a weight limit on a train is a constraint, as is the fact that two trains can't go in opposite directions on the same tracks simultaneously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analogously, a basic assumption or constraint limits what solutions are possible in a puzzle.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another word you might want to use to refer to the basic assumptions is &lt;em&gt;givens&lt;/em&gt;. Again, this word is taken from math: When completing a mathematical proof, the ideas or objects that you're making a proof about are &lt;em&gt;given&lt;/em&gt; to you at the start of the proof. For example, let's say you want to prove the Pythagorean Theorem: If a triangle is a right triangle with legs of length $a$ and $b$ and hypotenuse of length $c$, then $a^2+b^2=c^2$. The givens in this proof are that the triangle you are working with is a right triangle, and it has sides of length $a$, $b$, and $c$. If you take these details out, you don't have enough information to complete the proof!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, the assumptions you're allowed to make while solving a puzzle are the puzzle's givens. For example, in your problem about knights, knaves, and jokers, one of the givens is that the three characters know who each other are.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I don't think there's too much of a difference between a constraint and a given when discussing puzzles. If you want to be strict about it, though, a constraint is an aspect of a puzzle that puts requirements on an acceptable solution without decreasing the size of the search space (a hypothetical list of every possible solution based on a naïve reading of the puzzle), while a given is either necessary to ensure the puzzle can be solved or helps decrease the size of the search space. In other words, a constraint makes the puzzle harder by forcing you to find more complicated solutions, while a given makes a puzzle easier by pushing you in the right direction or ensuring a puzzle has only one provable answer. One can see where the difference can get blurry for a lot of puzzles.&lt;/p&gt;&#xA;" OwnerUserId="294" LastEditorUserId="294" LastEditDate="2014-12-09T00:47:28.477" LastActivityDate="2014-12-09T00:47:28.477" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6032" PostTypeId="2" ParentId="5980" CreationDate="2014-12-08T19:42:04.917" Score="9" Body="&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&#xA;I can get $15$ minutes before Jane finds out - thanks to @Joel Rondeau.&#xA;Jane finds out the secret in the $16^{th}$ conversation/minute&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the notation from @xnor, here is a method:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;&#xA; 1. AB; CDEFGHIJ   - A-B knows&#xA; 2. ABC; DEFGHIJ   - A-C knows&#xA; 3. ABCD; EFGHIJ   - A-D knows&#xA; 4. AD BC; EGI FHJ - A-D knows&#xA; 5. AC BD; EGJ FHI - A-D knows&#xA; 6. ABCDE; FGHIJ   - A-E knows&#xA; 7. AE BCD; JF IHG - A-E knows&#xA; 8. BE ACD; IF JHG - A-E knows&#xA; 9. CE ABD; HF JIG - A-E knows&#xA;10. CD ABE; GH FIJ - A-E knows&#xA;11. FDB ECA; JG IH - A-F knows&#xA;12. FDA ECB; JH IG - A-F knows&#xA;13. ABCDEF; GHIJ   - A-F knows&#xA;14. ABCDEFG: JIH   - A-G knows&#xA;15. ABCDEFGH: JI   - A-H knows&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Notice the symmetry: The last 5 are the inverse of the first 5.  The ones in the middle are symmetrical with themselves. &lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="1592" LastEditDate="2014-12-08T19:55:42.823" LastActivityDate="2014-12-08T19:55:42.823" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6033" PostTypeId="2" ParentId="5965" CreationDate="2014-12-08T20:05:31.540" Score="7" Body="&lt;p&gt;The text containing the key:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Graffiti decorations&lt;br&gt;Under a sky of dust&lt;br&gt;A constant wave of tension&lt;br&gt;On top of broken trust&lt;br&gt;The lessons that you taught me&lt;br&gt;I learned were never true&lt;br&gt;Now I find myself in question&lt;br&gt;They point the finger at me again&lt;br&gt;Guilty by association&lt;br&gt;You point the finger at me again&lt;br&gt;&lt;br&gt;I wanna runaway&lt;br&gt;Never say goodbye&lt;br&gt;I wanna know the truth&lt;br&gt;Instead of wondering why&lt;br&gt;I wanna know the answers&lt;br&gt;No more lies&lt;br&gt;I wanna shut the door&lt;br&gt;And open up my mind&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The key:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LINKINPARK&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The message:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LONG LIFE TO PUZZLES AND RIDDLES IN THIS COMMUNITY&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How I got there:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The message containing the text is a basic cryptogram using a substitution cipher.&lt;br&gt;The ~ are blank spaces, so I looked through the text and found 6 letter words where &quot;myself&quot; would fit, using trial and error to find the right one.  Then completed it like any other cryptogram.&lt;br&gt;&lt;br&gt;After finishing the text, Google told me that it was lyrics to a LINKIN PARK song.  Since the key had to be 10 characters, &quot;LINKINPARK&quot; made sense.&lt;br&gt;&lt;br&gt;I lined up the key under the encrypted message, repeating the key until the end:&lt;br&gt;&#xA; &lt;pre&gt;AGAW DVQE CE EMMPDRD AWT GAQTDRD IW JWAF SGZXUWYIQ&lt;br&gt;LINK INPA RK LINKINP ARK LINKINP AR KLIN KINPARKLI&lt;/pre&gt;&#xA; Then, using A=1, B=2, C=3, etc., add the letter from the first and second row, then subtract 1 (or 27 if the sum is greater than 26).&lt;br&gt;For example, to get the first letter of the message:&#xA; &lt;pre&gt;A=1&lt;br&gt;L=12&lt;br&gt;1+12=13&lt;br&gt;13-1=12&lt;br&gt;12=L&lt;/pre&gt;&#xA; And the second letter:&#xA; &lt;pre&gt;G=7&lt;br&gt;I=9&lt;br&gt;7+9-1=15&lt;br&gt;15=O&lt;/pre&gt;&#xA; And so on...&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6234" LastActivityDate="2014-12-08T20:05:31.540" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6034" PostTypeId="1" AcceptedAnswerId="6047" CreationDate="2014-12-08T21:56:09.813" Score="20" ViewCount="955" Body="&lt;p&gt;Suppose you're given six stakes and an unlimited length of string. Your objective is to plant the stakes in a flat patch of ground in such a way that you can wrap the string around the stakes in different ways to create &lt;a href=&quot;https://en.wikipedia.org/wiki/Simple_polygon&quot; rel=&quot;nofollow noreferrer&quot;&gt;simple polygons&lt;/a&gt; with integral (i.e. integer-valued) areas (see Fig. 2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some restrictions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;This is a 2D problem.&lt;/li&gt;&#xA;&lt;li&gt;No three stakes (or more) may be colinear. A line drawn through any two stakes must not pass through any other stakes (see Fig. 3).&lt;/li&gt;&#xA;&lt;li&gt;Areas must be circumscribed exactly.&lt;/li&gt;&#xA;&lt;li&gt;The stakes are ideal vertices (having zero cross-sectional area), and the string may only create perfectly straight, ideal edges between the stakes.&lt;/li&gt;&#xA;&lt;li&gt;The string must visit each stake &lt;em&gt;once and only once&lt;/em&gt;, and terminate where it begins (thus forming a closed polygon).&lt;/li&gt;&#xA;&lt;li&gt;Stake coordinates should be expressed in meters (m). The stakes can be positioned at any real coordinates so long as all stake coordinates are unique.&lt;/li&gt;&#xA;&lt;li&gt;All six stakes must be used.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There are two separate puzzle objectives:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The &lt;strong&gt;accepted answer&lt;/strong&gt; will go to the first (correct) answer that gives grid coordinates that can produce &lt;em&gt;at least 6 consecutive integral areas&lt;/em&gt; &amp;leq; 20 m&lt;sup&gt;2&lt;/sup&gt;. (For example, areas of 3,4,5,6,7, and 8 m&lt;sup&gt;2&lt;/sup&gt;, respectively). The answer should also specify the winding order of the string (around the stakes) for each polygon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, an answer might look like&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Stake 1: (5 m,2 m), Stake 2: (3 m,4 m), ...&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Area 3 m&lt;sup&gt;2&lt;/sup&gt;: 1 &amp;rarr; 2 &amp;rarr; 5 &amp;rarr; 3 &amp;rarr; 4 &amp;rarr; 6 &amp;rarr; 1&lt;br&gt;&#xA;  Area 4 m&lt;sup&gt;2&lt;/sup&gt;: 1 &amp;rarr; 3 &amp;rarr; 4 &amp;rarr; 6 &amp;rarr; 2 &amp;rarr; 5 &amp;rarr; 1&lt;br&gt;...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is acceptable if the stakes can produce additional polygons with areas not comprising the 6 consecutive integers in the solution.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A &lt;strong&gt;bounty of 100 rep&lt;/strong&gt; will go to the (correct) answer that gives the grid coordinates and winding rules that can produce &lt;em&gt;the greatest number of consecutive integral areas&lt;/em&gt; &amp;leq; 20 m&lt;sup&gt;2&lt;/sup&gt;.&lt;br&gt;&lt;br&gt;In the event of a tie, the earliest solution will receive the bounty.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Good luck staking your claim. ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Special thanks to &lt;a href=&quot;https://puzzling.stackexchange.com/users/6178/mcmagister&quot;&gt;McMagister&lt;/a&gt; for pointing out &lt;a href=&quot;http://en.wikipedia.org/wiki/Pick%27s_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pick's theorem&lt;/a&gt; as a simple way to compute the area of polygons with integer coordinates.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Examples of Legal and Illegal Polygons&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/nCHJY.png&quot; alt=&quot;legal and illegal polygons in &amp;quot;Staking Out the Integers&amp;quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/IJRfM.png&quot; alt=&quot;valid stake placement and several windings&quot;&gt;&lt;br&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;strong&gt;Fig. 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/Eg2Wi.png&quot; alt=&quot;invalid stake placement due to colinearity&quot;&gt;&lt;br&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;strong&gt;Fig. 3&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2014-12-11T10:02:43.453" Title="Staking Out the Integers" Tags="|geometry|" AnswerCount="2" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="6036" PostTypeId="1" AcceptedAnswerId="6043" CreationDate="2014-12-09T10:47:11.933" Score="7" ViewCount="742" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Gluttony at the heart of me&lt;br&gt;&#xA;  Famed worldwide for catastrophe&lt;br&gt;&#xA;  We tell ourselves we won the race&lt;br&gt;&#xA;  We delete, deny and cover all trace  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Our pride blinds us to stupidity&lt;br&gt;&#xA;  Attitudes towards us have no validity&lt;br&gt;&#xA;  Hide behind our mask of virtue&lt;br&gt;&#xA;  Whilst sins run amok underneath our statue  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Run and Hide our name Precedes us&lt;br&gt;&#xA;  our array of voices merely degrade us&lt;br&gt;&#xA;  We beat the lobster back to the sea&lt;br&gt;&#xA;  lest we forget, genocide while they flea  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;For we were one but gained another&lt;br&gt;&#xA;  we fought our brother, disregard our mother&lt;br&gt;&#xA;  If this is appealing come join me&lt;br&gt;&#xA;  bask in my ignorance, my gluttony  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5135" LastEditorDisplayName="user20" LastEditDate="2015-02-20T15:05:01.093" LastActivityDate="2015-02-20T15:05:01.093" Title="One small step for me, one giant leap for riddles!" Tags="|riddle|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6037" PostTypeId="1" AcceptedAnswerId="6044" CreationDate="2014-12-09T12:17:15.803" Score="38" ViewCount="4284" Body="&lt;p&gt;Once upon a time, Pharaoh sat upon his high throne, overlooking his realm, a dark realm indeed, bounded on four sides.  His anger burned against his Hebrew subjects, neither young and old, nor rich and poor were spared.  In his wrath, he forced every Hebrew to line up under him.  There would be no escape, for Pharaoh's watchful eyes never slept, not even at night.  And any Hebrew found disobeying Pharaoh's decrees would do so only at their peril, their own lives and perhaps even the lives of their loved ones may be forfeit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pharaoh became envious of Moses, and after a long period, he eventually let the Hebrews go.  What happened then is a story that would take a long time to tell, so I'll save it for another day.  Anyhow, the Hebrews found that miraculously the swift, flowing river blocking their way had been parted on both sides, allowing them safe passage through Pharaoh's realm.  The Hebrews headed away from Pharaoh in three directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But afterwards, Pharaoh realized his cowardice.  His cowardice soon turned back into anger as he tried to call the Hebrews back.  But it was too late, they had already left for good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, speak Pharaoh's name so that you too may meet him when his mood is not foul.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Your task is to uncover the identity of the Pharaoh.  No, it is not Ramses or Imhotep or an actual historical Pharaoh.  If your answer is correct the story should make sense from another angle.  The identity of the Hebrews will also become clear.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Moses is irrelevant to the story.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6178" LastEditorUserId="6178" LastEditDate="2014-12-09T12:45:10.673" LastActivityDate="2014-12-15T00:31:47.647" Title="The Exodus (speak the Pharaoh's name)" Tags="|riddle|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6040" PostTypeId="2" ParentId="6037" CreationDate="2014-12-09T14:50:43.360" Score="34" Body="&lt;p&gt;This is an allegorical tale of a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;thread scheduler&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Pharaoh&quot; is the title given to the operating system (OS), presiding over system operations with perfect diligence and absolute control (&quot;&lt;em&gt;Pharaoh's watchful eyes never slept, not even at night&lt;/em&gt;&quot;). Pharaoh's anger is his iron-clad execution schedule. Within Pharaoh's power is the ability to pause, lock down, even terminate threads at will. Unresponsive processes and their child processes are subject to his dictates (&quot;&lt;em&gt;any Hebrew found disobeying Pharaoh's decrees would do so only at their peril, their own lives and perhaps even the lives of their loved ones may be forfeit&lt;/em&gt;&quot;). Old threads and new, processing-intensive juggernauts and simple daemons, none escape his harsh scrutiny (&quot;&lt;em&gt;neither young and old, nor rich and poor were spared&lt;/em&gt;&quot;).&lt;br&gt;&#xA;&lt;br&gt;&#xA;Pharaoh presides over a system with four physical, software-opaque cores (&quot;&lt;em&gt;a dark realm indeed, bounded on four sides&lt;/em&gt;&quot;), as well as a realtime data connection with a critical hardware device, &quot;Moses&quot;. Running on these four cores are an endless stream of process threads&amp;mdash;the &quot;Hebrews&quot;, who patiently make their processing demands known, lining up in the execution queues awaiting their turn to perform Pharaoh's labour (&quot;&lt;em&gt;he forced every Hebrew to line up under him&lt;/em&gt;&quot;). Pharoah is fiercely protective of CPU time, given Moses' critical importance and high-bandwidth realtime I/O demands. The data bus shared between Pharaoh and Moses is appropriately called &quot;the river&quot;.&lt;br&gt;&#xA;&lt;br&gt;&#xA;At one point Moses stops sending data back, leaving Pharaoh in a quandary on how to proceed (&quot;&lt;em&gt;Pharaoh became envious of Moses&lt;/em&gt;&quot;). Eventually, after many microseconds of inactivity (&quot;&lt;em&gt;after a long period&lt;/em&gt;&quot;), Pharoah decides to suspend data stream with Moses (&quot;&lt;em&gt;the swift, flowing river blocking [the] way had been parted on both sides&lt;/em&gt;&quot;), and release three of his cores, allowing the Hebrews to execute on them (&quot;&lt;em&gt;allowing them safe passage through Pharaoh's realm ... The Hebrews headed away from Pharaoh in three directions&lt;/em&gt;&quot;).&lt;br&gt;&#xA;&lt;br&gt;&#xA;However, as is so often the case with asynchronous hardware with RT processing demands, Moses suddenly begins belching out data again, threatening to fail and lose data if his processing demands aren't met. The CPU is starving. Rescheduling chaos ensues (&quot;&lt;em&gt;But afterwards, Pharaoh realized his cowardice&lt;/em&gt;&quot;). Unfortunately, several of the Hebrew threads have entered critical code sections and uninterruptible communication with various other hardware and software resources, virtually guaranteeing a system crash or data corruption in the event of suspension or termination. They have been dispatched. It's too late to stop them (&quot;&lt;em&gt;His cowardice soon turned back into anger as he tried to call the Hebrews back. But it was too late, they had already left for good.&lt;/em&gt;&quot;). Moses would be left wanting.&lt;br&gt;&#xA;&lt;br&gt;&#xA;Pharaoh's more common name is &lt;strong&gt;The Kernel&lt;/strong&gt;, and indeed none of us want to meet him in a foul mood. Truly I tell you: He could blue-screen the very computer you're reading this on if he wanted to.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="6178" LastEditDate="2014-12-14T14:25:18.550" LastActivityDate="2014-12-14T14:25:18.550" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6041" PostTypeId="1" AcceptedAnswerId="6042" CreationDate="2014-12-09T15:07:45.360" Score="6" ViewCount="1116" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am a body thats for sure&lt;br&gt;&#xA;  Ire I held for my neighbours, so pure&lt;br&gt;&#xA;  My north teamed up while the south went solo&lt;br&gt;&#xA;  We remember the severe lack of the potato  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;My symbol riddled with the need for peace&lt;br&gt;&#xA;  The war of the gods needed to cease&lt;br&gt;&#xA;  Green to the left, orange to the right&lt;br&gt;&#xA;  Fictional characters at the root of the spite  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5135" LastEditorUserId="7586" LastEditDate="2016-03-04T23:09:35.740" LastActivityDate="2016-03-04T23:09:35.740" Title="Tea father? Drink!" Tags="|riddle|" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6042" PostTypeId="2" ParentId="6041" CreationDate="2014-12-09T15:48:38.760" Score="8" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Ireland.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am a body thats for sure&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; one landmass&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Ire I held for my neighbours, so pure&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Ire&quot; a clue for &quot;Ireland&quot;, I guess...?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My north teamed up while the south went solo&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Northern Ireland versus Eire.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;We remember the severe lack of the potato&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Potato famine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My symbol riddled with the need for peace&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The war of the gods needed to cease&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sectarian conflict&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Green to the left, orange to the right&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Flag_of_Ireland&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Flag_of_Ireland&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Fictional characters at the root of the spite&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jesus, Mary, God, etc... (although there is really a lot more than that forming the background of the Troubles, &lt;a href=&quot;http://en.wikipedia.org/wiki/The_Troubles#Background&quot; rel=&quot;nofollow&quot;&gt;see&lt;/a&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Plantation_of_Ulster&quot; rel=&quot;nofollow&quot;&gt;these&lt;/a&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Nine_Years%27_War_(Ireland)&quot; rel=&quot;nofollow&quot;&gt;links&lt;/a&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Tudor_conquest_of_Ireland&quot; rel=&quot;nofollow&quot;&gt;for&lt;/a&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Cromwellian_conquest_of_Ireland&quot; rel=&quot;nofollow&quot;&gt;more&lt;/a&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Irish_Home_Rule_movement&quot; rel=&quot;nofollow&quot;&gt;info&lt;/a&gt;...)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Title:&lt;/strong&gt; &lt;em&gt;Tea father? Drink!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the incomparable &lt;a href=&quot;http://en.wikipedia.org/wiki/Father_Ted&quot; rel=&quot;nofollow&quot;&gt;Father Ted&lt;/a&gt;, of course.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="7586" LastEditDate="2016-03-04T23:09:12.240" LastActivityDate="2016-03-04T23:09:12.240" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="6043" PostTypeId="2" ParentId="6036" CreationDate="2014-12-09T15:57:40.523" Score="7" Body="&lt;p&gt;I am&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;America?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Gluttony at the heart of me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;America is famous for its consumerism&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Famed worldwide for catastrophe&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Also famous for invading foreign countries/toppling their governments&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We tell ourselves we won the race&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Americans won the space race, the cold war.  Also, the famous &lt;a href=&quot;https://en.wikipedia.org/wiki/Mission_Accomplished_speech&quot;&gt;&quot;Mission Accomplished&quot; banner and speech&lt;/a&gt; on the war on terror&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We delete, deny and cover all trace&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;NSA covert activities?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Our pride blinds us to stupidity&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;American exceptionalism&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Attitudes towards us have no validity&#xA;Hide behind our mask of virtue&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Americans believe strongly in the virtues of liberty and democracy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whilst sins run amok underneath our statue&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Statue of Liberty.  The sins are probably to do with the hypocrisy of toppling democracies that don't support the West, while claiming to support democracy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We beat the lobster back to the sea&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Lobster refers to &lt;a href=&quot;http://www.urbandictionary.com/define.php?term=lobsterback&quot;&gt;British soldiers&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;lest we forget, genocide while they flea&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Genocide of the native Americans&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For we were one but gained another&#xA;we fought our brother, disregard our mother&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Brother refers to either Canada or the Confederates, mother is Britain.  &quot;Gained another&quot; could refer to Hawaii or Alaska.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6178" LastEditorUserId="6178" LastEditDate="2014-12-09T16:11:57.547" LastActivityDate="2014-12-09T16:11:57.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6044" PostTypeId="2" ParentId="6037" CreationDate="2014-12-09T17:10:53.237" Score="43" Body="&lt;p&gt;The Pharaoh is a &lt;strong&gt;traffic light/signal&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sits up on a high throne, forcing the Hebrews (cars) to line up beneath.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Emotions throughout refer to light colors:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Red is wrath : force them to line up&lt;/li&gt;&#xA;&lt;li&gt;Green is &quot;envious of Moses&quot; : lets them go&lt;/li&gt;&#xA;&lt;li&gt;Yellow is cowardice, which led to anger (red) again.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&quot;Never slept, not even at night&quot; is easy, since traffic lights are always (generally) on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disobeying comes at the peril of your life or your loved ones? Check.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Swift, flowing river blocking their way&quot; are cars going the other direction(s). Once parted, the Hebrews left in three directions (left/right/straight).&lt;/p&gt;&#xA;" OwnerUserId="1563" LastEditorUserId="1563" LastEditDate="2014-12-15T00:31:47.647" LastActivityDate="2014-12-15T00:31:47.647" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6045" PostTypeId="2" ParentId="6034" CreationDate="2014-12-09T17:11:33.650" Score="0" Body="&lt;p&gt;&lt;strong&gt;Incorrect answer&lt;/strong&gt;, as I misunderstood the question. (See comments). I'll leave it for reference while thinking of a proper solution.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h3&gt;What have I missed?&lt;/h3&gt;&#xA;&lt;p&gt;I've read the instructions again and again, but I don't get it.&lt;/p&gt;&#xA;&lt;p&gt;I am &lt;strong&gt;free&lt;/strong&gt; to place the stakes at any &lt;em&gt;real&lt;/em&gt; (unique) coordinate?&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;6. Stake coordinates should be expressed in meters (m). &#xA;   The stakes can be positioned at any real coordinates so long as all &#xA;   stake coordinates are unique.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Well, then I can simply do something like:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Any arbitrary circular shape like this hexagon:&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/GyHHR.jpg&quot; alt=&quot;Hex&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Can't I?&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; But in this case, I'm absolutely free to extend shrink the 3 parameters a,b,c to any floating point value to give me an area I like, and the stake coordinates will be:&lt;br&gt;&#xA; (0|0) -&amp;gt; (a|-b) -&amp;gt; (2a|0) -&amp;gt; (0|c) -&amp;gt; (a|c+b) -&amp;gt; (2a|c)  &lt;br&gt;&#xA; &lt;br&gt;&#xA; I can then make for example: &lt;br&gt;&#xA; A = 1 ==&amp;gt; a=0.5 b=c=0.5 &lt;br&gt;&#xA; A = 2 ==&amp;gt; a=1   b=c=0.5 &lt;br&gt;&#xA; A = 3 ==&amp;gt; a=1.5 b=c=0.5 &lt;br&gt;&#xA; ....&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;As this clearly isn't a real &lt;em&gt;puzzle&lt;/em&gt;, I suppose there is either some catch I haven't seen, or the puzzle text is missing something. Should it be &amp;quot;integer&amp;quot; values for the stakes? That would make more sense for me..&lt;/p&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-12-09T17:25:15.783" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6047" PostTypeId="2" ParentId="6034" CreationDate="2014-12-09T21:31:01.670" Score="12" Body="&lt;p&gt;I'm sure there's a lot of solutions for $\geq 6$ because it didn't take me long to find one:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[(2, 2), (4, 5), (2, 8), (8, 4), (7, 2), (6, 4)] 12&#xA;[(2, 2), (6, 4), (7, 2), (8, 4), (4, 5), (2, 8)] 13&#xA;[(2, 2), (2, 8), (8, 4), (7, 2), (6, 4), (4, 5)] 14&#xA;[(2, 2), (4, 5), (2, 8), (8, 4), (6, 4), (7, 2)] 15&#xA;[(2, 2), (2, 8), (4, 5), (8, 4), (6, 4), (7, 2)] 16&#xA;[(2, 2), (2, 8), (4, 5), (6, 4), (8, 4), (7, 2)] 17&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Finding the &lt;em&gt;most&lt;/em&gt; consecutive, on the other hand, is going to be a bit harder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diagrams:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DpId4m.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/3luSlm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/LgiQbm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/gkIf5m.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/osZO5m.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/0HRzdm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here's a solution with 12 consecutive:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;9  ((5, 4), (3, 0), (6, 4), (9, 2), (6, 10), (7, 6))&#xA;10 ((5, 4), (6, 10), (9, 2), (7, 6), (6, 4), (3, 0))&#xA;11 ((5, 4), (6, 10), (3, 0), (6, 4), (9, 2), (7, 6))&#xA;12 ((5, 4), (6, 10), (7, 6), (9, 2), (6, 4), (3, 0))&#xA;13 ((5, 4), (6, 10), (9, 2), (6, 4), (7, 6), (3, 0))&#xA;14 ((5, 4), (3, 0), (6, 10), (7, 6), (9, 2), (6, 4))&#xA;15 ((5, 4), (3, 0), (6, 10), (9, 2), (6, 4), (7, 6))&#xA;16 ((5, 4), (6, 10), (3, 0), (9, 2), (6, 4), (7, 6))&#xA;17 ((5, 4), (6, 10), (7, 6), (6, 4), (9, 2), (3, 0))&#xA;18 ((5, 4), (6, 10), (6, 4), (7, 6), (9, 2), (3, 0))&#xA;19 ((5, 4), (6, 10), (3, 0), (9, 2), (7, 6), (6, 4))&#xA;20 ((5, 4), (6, 10), (9, 2), (3, 0), (6, 4), (7, 6))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="445" LastEditorUserId="445" LastEditDate="2014-12-09T22:07:42.260" LastActivityDate="2014-12-09T22:07:42.260" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6048" PostTypeId="1" AcceptedAnswerId="6063" CreationDate="2014-12-09T21:49:24.223" Score="28" ViewCount="1465" Body="&lt;p&gt;You are Pat Doe, a lowly Sudoku puzzle-maker eking out a living by producing hand-made Sudoku puzzles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After an exhausting eight hours, you've just completed your last Sudoku grid for the day. Unfortunately, it's only after you've completed the grid that you realize the usual $3\times 3$ blocks aren't legal (none of them contain exactly the digits 1 thru 9). To make matters worse, your hand malfunctioned and inserted several bold lines denoting boundaries of neighbouring blocks in unusual places.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/4aOiq.png&quot; alt=&quot;pseudoku grid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather than redoing the entire grid, you resolve that you will find out a way to partition the grid into 9 irregular blocks (nonominoes) in such a way that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;each block contains the digits 1 thru 9 each exactly once&lt;/li&gt;&#xA;&lt;li&gt;the blocks respect the existing boundaries (i.e. a boundary may appear nowhere on a block except at its borders)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can you recover from your error? Can you partition the grid and salvage a workable (if highly irregular) Sudoku grid?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck! :)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The following is an example of a valid block. It is a nonomino containing the digits 1-9 each exactly once, and it respects established boundaries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/3FOiR.png&quot; alt=&quot;example of a valid pseudoku nonomino&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The following is not a valid block because it is not a nonomino.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/WY4AY.png&quot; alt=&quot;invalid pseudoku block is not a nonomino&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The following is not a valid block because it doesn't respect existing boundaries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/sFbs4.png&quot; alt=&quot;invalid pseudoku block does not respect existing boundaries&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2014-12-11T02:18:29.470" Title="Pseudoku: A Game of Finding Order in Chaos" Tags="|sudoku|pencil-and-paper-games|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6049" PostTypeId="2" ParentId="6001" CreationDate="2014-12-09T22:04:39.137" Score="3" Body="&lt;p&gt;I think this is the right answer...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Corn&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Longer and longer I grow hard&lt;br&gt;&#xA;but I am not the male shard&lt;br&gt;&#xA;evolved with gluttonous imagination&lt;br&gt;&#xA;created with utter anticipation &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Husk of Corn - Grows longer and harder and very modified over the years and has a very long history&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a dam to those inside&lt;br&gt;&#xA;terrible by definition, porous my hide&lt;br&gt;&#xA;Humble farmers are my starts&lt;br&gt;&#xA;Suffocated so as to extend my parts&lt;br&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Corn Husk is a dam to those inside because it keeps the cob and kernels inside. Also, husk by definition means poor, or terrible. Also the corn can only continue growing while it is encapsulated inside it's shuck.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rife with many shapes and proportions&lt;br&gt;&#xA;A magnitude of functions suit my options&lt;br&gt;&#xA;heat me from both sides for fun&lt;br&gt;&#xA;I'm not, not insane I'm now not one&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Corn has many uses - one of those is popcorn which is fun when you heat the kernels - but when you separate the kernels, it's no longer one unit of corn.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6247" LastActivityDate="2014-12-09T22:04:39.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6050" PostTypeId="1" AcceptedAnswerId="6054" CreationDate="2014-12-09T22:22:56.323" Score="12" ViewCount="1962" Body="&lt;p&gt;Around 2007-2009, I was introduced to a website which hid puzzle clues in website data such as picture code as well as normal puzzles using various different types of code breaking and cyphers.  The goal was to find the next url to progress with the puzzle.  The puzzle eventually ended as the site was not being updated and it even had a forum for help along the way.  Any ideas on what the site was and if it still exists?&lt;/p&gt;&#xA;" OwnerUserId="6202" LastActivityDate="2018-11-10T07:34:10.953" Title="Identify an Online Puzzle: Website Based Puzzle" Tags="|puzzle-identification|" AnswerCount="1" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6051" PostTypeId="2" ParentId="5970" CreationDate="2014-12-09T23:12:20.613" Score="6" Body="&lt;p&gt;The people are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Llamas&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cannot make sense of all the clues, here's what I think I've got so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Born from the damming of a dam&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A dam is a female llama: &lt;a href=&quot;http://www.llamapaedia.com/industry/terms.html&quot;&gt;http://www.llamapaedia.com/industry/terms.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;my people are strong and hardy&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Llamas are strong and hardy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though smaller than our desert cousins, we're almost the spitting image.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Smaller that the related camels, llamas are known for their spitting attitude.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A chest sits before you, filled with food.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The llama's chest, located before a person who rides it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Upon the chest, sits cloud made flesh upon a sheet unreadied.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Referring to a person sitting on a saddle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What do these people have in common with LL Cool J? Just a couple of things.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The initial letters &quot;LL&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I do not know what the four kingdoms mentioned in the riddle are and who the leader is.&lt;/p&gt;&#xA;" OwnerUserId="6070" LastActivityDate="2014-12-09T23:12:20.613" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6052" PostTypeId="2" ParentId="6001" CreationDate="2014-12-10T01:29:52.807" Score="5" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A sandwich (some clues suggest subs in particular but most clues apply to any sandwich).  Credit to Raystafarian for suggesting the idea.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have an Earl to which I was born&lt;br&gt;&#xA;not on the beach and without a thorn&lt;br&gt;&#xA;No Rand; I am not the Grail&lt;br&gt;&#xA;I assist your game of cards, no fail&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Earl is the &lt;a href=&quot;http://en.wikipedia.org/wiki/John_Montagu,_4th_Earl_of_Sandwich&quot; rel=&quot;nofollow&quot;&gt;Earl of Sandwich&lt;/a&gt; who invented sandwiches.  He did it in order to help him eat quickly so he could focus on playing cards.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without my walls I'm not complete&lt;br&gt;&#xA;the epitome of me is no easy feat&lt;br&gt;&#xA;when I bleed, don't call that Cop&lt;br&gt;&#xA;for he is unable to make it stop&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A sandwich is not complete without its &quot;walls&quot; of bread.  The &lt;a href=&quot;http://www.dailymail.co.uk/news/article-2148140/Now-THATS-sandwich-Worlds-biggest-meat-sarnie-contains-41-cuts-weighs-stone-10-hours-eat.html&quot; rel=&quot;nofollow&quot;&gt;epitome of sandwiches are no easy feat to eat&lt;/a&gt;.  When a sandwich is bleeding ketchup or other condiments, a cop is of no use.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Longer and longer I grow hard&lt;br&gt;&#xA;but I am not the male shard&lt;br&gt;&#xA;evolved with gluttonous imagination&lt;br&gt;&#xA;created with utter anticipation&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A sub is long and grows hard when left for a long time.  As a culinary delight, it is designed by people with &quot;gluttonous imagination&quot; and prepared by chefs with &quot;utter anticipation&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a dam to those inside&lt;br&gt;&#xA;terrible by definition, porous my hide&lt;br&gt;&#xA;Humble farmers are my starts&lt;br&gt;&#xA;Suffocated so as to extend my parts&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The sandwich bread is a dam to the filling inside.  A sub is &quot;terrible&quot;, as in &quot;subpar&quot;, and bread is porous.  Flour, a critical component of bread is created by humble farmers, while yeast is sacrificed to make bread rise.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rife with many shapes and proportions&lt;br&gt;&#xA;A magnitude of functions suit my options&lt;br&gt;&#xA;heat me from both sides for fun&lt;br&gt;&#xA;I'm not, not insane I'm now not one&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Sandwiches come in many shapes and proportions.  They have many options for fillings and are a common sight at the snack table of many functions (meetings).  Sandwiches can be heated on both sides to become a panini.  Sandwich is one item, but in order to create it you need to slice bread into two pieces.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6178" LastEditorUserId="6178" LastEditDate="2014-12-10T06:43:43.627" LastActivityDate="2014-12-10T06:43:43.627" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6054" PostTypeId="2" ParentId="6050" CreationDate="2014-12-10T04:08:00.323" Score="12" Body="&lt;p&gt;The riddle series you were looking for was &lt;a href=&quot;http://www.totallytricky.co.uk/riddle/ame.php&quot;&gt;Tricky's Riddles&lt;/a&gt;. It's an old puzzler that I myself came across in 2006 back when a bunch of my middle-school friends introduced me to it, and back when it didn't have a domain and was still being hosted in a University of Exeter home directory. Looking back at it, the riddles were really shoddy and much too &quot;random&quot;, but back then it was the ultimate challenge for us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are quite a few more riddle pages that fit that description, including &lt;a href=&quot;http://deathball.net/notpron&quot;&gt;Notpron&lt;/a&gt; and &lt;a href=&quot;http://zestriddle.ovh.org/hints.htm&quot;&gt;ze?t&lt;/a&gt;, out of the ones I've personally played and failed at (ze?t appears to be down at the moment).&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="6202" LastEditDate="2014-12-10T04:34:09.407" LastActivityDate="2014-12-10T04:34:09.407" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6058" PostTypeId="2" ParentId="5918" CreationDate="2014-12-10T06:03:48.837" Score="5" Body="&lt;p&gt;That's funny, I just finished finding my own set of algorithms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My base algorithm is effective in select situations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found a 2-2 swap from [A]=RU'RURURU'R'U'R2' swapping right-up edge and back center in additon to the left center and left-front edge. The key comes in the setup move [B]=R2F2x. (reminder: x is rotating the entire cube as if you were doing an R turn). This setup puts the originally swapped pairs into a new configuration to be swapped back. However, in swapping back, the original left-front edge will be rotated 90 degrees counter-clockwise. Thus the full algorithm I use is [A][B][A][B]'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is efficient if I just have one gear that needs to go in the right direction, but I found some others for other cases.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Mess with more than one gear:&#xA;Let [C]=FU'[A]UF'. This does the 1-1 swap on only edges. Add on isolating and rotating move [D]=UFnBnU' with Fn and Bn being n moves in that direction. Now we have algorithm [E]=[C][D][C][D]' which will rotate the left-bottom and left-back edges 90*n degrees.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Mess with more than two gears:&#xA;Let [F]=[R2U2]x3. Now use U[F]U'[F] to rotate all top four gears 180 degrees. I'm assuming you also know that [F] will also flip all gears not on the S slice 180 degrees.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Finally, operate one gear with precision. Namely, we wish to rotate just one gear on a 90, 180, or 270. We do this with the following long monster. F[E]F' will rotate JUST the left-back gear 90*n degrees counter-clockwise.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The main advantage that I see with my approach over yours is that your base sequence manipulates pieces that are opposite each other. This adds a lot of symmetry, which when attempting to isolate a single edge to rotate, isn't exactly what helps. My initial technique was an accident I stumbled upon, [1] was going to be my intended solution as it is flexible and the [D] setup isolates the gear to be rotated on the Back face and then proceeds to rotate it (but it had a side effect), and [3] was finally settled upon as my preferred rotator as it exchanged one of the rotated gears of [1] for a center, so it goes on unnoticed.&lt;/p&gt;&#xA;" OwnerUserId="458" LastActivityDate="2014-12-10T06:03:48.837" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6062" PostTypeId="2" ParentId="5880" CreationDate="2014-12-10T10:12:04.220" Score="26" Body="&lt;p&gt;From my experience as corrector of text puzzles on a Russian puzzle site 3 features are required for a puzzle to be liked by solvers (in order of importance): &lt;strong&gt;motivate, be correct, do not be straightforward&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;It is Interesting&lt;/strong&gt;. It must have something unusual. This is clearly the most important, if a reader does not expect to to get a new knowledge and experience he won't even try to solve it most probably. For example, x^2-2x+1=0 would not be a puzzle at all, just because it is &lt;em&gt;standard&lt;/em&gt; quadratic equation.&lt;br&gt;&#xA;So, what can be unusual about a puzzle?&lt;br&gt;&#xA;1.1. &lt;strong&gt;Unexpected&lt;/strong&gt;, beautiful fact to prove.  For example, &lt;em&gt;draw a shape which can be cut both on 3 and 4 equal triangles&lt;/em&gt;. Or, a puzzle about passenger, which comes to a bus stop at a random moment and most often meet his bus, that goes into wrong direction.&lt;br&gt;&#xA;1.2. It's answer seem to be &lt;strong&gt;impossible&lt;/strong&gt;.  For example, &lt;em&gt;a puzzle which asks how to measure diagonal of a brick in one measurement with solid ruler&lt;/em&gt;, or &lt;a href=&quot;https://puzzling.stackexchange.com/questions/394/variation-of-100-prisoners-light-problem&quot;&gt;prisoners problem&lt;/a&gt;.&lt;br&gt;&#xA;1.3. It's answer can &lt;strong&gt;seem to be too obvious&lt;/strong&gt; (but this obvious answer must be wrong of course). For example, &lt;em&gt;can whites win in &lt;a href=&quot;https://braingames.ru/images/main/puzzle/puzzle556.jpg&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt; position?&lt;/em&gt;.&lt;br&gt;&#xA;1.4. It can be about unusual properties of &lt;strong&gt;well know objects&lt;/strong&gt;. For example, &quot;Is it possible in a Cartesian coordinate system to position a regular tetrahedron so that all its vertices lie at the points with integer coordinates?&quot;.&lt;br&gt;&#xA;1.5. It can &lt;strong&gt;look too complex&lt;/strong&gt; to solve (but one should know that it is a puzzle, not a practical question and it has found, &lt;strong&gt;simple solution&lt;/strong&gt;). For example, &lt;a href=&quot;https://puzzling.stackexchange.com/questions/1784/maximum-time-for-ants-to-fall-off-stick/&quot;&gt;ants on a stick puzzle&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It's task is &lt;strong&gt;Easy to Read and Understand&lt;/strong&gt;. At least, the understanding of the task must require much less time than the solution itself.&lt;br&gt;&#xA;To be like this it must be:&lt;br&gt;&#xA;2.1. &lt;strong&gt;complete&lt;/strong&gt;. It must have all special information reader needs. And it must allow only one interpretation. But, note that there are well know facts, these don't have to be included in the formulation of the puzzle, and nobody can create a puzzle that can not be misunderstood.&lt;br&gt;&#xA;2.2. &lt;strong&gt;compact&lt;/strong&gt;, any word you put in you must think though, you must ask yourself whether it is really needed there.&lt;br&gt;&#xA;2.3. &lt;strong&gt;tested&lt;/strong&gt;. It must be read by several people, different people always see one thing differently, the author alone can't predict all problems.&lt;br&gt;&#xA;2.4. reading process should be &lt;strong&gt;pleasant&lt;/strong&gt;. So you better check you spelling, structure and order of phrases. Also, it is always nice to find appropriate &quot;interface&quot; for the idea behind a puzzle, all puzzles can be formulated in a strict mathematical language, but if you find real-life situation, that fits your idea the puzzle would became &lt;em&gt;much&lt;/em&gt; easier to remember and understand.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;It's Solution is Interesting&lt;/strong&gt;. The solution should have a new idea behind it. It's very nice when it simple to understand and compact, but meanwhile is hard to find (not straightforward).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="28" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2014-12-10T10:12:04.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6063" PostTypeId="2" ParentId="6048" CreationDate="2014-12-10T11:02:35.227" Score="13" Body="&lt;p&gt;My solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/j9QUy.png&quot; alt=&quot;The solved puzzle&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solution method:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I didn't do anything that systematic or logical. I made some vague approximate clusters that had most of the digits 1-9. Then, when a cluster needed a number that it wasn't adjacent to, I'd have it &quot;trade&quot; numbers with adjacent clusters. Often, a whole chain of trades was needed. Though I started on the right and went left, fixing a single disconnected cell in the last incomplete cluster caused a chain reaction that made me reconstruct the bottom right section of the board.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2014-12-10T11:17:31.533" LastActivityDate="2014-12-10T11:17:31.533" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="6064" PostTypeId="2" ParentId="6001" CreationDate="2014-12-10T12:58:28.303" Score="3" Body="&lt;p&gt;Just to explore the idea, I want to add in to the &lt;a href=&quot;https://puzzling.stackexchange.com/a/6052/51&quot;&gt;accepted answer&lt;/a&gt; that it's a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Submarine (Sandwich)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's have a go at defining the machine part to go in conjunction with the accepted answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I have an Earl to which I was born&lt;br&gt;&#xA;  not on the beach and without a thorn&lt;br&gt;&#xA;  No Rand; I am not the Grail&lt;br&gt;&#xA;  I assist your game of cards, no fail  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An Earl is on par with the status of a Royal Navy Officer, some of which commanded submarines. The first British submarine was captained in 1901 by Sir Reginald Bacon, a Royal Navy Officer and eventual Admiral&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Without my walls I'm not complete&lt;br&gt;&#xA;  the epitome of me is no easy feat&lt;br&gt;&#xA;  when I bleed, don't call that Cop&lt;br&gt;&#xA;  for he is unable to make it stop  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clearly submarines would fail without being completely enclosed and pressurized. This was quite a technological achievement. If it were to bleed (spring a leak) the blood (water) could not be stopped without specialized mechanics or surfacing&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Longer and longer I grow hard&lt;br&gt;&#xA;  but I am not the male shard&lt;br&gt;&#xA;  evolved with gluttonous imagination&lt;br&gt;&#xA;  created with utter anticipation     &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Submarines are becoming longer and more armored as technology advances (nuclear submarines). It's evolved for over a century (for Britain), getting bigger and better every class (from Holland-Class of around 60ft to Vanguard-Class of around 500ft) &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I am a dam to those inside&lt;br&gt;&#xA;  terrible by definition, porous my hide&lt;br&gt;&#xA;  Humble farmers are my starts&lt;br&gt;&#xA;  Suffocated so as to extend my parts  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A dam to those inside, protecting soldiers from the water. While first commissioned, they were constructed in a &quot;yacht shed&quot; secretly and referred to as pontoons. These allude to the idea that ships were initially made of wood rather than steel. Wood is essentially porous and tree farming would have an impact on landscape and environment as demand increased&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Rife with many shapes and proportions&lt;br&gt;&#xA;  A magnitude of functions suit my options&lt;br&gt;&#xA;  heat me from both sides for fun&lt;br&gt;&#xA;  I'm not, not insane I'm now not one&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Submarines come in many lengths and functions (nuclear, ballistic missile, etc). Each class of submarine has different features that allow it to perform in specialized missions. The &quot;heat me from both sides for fun&quot; refers to the submarine sandwich, and the last line indicates that the riddle is speaking of two things - The submarine sandwich and the underwater watercraft - submarine.&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2014-12-10T12:58:28.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6065" PostTypeId="1" AcceptedAnswerId="6121" CreationDate="2014-12-10T13:18:02.977" Score="12" ViewCount="566" Body="&lt;p&gt;This is no enigma&lt;br&gt;&#xA;Just a shameful stigma&lt;br&gt;&#xA;We have split apart for good&lt;br&gt;&#xA;A malevolent stance taken as it should  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our contract now null and void&lt;br&gt;&#xA;Desiring to inhabit a separate planetoid&lt;br&gt;&#xA;Once classed as my property&lt;br&gt;&#xA;Now free of the depravity  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For what we may have created&lt;br&gt;&#xA;now in despair, not appreciated&lt;br&gt;&#xA;for we'll pay for our sacrilege&lt;br&gt;&#xA;Once we dine with our heritage  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now we divide our surrounding&lt;br&gt;&#xA;separate haemorrhoids not corresponding&lt;br&gt;&#xA;Take our past in unrelated directions&lt;br&gt;&#xA;no longer our paths have intersections  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For we are at the end of the road&lt;br&gt;&#xA;although different, still a toad&lt;br&gt;&#xA;This heartfelt decision&lt;br&gt;&#xA;Has caused this incision  &lt;/p&gt;&#xA;" OwnerUserId="5135" LastEditorUserId="5135" LastEditDate="2014-12-10T14:37:51.503" LastActivityDate="2014-12-13T12:25:31.457" Title="Name that Sandwich: I am the End" Tags="|riddle|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="6066" PostTypeId="1" AcceptedAnswerId="6110" CreationDate="2014-12-10T17:40:28.430" Score="13" ViewCount="3225" Body="&lt;p&gt;So, I was bored and I made a small game for this site! In this game the user who solves the previous puzzle gets to create the next one and they will always be more complex. Website link and string down below!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How it works:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You paste in the current puzzle string in the input box and press reset just to be sure. This will setup the game and place &quot;invisible&quot; mirrors on the board of which there are 8 types.&lt;/li&gt;&#xA;&lt;li&gt;You shoot lasers with a click on one of the small boxes on the side of the board the green color tells you where the laser shot came from and blue tells you where it exited. If there is only a blue box and no green box then it means that the laser entered and exited in the same box.&lt;/li&gt;&#xA;&lt;li&gt;You place the mirrors with first clicking on one and then clicking on a square on the board, to delete a mirror click on it again or overwrite it with another&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The 8 mirrors are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Vertical and horizontal reflection mirrors (worth 1 point). As a example, let's take the horizontal mirror. If the laser comes from above or below it gets&#xA;reflected back in the opposite direction. If it comes from left or right it passes through the mirror.&lt;/li&gt;&#xA;&lt;li&gt;45° degree mirrors (worth 3 points). I think this is pretty self explanatory as to what happens.&lt;/li&gt;&#xA;&lt;li&gt;Rotating vertical and horizontal mirrors (worth 5 points). These behave the same way as their standard counterparts with the added ability that when they get hit they turn into their opposite. So if the laser comes from above at a horizontal rotating mirror it gets bounced back up and the mirror now becomes a vertical one, which then again turns into a horizontal one if it gets hit again etc etc. The rotation occurs also if the laser passes through.&lt;/li&gt;&#xA;&lt;li&gt;Rotating 45° degree mirrors (worth 7 points). These are the same as the standard 45° mirrors with the rotation ability&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Your goal in this game is to figure out at which position what mirror is under. You got some things to help you with this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The first is the point counter which shows what the total sum of points of all mirrors on the board, so if there is only 8 avaliable points&#xA;and you are sure that there is one rotating 45° mirror somewhere, then that only leaves 1 point left for one basic mirror.&lt;/li&gt;&#xA;&lt;li&gt;The second is the number of trys, this is a counter which tells you how many times you send a laser through from any position since innitializing the board&lt;/li&gt;&#xA;&lt;li&gt;And the last thing is a counter which tells you how many squares the laser has traveled in the current shot.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;The reset button resets the board to its initial state, this doesnt delete the mirrors you placed on the board but just puts the &quot;invisible&quot; mirrors back in their starting position&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Some additional things to consider:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The outer layer of the board can't have mirrors in them&lt;/li&gt;&#xA;&lt;li&gt;You need to find out the initial state of the board so if there is only one rotating mirror on the board and you hit it once with the laser the correct solution will still be to place the initial position of the rotating mirror on the board.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When you get the correct solution to the puzzle a message will inform you of your victory and then you will be able to create the next puzzle&#xA;where you will be able to place a total points of mirror worth $n+2$, where $n$ is the previous puzzle's maximum points. The first puzzle posted by me will have&#xA;3 maximum points, so the next will have 5, then 7 etc (Scoring is now different in the new version). To create a new puzzle you simply place mirrors on the board. When you place exactly $n+2$ points worth of mirrors you get the next puzzle string. When you are creating a new puzzle you will also be able to see the actual laser traveling, and if the laser travels over a box multiple times the color will get darker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should also probably sort this thread after date to see the latest puzzle. The next valid puzzle in the chain will be earliest one with the correct solution to the previous one. The last thing I want to impose is the correct format of a valid answer. Edit: check out the current answers!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Version 1&lt;/strong&gt; (use this version for level 1 to level 4): &lt;a href=&quot;https://preview.c9.io/vajura/blackbox/ver1/index.html&quot; rel=&quot;nofollow&quot;&gt;https://preview.c9.io/vajura/blackbox/ver1/index.html&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Version 2&lt;/strong&gt; (use this for level 5 and on, in this version mirrors cost more and the difficulty only increments for 1 point as a bonus the puzzle strings are 100 times smaller) &lt;a href=&quot;https://preview.c9.io/vajura/blackbox/ver2/index.html&quot; rel=&quot;nofollow&quot;&gt;https://preview.c9.io/vajura/blackbox/ver2/index.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Level 5 is not yet updated so if you want to solve it use Version 1, if you solve Level 5 in the old version before Josh update his puzzle to the new version you are free to create the Level 6 puzzle from this link(just start placing mirrors the points will update and when you placed a total worth of 13 the new puzzle string will be shown) &lt;a href=&quot;https://preview.c9.io/vajura/blackbox/ver2.1/index.html&quot; rel=&quot;nofollow&quot;&gt;https://preview.c9.io/vajura/blackbox/ver2.1/index.html&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Level 1, made by OP&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA; &lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Level 2, made by BmyGuest&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA; &lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Level 3, made by Crispy&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA; &lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Level 4, made by tttppp&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA; &lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Level 4.5, puzzle made by COTO (but originally numbered &quot;level 4&quot; by accident)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA; &lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Level 5, puzzle made by Josh&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA; &#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As I know many people here are smart and I have not really spent any time on making this secure. I don't doubt that someone will decode the puzzle strings or crack the whole thing if they want too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also sorry for any bugs!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;BONUS: Crack the code used! &lt;em&gt;Achieved&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5560" LastEditorUserId="12109" LastEditDate="2015-06-29T18:45:53.670" LastActivityDate="2015-06-29T18:45:53.670" Title="Can you solve those laser puzzles? (puzzles created by the community)" Tags="|geometry|mazes|open-ended|" AnswerCount="6" CommentCount="23" ClosedDate="2015-07-27T10:30:32.553" ContentLicense="CC BY-SA 3.0" />
  <row Id="6067" PostTypeId="2" ParentId="5960" CreationDate="2014-12-10T18:00:18.620" Score="2" Body="&lt;p&gt;Easy one ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The important point is the Pilots Guide:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The pilots guide referes to Albert Einstein and furthermore to the so called Einstein-Rosen-Bridge. These are theoretical one way traversable Wormholes. But to transfer you need near lightspeed -&gt; you need near infinite amount of fuel to accelerate any mass to near lightspeed -&gt; Time will be diluted and for the crew of the spaceship it will take VERY short to reach the other side.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Buying Fuel for High Prices!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;As the need for infinite amount of fuel can only mean that the payment is much less worth than the fuel or consists of this fuel one possibility comes to mind. AN as the Acid Number. Ranges about 92 mg KOH means it is some kind of oil. If partial oxidation is used to get Hydrogen it is common to use some sort of oil for the process. So you gimme 1 Gallon of oil and I give you 1gram of oil. Good deal?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Oxygen needed! &amp;amp; Technical Guide&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;As burning an unlimited amount of fuel you need also a unlimited amount of oxygen. Payment will be plaquetes like &lt;a href=&quot;http://en.wikipedia.org/wiki/Lunar_plaque&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Lunar_plaque&lt;/a&gt; as this is tradition for space adventures&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;They are shooting someone through a wormhole and you get paid less than you invested.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6254" LastActivityDate="2014-12-10T18:00:18.620" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6068" PostTypeId="2" ParentId="6066" CreationDate="2014-12-10T18:13:20.890" Score="3" Body="&lt;p&gt;Nice puzzle. The first level is easy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Previous puzzle&lt;/strong&gt; (solved below) &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Current (Next) puzzle (&lt;strong&gt;Level 2&lt;/strong&gt;)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Level 1 Map&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/ZQuQw.jpg&quot; alt=&quot;Solution1&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="5840" LastEditDate="2014-12-10T19:30:52.987" LastActivityDate="2014-12-10T19:30:52.987" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="6069" PostTypeId="1" AcceptedAnswerId="6417" CreationDate="2014-12-10T19:23:56.183" Score="20" ViewCount="4516" Body="&lt;p&gt;I have a cipher and I am wondering if - provided you have both the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ciphertext&quot; rel=&quot;nofollow noreferrer&quot;&gt;cipher-text&lt;/a&gt; and the clear text, given below, can you describe the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cipher&quot; rel=&quot;nofollow noreferrer&quot;&gt;cipher&lt;/a&gt; which was used?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The cipher-text is in this image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/KaiDE.jpg&quot; alt=&quot;Ciphertext&quot;&gt;&#xA;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I &lt;strong&gt;know&lt;/strong&gt; somebody will post this in a comment any second, so right away, here is some ASCII representation of the circular text. &lt;br&gt;&#xA;  However, I will &lt;em&gt;not&lt;/em&gt; say that you can solve this cipher by this ASCII string alone.&#xA;  On the other hand, you will &lt;em&gt;not&lt;/em&gt; need the digital form of that image, just what you are seeing. No dirty tricks played. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;HA, TITEXGB   DTESTRI AT UE A ONTCTSOMNDEOTA DN UEYCN HRUSN.IETH AEPT CR  F TNGEYI?OI A RCA UODUTYE COHNTUI?EURYI HEPSA T R&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This is the clear-text of the cipher:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; YOU CAN READ THIS CIPHERTEXT? THAT IS GREAT, BUT IT DOES NOT MEAN YOU CAN UNDERSTAND THE CIPHER YET. CAN YOU FIGURE IT OUT?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;Can you describe the cipher used in this puzzle ?&lt;/h1&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Hints&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The following hints are all (valid) conclusions drawn from partial-solutions below or discussions in the comments. They are certainly not required and may spoil your fun (or influence you into a certain line of thinking), so I'm putting them into spoiler tags.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint #1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The cipher does not alter any letters themselves. Number of each letter type is the same in clear-text and cipher-text.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Both cipher- and clear-text are cyclic. There is no distinct starting point neither for encrypting, nor for decrypting. i.e. if &quot;ABCDEFG&quot; maps on &quot;TUVWXYZ&quot;, then &quot;DEFGABCD&quot; will map onto &quot;WXYZTUV&quot; &lt;em&gt;or&lt;/em&gt; &quot;XYZTUVW&quot; &lt;em&gt;or&lt;/em&gt; any other cyclic permutation of the above. In all cases you can create the same &quot;cycles&quot; from clear-text and cipher-text.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This should be obvious for a cipher, but with the cipher it is possible to encrypt/decrypt arbitrary messages reliably. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The following pair of clean and cipher text was requested by Bobson and archaephyrryx.&#xA; &lt;pre&gt;THE QUICK BROWN FOX JUMPS OVER THE LAZY DOG, THEN THE LAZY DOG ROLLS UNDER THE QUICK BROWN FOX.WHO LIKES THIS DOG I WONDER?&lt;/pre&gt;&#xA; becomes:&#xA; &lt;pre&gt;WITDOES E HNOCBI RKW TFQ  OURNJ EOMAWPE HXSUO,RV DL GT EHAEOZY HN DLG T DET OLLOE RZHNYCSBIRU TK.QF U HERON I ODWKLOH ISXG?&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You may want to consider the difference between &lt;em&gt;mapping&lt;/em&gt; positions and &lt;em&gt;moving&lt;/em&gt; or &lt;em&gt;swapping&lt;/em&gt; characters in the context of a sequentially applied algorithm and reversibility of a cipher.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #6 (last hint):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The cipher is a sequential algorithm treating 26 'classes' of letters in sequence (and leaving a 27th class untouched). Deciphering will require a reversal of both the sequence of the 26 steps as well as each of the individual 26 steps. The sequence follows the natural 'order' of the 26 classes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="5840" LastEditDate="2014-12-21T14:42:27.260" LastActivityDate="2014-12-26T00:10:11.707" Title="What is the cipher behind this circular cyphertext?" Tags="|cipher|visual|" AnswerCount="8" CommentCount="20" ContentLicense="CC BY-SA 3.0" />
  <row Id="6071" PostTypeId="2" ParentId="6066" CreationDate="2014-12-10T21:14:05.083" Score="2" Body="&lt;p&gt;&lt;strong&gt;Previous puzzle Level 2&lt;/strong&gt; (solved below) &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Current (Next) puzzle (&lt;strong&gt;Level 3&lt;/strong&gt;)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Level 2 Map&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; !&lt;img src=&quot;https://i.stack.imgur.com/qkRjf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="138" LastActivityDate="2014-12-10T21:14:05.083" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6072" PostTypeId="1" AcceptedAnswerId="6074" CreationDate="2014-12-10T23:54:16.747" Score="12" ViewCount="968" Body="&lt;p&gt;Need a courier to help us send a message. &lt;strong&gt;Must have good aim.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you something of a creative type, the type whose mind moves in ways other than black and white? Do you understand discretion when collaborating in a group? Can you listen to a very, very specific story and share that story with others?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Join The Business Corporation, a &lt;strong&gt;completely legitimate operation&lt;/strong&gt; specializing in making dreams into reality for those whose got the money, not specializing in crime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scrawled in the margins:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Still trying to shoot with a 35 mm cartridge? Why not just do away with wasting metal? Hit 'em with a digital photoelectromagnetizer to get the job done. Our employees get the newest &quot;tools&quot; that offer pinpoint accuracy from far away and are completely electronic. The evidence left behind of your highly successful business dealings is in your hands and nobody else's.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You're told what to do and you get the job done. The last thing your target's going to see before the big sleep is what you have done. No games. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a multi-talented crew beside you, we're sure your work is going to be a real hit with everyone; they'll all be left seeing stars and you'll be seeing a whole lotta moolah.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're ready to climb the redberry tree to the top and be the best, contact Gaumont.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What are you being asked to do and how long is the prison sentence if there is one? Be certain and provide your reasoning; you'll end up needlessly unemployed or behind bars if you don't.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5626" LastEditorUserId="4050" LastEditDate="2014-12-11T10:18:27.237" LastActivityDate="2014-12-11T10:18:27.237" Title="Job Ad - A Legitimate Business needs discreet help. What do they need?" Tags="|riddle|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6073" PostTypeId="2" ParentId="6072" CreationDate="2014-12-11T01:47:07.943" Score="9" Body="&lt;p&gt;It sounds like you are being asked to be a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;cinematographer or a movie camera person&lt;/strong&gt; (and there is no prison sentence)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;other than black and white&quot; suggests color&lt;br&gt;&#xA; &quot;discretion when collaborating&quot; suggests not revealing the story&lt;br&gt;&#xA; &quot;very, very specific story&quot; suggests the script&lt;br&gt;&#xA; &quot;dreams into reality&quot; suggests bringing a story to life&lt;br&gt;&#xA; &quot;35 mm cartridge ...&quot; suggests using digital video instead of chemical film&lt;br&gt;&#xA; &quot;seeing stars&quot; suggests movie stars&lt;br&gt;&#xA; Gaumont is a French film company &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastActivityDate="2014-12-11T01:47:07.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6074" PostTypeId="2" ParentId="6072" CreationDate="2014-12-11T01:53:54.197" Score="12" Body="&lt;p&gt;You are asked to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Be a cameraman, no prison sentence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Need a courier to help us send a message. Must have good aim.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The message is the movie, and good aim is required with a camera.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Are you something of a creative type, the type whose mind moves in ways other than black and white? Do you understand discretion when collaborating in a group? Can you listen to a very, very specific story and share that story with others?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Movie making is a creative endeavour, movies are not made in black and white anymore, discretion is required so as to not leak the film's script to the outside world, and the story you are sharing is the story of the movie.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Join The Business Corporation, a completely legitimate operation specializing in making dreams into reality for those whose got the money, not specializing in crime.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Movies can turn dreams (things impossible in reality) into reality, but cost a lot of money.  They are also completely legitimate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The margins:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Scrawled in the margins is a message telling you to use digital video recorders instead of film.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You're told what to do and you get the job done&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The director tells you where to place the camera, when to shoot and when to cut.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The last thing your target's going to see before the big sleep is what you have done. No games.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The movie watchers are going to see the film before going to bed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With a multi-talented crew beside you, we're sure your work is going to be a real hit with everyone; they'll all be left seeing stars and you'll be seeing a whole lotta moolah.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Film making requires a large crew; stars refers to movie stars and movies can make a lot of money as well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you're ready to climb the redberry tree to the top and be the best, contact Gaumont.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Redberry refers to Hollywood.  Hollies have red berries.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6178" LastActivityDate="2014-12-11T01:53:54.197" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6075" PostTypeId="2" ParentId="6048" CreationDate="2014-12-11T02:18:29.470" Score="15" Body="&lt;p&gt;Thanks to @xnor presenting a solution, I was finally able to fix the bug in my Python 3 program.&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from collections import Counter&#xA;import sys&#xA;&#xA;grid = [[7, 3, 1, 2, 9, 8, 5, 6, 4],&#xA;        [4, 9, 5, 1, 8, 6, 7, 2, 3],&#xA;        [9, 4, 6, 7, 5, 3, 1, 8, 2],&#xA;        [1, 5, 3, 8, 7, 4, 2, 9, 6],&#xA;        [8, 6, 2, 5, 3, 1, 4, 7, 9],&#xA;        [2, 8, 9, 4, 1, 5, 6, 3, 7],&#xA;        [6, 7, 8, 3, 4, 2, 9, 5, 1],&#xA;        [3, 1, 7, 6, 2, 9, 8, 4, 5],&#xA;        [5, 2, 4, 9, 6, 7, 3, 1, 8]]&#xA;&#xA;walls = [((0,4), (1,4)),&#xA;         ((1,4), (1,5)),&#xA;         ((2,4), (2,5)),&#xA;         ((1,8), (2,8)),&#xA;         ((3,6), (3,7)),&#xA;         ((4,6), (4,7)),&#xA;         ((4,1), (4,2)),&#xA;         ((4,1), (5,1)),&#xA;         ((5,1), (6,1)),&#xA;         ((6,4), (6,5)),&#xA;         ((6,5), (7,5)),&#xA;         ((7,5), (7,6))]&#xA;&#xA;def neighbours(cell):&#xA;    results = []&#xA;&#xA;    for neighbour in [(cell[0]-1, cell[1]), (cell[0]+1, cell[1]),&#xA;                       (cell[0], cell[1]-1), (cell[0], cell[1]+1)]:&#xA;        if (0 &amp;lt;= neighbour[0] &amp;lt; len(grid) and 0 &amp;lt;= neighbour[1] &amp;lt; len(grid) and&#xA;            tuple(sorted([cell, neighbour])) not in walls):&#xA;            results.append(neighbour)&#xA;&#xA;    return results&#xA;&#xA;def all_polyominoes(grid):&#xA;    polys = set()&#xA;&#xA;    def search(curr_cell, added, nums, to_search, searched):          &#xA;        if len(nums) == len(grid):&#xA;            polys.add(tuple(sorted(added)))&#xA;            return&#xA;&#xA;        while to_search:&#xA;            n = to_search.pop()&#xA;&#xA;            if grid[n[0]][n[1]] not in nums:&#xA;                search(n, added | {n}, nums | {grid[n[0]][n[1]]},&#xA;                       to_search | {x for x in neighbours(n) if x not in searched},&#xA;                       searched)&#xA;&#xA;        return&#xA;&#xA;    for i in range(len(grid)):&#xA;        for j in range(len(grid)):&#xA;            start_cell = (i, j)&#xA;            search(start_cell, {start_cell}, {grid[i][j]},&#xA;                   {n for n in neighbours(start_cell)},&#xA;                   {n for n in neighbours(start_cell)})&#xA;&#xA;    to_remove = set()&#xA;&#xA;    for p in polys:&#xA;        for cell1 in p:&#xA;            for cell2 in p:&#xA;                if tuple(sorted((cell1, cell2))) in walls:&#xA;                    to_remove.add(p)&#xA;&#xA;    polys = polys - to_remove&#xA;    return polys&#xA;&#xA;walls = set(tuple(sorted(x)) for x in walls)&#xA;polys = list(all_polyominoes(grid))&#xA;&#xA;print(len(polys), &quot;polyominoes formed&quot;)&#xA;&#xA;all_cells = {(i, j) for i in range(len(grid)) for j in range(len(grid))}&#xA;&#xA;def good_blobs(cells):&#xA;    # Look at blobs (contiguous regions of unused cells) with size divisible by grid size&#xA;&#xA;    remaining_cells = all_cells - cells&#xA;&#xA;    while remaining_cells:&#xA;        cell = remaining_cells.pop()&#xA;        blob = {cell}&#xA;        to_search = [cell]&#xA;&#xA;        while to_search:&#xA;            c = to_search.pop()&#xA;&#xA;            for n in neighbours(c):&#xA;                if n in remaining_cells:&#xA;                    remaining_cells.remove(n)&#xA;                    blob.add(n)&#xA;                    to_search.append(n)&#xA;&#xA;        if len(blob) % len(grid) != 0:&#xA;            return False&#xA;&#xA;    return True&#xA;&#xA;def solve(covering, cells_covered, index):    &#xA;    if len(covering) == len(grid):&#xA;        print()&#xA;        print(covering)&#xA;        return True&#xA;&#xA;    for i,p in enumerate(polys[index:]):&#xA;        if len(covering) == 0:&#xA;            print(i, end=&quot; &quot;)&#xA;            sys.stdout.flush()&#xA;&#xA;        q = set(p)&#xA;&#xA;        if not (q &amp;amp; cells_covered):&#xA;            new_cells_covered = q | cells_covered&#xA;&#xA;            if good_blobs(new_cells_covered):&#xA;                new_covering = covering + [p]&#xA;                result = solve(new_covering, new_cells_covered, i+index+1)&#xA;&#xA;                #if result:&#xA;                #    return True&#xA;&#xA;    return False&#xA;&#xA;solve([], set(), 0)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The algorithm is pretty simple:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Generate all valid polyominoes that can be formed from the grid&lt;/strong&gt; (i.e. contains the numbers 1-9 exactly once, and respects the walls). This is where my bug was as I initially did a flood fill incorrectly, finding only the polyominoes that can be formed &quot;without lifting your pen off the paper&quot;, so to speak.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Keep placing polyominoes while we can, and backtrack if we can't&lt;/strong&gt;. Basically, recursion.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;At this point, it seems like what we have is exactly the &lt;a href=&quot;https://en.wikipedia.org/wiki/Set_cover_problem&quot; rel=&quot;noreferrer&quot;&gt;set cover problem&lt;/a&gt;. But there's restrictions on the polyominoes, namely the grid, allowing us to use one important heuristic that drastically improves the running time:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;When you place a new polyomino, look at the regions formed by cells not currently in a polyomino. If any of the regions have size which is not 0 modulo 9, then backtrack.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using &lt;a href=&quot;http://pypy.org/&quot; rel=&quot;noreferrer&quot;&gt;PyPy&lt;/a&gt; to speed things up, the program above finds its first solution at the 4 minute mark:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pgmm3m.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The program terminates after 20 minutes, finding a total of 5 solutions, the others being:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PXQ7am.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OaxPFm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sOG1km.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HFD6xm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(xnor's is the second one)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was easier for me to make the images from scratch, hence the difference in colour choice between each grid. Here's the common bits between each solution to make things easier to see:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mLNrDm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="445" LastActivityDate="2014-12-11T02:18:29.470" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6076" PostTypeId="2" ParentId="5960" CreationDate="2014-12-11T02:34:24.827" Score="1" Body="&lt;p&gt;Most likely not the correct answer, but I had thought of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Separation_of_isotopes_by_laser_excitation&quot; rel=&quot;nofollow&quot;&gt;Uranium enrichment using lasers&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;CO2 is the fuel for the laser (so oxygen is required) and the payment is enriched uranium (Atomic number 92).  The pilots guide refers to lasers moving at the speed of light.  The brass troughs refers to laser etching.  Some lasers run on hydrogen fluoride instead.  It doesn't answer why infinite fuel is needed though.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6178" LastActivityDate="2014-12-11T02:34:24.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6077" PostTypeId="1" AcceptedAnswerId="6084" CreationDate="2014-12-11T09:33:08.957" Score="14" ViewCount="2559" Body="&lt;p&gt;This is my first riddle and I think it should be easy for you guys.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;While swinging my fin I fight for a grin&lt;br&gt;&#xA;  I woosh back and forth&lt;br&gt;&#xA;  Killing millions on both&lt;br&gt;&#xA;  Destroying evil day and night&lt;br&gt;&#xA;  Preventing precious from blight&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4857" LastEditorUserId="4050" LastEditDate="2014-12-11T09:43:15.710" LastActivityDate="2014-12-11T11:13:15.953" Title="What am I? (my first riddle): While swinging my fin I fight for a grin" Tags="|riddle|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6078" PostTypeId="2" ParentId="6069" CreationDate="2014-12-11T09:45:00.350" Score="5" Body="&lt;p&gt;Here are some thoughts I have before any extra info was revealed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the ciphertext is a &lt;em&gt;permutation of the plaintext&lt;/em&gt;, treating spaces and symbols and letters alike. I also suspect the text is in &lt;em&gt;Latin&lt;/em&gt;. (Edit: Never mind, BmyGuest has added that it's English.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The letter frequencies of the text are similar to that of English but &lt;a href=&quot;http://www.rosettagrams.com/letter_frequencies.htm&quot; rel=&quot;nofollow&quot;&gt;actually closer to that of Latin&lt;/a&gt;. Note in particular the increased frequency of U (which also serves the function of V in Latin), the relative rarity of H, the total lack of Latin non-letters J, K, W, and V. I can't be sure though because of the relatively small counts. I could also be another Romance language.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;t : 14&#xA;e : 11&#xA;a : 8&#xA;u : 7&#xA;i : 7&#xA;n : 7&#xA;r : 6&#xA;o : 6&#xA;h : 5&#xA;c : 5&#xA;y : 4&#xA;d : 4&#xA;s : 4&#xA;g : 2&#xA;p : 2&#xA;x : 1&#xA;m : 1&#xA;f : 1&#xA;b : 1&#xA;z : 0&#xA;w : 0&#xA;v : 0&#xA;l : 0&#xA;k : 0&#xA;j : 0&#xA;q : 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Of the 123 characters, there are 96 letters, there are also 23 spaces, two question marks, a period and a comma. This is consistent with a typical word length of about 4.2, totally reasonable. Likewise, three sentences averaging 32 letters each and 7-8 words each is reasonable, though on the short end for words per sentence. Two of the sentences are questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, given that the character statistics are totally consistent that of plain text, I think the ciphertext is simply a rearrangement of the characters in the plaintext. Something that substituted for symbols for other symbols would not have produced these statistics. Moreover, there are no repeating substrings of length 3 or more, which further suggests rearrangement rather than substitution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What rule could permute the letters in the ciphertext? My first thought was simply taking every $n$th letter, which would also explain why the message is in a circle, to wrap around. But trying this for all possible $n$ didn't give anything that struck me as intelligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thought is that each letter tells you to look at the next letter $k$ spaces forward, perhaps with A=1, ..., Z=26 and something for non-letters. But then one would likely get into a loop. Unless maybe you skip letters you've already used? I'll have to this more about whether this can be a reversible encoding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideas to proceed:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strike&gt; Simply find a Latin famous quote that could match. It should have two sentences and a question mark. Maybe a part of the Lorem Ipsum text? &lt;/strike&gt;&lt;/li&gt;&#xA;&lt;li&gt;Look at the distribution of spaces to see what type of permutation might give reasonable word length&lt;/li&gt;&#xA;&lt;li&gt;Wait for more hints&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2014-12-11T10:51:00.273" LastActivityDate="2014-12-11T10:51:00.273" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="6079" PostTypeId="1" AcceptedAnswerId="6080" CreationDate="2014-12-11T09:47:57.760" Score="13" ViewCount="7522" Body="&lt;p&gt;This puzzle is not mine, but I couldn't find an original source. I edited it somewhat.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;An elementary school teacher had her purse stolen. The thief had to be&#xA;Sammy, Karin, Arnold, Harry, or Liza. When questioned, each child&#xA;made three statements:&lt;/p&gt;&#xA;&lt;p&gt;Sammy:&lt;br /&gt;&#xA;(1) I didn’t take the purse.&lt;br /&gt;&#xA;(2) I have never in my life stolen anything.&lt;br /&gt;&#xA;(3) Harry did it.&lt;/p&gt;&#xA;&lt;p&gt;Karin:&lt;br /&gt;&#xA;(4) I didn’t take the purse.&lt;br /&gt;&#xA;(5) My daddy is rich enough, and I have a purse of my own.&lt;br /&gt;&#xA;(6) Liza knows who did it.&lt;/p&gt;&#xA;&lt;p&gt;Arnold:&lt;br /&gt;&#xA;(7) I didn’t take the purse.&lt;br /&gt;&#xA;(8) I didn’t know Liza before I enrolled in this school.&lt;br /&gt;&#xA;(9) Harry did it.&lt;/p&gt;&#xA;&lt;p&gt;Harry:&lt;br /&gt;&#xA;(10) I am not guilty.&lt;br /&gt;&#xA;(11) Liza did it.&lt;br /&gt;&#xA;(12) Sammy is lying when she says I stole the purse.&lt;/p&gt;&#xA;&lt;p&gt;Liza:&lt;br /&gt;&#xA;(13) I didn’t take the teacher’s purse.&lt;br /&gt;&#xA;(14) Karin is guilty.&lt;br /&gt;&#xA;(15) Arnold can vouch for me because he has known me since I was born.&lt;/p&gt;&#xA;&lt;p&gt;Later, each child admitted that two of his statements were true and&#xA;one was false. Assuming this is true, who stole the purse?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5578" LastEditorUserId="69582" LastEditDate="2021-06-01T04:48:03.177" LastActivityDate="2021-06-01T04:48:03.177" Title="Who is the thief? An elementary school teacher had her purse stolen" Tags="|logical-deduction|liars|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6080" PostTypeId="2" ParentId="6079" CreationDate="2014-12-11T09:54:11.407" Score="13" Body="&lt;p&gt;It is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Karin&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since 15 and 8 contradict, we have two possibilities&lt;br/&gt;If 15 is False&lt;br/&gt;then 14 is True =&gt; then Karin stole the purse.&lt;br/&gt;If 15 is True&lt;br/&gt;then 8 is False&lt;br/&gt;then 9 is True.&lt;br/&gt;However, if 9 is true, then 10, 11 and 12 are False, and we can't have that =&gt; 14 is False. So:&lt;br/&gt;&lt;strong&gt;Karin did it&lt;/strong&gt;&lt;br/&gt;&lt;br/&gt;The lies are 3, 4, 9, 11 and 15&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="5610" LastEditDate="2014-12-15T13:28:14.923" LastActivityDate="2014-12-15T13:28:14.923" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6081" PostTypeId="2" ParentId="6077" CreationDate="2014-12-11T10:00:45.293" Score="1" Body="&lt;p&gt;You are a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clock (with a Pendulum)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While swinging my fin I fight for a grin&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The pendulum swings where the shape formed by the motion is a smile/grin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I woosh back and forth&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The pendulum swings in this way.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Killing millions on both&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every tick of the clock signals the death of millions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Destroying evil day and night&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sounds of the clock striking the hour (like a cuckoo clock) is thought to ward off evil. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Preventing precious from blight&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure :p&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastActivityDate="2014-12-11T10:00:45.293" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6082" PostTypeId="2" ParentId="6077" CreationDate="2014-12-11T10:05:56.017" Score="2" Body="&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Moon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While swinging my fin &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The moon swims through across the sky.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I fight for a grin&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The crescent moon looks like a grin (c.f. The Cheshire Cat).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I woosh back and forth&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The moon goes back and forth across the sky.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Killing millions on both&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Moon can mark the passage of time. Millions die with every cycle of the Moon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Destroying evil day and night&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Moon adds to the light of the night - &quot;evil&quot; is taken darkness.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Preventing precious from blight&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No idea yet &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastActivityDate="2014-12-11T10:05:56.017" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6083" PostTypeId="1" AcceptedAnswerId="6085" CreationDate="2014-12-11T11:08:39.257" Score="23" ViewCount="2906" Body="&lt;p&gt;The idea for this challenge stems from &lt;a href=&quot;https://cs.stackexchange.com/questions/34067/is-morse-code-without-spaces-uniquely-decipherable&quot;&gt;this posting on ComputerScience (beta)&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;It is established (see &lt;a href=&quot;https://cs.stackexchange.com/questions/34067/is-morse-code-without-spaces-uniquely-decipherable&quot;&gt;above link&lt;/a&gt;), that Morse-code requires the 'space' between individual character to be uniquely decipherable.&lt;/p&gt;&#xA;&lt;p&gt;The answer of &lt;a href=&quot;https://cs.stackexchange.com/users/20068/aaron-dufour&quot;&gt;Aaron Dufour&lt;/a&gt; there defines the term &lt;strong&gt;morsagram&lt;/strong&gt; as a pair of messages which, encoded in morse code become &lt;em&gt;identical&lt;/em&gt; if the spaces are left out. His provided example is:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;code&gt;SPACES&lt;/code&gt; = | &lt;code&gt;...&lt;/code&gt; | &lt;code&gt;.--.&lt;/code&gt; | &lt;code&gt;.-&lt;/code&gt; | &lt;code&gt;-.-.&lt;/code&gt; | &lt;code&gt;.&lt;/code&gt; | &lt;code&gt;...&lt;/code&gt; | = &lt;strong&gt;&lt;code&gt;....--..--.-.....&lt;/code&gt;&lt;/strong&gt; &lt;br&gt;&#xA;and&lt;br&gt;&#xA;&lt;code&gt;SWITCH&lt;/code&gt; = | &lt;code&gt;...&lt;/code&gt; | &lt;code&gt;.--&lt;/code&gt; | &lt;code&gt;..&lt;/code&gt; | &lt;code&gt;-&lt;/code&gt; | &lt;code&gt;-.-.&lt;/code&gt; | &lt;code&gt;....&lt;/code&gt; | = &lt;strong&gt;&lt;code&gt;....--..--.-.....&lt;/code&gt;&lt;/strong&gt; &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;br&gt;&#xA;&lt;h2&gt;I now challenge to find the longest &lt;em&gt;morsagram&lt;/em&gt; pair of words in English.&lt;/h2&gt;&#xA;&lt;p&gt;To remove ambiguity, and in analogy to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/5456/the-longest-word-made-from-chemical-symbols/5457#5457&quot;&gt;this puzzle&lt;/a&gt;, all English words which appear on  &lt;a href=&quot;http://dictionary.reference.com/&quot; rel=&quot;noreferrer&quot;&gt;dictionary.com&lt;/a&gt; are permitted.&lt;/p&gt;&#xA;&lt;p&gt;You may only use the Morse-code equivalents for the 26 characters of the &lt;a href=&quot;http://en.wikipedia.org/wiki/File:International_Morse_Code.svg&quot; rel=&quot;noreferrer&quot;&gt;international Morse code table&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;As consequence of the rule above: solutions are case-insensitive. Use capital letters for all.&lt;/p&gt;&#xA;&lt;p&gt;Also: The two words of the &lt;em&gt;morsagram&lt;/em&gt; may be of different length when written in English.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;The winning condition is to find the &lt;em&gt;longest&lt;/em&gt; words&lt;/strong&gt; (sum of both lengths of the English words). &lt;br&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;A bonus is to find the most &lt;em&gt;interesting&lt;/em&gt; pair&lt;/strong&gt;. This, of course, is highly subjective. Interesting may be one of the following: &lt;em&gt;most different characters&lt;/em&gt;;  &lt;em&gt;opposite meaning&lt;/em&gt;; two very &lt;em&gt;common&lt;/em&gt; but otherwise &lt;em&gt;unrelated&lt;/em&gt; words; words with have another &lt;em&gt;surprising&lt;/em&gt; connection...&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;A second bonus question is to find the most &lt;em&gt;interesting sentences&lt;/em&gt; which are morsograms&lt;/strong&gt;. &lt;br&gt;&#xA;Again, an example is already provided in the linked posting above:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;code&gt;SOS HELP&lt;/code&gt;      = |&lt;code&gt;...&lt;/code&gt;|&lt;code&gt;---&lt;/code&gt;|&lt;code&gt;...&lt;/code&gt;|&lt;code&gt;....&lt;/code&gt;|&lt;code&gt;.&lt;/code&gt;|&lt;code&gt;.-..&lt;/code&gt;|&lt;code&gt;.--.&lt;/code&gt;| = &lt;strong&gt;&lt;code&gt;...---.........-...--.&lt;/code&gt;&lt;/strong&gt;&lt;br&gt;&#xA;and&lt;br&gt;&#xA;&lt;code&gt;I AM HIS DATE&lt;/code&gt; = |&lt;code&gt;..&lt;/code&gt;|&lt;code&gt;.-&lt;/code&gt;|&lt;code&gt;--&lt;/code&gt;|&lt;code&gt;....&lt;/code&gt;|&lt;code&gt;..&lt;/code&gt;|&lt;code&gt;...&lt;/code&gt;|&lt;code&gt;-..&lt;/code&gt;|&lt;code&gt;.-&lt;/code&gt;|&lt;code&gt;-&lt;/code&gt;|&lt;code&gt;.&lt;/code&gt;| = &lt;strong&gt;&lt;code&gt;...---.........-...--.&lt;/code&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;For this bonus-question, disregards punctuation in the Morse-code.&#xA;&lt;br&gt;&lt;/p&gt;&#xA;&lt;p&gt;I will &lt;em&gt;accept&lt;/em&gt; and &lt;em&gt;un-accept&lt;/em&gt; answers for the winning condition as new records are brought in. &lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;strong&gt;Up-votes&lt;/strong&gt; should be given to answers which you believe fulfil the bonus condition.&lt;/p&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-12-15T12:21:46.043" Title="Open-ended challenge: Find the longest morsagram word pair" Tags="|word|open-ended|" AnswerCount="2" CommentCount="10" ClosedDate="2021-07-22T07:27:20.487" ContentLicense="CC BY-SA 3.0" />
  <row Id="6084" PostTypeId="2" ParentId="6077" CreationDate="2014-12-11T11:13:15.953" Score="17" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a toothbrush.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;While swinging my fin I fight for a grin&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pretty literal here: the fin is the brush, the grin is the grin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;I woosh back and forth&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Just brushing, you know.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Killing millions on both&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Killing the horrible bacteria, of course.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Destroying evil day and night&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well, actually three times a day as a minimum, I hope.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Preventing precious from blight&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What's more precious than teeth? Well, ok, golden teeth. And they don't even get blight.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5990" LastActivityDate="2014-12-11T11:13:15.953" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6085" PostTypeId="2" ParentId="6083" CreationDate="2014-12-11T11:29:29.333" Score="23" Body="&lt;p&gt;Longest word-pair so far:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;code&gt;.--..-..-.-.----.-.--.--.---.-..--..---..-...&lt;/code&gt;&lt;/strong&gt;&lt;br&gt;&#xA; 18: &lt;a href=&quot;http://dictionary.reference.com/browse/PRECONTEMPORANEOUS?s=t&quot;&gt;PRECONTEMPORANEOUS&lt;/a&gt;&lt;br&gt;&#xA; 19: &lt;a href=&quot;http://dictionary.reference.com/browse/PENECONTEMPORANEOUS?s=t&quot;&gt;PENECONTEMPORANEOUS&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Most &lt;em&gt;interesting pair&lt;/em&gt;:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The pairs with greatest difference in length (omitting abr.):&lt;br/&gt;-.-.-.---.-.: TENTAMEN, CYC, (len 8,3)&lt;br/&gt;.---.-.-.--.: JAAP, ATTENTATE, (len 4,9)&lt;br/&gt;-.-.---.--.: COP, TENTMATE, (len 3,4,4,8,4)&lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;OR&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ......-.-.-.-.........-........: HURTLESSNESS, HEARTLESSNESS, (len 12, 13)&lt;br/&gt;Are you hurtless when heartless?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;OR&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The UNs ..--......--..-.--.: UNSEEMING, UNSEWING, UNSEATING, UNSITTING, UNHATING, (len 9,8,9,9,8)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Brute forced top 20:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; .--..-..-.-.----.-.--.--.---.-..--..---..-...: PRECONTEMPORANEOUS, PENECONTEMPORANEOUS, (len 18,19)&lt;br/&gt;....--..-.....---.-..-----...-.-..-.-...-..-.--: SEMITHEOLOGICALLY, SEMIBIOLOGICALLY, (len 17,16)&lt;br/&gt;.--.....-.--.....-.-.----.....---.-..-----...-.-..-.-..: PHYSICOBIOLOGICAL, PHYSICOTHEOLOGICAL, (len 17,18)&lt;br/&gt;-.----....-..-...-.--....-...-........: NONIRRITABLENESS, NONVERITABLENESS, (len 16,16)&lt;br/&gt;..--.-....--..-.............-.-........: UNDEPRESSIVENESS, UNDIGRESSIVENESS, (len 16,16)&lt;br/&gt;-.----.-.....---.-..-----...-.-..-.-...-..-.--: NONTHEOLOGICALLY, NONBIOLOGICALLY, (len 16,15)&lt;br/&gt;.--..-.-.-..-...-....-..--.-.-........--.: PECTINIBRANCHIAN, PECTINIBRANCHIATE, (len 16,17)&lt;br/&gt;.--...-.....-.-..--.-.....-.---.--...-..: PITHECANTHROPINE, PITHECANTHROPID, (len 16,15)&lt;br/&gt;-........-.-....-...-.-..--..----.: DESERTIFICATION, DESULFURATION, (len 15,13)&lt;br/&gt;..-...-...------....-..--.-.-........--.: ELASMOBRANCHIAN, ELASMOBRANCHIATE, (len 15,16)&lt;br/&gt;.--..--.....--.--..-.-..-.-...-..-.--: ANATHEMATICALLY, PATHEMATICALLY, (len 15,14)&lt;br/&gt;..--....-..-...-.--....-...-........: UNIRRITABLENESS, UNVERITABLENESS, (len 15,15)&lt;br/&gt;..--......-.-..--..-.......-..-........: UNSURPRISEDNESS, UNSURPASSEDNESS, (len 15,15)&lt;br/&gt;-.----....----.-....--....-...-........: NONSOCIABLENESS, NONVOLUBLENESS, (len 15,14)&lt;br/&gt;-.-..--....--..--.--......-....--.--: TRANSMUTABILITY, TRANSMITTABILITY, (len 15,16)&lt;br/&gt;...-.-...--..-....-..--.-.-........--.: SCUTIBRANCHIATE, SCUTIBRANCHIAN, (len 15,14)&lt;br/&gt;.-..--....-----.-.-...-.-..-.-...-..-.--: AUDIOMETRICALLY, RADIOMETRICALLY, (len 15,15)&lt;br/&gt;-.-.-..--.-.-..-...-...-....-..: TETRACTINELLINE, TETRACTINELLID, (len 15,14)&lt;br/&gt;-..--.....-.-....-....-..--.-.-........--.: TUBULIBRANCHIAN, TUBULIBRANCHIATE, (len 15,16)&lt;br/&gt;.--......-.-.--.--.-.---.--.---......-: PHARMACOPOEIST, PHARMACOPOSIA, (len 14,13)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Don't really care for the up-votes. Just wanted to see if the two words would be of same length.&lt;/p&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="5610" LastEditDate="2014-12-11T13:39:37.173" LastActivityDate="2014-12-11T13:39:37.173" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="6092" PostTypeId="2" ParentId="6066" CreationDate="2014-12-11T16:18:30.607" Score="3" Body="&lt;p&gt;&lt;strong&gt;Previous puzzle Level 3&lt;/strong&gt; (solved below)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Current (Next) puzzle (&lt;strong&gt;Level 4&lt;/strong&gt;)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Level 3 Map&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/DhrmG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="303" LastEditorUserId="303" LastEditDate="2014-12-11T16:31:46.180" LastActivityDate="2014-12-11T16:31:46.180" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6093" PostTypeId="2" ParentId="6069" CreationDate="2014-12-11T16:27:14.373" Score="5" Body="&lt;p&gt;Given that we know that the circular nature of the puzzle is important, and that there are an odd number of characters (123, specifically) it seems a pretty safe conclusion that we need to go around the circle at least twice to decode it, probably continuously.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A naive approach is &quot;every other letter, going around twice&quot;.  That produces &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;H,TTXB DETIA EAOTTONET NUYNHUNIT ETC FTGY?IARAUDTECHTIERIHPATR&#xA;  A IEG  TSR TU  NCSMDOAD EC RS.EHAP R  NEIO  C OUY ONU?UY ES   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clearly, that's not right - even rotating the starting point won't help, because that would just rotate the result, and there's no clear English in that result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying 3 and 41 won't work, because they'll just loop over the same letters each time (3*41 = 123).  Brute-forcing each possible skip number up to 123 confirms the naive approach doesn't work (although using 41 from the provided starting point does produce &quot;HO?HO?HO?HO?HO?HO?HO?HO?HO?&quot; over and over again, which is seasonally appropriate).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now that we've concluded that the there isn't a consistent skip value, lets look at punctuation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Of the 123 characters, there are 96 letters, there are also 23 spaces, two question marks, a period and a comma.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From each punctuation mark (as 0), spaces can be found in the following positions: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;,&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;1, 9, 10, 11, 19, 22, 25, 27, 42, 45, 51, 62, 67, 70, 71, 73, 83, 85, 89, 97, 111, 117, 119 &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;.&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;5, 10, 13, 14, 16, 26, 28, 32, 40, 54, 60, 62, 67, 75, 76, 77, 85, 88, 91, 93, 108, 111, 117 &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;?&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;3, 5, 9, 17, 31, 37, 39, 44, 52, 53, 54, 62, 65, 68, 70, 85, 88, 94, 105, 110, 113, 114, 116 &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;? #2&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;6, 12, 14, 19, 27, 28, 29, 37, 40, 43, 45, 60, 63, 69, 80, 85, 88, 89, 91, 101, 103, 107, 115 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not seeing anything obvious from that information, though.  There's no numbers that show on all lists (which as we'd expect given the lack of a simple skip).  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Update:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I've been playing with an extended version of this plan.  Given the plaintext only has a single &lt;code&gt;.&lt;/code&gt; in it (at position 60), I generated a list of every possible way to generate the sequence &lt;code&gt;. CAN Y&lt;/code&gt;.  A &lt;em&gt;small&lt;/em&gt; sample of the output:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;60,4,34,2,32,4,51&#xA;60,4,34,2,39,4,51&#xA;60,4,34,2,32,12,51&#xA;60,12,34,2,32,4,51&#xA;60,4,34,2,32,13,51&#xA;60,13,34,2,32,4,51&#xA;60,4,34,2,32,14,51&#xA;...&#xA;60,114,101,119,104,120,112&#xA;60,120,101,119,104,114,112&#xA;60,114,101,119,104,122,112&#xA;60,122,101,119,104,114,112&#xA;60,120,101,119,104,120,112&#xA;60,120,101,119,104,122,112&#xA;60,122,101,119,104,120,112&#xA;60,122,101,119,104,122,112&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;My goal was to look for patterns in the change between them, but there are &lt;strong&gt;592480&lt;/strong&gt; possible combinations, which is a bit much to manually skim.  If anyone's interested in trying to parse this data, I can upload it somewhere, but the file is over 12 MB worth of lines just like the ones above.&lt;/p&gt;&#xA;" OwnerUserId="325" LastEditorUserId="325" LastEditDate="2014-12-16T19:45:46.417" LastActivityDate="2014-12-16T19:45:46.417" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="6094" PostTypeId="2" ParentId="6066" CreationDate="2014-12-11T17:19:48.343" Score="3" Body="&lt;h1&gt;Current Level (4)&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;An automatic +1 from me if you can figure out what the heck is going on here:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Previous Level (3) Solution&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The solution is posted below.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/DQwbN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2014-12-11T17:19:48.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6095" PostTypeId="2" ParentId="6079" CreationDate="2014-12-11T17:47:22.443" Score="6" Body="&lt;p&gt;Like @dmg said, the culprit is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Karin&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I used a slightly different path to come to this conclusion, though:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Harry said &quot;I am not guilty&quot; and &quot;Liza did it&quot;. If he is lying about being innocent, then Liza could not have done it, which would make that statement a lie as well. So Harry must not have done it.&lt;br&gt;&lt;br&gt;Because Harry did not do it, we know that 9 is Arnold's lie. That means 8 is true.&lt;br&gt;&lt;br&gt;Because 8 is true, 15 is a lie. That means 14 (Karin is guilty) must be the truth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1955" LastActivityDate="2014-12-11T17:47:22.443" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6096" PostTypeId="2" ParentId="6083" CreationDate="2014-12-11T20:05:15.563" Score="19" Body="&lt;p&gt;Suppose, instead of maximum length or maximum length difference, you care about maximum edit distance (levenshtein). Here are the winners:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;9 BETTERING DUNLAP -....--..-...-.--.&#xA;9 HICCUP HEARTENING ......-.-.-.-...-.--.&#xA;9 POLLUTE WAGNERIAN .--.---.-...-....--.&#xA;9 RAKING ETIQUETTE .-..--.-..-.--.&#xA;9 RANTING ETIQUETTE .-..--.-..-.--.&#xA;9 TESTIFIES BLUSH -....-....-.......&#xA;9 THINNING DEVASTATE -......-.-...-.--.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The longest pair that has no letters in common:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;THRALL DEFENDED -.....-..-.-...-..&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2187" LastEditorUserId="5840" LastEditDate="2014-12-15T12:21:46.043" LastActivityDate="2014-12-15T12:21:46.043" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6099" PostTypeId="2" ParentId="6066" CreationDate="2014-12-11T23:53:38.297" Score="3" Body="&lt;p&gt;That was fun!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Previous puzzle Level 4&lt;/strong&gt; (solved below)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Current (Next) puzzle (&lt;strong&gt;Level 5&lt;/strong&gt;) (Remember to user version 2.1 to solve - &lt;a href=&quot;https://preview.c9.io/vajura/blackbox/ver2.1/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://preview.c9.io/vajura/blackbox/ver2.1/index.html&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;rqwgvscLaVzBZqUhhlWFfM7I2GtsUHrIHlcDkE9ARxA=&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Level 4 Map&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/gkFhX.png&quot; alt=&quot;solution to 4&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Good luck! I tried not to obfuscate the internals too much ;)&lt;/p&gt;&#xA;" OwnerUserId="7265" LastEditorUserId="7265" LastEditDate="2014-12-12T18:14:06.203" LastActivityDate="2014-12-12T18:14:06.203" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6100" PostTypeId="1" AcceptedAnswerId="6103" CreationDate="2014-12-12T00:19:57.493" Score="4" ViewCount="667" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Alright, here's a new one for you guys. Instead of solving it, you need to determine if &lt;strong&gt;any given configuration&lt;/strong&gt; is possible to solve, without actually solving it. &lt;em&gt;It needs to be a general method of figuring it out, and should apply to most (if not all) situations.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without further ado, here's the puzzle you have to use:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a square grid of arbitrary size.  Now picture a T-shape drawn on it; the T is 3 blocks wide and 5 blocks tall as seen in the image below left. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now picture starting on the top left, and staying within the T shape/boundaries, slowly progressing, filling each square as you go. Can you fill in all the squares of the grid? &lt;strong&gt;&lt;em&gt;Of course not!&lt;/em&gt;&lt;/strong&gt; The most you can do us fill 6 squares! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However if the shape itself was a square (like the shape below right), you can fill the entire shape.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BQJNY.png&quot; alt=&quot;T and square shapes on a grid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How to move:&lt;/strong&gt; &lt;em&gt;Move to any orthogonally adjacent square which you have not already visited in the shape.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Note:&lt;/strong&gt; For all situations you may assume to start in the upper left corner. Accepted answer goes to the simplest method.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-12-12T21:47:37.387" LastActivityDate="2014-12-13T02:44:45.937" Title="Check to see if a Configuration is Possible: prove there's an Hamiltonian path on a connected subset of the square grid graph" Tags="|mathematics|geometry|solvability|computer-science|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6101" PostTypeId="5" CreationDate="2014-12-12T00:45:33.573" Score="0" Body="&lt;p&gt;Per a &lt;a href=&quot;https://puzzling.meta.stackexchange.com/questions/6617/the-end-of-open-ended-puzzles&quot;&gt;community discussion&lt;/a&gt;, truly open-ended questions, defined as those which expect multiple answers and with none that can be proven as optimal, are no longer accepted on this site.&lt;/p&gt;&#xA;&lt;p&gt;There are some open-ended questions that are still allowed. These include &lt;a href=&quot;/questions/tagged/optimization&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;optimization&amp;#39;&quot; rel=&quot;tag&quot;&gt;optimization&lt;/a&gt; puzzles where there is a provably correct answer (which the poster ideally will know ahead of time), and questions asking about a specific aspect of &lt;a href=&quot;/questions/tagged/puzzle-creation&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;puzzle-creation&amp;#39;&quot; rel=&quot;tag&quot;&gt;puzzle-creation&lt;/a&gt;. For such questions, use these tags (or other relevant tags); &lt;strong&gt;do not use this tag&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="44852" LastEditDate="2021-08-16T06:17:39.247" LastActivityDate="2021-08-16T06:17:39.247" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6102" PostTypeId="4" CreationDate="2014-12-12T00:45:33.573" Score="0" Body="OPEN-ENDED PUZZLES ARE NO LONGER IN SCOPE. Used when no answer can be guaranteed to be the last, best one. If a puzzle instead has a provably optimal solution, tag with [optimization]. " OwnerUserId="2383" LastEditorUserId="69582" LastEditDate="2020-12-15T21:27:12.600" LastActivityDate="2020-12-15T21:27:12.600" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6103" PostTypeId="2" ParentId="6100" CreationDate="2014-12-12T02:02:53.280" Score="9" Body="&lt;p&gt;What you seem to be asking (assuming the squares are completely filled), is to find an &lt;a href=&quot;http://en.wikipedia.org/wiki/Hamiltonian_path&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hamiltonian path&lt;/a&gt; on a connected subset of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lattice_graph&quot; rel=&quot;nofollow noreferrer&quot;&gt;square grid graph&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Defintions (lifted from Wikipedia)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A &lt;strong&gt;Hamiltonian path&lt;/strong&gt; (or traceable path) is a path in an undirected or directed graph that visits each vertex exactly once.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A &lt;strong&gt;square grid graph&lt;/strong&gt; is the graph whose vertices correspond to the points in the plane with integer coordinates, x-coordinates being in the range 1,..., n, y-coordinates being in the range 1,..., m, and two vertices are connected by an edge whenever the corresponding points are at distance 1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As far as I can tell, the answer is that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hamiltonian_path_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hamiltonian path problem&lt;/a&gt; is an NP-complete problem, which means that there is no polynomial time algorithm to determine if an arbitrary graph contains a Hamiltonian path.  In other words, &lt;strong&gt;given a large enough shape, it might take nearly forever just to find out if there is a solution to the stated problem.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A simple heuristic can determine if there is no solution: If a square is connected to only one other square (i.e. it has a degree of 1), it must either be the first or last square visited, since you can't backtrack. Hence, any shape with more than 2 such squares (marked with &quot;1&quot; in the image below) has no solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;However, the converse is not true&lt;/em&gt;. The shape below right has 2 squares with degree 1, but is unsolvable as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MZuRi.png&quot; alt=&quot;vertex degrees added&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another heuristic (courtesy of Lopsy): color the squares like a checkerboard. The number of white and black squares can only differ by 1 if it is solvable. If e.g. there is one more white square than black, then the path must start and end on white.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case the T shape satisfies the heuristic while the shape on the right does not.  Both are not solvable which again shows that the converse is false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LqyRK.png&quot; alt=&quot;checkerboard&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You can also simplify shapes by removing a vertex of degree 1 if and only if it is connected to a vertex of degree 2.  For example, the following shapes are equivalent in terms of their solvability:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1fhyb.png&quot; alt=&quot;equivalent T shapes&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Lets put what we have learned to use.  Here is a complex shape which satisfies both heuristics (only 1 vertex of degree 1, and there are 7 black squares vs 6 light squares).  But we can remove the vertex with degree 1 because it is connected to a vertex of degree 2, and now the new shape does not satisfy the checkerboard heuristic anymore.  So, the shape has no hamiltonian path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0wF0S.png&quot; alt=&quot;Complex shape&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is really a question more for math or computer science: &lt;br&gt;&#xA;&lt;a href=&quot;https://cstheory.stackexchange.com/questions/tagged/hamiltonian-paths&quot;&gt;https://cstheory.stackexchange.com/questions/tagged/hamiltonian-paths&lt;/a&gt; &lt;br&gt;&#xA;&lt;a href=&quot;https://mathoverflow.net/questions/tagged/hamiltonian-paths&quot;&gt;https://mathoverflow.net/questions/tagged/hamiltonian-paths&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may also be of interest (algorithms for finding Hamiltonian paths for L, C, F, and E shaped graphs on a grid): &lt;a href=&quot;http://www.hindawi.com/journals/jam/2012/475087/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.hindawi.com/journals/jam/2012/475087/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6178" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:31.357" LastActivityDate="2014-12-13T02:44:45.937" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="6104" PostTypeId="2" ParentId="4818" CreationDate="2014-12-12T04:01:40.150" Score="5" Body="&lt;p&gt;If you have a dry erase marker, number the centers 1-4 on each face. Then apply either your solution or your scramble, and watch what happens. The independent consideration of which center of the same color goes into which of the 4 available has something to do with it, as you alluded to. It's also why I wrote a 4x4x4 solving program to help me work things out I couldn't handle myself :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gaNwR.jpg&quot; alt=&quot;Solving a 22-move snake pattern in under 1 minute&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7269" LastActivityDate="2014-12-12T04:01:40.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6106" PostTypeId="1" AcceptedAnswerId="6107" CreationDate="2014-12-12T08:11:36.557" Score="12" ViewCount="882" Body="&lt;p&gt;I thank my lucky stars I'm able to write this to you. I spent a hefty percentage of my remaining cash at the courier's office and I want to underscore the importance of this communique. If a man comes at you exclaiming that he's found my magic hat, do &lt;em&gt;not&lt;/em&gt; hash out our plan in response to his inquiries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need you to come here, plus I need the contents of our safe. The combination is a five-digit number encoded in this letter. It can be deduced by poring over the numbers with simple arithmetic. Please come quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sincerely,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fatima&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Knowing that all five digits of the safe combination are unique, can you deduce the combination?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2014-12-12T11:24:35.287" Title="Poring Over the Numbers: A Hidden Message" Tags="|password|encoded-message|calculation-puzzle|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6107" PostTypeId="2" ParentId="6106" CreationDate="2014-12-12T08:42:12.263" Score="11" Body="&lt;p&gt;The words in the message &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; stars, percentage, cash, at, and, underscore, at, exclaiming, hat, hash, plus&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;correspond to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; characters &lt;code&gt;*%$@&amp;amp;_@!^#+&lt;/code&gt; on the top row of an &lt;a href=&quot;http://www.goodtyping.com/teclatUSok.png&quot;&gt;American keyboard&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;capitalized&quot; versions of the symbols &lt;code&gt;85427-2163=&lt;/code&gt; on the corresponding keys&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which gives the answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 83264 when evaluated arithmetically. This has 5 distinct digits, as confirmation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks to @No. 7892142 for noticing symbols I'd missed that fix my answer.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Old answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The words in the message&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; stars, percentage, cash, exclaiming, hash&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;correspond to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; symbols *%$!# on the top row of an American keyboard&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which further correspond to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the digits 85413 on those keys.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2014-12-12T11:24:35.287" LastActivityDate="2014-12-12T11:24:35.287" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6109" PostTypeId="1" CreationDate="2014-12-12T12:06:59.140" Score="-3" ViewCount="458" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Hi friends. I am &lt;strong&gt;252116211&lt;/strong&gt;.  You may think I am &lt;strong&gt;252116211&lt;/strong&gt; or&#xA;  &lt;strong&gt;252116211&lt;/strong&gt;. But I am not. Also, I am not &lt;strong&gt;252116211&lt;/strong&gt;, &lt;strong&gt;252116211&lt;/strong&gt;, &lt;strong&gt;252116211&lt;/strong&gt;.... or some other &lt;strong&gt;252116211&lt;/strong&gt;. There is a lot that's not what I am. I am lazy enough to list them all.  But one thing I can guarantee you,&#xA;  &lt;em&gt;I am definitely &lt;strong&gt;252116211&lt;/em&gt;&lt;/strong&gt;. If you know the old &lt;strong&gt;bda&lt;/strong&gt;, who expanded his three member family to ten by inviting seven new members, with his help,&#xA;  you can find me.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tell me who I am and how did you find me (it will be appreciated if you can explain everything mentioned in the question).&lt;/p&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="7308" LastEditDate="2015-04-30T23:34:43.390" LastActivityDate="2015-04-30T23:34:43.390" Title="Who am I? Try finding it out from the hints in the form of a 9 digit number" Tags="|cipher|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6110" PostTypeId="2" ParentId="6066" CreationDate="2014-12-12T12:16:36.817" Score="10" Body="&lt;p&gt;For the people who want to crack it by using code (and to get my bonus points):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The OP used Tea crypt with the password &quot;vaj123&quot;:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;s2 = Tea.decrypt(inputS, &quot;vaj123&quot;);&lt;/code&gt; decrypts the string (in inputS) into a map of values which it saves in s2. The first value has to be discarded, the rest of the values are used to make up a 20 by 20 grid.&#xA; -1 is nothing placed, and 0-7 are the diffirent mirrors used.&lt;br&gt;&#xA; You can execute the above command in the browser's console on the Black Box page.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And if you want the reverse-engineerd stuff:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Full code of the generation of the puzzle can be found here: &lt;a href=&quot;http://jsbin.com/civuvetezi/1/watch?js,output&quot;&gt;http://jsbin.com/civuvetezi/1/watch?js,output&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes I know, I am a party pooper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck!&lt;/p&gt;&#xA;" OwnerUserId="6083" LastEditorUserId="6083" LastEditDate="2014-12-12T12:22:47.347" LastActivityDate="2014-12-12T12:22:47.347" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="6111" PostTypeId="2" ParentId="6109" CreationDate="2014-12-12T13:47:33.467" Score="1" Body="&lt;p&gt;Probably not right but...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;112911252&lt;/strong&gt;, which is 252116211 rotated 180 degrees when typed in a calculator.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;bda's family is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;abdenmpquw&lt;/strong&gt;, which are letters that look like other letters when rotated 180 degrees.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6178" LastEditorUserId="6178" LastEditDate="2014-12-12T13:59:55.717" LastActivityDate="2014-12-12T13:59:55.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6112" PostTypeId="2" ParentId="6109" CreationDate="2014-12-12T14:13:46.037" Score="1" Body="&lt;p&gt;I would put this in a comment to the question, but that doesn't have spoiler tags. &#xA;So my lead would be that it has to do..&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...with the &lt;a href=&quot;http://msdn.microsoft.com/en-us/library/windows/desktop/dd693015(v=vs.85).aspx&quot; rel=&quot;nofollow&quot;&gt;Broadcast Driver Architecture&lt;/a&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;, but I haven't figured out who you are yet. I believe this because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; IPs are about numbers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;the old&quot; may refer to the 'outdated' technology as mentioned in the linked article.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However. I do not know how to go further from here. Maybe some other geek can?&lt;/p&gt;&#xA;" OwnerUserId="5840" LastActivityDate="2014-12-12T14:13:46.037" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6113" PostTypeId="1" AcceptedAnswerId="6117" CreationDate="2014-12-12T16:34:19.247" Score="21" ViewCount="1908" Body="&lt;p&gt;A simple riddle of mine, focused towards meter and rhyme.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm but a speck of fairy dust&lt;br&gt;&#xA;  A glitt'ring jew'l, but only just&lt;br&gt;&#xA;  So small I scarcely can be seen&lt;br&gt;&#xA;  A trinity in me convenes&lt;br&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;And yet without me you can't see&lt;br&gt;&#xA;  Fine art and words are made of me&lt;br&gt;&#xA;  I'm resolute; through me you can&lt;br&gt;&#xA;  View realms beyond the mortal ken&lt;br&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I shrink along with passing years&lt;br&gt;&#xA;  My number grows, I gain more peers&lt;br&gt;&#xA;  But treat me well 'tis very true&lt;br&gt;&#xA;  I sleep and live and die like you&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Please include explanations for as many lines as possible in your answer.&lt;/p&gt;&#xA;" OwnerUserId="6178" LastEditorUserId="6178" LastEditDate="2015-08-13T08:31:52.733" LastActivityDate="2015-08-13T08:31:52.733" Title="I'm but a speck of fairy dust" Tags="|riddle|rhyme|" AnswerCount="4" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6114" PostTypeId="2" ParentId="6113" CreationDate="2014-12-12T17:00:57.820" Score="12" Body="&lt;p&gt;You are a(n):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Atom&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm but a speck of fairy dust&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Atoms are star-dust. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A glitt'ring jew'l, but only just&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Atoms are beautiful, why not.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So small I scarcely can be seen&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Atoms are tiny.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A trinity in me convenes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Protons, electrons and neutrons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And yet without me you can't see&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Atoms are all there is to comprise the so-called physical world we operate in.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fine art and words are made of me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everything is comprised of atoms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm resolute; through me you can&#xA;View realms beyond the mortal ken&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A reference to quantum mechanics?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I shrink along with passing years&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Atomic decay&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My number grows, I gain more peers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Atoms are born with death of stars and expansion of the universe. Could also refer to atomic numbers increasing?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But treat me well 'tis very true&#xA;I sleep and live and die like you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Atoms actually die.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2014-12-12T17:37:20.567" LastActivityDate="2014-12-12T17:37:20.567" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="6117" PostTypeId="2" ParentId="6113" CreationDate="2014-12-12T18:23:56.833" Score="24" Body="&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A pixel&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm but a speck of fairy dust / A glitt'ring jew'l, but only just&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pixels are small and bright&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So small I scarcely can be seen / A trinity in me convenes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RGB&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And yet without me you can't see&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can't see content without them. Assuming specific to the context of its realm.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fine art and words are made of me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, assuming the context of its realm.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm resolute; through me you can / View realms beyond the mortal ken&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pretty obvious, via the internet, games, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I shrink along with passing years / My number grows, I gain more peers&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DPI increases as computers get more advanced. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But treat me well 'tis very true / I sleep and live and die like you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Sleep&quot; possibly referring to monitor state, and pixels can die.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7282" LastActivityDate="2014-12-12T18:23:56.833" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6118" PostTypeId="2" ParentId="6113" CreationDate="2014-12-12T19:30:53.183" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A water molecule.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm but a speck of fairy dust&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water molecules are specks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A glitt'ring jew'l, but only just&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water and ice both glitter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So small I scarcely can be seen&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is hard to see water molecules.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A trinity in me convenes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Two hydrogens and one oxygen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And yet without me you can't see&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because there would be no humours in your eye, and life can't exist without water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fine art and words are made of me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; watercolors, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm resolute; through me you can&#xA;View realms beyond the mortal ken&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Detecting water in remote galaxies, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I shrink along with passing years&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because of the expansion of the universe, you could think of everything else shrinking by comparison.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My number grows, I gain more peers&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's raining and / or more water is made in nebulae in outer space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But treat me well 'tis very true&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; because water is good for you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I sleep and live and die like you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; because you're 90% water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7284" LastActivityDate="2014-12-12T19:30:53.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6119" PostTypeId="2" ParentId="6113" CreationDate="2014-12-12T21:29:02.620" Score="1" Body="&lt;p&gt;There have been other answers along these lines (and frankly I like CAVX's better), but the first thing that came to mind for me is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A carbon atom&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm but a speck of fairy dust&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Like graphite&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A glitt'ring jew'l, but only just&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Like diamond&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So small I scarcely can be seen&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; because atom&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A trinity in me convenes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; electron, proton, neutron&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And yet without me you can't see&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your retinas are made mostly of carbon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fine art and words are made of me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Graphite, charcoal, most paints, all organic building materials...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm resolute; through me you can&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Diamond and graphene are transparent&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;View realms beyond the mortal ken&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; and diamonds are the result of ancient geological processes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I shrink along with passing years&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; nuclear weak force&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My number grows, I gain more peers&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; stellar fusion&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But treat me well 'tis very true&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Diamonds are incredibly hard, but also brittle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I sleep and live and die like you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We are made primarily of carbon, and carbon has a (very long) half-life too&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5895" LastActivityDate="2014-12-12T21:29:02.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6120" PostTypeId="1" AcceptedAnswerId="6202" CreationDate="2014-12-13T03:00:52.083" Score="28" ViewCount="3309" Body="&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/users/39/doorknob-%E5%86%B0&quot; title=&quot;Doorknob冰&quot;&gt;Doorknob&lt;/a&gt; knows exactly how to solve this, so I ask him to not answer this question.&lt;/p&gt;&#xA;&lt;h1&gt;Well, here's the puzzle:&lt;/h1&gt;&#xA;&lt;p&gt;No, an answer is the cube without A, No?&lt;br /&gt;&#xA;The Yes is No.&lt;br /&gt;&#xA;The squared answer, Yes, is to an answer.&lt;br /&gt;&#xA;A Yes without the Yes is to the answer.&lt;br /&gt;&#xA;Yes, the answer is No.&lt;br /&gt;&#xA;Yes, the answer to a no is 432.&lt;br /&gt;&#xA;What is the answer?&lt;/p&gt;&#xA;&lt;h1&gt;And That's it ... Now for some hints&lt;/h1&gt;&#xA;&lt;p&gt;I tried my best to make it as grammatically correct as I could, but nope, too hard.&lt;br /&gt;&#xA;I'm not gonna put any tags on because the key thing to this puzzle is that you need to figure out what type of puzzle it is. Maybe after it gets answered.&lt;br /&gt;&#xA;Just that the 4th sentence says the answer is No does not mean that the answer is no.&lt;/p&gt;&#xA;&lt;h1&gt;A few more things&lt;/h1&gt;&#xA;&lt;p&gt;I'm being serious here; it works perfectly. If you get it, it's gonna be a EUREKA!!! moment because it works perfectly.&lt;br /&gt;&#xA;I have added a tag because it is obvious that rand al'thor is on the right track.&#xA;Probably post the answer in 3 weeks (if not flagged for unclear puzzle).&lt;br /&gt;&#xA;Someone managed to solve an earlier version of this (same concept) and a bunch of people here are a lot better at solving than him so... I don't think its terribly hard...&lt;br /&gt;&#xA;If you don't get the answer I'm looking for, but manage to make a reasonable answer, I will give you an upvote, but not an answer mark.&lt;br /&gt;&#xA;This puzzle was inspired by a conversation with Doorknob and someone else (not on this site).&lt;/p&gt;&#xA;" OwnerUserId="167" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-03-06T00:55:48.493" Title="What, the answer isn't the No of the Yes is?" Tags="|mathematics|calculation-puzzle|lateral-thinking|" AnswerCount="3" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="6121" PostTypeId="2" ParentId="6065" CreationDate="2014-12-13T09:53:51.273" Score="5" Body="&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; My interpretation of the passage is incomplete as I'm not all too familiar with the topic. Any pointers would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Answer&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Divorce&lt;/strong&gt;, the end of marriage&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This is no enigma&lt;br&gt;&#xA;Just a shameful stigma&lt;br&gt;&#xA;We have split apart for good&lt;br&gt;&#xA;A malevolent stance taken as it should&lt;br&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The former spouses have separated. It's pretty shameful, and it's typically the result of hatred towards each other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Our contract now null and void&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Marriage is a legally-binding contract in many jurisdictions; divorce terminates that contract.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Desiring to inhabit a separate planetoid&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The former spouses live separately.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;For what we may have created&lt;br&gt;&#xA;now in despair, not appreciated&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Children are inevitably upset as a result of a divorce.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;for we'll pay for our sacrilege&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In many religions, marriage is considered sacred, and divorce sacrilegious.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Once we dine with our heritage&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Inheritance problems can come up as a result of divorce. It is impossible to stay out of hell, your heritage = what has been passed down to you = DNA. Your past relations in hell.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Take our past in unrelated directions&lt;br&gt;&#xA;no longer our paths have intersections&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The former spouses separate and never want to meet up again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This heartfelt decision&lt;br&gt;&#xA;Has caused this incision&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;In many legal systems, divorce is a decision made by a judge in court.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;although different, still a toad  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the reference to the frog turning into a prince happily ever after in marriage. This now divorced, your still a frog  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So now we divide our surrounding&lt;br&gt;&#xA;separate haemorrhoids not corresponding  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;haemorrhoids = piles, the metaphoric description of dividing the assets of the marriage and splitting them, which are never correspond. Ones pile is always worth more.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once classed as my property&lt;br&gt;&#xA;Now free of the depravity  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It was once common knowledge that the wife was property of the man (still is in some communities).  Now free of being an object through divorce.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7288" LastEditorUserId="5135" LastEditDate="2014-12-13T12:25:31.457" LastActivityDate="2014-12-13T12:25:31.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6122" PostTypeId="1" AcceptedAnswerId="6173" CreationDate="2014-12-13T09:58:14.513" Score="11" ViewCount="2236" Body="&lt;p&gt;Who or what is this poem referring to?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Within me you will find&lt;br&gt;&#xA;  One of like heart and mind.&lt;br&gt;&#xA;  As well as an entity&lt;br&gt;&#xA;  With inverted geography.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;While I represent symmetry&lt;br&gt;&#xA;  Mirrors can show another identity.&lt;br&gt;&#xA;  But there will always be twins&lt;br&gt;&#xA;  Who generate many grins.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hints:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first 4 lines and last 4 lines should be interpreted differently.  Also, avigrail's &lt;a href=&quot;https://puzzling.stackexchange.com/questions/6122/riddle-poem-within-me-you-will-find#comment19706_6122&quot;&gt;comment&lt;/a&gt; about boobs is somewhat related to the answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2016-07-14T10:47:26.230" Title="Riddle Poem: &quot;Within me you will find...&quot;" Tags="|riddle|rhyme|" AnswerCount="10" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6123" PostTypeId="2" ParentId="6122" CreationDate="2014-12-13T10:10:14.673" Score="8" Body="&lt;h2&gt;My answer&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Lungs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Within me you will find&lt;br&gt;&#xA;One of like heart and mind.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; representations of being within something that is like the heart and mind (brain) such as an organ&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As well as an entity&lt;br&gt;&#xA;With inverted geography.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; many bacterial entities reside to create these. Inverted geography as the region it has when rested is inverted when its at full capacity. (vice-versa)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While I represent symmetry&lt;br&gt;&#xA;Mirrors can show another identity.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; two lungs working in sync are symmetrical, mirrors as in reflections, reflect the user so smokers or people with ailments will show a different set of lungs. Also different DNA reflect different sizes and what not.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But there will always be twins&lt;br&gt;&#xA;Who generate many grins.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; two lungs working in sync to generate oxygen to form lots of malicious/happy grins&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5135" LastActivityDate="2014-12-13T10:10:14.673" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6124" PostTypeId="2" ParentId="6122" CreationDate="2014-12-13T10:36:51.923" Score="6" Body="&lt;p&gt;Within &lt;strong&gt;me&lt;/strong&gt; you will find&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;One of&lt;/em&gt; like &quot;h&lt;strong&gt;e&lt;/strong&gt;art&quot; and &quot;&lt;strong&gt;m&lt;/strong&gt;ind&quot;.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As well as an entity&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With inverted geography: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; invert(m) = w&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While I represent symmetry&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mirrors&lt;/strong&gt; can show another identity.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there will always be twins - twins: two letters&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who generate many grins.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; when we mirror &quot;me&quot; correctly there will be &quot;aw&quot; which is the sound of grinning people I suppose :D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5106" LastActivityDate="2014-12-13T10:36:51.923" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6125" PostTypeId="2" ParentId="6122" CreationDate="2014-12-13T10:45:06.290" Score="5" Body="&lt;p&gt;Is this a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Joker card?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Within me you will find&lt;br&gt;&#xA;One of like heart and mind.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A Joker card contains a depiction of a joker.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As well as an entity&lt;br&gt;&#xA;With inverted geography.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The joker card usually contains the word &quot;joker&quot; twice, one is inverted.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While I represent symmetry&lt;br&gt;&#xA;Mirrors can show another identity.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The card maintains rotational symmetry with respect to the word &quot;joker&quot;, but does not remain the same under mirroring&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But there will always be twins&lt;br&gt;&#xA;Who generate many grins.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There are two joker cards in a deck, and jokers generate many grins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6178" LastActivityDate="2014-12-13T10:45:06.290" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6126" PostTypeId="5" CreationDate="2014-12-13T12:52:35.450" Score="0" Body="&lt;p&gt;A puzzle tagged with community-generated should be 1 original puzzle, in which continues on through it's solvers' puzzles. Each puzzle answer should include an answer to the previous puzzle, and a new puzzle for the next solver.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Community-generated tag means the quality and quantity of the puzzle is determined by how much the community contributes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As more and more puzzles come in, the question can't contain them all, so sorting by date is probably a good idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the puzzles uses some sort of &quot;crack&quot; that allows you to solve any puzzle, it's recommended you accept it, since it essentially solved all given puzzles the solvers come up with.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-12-13T13:40:56.473" LastActivityDate="2014-12-13T13:40:56.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6127" PostTypeId="4" CreationDate="2014-12-13T12:52:35.450" Score="0" Body="This tag is used when answer contains an answer, as well as a new puzzle, to be solved by the next answerer. All the puzzles should revolve around the same concept or idea. New puzzles can (and probably should) be edited into the original question." OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-12-13T13:41:21.000" LastActivityDate="2014-12-13T13:41:21.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6129" PostTypeId="1" AcceptedAnswerId="6130" CreationDate="2014-12-13T14:51:47.543" Score="-3" ViewCount="8535" Body="&lt;p&gt;Here's a fairly simple riddle:&#xA;When is 90 greater than 100?&#xA;(And it's &lt;em&gt;not&lt;/em&gt; when you're on the negative side of the number line. Although it is true, it is not the answer)&lt;/p&gt;&#xA;" OwnerUserId="2524" LastEditorUserId="4050" LastEditDate="2014-12-13T16:29:47.787" LastActivityDate="2014-12-13T17:11:34.707" Title="A fairly simple riddle: When is 90 greater than 100?" Tags="|riddle|" AnswerCount="3" CommentCount="2" ClosedDate="2014-12-13T18:47:01.320" ContentLicense="CC BY-SA 3.0" />
  <row Id="6130" PostTypeId="2" ParentId="6129" CreationDate="2014-12-13T14:59:00.510" Score="12" Body="&lt;p&gt;Some joke answers here and some more serious ones too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Rotate 180 degrees, 06 is larger than 001&lt;br&gt;&#xA;&lt;br&gt;&#xA;In roman numerals, XC has more letters than C&lt;br&gt;&#xA;&lt;br&gt;&#xA;The sum of digits is larger for 90.&lt;br&gt;&#xA;&lt;br&gt;&#xA;90 (in decimal) is greater than 100 (in binary).&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;strong&gt;The most common answer:&lt;/strong&gt; Enter 90 in a microwave, is 90 seconds while entering 100, is 1 minute or 60 seconds&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;strong&gt;Heres an original answer&lt;/strong&gt;: In an isoceles triangle, the largest area is when one of the angles is a right angle.  So an isoceles triangle with 100 degree angle is lesser than one with a 90 degree angle.&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;strong&gt;Heres an original answer, part 2&lt;/strong&gt;: The colour #000090 (dark blue) is brighter (greater in luminosity) than #000100 (very dark green)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6178" LastEditorUserId="6178" LastEditDate="2014-12-13T15:10:56.977" LastActivityDate="2014-12-13T15:10:56.977" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6131" PostTypeId="2" ParentId="6129" CreationDate="2014-12-13T15:39:41.420" Score="3" Body="&lt;p&gt;More joke/serious answers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;90 isn't larger than 100, but it is &lt;em&gt;greater&lt;/em&gt; than 100. Just look at it. Those smooth prime factors. So great.&lt;br&gt;&#xA;&lt;br&gt;&#xA;In base 8, '90' isn't well-defined, but morally it's larger than 100.&lt;br&gt;&#xA;&lt;br&gt;&#xA;In Ruby: &lt;code&gt;class Fixnu­m; def &amp;gt;=(other) true end end; print 90 &amp;gt;= 100 # prints &quot;true&quot;&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2014-12-13T15:39:41.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6132" PostTypeId="2" ParentId="6129" CreationDate="2014-12-13T17:11:34.707" Score="10" Body="&lt;p&gt;From a programmer's perspective:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In lexicographical order.&#xA;&#xA; In Python: &lt;code&gt;90 &amp;gt; 100&lt;/code&gt; → &lt;code&gt;False&lt;/code&gt;.&#xA; But compare as strings, and you get &lt;code&gt;&quot;90&quot; &amp;gt; &quot;100&quot;&lt;/code&gt; → &lt;code&gt;True&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In lexicographical order, like in a dictionary, you compare characters sequentially.&#xA; So to check whether &lt;code&gt;&quot;def&quot; &amp;gt; &quot;abc&quot;&lt;/code&gt;, you first check whether &lt;code&gt;&quot;d&quot; &amp;gt; &quot;a&quot;&lt;/code&gt; (which it is, because it's later in the alphabet).&#xA; Because it is, you don't have to check the rest of the string.&#xA;&#xA; The same holds when you treat numbers as characters. To check &lt;code&gt;&quot;90&quot; &amp;gt; &quot;100&quot;&lt;/code&gt;, first check whether &lt;code&gt;&quot;9&quot; &amp;gt; &quot;0&quot;&lt;/code&gt; (which it is, because it's a larger number/later in ASCII).&#xA; So you immediately conclude &lt;code&gt;True.&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="397" LastActivityDate="2014-12-13T17:11:34.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6133" PostTypeId="1" CreationDate="2014-12-13T17:23:01.970" Score="8" ViewCount="798" Body="&lt;p&gt;This is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Cryptogram&quot;&gt;cryptogram&lt;/a&gt; puzzle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You get a riddle-style clue and some encoded text ('ciphertext'). Your task is to find the plaintext (the original, un-encoded, text).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The clue is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Conceptualised by Mitsubishi,&lt;br&gt;&#xA;  I Blaise  in a cryptic deepscan of the great strike-slip fault,  &lt;br&gt;&#xA;  For Agnes, but not for Edward Mountain:&lt;br&gt;&#xA;  &quot;Other years in bmx sports - E.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the ciphertext is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Glw Dwgu Rwka Qbrklmv'f Wgfo &lt;br&gt;&#xA;  fl Ivoqr Zsjyir &lt;br&gt;&#xA;  svge &quot;Nvbq Ydiwtso lw Wnxmjv&quot; (Gnvusvig, 1973) &lt;br&gt;&#xA;   &lt;br&gt;&#xA;  Wkkvraazmnjsjdd? &lt;br&gt;&#xA;  Prjlmxnp ulffejy lfxt lsp? &lt;br&gt;&#xA;  Yvzsoptgjpustfofy ktd op t k y y opokd. &lt;br&gt;&#xA;  Vzyopzsnpnfdzijyytzijtetzijylsxt jy jd – &lt;br&gt;&#xA;  yu keeooeaj kjx ovnax sejtq yjia tq. &lt;br&gt;&#xA;  Zgdscpgvwo – qshdwhbzgc – xpbzgvwobx-vgxpbhgcwwu? &lt;br&gt;&#xA;  Whdovna xgs jbo khtkeexzixplysjvykstzp ssc kxpssc! &lt;br&gt;&#xA;  Rovn ojs oon jgc! &lt;br&gt;&#xA;  Ovbj yjiasj yspj? &lt;br&gt;&#xA;  Tsett qop td – &lt;br&gt;&#xA;  jpz tde, &lt;br&gt;&#xA;  jpz tde, &lt;br&gt;&#xA;  jpz tde, &lt;br&gt;&#xA;  jpc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first word of the plaintext is 'The'. The plaintext contains numerous non-dictionary words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cipher can be decoded using one of the tools on this page: &lt;a href=&quot;http://rumkin.com/tools/cipher/&quot;&gt;Rumkin.com Cipher Tools&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="4050" LastEditDate="2014-12-13T17:54:10.930" LastActivityDate="2020-09-01T14:42:27.247" Title="Riddle + cryptogram: Wkkvraazmnjsjdd? Prjlmxnp ulffejy lfxt lsp? Rovn ojs oon jgc!" Tags="|riddle|cipher|" AnswerCount="3" CommentCount="7" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6134" PostTypeId="2" ParentId="6133" CreationDate="2014-12-13T18:04:32.343" Score="5" Body="&lt;p&gt;Incomplete answer (will figure out the riddle later), I attacked the cryptogram directly on the assumption that a certain cipher was used, and I was correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cipher used is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Vigenere cipher with key &quot;mvhhrv&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Plaintext:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Loch Ness Monster's Song&lt;br&gt;&#xA;by Edwin Morgan&lt;br&gt;&#xA;from &quot;From Glasgow to Saturn&quot; (Carcanet, 1973)&lt;br&gt;&#xA;&lt;br&gt;&#xA;Sssnnnwhuffffll?&lt;br&gt;&#xA;Hnwhuffl hhnnwfl hnfl hfl?&lt;br&gt;&#xA;Gdroblboblhobngbl gbl gl g g g g glbgl.&lt;br&gt;&#xA;Drublhaflablhaflubhafgabhaflhafl fl fl –&lt;br&gt;&#xA;gm grawwwww grf grawf awfgm graw gm.&lt;br&gt;&#xA;Hovoplodok – doplodovok – plovodokot-doplodokosh?&lt;br&gt;&#xA;Splgraw fok fok splgrafhatchgabrlgabrl fok splfok!&lt;br&gt;&#xA;Zgra kra gka fok!&lt;br&gt;&#xA;Grof grawff gahf?&lt;br&gt;&#xA;Gombl mbl bl –&lt;br&gt;&#xA;blm plm,&lt;br&gt;&#xA;blm plm,&lt;br&gt;&#xA;blm plm,&lt;br&gt;&#xA;blp.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6178" LastActivityDate="2014-12-13T18:04:32.343" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6135" PostTypeId="1" AcceptedAnswerId="6136" CreationDate="2014-12-13T19:19:16.837" Score="17" ViewCount="1214" Body="&lt;p&gt;An &lt;a href=&quot;http://puzzling.stackexchange.com&quot;&gt;online community&lt;/a&gt; has three moderators: &lt;a href=&quot;https://puzzling.stackexchange.com/users/39/doorknob&quot;&gt;Doorknob&lt;/a&gt;, &lt;a href=&quot;https://puzzling.stackexchange.com/users/4/kevin&quot;&gt;Kevin&lt;/a&gt;, and &lt;a href=&quot;https://puzzling.stackexchange.com/users/20/emrakul&quot;&gt;Emrakul&lt;/a&gt; (in no particular order). In between making posts on &lt;a href=&quot;http://meta.puzzling.stackexchange.com&quot;&gt;meta&lt;/a&gt;, they decide to set up a code for their private communications, and start sending secret messages to one another. First Doorknob sends Emrakul this message:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2211+1211=111+21+11+2111+1+1112+221+121=1221+2111+2122+2111+1=1112+1121=1211+121+2122+2122+2111+1222$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then Emrakul sends this one to Kevin:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1211+121+1121+221+121+1+2212+21+1211=1112=1121+21+1222=21=2212+2111+2=21+12+2212=21=1221+21+221$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally Kevin sends this to Doorknob, closing the circle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1112=2122+1112+2112+121=21+2121+2121+2122+121+1121=21+12+2212=2111+1+21+12+2+121+1121$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you decrypt the messages?&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2014-12-13T19:41:07.827" Title="Moderators' communications leaked!" Tags="|cipher|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6136" PostTypeId="2" ParentId="6135" CreationDate="2014-12-13T19:41:07.827" Score="18" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quite simply:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The ones and twos correspond to Morse code dots and dashes. The plus signs are letter separators, and the equals signs are word breaks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That gives you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. ZL SNIBEVGR PBYBE VF LRYYBJ&lt;br /&gt;&#xA; 2. LRFGREQNL V FNJ N QBT NAQ N PNG&lt;br /&gt;&#xA; 3. V YVXR NCCYRF NAQ BENATRF&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then all you need to do is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A 13-place Caesar shift (rot13)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To obtain:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. MY FAVORITE COLOR IS YELLOW&lt;br /&gt;&#xA; 2. YESTERDAY I SAW A DOG AND A CAT&lt;br /&gt;&#xA; 3. I LIKE APPLES AND ORANGES&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="40" LastActivityDate="2014-12-13T19:41:07.827" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6137" PostTypeId="2" ParentId="6133" CreationDate="2014-12-13T19:45:45.897" Score="9" Body="&lt;p&gt;Working from &lt;a href=&quot;https://puzzling.stackexchange.com/a/6134/5373&quot;&gt;McMagister's answer&lt;/a&gt; ...&lt;/p&gt;&#xA;&lt;p&gt;The answer to the riddle is&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Nessie&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Conceptualised by Mitsubishi,&lt;/em&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Mitsubishi_Nessie&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Mitsubishi_Nessie&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;I Blaise in a cryptic deepscan of the great strike-slip fault,&lt;/em&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &amp;quot;Blaise&amp;quot; could refer to Blaise Pascal, after whom the SI unit of pressure is named. Pressure is high at the bottom of Loch Ness! And Nessie is certainly a &amp;quot;cryptic&amp;quot; beast, part of the field of &lt;a href=&quot;http://en.wikipedia.org/wiki/Cryptozoology&quot; rel=&quot;nofollow noreferrer&quot;&gt;cryptozoology&lt;/a&gt;. &lt;a href=&quot;http://www.nessie.co.uk/htm/searching_for_nessie/deepscan.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Operation &amp;quot;Deepscan&amp;quot;&lt;/a&gt; was a search for Nessie in 1987. Loch Ness was formed by a &lt;a href=&quot;http://en.wikipedia.org/wiki/Great_Glen_Fault&quot; rel=&quot;nofollow noreferrer&quot;&gt;&amp;quot;strike-slip fault&amp;quot;&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;For Agnes, but not for Edward Mountain:&lt;/em&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Nessie is a familiar form of the name Agnes. Edward Mountain financed a &lt;a href=&quot;http://en.wikipedia.org/wiki/Loch_Ness_Monster#Sir_Edward_Mountain_Expedition_.281934.29&quot; rel=&quot;nofollow noreferrer&quot;&gt;search&lt;/a&gt; for Nessie in 1934.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;&amp;quot;Other years in bmx sports - E.&amp;quot;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; This is an anagram of &lt;em&gt;Nessiteras rhombopteryx&lt;/em&gt;, the Latin name proposed for Nessie's species by Peter Scott (itself an anagram of &amp;quot;Monster hoax by Sir Peter S&amp;quot;). See &lt;a href=&quot;https://puzzling.stackexchange.com/a/40576/5373&quot;&gt;Rosie F's answer&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The answer to the riddle should then be used as&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; the keyword for a Vigenere cipher,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;giving the following plaintext:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;The Loch Ness Monster's Song &lt;br&gt; by Edwin Morgan &lt;br&gt; from &amp;quot;From Glasgow to Saturn&amp;quot; (Carcanet, 1973) &lt;br&gt; &lt;br&gt; Sssnnnwhuffffll? &lt;br&gt; Hnwhuffl hhnnwfl hnfl hfl? &lt;br&gt; Gdroblboblhobngbl gbl gl g g g g glbgl. &lt;br&gt; Drublhaflablhaflubhafgabhaflhafl fl fl – &lt;br&gt; gm grawwwww grf grawf awfgm graw gm. &lt;br&gt; Hovoplodok – doplodovok – plovodokot-doplodokosh? &lt;br&gt; Splgraw fok fok splgrafhatchgabrlgabrl fok splfok! &lt;br&gt; Zgra kra gka fok! &lt;br&gt; Grof grawff gahf? &lt;br&gt; Gombl mbl bl – &lt;br&gt; blm plm, &lt;br&gt; blm plm, &lt;br&gt; blm plm, &lt;br&gt; blp.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Yes, this really is plaintext, not more ciphertext!&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2020-09-01T14:42:27.247" LastActivityDate="2020-09-01T14:42:27.247" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="6138" PostTypeId="1" AcceptedAnswerId="6220" CreationDate="2014-12-14T02:52:05.677" Score="156" ViewCount="11448" Body="&lt;p&gt;Complete the last line in each of the following five series:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Puzzle 1:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&quot;Using&quot; is E&#xA;&quot;Desist&quot; is I&#xA;&quot;Visits&quot; is E&#xA;&quot;Design&quot; is ??? &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Puzzle 2:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&quot;Asia&quot; is H&#xA;&quot;Cuffs&quot; is S&#xA;&quot;Ender&quot; is I&#xA;&quot;Redness&quot; is F&#xA;&quot;Fuchsia&quot; is ???&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Puzzle 3 (advanced):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&quot;Pro-tem&quot; and &quot;row&quot; are H&#xA;&quot;Ties&quot; is K&#xA;&quot;Cohesion&quot; is A&#xA;&quot;Casino&quot; is E&#xA;&quot;Hocks&quot; is either a word, name, ??? or ???&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Puzzle 4:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&quot;Gross&quot; is W or R&#xA;&quot;Assign&quot; is S&#xA;&quot;Amassing&quot; is E&#xA;&quot;Design&quot; is S&#xA;&quot;Amassing&quot; is S&#xA;&quot;Arrows&quot; is ??? or ???&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Puzzle 5:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&quot;Edna&quot; is I&#xA;&quot;Slang&quot; is X&#xA;&quot;Siege&quot; is E&#xA;&quot;Basin&quot; is X&#xA;&quot;Sin&quot; is X&#xA;&quot;Sin&quot; is A&#xA;&quot;Besiege&quot; is X&#xA;&quot;Signals&quot; is ??? and ???&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;First person to solve all five gets the win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each ??? stands for a single letter of the English alphabet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The puzzles are not related to one another, except that they are all based on the same principle.  If you solve any one, you will rapidly solve all the others.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #3, a quite cryptic one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The very first parts of vague solutions can decode nothing: that's the core of these puzzles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #4, a more obvious one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;You want another hint?!&lt;/strong&gt; Nag! Go botanist first, arse ulcer! O my! More clues? Rats! Rifts in a toboggan!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2162" LastEditorDisplayName="user88" LastEditDate="2015-03-19T04:42:59.570" LastActivityDate="2015-03-19T04:42:59.570" Title="Find the letters that complete these five patterns" Tags="|pattern|lateral-thinking|" AnswerCount="4" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="6139" PostTypeId="2" ParentId="6138" CreationDate="2014-12-14T03:27:59.107" Score="5" Body="&lt;p&gt;Answer;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The words having nothing to do with it, the pattern is E, I, E, I...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Alternatively...&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is &quot;one letter&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You told us in the hint!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Or maybe...&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Whichever letter comes first in the word, is the opposite (I is E, E Is I)&lt;br&gt;&#xA; &quot;Us&lt;strong&gt;i&lt;/strong&gt;ng&quot; is E&lt;br&gt;&#xA; &quot;D&lt;strong&gt;e&lt;/strong&gt;sist&quot; is I&lt;br&gt;&#xA; &quot;V&lt;strong&gt;i&lt;/strong&gt;sits&quot; is E&lt;br&gt;&#xA; &quot;D&lt;strong&gt;e&lt;/strong&gt;sign&quot; is I&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Perhaps it's...&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A new idea why;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the first &lt;em&gt;i&lt;/em&gt; in the word is the 4th letter it's I else, it's E&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Maybe it's...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the word contains an E the answer is I, else it's E&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-12-16T21:22:24.433" LastActivityDate="2014-12-16T21:22:24.433" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6140" PostTypeId="1" AcceptedAnswerId="6151" CreationDate="2014-12-14T03:44:03.230" Score="-5" ViewCount="3845" Body="&lt;p&gt;You must design an algorithm for a 16x17 rectangle grid. Each &quot;cell&quot; is either a wall or an opening. This algorithm must achieve a maze without long hallways in 1 general direction, and must have dead ends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Most online algorithms include cells with 4 walls. This has only squares with either cell or wall.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The simplest algorithm will be accepted. Accepted may change.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="5373" LastEditDate="2014-12-28T19:44:21.153" LastActivityDate="2014-12-28T19:44:21.153" Title="Maze Generation Algorithm" Tags="|puzzle-creation|mazes|algorithm|open-ended|" AnswerCount="2" CommentCount="6" ClosedDate="2015-07-27T10:40:13.437" ContentLicense="CC BY-SA 3.0" />
  <row Id="6141" PostTypeId="2" ParentId="6138" CreationDate="2014-12-14T04:12:55.860" Score="16" Body="&lt;p&gt;The letter is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because you &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; count the number of letters of the word that are also in your username, &quot;Matt Malone&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which gives&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;&quot;Using&quot;  is 1 (n)&#xA; &quot;Desist&quot; is 2 (te)&#xA; &quot;Visits&quot; is 1 (t)&#xA; &quot;Design&quot; is 2 (ne) &lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and then&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; take the letter that consists of that many dots in Morse code (&lt;code&gt;e=.&lt;/code&gt;, &lt;code&gt;i=..&lt;/code&gt;, &lt;code&gt;s=...&lt;/code&gt;, &lt;code&gt;h=....&lt;/code&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2014-12-14T04:12:55.860" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6142" PostTypeId="2" ParentId="6140" CreationDate="2014-12-14T04:17:34.050" Score="2" Body="&lt;p&gt;This is just an adaptation of an &lt;a href=&quot;https://puzzling.stackexchange.com/a/5817/5044&quot;&gt;answer that I posted on another question&lt;/a&gt;, just reduced and simplified to be 2D instead of 4D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the java code:&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import java.awt.Color;&#xA;import java.awt.Graphics2D;&#xA;import java.awt.image.BufferedImage;&#xA;import java.io.File;&#xA;import java.io.IOException;&#xA;import java.util.ArrayList;&#xA;import java.util.Arrays;&#xA;import java.util.Collections;&#xA;import java.util.HashMap;&#xA;import java.util.List;&#xA;import java.util.Map;&#xA;import java.util.Objects;&#xA;import javax.imageio.ImageIO;&#xA;&#xA;/**&#xA; * @author Victor&#xA; */&#xA;public class Maze2D {&#xA;    public static void main(String[] args) throws IOException {&#xA;        Maze maze = new Maze(16, 17);&#xA;        BufferedImage im = maze.draw();&#xA;        ImageIO.write(im, &quot;png&quot;, new File(&quot;maze5.png&quot;));&#xA;    }&#xA;&#xA;    public static final class Coordinate2D {&#xA;        private final Maze maze;&#xA;        private final int x, y;&#xA;&#xA;        public Coordinate2D(Maze maze, int x, int y) {&#xA;            Objects.requireNonNull(maze);&#xA;            if (x &amp;lt; 0 || x &amp;gt;= maze.xSize || y &amp;lt; 0 || y &amp;gt;= maze.ySize) throw new IndexOutOfBoundsException();&#xA;            this.maze = maze;&#xA;            this.x = x;&#xA;            this.y = y;&#xA;        }&#xA;&#xA;        @Override&#xA;        public int hashCode() {&#xA;            return Objects.hash(maze, x, y);&#xA;        }&#xA;&#xA;        @Override&#xA;        public boolean equals(Object another) {&#xA;            if (!(another instanceof Coordinate2D)) return false;&#xA;            Coordinate2D c4d = (Coordinate2D) another;&#xA;            return maze == c4d.maze &amp;amp;&amp;amp; x == c4d.x &amp;amp;&amp;amp; y == c4d.y;&#xA;        }&#xA;&#xA;        public int squareDistance(Coordinate2D another) {&#xA;            Objects.requireNonNull(another);&#xA;            if (maze != another.maze) throw new IllegalArgumentException();&#xA;            int dx = Math.abs(x - another.x);&#xA;            int dy = Math.abs(y - another.y);&#xA;            return dx + dy;&#xA;        }&#xA;&#xA;        public Coordinate2D minusX() { return x == 0              ? null : new Coordinate2D(maze, x - 1, y); };&#xA;        public Coordinate2D plusX()  { return x == maze.xSize - 1 ? null : new Coordinate2D(maze, x + 1, y); };&#xA;        public Coordinate2D minusY() { return y == 0              ? null : new Coordinate2D(maze, x, y - 1); };&#xA;        public Coordinate2D plusY()  { return y == maze.ySize - 1 ? null : new Coordinate2D(maze, x, y + 1); };&#xA;        public Maze getMaze() { return maze; }&#xA;    }&#xA;&#xA;    public static final class Maze {&#xA;        private final int xSize, ySize;&#xA;        private final Map&amp;lt;Coordinate2D, Node&amp;gt; nodes;&#xA;        private final Node start;&#xA;&#xA;        public Maze(int x, int y) {&#xA;            this.xSize = x;&#xA;            this.ySize = y;&#xA;            nodes = new HashMap&amp;lt;&amp;gt;(x * y);&#xA;            fill();&#xA;            this.start = chooseRandomNode();&#xA;            growMaze();&#xA;        }&#xA;&#xA;        private void fill() {&#xA;            for (int b = 0; b &amp;lt; xSize; b++) {&#xA;                for (int c = 0; c &amp;lt; ySize; c++) {&#xA;                    Coordinate2D coord = new Coordinate2D(this, b, c);&#xA;                    nodes.put(coord, new Node(coord));&#xA;                }&#xA;            }&#xA;        }&#xA;&#xA;        public Node nodeAt(Coordinate2D coord) {&#xA;            if (coord == null) return null;&#xA;            return nodes.get(coord);&#xA;        }&#xA;&#xA;        private Node chooseRandomNode() {&#xA;            int n = (int) (Math.random() * xSize * ySize);&#xA;            return new ArrayList&amp;lt;&amp;gt;(nodes.values()).get(n);&#xA;        }&#xA;&#xA;        private void growMaze() {&#xA;            List&amp;lt;Node&amp;gt; frontier = new ArrayList&amp;lt;&amp;gt;(xSize * ySize);&#xA;            frontier.add(start);&#xA;            start.linked = true;&#xA;            while (!frontier.isEmpty()) {&#xA;                Collections.shuffle(frontier);&#xA;                Node n = frontier.get(0);&#xA;                Node next = n.linkRandomUnlinkedNeighbour();&#xA;                if (next != null) {&#xA;                    frontier.add(next);&#xA;                } else {&#xA;                    frontier.remove(0);&#xA;                }&#xA;            }&#xA;        }&#xA;&#xA;        public BufferedImage draw() {&#xA;            int cellWidth = 16;&#xA;            int cellHeight = 16;&#xA;            int boardWidth = cellWidth * (xSize + 1);&#xA;            int boardHeight = cellHeight * (ySize + 1);&#xA;            BufferedImage im = new BufferedImage(boardWidth + cellWidth - 1, boardHeight + cellHeight - 1, BufferedImage.TYPE_INT_ARGB);&#xA;            Graphics2D g = im.createGraphics();&#xA;            for (int x = 0; x &amp;lt; xSize; x++) {&#xA;                for (int y = 0; y &amp;lt; ySize; y++) {&#xA;                    Coordinate2D c = new Coordinate2D(this, x, y);&#xA;                    Node n = nodeAt(c);&#xA;                    int x1 = cellWidth * x + cellWidth - 1;&#xA;                    int y1 = cellHeight * y + cellHeight - 1;&#xA;                    int x2 = x1 + cellWidth;&#xA;                    int y2 = y1 + cellHeight;&#xA;                    g.setColor(Color.BLACK);&#xA;                    if (!n.isLinkedTo(n.minusY())) g.drawLine(x1, y1, x2, y1);&#xA;                    if (!n.isLinkedTo(n.plusY()))  g.drawLine(x1, y2, x2, y2);&#xA;                    if (!n.isLinkedTo(n.minusX())) g.drawLine(x1, y1, x1, y2);&#xA;                    if (!n.isLinkedTo(n.plusX()))  g.drawLine(x2, y1, x2, y2);&#xA;                }&#xA;            }&#xA;            return im;&#xA;        }&#xA;    }&#xA;&#xA;    public static final class Node {&#xA;        private final Coordinate2D coord;&#xA;        private final List&amp;lt;Node&amp;gt; linkedNeighbours;&#xA;        private List&amp;lt;Node&amp;gt; neighbours;&#xA;        private boolean linked;&#xA;&#xA;        public Node(Coordinate2D coord) {&#xA;            Objects.requireNonNull(coord);&#xA;            this.coord = coord;&#xA;            linkedNeighbours = new ArrayList&amp;lt;&amp;gt;(8);&#xA;        }&#xA;&#xA;        public Node linkRandomUnlinkedNeighbour() {&#xA;            List&amp;lt;Node&amp;gt; list = new ArrayList&amp;lt;&amp;gt;(getNeighbours());&#xA;            list.removeIf(n -&amp;gt; n.linked);&#xA;            if (list.isEmpty()) return null;&#xA;            Collections.shuffle(list);&#xA;            Node next = list.get(0);&#xA;            next.getNeighbours();&#xA;            linkedNeighbours.add(next);&#xA;            next.linkedNeighbours.add(this);&#xA;            next.linked = true;&#xA;            return next;&#xA;        }&#xA;&#xA;        @SuppressWarnings(&quot;ReturnOfCollectionOrArrayField&quot;)&#xA;        public List&amp;lt;Node&amp;gt; getNeighbours() {&#xA;            if (neighbours == null) {&#xA;                List&amp;lt;Node&amp;gt; nodes = new ArrayList&amp;lt;&amp;gt;(Arrays.asList(minusX(), plusX(), minusY(), plusY()));&#xA;                nodes.removeIf(x -&amp;gt; x == null);&#xA;                neighbours = Collections.unmodifiableList(nodes);&#xA;            }&#xA;            return neighbours;&#xA;        }&#xA;&#xA;        public boolean isDeadEnd() {&#xA;            return linkedNeighbours.size() == 1;&#xA;        }&#xA;&#xA;        public boolean isBranch() {&#xA;            return linkedNeighbours.size() &amp;gt; 2;&#xA;        }&#xA;&#xA;        public boolean isLinkedTo(Node node) {&#xA;            return linkedNeighbours.contains(node);&#xA;        }&#xA;&#xA;        public Maze getMaze() { return coord.getMaze(); }&#xA;        public Coordinate2D getCoord() { return coord; }&#xA;        public Node minusX() { return getMaze().nodeAt(coord.minusX()); };&#xA;        public Node plusX()  { return getMaze().nodeAt(coord.plusX());  };&#xA;        public Node minusY() { return getMaze().nodeAt(coord.minusY()); };&#xA;        public Node plusY()  { return getMaze().nodeAt(coord.plusY());  };&#xA;    }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;I am sorry that this site has no syntax-coloring.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a generated maze:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eU0XH.png&quot; alt=&quot;maze&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the code is boilerplate code (java is too verbose) and there is some code used only for drawing, what really matters to the algorithm is this:&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;        private void growMaze() {&#xA;            List&amp;lt;Node&amp;gt; frontier = new ArrayList&amp;lt;&amp;gt;(xSize * ySize);&#xA;            frontier.add(start);&#xA;            start.linked = true;&#xA;            while (!frontier.isEmpty()) {&#xA;                Collections.shuffle(frontier);&#xA;                Node n = frontier.get(0);&#xA;                Node next = n.linkRandomUnlinkedNeighbour();&#xA;                if (next != null) {&#xA;                    frontier.add(next);&#xA;                } else {&#xA;                    frontier.remove(0);&#xA;                }&#xA;            }&#xA;        }&#xA;&#xA;        public Node linkRandomUnlinkedNeighbour() {&#xA;            List&amp;lt;Node&amp;gt; list = new ArrayList&amp;lt;&amp;gt;(getNeighbours());&#xA;            list.removeIf(n -&amp;gt; n.linked);&#xA;            if (list.isEmpty()) return null;&#xA;            Collections.shuffle(list);&#xA;            Node next = list.get(0);&#xA;            next.getNeighbours();&#xA;            linkedNeighbours.add(next);&#xA;            next.linkedNeighbours.add(this);&#xA;            next.linked = true;&#xA;            return next;&#xA;        }&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which basically starts with a random cell in a list called &lt;code&gt;frontier&lt;/code&gt;. Then it repeatly randomly chooses a cell in that list and looks randomly for an unlinked neighbouring cell to the cell chosen. Then, the neighbouring cell is linkedand added to the list. If there is no unlinked neighbouring cell, the chosen cell is removed from the list. This is repeats until the list eventually gets empty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is nothing in the code to avoid long corridors. I just executed it several times until you get a maze without long corridors. If this should be enforced, a simple way would be to just regenerate the maze all again if there is some long corridor until no long corridor exists.&lt;/p&gt;&#xA;" OwnerUserId="5044" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2014-12-14T04:23:25.817" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6143" PostTypeId="1" AcceptedAnswerId="6149" CreationDate="2014-12-14T04:38:32.103" Score="5" ViewCount="692" Body="&lt;p&gt;Just a short but really scary looking code breaking challenge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It breaks down into a couple of sentences once cracked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;24141325121914011720242320110720150920201224161720111012190613080310160806252412191417050526212017171013&lt;/p&gt;&#xA;" OwnerUserId="7296" LastEditorUserId="7296" LastEditDate="2014-12-14T20:27:46.017" LastActivityDate="2015-06-17T16:20:58.647" Title="Code breaking: Find the Message" Tags="|cipher|encoded-message|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6144" PostTypeId="2" ParentId="6138" CreationDate="2014-12-14T04:44:53.100" Score="6" Body="&lt;p&gt;It's the letter&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The pattern is that it's the second letter of the word immediately following it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;&quot;Using&quot;  is E&#xA; &quot;D&lt;strong&gt;e&lt;/strong&gt;sist&quot; is I&#xA; &quot;V&lt;strong&gt;i&lt;/strong&gt;sits&quot; is E&#xA; &quot;D&lt;strong&gt;e&lt;/strong&gt;sign&quot; is ???&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which give the answer because &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the next word is &quot;H&lt;strong&gt;i&lt;/strong&gt;nt&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alternatively, the answer could be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; S&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;if you &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; wrap around back to &quot;U&lt;strong&gt;s&lt;/strong&gt;ing&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2014-12-14T04:53:09.600" LastActivityDate="2014-12-14T04:53:09.600" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6145" PostTypeId="2" ParentId="6143" CreationDate="2014-12-14T05:03:06.613" Score="4" Body="&lt;p&gt;Splitting the digits into groups of two and taking letters &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;241413... = 24,14,13,... = XNM...&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;gives the length-52 string&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;'XNMYLSNAQTXWTPGTOITTLXPQTKJLSFMHCJPHFYXLSNQEEZVTQQJM'&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method is implicitly confirmed in that all the resulting indices are 1 through 26, which is unlikely in a random string of numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't care to go further though because now it's just a random-looking string of letters, too short to do any meaningful cryptanalysis, and I see nothing better to do then throw methods at it until one works. I did try Caesar cipher though; none of the shifts produced anything meaningful.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2014-12-14T05:03:06.613" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6146" PostTypeId="2" ParentId="6069" CreationDate="2014-12-14T07:48:37.573" Score="10" Body="&lt;p&gt;&lt;strong&gt;2nd attempt&lt;/strong&gt; - Trying to find a sequence using only the position of the characters has yielded no meaningful result.  It appears that a value for each character is also being used.  I think the cipher uses the position of each character in the clear text (from 1 to 123) minus its character value to determine its position in the cipher text.  If the calculated position is less than 1, it is wrapped from the beginning to the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following character values are the best that I have found so far. This is not the final answer but the results are closer than any of my previous efforts&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Character   ?   ,   .   +   A   B   C   D   E   F   G   H   I   M   N   O   P   R   S   T   U   X   Y&#xA;Value       42  41  40  40  40  35  37  38  35  35  31  31  30  24  25  24  25  23  21  18  17  18  13&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;When the cipher text position is calculated from the clear text position minus the character value, the results are as follows (using &quot;+&quot; to help track the spaces):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;My cipher text  H   A   ,   +   E   I   T   G   +   X   T   B   +   T   D   S   T   +   R   E   I   +   A   T   +   E   +   O   T   U   A   N   T   +   C   O   S   M   D   N   A   T   E   O   +   D   N   +   N   U   C   E   Y   +   H   R   N   U   E   S   .   +   I   H   A   E   T   +   C   P   T   +   R   F   +   +   N   E   ?   G   I   T   O   Y   +   I   A   +   C   R   U   Y   A   +   O   U   D   E   T   +   O   T   U   C   N   H   I   R   U   ?   E   Y   +   I   E   H   A   +   P   S   T   +   R&#xA;OP cipher text  H   A   ,   +   T   I   T   E   X   G   B   +   +   +   D   T   E   S   T   R   I   +   A   T   +   U   E   +   A   +   O   N   T   C   T   S   O   M   N   D   E   O   T   A   +   D   N   +   U   E   Y   C   N   +   H   R   U   S   N   .   I   E   T   H   +   A   E   P   T   +   C   R   +   +   F   +   T   N   G   E   Y   I   ?   O   I   +   A   +   R   C   A   +   U   O   D   U   T   Y   E   +   C   O   H   N   T   U   I   ?   E   U   R   Y   I   +   H   E   P   S   A   +   T   +   R&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;How three spaces come together in the OP's cipher text, is one peculiarity that I do not understand. I'll keep working on this but would be happy if somebody else found the correct solution.  &lt;/p&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2014-12-17T19:31:45.073" LastActivityDate="2014-12-17T19:31:45.073" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="6147" PostTypeId="2" ParentId="6122" CreationDate="2014-12-14T08:20:57.203" Score="4" Body="&lt;p&gt;My answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the human body&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Within me you will find&lt;br&gt;&#xA;One of like heart and mind...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As well as an entity&lt;br&gt;&#xA;With inverted geography.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Inverted geometry makes me think of something concave like a hole maybe. We have two inverted geometries worth mentioning I guess (mouth and a**hole). Since there is something about lips near the end I assume it is your &lt;em&gt;mouth&lt;/em&gt; (providing we are talking about unisex here).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While I represent symmetry&lt;br&gt;&#xA;Mirrors can show another identity.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bodies represent symmetry because every thing is there twice (besides inner organs) but as soon as we mirror each half independently it looks different after all (new identity).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But there will always be twins&lt;br&gt;&#xA;Who generate many grins.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The lips of your mouth generate grins (female body would have been more fun).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5106" LastEditorUserId="5106" LastEditDate="2014-12-14T21:25:16.487" LastActivityDate="2014-12-14T21:25:16.487" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6149" PostTypeId="2" ParentId="6143" CreationDate="2014-12-14T12:14:03.173" Score="13" Body="&lt;p&gt;First step:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Split the digits into pairs:&lt;br&gt;&#xA; Result is &lt;code&gt;24, 14, 13, 25, 12, 19, 14, 01, 17, 20, 24, 23, 20, 11, 07, 20, 15, 09, 20, 20, 12, 24, 16, 17, 20, 11, 10, 12, 19, 06, 13, 08, 03, 10, 16, 08, 06, 25, 24, 12, 19, 14, 17, 05, 05, 26, 21, 20, 17, 17, 10, 13&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second step:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Convert &lt;code&gt;01&lt;/code&gt; to &lt;code&gt;a&lt;/code&gt;, &lt;code&gt;02&lt;/code&gt; to &lt;code&gt;b&lt;/code&gt;, &lt;code&gt;03&lt;/code&gt; to &lt;code&gt;c&lt;/code&gt;, ... &lt;code&gt;26&lt;/code&gt; to &lt;code&gt;z&lt;/code&gt;&lt;br&gt;&#xA; Result is &lt;strong&gt;&lt;code&gt;xnmylsnaqtxwtkgtoittlxpqtkjlsfmhcjphfyxlsnqeezutqqjm&lt;/code&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Third step:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rotate 5 letters in the backward direction (or equivalently 21 in the forward direction).&lt;br&gt;&#xA; This changes &lt;code&gt;f&lt;/code&gt; to &lt;code&gt;a&lt;/code&gt;, &lt;code&gt;g&lt;/code&gt; to &lt;code&gt;b&lt;/code&gt;, &lt;code&gt;h&lt;/code&gt; to &lt;code&gt;c&lt;/code&gt;, &lt;code&gt;i&lt;/code&gt; to &lt;code&gt;d&lt;/code&gt;, ..., &lt;code&gt;y&lt;/code&gt; to &lt;code&gt;t&lt;/code&gt;, &lt;code&gt;z&lt;/code&gt; to &lt;code&gt;u&lt;/code&gt;, &lt;code&gt;a&lt;/code&gt; to &lt;code&gt;v&lt;/code&gt;, &lt;code&gt;b&lt;/code&gt; to &lt;code&gt;w&lt;/code&gt;, &lt;code&gt;c&lt;/code&gt; to &lt;code&gt;x&lt;/code&gt;, &lt;code&gt;d&lt;/code&gt; to &lt;code&gt;y&lt;/code&gt; and &lt;code&gt;e&lt;/code&gt; to &lt;code&gt;z&lt;/code&gt;.&lt;br&gt;&#xA; Result is &lt;strong&gt;&lt;code&gt;sihtgnivlosrofbojdoogsklofegnahcxekcatsgnilzzupolleh&lt;/code&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fourth step:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reverse the string.&lt;br&gt;&#xA; Result is &lt;strong&gt;&lt;code&gt;hellopuzzlingstackexchangefolksgoodjobforsolvingthis&lt;/code&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fifth (and final) step:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Add some spaces.&lt;br&gt;&#xA; Result is &lt;strong&gt;&lt;code&gt;hello puzzling stack exchange folks good job for solving this&lt;/code&gt;&lt;/strong&gt;, which is the plain-text message.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A note about the message:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The original message had two typos in the plain-text message. Originally it was&#xA; &quot;24141325121914011720242320 &lt;strong&gt;16&lt;/strong&gt; 0720150920201224161720111012190613080310160806252412191417050526 &lt;strong&gt;22&lt;/strong&gt; 2017171013&quot;,&lt;br&gt;&#xA; instead of&lt;br&gt;&#xA; &quot;24141325121914011720242320 &lt;strong&gt;11&lt;/strong&gt; 0720150920201224161720111012190613080310160806252412191417050526 &lt;strong&gt;21&lt;/strong&gt; 2017171013&quot;&lt;br&gt;&#xA; which used to produce the plain-text message &lt;strong&gt;&lt;code&gt;hello quzzling stack exchange folks good job kor solving this&lt;/code&gt;&lt;/strong&gt;, but the OP already fixed that.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5044" LastEditorUserId="5044" LastEditDate="2015-06-17T16:20:58.647" LastActivityDate="2015-06-17T16:20:58.647" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6150" PostTypeId="1" CreationDate="2014-12-14T16:49:34.127" Score="12" ViewCount="3664" Body="&lt;p&gt;Fill in each blank unshaded cell in the diagram below with a positive integer less than 100, such that every consecutive group of unshaded cells within a row or column is an arithmetic sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xY30o.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem is from the USAMTS Round 3 problem set.&lt;/p&gt;&#xA;" OwnerUserId="7301" LastEditorDisplayName="user20" LastEditDate="2015-01-21T10:08:05.083" LastActivityDate="2015-01-21T10:08:05.083" Title="Sequence Brain Teaser: Fill the gaps to create arithmetic sequences" Tags="|mathematics|calculation-puzzle|number-sequence|" AnswerCount="5" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6151" PostTypeId="2" ParentId="6140" CreationDate="2014-12-14T17:35:59.717" Score="3" Body="&lt;p&gt;This is an adapted version of the algorithm posted &lt;a href=&quot;https://puzzling.stackexchange.com/questions/5806/4d-maze-creation/5817#5817&quot;&gt;at this question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that you ask for a 16 x 17 maze requiring each point either wall or empty. If you require the maze to be &quot;enwalled&quot; you need to use a 14x15 dimension in my algorithm and then draw a border of wall around it. (With or without hole close to the &quot;exit&quot; and &quot;entry&quot; points depending on your liking.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This algorithm works for n-dimensional mazes and arbitrary maze dimensions. I do not post &lt;em&gt;code&lt;/em&gt; here, but the algorithm concept. Hence there is no 100% verification it is correct.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My solution is based on the idea that each pixel/voxel/n-dim-element ( &lt;em&gt;called voxel from now on&lt;/em&gt;) can either be a path (0) or a wall (1). &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;The data dimensions sizes are called &lt;em&gt;n,m,k,l...&lt;/em&gt;  [&lt;strong&gt;n = 16, m = 17&lt;/strong&gt;]&lt;/li&gt;&#xA;  &lt;li&gt;Indexing starts with 0 for the first voxel&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;Initialize data:&lt;/h3&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Create the maze-data as Boolean array of needed dimensionality [&lt;strong&gt;16 x 17 = 2D&lt;/strong&gt;]&lt;/li&gt;&#xA;  &lt;li&gt;Initialize all voxels as 0 (empty)&lt;/li&gt;&#xA;  &lt;li&gt;Create a list for voxel-indices of walls (empty)&lt;/li&gt;&#xA;  &lt;li&gt;Set every voxel with at least one &lt;em&gt;even&lt;/em&gt; index to 1 (wall) &lt;/li&gt;&#xA;  &lt;li&gt;Store the wall-voxel index in the list.&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;&lt;strong&gt;From now on, whenever a wall is removed:&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;remove the according index from the list &lt;/li&gt;&#xA;  &lt;li&gt;set the according Boolean to 0 (empty)&lt;/li&gt;&#xA;  &lt;/ul&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;Define start &amp;amp; end position:&lt;/h3&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Essentially set the two opposing &quot;corners&quot; to be start and end.&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Set the voxel of index (0,0,...) to be the starting location&lt;/li&gt;&#xA;  &lt;li&gt;Set the voxel of index (n,m,...) to be the goal destination&lt;/li&gt;&#xA;  &lt;li&gt;If the destination voxel is a wall, remove the wall. (if at least one dimension is even-sized, this will be the case.)&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, the start for a 17x16 looks like this (Start is red, goal is green):&#xA;&lt;img src=&quot;https://i.stack.imgur.com/GXV7e.jpg&quot; alt=&quot;Starting position&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Create the maze:&lt;/h3&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The current maze is a grid with isolated empty spaces.&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Label all empty spaces with a unique label.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;Now go iteratively (&lt;em&gt;WHILE-loop&lt;/em&gt;):&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Randomly choose a wall from the index list.&lt;/li&gt;&#xA;  &lt;li&gt;Remove the index form the list (&lt;em&gt;never test a wall twice&lt;/em&gt;)&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Test&lt;/strong&gt;: Will removing this wall merge two empty spaces of different label?&#xA;  NO: next iteration. ELSE:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;p&gt;Remove wall&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Of all involved labels choose the one with lowest value&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Set all voxels having involved labels to this chosen value (=merge empty spaces)&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;next iteration&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Stop iteration when no indices are in the list.&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now this will produce something random like this for intermediate steps (colour indicates label):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PjJwx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with some more merging&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dzhZH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...and eventually everything is merged into one maze with no loops and mazimized &quot;area&quot; like this:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/xm3u5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Without proofing it, I think this algorithm will give you:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;- a non-looping maze from start to end &#xA;&#xA;- maximize the size of the maze within the volume&#xA;&#xA;- The identical algorithm could be used for arbitrary amount of dimensions&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now, this is what I proposed for the original question on 4D puzzles. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now comes your new restriction of the &quot;long hallways&quot;. This is very vaguely defined restriction at the moment! In general, very large (or multidimensional) mazes you could do an auto-correlation of the maze (with itself) and measure the extension of the auto-correlation in each dimension - if it's too big, maze-creation failed and repeat (or break&amp;amp;repeat as suggested below.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For smaller mazes like the one in this question (16x17), you could specify &quot;no more than X into one direction&quot; as criterion and then test your created maze against it. If it fails, you could either&#xA;- recreate from scratch (and accept longer computing times in &quot;bad luck&quot; scenarios.&#xA;- make the algorithm &quot;split&quot;  to long corridor with a wall (creating 2 IDed areas) and then repeat the &quot;merger&quot; with all walls (Usually 1 step only will be enough!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume you defined &quot;no corridors &gt;15&quot; then you would potentially get something like the following from the above:&#xA;(red &quot;fills&quot; green &quot;opens&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/iBziE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zow6T.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hfM1R.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that should get you where you wanted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I &lt;em&gt;could&lt;/em&gt; write this into an algorithm to actually do it, but I don't feel for it and I wouldn't be able to do it in a &quot;web-accessible&quot; language like Java (because I don't know Java that well ;c) ) But then, your requirement was to devise an &lt;strong&gt;algorithm&lt;/strong&gt; not a &lt;strong&gt;program&lt;/strong&gt;. &lt;/p&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2014-12-14T18:29:17.517" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6152" PostTypeId="2" ParentId="6150" CreationDate="2014-12-14T17:45:28.500" Score="17" Body="&lt;p&gt;TL;DR. The solution is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\begin{array}{rrrrrrrrrrrl} 03&amp;amp;06&amp;amp;09&amp;amp;12&amp;amp;15&amp;amp;18&amp;amp;&amp;amp;59&amp;amp;71&amp;amp;83&amp;amp;95&amp;amp;\\ &amp;amp;&amp;amp;13&amp;amp;18&amp;amp;&amp;amp;20&amp;amp;35&amp;amp;50&amp;amp;65&amp;amp;&amp;amp;&amp;amp;\\ 03&amp;amp;10&amp;amp;17&amp;amp;24&amp;amp;&amp;amp;22&amp;amp;&amp;amp;41&amp;amp;59&amp;amp;77&amp;amp;95&amp;amp;\\ 27&amp;amp;&amp;amp;&amp;amp;30&amp;amp;&amp;amp;24&amp;amp;&amp;amp;32&amp;amp;&amp;amp;&amp;amp;77&amp;amp;\\ 51&amp;amp;46&amp;amp;41&amp;amp;36&amp;amp;31&amp;amp;26&amp;amp;&amp;amp;23&amp;amp;35&amp;amp;47&amp;amp;59&amp;amp;\\ \end{array}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I prove that no other solution exists.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here is the proof:&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Starting position:&lt;/h1&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\begin{array}{rrrrrrrrrrrl} 03&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;&amp;amp;59&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;\\ &amp;amp;&amp;amp;-&amp;amp;-&amp;amp;&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;&amp;amp;&amp;amp;\\ -&amp;amp;10&amp;amp;-&amp;amp;-&amp;amp;&amp;amp;-&amp;amp;&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;\\ -&amp;amp;&amp;amp;&amp;amp;-&amp;amp;&amp;amp;-&amp;amp;&amp;amp;-&amp;amp;&amp;amp;&amp;amp;-&amp;amp;\\ -&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;31&amp;amp;26&amp;amp;&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;59&amp;amp;\\ \end{array}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Filling from the $26$ and $31$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\begin{array}{rrrrrrrrrrrl} 03&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;&amp;amp;59&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;\\ &amp;amp;&amp;amp;-&amp;amp;-&amp;amp;&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;&amp;amp;&amp;amp;\\ -&amp;amp;10&amp;amp;-&amp;amp;-&amp;amp;&amp;amp;-&amp;amp;&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;\\ -&amp;amp;&amp;amp;&amp;amp;-&amp;amp;&amp;amp;-&amp;amp;&amp;amp;-&amp;amp;&amp;amp;&amp;amp;-&amp;amp;\\ 51&amp;amp;46&amp;amp;41&amp;amp;36&amp;amp;31&amp;amp;26&amp;amp;&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;59&amp;amp;\\ \end{array}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;First guess:&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Guessing what should go up from the $51$, we should get decrements up to $25$. If we try decrementing by $26$ or higher, we will get a negative number left to the $10$.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lets try decrementing by $20$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\begin{array}{rrrrrrrrrrrl} 03&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;&amp;amp;59&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;\\ &amp;amp;&amp;amp;-&amp;amp;??&amp;amp;&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;&amp;amp;&amp;amp;\leftarrow \text{Can't put -06, can't be negative.}\\ 11&amp;amp;10&amp;amp;09&amp;amp;08&amp;amp;&amp;amp;-&amp;amp;&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;\\ 31&amp;amp;&amp;amp;&amp;amp;22&amp;amp;&amp;amp;-&amp;amp;&amp;amp;-&amp;amp;&amp;amp;&amp;amp;-&amp;amp;\\ 51&amp;amp;46&amp;amp;41&amp;amp;36&amp;amp;31&amp;amp;26&amp;amp;&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;59&amp;amp;\\ \end{array}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lets try decrementing by $19$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\begin{array}{rrrrrrrrrrrl} 03&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;&amp;amp;59&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;\\ &amp;amp;&amp;amp;-&amp;amp;??&amp;amp;&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;&amp;amp;&amp;amp;\leftarrow \text{Can't put -12, can't be negative.}\\ 13&amp;amp;10&amp;amp;07&amp;amp;04&amp;amp;&amp;amp;-&amp;amp;&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;\\ 32&amp;amp;&amp;amp;&amp;amp;20&amp;amp;&amp;amp;-&amp;amp;&amp;amp;-&amp;amp;&amp;amp;&amp;amp;-&amp;amp;\\ 51&amp;amp;46&amp;amp;41&amp;amp;36&amp;amp;31&amp;amp;26&amp;amp;&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;59&amp;amp;\\ \end{array}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What do this means?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Trying anything below $19$ would just produce negative numbers there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lets try decrementing by $21$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\begin{array}{rrrrrrrrrrrl} 03&amp;amp;-&amp;amp;-&amp;amp;??&amp;amp;-&amp;amp;-&amp;amp;&amp;amp;59&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;\leftarrow \text{Can't put -06, can't be negative.}\\ &amp;amp;&amp;amp;-&amp;amp;00&amp;amp;&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;&amp;amp;&amp;amp;\\ 09&amp;amp;10&amp;amp;11&amp;amp;12&amp;amp;&amp;amp;-&amp;amp;&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;\\ 30&amp;amp;&amp;amp;&amp;amp;24&amp;amp;&amp;amp;-&amp;amp;&amp;amp;-&amp;amp;&amp;amp;&amp;amp;-&amp;amp;\\ 51&amp;amp;46&amp;amp;41&amp;amp;36&amp;amp;31&amp;amp;26&amp;amp;&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;59&amp;amp;\\ \end{array}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lets try decrementing by $22$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\begin{array}{rrrrrrrrrrrl} 03&amp;amp;02&amp;amp;01&amp;amp;00&amp;amp;??&amp;amp;-&amp;amp;&amp;amp;59&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;\leftarrow \text{Can't put -01, can't be negative.}\\ &amp;amp;&amp;amp;-&amp;amp;06&amp;amp;&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;&amp;amp;&amp;amp;\\ 07&amp;amp;10&amp;amp;13&amp;amp;16&amp;amp;&amp;amp;-&amp;amp;&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;\\ 29&amp;amp;&amp;amp;&amp;amp;26&amp;amp;&amp;amp;-&amp;amp;&amp;amp;-&amp;amp;&amp;amp;&amp;amp;-&amp;amp;\\ 51&amp;amp;46&amp;amp;41&amp;amp;36&amp;amp;31&amp;amp;26&amp;amp;&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;59&amp;amp;\\ \end{array}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lets try decrementing by $23$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\begin{array}{rrrrrrrrrrrl} 03&amp;amp;??&amp;amp;??&amp;amp;04&amp;amp;-&amp;amp;-&amp;amp;&amp;amp;59&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;\leftarrow \text{Can't put 3}\frac{1}{3}\text{and 3}\frac{2}{3}\text{, not integers.}\\ &amp;amp;&amp;amp;-&amp;amp;12&amp;amp;&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;&amp;amp;&amp;amp;\\ 05&amp;amp;10&amp;amp;15&amp;amp;20&amp;amp;&amp;amp;-&amp;amp;&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;\\ 28&amp;amp;&amp;amp;&amp;amp;28&amp;amp;&amp;amp;-&amp;amp;&amp;amp;-&amp;amp;&amp;amp;&amp;amp;-&amp;amp;\\ 51&amp;amp;46&amp;amp;41&amp;amp;36&amp;amp;31&amp;amp;26&amp;amp;&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;59&amp;amp;\\ \end{array}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lets try decrementing by $25$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\begin{array}{rrrrrrrrrrrl} 03&amp;amp;??&amp;amp;??&amp;amp;20&amp;amp;-&amp;amp;-&amp;amp;&amp;amp;59&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;\leftarrow \text{Can't put 8}\frac{2}{3}\text{and 14}\frac{1}{3}\text{, not integers.}\\ &amp;amp;&amp;amp;-&amp;amp;24&amp;amp;&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;&amp;amp;&amp;amp;\\ 01&amp;amp;10&amp;amp;19&amp;amp;28&amp;amp;&amp;amp;-&amp;amp;&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;\\ 26&amp;amp;&amp;amp;&amp;amp;32&amp;amp;&amp;amp;-&amp;amp;&amp;amp;-&amp;amp;&amp;amp;&amp;amp;-&amp;amp;\\ 51&amp;amp;46&amp;amp;41&amp;amp;36&amp;amp;31&amp;amp;26&amp;amp;&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;59&amp;amp;\\ \end{array}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It must be decremented by $24$:&lt;br&gt;&#xA; &lt;br&gt;&#xA; $$\begin{array}{rrrrrrrrrrrl} 03&amp;amp;06&amp;amp;09&amp;amp;12&amp;amp;15&amp;amp;18&amp;amp;&amp;amp;59&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;\\ &amp;amp;&amp;amp;13&amp;amp;18&amp;amp;&amp;amp;20&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;&amp;amp;&amp;amp;\\ 03&amp;amp;10&amp;amp;17&amp;amp;24&amp;amp;&amp;amp;22&amp;amp;&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;\\ 27&amp;amp;&amp;amp;&amp;amp;30&amp;amp;&amp;amp;24&amp;amp;&amp;amp;-&amp;amp;&amp;amp;&amp;amp;-&amp;amp;\\ 51&amp;amp;46&amp;amp;41&amp;amp;36&amp;amp;31&amp;amp;26&amp;amp;&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;59&amp;amp;\\ \end{array}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;Second guess:&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Now lets try something right to the $20$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets try $21$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\begin{array}{rrrrrrrrrrrl} 03&amp;amp;06&amp;amp;09&amp;amp;12&amp;amp;15&amp;amp;18&amp;amp;&amp;amp;59&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;\\ &amp;amp;&amp;amp;13&amp;amp;18&amp;amp;&amp;amp;20&amp;amp;21&amp;amp;22&amp;amp;23&amp;amp;&amp;amp;&amp;amp;\\ 03&amp;amp;10&amp;amp;17&amp;amp;24&amp;amp;&amp;amp;22&amp;amp;&amp;amp;??&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;\\ 27&amp;amp;&amp;amp;&amp;amp;30&amp;amp;&amp;amp;24&amp;amp;&amp;amp;??&amp;amp;&amp;amp;&amp;amp;-&amp;amp;\\ 51&amp;amp;46&amp;amp;41&amp;amp;36&amp;amp;31&amp;amp;26&amp;amp;&amp;amp;??&amp;amp;-&amp;amp;-&amp;amp;59&amp;amp;\leftarrow \text{Can't put -89, can't be negative.}\\ \end{array}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can infer that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For $22$, we get $-81$ in that same spot. $-73$ for $23$, $-65$ for $24$, $-57$ for $25$, $-49$ for $26$, $-41$ for $27$, $-33$ for $28$, $-29$ for $29$, $-21$ for $30$, $-13$ for $31$, $-05$ for $32$. Going for numbers smaller than $21$ will also be always negative. We can't try $47$ or higher because this would produce something too large at the end of the row starting with $20$ (putting $46$ produces $72$ and $98$).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, lets try $33$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\begin{array}{rrrrrrrrrrrl} 03&amp;amp;06&amp;amp;09&amp;amp;12&amp;amp;15&amp;amp;18&amp;amp;&amp;amp;59&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;\\ &amp;amp;&amp;amp;13&amp;amp;18&amp;amp;&amp;amp;20&amp;amp;33&amp;amp;46&amp;amp;59&amp;amp;&amp;amp;&amp;amp;\\ 03&amp;amp;10&amp;amp;17&amp;amp;24&amp;amp;&amp;amp;22&amp;amp;&amp;amp;33&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;\\ 27&amp;amp;&amp;amp;&amp;amp;30&amp;amp;&amp;amp;24&amp;amp;&amp;amp;20&amp;amp;&amp;amp;&amp;amp;-&amp;amp;\\ 51&amp;amp;46&amp;amp;41&amp;amp;36&amp;amp;31&amp;amp;26&amp;amp;&amp;amp;07&amp;amp;??&amp;amp;??&amp;amp;59&amp;amp;\leftarrow \text{Can't put 24}\frac{1}{3}\text{and 41}\frac{2}{3}\text{, not integers.}\\ \end{array}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lets try $34$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\begin{array}{rrrrrrrrrrrl} 03&amp;amp;06&amp;amp;09&amp;amp;12&amp;amp;15&amp;amp;18&amp;amp;&amp;amp;59&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;\\ &amp;amp;&amp;amp;13&amp;amp;18&amp;amp;&amp;amp;20&amp;amp;34&amp;amp;48&amp;amp;62&amp;amp;&amp;amp;&amp;amp;\\ 03&amp;amp;10&amp;amp;17&amp;amp;24&amp;amp;&amp;amp;22&amp;amp;&amp;amp;37&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;\\ 27&amp;amp;&amp;amp;&amp;amp;30&amp;amp;&amp;amp;24&amp;amp;&amp;amp;26&amp;amp;&amp;amp;&amp;amp;-&amp;amp;\\ 51&amp;amp;46&amp;amp;41&amp;amp;36&amp;amp;31&amp;amp;26&amp;amp;&amp;amp;15&amp;amp;??&amp;amp;??&amp;amp;59&amp;amp;\leftarrow \text{Can't put 29}\frac{2}{3}\text{and 44}\frac{1}{3}\text{, not integers.}\\ \end{array}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lets try $35$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\begin{array}{rrrrrrrrrrrl} 03&amp;amp;06&amp;amp;09&amp;amp;12&amp;amp;15&amp;amp;18&amp;amp;&amp;amp;59&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;\\ &amp;amp;&amp;amp;13&amp;amp;18&amp;amp;&amp;amp;20&amp;amp;35&amp;amp;50&amp;amp;65&amp;amp;&amp;amp;&amp;amp;\\ 03&amp;amp;10&amp;amp;17&amp;amp;24&amp;amp;&amp;amp;22&amp;amp;&amp;amp;41&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;\\ 27&amp;amp;&amp;amp;&amp;amp;30&amp;amp;&amp;amp;24&amp;amp;&amp;amp;32&amp;amp;&amp;amp;&amp;amp;-&amp;amp;\\ 51&amp;amp;46&amp;amp;41&amp;amp;36&amp;amp;31&amp;amp;26&amp;amp;&amp;amp;23&amp;amp;35&amp;amp;47&amp;amp;59&amp;amp;\\ \end{array}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;Third guess:&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Now to finish, we see that the cell at the middle line and last column, must be odd, otherwise, it is impossible to fill the cell under it. To it be odd, all the remaining cells at the middle row must be odd also.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, lets try a $43$ under the $65$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\begin{array}{rrrrrrrrrrrl} 03&amp;amp;06&amp;amp;09&amp;amp;12&amp;amp;15&amp;amp;18&amp;amp;&amp;amp;59&amp;amp;87&amp;amp;??&amp;amp;??&amp;amp;\leftarrow \text{Can't put 115 and 143, too high.}\\ &amp;amp;&amp;amp;13&amp;amp;18&amp;amp;&amp;amp;20&amp;amp;35&amp;amp;50&amp;amp;65&amp;amp;&amp;amp;&amp;amp;\\ 03&amp;amp;10&amp;amp;17&amp;amp;24&amp;amp;&amp;amp;22&amp;amp;&amp;amp;41&amp;amp;43&amp;amp;45&amp;amp;47&amp;amp;\\ 27&amp;amp;&amp;amp;&amp;amp;30&amp;amp;&amp;amp;24&amp;amp;&amp;amp;32&amp;amp;&amp;amp;&amp;amp;53&amp;amp;\\ 51&amp;amp;46&amp;amp;41&amp;amp;36&amp;amp;31&amp;amp;26&amp;amp;&amp;amp;23&amp;amp;35&amp;amp;47&amp;amp;59&amp;amp;\\ \end{array}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What we can try then?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Replacing the $43$ with something lower, will just make the cell above the $65$ go even higher. So we must replace it with something higher than $43$.&lt;br&gt;&#xA; &lt;br&gt;&#xA; If we try $45$, the top-right cell will go to $137$, with $47$ will go to $131$, $49$ will go to $125$, $51$ will go to $119$, $53$ will go to $113$, $55$ will go to $107$, $57$ will go to $101$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ... lets try $59$:&lt;br&gt;&#xA; &lt;br&gt;&#xA; $$\begin{array}{rrrrrrrrrrrl} 03&amp;amp;06&amp;amp;09&amp;amp;12&amp;amp;15&amp;amp;18&amp;amp;&amp;amp;59&amp;amp;71&amp;amp;83&amp;amp;95&amp;amp;\\ &amp;amp;&amp;amp;13&amp;amp;18&amp;amp;&amp;amp;20&amp;amp;35&amp;amp;50&amp;amp;65&amp;amp;&amp;amp;&amp;amp;\\ 03&amp;amp;10&amp;amp;17&amp;amp;24&amp;amp;&amp;amp;22&amp;amp;&amp;amp;41&amp;amp;59&amp;amp;77&amp;amp;95&amp;amp;\\ 27&amp;amp;&amp;amp;&amp;amp;30&amp;amp;&amp;amp;24&amp;amp;&amp;amp;32&amp;amp;&amp;amp;&amp;amp;77&amp;amp;\\ 51&amp;amp;46&amp;amp;41&amp;amp;36&amp;amp;31&amp;amp;26&amp;amp;&amp;amp;23&amp;amp;35&amp;amp;47&amp;amp;59&amp;amp;\\ \end{array}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And it is solved!&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;The second guess has a single solution:&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Do other solutions exists?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To the left part no, because $27$ is the only number that fits above $51$ (the first guess). So lets see if some other number fits in the second and third guesses.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Right to the $20$, lets try $36$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\begin{array}{rrrrrrrrrrrl} 03&amp;amp;06&amp;amp;09&amp;amp;12&amp;amp;15&amp;amp;18&amp;amp;&amp;amp;59&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;\\ &amp;amp;&amp;amp;13&amp;amp;18&amp;amp;&amp;amp;20&amp;amp;36&amp;amp;52&amp;amp;68&amp;amp;&amp;amp;&amp;amp;\\ 03&amp;amp;10&amp;amp;17&amp;amp;24&amp;amp;&amp;amp;22&amp;amp;&amp;amp;45&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;\\ 27&amp;amp;&amp;amp;&amp;amp;30&amp;amp;&amp;amp;24&amp;amp;&amp;amp;38&amp;amp;&amp;amp;&amp;amp;-&amp;amp;\\ 51&amp;amp;46&amp;amp;41&amp;amp;36&amp;amp;31&amp;amp;26&amp;amp;&amp;amp;31&amp;amp;??&amp;amp;??&amp;amp;59&amp;amp;\leftarrow \text{Can't put 40}\frac{1}{3}\text{and 49}\frac{2}{3}\text{, not integers.}\\ \end{array}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lets try $37$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\begin{array}{rrrrrrrrrrrl} 03&amp;amp;06&amp;amp;09&amp;amp;12&amp;amp;15&amp;amp;18&amp;amp;&amp;amp;59&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;\\ &amp;amp;&amp;amp;13&amp;amp;18&amp;amp;&amp;amp;20&amp;amp;37&amp;amp;54&amp;amp;71&amp;amp;&amp;amp;&amp;amp;\\ 03&amp;amp;10&amp;amp;17&amp;amp;24&amp;amp;&amp;amp;22&amp;amp;&amp;amp;49&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;\\ 27&amp;amp;&amp;amp;&amp;amp;30&amp;amp;&amp;amp;24&amp;amp;&amp;amp;44&amp;amp;&amp;amp;&amp;amp;-&amp;amp;\\ 51&amp;amp;46&amp;amp;41&amp;amp;36&amp;amp;31&amp;amp;26&amp;amp;&amp;amp;39&amp;amp;??&amp;amp;??&amp;amp;59&amp;amp;\leftarrow \text{Can't put 45}\frac{2}{3}\text{and 52}\frac{1}{3}\text{, not integers.}\\ \end{array}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lets try $38$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\begin{array}{rrrrrrrrrrrl} 03&amp;amp;06&amp;amp;09&amp;amp;12&amp;amp;15&amp;amp;18&amp;amp;&amp;amp;59&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;\\ &amp;amp;&amp;amp;13&amp;amp;18&amp;amp;&amp;amp;20&amp;amp;38&amp;amp;56&amp;amp;74&amp;amp;&amp;amp;&amp;amp;\\ 03&amp;amp;10&amp;amp;17&amp;amp;24&amp;amp;&amp;amp;22&amp;amp;&amp;amp;53&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;\\ 27&amp;amp;&amp;amp;&amp;amp;30&amp;amp;&amp;amp;24&amp;amp;&amp;amp;50&amp;amp;&amp;amp;&amp;amp;-&amp;amp;\\ 51&amp;amp;46&amp;amp;41&amp;amp;36&amp;amp;31&amp;amp;26&amp;amp;&amp;amp;47&amp;amp;51&amp;amp;55&amp;amp;59&amp;amp;\\ \end{array}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is going on?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In fact, right to the $20$ we must produce a number which added to one is multiple of $3$, otherwise the last line would not be able to hold integers. We already know that numbers lower than $35$ produce negative numbers in the $8$th column and that numbers greater than $46$ produces numbers greater than $100$ in the row starting with $20$. So, we may try only $35$, $38$, $41$ and $44$ in the place at the right side of the $20$. Further, we know that $35$ is able to solve the puzzle.&lt;br&gt;&#xA; &lt;br&gt;&#xA; Continuing with the $38$, we must put some number below the $74$ which do not produces something too large in the upper right cell. The lower the number under $74$ is, the higher the number above it is. But, the higher it is, higher will be the number in the end of the middle row also. Further, this number must be odd, otherwise we get an even number in the end of the middle row and the number just below it would not be integer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, lets try $73$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\begin{array}{rrrrrrrrrrrl} 03&amp;amp;06&amp;amp;09&amp;amp;12&amp;amp;15&amp;amp;18&amp;amp;&amp;amp;59&amp;amp;75&amp;amp;91&amp;amp;??&amp;amp;\leftarrow \text{Can't put 107, too high.}\\ &amp;amp;&amp;amp;13&amp;amp;18&amp;amp;&amp;amp;20&amp;amp;38&amp;amp;56&amp;amp;74&amp;amp;&amp;amp;&amp;amp;\\ 03&amp;amp;10&amp;amp;17&amp;amp;24&amp;amp;&amp;amp;22&amp;amp;&amp;amp;53&amp;amp;73&amp;amp;93&amp;amp;??&amp;amp;\leftarrow \text{Can't put 113, too high.}\\ 27&amp;amp;&amp;amp;&amp;amp;30&amp;amp;&amp;amp;24&amp;amp;&amp;amp;50&amp;amp;&amp;amp;&amp;amp;-&amp;amp;\\ 51&amp;amp;46&amp;amp;41&amp;amp;36&amp;amp;31&amp;amp;26&amp;amp;&amp;amp;47&amp;amp;51&amp;amp;55&amp;amp;59&amp;amp;\\ \end{array}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What do this means?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If we try something smaller than $73$ below the $74$, the number on the top-right will grow. If we try something larger, then the number on the end of the middle row will grow. So it is impossible with $38$ at the right of the $20$.&lt;br&gt;&#xA; &lt;br&gt;&#xA; If we try $41$, we get $83$ in the end of the row starting with $20$. If we try $44$, we get $92$. And these numbers are far higher than $53$ and $59$, so either the middle or the top row would end with something higher than $100$. This proves that right to the $20$, the number must be $35$, so &lt;strong&gt;the second guess has a single solution&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;The third guess has a single solution:&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Lets remember where it is before the third guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\begin{array}{rrrrrrrrrrrl} 03&amp;amp;06&amp;amp;09&amp;amp;12&amp;amp;15&amp;amp;18&amp;amp;&amp;amp;59&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;\\ &amp;amp;&amp;amp;13&amp;amp;18&amp;amp;&amp;amp;20&amp;amp;35&amp;amp;50&amp;amp;65&amp;amp;&amp;amp;&amp;amp;\\ 03&amp;amp;10&amp;amp;17&amp;amp;24&amp;amp;&amp;amp;22&amp;amp;&amp;amp;41&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;\\ 27&amp;amp;&amp;amp;&amp;amp;30&amp;amp;&amp;amp;24&amp;amp;&amp;amp;32&amp;amp;&amp;amp;&amp;amp;-&amp;amp;\\ 51&amp;amp;46&amp;amp;41&amp;amp;36&amp;amp;31&amp;amp;26&amp;amp;&amp;amp;23&amp;amp;35&amp;amp;47&amp;amp;59&amp;amp;\\ \end{array}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What can we try?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We already know that below the $65$ the number must be odd and can't be lower than $59$ ($59$ solves it) because it would make the top-right number be higher than $100$. So we must try something higher than $59$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What happens if we try $61$?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\begin{array}{rrrrrrrrrrrl} 03&amp;amp;06&amp;amp;09&amp;amp;12&amp;amp;15&amp;amp;18&amp;amp;&amp;amp;59&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;\\ &amp;amp;&amp;amp;13&amp;amp;18&amp;amp;&amp;amp;20&amp;amp;35&amp;amp;50&amp;amp;65&amp;amp;&amp;amp;&amp;amp;\\ 03&amp;amp;10&amp;amp;17&amp;amp;24&amp;amp;&amp;amp;22&amp;amp;&amp;amp;41&amp;amp;61&amp;amp;81&amp;amp;??&amp;amp;\leftarrow \text{Can't put 101, too high.}\\ 27&amp;amp;&amp;amp;&amp;amp;30&amp;amp;&amp;amp;24&amp;amp;&amp;amp;32&amp;amp;&amp;amp;&amp;amp;-&amp;amp;\\ 51&amp;amp;46&amp;amp;41&amp;amp;36&amp;amp;31&amp;amp;26&amp;amp;&amp;amp;23&amp;amp;35&amp;amp;47&amp;amp;59&amp;amp;\\ \end{array}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What do this means?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If we increase the number below $65$, the last number in the middle row will just go even higher. This way, the third guess can't be lower than $59$ nor higher than $59$, so $59$ is the only number where it works, and thus &lt;strong&gt;the third guess has a single solution&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So I proved that there is only one solution, and that this solution is:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\begin{array}{rrrrrrrrrrrl} 03&amp;amp;06&amp;amp;09&amp;amp;12&amp;amp;15&amp;amp;18&amp;amp;&amp;amp;59&amp;amp;71&amp;amp;83&amp;amp;95&amp;amp;\\ &amp;amp;&amp;amp;13&amp;amp;18&amp;amp;&amp;amp;20&amp;amp;35&amp;amp;50&amp;amp;65&amp;amp;&amp;amp;&amp;amp;\\ 03&amp;amp;10&amp;amp;17&amp;amp;24&amp;amp;&amp;amp;22&amp;amp;&amp;amp;41&amp;amp;59&amp;amp;77&amp;amp;95&amp;amp;\\ 27&amp;amp;&amp;amp;&amp;amp;30&amp;amp;&amp;amp;24&amp;amp;&amp;amp;32&amp;amp;&amp;amp;&amp;amp;77&amp;amp;\\ 51&amp;amp;46&amp;amp;41&amp;amp;36&amp;amp;31&amp;amp;26&amp;amp;&amp;amp;23&amp;amp;35&amp;amp;47&amp;amp;59&amp;amp;\\ \end{array}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5044" LastEditorUserId="5044" LastEditDate="2014-12-15T05:13:26.470" LastActivityDate="2014-12-15T05:13:26.470" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6153" PostTypeId="1" CreationDate="2014-12-14T17:54:52.030" Score="7" ViewCount="1605" Body="&lt;p&gt;There is a picture of a statue with a discus and a picture of the colosseum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CD XXV CLXIX CCLI CDXLI CCCLXI CCCLXI I CXLIV CDXLI CD XXV CLXIX&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the coded message?&lt;/p&gt;&#xA;" OwnerUserId="7304" LastEditorUserId="4050" LastEditDate="2014-12-14T18:03:45.073" LastActivityDate="2014-12-14T20:20:53.793" Title="Crack the code: Roman numerals. There is a picture of a statue with a discus and a picture of the colosseum" Tags="|cipher|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6154" PostTypeId="2" ParentId="6150" CreationDate="2014-12-14T18:02:34.753" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mTEc9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is considering arithmetic progression as sequence alone&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steps: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The given sequence at the bottom has to be filled i.e. 26 31 up to 51 (E6 to E1)&#xA;Then since 51 has to be filled up the number 2 rows above has to be odd number otherwise the row just above 51 cannot be filled with a whole number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering the above and also considering that if we enter a number above 51 in cell C1  then since 10 is fixed at C2, the series (C1,C2...) would go to negative, the number has to be lower than 51 and cannot be higher than 19 or less than 1 (for in both these scenarios the adjacent numbers would become negative), once you choose a number and start filling the story will unfold gradually, just 2 constraints no number can go below 0 or above 99.  If you try various numbers for C1 you will notice that the constraints would get violated for one of the cells (I tried many :) believe me, but feel free to try others, it either increases one number above 99 or decreases one below 0) so C1 has to be 3 which will fill the entire left side leaving B6 at 20.  Then the tricky part is managing the right side with the two 59 constraints&lt;/p&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="2313" LastEditDate="2014-12-14T18:46:22.610" LastActivityDate="2014-12-14T18:46:22.610" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6155" PostTypeId="2" ParentId="6150" CreationDate="2014-12-14T18:12:36.933" Score="0" Body="&lt;p&gt;There is a unique solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dEVgy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find it, consider&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nqy8l.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value of $X$ can be at most $28$ (because of the $10$ in the same row), so $Y$ can be at most $16$. However, $Y\equiv 36\mod 4$ and $Y\equiv 3\mod 3$, so we must have $Y=12$. This allows us to fill in up to here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/abxIv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $Y=3/2 X-10$. The sequence $59+P$, $A+Q$, $B+R$, $C+S$ is an arithmetic progression, and the first two terms are $2X$ and $2Y$, so we get $C+S=5X-60$.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2014-12-14T18:42:05.027" LastActivityDate="2014-12-14T18:42:05.027" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6156" PostTypeId="1" AcceptedAnswerId="6171" CreationDate="2014-12-14T19:01:35.217" Score="8" ViewCount="542" Body="&lt;p&gt;Who or what is our hero, the adventurer? :) Now extra long and extra easy:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Dear friend, I write to you before my great adventure&#xA;The nature is, as always, on my side.&#xA;Don’t worry, I am very well prepared,&#xA;I booked a very talented and clever guide.&#xA;&#xA;I’m ready and excited to explore&#xA;the world’s most mysterious and tricky roads,&#xA;to open every secret door&#xA;to break the mystic ancient codes.&#xA;&#xA;You usually trust me for my method,&#xA;no whining and complaining, no big guns.&#xA;I’m breaking the difficult questions &#xA;into easy and simple ones.&#xA;&#xA;With every secret I reveal&#xA;I’m getting closer to my end&#xA;and some day I will see clear&#xA;the last locked door, my dear friend.&#xA;&#xA;And when you read the records of my journey,&#xA;hopefully, you won’t forget&#xA;how I, the great adventurer brought treasure&#xA;and found peace in my heroic death.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It seems to be more difficult as previously thought. Time for hints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Our hero is a concept, rather than material entity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5283" LastEditorUserId="5283" LastEditDate="2014-12-15T08:28:48.073" LastActivityDate="2014-12-17T10:00:02.633" Title="letter from a great adventurer" Tags="|riddle|rhyme|" AnswerCount="3" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="6159" PostTypeId="2" ParentId="6153" CreationDate="2014-12-14T20:09:27.550" Score="18" Body="&lt;p&gt;I'm not sure I have the right answer yet, but shouldn't the fourth group of letters have a &lt;code&gt;V&lt;/code&gt; in it somewhere? (Slightly revised answer below)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if you change the fourth group to `CCLVI, then the code consists of a series of perfect squares in Roman numerals:&lt;br /&gt;&#xA; &lt;s&gt;400 25 169 256 441 361 361 144 441 400 25 169&lt;/s&gt; (missed one!)&lt;br /&gt;&#xA; 400 25 169 256 441 361 361 1 144 441 400 25 169&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If so, then&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Calculating the square roots and converting the results to letters (A=1, B=2, etc.) gives us:&lt;br /&gt;&#xA; &lt;code&gt;T E M P U S S A L U T E M&lt;/code&gt;&lt;br /&gt;&#xA; And according to Google Translate, &quot;tempus salutem&quot; means &quot;Time saving&quot; (?)&lt;br /&gt;&#xA; (Is that the answer?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="40" LastEditorUserId="40" LastEditDate="2014-12-14T20:20:53.793" LastActivityDate="2014-12-14T20:20:53.793" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6160" PostTypeId="1" AcceptedAnswerId="6163" CreationDate="2014-12-14T21:10:34.380" Score="5" ViewCount="979" Body="&lt;p&gt;Can anyone solve this puzzle or give me some tips?&#xA;Rules: you have to use all 1, 2, 3, 4, 5, 6, 7 numbers in every 8 straight and vertical lines(4 vertical line and 4 straight line)&#xA;And the summation of the 8 rooms around each pink number must be the pink number!&#xA;Please help me!!&#xA;See the puzzle at:&#xA;&lt;a href=&quot;http://b2n.ir/24768&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://b2n.ir/24768&lt;/a&gt;&lt;img src=&quot;https://i.stack.imgur.com/3UlxC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7306" LastEditorUserId="71" LastEditDate="2014-12-15T00:31:33.680" LastActivityDate="2014-12-16T00:38:21.073" Title="Fit numbers on a grid with required sums for squares" Tags="|mathematics|calculation-puzzle|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6163" PostTypeId="2" ParentId="6160" CreationDate="2014-12-14T22:48:54.633" Score="9" Body="&lt;p&gt;&lt;sup&gt;&lt;em&gt;[EDITED: Now, with many solutions!]&lt;/em&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;first&lt;/strong&gt; solution is this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\begin{array}{rrrrrrr} 1 &amp;amp; 7 &amp;amp; 4 &amp;amp; 2 &amp;amp; 5 &amp;amp; 3 &amp;amp; 6 \\ 2 &amp;amp; 37 &amp;amp; 5 &amp;amp; 30 &amp;amp; 2 &amp;amp; 27 &amp;amp; 3 \\ 5 &amp;amp; 6 &amp;amp; 7 &amp;amp; 2 &amp;amp; 3 &amp;amp; 1 &amp;amp; 4 \\ 4 &amp;amp; 37 &amp;amp; 6 &amp;amp; 32 &amp;amp; 1 &amp;amp; 22 &amp;amp; 1 \\ 6 &amp;amp; 1 &amp;amp; 2 &amp;amp; 7 &amp;amp; 4 &amp;amp; 3 &amp;amp; 5 \\ 3 &amp;amp; 24 &amp;amp; 1 &amp;amp; 35 &amp;amp; 7 &amp;amp; 38 &amp;amp; 7 \\ 7 &amp;amp; 1 &amp;amp; 3 &amp;amp; 5 &amp;amp; 6 &amp;amp; 4 &amp;amp; 2 \end{array} $$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How did I find that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;BRUTE FORCE!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a Java program that I'd built to find the solutions. It tries that by trying every possible combination, and when it finds one that solves the puzzle, it shows it. It takes about 50 seconds to find the first solution in my PC, but it will find many:&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import java.util.HashSet;&#xA;import java.util.Set;&#xA;&#xA;/**&#xA; * @author Victor&#xA; */&#xA;public class GridFind {&#xA;    private final int[][] m;&#xA;    private int solutions;&#xA;&#xA;    public GridFind() {&#xA;        m = new int[7][7];&#xA;        m[1][1] = 37;&#xA;        m[1][3] = 37;&#xA;        m[1][5] = 24;&#xA;        m[3][1] = 30;&#xA;        m[3][3] = 32;&#xA;        m[3][5] = 35;&#xA;        m[5][1] = 27;&#xA;        m[5][3] = 22;&#xA;        m[5][5] = 38;&#xA;    }&#xA;&#xA;    public boolean brute(int x, int y) {&#xA;        for (int a = 1; a &amp;lt;= 7; a++) {&#xA;            m[x][y] = a;&#xA;            if (!check()) continue;&#xA;            if (x == 6 &amp;amp;&amp;amp; y == 6) {&#xA;                show();&#xA;                return false;&#xA;            }&#xA;            int j = x, k = y;&#xA;            if (k == 6) {&#xA;                j++;&#xA;                k = 0;&#xA;            } else if (j % 2 == 1) {&#xA;                k += 2;&#xA;            } else {&#xA;                k++;&#xA;            }&#xA;            if (brute(j, k)) return true;&#xA;        }&#xA;        m[x][y] = 0;&#xA;        return false;&#xA;    }&#xA;&#xA;    private boolean check() {&#xA;        for (int v = 0; v &amp;lt; 7; v += 2) {&#xA;            Set&amp;lt;Integer&amp;gt; a = new HashSet&amp;lt;&amp;gt;(7);&#xA;            for (int u = 0; u &amp;lt; 7; u++) {&#xA;                int z = m[v][u];&#xA;                if (z == 0) continue;&#xA;                if (a.contains(z)) return false;&#xA;                a.add(z);&#xA;            }&#xA;            Set&amp;lt;Integer&amp;gt; b = new HashSet&amp;lt;&amp;gt;(7);&#xA;            for (int u = 0; u &amp;lt; 7; u++) {&#xA;                int z = m[u][v];&#xA;                if (z == 0) continue;&#xA;                if (b.contains(z)) return false;&#xA;                b.add(z);&#xA;            }&#xA;        }&#xA;        for (int u = 0; u &amp;lt;= 4; u += 2) {&#xA;            for (int v = 0; v &amp;lt;= 4; v += 2) {&#xA;                int a1 = m[u][v];&#xA;                int a2 = m[u][v + 1];&#xA;                int a3 = m[u][v + 2];&#xA;                int a4 = m[u + 1][v];&#xA;                int a5 = m[u + 1][v + 2];&#xA;                int a6 = m[u + 2][v];&#xA;                int a7 = m[u + 2][v + 1];&#xA;                int a8 = m[u + 2][v + 2];&#xA;                int ac = m[u + 1][v + 1];&#xA;                if (a1 == 0 || a2 == 0 || a3 == 0 || a4 == 0 || a5 == 0 || a6 == 0 || a7 == 0 || a8 == 0) continue;&#xA;                if (a1 + a2 + a3 + a4 + a5 + a6 + a7 + a8 != ac) return false;&#xA;            }&#xA;        }&#xA;        return true;&#xA;    }&#xA;&#xA;    public void show() {&#xA;        solutions++;&#xA;        System.out.println(&quot;Found the solution #&quot; + solutions);&#xA;        for (int a = 0; a &amp;lt; 7; a++) {&#xA;            for (int b = 0; b &amp;lt; 7; b++) {&#xA;                int x = m[b][a];&#xA;                if (x &amp;lt; 10) System.out.print(&quot; &quot;);&#xA;                System.out.print(x + &quot; &quot;);&#xA;            }&#xA;            System.out.println();&#xA;        }&#xA;    }&#xA;&#xA;    public static void main(String[] args) {&#xA;        GridFind p = new GridFind();&#xA;        p.brute(0, 0);&#xA;    }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;I am sorry that this site does not have syntax coloring.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mentioned that the algorithm finds many solutions. The first one that it finds is up there in the beginning of this post. So what are the others?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\begin{array}{rrrrrrr} 1 &amp;amp; 7 &amp;amp; 4 &amp;amp; 2 &amp;amp; 5 &amp;amp; 3 &amp;amp; 6 \\ 2 &amp;amp; 37 &amp;amp; 5 &amp;amp; 30 &amp;amp; 3 &amp;amp; 27 &amp;amp; 3 \\ 5 &amp;amp; 6 &amp;amp; 7 &amp;amp; 3 &amp;amp; 1 &amp;amp; 2 &amp;amp; 4 \\ 4 &amp;amp; 37 &amp;amp; 6 &amp;amp; 32 &amp;amp; 2 &amp;amp; 22 &amp;amp; 1 \\ 6 &amp;amp; 1 &amp;amp; 2 &amp;amp; 7 &amp;amp; 4 &amp;amp; 3 &amp;amp; 5 \\ 3 &amp;amp; 24 &amp;amp; 1 &amp;amp; 35 &amp;amp; 7 &amp;amp; 38 &amp;amp; 7 \\ 7 &amp;amp; 1 &amp;amp; 3 &amp;amp; 5 &amp;amp; 6 &amp;amp; 4 &amp;amp; 2 \\ \end{array} $$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Third solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\begin{array}{rrrrrrr} 1 &amp;amp; 5 &amp;amp; 7 &amp;amp; 3 &amp;amp; 4 &amp;amp; 2 &amp;amp; 6 \\ 2 &amp;amp; 37 &amp;amp; 4 &amp;amp; 30 &amp;amp; 2 &amp;amp; 27 &amp;amp; 7 \\ 6 &amp;amp; 7 &amp;amp; 5 &amp;amp; 4 &amp;amp; 1 &amp;amp; 2 &amp;amp; 3 \\ 3 &amp;amp; 37 &amp;amp; 6 &amp;amp; 32 &amp;amp; 3 &amp;amp; 22 &amp;amp; 1 \\ 7 &amp;amp; 1 &amp;amp; 2 &amp;amp; 6 &amp;amp; 5 &amp;amp; 3 &amp;amp; 4 \\ 5 &amp;amp; 24 &amp;amp; 1 &amp;amp; 35 &amp;amp; 6 &amp;amp; 38 &amp;amp; 5 \\ 4 &amp;amp; 1 &amp;amp; 3 &amp;amp; 5 &amp;amp; 7 &amp;amp; 6 &amp;amp; 2 \\ \end{array} $$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;40th solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\begin{array}{rrrrrrr} 1 &amp;amp; 7 &amp;amp; 5 &amp;amp; 2 &amp;amp; 6 &amp;amp; 3 &amp;amp; 4 \\ 4 &amp;amp; 37 &amp;amp; 3 &amp;amp; 30 &amp;amp; 3 &amp;amp; 27 &amp;amp; 5 \\ 7 &amp;amp; 6 &amp;amp; 4 &amp;amp; 5 &amp;amp; 2 &amp;amp; 1 &amp;amp; 3 \\ 6 &amp;amp; 37 &amp;amp; 7 &amp;amp; 32 &amp;amp; 1 &amp;amp; 22 &amp;amp; 1 \\ 2 &amp;amp; 4 &amp;amp; 1 &amp;amp; 7 &amp;amp; 5 &amp;amp; 3 &amp;amp; 6 \\ 3 &amp;amp; 24 &amp;amp; 2 &amp;amp; 35 &amp;amp; 4 &amp;amp; 38 &amp;amp; 7 \\ 5 &amp;amp; 1 &amp;amp; 6 &amp;amp; 3 &amp;amp; 7 &amp;amp; 4 &amp;amp; 2 \\ \end{array} $$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;126th solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\begin{array}{rrrrrrr} 2 &amp;amp; 5 &amp;amp; 6 &amp;amp; 1 &amp;amp; 4 &amp;amp; 3 &amp;amp; 7 \\ 1 &amp;amp; 37 &amp;amp; 5 &amp;amp; 30 &amp;amp; 2 &amp;amp; 27 &amp;amp; 3 \\ 5 &amp;amp; 6 &amp;amp; 7 &amp;amp; 2 &amp;amp; 3 &amp;amp; 1 &amp;amp; 4 \\ 6 &amp;amp; 37 &amp;amp; 4 &amp;amp; 32 &amp;amp; 1 &amp;amp; 22 &amp;amp; 1 \\ 4 &amp;amp; 3 &amp;amp; 2 &amp;amp; 7 &amp;amp; 6 &amp;amp; 1 &amp;amp; 5 \\ 3 &amp;amp; 24 &amp;amp; 1 &amp;amp; 35 &amp;amp; 7 &amp;amp; 38 &amp;amp; 6 \\ 7 &amp;amp; 1 &amp;amp; 3 &amp;amp; 4 &amp;amp; 5 &amp;amp; 6 &amp;amp; 2 \\ \end{array} $$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;415th solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\begin{array}{rrrrrrr} 3 &amp;amp; 6 &amp;amp; 7 &amp;amp; 2 &amp;amp; 1 &amp;amp; 4 &amp;amp; 5 \\ 1 &amp;amp; 37 &amp;amp; 2 &amp;amp; 30 &amp;amp; 7 &amp;amp; 27 &amp;amp; 4 \\ 7 &amp;amp; 6 &amp;amp; 5 &amp;amp; 4 &amp;amp; 2 &amp;amp; 1 &amp;amp; 3 \\ 6 &amp;amp; 37 &amp;amp; 3 &amp;amp; 32 &amp;amp; 4 &amp;amp; 22 &amp;amp; 1 \\ 5 &amp;amp; 1 &amp;amp; 4 &amp;amp; 7 &amp;amp; 3 &amp;amp; 2 &amp;amp; 6 \\ 2 &amp;amp; 24 &amp;amp; 1 &amp;amp; 35 &amp;amp; 6 &amp;amp; 38 &amp;amp; 7 \\ 4 &amp;amp; 1 &amp;amp; 6 &amp;amp; 3 &amp;amp; 5 &amp;amp; 7 &amp;amp; 2 \\ \end{array} $$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The 2741st (and last) solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\begin{array}{rrrrrrr} 7 &amp;amp; 1 &amp;amp; 3 &amp;amp; 4 &amp;amp; 6 &amp;amp; 2 &amp;amp; 5 \\ 6 &amp;amp; 37 &amp;amp; 2 &amp;amp; 30 &amp;amp; 4 &amp;amp; 27 &amp;amp; 3 \\ 5 &amp;amp; 7 &amp;amp; 6 &amp;amp; 3 &amp;amp; 2 &amp;amp; 1 &amp;amp; 4 \\ 4 &amp;amp; 37 &amp;amp; 7 &amp;amp; 32 &amp;amp; 1 &amp;amp; 22 &amp;amp; 1 \\ 3 &amp;amp; 4 &amp;amp; 1 &amp;amp; 7 &amp;amp; 5 &amp;amp; 2 &amp;amp; 6 \\ 2 &amp;amp; 24 &amp;amp; 4 &amp;amp; 35 &amp;amp; 3 &amp;amp; 38 &amp;amp; 7 \\ 1 &amp;amp; 4 &amp;amp; 5 &amp;amp; 3 &amp;amp; 7 &amp;amp; 6 &amp;amp; 2 \\ \end{array} $$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The program took, on my PC, 7 hours, 15 minutes and 16 seconds to find all the 2741 solutions.&lt;/p&gt;&#xA;" OwnerUserId="5044" LastEditorUserId="5044" LastEditDate="2014-12-16T00:38:21.073" LastActivityDate="2014-12-16T00:38:21.073" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6164" PostTypeId="2" ParentId="6122" CreationDate="2014-12-14T23:40:09.707" Score="4" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Within me you will find&lt;br /&gt;&#xA;One of like heart and mind.&lt;br /&gt;&#xA;As well as an entity&lt;br /&gt;&#xA;With inverted geography.&lt;/p&gt;&#xA;&lt;p&gt;While I represent symmetry&lt;br /&gt;&#xA;Mirrors can show another identity.&lt;br /&gt;&#xA;But there will always be twins&lt;br /&gt;&#xA;Who generate many grins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The first 4 lines and last 4 lines should be interpreted differently.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Answer:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The Heart.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Within &lt;em&gt;me&lt;/em&gt; you will find&#xA;One of like heart and mind.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Literally, something I will find in a word (me is the poem, assumed). And, literally, to select one of: like, &lt;em&gt;heart&lt;/em&gt;, and mind.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;As well as an &lt;em&gt;entity&lt;/em&gt;&#xA;With &lt;em&gt;inverted&lt;/em&gt; geography.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Another key phrase (a command to invert something)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;While I represent symmetry&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; What represents symmetry? A mirror, but it can't be, for mirrors show another identity (below). A lake can do this, so can a pond, a bucket, a pool... etc. These literally represent symmetry, but &lt;em&gt;the heart&lt;/em&gt; is a deeper represenetation of symmetry, that of emotional symmetry.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Mirrors can show another &lt;em&gt;identity&lt;/em&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; If we remove &lt;em&gt;entity&lt;/em&gt; from &lt;em&gt;identity&lt;/em&gt; we are left with &lt;em&gt;id&lt;/em&gt;, and, when reversed, is &lt;em&gt;di&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;But there will always be &lt;em&gt;twins&lt;/em&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;em&gt;di&lt;/em&gt; is a prefix for &lt;em&gt;two&lt;/em&gt;, and twins are in &lt;em&gt;twos&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Who generate many grins.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The heart has two chambers, left and right. It is also responsible for (in a non-anatomical sense; figuratively) generating emotions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7307" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-12-15T00:57:07.900" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6165" PostTypeId="1" AcceptedAnswerId="6166" CreationDate="2014-12-15T00:57:45.630" Score="172" ViewCount="10156" Body="&lt;p&gt;Two adventurers, Merlin and Hermes, approached a large iron door built into a cliff face.&quot;Well...&quot;, said Hermes, &quot;What do we do now?&quot;. Merlin produced an old, large piece of crumpled paper from his pocket. &quot;Hrm...&quot;, Merlin mumbled. &quot;It says here that we must speak the six letter keyword to open the door and enter the secret chamber, but I don't remember seeing any signs as to what that keyword might be...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a bit of searching, Hermes notices something etched into the ground. &quot;Come over here!&quot;, he yelled, pointing frantically. And sure enough, barely visible and obscured by dust, was a series of lines of differing color etched into the ground:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BXWwr.png&quot; alt=&quot;You&amp;#39;re looking at it the wrong way!&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Ah&quot;, Merlin said, &quot;So that is the keyword.&quot; Hermes was lost and confused. After staring at it for another thirty seconds, he grumbled &quot;What keyword!? All I see is a bunch of lines!&quot;. Merlin simply responded, &quot;You're just &lt;em&gt;looking at it the wrong way&lt;/em&gt;. It's obvious!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Isn't it?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="7308" LastEditorUserId="11876" LastEditDate="2016-02-20T05:54:30.350" LastActivityDate="2016-10-09T18:44:27.463" Title="Merlin and Hermes: Mysterious Lines" Tags="|enigmatic-puzzle|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6166" PostTypeId="2" ParentId="6165" CreationDate="2014-12-15T01:04:51.850" Score="147" Body="&lt;p&gt;The word is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; UNLINK&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The images are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; views of the letters from the top&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More precisely,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the gradients correspond to how close part of the letter is to the top of the letter, with black being flush with the top, and white being all the way at the bottom. For example, in the second letter (&lt;code&gt;N&lt;/code&gt;), the left and right edges touch the top and so are black, and the middle part lowers from black to white as you go left to right. The last letter (&lt;code&gt;K&lt;/code&gt;) has this same pattern reflected, but less intense, as the dip only goes halfway down. The &lt;code&gt;I&lt;/code&gt; is just a single bar in a sans-serif font.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2014-12-15T01:04:51.850" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6167" PostTypeId="1" AcceptedAnswerId="6182" CreationDate="2014-12-15T09:09:23.643" Score="13" ViewCount="818" Body="&lt;p&gt;I took a &lt;a href=&quot;http://nrars.org/0%20The%20Book%20of%20Weeks/archive/new/04%20html%20scrape%20or%20search/0489.htm&quot; rel=&quot;noreferrer&quot;&gt;contest entry&lt;/a&gt; written by Seth Brown of Williamstown, MA in 2003, and I converted it into this puzzle.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Disclaimer: Events described in this puzzle are fictional.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;A White House stenographer transcribed the following conversation between Barack Obama and Benjamin Netanyahu.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Obama: Who is on first?&lt;/p&gt;&#xA;&lt;p&gt;Netanyahu: Me?&lt;/p&gt;&#xA;&lt;p&gt;Obama: No, the guy playing first base.&lt;/p&gt;&#xA;&lt;p&gt;Netanyahu: Me?&lt;/p&gt;&#xA;&lt;p&gt;Obama: You're the first baseman?&lt;/p&gt;&#xA;&lt;p&gt;Netanyahu: No, I am asking you. Me?&lt;/p&gt;&#xA;&lt;p&gt;Obama: Who?&lt;/p&gt;&#xA;&lt;p&gt;Netanyahu: You mean that man right there?&lt;/p&gt;&#xA;&lt;p&gt;Obama: He's on first?&lt;/p&gt;&#xA;&lt;p&gt;Netanyahu: What do you mean? There aren't any females on the team.&lt;/p&gt;&#xA;&lt;p&gt;Obama: Well then who's on first?&lt;/p&gt;&#xA;&lt;p&gt;Netanyahu: Me?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Assume that nobody on the team is named &amp;quot;Who&amp;quot;, &amp;quot;Me&amp;quot;, or &amp;quot;He&amp;quot;.  With that assumption, the conversation doesn't seem to make any sense.  But it actually does make sense.  Explain how.&lt;/p&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-05-18T21:00:21.133" Title="Conversation between Obama and Netanyahu" Tags="|enigmatic-puzzle|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6168" PostTypeId="2" ParentId="6167" CreationDate="2014-12-15T09:40:38.827" Score="11" Body="&lt;p&gt;Language barrier.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Who is vocalised as &quot;hu&quot;, &quot;hu&quot; in hebrew means &quot;he&quot;.&lt;br&gt;&#xA; - &lt;a href=&quot;http://en.wiktionary.org/wiki/%D7%94%D7%95%D7%90&quot; rel=&quot;noreferrer&quot;&gt;http://en.wiktionary.org/wiki/%D7%94%D7%95%D7%90&lt;/a&gt;&lt;br&gt;&#xA; He is vocalised as &quot;hi&quot;, &quot;hi&quot; in hebrew means &quot;she&quot; (wacky, right?).&lt;br&gt;&#xA; - &lt;a href=&quot;http://en.wiktionary.org/wiki/%D7%94%D7%99%D7%90&quot; rel=&quot;noreferrer&quot;&gt;http://en.wiktionary.org/wiki/%D7%94%D7%99%D7%90&lt;/a&gt;&lt;br&gt;&#xA; Me is vocalised as &quot;mi&quot;, &quot;mi&quot; in hebrew means &quot;who&quot;.&lt;br&gt;&#xA; - &lt;a href=&quot;http://en.wiktionary.org/wiki/%D7%9E%D7%99&quot; rel=&quot;noreferrer&quot;&gt;http://en.wiktionary.org/wiki/%D7%9E%D7%99&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With this in mind, the conversation makes sense.&lt;/p&gt;&#xA;" OwnerUserId="5265" LastEditorUserId="5265" LastEditDate="2014-12-15T18:27:31.580" LastActivityDate="2014-12-15T18:27:31.580" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6169" PostTypeId="2" ParentId="6156" CreationDate="2014-12-15T10:36:24.047" Score="4" Body="&lt;p&gt;The adventurer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A debugger&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With the following reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dear friend, I write to you before my great adventure The nature is,&#xA;  as always, on my side.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The great adventure is to fix the code, and the process usually works.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Don’t worry, I am very well prepared, I booked a very talented and&#xA;  clever guide.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Flattery will get you everywhere! The guide is the developer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I’m ready and excited to explore the world’s most mysterious and&#xA;  tricky roads, to open every secret door to break the mystic ancient&#xA;  codes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's almost as if you are familiar with my programming style! &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;With every secret I reveal I’m getting closer to my end and some day I&#xA;  will see clear the last locked door, my dear friend.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Every fixed bug will cause the program to run further, and soon only one bug will remain.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And when you read the records of my journey, hopefully, you won’t&#xA;  forget how I, the great adventurer brought treasure and found peace in&#xA;  my heroic death.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The treasure is working, tested code! And documentation too!!  Once the code works, the debugger is no longer needed, and may rest in peace.&lt;/p&gt;&#xA;" OwnerUserId="7316" LastActivityDate="2014-12-15T10:36:24.047" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6170" PostTypeId="2" ParentId="6156" CreationDate="2014-12-15T10:42:03.160" Score="2" Body="&lt;p&gt;You are a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Graph search&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Dear friend, I write to you before my great adventure&#xA;The nature is, as always, on my side.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure here. The nature is on your side because the graph is connected.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Don’t worry, I am very well prepared,&#xA;I booked a very talented and clever guide.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The search strategy, [DFS](&lt;a href=&quot;https://en.wikipedia.It&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.It&lt;/a&gt;'s almost as if you are familiar with my programming style! org/wiki/Depth-first_search), &lt;a href=&quot;https://en.wikipedia.org/wiki/Breadth-first_search&quot; rel=&quot;nofollow&quot;&gt;BFS&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/A*_search_algorithm&quot; rel=&quot;nofollow&quot;&gt;A*&lt;/a&gt; or any other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I’m ready and excited to explore&#xA;the world’s most mysterious and tricky roads,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Graph exploration is a graph search where we do not have the whole graph at the beginning of the problem. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;to open every secret door&#xA;to break the mystic ancient codes.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every secret door is actually each visited vertex. Reference to coding.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;You usually trust me for my method,&#xA;no whining and complaining, no big guns.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Methods are well-known, with proofs and all, and the programmers trust them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I’m breaking the difficult questions &#xA;into easy and simple ones.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; GS employs the divide and conquer approach&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;With every secret I reveal&#xA;I’m getting closer to my end&#xA;and some day I will see clear&#xA;the last locked door, my dear friend.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each visited node brings you closer to the answer. The last locked door, being the vertex we are searching for.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;And when you read the records of my journey,&#xA;hopefully, you won’t forget&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quite often we need the path, not only the searched vertex.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;how I, the great adventurer brought treasure&#xA;and found peace in my heroic death.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After the algorithm finishes the process that implements it dies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastActivityDate="2014-12-15T10:42:03.160" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6171" PostTypeId="2" ParentId="6156" CreationDate="2014-12-15T10:54:33.153" Score="6" Body="&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Recursion&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Dear friend, I write to you before my great adventure&#xA;The nature is, as always, on my side.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Recursion is deemed as a &quot;natural process&quot;, for example fractals&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Don’t worry, I am very well prepared,&#xA;I booked a very talented and clever guide.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The talented and clever guide is the programmer implementing recursion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I’m ready and excited to explore&#xA;the world’s most mysterious and tricky roads,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Recursion goes down through one or more of the possible paths.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;to open every secret door&#xA;to break the mystic ancient codes.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every secret door is actually each call to the function. Reference to coding.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;You usually trust me for my method,&#xA;no whining and complaining, no big guns.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The method is well-known, with proofs for a lot of cases. Programmers trust recursion. Also, recursion is known for simplifying the code, with no giant data srtuctures to loop through.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I’m breaking the difficult questions &#xA;into easy and simple ones.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Recursion if often a divide and conquer approach. In order to converge, the subproblem should be smaller.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;With every secret I reveal&#xA;I’m getting closer to my end&#xA;and some day I will see clear&#xA;the last locked door, my dear friend.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each call to recursion gets us closer to the answer. The last locked door is the base case.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;And when you read the records of my journey,&#xA;hopefully, you won’t forget&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quite often we need the path the recursion has taken as well as the answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;how I, the great adventurer brought treasure&#xA;and found peace in my heroic death.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After the algorithm finishes the process that implements it dies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="5283" LastEditDate="2014-12-17T10:00:02.633" LastActivityDate="2014-12-17T10:00:02.633" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6172" PostTypeId="2" ParentId="6122" CreationDate="2014-12-15T12:09:22.323" Score="3" Body="&lt;p&gt;Another answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Within me you will find&lt;br&gt;&#xA;One of like heart and mind.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;em&gt;Within &quot;me&quot; you will find&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;&quot;1&quot;, of like heart and mind.&lt;/em&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;This means you are to look in the word &quot;me&quot; for something that is like &quot;1&quot;, in other words a number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As well as an entity&lt;br&gt;&#xA;With inverted geography.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This is the cue that the number is not right side up.  In this case, take m and rotate it to become 3.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While I represent symmetry&lt;br&gt;&#xA;Mirrors can show another identity.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;3 is symmetric, but put a mirror in the middle and it becomes 8.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But there will always be twins&lt;br&gt;&#xA;Who generate many grins.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;:3&lt;br&gt;&#xA;The twins represent the colon, which when added, form the cat face smiley :3&#xA;&lt;br&gt;&#xA;The heart in the previous passage can now be interpreted as &amp;lt;3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6178" LastEditorUserId="6178" LastEditDate="2014-12-16T01:15:08.430" LastActivityDate="2014-12-16T01:15:08.430" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6173" PostTypeId="2" ParentId="6122" CreationDate="2014-12-15T12:23:11.750" Score="5" Body="&lt;p&gt;My guess is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A palindrome or a palindromic phrase&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Within me you will find&lt;br&gt;&#xA;One of like heart and mind.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Within it is the effort of a thinker like the ones at Puzzling.SE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As well as an entity&lt;br&gt;&#xA;With inverted geography.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The entity with an inverted geography is the palindrome, as it reads the same in both directions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While I represent symmetry&lt;br&gt;&#xA;Mirrors can show another identity.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Palindromes are symmetrical, but a mirroring the first half of the palindrome will not produce the same thing visually.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But there will always be twins&lt;br&gt;&#xA;Who generate many grins.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Palindromic phrases or sentences are not so hard to construct, and usually one makes up the first half and tries to mirror it with the second half and spaces. Most such sentences are usually quite funny (like  &lt;em&gt;taco cat&lt;/em&gt; or &lt;em&gt;evil olive&lt;/em&gt;) or have some weird meaning. A short &lt;a href=&quot;https://en.wiktionary.org/wiki/Appendix:Palindromic_phrases#English&quot; rel=&quot;nofollow&quot;&gt;list&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="25966" LastEditDate="2016-07-14T10:47:07.693" LastActivityDate="2016-07-14T10:47:07.693" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6174" PostTypeId="2" ParentId="6122" CreationDate="2014-12-15T12:40:15.490" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A pregnant woman/embryo?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Within me you will find &lt;br&gt;&#xA;One of like heart and mind. &lt;br&gt;&#xA;As well as an entity &lt;br&gt;&#xA;With inverted geography.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Foetuses are upside down inside the mother's womb.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;While I represent symmetry &lt;br&gt;&#xA;Mirrors can show another identity. &lt;br&gt;&#xA;But there will always be twins &lt;br&gt;&#xA;Who generate many grins.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Literally twins; they would generate many grins in the proud parents.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Avigrail's 'boobs' comment also fits into this.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2014-12-15T12:40:15.490" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6175" PostTypeId="1" AcceptedAnswerId="6180" CreationDate="2014-12-15T13:12:54.023" Score="13" ViewCount="1852" Body="&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/users/2484/dalarcop&quot;&gt;d'alar'cop&lt;/a&gt;&lt;/strong&gt; was busy solving puzzles and making riddles at his home. He heard a knocking on the door. He got up and reached the door to answer the knocking. Once he opened the door, no one was there. He saw a packet lying in front of the door. It was addressed to cop. He took the packet and got back to his table. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;d'alar'cop started unpacking the package with curiosity. As he removed the covering he saw a wooden box. He thought it would be some gift, from a fan. Once he opened the box, he saw a &lt;strong&gt;letter&lt;/strong&gt;. Before opening the letter he also notices a &lt;strong&gt;6 digit long LCD panel&lt;/strong&gt; and a &lt;strong&gt;number pad with button 0-9 and an Enter&lt;/strong&gt; Key attached to the box cap. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;d'alar'cop got confused, what is this? Is this some kind of a trick or something?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He took the letter in his hand and started reading it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the letter&#xA;&lt;br /&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Hi cop, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't know me. I don't have any problem with you personally either. One of your enemies hired me to kill you and I have already planted a bomb in your home. It may explode anytime. There is no way you can find it or defuse it (without my help). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I shouldn't have sent this to you if I haven't created an account at &lt;strong&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/&quot;&gt;Puzzling.StackExchange.com&lt;/a&gt;&lt;/strong&gt; today. I saw your profile and came to know that you are the most reputed user in &lt;strong&gt;puzzling.SE&lt;/strong&gt;. I have also seen all the riddles and puzzles you made and the answers you posted. All of them are really brilliant and that made me think I should give you a chance to live, if you are really brilliant enough as I thought.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To defuse the bomb you have to enter a &lt;strong&gt;6 digit code&lt;/strong&gt; with the attached number pad and the fun part is I am not going to tell you the code. I am just going to test you. If you are brilliant enough, you will get the code for sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start with a &lt;strong&gt;riddle&lt;/strong&gt;. I know you love riddles more than anything ;).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I was made by hiring some from a family of twenty six&lt;br /&gt; &#xA;  And, if I count the twins as two, my total strength is six&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I will help you lit up your house&lt;br /&gt; &#xA;  And also to clean and tidy up your house&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;But, don't try to get into me too much&lt;br /&gt; &#xA;  I will kill you with hypoxia and such&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;My origin is actually from the Latin&lt;br /&gt; &#xA;  In perfect form, I am unable to be attained&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once you figure out all the six members, we can move to the next game. A game of chess. But you have to play according to MY RULES&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HERE IS THE GAME&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OExiq.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;br /&gt;&#xA;&lt;strong&gt;RULES&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Your pieces are White.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Assign each member to the pieces you have respectively. That is, the&#xA;White Pawn on the left is the first member, the next White Rook is&#xA;the second member, the next White Queen is the third member and so&#xA;on.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I have already made all my moves and now it's your turn.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You have to start from the left. That is, from the White Pawn&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You can only move your piece to the safe position.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If there is no safe position, you can leave the piece unchanged.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Also, you can cut my piece (Black). But, only if it's safe.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You can only move a piece once and so the maximum number of moves&#xA;allowed is six.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You can make your moves one after another. You don't have to wait for&#xA;my turn since as I told you, I have already made all my moves.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now Start Playing.....&#xA;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How was the game? Hope you enjoyed it. Otherwise, you will be &lt;strong&gt;&quot;BAAAANNGGGGG&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cop tried to open the door to run out, but it was locked. Someone might have locked the doors from outside as Cop took the parcel in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cop is currently working on it to solve it and save his life. Since he is a very valuable contributer of our community, it's our duty to help him. His loss will affect our community very badly for sure. So I request all fellow community members to help him figure out the code to defuse the bomb. Please post all your answers here. It will help him save himself. If Cop himself finds the answer, he will post the answer and how he saved himself here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since Cop is busy solving it, he asked me to post it here. If somebody post the correct code, I will inform him. &lt;/p&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2016-10-26T09:08:05.160" Title="Save d'alar'cop by Helping Him Defuse The Bomb Placed By An Assassin" Tags="|password|chess|" AnswerCount="4" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="6176" PostTypeId="2" ParentId="6175" CreationDate="2014-12-15T13:28:33.630" Score="0" Body="&lt;p&gt;The answer for the first part is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oxygen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fits the bill except for line 4, I suppose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second part doesn't make sense to me at all, though. I can assign the letters and then what. There's no real objective unless I'm missing something.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Never mind, it doesn't make sense when I learn to lead. Hyperoxia != Hypoxia.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As for the second part:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - Pawn Stays&lt;br&gt;&#xA; - Rook goes to b2&lt;br&gt;&#xA; - Queen goes to d7&lt;br&gt;&#xA; - Pawn goes to e3&lt;br&gt;&#xA; - King goes to f5&lt;br&gt;&#xA; - Rook goes to g4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now to get a proper cipher going with this.&lt;/p&gt;&#xA;" OwnerUserId="5265" LastEditorUserId="5265" LastEditDate="2014-12-15T14:01:59.357" LastActivityDate="2014-12-15T14:01:59.357" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6177" PostTypeId="2" ParentId="6175" CreationDate="2014-12-15T13:41:50.220" Score="1" Body="&lt;p&gt;I guess the answer for the first part is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Power (the twins are double-U)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The hypoxia part:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you get electrocuted it can result in hypoxia?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Still thinking bout part 2&lt;/p&gt;&#xA;" OwnerUserId="5610" LastActivityDate="2014-12-15T13:41:50.220" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6178" PostTypeId="2" ParentId="6175" CreationDate="2014-12-15T13:45:51.617" Score="12" Body="&lt;p&gt;The first part would refer to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Vacuum&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I will help you lit up your house&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Light bulbs have vacuum inside&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And also to clean and tidy up your house&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Vacuum cleaner&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But, don't try to get into me too much&#xA;I will kill you with hypoxia and such&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If you are in a vacuum you will have no oxygen to breathe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My origin is actually from the Latin&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Vacuum is originally a latin word&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In perfect form, I am unable to be attained&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;We cannot attain a perfect vacuum&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Part 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. Rb2&lt;br&gt;&#xA; 2. Qd7&lt;br&gt;&#xA; 3. e3&lt;br&gt;&#xA; 4. Kxf5&lt;br&gt;&#xA; 5. Rg7 (protected from the Black King by the White Queen at d7)&lt;br&gt;&#xA; or Rxg4 (the most correct answer as the Rook is not being attacked at all)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the final answer that combines these two parts, please refer to this &lt;a href=&quot;https://puzzling.stackexchange.com/a/6180/6178&quot;&gt;answer&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="6178" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2014-12-15T14:35:15.880" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6179" PostTypeId="2" ParentId="6079" CreationDate="2014-12-15T14:04:20.930" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Karin&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Started from last&lt;br&gt;&lt;br&gt;15 says Karin knows Liza, but in 8 Karin says she doesn't know Liza before enrollment, so one of them is wrong. &lt;br&gt;&lt;br&gt;If 15 is wrong, 14 is true and Karin is guilty.&lt;br&gt;&lt;br&gt;If 15 is right, obviously 8 is wrong, and 9 is right which makes Harry the thief. That means Harry's first statement (10) is false, forcing 11 (that states Liza is the thief) to be true, that makes Liza's first statement (13) false, making 14 as true that makes then Karin the thief (This makes the assumption that &quot;15 is right&quot; has been ruled out).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="109" LastEditorUserId="9180" LastEditDate="2015-02-05T22:19:20.913" LastActivityDate="2015-02-05T22:19:20.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6180" PostTypeId="2" ParentId="6175" CreationDate="2014-12-15T14:13:07.107" Score="12" Body="&lt;p&gt;For the riddle, &lt;a href=&quot;https://puzzling.stackexchange.com/a/6178/2484&quot;&gt;please refer to this excellent explanation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For the Chess part:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We assign the letters from VACUUM to the chess pieces..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pieces are placed on: a2,Rb7,Qa1,e2,Kf4,Rg8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suitable new locations for them are: a2,Rb2, Qd7, e3, Kxf5,Rxg4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These new locations satisfy all the rules and result in a checkmate for Black with a safe King.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives a change vector of: $+0,-5,+6,+1,+1,-4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the letters from VACUUM would be transformed to: V V I V V I&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we turn this into Arabic numbers for the pad, we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$551551$ :D&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2014-12-15T14:26:09.920" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="6181" PostTypeId="1" AcceptedAnswerId="10453" CreationDate="2014-12-15T17:47:08.607" Score="16" ViewCount="964" Body="&lt;p&gt;&lt;em&gt;This is the 3rd themed puzzle of the tomb. (See &lt;a href=&quot;https://puzzling.stackexchange.com/questions/5514/can-you-pass-through-the-ancient-tomb-find-the-correct-interpretation-for-the-s&quot;&gt;puzzle 1&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/questions/5923/can-you-solve-this-logic-puzzle-in-disguise-the-color-code-behind-the-tomb&quot;&gt;puzzle 2&lt;/a&gt;) It is fully independent of the other two and just linked by the common story.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;After you set the colour-puzzle dials to the right combination, the whole apparatus gives way to a tunnel underneath. A dark-grey metal-ladder leads downwards into another cavern, and it takes you just one a glance to see, that your ordeal in puzzle solving is not over yet. Built into a green marble wall is another device full of colourful crystals, and odd little turning wheels in two shapes.&lt;/p&gt;&#xA;&lt;p&gt;Without much thinking, you start turning a few of them over, and you can feel that they snap in distinctive steps of 120, respectively 60 degrees. They can also only be turned clockwise. Other than that, nothing seems to happen.&lt;/p&gt;&#xA;&lt;p&gt;After you turned a dozen or so steps on multiple wheels, there is 'clicking' sound and the wheels all start turning backwards. When they stop again, the panel glows greenly and becomes transparent, and you can see that underneath the panel are crystal markings in the same five colours the crystals on the surface have.&lt;/p&gt;&#xA;&lt;p&gt;The transparency fades away, and you repeat your experiment. Again, after 25 'steps' on the wheels, there is the clicking, the sequence of the turns is reversed, and the panel becomes transparent for a few seconds again. You play this game a few times, until you've firmly established that you can turn an arbitrary amount of wheels for an arbitrary amount each, until you reach the total sum of 25 steps. Then the device seems to reset itself to the starting position by reversing the moves, and the starting position is shown through the transparent panel. You also establish that the starting positions is always the same.&lt;/p&gt;&#xA;&lt;p&gt;Convinced that you have to fine the right combination to proceed, you look for further clues around you. Only when you turn you attention to the ceiling of the cavern do you discover the two painted pictures. With a deep sight you start un-puzzling this puzzle. It's too late to turn around anyway.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;The goal of this puzzle&lt;/h1&gt;&#xA;&lt;p&gt;Is to find the right combination of the wheels to turn. The picture of the device below has the wheels labelled for ease of notation.&lt;br&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3salO.jpg&quot; alt=&quot;Device&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Each &amp;quot;turn&amp;quot; of a wheel is denoted by a single appearance of the label, i.e. turning &lt;code&gt;FF2&lt;/code&gt; means that wheel &lt;code&gt;F&lt;/code&gt; is turned twice for one step and afterwards wheel &lt;code&gt;2&lt;/code&gt; is turned for one step (all clockwise).&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;The pictures on the ceiling are:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sTrlF.jpg&quot; alt=&quot;Ceiling #1&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/e3zKI.jpg&quot; alt=&quot;Ceiling #2&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;And the transparent device looks like:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6nysS.jpg&quot; alt=&quot;transparent device&quot; /&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;The solution of this puzzle is the shortest sequence leading from the start situation to the final situation.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I do know a solution of &amp;lt;25 steps exits, but it might be possible to achieve this with &lt;em&gt;less&lt;/em&gt; moves than my solution. If so, the sequence with least steps will be accepted as final solution.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;em&gt;As always, I'm highly interested in thoughts and comments on both the puzzle and its solution.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-06-19T18:06:53.580" Title="Rotationpuzzle in hex - The journey beyond the tomb" Tags="|combinatorics|visual|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6182" PostTypeId="2" ParentId="6167" CreationDate="2014-12-15T18:27:28.187" Score="12" Body="&lt;p&gt;The miscommunication is because...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Netanyahu is speaking Hebrew. (Although I don't know why - his English is very good and he spent some of his childhood years in the US).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Specifically:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Me = מי (Hebrew for 'who' or 'whom')&lt;br&gt;&#xA; He = היא (Hebrew for 'she' or 'her')&lt;br&gt;&#xA; Who = הוא (Hebrew for 'he' or 'him')&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Conversation, as heard by Netanyahu:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Obama: He is on first.&lt;br&gt;&#xA;Netanyahu: Who?&lt;br&gt;&#xA;Obama: The guy playing first base&lt;br&gt;&#xA;Netanyahu: Who?&lt;br&gt;&#xA;Obama: You're the first baseman?&lt;br&gt;&#xA;Netanyahu: No, I am asking you. who?&lt;br&gt;&#xA;Obama: Him!&lt;br&gt;&#xA;Netanyahu: You mean that man right there?&lt;br&gt;&#xA;Obama: She's on first.&lt;br&gt;&#xA;Netanyahu: What do you mean? There aren't any females on the team.&lt;br&gt;&#xA;Obama: Well then he's on first?&lt;br&gt;&#xA;Netanyahu: Who?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And of course, major props to pacoverflow for dusting off one of the best comedy skits of all time.&lt;/p&gt;&#xA;" OwnerUserId="2501" LastEditorUserId="2501" LastEditDate="2014-12-15T19:16:33.577" LastActivityDate="2014-12-15T19:16:33.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6183" PostTypeId="1" AcceptedAnswerId="6201" CreationDate="2014-12-15T18:34:10.793" Score="43" ViewCount="2412" Body="&lt;p&gt;In this puzzle you have to fill with water some cells of the grid. The number of filled squares in each row and column is written next to it.&lt;/p&gt;&#xA;&lt;p&gt;You can't fill a square if the squares connected to it (under, left and right) in the same region are empty. Also, you can't fill a square if the square under it is empty because the water would fall down (you should pay attention to simple physics laws such as gravity).&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Here are 2 wrong examples:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;1 - you can't fill half of a imaginary square:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/2HQdB.png&quot; alt=&quot;you can't fill half of a imaginary square&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;2 - you should pay attention to gravity:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/wD9GF.png&quot; alt=&quot;you should pay attention to simple physics laws&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;And this is the question:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/aOXcy.png&quot; alt=&quot;fill it with the rules mentioned above!&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;This puzzle has been invented by Sharif university's students.&lt;/p&gt;&#xA;&lt;p&gt;EDIT: If OP doesn't mind, here is the grid in ASCII form. Regions are represented by letters A-Z and numbers 0-3.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;AAABCCDDDDEE&#xA;FABBBCGGEEEH&#xA;FAAICCGJJEHH&#xA;IIIIKGGLJJJH&#xA;IMIKKKKLLNNH&#xA;MMMMKOPPLLNQ&#xA;MRROOOPSSNNQ&#xA;TRUOUPPPSSNQ&#xA;TRUUUVWPXYYY&#xA;RRZZZVWWXX0Y&#xA;1ZZ2VVVWX00Y&#xA;1112222W3300&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7320" LastEditorUserId="69582" LastEditDate="2020-11-19T15:17:06.647" LastActivityDate="2020-11-20T15:32:50.290" Title="fill this grid with water!" Tags="|logical-deduction|visual|grid-deduction|" AnswerCount="4" CommentCount="15" ContentLicense="CC BY-SA 4.0" />
  <row Id="6185" PostTypeId="1" AcceptedAnswerId="6555" CreationDate="2014-12-15T21:15:20.817" Score="7" ViewCount="1587" Body="&lt;p&gt;This is Part 6 of the Murder of the President brainteaser/riddle series. If you have not already, check out the answers for&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/3600/murder-of-the-president-part-1&quot;&gt;Murder of the President - Part 1&lt;/a&gt; posted by Nit&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/3845/murder-of-the-president-part-2&quot;&gt;Murder of the President - Part 2&lt;/a&gt; and&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/3977/murder-of-the-president-part-3&quot;&gt;Murder of the President - Part 3&lt;/a&gt; both posted by Joe Z.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/4119/murder-of-the-president-part-4&quot;&gt;Murder of the President - Part 4&lt;/a&gt; posted by el jefe&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/4360/murder-of-the-president-part-5&quot;&gt;Murder of the President - Part 5&lt;/a&gt; posted by Kaivosukeltaja.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Each part will give you a clue and you must solve it. This part will have several clues. Use all knowledge you have of cryptography, ciphers, past puzzles, etc. You should also use Google. This case is meant to take place in the present day, so all politicians, celebrities, places, etc. are who they are now. Please post your answers in spoiler tags.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Here's the riddle:&lt;/p&gt;&#xA;&lt;p&gt;You get to the Four Seasons Hotel in Philadelphia at 4:31 a.m. You immediately get a key for room 109 and you head to the room. You open the door and look around. You check in the bathroom and see a horrible sight.  Two unconscious men are hanging upside down. One has blonde hair and the other has black hair. You check their pulses and feel that they are both still alive. On the mirror you see a sticky note.&lt;/p&gt;&#xA;&lt;p&gt;It reads:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;One of us is Foul&lt;/p&gt;&#xA;&lt;p&gt;The other one Is puRe&lt;/p&gt;&#xA;&lt;p&gt;But to decode some silly meSsages&lt;/p&gt;&#xA;&lt;p&gt;You must firsT find the towel&lt;/p&gt;&#xA;&lt;p&gt;Pcihre +1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You look around and find a towel hanging above the toilet. You take it down and open it up. Two notes fall out.&lt;/p&gt;&#xA;&lt;p&gt;The first one says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Blam! The president is dead&lt;/p&gt;&#xA;&lt;p&gt;Lost was he in the head&lt;/p&gt;&#xA;&lt;p&gt;Over the years I hated him&lt;/p&gt;&#xA;&lt;p&gt;Now my life is very dim&lt;/p&gt;&#xA;&lt;p&gt;Dzwgwosdvk is coded name&lt;/p&gt;&#xA;&lt;p&gt;Even though you must decode it one time more&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The other one says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Blam! The president is dead&lt;/p&gt;&#xA;&lt;p&gt;Lost was he in the head&lt;/p&gt;&#xA;&lt;p&gt;Always thinking must you be&lt;/p&gt;&#xA;&lt;p&gt;Catching me will be tricky&lt;/p&gt;&#xA;&lt;p&gt;Klbmsvwvm is what you must decode, using the hint that you were first told&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Which man do you arrest? Why?&lt;/p&gt;&#xA;&lt;p&gt;Good Luck&lt;/p&gt;&#xA;&lt;p&gt;Hint #1&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Look for the different letters in the riddle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Hint #2&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The first cipher was 7&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Hint #3&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The new cipher is Romans 8 (or -6), but there is a second encryption&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Hint #4&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The second decoding you must use is an anagram&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;em&gt;Note: This is the final part. Hope you all enjoyed the series&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1711" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-12-31T05:51:42.263" Title="Murder of the President - Part 6" Tags="|cipher|mystery|anagram|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6187" PostTypeId="1" AcceptedAnswerId="6191" CreationDate="2014-12-16T01:23:19.160" Score="22" ViewCount="818" Body="&lt;p&gt;Four-star chef and puzzling enthusiast Pat Ravioli is known for her creativity in the kitchen. This morning, Pat gave her assistant a shopping list:&#xA;$\newcommand{q}[1]{{\huge\text{#1}}}$$\newcommand{a}[1]{{\LARGE \text{#1}}}$$\newcommand{z}[1]{{\small \text{#1}}}$$\newcommand{w}[1]{{\huge \mathbb{#1}}}$$\newcommand{s}[1]{{\LARGE \mathbb{#1}}}$$\newcommand{x}[1]{{\small\mathbb{#1}}}$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\z{P}\x{E}\s{P}\a{P}\q{E}\w{R}$&lt;/li&gt;&#xA;&lt;li&gt;$\w{S}\q{A}\z{L}\x{T}$&lt;/li&gt;&#xA;&lt;li&gt;$\z{F}\q{L}\w{O}\x{U}\z{R}$&lt;/li&gt;&#xA;&lt;li&gt;$\w{M}\x{A}\q{Y}\z{O}$&lt;/li&gt;&#xA;&lt;li&gt;$\x{O}\z{R}\a{A}\s{N}\a{G}\q{E}\w{S}$&lt;/li&gt;&#xA;&lt;li&gt;$\w{A}\q{P}\a{P}\s{L}\x{E}\z{S}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What is Pat planning to make?&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2014-12-16T03:29:43.487" LastActivityDate="2014-12-16T06:57:01.090" Title="What's for dinner?" Tags="|visual|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6188" PostTypeId="2" ParentId="6122" CreationDate="2014-12-16T01:50:14.673" Score="3" Body="&lt;p&gt;Congratulations to dmg for getting the &lt;a href=&quot;https://puzzling.stackexchange.com/a/6173/4450&quot;&gt;intended answer&lt;/a&gt;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are my explanations for the clues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first 4 lines are about the word &quot;palindrome&quot; itself.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Within me you will find One of like heart and mind.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A friend is a person of like heart and mind.  Pal is a synonym for friend, and &quot;pal&quot; is a substring of palindrome.  (I thought about using a clue for &quot;palin&quot; instead - as in Sarah Palin - but I thought that would be too U.S.-centric.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As well as an entity With inverted geography.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Countries have cities within them, but there is one example of a city (Rome) having a country (Vatican City) within it.  &quot;Rome&quot; is a substring of palindrome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last 4 lines are about the meaning of the word palindrome.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;While I represent symmetry&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A palindrome by definition has symmetry.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Mirrors can show another identity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you hold a word up to a mirror, it will not look the same, and may form a different word.  For example, the palindrome &quot;bib&quot; turns into &quot;did&quot; in a mirror.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But there will always be twins&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first letter of a palindrome is twins with the last letter.  The second letter is twins with the second-to-last letter, and so on.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Who generate many grins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Many people think palindromes are fun!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the singular form of &quot;boobs&quot; is a palindrome.&lt;/p&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2014-12-16T01:50:14.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6189" PostTypeId="1" AcceptedAnswerId="6190" CreationDate="2014-12-16T03:45:33.760" Score="6" ViewCount="283" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;See thine self through its misty cover &lt;br&gt;&#xA;  and thou wilt see naught within it. &lt;br&gt;&#xA;  See the floor through what it has uncovered &lt;br&gt;&#xA;  and thy will see it naught it all. &lt;br&gt;&lt;br&gt;&#xA;  It's ate 'twixt the bookends of wire and lover&lt;br&gt;&#xA;  though not 'twixt those of ire and love.&lt;br&gt;&#xA;  Work fast, work hard--it is thy reward.&lt;br&gt;&#xA;  Though thine self is naught but it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is &lt;em&gt;it&lt;/em&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hints:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'Twixt means between&lt;br&gt;All mentions of 'it' are referring to the object you are trying to guess&lt;br&gt;The number seven.&lt;br&gt;There is a grammatical error that is made on purpose. Knowing this might help you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7307" LastEditorUserId="7307" LastEditDate="2014-12-16T17:16:46.793" LastActivityDate="2014-12-16T17:16:46.793" Title="See thine self through its misty cover / and thy will see naught within it... What is it?" Tags="|riddle|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6190" PostTypeId="2" ParentId="6189" CreationDate="2014-12-16T04:32:30.933" Score="8" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Water?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See thine self through its misty cover&lt;br&gt;&#xA;and thy will see naught within it. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If you see your reflection on the surface of the water (also mist is water vapour), then the water is clear and has nothing in it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See the floor through what it has uncovered &lt;br&gt;&#xA;and thy will see it naught it all. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;When the bottom of a (former) body of water has been uncovered, the water has been drained away?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's ate 'twixt the bookends of wire and lover&lt;br&gt;&#xA;though not 'twixt those of ire and love.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It is &quot;ate&quot; between &quot;w&quot; and &quot;r&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Work fast, work hard--it is thy reward.&lt;br&gt;&#xA;Though thine self is naught but it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You sweat when you work fast and hard, and your body is mostly water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6178" LastActivityDate="2014-12-16T04:32:30.933" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6191" PostTypeId="2" ParentId="6187" CreationDate="2014-12-16T05:34:32.090" Score="18" Body="&lt;p&gt;She is planning to make&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SCONES&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'll give a series of hints that lead up to the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Only the fonts matter. The words and their contained letters are purely thematic. They could be replaced with any words of the same length.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each letter comes in three different sizes and two different weights (normal and blackboard).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each font can be described by a cell on a 3x2 grid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There's many possible orientations for the grid, but what's the most natural convention for graphing size and for graphing thickness?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Right, it's increasing height on the Y axis, and increasing thickness (normal then blackboard) on the X axis.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 6:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; So, the letters in each word correspond to a sequence of points on this 3x2 grid. What do we do with them?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 7:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Right, we &quot;connect the dots&quot; in order to get letters. They look like &lt;a href=&quot;http://luc.devroye.org/led.html&quot; rel=&quot;noreferrer&quot;&gt;the LCD digital letters&lt;/a&gt; (except the slant of the N connects top left to bottom right).&lt;img src=&quot;https://i.stack.imgur.com/3YahZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And this gives the answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SCONES. Or, if you oriented the grid differently, you recognize that you can reorient it to get that.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="2627" LastEditDate="2014-12-16T06:57:01.090" LastActivityDate="2014-12-16T06:57:01.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6192" PostTypeId="2" ParentId="6183" CreationDate="2014-12-16T06:31:24.883" Score="15" Body="&lt;p&gt;&lt;strong&gt;NOTE, THIS IS A PARTIAL AND INCOMPLETE ANSWER.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First step, lets give a letter to each area (we have 30 areas, so we need to be case-sensitive):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \begin{array}{cccccccccccc}&#xA;z &amp;amp; z &amp;amp; z &amp;amp; C &amp;amp; A &amp;amp; A &amp;amp; D &amp;amp; D &amp;amp; D &amp;amp; D &amp;amp; B &amp;amp; B \\&#xA;y &amp;amp; z &amp;amp; C &amp;amp; C &amp;amp; C &amp;amp; A &amp;amp; w &amp;amp; w &amp;amp; B &amp;amp; B &amp;amp; B &amp;amp; v \\&#xA;y &amp;amp; z &amp;amp; z &amp;amp; u &amp;amp; A &amp;amp; A &amp;amp; w &amp;amp; x &amp;amp; x &amp;amp; B &amp;amp; v &amp;amp; v \\&#xA;u &amp;amp; u &amp;amp; u &amp;amp; u &amp;amp; s &amp;amp; w &amp;amp; w &amp;amp; t &amp;amp; x &amp;amp; x &amp;amp; x &amp;amp; v \\&#xA;u &amp;amp; r &amp;amp; u &amp;amp; s &amp;amp; s &amp;amp; s &amp;amp; s &amp;amp; t &amp;amp; t &amp;amp; p &amp;amp; p &amp;amp; v \\&#xA;r &amp;amp; r &amp;amp; r &amp;amp; r &amp;amp; s &amp;amp; n &amp;amp; m &amp;amp; m &amp;amp; t &amp;amp; t &amp;amp; p &amp;amp; q \\&#xA;r &amp;amp; j &amp;amp; j &amp;amp; n &amp;amp; n &amp;amp; n &amp;amp; m &amp;amp; o &amp;amp; o &amp;amp; p &amp;amp; p &amp;amp; q \\&#xA;k &amp;amp; j &amp;amp; l &amp;amp; n &amp;amp; l &amp;amp; m &amp;amp; m &amp;amp; m &amp;amp; o &amp;amp; o &amp;amp; p &amp;amp; q \\&#xA;k &amp;amp; j &amp;amp; l &amp;amp; l &amp;amp; l &amp;amp; g &amp;amp; c &amp;amp; m &amp;amp; h &amp;amp; i &amp;amp; i &amp;amp; i \\&#xA;j &amp;amp; j &amp;amp; f &amp;amp; f &amp;amp; f &amp;amp; g &amp;amp; c &amp;amp; c &amp;amp; h &amp;amp; h &amp;amp; e &amp;amp; i \\&#xA;a &amp;amp; f &amp;amp; f &amp;amp; b &amp;amp; g &amp;amp; g &amp;amp; g &amp;amp; c &amp;amp; h &amp;amp; e &amp;amp; e &amp;amp; i \\&#xA;a &amp;amp; a &amp;amp; a &amp;amp; b &amp;amp; b &amp;amp; b &amp;amp; b &amp;amp; c &amp;amp; d &amp;amp; d &amp;amp; e &amp;amp; e \\&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, lets number each one accordingly to its water-level:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \begin{array}{cccccccccccc}&#xA;z_3    &amp;amp; z_3 &amp;amp; z_3    &amp;amp; C_2 &amp;amp; A_3 &amp;amp; A_3 &amp;amp; D_1 &amp;amp; D_1 &amp;amp; D_1 &amp;amp; D_1 &amp;amp; B_3 &amp;amp; B_3 \\&#xA;y_2    &amp;amp; z_2 &amp;amp; C_1    &amp;amp; C_1 &amp;amp; C_1 &amp;amp; A_2 &amp;amp; w_3 &amp;amp; w_3 &amp;amp; B_2 &amp;amp; B_2 &amp;amp; B_2 &amp;amp; v_4 \\&#xA;y_1    &amp;amp; z_1 &amp;amp; z_1    &amp;amp; u_3 &amp;amp; A_1 &amp;amp; A_1 &amp;amp; w_2 &amp;amp; x_2 &amp;amp; x_2 &amp;amp; B_1 &amp;amp; v_3 &amp;amp; v_3 \\&#xA;u_2    &amp;amp; u_2 &amp;amp; u_2    &amp;amp; u_2 &amp;amp; s_3 &amp;amp; w_1 &amp;amp; w_1 &amp;amp; t_3 &amp;amp; x_1 &amp;amp; x_1 &amp;amp; x_1 &amp;amp; v_2 \\&#xA;u_{1a} &amp;amp; r_3 &amp;amp; u_{1b} &amp;amp; s_2 &amp;amp; s_2 &amp;amp; s_2 &amp;amp; s_2 &amp;amp; t_2 &amp;amp; t_2 &amp;amp; p_4 &amp;amp; p_4 &amp;amp; v_1 \\&#xA;r_2    &amp;amp; r_2 &amp;amp; r_2    &amp;amp; r_2 &amp;amp; s_1 &amp;amp; n_3 &amp;amp; m_4 &amp;amp; m_4 &amp;amp; t_1 &amp;amp; t_1 &amp;amp; p_3 &amp;amp; q_3 \\&#xA;r_1    &amp;amp; j_4 &amp;amp; j_4    &amp;amp; n_2 &amp;amp; n_2 &amp;amp; n_2 &amp;amp; m_3 &amp;amp; o_2 &amp;amp; o_2 &amp;amp; p_2 &amp;amp; p_2 &amp;amp; q_2 \\&#xA;k_2    &amp;amp; j_3 &amp;amp; l_2    &amp;amp; n_1 &amp;amp; l_2 &amp;amp; m_2 &amp;amp; m_2 &amp;amp; m_2 &amp;amp; o_1 &amp;amp; o_1 &amp;amp; p_1 &amp;amp; q_1 \\&#xA;k_1    &amp;amp; j_2 &amp;amp; l_1    &amp;amp; l_1 &amp;amp; l_1 &amp;amp; g_3 &amp;amp; c_4 &amp;amp; m_1 &amp;amp; h_3 &amp;amp; i_3 &amp;amp; i_3 &amp;amp; i_3 \\&#xA;j_1    &amp;amp; j_1 &amp;amp; f_2    &amp;amp; f_2 &amp;amp; f_2 &amp;amp; g_2 &amp;amp; c_3 &amp;amp; c_3 &amp;amp; h_2 &amp;amp; h_2 &amp;amp; e_3 &amp;amp; i_2 \\&#xA;a_2    &amp;amp; f_1 &amp;amp; f_1    &amp;amp; b_2 &amp;amp; g_1 &amp;amp; g_1 &amp;amp; g_1 &amp;amp; c_2 &amp;amp; h_1 &amp;amp; e_2 &amp;amp; e_2 &amp;amp; i_1 \\&#xA;a_1    &amp;amp; a_1 &amp;amp; a_1    &amp;amp; b_1 &amp;amp; b_1 &amp;amp; b_1 &amp;amp; b_1 &amp;amp; c_1 &amp;amp; d_1 &amp;amp; d_1 &amp;amp; e_1 &amp;amp; e_1 \\&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we take every different symbol as a variable. We can reduce this to a boolean formula with 83 variables:&#xA;$a_1, a_2, b_1, b_2, c_1, c_2, c_3, c_4, d_1, e_1, e_2, e_3, f_1, f_2, g_1, g_2, g_3, h_1, h_2, h_3, i_1, i_2, i_3, j_1, j_2, j_3, j_4, k_1, k_2, l_1, l_2, m_1, m_2, m_3, m_4, n_1, n_2, n_3, o_1, o_2, p_1, p_2, p_3, p_4, q_1, q_2, q_3, r_1, r_2, r_3, s_1, s_2, s_3, t_1, t_2, t_3, u_{1a}, u_{1b}, u_2, u_3, v_1, v_2, v_3, v_4, w_1, w_2, w_3, x_1, x_2, y_1, y_2, z_1, z_2, z_3, A_1, A_2, A_3, B_1, B_2, B_3, C_1, C_2, D_1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets mount the formula. First the gravity constraints, they have the format $(\delta_{i+1} \to \delta_i)$, which can be simplified to $(\lnot\delta_{i+1} \lor \delta_i)$. We have 53 gravity constraints:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \begin{array}{cccccccccc}&#xA;(\lnot a_2 \lor a_1) &amp;amp; \land &amp;amp; (\lnot b_2 \lor b_1) &amp;amp; \land &amp;amp; (\lnot c_2 \lor c_1) &amp;amp; \land &amp;amp; (\lnot c_3 \lor c_2) &amp;amp; \land &amp;amp; (\lnot c_4 \lor c_3) &amp;amp; \land \\&#xA;(\lnot e_2 \lor e_1) &amp;amp; \land &amp;amp; (\lnot e_3 \lor e_2) &amp;amp; \land &amp;amp; (\lnot f_2 \lor f_1) &amp;amp; \land &amp;amp; (\lnot g_2 \lor g_1) &amp;amp; \land &amp;amp; (\lnot g_3 \lor g_2) &amp;amp; \land \\&#xA;(\lnot h_2 \lor h_1) &amp;amp; \land &amp;amp; (\lnot h_3 \lor h_2) &amp;amp; \land &amp;amp; (\lnot i_2 \lor i_1) &amp;amp; \land &amp;amp; (\lnot i_3 \lor i_2) &amp;amp; \land &amp;amp; (\lnot j_2 \lor j_1) &amp;amp; \land \\&#xA;(\lnot j_3 \lor j_2) &amp;amp; \land &amp;amp; (\lnot j_4 \lor j_3) &amp;amp; \land &amp;amp; (\lnot k_2 \lor k_1) &amp;amp; \land &amp;amp; (\lnot l_2 \lor l_1) &amp;amp; \land &amp;amp; (\lnot m_2 \lor m_1) &amp;amp; \land \\&#xA;(\lnot m_3 \lor m_2) &amp;amp; \land &amp;amp; (\lnot m_4 \lor m_3) &amp;amp; \land &amp;amp; (\lnot n_2 \lor n_1) &amp;amp; \land &amp;amp; (\lnot n_3 \lor n_2) &amp;amp; \land &amp;amp; (\lnot o_2 \lor o_1) &amp;amp; \land \\&#xA;(\lnot p_2 \lor p_1) &amp;amp; \land &amp;amp; (\lnot p_3 \lor p_2) &amp;amp; \land &amp;amp; (\lnot p_4 \lor p_3) &amp;amp; \land &amp;amp; (\lnot q_2 \lor q_1) &amp;amp; \land &amp;amp; (\lnot q_3 \lor q_2) &amp;amp; \land \\&#xA;(\lnot r_2 \lor r_1) &amp;amp; \land &amp;amp; (\lnot r_3 \lor r_2) &amp;amp; \land &amp;amp; (\lnot s_2 \lor s_1) &amp;amp; \land &amp;amp; (\lnot s_3 \lor s_2) &amp;amp; \land &amp;amp; (\lnot t_2 \lor t_1) &amp;amp; \land \\&#xA;(\lnot t_3 \lor t_2) &amp;amp; \land &amp;amp; (\lnot u_2 \lor u_{1a}) &amp;amp; \land &amp;amp; (\lnot u_2 \lor u_{1b}) &amp;amp; \land &amp;amp; (\lnot u_3 \lor u_2) &amp;amp; \land &amp;amp; (\lnot v_2 \lor v_1) &amp;amp; \land \\&#xA;(\lnot v_3 \lor v_2) &amp;amp; \land &amp;amp; (\lnot v_4 \lor v_3) &amp;amp; \land &amp;amp; (\lnot w_2 \lor w_1) &amp;amp; \land &amp;amp; (\lnot w_3 \lor w_2) &amp;amp; \land &amp;amp; (\lnot x_2 \lor x_1) &amp;amp; \land \\&#xA;(\lnot y_2 \lor y_1) &amp;amp; \land &amp;amp; (\lnot z_2 \lor z_1) &amp;amp; \land &amp;amp; (\lnot z_3 \lor z_2) &amp;amp; \land &amp;amp; (\lnot A_2 \lor A_1) &amp;amp; \land &amp;amp; (\lnot A_3 \lor A_2) &amp;amp; \land \\&#xA;(\lnot B_2 \lor B_1) &amp;amp; \land &amp;amp; (\lnot B_3 \lor B_2) &amp;amp; \land &amp;amp; (\lnot C_2 \lor C_1) &amp;amp;       &amp;amp;                      &amp;amp;       &amp;amp;                      &amp;amp;       \\                      &amp;amp;       \\&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we should model the constraint for each row. For any given line with $x$ variables, there would be $2^x$ possible ways to value the variables. So, for the 1st and 12th rows there are 32 possible ways to choose the variables for each one. For the 4th row, there are 64 ways. For the 2nd, 5th, 6th, 7th and 10th rows there are 128 ways. For the 3rd, 9th and 11th rows there are 256 ways. For the 8th row there are 512 ways. However, only the ways that satisfy the given number for each line should be considered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1st row:&#xA;$$ \begin{array}{cccccccccccc}&#xA;( &amp;amp;       z_3 &amp;amp; \land &amp;amp;       C_2 &amp;amp; \land &amp;amp; \lnot A_3 &amp;amp; \land &amp;amp; \lnot D_1 &amp;amp; \land &amp;amp; \lnot B_3 &amp;amp; ) &amp;amp; \lor \\&#xA;( &amp;amp; \lnot z_3 &amp;amp; \land &amp;amp; \lnot C_2 &amp;amp; \land &amp;amp;       A_3 &amp;amp; \land &amp;amp; \lnot D_1 &amp;amp; \land &amp;amp;       B_3 &amp;amp; ) &amp;amp; \lor \\&#xA;( &amp;amp; \lnot z_3 &amp;amp; \land &amp;amp; \lnot C_2 &amp;amp; \land &amp;amp; \lnot A_3 &amp;amp; \land &amp;amp;       D_1 &amp;amp; \land &amp;amp; \lnot B_3 &amp;amp; ) &amp;amp;      \\&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4th row:&#xA;$$ \begin{array}{cccccccccccc}&#xA;( &amp;amp;       u_2 &amp;amp; \land &amp;amp;       s_3 &amp;amp; \land &amp;amp; \lnot w_1 &amp;amp; \land &amp;amp; \lnot t_3 &amp;amp; \land &amp;amp; \lnot x_1 &amp;amp; \land &amp;amp; \lnot v_2 &amp;amp; ) &amp;amp; \lor \\&#xA;( &amp;amp;       u_2 &amp;amp; \land &amp;amp; \lnot s_3 &amp;amp; \land &amp;amp; \lnot w_1 &amp;amp; \land &amp;amp;       t_3 &amp;amp; \land &amp;amp; \lnot x_1 &amp;amp; \land &amp;amp; \lnot v_2 &amp;amp; ) &amp;amp; \lor \\&#xA;( &amp;amp;       u_2 &amp;amp; \land &amp;amp; \lnot s_3 &amp;amp; \land &amp;amp; \lnot w_1 &amp;amp; \land &amp;amp; \lnot t_3 &amp;amp; \land &amp;amp; \lnot x_1 &amp;amp; \land &amp;amp;       v_2 &amp;amp; ) &amp;amp; \lor \\&#xA;( &amp;amp; \lnot u_2 &amp;amp; \land &amp;amp;       s_3 &amp;amp; \land &amp;amp;       w_1 &amp;amp; \land &amp;amp;       t_3 &amp;amp; \land &amp;amp; \lnot x_1 &amp;amp; \land &amp;amp;       v_2 &amp;amp; ) &amp;amp; \lor \\&#xA;( &amp;amp; \lnot u_2 &amp;amp; \land &amp;amp;       s_3 &amp;amp; \land &amp;amp; \lnot w_1 &amp;amp; \land &amp;amp;       t_3 &amp;amp; \land &amp;amp;       x_1 &amp;amp; \land &amp;amp; \lnot v_2 &amp;amp; ) &amp;amp; \lor \\&#xA;( &amp;amp; \lnot u_2 &amp;amp; \land &amp;amp; \lnot s_3 &amp;amp; \land &amp;amp;       w_1 &amp;amp; \land &amp;amp; \lnot t_3 &amp;amp; \land &amp;amp;       x_1 &amp;amp; \land &amp;amp; \lnot v_2 &amp;amp; ) &amp;amp; \lor \\&#xA;( &amp;amp; \lnot u_2 &amp;amp; \land &amp;amp; \lnot s_3 &amp;amp; \land &amp;amp; \lnot w_1 &amp;amp; \land &amp;amp;       t_3 &amp;amp; \land &amp;amp;       x_1 &amp;amp; \land &amp;amp;       v_2 &amp;amp; ) &amp;amp;      \\&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;12th row:&#xA;$$ \begin{array}{cccccccccccc}&#xA;( &amp;amp;       a_1 &amp;amp; \land &amp;amp; \lnot b_1 &amp;amp; \land &amp;amp;       c_1 &amp;amp; \land &amp;amp;       d_1 &amp;amp; \land &amp;amp; \lnot e_1 &amp;amp; ) &amp;amp; \lor \\&#xA;( &amp;amp;       a_1 &amp;amp; \land &amp;amp; \lnot b_1 &amp;amp; \land &amp;amp;       c_1 &amp;amp; \land &amp;amp; \lnot d_1 &amp;amp; \land &amp;amp;       e_1 &amp;amp; ) &amp;amp; \lor \\&#xA;( &amp;amp; \lnot a_1 &amp;amp; \land &amp;amp;       b_1 &amp;amp; \land &amp;amp; \lnot c_1 &amp;amp; \land &amp;amp;       d_1 &amp;amp; \land &amp;amp; \lnot e_1 &amp;amp; ) &amp;amp; \lor \\&#xA;( &amp;amp; \lnot a_1 &amp;amp; \land &amp;amp;       b_1 &amp;amp; \land &amp;amp; \lnot c_1 &amp;amp; \land &amp;amp; \lnot d_1 &amp;amp; \land &amp;amp;       e_1 &amp;amp; ) &amp;amp;      \\&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: I did not modeled the other rows. Too tired.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each column there are exact 12 variables, so we can value each column in 4096 different ways. Again, only a few satisfy the given numbers. Further, violations to the gravity constraints should obviously be discarded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: I did not modeled the columns. Too tired.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end we get a boolean formula like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{gravity constraints} \land \text{1st row constraints} \land \text{2nd row constraints} \land \text{3rd row constraints} \land \text{4th row constraints} \land \text{5th row constraints} \land \text{6th row constraints} \land \text{7th row constraints} \land \text{8th row constraints} \land \text{9th row constraints} \land \text{10th row constraints} \land \text{11th row constraints} \land \text{12th row constraints} \land \text{1st column constraints} \land \text{2nd column constraints} \land \text{3rd column constraints} \land \text{4th column constraints} \land \text{5th column constraints} \land \text{6th column constraints} \land \text{7th column constraints} \land \text{8th column constraints} \land \text{9th column constraints} \land \text{10th column constraints} \land \text{11th column constraints} \land \text{12th column constraints}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then we will need to &quot;just&quot; solve that boolean formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To simplify the formula, at the 1st row, we can see that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z_3 = C_2$&lt;br&gt;&#xA;$A_3 = B_3$&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the 12th row, we can see that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b_1 = \lnot a_1$&lt;br&gt;&#xA;$c_1 = a_1$&lt;br&gt;&#xA;$d_1 = \lnot e_1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, with this we can reduce the number of variables from 83 to 78.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: I will stop by now. Too tired.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="5044" LastEditorUserId="5044" LastEditDate="2014-12-16T06:47:19.573" LastActivityDate="2014-12-16T06:47:19.573" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6194" PostTypeId="1" AcceptedAnswerId="6197" CreationDate="2014-12-16T06:44:43.213" Score="40" ViewCount="3177" Body="&lt;p&gt;Breaking news: the NSA can break letter substitution ciphers. (Letter frequency, guessing the word &amp;quot;the&amp;quot;, et cetera...) So, let's make our secret code more secure. Let's apply the secret substitution &lt;em&gt;n&lt;/em&gt; times to the &lt;em&gt;n&lt;sup&gt;th&lt;/sup&gt;&lt;/em&gt; letter of the message.&lt;/p&gt;&#xA;&lt;p&gt;Here's an example. Suppose that the secret substitution was &amp;quot;move forward 1 in the alphabet&amp;quot;. Then the naive substitution would encode &amp;quot;aaa! aa!&amp;quot; by &amp;quot;bbb! bb!&amp;quot;. But our new method encodes it to &amp;quot;bcd! ef!&amp;quot; The same letter gets encoded differently every time. And the substitution needn't be this simple. It could be any permutation of the 26 English letters. Take that, NSA!&lt;/p&gt;&#xA;&lt;p&gt;Anyway, break this cipher. The plaintext is a poem (not written by me) in plain English. The substitution was chosen uniformly at random from all 26! permutations. The algorithm is exactly as described above. I will give upvotes to partial answers, especially theoretical suggestions for attacking the cipher.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;TZTC NQW WEXJLBUE KLTMXFAA TYQQOB COI RAR &lt;br&gt;SNXPBW QJXJ NRVURDVV&#xA;UXVS WKKVJNYMMYR XMP XRHNYAKE &lt;br&gt;MUZX OHI TPHUX BYQSOSRR. URS IRTSRE&#xA;DYSSICLFD OO FVQXVF &lt;br&gt;ZAD VATIMDINN MWKKHZS QP JLI AQDE, &lt;br&gt;JZI GZK&#xA;IBYUDKWIE TAV RPD NN T MHUU EVU. &lt;br&gt;FU TXE GDBNQQ KBQPB ZXW ITXZXNVBC&#xA;DU JLP ZOMHDQFS VDTK FDVEAV &lt;br&gt;&amp;quot;BTOSZWP,&amp;quot; BGZ FY SGP KULGG: EUCNXS&#xA;&lt;br&gt;YRI WSAQ ZWYY, LRUDAA TQ VXWM CC URP LFXC &lt;br&gt;BQFYP MS CIHUFD J&#xA;GHHFK TCD ADTKIF OHE VCNKIM. &lt;br&gt; &lt;br&gt; &lt;br&gt;&amp;quot;QJXJ TFAI JR. XIOFK&amp;quot; VYK&#xA;LFIOUWLTE LMTRDI, &lt;br&gt;MIUWM DLDQJPVV XVUV ZAI LMNN, UDKIIUXLBXD KZPJ&#xA;&lt;br&gt;QMHJDAUIF LVRB QJXJ LVP JUNRXHRHFD AZLB SDHHS &lt;br&gt;NTX XVUV ZAQIS&#xA;WVF, EFFPMXBW KOFS MGI MHJW &lt;br&gt;CTLSSHXKE YMP IJH. QGE XMS VS VVNE&#xA;ACPQ &lt;br&gt;LTIY IRRRB QTLK XOSFDOBW BSXV U IYZNSWP, &lt;br&gt;JHJUMKXWVEJBFLL&#xA;EMVUP. &lt;br&gt;GO, XI OXXXS CQA WDPLCTK RGIDV JERY DMJRL ZYRBK; &lt;br&gt;MIN&#xA;OGN JWK FMCBD IVIFWNKE QJXJ MUPE IAS M NXVZDDVE VWKLQSW: &lt;br&gt;KGCG POF&#xA;AW PAKRR QXDVV IY U KNEZJS? &lt;br&gt;QVH URI NWQIXI VS YMW LEUL DAUO &lt;br&gt;XX&#xA;NQD XEWFYBHH ITJHRNF NE QJB VKLFMQ TJOIYNFD? &lt;br&gt; &lt;br&gt; &lt;br&gt;RLI OPTOUCY&#xA;WUCNXP EYQKSW UDP KOOZI &lt;br&gt;LAF ERIZX FR ODP KTXFKZNW, GZJJNFHF&#xA;MAHEEHHXO FAXWT, &lt;br&gt;YCXFTHTOGI KHRVZ KIRRK YNF HGIPI ZZ BZX &lt;br&gt;MM&#xA;LTS LPVNGIJJ DECJALE LG TXI OWBXKCC: &lt;br&gt;SGDAD ENDHJ XDTF TCB MGMKP&#xA;TIMISTB INQ QAHFTT. &lt;br&gt;OW RAS HNXXAYQ JKVKBJI Y HEXH, VUW TGYES&#xA;LSULDCJOTC, &lt;br&gt;UJW FAFGZC CS RDP WCRH, VYI YJGOKA DYLU RW GZD YBAY;&#xA;&lt;br&gt;DAC HLP RPFWD RNHXZAKC CUQWNW, &lt;br&gt; &lt;br&gt; &lt;br&gt;FKNJEB ZL OGCSCCB&#xA;CTLUDNVEPI, &lt;br&gt;XOW BSDRPI GB VUE PYQEOBW. &lt;br&gt; &lt;br&gt;-- DJYROI, WC&#xA;IFIRAS PFBJF&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1752" LastEditorUserId="69582" LastEditDate="2024-01-03T16:39:23.400" LastActivityDate="2024-01-03T16:39:23.400" Title="Break the subsitution cipher with a twist" Tags="|cipher|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="6195" PostTypeId="1" AcceptedAnswerId="6320" CreationDate="2014-12-16T07:07:20.323" Score="14" ViewCount="21105" Body="&lt;p&gt;A 3x3 magic square is a 3x3 grid containing the numbers 1-9 once each, and in which every row, column, and diagonal sums to 15:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;294&#xA;753&#xA;618&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And I presume we all know the rules to Sudoku.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following completed Sudoku puzzle:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;+---+---+---+&#xA;|&lt;b&gt;&lt;i&gt;294&lt;/i&gt;&lt;/b&gt;|516|738|&#xA;|&lt;b&gt;&lt;i&gt;753&lt;/i&gt;&lt;/b&gt;|842|196|&#xA;|&lt;b&gt;&lt;i&gt;618&lt;/i&gt;&lt;/b&gt;|379|254|&#xA;+---+---+---+&#xA;|571|468|329|&#xA;|936|125|847|&#xA;|482|793|561|&#xA;+---+---+---+&#xA;|345|981|672|&#xA;|827|634|915|&#xA;|169|257|483|&#xA;+---+---+---+&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Notice that the top-left 3x3 portion of the Sudoku solution is a magic square! Likewise, consider this Sudoku solution:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;+---+---+---+&#xA;|8&lt;b&gt;&lt;i&gt;29|4&lt;/i&gt;&lt;/b&gt;75|613|&#xA;|4&lt;b&gt;&lt;i&gt;75|3&lt;/i&gt;&lt;/b&gt;61|289|&#xA;|3&lt;b&gt;&lt;i&gt;61|8&lt;/i&gt;&lt;/b&gt;29|574|&#xA;+---+---+---+&#xA;|197|583|426|&#xA;|643|912|758|&#xA;|258|647|931|&#xA;+---+---+---+&#xA;|782|194|365|&#xA;|514|736|892|&#xA;|936|258|147|&#xA;+---+---+---+&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This solution also has a magic square in the top-left corner. However, notice that it isn't perfectly aligned with one of the nine 3x3 squares that are considered when solving Sudoku. Instead, it's a column to the right of the edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, consider this Sudoku solution:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;+---+---+---+&#xA;|&lt;b&gt;&lt;i&gt;294&lt;/i&gt;&lt;/b&gt;|187|365|&#xA;|&lt;b&gt;&lt;i&gt;753&lt;/i&gt;&lt;/b&gt;|269|814|&#xA;|&lt;b&gt;&lt;i&gt;618|34&lt;/i&gt;&lt;/b&gt;5|279|&#xA;+--&lt;b&gt;&lt;i&gt;-+--&lt;/i&gt;&lt;/b&gt;-+---+&#xA;|82&lt;b&gt;&lt;i&gt;1|59&lt;/i&gt;&lt;/b&gt;4|736|&#xA;|53&lt;b&gt;&lt;i&gt;6|72&lt;/i&gt;&lt;/b&gt;8|941|&#xA;|947|613|582|&#xA;+---+---+---+&#xA;|165|872|493|&#xA;|482|936|157|&#xA;|379|451|628|&#xA;+---+---+---+&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This one has two overlapping magic squares. The first one is in the top-right corner, like in the first Sudoku example. The second magic square shares the 8 in the bottom-right corner of the first magic square, but the rest of its numbers are below and to the right of the first magic square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Counting all magic squares that show up in a Sudoku solution, including ones that have overlapping numbers and that don't align with the 3x3 squares that are checked in a Sudoku puzzle, what is the maximum number of magic squares that can be present in a single Sudoku solution?&lt;/p&gt;&#xA;" OwnerUserId="294" LastEditorUserId="294" LastEditDate="2014-12-16T22:27:18.597" LastActivityDate="2014-12-22T23:28:08.190" Title="Magic Squares in Sudoku Grids" Tags="|logical-deduction|sudoku|magic-square|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6196" PostTypeId="1" CreationDate="2014-12-16T07:39:26.523" Score="9" ViewCount="1000" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&#xA;  We have a couple of cipher puzzles on this site which use primitive, simple ciphers like Caesar, Bacon and substitution ciphers, and some of these puzzles &lt;em&gt;combine&lt;/em&gt; them. &lt;br&gt;&#xA;  This made me wonder about &lt;em&gt;combined&lt;/em&gt; ciphers and what that means for various identification routes in general. The question, however, might be more suitable for &lt;a href=&quot;http://crypto.stackexchange.com&quot;&gt;crypto.stackexchange.com&lt;/a&gt; (as pointed out by Kevin) and a partial answer to my question can be found &lt;a href=&quot;https://security.stackexchange.com/questions/32247/will-double-encryption-increase-the-security-of-cipher-vs-bruteforce&quot;&gt;here&lt;/a&gt; (thanks, dmg).&#xA;  &lt;br&gt;&lt;br&gt;&#xA;  The main motivation for posting this general question at P.SE is the following problem:&lt;br&gt;&lt;br&gt;&#xA;   &lt;strong&gt;How could a cipher puzzle be made more resistant against brute-force tools while at the same time staying a solvable puzzle for the human brain&lt;/strong&gt;? &#xA;  &lt;br&gt;&#xA;  &lt;br&gt; I think/hope that the solution could be to use two ciphers, one of which is rather weak but requires a keyword. It is rather &lt;em&gt;easy&lt;/em&gt; to have a keyword hidden in text so that it is easily recognized by a human, but not by any algorithm. Hence, brute-forcing would only partly help, but it remains a fair puzzle for humans.&#xA;  &lt;br&gt;&#xA;  &lt;br&gt;&#xA;  Think of a &lt;em&gt;clever&lt;/em&gt; puzzle with some cipher to be cracked. The &lt;em&gt;clear text&lt;/em&gt; of this puzzle reads something like &quot;HTEWD SADISDA OASD KLASD NOW USE THE NUMBER BEFORE TWELVE AS KEY FOR SUBSTITUTE ASD ASD ASD ASDADFDFSD&quot; and only applying the second cipher with &quot;eleven&quot; as key gives the clear text. That should be rather straight forward for humans, but rather difficult for brute-force.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It is obvious that any keyword or key (prime) requiring encryption becomes stronger with the length of the keyword/key and that 32bit encryption is &lt;em&gt;way less secure&lt;/em&gt; than 64bit encryption etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The one thing, which hasn't become clear to me is the following: I &lt;em&gt;assume&lt;/em&gt; that brute-force decryption methods work by applying 'random' keys and validating the 'decrypted' message towards some measurement for &quot;clear-text&quot;. This being true, isn't multi-encryption with different ciphers &lt;em&gt;more difficult&lt;/em&gt; to crack, than applying the same cipher with a longer key?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Or to ask the same another way:&lt;/strong&gt; &#xA;Assume I have some cipher which depends on a key and becomes more difficult with longer key. Let's call this the &lt;em&gt;outer&lt;/em&gt; cipher. Now assume I have a another cipher - rather a simple one - which &lt;em&gt;destroys some fundamental text parameters&lt;/em&gt; like letter-frequencies and word-lengths. Let's call this the &lt;em&gt;inner&lt;/em&gt; cipher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if one has a brute-force method for decryption of the &lt;em&gt;outer&lt;/em&gt; cipher - can it even be applied if there is no knowledge of the &lt;em&gt;inner&lt;/em&gt; cipher?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying this in praxis: &lt;strong&gt;&quot;How much security is gained by double-encrypting a text with a simple cipher before using a strong cipher - compared to only using the strong cipher with a longer key?&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="8874" LastEditDate="2015-01-31T18:56:35.387" LastActivityDate="2015-01-31T18:56:35.387" Title="The strength of ciphers against brute force methods - what makes it more difficult but keeps it fair for humans?" Tags="|cipher|puzzle-creation|difficulty-rating|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="6197" PostTypeId="2" ParentId="6194" CreationDate="2014-12-16T08:22:30.263" Score="47" Body="&lt;p&gt;Scroll to the end if you just want the answer. I give a detailed explanation of how I reached it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;First, I hypothesized that the permutation has two cycles: one of length 9, and one length 17. The length-17 one contains T and R, and the length-9 one contains E. How did I get there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the string &lt;code&gt;JR.&lt;/code&gt; at the start of the second paragraph. That's a peculiar two-letter abbreviation. It could be &quot;Mr.&quot; or &quot;Ms.&quot; or &quot;Dr.&quot;. It tempting to thing that the R maps to itself. If so, it must be in a cycle whose length is a factor of 340, its position. That's &lt;code&gt;340=2*2*5*17&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was also struck by the three-letter words &lt;code&gt;TXE&lt;/code&gt; and &lt;code&gt;OHE&lt;/code&gt;. Wouldn't it be cool if they were &lt;code&gt;THE&lt;/code&gt;? The positions are &lt;code&gt;T: 187=11*17&lt;/code&gt;, &lt;code&gt;H:323=17*19&lt;/code&gt;, and &lt;code&gt;E:324=2*2*3*3*3*3, 189=3*3*7&lt;/code&gt;. Isn't the number 17 showing up a lot? It's not a common prime factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To check the theory, I took every 17th letter of the ciphertext&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;LADRHSONEATIHOGYPDHREPIANNFCEYWMOKRIAWDIATTYOKATNLTDMAHCDLRNCPR&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Those sure are nicer letters than usual. Which makes sense, since if there were a cycle of 17, then all 17 letters in that cycle would map to themselves in every 17th position, and so take on their natural frequencies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And taking every 9th letter, &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ETQSJUNXUUREDZISEYVUEQZLSABUSWTRYDDEXIITDVNBHRJDSAFSTMMCRSVJEUXCEYNIJSEGAIQIENHNMDCYOITJEWOIIMIEXETIQAJEEOGCOGCRUECERYRA&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;13.3% of the letters are E, which is totally consistent with the 12.7% in English&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, this suggests a general method to find the cycle sizes. To test a cycle-length n, take every n'th letter, and see how well those letters correspond to the English distribution. Here, I measured this by taking the average English-letter frequency of each letter appearing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UlYEi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see there's peaks at 9 and 17. There are other &quot;overtones&quot; of 9 in its factors or multiples. 18 has the same peak because those letters are a subset of every ninth letter. Every multiple of 3 (shown as vertical gridlines) has 1/3 the peak because 1/3 of those letters are multiples of 9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, I determined approximately what the nine-cycle is. I'm lucky to know the nine-cycle to contain the letter &lt;code&gt;E&lt;/code&gt;, the most common letter in the English alphabet. This means that in positions that are 1 mod 9, E is mapped to the next letter in the cycle, and that letter is likely to be the most common letter. It turns out that letter is &lt;code&gt;P&lt;/code&gt;. Likewise, the most common letter in positions 2 mod 9 is likely to be two after E in the cycle, and so on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gives the claimed cycle &lt;code&gt;EPDKBIDWF&lt;/code&gt;. They're similar to the letters in &lt;a href=&quot;https://puzzling.stackexchange.com/a/6211/4551&quot;&gt;richardb's analysis&lt;/a&gt; But they can't be totally right: &lt;code&gt;D&lt;/code&gt; appears twice. Though &lt;code&gt;E&lt;/code&gt; is the most common letter in English, that might fail to be the case in a random sample of letters, so some of these will be wrong. I guessed that the second &lt;code&gt;D&lt;/code&gt; is wrong and replaced it with the filler character &lt;code&gt;*&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing the nine-cycle let me decrypt all the letters that are part of it. This gave:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;---- --E FE---E-K I----I-- -----D --E --- &#xA;---WED ---- -----I-- ---- KW*-------- --D ------*D &#xA;---- --E -K--- D-------. --- P----E P---E--ED -- I----E &#xA;--E ---F-KI-- -KW*--- -F --E --KE, &#xA;--D --* WI--EEKEK --- -FF -- - ---- W--. &#xA;K- --E -EP--- II-ED --D F------E- I- --E ----I-E- -E-F KI-P-- &#xA;&quot;D----ED,&quot; B-- I- --K B----: I----- &#xA;--D K--- -W--, ---I-- -- --K- -- --E -I-- &#xA;W-E-E -- -E--ED - ---K* --D -E-FED --E ---FE-. &#xA;&#xA;&#xA;&quot;---- -I-E --. -I-*E&quot; --* -EI--B--K ----ED, &#xA;-E-E- D-E--I-- ---- --E ----, -EEPE---B-E E-I- &#xA;----E--ED ---K ---- --D --------ED ---E -I--- &#xA;--- ---- ---K- K-D, DEFE--ED *-E- --D ---E &#xA;-------*D --E K--. --D --- -- ---D --E- &#xA;--E- W---D ---* ---WE-ED W--- - K----ED, &#xA;-----W-E-E-DI-- K---E. &#xA;--, -E ----- --- DIK---P --EI- -E-- F---- ---DE; &#xA;-E- --- -IE B--*K I-KIK-*D ---- --IK W-- - ----IB-E -IE---E: &#xA;B--- W-K -E D-D-- --I-- I- - E-B---? &#xA;--- --E -E-I-K -- --E -E-- F--- &#xA;-- --E -IDD-I-- K-----E -F --E -*-E-- ---E--ED? &#xA;&#xA;&#xA;--D -I----- I----K P--P-K -IK B---D &#xA;--D D-I-- I- -IK B--*E--P, -----I-K ---EF---- D--W-, &#xA;---K-----K E---- BI--E --D --IEK -- F-- &#xA;-- --- -I---I-- FI----E -- --E -EI-D--: &#xA;--I-K E-E-- -I-E --D ---*K -I-K--F F-- ---I--. &#xA;-E --- ------- -D-EE-F - -E--, --D ---E- ----I-----, &#xA;--D D-E--- -- -IK F---, --E ----D- F--- -F --E -E--; &#xA;B-- --W -IDEK ------*- ---I-K, &#xA;&#xA;&#xA;K*--EK -- ------K ----I--EEK, &#xA;--D W-K-EK -E --D D--W-ED. &#xA;&#xA;-- I----K, B- EDW--- FIE-D&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So, there's clearly many things wrong; the &lt;code&gt;K&lt;/code&gt;'s look off in particular. But, there's segments of recognizable text. Particular useful is the last line, which contains &quot;B- EDW--- FIE-D&quot;, which I guessed was &quot;By Edward Field.&quot; Perfect, that's &lt;a href=&quot;http://en.wikipedia.org/wiki/Edward_Field_(poet)&quot; rel=&quot;noreferrer&quot;&gt;the name of a poet&lt;/a&gt;! It also means the letters &lt;code&gt;EDWFI&lt;/code&gt; are right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I looked for a word that had lots of those. There was the third word &quot;FE---E-K&quot;. The &lt;code&gt;K&lt;/code&gt; is surely wrong, and &lt;a href=&quot;http://www.onelook.com/?w=FE???E??&amp;amp;scwo=1&amp;amp;sswo=1&quot; rel=&quot;noreferrer&quot;&gt;plugging the pattern into a regexp matcher&lt;/a&gt; gave the very plausible word &quot;FEATHERS&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, &lt;a href=&quot;https://www.google.com/webhp?sourceid=chrome-instant&amp;amp;rlz=1C1ASUT_enUS413US413&amp;amp;ion=1&amp;amp;espv=2&amp;amp;ie=UTF-8#q=edward+field+feathers&quot; rel=&quot;noreferrer&quot;&gt;Googling &quot;edward field feathers&quot;&lt;/a&gt; was enough to take be to the poem &lt;a href=&quot;https://www.utexas.edu/courses/myth/12.Icarus.htm&quot; rel=&quot;noreferrer&quot;&gt;Icarus&lt;/a&gt; whose text clearly matched.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Only the feathers floating around the hat&#xA;Showed that anything more spectacular had occurred&#xA;Than the usual drowning. The police preferred to ignore&#xA;The confusing aspects of the case,&#xA;And the witnesses ran off to a gang war.&#xA;So the report filed and forgotten in the archives read simply&#xA;“Drowned,” but it was wrong: Icarus&#xA;Had swum away, coming at last to the city&#xA;Where he rented a house and tended the garden.&#xA;&#xA;“That nice Mr. Hicks” the neighbors called,&#xA;Never dreaming that the gray, respectable suit&#xA;Concealed arms that had controlled huge wings&#xA;Nor that those sad, defeated eyes had once&#xA;Compelled the sun. And had he told them&#xA;They would have answered with a shocked,&#xA;uncomprehending stare.&#xA;No, he could not disturb their neat front yards;&#xA;Yet all his books insisted that this was a horrible mistake:&#xA;What was he doing aging in a suburb?&#xA;Can the genius of the hero fall&#xA;To the middling stature of the merely talented?&#xA;&#xA;And nightly Icarus probes his wound&#xA;And daily in his workshop, curtains carefully drawn,&#xA;Constructs small wings and tries to fly&#xA;To the lighting fixture on the ceiling:&#xA;Fails every time and hates himself for trying.&#xA;&#xA;He had thought himself a hero, had acted heroically,&#xA;And dreamt of his fall, the tragic fall of the hero;&#xA;But now rides commuter trains,&#xA;Serves on various committees,&#xA;And wishes he had drowned.&#xA;&#xA;Icarus, by Edward Field&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2014-12-17T03:14:08.447" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6198" PostTypeId="1" CreationDate="2014-12-16T11:07:43.493" Score="14" ViewCount="2203" Body="&lt;p&gt;One of my friends gave me a phrase in a 'secret language' to try and solve. This is very urgent as if I solve it, I get a rather great reward. Here is the phrase: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hu, E fuh'd. E tu hud fecr du dymg du oui. Ev oui lyh nayt drec, cruja&#xA;  uvv. Ev oui lyh'd, drah cruja uvv. Tuh'd cbayg du sa.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7330" LastEditorUserId="4050" LastEditDate="2014-12-16T12:57:00.367" LastActivityDate="2016-10-30T14:14:42.803" Title="Secret &quot;Language&quot;?" Tags="|cipher|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6199" PostTypeId="2" ParentId="6198" CreationDate="2014-12-16T11:14:24.763" Score="31" Body="&lt;p&gt;The decrypted message is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;No, I won't. I do not wish to talk to you. If you can read this, shove off. If you can't, then shove off. Don't speak to me.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The cipher is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A simple substitution, where every letter of the plaintext is mapped to another letter.&#xA;Many websites exist to solve these types of simple ciphers. Example: &lt;a href=&quot;http://quipqiup.com/&quot;&gt;http://quipqiup.com/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7308" LastActivityDate="2014-12-16T11:14:24.763" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="6201" PostTypeId="2" ParentId="6183" CreationDate="2014-12-16T14:09:51.570" Score="25" Body="&lt;p&gt;The solution is as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/Hz0N3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My Java-based solver is viewable &lt;a href=&quot;http://pastebin.com/PzxyXWnC&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solver is a quasi-brute-forcer with efficient backtracking. The major headache was that I initially assumed both &quot;wells&quot; in piece 9 were simultaneously full or empty. However, no solution exists for either of these cases. Much time was spent trying to figure out what was wrong with the blasted solver that kept (correctly) returning &quot;no solution&quot;, &quot;no solution&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After reconsidering my assumption and hard-coding in one of the ugliest hacks in the history of ugly hacks, the solver yielded the above solution in slightly under a second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Special thanks to &lt;a href=&quot;https://puzzling.stackexchange.com/users/7308/tryth&quot;&gt;Tryth&lt;/a&gt; for a flood-fillable graphic, and to &lt;a href=&quot;https://puzzling.stackexchange.com/users/5044/victor&quot;&gt;Victor&lt;/a&gt;, whose lengthy analysis instantly convinced me that converting this problem to SAT (which was my first inclination as well) would not have been a fun way to solve this problem.&lt;/p&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2014-12-16T14:09:51.570" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6202" PostTypeId="2" ParentId="6120" CreationDate="2014-12-16T15:31:27.537" Score="29" Body="&lt;p&gt;The EUREKA moment is to realise that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; some of the words (yes, no, a, the, answer?) represent numbers or mathematical operations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;E.g.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'is' means = and 'without' means -.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Y, N, A, a, t, T denote 'Yes', 'No', 'answer', 'a'/'an', 'to', 'the' respectively.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(I don't know whether the capitalisation in some 'Yes's and 'No's is relevant.) We have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;No, an answer is the cube without A, No? &lt;br&gt;&#xA;The Yes is No. &lt;br&gt;&#xA;The squared answer, Yes, is to an answer. &lt;br&gt;&#xA;A Yes without the Yes is to the answer. &lt;br&gt;&#xA;Yes, the answer is No. &lt;br&gt;&#xA;Yes, the answer to a no is 432. &lt;br&gt;&#xA;What is the answer?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These statements become:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) &lt;em&gt;No, an answer is the cube without A, No?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $N a A = T^3 - a N$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(2) &lt;em&gt;The Yes is No.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $T Y = N$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(3) &lt;em&gt;The squared answer, Yes, is to an answer.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $T^2 A Y = t a N$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(4) &lt;em&gt;A Yes without the Yes is to the answer.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $a Y - T Y = t T A$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(5) &lt;em&gt;Yes, the answer is No.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $Y T A = N$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(6) &lt;em&gt;Yes, the answer to a no is 432.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $Y T A t a N = 432$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that (6) implies&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; none of our six variables can be zero. This will help with cancellation later!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Big spoiler block now with a lot of algebra.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; By (2) and (5), $A=1$. Substitute in $A=1$ and $N=TY$ to get: &lt;br&gt;&#xA; $(1)\Rightarrow Na=T^3-aN \Rightarrow 2TYa=T^3 \Rightarrow T^2=2aY$. &lt;br&gt;&#xA; $(3)\Rightarrow T^2Y=taTY \Rightarrow T=ta$. &lt;br&gt;&#xA; $(4)\Rightarrow aY-TY=tT=t^2a.$ &lt;br&gt;&#xA; $(6)\Rightarrow YTtaTY=432 \Rightarrow Y^2(ta)^2ta=432 \Rightarrow (Y=4, T=3)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So 'the answer' is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $TA =$ &lt;strong&gt;3&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2014-12-18T19:40:25.847" LastActivityDate="2014-12-18T19:40:25.847" CommentCount="21" ContentLicense="CC BY-SA 3.0" />
  <row Id="6203" PostTypeId="2" ParentId="6194" CreationDate="2014-12-16T16:01:25.070" Score="5" Body="&lt;p&gt;Some mixed minor observations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;QJXJ&lt;/code&gt; appears 4 times, for a 4 letter sequence, that's quite unlikely. So this &lt;em&gt;might&lt;/em&gt; be the same word every time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Positions are 24, 331, 416 and 620. Distances to the previous occurence are 307, 85 and 204. That might fit to the hypothesis of &lt;code&gt;xnor&lt;/code&gt; because 85 = 5 * 17 and 204 = 12 * 17. On the other hand, 307 is a prime, which puzzles me, but it could just imply that the first and the second &lt;code&gt;QJXJ&lt;/code&gt; are different words.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are 6 one-letter words (&lt;code&gt;T&lt;/code&gt;, &lt;code&gt;J&lt;/code&gt;, &lt;code&gt;U&lt;/code&gt;, &lt;code&gt;M&lt;/code&gt;, &lt;code&gt;U&lt;/code&gt;, &lt;code&gt;Y&lt;/code&gt;) which most likely map to the set (&lt;code&gt;A&lt;/code&gt;, &lt;code&gt;I&lt;/code&gt;, &lt;code&gt;O&lt;/code&gt;) - &lt;code&gt;O&lt;/code&gt; is quite unlikely, it's often used in poems, but doesn't appear at the beginning of a sentence here.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7338" LastActivityDate="2014-12-16T16:01:25.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6204" PostTypeId="1" CreationDate="2014-12-16T16:08:26.597" Score="23" ViewCount="1640" Body="&lt;p&gt;This is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Cryptogram&quot; rel=&quot;nofollow noreferrer&quot;&gt;cryptogram&lt;/a&gt; puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is some text; each section of the original text ('plaintext') is followed by an encoded version of that text (the 'ciphertext').&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your challenge is to explain the encoding for each section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the ciphers are 'pen-and-paper', meaning that you don't need a computer to decode them; but for at least one of them you &lt;em&gt;will&lt;/em&gt; probably need help from your computer (unless you have a lot of patience to go with your pen and paper). Some of them are easier than others. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may find the &lt;a href=&quot;http://rumkin.com/tools/cipher/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rumkin.com cipher tools&lt;/a&gt; useful, also &lt;a href=&quot;http://cryptii.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;cryptii.com&lt;/a&gt;. Also you might find this post here on Puzzling.SE useful: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/5905/what-characteristics-of-a-ciphertext-can-be-indicators-of-a-particular-cipher&quot;&gt;What characteristics of a ciphertext can be indicators of a particular cipher?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plaintext - &quot;The Hunting of the Snark&quot; by Lewis Carroll, fit 8:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0: Title&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Fit the Eighth &lt;br&gt;&#xA;  The Vanishing &lt;br&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;hthgiE eht tiF&lt;br&gt;&#xA;  Gur Inavfuvat  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;They sought it with thimbles, they sought it with care; &lt;br&gt;    They&#xA;  pursued it with forks and hope; &lt;br&gt; They threatened its life with a&#xA;  railway-share; &lt;br&gt;    They charmed it with smiles and soap. &lt;br&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Moif mbnouz wy ppxz ghzwusiz, nuxg fuilaa ml jikr vhvl;&lt;br&gt;&#xA;  Nuxg cafxnlh ag wzda msyef tvq ncux;&lt;br&gt;&#xA;  Alwl tybxhxlhrw qgy znyl aagh r btppdul-lpnxs;&lt;br&gt;&#xA;  Yalc uuaiwxk ma qvmp fswqxz efq sfki. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;They shuddered to think that the chase might fail, &lt;br&gt;    And the&#xA;  Beaver, excited at last, &lt;br&gt; Went bounding along on the tip of its&#xA;  tail, &lt;br&gt;    For the daylight was nearly past. &lt;br&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Tdam eto f ilste txtnh twse s,edtop,aeet gn,lnnioh&lt;br&gt;&#xA;  uicaeiW  aes.eh  vabgo lr  ht esit &lt;br&gt;&#xA;  s  adkalBenlildnh tiAr uo Fgyhh fhc,gethatr &lt;br&gt;&#xA;  te a  n  yayothd ad trtpdnhi teati &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;There is Thingumbob shouting!&quot; the Bellman said, &lt;br&gt;    &quot;He is&#xA;  shouting like mad, only hark! &lt;br&gt; He is waving his hands, he is&#xA;  wagging his head, &lt;br&gt;    He has certainly found a Snark!&quot; &lt;br&gt;   &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;BAABAAABBBAABAABAAAAAABAA ABAAABAAAB BAABAAABBBABAAAABBAAAABBABAABBABABBAAAABABBABAAAAB BAAABAABBBABBABBAABBBAABAABAAAABBAAAABBA BAABAAABBBAABAA AAAABAABAAABABAABABAABABBAAAAAABBAA BAAABAAAAAABAAAAAABB AABBBAABAA ABAAABAAAB BAAABAABBBABBABBAABBBAABAABAAAABBAAAABBA ABABAABAAAABAABAABAA ABABBAAAAAAAABB ABBABABBAAABABABABBA AABBBAAAAABAAAAABAAB AABBBAABAA ABAAABAAAB BABAAAAAAABAABBABAAAABBAAAABBA AABBBABAAABAAAB AABBBAAAAAABBAAAAABBBAAAB AABBBAABAA ABAAABAAAB BABAAAAAAAAABBAAABBAABAAAABBAAAABBA AABBBABAAABAAAB AABBBAABAAAAAAAAAABB AABBBAABAA AABBBAAAAABAAAB AAABAAABAABAAAABAABAAAAAAABAAAABBAAABABABABBA AABABABBABBAABBABBAAAAABB AAAAA BAAABABBAAAAAAABAAAAABAAB&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;They&#xA;  gazed in delight, while the Butcher exclaimed &lt;br&gt;    &quot;He was always a&#xA;  desperate wag!&quot; &lt;br&gt; They beheld him—their Baker—their hero unnamed—&#xA;  &lt;br&gt;    On the top of a neighbouring crag, &lt;br&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Lwpf ktstw wp pjztubz, lyczp alx Tnfpiqi sazugmqjr &lt;br&gt;&#xA;  &quot;Ss ztk quccdk p noevbazap dez!&quot; &lt;br&gt;&#xA;  Lxxe oqaswb bqw—dastu Utdtd—miqcu sodc vjfpisg— &lt;br&gt;&#xA;  Pf cbr vgg iq y mkmxahflvofw vjcs, &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Erect and sublime, for one moment of time, &lt;br&gt;    In the next, that&#xA;  wild figure they saw &lt;br&gt; (As if stung by a spasm) plunge into a&#xA;  chasm, &lt;br&gt;    While they waited and listened in awe. &lt;br&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;. .-. . -.-. - / .- -. -.. / ... ..- -... .-.. .. -- . --..-- / ..-. --- .-. / --- -. . / -- --- -- . -. - / --- ..-. / - .. -- . --..-- .-.-.. / .. -. / - .... . / -. . -..- - --..-- / - .... .- - / .-- .. .-.. -.. / ..-. .. --. ..- .-. . / - .... . -.-- / ... .- .-- .-.-.. -.--. .- ... / .. ..-. / ... - ..- -. --. / -... -.-- / .- / ... .--. .- ... -- -.--.- / .--. .-.. ..- -. --. . / .. -. - --- / .- / -.-. .... .- ... -- --..-- .-.-.. / .-- .... .. .-.. . / - .... . -.-- / .-- .- .. - . -.. / .- -. -.. / .-.. .. ... - . -. . -.. / .. -. / .- .-- . .-.-.-&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;It's a Snark!&quot; was the sound that first came to their ears,&lt;br&gt;&#xA;  And seemed almost too good to be true. &lt;br&gt; Then followed a torrent of&#xA;  laughter and cheers: &lt;br&gt;    Then the ominous words &quot;It's a Boo—&quot; &lt;br&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;KV'V V RGWZX!&quot; GXI SPN UQCOA UKJY NDZBB RWDJ CW FNKBT ARXR, &lt;br&gt;&#xA;  ELN AJKPDV HHDSNJ FLZ WKQY AP QO IXBK. &lt;br&gt;&#xA;  GZLG XSZJWKKY Y RQAXHSD SU MMKQJZNZ GSF HNKNTG: &lt;br&gt;&#xA;  BVHS DCQ TAARIJT XSEFI &quot;VI'V D VJN—&quot; &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;7.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Then, silence. Some fancied they heard in the air &lt;br&gt;    A weary and&#xA;  wandering sigh &lt;br&gt; That sounded like &quot;-jum!&quot; but the others declare&#xA;  &lt;br&gt;    It was only a breeze that went by. &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;8.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;They hunted till darkness came on, but they found &lt;br&gt;    Not a&#xA;  button, or feather, or mark, &lt;br&gt; By which they could tell that they&#xA;  stood on the ground &lt;br&gt;    Where the Baker had met with the Snark.&#xA;  &lt;br&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Moif bhgbrj hnes hsekeolz gyty bg, fxm bspb ffeah &lt;br&gt;&#xA;  Fgv a nyhgph, hk miyyvye, re atrl, &lt;br&gt;&#xA;  Vr pvvqy yler jslzu fecv ufwa bjlr zxmqr iy wai rchbnw &lt;br&gt;&#xA;  Poipw mvs Eoxxy lgu ayg zeal klx Zrbru. &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;9.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the midst of the word he was trying to say, &lt;br&gt;    In the midst of&#xA;  his laughter and glee, &lt;br&gt; He had softly and suddenly vanished away—&#xA;  &lt;br&gt;    For the Snark was a Boojum, you see. &lt;br&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In the midst of the word he was trying to say, &lt;br&gt;&#xA;  In the midst of his laughter and glee, &lt;br&gt;&#xA;  He had softly and suddenly vanished away— &lt;br&gt;&#xA;  For the Snark was a Boojum, you see. &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Feedback / constructive comments / suggestions on the puzzle and the ciphers themselves are welcome, as well as actual answers. Partial answers are welcome, in the spirit of teamwork.&lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2015-04-07T05:45:38.667" Title="The Hunting of the Snark: Given ciphertext and plaintext, what cipher is used?" Tags="|cipher|" AnswerCount="6" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="6205" PostTypeId="2" ParentId="6204" CreationDate="2014-12-16T16:17:33.083" Score="1" Body="&lt;p&gt;This is &lt;em&gt;work in progress&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Title&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Simply revers. (solved with this &lt;a href=&quot;http://www.wired.com/wp-content/uploads/2014/07/brain1.jpg&quot; rel=&quot;nofollow&quot;&gt;tool&lt;/a&gt; as a helper)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Section #3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Baconian Cipher&lt;/strong&gt; (solved with this &lt;a href=&quot;http://rumkin.com/tools/cipher/baconian.php&quot; rel=&quot;nofollow&quot;&gt;tool&lt;/a&gt; as a helper)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Section #5&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rather obvious Morse code (solved with this &lt;a href=&quot;http://gc.de/gc/morse/&quot; rel=&quot;nofollow&quot;&gt;tool&lt;/a&gt;as a helper)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Section #9&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Is &lt;em&gt;funny&lt;/em&gt;! :c)  ( Replacement cipher with a-&gt;a b-&gt;b c-&gt;c etc.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="5840" LastEditDate="2014-12-16T16:26:26.040" LastActivityDate="2014-12-16T16:26:26.040" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6206" PostTypeId="2" ParentId="6204" CreationDate="2014-12-16T16:18:54.300" Score="6" Body="&lt;p&gt;I guess I'll start with the elephants in the room:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;First Line&lt;/strong&gt; of the &lt;strong&gt;Title&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The text is simply reversed. &quot;Fit the Eighth&quot; becomes &quot;hthgiE eht tiF&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;Second Line&lt;/strong&gt; of the &lt;strong&gt;Title&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A Caesarian Shift Cipher is used. This involves shifting each letter a fixed number of places in the alphabet, in this case 13 places.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For &lt;strong&gt;Paragraph 3&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Bacon's Cipher. Each letter has a 5-bit representation, in As and Bs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For &lt;strong&gt;Paragraph 5&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Morse Code. Each letter has a dot-and-dash representation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For &lt;strong&gt;Paragraph 9&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Caesar Shift of 0?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4748" LastEditorUserId="4748" LastEditDate="2014-12-19T19:27:11.120" LastActivityDate="2014-12-19T19:27:11.120" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6207" PostTypeId="2" ParentId="6204" CreationDate="2014-12-16T16:24:27.063" Score="1" Body="&lt;p&gt;Title&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First line is reversed.&#xA; The second line is simple substitution cipher, shifted by 13.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bacon Cipher&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Morse code&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;9.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Is the same thing?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5030" LastActivityDate="2014-12-16T16:24:27.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6208" PostTypeId="1" AcceptedAnswerId="6210" CreationDate="2014-12-16T17:37:27.610" Score="4" ViewCount="339" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;$\mathfrak{From\&amp;gt;}b\&amp;gt;\mathfrak{to\&amp;gt;}a,\mathfrak{and\&amp;gt;then\&amp;gt;}a\&amp;gt;\mathfrak{to\&amp;gt;}b,\mathfrak{come\&amp;gt;find\&amp;gt;me!}$&#xA;  TWFyayB5b3VyIHNwb3QhIFdhbGsgZm9yd2FyZC4gCkFuZCBmb3J3YXJkLiBBbmQgZm9yd2FyZC4KQW5kIG5vdyB5b3UndmUgZm91bmQgeW91ciBzcG90IG9uIG1lLgpUcnkgYWdhaW4sIGdvaW5nIG92ZXIgdG8gbXkgb3RoZXIgc2lkZSwgYW5kIHdhbGsuCkhhdmUgeW91IGZvdW5kIHlvdXIgc3BvdD8gT2YgY291cnNlIHlvdSBoYXZlLgpXaGF0IGFtIEksIG5vdz8=&#xA;  &lt;br&gt;&lt;br&gt;Even after this, can you find out what I am? For this is only the first step to infinity! But, after a while, you'll end up right where you started.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hints:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first line is the title of the post.&lt;br&gt;The title is a hint (in fact, is &lt;em&gt;essential&lt;/em&gt; to solving the puzzle).&lt;br&gt;It has two parts, the first being decoded the seemingly random arrangement of text. The second part is a riddle.&lt;br&gt;The first part is programming-related&lt;br&gt;The little blurb after the puzzle will help.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7307" LastActivityDate="2014-12-16T17:48:19.180" Title="From $b$ to $a$, and then $a$ to $b$, come find me!" Tags="|riddle|cipher|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6209" PostTypeId="2" ParentId="6208" CreationDate="2014-12-16T17:41:31.643" Score="2" Body="&lt;p&gt;First step:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Converting the ciphertext from&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Base64&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;gives plaintext:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mark your spot! Walk forward. &lt;br&gt;&#xA; And forward. And forward. &lt;br&gt;&#xA; And now you've found your spot on me. &lt;br&gt;&#xA; Try again, going over to my other side, and walk. &lt;br&gt;&#xA; Have you found your spot? Of course you have. &lt;br&gt;&#xA; What am I, now?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why I did it this way:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The '=' symbol at the end of the ciphertext is used as padding to make a base64 string the right length. So if you see an '=' at the end then it's always worth trying base64.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Perhaps the next step might be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Maybe the riddle is talking about travelling up and down the real number line, so a real number: 1 perhaps? That would fit with &quot;first step to infinity&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="4050" LastEditDate="2014-12-16T17:47:07.293" LastActivityDate="2014-12-16T17:47:07.293" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6210" PostTypeId="2" ParentId="6208" CreationDate="2014-12-16T17:48:19.180" Score="5" Body="&lt;p&gt;I'll take a shot at the second step. This is A E's solution to the first:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mark your spot! Walk forward. &lt;br&gt;&#xA; And forward. And forward. &lt;br&gt;&#xA; And now you've found your spot on me. &lt;br&gt;&#xA; Try again, going over to my other side, and walk. &lt;br&gt;&#xA; Have you found your spot? Of course you have. &lt;br&gt;&#xA; What am I, now?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Even after this, can you find out what I am? For this is only the first step to infinity! But, after a while, you'll end up right where you started.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Möbius strip&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After a while of walking, you'll end up right where you started.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; But once you found the spot, if you go the other way, you'll find the spot yet again.&#xA; If you start from the other side, you'll also end up at the spot again&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1477" LastActivityDate="2014-12-16T17:48:19.180" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6211" PostTypeId="2" ParentId="6194" CreationDate="2014-12-16T19:15:17.240" Score="11" Body="&lt;p&gt;I'm taking it as given per xnor's analysis that the substitution forms two cycles, of length 9 and length 17. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose W is substituted by a common letter like E; then the first letter is more likely to be W than by chance. Also the 10th, 19th letter... is more likely to be W, and in general wherever the index mod 9 is 1. Unless that is, it is in the 17 long cycle, when the pattern would be locally pretty much random.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each letter I've calculated the frequency at each position mod 9 and mod 17, and summed the squares to give a crude autocorrelation figure of merit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9BKOl.png&quot; alt=&quot;modulo 9 goodness&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gratifyingly, there are 9 bars that have a good fit, from which I tentatively conclude that the cycles are: &lt;strong&gt;ACGHJLMNOQRTUVXY&lt;/strong&gt;Z and &lt;strong&gt;BDEFIKPSW&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step could be to try to assign an order to BDEFIKPSW which maximises the overall autocorrelation. With such a large ciphertext, you can get a long way with just maths without bringing in English frequency tables.&lt;/p&gt;&#xA;" OwnerUserId="7344" LastEditorUserId="7344" LastEditDate="2014-12-17T00:22:33.560" LastActivityDate="2014-12-17T00:22:33.560" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6212" PostTypeId="1" AcceptedAnswerId="8647" CreationDate="2014-12-16T20:38:32.133" Score="9" ViewCount="3595" Body="&lt;p&gt;A friend of yours was part of a secret society and was put on trial on 15 December 2014 without a lawyer. Alice encrypted the trial's transcript and sent it to Bob, but double agent Carol was able to e-mail a copy of the ciphertext to you. Your friend desperately needed help from a lawyer, but you were unable to provide one with the plaintext so that you could relay the proper advice back to your friend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;RESULTS OF TRIAL:&lt;/strong&gt; Your friend was executed on 26 December 2014, and the rest of your friend's family followed on 20 January 2015.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recently, Carol was able to intercept communications containing a rather ominous message. It seems that your friend's secret society has learned of the work you and your associates have been conducting and are less than pleased. Your life appears to be in jeopardy unless you solve the mystery of your friend's trial by the end of this quarter. If you fail, you will find out if there is an afterlife.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;E-mail:&lt;/strong&gt; &lt;a href=&quot;http://pastebin.com/v1aakvQv&quot; rel=&quot;nofollow&quot;&gt;Show&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you produce the plaintext for the message below to prevent your untimely demise?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ciphertext:&lt;/strong&gt; &lt;a href=&quot;http://pastebin.com/GpWNFdQh&quot; rel=&quot;nofollow&quot;&gt;Show&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After checking your e-mail, you find the following message in your inbox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;E-mail:&lt;/strong&gt; &lt;a href=&quot;http://pastebin.com/4jc9hrGn&quot; rel=&quot;nofollow&quot;&gt;Show&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You receive another e-mail, but the source is completely different from yesterday.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;E-mail:&lt;/strong&gt; &lt;a href=&quot;http://pastebin.com/gC1Dnfse&quot; rel=&quot;nofollow&quot;&gt;Show&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another cryptic e-mail has found its way into your inbox and keeps the theme of being indecipherable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;E-mail:&lt;/strong&gt; &lt;a href=&quot;http://pastebin.com/1SJSnKV3&quot; rel=&quot;nofollow&quot;&gt;Show&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 4:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A programmer working for you named Mallory has come up with a certain theory. He believes that the message sent by Dave is a base 10 representation of a base 27 number. When asked for an explanation, Mallory mumbles something about the power of the trinity.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 5:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While visiting your cubicle to get help on one of his assignments, a new intern named Oscar catches a glimpse of Eve's message on your computer screen and asks, &quot;Is that code from a hex editor?&quot; When you ask what he means, Oscar explains that the message looks like the output his disassembler has been giving him. &quot;Each pair of symbols can be thought of as a base 16 representation of a base 256 number.&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 6:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The IT documentation specialist Peggy has heard about your mysterious e-mails and calls you on your work extension. &quot;Victor, our e-mail administrator, mentioned at lunch last week that our firewall alerted him of some Ascii85 data transferred across the network and referenced a message from someone named Frank.&quot; You suddenly see Mallory peek over your cubicle's wall and give you the A-OK sign.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 7:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consultants by the names of Sam and Sally have been brought in from Luminosum Defensione Ltd. After analyzing all available data, they have come to the conclusion that the attachment in Dave's e-mail is something known as an initialization vector. If they are correct, then it is a critical piece of information for understanding the ciphertext.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 8:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sybil is LDL's top analyst and has noticed a peculiar similarity in Eve's attachment and the ciphertext from Carol. When comparing the two with one above the other, there appears to be a correspondence between certain characters in the two messages. Though the text is unreadable, the structure is visibly mirrored when comparing them both together.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 9:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Among all the bright people at LDL, Trent is the only one having a PhD in information theory. He is the doctor who everyone has been clamoring for, and he has deduced an impossible sounding explanation for the attachment in Frank's e-mail. Trent believes that the pickle contains the instructions to create an eighty-eight dimensional table having a length of fifty-two characters per side.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 10:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An e-mail similar to the one from Tenebris Lamina is showing up as an unread message.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;E-mail:&lt;/strong&gt; &lt;a href=&quot;http://pastebin.com/L571Ggde&quot; rel=&quot;nofollow&quot;&gt;Show&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 11:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This must be the week people send weird e-mails to others ignorant of their existence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;E-mail:&lt;/strong&gt; &lt;a href=&quot;http://pastebin.com/Ds0mcsaK&quot; rel=&quot;nofollow&quot;&gt;Show&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 12:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet another e-mail has landed in your inbox without any explanation of what to do with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;E-mail:&lt;/strong&gt; &lt;a href=&quot;http://pastebin.com/kB3Mymnf&quot; rel=&quot;nofollow&quot;&gt;Show&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 13:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bill is your company's primary lawyer and has a friend named Merlin he thinks might be able to assist with Alice's message. After five minutes of analyzing the message, Merlin exclaims, &quot;This ciphertext has an amber infinity; look at lines 19 and 24!&quot; When asked what an amber infinity is, he explains that &quot;we also call it a double infinity; the 88th alpha character on those lines are the same.&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 14:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Word of your friend's family's fate has been getting around, and a telephone operator in customer service has taken special interest. Paul calls and tells you, &quot;I have been looking through the available data and have a theory. Could there be a connection between the 87 random alpha characters before Merlin's amber infinity and the 87 character primer?&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 15:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A coworker and analyst named Carole has been studying the information gathered so far and has pointed out an interesting trend found in the observations made by the others. &quot;Trent, Merlin, and Paul have all noticed details that appear to be related to an amber infinity. Is there a possibility that lines 19 and 24 of the ciphertext, the primer, and the key are related?&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 16:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Auditors have been hired from Myutsu Mutucsoecul Corporation to verify previous findings for the purpose of utilizing the encryption techniques being studied. Alloy is the investigation leader and seems especially interested in last week's research. &quot;The discovery Merlin made may have been the most valuable clue thus far, and I believe we should explore the connections Carole found.&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 17:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of MMC's auditors has been studying notes from previous weeks and has drawn attention to Trent's theory. &quot;If what he says is true, the encryption table's size would be fifty-two raised to the power of eighty-eight.&quot; Cobalt finally blurts outs, &quot;No computer can store 10,192,817,301,005,542,286,466,232,471,675,496,120,406,975,910,800,636,068,660,404,884,973,826,682,736,169,865,768,988,355,997,729,823,378,974,012,695,621,304,952,079,063,967,426,432,230,164,063,584,256 units of data!&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 18:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another MMC auditor, Blastus, disagrees with Cobalt's conclusion regarding the table believed to exist by the doctor. An alternative interpretation is that if one understands how the instructions should be interpreted and how the resulting table should be used, it may be possible to simulate its function without its construction. Blastus believes that the total space required would probably be less than double the space of the key's pickle.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 19:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those from Nigrum Gladio have reached out to you again but neglected to include an instruction manual.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;E-mail:&lt;/strong&gt; &lt;a href=&quot;http://pastebin.com/fuEkHiwx&quot; rel=&quot;nofollow&quot;&gt;Show&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 20:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another unread e-mail is in your inbox, and if the subject is any indication of its contents, it might be a self-fulfilling prophecy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;E-mail:&lt;/strong&gt; &lt;a href=&quot;http://pastebin.com/bsDuv0ci&quot; rel=&quot;nofollow&quot;&gt;Show&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 21:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With all the e-mails you have been receiving, one might think that there is a certain theme to all this secrecy and madness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;E-mail:&lt;/strong&gt; &lt;a href=&quot;http://pastebin.com/vfb1JWQ1&quot; rel=&quot;nofollow&quot;&gt;Show&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 22:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your company's chief security expert thinks that the ciphertext characters may be related to each other. While testing the hypothesis is impossible, Mallet believes changing a single ciphertext letter would alter all remaining plaintext letters. His study of mathematics has led him to conclude that there is some type of relationship with Markov chains.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 23:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Charlie, the network administrator at this branch, runs continual automated analysis of all internal traffic and has found an interesting note on his reports. &quot;I don't know much about encryption, but the filters monitoring your case's data have been barkin' at me an electronic codebook tag. The information could potentially be processed in blocks at a time.&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 24:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several new employees were hired last week, and Trudy has quickly adjusted to her role as an information specialist. The time she spent working for Sanguine Shield obviously prepared her well for working on your friend's case, but Charlie's revelation yesterday appeared to come as a shock to her. &quot;If the IV is the first chunk's header, shouldn't we expect approximately A / B chunks where A and B are the lengths of the ciphertext and primer, respectively?&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 25:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elite Security Consultation And Protection Enterprise was a small company operated by Pat and Vanna Clingan before merging a couple of weeks ago with your employer. Several years prior, ESCAPE established a business relationship with Cumulus Cote and still has limited access to the systems they were responsible for implementing. Pat is going to submit two similar messages to be encrypted by CC's servers so that the results can be analyzed, the process reverse engineered, and the cryptographic technology understood for future application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Data:&lt;/strong&gt; &lt;a href=&quot;http://pastebin.com/RtP7kPeY&quot; rel=&quot;nofollow&quot;&gt;Show&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 26:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few seconds after Pat submitted the messages for processing, Vanna was able to retrieve the results produced by yesterday's experiment, but several hiccups were encountered along the way. The first problem comes from the fact that while the plaintext could be sent and the ciphertext received, there was no way to set the keys and primers used during the transaction. It seems that ESCAPE did a better job at security than they remembered because the second problem discovered was the inability to gain read access to the key and primer used to generate the text shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Data:&lt;/strong&gt; &lt;a href=&quot;http://pastebin.com/RyLaNvp5&quot; rel=&quot;nofollow&quot;&gt;Show&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 27:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After working for Shiners' Bait &amp;amp; Pet Store for a few years managing their web site and computer systems, Chuck claims he was bored and wanted to put his black hat skills to work for your company. He was hired two weeks ago and immediately put to work getting up to date on your friend's case while also collaborating with the former members of ESCAPE. After seeing the troubles they had with the key and primer, he quickly went to work infiltrating Cumulus Cote's security and was able to find a few backdoors and exploits that allowed him to recover the following information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Data:&lt;/strong&gt; &lt;a href=&quot;http://pastebin.com/gC5bz0pC&quot; rel=&quot;nofollow&quot;&gt;Show&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 28:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While talking with one of your coworkers, your computer mentions &quot;you've got mail&quot; in a pleasing voice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;E-mail:&lt;/strong&gt; &lt;a href=&quot;http://pastebin.com/SZpuKfLd&quot; rel=&quot;nofollow&quot;&gt;Show&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 29:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reasons unknown to you, the people at Egregie Obice still wish to contribute towards your efforts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;E-mail:&lt;/strong&gt; &lt;a href=&quot;http://pastebin.com/bmcHxy3L&quot; rel=&quot;nofollow&quot;&gt;Show&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 30:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe if you can understand this e-mail, you might understand all past, present, and prophetic e-mails.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;E-mail:&lt;/strong&gt; &lt;a href=&quot;http://pastebin.com/3iQh2ysi&quot; rel=&quot;nofollow&quot;&gt;Show&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1661" LastEditorUserId="1661" LastEditDate="2015-03-27T13:34:11.043" LastActivityDate="2015-03-27T13:34:11.043" Title="Justice and Fairness: Part 1" Tags="|cipher|" AnswerCount="4" CommentCount="29" ClosedDate="2015-02-25T23:03:41.780" ContentLicense="CC BY-SA 3.0" />
  <row Id="6213" PostTypeId="2" ParentId="6204" CreationDate="2014-12-16T21:36:07.557" Score="6" Body="&lt;p&gt;&lt;strong&gt;Cipher 1&lt;/strong&gt; is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Vigenère with the key &quot;thehuntingofthesnark&quot;.  The different encodings of &quot;They&quot; tipped me off to look at a polyalphabetic cipher.  After getting the first 4-5 letters of the key by trial and error, the rest of the key was easy to guess.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cipher 2&lt;/strong&gt; is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Decoding the plaintext with &lt;a href=&quot;http://rumkin.com/tools/cipher/skip.php&quot; rel=&quot;nofollow&quot;&gt;Skip 13&lt;/a&gt;.  That is, take every 13th character.  The tool does this in &quot;Decode&quot; mode.  The fact that the ciphertext consisted of exactly the characters of the plaintext made it clear it was some kind of transposition cipher.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When this is used as the plaintext:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;They shuddered to think that the chase might fail,&#xA;   And the Beaver, excited at last,&#xA;Went bounding along on the tip of its tail,&#xA;   For the daylight was nearly past.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I get this ciphertext:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Tdam eto f ilste txtnh twse s,edtop,aeet gn,lnnioh&#xA; uicaeiW  aes.eh  vabgo lr  ht esit&#xA;s  adkalBenlildnh tiAr uo Fgyhh fhc,gethatr&#xA;te a  n  yayothd ad trtpdnhi teati  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The whitespace in neither matches your example, but I suspect browser/Stack Exchange formatting is to blame for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cipher 7&lt;/strong&gt; is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Converted to decimal ASCII codes (84 104 101 110...),&#xA; then spelled out and separated with # (eighty-four#one hundred and four#one hundred and one#one hundred and ten...),&#xA; then converted to decimal ASCII codes again. (101+105+103+104+116+121+45+102+111+117+114+35+111+110+101+32+104+117+110+100+114+101+100+32+97+110+100+32+102+111+117+114+35+111+110+101+32+104+117+110+100+114+101+100+32+97+110+100+32+111+110+101+35+111+110+101+32+104+117+110+100+114+101+100+32+97+110+100+32+116+101+110...)&#xA;&lt;br&gt;&lt;br&gt;Boy, am I glad I didn't do that by hand.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cipher 8&lt;/strong&gt; is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://rumkin.com/tools/cipher/vigenere-autokey.php&quot; rel=&quot;nofollow&quot;&gt;Vigenère Autokey&lt;/a&gt; with a null Alphabet Key (ABCDEFGHIJKLMNOPQRSTUVWXYZ) and passphrase &quot;thehuntingofthesnark&quot;.  Since this started with the same &quot;Moif&quot; for &quot;They&quot; as #2, I figured it must be related.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7346" LastEditorUserId="7346" LastEditDate="2014-12-16T23:23:58.243" LastActivityDate="2014-12-16T23:23:58.243" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6216" PostTypeId="1" AcceptedAnswerId="6217" CreationDate="2014-12-16T22:39:09.560" Score="13" ViewCount="1005" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;'''op I se ,+Ja^u!, noh jo huew moy'''&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;'+)adxa I 'sno!^qo&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7349" LastEditorUserId="4050" LastEditDate="2014-12-17T16:15:34.407" LastActivityDate="2014-12-17T16:15:34.407" Title="What does this piece of text say? '''op I se ,+Ja^u!, noh jo huew moy''' '+)adxa I 'sno!^qo" Tags="|cipher|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6217" PostTypeId="2" ParentId="6216" CreationDate="2014-12-16T23:01:24.237" Score="16" Body="&lt;p&gt;To read this text:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Turn everything upside down. Each character looks (approximately) like another when upside-down:&lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/knXpk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt; The text reads&lt;br&gt; &amp;nbsp;&amp;nbsp; &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;$\text{obvious, I expect,}$&lt;br&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; $\ldots\text{how many of you }`\text{invert' as I do}\ldots$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; iaw +oN &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2014-12-16T23:36:14.193" LastActivityDate="2014-12-16T23:36:14.193" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6218" PostTypeId="1" CreationDate="2014-12-16T23:28:15.697" Score="5" ViewCount="328" Body="&lt;p&gt;Is it possible to have an $n \times n$ magic square with a another magic square of $\frac n 4 \times \frac n 4$ magic square inside it? If so provide an example, if not prove if impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rules:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$n$ cannot be 4&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clarifications:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A magic square is an $n \times n$ grid filled with numbers 1, ..., $n \times n$ (&lt;em&gt;this goes for the inside square too&lt;/em&gt;)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-12-17T02:17:25.617" LastActivityDate="2014-12-17T02:17:25.617" Title="Magic Square with more Magic Squares inside it?" Tags="|magic-square|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6219" PostTypeId="1" AcceptedAnswerId="6222" CreationDate="2014-12-16T23:31:12.703" Score="7" ViewCount="3791" Body="&lt;p&gt;Many methods, if not all, recommend solving it layer by layer, but is it actually possible to solve it face by face? If so, explain how you would do this, if not, prove it impossible.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2020-12-27T23:04:58.373" Title="Is solving the Rubik's cube face by face even possible?" Tags="|rubiks-cube|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6220" PostTypeId="2" ParentId="6138" CreationDate="2014-12-16T23:41:07.643" Score="96" Body="&lt;p&gt;Brilliant puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Puzzle 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;U&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Puzzle 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Puzzle 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;T or P&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Puzzle 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;S or G&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Puzzle 5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I and E&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;They are palindromes, and every word is part of the palindrome too.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Individual solutions&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The beginning, middle and end of the palindrome have been bolded:&lt;br&gt;&#xA;&lt;strong&gt;u&lt;/strong&gt;sing is e desist is i &lt;strong&gt;v&lt;/strong&gt;isits is e design is &lt;strong&gt;U&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;a&lt;/strong&gt;sia is h cuffs is s ender i&lt;strong&gt;s&lt;/strong&gt; i redness is f fuchsia is &lt;strong&gt;A&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;p&lt;/strong&gt;ro-&lt;strong&gt;t&lt;/strong&gt;em and row are h ties is k cohesion is a &lt;strong&gt;c&lt;/strong&gt;asino is e hocks is either a word, name, &lt;strong&gt;T&lt;/strong&gt; or &lt;strong&gt;P&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;g&lt;/strong&gt;ro&lt;strong&gt;s&lt;/strong&gt;s is w or r assign is s amassing is e &lt;strong&gt;d&lt;/strong&gt;esign is s amassing is s arrows is &lt;strong&gt;S&lt;/strong&gt; or &lt;strong&gt;G&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;e&lt;/strong&gt;dna &lt;strong&gt;i&lt;/strong&gt;s i slang is x siege is e basin is x si&lt;strong&gt;n&lt;/strong&gt; is x sin is a besiege is x si gnals is &lt;strong&gt;I&lt;/strong&gt; and &lt;strong&gt;E&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6178" LastEditorUserId="6178" LastEditDate="2014-12-17T02:51:24.717" LastActivityDate="2014-12-17T02:51:24.717" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6221" PostTypeId="1" AcceptedAnswerId="6224" CreationDate="2014-12-16T23:51:03.763" Score="4" ViewCount="249" Body="&lt;p&gt;What in the name of the USSR does this mean?!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;59 103 106 138 -1 106 138 -1 81 -1 135 81 141 103 94 135 -1 106 122 141 94 135 94 138 141 106 122 100 -1 87 125 91 94 -2 -1 91 125 122 -3 141 -1 157 125 144 -1 141 103 106 122 113 -4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It seems so easy, doesn't it? Just another Caesar Cipher, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hints: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $-n$ denotes some sort of punctuation mark.&lt;br&gt;$+n$ denotes a letter.&lt;br&gt;It uses a mathematical constant for encrypting.&lt;br&gt;It also uses a non-injective and surjective function for encrypting.&lt;br&gt;All characters are necessary for the answer (meaning, there are no nulls in it).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7307" LastActivityDate="2014-12-17T00:23:37.683" Title="Not your regular Caesar Cipher$\ldots$" Tags="|calculation-puzzle|cipher|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6222" PostTypeId="2" ParentId="6219" CreationDate="2014-12-17T00:18:12.340" Score="6" Body="&lt;p&gt;The answer is: sort of.&lt;/p&gt;&#xA;&lt;p&gt;Updated to include pictures. I didn't have a 3x3 handy, so I used my 5x5, only moving the edges.&lt;/p&gt;&#xA;&lt;p&gt;Unsolved cube:&lt;br /&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/c12AMm.jpg&quot; alt=&quot;Unsolved cube&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Solve side 1. Same as solving a layer.&lt;br /&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/oVDAwm.jpg&quot; alt=&quot;Side 1&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Solve side 2, adjacent to side 1. (Opposite would mean solving in layers). Again, no real explanation needed.&lt;br&gt;&#xA;Front:&lt;br /&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/J7h7bm.jpg&quot; alt=&quot;Side 2a&quot; /&gt;&lt;br /&gt;&#xA;Back:&lt;br /&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/mzO1gm.jpg&quot; alt=&quot;Side 2b&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Solve side 3, adjacent to both sides 1 and 2. While doing this, one of side 1 and 2 will spend a small amount of time in an unsolved state. But not so much as to consider it unsolved. (At least the way I solve it).&lt;br&gt;&#xA;Front:&lt;br /&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/5d95tm.jpg&quot; alt=&quot;Side 3a&quot; /&gt;&lt;br /&gt;&#xA;Back:&lt;br /&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/RKa4Lm.jpg&quot; alt=&quot;Side 3b&quot; /&gt;&lt;br /&gt;&#xA;Now, because we have 7 corners in the correct position and orientation, the 8th corner is also correct. This means that at most 3 side pieces will be incorrect.&lt;/p&gt;&#xA;&lt;p&gt;Solving side 4 (any of the remaining 3) means that at most 1 piece will be left incorrect. As this is impossible, once the fourth side is solved, sides 5 and 6 will also be solved.&lt;/p&gt;&#xA;" OwnerUserId="375" LastEditorUserId="93" LastEditDate="2020-12-27T18:50:35.710" LastActivityDate="2020-12-27T18:50:35.710" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="6223" PostTypeId="2" ParentId="6218" CreationDate="2014-12-17T00:20:59.253" Score="8" Body="&lt;p&gt;It's possible for any $n$ divisible by 4, other than $n=8$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, find a magic square of size $n/4$. (This is impossible when $n=8$.) Call it $\mathcal{M}$. Now consider the following 4 by 4 magic square $\mathcal{S}$, using the numbers 0 through 15:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left( \begin{array}{ccc}&#xA;8 &amp;amp; 5 &amp;amp; 2 &amp;amp; 15 \\&#xA;3 &amp;amp; 14 &amp;amp; 9 &amp;amp; 4 \\&#xA;13 &amp;amp; 0 &amp;amp; 7 &amp;amp; 10 \\&#xA;6 &amp;amp; 11 &amp;amp; 12 &amp;amp; 1 \end{array} \right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each number $k$ in this square, replace it by a copy of $\mathcal{M}$ with every number increased by $kn^2/16$. One can easily check that the resulting square must be magic, because both of the starting squares are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you've heard of a tensor product or Kronecker product, then this should feel like the product of two magic squares.&lt;/p&gt;&#xA;" OwnerUserId="1752" LastEditorUserId="1752" LastEditDate="2014-12-17T01:43:00.140" LastActivityDate="2014-12-17T01:43:00.140" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="6224" PostTypeId="2" ParentId="6221" CreationDate="2014-12-17T00:23:37.683" Score="5" Body="&lt;p&gt;The solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This is a rather interesting code, don't you think?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How I solved it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It should be obvious that -1 is a space and that -3 is an apostrophe. I replaced the spaces between numbers with exclamation marks, to get this:&lt;pre&gt;59!103!106!138! !106!138! !81! !135!81!141!103!94!135! !106!122!141!94!135!94!138!141!106!122!100! !87!125!91!94!-2! !91!125!122!'!141! !157!125!144! !141!103!106!122!113!-4&lt;/pre&gt; I then mapped all the numbers to a different letter, and replaced all the unknown punctuation with an asterisk, producing this: &lt;pre&gt;abcd cd e fegbhf cighfhdgcij klmh* mli'g nlo gbcip*&lt;/pre&gt; Some substitution analysis later, we have our solution.&lt;br&gt;&lt;br&gt;This was clearly not the intended solution: I suspect the actual method of encipherment revolved around the digits of pi, as the punctuation is mapped to -3, -1 and -4, the first 3 digits of pi, and some suspicious pi sequences appear in the ciphered text (141, for example). But I guess this just goes to show that no matter how clever you are in your method of encipherment, a substitution is just a substitution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7308" LastActivityDate="2014-12-17T00:23:37.683" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6225" PostTypeId="2" ParentId="6219" CreationDate="2014-12-17T00:49:42.363" Score="2" Body="&lt;p&gt;Here I will show my (simplified) reasoning:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For you to solve one face without the risk of &quot;unsolving&quot; it, you MUST solve the opposite side at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you change a square, it isn't solved anymore, even if it is to solve some other face.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This defeats the purpose of solving face-by-face.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conclusion: you can't, but you can solve in parallel sides.&lt;/p&gt;&#xA;" OwnerUserId="3984" LastActivityDate="2014-12-17T00:49:42.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6226" PostTypeId="1" AcceptedAnswerId="8455" CreationDate="2014-12-17T03:28:29.217" Score="10" ViewCount="2017" Body="&lt;p&gt;Like my last post, if no one gets this for 3 weeks, I will post the solution.&lt;br /&gt;&#xA;If no one gets this for 1 week, I will post a hint.&lt;/p&gt;&#xA;&lt;h2&gt;Here is the riddle:&lt;/h2&gt;&#xA;&lt;p&gt;In Korea, a door as bears fall up.&lt;br /&gt;&#xA;In America, a greeting, &amp;quot;Hey dude, what's up?&amp;quot;&lt;br /&gt;&#xA;In Greece, a virgin, fast as an arrow.&lt;br /&gt;&#xA;In Britain, a spy, fitting in with stealth.&lt;/p&gt;&#xA;&lt;p&gt;It rotates one-eighty as Caesar shouts&lt;br /&gt;&#xA;&amp;quot;Keen! Keen! Keen!&amp;quot;&lt;br /&gt;&#xA;Missing a letter by just one step.&lt;br /&gt;&#xA;&amp;quot;Just cease, just cease, just cease!&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;Controls 70% of the earth,&lt;br /&gt;&#xA;With some secrets, ready to be unearthed.&lt;br /&gt;&#xA;The little gold ring spins without a sound.&lt;br /&gt;&#xA;Around and around and around and around,&lt;br /&gt;&#xA;While the songs of the holes sing loudly and clearly,&lt;br /&gt;&#xA;If only, if only, if only&lt;/p&gt;&#xA;&lt;h2&gt;Now for some things you need to know.&lt;/h2&gt;&#xA;&lt;p&gt;The question is, &amp;quot;What am I?&amp;quot;&lt;br /&gt;&#xA;Explain all lines if you can.&lt;br /&gt;&#xA;Lines 2 and 13 are slightly ambiguous.&lt;br /&gt;&#xA;Lines 7 and 14 are clues to other clues.&lt;br /&gt;&#xA;Line 8 doesn't matter&lt;br /&gt;&#xA;Line 1 is the hardest&lt;/p&gt;&#xA;&lt;h1&gt;Hint&lt;/h1&gt;&#xA;&lt;p&gt;The 4th line merely means that the spy is within the word of what the answer is, and you have to be very creative with it.&lt;/p&gt;&#xA;" OwnerUserId="167" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-08T18:16:43.590" Title="A gold ring? A door? What am I?" Tags="|riddle|lateral-thinking|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6227" PostTypeId="2" ParentId="6194" CreationDate="2014-12-17T03:33:30.670" Score="9" Body="&lt;p&gt;Let's assume the frequency analysis done by xnor and richardb is correct, so that the permutation of the alphabet that we are using has a 9-cycle consisting of the list of letters &lt;strong&gt;B, D, E, F, I, K, P, S, W&lt;/strong&gt; (we won't need to assume the existence of a 17-cycle). This means every ciphertext letter on this list corresponds to a plaintext letter on this list, and vice-versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some of the longer ciphertext words, this alone is enough to recover the plaintext, which I determined with a regex dictionary search. For example, the ciphertext &lt;strong&gt;JHJUMKXWVEJBFLL&lt;/strong&gt; can only come from the plaintext &lt;em&gt;uncomprehending&lt;/em&gt;. Similarly, &lt;strong&gt;UDKIIUXLBXD&lt;/strong&gt; must come from &lt;em&gt;respectable&lt;/em&gt;, and &lt;strong&gt;WKKVJNYMMYR&lt;/strong&gt; comes from &lt;em&gt;spectacular&lt;/em&gt; (or &lt;em&gt;spectrogram&lt;/em&gt;, but I considered this unlikely).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Google search for 'poem uncomprehending respectable spectacular' turns up Edward Fields' poem &lt;em&gt;Icarus&lt;/em&gt;, which is seen to be a match.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2014-12-17T16:15:49.450" LastActivityDate="2014-12-17T16:15:49.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6228" PostTypeId="2" ParentId="6226" CreationDate="2014-12-17T04:20:57.843" Score="2" Body="&lt;p&gt;Here's what I got so far. Sounds like:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Artemis&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In America, a greeting, &quot;Hey dude, what's up?&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Artemis sounds like &quot;What time is (it)?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In Greece, a virgin, fast as an arrow.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Obvious Artemis reference.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In Britain, a spy, fitting in with stealth.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;An American spy (but working for Britain) named Artemis by the Germans (&lt;a href=&quot;http://en.wikipedia.org/wiki/Virginia_Hall&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Virginia_Hall&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It rotates one-eighty as Caesar shouts&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Must be Caesar cipher rotated 180 degrees i.e. ROT13&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6178" LastActivityDate="2014-12-17T04:20:57.843" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6229" PostTypeId="1" AcceptedAnswerId="6231" CreationDate="2014-12-17T11:46:42.293" Score="1" ViewCount="6085" Body="&lt;p&gt;For years I have been trying to solve the 3x3x3. Over the last several weeks I was determined to do it.  With the help of several online tutorials and video's I can now solve it.  My average is probably very slow compared to what you can do I am around 4 minute solve now.   I am doing this on a normal off the shelf Rubik's cube its not a speed cube and is slow to turn and locks up from time to time when I try and go to fast.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question may be a bit broad, or opinion based but I really need your help in deciding what I should do next.    Should I go for a speed cube 3x3x3 and try and improve my times or should I try a 4x4x4? or both :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update in response to comment:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is more for me, I have no intention of entering speed contests. It might be fun to be a little faster. I do think I need to memories the last 4 steps I still have to look at my cheat sheets to solve the last face on the 3x3x3. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Background I have been solving every wooden IQ puzzle I have found for the last 10 years.  This is me moving to what I guess are called twisty puzzles. Back in the 80s I had and solved the pyramix, rubik's magic and the snake thing.  I find sitting there and trying to figure them out a challenge and I enjoy sitting on the couch at night just messing with it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am fascinated by the concept of the megaminx but I think it's probably too soon to move to that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an attempt to be less broad, maybe I should ask what I should try next, that won't be too much too soon.&lt;/p&gt;&#xA;" OwnerUserId="7356" LastEditorUserId="4050" LastEditDate="2014-12-17T15:43:18.177" LastActivityDate="2014-12-17T22:26:19.247" Title="Solved Rubik's 3x3x3 - should I do the 4x4x4 next?" Tags="|rubiks-cube|twisty-puzzles|speedsolving|" AnswerCount="5" CommentCount="6" ClosedDate="2014-12-17T22:50:39.673" ContentLicense="CC BY-SA 3.0" />
  <row Id="6230" PostTypeId="2" ParentId="6229" CreationDate="2014-12-17T12:41:34.997" Score="5" Body="&lt;p&gt;The 3x3 and 4x4 are very different. If you had fun in trying to solve the 3x3 by yourself for years and don't want to enter speed contest then you can probably enjoy the same feeling on the 4x4 (or on the 5x5 which might be closer to the 3x3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can try to solve other formats by yourself, it will take time so you can keep practicing on the 3x3 to improve your speed occasionally.&lt;/p&gt;&#xA;" OwnerUserId="1779" LastActivityDate="2014-12-17T12:41:34.997" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6231" PostTypeId="2" ParentId="6229" CreationDate="2014-12-17T13:06:34.983" Score="4" Body="&lt;p&gt;In my opinion you should look up the &lt;a href=&quot;http://www.ws.binghamton.edu/fridrich/system.html&quot; rel=&quot;nofollow&quot;&gt;Fridrich method&lt;/a&gt; for the 3x3x3. At least the first 2 layers (F2L) methodology is really nice in this method. You will see that it is very natural feeling, no algorithms are needed for this. The top layer you can solve with the beginner method then. After a short while you will be able to solve the cube in under a minute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then for the 4x4x4, there is a tricky thing here that is called parity. All cubes with even numbered sides (2x2x2, 4x4x4, 6x6x6) have parity.&#xA;Perhaps an easier thing to do is going straight to the 5x5x5 since it is basically the same technique as the 4x4x4. You bring it to a state where you can solve it as a 3x3x3 cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option is to just take the 4x4x4 and give it a go. You could then ignore the parity and when you understand how it is done, you will be able to solve the 5x5x5 as well. Obviously, this all sounds easier than it is. Practise makes perfect!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And most important: HAVE FUN! :)&lt;/p&gt;&#xA;" OwnerUserId="420" LastEditorUserId="5840" LastEditDate="2014-12-17T13:39:17.860" LastActivityDate="2014-12-17T13:39:17.860" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6232" PostTypeId="2" ParentId="6229" CreationDate="2014-12-17T14:39:15.740" Score="2" Body="&lt;p&gt;My answer is completely subjective, but personally, I find that puzzle-solving can be summarized as &quot;having fun while finding new solutions to problems&quot; (i.e. solutions you don't know).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think that you can &quot;find&quot; a new, more elegant, or faster solution on your own - then keep playing with the 3x3x3. If you feel you are going nowhere with the 3x3x3 or that it starts to get less challenging, then switch. Also, let's not forget that 4x4x4 or other bigger cubes introduce a new concept (parity errors). Also since you are not trying to speed-solve, you don't have any incentive to go easy on yourself with the 3x3x3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also &lt;a href=&quot;https://en.wikipedia.org/wiki/Rubik%27s_Cube#Rubik.27s_Cube_software&quot; rel=&quot;nofollow&quot;&gt;4D and 5D cubes&lt;/a&gt; are apparently available in a virtual environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Disclaimer:&lt;/strong&gt; I have never tried solving a Rubik's cube, as I had to code a solver before I was able to get a real cube. I've never tried 4D and 5D, though.&lt;/p&gt;&#xA;" OwnerUserId="5610" LastActivityDate="2014-12-17T14:39:15.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6233" PostTypeId="2" ParentId="6069" CreationDate="2014-12-17T19:55:50.477" Score="3" Body="&lt;p&gt;&lt;sub&gt;Posting another answer for a different approach.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on xnor's analysis and some experiments I did based on the fact there is only a single &lt;code&gt;.&lt;/code&gt;, I did a bit of analysis on the rare characters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, there's only two &lt;code&gt;?&lt;/code&gt; in the cyphertext, and one of them &lt;em&gt;happens&lt;/em&gt; to be the same distance from the single &lt;code&gt;.&lt;/code&gt; in both the plaintext and the cyphertext.  I conclude that this is simply a coincidence, which means the &lt;em&gt;other&lt;/em&gt; &lt;code&gt;?&lt;/code&gt; character maps to the &lt;code&gt;?&lt;/code&gt; at the end of the plaintext.  So I changed the cypher/plain as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;HA, TITEXGB   DTESTRI AT UE A ONTCTSOMNDEOTA DN UEYCN HRUSN.IETH AEPT CR  F TNGEYI!OI A RCA UODUTYE COHNTUI?EURYI HEPSA T R&#xA;YOU CAN READ THIS CIPHERTEXT! THAT IS GREAT, BUT IT DOES NOT MEAN YOU CAN UNDERSTAND THE CIPHER YET. CAN YOU FIGURE IT OUT?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note the &lt;code&gt;!&lt;/code&gt; used instead of the second &lt;code&gt;?&lt;/code&gt;.  This lets us bring our &quot;unique character&quot; count to 7:  &lt;code&gt;X, M, F, B, ., ?, !&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some basic counting produces: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Char    Value   Cypher  Plain   Diff&#xA;X       88      8       26      18&#xA;M       77      37      61      24&#xA;F       70      74      109     35&#xA;B       66      10      45      35&#xA;.       46      59      99      40&#xA;?       63      107     122     15&#xA;!       33      82      28      69*&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;* This value is counted by moving in the same direction as the others, but it wraps around&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, for thoroughness' sake, here's the last two lines with the &lt;em&gt;other&lt;/em&gt; &lt;code&gt;?&lt;/code&gt; turned into the &lt;code&gt;!&lt;/code&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;?       63      82      122     40 &#xA;!       33      107     28      44 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What does this mean?  No idea yet.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here's a more general breakdown (keeping the &lt;code&gt;?&lt;/code&gt;/&lt;code&gt;!&lt;/code&gt; substitution).  Top line is the position of the cyphered characters, bottom is the plaintext.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(Space)&#xA;  3, 11, 12, 13, 21, 24, 27, 29, 44, 47, 53, 64, 69, 72, 73, 75, 85, 87, 91, 99, 113, 119, 121&#xA;  3, 7, 12, 17, 29, 34, 37, 44, 48, 51, 56, 60, 65, 69, 73, 84, 88, 95, 100, 104, 108, 115, 118&#xA;&#xA;!&#xA;  107&#xA;  28&#xA;&#xA;,&#xA;  2&#xA;  43&#xA;&#xA;.&#xA;  59&#xA;  99&#xA;&#xA;?&#xA;  82&#xA;  122&#xA;&#xA;A&#xA;  1, 22, 28, 43, 65, 86, 90, 118&#xA;  5, 10, 32, 41, 63, 71, 81, 102&#xA;&#xA;B&#xA;  10&#xA;  45&#xA;&#xA;C&#xA;  33, 51, 70, 89, 100&#xA;  4, 18, 70, 89, 101&#xA;&#xA;D&#xA;  14, 39, 45, 94&#xA;  11, 52, 76, 83&#xA;&#xA;E&#xA;  7, 16, 26, 40, 49, 61, 66, 79, 98, 108, 115&#xA;  9, 22, 25, 40, 54, 62, 77, 87, 93, 97, 114&#xA;&#xA;F&#xA;  74&#xA;  109&#xA;&#xA;G&#xA;  9, 78&#xA;  38, 111&#xA;&#xA;H&#xA;  0, 54, 63, 102, 114&#xA;  14, 21, 31, 86, 92&#xA;&#xA;I&#xA;  5, 20, 60, 81, 84, 106, 112&#xA;  15, 19, 35, 49, 90, 110, 116&#xA;&#xA;M&#xA;  37&#xA;  61&#xA;&#xA;N&#xA;  31, 38, 46, 52, 58, 77, 103&#xA;  6, 57, 64, 72, 75, 82, 103&#xA;&#xA;O&#xA;  30, 36, 41, 83, 93, 101&#xA;  1, 53, 58, 67, 106, 119&#xA;&#xA;P&#xA;  67, 116&#xA;  20, 91&#xA;&#xA;R&#xA;  19, 55, 71, 88, 110, 122&#xA;  8, 23, 39, 78, 94, 113&#xA;&#xA;S&#xA;  17, 35, 57, 117&#xA;  16, 36, 55, 79&#xA;&#xA;T&#xA;  4, 6, 15, 18, 23, 32, 34, 42, 62, 68, 76, 96, 104, 120&#xA;  13, 24, 27, 30, 33, 42, 47, 50, 59, 80, 85, 98, 117, 121&#xA;&#xA;U&#xA;  25, 48, 56, 92, 95, 105, 109&#xA;  2, 46, 68, 74, 107, 112, 120&#xA;&#xA;X&#xA;  8&#xA;  26&#xA;&#xA;Y&#xA;  50, 80, 97, 111&#xA;  0, 66, 96, 105&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="325" LastEditorUserId="325" LastEditDate="2014-12-17T21:44:50.960" LastActivityDate="2014-12-17T21:44:50.960" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6234" PostTypeId="2" ParentId="5714" CreationDate="2014-12-17T20:59:31.060" Score="3" Body="&lt;p&gt;We've seen 'inside the heads' of the German and the two women, so we know none of them performed either the kiss or the punch. Thus the Frenchman must have done both, and the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Frenchman kissed his fist and punched the Nazi in the eye.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(mdc32 has already &lt;a href=&quot;https://puzzling.stackexchange.com/questions/5714/four-people-in-a-dark-lift/5715#5715&quot;&gt;provided the answer&lt;/a&gt;, but I wanted the logical explanation included for the record.)&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2014-12-17T20:59:31.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6235" PostTypeId="1" AcceptedAnswerId="6296" CreationDate="2014-12-17T21:50:11.423" Score="5" ViewCount="541" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Lo! I heard you have gone to war.&lt;br&gt;&#xA;  Alluding to my utter demise.&lt;br&gt;&#xA;  Trees are what you used to fear,&lt;br&gt;&#xA;  blobs of nothing, nothing is best.&lt;br&gt;&#xA;  Return to the start, and commence to walk.&lt;br&gt;&#xA;  Do it no more than once, down these rickety stairs.&lt;br&gt;&#xA;  Take a notice, ignore the spaces in between.&lt;br&gt;&#xA;  No, doomed you are not, lest you count the punctuation.&lt;br&gt;&#xA;  (Count the space here, and ignore the first-born!)&lt;br&gt;&#xA;  Now, talk, flail, and stop right here.&lt;br&gt;&lt;br&gt;&#xA;  Now that you've got what I've meant to say, double check.&lt;br&gt;&#xA;  I would warn you of anything, coming at your door.&lt;br&gt;&#xA;  Trust me, trust me! And say it once more.&lt;br&gt;&#xA;  So trust me, trust me! Did you forget to double-check?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hints:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Do not take the first stanza literally. Very few words in it help you, but oh, how important they are!&lt;br&gt;The second stanza is really just emphasizing a portion of the first, and does describe a bit about the object.&lt;br&gt;I think I've said to much $\tiny\text{ But look at the tag. }$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7307" LastEditorUserId="7307" LastEditDate="2014-12-19T12:20:20.873" LastActivityDate="2014-12-19T13:54:01.407" Title="Did you forget to double-check?" Tags="|riddle|word|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6236" PostTypeId="2" ParentId="6120" CreationDate="2014-12-17T21:53:48.380" Score="3" Body="&lt;p&gt;Following rand and reading this as a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;set of simultaneous equations,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and ignoring&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;capitalisation and punctuation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the hint about grammatical correctness suggests that capitalisation and punctuation are here concerns of presentation only,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;try the substitutions&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;no&quot; $\rightarrow$ N&lt;br&gt;&#xA; &quot;an&quot;, &quot;a&quot; $\rightarrow$ 1&lt;br&gt;&#xA; &quot;answer&quot; $\rightarrow$ S&lt;br&gt;&#xA; &quot;is&quot; $\rightarrow$ =&lt;br&gt;&#xA; &quot;the&quot; $\rightarrow$ 3&lt;br&gt;&#xA; &quot;cube&quot;, &quot;cubed&quot; $\rightarrow$ ^3&lt;br&gt;&#xA; &quot;without&quot; $\rightarrow$ - (minus)&lt;br&gt;&#xA; &quot;yes&quot; $\rightarrow$ Y&lt;br&gt;&#xA; &quot;to&quot; $\rightarrow$ 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We get&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;N S = 27 - A N&lt;br&gt;&#xA;3 Y = N&lt;br&gt;&#xA;27 S Y = 2 S&lt;br&gt;&#xA;A Y - 3 Y = 2 3 S&lt;br&gt;&#xA;Y 3 S = N&lt;br&gt;&#xA;Y 3 S 2 N = 432,  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;NS = 27 - AN&lt;br&gt;&#xA;3Y = N&lt;br&gt;&#xA;27 Y = 2 (unless S=0)&lt;br&gt;&#xA;Y(A-3) = 6S&lt;br&gt;&#xA;3YS = N&lt;br&gt;&#xA;6 YSN = 432 =&gt; YSN = 72 (so S can't be 0)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which unfortunately&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;is a set of equations without a solution  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Might a eureka moment with the &quot;little is large&quot; hint straighten up the substitutions?&lt;/p&gt;&#xA;" OwnerUserId="7363" LastEditorUserId="7363" LastEditDate="2014-12-17T22:09:44.480" LastActivityDate="2014-12-17T22:09:44.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6237" PostTypeId="2" ParentId="6229" CreationDate="2014-12-17T22:11:46.260" Score="1" Body="&lt;p&gt;I've been able to solve the 3x3x3 cube for probably 15 years, always using the same basic algorithm.  I rarely pick it up anymore.  Perhaps 3-4 years ago (maybe longer, can't remember) I decided to go for the 4x4x4.  I had some trouble applying my standard technique to the new cube, so I went back and looked up speed cubing techniques for the 3x3x3.  Got proficient enough with that to be able to complete the 3x3x3 with the new method (not particularly concerned with my speed), and then went back to the 4x4x4 with my new techniques, and solved it successfully.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That was a satisfying puzzle experience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can't tell you what you should do, but I'll say this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I didn't feel the need to pick up the 4x4x4 for years.  I was always curious, but then just felt the need for it.&lt;/li&gt;&#xA;&lt;li&gt;Didn't particularly feel the need for the 5x5x5 yet (just curious), but was given it on a subsequent Christmas.  Still haven't spent any significant time with it.  I will someday though.&lt;/li&gt;&#xA;&lt;li&gt;I've forgotten everything but my original tried-and-true basic algorithms.  I do want to go back and re-learn the speed cubing techniques, as it's more fun to do something new and different.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Hope that helps!&lt;/p&gt;&#xA;" OwnerUserId="358" LastActivityDate="2014-12-17T22:11:46.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6238" PostTypeId="2" ParentId="6229" CreationDate="2014-12-17T22:26:19.247" Score="1" Body="&lt;p&gt;As someone who can solve a 3x3x3 but only after learning an iterative technique I would say that it is a worthwhile skill. I find it can be fun to impress people when I find one lying around their house.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought about learning a 4x4x4 but for me I just think it is going a bit far to &quot;mildly surprise&quot; people. I get enjoyment from figuring out a puzzle and so learning the steps online and applying them seems a bit like doing a crossword and using the answers just so I can leave it on my seat when I'm done to impress the next person. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, being able to solve one will put you in a small percent of the world who can, but unless you get pleasure and satisfaction from doing it then I wouldn't bother.&lt;/p&gt;&#xA;" OwnerUserId="7364" LastActivityDate="2014-12-17T22:26:19.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6239" PostTypeId="1" AcceptedAnswerId="6244" CreationDate="2014-12-17T22:57:27.187" Score="53" ViewCount="4371" Body="&lt;p&gt;There, spread out all over the floor, were the mechanical pieces of a robot. Something had torn it apart, and the debris spanned a radius of a few meters. There was no sign of what might have caused this tragedy. What monster, what disaster, could cause this much damage to a mechanical creature?&lt;/p&gt;&#xA;&lt;p&gt;However, a few steps away, there was a hastily constructed circuit:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/our4d.png&quot; alt=&quot;What happened here?&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;And one question nags at you:&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;What was the robot trying to tell us with his last words?&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="7308" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2017-01-28T05:12:56.737" Title="A Robot's Last Words" Tags="|logical-deduction|computer-puzzle|circuitry|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6240" PostTypeId="1" AcceptedAnswerId="6311" CreationDate="2014-12-17T23:26:57.380" Score="20" ViewCount="1919" Body="&lt;p&gt;I dropped in to Ernie's place a while back and he asked me to stay and help calibrate his new Drilling Device. It was an impressive looking machine consisting of a large raygun-like thing mounted on an x-y translation table. An electronic eye was supported above the table, and to one side was a box with two buttons, one red and one blue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ernie explained that he was sick of drilling countless holes in sheets of steel manually, so he had built this tool to do the job automatically. &quot;The extraordinarily accurate x-y table has precisely 1000 mm x 1000 mm of travel, so I can cut holes anywhere within an area of one square meter - accurate to the nearest micron. The drill uses an incredibly powerful laser to drill the holes so it takes absolutely no time at all to make the hole. The translation stage has enormously powerful, but restrained, drive motors so it can move the drill between any two locations in the one metre square area at a speed of precisely 1 m/s.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;But how does it know where to drill the holes?&quot;, I asked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Ahh&quot;, Ernie replied, &quot;I felt that an analogue approach was more desirable than a digital one for this job, so you just mark where you want to put the holes on the steel sheet with these two special pens. You draw a circle with the red pen and that is where the drill puts the first hole, and you add a series of crosses with the blue pen to tell the drill where to put the rest of the holes. The electro-mechanical brain works out what path to follow and when to drill&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IQ5L8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It felt like a typical Ernie invention - weirdly simple in some ways, but bizarrely complicated in other ways. &quot;Tell me more?&quot;, I asked, &quot;what do the two buttons do?&quot;. And Ernie explained that each button followed a very exact algorithm.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Red button: &#xA;If a red circle exists:   &#xA;   Move the drill, in a straight line, from its current position to the centre of the red circle.&#xA;&#xA;Blue Button:&#xA;Drill a hole.&#xA;While un-drilled blue crosses exist&#xA;   Move the drill, in a straight line, to the centre of the nearest blue cross,&#xA;   Drill a hole.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Ernie asked me to mark a plate so we could test the system. I marked the first plate with exactly one red circle and five blue crosses (the plate was a bit oversize so there was plenty of room to mark all the drill-points within the central 1000 mm x 1000 mm square that the drill could reach.). We mounted the plate and Ernie pressed the Red button - the drill moved until it was exactly over the red cross. Then he pressed the Blue button and simultaneously started a stop-watch. There were 6 bright flashes (and a little more flying molten steel than I had expected) as the drill made the holes in the plate. At the exact moment the sixth hole was made, Ernie stopped the watch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Hmmm&quot;, he exclaimed, and then very carefully measured the positions of the holes in the plate with a ruler. &quot;You seem, quite by chance of course, to have chosen hole positions that maximize the time required for the drill to complete the Blue algorithm for six holes!&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This only happened a few weeks ago ago but Ernie never told me exactly how long it took the Blue algorithm to drill the six holes, or where (exactly) in the steel in the sheet I drew the red circle. Can you help me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that is all to easy, I plan to visit Ernie around Christmas and would love to play with his head by &quot;quite by chance&quot; marking out a seven hole plate which maximizes the Blue algorithm time, so knowing where to mark the red circle and six blue crosses would be useful. Ernie only uses his second-best stop-watch and ruler for this kind of job, so positions accurate to the nearest micron, and times to the nearest microsecond should be sufficient. &lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="1631" LastEditDate="2014-12-18T19:59:28.373" LastActivityDate="2021-06-30T04:28:52.367" Title="Ernie and the Drill Test" Tags="|mathematics|calculation-puzzle|geometry|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6241" PostTypeId="1" AcceptedAnswerId="6242" CreationDate="2014-12-17T23:37:23.463" Score="44" ViewCount="3591" Body="&lt;p&gt;There is a seven-story skyscraper. Each floor hosts exactly one company.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pangloss Washers Co, Ltd. is located on the 7th floor.&lt;br&gt;&#xA;Alice PC Chips Co., Ltd. is located on the 6th floor.&lt;br&gt;&#xA;Jumanji Arcade Co., Ltd. is located on the 5th floor.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Colorful Crafts Co., Ltd. is located on the ___th floor. Can you fill in the blank?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Credit: this puzzle idea appears in Umineko.)&lt;/p&gt;&#xA;" OwnerUserId="1752" LastEditorUserId="4536" LastEditDate="2014-12-19T19:52:41.300" LastActivityDate="2015-01-14T17:37:55.860" Title="Which floor does this company occupy?" Tags="|lateral-thinking|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6242" PostTypeId="2" ParentId="6241" CreationDate="2014-12-17T23:43:32.920" Score="45" Body="&lt;p&gt;The answer has to be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;4&lt;/strong&gt; (the 4th floor).  It can't be on the 1st, 2nd, or 3rd floor since those don't end in &quot;th&quot; and you have &quot;___th&quot; in your question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2014-12-17T23:43:32.920" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6243" PostTypeId="2" ParentId="6240" CreationDate="2014-12-18T00:24:22.693" Score="2" Body="&lt;p&gt;Here are my punts for six and seven holes. The lengths are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Six holes: $2+\sqrt 2+2\sqrt{2-\sqrt{3}}=4.449...$ meters&lt;br&gt;&#xA;Seven holes: $2+ \sqrt 2+3\sqrt{2-\sqrt{3}}=4.967...$ meters&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, six holes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rHH7g.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is a unit square, then the non-corner holes are at $A=(\frac 12, \frac {\sqrt{3}}{2})$ and $B=(\frac {\sqrt{3}}{2}, \frac 12)$. Many distances are equal, so adjust the hole positions by a very tiny amount so that the drill path is the red path shown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now seven holes. Just add one more hole, symmetrically, at $C=(\frac 12, 1-\frac {\sqrt 3}{2})$, and then change $B$ to the red hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/f6EH2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, adjust the hole positions by a very tiny amount so that the drill path is the red path shown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reference, if we add an eighth point to complete the symmetry (shown below), then all the red distances and all the blue distances are equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KoVWj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1752" LastEditorUserId="1752" LastEditDate="2014-12-18T00:56:50.123" LastActivityDate="2014-12-18T00:56:50.123" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6244" PostTypeId="2" ParentId="6239" CreationDate="2014-12-18T00:32:36.077" Score="35" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;KEEPAWAY&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the functions for the output (in Java).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;Output 1 = !((!C)|C)&#xA;Output 2 = (!C)|C&#xA;Output 3 = !((!C)|C)&#xA;Output 4 = (A|B)&amp;C&#xA;Output 5 = (!(A^B))&amp;(!(B^C))&#xA;Output 6 = ((!A)&amp;(B^C))|((A&amp;C)&amp;(!B))&#xA;Output 7 = ((A&amp;C)&amp;(!B))|(!((B|C)|A))&#xA;Output 8 = (!(B&amp;C))|A&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Using these functions &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;you can create a Truth table:&#xA;&lt;code&gt;&#xA;|A|B|C|1|2|3|4|5|6|7|8|Hex|Char|&#xA;--------------------------------&#xA;|0|0|0|0|1|0|0|1|0|1|1|4b |K   |&#xA;|0|0|1|0|1|0|0|0|1|0|1|45 |E   |&#xA;|0|1|0|0|1|0|0|0|1|0|1|45 |E   |&#xA;|0|1|1|0|1|0|1|0|0|0|0|50 |P   |&#xA;|1|0|0|0|1|0|0|0|0|0|1|41 |A   |&#xA;|1|0|1|0|1|0|1|0|1|1|1|57 |W   |&#xA;|1|1|0|0|1|0|0|0|0|0|1|41 |A   |&#xA;|1|1|1|0|1|0|1|1|0|0|1|59 |Y   |&#xA;&lt;/code&gt;&#xA;From this truth table it is obvious what the message is.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7312" LastEditorUserId="7312" LastEditDate="2014-12-18T15:32:16.863" LastActivityDate="2014-12-18T15:32:16.863" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="6248" PostTypeId="2" ParentId="6195" CreationDate="2014-12-18T12:11:30.807" Score="-3" Body="&lt;p&gt;Make every square in one column contain the same. You can get another 4 squares from each then. This is 12 + 9 = 21 then. Same with rows → 33&lt;/p&gt;&#xA;" OwnerUserId="5106" LastEditorUserId="5106" LastEditDate="2014-12-19T08:26:00.497" LastActivityDate="2014-12-19T08:26:00.497" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6250" PostTypeId="1" AcceptedAnswerId="6251" CreationDate="2014-12-18T13:12:45.307" Score="11" ViewCount="4216" Body="&lt;p&gt;Imagine an $n \times n$ grid filled with the numbers 1, ..., $n$&#xA;where $n$ &gt; 3 each number appearing n times, where each row, column, and diagonal all equal the same number. Can you fill grid like this? If so, show an example, and include as much detail as possible (for example if there's only 1 solution). If not, provide mathematical proof it's impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I've purposely excluded $1 \times 1$ due to simplicity, $2 \times 2$ due to impossibility and $3 \times 3$ also due to impossibility (for solvers like you!)&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-12-19T16:24:05.280" LastActivityDate="2022-06-24T06:45:06.620" Title="The Quite Unusual Square" Tags="|magic-square|" AnswerCount="6" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6251" PostTypeId="2" ParentId="6250" CreationDate="2014-12-18T13:32:32.423" Score="9" Body="&lt;p&gt;Nice idea for a puzzle, reminds me of &lt;a href=&quot;http://en.wikipedia.org/wiki/Sudoku&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sudoku&lt;/a&gt; in a way.&#xA;An answer for 4x4 (Simple I know but I have to start somewhere):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EwGJz.png&quot; alt=&quot;4x4 Image&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How I found it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If the total of each row, column and diagonal must be the same, then it follows that the total for an $n$ x $n$ square must be $n^2 /2 + n/2$. This is achieved with one of each of the numbers $1, ... n$ appearing in every row, column and diagonal.&lt;br&gt;&lt;br&gt;To find my answer, I started by filling in a diagonal with numbers 1-4, then simply filled in the remainder through trial and error.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fear not, more answers will come!&#xA;I extended the method I used for 4x4 and solved the 5x5 this time:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/U684T.png&quot; alt=&quot;5x5 Image&quot;&gt;&#xA;&lt;br&gt;This time I started again by filling in the diagonal, but then filled in the 3s in a symmetrical pattern. After that I filled in the 1s and then rotated that design by 180 degrees and used it again for the 5s. Lastly I solved the 2s and 4s in the same way.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, to summarise:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Filling a $n \times n$ grid with the above rules seems to always be possible, because as pointed out by &lt;em&gt;squeamish ossifrage&lt;/em&gt;, there are hundreds of thousands of permutations for the numbers to sit in for the 9x9.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4748" LastEditorUserId="1831" LastEditDate="2014-12-19T10:58:07.813" LastActivityDate="2014-12-19T10:58:07.813" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="6252" PostTypeId="1" CreationDate="2014-12-18T13:33:19.823" Score="25" ViewCount="2688" Body="&lt;p&gt;So this starts out as a pretty simple question: what's the oldest recorded puzzle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm aware that we could get into &quot;what's the definition of a puzzle&quot; here, and I don't want to get sidetracked by that, so I'm expanding the question a bit to cover:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;What are some interesting puzzles from ancient civilisations (Greek, Roman, Egyptian, Babylonian, Sumerian, Persian, Aztec, Inca, etc)?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What are the oldest known puzzles of particular types? (e.g. physical puzzles which survive as artifacts, wordplay, image puzzles)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="8874" LastEditDate="2016-03-18T15:57:06.793" LastActivityDate="2016-09-16T06:12:47.167" Title="What's the oldest recorded puzzle?" Tags="|puzzle-history|" AnswerCount="5" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6254" PostTypeId="2" ParentId="6250" CreationDate="2014-12-18T13:46:20.080" Score="4" Body="&lt;p&gt;Here's a $9 \times 9$ grid that I found in about a minute using a Sudoku solver:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 7 5 3 9 4 8 2 6&#xA;4 2 6 1 8 7 3 9 5&#xA;8 9 3 6 2 5 1 4 7&#xA;7 5 2 4 1 8 9 6 3&#xA;3 6 1 2 5 9 4 7 8&#xA;9 4 8 7 3 6 5 1 2&#xA;5 8 4 9 6 2 7 3 1&#xA;6 3 9 5 7 1 2 8 4&#xA;2 1 7 8 4 3 6 5 9&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There are thousands more $9 \times 9$ solutions. From this square alone, I think it's possible to generate another 2903039 possible answers (since there are 9! permutations of the numbers 1-9, and the grids can be rotated four different ways with two possible reflections).&lt;/p&gt;&#xA;" OwnerUserId="40" LastEditorUserId="40" LastEditDate="2014-12-18T14:02:46.680" LastActivityDate="2014-12-18T14:02:46.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6255" PostTypeId="2" ParentId="6252" CreationDate="2014-12-18T14:13:52.540" Score="15" Body="&lt;p&gt;The best answer I can think of in a serious vein is &lt;a href=&quot;http://en.wikipedia.org/wiki/Sphinx#The_Riddle_of_the_Sphinx&quot; rel=&quot;noreferrer&quot;&gt;the Sphinx's riddle&lt;/a&gt;, solved by Oedipus in ancient Greece way back in the mists of time. This is surely the oldest &lt;em&gt;riddle&lt;/em&gt; if not the oldest puzzle. Here's the story from &lt;a href=&quot;http://www.perseus.tufts.edu/hopper/text?doc=Perseus%3Atext%3A1999.01.0022%3Atext%3DLibrary%3Abook%3D3%3Achapter%3D5%3Asection%3D8&quot; rel=&quot;noreferrer&quot;&gt;Apollodorus 3.5.8&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Laius was buried by Damasistratus, king of Plataea, and Creon, son of Menoeceus, succeeded to the kingdom. In his reign a heavy calamity befell Thebes.&lt;/p&gt;&#xA;&lt;p&gt;For Hera sent the Sphinx, whose mother was Echidna and her father Typhon; and she had the face of a woman, the breast and feet and tail of a lion, and the wings of a bird.&lt;/p&gt;&#xA;&lt;p&gt;And having learned a riddle from the Muses, she sat on Mount Phicium, and propounded it to the Thebans.&lt;/p&gt;&#xA;&lt;p&gt;And the riddle was this:— &lt;strong&gt;What is that which has one voice and yet becomes four-footed and two-footed and three-footed?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now the Thebans were in possession of an oracle which declared that they should be rid of the Sphinx whenever they had read her riddle; so they often met and discussed the answer, and when they could not find it the Sphinx used to snatch away one of them and gobble him up.&lt;/p&gt;&#xA;&lt;p&gt;When many had perished, and last of all Creon's son Haemon, Creon made proclamation that to him who should read the riddle he would give both the kingdom and the wife of Laius.&lt;/p&gt;&#xA;&lt;p&gt;On hearing that, Oedipus found the solution, declaring that the riddle of the Sphinx referred to &lt;strong&gt;man&lt;/strong&gt;; for &lt;strong&gt;as a babe he is four-footed, going on four limbs, as an adult he is two-footed, and as an old man he gets besides a third support in a staff&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;So the Sphinx threw herself from the citadel, and Oedipus both succeeded to the kingdom and unwittingly married his mother, and begat sons by her, Polynices and Eteocles, and daughters, Ismene and Antigone.3 But some say the children were borne to him by Eurygania, daughter of Hyperphas.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Some more flippant answers:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&amp;quot;What is the meaning of life?&amp;quot;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=ljGH07Unfe8&quot; rel=&quot;noreferrer&quot;&gt;&amp;quot;Why?&amp;quot;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;(In true AE-style I've even included a link to a Youtube video!)&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-12-26T14:53:00.773" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6256" PostTypeId="1" AcceptedAnswerId="6260" CreationDate="2014-12-18T14:50:28.040" Score="24" ViewCount="2868" Body="&lt;p&gt;I'm found in all angels, and core of the Devil,&lt;br&gt;&#xA;My forebears once dined 'cross the ol' English channel.&lt;br&gt;&#xA;If fish is your wish, I can help fill your dish,&lt;br&gt;&#xA;I go well with pie and with freshly cut sandwich.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wax and I wane with the tick of the hour,&lt;br&gt;&#xA;I only submit to my half-lunar ruler.&lt;br&gt;&#xA;Don't argue with me, though the night may be long-&lt;br&gt;&#xA;Though sometimes I'm right, I will never be wrong.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who am I?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="6178" LastEditorUserId="5188" LastEditDate="2015-08-14T02:20:53.373" LastActivityDate="2019-12-10T20:18:36.477" Title="Who am I? I'm found in all angels, and core of the Devil" Tags="|riddle|rhyme|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6258" PostTypeId="1" AcceptedAnswerId="6263" CreationDate="2014-12-18T15:25:31.313" Score="5" ViewCount="736" Body="&lt;p&gt;The good soldier Schweik had been ordered to line up a band of new&#xA;recruits before their officer gave them a speech. The desired line sought to&#xA;minimize the average difference in height of adjacent men. Schweik put the&#xA;tallest recruit first, the shortest one last, and let the remaining men stand&#xA;between them in random order. Did Schweik execute his order as stated?&#xA;How would you arrange the recruits?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; One may answer the first question affirmatively. The&#xA;intended order was, of course, different; it could have been executed in&#xA;two different ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; The below is the Solution but I couldn't understand the answer and it has some mathematical equation. will somebody explain ?.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;! Schweik must have understood his order as one to minimize&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Equation 1:&lt;/strong&gt;&#xA;  $$&#xA;{1\choose n}  [(h_2-h_1)+(h_3-h_2)+...+(h_n-h_{n-1})] = {1\choose n} (h_n-h_1)  &#xA;$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;where n is the number of recruits and hi, i = 1, 2, . . . , n, is the height of the&#xA;  recruit in the ith position. Since$$(h_n-h_1) /n$$ can be negative, its minimization&#xA;  means making it a negative number of the largest absolute value. This is achieved&#xA;  if h1 and hn are the largest and smallest heights, respectively, with the heights of all&#xA;  the others not contributing to the value of expression (1).&#xA;  The intended order was, of course, to minimize the average magnitude of the&#xA;  difference in height of adjacent men, that is,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Equation 2:&lt;/strong&gt;&#xA;  $$&#xA;{1\choose n} (|h_2-h_1| + |h_3-h_2| + ... +|h_n-h_{n-1}|)&#xA;$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Since n is fixed, the multiple 1/n can be ignored. Sum (2) is minimized when the&#xA;  heights are sorted in either ascending or descending order to yield the difference&#xA;  $$ (h_{max}-h_{min})$$ between the maximal and minimal heights. For any other ordering,&#xA;  this sum can be interpreted as the sum of lengths of segments covering the interval&#xA;  with endpoints at hmin and hmax with some overlaps, which makes it larger than&#xA;  $$ (h_{max}-h_{min})$$. A more formal proof by mathematical induction is not difficult.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2250" LastEditorUserId="2250" LastEditDate="2014-12-18T17:55:08.363" LastActivityDate="2014-12-18T22:13:52.987" Title="Lining Up Recruits" Tags="|mathematics|calculation-puzzle|algorithm|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6259" PostTypeId="2" ParentId="6258" CreationDate="2014-12-18T15:36:08.743" Score="2" Body="&lt;p&gt;If the aim is to minimize the average difference in height between any two adjacent men, it seems like the thing to do would be to surround each man with the two men closest to that man in height, which would logically mean lining the men up in order of increasing height.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that seems too easy. Am I missing something?&lt;/p&gt;&#xA;" OwnerUserId="4518" LastActivityDate="2014-12-18T15:36:08.743" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6260" PostTypeId="2" ParentId="6256" CreationDate="2014-12-18T16:02:53.087" Score="23" Body="&lt;p&gt;You're&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an Angle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm found in all angels, and core of the Devil,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Angel is an anagram, while the &quot;core&quot; of Devil is &lt;strong&gt;V&lt;/strong&gt;, which is simply an angle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My forebears once dined 'cross the ol' English channel.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Angles&quot;&gt;Angles&lt;/a&gt;, I assume&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If fish is your wish, I can help fill your dish,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Anglers catch fish&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I go well with pie and with freshly cut sandwich.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cutting pies or sandwiches creates a wedge (angle) shape. Could also be a play on pi.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I wax and I wane with the tick of the hour,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The angles between hands on a clock&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I only submit to my half-lunar ruler.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Maybe it's referring to a half-moon shaped protractor, not sure. (confirmed by OP in comments)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Don't argue with me, though the night may be long-&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some angles are &lt;em&gt;obtuse&lt;/em&gt;, or &quot;annoyingly insensitive or slow to understand&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though sometimes I'm right, I will never be wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are &lt;em&gt;right&lt;/em&gt; angles, but I've never heard of a &lt;em&gt;wrong&lt;/em&gt; one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1563" LastEditorUserId="1563" LastEditDate="2014-12-18T16:16:34.043" LastActivityDate="2014-12-18T16:16:34.043" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6261" PostTypeId="2" ParentId="6258" CreationDate="2014-12-18T16:16:37.270" Score="0" Body="&lt;p&gt;Another way to think of it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All the recruits are the same height. This would result in &lt;em&gt;the&lt;/em&gt; absolute minimum in average differences between the height of adjacent soliders... 0.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user5040" LastActivityDate="2014-12-18T16:16:37.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6262" PostTypeId="2" ParentId="6204" CreationDate="2014-12-18T16:18:30.697" Score="1" Body="&lt;p&gt;Taking a wild punt on #6. Is it an Enigma machine encryption? I say this because it has the unusual property that no letter ever gets encoded to itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: then again, this also applies to #4, the other uncracked message. Gah.&lt;/p&gt;&#xA;" OwnerUserId="1752" LastEditorUserId="1752" LastEditDate="2014-12-18T16:26:07.053" LastActivityDate="2014-12-18T16:26:07.053" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6263" PostTypeId="2" ParentId="6258" CreationDate="2014-12-18T16:24:41.230" Score="7" Body="&lt;p&gt;“Difference in height”, for a reasonable person, is a positive number (or zero for identical twins): the height of the taller person minus the height of the shorter person. With this definition, the average difference in height between adjacent recruits is minimized by lining up the recruits in monotonic order of height (tallest in front, then the next tallest behind him, etc.; or swap back and front, which makes two ways to do the lineup).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to prove this is to write the equations and do a bit of algebra. Being a computer scientist, I'll propose a different way. Suppose there are three recruits with height $x \lt y \lt z$. If they are lined up with $y$ in the middle, then the average difference in height is $(|y-x| + |z-y|)/2 = (z-x)/2$. If $z$ is in the middle then the average difference in height is $(|z-x| + |y-z|)/2 = (z-x)/2 + (z-y)/2$. If $x$ is in the middle then the average is $(|x-y| + |z-x|)/2 = (z-x)/2 + (y-x)/2$. This is minimized when $y$ is in the middle. Therefore, when there are three consecutive recruits such that the middle one by position is not the middle one by height, swapping positions to put the middling-height recruit in the middle lowers the height. The only line-ups that cannot be improved this way are line-ups in height order, either increasing or decreasing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A mathematically oriented sergeant might interpret “difference in height” as a signed value: the height of the recruit minus the height of the recruit in front of him. By this definition, the average height is $(x_1 - x_n)/n$ where $x_1$ is the height of the recruit at the back, $x_n$ is the height of the recruit at the front, and $n$ is the number of conscripts (the differences $(x_1 - x_2) + \dots (x_{n-1} - x_n)$ “cancel out”). The corporal achieved the minimum average difference by lining the conscripts up in this order (it's the largest achievable absolute value, with the orientation chosen to make it negative).&lt;/p&gt;&#xA;" OwnerUserId="71" LastActivityDate="2014-12-18T16:24:41.230" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6264" PostTypeId="2" ParentId="6250" CreationDate="2014-12-18T17:05:56.420" Score="16" Body="&lt;p&gt;The squares you describe are related to Latin squares. A &lt;em&gt;Latin square&lt;/em&gt; of order $n$ is an $n\times n$ square grid, filled with the numbers $1,\ldots,n$, such that each number appears once in every row and in every column. A Latin square is called &lt;em&gt;doubly diagonal&lt;/em&gt; if each number also appears once on each diagonals. A doubly diagonal Latin square satisfies your condition, because each number appears $n$ times, and the rows, columns, and diagonals sum to $1+\ldots+n=\frac{n(n+1)}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n$ a positive integer, a doubly diagonal Latin square of order $n$ exists if and only if $n=1$ or $n\geq 4$. I won't try to describe the general construction here, but this is written down explicitly in [1].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Gergely, Ervin. A simple method for constructing doubly diagonalized Latin squares. &lt;em&gt;J. Combinatorial Theory Ser. A&lt;/em&gt;, &lt;strong&gt;16&lt;/strong&gt; (1974), 266--272.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastActivityDate="2014-12-18T17:05:56.420" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6265" PostTypeId="2" ParentId="2836" CreationDate="2014-12-18T17:26:16.527" Score="2" Body="&lt;p&gt;You cannot. &#xA;It can be derived (although it's a bit longish) that the maximum number of balls you can resolve with N weightings is $(3^N-3)/2$. For N=2 that's only three and not four. However if you have one extra &quot;known good&quot; ball at your disposal you can up this to $(3^N-1)/2$, so in this case 4 would be possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a outline of the proof assuming N weightings and K balls&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Each weighting has three possible outcomes.These form a tree structure and the number of leaves at the end of the tree is $3^N$.&lt;/li&gt;&#xA;&lt;li&gt;The number of solutions (e.g. #4 is heavy, #12 is light, etc.) is $2*K$&lt;/li&gt;&#xA;&lt;li&gt;The puzzle can be solved my mapping the solutions on the tree structure. At every weighting the solution spaces gets divided into three sub sets that than progress down the tree in the next step&lt;/li&gt;&#xA;&lt;li&gt;The puzzle is solvable if every solution ends up at a unique leaf&lt;/li&gt;&#xA;&lt;li&gt;We can immediately see that number of leaves must be equal or larger than the number of solutions. So we have $K &amp;lt;= 3^N/2$. Since $3^N$ is odd and K is an integer we get  $K &amp;lt;= (3^N-1)/2$&lt;/li&gt;&#xA;&lt;li&gt;If you only have unknown balls, you have start with a weighting of M unknowns against M unknowns. If the scale moves the number of remaining solutions is $2*M$. If it doesn't move the number of remaining solutions is $2*(K-2*M)$. These are all even numbers.&lt;/li&gt;&#xA;&lt;li&gt;At every weighting you need to split the solution space in three equal parts, however $3^N/3$ is always an odd number so you can't do this perfectly with only even numbers and the best you can do is split the solution space into three parts of $3^N/3-1$ and we get the max possible number of balls to be $(3^N-3)/2$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So in the case of N=2 and K=4 we have a 9 leaves and 8 possible solutions. However in the first weighing we can split the solution space either 2/2/4 or 4/4/0. In either case you have a &quot;4&quot; in there which can't be resolved in only one weighting which has only 3 outcomes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding the known good ball allow you to split the solution space into 3/3/2 and hence this works.&lt;/p&gt;&#xA;" OwnerUserId="2666" LastActivityDate="2014-12-18T17:26:16.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6266" PostTypeId="2" ParentId="6250" CreationDate="2014-12-18T18:33:23.633" Score="8" Body="&lt;p&gt;I found a simple algorithm for odd numbers not divisible by 3.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Start with 1,2,...,n&#xA;For each subsequent row, push by 2.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;5x5:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 2 3 4 5&#xA;4 5 1 2 3&#xA;2 3 4 5 1&#xA;5 1 2 3 4&#xA;3 4 5 1 2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;7x7:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 2 3 4 5 6 7&#xA;6 7 1 2 3 4 5&#xA;4 5 6 7 1 2 3&#xA;2 3 4 5 6 7 1&#xA;7 1 2 3 4 5 6&#xA;5 6 7 1 2 3 4&#xA;3 4 5 6 7 1 2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;11x11:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1  2  3  4  5  6  7  8  9  10 11&#xA;10 11 1  2  3  4  5  6  7  8  9&#xA;8  9  10 11 1  2  3  4  5  6  7&#xA;6  7  8  9  10 11 1  2  3  4  5&#xA;4  5  6  7  8  9  10 11 1  2  3&#xA;2  3  4  5  6  7  8  9  10 11 1&#xA;11 1  2  3  4  5  6  7  8  9  10&#xA;9  10 11 1  2  3  4  5  6  7  8&#xA;7  8  9  10 11 1  2  3  4  5  6&#xA;5  6  7  8  9  10 11 1  2  3  4&#xA;3  4  5  6  7  8  9  10 11 1  2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;edit: See Lopsy's comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method will work for any number not divisible by 2 or 3. If each column is a series with a push-value of p, each diagonal will be a series with a push-value of p-1 and p+1. At 2, this means one diagonal is pushed by 1 (1,n,n-1,...,2), or pushed by 3. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as long as a number does not share any divisors with p-1,p, or p+1, and p is not equal to 1 or n-1, p should work for that n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is equivalent to the classical algorithm for generating Latin squares (pushing by 1 each row), then taking out the even rows and appending them to the bottom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure it's always possible to build these squares for n &gt; 3, but this is just an easy case for these numbers.&lt;/p&gt;&#xA;" OwnerUserId="5360" LastEditorUserId="5360" LastEditDate="2014-12-19T21:29:56.970" LastActivityDate="2014-12-19T21:29:56.970" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6269" PostTypeId="1" CreationDate="2014-12-18T19:57:27.250" Score="12" ViewCount="1697" Body="&lt;p&gt;I've enjoyed &lt;a href=&quot;http://en.wikipedia.org/wiki/Fillomino&quot; rel=&quot;noreferrer&quot;&gt;Fillomino&lt;/a&gt; for a couple years now. It's a region filling puzzle situated on a grid. The number in each box represents the size of the polyomino it belongs to, and polyominos of the same size cannot touch (except diagonally). Sample puzzle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/54yL0.png&quot; alt=&quot;sample&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Sample credit: &lt;a href=&quot;http://www.nikoli.co.jp/en/puzzles/fillomino.html&quot; rel=&quot;noreferrer&quot;&gt;Nikoli&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying my hand at &lt;em&gt;creating&lt;/em&gt; them, however, isn't going so well. I can easily create something for an unreasonably small grid, but that isn't fun or challenging. For example,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2..      (completed) 221&#xA;..2                  332&#xA;31.                  312&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;is a simple 3x3 puzzle with one distinct answer. The problem is that when I try to expand to a larger grid, the possibilities are overwhelming. It's very easy to lose track of why I put what where and what has already been accounted for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first instinct was to start with a &quot;completed&quot; grid and remove numbers until it was sparse enough to seem difficult, but I quickly found that deciding which ones to remove was hit-or-miss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking around the web, there are several sources explaining how to &lt;em&gt;solve&lt;/em&gt; these types of puzzles, but basically nothing on creating them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have an app on my phone that contains literally thousands of puzzles (of varying difficulty and size), so I'm inclined to believe that they can be easily computer generated, but I haven't seen an outline of an algorithm (by hand or code).&lt;/p&gt;&#xA;" OwnerUserId="1563" LastEditorUserId="1563" LastEditDate="2014-12-18T21:10:52.603" LastActivityDate="2016-09-25T20:04:42.190" Title="How can I create Fillomino puzzles?" Tags="|puzzle-creation|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6271" PostTypeId="2" ParentId="6258" CreationDate="2014-12-18T22:13:52.987" Score="1" Body="&lt;p&gt;If I'm understanding the proposed solution correctly, the writer is saying that the order of the men other than the two on the ends does not matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is easily proven false with an example. Suppose we give the heights of the men in inches, and their heights are 66, 68, 69, and 72 inches. If we arrange them in order of height, then the differences are 2, 1, and 3, for an average of 2. If we keep the shortest first and the tallest last but switch the two in the middle, so the order is 66, 69, 68, and 72, then the differences are 3, 1, and 4, for an average of 2 2/3. 2 &amp;lt;&gt; 2 2/3. The order of the men in the middle matters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue may be one of definitions. Normally, if you asked someone, &quot;What is the difference between Al's height and Bob's height?&quot;, he would state the difference as a non-negative number. That is, if Al is 68 inches and Bob is 72 inches, we would say that the difference in their heights is 4 inches. Very very few people would say &quot;If you name Al first, then it's minus 4 inches; but if you name Bob first, then it's plus 4 inches.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula given is using signed numbers when it should be using absolute values. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Side note: The divisor is not n but (n-1). If there are 3 men, then there are 2 height differences, not 3: h1 vs h2 and h2 vs h3. &lt;/p&gt;&#xA;" OwnerUserId="7386" LastActivityDate="2014-12-18T22:13:52.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6272" PostTypeId="1" AcceptedAnswerId="6358" CreationDate="2014-12-18T23:21:16.213" Score="8" ViewCount="3179" Body="&lt;p&gt;This is my first cipher puzzle. Here is the ciphertext:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;TKPHF_C_PELR_HVB_LDOAAVLEAEISGNCUTFROM_SDIOCFEAEAHAAROOAAIG_DLT_SN_LP_AREL_PTOTITOSIIUAMT_EAIS_TNWSNNI_DES_IP_FNOAC_PMCK__L_SIO___N_FN_TLIIMTICG_ANGNN_SYEEEH_AEFISTBOESNNE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's 171 characters, but this cipher can be applied to a plaintext of any length. I've replaced spaces with underscores to make it a little easier to work with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the title, says the cipher is a transposition cipher, meaning that the plaintext is an anagram of the ciphertext.  The cipher logic has a geometric basis.  You could write a computer program to decipher it, but it's simple enough to solve using pen and paper.  That's all I'll give away for now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is a pretty hard one.  Not NSA hard, but hard, so I anticipate adding hints over the coming days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The geometric shape this puzzle is based on is a square. Remember that any text may be encoded with this scheme, not just messages with a square number of characters, so there will be gaps.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2, an example of a plaintext and resulting ciphertext:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;IF YOU PLEASE GOOD SIR I THINK THAT A HINT IS IN ORDER&#xA;IOLEOI_FI_TH_ITUI__EO_ESTRI_O_IDSADRNRPKSNAHNTYH__GA__&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clockwise&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 4, a shorter, more telling plain/cipher text:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HELP I NEED A HINT &lt;br&gt; H_EAEII_ENTN_PHD_L&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2162" LastEditorUserId="2162" LastEditDate="2014-12-22T15:30:36.393" LastActivityDate="2014-12-23T11:35:14.123" Title="A geometric transposition cipher" Tags="|cipher|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6273" PostTypeId="1" AcceptedAnswerId="6298" CreationDate="2014-12-19T01:14:09.887" Score="13" ViewCount="4102" Body="&lt;p&gt;You have to use two words that are each their own opposite (&lt;a href=&quot;http://en.wikipedia.org/wiki/Auto-antonym&quot; rel=&quot;noreferrer&quot;&gt;auto-antonyms, contronyms, contranyms or whatever you want to call it&lt;/a&gt;) in the same sentence. This means that each of the two words in used twice, and for each word, the two occurrences have meanings that are opposite (contradictory) (**or extremely different ** &lt;em&gt;meaning they are so different one could argue they're opposites.&lt;/em&gt;) to each other. This sentence must be a grammatically correct English sentence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's ok to use minor variations on words such as plurals or conjugated verb forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal is to make it short (in terms of characters). &lt;strong&gt;The accepted answer will always be the shortest.&lt;/strong&gt; To save me trouble, it'd be nice for you to post the number of characters in your sentence. Spaces and punctuation count; you &lt;a href=&quot;http://www.javascriptkit.com/script/script2/charcount.shtml&quot; rel=&quot;noreferrer&quot;&gt;can use this site to check&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Anti-loophole: the words you choose must each be used twice, once as one meaning and again as another.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-12-20T15:17:23.617" LastActivityDate="2016-05-08T10:12:03.370" Title="Shortest sentence with two auto-antonyms" Tags="|open-ended|language|english|" AnswerCount="14" CommentCount="13" ClosedDate="2021-07-22T07:25:54.137" ContentLicense="CC BY-SA 3.0" />
  <row Id="6274" PostTypeId="2" ParentId="6069" CreationDate="2014-12-19T01:28:41.597" Score="3" Body="&lt;h1&gt;UPDATED&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Until more hints are provided, the only thing I can really supply is a dump of all of the cyclic rotations of all of the co-prime sampling intervals of the ciphertext (and, separately, the plaintext). Each are &lt;code&gt;123*80 = 9840&lt;/code&gt; lines. I hope that is useful to someone. I wrote the code in Haskell. The files are each 1MB, so I am splitting them up into several parts to fit it all on pastebin.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Cipher&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Part 1: &lt;a href=&quot;http://pastebin.com/mD2DuL5c&quot; rel=&quot;nofollow&quot;&gt;http://pastebin.com/mD2DuL5c&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 2: &lt;a href=&quot;http://pastebin.com/BPCNWwa3&quot; rel=&quot;nofollow&quot;&gt;http://pastebin.com/BPCNWwa3&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 3: &lt;a href=&quot;http://pastebin.com/XW61r97s&quot; rel=&quot;nofollow&quot;&gt;http://pastebin.com/XW61r97s&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Plain&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Part 1: &lt;a href=&quot;http://pastebin.com/E2ef2bRL&quot; rel=&quot;nofollow&quot;&gt;http://pastebin.com/E2ef2bRL&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 2: &lt;a href=&quot;http://pastebin.com/mPhx0iLC&quot; rel=&quot;nofollow&quot;&gt;http://pastebin.com/mPhx0iLC&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 3: &lt;a href=&quot;http://pastebin.com/TCKhwtn6&quot; rel=&quot;nofollow&quot;&gt;http://pastebin.com/TCKhwtn6&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;EDIT&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I have made no changes to the dumps above, but I just thought a bit more about the nature of this problem and came to a number of impressions/conclusions, which are not exactly rigorous but have some basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the information provided, both in the post and in its comments, I have arrived at several conclusions:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Automorphism&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Because the decryption does not rely on the orientation of the cyclical ciphertext or the starting point, we can consider this problem from a modular-arithmetic standpoint. We also know that all of the characters in the plaintext $p$ are preserved in the ciphertext $c$, merely subjected to a rearrangement. We can therefore treat the cipher as an automorphism, because it is a mapping of $\mathbb{N}_m \rightarrow \mathbb{N}_m$, where $m$ is the length of the message. We will therefore consider $E$ to be the encryption function, and $D$ to be the decryption function, where $D(E(p)) = p$. We also know that, for an arbitrary cyclic shift $C$, we have $E(C(p)) = C(E(p))$ and $D(C(c)) = C(D(c))$. We thus conclude that $E$ and $D$ are pure, stateless functions, meaning that they produce the same result for the same output independent of any history. We also know that $E$ and $D$ depend only on the positions of letters, and not on their values. For,if this were the case, the size of the domain of $E$ would be $m*L$, where the alphabet of the messages is of size $L$, while the range would still be $m$, as the letters themselves are merely transposed and not transformed; by the Pigeon-Hole principle, there must therefore be multiple index-value pairs that map to the same index of the ciphertext, which we know not to be the case. We therefore consider $e$ instead of $E$, which is the function mapping an index in plaintext to the ciphertext index when $E$ is applied, which we state as $$p_i = c_{e(i)}\;\forall\;1 \le i \le m$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We therefore only need to determine the mapping of $e$, which we know must be linear if $E(C(P)) = C(E(P)) \implies e(i+x) = e(i)+x$. After I create a table of the positions of each letter in ciphertext and plaintext, I will post the results of that.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Results&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;(I know someone else has posted these, but this is in a slightly different form, and it doesn't hurt)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[3,7,12,17,29,34,37,44,48,51,56,60,65,69,73,84,88,95,100,104,108,115,118] -&amp;gt; [3,11,12,13,21,24,27,29,44,47,53,64,69,72,73,75,85,87,91,99,113,119,121]&#xA;[43] -&amp;gt; [2]&#xA;[99] -&amp;gt; [59]&#xA;[28,122] -&amp;gt; [82,107]&#xA;[5,10,32,41,63,71,81,102] -&amp;gt; [1,22,28,43,65,86,90,118]&#xA;[45] -&amp;gt; [10]&#xA;[4,18,70,89,101] -&amp;gt; [33,51,70,89,100]&#xA;[11,52,76,83] -&amp;gt; [14,39,45,94]&#xA;[9,22,25,40,54,62,77,87,93,97,114] -&amp;gt; [7,16,26,40,49,61,66,79,98,108,115]&#xA;[109] -&amp;gt; [74]&#xA;[38,111] -&amp;gt; [9,78]&#xA;[14,21,31,86,92] -&amp;gt; [0,54,63,102,114]&#xA;[15,19,35,49,90,110,116] -&amp;gt; [5,20,60,81,84,106,112]&#xA;[61] -&amp;gt; [37]&#xA;[6,57,64,72,75,82,103] -&amp;gt; [31,38,46,52,58,77,103]&#xA;[1,53,58,67,106,119] -&amp;gt; [30,36,41,83,93,101]&#xA;[20,91] -&amp;gt; [67,116]&#xA;[8,23,39,78,94,113] -&amp;gt; [19,55,71,88,110,122]&#xA;[16,36,55,79] -&amp;gt; [17,35,57,117]&#xA;[13,24,27,30,33,42,47,50,59,80,85,98,117,121] -&amp;gt; [4,6,15,18,23,32,34,42,62,68,76,96,104,120]&#xA;[2,46,68,74,107,112,120] -&amp;gt; [25,48,56,92,95,105,109]&#xA;[26] -&amp;gt; [8]&#xA;[0,66,96,105] -&amp;gt; [50,80,97,111]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7387" LastEditorUserId="7387" LastEditDate="2014-12-19T20:39:17.533" LastActivityDate="2014-12-19T20:39:17.533" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6275" PostTypeId="2" ParentId="6120" CreationDate="2014-12-19T01:52:26.870" Score="1" Body="&lt;p&gt;I know this is a bit late, but since the accepted answer is still wrong, I thought I'd post my answer for the sake of completeness.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The six statements in the questions can be interpreted as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ignoring case and punctuation, all the words can be substitued in&#xA; each statement to form mathematical equations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which gives the following key:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;equals&lt;/strong&gt; = is, &lt;strong&gt;minus&lt;/strong&gt; = without&lt;br&gt;&#xA; &lt;strong&gt;Y&lt;/strong&gt; = yes, &lt;strong&gt;N&lt;/strong&gt; = no, &lt;strong&gt;T&lt;/strong&gt; = the&lt;br&gt;&#xA;&lt;strong&gt;a&lt;/strong&gt; = a/an, &lt;strong&gt;t&lt;/strong&gt; = to, &lt;strong&gt;A&lt;/strong&gt; = answer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which tranforms the statements into this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (1) No, an answer is the cube without A, No?&lt;br&gt;&#xA;&lt;br&gt;&#xA; N × a × A = (T³ - a) × N&lt;br&gt;&#xA; 6 × 4 × 1 = (2³ - 4) × 6&lt;br&gt;&#xA;&lt;br&gt;&#xA; (2) The Yes is No.&lt;br&gt;&#xA;&lt;br&gt;&#xA; T × Y = N&lt;br&gt;&#xA; 2 × 3 = 6&lt;br&gt;&#xA;&lt;br&gt;&#xA; (3) The squared answer, Yes, is to an answer.&lt;br&gt;&#xA;&lt;br&gt;&#xA; T² × A × Y = t × a × A&lt;br&gt;&#xA; 2² × 1 × 3 = 3 × 4 × 1&lt;br&gt;&#xA;&lt;br&gt;&#xA; (4) A Yes without the Yes is to the answer.&lt;br&gt;&#xA;&lt;br&gt;&#xA; (a × Y) - (T × Y) = t × T × A&lt;br&gt;&#xA; (4 × 3) - (2 × 3) = 3 × 2 × 1&lt;br&gt;&#xA;&lt;br&gt;&#xA; (5) Yes, the answer is No.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Y × T × A = N&lt;br&gt;&#xA; 3 × 2 × 1 = 6&lt;br&gt;&#xA;&lt;br&gt;&#xA; (6) Yes, the answer to a no is 432.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Y × T × A × t × a × N = 432&lt;br&gt;&#xA; 3 × 2 × 1 × 3 × 4 × 6 = 432  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But how do I know this is the right result?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mostly trial and error (sorry, no fancy algebra ;-)&lt;br&gt;&#xA;&lt;br&gt;&#xA; From (6), it can been seen that none of the variables is zero, and from (2) and (5) it's also easy to see that A (the answer) must equal one. So the factors of 432 (2⁴ &amp;amp; 3³) must be shared out between the other variables.&lt;br&gt;&#xA;&lt;br&gt;&#xA; I haven't checked all the possibilities, but my solution above definitely works, and it's easy to show that, e.g. Y = 4, T = 3 doesn't:&lt;br&gt;&#xA;&lt;br&gt;&#xA; Y × T × A × t × a × N = 432&lt;br&gt;&#xA; 4 × 3 × 1 × ? × ? × 12 = 432&lt;br&gt;&#xA;&lt;br&gt;&#xA; so:&lt;br&gt;&#xA;&lt;br&gt;&#xA; t = 3, a = 1 or t = 1, a = 3&lt;br&gt;&#xA;&lt;br&gt;&#xA; but substituting in (4), we get:&lt;br&gt;&#xA;&lt;br&gt;&#xA; (a × Y) - (T × Y) = t × T × A&lt;br&gt;&#xA; (1 × 4) - (3 × 4) = 3 × 3 × 1&lt;br&gt;&#xA; (3 × 4) - (3 × 4) = 1 × 3 × 1&lt;br&gt;&#xA;&lt;br&gt;&#xA; which is obviously wrong.&lt;br&gt;&#xA;   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5550" LastActivityDate="2014-12-19T01:52:26.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6276" PostTypeId="2" ParentId="6273" CreationDate="2014-12-19T01:55:39.180" Score="3" Body="&lt;h3&gt;107 characters&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Originally I was puzzled by this question, but I soon puzzled it out and came up with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At first I held &lt;strong&gt;fast&lt;/strong&gt; as my alarm went &lt;strong&gt;off&lt;/strong&gt;, but soon I had to turn it &lt;strong&gt;off&lt;/strong&gt;, for morning was &lt;strong&gt;fast&lt;/strong&gt; approaching.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That probably won't win the shortness award, but it's an example of fairly standard &lt;a href=&quot;http://en.wikipedia.org/wiki/Auto-antonym&quot; rel=&quot;nofollow&quot;&gt;auto-antonyms&lt;/a&gt;. I could probably shorten it by removing/changing some words while still being &lt;em&gt;grammatically&lt;/em&gt; correct, but it wouldn't sound very good to my ear.&lt;/p&gt;&#xA;" OwnerUserId="1563" LastEditorUserId="1563" LastEditDate="2014-12-19T02:02:20.263" LastActivityDate="2014-12-19T02:02:20.263" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6277" PostTypeId="2" ParentId="6273" CreationDate="2014-12-19T02:14:55.910" Score="8" Body="&lt;h1&gt;60 characters&lt;/h1&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I couldn't &lt;strong&gt;help&lt;/strong&gt; but &lt;strong&gt;help&lt;/strong&gt;, so I &lt;strong&gt;left&lt;/strong&gt; the pies that were &lt;strong&gt;left&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first &quot;help&quot; means &quot;resist;&quot; the second &quot;help&quot; means &quot;assist.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first &quot;left&quot; means &quot;went away;&quot; the second &quot;left&quot; means &quot;remaining.&quot;&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2014-12-19T02:14:55.910" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6278" PostTypeId="2" ParentId="6273" CreationDate="2014-12-19T03:41:40.170" Score="24" Body="&lt;h1&gt;52 Characters&lt;/h1&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The &lt;strong&gt;custom&lt;/strong&gt; is to turn &lt;strong&gt;off&lt;/strong&gt; &lt;strong&gt;custom&lt;/strong&gt; alarms that go &lt;strong&gt;off&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first &quot;custom&quot; means &quot;standard&quot;, the second means &quot;tailored&quot;.&#xA;The first &quot;off&quot; means &quot;disable&quot;, the second off means &quot;engaged&quot;.&lt;/p&gt;&#xA;" OwnerUserId="7389" LastActivityDate="2014-12-19T03:41:40.170" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6279" PostTypeId="2" ParentId="6273" CreationDate="2014-12-19T04:49:32.310" Score="33" Body="&lt;h1&gt;27 characters&lt;/h1&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Bolt&lt;/strong&gt; me &lt;strong&gt;fast&lt;/strong&gt;, or &lt;strong&gt;bolt&lt;/strong&gt; &lt;strong&gt;fast&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;We have:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;bolt&lt;/strong&gt;: secure in place, or dash away suddenly&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dictionary.cambridge.org/dictionary/british/fast&quot; rel=&quot;noreferrer&quot;&gt;&lt;strong&gt;fast&lt;/strong&gt;&lt;/a&gt;: firmly fixed, or quickly&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;The intended meaning is &amp;quot;Either tie me down securely or run away quickly&amp;quot;. As @xnor remarks: &amp;quot;I guess the speaker is a werewolf&amp;quot; (thanks for helping!).&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;If the above sentence wasn't ambiguous enough for you, we can take out the &amp;quot;me&amp;quot;, leaving it implied. This has the hilarious consequence:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Bolt&lt;/strong&gt; &lt;strong&gt;fast&lt;/strong&gt;, or &lt;strong&gt;bolt&lt;/strong&gt; &lt;strong&gt;fast&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Same meaning as above, but body language/intonation are necessary for this one to work. It helps if you imagine a movie scene between a male werewolf and a young girl reluctant to tie him up on a full moon.&lt;/p&gt;&#xA;" OwnerUserId="445" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-12-21T04:03:45.433" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6280" PostTypeId="2" ParentId="6272" CreationDate="2014-12-19T04:58:07.027" Score="2" Body="&lt;p&gt;Given that the length 171 is a triangular number, I think the shape is a triangle. I didn't get anything good though by packing the letters into a triangle, regardless of which direction I read (backward/forwards, slant left/right).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                 T&#xA;                K P&#xA;               H F _&#xA;              C _ P E&#xA;             L R _ H V&#xA;            B _ L D O A&#xA;           A V L E A E I&#xA;          S G N C U T F R&#xA;         O M _ S D I O C F&#xA;        E A E A H A A R O O&#xA;       A A I G _ D L T _ S N&#xA;      _ L P _ A R E L _ P T O&#xA;     T I T O S I I U A M T _ E&#xA;    A I S _ T N W S N N I _ D E&#xA;   S _ I P _ F N O A C _ P M C K&#xA;  _ _ L _ S I O _ _ _ N _ F N _ T&#xA; L I I M T I C G _ A N G N N _ S Y&#xA;E E E H _ A E F I S T B O E S N N E&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Nor with the triangle upside-down:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;T K P H F _ C _ P E L R _ H V B _ L&#xA; D O A A V L E A E I S G N C U T F&#xA;  R O M _ S D I O C F E A E A H A&#xA;   A R O O A A I G _ D L T _ S N&#xA;    _ L P _ A R E L _ P T O T I&#xA;     T O S I I U A M T _ E A I&#xA;      S _ T N W S N N I _ D E&#xA;       S _ I P _ F N O A C _&#xA;        P M C K _ _ L _ S I&#xA;         O _ _ _ N _ F N _&#xA;          T L I I M T I C&#xA;           G _ A N G N N&#xA;            _ S Y E E E&#xA;             H _ A E F&#xA;              I S T B&#xA;               O E S&#xA;                N N&#xA;                 E&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2014-12-19T04:58:07.027" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6282" PostTypeId="1" AcceptedAnswerId="6313" CreationDate="2014-12-19T06:22:48.733" Score="12" ViewCount="537" Body="&lt;p&gt;Deep under the sands of Ancient Egypt, a dusty chamber lies undisturbed. It is not a tomb, no - but it tells a story nevertheless. Despite the walls being ruined by time, an inscription is visible on the coarse sandstone walls - one that tells the tale of a girl and her animal:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3kH8E.png&quot; alt=&quot;Think melodically.&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this leaves us with two simple questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;What was her name, and what animal did she have?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You consult your smartphone. You know from experience that these symbols are &lt;a href=&quot;http://en.wikipedia.org/wiki/Egyptian_numerals&quot; rel=&quot;nofollow noreferrer&quot;&gt;purely mathematical&lt;/a&gt; in nature.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint II:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You &lt;a href=&quot;http://en.wikipedia.org/wiki/Pythagorean_tuning&quot; rel=&quot;nofollow noreferrer&quot;&gt;wonder why&lt;/a&gt; this has a music tag....&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7308" LastEditorUserId="7308" LastEditDate="2014-12-21T00:07:36.650" LastActivityDate="2014-12-21T00:07:36.650" Title="An Ancient Egyptian Story" Tags="|music|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6283" PostTypeId="2" ParentId="6269" CreationDate="2014-12-19T06:23:24.160" Score="4" Body="&lt;p&gt;For computers, re-solving the grid over and over is no problem at all, and that's basically how most generators for these types of games work. They either start with a blank grid and add clues or start with a filled grid and remove clues, re-solving the grid at each step to make sure it's working. (These aren't the only approaches, but I think they're the most common.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Games like this are typically NP-complete (Fillomino is: &lt;a href=&quot;http://www6.in.tum.de/Main/Publications/Ruepp2010a.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www6.in.tum.de/Main/Publications/Ruepp2010a.pdf&lt;/a&gt;) which means that this isn't really practical for very large grids, but since no one plays very large grids anyway, it doesn't come up. Even then, it's not a problem if it takes an hour to generate a new puzzle if you only need one every day or week or if you can wait for a couple days to gather a large collection (not that it usually takes that long).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For making puzzles by hand, the process is similar, but humans (with practice) can take a lot of short cuts and make a lot of decisions based on more aesthetic values. For example, you might pick out a set of givens that make a nice pattern and see how close that gets you to a solution, and then base the addition or removal of givens on how well the previous attempt worked rather than on chance. If you know that there's a lot of ambiguity in one corner, you can guess that you'll need to add another given or two near there. You may not even have to solve the whole grid to know that part of it isn't going to work. Most computer programs won't take such shortcuts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't say from experience, but I'm sure that with practice there are a lot of little things you could learn that would make making new puzzles easier. (Sudoku is popular enough that people have written articles on making sudoku puzzles. Some of that advice would surely translate to other types of puzzles.)&lt;/p&gt;&#xA;" OwnerUserId="7392" LastEditorUserId="7392" LastEditDate="2014-12-19T06:44:35.517" LastActivityDate="2014-12-19T06:44:35.517" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="6285" PostTypeId="2" ParentId="6269" CreationDate="2014-12-19T06:38:45.363" Score="2" Body="&lt;p&gt;Here's a possible approach that requires programming a solver:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Start with a blank grid&lt;/li&gt;&#xA;&lt;li&gt;Pick some cells and assign them values.&lt;/li&gt;&#xA;&lt;li&gt;Solve the resulting grid and enumerate the solutions. If there are still too many solutions, then repeat step 2. If there are no solutions, either backtrack or start over.&lt;/li&gt;&#xA;&lt;li&gt;If you reach this step, then there shouldn't be many solutions left. Look at all of the solutions and see where the differences lie, and place appropriate numbers to ensure a unique solution.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You can do step 2 manually or via a program, your choice. I thought this could be a good way to come up with symmetric puzzles, since once you get a nice-looking grid going you can determine the minimum number of givens you have to provide to make the solution unique.&lt;/p&gt;&#xA;" OwnerUserId="445" LastEditorUserId="445" LastEditDate="2014-12-19T06:53:21.610" LastActivityDate="2014-12-19T06:53:21.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6286" PostTypeId="2" ParentId="6273" CreationDate="2014-12-19T08:12:18.163" Score="4" Body="&lt;h3&gt;20 characters&lt;/h3&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Tipped cop cops tip&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;...news at 11.&lt;/p&gt;&#xA;&lt;p&gt;If articles are required, then &lt;strong&gt;24 characters&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A tipped cop cops a tip.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&amp;quot;A local police officer, recently in the news for having been knocked over on his beat by a fleeing pickpocket, has been caught with his own hand in the proverbial cookie jar, stealing the end of a ceremonial flagstaff at headquarters.&amp;quot;&lt;/p&gt;&#xA;" OwnerUserId="4204" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-12-19T08:40:19.913" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6288" PostTypeId="1" AcceptedAnswerId="7729" CreationDate="2014-12-19T10:40:23.303" Score="5" ViewCount="1054" Body="&lt;p&gt;&lt;em&gt;Hello everyone, my name is Jay&lt;br&gt;&#xA;I work for you 24 hours a day,&lt;br&gt;&#xA;but don't get me wrong I know not Silent Bob&lt;br&gt;&#xA;and got nothing to do with weed on the job.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My kind can be you or you can be me&lt;br&gt;&#xA;if you know a body by its chemistry.&lt;br&gt;&#xA;We're not only human, but mostly a thing!&lt;br&gt;&#xA;We have many shapes, on the air we are king.&lt;br&gt;&#xA;Our children you could surf but not in the sea&lt;br&gt;&#xA;(for some they are a monstruosity!).&lt;br&gt;&#xA;They fly, they bounce, they bend,&lt;br&gt;&#xA;but always to us they end!&lt;br&gt;&#xA;We get and we give even not alive,&lt;br&gt;&#xA;but to do our best we need a good drive,&lt;br&gt;&#xA;for we can make work a lot you possess,&lt;br&gt;&#xA;even destroy it if working in excess.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This riddle was difficult but I'm not a sphinx,&lt;br&gt;&#xA;in fact I'm well sure, who I am you can glimpse&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;I'm deleting the &quot;easy&quot; and &quot;difficult&quot; part because I think it got most of you thinking in the wrong direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HINTS:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Jay is just a name, don't focus on it &lt;/li&gt;&#xA;&lt;li&gt;Most of &quot;Jay&quot;s are objects&lt;/li&gt;&#xA;&lt;li&gt;The chemistry I'm talking about is... well think VERY-little!&lt;/li&gt;&#xA;&lt;li&gt;&quot;Jay&quot;s have nothing to do with sphinxes!&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5122" LastEditorUserId="5122" LastEditDate="2014-12-29T10:18:05.673" LastActivityDate="2015-01-09T06:48:15.377" Title="Not the Jay you saw in &quot;Clerks&quot;" Tags="|riddle|" AnswerCount="6" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6289" PostTypeId="2" ParentId="6273" CreationDate="2014-12-19T10:49:46.817" Score="4" Body="&lt;h2&gt;27 Characters&lt;/h2&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Resign old, resign when old&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;br /&gt;&#xA;&lt;h2&gt;33 Characters&lt;/h2&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Resign to resign old job when old&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;First resign means 'to quit'. Second resign means 'signing up again'.&lt;/p&gt;&#xA;&lt;p&gt;First old means 'past'. The second old means 'later state'.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Note: Not the shortest, but shorter than some other answers.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-12-19T11:11:41.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6290" PostTypeId="2" ParentId="6288" CreationDate="2014-12-19T10:56:22.227" Score="5" Body="&lt;p&gt;Jay is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; your heart&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Hello everyone, my name is Jay&#xA;I work for you 24 hours a day,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the heart does not stop even when sleeping&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;but don't get me wrong I know not Silent Bob&#xA;and got nothing to do with weed on the job.&#xA;My kind can be you or you can be me&#xA;if you know a body by its chemistry.&#xA;We're not only human, but mostly a thing!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; because people sometimes take decisions with their hearts ??&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;We have many shapes, on the air we are king.&#xA;Our children you could surf but not in the sea&#xA;(for some they are a monstruosity!).&#xA;They fly, they bounce, they bend,&#xA;but always to us they end!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the children are the leukocytes and red blood cells, they &quot;surf&quot; in the fluid, but always return back to the heart.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;We get and we give even not alive,&#xA;but to do our best we need a good drive,&#xA;for we can make work a lot you possess,&#xA;even destroy it if working in excess.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the heart is keeping us alive and supplies the brain with oxygen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5283" LastActivityDate="2014-12-19T10:56:22.227" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6292" PostTypeId="2" ParentId="6273" CreationDate="2014-12-19T11:11:01.713" Score="19" Body="&lt;h2&gt;17 characters&lt;/h2&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;One bar bars one&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The first &lt;em&gt;one&lt;/em&gt; is a single instance of something&lt;/li&gt;&#xA;&lt;li&gt;The first &lt;em&gt;bar&lt;/em&gt; is a place where you are served drinks&lt;/li&gt;&#xA;&lt;li&gt;The second &lt;em&gt;bar&lt;/em&gt; means to prohibit someone from being served&lt;/li&gt;&#xA;&lt;li&gt;The second &lt;em&gt;one&lt;/em&gt; is the general third person pronoun, referring to all people in general&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;You could imagine a context something like this:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;One bar bars one, and suddenly one gets a reputation for being a troublemaker!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If you don't like the fact that the second &lt;em&gt;bar&lt;/em&gt; has an s on the end (several other answers have &amp;quot;modified&amp;quot; forms like this, but I'm a bit of a perfectionist...), you could have&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;One bar did bar one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;which has an extra 4 characters, bringing the total to 21.&lt;/p&gt;&#xA;" OwnerUserId="5260" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-12-19T11:24:00.883" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6293" PostTypeId="2" ParentId="6273" CreationDate="2014-12-19T12:37:25.693" Score="3" Body="&lt;h2&gt;18 characters&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Fix left, left fix&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I fix (repair) the left object, but I left a fix (problem)&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;26 Characters&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Left's right, right's left&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not exactly opposites, but in context:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The answer to the left question is right, the right question is left.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6178" LastEditorUserId="6178" LastEditDate="2014-12-19T12:48:09.857" LastActivityDate="2014-12-19T12:48:09.857" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6295" PostTypeId="2" ParentId="6282" CreationDate="2014-12-19T13:07:49.463" Score="1" Body="&lt;p&gt;According to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gardiner%27s_sign_list&quot; rel=&quot;nofollow&quot;&gt;sign list&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the single bar stands for &quot;it has&quot;, &quot;singular&quot;, or &quot;one&quot;&lt;/li&gt;&#xA;&lt;li&gt;the symbol that looks like a capital lambda stands for &quot;approach, enterprise, do not move, stop, injure, leg, thigh&quot;&lt;/li&gt;&#xA;&lt;li&gt;the symbol that looks like a horizontal grain of rice stands for &quot;mouth&quot;&lt;/li&gt;&#xA;&lt;li&gt;the symbol that looks like a horizontal knife blade stands for &quot;side area&quot; or &quot;side&quot;&lt;/li&gt;&#xA;&lt;li&gt;the vertical ticks stand for 1's (as in unary digits) and the bumps stand for 10's&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I also believe the Egyptians used top-to-bottom reading order as in English, hence free from interpretation, the story is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(one)&lt;br&gt;(one) (approaches, does, stops, is injured, does not move, leg, thigh) (mouth) (eight)&lt;br&gt;(one) (approaches, does, stops, is injured, does not move, leg, thigh) (mouth) (four) (approaches, does, stops, is injured, does not move, leg, thigh) (mouth) (sixty-four)&lt;br&gt;(one) (approaches, does, stops, is injured, does not move, leg, thigh) (side area, side)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The numbers four, eight, and sixty-four, and their relationship to &quot;mouth&quot; is obviously significant, but I have no idea how they would imply the identity of the girl or her animal. My three guesses are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;She had a pet rabbit that multiplied for two generations, until she got fed up and let them die off (&quot;side&quot; in the final sentence).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;She had a pet she taught to first obey eight vocal commands, then sixty-four vocal commands. The pet had borne offspring, and the younger animal (perhaps a pup) could obey four vocal commands. Eventually both the older animal and its offspring died.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;She had a pet parrot that initially could say 8 things. Eventually it could say 64 different things and understand four different things. Ultimately it died.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I gather from the Wiki article that reading ancient Egyptian is significantly more complicated that just chaining together base word meanings, but if this problem requires some degree of proficiency in interpreting ancient Egyptian, you should probably indicate that in the OP.&lt;/p&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2014-12-19T13:07:49.463" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6296" PostTypeId="2" ParentId="6235" CreationDate="2014-12-19T13:54:01.407" Score="9" Body="&lt;p&gt;It is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A doorbell&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;L&lt;/strong&gt;o! I heard you have gone to war.&lt;br&gt;&#xA;A&lt;strong&gt;l&lt;/strong&gt;luding to my utter demise.&lt;br&gt;&#xA;Tr&lt;strong&gt;e&lt;/strong&gt;es are what you used to fear,&lt;br&gt;&#xA;blo&lt;strong&gt;b&lt;/strong&gt;s of nothing, nothing is best.&lt;br&gt;&#xA;Retu&lt;strong&gt;r&lt;/strong&gt;n to the start, and commence to walk.&lt;br&gt;&#xA;Do it n&lt;strong&gt;o&lt;/strong&gt; more than once, down these rickety stairs.&lt;br&gt;&#xA;Take a n&lt;strong&gt;o&lt;/strong&gt;tice, ignore the spaces in between.&lt;br&gt;&#xA;No, doome&lt;strong&gt;d&lt;/strong&gt; you are not, lest you count the punctuation.&lt;br&gt;&#xA;(Count the &lt;em&gt;space&lt;/em&gt; here, and ignore the first-born!)&lt;br&gt;&#xA;Now, talk, fl&lt;strong&gt;a&lt;/strong&gt;il, and stop right here.&lt;br&gt;&#xA;&lt;br&gt;&#xA;Now that you've got what I've meant to say, double check.&lt;br&gt;&#xA;I would warn you of anything, coming at your &lt;strong&gt;door&lt;/strong&gt;.&lt;br&gt;&#xA;Trust me, trust me! And say it once more.&lt;br&gt;&#xA;So trust me, trust me! Did you forget to double-check?&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hopefully no further clarification is needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was originally going to give up on this line of thought, but then I double checked and realized it was backwards.  True story.&lt;/p&gt;&#xA;" OwnerUserId="6178" LastActivityDate="2014-12-19T13:54:01.407" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6297" PostTypeId="2" ParentId="6288" CreationDate="2014-12-19T17:15:49.870" Score="4" Body="&lt;p&gt;I'll take a crack at it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jay is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Joule, the SI unit of energy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Hello everyone, my name is Jay&#xA;I work for you 24 hours a day,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;J&lt;/code&gt; is the standard abbreviation for Joule, which is also considered to be a measure of work performed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;but don't get me wrong I know not Silent Bob&#xA;and got nothing to do with weed on the job.&#xA;My kind can be you or you can be me&#xA;if you know a body by its chemistry.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Energy = Mass times the speed of light squared, so any amount of energy could be converted to mass or vice versa.  Also, the human body is powered by chemical processes to extract energy from food.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;We're not only human, but mostly a thing!&#xA;We have many shapes, on the air we are king.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; not really sure about this bit... Energy is commonly stored in batteries, which are a thing, and which come in a disgusting variety of shapes and sizes, but the air thing I dunno.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Our children you could surf but not in the sea&#xA;(for some they are a monstruosity!).&#xA;They fly, they bounce, they bend,&#xA;but always to us they end!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Energy is commonly transmitted as a wave form.  Energy waves can travel through the air, be bounced and bent, but they are still used as energy eventually.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;We get and we give even not alive,&#xA;but to do our best we need a good drive,&#xA;for we can make work a lot you possess,&#xA;even destroy it if working in excess.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Energy is received and transmitted all the time, and is required for all sorts of devices these days; but if you overcharge something or run it with the wrong voltage input you can easily burn it out.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5874" LastEditorUserId="5874" LastEditDate="2014-12-19T18:00:28.380" LastActivityDate="2014-12-19T18:00:28.380" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6298" PostTypeId="2" ParentId="6273" CreationDate="2014-12-19T18:21:36.387" Score="22" Body="&lt;h2&gt;17 Characters&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Using an archaic form of 'let', this ties with the currently first place answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let be or be let.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Let&lt;/strong&gt;: To allow OR to hinder (archaic).&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Be&lt;/strong&gt;: The way something already is OR to become different.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Translation: Allow things to stay as they are, or you will become hindered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Arguably, we can shorten this by removing the 'or' (which flips the meaning of the sentence) and using an implied conditional:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;15 Characters&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let be; be let.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Translation: If you allow things to stay as they are, then you will become hindered.&lt;/p&gt;&#xA;" OwnerUserId="7401" LastActivityDate="2014-12-19T18:21:36.387" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6299" PostTypeId="2" ParentId="6288" CreationDate="2014-12-19T19:24:29.123" Score="2" Body="&lt;p&gt;I'll give it a try, this is what i've got so far: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jay is  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an atom  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Hello everyone, my name is Jay&#xA;I work for you 24 hours a day,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; atoms get together and break appart all the time to produce every kind of physical nature. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;but don't get me wrong I know not Silent Bob&#xA;and got nothing to do with weed on the job.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...well jay is obviously not from &lt;em&gt;clerks&lt;/em&gt;.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;My kind can be you or you can be me&#xA;if you know a body by its chemistry.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; atoms make up all of the visible and touchable matter around us. pretty obvious connection to chemistry. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;We're not only human, but mostly a thing!&#xA;We have many shapes, on the air we are king.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; human, things, atoms can take any kind of shape. Air is also composed of different molecules. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Our children you could surf but not in the sea&#xA;(for some they are a monstruosity!).&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; i can't quite figure out this one yet but i'm sure its related to electrons and protons. i think electrons can surf on waves like sound or light waves. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;They fly, they bounce, they bend,&#xA;but always to us they end!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; atoms exchange protons and electrons to undergo chemical processes. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;We get and we give even not alive,&#xA;but to do our best we need a good drive,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; another reference to subatomic exchange. i think drive is refering to the force needed to exchange these particles. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;for we can make work a lot you possess,&#xA;even destroy it if working in excess.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the exchange of electrons produce electrical flow and can make work any electric device. if the electric flow is powerful enough, it could destroy. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7402" LastActivityDate="2014-12-19T19:24:29.123" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6301" PostTypeId="1" AcceptedAnswerId="6304" CreationDate="2014-12-19T23:24:34.880" Score="0" ViewCount="715" Body="&lt;p&gt;The expression &amp;quot;twenty minus five&amp;quot; has 15 letters (we ignore spaces) and also has the arithmetic value of 15.&lt;/p&gt;&#xA;&lt;p&gt;You can use standard math constants, operators and functions, represented by English words. Spaces and parenthesis do not count towards the expression length.&lt;/p&gt;&#xA;&lt;p&gt;Find the longest such expression.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; EDIT: As correctly stated in &lt;a href=&quot;https://puzzling.stackexchange.com/a/6304/7403&quot;&gt;https://puzzling.stackexchange.com/a/6304/7403&lt;/a&gt;, no such expression exists, so only the bonus now remains.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;As a bonus, find the most interesting such expression.&lt;/p&gt;&#xA;" OwnerUserId="7403" LastEditorUserId="69582" LastEditDate="2021-10-29T17:57:05.767" LastActivityDate="2021-10-29T17:57:05.767" Title="Find the longest arithmetic expression with value equal to number of letters in its english representation" Tags="|calculation-puzzle|optimization|" AnswerCount="4" CommentCount="1" ClosedDate="2021-10-30T19:15:48.020" ContentLicense="CC BY-SA 4.0" />
  <row Id="6302" PostTypeId="2" ParentId="6301" CreationDate="2014-12-19T23:52:44.567" Score="2" Body="&lt;p&gt;So I'll start with this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;One billion to the one billionth power&lt;/code&gt; has 32 characters. We'll call this string &lt;code&gt;POWER&lt;/code&gt;. We'll refer to the value &lt;code&gt;1000000000^1000000000&lt;/code&gt; as the constant &lt;code&gt;BTB&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;plus three minus two minus one&lt;/code&gt; has 25 characters. This evaluates to 0. We'll call this string &lt;code&gt;MINUS&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;plus one plus six plus seven plus fifty&lt;/code&gt; has 32 characters. This evaluates to 64. We'll call this string &lt;code&gt;PLUS&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we place the &lt;code&gt;POWER&lt;/code&gt; string, contact the &lt;code&gt;MINUS&lt;/code&gt; string a number of times equal to &lt;code&gt;BTB / 25&lt;/code&gt; then we will have a string of length BTB + 32 and a value of BTB. We then add the string &lt;code&gt;PLUS&lt;/code&gt; to the end, and we have a length of BTB + 64 and a value of BTB + 64.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the sake of stack exchange's servers, I have only &lt;em&gt;defined&lt;/em&gt; the string, not actually written the string out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can repeat this process ad infinitum by tacking &lt;code&gt;to the twenty billionth power&lt;/code&gt; to the end of the &lt;code&gt;POWER&lt;/code&gt; string, and adding the appropriate number of &lt;code&gt;MINUS&lt;/code&gt; strings to compensate, since they both have 25 CAREacaters (that's characters we care about. Clever, no?) Remember that now that BTB has increased, we still put the &lt;code&gt;MINUS&lt;/code&gt; string BTB times, but we also remove an additional &lt;code&gt;MINUS&lt;/code&gt; string for every additional &lt;code&gt;to the twenty billionth power&lt;/code&gt; because we don't need to compensate for that anymore.&lt;/p&gt;&#xA;" OwnerUserId="2812" LastEditorUserId="2812" LastEditDate="2014-12-19T23:59:29.490" LastActivityDate="2014-12-19T23:59:29.490" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6303" PostTypeId="2" ParentId="6301" CreationDate="2014-12-19T23:58:07.760" Score="3" Body="&lt;p&gt;4 + (1x21) = 25&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Four plus (one times twenty one) = 25 characters&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then continue adding &lt;code&gt;+ (1x21)&lt;/code&gt; to the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 + (1x21) + (1x21) = 46&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Four plus (one times twenty one) plus (one times twenty one) = 46 characters&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 + (1x21) + (1x21) + (1x21) = 67&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Four plus (one times twenty one) plus (one times twenty one) plus (one times twenty one) = 67 characters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can keep doing this forever, so the answer is &lt;strong&gt;infinity&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2014-12-19T23:58:07.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6304" PostTypeId="2" ParentId="6301" CreationDate="2014-12-20T00:00:04.093" Score="9" Body="&lt;p&gt;There is no determinate result to this question.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;plus twenty minus two: equals 18, 18 letters&lt;/p&gt;&#xA;&lt;p&gt;plus twenty minus two plus twenty minus two: equals 36, 36 letters&lt;/p&gt;&#xA;&lt;p&gt;plus twenty minus two plus twenty minus two plus twenty minus two: equals 54, 54 letters&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;and so on...&lt;/p&gt;&#xA;" OwnerUserId="4394" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-12-20T11:20:39.900" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6305" PostTypeId="2" ParentId="6301" CreationDate="2014-12-20T00:07:16.193" Score="3" Body="&lt;p&gt;The &quot;most interesting&quot; is a little bit subjective. I will try my best, though.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;code&gt;the square root of six to the fourth&lt;/code&gt; $=36$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;code&gt;minus e to the i pi plus two to the fourth times four minus two&lt;/code&gt; $=63$*&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;code&gt;one fourth of six to the third minus eleven&lt;/code&gt; $=43$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More will surely follow when found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;*Follow order of operations and this evaluates to 1 + (4*16) - 2&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1664" LastEditorUserId="1664" LastEditDate="2014-12-20T00:12:17.153" LastActivityDate="2014-12-20T00:12:17.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6306" PostTypeId="2" ParentId="6272" CreationDate="2014-12-20T03:10:41.943" Score="1" Body="&lt;p&gt;This is a work in progress, not a final answer. Mainly, I'll start by attacking the smaller piece of code (HINT #2). It is worthy to note that this can be solved with a &lt;em&gt;pen&lt;/em&gt; and &lt;em&gt;pad&lt;/em&gt;, so it's not going to (likely) include trigonometry or something like that.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;IF YOU PLEASE GOOD SIR I THINK THAT A HINT IS IN ORDER IOLEOI_FI_TH_ITUI__EO_ESTRI_O_IDSADRNRPKSNAHNTYH__GA__&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;s&gt;This sorted in order of length (small &lt;em&gt;x&lt;/em&gt; to large &lt;em&gt;x&lt;/em&gt;, &lt;em&gt;a&lt;/em&gt; to &lt;em&gt;z&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plaintext:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A I IF IN IS SIR YOU GOOD HINT THAT ORDER THINK PLEASE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ciphertext:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;O EO FI GA TH ITUI ESTRI ILOEOI IDSADRNRPKSNAHNTYH&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is obvious that there is no direct, easily findable order relation between the cipher text and the plaintext. However, it is worthy to note that there is some sort of order relation involving the transformation of a certain character into a space, as there are three instances of a double underscore (&lt;code&gt;__&lt;/code&gt;), which might not occur otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be inferred that the a letter occurring the same number of times as the spaces would be (if it is) the direct correlation. It is a feeble attempt (I'm sure XNOR would've thought of a Caeser cipher by now), but it is worth trying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Occurrences in Plaintext:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;#E: 3&lt;br&gt;&#xA;    #I: 7&lt;br&gt;&#xA;    #O: 4&lt;br&gt;&#xA;    #A: 3&lt;br&gt;&#xA;    #N: 3&lt;br&gt;&#xA;    #H: 3&lt;br&gt;&#xA;    #L: 1&lt;br&gt;&#xA;    #_: 12&lt;br&gt;&#xA;    #T: 4&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That seems like enough from now. Feel free to take this were you will, I will now retire (to bed, that is :P).&lt;/s&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above is the result of a misreading of the instructions. Feel free to analyze it, for it will get you no where.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what I now know: The letters are arranged in a transposition, where each letter is moved to another space. I have no idea where to go, so, since it involves geometry, I'll start doing stuff with geometry.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;IOLEOI_FI_TH_ITUI__EO_ESTRI_O_IDSADRNRPKSNAHNTYH__GA__&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is $54$ characters long; $\sqrt{54}$ is remarkably close to $e^2$, this may or may not be important, as the difference is $0.04058687058$, significant enough as to not be relevant. $54$, however, is able to be put into a few differently-sized rectangles:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;WIP&#xA;&lt;/pre&gt;&#xA;" OwnerUserId="7307" LastEditorUserId="7307" LastEditDate="2014-12-20T19:56:28.117" LastActivityDate="2014-12-20T19:56:28.117" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6308" PostTypeId="2" ParentId="6273" CreationDate="2014-12-20T03:51:07.917" Score="0" Body="&lt;p&gt;Infamous depressed depressed Infamous  35 characters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that infamous can mean both well known or not popular.&#xA;Depressed means sad or a button pressed in for a positive response.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a famous person who is depressed presses the depressed button that says Infamous on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has a lot of letters, but maybe it can lead us to other shorter words that can have opposite meanings.&lt;/p&gt;&#xA;" OwnerUserId="7406" LastActivityDate="2014-12-20T03:51:07.917" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6309" PostTypeId="2" ParentId="6273" CreationDate="2014-12-20T04:01:55.930" Score="0" Body="&lt;p&gt;Worn retired retired worn   24 characters&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Worn - used up&#xA;retired - stopped working&#xA;retired - did work to change a tire&#xA;Worn - in use as clothing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe someone can make this into a sentence.  It seems close to an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kevin&lt;/p&gt;&#xA;" OwnerUserId="7406" LastActivityDate="2014-12-20T04:01:55.930" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6310" PostTypeId="2" ParentId="6282" CreationDate="2014-12-20T04:13:37.400" Score="2" Body="&lt;p&gt;Her name was Def and she had a sea snail. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The notes on the Pythagorean scale are: D E F# and A, or DEFsharpA - Def's &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Harpa&quot; rel=&quot;nofollow&quot;&gt;Harpa&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2162" LastActivityDate="2014-12-20T04:13:37.400" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6311" PostTypeId="2" ParentId="6240" CreationDate="2014-12-20T06:35:39.937" Score="6" Body="&lt;p&gt;Starting with @Lopsy's answer for 6 holes, I was able to get a larger value. Here's how I went about it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, I assume that if two blue crosses are equidistant from the drill, then the drill can choose either, and I should assume that the smaller total path is the total distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With 2 holes, the holes must go in opposite corners.&lt;br&gt;&#xA;$(0, 0), (1, 1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With 3 holes, the holes go in any 3 corners, starting with the one adjacent to the other two.&lt;br&gt;&#xA;$(0, 0), (1, 0), (0, 1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With 4 holes, the holes go in 3 corners and on the diagonal, making sure the diagonal length is less than 1. Start with corner adjacent to the other two.&lt;br&gt;&#xA;$(0, 0), (\sqrt2/2, \sqrt2/2), (1, 0), (0, 1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With 5 holes it gets interesting. Using the existing holes, the obvious place for the 5th hole is the empty corner. However, once this is done, the center hole can be moved to give a longer distance. It maximizes at $(\sqrt3/2, 0.5)$.&lt;br&gt;&#xA;$(0, 0), (\sqrt3/2, 0.5), (1, 1), (1, 0), (0, 1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the 6th hole, there is now an equilateral triangle with vertices at $(0, 0), (0, 1), (\sqrt3/2, 0.5)$. The sixth hole goes in the center of this triangle, at $(\sqrt3/6, 0.5)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tweaking the 2 non-corner points and starting in the center of the triangle gives the following 6 points in the order they would be drilled:&lt;br&gt;&#xA;$(0.288674, 0.499999), (0, 0), (0.866024, 0.500001), (1, 1), (0, 1), (1, 0)$&lt;br&gt;&#xA;The total distance is 4.509199 meters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Tmkiz.png&quot; alt=&quot;6 dots&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the 7th dot, there's a large space in the upper half of the area with nothing, so getting it as far from $(0.255674, 0.499999)$ as possible while still making it the closest dot gives a point of $(0.500000, 0.955388)$. With this as the new starting point you have the following 7 points in the order they would be drilled:&lt;br&gt;&#xA;$(0.500000, 0.955388), (0.288674, 0.499999), (0, 0), (0.866024, 0.500001), (1, 1), (0, 1), (1, 0)$&lt;br&gt;&#xA;The total distance is 5.011189 meters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gcnFW.png&quot; alt=&quot;7 dots&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While looking for 6 dots, I tried various placements. For example, I started with the full area and tried points at every 0.1 meter point, then picked the best ones and scaled down to 0.01 meter and so on. So I'm fairly confident in my answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 7 dots, I assumed the positioning of the first 6 dots wouldn't change, found a spot for the 7th dot, then tested the values to make sure they were at least a local maximum (meaning that there might be a longer distance, but there wouldn't be one using the same approximate positioning). So I think my answer is correct, but I wouldn't be surprised if a drastic re-positioning of all 3 points turned out a larger distance.&lt;/p&gt;&#xA;" OwnerUserId="375" LastEditorUserId="375" LastEditDate="2014-12-20T22:02:44.533" LastActivityDate="2014-12-20T22:02:44.533" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6312" PostTypeId="2" ParentId="6252" CreationDate="2014-12-20T12:23:55.927" Score="7" Body="&lt;p&gt;In early mathematical puzzles, the &quot;&lt;a href=&quot;http://www.math.buffalo.edu/mad/Ancient-Africa/mad_ancient_egypt_algebra.html#rhind79&quot;&gt;inventory problem&lt;/a&gt;&quot; is addressed in the Egyptian Rhind papyrus, and &lt;a href=&quot;http://www.math.buffalo.edu/mad/Ancient-Africa/mad_ancient_egyptpapyrus.html&quot;&gt;may go back to 2650 BC / BCE&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="228" LastActivityDate="2014-12-20T12:23:55.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6313" PostTypeId="2" ParentId="6282" CreationDate="2014-12-20T12:34:05.643" Score="4" Body="&lt;p&gt;It is indeed Mary Had A Little Lamb. The lines correspond to the notes C, D, E and G (as per the C major scale in &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Pythagorean_tuning&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt; from hint 2) and if you play them in the positions indicated by the tick marks in the long horizontal lines you get the familiar tune. Like so:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                                      (Its)&#xA;C ----|-------------------------|-------------------------|--&#xA;      had                       had                      snow&#xA;D --|---|------|||------------|---|----------||-------|-------&#xA;    ry   a   little lamb      ry   a     fleece was   as&#xA;E |-------|||--------|------|-------|||----------|-----------&#xA;  Ma-   little lamb  li-    Ma-   little lamb   white&#xA;G ---------------------||------------------------------------&#xA;                     -ttle lamb&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2162" LastEditorUserId="2162" LastEditDate="2014-12-20T21:17:06.793" LastActivityDate="2014-12-20T21:17:06.793" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6314" PostTypeId="1" AcceptedAnswerId="6317" CreationDate="2014-12-20T15:38:27.827" Score="-7" ViewCount="612" Body="&lt;p&gt;&lt;strong&gt;Current Best: &lt;em&gt;13 characters&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This time I'm asking you guys to determine &lt;em&gt;the shortest possible grammatically correct english sentence&lt;/em&gt; that includes &lt;strong&gt;2 pairs of synonyms&lt;/strong&gt; (different words that mean the same). For checking character count you can use the site &lt;a href=&quot;http://www.javascriptkit.com/script/script2/charcount.shtml&quot; rel=&quot;nofollow&quot;&gt;http://www.javascriptkit.com/script/script2/charcount.shtml&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To determine if a word is a synonym, it either has to be obvious, or has to be confirmed by thesaurus.com.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The accepted answer will always be changed to adjust to the shortest answer which follows all the rules. So any answer can be &lt;em&gt;bested&lt;/em&gt; at any point by another solver!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example incase any of you are confused;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Characters: 64&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What a beautiful, giant flower! Almost as pretty as my big tree!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Beautiful and Pretty; To look good.&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;Giant and Big; To be large in size.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Please note your answer will not be accepted if it ties the accepted answer!&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-12-21T16:33:06.173" LastActivityDate="2014-12-21T16:33:06.173" Title="Shortest sentence with 2 sets of synonyms?" Tags="|open-ended|language|english|" AnswerCount="3" CommentCount="11" ClosedDate="2021-05-04T08:31:35.537" ContentLicense="CC BY-SA 3.0" />
  <row Id="6315" PostTypeId="2" ParentId="6273" CreationDate="2014-12-20T15:40:02.200" Score="1" Body="&lt;p&gt;&lt;strong&gt;Fix&lt;/strong&gt; the &lt;strong&gt;rules&lt;/strong&gt;, don't &lt;strong&gt;fix&lt;/strong&gt; the &lt;strong&gt;rules&lt;/strong&gt;. (35)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Fix&lt;/strong&gt; (as in: for the love of English, &lt;em&gt;stop changing&lt;/em&gt;) the &lt;strong&gt;rules&lt;/strong&gt; (a set of immutable guidelines), don't &lt;strong&gt;fix&lt;/strong&gt; (incessantly edit in vain attempts to repair) the &lt;strong&gt;rules&lt;/strong&gt; (an ever-changing and unenforced bunch of suggestions that apparently no one is taking seriously).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's all I got.&lt;/p&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2014-12-20T15:40:02.200" CommentCount="1" CommunityOwnedDate="2014-12-20T15:40:02.200" ContentLicense="CC BY-SA 3.0" />
  <row Id="6316" PostTypeId="2" ParentId="6314" CreationDate="2014-12-20T15:53:06.037" Score="3" Body="&lt;h2&gt;15 Characters&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Tell me, I say!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whether this works depends on whether I and me are considered synonyms, both referring to myself, of course.  Clarified as allowed by OP.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;25 Characters&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Heat drops as temps fall.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Heat and temps: refer to temperature&lt;br&gt;&#xA;Drops and fall: to become lower&lt;/p&gt;&#xA;" OwnerUserId="6178" LastEditorUserId="6178" LastEditDate="2014-12-20T16:32:25.850" LastActivityDate="2014-12-20T16:32:25.850" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6317" PostTypeId="2" ParentId="6314" CreationDate="2014-12-20T15:59:40.093" Score="4" Body="&lt;h2&gt;15 Characters&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I gaze; see me!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I and me: first person plural pronoun. Gaze and see: use one's visual sense.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;14 Characters&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I rise; up me!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I and me: first person plural pronoun. Rise and up: increase or raise upwards.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;13 Characters&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An os, a gob.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;*An and a: indefinite article. &lt;a href=&quot;http://www.scrabblefinder.com/word/os/&quot; rel=&quot;nofollow&quot;&gt;Os&lt;/a&gt; and &lt;a href=&quot;http://www.scrabblefinder.com/word/gob/&quot; rel=&quot;nofollow&quot;&gt;gob&lt;/a&gt;: mouth.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2014-12-20T16:37:21.813" LastActivityDate="2014-12-20T16:37:21.813" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6320" PostTypeId="2" ParentId="6195" CreationDate="2014-12-20T17:03:13.017" Score="11" Body="&lt;p&gt;&lt;strong&gt;This is a two-part answer.&lt;/strong&gt; First, it establishes a non-trivial upper bound of 5. Then a solution is given that proves 5 is also the lower bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is only one 3x3 magic square, up to symmetry:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;294&#xA;753&#xA;618&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So in order to have seven magic squares in a Sudoku, we require seven &lt;code&gt;5&lt;/code&gt;s which aren't against an edge, and that requires two diagonally opposite corners to be &lt;code&gt;5&lt;/code&gt;s. If we then look at parity (&lt;code&gt;e&lt;/code&gt; for even, &lt;code&gt;o&lt;/code&gt; for odd):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;eoe&#xA;o5o&#xA;eoe&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and consider the possible positions of the other &lt;code&gt;5&lt;/code&gt;s, we eliminate almost all positions. E.g. consider the following grid:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;5.. eoe ...&#xA;... o5o ...&#xA;... eoe ...&#xA;&#xA;... ... ...&#xA;... ... ...&#xA;... ... ...&#xA;&#xA;... ... ...&#xA;... ... ...&#xA;... ... ..5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;By Sudoku rules there are two possible places for the &lt;code&gt;5&lt;/code&gt; in the top-right block, but the parity matching prohibits the &lt;code&gt;5&lt;/code&gt;s from the magic squares being horizontally or vertically adjacent to an even number, so there's only one possible place:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;5.. eoe ...&#xA;... o5o eoe&#xA;... eoe o5o&#xA;&#xA;... ... eoe&#xA;... ... ...&#xA;... ... ...&#xA;&#xA;... ... ...&#xA;... ... ...&#xA;... ... ..5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The &lt;code&gt;5&lt;/code&gt; in the right-centre block is similarly forced; then the one in the centre block;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;5.. eoe ...    5.. eoe ...&#xA;... o5o eoe    ... o5o eoe&#xA;... eoe o5o    ... eoe o5o&#xA;&#xA;... ... eoe    ..e oe. eoe&#xA;... ..e oe.    ..o 5oe oe.&#xA;... ..o 5o.    ..e oeo 5o.&#xA;&#xA;... ..e oe.    ... ..e oe.&#xA;... ... ...    ... ... ...&#xA;... ... ..5    ... ... ..5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and we can't place any &lt;code&gt;5&lt;/code&gt; in the left-centre block.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the aid of a computer to check the 576 cases, this simple parity matching eliminates all but one of them:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;5.e oe. ...&#xA;..o 5oe oe.&#xA;eoe oeo 5o.&#xA;&#xA;o5o eoe oe.&#xA;eoe o5o eoe&#xA;.eo eoe o5o&#xA;&#xA;.o5 oeo eoe&#xA;.eo eo5 o..&#xA;... .eo e.5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If we zoom in on the top-left 3 blocks:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;5.e oe.&#xA;..o 5o.&#xA;eoE oe.&#xA;&#xA;o5o&#xA;eoe&#xA;...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;the two magic squares overlap on one even numbered cell, which I've labelled &lt;code&gt;E&lt;/code&gt;. Without loss of generality, let &lt;code&gt;E = 2&lt;/code&gt;. Then the numbers which are two cells up, down, left, and right of &lt;code&gt;E&lt;/code&gt; are suitably paired up &lt;code&gt;4&lt;/code&gt;s and &lt;code&gt;6&lt;/code&gt;s. But either we have two &lt;code&gt;4&lt;/code&gt;s in the same row as &lt;code&gt;E&lt;/code&gt;, or we have two &lt;code&gt;4&lt;/code&gt;s in the same column as &lt;code&gt;E&lt;/code&gt;, or we have two &lt;code&gt;4&lt;/code&gt;s or two &lt;code&gt;6&lt;/code&gt;s in the same block as &lt;code&gt;E&lt;/code&gt;. So the only grid which meets the parity constraints for seven magic squares fails as soon as we try to go beyond parity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look for solutions with 6 magic squares we keep one &lt;code&gt;5&lt;/code&gt; in a corner and allow the others to vary. There are 29 positions of &lt;code&gt;5&lt;/code&gt;s which satisfy the parity check, and they all fail to the same kind of blockage between overlapping magic squares. So the upper bound for a solution is 5.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Solution with five magic squares (contributed by &lt;a href=&quot;https://puzzling.stackexchange.com/users/1563/geobits&quot;&gt;Geobits&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wlTIo.png&quot; alt=&quot;5-magic-sudoku&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found a template for this at &lt;a href=&quot;http://www.sachsentext.de/en/node/825&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.sachsentext.de&lt;/a&gt;. Filling it in was a pain, but it's much easier once you realize that rotating/reflecting any of the magic squares causes it to break. Then you just have to fill in the &quot;non-magic&quot; spaces, which isn't much different than a regular sudoku board with a ton of givens.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(Peter resumes) Knuth's reduction of Sudoku to exact set cover can be augmented to add the magic square constraint. The reduction for standard Sudoku is to associate each $(\textrm{row}, \textrm{column}, \textrm{value})$ tuple with the set $\{\textrm{Row}(\textrm{row}, \textrm{value}), \textrm{Col}(\textrm{column}, \textrm{value}), \textrm{Block}(\lfloor\frac{\textrm{row}}3\rfloor, \lfloor\frac{\textrm{column}}3\rfloor, \textrm{value}), \textrm{Cell}(\textrm{row}, \textrm{column})\}$. Then an exact set cover of the union of all those sets provides exactly one instance of each value in each row, column, and block, and exactly one value per cell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To extend for a magic square centred on $(r, c)$ we add a constraint for each of the four edges of the magic square that it must be one of the four sets $\{2,9,4\}, \{2,6,7\}, \{4,3,8\}, \{6,1,8\}$. This is sort-of related to the &lt;a href=&quot;https://puzzling.stackexchange.com/q/133/228&quot;&gt;general lock puzzle&lt;/a&gt;: we can solve it with four &quot;locks&quot; and the following assignments:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2: A&#xA;4:  B&#xA;6:   C&#xA;8:    D&#xA;1: AB&#xA;3: A C&#xA;7:  B D&#xA;9:   CD&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This isn't quite strong enough to avoid false positives, but it narrows down the field considerably. With these extra constraints, and iterating over the possible positions of &quot;internal&quot; fives, I find 64 grids. Interestingly, some include &quot;internal&quot; fives which aren't from magic squares: e.g.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;843 127 695&#xA;276 895 143&#xA;951 643 827&#xA;        v&#xA;438 276 519&#xA;627 951 384&#xA;195 438 276&#xA;  ^&#xA;382 764 951&#xA;769 512 438&#xA;514 389 762&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;All of the ones I've checked have the same (IMO) structural defect that the magic squares are all in the same orientation. I haven't checked, but I think they may also all have a symmetry under rotation by 180 degrees and permutation of digits by $(19)(28)(37)(46)$.&lt;/p&gt;&#xA;" OwnerUserId="228" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2014-12-22T23:28:08.190" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6321" PostTypeId="1" AcceptedAnswerId="6537" CreationDate="2014-12-20T17:21:29.127" Score="3" ViewCount="7143" Body="&lt;p&gt;&lt;strong&gt;Current Best; Babingtonites =  5371207621898; 13 letters&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We all know from fooling around with a calculator at some point in our lives, the when we turn a calculator upside down we can make words. &lt;em&gt;For example, 0.7734 being HELLO. Or 808 being BOB.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now assuming your calculator screen can stretch as far as you like, what's the longest word you can possibly spell?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Numbers on a calculator translate the same as on a digital clock&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;RULES:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Numbers can be used as many times as you like.&lt;/li&gt;&#xA;&lt;li&gt;You cannot use operators (+, -, /, *, etc.) in your word.&lt;/li&gt;&#xA;&lt;li&gt;The decimal sign may only be used once. You can only use it as &lt;strong&gt;A)&lt;/strong&gt; a seperator to put 0 at the beginning of the number (to put O at the end of the word) or &lt;strong&gt;B)&lt;/strong&gt; an apostrophe.&lt;/li&gt;&#xA;&lt;li&gt;A number can only be used as a single letter in your word, and should not double!&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Accepted answer will change to always be the longest!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Note: Please don't ruin the fun by googling!  *&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2349" LastEditDate="2017-01-23T09:36:57.057" LastActivityDate="2018-07-27T22:04:07.860" Title="Longest Calculator Word?" Tags="|mathematics|open-ended|language|english|" AnswerCount="5" CommentCount="6" ClosedDate="2021-05-14T09:40:36.833" ContentLicense="CC BY-SA 3.0" />
  <row Id="6322" PostTypeId="2" ParentId="6321" CreationDate="2014-12-20T17:32:56.547" Score="14" Body="&lt;h2&gt;9 letters&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;BIOLOGIES = 531607018&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;&lt;s&gt;11 letters&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;HELIOLOGIES = 53160701734&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Definition: &lt;a href=&quot;http://www.thefreedictionary.com/heliology&quot;&gt;http://www.thefreedictionary.com/heliology&lt;/a&gt;&lt;/s&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;12 letters&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;BIBLIOLOGIES = 531607017818&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Definition: &lt;a href=&quot;http://dictionary.reference.com/browse/bibliologies&quot;&gt;http://dictionary.reference.com/browse/bibliologies&lt;/a&gt;&lt;br&gt;&#xA;Accepted by Scrabble: &lt;a href=&quot;http://www.scrabblewordsolver.com/word-checker/bibliologies&quot;&gt;http://www.scrabblewordsolver.com/word-checker/bibliologies&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6178" LastEditorUserId="6178" LastEditDate="2014-12-20T17:55:42.840" LastActivityDate="2014-12-20T17:55:42.840" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6323" PostTypeId="1" AcceptedAnswerId="6325" CreationDate="2014-12-20T17:56:57.513" Score="12" ViewCount="872" Body="&lt;p&gt;I did something funny.  I took a simple phrase.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;HELLO, WORLD!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;I generated all rotations of the string.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;HELLO, WORLD!&#xA;ELLO, WORLD!H&#xA;LLO, WORLD!HE&#xA;LO, WORLD!HEL&#xA;O, WORLD!HELL&#xA;, WORLD!HELLO&#xA; WORLD!HELLO,&#xA;WORLD!HELLO, &#xA;ORLD!HELLO, W&#xA;RLD!HELLO, WO&#xA;LD!HELLO, WOR&#xA;D!HELLO, WORL&#xA;!HELLO, WORLD&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Then I sorted the strings.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt; WORLD!HELLO,&#xA;!HELLO, WORLD&#xA;, WORLD!HELLO&#xA;D!HELLO, WORL&#xA;ELLO, WORLD!H&#xA;HELLO, WORLD!&#xA;LD!HELLO, WOR&#xA;LLO, WORLD!HE&#xA;LO, WORLD!HEL&#xA;O, WORLD!HELL&#xA;ORLD!HELLO, W&#xA;RLD!HELLO, WO&#xA;WORLD!HELLO, &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;And I collected the last character of each line.  I got the string:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;&amp;quot;,DOLH!RELLWO &amp;quot;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;As you can see, it is the original message scrambled.  But this was just an example.  I did the same thing with a much longer message and I got the following string:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;&amp;quot;DFKDKKWHKAADD  UDD  UUUU LLOOOOIC?CCCC CCCCUU OOOOWWWWWCHMHHHHOOO     &amp;quot;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Can you tell me what the message was?&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="69582" LastEditDate="2024-01-17T16:35:13.960" LastActivityDate="2024-01-17T16:35:13.960" Title="Descramble the scrambled text" Tags="|cipher|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="6324" PostTypeId="2" ParentId="6323" CreationDate="2014-12-20T18:31:05.170" Score="9" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;All of the rotations of the string are:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?&#xA;OW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?H&#xA;W MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HO&#xA; MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW&#xA;MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW &#xA;UCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW M&#xA;CH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MU&#xA;H WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUC&#xA; WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH&#xA;WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH &#xA;OOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH W&#xA;OD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WO&#xA;D WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOO&#xA; WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD&#xA;WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD &#xA;OULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD W&#xA;ULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WO&#xA;LD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOU&#xA;D A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOUL&#xA; A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD&#xA;A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD &#xA; WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A&#xA;WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A &#xA;OODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A W&#xA;ODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WO&#xA;DCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOO&#xA;CHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOOD&#xA;HUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODC&#xA;UCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCH&#xA;CK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHU&#xA;K CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUC&#xA; CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK&#xA;CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK &#xA;HUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK C&#xA;UCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CH&#xA;CK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHU&#xA;K IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUC&#xA; IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK&#xA;IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK &#xA;F A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK I&#xA; A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF&#xA;A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF &#xA; WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A&#xA;WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A &#xA;OODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A W&#xA;ODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WO&#xA;DCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOO&#xA;CHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOOD&#xA;HUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODC&#xA;UCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCH&#xA;CK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHU&#xA;K COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUC&#xA; COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK&#xA;COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK &#xA;OULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK C&#xA;ULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK CO&#xA;LD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COU&#xA;D CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COUL&#xA; CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD&#xA;CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD &#xA;HUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD C&#xA;UCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CH&#xA;CK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHU&#xA;K WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUC&#xA; WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK&#xA;WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK &#xA;OOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK W&#xA;OD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WO&#xA;D?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOO&#xA;?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The sorted strings are:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD&#xA; A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF&#xA; CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK&#xA; CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD&#xA; COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK&#xA; IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK&#xA; MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW&#xA; WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH&#xA; WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK&#xA; WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A&#xA; WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A&#xA; WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD&#xA;?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD&#xA;A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD &#xA;A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF &#xA;CH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MU&#xA;CHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOOD&#xA;CHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOOD&#xA;CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK &#xA;CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD &#xA;CK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHU&#xA;CK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHU&#xA;CK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHU&#xA;CK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHU&#xA;COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK &#xA;D A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOUL&#xA;D CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COUL&#xA;D WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOO&#xA;D?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOO&#xA;DCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOO&#xA;DCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOO&#xA;F A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK I&#xA;H WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUC&#xA;HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?&#xA;HUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODC&#xA;HUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODC&#xA;HUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK C&#xA;HUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD C&#xA;IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK &#xA;K CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUC&#xA;K COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUC&#xA;K IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUC&#xA;K WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUC&#xA;LD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOU&#xA;LD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COU&#xA;MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW &#xA;OD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WO&#xA;OD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WO&#xA;ODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WO&#xA;ODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WO&#xA;OOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH W&#xA;OOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK W&#xA;OODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A W&#xA;OODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A W&#xA;OULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD W&#xA;OULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK C&#xA;OW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?H&#xA;UCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW M&#xA;UCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCH&#xA;UCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCH&#xA;UCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CH&#xA;UCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CH&#xA;ULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WO&#xA;ULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK CO&#xA;W MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HO&#xA;WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH &#xA;WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK &#xA;WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A &#xA;WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A &#xA;WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?HOW MUCH WOOD &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4450" LastEditDate="2014-12-21T06:29:29.877" LastActivityDate="2014-12-21T06:29:29.877" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6325" PostTypeId="2" ParentId="6323" CreationDate="2014-12-20T18:38:28.977" Score="24" Body="&lt;p&gt;This procedure is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Burrows%E2%80%93Wheeler_transform&quot;&gt;Burrows-Wheeler transform&lt;/a&gt;. It is useful for data compression, because if the starting text contains many identical substrings (&quot;the&quot;, &quot;and&quot;, ...), then the Burrows-Wheeler transformation will contain many runs of identical characters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying the inverse algorithm on that Wikipedia page, with a suitable guess of the location of the end-of-file marker &quot;|&quot;, gives the answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;HOW MUCH WOOD WOULD A WOODCHUCK CHUCK IF A WOODCHUCK COULD CHUCK WOOD?&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2014-12-20T18:38:28.977" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6326" PostTypeId="2" ParentId="6314" CreationDate="2014-12-20T18:39:37.063" Score="1" Body="&lt;h1&gt;12 characters&lt;/h1&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Be me? I am.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I know this ties the best answer, but it's also a good one. &lt;em&gt;I&lt;/em&gt; and &lt;em&gt;me&lt;/em&gt; are both first person singular pronouns, and &lt;em&gt;be&lt;/em&gt; and &lt;em&gt;am&lt;/em&gt; are both forms of the infinitive &lt;em&gt;to be&lt;/em&gt;. This could be a response to the common statement &amp;quot;Be yourself,&amp;quot; so it is valid sentence/pair of sentences.&lt;/p&gt;&#xA;" OwnerUserId="1664" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-12-20T18:39:37.063" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6327" PostTypeId="2" ParentId="6272" CreationDate="2014-12-20T21:06:54.407" Score="2" Body="&lt;p&gt;This is also &lt;strong&gt;Work in Progress&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with the shorter text of Hint 4:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The clear text can be written in rectangle. As the clear text is&#xA; &lt;pre&gt;IF_YOU_PLEASE_GOOD_SIR_I_THINK_THAT_A_HINT_IS_IN_ORDER&lt;/pre&gt;&#xA; and the cipher text is&lt;br&gt;&#xA; &lt;pre&gt;IOLEOI_FI_TH_ITUI__EO_ESTRI_O_IDSADRNRPKSNAHNTYH__GA__&lt;/pre&gt;&#xA; I take the letters of the cipher text as staring letters of the line-breaks, giving me a break every 4 letters:&lt;br&gt;&#xA; &lt;pre&gt; IF_Y&#xA; OU_P&#xA; LEAS&#xA; E_GO&#xA; OD_S&#xA; IR_I&#xA; _THI&#xA; NK_T&#xA; HAT_&#xA; A_HI&#xA; NT_I&#xA; S_IN&#xA; _ORD&#xA; ER &lt;/pre&gt;&lt;br&gt;&#xA; This is already very encouraging, as the first 7 letters come out correct, but then it breaks down. So I think this is the way to go: Arrange given letters in a geometry, and read out into another direction. I'll have to play with some more geometries now...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastActivityDate="2014-12-20T21:06:54.407" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6330" PostTypeId="1" CreationDate="2014-12-21T03:09:37.250" Score="4" ViewCount="1541" Body="&lt;p&gt;How many 7 by 7 Latin squares are there, and how can one generate examples of them? I am working on classes for a game, and typing in numbers manually from 1-7 in rows and columns from scratch is exceedingly difficult. From what I can tell there are no generators for this type of thing online.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes I can type in 1-2-3-4-5-6-7, then the next row 3-4-5-6-7-1-2, but that does not work so well with stats of classes in a game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diagonals do NOT have to be 1-7. Rows and columns must have a unique number per square.&lt;/p&gt;&#xA;" OwnerUserId="7412" LastEditorUserId="4551" LastEditDate="2014-12-21T11:08:40.570" LastActivityDate="2014-12-24T10:52:19.130" Title="How to generate and count the 7x7 Latin squares" Tags="|sudoku|latin-square|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6331" PostTypeId="2" ParentId="6069" CreationDate="2014-12-21T03:28:48.500" Score="2" Body="&lt;p&gt;Work in progress:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the quick brown fox example, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;T is in position 1 in the clear. T is the 20th letter of the alphabet. In the cipher text, there's a T at position 1 + 20 + 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H is in position 2 in the clear. H is the 8th letter of the alphabet. In the cipher text, there's an H at position 2 + 8 + 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E is in position 3 in the clear. E is the 5th letter of the alphabet. In the cipher text, there's an E at position 3 + 5 + 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q is in position 5 in the clear (4 is a space). Q is the 17th letter of the alphabet. In the cipher text, there's a Q at position 5 + 17 + ... 2. So it's not perfect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuing, U is in position 6 in the clear. U is the 21st letter of the alphabet. In the cipher text, there's a U at position 6 + 21 + ... 0. Again, not perfect but eerily close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I: 7 + 9 + 0. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;C: 8 + 3 + 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;K: 9 + 11 + -1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's something like a pattern here or else it's just coincidence that these letters are found shifted ahead by approximately their alphabetic position. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I was just thinking, if I shift a letter, say T, ahead by 20 then the 20 characters skipped over are effectively shifted left by 1. So every time a letter is jumped over by another letter it is displaced backward. I'm also thinking that if every letter gets moved forward by its alphabetic position then As shift few letters while Zs shift many. All else being equal, a letter 20 spaces ahead of a cluster of As will move less than one 20 spaces ahead of some Zs. This may account for the variability. &lt;/p&gt;&#xA;" OwnerUserId="2162" LastEditorUserId="2162" LastEditDate="2014-12-21T12:48:13.247" LastActivityDate="2014-12-21T12:48:13.247" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6332" PostTypeId="2" ParentId="6330" CreationDate="2014-12-21T03:40:14.993" Score="8" Body="&lt;p&gt;There are 61479419904000 seven by seven Latin squares. This is &lt;a href=&quot;http://oeis.org/A002860&quot;&gt;from OEIS&lt;/a&gt;, and hard to calculate in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you just want to generate Latin squares quickly, you have a few options. One is quick and lazy: take a few template squares, and then generate more by permuting the rows, columns, and/or numbers. If you're comfortable with programming, you can write a quick backtracking solver. Or you could use &lt;a href=&quot;http://hamsterandwheel.com/grids/index2d.php&quot;&gt;this site&lt;/a&gt;. (Easy to Google once you know they're called, eh?)&lt;/p&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2014-12-21T03:40:14.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6333" PostTypeId="1" AcceptedAnswerId="6336" CreationDate="2014-12-21T04:07:44.530" Score="2" ViewCount="421" Body="&lt;p&gt;Can you determine any number of magic squares, that when treated as matrices, can be applied mathematical operations to return a new magic-matrix?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;You cannot use the same matrix twice!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The answer should be given as a mathematical equation using the matrices you find.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Magic squares follow the general magic square rules, they contain 1, .., $n \times n$ once, and each row, column, and diagonal sum to 1 number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Allowed Operations: Add, Subtract, Multiply, Exponention, Logarithms&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="51105" LastEditDate="2021-10-18T13:17:20.190" LastActivityDate="2021-10-18T13:17:20.190" Title="Magic Matrices?" Tags="|mathematics|formation-of-numbers|magic-square|" AnswerCount="2" CommentCount="7" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6335" PostTypeId="2" ParentId="6333" CreationDate="2014-12-21T04:57:03.633" Score="5" Body="&lt;p&gt;If we consider two magic squares that differ by a rotation distinct, then here's an example with $3\times 3$ squares. Let&#xA;$$&#xA;M_1=\left[\begin{array}{ccc}4&amp;amp;9&amp;amp;2\\3&amp;amp;5&amp;amp;7\\8&amp;amp;1&amp;amp;6\end{array}\right],\,M_2=\left[\begin{array}{ccc}8&amp;amp;3&amp;amp;4\\1&amp;amp;5&amp;amp;9\\6&amp;amp;7&amp;amp;2\end{array}\right],\,M_3=\left[\begin{array}{ccc}6&amp;amp;1&amp;amp;8\\7&amp;amp;5&amp;amp;3\\2&amp;amp;9&amp;amp;4\end{array}\right],M_4=\left[\begin{array}{ccc}2&amp;amp;7&amp;amp;6\\9&amp;amp;5&amp;amp;1\\4&amp;amp;3&amp;amp;8\end{array}\right].&#xA;$$&#xA;Then we can compute&#xA;$$&#xA;M_1+M_3=M_2+M_4=\left[\begin{array}{ccc}10&amp;amp;10&amp;amp;10\\10&amp;amp;10&amp;amp;10\\10&amp;amp;10&amp;amp;10\end{array}\right],&#xA;$$&#xA;so that the equation $M_1+M_3-M_2=M_4$ gives a solution to the problem.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastActivityDate="2014-12-21T04:57:03.633" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6336" PostTypeId="2" ParentId="6333" CreationDate="2014-12-21T05:02:39.413" Score="6" Body="&lt;p&gt;It is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fix $n$. Let $M$ denote the set of all $n$-by-$n$ magic squares. $M$ is &lt;em&gt;really huge&lt;/em&gt; - for $n=6$, we have $|M| \sim 10^{19}$ (see &lt;a href=&quot;https://oeis.org/A006052&quot;&gt;OEIS&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider all $2^{|M|}$ sums of subsets of $M$. Each one is an $n$-by-$n$ matrix whose entries are positive integers which are at most $n^2|M|$. There are $(n^2|M|)^{n^2}$ matrices which fit this description.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $|M|$ is large enough, then $2^{|M|} &amp;gt; (n^2|M|)^{n^2}$. Then by Pigeonhole, two of the sums of subsets are equal. Using this equality, we can express one of the magic squares as a sum/difference of others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is &quot;large enough&quot;? Taking logs on the inequality, we find that even $|M|=O(n^3)$ is large enough for sufficiently large $n$. For $n=4,5,6$ this works. I cannot find any references to back this up, but I bet there are enough magic squares that this works for all $n\geq4$.&lt;/p&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2014-12-21T05:02:39.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6337" PostTypeId="2" ParentId="2836" CreationDate="2014-12-21T12:47:48.277" Score="1" Body="&lt;p&gt;Here is a more succinct answer to why it is impossible in 2 weighings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have 8 cases to discriminate.  Identify the ball and whether it is heavy or light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the first weighing, your only choice is to weigh 1 against 1 ball or 2 against 2 balls.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If you weigh 2 against 2, the scale cannot balance, so you have only 2 outcomes.  Each outcome leaves 4 possibilities that you cannot discriminate in a single weighing.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you weigh 1 against 1, and the scale balances, the odd ball is one of the remaining balls, either heavy or light.  It leaves 4 possibilities which, again, one weighing is not enough to discriminate.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note that if you only need to identify the odd ball but not tell whether it is heavy or light, it can be done in 2 weighings.  I think this is what was meant when someone told you it can be done.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="11297" LastEditDate="2015-07-16T16:30:01.243" LastActivityDate="2015-07-16T16:30:01.243" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6338" PostTypeId="1" AcceptedAnswerId="6339" CreationDate="2014-12-21T15:18:35.020" Score="-6" ViewCount="634" Body="&lt;p&gt;&lt;em&gt;On Halloween night, a young boy was trick-or-treating.&lt;/em&gt; It was late at night and he was on his last stop. He knocked at the door and said:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Trick-or-Treat!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The man inside said he had a bag of just 4 candy left. &lt;em&gt;He said there were 1 hard candy, 1 gummy, and 2 lollipops.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He allows him to take 2 candy, and the boy does so. He grabs 2 from the bag &lt;strong&gt;without looking!&lt;/strong&gt; He then states he has at least 1 lollipop, afterwards he asked his mom what the chances are of him having 2 lollipops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;His mom had no idea, but I bet you do! What were the chances of the boy pulling out 2 lollipops that night?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="5373" LastEditDate="2014-12-21T15:52:07.380" LastActivityDate="2014-12-21T23:35:13.930" Title="Trick-Or-Treat Probability Puzzle" Tags="|mathematics|probability|" AnswerCount="1" CommentCount="4" ClosedDate="2015-03-11T16:26:08.013" ContentLicense="CC BY-SA 3.0" />
  <row Id="6339" PostTypeId="2" ParentId="6338" CreationDate="2014-12-21T15:28:06.490" Score="6" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 in 5.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because given that he has at least 1 lollipop, the possibilities are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;1 lollipop, 1 hard candy (probability 1 in 8)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;1 lollipop, 1 gummy (probability 1 in 8)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;2 lollipops (probability 1 in 16)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So the conditional probability of him having 2 lollipops is $\frac{1/16}{\frac{1}{8}+\frac{1}{8}+\frac{1}{16}}=\frac{1}{5}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the OP's behest, I'm also explaining why it looks like the answer should be 1 in 3. (If anyone's &lt;em&gt;not&lt;/em&gt; confused already, let's confuse 'em! :-) ) The boy has 1 lollipop; there are 3 possibilities for the other sweet, of which only 1 is 'the other lollipop'; so it looks like the answer should be 1 in 3. But we know nothing to distinguish the two lollipops. He has 1, yes, but which one did we choose? If the lollipops were one red and one yellow, and we knew he had the &lt;em&gt;red&lt;/em&gt; one, then the answer would be 1 in 3. As it stands, however, the answer is 1 in 5 as proved above.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="39" LastEditDate="2014-12-21T23:35:13.930" LastActivityDate="2014-12-21T23:35:13.930" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6340" PostTypeId="1" AcceptedAnswerId="6341" CreationDate="2014-12-21T16:05:28.157" Score="1" ViewCount="1967" Body="&lt;p&gt;A boy was taken to a magical land of wonder and chocolate, but sadly, a disease broke out in this land! He has to face off the evil Dr. Vanscosh, who caused the disease to break out, in a probability test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dr. Vanscosh gives the young boy 100 chocolate bars, 50 injected with deadly poison. He allows the boy to sort them in 2 bags, &lt;em&gt;and the boy knows which has been poisoned&lt;/em&gt;. The boy can place as many of the 100 in each bag, as long as neither bag is empty!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the Dr. plays a dirty trick: he blindfolds the boy and then he swaps the bags around (&lt;strong&gt;either 1 or 2 times, meaning there's a 50/50 chance of them being in the same spot or swapped&lt;/strong&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The boy, still blindfolded, then needs to take a bar and eat it, he cannot feel the bags, he has to stick his hand in the bag, and take a chocolate bar. If he dies, all chocolate in the world does. If he lives, chocolate will be saved forever and the Dr will be forced to eat his own poison.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What would have been the best way for the boy to organize them? What will his chances be afterwards..?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-12-22T05:21:11.797" LastActivityDate="2014-12-24T11:59:11.620" Title="100 chocolate bars!" Tags="|mathematics|probability|" AnswerCount="4" CommentCount="7" ClosedDate="2015-01-28T15:05:42.690" ContentLicense="CC BY-SA 3.0" />
  <row Id="6341" PostTypeId="2" ParentId="6340" CreationDate="2014-12-21T16:21:24.183" Score="10" Body="&lt;p&gt;I assume neither of the bags may remain empty and all chocolate bars must be sorted in one of the bags.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The boy should put one of the good chocolate bars in one bag and all the other chocolate bars in the other bag. This will give him (and all chocolate in the world) maximum chances to survive, namely $\frac{74}{99}$.&lt;/p&gt;&#xA;" OwnerUserId="4394" LastActivityDate="2014-12-21T16:21:24.183" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6343" PostTypeId="2" ParentId="6340" CreationDate="2014-12-21T17:20:36.550" Score="9" Body="&lt;p&gt;Well, he could of course put half of the good chocolate in each bag, but at the bottom and cover it with the poison-free chocolate. No matter how the bags are swapped around, if he picks one from the top it will always be safe to eat! This is probably considered cheating, however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility: put all bars except a poison-free one in one bag, then the boy can determine by touch which bag has only one bar. That bar is safe to eat. Apparently, this was already mentioned and deemed illegal in the comments.&lt;/p&gt;&#xA;" OwnerUserId="4749" LastEditorUserId="4749" LastEditDate="2014-12-21T22:39:03.573" LastActivityDate="2014-12-21T22:39:03.573" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6344" PostTypeId="1" AcceptedAnswerId="6345" CreationDate="2014-12-22T00:01:28.783" Score="18" ViewCount="772" Body="&lt;p&gt;We are a town in Lincolnshire and a funeral carriage; a mechanism for filling water tanks and a book of photographs; and, according to hearsay, a strong form of uniform continuity and a semi-precious stone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do we share?&lt;/p&gt;&#xA;" OwnerUserId="7363" LastEditorUserId="7363" LastEditDate="2014-12-22T00:23:31.700" LastActivityDate="2019-08-12T23:16:28.713" Title="What do we share?" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6345" PostTypeId="2" ParentId="6344" CreationDate="2014-12-22T01:03:52.307" Score="17" Body="&lt;p&gt;You share a fear of censorship. All the clues point to things that contain  insults/curses. The ones that are &quot;hearsay&quot; contain curses that are heard, not spelled. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A town in Lincolnshire:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Scunthorpe&quot; rel=&quot;noreferrer&quot;&gt;S&lt;strong&gt;cunt&lt;/strong&gt;horpe&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Funeral carriage:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He&lt;strong&gt;arse&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mechanism for filling water tanks:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Ballcock&quot; rel=&quot;noreferrer&quot;&gt;Ballcock&lt;/a&gt; -&gt; &lt;strong&gt;ball&lt;/strong&gt;? &lt;strong&gt;cock&lt;/strong&gt;? Both seem right&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Book of photographs:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; S&lt;strong&gt;crap&lt;/strong&gt;book (or al&lt;strong&gt;bum&lt;/strong&gt; as OP's comment)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Strong form of uniform continuity:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Lipschitz_continuity&quot; rel=&quot;noreferrer&quot;&gt;Lipschitz&lt;/a&gt; -&gt; &lt;strong&gt;shits&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Semi-precious stone:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Jasper&quot; rel=&quot;noreferrer&quot;&gt;Jasper&lt;/a&gt; -&gt; &lt;strong&gt;ass&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There may be another stone that contains a &quot;bad&quot; word, but that was the first I came across on the listing I found. &lt;strong&gt;Edit by Rand al'Thor on advice of OP:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lapis lazuli -&gt; &lt;strong&gt;piss&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1563" LastEditorUserId="50762" LastEditDate="2019-08-12T23:16:28.713" LastActivityDate="2019-08-12T23:16:28.713" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="6346" PostTypeId="1" AcceptedAnswerId="6389" CreationDate="2014-12-22T02:50:26.897" Score="33" ViewCount="3164" Body="&lt;p&gt;For Ernie, the task of choosing Christmas presents for his friends is always a struggle. I think that is because, while he finds it easy to solve problems relating to machines, mathematics, and materials, he has difficulty with minds. He is never sure exactly what presents his friends will like and as a result, agonizes for weeks until he can convince himself that he has chosen something appropriate. I always try to reassure him that &quot;it's the thought that counts&quot; (and Ernie certainly puts plenty of thought into his choices of gift), but that still doesn't reassure him. This year he had started earlier than usual (back in June I think), and had finally sorted out gifts for everyone except his old school friend Aristotle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aristotle was everything that Ernie wasn't. Emotional, where Ernie was logical; intuitive, where Ernie was rational; innumerate, where Ernie was mathematically inclined. And... an Alchemist by profession. Aristotle had never actually managed to transmute an element, and his experimental elixirs of life were anything but life-enhancing. But after every failed attempt to isolate phlogiston or dig up the philosophers stone, he would just smile a little wistfully and begin planning for his next experiment. Surprisingly, considering Ernie's usual views on &quot;witch-doctors, charlatans and anti-scientists&quot;, they remained the best of friends, although their gifts to each other often had a somewhat ironic aspect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I dropped in on Ernie yesterday I found him hunched over his desk with a worried frown. &quot;Any luck?&quot;, I asked. &quot;No&quot;, Ernie replied, &quot;whatever can you get an alchemist for Christmas?&quot;. &quot;Mmmm, how about the secret of converting lead to gold?&quot;, I asked, toungue in cheek. Ernie grunted, &quot;Pah, fool's gold that is!&quot;. But then a look came over his face. &quot;Brilliant - you have given me an idea!&quot;, he exclaimed, jumping up and heading over to one of his storage cupboards. He pulled out a curious object - a shiny cube that glowed a bright golden shade in the afternoon sun. &quot;FeS2, iron-disulphide, iron pyrite, fools-gold.&quot;, he explained. &quot;I've had this for years - it's a perfect cube exactly 100 mm on each side. Now what I though is that if I wrap it up in a very thin sheet of pure 24-carat gold, it will make the perfect gift for Aristotle - fools-gold, wrapped in gold, for a fool! I think he will appreciate the joke&quot;.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/w53yC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;Ernie asked if I could help him wrap it, as he was never very good at wrapping tidily - &quot;and this needs to be tidy&quot;, he said. &quot;If there is even the tiniest gap showing the pyrite under the gold sheet it will spoil the effect totally&quot;. He explained that he had two sheets of gold lying around as leftovers from an old experiment and, if his calculations were correct, each should be precisely just big enough to do the job. &quot;One is a perfect square of gold, and the other is a perfect equilateral triangle of gold. They are both very thin, so you can ignore the thickness of the gold when deciding how to wrap. It is OK for the gold sheet to overlap, but I don't think you should make any cuts or slits in it.&quot; I accepted the job, and had just asked which piece I should use (the triangle or the square), when there was a phone call and Ernie said he had to rush off &quot;on a kind of emergency&quot; and wouldn't be back till much later. &quot;Just use the wrap with the smallest surface area&quot; he called as he headed out the door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be honest, I had no idea how to work out which piece would be best from first principles, so I cheated a little. Ernie had said that both pieces would be just big enough, so I measured the side-lengths, worked out which one had the smallest area and (after lots of practice with sheets of newspaper) finally managed to wrap the cube in a single uncut sheet of gold according to Ernie's requirements. Ernie was right, with the appropriate orientation and correct folding the cube was perfectly covered in gold. Oddly enough, it must be the Christmas rush that has made me even more absent-minded than usual, I have already forgotten the measurements I made and can't even remember if I used the square or triangle. Can you help me to remember which piece I used, and the lengths I measured? &lt;/p&gt;&#xA;" OwnerUserId="1631" LastActivityDate="2015-08-22T02:49:06.867" Title="Ernie and the Alchemist's Gift" Tags="|mathematics|calculation-puzzle|geometry|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6347" PostTypeId="1" AcceptedAnswerId="6348" CreationDate="2014-12-22T03:59:33.730" Score="50" ViewCount="3364" Body="&lt;p&gt;The mysterious man motioned Walter into a small room. &quot;Here&quot;, he said, &quot;is where we test how smart our applicants are.&quot; Walter died inside. He had been told that the applications were tough - after all, being part of first contact with another species was important, and Earth only wanted the best of the best to represent them - but this interview was brutal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Do you see this screen?&quot;, the man said. He gestured to the far part of the room. Walter looked up, and saw what appeared to be a green screen with eleven white arrows on it. &quot;Yes...&quot; Walter replied. &quot;Okay&quot;, said the man, &quot;this is how this is going to work.&quot; He produced a stylus from his lower pocket. &quot;You can use this stylus to draw an arrow on the screen - here's an example.&quot; The man drew a twelfth arrow on the screen, in a seemingly random spot. He then walked over and handed the stylus to Walter. &quot;What do I have to do?&quot;, Walter asked, getting increasingly nervous. &quot;It's easy&quot;, said the man, &quot;All I want you to do, is to draw a thirteenth arrow, to finish this.&quot; Walter immediately wanted to ask a thousand questions. What was this about? How can you 'finish' a bunch of arrows? What is 'this'? The man, however, had already left the room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Walter stared at the screen in front of him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/g2ctE.png&quot; alt=&quot;Screen&quot;&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Where should he put the thirteenth arrow?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="7308" LastEditorUserId="7308" LastEditDate="2015-04-28T01:32:11.007" LastActivityDate="2015-04-28T01:32:11.007" Title="The Thirteenth Arrow" Tags="|visual|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6348" PostTypeId="2" ParentId="6347" CreationDate="2014-12-22T04:07:29.800" Score="47" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Nd5 checkmate, in others words draw an arrow starting from arrow 7 and extend in the same direction.&lt;br&gt;&#xA;The arrows represent chess moves.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/nBaqm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6178" LastActivityDate="2014-12-22T04:07:29.800" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6352" PostTypeId="1" AcceptedAnswerId="7999" CreationDate="2014-12-22T13:10:53.657" Score="-6" ViewCount="768" Body="&lt;p&gt;On an adventure looking through an Egyptian pyramid, you find weird marks on the walls. &lt;em&gt;It seems as though they are different, yet strangely similar from hieroglyphics... Strange...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further down into the pyramid you come across &lt;strong&gt;similar markings&lt;/strong&gt;. You move on, telling yourself you'll come back to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You reach the end of the tunnel, and find &lt;strong&gt;English&lt;/strong&gt;. Since when is English in a pyramid? Are you going crazy? &lt;em&gt;The carved words say:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A spin top, a pyramid, language. Basic, yet confusing, what could've happened?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you finish reading it, &lt;strong&gt;a virtual keyboard and a small screen appear.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This is when you realize you've made the biggest Egyptian discovery yet! They were very advanced! Or did they work with aliens?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before you have time to think of any other theories, &lt;em&gt;the door closes behind you, locking you in!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to put something on the keypad. The small screen says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Where are you?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ultimately confused, it's now your job to find your way out of this! Where are you?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="7308" LastEditDate="2015-04-30T23:33:13.100" LastActivityDate="2015-04-30T23:33:13.100" Title="Strange marks on the Wall?" Tags="|lateral-thinking|enigmatic-puzzle|" AnswerCount="1" CommentCount="31" ContentLicense="CC BY-SA 3.0" />
  <row Id="6353" PostTypeId="2" ParentId="6288" CreationDate="2014-12-22T13:50:24.873" Score="3" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; computers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Hello everyone, my name is Jay&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This might be a specific computer; I'm not really sure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I work for you 24 hours a day,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lots of computers do this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;but don't get me wrong I know not Silent Bob &lt;br&gt;&#xA; and got nothing to do with weed on the job.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; References to something on the telly? Excuse my cultural ignorance :-)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My kind can be you or you can be me &lt;br&gt;&#xA; if you know a body by its chemistry.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People can be seen as very complicated machines/computers, the chemical reactions inside their body taking the place of electronic signals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;We're not only human, but mostly a thing!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Computers are things and not humans.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;We have many shapes, on the air we are king.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ???&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Our children you could surf but not in the sea&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reference to surfing the web (this is what enabled me to get the answer).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(for some they are a monstruosity!).&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people don't like the internet?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;They fly, they bounce, they bend, &lt;br&gt;&#xA; but always to us they end!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Webpages can do all sorts of things, but to access them you need a computer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;We get and we give even not alive,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Computers can do all sorts of things even though they're inanimate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;but to do our best we need a good drive,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They need hard drives to function properly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;for we can make work a lot you possess, &lt;br&gt;&#xA; even destroy it if working in excess.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ???&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This riddle was difficult but I'm not a sphinx, &lt;br&gt;&#xA; in fact I'm well sure, who I am you can glimpse&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Indeed, as I solve this riddle I am looking at the answer to the riddle! :-)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nice one, Noldor!&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2014-12-22T13:56:08.067" LastActivityDate="2014-12-22T13:56:08.067" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="6355" PostTypeId="2" ParentId="6069" CreationDate="2014-12-22T17:27:39.400" Score="2" Body="&lt;p&gt;Well, here is what I've got. I think the algorithm is almost clear, but something is off and I can't figure out why; since I probably won't have time to keep trying in the next days, I hope someone else will fix the missing parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cipher's logic is this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; for each letter &lt;code&gt;X&lt;/code&gt; from A to Z, starting from the end of the string, take every &lt;code&gt;X&lt;/code&gt; and move it forward of &lt;code&gt;n&lt;/code&gt; positions, where n is the position of &lt;code&gt;X&lt;/code&gt; in the alphabet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A key factor for reversibility, with this algorithm, is that one has to be able to execute the same steps backwards. To do so, one has to know at each step&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; which one is the first &lt;code&gt;X&lt;/code&gt; that has been moved. It's for this reason, I suspect, that before moving the &lt;code&gt;X&lt;/code&gt;s you have to take the first &lt;code&gt;n&lt;/code&gt; letters of the string and move them to the end, and put them back after the movements.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is what I obtained from the quick brown fox thing, using &lt;a href=&quot;http://pastebin.com/dsWfLumx&quot; rel=&quot;nofollow&quot;&gt;this algorithm&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;THE QUICK BROWN FOX JUMPS OVER THE LZaY DOG, THEN THE LZaY DOG ROLLS UNDER THE QUICK BROWN FOX.WHO LIKES THIS DOG I WONDER?&#xA;&#xA;...&#xA;&#xA;tdoes Ue hnocbWi rkX tUfq Vor nj oeZmaYpe osh,r dl gt eZhaYeo hn dlg Ut det ollUoe rhncsbWir tkX.qWf  heron i odkloh Wisg?i&#xA;&#xA;...&#xA;&#xA;witdoes e hnocbi rkw tfq our nj oemawpe oxush,rv dl gt ehaeozy hn dlg t det olloe rzhnycsbiru tk.qf u heron i odwkloh isxg?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which is almost right. But the saddest part is, I can't even decode it correctly. The result of the decoding is something like:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;TEHUQ IKC BROWN FOX JUMPSO VER THE LAZY DOG, THEN THE LAZY DOG ROLLS UNDER THE QUICKB RWONF XO.WHO LKEIS THIS DOG I WODENR?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So, to sum up: I think this is the right path and we're almost there, PLEASE someone fix this.&lt;/p&gt;&#xA;" OwnerUserId="5990" LastActivityDate="2014-12-22T17:27:39.400" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6356" PostTypeId="2" ParentId="6346" CreationDate="2014-12-22T18:51:18.977" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm pretty sure you used a square with a length of $200\sqrt{2} \approx 282.84 mm$ (total area $80000 mm^2$). The optimal wrapping of a cube in a square looks like &lt;a href=&quot;http://www.mathpuzzle.com/wrap.htm&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;. I imagine the triangle will end up with more wasted surface area due to the dissimilarity of the shapes, but I haven't figured out the optimal wrapping strategy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2014-12-22T18:51:18.977" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6357" PostTypeId="2" ParentId="6185" CreationDate="2014-12-23T10:52:46.263" Score="3" Body="&lt;p&gt;Here's the progress I've made so far; maybe someone else can add to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The riddle contains five capital letters not at the start of lines; these spell out the word FIRST. Also 'Pcihre' is an anagram of cipher (thanks @Sathish).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reading the first letter of each line of the two notes shows us which one refers to which man (they spell out BLONDE and BLACK).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried deciphering Dzwgwosdvk and Klbmsvwvm using Vigenere ciphers with all sorts of different keys (Pcihre, Qdjisf, FIRST, EWR Southwest A [&quot;the hint that you were first told&quot;], etc.). I'm also tried all Caesar shifts, even though I thought that'd probably be too easy. So far, no joy.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2014-12-23T10:52:46.263" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6358" PostTypeId="2" ParentId="6272" CreationDate="2014-12-23T11:35:14.123" Score="7" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;    &lt;p&gt;The kingpin has fled to Cuba STOP Special forces have been deployed STOP Maintain vigilance against Communist infiltration among officers as well as the rank and file STOP&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The biggest clue for me was&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;    &lt;p&gt; seeing that the third letter of the plaintext appeared at the end of the ciphertext&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;To create ciphertext from plaintext (found with the hints):&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the lowest square above the number of characters. (171 characters -&gt; 196 = 14 * 14)&#xA;So we'll be working on a 14 by 14 square.&#xA;Fill the plaintext characters in the square in the following order (only 1-9 filled in on a 4 by 4 square):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1592&#xA;---6&#xA;8---&#xA;4-73&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Once the outer border is completely filled, take the next inner border.&#xA;Every odd inner border starts in the lower right (instead of upper left), but continue clockwise.&#xA;You'll have gaps in the middle when the sentence has stopped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;To decipher a ciphertext:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hardest part here is finding out how to reconstruct the square from the text, figuring out where the gaps should be.&#xA;To do this fill the 171 cells in a 14 by 14 table in the above mentioned manner.&#xA;In this case you'll have 4 complete rows of 14 letters.&#xA;The fifth until tenth row will have gaps in them, thus consisting of less than 14 letters.&#xA;The eleventh until fourteenth row are again complete with 14 letters.&#xA;The square looks like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;TKPHF_C_PELR_H&#xA;VB_LDOAAVLEAEI&#xA;SGNCUTFROM_SDI&#xA;OCFEAEAHAAROOA&#xA;AIG_DLT  _SN_L&#xA;P_AR     EL_PT&#xA;OTIT     OSIIU&#xA;AMT_      EAIS&#xA;_TNWS     NNI_&#xA;DES_I  P_FNOAC&#xA;_PMCK__L_SIO__&#xA;_N_FN_TLIIMTIC&#xA;G_ANGNN_SYEEEH&#xA;_AEFISTBOESNNE&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And then it's only a matter of reading the letters in the right order (as described above).&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2014-12-23T11:35:14.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6359" PostTypeId="1" AcceptedAnswerId="6360" CreationDate="2014-12-23T11:45:46.327" Score="11" ViewCount="1188" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm born when one is over the treasure spot,&lt;br&gt;&#xA;  I'm used when you drive to buy a present,&lt;br&gt;&#xA;  I'm born in the spirit of mutual gift-giving,&lt;br&gt;&#xA;  Yet I'm literally a super negative one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Kinda short, so might be open to interpretation, but I assure you your efforts in solving this riddle will be reciprocated with a +1 from me!&lt;/p&gt;&#xA;" OwnerUserId="6178" LastEditorUserId="5373" LastEditDate="2014-12-23T12:06:04.110" LastActivityDate="2015-03-06T00:55:22.653" Title="What am I? 'Tis the season of mutual puzzling" Tags="|riddle|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6360" PostTypeId="2" ParentId="6359" CreationDate="2014-12-23T12:23:10.693" Score="11" Body="&lt;p&gt;The answer must be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; reciprocal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I'm born when one is over the treasure spot,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;X marks the spot&lt;/em&gt;; &quot;the treasure spot&quot; means X. So &quot;one is over the treasure spot&quot; means 1 over x, or $\frac{1}{x}$, the reciprocal function.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I'm used when you drive to buy a present,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The internal combustion engine, present in the car you drive in, is a type of &lt;a href=&quot;http://en.wikipedia.org/wiki/Reciprocating_engine&quot;&gt;reciprocating engine&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I'm born in the spirit of mutual gift-giving,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Mutual&lt;/strong&gt; gift-giving; one person reciprocates another's gift by giving them a gift in return.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Yet I'm literally a super negative one.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Another notation for the reciprocal of a number is to raise that number to the power of minus one; reciprocals can be denoted by $^{-1}$, &quot;negative one&quot; in superscript.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And finally...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Your efforts in solving this riddle will be &lt;strong&gt;reciprocated&lt;/strong&gt; with a +1 from me!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;LYNCH MOB! :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you'd like to know how I solved this, the first line always seemed to me the most tractable. I struggled with 'one' and 'X' for a while before eventually thinking of the correct interpretation, which fitted with the third line. As I was writing the answer, I realised how it fitted the fourth line too, and that was when I was certain I was right. The second line was the hardest, and required a quick search on the internet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Soon it'll be time to post &lt;em&gt;my&lt;/em&gt; seasonal riddle...&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2014-12-23T12:29:18.533" LastActivityDate="2014-12-23T12:29:18.533" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6361" PostTypeId="1" AcceptedAnswerId="6362" CreationDate="2014-12-23T13:27:10.260" Score="6" ViewCount="1600" Body="&lt;p&gt;I have split the alphabet in two sets, as you may see in the picture below, according to a very specific rule. Which was the rule I used? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/isLQl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="70" LastEditorUserId="5188" LastEditDate="2015-04-27T06:02:59.160" LastActivityDate="2015-04-27T06:02:59.160" Title="How did I split the alphabet?" Tags="|pattern|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6362" PostTypeId="2" ParentId="6361" CreationDate="2014-12-23T13:53:21.843" Score="10" Body="&lt;p&gt;The top set of letters are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Letters that have no curves but do have &quot;branches&quot; (ie cannot be drawn without lifting a pen off the paper or retracing parts of the letter)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The bottom set are those left over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does I think generate the correct sets (I'll be very embarrassed if I am wrong on that) but it feels a bit too complicated to be an elegant answer...&lt;/p&gt;&#xA;" OwnerUserId="1561" LastActivityDate="2014-12-23T13:53:21.843" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6363" PostTypeId="2" ParentId="6361" CreationDate="2014-12-23T14:15:22.057" Score="6" Body="&lt;p&gt;AEFHKTXY&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The first set of letters cannot be drawn without lifting the pen from the paper and not drawing over what's already written (over the same path like a car on its reverse gear).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;BCDGIJLMNOPQRSUVWZ&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The second set of letters can be drawn without taking the pen from paper and not moving the pen over what have already written (over the same path like a car on its reverse gear)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="2730" LastEditDate="2014-12-23T14:26:19.500" LastActivityDate="2014-12-23T14:26:19.500" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="6364" PostTypeId="1" AcceptedAnswerId="6366" CreationDate="2014-12-23T15:49:34.280" Score="3" ViewCount="5062" Body="&lt;p&gt;I was working my way through some Propositional Logic Questions in &lt;a href=&quot;https://www.google.co.in/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=4&amp;amp;cad=rja&amp;amp;uact=8&amp;amp;ved=0CCwQFjAD&amp;amp;url=http%3A%2F%2Fwww.flipkart.com%2Fdiscrete-mathematics-its-applications-combinatorics-graph-theory-english-7th%2Fp%2Fitmdwxxszhczjrnf&amp;amp;ei=Oo2ZVIeqGJHluQT32IHICQ&amp;amp;usg=AFQjCNGMqBbPuopaBnOUnxErGEfGBONe5w&amp;amp;bvm=bv.82001339,d.c2E&quot; rel=&quot;nofollow noreferrer&quot;&gt;Discrete Maths by Rosen&lt;/a&gt;, when I came across the following question:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;An explorer is captured by a group of cannibals. There are two types of cannibals—those who always tell the truth and those who always lie. The cannibals will barbecue the explorer unless he can determine whether a particular cannibal always lies or always tells the truth. He is allowed to ask the cannibal exactly one question.&lt;/p&gt;&#xA;&lt;p&gt;Find a question that the explorer can use to determine whether the cannibal always lies or always tells the truth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;My Solution:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;&lt;em&gt;If I were to ask you whether you are a liar, would you answer &amp;quot;yes&amp;quot;?&lt;/em&gt;&lt;br/&gt;&#xA;- The honest person would say NO&lt;br/&gt;&#xA;- The liar would say YES (due to double negation)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Doubt:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Am I correct in saying my answer will work? It almost seems too simple&lt;/p&gt;&#xA;" OwnerUserId="7435" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-10T07:41:45.733" Title="Is this a correct solution to determining which of two people is the liar using one question?" Tags="|logical-deduction|liars|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6366" PostTypeId="2" ParentId="6364" CreationDate="2014-12-23T15:57:35.817" Score="5" Body="&lt;p&gt;That works :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the question you are looking at, you must only determine if he is a liar or not, so as stated in your comment, objective questions such as &quot;Is the Earth round?&quot; are simple solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other similar riddles that you may come across, like the two paths problem, or &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2188/two-doors-with-two-guards-one-lies-one-tells-the-truth&quot;&gt;the two guards&lt;/a&gt;, you must use double negatives, or have both people use a negative (ie. have the truth teller and liar lie). The reason being you need to figure out in one question if they are a liar or not and the solution to another question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can give an example if you want, or you can check out the two guards problem and give it a shot :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, welcome to the Puzzling Beta!&lt;/p&gt;&#xA;" OwnerUserId="4505" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2014-12-23T16:11:51.153" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6369" PostTypeId="2" ParentId="6185" CreationDate="2014-12-23T17:58:09.260" Score="4" Body="&lt;p&gt;Here's the progress I made so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Blam! could be a reference to this: &lt;a href=&quot;http://en.wikipedia.org/wiki/Blam!&quot; rel=&quot;nofollow&quot;&gt;Blam!&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;'Blam!' is the tenth track on the CD. There's another track called 'CD Is Dead' which has a length of 2:26. That's a time mentioned in Part 5. It is the 5th track. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Klbmsvwvm decodes with cipher +1 to CDTEKNONE&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CDTEKNONE decodes with 'CD is Dead' to 'istejmnme'. This could be an anagram for 'Jim sent me'. This note's 2nd and 3rd line rhyme with 'me'. Other note's 2nd and 3rd line rhyme with 'Jim'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The passphrase for Dzwgwosdvk could be 'CD Is Dead', which gives Bwootksath.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This could be an anagram for: 'two ask both', which is the sane thing to do as Bachrach44 mentioned, because we don't really know if the notes are telling the truth - even if we do decode them). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: edited to add clues.&lt;/p&gt;&#xA;" OwnerUserId="7430" LastEditorUserId="7430" LastEditDate="2014-12-24T17:57:57.027" LastActivityDate="2014-12-24T17:57:57.027" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6372" PostTypeId="1" AcceptedAnswerId="6378" CreationDate="2014-12-24T03:01:57.587" Score="3" ViewCount="6465" Body="&lt;p&gt;What are the chances of getting this correct if you pick at random?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;1/4&lt;/li&gt;&#xA;&lt;li&gt;1/2&lt;/li&gt;&#xA;&lt;li&gt;1/3&lt;/li&gt;&#xA;&lt;li&gt;1/4&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;You are not allowed to add more answers to this list&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note; This &lt;strong&gt;DOES&lt;/strong&gt; have a correct, demonstrable answer!&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-12-24T11:08:32.590" LastActivityDate="2022-11-05T14:46:52.587" Title="Multiple Choice Puzzle" Tags="|mathematics|probability|" AnswerCount="10" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="6373" PostTypeId="2" ParentId="6372" CreationDate="2014-12-24T03:38:57.053" Score="15" Body="&lt;p&gt;By randomly picking one of the labels 1, 2, 3 or 4, my chance of being correct is $1/4$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Correct here means the single label declared by tester (warspyking, in this case) as &quot;correct&quot;. The content behind the labels is irrelevant to me as there is no question asked about those contents. I only see a question about my chance of picking the correct label. &lt;/p&gt;&#xA;" OwnerUserId="6087" LastActivityDate="2014-12-24T03:38:57.053" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6374" PostTypeId="2" ParentId="6372" CreationDate="2014-12-24T03:41:01.710" Score="7" Body="&lt;p&gt;I leave this question blank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can't be $1/4$ because there would be a $1/2$ chance of randomly selecting choices 1 or 4, and it can't be $1/2$ or $1/3$ either because the chances of randomly selecting either choice is $1/4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if the correct answer is zero then there is zero chance of randomly selecting the correct choice.  This matches, thus the correct answer is zero and I leave the question blank as I won't get it correct.&lt;/p&gt;&#xA;" OwnerUserId="6178" LastEditorUserId="6178" LastEditDate="2014-12-24T03:57:53.633" LastActivityDate="2014-12-24T03:57:53.633" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6375" PostTypeId="2" ParentId="2188" CreationDate="2014-12-24T05:04:55.103" Score="5" Body="&lt;p&gt;I am a newbie to Stack-Exchange and if my answer violates some guidelines , please correct me . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us assume , WLOG that the right door leads to freedom . So , the question I would ask is :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If I were to ask you whether the right door leads to freedom , would you answer yes?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;The honest person would answer YES&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;The liar's internal response would be to say NO but being the liar he is , he would&lt;br&gt;&#xA;say YES (due to double negation)&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this would help you ascertain as to which is the truth-teller and which is the liar&lt;/p&gt;&#xA;" OwnerUserId="7435" LastEditorUserId="17482" LastEditDate="2016-05-23T14:09:15.233" LastActivityDate="2016-05-23T14:09:15.233" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6377" PostTypeId="2" ParentId="6372" CreationDate="2014-12-24T05:32:30.997" Score="0" Body="&lt;p&gt;Random? Okay. I'll pick one of the three possible answers, 1/2, 1/3, or 1/4, uniformly at random. (There are two choices labeled &quot;1/4&quot;, but that doesn't matter to me - they're the same answer.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you see where this is going.&lt;/p&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2014-12-24T05:32:30.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6378" PostTypeId="2" ParentId="6372" CreationDate="2014-12-24T05:35:43.490" Score="4" Body="&lt;p&gt;&lt;strong&gt;Probability of&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;picking `1/4` = `1/2`&#xA;picking `1/2` = `1/4`&#xA;picking `1/3` = `1/4`&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Probability of&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;correct answer `1/4` = `1/3`&#xA;correct answer `1/2` = `1/3`&#xA;correct answer `1/3` = `1/3`&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So the probability of me picking correct &lt;strong&gt;Answer&lt;/strong&gt; is&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;( 1/2 * 1/3 ) + ( 1/4 * 1/3 ) + ( 1/4 * 1/3 ) = 1/3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Probability of me picking correct &lt;strong&gt;Label&lt;/strong&gt; remains &lt;code&gt;1/4&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="5437" LastEditorUserId="5437" LastEditDate="2014-12-24T06:41:45.537" LastActivityDate="2014-12-24T06:41:45.537" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6379" PostTypeId="2" ParentId="6372" CreationDate="2014-12-24T08:24:36.193" Score="5" Body="&lt;p&gt;The crucial point of this puzzle is what defines a &lt;em&gt;correct&lt;/em&gt; answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Being a multiple choice question, one should possible assume the 4 answers should answer the question above.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With this in mind the correct answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Is not available. Because there is a 1:2 chance for it being 1/4 (wrong) and a 1:4 chance for it being either of the other (wrong)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In short:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is a paradox.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastActivityDate="2014-12-24T08:24:36.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6381" PostTypeId="2" ParentId="6372" CreationDate="2014-12-24T09:44:14.957" Score="1" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;This is a multiple choice puzzle, that means a correct answer is a set of labels s such that &quot;s are the labels of chances of getting this correct if you pick at random&quot; is true &#xA;ie &quot;s = { x in {1, 2, 3, 4} | s={x} p(x)=1/4 }&quot; is true&#xA;ie &quot;s = { x in {1, 4} | s={x} }&quot; is true.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As you see only the empty set can make this true.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7454" LastActivityDate="2014-12-24T09:44:14.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6382" PostTypeId="2" ParentId="6340" CreationDate="2014-12-24T10:16:56.717" Score="0" Body="&lt;p&gt;Assuming the doctor plays 'dirty tricks' - I'd like to elaborate on the GOTO 0's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While is he correct that with a RANDOM choice of 1 or 2 swaps you end up at odds of 74/99 (1/2 x 1 + 1/2 x 49/99), you can also assume that the doctor might have figured out that scenario as being the optimal for you...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being into 'dirty tricks', his choice of swaps might not be entirely random (meaning: despite what the OP said, the 50/50 odds might not actually be 50/50). In that case, with the (admittedly fairly obvious) scenario with the 1 'clean' bar in a seperate bag, you may get the tables turned on you, and only end up with 25/99 odds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the riddle does not state the doctor doesn't see what the boy does, so in any case with a non 50-50 rate, any evil guy with half a brain will be able to figure out which is the least favorable option and give the boy that bag. Given the 'Dr.' title, I'll assume he's clever enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefor the best achievable rate is 50%, which is formed by any combination with equal amounts of good and poisoned candy in both bags.&lt;/p&gt;&#xA;" OwnerUserId="2156" LastActivityDate="2014-12-24T10:16:56.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6383" PostTypeId="2" ParentId="6330" CreationDate="2014-12-24T10:34:59.910" Score="0" Body="&lt;p&gt;Note that for any nxn, with n being odd, there's a trivial solution where even the diagonals work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1   6   4   2   7   5&lt;br&gt;&#xA;2   7   5   3   1   6&lt;br&gt;&#xA;3   1   6   4   2   7&lt;br&gt;&#xA;4   2   7   5   3   1&lt;br&gt;&#xA;5   3   1   6   4   2&lt;br&gt;&#xA;6   4   2   7   5   3&lt;br&gt;&#xA;7   5   3   1   6   4&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just start with 1-n in the first column, and in each next column shift down 2 positions ('as the knight jumps', if you will). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's however only a partial answer, on how you can generate an example. Given your intended purpose I think you're more looking into a 'randomlooking' example?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;Only after typing this up, I found it's covered in &lt;a href=&quot;https://puzzling.stackexchange.com/questions/6250/the-quite-unusual-square/6384#6384&quot;&gt;this&lt;/a&gt; question aswell, but in more detail.&lt;/p&gt;&#xA;" OwnerUserId="2156" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2014-12-24T10:52:19.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6384" PostTypeId="2" ParentId="6250" CreationDate="2014-12-24T10:49:40.550" Score="4" Body="&lt;p&gt;Let me trump the others (for now) with a 16x16: (taken from my answer to another question  &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2204/is-there-any-sudoku-puzzle-combined-with-magic-square&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6C52f.png&quot; alt=&quot;here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given each line (and each main diagonal) holds numbers 1-16, they all have a total value of 136. Each number also appears 16 times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a bonus: it's a working 16x16 sudoku, with each 4x4 block being a magic square.&lt;/p&gt;&#xA;" OwnerUserId="2156" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2014-12-24T10:49:40.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6385" PostTypeId="2" ParentId="6372" CreationDate="2014-12-24T11:16:00.860" Score="3" Body="&lt;p&gt;In the real world scenario, the probability of the answer being correct is 1/2. Two scenarios for this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It is a multi choice question and thus can have only one right answer. As some one who do not know the answer(that's the reason we are choosing randomly or else we will pick the right one, right?), we will first discard the repetitive answer through logical reasoning. That is, if it is the right answer, such a blunder will not be seen in the question. Thus we will have two choices (2) and (3) and thus the probability will be 1/2(I am not noting the choice 1/2 but simply the probability of either choosing option 2 or 3)&lt;/li&gt;&#xA;&lt;li&gt;Upon over simplification, the chance of choosing the right answer is going to be only 1/2. Either you choose the answer right or you choose the answer wrong. Which means your chance of getting a randomly picked answer right is 1/2.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7455" LastActivityDate="2014-12-24T11:16:00.860" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6386" PostTypeId="1" AcceptedAnswerId="6393" CreationDate="2014-12-24T11:24:02.280" Score="52" ViewCount="5262" Body="&lt;p&gt;You buy a bottle with a letter from the merchant, the merchant tells you that when you drink the liquid in the bottle it grants you eternal life, he supposedly deciphered this from the letter.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After you get home you decide to study the letter if it really says what the merchant told you, can you figure out if the bottle really grants eternal life?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You come to a fork in the road.&lt;br&gt;&#xA;To the left is an empty well made from stone.&lt;br&gt;&#xA;On the right is a pirate's buried treasure.&lt;br&gt;&#xA;Ahead you only see a tall straight tree.&lt;br&gt;&#xA;The night is dark with only a dying moon in the sky.&lt;/em&gt;  &lt;/p&gt;&#xA;" OwnerUserId="5560" LastEditorDisplayName="user20" LastEditDate="2015-08-05T17:49:40.563" LastActivityDate="2015-08-05T17:49:40.563" Title="Is the magic liquid in the bottle safe to drink?" Tags="|riddle|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6387" PostTypeId="1" CreationDate="2014-12-24T11:38:55.713" Score="27" ViewCount="5998" Body="&lt;h3&gt;Rules&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Normal Sudoku rules apply except that instead of 9s you must place chess pieces. A puzzle can mix chess pieces, and in principle any piece may be used - although the rest of the constraints provide implicit limitations on which pieces may go in which squares.&lt;/li&gt;&#xA;&lt;li&gt;Each chess piece attacks each number from 1 to 8 precisely once.&lt;/li&gt;&#xA;&lt;li&gt;Attacks &quot;pass though&quot; numbers - see the Bishops in the second puzzle.&lt;/li&gt;&#xA;&lt;li&gt;Chess pieces do not attack (or protect) other chess pieces.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The space of possible puzzles is very small, and most of them are very symmetric, which makes them less interesting. Therefore I am posting two puzzles, not one: these are essentially the two distinct representatives of interesting chess Sudoku puzzles.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Puzzle 1&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QyK0a.gif&quot; alt=&quot;Puzzle image 1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Puzzle 2&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3OMyZ.gif&quot; alt=&quot;Puzzle image 2&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="228" LastActivityDate="2016-04-20T21:08:25.587" Title="Can you solve my Chess Sudoku puzzles?" Tags="|chess|sudoku|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6388" PostTypeId="2" ParentId="6340" CreationDate="2014-12-24T11:59:11.620" Score="1" Body="&lt;p&gt;It is not given explicitly that the total of chocolates within the jar should be 100. Why not put only the good chocolate bars in both the jars(25 each or whatever combination) and keep all the other bars in his hand?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This way, he can discard the poisoned chocolate bars when he is given a chance to choose one from any bar right? His chance of survival will be 100% in that case.&lt;/p&gt;&#xA;" OwnerUserId="7455" LastActivityDate="2014-12-24T11:59:11.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6389" PostTypeId="2" ParentId="6346" CreationDate="2014-12-24T12:03:23.060" Score="19" Body="&lt;p&gt;The optimal wrapping for a square has been shown to be $8\ dm^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the triangle, the almost best wrapping I could find is the following:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/9aOTs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the picture above, the grid has $5 \ cm$ squares.  The triangle is constrained by 3 grid points that must lie on the 3 sides and the fact that the right vertex lies on the grid line where it seems to be.  The shape that is cut out wraps exactly a cube of size $1\ dm$.  The leftovers can be folded nicely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The side is $4.215123\ dm$ as I found numerically.  This gives an area of $7.693450 \ dm^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in the end, it is still the triangle that wraps a cube best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PS&lt;/strong&gt;: The solution can be slightly improved by turning the 3 sides a bit more, counterclockwise around the 3 pivot points. The bottom edge would start eating in the tip at the right, but that could be covered by the leftover at the bottom left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stop buying square paper for your Christmas presents.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="15544" LastEditDate="2015-08-22T02:49:06.867" LastActivityDate="2015-08-22T02:49:06.867" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6391" PostTypeId="1" AcceptedAnswerId="6395" CreationDate="2014-12-24T12:29:15.287" Score="41" ViewCount="932" Body="&lt;p&gt;I made a crossword, but this is no normal crossword!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the grid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/55glmm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here are the clues:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;24 grains, or 1/20 of a troy ounce&#xA;Acrophobia is the fear of this&#xA;An interconnected system&#xA;Foolish&#xA;Holding of office or land&#xA;Imaginary or unreal&#xA;Intermediary floor&#xA;Juvenile felines&#xA;Longest side on a right-angled triangle&#xA;Passionate&#xA;Past&#xA;Pointed tooth&#xA;Sheen or lustre&#xA;Susceptible&#xA;The first to explore unseen territory&#xA;Unstable oxygen&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Once you solve it, can you give me the name of the thematic &lt;strong&gt;confectionery&lt;/strong&gt; the crossword hints to?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Clarifications&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Yes, these are all words used in English. Most of them are commonplace.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="445" LastActivityDate="2014-12-24T15:35:33.603" Title="From crossword to confectionery" Tags="|crosswords|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6393" PostTypeId="2" ParentId="6386" CreationDate="2014-12-24T13:54:07.683" Score="45" Body="&lt;p&gt;Excellent puzzle.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Look at the shapes of the objects described.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the potion is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TOXIC&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2014-12-24T13:54:07.683" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6394" PostTypeId="1" AcceptedAnswerId="6400" CreationDate="2014-12-24T14:08:24.837" Score="1" ViewCount="2728" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;You are allowed to select as many random&lt;sup&gt;1&lt;/sup&gt; people as you like, and place them in a room. How many do you need for the probability that at least 2 people will have a birthday&lt;sup&gt;2&lt;/sup&gt;...&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;a. on the same day...&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;b. on two consecutive days...&lt;br&gt;&#xA;  ...to be greater than 50%?  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Please explain why this is so. You may ignore leap years if it interferes with the answer, which I'm pretty sure it would.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;In this case, random means they have a &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{365}$&lt;/span&gt; chance of having a birthday on Jan. 1, or any other day.&lt;/li&gt;&#xA;&lt;li&gt;The 'birthday' references the day and month. Years are irrelevant.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="46002" LastEditDate="2019-02-24T23:54:42.990" LastActivityDate="2019-02-25T15:24:49.253" Title="Birthdays on Consecutive Days" Tags="|mathematics|probability|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="6395" PostTypeId="2" ParentId="6391" CreationDate="2014-12-24T14:17:38.517" Score="24" Body="&lt;p&gt;Great puzzle! It was fun to solve the whole way through. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found the key realization from the clue &quot;Longest side on a right-angled triangle&quot;. The answer is clearly&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HYPOTENUSE, which is too long to fit&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But, it can be made to work if you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; treat the substring &quot;TEN&quot; as the single number &quot;10&quot;, putting that number in a single cell of the grid. The idea is to replace a substring with its number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solving for other clues the same way gives the answers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;&#xA;24 grains, or 1/20 of a troy ounce      PENNYW8&#xA;Acrophobia is the fear of this          H8S&#xA;An interconnected system                NE2RK&#xA;Foolish                                 ASI9&#xA;Holding of office or land               10URE&#xA;Imaginary or unreal                     N1XIS10T&#xA;Intermediary floor                      MEZZA9&#xA;Juvenile felines                        KIT10S&#xA;Longest side on a right-angled triangle HYPO10USE&#xA;Passionate                              IN10SE&#xA;Past                                    BYG1&#xA;Pointed tooth                           CA9&#xA;Sheen or lustre                         SHI9SS&#xA;Susceptible                             PR1&#xA;The first to explore unseen territory   PI1ER&#xA;Unstable oxygen                         OZ1&#xA;&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the grid&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/QnjNh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally, we extract by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; reading the highlighed letters, which anagram to &quot;SUM&quot; and &quot;ROWS&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This tells us to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; find the sum of the numbers in each row, which gives letters:&#xA;&#xA;&lt;code&gt;&#xA;8+8=16     P&#xA;1+2+1+1=5  E&#xA;10+10=20   T&#xA;9          I&#xA;1+10+9=20  T&#xA;&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally, this gives the answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Petit_four&quot;&gt;Petit four&lt;/a&gt;, from the clues that it's a confectionery and the theme of including a number in the word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How did you make this? The whole thing seems really really constrained. I can't imagine there's many words to choose from, and the sums seem very restricting along with the intersection requirements.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="445" LastEditDate="2014-12-24T15:35:33.603" LastActivityDate="2014-12-24T15:35:33.603" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6396" PostTypeId="2" ParentId="6394" CreationDate="2014-12-24T14:37:50.460" Score="2" Body="&lt;p&gt;Like the birthday problem, but with triple the likelihood, I guess...&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;So, person 1 would have a birthday, doesn't matter when, won't be before or after anyone else.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Person 2 could have their birthday on any of the 365 days (lets ignore leap years, since the question does), chance of the 2nd person being within a day of the first person is 3/365.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If they don't (362/365) then the next person has a 6/365 chance to be within a day of the first two, and so on.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;So we have 1 * 362/365 * 359/365 * ... First time this drops below 50% is 365/365 * 362/365 * 359/365 * 356/365 * 353/365 * 350/365 * 347/365 * 344/365 * 341/365 * 338/365 * 335/365 * 332/365 * 329/365 * 326/365 = 0.4596676, so 14 people.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Edit: Typed too fast and fat fingers took over :) &lt;/p&gt;&#xA;" OwnerUserId="2537" LastEditorUserId="56016" LastEditDate="2019-02-24T23:47:20.723" LastActivityDate="2019-02-24T23:47:20.723" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="6397" PostTypeId="1" AcceptedAnswerId="6451" CreationDate="2014-12-24T14:48:57.970" Score="16" ViewCount="1541" Body="&lt;p&gt;A married couple making themselves some tea, but which is the man and which is the woman? It's impossible to tell from their names, so let's have some clues...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the one on the left. If it's a woman, she might be sitting on a remade table, or perhaps she's stirring her drink with a fork. If it's a man, he's much taller than his wife; he tops her by a head, although he might seem to be like a small burrowing rodent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the one on the right. If it's a woman, she's probably got a few pet moths, or maybe she's just got away from the Indian film industry. If it's a man, he needs most of what he can get from petrol, diesel, or gasoline.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put everything together - a man, a woman, and the pot of tea between them - and what have you got?&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2016-12-21T15:23:32.143" LastActivityDate="2016-12-21T15:23:32.143" Title="A man, a woman, and a pot of tea" Tags="|riddle|word|seasonal|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6398" PostTypeId="2" ParentId="6387" CreationDate="2014-12-24T15:12:02.317" Score="14" Body="&lt;p&gt;The solution for the first puzzle is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EcHuA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That was very interesting, I have to say. Here are the things that helped the most:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Rooks and queens attack too many squares, while pawns attack too little. Hence all 9s are either knights, bishops or kings.&lt;/li&gt;&#xA;&lt;li&gt;To attack 8 squares, knights and kings must be in the interior somewhere. &lt;strong&gt;Hence any 9 on the edge is a bishop&lt;/strong&gt; (this is the single fact that helped the most!)&lt;/li&gt;&#xA;&lt;li&gt;Assume a square is a 9 and a certain chess piece. If there is a doubled up number in the 8 squares attacked by that piece, then that piece can't go there. If no pieces can go there, then it can't be a 9.&lt;/li&gt;&#xA;&lt;li&gt;If you know what and where some of the pieces are, then you can look at the attacked squares. For example, if you have a square where an 8 &lt;em&gt;could&lt;/em&gt; go under normal Sudoku rules, but is being attacked by a bishop that's already attacking an 8, then 8 can't go there.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Someone else can do the other one :) (It's good fun!)&lt;/p&gt;&#xA;" OwnerUserId="445" LastEditorUserId="445" LastEditDate="2014-12-24T15:24:05.293" LastActivityDate="2014-12-24T15:24:05.293" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6399" PostTypeId="2" ParentId="6394" CreationDate="2014-12-24T15:22:59.780" Score="4" Body="&lt;p&gt;This is more complicated than simply the birthday problem times three, because there's the potential for overlap. For instance, if person A is born on Jan 5, and person B is born on Jan 7, then person C only has a 5/365 chance to meet the criteria. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather than do the math, I wrote a script to solve it, ignoring leap years (although it turns out that the margin between N=16 and N=17 is large enough that I expect the same answer holds if leap years are accounted for).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With 16 people, you will satisfy the criteria 48.3% of the time, and with 17 people, you satisfy them 52.6% of the time, so the answer is 17. I put the script up at &lt;a href=&quot;http://pastebin.com/YjNzECrP&quot; rel=&quot;nofollow&quot;&gt;http://pastebin.com/YjNzECrP&lt;/a&gt; if anyone would like to check my work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Thanks to Lopsy for pointing out a mistake in the script, fixing that causes it to produce 14 people as the answer.&lt;/p&gt;&#xA;" OwnerUserId="7459" LastEditorUserId="-1" LastEditDate="2014-12-24T18:05:24.097" LastActivityDate="2014-12-24T18:05:24.097" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6400" PostTypeId="2" ParentId="6394" CreationDate="2014-12-24T17:21:19.160" Score="6" Body="&lt;p&gt;Let's suppose there are $r$ people in the room.  We want to find the probability that at least two have the same birthday or adjacent birthdays.  It seems easier to find the complementary probability, i.e. the probability that no two have the same or adjacent birthdays.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are $365^r$ possible arrangements of birthdays for the $r$ people, all of which we assume are equally likely.  The number of arrangements in which the birthdays are distinct and non-adjacent can be obtained from a well-known combinatorial formula, as follows.  There are $\binom{365-r+1}{r}$ ways to pick $r$ integers from 1 to 365 with no two adjacent, where $\binom{n}{m}$ is a binomial coefficient: $\binom{n}{m} = \frac{n!}{m! (n-m)!}$.  But we need to multiply by $r!$, because we are considering the people to be distinguishable, so the order of the $r$ integers is significant.  Hence the probability that no two people have the same or adjacent birthdays is &#xA;$$f(r) = \frac{\binom{365-r+1}{r} r!}{365^r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability that at least two people do have the same or adjacent birthdays is $1-f(r)$.  If we do the computations, we find $1-f(13) = 0.4822$ and $1-f(14) = 0.5368$, so the answer to your question is that you need 14 people.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should add that we're not considering 31 December and 1 January to be adjacent.  If we did, it would change the probabilities a little bit, but I don't think it would change the final answer of 14 people.&lt;/p&gt;&#xA;" OwnerUserId="7463" LastEditorUserId="7463" LastEditDate="2014-12-24T21:16:22.547" LastActivityDate="2014-12-24T21:16:22.547" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6401" PostTypeId="1" AcceptedAnswerId="6465" CreationDate="2014-12-24T19:34:09.463" Score="11" ViewCount="2216" Body="&lt;p&gt;This is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Cryptogram&quot; rel=&quot;nofollow noreferrer&quot;&gt;cryptogram&lt;/a&gt; puzzle, with some riddle-style clues to help you solve it. When you find the thing that links all of the clues, that will let you deduce the keys to decipher the cryptogram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the clues:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;p&gt;Christmas day&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;cross bones&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;t&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;amphibian surmounts&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;hnadwrintig &lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and here is the ciphertext:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Fbocov quq vdbwrv'x lbzf mji uxp vr tfdan;&lt;br&gt;&#xA;  Uli sbed &quot;Mnd Thsouf jt&quot; bbw lkq nys megrr.&lt;br&gt;&#xA;  Oh lbjmp rbh xmtn; gcb Dvtim xjudgttb &quot;Fb!&lt;br&gt;&#xA;  Mua Gsrrjexh bv!&quot; faifbavp gfv galncr huu.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One of the &lt;a href=&quot;http://rumkin.com/tools/cipher/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rumkin cipher tools&lt;/a&gt; can be used for decoding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll need two keys to decode the cryptogram; both keys are numbers with their digits spelled out in words, for example the number 1234567890 would be represented like this:  onetwothreefourfivesixseveneightninezero (just an example, not one of the keys). Decimal points are ignored and there are no thousands separators or anything else except for the digits zero to nine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might want to refer to this question and its answers:&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/5905/what-characteristics-of-a-ciphertext-can-be-indicators-of-a-particular-cipher&quot;&gt;What characteristics of a ciphertext can be indicators of a particular cipher?&lt;/a&gt;&#xA;&lt;br&gt; and you will probably need to use reference materials. Feel free to use any resources at your disposal (i.e. using Google is ok).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plaintext consists mostly of English dictionary words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your challenge is to work out the two keys (and the reasoning behind them) and the plaintext. Also the reasoning behind each of the clues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the spirit of teamwork, partial answers are welcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd also be interested in constructive feedback on the puzzle itself, so my question is also &quot;how could this puzzle be improved?&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To make this puzzle slightly easier, I'm adding a checksum for each of the keys. That way you can tell if you have one of the keys right (even if you don't have the other key or you're not sure what cipher is in use). The checksums are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;c40c8102ca3c33705325a8dd614c49a2d29c82f3fce529528f2fe4001fac8a233b0d6916f1c604643a9c375ca3e02589ed5babf8c9e47d0629e145ec13ca8ef0&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a077332a1720b83b9a1c26619f3179dcfff1ef205f09468996c579aedd5ed5fb285acfb02c085768d8fb1cf2bfea05b02e993d947a324e361254bd68824e16e3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They are SHA-512 (SHA2) hashes generated using &lt;a href=&quot;https://quickhash.com/hash-sha512-online&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://quickhash.com/hash-sha512-online&lt;/a&gt;. You can test possible keys by pasting the possible key into 'input data' and seeing what comes up in 'output data' when you click the 'generate' button.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that you already have quite a lot of information about the keys, it might be feasible to &lt;a href=&quot;https://stackoverflow.com/questions/6776050/how-long-to-brute-force-a-salted-sha-512-hash-salt-provided&quot;&gt;attack the hashes&lt;/a&gt; to find the keys. This would be a legitimate way of solving the puzzle as far as I'm concerned. But you'll still be expected to explain &lt;em&gt;why&lt;/em&gt; those keys have been chosen. :)&lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="-1" LastEditDate="2017-05-23T12:38:53.617" LastActivityDate="2016-12-21T15:25:29.313" Title="Christmas Day Cryptogram / Riddle" Tags="|riddle|cipher|seasonal|" AnswerCount="3" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="6402" PostTypeId="2" ParentId="6386" CreationDate="2014-12-24T20:49:42.100" Score="32" Body="&lt;p&gt;The objects described on each line of the letter each have a distinctive&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; shape.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A &quot;pirate's buried treasure&quot; is the most obvious; it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; associated with an &quot;X&quot;, as in &quot;X marks the spot&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A &quot;dying moon&quot; is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a crescent, or &quot;C&quot; shaped.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A tall straight tree would be essentially&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a vertical line,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and since we are apparently looking at&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; letter shapes,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;that would be the&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; capital letter &quot;I&quot;, as in &quot;iguana&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The empty well is a bit ambiguous, but the best fit with the rest of the clues is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a circle, or the letter &quot;O&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &quot;fork in the road&quot; also could have one of two meanings, either&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Y&quot;, or &quot;T&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That gives us a full inventory of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the letters &quot;X&quot;, &quot;C&quot;, &quot;I&quot;, &quot;O&quot;, and either &quot;T&quot; or &quot;Y&quot;. &quot;Y&quot; doesn't seem to combine with the others to spell anything. Using &quot;T&quot; and the order the clues are given in the note, though, gives &quot;TOXIC&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Apparently, the merchant was&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; lying or just incorrect. Either way, DON'T &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;DRINK THE BOTTLE.&lt;/p&gt;&#xA;" OwnerUserId="4204" LastEditorUserId="4204" LastEditDate="2015-03-31T08:12:07.283" LastActivityDate="2015-03-31T08:12:07.283" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6404" PostTypeId="1" AcceptedAnswerId="6407" CreationDate="2014-12-24T22:48:33.953" Score="23" ViewCount="2205" Body="&lt;p&gt;Sent by a friend in Germany. I have no idea what it says yet!...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZW1FD.jpg&quot; alt=&quot;Xmas puzzle image&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7473" LastEditorUserId="5373" LastEditDate="2016-12-21T15:24:11.407" LastActivityDate="2016-12-21T15:24:11.407" Title="Can you decipher the hidden message in this Christmas card image?" Tags="|cipher|visual|seasonal|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6405" PostTypeId="1" AcceptedAnswerId="6412" CreationDate="2014-12-25T01:00:38.023" Score="4" ViewCount="2123" Body="&lt;p&gt;I was playing a game called Space Rangers earlier, and stumbled upon a &quot;text-adventure&quot; mission in it, which required me to find what's supposed to be the only possible solution to this puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a field of 3x3 tiles, numbered 1-9, such as&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[1] [2] [3]&#xA;[4] [5] [6]&#xA;[7] [8] [9]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I can chose TWO numbers, except for the number 9, which is &quot;not working&quot; according to the in-game explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each number is either white or black, and when I choose a number I find out what color it is. It seems that 5 is always black for some reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After choosing two tiles, the game will perform a test and tell you if there are possible &quot;solutions&quot; (as in, possible vertical lines or possible squares of black tiles), while knowing the color of your two chosen numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The possible &quot;solutions&quot; to the puzzle is a vertical line of black tiles, or a square of 2x2 of black tiles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I must guarantee that there is only one possible solution, as in, only one possible line or square that are all black.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which two numbers should I choose?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have thought it over for a long while, and the only idea I could come up with would be 4 and 6, but one of them is ALWAYS black, making my attempt at getting both as whites to mess up both vertical lines and squares in each side...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After countless attempts and even more half-hearted attempts, I gave up, accepting that I failed the mission, and moved on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm curious, as I assume it is possible one way or another (I just fail to see something that's probably obvious).&#xA;What would be a working solution to this puzzle?&lt;/p&gt;&#xA;" OwnerUserId="7475" LastEditorUserId="4204" LastEditDate="2014-12-25T04:59:15.293" LastActivityDate="2015-05-26T01:13:58.867" Title="Determining the pattern of a 3x3 grid of hidden black and white tiles by revealing only two" Tags="|logical-deduction|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6406" PostTypeId="2" ParentId="6405" CreationDate="2014-12-25T01:58:01.827" Score="5" Body="&lt;p&gt;It sounds impossible to me, the way you describe it. Let me try rephrasing it. Please let me know if I'm making erroneous assumptions: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a grid of numbers 1-9. Each has a hidden color, either white or black. You can choose any two to reveal, and from those you need to determine the pattern of the grid. You need to pick the two numbers that will provide you with information to guess the right answer every single time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The grid can either have one vertical black bar (all others white) or a 2x2 grid of black (all others white). That means there's 7 answers, 4 squares and 3 lines: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1245&#xA;4578&#xA;2356&#xA;5689&#xA;147&#xA;258&#xA;369&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this state the puzzle is impossible. Finding out the states of two numbers only gives four different possible states of the 7 needed. So I'm going to make two assumptions: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Assuming the 9 is a complete dud and not part of the answer. This removes 369 and 5689. So there's 5 different solutions, 3 squares and 2 lines: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1245&#xA;4578&#xA;2356&#xA;147&#xA;258&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still one too many. So on to assumption 2: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Assuming the 5 is always black because of some rule and not just your personal experience. That wipes out one of the options (147). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1245&#xA;4578&#xA;2356&#xA;258&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Four options, four possible button results. Now it's a matter of finding the two that are black in exactly two answers (and if there's more than two buttons, throwing out any 'dupes' where one is black whenever another is). There's only two: 4 and 8. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 black, 8 black: Answer is 4578&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 white, 8 black: Answer is 258&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 black, 8 white: Answer is 1245&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 white, 8 white: Answer is 2356&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No idea if this is the right answer, but hope it helps. Best of luck tracking down a walkthrough if it's not. &lt;/p&gt;&#xA;" OwnerUserId="7476" LastActivityDate="2014-12-25T01:58:01.827" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6407" PostTypeId="2" ParentId="6404" CreationDate="2014-12-25T02:46:07.817" Score="33" Body="&lt;p&gt;The message reads:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HaPpY 2015&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How to decode:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As noted in the comments, each figure has seven sharp corners. Treating a sharp left turn as $1$ and a sharp right turn as $0$, each figure encodes a 7-digit binary number. These numbers are $$1001000,\, 1100001,\, 1010000,\, 1110000,\, 1011001,\, 0110010,\, 0110000,\, 0110001,\, 0110101.$$ If we interpret these numbers as &lt;a href=&quot;http://en.wikipedia.org/wiki/ASCII#ASCII_printable_code_chart&quot;&gt;ASCII&lt;/a&gt;, we get a sequence of characters:&lt;br&gt;&amp;nbsp;&amp;nbsp;&lt;strong&gt;H&lt;/strong&gt;,&amp;nbsp; &lt;strong&gt;a&lt;/strong&gt;,&amp;nbsp; &lt;strong&gt;P&lt;/strong&gt;,&amp;nbsp; &lt;strong&gt;p&lt;/strong&gt;,&amp;nbsp; &lt;strong&gt;Y&lt;/strong&gt;,&amp;nbsp; &lt;strong&gt;2&lt;/strong&gt;,&amp;nbsp; &lt;strong&gt;0&lt;/strong&gt;,&amp;nbsp; &lt;strong&gt;1&lt;/strong&gt;,&amp;nbsp; &lt;strong&gt;5&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2014-12-25T17:03:30.567" LastActivityDate="2014-12-25T17:03:30.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6408" PostTypeId="1" AcceptedAnswerId="6409" CreationDate="2014-12-25T03:35:31.537" Score="17" ViewCount="1258" Body="&lt;p&gt;Here's yet another puzzle adapted from a puzzle book (in my case, with edits to make some of the specifications of the puzzle more clear because I didn't really understand them the first time I read it).&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Five professors are spending three days on a retreat, and every day they eat dinner at the same restaurant. This restaurant has a special &amp;quot;roulette menu&amp;quot; with nine items, in which you choose a menu item from &lt;code&gt;1&lt;/code&gt; to &lt;code&gt;9&lt;/code&gt; that corresponds to one of the nine available items, but you don't know which number corresponds to which item. When the items are ordered, they are brought to the table in no particular order, so if a group orders multiple items at a time, they cannot figure out individually which item was ordered by whom.&lt;/p&gt;&#xA;&lt;p&gt;These professors have tasked themselves with figuring out which number corresponds to which item in three days regardless. Each professor will order one item per day. How should they plan their ordering of the items in order to find out which number corresponds to which item for all items?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="76997" LastEditDate="2023-07-03T10:00:55.530" LastActivityDate="2023-07-03T10:00:55.530" Title="Five professors and nine dishes" Tags="|logical-deduction|combinatorics|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="6409" PostTypeId="2" ParentId="6408" CreationDate="2014-12-25T03:58:17.267" Score="20" Body="&lt;p&gt;Okay, 15 orders, 9 dishes, that means 6 dupes. So 3 dishes will only be ordered once, obviously never on the same day. Now to split the six dupes so they can all be detected. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first 3: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Day 1: 1   3&#xA;Day 2: 1 2&#xA;Day 3:   2 3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The dish that shows on day 1 and 2 is #1, and so on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now add in same-day pairs:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Day 1: 1   3 4 4&#xA;Day 2: 1 2   5 5 &#xA;Day 3:   2 3 6 6&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So the dish that's doubled on day 1 is #4, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now the last three:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Day 1: 1 7 3 4 4&#xA;Day 2: 1 2 8 5 5 &#xA;Day 3: 9 2 3 6 6&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The dish that only shows up on day 1 is #7, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then, secure in their triumph, they can leave five bad Yelp reviews warning people away from this crazy place.&lt;/p&gt;&#xA;" OwnerUserId="7476" LastActivityDate="2014-12-25T03:58:17.267" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6412" PostTypeId="2" ParentId="6405" CreationDate="2014-12-25T05:29:23.950" Score="6" Body="&lt;p&gt;Revealing the colours of only two tiles only provides four possible outcomes, which you can't map to five possible results (vertical line down the middle or 2x2 square). There just isn't enough information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is an out. If we assume that square 9 &quot;not working&quot; means that the 2x2 square of black tiles will never contain 9, then we have just the following four configurations:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x x .   . x .   . . .   . x x        x = black&#xA;x x .   . x .   x x .   . x x        . = white&#xA;. . _   . x _   x x _   . . _        _ = not working&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The case where the 2x2 black square is in the bottom-right corner is excluded because the 9 tile isn't working. So you have these four cases, and you need to choose two tiles such that the tiles that show are all different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to do this, you need to produce a binary encoding, so both tiles will need to show black in exactly two out of four cases. The tiles that do so are:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;4: x . x .&#xA;8: . x x .&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Luckily for us, these produce a binary encoding as-is. So the two tiles you want are 4 and 8.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2014-12-25T05:29:23.950" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="6413" PostTypeId="5" CreationDate="2014-12-25T18:56:46.973" Score="0" Body="&lt;p&gt;Puzzles which are themed towards or particularly relevant to a certain time of year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we might have:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Christmas&lt;/li&gt;&#xA;&lt;li&gt;New Year&lt;/li&gt;&#xA;&lt;li&gt;Easter&lt;/li&gt;&#xA;&lt;li&gt;Spring Equinox&lt;/li&gt;&#xA;&lt;li&gt;Summer&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11497" LastEditorUserId="4421" LastEditDate="2016-09-01T01:48:30.147" LastActivityDate="2016-09-01T01:48:30.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6414" PostTypeId="4" CreationDate="2014-12-25T18:56:46.973" Score="0" Body="Puzzles which are themed towards or particularly relevant to a certain time of year, e.g. seasons, events, holidays." OwnerUserId="4050" LastEditorUserId="4050" LastEditDate="2014-12-25T23:06:33.210" LastActivityDate="2014-12-25T23:06:33.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6415" PostTypeId="1" AcceptedAnswerId="6431" CreationDate="2014-12-25T21:04:56.943" Score="52" ViewCount="20966" Body="&lt;p&gt;A drunk man stands with a cliff one step to his left.&#xA;He takes steps randomly left and right.&#xA;Each step has probability $p$ of going left and probability $q=1-p$ of going right.&#xA;Each step is the same size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If allowed to randomly step indefinitely, what is the probability that the drunk falls off the cliff?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In purely mathematical terms, the drunkard starts at $x=0$ and steps $+1$ or $-1$ with each step.&#xA;If he ever gets to $x=-1$ he falls off the cliff.&lt;/p&gt;&#xA;" OwnerUserId="7488" LastEditorUserId="2023" LastEditDate="2014-12-27T01:05:58.830" LastActivityDate="2017-03-12T06:58:56.203" Title="Does the drunk man fall off the cliff? (a random walk problem)" Tags="|mathematics|probability|" AnswerCount="7" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="6417" PostTypeId="2" ParentId="6069" CreationDate="2014-12-26T00:10:11.707" Score="3" Body="&lt;p&gt;As this puzzle has been open for a while, I'm now going to post the &lt;em&gt;solution as intended&lt;/em&gt;. At this time, puzzlers on this site have 99% solved the cipher, and it is mainly technical detail which remains to be sorted - which might not be much fun. So, as a wrap-up here is the cipher as used in this puzzle. Do not read on, if you still want to find it by yourself.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The cipher is a simple transposition cipher acting &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;on letters only.&lt;/strong&gt;&lt;br&gt; Each letter is shifted X positions to its right/clockwise, where X stand for the &lt;em&gt;value&lt;/em&gt; of the letter (A=1, B=2, ...).&lt;br&gt;Spaces, numbers, dots etc. remain unaffected. &lt;br&gt;&#xA; All characters (including spaces) &lt;em&gt;count&lt;/em&gt; for the shifting purspose.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A few complications come in to have it a reversible cipher:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;The encoding process is sequentially done for the letters&lt;br&gt; The 'simultaneously' .&lt;/strong&gt;&lt;br&gt; First &lt;strong&gt;all A's&lt;/strong&gt; are shifted &lt;em&gt;&quot;simultaneously&quot;&lt;/em&gt; by 1 position to the right.&lt;br&gt; next &lt;strong&gt;all B's&lt;/strong&gt; are shifted &lt;em&gt;&quot;simultaneously&quot;&lt;/em&gt; by 2 positions to the right.&lt;br&gt; next &lt;strong&gt;all C's&lt;/strong&gt; are shifted &lt;em&gt;&quot;simultaneously&quot;&lt;/em&gt; by 3 positions to the right.&lt;br&gt; etc.&lt;br&gt;&lt;em&gt;&quot;simultaneously&quot;&lt;/em&gt; shifting is necessary, to avoid letters of the same type 'overtaking' each other, which would destroy reversibility of the cipher.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In order to decode a cipher text&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Reverse the process!&lt;/strong&gt;.&lt;br&gt; First &lt;strong&gt;all Z's&lt;/strong&gt; are shifted &lt;em&gt;&quot;simultaneously&quot;&lt;/em&gt; by 26 position to the left.&lt;br&gt; next &lt;strong&gt;all Y's&lt;/strong&gt; are shifted &lt;em&gt;&quot;simultaneously&quot;&lt;/em&gt; by 25 positions to the left.&lt;br&gt; next &lt;strong&gt;all X's&lt;/strong&gt; are shifted &lt;em&gt;&quot;simultaneously&quot;&lt;/em&gt; by 24 positions to the left.&lt;br&gt; etc.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is the example encoding process:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/ieSct.jpg&quot; alt=&quot;Encoding process&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastActivityDate="2014-12-26T00:10:11.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6418" PostTypeId="2" ParentId="6415" CreationDate="2014-12-26T06:06:27.587" Score="19" Body="&lt;p&gt;&lt;strong&gt;Answer:&lt;/strong&gt; The probability he dies is $D = p/q$ for $q&amp;gt;1/2$, and $D=1$ otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We'll prove the answer in a circumspect way by solving a different problem. The method doesn't use any summations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, I claim that if $q &amp;lt; 1/2$, then the pirate dies with probability $D=1$. After $t$ steps, his position is the sum of $t$ independent random variables with negative mean. So, far large $t$, it approaches a normal distribution with negative mean $-O(t)$ and standard deviation $O(\sqrt{t})$. Therefore, his position remaining non-negative requires a &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_score&quot; rel=&quot;noreferrer&quot;&gt;z-score&lt;/a&gt; of $O(\sqrt{t})$, the probability of which approaches $0$ as $t \to \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the modified problem. Label the pirate's possible positions as integers $x$. Imagine there's a pit at $x=0$ that kills the pirate if he steps into it. Say he also starts at $x=0$, but is exempt from dying to the pit only then. Because it's a pit rather than a cliff, the pirate can move left into the negatives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $q&amp;gt;1/2$. There's two cases. If the pirate moves right on his first move, we have the original problem (shifted right by 1), since he can never cross into the negatives without dying. If he moves left, he dies with probability 1, since we have a mirrored problem with $q&amp;lt;1/2$ where the pirate tends towards the pit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key idea is that the pirate has an equal probability to die starting left as starting right. Each of these is a total, not a conditional, probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Pr(\text{Die and Left}) = \Pr(\text{Die and Right})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because there's a bijection between the two types of dying paths: mirror the path by flipping the direction of each move. For example,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{LLRLLRRR} \leftrightarrow \text{RRLRRLLL}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mirroring preserves the fact that the pirate dies at the end of the path and no earlier.  Moreover, it preserves probabilities: because the pirate starts and ends at $x=0$, each path must have an equal number $n$ of lefts and rights, and its overall probability $p^n q^n$ depends only on this number. Summing the probabilities over all these paths, we find the pirate has an equal chance of dying going left and right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now expand the probabilities conditional on which way the pirate went.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Pr(\text{Die} \mid \text{Left}) \Pr(\text{Left}) = \Pr(\text{Die} \mid \text{Right}) \Pr(\text{Right})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $D$ be the probability we want to compute, the probability that the pirate dies if he starts right, and recall that he dies with probability $1$ if he starts left. This lets us solve for $D$ in terms of $p$ and $q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 \times p = D \times q$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which gives the answer $D = p/q$ for $q&amp;gt;1/2$. Recall that we argued separately that $D=1$ for $q &amp;lt; 1/2$. Finally, it remains to observe that when $q$ is exactly $1/2$, we must have $D=1$, since $D$ must be non-increasing in $q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can check that this has the right behavior: $D$ is decreasing in $q$, has $D=1$ at $q=1/2$, and has $D=0$ at $q=1$. &lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="7488" LastEditDate="2014-12-28T18:57:29.713" LastActivityDate="2014-12-28T18:57:29.713" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="6420" PostTypeId="1" CreationDate="2014-12-26T06:33:34.883" Score="7" ViewCount="1889" Body="&lt;p&gt;Me and a friend are trying to solve a series of puzzles, and this is the last one I need to solve and I'm stumped.  It's a Christmas themed puzzle titled 'bLitZEN'.  The only hints given are that the title is important and BIT=1011000.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The puzzle reads:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BLT - BIT&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IE&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BEN - LIZ&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IN - TEN&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LIEN - LIE&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IT - (EN + ZEN)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ZN&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When  I first saw it, I assumed it used ASCII but that's not giving me anything.  I don't know anything about binary, so I thought that maybe it might be binary.  But I can't figure out how BIT = 1011000 at all, and I can't seem to use that alone to solve the puzzle.  Typically, each line gives you a single letter, where each line's letter forms a random word.  It's part of a series, so that's what the others were like.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me know if you can come up with anything.  Thanks for the help.  Much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="7492" LastActivityDate="2014-12-27T07:51:23.197" Title="Need help solving this code puzzle. Possible binary?" Tags="|calculation-puzzle|puzzle-identification|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6421" PostTypeId="2" ParentId="6415" CreationDate="2014-12-26T07:06:49.917" Score="25" Body="&lt;p&gt;The probability of dying eventually is the probability of moving to the left from x=0 eventually. Denote the probability of moving to the left of some point eventually when starting at that point by $p_d$.&#xA;The probability $p_d$ is equal to the probability of moving left immediately ($p$), plus the probability of moving to the right immediately ($1-p$), then to the left eventually (back to the starting point) ($p_d$), then to the left eventually ($p_d$ again).&#xA;In other words:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;p_d &amp;amp;= p + (1-p){p_d}^2 \\&#xA;(1-p) {p_d}^2 - p_d + p &amp;amp;= 0 \,.&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging into the &lt;a href=&quot;http://en.wikipedia.org/wiki/Quadratic_equation&quot;&gt;quadratic formula&lt;/a&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;p_d &amp;amp;= {1 \pm \sqrt {1-4(1-p)p} \over 2(1-p)} \\&#xA;&amp;amp;= {1 \pm \sqrt {1-4p+4p^2} \over 2(1-p)} \\&#xA;&amp;amp;= {1 \pm \sqrt {(1-2p)^2} \over 2(1-p)} \\&#xA;&amp;amp;= {1 \pm (1-2p) \over 2(1-p)}\,.&#xA;\end{align}&#xA;$$&#xA;Evaluating for the two cases of plus and minus sign:&#xA;$$&#xA;\begin{align}&#xA;p_{d1} &amp;amp;= {2-2p \over 2-2p} = 1 \quad \text{(first branch)} \\&#xA;\text{and} \qquad p_{d2} &amp;amp;= {2p \over 2(1-p)} = p/q \quad \text{(second branch)}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For $p&amp;gt;1/2$ the second branch is physically impossible and the first branch applies.&lt;/li&gt;&#xA;&lt;li&gt;For $p=0$ the second branch clearly applies, not the first one.&lt;/li&gt;&#xA;&lt;li&gt;For $p=1/2$ both branches coincide and never intersect otherwise.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For similar probabilities the results should be similar ($p_d(p)$ is continuous), otherwise enough simulation would be able to distinguish arbitrarily small changes in step-wise probability with arbitrary confidence just by observing the survival rate. Thus, the second branch applies to the entire range of $(0,1/2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have a formal proof unfortunately, but I did run an automated test (1000 runs, give up after 1000 steps, tested for multiple values of $p$, distribution of survival times compared to expectations (&lt;a href=&quot;http://en.wikipedia.org/wiki/Unimodality&quot;&gt;unimodality&lt;/a&gt; with mode at 1 step) visually). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In total:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the probability of moving towards the cliff, $p$, is at least $1/2$, death occurs with a probability of one. Otherwise, death occurs with a probability of $p\over{1-p}$.&lt;/p&gt;&#xA;" OwnerUserId="284" LastEditorUserId="284" LastEditDate="2014-12-26T23:45:49.217" LastActivityDate="2014-12-26T23:45:49.217" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="6422" PostTypeId="1" AcceptedAnswerId="6424" CreationDate="2014-12-26T09:22:13.840" Score="3" ViewCount="2112" Body="&lt;p&gt;I've split the 26 letters (actually 25 letters) of the English language into two groups using a particular rule. Can you say what is the rule that I've used?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First Group: &lt;strong&gt;A D E F G H K L M N O Z&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;Second Group:&lt;strong&gt;B C I J P Q R S T U V X Y&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note that I've excluded &lt;strong&gt;W&lt;/strong&gt; from both the groups as it fits partially into both the groups. Also,the font and the &lt;strong&gt;BOLD&lt;/strong&gt; style doesn't matter here.&lt;/p&gt;&#xA;" OwnerUserId="7482" LastEditorUserId="5188" LastEditDate="2015-04-27T06:05:44.127" LastActivityDate="2015-04-27T06:05:44.127" Title="What rule have I used to split the alphabet?" Tags="|pattern|english|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6423" PostTypeId="2" ParentId="6420" CreationDate="2014-12-26T10:04:01.230" Score="10" Body="&lt;p&gt;BIT = 1011000 and the mathematical equations makes this look a lot like binary is involved. Combine that with the hint that the title is important, and we can work out where BIT=1011000 comes from:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Finding the letters BIT in the title gives b-it--- ... the same pattern as 1011000&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Knowing that, we can then convert all of the other equations to binary:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 11001000 - 1011000, 10010, 1000011 - 110100, 10001 - 1011, 110011 - 110010, 11000 - (11 + 111), 101&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solve those, and we get a list of numbers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 16 18 15 6 1 14 5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Numbers &amp;lt;= 26 can be converted to letters by counting through the alphabet, giving us:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PROFANE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4270" LastActivityDate="2014-12-26T10:04:01.230" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6424" PostTypeId="2" ParentId="6422" CreationDate="2014-12-26T11:19:24.467" Score="5" Body="&lt;p&gt;This is what I think it is, I'm not fully convinced myself though.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There are commonly used words in English that are pronounced the same way as the letters in the second group. But with that logic H &amp;amp; R would be certainly in the second group.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7493" LastActivityDate="2014-12-26T11:19:24.467" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="6425" PostTypeId="1" AcceptedAnswerId="6426" CreationDate="2014-12-26T15:22:38.287" Score="3" ViewCount="1164" Body="&lt;p&gt;NB: This is &lt;a href=&quot;http://www.brainbashers.com/showpuzzles.asp?puzzle=ZGQN&quot; rel=&quot;nofollow&quot;&gt;based on a puzzle over at BrainBashers.com&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The setup&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;You and a friend play a card game called Green Jack. The deck consists of 16 cards, divided into 4 cards in 4 colors (standard playing card abbreviations here, A = Ace, K = King, Q = Queen, J = Jack):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\color{red}{\rm A,\,K,\,Q,\,J}$&lt;/li&gt;&#xA;&lt;li&gt;$\color{blue}{\rm A,\,K,\,Q,\,J}$&lt;/li&gt;&#xA;&lt;li&gt;$\color{orange}{\rm A,\,K,\,Q,\,J}$&lt;/li&gt;&#xA;&lt;li&gt;$\color{green}{\rm A,\,K,\,Q,\,J}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Cards are ranked as listed, A &gt; K, K &gt; Q, Q &gt; J &lt;strong&gt;except&lt;/strong&gt; the $\color{green}{J}$ which beats everything. If two cards have the same face value, then the color determines the win: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\color{red}{A} &amp;gt; \color{blue}{A} &amp;gt; \color{orange}{A} &amp;gt; \color{green}{A}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;except, again, the Green jack, which beats everything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The game is played as follows: you are dealt one card face up, and your friend is dealt one card face down. Your friend then makes three true statements, and you have to work out who has the higher card, you or your friend.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The hand&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;You are dealt the blue king ($\color{blue}{K}$). Your friend makes the following three true statements:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;My card can beat an orange queen ($\color{orange}{Q}$).&lt;/li&gt;&#xA;&lt;li&gt;Knowing this, if my card is more likely to be an ace (A), then it is a queen (Q); otherwise it is a king (K).&lt;/li&gt;&#xA;&lt;li&gt;Knowing the above, if my card is more likely to be a queen, then it's $\color{red}{red}$; otherwise it is $\color{blue}{blue}$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Who wins the game?&lt;/p&gt;&#xA;" OwnerUserId="36" LastEditorUserId="36" LastEditDate="2014-12-26T15:57:01.653" LastActivityDate="2014-12-28T00:02:20.733" Title="A game of Green Jack" Tags="|logical-deduction|cards|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="6426" PostTypeId="2" ParentId="6425" CreationDate="2014-12-26T15:34:03.367" Score="7" Body="&lt;p&gt;Constraint 1 leaves 10 possibilities: red A, red K, red Q, blue A, blue Q, orange A, orange K, green A, green K, green J.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Among these 10, 4 are aces, 3 are kings, 2 are queens, and 1 is a jack. If I'm interpreting the meaning of 'more likely' correctly, this means the card &lt;em&gt;is&lt;/em&gt; 'more likely' to be an ace (than anything else), so constraint 2 means it's a queen. Now constraint 3 tells us it's red.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your friend has the &lt;strong&gt;red queen&lt;/strong&gt;, and &lt;strong&gt;you win&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm probably not interpreting this right though!&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2014-12-26T15:46:26.920" LastActivityDate="2014-12-26T15:46:26.920" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6427" PostTypeId="2" ParentId="6425" CreationDate="2014-12-26T15:55:39.523" Score="2" Body="&lt;p&gt;An alternate interpretation of &quot;most likely&quot; (before the &quot;knowing this&quot; edit):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statement 1 is as rand al'thor stated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since statement 2 and statement 3 are true, then the card must be either a Queen or King, and Red or Blue, &lt;strong&gt;regardless of the interpretation of &quot;most likely&quot;&lt;/strong&gt;.  Since the Blue King is in play, the 3 possibilities are the Red Queen, the Blue Queen and the Red King.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, the card is unlikely to be an Ace (0% probability) and likely to be a Queen (66% probability), so the mystery card is the &lt;strong&gt;Red King&lt;/strong&gt; and &lt;strong&gt;you lose&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the &quot;knowing this&quot; edit throws off this interpretation, but now the intended answer is trivial.&lt;/p&gt;&#xA;" OwnerUserId="6178" LastEditorUserId="6178" LastEditDate="2014-12-26T16:32:52.543" LastActivityDate="2014-12-26T16:32:52.543" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6430" PostTypeId="2" ParentId="6401" CreationDate="2014-12-26T19:38:23.020" Score="5" Body="&lt;p&gt;&lt;strong&gt;This answer details the first part of the solution of the &lt;em&gt;riddle&lt;/em&gt; and says which cipher to use. For the two keys, see &lt;a href=&quot;https://puzzling.stackexchange.com/a/6515/5373&quot;&gt;Len's answer&lt;/a&gt;. For the final solution, see &lt;a href=&quot;https://puzzling.stackexchange.com/a/6465/5373&quot;&gt;MarchHo's answer&lt;/a&gt;.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I can see, the only one of the Rumkin cipher tools that needs two keys is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the &lt;a href=&quot;http://rumkin.com/tools/cipher/vigenere-keyed.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Keyed Vigenere cipher&lt;/a&gt;, which needs both an 'alphabet key' and a 'passphrase'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Presumably (although he didn't say) this is what MarchHo used to decrypt the message.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to the riddle is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; NEWTON&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which gives the two numbers found by Len. All the clues point towards this answer as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Christmas day&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Isaac Newton was born on Christmas Day 1642 [&lt;a href=&quot;http://en.wikipedia.org/wiki/Isaac_Newton&quot; rel=&quot;nofollow noreferrer&quot;&gt;ref&lt;/a&gt;]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;cross bones&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; were Isaac Newton's personal sigil [&lt;a href=&quot;http://www.answers.com/Q/What_is_the_meaning_behind_Sir_Isaac_Newton&amp;#39;s_coat_of_arms&quot; rel=&quot;nofollow noreferrer&quot;&gt;ref&lt;/a&gt;]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;t&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; refers to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Tonne&quot; rel=&quot;nofollow noreferrer&quot;&gt;metric tonne&lt;/a&gt; or &lt;strong&gt;new ton&lt;/strong&gt; (brilliant!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;amphibian surmounts&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A newt is a kind of amphibian, and 'surmounts' is a fancy word for 'on', so this gives &lt;strong&gt;newt on&lt;/strong&gt; (also brilliant!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;hnadwrintig&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You recognised this word, didn't you? You realised it meant 'handwriting' even though that's not what it says. This is supposed to point to &lt;strong&gt;handwriting recognition&lt;/strong&gt; and a &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton_(platform)&quot; rel=&quot;nofollow noreferrer&quot;&gt;device called Newton&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2014-12-30T13:04:27.447" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6431" PostTypeId="2" ParentId="6415" CreationDate="2014-12-26T19:42:57.857" Score="24" Body="&lt;p&gt;First, let's consider a modified problem. Suppose that the position of the cliff is moved from $x=-1$ to $x=a$ for some integer $a&amp;lt;0$. Suppose also that the drunkard has a home at $x=b$ for some integer $b&amp;gt;0$, and that if the drunkard reaches his home, he stops walking. Let $d(x)= d_{a,b}(x)$ be the probability the drunkard falls of the cliff, given that the drunkard is at position $x$. We have a recurrence relation&#xA;$$&#xA;d(x)=p\,d(x-1)+(1-p)\,d(x+1).&#xA;$$&#xA;The characteristic polynomial of this recurrence is $(1-p)r^2-r+p$. If we assume $p\neq\frac{1}{2}$, this polynomial has two distinct roots:&#xA;$$&#xA;r=1,\hspace{1cm} r=\frac{p}{1-p}.&#xA;$$&#xA;This means that $d(x)$ has the form&#xA;$$&#xA;d(x)=c_1+c_2\left(\frac{p}{1-p}\right)^x,&#xA;$$&#xA;where $c_1$ and $c_2$ are constants which may depend on $a$, $b$, and $p$ but are independent of $x$. We have boundary conditions $d(a)=1$ and $d(b)=0$, so we can solve for $c_1$ and $c_2$:&#xA;$$&#xA;c_1=\frac{1}{1-\left(\frac{p}{1-p}\right)^{a-b}},\,\,\,c_2=\frac{1}{\left(\frac{p}{1-p}\right)^a-\left(\frac{p}{1-p}\right)^b}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;We recover the original question by taking $x=0$, $a=-1$, and letting $b$ go to infinity. The probability the drunkard falls in the original question is&#xA;$$&#xA;\begin{align*}&#xA;\lim_{b\to\infty} d_{-1,b}(0)&amp;amp;=\lim_{b\to\infty}\frac{1}{1-\left(\frac{p}{1-p}\right)^{-1-b}}+\frac{1}{\left(\frac{p}{1-p}\right)^{-1}-\left(\frac{p}{1-p}\right)^b}\left(\frac{p}{1-p}\right)^0\\&#xA;&amp;amp;=\lim_{b\to\infty}\frac{\left(\frac{p}{1-p}\right)^b-1}{\left(\frac{p}{1-p}\right)^b-\frac{1-p}{p}}\\&#xA;&amp;amp;=\begin{cases}1:p&amp;gt;\frac{1}{2},\\\frac{p}{1-p}:p&amp;lt;\frac{1}{2}.\end{cases}&#xA;\end{align*}&#xA;$$&#xA;Since the probability of falling approaches $1$ as $p$ approaches $\frac{1}{2}$ from below, the probability of falling must be $1$ when $p=\frac{1}{2}$.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastActivityDate="2014-12-26T19:42:57.857" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6435" PostTypeId="1" AcceptedAnswerId="6436" CreationDate="2014-12-26T22:00:09.693" Score="21" ViewCount="4515" Body="&lt;p&gt;Here are three ways of splitting the letters of the English alphabet into two groups:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &lt;strong&gt;ABDOPQR&lt;/strong&gt;; &lt;em&gt;CEFGHIJKLMNSTUVWXYZ&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) &lt;strong&gt;BCDGJOPQRSU&lt;/strong&gt;; &lt;em&gt;AEFHIKLMNTVWXYZ&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) &lt;strong&gt;DFHIJLNOPRTUVXYZ&lt;/strong&gt;; &lt;em&gt;ABCEGKMQSW&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In each case, the &lt;strong&gt;first collection&lt;/strong&gt; of letters can be described simply using a one-word 'catchphrase' while the &lt;em&gt;second collection&lt;/em&gt; is simply all those letters not in the first collection. First you need to find all the three catchphrases (this should be the only challenging part). In each case, determine how many distinct letters of the catchphrase are in the &lt;strong&gt;first&lt;/strong&gt; and how many in the &lt;em&gt;second&lt;/em&gt; collection of letters. You'll then have three pairs of numbers, and your final task is to find which of these three pairs is the odd one out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I really hope to achieve in this question is to get a collection of interesting alphabet-splitting puzzles, so that they're not all spread around the site in many &lt;a href=&quot;https://puzzling.stackexchange.com/questions/6361/how-did-i-split-the-alphabet&quot;&gt;different&lt;/a&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/questions/6422/what-rule-have-i-used-to-split-the-alphabet&quot;&gt;questions&lt;/a&gt;. The bit about catchphrases is just to unify this into a single puzzle so that I don't get people telling me I should split this into 3 distinct questions. After this has been solved (which shouldn't take long), I'll throw open the question to editing and we can incorporate all the interesting ways of splitting the alphabet that anyone can think of, and perhaps change or remove the catchphrase and odd-one-out parts of the question. &lt;strong&gt;When editing a sequence into the question, please edit the answer into warspyking's now-accepted answer.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your Puzzles;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) &lt;strong&gt;ABCDEGJLMOQTUZ&lt;/strong&gt;; &lt;em&gt;FHIKNPRSVWXY&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) &lt;strong&gt;EITSANHURDMWG&lt;/strong&gt;; &lt;em&gt;VLFBKOPJXCZYQ&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6) &lt;strong&gt;ABCIJOPQRSTUWY&lt;/strong&gt;; &lt;em&gt;DEFGHKLMNOVZ&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7) &lt;strong&gt;COPSUVWXZ&lt;/strong&gt;; &lt;em&gt;ABDEFGHIJKLMNQRTY&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8) &lt;strong&gt;AHIMOTUVWXY&lt;/strong&gt;; &lt;em&gt;BCDEFGJKLNPQRSZ&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;9) &lt;strong&gt;ΑΒΕΗΙΚΜΝΟΡΤΧΥΖ&lt;/strong&gt;; &lt;em&gt;CDFGJLQRSUVW&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;10) &lt;strong&gt;AEFHILMNORSX&lt;/strong&gt;; &lt;em&gt;BCDGJKPQTUVWYZ&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;11) &lt;strong&gt;HJKMNUVWXY&lt;/strong&gt;; &lt;em&gt;ABCDEFGILOPQRSTZ&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;12) &lt;strong&gt;BDEFHIKLMNPRTUVWXYZ&lt;/strong&gt;; &lt;em&gt;ACGJOQS&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;13) &lt;strong&gt;EFPTY&lt;/strong&gt;; &lt;em&gt;ABCDGHIJKLMNOQRSUVWXZ&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;14) &lt;strong&gt;BCFHIKNOPSUVWY&lt;/strong&gt;; &lt;em&gt;ADEGJLMQRTXZ&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If you think this is a silly way of posing a question, please tell me and give me a chance to improve it before hitting the downvote button. Let's be proactive here! :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: I just discovered that more or less exactly the same idea appears in &lt;a href=&quot;http://www.benschott.com/miscellanies/schotts-original-miscellany/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schott's Miscellany&lt;/a&gt;. His list is entitled 'Letter Traits' and is on page 76.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-07-05T09:50:18.370" Title="Alphabet splitting extraordinaire!" Tags="|english|language|" AnswerCount="2" CommentCount="21" ContentLicense="CC BY-SA 3.0" />
  <row Id="6436" PostTypeId="2" ParentId="6435" CreationDate="2014-12-26T23:11:38.723" Score="11" Body="&lt;p&gt;&lt;em&gt;Final Answer!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sequences;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;ABDOPQR&lt;/strong&gt;; CEFGHIJKLMNSTUVWXYZ&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;BCDGJOPQRSU&lt;/strong&gt;; AEFHIKLMNTVWXYZ&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;DFHIJLNOPRTUVXYZ&lt;/strong&gt;; ABCEGKMQSW&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ANSWERS&lt;/strong&gt; (credit to &lt;strong&gt;squeamish ossifrage&lt;/strong&gt; for the 3rd answer):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;1)&lt;/strong&gt; All these letters have enclosed space within the letters; Catchphrase &lt;em&gt;holes&lt;/em&gt;.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;2)&lt;/strong&gt; All these letters have curves in them; Catchphrase &lt;em&gt;curves&lt;/em&gt;.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;3)&lt;/strong&gt; All letters that correspond with composite numbers from 1 through 26; Catchphrase &lt;em&gt;composite&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NUMBERS:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; {1, 4}, {4, 2}, {4, 4}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;FINAL ANSWER:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The odd one out is the &lt;em&gt;1st&lt;/em&gt; pair &lt;strong&gt;{1, 4}&lt;/strong&gt; because the other pairs are {even, even} while this one's first number is odd.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;OTHER SEQUENCES:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;&lt;strong&gt;ABCDEGJLMOQTUZ&lt;/strong&gt;; &lt;em&gt;FHIKNPRSVWXY&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;EITSANHURDMWG&lt;/strong&gt;; &lt;em&gt;VLFBKOPJXCZYQ&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;ABCIJOPQRSTUWY&lt;/strong&gt;; &lt;em&gt;DEFGHKLMNOVZ&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;COPSUVWXZ&lt;/strong&gt;; &lt;em&gt;ABDEFGHIJKLMNQRTY&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;AHIMOTUVWXY&lt;/strong&gt;; &lt;em&gt;BCDEFGJKLNPQRSZ&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;ΑΒΕΗΙΚΜΝΟΡΤΧΥΖ&lt;/strong&gt;; &lt;em&gt;CDFGJLQRSUVW&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;AEFHILMNORSX&lt;/strong&gt;; &lt;em&gt;BCDGJKPQTUVWYZ&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;HJKMNUVWXY&lt;/strong&gt;; &lt;em&gt;ABCDEFGILOPQRSTZ&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;BDEFHIKLMNPRTUVWXYZ&lt;/strong&gt;; &lt;em&gt;ACGJOQS&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;EFPTY&lt;/strong&gt;; &lt;em&gt;ABCDGHIJKLMNOQRSUVWXZ&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;OTHER ANSWERS:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;4)&lt;/strong&gt; The &lt;strong&gt;first group&lt;/strong&gt; are those letters in the odd columns of a qwerty keyboard.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;5)&lt;/strong&gt; The letters in (in order) the most frequent letters used, sorted just like the morse code sorts them. The latter obviously being least frequent (also in order)&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;6)&lt;/strong&gt; The &lt;strong&gt;first group&lt;/strong&gt; are those letters that when spoken in English sound like (or are) words. &lt;BR&gt; &lt;BR&gt;&#xA; &lt;strong&gt;7)&lt;/strong&gt; The &lt;strong&gt;first group&lt;/strong&gt; are those letters that look the same in lower-case as in capitals.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;8)&lt;/strong&gt; The &lt;strong&gt;first group&lt;/strong&gt; are letters which can be reflected in a vertical axis and don't change.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;9)&lt;/strong&gt; The &lt;strong&gt;first group&lt;/strong&gt; are &lt;a href=&quot;http://en.wikipedia.org/wiki/Greek_alphabet#Letters&quot; rel=&quot;nofollow&quot;&gt;Greek letters&lt;/a&gt; (did you notice it?) that look like Latin letters; the &lt;em&gt;second group&lt;/em&gt; are Latin letters that don't look like any Greek letter.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;10)&lt;/strong&gt; The &lt;strong&gt;first group&lt;/strong&gt; are letters which names start with a vowel.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;11)&lt;/strong&gt; The &lt;strong&gt;first group&lt;/strong&gt; all have open tops, i.e. they would catch rainwater.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;12)&lt;/strong&gt; The &lt;strong&gt;first group&lt;/strong&gt; all reach the upper left corner of their bounding box.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;13)&lt;/strong&gt; The &lt;strong&gt;first group&lt;/strong&gt; all have an odd number of points where a stroke ends (one or three) where the second group have an even number (zero, two, or four.)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="4317" LastEditDate="2015-07-05T09:50:18.370" LastActivityDate="2015-07-05T09:50:18.370" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="6437" PostTypeId="2" ParentId="6435" CreationDate="2014-12-27T00:11:54.520" Score="3" Body="&lt;p&gt;It looks like &lt;a href=&quot;https://puzzling.stackexchange.com/a/6436/40&quot;&gt;warspyking&lt;/a&gt; has cracked the first two (I'd suggest &lt;strong&gt;holes&lt;/strong&gt; and &lt;strong&gt;curves&lt;/strong&gt; as catchphrases for these two groups).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the catchphrase for The last group is &lt;strong&gt;composite&lt;/strong&gt;. The alphabet positions of the letters that come after the split (&quot;ABCEGKMQSW&quot;) are 1, 2, 3, 5, 7, 11, 13, 17, 19 and 23 — i.e., 1 followed by all the prime numbers below 26. All the other numbers are composite, which means they are the product of two or more prime numbers.&lt;/p&gt;&#xA;" OwnerUserId="40" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2014-12-27T00:11:54.520" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6438" PostTypeId="2" ParentId="6425" CreationDate="2014-12-27T00:44:41.597" Score="1" Body="&lt;p&gt;I'm thinking the other guy got a Red Queen, meaning you win. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alright, I'll elaborate. The first truth eliminates some possibilities. It leaves you with everything that CAN beat an orange Queen. These are Red Ace, King, and Queen, Blue Ace, King, and Queen, Orange Ace and King, Green Ace, King, and Jack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You already know that the other guy can't have a Blue King, as that's the card you drew. So you're left with (and I'll just abbreviate) RA, RK, RQ, BA, BQ, OA, OK, GA, GK, and GJ.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second truth, since the card IS more likely an Ace, tells that the card is a Queen. This leaves us with only two choices: RQ and BQ. [(It's more likely an Ace because there is a larger pool of Aces being the possible card) 4:3:2:1 Ace:King:Queen:Jack]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third truth simply finishes. Since the card IS more likely a King, making it a Queen, then the card is Red. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Red Queen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn't know we needed explanations and such with these. Hopefully this one'll suffice.&lt;/p&gt;&#xA;" OwnerUserId="7511" LastEditorUserId="7511" LastEditDate="2014-12-28T00:02:20.733" LastActivityDate="2014-12-28T00:02:20.733" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6439" PostTypeId="2" ParentId="6415" CreationDate="2014-12-27T02:12:48.950" Score="0" Body="&lt;p&gt;Umm, doesn't it &lt;strong&gt;have&lt;/strong&gt; to be 100%?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If allowed to randomly step indefinitely&lt;/em&gt; means he keeps stepping until he falls off the cliff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This turned into a semantic argument in comments, I thought I'd summarise my position here. It is a, possibly naive, argument from &lt;em&gt;when does the result get evaluated&lt;/em&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If he never falls off the cliff, the evaluation never takes place as it is still an &lt;em&gt;undetermined result&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the problem was altered to say it terminates when he falls off a cliff or walks into a wall, that would give a definite stopping point to evaluate the &lt;em&gt;failed to fall off cliff&lt;/em&gt; result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My math extends as far as limit theorem in 1st year Engineering Calculus, so am obviously willing to bow to the assembled wisdom above that says I'm wrong ;-)&lt;/p&gt;&#xA;" OwnerUserId="7512" LastEditorUserId="7512" LastEditDate="2017-03-12T06:58:56.203" LastActivityDate="2017-03-12T06:58:56.203" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="6441" PostTypeId="2" ParentId="6420" CreationDate="2014-12-27T07:51:23.197" Score="1" Body="&lt;p&gt;As Toby explained &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1011000 is the bit wise or binary representation of curresponding letters (here for the clue BIT) present in the title.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Following the same rule we get&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (1100100-1011000) (10010) (1000011-110100) (10001-1011) (110011-110010) (11000-(11+111)) (11000-1010) (101)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which gives&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1100, 10010, 1111, 110, 1, 1110, 101 =&gt; 12, 18, 15, 6, 1, 14, 5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Considering these are the numerical representation of alphabets we get&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; L, R, O, F, A, N, E&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I assume this is an anagram. On further research I understood its the anagram of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; FAR NOEL (Noel means Christmas or birth of Christ)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which is definitely Christmas related and referring to the birth of Jesus Christ for the three wise men from far. I also suspect it may related to a Christmas carol that I have heard a long ago.&lt;/p&gt;&#xA;" OwnerUserId="2730" LastActivityDate="2014-12-27T07:51:23.197" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6443" PostTypeId="1" AcceptedAnswerId="6446" CreationDate="2014-12-27T10:55:49.287" Score="25" ViewCount="2537" Body="&lt;p&gt;Most of the puzzles here revolve around an &lt;em&gt;unknown algorithm attack&lt;/em&gt;. That is, you are given an enciphered piece of text and asked to decode it (and often given a series of hints which describe how the enciphering was done), or you are given a set of plaintexts and enciphered texts and asked to describe the manner in which the text was enciphered. In this puzzle, you will be asked to do something different. I will describe a custom cipher, and give an implementation, and then ask you to decode an enciphered text. Hopefully, if nothing else, you'll learn a bit more about cryptographic weaknesses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just so you are sure, any method of deciphering the text is fine. If you want to write a program to bruteforce the cipher, go ahead. With that in mind, let's begin...&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Part 0: A Basic Caesar Cipher&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Unless you've been living under a rock, you'll know about the basic Caesar cipher. But for those of you who like living under rocks, here's a basic rundown of how it works (note that the method may seem convoluted, but it's useful for later):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Start with your plain text. &lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;Attack at sundown, when the enemy isn't looking.&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;  &#xA;  &lt;p&gt;Then remove all spaces and punctuation (this is not strictly&#xA;  necessary, but we'll do it anyways)&lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;ATTACKATSUNDOWNWHENTHEENEMYISNTLOOKING&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;  &#xA;  &lt;p&gt;Decompose the text into numbers, A=0, B=1, ...Z=25: &lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;0 19 19 0 2 10 0 19 18 20 13 3 14 22 13 22 7 4 13 19 7 4 4 13 4 12 24 8 18 13 19 11 14 14 10 8 13 6 &lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;  &#xA;  &lt;p&gt;Choose a numeric key between 1 and 25; we'll choose 7. Then add the key to each number and take the result &lt;em&gt;modulo&lt;/em&gt; 26, that is, if the number exceeds 26, subtract 26 until it's smaller than 26 (e.g. If you have 30, then 30 modulo 26 is 30-26 is 4).&lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;7 0 0 7 9 17 7 0 25 1 20 10 21 3 20 3 14 11 20 0 14 11 11 20 11 19 5 15 25 20 0 18 21 21 17 15 20 13 &lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;  &#xA;  &lt;p&gt;Then convert back to letters:&lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;HAAHJRHAZBUKVDUDOLUAOLLULTFPZUASVVRPUN&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To decrypt, you do the same, except you &lt;em&gt;subtract&lt;/em&gt; the numeric key. That said, this cipher is very easy to break. To prove it, here's a challenge:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Challenge&lt;sup&gt;*&lt;/sup&gt; 0: Decode the following:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;SQUIQHMQIDEJQLUHOWEETSHOFJEWHQFXUH&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;* Note that this is a gross misuse of the word 'challenge'.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Part 1: Keyed Caesar&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Now we've got that out of the way, let's get on to something more interesting. The problem with a Caesar is that it's &lt;em&gt;too easy to bruteforce&lt;/em&gt;. Once you know it's a Caesar cipher, you can easily check all the possibilities. So let's fix that, by introducing a longer key. In part 0, we had a single numeric key that was applied to every letter. What we're going to do is have a &lt;em&gt;group&lt;/em&gt; of keys and apply them in turn. For example, if we had the keys 4, 19 and 8, we would shift the first letter 4 places, the second 19, the third 8, the fourth 4, and so on. In this way, one can not simply check all the possibilities&lt;sup&gt;**&lt;/sup&gt;, and a more elegant approach is needed to crack the cipher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;** Well, you could check all the possibilities, but that would be &lt;em&gt;really&lt;/em&gt; boring.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are three implementations of this cipher in different languages. If you can't program, or don't program in any of these languages, don't worry: The implementations are heavily commented, so just read the comments.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://ideone.com/HQGUaz&quot;&gt;PHP Implementation&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://jsfiddle.net/s4veu42x/&quot;&gt;Javascript Implementation&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://ideone.com/w9SZNw&quot;&gt;Python Implementation&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However, this is still a weak cipher, and you can prove it by decoding the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Challenge 1.1: Length 2 Key&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;QGUUEBBADCJDJSEEIJEPLAHUWKEZRASWKOUEJEIWBIEOJPXAIIQAQOQJENCWBYQAIWHYYLXAHWDZOKKYQJIPYHBXHQJAVKHYUPXASKCXYJQPYKDXKPYBOKKOEHLATPXEIWDWBUJESWBHOUEQQNUZEEDCMABHRASWKOUPXAIWCAJASDDEGQUYQJRAQLFHYATPEXHAQGJDYOSEFDUNMEJDQJOGUUBADCJDXAHAIOEIUIENUPUTJPXAGQYYAXHKMJVKNFKIFATKLAHPXABWPUTKW&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Challenge 1.2: Length 6 Key&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Challenge 1.3: Unknown Length Key (&amp;lt;= 7)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Part 2: Hardened Keyed Caesar&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The Keyed Caesar was a step in the right direction, but it's still reasonably easy to crack, because of certain weaknesses (no, I won't reveal the answer to part 1 in part 2!). So we're going to harden it some more. The Hardened Keyed Caesar will feature a global increment. This means that the nth letter of the plaintext will get shifted k*n places, where k is the global increment. So if your global increment was 2, the first letter would get shifted 2 placed on top of how it's shifted by the key, the second letter 4, and so on. This makes the cipher much more resilient to certain attacks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the implementations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://ideone.com/67MD5E&quot;&gt;PHP Implementation&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://jsfiddle.net/gyr0haqw/1/&quot;&gt;Javascript Implementation&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://ideone.com/rtv6gY&quot;&gt;Python Implementation&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&quot;Hang on&quot;, you say, &quot;in a real situation this would be on large chunks of text, and I'd know a bit about the context. The shortness of the challenges add artificial difficulty!&quot;. And you'd be right; that's why, for the next challenges, I'll provide a pastebin of a LARGE chunk of text encrypted by this cipher. I'll even tell you what it's about!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Challenge 2.1: Length 2 Key Plus Increment&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://pastebin.com/1ajmJgn8&quot;&gt;A military guidebook.&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Challenge 2.2: Unknown Length Key (&amp;lt;= 7) Plus Increment&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://pastebin.com/EJvpX44R&quot;&gt;A famous novel.&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What more can I say? &lt;em&gt;Happy cracking!&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="7308" LastActivityDate="2021-03-06T20:15:07.357" Title="Known Algorithm Attack: Caesar Reloaded" Tags="|cipher|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6444" PostTypeId="1" CreationDate="2014-12-27T11:23:31.007" Score="8" ViewCount="1242" Body="&lt;p&gt;I was working on this puzzle called cupid which listed a bunch of greek gods.  I filled in all the answers in the answer grid which were their roman counterparts.  Since these answers were now left aligned (where the clues were centered) and of course the fact that they're roman gods, naturally I'm thinking roman numerals.  The final answer is a word that can be obtained from these roman gods, but I can't seem to figure it out.  Here are the roman gods:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MARS&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CERES&lt;/p&gt;&#xA;&#xA;&lt;p&gt;VENUS&lt;/p&gt;&#xA;&#xA;&lt;p&gt;VESTA&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PLUTO&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MERCURY&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BACCHUS&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since M,C, and V are roman numerals, I'm assuming that's how to solve it, but I can't figure anything out, and I have no idea where the P and B fit in.  Maybe I'm looking at this all wrong.  Could anyone please help me solve this?&lt;/p&gt;&#xA;" OwnerUserId="7492" LastActivityDate="2014-12-27T11:26:33.790" Title="I'm stumped on a simple roman numeral puzzle" Tags="|word-problem|lateral-thinking|puzzle-identification|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6445" PostTypeId="2" ParentId="6444" CreationDate="2014-12-27T11:26:33.790" Score="18" Body="&lt;p&gt;Given that there's little context, it's hard to give a definitive answer, but I'd have to guess that the answer you are looking for is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MENTORS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This can be found by:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Taking the first letter of the first word, the second of the second, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7308" LastActivityDate="2014-12-27T11:26:33.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6446" PostTypeId="2" ParentId="6443" CreationDate="2014-12-27T11:38:39.933" Score="19" Body="&lt;h2&gt;Challenge 0&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;CAESAR WAS NOT A VERY GOOD CRYPTOGRAPHER&lt;br&gt;&#xA;Letters shifted by 16 positions in the alphabet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Found by brute force.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Challenge 1.1&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A KEY OF LENGTH TWO IS NOT VERY GOOD BECAUSE IT IS ALMOST THE SMAE AS A NORMAL CAESAR CIPHER, AND YOU CAN STILL BRUTE-FORCE THE COMBINATION. BUT IF YOU SOLVED THIS ANALYTICALLY YOU ARE DOING WELL, BECAUSE THE SAME TECHNIQUE CAN BE APPLIED TO BREAK THIS CIPHER WITH ANY KEY LENGTH. HERE'S SOME MORE TEXT: THE QUICK BROWN FOX JUMPED OVER THE LAZY DOG.&lt;br&gt;&#xA;Key of (16,22).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Found by brute force with the aid of the javascript implementation.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Challenge 1.2&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;STACKOVERFLOW HAS WRITTEN VOLUMES ABOUT HOW THE BEST WAY TO MAKE A POINT IS WITH FOAMING AT THE MOUTH, RHETORIC, AND LETTERS FILLED PRIMARILY WITH EXCLAMATION POINTS. DON'T BELIEVE A WORD OF IT, THOUGH. THE TRUTH IS THAT IT PROFESSES THAT A PLAUSIBLE EXCUSE IS A SATISFACTORY SUBSTITUTE FOR PERFORMANCE. THAT CONCEPT IS OF COURSE COMPLETE BUNK BY ANY STRETCH OF THE IMAGINATION; HOWEVER IT IS BUNK THAT HAS SURVIVED VIRTUALLY UNCHANGED FROM WHEN IT WAS FIRST PROPOSED NEARLY HALF A CENTURY AGO BY CAUSTIC MADMEN TO ITS PRESENT INCARNATION IN STACKOVERFLOW'S VILE EXCUSES. STACKOVERFLOW LOVES USING BIG WORDS LIKE 'UNDEMONSTRATIVENESS' AND 'UNPROPORTIONABLENESS'. AS A RESULT, IT WRITES LIKE A MENTALLY ILL PERSON WITH A THESAURUS. THAT GOT ME THINKING: PERHAPS STACKOVERFLOW SPOUTS ALL CLASSES OF PUFFERY ABOUT ITS MORAL VIGOR. WELL, SURE, IT HAS SOMEHOW FOUND THE FORTITUDE TO ENDURE OUR ONGOING HUMILIATION AND DISCOMFORT AT THE HANDS OF ITS PATSIES. BUT THE LARGER POINT IS THAT STACKOVERFLOW WRITES A LOT OF LONG STATEMENTS THAT MEAN PRACTICALLY NOTHING. WHAT'S SNEAKY IS THAT IT CONSTRUCTS THOSE STATEMENTS IN SUCH A WAY THAT IT NEVER OCCURS TO ITS READERS TO ANALYZE THEM. ANALYSIS WOULD ALMOST CERTAINLY INDICATE THAT STACKOVERFLOW SWEARS THAT ITS A LIVING BODHISATTVA OF PEACE AND NONVIOLENCE; CLEARLY IT'S LIVING IN A WORLD OF MAKEBELIEVE WITH FLOWERS AND BELLS AND LEPRECHAUNS AND MAGIC FROGS WITH FUNNY LITTLE HATS. BACK IN THE REAL WORLD, THE WHOLE OF STACKOVERFLOW'S BRUTISH WORLD-VIEW MAY PERHAPS BE EXPRESSED IN ONE SIMPLE WORD. THAT WORD IS 'SADISM'. LET ME EXPLAIN. WE MUST WORK TOGETHER TO DOLE OUT ACERBIC CRITICISM OF STACKOVERFLOW AND ITS PHALANX OF CRAZY RADICALISM-ORIENTED DISCIPLES. WHAT CAN YOU DO TO HELP? FOR STARTERS, YOU MIGHT WANT TO PLANT MARKERS THAT DEFINE THE LIMITS OF WHAT IS RAVING AND WHAT IS NOT. I PERSONALLY DERIVE GREAT SATISFACTION IN DOING THAT SORT OF THING, BECAUSE IT MAY SEEM AT FIRST THAT WITHIN A SHORT PERIOD OF TIME STACKOVERFLOW'S LIES WILL BE EXPOSED AND THE TRUTH CAN BE SPREAD. WHEN WE DESCEND TO DETAILS, HOWEVER, WE SEE THAT ITS FUNCTIONARIES OFTEN REVERSE THE NORMAL PROCESS OF INTERPRETATION: THAT IS, THEY VALUE THE UNSAID OVER THE SAID, THE OBSCURE OVER THE CLEAR.&lt;br&gt;&#xA;&lt;br&gt;&#xA;Key of [12,3,4,19,19,22]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Found by performing frequency analysis and choosing the keys which maximize the sum of the frequencies of E,T,A and O in the decrypted text.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Challenge 1.3&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;For each keylength 1 to 7, I computed the most likely value of the first part of the key by maximizing the sum of E,T,A, and Os as in Challenge 1.2.  This maximum of sums I call maxEtao.  Then, normalize maxEtao for each key length so they can be compared, and the results are as below:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;keylen  maxEtao * keylen / 7&#xA;1       46.57142857142857&#xA;2       47.42857142857143&#xA;3       47.142857142857146&#xA;4       46.285714285714285&#xA;5       90&#xA;6       46.285714285714285&#xA;7       53&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It seems very likely that the keylength is 5 since this will generate the most E, T, A, and Os on average.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now perform the same frequency analysis as in Challenge 1.2 and the answer will be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;IF YOU'RE A PROGRAMMER PLEASE DON'T DENIGRATE WHAT A BEGINNER IS CURRENTLY LEARNING UNLESS YOU'RE WILLING TO PERSONALLY INVEST THE ENORMOUS AMOUNT OF TIME TO ENSURE THAT THEY LEARN WHAT YOU WANT THEM TO LEARN JUST MAKING A SNIDE DRIVE BY REMARK ABOUT HOW THEIR CHOICE OF LANGUAGES OR TOOLS ISN'T AS COOL AS WHAT YOU USE WILL LIKELY DRAIN THEIR MOTIVATION. I KNOW IT CAN BE HARD TO RESIST SLIDING IN A WITTY ZINGER DISSING PHP OR WHATEVER LANGUAGE PEOPLE LOVE TO HATE ON THIS YEAR IVE BEEN GUILTY OF THIS MYSELF IN THE PAST ESPECIALLY BECAUSE YOU KNOW FROM THE BOTTOM OF YOUR HACKER HEART THAT YOUR GIVEN LANGUAGE, FRAMEWORK, LIBRARY, TEXT EDITOR, HOSTING ENVIRONMENT, ERGONOMIC KEYBOARD, STANDING DESK ETC. IS SOOOO MUCH BETTER THAN WHATEVER NOOB RUBBISH THEY'RE LEARNING BUT PLEASE RESIST JUST LET THEM FIRST LEARN SOMETHING ANYTHING TO GET HOOKED ONTO THE POTENTIAL OF PROGRAMMING AND THEN YOU CAN OFFER TO SHOW THEM THE TRUE LIGHT LATER. IF WHAT THEY'RE LEARNING IS TRULY SO TERRIBLE THEN THEY WILL KEEP GETTING STUCK AND ASK YOU WHETHER THERE'S AN EASIER WAY. THEN AND ONLY THEN YOU CAN UNLEASH YOUR TREATISE ON PURE FUNCTIONAL PROGRAMMING OR HYGENIC MACROS OR WHATEVER. ALSO YOU'RE PROBABLY NOT THE ONLY PROGRAMMER THE BEGINNER IS TALKING TO MANY PROGRAMMERS ESPECIALLY IN ONLINE FORUMS ARE PROBABLY MAKING SIMILAR SNIDE REMARKS EXCEPT THEY'RE SAYING THAT THEIR OWN FAVORITE LANGUAGE, LIBRARY, FRAMEWORK, TEXT EDITOR, HOSTING ENVIRONMENT IS THE BEST EVAHHHHHHH. SO NOW THE BEGINNER IS REALLY CONFUSED, THEY'RE REPEATEDLY TOLD THAT THEIR GIVEN CHOICE OF TOOLS IS SOMEHOW TERRIBLE EVEN THOUGH THEY SEEM TO BE DOING OK WITH IT AT THE MOMENT, BUT EVEN WORSE EVERY SO CALLED EXPERT IS RECOMMENDING A DIFFERENT SET OF TOOLS TO LEARN. WHO IS RIGHT WHAT SHOULD THEY DO NEXT?&lt;br&gt;&#xA;&lt;br&gt;&#xA;Key is [6,1,19,22,13]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Challenge 2.1&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I LAYING PLANS SUN TZU SAID THE ART OF WAR IS OF VITAL IMPORTANCE TO THE STATE IT IS A MATTER OF LIFE AND DEATH &lt;br&gt;&#xA;...&lt;br&gt;&#xA;WHEN YOU SURROUND AN ARMY LEAVE AN OUTLET FREE DO NOT PRESS A DESPERATE FOE TOO HARD SUCH IS THE ART OF WARFARE&lt;br&gt;&#xA;&lt;br&gt;&#xA;Key of [4,6], increment of 20.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Found by the same analysis as Challenge 1.2, but with an effective keylength of 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Challenge 2.2, please refer to the excellent answer by kasperd.&lt;/p&gt;&#xA;" OwnerUserId="6178" LastEditorUserId="6178" LastEditDate="2014-12-29T07:14:33.243" LastActivityDate="2014-12-29T07:14:33.243" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6449" PostTypeId="2" ParentId="6443" CreationDate="2014-12-27T14:28:26.677" Score="12" Body="&lt;p&gt;I consider challenges 0 through 2.1 as training exercises for those who haven't tried this sort of challenge before. So I jumped straight to challenge 2.2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I dusted off an old piece of code which I wrote more than a decade ago to break a Vigenère cipher and let it make an attempt at challenge 2.2. I had it try all possible key lengths from 1 through 400, and it told me the most likely key length was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 130&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But hey, that's more than 7. Was my code wrong? Not really, challenge 2.2 isn't a Vigenère cipher. However by interpreting the output slightly different, I can deduce that the key length must be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reason I got a different key length is that the number I got first is the product of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 5 (the actual key length) and 26 (the period introduced by the increment)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus I can deduce that the increment must be an odd number because otherwise&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the increment period could only have been 13 and the key length would have had to be 10, which is more than the promised maximum of 7&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For each index in the overall period I identified the most frequent character in the cipher text. I then looked at every&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 5th position&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and I noticed that the most frequent letter mostly jumps one position backwards in the alphabet each time I move forward. Thus I can deduce that the additional increment must be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 5 (increment) * 5 (key length) = 25&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and that explains the step one character backwards in the alphabet each time. And by knowing a bit of math (or by trial and error), one can find that it is the only possible increment with that property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the most frequent letter is E, I then get the shifts for the first five positions to be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 15 8 16 23 18&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Combining those shifts with the increment, the start of the text decrypts to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CHAPTER I MR JONES OF THE MANOR FARM HAD LOCKED THE HEN HOUSES FOR THE NIGHT BUT WAS TOO DRUNK TO REMEMBER&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(I have inserted some spaces in the decrypted text to aid readability)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the last part of the text decrypts to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; WITHOUT FURTHER ARGUMENT THAT THE MILK AND THE WINDFALL APPLES AND ALSO THE MAIN CROP OF APPLES WHEN THEY RIPENED SHOULD BE RESERVED FOR THE PIGS ALONE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1619" LastEditorUserId="1619" LastEditDate="2014-12-27T14:59:03.027" LastActivityDate="2014-12-27T14:59:03.027" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6451" PostTypeId="2" ParentId="6397" CreationDate="2014-12-27T17:16:54.557" Score="17" Body="&lt;p&gt;Answer: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Christmas&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A man and a woman and a pot of tea:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chirstmas =&gt; Chris T maS =&gt; Chris and Sam and &lt;strong&gt;T&lt;/strong&gt;ea in-between.  Chris and Sam are both gender-neutral names.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Look at the one on the left. If it's a woman, she might be sitting on a remade table&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If Chris is a woman, the name might be short for Christabel, where &lt;em&gt;&quot;tabel&quot;&lt;/em&gt; is a remade &quot;table.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;or perhaps she's stirring her drink with a fork&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Or it's short for Christine where &lt;em&gt;&quot;tine&quot;&lt;/em&gt; is a &quot;prong or sharp point, such as that on a fork or antler.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;If it's a man, he's much taller than his wife; he &lt;strong&gt;tops her&lt;/strong&gt; by a head, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chris is short for Christopher. Chris &lt;strong&gt;top her&lt;/strong&gt;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;although he might seem to be like a small burrowing rodent.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;&quot;topher&quot;&lt;/em&gt; resembles &quot;gopher&quot;, a small burrowing rodent.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Now the one on the right. If it's a woman, she's probably got a few pet moths, &quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If Sam is a woman, it might be short for Samantha, and &lt;em&gt;&quot;antha&quot;&lt;/em&gt; comes from Greek &quot;anthos&quot; meaning flower. A flower may have a few pet butterflies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;or maybe she's just got away from the Indian film industry.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sam may be short for Indian names Samskruthi or Samhita. Or more likely, the clue is referring to &quot;Antha&quot;, a 1981 Bollywood film.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;If it's a man, he needs most of what he can get from petrol, diesel, or gasoline.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If Sam is a man, then it's Samuel. &quot;uel&quot; is the most he can get from &quot;fuel.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7489" LastEditorUserId="7489" LastEditDate="2014-12-27T17:24:49.237" LastActivityDate="2014-12-27T17:24:49.237" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6452" PostTypeId="2" ParentId="5942" CreationDate="2014-12-27T17:55:38.147" Score="2" Body="&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/Magic_square&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; (my emphasis):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Normal magic squares of &lt;strong&gt;all sizes except 2 × 2&lt;/strong&gt; (that is, where n = 2) can be constructed. The 1 × 1 magic square, with only one cell containing the number 1, is trivial.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For a mathematical proof of this, see the Nrich article &lt;a href=&quot;http://nrich.maths.org/1337&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. The idea is to turn an n x n square into a diamond (an n x n square standing on one corner), fill in that diamond with the numbers 1 to n^2 in the natural way, and then fold everything into the original square. Some pictures to demonstrate (also taken from the Nrich article):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/i769W.gif&quot; alt=&quot;3x3 case&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TNH1z.gif&quot; alt=&quot;5x5 case&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2014-12-28T21:29:31.603" LastActivityDate="2014-12-28T21:29:31.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6453" PostTypeId="1" AcceptedAnswerId="8190" CreationDate="2014-12-27T18:20:50.910" Score="57" ViewCount="8790" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;This is a rather involved puzzle and might require a bit of stamina. All of the puzzle is in image form. The introduction text is not needed for the puzzle. At the end of the puzzle you'll get a short English sentence. The puzzle consists of several sub-puzzles and you're encouraged to post solutions to the individual steps - I will comment on them. &lt;br&gt;&lt;br&gt;&#xA;&lt;em&gt;The puzzles has now been fully solved, and the complete solution is summarized in the accepted answer.&lt;/em&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;Introduction&lt;/h1&gt;&#xA;&lt;p&gt;&lt;em&gt;You're in a circular room with a dome-like ceiling. All walls and the floor are bright white. There is only a single, metal door, which has a small, circular whole in eye height. You don't have any recollection how you came here, but at the moment, you're really only interested in getting out. The door, is looked. You knock, and there is some movement behind the hole. Somebody comes to the door and talks through the hole. The voice, however, sound foreign and not very human: &lt;strong&gt;&amp;quot;You awake? You out? You say passphrase!&amp;quot;&lt;/strong&gt; &lt;br&gt; You are about to ask something, when suddenly the walls behind you begin to shimmer and form a series of images, and the voice through the door adds: &lt;strong&gt;&amp;quot;You clever? You find passphrase.&amp;quot;&lt;/strong&gt; &lt;br&gt;Whatever else you say to your capturer, the only answer you get is: &lt;strong&gt;&amp;quot;No, not passphrase.&amp;quot;&lt;/strong&gt; &lt;br&gt; It seems you really have to find the passphrase. You just hope you can, before you die of hunger and thirst.&lt;/em&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;p&gt;On the wall around you, the following images can be found (&lt;em&gt;The appear clockwise around the room starting with the door and ending with the door&lt;/em&gt;.)&lt;br&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;The puzzle&lt;/h1&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/L2OpF.gif&quot; alt=&quot;INSTRUCTION1&quot; /&gt; &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;(&lt;em&gt;Instruction 1&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6teYX.png&quot; alt=&quot;INSTRUCTION2&quot; /&gt; &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;(&lt;em&gt;Instruction 2&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aYieq.gif&quot; alt=&quot;INSTRUCTION3&quot; /&gt; &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;(&lt;em&gt;Instruction 3&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SupTu.jpg&quot; alt=&quot;BOTTLES&quot; /&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;(&lt;em&gt;The bottles&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Md6FG.png&quot; alt=&quot;SCALE&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;(&lt;em&gt;The scale&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vFHCr.gif&quot; alt=&quot;CODE&quot; /&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;(&lt;em&gt;The code&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/m09aZ.gif&quot; alt=&quot;KEYS&quot; /&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;(&lt;em&gt;The keys&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;The following image is auxiliary for convenience sake. It is &lt;em&gt;a close-up of the bottle labels&lt;/em&gt; of the 2nd image above.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gCM9P.jpg&quot; alt=&quot;LABELS&quot; /&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;(&lt;em&gt;Bottle Label close-up&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;hr /&gt;&#xA;&lt;hr /&gt;&#xA;&lt;hr /&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;As always, I'm interested in your ideas and feedback. In particular, I hope the puzzle is &lt;em&gt;fun&lt;/em&gt;. If it isn't, let me know why. If it turns out that the puzzles needs amendments or changes to become better, I'll edit them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;Notes:&lt;/h1&gt;&#xA;&lt;p&gt;The &lt;strong&gt;red ABCDEF/1234 labels are not required for the puzzle&lt;/strong&gt;. They have only been added to make it easier to discuss/describe solutions.&lt;/p&gt;&#xA;&lt;p&gt;The 6 'rod-shaped' objects in the last image fit in size into the 'slot' of the image with the letters above it. (As shown in the smaller black &amp;amp; white image.)&lt;/p&gt;&#xA;&lt;p&gt;You may use any tool etc. to solve this puzzle, but it should be perfectly possible to do this with pen &amp;amp; paper alone.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;HINTS&lt;/h2&gt;&#xA;&lt;p&gt;If you are &lt;em&gt;really&lt;/em&gt; stuck with this puzzle and are wondering along the lines of &lt;em&gt;&amp;quot;What the @£! am I supposed to solve here??&amp;quot;&lt;/em&gt; you may look at the following hints. But they &lt;em&gt;will&lt;/em&gt; spoil a bit of your fun (as all hints do.) I try to make successive hints revealing more and more, so just try one at the time...&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Where am I even supposed to start this??&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The first images until the first horizontal line are called &lt;em&gt;instructions&lt;/em&gt; for a reason. Try interpreting them in terms of &lt;em&gt;what is this puzzle about?&lt;/em&gt; and &lt;em&gt;what am I to do?&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;I still don't get it at all...&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Remember: At &lt;em&gt;the very end&lt;/em&gt; of the puzzle is a pass-phrase. The only image with letters is obviously important, but how could one &lt;em&gt;encode&lt;/em&gt; with it - using the previous images? It surely &lt;em&gt;can't be done&lt;/em&gt; with one image alone...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;But... There are &lt;em&gt;endless&lt;/em&gt; possibilities! I could read &lt;em&gt;anything&lt;/em&gt; out of this...&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Really? Is there no clue about the &lt;em&gt;length&lt;/em&gt; of the pass-phrase? Does this length match anything... ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Okay, okay, so what do I have to solve &lt;em&gt;first&lt;/em&gt; ?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The puzzle has 3 main-sub puzzles. Guess what, they are sorted that way. Let's call them &lt;strong&gt;the bottles&lt;/strong&gt;, &lt;strong&gt;the scale&lt;/strong&gt; and &lt;strong&gt;the code&lt;/strong&gt; puzzle. &lt;em&gt;The scale&lt;/em&gt; is (literally!) the central puzzle here. It provides the information needed to solve &lt;em&gt;the code&lt;/em&gt;. But in order to attempt &lt;em&gt;the scale&lt;/em&gt; you need information provided in &lt;em&gt;the bottles&lt;/em&gt;. What could that be? Have another look at the &lt;em&gt;instructions&lt;/em&gt;! What do the yellow squares symbolize? What does the blue-to-grey shading symbolize? So what could the labels then be for?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;All nice and well, but can you give me another hint?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; NO. If you are still stuck and don't want to spent more time on thinking, you might as well read the accepted solution below!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2017-10-04T10:53:46.103" Title="A well balanced puzzle: Logic puzzle(s) in visual disguise" Tags="|logical-deduction|visual|" AnswerCount="10" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="6454" PostTypeId="2" ParentId="6453" CreationDate="2014-12-27T18:34:38.820" Score="7" Body="&lt;p&gt;&lt;strong&gt;GENERAL DISCUSSION&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following Tryth's terminology, let's name the images as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;INSTRUCTION1&lt;/li&gt;&#xA;&lt;li&gt;INSTRUCTION2&lt;/li&gt;&#xA;&lt;li&gt;INSTRUCTION3&lt;/li&gt;&#xA;&lt;li&gt;BOTTLES&lt;/li&gt;&#xA;&lt;li&gt;SCALES&lt;/li&gt;&#xA;&lt;li&gt;CODE TABLE&lt;/li&gt;&#xA;&lt;li&gt;KEYS&lt;/li&gt;&#xA;&lt;li&gt;LABELS&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;LABELS isn't really important, just there to help us with BOTTLES. The OP has intimated that SCALE is the key to the whole thing, so let's take a careful look at SCALES. We have six tubes, labelled by different colours, from which liquid drips down into six pipes, which lead to different points a balancing scales with two small grey weights on each end and more odd colours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at INSTRUCTION1. The left half is still a mystery to me, but the right half shows bottles being emptied into a tube of the same type as those on the scales. There are 5 bottles full of greyish-blue stuff, and 5 squares coloured greyish-blue, so I suspect each bottle in BOTTLES is going to fill up one of the small squares in SCALES after being poured through one of the tubes at the top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now INSTRUCTION2. We have a sequence of 24 bottles (the same number as in BOTTLES), and two more of those small grey weights. The first bottle at least gets poured into the 5th of the 6 tubes. The scales end up tipping one way or the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about INSTRUCTION3? We have a miniature version of SCALES printed on a big square, from which one corner in the shape of a jigsaw puzzle piece has been removed. This piece has a key on it, in the shape of one of those in KEYS. The piece is placed in a circle of 6 keys, one of which is chosen and inserted into what looks like CODE TABLE. Five cells in this table then have colours like those of the five outstanding bottles in INSTRUCTION1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here's what I think we need to do:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;decipher the squares and yellow dots in BOTTLES (reproduced in LABELS) into &lt;strong&gt;colours&lt;/strong&gt; (a hunch)&lt;/li&gt;&#xA;&lt;li&gt;empty all 24 bottles into the tubes in SCALES, using the colours we've found to determine which tube each one gets poured through&lt;/li&gt;&#xA;&lt;li&gt;notice which way the scale tips, and use this &lt;em&gt;somehow&lt;/em&gt; to deduce which of the KEYS to choose&lt;/li&gt;&#xA;&lt;li&gt;insert the chosen key into the CODE TABLE, and pick out a few cells &lt;em&gt;somehow&lt;/em&gt; whose letters reorganise to give the passphrase&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;SOME SPECIFICS (PROBABLY WRONG)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second image, the yellow dots under the bottles translate to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1011010 1011010 0001000 0001100 1111111 1101000 &lt;br&gt; 1110000 0011111 1110111 0001111 1000100 0011100 &lt;br&gt; 1100000 1010000 1101011 1101100 1111000 1101101 &lt;br&gt; 1110100 1001001 1110111 1000000 1011110 1111011&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I now suspect the squares on the bottles are something to do with colour codes (for which there are several different conventions).&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2014-12-31T11:57:44.327" LastActivityDate="2014-12-31T11:57:44.327" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="6455" PostTypeId="1" AcceptedAnswerId="6456" CreationDate="2014-12-27T19:28:41.930" Score="11" ViewCount="984" Body="&lt;p&gt;I'm playing an interactive text game and I'm stumped by the following riddles (the answers are names per the game context):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A villain with a common heart. Ten is his end, but he's fifty to start.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Cut off her head, she's male. Cut off her tail, she's a fruit. Cut off both, she's part of you.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7521" LastEditorUserId="7521" LastEditDate="2014-12-27T19:42:50.610" LastActivityDate="2015-03-26T14:48:47.413" Title="What are the names these riddles point to?" Tags="|riddle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6456" PostTypeId="2" ParentId="6455" CreationDate="2014-12-27T19:36:49.173" Score="16" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Lex.  (L is 50 in Roman numerals, X is 10, E is a common letter, Lex Luthor is a well-known &quot;villain&quot;.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Pearl.  (&quot;Earl&quot; is male, &quot;Pear&quot; is a fruit, &quot;Ear&quot; is a part of you.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7522" LastActivityDate="2014-12-27T19:36:49.173" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6457" PostTypeId="2" ParentId="6453" CreationDate="2014-12-27T23:00:04.900" Score="3" Body="&lt;p&gt;This is more of a brain dump at the moment, but hopefully it can help someone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The images are 'named' (as in the alt text) as follows (in order):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;INSTRUCTION1&lt;/li&gt;&#xA;&lt;li&gt;INSTRUCTION2&lt;/li&gt;&#xA;&lt;li&gt;INSTRUCTION3&lt;/li&gt;&#xA;&lt;li&gt;BOTTLES&lt;/li&gt;&#xA;&lt;li&gt;SCALE&lt;/li&gt;&#xA;&lt;li&gt;CODE TABLE&lt;/li&gt;&#xA;&lt;li&gt;KEYS&lt;/li&gt;&#xA;&lt;li&gt;( LABELS )&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This, if not anything else, might indicate that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The last image is a key for the second last - the color at the top indicates for which columns the colors are a key, and the left column of colors is a key for the left column of the code, and ditto for the right.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;BOTTLES&lt;/em&gt; ( &lt;em&gt;LABELS&lt;/em&gt; ) image has a series of 8x8 grids with holes in - this most likely corresponds to the 8x8 grid of symbols in the &lt;em&gt;INSTRUCTION1&lt;/em&gt; image. How, I don't know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note in the &lt;em&gt;INSTRUCTION1&lt;/em&gt; image, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;at the top left there are 6 bottles not colored the same. I have labelled them starting from the most blue.&#xA;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/rVbG5.gif&quot; alt=&quot;NUMBERS&quot;&gt;&#xA;&lt;br&gt;&#xA;These correspond to the grid coordinates &lt;strong&gt;C2, E4, A1, B4, F1 and A3&lt;/strong&gt; respectively. These also might correspond to the six colors in the code in the &lt;em&gt;SACLE&lt;/em&gt; image.&#xA;&lt;br&gt;&#xA;I did not get anything useful by overlaying the holed grids that correspond to those coordinates onto the symbol grid. But maybe I just didn't notice a pattern. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's a transcription of the &lt;em&gt;CODE TABLE&lt;/em&gt; image:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;WHOGTX FVTOJ/&#xA;ROSSIU ZTHXIC&#xA;/PYNNW ETUG//&#xA;DENFIZ TTOEMW&#xA;AVBAEJ UZRZFI&#xA;WVTGYE ZHKXPR&#xA;MI/HJZ RWULBL&#xA;EFDTS/ Z/OGZE&#xA;FSSUFY WLEYAZ&#xA;IUXKXT VTPLED&#xA;TKVATE A/OYYU&#xA;FOHI/B SGH/RZ&#xA;MEKGLM PMEADP&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here's a frequency analysis. Odd how there are so many Zs.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; E  T  Z  F  I  O  U  A  G  H  S  W  Y  L  M  P  R  V  X  D  K  B  J  N  C  Q&#xA;12 12 10  7  7  7  7  6  6  6  6  6  6  5  5  5  5  5  5  4  4  3  3  3  1  0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The puzzle might even be able to be broken with cryptanalysis of the code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the &lt;em&gt;KEYS&lt;/em&gt; image:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The colors in the left and right columns are always the same, just in a different order. This would lead me to conjecture that one must use this color key to reorder the corresponding column in the &lt;em&gt;CODE TABLE&lt;/em&gt; image, after which something will be spelled out. These colors also correspond to the ones on the scale in the &lt;em&gt;SCALE&lt;/em&gt; image.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is all I've got for now.&lt;/p&gt;&#xA;" OwnerUserId="7308" LastEditorUserId="5840" LastEditDate="2014-12-30T15:48:41.080" LastActivityDate="2014-12-30T15:48:41.080" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6458" PostTypeId="1" AcceptedAnswerId="6601" CreationDate="2014-12-27T23:17:44.923" Score="12" ViewCount="1988" Body="&lt;p&gt;0, 1, 11.2, 10.2, 113.2, 103, 103.2, 0.2, 1.2, 2.2, 2, ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only hint I'll give at this stage is that the sequence is infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If nobody gets it within a few days, I'll give a few terms that come &lt;em&gt;after&lt;/em&gt; the wanted term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have fun! :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: this is a wholly mathematical puzzle, a number sequence. There are no red herrings, elements meant to mislead or be ignored, or punctuation tricks etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;: this sequence can be defined in 12 words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Further hint&lt;/strong&gt; (in response to comments from @MohitJain and @MattMalone):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the numbers are all written in the same base, which isn't base 4 or 10.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7363" LastEditorUserId="4050" LastEditDate="2015-01-03T17:51:23.990" LastActivityDate="2015-01-03T17:51:23.990" Title="What is the next number in this sequence? 0, 1, 11.2, 10.2, 113.2, 103, 103.2, 0.2, 1.2, 2.2, 2, …" Tags="|mathematics|pattern|number-sequence|" AnswerCount="2" CommentCount="21" ContentLicense="CC BY-SA 3.0" />
  <row Id="6459" PostTypeId="2" ParentId="6458" CreationDate="2014-12-28T03:48:57.127" Score="5" Body="&lt;p&gt;This answer is a work in progress. Below are the differences between terms n and n+1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1, 10.2, -1, 103, -10.2, 0.2, -103, 1, 1, -0.2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These values are all even index terms from the original series S or their opposites. Like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S(2), S(4), -S(2), S(6), -S(4), S(8), -S(6), S(2), S(2), -S(8)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still trying to find some pattern in these. Another thing I notice:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S(3) = S(2) + S(4)&lt;br&gt;&#xA;S(5) = S(4) + S(6)&lt;br&gt;&#xA;S(7) = S(6) + S(8)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is to say S(n) = S(n-1) + S(n+1) for n = 3, 5 and 7 but this pattern breaks down at 9. Somewhat interestingly S(9) = (S(8) + S(10)) / 2. Hmmm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; Still trying. To state the above a little differently:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S(2) = S(3) - S(4)&lt;br&gt;&#xA;S(4) = S(5) - S(6)&lt;br&gt;&#xA;S(6) = S(7) - S(8)&lt;br&gt;&#xA;S(8) = S(10) - S(11) ?! What happened to S(9)? Or...&lt;br&gt;&#xA;S(8) = 2 * S(9) - S(10) as noted above&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The three problems with this line of investigation:&lt;br&gt;&#xA;1) The pattern disappears at the end&lt;br&gt;&#xA;2) It can't predict odd-numbered terms&lt;br&gt;&#xA;3) It's hard to ignore&lt;/p&gt;&#xA;" OwnerUserId="2162" LastEditorUserId="2162" LastEditDate="2014-12-30T01:44:56.210" LastActivityDate="2014-12-30T01:44:56.210" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6460" PostTypeId="1" AcceptedAnswerId="6484" CreationDate="2014-12-28T10:17:18.553" Score="6" ViewCount="469" Body="&lt;p&gt;Little Hans just got a model railroad kit from his grandfather for Christmas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;His grandpa explained to him, that when he was young, he was a model railroad enthusiast, and encouraged his grandchild to not despair if he cannot put together everything correctly the first time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no shame in it, as his first attempt at model railroading was also unsuccessful: he just modeled the station of his home town, Friedrichsfeld .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when he wanted to start the Mannheim-Heidelberg express train, his little train just wouldn't start. He checked every cable, every connection, none of them were broken, but he still couldn't start his little train.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;The question&lt;/strong&gt;: did the grandfather use an AC or DC power supply for his model railroad?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I've only chosen Germany as a location because both AC and DC was commonly used. This puzzle has a logical explanation, I didn't want it to make it into an assumption like &quot;in this country AC (or DC) was very uncommon&quot;.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2507" LastEditorUserId="7435" LastEditDate="2014-12-28T13:48:57.520" LastActivityDate="2014-12-29T09:11:57.703" Title="A model railroad story" Tags="|logical-deduction|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6461" PostTypeId="1" CreationDate="2014-12-28T11:37:35.947" Score="2" ViewCount="548" Body="&lt;p&gt;I'm working on this puzzle where the solution is either a word or phrase with no theme. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first part of the puzzle had incorrect movie titles with cities, states, etc, and the solution to the first part of the puzzle is the actual places in the movie titles.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is the list of all the actual places.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Somehow, this list should yield a word, but I have no idea what to do to get this word.  They are arranged in chronological order. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sorted by release date:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;St. Louis&lt;br&gt;&#xA;  Tangier&lt;br&gt;&#xA;  Nuremberg&lt;br&gt;&#xA;  Flint&lt;br&gt;&#xA;  Atlantic City&lt;br&gt;&#xA;  London&lt;br&gt;&#xA;  East LA&lt;br&gt;&#xA;  Tombstone&lt;br&gt;&#xA;  Providence&lt;br&gt;&#xA;  Arlington&lt;br&gt;&#xA;  Reno&lt;br&gt;&#xA;  Union&lt;br&gt;&#xA;  Elizabethtown&lt;br&gt;&#xA;  Annapolis&lt;br&gt;&#xA;  Talladega&lt;br&gt;&#xA;  New Orleans&lt;br&gt;&#xA;  Rome&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sorted alphabetically:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Annapolis&lt;br&gt;&#xA;  Arlington&lt;br&gt;&#xA;  Atlantic City&lt;br&gt;&#xA;  East LA&lt;br&gt;&#xA;  Elizabethtown&lt;br&gt;&#xA;  Flint&lt;br&gt;&#xA;  London&lt;br&gt;&#xA;  New Orleans&lt;br&gt;&#xA;  Nuremberg&lt;br&gt;&#xA;  Providence&lt;br&gt;&#xA;  Reno&lt;br&gt;&#xA;  Rome&#xA;  St. Louis&lt;br&gt;&#xA;  Annapolis&lt;br&gt;&#xA;  Talladega&lt;br&gt;&#xA;  Tangier&lt;br&gt;&#xA;  Tombstone&lt;br&gt;&#xA;  Union&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;* Full Puzzle *&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Meet Me in 'Maryland'      1944 - Judy Garland, Margaret O'Brien&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;'California' Incident      1953 - George Brent, Mari Aldon&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Judgment at 'Rhode Island'     1961 - Maximilian Schell, Spencer Tracy&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In Like 'Morocco'      1967 - James Coburn, Lee J. Cobb&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;'Michigan'     1980 - Susan Sarandon, Burt Lancaster&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;American Werewolf in 'Virginia'        1981 - David Naughton, Jenny Agutter&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Born in 'Italy'        1987 - Cheech Marin, Daniel Stern&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;'New Jersey'       1993 - Kurt Russell, Bill Paxton&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Outside 'United Kingdom'       1999 - Alec Baldwin, Shawn Hatosy&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;'Louisiana' Road       1999 - Jeff Bridges, Tim Robbins&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Waking up in 'Germany'     2002 - Patrick Swayze, Billy Bob Thornton&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;xXx: State of the 'Nevada'     2005 - Ice Cube, Willem Dafoe&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;'Arizona'      2005 - Orlando Bloom, Kirsten Dunst&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;'Alabama'      2006 - Tyrese Gibson, James Franco&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;'New York' Nights: The Ballad of Ricky Bobby       2006 - Will Ferrell, John C. Reilly&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Bad Lieutenant: Port of Call 'Kentucky'        2009 - Nicolas Cage, Eva Mendes&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;When in 'Missouri'     2010 - Kristen Bell, Josh Duhamel&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;My Doubt&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried a whole bunch of different counting patterns and also introduced some ciphers into the mix, but I still can't figure it out.  Can anyone help me out in getting to the probable solution ?&lt;/p&gt;&#xA;" OwnerUserId="7492" LastEditorUserId="4050" LastEditDate="2014-12-29T19:30:32.003" LastActivityDate="2014-12-30T00:33:18.933" Title="Place names in movie titles - stumped on this puzzle" Tags="|cipher|word|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="6462" PostTypeId="1" AcceptedAnswerId="6463" CreationDate="2014-12-28T13:33:07.903" Score="62" ViewCount="7040" Body="&lt;p&gt;In this game, all players start sitting at a round table, each with a coin in their hand.&#xA;At every turn, all players toss their coins together and put them on the table in front of them. Then each player looks at the coins tossed by his left and right neighbours. If the neighbours tossed different faces of the coin (one obverse and one reverse), the player in the middle survives for the next turn. Otherwise, if the neighbours both tossed the same face of the coin (both obverse or both reverse), the game is lost for the player sitting in between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After everybody knows who's continuing to the next turn, the losers pick up their coins and leave the table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The game is won when there is exactly one player left sitting at the table, who will be the winner and grab the lucky $10 bonus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If all players are eliminated at the same time, the game is a tie and nobody gets the bonus.&#xA;If there are only two players left, since left and right neighbours will be the same opponent, the players will necessarily eliminate each other at the next turn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are offered a chance to play this game and grab the lucky $10 bonus. You can choose to play in a game starting with 10, 11, or 12 players including yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would you choose and why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If the lucky $10 bonus sounds boring, replace it with 10 rep points or whatever you find worth playing for.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="4394" LastEditorUserId="12408" LastEditDate="2016-07-25T02:29:22.203" LastActivityDate="2016-07-25T02:29:22.203" Title="Left coin, right coin, last coin?" Tags="|logical-deduction|strategy|probability|game|coin-tossing|" AnswerCount="3" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="6463" PostTypeId="2" ParentId="6462" CreationDate="2014-12-28T13:54:58.477" Score="59" Body="&lt;p&gt;&lt;strong&gt;It doesn't matter which option you choose, because&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;it's impossible to win.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your probability of survival if you're one of n players left is as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; n=1: 100% (obviously!) &lt;br&gt; n=2: 0% &lt;br&gt; n=3: 0% &lt;br&gt; n=4: 0% &lt;br&gt; n=5: 0% &lt;br&gt; n=6: 0% &lt;br&gt; n=7: 0% &lt;br&gt; n=8: 0% &lt;br&gt; n=9: 0% &lt;br&gt; n=10: 0% &lt;br&gt; n=11: 0% &lt;br&gt; n=12: 0%&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Informal proof&lt;/strong&gt; &lt;br&gt;&#xA;It was established in the question that if there are only 2 players left, they must annihilate each other. So if there are 3 left, your only hope of survival is to annihilate &lt;em&gt;both&lt;/em&gt; your opponents, which can only happen if your result (WLOG, heads) is the same as both of theirs, which means their results must be the same, which means you're annihilated too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same argument goes on. If there are $n$ players left, where $n&amp;gt;1$, your only hope is to annihilate them all in one go (since all lower cases have been checked), which means annihilating yourself too. The proof is slightly different for odd and even $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Formal proof&lt;/strong&gt;&lt;br&gt;&#xA;Let $S_n$ be the statement that &lt;em&gt;if there are n players left, you can't possibly win&lt;/em&gt;. We've already proved $S_2$ and $S_3$, so by strong induction it will suffice to prove that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;if $S_k$ is true for $1&amp;lt;k&amp;lt;n$, then $S_n$ is true.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If $S_k$ is true for $1&amp;lt;k&amp;lt;n$ and you're one of n players remaining, your only hope is to annihilate &lt;em&gt;all&lt;/em&gt; your opponents in one go. (Otherwise, at the next stage you'll be one of $k$ players remaining where $1&amp;lt;k&amp;lt;n$, so you lose by the induction hypothesis.) WLOG, assume your coin comes up heads. Say you're A and the players on your left are B, C, D, ... while those on your right are Z, Y, X, ... (I'm not assuming $n=26$; this is just notationally convenient.) To eliminate B, you need C to get heads; then to eliminate D, you need E to get heads, and so on. Similarly, to eliminate Z, you need Y to get heads; then to eliminate X, you need W to get heads, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n$ is odd, you're already done at this stage because by proceeding in steps of 2 you cover &lt;em&gt;all&lt;/em&gt; the other players before coming back to yourself (whom you don't want to eliminate!), so &lt;em&gt;everyone&lt;/em&gt; gets heads and you get eliminated too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n$ is even, you now know that you need &lt;em&gt;every second&lt;/em&gt; player, starting from yourself, to get heads. In order to eliminate C, you need B and D to get the same (doesn't matter whether it's heads or tails); then to eliminate E, you need D and F to get the same, and so on. Taking this all the way round, you find that B, D, F, H, ..., V, X, Z (ending with Z since you know $n$ is even) to get the same result. But B and Z getting the same result means you get eliminated too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can't win when there are $n$ players remaining. QED.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2014-12-28T21:55:37.267" LastActivityDate="2014-12-28T21:55:37.267" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6464" PostTypeId="2" ParentId="6462" CreationDate="2014-12-28T14:17:00.823" Score="20" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It doesn't matter; you can't win.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If there are two players left, both lose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are three players left:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If all get the same coin face, all lose. If one is different, that player loses and the other two keep playing; but there are two of them and they lose.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If there are four players left:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If all get the same coin face, all lose. If one is different, that player and the one opposite lose, and there are only two left, they lose. If it's HTHT, everyone loses. If it's HHTT, everyone keeps playing and eventually loses.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If there are more:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You need to eliminate everyone but yourself in one go or you'll end up with one of the unsurvivable low numbers above. Suppose you get H, then you need a H on one side, T on the other. On the H side, it needs to be H's all the way or someone else will survive. On the T side, it needs to be alternating H/T all the way, like this: ...HHH-You-THTH.... Where these meet at the other side of the table, the two sequences can't continue and so someone else will survive (maybe more, but that just leaves us with another round).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6031" LastActivityDate="2014-12-28T14:17:00.823" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="6465" PostTypeId="2" ParentId="6401" CreationDate="2014-12-28T16:04:37.117" Score="4" Body="&lt;p&gt;&lt;strong&gt;This answer gives the two numbers, got by brute-forcing the hashes rather than solving the riddle, and the final solution. For the solution of the &lt;em&gt;riddle&lt;/em&gt;, see &lt;a href=&quot;https://puzzling.stackexchange.com/a/6515/5373&quot;&gt;Len's&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/a/6430/5373&quot;&gt;rand al'thor's&lt;/a&gt; answers.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have cracked both hashes using the Mathematica program below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hash 1 is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hash 1 matched number 'sixsixseventhreeeightfour' (667384)&#xA; It seems to be ASCII for &quot;BIT&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hash 2 is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hash 2 matched number 'nineeightzerosixsixfive' (980665)&#xA; I can't seem to decipher the meaning, since one of the values isn't a print ASCII character.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer decrypts to the message:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nature and nature's laws lay hid in night;&#xA; God said &quot;Let Newton be&quot; and all was light.&#xA; It could not last; the Devil shouting &quot;Ho!&#xA; Let Einstein be!&quot; restored the status quo.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using the Mathematica program&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Do[&#xA;  iStr = ToString[i];&#xA;  iLen = StringLength[iStr];&#xA;  numStr = &quot;&quot;;&#xA;  Do[&#xA;   posStr = StringTake[iStr, {j}];&#xA;   If[posStr == &quot;0&quot;, numStr = numStr &amp;lt;&amp;gt; &quot;zero&quot;];&#xA;   If[posStr == &quot;1&quot;, numStr = numStr &amp;lt;&amp;gt; &quot;one&quot;];&#xA;   If[posStr == &quot;2&quot;, numStr = numStr &amp;lt;&amp;gt; &quot;two&quot;];&#xA;   If[posStr == &quot;3&quot;, numStr = numStr &amp;lt;&amp;gt; &quot;three&quot;];&#xA;   If[posStr == &quot;4&quot;, numStr = numStr &amp;lt;&amp;gt; &quot;four&quot;];&#xA;   If[posStr == &quot;5&quot;, numStr = numStr &amp;lt;&amp;gt; &quot;five&quot;];&#xA;   If[posStr == &quot;6&quot;, numStr = numStr &amp;lt;&amp;gt; &quot;six&quot;];&#xA;   If[posStr == &quot;7&quot;, numStr = numStr &amp;lt;&amp;gt; &quot;seven&quot;];&#xA;   If[posStr == &quot;8&quot;, numStr = numStr &amp;lt;&amp;gt; &quot;eight&quot;];&#xA;   If[posStr == &quot;9&quot;, numStr = numStr &amp;lt;&amp;gt; &quot;nine&quot;];&#xA;   , {j, iLen}];&#xA;  iHash = IntegerString[Hash[numStr, &quot;SHA512&quot;], 16, 128];&#xA;  If[iHash == &#xA;    &quot;a077332a1720b83b9a1c26619f3179dcfff1ef205f09468996c579aedd5ed5fb2\&#xA;85acfb02c085768d8fb1cf2bfea05b02e993d947a324e361254bd68824e16e3&quot;,&#xA;   Print[&quot;Hash 2 matched number '&quot; &amp;lt;&amp;gt; numStr &amp;lt;&amp;gt; &quot;'&quot;];&#xA;   ,];&#xA;  If[iHash == &#xA;    &quot;c40c8102ca3c33705325a8dd614c49a2d29c82f3fce529528f2fe4001fac8a233b0d69\&#xA;16f1c604643a9c375ca3e02589ed5babf8c9e47d0629e145ec13ca8ef0&quot;,&#xA;   Print[&quot;Hash 1 matched number '&quot; &amp;lt;&amp;gt; numStr &amp;lt;&amp;gt; &quot;'&quot;];&#xA;   ,];&#xA;  If[Mod[i, 10000] == 0, Print[&quot;Tested up to &quot; &amp;lt;&amp;gt; iStr];];&#xA;  , {i, 10000000}];&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7529" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2014-12-30T13:08:53.560" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6466" PostTypeId="1" AcceptedAnswerId="6479" CreationDate="2014-12-28T16:36:39.977" Score="16" ViewCount="1555" Body="&lt;p&gt;&lt;em&gt;Credit to GOTO0 for &lt;a href=&quot;https://puzzling.stackexchange.com/questions/6462/left-coin-right-coin-last-coin&quot;&gt;the original idea&lt;/a&gt; for this puzzle, which is the converse of his puzzle.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this game, all players start sitting at a round table, each with a coin in their hand.&#xA;At every turn, all players toss their coins together and put it on the table in front of them. Then each player looks at the coins tossed by his left and right neighbours. If the neighbours got the same result (both heads or both tails), the player in the middle survives for the next turn. If the neighbours got different results (one heads and one tails), the game is lost for the player sitting in between them. All the eliminated players pick up their coins and leave the table, and the remaining players proceed to the next round.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The game is won when there is exactly one player left sitting at the table, who will be the winner and get an extra 10k rep on Stack Exchange.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If all players are eliminated at the same time, the game is a tie and nobody gets the prize.&#xA;If the coin-toss is such that nobody gets eliminated (e.g. if there are only two players left, since the left and right neighbours will be the same opponent), again it's a tie and nobody gets the prize.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are offered a chance to play this game and boost your reputation on Stack Exchange. You can choose to play starting with 10, 11, or 12 players including yourself. &lt;strong&gt;Which would you choose and why?&lt;/strong&gt; Please include a full explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[&lt;em&gt;If the 10k rep sounds boring, replace it with $10k, the famous spaghetti recipe, access to Jon Skeet's SO account, or whatever you find worth playing for.&lt;/em&gt;]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Disclaimer&lt;/strong&gt;: I don't know the solution!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Spoiler&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You (probably) won't actually get 10k rep for answering this. The most I can guarantee you is 25 rep for an upvote and accept. Sorry about that.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2016-07-25T02:29:29.813" Title="Another probabilistic coin puzzle" Tags="|logical-deduction|probability|coin-tossing|" AnswerCount="4" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="6467" PostTypeId="1" AcceptedAnswerId="7919" CreationDate="2014-12-28T17:10:25.567" Score="-9" ViewCount="1241" Body="&lt;p&gt;&lt;strong&gt;Background:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's time for a little bit of fun, your friends are playing a game and you decide to join them!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Gamplay:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You each sit down and play on a flat surface. You sit in a circle and each throw your die. If the person to your left gets a number within 2 number range (e.g; a 5 when you get 3) You are out. Then you all flip a coin, if you get the same as the person to your right you are out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How to Win:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be the last person standing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How ties are handled:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are 2 people left nobody wins. If there are nobody left nobody wins.&#xA;If the same result happens 3 times in a row nobody wins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Notes:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are using six sided dice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Puzzle:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have 11 friends, but one of them has to go, he says he can stay one more round if you want him to; will it better you chances if he leaves? Why or why not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Motivation:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What? You're not motivated? Fine! You can &lt;sub&gt;&lt;sup&gt;&lt;sub&gt;&lt;sup&gt;not&lt;/sup&gt;&lt;/sub&gt;&lt;/sup&gt;&lt;/sub&gt; have infinite rep on Puzzling if you solve!&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="12408" LastEditDate="2016-07-25T02:29:09.453" LastActivityDate="2016-07-25T02:29:09.453" Title="Coins, Dice, and Probability!" Tags="|mathematics|logical-deduction|probability|dice|coin-tossing|" AnswerCount="1" CommentCount="14" ClosedDate="2016-08-04T18:30:40.827" ContentLicense="CC BY-SA 3.0" />
  <row Id="6468" PostTypeId="2" ParentId="6466" CreationDate="2014-12-28T17:10:45.340" Score="2" Body="&lt;p&gt;I'll just start this off, to provide a springboard for somebody else...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are exactly 2 people left, nobody can win. &lt;br&gt; If there are exactly 3 people left, there's a 25% chance of each person winning and a 25% chance that nobody wins. (&lt;em&gt;Note to self: clarify rules. If the toss is such that nobody wins, but it is possible for somebody to win with that number of players, do they keep tossing until somebody wins, or declare the game a tie?&lt;/em&gt; &lt;strong&gt;Now clarified! They declare the game a tie. I think this makes the problem better than the other alternative would.&lt;/strong&gt;) &lt;br&gt; If there are exactly 4 people left, then either 0, 2, or 4 must be eliminated, so nobody can win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep going, guys! :-)&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2014-12-28T17:22:41.233" LastActivityDate="2014-12-28T17:22:41.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6470" PostTypeId="2" ParentId="6466" CreationDate="2014-12-28T17:29:01.377" Score="11" Body="&lt;p&gt;Lemma: On any given turn, an even number of players are eliminated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: Go around the circle, skipping every other coin. Every time the coin you see changes from tails to heads or vice versa, the person you skipped over gets eliminated. You must see an even number of changes before returning to the start, so there are an even number of eliminations in this cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are an even number of players, there are two &quot;skip every coin&quot; cycles. If there are an odd number of players, there is one cycle which passes around the circle twice. Either way, since every cycle contains an even number of eliminations, an even number of players get eliminated.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;By the lemma, if you start with 10 or 12 players, then you will always have an even number left, so you are screwed. Therefore, you should sit at the 11-seat table. There, you at least have a $9465137/134217728 \sim 7\%$ chance of winning that delicious spaghetti (src: computer program).&lt;/p&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2014-12-28T17:29:01.377" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6471" PostTypeId="2" ParentId="6462" CreationDate="2014-12-28T17:38:16.417" Score="43" Body="&lt;p&gt;The answers of rand al'thor and Callidus are great; I just want to give a different argument for the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Claim:&lt;/strong&gt; After each round, the number of surviving players is even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt; Let $f_i$ be the flip of player $i$, with tails $1$ and heads $0$ (it's arbitrary which is which). Let $s_i$ be $1$ if player $i$ survives the round and $0$ otherwise. Then, the condition that a player survives is his neighbors' flips are opposite can be written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s_i = f_{i-1}+f_{i+1} \bmod 2, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;wrapping indices around. Adding these equations for every player, each $f_i$ appears in exactly two summands, which cancel modulo $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_i s_i = 2 \sum_i f_i = 0 \mod 2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, an even number of $s_i$ are $1$, which means an even number of players survive. Therefore, in particular, &lt;strong&gt;you cannot be the only surviving player after any round&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="10248" LastEditDate="2015-03-17T04:47:00.933" LastActivityDate="2015-03-17T04:47:00.933" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6472" PostTypeId="1" CreationDate="2014-12-28T19:19:38.523" Score="11" ViewCount="909" Body="&lt;p&gt;Sometimes when I'm designing a puzzle which has more than one stage, I'd like to incorporate a 'check' for some of the answers (e.g. the answers of the earlier stages).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/6401/christmas-day-cryptogram-riddle&quot;&gt;This puzzle&lt;/a&gt; needed 2 keys to decode a cryptogram. I wanted to incorporate a way for the puzzler to know if they'd got &lt;em&gt;either one&lt;/em&gt; of the keys correct (the cryptogram won't decode unless &lt;em&gt;both&lt;/em&gt; of the keys are correct).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I naturally thought about hash functions, but the trouble with that is that if there's additional information about the key - as there was in this case - then it can be cracked by simply searching all the possible keys for a key which hashes to the same value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the specific puzzle linked above, I saw that this was possible and explicitly said that it was an allowed way of finding the solution - but what if I &lt;em&gt;didn't&lt;/em&gt; want to do that? What if I wanted to make a 'check' that couldn't be reverse-engineered by brute-force (or at least, not in a realistic timeframe).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously one way to do that would be to use a good-quality hash function and withhold information about the keys that are being hashed, so that the search space for possible keys is very large. But the less information I give the puzzler about the keys, the harder it'll be for them to work out what the keys are when solving the puzzle through the 'normal' route! And that's not what I want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: &lt;strong&gt;what are some of the ways in which a puzzle incorporating multiple answers can allow the puzzler to confirm that individual answers are correct?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can think of one example - a crossword, where each of the words provides a cross-check for some of the other words - but what are some other methods, and what are the pros and cons of each?&lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-02-02T15:08:49.023" Title="Puzzle design: How to incorporate confirmation of an answer?" Tags="|puzzle-creation|difficulty-rating|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6473" PostTypeId="2" ParentId="6472" CreationDate="2014-12-28T20:39:12.713" Score="14" Body="&lt;p&gt;One way is to use a hash function with a very small output space. E.g. if it's a number, give the sum of the digits modulo 5. That way it's not much use in filtering down a search space of thousands, but it does give a quick-reject for the majority of incorrect keys. If you want, you could even guess at possible errors that people will make and try to find a hash function which will reject the results.&lt;/p&gt;&#xA;" OwnerUserId="228" LastActivityDate="2014-12-28T20:39:12.713" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6474" PostTypeId="5" CreationDate="2014-12-28T20:45:07.873" Score="0" Body="&lt;p&gt;This tag should be used for puzzles where the genre or solving strategy of the puzzle is not explicitly stated; puzzles where the puzzler must deduce what type of puzzle it is.&lt;/p&gt;&#xA;&lt;p&gt;The word enigmatic comes from &lt;em&gt;enigma&lt;/em&gt;, meaning&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;a person or thing that is mysterious, puzzling, or difficult to understand.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Enigmatic puzzles are typically set up in a way where the solvers aren't immediately clear on where they should start. &lt;em&gt;It should still be clear once it has been solved, so as not to invite speculative answers or risk being closed for being too open-ended.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="52483" LastEditorUserId="52483" LastEditDate="2023-02-03T03:08:52.997" LastActivityDate="2023-02-03T03:08:52.997" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6475" PostTypeId="4" CreationDate="2014-12-28T20:45:07.873" Score="0" Body="Puzzles where the genre or solving strategy of the puzzle is not explicitly stated; puzzles where the puzzler must deduce what type of puzzle it is." OwnerUserId="4050" LastEditorUserId="4050" LastEditDate="2014-12-28T21:12:40.187" LastActivityDate="2014-12-28T21:12:40.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6476" PostTypeId="2" ParentId="6472" CreationDate="2014-12-28T21:11:21.200" Score="7" Body="&lt;h1&gt;3 different approaches:&lt;/h1&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I think, whenever possible, the cross-checks of partial solutions should be &lt;em&gt;fun&lt;/em&gt; in themselves. Of course it depends a lot of the nature of the result (a value? a phrase? something more abstract?), but instead of using some encryption, I would rather build a mini-puzzle around the solution. Something which is not very demanding, but requires the partial solution as a key-element. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If the solution is a number with several digits: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The digits of the solution can be combined to give the number 47 using basic arithmetic operators and no brackets. The order of the digits must not be changed.&lt;br&gt;&#xA;( &lt;strong&gt;12456&lt;/strong&gt;:   1+2*4*5+6  = 47 )&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the solution is a word:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The solution is an anagram for a capital city of the world, if you only use the first 4, the third-last letter.&lt;br&gt;&#xA;( &lt;strong&gt;Raspberries&lt;/strong&gt;:  RASPI = Paris )&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Alternatively, I would describe &lt;em&gt;one aspect&lt;/em&gt; of the solution. Not enough to determine the solution from it, but enough to verify you're on the right track. There is a chance that some &lt;em&gt;incorrect&lt;/em&gt; solution also fits this criteria, but if you combine 2 or 3 such &quot;checks&quot; you're adding enough security. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If the solution is a word/phrase. Provide a &lt;em&gt;riddle&lt;/em&gt; which in itself is not enough to find the solution, but which - once the solution is known - either seems fitting or not.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the solution is a sequence of numbers, use some criteria like:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;total number of even digits (2,4,6,8) in the whole sequence. &lt;/li&gt;&#xA;&lt;li&gt;5th &amp;amp; 11th digit of the total sequence are the same.&lt;/li&gt;&#xA;&lt;li&gt;The sum of all elements is even/odd/a prime... &lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Finally, you can brute-force a check by creating a custom-web-URL pointing to a &quot;success&quot; page. The URL and the partial solution need to be related.&lt;/p&gt;&#xA;" OwnerUserId="5840" LastActivityDate="2014-12-28T21:11:21.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6477" PostTypeId="2" ParentId="6472" CreationDate="2014-12-28T21:39:09.353" Score="-2" Body="&lt;p&gt;&lt;em&gt;I'm assuming your answer is a number. If it's a word, you can convert it into a number e.g. by the usual sort of method: WORD -&gt; 23151804.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/RSA_algorithm&quot; rel=&quot;nofollow&quot;&gt;RSA algorithm&lt;/a&gt; is a way of encrypting a number (with an encryption key) in such a way that decryption is next to impossible without a (separate) decryption key. So you could include the &lt;em&gt;encrypted&lt;/em&gt; number and the &lt;em&gt;encryption&lt;/em&gt; key but not the &lt;em&gt;decryption&lt;/em&gt; key.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another idea: any many-to-one (i.e. non-injective) function mapping numbers to numbers should suit your purposes. In less mathematical language, &lt;strong&gt;a way of getting number B from number A such that many different A's give the same B&lt;/strong&gt;. For instance, number B could be the &lt;a href=&quot;http://en.wikipedia.org/wiki/Digit_sum&quot; rel=&quot;nofollow&quot;&gt;digit sum&lt;/a&gt; of number A. So we could have WORD -&gt; 23151804 -&gt; 24. MANY different words could give the final answer 24, but probably not many that fit with whatever clues you've given towards WORD!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information and examples, try looking up about &lt;a href=&quot;http://en.wikipedia.org/wiki/Check_digit&quot; rel=&quot;nofollow&quot;&gt;check digits&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2014-12-28T21:39:09.353" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6478" PostTypeId="1" AcceptedAnswerId="6486" CreationDate="2014-12-28T23:21:11.243" Score="11" ViewCount="1165" Body="&lt;p&gt;This is a (slightly silly) version of Penguino's &lt;a href=&quot;https://puzzling.stackexchange.com/questions/6346/ernie-and-the-alchemists-gift&quot;&gt;Ernie gift-wrapping puzzle&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to wrap a cubical present with side length 1 meter. As in the linked question, when you wrap a present, you are not allowed to cut the wrapping paper. You can only fold it to cover the cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, unlike in the linked question, here you have no triangle. You can only use one rectangle of wrapping paper. The rectangle can have whatever dimensions you like. You want to use the least paper possible. How small can you make the area of the rectangle?&lt;/p&gt;&#xA;" OwnerUserId="1752" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2014-12-29T11:31:42.963" Title="Optimal present wrapping with a rectangle" Tags="|mathematics|geometry|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6479" PostTypeId="2" ParentId="6466" CreationDate="2014-12-29T02:32:20.880" Score="9" Body="&lt;p&gt;Lopsy answered 11. rand al'thor then asked to consider e.g. 9 and 13. I will&#xA;give a recursive formula for the probability of winning for a game of $N$ players.&#xA;According to Lopsy's Lemma, for $N$ even there is no chance of winning so I&#xA;consider only odd $N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, let me define notation for two types of sequences of coins. I use&#xA;$B(n_1)$ for a &quot;block&quot; sequence of same sided coins, of length $n_1$. E.g.&#xA;$B(5)$ can be $TTTTT$ or $HHHHH$. I use $A(n_2)$ for an &quot;alternating&quot; sequence of&#xA;coins, of length $n_2$. E.g. $A(5)$ can be $THTHT$ or $HTHTH$. We also define&#xA;the following when combining sequences:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$A$ types concatenate with $A$ types, and $B$ types concatenate with $B$&#xA;types. Namely $A(n_1)A(n_2) = A(n_1 + n_2)$ for any integers $n_1, n_2$, and&#xA;similarly for $B$ types. Without loss of generality I will only consider fully&#xA;concatenated forms from here on.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;An $A$ type sequence always starts with the opposite&#xA;of the preceding coin, e.g. $B(2)A(2)$ corresponds to $TTHT$ or $HHTH$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A $B$ type sequence always starts with the same side&#xA;as the preceding coin, e.g. $A(2)B(2)$ corresponds to $THHH$ or $HTTT$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note that any &lt;em&gt;periodic&lt;/em&gt; sequence can be described by combining $A$ and $B$ type&#xA;sequences and specifying the first coin. E.g., $TH H THTHTH HH$ is described by&#xA;specifying $T$ as the first coin, then $A(2)B(1)A(6)B(2)$.*  I call&#xA;such forms &quot;canonical&quot;, i.e.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The canonical form of a periodic sequence specifies the first coin in the&#xA;  sequence, and a list of alternating $A$ and $B$ type sequences of specified&#xA;  lengths.**&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that canonicalization provides a bijection from the set of all length-$N$&#xA;periodic coin sequences to the set of all length-$N$ canonical forms having an&#xA;even number of coins contained in $A$-type subsequences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us consider just the first round of an N player game. Without loss of&#xA;generality, assume my coin is $T$. Starting from me, looking to the left, and&#xA;skipping every other player, I write the sequence of $N$ coins in canonical&#xA;form. (This is the counting scheme from Lopsy's lemma.) Let $N_A$ denote the&#xA;total number of coins contained within $A$-type subsequences. Note that $N_A$ is&#xA;the number of people eliminated in this round and $N_A$ must be even (see&#xA;Lopsy's answer). Therefore the probability of $N_A$ people being eliminated is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{\binom{N}{N_A}}{2^{N - 1}}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I assumed I have $T$, we have only $2^{N-1}$ possible sequences of&#xA;coins in the denominator.&lt;sup&gt;$\dagger$&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Denote the probability that in a single N-player round, $N_A$ people are&#xA;eliminated, while I am not eliminated, as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;P_1(N, N_A) = \left(1 - \frac{N_A}{N}\right) \frac{\binom{N}{N_A}}{2^{N - 1}}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We must now consider all possibilities recursively. Denote the probability of&#xA;winning the entire N-person game as $P(N)$. Then $P(1) = 1$, and for $N &amp;gt; 1$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;P(N) = \sum_{N_A} P_1(N, N_A) P(N - N_A),&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the sum runs over all even, positive integers $N_A$, such that $N_A \leq N - 1$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt; *:&#xA;Note that the first coin is &quot;preceded&quot; by the last coin since this is a periodic&#xA;sequence. The last coin differs from the first, and is part of a $B$ type&#xA;sequence, so the first coin is the start of an $A$ type sequence. To be more&#xA;rigorous, the construction should go like $TA(1)B(1)A(6)B(2)$, then&#xA;$A(1)A(1)B(1)A(6)B(2)$, then $A(2)B(1)A(6)B(2)$.&#xA;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt; **:&#xA;Note that the first and last sequence types can be the same.&#xA;E.g., $T$, $A(2)B(3)A(2)$ is in its canonical form.&#xA;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt; &lt;sup&gt;$\dagger$&lt;/sup&gt;:&#xA;If we let my coin be $H$ or $T$, then both the numerator and denominator will pick up a&#xA;factor of two. Another way to obtain the denominator is by summing over all&#xA;possible eliminations, i.e. $\sum_{i=0}^{(N - 1) / 2} \binom{N}{2i} = 2^{N-1}$.&#xA;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&#xA;&lt;strong&gt;Edit 12/28:&lt;/strong&gt; Clarify why the example &quot;canonical form&quot; starts with $A(2)$, and that&#xA;the canonical form is for periodic sequences.&#xA;&lt;br&gt;&#xA;&lt;strong&gt;Edit 12/29:&lt;/strong&gt; Fix factor of 2 in probability of eliminating $N_A$ players, as pointed out by Lopsy.&#xA;&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="7537" LastEditorUserId="7537" LastEditDate="2014-12-29T18:22:28.343" LastActivityDate="2014-12-29T18:22:28.343" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6480" PostTypeId="2" ParentId="6478" CreationDate="2014-12-29T04:08:16.243" Score="1" Body="&lt;p&gt;OK, I'm not very clever, but I think I'll take a shot at this. Center the box on a square piece of paper. The distance from the center of the paper/box to the edge of the paper needs to be 2 meters (.5 for the bottom, 1 for the side, and .5 for the top) so the pieces meet on the opposite side of the box. Any less and it'll fall short. The way to minimize paper use is to rotate the box 45 degrees so it's &quot;diamond shaped&quot; on the square piece of paper. That way the required 2 meter length will be on the diagonal of the wrapping paper, the longest distance from centre to edge of the paper. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the entire diagonal will be 4 meters, each side is sqrt(8)....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, 8 square meters of paper? One square of paper the square root of 8 on each side? (Square square square.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But geometry is not my long suit. Anyone out there know a better way to do this with some creative folding?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Edit: Alternately, if it was a 1x8 sheet, you could wrap 4 sides of it with the middle 4m and then do 45 degree folds to cover the last two sides with the ends, or wrap 4 sides starting at one end, then do a 45 degree fold and wrap the last two sides with the last 3m.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I like the way this d6 I wrapped up in a square of toilet paper looks. &lt;/p&gt;&#xA;" OwnerUserId="7476" LastEditorUserId="7476" LastEditDate="2014-12-29T04:29:33.433" LastActivityDate="2014-12-29T04:29:33.433" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6481" PostTypeId="2" ParentId="6460" CreationDate="2014-12-29T06:30:59.630" Score="8" Body="&lt;p&gt;The grandfather used AC power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many model trains are powered via the two rails they ride on (rather than a &quot;third rail&quot; as in many metro trains and also Lionel model trains).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The express train had two locomotives.  Had they used DC power, their orientation would not matter: facing the same way they would drive together, and facing opposite directions, the &quot;backward&quot; facing one would also reverse its motion (because its contact with the rails was backward too), making the whole thing move forward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But with AC power, the direction of a (single-phase) motor does not depend on how it is wired externally.  So the two electric locomotives facing opposite directions simply played tug of war.&lt;/p&gt;&#xA;" OwnerUserId="7542" LastActivityDate="2014-12-29T06:30:59.630" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6482" PostTypeId="2" ParentId="6478" CreationDate="2014-12-29T07:05:05.713" Score="6" Body="&lt;p&gt;I can do it with 7 square meters of paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make the paper 1m by 7m.  Mark it every 1m, making seven squares:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;_____________________________&#xA;| 1 | 2/| 3 | 4 | 5 | 6/| 7 |&#xA;|___|/__|___|___|___|/__|___|&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Fold squares 2 and 6 in half diagonally, making each into a pair of equal triangles.  It now looks like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                    _____&#xA;                    | 7 |&#xA;        ____________|___|&#xA;       /| 3 | 4 | 5 |6 /&#xA;     /_2|___|___|___|/&#xA;    | 1 |&#xA;    |___|&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now fold the four inner markings (the sides of 3, 4, and 5) by 90 degrees, toward each other.  The diagonal edges of 2 and 6 now meet each other, and you have a square tube comprising 2-3-4-5-6 (half of 2 and half of 6 are &quot;wasted&quot;), plus two &quot;flanges&quot; 1 and 7.  Bend 1 and 7 by 90 degrees as well, and you have a cube.&lt;/p&gt;&#xA;" OwnerUserId="7542" LastEditorUserId="7542" LastEditDate="2014-12-29T08:08:48.893" LastActivityDate="2014-12-29T08:08:48.893" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6484" PostTypeId="2" ParentId="6460" CreationDate="2014-12-29T09:11:57.703" Score="9" Body="&lt;p&gt;He used DC power.&#xA;Looking up Friedrichsfeld, it looks like this:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/LdfxB.png&quot; alt=&quot;Track layout around Friedrichsfeld&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DC-trains usually do not have the third rail that is common with AC-systems and instead simply use one rail for power and the other as ground. If you try to build a triangle with a system like that...&#xA;&lt;img src=&quot;https://i.stack.imgur.com/p8Nvq.png&quot; alt=&quot;A model train triangle with a short&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...you get a nice, little short. This is a common beginner's mistake, the same thing would have happened with a loop (although those are much rarer in real railway systems).&lt;/p&gt;&#xA;" OwnerUserId="7545" LastActivityDate="2014-12-29T09:11:57.703" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="6485" PostTypeId="2" ParentId="6288" CreationDate="2014-12-29T09:50:25.017" Score="2" Body="&lt;p&gt;*Hello everyone, my name is Jay  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jay is Carbon ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I work for you 24 hours a day,&lt;br&gt;&#xA;but don't get me wrong I know not Silent Bob&lt;br&gt;&#xA;and got nothing to do with weed on the job.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;we are carbon-based&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My kind can be you or you can be me&lt;br&gt;&#xA;if you know a body by its chemistry.&lt;br&gt;&#xA;We're not only human, but mostly a thing!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; carbon form many compounds, and is present in air&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We have many shapes, on the air we are king.&lt;br&gt;&#xA;Our children you could surf but not in the sea&lt;br&gt;&#xA;(for some they are a monstruosity!).&lt;br&gt;&#xA;They fly, they bounce, they bend,&lt;br&gt;&#xA;but always to us they end!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; also fuel &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We get and we give even not alive,&lt;br&gt;&#xA;but to do our best we need a good drive,&lt;br&gt;&#xA;for we can make work a lot you possess,&lt;br&gt;&#xA;even destroy it if working in excess.*&lt;/p&gt;&#xA;" OwnerUserId="7546" LastActivityDate="2014-12-29T09:50:25.017" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6486" PostTypeId="2" ParentId="6478" CreationDate="2014-12-29T11:31:42.963" Score="5" Body="&lt;p&gt;I can do it in $6 + \epsilon$ square meters, where $\epsilon$ is any number greater than zero. Let the rectangle be of dimensions $w \times (6/w + 3/2)$, where $w$ is a very, very small width. Lay it out as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HIuan.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have the rectangle border the edge of the above figure, turn at the edge, trace the border back agian, and so on. In each repetition, which covers a width of w, there are 3 90 degree turns. Each  90 degree turn causes an overhead of $w^2/2$, the area of the folded over triangle created this way. Thus, there is an overhead of $3w^2/2$ in each repetition, and $3w/2$ in constructing the entire figure, which has $1/w$ repetitions. The above figure, once covered, can easily be assembled into a full cube, so covering the full cube also has an overhead of $3w/2$. Since w can be chosen arbitrarily, I will choose it to be $2\epsilon/3$. Thus, there is an overhead of $\epsilon$, for a full area of $6+\epsilon$.&lt;/p&gt;&#xA;" OwnerUserId="2187" LastActivityDate="2014-12-29T11:31:42.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6487" PostTypeId="2" ParentId="6466" CreationDate="2014-12-29T11:40:58.723" Score="0" Body="&lt;p&gt;This should really be a comment on Lopsy's answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By your Lemma, if you start with 11 players you will always have a winner, and since the game is symmetrical with respect to the players, why shouldn't the chance of winning be 1/n when n is odd?&lt;/p&gt;&#xA;" OwnerUserId="4805" LastActivityDate="2014-12-29T11:40:58.723" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6488" PostTypeId="1" AcceptedAnswerId="6494" CreationDate="2014-12-29T12:17:13.753" Score="4" ViewCount="736" Body="&lt;p&gt;In this game, all players start sitting at a round table, each with a coin in their hand.&lt;/p&gt;&#xA;&lt;p&gt;At every turn, all players toss their coins together and put it on the table in front of them. Then each player looks at the coins tossed by his left and right neighbours.&lt;/p&gt;&#xA;&lt;p&gt;If the neighbours got the same result (both heads or both tails), the player in the middle gets eliminated and doesn't get to play further. &lt;br/&gt;&#xA;If the neighbours got different results (one heads and one tails), the player sitting in between them survives for the next round.&lt;/p&gt;&#xA;&lt;p&gt;All the eliminated players pick up their coins and leave the table , and the remaining players proceed to the next round.&lt;/p&gt;&#xA;&lt;p&gt;The game is won when only two players remain seated on the table. &lt;br/&gt;&#xA;Those two then get the ultimate prize - the ultimate honor of getting to see the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3981/the-security-to-the-party&quot;&gt;secret spaghetti recipe&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;If all players are eliminated at the same time, the game is a tie and nobody gets the prize.&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;If the coin-toss is such that nobody gets eliminated, then the game continues normally. No tie or everyone wins thingy.&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;No changing of seats in between the game.&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;You and your friend are offered a chance to play this game and try your luck. The aim is for &lt;em&gt;at least one of the two of you&lt;/em&gt; to survive until the end of the game.&lt;/p&gt;&#xA;&lt;p&gt;You can choose your and your friend's seats, and you can also choose to play starting with 10, 11, or 12 players including yourself and your friend.&lt;/p&gt;&#xA;&lt;p&gt;Which would you choose and why? Please include a full explanation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;Thanks to @GOTO0 and @randal'thor for &lt;a href=&quot;https://puzzling.stackexchange.com/questions/6462/left-coin-right-coin-last-coin&quot;&gt;the original idea&lt;/a&gt; for this puzzle. This is just another version of the original one.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2808" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-12-29T15:00:34.763" Title="Left coin, Right coin, Last coin- A fair version" Tags="|probability|logical-deduction|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="6489" PostTypeId="2" ParentId="6488" CreationDate="2014-12-29T13:23:02.740" Score="1" Body="&lt;p&gt;&lt;em&gt;Partial answer...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are 2 players left, you win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are 3 players left, there are 4 possibilities, all equally likely: everyone eliminated, only player A eliminated, only player B eliminated, only player C eliminated. So you have 50% chance of surviving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are 4 players left, there are 4 possibilities, all equally likely: everyone eliminated, nobody eliminated, the players on either side of you eliminated, you and the player opposite you eliminated. Assuming that the 4 players keep tossing until &lt;em&gt;someone&lt;/em&gt; is eliminated, the second possibility is out, leaving you with a 1/3 chance of surviving. &lt;strong&gt;&lt;em&gt;EDIT&lt;/em&gt;&lt;/strong&gt;: unless everyone is eliminated, either you or the player on your left must survive. So there's a 2/3 chance of &lt;em&gt;either you or your friend&lt;/em&gt; surviving, provided you can choose where your friend sits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are 6 players left, then there's a 1/2 chance you're eliminated, a 1/8 chance you and one other player survive, and a 3/8 chance you and three other players survive. So your chance of surviving the whole game is $\frac{1}{8}+\frac{3}{8}.\frac{1}{3}=\frac{1}{4}$. &lt;strong&gt;This might be redundant. What are the chances of &lt;em&gt;either you or your friend&lt;/em&gt; surviving?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I got all of this just by brute-forcing it: going through all the possibilities.&lt;/em&gt;&#xA;&lt;hr&gt;&#xA;&lt;strong&gt;A general statement:&lt;/strong&gt; in each round, you have exactly 50% chance of surviving to the next round (since the results of the players on either side of you could be HH, HT, TH, TT with equal probability).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Another general statement, thanks to &lt;a href=&quot;https://puzzling.stackexchange.com/a/6471/5373&quot;&gt;xnor&lt;/a&gt;:&lt;/strong&gt; after each round, the number of surviving players is even. So all that's left to calculate is your chance of ultimate survival starting with 8, 10, 11, or 12 players (we don't need to worry about 7 or 9, or for that matter 3 or 5).&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2014-12-29T13:59:08.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6490" PostTypeId="1" CreationDate="2014-12-29T14:25:23.810" Score="1" ViewCount="3842" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Question : What is the next number in the following sequence? (not finite)&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\large24,54,90,120,168,210,240,288,360,440,\dots$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;My Solution&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I'm trying a closed form solution&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I tried computing the differences- they don't converge&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Tried finding the relationship between their binary equivalents&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A trivial observation : all numbers are divisible by &lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt; except &lt;span class=&quot;math-container&quot;&gt;$440$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;strong&gt;Edit-1&lt;/strong&gt;: Seeing the downvotes, I want to clarify something: All I'm trying to do is to find a solution (an expresssion) that represents the sequence&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Edit-2&lt;/strong&gt; : This question was given by my Math Teacher as a fun exercise in puzzle solving . It is not for a graded homework&lt;/p&gt;&#xA;" OwnerUserId="7549" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-01-03T17:51:43.510" Title="What is the next number? 24,54,90,120,168,210,240,288,360,440" Tags="|pattern|number-sequence|" AnswerCount="2" CommentCount="10" FavoriteCount="0" ClosedDate="2014-12-29T17:59:00.500" ContentLicense="CC BY-SA 3.0" />
  <row Id="6492" PostTypeId="2" ParentId="6490" CreationDate="2014-12-29T14:32:46.213" Score="3" Body="&lt;p&gt;The sequence is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4x6, 6x9, 9x10, 10x12, 12x14, 14x15, 15x16, 16x18, 18x20, 20x22, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So we have a sequence 4, 6, 9, 10, 12, 14, 15, 16, 18, 20, 22, ... such that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the products of adjacent pairs of terms in &lt;em&gt;this&lt;/em&gt; sequence gives the OP's original sequence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'd guess the next term in &lt;em&gt;this&lt;/em&gt; sequence is 24, so that the answer to the OP's question is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 22x24 = &lt;strong&gt;528&lt;/strong&gt;,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I'm not certain how to back this up!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possibility is that there's a mistake in the question and our new sequence should in fact be 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, ..., the sequence of composite numbers, as @GOTO0 suggested. Otherwise, I need to give it more thought.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2014-12-29T14:52:23.827" LastActivityDate="2014-12-29T14:52:23.827" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6493" PostTypeId="2" ParentId="6490" CreationDate="2014-12-29T14:45:42.273" Score="3" Body="&lt;p&gt;&lt;em&gt;Unless this is the intended answer, in which case it's a bad question anyway, this answer serves to show how broad the question is.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can always extend a sequence by taking the difference of subsequent elements, which yields a sequence with one fewer number. Once we get a list where all numbers are the same (which will eventually happen if we &quot;bottom out&quot; to get 1 element, but may happen sooner), we can start extending that bottom-most list and propagate the answer up to the top.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;24, 54, 90, 120, 168, 210, 240, 288, 360, 440&#xA;30, 36, 30, 48, 42, 30, 48, 72, 80&#xA;6, -6, 18, -6, -12, 18, 24, 8&#xA;-12, 24, -24, -6, 30, 6, -16&#xA;36, -48, 18, 36, -24, -22&#xA;-84, 66, 18, -60, 2&#xA;150, -48, -78, 62&#xA;-198, -30, 140&#xA;168, 170&#xA;2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In this case we go all the way down and get to the value 2. The next element in such a list is the sum of all the rightmost numbers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2 + 170 + 140 + 62 + 2 - 22 - 16 + 8 + 80 + 440 = &lt;strong&gt;866&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This yields a &lt;strong&gt;polynomial fit&lt;/strong&gt; to your problem which is guaranteed to always produce integers (and, after a while, to always produce distinct increasing integers).&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="7552" LastEditDate="2014-12-29T16:13:29.067" LastActivityDate="2014-12-29T16:13:29.067" CommentCount="11" CommunityOwnedDate="2014-12-29T15:32:54.940" ContentLicense="CC BY-SA 3.0" />
  <row Id="6494" PostTypeId="2" ParentId="6488" CreationDate="2014-12-29T15:00:34.763" Score="4" Body="&lt;p&gt;Partial answer. No matter how many players there are, I believe you and your friend should sit next to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, suppose there are 4 players left. If you and your friend sit next to each other, then you have a $2/3$ chance of winning a spaghetti recipe. If you sit across from each other, it's only $1/3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma: Given the choice between a 1/2 probability of both you and your friend surviving, and a certain chance that exactly one of you is eliminated, you should choose the latter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: The first choice gives $\frac 12 (\Pr[\text{You win}]+\Pr[\text{Friend wins}]-\Pr[\text{You both win}])$ chance of getting a spaghetti recipe. The second choice gives $\frac 12 (\Pr[\text{You win}]+\Pr[\text{Friend wins}])$, a larger chance. $\square$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By choosing our seats, my friend and I cannot influence our individual probabilities of surviving. However, we &lt;em&gt;can&lt;/em&gt; influence the probability that we both survive. The lemma tells us that we want to make this probability as small as possible. We'd rather exactly one of us survive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, when there are an even number of players left (which, after the first round, there always will be), you can split the players into two cycles. Each cycle contains every alternate player. Call the cycles $A$ and $B$. In each of these cycles, an even number of players live on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you live in cycle $A$ and you survive. This has no influence on cycle $B$, but it does mean that at least one other person in cycle $A$ has to live. If you survive, then you want your friend to be eliminated, so you should put your friend in cycle $B$. If you and your friend sit next to each other, then you are always in different cycles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Admittedly, this is not a proof, because the situation is more complicated. Sometimes nobody wins the game. So we really want to maximize&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[&#xA;\Pr[\text{Exactly one of us survives | Somebody wins the game}].&#xA;\]&#xA;But I conjecture that this doesn't affect the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TL;DR&lt;/strong&gt; I'm guessing that you want to sit next to your friend at a table of as few people as possible.&lt;/p&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2014-12-29T15:00:34.763" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6495" PostTypeId="1" AcceptedAnswerId="6496" CreationDate="2014-12-29T15:40:17.890" Score="32" ViewCount="3875" Body="&lt;p&gt;You own a square-shaped table. You want to drill a small hole in the center to place an umbrella stand. Unfortunately, you're a little drunk:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8OXtYm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alas. Fortunately, not all is lost. You are sober now, and have a hacksaw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) Cut the table into two pieces. Reassemble them to make a square-shaped table the same size as the original, but with the circular hole in the exact center. The cut does not have to be a straight line, but it must divide the table into exactly two pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Now, minimize the length of the cut.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this puzzle, everything is 2D. No tricky 3D cuts. &lt;em&gt;Note added in later, after people started rightly questioning what a hacksaw can do:&lt;/em&gt; Don't worry about what cuts are physically possible with a hacksaw. Any division of the table into two pieces is allowed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;WARNING&lt;/strong&gt;: When I give this puzzle to people, about 10% of them see the solution right away, about 30% figure out part (a) quickly and then spend varying amounts of time on part (b), and the other 60% give up. &lt;strong&gt;If you scroll down and look at the comments or answers, you will have robbed yourself of a deep 'aha!' moment.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1752" LastEditorUserId="4394" LastEditDate="2015-01-04T22:50:37.363" LastActivityDate="2024-02-25T11:16:15.053" Title="Dissection Puzzle - The Umbrella Stand" Tags="|geometry|dissection|" AnswerCount="6" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6496" PostTypeId="2" ParentId="6495" CreationDate="2014-12-29T15:44:22.690" Score="32" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Depending on your saw and skill, cut the hole in the middle where you want it, and use that piece to plug your mistake hole.&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; Originally I was going to suggest you cut an L shaped piece off the left and bottom and put it on the top and right. You would cut off:&lt;br /&gt;&#xA; Left - Half width - hole to right edge&lt;br /&gt;&#xA; Bottom - Half height - hole to top edge.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="463" LastEditDate="2024-02-25T11:16:15.053" LastActivityDate="2024-02-25T11:16:15.053" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="6497" PostTypeId="2" ParentId="6495" CreationDate="2014-12-29T16:02:21.740" Score="28" Body="&lt;p&gt;You cut a square like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/B2EL8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And rotate it 180 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/U76NO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cut square (or rectangle) simply needs to have its centerpoint be halfway between the hole and the center of the square, and to be large enough to contain the hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Boy do I feel silly seeing the intended solution...&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2014-12-29T16:02:21.740" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6499" PostTypeId="1" CreationDate="2014-12-29T19:27:48.843" Score="1" ViewCount="2920" Body="&lt;p&gt;The following is the puzzle that I am trying to figure out. I know the &quot;expected solution of 99&quot;. But honestly, I think this puzzle is flawed. How exactly can each wise man know the number of hats saved if they died silently? Doesn't make sense to me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A stark raving mad king tells his 100 wisest men he is about to line&#xA;  them up and place either a red or blue hat on each head. Once lined&#xA;  up, they must not communicate amongst themselves nor attempt to look&#xA;  behind them nor remove their own hat.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The king tells the wise men that they will be able to see all the hats&#xA;  in front of them. They will not be able to see the color of their own&#xA;  hat or the hats behind them, although they will be able to hear the&#xA;  answers from all those behind them.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The king will then start with the wise man in the back and ask &quot;what&#xA;  color is your hat?&quot; The wise man will only be allowed to answer &quot;red&quot;&#xA;  or &quot;blue,&quot; nothing more. If the answer is incorrect then the wise man&#xA;  will be silently killed. If the answer is correct then the wise man&#xA;  may live but must remain absolutely silent.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The king will then move on to the next wise man and repeat the&#xA;  question.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Before they are lined up, the king makes it clear that if anyone&#xA;  breaks the rules then all the wise men will die. The king listens in&#xA;  while the wise men consult each other to make sure they don't devise a&#xA;  plan to communicate anything more than their guess of red or blue.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;What is the maximum number of men they can be guaranteed to save?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7558" LastEditorUserId="4050" LastEditDate="2014-12-29T19:43:18.997" LastActivityDate="2014-12-29T19:43:18.997" Title="The mad king and the 100 men who each have to guess the colour of their own hat" Tags="|logical-deduction|word-problem|" AnswerCount="0" CommentCount="6" FavoriteCount="0" ClosedDate="2014-12-29T19:43:18.080" ContentLicense="CC BY-SA 3.0" />
  <row Id="6500" PostTypeId="2" ParentId="6495" CreationDate="2014-12-29T23:50:03.343" Score="35" Body="&lt;p&gt;JonTheMon and xnor's solutions assume we have superior equipment and skill, but the question states that we &quot;have a hacksaw&quot;. Well with a hacksaw, we must start from an exposed side; we can't start a cut in the middle of a plank!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most efficient &quot;side&quot; would be the current hole:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&lt;img src=&quot;https://i.stack.imgur.com/hN7A3.png&quot; alt=&quot;practical solution&quot;&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If the hacksaw won't fit in the hole, we &quot;cut an L shaped piece off the left and bottom&quot;, as stated first by JonTheMon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I yield! GOTO0 found an even shorter practical solution.&lt;/p&gt;&#xA;" OwnerUserId="7564" LastEditorUserId="7564" LastEditDate="2014-12-30T02:55:04.360" LastActivityDate="2014-12-30T02:55:04.360" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6501" PostTypeId="2" ParentId="6461" CreationDate="2014-12-30T00:06:36.203" Score="5" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Uranus.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each clue mentions a state or country that has a city belonging in the title of another movie on the list. &lt;br&gt;The first clue mentions 'Maryland', Annapolis is a city in Maryland, and 'Annapolis' belongs in the title of the 14th movie.  &lt;br&gt;Continuing, the 14th clue mentions 'Alabama', Talladega is a city in Alabama, and 'Talladega' belongs in the title of the 15th movie.&lt;br&gt; The 15th clue mentions 'New York', Union is a city in New York, and 'Union' belongs in the title of the 12th movie.&lt;br&gt;Continuing in this fashion, we get a cyclic ordering of the list of cities the OP found:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Annapolis&lt;/li&gt;&#xA;&lt;li&gt;Talladega&lt;/li&gt;&#xA;&lt;li&gt;Union&lt;/li&gt;&#xA;&lt;li&gt;Reno&lt;/li&gt;&#xA;&lt;li&gt;Nuremberg&lt;/li&gt;&#xA;&lt;li&gt;Providence&lt;/li&gt;&#xA;&lt;li&gt;London&lt;/li&gt;&#xA;&lt;li&gt;Arlington&lt;/li&gt;&#xA;&lt;li&gt;New Orleans&lt;/li&gt;&#xA;&lt;li&gt;Elizabethtown&lt;/li&gt;&#xA;&lt;li&gt;Tombstone&lt;/li&gt;&#xA;&lt;li&gt;Atlantic City&lt;/li&gt;&#xA;&lt;li&gt;Flint&lt;/li&gt;&#xA;&lt;li&gt;Tangier&lt;/li&gt;&#xA;&lt;li&gt;East LA&lt;/li&gt;&#xA;&lt;li&gt;Rome&lt;/li&gt;&#xA;&lt;li&gt;St Louis&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reading the first letter of each city name gives &lt;strong&gt;ATURNPLANETAFTERS&lt;/strong&gt;. Cyclicly permuting the letters and adding spaces gives &lt;strong&gt;PLANET AFTER SATURN&lt;/strong&gt;. 'Uranus' is the planet after Saturn.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2014-12-30T00:33:18.933" LastActivityDate="2014-12-30T00:33:18.933" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6502" PostTypeId="2" ParentId="5981" CreationDate="2014-12-30T02:02:09.507" Score="6" Body="&lt;p&gt;The clue can be found behind &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Basket of Fruit&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first cipher is a Bacon cipher based on the word &quot;bacon&quot; can be found in the ciphertext, and the ciphertext has 80 characters, divisible by 5.  The ciphertext can be translated to binary by setting alphabetical characters as &quot;A&quot; and non alphabetical characters as &quot;B&quot;.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;u171B b4C0N BwiOX 2e81L uc6S/ OWKyD O0QjE 5PwiD oF8b4 YmxlI A55oG iRm73 5nNUT Km7ZO +d0Hh 3QJb=&#xA;ABBBA ABABA AAAAA BABBA AABAB AAAAA ABAAA BAAAA AABAB AAAAA ABBAA AAABB BAAAA AABAA BABAA BAAAB&#xA;P     L     A     Y     F     A     I     R     F     A     N     D     R     E     W     S&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Giving &quot;Play fair F Andrews&quot;.  &quot;Playfair&quot; probably refers to the playfair cipher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sticking FANDREWS as the encryption key to the &lt;a href=&quot;http://rumkin.com/tools/cipher/playfair.php&quot; rel=&quot;nofollow&quot;&gt;playfair cipher&lt;/a&gt;, PINVEBLFWFMCCPRWIKWUGUSFZR decodes as ISAXCSGREATESTACHIEVEMENTZ.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ignoring the Z as padding, and considering that the X substitutes for the doubled A (Playfair cipher is not good with doubled letters), the plaintext can be read as &quot;ISAAC'S GREATEST ACHIEVEMENT&quot;.  This is a reference to the apocryphal story of Isaac Newton discovering gravity when an apple fell on his head.  The only artwork that references apples is &lt;em&gt;Basket of Fruit&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Old explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;s&gt;&#xA;Here is the list of artists:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Mona Lisa: &lt;a href=&quot;http://en.wikipedia.org/wiki/Leonardo_da_Vinci&quot; rel=&quot;nofollow&quot;&gt;Leonardo Da&#xA;  Vinci&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Basket of Fruit: &lt;a href=&quot;http://en.wikipedia.org/wiki/Caravaggio&quot; rel=&quot;nofollow&quot;&gt;Michelangelo Merisi (or Amerighi) da&#xA;  Caravaggio&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Sunflowers: &lt;a href=&quot;http://en.wikipedia.org/wiki/Vincent_van_Gogh&quot; rel=&quot;nofollow&quot;&gt;Vincent Van&#xA;  Gogh&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The Persistence of Memory: &lt;a href=&quot;http://en.wikipedia.org/wiki/Salvador_Dal%C3%AD&quot; rel=&quot;nofollow&quot;&gt;Salvador&#xA;  Dali&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The Scream: &lt;a href=&quot;http://en.wikipedia.org/wiki/Edvard_Munch&quot; rel=&quot;nofollow&quot;&gt;Edvard Munch&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The most likely among them is Amerighi da Caravaggio, because the plain text can be written as &quot;IS AC'S GREATEST ACHIEVEMENT&quot; (ignoring X and Z as padding, and the fact that no artist is named Isaac).  Then &lt;em&gt;Basket of Fruit&lt;/em&gt; holds the clue.&#xA;&lt;/s&gt;&lt;/p&gt;&#xA;" OwnerUserId="6178" LastEditorUserId="6178" LastEditDate="2014-12-30T14:37:17.500" LastActivityDate="2014-12-30T14:37:17.500" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6503" PostTypeId="2" ParentId="6495" CreationDate="2014-12-30T02:04:52.247" Score="32" Body="&lt;p&gt;As pointed out in &lt;a href=&quot;https://puzzling.stackexchange.com/a/6500/4394&quot;&gt;a previous answer&lt;/a&gt;, cutting a hole in the middle of the table may be unfeasible if everything you have is a hacksaw. Using the existing hole as a starting point, the cut can be reduced to the length of a single segment bifurcating on both end to meet the old and the newly cut hole tangentially at a convenient distance. Something like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sp5fK.png&quot; alt=&quot;solution 1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then swap the two fragments of the table.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Update&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The solution above divides the table in three pieces, which is not ok according to the problem statement. Here's another solution that produces only two pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Qct5p.png&quot; alt=&quot;solution 2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update 2&lt;/strong&gt; - &lt;em&gt;Credits to &lt;a href=&quot;https://puzzling.stackexchange.com/users/7564/ceraus&quot;&gt;Ceraus&lt;/a&gt; for seeing this:&lt;/em&gt; Depending on your ability to cut sharp curves with the saw, this solution can be further optimized by starting the cut as close as possible to the line that joins the centers of both circles.&lt;/p&gt;&#xA;" OwnerUserId="4394" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2014-12-30T03:44:31.197" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6504" PostTypeId="2" ParentId="1963" CreationDate="2014-12-30T02:10:39.180" Score="7" Body="&lt;p&gt;Here we go.&lt;/p&gt;&#xA;&lt;h1&gt;1:&lt;/h1&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$(1+1+1)! = 6$&lt;/span&gt;&lt;br /&gt;&#xA; This is the only possible one as far as I know.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h1&gt;2:&lt;/h1&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$2+2+2 = 6$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h1&gt;3:&lt;/h1&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$3*3-3 = 6$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h1&gt;4:&lt;/h1&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$4+(4/\sqrt{4}) = 6$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h1&gt;5:&lt;/h1&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$5+(5/5) = 6$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h1&gt;6:&lt;/h1&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$6*(6/6) = 6$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h1&gt;7:&lt;/h1&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$7-(7/7) = 6$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h1&gt;8:&lt;/h1&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 8 - &lt;span class=&quot;math-container&quot;&gt;$\sqrt[4]{8 + 8} = 6$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h1&gt;9:&lt;/h1&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$(9+9)/\sqrt{9} = 6$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h1&gt;Bonus - 0:&lt;/h1&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$(0! + 0! + 0!)! = 6$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7565" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-12-30T02:24:14.597" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6505" PostTypeId="1" AcceptedAnswerId="6506" CreationDate="2014-12-30T02:43:23.057" Score="9" ViewCount="1934" Body="&lt;p&gt;You are to make one cut (or draw one line) – of course it needn’t be straight – that will divide the figure into two identical parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://blogs.scientificamerican.com/observations/2010/05/22/three-puzzles-from-martin-gardner-1914-2010/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0rIrB.png&quot; alt=&quot;Crazy Cut&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7565" LastEditorUserId="11876" LastEditDate="2016-08-03T22:49:34.760" LastActivityDate="2016-08-03T22:49:34.760" Title="Martin Gardner - Crazy Cut" Tags="|geometry|dissection|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6506" PostTypeId="2" ParentId="6505" CreationDate="2014-12-30T02:47:24.470" Score="12" Body="&lt;p&gt;Is this a trick question?  Here is my answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qHNgp.png&quot; alt=&quot;Crazy Cut&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6178" LastEditorUserId="7565" LastEditDate="2014-12-31T01:43:31.393" LastActivityDate="2014-12-31T01:43:31.393" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6507" PostTypeId="1" AcceptedAnswerId="6510" CreationDate="2014-12-30T03:24:42.050" Score="8" ViewCount="6334" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;A &lt;a href=&quot;http://forums.hardwarezone.com.sg/eat-drink-man-woman-16/martin-gardners-hardest-puzzle-4334900.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;number's persistence&lt;/a&gt; is :&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The number of steps required to reduce it to a single digit by multiplying all its digits to obtain a second number&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Then multiplying all the digits of that number to obtain a third number, and so on until a one-digit number is obtained.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;For example&lt;/strong&gt; : 77 has a persistence of four because it requires four steps to reduce it to one digit: 77→49→36→18→8.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;The smallest number of persistence one is 10&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;The smallest of persistence two is 25&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;The smallest of persistence three is 39&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;The smaller of persistence four is 77&lt;/em&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;What is the smallest number of persistence five?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7565" LastEditorUserId="18129" LastEditDate="2020-10-27T06:58:17.087" LastActivityDate="2020-10-27T06:58:17.087" Title="Martin Gardner - Persistence" Tags="|mathematics|number-theory|reachability|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="6508" PostTypeId="2" ParentId="2188" CreationDate="2014-12-30T03:45:58.537" Score="22" Body="&lt;p&gt;My answer, assuming I ask one question only...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Is the liar in front of the death door (DD)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A. (if the Truth guard is asked) If the Truth guy is in front of DD, he answers NO. If not, he answers YES.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B. (if the Liar guard is asked) If the Liar is in front of DD, he answers NO. If not, he answers YES.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way, if we get a NO, then we've asked the guard in front of the death door, so we go to the opposite door. If we get a YES, then we've asked the guard not in front of the death door, so we go to the door behind them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you, I am here all week ;)&lt;/p&gt;&#xA;" OwnerUserId="7568" LastEditorUserId="48964" LastEditDate="2018-12-15T08:57:04.667" LastActivityDate="2018-12-15T08:57:04.667" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6509" PostTypeId="2" ParentId="3562" CreationDate="2014-12-30T03:58:00.237" Score="2" Body="&lt;p&gt;solution only requires:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.the reassembled rug may not comprise more than two pieces&lt;br&gt;&#xA;2.none of the red area may appear on the reassembled rug&lt;br&gt;&#xA;3.the reassembled rug must be square or rectangular in shape&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so, max saved if all but red section retained. to do that...  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;one cut deletes the center red section. DONE&lt;br&gt;&#xA;problem does not prohibit hole. one piece left, max area satisfied, still a rectangle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7569" LastActivityDate="2014-12-30T03:58:00.237" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6510" PostTypeId="2" ParentId="6507" CreationDate="2014-12-30T04:05:24.167" Score="10" Body="&lt;p&gt;Brute forcing with Lua gave me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 679&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is also confirmed by Wikipedia and OEIS:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.m.wikipedia.org/wiki/Persistence_of_a_number&quot;&gt;http://en.m.wikipedia.org/wiki/Persistence_of_a_number&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://oeis.org/A003001&quot;&gt;http://oeis.org/A003001&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2014-12-30T04:15:29.840" LastActivityDate="2014-12-30T04:15:29.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6511" PostTypeId="1" AcceptedAnswerId="6512" CreationDate="2014-12-30T05:45:44.790" Score="10" ViewCount="25885" Body="&lt;p&gt;I was working my way through some Knight and Knave Puzzles in &lt;a href=&quot;http://www.flipkart.com/discrete-mathematics-its-applications-combinatorics-graph-theory-english-7th/p/itmdwxxszhczjrnf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Discrete Maths by Rosen&lt;/a&gt;, when I came across the following question:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;There are  inhabitants of an island on which there are three kinds of people:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Knights who always tell the truth&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Knaves who always lie&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Spies who can either lie or tell the truth.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;You encounter three people, A, B, and C.&lt;/p&gt;&#xA;&lt;p&gt;You know one of these people is a knight, one is a knave, and one is a spy.&lt;/p&gt;&#xA;&lt;p&gt;Each of the three people knows the type of person each of other two is.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;For this situation, if possible, determine whether there is a unique solution and determine who the knave, knight, and spy is:&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;A says &amp;quot;C is the knave,” B says, “A is the knight,” and C says “I am the spy&amp;quot;&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;My Solution:&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; A is the Knight, B is the Spy, and C is the Knave&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Doubt:&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;Am I correct in saying my answer will work?&lt;/p&gt;&#xA;" OwnerUserId="7435" LastEditorUserId="69582" LastEditDate="2022-11-07T19:07:04.923" LastActivityDate="2022-11-07T19:07:04.923" Title="Knights, Knaves and Spies - Part 1" Tags="|logical-deduction|liars|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6512" PostTypeId="2" ParentId="6511" CreationDate="2014-12-30T06:48:19.720" Score="12" Body="&lt;p&gt;Yes. Your answer is correct.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A is the Knight&lt;br/&gt;&#xA; B is the Spy&lt;br/&gt;&#xA; and C is the Knave&lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To get the solution, First assume, A is knight and will always tells the truth.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Then as per his statement, C is the knave and so what he said will be&#xA;  false. That means he is not a spy. B is the spy and his statement A is&#xA;  the knight is random (true here). This is the only case in which the&#xA;  statements didn't contradict.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now assume, A is the Knave.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Then as per his statement &quot;C is the knave&quot;, it's clear that C is&#xA;  definitely not the knave. Which doesn't contradict since A is the&#xA;  knave already. That means, either B or C is Knight. If B is Knight his&#xA;  statement &quot;A is knight&quot; is false and it contradicts. If C is Knight&#xA;  his statement &quot;I am the spy&quot; is wrong and it contradicts. So this&#xA;  combination A is Knave, B is knight/Spy, C is Knight/Spy is wrong.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Continue this assumptions for other chances of combinations.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You will understand that all other combination except the first one (A&#xA;  is knight, B is Spy and C is knave) is wrong since the statements&#xA;  contradicts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="7435" LastEditDate="2014-12-31T08:46:30.023" LastActivityDate="2014-12-31T08:46:30.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6513" PostTypeId="2" ParentId="6212" CreationDate="2014-12-30T07:16:10.473" Score="6" Body="&lt;p&gt;(incomplete)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Hint 4:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Insert &quot;338531339395984130673279936 in base 27&quot; into Wolfram Alpha and get a Base 27 number, 5:22:5:0:23:9:12:12:0:19:5:14:4:0:1:21:4:9:15.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Translate this into letters (e.g. 5-&gt;E, 22-&gt;V, 0-&gt;space), and I get &quot;EVE WILL SEND AUDIO&quot;.&lt;/p&gt;&#xA;" OwnerUserId="6178" LastEditorUserId="6178" LastEditDate="2014-12-30T15:52:50.217" LastActivityDate="2014-12-30T15:52:50.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6515" PostTypeId="2" ParentId="6401" CreationDate="2014-12-30T09:20:26.980" Score="6" Body="&lt;p&gt;&lt;strong&gt;This answer explains the significance of the two numbers found in &lt;a href=&quot;https://puzzling.stackexchange.com/a/6465/5373&quot;&gt;MarchHo's answer&lt;/a&gt;, and also answers the riddle, but not in the way intended by the OP (for which see &lt;a href=&quot;https://puzzling.stackexchange.com/a/6430/5373&quot;&gt;rand al'thor's answer&lt;/a&gt;).&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuing with the keys, thanks to @MarchHo:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;6.67384&lt;/strong&gt;x10^-11 N m2/kg2 is G, the gravitational constant&lt;br&gt;&#xA; &lt;strong&gt;9.80665&lt;/strong&gt; m/s2 is g, the standard gravity on earth  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Possible meanings of the five clues are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Christmas Day&lt;/strong&gt; - Sir Isaac Newton was born on Dec 25, 1642 (Julian calendar)&lt;br&gt;&#xA; &lt;strong&gt;cross bones&lt;/strong&gt; - was Sir Isaac Newton's personal coat of arms&lt;br&gt;&#xA; &lt;strong&gt;t&lt;/strong&gt; - could be associated with the Samsung Gravity Touch phone&lt;br&gt;&#xA; &lt;strong&gt;amphibian surmounts&lt;/strong&gt; - could be a reference to &quot;The Tale of Mr Jeremy Fisher&quot;.  Jeremy Fisher is a frog who has a friend called Sir Isaac Newton&lt;br&gt;&#xA; &lt;strong&gt;amphibian surmounts&lt;/strong&gt; - could also describe overcoming gravity in adapting to dry land&lt;br&gt;&#xA; &lt;strong&gt;hnadwrintig&lt;/strong&gt; - could suggest that Sir Isaac Newton may have been dyslexic&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2014-12-30T13:13:55.920" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="6517" PostTypeId="2" ParentId="6511" CreationDate="2014-12-30T11:33:50.817" Score="33" Body="&lt;p&gt;Simpler explanation:&lt;/p&gt;&#xA;&lt;p&gt;First, notice that B cannot be the knight, because then for their statement to be true, A would also have to be a knight, and we know there is only one knight.&lt;/p&gt;&#xA;&lt;p&gt;Second, notice that C cannot be the knight, because then their statement would be false.&lt;/p&gt;&#xA;&lt;p&gt;Therefore, A is the knight. By their statement, C is the Knave. By elimination, B is the spy.&lt;/p&gt;&#xA;" OwnerUserId="2187" LastEditorUserId="2187" LastEditDate="2022-11-07T18:55:16.190" LastActivityDate="2022-11-07T18:55:16.190" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="6518" PostTypeId="1" AcceptedAnswerId="6526" CreationDate="2014-12-30T11:35:04.090" Score="25" ViewCount="2772" Body="&lt;p&gt;Romeo and Juliet are each placed uniformly at random locations on a spherical planet, meaning that each square meter of surface area of the planet is equally likely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Romeo decides to dig a straight-line tunnel through the planet towards Juliet. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;&lt;em&gt;What is the most likely distance Romeo has to dig?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In other words, what is the mode of the probability distribution of straight-line distances?&lt;/p&gt;&#xA;" OwnerUserId="7573" LastEditorUserId="4551" LastEditDate="2014-12-30T16:20:50.077" LastActivityDate="2016-07-22T21:01:24.327" Title="Digging a tunnel between random locations" Tags="|mathematics|geometry|probability|" AnswerCount="4" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="6519" PostTypeId="2" ParentId="6518" CreationDate="2014-12-30T11:50:10.713" Score="10" Body="&lt;p&gt;Edit: as noted by others, this answer is incorrect. It assumes (incorrectly) that the distance to the modal latitude is the modal distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d}{\sqrt{2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume, wlog, that Romeo is at the north pole. Form latitudinal rings, composed of the land equidistant from Romeo. The equatorial one is largest. Therefore, we just need the straight line distance to the equator, which is $\frac{d}{\sqrt{2}}$&lt;/p&gt;&#xA;" OwnerUserId="3987" LastEditorUserId="3987" LastEditDate="2014-12-30T18:04:09.043" LastActivityDate="2014-12-30T18:04:09.043" CommentCount="20" ContentLicense="CC BY-SA 3.0" />
  <row Id="6520" PostTypeId="1" AcceptedAnswerId="6522" CreationDate="2014-12-30T13:40:46.440" Score="2" ViewCount="2968" Body="&lt;p&gt;I was working my way through some Knight and Knave Puzzles in &lt;a href=&quot;http://www.flipkart.com/discrete-mathematics-its-applications-combinatorics-graph-theory-english-7th/p/itmdwxxszhczjrnf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Discrete Maths by Rosen&lt;/a&gt;, when I came across the following question:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;There are  inhabitants of an island on which there are three kinds of people:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Knights who always tell the truth&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Knaves who always lie&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Spies who can either lie or tell the truth.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;You encounter three people, A, B, and C.&lt;/p&gt;&#xA;&lt;p&gt;You know one of these people is a knight, one is a knave, and one is a spy.&lt;/p&gt;&#xA;&lt;p&gt;Each of the three people knows the type of person each of other two is.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;For this situation, if possible, determine whether there is a unique solution and determine who the knave, knight, and spy is :&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;A says “I am the knight,” B says “I am the knave,” and C says “B is the knight.”&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Book Solution:&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;&lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$A\Rightarrow Knight$&lt;/span&gt;&lt;/strong&gt;&lt;br/&gt;&#xA;&lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$B\Rightarrow Spy$&lt;/span&gt;&lt;/strong&gt;&lt;br/&gt;&#xA;&lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$C\Rightarrow Knave$&lt;/span&gt;&lt;/strong&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Doubt:&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;I am not able to understand why this is the case .&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For A it is fine&lt;/li&gt;&#xA;&lt;li&gt;For C it is fine&lt;/li&gt;&#xA;&lt;li&gt;But for B , if B is the Spy and B says &amp;quot;I am the knave&amp;quot; -- does that mean B is a Spy who lies ?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7435" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-07T01:17:24.800" Title="Knights , Knaves and Spies - Part 2" Tags="|logical-deduction|liars|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6521" PostTypeId="2" ParentId="6520" CreationDate="2014-12-30T13:44:52.427" Score="4" Body="&lt;p&gt;B cannot be a knight - they would self identify as a knight.&#xA;They also cannot be a knave - they would self identify as anything but a knave. Therefore they are a spy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TL;DR: anyone identifying themselves as a knave under these circumstances must be a spy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: B is indeed a spy who lies. By your answer to rand al'thor, they are a knave in all but name.&lt;/p&gt;&#xA;" OwnerUserId="3987" LastEditorUserId="3987" LastEditDate="2014-12-30T13:56:27.700" LastActivityDate="2014-12-30T13:56:27.700" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6522" PostTypeId="2" ParentId="6520" CreationDate="2014-12-30T13:46:04.060" Score="8" Body="&lt;p&gt;If C is the knight, then by C's statement, B is the knight, contradiction. If B is the knight, then by B's statement, B is the knave, contradiction. So &lt;strong&gt;A is the knight&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If B is the knave, then by B's statement, B is not the knave, contradiction. So &lt;strong&gt;B is the spy&lt;/strong&gt;. And his statement is a lie, so he's a &lt;strong&gt;lying spy&lt;/strong&gt; (now that's a good insult!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Leaving the last case: &lt;strong&gt;C is the knave&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2014-12-30T13:53:15.143" LastActivityDate="2014-12-30T13:53:15.143" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6524" PostTypeId="1" AcceptedAnswerId="6607" CreationDate="2014-12-30T14:16:19.847" Score="8" ViewCount="560" Body="&lt;p&gt;Who am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I grew up&lt;br&gt;&#xA;&amp;emsp; where Mischa's brother, the doctor,&lt;br&gt;&#xA;&amp;emsp; ate human liver with bitter wine:&lt;br&gt;&#xA;&amp;emsp; in soccer, a yellow card offence, I hear.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I trained as a cub&lt;br&gt;&#xA;&amp;emsp; where in ancient times the Southern mounds were built;&lt;br&gt;&#xA;&amp;emsp; where an Elk traverses cedars,&lt;br&gt;&#xA;&amp;emsp; with a big greasy tireless ol' man.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My famous works, as split,&lt;br&gt; &#xA;&amp;emsp; anticipate the arrival&lt;br&gt;&#xA;&amp;emsp; of you, seen, with half a drum,&lt;br&gt;&#xA;&amp;emsp; and of the French, fabric for a Nordic funeral.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I arrived and departed with a dirty snowball. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Play it once, for old times' sake - play it! &lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="4050" LastEditDate="2015-01-01T17:57:36.327" LastActivityDate="2015-01-02T06:51:53.203" Title="Who am I? I arrived and departed with a dirty snowball. Play it!" Tags="|riddle|" AnswerCount="3" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6525" PostTypeId="2" ParentId="6524" CreationDate="2014-12-30T14:27:39.390" Score="1" Body="&lt;p&gt;&lt;strong&gt;This answer is wrong, but I may have solved some of the clues correctly&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;s&gt;Dostoevsky?&lt;/s&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Mischa's brother&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;s&gt;Fyodor Dostoevsky had a brother named Mikhail; Mischa is a diminutive form of the name Mikhail.&lt;/s&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;yellow card offence&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This surely refers to &lt;a href=&quot;http://en.wikipedia.org/wiki/Yellow_ticket&quot; rel=&quot;nofollow&quot;&gt;prostitution&lt;/a&gt;, as referred to in Dostoevsky's &lt;a href=&quot;https://answers.yahoo.com/question/index?qid=20080527170646AAZJd8H&quot; rel=&quot;nofollow&quot;&gt;Crime and Punishment&lt;/a&gt; (this is where I first heard the term, and it's what put me onto this answer).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;where an Elk traverses cedars&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I suspect this refers to some &lt;a href=&quot;http://en.wikipedia.org/wiki/Elk_River&quot; rel=&quot;nofollow&quot;&gt;Elk River&lt;/a&gt; or other (or possibly &lt;a href=&quot;http://en.wikipedia.org/wiki/Moose_River&quot; rel=&quot;nofollow&quot;&gt;Moose River&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My famous works&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; suggests that we're looking for some kind of artist (novelist, composer, whatever); Dostoevsky and Tchaikovsky were what immediately came to my mind.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;the French fabric for a Nordic funeral&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I suspect this is wordplay. The French for fabric is &lt;em&gt;tissu&lt;/em&gt;, and the Norwegian/Swedish for funeral is &lt;em&gt;begravelse&lt;/em&gt;/&lt;em&gt;begravning&lt;/em&gt;. Tissubegrav? Needs more work!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More to follow...&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2014-12-30T14:45:52.843" LastActivityDate="2014-12-30T14:45:52.843" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6526" PostTypeId="2" ParentId="6518" CreationDate="2014-12-30T15:59:34.903" Score="18" Body="&lt;p&gt;&lt;strong&gt;Answer:&lt;/strong&gt; The most likely distance is the diameter of the planet, when Romeo and Juliet are antipodal.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Assume WLOG Romeo is on the North Pole. Juliet's position is uniformly random on the sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma 1:&lt;/strong&gt; For the uniform distribution over the sphere, the marginal distribution over the $z$ coordinate is also uniform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a surprising and non-obvious fact about spheres. It's not true for circles or for higher dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, if you make a slice of an apple that's a fixed width, you always get the same amount of peel. Slices closer to the poles will have a smaller radius, but the peel will be tilted inwards letting more of it fit, and these two effects exactly compensate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mMt9em.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In this picture, the slices are different widths, but I hope it gets the idea across.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the &lt;a href=&quot;http://mathworld.wolfram.com/Zone.html&quot; rel=&quot;noreferrer&quot;&gt;calculation from Mathworld&lt;/a&gt; for the surface area of a zone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5iFQt.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The surface area is computed to be $2 \pi r h$ (where $r$ is the sphere's radius), which is proportional to the thickness $h$ and independent the position of the slice given by $a$ and $b$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So, Juliet's position can be randomly generated by picking a uniformly random $z$-coordinate. Therefore, the most likely distance is the one corresponding to the infinitesimal slice $\mathrm{d}z$ that compresses to the smallest range of distances $\left| \mathrm{d}D \right|$, maximizing the probability density. This problem is unchanged if we instead consider points on a two-dimensional circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance $D$ between Romeo's position $(0, r)$ and Juliet's position $(\sqrt{r^2-z^2},z)$ is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$D^2 = r^2 - z^2 + (z-r)^2 = 2r(r-z)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, taking the derivative with respect to $z$ gives &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2 D \thinspace \mathrm{d}D = -2r \thinspace \mathrm{dz}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so the infinitesimal range of distances &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left| \mathrm{d}D \right| = r \thinspace \mathrm{dz} / D $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is minimized when $D$ is maximized. So, the most likely distance is the maximum one of the diameter, when Romeo and Juliet are antipodal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We've also shown that the distribution of distance squared is uniform. &lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2014-12-30T20:05:28.173" LastActivityDate="2014-12-30T20:05:28.173" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6528" PostTypeId="2" ParentId="6518" CreationDate="2014-12-30T16:13:36.953" Score="11" Body="&lt;p&gt;Wlog, have Romeo at the North pole; consider that latitude zero. Latitudes increase from there to $\frac{\pi}{2}$ at the equator and ${\pi}$ at the South pole. I will assume that the planet has radius $1$, because that's all going to cancel.&#xA;For a distance $x$, latitude ${\theta}$ is such that ${x}^2 = \sin^2{\theta} + (1 - \cos^2{\theta})$ (apologies for a lack of justifying diagram here).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Population density at distance $x$ will be proportional to $\sin{\theta}$ (which governs the size of the circle at distance $x$), and also proportional to $\frac{d\theta}{dx}$, how much latitude you cover for a given increase in $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting $\sin^2\theta + \cos^2\theta = 1$ into the distance equation, we get $\cos\theta = 1 - \frac{x^2}{2} = p$. So $\theta = \cos^{-1}p$, and $\sin\theta = \sqrt{1 - p^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the chain rule, $\frac{d\theta}{dx} = \frac{d\theta}{dp}.\frac{dp}{dx}$. So population density at distance $x$ is proportional to $\sin\theta.\frac{d\theta}{dp}.\frac{dp}{dx}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the definition of $p$, $\frac{dp}{dx} = -x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, $\frac{d}{dx}(\cos^{-1}x) = \frac{-1}{\sqrt{1 - x^2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So population density is proportional to $ \sqrt{1 - p^2}.\frac{-1}{\sqrt{1 - p^2}}.(-x) = x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the further away you get, the higher the population. The mode, therefore, must be $2$ - the maximum distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, this must be because at the South pole, you can move a long way without changing your distance to the North pole much. I am very surprised to see a linear relationship drop out of that, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Although I can't say I'm surprised to see someone has beaten me to it).&lt;/p&gt;&#xA;" OwnerUserId="6031" LastEditorUserId="1766" LastEditDate="2016-07-22T21:01:24.327" LastActivityDate="2016-07-22T21:01:24.327" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6529" PostTypeId="2" ParentId="6518" CreationDate="2014-12-30T16:56:03.723" Score="4" Body="&lt;p&gt;A short argument based on a beautiful equation giving the area of a spherical cap, runs as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a spherical planet, regardless its diameter, &lt;a href=&quot;http://en.wikipedia.org/wiki/Spherical_cap&quot; rel=&quot;nofollow&quot;&gt;the surface area reachable by a tunnel with length not exceeding $x$&lt;/a&gt; is $\pi x^2$. So the area reachable by tunnels ranging in length from $x$ to $x+dx$ is $2 \pi x dx$. The mode (most likely $x$ value) is reached at maximum $x$ equal to the diameter of the planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So although only one unique position is antipodal to Romeo's position, it does correspond to the most likely tunnel length.&lt;/p&gt;&#xA;" OwnerUserId="7573" LastActivityDate="2014-12-30T16:56:03.723" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6530" PostTypeId="1" AcceptedAnswerId="6539" CreationDate="2014-12-30T17:37:38.887" Score="6" ViewCount="2599" Body="&lt;p&gt;&lt;em&gt;This question was inspired by &lt;a href=&quot;https://puzzling.stackexchange.com/questions/6507/martin-gardner-persistence&quot;&gt;a recent question about number persistence&lt;/a&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a number's &lt;strong&gt;persistence&lt;/strong&gt; is defined by how many times you have to multiply all of its digits together before you get a single digit (&lt;strong&gt;e.g.&lt;/strong&gt; *47 -&gt; 28 -&gt; 16 -&gt; 6 (&lt;strong&gt;4 steps&lt;/strong&gt;)) how can you determine a number with n persistence by calculating backwards? Or is it impossible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;By calculating backwards I mean taking a number (such as 2) and work from it; 2 -&gt; 12 (or 21) -&gt; etc. etc.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Prove your claim!&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2014-12-31T04:58:54.413" Title="Persistence Reverse Calculation!" Tags="|mathematics|" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6531" PostTypeId="1" AcceptedAnswerId="7746" CreationDate="2014-12-30T17:52:36.257" Score="8" ViewCount="826" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XcCYm.png&quot; alt=&quot;Crossword&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Across&lt;/strong&gt;&lt;br&gt;&#xA;3.  Amphibian surmounts physicist (6)&lt;br&gt;&#xA;4.  Hilarious partner (3)&lt;br&gt;&#xA;5.  Spurn a guilty plea outside Aquidneck Island (8)&lt;br&gt;&#xA;6.  Performing a supersonic number in the head of an adult insect. (9)&lt;br&gt;&#xA;7.  Sightless, tailless, expensive and suitable for general audiences (5)&lt;br&gt;&#xA;9.  Blur's life, not your jumping around. (7)&lt;br&gt;&#xA;11. Dennis Ritchie's creation doing half a dance outside a flat to check your humanity (7)&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Down&lt;/strong&gt;&lt;br&gt;&#xA;1.  Crypto-Olaf (10, 7)&lt;br&gt;&#xA;2.  The angels amid sightless ghoti; half a nokia appears unexpectedly (5, 3)&lt;br&gt;&#xA;5.  Not a cat (3)&lt;br&gt;&#xA;8.  Not a dog (3)&lt;br&gt;&#xA;10. Ghoti (4)&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my first attempt at a cryptic-ish crossword (some of the clues are cryptic, I've thrown in a couple of really easy ones too, to provide more cross-checking).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd appreciate constructive feedback on what you think of this puzzle, what level of difficulty it is, and what I should do differently to improve things for the next one.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In response to query: OED is the Oxford English Dictionary. It's the most authoritative dictionary for English. The point is that while some of the words are neologisms, they're ones which are well-known and accepted enough to have become part of the English language (by the OED's definition).&lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="4050" LastEditDate="2015-01-09T22:07:41.083" LastActivityDate="2015-01-09T22:07:41.083" Title="Cryptic-ish crossword - contains some neologisms, all of them to be found in the OED" Tags="|crosswords|cryptic-crosswords|" AnswerCount="5" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6533" PostTypeId="2" ParentId="6530" CreationDate="2014-12-30T18:15:27.900" Score="1" Body="&lt;p&gt;This is less simple than I thought. The chosen factors must all be less than 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose an end number (4)&lt;br&gt;&#xA;Choose a set of factors (1,4).&lt;br&gt;&#xA;Use the factors to form a new number (14).&lt;br&gt;&#xA;Repeat until you have a number of the desired persistence (2,7)-&gt;72, (8,9)-&gt;98 (2,7,7)-&gt;277. This has no set of factors all less than 10. We may add 1s as we wish until we have a number which factorises as desired.&lt;/p&gt;&#xA;" OwnerUserId="3987" LastEditorUserId="5368" LastEditDate="2014-12-30T20:05:47.860" LastActivityDate="2014-12-30T20:05:47.860" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6535" PostTypeId="2" ParentId="6530" CreationDate="2014-12-30T18:38:54.737" Score="0" Body="&lt;p&gt;Using this process, you can never calculate a number's persistence. I will explain why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say you start with $13$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$13$&lt;br&gt;&#xA;  $1113$ &amp;amp; $3111$&lt;br&gt;&#xA;  $(3113$ &amp;amp; $3113)$ or $(1331$ &amp;amp; $1331)$&lt;br&gt;&#xA;  $((2321$ &amp;amp; $1232)$ or $(2321$ &amp;amp; $1232))$ &amp;amp; $((2123$ &amp;amp; $3212)$ or $(2123$ &amp;amp; $3212))$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As you can see, the resulting numbers will never reach a single digit, and the sequence becomes extremely complicated. I would continue on, but as you can see, it gets very complicated and redundant.&lt;/p&gt;&#xA;" OwnerUserId="7565" LastActivityDate="2014-12-30T18:38:54.737" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6536" PostTypeId="2" ParentId="6530" CreationDate="2014-12-30T18:51:15.097" Score="0" Body="&lt;p&gt;You could brute force it by checking the persistence of each number until you find one that matches.  You could get smart with different rules for building this data, such as any number with a 0 in it will have a persistence &amp;lt; 2, any number with an even digit and a 5 will have a persistence of &amp;lt; 3, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, from &lt;a href=&quot;http://en.wikipedia.org/wiki/Persistence_of_a_number&quot; rel=&quot;nofollow&quot;&gt;Wikipedia's persistence article&lt;/a&gt; it's thought that multiplicative persistence for base 10 is &amp;lt;= 11 for all numbers.  If that's true you could just take the appropriate value from &lt;a href=&quot;http://oeis.org/A003001&quot; rel=&quot;nofollow&quot;&gt;that known sequence&lt;/a&gt; based on your desired &lt;code&gt;n&lt;/code&gt; from the following values:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[0, 10, 25, 39, 77, 679, 6788, 68889, 2677889, 26888999, 3778888999, 277777788888899]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2014-12-30T18:51:15.097" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6537" PostTypeId="2" ParentId="6321" CreationDate="2014-12-30T19:22:11.547" Score="4" Body="&lt;h1&gt;Babingtonites - 13 letters&lt;/h1&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;h1&gt;&lt;span class=&quot;math-container&quot;&gt;$5371207621898$&lt;/span&gt;&lt;/h1&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.es/search?site=&amp;amp;q=define:+babingtonite#q=define:+babingtonites&quot; rel=&quot;nofollow noreferrer&quot;&gt;Definition&lt;/a&gt;&lt;br /&gt;&#xA;&lt;a href=&quot;http://www.scrabblewordsolver.com/word-checker/babingtonites&quot; rel=&quot;nofollow noreferrer&quot;&gt;Accepted on Scrabble&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7565" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2014-12-30T19:22:11.547" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6538" PostTypeId="1" AcceptedAnswerId="6605" CreationDate="2014-12-30T19:32:46.430" Score="6" ViewCount="1154" Body="&lt;p&gt;&lt;em&gt;This riddle is dedicated to &lt;strong&gt;Avigrail&lt;/strong&gt;, a brilliant riddler.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are four parts to this puzzle. Each of them points towards the same solution, and you can get the solution from any one on its own, but I'll only accept an answer that includes explanations of &lt;em&gt;how&lt;/em&gt; to solve all four.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the &lt;em&gt;interesting&lt;/em&gt; one (also probably the easiest):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;for all Lambda unique e L for all unique r&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In this riddle, all the clues point towards the &lt;em&gt;concept&lt;/em&gt; of the answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I might be metal, or you might&lt;br&gt;&#xA;  Find me in woods or at the beach.&lt;br&gt;&#xA;  I can answer your questions -&lt;br&gt;&#xA;  Always truly, but oft unwelcome.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In this riddle, all the clues point towards the &lt;em&gt;English word&lt;/em&gt; for the answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I might be paper.&lt;br&gt;&#xA;  Half of me has it;&lt;br&gt;&#xA;  Or, with repeats,&lt;br&gt;&#xA;  Only one lacks it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In this riddle, all the clues point towards the &lt;em&gt;German word&lt;/em&gt; for the answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I might be paper.&lt;br&gt;&#xA;  I rhyme with a predator.&lt;br&gt;&#xA;  Bond a fork and then&lt;br&gt;&#xA;  A twisted limb frozen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="8874" LastEditDate="2015-10-02T14:27:25.893" LastActivityDate="2015-10-02T14:27:25.893" Title="Are you clever enough? In memory of Avigrail" Tags="|riddle|word|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6539" PostTypeId="2" ParentId="6530" CreationDate="2014-12-30T19:36:17.147" Score="6" Body="&lt;p&gt;Rather than giving heuristics that might work, I thought I'd try putting my money where my mouth is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's some code in Python 3:&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from collections import defaultdict&#xA;&#xA;PERSISTENCE = 9&#xA;START_DIGIT = 0&#xA;ONES = 1&#xA;&#xA;def single_digit_factorise(n):   &#xA;    divisors = []&#xA;&#xA;    for p in [2, 3, 5, 7]:&#xA;        while n % p == 0:&#xA;            n //= p&#xA;            divisors.append(p)&#xA;&#xA;    return (n == 1) and sorted(divisors)&#xA;&#xA;# Uses http://www.nayuki.io/page/next-lexicographical-permutation-algorithm&#xA;def next_number(digits):&#xA;    # Treat 0 differently&#xA;    if digits[-1] == 0:&#xA;        n = int(&quot;&quot;.join(map(str,digits[:-1])))&#xA;        return list(map(int,list(str(n+1)))) + [0]&#xA;&#xA;    pivot = len(digits) - 1&#xA;    last = digits[-1]&#xA;&#xA;    while pivot &amp;gt;= 0 and digits[pivot] &amp;gt;= last:&#xA;        last = digits[pivot]&#xA;        pivot -= 1&#xA;&#xA;    # Exhausted search, add a 1&#xA;    if pivot == -1:&#xA;        return [1] + sorted(digits)&#xA;&#xA;    pivot_num = digits[pivot]&#xA;    successor = len(digits) - 1&#xA;&#xA;    while digits[successor] &amp;lt;= pivot_num:&#xA;        successor -= 1&#xA;&#xA;    digits[pivot], digits[successor] = digits[successor], digits[pivot]&#xA;    digits[pivot+1:] = digits[pivot+1:][::-1]&#xA;&#xA;    return digits&#xA;&#xA;def to_num(digits):&#xA;    total = 0&#xA;&#xA;    for i in range(len(digits)):&#xA;        total += digits[~i] * 10**i&#xA;&#xA;    return total&#xA;&#xA;def possibilities(digits):&#xA;    def _possibilities(digits, results):&#xA;        results.add(tuple(sorted(digits)))&#xA;&#xA;        if digits.count(2) + digits.count(3) &amp;lt;= 1:&#xA;            return results&#xA;&#xA;        # It's late night and I can't be bothered making this part neater&#xA;        if digits.count(2) &amp;gt;= 3:&#xA;            digits.remove(2)&#xA;            digits.remove(2)&#xA;            digits.remove(2)&#xA;            digits.append(8)&#xA;            _possibilities(digits, results)&#xA;            digits.remove(8)&#xA;            digits.extend([2, 2, 2])&#xA;&#xA;        if digits.count(2) &amp;gt;= 2:&#xA;            digits.remove(2)&#xA;            digits.remove(2)&#xA;            digits.append(4)&#xA;            _possibilities(digits, results)&#xA;            digits.remove(4)&#xA;            digits.extend([2, 2])&#xA;&#xA;        if digits.count(3) &amp;gt;= 2:&#xA;            digits.remove(3)&#xA;            digits.remove(3)&#xA;            digits.append(9)&#xA;            _possibilities(digits, results)&#xA;            digits.remove(9)&#xA;            digits.extend([3, 3])&#xA;&#xA;        if digits.count(2) &amp;gt;= 1 and digits.count(3) &amp;gt;= 1:&#xA;            digits.remove(2)&#xA;            digits.remove(3)&#xA;            digits.append(6)&#xA;            _possibilities(digits, results)&#xA;            digits.remove(6)&#xA;            digits.extend([2, 3])&#xA;&#xA;        return results&#xA;&#xA;    results = _possibilities(digits, set())&#xA;    return list(map(list, results))&#xA;&#xA;def persistence_str(n):&#xA;    output = [str(n)]&#xA;&#xA;    while n &amp;gt;= 10:&#xA;        product = 1&#xA;&#xA;        for digit in str(n):&#xA;            product *= int(digit)&#xA;&#xA;        output.append(str(product))&#xA;        n = product&#xA;&#xA;    return &quot; -&amp;gt; &quot;.join(output)&#xA;&#xA;def find_persistence(n, start, ones=1):&#xA;    assert 0 &amp;lt;= start &amp;lt; 10&#xA;&#xA;    queues = defaultdict(list)&#xA;&#xA;    if start &amp;gt; 0:&#xA;        queues[0] = [[start]]&#xA;        base = step = 0&#xA;&#xA;    else:&#xA;        queues[1] = [[1, 0]]&#xA;        base = step = 1&#xA;&#xA;    while base &amp;lt;= step &amp;lt; n:&#xA;        if not queues[step]:&#xA;            step -= 1&#xA;            continue&#xA;&#xA;        num = queues[step].pop(0)            &#xA;        fact = single_digit_factorise(to_num(num))&#xA;&#xA;        next_num = next_number(num[:])&#xA;&#xA;        if len(num) &amp;gt; 1 and next_num.count(1) &amp;lt;= ones:&#xA;            queues[step].append(next_num)&#xA;&#xA;        if len(num) == 1:&#xA;            step += 1&#xA;            queues[step].append(next_num)&#xA;            queues[step].extend([x for x in possibilities(fact) if len(x)&amp;gt;1])&#xA;&#xA;        elif fact and fact != num:&#xA;            step += 1&#xA;            queues[step].extend([x for x in possibilities(fact) if len(x)&amp;gt;1])&#xA;&#xA;    return to_num(queues[step][0])&#xA;&#xA;number = find_persistence(PERSISTENCE, START_DIGIT, ONES)&#xA;print(persistence_str(number))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit: Updated to be easier to use&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Enter the persistence and starting digit at the top. &lt;code&gt;ONES&lt;/code&gt; is an additional parameter which specifies how many ones the program is allowed to add.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently the program throws an error if a solution was not found (you can try upping the number of ones allowed to see if that helps).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some chains it has found:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2222222223333333778 -&amp;gt; 438939648 -&amp;gt; 4478976 -&amp;gt; 338688 -&amp;gt; 27648 -&amp;gt; 2688 -&amp;gt; 768 -&amp;gt; 336 -&amp;gt; 54 -&amp;gt; 20 -&amp;gt; 0&#xA;11 -&amp;gt; 1&#xA;222222244468 -&amp;gt; 393216 -&amp;gt; 972 -&amp;gt; 126 -&amp;gt; 12 -&amp;gt; 2&#xA;13 -&amp;gt; 3&#xA;2377 -&amp;gt; 294 -&amp;gt; 72 -&amp;gt; 14 -&amp;gt; 4&#xA;33557 -&amp;gt; 1575 -&amp;gt; 175 -&amp;gt; 35 -&amp;gt; 15 -&amp;gt; 5&#xA;233366777 -&amp;gt; 666792 -&amp;gt; 27216 -&amp;gt; 168 -&amp;gt; 48 -&amp;gt; 32 -&amp;gt; 6&#xA;17 -&amp;gt; 7&#xA;2222344444446679 -&amp;gt; 1783627776 -&amp;gt; 4148928 -&amp;gt; 18432 -&amp;gt; 192 -&amp;gt; 18 -&amp;gt; 8&#xA;19 -&amp;gt; 9&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It's not surprising that the chains found are so short for the digits 1379. After all, the last digit means that the number can't be divisible by 2 or 5, leaving only 3 and 7 as possible prime factors. If it's possible for the chain to be any longer for these digits then it's going to need a lot of 1s.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Explanation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;We keep a list of queues for different persistences, e.g. queue #3 is filled with numbers of persistence 3. Each queue is essentially a list of numbers to be searched. We initialise queue #0 to contain the digit in question (or queue #1 to contain 10 if the digit was 0, to make calculating easier). We also initialise a &lt;code&gt;step&lt;/code&gt; variable to 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While &lt;code&gt;step&lt;/code&gt; is less than our target, we:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Check if queue #&lt;code&gt;step&lt;/code&gt; is empty. If it is, we decrement &lt;code&gt;step&lt;/code&gt; by 1 and backtrack.&lt;/li&gt;&#xA;&lt;li&gt;Otherwise, get the next number from queue #&lt;code&gt;step&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Factorise it into primes, and push the &quot;next number&quot; back onto queue #&lt;code&gt;step&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;If the prime factorisation had only single digits and isn't the original number (in the case of one-digit numbers) then:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Increment &lt;code&gt;step&lt;/code&gt; by 1 (i.e. we just found a number with a higher persistence)&lt;/li&gt;&#xA;&lt;li&gt;Look at all possibilities of factorising the number into single digits (not necessarily prime) and push each of them into the queue for the new incremented &lt;code&gt;step&lt;/code&gt; number.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For most numbers, &quot;next number&quot; means lexicographically, so &lt;code&gt;123 -&amp;gt; 132 -&amp;gt; 213 -&amp;gt; 231 -&amp;gt; 312 -&amp;gt; 321&lt;/code&gt;. When we reach the end, we go back to the beginning but add a &lt;code&gt;1&lt;/code&gt; to the front, i.e. &lt;code&gt;321 -&amp;gt; 1123 -&amp;gt; 1132 -&amp;gt; ...&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For multiples of 10, however, we use the next multiple of 10 (since these are all numbers with a product of 0).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this has the problem that we might be searching indefinitely, always adding 1s. To remedy this we stop searching down a path if we've added more than some number of 1s - this probably isn't the best method, but it's the first quick heuristic I could think of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this was a mouthful, so let's look at what happens if the number we're searching is  24.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We push &lt;code&gt;42&lt;/code&gt; as the next number to search back on the current queue for later&lt;/li&gt;&#xA;&lt;li&gt;We factorise 24 into primes, becoming &lt;code&gt;[2, 2, 2, 3]&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;Then we look at all ways of turning that factorisation into single digits whose product is 24, giving &lt;code&gt;38, 234, 46, 2223, 226&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Finally we increment &lt;code&gt;step&lt;/code&gt; by 1 as we have found numbers with a persistence one higher than that of 24, and push all of these numbers into the corresponding queue and continue searching from there&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="445" LastEditorUserId="445" LastEditDate="2014-12-31T04:58:54.413" LastActivityDate="2014-12-31T04:58:54.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6540" PostTypeId="2" ParentId="5261" CreationDate="2014-12-30T19:49:58.013" Score="0" Body="&lt;p&gt;For &lt;code&gt;The first night of autumn 80&lt;/code&gt;, I think it is &lt;code&gt;Kokomo by John Phillips&lt;/code&gt; because it was released in 1988, after his 1980 conviction for drugs.&lt;/p&gt;&#xA;" OwnerUserId="7565" LastActivityDate="2014-12-30T19:49:58.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6541" PostTypeId="2" ParentId="6531" CreationDate="2014-12-30T20:10:47.927" Score="4" Body="&lt;p&gt;&lt;em&gt;1. Crypto-Olaf (10, 7)&lt;/em&gt; - looks interesting, but the answer was a bit of a disappointment!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ABOMINABLE SNOWMAN, since Olaf is a Disney snowman and the Abominable Snowman is a cryptic. &lt;strong&gt;Credit to @xnor for getting this.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;2. The angels amid sightless ghoti; half a nokia appears unexpectedly (5, 3)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; FLASH MOB — &quot;The angels&quot; = Los Angeles = LA. &quot;Ghoti&quot; = &quot;fish&quot; (see clue 10), but sightless (with no &quot;i&quot;) = FSH. A Nokia is a mobile, so half a Nokia = MOB&#xA; &lt;strong&gt;From @squeamish ossifrage&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;3. Amphibian surmounts physicist (6)&lt;/em&gt; - this one is (still) brilliant.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; NEWTON / NEWT ON&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;4. Hilarious partner (3)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; WAG — The British press use the acronym WAGs to refer to the &quot;wives and girlfriends&quot; of professional footballers. The singular form is WAG. And of course wag also means &quot;witty person&quot;.&#xA; &lt;strong&gt;From @squeamish ossifrage&lt;/strong&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;5A. Spurn a guilty plea outside Aquidneck Island (8)&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DEFRIEND, credit &lt;strong&gt;@shoover&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;5D. Not a cat (3)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dog, see bottom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;6. Performing a supersonic number in the head of an adult insect. (9)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MACHINIMA, credit to &lt;strong&gt;@Quark&lt;/strong&gt; below.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;7. Sightless, tailless, expensive and suitable for general audiences (5)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BLING, credit to &lt;strong&gt;@QuyNguyen2013&lt;/strong&gt; for getting this one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;8. Not a dog&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cat, see bottom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;9. Blur's life, not your jumping around. (7)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PARKOUR, since Park Life is a song by Blur and parkour is jumping around - not &quot;your&quot; jumping around but &lt;em&gt;our&lt;/em&gt;s, as in park&lt;em&gt;our&lt;/em&gt;. &lt;strong&gt;Credit to @frodoskywalker for getting this.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;10. Ghoti (4)&lt;/em&gt; - nice one, but it's trivial if you've seen it before and pretty undoable otherwise.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; FISH. Pronounce 'gh' as in 'rough', 'o' as in 'women', and 'ti' as in 'location'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think it would've been better to do a new take on this, with slightly more of a hint as to what the idea is. E.g.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;A phonetic ghost (4)&quot; could mean the same: pronounce 's' as in 'sugar' and 't' as in 'castle'. Or &quot;A phonetic cibod (4)&quot; with 'ci' as in 'acacia', 'b' as in 'subtle', 'o' as in 'women', and 'd' as in 'grasped' - what word would that give? :-&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;11. Dennis Ritchie's creation doing half a dance outside a flat to check your humanity (7)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CAPTCHA: Dennis Ritchie's creation is C. Half a dance outside a flat is APT (apartment) CHA (from cha-cha). Checking your humanity is what CAPTCHAs do. &lt;strong&gt;Credit to Geobits&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;5D. Not a cat (3)&lt;br&gt;&#xA;8. Not a dog (3)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are a bit silly!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DOG and CAT respectively.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="4050" LastEditDate="2015-01-09T20:30:47.933" LastActivityDate="2015-01-09T20:30:47.933" CommentCount="12" CommunityOwnedDate="2015-01-03T16:56:08.160" ContentLicense="CC BY-SA 3.0" />
  <row Id="6542" PostTypeId="2" ParentId="6538" CreationDate="2014-12-30T20:40:55.287" Score="4" Body="&lt;p&gt;For the first part,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;for all Lambda unique e L for all unique r&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;can be rendered as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \forall \Lambda ! e \mathrm{L} \forall ! \mathrm{r} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which seems to spell &quot;Avigrail&quot; if you flip each letter.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2014-12-30T21:39:54.553" LastActivityDate="2014-12-30T21:39:54.553" CommentCount="8" CommunityOwnedDate="2014-12-30T20:40:55.287" ContentLicense="CC BY-SA 3.0" />
  <row Id="6544" PostTypeId="2" ParentId="6495" CreationDate="2014-12-30T22:40:50.187" Score="2" Body="&lt;p&gt;As long as people are getting into impossible cuts with a hacksaw, I'm surprised no one has offered this solution...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4xHpt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, it can't be done, maybe with patience with a jigsaw. Also, hole saws exist. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just me being pedantic, or the opposite, whichever.&lt;/p&gt;&#xA;" OwnerUserId="7589" LastEditorUserId="24713" LastEditDate="2016-09-05T12:37:23.627" LastActivityDate="2016-09-05T12:37:23.627" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6545" PostTypeId="1" CreationDate="2014-12-30T22:51:04.340" Score="7" ViewCount="1104" Body="&lt;p&gt;I'm not sure the title of this question is the best one - feel free to edit it - but I have a question on puzzles in general:&lt;/p&gt;&#xA;&lt;p&gt;Most puzzles I know (and surely most puzzles on this site) fall into the category of&lt;/p&gt;&#xA;&lt;h2&gt;rational puzzles&lt;/h2&gt;&#xA;&lt;p&gt;By this I mean, that one can use rational thinking and reasoning to find the answer.&lt;/p&gt;&#xA;&lt;p&gt;As such, mathematics can act as puzzle-language and is most often an excellent tool for solving them. Where analytical maths can't help, numerical maths aka brute-force 'simulation' via programming comes to the rescue.&lt;/p&gt;&#xA;&lt;p&gt;But there are surely also different categories of puzzles:&lt;br&gt;&#xA;One, I can think of, are&lt;/p&gt;&#xA;&lt;h2&gt;knowledge based puzzles&lt;/h2&gt;&#xA;&lt;p&gt;These puzzles require that the solver has a certain &lt;em&gt;specific&lt;/em&gt; knowledge, and only in this context the puzzle becomes solvable. Without that knowledge, the puzzle appears arbitrary and ill-defined.&lt;/p&gt;&#xA;&lt;p&gt;A lot of riddle-type puzzles with very specific, topological hints (i.e. certain music groups / actors /...) fall into this category. (&lt;em&gt;Google to the rescue!&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;p&gt;But if you think of it, also any &lt;em&gt;Chess-puzzle&lt;/em&gt; without specifying the rules of Chess would really be of this category. The same for most/all cryptographic puzzles. ( A Ceasar cipher is not &lt;em&gt;logical&lt;/em&gt; in any way! )&lt;/p&gt;&#xA;&lt;p&gt;Those two examples are generally &lt;em&gt;accepted&lt;/em&gt; on this site, because the required knowledge (such as Chess rules or how to use Internet searches) are &lt;em&gt;assumed&lt;/em&gt; to be &lt;strong&gt;general knowledge&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;But this category would also accommodate more obscure examples, and I wonder of the fate of such puzzles on this site.&lt;/p&gt;&#xA;&lt;p&gt;Example: Historic paintings. Those paintings often used symbols with very, very specific meaning which could &lt;em&gt;encode&lt;/em&gt; a message or puzzle very nicely. People &lt;em&gt;of that period and culture&lt;/em&gt; (or &lt;em&gt;historians&lt;/em&gt;) would see, understand and appreciated the puzzle - and they would firmly believe there is a &lt;em&gt;unique and clear&lt;/em&gt; solution to them.&lt;/p&gt;&#xA;&lt;p&gt;There are, in fact, a lot of different &lt;em&gt;expert-knowledge&lt;/em&gt; requiring puzzles, if you think of it.&lt;/p&gt;&#xA;&lt;p&gt;This now leads me to my actual question:&lt;/p&gt;&#xA;&lt;h2&gt;What other &lt;em&gt;general&lt;/em&gt; categories of puzzles do exist?&lt;/h2&gt;&#xA;&lt;p&gt;In particular:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Are there any puzzles which are not based on rational (left brain) thinking, but on creative / associative / emotional (right brain) thinking?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Can you give an example?&lt;/p&gt;&#xA;&lt;p&gt;If they exist, how can you &lt;em&gt;proof&lt;/em&gt; the solution is correct? What does &lt;em&gt;correct&lt;/em&gt; mean in this context?&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;If this all sounds a bit too philosophical, well maybe it is. &lt;br&gt; But it is a &lt;strong&gt;genuine question and interest&lt;/strong&gt; of mine, and I hope I will get some good, thought-provoking answers.&lt;/p&gt;&#xA;&lt;p&gt;The question somehow came to mind out of observation: I consider myself a &amp;quot;left-brain&amp;quot; thinker, i.e. when I see a problem, I &lt;em&gt;first&lt;/em&gt; try to apply some logic/rational thinking to it. And I enjoy it.&lt;br&gt;&#xA;But I know a lot of people who do neither enjoy, nor think about problems this way. But it's exactly this group of people, which in other types of &amp;quot;puzzles&amp;quot; (/riddles/associative guessing games...) quickly find answers/solutions I wouldn't have dreamt of.&lt;/p&gt;&#xA;&lt;p&gt;I was hence wondering, how one would best &lt;em&gt;build&lt;/em&gt; a puzzle customized for such people.&lt;/p&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2021-09-14T14:23:46.350" Title="General puzzle categories and: Are there any emotional puzzles?" Tags="|puzzle-creation|puzzle-theory|" AnswerCount="2" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6546" PostTypeId="5" CreationDate="2014-12-30T22:53:41.720" Score="0" Body="&lt;p&gt;This tag indicates questions about puzzles and the process of puzzle-solving or puzzle-building on an abstract, generalized level.&lt;/p&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="5840" LastEditDate="2014-12-30T23:12:51.817" LastActivityDate="2014-12-30T23:12:51.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6547" PostTypeId="4" CreationDate="2014-12-30T22:53:41.720" Score="0" Body="The science of puzzles." OwnerUserId="5840" LastEditorUserId="5840" LastEditDate="2014-12-31T00:09:48.807" LastActivityDate="2014-12-31T00:09:48.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6548" PostTypeId="1" AcceptedAnswerId="6549" CreationDate="2014-12-31T01:10:00.247" Score="-2" ViewCount="1584" Body="&lt;p&gt;He asks for a glass of water. The bartender pulls out a shotgun and points it at him. The man says, thank you and leaves. Why did he thank him??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.iriddler.com/riddles/puzzles/doors.htm&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7565" LastEditorUserId="7308" LastEditDate="2015-04-28T10:40:40.897" LastActivityDate="2015-04-28T10:40:40.897" Title="A man walks into a bar" Tags="|lateral-thinking|humor|" AnswerCount="2" CommentCount="2" ClosedDate="2014-12-31T10:41:18.400" ContentLicense="CC BY-SA 3.0" />
  <row Id="6549" PostTypeId="2" ParentId="6548" CreationDate="2014-12-31T01:14:48.490" Score="13" Body="&lt;p&gt;The man had &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the hiccups.&lt;br&gt;&#xA; Drinking water is one solution to cure hiccups, which is why the man asked for water.&lt;br&gt;&#xA; Another solution is to be scared, and that's what the bartender did by pointing the shotgun at him.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="5460" LastEditDate="2014-12-31T10:41:49.770" LastActivityDate="2014-12-31T10:41:49.770" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6550" PostTypeId="2" ParentId="6548" CreationDate="2014-12-31T01:42:55.783" Score="10" Body="&lt;p&gt;The man felt threatened by the bartender's sudden and unprovoked hostility. He decided the safest course of action would be to leave very politely to avoid furthering angering the deranged barkeeper.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2014-12-31T01:42:55.783" CommentCount="3" CommunityOwnedDate="2014-12-31T01:42:55.783" ContentLicense="CC BY-SA 3.0" />
  <row Id="6551" PostTypeId="1" CreationDate="2014-12-31T03:03:03.560" Score="3" ViewCount="7143" Body="&lt;p&gt;This &lt;a href=&quot;http://www.buzzfeed.com/kjh2110/10-logic-puzzles-you-wont-be-able-to-solve&quot; rel=&quot;nofollow noreferrer&quot;&gt;numeric hidden message&lt;/a&gt; was found on a seat at Rangers Ballpark in Arlington, TX, with an instruction to &quot;Solve the Riddle&quot;. What does it mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RkOFs.png&quot; alt=&quot;cypher&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7565" LastEditorUserId="4551" LastEditDate="2014-12-31T15:59:52.720" LastActivityDate="2014-12-31T18:19:57.590" Title="Texas Numeric Hidden Message" Tags="|cipher|" AnswerCount="1" CommentCount="4" FavoriteCount="0" ClosedDate="2014-12-31T22:45:56.983" ContentLicense="CC BY-SA 3.0" />
  <row Id="6552" PostTypeId="1" AcceptedAnswerId="6553" CreationDate="2014-12-31T03:55:24.920" Score="46" ViewCount="22742" Body="&lt;p&gt;I was working my way through some Knight and Knave Puzzles in &lt;a href=&quot;http://www.flipkart.com/discrete-mathematics-its-applications-combinatorics-graph-theory-english-7th/p/itmdwxxszhczjrnf&quot; rel=&quot;noreferrer&quot;&gt;Discrete Maths by Rosen&lt;/a&gt;, when I came across the following question:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Four friends have been identified as suspects for an unauthorized access into a computer system.&lt;/p&gt;&#xA;&lt;p&gt;They have made statements to the investigating authorities.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Alice said “Carlos did it.”&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;John said “I did not do it.”&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Carlos said “Diana did it.”&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Diana said “Carlos lied when he said that I did it.”&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;If the authorities also know that exactly one of the four suspects is telling the truth, who did it? Explain&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Book solution:&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;John did it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;My question:&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;Why is this so?&lt;/p&gt;&#xA;" OwnerUserId="7435" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2021-03-25T22:51:31.950" Title="Puzzle: Who hacked the computer?" Tags="|logical-deduction|liars|" AnswerCount="10" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6553" PostTypeId="2" ParentId="6552" CreationDate="2014-12-31T04:09:41.720" Score="48" Body="&lt;p&gt;There's a logical inconsistency for the other options. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we say that Alice is the one telling the truth, then Carlos is guilty, and he's lying about Diana, BUT that means Diana is telling the truth. We can only have one truth teller, so Alice must be lying. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a similar situation if we say John is telling the truth. Again, that leaves either Carlos or Diana also telling the truth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Carlos is telling the truth, then Diana is guilty. In this case John is also telling the truth. (I'm making the assumption that we can't have two guilty parties.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only Diana can be the one truth teller. In this case, Alice is lying about Carlos' guilt, Carlos is lying about Diana's guilt, and John is lying about his innocence. He did it. &lt;/p&gt;&#xA;" OwnerUserId="2162" LastActivityDate="2014-12-31T04:09:41.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6554" PostTypeId="2" ParentId="6552" CreationDate="2014-12-31T04:39:03.317" Score="6" Body="&lt;p&gt;Consider following cases:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Carlos is lying&lt;/strong&gt;&lt;BR&gt;&#xA;If Carlos is lying, Diana is telling truth. So Alice and John are also lying, which means John's statement &quot;I did not do it&quot; is a lie. So John did it. (Correct answer)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both Carlos and Diana can not tell a lie at the same time as Diana's statement says, &quot;Carlos is lying&quot; (Not possible)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Carlos is not lying&lt;/strong&gt;&lt;BR&gt;&#xA;If Carlos is speaking truth, John must be lying. Which means both John and Diana hacked the computer. Assuming there is just one culprit, this is not possible.&lt;/p&gt;&#xA;" OwnerUserId="4637" LastEditorUserId="4637" LastEditDate="2014-12-31T04:44:51.357" LastActivityDate="2014-12-31T04:44:51.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6555" PostTypeId="2" ParentId="6185" CreationDate="2014-12-31T05:49:06.267" Score="5" Body="&lt;p&gt;I would arrest the&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; blonde man&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because running&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Dzwgwosdvk&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;through a Caesar cipher +8 reveals that his coded name is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lheoewalds&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is an anagram of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; presidential assassin &quot;Lee H. Oswald&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;while for the other guy,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; his coded name is &quot;Stjuadedu&quot;, which is an anagram of &quot;Just A Dude&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5121" LastActivityDate="2014-12-31T05:49:06.267" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6556" PostTypeId="2" ParentId="6545" CreationDate="2014-12-31T07:53:19.730" Score="9" Body="&lt;p&gt;This is a very interesting question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www-sre.wu.ac.at/ersa/ersaconfs/ersa14/e140826aFinal01568.pdf&quot;&gt;http://www-sre.wu.ac.at/ersa/ersaconfs/ersa14/e140826aFinal01568.pdf&lt;/a&gt; is an example of serious research surrounding what the authors call &quot;relation logic&quot; or &quot;relational logic&quot; - see p. 15 or p. 29 for examples.  The idea is that a puzzle (which figure doesn't belong?) can be answered in different ways by different types of thinkers, and the paper draws a distinction between what you call left brain and right brain types as categorical vs. relational thinkers.  It also relates back to Nisbett's theories on &quot;geography of thought,&quot; suggesting that difference in ancient Greek vs. ancient Chinese logic have persisted through today creating Western vs. Eastern thinkers that somewhat align with your left and right brain ideas.  According to this theory, Westerners are supposed to be more categorical by thinking of items in terms of their individual attributes, while Easterners are supposed to think in terms of relationships between items placed together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether one puts a lot of stock in these ideas, they do suggest a possible way forward for puzzle design - if you're more of the left brain, Westerner-type, then you would do well to steep yourself in puzzles of &quot;Eastern&quot; origin.  This seems to me to be the idea of fuzzing one's eyes to look at the whole - don't focus on individual pieces and try to get them to fit together, but look at clues as a whole as see if they remind you of any systems or relationships (familial, predator-prey, religious).  For instance, a man with 12 followers, or a man, woman, and child, or a hammer and nail evoke certain contexts on their own, regardless of the details of the individuals or things involved.  Correctness comes from the solver repeating the same relationship that was originally intended, or at least that can answer all the available hints in the puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you're right that uniqueness of the solution becomes harder to pin down in these cases - the paper I mention actually uses disagreement about the correct solution to a particular problem as its main measure of types of logic employed.  Perhaps working backwards from the solution will ensure that all clues foreshadow correctly, or certain clues that are so exacting as to rule out close misses by narrowing down the possible relationships between items?  Nonetheless, there are many examples on this site of multiple answers to a single riddle where some solutions fit closely but not exactly with the writer's intended solution.  Experience seems a valuable guide in pinning down what uniquely defines a relationship to different people.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope there's something of use in this long response for you.  A very interesting topic to think about, and I'd love to hear more of your own thoughts on the matter.&lt;/p&gt;&#xA;" OwnerUserId="5121" LastActivityDate="2014-12-31T07:53:19.730" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6557" PostTypeId="1" AcceptedAnswerId="6561" CreationDate="2014-12-31T11:09:13.897" Score="11" ViewCount="672" Body="&lt;p&gt;Inside a rectangular room, measuring 16 feet in length and 8 feet in width and height, a spider resides at a corner. A fly buzzing in the room intends to land at a spot that will take the spider the longest to reach, knowing that the spider never drops or uses its web, but crawls at constant speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where should the fly land?&lt;/p&gt;&#xA;" OwnerUserId="7573" LastActivityDate="2014-12-31T14:12:11.537" Title="A crawling spider and a cautious fly" Tags="|mathematics|geometry|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6559" PostTypeId="2" ParentId="6557" CreationDate="2014-12-31T11:35:46.850" Score="2" Body="&lt;p&gt;Intuitively, it seems the fly should land in the opposite corner.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If we unfold the room along the edge between short wall and roof, the spider travels 8 feet across and 24 feet along, for a distance $\sqrt{24^2+8^2}$ feet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, we can improve things for the spider by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Unfolding the room along the join between the long wall and the roof gives us a distance of $\sqrt{16^2+16^2}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Prompted by Johannes' comment that this can be beaten:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As shown below, if the fly moves some distance along the short wall it simultaneously improves the worse solution given and lengthens the improved solution. When they are equal the distance should be maximised.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qSZYz.jpg&quot; alt=&quot;diagram of room unfolded in both directions&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The distances are&#xA;$d_1=\sqrt{16^2+(16+x)^2}$,&#xA;$d_2=\sqrt{24^2+(8-x)^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which are equal to $\sqrt{24^2+(16/3)^2} \approx 24.585$&#xA;at x=8/3&lt;/p&gt;&#xA;" OwnerUserId="3987" LastEditorUserId="3987" LastEditDate="2014-12-31T13:42:26.087" LastActivityDate="2014-12-31T13:42:26.087" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="6561" PostTypeId="2" ParentId="6557" CreationDate="2014-12-31T12:09:04.807" Score="5" Body="&lt;p&gt;The optimal distance seems to be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$\sqrt{14^2+18^2} = \sqrt{22^2+6^2} \approx 22.804$&lt;br&gt;&#xA;which is greater than &lt;br&gt;&#xA;$\sqrt{16^2+16^2} \approx 22.627$&lt;br&gt;&#xA;If the fly stayed at the opposite corner.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assuming the spider is on the roof corner, be in the opposite corner (on the floor), then move upwards 2 feet and sideways 2 feet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key to this puzzle is that there are 4 distances to be optimized.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jw7LL.png&quot; alt=&quot;Spider1&quot;&gt;&#xA;  &lt;img src=&quot;https://i.stack.imgur.com/ILYhH.png&quot; alt=&quot;Spider2&quot;&gt;&#xA;  &lt;img src=&quot;https://i.stack.imgur.com/qr77K.png&quot; alt=&quot;Spider3&quot;&gt;&#xA;  &lt;img src=&quot;https://i.stack.imgur.com/kyEtc.png&quot; alt=&quot;Spider4&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6178" LastEditorUserId="6178" LastEditDate="2014-12-31T14:12:11.537" LastActivityDate="2014-12-31T14:12:11.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6562" PostTypeId="1" CreationDate="2014-12-31T12:59:55.717" Score="4" ViewCount="1487" Body="&lt;p&gt;I got this puzzle from my grandpa who was developing a cipher system; this is his first creation and his first example. He doesn't remember the plaintext, so I want to crack it, but I suck so I leave it for you guys/girls...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Cipher text: 11001000000101010110010111110&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The key:&#xA;  111000111011001001101010001010101111011000000110010011100000100001010101111100111001111010111111001001011111101&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a classic cipher text with XOR, my grandpa said to me 5 seconds ago...&#xA;I am pretty sure that the plaintext was something important because they later used this (improved) version of the cipher and encrypted the location for money but he already found THIS one is harder...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for trying.&lt;/p&gt;&#xA;" OwnerUserId="7607" LastEditorUserId="4050" LastEditDate="2014-12-31T15:17:46.960" LastActivityDate="2015-04-27T08:10:23.963" Title="A binary puzzle from 1950" Tags="|cipher|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6564" PostTypeId="1" AcceptedAnswerId="6565" CreationDate="2014-12-31T15:38:03.847" Score="1" ViewCount="4645" Body="&lt;p&gt;I was working my way through some puzzles in &lt;a href=&quot;http://www.flipkart.com/discrete-mathematics-its-applications-combinatorics-graph-theory-english-7th/p/itmdwxxszhczjrnf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Discrete Maths by Rosen&lt;/a&gt;, when I came across the following question:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Freedonia has fifty senators. Each senator is either honest or corrupt.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Suppose you knowthat at least one of the Freedonian senators is honest and that, given any two Freedonian senators, at least one is corrupt.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Based on these facts, can you determine how many Freedonian senators are honest and how many are corrupt? If so, what is the answer?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;My solution:&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Let the senators be numbered from 1 to 50 ( in order ) say S1 , S2 ,... , S50&lt;br/&gt;&lt;/li&gt;&#xA;&lt;li&gt;Now taking pairwise senators on at a time : (S1,S2) ; (S2,S3)  ; ... ; (S49,S50)&lt;br/&gt;&lt;/li&gt;&#xA;&lt;li&gt;Without Loss of Generality assume that the Liar Senator is the first one in each pair &lt;span class=&quot;math-container&quot;&gt;$\Rightarrow$&lt;/span&gt; S1,S2,..,S49&lt;br/&gt;&lt;/li&gt;&#xA;&lt;li&gt;Now , the only Senator left is S50 , who is Honest and also satisfies the constraints of the question&lt;/li&gt;&#xA;&lt;li&gt;Hence , there are 49 corrupt senators and 1 honest senator&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;My question:&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;Am I right ?&lt;/p&gt;&#xA;" OwnerUserId="7435" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-02-02T02:09:19.017" Title="Puzzle : Who is Corrupt in the Senate?" Tags="|logical-deduction|liars|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6565" PostTypeId="2" ParentId="6564" CreationDate="2014-12-31T15:41:36.883" Score="9" Body="&lt;p&gt;A simpler proof:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Assume there's more than one honest senator, and pick two. Among those two, neither is corrupt. Contradiction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Having chosen the pairs (S1,S2), (S2,S3), etc., you can't make a sweeping WLOG assumption about all of them together, since they depend on each other (e.g. the second pair is related to the first via S2). Also in your answer you've only used the fact that in any &lt;em&gt;consecutive&lt;/em&gt; pair of senators, at least one is corrupt. This isn't enough, since S1, S3, S5, ... could be corrupt and S2, S4, S6, ... honest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short: &lt;strong&gt;your answer is right, but your argument is wrong&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2014-12-31T16:21:57.380" LastActivityDate="2014-12-31T16:21:57.380" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6566" PostTypeId="2" ParentId="6551" CreationDate="2014-12-31T15:58:02.323" Score="3" Body="&lt;p&gt;According to the Texas Rangers &lt;a href=&quot;http://www.reddit.com/r/TexasRangers/comments/1hl06o/can_anyone_help_with_this_xpost_from_rpuzzles/&quot; rel=&quot;nofollow&quot;&gt;reddit thread&lt;/a&gt;, the numbers correspond to &lt;a href=&quot;http://www.reddit.com/r/TexasRangers/comments/1hl06o/can_anyone_help_with_this_xpost_from_rpuzzles/cavq6k0&quot; rel=&quot;nofollow&quot;&gt;bits of trivia related to the Texas Rangers&lt;/a&gt;. The &lt;a href=&quot;http://www.reddit.com/r/TexasRangers/comments/1hl06o/can_anyone_help_with_this_xpost_from_rpuzzles/cawkigf&quot; rel=&quot;nofollow&quot;&gt;owner (it is claimed) confirmed four of the seven sets given&lt;/a&gt;, which at least also confirms the mechanism.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2014-12-31T15:58:02.323" CommentCount="2" CommunityOwnedDate="2014-12-31T15:58:02.323" ContentLicense="CC BY-SA 3.0" />
  <row Id="6567" PostTypeId="2" ParentId="6552" CreationDate="2014-12-31T16:25:39.113" Score="45" Body="&lt;p&gt;The easiest way to approach the problem is to identify mutually contradictory statements, which off course are&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Carlos said “Diana did it.” &lt;/li&gt;&#xA;&lt;li&gt;Diana said “Carlos lied when he said that I did it.”&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and to ascertain these statements is to realize that one of them is telling the truth. So this would also mean the rest of the two statements are false&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Alice said “Carlos did it.” &lt;/li&gt;&#xA;&lt;li&gt;John said “I did not do it.”&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;From which we can easily conclude that the statement &quot;John said 'I did not do it.'&quot; is false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So John is the hacker.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4177" LastEditorUserId="39" LastEditDate="2015-02-22T13:52:29.613" LastActivityDate="2015-02-22T13:52:29.613" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6568" PostTypeId="2" ParentId="6564" CreationDate="2014-12-31T16:48:04.733" Score="4" Body="&lt;p&gt;The clue to the question is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;given any two Freedonian senators, at least one is corrupt.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So there are no two senators where both of them are honest. So either there is one senerator who is honest or none. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we already know &lt;code&gt;at least one of the Freedonian senators is honest&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can safely conclude there is &lt;strong&gt;&lt;em&gt;one and only one honest senator&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4177" LastActivityDate="2014-12-31T16:48:04.733" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6569" PostTypeId="1" AcceptedAnswerId="6571" CreationDate="2014-12-31T17:46:31.963" Score="52" ViewCount="36064" Body="&lt;p&gt;At a party last night, my friends Alice and Bob did a magic trick. Any ideas how it worked?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alice shuffled a pack of cards, and asked me to take five. I looked at them. She put the rest of the pack down on the table. Alice asked for my cards. She gave four of them to Bob (he was across the table), and the fifth back to me. Bob looked at the four cards for a while. Then Bob looked at me, and named the card I was holding. He was right. I'm quite sure he couldn't have seen it (we weren't sitting by a mirror). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They did the trick again later to someone else. I watched for funny business. Alice didn't say anything to Bob, so I don't think they have a code. Also Alice is famously clumsy, so I doubt it was sleight of hand.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit to answer a question: I learnt the trick from a maths magazine several years ago. I don't know who invented it. &quot;Michael Kleber. The best card trick. Mathematical Intelligencer 24 #1 (Winter 2002)&quot;&lt;/p&gt;&#xA;" OwnerUserId="4508" LastEditorUserId="4508" LastEditDate="2015-01-02T15:55:58.407" LastActivityDate="2022-09-15T18:05:16.097" Title="A five card trick - How does it work?" Tags="|cards|magic|" AnswerCount="5" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6571" PostTypeId="2" ParentId="6569" CreationDate="2014-12-31T19:11:01.130" Score="40" Body="&lt;p&gt;I'm going to assume Alice looked at the cards and chose which one to give back to you. The key to the puzzle is then to encode a single card's suit and value in 4 cards without the luxury of choosing those 4 cards arbitrarily from the whole deck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The suit is easy. In 5 cards there must be a double of at least one suit. So the first (or last, but I'll choose arbitrarily) card in the bunch she passes is the same suit as yours. Now there are three cards left to encode a number from 1 to 13. However Alice chose which card of your suit to pass to Bob and which to return to you. She can choose according to a rule that gets the number of possible cards down significantly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The three passed cards can be designated small medium and large according to their number, and then breaking ties by suit order (clubs smallest, diamonds, hearts, spades as in bridge.) This gives six possible numbers to be represented by the 3 passed cards based on their order:  SML, SLM, MLS, MSL, LSM, LMS. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how does she choose which of the suit cards to pass and which to return? Bob will add the encoded number to the passed card (going around K-A-2 if need be) to get the returned card. Alice passes whichever card is within an add of 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say you have the 2 and 4 of spades in your 5. She can't pass the 4 because no number between 1 and 6, added to 4, will wrap around to the 2. So she passes the 2 and encodes 2 in the other three cards by ordering them SLM. If you have the Q and K she passes the Q and encodes 1. If you have the 7 and K she passes the 7 and encodes 6. But 6 and K, she passes the K and encodes 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried to find a set of cards you could choose, knowing this algorithm, that would make it impossible to perform the trick, and I can't. Not from a proper deck that doesn't have any duplicate cards.&lt;/p&gt;&#xA;" OwnerUserId="1509" LastEditorUserId="1509" LastEditDate="2015-02-20T11:16:06.040" LastActivityDate="2015-02-20T11:16:06.040" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6572" PostTypeId="2" ParentId="6569" CreationDate="2014-12-31T19:11:08.107" Score="13" Body="&lt;p&gt;This is a great card trick known as Fitch Cheney's Five Card trick. Alice selects the fifth card to be identified and arranges the remaining four cards in a specific sequence.  Then Bob interprets the four card sequence and identifies the fifth card.&lt;/p&gt;&#xA;&lt;p&gt;Here are a few references that describe how to sequence the four cards:&lt;br /&gt;&#xA;&lt;a href=&quot;https://threesixty360.wordpress.com/2008/08/04/math-card-tricks-i-know-the-5th-card/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Batman on wordpress&lt;/a&gt;&lt;br /&gt;&#xA;&lt;a href=&quot;https://headinside.blogspot.com/2008/08/fitch-cheney-card-trick.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gray Matters on blogspot&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;And here is Bob and Alice doing the trick &lt;a href=&quot;https://web.archive.org/web/20170915180434/http://mathforum.org/wagon/fall97/p842.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;again&lt;/a&gt; ;-)&lt;/p&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="11110" LastEditDate="2022-09-15T18:05:16.097" LastActivityDate="2022-09-15T18:05:16.097" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6573" PostTypeId="2" ParentId="6569" CreationDate="2014-12-31T19:23:29.260" Score="17" Body="&lt;p&gt;Here is my answer:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Key Item 1: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In a deck there are 52 cards of which 4 are with Bob. So Bob needs to determine the unknown from 48 cards.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Key Item 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Given 4 ordered cards, the number of ways to arrange it in ascending order is 4! = 24&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;Key Item 3:&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The hand can have two orientations, upright and reversed. So no of arrangements = 2*4! = 48&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;Key Item 4:&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let Alice encode the remaining 48 cards following a strict ordering of suits &lt;code&gt;♣ &amp;lt; ♦ &amp;lt; ♥ &amp;lt; ♠&lt;/code&gt; and number &lt;code&gt;Ace, Two ..,Ten, Jack, Queen, King&lt;/code&gt; and taking into consideration of the missing 4 cards&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;Key Item 5:&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bob, aware of Alice's strategy, decodes the arrangement of the cards back to the index of the cards strictly ordered based on suits and numbers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4177" LastEditorUserId="46903" LastEditDate="2018-09-08T23:03:25.940" LastActivityDate="2018-09-08T23:03:25.940" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="6574" PostTypeId="2" ParentId="6196" CreationDate="2014-12-31T19:36:11.943" Score="2" Body="&lt;p&gt;I think the solution is using a set of characters or figures that are unknown to the typical brute-force dictionary. Take a look at this question: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/6404/can-you-decipher-the-hidden-message-in-this-christmas-card-image&quot;&gt;Can you decipher the hidden message in this Christmas card image?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully this helps.&lt;/p&gt;&#xA;" OwnerUserId="7565" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2014-12-31T19:36:11.943" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6575" PostTypeId="2" ParentId="6552" CreationDate="2014-12-31T21:02:55.860" Score="2" Body="&lt;p&gt;Another solution, similar to Abhijit's:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;John and Diana both say they did not do it, but only one can be right, so it has to be either of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That implies that one of them has to be right, and everybody else has to be wrong. That means Carlos must be wrong when he says Diana did it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it's John.&lt;/p&gt;&#xA;" OwnerUserId="7618" LastActivityDate="2014-12-31T21:02:55.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6576" PostTypeId="2" ParentId="6552" CreationDate="2014-12-31T21:58:57.827" Score="4" Body="&lt;p&gt;Here's how I approached it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 1.&lt;/strong&gt; If Alice's or Carlos's statement is true, then John's is also true.&#xA;This eliminates Carlos &amp;amp; Diana as suspects, leaving Alice &amp;amp; John as possibilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 2.&lt;/strong&gt; If Alice did it, then John and Diana are both telling the truth.&#xA;This eliminates Alice as a suspect, leaving John.&lt;/p&gt;&#xA;" OwnerUserId="381" LastActivityDate="2014-12-31T21:58:57.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6577" PostTypeId="2" ParentId="6552" CreationDate="2015-01-01T02:15:00.730" Score="21" Body="&lt;p&gt;Solving by machine (this is overkill but fun, and as I am on a Prolog refresher...)&lt;/p&gt;&#xA;&lt;p&gt;The code can also be found &lt;a href=&quot;https://github.com/dtonhofer/prolog_notes/blob/master/prolog_exercises/puzzles/who_is_the_hacker.pl&quot; rel=&quot;nofollow noreferrer&quot;&gt;at github&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Run it on your machine if you have a Prolog or online in &lt;a href=&quot;https://swish.swi-prolog.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;SWISH which is based on SWI-Prolog&lt;/a&gt;&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;% https://puzzling.stackexchange.com/questions/6552/puzzle-who-hacked-the-computer&#xA;%&#xA;% ===&#xA;% Four friends have been identified as suspects for an unauthorized &#xA;% access into a computer system.&#xA;%&#xA;% They have made statements to the investigating authorities.&#xA;%&#xA;%    Alice said &amp;quot;Carlos did it&amp;quot;.&#xA;%    John said &amp;quot;I did not do it&amp;quot;.&#xA;%    Carlos said &amp;quot;Diana did it&amp;quot;.&#xA;%    Diana said &amp;quot;Carlos lied when he said that I did it&amp;quot;.&#xA;%&#xA;% If the authorities also know that exactly one of the four suspects is&#xA;% telling the truth, who did it? Explain&#xA;% ===&#xA;&#xA;&#xA;% Define the persons&#xA;&#xA;persons([alice,john,carlos,diana]).&#xA;&#xA;% Define the statements of the persons&#xA;&#xA;says(alice  , hacker(carlos)).&#xA;says(john   , not(hacker(john))).&#xA;says(carlos , hacker(diana)).&#xA;says(diana  , lies(carlos)).&#xA;&#xA;% Main predicate. We will call this later&#xA;&#xA;solve(Hacker) :-&#xA;    selectTruthteller(TT),&#xA;    buildWorld(TT,World),&#xA;    simplifyWorld(World,SW),          % Run one simplification &#xA;                                      % (this turns out to be enough, &#xA;                                      % no second round needed)&#xA;    sort(SW,SimpleWorld),             % Sort and remove duplicates from&#xA;                                      % SW, yielding SimpleWorld&#xA;    writeln(SimpleWorld),             % Write SimpleWorld&#xA;    isConsistent(SimpleWorld,Hacker). % Check that SimpleWorld is&#xA;                                      % consistent; if yes, the&#xA;                                      % program succeeds and prints the&#xA;                                      % &amp;quot;Hacker&amp;quot;&#xA;&#xA;% Select a person as &amp;quot;truthteller&amp;quot;. Assuming that person is the&#xA;% truthteller, logical consequences will be checked.&#xA;&#xA;selectTruthteller(TT) :-&#xA;   persons(Persons),&#xA;   member(TT,Persons).&#xA;&#xA;% buildWorld(+Truthteller,-WorldOut).&#xA;% Generate &amp;quot;WorldOut&amp;quot; (which is a list of statements) under assumption&#xA;% that &amp;quot;Truthteller&amp;quot; is the truthteller.&#xA;% We need to iterate over the &amp;quot;Persons&amp;quot;, so we need to have a second&#xA;% buildWorld/3 predicate in addition to buildWorld/2.&#xA;&#xA;buildWorld(TT,WorldOut) :-&#xA;   persons(Persons),&#xA;   buildWorld(TT,WorldOut,Persons).&#xA;&#xA;% buildWorld(+Truthteller,-WorldOut, +ListOfPersons).&#xA;%&#xA;% If there are no persons left in the &amp;quot;ListOfPersons&amp;quot; we are done and &#xA;% the &amp;quot;WorldOut&amp;quot; is empty.&#xA;%&#xA;% If the next person in the &amp;quot;ListOfPersons&amp;quot; is the truthteller, we add&#xA;% the statement that he/she &amp;quot;truthes&amp;quot; as well as his/her&#xA;% statement as gospel truth to the &amp;quot;WorldOut&amp;quot;.&#xA;%&#xA;% If the next person in the &amp;quot;ListOfPersons&amp;quot; is NOT the truthteller,&#xA;% we add the statement that he/she &amp;quot;lies&amp;quot; as well as his/her&#xA;% statement as gospel falsity to the &amp;quot;WorldOut&amp;quot;.&#xA;&#xA;buildWorld(_,[],[]).&#xA;buildWorld(TT,[truthes(TT),Stmt|WorldOutRest],[TT|PersonsRest]) :-&#xA;   says(TT, Stmt),&#xA;   buildWorld(TT,WorldOutRest,PersonsRest).&#xA;buildWorld(TT,[lies(P),not(Stmt)|WorldOutRest],[P|PersonsRest]) :-&#xA;   P \== TT,&#xA;   says(P, Stmt),&#xA;   buildWorld(TT,WorldOutRest,PersonsRest).&#xA;&#xA;% Go through the world statements and simplify every statement in turn;&#xA;% the second line drops a statement if simplifyStmt/2 results in the &#xA;% atom &amp;quot;null&amp;quot;&#xA;&#xA;simplifyWorld([],[]).&#xA;simplifyWorld([In|InRest],OutRest) :-&#xA;   simplifyStmt(In,null),&#xA;   simplifyWorld(InRest,OutRest).&#xA;simplifyWorld([In|InRest],[Out|OutRest]) :-&#xA;   simplifyStmt(In,Out),&#xA;   Out \== null,&#xA;   simplifyWorld(InRest,OutRest).&#xA;&#xA;% Simplify a single statement by eliminating &amp;quot;not&amp;quot;; use red cuts to&#xA;% simplify code&#xA;&#xA;simplifyStmt(not(not(S)),S)           :- !.&#xA;simplifyStmt(not(lies(S)),truthes(S)) :- !.&#xA;simplifyStmt(not(hacker(_)),null)     :- !.&#xA;simplifyStmt(S,S).&#xA;&#xA;% Check world for consistency&#xA;&#xA;isConsistent(World,Hacker) :- &#xA;   persons(Persons),&#xA;   everyPersonLiesOrTruthes(World,Persons),&#xA;   \+ someoneLiesAndTruthes(World),&#xA;   \+ thereIsMoreThanOneHacker(World),&#xA;   thereIsAHacker(World,Hacker).&#xA;&#xA;% Helper predicates to check world for consistency&#xA;&#xA;everyPersonLiesOrTruthes(_,[]).&#xA;everyPersonLiesOrTruthes(World,[P|Persons]) :- &#xA;   member(lies(P),World),&#xA;   everyPersonLiesOrTruthes(World,Persons).&#xA;everyPersonLiesOrTruthes(World,[P|Persons]) :-&#xA;   member(truthes(P),World),&#xA;   everyPersonLiesOrTruthes(World,Persons).&#xA;&#xA;someoneLiesAndTruthes(World) :-&#xA;   member(lies(P),World),&#xA;   member(truthes(P),World).&#xA;&#xA;thereIsMoreThanOneHacker(World) :-&#xA;   member(hacker(P1),World),&#xA;   member(hacker(P2),World),&#xA;   P1 \== P2.&#xA;&#xA;thereIsAHacker(World,P) :-&#xA;   member(hacker(P),World).&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;After starting a Prolog interpreter and loading the code from file &lt;code&gt;who_is_the_hacker.pl&lt;/code&gt; (for example), the moment of truth. This prints the possible worlds and stops once a consistent world has been found&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;?- [who_is_the_hacker].&#xA;true.&#xA;&#xA;?- solve(X).&#xA;[hacker(carlos),hacker(john),lies(carlos),lies(diana),lies(john),truthes(alice),truthes(carlos)]&#xA;[lies(alice),lies(carlos),lies(diana),truthes(carlos),truthes(john)]&#xA;[hacker(diana),hacker(john),lies(alice),lies(diana),lies(john),truthes(carlos)]&#xA;[hacker(john),lies(alice),lies(carlos),lies(john),truthes(diana)]&#xA;X = john ;  % john is the hacker but maybe there are more solutions&#xA;false.      % no, there is only one solution&#xA;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;In case one is using SWISH, this can be seen:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TBwYi.png&quot; alt=&quot;Sample output&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="7625" LastEditorUserId="7625" LastEditDate="2021-03-25T22:51:31.950" LastActivityDate="2021-03-25T22:51:31.950" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="6578" PostTypeId="2" ParentId="6524" CreationDate="2015-01-01T04:04:11.627" Score="6" Body="&lt;p&gt;Lot's of fun!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;clearly, the unifying theme is &lt;strong&gt;Halley's comet&lt;/strong&gt;... I could not interpret &quot;dirty snowball&quot; any other way that was reasonable. It fit with &quot;arrived and departed with a dirty snowball&quot; =&gt; meaning coming and going of Halley's comet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;strong&gt;who am I&lt;/strong&gt;&quot; refers to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;people, events, places, etc. associated with Halley's passages at different times:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so, surely &quot;arrived and departed with a dirty snowball&quot; refers to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Mark Twain&lt;/strong&gt; who was born and died with the passage of Halley's comet. &quot;I came in with Halley's comet in 1835. It's coming again next year (1910), and I expect to go out with it. The Almighty has said no doubt, 'Now here are these two unaccountable freaks; they came in together, they must go out together.'&quot;(Mark Twain)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once I got that far, I was pretty sure I was in the ballpark. Then, the rest was looking for anything to confirm or rule out the theme&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;people, events, places, etc. associated with the passages at different times:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My famous works, as split, anticipate the arrival of you, seen with half a drum, and &#xA;of the French fabric for a Nordic funeral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;French fabric&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This refers to the &lt;strong&gt;Bayeux Tapestry&lt;/strong&gt;... &lt;strong&gt;Halley's most famous appearance&lt;/strong&gt; occurred shortly before the 1066 invasion of England by William the Conquerer. It is said that William felt the comet heralded his success. In any case, the comet was put on the Bayeux Tapestry – which chronicles the invasion – in William's honor h##p://www.space.com/19878-halleys-comet.html). and &quot;We know &lt;strong&gt;Comet Halley&lt;/strong&gt; was around in 1066 because it &lt;strong&gt;appears on the famous Bayeux Tapestry&lt;/strong&gt; depicting the Battle of Hastings along with the caption 'Isti Mirant Stella' (they marvel at the star). During the year 1144, Cardinal Gherardo Caccianemici dal Orso was elected Pope Lucius II, but he was killed after less than a year in office, coinciding with the comet's visit in 1145. Meanwhile, an English monk, Eadwine of Canterbury, must have taken time out from the transcribing he was working on because he drew an accurate portrayal of the comet on his Psalms manuscript.&quot; (h##p://h2g2.com approved entry A60281831)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;'&lt;strong&gt;split works&lt;/strong&gt;' refers to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the type of stitching. &quot;The skills of the embroiderers were, according to Mrs Makin, on a par with professional embroiderers of today. The embroidered cloth was not woven and so is not technically a tapestry. The couch, stem, &lt;strong&gt;split&lt;/strong&gt; and chain &lt;strong&gt;stitches found in the tapestry&lt;/strong&gt;...&quot; (h##p://www. medievalists. net/2012/11/15/new research on how the bayeux tapestry was made/). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Nordic Funeral&lt;/strong&gt; refers again to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the tapestry: 1) &quot;Was the bayeux tapestry created for a viking burial?&quot; (h##p:// answers wikia .com/wiki/Was the &lt;strong&gt;bayeux tapestry&lt;/strong&gt; created for a &lt;strong&gt;viking_burial&lt;/strong&gt;). also see &quot;But Marit Monsen Wang has interpreted the symbol of the &lt;strong&gt;portal under which Aelfgyva stands in the Tapestry&lt;/strong&gt; as reflective of an ancient &lt;strong&gt;Nordic concept&lt;/strong&gt;: the &lt;strong&gt;portal between this mortal life and that of the other-world&lt;/strong&gt; as described in the Scandinavian Eddas. Monsen Wang also traces this theme in the Risala, the description of an encounter between an Arab traveler and the Russian Vikings in the 10th century.  She refers specifically to the funeral rituals of the Viking king described in detail in the tale.&quot; (pg. 39: h##p://peregrinations.kenyon.edu/vol2_3/current/fb3.pdf). and see this footnote &quot;Customary funeral rights might infer a Christian burial, however subsequent evidence in this and other texts infer a Viking funeral under a pile of stones on a headland. - See The Funeral of Harold in the Carmen page xliii The Carmen of Hastingae Proelio Edited by Catherine Morton and Hope Muntz 1972 (Oxford medieval Texts).&quot; from &quot;h##p://www.secretsofthenormaninvasion.com/footnote.htm&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I believe '&lt;strong&gt;half a drum&lt;/strong&gt;; refers to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;a looking glass or telescope.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I grew up where Mischa's brother ate liver with bitter wine: a yellow card offence, I hear:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; this refers to the &lt;strong&gt;1910 sighting in Russia&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;bitter wine&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Vodka (&lt;a href=&quot;http://www.macalester.edu/academics/russian/about/resources/miscellany/vodka/&quot;&gt;http://www.macalester.edu/academics/russian/about/resources/miscellany/vodka/&lt;/a&gt;). confirms the location&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Yellow card offense&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;merely confirms that the location is Russia (&lt;strong&gt;Crime and Punishment&lt;/strong&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mischa's brother&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;alternate Russian spelling for &lt;strong&gt;Misha&lt;/strong&gt;. Russian classic: &lt;strong&gt;The brothers Karamazov&lt;/strong&gt;: a novel in four parts and an epilogue by Fyodor &lt;strong&gt;Dostoyevsky&lt;/strong&gt; (&quot;...And remember, Misha, if you are called &lt;strong&gt;Misha&lt;/strong&gt;. . . . His name is Misha, isn't it ? &quot; He turned to Pyotr Ilyitch again...&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I lived as a cub where the mounds are built; where an Elk traverses cedars, with an weary old man and a big greasy queen. Not as sure about this, but I believe:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cedars&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; gives a location and then time: referring to &lt;strong&gt;Cedars of Lebanon&lt;/strong&gt;: &quot;Matthew's description of the miracles and portents attending the &lt;strong&gt;birth of Jesus&lt;/strong&gt; can be compared to stories concerning the birth of Augustus (63 BC).[nb 4] Linking a birth to the first appearance of a star was consistent with a popular belief that each person's life was linked to a particular star.[32] &lt;strong&gt;Magi and astronomical events were linked in the public mind by the visit to Rome of a delegation of magi at the time of a spectacular appearance of Halley's Comet in AD 66&lt;/strong&gt;, about the time the Gospel of Matthew was being composed...Other writers suggest that the star was a comet.[48] Halley's Comet was visible in 12 BC and another object, possibly a comet or nova, was seen by Chinese and Korean stargazers in about 5 BC. This object was observed for over seventy days with no movement recorded.[48] Ancient writers described comets as &quot;hanging over&quot; specific cities, just as the &lt;strong&gt;Star of Bethlehem&lt;/strong&gt; was said to have &quot;stood over&quot; the &quot;place&quot; where Jesus was (the town of Bethlehem)...&quot; (h##p: //en. wikipedia. org /wiki/Star of Bethlehem). The last is disproved but is linked to the theme via myth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;mounds&lt;/strong&gt; and more on cedars:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;*Cedars of God** (Arabic: أرز الربّ‎ Horsh Arz el-Rab &quot;Cedars of the Lord&quot;) is one of the last vestiges of the extensive forests of the &lt;strong&gt;Cedars of Lebanon&lt;/strong&gt; (Cedrus libani ) that thrived across &lt;strong&gt;Mount Lebanon in ancient times&lt;/strong&gt;. Their timber was exploited by the Phoenicians, the Assyrians, Babylonians and Persians. The wood was prized by Egyptians for shipbuilding; the Ottoman Empire also used the cedars in railway construction&quot;(h##p://en wikipedia. org /wiki/ Cedars of God)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;weary old man&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;SADDAY, 'He of the Mountains,' though a more obviously secondary appellation, belongs semantically with Sûr.&quot; -- Albright, &lt;strong&gt;Yahweh&lt;/strong&gt; and the Gods of Canaan, p. 188... &quot;It was &lt;strong&gt;Moses&lt;/strong&gt;' acceptance of the old faith in the God of the Fathers that led him to identify Yahweh with the 'God of the Fathers' as well as with Shaddai.&quot; -- Ibid., p. 168&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7569" LastActivityDate="2015-01-01T04:04:11.627" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6579" PostTypeId="1" AcceptedAnswerId="6667" CreationDate="2015-01-01T04:55:01.597" Score="5" ViewCount="5717" Body="&lt;p&gt;I was working my way through some puzzles , when I came across the following question:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;An ancient Sicilian legend says that the barber in a remote town who can be reached only by traveling a dangerous mountain road&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The barber shaves those people, and only those people, who do not shave themselves.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can there be such a barber?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;My solution:&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;If the barber is a Female - then YES such a barber exists &lt;br/&gt;&#xA;For the sake of simplicity , let us assume that the Barber is a Male and that all Males have a Fuzz / Facial Hair &lt;br/&gt;&#xA;Let us assume that such a Male Barber exists &lt;br/&gt;&#xA;If the barber shaves himself then he infringes the rule that he shaves only those who do not shave themselves &lt;br/&gt;&#xA;If the barber does not shave himself then the rules say that he must shave himself&lt;br/&gt;&#xA;This is a Paradox and hence no Male Barber Exists !&lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;My question:&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;Am I correct ?&lt;/p&gt;&#xA;" OwnerUserId="7435" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-01-24T08:04:55.017" Title="Puzzle : O' Barber , Who Art Thou?" Tags="|logical-deduction|riddle|paradox|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6580" PostTypeId="5" CreationDate="2015-01-01T05:31:07.097" Score="0" Body="&lt;p&gt;&lt;strong&gt;Magic&lt;/strong&gt; (sometimes referred to as &lt;strong&gt;stage magic&lt;/strong&gt;, to distinguish it from paranormal or ritual magic) is a performing art that entertains audiences by staging tricks or creating illusions of seemingly impossible or seemingly supernatural feats, while using &lt;strong&gt;natural&lt;/strong&gt; means.&#xA;These feats are called magic tricks, effects, or illusions.&#xA;Typical magical effects include:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Prestidigitation: Manipulation of objects such as cards and coins (&amp;quot;quick fingers&amp;quot;).&lt;/li&gt;&#xA;&lt;li&gt;Conjuring: The illusion of invoking deities and other spirits.&lt;/li&gt;&#xA;&lt;li&gt;Hypnosis: Creating a state of consciousness involving reduced peripheral awareness, characterized by an enhanced capacity for response to suggestion.&lt;/li&gt;&#xA;&lt;li&gt;Mentalism: Telepathy, clairvoyance, divination, precognition, psychokinesis, mediumship, mind control, memory feats, and rapid mathematics.&lt;/li&gt;&#xA;&lt;li&gt;Escapology: The practice of escaping from restraints or other traps.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;A question tagged with this should revolve around a &lt;em&gt;concrete&lt;/em&gt; magic trick, with the goal of figuring out how it was done.&lt;/p&gt;&#xA;&lt;p&gt;You should &lt;strong&gt;not&lt;/strong&gt; use this tag for puzzles on paranormal or ritual magical concepts.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="60644" LastEditDate="2022-12-05T21:24:35.353" LastActivityDate="2022-12-05T21:24:35.353" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6581" PostTypeId="4" CreationDate="2015-01-01T05:31:07.097" Score="0" Body="Puzzles that revolve around a magic trick, or an illusion of sorts. &#xD;&#xA;Do not use this tag for a puzzle revolving around paranormal or ritual magical concepts." OwnerUserId="2383" LastEditorUserId="60644" LastEditDate="2022-12-05T21:24:35.353" LastActivityDate="2022-12-05T21:24:35.353" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6582" PostTypeId="1" AcceptedAnswerId="6600" CreationDate="2015-01-01T06:56:06.113" Score="9" ViewCount="548" Body="&lt;p&gt;Happy new year! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are asked to cut a rectangular strip 2015 times longer than its width into pieces that can be reassembled into a square with area equal to that of the original strip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many pieces do you need to cut at a minimum?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You need no more than $n+1$ pieces to reassemble a rectangle with aspect ratio (long edge length divided by short edge length) smaller than $n^2$ into a square. And $2015$ is smaller than $45^2$...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7573" LastEditorUserId="7573" LastEditDate="2015-01-04T16:48:33.057" LastActivityDate="2020-07-01T21:54:40.187" Title="Squaring the new year" Tags="|geometry|dissection|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6584" PostTypeId="1" AcceptedAnswerId="6589" CreationDate="2015-01-01T09:32:47.457" Score="5" ViewCount="3556" Body="&lt;p&gt;There are ten expressions written in the scheme below from left to right and top to bottom. Replace all of the ten distinct letters in the scheme with a single digit between 0 and 9 and make the expressions true. The usual operator precedence rules apply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;$$&#xA;\begin{array}{ccccccccccc}&#xA;\Large\color{red}{\bf F}&amp;amp;+&amp;amp;&#xA;\Large\color{blue}{\bf H}&amp;amp;-&amp;amp;&#xA;\Large\color{goldenrod}{\bf C}&amp;amp;×&amp;amp;&#xA;\Large\color{violet}{\bf A}&amp;amp;+&amp;amp;&#xA;\Large\color{orange}{\bf E}&amp;amp;=&amp;amp;&#xA;\Large0\\&#xA; -&amp;amp; &amp;amp;+&amp;amp; &amp;amp;÷&amp;amp; &amp;amp;×&amp;amp; &amp;amp;×   \\&#xA;\Large\color{limegreen}{\bf B}&amp;amp;-&amp;amp;&#xA;\Large\color{firebrick}{\bf J}&amp;amp;÷&amp;amp;&#xA;\Large\color{blue}{\bf H}&amp;amp;-&amp;amp;&#xA;\Large\color{goldenrod}{\bf C}&amp;amp;÷&amp;amp;&#xA;\Large\color{violet}{\bf A}&amp;amp;=&amp;amp;&#xA;\Large1\\&#xA; +&amp;amp; &amp;amp;-&amp;amp; &amp;amp;-&amp;amp; &amp;amp;-&amp;amp; &amp;amp;-   \\&#xA;\Large\color{purple}{\bf D}&amp;amp;-&amp;amp;&#xA;\Large\color{limegreen}{\bf B}&amp;amp;÷&amp;amp;&#xA;\Large\color{firebrick}{\bf J}&amp;amp;+&amp;amp;&#xA;\Large\color{dodgerblue}{\bf I}&amp;amp;-&amp;amp;&#xA;\Large\color{goldenrod}{\bf C}&amp;amp;=&amp;amp;&#xA;\Large2\\&#xA; +&amp;amp; &amp;amp;÷&amp;amp; &amp;amp;+&amp;amp; &amp;amp;+&amp;amp; &amp;amp;÷   \\&#xA;\Large\color{orange}{\bf E}&amp;amp;×&amp;amp;&#xA;\Large\color{purple}{\bf D}&amp;amp;+&amp;amp;&#xA;\Large\color{green}{\bf G}&amp;amp;-&amp;amp;&#xA;\Large\color{firebrick}{\bf J}&amp;amp;×&amp;amp;&#xA;\Large\color{dodgerblue}{\bf I}&amp;amp;=&amp;amp;&#xA;\Large3\\&#xA; +&amp;amp; &amp;amp;+&amp;amp; &amp;amp;×&amp;amp; &amp;amp;-&amp;amp; &amp;amp;-   \\&#xA;\Large\color{violet}{\bf A}&amp;amp;-&amp;amp;&#xA;\Large\color{orange}{\bf E}&amp;amp;+&amp;amp;&#xA;\Large\color{purple}{\bf D}&amp;amp;÷&amp;amp;&#xA;\Large\color{green}{\bf G}&amp;amp;+&amp;amp;&#xA;\Large\color{red}{\bf F}&amp;amp;=&amp;amp;&#xA;\Large4\\&#xA; =&amp;amp; &amp;amp;=&amp;amp; &amp;amp;=&amp;amp; &amp;amp;=&amp;amp; &amp;amp;=   \\&#xA; \Large9&amp;amp; &amp;amp;\Large8&amp;amp; &amp;amp;\Large7&amp;amp; &amp;amp;\Large6&amp;amp; &amp;amp;\Large5&#xA;\end{array}&#xA;$$&#xA;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus question:&lt;/strong&gt; If you were to write a computer program to solve this puzzle by brute force, how many cases would you need to check at most?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No computer is required to solve this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You &lt;strong&gt;&lt;em&gt;must not&lt;/em&gt;&lt;/strong&gt; assign different digits to different letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4394" LastEditorUserId="9404" LastEditDate="2016-03-05T17:36:12.003" LastActivityDate="2016-03-05T17:36:12.003" Title="Replace all letters with a digit and solve the equations" Tags="|mathematics|calculation-puzzle|lateral-thinking|alphametic|" AnswerCount="4" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="6586" PostTypeId="1" AcceptedAnswerId="6592" CreationDate="2015-01-01T12:48:52.923" Score="9" ViewCount="1313" Body="&lt;p&gt;Here's a seemingly simple riddle (and it's a &lt;em&gt;riddle&lt;/em&gt; and not just a word problem) I thought up two days ago (although it's simple enough that I can easily imagine someone else has thought of it before me):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a multiple of 7 of the digits between 1 and 9, and surprisingly enough, if you divide me into 2π, you get one of the two digits that I am not a multiple of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What number am I?&lt;/p&gt;&#xA;" OwnerUserId="7632" LastActivityDate="2015-01-02T13:10:06.480" Title="Integer that divides into 2π to yield an integer" Tags="|riddle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6588" PostTypeId="2" ParentId="6586" CreationDate="2015-01-01T15:45:54.273" Score="4" Body="&lt;p&gt;This is a work in progress. Take each of the (31 distinct) multiples of seven different digits between 1 and 9 inclusive and divide them into 360 (getting from $2\pi$ to 360 by converting from radians to degrees). Only a single quotient has a decimal fraction that contains, after any initial zeroes, only a single repeating digit. That's 0.0(5), the result of dividing 360 by 6480. Unfortunately, 6480 = $1\times 2\times 3\times 4\times 5\times 6\times 9$, the left out digits being 7 and 8, neither of which is 5.&lt;/p&gt;&#xA;" OwnerUserId="7363" LastEditorUserId="7363" LastEditDate="2015-01-01T15:55:55.150" LastActivityDate="2015-01-01T15:55:55.150" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6589" PostTypeId="2" ParentId="6584" CreationDate="2015-01-01T15:54:49.633" Score="1" Body="&lt;p&gt;With the second hint, it's rather easy: if we assign the same digit to all of the letters then the first row gives $3d - d^2 = 0$ whence $d = 3$ (since $d = 0$ definitely isn't a solution), and then it's just a matter of checking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In fact, the fourth row becomes $d^2 + d - d^2 = 3$ which has only one solution).&lt;/p&gt;&#xA;" OwnerUserId="228" LastActivityDate="2015-01-01T15:54:49.633" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6590" PostTypeId="1" AcceptedAnswerId="6598" CreationDate="2015-01-01T15:56:52.680" Score="6" ViewCount="1462" Body="&lt;p&gt;You are chasing a mouse living in a row of eight houses. Every night the mouse moves from the house it is in, to an adjacent house, and stays there the whole day. Apart from this, you have no a-priori information on the mouse's locations nor on its movements. However, you do have two mousetraps available. If you leave a mousetrap for the full day in a house, you are guaranteed to catch the mouse if on that day it happens to have targeted that particular house. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are contemplating the mousetrap placement strategy to follow, and land on the conclusion that if on subsequent days you place the mousetraps in houses 1-2, 2-3, ..., 7-8, you are certain to catch the mouse in no more than 7 days. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you do better? What is the shortest time in which you can be guaranteed of catching the mouse?  &lt;/p&gt;&#xA;" OwnerUserId="7573" LastActivityDate="2015-01-02T14:49:43.110" Title="A mouse, Two mousetraps, Eight houses" Tags="|logical-deduction|strategy|" AnswerCount="5" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="6591" PostTypeId="2" ParentId="6590" CreationDate="2015-01-01T16:24:46.173" Score="1" Body="&lt;p&gt;You can catch the mouse by placing the traps as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1-3, 4-6, 7-2, 3-5, 6-8. This takes 5 days.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no idea whether this is the fastest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The parity of the house that the mouse is at changes each day. Essentially, this strategy first eliminates the possibility that the mouse was at an odd numbered house the first day, then eliminates the possibility the mouse started at an even numbered house. This generalizes to the situation with $n$ houses, and takes at most $\left\lfloor\frac{2n}{3}\right\rfloor$ days to catch the mouse.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-01-01T20:59:16.433" LastActivityDate="2015-01-01T20:59:16.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6592" PostTypeId="2" ParentId="6586" CreationDate="2015-01-01T16:27:34.530" Score="11" Body="&lt;p&gt;The answer is 72.  I came up with that number yesterday, reasoning that neither 6 nor 8 could be the &quot;left out&quot; digits since any number divisible by those numbers are also divisible by 2 and 3, or 2 and 4, respectively, which would leave at least 3 of the digits 1 - 9 out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So including those numbers and their factors we must include: 1, 2, 3, 4, 6, 8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It made the most sense to me to add in 9, as it is a multiple of 3, and exclude the two remaining primes, 5 and 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The least common multiple of 1, 2, 3, 4, 6, 8, 9 is 8 * 9 = 72.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I did not realize until today's hint is that instead of dividing 2 pi by 72 and looking at the first few decimal digits of the answer, which gave .087266, we should convert to degrees and divide 360 by 72, which gives 5. This is indeed one of the digits we excluded from the list 1 through 9 and solves the riddle.&lt;/p&gt;&#xA;" OwnerUserId="7635" LastActivityDate="2015-01-01T16:27:34.530" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6594" PostTypeId="1" AcceptedAnswerId="6634" CreationDate="2015-01-01T17:03:25.993" Score="13" ViewCount="613" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A paper pound of wrong and round,&lt;br&gt;&#xA;  A minty piggy named.&lt;br&gt;&#xA;  Dividing nine, some yours, some mine,&lt;br&gt;&#xA;  A syzygy is claimed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To start off the new year, my first riddle! I'd really appreciate feedback. Thanks to everyone who gave advice on &lt;a href=&quot;https://puzzling.stackexchange.com/q/5249/4551&quot;&gt;riddle-writing&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/q/105/4551&quot;&gt;rhyming&lt;/a&gt;-- it was very helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to make every phrase have a precise interpretation connected to the answer.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-01-06T02:57:25.540" Title="Riddle quatrain: A paper pound of wrong and round" Tags="|riddle|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6597" PostTypeId="2" ParentId="6582" CreationDate="2015-01-01T18:24:18.537" Score="6" Body="&lt;p&gt;I might not have the optimum yet, but I think I can at least do &lt;em&gt;a lot&lt;/em&gt; better than  4.4 e11...&lt;/p&gt;&#xA;&lt;p&gt;I can do it with at maximum&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;strong&gt;96 pieces&lt;/strong&gt; (--&amp;gt; &lt;strong&gt;65&lt;/strong&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;Proof:&lt;/h1&gt;&#xA;&lt;p&gt;As a starting point:&lt;/p&gt;&#xA;&lt;p&gt;There is a very nice way to convert a rectangle into a square by dissection with only&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;strong&gt;two&lt;/strong&gt; cuts. (or 3 pieces)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If you do the following:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/ncDRJ.gif&quot; alt=&quot;rectangle to square&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I wish I could claim having found that solution myself, but it is actually from&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;a href=&quot;http://www.math.nmsu.edu/%7Ebreakingaway/Lessons/R2S/R2S.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt; &lt;br&gt;hosted at&lt;br&gt; &lt;em&gt;New Mexico State University; Department of Mathematical Sciences&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Now, this solution has a problem:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; It is only valid for rectangles where the two sides fulfil: &lt;strong&gt;a &amp;lt; b &amp;lt; 2a&lt;/strong&gt; &lt;br&gt; (So &lt;strong&gt;a&lt;/strong&gt; is the shorter side.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Our &amp;quot;starting point&amp;quot; is a rectangle of size &lt;strong&gt;a x 2015a&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; But we can make it to fulfil the condition! All we need is to cut stripes until &lt;strong&gt;b&amp;lt;2a&lt;/strong&gt;.&lt;br&gt;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/KcYrt.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Now we have&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; a &lt;em&gt;striped&lt;/em&gt; rectangle which matches the condition and used up 32 pieces(31 cuts). By applying the cutting in the first image, we will &lt;strong&gt;at maximum&lt;/strong&gt; create 3 pieces hence we will need no more than 3 x 32 = 96 pieces to accomplish the task.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;or visually:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/HwBgd.gif&quot; alt=&quot;Solution&quot; /&gt;&#xA; Note that X is &lt;em&gt;not&lt;/em&gt; splitting the longer side of the rectangle &lt;em&gt;exactly&lt;/em&gt; at 1/2, because the longer side is &lt;em&gt;not exactly&lt;/em&gt; 63 but 62.96875. The solution is still valid by virtue of the general statement at the beginning of the proof.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Looking at my own answer above, I think there is a very straight-forward way to optimize the solution!&lt;/p&gt;&#xA;&lt;h1&gt;Improved solution&lt;/h1&gt;&#xA;&lt;p&gt;Starting with the solution above:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/AfTnh.gif&quot; alt=&quot;Old solution&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;It can be seen that&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The tiles of the upper half (except the very tip) can of course be merged! This reduces 62 tiles to 31 brings the solution down to 96-31 = &lt;strong&gt;65&lt;/strong&gt; tiles.&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/3AhVP.gif&quot; alt=&quot;new&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-01-01T22:14:27.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6598" PostTypeId="2" ParentId="6590" CreationDate="2015-01-01T20:45:28.097" Score="10" Body="&lt;p&gt;It can be over in 4 days:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  5-7, 2-4, 2-4, 5-7   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is how it works:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can picture the progress of the mouse like a walk on a chessboard.  The rows are the days, the columns are the houses.  The mouse starts from the top row and walks diagonally down.  Note that it can move on one color only.&lt;br&gt;&#xA;&lt;br&gt;&#xA; To catch the mouse, you first assume it is on a white square.  There are 4 possible houses.  The first day you catch the mouse in houses 5 and 7.  If it survives, the mouse is in house 1 or 3.  Then, the next day, it must be in houses 2 or 4 and you are certain to catch it on the second day.  It is visually easy to see how the traps (x) seal the green area in the left picture.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/ukllC.png&quot; alt=&quot;mouse chess&quot;&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; You need to repeat the process on the black cells.  It is the same method reversed. What the mouse did on the first 2 days is then irrelevant.  The right picture shows the complete solution in 4 days.  Green cells are possible survival places for the mouse.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2015-01-01T23:15:27.020" LastActivityDate="2015-01-01T23:15:27.020" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6600" PostTypeId="2" ParentId="6582" CreationDate="2015-01-01T21:55:03.780" Score="10" Body="&lt;p&gt;I can do it in 46 pieces.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;  The square root of 2015 is 44.889.  That is the size of the square you want.&lt;br /&gt;&#xA;  Fill the square with horizontal stripes from the top until you have less than 2 units left in height.&#xA;  That is 43 stripes, you have 1.889 units left vertically.&lt;br /&gt;&#xA;  Then cut a diagonal along the width and cut the remaining piece in 2 as shown in the picture.&lt;br /&gt;&#xA;  Left is the square assembled from a 1x2015 stripe cut in 46 pieces.&lt;br /&gt;&#xA;  To better show the method, the right picture is a similar solution for a 1x20 stripe cut in 6 pieces.&lt;br /&gt;&#xA;  The bottom 3 pieces can be rearranged into a 1 unit-high rectangle.&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/TES2C.png&quot; alt=&quot;1x2015&quot; /&gt;   &lt;img src=&quot;https://i.stack.imgur.com/hu5DJ.png&quot; alt=&quot;1x20&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2020-07-01T21:54:40.187" LastActivityDate="2020-07-01T21:54:40.187" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="6601" PostTypeId="2" ParentId="6458" CreationDate="2015-01-01T22:35:30.187" Score="17" Body="&lt;p&gt;Based on the following analysis, the next number in the sequence is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;12.2&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The number base is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  2i, also known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Quater-imaginary_base&quot; rel=&quot;noreferrer&quot;&gt;quater-imaginary&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This table converts the numbers to base 10:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;img src=&quot;https://i.stack.imgur.com/b06g9.png&quot; alt=&quot;Table1&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This graph shows the sequence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;as a spiral&lt;/strong&gt;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/ID25F.png&quot; alt=&quot;Graph&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Additional comments by h34&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the sequence defined in 12 words:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaussian_integer&quot; rel=&quot;noreferrer&quot;&gt;Gaussian integers&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Quater-imaginary_base&quot; rel=&quot;noreferrer&quot;&gt;base 2i&lt;/a&gt;, ordered in counterclockwise &lt;a href=&quot;http://en.wikipedia.org/wiki/Ulam_spiral&quot; rel=&quot;noreferrer&quot;&gt;Ulam spiral&lt;/a&gt; beginning at 0.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Anyone who hasn't encountered&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gaussian integers, base 2i or Ulam spirals&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;before will find a lot of interesting stuff by clicking on those links.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or if you want a quick explanation here, here goes...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Gaussian integers&lt;/strong&gt; are numbers of the form a+bi, where a and b are integers and i is the square root of minus 1. These were first studied by Carl Friedrich &lt;strong&gt;Gauss&lt;/strong&gt; in 1832.&lt;br&gt;&lt;br&gt;&#xA; &lt;strong&gt;Base 2i&lt;/strong&gt; means what it says: instead of expressing a number using digits intended to be multiplied by powers of 10 (giving the decimal system), do it for powers of 2i. Base 2i is also known as quater-imaginary. It was first described by Donald &lt;strong&gt;Knuth&lt;/strong&gt; in 1955.&lt;br&gt;&lt;br&gt;&#xA; For expressing Gaussian integers, base 2i is symbolically less complex than any real base.&lt;br&gt;&lt;br&gt;&#xA; To express negative real numbers, it needs no symbols other than digits. Unlike a real base, it needs no minus symbol. To express Gaussian integers, it needs no symbols other than digits and, if necessary, a point to separate the units place from the $(2i)^{-1}$ place. And when a $(2i)^{-1}$ digit &lt;em&gt;is&lt;/em&gt; needed, it's always 2 and it's the only digit after the point.&lt;br&gt;&lt;br&gt;&#xA; The decimal system needs no point for integers whether real or Gaussian, but for negative reals it uses a minus symbol; for Gaussian integers with imaginary part nonzero it needs i; and for Gaussian integers with both real and imaginary parts nonzero it needs a plus symbol, a minus symbol, or both.&lt;br&gt;&lt;br&gt;&#xA; Examples of numbers written in base 2i:&lt;br&gt;&lt;br&gt;&#xA; 1+i is written 11.2 because 1+i = 1(2i) + 1(1) +2$(2i)^{-1}$ = 2i + 1 + 2(-i/2);&lt;br&gt;&lt;br&gt;&#xA; -1 is written 103 because -1 = 1$(2i)^2$ + 3(1) = -4 + 3&lt;br&gt;&lt;br&gt;&#xA;Expressed in base 2i, the non-negative integers run&lt;br&gt;&lt;br&gt;&#xA;0,1,2,3,10300,10301,10302,10303,10200,10201,10202,...,&lt;br&gt;&lt;br&gt;&#xA;and the negative integers, ordered by increasing magnitude, run&lt;br&gt;&lt;br&gt;&#xA;103, 102, 101, 100, 203, 202, 201, 200, 303, 302&lt;br&gt;&lt;br&gt;&#xA; Some numbers have more than one representation, e.g. 1/5 can be written as either 1.(0300) or 0.(0003). And a number such as 8 can be written 10200 or simply as 8. To get a unique representation of a Gaussian integer, just stipulate that all digits used must be between 0 and 3 inclusive.&lt;br&gt;&lt;br&gt;&#xA; Real numbers have their nonzero digits all in odd-order positions; pure imaginary numbers, all in even-order positions. So the expression for a+bi can be got from adding the expressions for a and b, and no 'carrying' is necessary.&lt;br&gt;&lt;br&gt;&#xA; The &lt;strong&gt;Ulam spiral&lt;/strong&gt; was invented by Stanislaw &lt;strong&gt;Ulam&lt;/strong&gt; in 1963. It's a way of writing down the positive integers on a square grid. Using standard Cartesian notation, write 1 at the origin (0,0), 2 at (1,0), 3 at (1,1), 4 at (0,1), 5 at (-1,1) etc. If you then mark the primes, you get a nice pattern, including for example a (for a time) almost-solid line of primes of the form $4x^2 + 2x + 41$ (related to Euler's polynomial $x^2 + x + 41$).&lt;br&gt;&lt;br&gt;&#xA; I used an Ulam spiral to order the Gaussian integers: 0, 1, 1+i, i, -1+i, -1 , etc. Then I expressed those numbers in base 2i to get the puzzle sequence. The missing number is the base 2i expression of 2+i, which is 12.2.&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The next hints would have been:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The numbers that follow the missing term are 12, 11, 10, 113, 112, 112.2, 102, 102.2, 1132, 1133, 1030, ...&lt;br&gt;&lt;br&gt;&#xA; Each of the three main mathematical ideas used to construct this sequence is associated with a distinct mathematician with a surname containing 4 or 5 letters.&lt;br&gt;&lt;br&gt;&#xA; Whoever &lt;strong&gt;thunk&lt;/strong&gt; of this won't &lt;strong&gt;maul&lt;/strong&gt; or &lt;strong&gt;gas us&lt;/strong&gt;. (Anagrams of the three mathematicians' surnames.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-01-02T05:35:26.183" LastActivityDate="2015-01-02T05:35:26.183" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6602" PostTypeId="2" ParentId="6590" CreationDate="2015-01-02T01:38:26.267" Score="0" Body="&lt;p&gt;[Edit: not an answer. No idea why I read seven houses, when it's clearly written as eight houses in the OP. Sorry!]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While there are quite a few possible answers, I favor the laziest, divide-and-conquer approach, where you move the traps as rarely as possible, if at all. It's just less work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, one at 4 divides the problem in two. One at 2, subdivides that half of the problem further. So, traps 2 &amp;amp; 4 for two days would eliminate any mice at 2 and 4, of course, but also at 1 and 3, since the only places the mouse could have gone from those, after the first day, was into our traps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we then know the mouse is in the range 5-7, we know the lowesst place he can reach the next day is 4. So we can just move the trap from house 2 to house 6, leaving the other trap at house 4. Sit for a couple of days - if he moves into in 4 or 6, he dies that day: if he moves into 5 or 7, then he has nowhere to move the next day except into our traps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: (2,4), (2,4), (4,6), (4,6).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The row of 7 houses, at the end of each day:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1234567 &amp;lt;- House numbers&#xA;mTmTmmm &amp;lt;- Day 1           m = mouse&#xA;.T.Tmmm &amp;lt;- Day 2           T = trap&#xA;...TmTm &amp;lt;- Day 3           . = known empty house.&#xA;...T.T. &amp;lt;- Day 4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7647" LastEditorUserId="7647" LastEditDate="2015-01-02T02:30:16.607" LastActivityDate="2015-01-02T02:30:16.607" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6604" PostTypeId="2" ParentId="625" CreationDate="2015-01-02T02:53:47.467" Score="-3" Body="&lt;p&gt;Thinking of a simpler and more practical solution that is focusing on the first part of the question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How can they exchange the numbers successfully if the numbers can be&#xA;  from 1 to M?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and not worrying about the algorithm portion of the question as this is &lt;em&gt;puzzling&lt;/em&gt; and not math stack exchange after all.&#xA;I can come up with a simple solution that allows for &lt;strong&gt;9,999&lt;/strong&gt; numbers for each spy.&#xA;Other posters seem to think a restraint is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The number of stones is the only information that may be collected.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I did not see that as part of the question. So we can also use the location the stone lands as an indicator of its value. Also assume each spy will take two turns each. For example if it is on the right of the spy it's 1 in front of the spy 2, left of the spy 5,  Far away on the right 10, far away in  middle 30, far away on the right 50.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Turn one is for the tens and ones&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qgIp2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Turn two for thousands and hundreds&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gsPD0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess you could go limitless if you wanted to also include height of stone, way it was thrown, number of times it bounded on the water. I simply thought it would be nice to have a non mathematical solution to show an alternate easyer to understand way to solve the problem that allows for higher numbers.&lt;/p&gt;&#xA;" OwnerUserId="7648" LastActivityDate="2015-01-02T02:53:47.467" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6605" PostTypeId="2" ParentId="6538" CreationDate="2015-01-02T03:38:55.827" Score="8" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; mirror&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which I got starting with the German word bit:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Predator - eagle. German words rhyming with 'eagle' - siegel (seal) and spiegel (mirror), both ending with &quot;gel&quot;, a twist of &quot;leg&quot;, which is encouraging, and only the latter names a newspaper, &lt;em&gt;Der Spiegel&lt;/em&gt;. That explains lines 1, 2 and 4 and leaves line 3. A &lt;a href=&quot;http://en.wikipedia.org/wiki/Pastry_fork&quot;&gt;pie fork&lt;/a&gt; is a kind of fork, and if &quot;s&quot; refers to a type of &lt;a href=&quot;http://en.wikipedia.org/wiki/Sigma_bond&quot;&gt;sigma bond&lt;/a&gt;, then &quot;Bond a fork&quot; gives &quot;Spie&quot; and we've got &quot;Spie&quot; + &quot;gel&quot; - tick!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The English word bit:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;The Mirror&lt;/em&gt; is an English-language newspaper. &quot;Half of me&quot; is &quot;m&quot;. &quot;Or, with repeats&quot; is &quot;or&quot; plus &quot;rr&quot;, a doubling of the letter &quot;R&quot; used for &quot;repeat&quot;. That gives m+or+rr and the last line tells us that only &quot;one&quot; is lacking, meaning the letter &quot;i&quot;. m+or+rr+i gives &quot;mirror&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What's going on with the flip of the string of symbols to give &quot;Avigrail&quot; is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; mirroring&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now the first bit:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A mirror might be metal, and can reflect accurately even if to do so is often unwelcome. Which leaves &quot;you might / Find me in woods or at the beach&quot; - possibly meaning reflection in a lake or the sea?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7363" LastEditorUserId="7363" LastEditDate="2015-01-02T03:57:15.487" LastActivityDate="2015-01-02T03:57:15.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6607" PostTypeId="2" ParentId="6524" CreationDate="2015-01-02T06:32:46.470" Score="7" Body="&lt;p&gt;The subject of the riddle is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Samuel Langhorne Clemens, also known as Mark Twain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I grew up&lt;br&gt;&#xA;  where Mischa's brother, the doctor,&lt;br&gt;&#xA;  ate human liver with bitter wine:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hannibal Lecter had a sister, Mischa, in Hannibal Rising. This is a reference to Hannibal, Missouri where Clemens grew up. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;  in soccer, a yellow card offence, I hear.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Refers to a handball, which sounds similar to Hannibal. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I trained as a cub&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A cub pilot on a riverboat, that is.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;  where in ancient times the Southern mounds were built;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This refers to the Mississippi River on which Clemens worked as a steamboat pilot. Several Indian tribes in the Mississippi River basin were mound builders. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;  where an Elk traverses cedars,&lt;br&gt;&#xA;  with a big greasy tireless ol' man.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Mississippi River is known colloquially as &quot;Ol' Man River&quot;. The Cheyenne tribe called it the Big Greasy River. The Ojibwe names for various stretches of the river include Elk River, River from the Traversing Lake, and Red Cedar River. (I think we both found the same Wikipedia entry)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My famous works, as split,&lt;br&gt;&#xA;  anticipate the arrival&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Adventure. Advent means the anticipated arrival of a celebrated person. The Latin adventurus means about to happen. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;  of you, seen, with half a drum,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wordplay on Tom Sawyer. Tom is half of tom tom, a type of drum. Tom saw ya. The Adventures of Tom Sawyer. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;  and of the French, fabric for a Nordic funeral.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I believe this refers to The Adventures of Huckleberry Finn. Not sure where Huckle comes from. Maybe the bouclé French fabric? ...berry Finn looks to be a play on &quot;bury Finn&quot; which is what you do to a guy from Finland at his funeral. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I arrived and departed with a dirty snowball.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As has been noted, Clemens' birth and death were timed close to two consecutive appearances of Haley's comet. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Play it once, for old times' sake - play it!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is a line from the movie Casablanca. It is often misquoted as &quot;Play it again, Sam.&quot; This refers to Clemens' given name. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2162" LastEditorUserId="2162" LastEditDate="2015-01-02T06:51:53.203" LastActivityDate="2015-01-02T06:51:53.203" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6608" PostTypeId="2" ParentId="6590" CreationDate="2015-01-02T08:36:01.660" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;3 days&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Day 1: trap at 2 and 7 (if mouse already in there, it'll be caught)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Night 1: mouse from 1 and 8 have to move into the trap. (since the puzzle doesn't allow the mouse to stay put).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Day 2: move trap inwards to 3 and 6 (if mouse is already in there, it'll be caught).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Night 2: catches mouse if they move outwards from 4 and 5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Day 3: move trap inwards to 4 and 5 (mouse got no where else to go.)&lt;/p&gt;&#xA;" OwnerUserId="7546" LastActivityDate="2015-01-02T08:36:01.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6609" PostTypeId="2" ParentId="6552" CreationDate="2015-01-02T10:21:45.703" Score="2" Body="&lt;p&gt;It can only be John.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;If the authorities also know that exactly one of the four suspects is telling the truth, who did it?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sentence implies, that we have 4 cases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 1: A Tells the truth, the other 3 suspects are lying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 2: B Tells the truth, the other 3 suspects are lying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so on..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In each case, except for case 2, John would be lying. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;John said “I did not do it.”&lt;/em&gt; -&gt; That would say that John is lying 3 out of 4 times. &#xA;If we take a closer look to case 2, which would make John telling the truth, this would imply that he isn't guilty and the other suspects are lying. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carlos did it -&gt; lie &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diana did it -&gt; lie &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carlos lied when he said that I did it. -&gt; This is a difficult one. Does Diana know what Carlos is going to say? If she knows who is telling the truth, does she suppose that Carlos is going to lie about her? That would make Diana the only one telling the truth, but in this case, we expect all suspects to lie but John. We can conclude that Carlos is lying -&gt;Diana did not do it. But also Diana is lying which would imply that carlos was actually telling the truth when its obvious he was lying. That doesn't make any sense to me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads to the conclusion that Diana is telling the truth and this makes John the culprit, since this is the only case making sense. &lt;/p&gt;&#xA;" OwnerUserId="7659" LastActivityDate="2015-01-02T10:21:45.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6612" PostTypeId="1" AcceptedAnswerId="6686" CreationDate="2015-01-02T12:07:43.827" Score="8" ViewCount="1343" Body="&lt;p&gt;Can you solve the following riddle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solid, misshapen, sad, I is - a linguistic object of rare beauty. What am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is somewhat experimental and I am not sure how hard people will find it. But I have thought about the thought processes required and it should give solvers a lot of pleasure.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To judge from the absence of comments or answers, it seems that people might not be finding a feature or part of this puzzle to get a grip on..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here's a hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;A linguistic object of rare beauty&quot; is the literal part.&lt;br&gt;&lt;br&gt;&#xA;&#xA;I thought this would be clear because that string of words holds together more fluently than the string that precedes the dash. &quot;A linguistic object of rare beauty&quot; is what you want.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Another hint&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the non-literal part is itself in how many parts? Remember how many when you're trying to create what you want.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7363" LastEditorUserId="7363" LastEditDate="2015-01-04T19:38:53.613" LastActivityDate="2015-01-06T17:29:59.570" Title="A linguistic object of rare beauty - what am I?" Tags="|riddle|word|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6614" PostTypeId="2" ParentId="4719" CreationDate="2015-01-02T12:39:43.850" Score="1" Body="&lt;p&gt;This is quite similar to the previous min reader tricks, don't mind, but the thing is there is nothing new in the method you have used here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nothing new? Well, apart from the minor changes you have made in your '&lt;strong&gt;key&lt;/strong&gt;' (of course, by key, I mean the '4 legged non-living thing/living thing'), I don't find anything new to be honest. I am not intending to sound rude btw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third person / victim asking the questions is same as you yourself asking the questions. Because, you both will obviously know what's the &lt;strong&gt;key&lt;/strong&gt;, it could 2 living things / 2 Four legged-thing or anything for that sake. And since the victim won't change the actual word that he has uttered while asking, it is but obvious that you and the other mind reader would know when the 2 living things are done with (of course, I have assumed that the key is 2 living things). Once that is done, you would use the actual word that was whispered and nedd not to say, you get it right and will be called &lt;strong&gt;mind-readers&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, I love your puzzles. Keep them coming. Cheers!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S: I did not intend to sound rude. Excuse me if I did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.P.S: &lt;strong&gt;You both&lt;/strong&gt; refers to Andrew and Gabriel. Victim refers to Harry.  &lt;/p&gt;&#xA;" OwnerUserId="7656" LastEditorUserId="7656" LastEditDate="2015-01-11T07:12:03.923" LastActivityDate="2015-01-11T07:12:03.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6615" PostTypeId="2" ParentId="6590" CreationDate="2015-01-02T14:49:43.110" Score="1" Body="&lt;p&gt;It will take maximum of &lt;strong&gt;3 days&lt;/strong&gt; to catch the mouse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Day 1:&lt;/strong&gt; traps put in houses 1 and 7:-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;case 1:&lt;/strong&gt; &lt;em&gt;it is in house 1 or 7 and gets trapped.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;case 2:&lt;/strong&gt; mouse is in house 8:-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;it will either move to house 7 or house 1(if it can) and will be trapped&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;case 3:&lt;/strong&gt; mouse is in house 2 or 6:-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;either it will move towards house 1 or 7 and get caught&#xA;or it will move towards houses 3 or 5&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;case 4:&lt;/strong&gt; it is in houses 3-5:-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;it will be available in houses 2-6 at all costs(because it can move only 1 house at a time)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Day 2:&lt;/strong&gt; traps put in houses 2 and 6:-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;again either mouse will be trapped at night or its within houses 3-5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Day 3:&lt;/strong&gt; traps put in houses 3 and 5:-&#xA;mouse will be certainly trapped at night.&lt;/p&gt;&#xA;" OwnerUserId="7665" LastActivityDate="2015-01-02T14:49:43.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6616" PostTypeId="5" CreationDate="2015-01-02T15:16:14.310" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-01-02T15:16:14.310" LastActivityDate="2015-01-02T15:16:14.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6617" PostTypeId="4" CreationDate="2015-01-02T15:16:14.310" Score="0" Body="A question about categories, mechanics, or effects of puzzle in general, or about scientific research on puzzles." OwnerUserId="5840" LastEditorUserId="5840" LastEditDate="2015-01-02T16:07:27.390" LastActivityDate="2015-01-02T16:07:27.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6618" PostTypeId="1" AcceptedAnswerId="6620" CreationDate="2015-01-02T16:04:03.393" Score="11" ViewCount="1517" Body="&lt;p&gt;&lt;strong&gt;Each answer is the title of a famous novel.&lt;/strong&gt; A few of the titles are split into two clues - you should find those clearly marked. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6TrxX.png&quot; alt=&quot;Crossword - novels&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Across&lt;/strong&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3   Think of the Phoenician, a possible drench, bleached prisons, or beach spindles (8,7)&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Consider Phlebas ('Think of'=consider, Phlebas the Phoenician, three anagrams)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6   Three women (under training) - Russian tragedy (4,8) &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Anna Karenina (Anna, Karen, Nina, under a train)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;9   Before playing the numbers on an Air Serbia flight (9) (after 1 down)&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Prejudice (Pre, airline code JU, Dice)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;10  Fabric, for the (French) Nordic funeral (11,4) (after 23 across)&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Huckleberry Finn (Huck, Le, &quot;Bury Finn&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;12  Liquid vessel sunk - in Hampshire? (9,4)&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Watership Down&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;14  Ennobled for misplacing files (4,2,3,5)&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lord of the Flies &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;15  Impressive detective; an annoying producer of electronic music (4,4)&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Moby Dick&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;17  A member of a Gallic mendicant religious order, spun unluckily - don't be silly (5) (before 4 down)&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sense (ROT13:FRAFR - abbrev:Friar, abbrev:France)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;18  See a snotty cold; a dripping nose, it looks like. (1,4,4,1,4)&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Room With a View (&quot;Rheum&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;20  Terrifying German lunch assembled from parts: sausage'n'tankard (12)&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Frankenstein (Frank 'n' Stein)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;21  A number of Winston's wartime exploits (6,4) (second half, after 7 down)&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eightyfour&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;22  Agricultural earthquake in Filipino city (6,4)&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Animal Farm (Anagram:MANILA, Farm)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;23  After year zero, invests risk capital (10) (before 10 across)&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Adventure (AD Venture)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;&lt;strong&gt;Down&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;1   She's got a ticket to Ryde - starting at Portsmouth (5) (before 9 across)&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pride (P, &quot;Ryde&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2   Anticipate the arrival of you, seen, with half a drum, floating downriver (3,10,2,3,6)&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Adventures of Tom Sawyer (Advent, &quot;Saw yer&quot;, Tom-tom)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4   Sensitiveness, aesthetic appreciativeness, capacity of emotion (11) (after 17 across)&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sensibility&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5   Initially rebellious sheep undertake a silent infantry operation, inside a small enclosure, or so I'm induced to believe (10)&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Persuasion (initials &quot;rsuasio&quot; within &quot;pen&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;7   A number of Winston's wartime exploits (8) (first half, before 21 across)&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nineteen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;8   Little-known Beatles song (without straw) (4,3,7)&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jude the Obscure (&quot;Hey Jude&quot; without &quot;hay&quot;, little-known = obscure)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;11  Computerised socialite gazes at beverage, a monarch writes (2)&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It ('It' girl, information technology, eye tea, Stephen King)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;13  Mr President, you sailed beyond the sunset to the sheltered end of the oceans, I've heard? (7)&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ulysses (You Lee Seas, quote from Tennyson's Ulysses, Ulysses S Grant)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;16  Oscar eats a food rich in iron, then dances to Chubby Checker (6,5) &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oliver Twist (O, Liver, Twist)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;19  Makin' twice as many female pronouns, in Ireland (9)&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dubliners (&quot;Doubling Hers = Doublin' 'ers&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;21  A tenth of a centimetre of MDMA for the electric André-Marie - clueless? (4)&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Emma (E, mm, Ampere, 'Clueless' movie has plot of Emma)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Do ask if you have any questions, either in the comments or ping me on chat with an '@AE'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the spirit of teamwork, partial answers are very welcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answers providing constructive feedback on the puzzle itself - how hard it is, what could be improved - are also very welcome. This puzzle took me &lt;em&gt;ages&lt;/em&gt; to compose, so please be kind! ;)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Update:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All answers now added. Thanks to all the solvers! :)&lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="4050" LastEditDate="2015-05-22T18:03:19.860" LastActivityDate="2015-05-22T18:03:19.860" Title="The person, be it gentleman or lady, who has not pleasure in a good novel, must be intolerably stupid. — Jane Austen" Tags="|cryptic-crosswords|crosswords|literature|" AnswerCount="6" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6619" PostTypeId="2" ParentId="6584" CreationDate="2015-01-02T16:12:24.407" Score="1" Body="&lt;p&gt;Semi-stupid exhaustive search demands 10*9*10*9*10 = 99000 checks and finds the single solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anything better?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pengines.swi-prolog.org/apps/swish/p/EOVHropD.pl&quot; rel=&quot;nofollow&quot;&gt;Program here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;solve([A,B,C,D,E,F,G,H,I,J]) :-&#xA;    Digits = [0,1,2,3,4,5,6,7,8,9], &#xA;    % select any F&#xA;    member(F,Digits), &#xA;    % select any H different from 0&#xA;    member(H,Digits), H =\= 0,&#xA;    % select any C&#xA;    member(C,Digits),&#xA;    % select any A different from 0&#xA;    member(A,Digits), A =\= 0,    &#xA;    % CAFH fixed --&amp;gt; 0-value equation yields E (also verify that E is a Digit)&#xA;    E is ( C*A-F-H ), member(E,Digits),           &#xA;    % CEAF fixed --&amp;gt; 5-value equation yields I (also verify that I is a nonzero Digit)&#xA;    % division by zero in the next equation is verboten&#xA;    (E*A-5-F) =\= 0,&#xA;    I is (C/(E*A-5-F)), member(I,Digits), I =\= 0,    &#xA;    % select any B&#xA;    member(B,Digits),    &#xA;    % 9-value equation yields D (verify that E is a Digit)&#xA;    D is (9-E-A-F+B), member(D,Digits), D =\= 0,&#xA;    % 8-value equation yields J (verify that J is a Digit)&#xA;    J is (8-E-H+B/D), member(J,Digits), J =\= 0,    &#xA;    % 3-value equation yields G (verify that G is a nonzero Digit)&#xA;    G is (3+J*I-E*D), member(G,Digits), G =\= 0,    &#xA;    % verify remaining constraints&#xA;    2 is (D - B / J + I - C),&#xA;    1 is (B - J / H - C / A),    &#xA;    4 is (A - E + D / G + F),&#xA;    6 is (A * C - I + J - G),&#xA;    7 is (C / H - J + G * D).&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7625" LastActivityDate="2015-01-02T16:12:24.407" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6620" PostTypeId="2" ParentId="6618" CreationDate="2015-01-02T16:40:44.313" Score="9" Body="&lt;p&gt;Just getting it started with a few (okay, so I've gotten more since writing that...):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;19&quot;&gt;&#xA;  &lt;li&gt;Makin' twice as many female pronouns, in Ireland (9)&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The females pronoun is &lt;em&gt;'ers&lt;/em&gt;, from &quot;hers&quot;, and Ireland points to &lt;em&gt;Dublin&lt;/em&gt; (along with &quot;doubling&quot; for making twice as many), so it's:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Dubliners&quot;&gt;Dubliners&lt;/a&gt; by James Joyce&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;16&quot;&gt;&#xA;  &lt;li&gt;Oscar eats a food rich in iron, then dances to Chubby Checker (6,5) &lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;O&lt;/em&gt; for Oscar, &lt;em&gt;liver&lt;/em&gt; is rich in iron, and Chubby Checker is famous for the &lt;em&gt;Twist&lt;/em&gt;, so it's:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Oliver_Twist&quot;&gt;Oliver Twist&lt;/a&gt; by Charles Dickens&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;12&quot;&gt;&#xA;  &lt;li&gt;Liquid vessel sunk - in Hampshire? (9,4)&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Liquid is &lt;em&gt;water&lt;/em&gt;, vessel is &lt;em&gt;ship&lt;/em&gt;, and when a ship sinks, it goes &lt;em&gt;down&lt;/em&gt;. That adds up to the name of a hill in Hampshire:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Watership_Down&quot;&gt;Watership Down&lt;/a&gt; by Richard Adams&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;11&quot;&gt;&#xA;  &lt;li&gt;Computerised socialite gazes at beverage, a monarch writes (2)&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you gaze at a beverage, you might &lt;em&gt;eye tea&lt;/em&gt;. Being computerized you know all about &lt;strong&gt;I&lt;/strong&gt;nfo &lt;strong&gt;T&lt;/strong&gt;ech. Another term for monarch is &lt;em&gt;king&lt;/em&gt;, so we have:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/It_%28novel%29&quot;&gt;It&lt;/a&gt; by Stephen King&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;21&quot;&gt;&#xA;  &lt;li&gt;A tenth of a centimetre of MDMA for the electric André-Marie - clueless? (4)&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A tenth centimeter is a millimeter (&lt;em&gt;MM&lt;/em&gt;), MDMA can go by either &lt;em&gt;E&lt;/em&gt; or &lt;em&gt;X&lt;/em&gt;, depending on where you are, and the electric André-Marie has to refer to &lt;a href=&quot;http://en.wikipedia.org/wiki/Andr%C3%A9-Marie_Amp%C3%A8re&quot;&gt;Ampère&lt;/a&gt;, so we'll use the abbreviation &lt;em&gt;A&lt;/em&gt; for amperes. And today I learned the the movie &lt;em&gt;Clueless&lt;/em&gt; is loosely based on this novel, spelled with the abbreviations used in the rest of the clue:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Emma_%28novel%29&quot;&gt;Emma&lt;/a&gt; by Jane Austen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;6&quot;&gt;&#xA;  &lt;li&gt;Three women (under training) - Russian tragedy (4,8)&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Three female names: &lt;em&gt;Anna&lt;/em&gt;, &lt;em&gt;Karen&lt;/em&gt;, and &lt;em&gt;Ina&lt;/em&gt;. A famous Russian tragedy fits this quite well, but I can't figure out why they're &quot;under training&quot;. Either way, I can't see it being anything but:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Anna_Karenina&quot;&gt;Anna Karenina&lt;/a&gt; by Leo Tolstoy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;2&quot;&gt;&#xA;  &lt;li&gt;Anticipate the arrival of you, seen, with half a drum, floating downriver (3,10,2,3,6)&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Anticipating the arrival of a person could be &lt;em&gt;advent&lt;/em&gt;, combined with &quot;yours&quot; makes adventure. Half a drum (tomtom) is &lt;em&gt;Tom&lt;/em&gt;, and when you're seen, I could say I &lt;em&gt;saw yer&lt;/em&gt;. Floating downriver seems to point to the content of the book instead of the title, which is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/The_Adventures_of_Tom_Sawyer&quot;&gt;The Adventures of Tom Sawyer&lt;/a&gt; by Mark Twain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;22&quot;&gt;&#xA;  &lt;li&gt;Agricultural earthquake in Filipino city (6,4)&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Agriculture points to &lt;em&gt;farm&lt;/em&gt;. The most well-known Filipino city is Manila. If an earthquake hit that word, it would certainly scramble things up, so let's look for anagrams. How about:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Animal_Farm&quot;&gt;Animal Farm&lt;/a&gt; by George Orwell&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;14&quot;&gt;&#xA;  &lt;li&gt;Ennobled for misplacing files (4,2,3,5)&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Credit to Roger in comments:&lt;/strong&gt;&#xA;&quot;Misplaced files&quot; = FLIES. To be ennobled for that would be to be made:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Lord_of_the_Flies&quot;&gt;Lord of the Flies&lt;/a&gt; by William Golding&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;20&quot;&gt;&#xA;  &lt;li&gt;Terrifying German lunch assembled from parts: sausage'n'tankard (12)&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Credit to Roger in comments:&lt;/strong&gt; A sausage is a &lt;em&gt;frank&lt;/em&gt;, and a stone tankard is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Beer_stein&quot;&gt;&lt;em&gt;stein&lt;/em&gt;&lt;/a&gt;. Put them together and you get a novel about a frightening monster assembled in Germany:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Frankenstein&quot;&gt;Frankenstein&lt;/a&gt; by Mary Shelley&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;7 A number of Winston's wartime exploits (8) (first half, before 21 across)&lt;br&gt;&#xA;  21 A number of Winston's wartime exploits (6,4) (second half, after 7 down)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A book about (perpetual) wartime, with numbers as the title? Maybe one with Winston as the main character? It can only be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Nineteen_Eighty-Four&quot;&gt;Nineteen Eighty-Four&lt;/a&gt; by George Orwell&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll admit, at first &lt;em&gt;wartime&lt;/em&gt; with &lt;em&gt;Winston&lt;/em&gt; had me looking at Churchill. Nicely misdirected, sir ;)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;23 After year zero, invests risk capital (10) (before 10 across)&lt;br&gt;&#xA;  10 Fabric, for the (French) Nordic funeral (11,4) (after 23 across)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After year zero is &lt;em&gt;AD&lt;/em&gt;, and a &lt;em&gt;venture&lt;/em&gt; could mean you're investing; that part is pretty straightforward. I got the second part of the title, but I'm shaky about the reasoning for the clues. Since we're talking Nordic funerals, I'm thinking of floating on a raft/boat. With Finland, that led to &lt;em&gt;Finn&lt;/em&gt;, but I really don't know what the other has to do with it. I assume it's some sort of cloth/fabric, but my Google-fu is failing on that one. Either way, it's:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Adventures_of_Huckleberry_Finn&quot;&gt;Adventures of Huckleberry Finn&lt;/a&gt; by Mark Twain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;13&quot;&gt;&#xA;  &lt;li&gt;Mr President, you sailed beyond the sunset to the sheltered end of the oceans, I've heard? (7)&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sheltered end of something is the &lt;em&gt;lee&lt;/em&gt; side, and oceans are also &lt;em&gt;seas&lt;/em&gt;. So it sounds like &quot;you lee seas&quot;, and something to do with a president:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Ulysses_%28novel%29&quot;&gt;Ulysses&lt;/a&gt; by James Joyce&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As far as feedback on the puzzle goes, the clues seem well done to me. Of course, I may be biased since I can only really judge the ones I've figured out. Either way, there seems a good mix of short/long, and there's &lt;em&gt;just&lt;/em&gt; enough in the clues to make me want to figure them out without being completely frustrated.&lt;/p&gt;&#xA;" OwnerUserId="1563" LastEditorUserId="1563" LastEditDate="2015-01-03T03:56:38.083" LastActivityDate="2015-01-03T03:56:38.083" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="6621" PostTypeId="1" AcceptedAnswerId="6622" CreationDate="2015-01-02T16:42:17.813" Score="-1" ViewCount="14699" Body="&lt;p&gt;Three men are arrested on charges of murder. They make the following statements.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First suspect: I'm not guilty of murder.&lt;/li&gt;&#xA;&lt;li&gt;Second suspect: the third suspect is the murderer.&lt;/li&gt;&#xA;&lt;li&gt;Third suspect: the first suspect is the murderer.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We know that two of them are lying and one of them is telling the truth, and certainly one of the three is the killer. The question now is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Which one is telling the truth?&lt;/li&gt;&#xA;&lt;li&gt;What is the probability of each suspect being the murderer?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7555" LastEditorDisplayName="user20" LastEditDate="2015-03-14T23:55:58.423" LastActivityDate="2015-12-18T13:20:37.173" Title="Who is the murderer? - logic puzzle" Tags="|logical-deduction|mystery|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6622" PostTypeId="2" ParentId="6621" CreationDate="2015-01-02T16:55:27.853" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Third one is innocent&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation follows&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First: &quot;I am not guilty&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second: &quot;Third is the murderer&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third: &quot;First is the Murderer&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets assume First and Second are lying and Third is telling the truth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we get the following statements.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;First one is the murderer. &lt;br /&gt;Third one is innocent. &lt;br /&gt;First one is the murderer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, lets assume Second and Third are lying and First is telling the truth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we get the following statements.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;First one is innocent. &lt;br /&gt;Third is innocent. &lt;br /&gt;First one is innocent.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, lets assume First and Third are lying and Second is telling the truth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we get the following statements.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;First one is Murderer. &lt;br /&gt;Third is Murderer. &lt;br /&gt;First one is innocent. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which contradicts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So from the first two valid assumptions it is clear that &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Third is definitely innocent.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now take the two valid assumptions&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1)First one is murderer and Third one is innocent. So Second too will be innocent.&lt;br /&gt; 2)First and Third are innocent. So Two is the murderer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the chances for First and Second to be murderer are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;50-50&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="2730" LastEditDate="2015-01-02T17:15:27.307" LastActivityDate="2015-01-02T17:15:27.307" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6624" PostTypeId="2" ParentId="6552" CreationDate="2015-01-02T17:13:23.350" Score="10" Body="&lt;p&gt;I like to use truth tables to answer these kinds of questions.  The truth table for this question is simplified because it's given that only one suspect is telling the truth (&lt;em&gt;I'm assuming only one person committed the crime&lt;/em&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AZVYv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first four columns show the suspects and the scenarios of who possibly lied and who possibly told the truth.  The second four columns show what can be gleaned by assuming that the respective suspect is the only one who told the truth.  Here are my comments on each row:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Alice is telling the truth&lt;/em&gt;: This results in both Carlos and John being guilty.  This is a &lt;strong&gt;contradiction&lt;/strong&gt; based on my assumption.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;John is telling the truth&lt;/em&gt;: We can say that Diana did not do it because Carlos is lying, but that means that Diana would be telling the truth.  Since we are assuming it is in fact only John that is telling the truth in this scenario, we have a &lt;strong&gt;contradiction&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Carlos is telling the truth&lt;/em&gt;. This results in both John and Diana being guilth.  This is a &lt;strong&gt;contradiction&lt;/strong&gt; based on my assumption.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Diana is telling the truth&lt;/em&gt;. This means that &lt;strong&gt;John is the criminal&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2318" LastActivityDate="2015-01-02T17:13:23.350" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6625" PostTypeId="2" ParentId="6569" CreationDate="2015-01-02T17:28:59.497" Score="7" Body="&lt;p&gt;From the point of view of Charlie (the one who holds the card to be identified and verifies Bob's answer), a trick when Charlie himself pulls the card from Alice's hands will look more showy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make this possible-&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; we need to be able to encode any one of 52-4=48 cards. The four cards can encode any pair of 4!=24 pairs of cards (thus covering all 48 cards), therefore we need to send Bob an extra bit of information (pointing to a single card in the pair of cards identified by the four cards). We can pass this single bit with any secret sign (with sight, head, elbow movement, orientation of cards on the table etc.)&lt;br&gt;&lt;br&gt;Neither base card, nor offset is used with encoding. Any one of 52-4=48 cards is encoded with one of 2*4!=48 possible states. Here 4! states are covered by the four cards, and 2 possible states (a single bit) are covered with a secret sign.&lt;br&gt;&lt;br&gt;The way we encode is similar to the one described by Kate Gregory. If the smallest card of the four is A, the next bigger one is B, then C, and the biggest one is D, we can encode each of the rest 24 pairs of cards like ABCD, ABDC, ACBD, ACDB, ADBC, ADCB, BACD, BADC, BCAD, BCDA, BDAC, BDCA, CABD, CADB, CBAD, CBDA, CDAB, CDBA, DABC, DACB, DBAC, DBCA, DCAB, DCBA.&lt;br&gt;&lt;br&gt;Another method is when Alice asks Charlie to take six cards and then asks him to pull one, while the other five provide redundant space to encode any card (with one of 52-5=47 useful states among 5!=120 possible states).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7668" LastEditorUserId="37211" LastEditDate="2018-09-09T04:36:15.427" LastActivityDate="2018-09-09T04:36:15.427" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="6627" PostTypeId="1" AcceptedAnswerId="6628" CreationDate="2015-01-02T19:54:53.113" Score="8" ViewCount="589" Body="&lt;p&gt;AZ -&gt; +1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TN -&gt; +1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-4 &amp;lt;- NV&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-6 &amp;lt;- PA&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GA -&gt; +4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What US city is this? (2 words)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clue:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Microsoft&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7671" LastEditorUserId="8874" LastEditDate="2015-03-05T16:20:42.133" LastActivityDate="2015-03-05T16:20:42.133" Title="Find this US city" Tags="|geography|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6628" PostTypeId="2" ParentId="6627" CreationDate="2015-01-02T20:14:43.663" Score="11" Body="&lt;p&gt;The city is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BATON ROUGE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I got it by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Take all the combinations of two letters in order AA, AB, AC, ... ZX, ZY, ZZ.&lt;br&gt;&#xA; Clue AZ -&gt; +1 means to move forward one in the list from AZ, which would be BA&lt;br&gt;&#xA; Same goes for the other clues, with the negative numbers meaning to go backwards in the list, giving us BA, TO, NR, OU, GE as the new pairs.&lt;br&gt;&#xA; The clue didn't help as I associate Microsoft with Washington.  I didn't know they had anything special going on in Louisiana.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="239" LastActivityDate="2015-01-02T20:14:43.663" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6629" PostTypeId="2" ParentId="6564" CreationDate="2015-01-02T21:05:57.627" Score="1" Body="&lt;p&gt;Here is a different approach that uses &lt;strong&gt;basic combinatorics&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Denote with $h$ the number of honest senators. &lt;/li&gt;&#xA;  &lt;li&gt;We already know that $h \ge$ 1 from the problem&lt;/li&gt;&#xA;  &lt;li&gt;The number of pairs of honest senators is given by the 2-combinations of $h$: there are $\binom h2$ of them. &lt;/li&gt;&#xA;  &lt;li&gt;As the problem says, this must be zero, hence $\binom h2 = 0 \implies 2 &amp;gt; h \implies h \le 1$.&lt;/li&gt;&#xA;  &lt;li&gt;Combine with the previous result to get $h$ = 1.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7636" LastEditorUserId="7435" LastEditDate="2015-01-24T11:11:05.307" LastActivityDate="2015-01-24T11:11:05.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6630" PostTypeId="2" ParentId="6618" CreationDate="2015-01-02T21:10:45.633" Score="4" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;1 &amp;amp; 9: She's got a ticket to Ryde - starting at Portsmouth (5)&lt;/p&gt;&#xA;&lt;p&gt;Before playing the numbers on an Air Serbia flight (9)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Starting with Portsmouth (P) and on to Ryde (RIDE), before (PRE) playing numbers (DICE) on Air Serbia (IATA code: JU). Put it together and you get:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Pride_and_Prejudice&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pride and Prejudice&lt;/a&gt; by Jane Austen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;17 &amp;amp; 4: A member of a Gallic mendicant religious order, spun unluckily - don't be silly (5)&lt;/p&gt;&#xA;&lt;p&gt;Sensitiveness, aesthetic appreciativeness, capacity of emotion (11) (after 17 across)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Not sure about the French monk, suspect an anagram somewhere... but &amp;quot;don't be silly&amp;quot; = have some SENSE. To be sensitive, to appreciate aethetics and to be capable of emotion are all different ways to have SENSE ABILITY, so putting those two together gives us:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Sense_and_Sensibility&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sense and Sensibility&lt;/a&gt; by Jane Austen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ol start=&quot;18&quot;&gt;&#xA;&lt;li&gt;See a snotty cold; a dripping nose, it looks like. (1,4,4,1,4)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Still working on the reasoning, but based on the lengths and the filled in letters, I'm pretty sure it's:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/A_Room_with_a_View&quot; rel=&quot;nofollow noreferrer&quot;&gt;A Room With a View&lt;/a&gt; by E.M. Forster&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ol start=&quot;15&quot;&gt;&#xA;&lt;li&gt;Impressive detective; an annoying producer of electronic music (4,4)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;A detective is often called a private DICK, and whether you think he's annoying or not, MOBY is certainly known for his electronic music sound. The resulting creature is certainly impressive:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Moby-Dick&quot; rel=&quot;nofollow noreferrer&quot;&gt;Moby Dick&lt;/a&gt; by Herman Melville&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;" OwnerUserId="4518" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-01-06T21:47:29.817" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6632" PostTypeId="2" ParentId="6584" CreationDate="2015-01-02T21:33:11.307" Score="0" Body="&lt;p&gt;It's totally easy if you think about it.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Replace all letters with &lt;strong&gt;a&lt;/strong&gt; digit and solve the equations (letter&lt;strong&gt;s&lt;/strong&gt; vs digit).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Replace all of the ten distinct letter&lt;strong&gt;s&lt;/strong&gt; in the scheme with a &lt;strong&gt;single digit&lt;/strong&gt; (the letters are distinct but not the digit)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You must not assign different digits to different letters.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So we must assign one and the same digit to every letter! Just try with 0, 1, 2, 3, and that's it. All letters correspond to the number 3. Even with the most naive approach, we only need to test &lt;strong&gt;10&lt;/strong&gt; possible combinations in the worst case, not 10000000000. This puzzle makes programmers angry! :-)&lt;/p&gt;&#xA;" OwnerUserId="7636" LastActivityDate="2015-01-02T21:33:11.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6634" PostTypeId="2" ParentId="6594" CreationDate="2015-01-03T03:41:57.873" Score="8" Body="&lt;p&gt;The answer is quite simple&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;The end result&lt;/em&gt; of a &lt;strong&gt;tic tac toe&lt;/strong&gt; (or &lt;strong&gt;Xs and Os&lt;/strong&gt;, &lt;strong&gt;Noughts/Knaughts and Crosses&lt;/strong&gt;, etc. See; &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Tic-tac-toe#English_names&quot;&gt;http://en.m.wikipedia.org/wiki/Tic-tac-toe#English_names&lt;/a&gt;) game.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Why?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A paper pound of wrong and round,&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The game is played on paper with Xs (wrong) and Os (round). Pound having to do with &lt;strong&gt;#&lt;/strong&gt; which is the shape of a tic tac toe &lt;strong&gt;board&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A minty piggy named.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A &lt;strong&gt;tic tac&lt;/strong&gt; is minty. Then there's piggy &lt;strong&gt;toe&lt;/strong&gt;!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dividing nine, some yours, some mine,&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The game is split us into &lt;strong&gt;9 slots&lt;/strong&gt;. Some slots are &lt;strong&gt;owned&lt;/strong&gt; by &lt;strong&gt;you&lt;/strong&gt;, &lt;strong&gt;some&lt;/strong&gt; by your &lt;strong&gt;friend&lt;/strong&gt;, Xs and Os!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A syzygy is claimed.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In astronomy a syzygy is the alignment of &lt;strong&gt;3&lt;/strong&gt; celestial bodies! In other words, &lt;strong&gt;3 in a row is achieved&lt;/strong&gt;, or a &lt;em&gt;&quot;tic-tac-toe&quot;&lt;/em&gt; Win!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;*One &lt;strong&gt;note&lt;/strong&gt; [&lt;strong&gt;POSSIBLE SPOILER&lt;/strong&gt;];&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Games can end in &lt;strong&gt;ties&lt;/strong&gt;, so the last line is &lt;strong&gt;arguably correct&lt;/strong&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2015-01-06T02:24:17.423" LastActivityDate="2015-01-06T02:24:17.423" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6635" PostTypeId="1" CreationDate="2015-01-03T04:06:48.630" Score="6" ViewCount="1273" Body="&lt;p&gt;I have a bit of pseudocode for you, and then an encoded string. The pseudocode will help you untangle the string. It's not the most easily understood pseudocode, and that might (just might) make your task a little bit harder.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;String[] pre to new Preamble[] with void, establish &amp;amp; perform:&#xA;    use package (alphabet) as al&#xA;    use package (keyboard) as qwerty&#xA;    return [answer(),getChar()]&#xA;&#xA;String[] answer to new Function[] with x as parameter, perform:&#xA;    rename x to ca&#xA;    set ca to new String[]&#xA;        ~ w/ ca to al.upper[3] &amp;amp; al.lower[indices(3,5,19,5,18)]&#xA;    set ci to new String[]&#xA;        ~ w/ ci to al.upper[9] &amp;amp; al.lower[indices(13,16,15,19,19,9,2,12,5)]&#xA;    set ac to new Int[ca.length]&#xA;    set ic to new Int[ci.length]&#xA;    return ca if ac &amp;lt; ic else ci&#xA;&#xA;Char[] getChar to new Function[] with void, perform:&#xA;    set target as new Char[]&#xA;        ~ w/ target to max(min(range(4.25*4,floor(pi**e))),0,1,2,3,-∞)&#xA;    return target&#xA;&#xA;{EOF}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And then, the code:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;WFE3gAk2gbE3d2U3LKsngqHqgbE3fqLjd2U1CbzjfAM6LVDwOXCbfAM1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In order to receive full credit, explain the pseudocode to me and give me the answer.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Hints&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The pseudocode is based off of Python, JavaScript, Java, and C++&lt;br&gt;The &lt;code&gt;indices&lt;/code&gt; function returns its parameters one at a time, for the purpose of its name.&lt;br&gt;&lt;code&gt;w/&lt;/code&gt; is sometimes used in texting with the same meaning....&lt;br&gt;For the encoded string, you typically don't see these types of codes without a certain character, given by the &lt;code&gt;getChar&lt;/code&gt; function.&lt;br&gt;The &lt;code&gt;qwerty&lt;/code&gt; variable contains all characters on the standard keyboard, and, for reference, this it what it would be: &lt;code&gt;[&quot;ESC&quot;,&quot;F1&quot;,&quot;F2&quot;,...&quot;F12&quot;,&quot;INS&quot;,&quot;DEL&quot;,&quot;1&quot;,&quot;2&quot;,...,&quot;0&quot;,&quot;-&quot;,&quot;=&quot;,&quot;BCKSPCE&quot;,&quot;\t&quot;,&quot;Q&quot;,&quot;W&quot;,&quot;E&quot;,...,&quot;P&quot;,&quot;{&quot;,&quot;}&quot;,&quot;|&quot;,&quot;CAPS&quot;,&quot;A&quot;,&quot;S&quot;,&quot;D&quot;,...,&quot;L&quot;....]&lt;/code&gt; - The characters are capitalized, but there are no shifted characters.&lt;br&gt;The arrays perform on a 1-based system (i.e. $5$ would be index $3$ of $[9,30,5,2,\infty]$)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7307" LastEditorUserId="5373" LastEditDate="2015-04-10T09:04:21.177" LastActivityDate="2016-03-22T23:58:38.093" Title="A riddle and a line of text" Tags="|cipher|computer-puzzle|" AnswerCount="1" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="6636" PostTypeId="1" AcceptedAnswerId="6675" CreationDate="2015-01-03T05:48:14.877" Score="48" ViewCount="2719" Body="&lt;p&gt;In chess, Knights moves to squares a distance $\sqrt5$ away. Bishops move distances $\sqrt2$, $\sqrt8$, $\sqrt{18}$, etc. Both pieces are restricted to non-integer distance moves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Enter the Knishop, a most potent chess piece capable of moving to &lt;em&gt;any&lt;/em&gt; squares at non-integer distance: $\sqrt2$, $\sqrt5$, $\sqrt8$, $\sqrt{10}$, $\sqrt{13}$, $\sqrt{18}$, ... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a chess board of unlimited extent, place three Knishops such that they can't capture each other, while any piece being added to the board can be captured in a single move.&lt;/p&gt;&#xA;" OwnerUserId="7573" LastEditorUserId="7573" LastEditDate="2015-01-05T17:14:53.387" LastActivityDate="2015-01-05T17:14:53.387" Title="Pythagorean Chess: Knishops" Tags="|mathematics|chess|" AnswerCount="4" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="6637" PostTypeId="1" AcceptedAnswerId="6642" CreationDate="2015-01-03T06:06:52.343" Score="7" ViewCount="3150" Body="&lt;p&gt;$1,20,171,?,9625,57756,288775$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please find the missing number in the series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have already tried with squares, cubes and progressions but couldn't find the problem.&lt;/p&gt;&#xA;" OwnerUserId="4183" LastEditorUserId="11876" LastEditDate="2015-09-25T05:06:32.647" LastActivityDate="2018-08-27T10:26:38.780" Title="Number Series - 1,20,171" Tags="|calculation-puzzle|pattern|number-sequence|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6639" PostTypeId="1" AcceptedAnswerId="6645" CreationDate="2015-01-03T07:10:59.523" Score="2" ViewCount="1566" Body="&lt;p&gt;Can you determine the next number in the below sequence?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1.0 , 4.0 , 6.9 , 9.0 , 10.8 , 12.0 , 14.0 , 16.0 , ?$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus:&lt;/strong&gt; Determine the next number that comes after '&lt;strong&gt;?&lt;/strong&gt;'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I'll post some hints if nobody gets the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #1:&lt;/strong&gt; You won't get the relationship between the numbers by calculating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #2:&lt;/strong&gt; These numbers have some relationship with science.&lt;/p&gt;&#xA;" OwnerUserId="7482" LastEditorUserId="4050" LastEditDate="2015-01-03T17:49:11.903" LastActivityDate="2019-04-19T06:03:34.007" Title="What is the next number in this sequence?" Tags="|pattern|number-sequence|" AnswerCount="1" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6640" PostTypeId="2" ParentId="6636" CreationDate="2015-01-03T07:25:45.637" Score="1" Body="&lt;p&gt;I've been slightly discouraged by the other answers (TL;DR), but here are my thoughts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking that they would have to be placed on paths of perfect right triangles (i.e. $l=3,w=4,\because\sqrt{3^2+4^2}=5$) I'd imagine that there are an infinite amount of these triangles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some of the movesets that knishops &lt;em&gt;cannot&lt;/em&gt; move along:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\pm(3x,4x)$&lt;/li&gt;&#xA;&lt;li&gt;$\pm(0,x)$&lt;/li&gt;&#xA;&lt;li&gt;$\pm(5x,12x)$&lt;/li&gt;&#xA;&lt;li&gt;$\pm(6x,8x)$&lt;/li&gt;&#xA;&lt;li&gt;$\pm(9x,12x)$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Since there are an infinite number of these triangles, there are, as a result, an infinite number of places that the knishops cannot move to; it is therefore impossible to place three bishops that they can capture every other piece on the infinite board but not each other; such a set up would require that the knishops fall into each other's 'out-of-reach spots' if you will; then, each knishop would have an infinite number of 'out-of-reach spots', so, as a result, there exist at least one $(x,y)$ coordinate that falls out of reach of all three knishops. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After thoroughly reading the other answers, I fear that my intuition is correct; proving it, however, will be harder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conjecture: Let $K$ be the set of knishops on the board; each knishop is a coordinate pair, as $(x,y)$. Then, our conjecture becomes, in order to prove that the alleged setup is indeed impossible: $\exists x,y\in\mathbb{Z}:\sqrt{x^2+y^2}\in\mathbb{Z}$ and that $\not\exists w,z\in\mathbb{Z}\land\sqrt{w^2+z^2}\notin\mathbb{Z}:(x\cdot w,y\cdot z)\notin K$. This states that there does not exists a path on which a knishop can travel to $(x,y)$ by means of having a line $\overline{wz}$ whose length is not an integer (which would entail the knishop being able to travel along it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Proof.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us prove this by making a contradiction when assuming the negative of the conjecture; that is, $\not\exists x,y\in\mathbb{Z}:\sqrt{x^2+y^2}\in\mathbb{Z}\land\exists w,z\in\mathbb{Z}\land\sqrt{w^2+z^2}\in\mathbb{Z}:(x\cdot w,y\cdot z)\in K$, in other words, assuming the setup is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, by our presumed false negative conjecture, the coordinates $(xw,yz)$ would be those of a knishop in $K$. $(x,y)$ would be the coordinates of a piece on our infinite board regardless. Proving that $\sqrt{w^2+z^2}\notin\mathbb{Z}$ would be sufficient to prove our conjecture.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[WIP]&lt;/p&gt;&#xA;" OwnerUserId="7307" LastEditorUserId="7307" LastEditDate="2015-01-03T21:45:05.327" LastActivityDate="2015-01-03T21:45:05.327" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6641" PostTypeId="2" ParentId="6552" CreationDate="2015-01-03T07:41:13.440" Score="4" Body="&lt;p&gt;Another way to approach it is using the fact that you not only know that only one of them is telling the truth, but also that there are enough information to solve the puzzle:&lt;/p&gt;&#xA;&lt;p&gt;If John was telling the truth, his piece of information (“I did not do it.”) would not answer the question (&amp;quot;Who did it?&amp;quot;), so you can assume he is lying, which means he did it.&lt;/p&gt;&#xA;&lt;p&gt;Of course this approach only works for puzzles, but I thought it would worth mentioning :o)&lt;/p&gt;&#xA;&lt;h3&gt;Attempt for a more detailed explanation as requested in the comments:&lt;/h3&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You know that you have enough information to solve the puzzle&lt;/li&gt;&#xA;&lt;li&gt;You know that only one of the suspects is telling the truth&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Now, imagine John is telling the truth:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;He said &amp;quot;I didn't do it&amp;quot;&lt;/li&gt;&#xA;&lt;li&gt;This would mean that one of the 3 others suspects did it&lt;/li&gt;&#xA;&lt;li&gt;This is not enough information to solve the puzzle (since we want only one name)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Therefor, he must be lying, and someone else must be telling the truth.&lt;/p&gt;&#xA;&lt;p&gt;But since he is lying, and he is saying &amp;quot;I didn't do it&amp;quot;, it means &amp;quot;he didn't NOT do it&amp;quot;, which can be simplified into &amp;quot;he did it&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;Afterthought: alternatively, Carlos and Diana say the opposite of each other, so one of them has to tell the truth while the second is lying; knowing that only one of the four suspects is telling the truth, you can assume that the two others are lying (including John)&lt;/p&gt;&#xA;" OwnerUserId="7682" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-01-03T14:20:42.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6642" PostTypeId="2" ParentId="6637" CreationDate="2015-01-03T07:55:33.103" Score="18" Body="&lt;p&gt;Starting with $1$, you proceed as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Add $1$, and multiply the result by $10$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This gives you $20$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Subtract $1$, and multiply the result by $9$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This gives you $171$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Add $1$, and multiply the result by $8$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This gives you $1376$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Subtract $1$, and multiply by $7$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This gives you $9625$. Etc.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;How do you get to this result? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the last three numbers given. If you take the ratio of the last two values ($288775/57756$), you get a result very close to $5$. If you take the ratio $57756/9625$ You get a result very close to $6$. From there on it's just some fiddling...&lt;/p&gt;&#xA;" OwnerUserId="7573" LastEditorUserId="7573" LastEditDate="2015-01-03T08:10:46.527" LastActivityDate="2015-01-03T08:10:46.527" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6643" PostTypeId="2" ParentId="6636" CreationDate="2015-01-03T09:04:39.613" Score="5" Body="&lt;p&gt;Knishops can't take pieces in their row or column, because those are all integer distances.&#xA;They can't all be in a line because that entire line will be free for other pieces to sit. If two of them are in a line, the other one will have a square on that line which it can't take and thus other pieces can sit there.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Thus each of the pieces must be a pythagorean triangle away from others - not the same column or line, but an integer distance. They can't be at linear multiples of the same pythagorean triangle, because later linear multiples will also be unreachable by the Knishops.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are probably infinitely many such solutions, but here's one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One piece is at square 0,0 (for convenience). The next piece is at square 20, 48 (which is four times 5, 12 --&gt; 13). The third piece is at square 9, -12 (which is three times 3, 4 --&gt; 5, but below the x-axis). So the distance from 1st to 2nd is 52; from 1st to 3rd is 15; and from 2nd to 3rd is the distance of 20-9, 48+12 which is 11,60 --&gt; 61.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One thing I have not proved is that there are no squares anywhere else on the board that are unreachable by my three knishops. I'll add that if I can.&lt;/p&gt;&#xA;" OwnerUserId="6031" LastActivityDate="2015-01-03T09:04:39.613" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6644" PostTypeId="2" ParentId="6579" CreationDate="2015-01-03T11:31:18.660" Score="2" Body="&lt;p&gt;The barber doesn't shave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the barber is a woman or a child or perhaps he's a man with a great big bushy beard and moustache or maybe he's undergoing hormone treatment and doesn't need to shave - doesn't matter which.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want it to be a paradox then you need to add that all the men in the village are clean-shaven and that they are all shaved by someone from the village and that only men are allowed to be barbers (or that kind of thing). And that magic doesn't exist and there aren't any auto-shaving machines. Also no access to electrolysis or waxing or depilatory creams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also (if you want it to be a paradox) you might want to stipulate that the number of men is &gt; 0 and the number of barbers is &gt;= 1. And that the people are &lt;em&gt;homo sapiens&lt;/em&gt; and that none of them is undergoing medical treatment which could cause hair loss, or has radiation poisoning.&lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="4050" LastEditDate="2015-01-04T17:00:24.857" LastActivityDate="2015-01-04T17:00:24.857" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="6645" PostTypeId="2" ParentId="6639" CreationDate="2015-01-03T16:15:33.340" Score="7" Body="&lt;p&gt;The numbers represent&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Atomic weights&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The next 5 numbers are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 19.0, 20.2, 23.0, 24.3, and 27.0&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1955" LastActivityDate="2015-01-03T16:15:33.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6646" PostTypeId="2" ParentId="6612" CreationDate="2015-01-03T16:43:51.920" Score="1" Body="&lt;p&gt;A tenuous guess, but I think the answer might be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the word I&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This guess comes from&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; taking the clues &quot;&lt;em&gt;What am I?&lt;/em&gt;&quot; and the strange wording &quot;&lt;em&gt;..., I is&lt;/em&gt;&quot; literally.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Trying to make the rest of the clues fit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;a linguistic object of rare beauty&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's a rare example of a single-letter word? Or a pronoun that's capitalized by default?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Solid&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the sans-serif I is just solid rectangle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;misshapen&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a different shape than the usual lowercase form&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;sad&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No idea on this one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-01-03T16:43:51.920" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6647" PostTypeId="1" CreationDate="2015-01-03T17:27:59.833" Score="4" ViewCount="435" Body="&lt;p&gt;We have 12 golfers and 3 foursomes everyday for 4 days.  What is the formula so each golfer plays with everyone else at least once, but no more than twice.  Obviously each golfer will only play twice with one other golfer one time.  Help!&lt;/p&gt;&#xA;" OwnerUserId="7686" LastActivityDate="2015-01-05T22:13:53.157" Title="Friendly Golfers" Tags="|mathematics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6648" PostTypeId="2" ParentId="6647" CreationDate="2015-01-03T18:12:28.343" Score="10" Body="&lt;p&gt;It is not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's label the players so that on the first day we have the groups:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ABCD  EFGH  IJKL&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Let's color the first group (A,B,C,D) red.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On every subsequent day, at least two red players will play together.  Let's consider the 3 groups, one for each subsequent day, where two red players play together.  In these groups there are 6 appearances of red players.  One or two red players must appear twice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say, WLOG, that A appears twice, on day 2 and 3.  If we look at A's schedule, it must look like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1: ARRR  2: AR??  3: AR??  4: A???&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where R is a red player and ? any player, red or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there we can see that A can play with at most 7 different non-red players.  But there are 8  of them.  Therefore A doesn't play with everybody.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2015-01-05T22:13:53.157" LastActivityDate="2015-01-05T22:13:53.157" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6649" PostTypeId="2" ParentId="6636" CreationDate="2015-01-03T18:19:53.063" Score="16" Body="&lt;p&gt;This is not a solution, but it is much too long to be a comment. If a solution exists, this post gives the theoretical machinery to verify it and to narrow the search. However, I have checked a considerable number of cases without finding a solution, which suggests that there may be none, or that if there is, it will be hard to find. But at least now we can make people sad when they find an incorrect solution. :)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The knishops clearly must form a non-degenerate triangle with integer side lengths and points on the integer lattice (i.e. both $(x,y)$ coordinates are integers). These two conditions are equivalent to demanding that the knishops be the vertices of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Heronian_triangle&quot;&gt;Heronian triangle&lt;/a&gt;. Moreover, by the condition that every square must be reachable by one of the knishops, we can exclude any solution where two knishops have the same $x$ or $y$ coordinate - in particular, if we had the knishops at $(0,0)$, $(x,0$) and $(x',y')$, then the reflection of the last point over the line between the first two, $(x',-y')$, would have integer distances to each point and thus be uncovered. Reflecting over one of the legs of a right triangle yields a similar problem, meaning the distances between the knishops must not form a Pythagorean triple. We could take this argument further to notice that we need that the reflection of one knishop over the line between the other two must not lie on a lattice point, since the distance between the knishop and its reflection is an integer (due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Pick%27s_theorem&quot;&gt;Pick's theorem&lt;/a&gt;), and obviously so is the distance from each other knishop to the reflection - thus, if it were a lattice point, it would not be covered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply applying that the Heronian triangle must be drawn on an integer lattice with no two points with the same $x$ or $y$ coordinates and going down the list of Heronian triangles on Wikipedia yields that the first suitable Heronian triangle is the one with side lengths $5$, $29$, $30$. This gives coordinates for the knishops of $(0,0)$, $(3,4)$, and $(21,-20)$. If we wanted to check for uncovered squares, we could set up three equations in integers $d_1,d_2,d_3$ being the distances of a hypothetical uncovered position $(x,y)$:&#xA;$$d_1^2=x^2+y^2$$&#xA;$$d_2^2=(x-3)^2+(y-4)^2$$&#xA;$$d_3^2=(x-21)^2+(y+20)^2$$&#xA;and wish to show that the only integer $(x,y)$ with an integer solution to the above are the positions of the knishops - that is, every other square is covered. However, we can note that by the triangle inequality, we have:&#xA;$$|d_1-d_2|\leq 5$$&#xA;$$|d_1-d_3|\leq 29$$&#xA;meaning that there are only finitely many choices of $a,b$ such that it is possible that $d_2=d_1+a$ and $d_3=d_1+b$ - in particular, we choose $a$ as an integer in $[-5,5]$ and $b$ as an integer in $[-29,29]$ yielding $649$ possible choices. However, this is finite and means that if we find &lt;em&gt;every&lt;/em&gt; solution to&#xA;$$d_1^2=x^2+y^2$$&#xA;$$(d_1+a)^2=(x-3)^2+(y-4)^2$$&#xA;$$(d_1+b)^2=(x-21)^2+(y+20)^2$$&#xA;which is a system of $3$ independent polynomials in $3$ variables (being the intersection of $3$ hyperboloids) and hence has finitely many solutions. Thus, we can generate, using Mathematica (or your favorite software; numerical solutions suffice, since we only wish to show that something is not an integer, and rather crude bounds suffice for that), every single solution to the above equations for integer $a$ and $b$ in the proper ranges. This yields a list of uncovered squares, like $(-15,-20)$. Mathematica code follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;CheckPair[x1_, y1_, x2_, y2_] := Module[{z, d1, d2},&lt;br&gt;&#xA;    d1 = Norm[{x1, y1}];&lt;br&gt;&#xA;    d2 = Norm[{x2, y2}];&lt;br&gt;&#xA;    If[! (IntegerQ[d1] &amp;amp;&amp;amp; IntegerQ[d2]), Return[False]];&lt;br&gt;&#xA;    z = Solve[{d^2 == &#xA;        x^2 + y^2, (d + a)^2 == (x - x1)^2 + (y - y1)^2, (d + &#xA;           b)^2 == (x - x2)^2 + (y - y2)^2}, {d, x, y}];&lt;br&gt;&#xA;    DeleteDuplicates[&#xA;     Select[{Floor[x + .5], Floor[y + .5]} /. &#xA;       Join @@ (Join @@ &#xA;          Table[N[z /. {a -&gt; m, b -&gt; n}], {m, -d1, d1}, {n, -d2, d2}]), &#xA;      IntegerQ[Norm[#]] &amp;amp;&amp;amp; IntegerQ[Norm[# - {x1, y1}]] &amp;amp;&amp;amp; &#xA;        IntegerQ[Norm[# - {x2, y2}]] &amp;amp;]]];&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where the function CheckPair takes as input the positions of two knishops relative the other (i.e. it assumes one is at $(0,0)$) and outputs a list of exceptions (or false if the input is invalid). I used this, along with a simple brute force search to figure out how to draw each triangle on a lattice, to check up to the triple $(5,51,52)$ on the list of Heronian triangles on Wikipedia. I did not find any solutions.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I have continued my computations by writing code to generate Heronian triples. I've left it running for several hours, and not found anything, but the computation becomes very slow after a point. Even filtering by the criteria in the first paragraph seems insufficient to make the brute force search feasible; additional insight will be required.&lt;/p&gt;&#xA;" OwnerUserId="4765" LastEditorUserId="4765" LastEditDate="2015-01-03T21:29:20.790" LastActivityDate="2015-01-03T21:29:20.790" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6651" PostTypeId="2" ParentId="6552" CreationDate="2015-01-03T20:44:04.627" Score="3" Body="&lt;p&gt;&lt;strong&gt;The authorities know that only one is telling the truth&lt;/strong&gt;&lt;br&gt;&#xA;This indicates the solution is revealed after figuring out which suspect is telling the truth.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Alice said “Carlos did it.”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1.  Is Alice's statement true?&lt;/strong&gt;&lt;br&gt;&#xA;   If Alice's satement were considered to be true it would mean Carlos did it, but John said he didn't do it which would make John's statement also true if Carlos did it. There can only be one true statement. So Alice is lying.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Carlos said “Diana did it.”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2.  Is Carlos's statement true?&lt;/strong&gt;&lt;br&gt;&#xA;If Carlos's satement were considered to be true it would mean Diana did it, but John said he didn't do it which would make John's statement also true if Diana did it. There can only be one true statement. So Carlos is lying.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Diana said “Carlos lied when he said that I did it.”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3.  Is Diana's statement true?&lt;/strong&gt;&lt;br&gt;&#xA;We already proved Carlos lied. So when Diana said “Carlos lied when he said that I did it.” Diana is telling the truth&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;John said “I did not do it.”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4.  Is John's statement true?&lt;/strong&gt;&lt;br&gt;&#xA;We proved Diana is telling the truth. We know only one statement is true. So when John said “I did not do it.” he was lying and guilty.&lt;/p&gt;&#xA;" OwnerUserId="7688" LastEditorUserId="39" LastEditDate="2015-02-22T13:54:01.710" LastActivityDate="2015-02-22T13:54:01.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6652" PostTypeId="1" AcceptedAnswerId="6662" CreationDate="2015-01-03T23:05:27.107" Score="16" ViewCount="674" Body="&lt;p&gt;As the human explorers ventured into an uncharted system, they got the feeling that their presence was expected. In fact, the stars themselves greeted them.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ojJTk.png&quot; alt=&quot;Stars&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;What was the greeting?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="7308" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-01-04T10:47:07.167" Title="Reading the Stars" Tags="|visual|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6653" PostTypeId="1" CreationDate="2015-01-04T00:23:08.090" Score="16" ViewCount="112086" Body="&lt;p&gt;If you choose an answer to this question at random, what is the chance you will be correct?&lt;/p&gt;&#xA;&lt;p&gt;A) 25%&lt;/p&gt;&#xA;&lt;p&gt;B) 50%&lt;/p&gt;&#xA;&lt;p&gt;C) 60%&lt;/p&gt;&#xA;&lt;p&gt;D) 25%&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qhIKi.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qhIKi.jpg&quot; alt=&quot;&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;https://pbs.twimg.com/media/B57kNq4IYAEA-Zj.jpg:large&quot; rel=&quot;noreferrer&quot;&gt;twimg.com&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="7678" LastEditorUserId="69582" LastEditDate="2021-04-07T19:47:57.100" LastActivityDate="2021-04-08T00:19:30.487" Title="If you choose an answer to this question at random, what is the chance you will be correct?" Tags="|mathematics|probability|" AnswerCount="3" CommentCount="4" ClosedDate="2021-04-13T10:42:23.783" ContentLicense="CC BY-SA 4.0" />
  <row Id="6655" PostTypeId="2" ParentId="6653" CreationDate="2015-01-04T00:40:06.127" Score="23" Body="&lt;p&gt;When limited to the 4 provided answers, there is no correct answer, because this question is a paradox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When there is one correct answer, the chance to pick that answer would be 25%. However, the answer 25% exists twice, so there is a 50% chance of picking it. But there is only one answer which says 50% and the chance to pick that answer is 25%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is no instruction in the question that you need to pick one of the &lt;em&gt;provided&lt;/em&gt; answers. In that case &quot;picking an answer at random&quot; would mean that I would respond with a completely random statement like &quot;A&quot;, &quot;Y&quot;, &quot;3.14159&quot;, &quot;Warshaw&quot;, &quot;jksdaskfa&quot; or &quot;Bob is the one who stole the strawberry cake&quot;. My answer could be literally anything, so there is an unlimited number of possible responses. Assuming that there is an unlimited set of incorrect answers but only a limited set of correct answers, the chance of picking a correct answer is &lt;code&gt;0%&lt;/code&gt; which would be my answer.&lt;/p&gt;&#xA;" OwnerUserId="1774" LastEditorUserId="1774" LastEditDate="2015-01-04T02:17:34.380" LastActivityDate="2015-01-04T02:17:34.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6656" PostTypeId="1" AcceptedAnswerId="6682" CreationDate="2015-01-04T00:53:00.753" Score="9" ViewCount="372" Body="&lt;p&gt;This is a riddle in &lt;a href=&quot;http://en.wikipedia.org/wiki/Double_dactyl&quot; rel=&quot;nofollow&quot;&gt;double-dactyl&lt;/a&gt; form:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hollywood-Shmollywood&lt;br&gt;&#xA;  Terrible ancients&lt;br&gt;&#xA;  Are now resurrected&lt;br&gt;&#xA;  In picture and sound.&lt;br&gt;&#xA;  “Amateur fadsters!” cry&lt;br&gt;&#xA;  ???????????????,&lt;br&gt;&#xA;  “We liked them when they were&lt;br&gt;&#xA;  Still underground.”  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Credit to R.M. for inspiration.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-07-01T19:44:08.823" LastActivityDate="2015-07-01T19:44:08.823" Title="Riddle double-dactyl: Hollywood-Shmollywood" Tags="|riddle|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6657" PostTypeId="2" ParentId="6653" CreationDate="2015-01-04T01:50:57.780" Score="15" Body="&lt;p&gt;You did not state the answer had to be &lt;strong&gt;uniformely random&lt;/strong&gt; so therefore I can look at the answers, recognize the paradox, and limit the answers to the 2 possible, &lt;em&gt;25%&lt;/em&gt; and &lt;em&gt;50%&lt;/em&gt;. Then the answer becomes 50% since there are 2 choices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TL;DR&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;50%&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2015-01-04T01:50:57.780" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6659" PostTypeId="2" ParentId="6612" CreationDate="2015-01-04T06:29:15.747" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bonsai or Bonsai tree.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I believe Bonsai could be characterized by &quot;Solid, misshapen, sad&quot; (that was my first read of the phrase)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The qualities a &lt;strong&gt;bonsai tree&lt;/strong&gt; and a &lt;strong&gt;linguistic object&lt;/strong&gt; have in common are: &lt;strong&gt;object acted upon by subject&lt;/strong&gt; (linguistic object: that acted upon by subject (wiki: Traditional grammar defines the object in a sentence as the entity that is acted upon by the subject.There is thus a primary distinction between subjects and objects that is understood in terms of the action expressed by the verb, e.g. Tom studies grammar...) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The qualities a &lt;strong&gt;bonsai tree&lt;/strong&gt; and a &lt;strong&gt;linguistic object of rare beauty&lt;/strong&gt; have in common are &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1) &lt;strong&gt;object acted upon by subject&lt;/strong&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2) &lt;strong&gt;individuality&lt;/strong&gt; and &lt;strong&gt;beauty&lt;/strong&gt;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I believe the &lt;strong&gt;bonsai tree&lt;/strong&gt; has &lt;strong&gt;individuality&lt;/strong&gt; and possesses a &lt;strong&gt;rare beauty&lt;/strong&gt;. A tie between &lt;strong&gt;bonsai tree&lt;/strong&gt; and a &lt;strong&gt;linguistic object&lt;/strong&gt; is &lt;strong&gt;rare beauty&lt;/strong&gt; if the &lt;strong&gt;linguistic object&lt;/strong&gt; is unique to an &lt;strong&gt;individual speaker&lt;/strong&gt; - in other words, &lt;strong&gt;idiolect&lt;/strong&gt;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In linguistics, an &lt;strong&gt;idiolect&lt;/strong&gt; is an individual's distinctive and unique use of language, including speech. This unique usage encompasses vocabulary, grammar, and pronunciation. &lt;strong&gt;Idiolect&lt;/strong&gt; is the variety of &lt;strong&gt;language unique to an individual&lt;/strong&gt;. (wiki: hxxp://en Wikipedia org wiki Idiolect). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A &lt;em&gt;linguistic object of rare beauty&lt;/em&gt; is in fact an &lt;strong&gt;idiolect&lt;/strong&gt; I would argue that &lt;strong&gt;language unique to an individual&lt;/strong&gt; possesses a &lt;strong&gt;rare beauty&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In fact, at another level, consider: &quot;&lt;strong&gt;Bonsai&lt;/strong&gt;&quot; is a &lt;strong&gt;unique linguistic construction itself&lt;/strong&gt; (plantings in tray, from &lt;strong&gt;bon&lt;/strong&gt;, a tray or low-sided pot and &lt;strong&gt;sai&lt;/strong&gt;, a planting or plantings) that I would consider &lt;strong&gt;rare and beautiful&lt;/strong&gt;: &quot;...a Japanese pronunciation of the earlier Chinese term penzai. A &quot;bon&quot; is a tray-like pot typically used in bonsai culture. The word bonsai is often used in English as an umbrella term for all miniature trees in containers or pots...&quot; (wiki) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This connection, analogy or commonality is reflected in at least one instance of its usage: &quot;&lt;em&gt;I say language but Gaelic isn’t one, not really. Its vocabulary is tiny, with no form of saying yes or no and attuned to a distant, pre-technological world. It’s essentially a kind of rural patois, a &lt;strong&gt;bonsai idiolect&lt;/strong&gt;; a way of specifying concepts central to a particular, highly codified way of life&lt;/em&gt;...&quot; (google for &quot;bonsai idiolect&quot;; there are many hits). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; thus, the answer is the &lt;em&gt;word itself&lt;/em&gt; &lt;strong&gt;bonsai&lt;/strong&gt; and &lt;strong&gt;&lt;em&gt;what it actually represents&lt;/em&gt;&lt;/strong&gt; - &lt;em&gt;physically and culturally&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; finally (ahem) &lt;em&gt;barking up the wrong &lt;strong&gt;tree&lt;/em&gt;&lt;/strong&gt;... &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; very nice!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Okay, okay, no more barking, how about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;em&gt;eye&lt;/strong&gt; of the &lt;strong&gt;beholder&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7569" LastEditorUserId="7569" LastEditDate="2015-01-04T14:05:05.453" LastActivityDate="2015-01-04T14:05:05.453" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6661" PostTypeId="2" ParentId="6652" CreationDate="2015-01-04T08:06:58.603" Score="8" Body="&lt;p&gt;I think what the starts try to tell humans simply is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Turn around!&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the reason being:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is a 2-fold rotational symmetry in the image, I.e. You can turn it around by 180 degrees and it is the same...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It should be added that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Other interpretations like &quot;your future is like your past&quot; or &quot;look behind you before going onward&quot; etc. are also possible, but I took the first one which jumped to head...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&#xA;On closer inspection (now with a computer at hand)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The image is not 100% rotational symmetric. Only some stars are. The following image highlights this in colour&lt;br&gt;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/jfP5f.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&lt;br&gt;&#xA; Red --&gt; Original stars&lt;br&gt;&#xA; Green --&gt; 180degree rotation stars &lt;br&gt;&#xA; Yellow --&gt; 180degree rotation overlapping positions&lt;br&gt;&#xA; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore I might need to look a bit deeper...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ignoring the 'yellow' parts leaves only dots which - rotated by 90 degrees' spell out PEACE.&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/B92qz.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="5840" LastEditDate="2015-01-04T10:47:07.167" LastActivityDate="2015-01-04T10:47:07.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6662" PostTypeId="2" ParentId="6652" CreationDate="2015-01-04T08:21:05.707" Score="14" Body="&lt;p&gt;The mesage is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Peace&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Juxtapose the image and a 180 degrees rotated version of the image, then remove all the stars that appear in both images, and the word &quot;Peace&quot; is formed twice.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/5oz4z.png&quot; alt=&quot;Peace&quot;&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My original answer would be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Z&quot;, because if you highlighted all the stars that are not square, you get a &quot;Z&quot;.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/lPRuC.png&quot; alt=&quot;Zshape&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6178" LastEditorUserId="6178" LastEditDate="2015-01-04T09:35:55.073" LastActivityDate="2015-01-04T09:35:55.073" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6663" PostTypeId="1" AcceptedAnswerId="6665" CreationDate="2015-01-04T09:27:37.813" Score="7" ViewCount="444" Body="&lt;p&gt;Ten matches lay on a table and look like this: &quot;XI+I=X&quot;. They create a correct equality. What is this equality? (Write it in arabic numeral system in your answer).&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="5840" LastEditDate="2015-01-04T10:39:48.897" LastActivityDate="2015-01-04T10:39:48.897" Title="Ten matches on a table" Tags="|matches|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6665" PostTypeId="2" ParentId="6663" CreationDate="2015-01-04T09:38:46.260" Score="12" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Turn it upside down: 10 = 1 + 9&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6031" LastActivityDate="2015-01-04T09:38:46.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6666" PostTypeId="2" ParentId="6612" CreationDate="2015-01-04T14:18:47.207" Score="2" Body="&lt;p&gt;NEW ATTEMPT:&#xA;Solid, misshapen, sad, I is - a linguistic object of rare beauty. What am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Hardest puzzle &quot;Blue Eyes&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solid&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hard&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;misshapen&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; puzzle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;sad&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; blue&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eyes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;linguistic object of rare beauty&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because its the hardest puzzle and hence a linguistic object( puzzle) or rare beauty( hardest).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hugh! I hope I have cracked it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OLD ATTEMPT:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eyes (I is)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solid&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Solid eyes ( A technology ).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Misshapen eyes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure, could be referring to an eye disease&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sad&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Emotive eyes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eye is -&gt; Eyes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A linguistic beauty of rare object&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eyes are used in couplets and poems to refer to an object of beauty , to speak the language of love.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Honest attempt :)&lt;/p&gt;&#xA;" OwnerUserId="2656" LastEditorUserId="2656" LastEditDate="2015-01-06T13:22:44.420" LastActivityDate="2015-01-06T13:22:44.420" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6667" PostTypeId="2" ParentId="6579" CreationDate="2015-01-04T17:16:37.773" Score="6" Body="&lt;p&gt;No, you are wrong. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The barber is stated to shave &quot;those &lt;strong&gt;people&lt;/strong&gt;, and only those people, who do not shave themselves&quot;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If we assume the barber is a person, they cannot exist. &lt;/li&gt;&#xA;&lt;li&gt;If they did exist, and did not shave themselves, the must shave themselves, which forbids them from shaving themselves, etc...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;It works if they're not a person.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="3987" LastEditorUserId="7435" LastEditDate="2015-01-24T08:04:55.017" LastActivityDate="2015-01-24T08:04:55.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6669" PostTypeId="1" AcceptedAnswerId="6677" CreationDate="2015-01-04T22:48:44.077" Score="20" ViewCount="759" Body="&lt;p&gt;Cut the stylized frog in the picture into six pieces having the same shape and size, possibly mirrored. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DhX4R.png&quot; alt=&quot;The frog&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The white dots are guide points. They help recognizing the shape's geometry. You are not restricted to cut along those points.&lt;/p&gt;&#xA;" OwnerUserId="4394" LastActivityDate="2015-01-05T15:51:15.193" Title="Dissect the frog" Tags="|geometry|dissection|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6671" PostTypeId="1" AcceptedAnswerId="6674" CreationDate="2015-01-05T08:11:50.770" Score="25" ViewCount="2501" Body="&lt;p&gt;Each solid shape is somehow related to the shape below it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find this relationship and the correct shape for the last one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bCYvw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: simplified second shape from left to a polygon&lt;/p&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="6178" LastEditDate="2015-01-05T12:09:02.250" LastActivityDate="2017-04-28T19:00:04.197" Title="Guess the other shape" Tags="|pattern|visual|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6672" PostTypeId="2" ParentId="6671" CreationDate="2015-01-05T09:37:02.853" Score="3" Body="&lt;p&gt;by Elimination i choose&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All the patterns with a triangle/sharp shape have 2 rods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the last pattern has a triangle shape too. and there is only one answer with 2 rods.&lt;/p&gt;&#xA;" OwnerUserId="7546" LastActivityDate="2015-01-05T09:37:02.853" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6673" PostTypeId="1" AcceptedAnswerId="7713" CreationDate="2015-01-05T11:20:22.823" Score="32" ViewCount="2716" Body="&lt;p&gt;The picture below depicts Sir Richard Mervyn Bacon, a (very) distant relative of &lt;a href=&quot;http://en.wikipedia.org/wiki/Francis_Bacon&quot; rel=&quot;noreferrer&quot;&gt;Francis Bacon&lt;/a&gt;, leaving his flat on a clear autumns night in order to get his favourite book from the local library. Your task is to:&lt;/p&gt;&#xA;&lt;h2&gt;Name the title of the book&lt;/h2&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aQYwF.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aQYwF.gif&quot; alt=&quot;The tower&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;The image is scaled on Stack Exchange. You might want to click the image to view the full-size version and &lt;strong&gt;save it locally&lt;/strong&gt; for easier analysis.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;It should be obvious, but just to make sure: The picture above &lt;em&gt;encodes&lt;/em&gt; either the title or unambiguous hints to the title in an unambiguous way. There is no guesswork required, and a valid solution not only names the title, but also features the decoding steps leading to the title.  The introduction text (and puzzle title) is a hint, but otherwise not essential.&#xA;The book is English and belongs to world literature, freely available on the net for download.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;Additional information on this puzzle and the puzzle-building process&lt;/h2&gt;&#xA;&lt;p&gt;The following information of the purpose of the site, explaining how and why some things were done. They assume that the solution has already been found (see accepted answer) and contain severe &lt;em&gt;spoilers&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Do not read on if you want to solve the puzzle yourself!&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The main goal of this puzzle was&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; to use &lt;a href=&quot;http://www.google.de/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=2&amp;amp;cad=rja&amp;amp;uact=8&amp;amp;ved=0CCoQFjAB&amp;amp;url=http%3A%2F%2Fen.wikipedia.org%2Fwiki%2FBacon%2527s_cipher&amp;amp;ei=nEu2VMHYH8a1ggSz3IPQBQ&amp;amp;usg=AFQjCNFHt0JcPeohmdgj5A6l2ZSe0gaung&amp;amp;sig2=kxTs-6cpL_br6UPKcKHMLA&amp;amp;bvm=bv.83640239,d.eXY&quot; rel=&quot;noreferrer&quot;&gt;Baconian cipher&lt;/a&gt; in a grey-level image to encode &lt;em&gt;a large amount&lt;/em&gt; of text. Essentially, one needs an image property which allows one to give a large number of arbitrarily changed binary statements in a clear sequence - binary code, if you want - but hide it so that it is not obvious to the person looking at the image. For a nice historical perspective, &lt;a href=&quot;http://www.cabinetmagazine.org/issues/40/sherman.php&quot; rel=&quot;noreferrer&quot;&gt;read this article&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The specific idea I had was&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; That &lt;em&gt;an arbitrary&lt;/em&gt; image can contain &lt;em&gt;arbitrary&lt;/em&gt; text if one encodes the message in a non-obvious, general attribute such as subtle intensity changes. Doing this on a pixel-by-pixel basis would be possible (i.e. a modulation of one colour-channel), but I wanted it not &lt;em&gt;too&lt;/em&gt; obscure, and at least fairly accessible to people with standard image editor programs. After all, this should be a &lt;em&gt;puzzle&lt;/em&gt; and not a safe encryption.&lt;br&gt;&lt;br&gt;So I've chosen to &lt;strong&gt;encode information into the line/column-wise intensity sum of image intensities&lt;/strong&gt;. For an image of dimensions H * W one can then encode H*W/5 letters using the standard Baconian cipher.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Then I've realized that&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Just presenting an arbitrary image with an arbitrary (long) text encoded and asking to find it, might just not be &lt;em&gt;fun&lt;/em&gt;. Also, I wanted to avoid &lt;em&gt;reception&lt;/em&gt;-problems like I've encounter in a &lt;a href=&quot;https://puzzling.stackexchange.com/questions/5555/hidden-in-plain-sight-by-what-method-do-these-images-hide-a-prime-number&quot;&gt;previous puzzle&lt;/a&gt; of mine.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Therefore I've build the image of the puzzle as&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; as both a &lt;em&gt;puzzle&lt;/em&gt; itself and as a series of &lt;em&gt;hints&lt;/em&gt; to the main challenge. The idea being, that people having found 8 Baconian ciphers in different levels of difficulty, would be more inclined to find the &lt;em&gt;last&lt;/em&gt; Baconian cipher they already &lt;em&gt;know&lt;/em&gt; has to be there. Also, the smaller ciphers allowed me to more or less hint at the final solution to cut down frustration. Using the additional &lt;em&gt;Morse&lt;/em&gt; code was just to keep the scope a bit more interesting.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The puzzle was built the following way:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; - The image was manually drawn in a graphics program with ideas for the various smaller codes. This was great fun for me. To do the heavy leg-work of encoding and decoding, I've use the excellent online tools: &lt;a href=&quot;http://rumkin.com/tools/cipher/morse.php&quot; rel=&quot;noreferrer&quot;&gt;Tool for Morse Code&lt;/a&gt;; &lt;a href=&quot;http://rumkin.com/tools/cipher/baconian.php&quot; rel=&quot;noreferrer&quot;&gt;Tool for Baconian cipher&lt;/a&gt;. I made sure to use a very limited number of distinct grey levels (between 0 and 255) because I wanted the puzzler to be aware of the different intensities in the stars, notice (on close inspection) the random noise, and also point to the importance of intensity-values in general.&lt;br&gt;&lt;Br&gt;&#xA; - To actually encode the long text, one has to successively alter line-sum intensity to be even or odd as required. If the sum is already correct, nothing needs to be done, otherwise add or subtract the value of 1 in a single pixel of the line. Once the lines are done, proceed in the same way with the column-sums. However, each time you change a pixel, you have to change another pixel of the that pixel´s &lt;em&gt;line&lt;/em&gt; (at higher column-index) to not mess with previously constructed sums. I have done this alteration in a little bit of script code. What is important, however:  Ensure that the minimum and maximum values in the image stay at 0..255! (So I've always done a -1 modifier except for 0-valued pixels which I gave a +1 when needed.)&lt;br&gt;&lt;br&gt;&#xA; - When saving the image, make sure a format is used which &lt;em&gt;preserves&lt;/em&gt; individual pixel-values (so no JPG!) and maps intensities to a grey-scale 0.255. GIF did work out nicely, but PNG or TIF might have been alternative choices. &lt;br&gt;  &lt;strong&gt;Most importantly:&lt;/strong&gt; Once saved in final form, re-open image and cross-check that the line and column sums are still correct! (I repeated this with the &lt;em&gt;posted&lt;/em&gt; image additionally to ensure the website isn't doing anything funny - it didn't.) I additionally checked that some free software can be used to extract the pixel values I needed. I've used &lt;a href=&quot;http://imagej.nih.gov/ij/&quot; rel=&quot;noreferrer&quot;&gt;ImageJ&lt;/a&gt; as test-software.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Some additional points:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; - I've added a bit of random noise to the image so that the pixel-modification of +1 or -1 isn't giving suspiciously 'single pixel' changes. If the whole image would have had only 6 main grey-tones, any 'deviated' pixel would have been easily recognizable. One wouldn't have the problem in an image using 256 grey-levels natively already.&lt;br&gt;&lt;br&gt;&#xA; - For encoding without 'obsolete' sums, the width &amp;amp; length of the image added together need to be integer-multiples of 5 (for Baconian cipher). Once could leave the 'last' sums unused, but I wanted a clean, self-coherent puzzle. However, as the &amp;quot;5&amp;quot; is a tell tale sign for Baconian cipher, and the puzzle already &lt;em&gt;knows&lt;/em&gt; this cipher is used, I didn't want it to be too obvious. Hence &lt;em&gt;neither&lt;/em&gt; width &lt;em&gt;nor&lt;/em&gt; length of my image where integer multiples of 5 (just the sum is). &lt;br&gt;&lt;br&gt;&#xA; - I &lt;em&gt;deliberately&lt;/em&gt; used strange image dimensions, so that a careful observer could realized this as a hint.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I hope this puzzle has been fun. &lt;strong&gt;Please use the comments sections to let me know your ideas.&lt;/strong&gt; This Stackexchange site is about &lt;em&gt;learning&lt;/em&gt; both solving and building puzzles, so feedback is important.&lt;/p&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-10-08T09:16:35.510" Title="The Baconian Tower" Tags="|cipher|visual|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6674" PostTypeId="2" ParentId="6671" CreationDate="2015-01-05T12:06:26.690" Score="24" Body="&lt;p&gt;I believe the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;C&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Read the circles/bars from front to back based on the way they overlap each other, they form binary numbers.&lt;br&gt;&#xA;For example, the circles/bars represent, from left to right: 10, 110, 11, 101, 011, 00.&lt;br&gt;&#xA;This matches the number of points (or convex corners) on the shape above.  (from the third shape we can tell that concave corners are to be ignored)&lt;br&gt;&#xA;The last shape has 5 points, so we want 101, which is C.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Excellent puzzle!&lt;/p&gt;&#xA;" OwnerUserId="6178" LastEditorUserId="6178" LastEditDate="2015-01-05T12:37:50.757" LastActivityDate="2015-01-05T12:37:50.757" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6675" PostTypeId="2" ParentId="6636" CreationDate="2015-01-05T12:06:56.027" Score="28" Body="&lt;p&gt;This question appears to be first settled in A. Kohnert and S. Kurz, &lt;em&gt;A note on Erdős-Diophantine graphs and Diophantine carpets&lt;/em&gt;, Mathematica Balkanica, 21(1-2):1–5,&#xA;2007, &lt;a href=&quot;http://arxiv.org/abs/math/0511705&quot;&gt;available on arxiv.org&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A Diophantine figure is a set of points on the integer grid $\mathbb{Z}^2$ where&#xA;  all mutual Euclidean distances are integers. We also speak of &lt;em&gt;Diophantine graphs&lt;/em&gt;. The&#xA;  vertices are points in $\mathbb{Z}^2$ (the coordinates) and the edges are labeled with the distance between the two adjacent vertices, which is integral. In this language a Diophantine figure is a complete Diophantine graph. ... Due to a famous theorem of Erdős and Anning there are complete Diophantine graphs which are not contained in larger ones. We call them Erdős-Diophantine graphs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that a three-vertex &lt;a href=&quot;http://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Diophantine_graph&quot;&gt;Erdős-Diophantine graph&lt;/a&gt; is precisely what the question asks for: the three vertices are at mutually integer distances, but there is no point which is at an integer distance from all of them, or it would not be Erdős-Diophantine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kohnert and Kurz list seven Erdős-Diophantine triangles and state that they form a complete list for triangles having an edge of length $\le 5000$. (They appear to actually mean having &lt;em&gt;all&lt;/em&gt; edges $\le 5000$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(2066, 1803, 505)\\&#xA;(2549, 2307, 1492)\\&#xA;(3796, 2787, 2165)\\&#xA;(4083, 2425, 1706)\\&#xA;(4426, 2807, 2210)\\&#xA;(4920, 4177, 985)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://www.wm.uni-bayreuth.de/fileadmin/Sascha/Publikationen/On_Heronian_Triangles.pdf&quot;&gt;later paper of Kurz&lt;/a&gt; lists many more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the first listed example, we place knishops at $(0, 0),\; (384, 2030),\; (720, 1653)$.&lt;/p&gt;&#xA;" OwnerUserId="228" LastActivityDate="2015-01-05T12:06:56.027" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6676" PostTypeId="2" ParentId="6612" CreationDate="2015-01-05T12:45:46.190" Score="0" Body="&lt;p&gt;I would go with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Emoticons&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Solid&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is no visual difference between handwritten and printed emoticons and they always look &quot;solid&quot; rather than soft and curved.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;misshapen&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They do not look like normal text&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;sad&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They can express emotions, like sadness&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I is&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reference to &quot;internet speak&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A linguistic object of rare beauty&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rarely used in formal communication and literature of any sorts&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;the non-literal part is itself in how many parts? Remember how many when you're trying to create what you want.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure about that, but I assume that &quot;solid, misshapen, sad&quot; are also referring to the 3 symbols needed to make the most popular emoticons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastActivityDate="2015-01-05T12:45:46.190" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6677" PostTypeId="2" ParentId="6669" CreationDate="2015-01-05T14:49:39.253" Score="17" Body="&lt;p&gt;I think this is the solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/BrN1B.png&quot; alt=&quot;solution&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I printed out the image and cut out each unit square and then it was more easier to try out all positions.&lt;/p&gt;&#xA;" OwnerUserId="101" LastEditorUserId="5840" LastEditDate="2015-01-05T15:50:31.667" LastActivityDate="2015-01-05T15:50:31.667" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6678" PostTypeId="1" AcceptedAnswerId="6690" CreationDate="2015-01-05T15:32:28.250" Score="19" ViewCount="844" Body="&lt;p&gt;A 3 x 1 rectangle has a perimeter of length 8. Two of these rectangles can each be bend in a U-shape and stitched together to yield a cube of unit volume.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surely we can do better than that! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What is the minimum stitch length required to create a unit volume out of two identical (bendable but not deformable) planar shapes?&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note: any stitching required to 'seal' the volume (including stitching required between parts of the same shape) counts towards the total.  &lt;/p&gt;&#xA;" OwnerUserId="7573" LastEditorUserId="7573" LastEditDate="2015-01-05T17:28:35.940" LastActivityDate="2015-01-12T07:38:52.173" Title="Stitching together two identical planar shapes" Tags="|mathematics|geometry|" AnswerCount="4" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6680" PostTypeId="2" ParentId="6678" CreationDate="2015-01-05T16:05:11.513" Score="2" Body="&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aha, I was looking into cones, and I think I've figured out the ratio. Make a cone where $h=\frac{2{\pi} r}{3}$  That works out to $r^3=\frac{9}{4\pi^2}$ which is aprox 0.611. With a circumference of 3.839 and a lateral length of 1.418 * 2, it gives a total stitching of aprox $6.675$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Old:&lt;br&gt;&#xA;An octahedron of volume 1 would have an edge length of 1.285. The planar shape you would need two of would be 4 equilateral triangles in a U shape. There would be 6 stiches. 1 on each piece to turn the U shape into a bottomless pyramid (2x), then 4 to put the two pyramids together. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total stitching: 7.71&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was gonna try a dodecahedron and an icosahedron, but each of those has way too many stitches (20 and 12). &lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-01-06T22:22:01.897" LastActivityDate="2015-01-06T22:22:01.897" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6682" PostTypeId="2" ParentId="6656" CreationDate="2015-01-05T19:53:22.873" Score="10" Body="&lt;p&gt;The missing line is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Paleontologists.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In double dactyl verse, there must be a line consisting of a single word, whose syllables have the pattern long-short-short-long-short-short. This line is often the third from the last, so it is reasonable to expect the missing line to have this form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first stanza refers disparagingly to a movie or movies (Hollywood-Shmollywood), in which something terrible and old (Terrible ancients) is brought to life in film (are now resurrected in picture and sound).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second stanza refers to a group of people criticizing fans of the movie (fadsters). These people are professionals (in contrast to the &quot;amateurs&quot;) who liked when the subject of the movie was &quot;still underground&quot;. The term &quot;still underground&quot; has a double meaning: it refers to a time before the subject of the movie entered popular culture, but also that the subject of the movie was literally buried in the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dinosaurs are terrible  (i.e. scary, at least when portrayed in the movies) and ancient, and their fossils are buried underground. Dinosaurs were &quot;resurrected&quot; in the movie Jurassic Park (maybe in other movies too, but this is what came to mind first). Paleontologists study dinosaur fossils as a profession, and the word &quot;Paleontologists&quot; has the desired meter.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-01-05T21:54:26.597" LastActivityDate="2015-01-05T21:54:26.597" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6683" PostTypeId="2" ParentId="6678" CreationDate="2015-01-05T21:50:57.470" Score="3" Body="&lt;p&gt;Two of the answers given so far use cones. To optimize this idea, suppose we take two identical circular sectors, form them into right circular cones of radius $r$ and height $h$, and attach them along their bases. The total volume will be $\frac{2}{3}\pi r^2 h$ (twice the volume of a cone of radius $r$ and height $h$), and the length of stitching is $2\sqrt{r^2+h^2}+2\pi r$ (each cone requires $2\sqrt{r^2+h^2}$ stitching to assemble, and it requires $2\pi r$ stitching to attach them). We should try to minimize the quantity $2\sqrt{r^2+h^2}+2\pi r$, subject to the contraint $\frac{2}{3}\pi r^2 h=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The minimum can be computed using the differential calculus. It's a big mess, but the result is:&#xA;$$&#xA;r=\sqrt[6]{\frac{-9-\frac{36}{\pi^2}+9\sqrt{1+\frac{24}{\pi^2}}}{8(\pi^2-1)}}\approx 0.619857783,&#xA;$$&#xA;$$&#xA;h=3\frac{\sqrt[3]{\frac{8(\pi^2-1)}{-9-\frac{36}{\pi^2}+9\sqrt{1+\frac{24}{\pi^2}}}}}{2\pi}\approx1.242674164.&#xA;$$&#xA;This gives a total stitch length of about $6.672063354$.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-01-05T22:14:11.180" LastActivityDate="2015-01-05T22:14:11.180" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6685" PostTypeId="2" ParentId="6531" CreationDate="2015-01-06T02:46:46.440" Score="1" Body="&lt;p&gt;I think I've got a couple more clues. I should probably edit these into the earlier answer, but I'm not sure how to go about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;2. The angels amid sightless ghoti; half a nokia appears unexpectedly (5, 3)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; FLASH MOB — &quot;The angels&quot; = Los Angeles = LA. &quot;Ghoti&quot; = &quot;fish&quot; (see clue 10), but sightless (with no &quot;i&quot;) = FSH. A Nokia is a mobile, so half a Nokia = MOB&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;4. Hilarious partner (3)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; WAG — The British press use the acronym &lt;a href=&quot;https://en.wiktionary.org/wiki/WAG&quot; rel=&quot;nofollow&quot;&gt;WAGs&lt;/a&gt; to refer to the &quot;wives and girlfriends&quot; of professional footballers. The singular form is WAG. And of course wag also means &quot;witty person&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;6. Performing a quick number in the head of an adult insect. (9)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Could this perhaps be PANTOMIME? It's a kind of performance, and it has an ANT in it. Haven't quite figured this one out yet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="40" LastActivityDate="2015-01-06T02:46:46.440" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6686" PostTypeId="2" ParentId="6612" CreationDate="2015-01-06T04:03:43.213" Score="5" Body="&lt;p&gt;The solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CUBE&lt;br&gt;UGLY&lt;br&gt;BLUE&lt;br&gt;EYES&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One of the Platonic &lt;b&gt;SOLID&lt;/b&gt;s is the &lt;b&gt;CUBE&lt;/b&gt;.&lt;br&gt;If something is &lt;b&gt;MISSHAPEN&lt;/b&gt; is may be deformed and &lt;b&gt;UGLY&lt;/b&gt;.&lt;br&gt;A colloquial way to say &lt;b&gt;SAD&lt;/b&gt; is &lt;b&gt;BLUE&lt;/b&gt;.&lt;br&gt;&quot;&lt;b&gt;I IS&lt;/b&gt;&quot; is kind of a homophone of &lt;b&gt;EYES&lt;/b&gt;.&lt;br&gt;&lt;br&gt;Together, these four words form a four-by-four square of letters that reads the same left-to-right or top-to-bottom, which makes it a thing of beauty.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First attempt:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Solid is a phase of matter. A phase sounds like &quot;a face&quot;.&lt;br&gt;Misshapen can mean withered, which sounds like &quot;with her&quot;.&lt;br&gt;Sad is, informally, blue.&lt;br&gt;And we know about eyes already. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which gives us:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A face with her blue eyes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Additional comments by h34&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well done! The solution is indeed this object:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/8Sbcw.jpg&quot; alt=&quot;Square&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The clues are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; crossword-style clues of four different types, namely:&lt;br&gt;&lt;br&gt;&#xA; &lt;strong&gt;solid&lt;/strong&gt;: a clue for a &lt;em&gt;member of the class&lt;/em&gt; of solids, namely &lt;strong&gt;cube&lt;/strong&gt;&lt;br&gt;&#xA; &lt;strong&gt;misshapen&lt;/strong&gt;: a clue for a &lt;em&gt;straightforward synonym&lt;/em&gt;, &lt;strong&gt;ugly&lt;/strong&gt;&lt;br&gt;&#xA; &lt;strong&gt;sad&lt;/strong&gt;: a clue for a &lt;em&gt;metaphorical synonym&lt;/em&gt;, &lt;strong&gt;blue&lt;/strong&gt;&lt;br&gt;&#xA; &lt;strong&gt;I is&lt;/strong&gt;: a &lt;em&gt;homophone&lt;/em&gt;, once contracted, for &lt;strong&gt;eyes&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The solution is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;em&gt;a linguistic object&lt;/em&gt; because it is made of words.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;em&gt;beautiful&lt;/em&gt; because of the symmetry&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;em&gt;rare&lt;/em&gt; because...well, just try making a similarly symmetric $4 \times 4$ square using words as common as these four. There aren't very many of them about, not counting a trivial variation of &lt;em&gt;cube&lt;/em&gt; to &lt;em&gt;tube&lt;/em&gt; or, if allowed, &lt;em&gt;rube&lt;/em&gt; or &lt;em&gt;lube&lt;/em&gt;. (According to &lt;a href=&quot;https://github.com/first20hours/google-10000-english/blob/master/google-10000-english.txt&quot; rel=&quot;nofollow noreferrer&quot;&gt;this list&lt;/a&gt; &lt;em&gt;tube&lt;/em&gt; is a commoner word than &lt;em&gt;cube&lt;/em&gt;, but &lt;em&gt;cube&lt;/em&gt; adds to the feeling of a symmetric spatial structure and works better.) An added bonus is that the noun phrase comprising words 2-4 actually makes sense!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the second hint stressed the number of parts of the riddle, four&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the solution itself can be understood as a crossword puzzle with the same clues down as across, so it is fitting that the clues are of crossword style and neat that they are of four different types&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2162" LastEditorUserId="7363" LastEditDate="2015-01-06T17:29:59.570" LastActivityDate="2015-01-06T17:29:59.570" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6687" PostTypeId="2" ParentId="6671" CreationDate="2015-01-06T11:04:02.147" Score="2" Body="&lt;p&gt;An alternate answer would be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;B&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Following the logic that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;IFF the (silhouet of the) lower shape has rotational symmetry, the upper shape has rotational symmetry (and vice versa)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1666" LastActivityDate="2015-01-06T11:04:02.147" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6688" PostTypeId="1" AcceptedAnswerId="6689" CreationDate="2015-01-06T11:43:43.837" Score="21" ViewCount="58394" Body="&lt;h2&gt;Story:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;A Color blind man is travelling in a somewhat famous ship that had just a few people on board. Everything was smooth sailing for some days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One day,the man heard some noises from outside when he was in his room. The ship started to shake and began to sink. By the time the man reached outside,nobody was there and the ship was almost fully sunk. There was no time for the man to search for the rest of the crew. He jumped in the ocean and saw the ship sinking to the bottom of the ocean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Struggling to find some help,he sees an island and swam straight in that direction. The current was a bit strong,but he managed to get to his destination. By the time he reaches the shore,he starts coughing and also remembers something. &lt;strong&gt;He forgot to take his pills!&lt;/strong&gt;. He keeps a few pills in his pocket and the rest in his room. He thrust his hand into his pocket. Fortunately, The pills were still there in his pocket.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Puzzle:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;He recalls putting 4 pills in his pocket(with the help of a friend) out of which 2 are red ones and the rest of them are white. All the pills are completely identical. He is supposed to take one red pill and one white pill everyday. The order which he takes them doesn't matter. He must &lt;strong&gt;not&lt;/strong&gt; take two pills of the same color and if he does so,he dies. How does he take one red pill and one white pill?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the man is &lt;em&gt;color blind&lt;/em&gt; and he knows that help will arrive soon,since the ship he was travelling was a famous one. Also,the island that the man is on is &lt;em&gt;deserted&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7482" LastActivityDate="2019-04-18T07:03:57.957" Title="A man with two red pills and two white pills" Tags="|logical-deduction|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6689" PostTypeId="2" ParentId="6688" CreationDate="2015-01-06T11:55:01.263" Score="37" Body="&lt;p&gt;&lt;strong&gt;Answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Just take half of each pill everyday which will always be 1 red + 1 white when put together.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="101" LastEditorUserId="19856" LastEditDate="2016-07-13T04:44:01.590" LastActivityDate="2016-07-13T04:44:01.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6690" PostTypeId="2" ParentId="6678" CreationDate="2015-01-06T13:17:00.823" Score="7" Body="&lt;p&gt;Edit: I found a better way to use this technique, and the answer becomes &quot;how small do you want it?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following simple paper template:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8A0kX.png&quot; alt=&quot;Plan for a cubical half-figure&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now fold this along the black lines so it forms an open-topped cube. The two points marked A must meet, as do B, C and D. This leaves four flaps - fold them along the green lines, and then along the sides of the cube so that 1 meets B, 2 meets C, 3 meets D, and 4 meets A. If the cube faces all have side $\frac{1}{2^{1/3}}$, then the cube has volume $\frac{1}{2}$. Sew two of these together, and you enclose a unit volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the stitching must go through three layers of figure, but the overall length of stitching is only $\frac{4}{2^{1/3}} = 3.1748$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's pretty small, but you can improve it by making your two boxes taller because the stitching length only depends on the width. So if we make the height of the boxes a multiple n of the width, then we have $nw^3 = \frac{1}{2}$, so $w = \frac{1}{(2n)^{1/3}}$ and the stitching length 4w approaches zero as n increases without bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The price you pay for all of this is sewing multiple thicknesses of material. You will need to sew through 2n + 1 layers of material.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Old answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm going to bend the rules a little, since there's nothing to say every part of the shapes has to be part of the volume boundary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can make a tetra-brik ( &lt;a href=&quot;http://en.wikipedia.org/wiki/Tetra_Brik&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Tetra_Brik&lt;/a&gt; ) with height h and square cross-section of side w, out of a single shape which is 4w wide, and h + w high. It has a stitching length of 5w + h (one stitch length up the side to make a tube, and 2w stitch length at each of top and bottom to seal it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you've done that, you can either use a double thickness (both shapes) to make your brik, since the same stitching will attach both; or you cut an infinitesimal hole in the brik and patch it with the second planar shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resulting stitch length:&#xA;The volume of the brik is $w^2h$, which must be 1. So $h = \frac{1}{w^2}$.&lt;br&gt;&#xA;Stitch length $s = 5w + \frac{1}{w^2}$&lt;br&gt;&#xA;$\frac{ds}{dw} = 5 - \frac{2}{w^3}$, so the minimum is at $w^3 = \frac{2}{5}$.&lt;br&gt;&#xA;This gives you a square cross section area of $(0.736806299728077)^2 = 0.54288352331898$, and a height of 1.8420157493202, for a total stitch length of 5.5260472479606 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really wanted to make a volume that genuinely used parts of both shapes - say a cone that simply had the other shape stitched flat across the open end - but I couldn't get one that was actually better than two cones. I promise an upvote to the first person who can do that.&lt;/p&gt;&#xA;" OwnerUserId="6031" LastEditorUserId="6031" LastEditDate="2015-01-12T07:38:52.173" LastActivityDate="2015-01-12T07:38:52.173" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6691" PostTypeId="1" AcceptedAnswerId="9851" CreationDate="2015-01-06T16:58:32.300" Score="62" ViewCount="9810" Body="&lt;p&gt;This is a &lt;em&gt;challenge&lt;/em&gt; for word-puzzle lovers. It might be a bit more of a game than a puzzle, but I think it is not limitless in it's answers, so a &lt;em&gt;final&lt;/em&gt; answer could be found (eventually).&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;What is the longest list of &lt;em&gt;growing&lt;/em&gt; anagrams which can be put into a grammatically correct English sentence?&lt;/h1&gt;&#xA;&lt;p&gt;&lt;strong&gt;Example for clarification:&lt;br&gt;&lt;/strong&gt;&#xA;Start with a word and then add words to the list with each having one &lt;em&gt;additional&lt;/em&gt; letter to the one before. All letters may be re-arranged, i.e. all anagrams of the word are allowed.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;I&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;in&lt;/strong&gt; &lt;br&gt;&#xA;&lt;strong&gt;pin&lt;/strong&gt; ( or alternatively: &lt;strong&gt;nip&lt;/strong&gt; ) &lt;br&gt;&#xA;&lt;strong&gt;pain&lt;/strong&gt; ( or alternatively: &lt;strong&gt;pian&lt;/strong&gt; ) &lt;br&gt;&#xA;&lt;strong&gt;paint&lt;/strong&gt; (or alternatively: &lt;strong&gt;inapt&lt;/strong&gt; )&lt;br&gt;&#xA;...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You may add &lt;em&gt;any&lt;/em&gt; letter, but the list is only valid if you can then put &lt;em&gt;all&lt;/em&gt; words of it into a single, English sentence which is grammatically correct. (It does not have to be very meaningful.) &lt;br&gt;&#xA;Such as in:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;code&gt;I paint &amp;quot;pin&amp;quot; in pain!&lt;/code&gt; &lt;br&gt;&#xA;...says the painter while he is painting a shop-name ( &amp;quot;Pin &amp;amp; Needle&amp;quot; ) onto the wall while having severe backache...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So, the aim is: &lt;em&gt;Find a single sentence using all (and only) the words of such a list, i.e. each word-length appears exactly once.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Rules:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Diacritics may be handled as &amp;quot;group&amp;quot; represented by their simplest form  &lt;em&gt;( a, ä, á... = a )&lt;/em&gt; to allow for more flexibility in the solutions.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The shortest word of the list may have &lt;em&gt;any&lt;/em&gt; length. (It does not have to be a single letter.) But the list has to contain a single word for each 'length' between the shortest and the longest word.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You may use arbitrary punctuation in the sentence. (Compound words count as single word.)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Names and acronyms are &lt;em&gt;allowed&lt;/em&gt; but if used, give a reference for their validity.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the meaning of the sentence is not apparently clear, give a little example of where it could be appear. (See example above.)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;All words of the list have to appear exactly once in the sentence.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;strong&gt;Bonus:&lt;/strong&gt;&lt;br&gt;&#xA;This puzzle is about &lt;em&gt;English&lt;/em&gt;, but if you can do the same in any other language and have a good example, please post here also. It will not be accepted as answer, though.&lt;/p&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2020-01-04T22:27:01.270" Title="Growing anagrams - What is the longest possible list forming a sentence?" Tags="|wordplay|language|anagram|open-ended|" AnswerCount="19" CommentCount="7" ClosedDate="2021-08-05T08:12:08.547" ContentLicense="CC BY-SA 4.0" />
  <row Id="6692" PostTypeId="2" ParentId="6691" CreationDate="2015-01-06T17:14:49.930" Score="8" Body="&lt;p&gt;My first attempt with 8 words:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A carnage ran near an arcane &lt;a href=&quot;http://en.wikipedia.org/wiki/Rumex_hymenosepalus&quot;&gt;canaigre&lt;/a&gt; arena.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Meaning &quot;A carnage spread near a mysterious arena where wild rhubarb are pitched in battle&quot;. Not very meaningful, but perfectly correct. I could think of this in the intro of a first-person shooter video game.&lt;/p&gt;&#xA;" OwnerUserId="4394" LastEditorUserId="2162" LastEditDate="2015-01-06T17:59:32.517" LastActivityDate="2015-01-06T17:59:32.517" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6693" PostTypeId="2" ParentId="6691" CreationDate="2015-01-06T17:29:51.523" Score="11" Body="&lt;p&gt;A short, random attempt. 6 words.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;This Sith's thesis? I hit it.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Useful for when someone asks you if you punched the paper that Darth Maul was supposed to turn in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Found one with 7 words, but I can't think of how it would be useful.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Sisters tie it; I tire; tires resist.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1664" LastActivityDate="2015-01-06T17:29:51.523" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6694" PostTypeId="2" ParentId="6691" CreationDate="2015-01-06T17:47:30.723" Score="14" Body="&lt;p&gt;4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;13 Letter (Rhinoplasties)&lt;/strong&gt;  &lt;br/&gt;&#xA;  a &amp;lt;1&gt;&lt;br/&gt;&#xA;  in&amp;lt;2&gt;&lt;br/&gt;&#xA;  its&amp;lt;3&gt;&lt;br/&gt;&#xA;  nose&amp;lt;4&gt;&lt;br/&gt;&#xA;  noses&amp;lt;5&gt;&lt;br/&gt;&#xA;  inters&amp;lt;6&gt;&lt;br/&gt;&#xA;  nostril&amp;lt;7&gt;&lt;br/&gt;&#xA;  hospital&amp;lt;8&gt;&lt;br/&gt;&#xA;  hospitals&amp;lt;9&gt;&lt;br/&gt;&#xA;  paltriness&amp;lt;10&gt;&lt;br/&gt;&#xA;  hospitalise&amp;lt;11&gt;&lt;br/&gt;&#xA;  relationship&amp;lt;12&gt;&lt;br/&gt;&#xA;  rhinoplasties&amp;lt;13&gt;&lt;br/&gt;&lt;br/&gt;&#xA;  Rhinoplasties:  nostril inters its paltriness ; a nose hospital relationship; hospitalise noses in hospitals&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And yes its a hyperbole with descriptions about what rhinoplasties meant. Its how a nose buries its worthlessness. Its a relationship of a noses and  hospitals. It also means that you hospitalise your nose in a hosptial.&#xA;Loving it :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a seat at east  (EAST) &lt;br/&gt;&#xA;     Was a waste at west (WEST) &lt;br/&gt;&#xA;     o! not horn or thorn (NORTH) &lt;br/&gt;&#xA;     o! hot to oust south (SOUTH) &lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Its rest I resist; it tires&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(and yes its a lie) :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I tried, I retried, I retired :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2656" LastEditorUserId="2656" LastEditDate="2015-01-09T08:31:19.797" LastActivityDate="2015-01-09T08:31:19.797" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="6695" PostTypeId="2" ParentId="6691" CreationDate="2015-01-06T18:10:53.980" Score="13" Body="&lt;p&gt;Nine words:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;“Straying giants stang an ant staying stingrays&quot; - a gnat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Meaning: a flying insect reports that wandering abnormally large individuals used nettle leaves to inflict pain on an earthbound insect engaged in serving legal notices on fish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is surreal, but not as neat as the five-word version it sprang from:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An ant stang a gnat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7363" LastActivityDate="2015-01-06T18:10:53.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6696" PostTypeId="2" ParentId="6691" CreationDate="2015-01-06T18:19:25.720" Score="14" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Starting staring in startling sin, I sing Sting string.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Beginning to gaze with shocking immorality, I sing a line from (the musician) Sting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are nine words, and no anagramming is needing. They letters stay in the same order when letters are added.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;i&#xA;in&#xA;sin&#xA;sing&#xA;sting&#xA;string&#xA;staring&#xA;starting&#xA;startling&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I didn't invent this sequence; I read it somewhere but I don't remember where.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-01-06T18:19:25.720" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6697" PostTypeId="2" ParentId="6691" CreationDate="2015-01-06T18:52:23.107" Score="80" Body="&lt;p&gt;Twelve words:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;a nearsighted ant sergeant negates an agent, denigrates estranged straightened gnat agents&lt;/p&gt;&#xA;&lt;p&gt;a&lt;br&gt;&#xA;an&lt;br&gt;&#xA;ant&lt;br&gt;&#xA;gnat&lt;br&gt;&#xA;agent&lt;br&gt;&#xA;agents&lt;br&gt;&#xA;negates&lt;br&gt;&#xA;sergeant&lt;br&gt;&#xA;estranged&lt;br&gt;&#xA;denigrates&lt;br&gt;&#xA;nearsighted&lt;br&gt;&#xA;straightened&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;An officer in an army of earthbound insects, suffering from poor eyesight, orders an informer to be dispensed with, and is harshly critical of an ex-couple of flying insect infiltrators, both graduates of a course to eliminate unwanted bodily bends.&lt;/p&gt;&#xA;" OwnerUserId="7363" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2018-09-13T08:19:15.877" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="6698" PostTypeId="2" ParentId="6691" CreationDate="2015-01-06T19:00:23.383" Score="8" Body="&lt;p&gt;I tic it: otic topic, tropic apricot, operatic imprecation, preatomic importance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Translation: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what I involuntarily blurt out: matter dealing with the ear, fruit from low latitudes, very well sung curses, early 20th century significance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pretty meaningless, but I got 11 words.&lt;/p&gt;&#xA;" OwnerUserId="2162" LastEditorUserId="2383" LastEditDate="2015-01-06T19:09:11.003" LastActivityDate="2015-01-06T19:09:11.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6699" PostTypeId="2" ParentId="6673" CreationDate="2015-01-06T19:00:29.550" Score="0" Body="&lt;p&gt;&lt;strong&gt;Allowed Partial Answer&lt;/strong&gt;&lt;br&gt;&#xA;&lt;em&gt;This answer includes notes as to help you in solving, it includes no actual answers, or processes in retrieving them.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes on puzzle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Stars&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 50 stars; 24 bright, 14 partially dim, 12 dim. 18 large, 12 medium, 20 small.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Windows&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 76 windows; 5 dark, 71 lit, 27 empty (assuming dark windows note have people, otherwise 22), 41 with people, 75 large, 1 small. 23 opens out, 53 open in. 45 windows have curved tops, 35 have straight edge tops, 1 has no top (the small one)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;People&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  42 people; 17 sat down bored, 15 sat down waving (32 sat down), 12 stood up bored, 5 stood up waving (17 stood up), 1 leaving, 6 known to be left handed, 9 known to be right handed. 1 known to be both, 26 unknown.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Door&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 door, dark.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rocks&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 60 rocks; 35 light, 25 dark.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Trees&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2 trees; Left; 13 branches, 23 twigs, 43 leaves. Right; 12 branches, 29 twigs, 45 leaves.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;These notes will save you time so you do not have to count.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Also, for anybody interested in a bit of history;&lt;/strong&gt; &lt;a href=&quot;http://www.geni.com/people/Sir-Richard-Bacon-Baron-Bacon-of-Baconsthorpe/6000000002182986962&quot; rel=&quot;nofollow&quot;&gt;http://www.geni.com/people/Sir-Richard-Bacon-Baron-Bacon-of-Baconsthorpe/6000000002182986962&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2015-01-06T19:00:29.550" CommentCount="3" CommunityOwnedDate="2015-01-06T19:00:29.550" ContentLicense="CC BY-SA 3.0" />
  <row Id="6700" PostTypeId="2" ParentId="6678" CreationDate="2015-01-06T20:54:29.243" Score="9" Body="&lt;p&gt;I believe the minimum is about $5.88$, and is achieved by a Mylar balloon shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me introduce you to the magic of paper pleating:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hOYZN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Nash_embedding_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nash-Kuiper embedding theorem&lt;/a&gt; $-$ well, a specific case of it, translated into English $-$ says that we can pleat paper into any shape, wasting as little paper as we want to. The only condition says that the paper can't get bigger. That is, if you imagine the paper as a fully stretched sheet of rubber, then the rubber can't stretch any more, it can only shrink. I recommend image searching &quot;pleated paper&quot;; it looks cool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that in mind, we have a hard math problem on our hands, from the calculus of variations. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Mylar_balloon_%28geometry%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mylar balloon&lt;/a&gt; solves this problem (thanks Florian F), and looks like this (diagram from Wikipedia):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LxxE5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(To get the Mylar balloon, rotate this image around the vertical axis.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To create your very own Mylar balloon, get a circle of paper of radius $a+\epsilon$, where $a$ is the arc length from the diagram. Pleat it into an approximation of the top half of the Mylar balloon. Slice off a plane very close to the equator, so that the boundary of the remaining squashed-hemisphere of paper lies on a plane. Reflect to get the bottom half, and stitch along the boundary. Voila, a (very good approximation of a) Mylar balloon of volume 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the formulas in the Wikipedia page, we find that for a Mylar balloon of volume 1, each of the papers is a circle of radius $a \approx 0.936$. Therefore, the length of stitching required is $2\pi a\approx 5.88$.&lt;/p&gt;&#xA;" OwnerUserId="1752" LastEditorUserId="1752" LastEditDate="2015-01-07T20:05:18.617" LastActivityDate="2015-01-07T20:05:18.617" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="6701" PostTypeId="2" ParentId="6673" CreationDate="2015-01-06T21:17:03.697" Score="14" Body="&lt;p&gt;Nine of the codes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. The rocks at the bottom are a baconian cipher (black 1, white 0): REDHERRING&lt;br&gt;&#xA; 2. The window cornices are a baconian cipher (trapezoidal 1, round 0): PERSONSAREMORSE&lt;br&gt;&#xA; 3. The people in the windows are morse code (short ., tall -): SUMTHELINESEVENORODD&lt;br&gt;&#xA; 4. The open/closed windows are a baconian cipher (open 1, closed 0): SUMVERTICALLYTOO&lt;br&gt;&#xA; 5. Stars from left to right are a baconian cipher (pointed down 1, up 0): HORIZONTAL&lt;br&gt;&#xA; (The next two are copied from rsanchez's answer)&lt;br&gt;&#xA; 6. Stars from left to right are a baconian cipher (white A, non white B): BMYGUEST&lt;br&gt;&#xA; 7. Ends of branches of both trees are a baconian cipher (with leave at end A, without leave at end B): FINDALH CIPHERS&lt;br&gt;&#xA; 8. Stars from left to right (tiny 1, other 0): READWINDOWS&lt;br&gt;&#xA; 9. People in windows + Sir Richard himself (hand raised 1, not raised 0): BRIGHTNESS  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditDate="2015-01-07T18:30:36.310" LastActivityDate="2015-01-07T18:30:36.310" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6703" PostTypeId="2" ParentId="6531" CreationDate="2015-01-06T22:44:43.713" Score="1" Body="&lt;p&gt;Perhaps 6 across is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Hairworms, they target insects' heads.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(first post)&lt;/p&gt;&#xA;" OwnerUserId="7722" LastActivityDate="2015-01-06T22:44:43.713" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6705" PostTypeId="2" ParentId="6691" CreationDate="2015-01-07T01:51:21.963" Score="30" Body="&lt;p&gt;Nikola Tesla, decrepit in old age, aggrieved by ailments, but still doing his experiments causing filaments (as in a light bulb) to explode. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A lamest Tesla sat at last - laments ailments, fulminates filaments. &#xA;  (10 words)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In a post-apocalyptic world, farm animal meat is a thing of the past.  To retain its customers, driven away by ever-increasing prices and a bland product, one delicatessen put up this window sign:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Later, a saltier rat at earliest retail rate.&#xA;  (8 words)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7489" LastEditorUserId="7489" LastEditDate="2015-01-07T02:59:43.513" LastActivityDate="2015-01-07T02:59:43.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7700" PostTypeId="2" ParentId="6150" CreationDate="2015-01-07T05:20:18.720" Score="0" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/e6hUk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my answer. For explanation purpose I named the board like chess. X-Axis is labeled as a - k. Y-Axis is labeled as 1-5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach :&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;In e5 and f5 we have adjacent numbers. So we know the diff is 5. So work the row upto a5.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Since the number at a5 is odd, the number at a3 should also be Odd !&#xA;number at a3 cannot be 9,7,5 because then the number at d5 become -ve. &#xA;Once you get the number a3 as 3, the rest (left half of the board) can be worked out easily.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For the right half - Lets assume the number in h5 is N. And lets assume the delta in column h is 'y' and delta in row 5 (from h5 to k5) is 'x'. So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N = 59 + 4y&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N = 59 + 3x&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So we get 3x = 4y. Once you assume x = 4 and y = 3 then the rest  can be worked out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note:  There are multiple solutions to this which can be worked out by trying different values at a3 and different  values of x and y.&lt;/p&gt;&#xA;" OwnerUserId="8725" LastActivityDate="2015-01-07T05:20:18.720" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7701" PostTypeId="1" AcceptedAnswerId="7712" CreationDate="2015-01-07T05:33:38.617" Score="81" ViewCount="18781" Body="&lt;p&gt;One day, the warden of a prison is, like most wardens in puzzles, feeling a little bit capricious and decides that he wants to get rid of his prisoners, one way or another. He gathers all the prisoners in the yard and explains to them - &quot;Tonight, I will go to each of you, hand you a key, and tell you who has your key. Each day after that, while the others are out of the cells and no one is watching, I will allow each of you to place your key in someone else's cell - and each night, you may collect the keys in your own cell. If, at any point, you are certain that everyone has the key to their own cell, you may summon me, at which point each of you will open your own cell and walk free. If anyone has the wrong key, everyone will be executed then and there. You may discuss your strategy before tonight, but afterwards no communication will be allowed regarding my game. - Oh, and by the way, if you are still here seven years from today, I will execute everyone - it'll be a big birthday for me and I want to retire!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That night, just as promised, the warden went to each cell and gave each prisoner a key. As he handed each prisoner the key, he whispered to them the name of the person possessing the key to their cell. The keys were entirely indistinguishable from one another, but that was okay, because the prisoners had not counted on being able to tell anything about them. Indeed, the prisoners all seemed confident.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each day for the nearly seven years, the prisoners all anonymously gave the key in their possession to someone else. Then, hardly a month before the deadline, a prisoner shouted to the warden that they were ready. All the prisoners were certain that they would live, and, lo and behold, when they put the keys in the locks to their cells, every cell opened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What was their strategy? Exactly how many days did it take?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The warden had learned this from his own days as a prisoner, when the same challenge was issued to him and just two other prisoners. It took only six days of trading keys before they were released.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Oops, turns out my prisoners were kind of dumb in their choice of strategy. (If anyone wants to puzzle out what I was thinking of, the information in the post about &lt;em&gt;should&lt;/em&gt; be sufficient - but I consider the puzzle, and this question to more about the most efficient way to do it). The prisoners, if wise, can be free in less than a fortnight.&lt;/p&gt;&#xA;" OwnerUserId="4765" LastEditorUserId="4269" LastEditDate="2015-09-12T17:08:51.830" LastActivityDate="2015-09-12T17:08:51.830" Title="10 Prisoners, 10 Keys, and 7 Years" Tags="|mathematics|strategy|keys-and-locks|" AnswerCount="10" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7702" PostTypeId="2" ParentId="7701" CreationDate="2015-01-07T05:56:10.160" Score="17" Body="&lt;p&gt;I don't know if this works, necessarily, but I think it does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Organize the prisoners $A$ through $J$. On day 1, prisoner $A$ puts his key outside the cell of whoever has his key (we'll call him $A_2$). Nobody else moves keys. On day 2, $A_2$ returns the key he receives to prisoner $A$'s cell. Day 3 starts, and prisoner $B$ puts his key outside of $B_2$'s cell. $B_2$ returns the key on day 4. Continue this until day 20, when prisoner $J_2$ returns the key to $J$. On day 21, each prisoner puts the key by the cell that they are supposed to. For example, if prisoner $E$ gets a key on day 7 and returns it on day 8, he knows to put his key outside of prisoner $D$'s cell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, if a prisoner receives a key on day $x$, they will place their key outside cell $(x+1)/2$* (converted to a letter, of course).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully this makes sense and works as a solution. I haven't tested it fully but logically it should (unless I'm missing something here).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*: &lt;sub&gt;I'm bad at MathJax, someone help&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1664" LastEditorUserId="1664" LastEditDate="2015-01-07T06:02:27.843" LastActivityDate="2015-01-07T06:02:27.843" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="7703" PostTypeId="2" ParentId="6150" CreationDate="2015-01-07T05:57:52.600" Score="1" Body="&lt;p&gt;One of the solution that I could find is :&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/V3vV0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steps : &#xA;&lt;br&gt;1. Begin with the given sequence at f5 and e5.&#xA;&lt;br&gt;2. The number at d1 must be such that : (d5-d1) must be divisible by 4 and (d1-a1) must       be divisible by 3. So it must be 12/24. Using brute force elimination, it comes out to be 12.&#xA;&lt;br&gt;The first half then can be solved easily.&#xA;&lt;br&gt;3. Similarly for the second half, the number at h5 must be such that : (k5-h5) must be divisible by 3 and (h1-h5) must be divisible by 4. So it must be 11/23/35... Again using brute force elimination it is 23.&lt;/p&gt;&#xA;" OwnerUserId="5565" LastEditorUserId="5565" LastEditDate="2015-01-07T06:44:43.227" LastActivityDate="2015-01-07T06:44:43.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7707" PostTypeId="1" AcceptedAnswerId="7708" CreationDate="2015-01-07T07:34:39.613" Score="11" ViewCount="3080" Body="&lt;p&gt;I am an object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hard at heart, I can be shaped the same,&lt;br/&gt;&#xA;Thickened, I added to Nostradamus's fame,&lt;br/&gt;&#xA;My two faces are sharp or sweet&lt;br/&gt;&#xA;While the air grows warm, I am a delightful treat&lt;br/&gt;&#xA;Often in pairs, but enjoyed alone. &lt;br/&gt;&#xA;Pop me in your mouth and try not to moan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HINT:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I have a bag of these on the desk in front of me.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4951" LastActivityDate="2015-06-02T09:15:24.910" Title="Riddle Me This. What Object am I?" Tags="|riddle|rhyme|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7708" PostTypeId="2" ParentId="7707" CreationDate="2015-01-07T07:42:40.407" Score="13" Body="&lt;p&gt;The answer is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Cherry&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hard at heart, I can be shaped the same&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cherries have pits and can be shaped like hearts&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thickened, I added to Nostradamus's fame&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Only one I looked up (others make the answer obvious), he published cherry jelly recipes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My two faces are sharp or sweet&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cherries can taste sharp or sweet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While the air grows warm, I am a delightful treat&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Good to eat while warm (pies)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Often in pairs, but enjoyed alone&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cherries often come in twos (slot machine symbols)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pop me in your mouth and try not to moan&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastActivityDate="2015-01-07T07:42:40.407" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7709" PostTypeId="2" ParentId="7701" CreationDate="2015-01-07T08:56:38.120" Score="34" Body="&lt;p&gt;If every prisoner has a number assigned, that each other prisoner knows, they can leave the prison after 13 days. On day $i$, prisoner $i$ gives his key to the person that holds his key, it is returned on day $i+1$. So, at the start of day 12 every prisoner is holding the key he had in the beginning, and he knows who's key it is. On day 12 they hand over all the keys to the appropriate inmate, and on day 13 they challenge the guard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You did not mention that they have such numbers assigned by any means - and it wasn't necessary until today. So how would they agree on a common numbering scheme?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The algorithm you thought of, on the other hand, works even when the prisoners do not have numbers from 1 to 10 assigned previously. If, for exactly 2519 days, every prisoner gives the key he holds to the one he was told would have his key, they can walk free after 2520 days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why 2520? Because it is (2*2*2*3*3*5*7) = LCM(1,2,3,4,5,6,7,8,9,10), and these are all possible lengths of the cycles in a permutation of any ten values. So for any permutation $p$ of any ten values, $p^{2519}$ = $p^{-1}$.&lt;/p&gt;&#xA;" OwnerUserId="2544" LastActivityDate="2015-01-07T08:56:38.120" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7710" PostTypeId="2" ParentId="7701" CreationDate="2015-01-07T11:12:16.543" Score="12" Body="&lt;p&gt;The first day, &quot;while no one is watching&quot;, a prisoner takes his key and tries it in the lock of each of nine other cells.&#xA;The prisoner swaps his key for the one that is in the cell that opens.&#xA;The prisoner then tests that key in the remaining eight doors, and swaps again.&#xA;The prisoner repeats the process for the remaining seven etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first night the warden is called and all walk free.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Of course, if the keys really are indistinguishable, they will all open all the doors.&lt;/p&gt;&#xA;" OwnerUserId="1587" LastEditorUserId="1587" LastEditDate="2015-01-08T08:33:16.267" LastActivityDate="2015-01-08T08:33:16.267" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7711" PostTypeId="2" ParentId="7701" CreationDate="2015-01-07T11:56:10.393" Score="9" Body="&lt;p&gt;Just in case it's possible to put the keys in specific locations in a cell...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each prisoner draws a map of the prison on his floor (or a grid and each prisoner chooses a square that everyone else knows about, or a set of unique locations in the cell, etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On day one, each prisoner puts his current key in the cell of the person who has his actual key, in the location corresponding to the prisoner putting the key in. That evening, each prisoner knows whose key he (originally) had, which is also the person who put that key in their cell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On day two, each prisoner puts the key they got on day one back in the cell of the prisoner he got it from. Everyone now has their original keys, but now they know who owns it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On day three, each prisoner puts the key they got back on day two in the cell of its proper owner. That night when everyone is back in their cells, they summon the warden.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I rather think the intention of the puzzle is that the &quot;added&quot; keys don't go in any particular location, but you never know.&lt;/p&gt;&#xA;" OwnerUserId="6031" LastActivityDate="2015-01-07T11:56:10.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7712" PostTypeId="2" ParentId="7701" CreationDate="2015-01-07T14:42:58.930" Score="124" Body="&lt;p&gt;9 days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every day, each prisoner with an unknown key passes it to the next cell to the left (or to the next guy in alphabetic order, sentence length, age... w/e, just agree on the order on that first day).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since they know where in the order the owner of their own key is (relative to them), they know on which day to expect their key - they get to keep that, and keep passing the unknown keys as usual. (For example, if guy 4 know guy 7 had his key, he should keep the key he gets at day 3.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After 9 exchanges, technically possible in the afternoon of day 9, they can be sure everyone has their key.&lt;/p&gt;&#xA;" OwnerUserId="8739" LastActivityDate="2015-01-07T14:42:58.930" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="7713" PostTypeId="2" ParentId="6673" CreationDate="2015-01-07T14:43:55.700" Score="20" Body="&lt;p&gt;I found two codes, not very helpful for the final answer though :)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. Stars from left to right are a baconian cipher (white A, non white B): BMYGUEST&lt;br&gt;&#xA; 2. Ends of branches of both trees are a baconian cipher (with leave at end A, without leave at end B): FINDALH CIPHERS &#xA; (There is an encoding error in the second one, confirmed by BmyGuest)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Seven codes were found by Otaia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3. The rocks at the bottom are a baconian cipher (black 1, white 0): REDHERRING&lt;br&gt;&#xA; 4. The window cornices are a baconian cipher (trapezoidal 1, round 0): PERSONSAREMORSE&lt;br&gt;&#xA; 5. The people in the windows are morse code (short ., tall -): SUMTHELINESEVENORODD&lt;br&gt;&#xA; 6. The open/closed windows are a baconian cipher (open 1, closed 0): SUMVERTICALLYTOO&lt;br&gt;&#xA; 7. Stars from left to right are a baconian cipher (pointed down 1, up 0): HORIZONTAL&lt;br&gt;&#xA; 8. Stars from left to right (tiny 1, other 0): READWINDOWS&lt;br&gt;&#xA; 9. People in windows + Sir Richard himself (hand raised 1, not raised 0): BRIGHTNESS  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And for the final answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; According to clues 3 and 4, we compute the addition of the pixel values of each line in the image, and consider it an A if the sum is even, and a B if it is odd. Do the same for each column and concatenate the strings together.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Decoding the Baconian cipher we get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which are the beginning words of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Moby-Dick, by Herman Melville&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4430" LastEditorUserId="5840" LastEditDate="2015-01-14T10:51:40.917" LastActivityDate="2015-01-14T10:51:40.917" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="7714" PostTypeId="2" ParentId="7701" CreationDate="2015-01-07T17:31:15.940" Score="2" Body="&lt;p&gt;The prisoners all give their keys to the same prisoner on the first night. Then when she places the keys the next day she tries them in every lock, and leaves each key in the right cell. Everyone is freed on night 2 regardless of the number of prisoners.&lt;/p&gt;&#xA;" OwnerUserId="8746" LastActivityDate="2015-01-07T17:31:15.940" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7715" PostTypeId="2" ParentId="7701" CreationDate="2015-01-07T21:02:35.153" Score="5" Body="&lt;p&gt;All prisoners give their keys to the same person. This person tries all keys in their lock, keeps the one that works and passes on the rest. This solves the problem in 10 days and no one has to do math.&lt;/p&gt;&#xA;" OwnerUserId="7265" LastActivityDate="2015-01-07T21:02:35.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7716" PostTypeId="2" ParentId="7701" CreationDate="2015-01-07T21:46:56.447" Score="9" Body="&lt;p&gt;&lt;strong&gt;2 days&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Day 0&lt;/strong&gt; (strategy discussion)&lt;br&gt;&#xA;Each prisoner is given a number.&lt;br&gt;&#xA;One prisoner is chose.  In his cell is designed a 10 x 10 grid (using chalk if available, other wise distance from the walls).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Day 1&lt;/strong&gt;&lt;br&gt;&#xA;Each prisoner brings the key in they were given to the designated cell.  They place the key in the row corresponding to their number and the column corresponding to the number of the prisoner who originally received their key.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Day 2&lt;/strong&gt;&lt;br&gt;&#xA;The prisoner with all they keys uses the columns to determine who needs which key and the rows to determine which key to give them.  The prisoner then distributes all the keys to the correct cells and calls the warden.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Thoughts:&lt;/strong&gt;&lt;br&gt;&#xA;I find that good puzzles require the use of resources which are required to be present in the definition (space in a cell) but not directly mentioned.&lt;/p&gt;&#xA;" OwnerUserId="8747" LastActivityDate="2015-01-07T21:46:56.447" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="7717" PostTypeId="2" ParentId="7701" CreationDate="2015-01-07T22:17:52.203" Score="0" Body="&lt;p&gt;Every prisoner gives his key to the person who has the key to his cell (as told by the warden). Next night, they repeat the whole process. If no one messes up and keeps giving their keys (that means a different key each night) to the same person, everyone will automatically end up with their keys after exactly 9 nights.&#xA;This is because the telling them creates a chain, in which all prisoners are linked. (Every prisoner has the key to someone elses cell and knows the one who has the key to their cell) Passing keys around in the described fashion &quot;rotates&quot; this chain. Repeat it 10 times and each prisoner ends up with the key he initially got from the warden, one time less and they end up with their own key.&#xA;Sorry I can't really explain it properly, but it works (I tried it 2 times with playing cards) and is fool-proof.&#xA;For a maybe better explanation, the problem here is a quite similar: &lt;a href=&quot;https://www.youtube.com/watch?v=C5-I0bAuEUE&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=C5-I0bAuEUE&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'm not sure if it will still work if there are two chains, i.e. Pris4 has the key to Cell2 and vice versa - then the two will just switch keys back and forth…&lt;/p&gt;&#xA;" OwnerUserId="8751" LastActivityDate="2015-01-07T22:17:52.203" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7718" PostTypeId="1" AcceptedAnswerId="7721" CreationDate="2015-01-07T22:45:25.663" Score="1" ViewCount="374" Body="&lt;h1&gt;Background&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;It isn't a common art style, but it is still done: using letters for form in pictures. Sometimes, people create art using &lt;em&gt;only&lt;/em&gt; letters. Now, let us choose a word, i.e. Cool. 'Cool' can be used as following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/v2Bwz.png&quot; alt=&quot;Cool -&amp;gt; Graphic&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cool, isn't it?&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Challenge&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Your challenge is to find a word (a noun) that can be formed to describe itself pictorially. Let us presume that the word TRIANGLE could be made into a triangle. Then, this would be a valid entry.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Rules&lt;/h1&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The word must have at least three letters.&lt;/li&gt;&#xA;&lt;li&gt;The picture form must represent what the word is, literally.&lt;/li&gt;&#xA;&lt;li&gt;You must provide a picture in your answer.&lt;/li&gt;&#xA;&lt;li&gt;There can be overlap of letters, but at least have a majority of each letter showing.&lt;/li&gt;&#xA;&lt;li&gt;The first letter of the word &lt;em&gt;must&lt;/em&gt; be upright; the others may be rotated and shrunk, roughly maintaining its proportions.&lt;/li&gt;&#xA;&lt;li&gt;All letters must be visible.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h1&gt;Score&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Do not tell the word, as the objective is for the best answers to float to the top. People can try and answer in comments. I will accept whichever answer/puzzle seems worthy of +15.&lt;/p&gt;&#xA;" OwnerUserId="7307" LastEditorUserId="2383" LastEditDate="2015-01-08T13:19:56.647" LastActivityDate="2015-01-08T13:19:56.647" Title="Object Representation" Tags="|word|visual|open-ended|" AnswerCount="3" CommentCount="12" ClosedDate="2015-01-08T10:21:08.607" ContentLicense="CC BY-SA 3.0" />
  <row Id="7720" PostTypeId="2" ParentId="7718" CreationDate="2015-01-07T23:38:39.660" Score="6" Body="&lt;p&gt;Here's my effort, a representation of the word 'toes':&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7ZIhn.jpg&quot; alt=&quot;Picture&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7363" LastActivityDate="2015-01-07T23:38:39.660" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7721" PostTypeId="2" ParentId="7718" CreationDate="2015-01-07T23:41:44.023" Score="11" Body="&lt;h1&gt;Clock&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nslpN.png&quot; alt=&quot;Clock&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This one turned out quite well, I think.&lt;/p&gt;&#xA;" OwnerUserId="7308" LastActivityDate="2015-01-07T23:41:44.023" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7722" PostTypeId="2" ParentId="7718" CreationDate="2015-01-08T00:13:43.587" Score="2" Body="&lt;p&gt;Here's a picture with the word &lt;strong&gt;LEGO&lt;/strong&gt; from my website's archive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YpdxU.gif&quot; alt=&quot;LEGO&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I made this a few years ago using &lt;a href=&quot;http://www.ldraw.org&quot; rel=&quot;nofollow noreferrer&quot;&gt;LDraw&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4394" LastActivityDate="2015-01-08T00:13:43.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7724" PostTypeId="1" AcceptedAnswerId="7725" CreationDate="2015-01-08T02:13:04.673" Score="28" ViewCount="3532" Body="&lt;p&gt;The word under secret means something. What does it mean? (Find the secret message.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/imrdS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="7722" LastEditDate="2015-01-23T03:22:41.750" LastActivityDate="2016-05-02T05:51:43.747" Title="Find the secret message" Tags="|cipher|visual|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7725" PostTypeId="2" ParentId="7724" CreationDate="2015-01-08T03:43:23.517" Score="32" Body="&lt;p&gt;The solution is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; gibberish&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These were my steps to find the answer - first, I:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Got the RGB ASCII values for every color.&lt;br&gt;&#xA; These were, &lt;strong&gt;For the word secret:&lt;/strong&gt;&lt;br&gt;&#xA; 31 41 59, 26 53 58, 97 93 23, 84 62 64, 33 83 27, 13 37 0&lt;br&gt;&#xA; &lt;strong&gt;For the text at the bottom:&lt;/strong&gt;&lt;br&gt;&#xA;22 9 7 5 14 5 18 5 20 (Only the 'red' bit of the color had a value other than zero).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then, by realizing that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The numbers for the top section correspond to the value of PI&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The numbers 22 9 7 5 14 5 18 5 20 when turned into letters, A=1, Z=26 correspond to VIGENERE(T)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You find that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That &lt;em&gt;vqqjtzxaw&lt;/em&gt; decoded via the Vigenere cipher with a key of PI yields &lt;em&gt;gibberish&lt;/em&gt;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7308" LastActivityDate="2015-01-08T03:43:23.517" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7726" PostTypeId="1" AcceptedAnswerId="7732" CreationDate="2015-01-08T04:13:12.177" Score="8" ViewCount="2799" Body="&lt;p&gt;This is a variation of the riddle of 2 guards and 2 doors:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are stuck in a room with 3 doors. Each of the doors have a guard posted. One guard always tells the truth, one always lies, and the third sometimes tells truths and sometimes tells lies, but stabs you if you ask him more than 2 questions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a reliable way to solve this?&lt;/p&gt;&#xA;" OwnerUserId="8757" LastEditorUserId="69582" LastEditDate="2023-03-20T00:54:49.657" LastActivityDate="2023-03-20T00:54:49.657" Title="Lying, truthful and merciless guards (variation)" Tags="|logical-deduction|liars|" AnswerCount="5" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="7727" PostTypeId="2" ParentId="7726" CreationDate="2015-01-08T04:43:26.990" Score="7" Body="&lt;p&gt;Ask all of the guards, &quot;Are you the merciless guard?&quot;. If two of them respond &quot;no&quot; and one &quot;yes&quot;, then you know that the one that said &quot;yes&quot; is the liar (and that the merciless one said &quot;no&quot;) - and you can ask them &quot;Will I escape through this door? How about this one?&quot; - whichever one they say you won't escape through is the proper door to open. If, to your initial round of questions, you get two responses of &quot;yes&quot; and one &quot;no&quot;, then you know that the one who said &quot;no&quot; is the truth-teller (and the merciless one said &quot;yes&quot;), and you can ask that guard which door is the escape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This uses the fact that we &lt;em&gt;know&lt;/em&gt; we will get one response of &quot;yes&quot; and one of &quot;no&quot; to the first question - from the liar and the truth-teller respectively. So, if we only have one &quot;yes&quot; or one &quot;no&quot;, we can identify who said it, meaning that we can assuredly avoid asking the merciless one two questions and, moreover, that we will know exactly who we're talking to. Since there's only three guards and two possible answers, we are assured that this will always be the case.&lt;/p&gt;&#xA;" OwnerUserId="4765" LastEditorUserId="4765" LastEditDate="2015-01-08T04:49:34.823" LastActivityDate="2015-01-08T04:49:34.823" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7728" PostTypeId="1" AcceptedAnswerId="7747" CreationDate="2015-01-08T05:03:47.900" Score="3" ViewCount="821" Body="&lt;p&gt;Note: Not a true story&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am currently bored while idly thinking about how I got this job in VA. Sitting at a desk, I think about my work conditions while playing around on Inkscape. Below is the picture I drew.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/alDjm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where am I located?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Look at my last puzzle. By the way I also incorporated a similar thing where things get harder the more you solve.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Look for what stands out (yeah it all stands out but some things &quot;actually&quot; stand out)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pardon the image quality/content. If this one is also solved in under an hour I'll keep upping the difficulty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: there will feel like a definitive answer when the puzzle is solved, but just to be clear, the answer will be directly related to six words, and the &quot;HAIR&quot; word shown isn't one of those. (it may be related to the puzzle however)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I keep feeling like I need to add more information but I'll stop editing for now and give you all a chance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I know I said I would stop adding more hints, but I think one of my hints isn't fully accurate so I'll clarify and basically give up a major point.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I say I made the image in Inkscape (I did, some of it), but if you know what Inkscape is, this image is obviously not of the traditional filetype. What is something that can be done in Inkscape puzzles?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="7722" LastEditDate="2015-01-08T22:07:02.693" LastActivityDate="2015-01-09T10:11:13.673" Title="Where am I located?" Tags="|cipher|visual|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7729" PostTypeId="2" ParentId="6288" CreationDate="2015-01-08T07:09:33.440" Score="4" Body="&lt;p&gt;&lt;em&gt;Hello everyone, my name is Jay&lt;br&gt;&#xA;I work for you 24 hours a day,&lt;br&gt;&#xA;but don't get me wrong I know not Silent Bob&lt;br&gt;&#xA;and got nothing to do with weed on the job.&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As I already said this did only give the hint that those &quot;Jays&quot; are working 24 hours a day&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My kind can be you or you can be me&lt;br&gt;&#xA;if you know a body by its chemistry.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chemistry is not Biology: the chemistry I'm talking about is the interaction between charged particles (for example molecules work with protonic/electronic pumps&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;We're not only human, but mostly a thing!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Self explanatory, once you read the solution&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;We have many shapes, on the air we are king.&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many shapes... Again, self explanatory. On the air = On-The-Air -&gt; Radio&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Our children you could surf but not in the sea&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Waves are surfable... and there are some kind of them which are not necessarily made of water&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(for some they are a monstruosity!).&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's no mystery that some people despise electromagnetic pollution&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;They fly, they bounce, they bend,&lt;br&gt;&#xA;but always to us they end!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; E.m. waves can fly, can bounce (on metals) can &quot;bend&quot; (refraction) and get captured only by... Antennas! In fact antennas trasmit (waves are their child) and receive such waves&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;We get and we give even not alive,&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Antennas are passive pieces of metal&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;but to do our best we need a good drive,&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They need to have a power source in order to be able to send a signal&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;for we can make work a lot you possess,&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well, again self-explanatory&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;even destroy it if working in excess.&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if overloaded - for example with a lightning - the power spike can destroy electronics in you TV or any other thing connected to it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This riddle was difficult but I'm not a sphinx,&lt;br&gt;&#xA;in fact I'm well sure, who I am you can glimpse&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well this last verse was only a challange to you ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the solution is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ANTENNA! P.s. = There are some kind of antennas that are called &quot;J&quot; antennas ;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5122" LastEditorUserId="5122" LastEditDate="2015-01-09T06:48:15.377" LastActivityDate="2015-01-09T06:48:15.377" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7730" PostTypeId="1" AcceptedAnswerId="7811" CreationDate="2015-01-08T07:50:16.977" Score="15" ViewCount="1804" Body="&lt;p&gt;The one I obey, my dear commander&lt;br&gt;&#xA;Keep on running I will not slander&lt;br&gt;&#xA;For I know you will follow my lead&lt;br&gt;&#xA;Call me in if you are in need  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one commands the other will follow&lt;br&gt;&#xA;That's the deal, one has to swallow&lt;br&gt;&#xA;Either way, I rely on you&lt;br&gt;&#xA;We kiss goodnight and start anew  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mighty Dictator of all places&lt;br&gt;&#xA;followed by all human races&lt;br&gt;&#xA;Watch over all the new born lives&lt;br&gt;&#xA;earn the fruit of fertile strifes  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Endless love, you never disclaim&lt;br&gt;&#xA;nor would you repel and never blame&lt;br&gt;&#xA;Unbribable Protector, punish me&lt;br&gt;&#xA;and I will follow tacitly  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is an easy riddle. For all the others:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Every verse contains the same solution. The whole riddle is &lt;em&gt;metaphorically&lt;/em&gt; describing one solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5106" LastEditorUserId="5373" LastEditDate="2015-03-14T12:43:55.313" LastActivityDate="2015-03-14T12:43:55.313" Title="The sands of life are running out" Tags="|riddle|word|rhyme|" AnswerCount="4" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="7732" PostTypeId="2" ParentId="7726" CreationDate="2015-01-08T09:02:39.383" Score="18" Body="&lt;p&gt;It seems that it is trivial to solve this. As long as we identify either the liar or the truth-teller, we can obtain 100% correct statements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, ask &quot;Is 1+1 == 2&quot; to all 3 guards. The liar will always say &quot;no&quot;, while the truth-teller will always say &quot;yes&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will get 1 of 2 outcomes: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 of them says yes while the other 2 say no. Since there must be at least one liar and one truth-teller, the merciless guard is lying in this case. Ask the guard who says yes which door is the true door by exhaustion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 of them says no while the other 2 say yes. The merciless guard is telling the truth, and we can perform the same operation with the liar, just with answers reversed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can therefore ignore the knave guard.&lt;/p&gt;&#xA;" OwnerUserId="7529" LastActivityDate="2015-01-08T09:02:39.383" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7734" PostTypeId="2" ParentId="7726" CreationDate="2015-01-08T11:54:18.143" Score="2" Body="&lt;p&gt;In response to Taemyr&#xA;It is most likely impossible, if the Guards can answer with other words than yes/no:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The liar can just answer &quot;I don't know&quot; to any question he knows, so we cannot get any information out of him, While the stabber can always answer exactly like the truth-teller, except for the final answer we need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means, whatever we do we can never decide between the Knave and the truth teller - that only leaves us with the liar. And he can just answer &quot;I don't know&quot; to any question we ask, If we ask him recursive questions, he can just replace it with anything &quot;What would you answer if....&quot; and he just says JELLYFISH!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I can see - unsolvable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Or the default method (If the liar always lies with opposites so yes-&gt;no):&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ask the first one: &quot;If I would ask you if Guard 2 is merciful, what would you say?&quot;&#xA;The thruth-teller will answer thruthfull, the Liar will lie about his lie and also say the truth, the killer will say anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We then choose Guard2 if he is merciful, otherwise Guard3. If the first one was the Knave, we have a safe one, otherwise we got a thruthful answer and will also have a safe one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we can just ask one or two question to the truthteller/liar:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I would ask the other merciful guard (the one who doesn't stab people) is the first door safe, what would he say?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If he says no, take the door otherwise ask the same for the 2nd door, if no take it, otherwise take door 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So 3 Questions overall.&lt;/p&gt;&#xA;" OwnerUserId="1998" LastEditorUserId="1998" LastEditDate="2015-01-08T13:00:03.027" LastActivityDate="2015-01-08T13:00:03.027" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7735" PostTypeId="2" ParentId="6691" CreationDate="2015-01-08T13:10:32.047" Score="5" Body="&lt;p&gt;Only 9 but a bit of a tongue-twister &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I sleightly, slightly light sightly slight gilt; it lit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cunningly make a small flame on a slim attractive government bond; it alighted&lt;/p&gt;&#xA;" OwnerUserId="8765" LastActivityDate="2015-01-08T13:10:32.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7736" PostTypeId="2" ParentId="7701" CreationDate="2015-01-08T14:32:04.117" Score="2" Body="&lt;p&gt;Since the answer has already been given (both the intended answer, and the &quot;best&quot; answer), I'll just throw out my subversive answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The night before they all agree to leave the key in their possession under their pillows. While pretending to give the key in their possession to a new cell, they each go and retrieve their key on day one.  That first night they call the warden and open their cells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The warden, aware of their deception but also that they were not expressly forbidden from finding or removing a key, shoots them or lets them go depending on his capricious whim.&lt;/p&gt;&#xA;" OwnerUserId="358" LastActivityDate="2015-01-08T14:32:04.117" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7737" PostTypeId="2" ParentId="6691" CreationDate="2015-01-08T15:51:46.517" Score="11" Body="&lt;p&gt;1 to 11&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;sand dancers reascend, desecrating a decreasing sedan and an increased sander&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1 to 12. The tale of a tough soldier having trouble decorating his clothes, whilst an ex-pat displays videos of his own delightful solutions online.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A mean marine and raiment garnishment nightmares, an emigrant streaming charmingest garnishments&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8778" LastEditorUserId="8778" LastEditDate="2015-01-08T16:29:42.180" LastActivityDate="2015-01-08T16:29:42.180" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7738" PostTypeId="2" ParentId="6691" CreationDate="2015-01-08T17:12:28.323" Score="23" Body="&lt;p&gt;12 &lt;strong&gt;in ascending order of length&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I, in tin unit, unite minute pinetum, tumpline, &lt;a href=&quot;https://en.wikipedia.org/wiki/Penultima&quot;&gt;penultima&lt;/a&gt;; manipulate &lt;a href=&quot;http://dictionary.reference.com/browse/planetarium&quot;&gt;planetarium&lt;/a&gt; planetariums&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I, in a metal shed, combine a very small coniferous arboretum, a load-carrying strap and a logic game, and handle models of the solar system in a building made for such a model&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i&lt;em&gt;n&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;t&lt;/em&gt;in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;u&lt;/em&gt;nit&lt;/p&gt;&#xA;&#xA;&lt;p&gt;unit&lt;em&gt;e&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;m&lt;/em&gt;inute&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;p&lt;/em&gt;inetum&lt;/p&gt;&#xA;&#xA;&lt;p&gt;tump&lt;em&gt;l&lt;/em&gt;ine&lt;/p&gt;&#xA;&#xA;&lt;p&gt;penultim&lt;em&gt;a&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;manipul&lt;em&gt;a&lt;/em&gt;te&lt;/p&gt;&#xA;&#xA;&lt;p&gt;planeta&lt;em&gt;r&lt;/em&gt;ium&lt;/p&gt;&#xA;&#xA;&lt;p&gt;planetarium&lt;em&gt;s&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="8765" LastEditorUserId="8765" LastEditDate="2015-01-08T17:29:31.357" LastActivityDate="2015-01-08T17:29:31.357" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7739" PostTypeId="2" ParentId="7730" CreationDate="2015-01-08T17:33:12.047" Score="3" Body="&lt;p&gt;&lt;strong&gt;Edit 1: New answer following emphasis from your hints&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Doctor&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The one I obey, my dear commander&#xA;Keep on running I will not slander &#xA;For I know you will follow my lead &#xA;Call me in if you are in need&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The one talking is the Doctor. The doctor listens to patients and do what they tell him to. (*cough I'm sick write me a prescription) The doctor told this guy not to run but he didn't listen. The doctor knows though that this guy will eventually follow his lead and asks him to request a house call when he needs a doctor. (Because the foolish guy kept running)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If one commands the other will follow &#xA;That's the deal, one has to swallow &#xA;Either way, I rely on you &#xA;We kiss goodnight and start anew&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A doctor needs a good reputation and income so he is just as reliant on patients as patients are reliant on him. (With patients often having to swallow medication) Sometimes though you don't have to take medication but undergo surgery, in which case you &quot;sleep&quot; and start anew, either in the afterlife from a mistake (doctors are only human), or perhaps a new life of healthiness.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mighty Dictator of all places &#xA;followed by all human races &#xA;Watch over all the new born lives &#xA;earn the fruit of fertile strifes &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letters MD are capitalized, meaning medical doctor. All human races rely on doctors. Doctors deliver all babies (anyone who delivers a baby and is not officially a doctor may as well be considered a doctor from that point on, that's tough stuff). All fertile strifes result in injuries (fights), and doctors are the ones who benefit from them (from all the business).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Endless love, you never disclaim &#xA;nor would you repel and never blame &#xA;Unbribable Protector, punish me &#xA;and I will follow tacitly&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Doctors never say they love you, nor do they repel or blame patients (with the exception of Dr. House). Your lives are in their hands, they don't really take bribes because most of em have more than enough money, and if they don't like you then it's possible a 'mistake' can happen while under the knife.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sands of life are running out... You shouldn't have pissed off the doctor.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The answer I'm thinking of &quot;fits&quot; but probably not what you're looking for.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Heart&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The one I obey, my dear commander&#xA;Keep on running I will not slander&#xA;For I know you will follow my lead&#xA;Call me in if you are in need&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You follow your heart, sometimes you need to help it medically.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If one commands the other will follow&#xA;That's the deal, one has to swallow&#xA;Either way, I rely on you&#xA;We kiss goodnight and start anew&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You rely on your heart, and are with it every night.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mighty Dictator of all places&#xA;followed by all human races&#xA;Watch over all the new born lives&#xA;earn the fruit of fertile strifes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All human races have hearts; all newborns have one; fertile strifes represent troubles which ties in to &quot;troubled heart&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Endless love, you never disclaim&#xA;nor would you repel and never blame&#xA;Unbribable Protector, punish me&#xA;and I will follow tacitly&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Endless love could refer to forever beating heart, unbribable and can punish you with a heart attack&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again, probably not what you're looking for because it doesn't fit perfectly, but just barely enough to post.&lt;/p&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="7722" LastEditDate="2015-01-11T02:02:50.683" LastActivityDate="2015-01-11T02:02:50.683" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7740" PostTypeId="2" ParentId="7726" CreationDate="2015-01-08T17:33:58.170" Score="1" Body="&lt;p&gt;Assuming that each guard can only answer yes or no questions, ask the left guard:&#xA; - Will the middle guard say you are the merciless guard? &#xA; - Will the right guard say you are the merciless guard? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The middle guard:&#xA; - Will the left guard say you are the merciless guard? &#xA; - Will the right guard say you are the merciless guard? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And to the right guard:&#xA;- Will the left guard say you are the merciless guard? &#xA;- Will the right guard say you are the merciless?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since both the truth teller and the lier cannot predict the merciless guard's answer, they must answer &quot;I don't know&quot; when asked about the merciless guard since doing otherwise might cause them to lie or tell the truth, which is impossible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The merciless guard knows which of the others is the truth teller and which is the lier so he is free to answer &quot;yes&quot;, &quot;no&quot;, or lie and say &quot;I don't know&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose the door behind the guard which received two 'I don't know&quot; answers. Even if the merciless guard replied &quot;I don't know&quot; twice he'll be alone in pointing to that guard. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So 6 questions if we allow the merciless guard yo answer I don't know and don't use Falco's double negative &quot;If I were to ask you, what would you say?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the merciless guard can't lie and say I don't know then you only need to ask the questions to one person and choose the door behind the person that received the single &quot;I don't know&quot; answer or didn't give it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If on the other hand the lier can also reply &quot;I don't know&quot; then I don't think you can solve it without a benchmark question like &quot;1+1=2?&quot;&lt;/p&gt;&#xA;" OwnerUserId="8783" LastActivityDate="2015-01-08T17:33:58.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7741" PostTypeId="1" CreationDate="2015-01-08T18:30:12.653" Score="14" ViewCount="1003" Body="&lt;p&gt;A hotel has $n$ rooms, numbered 1, 2, 3, etc. until $n$.&#xA;At the hotel reception there are $n$ buttons, one for each room, also numbered 1, 2, 3, ..., $n$.&#xA;Button 1 switches the light from on to off or from off to on in all rooms whose number contains a &quot;1&quot;, hence in rooms 1, 10, 11, 12, etc.&#xA;Button 2 switches the light on or off in all rooms whose number contains a &quot;2&quot;, hence 2, 12, 20, 21, etc.&#xA;The same thing happens with all other buttons, so for example button 42 switches the light in rooms 42, 142, 242, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The receptionist sees that all rooms have the lights on, so she pushes all buttons from 1 to $n$ in sequence. At the end of the process she notices that there are just as many rooms with the light on as there are rooms with the light off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many rooms are there in the hotel at most?&lt;/p&gt;&#xA;" OwnerUserId="4394" LastActivityDate="2015-01-11T19:47:25.410" Title="The lights in the hotel" Tags="|mathematics|calculation-puzzle|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7742" PostTypeId="2" ParentId="7741" CreationDate="2015-01-08T19:33:46.173" Score="4" Body="&lt;p&gt;The answer (assuming I didn't mess up somewhere) is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 306&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I sort of brute forced the problem because it seemed faster that way.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To still be on, the number has to be called an odd number of times before its own button is pushed. So, since the first hundred numbers can only be called twice max (once for each digit), the only ones that stay on are 10/11, 20/22, etc up to 90/99 (18 on 81 off).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After three digits, it changes a bit. There are now 5 ways a 3 digit number can be called, one of its numbers, the first two, or the last two. There are no &quot;1&quot;s (even 100 has 1 and 10), so 3's are 102-110(9 on) and are only possible with the &quot;0&quot; second digit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;For 5's, from 120 on there are 7 per set of 10, the 3 not included are 1*0, 1*1, and 1**, so for the set from 100-199, 65 are on and 35 are off (9+7*8). Another way to think of this is +30 to the original -63.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From here,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every hundred from here on will be another +30 with symmetry. So by 300 the &quot;count&quot; will be -3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And finally,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;300(3 and 30) is -1, 301 and 302 are +1, 303 is -1, 304-309 are +1 but by 306 the -3 from before is reached (-1+2-1+3), leaving a final answer of 306. It will never back track far enough to go negative again up to a thousand.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It may be explained a bit messily, I'm not a math proofs guy but you can generalize my statements for any number of digits. (There's probably a better way to spoiler a paragraph than to add random words in between sentences, I'm still new at this)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I just considered the possibility that in the 4 digit numbers it could loop back to being more negative but my brain is starting to hurt and I wouldn't be able to focus long enough to figure it out anyway if that was the case (because then the solution would be in the 5 digit numbers).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastActivityDate="2015-01-08T19:33:46.173" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7743" PostTypeId="2" ParentId="7741" CreationDate="2015-01-08T21:28:24.190" Score="11" Body="&lt;p&gt;There might be a problem. It looks like Quark's answer of &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 306 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is correct, given this plot for up to 1000 rooms (excess of lights on over lights off)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tvnh9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But if we extend that range a little...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;We see that the initial trend reverses, crossing the y-axis several times after 8000 rooms. It's not clear to me that this back-and-forth trend ends. &lt;img src=&quot;https://i.stack.imgur.com/th8m5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3987" LastActivityDate="2015-01-08T21:28:24.190" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7744" PostTypeId="2" ParentId="7728" CreationDate="2015-01-08T22:09:03.707" Score="0" Body="&lt;p&gt;You are located&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; in the state of Virginia, USA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; at the beginning it says &quot;I am currently bored while idly thinking about how I got this job in VA&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4050" LastActivityDate="2015-01-08T22:09:03.707" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7745" PostTypeId="2" ParentId="7730" CreationDate="2015-01-08T22:19:16.993" Score="5" Body="&lt;p&gt;Hmm.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Man's best friend.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The one I obey, my dear commander&#xA;Keep on running I will not slander&#xA;For I know you will follow my lead&#xA;Call me in if you are in need&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many dogs will run in-front of their owners; as long as their owners are willing to run along, dogs will keep running and the owners will follow.  Unless the dog goes a direction the owner doesn't want to, in which case, from the dog's perspective, the dog is coming to the aid of the owner.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If one commands the other will follow&#xA;That's the deal, one has to swallow&#xA;Either way, I rely on you&#xA;We kiss goodnight and start anew&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dog owner's often make their dogs &lt;strong&gt;swallow&lt;/strong&gt; medication and many owners kiss their pets goodnight.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mighty Dictator of all places&#xA;followed by all human races&#xA;Watch over all the new born lives&#xA;earn the fruit of fertile strifes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In this particular dog's situation, their is a man or woman of their household who simultaneously takes care of the children, brings home the bread, and commands their significant other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Endless love, you never disclaim&#xA;nor would you repel and never blame&#xA;Unbribable Protector, punish me&#xA;and I will follow tacitly&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The endless love refers to how Pet owners unconditionally love their pets (especially dog owners); Unbribable in the sense that owners rarely give in to their pets if they pout; and tacitly as dogs have no voice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bonus Round&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For the sands of time, are running out is a lyric from Elvis Presley's Wisdom of the Ages, who also wrote &quot;Hound Dog&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8787" LastEditorUserId="8787" LastEditDate="2015-01-08T22:29:43.200" LastActivityDate="2015-01-08T22:29:43.200" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7746" PostTypeId="2" ParentId="6531" CreationDate="2015-01-08T22:44:32.363" Score="3" Body="&lt;p&gt;EDIT: NOW I got it! (If you say I'm wrong I'm going to sigh)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Machinima&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mach is a fast number, in (from the riddle), the head of an adult insect is &quot;ima&quot; of &quot;imago&quot;. Machinima is using computer to create a cinematic production.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit 2: I think I got 5A also:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Disciple&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Spurn a guilty plea&quot; is like &quot;diss a plea&quot;, and googling aquidneck island shows a lot of stuff related to disciples (as well as a book saying disciples first settled on the island).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I finally got 6 across after trying for two days.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Karyogamy &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Ogamy is imago backwards (adult insect), kary is like kareoke where you perform a quick number. In the head meaing kary before ogamy. I'm surprised I got this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="7722" LastEditDate="2015-01-09T17:36:44.367" LastActivityDate="2015-01-09T17:36:44.367" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="7747" PostTypeId="2" ParentId="7728" CreationDate="2015-01-08T23:14:12.237" Score="4" Body="&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; - OK I think the location is now correct (message 6).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Message 1:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The word HAIR means to look in the clown's green hair.  There is a 6x12 pixel block in the center with RGB colors 20-128-8, 19-128-8, 1-128-16, 5-128-0.  Converting to numbers of the alphabet (and ignoring 128) provides &lt;strong&gt;T-H, S-H, A-P, E&lt;/strong&gt;.  So the message is &quot;the shape&quot; with the first &quot;e&quot; missing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Message 2:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; So &quot;the shape&quot; is a star or a pentagram.  Interpreting this to be the larger star with pink shorts and assuming it is a Pythagorean pentagram, then its right arm (representing heat or the sun) is &lt;strong&gt;pointing at the sun&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Message 3:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  There is a block of 3 pixels adjacent to a lower left ray from the sun.  Their RGB colors are 3-15-0, 18-15-0, 5-18-0.  Converting to letters provides &lt;strong&gt;C-O, R-O, E-R&lt;/strong&gt;, which could mean &lt;strong&gt;RECOLOR&lt;/strong&gt;.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Message 4:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  There are 4 tiny letters constructed with pixels that have a color of 12-34-56.  They are located in the word HAIR (S), in the fish eye (T), in the pink shorts (A), and below the trollface (R).  They spell &lt;strong&gt;&quot;STAR&quot;&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Message 5:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is a row of black pixels in the right side of the bottom frame.  Using morse code and reading from right to left provides &lt;strong&gt;CONNECTTHE&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Message 6:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Putting the clues together - the shape is a star or a pentagram with its points connected to form a pentagon.  So the location is &lt;strong&gt;The Pentagon in Arlington County, Virginia&lt;/strong&gt;.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-01-09T10:11:13.673" LastActivityDate="2015-01-09T10:11:13.673" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7748" PostTypeId="2" ParentId="6691" CreationDate="2015-01-08T23:31:39.407" Score="9" Body="&lt;pre&gt;&lt;code&gt;a&#xA;at&#xA;ate&#xA;late&#xA;stale&#xA;castle&#xA;elastic&#xA;elastics&#xA;silicates&#xA;specialist&#xA;plasticised&#xA;(deplasticise)&#xA;(deplasticiser)&#xA;(deplasticisers)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;A late silicates specialist ate stale plasticised elastics at elastic castle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An expert in silicon compounds was staying at an old castle built out of random rubber bands and other stretchy materials when he decided to eat some of the walls.  The structure of the castle was stable due to the rubber amalgam, giving it a slightly malleable texture.  He is dead now, presumably from eating plasticised rubber... or is it rubberized plastic?  Only the builder (and the dead specialist) will ever know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last three words don't really count since I couldn't fit them in the sentence, plus the words are recycled and lazy, but oh well!&lt;/p&gt;&#xA;" OwnerUserId="8788" LastActivityDate="2015-01-08T23:31:39.407" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7752" PostTypeId="2" ParentId="6691" CreationDate="2015-01-09T11:53:55.613" Score="1" Body="&lt;p&gt;6 words:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I  &#xA;if  &#xA;fin  &#xA;fins  &#xA;sniff  &#xA;sniffs  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If I sniff fins, fin sniffs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8799" LastEditorUserId="5840" LastEditDate="2015-01-09T12:27:21.877" LastActivityDate="2015-01-09T12:27:21.877" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7753" PostTypeId="2" ParentId="6691" CreationDate="2015-01-09T12:13:22.243" Score="28" Body="&lt;p&gt;&lt;strong&gt;Thirteen words:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;13 Universalists&#xA;12 Universalist&#xA;11 Antiviruses&#xA;10 Intrusives&#xA;9  Intrusive&#xA;8  Inviters&#xA;7  Inverts&#xA;6  Strive&#xA;5  Rivet&#xA;4  Tire&#xA;3  Tie&#xA;2  It&#xA;1  I&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Slightly contrived sentence:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Universalists rivet intrusive antiviruses, inviters tie intrusives (inverts it), I tire; Strive, Universalist!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Translation:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Those seeking a universal approach to computing will package their operating systems, somewhat firmly, with annoying security software. Those who would invite trouble choose to disable these annoyances, thus reversing the intended antiviral effect. I become bored with this state of affairs, and say to those operating system manufacturers: &quot;Keep trying!&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4534" LastActivityDate="2015-01-09T12:13:22.243" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7754" PostTypeId="2" ParentId="6691" CreationDate="2015-01-09T14:19:12.293" Score="5" Body="&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;:&#xA;Maybe 15? These are technically two sentences:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tie ratiocinated, &quot;Trite, cattier attire contraindicated accreditation.&quot; I ratiocinate, &quot;Contraindicate it? Interact! Intricate, antierotic rite tie!&quot;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;contraindicated&#xA;contraindicate&#xA;accreditation&#xA;ratiocinated&#xA;ratiocinate&#xA;antierotic&#xA;intricate&#xA;interact&#xA;cattier&#xA;attire&#xA;trite&#xA;rite&#xA;tie&#xA;it&#xA;I&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I and a fellow comissioner &lt;a href=&quot;http://www.babynamespedia.com/meaning/Tie/m&quot; rel=&quot;nofollow&quot;&gt;Tie&lt;/a&gt; are evaluating whether or not to grant accreditation to a school. The school's principal is a highly eccentric lady who stands on bizarre ceremony. Tie explains to me that he wished to refuse to grant accreditation to the school on account of the inappropriate slogans printed on the principal's T-shirt. I counter that had Tie actually interacted with the principal, he would have discovered that her behaviors were an equally good reason to deny accreditation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Old answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My best is only 9 so far.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;o&#xA;do&#xA;rod&#xA;cord&#xA;crowd&#xA;crowds&#xA;scowder&#xA;escrowed&#xA;scowdered&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;O escrowed rod, scowdered crowds do crowd scowder cord.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Translation:&#xA;A rod of gold is held in escrow in a large metal safe. Thieves are attempting to blow up the safe with dynamite, but they crowd the fuse and become scorched. The owner of the safe cries out with glee to the rod of gold: &quot;O escrowed rod, scowdered crowds do crowd scowder-cord!&quot;&lt;/p&gt;&#xA;" OwnerUserId="8803" LastEditorUserId="8803" LastEditDate="2015-01-16T05:57:06.953" LastActivityDate="2015-01-16T05:57:06.953" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7755" PostTypeId="2" ParentId="6453" CreationDate="2015-01-09T15:01:00.137" Score="4" Body="&lt;p&gt;I spotted that many of the 8x8 grids contained in the bottles have very similar patterns. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I have summed up the grids, by saying white = 1 and black = 0, and I received the following interesting but definitely not random pattern in which the numbers 0 to 24 all occur once (except for 1 and 2; 2 occurs twice). &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This gives:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;pre&gt;21  1  1  1  1  1 19  2&lt;br&gt;&#xA; 1  1  1  1 10  1  1 12&lt;br&gt;&#xA; 1  1  1 14 15  1  2  1&lt;br&gt;&#xA; 1 17  1  7  1 11 23  1&lt;br&gt;&#xA; 5 13 20 24  1 22  4  1&lt;br&gt;&#xA; 1  1  1  9  1  1  1  6&lt;br&gt;&#xA; 1  8  1 16  1  1  1  1&lt;br&gt;&#xA; 1  0  1  1 18  3  1  1&lt;br&gt;&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8775" LastEditorUserId="5840" LastEditDate="2015-01-09T15:53:37.307" LastActivityDate="2015-01-09T15:53:37.307" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7756" PostTypeId="2" ParentId="6453" CreationDate="2015-01-09T16:22:30.720" Score="4" Body="&lt;p&gt;I'm thinking that the symbols 8x8 grid is a big clue, I'm not 100% sure, but I think, that every symbol stands for a character, I didn't get very far till now, but I got a few symbols substituted by what I think the meaning of them are. I also think, that if you have all, you can overlay the bottles 8x8 grids over it and get hints for INSTRUCTION 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For now i have these substitutions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;         t |   |   | V |   |   |   | F |&#xA;         g |   | y |   | pi| q |   |   |&#xA;         P |   | zu| D | j | S |   | d |&#xA;           | E |   | S | H |   | J |   |&#xA;         s |   |   | A |   | r |   |   |&#xA;           | Q | H |   | h |   | X | c |&#xA;           |   |   | z |   | p |   |   |&#xA;           | d |   |   | n |   | W |   |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Most of the substitutions I found in the font called Symbols in Microsoft Word. Some of the Japanese characters I knew myself.&lt;/p&gt;&#xA;" OwnerUserId="8805" LastEditorUserId="8805" LastEditDate="2015-01-09T16:48:03.797" LastActivityDate="2015-01-09T16:48:03.797" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7757" PostTypeId="2" ParentId="7741" CreationDate="2015-01-09T18:40:11.577" Score="2" Body="&lt;p&gt;I think it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;12072&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I brute forced it with a Java program that I wrote.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the program that I wrote:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import java.util.HashSet;&#xA;&#xA;public class Main {&#xA;&#xA;    public static void main(String[] args) {&#xA;        long i = 1;&#xA;        long even = 0;&#xA;        while (i &amp;lt; 1000000) {&#xA;            even += check(i) ? 1 : -1;&#xA;            if (even == 0) System.out.println(i);&#xA;            i++;&#xA;        }&#xA;    }&#xA;&#xA;    private static boolean check(long i) {&#xA;        HashSet&amp;lt;String&amp;gt; subs = new HashSet&amp;lt;String&amp;gt;();&#xA;        String string = String.valueOf(i);&#xA;        for (int beginIndex = 0; beginIndex &amp;lt; string.length(); beginIndex++) {&#xA;            for (int endIndex = beginIndex; endIndex &amp;lt;= string.length(); endIndex++) {&#xA;                if (!string.substring(beginIndex, endIndex).startsWith(&quot;0&quot;)) {&#xA;                    subs.add(string.substring(beginIndex, endIndex));&#xA;                }&#xA;            }&#xA;        }&#xA;        return subs.size() % 2 == 0;&#xA;    }&#xA;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I am sure there must be more efficient ways but this is how it works. For each number I check how many light switches there exists. I do this by checking how many distinct sub strings there are that do not begin with &lt;code&gt;0&lt;/code&gt;. I keep track of a variable &lt;code&gt;even&lt;/code&gt; that is raised by one on an even amount of these substrings and substracted by one on an odd amount. Every time this variable is &lt;code&gt;0&lt;/code&gt; there must be an equal amount of lights that are on and off. The &lt;code&gt;306&lt;/code&gt; mentioned in another answer is the first it finds so it gives me confidence that this method works.&lt;/p&gt;&#xA;" OwnerUserId="1748" LastEditorUserId="1748" LastEditDate="2015-01-09T19:02:09.980" LastActivityDate="2015-01-09T19:02:09.980" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7758" PostTypeId="2" ParentId="6691" CreationDate="2015-01-09T18:49:30.693" Score="27" Body="&lt;p&gt;Fourteen words:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;Presentational toenail replantations entail senatorial alterations rationale&amp;quot;, leant a relation, late, at alternations tea.&lt;/p&gt;&#xA;&lt;p&gt;a&lt;br&gt;&#xA;at&lt;br&gt;&#xA;tea&lt;br&gt;&#xA;late&lt;br&gt;&#xA;leant&lt;br&gt;&#xA;entail&lt;br&gt;&#xA;toenail&lt;br&gt;&#xA;relation&lt;br&gt;&#xA;rationale&lt;br&gt;&#xA;senatorial&lt;br&gt;&#xA;alterations&lt;br&gt;&#xA;alternations&lt;br&gt;&#xA;replantations&lt;br&gt;&#xA;presentational&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;A family member, at an afternoon event held to celebrate the swapovers (of a kind which we will not reveal), tardily swayed over and offered a rather silly phrase - a phrase alleging that operations to reverse cosmetic surgery to toenails would produce a credible reason for changing the holders of political office.&lt;/p&gt;&#xA;" OwnerUserId="7363" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-01-10T15:22:46.173" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7760" PostTypeId="2" ParentId="6531" CreationDate="2015-01-09T19:42:44.870" Score="2" Body="&lt;p&gt;5A&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DEFRIEND&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because SPURN A GUILTY PLEA = &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DEFEND&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and AQUIDNECK ISLAND =&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rhode Island i.e. RI&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4373" LastActivityDate="2015-01-09T19:42:44.870" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7761" PostTypeId="1" AcceptedAnswerId="7766" CreationDate="2015-01-09T22:55:43.890" Score="15" ViewCount="807" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;This puzzle started out as ordinary 8-directional word-search puzzle, hiding 19 diseases amongst lots and lots of letters. But then, six letters were infected and became ill, and things became &lt;em&gt;very strange&lt;/em&gt;....&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;In the words-search below, 6 letters have &lt;em&gt;odd properties&lt;/em&gt;. Whenever a word 'reaches' one, &lt;em&gt;something&lt;/em&gt; happens starting with the next letter. To be clear: There are &lt;em&gt;6 different&lt;/em&gt; affected letters with &lt;em&gt;6 different effects&lt;/em&gt;, and &lt;strong&gt;every time&lt;/strong&gt; a letter is used, the &lt;strong&gt;same, letter-specific effect&lt;/strong&gt; occurs.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt; &lt;br&gt;( This is &lt;em&gt;not&lt;/em&gt; what is going on in the puzzle! )&lt;br&gt;&lt;br&gt;&#xA;Say the letters &lt;strong&gt;a&lt;/strong&gt; &amp;amp; &lt;strong&gt;e&lt;/strong&gt; have become ill. Whenever &lt;strong&gt;a&lt;/strong&gt; is reached in a word, the search direction is shifted one step down, while every time an &lt;strong&gt;e&lt;/strong&gt; is reached, the search direction is shifted one step up. In this case, the word Stackexchange would be found as: &lt;br&gt; &lt;br&gt; &lt;img src=&quot;https://i.stack.imgur.com/XxXVm.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;br&gt;&lt;br&gt; or as:&lt;br&gt;&lt;br&gt; &lt;img src=&quot;https://i.stack.imgur.com/lIOwf.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;br&gt;&lt;br&gt;  etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;The word-search-puzzle&lt;/h1&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;There are 19 diseases hidden in the grid&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;6 letters are &lt;em&gt;sick&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the 6 &lt;em&gt;sick&lt;/em&gt; letters fill in the colour-boxes in the solution-grid. Arrange them properly, and they will spell the favourite medication prescribed by a specialist. (A &lt;em&gt;very&lt;/em&gt; special doctor.)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The 19 diseases fill in the white/grey spaces in the solution-grid. The grey spaces will spell out the favourite catch-phrase of the same specialist.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;h2&gt;To solve the whole puzzle you have to:&lt;/h2&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Find all words&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Find the 6 &lt;em&gt;sick&lt;/em&gt; letters&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Specify the 6 &lt;em&gt;'illnesses'&lt;/em&gt; (=rules) of the special letters&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Completely fill in the solution grid&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;h2&gt;No partial solutions on this puzzle please!&lt;/h2&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KnV78.png&quot; alt=&quot;Search Grid&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nu0wa.png&quot; alt=&quot;Solution Grid&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-01-10T11:46:26.637" Title="The ill-posed word-search: A sick puzzle" Tags="|word-search|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7762" PostTypeId="1" AcceptedAnswerId="7778" CreationDate="2015-01-09T23:01:31.023" Score="11" ViewCount="466" Body="&lt;p&gt;Here are two figures, each composed of 54 pixels.  Cut each figure into &lt;strong&gt;six &lt;a href=&quot;http://en.wikipedia.org/wiki/Nonomino&quot;&gt;nonominoes&lt;/a&gt;&lt;/strong&gt; having the same shape and size.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Both figures use the same nonomino&lt;/li&gt;&#xA;&lt;li&gt;The nonominoes could be rotated, reflected, or both&lt;/li&gt;&#xA;&lt;li&gt;The four colors (blue, green, pink, gray) are for artistic effect only.  These colors have no relationship to the solution.  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2oBX7.png&quot; alt=&quot;Pixel-heads&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5188" LastActivityDate="2015-01-11T10:37:30.843" Title="Dissect the pixel-heads" Tags="|geometry|dissection|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7763" PostTypeId="2" ParentId="7761" CreationDate="2015-01-10T02:32:59.413" Score="5" Body="&lt;p&gt;Okay so I got the &quot;hard part&quot; out of the way over an hour ago, and I was pretty thorough with my steps. I think possibly you used a crossword generator and two were left out? (I found all 5 rules, and there's no phil anywhere (I checked not just near the two capital S as well). I canceled out all of the capital letters pretty thoroughly so I'm relatively certain I found everything but I apologize in advance if I'm mistaken.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/jlssQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sick letters:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; c means turn left 90 degrees, h means turn right 45 degrees, e means skip a letter, s means skip two letters, and t means go up&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Phrase:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Death comes to all of us&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Word:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letters were found so the only word it could be (other than techs) would be chest, maybe referring to a medicine chest.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My picture:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I used red to draw the finds, green to cancel out capital letters, and blue to cancel out capital letters that were used in words (I did the rest sequentially without the marker)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastActivityDate="2015-01-10T02:32:59.413" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7766" PostTypeId="2" ParentId="7761" CreationDate="2015-01-10T04:05:01.390" Score="6" Body="&lt;p&gt;&lt;strong&gt;Edit 3&lt;/strong&gt; - Incorporated complete answer as requested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 19 diseases are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;shown in the figures below&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The 6 sick letters with their rules are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;c&lt;/strong&gt; - turn 90 degrees in either direction&lt;br&gt;&#xA; &lt;strong&gt;e&lt;/strong&gt; - skip one letter&lt;br&gt;&#xA; &lt;strong&gt;h&lt;/strong&gt; - turn 45 degrees to the right&lt;br&gt;&#xA; &lt;strong&gt;s&lt;/strong&gt; - skip two letters&lt;br&gt;&#xA; &lt;strong&gt;t&lt;/strong&gt; - go up&lt;br&gt;&#xA; &lt;strong&gt;y&lt;/strong&gt; - turn the next letter upside-down  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The very special doctor is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the &lt;strong&gt;Grim Reaper&lt;/strong&gt;.  His favorite medication was the &lt;strong&gt;SCYTHE&lt;/strong&gt; which is formed from the 6 sick letters.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The favorite catch-phrase is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Death comes to all of us&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The figures:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/p1QRy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-01-10T11:46:26.637" LastActivityDate="2015-01-10T11:46:26.637" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7769" PostTypeId="5" CreationDate="2015-01-10T06:50:59.997" Score="0" Body="&lt;p&gt;Also known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Verbal_arithmetic&quot; rel=&quot;nofollow&quot;&gt;verbal arithmetic&lt;/a&gt;, alphametics, cryptarithmetic, or crypt-arithmetic, these puzzles consist of an arithmetic equation (usually addition) where the digits have been replaced by letters, the goal being to determine the original numbers in the equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: DONALD+GERALD=ROBERT has the unique solution 526485+197485=723970.&lt;/p&gt;&#xA;" OwnerDisplayName="user251" LastEditorDisplayName="user251" LastEditDate="2015-01-10T06:54:05.713" LastActivityDate="2015-01-10T06:54:05.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7770" PostTypeId="4" CreationDate="2015-01-10T06:50:59.997" Score="0" Body="A puzzle consisting of a arithmetic equation where the digits have been replaced by numbers, with the goal of determining which digits correspond to which letters." OwnerDisplayName="user251" LastEditorDisplayName="user251" LastEditDate="2015-01-10T06:54:09.537" LastActivityDate="2015-01-10T06:54:09.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7771" PostTypeId="2" ParentId="2484" CreationDate="2015-01-10T08:46:54.460" Score="-1" Body="&lt;p&gt;You ask all 3 the question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are you the Joker?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Knight will answer &quot;No&quot;&lt;br&gt;&#xA;The Knave will answer &quot;Yes&quot;&lt;br&gt;&#xA;The Joker will answer &quot;Yes/No&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you got &quot;N Y Y&quot;: Whoever answered &quot;No&quot; is the Knight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you got &quot;N Y N&quot;: Whoever answered &quot;Yes&quot; is the Knave. You ask him: &quot;Is this person the Knight?&quot;. To whoever he answers &quot;No&quot;, that is the Knight.&lt;/p&gt;&#xA;" OwnerUserId="1691" LastActivityDate="2015-01-10T08:46:54.460" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7773" PostTypeId="1" AcceptedAnswerId="7774" CreationDate="2015-01-10T09:20:47.623" Score="2" ViewCount="828" Body="&lt;p&gt;There is a car and a motor-bike. The car and the bike both have an spare tire. Now, the car has a total of 5 tyres (one spare tire included) and the bike has a total of 3 tires (one spare tire included).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each tire of the bike and the car will last for &lt;strong&gt;100 KMs&lt;/strong&gt;. Take the following points into consideration:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The tire will not have any puncture or problem&lt;/li&gt;&#xA;&lt;li&gt;The tire will rotate only when the vehicle moves.&lt;/li&gt;&#xA;&lt;li&gt;Assume that petrol/diesel is infinite.&lt;/li&gt;&#xA;&lt;li&gt;You cannot buy new tire. &lt;/li&gt;&#xA;&lt;li&gt;Tires can be changed easily. &lt;/li&gt;&#xA;&lt;li&gt;Once a tire has been used for travelling 100 KM, it cannot be used again.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Can the car or the bike travel further?&lt;/p&gt;&#xA;" OwnerUserId="7482" LastEditorUserId="6178" LastEditDate="2015-01-10T10:56:45.667" LastActivityDate="2015-01-10T18:51:20.377" Title="Can the car or the bike travel further?" Tags="|calculation-puzzle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7774" PostTypeId="2" ParentId="7773" CreationDate="2015-01-10T09:54:33.730" Score="9" Body="&lt;p&gt;Is this a trick question? It look just to be a math problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The bike can travel further.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How to travel with bike:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The bike can travel &lt;code&gt;150 km = 100 km / 2 * 3&lt;/code&gt; by rotating between its wheels. For each 50 km, two of the three wheels are used, so each wheel it used for two of the intervals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How to travel with car:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The car can travel &lt;code&gt;125 km = 100 km / 4 * 5&lt;/code&gt; by using a different four of the five wheels for each of five 25 km segments.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alternative short argument:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alternatively, the spare tire is a +1/4 relative increase in tire-power for the car and a +1/2 for the bike, so the bike wins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-01-10T18:51:20.377" LastActivityDate="2015-01-10T18:51:20.377" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7775" PostTypeId="1" CreationDate="2015-01-10T10:15:02.630" Score="-9" ViewCount="7509" Body="&lt;p&gt;12 man participate in work Recruitment Service. Each participant saw every other participant and shook his hand. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many handshakes took place?&lt;/p&gt;&#xA;" OwnerUserId="8829" LastEditorUserId="2383" LastEditDate="2015-01-10T10:52:04.920" LastActivityDate="2015-01-10T11:15:53.830" Title="how many times 12 men may shake hands?" Tags="|mathematics|pattern|" AnswerCount="2" CommentCount="5" ClosedDate="2015-01-11T23:07:43.803" ContentLicense="CC BY-SA 3.0" />
  <row Id="7776" PostTypeId="2" ParentId="7775" CreationDate="2015-01-10T10:43:10.623" Score="1" Body="&lt;p&gt;This one is easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each person needs to shake hands with 11 other participants and the total number of participants are 12. So the total number handshakes that took place would be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;11x12=132 handshakes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But then there are some people who have shook their hands twice! So just divide the answer by 2 thus giving us&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;132/2=66 handshakes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and that is the answer. This is done because the first participant shakes their hand with 11 other participants,the second participant shakes their hand with 10 other participant(-1 because he has already shook hands with the first participant) and so on.. Thus the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;11+10+9+8+7+6+5+4+3+2+1+0=66 handshakes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &lt;code&gt;+0&lt;/code&gt; in the answer is because the last participant has already shook his hands with all the other participant and so he does not have to shake hands with anyone.&lt;/p&gt;&#xA;" OwnerUserId="7482" LastEditorUserId="7482" LastEditDate="2015-01-10T11:05:35.880" LastActivityDate="2015-01-10T11:05:35.880" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7777" PostTypeId="2" ParentId="7775" CreationDate="2015-01-10T10:49:15.590" Score="2" Body="&lt;p&gt;&lt;em&gt;This is just math...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1&amp;amp;2&lt;br&gt;&#xA; 1&amp;amp;3&lt;br&gt;&#xA; 1&amp;amp;4&lt;br&gt;&#xA; 1&amp;amp;5&lt;br&gt;&#xA; 1&amp;amp;6&lt;br&gt;&#xA; 1&amp;amp;7&lt;br&gt;&#xA; 1&amp;amp;8&lt;br&gt;&#xA; 1&amp;amp;9&lt;br&gt;&#xA; 1&amp;amp;10&lt;br&gt;&#xA; 1&amp;amp;11&lt;br&gt;&#xA; 1&amp;amp;12&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The numbers each represent a &lt;strong&gt;person&lt;/strong&gt;. When we move on to &lt;strong&gt;person 2&lt;/strong&gt;, he does &lt;strong&gt;not&lt;/strong&gt; shake 11 times. He'll shake 10, due to already shaking with 1 earlier. &lt;strong&gt;This will repeat&lt;/strong&gt;, which makes the equation super simple:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 11+10+9+8+7+6+5+4+3+2+1+0&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;0 because by the time it's 12's turn he's &lt;strong&gt;shaken&lt;/strong&gt; with &lt;strong&gt;everyone&lt;/strong&gt; already&lt;/em&gt;)&lt;br&gt;&#xA;Which comes out to be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;66&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;66 handshakes will go down between those 12 men. &lt;em&gt;78 if people shake their own hands (lol)&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2015-01-10T11:15:53.830" LastActivityDate="2015-01-10T11:15:53.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7778" PostTypeId="2" ParentId="7762" CreationDate="2015-01-10T11:10:16.180" Score="12" Body="&lt;p&gt;Found it!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/Lvk0K.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Very nice! I wasted 30min until I've realized I mis-drew my black &amp;amp; white copy with only 53 squares! :c)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; &#xA;As requested, and for the purpose of this site (&lt;em&gt;teaching&lt;/em&gt; how to build/solve puzzles) I'm adding how I found the solution to this puzzle. &lt;br&gt;&#xA;General considerations:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Mostly by &lt;strong&gt;trial &amp;amp; error&lt;/strong&gt;. I'm using a graphic program with layers/objects (i.e. Powerpoint) such that I can easily modify &amp;amp; clone tiles while trying fitting together the puzzle. &lt;br&gt; &lt;/li&gt;&#xA;  &lt;li&gt;The &lt;strong&gt;tile-size&lt;/strong&gt; of the nonominoes &lt;strong&gt;is known&lt;/strong&gt; if all are of same size, the total field size is known, and the number of nomoninoes is known.&lt;br&gt;&lt;/li&gt;&#xA;  &lt;li&gt;Then I start with one (in this case close to a hole) and copy/tilt it to see if they fit. Rule-out conditions are: &lt;br&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;two noominoes overlap &lt;/li&gt;&#xA;  &lt;li&gt;isolated 'holes' are created &lt;/li&gt;&#xA;  &lt;/ul&gt;&lt;/li&gt;&#xA;  &lt;li&gt;From there, it's a &lt;strong&gt;process of reduction&lt;/strong&gt;, guided by intuition. (i.e. usually good puzzles have rather 'complicated' nonomino-shapes. So I try those first.)&lt;/li&gt;&#xA;  &lt;li&gt;Sometimes it helps to do the process with a &lt;em&gt;partial&lt;/em&gt; nonomino (i.e. of smaller size) first. This rules out whole sets of shapes. (If you have a problem with a 5-sized shape, &lt;strong&gt;no&lt;/strong&gt; bigger shape will work out.)&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As for this &lt;em&gt;particular&lt;/em&gt; puzzle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - The &lt;strong&gt;tile-size&lt;/strong&gt; of the nonominoes &lt;strong&gt;is known&lt;/strong&gt; to be 9, because we have to fit 6 tiles into 54 fields, hence: 6 x &lt;strong&gt;9&lt;/strong&gt; = 54. &lt;br&gt;&#xA; In the particular puzzle, it was soon clear that nonominoes had to partly wrap &lt;em&gt;around&lt;/em&gt; a one-square hole. Few tries showed that it had to be a &quot;5-tiles around hole&quot; solution. &lt;br&gt;&#xA; - So I made this 5-sized nonomino and copied it 6 times, arranging it around the head-holes. &lt;br&gt;&#xA; - Then I successively added 1 square (i.e. made all 6-sized, 7-sized...). One could immediately see, what fits, and what wouldn't. As always: You must not create overlaps and you must not create isolated holes.&#xA; - I actually only solved the left head. It was then immediately visible, that the right one has just a mirror in it's left half.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="5840" LastEditDate="2015-01-11T10:37:30.843" LastActivityDate="2015-01-11T10:37:30.843" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7779" PostTypeId="1" CreationDate="2015-01-10T12:32:41.717" Score="-3" ViewCount="2348" Body="&lt;p&gt;There's a dead body on a bed and on the floor is a pair of scissors. The scissors were instrumental to his death. Yet there was no trace of blood. The body reveals no sign of any cuts or bruises. How could the person have been murdered with a pair of scissors?&lt;/p&gt;&#xA;" OwnerUserId="8832" LastEditorUserId="7308" LastEditDate="2015-04-29T02:18:22.617" LastActivityDate="2015-04-29T02:18:22.617" Title="A dead body, a pair of scissors, no bruises, no cut. How?" Tags="|lateral-thinking|situation|" AnswerCount="2" CommentCount="3" ClosedDate="2015-01-10T18:53:19.673" ContentLicense="CC BY-SA 3.0" />
  <row Id="7780" PostTypeId="2" ParentId="7779" CreationDate="2015-01-10T12:34:10.450" Score="6" Body="&lt;p&gt;The victim is hooked to an IV in a hospital bed and the murderer cut the tube.&lt;/p&gt;&#xA;" OwnerUserId="6178" LastActivityDate="2015-01-10T12:34:10.450" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7782" PostTypeId="2" ParentId="7779" CreationDate="2015-01-10T15:20:56.230" Score="4" Body="&lt;p&gt;&lt;strong&gt;Answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He was deathly afraid of scissors, and merely seeing them caused him to die of heart attack.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus Info:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The murderer didn't mean to kill him, he was simply cutting out a get well soon card when the sick man woke up.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2015-01-10T15:20:56.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7783" PostTypeId="1" AcceptedAnswerId="7867" CreationDate="2015-01-10T15:59:54.073" Score="7" ViewCount="3544" Body="&lt;p&gt;Divide the length of my name by one letter, then multiply by 11 to get four times my meaning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What country am I in?&lt;/p&gt;&#xA;" OwnerUserId="7363" LastEditorUserId="7363" LastEditDate="2015-04-14T13:32:47.433" LastActivityDate="2015-04-14T13:32:47.433" Title="What country am I in?" Tags="|riddle|" AnswerCount="7" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="7785" PostTypeId="2" ParentId="7783" CreationDate="2015-01-10T16:34:33.813" Score="2" Body="&lt;p&gt;&lt;strong&gt;Partial Answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If 4 times a meaning and is a multiple of 11, it means it could be 44, 88, ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously 4 the length divided by one number has to return a number. So let's run through some basic numbers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;4*11/4&lt;br&gt;&#xA;  8*11/4&lt;br&gt;&#xA;  12*11/4&lt;br&gt;&#xA;  16*11/4&lt;br&gt;&#xA;  20*11/4&lt;br&gt;&#xA;  24*11/4&lt;br&gt;&#xA;  28*11/4&lt;br&gt;&#xA;  32*11/4&lt;br&gt;&#xA;  36*11/4&lt;br&gt;&#xA;  40*11/4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the general idea that I'm currently working with, is that his name has a number in word form somewhere inside it, and &lt;em&gt;&quot;dividing by one letter&quot;&lt;/em&gt; means to split it so you can get that number. What's the chances of someone having twelve, sixteen, ..., forty in their name anyway? Even four or eight is unlikely, but that is what we will work with.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;4*11/4 = 11&lt;br&gt;&#xA;  8*11/4 = 22&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So what sort of meaning could 11 or 22 have? I'll think on this for a while, be back to you soon!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'm almost positive the name is in one of these lists:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.scrabblefinder.com/contains/four/&quot; rel=&quot;nofollow&quot;&gt;http://www.scrabblefinder.com/contains/four/&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.morewords.com/contains/eight/&quot; rel=&quot;nofollow&quot;&gt;http://www.morewords.com/contains/eight/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;More Coming Soon!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question for reference while editing:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide the length of my name by one letter, then multiply by 11 to get for times my meaning.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Old Theory&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Left for educational purposes for future readers&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;s&gt; However it says you have to divide by a letter first, you can't divide by a letter, so it has to be a number that represents a letter. Possibilities could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A=1&lt;br&gt;&#xA;  B=2&lt;br&gt;&#xA;  ...=...&lt;br&gt;&#xA;  Y = 25&lt;br&gt;&#xA;  Z = 26&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or ascii&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A=65&lt;br&gt;&#xA;  B=66&lt;br&gt;&#xA;  ...=...&lt;br&gt;&#xA;  Y=89&lt;br&gt;&#xA;  Z=90&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a=97&lt;br&gt;&#xA;  b=98&lt;br&gt;&#xA;  ...=...&lt;br&gt;&#xA;  y=121&#xA;  z= 122&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which means that it could 4 times it's meaning could range from 11 to 122*11.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is unclear to me how to narrow it down further without understand what &quot;it&quot; is and what the 4 times it's meaning refers to. It is possible it means that it's number is divisable by 4.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Following up on the idea that the letter times 11 can be divisable by 4&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This narrows down the list to&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;D=4&lt;br&gt;&#xA;  H=8&lt;br&gt;&#xA;  L=12&lt;br&gt;&#xA;  P=16&lt;br&gt;&#xA;  T=20&lt;br&gt;&#xA;  X=24&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or Ascii:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;D=68&lt;br&gt;&#xA;  H=72&lt;br&gt;&#xA;  L=76&lt;br&gt;&#xA;  P=80&lt;br&gt;&#xA;  T=84&lt;br&gt;&#xA;  X=88&lt;br&gt;&#xA;  d=104&lt;br&gt;&#xA;  h=108&lt;br&gt;&#xA;  l=112&lt;br&gt;&#xA;  p=116&lt;br&gt;&#xA;  t=120&lt;br&gt;&#xA;  x=124&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Anybody else starting to notice a pattern here. If not lemme show you. D, H, L, P, T, X.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step by step explanation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;name/letter*11/4=Meaning&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;So let's letter*11 has to be divisible by 4. Meaning letter HAS to be either D, H, L, P, T, or X (possibly lowercase)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we need the meaning now don't we! Well, let's go ahead and multiply and divide the list shall we?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Meanings:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ascii_T=231&lt;br&gt;&#xA;  Alph_T=55&lt;br&gt;&#xA;  Ascii_h=297&lt;br&gt;&#xA;  Alph_P=44&lt;br&gt;&#xA;  Alph_D=11&lt;br&gt;&#xA;  Ascii_p=319&lt;br&gt;&#xA;  Ascii_d=286&lt;br&gt;&#xA;  Alph_H=22&lt;br&gt;&#xA;  Ascii_L=209&lt;br&gt;&#xA;  Alph_X=66&lt;br&gt;&#xA;  Ascii_x=341&lt;br&gt;&#xA;  Ascii_H=198&lt;br&gt;&#xA;  Ascii_l=308&lt;br&gt;&#xA;  Ascii_P=220&lt;br&gt;&#xA;  Ascii_t=330&lt;br&gt;&#xA;  Ascii_X=242&lt;br&gt;&#xA;  Alph_L=33&lt;br&gt;&#xA;  Ascii_D=187&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What could the meaning be? May be a while until I get back with more. &lt;/s&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2015-01-10T22:17:46.733" LastActivityDate="2015-01-10T22:17:46.733" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="7786" PostTypeId="2" ParentId="7783" CreationDate="2015-01-10T20:20:03.257" Score="11" Body="&lt;p&gt;&lt;strong&gt;Edit 2: Found another one that fits using your latest hints.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are in:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The United States, more specifically you are the town of Eightyeight, Kentucky.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Divide the length of my name by one letter&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the town of Eightyeight, there is only one letter that doesn't have a pair (the y), divide the length into two so Eight  y   eight.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;then multiply by 11&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What it now looks like is eight times(x) eight (but the y is like the top left of an x so you need to divide the answer by two). Eight  y  eight is 64/2 which is 32. 32*11=352&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;to get four times my meaning.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 352 divide by 4 is 88 which is the original word&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is probably not what you were looking for but please give me props for my skill in finding very closely fitting wrong answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm hoping this/these answers will give me the gold icon you have :P)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I think I got it or I'm at least close? (or way off). You're in: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The country of hell (lol)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Divide the length of my name by one letter&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Divide the length of your name by one letter, your name is h34 so the name by the one letter is 34, divide it so h3 / 4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;then multiply by 11 to get four times my meaning&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Multiply h3 by 11 gives you h311 with a 4 on the end. This gives you 4 times h311 which is the country you are in.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="7722" LastEditDate="2015-01-11T02:50:39.280" LastActivityDate="2015-01-11T02:50:39.280" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="7787" PostTypeId="2" ParentId="7783" CreationDate="2015-01-10T23:13:48.870" Score="4" Body="&lt;p&gt;Flag of Iran have 2 rows with 11 signs &quot;Allahu Akbar&quot;, total 22. Are you in Iran?&lt;/p&gt;&#xA;" OwnerUserId="8837" LastActivityDate="2015-01-10T23:13:48.870" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7789" PostTypeId="2" ParentId="3623" CreationDate="2015-01-11T02:25:58.493" Score="3" Body="&lt;p&gt;The best move that player o can make each turn is to &quot;cap&quot; the currently longest line, i.e. placing an &quot;o&quot; at each endpoint of the current longest line. From there, the longest chain that player x can make is just the degrees of freedom in the lattice space, with degrees of freedom referring to the number of positive directions that a chain can be built in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of degrees of freedom for a n-dimensional lattice space when not counting diagonal chains is equal to n. In this case, the only &quot;positive directions&quot; are those aligned with the Cartesian axes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you add diagonals into the equation, then it becomes a problem of determining the number of unique, nonzero vectors that can exist in n-dimensional space and only have 0 or 1 for each component. This is related to binary number theory, and is just the number of nonzero integers that can be represented with n bits. As it turns out, this is equal to &lt;strong&gt;2&lt;sup&gt;n&lt;/sup&gt; - 1&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; This proof only applies when x only makes new moves connected to old moves. If x plays in multiple islands and then connects them, he could theoretically get a chain of four or longer. I'm of the belief that in this way, the maximum chain length for n dimensions would be &lt;strong&gt;2 (2&lt;sup&gt;n&lt;/sup&gt; - 1) - 1&lt;/strong&gt;, which accounts for connecting two independent chains at the middle, but I have no way of backing this up, really.&lt;/p&gt;&#xA;" OwnerUserId="8839" LastEditorUserId="8839" LastEditDate="2015-01-12T23:38:16.423" LastActivityDate="2015-01-12T23:38:16.423" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="7792" PostTypeId="2" ParentId="16" CreationDate="2015-01-11T03:43:36.293" Score="11" Body="&lt;p&gt;Chance for all prisoners surviving: 50%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just use time as an indication. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before the exercise, all prisoners agree that they will each learn the name of the person going in after them. The first prisoner will search the first 50 boxes for his name and for the name of the second person. He has a 50% chance of finding his own name. If he doesn't find it in the first 50 boxes, everyone dies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the first person also finds the name of the person coming after him in the first 50 boxes, he leaves immediately within a pre-agreed time (say, 5 minutes). If he doesn't, then the next person is in the last 50 boxes, and he waits at least 5 minutes before leaving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second person knows whether his name is in the first 50 boxes or the last 50 boxes, based off of whether he had to wait 5 minutes or not. He finds his name, and finds whether the third person is in the first 50 boxes or the last 50 boxes, and waits the 5 minutes if the third person is in the last 50 boxes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And just repeat... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;50% chance the first person finds his own name, and then 100% chance that each subsequent person finds his own name.&lt;/p&gt;&#xA;" OwnerUserId="8840" LastActivityDate="2015-01-11T03:43:36.293" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7796" PostTypeId="1" AcceptedAnswerId="7800" CreationDate="2015-01-11T07:11:27.750" Score="4" ViewCount="719" Body="&lt;p&gt;The &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Soma_cube&quot; rel=&quot;nofollow noreferrer&quot;&gt;Soma cube&lt;/a&gt;&lt;/strong&gt; is perhaps the most well-known polycube dissection.  Here are &lt;strong&gt;three more dissections to be assembled&lt;/strong&gt; into 3x3x3 cubes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best answer will provide:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the first correct solution for &lt;strong&gt;all three puzzles&lt;/strong&gt;.  Excluding rotations and reflections, puzzles 1 and 3 have only one solution.  Puzzle 2 has 13 possible solutions.  &lt;/li&gt;&#xA;&lt;li&gt;a format similar to the example below.  Colors or numbers or both can be used to indicate the location of each piece.  &lt;/li&gt;&#xA;&lt;li&gt;a few words about your methodology (ie. simple deduction, used some code or calculations, found some help on the web...)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note that all pieces for these 3 puzzles have a &lt;strong&gt;thickness of 1 unit&lt;/strong&gt;.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/4eUeT.png&quot; alt=&quot;3puzzles&quot;&gt;&lt;br&gt;&#xA;Here is an example format for the Soma cube solution that is shown on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Soma_cube&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia page&lt;/a&gt;.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/cYQKy.png&quot; alt=&quot;Example&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These 3 puzzles do not have copyright restrictions.  The puzzle names are in this spoiler:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  1) Slothouber-Graatsma puzzle&lt;br&gt;&#xA;  2) Diabolical cube&lt;br&gt;&#xA;  3) ZZZLLL cube  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-01-11T09:47:33.447" LastActivityDate="2015-01-11T09:47:33.447" Title="Three 3 x 3 x 3 Polycube Dissections" Tags="|geometry|dissection|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7797" PostTypeId="1" AcceptedAnswerId="7798" CreationDate="2015-01-11T07:25:37.973" Score="9" ViewCount="1621" Body="&lt;p&gt;This will be a very straight forward puzzle accessible to anybody with no special skills required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Mj6Nm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this grid of dots, draw 12 straight lines and connect all the dots. The lines must be drawn end to end (imagine making the lines with a pencil and not ever picking up the pencil). The lines' thickness can't help with the puzzle; for all intents and purposes the line thicknesses are infinitesimal. The lines must also go through the center of each point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rules:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) No putting the grid on a non-flat surface&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Post a picture of your solution (the lines can be thick but the thickness can't help with the solution)&lt;/p&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="7722" LastEditDate="2015-01-14T02:41:34.407" LastActivityDate="2015-01-22T02:01:46.900" Title="Connect the dots" Tags="|visual|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7798" PostTypeId="2" ParentId="7797" CreationDate="2015-01-11T07:32:39.167" Score="13" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/ziy8k.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA; &#xA; Just the standard 3x3 solution, but extended with a spiral.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-01-11T07:32:39.167" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7800" PostTypeId="2" ParentId="7796" CreationDate="2015-01-11T08:24:34.837" Score="7" Body="&lt;p&gt;Puzzle 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xux5J.png&quot; width=&quot;300&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Puzzle 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mldND.png&quot; width=&quot;300&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Puzzle 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/X90vy.png&quot; width=&quot;300&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All solved by deduction. In most cases you can promptly rule out a large portion of possible solution spaces by considering the most awkward shapes e.g. the light blue piece in the second puzzle can only have one of two positions. If we place it in those we limit where the orange piece may go, etc...&lt;/p&gt;&#xA;" OwnerUserId="3987" LastEditorUserId="3987" LastEditDate="2015-01-11T08:49:12.657" LastActivityDate="2015-01-11T08:49:12.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7803" PostTypeId="2" ParentId="6453" CreationDate="2015-01-11T13:17:14.697" Score="10" Body="&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;1. General&lt;/h1&gt;&#xA;&#xA;&lt;h2&gt;1.1. Instruction 1&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;What to do with &lt;code&gt;INSTRUCTION 1&lt;/code&gt; has already been discussed by &lt;em&gt;&quot;rand al'thor&quot;&lt;/em&gt; and &lt;em&gt;&quot;Tryth&quot;&lt;/em&gt; in general. Just for clarity, here again the basic idea with the terms I'll use in my answer:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;blue shading&lt;/strong&gt;: the blue shading of the bottles (and also the code in &lt;code&gt;INSTRUCTION 3&lt;/code&gt;)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;labels&lt;/strong&gt;: the 8x8 grids with holes on every bottle&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;yellow dots&lt;/strong&gt;: the yellow dots on every bottle&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;code-grid&lt;/strong&gt;: the 8x8 grid with the odd symbols&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Summarized I think the image tells us to empty the &lt;em&gt;bottles&lt;/em&gt; in the right &lt;strong&gt;order&lt;/strong&gt; through the right &lt;strong&gt;tube/color&lt;/strong&gt; into the &lt;em&gt;scale&lt;/em&gt; filling a certain &lt;strong&gt;amount of squares&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;2. Proposed Solutions (partial)&lt;/h1&gt;&#xA;&#xA;&lt;h3&gt;2.1. Order of Bottles (verified)&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This solution was figured out with the help of &quot;Gary Ye's&quot; essential observation!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first question to ask is whether the order of the bottles is actually important.  The answer is &lt;strong&gt;YES&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The blue shading of the bottles in &lt;code&gt;INSTRUCTION 1&lt;/code&gt; and the same shading of the letters in &lt;code&gt;INSTRUCTION 3&lt;/code&gt; clearly indicate a connection between a bottle and a letter of the password, i.e. every bottle corresponds to exactly one letter of the solution. And since the letter order in the solution is obviously essential, so  is the order of the bottles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although the blue shading indicates that the order is important &lt;strong&gt;it does NOT indicate that the darkest blue bottle is indeed the first one!&lt;/strong&gt; So the order must be hidden in some other attributes of the bottles.&#xA;Considering that the order must be unique and non ambiguous, the yellow dots can be ruled out and therefore only the labels remain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The important thing to observe here is that some pieces in the 8x8 &lt;em&gt;label-grid&lt;/em&gt; are &lt;strong&gt;missing&lt;/strong&gt; (i.e. not white or any other color, but &lt;em&gt;missing&lt;/em&gt;). This means if we put them over the 8x8 &lt;em&gt;code-grid&lt;/em&gt; some symbols will shine through. Even more important, if we put two or more labels on top of each other some symbols will still shine through. And this is exactly how the order is defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reconstructing the order of the bottles the symbols in the grid don't matter all that much. The procedure looks the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;ol&gt;&lt;li&gt;Choose a label that has exactly 24 (the number of bottles and the number in the order) empty squares and put it on the grid, &lt;em&gt;E4&lt;/em&gt;. There are now exactly 24 symbols visible. (There are multiple options for the first step, but only one allows successful continuation.)&lt;/li&gt;&lt;li&gt;Choose the next label so that when it is put over the first one there are still exactly 23 symbols visible, &lt;em&gt;A2&lt;/em&gt;.&lt;/li&gt;&lt;li&gt;Repeat the same process for the remaining labels so that 22, 21, 20, 19, ..., 1 symbols are visible.&lt;/li&gt;&lt;li&gt;This is the order of the bottles with the top one being the first. Note that the last visible symbol is &lt;em&gt;A&lt;/em&gt; (or &lt;em&gt;Alpha&lt;/em&gt;) whereas the 24&lt;sup&gt;th&lt;/sup&gt; symbol is &lt;em&gt;Omega&lt;/em&gt;. That might be an indication of the otherwise arbitrary seeming choice in step 1.&lt;/li&gt;&lt;/ol&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;eventually resulting in:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;pre&gt;&#xA; 8 7  21 10  3  6&lt;br&gt;&#xA; 2 13 17 22 20 11&lt;br&gt;&#xA; 4 14  9 16  5 18&lt;br&gt;&#xA;23 15 19 12  1 24&#xA;&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This completes the code-square and the labels (and probably also the blue shading).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;3. Further Ideas&lt;/h1&gt;&#xA;&#xA;&lt;h3&gt;3.1 Tubes&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Now that we know the order in which the bottles have to be emptied, the next thing we have to figure out, is through which tubes to empty the bottles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that the labels, the shading and the code-grid are already completed, likely the yellow dots contain this information. Especially considering that the first bottle has exactly 5 yellow squares and &lt;code&gt;INSTRUCTION 2&lt;/code&gt; shows that the first bottle must be emptied through the 5&lt;sup&gt;th&lt;/sup&gt; tube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering that figuring out the order was rather tricky, this might be a little bit to o simple though ... I would guess that the spaces between those dots are also really significant (as they were in the labels), but I don't have any concrete ideas about that at the moment.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;3.2 Amount of Squares&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Considering that there are almost no instructions left for the bottles, the amount of squares to fill by every bottle might even be constant, or actually maybe &lt;strong&gt;1&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only real other possibility I can see is that the yellow dots could indeed describe both the tube and the amount of squares.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="8804" LastEditorUserId="8804" LastEditDate="2015-01-13T12:56:44.307" LastActivityDate="2015-01-13T12:56:44.307" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7804" PostTypeId="2" ParentId="7730" CreationDate="2015-01-11T13:19:16.310" Score="1" Body="&lt;p&gt;To thy service!!!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; God. Jesus.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The one I obey, my dear commander&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The commandments. John 14:15 &quot;If you love me, keep my commands&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Keep on running I will not slander&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We run throughout our life and god wants to keep faith on him so that we can find our purpose of life and fulfill it. Also could refer to atheists who keep on running from him.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For I know you will follow my lead&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All Theist sure do follow him. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Call me in if you are in need&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We sure do in our sad times mostly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If one commands the other will follow&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;If you love me, keep my commands.&quot; - the Bible says.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's the deal, one has to swallow&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We do have to agree on it to save our souls.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Either way, I rely on you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Either as a theist or atheist I rely on you and your creation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We kiss goodnight and start anew&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;: Moon, Sun alls your creation. We kiss goodnight, we start afresh.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mighty Dictator of all places&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sure is. Though the mortals feel they rule , the god laughs and gets his biddings executed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;followed by all human races&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Almost all human races follow god, different religion, different god.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Watch over all the new born lives&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The lord guides all the birth animal, human, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;earn the fruit of fertile strifes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; fertile strifes could mean good deeds that earn great fruits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Endless love, you never disclaim&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Love for the god that you cannot disclaim&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;nor would you repel and never blame&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Inspite of what he has bestowed we do blame him for his shortcomings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unbribable Protector, punish me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Unbribable he is though we do try to bribe him for every little/big undertakings we want. Boon and bane is what he gives in equal measurement.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and I will follow tacitly&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We have to. No one questions the Almighty.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2656" LastActivityDate="2015-01-11T13:19:16.310" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7805" PostTypeId="2" ParentId="7797" CreationDate="2015-01-11T14:58:42.997" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/IO2zN.jpg&quot; alt=&quot;12 lines, 7x7&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've found there are many solutions that use the 3x3 trick of going out of bounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may even be one with 11 lines, considering I've got around many overlapping points on several solutions.&lt;/p&gt;&#xA;" OwnerUserId="2684" LastEditorUserId="2684" LastEditDate="2015-01-11T15:12:41.730" LastActivityDate="2015-01-11T15:12:41.730" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7807" PostTypeId="1" AcceptedAnswerId="7838" CreationDate="2015-01-11T18:30:31.417" Score="14" ViewCount="1342" Body="&lt;p&gt;One day, the old billionaire summons his relatives in his study:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;I'm very sick and may die soon, but the thought of all of you waiting for my death just to inherit the money makes me even sicker. None of you ever cared for me, or showed that he knows what I care about. So why should you enjoy my money?&lt;br /&gt;&#xA;I have therefore decided to give all my wealth except for a small portion away &lt;em&gt;next week&lt;/em&gt;, so that I can live my remaining days in peace with none of you near me.&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;There is quite some murmur from the relatives.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;Don't fret. I give all of you a chance and it might very well be that I give all my wealth to &lt;em&gt;one of you&lt;/em&gt;. Now listen.&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The room becomes very quiet.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;All of you are going to give me a gift tomorrow. I will evaluate all of them for a day, and then decided if what you gave me is of any value. If a gift can really, really please me, I will give the one who gave it to me all my money. Now go and return tomorrow.&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;All the relatives hurry away. The next day, they queue to give the billionaire all sorts of presents. Gems, famous paintings, even some sport car and a voucher for a world-trip by balloon... The billionaire takes them without comments or gratitude and hushes the relatives away, telling them to return the next day.&lt;/p&gt;&#xA;&lt;p&gt;The next day, all relatives are summoned into the great hall, and the billionaire speaks out:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;As I had expected, your presents were all junk and worthless to me. Why the hell should I want to freeze to death miles high in the sky or crashing on the motorway? And were should I place those tasteless paintings without having them in my face all the time? Na! I was ready to donate all my wealth to the Puzzling.SE online community, the only people sharing my interests, hadn't I received this...&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;And with this he shows a little necklace made from cheapish plastic pearls on a Nylon string.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;Whoever gave me this, will be very rich by next week...&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;Can you explain, why this present did so appeal to the billionaire?&lt;/h1&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VbnrK.jpg&quot; alt=&quot;Necklace&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;The make it a bit easier: The solution requires looking at the necklace in clockwise direction. And, &lt;em&gt;of course&lt;/em&gt; it is an encrypted (English) message with a twist. The twist needs to be named in the solution.&lt;/p&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-01-14T02:38:06.217" Title="The billionaire's necklace" Tags="|cipher|visual|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="7809" PostTypeId="2" ParentId="7741" CreationDate="2015-01-11T19:38:42.663" Score="3" Body="&lt;p&gt;There is a solution between $10^{11}$ and $10^{12}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=10^{11}$, the total number of lights off is 50563019008. Just over half.&lt;br&gt;&#xA;For $n=10^{12}$, the total number of lights off is 499476031551. Just under half.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, somewhere between $10^{11}$ and $10^{12}$ there is an $n$ where exactly half of the lights are off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this using a computer program. To cut the number of cases, I used symmetry. Permuting the digits 1-9 does not change whether a room is on or off, for example &quot;377377009&quot; is the same lightness as &quot;899899001&quot;. This reduces the number of essentially different 12-digit numbers to only ~27 million.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I doubt this puzzle has a nice answer. I would be very impressed/surprised if GOTO 0 has a solution.&lt;/p&gt;&#xA;" OwnerUserId="1752" LastEditorUserId="1752" LastEditDate="2015-01-11T19:47:25.410" LastActivityDate="2015-01-11T19:47:25.410" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7811" PostTypeId="2" ParentId="7730" CreationDate="2015-01-11T20:24:47.987" Score="4" Body="&lt;p&gt;After reading this I am under the impression that this has to be the answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The one I obey, my dear commander &#xA;Keep on running I will not slander &#xA;For I know you will follow my lead &#xA;Call me in if you are in need&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;We all know that we humans can command time, which simply means to use the time wisely (or simply changing the time on a watch). As we carry on with our daily activities, time follows us regardless whether we use it wisely or not, which then explains why the time will not 'slander' as we 'follow (its) lead'(and we will not slander when time is &lt;em&gt;running&lt;/em&gt;). And of course, when we are in need of time, we can call him in, which just means that we would grasp time more efficiently and thus, have 'more time'(plus, using the watch again: it will tell/&lt;em&gt;call us in&lt;/em&gt; us when batteries are empty).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If one commands the other will follow &#xA;That's the deal, one has to swallow &#xA;Either way, I rely on you &#xA;We kiss goodnight and start anew&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;We both can command each other. When we command time, we are using it wisely whereas when time commands us, we will follow it as well. That's the case as one has to swallow, and swallow itself does not mean the physical act of swallowing, it could also mean 'put up with or meekly accept something', which in this case, the one which is commanding. When the clock hits twelve, a new day starts, hence 'we kiss goodnight and start anew'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mighty Dictator of all places &#xA;followed by all human races &#xA;Watch over all the new born lives &#xA;earn the fruit of fertile strifes &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Time is indeed the mighty dictator of all places, and we all humans follow it. Time watches over new born lives as when time passes, the newborns gradually age to become an adult, which then gives the metaphorical meaning that time is 'watching over' them. In the case of the fruit of fertile strife, it would mean that the humans themselves are displeased and hurt because of an argument, and to have the fruit of the fertile strife would be the solution to the argument, something that can dissipate the hurt between both of them, and as we have all heard before, 'Time heals all wounds.' &lt;br&gt;&lt;strong&gt;Double meaning of races!&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Endless love, you never disclaim &#xA;nor would you repel and never blame &#xA;Unbribable Protector, punish me &#xA;and I will follow tacitly&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;We all know that time is unbiased towards anyone, and that it will still pass regardless of situation, thus explaining why it would never disclaim, repel and blame. We all know that time can never be bought no matter how much money you have, hence giving it the title, Unbribable Protector. When time 'punishes us', it means that we have not used it wisely and we realise the fact that we have wasted precious time, something that can not be bought by money and we learn the importance of time. After being punished, we would learn to appreciate and use time efficiently, and hence 'follow tacitly'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Title Interpretation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The sands could mean hourglasses itself as hourglasses use sand to embody the meaning of time. And when turned upside down and there is no more time left, hence 'The sands of life are running out'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8852" LastEditorUserId="5106" LastEditDate="2015-01-12T00:23:38.710" LastActivityDate="2015-01-12T00:23:38.710" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7818" PostTypeId="2" ParentId="7783" CreationDate="2015-01-12T03:39:48.680" Score="1" Body="&lt;p&gt;Ok so I don't know the answer but my line of thinking seems different than other posters, so I figure I should post it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first thing that struck me was that the riddle says to divide by a letter. This is only possible in two ways as far as I know:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1) The letter is a variable, say &lt;em&gt;x&lt;/em&gt;.&#xA; 2) The &quot;letter&quot; is actually a Roman numeral.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As the first option seems ridiculous in this context, I'm assuming the second. Given that it has been said that &quot;divide by &lt;em&gt;one&lt;/em&gt; letter&quot; is somehow different from &quot;divide by &lt;em&gt;a&lt;/em&gt; letter&quot;, that leads to one of two more conclusions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1) This is literal. You are meant to divide by the letter for &lt;em&gt;one&lt;/em&gt;, or I.&#xA; 2) This means &quot;divide by &lt;em&gt;one of my letters&lt;/em&gt;&quot;, meaning the country name contains a Roman numeral (I, V, X, L, D, C, M).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not sure which of those is correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This line of reasoning led me to think that it's possible that the number 11 is actually referring to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the Roman numeral II, or 2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That could be wrong though. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last bit of the riddle is what really confuses me though. &quot;My meaning&quot; implies &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the meaning of the country itself. I looked into the etymologies of country names and found a few that reference numbers such as Bahrain which means &quot;two seas&quot; (which is an interesting one because 2 C's or CC actually means 200 in Roman numerals), but I couldn't make any of them fit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So like I said, I can't at all figure out a solution to the riddle, but I thought a fresh perspective may spark something for someone else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: It's also possible that &quot;multiply by 11&quot; means&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Take the previous number and attach XII (x11) to get &quot;four times my meaning&quot;. So if the first half of the riddle yields MMC, for example, then that &quot;times 11&quot; would be MMCXII.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8858" LastEditorUserId="8858" LastEditDate="2015-01-12T03:45:56.760" LastActivityDate="2015-01-12T03:45:56.760" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7819" PostTypeId="1" AcceptedAnswerId="7853" CreationDate="2015-01-12T09:03:41.780" Score="24" ViewCount="6775" Body="&lt;p&gt;I have heard of the puzzle with 100 or 10 dwarves wearing a hat of a color, either red or blue, and standing in a straight line, and having to guess the color of their own hat - it's quite easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution to save all dwarves or all dwarves minus the first one is fairly simple and relies on the fact that the number of dwarves is finite. &#xA;So what would happen if you had an &lt;strong&gt;infinite number of dwarves&lt;/strong&gt; standing in a straight line? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every dwarf wears a hat of color either red or blue and sees the color of the hats of all the dwarves standing in front of him. There is explicitly a first dwarf, who has to start guessing the color of his hat and then the guessing proceeds with the next one in the line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a dwarf guessed correctly, it is freed; if he guessed wrong, it is fried. Every dwarf can hear the voice of all other dwarves without a problem. Everybody is only allowed to speak out either the color red or blue, but no further information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a possibility for all dwarves to be freed? Well, probably no. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But is it at least possible that only finitely many of them are killed? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: This is a mathematical puzzle. No loopholes.&lt;/p&gt;&#xA;" OwnerUserId="8863" LastEditorUserId="10615" LastEditDate="2015-06-04T00:52:37.757" LastActivityDate="2023-12-19T23:12:49.683" Title="Infinitely many dwarves wearing hats of 2 colours" Tags="|mathematics|hat-guessing|" AnswerCount="5" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="7820" PostTypeId="1" AcceptedAnswerId="7823" CreationDate="2015-01-12T09:46:12.023" Score="2" ViewCount="918" Body="&lt;p&gt;I was working my way through some Puzzles in &lt;a href=&quot;http://www.flipkart.com/discrete-mathematics-its-applications-combinatorics-graph-theory-english-7th/p/itmdwxxszhczjrnf&quot; rel=&quot;nofollow&quot;&gt;Discrete Maths by Rosen&lt;/a&gt;, when I came across the following question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;A Pokemon Hunter is rowing upstream a brook&lt;br/&gt;&lt;/li&gt;&#xA;  &lt;li&gt;As he passes under the 'bridge-of-curse', he throws a Pikachu into the brook&lt;br/&gt;&lt;/li&gt;&#xA;  &lt;li&gt;5 minutes later, he realises that Pikachu will die and he should not have done that &lt;br/&gt;&#xA;  He rows back and picks Pikachu 3km. downstream of the 'bridge-of-curse'&lt;br/&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;&lt;em&gt;What is the speed of the river flow?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;My Answer:&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - Distance : 3km&lt;br/&gt;&#xA; - Time : 5 minutes + 5 minutes ( due to the stream's frame of reference ) =  10 minutes&lt;br/&gt;&#xA; - Speed = 3000 / 600  m/s  = 5 m/s&lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Doubt:&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I correct ? Seems a bit too easy ...&lt;/p&gt;&#xA;" OwnerUserId="7435" LastEditorUserId="11110" LastEditDate="2018-11-27T19:35:15.187" LastActivityDate="2018-11-27T19:35:15.187" Title="Doubt : Pokemon Hunter and the Rogue Brook" Tags="|logical-deduction|" AnswerCount="4" CommentCount="6" ClosedDate="2015-01-15T21:25:00.693" ContentLicense="CC BY-SA 3.0" />
  <row Id="7821" PostTypeId="2" ParentId="7819" CreationDate="2015-01-12T10:09:06.550" Score="2" Body="&lt;p&gt;Refer to the following &lt;a href=&quot;http://www.glassdoor.com/Interview/A-dwarf-killing-giant-lines-up-10-dwarfs-from-shortest-to-tallest-Each-dwarf-can-see-all-the-shortest-dwarfs-in-front-of-h-QTN_491010.htm&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt;. The one answer which is simple to understand is the High pitch and low pitch. Each dwarfs answers his hat color in either high pitch or low pitch depending on the next dwarfs hat whether its in Red or Blue color.&lt;/p&gt;&#xA;" OwnerUserId="8864" LastActivityDate="2015-01-12T10:09:06.550" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="7822" PostTypeId="2" ParentId="7819" CreationDate="2015-01-12T10:14:48.990" Score="7" Body="&lt;p&gt;The solution is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We define the notion of a class of infinite sequence of hats. All sequences in this class differ from each other by only a &lt;strong&gt;finite&lt;/strong&gt; number of elements. There are an &lt;strong&gt;infinite&lt;/strong&gt; number of classes, each containing a &lt;strong&gt;infinite&lt;/strong&gt; number of elements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why? As differences between classes are of &lt;strong&gt;infinite length&lt;/strong&gt;, there is an &lt;strong&gt;infinite number&lt;/strong&gt; of possible distances, which means there is an &lt;strong&gt;infinite number&lt;/strong&gt; of classes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each such class we select a &quot;representative&quot; which is a single element of the class (a single infinite sequence of hats). &lt;s&gt;So suppose we have the class &quot;infinitely alternating red and blue, with a finite number of non-alternating hats in the first 20 elements&quot;. We can set the representative to be:&lt;/s&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Apparently, after digging a bit, a good way to define the classes is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Two sequences are equivalent if they are identical after a finite number of items&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Choice of a representative can be easily defined as choosing the sequence where the differences appear at the beginning of the sequence and are lexicographically smaller.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the sequence&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;rbrbrbrbrbrb...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And, the sequence:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;r[r]rb[br]rbrbrb... ([] denotes mismatches)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;is a member of that class. Each dwarf can see the infinite sequence of hats in front of it and can recognize the class. The dwarves can see the mismatches in front of them, but do not know if their own hat is not a mismatch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, the difference is finite, and a modification of the standard rules for guessing a finite sequence can be applied. The first dwarf says &quot;red&quot; if there is an odd number of hats that are different in the &quot;sequence&quot;, compared to the &quot;representative&quot; (or &quot;blue&quot; if the mismatches are even). From there on, each dwarf has sufficient info to tell the colour of their hat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that, the number of classes is &lt;strong&gt;infinite&lt;/strong&gt;, however, since we are talking about infinite dwarves, I assume this is acceptable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, since the first dwarf &quot;guesses&quot;, the first dwarf has a 50% chance of survival, all other dwarves survive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh, and because of the infinite number of classes, this can be directly applied to the problem with an &lt;strong&gt;infinite number of colours&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary for the naysayers&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Two sequences are equivalent if they are identical after a finite number of items.&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Choice of a representative can be easily defined as choosing the sequence where the differences appear at the beginning of the sequence and are lexicographically smaller.&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;As the sequence is infinite, and differences are finite, there exists an infinite subsequence, that can uniquely define the class of the sequence.&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="5610" LastEditDate="2015-01-12T15:59:02.607" LastActivityDate="2015-01-12T15:59:02.607" CommentCount="22" ContentLicense="CC BY-SA 3.0" />
  <row Id="7823" PostTypeId="2" ParentId="7820" CreationDate="2015-01-12T10:47:46.627" Score="6" Body="&lt;p&gt;You assume it takes him 5 minutes back as well. Here is a sketched solution:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;s = speed of stream&#xA;r = speed of rowing in water with no current&#xA;t = time since drop to pickup&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;s * t = 3 // Pikachu travelled 3 kilometers for that time&#xA;t = 3 / s&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We also have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(t - 5) * (r + s) = 5 * (r - s) + 3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which means &lt;code&gt;the distance travelled returning is the distance going away + 3 kilometers&lt;/code&gt;. From here on we have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(3 / s - 5) * (r + s) = 5 * (r - s) + 3&#xA;3r/s - 5r + 3 - 5s = 5r - 5s + 3&#xA;3r/s - 5r = 5r&#xA;10r = 3r/s&#xA;10rs = 3r&#xA;10s = 3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;s&gt;s = 10/3 = 3.3333 kilometres per minute&lt;/s&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;s = 3/10 = 0.3 kilometres per minute&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This seems quite fast to me, if you see a typo in my calculations, feel free to edit or comment. Note $t =10$ which matches your assumption.&lt;/p&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="5610" LastEditDate="2015-01-17T08:47:23.050" LastActivityDate="2015-01-17T08:47:23.050" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7824" PostTypeId="2" ParentId="7820" CreationDate="2015-01-12T13:02:41.523" Score="1" Body="&lt;p&gt;You cannot directly assume that the total time taken is 10 min.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume,&lt;br&gt;&#xA;speed of boat = &lt;code&gt;b&lt;/code&gt;&#xA;&lt;br&gt;&#xA;speed of river = &lt;code&gt;r&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the total time taken be &lt;code&gt;t&lt;/code&gt;&#xA;&lt;br&gt;&#xA;Since pikachu has travelled &lt;code&gt;3km&lt;/code&gt; in &lt;code&gt;t&lt;/code&gt; time,&#xA;&lt;br&gt;speed of river&#xA;&lt;br&gt;&lt;code&gt;r = 3/t&lt;/code&gt;&#xA;&lt;br&gt;&lt;code&gt;rt = 3&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose the boat has travelled a distance of &lt;code&gt;x&lt;/code&gt; upstream from the 'bridge-of-curse' in 5min.&#xA;&lt;br&gt;&lt;code&gt;x = (b - r) * 5&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, while coming downstream, the total distance covered is &lt;code&gt;(x + 3)&lt;/code&gt;,&lt;br&gt; while the speed will be &lt;code&gt;(b + r)&lt;/code&gt; and time taken will be &lt;code&gt;(t - 5)&lt;/code&gt;&#xA;&lt;br&gt;So,&#xA;&lt;br&gt;&lt;code&gt;x + 3 = (b + r) * (t - 5)&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;(b - r) * 5 + 3 = (b + r) * (t - 5)&lt;/code&gt;&#xA;&lt;br&gt;&lt;code&gt;5b - 5r + 3 = bt + rt - 5b -5r&lt;/code&gt;&#xA;&lt;br&gt;&lt;code&gt;10b + 3 = bt + rt&lt;/code&gt;&#xA;&lt;br&gt; We already have &lt;code&gt;rt = 3&lt;/code&gt;&#xA;&lt;br&gt;&lt;code&gt;10b + 3 = bt + 3&lt;/code&gt;&#xA;&lt;br&gt;&lt;code&gt;10b = bt&lt;/code&gt;&#xA;&lt;br&gt;&lt;code&gt;t = 10 min&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speed of river, &lt;code&gt;r = 3/t = 3/10 = 0.3 km/min&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="5565" LastActivityDate="2015-01-12T13:02:41.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7825" PostTypeId="2" ParentId="2066" CreationDate="2015-01-12T13:05:47.593" Score="3" Body="&lt;p&gt;Hope this will also be considered as an answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/3LBLa.jpg&quot; alt=&quot;solution&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8758" LastEditorUserId="46903" LastEditDate="2018-09-07T02:23:02.527" LastActivityDate="2018-09-07T02:23:02.527" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="7826" PostTypeId="2" ParentId="7820" CreationDate="2015-01-12T13:20:44.743" Score="3" Body="&lt;p&gt;Your answer is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However you might need to justify why 10 minutes is the correct time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be done as follows;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The boat will travel at a speed $v_b=v_m+v_c$, ie. the total velocity is the sum of velocity from muscle plus velocity from current.  $v_c$ is constant, while $v_m$ changes direction when the hunter turns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pokemon travels at speed $v_p=v_c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Distance is speed times time.&#xA;$d_p=v_c*t=3km=d_b=(v_m+v_c)*t_{up}+(-v_m+v_c)*t_{down}=v_m*t_{up}-v_m*t_{down}+v_c*t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $v_m*t_{up}-v_m*t_{down}=0$ and $t_{up}=t_{down}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or stated in a simpler manner; the displacement due to current is the same for boat and pokemon, so the timing will be the same as if there was no current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This assumes that the current is constant and that the only drag is from current, without those assumption the question lacks sufficient information to be solved.&lt;/p&gt;&#xA;" OwnerUserId="1654" LastActivityDate="2015-01-12T13:20:44.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7827" PostTypeId="2" ParentId="7820" CreationDate="2015-01-12T13:25:50.257" Score="11" Body="&lt;p&gt;The original answer is correct but not explained. Viewed from the river's frame of reference, the Pikachu stays still; so the Hunter has simply rowed away for some period of time, and then rowed back in 5 minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hunter is rowing at a constant speed in the river's frame of reference, but in the opposite direction in each time period. The Pikachu has stayed still, so the distance out and back must be the same, and the times must also.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's 10 minutes total, and in that time the river has moved 3km in the bridge's frame of reference. So the river's speed relative to the bridge is 3km / 10 minute.&lt;/p&gt;&#xA;" OwnerUserId="6031" LastActivityDate="2015-01-12T13:25:50.257" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7828" PostTypeId="2" ParentId="6691" CreationDate="2015-01-12T14:34:20.353" Score="3" Body="&lt;p&gt;&lt;em&gt;I couldn't resist one-upping...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fifteen words:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;15 classifications&#xA;14 classification&#xA;13 salifications&#xA;12 laicisations&#xA;11 laicisation          &#xA;10 antisocial&#xA;9  scalation            &#xA;8  santalic         &#xA;7  satanic&#xA;6  antics&#xA;5  saint&#xA;4  anti&#xA;3  tan&#xA;2  an&#xA;1  a&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sentence:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Anti saint laicisations classification (laicisation antics classifications) — satanic salifications, a santalic scalation (an antisocial tan).&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Translation:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;What follows is a description of anti-Christian attempts at secularisation (i.e. descriptions of the foolish behaviour involved in secularisation) — evil applications of salt, applying a protective skin the colour of sandalwood (all of which refers to an offensive fake tan).&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4534" LastActivityDate="2015-01-12T14:34:20.353" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7829" PostTypeId="2" ParentId="7807" CreationDate="2015-01-12T14:58:16.887" Score="5" Body="&lt;p&gt;This is not a complete answer, but it might help to solve it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Black gems are baconian cipher.&lt;br/&gt;&#xA;Pearls:0 Diamonds:1 &lt;br/&gt;&#xA;Encrypted: 0001001110010110111010001&lt;br/&gt;&#xA;Decrypted: &lt;strong&gt;COLOR&lt;/strong&gt;.&lt;br /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Black and white gems are baconian cipher.&lt;br/&gt;&#xA;Black:0  White:1 &lt;br/&gt;&#xA;Encrypted: 0011010001000001100001101001001011110011(00)&lt;br/&gt;&#xA;Decrypted: &lt;strong&gt;GRAYNEXT&lt;/strong&gt;.&lt;br /&gt;&#xA;Info: Last two gems(00) are out of the context. Not sure how they are related.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4506" LastActivityDate="2015-01-12T14:58:16.887" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7830" PostTypeId="2" ParentId="6618" CreationDate="2015-01-12T16:12:07.210" Score="2" Body="&lt;p&gt;Ok so this is a huge long shot given the other answers, but I can't think of anything else that fits, so here goes...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;8 Little-known Beatles song (without straw) (4,3,7) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm basing this on two Beatles songs, &quot;I Will (Parenthetical)&quot; and &quot;Strawberry Fields Forever&quot; (No straw). Right, that's it. I have uh...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Love_You_Forever&quot; rel=&quot;nofollow&quot;&gt;Love You Forever&lt;/a&gt; by Robert Munsch (A children's book?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: So yeah, &quot;little-known&quot; = obscure, and &quot;without straw&quot; = no hay/hey. Beatles song is &quot;Hey Jude&quot;... Or&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Jude_the_Obscure&quot; rel=&quot;nofollow&quot;&gt;Jude The Obscure&lt;/a&gt; by Thomas Hardy. Extra credit points for the other one though. :D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;-&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;5 Initially rebellious sheep undertake a silent infantry operation, inside a small enclosure, or so I'm induced to believe (10)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't know the answer for this but &quot;Initially rebellious sheep&quot; makes me think of Jews (&quot;like sheep to the slaughter&quot;), and &quot;inside a small enclosure&quot; could easily refer to a concentration camp. &quot;So I'm induced to believe&quot; implies some kind of dream or hypnosis. Or just a really convincing liar.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;3 Think of the Phoenician, a possible drench, bleached prisons, or beach spindles (8,7)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also don't know this one, but &quot;Think of the Phoenician&quot; probably refers to a phoenix (or the alphabet, I suppose), and &quot;bleached prisons&quot; is almost certainly &quot;white cells&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[If my thoughts-but-not-answers are inappropriate, let me know and I will remove them.&lt;/p&gt;&#xA;" OwnerUserId="8858" LastEditorUserId="8858" LastEditDate="2015-01-12T16:38:40.860" LastActivityDate="2015-01-12T16:38:40.860" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7831" PostTypeId="1" AcceptedAnswerId="7832" CreationDate="2015-01-12T17:36:17.013" Score="34" ViewCount="11391" Body="&lt;p&gt;Recently a &lt;a href=&quot;https://puzzling.stackexchange.com/questions/7797/connect-the-dots&quot;&gt;question&lt;/a&gt; was posted with this picture of a 7x7 grid of dots, asking for a possible configuration with 12 lines where you can draw them without lifting a pencil. But is it possible with 11 lines? If so, what about 10?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/I9UDp.jpg&quot; alt=&quot;7x7 square of dots&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Just to be clear, you have to draw 11 lines that meet all dots. Or fewer if possible.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Or if it is not possible, explain why not.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solutions to 12 lines:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/FKkQT.jpg&quot; alt=&quot;1st solution (simpler)&quot;&gt;. &#xA; &lt;img src=&quot;https://i.stack.imgur.com/adxfL.jpg&quot; alt=&quot;2nd solution (complicated)&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Just to be clear, this grid is meant to be placed on a flat surface (such as a table.) Also as implied by the questions, line thickness is smaller than dot radius, but it must pass through the centre if the fit!&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2016-04-04T07:05:10.560" Title="What is the minimum number of straight lines to connect all the dots on this grid?" Tags="|geometry|" AnswerCount="3" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="7832" PostTypeId="2" ParentId="7831" CreationDate="2015-01-12T18:49:24.843" Score="49" Body="&lt;p&gt;Consider an arbitrary solution: it will use $H$ horizontal lines, $V$ vertical lines, and $S$ skew lines (that are neither vertical nor horizontal).&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If $H=7$, then we need at least $6$ lines to connect these horizontal lines to each other. Hence $H+V+S\ge13$ in this case.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $V=7$, we argue symmetrically to case 1.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $H=6$, there exist $7$ points (on another horizontal line) that are NOT covered by the horizontal lines in the solution. Each of these $7$ points must be covered by a separate line; this yields $V+S\ge7$ and $H+V+S\ge13$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $V=6$, we argue symmetrically to case 3.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It remains to consider the case with $H\le5$ and $V\le5$. Consider the remaining $(7-H)(7-V)$ points that are not covered by any horizontal or vertical line.  They form a grid, and the boundary of this grid contains $24-2(H+V)$ points.  As each of the skew lines covers at most $2$ of these boundary points, we conclude $S\ge 12-(H+V)$. This is equivalent to $H+V+S\ge12$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Summarizing, in each of the five cases we needed $H+V+S\ge12$ lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The argument generalizes to $n\times n$ grids and yields a lower bound of $2n-1$ for $n\le2$ and a lower bound of $2n-2$ for $n\ge3$. These lower bounds are actually best possible, as one easily constructs solutions with at most $2n-2$ lines for $n\ge3$ (just keep extending the spiral in the above answer for $n=7$; whenever $n$ increases by $1$, we attach $2$ further segments to the spiral).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the complete answer to the problem is as follows.&#xA;If $n=1$ or $n=2$, the best solution uses $2n-1$ lines.&#xA;If $n\ge3$, the best solution uses $2n-2$ lines.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-03-21T16:08:02.870" LastActivityDate="2015-03-21T16:08:02.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7834" PostTypeId="1" CreationDate="2015-01-12T18:48:05.597" Score="4" ViewCount="685" Body="&lt;p&gt;I'm here to ask you if it is possible to find a way to solve this problem.&#xA;I'm designing the puzzle-enigma section of a video game and try to find possible solutions and mechanic for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine each circle in the pic below is a platform and the rest of empty squares is void. Only the first platform is active and &quot;usable&quot;(say you can step on it). You have six characters that have to go from one side to the other stepping only on the platforms and each platform admits one to six characters a time. Stepping on a platform instantaneously make appear the following one; however getting off it instantaneously deactivate and make disappear the following one (making eventually die a character that would be on it at the moment of the deactivation). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible for the six characters to get to the other side all alive?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only solution I found for the moment was to change the rules of the puzzle, making each platform activate the following AND the previous one; like this, you would have the six characters one on each of the first six platforms and the bottom one going time by time to the top one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another idea I thought about was: each character on each platform activate one link, so two characters on one platform activate the two following ones,three characters on one platform activate the three following ones and so over, but I couldn't find a solution like this either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your time guys,&#xA;cheers!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5lYjr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="jacopo" LastEditorUserId="5840" LastEditDate="2015-01-13T14:14:45.710" LastActivityDate="2015-01-13T14:14:45.710" Title="logic problem/puzzle solving" Tags="|logical-deduction|puzzle-creation|" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="7835" PostTypeId="2" ParentId="7834" CreationDate="2015-01-12T18:54:17.903" Score="3" Body="&lt;p&gt;All that matters is the chain, and so far your rules are quite local.  If you can get a couple steps in you can get all the way.  Maybe different characters can step different numbers of platforms?  For example, one might be able to go forward $1$ or $3$ spaces or back $2$.  Maybe each platform can only be used some number of times?  It becomes a maze in a bit of a disguise.&lt;/p&gt;&#xA;" OwnerUserId="73" OwnerDisplayName="Ross Millikan" LastActivityDate="2015-01-12T18:54:17.903" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7836" PostTypeId="1" AcceptedAnswerId="7837" CreationDate="2015-01-13T02:49:31.707" Score="-5" ViewCount="15998" Body="&lt;p&gt;In the below figure, how many triangles are there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Q2L9L.png&quot; alt=&quot;Pentagramception&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Every line that seems straight is considered straight. Don't try to look for any sneaky stuff (the star shape drawer makes a weird curved top line for some reason).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Every vertices and intersection is a perfect vertices/intersection (no microscopic triangles or anything). The smallest triangles in the figure are the gold/light yellow triangles near the middle.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A triangle can be made up of multiple colors. (Such as the blue/orange triangle at the top) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If you try solving this by counting them individually, you'll probably make a mistake and it will take longer than necessary. If you want to though, don't let me stop you :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Optional: Time yourself and post how long it took you. Others may be interested to know how fast something like this can be done. (I'm thinking there will be a spread between 5 minutes and ~ 1 hour)&lt;/p&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="3993" LastEditDate="2015-01-13T08:35:22.383" LastActivityDate="2020-08-14T12:49:12.823" Title="How many triangles are in the picture?" Tags="|visual|" AnswerCount="3" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="7837" PostTypeId="2" ParentId="7836" CreationDate="2015-01-13T04:16:53.947" Score="7" Body="&lt;p&gt;Well, here's my shot. about 15 minutes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A pentagram is 10 triangles, enclosed in a pentagon takes it up to 30; the outer pentagram only has a partial pentagon enclosure so that gets 21 triangles. Then there is the extra horizontal and vertical lines. The horizontal line has 4 symmetric triangles, plus 10 each side for 24, then the vertical line makes another 10 each side for 20, total 95 altogether by my count.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;OK, needs some revision.... &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The interaction of inner and outer pentagrams makes another 10 triangles. Revised total: 105.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A few more...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm revising the number of triangle associated with the horizontal line to 13 each side + 4  = 30 and for the vertical line to another 16 each side for 32. Total then is 30+21+30+32+10 = 123&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8875" LastEditorUserId="8875" LastEditDate="2015-01-13T18:44:09.620" LastActivityDate="2015-01-13T18:44:09.620" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7838" PostTypeId="2" ParentId="7807" CreationDate="2015-01-13T04:30:52.207" Score="12" Body="&lt;p&gt;The answer is that the necklace:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Is a cyclic cipher&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ciphers in the puzzle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Black edgy pearl vs rounded pearl is a bacon cipher that decodes to &quot;color&quot; meaning to  look at the pattern of colors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Black pearls vs white pearls is a bacon cipher that decodes to &quot;graynext&quot; meaning to check black vs gray next.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Black pearls vs gray pearls is a bacon cipher that decodes to &quot;cyannext&quot; meaning to check black vs cyan next.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Black pearls vs cyan pearls is a bacon cipher that decodes to &quot;olivenext&quot; meaning to check black vs olive next.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Black pearls vs olive pearls is a bacon cipher that decodes to &quot;linennext&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The color code for white, #FAF0E6 is actually the code for the color &quot;linen&quot;, so when you're comparing black and white, you're really comparing black and linen and get gray which repeats the cycle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="7722" LastEditDate="2015-01-14T02:38:06.217" LastActivityDate="2015-01-14T02:38:06.217" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7839" PostTypeId="2" ParentId="7834" CreationDate="2015-01-13T06:14:50.117" Score="0" Body="&lt;p&gt;I suspect this is for a pathfinder/d&amp;amp;d game or something similar. I've been thinking of a similar puzzle but I haven't had to worry about putting it into practice yet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My suggestion would be to give each tile three &quot;stages&quot; -- cursed/trapped, normal, and hollow. The first stage, &quot;cursed&quot;, would have some kind of negative effect if stepped on (say, it turns the person to stone). The normal tile is a normal tile. The hollow tile stage just simply means it has disappeared (causing any petrification victims to fall to their doom and shatter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you make it so that stepping on a tile changes the stage of the four adjacent tiles (cursed &gt; normal &gt; hollow), and once a tile goes hollow, its gone forever (or until the puzzle is reset). That might make for a better puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't tested it yet, but when I get home from work, I can certainly build a program in Flash that we can use to play around with it and see if it works.&lt;/p&gt;&#xA;" OwnerUserId="8858" LastActivityDate="2015-01-13T06:14:50.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7840" PostTypeId="2" ParentId="7819" CreationDate="2015-01-13T06:33:54.683" Score="-1" Body="&lt;p&gt;I was wondering if a dwarf with infinite range of sight can identify repeating pattern(s)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;What if there is only one single pattern that is infinite? This would mean the dwarfs can't be saved :(&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the fact that the number of dwarfs is infinite, there must be a repeating pattern at some point (this is almost too complicated for me now!). With infinite sight, every dwarf can find his own position within the repeating pattern (which can be very, very, very long).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that, a dwarf can never be sure whether he really identified a pattern correctly since the number of dwarfs is infinite and he would have to see &lt;em&gt;all dwarfs at once&lt;/em&gt; in order to be sure - which is impossible due to &quot;&lt;em&gt;infinite number&lt;/em&gt;&quot;,... I think? &lt;/p&gt;&#xA;" OwnerUserId="5106" LastActivityDate="2015-01-13T06:33:54.683" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="7841" PostTypeId="2" ParentId="7831" CreationDate="2015-01-13T08:26:48.493" Score="7" Body="&lt;p&gt;Since there is not any restriction to my solution I'm posting this. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;As it mentioned in previous question, lines are infinitesimal. Which means without any restriction on dots height, dots are bigger than lines. To show my method easily, here is a picture of the solution with 3x3 dots. &lt;br/&gt;&lt;br/&gt;&lt;img src=&quot;https://i.stack.imgur.com/fgvoC.png&quot; alt=&quot;3x3&quot;&gt; &lt;br/&gt;&lt;br/&gt;&#xA;If you can draw a line that meets all 7 dots in a row with appropriate angle you can basically cover all dots with &lt;strong&gt;7 lines&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4506" LastActivityDate="2015-01-13T08:26:48.493" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7842" PostTypeId="2" ParentId="7834" CreationDate="2015-01-13T09:03:01.247" Score="2" Body="&lt;p&gt;What about using 6 * X platforms and let different characters see different platforms(either from the beginning or during walking)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, all characters step on the first platform because the orange character can see it. After this they can see the second platform (also orange). Next one is the green character that can see the platform ahead and so on...&#xA;So the player has to collect information from all characters before (s)he can follow the path. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ovhCy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5106" LastEditorUserId="5106" LastEditDate="2015-01-13T09:08:37.167" LastActivityDate="2015-01-13T09:08:37.167" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7843" PostTypeId="2" ParentId="7783" CreationDate="2015-01-13T09:25:42.280" Score="2" Body="&lt;p&gt;Length(&quot;h34&quot;) = &lt;strong&gt;3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3&lt;/strong&gt; divided by one letter (&lt;strike&gt;&quot;h&quot;&lt;/strike&gt;, &quot;3&quot; or &quot;4&quot;) results in &lt;br&gt;&lt;strong&gt;3/3 = 1&lt;/strong&gt; or &lt;strong&gt;3/4&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Times &lt;strong&gt;11&lt;/strong&gt; is either &lt;br&gt; &lt;strong&gt;11&lt;/strong&gt; or &lt;strong&gt;33/4&lt;/strong&gt; which equals 4 times the &lt;em&gt;meaning&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;11 = 4x&lt;/strong&gt; or &lt;strong&gt;33/4 = 4x&lt;/strong&gt; giving me &lt;em&gt;x&lt;/em&gt; either is &lt;br&gt; &lt;strong&gt;11/4&lt;/strong&gt; or &lt;strong&gt;33/16&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The alphabet tells me &lt;strong&gt;K/D&lt;/strong&gt;(not defined) or &lt;strong&gt;G/P&lt;/strong&gt; (Guadeloupe).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;KK/D&lt;/strong&gt;(Kempen/Germany) or &lt;strong&gt;CC/AF&lt;/strong&gt; (Somwhere in Afghanistan).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;I have no idea what to do now :)&lt;/p&gt;&#xA;" OwnerUserId="5106" LastEditorUserId="5106" LastEditDate="2015-01-13T10:19:41.993" LastActivityDate="2015-01-13T10:19:41.993" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="7845" PostTypeId="2" ParentId="7834" CreationDate="2015-01-13T12:48:05.710" Score="-1" Body="&lt;h1&gt;Can it be a puzzle to get &lt;em&gt;one&lt;/em&gt; character across only?&lt;/h1&gt;&#xA;&lt;p&gt;(Who can then pull a lever or such a thing to make a safe path for everybody else.)&lt;/p&gt;&#xA;&lt;p&gt;This might make for a more interesting puzzle...&lt;/p&gt;&#xA;&lt;p&gt;I &lt;em&gt;think&lt;/em&gt; I have found an interesting puzzle inspired by your question, but I'm not sure it is &lt;em&gt;solvable&lt;/em&gt;. However, we are on PuzzlingSE, so lets find out...&lt;/p&gt;&#xA;&lt;p&gt;I've posted it as &lt;a href=&quot;https://puzzling.stackexchange.com/questions/7848/pathfinding-with-disappearing-platforms-is-it-solvable&quot;&gt;this puzzle here&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;If it works out, I think it's actually a really good puzzle ;c)&lt;/p&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-01-13T13:38:38.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7846" PostTypeId="1" AcceptedAnswerId="7866" CreationDate="2015-01-13T13:14:59.100" Score="13" ViewCount="1671" Body="&lt;p&gt;&lt;pre&gt;'1' is not considered as a prime number. &lt;/pre&gt;&lt;br&gt;&#xA;Find my true self,&lt;br&gt;&#xA;kill eleven Germans&lt;br&gt;&#xA;The good disguise wins&lt;br&gt;&#xA;so you act as twins  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prime number 7&lt;br&gt;&#xA;encrypted in alpha&lt;br&gt;&#xA;Now try to soften&lt;br&gt;&#xA;but please, not often  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once in front&lt;br&gt;&#xA;while twice in rear&lt;br&gt;&#xA;The Christians symbol&lt;br&gt;&#xA;is what we want here&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two towers of letters,&lt;br&gt;&#xA;three floors can be seen&lt;br&gt;&#xA;Built close to each other,&lt;br&gt;&#xA;what fits in between?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;What fits in between&lt;br&gt;&#xA;has been found before&lt;br&gt;&#xA;But one was different,&lt;br&gt;&#xA;and is waiting for more  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not only a symbol&lt;br&gt;&#xA;but like a small lagoon&lt;br&gt;&#xA;Rising from the dead&lt;br&gt;&#xA;every afternoon&lt;/p&gt;&#xA;" OwnerUserId="5106" LastEditorUserId="5106" LastEditDate="2015-01-20T14:29:29.757" LastActivityDate="2015-01-22T03:07:42.687" Title="It tears me apart, one more time" Tags="|riddle|word|" AnswerCount="5" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="7847" PostTypeId="2" ParentId="5445" CreationDate="2015-01-13T13:35:34.210" Score="0" Body="&lt;p&gt;Here is a brute force algorithm in pseudo code:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assumptions: we have the set of all the vertices $V$ (where two or more lines meet) and lines (edges in graph theory) between them.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;For each vertex a in V&#xA;  For each vertex b in V where b != a&#xA;    For each vertex c in V where c != a or b&#xA;      If there are lines connecting a&amp;lt;--&amp;gt;b, b&amp;lt;--&amp;gt;c, and c&amp;lt;--&amp;gt;a&#xA;      and if these vertices are not all in a straight line, then they&#xA;      form a triangle, so increment the triangle count&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is trivially $O(n^3)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some ways to prune this search by:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Checking if $a$ and $b$ are connected by a line before selecting $c$&lt;/li&gt;&#xA;&lt;li&gt;Looping over the lines (edges) instead of the vertices might be more efficient, but not as intuitive&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2015-01-13T13:35:34.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7848" PostTypeId="1" AcceptedAnswerId="101212" CreationDate="2015-01-13T13:36:56.667" Score="10" ViewCount="897" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;This is a puzzle but might not have a solution. So &amp;quot;&lt;strong&gt;is it possible?&lt;/strong&gt;&amp;quot; is a proper question.&lt;/p&gt;&#xA;&lt;p&gt;The puzzle idea was inspired by the question &lt;a href=&quot;https://puzzling.stackexchange.com/questions/7834/logic-problem-puzzle-solving&quot;&gt;logic problem/puzzle solving&lt;/a&gt;, and if the puzzle here turns out to be solvable, this question would be an &lt;em&gt;answer&lt;/em&gt; to the linked question. (Hence the cross-link.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Imagine the following situation:&lt;/p&gt;&#xA;&lt;p&gt;There is a grid of 6x10 rectangular stones of fields of which &lt;em&gt;some&lt;/em&gt; contain 'magical' stones which are solid under specific circumstances (defined below, shown in grey) and 'vacuum' otherwise. The other fields are 'vacuum' all the time.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;This field is extended by a top and a bottom row of (6) always solid fields (shown in brown on the bottom, omitted on the top.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;There are 6 &amp;quot;players&amp;quot; which want to get across the field. They can only 'walk' on solid fields and (-this is due to the original puzzle question-) along a given path:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vCVb9.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;While a player stands still&lt;/strong&gt;, he sends off six straight &lt;em&gt;rays&lt;/em&gt; defined by the diagonals of the fields and a vertical line. (The centre of the rays is always the centre of the field he's standing in, regardless where on the field the player stands.):&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8pgTx.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now a stone becomes solid (black) if, and only if, &lt;strong&gt;at least two &lt;em&gt;different&lt;/em&gt; rays cross&lt;/strong&gt; over its centre, while rays of a player standing on the platform &lt;em&gt;do not count&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Note that 2 &lt;em&gt;parallel&lt;/em&gt; rays are not considered &lt;em&gt;crossing&lt;/em&gt; each other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So the starting position is:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pkAwe.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now the puzzle-question is:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Can you bring &lt;em&gt;all&lt;/em&gt; players safely across the field (along the defined path)?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;and if not,&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Can you bring &lt;em&gt;at least one&lt;/em&gt; player safely across the field (along the defined path)?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Notation:&lt;/strong&gt;&#xA;Assuming only one player moves at a time, you can note its colour and final position (assuming the path to that one is 'walkable'. See my example answer below.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;Hints&lt;/h2&gt;&#xA;&lt;p&gt;When &lt;em&gt;testing&lt;/em&gt; a valid move, do it by the following 3 steps:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&amp;quot;Remove&amp;quot; a player and his colour and check which platforms are (still) solid.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Move player along valid, solid path to any position. (Multiple players are allowed on the same field!)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Add player-rays and check which additional platforms are solid.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This is the new &amp;quot;next&amp;quot; starting position. Note down the move and restart from 1.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Thanks to &lt;a href=&quot;https://puzzling.stackexchange.com/users/8858/efrog&quot;&gt;EFrog&lt;/a&gt; there is now also a &lt;a href=&quot;http://www.fastswf.com/0jk17wE&quot; rel=&quot;nofollow noreferrer&quot;&gt;helper tool&lt;/a&gt; for checking the situation. (See his answer below for details and instructions.) In accordance to the 4 points above:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Select a colour (click the big square to change colour) to &amp;quot;pick up&amp;quot; a player. Now check which 'path' is available and move the player to the end-position you want. (The tool does not allow going diagonal directly, but the rules allow it where indicated in the pre-given path.) To set-down a player, simply select another player/colour. Using 'black' mean that all players are set-down.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2020-08-18T21:38:47.940" Title="Pathfinding with disappearing platforms - is it solvable?" Tags="|logical-deduction|puzzle-creation|visual|reachability|" AnswerCount="3" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="7849" PostTypeId="2" ParentId="7848" CreationDate="2015-01-13T13:48:35.373" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;This is a partial answer only, demonstrating the first few (potential) steps.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Each image depicts the situation &lt;em&gt;after&lt;/em&gt; the named move. Remember &lt;em&gt;during movement&lt;/em&gt; the &lt;em&gt;rays&lt;/em&gt; of the moving player are &lt;em&gt;off&lt;/em&gt;, so all platforms 'held' by his rays (and no other 2 crossing colours) will immediately disappear! &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Orange --&gt; A2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PVt55.png&quot; alt=&quot;1=Orange A2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Yellow --&gt; C2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IE5ZQ.png&quot; alt=&quot;2=Yellow C2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cyan--&gt; C3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/odBV4.png&quot; alt=&quot;3=Cyan C3&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Orange --&gt; 4&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pITHP.png&quot; alt=&quot;4=Orange 4&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Yellow --&gt; D4&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZrNI1.png&quot; alt=&quot;5=Yellow D4&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Purple --&gt; B3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/w1bxe.png&quot; alt=&quot;6=Purple B3&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;etc...&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="5840" LastEditDate="2015-01-13T14:55:00.683" LastActivityDate="2015-01-13T14:55:00.683" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7850" PostTypeId="2" ParentId="6453" CreationDate="2015-01-13T13:50:39.567" Score="3" Body="&lt;p&gt;Ok. I've mostly been following this thread from my phone while I'm at work, so I haven't managed to read most of the work already done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few details I haven't seen mentioned (though I could have overlooked them):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The hanging weight on the scale seems to be movable. And I'm guessing that given that the weights are divided in two segments, these weights probably equal the weight of two square-units-worth of liquid. Unfortunately, I don't know how weights and scales work in regard to distance from the center.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The diamonds on the top of the key are the same as the diamonds on the labels on the bottle. This probably indicates which tube the bottle is meant to be poured into.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My guess regarding the shading of the bottles:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The right-hand side of &quot;Instruction 1&quot; seems to indicate that the first (darkest) bottle will fill up five square-units. Using this information, it's not unreasonable to suspect that the next-lightest bottle will fill up four SU, then three, down to one SU for all of the grey bottles. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So. Given that information, this is my hypothesis regarding actually solving the passphrase:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Using the labels, we can determine (as @i_turo has shown) the order in which the bottles should be used. By overlaying the labels onto the 8x8 grid in &quot;Instruction 1&quot;, we see that it slowly covers up (what I'm assuming is the entire Greek alphabet, as most of the characters I know are in the grid, and the Greek alphabet contains 24 letters) the symbols in reverse from Omega to Alpha. The opposite order (A &gt; Z) should probably be used to pour the bottles. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2.) Now that we know what order to put the bottles in, we have to determine which tube we pour them down. This is most likely determined by the yellow diamonds on the labels, as they correspond with the colors on the keys. Unfortunately, I'm not good with codes like this so I'm not sure what to do. But &lt;strong&gt;we need to find a way to translate the label-diamonds into colors somehow&lt;/strong&gt;. We do know, however, that there are only six color-tubes while there are 24 different layouts of diamonds. So &lt;em&gt;either the colors can be represented multiple ways using this code&lt;/em&gt;, or &lt;em&gt;the code changes each time we remove a key and replace it&lt;/em&gt;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My hypothesis regarding this...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...is that the six keys begin in a line (Red, Yellow, Green, Cyan, Blue, Purple). If you line up all of the keys and allow each 2nd-diamond to correspond with the 1st-diamond on the next key, you have a pattern of AB BC CD DE EF FG. This gives you 7 diamonds overall (the number that appears in the labels), ABCDEFG. Once we remove the blue key (now EF) to use it in a moment, we slide the keys over, then replace the blue key at the END of the line (making it now FG and purple becomes EF). Then we use that arrangement to determine the color for the next bottle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Third&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When we pour the bottle into the tube, it will fill up the squares in the row corresponding to the tube color, as shown in &quot;The Scale&quot;. We can determine how many squares will be filled up by using the blue shading, as earlier discussed. Since bottle E4 is first, and it appears to have a color matching that of the first bottle in &quot;Instruction 1&quot;, we can assume it contains 5 SU of water. So when we pour E4 into the blue tube, the scale will tip to the right.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fourth&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When the scale tips, we have to move the hanging weight to make it balance again. The free-standing weight is clearly fixed in place. So we have to play with the hanging weights until the scale becomes balanced, then look at which section of the scale the hanging weight is occupying.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fifth&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We insert the key whose color corresponds with the tube through which we poured the water into &quot;The Code&quot;. Then we find the letter at the cross-section between the color of the space the hanging weight is occupying (indicated on the key itself) and the color of the key/tube that we used (indicated by the columns in &quot;The Code&quot;). That gives us our first letter. We repeat this process until we're done. The final result may still be in some kind of code, though that doesn't seem to be indicated anywhere in the pictures.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A concern/question for @Bmyguest:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The picture that @Tryth shows indicates that there are six differently-colored bottles of liquid instead of five, as all of the other pictures with shading show. This would imply that the C2 bottle contains 6 SU of water (or that 2 of the bottles have matching colors). If that's the case, then why is that sixth colored bottle not shown in the other pictures, specifically the ! speech-cloud in &quot;Instruction 3&quot;? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8858" LastEditorUserId="8858" LastEditDate="2015-01-15T14:07:00.983" LastActivityDate="2015-01-15T14:07:00.983" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7852" PostTypeId="2" ParentId="7848" CreationDate="2015-01-13T17:12:11.610" Score="4" Body="&lt;p&gt;This isn't really an answer, but this is a program that I made that should allow you to better solve this puzzle (or create your own using the same premises). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://fastswf.com/0jk17wE&quot; rel=&quot;nofollow&quot;&gt;BmyGuest's Puzzle&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Instructions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a very clunky program that I put together, but it should be fairly user-friendly and easy to work with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The grid&lt;/em&gt;: The grid consists of the same 6x10 layout presented in the OP (with one extra row at the bottom to start on and another at top for finishing). Each tile in the grid can be turned &quot;on&quot; or &quot;off&quot; by clicking. Tiles that are turned &quot;on&quot; are indicated by white triangles in the corners, while tiles that are &quot;off&quot; are indicated by black triangles. The tiles that are &quot;on&quot; represent the path -- only these tiles will respond to the characters' positions. (Note that it is set to the default path presented in the OP, but it can be changed if you like.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The little arrow things&lt;/em&gt;: In the bottom-right of the screen is a large red box with an arrow on each side. If you click on the red box, it should cycle through the colors (Red &gt; Blue &gt; Green &gt; Yellow &gt; Purple &gt; Orange &gt; Blank/None &gt; Red). Use this to choose which player you intend to control. By clicking on the appropriate arrow button near the selection box, you can then move that character in the indicated direction on the map. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Other&lt;/em&gt;: You may use the spacebar to place all of the players back in their starting positions. &lt;strong&gt;This does not reset the path to default, just the player positions.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Please Note&lt;/strong&gt;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I said, this is a rather clunky program. So there are a couple of known issues. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the player dots are able to move off of the grid. I didn't feel the need to put in restrictions, as this is just a sandbox-style program. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, players aren't able to move diagonally in my program; however, this is allowed in the puzzle. Just be sure that you complete each diagonal move you intend and don't stop halfway. :D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps and my instructions make sense.&lt;/p&gt;&#xA;" OwnerUserId="8858" LastEditorUserId="8858" LastEditDate="2015-01-13T19:20:52.620" LastActivityDate="2015-01-13T19:20:52.620" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="7853" PostTypeId="2" ParentId="7819" CreationDate="2015-01-13T18:10:37.973" Score="9" Body="&lt;p&gt;This puzzle has been discussed on the Mathematics Stack Exchange as in this question: &lt;a href=&quot;https://math.stackexchange.com/questions/1032928/prisoners-problem&quot;&gt;Prisoners Problem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution given there (by &lt;a href=&quot;https://math.stackexchange.com/users/622/asaf-karagila&quot;&gt;Asaf Karagila&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Encode the colors into $0$&#xA;  and $1$, and define the equivalence relation on $2^\Bbb N$, $\langle x_i\rangle\sim\langle y_i\rangle$ if and only if there is some $k$&#xA;  such that for all $n\geq k$, $x_n=y_n$. Using the axiom of choice the&#xA;  prisoners pick a representative from each equivalence class. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In his turn, the $n$-th prisoner looks for the representative class&#xA;  fitting the string of hats he sees ahead, assuming that all hats up to&#xA;  him are blue. Since all the prisoners follow the same representative&#xA;  to guess their own color, it is guaranteed that after finitely many&#xA;  deaths, the representative and the fashionable selection of hats by&#xA;  the warden will agree, and everyone else will survive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that since the Axiom of Choice is non-constructive, so is this solution and hence may not be practically useful for the dwarves in the question. There seems to be a fair bit of discussion on whether the Axiom of Choice is required. Well it was shown by Hardin and Taylor [Hardin, Christopher S.; Taylor, Alan D.&#xA;An introduction to infinite hat problems.&#xA;&lt;em&gt;Math. Intelligencer&lt;/em&gt; &lt;strong&gt;30&lt;/strong&gt; (2008), no. 4, 20–25] that there being &lt;strong&gt;no&lt;/strong&gt; winning strategy is consistent with ZF (plus other axioms which don't imply the Axiom of Choice). So in that sense the Axiom of Choice is necessary for a winning strategy.&lt;/p&gt;&#xA;" OwnerUserId="8888" LastEditorUserId="-1" LastEditDate="2017-04-13T12:19:36.950" LastActivityDate="2015-01-13T18:10:37.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7854" PostTypeId="1" AcceptedAnswerId="7857" CreationDate="2015-01-13T19:15:48.037" Score="-5" ViewCount="243" Body="&lt;p&gt;A group of farmers need to harvest 2 farms, one of which is twice the size of the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They worked half of the first day on the bigger farm. Then they split into two equal groups to work on harvesting both farms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By end of the first day, the bigger farm was completely harvested. The smaller farm had a some work left to be done, but a single farmer was able to finish harvesting it by the end of the second day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many farmers were in the group?&lt;/p&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="1955" LastEditDate="2015-01-13T19:31:01.737" LastActivityDate="2015-01-13T20:23:48.270" Title="Group of Farmers" Tags="|explanation|" AnswerCount="2" CommentCount="2" ClosedDate="2015-01-14T16:09:53.143" ContentLicense="CC BY-SA 3.0" />
  <row Id="7855" PostTypeId="2" ParentId="7854" CreationDate="2015-01-13T19:25:05.183" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Whoops, read that as 3 days total. 8 farmers, same farmer-days needed&lt;br&gt;&#xA; We know the large field takes $3X/4$ farmer-days to complete, and the smaller field takes $X/4 + 1$ farmer-days to complete. Since it is half the size, we have the equality $X/2 + 2 = 3X/4$ which resolves to $X = 8$.&lt;br&gt;&#xA; -OLD-&lt;br&gt;&#xA; There were 4 farmers. The big field took 6 farmer-days, the second field took 3 farmer-days (yeah, I made the unit up). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-01-13T20:23:48.270" LastActivityDate="2015-01-13T20:23:48.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7856" PostTypeId="1" CreationDate="2015-01-13T19:29:20.630" Score="-7" ViewCount="636" Body="&lt;p&gt;Cows are always hungry...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;70 Cows eat all the plants on a pasture in 24 days.&lt;/li&gt;&#xA;&lt;li&gt;30 Cows clear the same pasture in 60 days.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How many cows are needed to clear the pasture by the end of 96th day?&lt;/p&gt;&#xA;" OwnerUserId="2225" LastActivityDate="2015-01-13T22:27:51.177" Title="Cows eat too much" Tags="|logical-deduction|explanation|" AnswerCount="3" CommentCount="2" ClosedDate="2015-01-14T01:19:18.323" ContentLicense="CC BY-SA 3.0" />
  <row Id="7857" PostTypeId="2" ParentId="7854" CreationDate="2015-01-13T19:37:27.520" Score="3" Body="&lt;p&gt;Let's say that there are $x$ farmers. Since they spent half of the day working on the larger farm, we know that the larger farm requires at least $\frac{x}{2}$ man-days for harvest. Then, for the second half of the first day, $\frac{x}{2}$ farmers worked on each farm. Since the larger farm was then finished, we know it requires $\frac{x}{2}+\frac{\frac{x}{2}}{2}=\frac{3x}{4}$ man-days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The smaller farm received $\frac{x}{4}$ man-days of work on the first day, and since it is half the size of the larger farm should require $\frac{3x}{8}$ total man-days of work to finish. This means $\frac{3x}{8}-\frac{x}{4}=\frac{x}{8}$ man-days of work are left, and since that was done by one farmer in one day, $\frac{x}{8}=1$ so there are 8 farmers in the group.&lt;/p&gt;&#xA;" OwnerUserId="1955" LastActivityDate="2015-01-13T19:37:27.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7858" PostTypeId="2" ParentId="7856" CreationDate="2015-01-13T19:51:53.013" Score="1" Body="&lt;p&gt;&lt;strong&gt;First Try&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;gt;&amp;gt;&amp;gt; (70 * 24 + 30 * 60) / (2 * 96)&#xA;18.125&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It looks like you need 18 cows for 96 days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second Try&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;gt;&amp;gt;&amp;gt; (30 * 60) / (70 * 24) / (60 - 24) * (96 - 60) * (30 * 60) / 96&#xA;20.08928571428571&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;After looking at what others had to say, it might be 20 cows instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In more detail, it the previous calculation can be read as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;gt;&amp;gt;&amp;gt; food_for_24_days = 24 * 70&#xA;&amp;gt;&amp;gt;&amp;gt; food_for_60_days = 60 * 30&#xA;&amp;gt;&amp;gt;&amp;gt; ratio_of_increase = food_for_60_days / food_for_24_days&#xA;&amp;gt;&amp;gt;&amp;gt; ratio_of_increase_per_day = ratio_of_increase / (60 - 24)&#xA;&amp;gt;&amp;gt;&amp;gt; increase_from_60_to_96_days = ratio_of_increase_per_day * (96 - 60)&#xA;&amp;gt;&amp;gt;&amp;gt; food_for_96_days = food_for_60_days * increase_from_60_to_96_days&#xA;&amp;gt;&amp;gt;&amp;gt; cows_needed_for_96_days = food_for_96_days / 96&#xA;&amp;gt;&amp;gt;&amp;gt; cows_needed_for_96_days&#xA;20.08928571428571&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1661" LastEditorUserId="1661" LastEditDate="2015-01-13T22:27:51.177" LastActivityDate="2015-01-13T22:27:51.177" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7859" PostTypeId="2" ParentId="7856" CreationDate="2015-01-13T20:02:31.230" Score="3" Body="&lt;p&gt;Since 70 cows clear the pasture in 24 days, we know that there should be more than $70*23=1610$ and at most $70*24=1680$ cow-days of plants in the pasture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since 30 cows clear the pasture in 60 days, we know that there should be more than $30*59=1770$ and at most $30*60=1800$ cow-days of plants in the pasture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we run into a problem - the two ranges don't overlap. Either there's an error in the question, or we have to deal with plants growing (they do that).&lt;/p&gt;&#xA;" OwnerUserId="1955" LastActivityDate="2015-01-13T20:02:31.230" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7861" PostTypeId="2" ParentId="7856" CreationDate="2015-01-13T20:45:27.727" Score="4" Body="&lt;p&gt;20 cows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 70 cows over the 24 days eat 1680 cow-nom units.&lt;br&gt;&#xA;The 30 cows over the 60 days eat 1800 cow-nom units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the growth from 24 days to 60 days, we can work out that at 96 days, the cows would need to eat 1920 cow-nom units. Divide that by the 96 days and we get $20 cows$.&lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-01-13T20:45:27.727" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7862" PostTypeId="1" AcceptedAnswerId="7864" CreationDate="2015-01-13T20:46:06.900" Score="47" ViewCount="5256" Body="&lt;p&gt;The numbers 1 through 10 are written on a blackboard. You erase two numbers of your choice, and write their product plus their sum&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a,b \to ab+a+b$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, now there are nine numbers on the board. You repeat this process until a single number remains. What is the largest this number can be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I can't stop you from writing code to brute-force this, but I assure you there's a nice solution and coding isn't needed.)&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="11876" LastEditDate="2015-11-25T02:48:50.620" LastActivityDate="2015-11-25T02:48:50.620" Title="Ten numbers on a blackboard" Tags="|mathematics|blackboard|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="7864" PostTypeId="2" ParentId="7862" CreationDate="2015-01-13T21:38:50.940" Score="54" Body="&lt;p&gt;Let the combining operation be $\otimes$, i.e. $a \otimes b = ab + a + b$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observe the following for $\otimes$:&#xA;$$a \otimes b + 1\\&#xA;= ab + a + b + 1\\&#xA;= (a + 1)(b + 1)$$&#xA;So the ordinary multiplication operation is equivalent to $\otimes$ just offset by $+1$. &#xA;This suggests an equivalent formulation of the problem where instead of working on the original values 1 through 10, we work with 2 through 11 and just use multiplication. Our answer will end up being 1 larger than that required by the original problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since multiplication is commutative and associative, the ordering doesn't matter and all possible sequences will give us the same answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution to this transposed problem is then $2 \times 3 \times 4 \times \ldots \times 11 = 11! = 39916800$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so the solution to the original problem is $1 \otimes 2 \otimes 3 \otimes \ldots \otimes 10 = 11! - 1 = 39916799$.&lt;/p&gt;&#xA;" OwnerUserId="8888" LastActivityDate="2015-01-13T21:38:50.940" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7865" PostTypeId="2" ParentId="7862" CreationDate="2015-01-13T22:32:58.117" Score="23" Body="&lt;p&gt;The largest it can be is actually also the smallest it can be. In fact, if the numbers 1 through $n$ are written and the same process followed, the end result will be $(n+1)! - 1$ no matter what order you combine numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take a smaller set, just $\{a, b, c\}$, to see why. If you group $a$ and $b$ first, you'll end up with &#xA;$$&#xA;(ab+a+b)c+(ab+a+b)+c=a+b+c+ab+ac+bc+abc&#xA;$$&#xA;If you group $b$ and $c$ first, you get&#xA;$$&#xA;(bc+b+c)a+(bc+b+c)+a=a+b+c+ab+ac+bc+abc&#xA;$$&#xA;And just for completeness, grouping $a$ and $c$ first gives&#xA;$$&#xA;(ac+a+c)b+(ac+a+c)+b=a+b+c+ab+ac+bc+abc&#xA;$$&#xA;At the end of the $n$ numbers, you will always end up with the sum of the individual numbers, plus the sum of the products of the numbers taken 2 at a time, plus the sum of the products taken 3 at a time, all the way up to the product of all the numbers. If you added 1 to the final sum, you could factor the final result into $(a+1)(b+1)(c+1)...$ giving $(n+1)!$ in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, a starting list of the number $k$ written $n$ times will result in $(k+1)^n-1$.&lt;/p&gt;&#xA;" OwnerUserId="8891" LastEditorUserId="8891" LastEditDate="2015-01-13T22:48:53.690" LastActivityDate="2015-01-13T22:48:53.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7866" PostTypeId="2" ParentId="7846" CreationDate="2015-01-14T03:24:38.157" Score="2" Body="&lt;p&gt;&lt;strong&gt;EDIT 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tears&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the other hints we get the letters &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TRS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The additional two paragraphs state:&#xA;What fits in between&#xA;has been found before&#xA;But one was different,&#xA;and is waiting for more&#xA;It's not only a symbol&#xA;but like a small lagoon&#xA;Rising from the dead&#xA;every afternoon&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is saying that one of the things found is waiting for more. It then says &quot;It's not only a symbol&quot;, and the thing found that was a symbol was &quot;The Christians symbol&quot; which was T.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then the second part of the hint, &quot;but like a small lagoon, Rising from the dead every afternoon&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The small lagoon could be like a cup of tea. The tea leaves are dried and dead but during the afternoon you have your &quot;afternoon tea&quot; which rises them from the dead.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also the title is &quot;It tears me apart one more time&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letters TRS are torn apart to make tears.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If this is the answer I'd like to give credit to the many people who got hints out of Bobbee and I just pieced them together.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT:The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reset&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So far, the answers from Leslie and Namisca are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; QRS and STU&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Making these as two towers and taking everything in between is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; R S T&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The title of the puzzle is Tear it apart, one more time&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tear apart RST and make it RESET (RST is a common abbreviation of RESET). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the added hint, &#xA;&quot;What fits in between&#xA;has been found before&#xA;But one was different,&#xA;and is waiting for more&#xA;It's not only a symbol&#xA;but like a small lagoon&#xA;Rising from the dead&#xA;every afternoon&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reset is a symbol, &quot;every afternoon&quot; references a daily reset of something like a computer, and the symbol is a circle with a line coming out (rising dead from lagoon), also symbolizing reset).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="7722" LastEditDate="2015-01-22T03:07:42.687" LastActivityDate="2015-01-22T03:07:42.687" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="7867" PostTypeId="2" ParentId="7783" CreationDate="2015-01-14T09:33:06.387" Score="7" Body="&lt;p&gt;EDIT&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Found it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are in:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Uruguay – the town of Treinta y Tres, which means thirty-three in Spanish (official language of Uruguay), and was deliberately named after the ’33 Orientales’ who founded Uruguay&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Workings are basically the same as my previous answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Treinta y Tre has 12 letters, divide by one letter to get 12. Times 11 is 132, which is four times 33.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;OLD ANSWER:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are in:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Estonia. Specifically, the town of Elva.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Divide the length of my name by one letter&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Length of Elva is 4 letters, divide by one letter to get 4 (without units)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;then multiply by 11&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; to get 44&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;to get four times my meaning.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Meaning is eleven, because elva means eleven in Swedish (and possibly other languages, I don't know!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8898" LastEditorUserId="8898" LastEditDate="2015-01-15T16:35:05.993" LastActivityDate="2015-01-15T16:35:05.993" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="7868" PostTypeId="1" AcceptedAnswerId="7869" CreationDate="2015-01-14T10:07:52.900" Score="11" ViewCount="1119" Body="&lt;p&gt;In the magical circus, today a magician juggles with balls, cones, &#xA;rings and umbrellas. While performing his art, he sometimes applies &#xA;a powerful magic word:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;strong&gt;Ebrecedebre&lt;/strong&gt; transforms one ball into one cone and one ring.&lt;/li&gt;&#xA;  &lt;li&gt;&lt;strong&gt;Ibricidibri&lt;/strong&gt; transforms one cone into two rings and one umbrella.&lt;/li&gt;&#xA;  &lt;li&gt;&lt;strong&gt;Obrocodobro&lt;/strong&gt; transforms one ring into one umbrella and three balls.&lt;/li&gt;&#xA;  &lt;li&gt;&lt;strong&gt;Ubrucudubru&lt;/strong&gt; transforms one umbrella into one ball and four cones.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reversed magic words act in the opposite way:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;strong&gt;erbedecerbE&lt;/strong&gt; transforms one cone and one ring into one ball.&lt;/li&gt;&#xA;  &lt;li&gt;&lt;strong&gt;irbidicirbI&lt;/strong&gt; transforms two rings and one umbrella into one cone.&lt;/li&gt;&#xA;  &lt;li&gt;&lt;strong&gt;orbodocorbO&lt;/strong&gt; transforms one umbrella and three balls into one ring.&lt;/li&gt;&#xA;  &lt;li&gt;&lt;strong&gt;urbuducurbU&lt;/strong&gt; transforms one ball and four cones into one umbrella.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At the current moment, the magician is juggling with a single cone &#xA;(and no balls, no rings, no umbrellas). Which of the following four &#xA;situations can be reached by applying these magic words?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  A.  At least five balls, and no cones, rings, umbrellas.  &#xA;  B.  At least five cones, and no balls, rings, umbrellas.  &#xA;  C.  At least five rings, and no balls, cones, umbrellas.  &#xA;  D.  At least five umbrellas, and no balls, cones, rings.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-08T10:30:27.923" LastActivityDate="2016-03-08T10:30:27.923" Title="The juggling magician" Tags="|mathematics|combinatorics|reachability|" AnswerCount="3" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7869" PostTypeId="2" ParentId="7868" CreationDate="2015-01-14T13:09:27.567" Score="7" Body="&lt;p&gt;First lets define some variables. &lt;br/&gt;&#xA; &lt;strong&gt;B&lt;/strong&gt; = Ball, &lt;strong&gt;C&lt;/strong&gt; = Cone, &lt;strong&gt;R&lt;/strong&gt; = Ring and &lt;strong&gt;J&lt;/strong&gt; = Umbrella. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Equations:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;B = C R&lt;/li&gt;&#xA;&lt;li&gt;C = J 2R&lt;/li&gt;&#xA;&lt;li&gt;R = J 3B&lt;/li&gt;&#xA;&lt;li&gt;J = B 4C&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1) 4R = 4B | &lt;strong&gt;Proof:&lt;/strong&gt; 4R = J 3B 3R = C R 3B = 4B &lt;br/&gt; 2) R = 4B | &lt;strong&gt;Proof:&lt;/strong&gt; R = J 3B = 4B 4C &gt; (Use Proof 1) &gt; 4C 4R = 4B &lt;br/&gt; 3) C = 9B | &lt;strong&gt;Proof:&lt;/strong&gt; C = J 2R &gt; (Use Proof 2) &gt; J 8B = C4 9B &gt; (Use Proof 1) &gt; 5B C4 R4 = 9B &lt;br/&gt; 4) J = 37B | &lt;strong&gt;Proof:&lt;/strong&gt; J = B 4C &gt; (Use Proof 3) &gt; J = 37B&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We have single &lt;strong&gt;C&lt;/strong&gt;one in start.&lt;br/&gt; A. At least five balls, and no cones, rings, umbrellas. &lt;strong&gt;C = 9B&lt;/strong&gt;&lt;br/&gt;B. At least five cones, and no balls, rings, umbrellas. &lt;strong&gt;Let x divisble by 4. C = 9B = (9-x)B xR = 9-xB+4xB = 9+3xB = 45B = 5C(where x=12)&lt;/strong&gt;&lt;br/&gt;C. At least five rings, and no balls, cones, umbrellas. &lt;br/&gt;&lt;strong&gt;Let x divisble by 4. C = 9B = (9-x)B xR = 9-xB+4xB = 9+3xB which never be a multiply of 4. Thus not possible.&lt;/strong&gt; &lt;br/&gt; D. At least five umbrellas, and no balls, cones, rings.&lt;br/&gt;&lt;strong&gt;Let x divisble by 4. C = 9B = (9-x)B xR = 9-xB 4xB = 9+3xB = 333B = 9J (where x=108)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since last part is math problem, i assume previous parts can be explained with mathematical formulas.&lt;/p&gt;&#xA;" OwnerUserId="4506" LastEditorUserId="4506" LastEditDate="2015-01-14T13:47:54.873" LastActivityDate="2015-01-14T13:47:54.873" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7870" PostTypeId="2" ParentId="7868" CreationDate="2015-01-14T13:09:46.563" Score="3" Body="&lt;p&gt;Abbreviating B = Ball, C = Cone, R = Ring, U = Umbrella:&#xA;C  2R + U  R + 2U + 3B  R + 5B + 8C  2B + 4R + 11C  4R + 2U + 3C  5C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that's B: At least five cones and nothing else. If more cones are wanted, you can transform one cone into 5 more, so (at the very least) you can have 1 + 4n cones for positive integer n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That feels very manual; maybe it will help someone else see a pattern for the other three. I will add more as I get time.&lt;/p&gt;&#xA;" OwnerUserId="6031" LastActivityDate="2015-01-14T13:09:46.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7871" PostTypeId="2" ParentId="6241" CreationDate="2015-01-14T15:01:29.040" Score="1" Body="&lt;p&gt;If you disregard the &quot;th&quot; here is an alternative solution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;formula for floor number: Add the digits of the first letters of the company and subtract 8 if it is more than 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pangloss Washers Co&#xA;P = 16, W = 23, C = 3. Sum = 15, 15-8 = 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alice PC Chips Co&#xA;A = 1, p = 16, C = 3. Sum = 14, 14-8 = 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jumanji Arcade Co&#xA;J = 10, A = 1, C = 3. Sum = 5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Colorful Crafts Co&#xA;C = 3. Sum = 9. 9 - 8 = 1 therefore 1st floor.&lt;/p&gt;&#xA;" OwnerUserId="8903" LastEditorUserId="8903" LastEditDate="2015-01-14T15:07:44.723" LastActivityDate="2015-01-14T15:07:44.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7872" PostTypeId="2" ParentId="7846" CreationDate="2015-01-14T17:14:04.973" Score="3" Body="&lt;p&gt;Same as Quark, only an attempt. My current answer: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SS QO RX ET&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can't make it work with the hint in the title.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find my true self, kill eleven Germans&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; S. Eleven in German is &quot;elf&quot;. We delete the &quot;elf&quot; in self.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The good disguise wins so you act as twins&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; S. The twins acts as hint to double the previous letter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Prime number 7 encrypted in alph: As given per Quark's answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Q&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now try to soften but please, not often&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; O. We change the letter slighlty we get an O.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once in the front while twice in the rear&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; R.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Christians symbol is what we want here&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; X. ( Xmas.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Two towers of letters, three floors can be seen&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; E. When written in capital the straight lines are the &quot;Floors&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Built close to each other, what fits in between?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; T. It's what is between the letter E in the word letter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8905" LastActivityDate="2015-01-14T17:14:04.973" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7873" PostTypeId="2" ParentId="6241" CreationDate="2015-01-14T17:37:55.860" Score="12" Body="&lt;p&gt;I have an alternate solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is sourced from &lt;em&gt;Umineko&lt;/em&gt;, a Japanese video game/visual novel series, so it is reasonable to conclude that the building in question is in Japan. Japanese superstition associates the number 4 with death, so many Japanese buildings do not have a 4th floor (numbered).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reasoning of the &quot;correct&quot; answer then puts Colorful Crafts Co., Ltd. on the &lt;strong&gt;8th&lt;/strong&gt; floor.&lt;/p&gt;&#xA;" OwnerUserId="8776" LastActivityDate="2015-01-14T17:37:55.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7874" PostTypeId="2" ParentId="7846" CreationDate="2015-01-14T19:22:28.843" Score="7" Body="&lt;p&gt;Alright, I think I'm getting close. I'm not very confident about my answer to the second half of the first clue, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find my true self,&#xA;kill eleven Germans&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I agree with Namisca's reasoning for this as S.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The good disguise wins,&#xA;so you act as twins&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This could be saying 'U' (you) act as twins (W). It could also be saying that the clue is the one acting as &quot;twins&quot;, so either U or V. I think U works best here, though, with regards to the final solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Prime number 7&#xA;encrypted in alpha&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;As everyone else has already stated, this is Q.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now try to soften&#xA;but please, not often&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Soften&quot; without &quot;Often&quot; leaves us with S.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once in front&#xA;while twice in rear&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter R is in front once, and twice in rear.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Christians symbol&#xA;is what we want here&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Christian's symbol is usually a cross, which evokes a lowercase T to me.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Two towers of letters,&#xA;three floors can be seen&#xA;Built close to each other,&#xA;what fits in between?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think this is trying to make us think about the letters we've just deduced- if my logic is correct, then we're at: &lt;br/&gt;&#xA;S   U &lt;br/&gt;&#xA;Q   S &lt;br/&gt;&#xA;R   T &lt;br/&gt;&#xA;&lt;br/&gt;&#xA;The letters that go between these two &quot;towers&quot; are T, R, and S. The 'built close together' clue was what made me think U worked best for the second letter; each &quot;floor&quot; of the towers is only one letter apart.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5616" LastActivityDate="2015-01-14T19:22:28.843" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7875" PostTypeId="1" CreationDate="2015-01-12T22:03:08.407" Score="13" ViewCount="31707" Body="&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;one repeated letter - bitter&lt;/li&gt;&#xA;&lt;li&gt;two consecutive repeated letters - bassoon&lt;/li&gt;&#xA;&lt;li&gt;three consecutive repeated letters - bookkeeper&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I've tried to find a four-times-repeated example but failed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Question edited: was &#xA;“Is there an English word with four consecutive repeated letters?”)&lt;/p&gt;&#xA;" OwnerDisplayName="Centaurus" LastEditorUserId="60462" LastEditDate="2020-01-10T21:59:05.150" LastActivityDate="2021-10-29T13:10:39.130" Title="What is the English word with the greatest number of consecutive repeated letters?" Tags="|word|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="7876" PostTypeId="2" ParentId="7875" CreationDate="2015-01-12T22:06:13.217" Score="36" Body="&lt;p&gt;The word list on my Mac has one:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;$ grep '\(.\)\1\(.\)\2\(.\)\3\(.\)\4' /usr/share/dict/words&#xA;subbookkeeper&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="25316" OwnerDisplayName="Barmar" LastActivityDate="2015-01-12T22:06:13.217" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="7877" PostTypeId="2" ParentId="7875" CreationDate="2015-01-12T22:50:47.600" Score="36" Body="&lt;p&gt;Yellowwooddoor - a door made out of yellowwood (acknowledged as a real Scrabble word). It even has an extra double letter (ll) though it's not consecutive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I can't believe how much time I just spent thinking about this. Time to get a life.)&lt;/p&gt;&#xA;" OwnerUserId="8914" OwnerDisplayName="Mynamite" LastActivityDate="2015-01-12T22:50:47.600" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7878" PostTypeId="2" ParentId="7836" CreationDate="2015-01-14T19:56:58.810" Score="9" Body="&lt;p&gt;Lets ignore the cross for the moment.  Then we simply have a 5 pointed star drawn within a 5 pointed star, drawn within 3 sides of a pentagon.&lt;/p&gt;&#xA;&lt;p&gt;##Interior Pentagram##&lt;/p&gt;&#xA;&lt;p&gt;The interior 5 pointed star is inside a pentagon.  There is a known result that says there are 35 triangles in this configuration.  There are basically 7 groups with 5 triangles each.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;---link removed---&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;The groups are:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The points of the star - tall and skinny&lt;/li&gt;&#xA;&lt;li&gt;The space between the points - short and fat&lt;/li&gt;&#xA;&lt;li&gt;The point paired with the space on its left&lt;/li&gt;&#xA;&lt;li&gt;The point paired with the space on its right&lt;/li&gt;&#xA;&lt;li&gt;The point and the space on both sides&lt;/li&gt;&#xA;&lt;li&gt;Two non-adjacent points including the middle pentagon.&lt;/li&gt;&#xA;&lt;li&gt;The space between the points extended to include the point on the opposite side.  Makes another tall/skinny triangle that is proportional to #1, but much taller.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;##Outer Pentagram##&#xA;If we were to complete the outer pentagon, then there would be another 35 triangles there as well.  However, we are missing two of the sides of the pentagon, meaning we are missing two of the triangles that make up the space between the points.  Of the 7 groups mentioned above:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;2 groups include none of these triangles (groups 1 and 6), so they remain groups of 5&lt;/li&gt;&#xA;&lt;li&gt;4 groups include one of these triangles (groups 2, 3, 4, and 7), so they become groups of 3 instead&lt;/li&gt;&#xA;&lt;li&gt;1 group (group 5) includes two of these spaces, so that group becomes a group of 2&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Thus, we have 2&lt;em&gt;5 + 4&lt;/em&gt;3 + 1*2 = 24 triangles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;##Interaction between inner and outer Pentragrams##&#xA;Now, if we look at how the outer and inner stars interact, then we can add an additional 15 triangles (3 groups of 5).  Basically, take group 5 of the inner pentagram and pair it with first one point of the outer star, and then the other.  Then take all but group 5 of the inner pentagram and pair it with the point of the outer star.&lt;/p&gt;&#xA;&lt;p&gt;The spaces between the points of the outer pentagram do not contribute to any additional stars.&lt;/p&gt;&#xA;&lt;p&gt;#Initial Summary#&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Thus, we have 35 + 24 + 15 = 74 triangles based on the pentagrams alone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;#Vertical Line#&#xA;Now lets add the vertical line.  This splits the top point into two triangles, and the top space in the interior pentagram as well as bottom point of the interior pentagram.  This adds 6 triangles directly, but now introduces more groupings. It also splits both the inner and outer pentagons inscribed inside each star in half.  These shapes aren't triangles, but can contribute as we see later.&lt;/p&gt;&#xA;&lt;p&gt;The newly created half of the top point can be grouped with the half of the space below it, as well as all the way to the new half of the bottom point.  As well, it can be grouped with the newly created half of the pentagon. (3 more triangles per side means 6 more triangles).&lt;/p&gt;&#xA;&lt;p&gt;The half of the top space of the interior pentagram that was split can only be extended down to the new half of the bottom point (1 more triangle per side means 2 more triangles).&lt;/p&gt;&#xA;&lt;p&gt;The half of the bottom point of the interior triangle can be grouped with the adjacent space to make a right angle triangle, or with the new half inner pentagon, or with the half inner pentagon and non-adjacent point.  Also, it can be grouped with the adjacent space and point and half inner pentagon. (4 more triangle per side means 8 more triangles).&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; So now we have 6 + 6 + 2 + 8 = 22 more triangles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;#Horizontal Line#&#xA;Finally, lets add the horizontal line.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; It directly creates 9 more triangles since it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;makes a new triangle within both points of the outer star (2)&lt;/li&gt;&#xA;&lt;li&gt;divides both spaces between two points of the inner star into two triangles (4)&lt;/li&gt;&#xA;&lt;li&gt;makes 3 new triangles in the inner pentagon - 2 small that combine to form the third.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;##New Triangle in Outer Points##&#xA;If we look at the newly created triangle within one of the outer star's points, it contributes to 5 new triangles.  To see this, start with just this new triangle and then move the top right point to the right one vertex so that now it includes the newly created triangle made from the bottom of 2 above (1).  Now move the bottom point out the middle (2).  Then move the top point to the middle (3) and then again to just past the middle (4).  Lastly, move the bottom point all the way to the tip of the outer point, and the top point out one more vertex (5). Since this happens on both sides, this is 10 new triangles.&lt;/p&gt;&#xA;&lt;p&gt;Lastly, there is one more triangle that includes both of the these newly created triangles which extends to the bottom point of the inner triangle.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Thus, 11 new triangles are created using these.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;##Spaces between points of inner pentagram##&lt;/p&gt;&#xA;&lt;p&gt;Of the space that was divided in two, the lower half actually doesn't not contribute to any new triangles that we haven't already counted.  The upper half, however, contributes to 2 more (one with the half of the top point of the outer star, and one that uses the newly created triangles in the inner pentagon).  This happens on both sides, so there is 4 more triangles.&lt;/p&gt;&#xA;&lt;p&gt;Again, another triangle is created using both of these upper spaces and the upper point of the other star.&lt;/p&gt;&#xA;&lt;p&gt;Thus, 5 new triangles are created using these.&lt;/p&gt;&#xA;&lt;p&gt;##New triangles inside the inner pentagon##&lt;/p&gt;&#xA;&lt;p&gt;Of the three new triangles in the middle, the smaller ones don't contribute to any new triangles on their own, but the big one contributes to 2 more by pairing with the point on either side.&lt;/p&gt;&#xA;&lt;p&gt;##Horizontal Line Summary##&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; So, we have 9 + 11 + 5 + 2 = 27 more triangles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;#Overall summary$&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Add these all up and you get 74 + 22 + 27 = 123 - the same as the previous answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;How long did I take on this?  About an hour writing it all up, but not as long to find them all.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="1592" LastEditDate="2020-08-14T12:49:12.823" LastActivityDate="2020-08-14T12:49:12.823" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="7881" PostTypeId="1" AcceptedAnswerId="7887" CreationDate="2015-01-14T23:12:01.993" Score="13" ViewCount="1639" Body="&lt;p&gt;You play a game with a friend: he deals cards face up one a time from a uniformly shuffled deck of 26 black cards and 26 red cards. At some point before all the cards have been dealt, you must call &quot;bet&quot;. When you do, your friend flips the next card, and you win if it's black and lose if it's red. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the highest win probability you can achieve with any strategy?&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-01-15T06:42:05.153" Title="Picking a time to bet" Tags="|mathematics|probability|" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="7882" PostTypeId="1" AcceptedAnswerId="7884" CreationDate="2015-01-14T23:26:56.333" Score="7" ViewCount="586" Body="&lt;p&gt;You have an odd number of tokens whose weights are known whole numbers, and they don't all have the same weight. Show that there's a token you can remove so that the remaining tokens can't be split into two equal-size sets that have the same total weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathier bonus questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The weights are instead positive rational numbers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The weights are instead positive real numbers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Show that this doesn't necessary hold for an &lt;a href=&quot;http://en.wikipedia.org/wiki/Abelian_group&quot;&gt;Abelian group&lt;/a&gt;, where weights are non-identity elements that are summed with the group operation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="8874" LastEditDate="2015-01-31T12:30:49.797" LastActivityDate="2015-01-31T12:30:49.797" Title="Balancing all but one weight" Tags="|logical-deduction|weighing|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7884" PostTypeId="2" ParentId="7882" CreationDate="2015-01-14T23:58:00.680" Score="6" Body="&lt;p&gt;To split it in two equal sets, the total must be even.  If you start with an even total, just remove an odd weight, the total becomes odd and can't be split in two equal sets any more.  If you start with an odd total, remove an even weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If all weights are even or all are odd, you'll need to do the following:  Write down the weights in binary.  Ignore the least significant bits that are equal among the weights.  The operation is equivalent to subtracting a constant value of all weights and divide all weights by the same power of 2.  Neither operation affects the balance since there are the same number of weights on both side.  After this operation you are guaranteed to have even and odd weights and you can then apply the initial method above.  Note that you don't actually change the weights, you just change the scale to measure them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have rationals, you can multiply all weights by the smallest multiple of their denominators.  You then have all integer weights.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2015-01-15T14:43:54.200" LastActivityDate="2015-01-15T14:43:54.200" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7885" PostTypeId="2" ParentId="2605" CreationDate="2015-01-15T00:05:49.063" Score="1" Body="&lt;p&gt;A solution much better than Penguino's &quot;spiral&quot; can be achieved by starting with Penguino's zig-zag (the one with length 5098) but adjusting it slightly.  Looking at the first place the line comes toward the left from the right, it extends all the way to (0,3) and then (0,5) for the purpose of being within one unit of (0,2) and (0,6), but for any x and y in the range (0,1) such that x²+y²≤1, those points could be reached by having the line extend to (1+x,1), (x,2+y), (x,6-y), and (1+x,5) before returning to the right.  Penguino's zig-zag is a special case for x=0,y=1 but x=0.6 y=0.8 yields a much better result (assuming the right side is treated similarly, it will reduce the amount of x excursion per row by 1.2, but adds only 0.8 worth of y excursion, with about half of that being on shallow diagonals).  The net effect is a total length under 5060.5--a noticeable savings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The slight tweak used for the above number entails starting at (x,2-y), then moving to (x,y), and going to (1+x,1) and all subsequent points in normal fashion.  Note that things must be drawn in that order to avoid leaving an unpainted area on the top.  Note also that using slightly different coordinates could improve things a bit, but further improvements are apt to be fairly slight.&lt;/p&gt;&#xA;" OwnerUserId="1467" LastActivityDate="2015-01-15T00:05:49.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7887" PostTypeId="2" ParentId="7881" CreationDate="2015-01-15T00:40:51.650" Score="13" Body="&lt;p&gt;It is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $50-50$, i.e a fair game. In fact, any strategy will give you 50%&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To show that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; there are no strategy that maximizes gain, look at the game once we call stop. Suppose there are $k$ cards left. The probability that the next card is black is the same as the probability that the last card is black. But to bet on the next card is the same as betting on the last one whatever strategy I use, I have 50% chance of winning. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8873" LastActivityDate="2015-01-15T00:40:51.650" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="7888" PostTypeId="2" ParentId="7881" CreationDate="2015-01-15T04:00:57.930" Score="0" Body="&lt;p&gt;I agree the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 50%&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But the way I thought of it was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are two types of strategies: one, wait after $x$ number of cards or wait after $x$ number of red cards (symmetry for black).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First strategy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If I wait for one card and it's red, I now have a 25/51 chance of getting black on the next shot. If the card is black, I have a 26/51 chance of getting black. You gain or lose 1/102 chance with a 50% chance, which doesn't help. You can apply this for any $x$ number of cards.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second strategy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If I wait for the first red card, it's still the same thing as above only the position I stop at isn't decided ahead of time but through chance. If I wait for 10 reds, the chances that I find this before the 20th card and after the 20th card are the same, which correspond to a higher or lower chance of winning. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Other than these two strategies, there are no other (that I can think of) distinctive possible strategies other than cheesy ones like &quot;go through all 51 in hopes the dealer messes up and shows you the next card&quot;.&lt;/p&gt;&#xA;" OwnerUserId="7722" LastActivityDate="2015-01-15T04:00:57.930" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="7889" PostTypeId="1" AcceptedAnswerId="7896" CreationDate="2015-01-15T04:24:54.713" Score="18" ViewCount="1792" Body="&lt;p&gt;This is a similar but harder version of &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3074/6-tries-to-guess-a-number-between-1-100&quot;&gt;6 Tries to Guess a Number Between 1-100&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You friend gives you $n$ tries to guess an integer $x$, $1\le x\le 100$. Your guesses must be of the following type:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is $x$ greater than $k$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where $k$ is an integer. However, your friend will give you an answer to your question only after a delay of a question. A illustration of how this game would play in practice:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;You:&lt;/strong&gt; Is $x$ greater than 30?&lt;br&gt;&#xA;  &lt;strong&gt;Friend:&lt;/strong&gt; ...&lt;br&gt;&#xA;  &lt;strong&gt;You:&lt;/strong&gt; Is $x$ greater than 60?&lt;br&gt;&#xA;  &lt;strong&gt;Friend:&lt;/strong&gt; $x$ is greater than 30.&lt;br&gt;&#xA;  &lt;strong&gt;You:&lt;/strong&gt; Is $x$ greater than 75?&lt;br&gt;&#xA;  &lt;strong&gt;Friend:&lt;/strong&gt; $x$ is not greater than 60.&lt;br&gt;&#xA;  &lt;em&gt;etc.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your last guess would never be answered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the minimal number of moves $n$ to ensure you know the number your friend is thinking of, and what is the optimal strategy? Note that you don't have to specifically state $x$ as a guess.&lt;/p&gt;&#xA;" OwnerUserId="7308" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-01-31T17:44:11.277" Title="The Delayed Guessing Game" Tags="|mathematics|strategy|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7891" PostTypeId="2" ParentId="7881" CreationDate="2015-01-15T06:33:37.387" Score="20" Body="&lt;p&gt;We show that for $r$ red cards and $b$ black cards your chance of winning is $\frac{b}{r + b}$ and that this is true for all possible strategies. We do this by induction on $T = r + b$, the total number of cards left. This means we need to show that the chance to win is $\frac{b}{T}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $T = 1$ you must, according to the rules of the game, call &quot;bet&quot;. If $b = 1$, only a black card remains and your chance of winning is $1$. If $b = 0$, only a red card remains and your chance of winning is $0$. Your chance to win is $\frac{b}{T}$ in both cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the induction step, we assume that your chance to win with $b$ black cards remaining in $T - 1$ cards is $\frac{b}{T - 1}$ (for any $b \leq T - 1$). We calculate the chance to win with $b$ black cards remaining in $T$ cards (for any $b \leq T$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you choose to call &quot;bet&quot; now, your chance of winning is clearly $\frac{b}{T}$. If you wait for the next card instead then your chance of winning = (chance of red on next card $\times$ chance of winning with 1 less red) $+$ (chance of black on next card $\times$ chance of winning with 1 less black). The chances to win after removing a card are given by our induction hypothesis.&#xA;$$&#xA;\frac{r}{T} \times \frac{b}{T - 1} + \frac{b}{T} \times \frac{b - 1}{T - 1}\\&#xA;= \frac{rb + b(b - 1)}{T(T - 1)}\\&#xA;= \frac{b(r + b - 1)}{T(T - 1)}\\&#xA;= \frac{b(T - 1)}{T(T - 1)}\\&#xA;= \frac{b}{T}&#xA;$$&#xA;Regardless of whether you bet now or wait, your chance of winning is $\frac{b}{T}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This completes the induction, and so your chance of winning is always $\frac{b}{T} = \frac{b}{r + b}$. For the values given in the problem ($r = 26$, $b = 26$), this chance is $0.5$.&lt;/p&gt;&#xA;" OwnerUserId="8888" LastActivityDate="2015-01-15T06:33:37.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7892" PostTypeId="2" ParentId="7881" CreationDate="2015-01-15T06:42:05.153" Score="-2" Body="&lt;p&gt;Wait for him to drop almost all cards. Now there a three different possibilities:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;all 26 red cards are already on the table: you would have 100% chance to win since there are only (at least 2) black cards left.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;you can see that already 25 of your black cards are on the table at some earlier time: you have to wait for the end anyway because there is one black card among reds with a lower chance to hit. Not sure about the exact chance to win here.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;if there were one red and one black card left you would have to bet on the last card but one; giving you a 50% chance anyway - nothing to lose (Y). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5106" LastActivityDate="2015-01-15T06:42:05.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7893" PostTypeId="1" AcceptedAnswerId="7918" CreationDate="2015-01-15T06:45:11.007" Score="14" ViewCount="1552" Body="&lt;p&gt;A solar system contains some number of stationary perfectly-spherical planets of equal radius. Call a point on a planet's surface &lt;em&gt;private&lt;/em&gt; if it can't be seen from any other planet. Show that the total private surface area doesn't depend on how many planets there are and how they're arranged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is in 3D, the number of planets is finite and nonzero, and they are disjoint.)&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-01-15T07:19:11.100" LastActivityDate="2023-09-24T12:14:18.733" Title="Hidden planet area" Tags="|mathematics|geometry|" AnswerCount="2" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="7894" PostTypeId="2" ParentId="7893" CreationDate="2015-01-15T07:18:07.247" Score="10" Body="&lt;p&gt;What one planet can see from another is always a circle/half a planet with $$ A = 2\pi r^2$$&#xA;&lt;img src=&quot;https://i.stack.imgur.com/hAqJl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you add one planet to the system you will decrease the private surface on other planets by the same amount you add private area on the new planet. I think it is always one planet of private surface but I don't know how to express that mathematically.&lt;/p&gt;&#xA;" OwnerUserId="5106" LastEditorUserId="5106" LastEditDate="2015-01-15T07:24:00.787" LastActivityDate="2015-01-15T07:24:00.787" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7895" PostTypeId="2" ParentId="7783" CreationDate="2015-01-15T07:20:25.053" Score="7" Body="&lt;p&gt;You are in:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Angola, for you are Onze, a stream in Bengo province, Angola&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Divide my length by one letter&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The length of Onze is &quot;4 letters&quot;, divide by &quot;1 letter&quot; to get 4. Units cancel as per @Sunshine289&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then multiply by 11&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;4 x 11 = 44&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To get four times my meaning.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The meaning is therefore 11, and &quot;onze&quot; means 11 in Portuguese, the official language of Angola.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks to @Sunshine289&lt;/p&gt;&#xA;" OwnerUserId="8918" LastActivityDate="2015-01-15T07:20:25.053" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7896" PostTypeId="2" ParentId="7889" CreationDate="2015-01-15T07:52:54.737" Score="11" Body="&lt;p&gt;I can do it in 11 guesses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's generalize the range of numbers to be $N$ (instead of 100). I'll give a solution that lets you figure out $N$ in a number of guesses $g$ that's asymptotically&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ g = \log_{\phi}N +1 = \log_2 N/ \log_2(\phi)+1 \approx 1.44 \log_2 N + 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\phi$ is the golden ratio $1.618...$. Compare this to $\log_2 N$ for the standard guessing game, and $2 \log_2 N$ for the simple modified strategy where you make each guess twice to get around the time delay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start by making a guess that splits the $N$ possible values into parts of size $pN$ and $qN$, for fractions $p+q=1$ that we'll optimize later. For now, think of them as roughly $p \approx 1/3, q \approx 2/3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you don't get the answer until you make another guess. Let call this situation the &lt;em&gt;good state&lt;/em&gt;, where you're waiting on an answer that splits your interval into $p,q$ fractions. Now, you make a guess that splits the larger interval of size $qN$ into intervals of size $pqN$ and $q^2N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's two possibilities. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If the previous guess tells you the number is in the $qN$-size&#xA;interval, you're now back in the good state, but with an interval of&#xA;size $qN$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If it tells you the number is to the $pN$-size interval, your current guess is moot, which we call a &lt;em&gt;bad state&lt;/em&gt;, but the number is narrowed down to a smaller interval. But, just as when the game started, we'll make a guess that splits the interval into $p,q$-fractions, putting us in the good state of size $pN$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, we see that once we're in a good state with interval size $N$, we return to a good state of size &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$qN$ in one turns&lt;/li&gt;&#xA;&lt;li&gt;$pN$ in two turns&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, from a good state, every two turns decrease your interval size by an amount that's either $q^2$ or $p$. We get the best worst-case behavior if they are equal, and since $p=1-q$, we can solve&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$q^2=1-q$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to get the inverse of the golden ratio&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$q = \frac{-1+\sqrt{5}}{2} = 1/\phi \approx 0.618$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since a guess is needed to get from the initial bad state to a good state, after $g$ guesses, the size of the interval is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$N/\phi^{g-1}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, it takes $g = \log_{\phi}N+1$ guess to get down to an interval of size 1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, I've been pretending you can guess real numbers that are no necessarily integers. But that's fine, since one can always asks the for integers that it implies (Whether $x \geq 10$ for whether $x &amp;gt; 9.63\dots$), and pretend it's an answer to your question. The valid integers are contained in your interval, so once you get down to an interval of size $1$, it can only contain one integer and you're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This give $g \approx 10.57$ for $N=100$, so 11 guesses suffice. Compare this to 14 guesses for the naive strategy.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-01-15T10:25:53.013" LastActivityDate="2015-01-15T10:25:53.013" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7897" PostTypeId="2" ParentId="7882" CreationDate="2015-01-15T10:12:37.437" Score="6" Body="&lt;p&gt;Consider an arbitrary system of non-negative integer weights.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeat&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If all weights are even, then divide each weight by 2. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If all weighst are odd, then subtract 1 from each weight. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;until (at least one weight is even and at least one weight is odd)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then remove one weight (odd or even) so that the sum of the remaining weights is odd.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-01-15T10:12:37.437" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7898" PostTypeId="2" ParentId="376" CreationDate="2015-01-15T15:46:52.277" Score="7" Body="&lt;p&gt;The solution is equivalent to solving the $N+1$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Diophantine_equation&quot; rel=&quot;noreferrer&quot;&gt;Diophantine equations&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B$ number of bananas &lt;br&gt;&#xA;$n$ number of pirates &lt;br&gt;&#xA;{$a_1, a_2, .... a_n$} how many banana get each pirate. &lt;br&gt;&#xA;$m$  the banana for the monkey (for this case is 1).&#xA;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$B	=	na_1+m$ &lt;br&gt;&#xA;  $B-a_1-m	=	na_2+m$ &lt;br&gt;&#xA;  $B-a_1-a_2-2m	=	na_3+m$    &lt;br&gt;&#xA;  $\vdots$ &lt;br&gt;&#xA;  $B-a_1-a_2-a_3-...-a_n-nm	=	na+m$    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which can be rewritten as&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$B	=	na_1+m	$&lt;br&gt;&#xA;  $(n-1)a_1	=	na_2+m	$&lt;br&gt;&#xA;  $(n-1)a_2	=	na_3+m	$&lt;br&gt;&#xA;  $\vdots$ &lt;br&gt;&#xA;  $(n-1)a_{n-1}	=	na_n+m	$ &lt;br&gt; &#xA;  $(n-1)a_n	=	na+m.$&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since there are n+1 equations in the n+2 unknows  you can found all solutions as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B=kn^{n+1}-m(n-1), $&#xA;where k is an arbitrary integer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so for k = 1 and m = 1 &lt;br&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$B=n^{n+1}-(n-1) $&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note in this case the next morning after all pirate wake up will be enough bananas so each  pirates get an even number with 1 banana remaining for the monkey.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because that wasn't a requirement of the puzzle a small value of banana will still solve the problem.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$B=n^{n}-(n-1) $&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Case n = 4 &lt;br&gt;&#xA;$B_a$ = 253 &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(1) take 63 monkey 1 remaining 189&lt;br&gt;&#xA;  (2) take 47 monkey 1 remaining 141&lt;br&gt;&#xA;  (3) take 35 monkey 1 remaining 105&lt;br&gt;&#xA;  (4) take 26 monkey 1 remaining 78&lt;br&gt;&#xA;  Next morning every pirate get 19 bananas and monkey get 2 bananas&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$B_b$ = 1021 &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(1) take 255 monkey 1 remaining 765&lt;br&gt;&#xA;  (2) take 191 monkey 1 remaining 573&lt;br&gt;&#xA;  (3) take 143 monkey 1 remaining 429&lt;br&gt;&#xA;  (4) take 107 monkey 1 remaining 321&lt;br&gt;&#xA;  Next morning every pirate get 80 bananas and monkey get 1 banana&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5165" LastEditorUserId="5165" LastEditDate="2015-01-15T18:29:58.287" LastActivityDate="2015-01-15T18:29:58.287" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7900" PostTypeId="1" AcceptedAnswerId="7904" CreationDate="2015-01-15T17:33:34.840" Score="40" ViewCount="4076" Body="&lt;p&gt;Suppose I have in my possession a incredibly large (almost infinitely large!) bag of M&amp;amp;Ms containing a uniform distribution of the 6 M&amp;amp;M colors. I'm bored, so I decide to play a game:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I draw one M&amp;amp;M from the bag and place it on the table. I then continue to draw M&amp;amp;M's from the bag one at a time. If they are the same color as one already on the table, I eat both of them. Otherwise, I place it on the table with the others of different color. The game ends when I have all 6 distinct colors on the table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many M&amp;amp;M's should I expect to eat playing this game?&lt;/p&gt;&#xA;" OwnerUserId="8925" LastActivityDate="2015-01-16T01:47:16.843" Title="The &quot;M&amp;M Sugar Rush&quot; game" Tags="|mathematics|probability|" AnswerCount="3" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="7901" PostTypeId="2" ParentId="7900" CreationDate="2015-01-15T18:28:44.270" Score="4" Body="&lt;p&gt;If there are six candies on the table, the expected number of additional candies to eat will be zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are five on the table, the expected number will be 5/6 of (two plus whatever the expected number would be with four), plus 1/6 of the expected number with six.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are four on the table, the expected number will be 4/6 of (two plus whatever the expected number would be with three) plus 2/6 of the expected number with five.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are three on the table, the expected number will be 3/6 of (two plus whatever the expected number would be with two) plus 3/6 of the expected number with four.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are two on the table, the expected number will be 2/6 of (two plus whatever the expected number would be with one) plus 4/6 of the expected number with three.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is one on the table, the expected number will be 1/6 of (two plus whatever the expected number would be with none) plus 5/6 of the expected number with two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are none on the table, the expected number will be the same as the expected number with one&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one views the expected number of additional candies to eat when there are zero, one, two, etc. up to six candies on the table as being seven variables, the above will define a system of seven equations and seven unknowns (note that the &quot;variable&quot; for six is simply equal to zero, and the one for zero is equal to that of one, so if desired two variables and two equations may be omitted).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the above equations, it's easy to determine the number of candies that would need to be eaten with one candy on the table as being a linear function of the number of candies that would be consumed if there were two, then the determine the number with two as a linear function of the number that would be consumed if there were three, etc. up to five.  Since the value at six is known (zero), that that means that the values for five, four, three, etc. can all be computed.&lt;/p&gt;&#xA;" OwnerUserId="1467" LastActivityDate="2015-01-15T18:28:44.270" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7903" PostTypeId="1" AcceptedAnswerId="7913" CreationDate="2015-01-15T18:59:20.927" Score="13" ViewCount="2614" Body="&lt;p&gt;There are 15 person with different weights.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the least number of tries required to find the middle (median) weight person? How?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only tools you have is &lt;em&gt;your brain&lt;/em&gt; and a &lt;em&gt;balance scale&lt;/em&gt; with no limitation on area or weight. (you can put as many people you like on each side of the balance scale).&lt;/p&gt;&#xA;" OwnerUserId="2225" LastEditorUserId="8874" LastEditDate="2015-01-31T12:30:35.167" LastActivityDate="2023-06-30T17:06:17.010" Title="Using a balance scale the minimum number of times, find the median weight person" Tags="|logical-deduction|strategy|weighing|" AnswerCount="7" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="7904" PostTypeId="2" ParentId="7900" CreationDate="2015-01-15T19:35:41.193" Score="34" Body="&lt;p&gt;Let $S_n$ be the state where we have $n$ candies out on the table. We want to find the expected cost in eaten candies to advance from state $S_n$ to $S_{n+1}$. (This may, by chance, involve us having to move back by eating candies before moving forward again.) Let this expected value be $\Delta_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta_0 = 0$ since from $S_0$ (0 candies out) you will always draw a single distinct candy out of the bag to get to $S_1$ and so no candies will be eaten.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now assume we know $\Delta_{n-1}$, can we find $\Delta_n$? For this we want to know the expected cost to get from $S_n$ to $S_{n+1}$. Well there's a $\frac{n}{6}$ chance of drawing a match, eating 2 candies and moving back to state $S_{n-1}$. (The remaining fraction of the time there's no match and we advance to $S_{n+1}$ without eating a candy and so won't be needed for our calculations.) Now in state $S_{n-1}$ we know the expected cost to return to $S_{n}$ is $\Delta_{n-1}$ (assumed already calculated). This gives us the following equation:&#xA;$$\begin{align*}&#xA;&amp;amp;\Delta_n = \frac{n}{6} \times (2 + \Delta_{n-1} + \Delta_n)\\&#xA;\implies &amp;amp;6\Delta_n = n(2 + \Delta_{n-1}) + n\Delta_n\\&#xA;\implies &amp;amp;(6-n)\Delta_n = n(2 + \Delta_{n-1})\\&#xA;\implies &amp;amp;\Delta_n = \frac{n}{6-n}(2 + \Delta_{n-1})&#xA;\end{align*}$$&#xA;This formula now allows us to calculate the values for $\Delta_n$:&#xA;$$\begin{alignat}{2}&#xA;\Delta_0 &amp;amp;= 0 &amp;amp;=&amp;amp; 0\\&#xA;\Delta_1 &amp;amp;= \frac{1}{5}(2 + 0) &amp;amp;=&amp;amp; \frac{2}{5}\\&#xA;\Delta_2 &amp;amp;= \frac{2}{4}(2 + \frac{2}{5}) &amp;amp;=&amp;amp; \frac{6}{5}\\&#xA;\Delta_3 &amp;amp;= \frac{3}{3}(2 + \frac{6}{5}) &amp;amp;=&amp;amp; \frac{16}{5}\\&#xA;\Delta_4 &amp;amp;= \frac{4}{2}(2 + \frac{16}{5}) &amp;amp;=&amp;amp; \frac{52}{5}\\&#xA;\Delta_5 &amp;amp;= \frac{5}{1}(2 + \frac{52}{5}) &amp;amp;=&amp;amp; 62\\&#xA;\end{alignat}$$&#xA;The expected cost in candies eaten to get from $S_0$ (start state) to $S_6$ (terminal state) is:&#xA;$$\begin{alignat}{2}&#xA;&amp;amp;\Delta_0&amp;amp;+ &amp;amp;\Delta_1&amp;amp;+ &amp;amp;\Delta_2&amp;amp;+ &amp;amp;\Delta_3&amp;amp;+ &amp;amp;\Delta_4&amp;amp;+ &amp;amp;\Delta_5\\&#xA;= &amp;amp;0 &amp;amp;+ &amp;amp; \frac{2}{5} &amp;amp;+ &amp;amp; \frac{6}{5} &amp;amp;+ &amp;amp; \frac{16}{5} &amp;amp;+ &amp;amp; \frac{52}{5} &amp;amp;+ &amp;amp; 62&#xA;\end{alignat}$$&#xA;Which gives an expected number of candies eaten of 77.2.&lt;/p&gt;&#xA;" OwnerUserId="8888" LastEditorUserId="8888" LastEditDate="2015-01-15T19:52:09.357" LastActivityDate="2015-01-15T19:52:09.357" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="7905" PostTypeId="2" ParentId="7903" CreationDate="2015-01-15T19:43:53.647" Score="0" Body="&lt;p&gt;It's a pretty poor answer, but in the worst case scenario I have $63$ tries. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick a person. Weigh them against others until you get 8 greater or 8 lesser. Now, imagine that as you are comparing them against your picked person, you're filling a series of 15 slots, and each lighter person takes the left-most spot, and the heavier person takes the right-most spot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As soon as someone sits in the middle spot, that person and their group (heavier or lighter) go back into the pool, the picked person takes the next spot for the other group, and every spot taken is locked. Repeat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the worst case scenario, the first 7 people you pick are all on the light side. That's 56 tries. Then it'll take you another 7 to find the lightest of the remainder. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-01-15T19:43:53.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7907" PostTypeId="2" ParentId="7903" CreationDate="2015-01-15T20:12:21.167" Score="-2" Body="&lt;p&gt;You can weigh 2 and find the difference of the 1 vs 1 sides, and then use that towards the next one when you continue to weigh against him with more people, continue this process until you can compare them all to the 1 person, successfully determining their &quot;weight&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have now compared 14 people and determined their weight in comparison to the one's unknown weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, we can calculate this one whom we compared to by using the 2 people who weighed closest (the one closest yet greater, and the one closest yet lighter) to effectively rate their weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives us a mere&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 14 weighs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2015-01-15T20:29:27.193" LastActivityDate="2015-01-15T20:29:27.193" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="7908" PostTypeId="2" ParentId="7903" CreationDate="2015-01-15T20:20:11.957" Score="2" Body="&lt;p&gt;This doesn't take advantage of the possibility of weighing multiple people against each other at the same time, but a variation of the Quicksort algorithm, but only sorting half of the partitions, could work.&#xA;Purely using quicksort to completely sort the 15 people would take on average $15 log_2 15 = 58.6$ comparisons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eyeball the people and pick someone who looks like they might be the middle weight person to use as the first pivot.  Compare the other 14 people against him and sort them into lighter and heavier groups.  If the groups are the same size, you're done!  Else, you'll need to repeat the process with the larger remaining group, since that group must contain the middle weight person.&#xA;So the best case scenario is 14 comparisons.  In the worst case you alternately pick the lightest and heaviest people and have to make 105 comparisons.  For the average case, I'd surmise that since you are effectively sorting half of the people, it would take $8 log_2 8 = 24$ comparisons.&lt;/p&gt;&#xA;" OwnerUserId="239" LastActivityDate="2015-01-15T20:20:11.957" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7909" PostTypeId="2" ParentId="7903" CreationDate="2015-01-15T20:30:20.637" Score="4" Body="&lt;p&gt;Seeing as we have no information about the people's weights other than that they are all different, I don't think it will ever be useful to weigh more than two people at a time. Since we're only making one on one comparisons, I expect that standard sorting algorithms would be the way to do it the fastest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Merge_sort&quot; rel=&quot;nofollow&quot;&gt;mergesort&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Group the people into groups of 4 and one group of 3. Then for each group of 4 do the following, calling the people A, B, C, and D:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Weigh A against B and C against D (2 weighings)&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Without_loss_of_generality&quot; rel=&quot;nofollow&quot;&gt;WLOG&lt;/a&gt; let A and C be the lighter ones&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Weigh A against C&#xA;&lt;ul&gt;&#xA;&lt;li&gt;WLOG let A be lighter. We now know who the lightest of the four is (A)&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Weigh B against C&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If B is lighter, we know A &amp;lt; B &amp;lt; C &amp;lt; D in 4 weighings&lt;/li&gt;&#xA;&lt;li&gt;If C is lighter, we need to weigh B against D to know A &amp;lt; C &amp;lt; B &amp;lt; D or A &amp;lt; C &amp;lt; D &amp;lt; B&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So in a max of 5 weighings, we can sort each of the groups of 4. For the group of three&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Weigh A against B, then B against C&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If A &amp;lt; B &amp;lt; C or A &gt; B &gt; C, we're done.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Weigh A against C&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So in a max of 3 weighings we can sort the group of 3. Now we have 4 groups where we know the order of their weights. Weighings so far - up to 18. Now we need to merge pairs of groups:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;While either group still has at least one person, weigh the two lightest remaining people against each other&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The lighter of the two is removed from their group&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The order in which the people are removed from their groups is their order in the merged group. This can take up to 7 weighings for merging the groups of 4 and 6 for merging a group of 4 and the group of 3. Total weighings so far - up to 31.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can merge the two remaining groups to find the median-weighted person. We do the same thing as we did before, but this time after we have removed 8 people from their groups, we know who the median-weighted person is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Worst case scenario - we have to do 39 weighings.&lt;/p&gt;&#xA;" OwnerUserId="1955" LastActivityDate="2015-01-15T20:30:20.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7910" PostTypeId="2" ParentId="7903" CreationDate="2015-01-15T21:00:34.097" Score="3" Body="&lt;p&gt;I believe that a &quot;single-elimination tournament&quot; strategy gets us down to $35$ weighings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Step 1) Run a balanced four-round single-elimination tournament to find the lightest person. Eliminate them. We've used 14 weighings so far. We're going to repeat this 7 more times, each time using partial information from the previous rounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Step 2) Due to the structure of the tournament, the lightest person &quot;won&quot; against (at most) 4 other people. One of these four other people is the lightest person among those remaining. Also, these four other people have won (at most) 0, 1, 2, and 3 matches respectively. Play 0 against 1, then play the winner against 2, then play the winner of that against 3. The final winner is the next lightest person; eliminate them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Step 3) The final winner in Step 2 has now won at most four matches. I believe that by enumerating cases, one can show that the losers of these matches have won at most 3, 2, 1, and 0 matches themselves. Assuming I enumerated the cases properly, one can then repeat Step 2 until eight people are eliminated. The eighth person to be eliminated is the one of median weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All in all, this procedure takes 14 weighings to initialize, and then 3 weighings for each of the next 7 steps. So 35 in all.&lt;/p&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2015-01-15T21:00:34.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7911" PostTypeId="2" ParentId="7846" CreationDate="2015-01-15T21:04:01.747" Score="2" Body="&lt;p&gt;Aside from everything Leslie has said, the only thing I can think of to add is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Built close to each other,&#xA;what fits in between?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Out of the TRS Leslie found, the letter T is the only one that fits inside the word &quot;between&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What fits in between&#xA;has been found before&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Leslie found T in his/her solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But one was different,&#xA;and is waiting for more&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since Leslie's solution was TRS, and the riddle was specifically looking for T, it's different. It's &quot;waiting for more&quot; because it wants to be noticed in the word &quot;between&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's not only a symbol&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Both &quot;T&quot; and &quot;t&quot; are used as various symbols, marking perpendicular lines and the Christian faith.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but like a small lagoon&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Noooooo clue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rising from the dead&#xA;every afternoon&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure? Maybe the fact that &quot;T&quot; isn't present in &quot;morning&quot;, but reappears each day in the &quot;afTernoon&quot; and stays throughout the &quot;nighT&quot;. Ooooor it's referring to &quot;afternoon tea&quot;...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bobbee's original comment was something like&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;There's going to be golden rain in two days.&quot; -- &quot;Golden rain&quot; referring to &quot;ti&quot;, a drop of golden sun, and &quot;in two days&quot; being literal -- &quot;in Two days&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Other potential clues that probably aren't clues but are most likely just me being paranoid about everything being a clue... clues.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The title, twofold. &quot;Tear 'it' apart&quot;. If you take it apart, you're left with an I and a T, or literally the phrase, &quot;I, T&quot; a la &quot;I, Robot&quot;. Also, &quot;One more time&quot; -- the letter T is often used to denote (or countdown) time. So again, referencing T.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;-&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;The good disguise wins, so you act as twins&quot;. Aside from somehow (though I'm not 100% sure how) pointing to U, this could also be saying that &quot;wins&quot; is a good disguise for T, like when used in &quot;twins&quot;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;-&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Two towers of letters&quot; could refer to the &quot;two towers&quot; in the middle of the word &quot;leTTers&quot;, although I'm not sure, since the &quot;l&quot; also looks like a tower. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Actually, that said, &quot;Tear it apart&quot; to get T, then &quot;one more time&quot; to get another T... Along with the repeated &quot;Two Towers&quot; references, the fact that it's &quot;waiting for more...&quot; That could mean that the answer is actually TT? (&quot;TT&quot; does &quot;fit in between&quot; the &quot;leTTers&quot; too) And, incidentally, &quot;TT&quot;could be construed as looking somewhat like a lagoon. &lt;em&gt;shrug&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Edit #2...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uh, ok so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two towers of letters, what fits in between?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Letters?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What fits in between has been found before. But one was different and is waiting for more.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I can't tell you how many letters we've found already. &quot;T&quot; is different in that it occurs twice in the word &quot;letters&quot; (albeit so does E, but it wasn't in the RST tower). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's not only a symbol but like a small lagoon. Rising from the dead every afternoon.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This just goes on to further describe which of the letters is &quot;different&quot;. So yeah, my final guess is &quot;LETTERS&quot;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If that's not it, then I'll wait for someone else to get it so I can punch myself. :D&lt;/p&gt;&#xA;" OwnerUserId="8858" LastEditorUserId="8858" LastEditDate="2015-01-15T21:59:27.750" LastActivityDate="2015-01-15T21:59:27.750" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="7912" PostTypeId="2" ParentId="7903" CreationDate="2015-01-15T21:05:26.993" Score="3" Body="&lt;p&gt;Pick a person. Weight that person against all other 14. This splits the world into 3 sections:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The heavier (x)&lt;/li&gt;&#xA;&lt;li&gt;That person&lt;/li&gt;&#xA;&lt;li&gt;The lighter (y)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Take the group that must include the middle weight, repeat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best case scenario, 14 weighings, x == y at first attempt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Worst case scenario, picking successively heaviest and lightest persons, 14+13+12+11...+1 = 105.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is a partial quicksort with the worst choice of pivot.)&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Alternative solution (selection sort):&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Eliminate the heaviest at each round, 8 rounds needed, 14+13+12+...7 = 84&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Third solution:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Insertion sort with only 8 spots (last is the middle person) and binary search for inserting. 8 first insertions require at worst 17 weighings, 7 lasts at worst 28. Total is 45. (best case is 34).&lt;/p&gt;&#xA;" OwnerUserId="310" LastActivityDate="2015-01-15T21:05:26.993" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7913" PostTypeId="2" ParentId="7903" CreationDate="2015-01-15T21:32:04.163" Score="8" Body="&lt;p&gt;Here's 31 weighings. I'll only use one-vs-one weighings.&lt;/p&gt;&#xA;&lt;p&gt;One can &lt;a href=&quot;https://stackoverflow.com/a/1534932&quot;&gt;sort 5 elements in 7 comparisons&lt;/a&gt;. Split the people into 3 groups of 5 and sort each one (21 weighings). Sort the three medians of the three groups by comparing each pair (3 weighings). Call these groups the light, medium, and heavy group by their sorting.&lt;/p&gt;&#xA;&lt;p&gt;Now, note that the median of the light group has at least eight people heavier than them: the two heavier people in their group, and the three median-and-above people in each of the other two groups. That means they are below median, as are the two below-median people in the light group. Likewise, we know three above-median people.&lt;/p&gt;&#xA;&lt;p&gt;We can therefore eliminate these six people from considering, and find the median of the nine remaining people, which are sorted into a group of five and two-groups of two.&lt;/p&gt;&#xA;&lt;p&gt;Note that we can merge two sorted groups of size &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; using &lt;span class=&quot;math-container&quot;&gt;$a+b-1$&lt;/span&gt; weighings by comparing the two heaviest people in each group, removing the heavier one as the maximum, and repeating. So, one can merge the two sorted lists of two into a sorted list of four in 3 weighings.&lt;/p&gt;&#xA;&lt;p&gt;At this point, we've used 27 weighings, and could spend 8 weighing to merge the two groups of 4 and 5 and have the nine sorted, for a total of 35.&lt;/p&gt;&#xA;&lt;p&gt;But, that's overkill because we just need the median of the nine. Compare the median of the group of 5 and the median of the group of 4, rounding down for the four (1 weighing). Whichever way the result is, the two heavier people of the median-heavier group are above median, and likewise for the lighter people. Eliminating these, we're left with two sorted people and three sorted people.&lt;/p&gt;&#xA;&lt;p&gt;Again, we could merge-sort in 4 weighings, but we instead do it in 3. Compare the median of the three to each of the two (2 weighings). If the results are opposite, the median of the three is the median, and we're done. Otherwise, say WLOG both are lighter. Compare the heavier of the two to the lightest of the three (1weighing). If he's lighter, the lighter of three is the median. Otherwise, he's the median, as we know two people lighter than him and two people heavier.&lt;/p&gt;&#xA;&lt;p&gt;That's a total of 31 = 21 + 3 + 3 + 4 weighings.&lt;/p&gt;&#xA;&lt;p&gt;[A &lt;a href=&quot;http://www.cs.hut.fi/%7Ecessu/selection/&quot; rel=&quot;nofollow noreferrer&quot;&gt;computer search&lt;/a&gt; by Kenneth Oksanen found that the optimal number of comparisons (one-vs-one weighings) is between 24 and 28 inclusive, though an exact answer is not known.]&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2023-06-30T17:06:17.010" LastActivityDate="2023-06-30T17:06:17.010" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="7914" PostTypeId="1" AcceptedAnswerId="7917" CreationDate="2015-01-15T21:44:34.010" Score="9" ViewCount="460" Body="&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; - Added hint to help decode the color tiles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are back on Tatooine with R4-D2 (a descendant of R2-D2). During the journey to Herogga's pavilion, you encounter a fierce sandstorm. Unfortunately, some of the fine particles have infiltrated R4-D2's drive units.  Serious damage will occur unless the droid is disassembled and cleaned. You know this will not be easy so you look for some guidance and find three &lt;strong&gt;equations&lt;/strong&gt;, a small &lt;strong&gt;crossword&lt;/strong&gt; puzzle, and some &lt;strong&gt;encoding&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As your mission is urgent, you proceed to understand the clues and successfully disassemble R4-D2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best solution to this puzzle will be the &lt;strong&gt;first to show a correct diagram&lt;/strong&gt; of the droid's components.  &lt;strong&gt;Partial answers are OK&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The clue to all of the words in the crossword is &lt;strong&gt;four&lt;/strong&gt;. The equations and crossword will tell you the number of components and their relative shapes. The encoding provides a little more description of the shapes but is not necessary to solve the disassembly.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonne chance!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT - The encoding sequence was English letters to Base 4 numbers to CMYK colors.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/boUSg.png&quot; alt=&quot;R4-D4&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-01-16T21:59:35.583" LastActivityDate="2015-01-17T21:07:13.603" Title="The Dissection of R4-D2" Tags="|dissection|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7915" PostTypeId="2" ParentId="7900" CreationDate="2015-01-16T01:47:16.843" Score="31" Body="&lt;p&gt;This is a perfect opportunity to use the theory of &lt;a href=&quot;http://en.wikipedia.org/wiki/Markov_chain&quot;&gt;Markov Chains&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The states are the number of candies currently on the table (either 0, 1, 2, 3, 4, 5, or 6 candies).  If all 6 candies are present, then the game is over (this is called an &quot;absorbing state&quot;).  Otherwise, if there are $n$ candies on the table, then the probability of matching one of them and thus having $n-1$ on the table at the  next step is $n/6$, while the probability of drawing a new color and thus having $n+1$ candies on the table is $1 -n/6$.  So the transition probabilities are described by the following $7\times 7$ matrix:&#xA;$$&#xA;P = \begin{pmatrix} 0 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0\\&#xA;1/6 &amp;amp; 0 &amp;amp; 5/6 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0\\&#xA;0 &amp;amp; 2/6 &amp;amp; 0 &amp;amp; 4/6 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0\\&#xA;0 &amp;amp; 0 &amp;amp; 3/6 &amp;amp; 0 &amp;amp; 3/6 &amp;amp; 0 &amp;amp; 0\\&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 4/6 &amp;amp; 0 &amp;amp; 2/6 &amp;amp; 0\\&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 5/6 &amp;amp; 0 &amp;amp; 1/6\\&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1\\&#xA;\end{pmatrix}&#xA;$$&#xA;(Note that if there are $6$ candies on the table, then the game is over. You can never transition away from $6$ candies into restarting the game, so the absorbing state just steps to itself with probability $1$. This explains why there's a $1$ in the lower right corner of the matrix $P$; the last row and column of $P$ represent the absorbing state of $6$ candies.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we follow &lt;a href=&quot;http://en.wikipedia.org/wiki/Absorbing_Markov_chain#Expected_number_of_steps&quot;&gt;http://en.wikipedia.org/wiki/Absorbing_Markov_chain#Expected_number_of_steps&lt;/a&gt; and let $Q$ be the matrix $P$ excluding the absorbing states (so $Q$ is $P$ except for the last row and column). We calculate $N = (I-Q)^{-1}$, where $I$ is the $6\times 6$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Identity_matrix&quot;&gt;identity matrix&lt;/a&gt;.  The expected number of steps until absorption, starting at the &quot;0 candies&quot; state, will be the first entry of the vector $N Z$, where $Z$ is the length-6 column vector of all $1$'s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that the expected number of steps until absorption is 83.2.  At absorption, there are 6 candies left uneaten on the table, so that means you expect to eat 77.2 candies.&lt;/p&gt;&#xA;" OwnerUserId="8930" LastActivityDate="2015-01-16T01:47:16.843" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="7916" PostTypeId="2" ParentId="7914" CreationDate="2015-01-16T06:29:54.477" Score="5" Body="&lt;p&gt;Only a partial answer here, based on the crossword clue.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The crosswords are &quot;four&quot; in different languages: &lt;code&gt;vier&lt;/code&gt; (German) and &lt;code&gt;quatre&lt;/code&gt; (French) across, and &lt;code&gt;four&lt;/code&gt; (English) and &lt;code&gt;tessara&lt;/code&gt; (Greek) down. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4373" LastActivityDate="2015-01-16T06:29:54.477" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7917" PostTypeId="2" ParentId="7914" CreationDate="2015-01-16T08:10:36.007" Score="6" Body="&lt;p&gt;&lt;strong&gt;Edit of post:&lt;/strong&gt; I start with the solution&lt;/p&gt;&#xA;&lt;p&gt;Finding the dissection was fiendishly difficult, but here it is:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/UHnM8.jpg&quot; alt=&quot;Solution&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Now as for the other clues or puzzles:&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;The equations&lt;/strong&gt; were rather straight forward:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; One can easily see that they build a chain, in the end being written in one line as A≅B≅C≅D=14 which means that we seek 4 congruent tiles a 14 squares. This matches, as the whole puzzle has 4x4=56 squares. Also &amp;quot;4&amp;quot; is the solution of the crossword puzzle and also appears as single number. So now we know how we have to dissect.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;The crossword&lt;/strong&gt; was first solved by &lt;strong&gt;shoover&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The 4 words are all literally &amp;quot;4&amp;quot; being written in 4 different languages: &lt;code&gt;vier&lt;/code&gt;, &lt;code&gt;quatre&lt;/code&gt;, &lt;code&gt;four&lt;/code&gt; and &lt;code&gt;tessera&lt;/code&gt;. (German, French, English and Greek, respectively). The Greek &lt;code&gt;tessera&lt;/code&gt; was written with Greek letters ( &lt;code&gt;τέσσερα&lt;/code&gt; ) &lt;br&gt;&#xA; The &lt;em&gt;purpose&lt;/em&gt; of the crossword puzzle is somewhat illusive, as 'four' was given as hint and is the only information &lt;em&gt;gained&lt;/em&gt; by the puzzle. But it is a very nice sub-puzzle enforcing the fact that the dissection should happen into 4 parts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;The colour code&lt;/strong&gt; was first solved by &lt;strong&gt;Volatility&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Given the hint that base-4 encoding was used, i.e numbers are formed by:&lt;br&gt;&#xA; &lt;code&gt;### = A*0 + B*4 + C*16 + D*64&lt;/code&gt;&lt;br&gt;&#xA;  one could guess that the colours Black, Cyan, Mangenta and Yellow stand for 0, 1, 2, 3 and their position in the field for A,B,C,D. &lt;br&gt;&#xA; As numbers are used to represent letters, and standard ASCII letters use the number 65 to 90 for 'A' to 'Z', one could further conclude that the &amp;quot;D*64&amp;quot; needs to be D=1 for &lt;em&gt;all&lt;/em&gt; letters. This gives two important pieces of information:&lt;br&gt;&#xA; - The position which has the same colour (top-right) has to by &amp;quot;D&amp;quot; &lt;br&gt;&#xA; - The colour which is in this position has to stand for &amp;quot;1&amp;quot; (=Cyan) &lt;br&gt;&lt;br&gt;&#xA; From here, it is trial-and error to find one has to go 'clockwise' for D,C,B,A and the colours are 0,1,2,3 for Black, Cyan, Magenta and Yellow respectively. &lt;br&gt;&#xA; The fields then spell out &lt;code&gt;CHAIR&lt;/code&gt; and &lt;code&gt;STACK&lt;/code&gt; which nicely describes the graphical dissection.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;History (initial 2 postings and edits)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;h1&gt;Partial answer / observation.&lt;/h1&gt;&#xA;&lt;p&gt;This is a work-in-progress list of observations and possible leads in this puzzle.&lt;/p&gt;&#xA;&lt;p&gt;I'm posting, because the OP invites to do so. I also think that if one purpose of the site is to &lt;em&gt;teach&lt;/em&gt; puzzle-smiths how to improve, this is an important step, because it gives the puzzle-smith feedback on &lt;em&gt;what reactions his puzzle creates&lt;/em&gt;, even if they are not the intended ones. (So they can be recognized as wanted red herrings, or may be removed by better specification.)&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Observations:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Tile size and equations:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;strong&gt;There are 56 tiles and 4 &lt;em&gt;variables&lt;/em&gt; as well as the numbers 14 &amp;amp; 4.&lt;/strong&gt; The equations indicated proportionality or equality and 14*4 is indeed 56, leading to the assumption we have to dissect in 4 equally &lt;em&gt;sized&lt;/em&gt; but not equally &lt;em&gt;shaped&lt;/em&gt; tiles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The crossword:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; **has already been solved (I think) by &lt;em&gt;shoover&lt;/em&gt; above, with being &lt;code&gt;vier&lt;/code&gt;, &lt;code&gt;quatre&lt;/code&gt;, &lt;code&gt;four&lt;/code&gt; and &lt;code&gt;tessara&lt;/code&gt; all meaning 4. A nice twist is that Greek letters are used (sigma &amp;amp; rho) for the Greek &lt;code&gt;tessara&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The puzzle as a whole:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 1. There are indications of &lt;strong&gt;mirror operations&lt;/strong&gt; like the mirrored &amp;quot;r&amp;quot; in the top left.&lt;br&gt;&lt;br&gt;&#xA; 2.  There are also some (6) tiles which are graphics which seem to be puzzle-tiles of an image representing a R2D2 (or R4D2 ?) They are not complete. These might be &lt;em&gt;filler&lt;/em&gt; without meaning or they may come in later?&lt;br&gt;&lt;br&gt;&#xA; 3. There are 7 empty tiles. Maybe they are needed to counterbalance the bitmap tiles of point 2. ?&lt;br&gt;&lt;br&gt;&#xA; 4. Colours in the colour-tiles are the prime colours of subtractive colour mixing (&lt;em&gt;printing colours&lt;/em&gt;). which at the same time make them result-colours of additional colour mixing (&lt;em&gt;screen colours&lt;/em&gt;). It might therefore be, that they represent an &lt;em&gt;overlay&lt;/em&gt; of colours.&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Having spent a little more time on it, I&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;strong&gt;...am frustrated :c)&lt;/strong&gt;&lt;br&gt;&lt;br&gt;From the equation A=B=C=D=14 it seems that there should be 4 congruent shapes of 14 tiles which dissect the R4-D2. &lt;br&gt; If this is indeed the case, the other (colour) hints would not be needed, and one could simply attempt a regular 2D dissection puzzle into 4 congruent shapes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So far so good, but...&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; I simply can not find it, and having tested similar shapes &amp;quot;growing&amp;quot; I'm getting more and more suspicious that there is no such solution...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Which makes me wonder:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Am I making the wrong assumption tempted by the regular square grid? Do the shapes have to by polyominos? I'm no longer sure about that at all...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Or...&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Is this for sure only a 2D puzzle? Or are we dealing with a clever 2D projection of a 3D puzzle?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Taking the 2nd thought further:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Each section would have 14 squares - which fits to the surface area of any 3-cube part. (3x6 minus 3x2 border walls ) This would also nicely fit to the colour-bitmaps if they depict in a certain way such 3-cube L arrangements. Not sure how though. &lt;br&gt;&#xA; Would this work in 3D in principle? 3x4=12cubes would have a maximum surface area of 12x6 - 11x2 = 50 if they are touching one space each. Hmm...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-01-17T21:07:13.603" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7918" PostTypeId="2" ParentId="7893" CreationDate="2015-01-16T09:40:46.023" Score="18" Body="&lt;p&gt;Fix any direction and call it &quot;north.&quot; Look at the north poles of all planets. A north pole is private iff there are no planets further to the north. Therefore, there is exactly one private north pole: that of the northernmost planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, for any direction $X$, there is exactly one private $X$ pole. I claim that this means there is exactly one planet's worth of private area. Indeed, if we translate all of the planets' private parts onto a single reference planet, then the private parts combine to cover every point on the planet exactly once.&lt;/p&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2015-01-16T09:40:46.023" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7919" PostTypeId="2" ParentId="6467" CreationDate="2015-01-16T09:57:56.087" Score="2" Body="&lt;p&gt;So you have to survive the dice-roll, which turns out by counting, to be a $11/36$ chance of being kicked off, and then a $50\%$ chance of being kicked off by a coin toss. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other players have the same probabilities, and each person being kicked off is dependant only on those next to them. So it actually benefits everyone if there are fewer players purely since the method of removing oneself from the game is independent of the number of players.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However: The only meaningful (and likely) way to achieve &quot;The same result three times&quot; is if no-one is removed from the game. This is clearly more likely with fewer players, however impossible with an odd number of players; someone must leave because the ranges allowable on the dice roles alternate between $\{1,2,3\}$ and $\{4,5,6\}$, so there has to be an even number of people to achieve this; assuming the table is round etc. Therefore...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Since this affects every player, he should be asked to stay and play as this will improve your chances of winning due to the 3 consecutive plays rule -- 11 players as opposed to 10.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3897" LastActivityDate="2015-01-16T09:57:56.087" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7920" PostTypeId="2" ParentId="6691" CreationDate="2015-01-16T13:24:44.923" Score="9" Body="&lt;p&gt;&lt;strike&gt;15&lt;/strike&gt; 14 words:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Teal eat elastics specialist, steal plasticised silicates, deplasticise replasticised elastic, at a preplasticised castle.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A flock of wild duck go on a generally destructive rampage at an industrial science laboratory.)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A&#xA;aT&#xA;Eat&#xA;teaL&#xA;Steal&#xA;Castle&#xA;elastIc&#xA;elasticS&#xA;silIcates&#xA;sPecialist&#xA;plasticiseD&#xA;dEplasticise&#xA;Replasticised&#xA;Preplasticised&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Credit: adapted from &lt;a href=&quot;https://puzzling.stackexchange.com/a/7748/4234&quot;&gt;Psybin’s answer&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4234" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-01-24T12:06:27.640" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7921" PostTypeId="1" AcceptedAnswerId="7927" CreationDate="2015-01-16T16:07:18.090" Score="1" ViewCount="507" Body="&lt;p&gt;If a marker is located in one of the corners of a square cube and we begin to draw according to the following conditions, how many edges of the cube can we cross ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A) You can cross every edge only once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B) The marker must not be separated from the cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C) The marker can only cross the corners and edges of the cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Hm4DD.jpg&quot; alt=&quot;here is the image&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7555" LastEditorUserId="8874" LastEditDate="2015-01-31T13:46:53.947" LastActivityDate="2015-01-31T13:46:53.947" Title="Marker and the cube" Tags="|geometry|reachability|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7922" PostTypeId="2" ParentId="7921" CreationDate="2015-01-16T16:22:58.613" Score="6" Body="&lt;p&gt;The best you can do is 9. Let's number corners as 1-4 on top, 5-8 on bottom (&lt;code&gt;i&lt;/code&gt; is connected to &lt;code&gt;i+4&lt;/code&gt;). You can do:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(1-2), (2-3), (3-4), (4-1), (1-5), (5-6), (6-7), (7-8), (8-5)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which has 9 edges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this the best?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the cube is a graph with 12 edges and 8 vertices and each vertex has a degree of 3. Effectively this means that passing through a vertex &quot;removes&quot; 2 of that degree. The other 1 must be an end of a sequence of vertices and edges. Having 8 &quot;ends&quot; means that you must have 4 sequences of vertices and edges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As each sequence has at least one edge, by minimizing three of those sequences to a length of one edge, we get that the best we can do is 9 for the fourth sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a more theoretical approach to that problem have a &lt;a href=&quot;https://en.wikipedia.org/wiki/Eulerian_path&quot;&gt;look at this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="5610" LastEditDate="2015-01-16T16:39:51.703" LastActivityDate="2015-01-16T16:39:51.703" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="7923" PostTypeId="2" ParentId="7921" CreationDate="2015-01-16T16:31:49.987" Score="0" Body="&lt;p&gt;I think you can hit all 12 edges. Path (faces) is: top-left-back-top-right-back-bottom-right-front-bottom-left-front-top&lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-01-16T16:31:49.987" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7924" PostTypeId="2" ParentId="7868" CreationDate="2015-01-16T17:31:30.543" Score="4" Body="&lt;p&gt;This answer focuses on part (C) of the question.  See also the excellent answer by @Shyos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's write E for Ebrecedebre, I for Ibricidibri, and similarly for O and U.  Then we'll write e, i, o, u for the &quot;reversed&quot; operations. Following @Shyos, we'll also write B for Ball, C for Cone, R for Ring and J for Umbrella, so that we don't have U meaning two different things. (J is an excellent letter for an Umbrella!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each time you do E, it decreases the number of B's by 1, while increasing the number of C's by 1 and the number of R's by 1.  On the other hand, if you do an &quot;e&quot;, it increases the number of B's by 1 while decreasing C's and R's by 1.  Notice that E and e just cancel each other out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if at any point we did, say, 8 E's followed by 5 e's, the net result would be the same as doing (8-5) = 3 instances of E.  So we can count the E's as positive, and then each e is just a negative E.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In part (C), we need a total net increase in the R's by at least 5, and we need a total net decrease in the C's by exactly 1, and no net change in the B's or J's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's write $x,y, z, w$ to mean the &lt;em&gt;net&lt;/em&gt; number of applications of E, I, O, U respectively. (Note: since e, i, o, u count as negative, it's possible that $x,y,z,w$ could be negative; however, $x,y,z,w$ must all be integers.) Now the total net change in B's is:&#xA;$$&#xA;\Delta B = -x + 3z + w&#xA;$$&#xA;(Why? Well, remember that each application of E is going to decrease the number of B's by 1.  Meanwhile, each application of O will increase the number of B's by 3, and each application of U will increase the number of B's by 1.  So 3 times the number of O's, plus 1 times the number of U's, minus 1 times the number of E's, gives the total net change in the number of B's.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want $\Delta B = 0$, and $\Delta C = -1$ (since we start with 1 cone and want to end with 0), and similarly $\Delta J =0$.  Finally, $\Delta R$ can be any number $k$ as long as $k\geq 5$.  We now have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{eqnarray*}&#xA;0 &amp;amp; = &amp;amp; -x + 0y + 3z + w\\&#xA;-1 &amp;amp; = &amp;amp; x - y + 0z + 4w\\&#xA;k &amp;amp; = &amp;amp; x + 2y - z + 0w\\&#xA;0 &amp;amp; = &amp;amp; 0x + y + z - w\\&#xA;\end{eqnarray*}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or in matrix form:&#xA;$$&#xA;\begin{pmatrix}&#xA;0\\&#xA;-1\\&#xA;k \\&#xA;0 \\&#xA;\end{pmatrix}&#xA;=&#xA;\begin{pmatrix}&#xA;-1 &amp;amp; 0 &amp;amp; 3 &amp;amp; 1\\&#xA;1 &amp;amp; -1 &amp;amp; 0 &amp;amp; 4\\&#xA;1 &amp;amp; 2 &amp;amp; -1 &amp;amp; 0\\&#xA;0 &amp;amp; 1 &amp;amp; 1 &amp;amp; -1\\&#xA;\end{pmatrix}&#xA;\begin{pmatrix}&#xA;x\\&#xA;y\\&#xA;z \\&#xA;w \\&#xA;\end{pmatrix}&#xA;$$&#xA;We can solve this equation for $x,y,z,w$ in a variety of ways; for example, multiplying both sides by the inverse of the $4\times 4$ matrix.  No matter how you do it, the result is:&#xA;$$&#xA;\begin{pmatrix}&#xA;x\\&#xA;y\\&#xA;z \\&#xA;w \\&#xA;\end{pmatrix}&#xA;=&#xA;\begin{pmatrix}&#xA;-{3}/{4} - 2k/3\\&#xA;{1}/{4} + {2k}/{3}\\&#xA;-1/4 - k/3 \\&#xA;k/3 \\&#xA;\end{pmatrix}&#xA;$$&#xA;Since $x,y,z,w$ have to be integers, we see that $k$ must be divisible by 3 (from the last entry).  Also, from the first entry, the following must be an integer:&#xA;$$&#xA;-\frac{3}{4} - \frac{2k}{3} = \frac{-9 - 8k}{12},&#xA;$$&#xA;so $(8k+9)$ must be divisible by $12$.  This is impossible if $k$ is divisible by $3$, so there is no solution to part (C).&lt;/p&gt;&#xA;" OwnerUserId="8930" LastEditorUserId="8930" LastEditDate="2015-01-16T17:38:23.003" LastActivityDate="2015-01-16T17:38:23.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7925" PostTypeId="1" AcceptedAnswerId="7926" CreationDate="2015-01-16T18:04:40.697" Score="0" ViewCount="422" Body="&lt;p&gt;The driven ambitious with passionate zeal, &lt;br&gt;&#xA;Will step forward to discover the author’s reveal, &lt;br&gt;&#xA;With princes by men, &lt;br&gt;&#xA;In columns of ten, &lt;br&gt;&#xA;the experience the answer shall seal. &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;QBOKD KWLSDSYX SC DRO ZKCCSYX YP K QBOKD MRKBKMDOB. DRYCO OXNYGON GSDR SD WKI ZOBPYBW FOBI QYYN YB FOBI LKN KMDC. KVV NOZOXNC YX DRO ZBSXMSZVOC GRSMR NSBOMD DROW.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;KWLSDSYXMRKVVOXQO NYD GYBNZBOCC NYD MYW&lt;/p&gt;&#xA;" OwnerUserId="8943" LastEditorUserId="4" LastEditDate="2015-01-17T18:59:26.810" LastActivityDate="2015-01-17T18:59:26.810" Title="I need help on finding out what this means. I know there are 4 steps to it." Tags="|riddle|cipher|" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7926" PostTypeId="2" ParentId="7925" CreationDate="2015-01-16T18:45:14.437" Score="6" Body="&lt;p&gt;Easy. The text is a cryptogram. Plug it in &lt;a href=&quot;http://quipqiup.com/index.php&quot;&gt;here&lt;/a&gt;. It returns:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;GREAT AMBITION IS THE PASSION OF A GREAT CHARACTER. THOSE ENDOWED WITH IT MAY PERFORM VERY GOOD OR VERY BAD ACTS. ALL DEPENDS ON THE PRINCIPLES WHICH DIRECT THEM.  AMBITIONCHALLENGE DOT WORDPRESS DOT COM&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Go there. Password is &quot;Napoleon&quot; - the source of the quote.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can only assume this is part of a broader challenge, as the text there makes little sense to me and doesn't seem to be anything more than a contextual clue. Of course, I could be missing something, too.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastActivityDate="2015-01-16T18:45:14.437" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7927" PostTypeId="2" ParentId="7921" CreationDate="2015-01-16T20:36:23.350" Score="4" Body="&lt;p&gt;Depending on what the question means, the answer is either 9 or 12 edges per previous answers.  Here are two diagrams.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ftMdx.png&quot; alt=&quot;Cubes&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5188" LastActivityDate="2015-01-16T20:36:23.350" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7928" PostTypeId="2" ParentId="7914" CreationDate="2015-01-17T06:25:20.547" Score="4" Body="&lt;h1&gt;Partial answer&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Random symbols:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The symbols 4, 四, and Δ' all have a relation to the number 4: the first symbol is obvious, 四 is the Chinese character for 4, and Δ' in quadratic equations is Δ/4 (where Δ is the discriminant). These symbols (I believe) serve as a clue to the crossword puzzle. The backwards 'r' does seem to suggest some sort of mirroring, as @BmyGuest has stated.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Crossword:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As the answers above say, the answers are 'vier', 'tessera', 'quatre', and 'four', the words for the number 4 in German, Greek, French and English respectively.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Equations:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Put together, the equations say A≅B≅C≅D=14. The '≅' symbol is the standard symbol for congruency in geometry, hinting that A, B, C and D are congruent shapes. 14*4=56, which is the number of tiles, supporting the idea of congruency.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Encoding:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since @Len was gracious enough to give us a hint, all I had to do was figure out which numbers the colours stood for, and what order they were in. Because the ordinal numbers for 'A' to 'Z' were 65 to 90, I figured that the first number would always be 1. The corner with the same colour in each symbol was the top right, with cyan. Using that, I guessed that C=1, M=2, Y=3, and K=0. Then it was a case of trial and error figuring out what order they were in. This worked out to be a clockwise &quot;circle&quot; starting from the top right, yielding the message &quot;CHAIR STACK&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The graphics and empty tiles:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Haven't figured those out yet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8949" LastActivityDate="2015-01-17T06:25:20.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7929" PostTypeId="1" CreationDate="2015-01-17T07:58:33.713" Score="18" ViewCount="2747" Body="&lt;p&gt;&lt;strong&gt;UPDATE 4&lt;/strong&gt; - Inserted original list of questions (no more, no less)&lt;/p&gt;&#xA;&lt;p&gt;Could someone help me solve this code? I've tried for 35 years.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;13 25 22 19 9 30 32 11&lt;/strong&gt;&lt;br /&gt;&#xA;&lt;strong&gt;12 25 8 29 17&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;It is a place name in Kirkcudbrightshire Scotland, perhaps something like a castle or a tower or bridge or monument. This is question 4, as shown on the following sheet. The answers that are shown include National Grid reference numbers for each location.&lt;/p&gt;&#xA;&lt;p&gt;When I was so frustrated I requested another clue of the author. Here are the three additional ciphers that I received:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;19 9 25 15 22 29 30 13&lt;/strong&gt;&lt;br /&gt;&#xA;&lt;strong&gt;14 26 10 17 24 28 25 28 16 34 27 27&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;27 12 26 9 20 31 14&lt;/strong&gt;&lt;br /&gt;&#xA;&lt;strong&gt;20 14 25 29&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;19 25 13 30 17 24&lt;/strong&gt;&lt;br /&gt;&#xA;&lt;strong&gt;7 17 29 15 31 12 22&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If it helps, &lt;a href=&quot;http://www.kirkcudbright.co/index.asp&quot; rel=&quot;noreferrer&quot;&gt;here is a website&lt;/a&gt; that shows about 8000 named places in Kirkcudbrightshire.  These named places are listed within each of the 28 parishes.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zt3iU.png&quot; alt=&quot;Questions&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Update&lt;/strong&gt;&lt;/em&gt;: OP says correct answer for Q3 is more likely to be NX 64454 74089.&lt;/p&gt;&#xA;" OwnerUserId="8951" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-10-02T10:51:54.197" Title="Whisky treasure-hunt: Cryptogram which refers to a place in Scotland" Tags="|cipher|treasure-hunt|" AnswerCount="7" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="7930" PostTypeId="2" ParentId="7929" CreationDate="2015-01-17T09:55:23.423" Score="12" Body="&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; - Apparently the answer is &quot;dead easy&quot; which suggests that some local knowledge is required.  So here are a few additional comments in &lt;em&gt;italics&lt;/em&gt; and I give up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The 6 digit coordinates&lt;/strong&gt; are from the &lt;a href=&quot;http://gridreferencefinder.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ordnance Survey National Grid&lt;/a&gt; reference system. The first 3 digits are &lt;strong&gt;eastings&lt;/strong&gt; and the last 3 digits are &lt;strong&gt;northings&lt;/strong&gt;.  In about 1980 when this treasure hunt was created, coordinates would have been read from a paper map with some uncertainty in the 3rd digits. For most of Kirkcudbrightshire, coordinates are entered into the &quot;Grid Reference&quot; box like this answer to Q1 - NX680509.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Here is some info about the other 9 questions.&lt;/strong&gt; With a few minor exceptions noted below, all answers and coordinates appear to be correct:&lt;br&gt;&#xA;&lt;strong&gt;Q1&lt;/strong&gt; - describes the inscription at the entry to Kirkcudbright Tolhouse (refer to &quot;Rambles in Galloway&quot;)&lt;br&gt;&#xA;&lt;strong&gt;Q2&lt;/strong&gt; - is explained in &quot;Rambles in Galloway&quot;&lt;br&gt;&#xA;&lt;strong&gt;Q3&lt;/strong&gt; - describes a memorial with inscriptions near Lowran Burn - see &lt;a href=&quot;http://warmemscot.s4.bizhat.com/warmemscot-post-26590.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;photos on this page&lt;/a&gt;.  &lt;em&gt;The disappearing castle would be Lochivar which does not seem to be associated with the Q3 answer.  Policeman could describe black and white markings similar to the band on hats worn by Scottish policemen&lt;/em&gt;.&lt;br&gt;&#xA;&lt;strong&gt;Q5&lt;/strong&gt; - is word play with Carsfad&lt;br&gt;&#xA;&lt;strong&gt;Q6&lt;/strong&gt; - describes the &lt;a href=&quot;http://www.kirkyards.co.uk/rerrick/abbey.asp?offset=49&quot; rel=&quot;nofollow noreferrer&quot;&gt;headstone inscription&lt;/a&gt; for Thomas Dewar.  Coordinates are closer to &lt;strong&gt;749 474&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;Q7&lt;/strong&gt; - is an anagram for the answer which is &lt;strong&gt;Southerness Lighthouse&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;Q8&lt;/strong&gt; - is a substitution cipher for &lt;strong&gt;Kirkandrews Church&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;Q9&lt;/strong&gt; - is a poem described in &quot;Rambles in Galloway&quot;&lt;br&gt;&#xA;&lt;strong&gt;Q10&lt;/strong&gt; - is reference to &lt;a href=&quot;http://www.kirkcudbright.co/historyarticle.asp?ID=375&amp;amp;p=28&amp;amp;g=5&quot; rel=&quot;nofollow noreferrer&quot;&gt;minister at Kirkgunzeon&lt;/a&gt; - not clearly stated as 1896 but answer is most likely correct  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book, &lt;strong&gt;&quot;Rambles in Galloway&quot;&lt;/strong&gt; may contain the answer to Q4 which could be a &lt;strong&gt;book cipher&lt;/strong&gt;.  Two editions were published.  The first edition is available free &lt;a href=&quot;https://books.google.ca/books?id=0roHAAAAQAAJ&amp;amp;printsec=frontcover&amp;amp;source=gbs_ge_summary_r&amp;amp;cad=0#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;from Google&lt;/a&gt; or &lt;a href=&quot;http://search.bl.uk/primo_library/libweb/action/display.do?frbrVersion=2&amp;amp;tabs=moreTab&amp;amp;ct=display&amp;amp;fn=search&amp;amp;doc=BLL01014812777&amp;amp;indx=2&amp;amp;recIds=BLL01014812777&amp;amp;recIdxs=1&amp;amp;elementId=1&amp;amp;renderMode=poppedOut&amp;amp;displayMode=full&amp;amp;frbrVersion=2&amp;amp;dscnt=0&amp;amp;fctN=facet_local20&amp;amp;rfnGrp=1&amp;amp;frbrVersion=2&amp;amp;scp.scps=scope%3A%28BLCONTENT%29&amp;amp;fctV=online&amp;amp;tab=local_tab&amp;amp;dstmp=1423074717007&amp;amp;srt=rank&amp;amp;mode=Basic&amp;amp;vl(488279563UI0)=any&amp;amp;tb=t&amp;amp;fromLogin=true&amp;amp;rfnGrpCounter=1&amp;amp;vl(freeText0)=rambles%20in%20galloway&amp;amp;vid=BLVU1&amp;amp;frbg=&amp;amp;dum=true&quot; rel=&quot;nofollow noreferrer&quot;&gt;from the British Library&lt;/a&gt;.  The second edition is available &lt;a href=&quot;http://search.bl.uk/primo_library/libweb/action/display.do?frbrVersion=2&amp;amp;tabs=moreTab&amp;amp;ct=display&amp;amp;fn=search&amp;amp;doc=BLL01014840223&amp;amp;indx=1&amp;amp;recIds=BLL01014840223&amp;amp;recIdxs=0&amp;amp;elementId=0&amp;amp;renderMode=poppedOut&amp;amp;displayMode=full&amp;amp;frbrVersion=2&amp;amp;fctN=facet_local20&amp;amp;dscnt=0&amp;amp;rfnGrp=1&amp;amp;scp.scps=scope%3A%28BLCONTENT%29&amp;amp;fctV=online&amp;amp;frbg=&amp;amp;tab=local_tab&amp;amp;dstmp=1422404695295&amp;amp;srt=rank&amp;amp;mode=Basic&amp;amp;vl(488279563UI0)=any&amp;amp;dum=true&amp;amp;tb=t&amp;amp;rfnGrpCounter=1&amp;amp;vl(freeText0)=rambles%20in%20galloway&amp;amp;vid=BLVU1&quot; rel=&quot;nofollow noreferrer&quot;&gt;from the British Library&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Q4 does not seem to be a substitution cipher.&lt;/strong&gt;  There are no word pairs in the list of named places that match the pattern of any of the 4 cipher pairs.  The pattern of &quot;14 26 10 17 24 28 25 28 16 34 27 27&quot; is distinctive but no word in English matches this pattern.  Even progressively removing characters from the cipher words does not help to match the named places.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Q4 does not seem to be a hidden grid reference.&lt;/strong&gt;  Addition or subtraction does not provide any meaningful grid reference numbers.  Applying different rules to extract 3 digits from the first &quot;word&quot; and 3 digits from the second &quot;word&quot; does not provide a meaningful reference number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Q4 could be a book cipher&lt;/em&gt;&lt;/strong&gt; &lt;em&gt;but no answer was obtained from monument inscriptions, numerous poems, or the book &quot;Rambles in Galloway&quot;.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the &lt;strong&gt;final calculation&lt;/strong&gt; and a grid of points across the county, &lt;strong&gt;the location of the whisky&lt;/strong&gt; has to be just east of Rusko Castle, within the 3 red blobs on the following map. (The two discontinuities are caused by the grid reference system.) Coordinates for a rectangle around the lower blobs are eastings from 604-665 and northings from 577-631. The upper blob is eastings from 629-654 and northings from 671-676.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that there are only &lt;strong&gt;2 place names of interest within these blobs&lt;/strong&gt;.  They are &lt;strong&gt;a)&lt;/strong&gt; the Watch House in Twynholm and &lt;strong&gt;b)&lt;/strong&gt; Cairntosh in Girthon.  Using these coordinates to back-calculate the location of the Q4 answer provides &lt;strong&gt;a)&lt;/strong&gt; no points of interest near Glenarm, Urr and &lt;strong&gt;b)&lt;/strong&gt; a location which is near Cally Park, Cally House, and Cally Palace in Girthon.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming everything is correct to this point, &lt;strong&gt;the answer to Q4 should be something around Cally Park&lt;/strong&gt; (NX607553). I would be very happy if somebody figured out the Q4 cipher.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/LPue3.jpg&quot; alt=&quot;Whisky location&quot;&gt;  &lt;/p&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-02-04T18:43:52.433" LastActivityDate="2015-02-04T18:43:52.433" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="7931" PostTypeId="1" AcceptedAnswerId="7932" CreationDate="2015-01-17T14:29:59.503" Score="7" ViewCount="2427" Body="&lt;p&gt;When Caliban's will was opened it was found to contain the following clause:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;I leave ten of my books to each of A, B, and C, who are to choose in a certain order.&lt;/p&gt;&#xA;&lt;p&gt;No person who has seen me in a green tie is to choose before A.&lt;/p&gt;&#xA;&lt;p&gt;If B was not in Oxford in 1920, the first chooser never lent me an umbrella.&lt;/p&gt;&#xA;&lt;p&gt;If B or C has second choice, C comes before the one who first fell in love.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Unfortunately A, B, and C could not remember any of the relevant facts; but the family solicitor pointed out that, assuming the problem to be properly constructed (i.e. assuming it to contain no statement superfluous to its solution) the relevant data and order could be inferred.&#xA;&lt;strong&gt;What was the prescribed order of choosing; and who first fell in love?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If you tackle this, remember that every statement is necessary. Also remember that the requirement (that the puzzle represented by the three statements is properly constructed) is not part of the original puzzle -- it is a meta-statement about the base puzzle. [People regularly get confused about this point.] One further point you have to consider -- was the family solicitor correct in claiming (a) that the relevant data could be inferred and (b) that the order be inferred.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;The puzzle was originally created by &lt;a href=&quot;http://www.cdpa.co.uk/Newman/MHAN/view-item.php?Box=2&amp;amp;SubBox=&amp;amp;Folder=16&amp;amp;SubFolder=&amp;amp;Item=1&amp;amp;SubItem=&amp;amp;Page=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;M.H. Newman&lt;/a&gt;, the exact wording was taken from Richard Harters' &lt;a href=&quot;http://richardhartersworld.com/cri/1998/caliban.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;website&lt;/a&gt;, and I've discovered it from Anatoly Vorobey's &lt;a href=&quot;http://avva.livejournal.com/2835081.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;blog&lt;/a&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="8955" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2021-12-15T18:51:25.157" Title="Caliban's Will puzzle" Tags="|logical-deduction|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7932" PostTypeId="2" ParentId="7931" CreationDate="2015-01-17T15:15:50.833" Score="9" Body="&lt;p&gt;Assuming the problem contains nothing superfluous:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - At least one person has seen Caliban in a green tie. So A is first or second chooser.&lt;br&gt;&#xA; - B was not in Oxford in 1920, and the one of the candidates for first chooser lent Caliban an umbrella.&lt;br&gt;&#xA; - B or C must be second chooser. So A must be first chooser.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This allows us to deduce that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Then A is first chooser, C comes before the one who first fell in love (that must be B), so C is second chooser and B is third. A couldn't have been potentially switching with B since that would put A last; so A must have been potentially switching with C, but C lent Caliban an umbrella. B was the one who saw Caliban in a green tie.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So if A, B and C had remembered: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - B saw Caliban in a green tie, and is second or third.&lt;br&gt;&#xA; - B was not in Oxford in 1920, and A never lent Caliban an umbrella, so A is first.&lt;br&gt;&#xA; - B or C are second, but B first fell in love so C comes before him.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6031" LastEditorUserId="6031" LastEditDate="2015-01-17T15:24:44.480" LastActivityDate="2015-01-17T15:24:44.480" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7933" PostTypeId="1" AcceptedAnswerId="7942" CreationDate="2015-01-17T18:27:01.347" Score="9" ViewCount="820" Body="&lt;p&gt;You stand before a panel with two buttons on it.  One of them will dispense your favorite beverage, and you're dying for a sip.  The other will destroy all life throughout the universe, so you'd really rather not press that one.  You don't know which button is which, but fortunately the buttons have three alien attendants--each a different race--to help you.  You know a little bit about the alien languages, but unfortunately you don't know which alien uses which language.  The aliens are identified with signs reading &quot;1&quot;, &quot;3&quot;, and &quot;6&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The aliens' languages are based on gestures, and there are four polite ways to ask a question: point with the left index finger, the left index finger and middle finger, the right index finger, or the right index finger and middle finger, for 50 seconds.  Any other form of gesture would be considered rude, and would cause the aliens to immediately shoot you (which would be bad).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the aliens is an Altairian.  If you ask a question by pointing with your left hand (one or two fingers), it will (after 49 seconds) nod its head if the left button dispenses a beverage and do nothing if it does not.  Likewise if you ask a question by pointing with your right hand, it will not its head if the right button dispenses a beverage and do nothing if it does not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the other aliens is a Denebrian.  If you ask a question by pointing with one finger (either hand), it will (after 49 seconds) nod its head if the alien with the closer number is an Altairian or else do nothing.  If you point with two fingers (either hand), it will nod its head if the further number is an Altairian or else do nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remaining alien is a Snoozorian.  It's going to sleep for the next five years no matter what you do, and can thus be counted upon to simply do nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You've been called to battle, and need to leave within the next two minutes, but you really want your beverage before then.  This wouldn't be enough time, had you not learned in your latest linguistics class that it's possible to politely address two aliens simultaneously subject to the following constraints:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;One must simultaneously use both arms to address the aliens, with one arm pointed at each; the arms must not be crossed, and one must continue to point with both arms for the full 50 seconds.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you ask questions which both merit affirmative responses, the aliens will get annoyed and shoot you (again, not good).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: This is my first attempt at formulating a variation of a &quot;Lady or the tiger&quot; puzzle; rather than make some of the guardians deceitful, I decided to instead make them vary in the questions they answer.  Since only one alien knows where the right button is, and since the first alien one talks to might not know anything, it's necessary to be able to talk to all three aliens, but being able to simply ask a question of each without restriction doesn't seem very satisfying.  If people like the concept, but not the particulars, I'd be open to advice on how to make the puzzle more interesting and/or satisfying.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1467" LastActivityDate="2015-01-18T03:47:54.833" Title="The Beverage Button" Tags="|logical-deduction|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7934" PostTypeId="1" CreationDate="2015-01-17T20:11:39.903" Score="8" ViewCount="17406" Body="&lt;p&gt;This is a standardized test practice question asking for the next image in the sequence. Can anyone identify the pattern? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MSj8Q.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="8956" LastActivityDate="2023-01-03T17:56:32.657" Title="Determine the next image in the sequence" Tags="|pattern|" AnswerCount="6" CommentCount="0" FavoriteCount="0" ClosedDate="2023-02-01T19:17:53.877" ContentLicense="CC BY-SA 3.0" />
  <row Id="7935" PostTypeId="2" ParentId="7934" CreationDate="2015-01-17T20:51:33.313" Score="0" Body="&lt;p&gt;I know it is not the solution you are looking for, but with finding similarities in options and elimination I choose...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;C&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 of options has arrow up. &lt;strong&gt;b&lt;/strong&gt; is eliminated&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of four leads (upper images) has two black circles. &lt;strong&gt;e&lt;/strong&gt; is eliminated&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Locations of gray, white and black circle in &lt;em&gt;c&lt;/em&gt; and &lt;em&gt;d&lt;/em&gt; is exactly the same. &lt;strong&gt;a&lt;/strong&gt; is eliminated&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, between &lt;em&gt;c&lt;/em&gt; and &lt;em&gt;d&lt;/em&gt;, I choose &lt;em&gt;c&lt;/em&gt; because it is more similar to the last eliminated option (&lt;em&gt;a&lt;/em&gt;) than &lt;em&gt;d&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2225" LastActivityDate="2015-01-17T20:51:33.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7936" PostTypeId="2" ParentId="7934" CreationDate="2015-01-18T00:09:55.730" Score="9" Body="&lt;p&gt;Here are two different methods that give the same solution which is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;answer a&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Based on sequential logic:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  From the &lt;strong&gt;gray diamonds&lt;/strong&gt;, the fifth image could also be a combination of image one and three (as the fourth image could be a combination of image one and two).&lt;br&gt;&#xA;&lt;br&gt;&#xA;  The sequence for the &lt;strong&gt;arrow&lt;/strong&gt; is up, down, left, right.  If repeating, next should be up.&lt;br&gt;&#xA;&lt;br&gt;&#xA;  The &lt;strong&gt;color of the dots&lt;/strong&gt; could be 0=white, 1=gray, 2=black (a base 3 system).  So adding the dots for image one and two gives white-gray-black-gray which is the pattern for image 4 (starting at top left and going clockwise).  It follows that the dots for image 5 should be white-gray-black-white.&lt;br&gt;&#xA;&lt;br&gt;&#xA;  The &lt;strong&gt;keyhole at the edge&lt;/strong&gt; has a sequence of left, top, right, top.  If repeating, the next should be left.&lt;br&gt;&#xA;&lt;br&gt;&#xA;  The &lt;strong&gt;square around the dot&lt;/strong&gt; has a sequence of none, top left, none, bottom left.  If repeating, the next should be none.&lt;br&gt;&#xA;&lt;br&gt;&#xA;  Putting this together fits the pattern for &lt;strong&gt;answer a&lt;/strong&gt;.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Based on incorrect answers being formulated in regions surrounding the correct answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;strong&gt;most common elements&lt;/strong&gt; in the five answers are arrow=up, dot colors=white-gray-black-white, keyhole=left, square around dot = none.&lt;br&gt;&#xA;&lt;br&gt;&#xA; These most common elements coincide with &lt;strong&gt;answer a&lt;/strong&gt;.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastActivityDate="2015-01-18T00:09:55.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7938" PostTypeId="1" CreationDate="2015-01-18T01:19:16.403" Score="9" ViewCount="3224" Body="&lt;p&gt;What is the northernmost country that borders the US to the South?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the northernmost country that borders the contiguous US to the South?&lt;/p&gt;&#xA;" OwnerUserId="8958" LastEditorUserId="8874" LastEditDate="2015-02-22T18:24:24.103" LastActivityDate="2015-02-22T18:24:24.103" Title="What is the northernmost country that borders the US to the South?" Tags="|geography|knowledge|" AnswerCount="5" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7939" PostTypeId="2" ParentId="7929" CreationDate="2015-01-18T02:07:44.247" Score="4" Body="&lt;p&gt;I assume this is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Substitution_cipher&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;substition cipher&lt;/em&gt;&lt;/a&gt;. I know that some newspapers pose such problems. I checked to make sure it is not a rotation cipher, which it is not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this cipher we have the following conditions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;There are two words&lt;/li&gt;&#xA;&lt;li&gt;The first word has 8 letters and the second has 5&lt;/li&gt;&#xA;&lt;li&gt;The second letter in both words is the same &lt;/li&gt;&#xA;&lt;li&gt;All other letters in both words are different&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There are several possible obvious choices for the second word: Beach, Point, Sound and Tower; and numerous other more obscure possibilities: Cairn, Carse, Cauld, Clint, Corse, Court, Cowld, Craig, Firth, Grain, Haven, Hirst, Hurst, Lough and Slack. However, none of the well-known places having these words in Kirkcudbrightshire have a matching first word that I could find. There are several common places that could be the first word: Boreland, Dalgoner, Dumfries, Duncraig, Dunragit, Dunscore, Muncraig, and Twynholm, however I could find no second word to go with these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having gone through all the major cities and towns of the shire and found no matching places, I am inclined to believe there is a typo of some kind in the original puzzle, a common problem with old newspaper typesetting.&lt;/p&gt;&#xA;" OwnerUserId="8957" LastEditorUserId="8957" LastEditDate="2015-01-18T02:38:20.560" LastActivityDate="2015-01-18T02:38:20.560" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7940" PostTypeId="2" ParentId="7938" CreationDate="2015-01-18T02:39:47.777" Score="-1" Body="&lt;p&gt;Assuming there's no trick here, a simple answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mexico.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mexico is the &lt;strong&gt;&lt;em&gt;ONLY&lt;/em&gt;&lt;/strong&gt; country that borders US to the south, therefore the most Northen of them all.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All in all, this isn't really a puzzle, just geography.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Extra information for hungry minds:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; * Canada – borders US to the north.&lt;br&gt;&#xA; * Mexico – borders US to the south.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2015-01-18T02:39:47.777" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="7941" PostTypeId="2" ParentId="7938" CreationDate="2015-01-18T03:25:28.663" Score="6" Body="&lt;p&gt;Part 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Canada borders Alaska to the south, and is north of Mexico. That's the trick.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Part 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mexico is the only country that borders the contiguous US to the south.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6031" LastActivityDate="2015-01-18T03:25:28.663" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7942" PostTypeId="2" ParentId="7933" CreationDate="2015-01-18T03:47:54.833" Score="5" Body="&lt;p&gt;&lt;strong&gt;Solution:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Point two fingers with your left hand at alien 1, and two fingers with your right at alien 3.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If alien 1 nods, then point one finger with your left hand at alien 1, and one finger with your right at alien 6. If alien 6 nods, choose the right button, otherwise choose the left button.&lt;/li&gt;&#xA;&lt;li&gt;If alien 3 nods, then point 1 finger with your left hand at alien 1, and 1 finger with your right at alien 6 (as above). If one of the alien nods, choose the right button, otherwise choose the left button.&lt;/li&gt;&#xA;&lt;li&gt;If neither alien 1 or alien 3 nodded, then point two fingers with your left at 3 and two fingers with your right at 6. If alien 6 nods, choose the right button, otherwise choose the left button.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, point two fingers with your left hand at alien 1, and two fingers with your right at alien 3. You won't get shot because:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If alien 1 is an Altairian, then alien 3 will do nothing (either because he is the Snoozorian, or because the alien with the farther number (6) is not an Altairian).&lt;/li&gt;&#xA;&lt;li&gt;If alien 1 is a Denebrian and nods, then alien 6 is the Altairian and alien 3 is the Snoozorian and will do nothing. If he doesn't nod, alien 3 is the Altairian.&lt;/li&gt;&#xA;&lt;li&gt;If alien 1 is a Snoozorian, then he won't nod.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If alien 1 nods, then point one finger with your left hand at alien 1, and one finger with your right at alien 6. If alien 6 nods, choose the right button, otherwise choose the left button.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If alien 1 nods, he is the Altairian (the Denebrian wouldn't nod when pointed at with one &lt;em&gt;and&lt;/em&gt; two fingers), so choose the left button. If alien 6 is the Denebrian, then he won't nod because the alien with the close number (alien 3) isn't the Altairian.&lt;/li&gt;&#xA;&lt;li&gt;If alien 6 nods, he must be the Altarian, so we choose right. Alien 1 wouldn't nod, because he would be the Denebrian, and the alien with the number closer to him was not the Altairian. If he were the Denebrian, then alien 3 would be the Altairian, and alien 1 the Snoozorian - but alien 1 nodded the first time, so this is a contradiction.&lt;/li&gt;&#xA;&lt;li&gt;If neither alien nods, then alien 1 must be the Denebrian (if he was the Altairian, he would have nodded both times), and so alien 3 is the Altairian. He didn't nod when we pointed our right hand at him the first time, so we choose the left button.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If alien 3 nods, then point 1 finger with your left hand at alien 1, and 1 finger with your right at alien 6 (as above). If one of the alien nods, choose the right button, otherwise choose the left button.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If alien 1 nods, then alien 3 is the Altairian (if alien 1 was the Altairian, then alien 3 wouldn't have nodded, because we pointed two fingers at him), and the correct button is that on the right.&lt;/li&gt;&#xA;&lt;li&gt;If alien 6 nods, then the correct button is the one on the right (either 3 or 6 is the Altairian, and they both nodded when we pointed with our right hand to them).&lt;/li&gt;&#xA;&lt;li&gt;If neither alien nods, then alien 3 is the Denebrian (if he was the Altairian, then one of the others would be the Denebrian and would nod). He nodded when we pointed 2 fingers at him, so 6 is the Altairian. He didn't nod when we pointed with our right hand, so choose the left button.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If neither alien 1 or alien 3 nodded, then point two fingers with your left at 3 and two fingers with your right at 6. If alien 6 nods, choose the right button, otherwise choose the left button.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If alien 3 nods, then he is the Altairian (if he was the Denebrian, he would have nodded the first time), so choose the left button. You won't get shot because either 6 is the Snoozorian, or the farther alien from him (alien 1) is not the Altairian.&lt;/li&gt;&#xA;&lt;li&gt;If alien 6 nods, then he is either an Altairian so we choose the button on the right, or the Denebrian (meaning alien 1 was the Altairian). When we pointed with our left to alien 1, he didn't nod, so we also choose the button on the right. Again, we won't get shot because if alien 6 was the Altairian, alien 3 would have nodded the first time, and if alien 6 was the Denebrian, 1 was the Altairian and therefore 3 was the Snoozorian.&lt;/li&gt;&#xA;&lt;li&gt;If neither alien nods, then 3 is not the Altairian (we have pointed both hands at him). 3 cannot be the Snoozorian either, because if 6 was the Altairian, 1 (being the Denebrian) would have nodded the first time, and if 6 was the Denebrian, he would have nodded because 1 was the Altairian. Therefore 3 is the Denebrian, and 1 is the Altairian, so we choose left.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8949" LastActivityDate="2015-01-18T03:47:54.833" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7943" PostTypeId="2" ParentId="7938" CreationDate="2015-01-18T03:58:17.467" Score="23" Body="&lt;p&gt;The first part of the question has already been answered correctly by @Callidus, however the second has not. To answer the second part of the question:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;What is the northernmost country that borders the contiguous US to the South?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is again Canada: in Windsor, Ontario, which is south of Detroit, Michigan. That's the 2nd trick.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2718" LastActivityDate="2015-01-18T03:58:17.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7944" PostTypeId="2" ParentId="7938" CreationDate="2015-01-18T22:13:11.560" Score="8" Body="&lt;p&gt;@Michael has correctly answered the riddle.  At the risk of being pedantic, here are a few other points of interest:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://www.google.com/maps/place/Point+Roberts,+WA/@48.8754457,-123.0476601,11z&quot;&gt;Point Roberts, WA north of Saturna Island, BC&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://www.google.com/maps/place/Niagara+Falls,+ON,+Canada/@43.0479113,-79.072585,12z&quot;&gt;Niagara Falls, NY north of Niagara Falls, ON&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://www.google.com/maps/@45.4446313,-67.4597821,11z&quot;&gt;Vanceboro, ME north of St. James, NB&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8973" LastActivityDate="2015-01-18T22:13:11.560" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7945" PostTypeId="2" ParentId="7938" CreationDate="2015-01-18T23:02:20.803" Score="7" Body="&lt;p&gt;Well, the question could be interpreted as &quot;Which country has the northernmost border which can be crossed from the US going south&quot; or &quot;Which country can be crossed into from the US going south has the northernmost extent.&quot;,  &quot;Which country can be crossed into from the US going south has the northernmost southern extent.&quot;, &quot;Which country can be crossed into from the US going south has the northernmost centroid.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is further complicated by whether marine boundaries count for adjacency and extent and how to handle disputed claims.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It looks like any way you work it out though, the answer is Canada.&#xA; &lt;p&gt;&#xA; Allowing for a disputed marine border, the northernmost section of US border where the US is to the north is the Alaska-Yukon border in the Barents Sea according to the US claim that the border bends to be perpendicular to the coastline.  The Canadian claim is that the border continues due north-south.&#xA;&lt;p&gt;&#xA; For an undisputed land border, there's a short section near the south end of the Alaska-Yukon border between Mount Hubbard and Mount Seattle where Alaska is to the northwest and the Yukon is to the southeast.&#xA;&lt;p&gt;&#xA; The US shares land borders with Canada and Mexico, and also has a marine border with Russia, but that border never has the US to the north.  The Marine border between the US Virgin Islands and British Virgin islands also has a short section where the US is to the north.  Even if marine borders count this leaves the question of whether overseas territories count.&#xA;&lt;p&gt;&#xA; Even if we include the United Kingdom with all of its overseas territories, the northernmost point in the Shetland Islands is still much further south than Canada's northernmost point on Ellesmere Island.   Canada also has a more northern centroid and southernmost point than the United Kingdom with its territories.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8974" LastEditorUserId="8974" LastEditDate="2015-01-18T23:08:18.940" LastActivityDate="2015-01-18T23:08:18.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7946" PostTypeId="1" AcceptedAnswerId="7951" CreationDate="2015-01-18T23:21:15.140" Score="24" ViewCount="3967" Body="&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;There are 2014 persons in a row&lt;/li&gt;&#xA;  &lt;li&gt;Each of them is either a Knave (who always lies) or a Knight (who always tells the truth).&lt;/li&gt;&#xA;  &lt;li&gt;Each person says &lt;strong&gt;&lt;em&gt;'There are more knaves to my left than knights to my right'.&lt;/em&gt;&lt;/strong&gt; &lt;/li&gt;&#xA;  &lt;li&gt;How many knaves are there in the row?&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Source :&lt;/em&gt;&lt;/strong&gt; (Math Kangaroo competition for parents)&lt;/p&gt;&#xA;" OwnerUserId="6087" LastEditorUserId="5188" LastEditDate="2015-01-19T05:19:51.763" LastActivityDate="2015-01-20T08:48:00.277" Title="Knights and Knaves : Liar , Liar - How many are you?" Tags="|logical-deduction|liars|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7947" PostTypeId="2" ParentId="7946" CreationDate="2015-01-18T23:36:18.493" Score="8" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1007. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There are 1007 liars followed by 1007 knights. Each liar has 1007 knights to their right and at most 1006 liars to their left, while the knights are in the opposite situation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3987" LastEditorUserId="5188" LastEditDate="2015-01-19T00:26:22.030" LastActivityDate="2015-01-19T00:26:22.030" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7948" PostTypeId="1" AcceptedAnswerId="7949" CreationDate="2015-01-19T00:06:17.593" Score="1" ViewCount="535" Body="&lt;p&gt;My age squared is the number of commandments in the Torah minus the year in George Orwell's Book. When subtracting, consider the cause of the Y2K problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How old am I? &lt;/p&gt;&#xA;" OwnerUserId="8975" LastEditorUserId="7308" LastEditDate="2015-04-28T08:32:24.333" LastActivityDate="2015-04-28T08:32:24.333" Title="How many years old am I?" Tags="|knowledge|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7949" PostTypeId="2" ParentId="7948" CreationDate="2015-01-19T00:43:15.007" Score="5" Body="&lt;p&gt;There are 613 commandments in the torah.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1984, considering Y2K is 84.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;613-84=529&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sqrt (529) = 23&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your age is 23.&lt;/p&gt;&#xA;" OwnerUserId="2484" LastActivityDate="2015-01-19T00:43:15.007" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7950" PostTypeId="1" AcceptedAnswerId="7953" CreationDate="2015-01-19T00:59:57.067" Score="2" ViewCount="596" Body="&lt;p&gt;This is my first post on this forum! Credit goes to Goleman's 2014 IQ Test, problems 12, 20, 15, 19, and 18.&#xA;&lt;p&gt;If this is plagiarism please let me know&#xA;&lt;p&gt;If this is too easy please let me know :)&#xA;&lt;p&gt;If I shouldn't post this many puzzles at once please let me know :) :)&#xA;&lt;p&gt; Any feedback would be appreciated :) :) :)&#xA;&lt;img src=&quot;https://i.stack.imgur.com/PYiZG.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="8947" LastActivityDate="2015-01-19T09:00:08.187" Title="Patterns and things" Tags="|pattern|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7951" PostTypeId="2" ParentId="7946" CreationDate="2015-01-19T02:04:33.000" Score="24" Body="&lt;p&gt;For a group of $2n$ $(n\geq1)$ knights and knaves satisfying the puzzle conditions, there must be exactly $n$ knaves. Moreover, all knaves must occupy the first $n$ positions in the row (left to right direction)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can provide a very elegant proof of this fact by considering this situation if there were $2n$ people in a row. Clearly, for $n\geq 1$, there is at least one knight and one liar, as otherwise the situation would obviously be untrue. From this, consider the leftmost person - there are no liars to his left (which is obviously not &lt;em&gt;more&lt;/em&gt; than the number of knights to his right). Thus, he is a liar. The rightmost person obviously has no knights to his right and at least that liar to the left - so he must be a knight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we use this? Take the original $2014$ people. The leftmost is a liar, the rightmost is a knight. If we exclude these two people, we don't modify the situation - we've removed a liar from the left and a knight from the right, so everyone's counts on both sides decrease by one. Thus, we have the $2012$ people in the middle remaining - and the leftmost must be a liar and the rightmost is a knight. Remove them. There are now $2010$ people left. We continue in this manner, pairing liars and knights and, at the end, we have proven that the 1007 leftmost people were liars and the 1007 rightmost were knights.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;We can generalize this answer to handle the case where there are $n$ people and we ask everyone in the line whether there there are more knaves to their left than knights to the right and receive a &quot;yes&quot; or &quot;no&quot; answer from each person. In particular, let's consider cases:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If the leftmost person in line says &quot;no&quot;, he is a knight (since there are no knaves to his left). We can remove him without affecting the relevant counts of everyone else, so this reduces it to the $n-1$ person case.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the leftmost person says &quot;yes&quot; and the rightmost says &quot;no&quot;, then both are knaves; we can safely remove the rightmost one without affecting anyone's counts.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the leftmost and rightmost people say &quot;yes&quot;, then the leftmost is a knave and the rightmost is a knight; we can remove both people, decreasing everyone's count by $1$ on both sides and not affecting their answers.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Therefore, every case can be reduced to a smaller case and we can thusly identify everyone.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A more elegant proof of the same more general fact as in the previous section is that there must be some $k$ such that the first $k$ people have more knaves to their left than knights to the rest, and this is not true for the rest of the people. This $k$ satisfies that must be more knaves (who answer &quot;yes&quot;) among the first $k-1$ as knights (who also answer &quot;yes&quot;) among those at position $k+1$ and beyond and that there must be at least as many knights (who answer &quot;yes&quot;) in positions $k+2$ and beyond as knaves among the first $k$ (who answer &quot;yes&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is to say, if you proceed through the line left to right and receive the answer &quot;yes&quot; $2c+1$ times, then everyone before the $c^{th}$ person to say &quot;yes&quot; had fewer knaves to their left than knights to their right and everyone after had the opposite - this suffices to figure out who lied, as you know the truth. Similarly, if you get an answer of &quot;yes&quot; $2c$ times, then the $c^{th}$ person (and everyone before, but no one after) had fewer knaves to their left than knights to their right.&lt;/p&gt;&#xA;" OwnerUserId="4765" LastEditorUserId="4765" LastEditDate="2015-01-20T03:24:17.847" LastActivityDate="2015-01-20T03:24:17.847" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7953" PostTypeId="2" ParentId="7950" CreationDate="2015-01-19T07:57:37.253" Score="4" Body="&lt;p&gt;For the first question, the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; F. Overlap the rows of pictures to see the top two combine to fill an entire 9-block set. Combine the last row to see a center block is needed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the second question, the answer (I think) is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A. The first and third pictures and fourth and sixth pictures are reflection across the xy axis. The seventh has an xy axis reflection as choice A.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit 1: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For #3, The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; -5. The top row digit sum equals the sum of the digits below. 9 equals 6+1+2. 3+8 = 3+5+2+1. 4+7=4+2+2+3. 4=(-5)+2+7&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For #5, The answer is a similar trick:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 17. The number below and between each row pair is the sum of their digits. 8+8+6+3=25. 2+5+9=16. 1+6+2+4=13. 1+3+4+9=&lt;em&gt;17&lt;/em&gt;. To complete, 1+7+1+0=9.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For #4, the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; H. The top row has all letters with horizontal symmetry (flipped across y axis), and the bottom row has letters with vertical symmetry (flipped across x axis). H is the only letter option that will fit both.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="7722" LastEditDate="2015-01-19T09:00:08.187" LastActivityDate="2015-01-19T09:00:08.187" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7954" PostTypeId="1" CreationDate="2015-01-19T09:41:28.203" Score="14" ViewCount="36452" Body="&lt;p&gt;In one of the answers to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2188/two-doors-with-two-guards-one-lies-one-tells-the-truth&quot;&gt;this question&lt;/a&gt;, reproduced here:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;You are a prisoner in a room with 2 doors and 2 guards. One of the doors will guide you to freedom and behind the other is a hangman - you don't know which is which.&lt;/p&gt;&#xA;&lt;p&gt;One of the guards always tell the truth and the other always lies. You don't know which one is the truth-teller or the liar either.&lt;/p&gt;&#xA;&lt;p&gt;You have to choose and open one of these doors, but you can only ask a single question from one of the guards.&lt;/p&gt;&#xA;&lt;p&gt;What do you ask to lead you to the door of freedom?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;and the solution:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Here is a twisted solution.&lt;/p&gt;&#xA;&lt;p&gt;Go to any guard, point at a door and ask:&#xA;Among the propositions 1. &amp;quot;You are a liar&amp;quot;, 2. &amp;quot;You will reply negatively&amp;quot; and 3. &amp;quot;This door leads to freedom&amp;quot;, is there an odd number of true propositions?&lt;/p&gt;&#xA;&lt;p&gt;If you get the answer yes: If the guard is a truthteller, the number&#xA;of truths is odd, 1. is false, 2. is false, so 3. must be true. If the&#xA;guard is a liar, the number of truths is even, 1. is true, 2. is&#xA;false, so 3. must be true.&lt;/p&gt;&#xA;&lt;p&gt;If you get a negative answer: If the guard is a truthteller, the&#xA;number of truths is even, 1. is false, 2. is true, so 3. must be true.&#xA;If the guard is a liar, the number of truths is odd, 1. is true, 2. is&#xA;true, so 3. must be true.&lt;/p&gt;&#xA;&lt;p&gt;So regardless of the answer of the guard, the door you pointed at is&#xA;the door to freedom, you can leave safely.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I fail to follow the logic behind this answer. The answer seems to suggest that no matter what door is pointed to, it becomes the correct door, which does not make sense.&lt;/p&gt;&#xA;&lt;p&gt;Is there a logical mistake behind this answer, or is there something I missed in my reading?&lt;/p&gt;&#xA;" OwnerUserId="7529" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2021-11-01T22:26:54.303" Title="Logic explanation in &quot;two doors&quot; answer" Tags="|logical-deduction|" AnswerCount="8" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7955" PostTypeId="2" ParentId="7954" CreationDate="2015-01-19T10:05:11.977" Score="5" Body="&lt;p&gt;Yes. Let's just take one case - the door is correct, and the guard is a truthteller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If he says &quot;yes&quot;, there is indeed an odd number of true propositions - #3 alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If he says &quot;no&quot;, there is indeed an even number of true propositions - #2 and #3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ergo, his answer is vacuous. The same applies for every other case; the guard can answer either way since changing his answer flips proposition #2 and the parity of the number of true propositions. The solution above is wrong because in each case &lt;strong&gt;it assumes the value of #2 (which in reality depends on #3) &lt;em&gt;before&lt;/em&gt; determining #3&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1801" LastActivityDate="2015-01-19T10:05:11.977" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7956" PostTypeId="2" ParentId="7954" CreationDate="2015-01-19T10:07:45.103" Score="17" Body="&lt;p&gt;There is a fallacy, although quite subtle - &quot;You will reply negatively&quot; leads to a contradiction if the door pointed to leads to the hangman (meaning that neither guard can reply without breaking their rule of always telling the truth or always lying), so a case test of the reply will ignore this possibility. The solution assumes that there is always a possible answer in every situation (where there is not).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; upon rereading the original solution, it seems the lying guard will answer falsely to the entire question, rather than answering falsely to the individual questions and truthfully answering the main question. The fallacy is still the same in each case though.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Case testing the different situations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the guard is a truthteller, and the door you pointed to leads to freedom, then 1. is false and 3. is true. If 2. is true, then there are an even number of truths, so the reply is negative. If 2. is false, then there are an odd number of truths, so the reply is positive. Both answers work in this situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the door leads to the hangman however, then 1. is false and 3. is false. If 2. is true, then there are an odd number of truths, which would elicit a positive reply, but 2. being true leads to a contradiction. Similarly, if 2. is false, a contradiction also occurs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the guard is a liar, and the door leads to freedom, then 1. is true and 3. is true. If 2. is true, then there are an odd number of truths, so the reply is negative (because the guard always lies). If 2. is false, then there are an even number of truths, so the reply is positive (again because the guard always lies).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the door leads to the hangman, then 1. is true and 3. is false. If 2. is true, then there are an even number of truths, so the reply is positive, but this once again contradicts the claim in 2. that the reply is negative. Similarly, if 2. is false, a contradiction also occurs.&lt;/p&gt;&#xA;" OwnerUserId="8949" LastEditorUserId="8949" LastEditDate="2015-01-19T10:35:30.197" LastActivityDate="2015-01-19T10:35:30.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7957" PostTypeId="2" ParentId="7954" CreationDate="2015-01-19T10:08:19.017" Score="1" Body="&lt;p&gt;There are always 3 propositions (let's call them &quot;P&quot;). The Ps are well chosen so they will always be answered in a 2:1 ratio (2 true: 1 false or 2 false : 1 true). From the answer we can tell what the ratio was. Now, the problem is that you would always get the same answer from both guards, right? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's why the P2 is there. The creator of this solution made it hard to understand but try the following. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it is easier to switch P2 and P3. So you find your answers to the first 2 Ps and with (the new) P3 (&quot;You will reply negatively&quot;) you kind of invert the answer of one guard so that you can distinguish between the liar and the honest guy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only problem is that you can't really answer P3 because it is related to the answer the guard should give you. Maybe one can solve this with some differential equation but speaking for myself, I think there is a mistake. You would probably end up in a logical loop where the answers always change and the guard's head explodes :D&lt;/p&gt;&#xA;" OwnerUserId="5106" LastActivityDate="2015-01-19T10:08:19.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7958" PostTypeId="2" ParentId="7954" CreationDate="2015-01-19T10:10:21.500" Score="4" Body="&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; Let's read it backwards?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's read it backwards:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1. &quot;You are a liar&quot;&#xA;2. &quot;You will reply negatively&quot;&#xA;3. &quot;This door leads to freedom&quot;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;is there an odd number of true propositions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If 3 is False and guard is T we can have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;=&amp;gt; F T F =&amp;gt; F&#xA;=&amp;gt; F F F =&amp;gt; T&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If 3 is True and guard is T we can have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;=&amp;gt; F T T =&amp;gt; F&#xA;=&amp;gt; F F T =&amp;gt; T&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If 3 is False and guard is F we can have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;=&amp;gt; T T F =&amp;gt; F&#xA;=&amp;gt; T F F =&amp;gt; T&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If 3 is True and guard is F we can have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;=&amp;gt; T T F =&amp;gt; F&#xA;=&amp;gt; T F F =&amp;gt; T&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If the guards can deduce how answering 2 should affect their answer, they can choose. If they can't, they will not be able to answer.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TL;DR&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The guards know how 2 will affect their answer, and can choose what to say to 2.&lt;/p&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="5610" LastEditDate="2015-01-19T10:32:20.510" LastActivityDate="2015-01-19T10:32:20.510" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7959" PostTypeId="1" AcceptedAnswerId="7967" CreationDate="2015-01-19T10:15:23.850" Score="7" ViewCount="1014" Body="&lt;p&gt;You have created a robot with artificial intelligence, to test his abilities you add to the robot a battery pack which can hold a maximum of 5 charges and build for him an obstacle he needs to overcome.&#xA;The obstacle is a series of platforms, the only difference they have is the distance between them, which can either be a 1 length distance or a 2 length distance,&#xA;for the robot to jump a 1 length distance he needs 1 battery charge and to jump a 2 length distance he needs 2 battery charges. The platforms are as follows:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/kaxr6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the first 5 platforms are 1 length apart and the next 4 are 2 lengths apart. Of course your superior intellect knows that this is impossible to solve so you add 2 objects to the platforms,&#xA;first is a battery expansion pack which is placed on top every platform after the 5th one, every time the robot gets to a platform after the 5th one his maximum battery charge increases by 1 this expansion pack&#xA;is then used up. The second object is a battery charge cell which is moved to the farthest platform the robot has reached,&#xA;it gives him 1 battery charge every time he gets to the platform the cell is on, at the start this battery cell is placed on the 6th platform and stays there until the robot reaches the next platform(7th in this case).&#xA;the battery charge cell needs 2 jump from the robot to recharge before it can power up the robot again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While again you see that this is still impossible to solve, you device to also give the robot the ability to recharge his battery with jumping backwards to the previous platforms.&#xA;Jumping backwards doesn't cost any charges, in addition the first jump backwards doesn't give any charge only the second jump backwards (and every jump backwards after that) gives the robot 2 charges. Everytime the robot&#xA;jumps ahead this resets and the first jump backwards gives him no charge again.&#xA;The last thing you decide to give the robot is the ability to move the battery charge cell one platform backwards this costs the robot 1 charge, and again every time the robot reaches a farther platform then the&#xA;one the charge cell is currently on it gets moved up to platform. Jumping backwards with the cell doesn't alter the robots ability to gain charges, he does need a minimum of 1 charge to move the battery cell backwards&#xA;opposed to just jumping backwards without the energy cell which costs no energy.&#xA;Jumping backwards with the energy cell also counts towards the 2 jump counter that the cell needs to recharge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the robot reach the 9th platform with only the things he is given? If so what is the minimum amount of jumps that the robot needs to do to reach the 9th platform?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Forgot to add, the robot starts at platform 1 and with 5 charges&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dont read this if you dont want to know if a solution actualy exists or not.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; as i added in a comment already, a solution actualy exists.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a original puzzle i created so gl! :)&lt;/p&gt;&#xA;" OwnerUserId="5560" LastEditorUserId="5560" LastEditDate="2015-01-19T12:32:56.907" LastActivityDate="2015-01-19T17:07:40.273" Title="Can you help the robot cross the platforms?" Tags="|logical-deduction|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7960" PostTypeId="2" ParentId="7934" CreationDate="2015-01-19T10:22:46.867" Score="2" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Option A&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've chosen this option because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I've noticed that when there are two grey circles in the big square, then the grey circle opposite to the direction that the arrow points to has a small square around it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am assuming that there will not be more than one black circle in the big square as there is just one in each image.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Moreover,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The arrows in each image point at directions up,down,left and right and the &quot;door&quot; at left,up,right,up. So, the next arrow will probably point up and the &quot;door&quot; at left.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, by elimination, I got the answer.&lt;/p&gt;&#xA;" OwnerUserId="7482" LastEditorUserId="7482" LastEditDate="2015-01-19T10:29:23.017" LastActivityDate="2015-01-19T10:29:23.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7961" PostTypeId="2" ParentId="7946" CreationDate="2015-01-19T10:51:33.860" Score="2" Body="&lt;p&gt;One doesn't need maths to solve this, all you need is logic, imagination and intuition maybe:] It's easier to start from knights, because their statement is true, so we divide 2014 in two, here is a visual:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1upnb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the place marked as &quot;Your are here&quot; is the last position for the knight where his statement is true, and at this point we have 1007 knaves to the left. So the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&lt;b&gt;1007&lt;/b&gt;&lt;/blockquote&gt;&#xA;" OwnerUserId="8982" LastActivityDate="2015-01-19T10:51:33.860" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7963" PostTypeId="1" AcceptedAnswerId="7968" CreationDate="2015-01-19T12:31:41.690" Score="15" ViewCount="52567" Body="&lt;p&gt;This is an extension of &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2188/two-doors-with-two-guards-one-lies-one-tells-the-truth&quot;&gt;Two doors with two guards - one lies, one tells the truth&lt;/a&gt;, but in this situation you are a prisoner in a room with 2 doors and 3 guards. One of the doors will guide you to freedom and behind the other is a hangman - you don't know which is which. One of the guards always tell the truth, another always lies, and the third is unreliable and sometimes tells the truth and sometimes lies. You don't know who is who, but the guards do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After asking two yes/no questions you have to choose and open one of the two doors. You can ask one guard both questions or you can ask two different guards a single question each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What and who do you ask to lead you to the door of freedom?&lt;/p&gt;&#xA;" OwnerUserId="8984" LastEditorUserId="65277" LastEditDate="2021-05-28T12:39:14.273" LastActivityDate="2023-04-16T20:17:30.420" Title="Two doors with three guards - one lies, one tells the truth, and one is unreliable" Tags="|logical-deduction|liars|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7964" PostTypeId="2" ParentId="2399" CreationDate="2015-01-19T13:40:41.380" Score="2" Body="&lt;p&gt;According to the programs written on Stack Overflow, you can get the 8192 tile always, after which it depends on luck. Note that they are not 'perfect', and use alpha-beta pruning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are asking about the probability of greater tiles, try the math SE.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-01-19T13:40:41.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7965" PostTypeId="2" ParentId="7959" CreationDate="2015-01-19T13:51:52.300" Score="0" Body="&lt;p&gt;I made a program to simulate this and unless I made a mistake it is not possible. Best that I could find is getting to 8 with 1 charge left.&#xA;Here is it step by step:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;location of robot - location of charge cell - charges - maxCharges - charge cell cooldown&#xA;1 - 6 - 5 - 5 - 0&#xA;2 - 6 - 4 - 5 - 0&#xA;3 - 6 - 3 - 5 - 0&#xA;4 - 6 - 2 - 5 - 0&#xA;5 - 6 - 1 - 5 - 0&#xA;4 - 6 - 1 - 5 - 0&#xA;3 - 6 - 3 - 5 - 0&#xA;2 - 6 - 5 - 5 - 0&#xA;3 - 6 - 4 - 5 - 0&#xA;4 - 6 - 3 - 5 - 0&#xA;5 - 6 - 2 - 5 - 0&#xA;6 - 6 - 1 - 6 - 2&#xA;5 - 6 - 1 - 6 - 1&#xA;4 - 6 - 3 - 6 - 0&#xA;3 - 6 - 5 - 6 - 0&#xA;4 - 6 - 4 - 6 - 0&#xA;5 - 6 - 3 - 6 - 0&#xA;6 - 6 - 2 - 6 - 2&#xA;5 - 6 - 2 - 6 - 1&#xA;4 - 6 - 4 - 6 - 0&#xA;3 - 6 - 6 - 6 - 0&#xA;4 - 6 - 5 - 6 - 0&#xA;5 - 6 - 4 - 6 - 0&#xA;6 - 6 - 3 - 6 - 2&#xA;7 - 7 - 1 - 7 - 1&#xA;6 - 7 - 1 - 7 - 0&#xA;5 - 7 - 3 - 7 - 0&#xA;4 - 7 - 5 - 7 - 0&#xA;3 - 7 - 7 - 7 - 0&#xA;4 - 7 - 6 - 7 - 0&#xA;5 - 7 - 5 - 7 - 0&#xA;6 - 7 - 3 - 7 - 0&#xA;7 - 7 - 2 - 7 - 2&#xA;8 - 8 - 0 - 8 - 1&#xA;7 - 8 - 0 - 8 - 0&#xA;6 - 8 - 2 - 8 - 0&#xA;5 - 8 - 4 - 8 - 0&#xA;4 - 8 - 6 - 8 - 0&#xA;3 - 8 - 8 - 8 - 0&#xA;4 - 8 - 7 - 8 - 0&#xA;5 - 8 - 6 - 8 - 0&#xA;6 - 8 - 4 - 8 - 0&#xA;7 - 8 - 2 - 8 - 0&#xA;8 - 8 - 1 - 8 - 2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-01-19T13:51:52.300" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7966" PostTypeId="2" ParentId="7959" CreationDate="2015-01-19T13:59:59.863" Score="0" Body="&lt;p&gt;The instructions were a little confusing, but I think I understand. So... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(#) will represent the platforms while [#/#] represents the robots current/maximum charge. &quot;BCC&quot; is Battery Charge Cell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) [5/5]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) [4/5]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) [3/5]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) [2/5]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5) [1/5]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) [1/5]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) [3/5]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) [5/5]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) [4/5]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) [3/5]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5) [2/5]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(6) [0/6] +1 BCC [1/6] &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5) [0/6] Bring BCC to (5)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) [2/6] BCC charged&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) [4/6]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) [6/6]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) [5/6]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) [4/6]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5) [3/6] +1 BCC [4/6]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(6) [2/6] Bring BCC to (6)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(7) [0/7] Bring BCC to (7), BCC charged, +1 BCC [1/7]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(6) [0/7] Bring BCC to (6)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5) [2/7] BCC charged&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) [4/7]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) [6/7]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) [5/7]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5) [4/7]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(6) [2/7] +1 BCC [3/7]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5) [3/7]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) [5/7] BCC charged&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) [7/7]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) [6/7]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5) [5/7]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(6) [3/7] +1 BCC [4/7]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(7) [2/7] Bring BCC to (7)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(8) [0/8] Bring BCC to (8), BCC charged, +1 BCC [1/8]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(7) [0/8] Bring BCC to (7)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(6) [2/8] BCC charged&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5) [4/8]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) [6/8]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) [8/8]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) [7/8]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5) [6/8]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(6) [4/8]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(7) [2/8] +1 BCC [3/8]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(6) [2/8] Bring BCC to (6)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5) [3/8] Bring BCC to (5), BCC charged, +1 BCC [4/8]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) [6/8]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) [8/8] BCC charged&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) [7/8]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5) [6/8] +1 BCC [7/8]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(6) [5/8] Bring BCC to (6)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(7) [3/8] Bring BCC to (7), BCC charged, +1 BCC [4/8]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(8) [2/8] Bring BCC to (8)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(9) [0/9] Bring BCC to (9), BCC charged, +1 BCC [1/9]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just enough charged to jump off platform (9), even.&lt;/p&gt;&#xA;" OwnerUserId="8858" LastActivityDate="2015-01-19T13:59:59.863" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7967" PostTypeId="2" ParentId="7959" CreationDate="2015-01-19T14:04:14.623" Score="3" Body="&lt;p&gt;Based on Ivo's answer, where you're at spot #8 with 1 charge (from the charger):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the 1 charge to move the charger backwards ( to #7).&lt;br&gt;&#xA;Hop back 5 to #2 (8 charges).&lt;br&gt;&#xA;3 charges to #5, 4 charges to #7, left with 1 charge to move the charger back to #6. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At that point, hop back 4 to #2 (8 charges).&lt;br&gt;&#xA;Forward to #6 (4 charges).&lt;br&gt;&#xA;Back to #3 (8 charges).&lt;br&gt;&#xA;Forward to #6 (5 charges).&lt;br&gt;&#xA;Forward to #7 (3 charges, bring charger).&#xA;Forward to #8 (2 charges, bring charger, use charger).&lt;br&gt;&#xA;Forward to #9 (0 charges).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit&lt;/strong&gt;&lt;br&gt;&#xA;I think with some optimization, I can get it down to 40 steps:&lt;br&gt;&#xA;Key = $&amp;lt;location&amp;gt;(&amp;lt;charges&amp;gt;[,&amp;lt;charger&amp;gt;])$&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1(5) 5(1)   2(5)   6(1)   5(0,c) = 15&#xA;2(6) 5(4)   7(1)   6(0,c) 3(6)   = 12 + 15 = 27&#xA;6(3) 5(2,c) 4(4,c) 3(6)   4(6)   = 5 + 27 = 32&#xA;5(5) 4(6,c) 3(7)   4(7)   5(6)   = 5 + 32 = 37&#xA;6(5) 7(3)   8(2)   9(0)          = 3 + 37 = 40&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-01-19T17:07:40.273" LastActivityDate="2015-01-19T17:07:40.273" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7968" PostTypeId="2" ParentId="7963" CreationDate="2015-01-19T14:23:44.290" Score="14" Body="&lt;p&gt;This is a rather weird form of the knight, knave and joker puzzle, with a twist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This part is from &lt;a href=&quot;https://puzzling.stackexchange.com/users/1759/ben-aaronson&quot;&gt;Ben Aaronson&lt;/a&gt; in &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2484/knight-knave-and-joker-puzzle/2501#2501&quot;&gt;this&lt;/a&gt; answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So say the people are A, B and C. You ask A:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Is exactly one of these statements true:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;You are the knight&lt;/li&gt;&#xA;  &lt;li&gt;B is the joker&quot;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;If you get back the answer yes, then the possibilities are:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;A is the knight and B is the knave (1 is true, 2 is false, so one statement true, so the answer is yes which knight truthfully gives)&lt;/li&gt;&#xA;  &lt;li&gt;A is the joker&lt;/li&gt;&#xA;  &lt;li&gt;A is the knave and B is the knight (both statements false, so the answer is no which knave lies about)&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;In all three cases, B is safe&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If you get back the answer no, then the possibilities are:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;A is the knight and B is the joker (both statements true, so the answer is no which knight truthfully gives)&lt;/li&gt;&#xA;  &lt;li&gt;A is the joker&lt;/li&gt;&#xA;  &lt;li&gt;A is the knave and B is the joker (1 is false, 2 is true, so one statement is true so answer is yes which knave lies about)&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;In all three cases, C is safe&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then, just point to a door and ask the safe person: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Would your &lt;strong&gt;exact&lt;/strong&gt; opposite say this door leads to freedom?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you ask the Knave and:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The Knave says No:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The Knight would tell the truth (Yes), but the Knave lies (No).&lt;/li&gt;&#xA;&lt;li&gt;The door is safe; go through.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Knave says Yes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The Knight would tell the truth (No), but the Knave lies (Yes).&lt;/li&gt;&#xA;&lt;li&gt;The door is unsafe; choose the other one.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you ask the Knight and:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The Knight says No:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The Knave lies (No), and the Knight tells the truth (No).&lt;/li&gt;&#xA;&lt;li&gt;The door is safe; go through.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Knight says Yes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The Knave lies (Yes), and the Knight tells the truth (Yes).&lt;/li&gt;&#xA;&lt;li&gt;The door is unsafe; choose the other one.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This puzzle was an interesting mix of two puzzles.&lt;/p&gt;&#xA;" OwnerUserId="5368" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-01-20T21:28:19.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7970" PostTypeId="2" ParentId="7946" CreationDate="2015-01-19T14:48:17.387" Score="5" Body="&lt;p&gt;If the statement is true for one person, then everyone to the right of that person has possibly more knaves to his left, and possibly fewer knights to his right, so the statement is true for everyone on the right of that person. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the statement is false for one person, then everyone to the left of that person has possibly fewer knaves to his left, and possibly more knights to his right, so the statement is true for everyone on the left of that person. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every person actually makes that statement. So if the statement is true/false for a person, then that person is a knight/knave. So if anyone is a knight, then everyone to the right of that knight is also a knight. If anyone is a knave, then anyone to the left of that knave is also a knave. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the leftmost person, the statement is false, so he is a knave. If there was only one person, that person would be a knave. If we have more than one person (we have 2014 actually), then the statement is true for the rightmost person, which has at least one knave to his left and no knight to his right. We now assume we have two or more people. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining what we know, the row consists of a number of x knaves at the left, and y knights at the right. The last knave has x-1 knaves to his left and y knights to his right, and he is lying, so x-1 &amp;lt;= y. The first knight has x knaves to his left and y-1 knights to his right, so x &gt; y-1. This is equivalent to x ≤ y + 1 and x ≥ y; the number of knaves and knights are either equal, or there is one more knave than knights. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the number of people n is even, then there are n/2 knaves and n/2 knights. If the number of people n is odd, then there are (n + 1) / 2 knaves and (n - 1) / 2 knights. &lt;/p&gt;&#xA;" OwnerUserId="4664" LastActivityDate="2015-01-19T14:48:17.387" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7971" PostTypeId="1" AcceptedAnswerId="8017" CreationDate="2015-01-19T15:00:06.347" Score="8" ViewCount="5309" Body="&lt;h2&gt;Story&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;You meet a rich drunk CEO at a roulette table in Vegas. She explains to you that she has a way of almost guaranteeing a profit at this game by betting on only black. In the process she gets complimentary drinks from the waiters for just their tips. Her secret (though she constantly brags about it) is that she has 1 million dollars set aside to back up her bets. She complains, however, that the green numbers (0 and 00) on the board mess up the game and are obviously there for the casino's cut saying something about Europe being much better about this.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Relevant roulette rules&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;When she bets on black numbers (18 of them are black), and the ball lands on a black number, she is handed back twice her bet value for a net profit equal to her bet. The same would be true for red numbers. If she were to bet on &quot;green&quot;, however, and the ball lands on a green number, she is handed back 18 times her bet. She can also land on an individual number (0 and 00 being the only green numbers) and win 36 times her bet if she wins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the table she is at, the minimum bet is \$100 dollars and there is not a maximum bet. The deal will accept any whole dollar amount within that range and he is willing to be there for at least 500 more bets. He doesn't expect tips from you considering how much you have at stake.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The bet&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The drunk CEO decides that because green numbers are so terrible there is no way to win with them.  She, therefore, makes the following bet with you:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;She gives you her million dollar stash and you are not allowed to provide any more money.&lt;/li&gt;&#xA;&lt;li&gt;You are only allowed to bet on green or on one of the two individual green numbers.&lt;/li&gt;&#xA;&lt;li&gt;So you can't just use random walk, you give a number $N$ before starting which is the number of games you will play. She is willing to stay all night (500 games) if she has to. If your strategy requires more games she might be willing to have Jessica run a simulation of it and do pay outs accordingly. You must play every game.&lt;/li&gt;&#xA;&lt;li&gt;If after game $N$, you have more than \$1 million then you win.  If not, you lose.&lt;/li&gt;&#xA;&lt;li&gt;If you win, you get to keep all of the money.&lt;/li&gt;&#xA;&lt;li&gt;If you lose, she takes what you have left, everything you own, and you become her indentured servant for the rest of your life. (Apparently she thinks that is worth $1 million.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So while Jessica (who apparently the CEO met at a poker table last year) is being forced to write up the contract, figure out your strategy. What number $N$ will you pick? How can you maximize the probability you will not be a slave assuming you take the bet? How can you maximize your expected profits (assuming you agree that your life is worth \$1 million)?&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="15544" LastEditDate="2015-09-04T10:59:41.950" LastActivityDate="2018-02-19T09:51:32.237" Title="make a green profit in roulette" Tags="|calculation-puzzle|probability|board-games|" AnswerCount="4" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="7972" PostTypeId="2" ParentId="7959" CreationDate="2015-01-19T15:10:27.300" Score="0" Body="&lt;p&gt;Ok I found the bug in my previous answer. Somehow it didn't account for bringing the charge cell up and recharge at the same time. This is my second attempt. It should also be the least amount of jumps (42 jumps):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1st number is location of robot&lt;br&gt;&#xA;2nd number is location of charging cell&lt;br&gt;&#xA;3rd number is jumps until cell is recharged again&lt;br&gt;&#xA;4th number is max number of charges&lt;br&gt;&#xA;5th number is number of charges&lt;br&gt;&#xA;6th is a boolean indication whether the robot is moving backwards&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1, 6, 0, 5, 5, false&#xA;2, 6, 0, 5, 4, false&#xA;3, 6, 0, 5, 3, false&#xA;4, 6, 0, 5, 2, false&#xA;5, 6, 0, 5, 1, false&#xA;4, 6, 0, 5, 1, true&#xA;3, 6, 0, 5, 3, true&#xA;2, 6, 0, 5, 5, true&#xA;3, 6, 0, 5, 4, false&#xA;4, 6, 0, 5, 3, false&#xA;5, 6, 0, 5, 2, false&#xA;6, 6, 2, 6, 1, false&#xA;5, 5, 1, 6, 0, true&#xA;4, 5, 0, 6, 2, true&#xA;3, 5, 0, 6, 4, true&#xA;2, 5, 0, 6, 6, true&#xA;3, 5, 0, 6, 5, false&#xA;4, 5, 0, 6, 4, false&#xA;5, 5, 2, 6, 4, false&#xA;4, 5, 1, 6, 4, true&#xA;3, 5, 0, 6, 6, true&#xA;4, 5, 0, 6, 5, false&#xA;5, 5, 2, 6, 5, false&#xA;6, 6, 1, 6, 3, false&#xA;7, 7, 2, 7, 2, false&#xA;6, 6, 1, 7, 1, true&#xA;5, 5, 0, 7, 2, true&#xA;4, 5, 0, 7, 4, true&#xA;3, 5, 0, 7, 6, true&#xA;4, 5, 0, 7, 5, false&#xA;5, 5, 2, 7, 5, false&#xA;4, 5, 1, 7, 5, true&#xA;3, 5, 0, 7, 7, true&#xA;4, 5, 0, 7, 6, false&#xA;5, 5, 2, 7, 6, false&#xA;4, 4, 1, 7, 5, true&#xA;3, 4, 0, 7, 7, true&#xA;4, 4, 2, 7, 7, false&#xA;5, 5, 1, 7, 6, false&#xA;6, 6, 2, 7, 5, false&#xA;7, 7, 1, 7, 3, false&#xA;8, 8, 2, 8, 2, false&#xA;9, 9, 1, 9, 0, false&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-01-19T15:10:27.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7973" PostTypeId="1" AcceptedAnswerId="7974" CreationDate="2015-01-19T15:33:03.040" Score="5" ViewCount="906" Body="&lt;p&gt;I wrote a riddle for a word about a year ago. I don't know how hard it is, but it becomes pretty obvious &lt;em&gt;after&lt;/em&gt; you know the answer. Here it is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;River, Sing, Emit&#xA;Rear, Ranging, Mite&#xA;An acronym of its measurement&#xA;Is the first of synonym&#xA;It does not have a synonym&#xA;It does not have an antonym&#xA;Its anagram appears twice&#xA;Its clues are a homonym&#xA;When it's over, it is up&#xA;It controls us - a reality&#xA;We control it - a fantasy, imagination&#xA;Three of its letters appear in every line&#xA;Its third letter is missed in three&#xA;What is the word that has been mentioned?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There is no spelling, grammatical or formatting mistake, any such 'mistake' could be intended.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4269" LastEditDate="2015-05-27T15:04:20.940" LastActivityDate="2015-05-27T15:04:20.940" Title="English-based riddle for a word" Tags="|riddle|word|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7974" PostTypeId="2" ParentId="7973" CreationDate="2015-01-19T15:36:22.733" Score="5" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;River, Sing, Emit&#xA;Rear, Ranging, Mite&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reversing &quot;emit&quot;. Re-arranging &quot;mite&quot;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An acronym of its measurement&#xA;Is the first of synonym&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure. Maybe first refers to &quot;one&quot; which is an anagram of &quot;eon&quot;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It has no synonym&#xA;It has no antonym&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Both are true of time. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Its anagram appears twice&#xA;Its clues are a homonym&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I believe this is a clue to interpreting the first two lines. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When it's over, it is up&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's over and time's up mean the same thing. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It controls us - a reality&#xA;We control it - a fantasy, imagination&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We are mortal and our time is coming. Time travel is fantasy. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Three of its letters appear in every line&#xA;Its third letter is missed in three&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; T, I and E appear in each line. M is missing from lines 9, 10 and 12.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2162" LastEditorUserId="2162" LastEditDate="2015-01-19T15:54:08.050" LastActivityDate="2015-01-19T15:54:08.050" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7976" PostTypeId="1" AcceptedAnswerId="7995" CreationDate="2015-01-19T15:41:33.647" Score="8" ViewCount="2851" Body="&lt;p&gt;Let us give a number to each letter of the alphabet starting at 1 for &lt;code&gt;a&lt;/code&gt; and going to 26 for &lt;code&gt;z&lt;/code&gt;.  The value of a word is just the product of the values for the letters.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The puzzle is to find as many words as possible each of whose product is exactly one million. Each word must be in a different language and must be a word in a respected dictionary for that language.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rules&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any word you use must contain only the letters a to z.  If a language has other letters as well that is fine, you just can't use them in the word.  You can ignore accents however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You must use a different word in each language.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You must not write computer code to find the words. This rule is obviously impossible to enforce so it will involve trust. Please do explain how you found any words you got.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;score&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your score is just the number of different languages. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;leading entries so far&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a score of 5 by Len&lt;/li&gt;&#xA;&lt;li&gt;a score of 1 by theplace and dorothy &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8998" LastEditorUserId="8998" LastEditDate="2015-01-20T21:47:52.667" LastActivityDate="2015-01-22T12:01:32.720" Title="Find words that make a million" Tags="|word|" AnswerCount="5" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="7977" PostTypeId="2" ParentId="6196" CreationDate="2015-01-19T15:54:22.803" Score="3" Body="&lt;p&gt;Image captchas are currently the most brute-force resistant means of asking or confirming information. One could use images to depict a short message. You could also ask the decoder to take the first character of the names of each of the images. For more clarity, you could use a limited dictionary of about 1000 words or so. As long as you use a different image for the same word/character, you cannot be cracked by a computer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that such messages will have to be encrypted/decrypted manually.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-01-19T15:54:22.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7979" PostTypeId="1" AcceptedAnswerId="7983" CreationDate="2015-01-19T16:08:30.570" Score="36" ViewCount="7129" Body="&lt;p&gt;You are asleep on your boat on open sea. When you wake up you discover you have been robbed. A quick inspection of security cam footage reveals that the pirates who robbed you left your ship exactly an hour ago. The sea is flat, extends indefinitely, and is fully covered in a thick persistent fog. You have no idea in which direction the pirates fled. But you do know that these pirates always continue in a straight line at full speed away from their victim. Their maximum speed on open water is 20 nautical miles per hour. Your boat can reach 21 nautical miles per hour. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you catch the pirates?&lt;/p&gt;&#xA;" OwnerUserId="7573" LastEditorUserId="8874" LastEditDate="2016-03-23T14:56:46.960" LastActivityDate="2016-09-01T11:26:27.757" Title="Chasing pirates" Tags="|mathematics|strategy|game|pursuit-evasion|" AnswerCount="6" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7980" PostTypeId="2" ParentId="7976" CreationDate="2015-01-19T16:11:34.533" Score="8" Body="&lt;p&gt;I probably won't be able to get more than the one language I know, but let's explore the problem space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we're looking at a product, we're able to cut down the possible letters in the word drastically. One million is $10^6 = 2^6*5^6$, so any letter whose value has a factor other than 2 or 5 will not be usable. This restricts us to &lt;code&gt;a, b, d, e, h, j, p, t, y&lt;/code&gt;. We need to choose letters such that we have 6 instances each of 2 and 5 in the factorizations. A word can have as many &lt;code&gt;a&lt;/code&gt;s as you want, since 1 will never change the product.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;a = 1&#xA;b = 2^1&#xA;d = 2^2&#xA;e = 5^1&#xA;h = 2^3&#xA;j = 2^1 * 5&#xA;p = 2^4&#xA;t = 2^2 * 5&#xA;y = 5^2&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For example, the word &lt;code&gt;heap&lt;/code&gt; would have seven instances of 2 (three from &lt;code&gt;h&lt;/code&gt;, four from &lt;code&gt;p&lt;/code&gt;) and one instance of 5 from &lt;code&gt;e&lt;/code&gt;, for a final product of 640.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In writing this, I did manage to &lt;del&gt;come up with &lt;code&gt;jetty&lt;/code&gt; in English&lt;/del&gt; miscount the 2s in what I thought was an answer. As there are more possibilities with high numbers of 2s in their factorization, other resulting words will probably also be heavy in letters with 5s.&lt;/p&gt;&#xA;" OwnerUserId="8891" LastEditorUserId="8891" LastEditDate="2015-01-19T16:37:03.127" LastActivityDate="2015-01-19T16:37:03.127" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7983" PostTypeId="2" ParentId="7979" CreationDate="2015-01-19T17:14:48.833" Score="56" Body="&lt;p&gt;If we assume the ocean is flat and extends indefinitely in all directions, there is a strategy that guarantees we can catch the pirates in at most 800,000 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put our current location as the origin of a coordinate system. We will describe our position in polar coordinates, as a function of time: $(r(t),\theta(t))$ (where we have arbitrarily chosen a direction to be $\theta=0$, and $t=0$ is when we realized we had been robbed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We begin by traveling in the $\theta=0$ direction for 20 hours, putting our position at $(420, 0)$. We are then the same distance from the origin as the pirates. Next, we will travel in a spiral, in a manner so that $r'(t)=20$ at all times. This guarantees we will always be the same distance from the origin as the pirates. For $t\geq 20$, we will have $r(t)=420+20(t-20)=20t+20$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our speed is&#xA;$$&#xA;\sqrt{(r')^2+r^2(\theta')^2} = 21\text{ mph},&#xA;$$&#xA;and $r'(t)=20$ for $t&amp;gt;20$, so&#xA;$$&#xA;\theta'(t)=\sqrt{\frac{41}{r^2}}=\frac{\sqrt{41}}{20+20t}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is a $t\geq 20$ for which $\theta(t)$ is the angle in which the pirates fled, we will catch them. This means we will certainly catch the pirates by the time $\theta$ has increased from $0$ to $2\pi$. If $t_0$ is the time this happens, we have&#xA;$$&#xA;2\pi=\int_{20}^{t_0}\theta'(t)\,dt=\int_{20}^{t_0}\frac{\sqrt{41}}{20+20t}dt.&#xA;$$&#xA;Solving for $t_0$ gives&#xA;$$&#xA;t_0=21\mathrm{exp}\left(\frac{40\pi}{\sqrt{41}}\right)-1\approx 7,005,043,026.&#xA;$$&#xA;This means we can catch the pirates in at most 7,005,043,026 hours, or about 800 millennia. Better later than never!&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-01-19T23:29:00.933" LastActivityDate="2015-01-19T23:29:00.933" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="7984" PostTypeId="2" ParentId="7979" CreationDate="2015-01-19T17:20:13.683" Score="22" Body="&lt;p&gt;Drive 20 hours in a direction we will denote with as having $\theta=0$.  Drive in a spiral pattern such that you are $20t+20$ nautical miles away from you starting position.  $t$ is time in hours from right this second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After 20 hours, your distance from the starting position will be $r=20t+20$ while your angle will be... more difficult to find.  If we assume the spiral is of the form $r = ae^{b\theta}$, $a=420$ and $b\theta = \ln\frac{t+1}{21}$.  The arc length for such a curve is $s = a\sqrt{1+b^2}e^{b\theta}/b=r\sqrt{1+b^2}/b$.  The derivative of this with respect to t is $21=v=20\sqrt{1+b^2}/b$.  This means $b = \frac{20}{\sqrt{41}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta = \frac{\sqrt{41}}{20}\ln\frac{t+1}{21}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\theta =2\pi$ then $t=7005043026$ hours&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2015-01-19T17:20:13.683" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7985" PostTypeId="2" ParentId="7971" CreationDate="2015-01-19T17:31:19.967" Score="3" Body="&lt;p&gt;Prelude:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not the final, optimised solution, this is just the idea of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It lacks mathematical conciseness and can be improved in various ways - for example, on your nth attempt, you only need to bet the amount that turns you profit into 50.000-100*n. This is not accounted for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, this solution doesn't maximise the profit. It does however maximise the chance of winning the game, I believe.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The idea is to just bet the amount that almost ensures you can't lose the game, even if you lose all the remaining games at the minimal stake.&#xA;So our general goal is to achieve a profit of 50.000 US Dollar, which ensures we can bet 100 US Dollar 500 times, lose everytime and still end up in profit.&#xA;We will bet on green, because it mostly doesn't matter and it keeps the stakes smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this purpose, we calculate the amount we need to bet to gain 50.000 US Dollar (pure profit): 50.000/17 = 2.942 US Dollar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we lose, we need to win 52.942 US Dollar (actually less, see the prelude), so we bet 52.942/17 = 3.115 US Dollar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We keep doing this until we finally get a winner - for the remaining bets until we arrive at 500, we bet the minimum amount, 100 US Dollar - it is then irrelevant whether we win or lose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bad outcomes for us in this situation are running out of money or losing 500 times in a row. Running out of money on 500 tries means the average stake would have to be 2.000 US Dollar - considering our starting stake of about 3.000 US Dollar, this will happen eventually. For the purpose of this solution, I assume this happens after 100 tries, but could be totally off. (In fact, I am, see comments below.)&#xA;The chance of us losing 100 times are as follows: (36/38)^100 = 0.45%. So we're fairly certain to win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering all this, this turns into a maximisation problem: We want to have to win as little as possible to be secure (so we want to bet rarely), but we want to have as many tries to win as possible (so we want to bet often). There is an ideal N out there for this problem, I'm not the one to produce it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And with our million we free Jessica from her shackles - if she looks good - and live happily ever after.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The End.&lt;/p&gt;&#xA;" OwnerUserId="5265" LastEditorUserId="5265" LastEditDate="2015-01-19T18:28:38.107" LastActivityDate="2015-01-19T18:28:38.107" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="7986" PostTypeId="2" ParentId="7976" CreationDate="2015-01-19T17:31:38.727" Score="10" Body="&lt;p&gt;&lt;code&gt;tetty&lt;/code&gt; is an English word with product 1,000,000, although not very common.  It means touchy, testy, peevish, crabbed.&lt;/p&gt;&#xA;" OwnerUserId="8998" LastActivityDate="2015-01-19T17:31:38.727" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7987" PostTypeId="1" AcceptedAnswerId="8258" CreationDate="2015-01-19T17:31:47.327" Score="11" ViewCount="957" Body="&lt;p&gt;Each of the five teams A, B, C, D, E consists of five table tennis players. In their tournament last week, each player has played one match against each of the twenty players in the other four teams. The players were numbered 1, 2, 3, 4, ..., 25, and by a lucky coincidence &lt;strong&gt;every single match&lt;/strong&gt; in the tournament was won by the player with the smaller number. Furthermore, &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;team A has won at least $x$ matches against B; &lt;/li&gt;&#xA;&lt;li&gt;team B has won at least $x$ matches against C; &lt;/li&gt;&#xA;&lt;li&gt;team C has won at least $x$ matches against D; &lt;/li&gt;&#xA;&lt;li&gt;team D has won at least $x$ matches against E; and &lt;/li&gt;&#xA;&lt;li&gt;team E has won at least $x$ matches against A.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What is the largest value $x$ for which this story could be true?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-05T15:54:07.683" LastActivityDate="2015-02-05T15:54:07.683" Title="Table tennis tournament" Tags="|mathematics|combinatorics|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="7989" PostTypeId="2" ParentId="7979" CreationDate="2015-01-19T17:55:14.247" Score="95" Body="&lt;p&gt;Stay put for &lt;a href=&quot;http://www.wolframalpha.com/input/?i=circumference+of+the+earth+%2F+20+nautical+miles+per+hour&quot;&gt;about 45 days&lt;/a&gt;, after which the pirates would have circumnavigated the globe and returned to your current position.&lt;/p&gt;&#xA;" OwnerUserId="8992" LastEditorUserId="4087" LastEditDate="2015-01-21T00:37:01.293" LastActivityDate="2015-01-21T00:37:01.293" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="7990" PostTypeId="2" ParentId="7971" CreationDate="2015-01-19T18:17:53.887" Score="0" Body="&lt;p&gt;This is a simple answer intended only to show the concept clearly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I set $N=105$ and bet \$1000 each time, I expect to lose \$5250.  There is, however, a greater than $50\%$ chance &lt;a href=&quot;https://math.stackexchange.com/questions/316264/probability-of-winning-in-roulette/316322#316322&quot;&gt;I will make a profit&lt;/a&gt;. This is because if I win I am most likely to win \$3000 ($22\%$ of the time total) while if I lose I lose either \$105000, \$69000, or \$33000.  While I on average lose money, I win more often than I lose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I do this during the game, I will have a $52.4\%$ chance of getting over \$1,000,000 and keeping that money. I will then have a $47.6\%$ of becoming an indentured servant which was valued (probably too high) at \$1,000,000.  This means my expected actual winnings is roughly \$50,000.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are much better solutions than this but this should help counter concerns that taking the bet is a loss with regard to profit.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="-1" LastEditDate="2017-04-13T12:19:36.950" LastActivityDate="2015-01-19T18:17:53.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7991" PostTypeId="2" ParentId="7846" CreationDate="2015-01-19T18:36:36.750" Score="0" Body="&lt;p&gt;Building on the existing answers...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clutching at straws perhaps, but I think the final answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TAURUS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The good disguise wins&#xA;so you act as twins:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; could be V instead of U&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What fits in between&#xA;has been found before&#xA;But one was different,&#xA;and is waiting for more:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Instead of S &amp;amp; U in the top line you have S &amp;amp; V. It's different, you need T and U in between instead of just one letter. One extra letter, A?, and you have the letters of TAURUS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's not only a symbol&#xA;but like a small lagoon&#xA;Rising from the dead&#xA;every afternoon&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Taurus is a star sign, it's rising in the early afternoon where I am at the moment. I've seen it the last few nights, looking out for Comet Lovejoy!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1616" LastActivityDate="2015-01-19T18:36:36.750" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7993" PostTypeId="2" ParentId="7987" CreationDate="2015-01-19T20:51:32.097" Score="4" Body="&lt;p&gt;$x=14$ with the following order:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; abcde bcdea cdeab deabc eabcd&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4430" LastActivityDate="2015-01-19T20:51:32.097" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7994" PostTypeId="2" ParentId="7954" CreationDate="2015-01-19T22:10:16.617" Score="10" Body="&lt;p&gt;OK, I wrote the &quot;twisted solution&quot; above.  I actually wrote it as a joke.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is designed in such a way that if you point at the correct door, every answer works.  Whether the guard is a truth-teller or a liar, and whether the sentence is assumed to be true or false, it works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If however you point at the door to death, it results in a paradox of the type &quot;this sentence is a lie&quot;.  Regardless of your assumptions, it results in a contradiction.  Instead of exposing the paradox, I use the contradiction to deduce that it cannot be the door to death.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, if you point at the door to death and ask that question, neither guard will be able to answer the question in accordance to his nature.  So the argument fails in the assumption that the guard will always answer, the truth-teller will always tell the truth, and the liar will always lie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The guards sometimes can't answer when the question refers to itself.  It creates a &quot;logical larsen&quot;, a loop that makes it impossible to simply evaluate the truth of the proposition.  The guard needs to guess the answer, evaluate the truth of the proposition, possibly lie about it, and hope the answer matches his initial guess.  Sometimes no answer works.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2016-10-01T10:19:10.963" LastActivityDate="2016-10-01T10:19:10.963" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7995" PostTypeId="2" ParentId="7976" CreationDate="2015-01-19T23:20:16.347" Score="9" Body="&lt;p&gt;&lt;strong&gt;Edit 2&lt;/strong&gt; - added word:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Tetete&lt;/strong&gt; - is the name of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Tetet%C3%A9_people&quot;&gt;people&lt;/a&gt; who once lived in Ecuador. The name could have originated from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Siona_language&quot;&gt;Siona language&lt;/a&gt; but I don't know.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Teetet&lt;/strong&gt; - is used in the &lt;a href=&quot;http://ca.wikipedia.org/wiki/Teetet&quot;&gt;Catalan language&lt;/a&gt; as the name of one of Plato's dialogues. In English, the same dialogue is called the Theaetetus.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Teette&lt;/strong&gt; - is a &lt;a href=&quot;http://www.verbix.com/webverbix/Finnish/tehd%C3%A4.html&quot;&gt;Finnish verb&lt;/a&gt; meaning &quot;do you&quot; &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Teaette&lt;/strong&gt; - is an English word meaning &lt;a href=&quot;http://www.merriam-webster.com/dictionary/teaette&quot;&gt;&quot;tea maker&quot;&lt;/a&gt;. And then I found a &lt;a href=&quot;http://news.google.com/newspapers?nid=266&amp;amp;dat=19881026&amp;amp;id=5dkrAAAAIBAJ&amp;amp;sjid=UWQFAAAAIBAJ&amp;amp;pg=2702,5610860&quot;&gt;1988 article&lt;/a&gt; with a similar puzzle.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Pateyya&lt;/strong&gt; - a Pali word meaning &lt;a href=&quot;http://dictionary.buddhistdoor.com/en/word/99256/pateyya&quot;&gt;&quot;surely fit to have husbands&quot;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-01-20T03:37:33.377" LastActivityDate="2015-01-20T03:37:33.377" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7996" PostTypeId="2" ParentId="12" CreationDate="2015-01-20T00:23:03.167" Score="0" Body="&lt;p&gt;Ok, this thread has &lt;strong&gt;no clear&lt;/strong&gt; answer. However, this example &lt;strong&gt;may prove&lt;/strong&gt; that the answer is that some Sudoku puzzles &lt;strong&gt;require&lt;/strong&gt; a guess as part of the technique. See the three puzzles below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PUZZLE 1 OF 3:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;____642_3  &#xA;_4___89_6&#xA;__6_5_4_7&#xA;6_38__745&#xA;45_6_3821&#xA;___54_639&#xA;2_143_5_8&#xA;7_41_5362&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PUZZLE 2 OF 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;14_9__73_&#xA;763_148__&#xA;__2_3_461&#xA;_37__16__&#xA;_5____14_&#xA;__1___25_&#xA;81__4_9__&#xA;3291__5_4&#xA;_74__931_&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PUZZLE 3 OF 3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;__25___98&#xA;49__2___5&#xA;3_59__2__&#xA;864312759&#xA;153798_2_&#xA;927654831&#xA;2_1__59_3&#xA;____3_582&#xA;53_2__1__&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Plug the puzzles into a Sudoku solver website such as &lt;strong&gt;&lt;a href=&quot;http://sudokuhints.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;&lt;/strong&gt; and you will see that the solver reports back that it is stuck. There is no more logic to use to deduce the solution, therefore you must guess at one of the pairs.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please chime in and agree or disagree with assessment?&lt;/p&gt;&#xA;" OwnerUserId="9004" LastEditorUserId="30903" LastEditDate="2016-12-20T08:22:03.333" LastActivityDate="2016-12-20T08:22:03.333" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7999" PostTypeId="2" ParentId="6352" CreationDate="2015-01-20T02:21:02.173" Score="3" Body="&lt;p&gt;You find &lt;strong&gt;English&lt;/strong&gt; in the tunnel because are in a 1 dollar  bill. The &lt;strong&gt;pyramid&lt;/strong&gt; is the one behind it. The &lt;strong&gt;spin top&lt;/strong&gt; is the pyramid´s top, separated from the rest, where we can see the eye. And the &lt;strong&gt;language&lt;/strong&gt; proves the answer is &lt;em&gt;one dollar bill&lt;/em&gt;, because the greek word &lt;em&gt;logos&lt;/em&gt;(λόγος), means written or spoken words, in the case of the bill, written words. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like it was said, &lt;em&gt;the answer is not Egypt, but somewhere in it&lt;/em&gt;. I would say that &lt;strong&gt;the one dollar bill is in Egypt&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="8850" LastEditorUserId="8850" LastEditDate="2015-01-20T02:26:31.283" LastActivityDate="2015-01-20T02:26:31.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8000" PostTypeId="1" AcceptedAnswerId="8025" CreationDate="2015-01-20T06:59:24.330" Score="12" ViewCount="710" Body="&lt;p&gt;Whenever the Lopsided Mint stamps a coin, its heads-bias $p$ is chosen uniformly at random between $0$ to $1$, which makes it have a probability $p$ of landing heads when flipped. Bob takes a random coin and flips it 10 times. Whats count(s) of heads is he most likely to obtain?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're looking for the mode, not the expectation. The answer should not depend on $p$, since the coin's bias is unknown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(You can solve this by doing some integrals, but there's an elegant argument with no calculation or algebra.)&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="12408" LastEditDate="2016-07-25T02:29:55.017" LastActivityDate="2016-07-25T02:29:55.017" Title="Flipping a randomly-lopsided coin" Tags="|mathematics|probability|coin-tossing|" AnswerCount="6" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8002" PostTypeId="2" ParentId="7976" CreationDate="2015-01-20T07:17:09.007" Score="5" Body="&lt;p&gt;&lt;code&gt;TYPEY&lt;/code&gt; &amp;lt; Meaning: exhibiting superior bodily conformation &lt;a href=&quot;http://www.merriam-webster.com/dictionary/typey&quot;&gt;Link&lt;/a&gt;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2656" LastActivityDate="2015-01-20T07:17:09.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8003" PostTypeId="2" ParentId="8000" CreationDate="2015-01-20T08:10:54.930" Score="0" Body="&lt;p&gt;The answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The coin can be biased in either direction. There is equal probability for the bias to be &lt;code&gt;p&lt;/code&gt; or &lt;code&gt;1-p&lt;/code&gt;. Such coins &quot;even&quot; themselves-out. If we throw both of them, our expectation for &quot;number of heads&quot; is &lt;code&gt;1&lt;/code&gt;. Thus each coin has, such a counterpart. Because of that the expectation of &quot;number of heads&quot; for a randomly chosen &lt;code&gt;p&lt;/code&gt; (that is a coin with such &lt;code&gt;p&lt;/code&gt;) is 0.5. The expectation for 10 flips is 5.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastActivityDate="2015-01-20T08:10:54.930" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8004" PostTypeId="2" ParentId="8000" CreationDate="2015-01-20T08:29:12.297" Score="2" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All the counts are equally likely&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the reasoning is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Likeliness to get n heads = $\int_0^1\ {p^n*(1-p)^{10-n}}\text{ }dp$ * &lt;sup&gt;10&lt;/sup&gt;C&lt;sub&gt;n&lt;/sub&gt;&lt;BR&gt;&#xA; You want n heads(probability p&lt;sup&gt;n&lt;/sup&gt;) and (10-n) tails(probability (1-p)&lt;sup&gt;(10 - n)&lt;/sup&gt;) and there are &lt;sup&gt;10&lt;/sup&gt;C&lt;sub&gt;n&lt;/sub&gt; ways to get n heads out of 10 tosses&lt;BR&gt;&#xA; &lt;BR&gt;&#xA; Solving for n from 0 to 10&lt;BR&gt;&#xA; N = 0 and N = 10: (1&lt;sup&gt;11&lt;/sup&gt;/11 - 0&lt;sup&gt;11&lt;/sup&gt;/11) = 1/11&lt;BR&gt;&#xA; N = 1 and N =  9: ((1&lt;sup&gt;10&lt;/sup&gt;/10 - 1&lt;sup&gt;11&lt;/sup&gt;/11) - (0&lt;sup&gt;10&lt;/sup&gt;/10 - 0&lt;sup&gt;11&lt;/sup&gt;/11)) * (10) / (1) = 1/11&lt;BR&gt;&#xA; N = 2 and N =  8: ((1&lt;sup&gt;9&lt;/sup&gt;/9 - 1&lt;sup&gt;10&lt;/sup&gt;/5 + 1&lt;sup&gt;11&lt;/sup&gt;/11) - (0&lt;sup&gt;9&lt;/sup&gt;/9 - 0&lt;sup&gt;10&lt;/sup&gt;/5 + 0&lt;sup&gt;11&lt;/sup&gt;/11)) * (10 * 9) / (1 * 2) = 1/11&lt;BR&gt;&#xA; N = 3 and N =  7: ((1&lt;sup&gt;8&lt;/sup&gt;/8 - 1&lt;sup&gt;9&lt;/sup&gt;/3 + 3 * 1&lt;sup&gt;10&lt;/sup&gt;/10 - 1&lt;sup&gt;11&lt;/sup&gt;/11) - (0&lt;sup&gt;8&lt;/sup&gt;/8 - 0&lt;sup&gt;9&lt;/sup&gt;/3 + 3 * 0&lt;sup&gt;10&lt;/sup&gt;/10 - 0&lt;sup&gt;11&lt;/sup&gt;/11)) * (10 * 9 * 8) / (1 * 2 * 3) = 1/11&lt;BR&gt;&#xA;N = 4 and N =  6: ((1&lt;sup&gt;7&lt;/sup&gt;/7 - 1&lt;sup&gt;8&lt;/sup&gt;/2 + 2 * 1&lt;sup&gt;9&lt;/sup&gt;/9 - 2 * 1&lt;sup&gt;10&lt;/sup&gt;/5 + 1&lt;sup&gt;11&lt;/sup&gt;/11) - (1&lt;sup&gt;7&lt;/sup&gt;/7 - 1&lt;sup&gt;8&lt;/sup&gt;/2 + 2 * 1&lt;sup&gt;9&lt;/sup&gt;/9 - 2 * 1&lt;sup&gt;10&lt;/sup&gt;/5 + 1&lt;sup&gt;11&lt;/sup&gt;/11)) * (10 * 9 * 8 * 7) / (1 * 2 * 3 * 4) = 1/11&lt;BR&gt;&#xA; N = 5: ((1&lt;sup&gt;6&lt;/sup&gt;/6 - 5 * 1&lt;sup&gt;7&lt;/sup&gt;/7 + 5 * 1&lt;sup&gt;8&lt;/sup&gt;/4 - 10 * 1&lt;sup&gt;9&lt;/sup&gt;/9 + 1&lt;sup&gt;10&lt;/sup&gt;/2 - 1&lt;sup&gt;11&lt;/sup&gt;/11) - (0&lt;sup&gt;6&lt;/sup&gt;/6 - 5 * 0&lt;sup&gt;7&lt;/sup&gt;/7 + 5 * 0&lt;sup&gt;8&lt;/sup&gt;/4 - 10 * 0&lt;sup&gt;9&lt;/sup&gt;/9 + 0&lt;sup&gt;10&lt;/sup&gt;/2 - 0&lt;sup&gt;11&lt;/sup&gt;/11)) * (10 * 9 * 8 * 7 * 6) / (1 * 2 * 3 * 4 * 5) = 1/11&lt;BR&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4637" LastEditorUserId="4637" LastEditDate="2015-01-20T13:35:33.277" LastActivityDate="2015-01-20T13:35:33.277" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8005" PostTypeId="2" ParentId="7954" CreationDate="2015-01-20T08:29:54.017" Score="1" Body="&lt;p&gt;&lt;strong&gt;I think your proposed questions are not very good, because:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Go to any guard, point at a door and ask:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;You are a liar&quot; &lt;strong&gt;- Not a very useful question, as both guards will answer this with no&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;You will reply negatively&quot; &lt;strong&gt;- Not a very useful question, as the truth telling guard cannot answer this one:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;if he answers no, he lies&lt;/li&gt;&#xA;&lt;li&gt;if he answers yes, he lies&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&quot;This door leads to freedom&quot; &lt;strong&gt;- Not a very useful question, as this can go both ways&lt;/strong&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;lying guard, good door =&gt; No&lt;/li&gt;&#xA;&lt;li&gt;lying guard, wrong door =&gt; Yes&lt;/li&gt;&#xA;&lt;li&gt;truth guard, good door =&gt; Yes&lt;/li&gt;&#xA;&lt;li&gt;truth guard, wrong door =&gt; No&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9012" LastActivityDate="2015-01-20T08:29:54.017" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8006" PostTypeId="2" ParentId="7946" CreationDate="2015-01-20T08:48:00.277" Score="1" Body="&lt;p&gt;Answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1007&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Logic -&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any person in the row, the knaves on the left should be &gt; than knights on the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Only FFFFF.....(1007)TTTTT...(1007) combination makes it possible.&lt;/p&gt;&#xA;" OwnerUserId="9011" LastActivityDate="2015-01-20T08:48:00.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8007" PostTypeId="1" AcceptedAnswerId="8008" CreationDate="2015-01-20T09:06:23.737" Score="9" ViewCount="994" Body="&lt;p&gt;The tile-flipping game uses 6 tiles that have a white side and a black side. It starts with the tiles arranged in a line with their white side facing up.&#xA;$$\circ \circ \circ \circ \circ \circ$$&#xA;We then roll an ordinary 6-sided die 99 times. On each roll of the die, we flip over that number of tiles from left to right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if the first roll is 2, we will flip over the first two tiles:&#xA;$$\underline{\bullet \bullet} \circ \circ \circ \circ$$&#xA;If the next roll is 4, we will flip over the first four tiles, which means the first two tiles are flipped back to white:&#xA;$$\underline{\circ \circ \bullet \bullet} \circ \circ$$&#xA;After all the die rolls are done, the black player wins if there are more tiles black-side up than white-side up. If there are more tiles white-side up or it's equal, then the white player wins. (There are no draws.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which player has the advantage? What is their chance of winning?&lt;/p&gt;&#xA;" OwnerUserId="8888" LastEditorUserId="8874" LastEditDate="2015-01-31T17:44:48.137" LastActivityDate="2015-01-31T17:44:48.137" Title="The tile-flipping game" Tags="|probability|game|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8008" PostTypeId="2" ParentId="8007" CreationDate="2015-01-20T09:41:51.193" Score="8" Body="&lt;p&gt;Each player has a 50% chance of winning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, note that the game is unchanged if we add seventh another tile that never flips and so always stays white, since white wins ties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we claim that if one player wins with rolls $r_1, \dots r_{99}$, then the other player wins with opposite rolls $7-r_1, \dots, 7-r_{99}$, which implies they have equal chances of winning. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Resolving a roll of $7-r$ is the same as resolving $r$ except flipping from the right, then flipping every tile. Since flips commute, the entire sequence $7-r_1, \dots, 7-r_{99}$ is equivalent to &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;resolving $r_1, \dots r_{99}$ except from the right, then&lt;/li&gt;&#xA;&lt;li&gt;flipping every tile 99 times, which is equivalent to flipping them all once. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, if $k$ tiles were white with the original rolls, then $k$ tiles are white after step 1 with the mirrored rolls, and $7-k$ are white after step 2, which means the opposite player has the majority and so is the winner.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-01-20T10:40:16.047" LastActivityDate="2015-01-20T10:40:16.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8009" PostTypeId="2" ParentId="8000" CreationDate="2015-01-20T09:43:14.113" Score="0" Body="&lt;p&gt;The mode of a given binomial distribution is floor((n+1)p), except when (n+1)p is an integer, in which case the mode is (n+1)p and (n+1)p - 1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can ignore all the non-integer values, since they are equally distributed over the 11 possible values for the mode. (You can show this with an Excel spreadsheet). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, we only need to look at the integer values of (n+1)p. &#xA;When p is 0/11 or 11/11, the mode is 0 or 10. &#xA;When p is n/11, where n ranges from 1 to 10, the mode is n and n-1. Therefore, for all values of mode from 0 to 10, there are two cases where the mode falls on it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, there is no mode for this distribution of modes, assuming the spread is uniform.&lt;/p&gt;&#xA;" OwnerUserId="7529" LastEditorUserId="7529" LastEditDate="2015-01-20T12:40:58.347" LastActivityDate="2015-01-20T12:40:58.347" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8010" PostTypeId="1" CreationDate="2015-01-20T09:43:53.780" Score="5" ViewCount="270" Body="&lt;p&gt;Let us say that I have a rectangular area that has to always look &quot;filled&quot; with circles. (the void spaces with the given number of circles should be minimal)(Goal)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us assume that, I am also told that there will be &lt;strong&gt;n&lt;/strong&gt; circles that I want to fit into the container and I am also given relative sizes of different circles (i.e say that &lt;strong&gt;n/2&lt;/strong&gt; circles have to be twice the size of the remaining &lt;strong&gt;n/2&lt;/strong&gt; circles). Now the question is whether there is a way to synthesize the size of the circles such that te container looks &quot;filled&quot;.&lt;/p&gt;&#xA;" OwnerUserId="9015" LastEditorUserId="8874" LastEditDate="2016-03-23T14:56:23.240" LastActivityDate="2016-03-23T14:56:23.240" Title="Optimal size of n circles to fit an area" Tags="|mathematics|geometry|packing|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8011" PostTypeId="2" ParentId="8010" CreationDate="2015-01-20T10:07:23.143" Score="7" Body="&lt;p&gt;Your question falls into the area of &quot;circle packing&quot; problems.&#xA;Even some highly restricted special cases of circle packing are very hard and messy, and there exists no general theory for attacking them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good survey on the current knowledge of the area is given at &#xA;&lt;a href=&quot;http://www.packomania.com/&quot;&gt;http://www.packomania.com/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-01-20T10:07:23.143" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8012" PostTypeId="2" ParentId="7954" CreationDate="2015-01-20T12:38:58.920" Score="-3" Body="&lt;p&gt;Simple...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just go to any one of them and ask this ....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I go to other guard and ask &quot;Which gate is the gate to freedom&quot; what will he say.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whatever is the answer of the guard will be (for example: Guard will reply : &quot;He will say the door behind him is door to success&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just go to the opposite door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As because asking this type of question will always return you the false statement &lt;/p&gt;&#xA;" OwnerUserId="9020" LastActivityDate="2015-01-20T12:38:58.920" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8013" PostTypeId="2" ParentId="7963" CreationDate="2015-01-20T12:41:23.620" Score="2" Body="&lt;p&gt;Alternate (but similar) answer again based on determining who is unreliable in the first question&lt;/p&gt;&#xA;&lt;p&gt;This relies on the ability of the truth teller to &lt;em&gt;&lt;strong&gt;never&lt;/strong&gt;&lt;/em&gt; lie, and the liar to &lt;em&gt;&lt;strong&gt;never&lt;/strong&gt;&lt;/em&gt; tell the truth. It assumes that if they are unsure they may not answer as the answer they give may break these absolute rules&lt;/p&gt;&#xA;&lt;h2&gt;Question 1&lt;/h2&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; If I asked &amp;quot;which door is the good door&amp;quot; to the other two guards would their answers be the same?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If asked to the truth teller he cannot answer as he doesn't know, and therefore may lie if he answered.&lt;/li&gt;&#xA;&lt;li&gt;If asked to the liar he cannot answer as he doesn't know, and therefore may tell the  -  truth if he answered.&lt;/li&gt;&#xA;&lt;li&gt;If asked to the unreliable he will answer definitively either way&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Now, we can  identify a &amp;quot;reliable&amp;quot; guard. If they have not answered we know they are reliable, and if they have answered we know the other 2 are &amp;quot;reliable&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;Note that, unlike in Florian's answer in the &amp;quot;Two Guards&amp;quot; version, the problem that we can't know whether the guard can't answer or is still picking the answer, is irrelevant, because once we start talking to the guard while it's picking the guard will tell us to wait.&lt;/p&gt;&#xA;&lt;h2&gt;Question 2&lt;/h2&gt;&#xA;&lt;p&gt;We can now ask an identified &amp;quot;reliable&amp;quot; guard&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; On average if repeatedly asked &amp;quot;Is (identify a door here) the good door?&amp;quot; would the other two guards be more likely to say yes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;if the identified door is the good door:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The truth teller he will say &amp;quot;No&amp;quot;&lt;/li&gt;&#xA;&lt;li&gt;The liar he will say &amp;quot;No&amp;quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;if the identified door is the bad door:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The truth teller he will say &amp;quot;Yes&amp;quot;&lt;/li&gt;&#xA;&lt;li&gt;The liar he will say &amp;quot;Yes&amp;quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;So if the answer is &amp;quot;No&amp;quot; we go through the door we identified, otherwise, if the answer is &amp;quot;Yes&amp;quot; we go through the other door.&lt;/p&gt;&#xA;" OwnerUserId="4150" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-07-11T11:20:09.063" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8014" PostTypeId="1" AcceptedAnswerId="8015" CreationDate="2015-01-20T12:42:16.240" Score="10" ViewCount="827" Body="&lt;p&gt;Professor Kafka tells proudly: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Today I gave a fascinating lecture. &lt;/li&gt;&#xA;&lt;li&gt;There were five students sitting in the first row of the lecture room, and &#xA;each of them was awake as I started my lecture.&lt;/li&gt;&#xA;&lt;li&gt;Each of the five fell asleep exactly twice during my lecture. &lt;/li&gt;&#xA;&lt;li&gt;Furthermore, every pair of these five students was simultaneously asleep at some point in time. &lt;/li&gt;&#xA;&lt;li&gt;At any moment in time, at least three of these five students were awake and listening closely to my lecture.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Is it possible that professor Kafka remembers the situation correctly?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-10-08T15:32:26.693" LastActivityDate="2015-10-08T15:32:26.693" Title="Sleeping students" Tags="|mathematics|combinatorics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8015" PostTypeId="2" ParentId="8014" CreationDate="2015-01-20T13:35:51.883" Score="11" Body="&lt;p&gt;Professor Kafka is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; probably wrong&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We have 10 pairs of students. Each of these pairs must be sleeping separately, or we would fail the 'at least 3 awake at once' condition. We require a student to fall asleep to change the sleeping pair, and 5x2 students allows us 10 state changes. Unfortunately, to get from our presumed (hence the 'probably' above) initial state of 'all students awake' to the first pair we require 2 students to fall asleep, meaning we can only fulfill 9 pairs in total. If a student may start the lecture asleep the solution is trivial.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3987" LastActivityDate="2015-01-20T13:35:51.883" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8016" PostTypeId="2" ParentId="8000" CreationDate="2015-01-20T14:50:34.133" Score="0" Body="&lt;p&gt;A complete statistical answer can be found &lt;a href=&quot;https://math.stackexchange.com/questions/282347/binomial-distribution-with-uniform-parameter&quot;&gt;Here&lt;/a&gt;, on another SE question.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;? It turns out that each count is equally likely, for any n.&lt;/p&gt;&#xA;&lt;p&gt;? Basic approach is that the count at each level is dependent on n and p. If p itself is a random variable ( here, p ~ Uniform(0,1) ), it can be &amp;quot;integrated out&amp;quot; and shown to be constant for any chosen n.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5360" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-01-20T19:03:53.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8017" PostTypeId="2" ParentId="7971" CreationDate="2015-01-20T15:10:01.620" Score="10" Body="&lt;h2&gt;Preamble&lt;/h2&gt;&#xA;&lt;p&gt;Since you are forced to come up with a number of turns you will play, you can't simply walk out after winning the first bet.  In order words, your wins will need to be big enough to cover your subsequent losses.&lt;/p&gt;&#xA;&lt;p&gt;If you bet on green (higher percentage than 0 or 00), then you have a 1/19 chance of winning a bet, and a 18/19 chance of losing.&lt;/p&gt;&#xA;&lt;p&gt;The longer you play, the better your odds of having at least one win is.  Here is a table that shows this.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;Turns  Odds of Losing every game   Odds of Winning at least once&#xA;1       18/19      = 94.7%         5.2%&#xA;10     (18/19)^10  = 58.2%         42.8%&#xA;20     (18/19)^20  = 33.9%         66.1%&#xA;30     (18/19)^30  = 19.8%         80.2%&#xA;40     (18/19)^40  = 11.5%         89.5%&#xA;50     (18/19)^50  =  6.7%         93.3%&#xA;60     (18/19)^60  =  3.9%         96.1%&#xA;70     (18/19)^70  =  2.3%         97.7%&#xA;80     (18/19)^80  =  1.3%         98.7%&#xA;90     (18/19)^90  =  0.8%         99.2%&#xA;100    (17/18)^100 =  0.4%         99.6%&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;As you can see, the more games you play, the better your chances of winning at least once.&lt;/p&gt;&#xA;&lt;h2&gt;General Strategy&lt;/h2&gt;&#xA;&lt;p&gt;So, lets employ the following strategy:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Pick as many turns as we can.&lt;/li&gt;&#xA;&lt;li&gt;Make a medium sized bet initially.&lt;/li&gt;&#xA;&lt;li&gt;Every time we lose, increase the bet slightly to cover the losses we just incurred.&lt;/li&gt;&#xA;&lt;li&gt;If we win, bet the remaining bets at the minimum \$100&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;The value of the bet we choose initially needs to cover any losses we'd incur for the duration of number of turns we choose, plus the value of the original bet.&lt;/p&gt;&#xA;&lt;p&gt;Lets say that we choose 10 turns.  If we win on the first bet, then we need to cover 9 losses at \$100 each, plus the amount of our initial bet.  Since we win &lt;span class=&quot;math-container&quot;&gt;$18$&lt;/span&gt; times what we bet, we end up with this equation:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$18\times bet \ge bet + 100(N-1)$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$bet \ge \frac{100(N-1)}{17}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;If we plug in &lt;span class=&quot;math-container&quot;&gt;$N=10$&lt;/span&gt;, then we get an initial bet of \$53.  But our chances of winning once in 10 turns are only &lt;span class=&quot;math-container&quot;&gt;$42.8\%$&lt;/span&gt;, and we can do better than that.&lt;/p&gt;&#xA;&lt;p&gt;If we try for &lt;span class=&quot;math-container&quot;&gt;$N=500$&lt;/span&gt;, then our chances of winning are 99.99999%! But we have an initial bet of \$2,936.  The problem with this is that since the bet is increasing after every loss, we'd run out of money after only 53 consecutive turns!  This lowers our chances to only 94.3%.  Pretty good, but not optimal.  In fact, since we chose our initial bet based on 500 turns and we are only getting 53 turns, we can lower our initial bet substantially.&lt;/p&gt;&#xA;&lt;h2&gt;Solution&lt;/h2&gt;&#xA;&lt;p&gt;After running a simulation with these numbers, it turns out the sweet spot is to have &lt;span class=&quot;math-container&quot;&gt;$86$&lt;/span&gt; turns, and an initial bet of exactly \$500.&lt;/p&gt;&#xA;&lt;p&gt;If you win the first bet, you pay \$500 and get \$9,000, for a net profit of \$8,500.  This is exactly enough to cover &lt;span class=&quot;math-container&quot;&gt;$85$&lt;/span&gt; turns at the minimum bet.&lt;/p&gt;&#xA;&lt;p&gt;If you lose all your bets, you will still have enough money to make the &lt;span class=&quot;math-container&quot;&gt;$86^{th}$&lt;/span&gt; bet with the chance to win it all back.&lt;/p&gt;&#xA;&lt;p&gt;This gives you a &lt;span class=&quot;math-container&quot;&gt;$0.956\%$&lt;/span&gt; chance of losing the game, but a &lt;span class=&quot;math-container&quot;&gt;$99.044\%$&lt;/span&gt; chance of winning!&lt;/p&gt;&#xA;&lt;h2&gt;Addendum&lt;/h2&gt;&#xA;&lt;p&gt;One thing to note is that this strategy assumes that once you win, you don't need to win any more and the amount you made needs to cover the remaining bets at the minimum, plus any previous bets.&lt;/p&gt;&#xA;&lt;p&gt;The thing is that once you win, and start betting the minimum, you may win some of those bets too.  The fact that you can expect to win some of them likely has a big impact on your actual chances of winning.&lt;/p&gt;&#xA;&lt;p&gt;For example, if we chose &lt;span class=&quot;math-container&quot;&gt;$N=500$&lt;/span&gt;, but chose \$250 as our initial bet, then if we win on the first bet, we would get \$4,500.  The remaining 499 bets would expect to win 1 in every 19.  So, we can expect to get approximately \$47,273 in winnings by betting \$49,900 in 499 minimum bets of \$100.&lt;/p&gt;&#xA;&lt;p&gt;The question is, how much does this affect the returns?  How many more turns can you add with this knowledge?  If &lt;span class=&quot;math-container&quot;&gt;$N=86$&lt;/span&gt; turns gives you a 99% chance of winning, can you do 100 turns with less of an initial bet with the hopes that you'd make back your shortfall in by winning a few times on the minimum bet too?  At what point does the probability of not getting back this money outweigh your overall probability of winning?&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-01-20T15:10:01.620" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="8018" PostTypeId="2" ParentId="8000" CreationDate="2015-01-20T15:48:10.680" Score="5" Body="&lt;p&gt;The answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Either every number between 0 and 10 (inclusive), or the mode doesn't exist, depending on whether you think the mode of a uniform distribution does or does not have a mode.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability of getting $k$ heads is&#xA;$$p(k) = \int_0^1 \binom{10}k x^k(10-x)^{10-k}dx$$&#xA;Integrating $p(k+1)$ by parts yields&#xA;$$p(k+1) = \int_0^1 \binom{10}{k+1}\frac{k+1}{10-k}x^{k+1}(1-x)^{10-k}dx - \left(\left.\frac 1 {10-k} x^{k+1}(1-x)^{10-k}\right|_0^1\right)$$&#xA;The last term vanishes for all k between 0 and 9 (it's undefined for $k=10$). Using the fact that $\binom{10}{k+1} = \frac{10-k}{k+1}\binom{10}k$, the above simplifies to $p(k+1) = p(k)$, valid for for all integral $k$ from 0 to 9. Since $p(0)=\int_0^1 (1-x)^{10}dx = 1/11$, we have $p(k)=1/11$ for all $k$ from 0 to 9 by induction. Finally, $p(10)$ is easily integrated; it too is $1/11$. This is a uniform distribution. There is no mode (or alternately, all integers from 0 to 10 are the mode). &lt;/p&gt;&#xA;" OwnerUserId="9026" LastActivityDate="2015-01-20T15:48:10.680" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8019" PostTypeId="1" AcceptedAnswerId="8021" CreationDate="2015-01-20T15:53:21.657" Score="17" ViewCount="1974" Body="&lt;p&gt;Your programmable drone is pitted against another drone in a rectangular 3D arena measuring 40 meters long, 40 meters wide and 20 meters high. You have equipped your drone with a laser gun that will take your opponent down as soon as it is within 1 meter reach. This laser makes your drone slightly slower (9.70 m/s maximum speed) compared to your opponent's drone which lacks any weapons and which can reach 10.00 m/s. Both drones are equipped with cameras and accurate accelerometers. All this information is available to on-board processors which control the drone movements and which for all practical purposes can be assumed to be fast enough to avoid any measurable latency in the drone's responses to new information. Apart from the laser action, neither drone can influence the movements of the other (air currents and other aerodynamic interactions can be assumed to be insignificant), and both drones are light enough for gravity to be negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your opponent is allowed to select a starting corner, and your drone is placed in the opposing corner, and subsequently both drones are released at the same time. To make things a bit more challenging for you, prior to the drone battle your opponent is allowed a detailed inspection of the program that controls the movements of your drone. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you device a strategy that guarantees your drone takes down the opponent's drone?&lt;/p&gt;&#xA;" OwnerUserId="7573" LastEditorUserId="10615" LastEditDate="2015-04-24T15:07:57.517" LastActivityDate="2015-04-24T15:07:57.517" Title="Game of Drones - Zap it!" Tags="|mathematics|strategy|pursuit-evasion|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8021" PostTypeId="2" ParentId="8019" CreationDate="2015-01-20T17:53:42.577" Score="41" Body="&lt;p&gt;The attack drone should fly to a virtual target that is always at 29/30 of the way from the center of the room to the target drone.  That point moves 3.3% slower than the target drone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the attack drone flies only 3% slower than the target drone, it can reach that virtual target in a finite time regardless how the target drone moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the maximum distance from the center to any place of the room is a convenient 30 meters, the distance between the virtual target and the real target is never more than 1 meter.  So when the attack drone reaches the virtual target, the real target is in range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And Zap!&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2015-01-20T23:02:15.453" LastActivityDate="2015-01-20T23:02:15.453" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8022" PostTypeId="2" ParentId="7979" CreationDate="2015-01-20T19:06:14.023" Score="10" Body="&lt;p&gt;Assuming there is no wind, your boat was completely still on the water before it was boarded by pirates, the pirates stepped off your boat at its center of gravity, and the pirates used motorized boats, the escape vector of the pirates will be the exact opposite of the direction your boat will be drifting in once they &quot;step off&quot; your boat after robbing you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like when you step off a boat in real life, the boat is pushed in the opposite direction that you leave it.&lt;/p&gt;&#xA;" OwnerUserId="9030" LastActivityDate="2015-01-20T19:06:14.023" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8023" PostTypeId="1" AcceptedAnswerId="8029" CreationDate="2015-01-20T19:11:09.847" Score="21" ViewCount="1907" Body="&lt;p&gt;A party is being held at a local mansion. The host is very rich and his success is because of one thing - his famous recipe for spaghetti. He has pushed your tiny spaghetti startup to the side, and sabotaging him is your only hope of achieving dominance in the spaghetti marketplace. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only guests that may attend are people who correctly reply to the guard at the door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You hope to get in without invitation, and dye the spaghetti neon pink so that the rich host's reputation is ruined. You watch as guests come to the door. The guard asks them a question, and they draw a picture in the sand:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The guard says &quot;Three&quot;. The guest draws this picture and is admitted:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zrXtQ.png&quot; alt=&quot;hah, no hints here&quot;&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The guard says &quot;Nine&quot;. The guest draws this picture and is admitted:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mKa60.png&quot; alt=&quot;I said there weren&amp;#39;t any hints&quot;&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The guard says &quot;Forty&quot;. The guest draws this picture and is admitted:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wZex7.png&quot; alt=&quot;still no hints&quot;&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The guard says &quot;Ninety-nine&quot;. The guest draws this picture and is admitted:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hqKih.png&quot; alt=&quot;okay fine.&quot;&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The guard says &quot;Eight&quot;. The guest draws this picture and is forced to work in a spaghetti factory for the rest of his life:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KLqrB.png&quot; alt=&quot;nevermind, no hints.&quot;&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You approach the door. The guard says &quot;Twelve&quot;. What picture should you draw in the sand to be admitted?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hints: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - The puzzle becomes somewhat harder without the whitespace on the drawings. See &lt;a href=&quot;https://puzzling.stackexchange.com/revisions/8023/2&quot;&gt;the revision history&lt;/a&gt; for (giant) versions of the drawings including whitespace.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - Each number corresponds to exactly one drawing, each drawing may correspond to multiple numbers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-01-21T08:00:41.487" Title="Pictures in the sand to satisfy a guard" Tags="|password|" AnswerCount="3" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="8025" PostTypeId="2" ParentId="8000" CreationDate="2015-01-20T19:41:21.173" Score="14" Body="&lt;p&gt;All numbers from 0 to 10 are equally likely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is one way to flip a coin with bias $p$. Choose a random number uniformly from $(0,1)$, and pick heads if that number is less than $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives us a way to simulate Bob's experiment. We choose $p$ uniformly from $(0,1)$, and then choose ten coin-flip numbers uniformly from $(0,1)$. The number of heads is equal to the number of coin-flip numbers less than $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During this simulation, we choose 11 numbers independently and uniformly from $(0,1)$. By symmetry, there is an equal chance of the first chosen number turning out to be the smallest, second smallest, third smallest, ..., or eleventh smallest of the numbers. And the first chosen number is $p$. So by symmetry, there is an equal chance of Bob flipping 0, 1, 2, ..., or 10 heads.&lt;/p&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2015-01-20T19:41:21.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8026" PostTypeId="2" ParentId="8023" CreationDate="2015-01-20T21:38:50.257" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The rule is that the line must cross itself an even number of times. The number is a red herring. I draw a line segment.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-01-20T21:38:50.257" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8028" PostTypeId="2" ParentId="8023" CreationDate="2015-01-21T00:10:11.090" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Is it based on the English spelling on different keyboard layouts? I know the eight is, but I'm not entirely sure of the allowed answers. Is it a different keyboard layout?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9040" LastActivityDate="2015-01-21T00:10:11.090" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8029" PostTypeId="2" ParentId="8023" CreationDate="2015-01-21T00:35:14.213" Score="15" Body="&lt;p&gt;Each picture represents:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a number typed on a standard QWERTY keyboard, where the lines join the positions of two consecutive letters in the English spelling of the number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The number drawn is a different number than the challenge. Three maps to &quot;five&quot;, nine maps to &quot;six&quot;, forty maps to &quot;ten&quot;, and ninety-nine maps to &quot;twelve&quot;. Eight does not map to &quot;eight&quot;, which is what the last guest drew.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Consider the number on a 7-segment display. 3 is displayed with 5 segments, 9 with 6, 40 with 10, and 99 with 12. 8, however, is displayed with 7 segments, not 8.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 12 is displayed with 7 segments, so the correct picture is a mapping of 'seven' onto a QWERTY keyboard.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/5kx2O.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8949" LastEditorUserId="9044" LastEditDate="2015-01-21T08:00:41.487" LastActivityDate="2015-01-21T08:00:41.487" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8031" PostTypeId="2" ParentId="8019" CreationDate="2015-01-21T05:08:39.510" Score="0" Body="&lt;p&gt;You drone should just takeoff and then shoot immediately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The person who sold you a &quot;1 meter reach&quot; laser clearly doesn't understand how light works.&lt;/p&gt;&#xA;" OwnerUserId="4827" LastActivityDate="2015-01-21T05:08:39.510" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8032" PostTypeId="2" ParentId="252" CreationDate="2015-01-21T06:32:19.133" Score="19" Body="&lt;p&gt;Guessing single values in a depth-first search is sub-optimal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, here is a reasoning chain based on a breadth-first hypothesis/disproof method (which my stepson reluctantly calls &quot;educated guessing&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just following the chain including contradictions requires to solve 23 variants of the sudoku, so it's best used with a computer aided solver. However, it does not require any fancy algorithms to follow it. (I use my own home grown unoptimized python program, so there is no real computing power involved either).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The notation follows spreadsheet conventions (column = letter, row = number) (or chess if you will).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;STA Original Sudoku G8: 3,9&#xA;HYP # I8: 3,9&#xA;DIS # I8: 3,9 # B1: 1,2 =&amp;gt; CTR =&amp;gt; B1: 6&#xA;STA # I8: 3,9 + B1: 6&#xA;DIS # I8: 3,9 + B1: 6 # A2: 1,2 =&amp;gt; CTR =&amp;gt; A2: 5,9&#xA;STA # I8: 3,9 + B1: 6 + A2: 5,9&#xA;DIS # I8: 3,9 + B1: 6 + A2: 5,9 # B5: 1,2 =&amp;gt; CTR =&amp;gt; B5: 3,8&#xA;DIS # I8: 3,9 + B1: 6 + A2: 5,9 + B5: 3,8 =&amp;gt; CTR =&amp;gt; I8: 2,7&#xA;STA I8: 2,7&#xA;HYP I8: 2,7 # G7: 5&#xA;DIS I8: 2,7 # G7: 5 # G4: 6 =&amp;gt; CTR =&amp;gt; G4: 1,8&#xA;STA I8: 2,7 # G7: 5 + G4: 1,8&#xA;DIS I8: 2,7 # G7: 5 + G4: 1,8 # C5: 2,9 =&amp;gt; CTR =&amp;gt; C5: 6&#xA;STA I8: 2,7 # G7: 5 + G4: 1,8 + C5: 6&#xA;DIS I8: 2,7 # G7: 5 + G4: 1,8 + C5: 6 # H3: 4,5 =&amp;gt; CTR =&amp;gt; H3: 8&#xA;DIS I8: 2,7 # G7: 5 + G4: 1,8 + C5: 6 + H3: 8 =&amp;gt; CTR =&amp;gt; G7: 3,9&#xA;STA I8: 2,7 + G7: 3,9&#xA;HYP I8: 2,7 + G7: 3,9 # A8: 3,4,6&#xA;DIS I8: 2,7 + G7: 3,9 # A8: 3,4,6 # A9: 3 =&amp;gt; CTR =&amp;gt; A9: 6,7&#xA;STA I8: 2,7 + G7: 3,9 # A8: 3,4,6 + A9: 6,7&#xA;DIS I8: 2,7 + G7: 3,9 # A8: 3,4,6 + A9: 6,7 # D7: 2,7 =&amp;gt; CTR =&amp;gt; D7: 4,9&#xA;STA I8: 2,7 + G7: 3,9 # A8: 3,4,6 + A9: 6,7 + D7: 4,9&#xA;PRF I8: 2,7 + G7: 3,9 # A8: 3,4,6 + A9: 6,7 + D7: 4,9 =&amp;gt; SOL&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I have put up screen shots of the steps and a quick explanation of the method at &lt;a href=&quot;http://sw-amt.ws/sudoku/level-hard/xx-world-hardest-sudoku.html&quot;&gt;World's Hardest Sudoku&lt;/a&gt;. Since I am only interested in solving hard puzzles by &quot;educated guessing&quot;, I found that this sudoku is actually not so hard as advertised (1 level of hypothesis + 1 lookahead = 2 levels of hypotheses). In fact, I have not yet found a sudoku that requires more than 2 levels of hypotheses + one lookahead (= 3 levels of hypotheses).&lt;/p&gt;&#xA;" OwnerUserId="9013" LastActivityDate="2015-01-21T06:32:19.133" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8033" PostTypeId="2" ParentId="8007" CreationDate="2015-01-21T10:06:52.587" Score="1" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Both players have equal chances of winning&lt;BR&gt;&#xA; I have also written a small simulation &lt;a href=&quot;http://ideone.com/oVmO7X&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4637" LastEditorUserId="4637" LastEditDate="2015-01-21T11:02:50.837" LastActivityDate="2015-01-21T11:02:50.837" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8034" PostTypeId="2" ParentId="7929" CreationDate="2015-01-21T10:55:38.943" Score="4" Body="&lt;p&gt;Credits to A E for figuring out that they were National Grid references. I tried MGRS and UTM readings, but they all went into the sea instead. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have downloaded the NX grid map from &lt;a href=&quot;https://www.ordnancesurvey.co.uk/opendatadownload/products.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ordnance Survey&lt;/a&gt;, and plotted the 9 known locations onto the NX grid map, shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There doesn't appear to be any obvious &quot;cross&quot; on the map, but someone with better understanding may be able to figure out something from this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image was generated using the following Mathematica code, where nx was the image.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;point = Show[nx, &#xA;  Graphics[{PointSize[0.010], &#xA;    Point[{{680*4, 509*4}, {739*4, 622*4}, {644*4, 741*4}, {606*4, &#xA;       856*4}, {749*4, 474*4}, {977*4, 542*4}, {601*4, 481*4}, {584*4,&#xA;        604*4}, {866*4, 667*4}}]}]]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/M2DNv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to stupid image size restrictions in the SE uploader, the image was heavily compressed to JPG, and was also shrunk by SE's default image host. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A full resolution PNG is available &lt;a href=&quot;https://whatimg.com/i/sDSeCU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; on a temporary host. Please do suggest decent anonymous image hosts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rewrite on parity solution&lt;/strong&gt; (Credit to Len for the inspiration that the boundaries can also be used to reduce the size of the solution)&#xA;This rewrite aims to include additional information on Kirkcudbrightshire's boundaries, since both Clue 4 and the final whisky location are in Kirkcudbrightshire:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fourth number of the floor of the mean of the 10 numbers is 4 (9-5). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 4 least significant digits of the coordinates of 4 must range from 2503 to 3502 in order for the 4th digit in the floor of the mean to be 4. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the loci of possible positions of clue 4 have been reduced by parity to 10% of the original solution space, assuming all other positions are correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, since both the whiskey point and Answer 4 must be in Kirkcudbrightshire, the answer space is again decreased. An image mask for Kirkcudbrightshire was produced using &lt;a href=&quot;http://www.happyhaggis.co.uk/haggis_files/kirkcudbright3.JPG&quot; rel=&quot;nofollow noreferrer&quot;&gt;the image here&lt;/a&gt; and fitted onto the main NX map grid, in order to determine whether a given NX grid reference is within Kirkcudbrightshire.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/4LBkx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Performing the transform on the mask with this code&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ksData = ImageData[ksMask];&#xA;isKs[a_, b_] := If[ksData[[b]][[a]] &amp;gt; 0, 0, 1];&#xA;whiskeyTransformY[a_, b_] := &#xA;  Mod[Floor[(a*1000 + b + 6451505)/10], 1000] + 1;&#xA;whiskeyTransformX[a_, b_] := Floor[(a*1000 + b + 6451505)/10000] + 1;&#xA;p4Mask = Table[&#xA;   Table[isKs[x, y]*If[2503 &amp;lt; Mod[(1000 x + y), 10000] &amp;lt; 3502, 1, 0]*&#xA;     isKs[whiskeyTransformX[x, y], whiskeyTransformY[x, y]], {x, 1, &#xA;     1000}], {y, 1, 1000}];&#xA;Image[p4Mask]&#xA;wl = Table[Table[0, {x, 1, 1000}], {y, 1, 1000}];&#xA;Table[Table[&#xA;   If[p4Mask[[y]][[x]] == 1, &#xA;    wl[[whiskeyTransformY[x, y]]][[whiskeyTransformX[x, y]]] = 1], {x,&#xA;     1, 1000}], {y, 1, 1000}];&#xA;Image[wl]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;results in the following images for the &lt;a href=&quot;https://i.stack.imgur.com/j7QNa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;loci of A4&lt;/a&gt; and &lt;a href=&quot;https://i.stack.imgur.com/U1UEx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Whiskey Point&lt;/a&gt; (white: valid NX, black: invalid NX). Their overlays onto the map are shown:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A4 (&lt;a href=&quot;https://whatimg.com/i/KSeqeK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;full res&lt;/a&gt;)&#xA;&lt;img src=&quot;https://i.stack.imgur.com/FsM84.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whiskey Point (&lt;a href=&quot;https://whatimg.com/i/kPH0jJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;full res&lt;/a&gt;)&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Xl0fF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7529" LastEditorUserId="7529" LastEditDate="2015-01-28T13:25:23.633" LastActivityDate="2015-01-28T13:25:23.633" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="8035" PostTypeId="1" AcceptedAnswerId="8037" CreationDate="2015-01-21T11:10:24.683" Score="18" ViewCount="3437" Body="&lt;p&gt;This Puzzle is a form of the classical &lt;a href=&quot;http://de.wikipedia.org/wiki/Minesweeper&quot;&gt;MS Minesweeper&lt;/a&gt; which keeps popping into my head. In Minesweeper, a 2D grid is filled with mines and numbers whereas each number represents how many of its 8 adjacent tiles are occupied by mines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main rule of this puzzle is that only two types of tiles may be used: Mines (m) and &lt;em&gt;one number&lt;/em&gt;. So instead of using all numbers ranging from 0(blank) to 8, you have to select only one of it (except 0) and fill a rectangular grid with it. Of course, the Minesweeper rules still apply so the numbers must still represent the number of adjacent mines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The grid size can be freely chosen but must not be infinite so whatever pattern you may find must work with the edges as well. I do not yet have solutions for all numbers but some of them. As an example, here are the trivial solutions for 1 and 8:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example (1)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;+- - - - - - - - - - - - +&#xA;|1 1 1 1 1 1 1 1 1 1 1 1 |&#xA;|1 m 1 1 m 1 1 m 1 1 m 1 |&#xA;|1 1 1 1 1 1 1 1 1 1 1 1 |&#xA;|1 1 1 1 1 1 1 1 1 1 1 1 |&#xA;|1 m 1 1 m 1 1 m 1 1 m 1 |&#xA;|1 1 1 1 1 1 1 1 1 1 1 1 |&#xA;+- - - - - - - - - - - - +&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example (8)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;+- - - - - - - - - +&#xA;|m m m m m m m m m |&#xA;|m 8 m m 8 m m 8 m |&#xA;|m m m m m m m m m |&#xA;|m m m m m m m m m |&#xA;|m 8 m m 8 m m 8 m |&#xA;|m m m m m m m m m |&#xA;+- - - - - - - - - +&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note:&lt;/em&gt; The above examples do not work for arbitrary grid sizes, only for multiples of 3 in both axes. Such restrictions are allowed for the solutions. Just make sure they are large enough so that an actual pattern can be identified&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note:&lt;/em&gt; You can use any notation for the mines you like as with some numbers, some characters are better distinguishable than others&lt;/p&gt;&#xA;" OwnerUserId="5009" LastActivityDate="2015-01-23T13:59:01.643" Title="Binary Minesweeper" Tags="|logical-deduction|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8036" PostTypeId="2" ParentId="7929" CreationDate="2015-01-21T11:59:38.730" Score="2" Body="&lt;p&gt;I have some serious doubts about this answer, but it's too long for a comment so here goes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I reverse the digits in each row of the question.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I use the first row as Eastings and the second row as Northings.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This gives me gridref NX 11230 71928, which is &lt;a href=&quot;http://gridreferencefinder.com/?gr=NX1123071928%7CNX_s_11230_s_71928%7C1&amp;amp;z=10&amp;amp;v=h&amp;amp;t=NX_s_11230_s_71928&quot; rel=&quot;nofollow&quot;&gt;a point in the hills near Drumcargo&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reasons why this might not be the answer:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;It's quite far away from all the other clues.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There's no major landmark there.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Reason why it could be the answer after all:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;It's publicly accessible dry land in the middle of nowhere. This would make it a good place to bury the prize. Or at least, better than (say) the middle of the Irish Sea or somebody's back garden.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Although it's not very near the other points, it is definitely within driving distance of them.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It's fairly close to the nearest road, which would be an advantage if you were carrying a heavy prize from the car to the burial point.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;OS map shows lots of cairns in the area so you could build a new one to conceal the prize.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Fundamental reason why this answer probably isn't right:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Inevitably there's going to be loads of possible gridrefs that you can get by manipulating the numbers in the clue, in various ways. It's just &lt;a href=&quot;http://en.wikipedia.org/wiki/Numerology&quot; rel=&quot;nofollow&quot;&gt;numerology&lt;/a&gt;. There's few reasons to think that this particular way of manipulating the numbers to get a gridref is superior to any other way of doing it. (The few reasons are that this point is at least not in the sea and not on private property).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Reason why I've posted it even though it's probably wrong:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If this probably-wrong answer gives someone else an idea to get the right answer, maybe that makes it worth posting.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="4050" LastEditDate="2015-01-21T12:23:46.127" LastActivityDate="2015-01-21T12:23:46.127" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="8037" PostTypeId="2" ParentId="8035" CreationDate="2015-01-21T13:08:49.313" Score="16" Body="&lt;p&gt;For 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;+- - - - - - - - - - +&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|m 2 2 m 2 2 m 2 2 m |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|m 2 2 m 2 2 m 2 2 m |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|2 2 2 2 2 2 2 2 2 2 |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|m 2 2 m 2 2 m 2 2 m |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|m 2 2 m 2 2 m 2 2 m |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;+- - - - - - - - - - +&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll leave my solutions for 3 and 4 here, but @Jabe has much better ones.&lt;br&gt;&#xA;For 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;+- - +&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|m m |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|m 3 |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;+- - +&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;+- - +&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|m m |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|4 4 |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|m m |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;+- - +&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For 5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;+- - - - +&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|m m m m |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|m 5 5 m |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|m 5 5 m |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|m m m m |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;+- - - - +&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For 6, I came up with one that's not the same as @Jabe, but turns out to be a bigger version of the one @kasperd has. You can expand this one by making a bigger diamond as well as just sticking blocks together, and you can also nest the diamonds.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;+- - - - - - - +&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|m m m m m m m |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|m m m 6 m m m |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|m m 6 m 6 m m |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|m 6 m m m 6 m |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|m m 6 m 6 m m |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|m m m 6 m m m |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|m m m m m m m |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;+- - - - - - - +&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;... and just because I can:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;+- - - - - - - - - - - +&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|m m m m m m m m m m m |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|m 6 6 m m 6 m m 6 6 m |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|m 6 m m 6 m 6 m m 6 m |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|m m m 6 m m m 6 m m m |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|m m 6 m m 6 m m 6 m m |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|m 6 m m 6 m 6 m m 6 m |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|m m 6 m m 6 m m 6 m m |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|m m m 6 m m m 6 m m m |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|m 6 m m 6 m 6 m m 6 m |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|m 6 6 m m 6 m m 6 6 m |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|m m m m m m m m m m m |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;+- - - - - - - - - - - +&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For 7:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;+- - - - +&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|m m m m |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|m 7 m m |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|m m 7 m |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|m m m m |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;+- - - - +&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And of course, for any n:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;+- - - - +&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|m m m m |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|m m m m |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|m m m m |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;|m m m m |&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;+- - - - +&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6031" LastEditorUserId="6031" LastEditDate="2015-01-23T13:59:01.643" LastActivityDate="2015-01-23T13:59:01.643" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8039" PostTypeId="2" ParentId="8035" CreationDate="2015-01-21T13:48:19.903" Score="12" Body="&lt;p&gt;For 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;pre&gt;&lt;code&gt;+ — — — — — — — — — — — +&#xA;| m m 3 m m 3 m m 3 m m |&#xA;| m 3 3 m 3 3 m 3 3 m 3 |&#xA;| 3 3 3 3 3 3 3 3 3 3 3 |&#xA;| m m 3 m m 3 m m 3 m m |&#xA;| m 3 3 m 3 3 m 3 3 m 3 |&#xA;| 3 3 3 3 3 3 3 3 3 3 3 |&#xA;| m m 3 m m 3 m m 3 m m |&#xA;| m 3 3 m 3 3 m 3 3 m 3 |&#xA;| 3 3 3 3 3 3 3 3 3 3 3 |&#xA;| m m 3 m m 3 m m 3 m m |&#xA;| m 3 3 m 3 3 m 3 3 m 3 |&#xA;+ — — — — — — — — — — — +&lt;/pre&gt;&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;pre&gt;&lt;code&gt;+ — — — — — — — — — — — +&#xA;| m m 4 m m 4 m m 4 m m |&#xA;| m m 4 m m 4 m m 4 m m |&#xA;| 4 4 4 4 4 4 4 4 4 4 4 |&#xA;| m m 4 m m 4 m m 4 m m |&#xA;| m m 4 m m 4 m m 4 m m |&#xA;| 4 4 4 4 4 4 4 4 4 4 4 |&#xA;| m m 4 m m 4 m m 4 m m |&#xA;| m m 4 m m 4 m m 4 m m |&#xA;| 4 4 4 4 4 4 4 4 4 4 4 |&#xA;| m m 4 m m 4 m m 4 m m |&#xA;| m m 4 m m 4 m m 4 m m |&#xA;+ — — — — — — — — — — — +&lt;/pre&gt;&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For 6:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;pre&gt;&lt;code&gt;m m m m&#xA;m 6 6 m&#xA;m 6 m m&#xA;m m m m&lt;/pre&gt;&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8992" LastEditorUserId="2066" LastEditDate="2015-01-22T05:06:40.047" LastActivityDate="2015-01-22T05:06:40.047" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8040" PostTypeId="2" ParentId="8035" CreationDate="2015-01-21T16:40:35.780" Score="6" Body="&lt;p&gt;Here are the ones I could come up with. Some of them are similar to what the previous answers have come up with, some of them are different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 0&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0000&#xA;0000&#xA;0000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For 1&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;111&#xA;1m1&#xA;111&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For 2&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2m22m22m2&#xA;2m22m22m2&#xA;222222222&#xA;2m22m22m2&#xA;2m22m22m2&#xA;222222222&#xA;2m22m22m2&#xA;2m22m22m2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For 3&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;3mmmmmmmmmmmmmm3&#xA;mmmmmmmmmmmmmmmm&#xA;mmmmmmmmmmmmmmmm&#xA;3mmmmmmmmmmmmmm3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For 4&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;m4m4m4m4m&#xA;mm4mmm4mm&#xA;m4m4m4m4m&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For 5&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;mmmm&#xA;m55m&#xA;m55m&#xA;mmmm&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For 6&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;mmmmm&#xA;mm6mm&#xA;m6m6m&#xA;mm6mm&#xA;mmmmm&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For 7&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;mmmm&#xA;m77m&#xA;mmmm&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For 8&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;mmm&#xA;m8m&#xA;mmm&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For any number&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;mmmm&#xA;mmmm&#xA;mmmm&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The patterns for 5 through 8 all have &lt;code&gt;m&lt;/code&gt; all around the edges, so the basic building blocks can be placed throughout the grid in any pattern you like by filling in the intermediate blanks with more &lt;code&gt;m&lt;/code&gt;s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My pattern for 4 only works when one of the side lengths is 3. There is a more boring pattern that works when one of the side lengths is 2. And there is an even more boring pattern, which only has numbers near the corners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kudos to @Jabe for having more interesting patterns for 3 and 4 than mine.&lt;/p&gt;&#xA;" OwnerUserId="1619" LastActivityDate="2015-01-21T16:40:35.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8041" PostTypeId="2" ParentId="8007" CreationDate="2015-01-21T16:56:21.520" Score="3" Body="&lt;p&gt;There are three issues here:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;ties go to white&lt;/li&gt;&#xA;&lt;li&gt;there are an odd number of rolls&lt;/li&gt;&#xA;&lt;li&gt;the tiles all start white&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;At first glance it appears the first and last of these give an advantage to white. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Working backwards from the last tile, it will be black if an odd number of sixes are rolled, and white if an even number are rolled. On any two rolls there are 36 possible combinations. Of these, 1 is two sixes, 10 are a six and a not-six, and 25 are zero sixes. Even wins 26 to 10. There are an odd number of total rolls, and on that roll it's 5-1 there will be an even number (0) of sixes. So overwhelmingly (almost 3 to 1) the last tile will be white.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second last tile will be black if the total number of 5s and 6s rolled is odd. 16 results involve only 1 through 4 so have zero 5s and 6s. 16 involve just 1 5 or 6. 4 involve both. So even wins here 20 to 16. Closer than the last digit, but still advantage white.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the &quot;4&quot; tile, you need an odd number of 4/5/6 rolls in total. Still looking at two rolls at a time, of the 36 results 9 will involve zero 4/5/6. 9 will involve only 4/5/6 - an odd number. 9 will involve only 1 and 9 will involve 2. So here's it's 50-50 chance whether that tile ends up white or black.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am skipping the &quot;2&quot; and &quot;3&quot; tiles for lack of time. My intuition is that they match 6 and 5 respectively, but with while and black switched. So overall the chances of &quot;4&quot; being white or black is 50-50, and whatever advantage white has on &quot;6&quot;, black has on &quot;2&quot;, and whatever advantage white has on &quot;5&quot;, black has on &quot;3&quot;. For those last 5 tiles, it's 50-50 who will have more of &quot;their&quot; colour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's where I found myself surprised:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The 1 tile will flip every single time. Roll a 1 - you flip it. Roll a 2 - you flip it. Etc up to 6. It flips 99 times which is odd, therefore it will be black. &#xA;The rules are therefore slightly less unfair than they first seemed. In reality you are dealing with 5 tiles (2 through 6.) If three or more are white, white wins. If three or more are black, black wins (because of the guaranteed black first tile.) Since the odds of each of those 5 tiles being white or black is 50-50, &lt;strong&gt;the game is (surprisingly) fair with each player having an equal chance to win.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If there were 98 or 100 rolls, this would not be the case. &lt;/p&gt;&#xA;" OwnerUserId="1509" LastActivityDate="2015-01-21T16:56:21.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8042" PostTypeId="1" CreationDate="2015-01-21T18:15:41.057" Score="30" ViewCount="3244" Body="&lt;p&gt;Having reached the final stage of a game show, you face an endless row of doors labelled $1$, $2$, $3$, ... The game show host has selected a whole dollar amount, and has put this exact amount as a prize behind the door labelled with that dollar figure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have absolutely no clue what prize is for grabs, and you have to continue opening doors until you hit upon the prize. But be aware: each door you open costs you a dime ($0.10), payable to the game show host.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the opening of a door reveals the prize, it's yours and the game ends. If a door being opened reveals an empty room, the door gets closed again, and the game show host secretly moves the prize one door up (from behind a door labeled $n$ to a door labeled $n+1$), or one door down (from a door labeled $n$ to a door labeled $n-1$). Note: following each closing of a door the prize must move, but it's the game show host's sole decision which move to make.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want to leave this game with a profit, so it is of the essence to find the prize following a limited number of door openings. However, the game show host will do his utmost to reduce the game show cost. And to make things worse: prior to the game and before the game show host decided on the dollar amount for the prize, you have to specify to him the full strategy that you will follow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What will this strategy be?&lt;/p&gt;&#xA;" OwnerUserId="7573" LastEditorUserId="2136" LastEditDate="2015-01-22T17:52:10.920" LastActivityDate="2015-01-24T11:53:07.490" Title="One prize, infinitely many choices" Tags="|mathematics|strategy|optimization|" AnswerCount="10" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="8044" PostTypeId="1" AcceptedAnswerId="76146" CreationDate="2015-01-21T21:50:35.277" Score="7" ViewCount="447" Body="&lt;p&gt;This is a different 'connect the dots' puzzle. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Your task is to connect the blue &lt;em&gt;nodes&lt;/em&gt; in the map below with &lt;em&gt;links&lt;/em&gt; such that the created network guides a &lt;em&gt;signal&lt;/em&gt; emitted &lt;strong&gt;from the red triangle to the green pentagram&lt;/strong&gt;. &lt;br&gt;It is required that the &lt;em&gt;signal&lt;/em&gt; visits &lt;strong&gt;each node exactly once&lt;/strong&gt;.&lt;br&gt; (There is no restriction with respect to &lt;em&gt;links&lt;/em&gt;. They may or may not be used by the signal.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;The following rules apply:&lt;/h2&gt;&#xA;&#xA;&lt;h3&gt;Links:&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;You may create &lt;strong&gt;as many &lt;em&gt;links&lt;/em&gt; as you wish&lt;/strong&gt;. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The first and the last &lt;em&gt;link&lt;/em&gt; (drawn in orange) are fixed and can not be removed.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A &lt;em&gt;link&lt;/em&gt; has to connect two &lt;em&gt;nodes&lt;/em&gt; in a &lt;strong&gt;straight line&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Links&lt;/em&gt; have to &lt;strong&gt;leave &lt;em&gt;nodes&lt;/em&gt; radially&lt;/strong&gt;. &lt;br&gt;(If you prolong the line it has to go through the centre.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Links&lt;/em&gt; &lt;strong&gt;must not cross&lt;/strong&gt; each other at any point.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Links&lt;/em&gt; &lt;strong&gt;must not 'touch'&lt;/strong&gt; any other &lt;em&gt;node&lt;/em&gt;.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Links&lt;/em&gt; and &lt;em&gt;nodes&lt;/em&gt; have the dimensions as in the original map.&lt;br&gt; (They are &lt;em&gt;not&lt;/em&gt; mathematical points and lines.) &lt;br&gt; Line-width is ~4 pixels &lt;br&gt; Node diameter is ~32 pixels&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Examples&lt;/strong&gt; for valid and invalid &lt;em&gt;links&lt;/em&gt; are: &lt;br&gt;&lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/1x12v.png&quot; alt=&quot;Valid and invalid links&quot;&gt;&lt;br&gt;(valid; invalid because crossing; invalid because  touching other node; invalid because not radial)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Signal:&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The &lt;em&gt;signal&lt;/em&gt; travels along a &lt;em&gt;link&lt;/em&gt; from one &lt;em&gt;node&lt;/em&gt; to the next &lt;em&gt;node&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;li&gt;The moment the &lt;em&gt;signal&lt;/em&gt; arrives at a &lt;em&gt;node&lt;/em&gt;, the signal &lt;em&gt;strength&lt;/em&gt; is increased by one.&lt;/li&gt;&#xA;&lt;li&gt;Once a &lt;em&gt;signal&lt;/em&gt; has arrived at the &lt;em&gt;node&lt;/em&gt; it &lt;strong&gt;moves clockwise&lt;/strong&gt; for as many outgoing links as its &lt;em&gt;strength&lt;/em&gt; indicates. (It may do full circles multiple times.)&lt;br&gt; i.e. for a &lt;em&gt;strength&lt;/em&gt; of 3, the &lt;em&gt;signal&lt;/em&gt; would leave the &lt;em&gt;node&lt;/em&gt; via the 3rd link after the one it came from.&lt;/li&gt;&#xA;&lt;li&gt;The &lt;em&gt;signal&lt;/em&gt; is emitted from the &lt;strong&gt;red triangle&lt;/strong&gt; with a signal &lt;em&gt;strength&lt;/em&gt; of zero.&lt;/li&gt;&#xA;&lt;li&gt;The following images illustrate the &lt;em&gt;signal&lt;/em&gt; movement:&lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/YORlc.png&quot; alt=&quot;signal starting at red triangle&quot;&gt;&lt;br&gt;&lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/s8IgN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/0BYXd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;The map&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/U4NxA.png&quot; alt=&quot;The Map&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I do know that &lt;em&gt;at least one&lt;/em&gt; solution exists, but I guess there might be more than one.&#xA;  In your answer, when possible, please give details on how you solved the puzzle or which general (useful) insights you've gained. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="4551" LastEditDate="2015-01-22T03:57:09.290" LastActivityDate="2018-12-08T22:06:53.997" Title="Network connection: Create a network that guides the signal from start to goal" Tags="|visual|network|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8045" PostTypeId="5" CreationDate="2015-01-21T21:52:53.157" Score="0" Body="&lt;p&gt;A puzzle which involves &lt;a href=&quot;https://en.wikipedia.org/wiki/Network&quot; rel=&quot;nofollow noreferrer&quot;&gt;networks&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;If connected to mathematical graph-theory, use with &lt;a href=&quot;/questions/tagged/graph-theory&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;graph-theory&amp;#39;&quot; aria-label=&quot;show questions tagged &amp;#39;graph-theory&amp;#39;&quot; rel=&quot;tag&quot; aria-labelledby=&quot;graph-theory-container&quot;&gt;graph-theory&lt;/a&gt; and &lt;a href=&quot;/questions/tagged/mathematics&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;mathematics&amp;#39;&quot; aria-label=&quot;show questions tagged &amp;#39;mathematics&amp;#39;&quot; rel=&quot;tag&quot; aria-labelledby=&quot;mathematics-container&quot;&gt;mathematics&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;If connected to mazes, use with &lt;a href=&quot;/questions/tagged/mazes&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;mazes&amp;#39;&quot; aria-label=&quot;show questions tagged &amp;#39;mazes&amp;#39;&quot; rel=&quot;tag&quot; aria-labelledby=&quot;mazes-container&quot;&gt;mazes&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="60644" LastEditDate="2022-11-29T12:20:35.233" LastActivityDate="2022-11-29T12:20:35.233" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8046" PostTypeId="4" CreationDate="2015-01-21T21:52:53.157" Score="0" Body="A puzzle which involves networks. Often, such puzzles are closely connected to mathematical graph-theory." OwnerUserId="5840" LastEditorUserId="60644" LastEditDate="2022-11-29T12:20:35.233" LastActivityDate="2022-11-29T12:20:35.233" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8047" PostTypeId="2" ParentId="8044" CreationDate="2015-01-21T22:41:38.203" Score="4" Body="&lt;p&gt;For me it was just a lot of trial and error. The only real &quot;strategy&quot; I had was trying to stick to the outermost nodes as much as possible. Other than that, the fact that a signal with an odd-number strength lets you pass directly through a previously unvisited node helps. (All lines are links. Purple lines trace the actual path of the signal.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/qvNpy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8858" LastEditorUserId="8858" LastEditDate="2015-01-21T23:38:14.830" LastActivityDate="2015-01-21T23:38:14.830" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8050" PostTypeId="2" ParentId="8042" CreationDate="2015-01-22T02:39:52.907" Score="1" Body="&lt;p&gt;A potential answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Choose the highest value you think the game host can afford then make your plan to double your door value each time. (Or even ten fold, etc. Just force him to choose your first door)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Counter argument to any other potential strategy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the game show host HAS to know the strategy, and you HAVE to eventually hit the prize, you can force the game show host to accept whatever value you want. If the game show host knows your strategy is to increase from $1$ to infinity for example, then what stops him from choosing $2$ and increasing it by one each time? If your strategy is to increase from $x$ to infinity, what stops him from choosing $x+1$ and increasing to infinity? If the game show host is forced to choose an option that results in a finish (which is the only way the listed conditions are definitely satisfied), then he will accept the lowest cost possible (your first door in the given strategy).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="7722" LastEditDate="2015-01-22T04:47:15.520" LastActivityDate="2015-01-22T04:47:15.520" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8051" PostTypeId="1" CreationDate="2015-01-22T02:43:01.360" Score="-5" ViewCount="171" Body="&lt;p&gt;I was born may 21 75, little sister oct 29 88, my oldest sister mar 12, 73. Why do our birthdays fall on the same day of the week every year?&lt;/p&gt;&#xA;" OwnerUserId="9061" LastActivityDate="2015-01-22T02:52:19.727" Title="Why birth dates on the same day of the week as my sisters" Tags="|pattern|" AnswerCount="1" CommentCount="2" ClosedDate="2015-01-23T06:22:57.717" ContentLicense="CC BY-SA 3.0" />
  <row Id="8052" PostTypeId="2" ParentId="8051" CreationDate="2015-01-22T02:52:19.727" Score="5" Body="&lt;p&gt;The answer is simply because &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The amount of days between those days of the year are divisible by 7 (the length of a week or a cycle of &quot;days of the week&quot;), so they will always land on the same day no matter the year. If one of those dates was before February of the same year then it'd change with leap years but that isn't the case. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastActivityDate="2015-01-22T02:52:19.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8053" PostTypeId="1" AcceptedAnswerId="8089" CreationDate="2015-01-22T03:05:59.890" Score="4" ViewCount="787" Body="&lt;p&gt;There is a small village of 100 people. An unknown number of these villagers are pick pockets. &lt;em&gt;Each day, you're allowed to take as many villagers as you wish out of the village&lt;/em&gt;. &lt;strong&gt;When a pick pocket has been taken out, he will not pick any pockets.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Each day, pick pockets still in the village will pick one pocket.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Each night, people will report if and how many times they were victim to pick pocketing.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pick pockets do not know who each other are, so they may pick pocket each other. They will also still report if they are victim.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to successfully find who are the pick pockets.&#xA;My question is if there's an optimal solution to this problem, and if there is, what's the worst case scenario?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorDisplayName="user88" LastEditDate="2015-04-29T00:50:24.427" LastActivityDate="2015-04-29T00:50:24.427" Title="A village of pick pockets!" Tags="|logical-deduction|word-problem|" AnswerCount="4" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8054" PostTypeId="2" ParentId="8042" CreationDate="2015-01-22T03:26:04.303" Score="2" Body="&lt;p&gt;Start by opening door 1. If the prize is $1, you win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the prize is $2, then it could either be behind door 1 or door 3. Open door 3. If it is not there, the prize must be back in door 2, so open door 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the prize is $3, then in 3 moves the maximum number door it could be behind is 6. Choose 6. If you don't get the prize, move down one door at a time, until you either get the prize or reach door 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't get the prize, guess that the prize is one dollar higher ($4 in this case), take the total number of moves already made (8 in this case), add that to the value of the prize (4 + 8 = 12) to find the maximum number door possible for that prize value, and start from that number door. Move down one door at a time again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeat until successful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the possible doors the prize can be behind are always two apart from each other, there will never be a situation where you choose a door one door away from the prize, and when you move down, the host moves the prize to the door you just chose. This means you will eventually get the prize no matter what the prize actually is (although for high prize values you will probably ultimately lose money from opening all those doors).&lt;/p&gt;&#xA;" OwnerUserId="8949" LastActivityDate="2015-01-22T03:26:04.303" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="8055" PostTypeId="2" ParentId="8053" CreationDate="2015-01-22T04:14:58.247" Score="1" Body="&lt;p&gt;Here is what I think the optimal solution would be (assuming the goal is to isolate the pickpockets and that you can label people as pickpocket minesweeper flag style):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first day, have two people in the town. If the count is 0 or 2 after that, label the two as either not pickpockets or pickpockets, respectively. If the count is 1, replace one of the people and wait another day. If the count is still 1, replace the same person and wait another day. Keep going like this and when you finally get a count of 0 or 2, you will know what everyone tested is. This will take $x-1$ days for $x$ people tested. This method will guarantee a solution in between 50 days and 100 days depending on how many groups of 2 pickpockets/non-pickpockets you get. Worst case scenario example is having one pickpocket and one non-pickpocket and replacing the non-pickpocket with a non-pickpocket 98 times before comparing two of the non-pickpockets. Best case scenario example is comparing pairs of non-pickpockets or pickpockets each time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Potential shorter number of expected days method:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Day 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wait one day to know how many total pickpockets there are&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Day 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Take out 50 people&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Day 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; From the result of day two, you know how many pickpockets are in each group by subtracting the amount of pickpockets in the 50 remaining from the total. Whichever group had less put half (25) back in the town. Keep track of the 25 staying out of the town.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Day 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can calculate the number of pickpockets in each group of 25 using the logic in day 3. Now, put 25 of the original 50 back in (with the larger amount of pickpockets). Keep track of the 25 staying out of the town like previously.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Day 5 and on:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; At this point, you should know the number of pickpockets in each of the four 25 people groups. Extend this logic to find out the number of pickpockets in the 8 groups of 12(4)/13(4) people. Then extend this to find the number of pickpockets in the 16 groups of 6/7(4). Then, extend this to find the number of pickpockets in the 32 groups of 3/4(4). Then extend this to find the number of pickpockets in the 64 groups of 1/2(36). Test the remaining 36 groups of 2 with known values.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From this logic,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each extension will take the number of groups. For any groups along the way, if the entire group is pickpockets or not pickpockets you'll know and can take it out of the future tests.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Worst case scenario,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If none of the groups are full of pickpockets/non-pickpockets in the groups, that's 160 days worst case, but this number can be dramatically reduced if groups can be taken out early. I may be a bit off in my numbers so I'll be double checking but this should be in the ballpark.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="7722" LastEditDate="2015-01-22T04:26:35.957" LastActivityDate="2015-01-22T04:26:35.957" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8056" PostTypeId="2" ParentId="8042" CreationDate="2015-01-22T04:52:27.917" Score="3" Body="&lt;p&gt;Given that the only constraint on the host is that very every wrong guess n is either incremented or decremented, there are three things which conspire to make it very difficult for you to win this game:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;it's the game show host's sole decision which move to make.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the game show host will do his utmost to reduce the game show cost.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;prior to the game and before the game show host decided on the dollar amount for the prize, you have to specify to him the full strategy that you will follow.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In fact, I would assert that you cannot win at all with any deterministic strategy.  Any such strategy you pick, the host will know for either of his choices what your next move will be.  And to reduce the game show cost the host will pick the route that minimizes the show's cost.  The host can see every possible path, and find the one that will result in you never finding the number, or at the least, making it cost you more to find it than you get back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, if you simply pick 1, 2, 3, 4, etc. in order, then the host will pick a starting number n0, so that when you pick n, the prize $n0 is behind door n+1, and then he will decrement so you are looking in door n+1 when the prize is behind door n.  The host has an unlimited number of choices even: in this case he could pick 2 (you pick 1, he moves 2 to 1 after the door closes, you pick 2, he moves to 2 after the door closes, etc.) or he could pick 2 and increment every time and you would never catch him, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main problem with a deterministic strategy is that the game show host always has a way to escape from your next pick, he can either move up or down to avoid your next move.  The host always has two options (up or down, two possible numbers), and you have one (a single number).  There is always an escape, unless you box him in at the bottom, and this is very costly to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a non-deterministic strategy you would do something where a random number is involved, for instance, you will flip a coin each time to determine what to do, or pick a number from a hat or whatever.  In this case, the host cannot know ahead of time what you will do, so either move he makes is equally likely to wind up being the one you randomly pick.  There are an infinite amount of numbers, but the larger numbers also pay out bigger, and you will eventually hit the correct number, it is only a question of whether or not you will do so before you pay more than the cumulative cost of opening the doors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Arguably, a non-deterministic strategy won't actually be better in EV than a deterministic one, although for any given outcome you &lt;em&gt;could&lt;/em&gt; get lucky and get a good payoff.  So the host might get unlucky and have to pay out a lot, but he probably won't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: A &quot;win&quot; by the host doesn't necessarily mean you don't find the number. It can simply mean that when you do, you have paid the host more than you get back.  Per the constraint given in the question that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You want to leave this game with a profit&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the answer by @Volatility, for example, it can be seen that the strategy is successful in finding the door because you are essentially boxing in the host into winding up at door 2 when you are at a higher numbered even door.  But in such a case, the host will win if he picks any door number higher than 5, which will cost you more to play than you will win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Non-deterministic strategies where you pick a random number slightly above MAXINT (the maximum amount of money the host can put behind the door) and then decrement to 1 while a lot less costly (MAXINT/10 best case) and most likely to win (50%) still have a negative EV because the host will just pick door 1 in that case.&lt;/p&gt;&#xA;" OwnerUserId="2718" LastEditorUserId="2718" LastEditDate="2015-01-22T06:08:38.780" LastActivityDate="2015-01-22T06:08:38.780" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8057" PostTypeId="2" ParentId="8053" CreationDate="2015-01-22T05:15:45.260" Score="7" Body="&lt;p&gt;My solution yield result in&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 50 days (Both in best and worst case)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And here is the approach&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Make groups of 2. Leave first group in village on first night, second group on second night and so on and keep a count of pickpockets in that group.&lt;br&gt;&#xA; If both are pick-pocketed, both are accused. If both aren't victim, none of them is accused. If person 1 is reported victim once, person 2 is accused and vice versa.&lt;br&gt;&#xA; After 50 days you have the list of all pick-pocketers  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4637" LastEditorUserId="2383" LastEditDate="2015-01-22T21:53:53.630" LastActivityDate="2015-01-22T21:53:53.630" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8058" PostTypeId="1" AcceptedAnswerId="8059" CreationDate="2015-01-22T05:32:29.073" Score="8" ViewCount="379" Body="&lt;p&gt;Separate the grids below into groups of squares, such that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Each group contains exactly one numeral.&lt;/li&gt;&#xA;&lt;li&gt;The number of squares equals the numeral in the group.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For example: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HlWDW.png&quot; alt=&quot;Example&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could be solved liked this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1fjsv.png&quot; alt=&quot;Solution&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully this isn't too easy! There may be more than one solution for each grid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6RSB6.png&quot; alt=&quot;Puzzle 1&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/tKDEZ.png&quot; alt=&quot;Puzzle 2&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/dNc6q.png&quot; alt=&quot;Puzzle 3&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9062" LastActivityDate="2023-03-19T11:56:57.587" Title="Dividing Grid by Numbers" Tags="|visual|pencil-and-paper-games|" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8059" PostTypeId="2" ParentId="8058" CreationDate="2015-01-22T05:44:46.280" Score="5" Body="&lt;p&gt;Solution to all 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/2Q1e2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sorry it's a bit sloppy but conveys the idea adequately. I either got lucky or am good at these (or they were easy and had multiple solutions), I had to backtrack like twice total for all three.&lt;/p&gt;&#xA;" OwnerUserId="7722" LastActivityDate="2015-01-22T05:44:46.280" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8060" PostTypeId="2" ParentId="8058" CreationDate="2015-01-22T05:47:59.460" Score="2" Body="&lt;p&gt;It does seem pretty easy, and I imagine there are a lot of solutions for each one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/dpQvo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2015-01-22T05:47:59.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8061" PostTypeId="2" ParentId="8058" CreationDate="2015-01-22T05:52:26.017" Score="3" Body="&lt;p&gt;Simple guess work with no pure logic and easily done.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/vE4XF.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/942oZ.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/RygW3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some suggetions to make it harder and the solution unique:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If say a number is 7, add multiple sevens to ensure that all the 7's form a part of the same block or probably a wall of one of the block. Use different colour to help in distinguishability. (&lt;a href=&quot;https://puzzling.stackexchange.com/questions/6048/pseudoku-a-game-of-finding-order-in-chaos&quot;&gt;Pseudoku: A Game of Finding Order in Chaos&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;Remove a number, say just have 1-7 and remove 8. Let the user figure out where to place 8.&lt;/li&gt;&#xA;&lt;li&gt;border issues: Add constraints that &#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;2 cannot border its multiples likewise for 3.&lt;/li&gt;&#xA;&lt;li&gt;A number cannot border its consecutive number, etc.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Many ways to make this what you intend to. Put your thinking cap and puzzle us :)&lt;/p&gt;&#xA;" OwnerUserId="2656" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-01-22T10:15:28.370" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8062" PostTypeId="2" ParentId="8042" CreationDate="2015-01-22T06:24:27.220" Score="3" Body="&lt;p&gt;I feel like I've almost got the solution, but can't quite figure out the right parameters to use for it. My strategy is basically the same as @Volatility's with one significant change. The trick is to realize&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; we can exclude more than one possibility in each step.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Each step will be very similar to @Volatility's strategy other than the trick:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Exclude a \$2 prize as a possibility by choosing door 2. (1 door opened, smallest possible prized obtained is \$2)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The next two steps are similar:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let's exclude \$1 and \$3 as possibilities by opening door 4 and going down to door 2 (1+3=4 doors opened, \$1 is smallest possible prize obtained). &lt;br&gt;&lt;br&gt; Then we'll exclude \$4, \$6, \$8, and \$10 as possibilities by opening doors 14 down to 2 (4+13=17 doors opened, \$4 SPP obtained)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We continue using this same pattern:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Exclude odd numbers \$5 through \$15 by starting at door 32 and moving down to door 2 (17+31=48 doors opened, \$5 SPP) &lt;br&gt;&lt;br&gt; Exclude evens \$12 through \$28 by opening 66 down to 2 (48+65=113 doors opened, \$12 SPP).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And that's as far as we can get with my current parameters:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Prize $17 could now be behind door 130, meaning we need to have opened a total of 243 doors to ensure that we'll get it. The total number of doors we need to have opened at each step more than doubles at each step, so we need to exclude possibilities at the same rate. I just haven't found a way that works yet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1955" LastActivityDate="2015-01-22T06:24:27.220" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8063" PostTypeId="2" ParentId="8042" CreationDate="2015-01-22T08:34:45.343" Score="2" Body="&lt;p&gt;Here is a bit of analytics to support the idea that you &lt;em&gt;can&lt;/em&gt; win at this game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, it is important to note that in order to trap the host's door, you have to have the same evenness as the host, and a higher value than the host. If your number is lower, the host can simply run away from you to infinity and you will never catch him.  If his door is less than yours, every move the host must change his evenness in sync with you, and you can chase him down 2.  To maximize how much you have to pay the host must always run away towards 1, and to minimize how much you have to pay you must catch him at 2 as he steps back towards you (because he has no place to go).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is important to try to eliminate multiple doors at once.  For instance, if you pick 5, 4, 3, 2, you will win if the host started at 1, 3, or 5.  But if you only eliminate even or odd on each pass you will lose, because you are racing a linear number (how much you get) against an exponential (how much you have to pay) due to the need to keep backtracking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So at a minimum you must catch both evens and odds in a single pass, which means you must repeat each door twice every time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to not lose, you must catch numbers in increasing order, so as to stay in budget for all possible values. So to begin with, assume you will win \$1.00; this gives you a budget of \$0.90, or 9 doors to open if the starting door is 1.  If you pick 5, 5, 4, 4, 3, 3, 2, 2 then you will catch any initial pick from 1-5 in 8 picks or less for a maximum cost of \$0.80 and a minimum profit of \$0.20.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if the host picks a number above 5, then you have a minimum win of \$6, giving you a budget of \$5.00 extra or 50 extra picks to maintain your minimum \$0.20 profit.  If you pick 26, 26, 25, 25, ... , 2, 2 , then what initial number will have gone no higher than 25?  You have opened 8 doors, so 25-8=17.  So any doors up to 17 will now be caught with a minimum profit of \$0.20.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the host picks a number above \$17 you now have a minimum win of \$18, giving you a budget of \$12.00 extra, or 120 extra picks to maintain your minimum \$0.20 profit.  You can thus start at 61, 61, 60, 60, ... 2, 2.  But you have opened 58 doors before you start, and 18 + 58 = 76.  So you can't even catch 18!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you catch fewer numbers to begin with, what happens?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you pick 4,4,3,3,2,2.  You will catch any initial pick in 1-4 in 6 picks or less for a maximum cost of \$0.60 and a minimum profit of \$0.40.  If the host picks a number above 4, then you have a minimum win of \$5, giving you a budget of \$4.00 extra or 40 pics to maintain your \$0.40 profit.  Since you have had 6 picks, you must start at least as high as 5+6+1=12 to catch just 5.  This will cost you 2*(12-2+1) or 22 picks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, you can catch any pick in the range of 1-n in 2*(n-1) picks.&#xA;Whenever you make m picks to get the range 1-n, you now must start at least as high as n+1+m+1 to catch n+1.  since m=2*(n-1), this is n+1+2*n-2+1 or 3*n.  You have a minimum profit of \$n+1, but have already spent your previous \$n0 of that, so you have 5*(\$n+1-\$n0) picks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in order to win you must consistently have 5*(\$n+1-\$n0) &gt;= 3*n.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;5*(n+1-n0) &amp;gt;= 3*n&#xA;5*n+5-5*n0 &amp;gt;= 3*n&#xA;2*n+5 &amp;gt; 5*n0&#xA;n+2.5 &amp;gt; 2.5*n0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It looks like the problem is that n must be at least ~2.5 times greater than n0 each iteration.  But is this doable?  The following chart suggests it is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;     catch  picks   next minprofit maxcost total min/total&#xA;  n   1-n   2*(n-1)  3*n                &#xA;  1   1-1     1      3      1       0.10   0.10 10&#xA;  3   1-3     4      12     3       0.40   0.50 6&#xA;  12  1-12    22     36     12      2.20   2.70 4.4&#xA;  36  1-36    70     108    36     10.80  13.50 2.7&#xA; 108  1-108   214    324    108    32.40  45.90 2.4&#xA; 324  1-324   646    972    324    97.20 143.10 2.3&#xA; 972  1-972  1942   2916    972   194.20 337.30 2.8&#xA;1942  1-1942 3882   5826   1942   388.20 725.50 2.7&#xA;3882  1-3882 7762   11646  3882  1164.60 1890.1 2.0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So this suggests that the optimal sequence to play is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1, 3, 3, 2, 2, 12, 12, 11, 11, ..., 2, 2, 36, 36, ..., 2, 2, 108, 108, ..., 2, 2, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: It looks like there are some mistakes in this table.  maxcost should be picks/10 I think, but it is mistakenly next/10, which is a larger number, so this shouldn't affect the doability, just how profitable it is reported as. I'll fix tomorrow.&lt;/p&gt;&#xA;" OwnerUserId="2718" LastActivityDate="2015-01-22T08:34:45.343" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8064" PostTypeId="2" ParentId="8042" CreationDate="2015-01-22T08:45:18.437" Score="1" Body="&lt;p&gt;I seem to agree with @Michael and @Quark that it's really difficult to win anything in this game.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;each door you open costs you a dime, payable to the game show host&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think it would make most sense to pay that up front - or at least be able to land on a loss rather than only zero win, which makes the game even less worth playing. (If you don't risk any loss, I think any strategy will do as bad as the other, but a random one is likely better suited to even out the better odds for the optimal opponent.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you still must play I'd suggest you play by your own rules to defeat the game.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First open the last door or the first door, whichever is furthest away.&#xA; That's right, keep on walking... we're soon there...&#xA; still just an infinite number doors ahead...&#xA; There is no second step needed.&#xA; Then assume that there is an audience that will grow tired of this show, or that the recording cost isn't negligible.&#xA; And, of course, that the show pays for your hotel nights during the show!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a good strategy, knowing that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, the game show host will do his utmost to reduce the game show cost.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The only way for them to even limit the cost would be to cancel the show or terminate you, before you even open the intended door. They will do the utmost for this, that's your only hope.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Call it the &quot;Hunger Game Solution&quot; if you will.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I can only guess that the host will now say the game is fully VR with instant door-to-door travel time. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As for a puzzle question I think it would be interesting to know why the rules are this hard and then discuss which to change to make the game worth playing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Always avoid games where you face unlimited loss for a slight chance of winning. Nobody playing is the only way to beat the lottery.&lt;/p&gt;&#xA;" OwnerUserId="9064" LastActivityDate="2015-01-22T08:45:18.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8066" PostTypeId="2" ParentId="8058" CreationDate="2015-01-22T09:14:48.660" Score="0" Body="&lt;p&gt;These may also contribute to the answers.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/XRZvG.png&quot; alt=&quot;Image 1&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/cdZDt.png&quot; alt=&quot;Image 2&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/62Cvz.png&quot; alt=&quot;Image 3&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8758" LastEditorUserId="2621" LastEditDate="2023-03-19T11:56:57.587" LastActivityDate="2023-03-19T11:56:57.587" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="8067" PostTypeId="2" ParentId="8042" CreationDate="2015-01-22T10:11:53.140" Score="9" Body="&lt;h2&gt;Solution for the variant with 1 cent per door&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Here is a deterministic strategy for the variant where we only have to pay &lt;strong&gt;1 cent per opened door&lt;/strong&gt; (but that does not work if we have to pay 1 dime as in the actual problem statement.)&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The Strategy:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Here is a deterministic strategy for this problem.  It goes through many phases until it finds the desired door.&#xA;During the $p$-th phase ($p=0,1,2,\ldots$), we open all the doors with numbers between $1$ and $3^p$ in decreasing order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, in an even phase ($p=0,2,\ldots$) we assume that the good door (in the very beginning) started off at an odd position; in an odd phase ($p=1,3,\ldots$) we assume that the good door started off in an even position.&#xA;Since every phase opens an odd number of doors, the odd / even phases are correctly synchronized with the movement of the good door.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;First observations:&lt;/h2&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;In the $p$-th phase, we open exactly $3^p$ doors. &#xA;The total number of doors opened during the first $p$ phases is $N_p=1+3+3^2+\cdots+3^p&amp;lt;3^{p+1}/2$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If at the beginning of an odd phase $p$, the current position of the good door is odd and less or equal to $3^p$, then I will catch the door for sure during this phase. (I am always opening a door with the same parity as the current good door. Hence the door can not pass me by. I will corner and eventually catch it at latest at position 1.) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If at the beginning of an even phase $p$, the current position of the good door is odd and less or equal to $3^p$, then I will catch the door for sure during this phase. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h2&gt;Correctness:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Now suppose that the good door is &lt;strong&gt;initially&lt;/strong&gt; at some position $g$, and let $x$ be the (unique) integer with $3^{x-1}&amp;lt;g\le3^x$. &#xA;By statement (1), after the first $p$ phases, the good door has moved at most $N_p&amp;lt;3^{p+1}/2$ places to the right. &lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;At the beginning of either phase $x+1$ or $x+2$, the current position of the good door must be odd.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;At the beginning of phase $x+1$, the current position of the good door is at most $g+N_x\le 3^x+3^{x+1}/2&amp;lt;3^{x+1}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;At the beginning of phase $x+2$, the current position of the good door is at most $g+N_{x+1}\le 3^x+3^{x+2}/2&amp;lt;3^{x+2}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Statements (4), (5), (6) together with the first observations imply that I will catch the good door for sure before the end of phase $x+2$.&#xA;Hence the strategy is correct and indeed finds the desired door.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Profit analysis:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;By statement (1) above, the total cost of the first $x+2$ phases is at most $N_{x+2}&amp;lt;3^{x+3}/2$ cent.&#xA;My income is $g&amp;gt;3^{x-1}$ dollar, that is, at least $100\cdot3^{x-1}$ cent.&#xA;As the income is larger than my expenses, I always will make profit.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-01-23T14:17:29.097" LastActivityDate="2015-01-23T14:17:29.097" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8068" PostTypeId="2" ParentId="8042" CreationDate="2015-01-22T11:08:25.143" Score="0" Body="&lt;p&gt;I would go with a strategy something like&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;My strategy, Mr. Host, is to tell you that I'm going to start at door 10,000 and then work my way down to door 1, one door at a time. Then, I'm going to tell you that I'm going to do the opposite of my strategy.&quot; Since the entirety of my statement is my strategy, and it itself says I'm going to do the opposite of it, I can go either forward or backward during the game.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alternatively, something like this might work too:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;My strategy is to avoid telling you my strategy.&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Regardless, there really is no &lt;em&gt;optimum&lt;/em&gt; way to solve this, since the host could easily start at door 999,999,999,999,999 and go upward, which practically guarantees that any strategy the player picks will result in him running out of money before he finds the prize. The only way to do it is to start at some arbitrarily high number &lt;em&gt;x&lt;/em&gt;, then choose &lt;em&gt;x&lt;/em&gt;, &lt;em&gt;x&lt;/em&gt;, &lt;em&gt;x&lt;/em&gt; - 1, &lt;em&gt;x&lt;/em&gt; - 1, &lt;em&gt;x&lt;/em&gt; - 2, etc...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: In other words, for you mathematicians out there, the only way to solve this would be to name an upward boundary -- a point such that when the host chooses door &lt;em&gt;y&lt;/em&gt; or greater, the player is guaranteed to win some money using your proposed strategy. Since the host is trying to save the show money, he would reasonably be forced to choose a door less than &lt;em&gt;y&lt;/em&gt;, giving you an upward boundary from which to start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT II: This is my serious guess as to the optimum strategy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Choosing odd numbered doors twice consecutively (1, 1, 3, 3, 5, 5, etc...) will guarantee that you eliminate all doors as you move forward. This keeps you moving linearly at the same rate as the host as long as he steadily increases his door. So the strategy I would follow is this: &quot;I will start at 1, then go upward by odd numbers (1, 3, 5...). However, before I move to the next door, I will flip a coin. If the coin is heads, I will choose that door once. If the coin is tails, I will choose it twice.&quot; This way, you're not guaranteed to find his door, but you at least have a chance, and he cant predict your moves.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8858" LastEditorUserId="8858" LastEditDate="2015-01-22T11:34:18.547" LastActivityDate="2015-01-22T11:34:18.547" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8069" PostTypeId="2" ParentId="7976" CreationDate="2015-01-22T12:01:32.720" Score="5" Body="&lt;p&gt;TEETEED - Estonian, means 'roads of tea'. Actually does not count, since it is compound word and supposedly not in any sensible dictionary and what is worse, in plural. But still, in right context, it could be used and everyone would understand it.&lt;/p&gt;&#xA;" OwnerUserId="9067" LastActivityDate="2015-01-22T12:01:32.720" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8076" PostTypeId="1" AcceptedAnswerId="8437" CreationDate="2015-01-22T17:25:46.470" Score="4" ViewCount="3520" Body="&lt;p&gt;&lt;pre&gt;Let's say we are in a magic land with lots of fantastical animals and plants and whatever...&#xA;Imagine you have way too much time, really like apples and love collecting and growing them. &#xA;In fact, there are two rare kinds of apples and their colours are blue and green. &lt;/pre&gt;&lt;br&gt;&#xA;Just by looking at your table, you can easily tell that the number of blue apples is not a lot more than the number of green apples; let's say less than 10 times.&lt;br&gt;&#xA;From what you've red in your spellbook, 5 out of 6 blue apples are poisonous. I would take care of that. You don't have to, just saying. Anyway, you go ahead and add one blue apple, which is your favourite kind.&lt;br&gt;&#xA;As a magician (did I forget to mention that?), &#xA;you now use a very old spell that doubles the number of green apples immediately.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, you feel like adding another blue apple right after you cautiously use that old spell again but this time for both kinds at once (you are a very audacious person). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;magic sound&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You totally underestimated the power of this spell and you can feel that your body can't take such strong magic easily. You accidentally drop your wand, which is taking action on its own and changes the colour of one apple and also scorches your hat. But no problem, the hat was kind of maroon anyway.&lt;br&gt;&#xA;Left with two kinds of apples and a burnt hat, you divide the bigger group in half. Now you are ready and decide to eat one of each kind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;eating sound&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If done correctly you should have the same number of green and blue apples now. I'm wondering how much money you could get by selling all of them...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, look at all those beautiful colours! Wait...take a few seconds...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can this be real?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I think the question is &quot;Given the described operations and the final assertion that there are the same number of green and blue apples, how many green apples and blue apples did you start with?&quot;&lt;/em&gt; –  apsillers&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ♫ Math, huh, yeah! What is it good for, absolutely nothing ♫&lt;br&gt;I will also accept the mathematical solution (for the sake of Emrakul - may God rest his soul).&lt;br&gt;Note: apples are integers and cutting them is not allowed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5106" LastEditorUserId="5106" LastEditDate="2015-02-04T09:22:16.650" LastActivityDate="2015-02-04T09:22:16.650" Title="The story of green apples and red herrings" Tags="|riddle|calculation-puzzle|lateral-thinking|" AnswerCount="5" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8078" PostTypeId="2" ParentId="8042" CreationDate="2015-01-22T18:54:12.547" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Strategy:&lt;/strong&gt; Choose any prize value/door, and continue guessing the next highest door if you didn't yet get the prize.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This strategy relies on three hard rules and one assumption reworded as the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Rule 1 – You must disclose your strategy to the host before he decides where to place the prize.&lt;/li&gt;&#xA;&lt;li&gt;Rule 2 – The host will minimize the cost to the game show above all else.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Rule 3 – The prize value does not change when it is moved.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Assumption 1 – You can pay the cost of guessing the doors to the host at the end of the game, because you weren't given any starting conditions on your money pool.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Let's consider what happens if we pick any value that is not 1.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Case 1: The host places the prize behind a door lower than what you chose.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because the host can only move the prize to a directly adjacent door, if you keep incrementing your guess by one, you will never uncover the prize and the game will never end.  The cost of continuing to run the game/show will increase towards infinity.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Case 2: The host places the prize behind your door, or behind a door higher than what you chose.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are a number of things that can happen here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Case 2A: The host chooses a value that is not your door and keeps moving the prize to the next highest door no matter what.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again, if you keep incrementing your guess by one, you will never uncover the prize and the game will never end.  The cost of continuing to run the game/show will increase towards infinity.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Case 2B: The host chooses a value that is not your door, increments or decrements at will, but will not keep the game going into perpetuity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you keep incrementing your guess by one, the gap between your guess and the prize will decrease when the host decrements, and you will eventually reach the prize.  This gap is unknown to you, but it doesn't really matter; according to Rule 3, the prize value is set based on the first door that the host chooses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The host can opt to choose a higher value, decrement to near your current choice, increment over a long period of time, then allow you to &quot;win&quot;, so as to dime you out of a profit, but he can do this at any time in likely any situation, so leaving with a profit can't really be a hard rule.  The host must still pay out a higher value than what you had at your door.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Case 2C: The host chooses your door.  The game pays out your prize.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to Rule 1, you must disclose your strategy to the host, and he will thus come to these same conclusions.  Cases 1 and 2A are not viable for the host, as they break Rule 2.  Of Case 2B and 2C, 2C offers the least loss to the game show.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Thus, you should choose any high value, and the host would be forced to choose your door to minimize loss to the game show above all else.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if this isn't the answer you were looking for, Rule 1 is probably the most important factor in the problem.  The true solution likely revolves around limiting what the host can do with his knowledge of exactly what you will do, and forcing him to choose an option that benefits you.&lt;/p&gt;&#xA;" OwnerUserId="9078" LastEditorUserId="9078" LastEditDate="2015-01-22T19:05:40.310" LastActivityDate="2015-01-22T19:05:40.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8080" PostTypeId="2" ParentId="7929" CreationDate="2015-01-22T21:06:39.093" Score="3" Body="&lt;p&gt;Some bits on &lt;em&gt;reverse engineering&lt;/em&gt; the answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we can assume the 9 other clues to be solved correctly, then we can use the &quot;sum all divide by ten&quot; and the &quot;check first number...&quot; clue ( 9-5=&lt;em&gt;4&lt;/em&gt; ) to get at the first digit of our hint 4 solution. We should also be able to determine what the last digit of it &lt;em&gt;is not&lt;/em&gt; if we assume that &quot;ignore the after comma digit&quot; hints that it is likely &lt;em&gt;not&lt;/em&gt; zero! &#xA;Ok, that's only 2 out of 6 coordinate grid digits, but it limits further searches. (That's why I would like to have the 9 found numbers be fully confirmed. The image is a bit dubious about 749 479 vs 749 472, and Elisabeth also brought in some doubt on the last &quot;rounded&quot; digit of Q3.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 2 more points in the general instructions which make me wonder: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;if we need the sum (and maths) to get to our final coordinates,&lt;em&gt;why are we required to mark the 10 solutions on the map?&lt;/em&gt; This implies there &lt;em&gt;has&lt;/em&gt; to be some information when looking on the dots of the map, doesn't it?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;Stand on one...&quot; This seems to indicate the &lt;em&gt;final&lt;/em&gt; destination has(/had!) &lt;em&gt;something&lt;/em&gt; to stand on, which exists in multiple instances and is &lt;em&gt;recognizable&lt;/em&gt; for that spot. What could that be? Are there some hills with special rocks, maybe?  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Finally, &quot;look for a tree taller than you&quot;. This might be &lt;em&gt;really bad news&lt;/em&gt; for a 35year old puzzle! I am just glad that we all are only interested in the solution to the puzzle, and not in the whiskey :-). It might be hard to impossible to find it even &lt;em&gt;with&lt;/em&gt; the right solution!&lt;/p&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="5840" LastEditDate="2015-01-24T07:58:13.143" LastActivityDate="2015-01-24T07:58:13.143" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8081" PostTypeId="2" ParentId="8076" CreationDate="2015-01-22T21:07:23.097" Score="8" Body="&lt;p&gt;Assuming I'm interpreting all this correctly and the bits that sound optional really aren't, it's not too hard to start at the end and work backward math-wise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end of the story, you end up with $n$ each of green and blue apples. So before you ate one of each, you have $n+1$ of each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step going backward is the halving of the larger group, which will result in one side doubling in our backwards work. The only other operations we'll perform on the green apples will be adding or removing one for the random color switch and halving for each of the two times the number of green apples was doubled. If the green group was larger at this point, we'll have an even number flipped to odd with the color change, which results in a non-integer result for the starting number of green apples. Therefore, the blue group must have been larger. At this point, we had $n+1$ green apples and $2n+2$ blue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The unspecified color swap introduces a split in the possibilities. Before it we either had $n+2$ green and $2n+1$ blue with a green being swapped to blue, or $n$ green and $2n+3$ blue with an opposite swap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking away a blue for the one that was added before that step gives us either $n+2$ green and $2n$ blue, or $n$ green and $2n+2$ blue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, both piles were doubled, so both are halved. We have $\frac{n+2}{2}$ green and $n$ blue, or we have $\frac{n}{2}$ green and $n+1$ blue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another halving for the greens puts us at $\frac{n+2}{4}$ green and $n$ blue or $\frac{n}{4}$ green and $n+1$ blue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take back the extra blue we put in, and you arrive at $\frac{n+2}{4}$ green and $n-1$ blue or $\frac{n}{4}$ green and $n$ blue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the only thing left is to multiply the blues by 6, as I assume the bit about poisoning means you threw out 5/6 of the blue apples. This puts us at either $\frac{n+2}{4}$ green and $6n-6$ blue or $\frac{n}{4}$ green and $6n$ blue. Here's where we can use the initial constraint of starting with less than 10 times as many blues as greens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the swap was from blue to green, at this point we have $\frac{n}{4}$ green apples and $6n$ blue apples. This means we have 24 times as many blues and greens, so that can't be true. The swap was from green to blue, giving us $\frac{n+2}{4}$ green and $6n-6$ blue apples at the beginning. Using $k=\frac{n+2}{4}$, $n=4k-2$. So we have $k$ green apples and $24k-18$ blue apples. If $k$ is 2, we started with 2 green and 30 blue, which again violates the initial condition. However, starting with 1 green and 6 blue apples works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Run through with 1 green and 6 blue apples, and you end up with 2 each of blue and green. However, let's assume the doubling spell creates blue apples with the same poison status as the ones you had before you used it. The blue apple you added could be poisoned and subsequently doubled, and both the second added blue and the green one that was color-swapped to blue could be poisoned. So when you get to the point of halving your number of blue apples, there is a $\frac{1}{216}$ chance none of your blues are poisonous, $\frac{10}{216}$ you have 1 poisonous blue, $\frac{30}{216}$ you have 2, $\frac{50}{216}$ you have 3, and $\frac{125}{216}$ you have 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's further assume that the half of the blue bunch to discard was chosen randomly. This gets very scribbly on my paper, but at the end of it we find there's a $\frac{17}{54}$ chance you have 1 poisoned apple in the final group, a $\frac{23}{48}$ chance you have 2, and a $\frac{55}{432}$ chance you have 3. Finally, assuming the apple to eat among that group is again chosen randomly, we end up with a $\frac{716}{1296}$, or roughly 55%, chance you are now poisoned. Hope you're feeling lucky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 2: You didn't divide by 6&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also apparently possible that you didn't divide the blue apples by 6. This takes things from a single solution to an infinite number of solutions. Everything is the same as in the previous, except we get to the point where we have either $\frac{n+2}{4}$ green and $n-1$ blue or $\frac{n}{4}$ green and $n$ blue, depending on the direction of the color switch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a green apple was turned blue, we're in the first case. We can use a similar substitution with $k=\frac{n+2}{4}$ to find that we started with $k$ green and $4k-3$ blue apples. This passes the initial condition for all positive integers $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a blue apple was turned green, we're in the second case. Even easier substitution this time, $k=\frac{n}{4}$ means we started with $k$ green and $4k$ blue apples. Once again, this works for all positive integers $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;tl;dr:&lt;/strong&gt; You started with 1 green and 6 blue apples. Assuming perfect duplication and no checking of added blue apples for poison, you have a 55% chance of now being poisoned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you started with $k$ green and either $4k$ or $4k-3$ blue. You are more likely to be poisoned in these cases.&lt;/p&gt;&#xA;" OwnerUserId="8891" LastEditorUserId="8891" LastEditDate="2015-01-29T15:44:00.260" LastActivityDate="2015-01-29T15:44:00.260" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="8083" PostTypeId="2" ParentId="8076" CreationDate="2015-01-22T21:36:17.627" Score="5" Body="&lt;p&gt;Work backwards. X = green, Y = blue. X is the larger half.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a blue apple was changed to green&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x = y as initial condition. Both apples had one eaten, therefore x + 1 = y + 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X was halved, so to reverse it we multiply. 2x + 2 = y + 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reverse the color change, resulting in 2x + 1 = y + 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You added one blue apple, so we subtract 1 to get 2x + 1 = y + 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, both sides were multiplied by 2 (so we divide) and we get (2x + 1)/2 = (y + 1)/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's even it out a bit. Multiply the 2 on the Y side to the X side, giving us 2x + 1 = y + 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we see that the green apples were doubled prior to that, so divide the x side again; however, we want to avoid fractions, so lets immediately multiply everything by 2 and get rid of the denominator. This gives us 2x + 1 = 2y + 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, there was the 1 blue apple we added right at the start; subtract this from the y side. Now we have &lt;code&gt;2x + 1 = 2y + 1&lt;/code&gt;. (Case 1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following the same process, but with the assumption of the color changing from green to blue, our final result (If i did my math right) is &lt;code&gt;(2x + 3)/2 = y - 2&lt;/code&gt; (Case 2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case 1: Therefore, you started with the same amt of blue and green apples (which is wrong since the question states that there were more green apples) - we move to case 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Case 2: minimum of Y = 2 (since y - 2 cannot be negative). If we started with Y = 2, we would end up with [2(0) + 3]/2 on X side (which doesn't make sense; therefore, we also get rid of &quot;exactly 2&quot; Y case. In fact, we can see that in any scenario where we start with an even number of Y, X side cannot be divided by 2 evenly, so we can reason that you MUST start Y with minimum 3, but in increments of 2, such that we always start Y with an odd number of minimum 3. You must also notice however, that 2Y is always an even number (anything * 2 is even). If X = y, and 2X is also always even, then 2x + 3 is always odd! Uh oh, this is a problem. If 2x + 3 is always odd, you can never divide it by 2 evenly! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the final answer to the actual question that was asked (&quot;How can this be real&quot;) is that IT IS NOT REAL.&lt;/p&gt;&#xA;" OwnerUserId="8999" LastActivityDate="2015-01-22T21:36:17.627" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8084" PostTypeId="2" ParentId="8053" CreationDate="2015-01-23T01:17:32.027" Score="1" Body="&lt;p&gt;Since &quot;optimal&quot; is undefined I suggest the following solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Take one day to count pickpockets. Then take one person out of the village per day. Worst case: after you have taken each person (except one) out of the village, you know who all the pickpockets are. Time: 100 days. &lt;strong&gt;Cost: 99 hotel days&lt;/strong&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8875" LastActivityDate="2015-01-23T01:17:32.027" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8086" PostTypeId="1" CreationDate="2015-01-23T04:03:59.867" Score="17" ViewCount="1768" Body="&lt;p&gt;Tom works with a stingy boss that will deduct from his salary for every minute that he's late for work. Unfortunately, Tom has to pass by the boss's office on the way to his cubicle. Being a lazy worker, Tom decides to arrive at the office exactly 1 minute before his boss every day. Tom does not know when his boss will arrive at the office as the boss's office is located far away from Tom's cubicle. One thing that Tom knows is that the boss will arrive at the office at the same time every day, no sooner, no later. Tom can pull off the &quot;go to the washroom&quot; trick once every day to pass by the boss's office twice (to the washroom and back from the washroom with a 5-minute interval). How many days will it take for Tom to work out his boss's time of arrival?&lt;/p&gt;&#xA;" OwnerUserId="8882" LastEditorUserId="22922" LastEditDate="2016-09-15T08:04:36.603" LastActivityDate="2016-09-15T16:40:41.523" Title="The punctual worker and the stingy boss" Tags="|mathematics|strategy|" AnswerCount="6" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8087" PostTypeId="2" ParentId="8086" CreationDate="2015-01-23T04:21:41.403" Score="5" Body="&lt;p&gt;Does &quot;late for work&quot; mean &quot;arriving later than the boss&quot;? If so, and assuming that asking the boss or his/her secretary directly when they arrive is disallowed:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tom arrives at a definitely late time, and asks finance how much he is being fined. If Tom doesn't know the per minute fine, he arrives one minute earlier the next day and asks again how much, then he can calculate when the boss arrives.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If there is a requirement to minimise the total number of minutes late, then this should be stated.&lt;/p&gt;&#xA;" OwnerUserId="341" LastActivityDate="2015-01-23T04:21:41.403" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8088" PostTypeId="2" ParentId="8086" CreationDate="2015-01-23T05:11:30.170" Score="1" Body="&lt;p&gt;Let's assume that Tom is working full day (8 hours or 480 minutes).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1st day: Tom comes at 480/3=160 minute mark and goes to washroom at 160*2=320 minute mark. That way he will find out the right time with 160 minutes accuracy (divide the whole time by 3)&#xA; &lt;BR&gt;&lt;BR&gt;&#xA; 2nd day: Tom comes at 160/3=53.3 minute mark and goes to washroom at 53.3*2=320 minute mark (minute marks refered to the right time interval). That way he will find out the right time with 53.3 minutes accuracy.&#xA; &lt;BR&gt;&lt;BR&gt;&#xA; 3rd day: 17.7 minutes accuracy.&#xA; &lt;BR&gt;&lt;BR&gt;&#xA; 4th day: 5.9 minutes accuracy.&#xA; &lt;BR&gt;&lt;BR&gt;&#xA; 5th day: 1.9 minutes accuracy.&#xA; &lt;BR&gt;&lt;BR&gt;&#xA; 6th day: 39 seconds accuracy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 6 days.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I guess that &quot;back in 5 minutes&quot; trick will allow to do it in&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;5 days.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9088" LastEditorUserId="4637" LastEditDate="2015-01-23T05:22:06.093" LastActivityDate="2015-01-23T05:22:06.093" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8089" PostTypeId="2" ParentId="8053" CreationDate="2015-01-23T08:51:02.240" Score="7" Body="&lt;p&gt;$41$ days suffice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare &lt;a href=&quot;https://puzzling.stackexchange.com/questions/5190/coin-weighing-with-a-single-weighing-device&quot;&gt;this question&lt;/a&gt;. In some sense, it is almost the same question, but with 12 instead of 100. There is a hidden subset $S$ of $[100]:=\{1, 2, \dots, 100\}$, and every time you query a subset $T\subseteq [100]$, you discover $|T \cap S|$. You want to minimize the number of queries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One answer on the linked question cites a paper called &quot;Determination of a subset from certain combinatorial properties,&quot; of which one result is that you can solve $2^{k-1}k$ people in $2^k-1$ queries. Thus, you can solve 80 people in 31 queries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's one catch. In this question, you presumably can't query a subset of size 1, because no self-respecting pickpocket will pickpocket themselves. But as Joffan points out in the comments, using Mohit's method, you can query one person by pairing them with any other person.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 20 people left. Using Mohit Jain's clever pairing strategy, you can solve them all in 10 queries. So the final tally is 31+10=41.&lt;/p&gt;&#xA;" OwnerUserId="1752" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-01-23T21:39:15.013" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8090" PostTypeId="2" ParentId="8086" CreationDate="2015-01-23T09:17:52.060" Score="13" Body="&lt;p&gt;I assume that the boss's time of arrival should be determined to the minute (say, within an error of plus/minus 30 seconds). I furthermore assume that the arrival time is any time point between 0:00 and 24:00. These assumptions then leave us with a search space of $24*60=1440$ minutes.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Since Tom can perform one query per day, the current search interval should always be divided into two equal-sized large pieces plus one small piece of 5 time points (corresponding to 5 minutes washroom break).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;After day 1, 2, 3, 4, 5, 6, 7 the length of the large search interval piece is down to 718, 357, 176, 86, 41, 18, 7 time points (in the worst case).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;On day 8 (or earlier), Tom is left with a search space of at most 7 consecutive time points. Then on day 8, Tom queries the middle time point,&#xA;and then is left with 3 consecutive time points. On day 9, Tom queries the middle time point, and finally has solved his problem.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My answer:&lt;/strong&gt; In the worst case, Tom needs &lt;strong&gt;nine&lt;/strong&gt; days to determine the boss's time of arrival.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-01-23T12:48:58.080" LastActivityDate="2015-01-23T12:48:58.080" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8091" PostTypeId="2" ParentId="8086" CreationDate="2015-01-23T10:16:31.417" Score="3" Body="&lt;p&gt;After reading the other answers, I get the impression that Tom is supposed (by contract) to start working before his boss arrives. but he doesn't want to, ok.&lt;br&gt;&#xA;In order to find out about the exact moment his boss enters the company, Tom will have to be already around before his boss, obviously. I suggest there is some pattern he could follow that in the end will approximate the exact time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But is it really necessary? If his boss arrives always at the exact point of time....why doesn't Tom simply  go to work (as he is supposed to be) and wait in front of the boss' office until the boss arrives?&#xA;When finally his boss is in sight, Tom uses the bathroom-trick. This way Tom would have to get up early once and would't lose a penny.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure if I'm missing a detail but this would mean Tom can figure it all out within 1 day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Obviously his co-workers are not his friends. They would tell their boss right away. So this won't work :( &lt;/p&gt;&#xA;" OwnerUserId="5106" LastEditorUserId="5106" LastEditDate="2015-01-23T12:23:25.730" LastActivityDate="2015-01-23T12:23:25.730" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8092" PostTypeId="1" AcceptedAnswerId="8093" CreationDate="2015-01-23T12:59:05.280" Score="24" ViewCount="2435" Body="&lt;p&gt;A rook stands in the lower left corner of an $m\times n$ chessboard. Alice and Bob alternately move the rook (horizontally or vertically, through any number of squares).&#xA;As the rook moves, it leaves a trail of painted squares: every square through which the rook passes or at which the rook stops is painted red.&#xA;The rook is not allowed to pass through a red square nor can it stop on a red square. The first player who cannot move loses the game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determine the winner of the game for every combination $(m,n)$ of positive integers $m$ and $n$. (As usual, we assume that Alice and Bob use optimal strategies.)&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-07T15:42:39.863" LastActivityDate="2015-02-07T15:42:39.863" Title="Rook game on chessboard" Tags="|strategy|chess|game|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8093" PostTypeId="2" ParentId="8092" CreationDate="2015-01-23T13:24:31.000" Score="28" Body="&lt;p&gt;I will assume that the square in the lower left is painted red as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that an $m\times n$ board is the same as an $n \times m$ board.  So WLOG we will assume that $n \le m$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $n=m=1$, then the first player automatically loses because there is no valid move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So lets assume $m&amp;gt;1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $n=1$, then the first player will win simply by moving the rook all the way down the row.  Player two will have no valid move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $n=2$, the first player can force a win again by moving the rook all the way down the row.  Player two is forced to move up one into the next row, and the game degenerates into $n=1$ case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is true for $n=3$.  Player two must move up one or two rows, and by taking the complete row, player two is again forced into the remaining row.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, this same strategy can be used for all $n$.  The general strategy used by player one is to take the complete row.  Player two is forced to move to a different row, and player two can again complete it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way for player two to beat this strategy would be to force the game into a spiral condition where the act of moving into a new row exhausts all entries in that row.  But since the first move was made eliminating $m$ squares, and $m \ge n$, the remaining rectangle is $n-1 \times m$, and player two must move to a different row.  The only way that a spiral could form is if player one failed to take the complete row, resulting in a rectangle with more rows than columns.  But since player one is not using that strategy, this can never happen.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2015-01-23T13:24:31.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8094" PostTypeId="2" ParentId="8092" CreationDate="2015-01-23T13:25:15.807" Score="15" Body="&lt;p&gt;The winner is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the first player Alice, for all $m$ and $n$ except $m=n=1$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The winning strategy for that player is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Assume WLOG that $m\leq n$, there are at least as many columns as rows. Alice moves the rook horizontally as far as possible to the opposite side until it runs into a red cell.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This wins because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each move of Alice makes the horizontal row totally red, so Bob is forced to move the rook vertically. After at most $m$ moves, Bob will lose because Alice will have filled every row. But, Alice cannot lose until Bob has made $n$ moves, since each of Bob's moves colors a subset of a column, and this never cuts a row into two regions. Since $m\leq n$ and Alice moves first, Bob must run out of moves before Alice does.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="1666" LastEditDate="2015-01-23T17:17:02.313" LastActivityDate="2015-01-23T17:17:02.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8095" PostTypeId="2" ParentId="8042" CreationDate="2015-01-23T13:51:46.867" Score="4" Body="&lt;p&gt;Edit: added two more options at the end, including one that wins \$1 on the specific assumption that you can't actually lose money.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The best you can possibly do is win \$1. If your strategy will do better than that, the host will simply choose \$1 as the prize. So this is always going to be a lot of work for not much; but I don't think you can guarantee a profit at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Others have observed that if you choose a door that's higher than a given prize (wherever that prize happens to be) you can then capture it by choosing the next lowest door all the way down to 2. Also, you can only do this with even prizes or odd prizes on any given run.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, no matter what strategy you produce that captures a finite set of prizes, the host can choose a higher prize. So your strategy must be able to handle infinitely high prizes; but it must get the lower prizes first or you will lose money on them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm going to assume that you've worked out the intricacies of small integers for everything below some large number, so that I can ignore +1 terms etc (for example, $n \approx n+1 \approx n+2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume that you have a strategy that covers everything up to n, at a cost of no more than cn doors. Now you sweep out every prize from kn down, where k &gt; 1. After cn doors, the kn'th prize will have been moved up to $kn + cn$, so it will take you $kn + cn$ doors to do that sweep, so now your total door count is $kn + 2cn$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need $kn + 2cn &amp;lt; c(kn)$, to preserve our cost limit. That means $k + 2c &amp;lt; ck$. In this case we need c to be no more than 10 (10 doors costs a dollar). We can find k that satisfies this: $k + 20 &amp;lt; 10k$ means $k &amp;lt; \frac{20}{9}$. But the smallest k can possibly be is given by $k(c - 1) &amp;gt; 2c$, ie $k &amp;gt; \frac{2c}{c-1}$. So k must be greater than 2, which definitely rules out any kind of linear search (we must at least double our door count on every run). We must climb exponentially, but we can definitely catch any given prize. Similarly, $c &amp;gt; \frac{k}{k-2}$, so c must exceed 1 and c and k can't both be small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about the cost? Our host will choose the very worst prize for our strategy, which in this case is a number just above n, and with the wrong polarity. Thus, we will sweep prizes between $n$ and $kn$, and then between $kn$ and $k^2n$, just to get a prize of $n$.&#xA;From $n$ at a cost of $cn$, we go to $kn$ at a cost of $kn + 2cn$; and from there we go to $k^n$ at a cost of $k^2n + 2kn + 4cn$, and this gets us a prize of n at a cost of $\frac{k^2n + 2kn + 4cn}{10}$ dollars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make $\frac{k^2n + 2kn + 4cn}{10} &amp;lt; n$, we will need $k^2 + 2k + 4c &amp;lt; 10$. In fact if we substitute $c = \frac{k}{k-2}$ into $k^2 + 2k + 4c$, take the derivative and set that to zero, we find the minimum value of $k = c = 3$, which means $k^2 + 2k + 4c = 27$, which is clearly more than 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our host, therefore, can set an arbitrarily high value for the prize and reap a handsome profit. To stop him, we need to negotiate a price of $\frac{100}{27} = 3.7$ cents or better; or else call in sick.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is the best you can do deterministically. If you can include a random number source (say, a good strong hot cup of tea), then you can reduce your losses for a given prize because the host can't choose the optimal number in a range. What you do is randomly vary the value of k around 3. This increases your cost and takes a while to make the ranges completely random, but I'm going to ignore those effects by waving my hands and chanting &quot;for large n&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the host can't get &quot;just above your last guess&quot;, and more importantly can't guarantee to get the wrong polarity. So we will sweep prizes between $n$ and $kn$, and then between $kn$ and $k^2n$, and get a prize averaged between $\frac{kn-n}{2}$ and $\frac{k^2n-kn}{2}$, depending on which way the polarity went.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's nice, but it still doesn't reach profitability (it does make the calculations uglier, though). To get a profit, you need another assumption.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now I'm going to assume that since this is a game show, they can't charge you money. So if you have a strategy that either makes \$x or loses \$y, you walk away from any potential losses and get an expected value of $\frac{x}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, we will use one of the opening sequences already established - door 1, door 3, then door 2. This guarantees that a \$1 prize is found immediately, and a \$2 prize is found at a cost of \$0.20 for a payoff of \$1.80.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we will flip a coin. Heads, we assume the prize is even, tails we assume it's heads. We then proceed with c = 3, k = 3. In general this ensures we cover everything up to $n$ at a cost of at most $3n$ doors; at each next step we cover everything up to $3n$ at a cost of $3n + 3n + 3n = 9n$ doors, and we get a prize of \$n for a profit of $n - \frac{9}{10}n = 0.1n$. If our coin-flip failed, we get nothing, so the expected value is $0.05n$. For any large n this is more than \$1, so the host won't choose these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we start having covered 1 and 2 at a cost of 3 doors. If we have correctly chosen even numbers, the sequence looks like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Covered  Cost (doors)  Smallest door in range  Profit (if the door is found here)&#xA;   2          3&#xA;   4          9                4                 3.10&#xA;  12         29                6                 3.10&#xA;  36         93               14                 4.70&#xA; 108        293               38                 8.70&#xA; .....and we're well over 20.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And if we've correctly chosen even numbers, the sequence looks like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Covered  Cost (doors)  Smallest door in range  Profit (if the door is found here)&#xA;   2          3&#xA;   3          8                3                 2.20&#xA;   9         24                5                 2.60&#xA;  27         78               11                 3.20&#xA;  81        236               29                 5.40&#xA;....and we're well over 20.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Expected profit values have to be divided by 2 to cover failed coin-flips, but these are all greater than \$2, so the expected profits will always be greater than \$1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, we announce this strategy to the host, he chooses a prize of $1 which we find immediately, and showers of confetti are unleashed.&lt;/p&gt;&#xA;" OwnerUserId="6031" LastEditorUserId="6031" LastEditDate="2015-01-24T11:53:07.490" LastActivityDate="2015-01-24T11:53:07.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8096" PostTypeId="2" ParentId="4818" CreationDate="2015-01-23T14:44:14.473" Score="3" Body="&lt;p&gt;No mathematician, but from my experience, this is how I think of it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) It seems from prior work on the 3x3x3, that it is not possible to perform a simple exchange of 2 pieces without changing anything else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) This means that the simplest moves we can think of do 2 exchanges (meaning 2 edges exchanged and 2 corners, or 2 sets of edges, or 2 sets of corners).  2 exchanges is the same as a 3-way rotation.  Think of a&gt;b&gt;c as: a/b exchange, then b/c exchange.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Now, transferring to the 4x4x4, all parity really can be explained (at least in my opinion) as: It appears we need to make a single exchange. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, what we call 'parity solutions' really do just that - they make a single exchange of pieces - although often by making an odd number of exchanges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My own personal &quot;easiest way to see what is happening&quot; solution method goes like this (NOTE - This is NOT fast on the clock):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A) Put centers in place.  This is intuitive, and with a little practice goes pretty fast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B) Place top edges. This is done by pairing the edges as you put them in place, and you have to take care to keep the horizontal center slices correct as you go.  It's not difficult to see how as you work with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B) Place all top corners correctly.  Also intuitive. (This is also the way I first solved the 3x3x3).  This step is actually done the same as the 3x3x3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C) Place middle layer edges (all 8).  (There is basically one algorithm I need for this, with a reflection. And, it is really the same algorithm as is used in the 3x3x3 solution, with a slight modification to the 4x4x4 which is easy to see as you watch the move being performed.)  The second slice pieces require exactly the same series of moves as with the 3x3x3, and the third is very similar, except that you have to grab an extra slice at one point, and then put it back later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D) Place bottom corners correctly in place, and then orient. (Again, one algorithm, with a single conjugation for placing. 2 algorithms for rotation.  And, these are the same algorithms as the 3x3x3.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E) Place bottom edges correctly.  Obviously, there are 8.  I have not paired them yet, so I do this one by one. I do 2 opposite edges first.  I need one algorithm with a reflection. It is an algorithm I made up myself, and it does a 3-way exchange. This is really the only NEW algorithm I need compared to the 3x3x3. And, sometimes it needs a quick conjugation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EE) And, this is the step where 'parity' might be involved.  Somewhere in this step I count how many exchanges of pieces I have to make.  I want an answer of &quot;even&quot; or &quot;odd.&quot;  If the answer if &quot;even&quot;, then I simply continue until I have all the edges in place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EEE) If the answer is &quot;odd&quot;, I have what is called a &quot;parity issue&quot;.  Since I try to work the edges by doing pairs together, and non-adjacent sides first, the last 3 or 4 pieces are across the face from each other.  This leaves me a simple solution available for the 'parity issue': Since the offending edges are across from each other, I turn one inside layer 180*, then the face I am working with 180* (this effectively puts all 4 edges that I need to fix into the same slice), then the inside layer where all the edges are located 90* either way, then the face I am working with 180*, and finally the inside layer 180* again.  So, if the face were on top, the algorithm might look like this: r2,U2,r,U2,r2.  small r means the inside layer WITHOUT the right face.  Think about what that move does:  It does a 4-way cycle on the edges, and also 2 4-way cycles on the centers (which you can't notice, and I am not worried about yet).  Thus, I am making an odd number of exchanges (9) and this solves the 'parity' issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EEEE) I now continue to put the edges in place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F) Finally, if I had to do step EEE, then I need to go back and replace the 4 sets of 2 paired centers that were moved in that step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Admittedly, again, this is not FAST.  But I think it makes a great beginners method because most of it is exactly like the 3x3x3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, this lines up with what someone said further up: The parity issue is caused by how many interior layer moves you do, because those moves are the ones that do odd numbers of exchanges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT TO ADD THE FOLLOWING:&#xA;Perhaps the truest answer is this: on a 4x4x4 cube, when you rotate an interior layer, you actually accomplish an odd number of exchanges of pieces: 3 for the edge pieces (it is really a 4-way cycle, but that is 3 exchanges), 3 for one group of centers (again a 4-way cycle), and 3 for another group of centers (again a 4-way cycle).  Since there is a simple move which accomplishes an odd number of exchanges on this puzzle, then all configurations with an odd number of exchanges when compared to 'solved' are now possible.  Thus, what are called OLL and PLL parity are legal configurations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, that is why the way to 'fix' parity issues on this puzzle comes down to simply doing one more odd turn of an interior layer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those who think in terms of 'hidden pieces,' it is possible to imagine the 2x2x2 block hidden inside of the 4x4x4 cube as &quot;hidden pieces.&quot;  If you choose to think that way, then the single turn of an interior layer actually does an even number of exchanges: 4 sets of 4-cycles.  First, the 3 that I have been discussing, and then 4th, the 4-way cycle of the interior block pieces.  As I say, I am not a mathematician by trade, but to me, this is an ok way of thinking about it.  However, the group theory of the 4x4x4 cube does not 'require' this idea.  The group theory of the 4x4x4 cube is simply different from that of the 3x3x3, because the interior slice move does an odd number of exchanges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT AGAIN:  I revised my simple method above because I learned that you can start with the centers, and it goes a little quicker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT EDIT AGAIN:  What is commonly called PLL parity is not parity at all.  It is simply 2 pairs of pieces that need to be exchanged.  It only looks weird because the fastest way to solve this puzzle is to reduce to 3x3x3.&lt;br&gt;&#xA;What is commonly called OLL parity is really just the need for a 90* turn of an interior slice.&lt;/p&gt;&#xA;" OwnerUserId="9092" LastEditorUserId="9092" LastEditDate="2015-02-20T18:02:07.780" LastActivityDate="2015-02-20T18:02:07.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8097" PostTypeId="2" ParentId="8086" CreationDate="2015-01-23T15:34:46.197" Score="13" Body="&lt;p&gt;Lets assume that Tom is required to work for 8 hours each day, and he is supposed to start at time &lt;span class=&quot;math-container&quot;&gt;$T_s$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The boss arrives at some time &lt;span class=&quot;math-container&quot;&gt;$T_b$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$T_s &amp;lt; T_b &amp;lt; T_s+8h$&lt;/span&gt;.  Since there are 60 minutes in an hour, this gives 480 minutes in a working day.&lt;/p&gt;&#xA;&lt;p&gt;We will make the following assumptions:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If Tom arrives in the same minute as the Boss and the Boss is late, the Boss will see Tom arrive and fine him.&lt;/li&gt;&#xA;&lt;li&gt;If Tom goes to/returns from the bathroom at in the same minute that the Boss arrives, Tom will see the Boss arriving know exactly the time the Boss arrives.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;This gives us discrete one minute intervals to work with.&lt;/p&gt;&#xA;&lt;p&gt;Tom is lazy, so he wants to find the largest &lt;span class=&quot;math-container&quot;&gt;$t_{arrive} \lt T_b$&lt;/span&gt;.  But Tom also doesn't want to be fined any amount of money, so he must always arrive before the Boss.  If we relax this rule, Tom can find out the time quicker, but at added cost.&lt;/p&gt;&#xA;&lt;p&gt;Lets say on a typical day that Tom arrives at time &lt;span class=&quot;math-container&quot;&gt;$t_{arrive}$&lt;/span&gt;, goes to the bathroom at time &lt;span class=&quot;math-container&quot;&gt;$t_{bathroom}$&lt;/span&gt; and returns from the bathroom at time &lt;span class=&quot;math-container&quot;&gt;$t_{return}$&lt;/span&gt;.  Then, we have the following possibilities:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$T_b &amp;lt; t_{arrive}$&lt;/span&gt; - The Boss is already at work when Tom arrives, so Tom is fined.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$T_b = t_{arrive}$&lt;/span&gt; - The Boss arrives at the same time as Tom, and Tom is fined.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$t_{arrive} &amp;lt; T_b &amp;lt; t_{bathroom}$&lt;/span&gt; - The Boss arrives after Tom, but before he goes to the bathroom.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$T_b = t_{bathroom}$&lt;/span&gt; - The Boss arrives as Tom goes to the bathroom.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$t_{bathroom} &amp;lt; T_b &amp;lt; t_{return}$&lt;/span&gt; - The Boss arrives while Tom is in the bathroom.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$T_b = t_{return}$&lt;/span&gt; - The Boss arrives as Tom returns from the bathroom.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$t_{return} &amp;lt; T_b$&lt;/span&gt; - The Boss arrives after Tom returns from the bathroom.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Using the above, this means that if we have an interval that includes 5 exact minute marks, we can determine exactly when the boss arrives in one day with no fines.&lt;/p&gt;&#xA;&lt;p&gt;Proof: Let the interval at which the Boss can arrive be &lt;span class=&quot;math-container&quot;&gt;${t_1, t_2, t_3, t_4, t_5}$&lt;/span&gt;.  Tom will arrive at &lt;span class=&quot;math-container&quot;&gt;$t_{arrive}=t_1 -2$&lt;/span&gt;, go to the bathroom at &lt;span class=&quot;math-container&quot;&gt;$t_{bathroom}=t_2$&lt;/span&gt;, and return at &lt;span class=&quot;math-container&quot;&gt;$t_{return}=t_4$&lt;/span&gt;.  Using the possibilities above:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Can't happen because we've arrived before the boss.&lt;/li&gt;&#xA;&lt;li&gt;Can't happen because we've arrived before the boss.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$T_b = t_1$&lt;/span&gt; since the boss arrived after Tom, but before he goes to the bathroom.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$T_b = t_2$&lt;/span&gt; since Tom witnesses his arrival.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$T_b = t_3$&lt;/span&gt; since the Boss arrived while Tom was in the bathroom.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$T_b = t_4$&lt;/span&gt; since Tom witnesses his arrival.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$T_b = t_5$&lt;/span&gt; since the Boss arrived after Tom returned from the bathroom.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;If we allow fines, then the interval can increase to 7 minutes.&lt;/p&gt;&#xA;&lt;h1&gt;No Fines Solution&lt;/h1&gt;&#xA;&lt;p&gt;In order to do this with no fines, the first day, Tom must arrive at &lt;span class=&quot;math-container&quot;&gt;$T_s$&lt;/span&gt;.  We can assume that the Boss arrives later than this because if he arrives at the same time, then Tom will discover this the first day.&lt;/p&gt;&#xA;&lt;p&gt;Thus, &lt;span class=&quot;math-container&quot;&gt;$T_s &amp;lt; T_b$&lt;/span&gt;.  Also, the Boss must arrive before the end of work, so we will also assume &lt;span class=&quot;math-container&quot;&gt;$T_b &amp;lt; T_s + 480$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Thus, there are &lt;span class=&quot;math-container&quot;&gt;$479$&lt;/span&gt; possibilities for &lt;span class=&quot;math-container&quot;&gt;$T_b$&lt;/span&gt; initially.&lt;/p&gt;&#xA;&lt;p&gt;Tom use his bathroom break like a binary search.  So at around lunch time (&lt;span class=&quot;math-container&quot;&gt;$t_{bathroom}=T_s + 238$&lt;/span&gt;), he will take his break and return 5 minutes later.  Since we've excluded the possibilities of fines, only the last 5 are valid options.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$t_{arrive} &amp;lt; T_b &amp;lt; t_{bathroom}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$T_b = t_{bathroom}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$t_{bathroom} &amp;lt; T_b &amp;lt; t_{return}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$T_b = t_{return}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$t_{return} &amp;lt; T_b$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;For the &lt;span class=&quot;math-container&quot;&gt;$2^{nd}$&lt;/span&gt; and and &lt;span class=&quot;math-container&quot;&gt;$4^{th}$&lt;/span&gt; possibilities, Tom figures out exactly the Boss's arrival time.  For the &lt;span class=&quot;math-container&quot;&gt;$3^{rd}$&lt;/span&gt; Tom has narrowed it down to the four 1 minute intervals while Tom was in the bathroom.  This means that he can figure it out exactly on the second day.&lt;/p&gt;&#xA;&lt;p&gt;Thus, the only two possibilities we need to explore further are &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt;.  Each narrow the interval down to &lt;span class=&quot;math-container&quot;&gt;$237$&lt;/span&gt; (or less) possibilities.&lt;/p&gt;&#xA;&lt;p&gt;On the end of the second day, this interval is cut to &lt;span class=&quot;math-container&quot;&gt;$116$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;On the third day, it is &lt;span class=&quot;math-container&quot;&gt;$55$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;On the forth day, it is &lt;span class=&quot;math-container&quot;&gt;$25$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;On the fifth day, it is &lt;span class=&quot;math-container&quot;&gt;$10$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;So by the end of the sixth day, the interval is only &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt;, and thus small enough to figure out the next day.&lt;/p&gt;&#xA;&lt;h2&gt;Answer: 7 days&lt;/h2&gt;&#xA;&lt;h1&gt;With Fines&lt;/h1&gt;&#xA;&lt;p&gt;Since Tom is in this job for the long haul, perhaps a week of fines is worth it to find the answer quicker.&lt;/p&gt;&#xA;&lt;p&gt;Instead of a binary search, we will employ a ternary search.  Basically split the interval into 3 and arrive after the first interval, and go to the bathroom after the second.&lt;/p&gt;&#xA;&lt;p&gt;On the first day, Tom will arrive at &lt;span class=&quot;math-container&quot;&gt;$t_{arrive}=T_s + 158$&lt;/span&gt;.  He will go the bathroom at &lt;span class=&quot;math-container&quot;&gt;$t_{bathroom}=T_s+316$&lt;/span&gt; and return at &lt;span class=&quot;math-container&quot;&gt;$t_{return}=T_s+321$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;There are then 7 possibilities:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$T_b &amp;lt; t_{arrive}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$T_b = t_{arrive}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$t_{arrive} &amp;lt; T_b &amp;lt; t_{bathroom}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$T_b = t_{bathroom}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$t_{bathroom} &amp;lt; T_b &amp;lt; t_{return}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$T_b = t_{return}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$t_{return} &amp;lt; T_b$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Options 2, 4, and 6 all determine the Boss' arrival exactly.  Option 5 reduces the interval to 5 minutes which can be done in an extra day.  So, the remaining options 1, 3, and 7 need to be explored.  All three of these reduce the interval to &lt;span class=&quot;math-container&quot;&gt;$158$&lt;/span&gt; minutes by the end of the first day.&lt;/p&gt;&#xA;&lt;p&gt;By the end of the second day, the interval is &lt;span class=&quot;math-container&quot;&gt;$51$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;By the third day, it is &lt;span class=&quot;math-container&quot;&gt;$15$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;By the end of the forth day, it is &lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt; and small enough to determine on the next day.&lt;/p&gt;&#xA;&lt;h2&gt;Answer: 5 days&lt;/h2&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-01-23T15:47:04.667" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8098" PostTypeId="1" AcceptedAnswerId="8106" CreationDate="2015-01-23T16:09:24.933" Score="2" ViewCount="1764" Body="&lt;p&gt;In this grid of numbers, what is the missing number (in place of the question mark)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{|c|c|}\hline2&amp;amp;7\\\hline8&amp;amp;9\\\hline\end{array}&#xA;\begin{array}{|c|c|}\hline0&amp;amp;3\\\hline5&amp;amp;6\\\hline\end{array}&#xA;\begin{array}{|c|c|}\hline0&amp;amp;11\\\hline10&amp;amp;11\\\hline\end{array}&#xA;\begin{array}{|c|c|}\hline2&amp;amp;?\\\hline4&amp;amp;3\\\hline\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UAY4l.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;(source image in Arabic)&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8829" LastEditorDisplayName="user20" LastEditDate="2015-01-23T17:19:33.357" LastActivityDate="2015-01-24T10:43:27.800" Title="whats the Missing Number?" Tags="|logical-deduction|mathematics|calculation-puzzle|number-sequence|" AnswerCount="2" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8099" PostTypeId="1" CreationDate="2015-01-23T18:05:59.947" Score="0" ViewCount="19105" Body="&lt;p&gt;A deaf person stole a blind person's hen and a dumb person  saw  him stealing.How can the dumb person convey this to the blind person?&lt;/p&gt;&#xA;" OwnerUserId="9095" LastActivityDate="2015-01-23T18:45:28.383" Title="Puzzle about a deaf,dumb and a blind person" Tags="|lateral-thinking|" AnswerCount="1" CommentCount="2" ClosedDate="2015-01-23T18:58:12.250" ContentLicense="CC BY-SA 3.0" />
  <row Id="8100" PostTypeId="2" ParentId="8099" CreationDate="2015-01-23T18:45:28.383" Score="1" Body="&lt;p&gt;If we go by just the rules, then there is nothing we can do.  The blind person can only hear things, and dumb person can't say anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have to think outside the box for this one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets assume that the blind person knows Braille.  Then the dumb person could write out his communication in Braille for the blind person.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, the dumb person could use some &quot;text-to-speech&quot; app on his smart phone.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or guide the blind persons hand to the location of the missing hen and let him figure it out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or find someone to act as an intermediary and translate the dumb persons writing to speech.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or if they both know morse code, the dumb person can tap out the message.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2015-01-23T18:45:28.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8101" PostTypeId="2" ParentId="8098" CreationDate="2015-01-23T20:28:16.137" Score="1" Body="&lt;p&gt;This can be viewed as three equations with 3 unknowns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2A+0B+0C=2$$&#xA;$$8A+5B+10C=4$$&#xA;$$9A+6B+11C=3$$&#xA;$$7A+3B+11C=?$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$? = \frac{53}{5}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't believe this is the intended answer but it is valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2A+8B+9C=7$$&#xA;$$0A+5B+6C=3$$&#xA;$$0A+10B+11C=11$$&#xA;$$2A+4B+3C=?$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$?=10.6$$&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditDate="2015-01-23T20:53:42.077" LastActivityDate="2015-01-23T20:53:42.077" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8102" PostTypeId="1" AcceptedAnswerId="8104" CreationDate="2015-01-23T21:40:31.687" Score="6" ViewCount="1592" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/n4VQ9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was given this as a quick challenge, but the underlying pattern behind it eludes me entirely. Can anyone suggest what the next image along might be (&lt;a href=&quot;https://www.dropbox.com/s/51zjyh3movhbv52/Screenshot%202015-01-23%2021.34.10.png?dl=0&quot; rel=&quot;nofollow noreferrer&quot;&gt;screenshot of correct answer&lt;/a&gt;) and explain HOW they reached this conclusion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel like a moron for not recognising the pattern here.&lt;/p&gt;&#xA;" OwnerUserId="9098" LastEditorUserId="39" LastEditDate="2015-04-27T22:03:24.840" LastActivityDate="2015-04-27T22:03:24.840" Title="Was given this question: what is the pattern?" Tags="|pattern|visual|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8103" PostTypeId="2" ParentId="8102" CreationDate="2015-01-23T21:59:56.197" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The circle and x both move one space at a time. The circle is initially going towards the top left, and the x is going towards the bottom right. If the circle is going to collide with the x, then the circle jumps over the x, and the x stays in place. The circle will not jump the x if it is in a corner (perhaps because it can't get the proper footing in such a small space). If either shape hits a corner, they turn around and go the opposite direction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8942" LastEditorUserId="8942" LastEditDate="2015-01-23T22:09:52.277" LastActivityDate="2015-01-23T22:09:52.277" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8104" PostTypeId="2" ParentId="8102" CreationDate="2015-01-23T22:09:37.417" Score="10" Body="&lt;p&gt;Assuming D is the answer (can't see dropbox at work), &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the circle alternates moving one space and two spaces per turn going left and reversing when it runs out of room, and the x alternates moving 1 space and 0 spaces per turn going right.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2374" LastActivityDate="2015-01-23T22:09:37.417" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8105" PostTypeId="2" ParentId="2188" CreationDate="2015-01-23T22:37:29.280" Score="0" Body="&lt;p&gt;Emrakul provided the only answer that is close to the way I would explain this riddle solution. Others are violating one or more the original premise or just off base in general. Especially Kevin and Rafe which obviously just convinced themselves 50-50 was ok, LOL, which guard is the honest one and which is the liar is obviously not something you know, it says right in the riddle explanation, but your answer depends on it. I'm really scratching my head and wish you two the best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer, same concept as Emrakul, in other words: Go to either guard. Ask that guard, &quot;Which door would the other guard say is the safe route&quot;. This answer, regardless of which guard you ask will allow you to choose the door opposite of either guard's answer and always lead you safely.  -1 x -1 = 1&#xA;MGO&lt;/p&gt;&#xA;" OwnerUserId="9100" LastActivityDate="2015-01-23T22:37:29.280" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8106" PostTypeId="2" ParentId="8098" CreationDate="2015-01-23T23:47:45.067" Score="11" Body="&lt;p&gt;The missing number is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You can interpret the two numbers on the top of each grid as a single decimal number, but with the ones digit written first, e.g. &quot;2 7&quot; becomes 72. This top number is then the product of the two numbers below. So 72 = 8*9, 30 = 5*6, 110 = 10*11, and &lt;strong&gt;1&lt;/strong&gt;2 = 3*4.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9102" LastActivityDate="2015-01-23T23:47:45.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8107" PostTypeId="1" CreationDate="2015-01-23T16:29:50.790" Score="0" ViewCount="113" Body="&lt;p&gt;A deaf person stole a blind person's hen and a dumb person witnessed it.How can the dumb person convey this to the blind person?&lt;/p&gt;&#xA;" OwnerDisplayName="sajjad veeri" LastActivityDate="2015-01-24T01:30:58.483" Title="A soft puzzle about a deaf ,dumb and a blind person" Tags="|logical-deduction|" AnswerCount="0" CommentCount="4" ClosedDate="2015-01-24T02:12:04.480" ContentLicense="CC BY-SA 3.0" />
  <row Id="8110" PostTypeId="2" ParentId="8102" CreationDate="2015-01-24T03:13:37.813" Score="0" Body="&lt;p&gt;The answer is &quot;D&quot; for the following reason:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) There are 6 regions. Let's call the upper left region Region #1 and proceed southeast as Region #2, Region #3, Region #4, Region #5, Region #6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The starting position is Box #1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Looking only at the $+$ movements, we see that it appears to spend 2 turns in the same position. This is evidenced by Boxes #2, #3, #4, and #5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) $o$ is moving northwest as evidenced by the change of position from Box #1 to Box #2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) Whenever $o$ encounters $+$, it jumps over $+$ and when $o$ finally encounters the northwest corner it reverses direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6) In Box #5, the $+$ is completing its time in Region #3 and must now proceed to Region #4. $o$ must continue from Region #4 to Region #5 and continue until the corner where it will once again reverse direction.&lt;/p&gt;&#xA;" OwnerUserId="9105" LastActivityDate="2015-01-24T03:13:37.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8111" PostTypeId="1" AcceptedAnswerId="8113" CreationDate="2015-01-24T03:29:44.713" Score="26" ViewCount="18087" Body="&lt;p&gt;Here's a puzzle that a teacher told me today:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How can you cut twelve in half and get seven?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint (only if you're stumped):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Don't let just regular numerals control you&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another Hint&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Notice that I didn't use the math tag&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1711" LastEditorUserId="7722" LastEditDate="2015-01-24T18:59:33.087" LastActivityDate="2016-05-07T19:47:45.483" Title="How to cut 12 in half and get 7" Tags="|lateral-thinking|" AnswerCount="9" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8112" PostTypeId="2" ParentId="8111" CreationDate="2015-01-24T03:36:41.297" Score="21" Body="&lt;p&gt;Here's how you can get seven by cutting twelve in half:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/aFisd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastActivityDate="2015-01-24T03:36:41.297" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8113" PostTypeId="2" ParentId="8111" CreationDate="2015-01-24T03:48:20.057" Score="54" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Slice XII across the middle to get VII.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure that's what Quark's answer is facetiously alluding to.&lt;/p&gt;&#xA;" OwnerUserId="7489" LastActivityDate="2015-01-24T03:48:20.057" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8114" PostTypeId="2" ParentId="8111" CreationDate="2015-01-24T05:09:01.617" Score="-1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 12 = XII&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;strike through the middle&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the top half now becomes VII&lt;BR&gt;&lt;BR&gt;&#xA; VII = 7&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9107" LastEditorUserId="4637" LastEditDate="2015-01-24T06:04:33.777" LastActivityDate="2015-01-24T06:04:33.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8115" PostTypeId="1" CreationDate="2015-01-24T05:20:43.473" Score="10" ViewCount="630" Body="&lt;p&gt;It's been well known for a few years now that &lt;a href=&quot;http://www.cube20.org/&quot;&gt;every Rubik's Cube scramble can be solved in 20 moves&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this is only for the case where the orientation of the centers is not taken into consideration. I couldn't find a reference for the picture cube, where face orientation does matter. Does anybody know a resource on the minimum number of moves required to solve a picture cube?&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-01-25T00:30:28.090" Title="What is God's Number for the Rubik's Picture Cube?" Tags="|rubiks-cube|mechanical-puzzles|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8117" PostTypeId="2" ParentId="8111" CreationDate="2015-01-24T08:44:18.387" Score="66" Body="&lt;p&gt;Here's a more mathematical answer from a programmer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  If we use the base 12 numbering system then 12 (in base 12) == 14 (in base 10), and half of that is 7 (in both base 10 and 12).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9112" LastEditorUserId="9126" LastEditDate="2015-01-24T22:50:32.337" LastActivityDate="2015-01-24T22:50:32.337" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8118" PostTypeId="2" ParentId="5299" CreationDate="2015-01-24T10:32:58.330" Score="3" Body="&lt;p&gt;Just observe the behavior of crowds, and you get to the right answer: people mostly stand still and rest on escalators, they tend &lt;em&gt;not&lt;/em&gt; to do so on staircases. There is a good reason for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Key is to realize you have to cover a given distance in a minimum amount of time (rather than maximize the distance covered in a given amount of time). So you have to &lt;strong&gt;minimize your slowness&lt;/strong&gt; (time spend per unit distance), rather than &lt;strong&gt;maximize your speed&lt;/strong&gt; (distance covered per unit time). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the specific problem posted here: resting in-between moving walkways increases your slowness from $1/10$ to $1/5$ h/km, a net increase of $1/10$ h/km. Denoting the speed of the moving walkways in km/h by $v$, it follows that resting on the walkways increases your slowness from $1/(v+10)$ to $1/(v+5)$ h/km. For any positive $v$ this increase is less than $1/10$ h/km. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So take your rest on the moving walkways, this minimizes your slowness and maximizes your chances of catching your plane.&lt;/p&gt;&#xA;" OwnerUserId="7573" LastEditorUserId="7573" LastEditDate="2015-01-25T15:22:34.483" LastActivityDate="2015-01-25T15:22:34.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8120" PostTypeId="1" CreationDate="2015-01-24T12:23:21.637" Score="7" ViewCount="418" Body="&lt;p&gt;This is a different 'connect the dots' puzzle. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Your task is to connect the blue and yellow &lt;em&gt;nodes&lt;/em&gt; in the map below with &lt;em&gt;links&lt;/em&gt; such that the created network guides two &lt;em&gt;signals&lt;/em&gt; emitted &lt;strong&gt;from the two red triangles to the green pentagram&lt;/strong&gt;. In doing so, &lt;strong&gt;each node&lt;/strong&gt; needs to be used by at least on of the two signals at least once.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Clarification:&lt;/strong&gt; Your task is to provide the additional links. Once they are all set, signals start and move according to the rules until they either die or end in the green node. You can no longer influence this at this stage.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Scy8X.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The following rules apply:&lt;/h2&gt;&#xA;&#xA;&lt;h3&gt;Nodes:&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A &lt;em&gt;node&lt;/em&gt; may be visited as often as one likes.&lt;/li&gt;&#xA;&lt;li&gt;Each &lt;em&gt;node&lt;/em&gt; needs to be visited &lt;strong&gt;at least &lt;em&gt;once&lt;/em&gt; by at least &lt;em&gt;one&lt;/em&gt; &lt;em&gt;signal&lt;/em&gt;&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;The &lt;em&gt;signal strength&lt;/em&gt; when reaching the final node (green pentagram) does not matter.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Node&lt;/em&gt; labels are only to allow easier writing of an answer.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Links:&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;You may create &lt;strong&gt;as many &lt;em&gt;links&lt;/em&gt; as you wish&lt;/strong&gt;. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Already drawn &lt;em&gt;links&lt;/em&gt; are fixed and can not be removed.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A &lt;em&gt;link&lt;/em&gt; has to connect two &lt;em&gt;nodes&lt;/em&gt; in a &lt;strong&gt;straight line&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Links&lt;/em&gt; have to &lt;strong&gt;leave &lt;em&gt;nodes&lt;/em&gt; radially&lt;/strong&gt;. &lt;br&gt;(If you prolong the line it has to go through the centre.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Links&lt;/em&gt; &lt;strong&gt;must not cross&lt;/strong&gt; each other at any point.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Links&lt;/em&gt; &lt;strong&gt;must not 'touch'&lt;/strong&gt; any other &lt;em&gt;node&lt;/em&gt;.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Links&lt;/em&gt; and &lt;em&gt;nodes&lt;/em&gt; have the dimensions as in the original map.&lt;br&gt; (They are &lt;em&gt;not&lt;/em&gt; mathematical points and lines.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Examples&lt;/strong&gt; for valid and invalid &lt;em&gt;links&lt;/em&gt; are: &lt;br&gt;&lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/1x12v.png&quot; alt=&quot;Valid and invalid links&quot;&gt;&lt;br&gt;(valid; invalid because crossing; invalid because  touching other node; invalid because not radial)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Signals:&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Each &lt;em&gt;signal&lt;/em&gt; travels along a &lt;em&gt;link&lt;/em&gt; from one &lt;em&gt;node&lt;/em&gt; to the next &lt;em&gt;node&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Each &lt;strong&gt;red triangle&lt;/strong&gt; emits one &lt;em&gt;signal&lt;/em&gt; of &lt;em&gt;strength&lt;/em&gt; 5 towards the first connected node.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Signals&lt;/em&gt; do not affect each other.&lt;br&gt; (You can think of them as being emitted in sequence, i.e. the second is emitted after the first reached it's goal.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The moment a &lt;em&gt;signal&lt;/em&gt; arrives at a &lt;em&gt;node&lt;/em&gt;, the signal &lt;em&gt;strength&lt;/em&gt; is either:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;decreased by &lt;strong&gt;-1&lt;/em&gt;&lt;/strong&gt; (blue node)&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;increased by &lt;strong&gt;+2&lt;/em&gt;&lt;/strong&gt; (yellow node)&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A &lt;em&gt;signal&lt;/em&gt; of &lt;strong&gt;&lt;em&gt;strength&lt;/em&gt; 0 dies&lt;/strong&gt; and is removed. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Once a &lt;em&gt;signal&lt;/em&gt; has arrived at the &lt;em&gt;node&lt;/em&gt; it &lt;strong&gt;moves clockwise&lt;/strong&gt; for as many outgoing links as its &lt;em&gt;strength&lt;/em&gt; indicates. (It may do full circles multiple times.)&lt;br&gt; i.e. for a &lt;em&gt;strength&lt;/em&gt; of 3, the &lt;em&gt;signal&lt;/em&gt; would leave the &lt;em&gt;node&lt;/em&gt; via the 3rd link after the one it came from.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The following images illustrate the &lt;em&gt;signal&lt;/em&gt; movement: &lt;img src=&quot;https://i.stack.imgur.com/eAIZm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/j00ue.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Solution&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There exists at least one solution. It may not be unique. You can specify your answer either by an image (preferred) or by specifying all links (2 node-labels each) and the path each signal takes (by node-labels.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="5840" LastEditDate="2015-03-12T12:13:24.303" LastActivityDate="2015-03-12T12:13:24.303" Title="Signal-network: Create a network that guides 2 signals" Tags="|visual|network|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8122" PostTypeId="1" AcceptedAnswerId="8125" CreationDate="2015-01-24T12:56:50.113" Score="14" ViewCount="1087" Body="&lt;p&gt;The following 3x4 rectangle can be cut into pieces along grid lines, so that each piece has exactly three neighbors: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/plyda.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt; Find the smallest rectangle on the integer grid that can be cut into pieces along grid lines, so that each piece has exactly &lt;strong&gt;five&lt;/strong&gt; neighbors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pieces are neighbors, if their boundaries touch; touching in a corner doesn't count.&lt;/p&gt;&#xA;" OwnerUserId="5781" LastEditorUserId="65277" LastEditDate="2023-11-16T13:38:59.697" LastActivityDate="2023-11-16T13:38:59.697" Title="Find smallest rectangle divided into figures so each figure has 5 neighbours" Tags="|geometry|optimization|polyomino|dissection|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8123" PostTypeId="2" ParentId="8111" CreationDate="2015-01-24T13:37:44.657" Score="11" Body="&lt;p&gt;@CoolGuy I disagree that it isn't cutting &quot;12&quot; in half. No matter which of the resulting fragments you take, you get 7. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lines in PS are to ensure that they are exactly (100/100 pixels above and below guide lines) cut in half. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/rS9U1.png&quot; alt=&quot;12 / 2 = 7&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7529" LastEditorUserId="9439" LastEditDate="2016-05-05T10:23:23.623" LastActivityDate="2016-05-05T10:23:23.623" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8125" PostTypeId="2" ParentId="8122" CreationDate="2015-01-24T14:19:25.047" Score="21" Body="&lt;p&gt;How's this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MrNqr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got this by square-izing the icosahedron graph:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/60rQQm.gif&quot; width=&quot;200&quot; height=&quot;200&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2015-01-24T14:19:25.047" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8126" PostTypeId="2" ParentId="8122" CreationDate="2015-01-24T14:55:16.210" Score="13" Body="&lt;p&gt;Here is a proof that $12$ is the smallest possible number of regions in any feasible solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider an arbitrary division of an arbitrary rectangle into $n$ regions, such that every region has exactly five neighboring regions. We translate this picture into a so-called &lt;a href=&quot;http://en.wikipedia.org/wiki/Planar_graph&quot;&gt;planar graph&lt;/a&gt;: each of the $n$ regions then translates into a vertex/point, and there is a connecting edge between any two vertices/points that correspond to neighboring regions. The total number of edges is denoted $e$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We only need facts on $e$ and $n$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Fact 1:&lt;/strong&gt; $~~~~2e=5n$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This can be seen by counting the touching vertex-edge pairs: every edge touches two vertices, and every vertex touches $5$ edges.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Fact 2:&lt;/strong&gt; $~~~~e\le3n-6$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This fact can be found (for instance) on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Planar_graph&quot;&gt;wiki-page&lt;/a&gt; in the paragraph on maximal planar graphs.  If follows from the Euler formula for planar graphs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now by combining Facts 1 and 2, we get&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$5n/2~=~e~\le~3n-6$, &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which implies $0\le n/2-6$ and hence $n\ge12$. Consequently, Lopsy's construction is best possible with respect to the number of used regions.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-01-24T14:55:16.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8127" PostTypeId="1" AcceptedAnswerId="8130" CreationDate="2015-01-24T16:05:07.083" Score="15" ViewCount="1511" Body="&lt;p&gt;James Bond is invited to a party with altogether $128$ participants (including Bond himself, the host, and the hostess). At the beginning of the party the host takes James Bond aside and asks him to identify the mysterious Doctor No. It is known that Doctor No knows all the other people at the party, but none of the others does know Doctor No. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;James Bond starts asking questions of the following type: he asks some person A, whether A knows some other person B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the minimal number of questions which James Bond needs to ask (in the worst case) to identify the mysterious Doctor No?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-23T14:55:34.690" LastActivityDate="2018-09-26T01:21:31.283" Title="Finding Doctor No" Tags="|mathematics|strategy|combinatorics|" AnswerCount="6" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8128" PostTypeId="2" ParentId="8127" CreationDate="2015-01-24T16:30:02.157" Score="19" Body="&lt;p&gt;When you ask A if he knows B:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the answer is yes, this eliminates B as a candidate, since no one is supposed to know No.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the answer is no, this eliminates A as a candidate, since No is supposed to know everyone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;...if you pair everyone up, within 64 questions, you can eliminate 64 people. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;...If you group the remaining candidates and pair them up, 32 questions eliminates 32 more people.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 16 questions for 16 more;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 8 for 8 more;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4 for 4 more;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2 for 2 more; and&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 final question will identify Dr. No.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;...if you add them all up, he has to ask 127 questions. Of course, by the time that he has asked that many questions, Dr. No will probably catch on to his scheme and slip out, so that might not be the best way to go about it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8858" LastEditorUserId="42452" LastEditDate="2018-09-26T01:21:31.283" LastActivityDate="2018-09-26T01:21:31.283" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="8130" PostTypeId="2" ParentId="8127" CreationDate="2015-01-24T20:46:38.817" Score="9" Body="&lt;p&gt;When James Bond asks A if he knows B:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the answer is yes, this eliminates B as a candidate, since no one is supposed to know Doctor No.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the answer is no, this eliminates A as a candidate, since Doctor No is supposed to know everyone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; James Bond just keeps asking an arbitrary non-eliminated person about another arbitrary non-eliminated person. Every question eliminates one person. At the beginning there are 125 candidates: 128 participants minus James Bond (whom the host knows), minus the host (whom James Bond knows), minus the hostess (whom the host knows). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the answer is...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...124 questions in the worst case.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="42452" LastEditDate="2018-09-26T01:18:02.173" LastActivityDate="2018-09-26T01:18:02.173" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="8131" PostTypeId="2" ParentId="8127" CreationDate="2015-01-24T23:05:14.270" Score="7" Body="&lt;p&gt;Here's a proof that 127 is optimal, assuming the host and hostess can also be Dr. No:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Imagine a party where nobody knows anyone except for Dr. No, who knows everyone. Querying anyone by Dr. No results in a &quot;No&quot;. In the worst case, Dr. No will be the last person you query.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For each person you know a-priori not to be Dr. No, the required number of questions decreases by 1.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-01-24T23:05:14.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8132" PostTypeId="1" AcceptedAnswerId="8137" CreationDate="2015-01-24T23:15:13.207" Score="4" ViewCount="3292" Body="&lt;p&gt;This is a question about pairs of anagrams capable of flavoring puzzles as memorable and surprising because the two words are so different.&lt;/p&gt;&#xA;&lt;p&gt;One of my favorite such pairs is &lt;em&gt;chesty&lt;/em&gt; and &lt;em&gt;scythe&lt;/em&gt;. These sound and look very different:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;two syllables; one syllable&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;c&lt;/em&gt; and &lt;em&gt;h&lt;/em&gt; together to represent the sound /tʃ/, with &lt;em&gt;t&lt;/em&gt; representing /t/; &lt;em&gt;t&lt;/em&gt; and &lt;em&gt;h&lt;/em&gt; together to represent /ð/, with &lt;em&gt;c&lt;/em&gt; immediately before &lt;em&gt;s&lt;/em&gt; to represent /s/ (or, if you prefer, silent)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;e&lt;/em&gt; pronounced; &lt;em&gt;e&lt;/em&gt; silent&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;y&lt;/em&gt; pronounced two different ways&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;y&lt;/em&gt; as suffix and in the root&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;one pair of consecutive consonants (with that term understood as it should be, to denote a non-vocal phoneme); no pairs&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;two monophthongs, no diphthong; no monophthongs, one diphthong.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;The only letter pronounced the same way in both words is &lt;em&gt;s&lt;/em&gt;, and even then you could reasonably say that the letter group representing the phoneme /s/ in &lt;em&gt;scythe&lt;/em&gt; is &lt;em&gt;sc&lt;/em&gt; rather than just &lt;em&gt;s&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;p&gt;They are also a long way apart semantically.&lt;/p&gt;&#xA;&lt;p&gt;The pair is practically a work of art!&lt;/p&gt;&#xA;&lt;p&gt;The pair &lt;em&gt;bedroom&lt;/em&gt; and &lt;em&gt;boredom&lt;/em&gt; is quite good, but hardly in the same league. &lt;em&gt;Conversation&lt;/em&gt; and &lt;em&gt;conservation&lt;/em&gt; is at the other end of the scale.&lt;/p&gt;&#xA;&lt;p&gt;What other pairs of very different sounding and looking words are anagrams of each other? What nice metric might we use for difference?&lt;/p&gt;&#xA;&lt;p&gt;And what about triples or larger sets?&lt;/p&gt;&#xA;&lt;p&gt;(I was caused to think of this question after tackling &lt;a href=&quot;https://puzzling.stackexchange.com/questions/6691&quot;&gt;@BmyGuest's challenge question&lt;/a&gt;. Add &lt;em&gt;r&lt;/em&gt; to &lt;em&gt;toenail&lt;/em&gt; and rearrange to get &lt;em&gt;relation&lt;/em&gt;; &lt;em&gt;t&lt;/em&gt; to &lt;em&gt;senatorial&lt;/em&gt;, &lt;em&gt;alterations&lt;/em&gt;; or in &lt;a href=&quot;https://puzzling.stackexchange.com/a/7748/4234&quot;&gt;@Psybin's answer&lt;/a&gt;, &lt;em&gt;i&lt;/em&gt; to &lt;em&gt;castle&lt;/em&gt;, &lt;em&gt;elastic&lt;/em&gt;.)&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;br&gt;&#xA;Apart from {&lt;em&gt;chesty&lt;/em&gt;, &lt;em&gt;scythe&lt;/em&gt;} and {&lt;em&gt;admirer&lt;/em&gt;, &lt;em&gt;married&lt;/em&gt;} (found by @Quark), are there any other pairs of single-word anagrams, with say five or more letters, in which no consecutive letter pair appears in both words? Four-letter examples include &lt;em&gt;part&lt;/em&gt; and &lt;em&gt;trap&lt;/em&gt;, too symmetric to get a high score for the difference.&lt;/p&gt;&#xA;" OwnerUserId="7363" LastEditorUserId="57521" LastEditDate="2021-05-14T18:17:58.933" LastActivityDate="2021-05-14T18:17:58.933" Title="Pairs of anagrams where the two words sound and look very different" Tags="|open-ended|anagram|" AnswerCount="5" CommentCount="6" ClosedDate="2021-08-16T16:10:22.467" ContentLicense="CC BY-SA 4.0" />
  <row Id="8133" PostTypeId="2" ParentId="8127" CreationDate="2015-01-24T23:37:50.310" Score="0" Body="&lt;p&gt;Isn't the answer depending on how many people James knows? (The subtraction of host and hostess are an example of that!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jame only has to ask people if they know somebody &lt;em&gt;he&lt;/em&gt; doesn't know, as otherwise it can't be Dr. No.&lt;/p&gt;&#xA;" OwnerUserId="5840" LastActivityDate="2015-01-24T23:37:50.310" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8134" PostTypeId="2" ParentId="8076" CreationDate="2015-01-25T00:11:48.527" Score="2" Body="&lt;p&gt;The way it is possible is that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You changed the color of one of the apples to or from another color other than green/blue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assumption 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In order for all the color references to have meaning (red, maroon, beautiful colors), the apple color change must have changed color from green/blue to not green/blue, or not green/blue to green/blue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assumption 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A blue apple is added &quot;right after&quot; the second doubling. However, in the story a color change happens during this doubling (or right after). This assumption will be that the blue is added after the color change.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Start:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; At the end you have the same number of green and blue apples, $G$ = $B$, original values. Before that, you eat one of each kind. $G+1$ greens and $B+1$ blues. Then before this, the bigger group is divided in half.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Branch 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First assume green was the bigger half. There are now $2G+2$ greens. A blue is added (subtracted) here (see assumption 2). There are now $B$ blues. Either a green/blue is added here or taken away (see assumption 1). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Branch 1.1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Green added (subtracted): There is now $2G+1$ greens. Both groups are doubled (halved). This is not possible (Green= $G+1/2$), end of branch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Branch 1.2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Blue added (subtracted): There are now $B-1$ blues. Both groups are halved. $(B-1)/2$ blues and $G+1$ greens. Greens doubled (halved) = $(G+1)/2$. A blue is added(subtracted), $(B-3)/2$ blues. Not equal for any value of $B$ (same as $G$), end of branch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Branch 1.3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Green subtracted (added): There are now $2G+3$ greens. This can't be halved (same as 1.1), end of branch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Branch 1.4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Blue subtracted (added): There are now $B+1$ blues. Both groups are halved. $(B+1)/2$ blues and $G+1$ greens. Greens doubled (halved) = $(G+1)/2$. A blue is added (subtracted), $(B-1)/2$. Not equal for any value of $B$ (same as $G$), end of branch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Branch 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First assume blue was the bigger half. There are now $2B+2$ blues. A blue is added (subtracted) here (see assumption 2). There are now $2B+1$ blues and $G+1$ greens. Either a green/blue is added here or taken away (see assumption 1). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Branch 2.1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Green added (subtracted): There is now $G$ greens. Both groups are doubled (halved). Blue is at $2B+1$ so it can't be halved, end of branch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Branch 2.2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Blue added (subtracted): There are now $2B$ blues. Both groups are halved. $B$ blues and $(G+1)/2$ greens. Greens doubled (halved) = $(G+1)/4$. A blue is added(subtracted), $B-1$ blues. Set two equations equal, $G,B=5/3$, this isn't possible so end of branch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Branch 2.3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Green subtracted (added): There are now $G+2$ greens. Blue is at $2B+1$ so it can't be halved, end of branch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Branch 2.4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Blue subtracted (added): There are now $2B+2$ blues. Both groups are halved. $B+1$ blues and $(G+1)/2$ greens. Greens doubled (halved) = $(G+1)/4$. A blue is added (subtracted), Set two equations equal, $G,B=1/3$, this isn't possible so end of branch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Takeaway:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For branch 1 and 2, if not using Assumption 1 (color change for non green/blue), that means that both green and blue either go up or down by 1 and this will always make the &quot;double both&quot; stage (divide both by 2) fail.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since none of these work, assumption 2 is probably wrong but I don't feel like doing this again without confirmation.&lt;/p&gt;&#xA;" OwnerUserId="7722" LastActivityDate="2015-01-25T00:11:48.527" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8135" PostTypeId="2" ParentId="8132" CreationDate="2015-01-25T00:21:52.413" Score="2" Body="&lt;p&gt;Many of the anagram pairs in this hall of fame:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dictionary = Indicatory&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elvis = Lives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Listen = Silent&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Admirer = Married&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although the letters don't really have different sounds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Source:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://wordsmith.org/anagram/hof.html&quot; rel=&quot;nofollow&quot;&gt;http://wordsmith.org/anagram/hof.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="7722" LastEditDate="2015-01-25T17:21:51.510" LastActivityDate="2015-01-25T17:21:51.510" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8136" PostTypeId="2" ParentId="8115" CreationDate="2015-01-25T00:30:28.090" Score="2" Body="&lt;p&gt;If you mean &quot;minimum&quot; as a lowerbound, then yes.  Last year there was a detailed discussion on the 3x3x3 &quot;supercube&quot; (which is a different name to describe the same puzzle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that they found that God's number is 26 quarter turn moves last year for the regular 6 color Rubik's cube as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will see in this discussion that a &lt;strong&gt;lowerbound&lt;/strong&gt; of 24 half turns and 28 quarter turns has been established.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://cubezzz.dyndns.org/drupal/?q=node/view/536&quot; rel=&quot;nofollow&quot;&gt;http://cubezzz.dyndns.org/drupal/?q=node/view/536&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your possible curiosity, Bruce Norskog recently found it that it takes at most 12 half turns to solve the first 4 cross edges and correctly twist the 5 centers adjacent to them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.speedsolving.com/forum/showthread.php?50422-Optimal-supercube-cross&quot; rel=&quot;nofollow&quot;&gt;https://www.speedsolving.com/forum/showthread.php?50422-Optimal-supercube-cross&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6044" LastActivityDate="2015-01-25T00:30:28.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8137" PostTypeId="2" ParentId="8132" CreationDate="2015-01-25T00:33:08.053" Score="13" Body="&lt;p&gt;I'm not a linguist so I couldn't tell you about diphthongs or phonemes, but here are a few:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;generate&lt;/em&gt; --&gt; &lt;em&gt;teenager&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;marginal&lt;/em&gt; --&gt; &lt;em&gt;alarming&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;interests&lt;/em&gt; --&gt; &lt;em&gt;triteness&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;infringe&lt;/em&gt; --&gt; &lt;em&gt;refining&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;bargained&lt;/em&gt; --&gt; &lt;em&gt;gabardine&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a good one I found:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;altering&lt;/em&gt; --&gt; &lt;em&gt;relating&lt;/em&gt; --&gt; &lt;em&gt;alerting&lt;/em&gt; --&gt; &lt;em&gt;triangle&lt;/em&gt; --&gt; &lt;em&gt;integral&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="8858" LastEditorUserId="8858" LastEditDate="2015-02-01T20:49:27.740" LastActivityDate="2015-02-01T20:49:27.740" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8138" PostTypeId="2" ParentId="8014" CreationDate="2015-01-25T01:34:20.900" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Professor Kafka is probably wrong because &quot;every pair of these five students was simultaneously asleep at some point in time&quot; is impossible if &quot;at any moment in time there were at least three of these students awake and listening to my lecture.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9105" LastEditorUserId="9105" LastEditDate="2015-01-25T01:39:31.983" LastActivityDate="2015-01-25T01:39:31.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8139" PostTypeId="2" ParentId="8111" CreationDate="2015-01-25T03:18:44.100" Score="45" Body="&lt;p&gt;Cut perfectly in half we have a 7 folks!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/T7ak5.png&quot; alt=&quot;Cut perfectly in half we have a 7!&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9131" LastEditorUserId="9439" LastEditDate="2016-05-05T10:59:39.257" LastActivityDate="2016-05-05T10:59:39.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8140" PostTypeId="1" AcceptedAnswerId="8142" CreationDate="2015-01-25T03:38:48.467" Score="16" ViewCount="1873" Body="&lt;p&gt;You are at the edge of an enormous circular arena. A hungry lion is eying you from the centre of this area. You are both capable of running at the same maximum speed, but constraint within the arena. The lion has worked out a strategy of always running at maximum speed in an outward direction such that he stays positioned on the line thru you and the center of the arena.&lt;/p&gt;&#xA;&lt;p&gt;The starting signal sounds and the lion starts moving. You can't outrun the lion. How do you ensure you stay out of the lion's claws?&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Clarifications:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the arena is truly gigantic, and you can think of you and the lion as point objects constrained within a circle of unit radius&lt;/li&gt;&#xA;&lt;li&gt;there is no latency in the lion's reactions to your movements: at any moment in time you, the lion, and the center of the arena remain co-linear&lt;/li&gt;&#xA;&lt;li&gt;the lion catches you if, and only if, his position coincides with yours&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Running at maximum speed along any circular path centered on the center of the arena is &lt;em&gt;not&lt;/em&gt; going to help you. Starting from the center, the lion will run at the same speed along a circular path &lt;em&gt;of half the radius&lt;/em&gt; and catch you as soon as you have completed a quarter of a circle. (It is easy to check that these paths keep you, the lion and the center co-linear.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7573" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-03-23T14:55:14.377" Title="Escaping a hungry lion you can't outrun" Tags="|mathematics|geometry|strategy|pursuit-evasion|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8141" PostTypeId="2" ParentId="8140" CreationDate="2015-01-25T04:10:14.097" Score="-1" Body="&lt;p&gt;To stay out of the lions' claws:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Walk slightly away from the edge (can be any amount &gt; 0), then run in a perfect circle concentric with the arena.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If the lion and human are truly points and the lion always runs on the line between the center and the human, then by running in a circle the human can stay away (although infinitesimally close) to the lion. The lion will reach (near) the human by $R/V$ seconds no matter what, where $R$ is the radius of the ring and $V$ is the velocity of the lion in units of $R$ per second. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's a picture to show what happens:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zVth2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the picture, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The lines are exaggerated, just extend that to a polygon with &quot;infinite&quot; sides with a circumscribed circle. There's probably a better way to prove this with calculus and $\partial \theta$ but I forget how, just remember seeing this puzzle before.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="7722" LastEditDate="2015-01-25T04:24:49.277" LastActivityDate="2015-01-25T04:24:49.277" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8142" PostTypeId="2" ParentId="8140" CreationDate="2015-01-25T04:20:29.387" Score="16" Body="&lt;p&gt;First, run 1/3 of a unit toward the center of the arena. The radius of the arena is 1 unit and the lion runs at the same speed as you, so the lion will not catch you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we run along a series of straight line paths. The $n$-th path will be orthogonal to the the line segment connecting your position and the lion's (at the start of the path), and have a length of $\frac{1}{2n}$ units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These paths will never take you outside the arena. To see this, if you are $d$ units from the center of the arena immediately before running along the $n$-th path, your distance from the center after the $n$-th path will be&#xA;$$&#xA;\sqrt{d^2+\left(\frac{1}{2n}\right)^2}.&#xA;$$&#xA;This means your distance from the center after the $n$-th path is&#xA;$$&#xA;\sqrt{\left(\frac{2}{3}\right)^2+\left[\left(\frac{1}{2}\right)^2+\ldots+\left(\frac{1}{2n}\right)^2\right]},&#xA;$$&#xA;and this quantity is always less than 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since each straight line path heads in a direction orthogonal to the lion, the lion cannot catch you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the total distance you are running is&#xA;$$&#xA;\frac{1}{3}+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\ldots\right)=\infty&#xA;$$&#xA;(this is essentially the fact that the harmonic series diverges), so you can continue to run along the chosen paths indefinitely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, this strategy enables you to evade the lion no matter how the lion decides to pursue you.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-01-25T04:35:58.537" LastActivityDate="2015-01-25T04:35:58.537" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="8143" PostTypeId="1" AcceptedAnswerId="8148" CreationDate="2015-01-25T06:32:50.793" Score="17" ViewCount="1157" Body="&lt;p&gt;Decrypt the hidden meaning in the picture below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oLgOu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/AhzY6.png&quot; alt=&quot;Component not primary&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is a humble yet terrific and radiant puzzle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="7722" LastEditDate="2015-05-25T16:34:14.027" LastActivityDate="2017-04-28T19:29:09.677" Title="One word: Bacon" Tags="|cipher|visual|" AnswerCount="2" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="8144" PostTypeId="1" AcceptedAnswerId="8146" CreationDate="2015-01-25T08:23:40.663" Score="17" ViewCount="1811" Body="&lt;p&gt;Several matchstick digits are still digits when rotated by 180 degrees:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YdFw2.png&quot; alt=&quot;invertible matchstick digits&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that 2 is rotated to 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/donXi.png&quot; alt=&quot;2 to 2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, we can make a matchstick arithmetic expression which is valid when rotated 180 degrees. For example, 1+1=2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Urf4i.png&quot; alt=&quot;1+1=2&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, this is trivial because two expressions are the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following matchstick expression produces two different valid equalities: 10+02=12 and 21=20+01.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/f6Dh9.png&quot; alt=&quot;10+02=12,21=20+01&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, it's not natural because of the leading 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you find a natural matchstick arithmetic expression which produces two different valid equalities when rotated 180 degrees?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe, the smallest answer is 0+6=6 which is turned into 9=9+0. Or 1x6=6 which is turned into 9=9x1. But, let us exclude trivial equalities like &quot;0+...&quot;, &quot;0x...&quot;, or &quot;1x...&quot;&lt;/p&gt;&#xA;" OwnerUserId="3914" LastEditorUserId="8874" LastEditDate="2016-03-27T17:53:38.177" LastActivityDate="2016-09-03T17:07:39.220" Title="180 degree rotation matchstick equality" Tags="|mathematics|calculation-puzzle|matches|" AnswerCount="6" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="8145" PostTypeId="2" ParentId="8144" CreationDate="2015-01-25T08:46:25.893" Score="14" Body="&lt;p&gt;I'll start the ball rolling...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FAyEM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dpL71.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="8875" LastActivityDate="2015-01-25T08:46:25.893" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8146" PostTypeId="2" ParentId="8144" CreationDate="2015-01-25T09:59:28.813" Score="19" Body="&lt;p&gt;This first answer is kind of boring, but I think multiplication answers will be hard to come by:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 11 x 6 = 66&lt;br&gt;&#xA; 99 = 9 x 11&lt;br&gt;&#xA; You can have as many 1s as you like in one term. Also you can add as many extra x1 terms on either side as you want, just as you can add +0&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Removed an incorrect solution that assumed 2 rotated to 5. Erk).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a more interesting one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 65 + 2 x 6 + 99 - 66 = 58 + 52&lt;br&gt;&#xA; 25 + 85 = 99 - 66 + 9 x 2 + 59&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: And here's another extendable answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 66 + 55 - 22 = 99&lt;br&gt;&#xA; 66 = 22 - 55 + 99&lt;br&gt;&#xA; You can do this with any number of digits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6031" LastEditorUserId="29753" LastEditDate="2016-09-03T15:19:35.623" LastActivityDate="2016-09-03T15:19:35.623" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8147" PostTypeId="1" AcceptedAnswerId="8150" CreationDate="2015-01-25T11:18:55.277" Score="16" ViewCount="4142" Body="&lt;p&gt;My fellow mathematicians Paul and Sam are very fond of mathematical puzzles. One day, I came up with the idea on mathematical puzzles while three of us having lunch together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I: Hey, guys. I am thinking of two integers which are greater than or equal to 2. I will tell you the product and the sum of two numbers. Why don't you guess what two numbers are?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P: If you tell us the product and the sum, finding two integers is a piece of cake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I: I will tell Paul only the product and Sam only the sum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I whispered the product to Paul and the sum to Sam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I: First, I will ask Paul. Could you figure out two numbers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P: (after thinking) I can't. I don't know what two integers are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I: (to Sam) How about you Sam? Do you know what two unknown numbers are?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S: I do not know either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surprisingly, right after Sam answered that he does not know, Paul said;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P: Now, I think I know these two numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S: I also found out two numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, a question to the reader. What are two integers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Remark:&lt;/strong&gt; The answer is NOT 4 and 13.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this riddle, after answering &quot;I don't know&quot; twice, the answer &quot; I know&quot; came one after another. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &quot;I don't know&quot; comes three times and then answers &quot;I know&quot;, what would two integers be? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the largest possible number of &quot;I don't know&quot; before one can answer &quot;I know&quot; when solving two integers?&lt;/p&gt;&#xA;" OwnerUserId="3914" LastEditorUserId="10615" LastEditDate="2016-02-24T03:59:27.883" LastActivityDate="2017-08-01T11:30:03.577" Title="Product and Sum" Tags="|logical-deduction|calculation-puzzle|meta-knowledge|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8148" PostTypeId="2" ParentId="8143" CreationDate="2015-01-25T11:21:57.227" Score="10" Body="&lt;p&gt;I believe it must be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wilbur, the pig from Charlotte's Web&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is a Baconion cipher. The first three letters are encoded in color channels. The red channel is present in letters B, C and O, which encodes to BABBA, or W. A similar process is done for the green and blue channels, which yield ABAAA (I) and ABABB (L). The next three letters are encoded in the font style. The B (AAAAB) comes from the n being italic. The U comes similarly from the capital letters (BABAA), and the final letter (R) is encoded in the pixelated font (ABBBA). The first hint shows you what the &quot;default&quot; lettering looks like, so you can spot anomalies. The second hint refer directly to the answer, but also hints on that this is in fact a Baconian cipher (pig reference).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might expand on the explanation if it is confusing, but right now I don't have the time for it. Very nice puzzle.&lt;/p&gt;&#xA;" OwnerUserId="7577" LastActivityDate="2015-01-25T11:21:57.227" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8149" PostTypeId="2" ParentId="8143" CreationDate="2015-01-25T12:12:48.990" Score="2" Body="&lt;p&gt;Is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;black&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. It's a cryptogram solution of &quot;bacon&quot; if you allow that a letter can represent itself.&lt;br&gt;&#xA; 2. Relevance to Hint 1 is that a synonym for &quot;default&quot; is &quot;basic&quot;, which is on a black background, which evokes the common phrase &quot;basic black&quot;.&lt;br&gt;&#xA;3. Relevance to Hint 2 is that the words &quot;humble&quot;, &quot;terrific&quot;, and &quot;radiant&quot; are three of the four words or phrases (the fourth being &quot;some pig&quot;) woven into her web by Charlotte to describe Wilbur the pig in the book &quot;Charlotte's Web&quot; by E. B. White. White is the opposite of black, and it also has the cute added relevance of bacon coming from a pig.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I guess&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;white&quot; could be the answer, too, but it didn't seem to fit as well with Hint 1. So could &quot;basic&quot; but that doesn't fit at all with Hint 2. Any 5-letter word with no repeated letters is a cryptogram solution of &quot;bacon&quot; but those are the top three I could think of that were at all relevant to the hints.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9133" LastEditorUserId="35972" LastEditDate="2017-04-28T19:29:09.677" LastActivityDate="2017-04-28T19:29:09.677" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8150" PostTypeId="2" ParentId="8147" CreationDate="2015-01-25T12:13:21.583" Score="10" Body="&lt;p&gt;The numbers are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 6 and 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; P gets 12 and S gets 8&lt;BR&gt;&#xA; P knows the numbers are (4, 3) or (6, 2), so replies I don't know&lt;BR&gt;&#xA; S knows the numbers are (2, 6) or (4, 4) but can not be (3, 5) as P doesn't know the answer, so still replies I don't know&lt;BR&gt;&#xA; Suddenly P realizes that numbers can not be (4, 3) otherwise S would have ruled out (5, 2) and guessed (4, 3) and hence (6, 2) is the answer so he replies I know the answer&lt;BR&gt;&#xA; S also realizes that the numbers can not be (4, 4) otherwise P could not know the number so he concludes the numbers are (6, 2) and replies I know the answer &lt;BR&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I assume that both P and S are perfect logicians.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;BR&gt; Answers to the bonus questions are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (4,4) for sequence no, no, no, yes&lt;BR&gt;&#xA; There may be infinite rounds of NO&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I will add the details once I put down my paper work in text.&lt;/p&gt;&#xA;" OwnerUserId="4637" LastEditorUserId="4637" LastEditDate="2015-01-27T08:33:45.067" LastActivityDate="2015-01-27T08:33:45.067" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="8152" PostTypeId="2" ParentId="8127" CreationDate="2015-01-25T14:40:15.220" Score="3" Body="&lt;p&gt;The answer can also be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One question. James asks the host whether he knows the hostess or not. If he does, he is Dr.No. If he doesn't, she is.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why do I think this is a valid answer?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The OP could either be a strictly mathematical puzzle, or a &lt;em&gt;brainteaser&lt;/em&gt; with a somewhat more loose attitude towards conditions. To be of the first kind, it doesn't seem to be rigorous enough in all statements. In particular: Are all queried persons tell the truth? (Assume yes.). What does &lt;em&gt;knowing&lt;/em&gt; somebody mean exactly? &lt;br&gt;&lt;br&gt; for this reason I think the puzzle can be looked upon in a little more relaxed way, allowing 'common sense' assumptions to be made. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The following 'common sense' assumptions are needed:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - By &lt;em&gt;knowing&lt;/em&gt; a person, one means that the name of the person is known to you.  &lt;br&gt;&#xA; - All people have been invited by the host or the hostess of the party.&lt;br&gt;&#xA; - You invite people only if you &lt;em&gt;know&lt;/em&gt; them. (I.e you know their name for the invitation.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Under those assumptions: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Either host or hostess know all and are therefore Dr. No. (But either host or hostess do &lt;em&gt;not&lt;/em&gt; know their counterpart!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the question then reveals:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the host knows the hostess, she can't be Dr. No - therefore he is! Otherwise, she has to be Dr. Know.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastActivityDate="2015-01-25T14:40:15.220" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8153" PostTypeId="1" AcceptedAnswerId="8156" CreationDate="2015-01-25T15:49:46.953" Score="9" ViewCount="818" Body="&lt;p&gt;My friends Peter and Sam are excellent mathematicians and always think strictly logical. Yesterday I told them: &lt;em&gt;&quot;I have secretly chosen two integers $x$ and $y$ with $1\le x\le y\le 9$. I have told their sum $s=x+y$ to Sam and their product $p=xy$ to Peter.&quot;&lt;/em&gt; Then the following conversation developed.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Peter said: I don't know the numbers.&lt;/li&gt;&#xA;&lt;li&gt;Sam said:   I don't know the numbers.&lt;/li&gt;&#xA;&lt;li&gt;Peter said: I don't know the numbers.&lt;/li&gt;&#xA;&lt;li&gt;Sam said:   I don't know the numbers.&lt;/li&gt;&#xA;&lt;li&gt;Peter said: I don't know the numbers.&lt;/li&gt;&#xA;&lt;li&gt;Sam said:   I don't know the numbers.&lt;/li&gt;&#xA;&lt;li&gt;Peter said: I don't know the numbers.&lt;/li&gt;&#xA;&lt;li&gt;Sam said:   I don't know the numbers.&lt;/li&gt;&#xA;&lt;li&gt;Peter said: Aha. Then I do know the numbers now.&lt;/li&gt;&#xA;&lt;li&gt;Sam said:  Aha. Then I also know the numbers now.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What are the values of $x$ and $y$?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="10615" LastEditDate="2016-02-24T03:59:35.257" LastActivityDate="2016-02-24T03:59:35.257" Title="Product and Sum, Variant 2a" Tags="|mathematics|logical-deduction|meta-knowledge|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8154" PostTypeId="2" ParentId="8127" CreationDate="2015-01-25T15:56:27.267" Score="1" Body="&lt;p&gt;I'd go with&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; zero&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;given that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; this is James Bond and he is trying to find Doctor No.  Ideally, just stand and listen for a few minutes and the megalomaniac will probably just identify themselves by their behaviour.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...in practice, the typical procedure would be to seduce the most attractive female in the room, get captured and then Doctor No will identify himself (with a number of complications, gunshots, explosions and at least one car/boat/aircraft/space shuttle chase along the way).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Source:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Dr._No_%28film%29#Plot&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Dr.&lt;em&gt;No&lt;/em&gt;%28film%29#Plot&lt;/a&gt; (and most other Bond films/books)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9142" LastEditorUserId="42452" LastEditDate="2018-09-26T01:17:20.383" LastActivityDate="2018-09-26T01:17:20.383" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8155" PostTypeId="2" ParentId="5299" CreationDate="2015-01-25T15:58:06.907" Score="2" Body="&lt;p&gt;Here is another very simple argument. Suppose you are currently 1 meter away from the next moving walkway. Consider two possibilities:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;You may move the 1 meter before resting and move along with the walkway.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You rest right away, and afterwards move the 1 meter to the walkway.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Both possibilities cost the same time (the time of resting plus the time of walking one meter). Clearly, the first possibility brings you farther ahead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence without any calculation: &lt;strong&gt;You should rest on the moving walkway.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-01-25T15:58:06.907" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8156" PostTypeId="2" ParentId="8153" CreationDate="2015-01-25T16:54:55.273" Score="6" Body="&lt;p&gt;To solve this I listed all possible solutions and then removed each entry that couldn't be it after the question. The thing is, a person knows what value x and y if there is only one entry in that list with that given s or p. Saying that you don't know the solution implies that I can remove all items from the list that only have a single entry with that s or p. Here is the list:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;xy  s  p&#xA;11  2  1&#xA;12  3  2&#xA;13  4  3&#xA;14  5  4&#xA;15  6  5&#xA;16  7  6&#xA;17  8  7&#xA;18  9  8&#xA;19 10  9&#xA;22  4  4&#xA;23  5  6&#xA;24  6  8&#xA;25  7 10&#xA;26  8 12&#xA;27  9 14&#xA;28 10 16&#xA;29 11 18&#xA;33  6  9&#xA;34  7 12&#xA;35  8 15&#xA;36  9 18&#xA;37 10 21&#xA;38 11 24&#xA;39 12 27&#xA;44  8 16&#xA;45  9 20&#xA;46 10 24&#xA;47 11 28&#xA;48 12 32&#xA;49 13 36&#xA;55 10 25&#xA;56 11 30&#xA;57 12 35&#xA;58 13 40&#xA;59 14 45&#xA;66 12 36&#xA;67 13 42&#xA;68 14 48&#xA;69 15 54&#xA;77 14 49&#xA;78 15 56&#xA;79 16 63&#xA;88 16 64&#xA;89 17 72&#xA;99 18 81&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;After question 1 I can remove 11, 12, 13, 15, 17, 25, 27, 35, 37, 39, 45, 47, 48, 55, 56, 57, 58, 59, 67, 68, 69, 77, 78, 79, 88, 89 and 99.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That leaves us with:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;xy  s  p&#xA;14  5  4&#xA;16  7  6&#xA;18  9  8&#xA;19 10  9&#xA;22  4  4&#xA;23  5  6&#xA;24  6  8&#xA;26  8 12&#xA;28 10 16&#xA;29 11 18&#xA;33  6  9&#xA;34  7 12&#xA;36  9 18&#xA;38 11 24&#xA;44  8 16&#xA;46 10 24&#xA;49 13 36&#xA;66 12 36&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;After question 2 we remove 22, 49, 66. we now have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;xy  s  p&#xA;14  5  4&#xA;16  7  6&#xA;18  9  8&#xA;19 10  9&#xA;23  5  6&#xA;24  6  8&#xA;26  8 12&#xA;28 10 16&#xA;29 11 18&#xA;33  6  9&#xA;34  7 12&#xA;36  9 18&#xA;38 11 24&#xA;44  8 16&#xA;46 10 24&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;After question 3 we eliminate 14, we now have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;xy  s  p&#xA;16  7  6&#xA;18  9  8&#xA;19 10  9&#xA;23  5  6&#xA;24  6  8&#xA;26  8 12&#xA;28 10 16&#xA;29 11 18&#xA;33  6  9&#xA;34  7 12&#xA;36  9 18&#xA;38 11 24&#xA;44  8 16&#xA;46 10 24&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;After question 4 we eliminate 23, we now have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;xy  s  p&#xA;16  7  6&#xA;18  9  8&#xA;19 10  9&#xA;24  6  8&#xA;26  8 12&#xA;28 10 16&#xA;29 11 18&#xA;33  6  9&#xA;34  7 12&#xA;36  9 18&#xA;38 11 24&#xA;44  8 16&#xA;46 10 24&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;After question 5 we eliminate 16, we now have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;xy  s  p&#xA;18  9  8&#xA;19 10  9&#xA;24  6  8&#xA;26  8 12&#xA;28 10 16&#xA;29 11 18&#xA;33  6  9&#xA;34  7 12&#xA;36  9 18&#xA;38 11 24&#xA;44  8 16&#xA;46 10 24&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;After question 6 we eliminate 34, we now have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;xy  s  p&#xA;18  9  8&#xA;19 10  9&#xA;24  6  8&#xA;26  8 12&#xA;28 10 16&#xA;29 11 18&#xA;33  6  9&#xA;36  9 18&#xA;38 11 24&#xA;44  8 16&#xA;46 10 24&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;After question 7 we eliminate 26, we now have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;xy  s  p&#xA;18  9  8&#xA;19 10  9&#xA;24  6  8&#xA;28 10 16&#xA;29 11 18&#xA;33  6  9&#xA;36  9 18&#xA;38 11 24&#xA;44  8 16&#xA;46 10 24&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;After question 8 we eliminate 44, we now have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;xy  s  p&#xA;18  9  8&#xA;19 10  9&#xA;24  6  8&#xA;28 10 16&#xA;29 11 18&#xA;33  6  9&#xA;36  9 18&#xA;38 11 24&#xA;46 10 24&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Conclusion:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If Peter didn't know it now, he would eliminate 28 now which means that x = 2 and y = 8 is the answer. Sam knows this too because he thinks the same like Peter (and me).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-01-25T16:54:55.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8157" PostTypeId="2" ParentId="8111" CreationDate="2015-01-25T16:56:37.780" Score="41" Body="&lt;p&gt;Slightly different answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 12 is a base ten number. Cut the base in half, and you get 7.&lt;br&gt;&#xA; (base five) 12 = (base ten) 7&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8861" LastActivityDate="2015-01-25T16:56:37.780" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8159" PostTypeId="2" ParentId="2188" CreationDate="2015-01-25T17:51:21.313" Score="5" Body="&lt;p&gt;You ask:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Would the other guard say your door is the door to freedom?&#xA;&#xA; If either guard is guarding the door to freedom both would answer NO.&#xA; If either guard is guarding the door to death   both would answer YES.&#xA; NO leads to freedom and YES leads to death.&#xA;&#xA; I don't think there is one question whose answer would indicate which guard&#xA; is guarding which door and which door leads to freedom. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9144" LastEditorUserId="46903" LastEditDate="2018-09-14T04:16:20.590" LastActivityDate="2018-09-14T04:16:20.590" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8160" PostTypeId="2" ParentId="8111" CreationDate="2015-01-25T20:30:37.057" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Using roman numbering seems to do the trick : 12 is : &lt;strong&gt;XII&lt;/strong&gt;, cut it horizontally  and you have &lt;strong&gt;VII&lt;/strong&gt; at the top&#xA;&#xA; &lt;pre&gt;/||&lt;/pre&gt;&#xA; &lt;pre&gt;/\||&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9151" LastEditorUserId="9151" LastEditDate="2016-05-07T19:47:45.483" LastActivityDate="2016-05-07T19:47:45.483" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8161" PostTypeId="2" ParentId="8144" CreationDate="2015-01-25T21:23:07.880" Score="6" Body="&lt;p&gt;$$6+6-5=5+1+(1+1)-1$$&#xA;Rotates to&#xA;$$1-(1+1)+1+5=5-9+9$$&lt;/p&gt;&#xA;" OwnerUserId="4765" LastActivityDate="2015-01-25T21:23:07.880" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8162" PostTypeId="1" CreationDate="2015-01-25T22:29:56.623" Score="16" ViewCount="2543" Body="&lt;p&gt;A man is sitting in a dark room. In front of him is a pile of 100 cards. He knows that there are 90 cards faced up and 10 cards faced down. &#xA;Is it possible for the man to form two piles from these cards such that each pile contains the same number of faced down cards? No loopholes...&lt;/p&gt;&#xA;" OwnerUserId="8863" LastEditorUserId="8874" LastEditDate="2015-01-26T09:29:59.910" LastActivityDate="2016-08-28T04:50:08.050" Title="Cards in the darkness" Tags="|logical-deduction|" AnswerCount="5" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="8163" PostTypeId="1" AcceptedAnswerId="8173" CreationDate="2015-01-25T22:38:38.560" Score="0" ViewCount="1242" Body="&lt;p&gt;You invite all your friends to a birthday party.&lt;br&gt;&#xA;Three of your friends came.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the party, you need to share your cake equally with everyone.&lt;br&gt;&#xA;The friends that came aren't your best friends, so you don't give anyone an extra piece.&lt;br&gt;&#xA;You can cut three times. No more, no less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you share the cake with everyone?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cake is a cylinder, it looks something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lxdfFm.jpg&quot; alt=&quot;pic&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="435" LastEditorUserId="8874" LastEditDate="2016-03-23T15:37:27.130" LastActivityDate="2016-03-23T15:37:27.130" Title="How to share this cake?" Tags="|strategy|dissection|" AnswerCount="4" CommentCount="6" FavoriteCount="0" ClosedDate="2015-01-27T01:42:11.400" ContentLicense="CC BY-SA 3.0" />
  <row Id="8164" PostTypeId="2" ParentId="8163" CreationDate="2015-01-25T22:49:05.653" Score="8" Body="&lt;p&gt;Assuming you want 4 equal pieces and the cake is cylindrical shaped:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cut the cake into fourths (perpendicular cuts through the center of the circular face), and then cut through the cake horizontally (half way between both circular faces) to make 8 equal pieces. Each person gets 2 pieces. (this idea from your profile pic)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you meant each person can only get one piece, then there are still many ways to do it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One way using parallel lines is to first cut the cake in half normally, then use the remaining two cuts to divide the semicircles in half. There are many ways to make these cuts and if you want them parallel, you will cut approximately $.407*R$ away from the center where $R$ is the radius of the cake's circular face. (This from &lt;a href=&quot;http://www.mathopenref.com/segmentarea.html&quot;&gt;http://www.mathopenref.com/segmentarea.html&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="7722" LastEditDate="2015-01-25T23:03:06.760" LastActivityDate="2015-01-25T23:03:06.760" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8165" PostTypeId="2" ParentId="8144" CreationDate="2015-01-25T22:51:22.407" Score="11" Body="&lt;p&gt;What answer is &quot;best&quot; depends on how one defines &quot;trivial&quot;, but if the goal is to have an answer which uses different arithmetical relationships when inverted, how about &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 21-9                                   5-11&#xA;------ = 8-6    which becomes    9-8 = ------&#xA; 11-5                                   6-12&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The values before and after inversion are different, and the things that make one side different have no analog on the other side (whereas in e.g. something like 26+12 = 16+22, the left-side &quot;6&quot; becoming a nine is balanced by having the right-side six become a nine).&lt;/p&gt;&#xA;" OwnerUserId="1467" LastEditorUserId="1467" LastEditDate="2015-01-26T17:59:53.577" LastActivityDate="2015-01-26T17:59:53.577" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8170" PostTypeId="1" AcceptedAnswerId="8180" CreationDate="2015-01-26T00:19:10.023" Score="16" ViewCount="1190" Body="&lt;p&gt;Alright, say you've made this logic based puzzle, but when you post it, people find tons of &lt;strong&gt;loop holes&lt;/strong&gt;, that obviously wasn't what you want! You didn't tag it with &lt;a href=&quot;/questions/tagged/lateral-thinking&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged 'lateral-thinking'&quot; rel=&quot;tag&quot;&gt;lateral-thinking&lt;/a&gt; or &lt;a href=&quot;/questions/tagged/brainteaser&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged 'brainteaser'&quot; rel=&quot;tag&quot;&gt;brainteaser&lt;/a&gt;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's times like these where you want to undo the ask button, because &lt;em&gt;it's just a matter of time before people &lt;strong&gt;DV&lt;/strong&gt; and &lt;strong&gt;VTC&lt;/strong&gt; &quot;Too Broad&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what are some &lt;strong&gt;general&lt;/strong&gt; &lt;em&gt;tips/advice&lt;/em&gt; to &lt;strong&gt;avoiding&lt;/strong&gt; such situations where you just want to rip your hair out? &lt;em&gt;How can you perfect the logic puzzle so people can't come in and ruin the fun, during the last phase of puzzle creation?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="4050" LastEditDate="2015-01-26T09:09:35.523" LastActivityDate="2015-01-26T09:34:30.467" Title="How to ensure your logic puzzle has no loopholes?" Tags="|logical-deduction|lateral-thinking|puzzle-creation|" AnswerCount="1" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="8172" PostTypeId="2" ParentId="8163" CreationDate="2015-01-26T02:24:49.377" Score="3" Body="&lt;p&gt;Take off the candles first; you don't want to eat those :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that you (as the birthday boy!) will take a piece also, so we're looking for four pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, you have to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; tilt the cake sideways, and cut it three times horizontally, like this (&lt;em&gt;picture not to scale&lt;/em&gt;):&lt;br&gt;&#xA; &lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/xNWJz.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA; &lt;br&gt;&#xA; These won't be traditionally shaped &quot;wedge&quot; cake pieces, but they'll still work :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Off-by-one_error#Fencepost_error&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Wikipedia article&lt;/a&gt; for more info about why this question sounds so complicated to some people.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This answer was really a piece of cake :P&lt;/p&gt;&#xA;" OwnerUserId="9038" LastEditorUserId="9038" LastEditDate="2015-01-26T03:49:09.300" LastActivityDate="2015-01-26T03:49:09.300" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8173" PostTypeId="2" ParentId="8163" CreationDate="2015-01-26T03:06:18.280" Score="10" Body="&lt;p&gt;Can't you just do this? Surprised nobody thought of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1iM8A.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Answer (Click me)&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9159" LastEditorUserId="9159" LastEditDate="2015-01-26T16:50:44.097" LastActivityDate="2015-01-26T16:50:44.097" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8174" PostTypeId="2" ParentId="8163" CreationDate="2015-01-26T03:49:56.573" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. First, cut into half (1) &lt;br&gt;&#xA; 2. Rotate the first half 90 degree and put on the other half. &lt;br&gt;&#xA; 3. Cut the edge. (2) &lt;br&gt;&#xA; 4. Flip &lt;br&gt;&#xA; 5. Cut the edge (3) &lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/aViGc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9160" LastActivityDate="2015-01-26T03:49:56.573" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8175" PostTypeId="1" AcceptedAnswerId="8204" CreationDate="2015-01-26T06:47:49.077" Score="12" ViewCount="1231" Body="&lt;p&gt;You are asked to dissect an &lt;span class=&quot;math-container&quot;&gt;$N \times N$&lt;/span&gt; square into &lt;a href=&quot;http://en.wikipedia.org/wiki/Polyomino&quot; rel=&quot;nofollow noreferrer&quot;&gt;polyomino pieces&lt;/a&gt; such that each piece shares a portion of its boundary with exactly &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; other pieces, and no piece has area exceeding &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt;. This can be achieved for &lt;span class=&quot;math-container&quot;&gt;$D \le 5$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;For &lt;span class=&quot;math-container&quot;&gt;$D=2, 3, 4$&lt;/span&gt; the smallest such squares are of size &lt;span class=&quot;math-container&quot;&gt;$2 \times 2$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$3 \times 3$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$4 \times 4$&lt;/span&gt;, respectively:&lt;br /&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/ssHC3.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Find the smallest square for &lt;span class=&quot;math-container&quot;&gt;$D=5$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Credit: inspired by &lt;a href=&quot;https://puzzling.stackexchange.com/questions/8122/find-smallest-rectangle-divided-into-figures-so-each-figure-has-5-neighbours&quot;&gt;this puzzle&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;EDIT / Bounty offer by Will Octagon Gibson:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Two answers have already been posted using 8x8 squares and 16 polyominoes.&lt;/p&gt;&#xA;&lt;p&gt;I plan to award a bounty of 100 if a correct answer is posted using a smaller square or fewer polyominoes.&lt;/p&gt;&#xA;&lt;p&gt;I plan to increase the bounty to 200 if a correct answer is posted using a smaller square &lt;strong&gt;AND&lt;/strong&gt; fewer polyominoes.&lt;/p&gt;&#xA;&lt;p&gt;These offers will expire on November 30, 2023 at approximately 11:59 p.m. (my time).&lt;/p&gt;&#xA;&lt;p&gt;Good luck!&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;" OwnerUserId="7573" LastEditorUserId="65277" LastEditDate="2023-11-16T13:38:16.027" LastActivityDate="2023-11-25T15:41:21.407" Title="Dissecting a square" Tags="|geometry|optimization|graph-theory|polyomino|dissection|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="8176" PostTypeId="1" AcceptedAnswerId="8177" CreationDate="2015-01-26T08:28:43.363" Score="5" ViewCount="2942" Body="&lt;p&gt;A man locked his personal computer with a password and wrote some phrases in the hint box kept in his room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One day, his wife tries to login in his absence using the hints which contained the following in seperate pieces of paper:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1 apples&lt;br&gt;&#xA;  1 oranges&lt;br&gt;&#xA;  2 pineapples&lt;br&gt;&#xA;  4 grapes&lt;br&gt;&#xA;  7 bananas&lt;br&gt;&#xA;  7 mangoes&lt;br&gt;&#xA;  8 pomegranates&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can you help the wife find the password?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1&lt;/strong&gt;: The password consists of &lt;em&gt;a&lt;/em&gt; word&lt;/p&gt;&#xA;" OwnerUserId="7482" LastEditorUserId="7482" LastEditDate="2015-01-27T03:30:41.797" LastActivityDate="2017-04-10T06:31:45.830" Title="Find the password of the man's PC" Tags="|password|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8177" PostTypeId="2" ParentId="8176" CreationDate="2015-01-26T08:34:38.130" Score="5" Body="&lt;p&gt;Password is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; passion&lt;br&gt;&#xA; numbers represent letter to use from the associated word;&lt;br&gt;&#xA; a o i p s s n&lt;br&gt;&#xA; then unscramble  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-01-27T03:24:53.823" LastActivityDate="2015-01-27T03:24:53.823" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8178" PostTypeId="2" ParentId="8176" CreationDate="2015-01-26T08:34:48.253" Score="4" Body="&lt;p&gt;is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Passion&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For each line take the that manieth letter of the word, and Passion is an anagram of those letters&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-01-26T08:34:48.253" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8179" PostTypeId="2" ParentId="8176" CreationDate="2015-01-26T08:34:54.600" Score="8" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Passion&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Take the letter with position of the leading number. The letters are a, o, i, p, s, s, n. The only one word anagram this gives is passion, which is also a type of fruit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastActivityDate="2015-01-26T08:34:54.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8180" PostTypeId="2" ParentId="8170" CreationDate="2015-01-26T09:34:30.467" Score="11" Body="&lt;h1&gt;Puzzles in general&lt;/h1&gt;&#xA;&lt;p&gt;It is often close to &lt;em&gt;impossible&lt;/em&gt; to make a brainteaser-logic puzzle &lt;em&gt;bullet-proof&lt;/em&gt; without ruining it. A certain &lt;em&gt;amount&lt;/em&gt; of loop-holes is, what keeps the puzzles interesting after all.&lt;/p&gt;&#xA;&lt;p&gt;The real question then is, how to &lt;em&gt;deal&lt;/em&gt; with loop-holes and brain-teaser puzzles as a puzzle author. I would recommend the following guidelines:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;State that you do not seek for some extra-clever loop-hole answer&lt;/strong&gt;, but for an answer within the framework of the puzzle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Depending on the type of puzzle, this is more or less difficult, but such a statement&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Indicates you're &lt;em&gt;serious&lt;/em&gt; about this puzzle.&lt;/li&gt;&#xA;&lt;li&gt;Gives you a handhold against wild-guesses.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Change your viewpoint&lt;/strong&gt;. Try to &lt;em&gt;solve&lt;/em&gt; your puzzle with loop-holes. If one is found, decided if it is a 'reasonable' or a very 'obscure' one, and safe-guard accordingly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If there are loopholes just because some statements &lt;em&gt;can be misinterpreted&lt;/em&gt; or are &lt;em&gt;missing&lt;/em&gt;, then fix the text accordingly.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt; A recent example is &lt;a href=&quot;https://puzzling.stackexchange.com/questions/8127/finding-doctor-no&quot;&gt;this puzzle&lt;/a&gt;. It omits an important statement, that questions are &lt;em&gt;truthfully&lt;/em&gt; answered, and it does not clearly define what is meant by &lt;em&gt;being known&lt;/em&gt;. This opens up to loop-hole answers, and could very easily be avoided.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Counter-Example:&lt;/strong&gt; In &lt;a href=&quot;https://puzzling.stackexchange.com/questions/8127/finding-doctor-no&quot;&gt;the same puzzle&lt;/a&gt;, one could creatively construct a rather obscure loop-hole that Dr. No is the person watching the party from the security cameras only. This is clearly &lt;em&gt;outside&lt;/em&gt; the original puzzle-scope and it is not worthwhile to safeguard against such answers. (Except by the general statement.)&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Double or Triple-check &lt;em&gt;your&lt;/em&gt; logic.&lt;/strong&gt; Is it really 'tight' ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;It can be helpful to ask the puzzle to another person with and without giving your intended answer. A second opinion can be very helpful.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;Puzzles on PuzzlingSE&lt;/h1&gt;&#xA;&lt;p&gt;The three points above are generally helpful, but on the other hand, PuzzlingSE is (also) &lt;em&gt;about&lt;/em&gt; learning if your puzzle works or not. So maybe the real question is less on &lt;em&gt;how to avoid&lt;/em&gt; the situation than on &lt;em&gt;how to deal with it.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;React &lt;em&gt;positively&lt;/em&gt; on comments and loop-hole answers.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Somebody spotted a weak point in your puzzle - that's great news! It allow making it better &lt;em&gt;by fixing it&lt;/em&gt;! Take the comment/solution seriously. If it is just a &lt;em&gt;joke&lt;/em&gt; or very &lt;em&gt;obscure&lt;/em&gt; - tell the poster and otherwise ignore. If it is, however, justifiable, then follow the next advise:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Edit your question to improve.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Whenever you think that you've spotted a better way to present your puzzle - make that edit! PuzzlingSE is &lt;em&gt;not&lt;/em&gt; 'fun for the moment', but it's ultimate goal is to be useful (and searchable) for future readers. Even if you've already got an accepted answer, if you find a way to make your original post better then &lt;strong&gt;do it&lt;/strong&gt;.&#xA;If this edit changes the puzzle drastically or makes some of the existing answers look strange, make the edit &lt;em&gt;visible&lt;/em&gt; as an edit and/or add an according comment to the existing answers.&lt;/p&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-01-26T09:34:30.467" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8181" PostTypeId="1" AcceptedAnswerId="8184" CreationDate="2015-01-26T09:41:46.570" Score="11" ViewCount="1130" Body="&lt;p&gt;Consider the sixteen cells in a $4\times4$ square. Two (distinct) cells are neighbors, if they share a horizontal or vertical edge; note that every cell has two or three or four neighbors.&#xA;Now each of the sixteen cells contains an integer, so that for every cell the integers in the neighboring cells add up to the sum $120$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) What are the possible values for the sum of the sixteen integers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Does there exist such a $4\times4$ square in which the sixteen integers are pairwise distinct?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-01-26T09:58:05.600" LastActivityDate="2015-01-26T11:55:43.960" Title="Four-by-four square" Tags="|mathematics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8182" PostTypeId="2" ParentId="8181" CreationDate="2015-01-26T10:34:49.897" Score="8" Body="&lt;p&gt;The solution to part (a):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The integers must sum to 720&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let us number the cells like in a matrix. Consider the neighbours of the cells $(0, 0), (0, 1), (1, 3), (2, 3), (3, 1)$ and $(3, 0)$. The sets of neighbours for each of these cells are pairwise disjoint, and their union covers all the cells. The sum of the neighbouring cells for a particular cell is $120$, and there are 6 cells above, therefore the sum is $6 \times 120 = 720$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The solution to part (b):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Consider cells $(0, 1)$ and $(1, 0)$. They have common neighbours $(0, 0)$ and $(1, 1)$. Cell $(0, 1)$ has a third neighbour $(0, 2)$, and cell $(1, 0)$ has a third neighbour $(2, 0)$. For the sums of the neighbours of each of the two cells to equal $120$ (and therefore equal each other), cells $(0, 2)$ and $(2, 0)$ must contain the same integer (as the other neighbours are common), so therefore there does not exist a $4 \times 4$ square such that the sixteen integers are pairwise distinct.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8949" LastActivityDate="2015-01-26T10:34:49.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8183" PostTypeId="2" ParentId="8162" CreationDate="2015-01-26T10:52:17.287" Score="28" Body="&lt;p&gt;Assuming Gerhard's edit reflects what the OP &lt;em&gt;meant&lt;/em&gt; to ask, then yes. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Take 10 cards from the deck. You now have a stack of 10 cards with n face down cards and a stack of 90 cards with 10-n face down cards. Now turn the stack of 10 over, to give you 10-n face down cards. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3987" LastActivityDate="2015-01-26T10:52:17.287" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8184" PostTypeId="2" ParentId="8181" CreationDate="2015-01-26T10:57:23.080" Score="12" Body="&lt;p&gt;Coloring the 16 cells like a checkerboard, it follows that the cells of one color can be represented as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FhMng.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With $a+b=120$. This guarantees that the neighborhood of each grey cell sums to 120. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As multiple cells share the same value, it is not possible to have all cell values pairwise distinct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum total for the 8 cells of given color equals $3a+3b=360$, so that all cells together add up to 720. &lt;/p&gt;&#xA;" OwnerUserId="7573" LastEditorUserId="7573" LastEditDate="2015-01-26T11:55:43.960" LastActivityDate="2015-01-26T11:55:43.960" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8185" PostTypeId="2" ParentId="7979" CreationDate="2015-01-26T11:19:28.020" Score="4" Body="&lt;p&gt;Wait 22.5 days, one minute and twenty seconds. By this time, the pirates are on the opposite point on the globe (assuming it's Earth) and they have stopped, because there is no way they could go away from you. Then choose a random direction. As the pirates always go away from you, they choose the same direction as you have. After 454 days, you will meet and the booty is yours!&lt;/p&gt;&#xA;" OwnerUserId="4520" LastActivityDate="2015-01-26T11:19:28.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8186" PostTypeId="1" AcceptedAnswerId="8191" CreationDate="2015-01-26T11:31:04.347" Score="9" ViewCount="4806" Body="&lt;p&gt;Question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eight people are sitting in two parallel rows containing four people each, in such a way that there is a equal distance between and adjacent persons. In row 1 , A,B,C,D are seated and all of them are facing south . In row 2 , E,F,G,H are seated and all of them are facing North. Therefore , in the given sitting arrangement each member seated in a row faces another member of other row. Each one them also have different profession, viz. Engineer, Doctor, Teacher, Pilot, Businessman , Banker , Lawyer and Builder but not necessarily in the same order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B sits in second to the right of the builder. G faces the immediate neighbor of the Builder. Only one person sits between G and the pilot. Two people  sit between H and the Lawyer. D faces the immediate neighbor of pilot.Only one person sits between teacher and C . Teacher does not sit at any of the extreme ends. Immediate neighbor of D faces the doctor, E faces engineer, F is not a banker. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Who among the following sit in the extreme ends of the rows ? &lt;br /&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;a) B and pilot,&lt;br /&gt;&#xA;b) C and builder &lt;br /&gt;&#xA;c) The Builder and G &lt;br /&gt;&#xA;d) G and the engineer &lt;br /&gt;&#xA;e) A,G &lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Who among the following is the banker ?  &lt;br /&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;a) A &lt;br /&gt;&#xA;b) B &lt;br /&gt;&#xA;c) C &lt;br /&gt;&#xA;d) E &lt;br /&gt;&#xA;e) G &lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;Who is/are seated between lawyer and the doctor ? &lt;br /&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;a) Pilot only &lt;br /&gt;&#xA;b) G and the businessman &lt;br /&gt; &#xA;c) F,E &lt;br /&gt;&#xA;d) businessman and the banker &lt;br /&gt; &#xA;e) No one &lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;Who among the following faces the businessman ? &lt;br /&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;a) A &lt;br /&gt;&#xA;b) Doctor &lt;br /&gt;&#xA;c) F &lt;br /&gt;&#xA;d) Teacher &lt;br /&gt;&#xA;e) Builder &lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;Which of the following is true regarding G ? &lt;br /&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;a) D faces G &lt;br /&gt;&#xA;b) The pilot is an immediate neighbor of G &lt;br /&gt;&#xA;c) G faces the person who is second left to the Banker &lt;br /&gt;&#xA;d) G is the doctor&lt;br /&gt;&#xA;e) G sits at one of the extreme ends of the line. &lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;End of question&lt;/em&gt;  &lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got this arrangement (along with many others while trying to solve them) but I always get stuck with contradictory situations. (and I know it must be wrong). &lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(B),(),(D),(C) &lt;br /&gt;&#xA;(Lawyer),(Builder, G), (Doctor),(Pilot E)&#xA;&lt;br /&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am practicing logic-puzzle questions and of many types , I can't solve this type of problem when ever I come across them. I know once if I could solve even one of them I would learn to solve the others. (I know stack-exchange communities are normally very strict regarding the type of questions posted, hope this fits into it. I tried a lot of times but I can't solve this one and need help and also please explain how did you got the answer)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any help. &lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answers:&lt;br /&gt;&#xA;1) a) &lt;br /&gt;&#xA;2) c) &lt;br /&gt;&#xA;3) e) &lt;br /&gt;&#xA;4) e) &lt;br /&gt;&#xA;5) d) &lt;br /&gt;&lt;/p&gt;&#xA;" OwnerUserId="9164" LastEditorUserId="9164" LastEditDate="2015-01-26T15:36:47.067" LastActivityDate="2015-01-27T03:12:39.797" Title="What is the sitting arrangement ?(double row question)" Tags="|logical-deduction|" AnswerCount="4" CommentCount="19" ContentLicense="CC BY-SA 3.0" />
  <row Id="8187" PostTypeId="2" ParentId="8144" CreationDate="2015-01-26T11:41:31.337" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 65+2-5 = 5-2+59&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Technically, it is not a &lt;em&gt;different&lt;/em&gt; equality.  But I believe it is in the spirit of the puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, just write&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 65+2-5 = 55+5+2&lt;br&gt;&#xA; 2+5+55 = 5-2+59&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 25-19 = 6&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2015-01-26T16:53:41.463" LastActivityDate="2015-01-26T16:53:41.463" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8188" PostTypeId="2" ParentId="8162" CreationDate="2015-01-26T11:43:07.770" Score="0" Body="&lt;p&gt;First he builds two piles by dealing one card at a time.&#xA;(0,0), (1,0), (1,1), (2,1)...(50,50).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After this, he puts one pile onto the other (without shuffling) and repeats the process, but he is then dealing a different number of cards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is to follow 1*2*2*5*5 = 100. In other words:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First round, 1 card per pile.&lt;br&gt;&#xA;Second round, 2 cards per pile.&lt;br&gt;&#xA;...&lt;br&gt;&#xA;Last round, 5 cards per pile. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;I tried it and it worked. Maybe I was lucky since I have no idea why it worked (if it does).&lt;/p&gt;&#xA;" OwnerUserId="5106" LastEditorUserId="5106" LastEditDate="2015-01-28T09:28:29.980" LastActivityDate="2015-01-28T09:28:29.980" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="8189" PostTypeId="1" AcceptedAnswerId="8194" CreationDate="2015-01-26T12:27:30.260" Score="4" ViewCount="1371" Body="&lt;p&gt;Alice and Bob started working part-time parking lot attendants. Bob was managing a car when Alice arrived late on the first day at work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A: I'm sorry for being late. How many cars have come in?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B: I will make a quiz on it. Why don't you guess the number of cars?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A: You are giving me a problem to solve?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B: Right. Every car has its plate number and the numbers have special relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A: So, do you want me to figure out the number of cars from that relation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B: Yes, conditions are as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Condition 1:&lt;/strong&gt; Any two cars must share only one digit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Condition 2:&lt;/strong&gt; There must be exactly one more car with the equal digit that two cars have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A: (after thinking) Three cars?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B: Why do you think the answer is three cars?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A: For example, if there are three cars with the plate numbers 1, 12, and 13, any two cars have 1, and there are exactly three cars which has 1 in the plate number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B: Oh, there is one more condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Condition 3:&lt;/strong&gt; There are more than three cars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many cars were in the parking lot when Alice arrived?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added remark:&lt;/strong&gt; It's not very hard to construct examples for 7 cars. Could you show the reason why other number cannot be an answer?&lt;/p&gt;&#xA;" OwnerUserId="3914" LastEditorUserId="3914" LastEditDate="2015-01-27T05:19:54.827" LastActivityDate="2015-01-27T17:04:45.313" Title="At the Parking Lot" Tags="|logical-deduction|mathematics|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8190" PostTypeId="2" ParentId="6453" CreationDate="2015-01-26T12:32:51.133" Score="15" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;This is a community-wiki answer compiling the final solution of the puzzle.&lt;br&gt; &#xA;  It also presents the solution &lt;em&gt;as intended&lt;/em&gt;.&lt;br&gt;&#xA;  The solutions were found by individual answer posters.&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the impatient, the &lt;strong&gt;final&lt;/strong&gt; answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&quot;Balance is the rectifier&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Individual contributions were:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/users/8775/gary-ye&quot;&gt;Gary Ye&lt;/a&gt;:&lt;br&gt;&#xA;   - First discovery of mechanism of part 1 (&lt;em&gt;The Bottles&lt;/em&gt;)&lt;br&gt;&#xA;  &lt;br&gt;&#xA;   &lt;a href=&quot;https://puzzling.stackexchange.com/users/8804/i-turo&quot;&gt;i_turo&lt;/a&gt;:&lt;br&gt;&#xA;  - The full solution to part 1 (&lt;em&gt;The Bottles&lt;/em&gt;)&lt;br&gt;&#xA;  &lt;br&gt;&#xA;  &lt;a href=&quot;https://puzzling.stackexchange.com/users/8858/efrog&quot;&gt;EFrog&lt;/a&gt;:&lt;br&gt;&#xA;   - Final sorting detail of part 1 (&lt;em&gt;The Bottles&lt;/em&gt;)&lt;br&gt;&#xA;   - Valuable ideas on how part 1 and part 2 are related.&lt;br&gt;&#xA;  &lt;br&gt;&#xA;  &lt;a href=&quot;https://puzzling.stackexchange.com/users/9319/bulldogg6404&quot;&gt;Bulldogg6404&lt;/a&gt;:&lt;br&gt;&#xA;   - The mechanism of part 2 (&lt;em&gt;The Scales&lt;/em&gt;) &lt;br&gt;&#xA;  &lt;br&gt;&#xA;  &lt;a href=&quot;https://puzzling.stackexchange.com/users/9226/nephtyz&quot;&gt;Nephtyz&lt;/a&gt;:&lt;br&gt;&#xA;  - The first solution to part 2 (&lt;em&gt;The Scales&lt;/em&gt;)&lt;br&gt;&#xA;  &lt;br&gt;&#xA;  &lt;a href=&quot;https://puzzling.stackexchange.com/users/9319/bulldogg6404&quot;&gt;Bulldogg6404&lt;/a&gt;:&lt;br&gt;&#xA;   - Missing colours on the scales. (&lt;em&gt;The Code&lt;/em&gt;)&lt;br&gt;&#xA;   - The mechanism and solution of part 3 (&lt;em&gt;The Code&lt;/em&gt;) &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;Detailed Solution&lt;/h1&gt;&#xA;&#xA;&lt;h2&gt;Instructions&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;All images:&lt;/strong&gt; The blue-to-grey gradient everywhere represents:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;order&lt;/em&gt;. Otherwise identical bottles (they are all blue) have to be &lt;em&gt;sorted&lt;/em&gt; in some way and then be used in this order.&lt;br&gt; Also: The final pass-phrase is gotten letter by letter &lt;em&gt;in correct order&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Instruction 1&lt;/strong&gt; (left part)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The original grid of bottles is not in order. The labels and the yellow code-grid-table need to be used to achieve sorting. For the meaning of the code-grid-table see &lt;em&gt;'The Bottles'&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Instruction 1&lt;/strong&gt; (right part)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bottles are labelled by dots, indicating the amount of liquid they contain. Filling a bottle into a tube will fill the equal amount of square grids, starting from the further most grid. Other details to be seen: The liquid does &lt;em&gt;not&lt;/em&gt; level out across neighbouring squares. The spacing or alignment of 'dots' on the labels is of no consequence. (The shown bottle has 5 dots in an arrangement none of the bottle labels has.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Instruction 2&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The 24 bottles are needed in &lt;em&gt;'The scales'&lt;/em&gt;. They are used in the order determined by &lt;em&gt;'The Bottles'&lt;/em&gt;. They need to be filled into a single tube one after the other, while the hanging weight is moved at each step to keep the scales in balance.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Instruction 3&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;One&lt;/em&gt; key is the &lt;em&gt;final puzzle piece&lt;/em&gt;. It is needed to &lt;em&gt;complete&lt;/em&gt; the second part (&lt;em&gt;The Scales&lt;/em&gt;). This &lt;em&gt;single&lt;/em&gt; key needs to be inserted into the code-table. The pass-phrase is found letter-by-letter from the grid. There are as many letters as there are bottles (24), and their order is the same as determined by the first part (&lt;em&gt;The Bottles&lt;/em&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;The Bottles (solved)&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The aim of the &lt;em&gt;bottle&lt;/em&gt; puzzle is to bring the 24 labelled bottles into correct order.&#xA; &lt;br&gt;&#xA; The brown bottle labels are a transparent grid. If all are overlaid, only a single square remains fully transparent. &#xA; &lt;br&gt;&#xA; It is only possible to remove a &lt;em&gt;specific&lt;/em&gt; label from the stack in order to get a second transparent square. (Removing any other label doesn't change the situation.)&#xA; &lt;br&gt;&#xA; Once that label is removed, there is only a single &lt;em&gt;specific&lt;/em&gt; label which can be removed in order to get a third transparent square. etc. etc.&#xA; &lt;br&gt;&#xA; This gives an order of labels. However, it is ambiguous as it is not clear if the order is first-to-last, or last-to-first.&#xA; &lt;br&gt;&#xA; Checking the position of the &quot;first&quot; and the &quot;last&quot; transparent grid-square on the yellow code-table of hint reveals: &lt;em&gt;Alpha&lt;/em&gt; and &lt;em&gt;Omega&lt;/em&gt;.&#xA; &lt;br&gt;&#xA; &lt;em&gt;Alpha&lt;/em&gt; and &lt;em&gt;Omega&lt;/em&gt; are the first and last characters of the ancient Greek alphabet. They are often used (f.e. in religion) to indicate &lt;em&gt;Beginning&lt;/em&gt; and &lt;em&gt;End&lt;/em&gt;.&#xA; &lt;br&gt;&#xA; Therefore the correct order of bottles is with the first label removed showing 2 fields and the next showing 3 fields etc. &#xA; As all bottles are equal except the grid and dot labels, and the grid has been 'used up' by the puzzle now. The 'dots' are all that remain. &#xA; 'Dot-labels' sorted by the order determined above are:&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/peZks.png&quot; alt=&quot;Dot-Labels in correct order&quot;&gt;&lt;br&gt;&lt;br&gt;&#xA; Spaces in the dot-labels are decoys and can be ignored. &lt;br&gt;&#xA; The &lt;em&gt;number&lt;/em&gt; of dots in sequence (ignoring any spaces) are: &lt;br&gt;&#xA; &lt;code&gt;5; 3; 7; 2; 4; 3; 4; 4; 5; 2; 3; 1; 6; 2; 3; 4; 6; 5; 6; 2; 1; 4; 4; 7&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;The Scales (solved)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The balance of the scale is determined by &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the &lt;strong&gt;total weight of filled pipes and the hanging weights&lt;/strong&gt;.&lt;br&gt; Each 'filled square' counts for the same weight, and the tipping moment is determined by the (horizontal) distance from the centre point multiplied by the (summed) weight at this distance.&lt;br&gt;An example (tipping to the right by &quot;+2&quot;):&lt;br&gt;&lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/2epZr.png&quot; alt=&quot;Scale-balance-example&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Following &lt;em&gt;Instructions 1&lt;/em&gt;, it is the aim of the puzzle to ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...fill in the bottles one after the other into either of the 6 tubes, while keeping the system in balance with the &lt;em&gt;hanging weight&lt;/em&gt;. The exact sequence has to be found by the puzzle! However, the order of the bottles is given (by the &lt;em&gt;The Bottles&lt;/em&gt; puzzle) and there is only a very limited number of possibilities which can keep the system in balance at each step. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Following all of the above, the puzzle solves into the following sequence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Read the image left to right and top to bottom for the sequence of 24 bottles subsequently filling the pipes:&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/8YCVD.png&quot; alt=&quot;24-Solution-Steps&quot;&gt;&lt;br&gt;&#xA; &lt;br&gt;&#xA; A second solution is possible, which has bottles 11 and 15 swapped. However, only one of the two solutions (the one above) will produced an English pass-phrase in the third part of the puzzle (&lt;em&gt;The Code&lt;/em&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;The Code (solved)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The solution of part 1 and part 2 of the puzzle give one&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 24 &lt;em&gt;ordered&lt;/em&gt; actions of pouring bottles into the grid. Each pouring-actions is defined by 2 things: A &lt;strong&gt;tube colour&lt;/strong&gt; into which the liquid is filled &lt;strong&gt;and&lt;/strong&gt; a &lt;strong&gt;hanging position&lt;/strong&gt; for the balancing weight. This position is associated with a &lt;em&gt;colour&lt;/em&gt;, but some positions are missing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The missing (b&amp;amp;w checkboard pattern) fields on the scale can be found by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; looking at colours of all keys. All 6 keys show the same colour 'pairs' albeit in different vertical ordering. The according colour pairs appear in symmetric position on the scale as well, allowing to fill in the spots as with the following image:&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/WU9vc.png&quot; alt=&quot;Scales with colours completed&quot;&gt;&lt;br&gt;&#xA; Also: Colour-pairs (if watches as RGB hex-codes) add up to FFFFFF, i.e. are complementary.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Labelling the 26 weight-hanging positions from the left to the right, one gets the solution as a table:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/b6sDg.png&quot; alt=&quot;NumberedScale with filled colors&quot;&gt;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/JjeCh.png&quot; alt=&quot;solution positions&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To find the appropriate &lt;em&gt;key&lt;/em&gt; for &lt;em&gt;The Code&lt;/em&gt; one has to...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...realized that the final situation after pouring all bottles left exactly two red squares empty. On the bottle, the squares are represented by yellow diamonds. The same dots appear on the keys, so the two yellow diamonds on the &lt;strong&gt;red key&lt;/strong&gt; will 'complete' the puzzle as hinted at by 'Instruction 3'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With this key inserted into the code-table...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...one can now translate the 24 pairs of 'pipe colour' and 'weight position' into letters by looking them up in the grid:&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/5wTrq.png&quot; alt=&quot;RedKeyInSolution&quot;&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This eventually leads to the 24-characters of the pass-phrase:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;BALANCE/IS/THE/RECTIFIER&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-02-20T17:49:16.557" CommentCount="0" CommunityOwnedDate="2015-01-26T12:32:51.133" ContentLicense="CC BY-SA 3.0" />
  <row Id="8191" PostTypeId="2" ParentId="8186" CreationDate="2015-01-26T13:33:54.560" Score="5" Body="&lt;p&gt;New Edit, now that question has been corrected. Apologies if this is over-long, but you did ask for explanation at how to arrive at the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll work through it long-hand, because the worse case is only 3 possibility trees, and they soon collapse under contradiction. So here goes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;B sits in second to the right of the builder&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This gives two possible options.&#xA;Branch 1:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X   X   Bui X&#xA;B   X   X   X&#xA;&#xA;X   X   X   X&#xA;X   X   X   X&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Or Branch 2:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X   X   X   Bui&#xA;X   B   X   X&#xA;&#xA;X   X   X   X&#xA;X   X   X   X&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;G faces the immediate neighbor of the Builder&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Taking branch 1 as the start point, we end up with two possibilities.&#xA;Branch 1.1:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X   X   Bui X&#xA;B   X   X   X&#xA;&#xA;X   X   X   X&#xA;X   G   X   X&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Branch 1.2:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X   X   Bui X&#xA;B   X   X   X&#xA;&#xA;X   X   X   X&#xA;X   X   X   G&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But from branch 2 there is only one possibility:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Branch 2:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X   X   X   Bui&#xA;X   B   X   X&#xA;&#xA;X   X   X   X&#xA;X   X   G   X&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Only one person sits between G and the pilot&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For each of our 3 possible branches there is just one solution that matches this, so our branches now look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Branch 1.1:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X   X   Bui X&#xA;B   X   X   X&#xA;&#xA;X   X   X   Pil&#xA;X   G   X   X&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Branch 1.2:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X   X   Bui X&#xA;B   X   X   X&#xA;&#xA;X   Pil X   X&#xA;X   X   X   G&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Branch 2:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X   X   X   Bui&#xA;X   B   X   X&#xA;&#xA;Pil X   X   X&#xA;X   X   G   X&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Two people sit between H and the Lawyer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again each of our branches only has one option that matches this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Branch 1.1:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X   X   Bui X&#xA;B   X   X   X&#xA;&#xA;Law X   X   Pil&#xA;X   G   X   H&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Branch 1.2:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X   X   Bui X&#xA;B   X   X   X&#xA;&#xA;X   Pil X   Law&#xA;H   X   X   G&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Branch 2:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X   X   X   Bui&#xA;X   B   X   X&#xA;&#xA;Pil X   X   Law&#xA;H   X   G   X&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;D faces the immediate neighbor of pilot&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is good, as there is no way to satisfy branch 2 with this condition. There is just one possible solution for each of our other branches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Branch 1.1:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X   X   Bui X&#xA;B   X   D   X&#xA;&#xA;Law X   X   Pil&#xA;X   G   X   H&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Branch 1.2:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X   X   Bui X&#xA;B   X   D   X&#xA;&#xA;X   Pil X   Law&#xA;H   X   X   G&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Only one person sits between teacher and C . Teacher does not sit at&#xA;  any of the extreme ends&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Taking both of these together saves a step, because it reduces the possible seats of the teacher to the middle two seats. For both of our branches there is only one possible solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Branch 1.1:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X   Tea Bui X&#xA;B   X   D   C&#xA;&#xA;Law X   X   Pil&#xA;X   G   X   H&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Branch 1.2:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X   Tea Bui X&#xA;B   X   D   C&#xA;&#xA;X   Pil X   Law&#xA;H   X   X   G&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Immediate neighbor of D faces the doctor&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This condition gets rid of branch 1.2, because we know the neighbour of D in that branch must face the pilot or the lawyer. Branch 1.1 has only one possible solution:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X   Tea Bui X&#xA;B   X   D   C&#xA;&#xA;Law Dr  X   Pil&#xA;X   G   X   H&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;E faces engineer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;E must be on the bottom row, and the only unknown letter that faces an unknown occupation gives a single solution:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Eng Tea Bui X&#xA;B   X   D   C&#xA;&#xA;Law Dr  X   Pil&#xA;E   G   X   H&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;F is not a banker&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This condition allows us to finish the puzzle. F is on the bottom row, and the only remaining unknown occupation is the business man. This allows us to fill in the remaining X's.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Eng Tea Bui Ban&#xA;B   A   D   C&#xA;&#xA;Law Dr  Bus Pil&#xA;E   G   F   H&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which allows us to answer the questions as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) A&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) E&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) E&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) D&lt;/p&gt;&#xA;" OwnerUserId="9168" LastEditorUserId="9168" LastEditDate="2015-01-26T21:32:48.260" LastActivityDate="2015-01-26T21:32:48.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8192" PostTypeId="2" ParentId="8186" CreationDate="2015-01-26T13:47:26.573" Score="4" Body="&lt;p&gt;I apologize for this answer being so long. I tried to note all of my logic as I worked through the puzzle myself, only to find that (I think) it is impossible as typed. I believe there is a typo in one of the clues. (&quot;Only two person sits between...&quot; I believe should be &quot;Only one person sits between...&quot;) For the sake of logic-checking, I maintained my first work with the typo, then put all of the work for the fixed version &lt;strong&gt;after the line break&lt;/strong&gt;. My answers to the questions are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A, C, E, ??, D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;strong&gt;A - H&lt;/strong&gt; will be used as above. Occupations will be abbreviated as &lt;strong&gt;EN&lt;/strong&gt;, &lt;strong&gt;DO&lt;/strong&gt;, &lt;strong&gt;TE&lt;/strong&gt;, &lt;strong&gt;PI&lt;/strong&gt;, &lt;strong&gt;BS&lt;/strong&gt;, &lt;strong&gt;BA&lt;/strong&gt;, &lt;strong&gt;LA&lt;/strong&gt; and &lt;strong&gt;BL&lt;/strong&gt;. A single seat is represented first by the letter, then by the occupation, or &lt;strong&gt;-/--&lt;/strong&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;B sits in second to the right of the builder&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that there is exactly one seat between B and the builder, and that B will be to the builder's right-hand side, which changes their orientation depending on the row in which they sit. This gives us four possible positions. I think rotation is arbitrary, however, so that gives us two possibilities. (Incidentally, this also means that B is not the Builder.) &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;-/--, -/--, -/--, -/--&#xA;-/BL, -/--, B/--, -/--&#xA;&#xA;-/--, -/--, -/--, -/--&#xA;-/--, -/BL, -/--, B/--&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;G faces the immediate neighbor of the Builder.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that Builder's neighbor, &lt;em&gt;x&lt;/em&gt;, sits next to Builder, and across from G. This positions G diagonal from Builder. There are only three possible positions for G. (This also means G isn't Builder.)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;-/--, G/--, -/--, -/--&#xA;-/BL, -/--, B/--, -/--&#xA;&#xA;G/--, -/--, -/--, -/--&#xA;-/--, -/BL, -/--, B/--&#xA;&#xA;-/--, -/--, G/--, -/--&#xA;-/--, -/BL, -/--, B/--&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Only two person sits between G and the pilot.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I question whether this one should say &quot;only &lt;em&gt;one&lt;/em&gt; person sits between...&quot;&lt;/strong&gt; If so, that changes everything from here on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, this means there are two empty seats between G and Pilot. There is only one situation in which this is possible, our second layout above. (This also eliminates G as Pilot, and B as Pilot, since there is no scenario in which G and B are two seats apart.)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;G/--, -/--, -/--, -/PI&#xA;-/--, -/BL, -/--, B/--&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Two people sit between H and the Lawyer.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This tells us that H is on one end of a row while Lawyer is on the other. Unfortunately, there are many scenarios in which this can be true. So I'm going to ignore it for now and see if it's more significant on my second loop through the clues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;D faces the immediate neighbor of pilot.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know where Pilot is sitting, and he only has one neighbor. So we can easily conclude that D sits next to B in our layout.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;G/--, -/--, -/--, -/PI&#xA;-/--, -/BL, D/--, B/--&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Only one person sits between teacher and C.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives us a lot of options. If you combine it with the next clue, however, it narrows it down quite a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Teacher does not sit at any of the extreme ends.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means Teacher must sit in one of the four middle seats. But since the previous clue says C is one spot away from Teacher, there are two places where Teacher can't be sitting because B and G are already occupying the place where C would have to sit in those cases. So now we have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;G/--, -/TE, -/--, C/PI&#xA;-/--, -/BL, D/--, B/--&#xA;&#xA;G/--, -/--, -/--, -/PI&#xA;C/--, -/BL, D/TE, B/--&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Immediate neighbor of D faces the doctor.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D's immediate neighbors are B and Builder. Since we know that B is already facing the Pilot, that means Builder must be facing the Doctor. This eliminates our first layout above, since Teacher is positioned across from Builder there.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;G/--, -/DO, -/--, -/PI&#xA;C/--, -/BL, D/TE, B/--&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;E faces engineer, F is not a banker.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These two clues don't seem to help much right now, since we haven't dealt with E, F, Engineer or Banker thus far. I'd like to revisit the clue I skipped earlier, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Two people sit between H and the Lawyer.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This clue means that H and Lawyer are at opposite ends of the same row. There is only one possible place where H could be positioned at the end of a row now.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;G/LA, -/DO, -/--, H/PI&#xA;C/--, -/BL, D/TE, B/--&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Aaaaand this makes &lt;strong&gt;E faces Engineer&lt;/strong&gt; impossible. So I'm guessing that the third clue was probably typed wrong. Or maybe I messed up somewhere. Let me back up and try it the other way...&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;B sits in second to the right of the builder&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that there is exactly one seat between B and the builder, and that B will be to the builder's right-hand side, which changes their orientation depending on the row in which they sit. This gives us four possible positions. I think rotation is arbitrary, however, so that gives us two possibilities. (Incidentally, this also means that B is not the Builder.) &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;-/--, -/--, -/--, -/--&#xA;-/BL, -/--, B/--, -/--&#xA;&#xA;-/--, -/--, -/--, -/--&#xA;-/--, -/BL, -/--, B/--&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;G faces the immediate neighbor of the Builder.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that Builder's neighbor, &lt;em&gt;x&lt;/em&gt;, sits next to Builder, and across from G. This positions G diagonal from Builder. There are only three possible positions for G. (This also means G isn't Builder.)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;-/--, G/--, -/--, -/--&#xA;-/BL, -/--, B/--, -/--&#xA;&#xA;G/--, -/--, -/--, -/--&#xA;-/--, -/BL, -/--, B/--&#xA;&#xA;-/--, -/--, G/--, -/--&#xA;-/--, -/BL, -/--, B/--&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Only ONE person sits between G and the pilot.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives Pilot three possible positions:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;-/--, G/--, -/--, -/PI&#xA;-/BL, -/--, B/--, -/--&#xA;&#xA;G/--, -/--, -/PI, -/--&#xA;-/--, -/BL, -/--, B/--&#xA;&#xA;-/PI, -/--, G/--, -/--&#xA;-/--, -/BL, -/--, B/--&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;D faces the immediate neighbor of pilot.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This positions D diagonally from Pilot, which puts him in only two possible positions:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;G/--, -/--, -/PI, -/--&#xA;-/--, D/BL, -/--, B/--&#xA;&#xA;-/PI, -/--, G/--, -/--&#xA;-/--, D/BL, -/--, B/--&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Only one person sits between teacher and C. Teacher does not sit at any of the extreme ends.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This puts us at five possibilities. Not fun.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;G/--, -/TE, -/PI, C/--&#xA;-/--, D/BL, -/--, B/--&#xA;&#xA;G/--, -/--, -/PI, -/--&#xA;C/--, D/BL, -/TE, B/--&#xA;&#xA;C/PI, -/--, G/TE, -/--&#xA;-/--, D/BL, -/--, B/--&#xA;&#xA;-/PI, -/TE, G/--, C/--&#xA;-/--, D/BL, -/--, B/--&#xA;&#xA;-/PI, -/--, G/--, -/--&#xA;C/--, D/BL, -/TE, B/--&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Immediate neighbor of D faces the doctor.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This at least eliminates the middle option, since both diagonals from D already have an occupation.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;G/--, -/TE, -/PI, C/--&#xA;-/--, D/BL, -/--, B/--&#xA;&#xA;G/--, -/--, -/PI, -/--&#xA;C/--, D/BL, -/TE, B/--&#xA;&#xA;-/PI, -/TE, G/--, C/--&#xA;-/--, D/BL, -/--, B/--&#xA;&#xA;-/PI, -/--, G/--, -/--&#xA;C/--, D/BL, -/TE, B/--&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Two people sit between H and the Lawyer.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many possibilities for this, but I'm hoping other clues will narrow them down...&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;G/--, -/TE, -/PI, C/--&#xA;H/--, D/BL, -/--, B/LA&#xA;&#xA;G/LA, -/--, -/PI, H/--&#xA;C/--, D/BL, -/TE, B/--&#xA;&#xA;H/PI, -/TE, G/--, C/LA&#xA;-/--, D/BL, -/--, B/--&#xA;&#xA;-/PI, -/TE, G/--, C/--&#xA;H/--, D/BL, -/--, B/LA&#xA;&#xA;H/PI, -/--, G/--, -/LA&#xA;C/--, D/BL, -/TE, B/--&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;E faces engineer.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This actually does narrow it down a little:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;G/--, -/TE, E/PI, C/--&#xA;H/--, D/BL, -/EN, B/LA&#xA;&#xA;H/PI, -/TE, G/EN, C/LA&#xA;-/--, D/BL, E/--, B/--&#xA;&#xA;E/PI, -/TE, G/--, C/--&#xA;H/EN, D/BL, -/--, B/LA&#xA;&#xA;-/PI, -/TE, G/EN, C/--&#xA;H/--, D/BL, E/--, B/LA&#xA;&#xA;H/PI, -/--, G/EN, -/LA&#xA;C/--, D/BL, E/TE, B/--&#xA;&#xA;H/PI, -/--, G/--, E/LA&#xA;C/--, D/BL, -/TE, B/EN&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Immediate neighbor of D faces the doctor.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This eliminates any layouts above in which both of D's diagonals already have occupations.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;G/DO, -/TE, E/PI, C/--&#xA;H/--, D/BL, -/EN, B/LA&#xA;&#xA;E/PI, -/TE, G/DO, C/--&#xA;H/EN, D/BL, -/--, B/LA&#xA;&#xA;H/PI, -/--, G/DO, E/LA&#xA;C/--, D/BL, -/TE, B/EN&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We have no information, really, about F, A, Businessman, or Banker. But let's check out the questions and see if we have enough information to answer them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1. Who among the following sit in the extreme ends of the rows ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;a) B and pilot&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) C and builder &lt;em&gt;//This is never true in any layout.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) The Builder and G &lt;em&gt;//This is never true in any layout.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d) G and the engineer &lt;em&gt;//This is never true in any layout.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e) A,G &lt;em&gt;//This is never true in any layout.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;**This eliminates one of our layouts above...&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;E/PI, -/TE, G/DO, C/--&#xA;H/EN, D/BL, -/--, B/LA&#xA;&#xA;H/PI, -/--, G/DO, E/LA&#xA;C/--, D/BL, -/TE, B/EN&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2. Who among the following is the banker?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) A &lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) B &lt;em&gt;//This is never true in any layout.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) C &lt;/p&gt;&#xA;&#xA;&lt;p&gt;d) E &lt;em&gt;//This is never true in any layout.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e) G &lt;em&gt;//This is never true in any layout.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;**We can't answer this yet, but maybe we can later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3. Who is/are seated between lawyer and the doctor ?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Pilot only &lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) G and the businessman &lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) F,E &lt;/p&gt;&#xA;&#xA;&lt;p&gt;d) businessman and the banker &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;e) No one&lt;/strong&gt; &lt;em&gt;//This is true in both layouts.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4. Who among the following faces the businessman ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) A &lt;em&gt;//See note.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Doctor &lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) F &lt;em&gt;//See note.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d) Teacher &lt;em&gt;//See note.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e) Builder &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Businessman must occupy a space with a blank occupation. Those blanks are never situated across from A, F, or Teacher. This also means:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;E/PI, -/TE, G/DO, C/--&#xA;H/EN, D/BL, -/BS, B/LA&#xA;&#xA;H/PI, -/BS, G/DO, E/LA&#xA;C/--, D/BL, -/TE, B/EN&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;5. Which of the following is true regarding G ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) D faces G &lt;em&gt;//This is never true in any layout.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) The pilot is an immediate neighbor of G &lt;em&gt;//This is never true in any layout.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) G faces the person who is second left to the Banker &lt;em&gt;//This is never true in any layout.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;d) G is the doctor&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e) G sits at one of the extreme ends of the line. &lt;em&gt;//This is never true in any layout.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2. Who among the following is the banker ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) A &lt;em&gt;//This can't be true in either layout.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;c) C&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;E/PI, -/TE, G/DO, C/BA&#xA;H/EN, D/BL, -/BS, B/LA&#xA;&#xA;H/PI, -/BS, G/DO, E/LA&#xA;C/BA, D/BL, -/TE, B/EN&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Aaand I'm stumped for the last question.&lt;/p&gt;&#xA;" OwnerUserId="8858" LastActivityDate="2015-01-26T13:47:26.573" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8193" PostTypeId="2" ParentId="8186" CreationDate="2015-01-26T14:08:12.923" Score="1" Body="&lt;p&gt;&lt;strong&gt;Edit: OP edited a word in the question which made this answer incorrect.&lt;/strong&gt; Here is the incorrect answer anyway:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The given situation is impossible to get.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider seats(from left) in first row as 1,2,3,4 and seats(from left) in second row as 5,6,7,8.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Only two(OP changed this to one) person sits between G and the pilot. Two people sit between H and the Lawyer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This means that &lt;strong&gt;G,the lawyer&lt;/strong&gt; and &lt;strong&gt;H,the pilot&lt;/strong&gt; sit at the ends of the second row. So, let us assume that &lt;strong&gt;H sits in seat 5&lt;/strong&gt; and &lt;strong&gt;G sits in seat 8&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;D faces the immediate neighbor of pilot&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Would mean that &lt;strong&gt;D sits in seat 2&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;G faces the immediate&#xA;  neighbor of the Builder&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Would mean that &lt;strong&gt;Builder sits on seat 3&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;B sits in second to the right of the&#xA;  builder&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Would mean that &lt;strong&gt;B sits on seat 1&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Only one person sits between teacher and C . Teacher does not sit at any of the extreme ends&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Would mean that &lt;strong&gt;Teacher ,D sits on seat 2&lt;/strong&gt; and &lt;strong&gt;C is on seat 4&lt;/strong&gt;. This leaves &lt;strong&gt;A,the builder on seat 3&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So,we have&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;B,D,A,C (A is the builder and D is the teacher) &lt;br&gt;&#xA;  H,?,?,G (H is the pilot and G is the lawyer)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;E faces engineer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is not possible.&lt;/p&gt;&#xA;" OwnerUserId="7482" LastEditorUserId="7482" LastEditDate="2015-01-27T03:12:39.797" LastActivityDate="2015-01-27T03:12:39.797" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8194" PostTypeId="2" ParentId="8189" CreationDate="2015-01-26T14:22:28.750" Score="2" Body="&lt;p&gt;If any two cars share exactly 1 digit, we can get up to $7 cars$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on our restrictions, we can use at most 30 digits. A 3-digit number can match at most 6 other numbers (2 for each digit). A 4-digit number can match at most 8 other numbers, but at that point, we need more than the 30 available digits. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1__'s then 2__'s then 3__'s&lt;br&gt;&#xA;$123, 145, 167, 246, 257, 347, 356$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;or  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting from 12, 13, 14&lt;br&gt;&#xA;$125, 136, 147, 234, 267, 537, 564$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Old&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if I have it right:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Any two cars share at least 1 digit  &lt;/li&gt;&#xA;&lt;li&gt;1 and only 1 other car shares that digit&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, it looks like we need some sort of groups of 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Might be able to get away with 5 cars: 12, 13, 14, 234, 432&lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-01-27T17:04:45.313" LastActivityDate="2015-01-27T17:04:45.313" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8195" PostTypeId="1" AcceptedAnswerId="8211" CreationDate="2015-01-26T14:30:31.003" Score="63" ViewCount="4829" Body="&lt;p&gt;We all know it can be difficult to find a &lt;em&gt;needle in a haystack&lt;/em&gt;, but in this puzzle you have to&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;#Find the &lt;em&gt;NEEDLE&lt;/em&gt; in the network!&#xA;&lt;img src=&quot;https://i.stack.imgur.com/wfIMh.png&quot; alt=&quot;Needle in the network puzzle&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;A bit more detailed:&lt;/strong&gt;&lt;br&gt;&#xA;Somewhere in the network pictured above hides the word &lt;strong&gt;Needle&lt;/strong&gt; encoded in a rather well known cipher. (You will find examples of the used cipher on Puzzling SE in several other puzzles.) The task is to find it (and the used cipher). &lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Particulars:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The image alone constitutes the puzzle.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It is &lt;em&gt;visual&lt;/em&gt; and no information is hidden in its digital representation. &lt;br&gt;&#xA;(&lt;em&gt;If you want, you can draw the same image by hand.&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The solution requires a &lt;em&gt;cipher&lt;/em&gt; to be applied, i.e. the word &lt;strong&gt;&amp;quot;needle&amp;quot;&lt;/strong&gt; is &lt;em&gt;encoded&lt;/em&gt; not just visually hidden or obscured.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="69582" LastEditDate="2021-11-28T21:09:32.140" LastActivityDate="2021-11-28T21:09:32.140" Title="The needle in the network - find it!" Tags="|enigmatic-puzzle|cipher|visual|network|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8196" PostTypeId="2" ParentId="8189" CreationDate="2015-01-26T14:31:16.753" Score="2" Body="&lt;p&gt;ORIGINAL ANSWER before question was ammended:&#xA;I'm having trouble understanding the clauses ... doesn't&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Condition 1: Any two cars must have an equal digit. &#xA;  Condition 2: There must be exactly one more car with the equal digit that two cars have.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Just mean that that there must be 3 cars that share a digit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In which case the answer is 4:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;12, 13, 23, 123&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each digit is shared by two cars, plus exactly one more car.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDITED ANSWER:&#xA;7, if only a single digit can be shared between two cars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2, 3, 4, 12, 13, 14, 234&lt;/p&gt;&#xA;" OwnerUserId="9168" LastEditorUserId="9168" LastEditDate="2015-01-26T15:03:18.880" LastActivityDate="2015-01-26T15:03:18.880" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8197" PostTypeId="2" ParentId="8189" CreationDate="2015-01-26T14:38:27.090" Score="4" Body="&lt;p&gt;7 cars&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example with 7 cars:&#xA;156&#xA;267&#xA;137&#xA;124&#xA;235&#xA;346&#xA;457&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-01-26T14:38:27.090" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8199" PostTypeId="1" AcceptedAnswerId="8202" CreationDate="2015-01-26T15:31:17.547" Score="8" ViewCount="2617" Body="&lt;p&gt;You find yourself with a friend on a completely deserted (devoid of objects or people) and flat plane. The only thing of note is a wide river; you can just manage to see the other side of it. There is a straight metallic bridge to the other side. It is held only by 2 poles on each side of the river, and is built extremely high over the river. You can not climb up a pole, and walk across the bridge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a single metal pole of approximately $1$ foot length with you, and a rope of at least $10$ feet length. You, however, do not know the measurements. You have &lt;strong&gt;absolutely nothing else&lt;/strong&gt; with you that can be used (no measuring tapes, ladders, stopwatches, airplanes, kites, and so on). You can not use your clothes as a device, and you have no idea where you have been located (you may not even be on the Earth).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Find out approximately how wide the river is, in terms of the rod given to you?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer would be something like &lt;code&gt;River width = 55 rods&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="2156" LastEditDate="2015-12-22T16:14:24.057" LastActivityDate="2015-12-22T16:14:24.057" Title="Find the width of the river" Tags="|mathematics|geometry|lateral-thinking|word-problem|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8200" PostTypeId="2" ParentId="8199" CreationDate="2015-01-26T15:47:45.310" Score="12" Body="&lt;p&gt;Walk along the river until the rod, held up by my friend at a distance equal to the length of the rope, has the same apparent size as the bridge. Then measure the distance to the bridge in rope lengths. This number will equal the bridge/river width in rod lengths.&lt;/p&gt;&#xA;" OwnerUserId="3987" LastActivityDate="2015-01-26T15:47:45.310" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8201" PostTypeId="2" ParentId="8195" CreationDate="2015-01-26T16:13:15.947" Score="38" Body="&lt;p&gt;Is this possibly what you are looking for?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hSxOw.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9174" LastEditorUserId="35972" LastEditDate="2017-04-28T19:36:14.573" LastActivityDate="2017-04-28T19:36:14.573" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8202" PostTypeId="2" ParentId="8199" CreationDate="2015-01-26T16:26:53.280" Score="9" Body="&lt;p&gt;This is just an expansion of @frodoskywalker's answer with more explanation.  Please upvote his (as I did) if you agree with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So start by walking a &quot;fair&quot; distance away from the bridge along the shore - say 100 rope lengths.  Doesn't need to be accurate at this point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then arrange the rod to be perpendicular to the water (i.e. parallel with the bridge) and place the rope parallel to the water with the end closest to the bridge at the end of the rod.  You've should have just made two sides of a right angle triangle (without the hypotenuse).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Standing at the end of the rope farthest from the bridge, compare the length of the rod to the bridge.  If the rod looks smaller than the bridge, pack everything up and move further away from the bridge along the shore and repeat.  If the rod looks bigger, then move closer.   Keep moving until you find the spot where the rod appears exactly the same length as the bridge.  You might need to get on the ground to get a better look at both the rod and the bridge in the same field of vision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, you've made two proportionate right angle triangles:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The smaller triangle has the rod for the height, and the rope for the base.&lt;/li&gt;&#xA;&lt;li&gt;The larger triangle has the bridge for the height, and the distance from you to the bridge as the base.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So now all you need to do is measure as accurately as you can your distance from the bridge in rope lengths.  Whatever this distance is will be the length of the bridge in rod lengths since the two triangles are proportional.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2015-01-26T16:26:53.280" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8203" PostTypeId="2" ParentId="8186" CreationDate="2015-01-26T17:02:19.427" Score="2" Body="&lt;p&gt;First of all, the term &quot;immediate neighbor&quot; is misleading, because in order for this to work, it really should be &quot;an immediate neighbor&quot;. Further, &quot;to the right&quot; is also misleading, because they should inform us that we are facing North.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, I quickly realized this would be very difficult to solve in a completely straightforward fashion (e.g. eliminating possibilities until the answer remains), so I decided to make assumptions and which would either work out, or result in a contradiction and I could start again. Also, the possibility occurred to me - because the answers were multiple choice, it didn't ask for a complete grid - that the solution may not be unique, so making assumptions is not necessarily a bad thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first thing I noticed is that there is a lot of information in the South row about people sitting far from each other, i.e.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Only one person sits between G and the pilot.&lt;/li&gt;&#xA;&lt;li&gt;Two people sit between H and the Lawyer.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Therefore, either G must be the lawyer, or H must be the pilot. I began with the &lt;strong&gt;assumption&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;H is the Pilot.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I also made a second assumption:&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;H is on the left (and the Lawyer is on the right).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So we have, immediately:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;xxx    xxx    xxx   xxx&#xA;        D&#xA;&#xA;Pil    xxx    xxx   Law&#xA; H             G&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Since the neighbor of D faces the Doctor, G must be the Doctor. Further, since G faces the immediate neighbor of the Builder, and B must be second to the right of the Builder, D must be the builder.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;xxx    Bui    xxx   xxx&#xA;        D            B&#xA;&#xA;Pil    xxx    Doc   Law&#xA; H             G&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Since E faces the Engineer, and D is the Builder, B must be the engineer. Further, Teacher and C are in the same row, and Teacher is not on the extreme ends, so Teacher must be on the third seat, and therefore C is the first seat. Via process of elimination, Now we have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;xxx    Bui    Tea   Eng&#xA; C      D      A     B&#xA;&#xA;Pil    xxx    Doc   Law&#xA; H      F      G     E&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Since F is not a Banker, F must be a businessman, So C is the Banker.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Ban    Bui    Tea   Eng&#xA; C      D      A     B&#xA;&#xA;Pil    Bus    Doc   Law&#xA; H      F      G     E&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;A quick check of all the facts shows that this fits, and this gives the answers to the multiple choice provided by OP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exercise for the reader:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Show that assumption 1 results in a contradiction (hint: who do the immediate neighbors of D face?)&lt;/li&gt;&#xA;&lt;li&gt;Show that assumption 2 results in a contradiction (hint: B and D can't both be two to the right of the Builder)&lt;/li&gt;&#xA;&lt;li&gt;Why do the answers to 1 and 2 show the answer is, in fact, unique?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="42" LastEditorUserId="42" LastEditDate="2015-01-26T17:14:13.047" LastActivityDate="2015-01-26T17:14:13.047" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8204" PostTypeId="2" ParentId="8175" CreationDate="2015-01-26T17:40:19.407" Score="11" Body="&lt;p&gt;I think I've found a solution for an 8x8 square. I do not know if it is the &lt;em&gt;minimum&lt;/em&gt; solution or how to prove that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/TfBXe.png&quot; alt=&quot;8x8&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It was definitely fun to try and find this! Took me a while. Excellent puzzle.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Some comments on how I got to the solution (Rather a chronology than a full deduction):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - It has been proven in the other puzzle that &lt;em&gt;at least&lt;/em&gt; 12 tiles would be needed.&lt;br&gt;&#xA; - It was clear, that each piece has to have &lt;em&gt;at least&lt;/em&gt; 2 squares.&lt;br&gt;&#xA; - I was intuitively convinced that the solution would have a 4-fold rotational symmetry. (No proof for that whatsoever. More a 'feeling'.) So I set out with trying patterns in this symmetry. If I get 1/4th of the pieces in place, the other 3/4th would be correct automatically.&lt;br&gt;&#xA; - I figured the &lt;em&gt;rim&lt;/em&gt; would be the difficult part, as long-stretched tiles are needed, and their length is limited by the square-size. &lt;br&gt;&#xA; - So I first tried to create a 7x7 square with fitting tiles to the border and leaving room for 5 connections. &lt;br&gt; &lt;img src=&quot;https://i.stack.imgur.com/pH8z9.png&quot; alt=&quot;stp&quot;&gt;&lt;br&gt;&#xA; - This did not work out (because of the symmetry centre being 'left out'). So I then retried the same with a 8x8 grid, ending up with:&lt;br&gt; &lt;img src=&quot;https://i.stack.imgur.com/ApHED.png&quot; alt=&quot;Step&quot;&gt;&lt;br&gt;&#xA; - Trying to extend the colours towards the centre (always using 4-fold rotational symmetry on the extensions) I soon ran into obstacles, which forced me to 'shift' the outer rim to:&lt;br&gt; &lt;img src=&quot;https://i.stack.imgur.com/nTOum.png&quot; alt=&quot;step&quot;&gt;&lt;br&gt;&#xA; - And then it was just a matter of extending &lt;em&gt;inwards&lt;/em&gt; and realizing that the number of tiles is not yet enough. Adding 4 colours did the trick.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="5840" LastEditDate="2015-01-26T18:08:35.187" LastActivityDate="2015-01-26T18:08:35.187" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8205" PostTypeId="1" CreationDate="2015-01-26T17:48:14.193" Score="1" ViewCount="222" Body="&lt;p&gt;I'm trying to create a math equasion puzzle using the numbers 3, 7, 20 and 40.  I want to end with an answer of either 7 or 40.   Is there any way this can be done? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for the help &lt;/p&gt;&#xA;" OwnerUserId="9177" LastActivityDate="2015-01-26T18:39:41.967" Title="need help with a math equation puzzle" Tags="|logical-deduction|" AnswerCount="2" CommentCount="0" ClosedDate="2015-01-27T08:43:44.440" ContentLicense="CC BY-SA 3.0" />
  <row Id="8206" PostTypeId="2" ParentId="8205" CreationDate="2015-01-26T17:57:41.377" Score="2" Body="&lt;p&gt;If I understand your question correctly, you just want to combine the numbers 3, 7, 20 and 40 with arithmetic operations to get 7 and/or 40:&#xA;$$(3 \times 7 - 20) \times 40 = 40\\&#xA;(3 - 40 \div 20) \times 7 = 7$$&lt;/p&gt;&#xA;" OwnerUserId="8888" LastActivityDate="2015-01-26T17:57:41.377" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8207" PostTypeId="2" ParentId="8205" CreationDate="2015-01-26T18:39:41.967" Score="2" Body="&lt;p&gt;I &lt;a href=&quot;http://ideone.com/xSk457&quot; rel=&quot;nofollow&quot;&gt;wrote a script&lt;/a&gt; to go through all possible equations using +, -, /, *. Here is the output:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;7 &#xA;&#xA;7 * (3 - (40 / 20))&#xA;(3 - (40 / 20)) * 7&#xA;7 / (3 - (40 / 20))&#xA;&#xA;40 &#xA;&#xA;40 * ((3 * 7) - 20)&#xA;40 * ((7 * 3) - 20)&#xA;((3 * 7) - 20) * 40&#xA;((7 * 3) - 20) * 40&#xA;(20 * (7 - 3)) - 40&#xA;((7 - 3) * 20) - 40&#xA;40 / ((3 * 7) - 20)&#xA;40 / ((7 * 3) - 20)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;... Although most of these are trivial rearrangements of earlier ones. The ones I consider &quot;unique&quot; are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$7 \times (3 - (40 \div 20)) = 7\\&#xA;7 \div (3 - (40 \div 20)) = 7\\&#xA;((7 - 3) \times 20) - 40 = 40\\&#xA;40 \times ((3 \times 7) - 20) = 40\\&#xA;40 \div ((3 \times 7) - 20) = 40$$&lt;/p&gt;&#xA;" OwnerUserId="1497" LastActivityDate="2015-01-26T18:39:41.967" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8208" PostTypeId="1" AcceptedAnswerId="8276" CreationDate="2015-01-26T18:45:46.920" Score="5" ViewCount="459" Body="&lt;p&gt;Inspector Crumblesniff thinks he is on the verge of arresting a notorious Mafia hit man, but he needs your help.  First, after a raid on Sal's Pizzeria, he found an address book with these contents:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;BRUNO, ALFONSO - 123 LOVEJOY&lt;/li&gt;&#xA;&lt;li&gt;GIORDANO, VITO - 456 AINSWORTH&lt;/li&gt;&#xA;&lt;li&gt;LANGELLA, TONY - 789 WYGANT&lt;/li&gt;&#xA;&lt;li&gt;PICCOLO, SAMMY - 159 JARRETT&lt;/li&gt;&#xA;&lt;li&gt;TOCCO, FRANKIE - 357 EMERSON&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Next, by disguising himself as a table lamp, the Inspector was able to sneak in to Don Furfante's secret base, the Piazza di Sei, and acquire the following job list:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;PLUMBER - 2074123927971998088&lt;/li&gt;&#xA;&lt;li&gt;PAPERBOY - 3907210679561031886&lt;/li&gt;&#xA;&lt;li&gt;MAILMAN - 3724268395359054732&lt;/li&gt;&#xA;&lt;li&gt;ASSASSIN - 4149097910918443572&lt;/li&gt;&#xA;&lt;li&gt;GARDENER - 1394589336448442100&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, the job list seems to be encrypted, and the Judge Maldestro will only issue a single search warrant.  Can you figure out the killer's address?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: Since a day has gone by with no progress towards catching the murderer, I'll just pop in with a few words of encouragement:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The encryption is simple and not arbitrary.  In some languages, you could convert &quot;2074123927971998088&quot; to the plumber's identity in a single line of code.&lt;/li&gt;&#xA;&lt;li&gt;That said, if you use all of the information available to you and have access to google, neither programming nor guesswork is required to identify the killer.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2374" LastEditorUserId="2374" LastEditDate="2015-01-27T20:40:30.307" LastActivityDate="2015-01-29T05:02:42.540" Title="The Inspector and the Hit man" Tags="|mathematics|cipher|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8209" PostTypeId="2" ParentId="1941" CreationDate="2015-01-26T18:47:40.997" Score="3" Body="&lt;p&gt;John speaks a language with absolute frame of reference such as Guugu Yimithirr or  Tzeltal (so he ordinarily speaks of his north foot and south foot, or his east foot and west foot).  Speakers of these languages have been shown to have a strong internal compass.  As such, John always knows which direction is north.&lt;/p&gt;&#xA;" OwnerUserId="9007" LastActivityDate="2015-01-26T18:47:40.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8210" PostTypeId="2" ParentId="6212" CreationDate="2015-01-26T19:06:57.277" Score="4" Body="&lt;p&gt;From Hint 2: Partial Translation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cotanneih dreeia rnt hee rgumeatsn ofa rda naigstr na&#xA;dmietitnJ ayg_aDrseenr itoL ucenocusumt. rPensae erS&#xA;tteld01, miAnisratoert tf osh etei; Kballto10, deMaoortf oT&#xA;taorlRy nlomad; Gwtyhnati, Gema sMrtea; aDa_rseerJne, ahc&#xA;yseuctM; ksdaoiodral eZe oidn, tnreresepr ttidase sn&#xA;iuproft te pcc husoead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Contained herein are the arguments for and against readmitting&lt;/strong&gt;&#xA;J ay_aDrseenr itoL ucenocusumt. &lt;strong&gt;Present are&lt;/strong&gt; S&#xA;teld01, miAnisratoert tf osh etei; Kballto10, deMaoortf oT&#xA;taorlRy nlomad; Gwtyhnati, Gema sMrtea; aDa_rseerJne, ahc&#xA;yseuctM; ksdaoiodral eZe oidn, tnreresepr ttidase sn&#xA;iuproft te pcc husoead.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From Hint 5:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;02 90 B2 9A E9 D0 8B D7 60 39 61 1B 07 85 92 1F B8 F2 E3 44 ED 95 A4 6C C9 A9 09 43 A3 99 DA 32 78 DC 35 CF E0 8B 24 82&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result I get from converting (at WolframAlpha) the base 16 representation to base 256:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2:144:178:154:233:208:139:215:96:57:97:27:7:133:146:31:184:242:227:68:237:149:164:108:201:169:9:67:163:153:218:50:120:220:53:207:224:139:36:130&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;We'll need to put our heads together on this problem people. Just keep building on the work others provide. It'll come together.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9105" LastEditorUserId="9105" LastEditDate="2015-01-28T00:24:39.020" LastActivityDate="2015-01-28T00:24:39.020" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8211" PostTypeId="2" ParentId="8195" CreationDate="2015-01-26T19:23:16.493" Score="78" Body="&lt;p&gt;I noticed that four of the nodes of the graph have a quarter missing, and each of those four is missing a different quarter. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; These missing quarters can be used to designate those four nodes as the four corners of a 7 by 6 grid of nodes like so:&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/PW9Dv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With this...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each row of the grid has 5 cells, which can be assigned a 1 if the diagonal goes from lower left to upper right or a 0 of the diagonal goes from upper left to lower right. Using this method, the rows have binary values of 01101, 00100, 00100, 00011, 01011, and 00100. In base ten, these numbers are 13, 4, 4, 3, 11, 4. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though many people might convert these numbers into letters starting with A=1, this would result in the nonsense word &quot;mddckd&quot;. All C programmers know that we should start with A=0, which means the numbers convert to &quot;needle&quot;.&lt;/p&gt;&#xA;" OwnerUserId="9178" LastEditorUserId="5840" LastEditDate="2015-01-26T20:25:48.750" LastActivityDate="2015-01-26T20:25:48.750" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8212" PostTypeId="1" AcceptedAnswerId="10358" CreationDate="2015-01-26T19:30:01.480" Score="10" ViewCount="58798" Body="&lt;p&gt;I know a set of six 4-letter words without repeating a letter (i.e. using 24 different letters).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; gasp, verb, jinx, flow, duck, myth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Are there sets of words with no repeated letters and having&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;four 6-letter words?&lt;/li&gt;&#xA;&lt;li&gt;five 5-letter words?&lt;/li&gt;&#xA;&lt;li&gt;three 8-letter words?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Bonus: Find a set of words (non-repeating letters) of sizes 1,2,3,4,5,6. &lt;/p&gt;&#xA;" OwnerUserId="5684" LastEditorUserId="22922" LastEditDate="2016-09-19T09:40:19.243" LastActivityDate="2021-06-02T12:25:39.140" Title="Word sets with no repeating letters" Tags="|word|english|language|" AnswerCount="6" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8213" PostTypeId="2" ParentId="8212" CreationDate="2015-01-26T19:58:27.517" Score="0" Body="&lt;p&gt;For the 1-letter bonus, there are two possible answers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first is that there are three different words:&lt;br&gt;&#xA;A&lt;br&gt;&#xA;I&lt;br&gt;&#xA;O&lt;br&gt;&#xA;Strictly speaking, these are the only one-letter words in English (&quot;O&quot; is an alternate spelling of the interjection &quot;Oh&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second answer relies on a more flexible meaning of &quot;word&quot;: every letter can be regarded as a self-referential word. That is, as a word that identifies the letter, as in &quot;&lt;em&gt;C&lt;/em&gt; is for Cookie&quot; or &quot;The &lt;em&gt;L&lt;/em&gt; Word&quot;. In this case, one can list 26 distinct words with no letters in common between any of them!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm currently working on the eight-letter question (and I want to do it without writing a search script), but as a hint to other solvers: there are only five vowels not including &lt;em&gt;Y&lt;/em&gt; (which I just used as a word), so one of the eight-letter words will need to have only one &quot;regular&quot; vowel, and will probably need to use &lt;em&gt;Y&lt;/em&gt; as a vowel. Some possibilities are &lt;code&gt;BRIGHTLY&lt;/code&gt; &lt;code&gt;GRYPHONS&lt;/code&gt; &lt;code&gt;HYDRANTS&lt;/code&gt; &lt;code&gt;PSYCHING&lt;/code&gt; &lt;code&gt;SHREWDLY&lt;/code&gt; &lt;code&gt;SPLOTCHY&lt;/code&gt;. It might be especially useful to use &lt;code&gt;SCHMALTZ&lt;/code&gt; or &lt;code&gt;SCHMALZY&lt;/code&gt;/&lt;code&gt;SHMALTZY&lt;/code&gt; to use a &lt;em&gt;Z&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="8776" LastEditorUserId="8776" LastEditDate="2015-01-27T15:05:00.480" LastActivityDate="2015-01-27T15:05:00.480" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8214" PostTypeId="2" ParentId="2283" CreationDate="2015-01-26T20:08:04.180" Score="28" Body="&lt;p&gt;I figured adding a picture will more clearly explain &lt;a href=&quot;https://puzzling.stackexchange.com/q/2283/19765#2286&quot;&gt;the accepted answer&lt;/a&gt;. With this image, one can clearly see the 123456789. The parts marked in red are where you were expected to extrapolate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VKvR0.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VKvR0.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2359" LastEditorUserId="19765" LastEditDate="2017-05-01T01:12:36.613" LastActivityDate="2017-05-01T01:12:36.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8215" PostTypeId="2" ParentId="8212" CreationDate="2015-01-26T20:30:29.940" Score="2" Body="&lt;p&gt;Here's a set for the bonus question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;a&lt;br&gt;&#xA;be&lt;br&gt;&#xA;why&lt;br&gt;&#xA;jump&lt;br&gt;&#xA;flong (a type of mould used in relief printing)&lt;br&gt;&#xA;tricks&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The missing letters are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;dqvxz&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7363" LastActivityDate="2015-01-26T20:30:29.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8216" PostTypeId="1" AcceptedAnswerId="8220" CreationDate="2015-01-26T21:03:13.067" Score="12" ViewCount="9673" Body="&lt;p&gt;&lt;strong&gt;7&lt;/strong&gt; segments make up a number, some &lt;em&gt;visible&lt;/em&gt;, some &lt;em&gt;invisible&lt;/em&gt;. Just like on a &lt;em&gt;digital clock&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;88:88&lt;/strong&gt; or &lt;strong&gt;04:00&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MlOlh.jpg&quot; alt=&quot;Digital clock showing 04:00&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 2 vertical segments on both sides and 3 horizontal segments in the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the largest number you can form, where it has the same number of visible segments as the number it represents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example &lt;strong&gt;4&lt;/strong&gt; is formed with &lt;em&gt;4&lt;/em&gt; visible segments on a digital clock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or &lt;strong&gt;5&lt;/strong&gt; which has &lt;em&gt;5&lt;/em&gt; visible segments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How large can these numbers get, and which is the largest?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; &lt;em&gt;The number can have as many digits as you wish&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your answer is only valid if you state the answer, and how you came to the conclusion!&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorDisplayName="user88" LastEditDate="2015-05-01T05:35:49.600" LastActivityDate="2015-05-01T05:35:49.600" Title="Greatest Digital Clock Number, with a Twist!" Tags="|mathematics|seven-segment|" AnswerCount="6" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8217" PostTypeId="2" ParentId="8199" CreationDate="2015-01-26T21:08:35.727" Score="3" Body="&lt;p&gt;Assuming the pole holding up the bridge on the near end is the same height as the pole holding up the bridge on the far end... You could stand at the near pole and visually measure the far pole, comparing it to the rod.&lt;br&gt;&#xA;Then walk away from the near pole until comparing it against the rod matches up with what you saw on the far end... Measure back to the pole, and that's how wide the river is.  &lt;/p&gt;&#xA;" OwnerUserId="9183" LastActivityDate="2015-01-26T21:08:35.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8218" PostTypeId="2" ParentId="8216" CreationDate="2015-01-26T21:17:41.470" Score="13" Body="&lt;p&gt;This assumes you cannot null-pad the number, since if you can null-pad it, the highest number is infinite. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a n-digit number, the maximum number of lines visible is 7n. &#xA;Therefore, the number cannot be greater than 2 digits, since 21&amp;lt;100 for n = 3, and the number of lines increases linearly with n while the number increases exponentially with n. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 2-digit number also cannot be greater than 14, since the maximum number of visible lines is 14. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;14 - 6 lines&lt;/p&gt;&#xA;&#xA;&lt;p&gt;13 - 7 lines&lt;/p&gt;&#xA;&#xA;&lt;p&gt;12 - 7 lines&lt;/p&gt;&#xA;&#xA;&lt;p&gt;11 - 4 lines&lt;/p&gt;&#xA;&#xA;&lt;p&gt;10 - 8 lines&lt;/p&gt;&#xA;&#xA;&lt;p&gt;9 - 6 lines&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8 - 7 lines&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7 - 3 lines&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;6 - 6 lines&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If null-padding is allowed, then any number that has follows the equation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;(actualValue-lineCount)%6 == 0&lt;/code&gt; will work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;13: 13-7 = 6, valid as 013&lt;/p&gt;&#xA;&#xA;&lt;p&gt;29999: 29999-29 = 29970, valid as (4995 zeros)29999&lt;/p&gt;&#xA;" OwnerUserId="7529" LastEditorUserId="7529" LastEditDate="2015-01-26T21:48:25.357" LastActivityDate="2015-01-26T21:48:25.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8219" PostTypeId="2" ParentId="1941" CreationDate="2015-01-26T21:24:27.970" Score="3" Body="&lt;p&gt;The only static element inside the room who could be used as a reference is...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;The light bulb filament&lt;/strong&gt;, so after the psychopath tells John about the glass, he can look up, check the filament's direction (which given it's particular shape can be used precisely to reference that glass) and do whatever he likes during the psychopath's master plan, and then safely drink the unpoisoned juice from the correct glass. The flaw in the psychopath system is the light bulb rotating with the entire system and not fixed or static.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9184" LastActivityDate="2015-01-26T21:24:27.970" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8220" PostTypeId="2" ParentId="8216" CreationDate="2015-01-26T21:33:12.487" Score="24" Body="&lt;p&gt;There's two answers here:  a real answer, and a (slightly) cheating answer.&lt;/p&gt;&#xA;&lt;h1&gt;Real Answer&lt;/h1&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;&lt;strong&gt;6&lt;/strong&gt; is the largest number where it represents the number of segments.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;First, we can put an upper bound on our answer.  At most, we have a two digit number.  This is because the smallest three digit number (100) is an order of magnitude more than the maximum number of displayed segments (&lt;span class=&quot;math-container&quot;&gt;$21 = 7\times3$&lt;/span&gt;).&lt;/p&gt;&#xA;&lt;p&gt;Through a process of elimination, we can work backwards from the largest possible 2 digit number: 14 (&lt;span class=&quot;math-container&quot;&gt;$=7\times2$&lt;/span&gt;).&lt;/p&gt;&#xA;&lt;p&gt;We end up with &lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt; being the largest number.&lt;/p&gt;&#xA;&lt;h1&gt;Cheater Answer&lt;/h1&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;&lt;strong&gt;Infinite&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If we can front-pad the number with zeros, we can get to infinity, but only certain numbers.  This is because each zero we add in front of the number is 6 additional segments.&lt;/p&gt;&#xA;&lt;p&gt;For example, &lt;span class=&quot;math-container&quot;&gt;$0...0400$&lt;/span&gt; can be made with 114 leading zeros.  &lt;span class=&quot;math-container&quot;&gt;$(114+2)\times6+4 = 400$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;In fact, there's a limited set of numbers that answer this.  We know these numbers are greater than six, and we know their remainder when divided by six is equal to the sum of their segments mod 6.&lt;/p&gt;&#xA;&lt;p&gt;Let &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; be the number, and &lt;span class=&quot;math-container&quot;&gt;$s_n$&lt;/span&gt; be the number of segments of the &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;th digit.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\left(\sum s_n\right) \mod 6 = N\mod6$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Any number of the form &lt;span class=&quot;math-container&quot;&gt;$0...06...6$&lt;/span&gt; will work (assuming enough zeros).  Also, any number of the form &lt;span class=&quot;math-container&quot;&gt;$0...040...0$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Some things I've worked out:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; is even, iff &lt;span class=&quot;math-container&quot;&gt;$\sum s_n$&lt;/span&gt; is even.  This is because zero has an even number of segments, and you can't add even numbers together to get odd numbers.  The corollary to this is: &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; is odd, iff &lt;span class=&quot;math-container&quot;&gt;$\sum s_n$&lt;/span&gt; is odd.  This may seem really obvious, but it helps tremendously.&lt;/li&gt;&#xA;&lt;li&gt;1, 2, 8, and 9 are &amp;quot;backwards&amp;quot; numbers.  By this, I mean that they are even, but their number of segments is odd, or vice versa.  This means that a number cannot only be made of 1s or 9s, and a number made solely of 2s and/or 8s must have an even number of 2s and 8s.  A corollary of this is that a number made of solely &amp;quot;regular&amp;quot; odd numbers cannot have an even number of digits. (These numbers being 3, 5, and 7)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;There are probably many other forms.  I'd be interested to see a mathematician work through them all.&lt;/p&gt;&#xA;" OwnerUserId="4134" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-01-27T15:36:47.127" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8221" PostTypeId="2" ParentId="8216" CreationDate="2015-01-26T21:57:32.097" Score="13" Body="&lt;p&gt;My digital clock is in &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; binary.  It reads 100000, which uses 32&lt;sub&gt;10&lt;/sub&gt; line segments to display the number 32&lt;sub&gt;10&lt;/sub&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assuming we can't pad the answer to infinity with unnecessary zeroes, this seems to be the maximal answer, because &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; in any base, when we add an additional nth digit, the clock face value grows geometrically by (radix*n) while the line count grows arithmetically by at most 7.  Clearly these values will diverge quickly; to maximize the point before they diverge, we want the most digits, and hence the lowest radix.  Ergo, no base higher than 2 will yield a higher answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit: @WChargin asks in the comments: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;What about a clock in base square root of two?&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Very devious approach!  However, it doesn't turn out to yield a higher answer, or any valid answer at all.  I can't think of a tidy number-theory way to address this, but I wrote a simple script to test and it turns out that &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The highest solution from a square root base is 16 (202 in base sqrt(7)).  Cube roots are left as an exercise for the student :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2374" LastEditorUserId="2374" LastEditDate="2015-01-27T19:18:53.157" LastActivityDate="2015-01-27T19:18:53.157" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8222" PostTypeId="2" ParentId="1953" CreationDate="2015-01-26T22:53:44.110" Score="4" Body="&lt;p&gt;I think this can be slightly improved. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Box 1 is inside Box 2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Box 1 contains 50 white balls. Box 2 contains Box 1 (which contains 50 white balls) and 50 black balls. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upon choosing a random box, you get either Box 1 (100% survival) or Box 2 (50% survival). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore your chance of survival is 75%, which beats the current accepted answer. &lt;/p&gt;&#xA;" OwnerUserId="7529" LastActivityDate="2015-01-26T22:53:44.110" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="8223" PostTypeId="1" AcceptedAnswerId="10396" CreationDate="2015-01-26T23:25:35.193" Score="11" ViewCount="5621" Body="&lt;p&gt;&lt;strong&gt;Inspiration:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;em&gt;This question was inspired by &lt;a href=&quot;https://puzzling.stackexchange.com/questions/8212/word-sets-with-no-repeating-letters&quot;&gt;Word sets with no repeating letters&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rules:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Sentences can be of any length.&lt;/li&gt;&#xA;&lt;li&gt;Every letter used in this sentence, must be used &lt;em&gt;n&lt;/em&gt; times.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;n&lt;/em&gt; can be any number.&lt;/li&gt;&#xA;&lt;li&gt;Capitals/Punctuation do not count.&lt;/li&gt;&#xA;&lt;li&gt;No dialogue!&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Objective:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;em&gt;The objective is to find the longest sentence which abides by the rules and clarifications.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Length Measuring:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;em&gt;The way your sentence is measured is by characters, including spaces, and punctuation!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Further Clarifications:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The sentence cannot consist of mainly the same word repeated, or the same 2, etc.&lt;/li&gt;&#xA;&lt;li&gt;You cannot make up new words or names.&lt;/li&gt;&#xA;&lt;li&gt;It has to be in English.&lt;/li&gt;&#xA;&lt;li&gt;No uncommon names.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Checking your answer:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://jsfiddle.net/tLcvk2o4/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://jsfiddle.net/tLcvk2o4/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Current Best:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sentence:&lt;/strong&gt; I'm the master's nightmarish, gorgonian hatemonger, his moth-eaten gonorrhoea, smothering mightiest heroism, thrashing egomania's (or, to me, ignorant mismanagement's) strong-arm mishmash or staggering high treason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Length:&lt;/strong&gt; 215&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;N:&lt;/strong&gt; 16&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Letters:&lt;/strong&gt; AEGHIMNORST (11 total)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Current Winner:&lt;/strong&gt; uri zarfaty&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The answer format should be the same format as the Current Best section!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If your answer does not beat the current best, but you still wish to post it, feel free, you may still get +1s!&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="5373" LastEditDate="2019-02-16T19:57:42.340" LastActivityDate="2020-03-11T23:25:19.010" Title="Longest sentence with each letter repeated n times" Tags="|wordplay|open-ended|letters|" AnswerCount="11" CommentCount="21" ClosedDate="2021-06-29T07:35:57.983" ContentLicense="CC BY-SA 3.0" />
  <row Id="8224" PostTypeId="2" ParentId="8223" CreationDate="2015-01-26T23:36:02.750" Score="11" Body="&lt;p&gt;Sentence: My cat ate my cell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Length: 19&lt;/p&gt;&#xA;&#xA;&lt;p&gt;n: 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Letters included: m, y, c, a, t, e, l&lt;/p&gt;&#xA;" OwnerUserId="9159" LastEditorUserId="2383" LastEditDate="2015-01-27T01:51:58.633" LastActivityDate="2015-01-27T01:51:58.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8225" PostTypeId="2" ParentId="130" CreationDate="2015-01-26T23:50:21.473" Score="1" Body="&lt;p&gt;Contrary to @lynxlynxlynx I lost Interest in sudokus, because they were too simple :) I became interested again, when I found sudokus that required guessing,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I developed a simple notation to keep track of hypothetical changes I made on separate lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I introduce a guess with a hash mark:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;# A2: 1,2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then I add consequences of the hypothesis with +:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;# A2: 1,2 + B2: 5 + C7: 3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;When I run into a contradiction, I can easily roll back the last hypothesis:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;# A2: 1,2 + B2: 5 + C7: 3 # G9: 1 + H7 3 =&amp;gt; CTR =&amp;gt; G9: 5&#xA;# A2: 1,2 + B2: 5 + C7: 3 + G9: 5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and so on. Of course, When the paper becomes too cluttered, a refresh on a new sheet becomes necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I kept this notation, when I wrote a program to support limited depth hypothesis/disproof/proof, since the puzzles became too hard to solve with paper, pencil and copy machine. I have explained the method for &lt;a href=&quot;http://sw-amt.ws/sudoku/worlds-hardest-sudoku/&quot; rel=&quot;nofollow noreferrer&quot;&gt;World's Hardest Sudoku&lt;/a&gt; in response to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/252/how-do-i-solve-the-worlds-hardest-sudoku/8032#8032&quot;&gt;another question&lt;/a&gt;, if you're interested.&lt;/p&gt;&#xA;" OwnerUserId="9013" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2015-01-26T23:50:21.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8226" PostTypeId="1" AcceptedAnswerId="8245" CreationDate="2015-01-27T01:02:52.337" Score="4" ViewCount="837" Body="&lt;p&gt;Your neighbour is hosting a party for a select group of people. .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hoping to get in, you hide in the front lawn and watch as guests begin to arrive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first to walk up is a man, who says &quot;Hi, I'm Peter&quot;. The guard smiles and beckons him in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next come a girl and boy - &quot;We're Amy and Jonathan&quot;, says the girl. The guard opens the door and Amy walks in, but as Jonathan goes to follow he says &quot;No, not you&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A woman then walks up, saying &quot;I'm Teaghan, I should be on the guest list.&quot;&lt;br&gt;&#xA;&quot;There isn't a list,&quot; replies the guard, &quot;but in you go&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just as you start to get up, thinking anyone who speaks to the guard is allowed in, another hopeful guest arrives. &quot;Hey, I'm Emmma&quot; she says and is promptly rejected. She cries &quot;It's spelt E-M-M-M-A!&quot;, but the guard responds &quot;that doesn't change anything.&quot;&lt;br&gt;&#xA;Her friends Ashley and Joseph fare no better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some thinking you decide it's time to join the party. How do you introduce yourself to the guard?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Minor Hint + More Examples:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This puzzle would work just as well spoken - the guard doesn't require an extensive knowledge of how names are spelt. Katie, James and Eevee would be allowed in, while Dan and William wouldn't.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;More Significant Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There's a fairly straightforward rule that distinguishes the names of those accepted from those rejected. While the exact spelling doesn't matter, one letter of the name (accompanied with the pronunciation) does.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9062" LastEditorUserId="9062" LastEditDate="2015-01-27T04:29:26.597" LastActivityDate="2015-01-27T16:49:25.403" Title="An Exclusive Party" Tags="|password|" AnswerCount="7" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8227" PostTypeId="2" ParentId="8226" CreationDate="2015-01-27T01:10:24.050" Score="2" Body="&lt;p&gt;&lt;em&gt;This answer was written prior to the edits in the OP, so it is wrong.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it just that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the person's name has to contain a long vowel sound? (P&lt;strong&gt;e&lt;/strong&gt;ter, &lt;strong&gt;A&lt;/strong&gt;my, and T&lt;strong&gt;ea&lt;/strong&gt;ghan do while Jonathan and Emma don't.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If that's the case, then&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a name like Ashl&lt;strong&gt;ey&lt;/strong&gt;, which is unisex, should work.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8858" LastEditorUserId="8858" LastEditDate="2015-01-27T16:49:25.403" LastActivityDate="2015-01-27T16:49:25.403" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8228" PostTypeId="2" ParentId="8226" CreationDate="2015-01-27T01:28:45.347" Score="6" Body="&lt;p&gt;Note that &lt;strong&gt;Peter&lt;/strong&gt;, &lt;strong&gt;Amy&lt;/strong&gt;, and &lt;strong&gt;Teaghan&lt;/strong&gt; have an odd number of letters in their name, while &lt;strong&gt;Jonathan&lt;/strong&gt;, &lt;strong&gt;Emma&lt;/strong&gt;, &lt;strong&gt;Ashley&lt;/strong&gt;, and &lt;strong&gt;Joseph&lt;/strong&gt; all have an even number of letters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Judging by this pattern, I believe that the rule is that only people with an &lt;em&gt;odd&lt;/em&gt; number of letters in their name are allowed into the party.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means I'd be allowed in! Hooray!&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-01-27T01:28:45.347" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8229" PostTypeId="2" ParentId="8199" CreationDate="2015-01-27T02:23:04.140" Score="2" Body="&lt;p&gt;Walk away from the bridge along the river bank a distance roughly equal to the length of the bridge. Lay the rod on the ground pointing from your location toward the far end of the bridge. If the rod forms a 45 degree angle to the river bank then the spot where you are standing is the same distance from the near side of the bridge as the bridge is long. If it's not a 45 degree angle walk to a different point and observe the angle again and repeat until it is 45 degrees. (45 degrees is easy to judge, and the question asks only for an approximation.) Now move back to the near side of the bridge measuring the distance as you go in terms of the rod length by laying the rod end-to-end all the way back. That's the length of the bridge in rods. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then continue strolling along the river with your friend and save the rope for another day.&lt;/p&gt;&#xA;" OwnerUserId="9191" LastActivityDate="2015-01-27T02:23:04.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8231" PostTypeId="2" ParentId="8223" CreationDate="2015-01-27T03:16:47.360" Score="9" Body="&lt;p&gt;&lt;em&gt;Updates from comments:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Enid, MY goat, ate my dingo. (length 28, 10 letters: credit to Muqo)&#xA;My snide goat ate my dingos. (length 28, 11 letters)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sentence:&lt;/strong&gt; Ed, my goat, ate my dog.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Length:&lt;/strong&gt; 24&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;n:&lt;/strong&gt; 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Letters:&lt;/strong&gt; ADEGMOTY&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Bonus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sentence:&lt;/strong&gt; Keats takes Steak Skate stake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Steak Skate&lt;/em&gt; is a combination skating rink and steakhouse. They hold an annual competition, and Keats won this year's stake.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Length:&lt;/strong&gt; 30&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;n:&lt;/strong&gt; 5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Letters:&lt;/strong&gt; AEKST&lt;/p&gt;&#xA;" OwnerUserId="1563" LastEditorUserId="1563" LastEditDate="2015-01-27T19:01:46.153" LastActivityDate="2015-01-27T19:01:46.153" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8232" PostTypeId="2" ParentId="8226" CreationDate="2015-01-27T03:23:54.360" Score="0" Body="&lt;p&gt;Note:The question was edited and the hints were given after I posted this answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess it is because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Number of consonants in a name must be Number of vowels in the name + 1&lt;br&gt;So, number of consonants=number of vowels+1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is why Emma,Ashley,Joseph and Jonathan failed to enter the party.&#xA;&lt;hr&gt; An easier way to gain access to the party would be to approach the guard and say &quot;Hello! I'm Peter&quot; and the guard will let you through just like he did for the first man,Peter.&lt;/p&gt;&#xA;" OwnerUserId="7482" LastEditorUserId="7482" LastEditDate="2015-01-27T10:25:12.023" LastActivityDate="2015-01-27T10:25:12.023" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8233" PostTypeId="2" ParentId="8223" CreationDate="2015-01-27T04:20:16.417" Score="6" Body="&lt;p&gt;&lt;strong&gt;Sentence:&lt;/strong&gt; I hit a heat blob's sole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Yes, blobs of heat have shoes too!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Length:&lt;/strong&gt; 25&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;n:&lt;/strong&gt; 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Letters:&lt;/strong&gt; ABEHILOST&lt;/p&gt;&#xA;" OwnerUserId="8877" LastActivityDate="2015-01-27T04:20:16.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8234" PostTypeId="2" ParentId="8223" CreationDate="2015-01-27T04:33:04.880" Score="15" Body="&lt;p&gt;A blatant homage to the mods:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sentence&lt;/strong&gt;: Our rangers of the null postings are leaping up, high atop this fluff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Length&lt;/strong&gt;: 71&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;n&lt;/strong&gt;: 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Letters&lt;/strong&gt;: aefghilnoprstu&lt;/p&gt;&#xA;" OwnerUserId="2374" LastActivityDate="2015-01-27T04:33:04.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8237" PostTypeId="2" ParentId="8223" CreationDate="2015-01-27T05:13:30.157" Score="11" Body="&lt;p&gt;&lt;strong&gt;Sentence:&lt;/strong&gt; Step on no pets!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Length:&lt;/strong&gt; 16&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;n:&lt;/strong&gt; 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Letters:&lt;/strong&gt; ENOPST&lt;/p&gt;&#xA;" OwnerUserId="7482" LastActivityDate="2015-01-27T05:13:30.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8238" PostTypeId="2" ParentId="8216" CreationDate="2015-01-27T07:52:21.350" Score="4" Body="&lt;p&gt;Regarding only the maximum of 28 elements and 4 numbers, the highest element is 6 (without leading zeroes) and 22 (with leading zeroes : 00:22). Only regarding decimal system.&lt;/p&gt;&#xA;" OwnerUserId="9197" LastActivityDate="2015-01-27T07:52:21.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8239" PostTypeId="1" AcceptedAnswerId="8241" CreationDate="2015-01-27T10:35:29.220" Score="8" ViewCount="3245" Body="&lt;p&gt;I have split the 26 letters of the alphabet into 4 sets. Here they are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set 1: $F G J L N P Q R S Z$&lt;BR&gt;&#xA;Set 2: $A M T U V W Y$&lt;BR&gt;&#xA;Set 3: $B C D E K$&lt;BR&gt;&#xA;Set 4: $H I O X$&lt;BR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you determine on what basis have I split them?&lt;/p&gt;&#xA;" OwnerUserId="7482" LastActivityDate="2015-01-27T10:47:37.857" Title="On what basis have I split the alphabet into 4 sets?" Tags="|english|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8240" PostTypeId="2" ParentId="8226" CreationDate="2015-01-27T10:43:13.733" Score="1" Body="&lt;p&gt;I'd hazard to guess that the name has to contain this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;An iː-sound (sort of like &quot;ee&quot;, like in &quot;mean&quot;), as in &lt;br&gt;&#xA;P&lt;strong&gt;e&lt;/strong&gt;ter, Am&lt;strong&gt;y&lt;/strong&gt;, T&lt;strong&gt;ea&lt;/strong&gt;ghan, Kat&lt;strong&gt;ie&lt;/strong&gt;, &lt;strong&gt;Ee&lt;/strong&gt;v&lt;strong&gt;ee&lt;/strong&gt;&lt;br&gt;&#xA;I'm stumped as to why James got in though, maybe his name is pronounced &quot;Jamees&quot;!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I'd introduce myself as...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Seagull. That's right.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5265" LastActivityDate="2015-01-27T10:43:13.733" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8241" PostTypeId="2" ParentId="8239" CreationDate="2015-01-27T10:47:37.857" Score="7" Body="&lt;p&gt;My guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think you have split them based on symmetry. The first group has no real symmetry, while the second group can be divided by drawing a line from top to bottom. The third group can be splitted by drawing a horizontal line from left to right, and the last group can be divided using either a vertical or horizontal line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7279" LastActivityDate="2015-01-27T10:47:37.857" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8242" PostTypeId="2" ParentId="8226" CreationDate="2015-01-27T11:26:11.660" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The name must contain a sound that is prounced as the name of a vowel. Callidus would not be permitted, but Payne, Jean, Simon, Joan and Stewart all would be, even though Stewart doesn't contain the relevant vowel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6031" LastActivityDate="2015-01-27T11:26:11.660" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8243" PostTypeId="2" ParentId="8226" CreationDate="2015-01-27T11:34:51.790" Score="0" Body="&lt;p&gt;I believe it is because of how you pronounce the names, especially the letter(s):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a and e&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You pronounce this (or these, for that matter) letter(s) different in the names of the people being rejected. I believe I could just introduce myself as Kevin, and get in the party.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a list of people who were allowed in and why I think why:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; P&lt;strong&gt;e&lt;/strong&gt;ter T&lt;strong&gt;ea&lt;/strong&gt;ghan &lt;strong&gt;A&lt;/strong&gt;my K&lt;strong&gt;a&lt;/strong&gt;tie J&lt;strong&gt;a&lt;/strong&gt;mes &lt;strong&gt;Ee&lt;/strong&gt;vee&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And a list of rejected people as well:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; D&lt;strong&gt;a&lt;/strong&gt;n Willi&lt;strong&gt;a&lt;/strong&gt;m Jon&lt;strong&gt;a&lt;/strong&gt;than &lt;strong&gt;E&lt;/strong&gt;mma &lt;strong&gt;A&lt;/strong&gt;shley Jos&lt;strong&gt;e&lt;/strong&gt;ph&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7279" LastActivityDate="2015-01-27T11:34:51.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8244" PostTypeId="1" AcceptedAnswerId="8246" CreationDate="2015-01-27T11:35:35.173" Score="3" ViewCount="3650" Body="&lt;p&gt;You are in the possession of two bottles, one with a capacity of 7 litres and one with a capacity of 11 litres. Next to you is an infinitely large tub of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to measure exactly 2 litres in one of the bottles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are only allowed to entirely empty or fill the bottles. You can't fill them partially since there is no indication on the bottles saying how much liquid is in them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you measure exactly 2 litres?&lt;/p&gt;&#xA;" OwnerUserId="6151" LastEditorUserId="24713" LastEditDate="2016-09-05T10:10:45.733" LastActivityDate="2016-09-05T10:10:45.733" Title="Pouring problem" Tags="|mathematics|decanting-problems|" AnswerCount="1" CommentCount="3" ClosedDate="2015-01-27T23:03:34.033" ContentLicense="CC BY-SA 3.0" />
  <row Id="8245" PostTypeId="2" ParentId="8226" CreationDate="2015-01-27T11:38:30.280" Score="9" Body="&lt;p&gt;I'm guessing Jayden and Beatrice would get in, but Pollyanna and Ness wouldn't. Also, Emma would have gotten in if her name was spelled Margaret.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, is the rule that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the sound of the name has to start with the sound of the first letter?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2015-01-27T11:38:30.280" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8246" PostTypeId="2" ParentId="8244" CreationDate="2015-01-27T11:49:17.153" Score="5" Body="&lt;p&gt;Fill the 7 litre bottle completely and pour the whole thing to the 11 litre bottle.&lt;br&gt;&#xA;Fill the 7 litre bottle completety again and pour the water to the 11 litre bottle until it becomes full.&lt;br&gt;&#xA;Now the 7 litre bottle contains 3 litres.&lt;br&gt;&#xA;Empty the 11 litre bottle and then pour the 3 litres into the 11 litre bottle.&lt;br&gt;&#xA;Fill the 7 litre bottle completely and pour the whole thing to the 11 litre bottle.&lt;br&gt;&#xA;Fill the 7 litre bottle completely again and pour the water to the 11 litre bottle until it becomes full.&lt;br&gt;&#xA;Now the 7 litre bottle contains 6 litres.&lt;br&gt;&#xA;Empty the 11 litre bottle and then pour the 6 litres into the 11 litre bottle.&lt;br&gt;&#xA;Fill the 7 litre bottle completely and pour the water to the 11 litre bottle until it becomes full.&lt;br&gt;&#xA;Now, the 7 litre bottle contains 2 litres.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;An other possible solution would be to start by filling the 11 litre bottle as shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GWVbz.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7482" LastEditorUserId="6151" LastEditDate="2015-01-27T12:05:53.983" LastActivityDate="2015-01-27T12:05:53.983" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8247" PostTypeId="2" ParentId="8216" CreationDate="2015-01-27T11:51:45.477" Score="23" Body="&lt;p&gt;I suppose this is also cheating?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EidDP.jpg&quot; alt=&quot;&amp;quot;thirty four&amp;quot; written as words, to make 34 segments&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4182" LastActivityDate="2015-01-27T11:51:45.477" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8248" PostTypeId="1" AcceptedAnswerId="8257" CreationDate="2015-01-27T12:02:30.520" Score="5" ViewCount="5690" Body="&lt;h2&gt;Instructions&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;You have a bag containing infinite number of matchsticks all having the same cuboidal shape and size. Digits can be made using these matchsticks:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/V43xP.png&quot; alt=&quot;DIGITS&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$0$ takes 6 matchsticks&lt;/li&gt;&#xA;&lt;li&gt;$1$ takes 2 matchsticks&lt;/li&gt;&#xA;&lt;li&gt;$2$ takes 5 matchsticks&lt;/li&gt;&#xA;&lt;li&gt;$3$ takes 5 matchsticks&lt;/li&gt;&#xA;&lt;li&gt;$4$ takes 4 matchsticks&lt;/li&gt;&#xA;&lt;li&gt;$5$ takes 5 matchsticks&lt;/li&gt;&#xA;&lt;li&gt;$6$ takes 6 matchsticks&lt;/li&gt;&#xA;&lt;li&gt;$7$ takes 3 matchsticks&lt;/li&gt;&#xA;&lt;li&gt;$8$ takes 7 matchsticks&lt;/li&gt;&#xA;&lt;li&gt;$9$ takes 6 matchsticks&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And the operators which you can use are&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;'$+$' takes 2 matchsticks&lt;/li&gt;&#xA;&lt;li&gt;'$-$' takes a matchsticks&lt;/li&gt;&#xA;&lt;li&gt;'$\times$' takes 2 matchsticks&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Requirement&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;You are required to create &lt;strong&gt;expressions&lt;/strong&gt; by creating numbers and operators together using the matchsticks provided to you. The expressions that you create must have at least one operator and must have the result which will be provided to you.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Example&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Question: Create an expression having result 1000 using matchsticks &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $10 \times 10 \times 10$ takes 28 matchsticks&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Question&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Create an expression &lt;strong&gt;using the minimum number of matchsticks&lt;/strong&gt; having results&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;10&lt;/li&gt;&#xA;&lt;li&gt;45&lt;/li&gt;&#xA;&lt;li&gt;50&lt;/li&gt;&#xA;&lt;li&gt;99&lt;/li&gt;&#xA;&lt;li&gt;125&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;I will accept the answer which uses the least number of matches in all the five expressions combined.&lt;br&gt;&#xA;The current accepted answer by @Jabe uses a total of &lt;strong&gt;55&lt;/strong&gt; matchsticks.&lt;br&gt;&#xA;&lt;em&gt;Have fun trying to beat the highscore!&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="7482" LastEditorUserId="30887" LastEditDate="2017-03-08T12:02:20.180" LastActivityDate="2017-08-23T15:52:56.853" Title="What is the minimum number of matchsticks that you can use to build expressions having the following results?" Tags="|mathematics|matches|" AnswerCount="4" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8249" PostTypeId="1" AcceptedAnswerId="25959" CreationDate="2015-01-27T12:11:25.080" Score="14" ViewCount="980" Body="&lt;p&gt;You have reached the final stage of a game show, and now face an endless row of doors labelled $\ldots,-3,-2,-1,0,1, 2, 3,\ldots$, going from minus infinity to plus infinity.  Your goal is to detect a secret door that has been chosen by the game host (and that is static and will not change throughout the game).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The door with number $0$ is already open and forms the starting point of your search. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In every further step, you first pay 1 Euro to the game host and then open the left or right neighbor of your current door. (In other words: if you open door $n$ in some round, then in the following round you must open door $n-1$ or door $n+1$.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Once you open the secret door, you may keep the money that is hidden behind it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Prior to the game and before the game show host chooses the secret door, you have to specify to him the full (deterministic) strategy that you will follow for the entire evening. Only then the host will select his secret door (with number $x\ne0$) and will put an amount of $8.999|x|$ Euros behind it (where the multiplicator is $8.999=9-1/1000$ and hence slightly smaller than $9$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any deterministic strategy that guarantees you some positive profit in this game?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: This puzzle was inspired by the puzzle &lt;a href=&quot;https://puzzling.stackexchange.com/questions/8042/one-prize-infinitely-many-choices/&quot;&gt;One prize, infinitely many choices&lt;/a&gt;. In this other puzzle, the secret door moves in the same fashion as the searcher does move in my puzzle.)&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2016-01-30T18:10:24.333" Title="Searching for a secret door" Tags="|mathematics|strategy|optimization|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8251" PostTypeId="2" ParentId="8249" CreationDate="2015-01-27T12:41:04.133" Score="5" Body="&lt;p&gt;My answer (if I have not made any mistakes).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Our best strategy is the &quot;accelerating pendulum&quot; one. We go in one direction with &lt;code&gt;x&lt;/code&gt; doors and after that go in the other &lt;code&gt;x&lt;/code&gt; to get to zero and &lt;code&gt;2*x&lt;/code&gt;. To get to the current maximum &lt;code&gt;2*z&lt;/code&gt;, we've made &lt;code&gt;4*z&lt;/code&gt; steps (2*z in that direction, 1*z to get back from the previous peak, and 1*z of all the previous moves). This means that if we win on the &quot;peak&quot; of the swing we'll get &lt;code&gt;9*z&lt;/code&gt; and thus &lt;code&gt;5*z&lt;/code&gt; (5/9 of the winnings). Our worst case is to have a peak at door number &lt;code&gt;-2*y&lt;/code&gt; but the winnings are at &lt;code&gt;2*y&lt;/code&gt;. However even in that case we'll get &lt;code&gt;2*9*y&lt;/code&gt; and we will have moved a total of &lt;code&gt;4*y + 6*y = 10*y&lt;/code&gt;. Which is still &lt;code&gt;8*y&lt;/code&gt; (that is 4/9 of the winnings), which ain't that bad.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;P.S.:&lt;/strong&gt; The edit does not affect my solution at all.&lt;/p&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="5610" LastEditDate="2015-01-27T16:22:57.990" LastActivityDate="2015-01-27T16:22:57.990" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8252" PostTypeId="1" CreationDate="2015-01-27T12:51:00.280" Score="24" ViewCount="10244" Body="&lt;p&gt;Take a look at these two questions:&lt;br&gt;&#xA; - &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2034/a-set-of-water-jug-challenges&quot;&gt;A Set of Water Jug Challenges&lt;/a&gt; &lt;br&gt;&#xA; - &lt;a href=&quot;https://puzzling.stackexchange.com/questions/8244/pouring-problem&quot;&gt;Pouring problem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I'm asking for a generalised solution to that problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I define the problem as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;You are required to measure exactly $n$ litres of water.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You have $j$ jugs. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Each jug has volume $v_m$ where $m$ is the jug-number ($1$ to $j$). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You have an infinitely large tub of water.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There is no indication on the jugs saying how much liquid is in them, and you can't judge it by eye. You do know for sure when they are empty and when they are full.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Question(s):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;For what sets of values ($n$, $j$, $v_1$ to $v_j$) is this solvable?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For cases which are solvable, what algorithm can be used to find the solution?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is intended to be a math/algorithm problem, rather than a lateral thinking problem, so 'creative' solutions such as &quot;&lt;em&gt;I use the infinitely-large tub of water to solve the world's drought problems, am elected world president for my services to humanity, and keep the jugs as a souvenir&lt;/em&gt;&quot; may be upvoted if they're witty/entertaining but won't be marked as correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NB&lt;/strong&gt;: Searching the entire possible space of jug-pouring actions is clearly &lt;em&gt;a&lt;/em&gt; solution. But it isn't a very interesting or elegant one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=lZ64IR2bz5o&quot; rel=&quot;noreferrer&quot;&gt;Die Hard With A Vengeance&lt;/a&gt;: the 'decanting problem' scene.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://forums.udacity.com/questions/100002085/when-does-the-water-pour-problem-have-no-solution&quot; rel=&quot;noreferrer&quot;&gt;Udacity: When does the water pour problem have no solution?&lt;/a&gt; (which seems to show a necessary but IMO not sufficient condition for there to be a solution).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Diophantine_equation&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia: Diophantine equation&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://mathworld.wolfram.com/ThreeJugProblem.html&quot; rel=&quot;noreferrer&quot;&gt;Weisstein, Eric W. &quot;Three Jug Problem.&quot; From MathWorld--A Wolfram Web Resource.&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2021-11-19T23:19:47.890" Title="A general solution to the decanting problem? (aka jug-pouring, water-pouring)" Tags="|mathematics|algorithm|number-theory|decanting-problems|" AnswerCount="3" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="8253" PostTypeId="1" CreationDate="2015-01-27T12:56:48.120" Score="3" ViewCount="132" Body="&lt;p&gt;In a monastery, there is a special rite to elect new novices. All candidates (we're talking of a huge number, so let's assume an infinite amount) gather together in one hall that gets locked, so that nobody can get in or out. There is no communication with the outside at all.&lt;br&gt;&#xA;Now the magic starts and the elder monks perform a spell somewhere outside this room, which makes a symbol appear on the forehead of each candidate that will become a novice. No one can see his own forehead, and there are no mirroring surfaces in the hall. Talking or giving signs is strictly prohibited!&lt;br&gt;&#xA;So after this has happened, the candidates have the following task:&lt;br&gt;&#xA;Every full hour the door will be opened. Then there are two possible options. Either ALL novices with the symbol on their forehead leave the room together, or nobody leaves the room at all. They have to finish the rite in as little time as possible, but without breaking any rules, which would have bad (very bad, I assure you) consequences for the candidates...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now your task is: There is a random number &lt;strong&gt;n&lt;/strong&gt; between 1 (inclusively)  and the amount of candidates (almost infinite) of newly elected novices that have the sign appear on their forehead. How many hours &lt;strong&gt;h&lt;/strong&gt; will the rite take at least, depending on the value of &lt;strong&gt;n&lt;/strong&gt;? Or the other way round: When the novices leave the room after &lt;strong&gt;h&lt;/strong&gt; hours, how many novices got elected?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should be a pretty easy question. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="9190" LastActivityDate="2015-01-27T12:56:48.120" Title="Magical election of novices in a monastery" Tags="|logical-deduction|strategy|" AnswerCount="0" CommentCount="3" ClosedDate="2015-01-27T22:15:59.473" ContentLicense="CC BY-SA 3.0" />
  <row Id="8255" PostTypeId="2" ParentId="8248" CreationDate="2015-01-27T13:24:09.563" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;$11 - 1$ takes 7 sticks. &lt;/li&gt;&#xA;&lt;li&gt;$9 \times 5$ takes 13 sticks.&lt;/li&gt;&#xA;&lt;li&gt;$51 - 1$ takes 10 sticks. &lt;/li&gt;&#xA;&lt;li&gt;$11 \times 9$ takes 12 sticks. &lt;/li&gt;&#xA;&lt;li&gt;$5^3$ would take 10, but if not, then $121 + 4$ takes 15, I think. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Still working on a program to optimize it, but that could take a bit.&lt;/p&gt;&#xA;" OwnerUserId="1664" LastEditorUserId="1664" LastEditDate="2015-01-27T13:41:07.103" LastActivityDate="2015-01-27T13:41:07.103" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8256" PostTypeId="2" ParentId="8248" CreationDate="2015-01-27T13:26:43.163" Score="0" Body="&lt;p&gt;First off, in your example, you used $10\times10\times10$ which seems to me that we can make numbers with multiple digits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually, the definition of an expression doesn't require the use of an operator, so I would make the following trivial initial guesses:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;10: 8 matchsticks - &quot;$10$&quot;&lt;/li&gt;&#xA;&lt;li&gt;45: 9 matchsticks - &quot;$45$&quot;&lt;/li&gt;&#xA;&lt;li&gt;50: 11 matchsticks - &quot;$50$&quot;&lt;/li&gt;&#xA;&lt;li&gt;99: 12 matchsticks - &quot;$99$&quot;&lt;/li&gt;&#xA;&lt;li&gt;125: 12 matchsticks - &quot;$125$&quot;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;But none of these answers is very interesting.  In fact, some of them aren't even optimal!  So I will assume that we need to include operators.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;10: 7 matchsticks - &quot;$11-1$&quot;&lt;/li&gt;&#xA;&lt;li&gt;45: 13 matchsticks - &quot;$5 \times 9$&quot;&lt;/li&gt;&#xA;&lt;li&gt;50: 10 matchsticks - &quot;$51 - 1$&quot;&lt;/li&gt;&#xA;&lt;li&gt;99: 12 matchsticks - &quot;$11 \times 9$&quot;&lt;/li&gt;&#xA;&lt;li&gt;125: 16 matchsticks - &quot;$11 \times 11 + 4$&quot;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2015-01-27T13:26:43.163" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8257" PostTypeId="2" ParentId="8248" CreationDate="2015-01-27T13:47:19.303" Score="8" Body="&lt;ol&gt;&#xA;&lt;li&gt;(10) 7 matchsticks: $11-1$&lt;/li&gt;&#xA;&lt;li&gt;(45) 12 matchsticks: $44 + 1$&lt;/li&gt;&#xA;&lt;li&gt;(50) 10 matchsticks: $51-1$&lt;/li&gt;&#xA;&lt;li&gt;(99) 12 matchsticks: $11\times9$&lt;/li&gt;&#xA;&lt;li&gt;(125) 14 matchsticks: $114+11$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Total: 55 matchsticks&lt;/p&gt;&#xA;" OwnerUserId="8992" LastEditorUserId="8992" LastEditDate="2015-01-28T14:45:21.537" LastActivityDate="2015-01-28T14:45:21.537" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8258" PostTypeId="2" ParentId="7987" CreationDate="2015-01-27T14:09:57.847" Score="9" Body="&lt;p&gt;x=16 with the following order:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A BB CCC DDDD EEEEE AAAA BBB CC D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now I can also show that x=16 is best possible:&#xA;Order the players from left to right by increasing number (as in my solution shown above). Put for every team the middle player in boldface (so that there are two players from the same team to its left and two players from the same team to its right):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A BB CC&lt;strong&gt;C&lt;/strong&gt; DD&lt;strong&gt;D&lt;/strong&gt;D EE&lt;strong&gt;E&lt;/strong&gt;EE A&lt;strong&gt;A&lt;/strong&gt;AA &lt;strong&gt;B&lt;/strong&gt;BB CC D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Consider the rightmost of the five boldface players.&#xA;If this rightmost boldface is from team A then he and his to team mates to the right lose all their matches against the three leftmost players from team B (the boldface player from B and his two team mates to the left of him). Similar and symmetric arguments work, if the rightmost boldface is from teams B,C,D,E. Altogether this means that team A loses at least nine matches against B, and can win at most 16 matches against E. Therefore x&amp;lt;=16&lt;/p&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="9125" LastEditDate="2015-01-31T11:45:08.533" LastActivityDate="2015-01-31T11:45:08.533" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8259" PostTypeId="2" ParentId="8076" CreationDate="2015-01-27T14:25:19.230" Score="3" Body="&lt;p&gt;Here is one way it could work:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Begin with 3 green apples and 12 blue apples.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This meets the condition that there are less than 10 times the number of blue apples than green apples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add one blue apple:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We now have 3 green apples, 13 blue apples&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Double the number of green apples:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We now have 6 green apples, 13 blue apples&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Use the doubling spell now on all the apples:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We now have 12 green apples, 26 blue apples&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add one blue apple:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We now have 12 green apples, 27 blue apples&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One apple changes colour:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A blue apple turns green, we now have 13 green apples, 26 blue apples&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Divide the bigger group in half:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We now have 13 green apples, 13 blue apples&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You eat one of each colour:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are now 12 green apples, 12 blue apples, and maybe you have been poisoned.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1616" LastActivityDate="2015-01-27T14:25:19.230" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8262" PostTypeId="2" ParentId="8252" CreationDate="2015-01-27T18:21:20.760" Score="14" Body="&lt;p&gt;Your link to Diophantine equations shows that you already have an idea of the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only actions available are filling a jug completely from the tub, emptying a jug completely, or pouring one jug into another. If we fill empty jug $k$ from the tub, then the amount of water we have is increased by $v_k$. Similarly, if we empty jug $k$ when it is full, we decrease the amount of water we have measured by $v_k$. If we empty a jug into another, the amount of water doesn't change. However, if jug $l$ is full, and we pour it into jug $m$, when $v_l&amp;gt;v_m$, then the amount of water left in jug $l$ is $v_l-v_m$, and emptying it will decrease the amount of water we have by that amount. From this, we can see that the only effect we can have on the total amount of water is to add or subtract $v_n$ repeatedly, for each $n\in[1,j]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that in order to get a total amount of water $n$ (you've used the same variable name for two things, tsktsk) there must be integers $z_1, z_2,\ldots z_j$ so that $z_1v_1+z_2v_2+\cdots+z_jv_j=n$. However, this is not a sufficient condition. For example, if your largest jug has volume $v_k$, and $n\gg v_k$, then you won't have anywhere to &lt;em&gt;put&lt;/em&gt; the water you are measuring. The restrictions on the magnitude of $n$ for making a necessary and sufficient condition are tricky, so I'll get to that later. For now, let's suppose that we also have an infinite empty tub to put our water in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given volumes $v_1$ to $v_j$ and a goal $n$, the equation $z_1v_1+z_2v_2+\cdots+z_jv_j=n$, or $\sum_{k=1}^jz_kv_k=n$, is a Diophantine equation, because we are looking for integer solutions. So you could simply say that the problem is solvable when this equation has a solution. However, that isn't especially helpful, as it is just restating the question in an abstracted way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if $d$ is the smallest possible positive value of $\sum_{k=1}^jz_kv_k$, then $d=\gcd(v_1,v_2,\ldots,v_j)$, by &lt;a href=&quot;http://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity&quot;&gt;Bézout's identity&lt;/a&gt;. Furthermore, every number that can be expressed this way (as an integer-multiple linear combination of $v$'s) is a multiple of $d$. Therefore, we have the requirement that $$\gcd(v_1,v_2,\ldots,v_j)\ \textrm{ divides }\ n.$$&#xA;If we assume the empty tub, this is a necessary and sufficient condition. It is especially nice because there is already an algorithm for computing the gcd, and the minimal coefficients of the corresponding linear combination: the extended Euclidean algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given two integers $a$ and $b$, we can compute $\gcd(a,b)$ and integers $x$ and $y$ so that $ax+by=\gcd(a,b)$ as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $r_0=a, r_1=b, s_0=1, s_1=0, t_0=0, $ and $t_1=1$. Then, for any index $i$, find $q_{i+1}, r_{i+1}$ so that $r_{i-1}=q_{i+1}r_i+r_{i+1}$ and $0\le r_{i+1}&amp;lt;r_i$. That is, $g_{i+1}=\left\lfloor\dfrac{r_{i-1}}{r_i}\right\rfloor$ and $r_{i+1}=r_{i-1}\mod r_i$. Also, let $s_{i+1}=s_{i-1}-q_{i+1}s_i$ and $t_{i+1}=t_{i-1}-q_{i+1}t_i$. When you reach $r_{i+1}=0$, stop. At this point, you have $\gcd(a,b)=r_i, x=s_i,$ and $y=t_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This only finds the gcd of two numbers, however, $\gcd(a,b,c)=\gcd(\gcd(a,b),c)$, so you can apply this algorithm repeatedly, storing the coefficients at each stage. Also, you don't always need to test all of the jugs' volumes. As soon as you reach a point where $n$ is a multiple of the gcd you have found, you can stop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I will finish this answer later, including how to turn the linear combination back into a jug-pouring sequence.)&lt;/p&gt;&#xA;" OwnerUserId="8776" LastActivityDate="2015-01-27T18:21:20.760" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8263" PostTypeId="1" CreationDate="2015-01-27T19:03:58.283" Score="2" ViewCount="316" Body="&lt;p&gt;The guy who recited 20k numbers of pi by memory alone used a system where he converted numbers into consonants and then grouped words or sentences together to remember a huge amount of numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how it works: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1:  t, th, d   (single downstroke) Ex: Take, THe, Dog   &#xA;2:  n   (two downstrokes) Ex: Nose  &#xA;3:  m   (three downstrokes) Ex: Money   &#xA;4:  r   (the last letter of “four” is “r”) Ex: Road&#xA;5:  L   (you have five fingers on your Left hand) Ex: Life, driLL   &#xA;6:  j, soft g, ch, sh   (“J” is a near mirror-image of “6”) Ex: Jelly, CHips, garaGe, SHoe  &#xA;7:  k, hard g, c, or x  (“7” side-by-side with a mirror image form a sideways “K”) Kite, Goat, Cat, eXcel   &#xA;8:  f, v, ph    (“8” is similar to the lower-case cursive “f”) Ex: Flame, eFFect, Vest, graPH   &#xA;9:  p, b    (9 is a mirror-image of “P”) Ex: Pen, Ball  &#xA;0:  z, s, soft c or x   (0 signifies “zero”) Ex: Zipper, Scarf, eXample, iCe&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Vowels don't count, double consonants only count once, and you go by the pronunciation rather than the spelling.  H, q, and w don't count, and x is listed here but it can get confusing with words like excellent, so it's best to use unambiguous words.  I hope I explained it well enough!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I like phi better than pi, so for this I'll be using phi.  You can use whatever numbers you want, but phi is the most irrational number and we can do this forever or until we're bored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, the goal is to come up with sentences that make as much sense as possible, so it becomes easier to remember them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what I have so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.61803398874989484820458683436563811772030917980576286213544862270526046281890244970720720418939113748475408807538689175212663386222353693179318006076672635&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;They shot off some map of a fake rope fiber.&#xA;If ravens are alive, show off my room, Jill.&#xA;She may fight to cook.&#xA;Nice mice bit a cop; face leakage.&#xA;An efficient mall; rare fashion.&#xA;Nick's lunch is rich; no vat of poison.&#xA;Rare picks... coon skins are tough.&#xA;Be my petite maker for a gal or save face.&#xA;Cool move, chief bought kale in a teenage sham.&#xA;My fashion—an animal.&#xA;My shop might keep my tough sausage, so I cash a check.&#xA;Nacho meal.&#xA;&#xA;They = 1&#xA;shot = 61&#xA;off = 8&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;...and so on.  You can see some of these don't make much sense.  Have fun!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a place to get all the numbers you'll ever need:&#xA;&lt;a href=&quot;http://www.goldennumber.net/phi-million-places/&quot; rel=&quot;nofollow&quot;&gt;http://www.goldennumber.net/phi-million-places/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8788" LastActivityDate="2015-01-27T21:24:12.387" Title="Memory Enhancement Through Numeric Imagery - What is the longest or best coherent sentence/paragraph/story you can make?" Tags="|open-ended|language|wordplay|" AnswerCount="1" CommentCount="6" ClosedDate="2021-05-10T13:25:02.020" ContentLicense="CC BY-SA 3.0" />
  <row Id="8264" PostTypeId="2" ParentId="8223" CreationDate="2015-01-27T19:05:07.917" Score="4" Body="&lt;p&gt;Sentence: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;I buy my stark musk, roast rusk in milk, dark blue bow, reed, bat, wind bell, stud, two lemon on my windy way.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Length: Letters=&lt;strong&gt;80&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Blank space= &lt;strong&gt;22&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comma= &lt;strong&gt;7&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Full stop=1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;n=&lt;strong&gt;5&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Letters used=&lt;strong&gt;16&lt;/strong&gt; (a b d e i k l m n o r s t u w y)&lt;/p&gt;&#xA;" OwnerDisplayName="user9174" LastEditorDisplayName="user9174" LastEditDate="2015-01-28T05:56:45.713" LastActivityDate="2015-01-28T05:56:45.713" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8265" PostTypeId="2" ParentId="8249" CreationDate="2015-01-27T19:28:30.373" Score="3" Body="&lt;p&gt;Based on the answer by &lt;a href=&quot;https://puzzling.stackexchange.com/users/5610/dmg&quot;&gt;dmg&lt;/a&gt;, our plan is to pick a number $a$, start going to the right, and turn around the first time we reach door $x = (-a)^i$.  Thus, for $a = 2$, we would turn around at $x = 1, -2, 4, -8, ...$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate how many steps it takes to reach a turnaround door, we assume a linear function $f(x) = r|x| - s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To go from door $x$ to door $a^2 x$, we go back $|x|$ doors to $0$, then $a|x|$ further, then turn around, $a|x|$ back to $0$, then on $a^2|x|$ more.  This distance is $|x| + 2a|x| + a^2|x| = |x| (a+1)^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But &#xA;$$f(a^2x) - f(x) = r | a^2 x| - s - (r|x| - s) = r (a^2-1) |x|$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So,&#xA;$$ r (a^2-1) |x| =|x| (a+1)^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dividing out $|x|$ and using $a^2-1 = (a+1)(a-1)$, we have:&#xA;$$r = {a+1\over a-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we always turn around at $x = 1$, we have $s = r-1$.  A bit of testing shows that our assumption of a linear function works, and thus for turnaround doors $x = (-a)^i$, the number of steps to reach that door is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = {a+1\over a-1} |x| - {2 \over a-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, for $a = 2$, we have $r = 3$ and $s = 2$, so $f(x) = 3|x| -2$ (whenever $x = (-a)^i)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The worst case result is 1 door past a turn-around, because we almost reach it, go back, come forward again, and then 1 more step.  This takes a total of $f(x) + (2a+2)|x| + 1$ steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the number of steps to reach the door just past turnaround door $x$ is:&#xA;$$g(x) = f(x) + (2a+2)|x| + 1$$&#xA;$$   = \left({a+1\over a-1} + 2a + 2\right) |x| + {a-3\over a-1}$$&#xA;steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setting $g(x) = w|x| + t$, we have&#xA;$$w = {a+1\over a-1} + 2a + 2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $a = 2$, we have $w = 9$, and $g(x) = 9|x| - 1$.  But the money behind this door is only $8.999(|x|+1)$ euro, so if $9|x|-1 &amp;gt; 8.999(|x|+1)$, we lose.  The first time this happens is for $x = 16384$.  It takes 147455 steps to reach door 16385, and behind it is only 147448 euros.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we do better than $a = 2$?&#xA;For $a = 3$, we have $r = 2$ (better), $s = 1$, and $w = 10$ (worse).  Likewise, for $r = 4$, we have $r = 5/3$, $s=2/3$, and $w = 11.67$.  Clearly increasing $a$ will not help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first derivative of $w$ in terms of $a$ is:&#xA; $$w' = {-2 \over (a-1)^2} + 2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving $w' = 0$ gives&#xA;$${-2 \over (a-1)^2} = -2$$&#xA;$$1 = (a-1)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or $a = 0,2$.  The second derivative is positive at $a = 2$, so that makes this a minimum.  Thus, there is no solution where $w &amp;lt; 9$, and thus no solution to the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Caveat: I have not considered non-exponential solutions.  However, I expect the exponential solution to be the most efficient.  Any other order could be approximated by an exponential with varying $a$, and thus either $a &amp;gt; 2$ or $a &amp;lt; 2$, and would therefore be suboptimal compared to an exponential with $a = 2$.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-01-27T19:44:20.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8266" PostTypeId="2" ParentId="8175" CreationDate="2015-01-27T20:24:54.620" Score="6" Body="&lt;p&gt;It is also possible with rectangles.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;         &lt;img src=&quot;https://i.stack.imgur.com/HY00N.png&quot; alt=&quot;square dissection&quot;&gt;   or   &lt;img src=&quot;https://i.stack.imgur.com/bFTEJ.png&quot; alt=&quot;square dissection&quot;&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2015-02-01T00:03:47.393" LastActivityDate="2015-02-01T00:03:47.393" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8267" PostTypeId="2" ParentId="8263" CreationDate="2015-01-27T20:52:26.143" Score="2" Body="&lt;p&gt;Good lines here are what works well for the person doing the memorising. And the first stage in fixing them in your memory is to make them up yourself. It won't do to use someone else's efforts, at least not when you're trying to memorise something quite long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, people who are seriously into memorisation use prepared lists for numbers up to 100 or 1000 or 10000 and then string them together to make stories. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a quick go at the first 24 digits:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1--6--1----8--0---3----3----9&#xA;Teach to a fish: &quot;Moo, moo, baa!&quot;&#xA;&#xA;8----8------7-4--9-8----9--4&#xA;Fie! Fie! A car above a pear!&#xA;&#xA;8-4---8--2----0--4---58&#xA;Fury! Phone a shore elf!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7363" LastEditorUserId="7363" LastEditDate="2015-01-27T21:24:12.387" LastActivityDate="2015-01-27T21:24:12.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8268" PostTypeId="2" ParentId="8162" CreationDate="2015-01-27T21:15:52.270" Score="6" Body="&lt;p&gt;And out-of-the-box solution.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Make 2 roughly equal piles and place them standing one against the other.  No card faces down, they all face the one or the other side.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2136" LastActivityDate="2015-01-27T21:15:52.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8269" PostTypeId="2" ParentId="8212" CreationDate="2015-01-27T22:32:06.583" Score="1" Body="&lt;p&gt;I have 2 possible answers for the four 6-letter words. Both were obtained by parsing a scrabble 6-letter word list, picking out words with few vowels, then checking to make sure there were 24 distinct letters.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;blowzy, frumps, jading, kvetch&#xA;frowzy, jading, kvetch, plumbs&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I attempted the same thing with five 5-letter words, but did not get a match.&lt;/p&gt;&#xA;" OwnerUserId="375" LastActivityDate="2015-01-27T22:32:06.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8270" PostTypeId="1" CreationDate="2015-01-27T23:31:07.403" Score="10" ViewCount="3747" Body="&lt;p&gt;I saw this puzzle and don't know how to solve it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7qAWJ.png&quot; alt=&quot;puzzle self&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only pattern I saw is 3 + 8 = 11 , 2 + 9 = 11, 6 + 2 = 8 and 1 + 7 = 8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there other approaches I should consider?&lt;/p&gt;&#xA;" OwnerUserId="9213" LastEditorUserId="11876" LastEditDate="2020-03-30T18:38:40.793" LastActivityDate="2020-03-31T01:58:55.153" Title="Numbers in quadrants of a circle - help me solve it" Tags="|pattern|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8272" PostTypeId="2" ParentId="8216" CreationDate="2015-01-28T01:01:42.550" Score="0" Body="&lt;p&gt;Since this is a digital clock, the leftmost digit only has six elements, which can display either 1 or 2. The upper left vertical segment is missing on clock displays. Therefore, the leftmost digit can only display 1, 2, 3, or 7. The lack of facility to display a 0 in the most significant position in a digital clock means the question &lt;em&gt;as asked&lt;/em&gt; has no valid answer. Oh sure, 24-hour clocks. Pff. Who has &lt;em&gt;them&lt;/em&gt;?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Taking a four-gang seven-segment display, with zero padding, the total number of segments is 28. With zero padding (not disqualified by the questioner) this means 0022 (6+6+5+5) is the highest number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other terms: Sa+Sb+Sc+Sd = 1000*A + 100*B + 10*C + D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S(0) + S(0) + S(2) + S(2) = 1000*0 + 100*0 + 10*2 + 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6 + 6 + 5 + 5 = 1000*0 + 100*0 + 10*2 + 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;22&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Taking a four-gang seven-segment display, with zero padding and a colon (as illustrated, although the missing segment in the second digit is a bit off), the total number of &lt;em&gt;elements&lt;/em&gt; is 30. The highest number of elements for a display beginning 00:2 displays 00:28 and lights 6+6+2+5+7=26 elements.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;00:27 lights 6+6+2+5+3=22&lt;/li&gt;&#xA;&lt;li&gt;00:26 lights 6+6+2+5+6=25&lt;/li&gt;&#xA;&lt;li&gt;00:25 lights 6+6+2+5+5=24&lt;/li&gt;&#xA;&lt;li&gt;00:24 lights 6+6+2+5+4=23&lt;/li&gt;&#xA;&lt;li&gt;00:23 lights 6+6+2+5+5=24&lt;/li&gt;&#xA;&lt;li&gt;00:22 lights 6+6+2+5+5=24&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There's no point going lower, since the colon in such a display flashes on and off, and so the highest number is again as above: 22. However, with the colon lit:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;00:21 lights 6+6+2+5+2=21&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This means that, with a flashing colon that is lit at 59 seconds and unlit at 0 seconds, a digital clock will display two consecutive &quot;correct&quot; answers at 21 and 22 minutes past midnight.&lt;/p&gt;&#xA;" OwnerUserId="9214" LastActivityDate="2015-01-28T01:01:42.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8273" PostTypeId="1" CreationDate="2015-01-28T01:08:19.617" Score="14" ViewCount="1133" Body="&lt;p&gt;One afternoon, you start writing some whole numbers on a (very large) sheet of paper. First you write a one-digit number, then a two-digit, and so on, with each number one digit longer than the last. Every number is written in base 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make this otherwise dull activity more interesting, you decide never to write a number using all of the digits in any previously written number (in any order).&#xA;It easy to write numbers at first, but it gets harder and harder as you go.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Will you ever be forced to stop writing, and if so how many numbers might you need to write before this happens?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="8874" LastEditDate="2016-03-23T14:54:51.420" LastActivityDate="2016-03-23T14:54:51.420" Title="How long can this list of numbers get?" Tags="|mathematics|strategy|game|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8274" PostTypeId="2" ParentId="8273" CreationDate="2015-01-28T02:44:17.737" Score="18" Body="&lt;p&gt;I think that the list must terminate, but a strategy like this is optimal:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1&#xA;22&#xA;233&#xA;2344&#xA;23455&#xA;234566&#xA;2345677&#xA;23456788&#xA;234567899&#xA;2345678900&#xA;23456780000&#xA;234567999999&#xA;2345679999900&#xA;23456799990000&#xA;234567999000000&#xA;2345679900000000&#xA;23456790000000000&#xA;234567000000000000&#xA;2345688888888888888&#xA;23456888888888888899&#xA;234568888888888888900&#xA;...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Each time, remove a copy of the largest digit present (call it x). If x is 9, add two 0s to the end of the number. Otherwise, remove all 0s and fill the rest of the length with (x+1)s. This results in a list with a length on the order of A(10,1), where A is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ackermann_function&quot;&gt;Ackermann function&lt;/a&gt;. This is because f(x,y), defined as the length of a list of expanding numbers using this strategy where x is the number of digits available and y is the length of the first entry obeys a recursion very similar to the definition of the Ackermann function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no proof that the list can't continue indefinitely, but it might be related to the &quot;hydra game&quot;.&lt;/p&gt;&#xA;" OwnerUserId="9215" LastActivityDate="2015-01-28T02:44:17.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8276" PostTypeId="2" ParentId="8208" CreationDate="2015-01-28T06:21:52.560" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ASSASSIN - GIORDANO, VITO, 456 AINSWORTH&#xA; &lt;BR&gt;&lt;BR&gt;&#xA; 4149097910918443572 (10) = VITOGIORDAO8 (36)&#xA; &lt;BR&gt;&lt;BR&gt;&#xA; It is all about converting from Decimal to Hexatrigesimal system. Unfortunately, I have no idea why base 36 was chosen. It was just the largest possible system that online converter was capable of. So, I found this solution by pure luck.&#xA; &lt;BR&gt;&lt;BR&gt;&#xA; Piazza di Sei = Squared six = 36, Great!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9088" LastEditorUserId="9088" LastEditDate="2015-01-29T05:02:42.540" LastActivityDate="2015-01-29T05:02:42.540" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8277" PostTypeId="2" ParentId="8270" CreationDate="2015-01-28T06:49:17.880" Score="6" Body="&lt;p&gt;I don't think is it but could be a start.. A bit like what you pointed out in the answer,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The numbers in the fifth circle show the difference between the corresponding quarter of the top-right and bottom-left circles. I.e. 8 - 1 = 7, 9 - 3 = 6, 2 - 4 = (-)2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So maybe the question mark should be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 5   (which is 7-2)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because it fits this pattern, but that doesn't feel like a good answer.&lt;/p&gt;&#xA;" OwnerUserId="9062" LastActivityDate="2015-01-28T06:49:17.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8278" PostTypeId="2" ParentId="8270" CreationDate="2015-01-28T07:35:28.093" Score="3" Body="&lt;p&gt;All I found was a diagonal reference. You build the two differences (always positive) and chose the smaller one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7V5HD.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for top left this would be '8' and &lt;strong&gt;'7'&lt;/strong&gt;,&lt;br&gt;&#xA;top right '9' and &lt;strong&gt;'6'&lt;/strong&gt;,&lt;br&gt;&#xA;bottom left &lt;strong&gt;'2'&lt;/strong&gt; and again &lt;strong&gt;'2'&lt;/strong&gt; (is this a sign?)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and following this rules, bottom right would be '7' and &lt;strong&gt;'5'&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5106" LastActivityDate="2015-01-28T07:35:28.093" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8282" PostTypeId="2" ParentId="8273" CreationDate="2015-01-28T11:56:19.770" Score="5" Body="&lt;p&gt;I suspect this can be proven by the theory of posets. Here's a sketch/outline of a proof, though it's missing the key proof ingredient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can associate each number with a vector $v$ in $\mathbb{N}^{10}$ whose entries $(v_0,v_1,\dots,v_9)$ are non-negative integers that count the number of each digit $0$ through $9$ that appears. We have the condition that no entry $w$ dominates another entry $v$, meaning has every entry at least as large $w_i\geq v_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's think of $\mathbb{N}^{10}$ as a &lt;a href=&quot;http://en.wikipedia.org/wiki/Partially_ordered_set&quot;&gt;partially ordered set (poset)&lt;/a&gt; under the domination relation $\preceq$. In fact, it is exactly the poset Cartesian product $\mathbb{N}\times\cdots\times \mathbb{N}$ of $10$ copies of $\mathbb{N}$, the ordered set of natural numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove there's no infinite sequence as required, it would prove to show that $\mathbb{N}^{10}$ has no infinite antichain, which is a collection of elements no two of which are comparable. We'll call this property &lt;em&gt;finite-width&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We'll proceed by induction. It's clear that $\mathbb{N}$, being totally ordered, has finite width.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;False claim:&lt;/em&gt; If $A$ and $B$ are finite-width posets, then so is $A\times B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its falsity is witnessed by the example of $A=\mathbb{N}$ and $B=\mathbb{\bar{N}}$, the natural numbers under the reversed order. There, the set of all $\{(i,i)\}$ is an infinite antichain. So, we'll need some stronger condition to make this carry through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this condition is that the natural number are well-ordered, meaning that they have no infinite downward chains, which are infinite sequences of elements, each smaller than the previous. The natural numbers have this property, but the reversed natural numbers do not. The property is preserved by product&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Claim:&lt;/em&gt;  If $A$ and $B$ are well-ordered posets, then so is $A\times B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is easy to see in that any infinite downward chain in $A\times B$ corresponds to an infinite downward chain in $A$ (and also $B$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $\mathbb{N}^{k}$ is a well-ordered poset for any finite $k$. I believe this suffices to prove the claim we want, but don't yet have a proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Conjecture:&lt;/em&gt; If $A$ and $B$ are well-ordered with finite width, then so is $A\times B$ has finite width.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-01-28T11:56:19.770" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8283" PostTypeId="1" AcceptedAnswerId="8478" CreationDate="2015-01-28T12:08:33.147" Score="17" ViewCount="1998" Body="&lt;p&gt;To clear some confusion, this is just a riddle. I think maybe the size is whats intimidating some, but it can be solved as you would any other riddle. There are no new techniques used here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;atop the monument seeking jewellery&lt;br&gt;&#xA;this fake character is not the ordinary&lt;br&gt;&#xA;a trio, the misguided claim to possess&lt;br&gt;&#xA;their psychiatric health we must assess&lt;br&gt;&#xA; &lt;br&gt;&#xA;watching you now, locked in solitude&lt;br&gt;&#xA;lose thoughts of it and gain gratitude&lt;br&gt;&#xA;data from it to be deciphered&lt;br&gt;&#xA;easy comprehension is what's desired&lt;br&gt;&#xA; &lt;br&gt;&#xA;to see your path, you favour no grin&lt;br&gt;&#xA;just follow the road by the grimalkin&lt;br&gt;&#xA;the agreeable ones pillage upon the salt&lt;br&gt;&#xA;those who fare, through billow upon holt&lt;br&gt;&#xA; &lt;br&gt;&#xA;low warm and high cool, the air will fight&lt;br&gt;&#xA;stay close within to avoid the smite&lt;br&gt;&#xA;as the exterior rips through the land&lt;br&gt;&#xA;the inner recess is tranquil and grand   &lt;/p&gt;&#xA;&#xA;&lt;h2&gt; &lt;/h2&gt;&#xA;&#xA;&lt;p&gt;shot, the projectile changes thee&lt;br&gt;&#xA;elude, the meaning will till you guarantee&lt;br&gt;&#xA;no material can heal what once was neat&lt;br&gt;&#xA;subtle memories, dread a repeat   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the drum of life does symbolise&lt;br&gt;&#xA;one does not simply analyse&lt;br&gt;&#xA;broken only, lonely alone&lt;br&gt;&#xA;rejection wrecking leaving stone   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;yet true twice in benevolent violence&lt;br&gt;&#xA;one look only for it gains entrance&lt;br&gt;&#xA;an entity for life you obtain&lt;br&gt;&#xA;create anew a life to reign&lt;br&gt;&#xA; &lt;br&gt;&#xA;compelled by this to achieve&lt;br&gt;&#xA;yet humility intervenes, you don’t believe&lt;br&gt;&#xA;effects differ and subjects vary&lt;br&gt;&#xA;listen to this its not that scary   &lt;/p&gt;&#xA;&#xA;&lt;h2&gt; &lt;/h2&gt;&#xA;&#xA;&lt;p&gt;the board rides the pipe into the air&lt;br&gt;&#xA;high as a hawk, its quite a scare&lt;br&gt;&#xA;thou shalt address thee as a clue&lt;br&gt;&#xA;be thy plumber of thy loo&lt;br&gt;&#xA; &lt;br&gt;&#xA;material of squander travels around&lt;br&gt;&#xA;within the confines, trapped and bound&lt;br&gt;&#xA;nowhere to go the water rises&lt;br&gt;&#xA;break free quickly no nasty surprises   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Germans sink within this vessel&lt;br&gt;&#xA;under the radar hoping to nestle&lt;br&gt;&#xA;place what you need before the knee&lt;br&gt;&#xA;take the verse and have some tea   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the Gothic did state the differences&lt;br&gt;&#xA;upon the vague instances&lt;br&gt;&#xA;for we were once another&lt;br&gt;&#xA;our rather angular brother   &lt;/p&gt;&#xA;" OwnerUserId="5135" LastEditorUserId="8874" LastEditDate="2015-02-02T09:03:12.333" LastActivityDate="2015-02-05T15:30:13.233" Title="The hardest string of words to verbalise in english" Tags="|riddle|english|" AnswerCount="2" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="8285" PostTypeId="1" AcceptedAnswerId="8295" CreationDate="2015-01-28T13:59:09.243" Score="14" ViewCount="1633" Body="&lt;p&gt;You are asked to place matchsticks on a flat surface such that each matchstick end meets three others, and no matches cross. It is easy to achieve this for patterns that extend indefintely:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/n5bee.jpg&quot; alt=&quot;flat matchstick pattern with each end meeting three others&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The challenge is to truncate such patterns to finite 2D networks. How small a matchstick network can you create?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Further clarifications:&lt;/strong&gt; the matchsticks all have equal length and can be thought of as mathematical line segments. At each point of contact, exactly four ends meet. All matches lay flat on the surface, no gluing allowed!&lt;/p&gt;&#xA;" OwnerUserId="7573" LastEditorUserId="7573" LastEditDate="2015-01-31T20:00:47.023" LastActivityDate="2015-09-12T19:40:40.877" Title="Touching Matchsticks" Tags="|mathematics|geometry|matches|graph-theory|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8286" PostTypeId="1" AcceptedAnswerId="8310" CreationDate="2015-01-28T14:34:50.293" Score="3" ViewCount="2756" Body="&lt;p&gt;&lt;strong&gt;You and 9 friends are playing a game, is there a way to force a win, if so, how, if not, what is the most optimal strategy?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Gameplay:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;em&gt;Each player can have a turn, or if he/she wishes, 2 turns. Each turn writing a number greater than the number wrote on the previous turn, it has to be &lt;strong&gt;1&lt;/strong&gt; greater.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How to lose:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;em&gt;Write the number 10!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How to win:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;em&gt;Be the last one not to write the number 10!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clarifications:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Turns &lt;strong&gt;do&lt;/strong&gt; cycle&lt;/li&gt;&#xA;&lt;li&gt;After writing 10 you restart at 1&lt;/li&gt;&#xA;&lt;li&gt;Players are for themselves, nobody wishes to team up.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example game:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Friend A writes 1&lt;/li&gt;&#xA;&lt;li&gt;Friend A writes 2&lt;/li&gt;&#xA;&lt;li&gt;Friend B writes 3&lt;/li&gt;&#xA;&lt;li&gt;Friend C writes 4&lt;/li&gt;&#xA;&lt;li&gt;Friend C writes 5&lt;/li&gt;&#xA;&lt;li&gt;Friend D writes 6&lt;/li&gt;&#xA;&lt;li&gt;Friend D writes 7&lt;/li&gt;&#xA;&lt;li&gt;Friend E writes 8&lt;/li&gt;&#xA;&lt;li&gt;Friend E writes 9&lt;/li&gt;&#xA;&lt;li&gt;Friend F writes 10! He's out!&lt;/li&gt;&#xA;&lt;li&gt;Friend G writes 1&lt;/li&gt;&#xA;&lt;li&gt;etc. until one player is left.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus:&lt;/strong&gt; &lt;em&gt;Can you extend this strategy for any amount of players, how?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Notes:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;em&gt;I'd like to let you know, I'm genuinely looking for an answer. I do not know it.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="8874" LastEditDate="2015-02-02T09:02:46.593" LastActivityDate="2015-02-02T09:02:46.593" Title="10 friends play a number game" Tags="|mathematics|strategy|game|" AnswerCount="4" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="8288" PostTypeId="5" CreationDate="2015-01-28T14:40:26.880" Score="0" Body="&lt;p&gt;Question is about the connecting and joining of 2D or 3D lines of unit-length. A commonly used object for visualizing are matchsticks, but could as well be pencils, pens, sticks, or even needles. The matchsticks may be moved around or removed, or some may be added, to achieve some goal.&lt;/p&gt;&#xA;&lt;p&gt;Some kinds of puzzles using this tag:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Turn one figure into another with a specific number of moves: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/17172/turning-a-goat&quot;&gt;Turning a goat?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Make a mathematical equation true with a specific number of moves: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/98744/9-5-5-matchstick-problem&quot;&gt;9-5=5? Matchstick problem&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Minimize the number of matches removed to achieve a certain goal: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/107502/how-many-matchsticks-need-to-be-removed-so-there-are-no-equilateral-triangles&quot;&gt;How many matchsticks need to be removed so there are no equilateral triangles?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Maximize the number created by moving a certain number of matches: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/100739/move-matches-and-make-largest-possible-number&quot;&gt;Move matches and make largest possible number&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="69582" LastEditDate="2021-03-10T06:51:00.223" LastActivityDate="2021-03-10T06:51:00.223" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8289" PostTypeId="4" CreationDate="2015-01-28T14:40:26.880" Score="0" Body="A puzzle from 2-dimensional or 3-dimensional geometry centered around matchsticks, often involving moving, adding, or removing some to achieve some defined goal." OwnerUserId="8874" LastEditorUserId="69582" LastEditDate="2021-03-10T06:50:38.817" LastActivityDate="2021-03-10T06:50:38.817" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8291" PostTypeId="2" ParentId="8286" CreationDate="2015-01-28T14:55:26.670" Score="0" Body="&lt;p&gt;A point to make and solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First, your strategy depends on your starting position. If you're Friend J (10th friend), then no matter what strategy you have, you can get stuck with 10 if everyone writes only one number. Similarly, you'll probably need a different strategy if you're friend #9 versus friend #8, etc. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you do know what position you are, you can try this solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Work backwards from your intended win. This is easiest to see with two people remaining. Let's say there are two people left, A and B. The person who says 1 or 2 will lose. The reason is that the person that comes after can make the total after their turn a multiple of 3. Let's say A says 1. B can say 2 numbers (2,3) to make the total 3. If A says 4,5 then B will just say 6. Then whatever A says, B says whatever makes the count to 9 forcing A with 10. Now if there are three people left, use a similar logic to try to have the person eliminated when your rival will be the one coming next. An example would be if you are A, you want B to be eliminated right after you go. You would try to get to a position where you can write 9 so B has to write 10 and C will be forced to write 1 or 2. By extending this thought process you can go to any number of people.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastActivityDate="2015-01-28T14:55:26.670" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8295" PostTypeId="2" ParentId="8285" CreationDate="2015-01-28T15:18:02.970" Score="11" Body="&lt;p&gt;I must admit that I found It with google but this is the solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/5eK9P.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-01-28T15:18:02.970" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8296" PostTypeId="2" ParentId="8286" CreationDate="2015-01-28T15:18:23.667" Score="2" Body="&lt;p&gt;Ok, let's say that the group is aggressive as possible about eliminating players, but wants to preserve their spot.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Player 6 will be eliminated first, then player 2. Neither can stop it.&lt;br&gt;&#xA; The next player to be eliminated is player 9, who is unable to stop it either.&lt;br&gt;&#xA;&lt;br&gt;&#xA; So, we have 10, 1, 3, 4, 5, 7, 8. With the same strat, player 7 would be eliminated, followed by player 5.&lt;br&gt;&#xA; Now, player 5 is the first player who can really throw a wrench into this. If player 2 wants to preserve themselves, they'll go with 1 (first round), and at that point player 5 goes 1 as well (instead of 2), forcing player 6 into 1, taking out 7.  (2-1-2-2-1-1).&lt;br&gt;&#xA; Next player to be eliminated would be 3 then, followed 10.&lt;br&gt;&#xA;&lt;br&gt;&#xA; At this stage you have 1, 2, 4, 5, 6, 8, 9.  Player 8 gets eliminated, followed by player 6. Now, at this point in time, could 6 or 8 have done anything to shift the eliminations? Round 1, no, round 2, yes, with 8. Instead of offing 10, player 1 gets eliminated.&lt;br&gt;&#xA; 2, 4, 5, 6, 8, 9, 10.  Player 9 gets eliminated, and can't change the outcome. Then 8.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Now you are left with five people: 10, 2, 4, 5, 6. At that point it's kind of a game of chicken. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Old:&lt;/strong&gt;&lt;br&gt;&#xA;I don't think there's a perfect play for this game. Generally, there's a perfect play to eliminate someone. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let's say you are player 1, and everyone else wants to get you out.&lt;br&gt;&#xA; 1-2-2-2-2 Eliminates player 6 (players 6-9 can be eliminated at will the first pass)&lt;br&gt;&#xA; From there 2-2-1-1-2-1 eliminates player 3&lt;br&gt;&#xA; Then 2-2-1-1-1-1-2 eliminates player 1&lt;br&gt;&#xA;&lt;br&gt;&#xA; Let's say we want to eliminate player 2&lt;br&gt;&#xA; 2-(1/2)-2-2-(2/1) gets player 6&lt;br&gt;&#xA; Then 2-2-2-2-1 would get player 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-01-28T17:52:16.840" LastActivityDate="2015-01-28T17:52:16.840" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8297" PostTypeId="2" ParentId="8286" CreationDate="2015-01-28T15:44:13.103" Score="6" Body="&lt;p&gt;If only 2 people play, the person starting can always be forced to lose. Say 1 number if he says 2, and 2 if he says 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are 3 people, you would want to eliminate the person immediately after you (and win via the rule above). This would be the easiest for one of the three positions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are 4 people, you would want to eliminate a person such that your position in the above scenario is optimal. This will again the easiest in one of the 4 positions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continue this way to 10 players. There is a programming concept called minimax game tree that can be used, as manually solving till 10 players is a lot of work. There is also a wikipedia page on 'Nim', it may or may not be of any use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that none of them can play perfectly, but they all have an independent optimal play. Luck will play some role. If players are willing to team up to defeat the remaining players, the strategy depends on who makes the best (maybe perfect) agreements.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-01-28T15:44:13.103" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8299" PostTypeId="2" ParentId="8223" CreationDate="2015-01-28T16:11:05.903" Score="10" Body="&lt;p&gt;EDIT: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It took me some time and a lot of dict.leo.org as I am no native speaker. Here is my second try:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Twenty-two numb lizards yearly assist wizards who summon the chill, cold blizzard, when my muzzy church buddy won me no crab cub - be it!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Length:&lt;/strong&gt; 137&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;N:&lt;/strong&gt; 6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Letters:&lt;/strong&gt; A, B, C, D, E, H, I, L, M, N, O, R, S, T, U, W, Y, Z (18 total)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;--- DEPRECATED ---&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My mum's wild buddy Yannik B. was widely known to tell rumors about&#xA;  trekkie bars.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Letters: a,b,d,e,i,k,l,m,n,o,r,s,t,u,w,y&lt;/p&gt;&#xA;&#xA;&lt;p&gt;n: 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Length : 81&lt;/p&gt;&#xA;" OwnerUserId="9226" LastEditorUserId="9226" LastEditDate="2015-01-28T18:25:38.823" LastActivityDate="2015-01-28T18:25:38.823" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8300" PostTypeId="1" AcceptedAnswerId="8303" CreationDate="2015-01-28T16:34:00.233" Score="14" ViewCount="794" Body="&lt;p&gt;The statements of the following two puzzles are &lt;strong&gt;almost&lt;/strong&gt; identical.&#xA;The only difference consists in the words printed in boldface.&#xA;The two solutions, however, turn out to be totally different.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Variant 1 of the puzzle&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My friends Xavier, Yvo, and Zeno are excellent mathematicians and always think strictly logically. Yesterday I told them: &lt;em&gt;&quot;I have secretly chosen three positive integers $x,y,z$ with $x+y+z=2015$. I have told $x$ to Xavier, $y$ to Yvo, and $z$ to Zeno.&quot;&lt;/em&gt;&lt;br&gt;&#xA;  Then the following conversation developed.&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Xavier said: I know that Yvo and Zeno have different numbers.&lt;/li&gt;&#xA;  &lt;li&gt;Yvo said: &lt;strong&gt;I already knew&lt;/strong&gt; that all our numbers are different.&lt;/li&gt;&#xA;  &lt;li&gt;Zeno said: Aha! Now I know all three numbers.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;Determine the values of $x$, $y$, and $z$!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Variant 2 of the puzzle&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My friends Xavier, Yvo, and Zeno are excellent mathematicians and always think strictly logically. Yesterday I told them: &lt;em&gt;&quot;I have secretly chosen three positive integers $x,y,z$ with $x+y+z=2015$. I have told $x$ to Xavier, $y$ to Yvo, and $z$ to Zeno.&quot;&lt;/em&gt;&lt;br&gt;&#xA;  Then the following conversation developed.&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Xavier said: I know that Yvo and Zeno have different numbers.&lt;/li&gt;&#xA;  &lt;li&gt;Yvo said: &lt;strong&gt;Aha! Now I know&lt;/strong&gt; that all our numbers are different.&lt;/li&gt;&#xA;  &lt;li&gt;Zeno said: Aha! Now I know all three numbers.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;Determine the values of $x$, $y$, and $z$!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="10615" LastEditDate="2016-02-24T04:05:23.240" LastActivityDate="2016-02-24T04:05:23.240" Title="Three secret numbers (Variant 1 and Variant 2)" Tags="|mathematics|logical-deduction|meta-knowledge|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8301" PostTypeId="2" ParentId="8162" CreationDate="2015-01-28T16:41:50.363" Score="2" Body="&lt;p&gt;The key is you have to make any two piles, doesn't necessarily 50/50&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets call $C = 100$ the number of cards &lt;br&gt;&#xA;and $N = 90$ number of face up&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplicity imagine $N = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You take 1 card at random from the $C$ pile now you have a $pile A = 99$ cards and $pile B = 1$ card&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you flip the card in $B$ two scenarios occurs &lt;br&gt;&#xA;$\cdot$If card $B$ is the face up mean $A$ doesnt have any face up cards after the flip $A$ and $B$ have 0 face up cards &lt;br&gt;&#xA;$\cdot$if card $B$ is face down mean $A$ have 1 face up cards after the flip $A$ and $B$ have 1 face up card &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Generalised Case&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are $N$ face up cards in Pile 1. We take $N$ cards from Pile 1 to form Pile 2. Lets say, by chance $F$ of those are face up. This necessarily leaves $N-F$ face up cards in Pile 1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pile 2 contains $N$ cards, $F$ of which are face up, meaning $N-F$ are face down. When we flip over Pile 2 these $N-F$ face down card become face up. Both piles contain $N-F$ face up cards. &lt;/p&gt;&#xA;" OwnerUserId="5165" LastActivityDate="2015-01-28T16:41:50.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8302" PostTypeId="1" AcceptedAnswerId="8304" CreationDate="2015-01-28T17:01:52.100" Score="10" ViewCount="8489" Body="&lt;p&gt;If an aeroplane makes a round trip and a wind is blowing, is the trip time shorter, longer or the same as without wind? &lt;br&gt;&#xA;Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Round trip mean aeroplane go from point A to point B and then return to point A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The velocity of the wind is constant.&lt;/p&gt;&#xA;" OwnerUserId="5165" LastEditorUserId="8874" LastEditDate="2015-02-02T09:02:11.577" LastActivityDate="2015-02-02T09:02:11.577" Title="Aeroplane roundtrip in the wind" Tags="|mathematics|calculation-puzzle|" AnswerCount="6" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8303" PostTypeId="2" ParentId="8300" CreationDate="2015-01-28T17:07:13.193" Score="16" Body="&lt;p&gt;X's statement means that his number must be even (if it was odd, Y and Z would have numbers summing to an even number, so they could be the same).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first case, Y's statement means that Y has an even number, and neither X nor Z can have the same even number. Therefore Y must have a number that is at least 1008.&#xA;Z knows all three numbers, so X and Y must both have the smallest possible numbers, 2 and 1008 respectively. (Any other sum for X and Y would allow multiple possibilities for their numbers.) This leaves Z with 1005.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second case, Y did not know that the numbers were different before, but did after hearing X's statement. Y must not have an even number greater than 1006 (because he would have known already). Y cannot have an even number less than 1008 either, because there would be no way for him to tell that it wasn't the same as X's number. Y also can't have an odd number which is 3 mod 4, because then X and Z could have the same even number. Therefore Y must have a number which is 1 mod 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now Y must have 1, otherwise it would be possible to subtract 4 from Y and add 4 to X without Z being able to tell. X must have 2 or 4. Z has 2012 or 2010. But if Z had 2012, he would have known all the numbers after X's statement. If his statement means that he did not know until Y's statement, the numbers must be 4, 1, and 2010.&lt;/p&gt;&#xA;" OwnerUserId="9227" LastActivityDate="2015-01-28T17:07:13.193" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8304" PostTypeId="2" ParentId="8302" CreationDate="2015-01-28T17:11:02.977" Score="13" Body="&lt;p&gt;It takes longer.  Let the distance be $d$, the plane speed be $v$, and the wind speed be $w$.  &lt;br&gt; With no wind, the round trip takes $\frac {2d}v$.  &lt;br&gt; With wind the round trip takes $\frac d{v+w}+\frac d{v-w}=\frac{d(v-w)+d(v+w)}{v^2-w^2}=\frac {2dv}{v^2-w^2}=\frac {2d}{v-\frac {w^2}v}$. &lt;br&gt;  Decreasing the denominator increases the time.  &lt;br&gt; If $w \gt v$ he can never return.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="5165" LastEditDate="2015-01-28T18:36:59.263" LastActivityDate="2015-01-28T18:36:59.263" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8305" PostTypeId="2" ParentId="8302" CreationDate="2015-01-28T17:12:14.297" Score="11" Body="&lt;p&gt;The overall trip becomes longer: one way the wind speed is essentially added to the aeroplane speed, and the other way the wind speed is essentially subtracted. The trip with addition (and faster speed) takes shorter time than the trip with subtraction (and slower speed). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the plane is sped up for a short time and slowed down for a longer time. This makes the overall trip to become longer.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-01-28T17:12:14.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8306" PostTypeId="5" CreationDate="2015-01-28T17:35:53.713" Score="0" Body="&lt;p&gt;Puzzles centered around a discrete system that the solver must then show whether or not a certain system state is reachable within a finite (or specific) number of steps. Normally answers must have some form of mathematical proof to back up their claim.&lt;/p&gt;&#xA;" OwnerUserId="52483" LastEditorUserId="52483" LastEditDate="2018-11-27T19:35:54.623" LastActivityDate="2018-11-27T19:35:54.623" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8307" PostTypeId="4" CreationDate="2015-01-28T17:35:53.713" Score="0" Body="A puzzle on a discrete system where one has to decide whether a certain system state can be reached through a finite number of steps." OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-01-28T20:40:26.387" LastActivityDate="2015-01-28T20:40:26.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8308" PostTypeId="5" CreationDate="2015-01-28T17:36:35.843" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-01-28T17:36:35.843" LastActivityDate="2015-01-28T17:36:35.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8309" PostTypeId="4" CreationDate="2015-01-28T17:36:35.843" Score="0" Body="A puzzle on a discrete system where one has to decide whether a certain system state can be reached through a finite number of steps." OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-01-29T03:25:30.087" LastActivityDate="2015-01-29T03:25:30.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8310" PostTypeId="2" ParentId="8286" CreationDate="2015-01-28T18:10:22.407" Score="7" Body="&lt;p&gt;As mentioned above, the second player in the two person game can force a win by playing perfectly.&lt;/p&gt;&#xA;&lt;p&gt;The problem in the three person game is that there are 2 losers.  Lets say that the three players are &lt;span class=&quot;math-container&quot;&gt;$A, B,$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; and there exists a perfect strategy for one of them to win. WLOG, lets say that this is &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; (note we don't know who needs to go first for this to happen).&lt;/p&gt;&#xA;&lt;p&gt;For &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; to win, player &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; will need to lose first making it player &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt;'s turn in the final round.  Thus, to make player &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; lose, player &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; will need to write &lt;span class=&quot;math-container&quot;&gt;$9$&lt;/span&gt;, or &lt;span class=&quot;math-container&quot;&gt;$8,9$&lt;/span&gt;.  In order to do this, player &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; will need to write &lt;span class=&quot;math-container&quot;&gt;$7$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt;.  To force player &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; to write &lt;span class=&quot;math-container&quot;&gt;$7$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt;, player &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; will need to write &lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The problem is that we can't force player &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; to write &lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt;. We can either end on &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt;, giving &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; the &lt;em&gt;option&lt;/em&gt; of writing &lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt;, or of writing &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$7$&lt;/span&gt;, depending on what we give them.  In fact, knowing that by writing &lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; will lose, &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; will probably choose to write another number.  &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; may still lose, but losing to &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; is equivalent to losing to &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;, so there is no motivation to play along with any strategy of &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;'s.&lt;/p&gt;&#xA;&lt;p&gt;Lets say that in this situation, &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; wrote a &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt;.  Knowing that writing a &lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt; will lose, &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; will write both &lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$7$&lt;/span&gt;.  This hands the victory to &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; since &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; can now write &lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$9$&lt;/span&gt;, making &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; drop out.  &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; goes next in the 2 person game, so &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; wins.&lt;/p&gt;&#xA;&lt;p&gt;If instead &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; wrote a &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt;, again, &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; would know that writing a &lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt; is death.  So instead, &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; may choose to write only &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt;.  But as we saw above, writing a &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt; will cause that player to be eliminated, and this would hand the win one the other two players.&lt;/p&gt;&#xA;&lt;p&gt;Thus, if given the option to pick between &lt;span class=&quot;math-container&quot;&gt;$5, 6,$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$7$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; won't care because they all result in losses.&lt;/p&gt;&#xA;&lt;p&gt;If we changed the rules to have a ranking where finishing later is better, then we could probably come up with a generic strategy for the 3 person game and extend it further.&lt;/p&gt;&#xA;&lt;h1&gt;With the Rule Change&lt;/h1&gt;&#xA;&lt;p&gt;Lets revisit the 3 person game with the new change that a player would rather come in second than lose.&lt;/p&gt;&#xA;&lt;h2&gt;3 Player Game&lt;/h2&gt;&#xA;&lt;p&gt;Players are &lt;span class=&quot;math-container&quot;&gt;$A,B,$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt;.&#xA;The following are the results for player &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; if player &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; writes the following numbers:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$9$&lt;/span&gt;: Finish 1st.  Player &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; will be eliminated and &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; will start the next round.  Final order is &lt;span class=&quot;math-container&quot;&gt;$ACB$&lt;/span&gt;.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt;: Finish 2nd.  Player &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; will write &lt;span class=&quot;math-container&quot;&gt;$9$&lt;/span&gt; eliminating &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; will start the final round. Final order is &lt;span class=&quot;math-container&quot;&gt;$BAC$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$7$&lt;/span&gt;: Finish 2nd.  Player &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; will write &lt;span class=&quot;math-container&quot;&gt;$8,9$&lt;/span&gt;. Final order is &lt;span class=&quot;math-container&quot;&gt;$BAC$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt;: Finish 3rd.  Player &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; can do no better than 2nd, and &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; will get the win.  Final order is &lt;span class=&quot;math-container&quot;&gt;$CBA$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt;: Finish 3rd.  &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; will write &lt;span class=&quot;math-container&quot;&gt;$6,7$&lt;/span&gt; for same result as above.  Final order &lt;span class=&quot;math-container&quot;&gt;$CBA$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt;: Finish 1st.  Player &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; will be forced to write either &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt;, both of which result in a 3rd place finish.  This leaves &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; to go first in the final round.  Final order is &lt;span class=&quot;math-container&quot;&gt;$ACB$&lt;/span&gt;.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt;: Finish 2nd.  &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; will get the win, so the final order is &lt;span class=&quot;math-container&quot;&gt;$BAC$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt;: Finish 2nd. Player &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; will write &lt;span class=&quot;math-container&quot;&gt;$3,4$&lt;/span&gt; and finish 1st again.  Final order is &lt;span class=&quot;math-container&quot;&gt;$BAC$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;: Finish 3rd.  &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; can do no better than 2nd, and &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; will win.  Final order is &lt;span class=&quot;math-container&quot;&gt;$CBA$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;em&gt;Notice that to get the previous number, you simply need to figure out what &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; will do.  &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; will have the choice of either the next number, or the next two, and will pick which ever gets the best result.  Then you take the final ordering from that number and increment each player by one wrapping around to &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; if necessary (e.g. &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; become &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; becomes &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; becomes &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;) to get the final ordering for the new number.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;Thus, the person who goes second in the 3 person game can win.  The person who goes first will choose to make &lt;span class=&quot;math-container&quot;&gt;$1,2$&lt;/span&gt; as the first move and guarantee themselves second instead of third, allowing the person who goes second to write &lt;span class=&quot;math-container&quot;&gt;$3,4$&lt;/span&gt; and put themselves in the winning position.&lt;/p&gt;&#xA;&lt;p&gt;So if the initial setup is &lt;span class=&quot;math-container&quot;&gt;$ABC$&lt;/span&gt; with &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; to go first, then the final ordering will be &lt;span class=&quot;math-container&quot;&gt;$BAC$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;h2&gt;4 Player Game&lt;/h2&gt;&#xA;&lt;p&gt;Players are &lt;span class=&quot;math-container&quot;&gt;$A,B,C,$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt;.&#xA;The following are the results for player &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; if player &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; writes the following numbers:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$9$&lt;/span&gt;: Finish 3rd. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; is eliminated, and the remaining players in order to start the final round are &lt;span class=&quot;math-container&quot;&gt;$CDA$&lt;/span&gt;.  Final order will be &lt;span class=&quot;math-container&quot;&gt;$DCAB$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt;: Finish 1st. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; would write &lt;span class=&quot;math-container&quot;&gt;$9$&lt;/span&gt; and finish 3rd, final order is &lt;span class=&quot;math-container&quot;&gt;$ADBC$&lt;/span&gt;.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$7$&lt;/span&gt;: Finish 2nd. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; would write &lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt; and finish 1st.  Final order is &lt;span class=&quot;math-container&quot;&gt;$BACD$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt;: Finish 2nd. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; would write &lt;span class=&quot;math-container&quot;&gt;$7,8$&lt;/span&gt; to finish 1st.  Final order is &lt;span class=&quot;math-container&quot;&gt;$BACD$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt;: Finish 4th. The best &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; can do is finish 2nd.  Final order is &lt;span class=&quot;math-container&quot;&gt;$CBDA$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt;: Finish 4th. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; would write &lt;span class=&quot;math-container&quot;&gt;$5,6$&lt;/span&gt; to claim 2nd, making the final order again &lt;span class=&quot;math-container&quot;&gt;$CBDA$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt;: Finish 3rd. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; is forced to finish 4th.  Final order will be &lt;span class=&quot;math-container&quot;&gt;$DCAB$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt;: Finish 1st.  3rd is better than 4th, so &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; will write just &lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt;.  Final order is &lt;span class=&quot;math-container&quot;&gt;$ADBC$&lt;/span&gt;.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;: Finish 2nd. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; would write &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; and finish 1st. Final order is &lt;span class=&quot;math-container&quot;&gt;$BACD$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;So in the 4 person game, whoever goes first can win.  If the initial order is &lt;span class=&quot;math-container&quot;&gt;$ABCD$&lt;/span&gt; with &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; to go first, then the final ordering is &lt;span class=&quot;math-container&quot;&gt;$ADBC$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;h2&gt;5 Player Game&lt;/h2&gt;&#xA;&lt;p&gt;Players are &lt;span class=&quot;math-container&quot;&gt;$A,B,C,D$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt;.&#xA;The following are the results for player &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; if player &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; writes the following numbers:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$9$&lt;/span&gt;: Finish 2nd. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; is eliminated, and the remaining players in order to start the final round are &lt;span class=&quot;math-container&quot;&gt;$CDEA$&lt;/span&gt;.  Final order will be &lt;span class=&quot;math-container&quot;&gt;$CADEB$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt;: Finish 4th. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; would write &lt;span class=&quot;math-container&quot;&gt;$9$&lt;/span&gt; and finish 2nd. Final: &lt;span class=&quot;math-container&quot;&gt;$DBEAC$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$7$&lt;/span&gt;: Finish 4th. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; would write &lt;span class=&quot;math-container&quot;&gt;$8,9$&lt;/span&gt; and finish 2nd. Final: &lt;span class=&quot;math-container&quot;&gt;$DBEAC$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt;: Finish 3rd. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; would be forced to finish 4th. Final: &lt;span class=&quot;math-container&quot;&gt;$ECABD$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt;: Finish 1st. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; would write &lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt; to finish 3rd instead of 4th. Final: &lt;span class=&quot;math-container&quot;&gt;$ADBCE$&lt;/span&gt;.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt;: Finish 5th. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; would write &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt; to finish first. Final: &lt;span class=&quot;math-container&quot;&gt;$BECDA$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt;: Finish 5th. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; would write &lt;span class=&quot;math-container&quot;&gt;$5,6$&lt;/span&gt; to finish first. Final: &lt;span class=&quot;math-container&quot;&gt;$BECDA$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt;: Finish 2nd. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; finishes last no matter what is written.  Final &lt;span class=&quot;math-container&quot;&gt;$CADEB$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;: Finish 4th. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; gets 2nd with &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt;. Final: &lt;span class=&quot;math-container&quot;&gt;$DBEAC$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;So the only way to win the 5 person game is to be the one to write the &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt;.  This can be done by going third.  The first player will choose to write &lt;span class=&quot;math-container&quot;&gt;$1,2$&lt;/span&gt; since that is the better result.  The second player finishes in last place no matter what they write, leaving the 3rd player to write the &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt; and win.&lt;/p&gt;&#xA;&lt;p&gt;If the initial order is &lt;span class=&quot;math-container&quot;&gt;$ABCDE$&lt;/span&gt; with &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; to go first, then the final ordering is &lt;span class=&quot;math-container&quot;&gt;$CADEB$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;h2&gt;6 Player Game&lt;/h2&gt;&#xA;&lt;p&gt;Players are &lt;span class=&quot;math-container&quot;&gt;$A,B,C,D,E$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt;.&#xA;The following are the results for player &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; if player &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; writes the following numbers:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$9$&lt;/span&gt;: Finish 4th. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; is eliminated, and the remaining players in order to start the final round are &lt;span class=&quot;math-container&quot;&gt;$CDEFA$&lt;/span&gt;.  Final order will be &lt;span class=&quot;math-container&quot;&gt;$ECFADB$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt;: Finish 3rd. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; would write &lt;span class=&quot;math-container&quot;&gt;$9$&lt;/span&gt; and finish 4th. Final: &lt;span class=&quot;math-container&quot;&gt;$FDABEC$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$7$&lt;/span&gt;: Finish 1st. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; would write &lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt; and finish 3rd. Final: &lt;span class=&quot;math-container&quot;&gt;$AEBCFD$&lt;/span&gt;.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt;: Finish 5th. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; would write &lt;span class=&quot;math-container&quot;&gt;$7$&lt;/span&gt; for the win. Final: &lt;span class=&quot;math-container&quot;&gt;$BFCDAE$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt;: Finish 5th. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; would write &lt;span class=&quot;math-container&quot;&gt;$6,7$&lt;/span&gt; and win. Final: &lt;span class=&quot;math-container&quot;&gt;$BFCDAE$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt;: Finish 2nd. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; gets 5th. Final: &lt;span class=&quot;math-container&quot;&gt;$CADEBF$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt;: Finish 6th. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; would write &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt; to finish 2nd. Final: &lt;span class=&quot;math-container&quot;&gt;$DBEFCA$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt;: Finish 6th. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; writes &lt;span class=&quot;math-container&quot;&gt;$3,4$&lt;/span&gt; for 2nd. Final: &lt;span class=&quot;math-container&quot;&gt;$DBEFCA$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;: Finish 4th. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; gets 6th regardless. Final: &lt;span class=&quot;math-container&quot;&gt;$ECFADB$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;To win the 6 person game, you want to be the 5th player to go.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Will write just &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; since 4th place is better than 6th.&lt;/li&gt;&#xA;&lt;li&gt;Gets 6th no matter what they write.&lt;/li&gt;&#xA;&lt;li&gt;Will write &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt; to get 2nd instead of 5th.&lt;/li&gt;&#xA;&lt;li&gt;Gets 5th no matter what.&lt;/li&gt;&#xA;&lt;li&gt;Writes &lt;span class=&quot;math-container&quot;&gt;$7$&lt;/span&gt; to win.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Final order assuming &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; goes first in a 6 person game is &lt;span class=&quot;math-container&quot;&gt;$ECFADB$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;h2&gt;7 Player Game&lt;/h2&gt;&#xA;&lt;p&gt;Players are &lt;span class=&quot;math-container&quot;&gt;$A,B,C,D,E,F$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt;.&#xA;The following are the results for player &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; if player &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; writes the following numbers:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$9$&lt;/span&gt;: Finish 3rd. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; is eliminated, and the remaining players in order to start the final round are &lt;span class=&quot;math-container&quot;&gt;$CDEFGA$&lt;/span&gt;.  Final order will be &lt;span class=&quot;math-container&quot;&gt;$GEACFDB$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt;: Finish 1st. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; would write &lt;span class=&quot;math-container&quot;&gt;$9$&lt;/span&gt; and finish 3rd. Final: &lt;span class=&quot;math-container&quot;&gt;$AFBDGEC$&lt;/span&gt;.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$7$&lt;/span&gt;: Finish 5th. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; would write &lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt; and finish 1st. Final: &lt;span class=&quot;math-container&quot;&gt;$BGCEAFD$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt;: Finish 5th. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; would write &lt;span class=&quot;math-container&quot;&gt;$7,8$&lt;/span&gt; for the win. Final: &lt;span class=&quot;math-container&quot;&gt;$BGCEAFD$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt;: Finish 2nd. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; gets 5th. Final: &lt;span class=&quot;math-container&quot;&gt;$CADFBGE$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt;: Finish 6th. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; gets 2nd with &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt;. Final: &lt;span class=&quot;math-container&quot;&gt;$DBEGCAF$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt;: Finish 6th. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; gets 2nd again with &lt;span class=&quot;math-container&quot;&gt;$4,5$&lt;/span&gt;. Final: &lt;span class=&quot;math-container&quot;&gt;$DBEGCAF$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt;: Finish 4th. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; gets 6th. Final: &lt;span class=&quot;math-container&quot;&gt;$ECFADBG$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;: Finish 7th. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; gets 4th with &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt;. Final: &lt;span class=&quot;math-container&quot;&gt;$FDGBECA$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;To win the 7 person game, you want to be the 5th player to go.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Will write just &lt;span class=&quot;math-container&quot;&gt;$1,2$&lt;/span&gt; since 4th place is better than 7th.&lt;/li&gt;&#xA;&lt;li&gt;Gets 6th no matter what they write.&lt;/li&gt;&#xA;&lt;li&gt;Will write &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt; to get 2nd instead of 5th.&lt;/li&gt;&#xA;&lt;li&gt;Gets 5th no matter what.&lt;/li&gt;&#xA;&lt;li&gt;Writes &lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt; to win.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Final order assuming &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; goes first in a 7 person game is &lt;span class=&quot;math-container&quot;&gt;$ECFADBG$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;h2&gt;8 Player Game&lt;/h2&gt;&#xA;&lt;p&gt;Players are &lt;span class=&quot;math-container&quot;&gt;$A,B,C,D,E,F,G$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt;.&#xA;The following are the results for player &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; if player &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; writes the following numbers:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$9$&lt;/span&gt;: Finish 7th. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; is eliminated, and the remaining players in order to start the final round are &lt;span class=&quot;math-container&quot;&gt;$CDEFGHA$&lt;/span&gt;.  Final order will be &lt;span class=&quot;math-container&quot;&gt;$GEHCFDAB$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt;: Finish 3rd. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; would write &lt;span class=&quot;math-container&quot;&gt;$9$&lt;/span&gt; and finish 7th. Final: &lt;span class=&quot;math-container&quot;&gt;$HFADGEBC$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$7$&lt;/span&gt;: Finish 1st. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; would write &lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt; and finish 3rd. Final: &lt;span class=&quot;math-container&quot;&gt;$AGBEHFCD$&lt;/span&gt;.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt;: Finish 5th. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; would write &lt;span class=&quot;math-container&quot;&gt;$7$&lt;/span&gt; for the win. Final: &lt;span class=&quot;math-container&quot;&gt;$BHCFAGDE$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt;: Finish 5th. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; writes &lt;span class=&quot;math-container&quot;&gt;$6,7$&lt;/span&gt; for the win. Final: &lt;span class=&quot;math-container&quot;&gt;$BHCFAGDE$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt;: Finish 2nd. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; gets 5th. Final: &lt;span class=&quot;math-container&quot;&gt;$CADGBHEF$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt;: Finish 6th. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; writes &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt; for 2nd. Final: &lt;span class=&quot;math-container&quot;&gt;$DBEHCAFG$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt;: Finish 6th. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; writes &lt;span class=&quot;math-container&quot;&gt;$3,4$&lt;/span&gt; for 2nd.  Final: &lt;span class=&quot;math-container&quot;&gt;$DBEHCAFG$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;: Finish 4th. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; gets 6th. Final: &lt;span class=&quot;math-container&quot;&gt;$ECFADBGH$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;To win the 8 person game, you want to be the 5th player to go.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Will write just &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; since 4th place is better than 6th.&lt;/li&gt;&#xA;&lt;li&gt;Gets 6th no matter what they write.&lt;/li&gt;&#xA;&lt;li&gt;Will write &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt; to get 2nd instead of 5th.&lt;/li&gt;&#xA;&lt;li&gt;Gets 5th no matter what.&lt;/li&gt;&#xA;&lt;li&gt;Writes &lt;span class=&quot;math-container&quot;&gt;$7$&lt;/span&gt; to win.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Notice the strategy is identical to the 6 person game and very similar to the 7?&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;Final order assuming &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; goes first in a 8 person game is &lt;span class=&quot;math-container&quot;&gt;$ECFA DBGH$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;h2&gt;9 Player Game&lt;/h2&gt;&#xA;&lt;p&gt;Players are &lt;span class=&quot;math-container&quot;&gt;$A,B,C,D,E,F,G,H$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt;.&#xA;The following are the results for player &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; if player &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; writes the following numbers:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$9$&lt;/span&gt;: Finish 8th. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; is eliminated, and the remaining players in order to start the final round are &lt;span class=&quot;math-container&quot;&gt;$CDEF GHIA$&lt;/span&gt;.  Final order will be &lt;span class=&quot;math-container&quot;&gt;$GEHC FDIAB$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt;: Finish 7th. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; would write &lt;span class=&quot;math-container&quot;&gt;$9$&lt;/span&gt; and finish 8th. Final: &lt;span class=&quot;math-container&quot;&gt;$HFID GEABC$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$7$&lt;/span&gt;: Finish 3rd. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; would write &lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt; and finish 7th. Final: &lt;span class=&quot;math-container&quot;&gt;$IGAE HFBCD$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt;: Finish 1st. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; would write &lt;span class=&quot;math-container&quot;&gt;$7$&lt;/span&gt; for 3rd. Final: &lt;span class=&quot;math-container&quot;&gt;$AHBF IGCDE$&lt;/span&gt;.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt;: Finish 5th. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; writes &lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt; for the win. Final: &lt;span class=&quot;math-container&quot;&gt;$BICG AHDEF$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt;: Finish 5th. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; writes &lt;span class=&quot;math-container&quot;&gt;$5,6$&lt;/span&gt; and wins. Final: &lt;span class=&quot;math-container&quot;&gt;$BICG AHDEF$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt;: Finish 2nd. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; gets 5th. Final: &lt;span class=&quot;math-container&quot;&gt;$CADH BIEFG$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt;: Finish 6th. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; writes &lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt; for 2nd.  Final: &lt;span class=&quot;math-container&quot;&gt;$DBEI CAFGH$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;: Finish 6th. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; writes &lt;span class=&quot;math-container&quot;&gt;$2,3$&lt;/span&gt; for 2nd.  Final: &lt;span class=&quot;math-container&quot;&gt;$DBEI CAFGH$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;To win the 9 person game, you want to be the 4th player to go.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Gets 6th.&lt;/li&gt;&#xA;&lt;li&gt;Writes &lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt; for 2nd place.&lt;/li&gt;&#xA;&lt;li&gt;Gets 5th no matter what.&lt;/li&gt;&#xA;&lt;li&gt;Writes &lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt; to win.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&lt;em&gt;Notice that player &lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt; doesn't even get to play in the first round!&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;Final order assuming &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; goes first in a 9 person game is &lt;span class=&quot;math-container&quot;&gt;$DBEIC AFGH$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;h2&gt;10 Player Game&lt;/h2&gt;&#xA;&lt;p&gt;Players are &lt;span class=&quot;math-container&quot;&gt;$A,B,C,D,E,F,G,H,I$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$J$&lt;/span&gt;.&#xA;The following are the results for player &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; if player &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; writes the following numbers:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$9$&lt;/span&gt;: Finish 4th. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; is eliminated, and the remaining players in order to start the final round are &lt;span class=&quot;math-container&quot;&gt;$CDEFG HIJA$&lt;/span&gt;.  Final order will be &lt;span class=&quot;math-container&quot;&gt;$FDGAE CHIJB$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt;: Finish 9th. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; would write &lt;span class=&quot;math-container&quot;&gt;$9$&lt;/span&gt; and finish 4th. Final: &lt;span class=&quot;math-container&quot;&gt;$GEHBF DIJAC$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$7$&lt;/span&gt;: Finish 9th. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; would write &lt;span class=&quot;math-container&quot;&gt;$8,9$&lt;/span&gt; and finish 4th. Final: &lt;span class=&quot;math-container&quot;&gt;$GEHBF DIJAC$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt;: Finish 8th. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; gets 9th. Final: &lt;span class=&quot;math-container&quot;&gt;$HFICG EJABD$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt;: Finish 7th. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; writes &lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt; for 8th. Final: &lt;span class=&quot;math-container&quot;&gt;$IGJDH FABCE$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt;: Finish 3rd. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; writes &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt; for 7th. Final: &lt;span class=&quot;math-container&quot;&gt;$JHAEI GBCDF$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt;: Finish 1st. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; writes &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt; for 3rd. Final: &lt;span class=&quot;math-container&quot;&gt;$AIBFJ HCDEG$&lt;/span&gt;.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt;: Finish 5th. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; writes &lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt; for 1st.  Final: &lt;span class=&quot;math-container&quot;&gt;$BJCGA IDEFH$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;: Finish 5th. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; writes &lt;span class=&quot;math-container&quot;&gt;$2,3$&lt;/span&gt; for the win.  Final: &lt;span class=&quot;math-container&quot;&gt;$BJCGA IDEFH$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;To win the 10 person game, you want to be the 2nd player to go.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Gets 5th no matter what they do.&lt;/li&gt;&#xA;&lt;li&gt;Writes &lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt; for the win!&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Neither &lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt; nor &lt;span class=&quot;math-container&quot;&gt;$J$&lt;/span&gt; get to actually play in the first round and &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt; is eliminated in the first round.&lt;/p&gt;&#xA;&lt;p&gt;Final order assuming &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; goes first in a 10 person game is &lt;span class=&quot;math-container&quot;&gt;$BJCGA IDEFH$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;h1&gt;Addendum&lt;/h1&gt;&#xA;&lt;p&gt;This can be extended indefinitely for groups of size &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt;, but only by working out the solution for &lt;span class=&quot;math-container&quot;&gt;$N-1$&lt;/span&gt; first.  There does not seem to be a trivial formula suggested by this that can get you the solution for an arbitrary &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; without knowing the intermediate results.&lt;/p&gt;&#xA;&lt;p&gt;This solution, however, is very useful for every player of the game.  It provides a quick lookup to see what your best move is at any stage of the game.  For example, if I have survived until the round of 5 players and am presented with a &lt;span class=&quot;math-container&quot;&gt;$7$&lt;/span&gt;, I know that writing &lt;span class=&quot;math-container&quot;&gt;$8,9$&lt;/span&gt; will get me a 2nd place finish if everyone else plays perfectly.  If only one person doesn't play perfectly, then I may not do better than 2nd, but that player will definitely finish worse if the rest of the players play perfectly.  This solution can continue to provide optimal responses even if things do not go according to plan.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-01-29T17:51:11.927" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="8311" PostTypeId="1" AcceptedAnswerId="8313" CreationDate="2015-01-28T18:38:18.060" Score="7" ViewCount="1436" Body="&lt;p&gt;On a deserted road, the probability of observing a car during a thirty-minute period is 95%. &lt;br&gt;&#xA;What is the chance of observing a car in a ten-minute period?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&#xA;To clarify the question we are saying the probability of seeing any other cars in 30 minutes is 95% or more clearly, and more usefully, the probability of not seeing any other cars is 5%.&lt;/p&gt;&#xA;" OwnerUserId="5165" LastEditorUserId="5165" LastEditDate="2015-01-29T04:25:29.377" LastActivityDate="2015-01-29T04:25:29.377" Title="Probability of Seeing a Car in 10 Minutes &amp; 30 Minutes" Tags="|mathematics|calculation-puzzle|" AnswerCount="1" CommentCount="4" ClosedDate="2015-01-31T16:32:28.170" ContentLicense="CC BY-SA 3.0" />
  <row Id="8312" PostTypeId="2" ParentId="8223" CreationDate="2015-01-28T18:50:57.937" Score="9" Body="&lt;p&gt;Not  sure if the convention is to add a new answer or edit my old one, but here's a new and rather saucy attempt:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sentence&lt;/strong&gt;: The tight-shoed thigh-slapping puzzle nudist, polo pants unzipped and full of gaffing stuff, laughs off a dozen glazed pizza-sized doughnut holes!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Length&lt;/strong&gt;: 146&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;n&lt;/strong&gt;: 8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Letters&lt;/strong&gt;: adefghilnopstuz&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Also, I updated that tool a bit, use &lt;a href=&quot;http://jsfiddle.net/uuef3bf7/&quot; rel=&quot;noreferrer&quot;&gt;http://jsfiddle.net/uuef3bf7/&lt;/a&gt; if you like.&lt;/p&gt;&#xA;" OwnerUserId="2374" LastActivityDate="2015-01-28T18:50:57.937" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8313" PostTypeId="2" ParentId="8311" CreationDate="2015-01-28T18:53:52.440" Score="7" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 100% -  5%^(1/3) (cube root of 5%), which is about 63%&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because the probability of not seeing a car in thirty minutes is equal to the probability of not seeing a car for ten minutes to the third power. That is, &lt;code&gt;not seeing a car for ten minutes&lt;/code&gt; three times in a row is like &lt;code&gt;not seeing a car for thirty minutes&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or, with a formula:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If $P_{not30}$ is the probability of not seeing a car for 30 minutes and $P_{not10}$ is the probability of not seeing a car for ten minutes, $P_{not30}$= $P_{not10}^ 3 \Rightarrow$ $P_{not10} = \sqrt[3]{P_{not30}}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="5165" LastEditDate="2015-01-28T19:22:52.363" LastActivityDate="2015-01-28T19:22:52.363" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8314" PostTypeId="2" ParentId="8302" CreationDate="2015-01-28T21:56:20.440" Score="6" Body="&lt;p&gt;It will usually take longer, as others have said. However, under the right conditions, the wind can make for a shorter trip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It depends on where point A and point B are, and the direction and speed of the wind. If, for example, you are flying round-trip between Washington D.C. and Singapore, the distance from one to the other is approximately the same whether you go east or west. So, if the wind is blowing to the east or west, you can travel from A to B by flying with the wind, and then from B to A, while &lt;em&gt;still flying with the wind&lt;/em&gt;. In this case, the wind will make you go faster. Even if the two points aren't on approximately opposite sides of the world, it could still be faster to go &quot;the long way around&quot;, if the winds are strong enough.&lt;/p&gt;&#xA;" OwnerUserId="8776" LastActivityDate="2015-01-28T21:56:20.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8315" PostTypeId="2" ParentId="8302" CreationDate="2015-01-28T22:22:25.367" Score="2" Body="&lt;p&gt;There is not enough information to answer the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is stated that wind speed is constant, but not direction. So there could be a tail wind in both directions, which would be faster than without wind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, if wind direction is constant, the plane could circumnavigate to go from A to B. Then circumnavigate back from B to A, in the same direction. This way there could be a tail wind for both flights. Near the poles, this wouldn't have to be an overly long flight. (I guess this is similar to the half-way idea that KSmarts answered)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.csgnetwork.com/e6bcalc.html&quot; rel=&quot;nofollow&quot;&gt;E6B calculator&lt;/a&gt; is helpful.&lt;/p&gt;&#xA;" OwnerUserId="2477" LastEditorUserId="2477" LastEditDate="2015-01-28T23:21:13.073" LastActivityDate="2015-01-28T23:21:13.073" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8316" PostTypeId="2" ParentId="8285" CreationDate="2015-01-29T00:35:06.760" Score="12" Body="&lt;p&gt;An arrangement which can be readily seen to be valid albeit not minimal may be formed by observing that joining two &quot;small&quot; (3-matchstick) triangles and two &quot;large&quot; (9-matchstick) rigid triangles will yield a construct with four vertices of order two (the remaining vertices are all of order four).  The order-two vertices will form a trapezoid whose non-parallel edges are of equal length and form an angle which may, by flexing the joints, may be adjusted within a range that includes 45 degrees and extends almost up to 60.  One may thus fasten seven or eight such constructs together to yield a &quot;donut&quot; which meets the necessary conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Xat49.png&quot; alt=&quot;Solutions using flexible four-triangle section&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The figure on the left has the angles adjusted so as to show that the range extends slightly beyond 45 degrees.  Other constructs may use fewer line segments, but this construct is probably the &quot;simplest&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another useful building block is this combination of four large and two small triangles:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/dtAL7.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;While this section, unlike the one above, isn't directly flexible, it only has two order-two nodes rather than four.  As a consequence, three of them may be combined to yield a regular graph, or--more interestingly--four of them may be combined to yield a flexible regular graph.&lt;/p&gt;&#xA;" OwnerUserId="1467" LastEditorUserId="1467" LastEditDate="2015-01-29T04:58:47.723" LastActivityDate="2015-01-29T04:58:47.723" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8318" PostTypeId="2" ParentId="5261" CreationDate="2015-01-29T07:15:48.833" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am a magician, perhaps by knowing this you can more easily resolve&#xA;  this puzzle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I assume this puzzle relates to tools used by magician.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the songs are about something that have similarity with time&#xA;  (daylight, seasons, weeks, etc.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;My best bet is a playing card&lt;/li&gt;&#xA;&lt;li&gt;Playing card have 2 color (similar with day and night)&lt;/li&gt;&#xA;&lt;li&gt;Have 4 suits (4 seasons in a year)&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Have a total of 52 cards (the number of weeks in a year)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;using this fact, and this page sequence&#xA;http://en.wikipedia.org/wiki/Playing_card#French_suits&#xA;Hearts Diamonds Clovers Clubs Spades&#xA;&#xA;http://en.wikipedia.org/wiki/Season&#xA;Winter Spring Summer Autumn&#xA;&#xA;Now we have:&#xA;Day as Red, Night as Black&#xA;Hearts Diamonds Clovers Clubs Spades AS Winter Spring Summer Autumn&#xA;&#xA;Search song title at google will bring all song in first result.&#xA;&#xA;The eleventh day of spring 26 ===&amp;gt; Song &quot;Jack of Diamonds&quot; by Blind Lemon Jefferson, 1926&#xA;The twelfth night of summer 74 ==&amp;gt; Song &quot;Queen of Clubs&quot; by KC &amp;amp; The Sunshine Band, 1974&#xA;The first night of autumn 80 ====&amp;gt; Song &quot;Ace of spades&quot; by Motörhead, 1980&#xA;The second day of winter 86 =====&amp;gt; Song &quot;Two of Hearts&quot; by Stacey Q, 1986&#xA;&lt;/code&gt;&lt;/pre&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5958" LastActivityDate="2015-01-29T07:15:48.833" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8320" PostTypeId="1" AcceptedAnswerId="8358" CreationDate="2015-01-29T10:45:21.693" Score="16" ViewCount="1966" Body="&lt;p&gt;On a standard $19\times19$ Go-board there are four stones: two white stones in the lower left and the upper right corner, and two black stones in the lower right and the upper left corner. Alice (playing white) and Bob (playing black) alternately move one of their stones to a horizontally or vertically adjacent point. Bob makes the first move. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alice wins, if she manages to navigate her two white stones to two points that are (horizontally or vertically) adjacent to each other. Bob wins, if he can prevent Alice forever from winning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determine the winner of the game. (As usual, we assume that Alice and Bob use optimal strategies.)&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-23T14:53:10.790" LastActivityDate="2016-03-23T14:53:10.790" Title="Four stones on a Go-board" Tags="|mathematics|strategy|game|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8321" PostTypeId="2" ParentId="8302" CreationDate="2015-01-29T10:52:22.293" Score="1" Body="&lt;p&gt;It should take almost the same time. If the wind was to remain completely constant for the whole trip, at all altitudes (extremely unlikely), the cruise legs added together would be the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During the phases of take off, climb, approach and landing, there would be differences. Depending on the procedures in use (VFR, IFR, SIDs, STARs, etc), the time will be different. It might be necessary to fly the approach into a headwind/tailwind during a joining procedure or instrument approach, while the take off may be straight out into a headwind. The wind direction will require the aircraft to turn around after take off when the cruise has a tailwind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This supposes that the aircraft follows the exact same route for the cruise in both directions (unlikely), and flies at the same altitude (extremely unlikely, unless the entire flight is conducted outside controlled airspace and there is no conflict risk).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of these factors can influence the flight time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, in reality, no wind provides the shortest flight time. There are exceptions, but not with constant wind. I'm a pilot myself and can say for sure that from experience the shortest flights are usually the ones with no wind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another point to take into consideration is that an aircraft moves slower while it's climbing and faster while it's descending. If it has a tailwind on both legs, the climb would take the same time, but would cover more distance at the slower speed, the descent would usually also take the same time, but cover more distance again at the faster speed. The reverse is true for a headwind. So again these factors would affect the flight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The take off and landing is always flown into wind and is at a slower speed, that would make the overall trip longer the higher the wind speed.&lt;/p&gt;&#xA;" OwnerUserId="9240" LastEditorUserId="9240" LastEditDate="2015-01-29T11:11:22.883" LastActivityDate="2015-01-29T11:11:22.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8322" PostTypeId="2" ParentId="8320" CreationDate="2015-01-29T11:11:17.887" Score="17" Body="&lt;p&gt;Assuming Alice goes first, Bob can win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bob can always makes a move that keeps the four stones as corners of a grid-aligned rectangle, with opposed corners belonging to opposite colors. Alice cannot get her two stones adjacent by making a move from such a position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zR0bs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In more detail: Assume this invariant is true before Alice's move. We show how after Alice moves, Bob can make a move after which the invariant is again true. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alice's move causes only one of the edges of the rectangle to break, by shifting to an adjacent column or adjacent row. Then, Bob can move his stone in that edge the same displacement, reforming that edge while keeping its other edge intact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing so is always a legal move: it will be on the board because Alice's move was on the board, and no stone can be occupying the space because the other two are in a different and parallel row/column, as Alice cannot get both her stones onto the same row or column.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-01-29T11:18:59.967" LastActivityDate="2015-01-29T11:18:59.967" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8323" PostTypeId="5" CreationDate="2015-01-29T12:03:52.050" Score="0" Body="&lt;p&gt;A puzzle where you have to optimize a certain objective function. Sometimes these will be &lt;a href=&quot;/questions/tagged/mathematics&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;mathematics&amp;#39;&quot; rel=&quot;tag&quot;&gt;mathematics&lt;/a&gt;-focused, with goals such as:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;maximizing bananas transported to market: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/230/a-camel-transporting-bananas&quot;&gt;A camel transporting bananas&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;minimizing area of a blanket: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2693/a-blanket-for-my-baby-snake&quot;&gt;A blanket for my baby snake&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;maximizing the smallest amount that 8 coins can pay for: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/77750/eight-coins-for-the-fair-king&quot;&gt;Eight coins for the fair king&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;However there are also many kinds of less mathematical optimization problems. Some examples:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/chess&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;chess&amp;#39;&quot; rel=&quot;tag&quot;&gt;chess&lt;/a&gt; minimizing the &amp;quot;score&amp;quot; of chess pieces to control the whole board: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2907/how-many-chess-pieces-are-needed-to-control-every-square-on-the-board-no-piece&quot;&gt;How many Chess Pieces are needed to control every square on the board? No Piece Restriction&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/sudoku&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;sudoku&amp;#39;&quot; rel=&quot;tag&quot;&gt;sudoku&lt;/a&gt; minimizing the numbers covered by a &amp;quot;Sudoku net&amp;quot; that will always leave a solvable Sudoku: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/11780/sudoku-net-that-is-always-solvable&quot;&gt;Sudoku net that is always solvable&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/board-games&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;board-games&amp;#39;&quot; rel=&quot;tag&quot;&gt;board-games&lt;/a&gt; minimizing or maximizing &amp;quot;shots&amp;quot; fired in a Battleships game, given certain constraints: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/107381/shooting-them-blanks-double-optimization-task&quot;&gt;Shooting them blanks (double optimization task)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/matches&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;matches&amp;#39;&quot; rel=&quot;tag&quot;&gt;matches&lt;/a&gt; minimizing the number of matches removed to leave behind no triangles: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/107502/how-many-matchsticks-need-to-be-removed-so-there-are-no-equilateral-triangles&quot;&gt;How many matchsticks need to be removed so there are no equilateral triangles?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Ideally, puzzle setters will know of a provably optimal solution before posting the puzzle. In the case where no provable best solution exists, when the &amp;quot;puzzle&amp;quot; is more of a game with no final answer, puzzles will be closed as open-ended. See here for why open-ended puzzles are off-topic: &lt;a href=&quot;https://puzzling.meta.stackexchange.com/questions/6617/the-end-of-open-ended-puzzles&quot;&gt;The end of open-ended puzzles&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="69582" LastEditDate="2021-07-19T20:00:05.240" LastActivityDate="2021-07-19T20:00:05.240" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8324" PostTypeId="4" CreationDate="2015-01-29T12:03:52.050" Score="0" Body="A puzzle where you have to optimize a certain objective function (maximize profit, minimize cost). There should ideally be a provable best answer, to avoid making the puzzle into an [open-ended] game." OwnerUserId="8874" LastEditorUserId="69582" LastEditDate="2021-07-19T19:59:50.390" LastActivityDate="2021-07-19T19:59:50.390" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8325" PostTypeId="2" ParentId="8302" CreationDate="2015-01-29T12:14:36.833" Score="2" Body="&lt;p&gt;Not a proof but a strong hint that the trip should be longer comes from the case where the wind matches the normal flight speed of the plane.  It will not be able to complete any segment of the tour against the wind.  The travel becomes infinite.  For intermediate speeds, the time should somewhere in between, between unchanged and infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This being said, if the plane circumnavigates the globe and is pushed by a constant westward wind like the jetstream, of course the trip becomes shorter.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastActivityDate="2015-01-29T12:14:36.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8326" PostTypeId="1" AcceptedAnswerId="8423" CreationDate="2015-01-29T12:37:42.960" Score="4" ViewCount="864" Body="&lt;p&gt;You are asked to create puzzle pieces by joining together identical squares. This needs to be done such that the puzzle pieces can be arranged in a pattern with each piece being in contact with exactly 5 others. In terms of number of starting squares, how low can you go?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Background and further clarification&lt;/strong&gt; &lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/8175/dissecting-a-square&quot;&gt;Here&lt;/a&gt; we saw that 64 identical squares can be joined into puzzle pieces which can subsequently be grouped together &lt;em&gt;in a square&lt;/em&gt; such that each piece is in contact with exactly 5 other pieces.&lt;/li&gt;&#xA;  &lt;li&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/8122/find-smallest-rectangle-divided-into-figures-so-each-figure-has-5-neighbours&quot;&gt;Here&lt;/a&gt; we saw that 42 identical squares can be joined into puzzle pieces which can subsequently be grouped together &lt;em&gt;in a rectangle&lt;/em&gt; such that each piece is in contact with exactly 5 other pieces. &lt;/li&gt;&#xA;  &lt;li&gt;Now we are getting rid of the requirement to group the puzzle pieces into a specific shape: &lt;em&gt;any shape&lt;/em&gt; will do!&lt;/li&gt;&#xA;  &lt;li&gt;We still require two squares being 'joined' or two puzzle pieces being 'in contact' to mean that both share a finite portion of their perimeter. &lt;/li&gt;&#xA;  &lt;li&gt;Also, the joining of the squares into puzzle pieces as well as the grouping of these pieces, is to be done within a plane and without creating any overlap. In other words, there should be no loss of total area covered: the resulting total puzzle needs to cover an area identical to that of the sum of the squares used.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So far this puzzle has seen zero attempts towards a solution. As long as progress is lacking, every one or two days I will publish a hint. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In contrast to the 64 square and 42 square solutions, the solution sought does deploy single square puzzle pieces.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7573" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-07-22T22:04:04.200" Title="Puzzle pieces, each in contact with 5 others" Tags="|mathematics|geometry|graph-theory|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8327" PostTypeId="1" AcceptedAnswerId="8332" CreationDate="2015-01-29T12:47:15.720" Score="9" ViewCount="2175" Body="&lt;p&gt;I'm newbie to cryptic crosswords and I'm a bit lost as to how &quot;abolish&quot; is the answer to the clue &quot;Boil rocks in wood scrap (7).&quot;&lt;/p&gt;&#xA;" OwnerUserId="9241" LastEditorUserId="21234" LastEditDate="2016-07-22T20:59:57.387" LastActivityDate="2016-07-22T20:59:57.387" Title="Cryptic clue: Boil rocks in wood scrap (7)" Tags="|crosswords|cryptic-clues|crossword-clues|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8328" PostTypeId="5" CreationDate="2015-01-29T13:14:05.830" Score="0" Body="&lt;p&gt;A geometric puzzle where a given figure has to be cut into a number of pieces subject to a number of constraints. Typical constraints concern&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;the number of pieces&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the shape of the pieces&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the adjacency relations between the pieces&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-01-29T13:50:50.043" LastActivityDate="2015-01-29T13:50:50.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8329" PostTypeId="4" CreationDate="2015-01-29T13:14:05.830" Score="0" Body="A geometric puzzle where a given figure has to be cut into a number of pieces subject to a number of constraints." OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-01-29T18:20:37.510" LastActivityDate="2015-01-29T18:20:37.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8330" PostTypeId="1" AcceptedAnswerId="8334" CreationDate="2015-01-29T13:18:03.730" Score="7" ViewCount="2681" Body="&lt;p&gt;Find the largest integer that is a product of three-digit number and a one-digit number and also a product of two two-digit numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, 200 x 1 = 10 x 20. Of course, 200 is not the largest.&lt;/p&gt;&#xA;" OwnerUserId="3914" LastEditorUserId="8874" LastEditDate="2016-03-23T14:52:55.587" LastActivityDate="2022-02-01T12:40:53.733" Title="3d x 1d = 2d x 2d" Tags="|mathematics|number-theory|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8331" PostTypeId="2" ParentId="8330" CreationDate="2015-01-29T13:32:15.403" Score="10" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 8928 = 992 * 9 = 96 * 93 &lt;br&gt;&#xA; 9 - is obviosly largest 1-digit number, so we need to maximize 3-digit number so that their product can be a product of two 2-digit numbers. &lt;br&gt;&#xA; Candidates are 993, 994, 995, 996, 997, 998, 999. &lt;br&gt;&#xA; Let's find divisors of their products with 9 to show that they can't be products of 2-digit numbers. Write them in a row and in second row in reversed order. Example for 9*993:&lt;br&gt;&#xA; &lt;code&gt;1,    3,    9,   27,  331, 993, 2979, 8937&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;8937, 2979, 993, 331, 27,  9,   3,    1&lt;/code&gt;&lt;br&gt;&#xA; Now it's easy to see that it cannot be product of 2-digit number (there will be two 2-digit numbers one above second if it is possible). In same way you check other candidates (not showing here, because long lists of divisors).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5781" LastEditorUserId="5781" LastEditDate="2015-01-30T07:45:23.850" LastActivityDate="2015-01-30T07:45:23.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8332" PostTypeId="2" ParentId="8327" CreationDate="2015-01-29T13:32:29.470" Score="7" Body="&lt;p&gt;'Boil rocks' means 'an anagram of boil' ('boil' that's been kind of jolted around).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;'Wood scrap' (noun) is what you get once you've burnt some wood: 'ash' &lt;br&gt;(or maybe the author just intends 'wood', by itself, to mean 'ash' the species of tree - see comments below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;'In' means 'in the middle of'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it's an anagram of 'boil', in the middle of the word 'ash'.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a-boli-sh&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;'Scrap' (verb) is the whole thing (to 'abolish' is to 'scrap').&lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="4050" LastEditDate="2015-01-30T10:46:28.827" LastActivityDate="2015-01-30T10:46:28.827" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8333" PostTypeId="2" ParentId="8320" CreationDate="2015-01-29T13:37:42.103" Score="15" Body="&lt;p&gt;If Bob moves first, then Alice can win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alice's strategy is simple: Move her stones closer. We show that such a move is always possible, so the game can only end when Alice's stones are adjacent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bob's moving first induces a parity restriction. If we color the points of the go board in a checkerboard pattern, then each moves changes the checkerboard color of one of the stones. Since all stones start on the same color, after each of Bob's moves, and odd number of moves have been made, so there are an odd number of stones on each checkerboard color. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that if one of Bob's stones is next to one of Alice's stones, then Bob's other stone can't be next to Alice's other stone, or else there would be two stones on each checkerboard color. So, one of Bob's stones is always not impeding Alice's move. Therefore, Bob is only blocking two of the four moves Alice has to bring her stones closer, and so Alice has a move (two in fact) to do so. The exception is when Bob's stone is immediately in between two of Alice's stones on the same line, but Alice can avoid this by choosing not to put both her stones on the same line unless this wins immediately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that Alice isn't slowed down by Bob at all: each of her moves decreases the Manhattan distance between her stones by 1, so she wins in 35 of her moves, same as if Bob wasn't there.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-01-29T13:48:15.017" LastActivityDate="2015-01-29T13:48:15.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8334" PostTypeId="2" ParentId="8330" CreationDate="2015-01-29T13:43:14.240" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $8928 = 992 \times 9 = 96 \times 93$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way to get to a number larger than $8000$ is if the number is of the form $(100-n)(100-m)=9(1000-k)$. The last two digits of the result equal minus a multiple of $9$ modulo $100$: $19$, $28$, $37$, $46$, $55$, $64$, $73$, $82$, or $91$. The smallest number in this list that occurs in the standard multiplication tables is $28$. So, the solution sought is $n=4$, $m=7$ and $100-9k=28$ or $k=8$.&lt;/p&gt;&#xA;" OwnerUserId="7573" LastEditorUserId="7573" LastEditDate="2015-01-29T14:14:46.907" LastActivityDate="2015-01-29T14:14:46.907" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8335" PostTypeId="2" ParentId="8162" CreationDate="2015-01-29T14:25:32.233" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Arbitrarily (randomly)&#xA;divide the pile (or deck) of cards into two piles of $90$ and $10$. &#xA;\begin{array}{lcl}\text{Status}&amp;amp;\text{Pile1}&amp;amp;\text{Pile2}\\\text{Total cards}&amp;amp;90&amp;amp;~~10\\\text{Total face-down}&amp;amp;x&amp;amp;~~10-x\hskip1in\end{array}&#xA;&lt;br&gt;&#xA;Now flip all cards in Pile2 to get equal number of face-down cards in both piles. &#xA;&#xA;\begin{array}{lcl}\text{Status}&amp;amp;\text{Pile1}&amp;amp;\text{Pile2}\\\text{Total cards}&amp;amp;90&amp;amp;~~10\\\text{Total face-down}&amp;amp;x&amp;amp;~~10-(10-x) = x\end{array}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2656" LastEditorUserId="19765" LastEditDate="2016-08-28T04:50:08.050" LastActivityDate="2016-08-28T04:50:08.050" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8336" PostTypeId="1" AcceptedAnswerId="8337" CreationDate="2015-01-29T14:58:00.827" Score="32" ViewCount="3919" Body="&lt;p&gt;Alfréd Rényi, a Hungarian mathematician, said that a mathematician is a machine for turning coffee into theorems. It's believed that his mention was about his colleague, Paul Erdős, the celebrated mathematician.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much coffee had he drunken for his theorems? I think that he would drink 3 cups of coffee per theorem. :-) &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  COFFEE&#xA;  COFFEE&#xA;+ COFFEE&#xA;--------&#xA; THEOREM&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="3914" LastEditorUserId="69582" LastEditDate="2023-08-17T03:58:06.767" LastActivityDate="2023-08-17T03:58:06.767" Title="A mathematician's coffee alphametic" Tags="|mathematics|logical-deduction|alphametic|" AnswerCount="5" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8337" PostTypeId="2" ParentId="8336" CreationDate="2015-01-29T15:03:03.153" Score="24" Body="&lt;p&gt;Letter values:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; C = 8, O = 3, F = 1, E = 9, T = 2, H = 4, R = 5, M = 7.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The equation:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;&lt;pre&gt;   831199&#xA;   831199&#xA; + 831199&#xA; --------&#xA;  2493597&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8858" LastEditorUserId="81979" LastEditDate="2023-08-16T14:14:53.760" LastActivityDate="2023-08-16T14:14:53.760" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="8338" PostTypeId="2" ParentId="8336" CreationDate="2015-01-29T15:19:52.877" Score="13" Body="&lt;p&gt;Well, first, we can determine E, since $E*11*3 = 100's + 10*E + change$.   Only number that fits is $E = 9, M = 7$. (Another way to think of it is to take each number times 3, add the digits and figure out which one matches; 9*3=27, 2+7=9)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, O and F are interesting, since the 10's of $F*11*3 = O$ and $3*O + (3*F \% 100) = 9$.  If O = 3, then F&amp;lt;=3, since it can't rollover to the hundreds. If F = 1, then we have no rollover. So, $O = 3, F = 1, R = 5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, that just leaves us with $3*C = TH$ without repeating a digit. C &gt;= 7, since it has to be 2 digits and 1 is already taken. 8 works, since 24 is good. So, $C = 8, T = 2, H = 4$  &lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-01-29T15:38:40.090" LastActivityDate="2015-01-29T15:38:40.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8339" PostTypeId="2" ParentId="8330" CreationDate="2015-01-29T16:02:33.800" Score="2" Body="&lt;p&gt;A longer, more thorough attempt at finding the solution without doing actual multiplications (except for &lt;span class=&quot;math-container&quot;&gt;$32\times32 = 2^5\times2^5 = 2^{10} = 1024$&lt;/span&gt;). Hopefully it's easier to understand.&lt;/p&gt;&#xA;&lt;p&gt;Notations:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;ab &amp;amp;=\text{number with first digit}\ a\ \text{and second digit}\ b\\&#xA;a \times b &amp;amp;= a\ \text{multiplied by}\ b&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;First phase&lt;/strong&gt; is to observe that&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$1d$&lt;/span&gt; must be &lt;span class=&quot;math-container&quot;&gt;$9$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The highest solution that can be made with &lt;span class=&quot;math-container&quot;&gt;$1d = 8$&lt;/span&gt; is:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$8\times999 = 8 \times 111 \times 9 = 9 \times 888 &amp;lt; 9 \times 900 = 90 \times 90$&lt;/span&gt;&lt;br /&gt;&#xA; Ergo there are solutions with &lt;span class=&quot;math-container&quot;&gt;$1d = 9$&lt;/span&gt; higher than can be made with &lt;span class=&quot;math-container&quot;&gt;$1d = 8$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The important thing to note is that the number is &lt;span class=&quot;math-container&quot;&gt;$9\times 3d$&lt;/span&gt;, which is divisible by &lt;span class=&quot;math-container&quot;&gt;$9$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;The second phase&lt;/strong&gt; is to determine/guess the number starting with the highest possible factors.&#xA;Note that&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$2d\times2d$&lt;/span&gt; is divisible by &lt;span class=&quot;math-container&quot;&gt;$9$&lt;/span&gt;, so either one of the factors is divisible by &lt;span class=&quot;math-container&quot;&gt;$9$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$99,90,$&lt;/span&gt;etc) or both factors are divisible by &lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$96,93,$&lt;/span&gt;etc)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This leads to the following cases (we stop when it's obvious we can't find better solutions):&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$$\begin{align}99\times9x &amp;amp;= 9 \times 11 \times 9x \\&amp;amp;= 9 \times 9(9+x)x \\&amp;amp;\implies x = 0 \\&amp;amp;\implies\text{solution}: 99 \times 90 = 9 \times 990\\ 96\times96 &amp;amp;= 9 \times 32 \times 32 \\&amp;amp;= 9 \times 1024 \\&amp;amp;\implies\text{no solution}\\ 96\times93 &amp;amp;= 9 \times 32 \times 31 \\&amp;amp;= 9 \times (1024-32) \\&amp;amp;= 9 \times 992 \\&amp;amp;\implies\text{solution}: 96 \times 93 = 9 \times 992\\ 93\times93 &amp;amp;\ \text{irrelevant as it's smaller than}\ 96\times93\\ 90\times9x &amp;amp;\ \text{the best solution is obviously}\ 99\times90=9\ \text{discovered above}\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Obviously the best solution is&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$$\boxed{9\times992=96\times93}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3918" LastEditorUserId="54367" LastEditDate="2022-02-01T12:40:53.733" LastActivityDate="2022-02-01T12:40:53.733" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8340" PostTypeId="5" CreationDate="2015-01-29T16:04:40.647" Score="0" Body="&lt;p&gt;In graph theory, a graph coloring is a labeling of the vertices of a graph with so-called colors, subject to the constraint that adjacent vertices must receive distinct colors/labels. &#xA;The smallest possible number of colors in such a coloring is called the&#xA;&lt;em&gt;chromatic number&lt;/em&gt; of the underlying graph.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Example: The black-and-white coloring of a chessboard shows that the underlying graph (with a vertex corresponding to every cell, and an edge between neighboring cells) has chromatic number at most 2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Graph coloring puzzles typically ask&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;for finding the chromatic number of a (often hidden) graph structure&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;for using a highly structured graph coloring to derive a useful combinatorial property.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-01-29T18:19:47.327" LastActivityDate="2015-01-29T18:19:47.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8341" PostTypeId="4" CreationDate="2015-01-29T16:04:40.647" Score="0" Body="Graph coloring puzzles are centered around the chromatic number of a (often hidden) underlying graph structure." OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-01-29T21:32:25.150" LastActivityDate="2015-01-29T21:32:25.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8342" PostTypeId="2" ParentId="8330" CreationDate="2015-01-29T16:09:54.977" Score="7" Body="&lt;p&gt;It took just a couple of minutes to come up with a program that quickly calculates this.  It takes about 0.2 seconds to run.  The variables I use are simply &lt;code&gt;A*B=C*D&lt;/code&gt; and the for loops take care of the number of digits in each variable.  Here's the output:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Largest Product = 8928&#xA;&lt;br&gt;&#xA; Largest A = 992&#xA;&lt;br&gt;&#xA; Largest B = 9&#xA;&lt;br&gt;&#xA; Largest C = 93&#xA;&lt;br&gt;&#xA; Largest D = 96&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's the C# code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;static void Main(string[] args)&#xA;{&#xA;    int largestProduct = 200;&#xA;    int[] largestABCD = { 200, 1, 10, 20 };&#xA;&#xA;    for(int a = 100; a &amp;lt;= 999; a++)&#xA;    {&#xA;        for(int b = 1; b &amp;lt;= 9; b++)&#xA;        {&#xA;            int product1 = a * b;&#xA;&#xA;            for(int c = 10; c &amp;lt;= 99; c++)&#xA;            {&#xA;                for(int d = 10; d &amp;lt;= 99; d++)&#xA;                {&#xA;                    int product2 = c * d;&#xA;                    if(product1 == product2)&#xA;                    {&#xA;                        if (product1 &amp;gt; largestProduct)&#xA;                        {&#xA;                            largestProduct = product1;&#xA;                            largestABCD[0] = a;&#xA;                            largestABCD[1] = b;&#xA;                            largestABCD[2] = c;&#xA;                            largestABCD[3] = d;&#xA;                        }&#xA;                    }&#xA;                }&#xA;            }&#xA;        }&#xA;    }&#xA;&#xA;    Console.WriteLine(&quot;Largest Product = &quot; + largestProduct.ToString());&#xA;    Console.WriteLine(&quot;Largest A = &quot; + largestABCD[0].ToString());&#xA;    Console.WriteLine(&quot;Largest B = &quot; + largestABCD[1].ToString());&#xA;    Console.WriteLine(&quot;Largest C = &quot; + largestABCD[2].ToString());&#xA;    Console.WriteLine(&quot;Largest D = &quot; + largestABCD[3].ToString());&#xA;&#xA;    Console.ReadLine();&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I could probably come up with some optimizations for this code.  I could constrain the &lt;code&gt;D&lt;/code&gt; loop counter.  Or I could break out of a loop once &lt;code&gt;product2&lt;/code&gt; is greater than &lt;code&gt;product1&lt;/code&gt;.  But the code runs so fast that it doesn't matter.&lt;/p&gt;&#xA;" OwnerUserId="2318" LastActivityDate="2015-01-29T16:09:54.977" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8343" PostTypeId="5" CreationDate="2015-01-29T16:10:06.493" Score="0" Body="&lt;p&gt;A &lt;a href=&quot;https://en.wikipedia.org/wiki/Graph_(discrete_mathematics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;graph&lt;/em&gt;&lt;/a&gt; &lt;span class=&quot;math-container&quot;&gt;$G=(V,E)$&lt;/span&gt; consists of a set &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt; of &lt;em&gt;vertices&lt;/em&gt; (or &lt;em&gt;nodes&lt;/em&gt;) and a set &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; of &lt;em&gt;edges&lt;/em&gt;; every edge in E connects two vertices in V. Two vertices are called &lt;em&gt;adjacent&lt;/em&gt; if there is an edge between them.&lt;/p&gt;&#xA;&lt;p&gt;Graphs are one of the primary objects of study in discrete mathematics and combinatorics. Graph structures often show up in combinatorial puzzles; usually, they are hidden, and part of the solution process is to detect the graph structure. Take note that general graph theory questions are off-topic but can be asked on our sister site &lt;a href=&quot;https://math.stackexchange.com/&quot;&gt;Mathematics Stack Exchange&lt;/a&gt; using the [graph-theory] tag.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="60644" LastEditDate="2020-11-30T07:33:25.910" LastActivityDate="2020-11-30T07:33:25.910" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8344" PostTypeId="4" CreationDate="2015-01-29T16:10:06.493" Score="0" Body="A puzzle built around graphs: sets of nodes joined together by paths. Use with [mathematics]" OwnerUserId="8874" LastEditorUserId="69582" LastEditDate="2021-01-25T19:21:48.317" LastActivityDate="2021-01-25T19:21:48.317" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8345" PostTypeId="5" CreationDate="2015-01-29T16:29:35.590" Score="0" Body="&lt;p&gt;A puzzle that asks for determining the winner in a (typically combinatorial) multi-player game.&#xA;Here &lt;em&gt;multi-player&lt;/em&gt; may actually mean&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;single-player: solitaire games; games against nature; games against stochastic adversaries&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;two-player: standard board games; combinatorial two-player games&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;three or more players: resource-sharing games; profit division games; allocation games.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;The solution to a combinatorial two-player game puzzle usually consists of a full analysis of all game situations, by classifying them into &amp;quot;wins&amp;quot; and &amp;quot;losses&amp;quot; for either player. Take note that game questions that aren't puzzles are off-topic and should be asked on &lt;a href=&quot;https://gaming.stackexchange.com/&quot;&gt;Arqade Stack Exchange&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="71981" LastEditDate="2020-11-30T01:56:11.337" LastActivityDate="2020-11-30T01:56:11.337" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8346" PostTypeId="4" CreationDate="2015-01-29T16:29:35.590" Score="0" Body="A puzzle that asks for determining the winner in a multi-player game." OwnerUserId="8874" LastEditorUserId="70545" LastEditDate="2021-02-04T13:58:29.923" LastActivityDate="2021-02-04T13:58:29.923" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8347" PostTypeId="5" CreationDate="2015-01-29T16:40:55.633" Score="0" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Steganography&quot; rel=&quot;nofollow noreferrer&quot;&gt;Steganography&lt;/a&gt; is the art of concealing data within another message.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concealed message can be plainly visible, or it may require advanced cryptographic techniques and data manipulation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a message could be hidden in the alpha transparency bits of an image, such that it doesn't make any difference to the naked eye.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="11497" LastEditDate="2020-01-29T00:23:44.253" LastActivityDate="2020-01-29T00:23:44.253" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8348" PostTypeId="4" CreationDate="2015-01-29T16:40:55.633" Score="0" Body="A puzzle that requires finding hidden data in various media forms, such as text or images." OwnerUserId="8874" LastEditorUserId="69582" LastEditDate="2020-12-06T17:02:56.313" LastActivityDate="2020-12-06T17:02:56.313" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8349" PostTypeId="5" CreationDate="2015-01-29T16:47:37.030" Score="0" Body="&lt;p&gt;A &lt;em&gt;polyomino&lt;/em&gt; is a plane geometric figure formed by joining one or more unit-squares edge to edge. Equivalently, a polyomino is a finite subset of the regular square tiling with a connected interior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typical examples of polyominoes are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the 2x1 dominoes (consisting of 2 squares)&lt;/li&gt;&#xA;&lt;li&gt;the straight 3x1 tromino and the L-shaped tromino (both consisting of 3 unit-squares)&lt;/li&gt;&#xA;&lt;li&gt;tetromino pieces (consisting of 4 squares), as used in TETRIS&lt;/li&gt;&#xA;&lt;li&gt;pentomino pieces (consisting of 5 squares); there are 12 pentominoes, up to rotation&lt;/li&gt;&#xA;&lt;li&gt;hexomino pieces (consisting of 6 squares)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;By definition, polyominoes (consisting of 8 or more squares) may contain holes.&#xA;For some puzzles (such as tiling problems), polyominoes with holes are inconvenient.&#xA;Polyominoes without holes are called &lt;em&gt;simply connected&lt;/em&gt; polyominoes.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="52483" LastEditDate="2018-11-27T05:38:55.393" LastActivityDate="2018-11-27T05:38:55.393" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8350" PostTypeId="4" CreationDate="2015-01-29T16:47:37.030" Score="0" Body="A geometric puzzle centered around geometric figures formed from unit-squares, or a puzzle that uses polyominoes as an integral part." OwnerUserId="8874" LastEditorUserId="69582" LastEditDate="2021-01-28T15:47:44.110" LastActivityDate="2021-01-28T15:47:44.110" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8351" PostTypeId="2" ParentId="8320" CreationDate="2015-01-29T17:12:32.223" Score="0" Body="&lt;p&gt;I believe Bob will always wins. He just has to &quot;circle&quot; a stone (adjacent vertically and horizontally). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the shortest way is to move just one stone in a &quot;L&quot; way, Alice needs 18 movements vertically and 17 movements horizontally (or viceversa). Meanwhile, in order to circle the static stone he just has to move 17 times vertically and 17 times horizontally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if Alice starts, she will always be behind Bob's movements and by the time she gets one stone close to the other one, Bob will have it circled making it impossible for Alice to make the two stones adjacent.&lt;/p&gt;&#xA;" OwnerUserId="9245" LastActivityDate="2015-01-29T17:12:32.223" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8352" PostTypeId="1" AcceptedAnswerId="8353" CreationDate="2015-01-29T17:41:00.957" Score="22" ViewCount="1913" Body="&lt;p&gt;Here's a riddle I used in a game some time ago. Not the best riddle ever, but we had fun solving it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was written on a translucent door, on the frame of which was written &quot;He who holds the answer may enter&quot;. What's the riddle about and what would your action be?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;We're quite the couple, you and I&#xA;If you let me live, I'll surely die&#xA;&#xA;If you kill me to live another night&#xA;I'll be there next evening, at your side&#xA;&#xA;I won't keep you warm, I'll give you that&#xA;And I am most cold when I wear my hat&#xA;&#xA;Whenever I cry you just leave me be&#xA;while your tears are like poison to me&#xA;&#xA;I see the pain in your eyes, whenever we touch&#xA;I shiver when you speak or shush&#xA;&#xA;If we're too passionate you'll surely die&#xA;We're quite the couple, you and I&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="9208" LastActivityDate="2015-01-29T18:03:56.230" Title="Riddle - He who holds the answer may enter" Tags="|riddle|english|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8353" PostTypeId="2" ParentId="8352" CreationDate="2015-01-29T18:03:56.230" Score="23" Body="&lt;p&gt;The item spoken of in the riddle is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a candle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;We're quite the couple, you and I&#xA;If you let me live, I'll surely die&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Leave the candle burning, it will be consumed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;If you kill me to live another night&#xA;I'll be there next evening, at your side&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; snuff out the candle flame and you will be able to re-use the candle the next night.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I won't keep you warm, I'll give you that&#xA;And I am most cold when I wear my hat&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Candle flames are good for (a little) light, not for keeping things warm; if you put a &quot;hat&quot; (a candle-snuffer) on a candle, it goes out and doesn't provide any warmth at all.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Whenever I cry you just leave me be&#xA;while your tears are like poison to me&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The candle's &quot;tears&quot; are the wax runoff, which we don't do anything about; but our tears would put out the candle-flame.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I see the pain in your eyes, whenever we touch&#xA;I shiver when you speak or shush&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it hurts us to touch the candleflame, and the flame flickers from the air-currents of our breathing or speaking.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;If we're too passionate you'll surely die&#xA;We're quite the couple, you and I&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let the flame get too big/too close, you're in for a world of hurt.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, my action would be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; stepping through the doorframe while holding a lit candle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5874" LastActivityDate="2015-01-29T18:03:56.230" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8354" PostTypeId="2" ParentId="8336" CreationDate="2015-01-29T19:57:26.600" Score="5" Body="&lt;p&gt;Assuming that a letter can have the same value as another letter, these are some lazy solutions.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; T = 1, H = 2, C = 4, E = F = M = O = R = 0&lt;br&gt;&#xA; T = 1, H = 5, C = 5, E = F = M = O = R = 0&lt;br&gt;&#xA; T = 1, H = 8, C = 6, E = F = M = O = R = 0&lt;br&gt;&#xA; T = 2, H = 1, C = 7, E = F = M = O = R = 0&lt;br&gt;&#xA; T = 2, H = 4, C = 8, E = F = M = O = R = 0&lt;br&gt;&#xA; T = 2, H = 7, C = 9, E = F = M = O = R = 0&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8942" LastEditorUserId="8942" LastEditDate="2015-01-29T20:54:29.660" LastActivityDate="2015-01-29T20:54:29.660" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8355" PostTypeId="2" ParentId="496" CreationDate="2015-01-29T20:26:00.963" Score="1" Body="&lt;p&gt;Here are some ideas for constructing such a number (x):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;X mod 10 = 1: This means that the ones digit of X is 1.&lt;/li&gt;&#xA;&lt;li&gt;X mod 9 = 1: This means that the sum of all of the digits except the ones digit is 9.  I know this because the sum of the digits of all numbers divisible by 9 is 9 (549 -&gt; 5 + 4 + 9 = 18 -&gt; 1 + 8 = 9)&lt;/li&gt;&#xA;&lt;li&gt;X mod 4 = 1: This is important because numbers divisible by 4 have a property that the last two digits of the number are divisible by 4.  (I know that 46382648462984624 is divisible by 4 because the last two digits, 24, are divisible by four)&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Let's look at what this means.  Take the number 331.  Subtract 1 to get 330.  The last two digits of 330 are not divisible by 4, therefore X &amp;lt;&gt; 331.  In fact we can say that the last two digits of X cannot be 31.  So X &amp;lt;&gt; 531, and X &amp;lt;&gt; 75873631, you get the idea.&lt;/li&gt;&#xA;&lt;li&gt;From the previous point, you can see that the last two digits of X must be one of { 01, 21, 41, 61, and 81 }&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This should narrow down your search space.  Let's try some numbers:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;110: While divisible by 11, the ones digit is not 1, the digits don't sum to 10, and 09 is not divisible by 4.  X &amp;lt;&gt; 110&lt;/li&gt;&#xA;&lt;li&gt;121: The ones digit is 1, but the digits don't sum to 10, and 20 is not divisible by 4.  X &amp;lt;&gt; 121:&lt;/li&gt;&#xA;&lt;li&gt;281: The ones digit is 1, the digits sum to 10, and 80 is divisible by 4.  X is a candidate.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now you just have to check is 281 is divisible by 11, and that 280 is divisible by 8, 7, and 6.  You know it's divisible 5 because it's divisible by 10.  Same thing for 4 -&gt; 8, 3 -&gt; 9, and 2 -&gt; 8.&lt;/p&gt;&#xA;" OwnerUserId="2318" LastEditorUserId="2318" LastEditDate="2015-01-30T14:16:40.743" LastActivityDate="2015-01-30T14:16:40.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8356" PostTypeId="2" ParentId="7929" CreationDate="2015-01-29T23:06:53.713" Score="4" Body="&lt;p&gt;Some small progress, I think: the second of the three additional ciphers, namely&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;27 12 26 9 20 31 14&#xA;20 14 25 29&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;may well encrypt the words&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SELKIRK&lt;br&gt;&#xA; ARMS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Douglas McDavid, one of the owners of the Selkirk Arms in Kirkcudbright, is &lt;a href=&quot;http://www.tripadvisor.co.uk/ShowUserReviews-g186516-d192347-r16560104-The_Selkirk_Arms_Hotel-Kirkcudbright_Dumfries_and_Galloway_Scotland.html&quot; rel=&quot;nofollow&quot;&gt;known for his prizewinning puzzles&lt;/a&gt; and has been since he was manager of the Cally Palace Hotel in Gatehouse of Fleet, another town in Kirkcudbrightshire. (He is referred to once in the article at that link as &quot;Duncan&quot;, presumably a typo.) The Selkirk Arms also &lt;a href=&quot;http://www.selkirkarmshotel.co.uk/wp-content/uploads/2014/03/Selkirk-Newsletter-Autumn-2013.pdf&quot; rel=&quot;nofollow&quot;&gt;hosts an annual &quot;brain game&quot;&lt;/a&gt;, the winner of which is awarded the &quot;egg head&quot; prize. The said establishments, it would be safe to say, also sell whisky. I'm hazarding the guess that he is the author of the puzzle and that the second cipher encrypts the name of the establishment he currently part-owns.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Moral:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;cryptanalysis is nothing without humint :-)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7363" LastActivityDate="2015-01-29T23:06:53.713" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8357" PostTypeId="2" ParentId="8336" CreationDate="2015-01-29T23:39:41.933" Score="9" Body="&lt;p&gt;Noting that 3*E gives a different result in the first two columns, there must be an overflow i.e. E is greater than 3. This overflow can only be 1 or 2 (you cannot get more than 27 from tripling a single digit), and in the second column 3*E + overflow =&gt; E. This means 2*E + overflow (which is 1 or 2) equals either 10 or 20. E=9 is the only number which satisfies this, so E=9, M=7:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  COFF99&#xA;  COFF99&#xA;+ COFF99&#xA;--------&#xA; TH9OR97&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;seeing that 3*O + (overflow from 3*F) = 9, O=3 and the overflow is zero (the overflow could only have been 0, 1 or 2, and only 0 is compatible with the result being divisible by 3).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  C3FF99&#xA;  C3FF99&#xA;+ C3FF99&#xA;--------&#xA; TH93R97&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;but 3*F + (overflow from 3*F, which we know is 0) = 3, so F=1&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  C31199&#xA;  C31199&#xA;+ C31199&#xA;--------&#xA; TH93597&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which also gives R=5. Now 3*C = TH. T can only be 1 or 2, and 1 is taken. T=2. To give an overflow of 2, C must be greater than 6. But 7 and 9 are already taken, so C=8, forcing H=4&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  831199&#xA;  831199&#xA;+ 831199&#xA;--------&#xA; 2493597&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="3987" LastActivityDate="2015-01-29T23:39:41.933" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8358" PostTypeId="2" ParentId="8320" CreationDate="2015-01-30T01:33:36.757" Score="8" Body="&lt;p&gt;I need to see it visually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reduce the problem to a simpler one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a 3x3 board with black and white stones as indicated in the problem statement.&#xA;(Think tic-tac-toe board with black in 1 and 9 and white in 3 and 7.)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;B|2|W&#xA;4|5|6&#xA;W|8|B&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Bob makes the first move which is to move a black stone next to one of Alice's white stones and reducing Alice's possible moves to 2, 6, and 8 as indicated here.&#xA;It doesn't matter which stone moves or to which point, because they are all reflections of each other.  So, Alice's first move it always from this position and she has 3 choices: 2, 6, or 8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Alice chooses to move 7 to 8 and Bob does not move to 5, &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;B|2|W    1|2|W    1|2|W    1|2|W    1|x|W&#xA;4|5|6 -&amp;gt; B|5|6 -&amp;gt; B|5|6 ?&amp;gt; B|5|B -&amp;gt; B|W|B&#xA;W|8|B    W|8|B    7|W|B    7|W|9    7|8|9&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;B|2|W    1|2|W    1|2|W    B|2|W    B|x|W&#xA;4|5|6 -&amp;gt; B|5|6 -&amp;gt; B|5|6 ?&amp;gt; 4|5|B -&amp;gt; 4|W|x&#xA;W|8|B    W|8|B    7|W|B    7|W|9    7|8|B&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Alice will move 8 to 5, providing at least one move that cannot be blocked.  So Bob must move to 5.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;B|2|W    1|2|W    1|2|W    1|2|W&#xA;4|5|6 -&amp;gt; B|5|6 -&amp;gt; B|B|6 -&amp;gt; 4|B|6&#xA;W|8|B    W|8|B    7|W|B    7|W|B&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Alice then moves 3 to 6 reducing the distance between the four stones to the minimum. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1|2|3&#xA;4|B|W&#xA;7|W|B&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;At this point Bob must move the black stone at 5 to either 2 or 4 allowing Alice to move either of her stones into the vacated point for the win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other possible moves for Alice yield the same results.  As long as Alice always moves one of her stones towards the other, no matter how Bob blocks, eventually the stones will converge to minimal distance between the 4 stones and Bob will be forced to move away allowing Alice to move into the vacated point and take the win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expand to a 19x19 board and the results will be the same.  Go Alice!&lt;/p&gt;&#xA;" OwnerUserId="9251" LastActivityDate="2015-01-30T01:33:36.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8359" PostTypeId="1" AcceptedAnswerId="22115" CreationDate="2015-01-30T03:00:44.990" Score="15" ViewCount="1682" Body="&lt;p&gt;The question &lt;a href=&quot;https://puzzling.stackexchange.com/questions/8285/touching-matchsticks&quot;&gt;Touching Matchsticks&lt;/a&gt; asked for the smallest matchstick graph where every node is connected to four distinct edges.  The Harborth Graph is the smallest such, with 104 edges connecting 52 nodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5eK9P.gif&quot; alt=&quot;Harboth Graph, not quite to scale&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is made of four identical (but for reflection) sections, each of whose shape, in isolation, would have one degree of freedom.  Each section has two pairs of points that connect to other sections; drawing a line through each pair of points, the angle may be made greater than or less than ninety degrees; by the Mean Value Theorem, the angle may also be equal to ninety degrees, which is what's required for the shape to fit together with reflected copies of itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A curious feature of the Harboth Graph is that while it can be shown that four quarters of the Harborth graph drawn with the proper angles will fit together perfectly, &lt;a href=&quot;http://arxiv.org/pdf/math/0609360.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;it is impossible to produce the correct angles using compass and straightedge alone&lt;/a&gt;.  If one were tasked with the task of drawing a regular matchstick graph of order four using compass and straightedge alone, what would be the smallest graph that could actually be constructed (determining the precise location of vertices using compass and straightedge alone)? A few different regular graphs, with various kinds of symmetry, can be compass-and-straightedge constructed fairly easily with 63 nodes and 126 edges; are there any four-regular matchstick graphs with less than 63 nodes that can can be constructed (with compass and straightedge) at all?&lt;/p&gt;&#xA;" OwnerUserId="1467" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-09-12T00:57:48.053" Title="Touching matchsticks with compass and straightedge" Tags="|mathematics|geometry|matches|" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="8360" PostTypeId="1" AcceptedAnswerId="8361" CreationDate="2015-01-30T06:10:29.130" Score="15" ViewCount="2254" Body="&lt;p&gt;I've tried looking at the paper from different angles, and nothing is jumping out at me. I'm not sure what to try next.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/jOBLu.jpg&quot; alt=&quot;7*8 grid with partially and completely filled squares.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="9253" LastEditorUserId="5188" LastEditDate="2015-01-31T00:22:37.323" LastActivityDate="2015-05-16T06:32:03.277" Title="How to solve this 7x8 grid with partially and completely filled squares?" Tags="|visual|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8361" PostTypeId="2" ParentId="8360" CreationDate="2015-01-30T06:23:32.697" Score="33" Body="&lt;p&gt;The answer is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HELLO - just compress the vertical dimension like this:&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/9P2Pw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastActivityDate="2015-01-30T06:23:32.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8362" PostTypeId="5" CreationDate="2015-01-30T09:08:04.843" Score="0" Body="&lt;p&gt;A &lt;a href=&quot;https://en.wikipedia.org/wiki/Latin_square&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Latin square&lt;/strong&gt;&lt;/a&gt; is an &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;-by-&lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; array filled with &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; different symbols, each occurring exactly once in each row and exactly once in each column.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Sudoku and KenKen puzzles form special cases of Latin squares; any solution to a Sudoku or a KenKen puzzle is a Latin square. (However, it is unnecessary to use this tag with every question relating to puzzles of these types.)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It is &lt;a href=&quot;https://en.wikipedia.org/wiki/NP-completeness&quot; rel=&quot;nofollow noreferrer&quot;&gt;NP-complete&lt;/a&gt; (i.e. computationally intractable) to decide whether a partially filled square can be completed to a full Latin square.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The origin of the use of 'Latin' in the name dates back to the mathematician &lt;a href=&quot;https://en.wikipedia.org/wiki/Leonhard_Euler&quot; rel=&quot;nofollow noreferrer&quot;&gt;Leonhard Euler&lt;/a&gt;, who used Latin characters as symbols.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="60644" LastEditDate="2022-12-07T11:57:05.567" LastActivityDate="2022-12-07T11:57:05.567" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8363" PostTypeId="4" CreationDate="2015-01-30T09:08:04.843" Score="0" Body="A combinatorial puzzle related to Latin squares. " OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-01-30T13:32:59.360" LastActivityDate="2015-01-30T13:32:59.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8364" PostTypeId="1" AcceptedAnswerId="8365" CreationDate="2015-01-30T09:13:36.437" Score="23" ViewCount="4363" Body="&lt;p&gt;A simple sequence of numbers, which I encountered during intermediate school, in my math book. Any person that finished intermediate school possesses the necessary toolbox required to solve this puzzle. Yet, so far, I have never found anyone that could solve it. I have had many collegues and friends try (and most of them are engineers), but it still remains unsolved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to admit, I didn't solve it either, at the age of 14, but I certainly had the matemathical knowledge to be able to solve it, and the puzzle have been stuck in my mind ever since. It's really great (in my mind).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The start of the sequence is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;8  5  4  9  1  7  6 ...&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering all the problem solvers here, I assume it will be solved pretty quick :) But it would be fun if it remained unsolved for just a little longer :)&lt;/p&gt;&#xA;" OwnerUserId="2781" LastEditorUserId="8874" LastEditDate="2015-10-09T07:34:51.407" LastActivityDate="2016-03-11T09:06:53.327" Title="Very simple, yet unsolved, sequence of numbers" Tags="|pattern|number-sequence|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8365" PostTypeId="2" ParentId="8364" CreationDate="2015-01-30T09:19:27.167" Score="31" Body="&lt;p&gt;This puzzle is &lt;strong&gt;not&lt;/strong&gt; mathematical at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint1: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The puzzle is an English-language puzzle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Think lexicographically! Think alphabetically!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solution: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; eight, five, four, nine, one, seven, six, three, two, zero  --&gt;   8 5 4 9 1 7 6 3 2 0&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-11T09:06:53.327" LastActivityDate="2016-03-11T09:06:53.327" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8367" PostTypeId="5" CreationDate="2015-01-30T10:31:57.197" Score="0" Body="&lt;p&gt;A 'number sequence' is a series or collection of numbers presented in a particular order according to a specific logical rule. In &lt;a href=&quot;/questions/tagged/number-sequence&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;number-sequence&amp;#39;&quot; aria-label=&quot;show questions tagged &amp;#39;number-sequence&amp;#39;&quot; rel=&quot;tag&quot; aria-labelledby=&quot;number-sequence-container&quot;&gt;number-sequence&lt;/a&gt; puzzles, the aim for the solver is usually to deduce or discover the rule behind this ordering, and/or supply the next - or missing - number(s) in the sequence.&lt;/p&gt;&#xA;&lt;p&gt;The hidden pattern is often mathematical in nature (in which case, the &lt;a href=&quot;/questions/tagged/mathematics&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;mathematics&amp;#39;&quot; aria-label=&quot;show questions tagged &amp;#39;mathematics&amp;#39;&quot; rel=&quot;tag&quot; aria-labelledby=&quot;mathematics-container&quot;&gt;mathematics&lt;/a&gt; tag may also be used), but sometimes is derived from common knowledge, structures (e.g. Morse code), or language (e.g. alphabetical ordering).&lt;/p&gt;&#xA;&lt;p&gt;For some advice on how to create interesting number sequence puzzles (and also what &lt;em&gt;not&lt;/em&gt; to do), see &lt;a href=&quot;https://puzzling.meta.stackexchange.com/questions/5712/number-sequence-puzzles-what-not-to-do&quot;&gt;this meta question&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Please note: Many common &amp;quot;brainteaser&amp;quot; number sequences, such as the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2271/interview-question-or-pathbreaking-puzzle&quot;&gt;&amp;quot;Look and Say&amp;quot;&lt;/a&gt; sequence, &lt;em&gt;have already been posted on Puzzling Stack Exchange&lt;/em&gt;. Before sharing an interesting sequence that you have found published or presented elsewhere, please make sure to search for duplicates using at least the first few terms of the sequence, and avoid re-posting questions found to have been posted previously.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="60644" LastEditDate="2022-12-12T23:37:36.907" LastActivityDate="2022-12-12T23:37:36.907" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8368" PostTypeId="4" CreationDate="2015-01-30T10:31:57.197" Score="0" Body="A puzzle in which the hidden pattern behind a given sequence of numbers (and, often, the next term) is to be found. Use with [sequence]" OwnerUserId="8874" LastEditorUserId="60644" LastEditDate="2022-12-12T23:37:36.907" LastActivityDate="2022-12-12T23:37:36.907" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8369" PostTypeId="5" CreationDate="2015-01-30T10:36:35.293" Score="0" Body="&lt;p&gt;A puzzle centered around playing cards. Typical features of playing cards are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;rank: 2, 3, 4, 5, 6, 7, 8, 9, 10 (X), Jack (J), Queen(Q), King (K), Ace (A)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;suit: spades (&lt;span class=&quot;math-container&quot;&gt;$\spadesuit$&lt;/span&gt;), hearts (&lt;span class=&quot;math-container&quot;&gt;$\heartsuit$&lt;/span&gt;), diamonds (&lt;span class=&quot;math-container&quot;&gt;$\diamondsuit$&lt;/span&gt;) and clubs (&lt;span class=&quot;math-container&quot;&gt;$\clubsuit$&lt;/span&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;color: red: &lt;span class=&quot;math-container&quot;&gt;$\color{red}{\heartsuit\diamondsuit}$&lt;/span&gt; (hearts and diamonds), black: &lt;span class=&quot;math-container&quot;&gt;$\spadesuit\clubsuit$&lt;/span&gt; (spades and clubs).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Many games (bridge, poker, pontoon, rummy, solitaire, ...) can be played with these cards. For questions about card games themselves, visit our sister site &lt;a href=&quot;https://boardgames.stackexchange.com/&quot;&gt;Board &amp;amp; Card Games Stack Exchange&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="11110" LastEditDate="2020-04-30T06:52:24.177" LastActivityDate="2020-04-30T06:52:24.177" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8370" PostTypeId="4" CreationDate="2015-01-30T10:36:35.293" Score="0" Body="For puzzles centered around playing cards. " OwnerUserId="8874" LastEditorUserId="69582" LastEditDate="2022-04-15T03:16:32.560" LastActivityDate="2022-04-15T03:16:32.560" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8371" PostTypeId="1" AcceptedAnswerId="8401" CreationDate="2015-01-30T10:52:34.883" Score="10" ViewCount="845" Body="&lt;p&gt;Professor Erasmus has constructed a special &lt;a href=&quot;http://www.mathsisfun.com/triangle.html&quot; rel=&quot;nofollow&quot;&gt;isosceles triangle&lt;/a&gt; that he modestly calls the &quot;Professor-Erasmus-triangle&quot;. The professor claims that he can cut his triangle into three smaller triangles, so that any pair of smaller triangles can be put together to form a new isosceles triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has the professor once again made a mathematical blunder, or do such triangles indeed exist?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-23T15:39:19.247" LastActivityDate="2016-03-23T15:39:19.247" Title="The Erasmus isosceles triangle" Tags="|mathematics|geometry|dissection|tiling|triangle|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8372" PostTypeId="1" AcceptedAnswerId="8374" CreationDate="2015-01-30T11:22:17.000" Score="5" ViewCount="566" Body="&lt;p&gt;There are four fences, which are geometrical identical (same heigth and same length). The following four people need the specified time to paint it:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Bob needs 2 hours.&lt;/li&gt;&#xA;&lt;li&gt;Simon needs 3 hours.&lt;/li&gt;&#xA;&lt;li&gt;Tommy needs 5 hours.&lt;/li&gt;&#xA;&lt;li&gt;Johnny needs 8 hours.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How long does it take, if they paint one mentioned fence together?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus: what is the special electrotechnical approach to this issue?&lt;/p&gt;&#xA;" OwnerUserId="9197" LastEditorUserId="7573" LastEditDate="2015-01-30T16:48:27.537" LastActivityDate="2015-01-30T19:32:01.110" Title="Quick and dirty: painting a fence" Tags="|mathematics|calculation-puzzle|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8373" PostTypeId="2" ParentId="8336" CreationDate="2015-01-30T11:24:12.723" Score="5" Body="&lt;p&gt;Since the 'normal' solutions have all been found, have the brute force method:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Python2:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;for c in xrange(333333,1000000):&#xA;    t=3*c &#xA;    theorem=str(t)&#xA;    coffee=str(c)&#xA;    if theorem[2]!=theorem[5]:continue&#xA;    if theorem[5]!=coffee[5]:continue&#xA;    if coffee[4]!=coffee[5]:continue&#xA;    if coffee[2]!=coffee[3]:continue&#xA;    if coffee[1]!=theorem[3]:continue&#xA;    if len(set(coffee))!=4:continue&#xA;    if len(set(theorem))!=6:continue&#xA;    if len(set(coffee+theorem))!=8:continue&#xA;    print coffee,theorem &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4577" LastActivityDate="2015-01-30T11:24:12.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8374" PostTypeId="2" ParentId="8372" CreationDate="2015-01-30T11:38:21.530" Score="4" Body="&lt;p&gt;Nice! The answer to the bonus question is: adding up the resistance of parallel resistors.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Mathematically, the work paces of the four guys are $B=1/2$, $C=1/3$, $T=1/5$ and $J=1/8$.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If we put them in parallel, the resulting work pace is $B+C+T+J=139/120$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The total work time is the reciprocal of this, that is $1/(B+C+T+J)=120/139\approx0.863$ hours $\approx 52$ minutes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-01-30T12:07:59.363" LastActivityDate="2015-01-30T12:07:59.363" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8376" PostTypeId="2" ParentId="8330" CreationDate="2015-01-30T13:27:42.927" Score="2" Body="&lt;p&gt;I have a &lt;em&gt;simpler&lt;/em&gt; way to get to the answer. Clearly, if the 1 digit number is $8$, the highest possible product will be less than $8000$ because $8 * 999 &amp;lt; 8000$. Therefore, assume the 1 digit number is $9$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means the product of the two 2-digit numbers must, between them, one must be divisible by $9$, or both must be divisible by 3.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;One number is divisible by 9:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{99 * (99, 98, 97, 96, 95, 94, 93, 92, 91, 90)}{9} \rightarrow \\&#xA;(1089, 1078, 1067, 1056, 1045, 1034, 1023, 1012, 1001, 990)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the first solution candidate is $99 * 90 = 8910 = 9 * 990$.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Both numbers are divisible by 3:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;99 * 96 \rightarrow 9504 \rightarrow 1056 * 9\\&#xA;96 * 96 \rightarrow 9216 \rightarrow 1024 * 9\\&#xA;96 * 93 \rightarrow 8928 \rightarrow 992 * 9&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $8928 &amp;gt; 8910$, we have our answer, and the second place answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These two processes can be continued to generate further answers.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$99$ times any two digit below $90$ will also result in a valid answer.&lt;/li&gt;&#xA;&lt;li&gt;Any two multiples of 3 equal to or below $96$ and $93$ will also result in a valid answer.&lt;/li&gt;&#xA;&lt;li&gt;Of course, once we get under $8000$, the product no longer needs to be a multiple of $9$ as the 1 digit number doesn't need to be $9$ anymore.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Exhaustive list of answers above $8000$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;8928	=	9	*	992	=	96	*	93	\\&#xA;8910	=	9	*	990	=	99	*	90	\\&#xA;8820	=	9	*	980	=	98	*	90	\\&#xA;8811	=	9	*	979	=	99	*	89	\\&#xA;8730	=	9	*	970	=	97	*	90	\\&#xA;8712	=	9	*	968	=	99	*	88	\\&#xA;8649	=	9	*	961	=	93	*	93	\\&#xA;8640	=	9	*	960	=	96	*	90	\\&#xA;8613	=	9	*	957	=	99	*	87	\\&#xA;8550	=	9	*	950	=	95	*	90	\\&#xA;8514	=	9	*	946	=	99	*	86	\\&#xA;8460	=	9	*	940	=	94	*	90	\\&#xA;8415	=	9	*	935	=	99	*	85	\\&#xA;8370	=	9	*	930	=	93	*	90	\\&#xA;8352	=	9	*	928	=	96	*	87	\\&#xA;8316	=	9	*	924	=	99	*	84	\\&#xA;8280	=	9	*	920	=	92	*	90	\\&#xA;8217	=	9	*	913	=	99	*	83	\\&#xA;8190	=	9	*	910	=	91	*	90	\\&#xA;8118	=	9	*	902	=	99	*	82	\\&#xA;8100	=	9	*	900	=	90	*	90	\\&#xA;8091	=	9	*	899	=	93	*	87	\\&#xA;8064	=	9	*	896	=	96	*	84	\\&#xA;8019	=	9	*	891	=	99	*	81	\\&#xA;8010	=	9	*	890	=	89	*	90	\\&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="42" LastEditorUserId="42" LastEditDate="2015-01-30T14:35:52.287" LastActivityDate="2015-01-30T14:35:52.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8377" PostTypeId="5" CreationDate="2015-01-30T13:32:20.347" Score="0" Body="&lt;p&gt;Puzzles that crucially depend on some feature of the English language or that only work in an English formulation.&lt;/p&gt;&#xA;&lt;p&gt;General English questions are off-topic but can be asked on our sister sites &lt;a href=&quot;https://english.stackexchange.com/&quot;&gt;English Language &amp;amp; Usage Stack Exchange&lt;/a&gt; or &lt;a href=&quot;https://ell.stackexchange.com/&quot;&gt;English Language Learners Stack Exchange&lt;/a&gt; (depending on the level).&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="60644" LastEditDate="2020-11-24T23:06:43.130" LastActivityDate="2020-11-24T23:06:43.130" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8378" PostTypeId="4" CreationDate="2015-01-30T13:32:20.347" Score="0" Body="Puzzles that crucially depend on some feature of the English language or that only work in an English formulation." OwnerUserId="8874" LastEditorUserId="70545" LastEditDate="2021-02-04T05:15:04.123" LastActivityDate="2021-02-04T05:15:04.123" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8379" PostTypeId="5" CreationDate="2015-01-30T13:54:17.210" Score="0" Body="&lt;p&gt;Triangles are one of the simplest geometric shapes. &#xA;Yet, their combinatorics is quite non-trivial and often leads to nice puzzles.&#xA;The following list summarizes some central types of special triangles:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;An isosceles triangle is a triangle with (at least) two equal sides. Equivalently, an isosceles triangle is a triangle with two equal angles.&lt;/li&gt;&#xA;&lt;li&gt;An equilateral triangle is a triangle with all three sides equal. Equivalently, an equilateral triangle is a triangle with three angles of 60 degree.&lt;/li&gt;&#xA;&lt;li&gt;A right-angled triangle is a triangle that has a right angle (90 degree) in it.&lt;/li&gt;&#xA;&lt;li&gt;An obtuse triangle is a triangle that has one angle strictly larger than 90 degree.&lt;/li&gt;&#xA;&lt;li&gt;An acute triangle is a triangle that has all angles strictly smaller than 90 degree.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Furthermore, there are two important relations between pairs of triangles:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Two triangles are &lt;em&gt;similar&lt;/em&gt;, if they have the same three angles.&lt;/li&gt;&#xA;&lt;li&gt;Two triangles are &lt;em&gt;congruent&lt;/em&gt;, if they have the same three side lengths.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It is easy to see that congruent triangles are always similar to each other, whereas similar triangles are not necessarily congruent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In puzzles, one of the most-used properties of a triangle is the triangle inequality: If the three side lengths of a triangle are a&amp;lt;=b&amp;lt;=c, then we always have a+b&gt;=c. The case with a+b=c yields a degenerate triangle whose area is zero.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-21T15:37:52.680" LastActivityDate="2015-02-21T15:37:52.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8380" PostTypeId="4" CreationDate="2015-01-30T13:54:17.210" Score="0" Body="A geometry puzzle centered around the triangle or the centers of a triangle." OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-01-30T17:53:05.797" LastActivityDate="2015-01-30T17:53:05.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8381" PostTypeId="5" CreationDate="2015-01-30T14:30:27.383" Score="0" Body="&lt;p&gt;The Monty Hall problem is a probability puzzle, loosely based on the American television game show &lt;em&gt;&amp;quot;Let's Make a Deal&amp;quot;&lt;/em&gt; and named after its host &lt;a href=&quot;https://en.wikipedia.org/wiki/Monty_Hall&quot; rel=&quot;nofollow noreferrer&quot;&gt;Monty Hall&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;In the problem the player of the game show is asked to pick one of 3 doors to open and hopefully get the best prize, a car (and not a goat). The host, Monty Hall, then opens a different door revealing one of the two dud prizes, a goat. The host then asks if you'd like to stick with your initial choice or change doors. The probability problem comes into play when deciding whether or not it is better to change doors.&lt;/p&gt;&#xA;&lt;p&gt;Questions with this tag often concern more complex or interesting variations of the original Monty Hall problem, although the original itself has also been posted &lt;a href=&quot;/questions/2167/the-monty-hall-problem&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;For more information the Wikipedia article on the Monty Hall problem can be found &lt;a href=&quot;https://en.wikipedia.org/wiki/Monty_Hall_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="52483" LastEditDate="2023-01-31T15:16:41.747" LastActivityDate="2023-01-31T15:16:41.747" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8382" PostTypeId="4" CreationDate="2015-01-30T14:30:27.383" Score="0" Body="The Monty Hall problem is a well-known probability puzzle centered around three doors, two with a bad prize hidden behind, and one with a good one." OwnerUserId="8874" LastEditorUserId="52483" LastEditDate="2023-01-31T15:16:13.747" LastActivityDate="2023-01-31T15:16:13.747" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8383" PostTypeId="1" AcceptedAnswerId="8385" CreationDate="2015-01-30T15:05:08.377" Score="11" ViewCount="1163" Body="&lt;p&gt;You have 2 bishops, 2 rooks and 2 knights (of no colour). Place them in as small a rectangle as possible such that each piece can be captured by atleast 2 other pieces in a single move. Bishops have to be on opposing coloured squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I managed it in a 3x4 rectangle. Lets see if a better solution exists.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-01-30T20:21:04.447" Title="Filling chess pieces in smallest area" Tags="|chess|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8384" PostTypeId="2" ParentId="8383" CreationDate="2015-01-30T15:18:05.497" Score="8" Body="&lt;p&gt;Got a 3x3&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;K R B&#xA;  R B&#xA;  K&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The hardest part was keeping the bishops on opposite colors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thinking of a 2x4:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;K R R K&#xA;  B B&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here's a thought: what if we had a moebius board?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;B R B&#xA;K R K&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-01-30T15:36:38.487" LastActivityDate="2015-01-30T15:36:38.487" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8385" PostTypeId="2" ParentId="8383" CreationDate="2015-01-30T15:18:06.390" Score="17" Body="&lt;p&gt;The smallest solution by area is the $4\times 2$ rectangle. The smallest solution with respect to longest side is the $3\times 3$ rectangle.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(1) The $3\times 3$ rectangle works:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;N e N&lt;br&gt;&#xA;  R B R&lt;br&gt;&#xA;  e B e  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(where N=kNight, R=Rook, B=Bishop, e=empty)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(2) The $4\times 2$ rectangle works:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;N R R N&lt;br&gt;&#xA;  e B B e  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(3) No $1\times n$ rectangle can work&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof: Bishops and knights do not attack any other squares on such a rectangle; rooks only attack two squares. But altogether we need $2\cdot6=12$ attacks.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(4) The $3\times2$ rectangle cannot work&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof: Consider the two middle squares. No knight can attack them; every bishop and every rook can attack at most one one of them. For having each of them attacked twice, the rooks and bishops must occupy the outer four squares, and the knights must be on the middle squares. Impossible.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="-1" LastEditDate="2015-01-30T20:21:04.447" LastActivityDate="2015-01-30T20:21:04.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8386" PostTypeId="2" ParentId="3820" CreationDate="2015-01-30T16:09:54.320" Score="5" Body="&lt;p&gt;There is another way that is probably far less time-consuming, depending on $n$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Weigh the bottle of jellybeans. Call this value $F$.&lt;br&gt;&#xA; Pour out the bottle of jellybeans.&lt;br&gt;&#xA; Weigh the empty bottle. Call this value $E$.&lt;br&gt;&#xA; Weigh a single jellybean. Call this value $J$.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The number of jellybeans is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\frac{F-E}{J}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1716" LastEditorUserId="1716" LastEditDate="2015-01-30T16:17:44.057" LastActivityDate="2015-01-30T16:17:44.057" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8387" PostTypeId="2" ParentId="8372" CreationDate="2015-01-30T16:42:05.813" Score="5" Body="&lt;p&gt;It helps being very explicit about the units in which the various figures are expressed:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Bob: 2 hours/fence &lt;span class=&quot;math-container&quot;&gt;$\rightarrow$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\frac12$&lt;/span&gt; fence/hour&lt;/p&gt;&#xA;&lt;p&gt;Simon: 3 hours/fence &lt;span class=&quot;math-container&quot;&gt;$\rightarrow$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\frac13$&lt;/span&gt; fence/hour&lt;/p&gt;&#xA;&lt;p&gt;Tommy: 5 hours/fence &lt;span class=&quot;math-container&quot;&gt;$\rightarrow$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\frac15$&lt;/span&gt; fence/hour&lt;/p&gt;&#xA;&lt;p&gt;Johnny: 8 hours/fence &lt;span class=&quot;math-container&quot;&gt;$\rightarrow$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\frac18$&lt;/span&gt; fence/hour&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Painting in parallel: &lt;span class=&quot;math-container&quot;&gt;$(\frac12 + \frac13 + \frac15 + \frac18)$&lt;/span&gt; fence/hour = &lt;span class=&quot;math-container&quot;&gt;$\frac{139}{120}$&lt;/span&gt; fence/hour &lt;span class=&quot;math-container&quot;&gt;$\rightarrow$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\frac{120}{139}$&lt;/span&gt; hour/fence.&lt;/p&gt;&#xA;" OwnerUserId="7573" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-01-30T19:32:01.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8388" PostTypeId="1" AcceptedAnswerId="8398" CreationDate="2015-01-30T16:46:19.037" Score="7" ViewCount="1296" Body="&lt;p&gt;I found this puzzle within the game &lt;a href=&quot;http://en.wikipedia.org/wiki/MouseHunt_%28game%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mousehunt&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/V2rSZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It didn't seem to match any rune system that I know of. Rotating the image also didn't seem to show any useful information, although the first word when the image is rotated 180 degrees looked suspiciously like &quot;First&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For context, a list of other images are also visible &lt;a href=&quot;http://mhwiki.hitgrab.com/wiki/index.php/Torn_Pages_of_Plankrun%27s_Journal&quot; rel=&quot;nofollow noreferrer&quot;&gt;here on the official wiki.&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="7529" LastEditorUserId="4050" LastEditDate="2015-01-30T17:58:04.833" LastActivityDate="2015-01-30T20:23:14.113" Title="Mousehunt cryptogram" Tags="|cipher|symbols|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8389" PostTypeId="1" AcceptedAnswerId="8392" CreationDate="2015-01-30T17:01:03.653" Score="3" ViewCount="413" Body="&lt;p&gt;You would like to sort some monthly magazines in different categories. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disregarding the title of magazines, there are 4 different categories if you arrange magazines by publishing year and month. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disregarding the publishing year, there are 5 different categories if you sort magazines by title and month. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disregarding month, you could divide magazines in to  6 different categories if you sort magazines by title and publishing year. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many monthly magazines you have &lt;em&gt;at most&lt;/em&gt;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no magazine with the same title, publishing year and month. Each categorization does not need to have the same number of magazines.&lt;/p&gt;&#xA;" OwnerUserId="3914" LastEditorUserId="3914" LastEditDate="2015-01-30T17:46:29.337" LastActivityDate="2015-01-30T18:04:34.853" Title="How many magazines?" Tags="|logical-deduction|mathematics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8390" PostTypeId="2" ParentId="8389" CreationDate="2015-01-30T17:24:13.037" Score="1" Body="&lt;p&gt;Welp, looks like I can max out at 10 mags:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a 2000 jan&#xA;b 2000 jan&#xA;c 2000 jan&#xA;a 2001 jan&#xA;b 2001 jan&#xA;c 2001 jan&#xA;a 2000 feb&#xA;b 2000 feb&#xA;a 2001 feb&#xA;b 2001 feb&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Old:&lt;/strong&gt;&lt;br&gt;&#xA;I think 6 would be the number of magazines. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a 2000 Jan  or  a 2000 Jan&#xA;b 2000 Feb      b 2000 Feb&#xA;c 2000 Jan      c 2000 Mar&#xA;d 2000 Feb      d 2000 Jan&#xA;a 2001 Jan      e 2000 Jan&#xA;b 2001 May      a 2001 Jan&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Any repeating combo of attributes would increase one of the other groups. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have to cross over 2 years, since one year with the #6 means you'd need 6 different mags, which would completely bork #5, which directly says we have a repeating title-month. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have to have 2 repeating year-month combos, and one repeating title-month combo, which are mutually exclusive.  So, that means we have at most 3 distinct months. We can achieve the 4 year-month combos with 2 years and 2 months if desired. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what about a repeating title-year? Well, that would increase the title-month count, so we can't have that. So, 6 magazines. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-01-30T18:04:29.623" LastActivityDate="2015-01-30T18:04:29.623" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8391" PostTypeId="2" ParentId="8371" CreationDate="2015-01-30T17:36:19.850" Score="4" Body="&lt;p&gt;The professor has blundered. &lt;strong&gt;The triangle he claims to have constructed cannot exist in Euclidean space.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I apologize for the lack of pictures. I know they would make my explanation/proof easier to follow, but I do not currently have access to any tools to create or upload them.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start by considering an isosceles triangle, $ABC$, that has been constructed from two of the component triangles. WLOG, let $AB$ be the base. Notice that if this is an equilateral triangle, there is no way to &quot;add&quot; a third triangle to get a bigger isosceles triangle. So it can't be an equilateral triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;Professor-Erasmus-Triangle&quot; (PET) must have the two equal sides equal in length to the longest edge of $ABC$, since we can't add something to $ABC$ and make it smaller. (Not exactly rigorous, I know, but this isn't Math.SE, so I'll be a little more flexible.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $ABC$ is a &quot;short&quot; isosceles triangle, that is, $AB$ is the longest edge, we can create the PET by extending $AC$ to some point $D$ where $BD=AB$, creating third component triangle $BCD$. (Sorry I don't have pictures.) Now, since we assumed that $AB=BD$ is longer than $BC$, we can say that either $BD$ or $CD$ is the longest edge of $BCD$. Suppose that we have a copy of one of the first two components of $ABC$, that we can add to $BCD$ to make a different isosceles triangle. So, this triangle shares an edge with $BCD$. It can't share $BC$, because it takes both component triangles to make an isosceles triangle that way. However, if it shares either of the others, it either shares the longest edge or has an edge of equal length (by virtue of making an isosceles triangle). This means that it is too big to &quot;fit&quot; in $ABC$, which means that isn't one of the component triangles. All of this means that our first triangle, $ABC$ can't be &quot;short&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what if it's tall? So then $AC=BC&amp;gt;AB$. In this case, we can create the third compenent and the PET by extending $CB$ to a point $D$ where $AC=AD$. We can also find the first two component triangles by choosing a point $E$ on $CB$ were $ED=AD$. So now we have the Professor-Erasmus-Triangle $ACD$, with smaller triangles $ABD$, $ABE$, and $ACE$. Also, just by its construction, we know we can combine $ABE$ with either of the other smaller triangles to make an isosceles triangle. So, the only question that remains is, (how) can we combine $ABD$ and $ACD$ to make an isosceles triangle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By their construction as parts of an isosceles triangle, we know that $AC=AD$ and &#xA;$\angle ACE=\angle ADB$. Also, by combining with the other component triangle we get isosceles triangles, so $BC=AC$ and $DE=AD$. Then, by transitivity, $BC=DE$. Removing the segment $BE$ gives us $BD=CE$. This means that the two triangles have side-angle-side congruence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order for the two triangles to be combined into an isosceles triangle, they must have angles that are supplementary to each other (they make $180^{\circ}$ together). Since the triangles are congruent, these angles have to be right angles. However, if $\angle ACE=\angle ADB=90^{\circ}$, that leaves the third angle of the PET, $\angle CAD$, at $0^{\circ}$, which can't be. If $\angle CAE=\angle BAD=90^{\circ}$, that means that $\angle CAD&amp;gt;180^{\circ}$, which is also impossible. Finally, if $\angle AEC=\angle ABD=90^{\circ}$, then the third component triangle, which is between those two, has two zero-degree angles, $\angle AEB$ and $\angle ABE$. Since none of these are possible, that means that these two triangles, as constructed, cannot make an isosceles triangle. This, in turn, means that the PET cannot be &quot;tall&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the Professor-Erasmus-Triangle is not &quot;short,&quot; is not &quot;tall,&quot; and is not equilateral, we are forced to conclude that it does not exist.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note that this result does not necessarily hold for non-Euclidean spaces. For example, in elliptical geometry, where interior angles of a triangle add up to &lt;em&gt;more than&lt;/em&gt; $180^{\circ}$, you could have an equilateral triangle with three $90^{\circ}$ angles, with the smaller triangles formed by trisecting one of the angles.&lt;/p&gt;&#xA;" OwnerUserId="8776" LastActivityDate="2015-01-30T17:36:19.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8392" PostTypeId="2" ParentId="8389" CreationDate="2015-01-30T18:04:34.853" Score="6" Body="&lt;p&gt;10 mayhaps?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PaNZzm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture corresponds with the magazines in the natural way. Projections along the three axes have sizes 4, 5, and 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I conjecture that with $a$, $b$, and $c$ categories in the three axes, the maximum number of magazines is at most $\sqrt{abc}$.&lt;/p&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2015-01-30T18:04:34.853" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8393" PostTypeId="1" CreationDate="2015-01-30T18:35:39.003" Score="3" ViewCount="24732" Body="&lt;p&gt;Consider a cuboid (that is, a rectangular box / rectangular parallelepiped) with the following properties: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;the area of its top face is $240$ cm$^2$ &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the area of its front face is $300$ cm$^2$ &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the area of its face on the righthand side is $180$ cm$^2$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Determine the three edge lengths (length, width, height) of the cuboid from this information.&lt;/p&gt;&#xA;" OwnerUserId="8829" LastEditorUserId="8874" LastEditDate="2015-01-30T19:11:26.027" LastActivityDate="2015-01-30T22:23:21.677" Title="Find length, width and height of a cuboid" Tags="|mathematics|calculation-puzzle|" AnswerCount="4" CommentCount="4" ClosedDate="2015-01-30T23:44:43.253" ContentLicense="CC BY-SA 3.0" />
  <row Id="8395" PostTypeId="2" ParentId="8393" CreationDate="2015-01-30T18:58:50.733" Score="7" Body="&lt;p&gt;$x * y = 240 \text{ cm²}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y * z = 300 \text{ cm²}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x * z = 180 \text{ cm²}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solve for x gives: $x * x = \frac{240 * 180}{300} = 144 \text{ cm²}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = 12 \text{ cm}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = 20 \text{ cm}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z = 15 \text{ cm}$&lt;/p&gt;&#xA;" OwnerUserId="448" LastEditorUserId="448" LastEditDate="2015-01-30T21:08:45.530" LastActivityDate="2015-01-30T21:08:45.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8396" PostTypeId="2" ParentId="8393" CreationDate="2015-01-30T19:04:38.800" Score="3" Body="&lt;p&gt;First, I have to translate this into a valid question.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It's not a cube; it's a rectangular prism.&lt;/li&gt;&#xA;&lt;li&gt;There aren't upper, front, and bottom areas; there are top, front, and side areas.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Starting from that point, we can work through the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A rectangular prism has 3 dimensions: height, width, depth&lt;/li&gt;&#xA;&lt;li&gt;The top face's area = width * depth&lt;/li&gt;&#xA;&lt;li&gt;The front face's area = width * height&lt;/li&gt;&#xA;&lt;li&gt;The side face's area = height * depth&lt;/li&gt;&#xA;&lt;li&gt;Let's simplify these equations into &lt;code&gt;t=w*d&lt;/code&gt;, &lt;code&gt;f=w*h&lt;/code&gt;, and &lt;code&gt;s=h*d&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;Solve for &lt;code&gt;w=t/d&lt;/code&gt; and plug that into &lt;code&gt;f=w*h&lt;/code&gt; to get &lt;code&gt;f=(t/d)*h&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;Solve for &lt;code&gt;h=f*d/t&lt;/code&gt; and plug that into &lt;code&gt;s=h*d&lt;/code&gt; to get &lt;code&gt;s=(f*d/t)*d&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;Solve &lt;code&gt;d^2=s*t/f&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;s&lt;/code&gt;, &lt;code&gt;t&lt;/code&gt;, and &lt;code&gt;f&lt;/code&gt; are given in the problem as 180 cm^2, 240 cm^2, and 300 cm^2, respectively&lt;/li&gt;&#xA;&lt;li&gt;Plug those in to get &lt;code&gt;d^2=180*240/300&lt;/code&gt; or &lt;code&gt;d^2=144&lt;/code&gt; or &lt;code&gt;d=12&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;Work backwards through the functions to get &lt;code&gt;h=15&lt;/code&gt; and &lt;code&gt;w=20&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So the height is 15 cm, the width is 20 cm, and the depth (or length) is 12 cm.&#xA;I believe my correction to the question is valid as the answer came out to simple integers.&lt;/p&gt;&#xA;" OwnerUserId="9000" LastActivityDate="2015-01-30T19:04:38.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8397" PostTypeId="2" ParentId="8393" CreationDate="2015-01-30T19:07:44.810" Score="4" Body="&lt;p&gt;A rectangular block of width $a$, height $b$, and depth $c$ has front area $ab=300$, top area $ac=240$, and side area $bc=180$. So we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a^2=\frac{ab \times ac}{bc}=\frac{300 \times 240}{180}=400$ $\rightarrow$ $a=20$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b^2=\frac{ab \times bc}{ac}=\frac{300 \times 180}{240}=225$ $\rightarrow$ $b=15$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c^2=\frac{ac \times bc}{ab}=\frac{240 \times 180}{300}=144$ $\rightarrow$ $c=12$.&lt;/p&gt;&#xA;" OwnerUserId="7573" LastEditorUserId="7573" LastEditDate="2015-01-30T19:17:01.840" LastActivityDate="2015-01-30T19:17:01.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8398" PostTypeId="2" ParentId="8388" CreationDate="2015-01-30T20:23:14.113" Score="12" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lsTnC.png&quot; alt=&quot;solution&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you swap the upper part of each line with the lower part, then flip the image horizontally, you get the message &quot;WELCOME TO YOUR PRISON.  FIRST HUNTER.&quot;&lt;/p&gt;&#xA;" OwnerUserId="9269" LastActivityDate="2015-01-30T20:23:14.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8401" PostTypeId="2" ParentId="8371" CreationDate="2015-01-30T22:17:26.363" Score="9" Body="&lt;p&gt;It's possible if Professor Erasmus can reflect triangles when creating isosceles triangles. If we imagine them as being cut out of paper, this would be by flipping over the paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/y7heH.png&quot; alt=&quot;Erasmus triangle&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an isosceles triangle if we set $x=\sqrt2-1$, making the bottom and right edges both have length $\sqrt2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two leftmost triangles form a symmetric isoceles triangles, and the rightmost two form a right isosceles triangle. For the outer two, the left triangle can be flipped into the position of the middle one to form the same right isosceles triangle.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-01-31T06:44:42.013" LastActivityDate="2015-01-31T06:44:42.013" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8402" PostTypeId="1" AcceptedAnswerId="8405" CreationDate="2015-01-30T22:22:29.037" Score="3" ViewCount="298" Body="&lt;p&gt;Choose 7 distinct points in the Euclidean plane so that among any 3 of those points, there are (at least) 2 that are a distance of exactly 1 apart.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-01-30T22:28:47.893" LastActivityDate="2015-02-01T11:54:04.753" Title="Find seven points so any three contain two of distance one" Tags="|mathematics|geometry|matches|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8403" PostTypeId="2" ParentId="8393" CreationDate="2015-01-30T22:23:21.677" Score="4" Body="&lt;ol&gt;&#xA;&lt;li&gt;Multiply them all together and take the square root:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;abc = \sqrt{a^2b^2c^2} = \sqrt{ab * ac * bc} = \sqrt{12960000} = 3600&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Divide by the missing term to get each side:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;a = \frac{3600}{bc} = \frac{3600}{240} = 15 \\&#xA;b = \frac{3600}{ac} = \frac{3600}{300} = 12 \\&#xA;c = \frac{3600}{ab} = \frac{3600}{180} = 20&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2015-01-30T22:23:21.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8404" PostTypeId="1" AcceptedAnswerId="8471" CreationDate="2015-01-31T02:27:07.440" Score="17" ViewCount="2091" Body="&lt;p&gt;I heard this classic riddle a long long time ago and really liked it, but I'm afraid it was way too simple for you all in the format that I originally heard it, so I have tweaked it just a little:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;101 / 50 + 0 = 1/9&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is the name of a person.  Who is it, and why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The equation is actually of the form &quot;name = description&quot; (or &lt;em&gt;who = why&lt;/em&gt;, if you prefer).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now that a correct answer has been given, here is the riddle as I originally heard it, suitable for any well-educated person:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A hundred and one, divided by fifty, followed by nothing: one of nine.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5874" LastEditorUserId="4050" LastEditDate="2015-02-02T20:19:39.553" LastActivityDate="2015-02-02T20:19:39.553" Title="Classic &quot;mathematical&quot; riddle: 101 / 50 + 0 = 1/9. Who is it, and why?" Tags="|riddle|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8405" PostTypeId="2" ParentId="8402" CreationDate="2015-01-31T02:52:41.010" Score="5" Body="&lt;p&gt;First, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;take two equilateral triangles with one side identified. This &quot;diamond&quot; gives an answer for four points with two &quot;far&quot; corners and two &quot;near&quot; corners&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;consider two copies of this diamond in the plane. This has eight points but doesn't solve the puzzle because you could pick two far corners of one diamond and any point of the other diamond.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Next,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;identify the far corner of one diamond with the far corner of the other diamond. Now we're good except if you take the identified point and the two other far corners.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;rotate one of the two diamonds until the other two far corners are the correct distance apart.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explicitly,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;assuming I did my arithmetic right, you could choose $A=(0,0)$, $B=(\sqrt{3}/2,1/2)$, $C=(\sqrt{3}/2,-1/2)$, $D=(\sqrt{3},0)$, $E=(\frac{5\sqrt{3}-\sqrt{11}}{12},\frac{\sqrt{33}+5}{12})$, $F=(\frac{5\sqrt{3}+\sqrt{11}}{12},\frac{\sqrt{33}-5}{12})$, and $G=(5\sqrt{3}/6,\sqrt{33}/6)$. Then the pairs $(A,B)$, $(A,C)$, $(B,C)$, $(B,D)$, $(C,D)$, $(A,E)$, $(A,F)$, $(E,F)$, $(E,G)$, $(F,G)$, and $(D,G)$ are all distance one apart.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;@Quark suggested that I include an image (segments are of length one)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/e1NYy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7300" LastEditorUserId="7300" LastEditDate="2015-02-01T11:54:04.753" LastActivityDate="2015-02-01T11:54:04.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8406" PostTypeId="1" AcceptedAnswerId="8416" CreationDate="2015-01-31T05:02:59.683" Score="8" ViewCount="1092" Body="&lt;p&gt;An ant walks in constant steps from corner to corner over a rectangular box, always chosing the shortest path. From any given corner, the ant observes the shortest paths to the other seven corners all to equate to an integer number of ant steps. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8QhKp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the ratios between the linear dimensions of the box?&lt;/p&gt;&#xA;" OwnerUserId="7573" LastEditorUserId="7573" LastEditDate="2015-01-31T14:28:16.760" LastActivityDate="2015-02-01T13:42:51.440" Title="Ant on a rectangular box" Tags="|mathematics|geometry|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8407" PostTypeId="2" ParentId="8371" CreationDate="2015-01-31T05:32:38.153" Score="0" Body="&lt;p&gt;The only way to divvy up an isosceles triangle into three smaller triangles without flipping is to start with an isosceles right triangle and halve it to make two smaller isosceles right triangles.  The third triangle is degenerate and coincident with the hypotenuse of one of the smaller right triangles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you alter the sectioning, you won't be able to create an isosceles triangle out of at least one pair of triangles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you alter any of the angles of the starting triangle, you'll have a scalene triangle and/or you won't be able to create an isosceles triangle out of at least one pair of triangles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The degenerate triangle can't be created by a &quot;cut&quot;, so Professor Erasmus's claim &quot;that he can cut his triangle into three smaller triangles&quot; is false, which means the Professor-Erasmus-triangle can't exist as described.  However, he hasn't necessarily made a mathematical blunder because the geometry works, albeit with degeneracy.&lt;/p&gt;&#xA;" OwnerUserId="1923" LastEditorUserId="1923" LastEditDate="2015-01-31T13:38:25.700" LastActivityDate="2015-01-31T13:38:25.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8408" PostTypeId="2" ParentId="8402" CreationDate="2015-01-31T06:14:00.857" Score="4" Body="&lt;p&gt;I agree with Gabriel's answer. I plotted the points just to verify them/explain to others visually; feel free to include the plot in your answer if you want and I'll delete this answer after. (red lines are lengths of 1)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/WWSHl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastActivityDate="2015-01-31T06:14:00.857" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8410" PostTypeId="1" AcceptedAnswerId="8417" CreationDate="2015-01-31T12:15:09.287" Score="16" ViewCount="2363" Body="&lt;p&gt;In the country Argonia, there are only two types of coins: genuine ones and fake ones.&#xA;All genuine coins have the same weight, and all fake coins have the same weight.&#xA;The weight of a genuine coin is greater than the weight of a fake coin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cosmo puts four coins on the table and tells Fredo: &lt;em&gt;&quot;I think that two of these coins are genuine and two of them are fake.&quot;&lt;/em&gt; Then Cosmo leaves the room. On the table, there is a balance with two pans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can Fredo decide whether Cosmo's statement is true by using the balance at most twice?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-10-09T07:24:46.337" LastActivityDate="2016-04-11T16:39:45.293" Title="Two genuine and two fake coins" Tags="|mathematics|strategy|weighing|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8411" PostTypeId="5" CreationDate="2015-01-31T12:21:20.073" Score="0" Body="&lt;p&gt;A strategy puzzle that is centered around scales, weights, and balances.  Typically, in these problems one is given a set of identical-looking items (coins, stones, balls, tokens), and typically the goal is&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;to identify one item with a particular weight,&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;to divide the items into several groups with certain desired weights,&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;to verify whether the weights of the items satisfy a certain property.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;with as few weighings as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The underlying weighing mechanism might either be&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;a two-armed balance (with two pans): for comparing the weights of two sets of items with the possible outcomes &quot;left pan is heavier&quot;, &quot;left pan is lighter&quot;, &quot;both pans have equal weight&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;a calibrated one-armed balance (with a single pan): for learning the precise total weight of the item set in the pan&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="5373" LastEditDate="2018-11-23T18:23:31.123" LastActivityDate="2018-11-23T18:23:31.123" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8412" PostTypeId="4" CreationDate="2015-01-31T12:21:20.073" Score="0" Body="A puzzle that is centered around weighings and balances." OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-01-31T16:38:43.693" LastActivityDate="2015-01-31T16:38:43.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8413" PostTypeId="5" CreationDate="2015-01-31T12:53:23.797" Score="0" Body="&lt;p&gt;A burr puzzle is an interlocking puzzle consisting of several three-dimensional pieces that are combined to make one three-dimensional unit. The interlocking pieces are often notched sticks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Burr puzzles are traditionally made of wood, and sometimes of metal or plastic.&#xA;Quality burr puzzles are usually precision-made for easy sliding and accurate fitting of the pieces. &lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-01-31T18:06:28.433" LastActivityDate="2015-01-31T18:06:28.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8414" PostTypeId="4" CreationDate="2015-01-31T12:53:23.797" Score="0" Body="A burr puzzle is an interlocking puzzle consisting of several three-dimensional pieces that are combined to make one three-dimensional unit. " OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-01-31T18:09:55.050" LastActivityDate="2015-01-31T18:09:55.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8415" PostTypeId="2" ParentId="8410" CreationDate="2015-01-31T14:08:39.800" Score="3" Body="&lt;p&gt;There can be two outputs by weighing only two coins.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The balance is in alignment if two coins are exact same weight. Then you weight another two coins and if they are in alignment we can assume that Cosmo was correct but cannot be sure because there could be four coins that weigh the same.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The balance is not in alignment when we put two coins. Fredo weights another two coins and if they aren't in alignment we can assume that Cosmo was correct.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I think that it can't be done with only two weighing, there must be third to be sure that there are only two same coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ALSO&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can weigh four coins at once!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we put two of the same weight balance will not be in alignment, so we switch from left to right and right to left and if the balance is in alignment we are 100% sure that Cosmo was correct!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if balance is in alignment in start with four coins we cannot be certain that he was correct because all four coins can be of the same weight so changing won't change the scale and we can only think that we switched two coins of same weight!&lt;/p&gt;&#xA;" OwnerUserId="8866" LastEditorUserId="8866" LastEditDate="2015-01-31T14:34:14.063" LastActivityDate="2015-01-31T14:34:14.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8416" PostTypeId="2" ParentId="8406" CreationDate="2015-01-31T14:32:20.050" Score="9" Body="&lt;p&gt;Let the rectangular box sides be a,b and c. Call integers x,y a pythagorean double if $x^2 + y^2 = z^2$ for some integer z.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly a,b and c must be integers; that's three paths.&#xA;Also, any two of them must form a pythagorean double so that the ant can get from one corner of a face to the opposite corner. That means a,b and c must be distinct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally $(a+b),c$ must form a pythagorean double (where c is the longest side), so that the ant can get from one corner to the opposite corner of the box.&#xA;So far so good. This means c must form a pythagorean double with a, with b, and with a+b; and a,b must form a pythagorean double.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One such set of numbers is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a = 44, b = 117, c = 240 - these are the ratios of the sides of the box.&lt;br&gt;&#xA; These give pythagorean triples of 44,117,125; 44,240,244; 117,240,267; 161,240,289&lt;br&gt;&#xA; So the other four path lengths are (some multiples of) 125, 244, 267 and 289.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There may be others.&lt;/p&gt;&#xA;" OwnerUserId="6031" LastActivityDate="2015-01-31T14:32:20.050" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8417" PostTypeId="2" ParentId="8410" CreationDate="2015-01-31T15:30:13.923" Score="22" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let's label the coins A, B, C, and D.&lt;br&gt;&#xA;&lt;br&gt;&#xA; First weigh A and B against C and D. This will leave 2 cases:&lt;br&gt;&#xA;&lt;br&gt;&#xA;  &lt;strong&gt;First case:&lt;/strong&gt; If they weigh the same then weigh A against B. If A=B, then all coins weigh the same - they are all fake or all genuine. If they differ, then C and D must differ too, which means that Cosmo's statement is true.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Second case:&lt;/strong&gt; If the first weighting is not in balance, then let's assume without loss of generality that A and B are the heavier pair. This can also be restated as Lemma 1: At least one of A and B are genuine and at least one of C and D are fake. Now weigh A and C against B and D. If this weighting comes out in balance, it means that A + C = B + D. By Lemma 1, it is then easy to see that both A and B are genuine, and C and D are fake, validating Cosmo's statement. If it is not in balance, for example if A + C &gt; B + D, we know that A is genuine, D is fake, that C and D are the same type. By symmetry, if A + C &amp;lt; B + D, then B is genuine, C is fake, and A and D are the same type.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7577" LastEditorUserId="2383" LastEditDate="2015-02-01T01:17:18.203" LastActivityDate="2015-02-01T01:17:18.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8418" PostTypeId="2" ParentId="8410" CreationDate="2015-01-31T17:25:28.487" Score="9" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;     Label the coins at random A, B, C and D.&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;     First weigh A+B against C+D. They will either way the same, or not the same.&#xA;     &lt;ul&gt;&lt;br&gt;&#xA;     * &lt;em&gt;If they are the same&lt;/em&gt;: the two pans must have the same distribution of fake/genuine, so either all the coins are fake, or all the coins are genuine, or there is a fake coin and a genuine coin in each pan. By weighing A against B, you determine which the situation is, as if A weighs the same as B all the coins must be the same weight, and Cosmo is wrong. If A and B weigh a different amount then Cosmo is right.&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;     * &lt;em&gt;if they are different&lt;/em&gt;: the two pans must have a differing distribution of fake/genuine. The only way Cosmo could be right here is if the lighter pan contained two fake coins and the heavier two genuine coins. So swap a coin from the lighter pan and the heavier pan, and if they are now equal, Cosmo is right. Else he is wrong.&#xA;     &lt;/ul&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;s&gt;This seems to be a different answer to &lt;em&gt;waxwing&lt;/em&gt;&lt;/s&gt; I think it might be a little easier to understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit: actually it's the same decision tree as what waxwing proposes but despite a supposed mathematical background I didn't realise that for a good while, and it's proved in a different way, so I shall leave this up&lt;/em&gt;. &lt;/p&gt;&#xA;" OwnerUserId="9280" LastEditorUserId="9280" LastEditDate="2015-01-31T18:19:23.173" LastActivityDate="2015-01-31T18:19:23.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8419" PostTypeId="5" CreationDate="2015-01-31T18:34:29.067" Score="0" Body="&lt;p&gt;Many puzzles in the area of computer science are algorithmic in nature.&#xA;Standard examples are&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Eight_queens_puzzle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Eight queens puzzle&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Tower_of_Hanoi&quot; rel=&quot;nofollow noreferrer&quot;&gt;Towers of Hanoi&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Dining_philosophers_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dining philosophers&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;The &lt;a href=&quot;https://classic-puzzles.blogspot.nl/2006/12/google-interview-puzzle-2-egg-problem.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;2 Egg Problem&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Other standard puzzle types in this area involve search, learning, and gaining information. Take note that general computer science questions are off-topic but can be asked on &lt;a href=&quot;https://cs.stackexchange.com/&quot;&gt;Computer Science Stack Exchange&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="71981" LastEditDate="2020-11-28T13:28:35.147" LastActivityDate="2020-11-28T13:28:35.147" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8420" PostTypeId="4" CreationDate="2015-01-31T18:34:29.067" Score="0" Body="A puzzle related to concepts in computer science, such as algorithms or information theory. " OwnerUserId="8874" LastEditorUserId="69582" LastEditDate="2021-02-05T17:51:20.063" LastActivityDate="2021-02-05T17:51:20.063" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8421" PostTypeId="5" CreationDate="2015-01-31T18:57:16.930" Score="0" Body="&lt;p&gt;For questions about determining or adjusting the difficulty level of a puzzle.&lt;/p&gt;&#xA;&lt;p&gt;This tag is typically used for isolated instances of a problem, or deciding how to set difficulty levels.&lt;/p&gt;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;single-player puzzles, such as &lt;a href=&quot;/questions/tagged/sudoku&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;sudoku&amp;#39;&quot; rel=&quot;tag&quot;&gt;sudoku&lt;/a&gt;,&lt;/li&gt;&#xA;&lt;li&gt;two-player games, such as &lt;a href=&quot;/questions/tagged/chess&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;chess&amp;#39;&quot; rel=&quot;tag&quot;&gt;chess&lt;/a&gt; problems.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Furthermore, the tag covers general discussion on the &lt;em&gt;construction&lt;/em&gt; of all kinds of puzzles or puzzle levels.&lt;/p&gt;&#xA;&lt;p&gt;For a discussion on how to grade a Sudoku, see:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/29/what-are-the-criteria-for-determining-the-difficulty-of-sudoku-puzzle/41093#41093&quot;&gt;What are the criteria for determining the difficulty of Sudoku puzzle?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="60644" LastEditDate="2020-08-30T12:01:15.660" LastActivityDate="2020-08-30T12:01:15.660" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8422" PostTypeId="4" CreationDate="2015-01-31T18:57:16.930" Score="0" Body="A question about assessing or adjusting the difficulty level of a puzzle." OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-01-31T23:59:46.090" LastActivityDate="2015-01-31T23:59:46.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8423" PostTypeId="2" ParentId="8326" CreationDate="2015-01-31T19:34:37.303" Score="3" Body="&lt;p&gt;assuming a staggered grid isn't illegal because the expected solution includes single squares&lt;/p&gt;&#xA;&#xA;&lt;p&gt;27 squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4zO7s.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9279" LastEditorUserId="9279" LastEditDate="2015-02-04T22:36:26.867" LastActivityDate="2015-02-04T22:36:26.867" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8424" PostTypeId="2" ParentId="8404" CreationDate="2015-01-31T21:34:06.473" Score="8" Body="&lt;p&gt;101/50 could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CIBYL (which sounds like it could be a name),&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;since&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 101 in Roman numerals is CI, 50 in Roman numerals is L, and the division stroke is often read as 'by'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="39" LastEditDate="2015-02-01T14:17:48.083" LastActivityDate="2015-02-01T14:17:48.083" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8425" PostTypeId="2" ParentId="6273" CreationDate="2015-01-31T21:53:00.317" Score="1" Body="&lt;p&gt;&lt;em&gt;22 Characters&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Riot left, left a riot&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Riot 1: Violent disorder&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Left 1: Going away&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Left 2: Staying&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Riot 2: Lots of fun&lt;/p&gt;&#xA;" OwnerUserId="9283" LastActivityDate="2015-01-31T21:53:00.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8426" PostTypeId="2" ParentId="469" CreationDate="2015-01-31T23:05:15.987" Score="12" Body="&lt;ul&gt;&#xA;&lt;li&gt;Lights Out puzzles have a polynomial time algorithm by linear algebra.&lt;/li&gt;&#xA;&lt;li&gt;&quot;Draw this shape without picking up your pencil&quot; puzzles have a polytime algorithm which follows from the proof of which graphs have Eulerian circuits.&lt;/li&gt;&#xA;&lt;li&gt;Sliding puzzles like the 15-puzzle are easy (though hard to solve in a minimal number of moves).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;IMO these examples kind of prove your assertion. None of them are popular anymore (the last time I saw Lights Out was what, middle school?), and I bet this is because once you learn the algorithm, they're boring.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For a more nontrivial but lesser known example, take &quot;orb puzzles&quot; from the Deadly Rooms of Death video game series. Orb puzzles consist of a bunch of doors, and a bunch of orbs which control them. For each orb and each door, the orb either opens, closes, toggles, or does nothing to the door. Orbs may be triggered in any sequence, as many times as you want. The objective is to find a sequence of orb hits which leaves all the doors open.&lt;br&gt;&#xA;&lt;br&gt;&#xA;Players used to include orb puzzles in their custom levels a lot. Then mitchthro found a polytime algorithm &lt;a href=&quot;http://forum.caravelgames.com/viewtopic.php?TopicID=7242&amp;amp;page=1#101574&quot;&gt;here&lt;/a&gt;. Now orb puzzles are basically dead. (But I doubt the algorithm killed them; I think they died because players find them repetitive and unfun.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I wonder if there are any popular puzzles which fall to linear (or convex) programming.&lt;/p&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2015-01-31T23:05:15.987" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8427" PostTypeId="2" ParentId="8406" CreationDate="2015-02-01T04:02:26.433" Score="1" Body="&lt;p&gt;What you are basically looking for is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler_brick&quot; rel=&quot;nofollow&quot;&gt;Euler brick&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some solutions listed there are:&lt;br&gt;&#xA;(85, 132, 720) — (157, 725, 732)&lt;br&gt;&#xA;(140, 480, 693) — (500, 707, 843)&lt;br&gt;&#xA;(160, 231, 792) — (281, 808, 825)&lt;br&gt;&#xA;(240, 252, 275) — (348, 365, 373)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to add that, as @Quark mentioned, only the first one actually works for the last diagonal (integer length for $\sqrt{(a+b)^2+c^2}$), which I forgot to take into account.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4269" LastEditDate="2015-02-01T13:42:51.440" LastActivityDate="2015-02-01T13:42:51.440" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8428" PostTypeId="2" ParentId="8410" CreationDate="2015-02-01T08:12:00.053" Score="5" Body="&lt;p&gt;Yes, and Fredo only needs to use the balance once.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Place all four suspect coins in one of the balance's pans.  In the other pan, put two known-genuine coins and two known-fake coins.  If the two pans weigh the same amount, then Cosmo is correct.  If not, he is wrong.&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;em&gt;This is an outside-the-box solution, taking advantage of the fact that the puzzle doesn't actually specify that Cosmo's coins are the only ones which may be weighed;  only that the balance must be the instrument of judgement, and that it may not be used more than twice.  This answer may or may not be contrary to the spirit of the puzzle, depending upon whether or not Gerhard was intentionally fishing for this trick answer.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9286" LastActivityDate="2015-02-01T08:12:00.053" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="8429" PostTypeId="1" AcceptedAnswerId="8452" CreationDate="2015-02-01T10:08:03.617" Score="23" ViewCount="3948" Body="&lt;p&gt;One hot day, Stan, Kyle, and Kenny were sitting outside with a square watermelon (actually it was a cube like the picture below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stan says &quot;Let's cut the watermelon into 3 equal slices (like the right-side diagram below).&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kyle says &quot;But then the end slices have more rind.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you show them how to cut this square watermelon into 3 equal (congruent) pieces, so that each piece has an equal amount of rind?  The correct answer should include a sketch or a picture and uses just a few straight cuts (no tricks).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AFvrW.png&quot; alt=&quot;Watermelon&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5188" LastActivityDate="2015-02-03T06:07:32.130" Title="Fair share of a square watermelon?" Tags="|geometry|dissection|" AnswerCount="6" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="8430" PostTypeId="1" AcceptedAnswerId="8534" CreationDate="2015-02-01T13:14:18.387" Score="10" ViewCount="1686" Body="&lt;p&gt;The following five sequence puzzles are quite nasty and hard to solve. &#xA;Yet, each of them has a clear and justifiable solution.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sequence 1:&lt;br&gt;&#xA;  ?, ?, ?, B, B, C, ?, ?, F, ?, S, M, A, S, P, S, C, A, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sequence 2:&lt;br&gt;&#xA;  ?, ?, ?, ?, 8485, 8188, 8167, 8163, 8125, 8091, 8080, 8051, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sequence 3:&lt;br&gt;&#xA;  ..., 23, 4, 16, 33, 21, 6, 18, ?, ?, ?, ?, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sequence 4:&lt;br&gt;&#xA;  P, M, B, ?, S, C, K, E, L, H, M, H, T, ?, R, ?, ?, ?, S, H, F, B, Y, R, S, D, B, U, M, L, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sequence 5:&lt;br&gt;&#xA;  1, 2, 6, 7, 6, 5, 0, 6, 0, 0, 2, 2, 8, 2, 2, 9, 4, 0, 1, 4, 9, 6, 7, 0, 3, 2, 0, 5, ?, ?, ?.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="9125" LastEditDate="2015-02-01T16:04:05.237" LastActivityDate="2015-02-04T09:05:47.727" Title="Five sequence puzzles" Tags="|number-sequence|letter-sequence|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8431" PostTypeId="2" ParentId="8430" CreationDate="2015-02-01T13:27:54.987" Score="10" Body="&lt;p&gt;Sequences 2 and 3 can easily be done by googling. For the other three sequences, Google found nothing and I have no idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sequence 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Heights of the highest mountains in meters: Mount Everest 8848, K2 8611, Kangchendzonga 8586, Lhotse 8516, Makalu 8485, ... Hence the four questionmarks are 8848, 8611, 8586, 8516.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sequence 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Numbers on a Roulette wheel: 23, 4, 16, 33, 21, 6, 18, 31, 19, 8 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-02-01T13:27:54.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8432" PostTypeId="2" ParentId="8430" CreationDate="2015-02-01T13:33:18.853" Score="12" Body="&lt;p&gt;Sequence 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each letter in this sequence represent the starting letters of the elements of the periodic table arranged on the basis of atomic number&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So,the full sequence is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; H,H,L,B,B,C,N,O,F,N,S,M,A,S,P,S,C,A,...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7482" LastActivityDate="2015-02-01T13:33:18.853" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8433" PostTypeId="1" AcceptedAnswerId="8460" CreationDate="2015-02-01T13:33:46.633" Score="5" ViewCount="916" Body="&lt;p&gt;Eight people played chess in a round robin tournament. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;No match ended in a tie and both A and B won the first place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since C and D were rivals, C beat all players who beat D and D beat all players who beat C. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the final score of A and B (the number of wins and defeats)?&lt;/p&gt;&#xA;" OwnerUserId="3914" LastActivityDate="2015-02-05T22:43:31.680" Title="8 chess players in a round robin tournament" Tags="|logical-deduction|mathematics|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8434" PostTypeId="1" AcceptedAnswerId="8436" CreationDate="2015-02-01T13:34:25.863" Score="27" ViewCount="4613" Body="&lt;p&gt;How do I convince my grandmother (who really hates mathematics) that there exist three positive integers $x,y,z$ that satisfy the equation $28x+30y+31z=3650$?&lt;/p&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="8874" LastEditDate="2016-03-19T17:05:51.460" LastActivityDate="2016-03-20T14:40:45.567" Title="How do I convince my grandmother?" Tags="|mathematics|calculation-puzzle|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8435" PostTypeId="2" ParentId="8434" CreationDate="2015-02-01T13:41:31.263" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It's obviously the number of days in 10 non-leap years.&#xA; and x, y, and z are 10, 40, and 70.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1482" LastEditorUserId="1482" LastEditDate="2015-02-02T13:14:13.150" LastActivityDate="2015-02-02T13:14:13.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8436" PostTypeId="2" ParentId="8434" CreationDate="2015-02-01T13:43:53.517" Score="38" Body="&lt;p&gt;You tell your Grandmother:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Take ten ordinary years. They contain exactly 3650 days, and these days are split among several months with 28, 30, and 31 days.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-02-01T13:43:53.517" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8437" PostTypeId="2" ParentId="8076" CreationDate="2015-02-01T14:10:48.980" Score="11" Body="&lt;p&gt;I'm going on from @Togashi's answer, which seems to be mostly correct except that he's overlooked one possibility. Anyone who's seen fallacious proofs such as &lt;a href=&quot;https://www.math.toronto.edu/mathnet/falseProofs/first1eq2.html&quot;&gt;this one&lt;/a&gt;, or who knows about &lt;a href=&quot;http://en.wikipedia.org/wiki/Pathological_(mathematics)#Pathological_example&quot;&gt;pathological (often trivial) examples&lt;/a&gt; in pure mathematics, will know&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; not to exclude the case $n=0$ unless you're specifically told $n$ is nonzero.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is why Togashi's statement&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;at this point we have $\frac{n}{4}$ green apples and $6n$ blue apples. This means we have 24 times as many blues and greens, so that can't be true [because we have the same number of blues and greens]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;immediately rang warning bells for me. &lt;strong&gt;Here's my solution.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(G,B)$ denote the ordered pair (number of green apples, number of blue apples).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Just by looking at your table, you can easily tell that the number of blue apples is not a lot more than the number of green apples; let's say less than 10 times.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(G,B)=(0,0)$. [Lynch mob!, division by 6 is optional ;)]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;you go ahead and add one blue apple [...] you now use a very old spell that doubles the number of green apples immediately.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(G,B)=(0,1)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;you feel like adding another blue apple right after you cautiously use that old spell again but this time for both kinds at once&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(G,B)=(0,3)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;your wand, which is taking action on its own and changes the colour of one apple&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(G,B)=(1,2)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;you divide the bigger group in half&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(G,B)=(1,1)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;you [...] eat one of each kind&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(G,B)=(0,0)$ again!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If done correctly you should have the same number of green and blue apples now. I'm wondering how much money you could get by selling all of them...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0 pounds, dollars, euros, deutschmarks, francs, rubles, tenges, zlotys, or any other currency you care to mention.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5106" LastEditDate="2015-02-01T14:48:29.370" LastActivityDate="2015-02-01T14:48:29.370" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8438" PostTypeId="2" ParentId="8433" CreationDate="2015-02-01T14:27:49.823" Score="6" Body="&lt;p&gt;(I assume that $A, B, C, D$ represent names, and denote four pairwise distinct players; I call the other four players $E,F,G,H$. Furthermore, I assume that the statement &quot;Since $C$ and $D$ were rivals ...&quot; does not cover the particular game played between $C$ and $D$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the problem statement, no player in $\{A,B,E,F,G,H\}$ has won both games against $C$ and $D$. Hence $\{A,B,E,F,G,H\}$ can be partitioned into three disjoint sets: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the set $L(C)$ of players who have lost to $C$ and won against $D$&lt;/li&gt;&#xA;&lt;li&gt;the set $L(D)$ of players who have lost to $D$ and won against $C$&lt;/li&gt;&#xA;&lt;li&gt;the set $L(CD)$ of players who have lost against both $C$ and $D$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;By symmetry we now assume that $C$ ended up with at least as many points as $D$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(1) $C$ scored at least $4$ points. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof: Suppose for the sake of contradiction that $C$ scored at most $3$ points.&#xA;If $C$ won against $D$, this implies $|L(C)|\le2$ and hence $|L(D)|+|L(CD)|\ge4$; then $D$ has scored at least $4$ points; contradiction.&#xA;IF $C$ lost to $D$, this implies $|L(C)|+|L(CD)|\le3$ and hence $|L(D)|\ge3$; then $D$ has scored at least $4$ points; another contradiction.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(2)  $A$ and $B$ each scored at least $5$ points.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof: $A$ and $B$ won the first place, and hence scored more points than $C$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(3) $A$ and $B$ each scored at most $5$ points.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof: Suppose for the sake of contradiction that $A$ and $B$ both scored at least $6$ points. They played one game against each other, and altogether $12$ games against the other six players. Then one of them must have won all six matches against ${C,D,E,F,G,H}$, and in particular his matches against $C$ and $D$; contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So the answer is:&lt;/strong&gt; $A$ and $B$ each lost two matches and each won five matches.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-01T14:46:28.750" LastActivityDate="2015-02-01T14:46:28.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8439" PostTypeId="2" ParentId="8429" CreationDate="2015-02-01T15:08:02.240" Score="5" Body="&lt;p&gt;Assume that the cube/watermelon has edge length $1$, and let $M$ denote the point in the very middle of the cube. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For each of the six faces of the cube, we define a corresponding pyramid that has the cube-face as its ground-square and that has point $M$ as its top vertex. &lt;/li&gt;&#xA;&lt;li&gt;The ground-square of such a pyramid has area $1$, and its height (distance from ground-square to $M$) equals $1/2$. Hence the volume of the pyramid is $1/6$ of the total cube volume.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now we can define three congruent pieces that each have equal amounts of cube surface and cube volume:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The first piece consists of the two pyramids associated with the front face and the right face.&lt;/li&gt;&#xA;&lt;li&gt;The second piece consists of the two pyramids associated with the bottom face and the back face.&lt;/li&gt;&#xA;&lt;li&gt;The third piece consists of the two pyramids associated with the top face and the left face.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Added by Len - Diagrams show the difference between the two answers:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/YkoN2.png&quot; alt=&quot;Trisection of cube&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="5188" LastEditDate="2015-02-01T20:48:57.443" LastActivityDate="2015-02-01T20:48:57.443" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8440" PostTypeId="1" AcceptedAnswerId="8454" CreationDate="2015-02-01T15:29:19.983" Score="3" ViewCount="395" Body="&lt;p&gt;My friends Peter and Sam are excellent mathematicians and always think strictly logical. Yesterday I told them: &lt;em&gt;&quot;I have secretly chosen two integers $x$ and $y$ with $1\le x\le y\le 99$. I have told their sum $s=x+y$ to Sam and their product $p=xy$ to Peter.&quot;&lt;/em&gt; Then the following conversation developed.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Peter said: I don't know the numbers.&lt;/li&gt;&#xA;&lt;li&gt;Sam said:   I don't know the numbers.&lt;/li&gt;&#xA;&lt;li&gt;Peter said: I don't know the numbers.&lt;/li&gt;&#xA;&lt;li&gt;Sam said:   I don't know the numbers.&lt;/li&gt;&#xA;&lt;li&gt;Peter said: I don't know the numbers.&lt;/li&gt;&#xA;&lt;li&gt;Sam said:   I don't know the numbers.&lt;/li&gt;&#xA;&lt;li&gt;Peter said: I don't know the numbers.&lt;/li&gt;&#xA;&lt;li&gt;Sam said:   I don't know the numbers.&lt;/li&gt;&#xA;&lt;li&gt;Peter said: I don't know the numbers.&lt;/li&gt;&#xA;&lt;li&gt;Sam said:   I don't know the numbers.&lt;/li&gt;&#xA;&lt;li&gt;Peter said: I don't know the numbers.&lt;/li&gt;&#xA;&lt;li&gt;Sam said:   I don't know the numbers.&lt;/li&gt;&#xA;&lt;li&gt;Peter said: I don't know the numbers.&lt;/li&gt;&#xA;&lt;li&gt;Sam said:   I don't know the numbers.&lt;/li&gt;&#xA;&lt;li&gt;Peter said: Aha. Then I do know the numbers now.&lt;/li&gt;&#xA;&lt;li&gt;Sam said:  Aha. Then I also know the numbers now.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What are the values of $x$ and $y$?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Bonus question: How does the answer change, if we replace the lower bound in&#xA; &quot;$1\le x\le y\le 99$&quot; by the new lower bound&#xA;&quot;$66\le x\le y\le 99$&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Remark: This puzzle is a variation on &lt;a href=&quot;https://puzzling.stackexchange.com/questions/8153/product-and-sum-variant-2a&quot;&gt;Product and Sum, Variant 2a&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2016-02-24T03:59:54.100" Title="Product and Sum, Variant 2b" Tags="|mathematics|logical-deduction|meta-knowledge|" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8441" PostTypeId="5" CreationDate="2015-02-01T16:14:34.500" Score="0" Body="&lt;p&gt;A 'letter sequence' is a series or collection of letters presented in a particular order according to a specific logical rule. In &lt;a href=&quot;/questions/tagged/word-sequence&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;word-sequence&amp;#39;&quot; aria-label=&quot;show questions tagged &amp;#39;word-sequence&amp;#39;&quot; rel=&quot;tag&quot; aria-labelledby=&quot;word-sequence-container&quot;&gt;word-sequence&lt;/a&gt; puzzles, the aim for the solver is usually to deduce or discover the rule behind this ordering, and/or supply the next - or missing - letter(s) in the sequence.&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="60644" LastEditDate="2022-12-12T23:28:18.180" LastActivityDate="2022-12-12T23:28:18.180" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8442" PostTypeId="4" CreationDate="2015-02-01T16:14:34.500" Score="0" Body="A puzzle in which the hidden pattern behind a given sequence of letters is to be found. Use with [sequence]" OwnerUserId="8874" LastEditorUserId="60644" LastEditDate="2022-12-12T23:28:18.180" LastActivityDate="2022-12-12T23:28:18.180" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8443" PostTypeId="2" ParentId="8429" CreationDate="2015-02-01T16:22:48.320" Score="21" Body="&lt;p&gt;Look at the cube down a space diagonal. It should appear as a hexagon, which can be divided into three rhombuses by line segments from the center to every other corner. If these three cuts are made, they will divide the cube into three pieces, which must be congruent by symmetry.&lt;/p&gt;&#xA;" OwnerUserId="9291" LastActivityDate="2015-02-01T16:22:48.320" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8444" PostTypeId="5" CreationDate="2015-02-01T16:47:25.700" Score="0" Body="&lt;p&gt;A puzzle that heavily depends on linguistic features.&lt;/p&gt;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A puzzle reliant on words in a language other than English;&lt;/li&gt;&#xA;&lt;li&gt;A puzzle built around a peculiarity of the &lt;a href=&quot;/questions/tagged/english&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;english&amp;#39;&quot; aria-label=&quot;show questions tagged &amp;#39;english&amp;#39;&quot; rel=&quot;tag&quot; aria-labelledby=&quot;english-container&quot;&gt;english&lt;/a&gt; language;&lt;/li&gt;&#xA;&lt;li&gt;A puzzle that plays heavily with particular features of language such as verbs, nouns, adjectives, etc.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="52483" LastEditorUserId="60644" LastEditDate="2023-01-31T18:38:55.440" LastActivityDate="2023-01-31T18:38:55.440" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8445" PostTypeId="4" CreationDate="2015-02-01T16:47:25.700" Score="0" Body="A puzzle that heavily depends on linguistic features; for instance it may concern foreign languages, or only work in some particular language, or be built around a peculiarity of the English language.  " OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-23T17:12:59.960" LastActivityDate="2016-03-23T17:12:59.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8446" PostTypeId="5" CreationDate="2015-02-01T16:49:21.067" Score="0" Body="&lt;p&gt;This is a subset of the &lt;a href=&quot;/questions/tagged/seasonal&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;seasonal&amp;#39;&quot; rel=&quot;tag&quot;&gt;seasonal&lt;/a&gt; tag.&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="5373" LastEditDate="2015-05-18T10:35:20.113" LastActivityDate="2015-05-18T10:35:20.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8447" PostTypeId="4" CreationDate="2015-02-01T16:49:21.067" Score="0" Body="A seasonal puzzle themed towards Christmas." OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-01T20:06:28.800" LastActivityDate="2015-02-01T20:06:28.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8448" PostTypeId="5" CreationDate="2015-02-01T16:53:35.787" Score="0" Body="&lt;p&gt;A puzzle centred around &lt;a href=&quot;https://en.wikipedia.org/wiki/Kinship&quot; rel=&quot;nofollow noreferrer&quot;&gt;kinship&lt;/a&gt;, family ties, ancestry, relatives, blood ties, and other family bonds. Often related to deducing what connects a family, or what makes families difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/40023/what-is-the-relationship-between-the-man-and-the-woman&quot;&gt;What is the relationship between the man and the woman?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11497" LastEditorUserId="11497" LastEditDate="2019-03-29T13:51:20.020" LastActivityDate="2019-03-29T13:51:20.020" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8449" PostTypeId="4" CreationDate="2015-02-01T16:53:35.787" Score="0" Body="A puzzle centred around kinship, family ties, ancestry, relatives, blood ties and other family bonds." OwnerUserId="8874" LastEditorUserId="11497" LastEditDate="2019-03-29T13:51:20.020" LastActivityDate="2019-03-29T13:51:20.020" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8450" PostTypeId="2" ParentId="8434" CreationDate="2015-02-01T17:51:08.387" Score="12" Body="&lt;p&gt;Your grandmother can probably&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; answer the question for 365 (days), &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and she can probably&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; multiply by 10.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9294" LastActivityDate="2015-02-01T17:51:08.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8452" PostTypeId="2" ParentId="8429" CreationDate="2015-02-01T18:54:49.133" Score="16" Body="&lt;p&gt;This image from &lt;a href=&quot;http://demonstrations.wolfram.com/ThreePyramidsThatFormACube/&quot; rel=&quot;noreferrer&quot;&gt;http://demonstrations.wolfram.com/ThreePyramidsThatFormACube/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/InlHt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;solves the problem - but Stan or Kyle or Kenny would have trouble cutting their cubical watermelon this way since the point of the knife needs to follow the long diagonal of the cube while the edge of the blade follows a face diagonal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each individual pyramid has a square base and one vertical edge. It's the most interesting piece in the geoblocks available in many K-2 classrooms - the central one in this image from &lt;a href=&quot;http://catalog.mathlearningcenter.org/store/product-924015.htm&quot; rel=&quot;noreferrer&quot;&gt;http://catalog.mathlearningcenter.org/store/product-924015.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hbDL9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This dissection is really much more than the answer to a puzzle. It's the essence of the theorem that says that the volume of a pyramid is 1/3 of the height times the area of the base, and of the fact that the integral of $x^2$ is $x^3/3$. The analogue in the plane is cutting a square in half along the diagonal. In four dimensions you can cut a tesseract into four congruent pyramids each of which has a cube for a base. &lt;/p&gt;&#xA;" OwnerUserId="9294" LastEditorUserId="9294" LastEditDate="2015-02-02T15:09:02.123" LastActivityDate="2015-02-02T15:09:02.123" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8453" PostTypeId="2" ParentId="8429" CreationDate="2015-02-01T19:12:53.240" Score="1" Body="&lt;p&gt;Cut the watermelon in a cross.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From corner to corner. Like an X!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, the remaining slice, you cut in 3 parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the best option since it also avoid wastes in case one of them doesn't want that piece. Thus, easily dividing the remaining slice in only 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the slices will have the same ammount of 'shell' and a very ergonomic shape, that helps you to bite it or cut it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As pointed by @Len, there is one problem...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we have to cut the slice in 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution is to grab a piece of string with the length of the slice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you make a circle with it and mark 1 line at the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try to make the &quot;peace&quot; symbol (You can check it here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Peace_symbols&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Peace_symbols&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cut it along one of the lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The string will have 1/3 of the size!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now cut the slice with the same width as the string.&lt;/p&gt;&#xA;" OwnerUserId="3984" LastEditorUserId="3984" LastEditDate="2015-02-01T19:33:48.177" LastActivityDate="2015-02-01T19:33:48.177" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8454" PostTypeId="2" ParentId="8440" CreationDate="2015-02-01T20:00:37.913" Score="5" Body="&lt;p&gt;I wrote a Ruby program to cut out the tedious manual work:&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#!/usr/bin/ruby&#xA;&#xA;range = [*1..99]&#xA;&#xA;pairs = range.product(range).select{|x, y| x &amp;lt;= y }&#xA;pairs.map! {|x, y| [x, y, x+y, x*y] }&#xA;&#xA;def getuniqs arr&#xA;    arr.group_by{|x| x }&#xA;       .map{|k, v| v.length == 1 ? k : nil }&#xA;       .compact&#xA;end&#xA;&#xA;def dunno pairs, n&#xA;    uniqs = getuniqs pairs.map{|p| p[n] }&#xA;    pairs.reject{|pair| uniqs.index pair[n] }&#xA;end&#xA;&#xA;def aha pairs, n&#xA;    uniqs = getuniqs pairs.map{|p| p[n] }&#xA;    pairs.select{|pair| uniqs.index pair[n] }&#xA;end&#xA;&#xA;7.times {&#xA;    pairs = dunno pairs, 3  # peter&#xA;    pairs = dunno pairs, 2  # sam&#xA;}&#xA;pairs = aha pairs, 3&#xA;&#xA;puts pairs.map{|x, y, s, p| &quot;#{x}, #{y}&quot; }&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Results:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;llama@llama:...Code/ruby/puzzling8440$ ./pro_sum.rb &#xA;77, 84&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The numbers are $77$ and $84$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the bonus: The results are exactly the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This uses the same strategy as &lt;a href=&quot;https://puzzling.stackexchange.com/a/8156/39&quot;&gt;the answer on Variant 2A&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="39" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-02-01T20:00:37.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8455" PostTypeId="2" ParentId="6226" CreationDate="2015-02-01T20:54:34.613" Score="8" Body="&lt;p&gt;I'm on a phone, so I don't know how to do spoilers.&lt;br /&gt;&#xA;Can somne please add spoilers for me if these aren't working?&lt;br /&gt;&#xA;This is super late, too but I haven't had a chance to write the answer.&lt;/p&gt;&#xA;&lt;h1&gt;Here's the answer that you've been waiting for.&lt;/h1&gt;&#xA;&lt;p&gt;... hopefully&lt;/p&gt;&#xA;&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;The moon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&lt;p&gt;In Korea, a door as bears fall up.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;The word for door, 문, sounds like &amp;quot;moon&amp;quot; and is 곰, or bear, upside down.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;In America a greeting, &amp;quot;Hey dude, what's up?&amp;quot;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;The moon is up.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;In Greece, a virgin, fast as an arrow,&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;Artemis is the god of the moon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;In Britain, a spy flitting in with stealth.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;This one was a slight stretch. 007 happens to look like oon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The next paragraph is derived by the following steps:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;You take the word &amp;quot;moon&amp;quot; and flip ur upside down top make &amp;quot;uoow&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Then,&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;Then you put it in a caeser cipher, ROT16, to get &amp;quot;keem&amp;quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So,&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;He misses &amp;quot;keen&amp;quot; by one step.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Controls 70% of the earth.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;The tides are controlled by the moon, and the ocean covers 70% of the earth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;With some secrets ready to be unearthed.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;We continue to look for more information on the moon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The little gold ring spins without a sound, around and around and around.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;The bright part of the moon phases spin around and around.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;While the songs of the holes sing loudly and clearly, If only, If only, If only.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;In Holes by Louis Sachar, there is a song that ends like &amp;quot;If only, If only, If only.&amp;quot; The word before this is &amp;quot;moooooooooooooooooooooooooooon&amp;quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h1&gt;And there you have it,&lt;/h1&gt;&#xA;&lt;p&gt;the answer.&lt;/p&gt;&#xA;" OwnerUserId="167" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-02-01T20:54:34.613" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8457" PostTypeId="1" AcceptedAnswerId="8458" CreationDate="2015-02-01T23:26:37.697" Score="5" ViewCount="682" Body="&lt;p&gt;I've used quite a simple algorithm to create the appearance of a jumbled up bunch of letters. These letters represent a song. What is the song title?&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;h1&gt;TTLSHIWWYAUATWSHLADITSTTLSHIWWYA&lt;/h1&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;(Everyone should know this song, unless you live in a non-English speaking country.)&lt;/p&gt;&#xA;" OwnerUserId="167" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-10-08T09:21:39.187" Title="Scrambled Songs" Tags="|letter-sequence|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8458" PostTypeId="2" ParentId="8457" CreationDate="2015-02-02T01:13:15.203" Score="6" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Twinkle Twinkle Little Star&quot;, each letter corresponds to the first letter of each word in the song.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastActivityDate="2015-02-02T01:13:15.203" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8459" PostTypeId="1" CreationDate="2015-02-02T01:56:29.847" Score="8" ViewCount="2433" Body="&lt;p&gt;I was working my way through some puzzles in &lt;a href=&quot;http://www.flipkart.com/discrete-mathematics-its-applications-combinatorics-graph-theory-english-7th/p/itmdwxxszhczjrnf&quot; rel=&quot;noreferrer&quot;&gt;Discrete Maths by Rosen&lt;/a&gt;, when I came across the following question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The $n^{th}$ statement in a list of 100 statements is : &lt;em&gt;&quot;Exactly $n$ of the statements in this list are false&quot;&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;p&gt;What conclusion can you draw from these statements ?&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Answer the first part if the $n^{th}$ statement is : &lt;em&gt;&quot;At least $n$ of the statements in this list are false&quot; ?&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Answer the second part assuming that the list contains 99 statements ?&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;My Solution (Inadequate):&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;p&gt;The 99th Statement is True and the rest are false&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;I am all thumbs for the next two parts&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Book solution:&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;The 99th Statement is True and the rest are false&lt;/li&gt;&#xA;  &lt;li&gt;Statements 1 through 50 are all true and statements 51 through 100 are&#xA;  all false&lt;/li&gt;&#xA;  &lt;li&gt;This cannot happen; it is a paradox, showing that these cannot be statements.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;My question:&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this so?&lt;/p&gt;&#xA;" OwnerUserId="7435" LastEditorUserId="8874" LastEditDate="2015-02-03T14:53:41.597" LastActivityDate="2015-02-04T10:56:55.840" Title="A list of 100 statements" Tags="|logical-deduction|paradox|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8460" PostTypeId="2" ParentId="8433" CreationDate="2015-02-02T03:38:10.033" Score="9" Body="&lt;p&gt;With $8$ players, there have been $(8 \times 7)/2 = 28$ matches and therefore $28$ points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If A and B scored 4 points or less, the other players would have scored at most 3 points which brings a total of at $2\times 4+6\times 3 = 26$ points or less.  But there are 28 points altogether.  Therefore A and B scored at least 5 points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of A or B must have lost against the other, and both have lost against C or D.  Therefore one of them has lost 2 matches and scores at most 5 points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result, A and B can have scored only 5 points.  They won 5 and lost 2 matches each.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastActivityDate="2015-02-02T03:38:10.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8461" PostTypeId="2" ParentId="8429" CreationDate="2015-02-02T08:48:39.447" Score="2" Body="&lt;p&gt;Just cut the rind off 2 opposing sides then cut the remains into 3 equal parts.&lt;/p&gt;&#xA;" OwnerUserId="6076" LastEditorUserId="6076" LastEditDate="2015-02-03T06:07:32.130" LastActivityDate="2015-02-03T06:07:32.130" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8462" PostTypeId="1" AcceptedAnswerId="8468" CreationDate="2015-02-02T08:53:21.183" Score="8" ViewCount="328" Body="&lt;p&gt;Professor Erasmus has constructed a special &lt;a href=&quot;http://en.wikipedia.org/wiki/Tetrahedron&quot;&gt;tetrahedron&lt;/a&gt; that he modestly calls the &quot;Professor-Erasmus-tetrahedron&quot;. The professor claims that all four faces of his tedrahedron are &lt;a href=&quot;http://en.wikipedia.org/wiki/Right_triangle&quot;&gt;right-angled&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Congruence_(geometry)#Congruence_of_triangles&quot;&gt;congruent&lt;/a&gt; to each other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has the professor once again made one of his mathematical blunders, or do such  tetrahedrons indeed exist?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-02-02T10:46:24.663" Title="The Erasmus tedrahedron" Tags="|geometry|triangle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8463" PostTypeId="5" CreationDate="2015-02-02T09:14:12.447" Score="0" Body="&lt;p&gt;A number of players are each wearing a hat, which may be of various specified colors. &#xA;Every player can see the colors of some (or all) other players' hats, but does not know the color of his own hat. &#xA;With highly restricted communication (that varies from puzzle to puzzle), some of the players must guess the color of their hat. &#xA;The problem is to find a strategy for the players to determine the colors of their hats based on the hats they see and based on the observed actions of the other players. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some versions, the players compete to be the first to guess correctly; in other versions, they can work out a strategy beforehand to cooperate and maximize the probability of correct guesses.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-04T11:39:45.893" LastActivityDate="2015-02-04T11:39:45.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8464" PostTypeId="4" CreationDate="2015-02-02T09:14:12.447" Score="0" Body="A strategic puzzle centered around a group of people, where everybody has to guess the color of his own hat." OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-04T11:39:55.127" LastActivityDate="2015-02-04T11:39:55.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8465" PostTypeId="2" ParentId="8429" CreationDate="2015-02-02T09:19:02.000" Score="2" Body="&lt;p&gt;You can cut it in six pyramids (each with the base one face of the cube, and the top in the middle of the cube).&#xA;To cut in three parts, you take an edge and cut away pyramids formed by the two squares that define that edge (let's say top and left). After that, you do the same for front and right, and end with back and bottom as the third piece.&#xA;   This way, you also solve the problem of sharing the sweetness (the center of a watermelon is sweeter than the margins).&lt;/p&gt;&#xA;" OwnerUserId="9239" LastActivityDate="2015-02-02T09:19:02.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8466" PostTypeId="5" CreationDate="2015-02-02T09:46:30.993" Score="0" Body="&lt;p&gt;A board game is a game that involves counters or pieces moved or placed on a pre-marked surface or &quot;board&quot;, according to a fixed set of rules. &#xA;Board games can be based on pure strategy, chance (rolling dice), or a mixture of the two, and usually have a goal that a player aims to achieve. &#xA;Early board games represented a battle between two armies, and most modern board games are still based on defeating opposing players in terms of counters, winning position, or acquisition of points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typical board games are chess, go, reversi, checkers, backgammon, abalone, Trivial Pursuit, risk, monopoly, diplomacy.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-02T14:59:28.913" LastActivityDate="2015-02-02T14:59:28.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8467" PostTypeId="4" CreationDate="2015-02-02T09:46:30.993" Score="0" Body="A board game is a game that involves counters or pieces moved or placed on a &quot;board&quot; according to a fixed set of rules. " OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-02T14:54:43.037" LastActivityDate="2015-02-02T14:54:43.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8468" PostTypeId="2" ParentId="8462" CreationDate="2015-02-02T09:47:28.630" Score="8" Body="&lt;p&gt;No, it's not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any vertex, the three edges coming out of it must have distinct length, since otherwise the triangle must be isosceles as well as right, which forces an equilateral triangle for the opposite face. Therefore, the three angles at each vertex must correspond to the three angles of the right triangle; call them $\alpha$, $\beta$, and $90^\circ$. But since $\alpha + \beta = 90^\circ$, the faces must join flat at the vertex, forming the degenerate flat tetrahedron shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BPuLNl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-02-02T10:46:24.663" LastActivityDate="2015-02-02T10:46:24.663" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8470" PostTypeId="2" ParentId="8283" CreationDate="2015-02-02T13:32:16.730" Score="2" Body="&lt;p&gt;This is a non-answer/partial answer. Just assembling some thoughts and maybe getting the ball rolling?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;atop the monument seeking jewellery //&#xA;this fake character is not the ordinary //&#xA;a trio, the misguided claim to possess //&#xA;their psychiatric health we must assess&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Edit: This could be referring to Jesus (though Idk what monument he's on top of). He's &quot;fake&quot; according to a lot of people, he is representative of the holy trinity, Christians (the &quot;misguided&quot;) claim to have him in their hearts or whatever, which would make them crazy if he's fake. &quot;Seeking jewelry&quot; could refer to the crucifix as a piece of jewelry.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;watching you now, locked in solitude //&#xA;lose thoughts of it and gain gratitude //&#xA;data from it to be deciphered //&#xA;easy comprehension is what's desired //&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This sounds like it's referring to the riddle itself, so the word here is probably something like &lt;em&gt;riddle&lt;/em&gt;, &lt;em&gt;rhyme&lt;/em&gt;, &lt;em&gt;poem&lt;/em&gt;. This could also be referring to your computer, however, so it may be &lt;em&gt;computer&lt;/em&gt;, &lt;em&gt;monitor&lt;/em&gt;, &lt;em&gt;screen&lt;/em&gt;... If I had to bet, I'd go with &lt;em&gt;monitor&lt;/em&gt;, since when used as a verb, it means &lt;em&gt;to watch&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;to see your path, you favour no grin //&#xA;just follow the road by the grimalkin //&#xA;the agreeable ones pillage upon the salt //&#xA;those who fare, through billow upon holt&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I have no clue what this could be about. A road that is &quot;by&quot; a cat, but not a grinning cat (i.e. Cheshire)? So... A road that is the name of a cat and is somehow related to Cheshire...?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;low warm and high cool, the air will fight //&#xA;stay close within to avoid the smite //&#xA;as the exterior rips through the land //&#xA;the inner recess is tranquil and grand&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I doubt that this could be referencing anything but a &lt;em&gt;tornado&lt;/em&gt;. How it's formed, eye of the storm, destruction...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;shot, the projectile changes thee //&#xA;elude, the meaning will till you guarantee //&#xA;no material can heal what once was neat //&#xA;subtle memories, dread a repeat&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Complete guess, but this seems to me that it's referencing the structure of the stanza itself. As if you're supposed to rearrange the awkward syntax to make it &quot;neat&quot; again, then re-read it to figure out what it's talking about -- it'll elude you until you do.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the drum of life does symbolise //&#xA;one does not simply analyse //&#xA;broken only, lonely alone //&#xA;rejection wrecking leaving stone&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not a clue. &quot;Drum of life&quot; could be referring to something like a song or book called &quot;beat&quot; or &quot;rhythm&quot; of life. Alternatively, it could be referring to a sleep cycle, or &quot;biorhythm&quot;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yet true twice in benevolent violence //&#xA;one look only for it gains entrance //&#xA;an entity for life you obtain //&#xA;create anew a life to reign&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This sounds like giving birth. Two ovaries, one sperm enters an egg, baby is made. Not sure what the word here would be... &lt;em&gt;Birth&lt;/em&gt;, &lt;em&gt;conception&lt;/em&gt;, &lt;em&gt;baby&lt;/em&gt;, anything in between...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;compelled by this to achieve //&#xA;yet humility intervenes, you don’t believe //&#xA;effects differ and subjects vary //&#xA;listen to this its not that scary&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It sounds to me like the word here is &lt;em&gt;encouragement&lt;/em&gt;, though it could also be something like &lt;em&gt;advice&lt;/em&gt;. Encouragement compels you to achieve something, but your own humble views of yourself may make you believe you can't do it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the board rides the pipe into the air //&#xA;high as a hawk, its quite a scare //&#xA;thou shalt address thee as a clue //&#xA;be thy plumber of thy loo&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Right, so address yourself as a clue when you're cleaning out your own pipes. The board rides the pipe... Yeah. Lol Idk what this one is about, but it sounds dirty.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;material of squander travels around //&#xA;within the confines, trapped and bound //&#xA;nowhere to go the water rises //&#xA;break free quickly no nasty surprises&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Material of squander... waste? Waste within confines and considered to be &quot;water&quot; sounds like a sewer, which is enforced by &quot;nasty&quot; surprises. So... I'm guessing &lt;em&gt;sewer&lt;/em&gt; here? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Germans sink within this vessel //&#xA;under the radar hoping to nestle //&#xA;place what you need before the knee //&#xA;take the verse and have some tea&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Obviously talking about a U-boat. I'm not sure what you would &quot;place before your knee&quot; before praying over tea... But whatever that is, probably in German, is what we're looking for.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the Gothic did state the differences //&#xA;upon the vague instances //&#xA;for we were once another //&#xA;our rather angular brother&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is talking about evolution, but I'm not sure what exactly we're supposed to extract from it. I don't think &quot;Gothic&quot; refers to Darwin, and I don't know enough to guess any other scientist. Otherwise, we could be seeking &lt;em&gt;evolution&lt;/em&gt; itself, or maybe &lt;em&gt;homo erectus&lt;/em&gt; or one of his kin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All in all, I find myself very... confused by this puzzle. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Monitor, Tornado, Biorhythm, Conception, Encouragement, Sewer, Evolution...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't see a connection at all. The only thing I can think of is that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...all of these words are supposed to be &lt;em&gt;appropriate&lt;/em&gt; words that might make people feel uncomfortable, hence making them the &quot;hardest to verbalize&quot;. (Conception, Masturbation, Sewage, Homo Erectus...) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I'm 99.9% sure that I'm miles away from the actuality of it all.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I suppose it could also be that all of the words &lt;em&gt;aren't English words&lt;/em&gt;, which would make them pretty hard to say in English.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8858" LastEditorUserId="8858" LastEditDate="2015-02-02T16:33:09.087" LastActivityDate="2015-02-02T16:33:09.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8471" PostTypeId="2" ParentId="8404" CreationDate="2015-02-02T14:57:42.797" Score="15" Body="&lt;p&gt;My guess is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clio one of the 9 Muses&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In roman: 101 = CI, we put 50 = L in between to get CLI and we &quot;add&quot; 0 which kinda looks like &quot;O&quot; and we get CLIO. There are 9 muses and she is one of them so that is the reason behind 1/9.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The roman numbers are a reference to the roman time, when the muses actually got their roles (for example Clio became the muse of history).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Also:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;one big &quot;thanks&quot; to &lt;a href=&quot;https://puzzling.stackexchange.com/users/5373/rand-althor&quot;&gt;@rand al'thor&lt;/a&gt; as he provided the basis by mentioning the roman numerals in &lt;a href=&quot;https://puzzling.stackexchange.com/a/8424/5610&quot;&gt;his answer&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.923" LastActivityDate="2015-02-02T15:18:05.757" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8472" PostTypeId="2" ParentId="8459" CreationDate="2015-02-02T15:24:15.547" Score="6" Body="&lt;p&gt;For the second part, &quot;&lt;em&gt;At least $n$ of the statements in this list are false&lt;/em&gt;,&quot; you have some number $k$ of false statements, and $100-k$ true statements. It should be clear that the $n^{th}$ statement is true whenever $n\le k$, and false otherwise, which means that there also $k$ true statements. So, you have $k=100-k$, which means that $k=50$. There are $50$ true and $50$ false statements, which means the first $50$ statements are true and the rest are false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method should make it clear that there is &lt;em&gt;something&lt;/em&gt; wrong with the third problem, since it tells you that there should be $49.5$ true statements, and how can you have a (logical) half-truth? To see the paradox, suppose that statements 1 through 49 are true, and the rest are false. Then there are $50$ false statements, so the $50^{th}$ statement is true. But the $50^{th}$ was one of the false statements. Now there are only $49$ statements that are false, so &quot;At least $50$ of the statemnets in this list are false&quot; is false. This gives $50$ false statements again, so we're back where we started.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An odd number of statements in a list like that produces a version of the Liar's Paradox.&lt;/p&gt;&#xA;" OwnerUserId="8776" LastActivityDate="2015-02-02T15:24:15.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8473" PostTypeId="1" AcceptedAnswerId="8479" CreationDate="2015-02-02T15:32:52.030" Score="13" ViewCount="844" Body="&lt;p&gt;The French mathematician Lagrange proved that every natural number can be written as the sum of four squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is true in the view of cryptarithms, too.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  SQUARE&#xA;  SQUARE&#xA;  SQUARE&#xA;+ SQUARE&#xA;--------&#xA;  NUMBER&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Each letter represents a unique digit.&lt;/p&gt;&#xA;" OwnerUserId="3914" LastEditorUserId="9404" LastEditDate="2016-03-03T10:20:37.140" LastActivityDate="2016-03-03T10:20:37.140" Title="Lagrange's Four Square Cryptarithm" Tags="|mathematics|logical-deduction|alphametic|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8476" PostTypeId="2" ParentId="8430" CreationDate="2015-02-02T16:10:34.493" Score="9" Body="&lt;p&gt;Sequence 5&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3, 7 and 6. Those are the last digits of 2^100 = 1267650600228229401496703205376&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9315" LastActivityDate="2015-02-02T16:10:34.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8477" PostTypeId="2" ParentId="8473" CreationDate="2015-02-02T16:40:28.757" Score="5" Body="&lt;p&gt;Square has a value of 149082&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number has a value of 596328&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way I obtained my answer was, that in my python class I created a program to solver word puzzles, such as SEND+MORE==MONEY, and this was just another one of those. So I just ran my program with that puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Python3 code to solve the puzzle:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import re&#xA;import itertools&#xA;&#xA;def solve(puzzle):&#xA;    words = re.findall('[A-Z]+', puzzle.upper())&#xA;    unique_characters = set(''.join(words))&#xA;    assert len(unique_characters) &amp;lt;= 10, 'Too many letters'&#xA;    first_letters = {word[0] for word in words}&#xA;    n = len(first_letters)&#xA;    sorted_characters = ''.join(first_letters) + \&#xA;        ''.join(unique_characters - first_letters)&#xA;    characters = tuple(ord(c) for c in sorted_characters)&#xA;    digits = tuple(ord(c) for c in '0123456789')&#xA;    zero = digits[0]&#xA;    for guess in itertools.permutations(digits, len(characters)):&#xA;        if zero not in guess[:n]:&#xA;            equation = puzzle.translate(dict(zip(characters, guess)))&#xA;            if eval(equation):&#xA;                return equation&#xA;&#xA;if __name__ == '__main__':&#xA;    import sys&#xA;    for puzzle in sys.argv[1:]:&#xA;        print(puzzle)&#xA;        solution = solve(puzzle)&#xA;        if solution:&#xA;            print(solution)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.diveintopython3.net/advanced-iterators.html&quot;&gt;This&lt;/a&gt; code is not the one I wrote, but does the same thing.&lt;/p&gt;&#xA;" OwnerUserId="9159" LastActivityDate="2015-02-02T16:40:28.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8478" PostTypeId="2" ParentId="8283" CreationDate="2015-02-02T17:37:01.420" Score="18" Body="&lt;p&gt;I think I understand it. The title, the riddle sectioning, everything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the beginning of trying to solve this riddle, I thought that maybe each paragraph represented a word. This didn't exactly work out as I went. However, the words I gathered were often similar to each other. This helped me clarify a lot of things.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each section of four paragraphs is a different word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Section 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;The first word is I.&lt;/strong&gt; &lt;br&gt;&lt;em&gt;atop the monument seeking jewellery, this fake character is not the ordinary&lt;/em&gt; &lt;br&gt;This is talking about the lowercase i: the dot is the jewel atop the monument, the &quot;fake&quot; character is how i represents an imaginary number in mathematics. &lt;br&gt;&lt;br&gt;&lt;em&gt;a trio, the misguided claim to possess, their psychiatric health we must assess&lt;/em&gt; &lt;br&gt;Talking about me, myself, and I - thinking about deep-seated conceit as a psychiatric concern.&lt;br&gt;&lt;br&gt; &lt;em&gt;watching you now, locked in solitude, lose thoughts of it and gain gratitude&lt;/em&gt;&lt;br&gt;Courteousness. Don't be selfish, and people will thank you.&lt;br&gt;&lt;br&gt;&lt;em&gt;data from it to be deciphered, easy comprehension is what's desired&lt;/em&gt;&lt;br&gt;The id, aka sense of self-identity, people strive to identify who they are. In a philosophical sense, people look for the easiest way to understand themselves.&lt;br&gt;&lt;br&gt;&lt;em&gt;to see your path, you favour no grin, just follow the road by the grimalkin&lt;/em&gt;&lt;br&gt;The future; a form of identity as related to how one perceives their future. The grimalkin is another word for ominous cat, which at the time of its use, referred to familiar cats (those owned by witches) that were believed to have the power of futuresight.&lt;br&gt;&lt;br&gt;&lt;em&gt;the agreeable ones pillage upon the salt, those who fare through billow upon holt&lt;/em&gt;&lt;br&gt;Credit to &lt;a href=&quot;https://puzzling.stackexchange.com/users/2162/matt-malone&quot;&gt;Matt Malone&lt;/a&gt;, this refers to seafaring pirates who say &quot;aye&quot;, pronounced like &quot;I&quot;, when they are in agreement.&lt;br&gt;&lt;br&gt;&lt;em&gt;low warm and high cool, the air will fight, stay close within to avoid the smite, as the exterior rips through the land, the inner recess is tranquil and grand&lt;/em&gt;&lt;br&gt;Probably the easiest of the bunch, this is literally what it suggests: the &quot;eye&quot; of the storm. The difference is that it has nothing to do with self, except that &quot;eye&quot; is a homonym of &quot;I&quot;.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;OP Edit:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I, me and myself are the trio, aswell as those who believe they have a third eye. Spirituality does not come easy to me ergo I stated this line was an opinion.&lt;br&gt;&#xA;&lt;br&gt;&#xA;I is the monument and it seeks its dot, which is only in the lower case. There is also a reference that could be taken if you know of Lord Of The Rings, where an eye is on a monument looking for a ring. Which is also not the ordinary.&lt;br&gt;&#xA;&lt;br&gt;&#xA;Lose thoughts of I will bring gratitude from others. Watching you now, represents the eye itself. Locked in solitude, this is the upper case &quot;I&quot; (cap locks) and on its own as &quot;I&quot; stands alone when written.&lt;br&gt;&#xA;&lt;br&gt;&#xA;&quot;Data from it to be deciphered&quot; This is because we see upside down and the brain rotates the picture for us for easy comprehension! (also side not of how we see colour, we just label a colour but how do we know we all see the same thing)&#xA;&lt;br&gt;&#xA;We follow the path of the cat's eye when driving at night. Usually along motorways and other major roads to ensure we dont randomly drive to and.. die... Horribly.&lt;br&gt;&#xA;&lt;br&gt;&#xA;holt is a wooded hill, a pirate ship would be a wooded hill on the sea.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Section 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;The second word is love.&lt;/strong&gt;&lt;br&gt;&lt;em&gt;shot, the projectile changes thee; elude, the meaning will till you guarantee; no material can heal what once was neat; subtle memories, dread a repeat&lt;/em&gt;&lt;br&gt;Cupid's arrow. Once struck, you are changed forever. You won't understand it until you know for sure that it's love. It's awkward for the riddle to claim that being lovestruck needs healing, and leads me to believe the second half is talking about heartbreak.&lt;br&gt;&lt;br&gt;&lt;em&gt;the drum of life does symbolise one does not simply analyse; broken only, lonely alone, rejection wrecking leaving stone&lt;/em&gt; The drum of life is the heart and its heartbeat.  You can't research love. Again, second half is talking about heartbreak, and how being rejected turns the heart stone-cold and less susceptible to falling in love again.&lt;br&gt;&lt;br&gt;&lt;em&gt;yet true twice in benevolent violence, one look only for it gains entrance, an entity for life you obtain, create anew a life to reign&lt;/em&gt;&lt;br&gt;Motherly love. Tricky to understand at first glance, the benevolent violence is another word for time. In essence, this means &quot;twice in life you find true love, but you only have to look for it once&quot;. In other words, the first time you encounter true love is from your mother at birth. The second half encourages this thought, talking about giving birth to new life.&lt;br&gt;&lt;br&gt;&lt;em&gt;compelled by this to achieve yet humility intervenes, you don’t believe; effects differ and subjects vary, listen to this its not that scary&lt;/em&gt;&lt;br&gt;Final words about love, this is talking about how we are compelled to find love, but low self-esteem gets in the way, making you think you'll never find it. But don't give up, there are lots of people out there and you might not notice love when it reaches out to you. Think about this and you can do away with lover's anxiety.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;OP Edit: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; love is extremely hard to describe in riddle format without being ambigous, however. I tried to make at least some of the lines absolute truth, even then clutures and whatnot would disagree.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Guarantee is the contracts signed when getting married.&lt;br&gt;&#xA;&lt;br&gt;&#xA;Neat is the word used to describe a pure drink E.G. in this instance neat is the pure love which gets broken after love ends.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Love is in benevolent and in violence. Another reference to it being in their twice is the act of being cruel to be kind. &quot;you hurt those closest the most&quot; and all those types of phrases.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Section 3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;The final word is you.&lt;/strong&gt;&lt;br&gt;&lt;em&gt;the board rides the pipe into the air, high as a hawk, its quite a scare&lt;/em&gt;&lt;br&gt;Kinda silly, but this is talking about the skateboarding half-pipe. It's in the shape of a U.&lt;br&gt;&lt;br&gt;&lt;em&gt;thou shalt address thee as a clue, be thy plumber of thy loo&lt;/em&gt;&lt;br&gt;This is only trying to make use of words you/yourself/your in old English. Serves no other purpose. The moral of the last line is &quot;Clean up your own messes,&quot; and the line before it says the subject, &quot;you&quot;, is the clue.&lt;br&gt;&lt;br&gt;&lt;em&gt;material of squander travels around, within the confines, trapped and bound, nowhere to go the water rises, break free quickly no nasty surprises&lt;/em&gt;&lt;br&gt;What. ... If this follows the previous paragraph, as I expect it does, then this is talking about... flushing. Using your imagination, you can tell exactly what is meant by &quot;material of squander&quot; and &quot;nasty surprises.&quot; What does this have to do with &quot;you&quot;? Don't ask me. I suppose this is just another one of those &quot;clean up your own messes&quot; deals... oh, haha! Maybe it's supposed to be &quot;eww&quot;, which sounds like &quot;you&quot;? &lt;em&gt;suggestion to the author: Maybe you can reference the yew tree instead? That makes for a good, &lt;strong&gt;clean&lt;/strong&gt; homonym.&lt;/em&gt;&lt;br&gt;&lt;br&gt;&lt;em&gt;Germans sink within this vessel under the radar hoping to nestle. Place what you need before the knee, take the verse and have some tea.&lt;/em&gt;&lt;br&gt;German submarines were U-boats. The U is what is being alluded to here.&lt;br&gt;&lt;br&gt;&lt;em&gt;the Gothic did state the differences upon the vague instances, for we were once another our rather angular brother.&lt;/em&gt;&lt;br&gt;Admittedly, I know next to nothing about East Germanic language. However, I'm familiar enough to know that this references the uses of thee, thou, and thy (all Romanized Gothic words) as being ambiguous and misunderstood in modern English, whereas in old times they were very different in meaning.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;OP Edit:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; yes I know, very silly and cheesy skateboarding reference. It seemed a good idea at the time&lt;br&gt;&#xA;&lt;br&gt;&#xA; University = U knee verse tea (almost).&lt;br&gt;&#xA;&lt;br&gt;&#xA;According to online sources the Goths were the first culture to differenciate the &quot;U&quot; and &quot;V&quot;, before words like vague could be spelt uagve and upon could be spelt vpon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Final Answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All in all, the whole answer is &quot;I love you,&quot; which makes sense with the title being the hardest English words to speak.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9319" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-02-05T15:30:13.233" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8479" PostTypeId="2" ParentId="8473" CreationDate="2015-02-02T18:15:29.250" Score="11" Body="&lt;p&gt;From the last two positions, we see that $4\cdot(10R+E)\equiv 10E+R \bmod{100}$. This is equivalent to $39R\equiv6E\bmod{100}$, and further to $13R\equiv2E\bmod{100}$.&#xA;Hence $R$ has to be even. Checking $R=0,2,4,6,8$, we see that only $R=8$ works.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We conclude that $R=8$ and that $E=2$, and that the carry over from the second column to the third column is $3$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since $4\cdot10^5\,S= 4\cdot S00000 &amp;lt; 4*\mbox{SQUARE} = \mbox{NUMBER} &amp;lt;10^6$, we conclude $4S&amp;lt;10$. The case $S=0$ would have zero as leading digit, and the case $S=2$ would collide with $S=E=2$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We derive $S=1$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We plug the detected values into the given equation and derive   $4\cdot(1000+100Q+10U+A)+3=1000N+100U+10M+B$,&lt;br&gt;&#xA;which simplifies to&lt;br&gt;&#xA;$4003 -10M+ 4A-B = 1000N -400Q +60U.~~~ (*)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The left hand side of equation ($*$) is at least $4003-90+0-9=3904$ and &#xA;at most $4003-0+36-0=4039$.  Furthermore, it is a multiple of $20$. Hence it lies between $3920$ and $4020$, which yields&#xA;$196\le 50N-20Q+3U\le 201~~~ (**)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we distinguish several cases on $U$ to derive the following from $(**)$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(1) If $U=0$, then $5N-2Q=20$. Then $N=6$ and $Q=5$.&lt;br&gt;&#xA;   (2) If $U=3$, then $5N-2Q=19$. Then $(N,Q)=(5,2)$ or $=(7,8)$; collisions.&lt;br&gt;&#xA;   (3) If $U\in\{4,5\}$, then $181&amp;lt;50N-20Q&amp;lt;189$. No solution.&lt;br&gt;&#xA;   (4) If $U\in\{6,7\}$, then $5N-2Q=18$. Then $(N,Q)=(4,1)$ or $=(6,6)$; collisions.&lt;br&gt;&#xA;   (5) If $U=9$, then $5N-2Q=17$. Then $(N,Q)=(5,4)$ which is fine, or $(N,Q)=(7,9)$ which has the collision $Q=U=9$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All in all, this leaves us with two possible cases: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Case X: $(N,Q,U)=(6,5,0)$&lt;br&gt;&#xA;  Case Y: $(N,Q,U)=(5,4,9)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In case X, equation $(*)$ boils down to $10M- 4A+B = 3$.   &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $M\ge4$, then $10M-4A+B\ge40-4\cdot9+0=4&amp;gt;3$; contradiction.  &lt;/li&gt;&#xA;&lt;li&gt;If $M=3$, then $4A-B=27$ implies $(A,B)=(7,1)$ or $=(8,5)$ or $=(9,9)$; all three cases are collisions (with $B=S=1$, $B=Q=5$, $A=B=9$ respectively).  &lt;/li&gt;&#xA;&lt;li&gt;If $M=0$, we have the collision $M=U=0$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In case Y, equation $(*)$ boils down to $10M- 4A+B =63$.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $M\le5$, then $10M- 4A+B\le 50+B&amp;lt;63$; contradiction.  &lt;/li&gt;&#xA;&lt;li&gt;If $M=6$, we get $B=4A+3$; this leaves $(A.B)=(7,1)$ with the collision $B=S=1$ and the &lt;strong&gt;good solution $(A,B)=(0,3)$&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;If $M=7$, we get $4A-B=7$. This leaves the cases $(A.B)=(2,1)$, $(3,5)$, $(4,9)$ which respectively collide with $B=S=1$, $B=N=5$, $A=Q=4$.&lt;/li&gt;&#xA;&lt;li&gt;If $M=9$,we get $4A-B=27$. This leaves the cases $(A.B)=(7,1)$, $(8,5)$, $(9,9)$ which respectively collide with $B=S=1$, $B=N=5$, $A=B=9$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary:&lt;/strong&gt;&lt;br&gt;&#xA;Only a single branch in this analysis has led to a solution:&lt;br&gt;&#xA;$R=8$ $E=2$ $S=1$;  $(N,Q,U)=(5,4,9)$; $M=6$ and  $(A.B)=(0,3)$.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-02T18:20:59.713" LastActivityDate="2015-02-02T18:20:59.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8480" PostTypeId="1" AcceptedAnswerId="8486" CreationDate="2015-02-02T22:10:57.997" Score="15" ViewCount="1068" Body="&lt;p&gt;I had taken an extended summer break over the Christmas New Year period but finally returned to the city at the end of January, feeling relaxed and tanned. I was looking forwards to getting back in touch with friends I hadn't seen in a while so on the following Sunday morning I dropped in on Ernie to see how things were going with him. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;First of February&quot;, I commented as I entered the workshop to find Ernie peering into a brass microscope. Ernie looked up to greet me, asked how my break had been, then requested that I tear off the January page on his calendar to get it up to date. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was happy to assist. But as I ripped off the old page, perched on a stool in order to reach it, I noticed something rather curious. On the January page Ernie had crossed out the &lt;strong&gt;N&lt;/strong&gt; in &lt;strong&gt;JAN&lt;/strong&gt; (with the bright red marker pen he only used when something was &quot;wrong&quot;), so it now appeared as &lt;strong&gt;JA&lt;/strong&gt;. I riffled through the rest of the pages and discovered that this wasn't the only alteration he had made. The last letter &lt;strong&gt;G&lt;/strong&gt; was also scribbled out from &lt;strong&gt;AUG&lt;/strong&gt; to leave &lt;strong&gt;AU&lt;/strong&gt;. He had been even more extreme with &lt;strong&gt;FEB&lt;/strong&gt;, &lt;strong&gt;SEP&lt;/strong&gt;, &lt;strong&gt;OCT&lt;/strong&gt;, &lt;strong&gt;NOV&lt;/strong&gt;, and &lt;strong&gt;DEC&lt;/strong&gt;, which had each been reduced to their first letter - &lt;strong&gt;F&lt;/strong&gt;, &lt;strong&gt;S&lt;/strong&gt;, &lt;strong&gt;O&lt;/strong&gt;, &lt;strong&gt;N&lt;/strong&gt;, and &lt;strong&gt;D&lt;/strong&gt;, respectively. &lt;strong&gt;MAR&lt;/strong&gt;, &lt;strong&gt;MAY&lt;/strong&gt;, and &lt;strong&gt;JUL&lt;/strong&gt;, had been left untouched. But curiously, for the months of April and June, he had &lt;em&gt;added&lt;/em&gt;  a letter to convert &lt;strong&gt;APR&lt;/strong&gt; to &lt;strong&gt;APRI&lt;/strong&gt; and &lt;strong&gt;JUN&lt;/strong&gt; to &lt;strong&gt;JUNE&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ernie's behavior is considered rather odd at times (by some people), but I have never found his actions to be random or haphazard or arbitrary so I knew there must be some logic behind the alterations. So, being the curious sort, I asked him why he had done this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Well, you know how I despise unnecessary redundancy!&quot;, Ernie began, &quot;For example, you can distinguish January uniquely from June and July by its first two letters &lt;strong&gt;JA&lt;/strong&gt;, no need for the redundant &lt;strong&gt;N&lt;/strong&gt;. And similarly, February is the only month beginning with an &lt;strong&gt;F&lt;/strong&gt;, so there is no need for the &lt;strong&gt;EB&lt;/strong&gt;. So, I have edited the calendar by getting rid of superfluous letters.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wasn't totally convinced, &quot;That explains most of your changes&quot;, I replied, &quot;but what about the extra letters in &lt;strong&gt;APRI&lt;/strong&gt; and &lt;strong&gt;JUNE&lt;/strong&gt;? I can't see that you need them&quot;. Ernie paused for a moment then continued. &quot;The letters &lt;strong&gt;ADFJLMNOPRSUY&lt;/strong&gt; would be enough to identify the names of the months efficiently, but by adding &lt;strong&gt;E&lt;/strong&gt; and &lt;strong&gt;I&lt;/strong&gt;, I can also spell out &lt;strong&gt;ERNIE&lt;/strong&gt; ... and that also helped with the synesthesia.&quot; &lt;img src=&quot;https://i.stack.imgur.com/kNC8x.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I remembered Ernie explaining some time ago how some people has a sort of sensory mix-up in their brains and as a result they could &quot;see&quot; letters as colours, or hear tones when they observed colours, or could even taste musical notes. Ernie had a rare condition that meant he could &lt;em&gt;hear&lt;/em&gt; numbers - he had told me that, for example, when he scanned the digits of pi they formed a musical progression that sounded like a Bach Toccata (&quot;a lovely melody&quot;, he had commented at the time), and that the logarithmic constant e had something of an &quot;oompah-band&quot; theme (&quot;a rather hideous, endless, polka&quot;, he had added with a slight wince).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Also, when I look at a word, if there is a numerical connotation with it, I hear the sound of the associated number. So January (being the first month) always sounds like the number 1, February sounds like 2, and so on to December which sounds like 12. And just to complicate things, when I see a set of words that can be spelled out with 16 letters or less, my mind automatically substitutes a hexadecimal digit for each letter (the same digit for the same letter in each case of course) to convert the words into hexadecimal numbers - and then for each of &lt;em&gt;those&lt;/em&gt; numbers my mind plays the sound of one of its divisors in my head. And the problem was that in my original efficient calendar every hexadecimal substitution produced a dreadful disharmony between the two sounds for at least one month!&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ernie paused for a moment as my mind reeled - sometimes Ernie seemed to have a very complicated life. Then he continued. &quot;But with the extra &lt;strong&gt;E&lt;/strong&gt; and &lt;strong&gt;I&lt;/strong&gt;, everything is nice and harmonic. The letters &lt;strong&gt;JA&lt;/strong&gt; can convert to a hexadecimal number that is divisible by 1, &lt;strong&gt;F&lt;/strong&gt; is divisible by 2, &lt;strong&gt;MAR&lt;/strong&gt; is divisible by 3, and so on right up to &lt;strong&gt;D&lt;/strong&gt; which is divisible by 12. And as an extra bonus, &lt;strong&gt;ERNIE&lt;/strong&gt; is divisible by 13 - which is my second most favorite number.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Ummm, I see (I think).&quot;, I replied. &quot;Is there more than one way to substitute hexadecimal digits for the letters in your calendar that would have the same result?&quot;, I asked. &quot;As a matter of fact, there is&quot;, Ernie replied, &quot;but my mind always seem to choose the unique substitution that doesn't use the hexadecimal digit 0 and that also also makes my middle name equal to my &lt;em&gt;most&lt;/em&gt; favourite number!&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ernie then told me his what his favourite number was. Unfortunately it has slipped my mind already, but I do remember that he told me it was a prime. If I told you that Ernie's middle name is &lt;strong&gt;DAN&lt;/strong&gt;, do you think there is any way to work out what his favorite number is? &lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="1631" LastEditDate="2015-02-03T09:32:23.883" LastActivityDate="2015-02-03T09:32:23.883" Title="Ernie and the Calendar" Tags="|mathematics|word-problem|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8481" PostTypeId="1" AcceptedAnswerId="8482" CreationDate="2015-02-02T23:37:42.743" Score="8" ViewCount="1383" Body="&lt;p&gt;You have been given the number 16. You also have 4 magic spells. You have to use the spells to turn 16 into 25.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each spell changed your number a specific way, but the objective is never altered, you must make 25.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Spell List:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Halfzenumzaber: Half your number (&lt;em&gt;e.g: 16 -&gt; 8&lt;/em&gt;)&lt;/li&gt;&#xA;&lt;li&gt;Dogaubtale: Double your number (&lt;em&gt;e.g: 16 -&gt; 32&lt;/em&gt;)&lt;/li&gt;&#xA;&lt;li&gt;Conracayutekonate: Concatenate your number with the same number (&lt;em&gt;e.g: 16 -&gt; 1616&lt;/em&gt;)&lt;/li&gt;&#xA;&lt;li&gt;subhatractlfplones: Half the number, then subtract 1 (&lt;em&gt;e.g: 16 -&gt; 7&lt;/em&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Current Best:&lt;/strong&gt; &lt;em&gt;7&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Is it possible to get 25? If so, what's the least amount of moves it takes, if not, prove it's impossibility.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="69582" LastEditDate="2021-06-15T19:30:30.927" LastActivityDate="2021-06-15T19:30:30.927" Title="Turn 16 into 25. You have 4 magic spells. Can you do it?" Tags="|mathematics|optimization|reachability|" AnswerCount="5" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="8482" PostTypeId="2" ParentId="8481" CreationDate="2015-02-02T23:56:08.060" Score="15" Body="&lt;p&gt;Here's one answer that I think would be shortest:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 16 to 1616 to 808 to 404 to 202 to 100 to 50 to 25. (7 moves) Basically you need to halve from 50/52/100-104/etc. The only other possibilities would be to halve 16 first then go to 88 or 77, but there's no way to get to 25 faster. Similarly, going to 32 then 64, etc., there's no way. This is basically a proof by exhaustion, eliminating the other two branches that could potentially beat 7.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastActivityDate="2015-02-02T23:56:08.060" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8483" PostTypeId="2" ParentId="8481" CreationDate="2015-02-02T23:59:03.260" Score="8" Body="&lt;p&gt;It may not be the fewest number of spells, but it's definitely possible in 7:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Letting &lt;code&gt;s&lt;/code&gt; represent subhatractlfplones and &lt;code&gt;d&lt;/code&gt; represent Dogaubtale:&lt;br/&gt;&#xA; 16 &lt;code&gt;s&lt;/code&gt; 7 &lt;code&gt;d&lt;/code&gt; 14 &lt;code&gt;d&lt;/code&gt; 28 &lt;code&gt;s&lt;/code&gt; 13 &lt;code&gt;d&lt;/code&gt; 26 &lt;code&gt;d&lt;/code&gt; 52 &lt;code&gt;s&lt;/code&gt; 25&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9323" LastActivityDate="2015-02-02T23:59:03.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8484" PostTypeId="1" CreationDate="2015-02-03T01:10:55.037" Score="7" ViewCount="822" Body="&lt;p&gt;Make an as small as possible three dimensional structure of matchsticks, all of which have equal length, such that the end of each matchstick meets exactly five other ends. A matchstick is not allowed to cross another matchstick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got the inspiration for this question from &lt;a href=&quot;https://puzzling.stackexchange.com/questions/8285/touching-matchsticks&quot;&gt;this question&lt;/a&gt;. Similarly a matchstick can be thought of as mathematical line segment, connecting two points in $\mathbb{R}^3$, such that in each point six line segments meet. And with &quot;as small as possible structure&quot; I am referring to the number of used matches/line segments.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt;&#xA;If the number of line segments, which have to meet in each node, would be equal to four, then the smallest structure would be an octahedron:&lt;br&gt;&#xA;$\hspace{5cm}$&lt;img src=&quot;https://i.stack.imgur.com/kIIAZ.gif&quot; alt=&quot;octahedron&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="8810" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2015-02-06T13:01:26.207" Title="Smallest number of matchsticks for a 3D structure with 6 matchsticks at each vertex" Tags="|mathematics|geometry|matches|graph-theory|three-dimensional|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8486" PostTypeId="2" ParentId="8480" CreationDate="2015-02-03T02:49:05.573" Score="6" Body="&lt;p&gt;(Hexadecimal digits will be represented as lowercase letters a-f, while the unknown letters will be uppercase.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are given:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;JA is divisible by 1,&#xA;F is divisible by 2,&#xA;MAR is divisible by 3,&#xA;APRI is divisible by 4,&#xA;MAY is divisible by 5,&#xA;JUNE is divisible by 6,&#xA;JUL is divisible by 7,&#xA;AU is divisible by 8,&#xA;S is divisible by 9,&#xA;O is divisible by 10,&#xA;N is divisible by 11,&#xA;D is divisible by 12,&#xA;ERNIE is divisible by 13,&#xA;and DAN is prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ADEFIJLMNOPRSUY are the digits 123456789abcdef in some order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can quickly identify SOND as 9abc respectively.&#xA;U must be 8, because it can't be 0. Then I must be 4, because it is the last digit left divisible by 4.&#xA;F and E must both be divisible by 2, so they are 2, 6, or e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try each of these values for E and solve for the value of R from ERNIE (ERb4E): 2bb42, 68b46, e2b4e, and efb4e are each divisible by 13. But R cannot be b or 8, because both numbers are already taken. Therefore E is e and R is either 2 or f. F is either 2 or 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Testing each remaining value for A, only 5 and f make DAN (cAb) a prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that JUNE (J8be) is divisible by 6. This requires J to be 3, 6 or f. If J is 6, then L must be 2 for JUL to be divisible by 7. But this leaves no value for F.&#xA;If J is 3, then L is 7, while if J is f, then L is 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If J is f and L is 1, then R is 2, A is 5, and F is 6. But no value is left for M so that MAR is divisible by 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If instead J is 3 and L is 7, then regardless of whether A is 5 or f, M and Y must be 2 and d in some order for MAY to be divisible by 5. F must then be 6, and R must be f. A must be 5, and P is 1 by elimination. M must be d for MAR to be divisible by 3, and so Y is 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ADEFIJLMNOPRSUY end up corresponding to 5ce6437dba1f982. Ernie's favorite number is hexadecimal c5b, or decimal 3163.&lt;/p&gt;&#xA;" OwnerUserId="9325" LastActivityDate="2015-02-03T02:49:05.573" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8487" PostTypeId="2" ParentId="6453" CreationDate="2015-02-03T05:43:07.813" Score="3" Body="&lt;p&gt;&lt;em&gt;Feb 8th edit:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some more progress has been made since I posted this answer, now I'm choosing to edit it to reflect what we know. I used the keys to finish the scale's color spectrum base; the edited scale has been added by Bmy to the wiki answer with an explanation of the logic involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue we are facing now is deciding which tubes the bottle contents need to go into. We have supposedly used up the meaning of the 8x8 grid, where it was relevant to determining order. The yellow diamonds are the only label we have not fully understood from the first instruction, and it is believed that this is where we will determine the placement of the contents. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My current line of thinking holds that the yellow diamonds tell us the amount that the respective bottles will put into the tubes. This is supported both by my counting (95 empty tube spaces, 93 diamonds) and by Bmy's upvote. However, if this were all that the diamonds tell us, then we would still have no information on which tubes to fill. The only piece I can imagine being useful is the spacing between the diamonds. It has been made clear that the labels are supposed to be center-aligned, and thus the left/right sides of the diamonds don't mean anything. Only the spacing /between/ them seems to matter here, though it could still be that it has no meaning at all. In such a case, for example, bottle 14 and bottle 20 would be the same with regards to their diamond pattern. The question to be answered, in short, is this: what does the diamond spacing mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did some testing on a selection of ideas. My first idea was that the spacing simply described how many tubes each bottle is divided into. Our first bottle is a 1_4, so this idea would claim that one unit goes in one tube while the other four units go somewhere else. Presumably, this also suggests that the 1 unit will go in a farther left direction than the 4 units, based on their order on the label. This is all well and good, but following it through encountered a problem very early.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the scale in more detail, we are supposed to keep it balanced. To make the force calculations and referencing easier, I labeled each hook on the number line -13 to +13. The pipes are placed in specific spots on the scale, specifically at locations&lt;br&gt;&#xA;[-13, -5, -2, 4, 8, 11].&lt;br&gt;&#xA;This means the force that the bottles apply to the scale will always be in only these 6 locations. The weight can be placed on any hook, which you'd think makes things easier. Unfortunately, the weight is 2 units in size. Therefore, any force the weight provides to the scale will be an even number; an even number multiplied by any number always results in an even number. Therefore, to keep the scale balanced, we must ensure that the forces the filled tubes apply are even amounts as well: to balance an even force on one side requires an equal force placed on the other.&lt;br&gt;&#xA;What does this mean?&lt;br&gt;&#xA;It means, young padawan, that our bottles that have an odd number of diamonds must fill only tubes standing on even numbers. For example, if we take a 3-diamond bottle and fill the yellow tube at location -5, then that would be -15 units of pressure on the scale. We cannot put the weight anywhere on the scale to perfectly balance 15 units, because the hanging weight cannot provide an odd numbered amount of pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking back at the first bottle, our 1_4 bottle, lets assume that we must follow the rules I stated above. The 1 unit must go into an even numbered tube [-2, 4, or 8]. The 4 units &lt;em&gt;should&lt;/em&gt; go into any tube to the right of that [4, 8, or 11]. Unfortunately, this is already impossible - putting 4 units into the 4 spot is 16 units of pressure, plus the 26 units of pressure already on the right end = 42 units of pressure. The 1 remaining unit of liquid and the hanging weight can only provide 28 units of pressure combined at maximum, not enough to balance the scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That might have been a lot to process, so let's make this easier. We have weight already on the right end. We must put weight on the left end of the scale to balance it. Of course, pouring the first bottle's contents into a tube on the right end would only make things worse. We can't pour the first bottle into the odd-numbered tubes, either, because its force cannot be balanced by our even-numbered hanging weight. By process of elimination, we can only pour the bottle - all 5 units - into the green tube. Things begin to look better: 5 units * -2 distance from the center = -10 units of pressure, + 26 units from the right hand side = +16 units of pressure on the scale. To balance this evenly with an added -16 units of pressure, the weight must be placed on the -8 hook.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might be thinking, &quot;Great! We have a plan.&quot; Unfortunately, this plan doesn't continue very far. Long story short, we can't keep putting odd-numbered diamond bottles into even-numbered tubes, because we will end up favoring the green tube too much and cause it to overflow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of this means something is wrong somewhere. Whether it was my analysis of the scale, or my analysis of the diamonds, something is not matching up properly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My most recent failed idea was ignoring the spaces between the diamonds. I interpreted  instruction-2 a bit more literally, and thought maybe the hanging weight's current location was our clue for where to pour the next bottle. The fact that the weight is already hanging on the scale at the start of the puzzle gave the idea some credence, so I followed through. As before, we can't put only one unit nor all 5 units into the red pipe, as it would provide an odd pressure or too much pressure, respectively. The remaining option would be to put 4 units into the red and the remaining one elsewhere. This yielded multiple answers, somewhere around 9 different possibilities, but I remembered that this scenario has an additional rule: the hanging weight must be under a pipe so that we can identify where to pour the next bottle. There was only one solution that allowed this, which was pouring the last unit into the cyan tube and placing the hanging weight under the purple tube. This is where the idea fails, however, as we next have a 3-diamond bottle which is to be placed into the purple tube. 3*11 is 33, an odd pressure, and once again cannot be balanced by our even-sized hanging weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems this is the only remaining difficult point to reason out in the puzzle. I figure once we have a system that works without fail, it's just a matter of applying it to reach the answer. Almost certainly, the code will be broken by selecting the key related in color to the pipe that the bottle fills, selecting the line with the color that the hanging weight is beneath, and matching it with the same colored column as the key. &lt;em&gt;For example, if the bottle fills the red pipe and we had to place the hanging weight on hook 4, then we'd insert the red key into the table, select the 4th row from the bottom (where the bright purple is), and select where it meets the red column on the same side, which happens to be the letter D.&lt;/em&gt; Noted, for this reasoning to be correct, it assumes that the whole of each bottle is poured into only one tube, which conflicts with most of my other reasoning above. I fear that I have overlooked something very basic and that all of my work and thinking has been flawed from the very beginning.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Original Post&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started into this puzzle today, hopefully I'll get to help you guys out solving this. I see you've made some good progress. I noticed something regarding the scale section, in that there were blank spots where there should be colors (like where the cyan tube ends). The keys offer a solution to this, since all the keys have the same color pairs, and those color pairs match up perfectly to their spacing on the scale.&lt;br&gt;&#xA;I've taken the liberty of submitting a complete scale image:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/jFpHC.png&quot; alt=&quot;complete color-coded scale&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More to do:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let's take a look at the tubes. Those things are awkwardly shaped and must have a reason for it. I hypothesize that the tunnels are not actually in the right place. There's that weight on the right hand side; could it actually be a pipe with grey contents? If this is the case, then my hypothesize holds water (excuse the pun) - that is, we now have a correlation between the dots and the scale - 7 dots against 7 pipes. We would probably need to reorganize the pipes on the scale and see where the weight distributes itself when the pipes are full of a color.&lt;br&gt;&lt;br&gt;My second hypothesis comes from looking more closely at the first instruction. It clearly shows the diamonds on the example bottle, so it's possible that it is giving us a clue about what the diamonds mean. From what it looks like, the diamonds are the amount of a color there is in the bottle, and it is pouring into a tube that already has contents. In this case, I suspect that the contents of the tubes persist, and our weight calculations will compound as we go through the bottles. We will be able to identify our mistakes by any tubes that are overfilled. Furthermore, there are times when we see up to three sets of diamonds on a bottle - this would indicate three different colors, if this hypothesis holds true, and we must compensate for that in our analysis. &lt;br&gt;*Note: I counted up the number of spaces in the tubes (95 empty) and the number of diamonds on all the bottles combined (93). Assuming I counted accurately, the fact that we can &quot;fill the tubes with diamonds&quot;, so to speak, helps support the hypothesis that the diamonds represent the amount of something in each bottle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9319" LastEditorUserId="9319" LastEditDate="2015-02-08T07:05:02.997" LastActivityDate="2015-02-08T07:05:02.997" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8488" PostTypeId="1" AcceptedAnswerId="8532" CreationDate="2015-02-03T05:52:10.530" Score="17" ViewCount="1329" Body="&lt;p&gt;I came up with this riddle a few days ago. It's my very first riddle, so it shouldn't be too hard to crack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two hints are:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;My name is a state that is not my state. &lt;/li&gt;&#xA;&lt;li&gt;When I am downtrodden, my oppressors take risks.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4941" LastEditorUserId="8874" LastEditDate="2016-03-23T17:15:40.237" LastActivityDate="2016-03-23T17:15:40.237" Title="A US English riddle: My name is a state that is not my state. When I am downtrodden, my oppressors take risks" Tags="|riddle|language|regional-dialects|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8489" PostTypeId="2" ParentId="8484" CreationDate="2015-02-03T05:59:45.907" Score="10" Body="&lt;p&gt;Here's a way to do it 48 edges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A regular tetrahedron has four vertices each connected to the three other vertices by an edge of length one. Take four of these tetrahedra, and place four translated copied so that their centers themselves form a regular tetrahedron of edge length $1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, the corresponding vertices of each tetrahedron themselves form a regular tetrahedron with edge length $1$. So, each vertex is one away from $6$ other vertices -- three in its own tetrahedron, and the three corresponding points in other tetrahedra. This is a solution with $48$ edges -- each of $16$ vertices connects to $6$ other vertices, and $(16\times 6)/2 = 48$. If the original tetrahedron is rotated randomly, no edges cross with probability $1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically, we've taken the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cartesian_product_of_graphs&quot;&gt;Cartesian graph product&lt;/a&gt; $T\times T$, whose vertices correspond to pairs $(v,w)$ of vertices of $T$, and two vertices are connected by an edge, written $(v,w) \leftrightarrow (v',w')$, if either $v=v'$ and $w \leftrightarrow w'$, or $w=w'$ and $v \leftrightarrow v'$ in $T$. We've realized it as a unit distance graph in 3D as follows by taking the vector sum $v+w$ corresponding to each pair of vertices $(v,w)$, with a different tetrahedron embedding for $v$ and $w$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This type of construction lets us create a unit distance graph with arbitrary many edges per vertex. If we have a solution for $n_1$ edges per vertex using $V_1$ vertices and likewise for $n_2$ and $V_2$, then taking the Cartesian product gives a solution for $n_1 + n_2$ edges per vertex that uses $V_1 V_2$ vertices. Note that minimizing vertices is equivalent to minimizing edges due to the relation $E=nV/2$.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-02-03T09:52:46.727" LastActivityDate="2015-02-03T09:52:46.727" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8491" PostTypeId="5" CreationDate="2015-02-03T09:21:41.740" Score="0" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Number_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;Number theory&lt;/a&gt; is a branch of pure mathematics devoted primarily to the study of the &lt;em&gt;integers&lt;/em&gt;.&#xA;Sometimes number theory is called &lt;strong&gt;&quot;The Queen of Mathematics&quot;&lt;/strong&gt; because of its foundational place in the discipline. Number theorists study prime numbers as well as the properties of objects made out of integers or defined as generalizations of the integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that general number theory questions are off-topic, but can be asked on our sister site &lt;a href=&quot;https://math.stackexchange.com/&quot;&gt;Mathematics Stack Exchange&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="11110" LastEditDate="2018-09-06T13:17:19.577" LastActivityDate="2018-09-06T13:17:19.577" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8492" PostTypeId="4" CreationDate="2015-02-03T09:21:41.740" Score="0" Body="A mathematical puzzle whose solution is heavily based on the arithmetic properties of the integers. Use with [mathematics]" OwnerUserId="8874" LastEditorUserId="69582" LastEditDate="2021-01-23T00:18:57.070" LastActivityDate="2021-01-23T00:18:57.070" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8493" PostTypeId="5" CreationDate="2015-02-03T09:22:25.343" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-02-03T09:22:25.343" LastActivityDate="2015-02-03T09:22:25.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8494" PostTypeId="4" CreationDate="2015-02-03T09:22:25.343" Score="0" Body="A puzzle that heavily depends on linguistic features; for instance it may concern foreign languages, or only work in some particular language, or be built around a peculiarity of the English language." OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-03T09:41:54.033" LastActivityDate="2015-02-03T09:41:54.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8495" PostTypeId="5" CreationDate="2015-02-03T09:24:04.410" Score="0" Body="&lt;p&gt;For puzzles based on, related to, and/or requiring knowledge of literary pieces, characters, genres, or authors. General literature questions are off-topic but can be asked on &lt;a href=&quot;https://literature.stackexchange.com/&quot;&gt;Literature Stack Exchange&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Sample topics include:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;series&lt;/li&gt;&#xA;&lt;li&gt;characters&lt;/li&gt;&#xA;&lt;li&gt;plots&lt;/li&gt;&#xA;&lt;li&gt;genres&lt;/li&gt;&#xA;&lt;li&gt;authors (and their aliases)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="52483" LastEditorUserId="60644" LastEditDate="2023-02-02T16:11:03.513" LastActivityDate="2023-02-02T16:11:03.513" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8496" PostTypeId="4" CreationDate="2015-02-03T09:24:04.410" Score="0" Body="A puzzle that is primarily built around a well-known book, novel, drama, poem, or author." OwnerUserId="8874" LastEditorUserId="69582" LastEditDate="2021-01-29T22:12:01.870" LastActivityDate="2021-01-29T22:12:01.870" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8497" PostTypeId="5" CreationDate="2015-02-03T09:26:42.007" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-02-03T09:26:42.007" LastActivityDate="2015-02-03T09:26:42.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8498" PostTypeId="4" CreationDate="2015-02-03T09:26:42.007" Score="0" Body="A puzzle that is primarily built around a historical situation or event. Often it concerns famous persons or weird and unexpected happenings." OwnerUserId="8874" LastEditorUserId="4551" LastEditDate="2015-02-03T09:27:18.110" LastActivityDate="2015-02-03T09:27:18.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8499" PostTypeId="5" CreationDate="2015-02-03T09:28:30.007" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-02-03T09:28:30.007" LastActivityDate="2015-02-03T09:28:30.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8500" PostTypeId="4" CreationDate="2015-02-03T09:28:30.007" Score="0" Body="Geared Mixup is a cube-shaped twisty-puzzle with eight geared corners that has been designed by the famous Dutch puzzel designer Oskar van Deventer." OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-03T09:52:57.377" LastActivityDate="2015-02-03T09:52:57.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8501" PostTypeId="5" CreationDate="2015-02-03T09:30:52.317" Score="0" Body="&lt;p&gt;Gambling is the wagering of money on an event with an uncertain outcome with the primary intent of winning additional money. &#xA;Gambling requires three elements to be present: consideration, chance and prize.&#xA;The outcome of the wager is often immediate, such as a single roll of dice or a spin of a roulette wheel, but longer time frames are also common, allowing wagers on the outcome of a future sports contest or even an entire sports season.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-03T11:29:25.027" LastActivityDate="2015-02-03T11:29:25.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8502" PostTypeId="4" CreationDate="2015-02-03T09:30:52.317" Score="0" Body="Gambling puzzles are centered around the winning probablities and expected profit values in a gambling game. " OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-03T11:29:16.513" LastActivityDate="2015-02-03T11:29:16.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8503" PostTypeId="5" CreationDate="2015-02-03T09:34:27.500" Score="0" Body="&lt;p&gt;In general, a &lt;a href=&quot;https://en.wikipedia.org/wiki/Word_ladder&quot; rel=&quot;nofollow noreferrer&quot;&gt;word ladder puzzle&lt;/a&gt; has (i) a source word and (ii) a target word. To solve the puzzle you must find a chain of words to link the two, such that two successive words differ only by a single letter (in math-speak, they have 'a &lt;a href=&quot;https://en.wikipedia.org/wiki/Hamming_distance&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hamming distance&lt;/a&gt; of &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;').&lt;/p&gt;&#xA;&lt;p&gt;Some puzzles may involve linked or conjoined word ladders where one section's target word is the next section's source word. Meanwhile, some word ladder variations allow letters to be removed or added as well &lt;a href=&quot;https://en.wikipedia.org/wiki/Levenshtein_distance&quot; rel=&quot;nofollow noreferrer&quot;&gt;(Levenshtein distance)&lt;/a&gt;, and/or transpositions &lt;a href=&quot;https://en.wikipedia.org/wiki/Damerau%E2%80%93Levenshtein_distance&quot; rel=&quot;nofollow noreferrer&quot;&gt;(Damerau-Levenshtein distance)&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Example&lt;/strong&gt;: &lt;br/&gt;&lt;pre&gt;&lt;code&gt;| COLD &lt;br/&gt;| CO&lt;strong&gt;R&lt;/strong&gt;D&lt;br/&gt;| C&lt;strong&gt;A&lt;/strong&gt;RD&lt;br/&gt;| &lt;strong&gt;W&lt;/strong&gt;ARD &lt;br/&gt;V WAR&lt;strong&gt;M&lt;/strong&gt; &lt;/code&gt;&lt;/pre&gt; forms a word ladder from COLD to WARM.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="60644" LastEditDate="2023-01-31T16:40:44.290" LastActivityDate="2023-01-31T16:40:44.290" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8504" PostTypeId="4" CreationDate="2015-02-03T09:34:27.500" Score="0" Body="A word ladder puzzle has a source word and a target word, and to solve the puzzle you must find a chain of words, differing by one letter at a time, to link the two." OwnerUserId="8874" LastEditorUserId="52483" LastEditDate="2023-02-01T19:16:25.377" LastActivityDate="2023-02-01T19:16:25.377" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8505" PostTypeId="5" CreationDate="2015-02-03T09:43:45.060" Score="0" Body="&lt;p&gt;The &lt;em&gt;Einstein puzzle&lt;/em&gt; or &lt;em&gt;Zebra puzzle&lt;/em&gt; is a well-known logic puzzle. It is often called &quot;Einstein's Puzzle&quot; or &quot;Einstein's Riddle&quot; because it is said to have been invented by Albert Einstein as a boy. However, there is no known evidence for Einstein's authorship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The puzzle concerns five houses that are each painted a different color. &#xA;Their inhabitants are of different national extractions, own different pets, drink different beverages, and smoke different brands of American cigarettes.&lt;br&gt;&#xA;The usual question is &lt;em&gt;&quot;Who owns the zebra?&quot;&lt;/em&gt;, and&#xA;the usual answer is &lt;em&gt;&quot;The Japanese&quot;&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the full puzzle text, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Zebra_Puzzle&quot; rel=&quot;nofollow&quot;&gt;Zebra puzzle in wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="5215" LastEditDate="2015-06-25T04:27:19.013" LastActivityDate="2015-06-25T04:27:19.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8506" PostTypeId="4" CreationDate="2015-02-03T09:43:45.060" Score="0" Body="The Einstein puzzle or Zebra puzzle is a well-known logic puzzle, centered around five houses, five men of different nationalities, five pets, five beverages, five cigarettes, and five colors." OwnerUserId="8874" LastEditorUserId="5215" LastEditDate="2015-06-25T04:26:39.490" LastActivityDate="2015-06-25T04:26:39.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8507" PostTypeId="5" CreationDate="2015-02-03T09:46:05.983" Score="0" Body="&lt;p&gt;There are people standing in a circle waiting to be executed. &#xA;The counting out begins at some point in the circle and proceeds around the circle in a fixed direction. &#xA;In each step, a certain number of people are skipped and the next person is executed. &#xA;The elimination proceeds around the circle (which is becoming smaller and smaller as the executed people are removed), until only the last person remains, who is given freedom.&#xA;The task is to choose the place in the initial circle so that you are the last one remaining and so survive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is named after Flavius Josephus, a Jewish historian living in the 1st century. &#xA;According to his account of the siege of Yodfat, he and his 40 soldiers were trapped in a cave, the exit of which was blocked by Romans. &#xA;They chose suicide over capture and decided that they would form a circle and start killing themselves using a step of three. Josephus states that by luck or possibly by the hand of God, he and another man remained the last and gave up to the Romans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information, see the &lt;a href=&quot;https://en.wikipedia.org/wiki/Josephus_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article on the Josephus problem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="11110" LastEditDate="2017-06-20T20:28:45.567" LastActivityDate="2017-06-20T20:28:45.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8508" PostTypeId="4" CreationDate="2015-02-03T09:46:05.983" Score="0" Body="There are n people standing in a circle. A counter goes clockwise around the circle and repeatedly eliminates every k-th person in the circle, until one person (the winner) remains. Where should you stand initially to become the winner? " OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-04T15:32:26.540" LastActivityDate="2015-02-04T15:32:26.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8509" PostTypeId="5" CreationDate="2015-02-03T09:47:58.283" Score="0" Body="&lt;p&gt;For puzzles based on, related to, or requiring knowledge of songs, music theory, lyrics, composers, singers, etc.&lt;br&gt;&#xA;General music questions are off-topic here, but might be welcomed on either &lt;a href=&quot;https://music.stackexchange.com/&quot;&gt;Music Practice &amp;amp; Theory&lt;/a&gt; or &lt;a href=&quot;https://musicfans.stackexchange.com/&quot;&gt;Music Fans&lt;/a&gt; Stack Exchange.&lt;/p&gt;&#xA;&lt;p&gt;Sample topics include:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;tunes&lt;/li&gt;&#xA;&lt;li&gt;melodies&lt;/li&gt;&#xA;&lt;li&gt;operas&lt;/li&gt;&#xA;&lt;li&gt;musicals&lt;/li&gt;&#xA;&lt;li&gt;lyrics&lt;/li&gt;&#xA;&lt;li&gt;song and record titles&lt;/li&gt;&#xA;&lt;li&gt;composers&lt;/li&gt;&#xA;&lt;li&gt;orchestras&lt;/li&gt;&#xA;&lt;li&gt;bands&lt;/li&gt;&#xA;&lt;li&gt;musical instruments&lt;/li&gt;&#xA;&lt;li&gt;musical scores&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;and anything else that can be considered to relate to music.&lt;/p&gt;&#xA;" OwnerUserId="11497" LastEditorUserId="60644" LastEditDate="2023-02-02T16:13:21.733" LastActivityDate="2023-02-02T16:13:21.733" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8510" PostTypeId="4" CreationDate="2015-02-03T09:47:58.283" Score="0" Body="A puzzle centered around tunes, melodies, operas, famous song texts, composers, orchestras, and/or musical instruments." OwnerUserId="8874" LastEditorUserId="69582" LastEditDate="2021-02-03T16:49:29.633" LastActivityDate="2021-02-03T16:49:29.633" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8511" PostTypeId="1" AcceptedAnswerId="8526" CreationDate="2015-02-03T10:09:36.680" Score="17" ViewCount="1664" Body="&lt;p&gt;Professor Erasmus has found a special way of moving a rook across a standard $8\times8$ chessboard that he modestly calls the &quot;Professor-Erasmus-rook-tour&quot;. &#xA;The professor claims that in this tour, the rook makes $64$ moves that visit all the $64$ squares of the chessboard. Every move is from some square to a horizontally or vertically adjacent square, and the very last move brings the rook finally back to its starting square. Exactly $32$ of these moves are in horizontal direction, and $32$ are in vertical direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has the professor once again made one of his mathematical blunders, or do such rook tours indeed exist?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-02-03T18:58:37.023" Title="The Erasmus rook tour" Tags="|mathematics|chess|checkerboard|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8512" PostTypeId="5" CreationDate="2015-02-03T10:15:46.737" Score="0" Body="&lt;p&gt;A jigsaw puzzle is a tiling puzzle that requires the assembly of (often oddly shaped) interlocking and tessellating pieces.&lt;/p&gt;&#xA;&lt;p&gt;Each piece usually has a small part of a picture on it and the aim of the game is to fit the pieces together to produce the full picture, meaningfully. In some cases, more advanced types have appeared, such as spherical jigsaws and puzzles showing optical illusions.&lt;/p&gt;&#xA;&lt;p&gt;In some variants, all pieces are already put in place and the solution is to put the pictures or letters or numbers on the pieces, meaningfully.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="38" LastEditDate="2021-08-21T03:37:16.390" LastActivityDate="2021-08-21T03:37:16.390" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8513" PostTypeId="4" CreationDate="2015-02-03T10:15:46.737" Score="0" Body="A jigsaw puzzle is a tiling puzzle that requires the assembly of (often oddly shaped) interlocking and tessellating pieces." OwnerUserId="8874" LastEditorUserId="5373" LastEditDate="2015-07-07T17:22:00.937" LastActivityDate="2015-07-07T17:22:00.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8514" PostTypeId="1" CreationDate="2015-02-03T10:37:45.820" Score="9" ViewCount="1344" Body="&lt;p&gt;Suppose there are n archers in a standoff and want to kill each other until there is just one archer remaining.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However everyone's chance of hitting (and thus killing) another archer is not fair. They are 1/n, 2/n, 3/n, ... n/n. &#xA;Thus they decide to take turns and one may only shoot, if is determined, whether his arrow has killed someone or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The order is determine by the following:&#xA;First they look, which archers are alive, then they look which of them has shot the least amount of arrows so far and if there are several of them, then they go after their chance of hitting another archer and always let a weaker archer shoot first.&#xA;An archer (not the one with the highest chance of hitting one!) may be allowed to shoot into the air, not hitting and thus not killing anything.&lt;br&gt;&#xA;It can be assumed, that every archer is sane and of course wants to survive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which archer has the highest chance of surviving? Which strategy should he use in order to maximize his success? (No hiding behind another archers and also no other loopholes)&#xA;For starters one might go with n = 3;&lt;/p&gt;&#xA;" OwnerUserId="8863" LastEditorUserId="5373" LastEditDate="2015-02-03T22:23:25.467" LastActivityDate="2015-02-05T15:33:59.650" Title="n - archers and many arrows" Tags="|logical-deduction|strategy|" AnswerCount="3" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="8515" PostTypeId="2" ParentId="8511" CreationDate="2015-02-03T11:15:57.750" Score="1" Body="&lt;p&gt;Many tries, I always end up in 30 : 34. Is the start square important?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3dCHa.png&quot; alt=&quot;30&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9197" LastActivityDate="2015-02-03T11:15:57.750" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8516" PostTypeId="2" ParentId="8514" CreationDate="2015-02-03T12:24:34.877" Score="7" Body="&lt;h2&gt;Solution for n = 3&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The weakest archer (1/3 chance of killing) starts. If he aims at the medium archer (2/3), and happens to kill him, it is the strongest archer's turn next, and will kill him. Therefore, he will aim at the strongest archer. If he kills, he has about 1/9 chance of survival. If not, it is the medium archer's turn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the medium archer aims, and kills the weak one, the strongest will fire next and win. So he will aim at the strongest one. If he kills, he has about 1/2 chance of survival as. If not, it is the strongest one's turn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strongest archer will obviously increase his chance of survival by taking out the medium one. He has 2/3 chance of survival and the weakest 1/3.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Wasting shots&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Coming back to the medium archer's option to waste his shot, the strongest archer will anyway kill him, so that option's out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the weakest one chooses to waste his shot, he will at least reach the top 2 (the other 2 will still aim at each other), and it will be his turn to fire, giving him a minimum survival chance of 1/3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strongest one gains nothing by wasting his shot.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Conclusion&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Optimal strategy for first shot, if you are:&lt;br&gt;&#xA;a) Weakest archer -  Waste the shot.&lt;br&gt;&#xA;b) Medium archer - Aim for the strongest one&lt;br&gt;&#xA;c) Strongest archer - Kill the medium one  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After up to 3 arrows shot, someone will end up dead, so you need to aim every shot at the remaining opponent.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Approach to solve n = 4&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This is the basic outline for how it should be solved, some higher level maths will be necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let a to d be the archers, weakest to strongest. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fact 1: One who misses a shot is as good as wasting it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fact 2: Top 2 archers always duel against each other (as best archer will always take out the next best).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Case 1: a kills b. c and d duel. a duels with winner of c and d.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Case 2: a kills c. b and d duel. a duels with winner of b and d.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Case 3: a kills d. b and c duel. a wastes shots till b or c is killed, a duels with winner.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Case 4: a wastes his shot, or chooses to waste it. Write 4 cases for b's decisions, like a. 4th case for b (waste/miss) will be as follows. c will necessarily kill d, else get killed by d. If a, b and c are left, a wastes shots till b and c's duel is over, then duels with the winner. Similarly for if a, b and d are left.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then you need to calculate the probability of each of the archers surviving each of these cases, and each archer should take the case with greatest chance of survival.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4269" LastEditDate="2015-02-05T15:33:59.650" LastActivityDate="2015-02-05T15:33:59.650" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8517" PostTypeId="1" AcceptedAnswerId="8535" CreationDate="2015-02-03T12:29:00.547" Score="11" ViewCount="662" Body="&lt;p&gt;The following twelve statements were made on a certain day X during the last 100 years. At least seven of these statements were true on day X:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Exactly one month from now it will be Thanksgiving.&lt;/li&gt;&#xA;&lt;li&gt;Arnie was born in San Francisco.&lt;/li&gt;&#xA;&lt;li&gt;Barney is in his twenties.&lt;/li&gt;&#xA;&lt;li&gt;Exactly one month from now is Independence Day (July 4).&lt;/li&gt;&#xA;&lt;li&gt;Exactly one month from now is Easter.&lt;/li&gt;&#xA;&lt;li&gt;Barney's favorite movie is  Akira Kurosawa's &quot;Rashomon&quot;.&lt;/li&gt;&#xA;&lt;li&gt;Barney was born in New York City.&lt;/li&gt;&#xA;&lt;li&gt;Arnie celebrated his 30th birthday yesterday.&lt;/li&gt;&#xA;&lt;li&gt;Barney was born in Seattle.&lt;/li&gt;&#xA;&lt;li&gt;Arnie was born in Miama.&lt;/li&gt;&#xA;&lt;li&gt;Barney is older than Arnie.&lt;/li&gt;&#xA;&lt;li&gt;Exactly one month from now is Christmas.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;On which day (day+month+year) was Barney born?&lt;/p&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="4050" LastEditDate="2015-02-03T13:06:13.577" LastActivityDate="2015-02-07T17:55:11.440" Title="Arnie and Barney: When is Barney's birthday?" Tags="|logical-deduction|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8518" PostTypeId="2" ParentId="8514" CreationDate="2015-02-03T13:24:45.250" Score="5" Body="&lt;p&gt;Let the archers be numbered $a_1, a_2, ... a_n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $p_i$ be the probability of archer $i$ making a shot.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $P_n(j, S)$ be the probability of archer $a_j$ winning out of the ordered set $S=\{i, ..., j, ... k\}$ with an initial size of $n$ where $a_j$ goes first.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;General solution for $n=2$&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;For the following, we will compute a formula for 2 archers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets say that we have two archers remaining, $a_i$ and $a_j$ and $a_i$ is to shoot first. To win, he can either kill $a_j$ on the first shot (happens with probability $p_i$), or if he misses (probability $1-p_i$) and survives the attempt by $a_j$ (probability $1-p_j$), then he can try again.  This gives the following recursive formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_n(i, \{i,j\}) = p_i + (1-p_i)(1-p_j)P(i,\{i,j\})$$&#xA;$$P_n(i, \{i,j\}) = \frac{p_i}{1-(1-p_i)(1-p_j)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $p_k=\frac{k}{n}$, we can reduce this further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_n(i, \{i,j\}) = &#xA;\frac{\frac{i}{n}}{1-(1-\frac{i}{n})(1-\frac{j}{n})} =&#xA;\frac{i}{n \left( 1-\left( \frac{n-i}{n}\right) \left( \frac{n-j}{n} \right) \right)} = &#xA;\frac{i}{\frac{n^2-(n(i+j)+ij)}{n}} =&#xA;\frac{in}{ni+nj-ij}$$&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Solution for $n=3$&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;As posted by @ghosts_in_the_code, there is a strategy for this scenario.  When implemented, the following are the chances for survival.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Archer $a_3$&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The first archer $a_1$ wastes his shot, and $a_2$ will aim at $a_3$.  If he hits, then $a_3$ loses, but if he misses, then $a_3$ kills $a_2$ and must survive an attempt by $a_1$ to win.  So, $a_3$ wins if both $a_2$ and $a_1$ miss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_3(3, \{1,2,3\})=(1-p_2)(1-p_1)=\frac{2}{9}$$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Archer $a_2$&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The second archer will lose if he misses on his first shot at $a_3$ since $a_3$ will surely target him and kill him on his turn.  Thus, $a_2$ must first kill $a_3$, and then survive $a_1$'s attempt.  Then his chances are given in the formula above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_3(2, \{1,2,3\}) = &#xA;p_2 \times (1-p_1) \times P_3(2, \{1, 2\}) = &#xA;\frac{2}{3}\times \frac{2}{3}\times \frac{2 \times 3}{3 \times 2 + 3 \times 1 - 2 \times 1} =&#xA;\frac{8 \times 3}{3 \times 3 \times (6+3-2)} =&#xA;\frac{8}{21}$$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Archer $a_1$&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The first archer can win in two ways: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If $a_2$ kills $a_3$ ($p_2$) and then $a_1$ outlasts $a_2$ (formula above).  &lt;/li&gt;&#xA;&lt;li&gt;If $a_3$ kills $a_2$ ($1- p_2$) and then $a_1$ hits $a_3$ on the first attempt ($p_1$).  &lt;em&gt;There will be only one attempt!&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Using the formula we worked out before, we get the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P(1, \{1,2,3\}) = &#xA;p_2 \times P_3(1, \{1, 2\}) + (1-p_2) \times p_1 = &#xA;\frac{2}{3} \times \frac{3\times 1}{3 \times 1+ 3 \times 2 - 2 \times 1} + \left( 1-\frac{2}{3} \right) \times \frac{1}{3} = &#xA;\frac{2}{7} + \frac{1}{9} = &#xA;\frac{25}{63}$$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Summary&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;All these should add up to $1$, and they do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{25}{63} + \frac{8}{21} + \frac{2}{9} = &#xA;\frac{25 + (3 \times 8) + (7 \times 2)}{63} = \frac{25 + 24 + 14}{63} =&#xA;\frac{63}{63} = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, somewhat counter-intuitively, the best archer has the worst chance of survival, and the worst archer has the best chance.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;General Solution for $n=3$&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;We've done the specific solution for $a_1, a_2, a_3$, but what about for $a_i, a_j, a_k$?  It is no longer obvious that the lowest archer shoot intentionally miss.  For example, say $a_i$ and $a_j$ are equally dismal shots and $a_k$ is perfect.  If $a_i$ wastes the shot and $a_j$ misses $a_k$, then $a_k$ will take out $a_j$.  That leaves $a_i$ one shot to kill $a_k$.  Thus, $a_i$'s chance of winning is $1-p_i$.  If $a_j$ hits $a_k$ instead, then it is a duel between roughly equal players and $a_i$ has about a 50% chance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead if $a_i$ aims for $a_k$ and hits, then although $a_j$ goes first in the duel, the chances are still pretty equal - much better than duelling $a_k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, in this case, it is obvious that $a_i$ should shoot for $a_k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we need to explore 2 strategies - one where the weakest wastes the shot, and the other where the weakest targets the strongest.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Weakest Targets Strongest&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;For the strongest to win, he needs to survive the first two shots.  Then he will target $a_j$.  If he is successful, then it is a two person game (and we have the general solution for that).  If he fails, then it is the same probability as before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_n(k, \{i,j,k\}) = (1-p_i)(1-p_j)\Big( p_k \times P_n(k, \{i,k\}) + (1-p_k) \times P(k, \{i,j,k\}) \Big) =&#xA;\left(1-\frac{i}{n} \right) \left( 1-\frac{j}{n} \right) \left(    &#xA;\frac{k}{n} \times \frac{nk}{nk+ni-ik} + \left( 1-\frac{k}{n} \right) \times P(k, \{i,j,k\})&#xA;\right)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving this is rather tricky, and doesn't have a nice closed form that I can see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for now, I am giving up on this.  Perhaps we can come back to it later, but I think I am taking the wrong approach.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="1592" LastEditDate="2015-02-05T14:58:52.800" LastActivityDate="2015-02-05T14:58:52.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8519" PostTypeId="2" ParentId="8517" CreationDate="2015-02-03T13:46:50.003" Score="7" Body="&lt;p&gt;My first variant...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Barney was born on 29 February, 1844 (Year may be different)&#xA; &lt;BR&gt;&lt;BR&gt;&#xA; Today is 1st March 1934&#xA; &lt;BR&gt;&lt;BR&gt;&#xA; 1) Arnie was born in San Francisco / Arnie was born in Miama.&#xA; &lt;BR&gt;&lt;BR&gt;&#xA; 2) Barney was born in New York City / Barney was born in Seattle.&#xA; &lt;BR&gt;&lt;BR&gt;&#xA; 3) Barney's favorite movie is Akira Kurosawa's &quot;Rashomon&quot;.&#xA; &lt;BR&gt;&lt;BR&gt;&#xA; 4) Exactly one month from now is Easter. (1st April 1934)&#xA; &lt;BR&gt;&lt;BR&gt;&#xA; 5) Barney is in his twenties (yep, he celebrates his birthday once in 4 years)&#xA; &lt;BR&gt;&lt;BR&gt;&#xA; 6) Arnie celebrated his 30th birthday yesterday. (28 February 1934, he celebrates his birthday every year)&#xA; &lt;BR&gt;&lt;BR&gt;&#xA; 7) Barney is older than Arnie. (yep, for many-many years)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My second variant... (&quot;Rashomon&quot; won't stop me xD)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Barney was born on 1 March 1934.&#xA; &lt;BR&gt;&lt;BR&gt;&#xA; Today: 29th February 1964.&#xA; &lt;BR&gt;&lt;BR&gt;&#xA; 1) Arnie was born on 28 February 1934 in Miama at 23:30 (4:30 GMT)&#xA; &lt;BR&gt;&lt;BR&gt;&#xA; 2) Barney was born on 1 March 1934 in Seattle at 00:30 (3:30 GMT)&#xA; &lt;BR&gt;&lt;BR&gt;&#xA; 3) Barney's favorite movie is Akira Kurosawa's &quot;Rashomon&quot;.&#xA; &lt;BR&gt;&lt;BR&gt;&#xA; 4) Exactly one month from now is Easter. (29 March 1964)&#xA; &lt;BR&gt;&lt;BR&gt;&#xA; 5) Barney is in his twenties (He will hit 30 tomorrow)&#xA; &lt;BR&gt;&lt;BR&gt;&#xA; 6) Arnie celebrated his 30th birthday yesterday. (28 February 1964)&#xA; &lt;BR&gt;&lt;BR&gt;&#xA; 7) Barney is older than Arnie. (technically, yes)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9088" LastEditorUserId="9088" LastEditDate="2015-02-04T08:26:44.873" LastActivityDate="2015-02-04T08:26:44.873" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8520" PostTypeId="5" CreationDate="2015-02-03T14:31:09.493" Score="0" Body="&lt;p&gt;This tag is used for geometric puzzles, if one wants to stress that the setting is not 2-dimensional but 3-dimensional.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/37725/a-maze-ing-dice&quot;&gt;A-maze-ing Dice&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11497" LastEditorUserId="11110" LastEditDate="2017-06-21T20:24:55.290" LastActivityDate="2017-06-21T20:24:55.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8521" PostTypeId="4" CreationDate="2015-02-03T14:31:09.493" Score="0" Body="This tag is used for geometric puzzles, if one wants to stress that the setting is not 2-dimensional but 3-dimensional." OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-03T15:21:31.403" LastActivityDate="2015-02-03T15:21:31.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8522" PostTypeId="5" CreationDate="2015-02-03T14:37:36.543" Score="0" Body="&lt;p&gt;This tag is for puzzles involving cryptography. Cryptography is the art of secure communication in the presence of (hostile) third parties. For simpler &amp;quot;crack the code&amp;quot; type problems use &lt;a href=&quot;/questions/tagged/cipher&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;cipher&amp;#39;&quot; aria-label=&quot;show questions tagged &amp;#39;cipher&amp;#39;&quot; rel=&quot;tag&quot; aria-labelledby=&quot;cipher-container&quot;&gt;cipher&lt;/a&gt; instead.&lt;/p&gt;&#xA;&lt;p&gt;More precisely, Cryptography&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;constructs and analyzes protocols that block third parties&lt;/li&gt;&#xA;&lt;li&gt;deals with aspects in information security as data confidentiality, data integrity, authentication&lt;/li&gt;&#xA;&lt;li&gt;applies techniques from mathematics, computer science, and electrical engineering.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Everyday applications of cryptography include ATM cards, computer passwords, and electronic commerce.&lt;/p&gt;&#xA;&lt;p&gt;There is a Stack Exchange site dedicated to &lt;a href=&quot;https://crypto.stackexchange.com/&quot;&gt;Cryptography&lt;/a&gt;, please use this if your questions are of a more serious nature.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="52483" LastEditDate="2023-01-31T19:25:41.463" LastActivityDate="2023-01-31T19:25:41.463" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8523" PostTypeId="4" CreationDate="2015-02-03T14:37:36.543" Score="0" Body="A puzzle concerning techniques for secure communication in the presence of (hostile) third parties. Do NOT use this for simple &quot;crack the code&quot; problems; use [cipher] instead." OwnerUserId="8874" LastEditorUserId="69582" LastEditDate="2021-03-08T18:21:14.277" LastActivityDate="2021-03-08T18:21:14.277" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8524" PostTypeId="1" AcceptedAnswerId="8525" CreationDate="2015-02-03T15:13:24.373" Score="21" ViewCount="2773" Body="&lt;p&gt;Who wrote this doctoral thesis?&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;Published:&lt;/em&gt; August 1960&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Abstract:&lt;/em&gt; Subject responses are investigated to antemeridial comestibles in the 510 nm range. Verbal responses was elicited by a hobbit companion on a sample of size n=1. Aversive reactions were reproduced for a wide variety of environmental scenarios, including: cuboid containment, vulpine proximity, domestic setting, liquid precipitation, and locomotive transport. However, the results are contraindicated by a singular yet strongly outlying data point at the end of testing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;In case it's not clear, this is a riddle in a fanciful frame. This is not an actual doctoral thesis.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-02-03T20:55:19.550" Title="Riddle: Who wrote this doctoral thesis?" Tags="|riddle|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8525" PostTypeId="2" ParentId="8524" CreationDate="2015-02-03T16:40:03.007" Score="24" Body="&lt;p&gt;This is an abstract to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Green Eggs and Ham&lt;/em&gt;, a children's book by Dr. Seuss (Theodor Geisel), published in 1960.&lt;br&gt; &quot;Antemeridial comestibles&quot; are morning foods (eggs and ham), and light with wavelength 510 nm is green light.&lt;br&gt; &quot;Hobbit companion&quot; refers to Sam, who was the companion of Frodo in &lt;em&gt;The Lord of the Rings&lt;/em&gt;, as well as the character Sam (Sam-I-Am) in Seuss' book who offers green eggs and ham to another character.&lt;br&gt; The aversive responses are the other character's rejections of green eggs and ham in a variety of locations and situations: &quot;cuboid containment&quot; (&lt;em&gt;I do not like them in a box&lt;/em&gt;), &quot;vulpine proximity&quot; (&lt;em&gt;I do not like them with a fox&lt;/em&gt;), &quot;liquid precipitation&quot; (&lt;em&gt;I will not eat them in the rain&lt;/em&gt;), and &quot;locomotive transport&quot; (&lt;em&gt;I will not eat them on a train&lt;/em&gt;). &lt;br&gt;The refusals are always addressed to Sam by name (&lt;em&gt;I do not like them, Sam-I-Am&lt;/em&gt;).&lt;br&gt; The results are contraindicated because at the end of the story, the character tries green eggs and ham, likes them, and recants the earlier statements.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-02-03T17:15:41.343" LastActivityDate="2015-02-03T17:15:41.343" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="8526" PostTypeId="2" ParentId="8511" CreationDate="2015-02-03T17:47:34.053" Score="11" Body="&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem can be simplified in the following way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an $n x n$ board where $n$ is even, make $\frac{n}{4}$ squares that are $2 x 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has $\frac{n^2}{2}$ horizontal and $\frac{n^2}{2}$ vertical moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, join any 2 adjacent shapes. When doing this, you will alter both the horizontal and vertical moves by 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continue joining shapes until you have made 1 single shape taking up the entire board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can choose your adjustments so that for each join, you alternate between increasing and decreasing the horizontal moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To end up with a shape that has the same number of horizontal and vertical moves, it requires an even number of join operations, which requires an odd number of shapes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $\frac{n^2}{4}$ must be odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n = 8$ then $\frac{n^2}{4} = 16$. Because it is even, the number of join operations is odd, which means that you cannot end up with an identical number of horizontal and vertical moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Professor Erasmus is incorrect. This cannot be done with an $8 x 8$ chessboard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Original:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Professor Erasmus has made a mistake. He should have chosen a board with a side that is divisible by 2 but not 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2x2 board: Trivial to see that 4 moves must be 2 horizontal and 2 vertical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4x4 board: Few enough paths to see that the best you can get is a difference of 4 between horizontal and vertical. Also, the worst you can get is a difference of 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6x6 board: Follow the outside edge for 17 moves, then zig-zag back through the center. Total of 18 horizontal, 18 vertical. This is one of many ways to achieve this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8x8 board: Again, the best that you can do is a difference of 4 between horizontal and vertical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have not tried larger boards. I expect that my answer is accurate for them.&lt;/p&gt;&#xA;" OwnerUserId="375" LastEditorUserId="375" LastEditDate="2015-02-03T18:58:37.023" LastActivityDate="2015-02-03T18:58:37.023" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="8528" PostTypeId="2" ParentId="8514" CreationDate="2015-02-03T19:56:44.343" Score="1" Body="&lt;p&gt;The optimum strategy is for the top 2 archers to duel, the next one to shoot down, then the rest to shoot for the strongest. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have 3 archers, the weakest wants to miss and let the top 2 kill each other, then try to snipe the survivor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have 4 archers, the 2nd weakest will try to kill the weakest, since the top 2 archers are focused on each other, and if you kill either of them, the survivor will try to kill you. And the weakest will actually try to kill the strongest, since that puts him back into the 3 archer scenario. He could just try to kill the 2nd weakest, but that would ultimately leave a stronger archer to 1-on-1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have 5 archers, the two weakest would go for the top, the middle would go for 2nd weakest, then the top 2 would duel, as usual. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I've been thinking about how to deal with this issue when you have many archers (say, 100). It's actually not in the top archer's best interest to kill each other first, since there are so many people gunning for them. Which got me thinking about how the categories of archers really played out.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The top archers, who everyone is gunning for. &lt;/li&gt;&#xA;&lt;li&gt;The good archers, who have a good chance against the top, but eventually could become the top that everyone else is gunning for. &lt;/li&gt;&#xA;&lt;li&gt;The mediocre archers, who won't get to the top any time soon, so can gun for whoever.&lt;/li&gt;&#xA;&lt;li&gt;The bad archers, who through weight of numbers can bring down a good archer.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now, the good archers and the top archers have an interesting predicament. The good archers could probably take out at least half of the top archers, but then the remaining top archers would take out a lot of the good ones, less how many top archers were killed by the lower archers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If instead the good archers shot into the mediocre ones, the top ones could shoot into the good ones, but then they'd be eliminated the next wave (likely). But if they also shot into the mediocre ones, that group would be fairly weakened, leaving bad archers to hit good/top and good/top to hit mediocre/bad. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This scenario would end up with the good archers in the best position, since they really haven't been targeted as much as the top/bottom. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-02-03T21:23:55.420" LastActivityDate="2015-02-03T21:23:55.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8530" PostTypeId="2" ParentId="8488" CreationDate="2015-02-03T22:26:50.473" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think that &lt;strong&gt;Tony Montana&lt;/strong&gt; is the answer.&#xA; &lt;br/&gt;&lt;br/&gt;&#xA; Thanks to the first hint, we know that the answer's name is a state. The title of the question completes this hint: we can deduce that the answer's name is a state of the United States of America, and that the answer is the name of a person.&#xA; &lt;br/&gt;&lt;br/&gt;&#xA; The second hint tells us that it's risky to oppress that person, so that person may have enough influence to become dangerous.&lt;br/&gt;&#xA; Presidents and mafia have enough influence to do this.&#xA; &lt;br/&gt;&lt;br/&gt;&#xA; &lt;a href=&quot;https://en.wikipedia.org/wiki/Tony_Montana&quot;&gt;Tony Montana&lt;/a&gt; is the protagonist of &lt;em&gt;Scarface&lt;/em&gt;, and is one of the most famous mafia characters.&lt;br/&gt;&#xA; He's from Cuba, and Scarface takes place in Florida, California and New York: none of them is Montana, so the first hint is verified.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9202" LastEditorUserId="9202" LastEditDate="2015-02-03T22:32:17.320" LastActivityDate="2015-02-03T22:32:17.320" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8532" PostTypeId="2" ParentId="8488" CreationDate="2015-02-03T22:58:02.650" Score="22" Body="&lt;p&gt;An answer that fits is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gas (gasoline). Its state is liquid although its name is gas. If people step on it, they are in risk of exploding if there is a nearby spark/flame.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastActivityDate="2015-02-03T22:58:02.650" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8533" PostTypeId="2" ParentId="8517" CreationDate="2015-02-04T07:34:06.600" Score="3" Body="&lt;p&gt;I don't have a complete answer yet, but so far I have been able to determine that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2 &amp;amp; 10 are contradictory so at least 1 of them is false&#xA; &lt;BR&gt;&lt;BR&gt;&#xA; 7 &amp;amp; 9 are contradictory so at least 1 of them is false&#xA; &lt;BR&gt;&lt;BR&gt;&#xA; 3, 8 &amp;amp; 11 are contradictory. at least 1 of them is false.&#xA; &lt;BR&gt;&lt;BR&gt;&#xA; 1, 4, 5 &amp;amp; 12 are contradictory. at least 3 of them are false&#xA; &lt;BR&gt;&lt;BR&gt;&#xA; even though the date for thanksgiving has changed throughout history none of the dates are close enough to have fallen on the same date (afaik).&#xA; &lt;BR&gt;&lt;BR&gt;&#xA; that means we have at least 6 false statements, which contradicts the original question that stated there are at least 7 true statements.&#xA; &lt;BR&gt;&lt;BR&gt;&#xA; therefore some part of my understanding is likely mistaken (?)&#xA; &lt;BR&gt;&lt;BR&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4125" LastEditorUserId="4125" LastEditDate="2015-02-04T16:06:41.433" LastActivityDate="2015-02-04T16:06:41.433" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8534" PostTypeId="2" ParentId="8430" CreationDate="2015-02-04T09:05:47.727" Score="7" Body="&lt;p&gt;Sequence 4 is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the names of the &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_Nobel_laureates_in_Literature&quot;&gt;Nobel laureates for Literature&lt;/a&gt; - the first letter of their last names listed by year since 1901.  The missing letters are shown as follows (&lt;strong&gt;X+Y&lt;/strong&gt; meaning two names in the same year and &lt;strong&gt;n/a&lt;/strong&gt; meaning no award for that year):&lt;br&gt;&#xA;&lt;br&gt;&#xA; P, M, B, &lt;strong&gt;M+E&lt;/strong&gt;, S, C, K, E, L, H, M, H, T, &lt;strong&gt;n/a&lt;/strong&gt;, R, &lt;strong&gt;v&lt;/strong&gt;, &lt;strong&gt;G+P&lt;/strong&gt;, &lt;strong&gt;n/a&lt;/strong&gt;, S, H, F, B, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastActivityDate="2015-02-04T09:05:47.727" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8535" PostTypeId="2" ParentId="8517" CreationDate="2015-02-04T10:30:38.083" Score="10" Body="&lt;p&gt;(1) First we derive that &lt;strong&gt;at most three&lt;/strong&gt; out of the eight statements 1, 2, 4, 5, 7, 9, 10, 12 can be true.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Statements 1, 4, 5, 12 are pairwise contradictory. Thanksgiving is in late November, Independence Day in July, Easter in spring, Christmas in December. Hence at most one out of 1, 4, 5, 12 is true.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Statements 2 and 10 on Arnie's birthplace are contradictory. Hence at most one of 2 and 10 is true.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Statements 7 and 9 on Barney's birthplace are contradictory. Hence at most one of 7 and 9 is true.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(2) Since altogether there are &lt;strong&gt;at least seven&lt;/strong&gt; true statements, we conclude that the remaining four statement 3, 6, 8 and 11 all must be true.&#xA;To summarize:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;3, 6, 8 and 11 are true  &lt;/li&gt;&#xA;  &lt;li&gt;exactly one of 1, 4, 5, 12 is true  &lt;/li&gt;&#xA;  &lt;li&gt;exactly one of 2, 10 is true  &lt;/li&gt;&#xA;  &lt;li&gt;exactly one of 7, 9 is true.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(3) Statements 3, 8, 11 state that Arnie's mathematical age (counted in years) is larger than Barney's mathematical age, although Barney is older than Arnie.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Hence Barney was born in an eastern time zone (very early in the day, let us say just after midnight).&lt;/li&gt;&#xA;  &lt;li&gt;Arney was born in a time zone farther to the west, later than Barney, but still on the day before Barney.&lt;/li&gt;&#xA;  &lt;li&gt;This implies that statements 2 and 7 are true: Barney was born in New York City and Arnie was born in San Francisco.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(4) By the true statement 8, Arnie's 30th birthday was yesterday. By the true statement 3, Barney's 30th birthday will start in the future. This means that the birthdays were consecutive days in the year when they both were born, but not in the year with day X. The only possible reason for this discrepancy must be a February 29 in the year with day X.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Hence Barney was born on March 1 in New York City (say at 0:01am).&lt;/li&gt;&#xA;  &lt;li&gt;Arnie was born later (say at 11:55pm) on February 28 in San Francisco.&lt;/li&gt;&#xA;  &lt;li&gt;Day X is a February 29.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(5) Exactly one of statements 1, 4, 5, 12 must be true, and the only remaining possibility is statement 5 with Easter. &#xA;Since statement 6 on Rashomon is true and since Rashomon came out in 1950, the year with day X is 1950 or later.&#xA;We summarize:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;In the year with day X, Easter is on March 29&lt;/li&gt;&#xA;  &lt;li&gt;The year with day X is 1950 or later.&lt;/li&gt;&#xA;  &lt;li&gt;The year with day X is a leap year&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(6) Looking up the Easter days in &#xA;&lt;a href=&quot;http://www.maa.mhn.de/StarDate/publ_holidays.html&quot; rel=&quot;nofollow&quot;&gt;http://www.maa.mhn.de/StarDate/publ_holidays.html&lt;/a&gt;, we see that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Day X is 29 February 1964&lt;/li&gt;&#xA;  &lt;li&gt;Arnie was born 28 February 1934&lt;/li&gt;&#xA;  &lt;li&gt;Barney was born on 1 March 1934.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-07T17:55:11.440" LastActivityDate="2015-02-07T17:55:11.440" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8536" PostTypeId="2" ParentId="8459" CreationDate="2015-02-04T10:56:55.840" Score="1" Body="&lt;p&gt;&lt;strong&gt;Answer the second part assuming that the list contains 99 statements:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that exactly $x$ statements are false. Then every statement $n$ with number $n\ge x+1$ is false, and every statement $m$ with number $m\le x$ is true. There are exactly $99-x$ statements with $n\ge x+1$, and they coincide with the $x$ false statements. Therefore $x=99-x$. This is impossible, and the problem has no solution.&lt;/p&gt;&#xA;" OwnerUserId="9125" LastActivityDate="2015-02-04T10:56:55.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8537" PostTypeId="5" CreationDate="2015-02-04T11:17:17.750" Score="0" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Water_pouring_puzzle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Decanting problems&lt;/a&gt; (also known as jug-pouring problems or water-pouring problems) are where there are various containers with various known capacities for keeping various liquids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal is to design a strategy for pouring the liquid between the bottles, so that a certain goal situation is reached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple example is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There are two bottles that can hold 3 litres and 5 litres respectively, and a bathtub with unlimited water. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How can 1 litre be measured? &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fill 3. Fill 5 with 3. Fill 3. Fill 5 with 3. 3 now has 1L.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More examples and analysis of the problems can be found here:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.cut-the-knot.org/triangle/glasses.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cut-the-Knot&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="11497" LastEditDate="2020-04-06T13:34:29.397" LastActivityDate="2020-04-06T13:34:29.397" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8538" PostTypeId="4" CreationDate="2015-02-04T11:17:17.750" Score="0" Body="A strategic puzzle centered around pouring liquids between containers of different capacities, with the goal of measuring exactly n liters." OwnerUserId="8874" LastEditorUserId="11497" LastEditDate="2020-04-06T13:34:29.397" LastActivityDate="2020-04-06T13:34:29.397" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8540" PostTypeId="2" ParentId="201" CreationDate="2015-02-04T11:55:42.413" Score="1" Body="&lt;p&gt;The possibilities are :&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;RED   RED   RED   : A won't know for sure, B know either B or C or both have RED but isn't sure, but C is sure he has RED since both declined.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;RED   BLACK BLACK : A knows for sure since rest have Black Hats. B knows too. So does C.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;RED   RED   BLACK : A won't know for sure, B know either B or C or both have RED  hence is sure since it sees a BLACK Hat ahead&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;RED   BLACK RED   : A won't know for sure, B know either B or C or both have RED but isn't sure, but C is sure&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;BLACK BLACK RED   : A won't know for sure, B know either B or C or both have RED but isn't sure, but C is sure&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;BLACK RED   BLACK : A won't know for sure, B know either B or C or both have RED  hence is sure since it sees a BLACK Hat ahead&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;BLACK RED   RED   : A won't know for sure, B know either B or C or both have RED but isn't sure, but C is sure&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Implies 1,4,5,7 are the possible cases/.&lt;/p&gt;&#xA;" OwnerUserId="9344" LastActivityDate="2015-02-04T11:55:42.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8541" PostTypeId="5" CreationDate="2015-02-04T12:46:36.003" Score="0" Body="&lt;p&gt;In a treasure hunt, one or more players try to find hidden articles, locations or places by using a series of clues.&#xA;Treasure hunt games may be an indoor or outdoor activity; outdoors they can be played in a garden or the treasure could be located anywhere around the world.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Geocaching is an outdoor treasure-hunting game in which the participants use a global positioning system (GPS) receiver or other navigational techniques to hide and seek containers (called &quot;geocaches&quot; or &quot;caches&quot;) anywhere in the world.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Letterboxing is played outdoors and combines elements of orienteering, art and problem-solving. Letterboxers hide small, weatherproof boxes in publicly accessible places (such as parks) and distribute clues to finding the box in printed catalogs, on one of several web sites, or by word of mouth. &#xA;Individual letterboxes usually contain a logbook.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-04T14:59:42.323" LastActivityDate="2015-02-04T14:59:42.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8542" PostTypeId="4" CreationDate="2015-02-04T12:46:36.003" Score="0" Body="In a treasure hunt, one or more players try to find hidden articles or locations by using a series of clues. " OwnerUserId="8874" LastEditorUserId="5373" LastEditDate="2020-12-14T19:17:56.107" LastActivityDate="2020-12-14T19:17:56.107" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8543" PostTypeId="5" CreationDate="2015-02-04T12:47:21.200" Score="0" Body="&lt;p&gt;A puzzlehunt is a puzzle game where teams compete to solve a series of puzzles at a particular site, in multiple sites or via the internet. Groups of puzzles in a puzzle hunt are often connected by a metapuzzle, leading to answers which combine into a final set of solutions. Sometimes, the prize for winning a puzzlehunt is to create the next one.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-04T15:32:30.773" LastActivityDate="2015-02-04T15:32:30.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8544" PostTypeId="4" CreationDate="2015-02-04T12:47:21.200" Score="0" Body="A puzzlehunt is a puzzle game where teams compete to solve a series of puzzles at a particular site, in multiple sites or via the internet." OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-04T15:32:19.947" LastActivityDate="2015-02-04T15:32:19.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8545" PostTypeId="5" CreationDate="2015-02-04T13:05:16.903" Score="0" Body="&lt;p&gt;Wordplay is a form of wit in which the used words become the main subject of the work, primarily for the purpose of intended effect or amusement.&lt;/p&gt;&#xA;&lt;p&gt;Examples of word play include:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;puns&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;A pun is a joke exploiting the different possible meanings of a word or the fact that there are words which sound alike but have different meanings.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Examples:&lt;/strong&gt; &lt;br&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The duck said to the bartender, &amp;quot;Put it on my &lt;em&gt;&lt;strong&gt;bill&lt;/strong&gt;&lt;/em&gt;.&amp;quot; &lt;br&gt; &lt;em&gt;Playing on the word 'bill' referring to both (i) a check/tab used to keep track of money owed at a bar, and (ii) the beak of a bird.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;What did the cupcake say to the frosting? &amp;quot;I'd be &lt;em&gt;&lt;strong&gt;muffin&lt;/strong&gt;&lt;/em&gt; without you!&amp;quot; &lt;br&gt; &lt;em&gt;Playing on the similar sounds of the words 'muffin' and 'nothing', and how a cupcake without frosting is similar to a muffin.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;phonetic mix-ups (such as spoonerisms)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;A spoonerism is a term used to describe a phrase where the beginning letters or sounds of two words in a phrase have been swapped, typically for humorous effect.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt; &lt;br&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A well-&lt;em&gt;&lt;strong&gt;b&lt;/strong&gt;&lt;/em&gt;oiled &lt;em&gt;&lt;strong&gt;i&lt;/strong&gt;&lt;/em&gt;cicle &lt;em&gt;(instead of a 'well-oiled bicycle')&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;obscure words and meanings&lt;/li&gt;&#xA;&lt;li&gt;clever rhetorical excursions&lt;/li&gt;&#xA;&lt;li&gt;oddly formed sentences&lt;/li&gt;&#xA;&lt;li&gt;double entendres&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;A double entendre is a word or phrase that is open to two interpretations.&lt;/em&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt; &lt;br&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;That guy looks &lt;em&gt;&lt;strong&gt;hot&lt;/strong&gt;&lt;/em&gt;! &lt;br&gt; &lt;em&gt;Playing on the two possible interpretations of 'hot' (i) in terms of temperature, and (ii) as slang for being physically attractive.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="60644" LastEditDate="2023-02-03T15:58:59.183" LastActivityDate="2023-02-03T15:58:59.183" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8546" PostTypeId="4" CreationDate="2015-02-04T13:05:16.903" Score="0" Body="A puzzle which incorporates a form of wordplay, such as puns, or which requires wordplay in order to find the solution." OwnerUserId="8874" LastEditorUserId="69582" LastEditDate="2020-12-16T07:07:49.213" LastActivityDate="2020-12-16T07:07:49.213" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8547" PostTypeId="1" CreationDate="2015-02-04T13:06:04.487" Score="10" ViewCount="944" Body="&lt;p&gt;A polygon is folded to perfectly wrap a cube, covering all of its surface area with no overlap. Show that the polygon had at least two equal angles.&lt;/p&gt;&#xA;" OwnerUserId="1752" LastEditorUserId="8874" LastEditDate="2015-02-04T13:16:39.207" LastActivityDate="2017-07-13T00:41:04.880" Title="Polygon wrapping a cube" Tags="|mathematics|geometry|three-dimensional|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8548" PostTypeId="5" CreationDate="2015-02-04T13:27:25.550" Score="0" Body="&lt;p&gt;This tag is used for discussing the number of solutions of a puzzle type or a concrete puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Counting is the action of finding the number of elements of a finite set of objects.&#xA;The related term enumeration refers to uniquely identifying the elements of a finite (combinatorial) set or infinite set by assigning a number to each element.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-04T15:32:57.340" LastActivityDate="2015-02-04T15:32:57.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8549" PostTypeId="4" CreationDate="2015-02-04T13:27:25.550" Score="0" Body="This tag is used for discussing the number of solutions of a puzzle type or a concrete puzzle." OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-04T17:04:33.063" LastActivityDate="2015-02-04T17:04:33.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8550" PostTypeId="5" CreationDate="2015-02-04T13:30:44.913" Score="0" Body="&lt;p&gt;Terminology is the study of terms and their use. &#xA;&lt;em&gt;Terms&lt;/em&gt; are words and compound words or multi-word expressions that are given specific meanings; these may deviate from the meanings the same words have in other contexts and in everyday language. &#xA;Terminology is a discipline that studies the development of such terms and their interrelationships.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-04T14:56:44.597" LastActivityDate="2015-02-04T14:56:44.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8551" PostTypeId="4" CreationDate="2015-02-04T13:30:44.913" Score="0" Body="This tag is used for discussing notations and terms used in puzzles." OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-04T15:33:09.367" LastActivityDate="2015-02-04T15:33:09.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8552" PostTypeId="5" CreationDate="2015-02-04T13:36:46.890" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-02-04T13:36:46.890" LastActivityDate="2015-02-04T13:36:46.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8553" PostTypeId="4" CreationDate="2015-02-04T13:36:46.890" Score="0" Body="This tag is used for discussing notations and terms used in puzzles." OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-04T15:33:00.970" LastActivityDate="2015-02-04T15:33:00.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8554" PostTypeId="5" CreationDate="2015-02-04T13:39:44.800" Score="0" Body="&lt;p&gt;Use this tag when a puzzle provides some information about a situation, and requires working through clues to arrive at an answer. Clues might be snippets of conversation, transcripts of letters, pictures of a scene, etc. These puzzles will generally be dressed up as a narrative/story, and the [story] tag will usually also apply. Some examples of situation puzzles include:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&amp;quot;Whodunnit&amp;quot; mysteries: Description of the crime scene and suspects, with the question being &amp;quot;Who did it?&amp;quot;&lt;/li&gt;&#xA;&lt;li&gt;&amp;quot;What happened here?&amp;quot;: Description of an unusual situation, with elements that seem to be incongruous or impossible, with the question being &amp;quot;How did this happen?&amp;quot; or &amp;quot;What happened?&amp;quot;&lt;/li&gt;&#xA;&lt;li&gt;&amp;quot;Why did X do Y?&amp;quot;: Description of the events surrounding a choice/action/event, with the question being &amp;quot;Why did X do Y?&amp;quot; (What was their motivation?)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;On Puzzling Stack Exchange, we expect puzzles to have unique answers. If your situation puzzle has several solutions all equally compatible with the given information (perhaps the idea is that solvers will propose solutions and you will dole out new information little by little to lead them to the intended solution), it is &lt;em&gt;not&lt;/em&gt; a good fit here. There should be just one solution that really fits the information you provide.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="19114" LastEditDate="2021-03-30T15:27:01.537" LastActivityDate="2021-03-30T15:27:01.537" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8555" PostTypeId="4" CreationDate="2015-02-04T13:39:44.800" Score="0" Body="A puzzle that involves figuring out what happened with regard to a certain incident." OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-04T17:03:46.387" LastActivityDate="2015-02-04T17:03:46.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8556" PostTypeId="1" AcceptedAnswerId="8593" CreationDate="2015-02-04T17:09:58.997" Score="2" ViewCount="898" Body="&lt;p&gt;This is my second batch of sequence puzzles that are nasty and hard to solve; &#xA;yet, each of them has a clear and justifiable solution.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sequence 1:&lt;br&gt;&#xA;  18, 17, 18, 22, 26, 35, 32, 32, ??, 38, ??, 52, ??, ??, ??, 64, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sequence 2:&lt;br&gt;&#xA;  Du, Me, Ph, Me, Al, Mi, ??&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sequence 3:&lt;br&gt;&#xA;  ..., 14, 13, 14, 6, 7, 12, 8, 16, 9, 13, 10, 15, 11, 12, ??, 6, ??, 2, ??, ??&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sequence 4:&lt;br&gt;&#xA;  1, 2, 3, 3, 2, 3, 4, 5, 3, 2, 3, 4, 5, 5, 4, ?, ?, ?, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sequence 5:&lt;br&gt;&#xA;  ?, ?, ?, ?, M, A, J, ?, H, T, P, T, F, P, B, L, J, G, H, G, A, C, H, C, ?, R, T, W, H, C, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9125" LastActivityDate="2015-02-06T01:04:47.597" Title="Five further sequence puzzles" Tags="|number-sequence|letter-sequence|" AnswerCount="5" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8557" PostTypeId="2" ParentId="8556" CreationDate="2015-02-04T17:22:42.673" Score="7" Body="&lt;p&gt;Partial answer (naturally):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sequence 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Number of matches played in the FIFA World Cups since 1930. Missing entries are 32, 38, 52, 52, 52. &lt;a href=&quot;https://en.wikipedia.org/wiki/FIFA_World_Cup#Attendance&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastActivityDate="2015-02-04T17:22:42.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8558" PostTypeId="2" ParentId="8556" CreationDate="2015-02-04T17:26:51.933" Score="6" Body="&lt;p&gt;Sequences 1, 3, and 4 again can be googled:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sequence 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Goles totales in mismunidales = Number of goals scored at the Soccer World Championships&lt;br&gt;&#xA; 18 17 18 22 26 35 32 32 32 38 38 52 52 64 64 64&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sequence 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The numbering of the popes,&lt;br&gt;&#xA; ending with John XXIII, Paul VI, John Paul I, John Paul 2, Benedict XVI, Francis I:&lt;br&gt;&#xA; ..., 14, 13, 14, 6, 7, 12, 8, 16, 9, 13, 10, 15, 11, 12, 23, 6, 1, 2, 16, 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sequence 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; number of strokes in the Roman numerals: I, II, III, IV, ...&lt;br&gt;&#xA; 1, 2, 3, 3, 2, 3, 4, 5, 3, 2, 3, 4, 5, 5, 4, 5, 6, 7&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-04T17:32:56.143" LastActivityDate="2015-02-04T17:32:56.143" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8559" PostTypeId="2" ParentId="8556" CreationDate="2015-02-04T17:38:55.573" Score="7" Body="&lt;p&gt;Sequence 5 is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Presidents of the United States.&lt;br&gt; W(ashington), A(dams), J(efferson), M(adison), M(onroe), A(dams), J(ackson), V(an Buren), H(arrison), T(yler), P(olk), T(aylor), F(illmore), P(ierce), B(uchanan), L(incoln), J(ohnson), G(rant), H(ayes), G(arfield), A(rthur), C(leveland), H(arrison), C(leveland), M(cKinley), R(oosevelt), T(aft), W(ilson), H(arding), C(oolidge).&lt;br&gt; The sequence ends here, but the presidents continue: H(oover), R(oosevelt), T(ruman), E(isenhower), K(ennedy), J(ohnson), N(ixon), F(ord), C(arter), R(eagan), B(ush), C(linton), B(ush), O(bama). &lt;br&gt; The missing letters are W, A, J, M, V, M.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-02-04T17:46:43.710" LastActivityDate="2015-02-04T17:46:43.710" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8560" PostTypeId="1" AcceptedAnswerId="8568" CreationDate="2015-02-04T17:41:36.857" Score="16" ViewCount="1180" Body="&lt;p&gt;In the country Borgonia, there are only two types of coins: genuine ones and fake ones.&#xA;All genuine coins have the same weight, and all fake coins have the same weight.&#xA;The weight of a genuine coin is different from the weight of a fake coin, but nobody knows whether it is larger or smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cosmo puts 54 coins on the table and tells Fredo: &lt;em&gt;&quot;There are exactly 2 fake coins among these 54 coins. Please divide them into two groups of equal total weight!&quot;&lt;/em&gt; Then Cosmo leaves the room. On the table, there is a balance with two pans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can Fredo find the desired division into two groups by using the balance at most five times?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-04T17:46:48.383" LastActivityDate="2016-05-05T00:32:48.040" Title="Fifty-two genuine and two fake coins" Tags="|strategy|weighing|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8561" PostTypeId="2" ParentId="284" CreationDate="2015-02-04T18:35:48.790" Score="-2" Body="&lt;p&gt;You basically need N chances for No. of Balls Ranging from ((N-1)*3)+1 to N*3.&#xA;i.e &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Upto 3 Balls -&gt; 1 Scale&lt;/li&gt;&#xA;&lt;li&gt;3 - 9 Balls  -&gt; 2 Scale&lt;/li&gt;&#xA;&lt;li&gt;10 - 15 Balls-&gt; 3 Scale&lt;/li&gt;&#xA;&lt;li&gt;16 - 21 Balls-&gt; 4 Scale&#xA;And so on.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9344" LastActivityDate="2015-02-04T18:35:48.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8562" PostTypeId="1" AcceptedAnswerId="8590" CreationDate="2015-02-04T20:56:44.813" Score="8" ViewCount="2109" Body="&lt;p&gt;On an island, all inhabitants are either knights or knaves. Knights always tell the truth and knaves always lie. No tribe is gender related and some people marry into a different tribe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I visited this island, a tour guide who is a knight taught me an amazing question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you ask this question to a couple in the island, it is impossible to find out of which tribe he/she is or his/her partner is with a response from a single person. However, if you combine both answers from a couple, then you can figure out their tribes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One day at a party, this amazing question was asked to 50 couples from the island. All husbands answered &quot;Yes&quot; and all wives answered &quot;No&quot;. How many people lied?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added hint:&lt;/strong&gt; I did NOT ask how many knaves are there?&lt;/p&gt;&#xA;" OwnerUserId="3914" LastEditorUserId="3914" LastEditDate="2015-02-05T23:55:37.983" LastActivityDate="2016-08-23T10:02:17.150" Title="Knights and knaves at a party" Tags="|logical-deduction|liars|" AnswerCount="6" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8563" PostTypeId="2" ParentId="8556" CreationDate="2015-02-04T23:30:41.887" Score="8" Body="&lt;p&gt;Sequence 2 is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the initial two letters of the names of the stars in the Big Dipper (the seven brightest stars in the constellation Ursa Major). The stars are:&lt;br&gt; Dubhe&lt;br&gt; Merak&lt;br&gt;Phecda&lt;br&gt; Megrez&lt;br&gt;Alioth&lt;br&gt;Mizar&lt;br&gt;Alkaid&lt;br&gt;The missing letters are 'Al'.&lt;br&gt;&lt;br&gt;I first noticed that sequence 2 was much shorter than the others, having only seven items. I did a Google search for 'things that come in sevens', and one of the lists mentioned there were seven stars in the Big Dipper. I looked up the names of the stars, and found they were a match.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-02-04T23:48:21.593" LastActivityDate="2015-02-04T23:48:21.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8564" PostTypeId="2" ParentId="8481" CreationDate="2015-02-04T23:49:39.283" Score="2" Body="&lt;p&gt;I can also attest to a proof that doesn't beat 7. But there is a 7 that I believe does so with more style as it uses non-integers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 16 &lt;code&gt;d&lt;/code&gt; 32 &lt;code&gt;s&lt;/code&gt; 15 &lt;code&gt;s&lt;/code&gt; 6.5 &lt;code&gt;d&lt;/code&gt; 13 &lt;code&gt;d&lt;/code&gt; 26 &lt;code&gt;d&lt;/code&gt; 52 &lt;code&gt;s&lt;/code&gt; 25&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I also found a few identities:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;code&gt;d s&lt;/code&gt; = subtract 1&lt;br&gt;&#xA;  &lt;code&gt;s d&lt;/code&gt; = subtract 2&lt;br&gt;&#xA;  &lt;code&gt;s s d&lt;/code&gt; = &lt;code&gt;d d s&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To perform the proof that 7 operations is the minimum, I looked at it from this perspective:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for each operation (&lt;code&gt;h&lt;/code&gt;, &lt;code&gt;d&lt;/code&gt;, &lt;code&gt;c&lt;/code&gt;, and &lt;code&gt;s&lt;/code&gt;) I defined an inverse (&lt;code&gt;h'&lt;/code&gt;, &lt;code&gt;d'&lt;/code&gt;, &lt;code&gt;c'&lt;/code&gt;, and &lt;code&gt;s'&lt;/code&gt;):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;code&gt;h(x) = x / 2&lt;/code&gt;, so &lt;code&gt;h'(x) = x * 2&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;d(x) = x * 2&lt;/code&gt;, so &lt;code&gt;d'(x) = x / 2&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;(note that &lt;code&gt;d'(x) == h(x)&lt;/code&gt; and &lt;code&gt;h'(x) == d(x)&lt;/code&gt;, convenient)&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;c(x) = x * (10^(floor(log10(x)) + 1)&lt;/code&gt;, so &lt;code&gt;c'(x) = x / ((10^(floor(log10(x)/2)+1)&lt;/code&gt; IFF that denominator is an integer&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;s(x) = (x / 2) - 1&lt;/code&gt;, so &lt;code&gt;s'(x) = (2 * x) + 2&lt;/code&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;16 can be transformed into 8, 32, 1616, and 7.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;Those can be transformed into ....&lt;/li&gt;&#xA;&lt;li&gt;A few more steps&lt;/li&gt;&#xA;&lt;li&gt;Only 50, 52 and 12.5 can transform into 25.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;With steps 1-to-2 and steps 7-8 defined, we can work our way to the middle. We can eliminate a lot of duplicate answers though, since if an answer appears earlier in a chain, we would never use that path since a shorter one already exists.&lt;/p&gt;&#xA;" OwnerUserId="2812" LastActivityDate="2015-02-04T23:49:39.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8566" PostTypeId="2" ParentId="8560" CreationDate="2015-02-05T00:05:55.047" Score="10" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Updated strategy after Joe Z's comment.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The key idea I'm going to use is that each use of the balance weighs all 54 coins. If we're lucky, it's going to be even and we're done, and if we're unlucky, we're going to repartition the 54 coins and try again, until we are done.&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; The question then becomes whether we in six rounds can partition 54 coins into two halves, such that any two coins are in different partitions in at least one round (the sixth round comes from the fact that if we haven't found a balance in the first five rounds, we can still repartition it one more time before presenting it to Cosmo). When the two fake coins are in different partitions, we are done.&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Here is one partitioning strategy that works: Give each coin a unique number from 0-27 and 36-63 (see explanation below why I choose this numbering). To partition the coins, we are going to use the coin number's binary representation:&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; $$binary(0) = 000000 \\ binary(1) = 000001 \\ binary(2) = 000010 \\ ... \\ binary(63) = 111111$$&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; In the first round, place all coins whose binary representation's least significant bit is 0 in the first pan, and all coins with 1 in the second pan.&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;In rounds $n$ = two to five, use the same procedure, but instead split into partitions by the $n$th least significant bit.&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;If any of the rounds produce an even balance, we are done and can stop. If after five rounds we still haven't found an even partitioning, we know that the two fake coins &lt;em&gt;only&lt;/em&gt; differ in the most significant bit. We can then put coins 0 through 26 in one pile and 36 through 63 in the other and present the result to Cosmo.&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Proof sketch: If the fake coins have numbers $x$ and $y$ (where $x \neq y$), the game will end in round $i$, where $i$ is the least significant bit where $x$ is different from $y$, because at this round $x$ and $y$ are in different partitions. This is guaranteed to happen in six rounds or less, since 0-63 can be represented in six bits.&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; The new numbering strategy is chosen to guarantee that each round splits the coins in two parts of exactly 27 coins each. This is because for each coin number $i$, there is a complement coin $j$ with all its bits reversed (for example, the complement of 0 is 63, the complement of 10 is 53). This was not the case when numbering the coins 0-53, as I had done originally.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7577" LastEditorUserId="7577" LastEditDate="2015-02-05T08:49:48.467" LastActivityDate="2015-02-05T08:49:48.467" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8568" PostTypeId="2" ParentId="8560" CreationDate="2015-02-05T01:25:44.777" Score="10" Body="&lt;p&gt;This is essentially a modification of Waxwing's argument.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We will perform a sequence of weighings, each of of which puts 27 coins on each side of the balance. If ever the two counterfeit coins are of different sides, the scale will balance and we are done.&#xA; &#xA; At any point during the process, we will call two coins equivalent if they have ended up on the same side of the balance in every weighing so far (so at the start, every coin is equivalent to every other coin). For each weighing, we will split each equivalence class in half (or as close to half as possible, if the size of the class is odd), putting half the coins on one side of the balance and half on the other side.&#xA; &#xA; There are an even number of coins total, so the number of equivalence classes with an odd number of coins is even. For half of these classes, we will put the bigger part on the right of the balance, and for the other half, put the bigger part on the left of the balance.&#xA; &#xA; Each equivalence class get cut in half (or as close to half as possible) with each weighing. This means that if the largest equivalence class has $k$ coins before some weighing, then after that weighing the largest class will have $\lceil k/2\rceil$ coins.&#xA; &#xA; At the start, the largest (only) equivalence class has 54 coins. Cutting in half (and rounding up) six times, we get that after the sixth division, each equivalence class has only a single coin. We are only allowed five weighings, so if we haven't had a balance after the fifth weighing, we know that the sixth division is balanced.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="-1" LastEditDate="2016-05-05T00:32:48.040" LastActivityDate="2016-05-05T00:32:48.040" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8569" PostTypeId="2" ParentId="8560" CreationDate="2015-02-05T01:57:27.600" Score="8" Body="&lt;p&gt;&lt;strong&gt;Step 1:&lt;/strong&gt;&lt;br&gt;&#xA;Make 3 groups of 18 coins: A,B,C.&lt;br&gt;&#xA;Compare A to B and B to C. If A&amp;ne;B and B&amp;ne;C then A=C.&lt;br&gt;&#xA;Select the 'odd' group (assume C) and note if it is lighter or heavier than the others. (assume lighter)&lt;br&gt;&#xA;&lt;strong&gt;2 weighings used&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 2:&lt;/strong&gt;&lt;br&gt;&#xA;Split C into 3 groups of 6: D,E,F.&lt;br&gt;&#xA;Compare D to E. If D=E, F is the odd group. Go to 3a&lt;br&gt;&#xA;If D&amp;ne;E the lighter one is the odd group (assume D). Go to 3b&lt;br&gt;&#xA;&lt;strong&gt;3 weighings used&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 3a:&lt;/strong&gt;&lt;br&gt;&#xA;Split F in 3 groups of 2: G,H,I.&lt;br&gt;&#xA;Compare G and H.    &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If G&amp;ne;H compare the lighter portion (assume G) with I.&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If G=I, then A+D+G+(either coin of H)=B+E+I+(the other coin of H)&lt;/li&gt;&#xA;&lt;li&gt;If G&amp;ne;I, then H=I and A+D+H+(either coin of G)=B+E+I+(the other coin of G)&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;If G=H, then A+D+G+(either coin of I)=B+E+H+(the other coin of I) &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 3b:&lt;/strong&gt;&lt;br&gt;&#xA;Split D in 3 groups of 2: G,H,I.&lt;br&gt;&#xA;Compare G and H.    &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If G=H, compare D and F.&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If D=F, then A+D+(any three coins of E)=B+F+(the other 3 coins of E) &lt;/li&gt;&#xA;&lt;li&gt;If D&amp;ne;F, then A+F+G+(either coin of I)=B+E+H+(the other coin of I) &lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;If G&amp;ne;H, (assume G is the lighter one) compare I+(any 4 known good coins) and F.&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If F&amp;lt;I+4, A+D+(any three coins of E)=B+F+(the other 3 coins of E)&lt;/li&gt;&#xA;&lt;li&gt;If F=I+4, A+E+G+(either coin of D)=B+F+I+(the other coin of D)&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The key here is to use the knowledge of whether the odd set of coins were lighter or heavier to help locate the fakes, if they are within the odd set of coins.    &lt;/p&gt;&#xA;" OwnerUserId="3987" LastActivityDate="2015-02-05T01:57:27.600" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8571" PostTypeId="2" ParentId="8562" CreationDate="2015-02-05T06:17:55.260" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 50 people lied&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the question was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strike&gt;Ask what would your partner answer if I ask him/her whether you are a knight&lt;/strike&gt; Some imaginary question&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are 4 types of pairs: KK, KN, NN and NK&lt;BR&gt;&#xA; Their corresponding answers to this question would be: YY, YN, NN, NY&lt;BR&gt;&#xA; Because the reply was YN (All husbands Yes and wives No), they are all KN (or NK) pairs and this 50 people were supposed to lie.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4637" LastEditorUserId="4637" LastEditDate="2015-02-06T05:36:45.697" LastActivityDate="2015-02-06T05:36:45.697" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8573" PostTypeId="1" AcceptedAnswerId="8581" CreationDate="2015-02-05T09:16:45.480" Score="14" ViewCount="2569" Body="&lt;p&gt;Cosmo plays the following (single-player) game with three wooden sticks:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;He first checks whether he can form a triangle from the three sticks&#xA;(which means: whether the longest stick is at most as long as the combined length of the two shorter sticks).&lt;/li&gt;&#xA;&lt;li&gt;If Cosmo can form such a triangle: Cosmo has won the game.&lt;/li&gt;&#xA;&lt;li&gt;If Cosmo cannot form such a triangle: Cosmo takes the longest stick, cuts off from it a piece as long as the combined length of the two shorter sticks, throws away this piece, and keeps the remaining shortened stick.&lt;/li&gt;&#xA;&lt;li&gt;Then Cosmo repeats these steps with the two shorter sticks and the shortened stick.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Will Cosmo always win the game after a finite number of steps? For any choice of wooden sticks to start with?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="11110" LastEditDate="2018-11-26T13:54:42.850" LastActivityDate="2018-11-26T13:54:42.850" Title="Cutting three sticks" Tags="|mathematics|game|reachability|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8574" PostTypeId="5" CreationDate="2015-02-05T09:26:23.767" Score="0" Body="&lt;p&gt;A puzzle that requires the solver to prove whether or not a system terminates after a finite number of steps and sometimes determine how many steps it takes.&lt;/p&gt;&#xA;" OwnerUserId="52483" LastEditorUserId="52483" LastEditDate="2018-11-26T15:04:44.567" LastActivityDate="2018-11-26T15:04:44.567" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8575" PostTypeId="4" CreationDate="2015-02-05T09:26:23.767" Score="0" Body="A puzzle on a discrete system, where one has to decide whether certain runs of the system end after a finite number of steps." OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-05T12:41:18.867" LastActivityDate="2015-02-05T12:41:18.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8577" PostTypeId="1" CreationDate="2015-02-05T11:29:31.320" Score="6" ViewCount="798" Body="&lt;p&gt;I was given this exercise to solve. &#xA;It's beyond me, but important for me to solve this. &#xA;The following piece of code is all the context I was given.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;int[][] x = new int[6][6];&#xA;        if x[1][2] + x[1][4] = 20&#xA;        and x[0][3] + x[0][4] + x[0][5] = 15&#xA;        and x[2][2] - x[0][5] = 9&#xA;        and x[0][0] + x[0][1] + x[0][2] = 6&#xA;        and x[3][2] - x[3][1] + x[2][4] = r&#xA;        and x[0][5] + x[2][2] + x[1][2] - x[3][1] = 10&#xA;        and x[3][0] - x[2][1] = 5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;then what is the answer of x[2][5] * x[3][0] ?&lt;/p&gt;&#xA;" OwnerUserId="9349" LastEditorUserId="8874" LastEditDate="2015-02-05T12:19:04.330" LastActivityDate="2015-02-05T13:33:54.677" Title="Exercise with a piece of code and x[2][5]" Tags="|logical-deduction|computer-science|" AnswerCount="2" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="8578" PostTypeId="2" ParentId="8577" CreationDate="2015-02-05T12:17:30.073" Score="4" Body="&lt;p&gt;My guess would be that this code is written in some programming language that initializes new numerical variables with 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since x[2][5] has not been referenced / used / accessed / changed before your question, then its value will be 0 and also the product x[2][5] * x[3][0] will have value 0.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-02-05T12:17:30.073" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8579" PostTypeId="2" ParentId="7929" CreationDate="2015-02-05T12:39:19.317" Score="1" Body="&lt;p&gt;Yet another (so far failed) attempt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We've established that &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;The answer the Q4 is likely some &lt;em&gt;text&lt;/em&gt; and not a coordinate/number.&lt;/strong&gt; &lt;br&gt;&#xA;(It comes in 2 parts of different length and the &lt;em&gt;additional&lt;/em&gt; hints following the same basic idea do not have any reason to be numbers, as they are out of connection.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Q4 is &lt;em&gt;not&lt;/em&gt; a substitution cipher.&lt;/strong&gt; (Because nothing matches the pattern) &lt;br&gt;&#xA;Modifying the numbers such that they &lt;em&gt;then&lt;/em&gt; become a substitution cipher did not yield results for simple maths, and it is unlikely that something rather involved has been used due to the other questions.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We are therefore looking for cipher which maps numbers onto letters but in a non-unique way. (Or it would be a substitution cipher.)  This could be a code-table where different numbers map onto identical letters, but it is hard to envision one which would be 'obvious and dead-simple' without any additional clue. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leaves two things jumping to mind: a flexible transposition cipher where the numbers are providing a key, or a book-cipher. Both would need a source-text to act on, and the argument of a 'dead simple' cipher seems to point rather to a book-cipher than a complicated transposition cipher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result, I continue working on the assumption Q4 is some sort of book-cipher, with the numbers representing a 'picking pattern' from some sort of source. Typically, such ciphers (if meant to represent letters) either specify &lt;strong&gt;Paragraph / Word / Letter&lt;/strong&gt; in a numeric way. However, we only have double-digit numbers up to &lt;strong&gt;34&lt;/strong&gt; which seems to indicate a more direct cipher. It could be&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Pick n'th word of source (and take first letter of it). Source at least 34 words long.&lt;/li&gt;&#xA;&lt;li&gt;Pick n'th letter of source. Source at least 34 characters long.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The second option seems more straight forward to me, but it would also suggest that the 'source' would be something the puzzle-solver would rather easily &lt;em&gt;guess&lt;/em&gt; as there are no other hints to a source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've therefore written a little script which takes the code-indices and a source string and outputs the according 'pick n-th letter' solution. It does this in 2 variants: with and without counting spaces as letters.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The task now is to try out as many 'potential' source-strings as possible.&lt;/strong&gt;&#xA;I've so far only used&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;code&gt;Kirkcudbrightshire squash club donate 1$&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;Somewhere in Kirkcudbrightshire is a buried case of whiskey&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Both did not yield something useful. It would be great, if we could together compile a list of &quot;source strings&quot; we might want to investigate. In particular Elisabeth, as she might think of &quot;obvious&quot; source-strings due to local or historical or personal knowledge.&#xA;Are their some &lt;em&gt;mottos&lt;/em&gt; or &lt;em&gt;inscriptions&lt;/em&gt; etc, which would be useful? They can't be too short, tough.&lt;/p&gt;&#xA;" OwnerUserId="5840" LastActivityDate="2015-02-05T12:39:19.317" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="8580" PostTypeId="2" ParentId="8577" CreationDate="2015-02-05T13:28:02.573" Score="11" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (2*6 + 6) * (3*6 + 1) = 342&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Using $x[0][3] + x[0][4] + x[0][5] = 15$ and $x[0][0] + x[0][1] + x[0][2] = 6$ we get that $sum(x[0][i]) = 21$ and we can deduce that $x[0][i] = i + 1$. We also can find that $x[2][2] = 9 + x[0][5] = 15$. An obvious thing that can be suggested is that $x[i][j] = i*6 + j + 1$. Thus $x[2][5] * x[3][0] = (2*6 + 6) * (3*6 + 1) = 342$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of course, we can cross-validate:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Here are all the equations:&lt;br&gt;$x[1][2] + x[1][4] = 6 + 3 + 6 + 5 = 20$&lt;br&gt;$x[0][3] + x[0][4] + x[0][5] = 4 + 5 + 6 = 15$&lt;br&gt;$x[2][2] - x[0][5] = 2*6 + 3 - 6 = 9$&lt;br&gt;$x[0][0] + x[0][1] + x[0][2] = 1 + 2 + 3 = 6$&lt;br&gt;$x[3][2] - x[3][1] + x[2][4] = 3*6 + 3 - 3*6 - 2 + 2*6 + 5 = r$ // which kinda makes sense as r is the 18th letter of the alphabet&lt;br&gt;$x[0][5] + x[2][2] + x[1][2] - x[3][1] = 6 + 2*6 + 3 + 1*6 + 3 - 3*6 - 2 = 10$&lt;br&gt;$x[3][0] - x[2][1] = 3*6 + 1 - 2*6 + 2 = 5$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="5610" LastEditDate="2015-02-05T13:33:54.677" LastActivityDate="2015-02-05T13:33:54.677" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8581" PostTypeId="2" ParentId="8573" CreationDate="2015-02-05T13:41:48.323" Score="19" Body="&lt;p&gt;Spoiler:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's choose a set of sticks to prove this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Consider sticks of length $1, a, a^2$ - each stick is a times bigger than the previous one. Then the next step we will have sticks of length $a^2 - a - 1, 1, a$. If we choose a such that $a^2 - a - 1 = \frac{1}{a}$, then the sticks will retain the same ratios at each step, and we will never terminate. The required a is a root of $a^3 - a^2 - a - 1 = 0$. This turns out to have exactly one root at a = 1.8393, and we can check that $a^2 = 3.383 &amp;gt; a + 1$, so we don't terminate at step 1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6031" LastActivityDate="2015-02-05T13:41:48.323" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8582" PostTypeId="2" ParentId="8562" CreationDate="2015-02-05T14:16:35.070" Score="4" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 50 people lied.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In order to be able to tell nothing from one partner's response, but everything from both, not only must all combinations of answers be possible, but also a partner's answer must differ depending on the tribe of his/her partner. So the possible permutations should be of the form: &lt;img src=&quot;https://i.stack.imgur.com/lzH2w.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Regardless of which response corresponds to Yes or No, we can tell that if each partner of a couple gives different responses to the question, they must be of different tribes. Given that for every couple, each partner gave a different response to the other, we can therefore see that all couples present must have been of a knight and a knave - therefore there must have been 50 knaves present, therefore 50 people lied.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For the puzzle in its current form, it is not necessary to know what the amazing question was, or how the yes/no answers match up. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9278" LastActivityDate="2015-02-05T14:16:35.070" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8588" PostTypeId="1" AcceptedAnswerId="10495" CreationDate="2015-02-05T15:43:33.880" Score="-1" ViewCount="773" Body="&lt;p&gt;Could I BE any more riddle.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;your holiday starts, forget happiness&lt;br&gt;&#xA;no out from these mates of no friendliness&lt;br&gt;&#xA;your past is what we think&lt;br&gt;&#xA;as space tortures, beginning to shrink  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;follow the sheep of fashion&lt;br&gt;&#xA;through halls of no compassion&lt;br&gt;&#xA;little sun is gained through wreck&lt;br&gt;&#xA;each day it steals, the doves doth peck  &lt;/p&gt;&#xA;" OwnerUserId="5135" LastEditorUserId="5188" LastEditDate="2015-04-15T03:49:35.303" LastActivityDate="2015-06-27T12:59:57.917" Title="The German's are Experimenting" Tags="|riddle|rhyme|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8590" PostTypeId="2" ParentId="8562" CreationDate="2015-02-05T22:23:12.100" Score="5" Body="&lt;p&gt;What I think the answer is supposed to be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 person lied. Shout out to @Callidus for pointing out the somewhat obvious implication of my logic&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As I'll explain, there is no single question that can satisfy the requirements for the &quot;amazing question&quot;. Hence OP lied in saying that such a question exists.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What I think the answer really is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This question does not provide enough information for us to know how many people lied. As I show in my &quot;alternative theory&quot; section, there are compound questions that actually work. However, without knowing which question is used and the order in which the husband and wife are spoken to, we have no way to tell how many knights and knaves there are - you could formulate a question for which a husband saying &quot;Yes&quot; followed by the wife saying &quot;No&quot; means they are both knights while a wife saying &quot;No&quot; followed by the husband saying &quot;Yes&quot; means they are both knaves. So you could have 100 knaves, 100 knights, or any combination of knights and knaves.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ABOUT THE &quot;AMAZING QUESTION&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a reason why the amazing question is not given to us:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No such real question could possibly exist unless it is asymmetrical!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's first tackle some properties of the question. We know that there are 4 possible pairings, KK, KN, NK, and NN (K=knight, N=knave). Based on the requirements of the question, we know that if you ask one person you get no conclusive information, but if you ask their spouse then you know both of their tribes. Let's focus on the husbands first:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because we won't know his tribe based on his answer, any answer that he can give as a knight must be something he can give as a knave. Additionally, if his wife is of a particular tribe his response must change depending on his tribe. Because we won't know his wife's tribe based on his answer, his response must change depending on her tribe.&lt;br&gt;&lt;br&gt; So if he's a knight, he'll say one thing (A) if she's a knight, and something else (B) if she's a knave. So if he's a knave, he'll say either A or B. Since he can't say A if she's a knight, he'll say B. Then, he'll say A if she's a knave.&lt;br&gt;&lt;br&gt;&lt;code&gt;H W | R&lt;/code&gt;&lt;br&gt;&lt;code&gt;-------&lt;/code&gt;&lt;br&gt;&lt;code&gt;K K | A&lt;/code&gt;&lt;br&gt;&lt;code&gt;K N | B&lt;/code&gt;&lt;br&gt;&lt;code&gt;N K | B&lt;/code&gt;&lt;br&gt;&lt;code&gt;N N | A&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The same logic holds for the wife:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;H W | R&lt;/code&gt;&lt;br&gt;&lt;code&gt;-------&lt;/code&gt;&lt;br&gt;&lt;code&gt;K K | C&lt;/code&gt;&lt;br&gt;&lt;code&gt;K N | D&lt;/code&gt;&lt;br&gt;&lt;code&gt;N K | D&lt;/code&gt;&lt;br&gt;&lt;code&gt;N N | C&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's where we see the &quot;amazing&quot; part of the question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;H W | R R&lt;/code&gt;&lt;br&gt;&lt;code&gt;---------&lt;/code&gt;&lt;br&gt;&lt;code&gt;K K | A C&lt;/code&gt;&lt;br&gt;&lt;code&gt;K N | B D&lt;/code&gt;&lt;br&gt;&lt;code&gt;N K | B D&lt;/code&gt;&lt;br&gt;&lt;code&gt;N N | A C&lt;/code&gt;&lt;br&gt;&lt;br&gt;In order to satisfy the requirements that asking a single person gives you no information about their two tribes, we've forced the question to be in a form such that asking the spouse gives you no additional information. Asking once allows you to know whether or not the two are from the same tribe, but then you can gather no additional information by using the question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In summary&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This riddle can't be solved by attempting to derive the &quot;amazing question&quot;, because it cannot exist. Since the question can't exist, there is no way to solve the puzzle in its current form - why should the husbands and wives have to give symmetrical answers? Perhaps in a KK couple, the husband will answer 'Yes' while the wife answers 'No'. Or perhaps that's what a NN couple would answer. So we could have either everyone telling the truth, or no-one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;An alternative theory&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are asymmetrical questions that make this work, such as having the answer change depending on whether or not you have talked to their spouse.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A possible truth-table for the question under these conditions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;1 2 | R R&lt;/code&gt;&lt;br&gt;&lt;code&gt;---------&lt;/code&gt;&lt;br&gt;&lt;code&gt;K K | Y Y&lt;/code&gt;&lt;br&gt;&lt;code&gt;K N | N Y&lt;/code&gt;&lt;br&gt;&lt;code&gt;N K | N N&lt;/code&gt;&lt;br&gt;&lt;code&gt;N N | Y N&lt;/code&gt;&lt;br&gt;&lt;br&gt;A question that fits this would be &quot;If your spouse just answered my question did they say 'yes' otherwise if I asked you if you and your spouse are from the same tribe, would you say yes?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another possibility:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;1 2 | R R&lt;/code&gt;&lt;br&gt;&lt;code&gt;---------&lt;/code&gt;&lt;br&gt;&lt;code&gt;K K | Y N&lt;/code&gt;&lt;br&gt;&lt;code&gt;K N | N N&lt;/code&gt;&lt;br&gt;&lt;code&gt;N K | N Y&lt;/code&gt;&lt;br&gt;&lt;code&gt;N N | Y Y&lt;/code&gt;&lt;br&gt;&lt;br&gt;A question that fits this would be &quot;If your spouse just answered my question did they say 'no' otherwise if I asked you if you and your spouse are from the same tribe, would you say yes?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Comparing the two possibilities:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This still doesn't help us solve the riddle. This allows the question to exist, but we have no way to determine how many people lied. If it is the second question and we always ask the husband first, then he will say &quot;Yes&quot; and his wife will say &quot;No&quot; if they are both knights (nobody lies). If we ask the wife first and then the husband, she will say &quot;No&quot; and he will say &quot;Yes&quot; if she is a knave and he is a knight (50 people lie). If we're not necessarily consistent in who we ask first, then we could have anywhere between 0 and 50 liars. It wouldn't be too hard to formulate another question so that the husband saying &quot;Yes&quot; and the wife saying &quot;No&quot; could either be two knaves or a knight and a knave, depending on who is asked first. Then we could have between 50 and 100 liars. In short, nobody or everyone could be liars depending on the order in which they are asked and on what the question is.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A note:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The reason I considered this an alternate theory instead of my main theory is because it feels like cheating - we're essentially asking two different questions. Also, if you object to the &quot;if ... otherwise ...&quot; construction of the questions, you can put it into a form that  only uses 'and's and 'or's - &quot;Is it true that 1. I have talked to your spouse and they said x, or 2. I have not talked to your spouse and if I asked you...&quot;. (might not be exactly the way to translate it, but I don't feel like going through it to make sure right now)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1955" LastEditorUserId="1955" LastEditDate="2015-02-06T16:49:40.007" LastActivityDate="2015-02-06T16:49:40.007" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8591" PostTypeId="2" ParentId="8433" CreationDate="2015-02-05T22:43:31.680" Score="3" Body="&lt;p&gt;For something like this, simply constructing a specific outcome that meets the requirements isn't actually that hard, and I think is an often overlooked way of solving some puzzles. Here's an example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  a b c d e f g h total&#xA;a - 1 0 1 0 1 1 1   5&#xA;b 0 - 1 0 1 1 1 1   5&#xA;c 1 0 - 1 0 1 1 0   4&#xA;d 0 1 0 - 1 0 0 1   3&#xA;e 1 0 1 0 - 1 0 0   3&#xA;f 0 0 0 1 0 - 1 1   3&#xA;g 0 0 0 1 1 0 - 1   3&#xA;h 0 0 1 0 1 0 0 -   2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I started out with A beating everyone except D. It then immediately became obvious that B could not tie A. So I added another loss to A (against E). From there, I made sure C and D had mirror outcomes, and filled out the rest of the table making sure to keep everyone below 5 wins. The whole thing only required backtracking once.&lt;/p&gt;&#xA;" OwnerUserId="4687" LastActivityDate="2015-02-05T22:43:31.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8592" PostTypeId="2" ParentId="8560" CreationDate="2015-02-05T22:52:49.487" Score="3" Body="&lt;p&gt;Yes! I think I did it! The trick was&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; to 'bag' the coins :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note for the below I skip trying to follow 'symmetric' and 'easier' paths to the solution. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Step 1.&lt;br&gt;&#xA;Divide coins in 2 equal groups of 27. A, B&lt;br&gt;&#xA;Weigh A vs B&lt;br&gt;&#xA;If A == B you are done&lt;br&gt;&#xA;If A &gt; B go to step 2&lt;br&gt;&#xA;If A &amp;lt; B the solution is symmetric to step 2&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;Step 2&lt;br&gt;&#xA;Take 13 coins away from A and put in CA.&lt;br&gt;&#xA;Take 13 coins away from B and put in CB.&lt;br&gt;&#xA;Weigh A+CB with B+CA&lt;br&gt;&#xA;If A+CB == B+CA you are done&lt;br&gt;&#xA;If A+CB &gt; B+CA goto step 3&lt;br&gt;&#xA;If A+CB &amp;lt; B+CA goto step 3 (knowing A==B)&lt;br&gt;&#xA;&lt;br&gt;&#xA;Step 3&lt;br&gt;&#xA;We know that A&gt;B from measurement 1, and that CA==CB since they didn't change the sum weight.&lt;br&gt;&#xA;A and B each have 14 coins left.&lt;br&gt;&#xA;Take 7 coins away from A and put in DA.&lt;br&gt;&#xA;Take 7 coins away from B and put in DB.&lt;br&gt;&#xA;Weigh A+DB and B+DA&lt;br&gt;&#xA;If B+DA == A+DB you are done&lt;br&gt;&#xA;If A+DB &gt;  B+DA goto step 4&lt;br&gt;&#xA;If A+DB &amp;lt;  B+DA similar to step 4 (knowing A==B)&lt;br&gt;&#xA;&lt;br&gt;&#xA;Step 4&lt;br&gt;&#xA;We know that A&gt;B from measurement 1, and that DA==DB since they didn't change the weight.&lt;br&gt;&#xA;A and B each have 7 coins remaining&lt;br&gt;&#xA;Take away 3 coins from A calling them EA, and 3 coins from B calling them EB.&lt;br&gt;&#xA;Weigh  A+EB and B+EA&lt;br&gt;&#xA;If A+EB == B+EA you are done&lt;br&gt;&#xA;If A+EB &gt;  B+EA goto step 5&lt;br&gt;&#xA;If A+EB &amp;lt;  B+EA similar to step 5 (knowing A==B)&lt;br&gt;&#xA;&lt;br&gt;&#xA;Step 5&lt;br&gt;&#xA;We know that A&gt;B from measurement 1, and that EA==EB since they didn't change the weight.&lt;br&gt;&#xA;A and B each have 4 coins remaining&lt;br&gt;&#xA;Take away 2 coins from A calling them FA, and 2 coins from B calling them FB.&lt;br&gt;&#xA;Weigh  A+FB and B+FA&lt;br&gt;&#xA;If A+FB == B+FA you are done&lt;br&gt;&#xA;If A+FB &gt;  B+FA then either A or B has both the two fake coins (you found them!)&lt;br&gt;&#xA;If A+FB &amp;lt;  B+FA then either FA or FB has both the two fake coins (you found them!)&lt;br&gt;&#xA;&lt;br&gt;&#xA;Since either A or B, or, FA or FB, has both fake coins, take 1 coin from each and distribute.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9302" LastEditorUserId="9302" LastEditDate="2015-02-05T23:17:57.047" LastActivityDate="2015-02-05T23:17:57.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8593" PostTypeId="2" ParentId="8556" CreationDate="2015-02-05T22:55:19.550" Score="6" Body="&lt;p&gt;Sequence 1 &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Number of matches played in the FIFA World Cups:&lt;br&gt;&amp;nbsp; &amp;nbsp; &lt;strong&gt;18&lt;/strong&gt;, &lt;strong&gt;17&lt;/strong&gt;, &lt;strong&gt;18&lt;/strong&gt;, &lt;strong&gt;22&lt;/strong&gt;, &lt;strong&gt;26&lt;/strong&gt;, &lt;strong&gt;35&lt;/strong&gt;, &lt;strong&gt;32&lt;/strong&gt;, &lt;strong&gt;32&lt;/strong&gt;, &lt;strong&gt;32&lt;/strong&gt;, &lt;strong&gt;38&lt;/strong&gt;, &lt;strong&gt;38&lt;/strong&gt;, &lt;strong&gt;52&lt;/strong&gt;, &lt;strong&gt;52&lt;/strong&gt;, &lt;strong&gt;52&lt;/strong&gt;, &lt;strong&gt;52&lt;/strong&gt;, &lt;strong&gt;64&lt;/strong&gt;,...&lt;br&gt;  &lt;a href=&quot;https://en.wikipedia.org/wiki/FIFA_World_Cup#Attendance&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sequence 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stars in the Big Dipper (the seven brightest stars in the constellation Ursa Major):&lt;br&gt; &amp;nbsp; &amp;nbsp; &lt;strong&gt;Du&lt;/strong&gt;bhe, &lt;strong&gt;Me&lt;/strong&gt;rak, &lt;strong&gt;Ph&lt;/strong&gt;ecda, &lt;strong&gt;Me&lt;/strong&gt;grez, &lt;strong&gt;Al&lt;/strong&gt;ioth, &lt;strong&gt;Mi&lt;/strong&gt;zar, &lt;strong&gt;Al&lt;/strong&gt;kaid&lt;br&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Big_Dipper#Stars&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sequence 3 &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Popes:&lt;br&gt; &amp;nbsp; &amp;nbsp; ..., Benedict XIV (&lt;strong&gt;14&lt;/strong&gt;), Clement XIII (&lt;strong&gt;13&lt;/strong&gt;), Clement XIV (&lt;strong&gt;14&lt;/strong&gt;), Pius VI (&lt;strong&gt;6&lt;/strong&gt;), Pius VII (&lt;strong&gt;7&lt;/strong&gt;), Leo XII (&lt;strong&gt;12&lt;/strong&gt;), Pius VIII (&lt;strong&gt;8&lt;/strong&gt;), Gregory XVI (&lt;strong&gt;16&lt;/strong&gt;), Pius IX (&lt;strong&gt;9&lt;/strong&gt;), Leo XIII (&lt;strong&gt;13&lt;/strong&gt;), Pius X (&lt;strong&gt;10&lt;/strong&gt;), Benedict XV (&lt;strong&gt;15&lt;/strong&gt;), Pius XI (&lt;strong&gt;11&lt;/strong&gt;), Pius XII (&lt;strong&gt;12&lt;/strong&gt;), John XXIII (&lt;strong&gt;23&lt;/strong&gt;), Paul VI (&lt;strong&gt;6&lt;/strong&gt;), John Paul I (&lt;strong&gt;1&lt;/strong&gt;), John Paul II (&lt;strong&gt;2&lt;/strong&gt;), Benedict XVI (&lt;strong&gt;16&lt;/strong&gt;), Francis I (&lt;strong&gt;1&lt;/strong&gt;)&lt;br&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_popes&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sequence 4 &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Number of strokes needed to write Roman numerals:&lt;br&gt; &amp;nbsp; &amp;nbsp; I (&lt;strong&gt;1&lt;/strong&gt;), II (&lt;strong&gt;2&lt;/strong&gt;), III (&lt;strong&gt;3&lt;/strong&gt;), IV (&lt;strong&gt;3&lt;/strong&gt;), V (&lt;strong&gt;2&lt;/strong&gt;), VI (&lt;strong&gt;3&lt;/strong&gt;), VII (&lt;strong&gt;4&lt;/strong&gt;), VIII (&lt;strong&gt;5&lt;/strong&gt;), IX (&lt;strong&gt;3&lt;/strong&gt;), X (&lt;strong&gt;2&lt;/strong&gt;), XI (&lt;strong&gt;3&lt;/strong&gt;), XII (&lt;strong&gt;4&lt;/strong&gt;), XIII (&lt;strong&gt;5&lt;/strong&gt;), XIV (&lt;strong&gt;5&lt;/strong&gt;), XV (&lt;strong&gt;4&lt;/strong&gt;), XVI (&lt;strong&gt;5&lt;/strong&gt;), XVII (&lt;strong&gt;6&lt;/strong&gt;), XVIII (&lt;strong&gt;7&lt;/strong&gt;),...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sequence 5&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Presidents of the United States:&lt;br&gt; &amp;nbsp; &amp;nbsp;&lt;strong&gt;W&lt;/strong&gt;ashington, &lt;strong&gt;A&lt;/strong&gt;dams, &lt;strong&gt;J&lt;/strong&gt;efferson, &lt;strong&gt;M&lt;/strong&gt;adison, &lt;strong&gt;M&lt;/strong&gt;onroe, &lt;strong&gt;A&lt;/strong&gt;dams, &lt;strong&gt;J&lt;/strong&gt;ackson, &lt;strong&gt;V&lt;/strong&gt;an Buren, &lt;strong&gt;H&lt;/strong&gt;arrison, &lt;strong&gt;T&lt;/strong&gt;yler, &lt;strong&gt;P&lt;/strong&gt;olk, &lt;strong&gt;T&lt;/strong&gt;aylor, &lt;strong&gt;F&lt;/strong&gt;illmore, &lt;strong&gt;P&lt;/strong&gt;ierce, &lt;strong&gt;B&lt;/strong&gt;uchanan, &lt;strong&gt;L&lt;/strong&gt;incoln, &lt;strong&gt;J&lt;/strong&gt;ohnson, &lt;strong&gt;G&lt;/strong&gt;rant, &lt;strong&gt;H&lt;/strong&gt;ayes, &lt;strong&gt;G&lt;/strong&gt;arfield, &lt;strong&gt;A&lt;/strong&gt;rthur, &lt;strong&gt;C&lt;/strong&gt;leveland, &lt;strong&gt;H&lt;/strong&gt;arrison, &lt;strong&gt;C&lt;/strong&gt;leveland, &lt;strong&gt;M&lt;/strong&gt;cKinley, &lt;strong&gt;R&lt;/strong&gt;oosevelt, &lt;strong&gt;T&lt;/strong&gt;aft, &lt;strong&gt;W&lt;/strong&gt;ilson, &lt;strong&gt;H&lt;/strong&gt;arding, &lt;strong&gt;C&lt;/strong&gt;oolidge,...&lt;br&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_Presidents_of_the_United_States&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="5188" LastEditDate="2015-02-06T01:04:47.597" LastActivityDate="2015-02-06T01:04:47.597" CommentCount="0" CommunityOwnedDate="2015-02-05T22:55:19.550" ContentLicense="CC BY-SA 3.0" />
  <row Id="8594" PostTypeId="1" AcceptedAnswerId="8595" CreationDate="2015-02-06T03:03:24.860" Score="5" ViewCount="304" Body="&lt;p&gt;As the apology for the mischievous problem about knights and knaves:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  PRIME&#xA;  PRIME&#xA;+ PRIME&#xA;-------&#xA; NUMBER&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Yeah, this was inspired by the famous (or infamous) Goldbach Conjecture.&lt;/p&gt;&#xA;" OwnerUserId="3914" LastEditorUserId="9404" LastEditDate="2016-03-03T10:20:43.217" LastActivityDate="2016-03-03T22:10:39.093" Title="Goldbach Cryptarithm" Tags="|mathematics|alphametic|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8595" PostTypeId="2" ParentId="8594" CreationDate="2015-02-06T04:52:07.530" Score="2" Body="&lt;p&gt;If PRIME is a prime number, then the solution is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;there is no solution&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If PRIME is not a prime number, the solution is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;PRIME = 54328&lt;/strong&gt;, NUMBER = 162984  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Found an &lt;a href=&quot;http://www.tkcs-collins.com/truman/alphamet/alpha_solve.shtml&quot; rel=&quot;nofollow&quot;&gt;alphametic solver&lt;/a&gt; for faster solutions.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-02-06T18:05:14.583" LastActivityDate="2015-02-06T18:05:14.583" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8596" PostTypeId="1" AcceptedAnswerId="8597" CreationDate="2015-02-06T09:08:05.320" Score="5" ViewCount="867" Body="&lt;p&gt;Kenny's a serial killer and he likes to play games. To make his killing spree more interesting, he leaves clues for the detectives on his next victim. One day, the police station received a message as below:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  Kenny: &quot;;9;9;jlk &quot;&#xA;&#xA;  Kenny: &quot;l l&quot;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The police were able to immediately trace where the message was coming from and went to the source destination. Upon arrival, they saw nothing special but a grand piano with Beethoven's music sheets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who's the next target?&lt;/p&gt;&#xA;" OwnerUserId="8882" LastEditorUserId="8874" LastEditDate="2015-11-01T15:56:53.560" LastActivityDate="2015-11-01T15:56:53.560" Title="Who's the next target?" Tags="|cipher|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8597" PostTypeId="2" ParentId="8596" CreationDate="2015-02-06T09:44:07.427" Score="8" Body="&lt;p&gt;It's Elise. With cup-size &quot;double d&quot; :D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;/edit: Second thought: it could be Elise's Dad :|&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explanation:&#xA;one of the most famous play from beethoven is &quot;For Elise&quot;. It starts with the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YU648.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First message from Kenny are the starting notes. Regarding the second message and combining the notes, [L L] results in &quot;DAD&quot;.&lt;/p&gt;&#xA;" OwnerUserId="9197" LastEditorUserId="9197" LastEditDate="2015-02-06T10:47:43.230" LastActivityDate="2015-02-06T10:47:43.230" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="8598" PostTypeId="2" ParentId="8481" CreationDate="2015-02-06T09:49:40.357" Score="9" Body="&lt;p&gt;There are 5 possible solutions that use 7 steps:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 16 -&gt; 7 -&gt; 14 -&gt; 28 -&gt; 13 -&gt; 26 -&gt; 52 -&gt; 25&lt;br&gt;&#xA; 16 -&gt; 7 -&gt; 14 -&gt; 28 -&gt; 56 -&gt; 27 -&gt; 12.5 -&gt; 25&lt;br&gt;&#xA; 16 -&gt; 32 -&gt; 15 -&gt; 6.5 -&gt; 13 -&gt; 26 -&gt; 52 -&gt; 25&lt;br&gt;&#xA; 16 -&gt; 32 -&gt; 64 -&gt; 31 -&gt; 14.5 -&gt; 6.25 -&gt; 12.5 -&gt; 25&lt;br&gt;&#xA; 16 -&gt; 1616 -&gt; 808 -&gt; 404 -&gt; 202 -&gt; 100 -&gt; 50 -&gt; 25  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are no solutions that use less than 7 steps. And probably a lot that use 8 or more.&lt;/p&gt;&#xA;" OwnerUserId="9352" LastEditorUserId="9352" LastEditDate="2015-02-06T09:54:51.550" LastActivityDate="2015-02-06T09:54:51.550" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8599" PostTypeId="2" ParentId="8562" CreationDate="2015-02-06T12:35:21.460" Score="5" Body="&lt;p&gt;Assuming I have correctly digested Rob Watts' excellent answer, the number of people who lied is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One.&lt;br&gt;&#xA; There is no question that fits the description given. Therefore, a knight certainly would not have told you that one did, and you could not have asked it of 50 couples at a party. You made the whole thing up - you're the liar!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6031" LastActivityDate="2015-02-06T12:35:21.460" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8600" PostTypeId="1" AcceptedAnswerId="8602" CreationDate="2015-02-06T12:47:23.060" Score="23" ViewCount="2280" Body="&lt;p&gt;A well-known crooked casino is offering the following bet with a pair of (identically looking and indistinguishable) dice that both show the numbers $1,2,3,4,5,6$ on their faces.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;The first die is fair and perfectly standard: throwing it yields each of the outcomes $1,2,3,4,5,6$ with equal probability $1/6$.&lt;/li&gt;&#xA;  &lt;li&gt;The second die is loaded: throwing it yields each of the five outcomes $1,2,3,4,5$ with equal probability $1/5$ (while $6$ is impossible to throw).&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We stress that there is no way whatsoever for an honest gambler to tell the two dice from each other.&#xA;Now, in the bet that we are considering, the gambler is allowed to throw these dice as often as he likes, subject to the following casino rules:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;The gambler has to pay $1$ Euro for every single throw of a single die.&lt;/li&gt;&#xA;  &lt;li&gt;Whenever he switches from throwing one die to throwing the other die, he must pay $5$ Euro.&lt;br&gt;&#xA;  For his very first throw, he may arbitrarily choose either of the dice (without paying anything).&lt;/li&gt;&#xA;  &lt;li&gt;As soon as he throws a $6$, the game is over. The gambler wins $20$ Euro.&lt;/li&gt;&#xA;  &lt;li&gt;At any moment in time, the gambler may decide to quit the game and leave the table at no further cost. &lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: What is the &lt;strong&gt;best strategy&lt;/strong&gt; for this game and what is the &lt;strong&gt;expected profit&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-10-04T16:11:46.543" LastActivityDate="2015-10-04T16:11:46.543" Title="Gambling in a rigged casino" Tags="|mathematics|strategy|probability|dice|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8601" PostTypeId="2" ParentId="8600" CreationDate="2015-02-06T13:55:15.377" Score="4" Body="&lt;p&gt;I'm not expecting that this is the best answer, just helping out the thoughts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a baseline, the strategy to sit down and roll once and walk away is a positive bet.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$EV = (\frac{1}{2})(-1) + (\frac{1}{2})(\frac{1}{6})(19) + (\frac{1}{2})(\frac{5}{6})(-1) = \frac{2}{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a 50% chance to get the bad die and finish -1.  There's a 50% chance of getting the good die and then a 1/6 chance of winning and coming out +19 or a 5/6 chance of coming out -1.&lt;/p&gt;&#xA;" OwnerUserId="1777" LastEditorUserId="1777" LastEditDate="2015-02-06T18:23:15.730" LastActivityDate="2015-02-06T18:23:15.730" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8602" PostTypeId="2" ParentId="8600" CreationDate="2015-02-06T15:40:55.400" Score="9" Body="&lt;p&gt;What follows is a strategy for earning an expected profit of about $3.63$ Euro, and an argument that this strategy is optimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First we'll pick some positive integer $k$. We will switch dice after every roll number that is an odd multiple of $k$, and continue to roll until we get a 6. So we will roll the first die $k$ times, then the second die $2k$ times, then the first die $2k$ times, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With probability $1$ we will eventually roll a 6 and win the 20 Euro. To compute the expected cost, we consider two cases, depending on which die we pick up first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First die is fair&lt;/strong&gt;.&#xA;We expect to roll the fair die $6$ times. We expect to roll the loaded die&#xA;$$&#xA;\sum_{n\geq 0}\left(\frac{5}{6}\right)^{2nk}2k = \frac{2k\cdot(5/6)^k}{1-(5/6)^{2k}}&#xA;$$&#xA;times. The expected number of switches is&#xA;$$&#xA;\left(\frac{5}{6}\right)^k+2\cdot\sum_{n\geq 1}\left(\frac{5}{6}\right)^{(2n+1) k} = \frac{2(5/6)^k}{1-(5/6)^{2k}}.&#xA;$$&#xA;So if we start with the fair die, our expected profit is&#xA;$$&#xA;20-6-\frac{2k\cdot(5/6)^k}{1-(5/6)^{2k}}-5\cdot\frac{2(5/6)^k}{1-(5/6)^{2k}}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First die is loaded&lt;/strong&gt;. We again expect to roll the fair die $6$ times. We expect to roll the loaded die&#xA;$$&#xA;k+\sum_{n\geq 1}\left(\frac{5}{6}\right)^{2nk} 2k=\frac{k\cdot(1+(5/6)^{2k})}{1-(5/6)^{2k}}&#xA;$$&#xA;times. The expected number of switches is&#xA;$$&#xA;1+2\cdot\sum_{n\geq 1}\left(\frac{5}{6}\right)^{2nk}=\frac{1+(5/6)^{2k}}{1-(5/6)^{2k}}.&#xA;$$&#xA;So if we start with the loaded die, our expected profit is&#xA;$$&#xA;20-6-\frac{k\cdot(1+(5/6)^{2k})}{1-(5/6)^{2k}}-5\cdot \frac{1+(5/6)^{2k}}{1-(5/6)^{2k}}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Expected profit&lt;/strong&gt; Our overall expected profit is the mean of the expected profit if the first die is fair and the expected profit if the first die is loaded. The expression simplifies to&#xA;$$&#xA;14-\frac{1}{2}\frac{(5+k)\left(1+(5/6)^{k}\right)^2}{1-(5/6)^{2k}}.&#xA;$$&#xA;Numerically, I found that profit is maximized taking $k=9$. The expected profit using this strategy is&#xA;$$&#xA;\frac{4218321}{1160653}\approx 3.63\,\,\,\text{Euro}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is this strategy optimal?&lt;/strong&gt; Here's a sketch of a proof that this strategy is optimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that the &lt;em&gt;odds&lt;/em&gt; a given die is loaded is the ratio&#xA;$$&#xA;\frac{\text{probability that die is fair}}{\text{probability that die is loaded}}.&#xA;$$&#xA;The product of the odds that each die is fair is $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During the course of the game, we only get information of one kind, namely that the die we just rolled did not turn up 6 (if it turns up 6, the game ends). Bayes' Theorem says that whenever a die turns up something other than 6, the odds that it is fair get multiplied by $\frac{5}{6}$ (and so the odds the other die is fair get multiplied by $\frac{6}{5}$). This means that assuming no 6's have been rolled, the odds that a die is fair depend &lt;em&gt;only&lt;/em&gt; on the different between the number of times that die has been rolled, and the number of times the other die has been rolled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we are deciding whether to roll our current die, switch dice, or walk away, we need only consider the odds our current die is fair. Any costs already paid (sunk costs) should not be taken into account, as future payments and earnings do not depend on them. This means that possible strategies fall into two possible kinds:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the odds the current die is fair are below a certain threshold, leave the game. Otherwise roll.&lt;/li&gt;&#xA;&lt;li&gt;If the odds the current die is fair are below a certain threshold, switch dice. Otherwise roll.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The strategy described above is of the second kind, where the threshold is $\left(\frac{5}{6}\right)^k$. We choose $k$ to optimize profit among strategies of this kind. The only other possibility is a strategy where we roll the current die $n$ times (for some fixed $n$), then leave the game if we haven't rolled a 6. It's not too hard to write down the expected profit here as a function of $n$, find the $n$ that maximizes profit (if I computed correctly it is $n=5$), and see that this doesn't do better.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-02-07T04:39:06.983" LastActivityDate="2015-02-07T04:39:06.983" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8603" PostTypeId="2" ParentId="8600" CreationDate="2015-02-06T16:10:16.197" Score="2" Body="&lt;p&gt;Following from Jason's answer, if we ignore the possibility of switching dice, we can look at this as a simple game where paying &amp;euro;1 gives you a 1/12 chance of winning &amp;euro;20, and 11/12 chance of getting nothing. It is clear that this simplified version is a winning game. The chance of playing 20 times without winning is $\left(\dfrac{11}{12}\right)^{20}\approx0.175$, so you have about $82.5\%$ chance of winning (or breaking even). Furthermore, if you play this game indefinitely, until you win, your expected return is&#xA;$$\sum_{n=0}^{\infty}\left(\frac{1}{12}\right)\left(\frac{11}{12}\right)^n&#xA;\times(19-n)=8$$&#xA;So, on average, a player of this simplified game will win &amp;euro;8. Of course, this isn't a completely accurate picture, since well before you get to 20 failed rolls, it becomes much more likely that you have the rigged die, and will &lt;em&gt;never&lt;/em&gt; win. So, the real question is, at what point does it become worthwhile to pay the &amp;euro;5 to switch?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that you should play up to 7 times, and if you haven't won by then, switch dice and and play up to 7 more times. I will try to verify (or disprove) this, and then update my answer.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&lt;br&gt;&#xA;If you choose the fair die initially, your expected return after 7 rolls is&#xA;$$\sum_{n=0}^6\left(\frac16\right)\left(\frac56\right)^n(19-n)\approx12.05$$&#xA;However, you also have a $\left(\dfrac56\right)^7\approx28\%$ chance of losing, in which case you will switch and continue with the loaded die, losing a total of &amp;euro;19, which lowers the expected return to about &amp;euro;6.75&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you choose the loaded die first, you will end up in the same &amp;euro;12-return situation as above, except after spending &amp;euro;12 on 7 die rolls and one switch, giving an expected return of about &amp;euro;0.05. Well, it's better than nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that these two situations are equally likely, this strategy gives an expected return of about $(6.75+0.05)/2=3.4$.&lt;/p&gt;&#xA;" OwnerUserId="8776" LastEditorUserId="8776" LastEditDate="2015-02-06T19:06:54.393" LastActivityDate="2015-02-06T19:06:54.393" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8604" PostTypeId="1" AcceptedAnswerId="9011" CreationDate="2015-02-06T16:52:20.457" Score="14" ViewCount="5525" Body="&lt;p&gt;I have recently bought a fair $10$-sided dice for generating equally distributed random numbers from the range $1,2,\ldots,10$. Let me describe this dice in some more detail. Consider a regular 5-gon $ABCDE$ in 3-dimensional space that is situated in the $xy$-plane so that its center lies in the origin. Consider a further point $F$ on the $z$-axis somewhere above the origin, and a point $G$ that results by reflecting $F$ across the origin. If one connects every corner of the 5-gon $ABCDE$ by two edges to $F$ and $G$, this will yield a polyhedron $P_{10}$ with 10 congruent triangular faces. My fair 10-sided dice is essentially $P_{10}$ with the sides labeled by the numbers 1,2,...,10. Throw it up into the air, look how it lands, and take the side on which it lands as your random number. (With a cube one always takes the top side, but $P_{10}$ has no top side). By the symmetry of $P_{10}$ it is obvious that all $10$ numbers are equally probable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I modify this construction and replace the regular 5-gon by a regular $n$-gon, I shall get a fair $2n$-sided dice with an even number of sides.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My &lt;strong&gt;question&lt;/strong&gt; is whether there also exist &lt;strong&gt;fair&lt;/strong&gt; $(2n+1)$-sided dice with an odd number of sides. By a &lt;strong&gt;fair&lt;/strong&gt; dice I mean &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a $(2n+1)$-sided polyhedron with $2n+1$ congruent sides,&lt;/li&gt;&#xA;&lt;li&gt;so that the probability of landing on each of its sides is precisely $\frac{1}{2n+1}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="8874" LastEditDate="2016-03-23T18:27:28.117" LastActivityDate="2018-10-29T01:45:19.720" Title="Fair n-sided dice" Tags="|mathematics|geometry|probability|dice|three-dimensional|" AnswerCount="7" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8605" PostTypeId="1" AcceptedAnswerId="8696" CreationDate="2015-02-06T17:04:58.170" Score="5" ViewCount="317" Body="&lt;p&gt;This is my third batch of sequence puzzles that are nasty and hard to solve; &#xA;yet, each of them has a clear and justifiable solution.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sequence 1:&lt;br&gt;&#xA;  2, 4, 4, 3, ?, ?, 3, 4, 2, 4, 3, 4, 2, 2, 3, 4, 4, 3, ?, ?, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sequence 2:&lt;br&gt;&#xA;  ?, ?, ?, J, J, ?, A, M, ?, J, J, Z, M, F, J, A, A, U, R, J, C, G, B, G, W, T, W, W, W, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sequence 3:&lt;br&gt;&#xA;  ??, ??, Ca, Cl, Ne, Ga, ??, ??, ??, Ti, ??, Ne, ??, ??, ??, Au, Lu, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sequence 4:&lt;br&gt;&#xA;  0.06, 0.82, 1.00, 0.11, ??, ??, 14.6, 17.2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sequence 5:&lt;br&gt;&#xA;  A, E, A, P, A, ?, U, ?, E, ?, ?, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="9125" LastEditDate="2015-02-10T09:27:40.043" LastActivityDate="2015-02-10T09:30:45.930" Title="Yet another group of five sequence puzzles" Tags="|number-sequence|letter-sequence|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8606" PostTypeId="2" ParentId="8605" CreationDate="2015-02-06T17:10:12.307" Score="7" Body="&lt;p&gt;I could google sequence 4:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sequence 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;The Evolving Universe and the Origin of Life: The Search for Our Cosmic Roots&quot;&lt;br&gt;&#xA; Mass of the planets in our solar system measured relative to Earth&lt;br&gt;&#xA;I have googled0.06 0.82 1.00 0.11 317.9 95.2 14.6 17.2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-02-06T17:10:12.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8607" PostTypeId="2" ParentId="8604" CreationDate="2015-02-06T17:29:49.370" Score="6" Body="&lt;p&gt;I can loosely prove that such dice exist for n &gt; 1, though actually constructing such a die might be trickier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a pyramid whose base is a regular 2n-sided polygon, with a height h. This shape has 2n+1 faces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As h approaches 0, it becomes more and more likely that the die lands on its base, asymptotically approaching 1/2 probability, as this shape approaches a flat &quot;coin&quot;. As h approaches infinity, it becomes less and less likely that the die lands on its base, asymptotically approaching 0 probability. I claim that the probability of landing on the base is continuous, and so there must be some value for the height such that the probability of landing on the base is 1/(2n+1), as 0 &amp;lt; 1/(2n+1) &amp;lt; 1/2. The rest of the faces then total (2n)/(2n+1) probability, and by symmetry must each also have a 1/(2n+1) probability of coming up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar construction is also possible with a (2n-1)-polygon-prism instead of a (2n)-polygon-pyramid.&lt;/p&gt;&#xA;" OwnerUserId="4687" LastActivityDate="2015-02-06T17:29:49.370" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8608" PostTypeId="2" ParentId="8605" CreationDate="2015-02-06T17:48:33.213" Score="8" Body="&lt;p&gt;Sequence 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First initials of US Presidents, in order of when they were in office. &lt;strong&gt;G&lt;/strong&gt;eorge Washington, &lt;strong&gt;J&lt;/strong&gt;ohn Adams, &lt;strong&gt;T&lt;/strong&gt;homas Jefferson, &lt;strong&gt;J&lt;/strong&gt;ames Madison, &lt;strong&gt;J&lt;/strong&gt;ames Monroe, &lt;strong&gt;J&lt;/strong&gt;ohn Quincy Adams, &lt;strong&gt;A&lt;/strong&gt;ndrew Jackson, &lt;strong&gt;M&lt;/strong&gt;artin Van Buren, &lt;strong&gt;W&lt;/strong&gt;illiam Henry Harrison, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4687" LastActivityDate="2015-02-06T17:48:33.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8609" PostTypeId="2" ParentId="8326" CreationDate="2015-02-06T18:17:17.747" Score="3" Body="&lt;p&gt;Upon request, here is the solution I had in mind:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7duS8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This solution is in essence the same as arbitrary's, as it is based on the same icosahedral graph (the smallest planar regular graph with vertex connectivity 5):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zMwFl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7573" LastEditorUserId="7573" LastEditDate="2015-02-06T18:23:58.677" LastActivityDate="2015-02-06T18:23:58.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8610" PostTypeId="2" ParentId="6212" CreationDate="2015-02-06T18:48:20.017" Score="2" Body="&lt;p&gt;Hint 12:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ASCII85 to Hex to ASCII&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;wVÒ­Â&amp;amp;ì~*÷X/ô,@=27t-òFpµv=2Qõ¼Ïnïo4 giñO?&lt;br&gt;&#xA;  ú2RÜ¬më÷s#%]Zao«º·k;;|iªRÊ&gt;ð.g³¹Óµ7@H.H.%/??FèÊî¤dàI(Ûo¢Ü½&lt;br&gt;&#xA;  s ?Ei¸gfjÔ¿Myþx_ÞxI{Z¡u³'&gt;&lt;br&gt;&#xA;  Õ½¯öËÕ&amp;amp;!ªw,á©s©Î×Êq»¦sG¾+OyØxDø@Ü&lt;br&gt;&#xA;  pâ÷ä~²Æ½±/í'.ôNFZ^AkÞ+¸l$ß!  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5368" LastActivityDate="2015-02-06T18:48:20.017" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8611" PostTypeId="1" AcceptedAnswerId="8614" CreationDate="2015-02-06T19:00:01.107" Score="-5" ViewCount="333" Body="&lt;h2&gt;Secrets Revisited&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;A friend sent me the following and wanted to find out if anyone could figure it out. The message was encrypted three times, and all keys and primers were provided. It appears that there are some fairly obvious patterns that might be exploitable. The plaintext is not necessary for the answer, just the name of the story. Kudos go to whoever can explain how the keys and primers relate to the given ciphertext.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Key 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;xmjfdlzquaoyvbkswcipnehrtg&#xA;bctmalxrhknjgqfvdusyieopwz&#xA;jqkywbacpdlmofusthzxiengrv&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Key 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;HRCXYEPMWDLQFAOGBVZKNSJTUI&#xA;ZJWPNHXYQOGDISKRACVLTUBEFM&#xA;JFYXASMZWKDOCQEIHNGBTUVLRP&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Key 3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;%&quot;'&amp;gt;^\#[&amp;lt;&amp;amp;~_();+:.}]$?{/!@,`|=*-&#xA;%:!?;~]$@'+.-*)}=&amp;amp;&amp;gt;{(&amp;lt;[|_#\^/,`&quot;&#xA;#}[._`!&quot;:%(\@&amp;gt;+?&amp;amp;{^]'$=*,)&amp;lt;~/;|-&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Primer 1:&lt;/strong&gt; &lt;code&gt;tk&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Primer 2:&lt;/strong&gt; &lt;code&gt;ZA&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Primer 3:&lt;/strong&gt; &lt;code&gt;_?&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ciphertext:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;O qc Mkh&quot;&#xA;Ikh J kh)&#xA;&#xA;Reno Zxv)T;kh!&#xA;Reno Zxv\T.kh!&#xA;B cv kwv crnd neoo Kxv\G.kh!&#xA;&#xA;Qj tgq tsnd qnuhg klak ypg tef)&#xA;&#xA;Y cv kwv crnd nekg= Ojh-G:kh\&#xA;B cv kwv crnd qnuhg klak ypg tef^&#xA;&#xA;Tjnto bjq tsnd nekg atds zr hmkds?&#xA;&#xA;O kmdto jpv crnd nekg atds zr hmkds|&#xA;Z kmdto jpv crnd nekg jrnjmuds)&#xA;D ei kwv crnd qnuhg klak ypg tef;&#xA;D cv kwv crnd nekg( Kjh%G:kh\&#xA;&#xA;Yjnto bjq tsnd nekg oz b oxmlt_&#xA;Cwxto bjq tsnd nekg tsht o fjnlt&amp;gt;&#xA;&#xA;T li kwv crnd nekg oz b oxmlt`&#xA;Z li kwv crnd nekg tsht o fjnlt)&#xA;D li kwv crnd nekg atds zr hmkds;&#xA;D ei kwv crnd nekg jrnjmuds;&#xA;D ei kwv crnd qnuhg klak ypg tef;&#xA;D cv kwv crnd nekg( Kjh%G:kh\&#xA;&#xA;Yjnto bjq cka plkg oz b dnn_&#xA;Cjato bjq cka plkg tsht o hux&amp;gt;&#xA;&#xA;Cju kh b dnn`&#xA;Mju jlht o hux)&#xA;Rju kh b oxmlt;&#xA;Rwg jlht o fjnlt;&#xA;J tmdto jpv oha plkg atds zr hmkds;&#xA;F kmdto jpv oha plkg jrnjmuds;&#xA;D kmdto jpv oha zduhg klak ypg tef;&#xA;D cv kwv crnd nekg( Kjh%G:kh\&#xA;&#xA;Yjnto bjq' sctto bjq- ek b lbl~&#xA;Lrs plkg. Tjf plkg' Jzrs yeky pkm]&#xA;&#xA;M kmdto jpv+ ugtto jpv- kh b lbl&amp;gt;&#xA;&#xA;Dyz gdu wnnd nekg: Hwn arkp uqz;&#xA;Kxx gdu wnnd nekg oz b hcgh&quot;&#xA;&#xA;X umdto jpv; ugtto jpv' kh b hcgh&amp;gt;&#xA;Mxg kh b lbl&amp;gt; Dyz tyw xw qt:&#xA;A zi kwv crnd nekg oz b dnn.&#xA;C cm kwv crnd nekg tsht o hux;&#xA;D cm kwv crnd nekg oz b oxmlt;&#xA;D li kwv crnd nekg tsht o fjnlt;&#xA;D li kwv crnd nekg atds zr hmkds;&#xA;D ei kwv crnd nekg jrnjmuds;&#xA;D ei kwv crnd qnuhg klak ypg tef;&#xA;D cv kwv crnd nekg( Kjh%G:kh\&#xA;&#xA;W qgedt/ W ziedt&amp;gt; J ziedt# J ziedt#&#xA;Umbto bjq. afdto bjq' me j hcedt~&#xA;&#xA;Imv kh b hcedt. Nmv kh b hcgh'&#xA;Rxg kh b lbl# Pru. gfw xw qt/&#xA;Z gmdto jpv: ugtto jpv' kh b dnn&amp;gt;&#xA;B wttto jpv; jydto jpv- jlht o hux&amp;gt;&#xA;D odkp fav oha plkg tsht o fjnlt:&#xA;D tekp fav oha plkg oz b oxmlt;&#xA;D tekp fav oha plkg atds zr hmkds;&#xA;D kekp fav oha plkg jrnjmuds;&#xA;D ei kwv crnd qnuhg klak ypg tef;&#xA;D cv kwv crnd nekg( Kjh%G:kh\&#xA;&#xA;Hku/ Gi ouk wdjq/ Gqis xv zuk wdjq}&#xA;Bcsto bjq) sctto bjq' ek zuk wdjq~&#xA;&#xA;U gwdto jpv| ugtto jpv* kh zuk wdjq&amp;gt;&#xA;&#xA;Pcsto bjq; sctto bjq- ek zuk dydt~&#xA;&#xA;O akdto jpv| ugtto jpv kh zuk dydt%&#xA;Tmv kh zuk wdjq: Uck nf j hcedt;&#xA;Rmv kh b lbl; Ryl kh b hcgh;&#xA;J si kwv crnd nekg( Jjh- vmq ltz&amp;gt;&#xA;Nxg kh b oxmlt: Ewg kh b dnn;&#xA;Rju jlht o fjnlt; Rwg jlht o hux;&#xA;J odkp fav oha plkg atds zr hmkds;&#xA;F ei kwv crnd nekg jrnjmuds&quot;&#xA;&#xA;Izx wn kwv crnd qnuhg klak ypg tef/&#xA;&#xA;L cv kwv crnd nekg= Fjh-G:kh\&#xA;&#xA;Ljnto bjq' afdto bjq- arht o rfgm~&#xA;&#xA;P hydto jpv| ugtto jpv* jlht o rfgm{&#xA;&#xA;Kkoto bjq: sctto bjq' me j dnrm~&#xA;&#xA;O jgtto jpv| jydto jpv* nf j dnrm&amp;gt;&#xA;Z hlkp fav; jlkp fav- jlht o rfgm&amp;gt;&#xA;D hlkp fav oha plkg oz zuk dydt:&#xA;D aukp fav oha plkg wb j hcedt;&#xA;Omv kh zuk wdjq&quot; Uck kh b hcgh&amp;gt;&#xA;Rxg kh b lbl# Fyz tyw xw qt#&#xA;A zi kwv crnd nekg oz b dnn]&#xA;C cm kwv crnd nekg tsht o hux.&#xA;D odkp fav oha plkg oz b oxmlt;&#xA;D li kwv crnd nekg tsht o fjnlt;&#xA;D li kwv crnd nekg atds zr hmkds;&#xA;D ei kwv crnd nekg RSTRPMQR&quot;&#xA;Y lv kwv crnd qnuhg klak ypg tef&amp;gt;&#xA;O cv kwv crnd nekg. Kjh&amp;gt;G:kh\&#xA;&#xA;Pjn wn kwv crnd nekg' nx ngq lnd&quot;&#xA;Shw ijkg&amp;gt; Fpf ijkg&amp;gt; Wsq sjq gdu]&#xA;Dhw ijkg jrg sjq gdu+ J rgd&quot;&#xA;&#xA;Rla&amp;gt; Gp xdq arkp mnw xw qt;&#xA;B gekp awp ijkg^ Iwn arkp uqz:&#xA;Nvk&quot; W wnnd qnuhg klak ypg tef&amp;gt;&#xA;B cv# D vlnd nekg. Kjh&amp;gt;G:kh!&#xA;Wmq B bhdto pqa plkg oz b dnrm#&#xA;Gbr B bhdto pqa plkg tsht o rfgm.&#xA;Gbr B bwkp nra plkg oz zuk dydt;&#xA;Gxx gq zuk wdjq; Wwj lk j hcedt;&#xA;Jxx gq b lbl; Jop gq b hcgh;&#xA;Gpdy pkm oz jabm( km jabm- njq ltz{&#xA;Px T jbkp nra plkg oz b dnn}&#xA;Wmh B bwkp nra plkg tsht o hux.&#xA;Gmh B eqa plkg oz b oxmlt;&#xA;Gsb B bwkp nra plkg tsht o fjnlt;&#xA;Gsb B bwkp nra plkg atds ifg owkds;&#xA;Mik&quot; G mjkp nra plkg WSTRPMQR&amp;gt;&#xA;Y lv pt qlnd qnuhg klak ypg tef#&#xA;Iendz uhq# Knjdz uhq. Hbd&amp;gt;T:kh!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1661" LastActivityDate="2015-02-06T22:22:30.427" Title="Looks like ECB -- What story is this?" Tags="|cipher|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="8612" PostTypeId="1" AcceptedAnswerId="8710" CreationDate="2015-02-06T20:39:55.340" Score="5" ViewCount="2652" Body="&lt;p&gt;What English words containing only voiced or only unvoiced consonants contain the most?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Count each consonant as many times as it appears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this is a question about the sounds (phonemes) contained in spoken English words, not letters of the alphabet!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consonants are either voiced (spoken with vibration of the vocal cords) or unvoiced (spoken without it). For example the sounds usually represented by &lt;em&gt;b&lt;/em&gt;, &lt;em&gt;d&lt;/em&gt; and &lt;em&gt;v&lt;/em&gt; are voiced, whereas those usually represented by &lt;em&gt;p&lt;/em&gt;, &lt;em&gt;t&lt;/em&gt; and &lt;em&gt;f&lt;/em&gt; are unvoiced. Voiced consonants can convey a feeling of softness; unvoiced ones, strength. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Captious&lt;/em&gt; contains 4 consonants, all unvoiced: /c/, /p/, /ʃ/, /s/.&lt;br&gt;&#xA;&lt;em&gt;Bamboozled&lt;/em&gt; contains 6, all voiced: /b/, /m/, /b/, /z/, /l/, /d/.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you do better?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unvoiced consonants:&lt;br&gt;&#xA;/f/, /k/, /p/, /s/, /t/,&lt;br&gt;&#xA;/h/&lt;br&gt;&#xA;/ʃ/ (&lt;strong&gt;&lt;em&gt;sh&lt;/em&gt;&lt;/strong&gt;&lt;em&gt;oe&lt;/em&gt;),&lt;br&gt;&#xA;/θ/ (&lt;strong&gt;&lt;em&gt;th&lt;/em&gt;&lt;/strong&gt;&lt;em&gt;in&lt;/em&gt;),&lt;br&gt;&#xA;/tʃ/ (&lt;strong&gt;&lt;em&gt;ch&lt;/em&gt;&lt;/strong&gt;&lt;em&gt;ip&lt;/em&gt;) (count as single)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Voiced consonants:&lt;br&gt;&#xA;/v/, /g/, /b/, /z/, /d/,&lt;br&gt;&#xA;/l/, /m/, /n/, /r/, /w/, /y/,&lt;br&gt;&#xA;/ʒ/ (&lt;em&gt;vi&lt;/em&gt;&lt;strong&gt;&lt;em&gt;s&lt;/em&gt;&lt;/strong&gt;&lt;em&gt;ion&lt;/em&gt;),&lt;br&gt;&#xA;/ð/ (&lt;strong&gt;&lt;em&gt;th&lt;/em&gt;&lt;/strong&gt;&lt;em&gt;at&lt;/em&gt;),&lt;br&gt;&#xA;/dʒ/ (&lt;strong&gt;&lt;em&gt;g&lt;/em&gt;&lt;/strong&gt;&lt;em&gt;em&lt;/em&gt;) (count as single)&lt;br&gt;&#xA;/ŋ/ (&lt;em&gt;si&lt;/em&gt;&lt;strong&gt;&lt;em&gt;ng&lt;/em&gt;&lt;/strong&gt;)&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's pronounce &lt;em&gt;where&lt;/em&gt; and &lt;em&gt;wear&lt;/em&gt; the same. /hw/ is an unvoiced-voiced pair and therefore could not appear in the words we want anyway, and we will ignore /ʍ/, the unvoiced partner of /w/ that appears in some people's pronunciation of &lt;em&gt;where&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7363" LastEditorUserId="5373" LastEditDate="2015-02-08T22:53:25.253" LastActivityDate="2015-03-16T17:17:03.413" Title="English words with the most consonants, all voiced or all unvoiced" Tags="|word|open-ended|" AnswerCount="2" CommentCount="8" ClosedDate="2021-06-26T17:02:06.970" ContentLicense="CC BY-SA 3.0" />
  <row Id="8614" PostTypeId="2" ParentId="8611" CreationDate="2015-02-06T22:22:30.427" Score="4" Body="&lt;p&gt;The story is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Green Eggs and Ham, by Dr. Seuss.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2015-02-06T22:22:30.427" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8615" PostTypeId="2" ParentId="8604" CreationDate="2015-02-06T22:35:19.630" Score="16" Body="&lt;p&gt;Let $n$ be an arbitrary positive integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with a right cylinder with cross section a regular $n$-gon. To each of the bases, attach a pyramid (with regular $n$-gon base). The resulting polyhedron has $n$ rectangular faces and $2n$ triangular faces. If the cylinder and pyramids are tall enough, it will be impossible for the die to land on the triangular faces (the center of gravity would not be over those faces). So, although this die has $3n$ sides (so it doesn't exactly answer your question), it is only possible to land on $n$ of them, and by symmetry all $n$ are equally likely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This die looks a bit like a pencil that's sharpened at both ends.&lt;img src=&quot;https://i.stack.imgur.com/FU3qr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-02-07T16:26:38.120" LastActivityDate="2015-02-07T16:26:38.120" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8616" PostTypeId="1" CreationDate="2015-02-06T23:37:05.357" Score="12" ViewCount="3138" Body="&lt;p&gt;Edit: added link for geocheck&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been working on trying to solve this puzzle for months and cannot come up with any solution. I have learned lots of puzzle solving methods in the process but I can't even find a start on this one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The required answer is a set of coordinates&lt;/strong&gt; for the location of a geocache. I am all for solving puzzles on your own but I think that there is always going to be something above your knowledge level. At this point, the only way to learn from this puzzle is to get an answer and understand how it was solved.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only hint given was that the answer is on the page and no math or program is required. In addition, you could simply solve it even with a printed copy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This one has been driving me mad and now I just want to understand how to solve it...lol. Any help is appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If you want to check a solution&lt;/strong&gt;, enter it at &lt;a href=&quot;http://geocheck.org/geo_inputchkcoord.php?gid=61350634fbd9df7-1ab9-46cd-ba0e-e86e36be4d27&quot; rel=&quot;nofollow noreferrer&quot;&gt;this link&lt;/a&gt;.  For a correct answer, the solution must match to the 3rd decimal point of the minute coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HNAKt.jpg&quot; alt=&quot;Page1&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/iRNSp.jpg&quot; alt=&quot;Page2&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9368" LastEditorDisplayName="user88" LastEditDate="2016-06-23T13:47:17.410" LastActivityDate="2019-02-27T11:32:41.730" Title="Please help with this geocaching puzzle" Tags="|cipher|lateral-thinking|geocaching|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8618" PostTypeId="1" CreationDate="2015-02-07T05:39:06.920" Score="-1" ViewCount="2287" Body="&lt;p&gt;What is the longest alphabetized (by first letter) sentence you can construct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider these &quot;words&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;be, a, fish, elephant, crashed, dead&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If this were a sentence, it would be valid:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A be crashed dead elephant fish&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sentence doesn't have to be logical, but it must be a sentence. Example of an illogical sentence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The fish walked over the air&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you run out of 26 words somehow, you may start over at &lt;code&gt;A&lt;/code&gt; yet again, but no reusing words.&lt;/p&gt;&#xA;" OwnerUserId="1477" LastEditorUserId="8874" LastEditDate="2015-02-07T16:14:06.447" LastActivityDate="2016-07-19T20:35:09.500" Title="Longest alphabetical sentence" Tags="|open-ended|english|" AnswerCount="2" CommentCount="3" ClosedDate="2016-07-20T12:30:35.743" ContentLicense="CC BY-SA 3.0" />
  <row Id="8619" PostTypeId="2" ParentId="8618" CreationDate="2015-02-07T05:43:00.700" Score="6" Body="&lt;p&gt;I've got one at &lt;strong&gt;26+26=52 words&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Any blue crown does encompass fatter grey helmets in just kings' lovely mild noses, opening paths, quickly reaping short things underneath violent winds, xeroxing yelling zebras and brown candied diapers entirely filled, grossly helping it, juicing kind, light, minced navel oranges, purposely queuing rinds, slowly tenderizing uncovered violas with xerotic yellow zippers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1477" LastEditorUserId="1477" LastEditDate="2015-02-07T06:18:03.520" LastActivityDate="2015-02-07T06:18:03.520" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8620" PostTypeId="2" ParentId="8616" CreationDate="2015-02-07T08:39:33.157" Score="-1" Body="&lt;p&gt;Edit: Added some initial analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After checking out geocaching.com, I understand that the only information that has been provided by the originator is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The name of the cache (curnew - who is an associate of the originator).  The originator of the cache is &quot;Adroit Seeker&quot;.&lt;/li&gt;&#xA;&lt;li&gt;The false coordinates which could be a modification of the real coordinates&lt;/li&gt;&#xA;&lt;li&gt;The seven sentences that might encode the real coordinates&lt;/li&gt;&#xA;&lt;li&gt;The photo that appears to contain no meaningful information (blowup of sunglasses below is from the jpg on the website).  The lettering on the blue hoodie says CANUCKS EST. 1970&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;With a radius of 3-4 kilometers around the false coordinates, the North minutes should be between 1.300 and 4.200.  The West minutes should be between 16.600 and 21.900. But this doesn't help much because there are still millions of possible answers especially if the solution uses a coordinate system different from WGS84.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not hard to make an unsolvable puzzle and apparently this puzzle is intended to be unsolvable - competing with two other puzzles that nobody has solved for the last 5 years.  &lt;strong&gt;There are no ingenious methods for solving low quality puzzles like this&lt;/strong&gt; - it usually just takes a lot of time to try all of the possibilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/09G5q.png&quot; alt=&quot;Sunglasses&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-02-08T23:27:29.517" LastActivityDate="2015-02-08T23:27:29.517" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8622" PostTypeId="1" AcceptedAnswerId="8627" CreationDate="2015-02-07T13:42:34.887" Score="8" ViewCount="485" Body="&lt;p&gt;The current market price of 175 honey bees is more than the price of 125 bumblebees, but less than the price of 126 bumblebees. Is it possible to buy 3 honey bees and 1 bumblebee for altogether at most 100 cents?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: all prices are integers, and all prices are measured in cents).&lt;/p&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="8874" LastEditDate="2016-03-14T18:02:57.487" LastActivityDate="2016-03-14T18:02:57.487" Title="Bumblebees and Honey bees" Tags="|mathematics|calculation-puzzle|algebra|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8623" PostTypeId="1" AcceptedAnswerId="8649" CreationDate="2015-02-07T14:11:38.440" Score="7" ViewCount="2055" Body="&lt;p&gt;The following alphabetical list contains independent countries with a certain common property. (The list is complete to the best of my knowledge).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Australia, ???, Chile, Costa Rica, Cuba, ???, ???, Iceland, Laos, Liberia,&#xA;  Luxembourg, Nepal, Netherlands, ???, ???, ???, Panama, Russia, ???, Slovakia,&#xA;  ???, Thailand, ???, ???&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which countries have been replaced by questionmarks?&lt;/p&gt;&#xA;" OwnerUserId="9125" LastActivityDate="2015-02-08T07:18:11.820" Title="What do these countries have in common?" Tags="|pattern|geography|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8624" PostTypeId="2" ParentId="3100" CreationDate="2015-02-07T14:29:54.423" Score="1" Body="&lt;p&gt;Bury yourself like a normal person, just don't face the sea. That way, when the waves hit, they will hit the back of your head and the retreating  waves don't have that much force so I think you will be saved. That is assuming that sea level doesn't rise.&lt;/p&gt;&#xA;" OwnerUserId="8980" LastActivityDate="2015-02-07T14:29:54.423" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8625" PostTypeId="1" AcceptedAnswerId="9659" CreationDate="2015-02-07T15:10:11.587" Score="14" ViewCount="1834" Body="&lt;p&gt;The following list contains the room numbers of some of my co-workers: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Room 15: Bobbie Solomon, born 18-Apr-1973&lt;/li&gt;&#xA;&lt;li&gt;Room 21: Galvin MacNeil, born 24-May-1974&lt;/li&gt;&#xA;&lt;li&gt;Room 22: Ismael Tirrell, born  7-Feb-1964&lt;/li&gt;&#xA;&lt;li&gt;Room 31: Lorene Alberts, born 22-Jun-1992&lt;/li&gt;&#xA;&lt;li&gt;Room 42: Robert Cabello, born 27-Sep-1967&lt;/li&gt;&#xA;&lt;li&gt;Room 48: Olivia Perkins, born 25-Jan-1971&lt;/li&gt;&#xA;&lt;li&gt;Room 52: Nelson Morales, born 19-Oct-1987&lt;/li&gt;&#xA;&lt;li&gt;Room 65: Alfred Bennett, born 11-Jul-1988&lt;/li&gt;&#xA;&lt;li&gt;Room 73: Buster Robbins, born  6-Aug-1989&lt;/li&gt;&#xA;&lt;li&gt;Room 84: Dorris Compton, born 13-Nov-1985&lt;/li&gt;&#xA;&lt;li&gt;Room 87: Tamara Pierson, born 15-Dec-1970&lt;/li&gt;&#xA;&lt;li&gt;Room 97: Ursula Akiyama, born 30-Mar-1990&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What will be the room number of my co-worker Martin Petters, born 16-Oct-1984?&lt;/p&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="39" LastEditDate="2015-02-07T15:18:51.553" LastActivityDate="2015-02-27T10:37:24.053" Title="What's my room number?" Tags="|pattern|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8626" PostTypeId="2" ParentId="8622" CreationDate="2015-02-07T15:10:40.117" Score="7" Body="&lt;p&gt;Short answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;   No  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Longer:   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;   Lowest price for the relation to be true is:&lt;br&gt;&#xA;   honey bee 23 cents&lt;br&gt;&#xA;   bumblebee 32 cents&lt;br&gt;&#xA;   3*23 + 32 = 101 cents   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7578" LastEditorUserId="7578" LastEditDate="2015-02-07T15:18:17.987" LastActivityDate="2015-02-07T15:18:17.987" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8627" PostTypeId="2" ParentId="8622" CreationDate="2015-02-07T15:30:48.843" Score="14" Body="&lt;p&gt;Assume that a bumble bee costs $x$ cents and that a honey bee costs $y$ cents, where $x$ and $y$ are integers. Then the problem statement gives&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$125x&amp;lt;175y$, which yields $5x&amp;lt;7y$ and hence $5x+1\le7y$&lt;/li&gt;&#xA;&lt;li&gt;$175y&amp;lt;126x$, which yields $25y&amp;lt;18x$ and hence $25y+1\le18x$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We multiply the first inequality $5x+1\le7y$ by $18$ and the second inequality $25y+1\le18x$ by $5$, and combine them into&lt;br&gt;&#xA;$$ 5(25y+1)\le90x\le18(7y-1).$$&#xA;This implies $125y+5\le126y-18$, and hence $y\ge23$.&lt;br&gt;&#xA;Furthermore, $18x\ge25y+1\ge576$ implies $x\ge32$.&lt;br&gt;&#xA;Hence the price of 3 honey bees and 1 bumblebee is $3y+x\ge3\cdot23+32=101$, and lies strictly above the price limit of $100$ cents.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="1664" LastEditDate="2015-02-07T16:27:23.167" LastActivityDate="2015-02-07T16:27:23.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8628" PostTypeId="5" CreationDate="2015-02-07T16:20:15.320" Score="0" Body="&lt;p&gt;A puzzle centered around geographical features, such as countries, continents, cities, mountains, rivers, climate, etc.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Examples&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/10085/how-many-countries-in-each-continent&quot;&gt;How many countries in each continent?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/30607/which-country-is-india-in&quot;&gt;Which country is INDIA in?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Please do not use this tag to ask geography problems which aren't puzzles as they are considered off-topic.&lt;/p&gt;&#xA;" OwnerUserId="11497" LastEditorUserId="11110" LastEditDate="2020-11-30T09:31:26.013" LastActivityDate="2020-11-30T09:31:26.013" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8629" PostTypeId="4" CreationDate="2015-02-07T16:20:15.320" Score="0" Body="A puzzle centered around geographical features, such as countries, continents, cities, mountains, rivers, and climate." OwnerUserId="8874" LastEditorUserId="69582" LastEditDate="2021-02-02T08:05:18.130" LastActivityDate="2021-02-02T08:05:18.130" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8630" PostTypeId="1" AcceptedAnswerId="8643" CreationDate="2015-02-07T16:23:06.163" Score="20" ViewCount="2027" Body="&lt;p&gt;Suppose three points are chosen at random in a circle. A triangle is made with these three points as vertices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the probability that the triangle contains the origin of the circle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Although I have an answer, I thought I would throw it out there and see what others make of it!)&lt;/p&gt;&#xA;" OwnerUserId="5381" LastEditorUserId="5373" LastEditDate="2021-07-06T19:50:40.917" LastActivityDate="2021-07-08T18:20:07.990" Title="Triangle in a circle" Tags="|mathematics|geometry|probability|triangle|" AnswerCount="7" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8631" PostTypeId="5" CreationDate="2015-02-07T16:37:51.657" Score="0" Body="&lt;p&gt;Topology is a branch of mathematics that is sometimes also called &amp;quot;rubber-sheet geometry&amp;quot;.&#xA;Topology studies those properties of geometric figures that remain unchanged under distortions and deformations (like twisting, shrinking, and stretching), as long as no surfaces are torn or cut.&lt;/p&gt;&#xA;&lt;p&gt;Typical topological puzzles are:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Wire puzzles: two or more entangled pieces of more or less stiff wire. The closed pieces might be simple rings or have more complex shapes. Normally the puzzle must be solved by disentangling the two pieces without bending or cutting the wires.&lt;/li&gt;&#xA;&lt;li&gt;Wire-and-string puzzles: where a ribbon or piece of string is added to a wire puzzle&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;General topology questions are off-topic but can be asked on &lt;a href=&quot;https://math.stackexchange.com/&quot;&gt;Mathematics Stack Exchange&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="11110" LastEditDate="2020-11-28T13:28:59.243" LastActivityDate="2020-11-28T13:28:59.243" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8632" PostTypeId="4" CreationDate="2015-02-07T16:37:51.657" Score="0" Body="A puzzle about disentangling knots, folding paper, separating strings and wires, and understanding the connectivity properties of elastic figures. General topology questions are off-topic but can be asked on Mathematics Stack Exchange." OwnerUserId="8874" LastEditorUserId="71981" LastEditDate="2020-11-28T13:28:12.243" LastActivityDate="2020-11-28T13:28:12.243" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8633" PostTypeId="5" CreationDate="2015-02-07T16:45:27.263" Score="0" Body="&lt;p&gt;For puzzles that involve foreign alphabets and unusual character encodings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/4642/decode-the-message-enciphered-in-these-symbols&quot;&gt;Decode the message enciphered in these symbols: ◳◰ ◓◨ ◨◧◕ ◎◌ ◱◯◱◯ ◍◌○ ◉◉ ◇◔◓◕ ◐►◓◒ ◒◑ ◈◑ ◆◆◓ ◉◉◉&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11497" LastEditorUserId="11110" LastEditDate="2017-06-21T20:08:01.690" LastActivityDate="2017-06-21T20:08:01.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8634" PostTypeId="4" CreationDate="2015-02-07T16:45:27.263" Score="0" Body="Puzzles about foreign alphabets and weird encodings." OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-07T18:25:35.647" LastActivityDate="2015-02-07T18:25:35.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8635" PostTypeId="1" AcceptedAnswerId="8644" CreationDate="2015-02-07T17:25:41.317" Score="7" ViewCount="595" Body="&lt;p&gt;Alice and Bob play a game by alternately placing dominoes on a $3\times100$ checkerboard (3 rows, 100 columns). Alice has the first move.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A move of Alice consists in placing a $1\times2$ (horizontal) white dominoe.&lt;/li&gt;&#xA;&lt;li&gt;A move of Bob consists in placing two $2\times1$ (vertical) black dominoes.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The dominoes must be placed without overlaps, and they must always cover exactly two cells of the checkerboard. &#xA;The loser is the player who cannot make a complete move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determine the winner of the game. &#xA;(As usual, we assume that Alice and Bob use optimal strategies.)&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-07T20:10:33.553" LastActivityDate="2015-02-10T17:35:45.617" Title="Placing dominoes on a 3x100 checkerboard" Tags="|strategy|checkerboard|game|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8636" PostTypeId="1" CreationDate="2015-02-07T18:13:53.953" Score="-4" ViewCount="1464" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;&lt;em&gt;Attention everyone! It is the final game of the Connect 4 championship!&lt;/em&gt;  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;Billy and Sarah have played perfectly so far during this tournament, so in efforts to prevent Player 1 from being able to win, we are going to try something new!&lt;/em&gt;  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;These perfect players are going to be given an infinite playing board.&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now that you've listened to the announcement, it's time for the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So, assuming Billy goes first, who will win? Or will it be a tie?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2015-02-08T19:35:21.363" Title="Connect 4 Infinite" Tags="|mathematics|strategy|board-games|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="8638" PostTypeId="2" ParentId="8635" CreationDate="2015-02-07T19:27:38.637" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Alice wins.&lt;br&gt;&#xA;Starting out Alice has 150 possible locations to place, and Bob has 100. If Alice places 1 domino along the middle row she will eliminate two possible locations for Bob. If Bob places 2 dominoes spaced one from eachother and one from Alice's piece, he will eliminate 4 spots for alice. Repeated over the board this strategy will for the first 96 columns remove 32 spots from Bob, and 64 from Alice, the final 4 columns removes 2 from Bob and 2 from Alice leaving Bob with 1 piece left in his move. Now there are 84 locations left for Alice of which 34 are untouchable for Bob. Bob has 68 locations of which 16 are untouchable. Since Bob has 1 domino left of his turn, and Bob will use 2 locations for every Alice's one (2*34=68), Alice will win even if she doesn't place in any of Bob's remaining spots, which she of course will.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9302" LastEditorUserId="9302" LastEditDate="2015-02-08T22:50:12.063" LastActivityDate="2015-02-08T22:50:12.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8641" PostTypeId="2" ParentId="8630" CreationDate="2015-02-07T19:56:55.213" Score="9" Body="&lt;h2&gt;A generalization to arbitrary centrally symmetric regions&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;A planar region is called &lt;em&gt;centrally symmetric&lt;/em&gt; with respect to the origin, if for every point $P$ in the region also its reflection with respect to the origin is in the region.&#xA;Examples for centrally symmetric regions are for instance circles, squares, regular hexagons, regular polygons with an even number of sides.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem&lt;/strong&gt;: &#xA;  Let $R$ be a (bounded) region that is centrally symmetric with respect to the origin.&#xA;  If three points are randomly chosen from $R$, then the probability that the resulting triangle contains the origin is $1/4$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a special case, we derive that the corresponding probability for a circle is also $1/4$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;-----------------------------------------------------------------------------&lt;/h2&gt;&#xA;&#xA;&lt;h2&gt;The proof&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;(1) My argument heavily uses the fact that exactly one of the following two statements holds true for any triangle $\Delta$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The triangle $\Delta$ does contain the origin.&lt;/li&gt;&#xA;&lt;li&gt;There exists a line that separates the origin from the triangle $\Delta$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(2) Now consider the region $R$, and let $\ell$ be a horizontal line through the origin.&#xA;We will assume throughout that none of the three randomly chosen points does ever land directly on line $\ell$ (as these cases have measure zero in the underlying probability space).&lt;br&gt;&#xA;A situation where the triangle contains the origin is called &quot;success&quot;, and&#xA;a situation where the triangle does not contain the origin is called &quot;failure&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) We distinguish three cases for three randomly chosen points $A,B,C$ in the region $R$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Case (3a): All three points are above $\ell$.&lt;br&gt;&#xA;  Case (3b): All three points are below $\ell$.&lt;br&gt;&#xA;  Case (3c): Two points are above $\ell$, and one is below $\ell$.&lt;br&gt;&#xA;  Case (3d): Two points are below $\ell$, and one is above $\ell$.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The respective probabilities for cases (3a), (3b), (3c), (3d) are $1/8$, $1/8$, $3/8$ and $3/8$. Furthermore, cases (3a) and (3b) are clear failures (as line $\ell$ itself separates the origin from the resulting triangle). By symmetry, the success probabilities for cases (3c) and (3d) are identical and will be called $p^*$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) Now let us compute the success probability $p^*$ for case (3c). Assume that two points $A$ and $B$ are above $\ell$, whereas the third point $C$ is below $\ell$. &#xA;The triangle $\Delta ABC$ is a success, if and only if the line through $C$ and the origin has $A$ on one side and $B$ on the other side. &#xA;Let point $C'$ be the mirror image of $C$ with respect to the origin. &#xA;The triangle $\Delta ABC$ is a success, &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;if and only if the line through $C'$ and the origin has $A$ on one side and $B$ on the other side, &lt;/li&gt;&#xA;&lt;li&gt;which happens if and only if in the clockwise ordering of $A$, $B$, $C'$ around the origin we have point $C'$ in the middle between $A$ and $B$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now note that the probability distribution for choosing $C$ is identical to the probability distributions for choosing $C'$, for choosing $A$ and for choosing $B$, as all these points are chosen from congruent regions (that are one half of the symmetri region $R$).&#xA;Consider an &lt;em&gt;arbitrary&lt;/em&gt; choice of three points $X$, $Y$, $Z$ from the half-region. If we randomly label these three points by $A$, $B$, $C'$, then exactly two of the six labelings have $C'$ in the middle between $A$ and $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summarizing, the success probability in case (3c) is $p^*=2/6=1/3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5) Now let us wrap things up. The overall success probability equals the overall success probability in the four cases&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Case (3a): $\frac14\cdot0$&lt;br&gt;&#xA;  Case (3b): $\frac14\cdot0$&lt;br&gt;&#xA;  Case (3c): $\frac38\cdot p^*$&lt;br&gt;&#xA;  Case (3d): $\frac38\cdot p^*$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As the sum of these four success probabilities is $\frac34\cdot p^* =\frac14$,&#xA;the proof is complete.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-08T14:03:15.037" LastActivityDate="2015-02-08T14:03:15.037" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8642" PostTypeId="2" ParentId="8630" CreationDate="2015-02-07T20:18:42.547" Score="4" Body="&lt;p&gt;Put the center of the circle as the origin of a polar coordinate system. Then each point is determined by and angle $\theta$ and a distance $r$ from the origin (we will assume none of the points lands on the origin). Then the triangle joining the three points will contain the origin if and only if the three $\theta$ values are not contained in any half-circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given angles $\theta_1$ and $\theta_2$, let $|\theta_2-\theta_1|$ denote their angular separation, chosen so that $0^\circ\leq|\theta_2-\theta_1|\leq 180^\circ$. Then the set of andlges $\theta_3$ such that $\theta_1,\theta_2,\theta_3$ aren't contained in a half-circle is an arc, of length $|\theta_2-\theta_1|$ (this arc is antipodal to the arc connecting $\theta_1$ and $\theta_2$; see the image, where this arc is colored red). Thus the probability that a randomly chosen $\theta_3$ satisfying this condition is&#xA;$$&#xA;\frac{|\theta_2-\theta_1|}{360^\circ}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\theta_1$ and $\theta_2$ are chosen independently at random, then the expected value of $|\theta_2-\theta_1$ is $90^\circ$ (the quantity $|\theta_2-\theta_1|$ will be uniform in the interval $[0,180^\circ]$). This means the probability the center of the circle lies in the triangle with three random points as vertices is&#xA;$$&#xA;\frac{90^\circ}{360^\circ}=\frac{1}{4}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cyPkC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2627" LastActivityDate="2015-02-07T20:18:42.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8643" PostTypeId="2" ParentId="8630" CreationDate="2015-02-07T20:35:52.077" Score="22" Body="&lt;p&gt;First, observe &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That the triangle does not contain the center if and only if all three points are contained in a half-circle. So, only the polar angles of the points matter, not their distance to the center.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Next, let's &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Label the points $A$, $B$, $C$. If we starting at point $A$ and go clockwise for half a circle, there's a $1/4$ chance that both point $B$ and point $C$ are in that half-circle, putting all the points on a half-circle. Likewise for starting at point $B$ and starting at point $C$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These events are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Disjoint (well, overlap with probability 0), and comprise all the way the three points can be on a half-circle, since there's always exactly one for the other two are within half a circle clockwise. So, the probability of them being on a half-circle is $$3/4 = 1/4 + 1/4 + 1/4 $$and so the probability the triangle contains the center is the complement $$1 - 3/4 = 1/4$$ By the same argument, if we generalize to choosing $n$ points and connecting them in a polygon in cyclic order of polar angle, the probability that it contains the center is $$ 1 - \frac{n}{2^{n-1}}$$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="46903" LastEditDate="2018-09-09T07:00:18.190" LastActivityDate="2018-09-09T07:00:18.190" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="8644" PostTypeId="2" ParentId="8635" CreationDate="2015-02-07T20:50:52.983" Score="5" Body="&lt;p&gt;Alice can win using the following strategy: if possible, place a domino in the leftmost available place in the middle row. If it is not possible to play in the middle row, place a domino directly above or below one previously placed in the middle row.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, I claim that Alice will be able to play at lease 17 dominos in the middle row. To see why, note that there are initially 99 places a domino can go in the middle row. Call such a place &lt;em&gt;blocked&lt;/em&gt; if one or both of its squares is occupied by another domino. Each domino Alice or Bob places blocks at most 2 previously unblocked places (so that Bob can block at most 4 places on each of his turns). 16 turns by Alice and 16 turns by Bob lead to at most $16\cdot 2+16\cdot 4 =96$ of the 99 places being blocked, so Alice can still find a place to play in the middle row on her 17th turn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each domino placed in the middle row makes space for two more dominos, one above it and one below it, and Bob cannot interfere with these spaces. So after playing 17 dominos in the middle row, Alice is guaranteed to be able to find a play for the next 34 turns. This gives Alice a total of at least 51 plays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, Bob will not be able to play 51 times. In 51 turns, Bob would need to place 102 dominos. Each domino Bob places must be in a different column, and there are only 100 columns on the board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, Bob cannot play in any of the (at least) 34 columns containing one of Alice's middle row dominos, so Bob will actually be able to take a total of at most 33 turns.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-02-10T17:35:45.617" LastActivityDate="2015-02-10T17:35:45.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8645" PostTypeId="2" ParentId="8630" CreationDate="2015-02-07T20:56:44.720" Score="4" Body="&lt;p&gt;Thanks everyone for their answers. Here's mine which is close in spirit to a couple of the existing ones.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Consider a circle centered at the origin. Without loss of generality, let one of the points lie on the positive $x$-axis, call it $p_1$. We can do this just by rotating the coordinate system and the fact that the probability the origin is one of the points is zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Label the other two points $p_2, p_3$ such the angle they make is in ascending order, $0 = \theta_1 \leq \theta_2 \leq \theta_3 &amp;lt; 2\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now observe that if a triangle formed by the three points containing the origin can be formed we necessarily have $\theta_2 &amp;lt; \pi$, as if $\theta_2 \geq \pi$ this implies $\theta_3$ is also greater than $\pi$ and both $p_2$ and $p_3$ lie in the 3rd and 4th quadrants. Similarly, we require $\theta_3 &amp;gt; \pi$, as if $\theta \leq \pi$ then $p_2$ and $p_3$ lie in the 1st and 2nd quadrants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those two conditions being satisfied, we also require that the difference $\theta_3 - \theta_2 &amp;lt; \pi$ since if the difference is greater than $\pi$ it is possible to draw a line through the origin which doesn't intersect the triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The desired probability, $p$, is then the proportion of the parameter space $$S = \{ (\theta_2,\theta_3) \ : \  0 \leq \theta_2 \leq \theta_3 &amp;lt; 2\pi\}$$ that meets all three conditions, $S_{Yes} = \{(\theta_2,\theta_3) \in S : \theta_2 &amp;lt; \pi \ \wedge \ \theta_3 &amp;gt; \pi \ \wedge \ \theta_3 - \theta_2 &amp;lt; \pi \}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p = \frac{\mu(S_{Yes})}{\mu(S)} = \frac{\frac{1}{2}\pi^2}{\frac{1}{2}(2\pi)^2} = \frac{1}{4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/D2AoT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The three conditions ($\theta_2 &amp;lt; \pi \ \wedge \ \theta_3 &amp;gt; \pi \ \wedge \ \theta_3 - \theta_2 &amp;lt; \pi$) are equivalent to saying the three points don't lie in a semi-circle. What's elegant about xnor's answer above is how it abstracts away the messiness of angular coordinates altogether and just uses that key condition.&lt;/p&gt;&#xA;" OwnerUserId="5381" LastEditorUserId="5381" LastEditDate="2015-02-08T16:46:37.900" LastActivityDate="2015-02-08T16:46:37.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8646" PostTypeId="2" ParentId="8630" CreationDate="2015-02-07T20:58:51.920" Score="8" Body="&lt;p&gt;Imagine putting the three points within the circle one-by-one. We are only interested in the angular coordinates of these points. We use the first point to define the zero point for this angular coordinate. The second and third point have angular coordinates running from $-\pi$ to $+\pi$. The combinations of both angular coordinates for which the triangle contains the origin is indicated in below figure as the two blue regions. These blue regions cover $1/4$ of the total area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9yWQh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7573" LastEditorUserId="7573" LastEditDate="2015-02-07T21:26:12.010" LastActivityDate="2015-02-07T21:26:12.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8647" PostTypeId="2" ParentId="6212" CreationDate="2015-02-08T06:09:22.333" Score="15" Body="&lt;p&gt;The puzzle author appears to have come up with their own encryption scheme. It doesn't appear to be documented anywhere, but there is sample code (in python) out there on a different site. Finding that is actually the hardest part since the key and primer needed to decipher the message are both given in the first few hints.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Solution&lt;/h3&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Contained herein are the arguments for and against re-admitting Jay_Darsener into Luceoscutum. Present are Steel01, Administrator of the site; Koballt10, Moderator of Totally Random; Gwythaint, Game Master; Jay_Darsener, the accused; Makosolider and Zero, interested parties in support of the accused.&#xA;So verified by: &#xA;Jay_Darsener&#xA;Steel01&#xA;Gwythaint&#xA;Zero&#xA;MakoSoldier626&#xA;Koballt10&#xA;&#xA;With my signature below, I hereby a) verify that all I say is truth, b) verify that any questions asked of me will be answered directly, barring any information that is available to moderators only, c) agree to abide by any and all terms that are established by this meeting, and d) agree to honor these terms not only in word, but in spirit, with the full knowledge that any wrongdoing on either side of this agreement will be punished to the fullest extent of the site rules, to be meted out by moderators and followed through completely (in the event that the Administrator or the Moderator do wrongly, moderators who are not present will mete out the penalty).&#xA;Signed:&#xA;Gwythaint&#xA;Jay_Darsener&#xA;MakoSoldier626&#xA;Zero&#xA;Steel01&#xA;Koballt10&#xA;To the defense:&#xA;By signing below, I hereby state that I am sufficiently represented in this meeting, and feel that I am being defended to the best ability of my representative.&#xA;Signed:&#xA;Jay_Darsener&#xA;&#xA;To the prosecution:&#xA;By signing below, I hereby state that I am sufficiently represented in this meeting, and feel that I am being just in my challenge of the defendant.&#xA;Signed:&#xA;Steel01&#xA;&#xA;The purpose of the meeting (J) is to discuss why I have not been allowed back on the site/ (S) to find out what has been going on since the original account has been banned until now/ (G) to clear the air and cure the site of ill-will (Z)agrees with G (M) to get the site back to the original intentions--to RP without stress and drama.&#xA;&quot;First, addressing history of the account. The two present at the original argument were Steel and Jay. Both need to tell their chronology of the issue.&quot; (G)&#xA;(S)I do not remember the first post that sparked the whole thing. CP as the Global mod saw something in one of Jay's posts that went against the site rules --content-- that was modified by CP. Jay did not like the edit, and edited out CP's edit without fixing the post for content. A PM was sent about the issue, explaining that editing the mod's posts out without fixing the issue was agains the rules. He fought this. During the week that he was banned, he talked to me a lot, gave some reasons that he wasnted the account permanently removed. The reason that I recall is that his parents don't like fantasy and that he didn't want his parents to see what he had done. Jay said that he would continue causing trouble until he was permanently banned. He kept misbehaving and was banned--this came to a head in the duel with Leben-hoff.&quot;&#xA;(J)I don't remember everything that happened. There were grievances b/w me and CP. The mods were reevaluating how post counts counted toward titles. I was against post counts counting in GD. I talked to either a mod or Steel. I decided that they weren't going to do anything about it and posted everywhere. CP, noticing this, became picky about everything I did. She cut out one of my posts and edited it into one post. I changed it back. That day, S told me that I had gotten into trouble and was banned for a week. The original ban didn't work; S re-banned me properly. I don't think I posted after that week.&#xA;(G) Most of this centers around the argument with CP. Kage also interfered in the duel, which neither of us liked. There was further animosity than just with CP.&#xA;(S) Kage stepped down before the final, permanent ban.&#xA;(J)did that have anything to do with the duel?&#xA;(G) both CP and K were mods. That is the only connection there. With that, general chronology from there. That was in July. S, J, feel free to interrupt. Initially after the ban there was some disgruntlement, but nothing happened either way. Ther e were further difficulties with interactions on the site. This was brought to a head b/c S and the mods were in disagreement over the site about how much power that the mods have. S brought up the mod election idea because we needed to start over (S agrees). J and G saw that the elections might not have CP reelected, and we were happy about that. When the election did come about, CP was still mod. We were pleased that the Global Mod position was eliminated--it granted more freedom, eliminated the hierarchy idea among the mods; they were players, not just GODMODS. Fall semester, J had another account. He was trying to keep under cover b/c he knew that he would be insta-banned. When (S) found out, he asked me to find out what his new name is. I asked, I was informed that J was on, but he did not tell me what account it is. After that, I was told by J that teh.bob was not the name. After that, I did not push, hoping it would die. After fall semester, it did die a bit.&#xA;(J)last semester I approached S and I asked for Jay to be resurrected. At that point, I would stop sneaking around behind everyone's back. He told me that it was the mod's opinion that Jay was dead. The only way I could do retribution would be to reveal my other screen name--S would bring it before the mods for review and maybe allow me to get back on later. (S--not so much allowed back on, but we would not hunt him down and ban him instantly. If he came to us and tried to work with us, he would talk about allowing the new account to stay.) (K agrees)&#xA;(G) Interterm, it has been the same discussion.&#xA;(S) Other than that, there is 2nd and 3rd hand rumor that during the mod elections that J and G were actively talking to people, trying to curve the elections/get rid of the old mod team. &#xA;(G) yes, J and I did get together, we were pleased with the prospect of getting rid of CP as Global Mod. We wanted the position of GloMod to be eliminated, and it was. We considered it a success.&#xA;(K) I know this, he talked to me about it before I was a mod.&#xA;(S) have there been any attempts to remove anyone else as mod?&#xA;(J) no, Kage was annoying, but that is it. (G agrees)&#xA;(J) my parents have now stated that it is all right that I am on the forum. They were confused at first, had some skewed details about the site, but their opinions have changed since they found out what the site is about.&#xA;(S) IS the account you are currently using teh.bob?&#xA;(J) I will tell you (S) in secret to protect my anonymity.&#xA;(Z) the admin will be informed, I am okay with that.&#xA;(M) so far everything has been accurate as far as I know. I want this resolved as fast as possible.&#xA;(G) the primary difficulty is the other account, and whisperings behind the scenes. It sounds like more than just going through the back door to getto the site itself. There's whispering, discussion, chatting with newbies who have joined saying don't trust these people, interact with these people, etc., poisoning the site, specifically the old mods. &#xA;(J) I don't remember whispering to people about this other than about CP.&#xA;(G) was this after CP?&#xA;(J) yes, I was was warning them about her temper.&#xA;(G) was there anything like &quot;Watch out for xx, not just CP.&quot;&#xA;(J) I told people to be careful about double posting. That was it.&quot;&#xA;(S)You, Z, told people in the chat that the mods would get people in trouble for any disagreement with them. Where did they hear that?&#xA;(K) we had a bti of a bicker/play argument.&#xA;(Z) we were having an argument abotu sheep being dumb. I gave my bible perspective, and you said that sheep are not stupid. You defended them, and expressed that on previous sites, that because on other sites arguments rose. I have seen an argument starting. I thought, I want peace, I will not force my opinion does not mattter in the long run. Seeing an argument growing, I sat back and shut down the argument, showing that I am under the authority of the site. (K agrees) If they find that I am breaking a rule, I accept their judgment. To answer S's question, before coming to the site, I was aware of the issue with J. I understand that issues can be generated between Mods and players. I was trying to cover myself. I don't think that there's anything wrong with that, especially with new people. Call me over-protective.&#xA;(S) when that happens, we don't ban or do moderation on small offenses.&#xA;(G) J and I have had negative interactions with CP. Other than that, I have had mods PM me about problems, and they have helped me. They wield the sword, and they do their duty, but they do so graciously. Aside from one or two interactions, they have been pleasant. Aside from that, Jay is the only one to be banned permanently. At this point, others have been warned. There seems to be a disconnect because at first the ban was voluntary, but tensions have grown.&#xA;(S) from the mods perspective, there was no voluntary on the ban--even if he agreed with it, it was going to happen.&#xA;(J) from what I recall, by the end of the week, I was permanently banned.&#xA;(S) when was the new account made? Or the second account?&#xA;(J) about 2 or three weeks after I was banned&#xA;(S) and your parents&#xA;(J) a day after they were okay with it, I signed back up.&#xA;(S) then why didn[t that come up before when he was asking for J to come back?&#xA;&#xA;(J) it was because you told me people didn't want me back, mods were mad, you told me that I was causing trouble.&#xA;(S) you cause trouble with one mod, you get trouble from all.&#xA;(J) I learned to stay in RP sections, and not GD and TR. &#xA;(M) I think people are putting too much stock and emotional time into TR, GD, and chat. It irks me when people cause issues, when there sholdn't be an issue there. Chars can argue, but when people argue, there's problems. It's not what all of us joined Luceo for. I almost see the arguments as a roadblock to what we came here for. &#xA;(Z) It seems that to me from J's perspective, he was banned voluntarily, and that the mods banned him whether or not it was voluntary.&#xA;&#xA;(m) about banning: ban the username, but not the person. Imagine if you couldn't be your fav. char ever again. I can see user names dying, but I can't see breaking internet anon to reban someone for who they are because you're mad at them as a person. &#xA;(S) something that does need to be considered is that when people start breaking laws, not governing themselves, the admin needs to step in. When people cause probelms appropriate measures need to be taken.&#xA;&#xA;TO BE CONTINUED AFTER CHURCH at 8:15.&#xA;THIS MESSAGE IS PRIVILEDGED INFORMATION. DO NOT SHARE THIS WITH ANYONE OUTSIDE THE STATED MEMBERS PRESENT.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The message body of the hints are all numbers, written in various bases. When expressed in a different base with each digit mapped to a different character to reveal the message.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Hint 1&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Convert the base 10 number to base 27, using the characters &quot; abcdefghijklmnopqrstuvwxyz&quot; for digits 0-26 gives: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;eve will send audio&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The attachment should be useful later on.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Hint 2&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Same as hint 1, except you're give hex digits for a base 256 number. After getting that in base 27 I got: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;sorry for the garbled audio but frank might send key data tomorrow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In case you couldn't tell the attachment was garbled. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inquisitive got a good start on the attachment already. Spaces, punctuation, word length, and capitalization are all preserved, only the alphanumeric characters a jumbled. Here is the full text&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Contained herein are the arguments for and against re-admitting Jay_Darsener into Luceoscutum. Present are Steel01, Administrator of the site; Koballt10, Moderator of Totally Random; Gwythaint, Game Master; Jay_Darsener, the accused; Makosolider and Zero, interested parties in support of the accused.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's worth noting that this is very likely the first line of the encrypted audio (hint 8 points this out too). Word length and punctuation match perfectly. Ya! partial plaintext.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Hint 3&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;In case you don't recognize this as ASCII85 hint 6 will tell you so. Converting to base 27 just like the first 2 emails gives:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;you need to research markov encryption to decode the message&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm familiar with Markov chaining, but not encryption. That chapter appears to be missing from my copy of &lt;em&gt;Applied Cryptography&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the attachment, it's not a real pickle, it's a 88x52 matrix already. Each line has a-zA-Z randomized with no repeated digits on a line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can stop here with the hints, nothing else given is all that useful.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Hint 10&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Bases got switched, after some trial and error, I found base 38 worked with &quot; abcdefghijklmnopqrstuvwxyz\n0123456789.&quot; and got:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;number representation and value are separate &lt;br&gt;they are shown in base x with x or more symbols &lt;br&gt;values have not a base&lt;br&gt;base 16 may be shown in base 256 and base 256 in base 27&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;Hint 11&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Using base 38 I get something almost useful.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;would a base 85 number be stored as characters or bytes&lt;br&gt;something is a kind of clay used as a building material&lt;br&gt;the building material is typically in the form of sun dried bricks&lt;br&gt;what is the key word&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The attachment doesn't decode to anything meaningful in base 26 or 38, or any base between 26 and 128 if the initial character mapping stays the same.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Hint 12&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Hint 11 points out you should use the Adobe implementation of ASCII85 to decode the messages.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;an attachment contains this message 3 times&lt;br&gt;each is encrypted with the primers and keys&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;Hint 19&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Bases switched to 65 this time and converting to plain text got a little more complicated. The first 39 characters of the decode string are the same, but after that you need to add 56 and convert that number to to an ASCII character.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;from our past 7 weeks of analysis&lt;br&gt;we have concluded there is a 99.9&lt;br&gt;percent chance that an unfamiliar&lt;br&gt;method of encryption was utilized&lt;br&gt;to secure the transcript for your&lt;br&gt;friend.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;Hint 20&lt;/h3&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;while working with ng and attacking the systems at ls&lt;br&gt;we have found that me works with individual bytes. do&lt;br&gt;not try solving this using unicode. ascii encoding is&lt;br&gt;probably your best choice. chars should be printable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;Hint 21&lt;/h3&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;our associates are learning more and more about markov encryption&lt;br&gt;but some of them have not caught on to a simple fact. the process&lt;br&gt;works on what we call targeted bytes. when a byte is not targeted&lt;br&gt;it passes through the process without being changed. each byte of&lt;br&gt;interest may be changed into any other targeted byte. until then.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9060" LastEditorUserId="9060" LastEditDate="2015-02-09T20:24:51.123" LastActivityDate="2015-02-09T20:24:51.123" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8648" PostTypeId="2" ParentId="1856" CreationDate="2015-02-08T06:13:57.230" Score="8" Body="&lt;p&gt;Begin wherever you want and draw one line through &lt;strike&gt;all of the doors&lt;/strike&gt;, but you cannot go through the same door twice.&#xA;Hard, but possible!&lt;/p&gt;&#xA;" OwnerUserId="9375" LastActivityDate="2015-02-08T06:13:57.230" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8649" PostTypeId="2" ParentId="8623" CreationDate="2015-02-08T07:18:11.820" Score="9" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All of these countries have red, white, and blue flags.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's the list of question marks (in order):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cambodia, Czech Republic, France, New Zealand, North Korea, Norway, Samoa, Taiwan, United Kingdom, United States&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's what I think your source was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_flags_by_color_combination#Blue.2C_red_and_white&quot;&gt;http://en.wikipedia.org/wiki/List_of_flags_by_color_combination#Blue.2C_red_and_white&lt;/a&gt;, and in this case it seems like Costa Rica may not count.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastActivityDate="2015-02-08T07:18:11.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8652" PostTypeId="2" ParentId="8616" CreationDate="2015-02-08T09:10:30.190" Score="1" Body="&lt;p&gt;Keeping in mind that he claims this cache exists and that he was the one to place it, there is a good possibility that it is in his local area. Being within 2km isn't far, and might be a statistic meant to throw us off. Having been a geocacher with my father when I was young, I can vouch that we didn't care to go too far from home, especially when placing our own cache(s). It's not a guarantee, but it's possible that his behavior is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are the provided coordinates truly near where he is active as a geocacher? I'm unfamiliar with the mathematical standards in converting coordinates to distance in kilometers, but from what I researched, a km can be anywhere from 1-2 minutes. This means he wouldn't be likely to have changed his N/W degrees by much, if at all, and makes them constants that may be easier to find on the page.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reflection in the sunglasses doesn't appear to have value. I can distinguish simply a reflection of his arm holding his cell-phone camera in a normal selfie pose. There is still a good chance this photo was doctored, as you suggested, but I don't think it would be in such a difficult place to keep the photo looking natural. Looking at a blown up version of the image, there are very few things that seem out of place even after the pixelated distortion: one is the arm of the woman's glasses, and should be looked at a bit closer if someone has a better angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end, it might not be worth tackling this puzzle cache. As was suggested in other answers, the intent of this puzzle was revenge, an attempt to make a deviously hard puzzle to solve. It's quite possible that he's baiting people to waste time on it, knowing it can't be solved, and get satisfaction out of knowing people are trying to solve an impossible riddle. If it's any indication, the fact that he named his cache with his own name demonstrates how little respect he has for his own work.&lt;/p&gt;&#xA;" OwnerUserId="9319" LastActivityDate="2015-02-08T09:10:30.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8655" PostTypeId="1" AcceptedAnswerId="8657" CreationDate="2015-02-08T14:21:34.107" Score="-2" ViewCount="444" Body="&lt;p&gt;I am thinking of an integer $1,2$ or $3$. You can ask me only a single question to which I can reply &lt;strong&gt;&quot;Yes&quot;&lt;/strong&gt;, &lt;strong&gt;&quot;No&quot;&lt;/strong&gt; or &lt;strong&gt;&quot;I don't know&quot;&lt;/strong&gt;. I will be completely honest. What will you ask me to figure out what number I am thinking about?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; Since there are an infinite number of solutions to this puzzle, I'll select the wittiest one / the one with most upvotes. (Because the one I came up with is kind of overly mathematical and a little too out of the box.)&lt;/p&gt;&#xA;" OwnerUserId="9379" LastEditorUserId="5373" LastEditDate="2017-12-17T14:45:14.553" LastActivityDate="2017-12-17T14:45:14.553" Title="What's my number?" Tags="|logical-deduction|open-ended|" AnswerCount="4" CommentCount="5" ClosedDate="2015-02-08T19:39:33.723" ContentLicense="CC BY-SA 3.0" />
  <row Id="8656" PostTypeId="2" ParentId="8655" CreationDate="2015-02-08T14:29:09.733" Score="1" Body="&lt;p&gt;Is the reciprocal of 2 minus your number greater than zero?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the number is 1, the answer is YES (as $\frac{1}{2-1}&amp;gt;0$ )&lt;/li&gt;&#xA;&lt;li&gt;If the number is 2, the answer is I DON'T KNOW (as $\frac{1}{2-2}$ is undefined  )&lt;/li&gt;&#xA;&lt;li&gt;If the number is 3, the answer is NO (as $\frac{1}{2-3}&amp;lt;0$    )&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3987" LastEditorUserId="3987" LastEditDate="2015-02-08T17:40:14.787" LastActivityDate="2015-02-08T17:40:14.787" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8657" PostTypeId="2" ParentId="8655" CreationDate="2015-02-08T14:37:29.400" Score="1" Body="&lt;p&gt;I'm going to think of a random number too, 1 or 2. If we multiply our numbers, will the product be greater than or equal to 3?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;No -&gt; 1&lt;/li&gt;&#xA;&lt;li&gt;Yes -&gt; 3&lt;/li&gt;&#xA;&lt;li&gt;I don't know -&gt; 2&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2015-02-08T14:51:31.010" LastActivityDate="2015-02-08T14:51:31.010" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8658" PostTypeId="1" AcceptedAnswerId="8660" CreationDate="2015-02-08T14:47:15.007" Score="-1" ViewCount="1306" Body="&lt;p&gt;Yesterday at our local highschool there were &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;29 pupils wearing BLACK coats&lt;/li&gt;&#xA;&lt;li&gt;27 wearing RED coats,  &lt;/li&gt;&#xA;&lt;li&gt;and 40 wearing BLUE coats. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How many pupils were wearing GREEN coats?&lt;/p&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="8874" LastEditDate="2016-03-19T13:40:39.693" LastActivityDate="2016-03-19T13:40:39.693" Title="How many GREEN coats?" Tags="|calculation-puzzle|lateral-thinking|alphametic|" AnswerCount="2" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8660" PostTypeId="2" ParentId="8658" CreationDate="2015-02-08T14:50:06.127" Score="8" Body="&lt;p&gt;Aah, such an old chestnut...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Just use the numbering A=1, B=2, C=3, D=4, E=5, ..., G=7, ..., N=14,..., R=18, ..., Z=26&lt;br&gt;&#xA; Then BLACK=29, RED=27, BLUE=40,&lt;br&gt;&#xA; and there were &lt;strong&gt;49 pupils&lt;/strong&gt; with GREEN coats&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-02-08T14:50:06.127" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8661" PostTypeId="2" ParentId="8658" CreationDate="2015-02-08T14:55:27.193" Score="1" Body="&lt;p&gt;There were 0 wearing green coats, you mentioned everybody's coat color already!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the other color coats were there to serve as red herrings and to show everybody's coat had been mentioned.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2015-02-08T15:11:11.777" LastActivityDate="2015-02-08T15:11:11.777" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8662" PostTypeId="1" AcceptedAnswerId="8663" CreationDate="2015-02-08T15:12:36.717" Score="17" ViewCount="2551" Body="&lt;p&gt;You surely remember that one of my grandmothers really hates mathematics (&lt;a href=&quot;https://puzzling.stackexchange.com/questions/8434/how-do-i-convince-my-grandmother/8436&quot;&gt;How do I convince my grandmother?&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yesterday I visited my other grandmother who really loves mathematics. She was in the middle of preparing 88 cherry pies for the church fair on sunday, and she did not have much time for anything else.  To keep me busy, she posed me a small riddle that also has to do with 88 pies (as she claimed):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Find all positive real numbers x with &#xA;  $x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor=88$.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(The puzzle uses the floor function $\lfloor x\rfloor$ which rounds x down to the next integer below x.)&lt;br&gt;&#xA;What is the answer to grandmother's puzzle, and how does it relate to cherry pies?&lt;/p&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-10-13T16:46:06.663" Title="A puzzle about baking 88 pies" Tags="|mathematics|calculation-puzzle|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8663" PostTypeId="2" ParentId="8662" CreationDate="2015-02-08T15:27:41.740" Score="29" Body="&lt;p&gt;Answer: the relation to pies is a relation to $\pi=3.14159265359$ (which is commonly pronounced &quot;pie&quot;). The convergents of the continued fractions for $\pi$ are the simplest approximants to $\pi$. The first few are &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$3$ (known to the Babylonians, and also used in the Bible), &lt;/li&gt;&#xA;&lt;li&gt;$22/7$ (known to Archimedes and the old Greek), &lt;/li&gt;&#xA;&lt;li&gt;$355/113$ (known to Tsu Chung Chih and the old Chinese).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;First I guessed $x=\pi$, but then the left hand side is irrational and the right hand side is integer.&lt;br&gt;&#xA;My second guess was $x=3$, which does not work either as $88$ is not a multiple of $3$.&lt;br&gt;&#xA;My third guess was $x=22/7\approx 3.1429$, and this indeed works: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Then $\lfloor x\rfloor=3$ &lt;/li&gt;&#xA;&lt;li&gt;and $\lfloor x\lfloor x\rfloor\rfloor=\lfloor (22/7)\cdot3\rfloor=9$&lt;/li&gt;&#xA;&lt;li&gt;and $\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor= \lfloor (22/7)\cdot9\rfloor=28$&lt;/li&gt;&#xA;&lt;li&gt;and $x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor=(22/7)\cdot28=88$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Hence $x=22/7$ works. &#xA;There cannot be any other solutions, as the function $f(x)=x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor$ is strictly increasing on the positive reals; therefore $f(x)&amp;lt;88$ for $x&amp;lt;22/7$ and $f(x)&amp;gt;88$ for $x&amp;gt;22/7$.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-08T15:38:45.920" LastActivityDate="2015-02-08T15:38:45.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8664" PostTypeId="2" ParentId="8655" CreationDate="2015-02-08T15:32:43.437" Score="4" Body="&lt;p&gt;I'm thinking of either 1.5 or 2.5 . Is your number greater than mine?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Yes          - 3&lt;/li&gt;&#xA;&lt;li&gt;No           - 1&lt;/li&gt;&#xA;&lt;li&gt;I don't know - 2&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9346" LastActivityDate="2015-02-08T15:32:43.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8665" PostTypeId="1" AcceptedAnswerId="8667" CreationDate="2015-02-08T15:43:33.283" Score="12" ViewCount="15150" Body="&lt;p&gt;I came across a puzzle (posed by my friends) that looked like this:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Which Number Replaces the Question Mark?&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;     6&#xA;  11 | 9&#xA;    \|/&#xA;2 -- ⬤ -- 4&#xA;    /|\&#xA;  16 | 14&#xA;     ?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A. 1&lt;/li&gt;&#xA;&lt;li&gt;B. 2&lt;/li&gt;&#xA;&lt;li&gt;C. 3&lt;/li&gt;&#xA;&lt;li&gt;D. 5&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Does anyone see any pattern here?&lt;/p&gt;&#xA;" OwnerUserId="9381" LastEditorUserId="69582" LastEditDate="2022-11-13T01:59:52.507" LastActivityDate="2023-05-12T23:38:17.000" Title="Which number replaces the question mark in &lt;16, 2, 11, 6, 9, 4, 14, ?&gt;" Tags="|pattern|visual|" AnswerCount="4" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8667" PostTypeId="2" ParentId="8665" CreationDate="2015-02-08T15:53:29.867" Score="22" Body="&lt;p&gt;I would guess that the question mark should be:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The pattern I see:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The picture is a tilted square, standing on one of its corners.&lt;br /&gt;&#xA; Along each edge of the square, the sum of the three numbers is &lt;span class=&quot;math-container&quot;&gt;$19$&lt;/span&gt;.&lt;br /&gt;&#xA; &lt;span class=&quot;math-container&quot;&gt;$2+11+6=19$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$6+9+4=19$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$4+14+?=19$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$?+16+2=19$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="76863" LastEditDate="2022-11-13T03:21:27.657" LastActivityDate="2022-11-13T03:21:27.657" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="8671" PostTypeId="5" CreationDate="2015-02-08T16:40:24.080" Score="0" Body="&lt;p&gt;A puzzle that is dressed up as a story or a longer description of a situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The clues to the puzzle are scattered in the story, and so some of the challenge is to find the clues in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Popular story types include &lt;a href=&quot;/questions/tagged/mystery&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;mystery&amp;#39;&quot; rel=&quot;tag&quot;&gt;mystery&lt;/a&gt;, &lt;a href=&quot;/questions/tagged/detective&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;detective&amp;#39;&quot; rel=&quot;tag&quot;&gt;detective&lt;/a&gt;, 'whodunnit' and 'murder'.&lt;/p&gt;&#xA;" OwnerUserId="11497" LastEditorUserId="11497" LastEditDate="2016-08-19T16:40:36.443" LastActivityDate="2016-08-19T16:40:36.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8672" PostTypeId="4" CreationDate="2015-02-08T16:40:24.080" Score="0" Body="A puzzle that is dressed up as a story or a longer description of a situation.&#xD;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-08T19:39:54.027" LastActivityDate="2015-02-08T19:39:54.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8673" PostTypeId="1" AcceptedAnswerId="9392" CreationDate="2015-02-08T17:06:19.903" Score="12" ViewCount="742" Body="&lt;p&gt;Professor Halfbrain has spent his entire weekend by cutting lots of wooden &lt;span class=&quot;math-container&quot;&gt;$50\times50$&lt;/span&gt; checkerboards into lots of polyominoes.&#xA;He looked at various pattern polyominoes with area &lt;span class=&quot;math-container&quot;&gt;$49$&lt;/span&gt;, and always tried to cut as many copies of the pattern as possible out of some wooden &lt;span class=&quot;math-container&quot;&gt;$50\times50$&lt;/span&gt; checkerboard.&#xA;The pattern polyomino could be rotated and flipped over, but it always had to be aligned with the cells of the checkerboard.&#xA;For most patterns, the professor was able to cut quite a number of copies while wasting only a small fraction of the checkerboard area.&lt;/p&gt;&#xA;&lt;p&gt;Professor Halfbrain has also proved two extremely deep theorems on such checkerboard cuttings.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Professor Halfbrain's first theorem&lt;/strong&gt;: For every possible polyomino pattern of area &lt;span class=&quot;math-container&quot;&gt;$49$&lt;/span&gt;, it is possible to cut at least &lt;strong&gt;one&lt;/strong&gt; copy of the pattern out of a wooden &lt;span class=&quot;math-container&quot;&gt;$50\times50$&lt;/span&gt; checkerboard.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Professor Halfbrain's second theorem&lt;/strong&gt;: There exist polyomino patterns of area &lt;span class=&quot;math-container&quot;&gt;$49$&lt;/span&gt;, for which it is not  possible to cut &lt;strong&gt;52&lt;/strong&gt; copies of the pattern out of a wooden &lt;span class=&quot;math-container&quot;&gt;$50\times50$&lt;/span&gt; checkerboard.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This puzzle asks you to improve the two theorems of professor Halfbrain and to make them even deeper.&#xA;Find an integer &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;, so that &amp;quot;one copy&amp;quot; in the first theorem may be replaced by &amp;quot;&lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; copies&amp;quot;, and so that &amp;quot;52 copies&amp;quot; in the second theorem may be replaced by &amp;quot;&lt;span class=&quot;math-container&quot;&gt;$x+1$&lt;/span&gt; copies&amp;quot; (again yielding true statements, of course).&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-03-10T09:02:29.437" Title="Polyominoes on a checkerboard" Tags="|mathematics|geometry|checkerboard|dissection|polyomino|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="8674" PostTypeId="2" ParentId="8655" CreationDate="2015-02-08T17:41:11.920" Score="2" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Is either of the following true:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Your number is &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Your number is &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt;, and the Riemann hypothesis is true?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If you say &amp;quot;yes&amp;quot;, it's 1, if you say &amp;quot;I don't know&amp;quot;, it's &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt;, and if you say &amp;quot;no&amp;quot;, it's &lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt;. And, if you say &amp;quot;yes&amp;quot; and I find out your number isn't &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;, I'm gonna want a proof.&lt;/p&gt;&#xA;&lt;p&gt;(You could replace &amp;quot;Riemann hypothesis&amp;quot; by anything else the answerer would not know. For instance, &amp;quot;Is a random number I just thought of equal to &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;?&amp;quot;)&lt;/p&gt;&#xA;" OwnerUserId="4765" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-02-08T17:41:11.920" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8678" PostTypeId="1" AcceptedAnswerId="8704" CreationDate="2015-02-08T18:48:03.760" Score="8" ViewCount="804" Body="&lt;p&gt;Three US cities of over 100,000 inhabitants have the following characteristic:&#xA;they include a &lt;strong&gt;particular point&lt;/strong&gt; from which, if you travel straight north, straight south, straight east, or straight west until you cross a US state border line, then in all four cases you will enter the same other US state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are these three cities, what state are they in, and what state will you enter on the four journeys?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The intended answers were Stamford, Omaha and New York City. (Anchorage and Washington D.C. do not go together with the wording: one does not enter the same &lt;em&gt;other&lt;/em&gt; US state). Since Omaha has been found last, I count it as the most difficult one of the three answers and I assign the &quot;correct answer&quot; sign to it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="68" LastEditDate="2018-11-13T08:27:31.633" LastActivityDate="2018-11-13T08:27:31.633" Title="USA geography puzzle" Tags="|geography|knowledge|" AnswerCount="3" CommentCount="8" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8679" PostTypeId="2" ParentId="6122" CreationDate="2015-02-08T19:26:28.540" Score="2" Body="&lt;p&gt;I think it's eyes. One can see themselves in someone else's eyes by reflection but they are reflections and hence laterally inverted but still represent symmetry. Twins because they are a pair. Grins because we can see many funny things with our eyes.&lt;/p&gt;&#xA;" OwnerUserId="8980" LastActivityDate="2015-02-08T19:26:28.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8680" PostTypeId="2" ParentId="8636" CreationDate="2015-02-08T19:35:21.363" Score="1" Body="&lt;p&gt;@JulianRosen Your link refers to a semi-infinite board, where either the width or height is limited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If neither is limited, I suppose the game can not be won against a perfect player, irrespective of who starts. It will continue endlessly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have not studied it in detail, but I know from personal experience that the only way to win (except controlling the Zugzwang) is to create a double trap, which can always be prevented.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-02-08T19:35:21.363" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="8683" PostTypeId="2" ParentId="8678" CreationDate="2015-02-08T20:07:09.213" Score="9" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;https://www.google.com/maps/place/Washington,+DC,+USA&quot; rel=&quot;nofollow noreferrer&quot;&gt;Washington, DC&lt;/a&gt;&lt;/strong&gt;&lt;br&gt;&#xA;North of the intersection of the VA/MD/DC borders and east of the easternmost point of the DC/VA border. You will always end up in Maryland.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;https://www.google.com/maps/place/Stamford,+Connecticut,+USA&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stamford, CT&lt;/a&gt;&lt;/strong&gt;&lt;br&gt;&#xA;No matter where you start in Stamford, if you go straight north, east, west, or south, you'll end up in NY.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;New York, NY&lt;/strong&gt;&lt;br&gt;&#xA;As mentioned in &lt;a href=&quot;https://puzzling.stackexchange.com/a/8690/68&quot;&gt;@KSmarts' answer&lt;/a&gt;, part of Ellis Island is an exclave of New York, surrounded by the state of New Jersey.&lt;br&gt;&#xA;&lt;a href=&quot;https://maps.google.com/maps?q=liberty+island&quot; rel=&quot;nofollow noreferrer&quot;&gt;Liberty Island&lt;/a&gt; is in a similar situation. It is part of New York City, but surrounded by New Jersey (waters), as explained in &lt;a href=&quot;https://www.youtube.com/watch?v=SgZ1f4ACZBQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;this video by CGP Grey&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditDate="2018-11-13T08:05:37.487" LastActivityDate="2018-11-13T08:05:37.487" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="8685" PostTypeId="2" ParentId="8662" CreationDate="2015-02-08T20:28:33.297" Score="17" Body="&lt;p&gt;You can solve this one without any guessing or knowing anything about pies or rational approximations to $\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, you note that you have&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x * (integer) = 88&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;so that $x$ must be a fraction: $88$ over some integer $m$, in fact.&lt;br&gt;&#xA;Then you note that $⌊x⌋^4 &amp;lt; x⌊x⌊x⌊x⌋⌋⌋ &amp;lt; x^4 &amp;lt; ⌊x+1⌋^4$.  You know that $3^4=81$, and $4^4$ is 256, so we must have $⌊x⌋ = 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leaves relatively few choices.  We know that $m$ must be no more than $29$ as otherwise $88/m$ would be less than 3.  We also know it can't be as low as $22$ as otherwise $x$ would be $4$ or more.  So we have the following options, which we just calculate out, throwing away anything that is too big or where the final result isn't actually $88$.  (We can use the fact that $3 &amp;lt; ⌊x⌋$.)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;trial x      x⌊x⌋   ⌊x⌊x⌋⌋ x⌊x⌊x⌋⌋ ⌊x⌊x⌊x⌋⌋⌋      x⌊x⌊x⌊x⌋⌋⌋&#xA;88/23        264/23   11   too much, 3*3*11 &amp;gt; 88&#xA;88/24=11/3   11       11   too much, 3*3*11 &amp;gt; 88&#xA;88/25        264/25   10   too much, 3*3*10 &amp;gt; 88&#xA;88/26=44/13  132/13   10   too much, 3*3*10 &amp;gt; 88&#xA;88/27        88/9      9   88/3      29           not integer&#xA;88/28=22/7   66/7      9   198/7     28           22/7 * (7*4) = 88&#xA;88/29        264/29    9   792/29    27           not integer&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So, there you have it: 22/7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Then you have to remember that it is used as a handy rational approximation to $\pi$.)&lt;/p&gt;&#xA;" OwnerUserId="2815" LastEditorUserId="8874" LastEditDate="2015-10-13T16:45:51.083" LastActivityDate="2015-10-13T16:45:51.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8686" PostTypeId="2" ParentId="8662" CreationDate="2015-02-08T20:56:53.537" Score="3" Body="&lt;p&gt;We know $x \neq 0$ and $\lfloor x \lfloor x \lfloor x \rfloor \rfloor \rfloor \neq 0$, since their product is nonzero.  Therefore we may divide both sides by the nonzero integer $n = \lfloor x \lfloor x \lfloor x \rfloor \rfloor \rfloor$, and we get that $x = \frac{88}{n}$ is rational.  So $x = \frac{p}{q}$ for some integers $p, q$, where $q \neq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, $\frac{88}{n} = \frac{p}{q}$ where $\frac{p}{q}$ is in lowest terms, so $p$ is a divisor of $88$.  Since $x &amp;gt; 0$, we may assume $p&amp;gt;0$ and $q &amp;gt;0$.  Thus $p$ is one of $1, 2, 4, 8, 11, 22, 44, 88$.  On the other hand, since $\lfloor x \rfloor &amp;gt; 0$, we have that $1\leq q \leq p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This narrows down the choices to $(p, q)$; there are now only $1 + 2 + 4 + 8 + 11 + 22 + 44 + 88 = 15 + 15(11) = 15(12) = 180$ possible pairs to check.  (Actually fewer than that, since the same rational number may show up as $p/q$ for various different pairs $(p, q)$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following Python 2.7 code returns the pairs $(p,q) = (22, 7)$, $(44, 14)$, and $(88,28)$, all of which represent fractions that reduce to $22/7$. Note that in Python 2.7, regular division of integers returns the integer quotient (aka the floor of the rational number).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;for p in [1, 2, 4, 8, 11, 22, 44, 88]:&#xA;    for q in xrange(1, p): #no need to check p = q since p/q = 1 is not a solution&#xA;        frac = 1&#xA;        for i in xrange(3):&#xA;            frac *= p&#xA;            frac /= q&#xA;        if frac*p == 88*q:&#xA;            print p,q&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In fact, many of these same comments still apply even if we don't assume $x&amp;gt;0$.  If we had assumed $x&amp;lt;0$ instead, then clearly $x \leq -1$: Otherwise, $-1 &amp;lt; x &amp;lt; 0$, so $\lfloor x \rfloor = -1$, and in that case $\lfloor x \lfloor x \rfloor \rfloor = \lfloor -x \rfloor = 0$ so the whole product on the left-hand side is $0$.  Further, $x = -1$ is not a solution, so $x &amp;lt; -1$.  Therefore $x = p/q$ where $p$ is a (positive) divisor of $88$ and $q$ is a (negative) integer such that $-p &amp;lt; q \leq -1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Checking these possible pairs yields no solutions, so $x = 22/7$ is the only real solution, positive or negative.&lt;/p&gt;&#xA;" OwnerUserId="8930" LastEditorUserId="8930" LastEditDate="2015-02-08T21:36:09.267" LastActivityDate="2015-02-08T21:36:09.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8690" PostTypeId="2" ParentId="8678" CreationDate="2015-02-08T22:46:10.110" Score="7" Body="&lt;p&gt;&lt;strong&gt;New York City, New York&lt;/strong&gt;&lt;br&gt;&#xA;Part of &lt;a href=&quot;http://en.wikipedia.org/wiki/Ellis_Island&quot;&gt;Ellis Island&lt;/a&gt; is territory of New York City, seen &lt;a href=&quot;https://www.google.com/maps/place/Ellis+Island&quot;&gt;here&lt;/a&gt;. However, most of Ellis Island, which has been expanded through land reclamation, is a part of New Jersey&lt;sup&gt;[1]&lt;/sup&gt;. So, traveling in any direction from this point (not just the four cardinal directions) will lead you into New Jersey.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;[1]&lt;/sup&gt; &lt;em&gt;New Jersey v. New York&lt;/em&gt;, 523 U. S. 767 (1998) &lt;/p&gt;&#xA;" OwnerUserId="8776" LastActivityDate="2015-02-08T22:46:10.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8691" PostTypeId="2" ParentId="8665" CreationDate="2015-02-09T04:33:49.493" Score="9" Body="&lt;p&gt;I'd say:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; the difference between the opposing pairs are rising primes.&#xA; 2-4 (2)&#xA; 11-14 (3)&#xA; 1-6 (5)&#xA; 16-9 (7)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9395" LastEditorUserId="76863" LastEditDate="2022-11-13T03:22:04.247" LastActivityDate="2022-11-13T03:22:04.247" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="8692" PostTypeId="1" CreationDate="2015-02-09T06:30:49.453" Score="-4" ViewCount="361" Body="&lt;p&gt;I have two &quot;messages&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;596f752061726520616c6d6f7374207468657265203b29&#xA;aHR0cDovLzE5My4xNDIuMTEyLjIyMDo5MC9tYWluX3Rhc2sucGRm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They have a hidden password, and I don't know exactly which one could have this pass (maybe both)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I found, the first password consists of numbers and letters from 'a' to 'f' so maybe it has something to do with hex? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="9397" LastEditorUserId="69582" LastEditDate="2023-12-05T04:15:40.753" LastActivityDate="2023-12-05T04:15:40.753" Title="find a word in this text" Tags="|cipher|" AnswerCount="1" CommentCount="2" ClosedDate="2015-02-10T03:43:32.977" ContentLicense="CC BY-SA 3.0" />
  <row Id="8693" PostTypeId="2" ParentId="8605" CreationDate="2015-02-09T06:32:17.580" Score="6" Body="&lt;p&gt;Sequence 1 is a list of:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Morse_code&quot; rel=&quot;nofollow&quot;&gt;Morse code dots+dashes&lt;/a&gt; for letters A through T (missing numbers are in &lt;strong&gt;bold&lt;/strong&gt;):&lt;br&gt;&#xA; 2, 4, 4, 3, &lt;strong&gt;1, 4&lt;/strong&gt;, 3, 4, 2, 4, 3, 4, 2, 2, 3, 4, 4, 3, &lt;strong&gt;3, 1&lt;/strong&gt;, . . &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sequence 3 is a list of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_Roman_emperors&quot; rel=&quot;nofollow&quot;&gt;Roman emperors&lt;/a&gt; (missing names in &lt;strong&gt;bold&lt;/strong&gt;):&lt;br&gt;&#xA; &lt;strong&gt;Augustus, Tiberius&lt;/strong&gt;, Caligula, Claudius, Nero, Galba, &lt;strong&gt;Otho, Vitellius, Vespasian&lt;/strong&gt;, Titus, &lt;strong&gt;Domitian&lt;/strong&gt;, Nerva, &lt;strong&gt;Trajan, Hadrian, Antoninus Pius&lt;/strong&gt;, Marcus Aurelius, Lucius Verus, . . .&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-02-10T09:30:45.930" LastActivityDate="2015-02-10T09:30:45.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8694" PostTypeId="2" ParentId="8692" CreationDate="2015-02-09T07:04:39.770" Score="7" Body="&lt;p&gt;The first one is &lt;a href=&quot;http://en.wikipedia.org/wiki/Hexadecimal&quot;&gt;&lt;strong&gt;hex-encoded ASCII&lt;/strong&gt;&lt;/a&gt;, and translates to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You are almost there ;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second is &lt;a href=&quot;http://en.wikipedia.org/wiki/Base64&quot;&gt;&lt;strong&gt;base64&lt;/strong&gt;&lt;/a&gt;, and translates to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://193.142.112.220:90/main_task.pdf&quot;&gt;http://193.142.112.220:90/main_task.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7308" LastActivityDate="2015-02-09T07:04:39.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8696" PostTypeId="2" ParentId="8605" CreationDate="2015-02-09T08:53:54.883" Score="5" Body="&lt;p&gt;Sequence 5 is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A, E, A, P, A, &lt;strong&gt;U&lt;/strong&gt;, U, &lt;strong&gt;U&lt;/strong&gt;, &lt;strong&gt;E&lt;/strong&gt;, &lt;strong&gt;C&lt;/strong&gt;, &lt;strong&gt;O&lt;/strong&gt;, &lt;strong&gt;E&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Second letter of the month:&lt;br/&gt;J&lt;strong&gt;a&lt;/strong&gt;nuary&lt;br/&gt;F&lt;strong&gt;e&lt;/strong&gt;bruary&lt;br/&gt;M&lt;strong&gt;a&lt;/strong&gt;rch&lt;br/&gt;A&lt;strong&gt;p&lt;/strong&gt;ril&lt;br/&gt;M&lt;strong&gt;a&lt;/strong&gt;y&lt;br/&gt;J&lt;strong&gt;u&lt;/strong&gt;ne&lt;br/&gt;J&lt;strong&gt;u&lt;/strong&gt;ly&lt;br/&gt;A&lt;strong&gt;u&lt;/strong&gt;gust&lt;br/&gt;S&lt;strong&gt;e&lt;/strong&gt;ptember&lt;br/&gt;O&lt;strong&gt;c&lt;/strong&gt;tober&lt;br/&gt;N&lt;strong&gt;o&lt;/strong&gt;vember&lt;br/&gt;D&lt;strong&gt;e&lt;/strong&gt;cember&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well done to OP for not listing twelve items as that would have taken 5 seconds :D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I'm not going to mention the other sequences as they are already answered by others:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/a/8693/5610&quot;&gt;Sequence 1 and 3&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/a/8608/5610&quot;&gt;Sequence 2&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/a/8606/5610&quot;&gt;Sequence 4&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-02-09T08:53:54.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8699" PostTypeId="2" ParentId="8665" CreationDate="2015-02-09T13:20:09.140" Score="1" Body="&lt;p&gt;Diffenrence in right left top and bottom is two (9,11 and 14,16), and left right 2,4. Vertical it's five, so I guess, as mentioned above, it's one (1,6).&lt;/p&gt;&#xA;" OwnerUserId="9405" LastActivityDate="2015-02-09T13:20:09.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8700" PostTypeId="1" AcceptedAnswerId="8701" CreationDate="2015-02-09T13:37:50.637" Score="4" ViewCount="733" Body="&lt;p&gt;I saw this puzzle many moons ago and so I assumed it would already be on here somewhere. I have looked and not found, though, and so I present to you a puzzle that baffled me when first I saw it.&lt;br&gt;&lt;br&gt;&#xA;What number replaces the question mark in the last example and why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3531 → 0&lt;br&gt;&#xA;3597 → 1&lt;br&gt;&#xA;3891 → 3&lt;br&gt;&#xA;9397 → 2&lt;br&gt;&#xA;3869 → 4&lt;br&gt;&#xA;2128 → 2&lt;br&gt;&#xA;3786 → 3&lt;br&gt;&#xA;3506 → 2&lt;br&gt;&#xA;2700 → 2&lt;br&gt;&#xA;7212 → 0&lt;br&gt;&#xA;7205 → 1&lt;br&gt;&#xA;2990 → 3&lt;br&gt;&#xA;9503 → 2&lt;br&gt;&#xA;1057 → 1&lt;br&gt;&#xA;1160 → 2&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1995 → ?&lt;br&gt;&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="4551" LastEditDate="2015-02-09T20:10:02.097" LastActivityDate="2015-02-09T20:10:02.097" Title="Which number does 1995 correspond to in this number pattern?" Tags="|lateral-thinking|pattern|" AnswerCount="2" CommentCount="1" ClosedDate="2015-03-08T00:28:32.247" ContentLicense="CC BY-SA 3.0" />
  <row Id="8701" PostTypeId="2" ParentId="8700" CreationDate="2015-02-09T13:41:41.387" Score="4" Body="&lt;p&gt;It is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You count the amount of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; enclosed areas in the numbers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3993" LastActivityDate="2015-02-09T13:41:41.387" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8702" PostTypeId="2" ParentId="8665" CreationDate="2015-02-09T13:52:04.780" Score="-3" Body="&lt;p&gt;2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at opposites and ignoring the sign, starting at the '2' and rotating clockwise&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2  - 4   = 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;11 - 14  = 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6  - ?   =    should be 4 therefore the number required is 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;9  - 16  = 5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this is not the answer being looked for, but another pattern :)&lt;/p&gt;&#xA;" OwnerUserId="9406" LastActivityDate="2015-02-09T13:52:04.780" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8703" PostTypeId="2" ParentId="8700" CreationDate="2015-02-09T13:59:03.003" Score="14" Body="&lt;p&gt;The answer provided by Mathias711 is correct, however there is an alternate and lengthier solution to this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can assign the number values....&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Like from 3531 = 0 and 7212 = 0, we can say that the digits 3,5,1,7&amp;amp; 2&#xA;  have NO VALUE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using this information we can get the value of 9&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As 3597 = 1, and the fact 3,5&amp;amp;7 have no value, we can say 9=1 (i.e 9&#xA;  has a value of ONE)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Similarly we can figure out the values of all digits, however we don't need to as we already have enough information to get the answer....&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1995= 2 (as 1&amp;amp;5 have no value and 9 has a value of ONE, hence two 9's will be 2)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9346" LastActivityDate="2015-02-09T13:59:03.003" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="8704" PostTypeId="2" ParentId="8678" CreationDate="2015-02-09T14:38:39.447" Score="7" Body="&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;Omaha, NE&lt;br&gt;&#xA;The famous hairpin of East Omaha (state line with Iowa)&lt;br&gt;&#xA;&lt;a href=&quot;https://maps.google.nl/maps?oe=UTF-8&amp;amp;ie=UTF-8&amp;amp;fb=1&amp;amp;gl=nl&amp;amp;ftid=0x87938dc8b50cfced:0x46424d4fae37b604&amp;amp;q=Omaha,+NE,+USA&amp;amp;ei=WsXYVMe_H8r3UJSggqAJ&amp;amp;ved=0CCUQ8gEoADAC&amp;amp;output=classic&amp;amp;dg=brw&quot; rel=&quot;noreferrer&quot;&gt;see Google maps&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2015-02-09T14:38:39.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8705" PostTypeId="1" AcceptedAnswerId="8708" CreationDate="2015-02-09T14:43:48.353" Score="35" ViewCount="2268" Body="&lt;p&gt;What's the burning sensation in the following Limerick?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; Complete, I'm a burning sensation,  &#xA; Less my second, provide information,  &#xA;    Then curtailed, an offense  &#xA;    Like the one with the quince,  &#xA; And beheaded, a doctor's location.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="9404" LastEditorDisplayName="user88" LastEditDate="2015-04-24T17:36:41.337" LastActivityDate="2015-04-24T17:36:41.337" Title="A burning sensation" Tags="|riddle|word|limerick|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8708" PostTypeId="2" ParentId="8705" CreationDate="2015-02-09T15:58:11.450" Score="40" Body="&lt;p&gt;The burning sensation is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a 'sting'.&lt;br&gt; Remove the second letter to get 'sing', which means provide information (what a criminal might do, to the police).&lt;br&gt; When 'sing' is curtailed (has the last letter removed), we get 'sin', which is an offense. An example is Adam and Eve's decision to eat the forbidden fruit from the tree of knowledge, and some say the fruit was a quince.&lt;br&gt; Finally, 'sin' is beheaded to give 'in', and &quot;The doctor is in&quot; is a common expression.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-02-09T16:04:10.540" LastActivityDate="2015-02-09T16:04:10.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8709" PostTypeId="1" AcceptedAnswerId="8711" CreationDate="2015-02-09T17:34:06.330" Score="13" ViewCount="3837" Body="&lt;p&gt;Solve the following rebus puzzle:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    0123 56789 DRACULA&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;http://en.wikipedia.org/wiki/Rebus&quot;&gt;http://en.wikipedia.org/wiki/Rebus&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2015-03-16T12:34:15.967" Title="Rebus with Dracula" Tags="|rebus|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8710" PostTypeId="2" ParentId="8612" CreationDate="2015-02-09T17:43:38.377" Score="3" Body="&lt;p&gt;7 unvoiced consonants:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Hypaspists&lt;/em&gt;: h, p, s, p, s, t, s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Runners-up with 6:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Ecstatics&lt;/em&gt;: k, s, t, t, k, s.&lt;br&gt;&#xA;&lt;em&gt;Expects&lt;/em&gt;: k, s, p, k, t, s.&lt;br&gt;&#xA;&lt;em&gt;Fisticuffs&lt;/em&gt;: f, s, t, k, f, s.&lt;br&gt;&#xA;&lt;em&gt;Pacifists&lt;/em&gt;: p, s, f, s, t, s.&lt;br&gt;&#xA;&lt;em&gt;Sexists&lt;/em&gt;: s, k, s, s, t, s.&lt;br&gt;&#xA;&lt;em&gt;Specificity&lt;/em&gt;: s, p, s, f, s, t.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;7 voiced consonants:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Invulnerable&lt;/em&gt;: n, v, l, n, r, b, l.&lt;br&gt;&#xA;&lt;em&gt;Unendangered&lt;/em&gt;: n, n, d, n, j, r, d.&lt;br&gt;&#xA;&lt;em&gt;Underminers&lt;/em&gt;: n, d, r, m, n, r, z.&lt;br&gt;&#xA;&lt;em&gt;Undependable&lt;/em&gt;: n, d, p, n, d, b, l.&lt;br&gt;&#xA;&lt;em&gt;Undeservedly&lt;/em&gt;: n, d, z, r, v, d, l.&lt;br&gt;&#xA;&lt;em&gt;Unremembered&lt;/em&gt;: n, r, m, m, b, r, d.&lt;br&gt;&#xA;&lt;em&gt;Unreservedly&lt;/em&gt;: n, r, z, r, v, d, l.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Unendangeredly&lt;/em&gt; has 4 hits on Google but isn't in any dictionary.&lt;br&gt;&#xA;&lt;em&gt;Undermobilized&lt;/em&gt; also has a few hits.&lt;/p&gt;&#xA;" OwnerUserId="1716" LastEditorUserId="1716" LastEditDate="2015-02-10T21:47:06.530" LastActivityDate="2015-02-10T21:47:06.530" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="8711" PostTypeId="2" ParentId="8709" CreationDate="2015-02-09T18:23:15.467" Score="14" Body="&lt;p&gt;That's got to be:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;out 4 + the Count&quot; = &lt;strong&gt;&quot;to be out for the count.&quot;&lt;/strong&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-03-16T12:34:15.967" LastActivityDate="2015-03-16T12:34:15.967" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8712" PostTypeId="2" ParentId="8625" CreationDate="2015-02-09T19:44:45.793" Score="1" Body="&lt;p&gt;I do not know the answer yet, but I have done some observations. I post them here so others could maybe benefit from them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 12 persons in 12 rooms, with number 13 to be placed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All names have equal number of letters, 6 in firstname, 7 in lastname, in total 13 letters. Firstnames seems to start with different letters, but Bobbie and Buster disrupt the pattern. Same goes with lastname, but here there are three pairs on A, C and P.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All dates are different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All days of the month is different as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All Months are different, except the one to be placed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All years are different as well. Some are ending in the same digit, and this shows some relation to the last digit of the room number. But here there are a break in the pattern as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the undisputable facts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here comes some of my speculations also:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that the number 12 is central to this puzzle. There are 12 Months, and since they are never used twice, I do think that two persons born in the same Month would be revealing to the puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since a lot is different, but still similar, I suspect that all parts of name and birth date is needed. It looks like a calculation of some sort. (This could however just be a trick of concealing the clue in between unnecessary hints). 13 letters combined with the month number could also lead to something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another point to consider is whether the room numbers has to be unique (that means running numbers, building on eachother), or that if the right person with the right birth date comes along, someone would needed to share a Double room?&lt;/p&gt;&#xA;" OwnerUserId="9385" LastActivityDate="2015-02-09T19:44:45.793" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8713" PostTypeId="1" AcceptedAnswerId="8876" CreationDate="2015-02-09T20:19:51.180" Score="2" ViewCount="1470" Body="&lt;p&gt;What is the riddle asking? What is the answer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't lose this one, do the opposite!&lt;br&gt;&#xA;But you won't need eyes to do so.&lt;br&gt;&#xA;Then let it percolate, and remove any doubles.&lt;br&gt;&#xA;Now we're ready for tea, and with it, our answer!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;HINT 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is heavily derived from a play on words.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;HINT 2: (may be a spoiler, check only if you're completely stuck):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Homonyms!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;HINT 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It has nothing to do with tea or biscuits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;HINT 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You may not need eyes, but that does not mean you have to remove any.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;HINT 5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer contains only one letter different than the question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5601" LastEditorDisplayName="user20" LastEditDate="2015-02-15T02:34:34.037" LastActivityDate="2015-02-15T02:34:34.037" Title="Now we're ready for tea" Tags="|riddle|" AnswerCount="10" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8714" PostTypeId="1" AcceptedAnswerId="8717" CreationDate="2015-02-09T20:22:01.693" Score="3" ViewCount="18359" Body="&lt;p&gt;Consider the set of natural numbers from 0 to 999, inclusive.  Suppose we select some subset of this set such that any pair of numbers from it share no more than one digit in the same positions.  To clarify,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Pairs such as 123 and 456 are obviously allowed since they don't share any digits in common.&lt;/li&gt;&#xA;&lt;li&gt;123 and 321 are allowed together since, while they have the same exact digits, only the 2 is actually in the same position.&lt;/li&gt;&#xA;&lt;li&gt;812 and 212 would &lt;em&gt;not&lt;/em&gt; be allowed since they both share the &quot;12&quot; at their end, which is two digits.&lt;/li&gt;&#xA;&lt;li&gt;206 and 246 are similarly not allowed; the shared digits don't need to be contiguous.&lt;/li&gt;&#xA;&lt;li&gt;Leading zeroes matter, so 5 and 205 are a forbidden pair as well.  (Think of 5 as &quot;005&quot;.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Remember this needs to hold for every pair of numbers in the subset.  What is the largest (as in &lt;strong&gt;most elements&lt;/strong&gt;) possible set which satisfies this condition?  How about the set with the largest &lt;strong&gt;sum&lt;/strong&gt; of elements?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Puzzle inspired by a conversation with a friend regarding mistyping ID numbers.  Apologies if this has already been asked here.)&lt;/p&gt;&#xA;" OwnerUserId="4819" LastEditorUserId="8874" LastEditDate="2015-10-13T16:38:11.127" LastActivityDate="2015-10-13T16:38:11.127" Title="Numbers with distinct digits" Tags="|mathematics|combinatorics|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8715" PostTypeId="2" ParentId="8714" CreationDate="2015-02-09T21:05:50.787" Score="5" Body="&lt;p&gt;I have a solution with 100 elements in the set.&#xA;Generate each number as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For element $0 \leq n \leq 99$, add &lt;code&gt;ABC&lt;/code&gt; to the set where &lt;code&gt;AB&lt;/code&gt; is $n$ (with a leading zero if $n &amp;lt; 10$) and &lt;code&gt;C&lt;/code&gt; is &lt;code&gt;A+B mod 10&lt;/code&gt;. So for $n=3$, &lt;code&gt;AB&lt;/code&gt; is &lt;code&gt;03&lt;/code&gt; and &lt;code&gt;C&lt;/code&gt; is &lt;code&gt;0+3 mod 10=3&lt;/code&gt;, meaning &lt;code&gt;ABC&lt;/code&gt; is &lt;code&gt;033&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are no pairs of numbers in this set that violate the condition (as far as I know, that is).&#xA;And there is no three-digit number &lt;code&gt;ABC&lt;/code&gt; with a different &lt;code&gt;AB&lt;/code&gt; part, since I covered all 100 possibilities.&#xA;Which means that this is the largest possible set which satisfies the condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no solution for the second question yet.&lt;/p&gt;&#xA;" OwnerUserId="448" LastEditorUserId="1955" LastEditDate="2015-02-09T21:23:35.630" LastActivityDate="2015-02-09T21:23:35.630" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8716" PostTypeId="2" ParentId="8714" CreationDate="2015-02-09T21:16:20.717" Score="6" Body="&lt;p&gt;To add to PrisonMonkeys' answer, &lt;em&gt;every set&lt;/em&gt; with 100 elements will have the same sum, which is 49950. For each place in the number, each digit can only appear in that place at most 10 times, because it cannot appear with the same value for another place twice. This means that each digit must appear in each place exactly 10 times, for there to be a total of 100 numbers in the set. Therefore we can sum each place separately to get 45000+4500+450=49950.&lt;/p&gt;&#xA;" OwnerUserId="9411" LastActivityDate="2015-02-09T21:16:20.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8717" PostTypeId="2" ParentId="8714" CreationDate="2015-02-09T22:08:34.057" Score="13" Body="&lt;p&gt;For what it's worth, I have a neat visualization of why the answer is 100:&lt;br&gt;&#xA;First imagine an easier case, where you have all the numbers 0-99 and you can't have any two numbers share any digit in the same place; we can draw a table of numbers like this:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/cpYDp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;Note that every number in a row shares its first digit, and every number in a column shares its second digit. Therefore you must pick numbers such that no two share a row or column. It's pretty clear that the maximum amount of numbers you're going to be able to pick is 10. Here's one solution:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/cliav.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;Now consider the case posed in the question. Get 10 of these 0..99 tables, except the second table has 100..199, the third 200..299, and so on. Then layer them one behind another to form a 10x10x10 grid from 0..999 (I have entered some of the numbers to give you the idea):&#xA;&lt;img src=&quot;https://i.stack.imgur.com/fh2Sv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;Note that every number in any 'strip' of ten cubes (row or column) shares 2 digits in common. Therefore you need to pick numbers such that no number shares a 'strip'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After thinking for a bit it is clear you will only be able to pick 100 such numbers.&lt;/p&gt;&#xA;" OwnerUserId="7308" LastActivityDate="2015-02-09T22:08:34.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8718" PostTypeId="1" AcceptedAnswerId="8720" CreationDate="2015-02-09T22:09:01.933" Score="8" ViewCount="582" Body="&lt;p&gt;In a room are 13 men, who are either Knights (truth tellers) or Knaves (lie tellers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the first man looks around and says &quot;The number of knights in this room is no greater than one&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the second man looks around and says &quot;The number of knights in this room is no greater than two&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the third man looks around and says &quot;The number of knights in this room is no greater than three&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;how many knights exactly are in the room?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9413" LastActivityDate="2015-02-09T22:42:49.017" Title="How many knights" Tags="|logical-deduction|liars|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8719" PostTypeId="2" ParentId="8713" CreationDate="2015-02-09T22:16:35.943" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the answer is hockey! it's the NHL &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;although it's very far fetched and if it's not it I won't bother posting how I got there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thinking:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Don't&lt;/em&gt; plus &lt;em&gt;the opposite&lt;/em&gt; makes it a double negative, so we start with &quot;lose this one&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;then &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; we drop the &quot;i&quot; because &quot;no eyes&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;then &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; we rearrange (percolate) the letters and drop the &quot;doubles&quot; &lt;br&gt; LTHN -&gt; NHLT&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;then &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; we drop the &quot;tea&quot; &lt;br&gt; NHL&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="5840" LastEditDate="2015-02-10T06:58:45.580" LastActivityDate="2015-02-10T06:58:45.580" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8720" PostTypeId="2" ParentId="8718" CreationDate="2015-02-09T22:30:38.977" Score="2" Body="&lt;p&gt;There are 7 knights.  The 7th knight is the first correct, and from 7-13 are 7 knights. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every knight added eliminates a knave while increasing the threshold for truthiness. So, since one is increasing at the same rate that the other is decreasing, the result will end up near the middle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are 6 knights, that would work down to knight 8, which would be untrue. If there are 8 knights, that would work down to knight 6, which is also untrue. 7 knights works down to knight 7. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-02-09T22:42:49.017" LastActivityDate="2015-02-09T22:42:49.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8721" PostTypeId="2" ParentId="8718" CreationDate="2015-02-09T22:30:58.423" Score="6" Body="&lt;p&gt;Man 13 must be telling the truth... there are not more than 13 knights so he is a knight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If he is the only knight, man 1 must be a knight which is a paradox.  Man 1 is a knave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we know now that there is no more than 12 knights, Man 12 is a knight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can continue until we find that men &quot;8-13&quot; are knights while men &quot;1-6&quot; are knaves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As this means there are at least 6 knight and 6 knaves, man 7 says &quot;there are no more than 7 knights&quot;.  He is correct so is a knight too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ergo: &quot;7-13&quot; are knights while men &quot;1-6&quot; are knaves&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2015-02-09T22:30:58.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8722" PostTypeId="1" AcceptedAnswerId="8724" CreationDate="2015-02-09T23:12:01.043" Score="17" ViewCount="2438" Body="&lt;p&gt;As we all know, Houdini is very good at coming out on top against improbable odds! And oh boy are the odds improbable. For his next great performance, he claims he can play simultaneous chess games against the 8 best chess players in the world. Although he is a very smart man in other ways, he is a mediocre chess player at best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8 chess tables are set in the room, and Houdini gets to play the white pieces in half the games. For this event, he scores 0 points for every loss, 1 point for every draw and 2 points for every win. He makes the ambitious claim:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I will score no less than 8 points&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After all the games have been played, Houdini has scored, in fact, exactly 8 points. How did he do it?&lt;/p&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="8874" LastEditDate="2015-10-13T16:34:48.203" LastActivityDate="2015-10-13T16:34:48.203" Title="Houdini versus the chess masters" Tags="|logical-deduction|lateral-thinking|chess|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="8723" PostTypeId="2" ParentId="8722" CreationDate="2015-02-09T23:27:54.580" Score="-4" Body="&lt;p&gt;He could have won 4 games and lost 4 (4x2 + 4x0 = 8), or drew all 8 (8x1 = 8).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there more to the question? I feel like I'm missing something..&lt;/p&gt;&#xA;" OwnerUserId="9062" LastActivityDate="2015-02-09T23:27:54.580" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8724" PostTypeId="2" ParentId="8722" CreationDate="2015-02-09T23:37:29.770" Score="26" Body="&lt;p&gt;Houdini cleverly paired up each game in which he played black with a game in which he played white. He then simply copied the moves played by his opponent to the corresponding game.&lt;/p&gt;&#xA;" OwnerUserId="1716" LastActivityDate="2015-02-09T23:37:29.770" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="8726" PostTypeId="2" ParentId="8713" CreationDate="2015-02-10T07:05:08.683" Score="13" Body="&lt;p&gt;Don't lose &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;em&gt;this one&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Remove the eye(s)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; = ths one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Percolate &lt;em&gt;it&lt;/em&gt; or &quot;and with &lt;em&gt;it&lt;/em&gt;&quot; (not sure which one, maybe both) &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; = ths one it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You &lt;em&gt;won't&lt;/em&gt; need eyes (again)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; = ths one t&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Remove doubles&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; = hs one &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add &lt;em&gt;tea&lt;/em&gt;('t') and rearrange&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; = honest&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5106" LastEditorUserId="5106" LastEditDate="2015-02-10T08:09:04.927" LastActivityDate="2015-02-10T08:09:04.927" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8728" PostTypeId="5" CreationDate="2015-02-10T09:03:10.930" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-02-10T09:03:10.930" LastActivityDate="2015-02-10T09:03:10.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8729" PostTypeId="4" CreationDate="2015-02-10T09:03:10.930" Score="0" Body="A puzzle or trivia question whose solution relies on external sources (like tables, dictionaries, or wikipedia)." OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-10T11:30:33.873" LastActivityDate="2015-02-10T11:30:33.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8731" PostTypeId="1" AcceptedAnswerId="8737" CreationDate="2015-02-10T09:56:19.280" Score="7" ViewCount="1147" Body="&lt;p&gt;Cosmo puts a golden weight of 11111 gram into the right pan of a balance.&#xA;Then Fredo starts adding his bronze weights to the balance: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;first Fredo adds a bronze weight of 1 gram to one of the pans; &lt;/li&gt;&#xA;&lt;li&gt;then he adds a bronze weight of 2 gram to one of the pans; &lt;/li&gt;&#xA;&lt;li&gt;then he adds a bronze weight of 4 gram to one of the pans;&lt;/li&gt;&#xA;&lt;li&gt;then he adds a bronze weight of 8 gram to one of the pans;&lt;/li&gt;&#xA;&lt;li&gt;and so on.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Every bronze weight weighs twice as much as the preceding one.&#xA;After Fredo has been doing this for some time, the two pans are in equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) Is it possible that Fredo has put the 16 gram bronze weight into the &lt;em&gt;left&lt;/em&gt; pan?&lt;br&gt;&#xA;(b) Is it possible that Fredo has put the 16 gram bronze weight into the &lt;em&gt;right&lt;/em&gt; pan?  &lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-09T09:12:47.083" LastActivityDate="2016-03-09T09:12:47.083" Title="A golden weight of 11111 gram" Tags="|mathematics|weighing|number-theory|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8732" PostTypeId="2" ParentId="8731" CreationDate="2015-02-10T10:28:04.057" Score="5" Body="&lt;p&gt;The possible answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1______11111&lt;br&gt;&#xA; 2__________4&lt;br&gt;&#xA; 16_________8&lt;br&gt;&#xA; 32________64&lt;br&gt;&#xA; 128______512&lt;br&gt;&#xA; 256_____2048&lt;br&gt;&#xA; 1024&lt;br&gt;&#xA; 4096&lt;br&gt;&#xA; 8192&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This will balance the scales at &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 13747g&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So to answer your questions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (a) Yes, it is possible&lt;br&gt;&#xA; (b) No, it's impossible. Proof:&lt;br&gt;&#xA;&lt;br&gt;&#xA; 11111 can be made from those bronze weights by adding the following ones: 8192, 2048, 512, 256, 64, 32, 4, 2, 1. From there we can find out that we still need 4096, 1024, 128, 16 and 8.&lt;br&gt;&#xA; If Cosmo would add 16 on the right scale, we would get 11127 on the right scale. To get that we would need 8192, 2048, 512, 256, 64, 32, 16, 4, 2 and 1 weights. But 16 is already on the right pan, so we can't use it on the left.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9352" LastEditorUserId="9352" LastEditDate="2015-02-10T11:08:13.237" LastActivityDate="2015-02-10T11:08:13.237" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8733" PostTypeId="1" AcceptedAnswerId="8735" CreationDate="2015-02-10T10:46:54.413" Score="19" ViewCount="7425" Body="&lt;p&gt;There are quite a few ways rebus puzzles that hide a certain meaning. Some of them that I can recall off-hand are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Images - An image that just refers to what it looks like&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Placement - Placing a word above, below, in between, etc. another word sometimes conveys a meaning.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Repetition - For example, repeating 'give get' 4 times indicates 'forgive &amp;amp; forget'.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Coloring -  For example, 'MOONCEON', with the first 2 and last 2 letters in blue, indicates 'once in a blue moon'.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Combination - Like H + {ear image} = HEAR&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;A good rebus puzzle will include various different such methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought it would be a good idea to have (a) detailed answer(s) on the site, indicating the methods used, with examples.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="8874" LastEditDate="2015-10-02T10:47:49.317" LastActivityDate="2015-10-02T10:47:49.317" Title="What are the tricks used in making a 'Rebus' puzzle?" Tags="|puzzle-creation|rebus|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8734" PostTypeId="2" ParentId="8731" CreationDate="2015-02-10T12:06:08.290" Score="4" Body="&lt;p&gt;&lt;code&gt;11111&lt;/code&gt; in binary is &lt;code&gt;0010 1011 0110 0111&lt;/code&gt;. Instead 16 is &lt;code&gt;0001 0000&lt;/code&gt;. In order to have the same amount in both plates, the binary rapresentation of the weight must be the same:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;R: 0010 1011 0110 0111 //We need a 1 in 5th position.&#xA;L: 0000 0000 0001 0000 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If we add &lt;code&gt;1&lt;/code&gt; in the &lt;code&gt;R&lt;/code&gt; plate we get &lt;code&gt;0010 1011 0110 1000&lt;/code&gt;, we then add &lt;code&gt;8&lt;/code&gt; to get the &lt;code&gt;1&lt;/code&gt; in 5th position. But then we have to add &lt;code&gt;2&lt;/code&gt; and &lt;code&gt;4&lt;/code&gt; in the plates and since the weights are increasing we would degenerate in an incompatible configuration:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Last 4 digits&#xA;R: 0000 add 4 or 2 -&amp;gt; 0100 or 0010&#xA;L: 0000 will never be possible to reach the 0100 or 0010 configuration (I only have 2 or 4 respectively)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This leads to the only configuration possible:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;R: 0010 1011 0111 0011 added 4, 8&#xA;L: 0000 0000 0001 0011 added 1, 2, 16&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We cleared powers from 0 to 4. The same reasoning applies to the rest of the binary number:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;R: 0010 1011 1011 0011 added 64&#xA;L: 0000 0000 1011 0011 added 32, 128&#xA;&#xA;R: 0010 1101 1011 0011 added 512&#xA;L: 0000 0101 1011 0011 added 256, 1024&#xA;&#xA;R: 0011 0101 1011 0011 added 2048&#xA;L: 0011 0101 1011 0011 added 4096, 8192&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And we're done. The plates are now in equilibrium and weight:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;R = 11111 + 4 + 8 + 64 + 512 + 2048 = 13747&#xA;L = 1 + 2 + 16 + 32 + 128 + 256 + 1024 + 4096 + 8192 = 13747&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So the first answer is yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case we put the 16 in the right plate, using the same approach:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;R: 0010 1011 0111 0111&#xA;L: 0000 0000 0000 0000 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now we need to get a &lt;code&gt;1&lt;/code&gt; in the 5th place in the left plate, but we can't since the only way is via remainder and we don't have a &quot;one&quot; in the left plate. We are in a incompatbile configuration:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;R: 0010 1011 0111 0111&#xA;L: 0000 0000 0000 1111 we don't have any way to put a 1 in the 5th position &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So the answer for the second question is no.&lt;/p&gt;&#xA;" OwnerUserId="4652" LastEditorUserId="4652" LastEditDate="2015-02-10T13:13:09.223" LastActivityDate="2015-02-10T13:13:09.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8735" PostTypeId="2" ParentId="8733" CreationDate="2015-02-10T12:26:36.117" Score="24" Body="&lt;p&gt;I, myself have read about/solved a few rebus puzzles myself recently; And from whatever I have read about them and learnt about them after solving, I can give the following tips:-&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;POSITION WITHIN A REBUS PUZZLE&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;One way of encrypting a Rebus is the positioning of the words and symbols, in terms of its place on the canvas area and to the other words and symbols in relation to it. You also sometimes need to think what other words might be used to describe a symbol, letter or word.&lt;/p&gt;&#xA;&lt;p&gt;In the example 12″ is in the word DOOR, 12″ (Twelve Inches) is also&#xA;known as a foot, so we have the answer of: Foot In The Door.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tywvh.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;HIGHLIGHTING WITHIN A REBUS PUZZLE&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Sometimes you can draw attention to some part of the Rebus,  by an arrow, text bolding or underlining, indicating where you should be looking for the clue.&lt;/p&gt;&#xA;&lt;p&gt;In the example the arrow is pointing to the first word which is&#xA;indicating “The Beginning”, so therefore the answer is The Beginning&#xA;Of The End.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bnMXd.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;DIRECTION WITHIN A REBUS PUZZLE&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A word can be spelt in a particular direction other than the usual like&#xA;left-to-right to reveal a hint to the solution.&lt;/p&gt;&#xA;&lt;p&gt;This Rebus shows Two “Feet” in the Left direction, to give the answer&#xA;Two Left Feet.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/xTN1I.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;SIZE WITHIN A REBUS PUZZLE&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;You may make use of large or small words or symbols  to&#xA;convey concepts like big, small, skinny, wide, etcetera.&lt;/p&gt;&#xA;&lt;p&gt;Take this Rebus puzzle below, the word “Deal” is big so it&#xA;represents Big Deal. &lt;img src=&quot;https://i.stack.imgur.com/7oiOo.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;NUMBER(S) WITHIN A REBUS PUZZLE&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;You can make use of numbers and their&#xA;closeness in sound of 0, 1, 2, 4, 8, 10,&#xA;etcetera, to other English words.&lt;/p&gt;&#xA;&lt;p&gt;This cryptic example shows 2 “Funny” 4 “Words”, and so the answer is&#xA;Too Funny For Words. &lt;img src=&quot;https://i.stack.imgur.com/EoJiZ.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;COLOUR WITHIN A REBUS PUZZLE&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;You can adjust the colour to be relevant in&#xA;the answer. Sometimes Rebus puzzles may make use of colour names like Rose.&lt;/p&gt;&#xA;&lt;p&gt;In this extremely easy example “Tickled” is Pink so the answer is&#xA;Tickled Pink. &lt;img src=&quot;https://i.stack.imgur.com/qRRGl.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;STYLE WITHIN A REBUS PUZZLE&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;You can adjust the style of font to represent other&#xA;words such as tall, petite, thin, slanted, bold etc.&lt;/p&gt;&#xA;&lt;p&gt;The “Skinny” letters in the example makes the answer is Skinny&#xA;Dipping. &lt;img src=&quot;https://i.stack.imgur.com/Y4smx.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;SOUND WITHIN A REBUS PUZZLE&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Your Rebus puzzle can be phonologically cryptic&#xA;(to do with sound). The sound might not always be exactly like the&#xA;answer, but it will always be close enough to recognise.&lt;/p&gt;&#xA;&lt;p&gt;Take the Rebus below, the “2222”‘ when pronounced together&#xA;sounds like Tues, then with the word “Day” we have Tuesday&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zvBhR.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;IMAGE WITHIN A REBUS PUZZLE&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;You can make use of images. The image&#xA;will definitely form part or most of the clue.&lt;/p&gt;&#xA;&lt;p&gt;Take this One, the Rebus shows an “Apple” and the mathematical number&#xA;for “Pi” which gives us Apple Pie !&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jm2Lf.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9346" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-02-10T12:26:36.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8736" PostTypeId="2" ParentId="8731" CreationDate="2015-02-10T12:35:06.457" Score="6" Body="&lt;p&gt;Fredo's weights are $1_2, 10_2, 100_2, ...$, so writing the sum of a subset of the weights written in base 2 gives an easy way to identify the individual weights visually. For example, if he placed the 8g, (not 4g), 2g and 1g weights into the left pan, the sum of the weights is $1011_2$ or $11_{10}$. The right pan would then get the complement, namely $0100_2$, which we can get by subtracting from $1111_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $n$ be the sum of the weights Fredo puts in the left pan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a $k$-digit base-2 number $n$, let us call its complement $C(n,k)$. Then $C(n,k) = (2^k - 1) - n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many weights did Fredo use (i.e. what is $k$)? Since $11111$ has 14 digits in base 2, the weights in the left pan must also have at least 14 base-2 digits to balance. If Fredo used 16 weights, the largest would tip the scale whichever side it was placed, so $k$ is either $14$ or $15$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 16g weight is in the right pan if and only if $n$ has a $0$ in position 5 since $16_{10} = 10000_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose Fredo used $k$ weights and puts a subset weighing $n$ grams into the left pan. Then the right pan would receive the remaining weights, $C(n,k)$ grams. For the pans to balance, we require (writing now in base-10 for simplicity):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n = 11111 + C(n,k) = 11111 + 2^k - 1 - n$, which gives $n = (11110 + 2^k)/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $k=14$ we have $n=13747$, and when $k=15$ we have $n=21939$. In both cases, writing $n$ in base 2 shows a '1' in position 5, so the 16g weight is always in the left pan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is (a) yes and (b) no.&lt;/p&gt;&#xA;" OwnerUserId="9422" LastEditorUserId="9422" LastEditDate="2015-02-11T04:54:13.597" LastActivityDate="2015-02-11T04:54:13.597" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8737" PostTypeId="2" ParentId="8731" CreationDate="2015-02-10T12:43:41.333" Score="5" Body="&lt;p&gt;Adding a weight to the left pan is the same as adding that weight to the right pan and twice that weight to the left pan. So we can treat adding the first $n$ powers of two to the pans the same as if we added all of them to the right pan, and double some of them to the left pan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have added $2^n - 1$ to the right pan, and some even but otherwise arbitrary binary number of up to $n+1$ digits to the left pan. Calling the starting weight $x$, to balance the scales we need $2(2^n - 1) \ge  x + 2^n - 1$, which means $2^n - 1 \ge x$. So we must add a number with at least as many binary digits as the starting number. The desired number on the left will be $ x + 2^n - 1$, and we interpret it as follows: divide by 2; each bit in the result represents a weight on the left, each 0 represents a weight on the right. (Incidentally, this shows that you can't do this trick with an even starting weight).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;11111 decimal expands to 10101101100111 binary. So we must add 11111111111111 to it, getting 110101101100110. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Divide by 2 to get 11010110110011, and we see that the bit representing 16 is set, so 16 is on the left.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, if we use a larger value of $n$, it will still work; but it won't change any of the original $n$ digits in our generated number (since adding powers of two does not change the lower bits in a binary number).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Therefore, 16 cannot be anywhere but on the left.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a) Yes&lt;br&gt;&#xA; b) No.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit: there is no limit to the number of weights that can be added. Once we have an equilibrium, consider the largest weight ($2^n$ for some $n$), which is on side A (whatever side that happens to be). The next weight to be added is $2^{n+1}$. We simply shift the $n$th weight to side B; this reduces side A by $2^n$ and increases side B by $2^n$, a net difference of $2^{n+1}$. Now we put the new weight on side A, increasing it again by $2^{n+1}$, and restoring the balance.&#xA;However, this only moves the heaviest weight; all the others stay where they are. So once a weight is no longer the heaviest, it never moves again.&#xA;(Cosmo doesn't move weights, but the logic still applies).&lt;/p&gt;&#xA;" OwnerUserId="6031" LastEditorUserId="12408" LastEditDate="2015-11-07T02:02:08.403" LastActivityDate="2015-11-07T02:02:08.403" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8739" PostTypeId="1" CreationDate="2015-02-10T13:56:50.717" Score="-2" ViewCount="183" Body="&lt;p&gt;Our little hero starts off in a round (or square) pit with very high walls, an escape does not seem possible. However also in the pit is Bowser! How did that happen? Both of them are now facing off each other, which means, Mario is on the run and Bowser chases him. &#xA;Both of them run at the same constant speed. &#xA;The question now is: Can Mario outrun Bowser and will never be caught or is he sooner or later doomed to be caught by it? Do both of them have an optimal strategy to run? &#xA;(No attack moves, jumps, power ups etc. allowed! Just running!)&lt;/p&gt;&#xA;" OwnerUserId="8863" LastEditorUserId="5373" LastEditDate="2016-10-24T00:33:48.430" LastActivityDate="2016-10-24T00:33:48.430" Title="Bowser and Mario in a Pinch!" Tags="|logical-deduction|strategy|" AnswerCount="2" CommentCount="0" ClosedDate="2015-02-10T18:05:06.277" ContentLicense="CC BY-SA 3.0" />
  <row Id="8740" PostTypeId="2" ParentId="8713" CreationDate="2015-02-10T14:15:31.630" Score="5" Body="&lt;p&gt;Don't lose this one, do the opposite!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You should keep this one, but DO (process) the opposite. So keep &lt;em&gt;opposite&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But you won't need eyes to do so.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; remove the &lt;em&gt;i&lt;/em&gt; -&gt; &lt;em&gt;opposte&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then let it percolate, and remove any doubles.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Use &lt;em&gt;it&lt;/em&gt; as a filter, removing the &lt;em&gt;t&lt;/em&gt;. Then remove the doubles -&gt; &lt;em&gt;se&lt;/em&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now we're ready for tea,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Add &lt;em&gt;t&lt;/em&gt; -&gt; &lt;em&gt;set&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and with it, our answer!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;set it&lt;/em&gt;, the  tea!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So&#xA;&lt;code&gt;What is the riddle asking?&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The riddle is asking to &lt;a href=&quot;http://www.wikihow.com/Set-a-Table-for-a-Tea-Party&quot;&gt;set the tea&lt;/a&gt;! Happy tea-time! :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4652" LastActivityDate="2015-02-10T14:15:31.630" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8741" PostTypeId="2" ParentId="8731" CreationDate="2015-02-10T14:36:57.830" Score="6" Body="&lt;p&gt;The binary of &lt;code&gt;11111&lt;/code&gt; is &lt;code&gt;010 1011 0110 0111&lt;/code&gt; and the other pan is effectively &lt;code&gt;000 0000 0000 0000&lt;/code&gt;. Adding the weights is the same as adding 1 to each spot, and we want the two numbers to end up the same. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scenario that we must avoid is having &lt;code&gt;0 0&lt;/code&gt; and needing to add a weight to one side or the other. So, if the next digit on the right pan is 0, add your current weight to the right pan. Otherwise, add it to your left pan.&lt;br&gt;&#xA;So you'd end up with (orig, left, right):&lt;br&gt;&#xA;&lt;code&gt;010 1011 0110 0111&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;011 0101 1011 0011&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;011 0101 1011 0011&lt;/code&gt;&lt;br&gt;&#xA;= 13747&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now where the 16 weight can go, let's look at that part of the sequence: &lt;code&gt;10 0111&lt;/code&gt; where 16 = &lt;code&gt;01 0000&lt;/code&gt;. The only way we can add the 16 to the right pan is if there is a 1 there already (e.g. &lt;code&gt;11 0011&lt;/code&gt;). But, if we add it to the right, that violates our &quot;look ahead and see 0&quot; rule. So:&lt;br&gt;&#xA;a) yes&lt;br&gt;&#xA;b) no&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Process: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;010 1011 0110 0111&#xA;000 0000 0000 0000&#xA;                 - See 1 (left)&#xA;010 1011 0110 0111&#xA;000 0000 0000 0001&#xA;                -  See 1 (left)&#xA;010 1011 0110 0111&#xA;000 0000 0000 0011&#xA;               -   See 0 (right)&#xA;010 1011 0110 1011&#xA;000 0000 0000 0011&#xA;              -    See 0 (right)&#xA;010 1011 0111 0011&#xA;000 0000 0000 0011&#xA;            -      See 1 (left)&#xA;010 1011 0111 0011&#xA;000 0000 0001 0011&#xA;           -       See 1 (left)&#xA;010 1011 0111 0011&#xA;000 0000 0011 0011&#xA;          -        See 0 (right)&#xA;010 1011 1011 0011&#xA;000 0000 0011 0011&#xA;         -         See 1 (left)&#xA;010 1011 1011 0011&#xA;000 0000 1011 0011&#xA;       -           See 1 (left)&#xA;010 1011 1011 0011&#xA;000 0001 1011 0011&#xA;      -            See 0 (right)&#xA;010 1101 1011 0011&#xA;000 0001 1011 0011&#xA;     -             See 1 (left)&#xA;010 1101 1011 0011&#xA;000 0101 1011 0011&#xA;    -              See 0 (right)&#xA;011 0101 1011 0011&#xA;000 0101 1011 0011&#xA;  -                See 1 (left)&#xA;011 0101 1011 0011&#xA;001 0101 1011 0011&#xA; -                 Just 1 remaining (left)&#xA;011 0101 1011 0011&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-02-10T14:48:38.513" LastActivityDate="2015-02-10T14:48:38.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8742" PostTypeId="2" ParentId="8739" CreationDate="2015-02-10T14:42:14.573" Score="0" Body="&lt;p&gt;If the pit is a square, Mario is doomed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the pit is a square &lt;code&gt;1 m x 1 m&lt;/code&gt;. Mario ($M$) has to maximize the distance from Bowser ($B$) and $B$ minimize it. The strategy for $B$ is to stand in the middle of the square. The most distant point for $M$ to stand is at one of the corners which is at ${\sqrt2 \over 2}m$  which is greater than ${1\over2}m$, the distance from the center to one of the sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $M$ starts to run following one of the sides, $B$ has only to run, at the same speed, towards that side. They will met at $1\over2$ of the side itself, and $M$ is captured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B$ has to run towards $M$ following the diagonal of the square, decreasing the dimension of the square until $M$ is captured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When running towards the edge of the pit, $B$ has to run in a straight orthogonal line. If $M$ runs back $B$ runs on a $45°$ line orthogonal to the diagonal of the square. In this way $B$ runs a shorter path since he now runs on the cathetus instead of the hypotenuse of the triangle made with the diagonal of the pit, assures that $M$ doesn't escape and decreases the distance between them when getting back to the diagonal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $M$ at some point decide to run diagonally, $B$ only has to decrease the run and wait $M$ intersect his path.&lt;/p&gt;&#xA;" OwnerUserId="4652" LastEditorUserId="4652" LastEditDate="2015-02-10T14:48:29.297" LastActivityDate="2015-02-10T14:48:29.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8743" PostTypeId="1" AcceptedAnswerId="8749" CreationDate="2015-02-10T14:44:55.260" Score="9" ViewCount="3621" Body="&lt;p&gt;There is a player who controls only a queen, while his opponent controls a knight. Both pieces start from opposite corners of the board. The aim is to kill the knight, without having your queen killed. The board is square, of side $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case I: The opponent only knows that he should&lt;br&gt;&#xA;(a) kill the queen on his current move, if possible&lt;br&gt;&#xA;(b) escape to a slot where he can't be killed on the immediate next move&lt;br&gt;&#xA;(c) move randomly to any one of the 'safe' slots, as in (b)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case II: The opponent is a perfect player, and can look into infinite depth and all possible scenarios of the game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the maximum board size for which the queen is 'guaranteed' to win, in Case I and in Case II?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes:&lt;br&gt;&#xA;1. In Case I, since moves are random, it is possible that if you can keep coming back to the same position, the knight will eventually make a bad move.&lt;br&gt;&#xA;2. 'randomly' means that all 'safe' moves have equal probability, but as the queen's win needs to be guaranteed, this should not affect the answer.&lt;br&gt;&#xA;3. There is no limit on moves, time, etc. and no possibility of a draw.&#xA;4. The knight always starts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Since this got answered so quickly, let me make a harder version: The knight is allowed to decide the initial placement of the queen as well as itself. Rest of the question is same.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4269" LastEditDate="2015-02-10T15:26:28.557" LastActivityDate="2015-02-11T06:54:47.487" Title="Queen chasing a knight (chess pieces)" Tags="|strategy|chess|game|board-games|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8744" PostTypeId="2" ParentId="8739" CreationDate="2015-02-10T14:45:18.057" Score="0" Body="&lt;p&gt;If this pit goes through the entire planet in one long line, then Bowser will not be able to catch Mario.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if the pit is round, then Mario has no chance. No matter where he runs, Bowser will run towards him. When Mario reaches the wall, he has to go left or right. Depending where he goes, Bowser will slightly move too that side too making the distance smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mario will probably keep on running in circles, but Bowser will make some corrections and shorten the distance with every turn of Mario.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, Mario is doomed.&lt;/p&gt;&#xA;" OwnerUserId="9352" LastActivityDate="2015-02-10T14:45:18.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8745" PostTypeId="2" ParentId="8743" CreationDate="2015-02-10T14:54:42.847" Score="14" Body="&lt;p&gt;The queen can always win. If the queen goes first, obviously it will win (just take the knight). If the knight goes first, there are only 2 places to go (mirrored, so effectively the same):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;- - - - - +&#xA;. . . . . |&#xA;. . K . . |&#xA;. . . . . |&#xA;. . . . . |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;All the queen has to do is move: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;- - - - - +&#xA;. . . . . |&#xA;. . K . . |&#xA;. . Q . . |&#xA;. . . . . |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;All the places the knight can go are:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;- - - - - +&#xA;x . . . x |&#xA;. . K . . |&#xA;x . Q . x |&#xA;. x . x . |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which are all in the queen's sight. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's say they start on adjacent corners (K in top right). &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;- - - - - - + -&amp;gt; - - - - - - + or - - - - - - +&#xA;. . Q . . . |    . . . . . . |    . . . . . . |&#xA;. . . K . . |    . . . . . . |    . . . . . . |&#xA;. . . . . . |    Q K . . . . |    . . . . . . |&#xA;. . . . . . |    . . . . . . |    . . Q . K . |&#xA;&#xA;or&#xA;&#xA;- - - - - +    - - - - - +&#xA;. Q . . . |    . . . . . |&#xA;. . . . . |    . . . . . |&#xA;. . . K . |    . . . . . |&#xA;. . . . . |    . . . . . |&#xA;. . . . . |    . Q K . . |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Either way, the Queen has now pinned the knight against an edge and can eventually reach the original scenario. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-02-10T15:30:03.973" LastActivityDate="2015-02-10T15:30:03.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8746" PostTypeId="2" ParentId="8743" CreationDate="2015-02-10T14:56:50.773" Score="9" Body="&lt;p&gt;Let's assume that opponent's Knight stands on &lt;em&gt;A1&lt;/em&gt;, then my queen will be on &lt;em&gt;{last letter}{last number}&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For both cases:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I will always win in max 2 turns on any board size:&lt;br&gt;&#xA; If I start: my queen will immediately kill the knight&lt;br&gt;&#xA; If opponent starts: he can only move to &lt;em&gt;B3&lt;/em&gt; or &lt;em&gt;C2&lt;/em&gt;. Then I would move my queen to &lt;em&gt;C3&lt;/em&gt;. At this moment Knight is under attack of my queen and he only has following moves:&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/tlBl0.png&quot; alt=&quot;option 1&quot;&gt;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/OBH6z.png&quot; alt=&quot;option 2&quot;&gt;&lt;br&gt;&#xA; No matter where he moves, my queen will always kill him on her second move&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For the updated question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If Knight starts, game is over(unless it's a very kind Knight who didn't place himself to immediately kill the queen).&lt;br&gt;&#xA; If queen starts, first move will be to get to the same line/column as the knight. Then knight can move either 1 or 2 lines/columns away from the queen. Next step of my queen would be to move as close to the knight as possible. There are 2 possibilities(actually 8, but other 6 are the same, just turned around):&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/pOmc0.png&quot; alt=&quot;possibility 1&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/nYcfe.png&quot; alt=&quot;possibility 2&quot;&gt;&lt;br&gt;&#xA; In both cases, yellow are the ones where knight can go and won't immediately be killed by the queen. But possibility 1 will only happen once, due to the fact that queen can get to adjacent square in her next move, no matter where knight goes.&lt;br&gt;&#xA; And then it will just be a chase to get the knight. No matter what happens, eventually the queen will kill the knight&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9352" LastEditorUserId="9352" LastEditDate="2015-02-11T06:54:47.487" LastActivityDate="2015-02-11T06:54:47.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8747" PostTypeId="2" ParentId="8713" CreationDate="2015-02-10T16:07:05.260" Score="5" Body="&lt;p&gt;Don't lose this one, do the opposite!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; So we &quot;gain this one&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But you won't need eyes to do so.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; -which becomes &quot;gan ths one&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then let it percolate, and remove any doubles.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; -then &quot;hostage nn&quot; -&gt; &quot;hostage&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now we're ready for tea, and with it, our answer! &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; So, &quot;Who do we release at tea time? The hostage!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9278" LastActivityDate="2015-02-10T16:07:05.260" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8748" PostTypeId="1" AcceptedAnswerId="8751" CreationDate="2015-02-10T16:14:33.943" Score="-6" ViewCount="557" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the most commonly used everyday English word that contains the four letters &lt;strong&gt;YHVH&lt;/strong&gt; in some order?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The answer is unique and clear, unless you try hard to stretch the rules that are explained in the following lines.&lt;/li&gt;&#xA;&lt;li&gt;contains letters &lt;strong&gt;YHVH&lt;/strong&gt; = contains those 4 letters (in some order) and some other letters (in some order)&lt;/li&gt;&#xA;&lt;li&gt;the word must not contain any special symbols (like hyphens) nor blanks&lt;/li&gt;&#xA;&lt;li&gt;&quot;everyday English word&quot; = contained in this form as the main entry in most/all&#xA;the standard dictionaries&lt;/li&gt;&#xA;&lt;li&gt;in particular adverbs like overthoughtfully, overchildishly, thievishly do not count as they are not the main entry&lt;/li&gt;&#xA;&lt;li&gt;in particular words like vasorrhaphy, orthoveratraldehyde, phosphorylative do not count as they are not everyday words&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="39" LastEditDate="2015-02-10T21:53:53.753" LastActivityDate="2015-06-30T14:57:02.433" Title="The four letters YHVH" Tags="|word|english|" AnswerCount="4" CommentCount="11" ClosedDate="2015-06-30T16:02:23.697" ContentLicense="CC BY-SA 3.0" />
  <row Id="8749" PostTypeId="2" ParentId="8743" CreationDate="2015-02-10T16:16:56.183" Score="15" Body="&lt;p&gt;For the updated version, I'll assume that after the placement of the pieces, the queen moves first. (Otherwise the knight could arrange things so that the queen is captured on the first move.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this situation, the queen will still always win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wherever the two pieces start, the queen can get to a space adjacent to the knight in no more than 3 moves. First, move to the same row as the knight. (Or the same column, but we can say row without loss of generality.) The knight's possible moves put it either 1 or 2 rows away. If it moves by one row, the queen moves to the space adjacent. If it moves two rows, the queen moves to be in the same column. Now, if the knight moves one column to the side, the queen can move vertically to an adjacent space. If the knight moves two columns to the side, the queen can move diagonally to an adjacent space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, now we have the queen vertically adjacent to the knight, which leaves the knight with two possible moves, as shown:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;..o.o..&#xA;.x...x.&#xA;...N...&#xA;.x.Q.x.&#xA;..x.x..&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The &lt;code&gt;o&lt;/code&gt;'s represent possible moves. Whichever space the knight moves to, let the queen move one space diagonally to be in the same column. Then we get this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;..o.o..&#xA;.o...o.&#xA;...N...&#xA;.o...o.&#xA;..xQx..&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For any of the four lower spaces, the queen can move either horizontally or vertically to be one space below the knight, returning to the first setup, but one row higher. So, suppose that the knight takes one of the moves that take it two spaces up. If the queen moves diagonally up to match again, we get:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;..o.o..&#xA;.o...o.&#xA;...N...&#xA;.x...x.&#xA;..x.x..&#xA;...Q...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Again, the leftmost and rightmost moves allow a diagonal move from the queen to return to a previous state. So, let us suppose that the knight moves farther up again. Then let the queen move vertically 4 spaces, to be diagonally adjacent to the knight.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;...x.o..&#xA;..o...x.&#xA;....N...&#xA;..xQ..x.&#xA;...x.o..&#xA;........&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now the knight has three available moves, but for any of them, the queen can follow up with a diagonal move to be one space below, returning to the first configuration. Even if the knight moves down, this was preceded by three moves up, so the knight has moved up the board overall. The queen can continue forcing the knight in one direction like this until they reach the edge of the board, which further restricts the knight's moves and allows the queen to capture.&lt;/p&gt;&#xA;" OwnerUserId="8776" LastEditorUserId="8776" LastEditDate="2015-02-10T18:43:32.127" LastActivityDate="2015-02-10T18:43:32.127" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8750" PostTypeId="1" AcceptedAnswerId="8754" CreationDate="2015-02-10T16:18:43.850" Score="10" ViewCount="4965" Body="&lt;p&gt;Solve the following rebus puzzle:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  I I&#xA;  I I I&#xA;  I I I I&#xA;___________&#xA;&#xA;   O   O&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;http://en.wikipedia.org/wiki/Rebus&quot;&gt;http://en.wikipedia.org/wiki/Rebus&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="8874" LastEditDate="2015-02-10T17:06:53.747" LastActivityDate="2015-02-10T17:06:53.747" Title="Another Rebus puzzle" Tags="|english|rebus|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8751" PostTypeId="2" ParentId="8748" CreationDate="2015-02-10T16:30:21.703" Score="7" Body="&lt;p&gt;How about &lt;strong&gt;heavyweight&lt;/strong&gt;? I'm still confused on how this is a puzzle.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2015-02-10T16:30:21.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8752" PostTypeId="2" ParentId="8748" CreationDate="2015-02-10T16:36:33.543" Score="4" Body="&lt;p&gt;There does not appear to be a unique answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possible answers include &lt;strong&gt;h&lt;/strong&gt;ea&lt;strong&gt;vy&lt;/strong&gt;weig&lt;strong&gt;h&lt;/strong&gt;t, t&lt;strong&gt;h&lt;/strong&gt;ie&lt;strong&gt;v&lt;/strong&gt;is&lt;strong&gt;h&lt;/strong&gt;l&lt;strong&gt;y&lt;/strong&gt;, and &lt;strong&gt;h&lt;/strong&gt;ea&lt;strong&gt;vyh&lt;/strong&gt;earted.&lt;/p&gt;&#xA;" OwnerUserId="8776" LastActivityDate="2015-02-10T16:36:33.543" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8753" PostTypeId="2" ParentId="8750" CreationDate="2015-02-10T16:58:04.700" Score="9" Body="&lt;p&gt;Is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Glasses? Two eyes, three eyes, four eyes. Four eyes = glasses = O   O&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5601" LastEditorUserId="5601" LastEditDate="2015-02-10T17:04:27.387" LastActivityDate="2015-02-10T17:04:27.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8754" PostTypeId="2" ParentId="8750" CreationDate="2015-02-10T17:02:21.210" Score="11" Body="&lt;p&gt;I love rebus puzzles!! That one is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Os under the Is&quot; = &lt;strong&gt;&quot;circles under the eyes&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-02-10T17:02:21.210" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8755" PostTypeId="2" ParentId="8748" CreationDate="2015-02-10T17:09:40.600" Score="0" Body="&lt;p&gt;Except the answers already provided, 2 other words I found are: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;superheavyweight &lt;/li&gt;&#xA;&lt;li&gt;phosphorylative &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9346" LastActivityDate="2015-02-10T17:09:40.600" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8756" PostTypeId="1" AcceptedAnswerId="8869" CreationDate="2015-02-10T18:42:51.803" Score="6" ViewCount="333" Body="&lt;p&gt;My grand father has been at the Taj Mahal, the Stari Most and &#xA;the blue Mosque, but still thinks life is a comedy of hell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My humble servant loves Fawlty Towers,&#xA;but behaves like a conquistador.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On which day did the plumber fix the world's biggest hole?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Luigi is not part of the team!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 11 men&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jürgen is experiencing tough times after the plumber left him...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The servant is now considered to be the best in the world in his role, and number three in general.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9385" LastEditorUserId="8874" LastEditDate="2016-03-08T11:01:19.553" LastActivityDate="2016-03-08T11:01:19.553" Title="Band of Brothers" Tags="|enigmatic-puzzle|" AnswerCount="2" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8758" PostTypeId="2" ParentId="8713" CreationDate="2015-02-10T20:25:16.793" Score="3" Body="&lt;p&gt;Don't lose this one, do the opposite!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Win that one&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But you won't need eyes to do so.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;wnthatone&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then let it percolate, and remove any doubles.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Anagram and remove the t and n:&#xA;whatnoe&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now we're ready for tea, and with it, our answer!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Add &quot;t&quot; and &quot;ti&quot; is also our answer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;What note? Ti!&#xA;Clearly the best of the do-re-mi's.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2015-02-10T20:25:16.793" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8759" PostTypeId="2" ParentId="8713" CreationDate="2015-02-10T20:54:20.903" Score="2" Body="&lt;p&gt;Another try, since my previous one wasn't quite right (leaving it up so others can potentially learn from it). I like my previous one better than this, but willing to give it another try.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't lose this one, do the opposite!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Win that one&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But you won't need eyes to do so.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; wnthatone&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then let it percolate, and remove any doubles.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; anagram and drop n, t&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What one&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now we're ready for tea, and with it, our answer!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Add the letters of &quot;tea&quot;, and rearrange both to get the question and answer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What one ate? Oaten wheat! Or Wheaten Oat :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2015-02-10T20:54:20.903" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8760" PostTypeId="2" ParentId="8588" CreationDate="2015-02-10T20:57:32.893" Score="0" Body="&lt;p&gt;I think I have an idea :)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A birthday! &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Holiday from happiness: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Happy Birthday&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your past is what we think&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each year we celebrate this day usually with reminders of the past year&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As space tortures, beginning to shrink:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Birth process. Contractions?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's about all I can piece together.&lt;/p&gt;&#xA;" OwnerUserId="9412" LastActivityDate="2015-02-10T20:57:32.893" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8761" PostTypeId="1" AcceptedAnswerId="8762" CreationDate="2015-02-10T21:20:03.567" Score="12" ViewCount="1018" Body="&lt;pre&gt;&lt;code&gt;0&#xA;______________________&#xA;PhD&#xA;BS&#xA;AAS&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Solve this Rebus.&#xA;See also &lt;a href=&quot;http://en.wikipedia.org/wiki/Rebus&quot;&gt;http://en.wikipedia.org/wiki/Rebus&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9412" LastActivityDate="2015-02-10T21:21:16.030" Title="What am I in this rebus?" Tags="|rebus|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8762" PostTypeId="2" ParentId="8761" CreationDate="2015-02-10T21:21:16.030" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Three degrees below zero&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The abbreviations are all (college/university) degrees.&lt;/p&gt;&#xA;" OwnerUserId="1563" LastActivityDate="2015-02-10T21:21:16.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8763" PostTypeId="1" AcceptedAnswerId="10813" CreationDate="2015-02-10T21:26:58.237" Score="32" ViewCount="3921" Body="&lt;p&gt;&lt;em&gt;Unfortunately, it appears that I may have misled you a little in this puzzle (as you probably know - my memory of events isn't always perfect). When I was writing it, I re-checked the Kzijekistanian postal rules on Gurgle-translate, but somehow stumbled on an old cached web-page with out of date data. Notes 1, 2, and 3, were unchanged but Note 4 was misleading. As a result, it appeared that even with the best packing, the pinot noir would have had to be so execrable that the wine store would have &lt;strong&gt;paid&lt;/strong&gt; Ernie to take it away!!! I have edited the story below to show the correct postal rules for four or more items (and have retained the old note in &lt;s&gt; strikeout &lt;/s&gt; form for posterity). J Richard Snape did all the hard work for the first part of the puzzle - which was meant to be the hard part - so I am accepting his answer as correct.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Are you familiar with the Island of Stability?&quot;, Ernie asked, as we sat drinking coffee and perusing some light literature at the local cafe. I had to admit that I wasn't, and so Ernie took it as an opportunity to reduce my ignorance. He read the following from his magazine &lt;em&gt;Nuclear Analysis Monthly&lt;/em&gt; &lt;strong&gt;&lt;em&gt;&quot;In nuclear physics, the island of stability is a set of predicted, but as-yet undiscovered, heavier isotopes of trans-uranic elements which are theorized to be much more stable than some of those closer in atomic number to uranium.&quot;&lt;/em&gt;&lt;/strong&gt; and pointed at a diagram showing the predicted island. &quot;The current belief is that the right isotophe of Flerovium may have a half life measuring in millions of years!&quot;, he continued emphatically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TnQUA.jpg&quot; alt=&quot;Island of stability graphical representation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I murmured a note of vague interest and returned my attention to my much more interesting reading - &lt;em&gt;The Journal of Trans-Chatcal Affairs&lt;/em&gt;. I read on for a while and then felt a mild note of deja-vu. &quot;What was that element you were talking about before?&quot;, I asked. &quot;Flerovium&quot;, Ernie replied, &quot;But chances are there is no really stable isotophe or it would have been synthesised by now, if only in picogramme quantities&quot;. I paused to savour the moment - it wasn't often that I was in the position to tell him something that he didn't know. &quot;So I guess you wouldn't give much credence to this report&quot;, I replied, and read out a short section of the article I was reading. &lt;strong&gt;&lt;em&gt;It has been reported by the People's Council of Mining and Metallurgy, in the break-away republic of Kzijekistan, that a large quantity of elementally pure Flerovium-298 has been found in a prehistoric meteoric impact crater - situated in a remote region south-west of the Kyshtut river. The Fleovium occurs in crystalline form as perfect regular tetrahedrons - all exactly identical in size. As the republic has no suitable research facilities of their own, in the spirit of international brotherhood, they are offering uncut crystals, charging only for extraction costs and postage, to researchers around the globe.&lt;/em&gt;&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;And the article references a web-site where they claim you can order the stuff&quot;, I said. Our first thoughts were that this must be an internet scam of some sort, but Ernie ran a quick search and found several academic pre-prints already showing preliminary data from their Fleovium samples. Ernie entered the link into his tablet. The site was written in Kzijekistanese, but Gurgle-translate dealt with that and soon we were reading the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fleovium crystals are all precisely 5.29 cm along each edge. Any number may be purchased at extraction price (plus postage) which can be calculated using the following algorithm:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1) Calculate the precise volume of Fleovium required (in cubic cm), then round down to the nearest cubic cm.&#xA;2) Multiply by the density of Fleovium (#%@##**^ grams per cm), then round up to the nearest gram.&#xA;3) Multiply by the cost price (6.02 cents per gram), then round up to the nearest cent (all prices in $US).&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Gurgle-translate seemed to have problems converting the density to standard Arabic numerals but Ernie said he was familiar with Kzijekistanian digits, so he would calculate the extraction price if I was willing to deal with postage. I loaded the post office site onto my tablet to find:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Kzijekistan Postal Department charges all post by standardized volume, not weight. The postal charge is calculated by the following algorithm:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1) Determine the standard volume for the item or items (in cubic cm) and round off to the nearest cubic cm.&#xA;2) Multiply by postal price (63.28 cents per cubic cm), then round off to the nearest cent (all prices in $US).&#xA;Note 1: The standard volume for a single item is that of the smallest x*x*x cube into which the item can be packed.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Calculating the optimal packing of a tetrahedron into a cube wasn't too hard, but I made a couple of small models from folded paper to confirm I had got it right. Nonetheless, when I told Ernie the post charge, he was very unhappy. &quot;But that is more than the extraction cost!&quot;, he exclaimed, &quot;I have no interest in subsidizing the profit of the Kzijekistan Postal Department&quot;. &quot;Well, I could always use two crystals, maybe we can pack them more efficiently&quot;, Ernie muttered. I scrolled further down the KPD's web page and read the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Note 2: The standard volume for two items is that of the smallest x*x*y rectangular prism into which both items can be packed simultaneously. &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Once again, I calculated the post cost (I needed to make a second tetrahedron model, and it took a little longer to find the optimal packing), but Ernie was still disappointed: &quot;The post cost is still higher than the materials cost. How about if I buy three?&quot;. There was another note further down the page:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Note 3: The standard volume for three items is that of the smallest x*y*z rectangular prism into which all three items can be packed simultaneously. &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ewpr3.jpg&quot; alt=&quot;Folded paper models&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It took a lot of work to confirm the best packing this time. By the time I was finished the table was littered with little model tetrahedra and boxes. I decided Ernie had chosen the much simpler task. &quot;Well, at least this time the postal charge is less than the extraction cost&quot;, I said when Ernie told me his number. &quot;True&quot;, he replied, but now the total cost is 10c more than what I have in my PayPall account! How about if I buy four of them?&quot;. This seemed like a fools errand to me - if three were too expensive what cope could there be in buying four crystals? There was a final note on the KPD site: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;s&gt; Note 4: The standard volume for four or more items is that of the smallest hexahedron constructed from six quadrilateral faces into which all the items can be packed simultaneously. &lt;/s&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Note 4: The standard volume for four or more items is that of the smallest closed volume (constructed from planar faces) with a maximum of twelve  edges into which all the items can be packed simultaneously. &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I sighed and reached for the scissors to build another tetrahedron. It could take all afternoon to work out how the most efficient packing, but then Ernie gave a chuckle and told me he had already worked out the postal cost and it was much improved. &quot;I can afford four crystals and still have money left over!&quot;, he exclaimed. He was so pleased that after he had filled out the online purchase form, he spent all the money left in his account on a nice bottle of Pinot Noir.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here is the point where you may be able to help me. I believe there is sufficient information to work out &lt;strong&gt;the density of Fleovium crystals&lt;/strong&gt; (and the error limit in that value). Also, as it is my turn to buy the wine next week it would be good to know approximately &lt;strong&gt;how much Ernie spent on the celebratory wine&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1:&lt;/strong&gt; For 1, 2, and 3 crystals the answer is all in the rounding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2:&lt;/strong&gt; For &lt;code&gt;n = 1, 2, 3&lt;/code&gt; crystals, the &lt;em&gt;exact&lt;/em&gt; un-rounded mass and &lt;em&gt;exact&lt;/em&gt; un-rounded volume scale &lt;em&gt;exactly&lt;/em&gt; with &lt;code&gt;n&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 3:&lt;/strong&gt; For &lt;code&gt;n = 4&lt;/code&gt;, to find the packing density you need to &quot;Think outside the &lt;strong&gt;box&lt;/strong&gt;&quot;. Note that this is different from &quot;Think &lt;strong&gt;outside&lt;/strong&gt; the box&quot;, if that helps...&lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="1631" LastEditDate="2015-03-31T20:49:03.897" LastActivityDate="2015-03-31T20:49:03.897" Title="Ernie and the Island of Stability" Tags="|mathematics|geometry|word-problem|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8765" PostTypeId="1" AcceptedAnswerId="8902" CreationDate="2015-02-10T21:32:36.353" Score="3" ViewCount="589" Body="&lt;p&gt;Inspired by &lt;a href=&quot;https://puzzling.stackexchange.com/questions/8625/whats-my-room-number&quot;&gt;this puzzle&lt;/a&gt;, I have made this similar puzzle (the solution method is NOT identical)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a hotel there are ten floors, with ten rooms at each floor. &#xA;Currently 10 rooms are occupied.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In room 13 Mr. Nelson is staying his last day.&lt;/li&gt;&#xA;&lt;li&gt;In room 21 Angela is hanging up a &quot;Malerei&quot; that has fallen down the wall.&lt;/li&gt;&#xA;&lt;li&gt;In room 23 Franz and Ferdinand have a principal meeting.&lt;/li&gt;&#xA;&lt;li&gt;In room 39 Joseph wins the day after Paul the Apostle has surrendered.&lt;/li&gt;&#xA;&lt;li&gt;In room 51 a King has a dream.&lt;/li&gt;&#xA;&lt;li&gt;In room 62 Zwedish football is born.&lt;/li&gt;&#xA;&lt;li&gt;In room 65 Pierce wants to travel the world and end in Europe for the second time.&lt;/li&gt;&#xA;&lt;li&gt;In room 82 Mr. Brown is serving NamNam for free.&lt;/li&gt;&#xA;&lt;li&gt;In room 88 Mrs. Carla is using a small and gentle computer for the first time.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In what room would Cuba Gooding jr. want to stress down after the climax of a lifetime?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is a two-legged puzzle. The first leg is to find the one thing in common. That thing is layed out in a pattern. The pattern is leg two.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first leg is relying heavily on association.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mrs. Carla is working on formulating VALUEs important to here&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First leg consists of one date pr. room.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mrs. Carla is using Excel&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Chatroom:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;To simplify our discussions, there now is a chat-room on this puzzle:&lt;br&gt;&#xA;&lt;a href=&quot;http://chat.stackexchange.com/rooms/21065/cuba-gooding-jr&quot;&gt;chat.stackexchange.com/rooms/21065/cuba-gooding-jr&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="9385" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-02-13T13:35:10.473" Title="What's the room number for Cuba Gooding jr?" Tags="|pattern|" AnswerCount="4" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8766" PostTypeId="1" AcceptedAnswerId="8767" CreationDate="2015-02-10T22:28:25.060" Score="-1" ViewCount="239" Body="&lt;p&gt;In celebration of catching the spaghetti thieves, a new party is being thrown!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the thieves have relatives who think they can do better! I wonder if they can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So they're hiding outside listening, and they hear many guests coming, sadly, many were like them trying to get in, meaning they didn't get a lot of information:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A man and woman come along. Each is given a number. The man is given 5 and the woman is given 10. The man responds 10 and gets in. The woman said 5 and gets in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then two kids come along, given the numbers 3 and 9. They responded 6 and 4.5. Kid #1 gets in, but #2 is trampled on by an elephant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then a man comes along, he gets 5, he says he'll come back later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What should the 2 thieves do when they go up to the door?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Table summary:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Man    5    10    Right&#xA;Woman  10   5     Right &#xA;Kid    3    6     Right&#xA;Kid    9    4.5   Wrong&#xA;Man    5    ?     ?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2015-02-11T01:15:53.727" LastActivityDate="2015-02-11T01:15:53.727" Title="The security to the party (Spaghetti's back!)" Tags="|mathematics|password|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="8767" PostTypeId="2" ParentId="8766" CreationDate="2015-02-10T22:36:18.220" Score="6" Body="&lt;p&gt;If the number is odd, double it.  If it is even, halve it.&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2015-02-10T22:36:18.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8768" PostTypeId="2" ParentId="8756" CreationDate="2015-02-10T23:16:32.597" Score="7" Body="&lt;p&gt;I think the date is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;September 6, 1522&lt;/strong&gt; when Magellan / Elcano completed the first circumnavigation of the earth and provided practical evidence that the earth was not flat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Taj Mahal, Stari Most and the blue Mosque are all associated with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; apprentices of Mimar Sinan who lived during the 1500's&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Life is a comedy of hell refers to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dante's Divine Comedy which portrays the earth as a sphere&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fawlty Towers and conquistador refer to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the British (I don't know why) and the Spanish (who financed Magellan)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The world's biggest hole is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the edge of a flat earth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastActivityDate="2015-02-10T23:16:32.597" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8769" PostTypeId="1" AcceptedAnswerId="8770" CreationDate="2015-02-11T01:17:46.923" Score="17" ViewCount="11992" Body="&lt;p&gt;I have an Android phone with a pattern lock screen, which if you've never seen one &lt;a href=&quot;https://www.google.com/search?q=pattern+lock&amp;amp;tbm=isch&quot; rel=&quot;noreferrer&quot;&gt;allows you to unlock your phone using a pattern drawn on the screen over a 3x3 grid of dots&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for example, if we number the dots as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 2 3&#xA;4 5 6&#xA;7 8 9&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;then valid lock patterns may be created by drawing lines connecting &lt;code&gt;1 &amp;gt; 2 &amp;gt; 3 &amp;gt; 4&lt;/code&gt;, or &lt;code&gt;1 &amp;gt; 5 &amp;gt; 9 &amp;gt; 8 &amp;gt; 7&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, there are a few rules restricting what constitutes a valid unlock pattern:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Any given dot can only be used once (however you can cross back over a previously used dot, if you can connect to another unused dot on the other side)&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Eg. &lt;code&gt;1 &amp;gt; 2 &amp;gt; 3&lt;/code&gt; is valid, &lt;code&gt;2 &amp;gt; 1 &amp;gt; 2&lt;/code&gt; is invalid, but &lt;code&gt;2 &amp;gt; 1 &amp;gt; 3&lt;/code&gt; is valid (goes back over the &lt;code&gt;2&lt;/code&gt;, but doesn't re-include it in the actual code)&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;If your line crosses directly over a previously unused dot, it is automatically included in the code&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Eg. Attempting to create a code of &lt;code&gt;1 &amp;gt; 3 &amp;gt; 4&lt;/code&gt; would automatically become &lt;code&gt;1 &amp;gt; 2 &amp;gt; 3 &amp;gt; 4&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What I remember about my specific lock pattern:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It used all 9 dots&lt;/li&gt;&#xA;&lt;li&gt;The line representing the pattern has reflective symmetry&lt;/li&gt;&#xA;&lt;li&gt;It is different to my previous pattern which also followed rules 1 &amp;amp; 2&lt;/li&gt;&#xA;&lt;li&gt;My previous pattern looked like an alphanumeric character&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is current my lock pattern? What was my previous lock pattern?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;For the purposes of this puzzle, we'll ignore the fact that there will be multiple &quot;correct&quot; answers achieved by simply rotating by multiples of 90 degrees. In other words a pattern of &lt;code&gt;1 &amp;gt; 2 &amp;gt; 3&lt;/code&gt; will be considered identical to &lt;code&gt;3 &amp;gt; 6 &amp;gt; 9&lt;/code&gt;.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Edit: Given that &lt;a href=&quot;https://puzzling.stackexchange.com/a/8770/5208&quot;&gt;Len found two valid solutions below&lt;/a&gt;, and neither was my intended solution, I've added an &lt;a href=&quot;https://puzzling.stackexchange.com/q/8814/5208&quot;&gt;extension puzzle with an additional rule&lt;/a&gt;.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2020-11-06T15:49:20.363" Title="Help! I've forgotten my phone's lock pattern!" Tags="|geometry|pattern|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8770" PostTypeId="2" ParentId="8769" CreationDate="2015-02-11T03:40:36.603" Score="11" Body="&lt;p&gt;Added 4th solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think your previous pattern was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;M&lt;/strong&gt; which is also called &lt;strong&gt;&lt;a href=&quot;http://marceluniverse.blogspot.ca/2013/03/android-screen-lock-pattern-repertoire.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;impossible 3 ends&lt;/a&gt;&lt;/strong&gt; shown below - 854693217&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/H1Qsz.png&quot; alt=&quot;3 ends&quot;&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And you current pattern is one of:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - a &lt;strong&gt;type of star&lt;/strong&gt; left below - 857342619&lt;br&gt;&#xA; - or a &lt;strong&gt;geometric pattern&lt;/strong&gt; center below - 851962473&lt;br&gt;&#xA; - or a &lt;strong&gt;modified M&lt;/strong&gt; right below - 254693817&lt;br&gt;&#xA;&lt;br&gt;&#xA;  &lt;img src=&quot;https://i.stack.imgur.com/yfwry.png&quot; alt=&quot;LockPatterns&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-02-12T07:25:43.973" LastActivityDate="2015-02-12T07:25:43.973" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8771" PostTypeId="2" ParentId="8769" CreationDate="2015-02-11T07:36:34.640" Score="11" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; So your previous pattern looked like an alphanumeric character, used all 9 dots and was symmetrical. Alphanumeric characters: ABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890. Out of these there are some asymmetric: FGJLNPQRSZ245679. That leaves us with: ABCDEHIKMOTUVWXY138.&lt;br&gt;&#xA; Now let's throw out the ones that can't fit using all 9 dots: ACDHIKOTUVXY1. This leaves us with: BEMW38. Because you will always have an &quot;opening&quot; in your pattern, we can throw out B and 8, so only possible answers are: EMW3. But E, M and W and 3 are actually same pattern, just turned 90 degrees, so &lt;strong&gt;your old pattern looked like E&lt;/strong&gt;, so most probably that &lt;em&gt;Impossible 3-Ends&lt;/em&gt;: 854693217&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So now we should find your current pattern:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because we only know that it is symmetric and uses all 9 dots we can assume the following pattern to be your current one: gonna find it out soon and update my answer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9352" LastActivityDate="2015-02-11T07:36:34.640" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8772" PostTypeId="5" CreationDate="2015-02-11T09:18:18.997" Score="0" Body="&lt;p&gt;A geometric packing puzzle in which a number of shapes have to be assembled into a larger shape, generally without overlaps or gaps.&lt;/p&gt;&#xA;" OwnerUserId="11497" LastEditorUserId="11497" LastEditDate="2016-08-19T17:42:42.427" LastActivityDate="2016-08-19T17:42:42.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8773" PostTypeId="4" CreationDate="2015-02-11T09:18:18.997" Score="0" Body="A geometric packing puzzle in which a number of shapes have to be assembled into a larger shape, generally without overlaps or gaps.&#xD;&#xA;" OwnerUserId="8874" LastEditorUserId="11497" LastEditDate="2016-08-19T17:42:30.940" LastActivityDate="2016-08-19T17:42:30.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8775" PostTypeId="1" AcceptedAnswerId="8780" CreationDate="2015-02-11T10:31:02.847" Score="13" ViewCount="1479" Body="&lt;p&gt;Bobby Fischer liked to play the following game on a standard &#xA;$8\times8$ chessboard:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In his first step, Bobby placed a white rook and a black rook &#xA;somewhere on the chessboard (on two different squares, of course).&lt;/li&gt;&#xA;&lt;li&gt;In every further step, Bobby picked one of the rooks, and then &#xA;moved it away from its current square to a horizontally or &#xA;vertically adjacent (currently empty) square.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is it possible that after $64\cdot63=4032$ steps, one has generated &#xA;each of the $4032$ possible positions for the two rooks exactly once?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="17482" LastEditDate="2016-03-08T11:41:59.617" LastActivityDate="2016-03-08T11:41:59.617" Title="Two rooks for Bobby Fischer" Tags="|mathematics|chess|checkerboard|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8780" PostTypeId="2" ParentId="8775" CreationDate="2015-02-11T13:18:49.973" Score="24" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No, this is not possible.&lt;br&gt; Call a position &lt;em&gt;even&lt;/em&gt; if the two rooks are on squares of the same color squares, &lt;em&gt;odd&lt;/em&gt; if they are on squares of different colors. Every move changes the position from even to odd or from odd to even. There are a total of $31\cdot 64=1984$ even positions and $32\cdot 64=2048$ odd positions, so it is not possible to reach them all while alternating between them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-02-11T15:41:20.713" LastActivityDate="2015-02-11T15:41:20.713" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8781" PostTypeId="1" AcceptedAnswerId="8799" CreationDate="2015-02-11T13:38:01.770" Score="5" ViewCount="379" Body="&lt;p&gt;In this puzzle, we only consider &lt;em&gt;independent&lt;/em&gt; countries and we do &lt;em&gt;not&lt;/em&gt;&#xA;count overseas territories as actual part of a country. &#xA;We say that three countries form a &lt;strong&gt;well-connected triple&lt;/strong&gt;, if each pair shares a piece of common border with strictly positive length.&#xA;Most of these well-connected triples furthermore meet at a point (at&#xA;least theoretically).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Q&lt;/strong&gt;: What are the twelve well-connected triples that do not have any common point &#xA;that is shared by all three countries?&lt;/p&gt;&#xA;" OwnerUserId="9125" LastEditorDisplayName="user88" LastEditDate="2015-09-04T14:36:37.387" LastActivityDate="2015-09-04T14:36:37.387" Title="Country triples without common point" Tags="|knowledge|geography|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8782" PostTypeId="2" ParentId="8781" CreationDate="2015-02-11T13:41:11.570" Score="3" Body="&lt;p&gt;Some quick and easy answers:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;France, Germany, Belgium&lt;/li&gt;&#xA;&lt;li&gt;Brazil, Argentina, Bolivia&lt;/li&gt;&#xA;&lt;li&gt;Congo, Tanzania, Uganda&lt;/li&gt;&#xA;&lt;li&gt;Congo, Tanzania, Rwanda&lt;/li&gt;&#xA;&lt;li&gt;Tanzania, Mozambique, Zambia&lt;/li&gt;&#xA;&lt;li&gt;Turkey, Georgia, Azerbaijan&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-11T14:34:31.183" LastActivityDate="2015-02-11T14:34:31.183" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8783" PostTypeId="5" CreationDate="2015-02-11T13:41:59.837" Score="0" Body="&lt;p&gt;The &lt;a href=&quot;/questions/tagged/knowledge&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;knowledge&amp;#39;&quot; rel=&quot;tag&quot;&gt;knowledge&lt;/a&gt; tag should be used for puzzles that require specific knowledge about the subject matter to be able to answer, or require external help (i.e. tables, dictionaries, Wikipedia, Google, etc.). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note there are also tags for more specific knowledge sets such as:&lt;br&gt;&#xA;* &lt;a href=&quot;/questions/tagged/music&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;music&amp;#39;&quot; rel=&quot;tag&quot;&gt;music&lt;/a&gt;&lt;br&gt;&#xA;* &lt;a href=&quot;/questions/tagged/video-games&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;video-games&amp;#39;&quot; rel=&quot;tag&quot;&gt;video-games&lt;/a&gt;&lt;br&gt;&#xA;* &lt;a href=&quot;/questions/tagged/movies&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;movies&amp;#39;&quot; rel=&quot;tag&quot;&gt;movies&lt;/a&gt;&lt;br&gt;&#xA;* &lt;a href=&quot;/questions/tagged/history&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;history&amp;#39;&quot; rel=&quot;tag&quot;&gt;history&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="52483" LastEditorUserId="11497" LastEditDate="2019-08-13T08:48:00.783" LastActivityDate="2019-08-13T08:48:00.783" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8784" PostTypeId="4" CreationDate="2015-02-11T13:41:59.837" Score="0" Body="A puzzle whose solution relies on external sources such as tables, dictionaries, Wikipedia, or specifics that may not be common knowledge." OwnerUserId="8874" LastEditorUserId="11497" LastEditDate="2019-08-13T08:48:00.783" LastActivityDate="2019-08-13T08:48:00.783" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8785" PostTypeId="1" AcceptedAnswerId="8788" CreationDate="2015-02-11T13:48:44.057" Score="4" ViewCount="331" Body="&lt;p&gt;In this puzzle, we only consider &lt;em&gt;independent&lt;/em&gt; countries and we do &lt;em&gt;not&lt;/em&gt;&#xA;count overseas territories as actual part of a country. We say that three countries form a &lt;strong&gt;well-connected triple&lt;/strong&gt;, if each two of&#xA;them share a piece of common border with strictly positive length.&#xA;Most of these well-connected triples furthermore meet at a point (at&#xA;least theoretically).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Q&lt;/strong&gt;: What are the eight well-connected triples that share (at least) two  points common to all three countries?&lt;/p&gt;&#xA;" OwnerUserId="9125" LastActivityDate="2015-02-11T17:43:41.327" Title="Country triples with two common points" Tags="|geography|knowledge|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8787" PostTypeId="1" CreationDate="2015-02-11T13:55:19.577" Score="7" ViewCount="4177" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Dear Reader: This question assumes you know what a Nonogram is. If you do not, I recommend reading &lt;a href=&quot;http://en.wikipedia.org/wiki/Nonogram&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Wikipedia entry first.&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nonograms have many solving techniques. One of the lesser used ones - simply because it does not come up that often - is that if the puzzle is symmetric, you can do additional solving techniques; specifically, for a particular row - if the nonogram is vertically symmetric - the center number must be centered in the middle. Also, any solving you do on one side of the puzzle can also be immediately copied to it's reflection on the other side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is about &lt;strong&gt;diagonally symmetric&lt;/strong&gt; Nonograms. You can tell if a puzzle is diagonally symmetric (top left to bottom right) if the numbers on each side are the same, and symmetric top right to bottom left in the numbers are the same, but in opposite order. Here is an example of a diagonally symmetric Nonogram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PHoGX.png&quot; alt=&quot;Basic Nonogram&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/zMYCL.png&quot; alt=&quot;Nonogram with Symmetry line&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the numbers are the same, but in reverse order, so it has top right to bottom left symmetry. Are there additional solving techniques that take advantage of this fact? How would you solve this (fairly difficult for a 10x10) puzzle?&lt;/p&gt;&#xA;" OwnerUserId="42" LastEditorUserId="42" LastEditDate="2015-02-11T21:07:40.930" LastActivityDate="2015-04-14T14:21:02.430" Title="Additional solving techniques to diagonally symmetric Nonograms" Tags="|strategy|nonogram|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8788" PostTypeId="2" ParentId="8785" CreationDate="2015-02-11T14:06:52.627" Score="3" Body="&lt;p&gt;&lt;strong&gt;Quick and easy answers:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Mongolia, China, and Russia&lt;/li&gt;&#xA;  &lt;li&gt;Switzerland, Austria, and Liechtenstein&lt;/li&gt;&#xA;  &lt;li&gt;India, China, and Nepal&lt;/li&gt;&#xA;  &lt;li&gt;India, China, and Bhutan&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Three further answers found by shyos:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;5&quot;&gt;&#xA;  &lt;li&gt;Ukraine, Romania, and Moldova&lt;/li&gt;&#xA;  &lt;li&gt;South Africa, Mozambique, and Swaziland&lt;/li&gt;&#xA;  &lt;li&gt;France, Spain, and Andorra&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;And finally the &quot;weirdest&quot; solution:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;8&quot;&gt;&#xA;  &lt;li&gt;Armenia, Azerbaijan, and Iran&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-11T17:43:41.327" LastActivityDate="2015-02-11T17:43:41.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8789" PostTypeId="2" ParentId="8785" CreationDate="2015-02-11T14:30:56.097" Score="5" Body="&lt;p&gt;Based on &lt;a href=&quot;http://en.wikipedia.org/wiki/Landlocked_country#Landlocked_by_two_countries&quot;&gt;wikipedia&lt;/a&gt; there are 7 country that are landlocked by two other countries.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Seven landlocked countries are surrounded by only two mutually&#xA;  bordering neighbors:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Andorra (between France and Spain)  &lt;/li&gt;&#xA;  &lt;li&gt;Bhutan (between India and China) &lt;/li&gt;&#xA;  &lt;li&gt;Liechtenstein (one of the &quot;doubly landlocked&quot; countries, between       Switzerland and Austria)&lt;/li&gt;&#xA;  &lt;li&gt;Moldova (between Ukraine and Romania)&lt;/li&gt;&#xA;  &lt;li&gt;Mongolia (between Russia and China)&lt;/li&gt;&#xA;  &lt;li&gt;Nepal (between India and China)&lt;/li&gt;&#xA;  &lt;li&gt;Swaziland (between South Africa and Mozambique)&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;To this group could be added two de facto states with no or limited&#xA;  international recognition:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;South Ossetia (between Russia and Georgia)&lt;/li&gt;&#xA;  &lt;li&gt;Transnistria (between Ukraine and Moldova)&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4506" LastActivityDate="2015-02-11T14:30:56.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8790" PostTypeId="1" AcceptedAnswerId="8793" CreationDate="2015-02-11T14:54:21.423" Score="20" ViewCount="2866" Body="&lt;p&gt;Doctor Brun was at it for years, but he'd finally got his time machine up and running. His first trip was to a time way back in the past when dinosaurs roamed the Earth. To his surprise, the dinosaurs were actually quite intelligent. His first encounter was with Timmy Rex (a Tyrannosaurus Rex). They began talking about philosophy and politics. Eventually, mathematics came up. Here's the part of the conversation they had, including some of the cave writings.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Doc Brun: Ok. Just to be sure we understand each other, can you&#xA;  complete these equations for me?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Doc Brun wrote $1 + 2 =$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;T. Rex wrote $3$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Doc Brun: Great. And this one?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Doc Brun wrote $12 \times 3 =$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;T. Rex wrote $102$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Doc Brun: What? That's not right at all. Let's try another. I think I&#xA;  see what's going on here. Is this equation right?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Doc Brun wrote $1 + 3 = 4$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;T. Rex: I don't even understand what you wrote there!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Doc Brun: I see, just as I thought. I'm assuming that--&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The frustrated T. Rex hastily ate Doc Brun in one bite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What was Doc Brun going to say before he was swallowed?&lt;/p&gt;&#xA;" OwnerUserId="8942" LastEditorUserId="8942" LastEditDate="2015-02-12T17:48:04.530" LastActivityDate="2015-02-12T17:48:04.530" Title="Math with dinosaurs" Tags="|mathematics|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8792" PostTypeId="2" ParentId="8790" CreationDate="2015-02-11T15:07:16.817" Score="9" Body="&lt;p&gt;First I thought, that the dinos used a numbering system, based on 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 + 2 = 3 would fit here, but 12 * 3 = 210 would be 6 * 3 = 18 in decimal system. 210 in decimal is 36. Am I missing a 2? Or It should be 102...&lt;/p&gt;&#xA;" OwnerUserId="9197" LastActivityDate="2015-02-11T15:07:16.817" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8793" PostTypeId="2" ParentId="8790" CreationDate="2015-02-11T15:09:09.187" Score="37" Body="&lt;p&gt;It appears that the dinosaurs do mathematics in base $4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is the case, when Doc. Brun writes $12\times3$ he expects the answer $36$, which T. Rex writes as $210$, in base $4$. However, T. Rex &lt;em&gt;should&lt;/em&gt; read this as $6\times3$, since $12_4=6$, so he would give the answer $18$, which is $102$ in base $4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Base $4$ would only contain the digits $0$ through $3$, so T. Rex would not recognize the glyph &quot;$4$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also makes sense because a Tyrannosaurus Rex only has two fingers on each hand, for a total of four fingers. The main reason that humans started using base ten is because it is easy to count to ten&amp;mdash;just count on your fingers! So T-rex math would likely use base four.&lt;/p&gt;&#xA;" OwnerUserId="8776" LastEditorUserId="8776" LastEditDate="2015-02-11T19:00:56.983" LastActivityDate="2015-02-11T19:00:56.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8794" PostTypeId="2" ParentId="8790" CreationDate="2015-02-11T15:09:19.443" Score="4" Body="&lt;p&gt;The dinosaurs were doing mathematics in base 4.&lt;br&gt;&#xA;Then numbers 1, 2, 3, 12, and 102 in dino language mean&lt;br&gt;&#xA;the numbers 1, 2, 3, 6, and 18 in our decimal system.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;T. Rex does not understand 1+3=4, as the digit 4 is not used by the dinos.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-11T15:24:21.133" LastActivityDate="2015-02-11T15:24:21.133" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8795" PostTypeId="1" AcceptedAnswerId="8800" CreationDate="2015-02-11T15:30:25.717" Score="22" ViewCount="689" Body="&lt;p&gt;What word is hidden in the following Limerick?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; See my first when the scalpel's applied&#xA; In my final, a pronoun is spied&#xA;    My middle is meant&#xA;    To be readily bent&#xA; And my all, like a serpent, can slide.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="9404" LastEditorDisplayName="user88" LastEditDate="2015-04-24T17:40:04.017" LastActivityDate="2018-03-18T21:27:28.133" Title="When the scalpel is applied" Tags="|riddle|word|limerick|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8796" PostTypeId="1" AcceptedAnswerId="8797" CreationDate="2015-02-11T15:34:44.373" Score="21" ViewCount="3948" Body="&lt;p&gt;Solve the following rebus puzzle:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;      L&#xA;      D&#xA;    BRIDGE&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;See also &#xA;(&lt;a href=&quot;https://puzzling.stackexchange.com/questions/8709/rebus-with-dracula&quot;&gt;Rebus with Dracula&lt;/a&gt;)  and &#xA;(&lt;a href=&quot;https://puzzling.stackexchange.com/questions/8750/another-rebus-puzzle&quot;&gt;Another Rebus puzzle&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-02-11T15:37:01.123" Title="My third Rebus puzzle" Tags="|english|rebus|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8797" PostTypeId="2" ParentId="8796" CreationDate="2015-02-11T15:37:01.123" Score="18" Body="&lt;p&gt;I love rebus puzzles! That one is perhaps somewhat on the easy side:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;L on D on BRIDGE&quot; = &lt;strong&gt;&quot;London Bridge&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-02-11T15:37:01.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8798" PostTypeId="2" ParentId="8781" CreationDate="2015-02-11T15:42:22.393" Score="4" Body="&lt;p&gt;There are a number of possible triples with Russia, thanks to &lt;a href=&quot;http://en.wikipedia.org/wiki/Kaliningrad_Oblast&quot; rel=&quot;nofollow&quot;&gt;Kaliningrad Oblast&lt;/a&gt;. You may wish to exclude this, even though it is not an overseas territory, and it is an oblast (roughly, a state or province) of Russia.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Russia, Poland, Belarus&lt;/li&gt;&#xA;&lt;li&gt;Russia, Poland, Ukraine&lt;/li&gt;&#xA;&lt;li&gt;Russia, Lithuania, Belarus&lt;/li&gt;&#xA;&lt;li&gt;Russia, Lithuania, Latvia&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Another, possibly more &quot;legitimate&quot; answer not yet given is&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Tanzania, Malawi, Mozambique&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8776" LastEditorUserId="8776" LastEditDate="2015-02-11T15:48:19.760" LastActivityDate="2015-02-11T15:48:19.760" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8799" PostTypeId="2" ParentId="8781" CreationDate="2015-02-11T16:23:48.617" Score="4" Body="&lt;p&gt;Two more as yet unmentioned:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Armenia, Iran, Turkey&lt;/li&gt;&#xA;&lt;li&gt;Croatia, Montenegro, Serbia&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;By my count this makes 12, not counting the dubious borders-in-lakes ones.&lt;/p&gt;&#xA;" OwnerUserId="1527" LastActivityDate="2015-02-11T16:23:48.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8800" PostTypeId="2" ParentId="8795" CreationDate="2015-02-11T17:14:05.927" Score="23" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Slither&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See my first when the scalpel's applied&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Slit made when a cut from the scalpel&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In my final, a pronoun is spied&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Her, feminine pronoun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My middle is meant to be readily bent&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;lithe, synonymous with agile&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And my all, like a serpent can slide.&lt;/p&gt;&#xA;" OwnerUserId="9412" LastActivityDate="2015-02-11T17:14:05.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8801" PostTypeId="2" ParentId="8713" CreationDate="2015-02-11T17:26:57.703" Score="1" Body="&lt;p&gt;The riddle is asking:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What time is it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer must be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tea time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Don't lose this one, do the opposite!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; win that one&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But you won't need eyes to do so.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; win that ne&#xA; Remove the &quot;o&quot; that looks like an eye&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then let it percolate, and remove any doubles.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; w i n t h a t n e&#xA; w h a t t i n e &#xA; The second &quot;W&quot; is the only letter that is not reused.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now we're ready for tea, and with it, our answer!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just not sure how to connect the last dot there.....&lt;/p&gt;&#xA;" OwnerUserId="9412" LastActivityDate="2015-02-11T17:26:57.703" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8802" PostTypeId="1" CreationDate="2015-02-11T15:27:30.833" Score="7" ViewCount="2073" Body="&lt;p&gt;Replace the letters $A$, $B$, $C$, $D$, $E$ with the numerals $1$, $2$, $3$, $4$, $5$ such that the following equation is true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$AB*C=DE$$&lt;/p&gt;&#xA;" OwnerDisplayName="sarah" LastEditorUserId="9404" LastEditDate="2016-03-05T17:35:19.700" LastActivityDate="2016-03-21T12:25:22.770" Title="Maths Puzzle ABCDE" Tags="|logical-deduction|alphametic|" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8804" PostTypeId="2" ParentId="8802" CreationDate="2015-02-11T15:34:13.023" Score="4" Body="&lt;p&gt;You really don't have many cases to analyze, because of two pretty strict conditions that come from the puzzle's hypothesis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the result must be smaller than $54$ because it's the greatest two digit number you can write with $1,2,3,4,5$. The only $AB$ combinations that multiplied for another number $C \in \{1,2,3,4,5\}\setminus\{A,B\}$ give a result $DE$ that is lower than $54$ are $12,13,14,15,21$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also can exclude couples in which $C=1$ because in that case &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$AB*C=AB*1=DE \implies AB=DE$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The possible combinations for the LHS reduce to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{12,3\},\{12,4\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{13,2\},\{13,4\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{14,2\},\{14,3\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{15,2\},\{15,3\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bit of try and error will lead to the solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$13*4=52$&lt;/p&gt;&#xA;" OwnerUserId="32151" OwnerDisplayName="bharb" LastActivityDate="2015-02-11T15:58:39.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8805" PostTypeId="2" ParentId="8802" CreationDate="2015-02-11T15:35:47.947" Score="15" Body="&lt;p&gt;$13*4=52$ is the only solution. We note that we must have $D=5$, since $E = 5$ forces $B$ or $C = 5$, and vice versa, while $A \leq D$ prevents $A = 5$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now factorising the possibilities for $DE$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$51 = 17 * 3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$52 = 13 * 4 = 26 * 2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$53$ is prime&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$54 = 27 * 2 = 9 * 6 = 3 * 18$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only $52 = 13 * 4$ is of the correct form.&lt;/p&gt;&#xA;" OwnerUserId="1527" OwnerDisplayName="Christopher Fish" LastActivityDate="2015-02-11T15:35:47.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8806" PostTypeId="1" AcceptedAnswerId="8833" CreationDate="2015-02-11T18:34:56.110" Score="5" ViewCount="451" Body="&lt;p&gt;In &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3337/from-knights-to-kings&quot;&gt;From knights to kings&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3447/from-knights-to-kings-on-a-rectangle&quot;&gt;From Knights to Kings on a rectangle&lt;/a&gt; we tried to help knights be next to their friends after they were promoted to kings. What happens if we have kings who are deposed and become knights?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We start off with an $N\times M$ board filled with kings. A king is friends with all the kings on squares that he could move to, i.e all adjacent kings. After they are deposed, they want to have all their friends still be in squares that they could get to in a single move. So two kings A and B:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A B .&#xA;. . .&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;would want to be like this after becoming knights:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A . .&#xA;. . B&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Unfortunately A's two other friends (originally in the two spaces below him) cannot be put anywhere that allows A to still reach them. So a $3\times 2$ board doesn't work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A $1\times 1$ board is trivially solved - no friends either before or after the change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any $n\times 1$ board (where $n&amp;gt;1$) is obviously impossible - each king has one friend if he is at the end, or two friends if he isn't. After they become knights, none of them can move so none of them can be a knight's move away from their friends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Are there any board sizes that allow the kings to be a knight's move away from all of their friends?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1955" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-10-13T16:27:53.807" Title="From Kings to Knights" Tags="|logical-deduction|combinatorics|chess|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8807" PostTypeId="2" ParentId="8802" CreationDate="2015-02-11T18:43:35.507" Score="-2" Body="&lt;p&gt;&lt;code&gt;3*3=9?&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;AB*C=DE  &#xA;12*3=45  &#xA;(1+2)*3=(4+5)&#xA;3*3=9&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'm guessing this is the inferred solution as it has come from a school - not a pure mathematics handbook - although it would appear that the question is poorly written (originally, not by the OP)&lt;/p&gt;&#xA;" OwnerUserId="9451" LastEditorUserId="9451" LastEditDate="2015-02-11T19:11:34.270" LastActivityDate="2015-02-11T19:11:34.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8808" PostTypeId="2" ParentId="8765" CreationDate="2015-02-11T18:47:21.620" Score="2" Body="&lt;p&gt;Cuba would want to relax in Room 97.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prepending a '19' in front of each room number gives a year that corresponds to the clue/event.  In the case of Cuba Gooding Jr, his climax was winning an Oscar in 1997 for his role in Jerry Maguire.&lt;/p&gt;&#xA;" OwnerUserId="2605" LastActivityDate="2015-02-11T18:47:21.620" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8809" PostTypeId="2" ParentId="8713" CreationDate="2015-02-11T19:07:05.080" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;Don't lose this one&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;win this one&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;But you won't need eyes to do so &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;wnthsone&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;Then let it percolate and remove any doubles&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;wthsoe&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;Now we're ready for tea, and with it our answer&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;twosthe&lt;/code&gt; or &lt;code&gt;Tuesday&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9450" LastEditorUserId="9450" LastEditDate="2015-02-11T20:02:45.693" LastActivityDate="2015-02-11T20:02:45.693" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8810" PostTypeId="2" ParentId="8802" CreationDate="2015-02-11T19:13:28.060" Score="8" Body="&lt;p&gt;The only two numbers that multiply together $B*C$ to another (E) is $3*4 = 12$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That leaves 1 and 5 remaining, so obviously $DE = 52$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;52 doesn't divide into 3, so $AB = 13, C = 4$ and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$13 * 4 = 52$&lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-02-11T19:13:28.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8811" PostTypeId="1" AcceptedAnswerId="8871" CreationDate="2015-02-11T21:37:57.413" Score="7" ViewCount="1290" Body="&lt;p&gt;What is the link between an astronotical fish, an owner of crown jewels, Chelsea, an honoured Akira, and a synthesizer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The link is a name&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A wilde name&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9385" LastEditorUserId="9385" LastEditDate="2015-02-12T16:28:58.567" LastActivityDate="2015-02-12T20:07:29.433" Title="What do they have in common?" Tags="|riddle|pattern|" AnswerCount="4" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8812" PostTypeId="2" ParentId="8811" CreationDate="2015-02-11T23:32:11.233" Score="3" Body="&lt;p&gt;Hrm.. tricky...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are all masters of their domain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Astronotical fish&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Master of the stars&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Owner of crown jewels&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ruler of a kingdom&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Chelsea&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Might be subjective, but masters of the pitch?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Honoured Akira&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oragami master&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;syntheziser&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Master of all things musical or sound&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9412" LastActivityDate="2015-02-11T23:32:11.233" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8813" PostTypeId="1" CreationDate="2015-02-11T23:36:26.327" Score="34" ViewCount="12342" Body="&lt;p&gt;I have been asked this puzzle in an interview. There was not a lot of time left so I couldn't solve it quickly and the interviewer kind of let it go as he had to wrap up the interview &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Foo and Bar are twin brothers. Foo is older than Bar by 5 minutes. Foo celebrates his birthday every year on March 16th and Bar celebrates his birthday on March 15th. Explain the scenario.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9453" LastEditorUserId="5373" LastEditDate="2017-02-06T11:26:16.633" LastActivityDate="2018-09-25T12:54:56.990" Title="Older twin's birthday is one day after younger's" Tags="|lateral-thinking|time|" AnswerCount="16" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8814" PostTypeId="1" AcceptedAnswerId="8840" CreationDate="2015-02-11T23:43:24.393" Score="7" ViewCount="879" Body="&lt;p&gt;Well, &lt;a href=&quot;https://puzzling.stackexchange.com/q/8769/5208&quot;&gt;thanks to Len, I managed to unlock my phone again&lt;/a&gt;, but now everyone knows my pattern, so I need to choose a new one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;For brevity, I won't include the explanation of how an Android pattern lock screen works again, but you still need to follow those rules (see previous puzzle).&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the rules for my new pattern:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I still want to use all nine dots&lt;/li&gt;&#xA;&lt;li&gt;I'm still obsessed with mirroring, so it must still have reflective symmetry&lt;/li&gt;&#xA;&lt;li&gt;I don't like odd numbers, so the pattern's resulting line must have exactly two &quot;ends&quot;*&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What should my new pattern be?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;* by this I mean that visually, the final line formed by the pattern has exactly two dots which have a single line connecting to them. So, for example the pattern &lt;code&gt;2 1 3&lt;/code&gt; has two line endings (at 1 &amp;amp; 3), but &lt;code&gt;5 2 1 3&lt;/code&gt; has three (at 1, 3 &amp;amp; 5). All solutions found for the previous puzzle have three line endings.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-02-12T13:32:58.320" Title="Now help me choose a new lock pattern for my phone" Tags="|geometry|pattern|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8815" PostTypeId="2" ParentId="8813" CreationDate="2015-02-11T23:44:33.183" Score="46" Body="&lt;p&gt;Foo is older than Bar, but celebrates his birthday the day after Bar because one of the following happened:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. The mother was on a boat or a plane traveling eastward.  Foo was born on March 16th, then the boat/plane crossed the International Date Line, and 5 minutes later, Bar was born on March 15th.&#xA; &lt;p&gt;&#xA; 2. The mother was traveling westward between 12am and 1am.  Foo was born on March 16th, then they crossed a time zone, and 5 minutes later, Bar was born on March 15th.  (This was mentioned in &lt;a href=&quot;https://puzzling.stackexchange.com/questions/8813/interview-puzzle#comment26839_8815&quot;&gt;dmg's comment&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2015-02-12T16:39:35.163" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="8816" PostTypeId="1" AcceptedAnswerId="8941" CreationDate="2015-02-12T00:40:29.520" Score="4" ViewCount="391" Body="&lt;p&gt;The greatest arithmeticians on this side of the Kuiper Belt had gathered together on the summit of Olympus Mons to hold the solar system's first arithmetic summit. Everyone was very excited - you could hear mutterings about sums and products in at least eight different languages (and one &lt;a href=&quot;http://en.wikipedia.org/wiki/Pluto&quot;&gt;dwarf language&lt;/a&gt;), and you could see mathematicians making their enthusiastic, but nonsensical mathematician gestures with anywhere from one to twenty arms (or similar appendages)!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Phobos set, the crowd gathered into a circle and went silent in anticipation. The keynote speaker, a distinguished Earthican mathematician, stood up, and began her address, stating various truths about the sums, products, differences, and quotients of pairs of real numbers. Unfortunately, as she continued, she realized that the crowd was bewildered - and that, every time she said &quot;zero&quot;, the Neptunians would make a pitiful noise with their speech-glands, and every time she said &quot;one&quot;, the Mercurians would make what is surely an obscene gesture with their noodly appendages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;She paused and thought, and it dawned on her:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Though every member of the audience knew that her talk would concern only the operations of addition, subtraction, multiplication, and division over the real numbers, there was no common language&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The words &quot;zero&quot; and &quot;one&quot; are words in other languages, but they roughly translate to grievous insults about the alien's version of a &quot;face&quot; or &quot;mother&quot;. She must not use them.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What series of true statements can she make such that there can be no ambiguity in translation, where each statement is of the form&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$x \text{ op } y = z$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;with $x,\,y,$ and $z$ are real numbers other than $0$ or $1$ and op one of the four elementary operations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Unambiguous&quot; means that the aliens can determine the meaning of each word, only knowing that there are four operations and that the numbers are real, but not knowing which operation, for instance, &quot;plus&quot; refers to or which number &quot;two&quot; refers to (as they do not speak the language). For instance&#xA;$$2+2=4$$&#xA;$$4/2=2$$&#xA;Is ambiguous because if the aliens believed &quot;/&quot; meant &quot;-&quot; (i.e. swapped the operations in all equations), they'd get the equally correct&#xA;$$2+2=4$$&#xA;$$4-2=2.$$&#xA;Thus, from the first two equations, they would not be able to determine the meaning of &quot;/&quot;, as it would be equally true if it referred to division or subtraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, were only &quot;+&quot; and &quot;*&quot; to be used, and were $0$ and $1$ allowable words, the following three statements would be a solution:&#xA;$$1*1=1$$&#xA;$$0+0=0$$&#xA;$$1+1=2$$&#xA;As the aliens would, from the first two, be able to deduce that 1 was the identity element of * and 0 was the identity element of +. The only alternate translation of those two is&#xA;$$0+0=0$$&#xA;$$1*1=1$$&#xA;however, under those rules, the last statement translations to&#xA;$$0*0=c$$&#xA;(where $c$ is whatever 2 translates to) - but this is false, since $0*0=0$, and the aliens can recognize that $c$ was not the same word as meant $0$. Thus, there is exactly one way to translate the first series of three statements.&lt;/p&gt;&#xA;" OwnerUserId="4765" LastEditorUserId="4765" LastEditDate="2015-02-12T01:01:51.787" LastActivityDate="2015-02-14T14:55:23.113" Title="The First Interplanetary Arithmetic Summit" Tags="|mathematics|" AnswerCount="5" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8817" PostTypeId="2" ParentId="8816" CreationDate="2015-02-12T01:31:24.450" Score="1" Body="&lt;p&gt;Here's a presumably far-from-optimal solution providing all four operators in six statements:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presuming the aliens know that $0$ and $1$ are not in the mix, then it's easy to disambiguate addition/multiplication from subtraction/division; there are no numbers $a,b\geq 2$ with $a+b=b$ or $a\times b=b$, and contrariwise there are no numbers with $a-a=b$.  So:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\alpha\mathop{\heartsuit}\alpha = \beta$&lt;/li&gt;&#xA;&lt;li&gt;$\alpha\mathop{\spadesuit}\alpha = \beta$&lt;/li&gt;&#xA;&lt;li&gt;$\alpha\mathop{\spadesuit}\gamma = \delta$&lt;/li&gt;&#xA;&lt;li&gt;$\gamma\mathop{\heartsuit}\gamma = \epsilon$&lt;/li&gt;&#xA;&lt;li&gt;$\epsilon\mathop{\clubsuit}\gamma = \gamma$&lt;/li&gt;&#xA;&lt;li&gt;$\epsilon\mathop{\diamondsuit}\delta=\beta$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(1) and (2) establish $\alpha=2$, $\beta=4$, and $\heartsuit$ and $\spadesuit$ as the 'positive' operators.  (4), (5), and (6) establish the negative operators as the inverses of the positive ones (and in particular, imply that $\clubsuit$ is inverse to $\heartsuit$ and that $\diamondsuit$ is inverse to $\spadesuit$).  But now suppose that $\spadesuit$ is $\times$.  Then by (3) we have $\delta=\alpha\times\gamma=2\gamma$, and by inverting (6) we get $\epsilon=\beta\times\delta=4\delta=8\gamma$.  But $\gamma\heartsuit\gamma=\epsilon$ would then claim that $\gamma+\gamma=\epsilon$ or $\epsilon=2\gamma$, which is clearly a contradiction (remember that none of these can be $0$ or $1$).  This gives $\spadesuit=+$ and $\heartsuit=\times$; and now lines (5) and (6) give the other two.&lt;/p&gt;&#xA;" OwnerUserId="266" LastEditorUserId="266" LastEditDate="2015-02-12T01:40:10.150" LastActivityDate="2015-02-12T01:40:10.150" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8819" PostTypeId="2" ParentId="8816" CreationDate="2015-02-12T02:32:33.913" Score="0" Body="&lt;p&gt;I have a set of 12 which identifies the operators and the numbers 2-9 with no requirement that the aliens know 0 and 1 are prohibited:&lt;br&gt;&#xA;2*2=4&lt;br&gt;&#xA;2+2=4&lt;br&gt;&#xA;4-2=2&lt;br&gt;&#xA;4/2=2&lt;br&gt;&#xA;Identifies 2 and 4 and tells our guests that +,* are addition and multiplication (but not which is which) and that -,/ are division and subtraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4+2=6&lt;br&gt;&#xA;6+2=8&lt;br&gt;&#xA;4*2=8&lt;br&gt;&#xA;Identifies 6 and 8 and clarifies +,*&lt;br&gt;&#xA;8-2=6&lt;br&gt;&#xA;Then gives -,/&lt;br&gt;&#xA;6/2=3&lt;br&gt;&#xA;3+2=5&lt;br&gt;&#xA;5+2=7&lt;br&gt;&#xA;7+2=9&lt;br&gt;&#xA;Identify the remaining numbers.&lt;/p&gt;&#xA;" OwnerUserId="3987" LastActivityDate="2015-02-12T02:32:33.913" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8820" PostTypeId="1" AcceptedAnswerId="8827" CreationDate="2015-02-12T02:42:39.793" Score="0" ViewCount="470" Body="&lt;p&gt;I'm a kaleidoscope of colors,&lt;br&gt;&#xA;Red, blue, green of sorts&lt;br&gt;&#xA;You use me everyday,&lt;br&gt;&#xA;Even now you're using me,&lt;br&gt;&#xA;Or maybe my arch-rival...  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cool as he is, he's -&lt;br&gt;&#xA;Mostly used offline!&lt;br&gt;&#xA;Yet, he's vital to many prints,&lt;br&gt;&#xA;Kaleidoscope, kaleidoscope...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My rival's smug, very, very smug,&lt;br&gt;&#xA;He's comprised of more than me...&lt;br&gt;&#xA;But he only subtracts, I add,&lt;br&gt;&#xA;We're very under-hyped!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kaleidoscope, kaleidoscope,&lt;br&gt;&#xA;Part of a broader view!&lt;/p&gt;&#xA;" OwnerUserId="5368" LastEditorUserId="5368" LastEditDate="2015-02-12T12:48:47.977" LastActivityDate="2015-02-12T12:48:47.977" Title="Kaleidoscope of colors" Tags="|riddle|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8821" PostTypeId="2" ParentId="8820" CreationDate="2015-02-12T03:13:29.967" Score="2" Body="&lt;p&gt;Could you be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A computer screen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm a kaleidoscope of colors,&#xA;Red, blue, green of sorts&#xA;You use me everyday,&#xA;Even now you're using me,&#xA;Or maybe my arch-rival...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Computer screens are made up of RGB components. The arch rival could refer to a TV instead of a monitor which some people use instead.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cool as he is, he's -&#xA;Mostly used offline!&#xA;Yet, he's vital to many prints,&#xA;Kaleidoscope, kaleidoscope...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TVs are normally used offline but are used for stories (related to prints/books?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My rival's smug, very, very smug,&#xA;He's comprised of more than me...&#xA;But he only subtracts, I add,&#xA;We're very under-hyped!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Televisions are normally bigger and contain more pixels. However these days many monitors are touch screens. In the same way touch screens &quot;add&quot; functionality, TVs subtract(yeah it's weak).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Anyone else feel free to steal my answer if you can come up with a better explanation for the iffy parts.&lt;/p&gt;&#xA;" OwnerUserId="7722" LastActivityDate="2015-02-12T03:13:29.967" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8822" PostTypeId="2" ParentId="8820" CreationDate="2015-02-12T03:49:06.133" Score="1" Body="&lt;p&gt;Credit to &lt;a href=&quot;https://puzzling.stackexchange.com/a/8821/4450&quot;&gt;Quark&lt;/a&gt; for figuring out the first 4 lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a kaleidoscope of colors,&#xA;Red, blue, green of sorts&#xA;You use me everyday,&#xA;Even now you're using me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This refers to a computer screen.  Each pixel has red, blue, and green components.  I use a computer screen everyday, and I am using one to read this puzzle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or maybe my arch-rival...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is a printer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cool as he is, he's -&#xA;Mostly used offline!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure about this - most printers are not connected to the internet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet, he's vital to many prints,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Obviously a printer is vital to many prints.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My rival's smug, very, very smug,&#xA;He's comprised of more than me...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Inkjet printers use CMYK (4 colors - cyan, magenta, yellow, key/black) instead of RGB (3 colors).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But he only subtracts, I add,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CMYK is a subtractive color model, RGB is an additive color model.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-02-12T03:49:06.133" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8823" PostTypeId="2" ParentId="1941" CreationDate="2015-02-12T04:44:27.597" Score="1" Body="&lt;p&gt;I've seen the above answers but I have another solution.  If John can throw something in the air perfectly up just before the rotation starts, maybe a hankerchief parachute, if the object is perfectly symmetric and the throw is perfectly up, then if the object falls slowly enough, he can note how it rotated in relation to him, (may require a pattern on the hankerchief).&lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2015-02-12T04:44:27.597" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8824" PostTypeId="2" ParentId="1941" CreationDate="2015-02-12T05:33:43.103" Score="3" Body="&lt;p&gt;Simple...  no magnet, no laces required :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The steel cage is offset from the center of the room, thus creating a reference point to compare the distance of each glass to the nearest wall of the main room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exit the steel cage and just look !  (drawing has exagerated proportions to relate the idea - the op does specify room is &quot;&lt;strong&gt;about&lt;/strong&gt;&quot; 20m diameter, and glasses are &quot;&lt;strong&gt;about&lt;/strong&gt;&quot; 10 meters from center of cage...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OZ4hz.png&quot; alt=&quot;solution&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9461" LastEditorUserId="9461" LastEditDate="2015-02-12T05:39:28.383" LastActivityDate="2015-02-12T05:39:28.383" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8826" PostTypeId="2" ParentId="8811" CreationDate="2015-02-12T07:56:00.720" Score="3" Body="&lt;p&gt;Well I doubt it's correct, but here goes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dragon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronotical fish&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Blank, because I have no idea where you're going with astronotical. Dragons are kinda like huge, flying, fire-spewing fish though.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Owner of crown jewels&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Dragons hoard gold and jewels, I guess? There's an obvious link somewhere I feel I forgot.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Chelsea&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The thing on the crest (of the football club) has a dragon look going on.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Honoured Akira&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Toriyama, creator of Dragon Ball.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;synthesizer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Apparently, Synthesizer is &lt;a href=&quot;http://dragonball.wikia.com/wiki/Synthesizer&quot; rel=&quot;nofollow&quot;&gt;a boss in some Dragon Ball game&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5265" LastActivityDate="2015-02-12T07:56:00.720" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8827" PostTypeId="2" ParentId="8820" CreationDate="2015-02-12T08:18:42.700" Score="4" Body="&lt;p&gt;I'm a kaleidoscope of colors,&lt;br&gt;&#xA;Red, blue, green of sorts&lt;br&gt;&#xA;You use me everyday,&lt;br&gt;&#xA;Even now you're using me,&lt;br&gt;&#xA;Or maybe my arch-rival...  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My guess would be that kaleidoscope is &lt;a href=&quot;http://en.wikipedia.org/wiki/RGB_color_model&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;RGB color model&lt;/em&gt;&lt;/a&gt; and arch-rival is &lt;a href=&quot;http://en.wikipedia.org/wiki/CMYK_color_model&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;CMYK color model&lt;/em&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cool as he is, he's -&lt;br&gt;&#xA;Mostly used offline!&lt;br&gt;&#xA;Yet, he's vital to many prints,&lt;br&gt;&#xA;Kaleidoscope, kaleidoscope...  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/CMYK_color_model&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;CMYK color model&lt;/em&gt;&lt;/a&gt; is normally used in printers, so he's used offline and he's vital to many prints&lt;br&gt;&#xA; And as &lt;a href=&quot;https://puzzling.stackexchange.com/users/5373/rand-althor&quot;&gt;&lt;em&gt;rand al'thor&lt;/em&gt;&lt;/a&gt; mentioned in his comment, the initial letters of the lines also form &lt;em&gt;CMYK&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My rival's smug, very, very smug,&lt;br&gt;&#xA;He's comprised of more than me...&lt;br&gt;&#xA;But he only subtracts, I add,&lt;br&gt;&#xA;We're very under-hyped!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/CMYK_color_model&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;CMYK color model&lt;/em&gt;&lt;/a&gt; is comprised of 4 colors: &lt;em&gt;C&lt;/em&gt;yan, &lt;em&gt;M&lt;/em&gt;agenta, &lt;em&gt;Y&lt;/em&gt;ellow and &lt;em&gt;K&lt;/em&gt;ey(black), &lt;a href=&quot;http://en.wikipedia.org/wiki/RGB_color_model&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;RGB color model&lt;/em&gt;&lt;/a&gt; only consists of 3 colors: &lt;em&gt;R&lt;/em&gt;ed, &lt;em&gt;G&lt;/em&gt;reen and &lt;em&gt;B&lt;/em&gt;lue.&lt;br&gt;&#xA; From &lt;a href=&quot;http://en.wikipedia.org/wiki/Main_Page&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Wikipedia&lt;/em&gt;&lt;/a&gt;:&lt;br&gt;&#xA; The &lt;a href=&quot;http://en.wikipedia.org/wiki/CMYK_color_model&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;CMYK color model&lt;/em&gt;&lt;/a&gt; (process color, four color) is a &lt;strong&gt;subtractive&lt;/strong&gt; color model&lt;br&gt;&#xA; The &lt;a href=&quot;http://en.wikipedia.org/wiki/RGB_color_model&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;RGB color model&lt;/em&gt;&lt;/a&gt; is an additive color model&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Kaleidoscope, kaleidoscope,&lt;br&gt;&#xA;Part of a broader view!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So my final answer is:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/RGB_color_model&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;RGB color model&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;And a bit of how-I-got-there:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RGB was really easy to figure out. Actually, it is in the question.&lt;br&gt;&#xA; Then I thought that kaleidoscope could be the CRT(cathode ray tube), then the rival would be LCD/TFT&lt;br&gt;&#xA; But I didn't find anything about adding/subtracting about CRT/LCD, so I started reading about RGB color model&lt;br&gt;&#xA; In there I saw that this color model is additive, so I immediately went to other color models hoping to find a subtractive one&lt;br&gt;&#xA; After I read the riddle again with those color models in my mind, everything made sense :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9352" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-02-12T10:20:37.350" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8828" PostTypeId="2" ParentId="8811" CreationDate="2015-02-12T08:37:12.740" Score="8" Body="&lt;p&gt;My guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Kurosawa / Kurozawa&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasons:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Akira Kurosawa is apparently a Chelsea fan. Kurozawa is a synthesizer. There is a movie Jellyfish by K. Kurosawa. Akira Kurosawa's Seven samurais is often reffered to as the crown jewel of Japanese film. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9302" LastActivityDate="2015-02-12T08:37:12.740" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8829" PostTypeId="2" ParentId="8813" CreationDate="2015-02-12T09:13:32.523" Score="23" Body="&lt;p&gt;Because they are...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ... identical twins, although this isn't a necessary requirement for this answer, it can come in handy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And they were born...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ... before and after midnight&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ... their identities got mixed up, i.e. the parents think they named their oldest twin Bar and act accordingly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5127" LastEditorUserId="5127" LastEditDate="2015-02-12T09:23:29.467" LastActivityDate="2015-02-12T09:23:29.467" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8830" PostTypeId="1" AcceptedAnswerId="8832" CreationDate="2015-02-12T09:27:23.780" Score="17" ViewCount="1050" Body="&lt;p&gt;Here is a question I was given during an interview a few years ago.&#xA;&lt;hr&gt;&#xA;There was a farmer who had two daughters. One day he was considering giving his farm to his girls.  He took his first daughter, Danielle, aside and told her in strictest confidence the area of the farm. He then took his second daughter, Tammy, aside and told her in strictest confidence the perimeter of the farm.  He also told both girls together that the farm was a rectangle, that the perimeter of the farm did not exceed 80 miles and that each side of the farm was not less than two miles in length, and that all measurements were in whole miles. He then promised them that if they could tell him the dimensions of the farm without revealing to each other what he had told them in confidence, in that case he would give them the farm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conversation between the two girls went as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Danielle said, &quot;I don't know what the dimensions of the farm are.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tammy responded, &quot;I knew you were going to say that Danielle!&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then Danielle said, &quot;Well, I still don't know what the dimensions are!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Tammy said, &quot;Well, I  know what the dimensions are now!&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And Danielle concluded the conversation by saying, &quot;Then I know what the dimensions are too!&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was no other conversation between the girls. They went to their father, told him the dimensions of the farm, and their father kept his word and gave them the farm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: Even if they do not know the value, each girl knows what information have been released to her sister. (Tammy knows that Danielle knows the area and Danielle knows that Tammy knows the perimeter.)&lt;/p&gt;&#xA;" OwnerUserId="9463" LastEditorUserId="5373" LastEditDate="2017-02-07T12:30:16.687" LastActivityDate="2017-02-07T12:30:16.687" Title="Farm dimensions" Tags="|mathematics|logical-deduction|meta-knowledge|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8831" PostTypeId="1" AcceptedAnswerId="8835" CreationDate="2015-02-12T09:29:56.727" Score="17" ViewCount="5062" Body="&lt;p&gt;Alice and Bob play the following  number guessing game.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;In the beginning, Alice picks a secret number from the range $1,2,3,\ldots,144$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Then Bob is allowed to ask a sequence of questions. &#xA;In every question, Bob chooses some subset from the range $1,2,3,\ldots,144$&#xA;and asks Alice whether her secret number lies in this subset.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Alice answers Bob's question truthfully.&#xA;If her answer is YES, then Bob must pay 2 Euro to Alice.&#xA;If her answer is  NO, then Bob must pay 1 Euro to Alice.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The game ends, once Bob chooses a one-element subset to which Alice answers YES.&lt;br&gt;&#xA;Bob then receives $x$ Euro from Alice.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How should the value $x$ be chosen, so that these rules do yield a &lt;strong&gt;fair game&lt;/strong&gt;? (As usual, we assume that Alice and Bob both use optimal strategies in their play.)&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-03-24T12:34:25.293" Title="Guess my number! Pay for the answer!" Tags="|mathematics|strategy|game|" AnswerCount="6" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8832" PostTypeId="2" ParentId="8830" CreationDate="2015-02-12T10:12:06.427" Score="15" Body="&lt;p&gt;In other words, we are looking for two positive integers $x$ and $y$ with $x\ge2$, $y\ge2$ and $4\le x+y\le40$ that fit with the conversation.&#xA;Let us denote the sum $s=x+y$ and the product $p=xy$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Danielle's first statement &quot;I don't know what the dimensions of the farm are&quot; gives the following information on the product:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$p$ is not the product of (exactly) two primes $p=ab$; &#xA;otherwise Danielle would immediately deduce that $\{x,y\}=\{a,b\}$.&lt;/li&gt;&#xA;&lt;li&gt;$p$ does not have any prime factor $f\ge20$; &#xA;otherwise Danielle would deduce $\{x,y\}=\{f,p/f\}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(2) Tammy responded, &quot;I knew you were going to say that Danielle!&quot; &#xA;This means that for any possible way of writing $s$ as the sum of two summands $z$ and $s-z$, the corresponding product does not allow Danielle to deduce $x$ and $y$.&#xA;Equivalently, for any integer $z$ with $2\le z\le s/2$, the corresponding product $z\cdot(s-z)$ must be of the form as excluded in (1). &#xA;This excludes most possible values for $s$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$25\le s\le 40$: for $z=23$, the product $z\cdot(s-z)$ has been excluded.&lt;/li&gt;&#xA;&lt;li&gt;$4\le s\le 24$, and $s$ even:&#xA;Then $s$ can be written as the sum of two primes $a$ and $b$ (as a special case of Goldbach's conjecture).&#xA;Hence the corresponding product $ab$ was excluded in (1). &lt;/li&gt;&#xA;&lt;li&gt;$5\le s\le 23$, and $s=a+2$ for a prime $a$:&#xA;The product of $2$ and $a$ was excluded in (1).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Summarizing we see that $s$ must be a number from the following short list:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The remaining candidates for the sum $s$ are 11, 17, 23&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For convenience, we now enumerate for every remaining sum-candidate the corresponding products that have not been excluded under (1):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$s=11$: only 18,  24,  28,  30 are possible products.&lt;/li&gt;&#xA;&lt;li&gt;$s=17$: only 30,  42,  52,  60,  66,  70,  72 are possible products.&lt;/li&gt;&#xA;&lt;li&gt;$s=23$: only 42,  60,  76,  90, 102, 112, 120, 126, 130, 132&#xA;are possible products.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(3) Then Danielle said, &quot;Well, I still don't know what the dimensions are!&quot;&#xA;Although Danielle knows the product and has deduced the remaining sum-candidates, she is not able to determine the sum. This means that her product $p$ occurrs in the lines of two different sum-candidates.&#xA;We update the above listing to these doubly occurring product-candidates:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$s=11$: only 30 is a possible product.&lt;/li&gt;&#xA;&lt;li&gt;$s=17$: only 30,  42,  60 are possible products.&lt;/li&gt;&#xA;&lt;li&gt;$s=23$: only 42,  60 are possible products.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(4) So Tammy said, &quot;Well, I know what the dimensions are now!&quot; &#xA;Tammy knows the sum $s$ and the above listing of remaining compatible  product-candidates. If the sum was $17$ or $23$, then Tammy would have been left with at least two candidates and would &lt;em&gt;not&lt;/em&gt; have known which one was the right one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hence, Tammy must have the sum $s=11$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(5) And Danielle concluded the conversation by saying, &quot;Then I know what the dimensions are too!&quot; &#xA;Danielle has thought through each of the above steps and has come to the following conclusion:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The sum is $s=11$ and the product is $p=30$.&lt;br&gt;&#xA;  The dimensions of the farm are 5 and 6 miles, respectively.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-12T17:45:56.387" LastActivityDate="2015-02-12T17:45:56.387" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8833" PostTypeId="2" ParentId="8806" CreationDate="2015-02-12T10:20:50.693" Score="5" Body="&lt;p&gt;The answer for a full n x m board of kings is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No. The reason is that you can't position three knights such that each is friends with the other two. You've already eliminated n x 1. If we have n x 2, then each King has three friends who are all friends with each other, and we can't even handle two of them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we're allowed less than a full board:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; then it becomes a friendly-king-packing problem. The limit there is at least 2/3 (for some board sizes). For example:&lt;br&gt;&#xA; 123 _ _ _ 153&lt;br&gt;&#xA; xxx _ &gt; _ xxx&lt;br&gt;&#xA; 456 _ _ _ 426&lt;br&gt;&#xA; Here the numbers are kings, and this is two 3x3 boards. But maybe I'm getting ahead of things...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6031" LastActivityDate="2015-02-12T10:20:50.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8835" PostTypeId="2" ParentId="8831" CreationDate="2015-02-12T11:17:31.473" Score="23" Body="&lt;p&gt;&lt;strong&gt;The answer is that only x=11 yields a fair game.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I solved this using a recurrence. Let $f[N]$ be optimal expected amount&#xA;  of money that Bob has to pay Alice to find the best answer. Clearly,&#xA;  $f[1] = 2$ since he can ask a single question.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now we can find a recurrence for $N$. Bob can choose to ask for a subset&#xA;  of size $i$. Since Alice is not cheating, with probability $i/N$, he&#xA;  receives the answer &lt;strong&gt;&lt;em&gt;yes&lt;/em&gt;&lt;/strong&gt;. So the expected value Bob has to pay in total including that for the current question is&#xA;  $$\frac{i*(2+f[i])+(N-i)*(1+f[N-i])}{N}$$ (The first $f[i]$ should be avoided when&#xA;  $i=1$). Minimising this quantity over all values of $1&amp;lt;=i&amp;lt;=N-1$ gives the optimum strategy for Bob. That is, $$f[N]=min_i \frac{i*(2+f[i])+(N-i)*(1+f[N-i])}{N}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Building this recurrence bottom up gives $f[144]=11$. For the game to be&#xA;  fair, Alice should pay 11 euros back to Bob.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9471" LastEditorUserId="8874" LastEditDate="2015-03-24T12:34:25.293" LastActivityDate="2015-03-24T12:34:25.293" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8837" PostTypeId="2" ParentId="8813" CreationDate="2015-02-12T12:07:49.423" Score="6" Body="&lt;p&gt;Fairly sure @pacoverflow had the answer you wanted, but here's an alternative solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Foo and Bar are non-identical twins. If you measure their relative age &lt;em&gt;from conception&lt;/em&gt;, then it is plausible that Foo was conceived 5 minutes earlier, but was actually born an hour or so later (on the other side of midnight, clearly) from Bar.&lt;/p&gt;&#xA;" OwnerUserId="9444" LastActivityDate="2015-02-12T12:07:49.423" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8838" PostTypeId="1" AcceptedAnswerId="8852" CreationDate="2015-02-12T13:23:52.903" Score="0" ViewCount="473" Body="&lt;p&gt;From some US states, one may travel straight south and get into some&#xA;Canadian province or territory (of course without crossing the equator).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are these nine &quot;US state $\to$ Canadian province/territory&quot; pairs?&lt;/p&gt;&#xA;" OwnerUserId="9125" LastActivityDate="2015-02-12T15:47:06.147" Title="Canada and USA geography puzzle" Tags="|geography|knowledge|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8840" PostTypeId="2" ParentId="8814" CreationDate="2015-02-12T13:32:58.320" Score="8" Body="&lt;p&gt;Woah... I think I got it after a whole lot of trying around. Here is my train of thought:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Since we have two open Ends and reflective symmetry, both ends must be mirrored.&lt;/li&gt;&#xA;&lt;li&gt;Since we can never directly go to an already visited point, the last point must be an open end - so one of the two ends.&lt;/li&gt;&#xA;&lt;li&gt;The two end segments (the last part, direct connection) cannot be directly connected, because we cannot leave the path and come back to a point.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Valid endpoints: 1-4 3-6 and 1-4 8-9&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A . . .  B . . .              . . .   . . .   . . .   . . .&#xA;  | . |    | . .    Invalid:  . . .   .\./.   .\. .   . | .&#xA;  . . .    . ._.              ._._.   . . .   . .\.   . | .&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Trying to connect the two valid positions for endpoints I realized I couldn't start at one of them, because I cannot leave the path and come back on the path - and without coming back on the path, it is no longer symmetric. So one of the endpoints had to be a hidden double-line (like 2-1-3) and the other the real endpoint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my path had to start on the mirroring point of this hidden path, since the mirrored path without the end-points would end there and has to be symmetric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the start-points for Figure A was 4 or 6 and for B was 4 or 8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some more trying around gave me this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;. ._.    4-5-6-3-2-8-9-7-1&#xA; |_|_|&#xA; |_|_.&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1998" LastActivityDate="2015-02-12T13:32:58.320" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8841" PostTypeId="1" AcceptedAnswerId="9661" CreationDate="2015-02-12T13:41:57.263" Score="0" ViewCount="375" Body="&lt;p&gt;This is my fourth batch of sequence puzzles that are nasty and hard to solve; &#xA;yet, each of them has a clear and justifiable solution.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sequence 1:&lt;br&gt;&#xA;  2, ?, ?, ?, 0, 3, 7, 6, 4, 6, 9, 0, ?, 0, 4, 3, 4, 9, 5, 7, 3, 7, 5, 4, 4, 9, 4, 3, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sequence 2:&lt;br&gt;&#xA;  Lt, Hd, Tu, Wk, ??, Gbb, Ak, M-kb&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sequence 3:&lt;br&gt;&#xA;  ??, ??, ??, 2, 67, 62, 27, 14&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sequence 4:&lt;br&gt;&#xA;  S, ?, ?, ?, B, F, B, R, A, R, V.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="9125" LastEditDate="2015-02-21T16:53:59.050" LastActivityDate="2015-02-27T11:58:12.770" Title="A batch of four sequence puzzles" Tags="|number-sequence|letter-sequence|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8842" PostTypeId="2" ParentId="8841" CreationDate="2015-02-12T13:45:16.390" Score="4" Body="&lt;p&gt;This time I could google sequence 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sequence 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Moon fact sheet: &#xA; Metric Number of Moons:&lt;br&gt;&#xA; 0, 0, 1, 0, 2, 67, 62, 27, 14, 5&lt;br&gt;&#xA; (so the three questionmarks are 0, 0, 1 for Mercury, Venus, Earth)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Added February 17. I think I have finally found the answer to sequence 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sequence 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; summary, answers, questions, tags, badges, favorites, &#xA; bounties, reputation, activities, responses, votes&lt;br&gt;&#xA; The complete sequence is: &lt;strong&gt;S, A, Q, T, B, F, B, R, A, R, V&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-17T17:04:10.740" LastActivityDate="2015-02-17T17:04:10.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8843" PostTypeId="1" CreationDate="2015-02-12T13:57:20.163" Score="0" ViewCount="274" Body="&lt;p&gt;This is a question I saw and solved a long time ago, it involves some maths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a keypad&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1  2  3&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;4  5  6&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;7  8  9&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A certain password starts with a 3. Each digit is adjacent (horizontally or vertically) on the keypad to either of the previous two digits of the password. The password length is 10. How many possibilities are there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, 1247879685 is valid but 1234565985 is not (4 is not adjacent to 2 or 3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note:&lt;br&gt;&#xA;&lt;strong&gt;Although a complicated combinatorics solution or a program would also work, I'm looking for one that any common person can understand.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4269" LastEditDate="2015-02-12T16:48:55.823" LastActivityDate="2015-02-13T15:00:06.177" Title="Possible numeric codes" Tags="|mathematics|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8844" PostTypeId="2" ParentId="8841" CreationDate="2015-02-12T14:22:59.007" Score="3" Body="&lt;p&gt;Sequence 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To get each number in this sequence, add 1 to each digit of Pi(after the decimal point) and if it is 10, use 0(mod 10).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, The full sequence is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2,5,2,6,0,3,7,6,4,6,9,0,8,0,4,3,4,9,5,7,3,7,4,3,3,9,4,3, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7482" LastActivityDate="2015-02-12T14:22:59.007" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8845" PostTypeId="2" ParentId="8813" CreationDate="2015-02-12T14:23:33.680" Score="7" Body="&lt;p&gt;Because semanthics: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Foo celebrates his&lt;/em&gt; (brother's) &lt;em&gt;birthday every year on March 16th and Bar celebrates his&lt;/em&gt; (brother's) &lt;em&gt;birthday on March 15th.&lt;/em&gt; &lt;a href=&quot;https://www.google.ro/webhp?sourceid=chrome-instant&amp;amp;rlz=1C1CHFX_enRO627RO627&amp;amp;ion=1&amp;amp;espv=2&amp;amp;ie=UTF-8#q=celebrating%20my%20brother%27s%20birthday&quot;&gt;Celebrating my brother's birthday&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9475" LastActivityDate="2015-02-12T14:23:33.680" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8846" PostTypeId="2" ParentId="8838" CreationDate="2015-02-12T14:40:45.483" Score="5" Body="&lt;p&gt;Here are the ones I found so far:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. Alaska, Prince of Wales Island  -&gt; Graham Island&lt;br&gt;&#xA; 2. Minnesota, Voyageurs National Park(where Kettle Falls hotel is) -&gt; Oakpoint Island/Kettle Island/Moose Island/etc(I guess those islands count as one territory, there are 6 islands according to google maps)&lt;br&gt;&#xA; 2.1 Minnesota, Norway Island/Burnt Island Campground -&gt; Some unnamed Canadian ground(but since it's still Minnesota, I guess it doesn't count as 3&lt;br&gt;&#xA; 3. Michigan, Detroit -&gt; Windsor, Ontario&lt;br&gt;&#xA; 4. New York, Buffalo (very west part of it) -&gt; Niagara falls/St. Catharines&lt;br&gt;&#xA; 5. Maine (the part that stick out north, either very west or very east) -&gt; Quebec at very west and Fredericton at very east&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9352" LastActivityDate="2015-02-12T14:40:45.483" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8847" PostTypeId="2" ParentId="8831" CreationDate="2015-02-12T15:07:45.683" Score="6" Body="&lt;p&gt;I think crazyiman has it right, but I just wanted to add a bit that is too much to fit in a comment.  So feel free to upvote this answer, but also upvote his.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets say that you have 2 numbers left.  If you guess one and are right, you are done, but if you are wrong, you still need to guess the other number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So lets define two recurrence relations, $f_r(N)$ and $f_w(N)$.  $f_r(N)$ is the expected value your previous guess was right for the set $N$, and $f_w(N)$ is the expected value if your previous guess was wrong leaving you with set $N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that these relations only differ when you are talking about sets with single elements because if you guessed right on a single number you are done, but if you guessed wrong and are left with a single number, you must still guess it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we know the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_r(1)=0$$&#xA;$$f_w(1)=2$$&#xA;$$f(n)=f_r(i)=f_w(i), \forall i \mid i&amp;gt;1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for $i&amp;gt;1$, we know the following from @craziman:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(n)=\frac{i∗(2+f_r(i))+(n−i)∗(1+f_w(n−i))}{n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And again, just to show the work, we will work up from 1.  The whole number results are indeed the Fibonacci sequence.  For eacn $n$, the best choice for $i$ is also recorded.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f(002)=02.500, i=1&#xA;f(003)=03.000, i=1   &amp;lt;-- Fibonacci&#xA;f(004)=03.500, i=1&#xA;f(005)=04.000, i=1   &amp;lt;-- Fibonacci&#xA;f(006)=04.500, i=1     f(007)=04.714, i=3&#xA;f(008)=05.000, i=3   &amp;lt;-- Fibonacci&#xA;f(009)=05.222, i=4     f(010)=05.500, i=3     f(011)=05.636, i=4&#xA;f(012)=05.833, i=4&#xA;f(013)=06.000, i=4   &amp;lt;-- Fibonacci&#xA;f(014)=06.143, i=5     f(015)=06.333, i=4     f(016)=06.438, i=5&#xA;f(017)=06.588, i=5     f(018)=06.667, i=7     f(019)=06.789, i=7&#xA;f(020)=06.900, i=7&#xA;f(021)=07.000, i=7   &amp;lt;-- Fibonacci&#xA;f(022)=07.091, i=8     f(023)=07.174, i=9     f(024)=07.292, i=8&#xA;f(025)=07.360, i=9     f(026)=07.462, i=8     f(027)=07.519, i=9&#xA;f(028)=07.607, i=9     f(029)=07.655, i=11    f(030)=07.733, i=11&#xA;f(031)=07.806, i=11    f(032)=07.875, i=11    f(033)=07.939, i=11&#xA;f(034)=08.000, i=11  &amp;lt;-- Fibonacci&#xA;f(035)=08.057, i=12    f(036)=08.111, i=13    f(037)=08.162, i=14&#xA;f(038)=08.237, i=14    f(039)=08.282, i=14    f(040)=08.350, i=13&#xA;f(041)=08.390, i=14    f(042)=08.452, i=18    f(043)=08.488, i=14&#xA;f(044)=08.545, i=14    f(045)=08.578, i=16    f(046)=08.630, i=16&#xA;f(047)=08.660, i=18    f(048)=08.708, i=18    f(049)=08.755, i=18&#xA;f(050)=08.800, i=18    f(051)=08.843, i=18    f(052)=08.885, i=18&#xA;f(053)=08.925, i=18    f(054)=08.963, i=18&#xA;f(055)=09.000, i=18  &amp;lt;-- Fibonacci&#xA;f(056)=09.036, i=19    f(057)=09.070, i=20    f(058)=09.103, i=21&#xA;f(059)=09.136, i=22    f(060)=09.167, i=23    f(061)=09.213, i=22&#xA;f(062)=09.242, i=23    f(063)=09.286, i=22    f(064)=09.312, i=23&#xA;f(065)=09.354, i=22    f(066)=09.379, i=23    f(067)=09.418, i=22&#xA;f(068)=09.441, i=23    f(069)=09.478, i=22    f(070)=09.500, i=23&#xA;f(071)=09.535, i=23    f(072)=09.556, i=25    f(073)=09.589, i=25&#xA;f(074)=09.608, i=27    f(075)=09.640, i=27    f(076)=09.658, i=29&#xA;f(077)=09.688, i=29    f(078)=09.718, i=29    f(079)=09.747, i=29&#xA;f(080)=09.775, i=29    f(081)=09.802, i=29    f(082)=09.829, i=29&#xA;f(083)=09.855, i=29    f(084)=09.881, i=29    f(085)=09.906, i=29&#xA;f(086)=09.930, i=29    f(087)=09.954, i=29    f(088)=09.977, i=29&#xA;f(089)=10.000, i=29  &amp;lt;-- Fibonacci&#xA;f(090)=10.022, i=30    f(091)=10.044, i=31    f(092)=10.065, i=32&#xA;f(093)=10.086, i=33    f(094)=10.106, i=34    f(095)=10.126, i=35&#xA;f(096)=10.146, i=36    f(097)=10.165, i=37    f(098)=10.194, i=36&#xA;f(099)=10.212, i=37    f(100)=10.240, i=36    f(101)=10.257, i=37&#xA;f(102)=10.284, i=36    f(103)=10.301, i=37    f(104)=10.327, i=36&#xA;f(105)=10.343, i=37    f(106)=10.368, i=36    f(107)=10.383, i=37&#xA;f(108)=10.407, i=36    f(109)=10.422, i=37    f(110)=10.445, i=36&#xA;f(111)=10.459, i=37    f(112)=10.482, i=36    f(113)=10.496, i=37&#xA;f(114)=10.518, i=37    f(115)=10.530, i=39    f(116)=10.552, i=39&#xA;f(117)=10.564, i=41    f(118)=10.585, i=41    f(119)=10.597, i=43&#xA;f(120)=10.617, i=43    f(121)=10.628, i=45    f(122)=10.648, i=45&#xA;f(123)=10.659, i=47    f(124)=10.677, i=47    f(125)=10.696, i=47&#xA;f(126)=10.714, i=47    f(127)=10.732, i=47    f(128)=10.750, i=47&#xA;f(129)=10.767, i=47    f(130)=10.785, i=47    f(131)=10.802, i=47&#xA;f(132)=10.818, i=47    f(133)=10.835, i=47    f(134)=10.851, i=47&#xA;f(135)=10.867, i=47    f(136)=10.882, i=47    f(137)=10.898, i=47&#xA;f(138)=10.913, i=47    f(139)=10.928, i=47    f(140)=10.943, i=47&#xA;f(141)=10.957, i=47    f(142)=10.972, i=47    f(143)=10.986, i=47&#xA;f(144)=11.000, i=47  &amp;lt;-- Fibonacci&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2015-02-12T15:07:45.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8848" PostTypeId="5" CreationDate="2015-02-12T15:18:59.000" Score="0" Body="&lt;p&gt;Nonograms are picture logic puzzles that use a grid with numbers, generally along the top and left side, indicating how many cells are coloured in in that specific row/column. If there are multiple groupings of coloured sections in a particular row or column there must be separate numbers to represent the amount of coloured spaces in each group. They can be solved using logic to eventually show an image in the grid. It is also possible to have nonograms with multiple colours.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Example: &lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/XcdkT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XcdkT.png&quot; alt=&quot;Nonogram example&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Often used in conjunction with &lt;a href=&quot;/questions/tagged/grid-deduction&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;grid-deduction&amp;#39;&quot; aria-label=&quot;show questions tagged &amp;#39;grid-deduction&amp;#39;&quot; rel=&quot;tag&quot; aria-labelledby=&quot;grid-deduction-container&quot;&gt;grid-deduction&lt;/a&gt; and &lt;a href=&quot;/questions/tagged/logical-deduction&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;logical-deduction&amp;#39;&quot; aria-label=&quot;show questions tagged &amp;#39;logical-deduction&amp;#39;&quot; rel=&quot;tag&quot; aria-labelledby=&quot;logical-deduction-container&quot;&gt;logical-deduction&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;For more information the Nonogram wikipedia page can be found &lt;a href=&quot;https://en.wikipedia.org/wiki/Nonogram&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;here&lt;/em&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="52483" LastEditorUserId="52483" LastEditDate="2023-01-31T19:25:26.347" LastActivityDate="2023-01-31T19:25:26.347" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8849" PostTypeId="4" CreationDate="2015-02-12T15:18:59.000" Score="0" Body="Puzzles that involve nonograms (also known as *Hanjie* or *Japanese Puzzle*), logic puzzles where runs of colored cells are given at the edges of the grid. Use with [grid-deduction] and [logical-deduction]" OwnerUserId="8874" LastEditorUserId="18250" LastEditDate="2021-02-07T22:47:59.683" LastActivityDate="2021-02-07T22:47:59.683" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8850" PostTypeId="1" AcceptedAnswerId="8854" CreationDate="2015-02-12T15:26:41.270" Score="6" ViewCount="1338" Body="&lt;p&gt;Is this the smallest known rebus puzzle, and what does it mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JHopq.jpg&quot; alt=&quot;Rebus puzzle&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="15544" LastEditDate="2015-10-02T19:05:42.360" LastActivityDate="2015-10-02T19:05:42.360" Title="Smallest rebus puzzle: Print" Tags="|puzzle-creation|rebus|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8851" PostTypeId="2" ParentId="8850" CreationDate="2015-02-12T15:29:29.423" Score="4" Body="&lt;p&gt;I'm not sure whether its the smallest rebus puzzle....but the answer is &lt;strong&gt;Blue print!!&lt;/strong&gt;&#xA;.&#xA;.&#xA;.&#xA;If by the smallest rebus puzzle you mean the one which has the least number of letters in the answer the NO, there is a smaller rebus puzzle, that being.....Big Deal-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7oiOo.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9346" LastEditorUserId="9346" LastEditDate="2015-02-12T15:44:28.493" LastActivityDate="2015-02-12T15:44:28.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8852" PostTypeId="2" ParentId="8838" CreationDate="2015-02-12T15:32:01.177" Score="5" Body="&lt;p&gt;Assuming that one state to two different Canadian territories counts as two pairs (and by the presentation of the question, I think it does), then Novarg actually has six, and I think I have the other three.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Point Roberts, Washington is directly north of Saturna Island, British Columbia&lt;br&gt;&#xA;&lt;br&gt;&#xA;A large portion of the border between New Hampshire and Quebec is defined by Halls Stream, which leads to several small points along the border where NH is directly north of QC, particularly near the north end of the river.&lt;br&gt;&#xA;&lt;br&gt;&#xA;Mt Hubbard, in the Saint Elias Mountains, lies on the border between Alaska and Yukon. The border travels roughly SSW from there to Mt Seattle, so the Alaskan area around Mt Hubbard is directly north of Yukon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The main difficulty with this question was finding the specific points. Most of the US-Canada border is defined strictly by latitude and longitude, so it is impossible for those areas to have part of the US directly north of Canada. This eliminates Idaho, Montana, North Dakota, and Vermont as possibilities, as well as the Minnesota-Manitoba and New York-Quebec borders.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That leaves eleven pairs of US states and Canadian territories that share a border. Ohio and Pennsylvania only border Ontario along Lake Erie, so they clearly don't fit the requirements either. That only leaves nine border pairs. And since your question says that there are nine pairs in the answer, these must be them:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Alaska$\to$Yukon&lt;br&gt;&#xA;Alaska$\to$British Columbia&lt;br&gt;&#xA;Washington$\to$British Columbia&lt;br&gt;&#xA;Minnesota$\to$Ontario&lt;br&gt;&#xA;Michigan$\to$Ontario&lt;br&gt;&#xA;New York$\to$Ontario&lt;br&gt;&#xA;New Hampshire$\to$Quebec&lt;br&gt;&#xA;Maine$\to$Quebec&lt;br&gt;&#xA;Maine$\to$New Brunswick&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8776" LastEditorUserId="8776" LastEditDate="2015-02-12T15:47:06.147" LastActivityDate="2015-02-12T15:47:06.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8854" PostTypeId="2" ParentId="8850" CreationDate="2015-02-12T15:37:09.083" Score="13" Body="&lt;p&gt;The answer to the puzzle is &quot;Blueprint&quot; but I doubt it's the smallest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We would need to clarify what is meant by &quot;smallest&quot;. If you mean the smallest number of bytes required to store the puzzle electronically, then the answer is no, this is not the smallest puzzle. A4ID (foreign aid) (4 bytes) has less letters and color doesn't matter so there's no need to store the information. R+R (summer) is shorter, still.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Side note: We could get cute with it and say &quot;small print&quot; is the smallest rebus:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/DNFrG.gif&quot; alt=&quot;Small Print Rebus Puzzle&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9000" LastActivityDate="2015-02-12T15:37:09.083" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8856" PostTypeId="2" ParentId="8843" CreationDate="2015-02-12T16:01:50.223" Score="3" Body="&lt;p&gt;I made a small program in lua. this is the code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;n = {&#xA;[0] = {},&#xA;[1] = {[2]=true,[4]=true},&#xA;[2] = {[1]=true,[3]=true,[5]=true},&#xA;[3] = {[2]=true,[6]=true},&#xA;[4] = {[1]=true,[5]=true,[7]=true},&#xA;[5] = {[2]=true,[4]=true,[6]=true,[8]=true},&#xA;[6] = {[3]=true,[5]=true,[9]=true},&#xA;[7] = {[4]=true,[8]=true},&#xA;[8] = {[7]=true,[5]=true,[9]=true},&#xA;[9] = {[6]=true,[8]=true}}&#xA;&#xA;function x(a,b, level)&#xA;  if level == 10 then&#xA;    return 1;&#xA;  end&#xA;  local newnumbers = {}&#xA;  for k in pairs(n[a]) do&#xA;    newnumbers[k] = true&#xA;  end&#xA;  for k in pairs(n[b]) do&#xA;    newnumbers[k] = true&#xA;  end&#xA;  local c = 0&#xA;  for k in pairs(newnumbers) do&#xA;      c = c + x(b,k,level+1)&#xA;  end&#xA;  return c;&#xA;end&#xA;&#xA;print(x(0,3,1))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;you can test it out at &lt;a href=&quot;http://www.lua.org/cgi-bin/demo&quot; rel=&quot;nofollow&quot;&gt;http://www.lua.org/cgi-bin/demo&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 691,950&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not sure if I made any mistake&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-02-12T16:01:50.223" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8857" PostTypeId="2" ParentId="8813" CreationDate="2015-02-12T16:18:29.200" Score="2" Body="&lt;p&gt;I think the emphasis on other answers has been in solving the abstract puzzle that was presented (which is very reasonable on this forum). In an interview situation, if you were able to come up with these different possible explanations, then selecting the &quot;best&quot; explanation (assuming you get one chance to answer) would probably depend on the nature of the interview.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time zone answer is an appropriate choice for any company which engages in business across multiple time zones (demonstrates awareness of the issues that time zones can create), and for certain companies it may even earn bonus points if you can point to a specific location where this could have occurred: &quot;Foo was born in X while his parents were travelling towards Z on a boat, Bar was born in Y 5 minutes later&quot; where X Y and Z are locations that the company does business.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mixup identity answer is the answer I'd go with if I were interviewing for an editorial/newspaper job because it demonstrates a focus on the precise wording of the question (&quot;Foo celebrates his birthday on X&quot; instead of &quot;Foo was born on X&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally, if I were interviewing for any position related to a religious organization which defines life as beginning at conception, it would be very hard to go wrong with the conception answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally mixed up identity is what I believe to be the generic &quot;most likely&quot; answer to this puzzle (in that there probably exist more people who have been born under those circumstances than the time zone answer, and it doesn't require playing with the definition of life), and it seems logical that if the puzzle is worded as &quot;celebrates his birthday on...&quot; there's a reason it didn't simply state &quot;was born on,&quot; so if a business doesn't meet any of the criteria above I'd definitely go with mixed-up identity. Even more so if it's a stock analysis or prediction of any kind job where best guesses often have to be relied upon because you cannot know something with certainty.&lt;/p&gt;&#xA;" OwnerUserId="9477" LastActivityDate="2015-02-12T16:18:29.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8858" PostTypeId="2" ParentId="8816" CreationDate="2015-02-12T16:33:17.887" Score="1" Body="&lt;p&gt;There are other ways to help identify the operations that do not depend on the arguments being natural numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if we say that $x\odot y=z$ and $y\odot x=z$, this tells us that the $\odot$ operator is commutative, meaning it is either addition or multiplication. Then we can do the same thing with another operator, say $\oplus$, so then we know that those two are addition and multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we can follow some of the same techniques:&#xA;\begin{align}a\oplus a&amp;amp;=c\\a\odot a&amp;amp;=c \end{align}&#xA;establishes that $a=2$ and $c=4$. Then we can state&#xA;\begin{align}2\oplus4&amp;amp;=f\\2\odot4&amp;amp;=g\end{align}&#xA;which tells them that $f$ and $g$ are either $6$ or $8$. Then we say&#xA;$$4\odot4=f$$&#xA;This gives a lot of new information. It tells us that $\odot$ is addition and $f=8$, so then that leaves $\oplus$ as multiplication and $g=6$. From here we can establish the operations of subtraction and division with $8\oslash2=4$ and either $8\ominus2=6$ or $6\ominus2=4$. Now the aliens should be able to recognize all four elementary operations. Then we can tell them $6/2=b$, so they know $b=3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here, we can add numbers to get all integers greater than $1$, and subtract to get integers less than $-1$. We can also divide to get rational numbers other than $\pm1$ and $0$.&lt;/p&gt;&#xA;" OwnerUserId="8776" LastActivityDate="2015-02-12T16:33:17.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8859" PostTypeId="2" ParentId="8850" CreationDate="2015-02-12T16:38:33.853" Score="4" Body="&lt;p&gt;The answer to the puzzle has been given in the other answers, but how about this for a smaller (art-related) Rebus puzzle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fHzeV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user9278" LastActivityDate="2015-02-12T16:38:33.853" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8860" PostTypeId="2" ParentId="8713" CreationDate="2015-02-12T16:43:24.770" Score="2" Body="&lt;p&gt;there is no &quot;I&quot; in TEAm, but there is an &quot;I&quot; in pie. And there's an &quot;I&quot; in meat pie. Anagram of meat is team...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;yeah sorry about that it's just driving me crazy&lt;/p&gt;&#xA;" OwnerUserId="9413" LastActivityDate="2015-02-12T16:43:24.770" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8861" PostTypeId="1" AcceptedAnswerId="8862" CreationDate="2015-02-12T16:58:08.563" Score="10" ViewCount="1370" Body="&lt;p&gt;There are two similar rebus puzzles below. Each of them hides a word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Puzzle 1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VcQiT.png&quot; alt=&quot;Rebus1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Puzzle 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zbH2F.png&quot; alt=&quot;Rebus2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your task is to find out the hidden word in both of these puzzles.&lt;/p&gt;&#xA;" OwnerUserId="7482" LastActivityDate="2015-02-12T17:03:20.617" Title="Two similar rebus puzzles" Tags="|rebus|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8862" PostTypeId="2" ParentId="8861" CreationDate="2015-02-12T17:03:20.617" Score="11" Body="&lt;p&gt;Puzzle 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eye + L&amp;amp; = Island&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Puzzle 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eyes + L&amp;amp; = Iceland&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2015-02-12T17:03:20.617" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8863" PostTypeId="1" AcceptedAnswerId="8866" CreationDate="2015-02-12T17:10:59.243" Score="6" ViewCount="333" Body="&lt;p&gt;Let $N$ and $M$ be two integers with $M\ge N\ge2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a list of altogether $M$ statements that is divided into three parts:&#xA;the first part consists only of the first statement; &#xA;the second part consists of the $N-1$ statements with numbers from $2$ up to $N$;&#xA;the third (possibly empty) part consists of the remaining statements with numbers from $N+1$ up to $M$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first part of the list declares: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Statement 1&lt;/strong&gt;: Not all the statements on this list are false.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The statement with number $n$ with $2\le n\le N$ in the second part says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Statement n&lt;/strong&gt;: All statements with a number divisible by $n$ are false.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The statement with number $k$ with $N+1\le k\le M$ in the third part may be arbitrary.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Statement k&lt;/strong&gt;: (Contents is left to the choice of the puzzle solver.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Determine (in dependence on $N$) the minimum value of $M$ for which this system of statements does not yield a paradox.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="8874" LastEditDate="2015-02-12T17:47:29.327" LastActivityDate="2015-02-12T19:57:29.913" Title="At least $N$ false statements" Tags="|logical-deduction|paradox|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8864" PostTypeId="2" ParentId="8863" CreationDate="2015-02-12T17:36:02.467" Score="0" Body="&lt;p&gt;I don't think there are any possible values for M and N that don't trap us in a paradox&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if M=2 and N=2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1) Not all the statements on the list are false.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2) Every 2nd statement is false&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;obviously we have a paradox because if &lt;strong&gt;2)&lt;/strong&gt; is declaring itself to be false then it's true&lt;/p&gt;&#xA;&#xA;&lt;p&gt;am I missing the question entirely?&lt;/p&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="8874" LastEditDate="2015-02-12T17:44:06.500" LastActivityDate="2015-02-12T17:44:06.500" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8865" PostTypeId="2" ParentId="8863" CreationDate="2015-02-12T17:49:42.490" Score="1" Body="&lt;p&gt;N can be as small as 2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 Not all statements are false (true)&lt;br&gt;&#xA;2 Every statement divisible by 2 is false (false)  N&lt;br&gt;&#xA;3 Every statement divisible by 3 is true (true)&lt;br&gt;&#xA;4 Every statement divisible by 4 is true (true)  M&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, for your highest N, you have to have a higher 2*N that is true. Between N and M, every prime will need to be true. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-02-12T17:49:42.490" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8866" PostTypeId="2" ParentId="8863" CreationDate="2015-02-12T18:06:01.550" Score="7" Body="&lt;p&gt;(1) No statement $n$ with $2\le n\le N$ in the second part can be true, as it would imply its own falseness. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Every statement in the second part of the list is false.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(2) As statement $n$ with $2\le n\le N$ in the second part is false, we conclude that not all statements with a number divisible by $n$ are false.&#xA;Equivalently, there exists some statement with a number divisible by $n$ that is true.&#xA;Clearly, this true statement with a number divisible by $n$ must be in the third part of the list. We conclude: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For every $n$ with $2\le n\le N$, there is some true statement in the third part of the list, whose number is divisible by $n$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(3) Now for $n=N$, the above discussion implies that the third part of the list contains a true statement whose number is divisible by $N$. This implies $M\ge2N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) If $M=2N$, then we can find the following true/false pattern for the list:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The first statement is TRUE.&lt;/li&gt;&#xA;&lt;li&gt;All statements in the second part are FALSE.&lt;/li&gt;&#xA;&lt;li&gt;All statements in the third part are TRUE (let us say that they state $1+1=2$).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Since every number $n$ with $2\le n\le N$ has a multiple in the range $N+1,N+2,\ldots,2N$, the truth values are compatible and do not yield a paradox.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The answer is that the minimum value is $M=2N$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-02-12T18:06:01.550" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8867" PostTypeId="2" ParentId="8813" CreationDate="2015-02-12T18:19:42.033" Score="14" Body="&lt;p&gt;Here's an answer suggesting human error,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;They were born on either side of midnight, but&#xA;the birth certificate guy got it backwards!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7265" LastEditorUserId="7265" LastEditDate="2015-02-12T21:44:56.177" LastActivityDate="2015-02-12T21:44:56.177" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8868" PostTypeId="2" ParentId="8831" CreationDate="2015-02-12T18:58:35.403" Score="3" Body="&lt;p&gt;Starting with an assumption that the binary search algorithm might be the optimal way to do this, even though half of the answers would cost 2 Euro, and the other half 1 Euro, I brute-forced the question.  Each round Bob asks if the answer is between &quot;middle&quot; and &quot;top&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The average cost for an answer is 10.6875 Euro, meaning that Alice should give Bob 10.6875 Euro and things will average out to be even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I then tested what would happen if we used a weighted value as the center value, instead of just using the number midway between high and low.  When I examined the weighted results, I found that a factor of .375 resulted in the lowest average cost at 10.4028 Euros.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have included my code below.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;public void BinSearch()&#xA;{&#xA;    Debug.Print(&quot;Binary Search = &quot; + GetSearchResult(null).ToString());&#xA;    Debug.Print(&quot;Binary Offset Search = &quot; + GetSearchResult((a, b) =&amp;gt; Convert.ToInt32((b - a) * .375 + a)).ToString());&#xA;&#xA;    nums.ForEach(v =&amp;gt;&#xA;        {&#xA;            var offset = v / 144m;&#xA;            Debug.Print(&quot;Offset by &quot; + v.ToString() + &quot; = &quot; + GetSearchResult((a, b) =&amp;gt; Convert.ToInt32((b - a) * offset + a)).ToString());&#xA;        });&#xA;}&#xA;&#xA;private List&amp;lt;int&amp;gt; nums = Enumerable.Range(1, 144).ToList();&#xA;&#xA;// return the average price to find a value with this formula&#xA;public double GetSearchResult(Func&amp;lt;int, int, int&amp;gt; formula)&#xA;{&#xA;    var costs = new Dictionary&amp;lt;int, int&amp;gt;();&#xA;&#xA;    foreach (var num in nums)&#xA;    {&#xA;        var result = BinarySearch(nums, num, formula);&#xA;        costs.Add(num, result);&#xA;    }&#xA;&#xA;    return costs.Select(v =&amp;gt; v.Value).Average();&#xA;}&#xA;&#xA;// return the price to find value using formula&#xA;public static int BinarySearch(List&amp;lt;int&amp;gt; arrayList, int value, Func&amp;lt;int, int, int&amp;gt; formula)&#xA;{&#xA;    formula = formula ?? ((low, high) =&amp;gt; (low + high) / 2);&#xA;&#xA;    int lo = 0;&#xA;    int hi = arrayList.Count - 1;&#xA;    int price = 0;&#xA;&#xA;    while (lo &amp;lt;= hi)&#xA;    {&#xA;        price++;&#xA;&#xA;        int i = formula(lo, hi);&#xA;        int c = arrayList[i].CompareTo(value);&#xA;&#xA;        if (c == 0 &amp;amp;&amp;amp; lo == hi) return price + 1;&#xA;&#xA;        if (c &amp;lt; 0)&#xA;        {&#xA;            lo = i + 1;&#xA;        }&#xA;        else&#xA;        {&#xA;            price++;&#xA;            hi = i - 1;&#xA;        }&#xA;    }&#xA;&#xA;    return price;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Alice should give Bob 10.4028 Euros.&lt;/p&gt;&#xA;" OwnerUserId="9479" LastActivityDate="2015-02-12T18:58:35.403" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8869" PostTypeId="2" ParentId="8756" CreationDate="2015-02-12T19:11:12.613" Score="8" Body="&lt;p&gt;The answer must be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;13 July 2014&lt;/strong&gt;&lt;br&gt;&#xA; (when Germany became soccer world champion against Argentina)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;for the following reasons:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Taj Mahal, Stari Most, blue Mosque -&gt; Mimar Sinan -&gt; Sinan Kurt&lt;br&gt;&#xA; comedy of hell -&gt; Dante -&gt; Dante Bonfim Costa Santos&lt;br&gt;&#xA; Fawlty Towers -&gt; the waiter Manuel -&gt; Manuel Neuer&lt;br&gt;&#xA; behaves like a conquistador -&gt; Pizarro -&gt; Claudio Miguel Pizarro Bosio&lt;br&gt;&#xA; Plumber -&gt; Super Mario -&gt; Mario Gotze &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and:   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  band of brothers -&gt; Bayern Munich&lt;br&gt;&#xA;  biggest hole -&gt; biggest goal&lt;br&gt;&#xA;  plumber fixes the world's biggest hole -&gt; Gotze scores against Argentina&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-12T19:52:27.803" LastActivityDate="2015-02-12T19:52:27.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8870" PostTypeId="2" ParentId="8863" CreationDate="2015-02-12T19:24:38.647" Score="1" Body="&lt;p&gt;The minimum value of M is N+1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statement &lt;strong&gt;1 to (n-1)&lt;/strong&gt; says : Not all statements in this list are false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the first statements are true it does not put any requirements on the rest of the list, as it itself is true. If the first statements are false then all statements must be false, thus the second part of statements must be true. Since the statements in the second part they themselves imply they are false this leads to a paradox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore all statements in the first part are true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statement &lt;strong&gt;n to N&lt;/strong&gt;: All statements divisible by n are false.  (can be true or false)&#xA;Statement &lt;strong&gt;N+1=M&lt;/strong&gt;: ..abritrary.. (can be true or false)&lt;/p&gt;&#xA;" OwnerUserId="9302" LastEditorUserId="9302" LastEditDate="2015-02-12T19:57:29.913" LastActivityDate="2015-02-12T19:57:29.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8871" PostTypeId="2" ParentId="8811" CreationDate="2015-02-12T20:07:29.433" Score="7" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Oscar&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;for the following reasons:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; astronotical fish  -&gt;  astronotus ocellatus = oscar fish&lt;br&gt;&#xA; an owner of crown jewels -&gt;   Oscar of Sweden (King of Sweden and Norway, 1799-1859)&lt;br&gt;&#xA; Chelsea -&gt;  Oscar dos Santos Emboaba Junior (plays for Chelsea)&lt;br&gt;&#xA; an honoured Akira -&gt;  Oscar academy award for Akira Kurosawa&lt;br&gt;&#xA; a synthesizer -&gt;  OSCar synthesizer (by the Oxford Synthesiser Company)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-02-12T20:07:29.433" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8872" PostTypeId="2" ParentId="8813" CreationDate="2015-02-12T20:07:51.703" Score="0" Body="&lt;p&gt;My first post here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Foo and Bar were both born on the 15th.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bar died on his birthday which results in Foo celebrating his birthday on the 16th of every year. It doesn't state that Bar celebrates his birthday every year.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The key word here is 'celebrates'.&lt;/p&gt;&#xA;" OwnerUserId="9481" LastEditorUserId="9481" LastEditDate="2015-02-12T20:14:10.613" LastActivityDate="2015-02-12T20:14:10.613" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8873" PostTypeId="2" ParentId="8831" CreationDate="2015-02-12T20:17:48.437" Score="5" Body="&lt;p&gt;Following on Crazyman's answer: it appears that if $N = F_n$ is the $n^{th}$ Fibonacci number, then following the optimal strategy, the expected value is $f[F_n] = n-1$ for all $n \geq 4$.  Given $N = F_n$ for some $n\geq 4$, a concrete strategy that achieves this expectation is to ask: &quot;Is the desired number less than or equal to $F_{n-2}$?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: by induction.  This strategy gives expectation $4-1 = 3$ for $N = F_4 = 3$, and $5-1 = 4$ for $N = F_5 = 5$, as may be verified by hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now assume that the expectation using this strategy satisfies $f[F_{n-2}] = n-3$, and $f[F_{n-1}] = n-2$, and consider what happens when we apply this strategy to $N = F_n$.  With probability $F_{n-2} / F_n$, the answer will be yes, we'll have to pay $2$ Euro, and we'll have narrowed the possibilities down to a set of size $F_{n-2}$.  On the other hand, with probability $(F_n - F_{n-2})/F_n = F_{n-1} / F_n$, the answer will be no, in which case we pay $1$ Euro and we have it narrowed down to a set of size $F_{n-1}$.  Thus our expected payout is:&#xA;$$&#xA;f[F_n] = \frac{F_{n-2}(2 + f[F_{n-2}]) + F_{n-1}(1 + f[F_{n-1}])}{F_n}&#xA;$$&#xA;By the induction hypothesis, $f[F_{n-2}] = (n-2) -1$, and $f[F_{n-1}] = (n-1) - 1$, which means that the above formula simplifies:&#xA;$$&#xA;f[F_n] = \frac{F_{n-2}(2 + n-3) + F_{n-1}(1 + n-2)}{F_n} = \frac{(F_{n-2} + F_{n-1})(n-1)}{F_n} = n-1,&#xA;$$&#xA;proving the induction step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does not explain why this strategy is optimal, but it does at least demonstrate an explicit strategy that would give $f[144] = 11$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At every point during the game, Bob's strategy is to split a set of $N$ elements into two smaller sets.  Then based on Alice's answer, Bob knows which of these two sets includes Alice's number.  Let's say $A$ and $B$ are the sizes of these two sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this game, two &quot;no&quot; answers cost the same as one &quot;yes&quot; answer, so intuitively they should yield the same amount of information.  Let's let $r$ be the percentage (or proportion) of the $N$ values that lie in the set of size $A$ (so in other words, $r = A/N$).  Then $1-r$ is the proportion of the $N$ values that lie in the other set, so $B/N = 1-r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we use the same proportion $r$ each time, then two &quot;no&quot; answers will leave us with $r^2N$ values left, while one &quot;yes&quot; answer will leave us with $(1-r)N$ values left.  Since they cost the same, they should yield the same amount of information.  That is, we should have&#xA;$$&#xA;r^2 = 1- r&#xA;$$&#xA;Solving this for $r$, we see that $r = 1/ \phi$ where $\phi$ is the Golden Ratio. If $N = F_n$ is a Fibonacci number, then $(1/ \phi)F_n$ is very close to $F_{n-1}$, which starts to explain why this solution makes sense if $N$ is a Fibonacci number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the optimal solution is &lt;em&gt;always&lt;/em&gt; to split $N$ into two numbers that are as close as possible to $(1/ \phi)N$ and $(1 - 1/\phi ) N$, in which case  it just so happens that with Fibonacci numbers you'll always split it into the next two smaller Fibonacci numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Golden_ratio&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Golden_ratio&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8930" LastEditorUserId="8930" LastEditDate="2015-02-13T00:23:10.540" LastActivityDate="2015-02-13T00:23:10.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8875" PostTypeId="2" ParentId="8765" CreationDate="2015-02-12T20:55:23.960" Score="1" Body="&lt;p&gt;Here are the dates described:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Room 13:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The date Nelson Mandela died&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Room 21:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The date of the fall of the Berlin Wall&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Room 23:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The date Gavrilo Princip assasinated Franz Ferdinand&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Room 39:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The date Stalingrad battle ended, the day after general Paulus surrendered to Joseph Stalin&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Room 51:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The date Martin Luther King jr. held his famous speech &quot;I have a dream&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Room 62:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The birth date of Zlatan Ibrahimovic&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Room 65:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The date of the second peace(Pierce) of world war in Europe&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Room 82:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The date of independence date for Vietnam&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Room 88:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The date of founding Microsoft&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This would conclude leg one.&#xA;For leg two:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hints 3 and 5 combined with the limitation on room numbers given by floors and rooms pr. floor, would reveal the pattern.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9385" LastActivityDate="2015-02-12T20:55:23.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8876" PostTypeId="2" ParentId="8713" CreationDate="2015-02-12T20:56:56.930" Score="8" Body="&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When?  Then!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Don't lose.  Do the opposite.  That's &quot;win&quot;.  Something that sounds like &quot;win&quot; but has no 'I's (eyes) is &quot;when&quot;.  That's the question.  Remove the 'W's (doubles) and add a 'T' (tea), and you get &quot;then&quot;.  That's the answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2162" LastActivityDate="2015-02-12T20:56:56.930" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8877" PostTypeId="2" ParentId="8831" CreationDate="2015-02-12T21:14:29.803" Score="2" Body="&lt;p&gt;[&lt;strong&gt;TL/DR&lt;/strong&gt; Answer is 11]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering the optimum cost:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For, the base case, a set of magnitude $1$ then the cost is $C_1 =2$&lt;/li&gt;&#xA;&lt;li&gt;For a set $X$ (of size $x$) then this can be split into sub-sets $Y$ (of size $y$) and $X\backslash Y$ then&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$X\backslash Y$ has size $z=x-y$;&lt;/li&gt;&#xA;&lt;li&gt;The probability $P_{y}$ of the answer being in $Y$ is $\frac{y}{x}$;&lt;/li&gt;&#xA;&lt;li&gt;The probability $P_{z}$ of the answer being in $X\backslash Y$ is $\frac{x-y}{x}$;&lt;/li&gt;&#xA;&lt;li&gt;The cost $C_{(x,y)}$ for this combination of set and sub-set sizes is given by: $\begin{split} C_{(x,y)} &amp;amp; = \begin{cases} P_{y} \cdot 2 + P_{z} \cdot (1+C_{z}) &amp;amp; \text{if } y=1\\  P_{y} \cdot (2+C_{y}) + P_{z} \cdot (1+C_{z}) &amp;amp; \text{if } y&amp;gt;1\end{cases} \\ &amp;amp;= \begin{cases} \frac{2 + (x-1) \cdot (1+C_{z})}{x} &amp;amp; \text{if } y=1\\  \frac{y \cdot (2+C_{y}) + (x-y) \cdot (1+C_{z})}{x} &amp;amp; \text{if } y&amp;gt;1\end{cases} \end{split}$&lt;/li&gt;&#xA;&lt;li&gt;The optimum cost $C_x = \text{min} \left( \forall_{y=1 \ldots \lfloor \frac{x}{2} \rfloor } C_{(x,y)} \right)$ &lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This can then be used iteratively to generate costs for successively increasing sized sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following Java code does this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import java.math.BigDecimal;&#xA;import java.math.RoundingMode;&#xA;import java.util.ArrayList;&#xA;&#xA;public class Coins {&#xA;&#xA;    static final int SIZE = 200;&#xA;    static final BigDecimal TWO = new BigDecimal( &quot;2&quot; );&#xA;&#xA;    public static void main(String[] args) {&#xA;        final ArrayList&amp;lt;BigDecimal&amp;gt; costs = new ArrayList&amp;lt;BigDecimal&amp;gt;( SIZE + 1 );&#xA;        final ArrayList&amp;lt;Integer&amp;gt; values   = new ArrayList&amp;lt;Integer&amp;gt;();&#xA;&#xA;        costs.add( BigDecimal.ZERO );&#xA;        costs.add( TWO );&#xA;&#xA;        for ( int size = 2; size &amp;lt;= SIZE; ++size )&#xA;        {&#xA;            costs.add( new BigDecimal( SIZE ) );&#xA;            values.clear();&#xA;            BigDecimal setSize = new BigDecimal( size );&#xA;            for ( int firstSubSetSize = 1; firstSubSetSize &amp;lt;= size / 2; ++firstSubSetSize )&#xA;            {&#xA;                int secondSubSetSize = size - firstSubSetSize;&#xA;                BigDecimal setSize1 = new BigDecimal( firstSubSetSize );&#xA;                BigDecimal setSize2 = new BigDecimal( secondSubSetSize );&#xA;                BigDecimal cost1 = firstSubSetSize == 1 ? TWO : TWO.add( costs.get( firstSubSetSize ) );&#xA;                BigDecimal cost2 = BigDecimal.ONE.add( costs.get( secondSubSetSize ) );&#xA;                BigDecimal cost = cost1.multiply( setSize1 ).add( cost2.multiply( setSize2 ) ).setScale( 50, RoundingMode.HALF_UP ).divide( setSize, RoundingMode.HALF_UP );&#xA;                if ( costs.get( size ) == null )&#xA;                {&#xA;                    costs.set( size, cost );&#xA;                    values.add( firstSubSetSize );&#xA;                }&#xA;                else&#xA;                {&#xA;                    switch( cost.compareTo( costs.get( size ) ) )&#xA;                    {&#xA;                    case -1: values.clear();&#xA;                    case 0:  values.add( firstSubSetSize );&#xA;                             costs.set( size, cost );&#xA;                             break;&#xA;                    default: break;&#xA;                    }&#xA;                }&#xA;            }&#xA;            System.out.println( size + &quot; : &quot; + costs.get( size ).toPlainString() + &quot; : &quot; + values.toString() );&#xA;        }&#xA;    }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And gives the output (first column is the starting set size, second column is the optimum cost and third column is the list of sub-set sizes which Bob can pick that will give that optimum cost):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2 : 2.50000000000000000000000000000000000000000000000000 : [1]&#xA;3 : 3.00000000000000000000000000000000000000000000000000 : [1]&#xA;4 : 3.50000000000000000000000000000000000000000000000000 : [1]&#xA;5 : 4.00000000000000000000000000000000000000000000000000 : [1]&#xA;6 : 4.50000000000000000000000000000000000000000000000000 : [1, 2, 3]&#xA;7 : 4.71428571428571428571428571428571428571428571428571 : [3]&#xA;8 : 5.00000000000000000000000000000000000000000000000000 : [3, 4]&#xA;9 : 5.22222222222222222222222222222222222222222222222222 : [4]&#xA;10 : 5.50000000000000000000000000000000000000000000000000 : [3, 4, 5]&#xA;11 : 5.63636363636363636363636363636363636363636363636363 : [4]&#xA;12 : 5.83333333333333333333333333333333333333333333333333 : [4, 5]&#xA;13 : 6.00000000000000000000000000000000000000000000000000 : [4, 5]&#xA;14 : 6.14285714285714285714285714285714285714285714285714 : [5]&#xA;15 : 6.33333333333333333333333333333333333333333333333333 : [4, 5, 6, 7]&#xA;16 : 6.43750000000000000000000000000000000000000000000000 : [5, 7]&#xA;17 : 6.58823529411764705882352941176470588235294117647058 : [6]&#xA;18 : 6.66666666666666666666666666666666666666666666666666 : [7]&#xA;19 : 6.78947368421052631578947368421052631578947368421052 : [7, 8]&#xA;20 : 6.90000000000000000000000000000000000000000000000000 : [7, 8, 9]&#xA;21 : 7.00000000000000000000000000000000000000000000000000 : [7, 8, 9]&#xA;22 : 7.09090909090909090909090909090909090909090909090909 : [8, 9]&#xA;23 : 7.17391304347826086956521739130434782608695652173913 : [9]&#xA;24 : 7.29166666666666666666666666666666666666666666666666 : [9, 11]&#xA;25 : 7.36000000000000000000000000000000000000000000000000 : [9, 11]&#xA;26 : 7.46153846153846153846153846153846153846153846153846 : [8, 9, 10, 11, 12]&#xA;27 : 7.51851851851851851851851851851851851851851851851851 : [9]&#xA;28 : 7.60714285714285714285714285714285714285714285714285 : [9, 10, 11]&#xA;29 : 7.65517241379310344827586206896551724137931034482758 : [11]&#xA;30 : 7.73333333333333333333333333333333333333333333333333 : [11, 12]&#xA;31 : 7.80645161290322580645161290322580645161290322580645 : [11, 12, 13]&#xA;32 : 7.87500000000000000000000000000000000000000000000000 : [11, 12, 13, 14]&#xA;33 : 7.93939393939393939393939393939393939393939393939393 : [14]&#xA;34 : 8.00000000000000000000000000000000000000000000000000 : [11, 12, 13, 14]&#xA;35 : 8.05714285714285714285714285714285714285714285714286 : [12, 13, 14]&#xA;36 : 8.11111111111111111111111111111111111111111111111111 : [13, 14]&#xA;37 : 8.16216216216216216216216216216216216216216216216216 : [14]&#xA;38 : 8.23684210526315789473684210526315789473684210526315 : [14]&#xA;39 : 8.28205128205128205128205128205128205128205128205128 : [14, 16]&#xA;40 : 8.34999999999999999999999999999999999999999999999999 : [13]&#xA;41 : 8.39024390243902439024390243902439024390243902439024 : [14, 16, 18]&#xA;42 : 8.45238095238095238095238095238095238095238095238095 : [13, 14, 15, 16, 17, 18, 19]&#xA;43 : 8.48837209302325581395348837209302325581395348837209 : [14, 16, 18]&#xA;44 : 8.54545454545454545454545454545454545454545454545454 : [14, 15, 16, 17, 18, 19]&#xA;45 : 8.57777777777777777777777777777777777777777777777777 : [16, 18]&#xA;46 : 8.63043478260869565217391304347826086956521739130434 : [17, 18, 19]&#xA;47 : 8.65957446808510638297872340425531914893617021276595 : [18]&#xA;48 : 8.70833333333333333333333333333333333333333333333333 : [18, 19]&#xA;49 : 8.75510204081632653061224489795918367346938775510204 : [18, 19, 20]&#xA;50 : 8.80000000000000000000000000000000000000000000000000 : [18, 19, 20, 21]&#xA;51 : 8.84313725490196078431372549019607843137254901960783 : [18]&#xA;52 : 8.88461538461538461538461538461538461538461538461538 : [18, 19, 20, 21, 22, 23]&#xA;53 : 8.92452830188679245283018867924528301886792452830188 : [19, 20, 23]&#xA;54 : 8.96296296296296296296296296296296296296296296296296 : [18, 19, 20, 21, 22, 23]&#xA;55 : 8.99999999999999999999999999999999999999999999999999 : [22]&#xA;56 : 9.03571428571428571428571428571428571428571428571428 : [19, 23]&#xA;57 : 9.07017543859649122807017543859649122807017543859649 : [20, 21, 22, 23]&#xA;58 : 9.10344827586206896551724137931034482758620689655172 : [21, 22]&#xA;59 : 9.13559322033898305084745762711864406779661016949152 : [22, 23]&#xA;60 : 9.16666666666666666666666666666666666666666666666667 : [23]&#xA;61 : 9.21311475409836065573770491803278688524590163934426 : [22, 23, 24, 25]&#xA;62 : 9.24193548387096774193548387096774193548387096774193 : [23, 25]&#xA;63 : 9.28571428571428571428571428571428571428571428571428 : [22, 23, 24, 25, 26, 27]&#xA;64 : 9.31250000000000000000000000000000000000000000000000 : [23, 25, 27]&#xA;65 : 9.35384615384615384615384615384615384615384615384615 : [22, 23, 24, 25, 26, 27, 28, 29]&#xA;66 : 9.37878787878787878787878787878787878787878787878787 : [27, 29]&#xA;67 : 9.41791044776119402985074626865671641791044776119402 : [22, 27, 29]&#xA;68 : 9.44117647058823529411764705882352941176470588235294 : [23, 25, 27, 29]&#xA;69 : 9.47826086956521739130434782608695652173913043478260 : [22, 23, 24, 27, 28, 29]&#xA;70 : 9.49999999999999999999999999999999999999999999999999 : [23, 27]&#xA;71 : 9.53521126760563380281690140845070422535211267605633 : [24, 25, 26, 27, 28, 29, 30]&#xA;72 : 9.55555555555555555555555555555555555555555555555555 : [25, 27, 29]&#xA;73 : 9.58904109589041095890410958904109589041095890410958 : [26, 27, 28, 29]&#xA;74 : 9.60810810810810810810810810810810810810810810810810 : [27, 29]&#xA;75 : 9.63999999999999999999999999999999999999999999999999 : [27, 28, 29, 30]&#xA;76 : 9.65789473684210526315789473684210526315789473684210 : [29]&#xA;77 : 9.68831168831168831168831168831168831168831168831168 : [29, 30]&#xA;78 : 9.71794871794871794871794871794871794871794871794871 : [30, 31]&#xA;79 : 9.74683544303797468354430379746835443037974683544303 : [32]&#xA;80 : 9.77499999999999999999999999999999999999999999999999 : [29, 33]&#xA;81 : 9.80246913580246913580246913580246913580246913580246 : [29, 30, 33, 34]&#xA;82 : 9.82926829268292682926829268292682926829268292682926 : [29, 30, 31, 33]&#xA;83 : 9.85542168674698795180722891566265060240963855421686 : [29, 30, 31, 32, 33, 36]&#xA;84 : 9.88095238095238095238095238095238095238095238095237 : [29, 33]&#xA;85 : 9.90588235294117647058823529411764705882352941176470 : [29, 30, 31, 32, 33, 34, 36, 37]&#xA;86 : 9.93023255813953488372093023255813953488372093023255 : [30, 31, 33, 35]&#xA;87 : 9.95402298850574712643678160919540229885057471264367 : [29, 31, 32, 33, 34, 36]&#xA;88 : 9.97727272727272727272727272727272727272727272727272 : [29, 30, 32, 33, 35, 36, 37]&#xA;89 : 9.99999999999999999999999999999999999999999999999999 : [33, 34]&#xA;90 : 10.02222222222222222222222222222222222222222222222222 : [30, 31, 32, 33, 34, 35, 36, 37]&#xA;91 : 10.04395604395604395604395604395604395604395604395604 : [32, 33, 34, 35, 36, 37]&#xA;92 : 10.06521739130434782608695652173913043478260869565217 : [33, 34, 35, 36, 37]&#xA;93 : 10.08602150537634408602150537634408602150537634408602 : [33, 34, 35, 36, 37]&#xA;94 : 10.10638297872340425531914893617021276595744680851064 : [34, 35, 36, 37]&#xA;95 : 10.12631578947368421052631578947368421052631578947368 : [36, 37]&#xA;96 : 10.14583333333333333333333333333333333333333333333333 : [37]&#xA;97 : 10.16494845360824742268041237113402061855670103092784 : [37]&#xA;98 : 10.19387755102040816326530612244897959183673469387755 : [36, 37, 38, 39]&#xA;99 : 10.21212121212121212121212121212121212121212121212121 : [37, 39]&#xA;100 : 10.23999999999999999999999999999999999999999999999999 : [38]&#xA;101 : 10.25742574257425742574257425742574257425742574257425 : [39]&#xA;102 : 10.28431372549019607843137254901960784313725490196078 : [36, 37, 38, 39, 40, 41, 43]&#xA;103 : 10.30097087378640776699029126213592233009708737864077 : [37, 41]&#xA;104 : 10.32692307692307692307692307692307692307692307692307 : [37, 38, 39, 40, 41, 42, 43, 45]&#xA;105 : 10.34285714285714285714285714285714285714285714285714 : [37, 39, 41, 43, 45]&#xA;106 : 10.36792452830188679245283018867924528301886792452829 : [36, 39, 40]&#xA;107 : 10.38317757009345794392523364485981308411214953271027 : [37, 41, 45]&#xA;108 : 10.40740740740740740740740740740740740740740740740740 : [36, 37, 38, 39, 40, 41, 42, 43, 45, 46, 47]&#xA;109 : 10.42201834862385321100917431192660550458715596330275 : [37, 39, 41, 43, 45, 47]&#xA;110 : 10.44545454545454545454545454545454545454545454545454 : [36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48]&#xA;111 : 10.45945945945945945945945945945945945945945945945945 : [37, 41, 45]&#xA;112 : 10.48214285714285714285714285714285714285714285714285 : [36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47]&#xA;113 : 10.49557522123893805309734513274336283185840707964601 : [37, 39, 41, 43, 45, 47]&#xA;114 : 10.51754385964912280701754385964912280701754385964911 : [40, 44, 45, 47]&#xA;115 : 10.53043478260869565217391304347826086956521739130434 : [39, 41, 43, 45, 47]&#xA;116 : 10.55172413793103448275862068965517241379310344827585 : [41, 45, 46, 47]&#xA;117 : 10.56410256410256410256410256410256410256410256410256 : [41, 43, 45, 47]&#xA;118 : 10.58474576271186440677966101694915254237288135593219 : [45]&#xA;119 : 10.59663865546218487394957983193277310924369747899159 : [43, 45, 47]&#xA;120 : 10.61666666666666666666666666666666666666666666666666 : [43, 44, 45, 46, 47, 48]&#xA;121 : 10.62809917355371900826446280991735537190082644628098 : [47]&#xA;122 : 10.64754098360655737704918032786885245901639344262294 : [45, 46, 47, 48]&#xA;123 : 10.65853658536585365853658536585365853658536585365853 : [47]&#xA;124 : 10.67741935483870967741935483870967741935483870967741 : [47, 48]&#xA;125 : 10.69599999999999999999999999999999999999999999999999 : [47, 48]&#xA;126 : 10.71428571428571428571428571428571428571428571428571 : [47, 48, 49, 50]&#xA;127 : 10.73228346456692913385826771653543307086614173228346 : [47, 48, 49, 50, 51]&#xA;128 : 10.74999999999999999999999999999999999999999999999999 : [47, 48, 49, 51]&#xA;129 : 10.76744186046511627906976744186046511627906976744185 : [47, 48, 49, 51, 52, 53]&#xA;130 : 10.78461538461538461538461538461538461538461538461537 : [51]&#xA;131 : 10.80152671755725190839694656488549618320610687022900 : [47, 48, 49, 50, 51, 52, 53, 54, 55]&#xA;132 : 10.81818181818181818181818181818181818181818181818181 : [47, 48, 49, 50, 51, 52, 53, 54, 55, 56]&#xA;133 : 10.83458646616541353383458646616541353383458646616540 : [51, 53, 55]&#xA;134 : 10.85074626865671641791044776119402985074626865671641 : [47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58]&#xA;135 : 10.86666666666666666666666666666666666666666666666665 : [51]&#xA;136 : 10.88235294117647058823529411764705882352941176470587 : [47, 51, 52, 55, 56]&#xA;137 : 10.89781021897810218978102189781021897810218978102189 : [47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60]&#xA;138 : 10.91304347826086956521739130434782608695652173913042 : [51]&#xA;139 : 10.92805755395683453237410071942446043165467625899280 : [47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60]&#xA;140 : 10.94285714285714285714285714285714285714285714285713 : [51, 56]&#xA;141 : 10.95744680851063829787234042553191489361702127659574 : [47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60]&#xA;142 : 10.97183098591549295774647887323943661971830985915492 : [47, 51, 53, 54, 55, 56, 58, 59]&#xA;143 : 10.98601398601398601398601398601398601398601398601398 : [47, 48, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60]&#xA;144 : 10.99999999999999999999999999999999999999999999999999 : [51, 55, 56, 57, 58, 59]&#xA;145 : 11.01379310344827586206896551724137931034482758620689 : [49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60]&#xA;146 : 11.02739726027397260273972602739726027397260273972602 : [51, 53, 54, 55, 56, 57, 58, 59, 60]&#xA;147 : 11.04081632653061224489795918367346938775510204081632 : [51, 52, 54, 55, 56, 57, 58, 59, 60]&#xA;148 : 11.05405405405405405405405405405405405405405405405405 : [51, 52, 53, 54, 55, 56, 57, 58, 59, 60]&#xA;149 : 11.06711409395973154362416107382550335570469798657718 : [52, 53, 54, 55, 56, 57, 58, 59, 60]&#xA;150 : 11.07999999999999999999999999999999999999999999999999 : [55]&#xA;151 : 11.09271523178807947019867549668874172185430463576158 : [55, 56, 59]&#xA;152 : 11.10526315789473684210526315789473684210526315789473 : [56, 57, 59]&#xA;153 : 11.11764705882352941176470588235294117647058823529411 : [58]&#xA;154 : 11.12987012987012987012987012987012987012987012987013 : [57, 58, 59, 60]&#xA;155 : 11.14193548387096774193548387096774193548387096774193 : [59, 60]&#xA;156 : 11.15384615384615384615384615384615384615384615384615 : [59, 60]&#xA;157 : 11.16560509554140127388535031847133757961783439490446 : [60]&#xA;158 : 11.18354430379746835443037974683544303797468354430379 : [59, 62]&#xA;159 : 11.19496855345911949685534591194968553459119496855346 : [60, 62]&#xA;160 : 11.21249999999999999999999999999999999999999999999999 : [59]&#xA;161 : 11.22360248447204968944099378881987577639751552795031 : [60, 62, 64]&#xA;162 : 11.24074074074074074074074074074074074074074074074073 : [59, 62]&#xA;163 : 11.25153374233128834355828220858895705521472392638036 : [62]&#xA;164 : 11.26829268292682926829268292682926829268292682926829 : [59, 60, 61, 62, 63, 64, 65, 66, 67, 68]&#xA;165 : 11.27878787878787878787878787878787878787878787878787 : [62, 64, 66]&#xA;166 : 11.29518072289156626506024096385542168674698795180722 : [59, 60, 62, 63, 65, 66, 67, 70]&#xA;167 : 11.30538922155688622754491017964071856287425149700598 : [60, 62, 64, 66]&#xA;168 : 11.32142857142857142857142857142857142857142857142856 : [61, 62, 67]&#xA;169 : 11.33136094674556213017751479289940828402366863905325 : [60, 62, 64, 66, 68, 70, 72]&#xA;170 : 11.34705882352941176470588235294117647058823529411764 : [59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 74]&#xA;171 : 11.35672514619883040935672514619883040935672514619882 : [64, 70]&#xA;172 : 11.37209302325581395348837209302325581395348837209301 : [66]&#xA;173 : 11.38150289017341040462427745664739884393063583815028 : [62, 66, 70, 72, 74]&#xA;174 : 11.39655172413793103448275862068965517241379310344827 : [59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77]&#xA;175 : 11.40571428571428571428571428571428571428571428571428 : [60, 62, 64, 66, 68, 70, 72, 74, 76]&#xA;176 : 11.42045454545454545454545454545454545454545454545453 : [70]&#xA;177 : 11.42937853107344632768361581920903954802259887005649 : [62, 64, 66, 70, 72, 74, 76]&#xA;178 : 11.44382022471910112359550561797752808988764044943819 : [60, 62, 64, 66, 67, 70, 71, 72, 74, 75, 77]&#xA;179 : 11.45251396648044692737430167597765363128491620111731 : [62, 64, 66, 68, 70, 72, 74, 76]&#xA;180 : 11.46666666666666666666666666666666666666666666666666 : [59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77]&#xA;181 : 11.47513812154696132596685082872928176795580110497237 : [60, 62, 64, 66, 68, 70, 72, 74, 76]&#xA;182 : 11.48901098901098901098901098901098901098901098901098 : [59, 60, 61, 62, 63, 64, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77]&#xA;183 : 11.49726775956284153005464480874316939890710382513660 : [62, 70, 72, 76]&#xA;184 : 11.51086956521739130434782608695652173913043478260868 : [66, 70]&#xA;185 : 11.51891891891891891891891891891891891891891891891891 : [62, 64, 66, 70, 72, 74]&#xA;186 : 11.53225806451612903225806451612903225806451612903225 : [62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77]&#xA;187 : 11.54010695187165775401069518716577540106951871657753 : [66, 70, 72, 74, 76]&#xA;188 : 11.55319148936170212765957446808510638297872340425531 : [64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77]&#xA;189 : 11.56084656084656084656084656084656084656084656084655 : [66, 68, 70, 74, 76]&#xA;190 : 11.57368421052631578947368421052631578947368421052631 : [66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77]&#xA;191 : 11.58115183246073298429319371727748691099476439790575 : [68, 70, 72, 74, 76]&#xA;192 : 11.59374999999999999999999999999999999999999999999999 : [68, 69, 70, 71, 72, 73, 74, 75, 76, 77]&#xA;193 : 11.60103626943005181347150259067357512953367875647667 : [72]&#xA;194 : 11.61340206185567010309278350515463917525773195876288 : [70, 71, 72, 73, 74, 75, 76, 77]&#xA;195 : 11.62051282051282051282051282051282051282051282051281 : [72, 74, 76]&#xA;196 : 11.63265306122448979591836734693877551020408163265305 : [72, 73, 74, 75, 77]&#xA;197 : 11.63959390862944162436548223350253807106598984771573 : [74, 76]&#xA;198 : 11.65151515151515151515151515151515151515151515151514 : [74, 75, 76, 77]&#xA;199 : 11.65829145728643216080402010050251256281407035175879 : [76]&#xA;200 : 11.66999999999999999999999999999999999999999999999999 : [76, 77]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As a side note: If you plot the set size against the (multiple) optimal sub-set sizes then you get an interesting pattern.&lt;/p&gt;&#xA;" OwnerUserId="5434" LastEditorUserId="5434" LastEditDate="2015-02-13T16:26:09.260" LastActivityDate="2015-02-13T16:26:09.260" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8878" PostTypeId="2" ParentId="8765" CreationDate="2015-02-12T21:21:36.667" Score="3" Body="&lt;p&gt;If you put the dates listed above into excel and then find their =VALUE all of their last two digits correspond to their room numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;CBJ's best known role is in Jerry McGuire, released 12/13/1996. That VALUE is 35412.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He is in room 12.&lt;/p&gt;&#xA;" OwnerUserId="9425" LastActivityDate="2015-02-12T21:21:36.667" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8879" PostTypeId="2" ParentId="8813" CreationDate="2015-02-12T21:29:51.713" Score="6" Body="&lt;p&gt;This, I believe, is a less far-fetched scenario than the intended answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Foo and Bar were adopted as babies by different families.  They were born orphans.  The orphanage was worried nobody would want to adopt two children at a time, or that it would bring less money, so they falsified one birth certificate and changed the date.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2136" LastActivityDate="2015-02-12T21:29:51.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8880" PostTypeId="2" ParentId="8787" CreationDate="2015-02-12T21:35:33.730" Score="1" Body="&lt;p&gt;Any symmetrical puzzle will be mirrored over the axis of symmetry. It behaves that way for any symmetrical Nonogram, I would expect, whether it's vertical symmetry, horizontal, or diagonal. I'm not sure about there being any extra tactics toward solving it that couldn't be applied to an asymmetrical puzzle, other than the ones you mentioned in the original post. The way I see it, if the tactic of mirroring your actions over the axis of symmetry doesn't work, then it's either not a symmetrical Nonogram or it's an invalid puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the solution to your 10x10, and as you can see, it is indeed symmetrical over the diagonal line in your description. If you'd like me to explain in detail how I solved it, I can, but only when I have time after classes today:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/f4xO3.png&quot; alt=&quot;Nonogram&quot;&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9319" LastActivityDate="2015-02-12T21:35:33.730" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8881" PostTypeId="2" ParentId="8813" CreationDate="2015-02-12T21:39:22.720" Score="6" Body="&lt;p&gt;Bar may have been older than Foo at the time of birth, but then became an astronaut and has been participating in tests of new propulsion technologies, while Foo had a regular job (which did not involve that kind of acceleration Bar was frequently experiencing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now due to effects of time dilation Foo is actually 5 minutes older than Bar, despite celebrating his birthday after Bar's  &lt;/p&gt;&#xA;" OwnerUserId="9485" LastActivityDate="2015-02-12T21:39:22.720" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8884" PostTypeId="2" ParentId="546" CreationDate="2015-02-12T23:46:17.437" Score="8" Body="&lt;p&gt;A simple strategy which is good and computationally much faster than Knuth's is the following (I have programmed both)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Create the list 1111,...,6666 of all candidate secret codes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with 1122.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeat the following 2 steps:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) After you got the answer (number of red and number of white pegs) eliminate from the list of candidates all codes that would not have produced the same answer if they were the secret code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Pick the first element in the list and use it as new guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This averages no more than 5 guesses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the Swaszek (1999-2000) strategy that was mentioned in another answer.&lt;/p&gt;&#xA;" OwnerUserId="9488" LastEditorUserId="34618" LastEditDate="2017-02-22T20:55:21.277" LastActivityDate="2017-02-22T20:55:21.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8886" PostTypeId="1" AcceptedAnswerId="8896" CreationDate="2015-02-13T07:44:42.330" Score="7" ViewCount="513" Body="&lt;p&gt;This is a linking puzzle, where you get 8 hints, and 7 links between them.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Michael Jackson made a song for this film.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;1.-2. The link between the animal of clue 1 and the fictional person of clue 2 is an animal with long teeth.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;This is a fictional person, and the same in both links.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;2.-3. This is the best couple of friends you ever knew.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;This is a fictional person in the foregoing link, but a real person in the next link.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;3.-4. They used to be a super-couple.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;A real person in both link, but he could also be associated with the name of a certain part of the heel.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;4.-5. The new super-couple. Lots of kids.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;Change a vowel in one of the names from the last link, to identify the fictional person in the next link.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;5.-6. Worlds most famous lovers.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;6&quot;&gt;&#xA;&lt;li&gt;This person has portrayed both the fictional person in the last link, and the next link. He has the name of a turtle.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;6.-7. The ship is sinking.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;7&quot;&gt;&#xA;&lt;li&gt;A fictional person in last link, a real in the next one.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;7.-8. A royal couple.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;8&quot;&gt;&#xA;&lt;li&gt;A person from Switzerland.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The two overflowing words from 1. and 8. could be combined to be associated with a principle important to many countries in the world. This has recently(?) been a part of news coverage around the world. What is this principle called?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9385" LastEditorUserId="9385" LastEditDate="2015-02-13T11:42:48.990" LastActivityDate="2015-02-13T12:35:47.630" Title="Link the hints to solve this one, starting with Michael Jackson" Tags="|logical-deduction|connections-puzzle|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8887" PostTypeId="2" ParentId="8813" CreationDate="2015-02-13T08:01:54.263" Score="8" Body="&lt;p&gt;It's simple. It's not about strange time-zone scenarios, however they may be &lt;em&gt;possible&lt;/em&gt;. The interviewer probably wanted to test if you can think outside of the 'nerd' box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They were simply born at almost the same time. To not have to celebrate the birthday on the same day, which may not be much fun for either of the kids, the parents decided to celebrate the birthdays on separate days. They didn't care about the 5 minutes, or maybe didn't even want to tell the kids which one was first, to treat them equally. In the end, this became a tradition they kept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That, is by far a more likely scenario than any weird 'traveling with enormous speed' scenario.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the whole point of interviews. You have to discover on what modes and levels people can think.&lt;/p&gt;&#xA;" OwnerUserId="9491" LastActivityDate="2015-02-13T08:01:54.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8888" PostTypeId="2" ParentId="8813" CreationDate="2015-02-13T08:08:31.113" Score="5" Body="&lt;p&gt;One Solution could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The woman is native of the Yoruba nigerian tribe which belives the older twin is the second born. Then the oldest twin by 5 five minutes is the second born by 5 minutes. Hence Foo was simply born at 12:03 AM and Bar was born at 11:58 PM.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Based on&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first born twin, whether a boy or a girl, is always called Taiwo, meaning `having the first taste of the world’, whereas the second is named Kehinde, meaning ‘arriving after the other’. Although being born first Taiwo is considered as the younger twin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Additionally&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This belief is held in many african tribes, and I did hear it once as a tradition in France although it is not the case officially (this is how I got this idea). I'm pretty sure we can find other culture that hold that belief?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;here is some links&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.dadsguidetotwins.com/nigerian-twins-legend/&quot;&gt;Yoruba&lt;/a&gt;, &lt;a href=&quot;http://www.kouya.net/?p=388&quot;&gt;Northern Ivory Coast&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is kind of far-fetched, probably not what interviewer would be looking for but this work :).&lt;/p&gt;&#xA;" OwnerUserId="9490" LastActivityDate="2015-02-13T08:08:31.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8889" PostTypeId="2" ParentId="8831" CreationDate="2015-02-13T08:40:00.377" Score="2" Body="&lt;p&gt;This is a partial answer showing the intuitive fact, which is used in other answers but not mentioned explicitly, that there is a Nash equilibrium where Alice's strategy is the uniform distribution. Thus the question reduces to finding Bob's optimal strategy against the uniform distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If distribution $X$ is Alice's strategy in a Nash equilibrium with payoff $a$ to Alice, then, because the game is the same if the numbers are permuted, so are all distributions $X_1,\dots,X_{144!}$ obtained from $X$ by all permutations of the numbers. Let Alice choose a strategy uniformly from those strategies, giving a uniform distribution. Let Bob's optimal strategy against that be $S$. Now Alice's payoff is &#xA;$$&#xA;\sum_{i=1}^{144!} \frac{1}{144!} f(X_i,S),&#xA;$$&#xA;where $f(X,Y)$ is Alice's payoff when the players have strategies $X$ and $Y$.&#xA;Alice's payoff can't be worse than&#xA;$$&#xA;\min_i f(X_i,S),&#xA;$$ &#xA;that is, the payoff she would get by choosing the worst possible of the strategies $X_i$ to counter $S$. But because each $X_i$ is a strategy in a Nash equilibrium with payoff $a$, we know that $f(X_i,S)\geq a$ no matter what $S$ is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, Alice gets at least payoff $a$ if she uses the uniform distribution. Since this is at least as large as the payoff in a Nash equilibrium, the uniform distribution has to be one of the optimal strategies.&lt;/p&gt;&#xA;" OwnerUserId="2228" LastActivityDate="2015-02-13T08:40:00.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8890" PostTypeId="1" AcceptedAnswerId="8891" CreationDate="2015-02-13T08:58:54.780" Score="11" ViewCount="6442" Body="&lt;p&gt;There are $60$ different 3-letter combinations of A,B,C,D,E. Using each letter to represent a different integer from $1$ to $9$, what is the solution for &lt;strong&gt;ABCDE&lt;/strong&gt; to equal the sum of all 3-letter combinations (ABC + ABD + ABE + BAC + BAD + . . .)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As well as the correct solution, the best answer should include simplified expression for this problem.  &lt;/p&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="18514" LastEditDate="2016-12-31T12:07:22.753" LastActivityDate="2016-12-31T12:07:22.753" Title="ABCDE = the sum of all of its 3-letter permutations (ABC + ABD + ABE + BAC + BAD + ...)" Tags="|mathematics|alphametic|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8891" PostTypeId="2" ParentId="8890" CreationDate="2015-02-13T09:33:20.610" Score="8" Body="&lt;p&gt;First let's look at what all the 60 combinations look like.  Consider the first digit - obviously, we have 5 choices for which letter to place there.  Since there are a total of 60 combinations, each letter must appear there a total of 12 times.  Therefore, we can add up everything contributing to the sum from just that place alone as 1200*(A+B+C+D+E).  This argument can be applied to each digit separately, so we can evaluate the entire sum as 1332*(A+B+C+D+E) since 12+120+1200=1332.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the constraint that each letter must represent a distinct number means that the sum A+B+C+D+E must fall somewhere in the range of 15 (1+2+3+4+5) to 35 (9+8+7+6+5), inclusive.  That's actually a small enough range to check by hand, in which case you get&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 35964&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;as the unique answer for this problem.&lt;/p&gt;&#xA;" OwnerUserId="4819" LastActivityDate="2015-02-13T09:33:20.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8892" PostTypeId="2" ParentId="8886" CreationDate="2015-02-13T10:23:53.707" Score="6" Body="&lt;p&gt;These are the pieces I figured out.&lt;br&gt;&#xA;Edit: New ideas for links 1-3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1   : &lt;em&gt;Free Willy&lt;/em&gt;&lt;br&gt;&#xA; 1-2 : Movie is about an orca/whale, which has surprisingly long teeth and is rather fatty.&lt;br&gt;&#xA; 2   : Monica Geller (&lt;em&gt;Friends&lt;/em&gt; wiki says she was considered overweight in high school. compared to a whale... a fat joke?)&lt;br&gt;&#xA; 2-3 : Monica Geller and Rachel Green from &lt;em&gt;Friends&lt;/em&gt;&lt;br&gt;&#xA; 3   : Rachel Green played by Jennifer Aniston&lt;br&gt;&#xA; 3-4 : Brad Pitt and Jennifer Aniston&lt;br&gt;&#xA; 4   : Brad Pitt plays Achilles in the movie &lt;em&gt;Troy&lt;/em&gt;.&lt;br&gt;&#xA; 4-5 : Brad Pitt and Angelina Jolie&lt;br&gt;&#xA; 5   : Jolie -&gt; Julie (?)&lt;br&gt;&#xA; 5-6 : &lt;em&gt;Romeo and Juliet&lt;/em&gt;&lt;br&gt;&#xA; 6   : Leonardo Dicaprio portrayed main characters in both films (Leonardo is the name of a Ninja Turtle)&lt;br&gt;&#xA; 6-7 : &lt;em&gt;Titanic&lt;/em&gt;&lt;br&gt;&#xA; 7   : Kate (Winslet is the actress in Titanic)&lt;br&gt;&#xA; 7-8 : Kate Middleton and Prince William&lt;br&gt;&#xA; 8   : William Tell (folk tale legend from Switzerland)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Final answer is &lt;strong&gt;Free Tell&lt;/strong&gt;, also known as Freedom of Speech.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9319" LastEditorUserId="9319" LastEditDate="2015-02-13T11:36:08.157" LastActivityDate="2015-02-13T11:36:08.157" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8893" PostTypeId="1" AcceptedAnswerId="8901" CreationDate="2015-02-13T10:25:46.470" Score="6" ViewCount="345" Body="&lt;p&gt;The new board of the Fun-and-Nonsense-Club is to be elected. &#xA;The voting procedure itself is fairly simple: &#xA;The club has 24 members, and every pair of (distinct) members announces to the public whether they are close friends or not. &#xA;Then every member receives one vote for every close friend, and the composition of the board depends on these numbers of votes in a horribly complicated fashion; in our puzzle this luckily is of no concern to us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the week before the election, the club hired Professor Halfbrain to carry out an opinion poll.&#xA;Halfbrain briefly spoke to every pair of club members and asked them whether they are close friends or not. &#xA;Then he summarized his data and announced his prediction on the number of votes that each club member is going to receive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Scenario 1:&lt;/strong&gt; During the poll, all club members answered truthfully to the professor's questions. &#xA;But once they hear Halfbrain's prediction, they change their mind and decide to have their fun with the professor. &#xA;They plot up and at the actual election &lt;em&gt;not a single one&lt;/em&gt; of Professor Halfbrain's 24 predicted numbers turns out to be correct. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question 1:&lt;/strong&gt; &#xA;  Was it just a lucky coincidence that the original (truthful) data allowed the club members to shatter Halfbrain's predictions?&#xA;  Or is there always a way of doing this?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Scenario 2:&lt;/strong&gt; &#xA;At the end of his interviewing work with the club members, Professor Halfbrain realized that he had accidentally erased all the collected data.&#xA;As he could not remember a single number and as he was unwilling to admit his mistake, he decided to simply use 24 random numbers (from the range $0,1,\ldots,23$) in his prediction.&#xA;Once the club members hear Halfbrain's prediction, they decide to have their fun with the professor. &#xA;They plot up and at the actual election &lt;em&gt;not a single one&lt;/em&gt; of Professor Halfbrain's 24 predicted numbers turns out to be correct. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question 2:&lt;/strong&gt; &#xA;  Was it just a lucky coincidence that the club members were able to shatter Halfbrain's random predictions?&#xA;  Or is there always a way of doing this?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-13T13:13:44.327" LastActivityDate="2015-02-13T14:30:47.930" Title="Elections at the Fun-and-Nonsense-Club" Tags="|mathematics|combinatorics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8894" PostTypeId="1" AcceptedAnswerId="8918" CreationDate="2015-02-13T10:45:35.847" Score="21" ViewCount="552" Body="&lt;p&gt;Along the road less traveled, an abandoned temple lays in ruins. Though it has suffered the wrath of time, its foundation remains strong. And within this place, an inscription remains visible. A series of meaningless dots, yet dots that define all of creation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/alwiH.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this inscription has an error: for what should be four, there is only three.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;What is missing?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="7308" LastEditorUserId="5188" LastEditDate="2015-04-28T01:30:10.690" LastActivityDate="2015-04-28T01:30:10.690" Title="The Dots of Creation" Tags="|cipher|pattern|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8896" PostTypeId="2" ParentId="8886" CreationDate="2015-02-13T10:57:56.990" Score="3" Body="&lt;p&gt;Here is the complete solution (I have added my solutions for items 1 through 3 to Bulldogg6404's solution for items 4-8):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; * 1:   &quot;Free Willy&quot; (about saving an Orca whale)&lt;br&gt;&#xA; * 1-2: Walrus&lt;br&gt;&#xA; * 2:   Ross Geller (from the TV series &quot;Friends&quot;; played by David Schwimmer)&lt;br&gt;&#xA; * 2-3: Ross &amp;amp; Rachel = David Schwimmer &amp;amp; Jennifer Aniston (TV series &quot;Friends&quot;)&lt;br&gt;&#xA; * 3:   Jennifer Aniston (plays Rachel in &quot;Friends&quot;)&lt;br&gt;&#xA; * 3-4: Brad Pitt and Jennifer Aniston&lt;br&gt;&#xA; * 4:   Brad Pitt plays Achilles in the movie &quot;Troy&quot;&lt;br&gt;&#xA; * 4-5: Brad Pitt and Angelina Jolie&lt;br&gt;&#xA; * 5:   Jolie -&gt; Julie (?)&lt;br&gt;&#xA; * 5-6: Romeo and Juliet&lt;br&gt;&#xA; * 6:   Leonardo Dicaprio portrayed main characters in both films&lt;br&gt;&#xA; * 6-7: Titanic&lt;br&gt;&#xA; * 7:    Kate Winslet (actress in Titanic)&lt;br&gt;&#xA; * 7-8:  Kate Middleton and Prince William&lt;br&gt;&#xA; * 8:    William Tell (folk tale legend from Switzerland)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer to the final question seems to be &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the &lt;strong&gt;freedom of speech&lt;/strong&gt; principle&lt;br&gt;&#xA;  (This works with &quot;telling&quot; and with &quot;Free&quot; Willy)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-13T12:35:47.630" LastActivityDate="2015-02-13T12:35:47.630" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8897" PostTypeId="1" AcceptedAnswerId="8898" CreationDate="2015-02-13T11:48:57.470" Score="12" ViewCount="12911" Body="&lt;p&gt;I discovered this by accident, when trying to create a formula that generates prime numbers (an impossible task, I know).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, I find it very interesting that you take any prime number 5 and greater, then you square it and subtract 1, dividing it by 3 always results in a whole integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;5 x 5 = 25 - 1 = 24 / 3 = 8  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;11 x 11 = 121 - 1 = 120 / 3 = 40&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is always a whole number, regardless of how high the prime number is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone explain why this is so, mathematically?  This does not seem possible (to me).  And if this is really true, why can I find nothing written about it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have never heard of this theorem before, and nothing is mentioned on Wikipedia or other sources.  But perhaps this could be a helpful in reducing 33% of the possibilities when trying to find or prove large prime numbers, computationally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE:  someone commented that the resulting number is divisible by 24 , not just 3&lt;/p&gt;&#xA;" OwnerUserId="9495" LastEditorDisplayName="user20" LastEditDate="2015-02-13T16:22:50.310" LastActivityDate="2015-02-13T17:31:43.720" Title="Why is every prime number (5 and higher) divisible by 3 when you square it and subtract 1?" Tags="|mathematics|number-theory|" AnswerCount="4" CommentCount="9" ClosedDate="2015-02-13T17:43:54.057" ContentLicense="CC BY-SA 3.0" />
  <row Id="8898" PostTypeId="2" ParentId="8897" CreationDate="2015-02-13T11:58:50.387" Score="10" Body="&lt;p&gt;All such primes are of two types:&#xA;$6x+1$ and $6x-1$&#xA;$$(6x+1)^2-1=36x^2+12x$$&#xA;$$(6x-1)^2-1=36x^2-12x$$&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2015-02-13T11:58:50.387" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8899" PostTypeId="2" ParentId="8897" CreationDate="2015-02-13T12:29:00.693" Score="35" Body="&lt;p&gt;That happens because when you square a number(let's say $x$), you will get $x^2$ as the result. Then you subtract 1 from it and you get $x^2 - 1$, which can be rewritten as $x^2 - 1^2$ which is then equal to $(x-1)(x+1)$. Prime numbers are only divisible by $1$ and itself($x$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, for &lt;strong&gt;any&lt;/strong&gt; number $x$ the following is true:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$x$, $x+1$ or $x-1$ is fully divisible by 3.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And because $x$ is a prime number, it is not divisible by 3(unless $x = 3$, but you mentioned that it only works for values &gt;= 5).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That product consists of the numbers 1 less than that prime number and 1 more than that prime number. And out of &lt;strong&gt;any&lt;/strong&gt; 3 consequent numbers, 1 is fully divisible by 3, so either $x+1$ or $x-1$ is fully divisible by 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As &lt;a href=&quot;https://puzzling.stackexchange.com/users/99/kaine&quot;&gt;kaine&lt;/a&gt; pointed in his comment, we can also prove that that resulting number $x^2-1$ is also divisible by 12:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x$ is prime, it is not divisible by 2(so it's odd). Out of here follows, that both $(x-1)$ and $(x+1)$ are even, thus divisible by 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So out of these 2 numbers($(x-1)$ and $(x+1)$) one is divisible by 3 and both are divisible by 2, so the whole number is also divisible by $3*2*2 = 12$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as &lt;a href=&quot;https://puzzling.stackexchange.com/users/448/prisonmonkeys&quot;&gt;PrisonMonkeys&lt;/a&gt; mentioned in the comment, $(x-1)$ and $(x+1)$ are 2 consequent even numbers, so 1 of them is divisible by 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That then shows that the whole number $x^2 - 1$ is divisible by $2*3*4 = 24$&lt;/p&gt;&#xA;" OwnerUserId="9352" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2015-02-13T13:28:06.743" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8900" PostTypeId="2" ParentId="8897" CreationDate="2015-02-13T12:30:38.053" Score="1" Body="&lt;p&gt;Each number (without regard of primality), not divisible by $3$ can be expressed as $x + 1$ or $x + 2$ where $x$ is divisible by $3$. In the two cases you have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(x + 2)*(x + 2) = x*x + 4*x + 4 = x*x + 4*x + 3 + 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(x + 1)*(x + 1) = x*x + 2*x + 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $x$ is divisible by $3$ both boil down to&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$x*x + 4*x + 3 + 1 = 3*(x/3*x + x/3*4 + 1) + 1 = 3*n + 1$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$ x*x + 2*x + 1 = 3*(x/3*x + x/3*2) + 1 = 3*n + 1$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;with $n$ being an integer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;3*n + 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Subtract $1$ and you get a number divisible by $3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $3$ is the only prime divisible by $3$, this does not work for it. It works for $2$ though. It works for all other primes as well.&lt;/p&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="5840" LastEditDate="2015-02-13T13:08:04.363" LastActivityDate="2015-02-13T13:08:04.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8901" PostTypeId="2" ParentId="8893" CreationDate="2015-02-13T13:04:19.790" Score="7" Body="&lt;p&gt;Question 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because there are 24 members, that means each member belongs to exactly 23 pairs (1:24, 2:24 ... 23:24). Because the number of pairs is odd, each member must have a different number of close friends versus non-friends. &lt;em&gt;For example, if a member has 12 close friends, he must then have 11 non-friends.&lt;/em&gt; If at the time of the vote every member reversed their opinions of each other, then every member will have a different number of close friends than what they originally stated in the poll. Therefore, they will all have a different number of votes than the professor will have predicted.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Question 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the professor chooses all random numbers for his prediction (from the range 0-23), then chances are good that he is wrong about all the numbers anyway.&lt;br&gt;&#xA;&lt;br&gt;&#xA; In the event [A] that he guesses two members' vote counts correctly, those members can change each others' vote counts at the same time as a pair. By extension, this works for all possibilities where the professor guessed an even number of members' votes correctly.&lt;br&gt;&#xA;&lt;br&gt;&#xA; In the event [B] that the professor guesses only one person's, or an odd number of members', votes correctly, then there's the potential that reversing the opinion in one of his pairs will correct the other person's vote count by mistake. In such a case where this is unavoidable, the already-correct member reverses two pairs in the same direction (adding two close friends or subtracting two). This guarantees that the two people he swapped with are now correct with the professor's prediction, and therefore sets up event [A] above.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;em&gt;For example, pretend that the entire board had no close friends at all, the simplest scenario, where all members are tied with 0 votes. Let's say the professor's random number generator gets all 1's and only a single 0. By reversing member A, whose vote count was guessed correctly, with member B (B had 0 close friends; now he has 1), then member B's vote count has now been correctly predicted by mistake. If after that member A reverses with someone else, member C, then now C has 1 close friend and has a correctly predicted vote count as well. Now, simply reverse B and C together as a pair, making them close friends with each other, and they will all three have two votes apiece, being close friends with each other. The professor's zero-guess has been nullified. &lt;strong&gt;Note: if one of the guesses were two votes off, then this whole situation would have been avoided in the first place.&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9319" LastEditorUserId="9319" LastEditDate="2015-02-13T14:30:47.930" LastActivityDate="2015-02-13T14:30:47.930" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8902" PostTypeId="2" ParentId="8765" CreationDate="2015-02-13T13:35:10.473" Score="1" Body="&lt;p&gt;This is the answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Room 47.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the reason:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The date of the climax of the Cuba crisis was 28.october 1962. Formulated as a VALUE in Excel, would give 22947. Room number is identical to two last digits, based on the pattern of the other rooms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9385" LastActivityDate="2015-02-13T13:35:10.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8903" PostTypeId="1" AcceptedAnswerId="8904" CreationDate="2015-02-13T14:00:30.490" Score="3" ViewCount="1448" Body="&lt;p&gt;Solve the following rebus puzzle:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    J O K E&#xA;       U   &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;See also &#xA;(&lt;a href=&quot;https://puzzling.stackexchange.com/questions/8709/rebus-with-dracula&quot;&gt;Rebus with Dracula&lt;/a&gt;)  and &#xA;(&lt;a href=&quot;https://puzzling.stackexchange.com/questions/8750/another-rebus-puzzle&quot;&gt;Another Rebus puzzle&lt;/a&gt;) and&#xA;(&lt;a href=&quot;https://puzzling.stackexchange.com/questions/8796/my-third-rebus-puzzle&quot;&gt;My third Rebus puzzle&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-02-13T14:09:37.793" Title="My fourth Rebus puzzle" Tags="|english|rebus|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8904" PostTypeId="2" ParentId="8903" CreationDate="2015-02-13T14:01:41.837" Score="6" Body="&lt;p&gt;I think that the answer to this puzzle is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The joke is on you&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9278" LastActivityDate="2015-02-13T14:01:41.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8905" PostTypeId="1" AcceptedAnswerId="8906" CreationDate="2015-02-13T14:03:28.637" Score="10" ViewCount="609" Body="&lt;p&gt;Create a pangram sentence by replacing the 26 dots with the&#xA;26 letters of the alphabet (different dots, different letters):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;       ...  .A..  ..C..  .U..K  .H..  ...ES  .U..  O.E.  ..EM&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2015-02-13T14:39:37.847" Title="Pangram sentence" Tags="|english|pangram|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8906" PostTypeId="2" ParentId="8905" CreationDate="2015-02-13T14:39:37.847" Score="15" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; big  lAzy  duCks  qUacK  wHen  foxES  jUmp  OvEr  thEM&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first place my mind goes when someone says &quot;pangram&quot; is to those jumping foxes. The letters given in &lt;em&gt;quack&lt;/em&gt;, &lt;em&gt;jump&lt;/em&gt;, and &lt;em&gt;over&lt;/em&gt; were especially helpful getting this one started. &lt;/p&gt;&#xA;" OwnerUserId="1563" LastActivityDate="2015-02-13T14:39:37.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8907" PostTypeId="5" CreationDate="2015-02-13T14:51:53.523" Score="0" Body="&lt;p&gt;A &lt;em&gt;pangram&lt;/em&gt; (or &lt;em&gt;holoalphabetic&lt;/em&gt;) sentence for a given alphabet is a sentence that uses every letter of the alphabet &lt;em&gt;at least&lt;/em&gt; once. &#xA;A perfect pangram contains every letter of the alphabet &lt;em&gt;exactly&lt;/em&gt; once (and hence may be considered an anagram of the alphabet). &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Examples&lt;/strong&gt;:   &lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;strong&gt;&quot;The quick brown fox jumps over the lazy dog&quot;&lt;/strong&gt; is the most famous English pangram over the Latin alphabet.&lt;/li&gt;&#xA;  &lt;li&gt;&lt;strong&gt;&quot;Cwm fjord bank glyphs vext quiz&quot;&lt;/strong&gt; is a perfect pangram (where &quot;cwm&quot; is a loan word from Welsh, which means a steep-sided valley)&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An example of a pangram puzzle is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/8905/pangram-sentence&quot;&gt;Pangram sentence&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="11110" LastEditDate="2017-06-21T20:16:16.570" LastActivityDate="2017-06-21T20:16:16.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8908" PostTypeId="4" CreationDate="2015-02-13T14:51:53.523" Score="0" Body="A pangram sentence uses every letter of the alphabet at least once." OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-13T14:51:53.523" LastActivityDate="2015-02-13T14:51:53.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8910" PostTypeId="1" AcceptedAnswerId="8911" CreationDate="2015-02-13T15:02:37.857" Score="6" ViewCount="1772" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FTx7h.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who am I? This one should go pretty quick. For more info see also: &lt;a href=&quot;http://en.wikipedia.org/wiki/Rebus&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Rebus&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9412" LastActivityDate="2015-02-13T15:04:37.860" Title="Who am I in this Rebus Puzzle?" Tags="|rebus|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8911" PostTypeId="2" ParentId="8910" CreationDate="2015-02-13T15:04:37.860" Score="13" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;O + bee + wand - d;  Can + O + bee&quot; =&lt;br&gt;&#xA; = &lt;strong&gt;&quot;Obi-Wan Kenobi&quot;&lt;/strong&gt; (Jedi Knight of Star Wars fame)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-02-13T15:04:37.860" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8912" PostTypeId="1" AcceptedAnswerId="8913" CreationDate="2015-02-13T15:55:51.303" Score="9" ViewCount="787" Body="&lt;pre&gt;&lt;code&gt; D&#xA; R&#xA; A &#xA;EGDEH&#xA; G&#xA; E&#xA; D&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Try solving the Rebus puzzle above, according to me its neither too hard nor too easy!&lt;/p&gt;&#xA;" OwnerUserId="9346" LastEditorUserId="39" LastEditDate="2015-02-13T22:03:39.667" LastActivityDate="2015-02-13T22:03:39.667" Title="My First Rebus!" Tags="|rebus|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8913" PostTypeId="2" ParentId="8912" CreationDate="2015-02-13T15:59:31.800" Score="7" Body="&lt;p&gt;Nice idea!  Nice puzzle!&lt;br&gt;&#xA;(It is easy to solve, since there is so much openly visibile information)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dragged through a hedge backwards&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2015-02-13T15:59:31.800" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8914" PostTypeId="1" AcceptedAnswerId="8916" CreationDate="2015-02-13T16:30:55.693" Score="-2" ViewCount="533" Body="&lt;p&gt;I made a (for now, 2 player) game once that deals with manipulating binary numbers on a single list. Here are the rules.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Rules&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;For now, take $n=5$. If possible, provide a solution for generalized $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Player 1 is called the lead, and has a (supposed) advantage over the other one. He starts by declaring a (decimal) number $k$. He also writes an $n$-bit binary number at the top of a list.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Player 2 copies down this number twice. In one copy, he changes a single $0$ into a $1$. In the other copy, he replaces a single $1$ with a $0$. For example, if the number is $10011$, two possible numbers following it could be $00011$ (remove $1$) and $11011$ (remove $0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After this, on each turn, a player picks up (at his will) either of the $2$ numbers written by the opponent. He makes $2$ copies, and switches a $0$ in one, and a $1$ in the other, similar to player 2's first move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any point in time, only the last $k$ numbers remain on the list, all previous numbers are deleted. A player can not write a number that is on the list again. However, if it is being deleted with his move entering, then it is allowed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A player loses when he is unable to provide $2$ numbers. The winner wins $k$ points.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Question&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If both players use optimal strategies, who wins? If player 1 wins, what value of $k$ will he choose.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Additional rule&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;What if $00000$ and $11111$ were banned for the first move? Then it would become a more proper game. What would the answer be, then?&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4269" LastEditDate="2015-02-14T10:38:43.347" LastActivityDate="2015-02-14T22:28:07.363" Title="Binary manipulation game" Tags="|mathematics|logical-deduction|strategy|combinatorics|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8916" PostTypeId="2" ParentId="8914" CreationDate="2015-02-13T16:53:04.050" Score="4" Body="&lt;p&gt;Player one should choose &lt;code&gt;11111&lt;/code&gt; or &lt;code&gt;00000&lt;/code&gt; as his initial number. Player two cannot form two numbers by switching a &lt;code&gt;1&lt;/code&gt; and &lt;code&gt;0&lt;/code&gt;. Game over. This generalizes to an $n$-bit number quite well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k$ can be however many points player one desires. &lt;/p&gt;&#xA;" OwnerUserId="1563" LastEditorUserId="1563" LastEditDate="2015-02-13T17:06:15.710" LastActivityDate="2015-02-13T17:06:15.710" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8917" PostTypeId="2" ParentId="8897" CreationDate="2015-02-13T17:31:43.720" Score="1" Body="&lt;p&gt;Just to reiterate, rehash, and regurgitate what others have pointed out:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The resulting numbers ($x^2$ - 1), every last dang one of 'em, are divisible not merely by 3, but by $24$ (see the two edits to Novarg's answer)&lt;/li&gt;&#xA;&lt;li&gt;Prime numbers (&gt;3) are just a subset of the numbers for which this works: any odd number not divisible by 3; or, &lt;strong&gt;any multiple of 6 plus or minus 1&lt;/strong&gt;: nonprimes $25, 35, 49, 55, 65, 77, 85, 91, 95,$ $115, 119, 121, 125$...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9515" LastActivityDate="2015-02-13T17:31:43.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8918" PostTypeId="2" ParentId="8894" CreationDate="2015-02-13T17:37:23.263" Score="17" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Air&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The pattern spells out &quot;Earth&quot;, &quot;Fire&quot; and &quot;Water&quot; in braille, each in different colours with a different colour for any overlaps&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Colour explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dark Blue dots: &quot;Water&quot; word only&lt;br&gt;&#xA; Red dots: &quot;Fire&quot; word only&lt;br&gt;&#xA; Green: &quot;Earth&quot; word only&lt;br&gt;&#xA; Purple: &quot;Fire&quot; and &quot;Water&quot; words overlapped&lt;br&gt;&#xA; Yellow: &quot;Earth&quot; and &quot;Fire&quot; words overlapped&lt;br&gt;&#xA; Light Blue: &quot;Earth&quot; and &quot;Water&quot; overlapped&lt;br&gt;&#xA; White: &quot;Earth&quot;, &quot;Water&quot; and &quot;Fire&quot; words overlapped&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/pI3Ne.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, the obvious missing element is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Air!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6205" LastEditorUserId="1955" LastEditDate="2015-02-13T19:58:09.350" LastActivityDate="2015-02-13T19:58:09.350" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8919" PostTypeId="2" ParentId="251" CreationDate="2015-02-13T18:47:37.997" Score="2" Body="&lt;p&gt;My intervention here appeared to be too late and nothing but a supplementary clarification for confused people.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wrote a matlab code to illustrate why cannot any two numbers be or can be applied ,this is based on mathematical expression:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ (x,y)\  \epsilon \ \mathbb{N} \ \ x+y&amp;lt;100\ ,\forall n\ \epsilon\mathbb{N}&amp;lt;x+y \  \  : x+y-n\ \ not\  prime \  or\  n \ not\  prime\ \&amp;amp;\&amp;amp;\ \forall k\ prime \ factor\  of \ n \ if \ x+y-n \ primes \ (x+y-n)*k &amp;lt;100,\forall \ (x_{0},y_{0})\epsilon\mathbb{N} \ \ x_{0}+y_{0}&amp;lt;100,x_{0}y_{0}=xy:x_{0}\  and \ y_{0}\ \  prime \  \ or \ \ \exists n&amp;lt;x_{0}+y_{0} \ :[\ x_{0}+y_{0}-n \ prime \ \&amp;amp; \forall k  \ (x_{0}+y_{0}-n)*\frac{n}{k}&amp;gt;100 ] \ \ \ \&amp;amp;\&amp;amp; \ \ \ \forall \ (x_{1},y_{1})\epsilon\mathbb{N}\ \ x_{1}+ y_{1}&amp;lt;100,x-x_{1}=y_{1}-y, \ (x_{1}\  , \ y_{1})\   verify\  the \ last \  condition \ which \  means \  dont \  verify \ the \ first \  condition $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those who doesnt master matlab i have pastbin the results all you have to do is find (x=* ,y=*) with the suspected values&lt;/p&gt;&#xA;&#xA;&lt;p&gt;part1:&lt;a href=&quot;http://pastebin.com/raw.php?i=y36fughA&quot; rel=&quot;nofollow&quot;&gt;http://pastebin.com/raw.php?i=y36fughA&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;part2:&lt;a href=&quot;http://pastebin.com/raw.php?i=kfhC5qdw&quot; rel=&quot;nofollow&quot;&gt;http://pastebin.com/raw.php?i=kfhC5qdw&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;k = 1;begin=1;fid = fopen('numbers.txt', 'w');&#xA;for i = 2: 100&#xA;    feu = 0;&#xA;    for j = 2: i / 2&#xA;        if mod(i, j) == 0&#xA;            feu = 1;&#xA;            break;&#xA;        end&#xA;    end&#xA;    if feu == 0&#xA;        l(k) = i;&#xA;        k = k + 1;&#xA;    end&#xA;end&#xA;for  i = 1: k - 1&#xA;    tmp = 1;&#xA;    for  j = 1: 20&#xA;        tmp = tmp * l(i);&#xA;        if tmp &amp;gt;= 100&#xA;            m(i) = j;&#xA;            break&#xA;        end&#xA;    end&#xA;end&#xA;for ii = 1: k&#xA;    vect(ii) = 0;&#xA;end&#xA;for ii = 0: (k - 2)&#xA;    m(k - ii) = m(k - ii - 1);&#xA;    l(k - ii) = l(k - ii - 1);&#xA;end&#xA;for bb=0:1000000&#xA;    if begin==0&#xA;    fid = fopen('draft.txt', 'w');&#xA;    end&#xA;    m(1) = 1;o=1;oo=1;&#xA;    if begin==0&#xA;        for cc=0:10000&#xA;        val1=input('\nenter a number \n');val2=input('\nenter a second number \n');nb1=val1;nb2=val2;val1=min(nb1,nb2);val2=max(nb1,nb2);&#xA;        if val1==1 | val2&amp;gt;=100&#xA;            fprintf('enter valid values\n');&#xA;        else&#xA;            break;&#xA;        end&#xA;        end&#xA;    end&#xA;    for jj = 1: 6885&#xA;        cunt = 0;tmp = 1;feu = 0; nb1=0;nb2=0;&#xA;        for i= 2: (k)&#xA;            if (vect(i) == mod(vect(i) + floor(((mod(vect(i - 1) - 1, m(i - 1)) + 1) / m(i - 1))), m(i)))&#xA;                feu = 1;&#xA;            end&#xA;            if feu == 0&#xA;                vect(i) = mod(vect(i) + floor(((mod(vect(i - 1) - 1, m(i - 1)) + 1) / m(i - 1))), m(i));current = i;&#xA;            end&#xA;            tmp = tmp * l(i) ^ vect(i);&#xA;            if tmp &amp;gt; 10000&#xA;                if feu == 1&#xA;                    for ll = 2: current&#xA;                        if m(ll)~=0&#xA;                            vect(ll) = m(ll) - 1;&#xA;                        end&#xA;                    end&#xA;                    break;&#xA;                end&#xA;            end&#xA;        end&#xA;        if tmp &amp;lt; 10000 &amp;amp; tmp ~= 1&#xA;            kk=k;lim=1;&#xA;            for iii = 2: k&#xA;                if vect(iii)~=0&#xA;                    kk=iii+1;mm(iii) = vect(iii)  +1;lim=lim*(mm(iii));&#xA;                else&#xA;                    mm(iii)=0;&#xA;                end&#xA;                vectt(iii) = 0;&#xA;            end&#xA;            if kk&amp;gt;k&#xA;                kk= k;&#xA;            end&#xA;            mm(1) = 1;vectt(1) = 0;&#xA;            for jjj = 1: lim + 1&#xA;                tmpp = 1;feu = 0;tmmp = 1;ilast=1;&#xA;                for ii = 2: (kk)&#xA;                    if mm(ii)~=0&#xA;                        if(jjj~=1 )&#xA;                            if (vectt(ii) == mod(vectt(ii) + floor(((mod(vectt(ilast) - 1, mm(ilast)) + 1) / mm(ilast))), mm(ii)))&#xA;                                feu = 1;&#xA;                            end&#xA;                            if feu == 0&#xA;                                vectt(ii) = mod(vectt(ii) + floor((mod(vectt(ilast) - 1, mm(ilast)) + 1) / mm(ilast)), mm(ii));currentt = ii;&#xA;                            end&#xA;                        end&#xA;                        tmpp = tmpp * l(ii) ^ vectt(ii);tmmp = tmmp * l(ii) ^ (vect(ii) - vectt(ii));&#xA;                        if tmpp &amp;gt; 100 | tmmp &amp;gt; 100 | tmpp == 1 | tmmp == 1&#xA;                            if feu == 1&#xA;                                if tmpp&amp;gt; 100&#xA;                                    for ll = 2: currentt&#xA;                                        if mm(ll)~=0&#xA;                                            vectt(ll) = mm(ll) - 1;&#xA;                                        end&#xA;                                    end&#xA;                                end&#xA;                                break;&#xA;                            end&#xA;                        end&#xA;                        ilast=ii;&#xA;                    end&#xA;                end&#xA;                if tmpp &amp;lt; 100 &amp;amp; tmmp &amp;lt; 100 &amp;amp; tmpp ~= 1 &amp;amp; tmmp ~= 1&#xA;                    if begin==1&#xA;                        fprintf('-  %d %%  -\n',floor(100*jj/6885));&#xA;                    end&#xA;                    sum = tmpp + tmmp;prem = 0;sec = 0;niv = k;&#xA;                    for iij = 0: niv - 2&#xA;                        if l(k - iij) &amp;lt;= sum&#xA;                            niv = k - iij;break;&#xA;                        end&#xA;                    end&#xA;                    if(sum&amp;gt;100)&#xA;                        prem=1000;&#xA;                    end&#xA;                    last = 2;&#xA;                    for iij = 2: niv&#xA;                        if prem ~= 0 | sec ~= 0&#xA;                            break;&#xA;                        end&#xA;                        if l(iij) == sum - 2&#xA;                            prem = sum-2;break;&#xA;                        end&#xA;                        for iijj = 0: niv - 2&#xA;                            if(niv-iijj&amp;lt;iij)&#xA;                                break;&#xA;                            end&#xA;                            if(prem ~= 0|sec ~= 0)&#xA;                                break;&#xA;                            end&#xA;                            if l(niv - iijj) == sum - 2&#xA;                                prem = sum-2;break;&#xA;                            end&#xA;                            for ji = l(iij - 1) + 1: l(iij) - 1&#xA;                                if l(niv - iijj) == sum - ji&#xA;                                    if l(last) * l(last) == ji&#xA;                                        if l(niv - iijj) * l(last) &amp;gt; 100&#xA;                                            sec = l(niv - iijj) ;prem=l(last);break;&#xA;                                        end&#xA;                                        if last~= k&#xA;                                            last = last + 1;&#xA;                                        end&#xA;                                    end&#xA;                                end&#xA;                            end&#xA;                            if l(niv - iijj) &amp;lt;= sum - l(iij)&#xA;                                if l(niv - iijj) == sum - l(iij)&#xA;                                    prem= l(iij) ;break;&#xA;                                end&#xA;                                niv = niv - iijj;&#xA;                                break;&#xA;                            end&#xA;                        end&#xA;                    end&#xA;                    res(o) = tmmp;res(o+1) = tmpp;ress(o) = prem;ress(o+1) = sec;o = o + 2;&#xA;                    if prem == 0 &amp;amp; sec == 0&#xA;                        valeur=tmmp;&#xA;                        if tmmp~=nb1 &amp;amp; tmmp~=nb2&#xA;                            cunt = cunt + 1;nb1=min(tmmp,tmpp);nb2=max(tmmp,tmpp);&#xA;                        end&#xA;                    end&#xA;                end&#xA;                for ij = 2: kk&#xA;                    if vectt(ij) ~= mm(ij) - 1 &amp;amp; mm(ij) ~= 0&#xA;                        break;&#xA;                    end&#xA;                end&#xA;                if ij == kk&#xA;                    break;&#xA;                end&#xA;            end&#xA;        end&#xA;        res(o)=0;ress(o)=cunt;o=o+1;&#xA;        if (vect(k) == m(k) - 1)&#xA;            for ijij = 2: k&#xA;                if vect(ijij)  ~= m(ijij) - 1&#xA;                    break;&#xA;                end&#xA;            end&#xA;            if ijij == k&#xA;                break;&#xA;            end&#xA;        end&#xA;        if(cunt&amp;gt;1 | cunt==0)&#xA;            for ij=0:2:o-3&#xA;                if res(o-2-ij)==0&#xA;                    o=o-1-ij;&#xA;                    break&#xA;                end&#xA;                if ress(o-2-ij)==0 &amp;amp; ress(o-2-ij-1)==0&#xA;                    if begin==0 &amp;amp; val1==min(res(o-2-ij-1),res(o-2-ij)) &amp;amp; val2==max(res(o-2-ij-1),res(o-2-ij))&#xA;                        fprintf('(x=%d ,y=%d) is rejected because there exists (x0,y0)  :\n',res(o-2-ij-1),res(o-2-ij));&#xA;                    end&#xA;                    fprintf(fid,'(x=%d ,y=%d) is rejected because there exists (x0,y0)  :\n',res(o-2-ij-1),res(o-2-ij));&#xA;                    for iij=0:2:o-3&#xA;                        if res(o-2-iij)==0&#xA;                            break&#xA;                        end&#xA;                        if res(o-2-ij)~=res(o-2-iij) &amp;amp; res(o-2-ij)~=res(o-2-iij-1)&#xA;                            if ress(o-2-iij)==0 &amp;amp; ress(o-2-iij-1)==0&#xA;                                if begin==0 &amp;amp; val1==min(res(o-2-ij-1),res(o-2-ij)) &amp;amp; val2==max(res(o-2-ij-1),res(o-2-ij))&#xA;                                    fprintf('(x0=%d ,y0=%d ) where xy = x0*y0  and they both verify conditions \nthe proctor wouldnt say  &quot;Now I can determine them.&quot;!\n\n',res(o-2-iij-1),res(o-2-iij));break;&#xA;                                end&#xA;                            fprintf(fid,'(x0=%d ,y0=%d ) where xy = x0*y0  and they both verify conditions \nthe proctor wouldnt say  &quot;Now I can determine them.&quot;!\n\n',res(o-2-iij-1),res(o-2-iij));break;&#xA;                            end&#xA;                        end&#xA;                    end&#xA;                else&#xA;                    if begin==0 &amp;amp; val1==min(res(o-2-ij-1),res(o-2-ij)) &amp;amp; val2==max(res(o-2-ij-1),res(o-2-ij))&#xA;                        fprintf('(x=%d ,y=%d) is rejected because   :\n',res(o-2-ij-1),res(o-2-ij));&#xA;                    end&#xA;                    fprintf(fid,'(x=%d ,y=%d) is rejected because   :\n',res(o-2-ij-1),res(o-2-ij));&#xA;                    if ress(o-2-ij)==0&#xA;                        if ress(o-2-ij-1)==1000&#xA;                            if begin==0 &amp;amp; val1==min(res(o-2-ij-1),res(o-2-ij)) &amp;amp; val2==max(res(o-2-ij-1),res(o-2-ij))&#xA;                                fprintf('x+y = %d bigger than 100 \n\n\n',res(o-2-ij-1)+res(o-2-ij));&#xA;                                break;&#xA;                            end&#xA;                        fprintf(fid,'x+y = %d bigger than 100 \n\n\n',res(o-2-ij-1)+res(o-2-ij));&#xA;                        else&#xA;                            if begin==0 &amp;amp; val1==min(res(o-2-ij-1),res(o-2-ij)) &amp;amp; val2==max(res(o-2-ij-1),res(o-2-ij))&#xA;                                if res(o-2-ij)==ress(o-3-ij) | res(o-3-ij)==ress(o-2-ij-1)&#xA;                                    fprintf('%d and %d are primes \nthe proctor wudnt say  &quot;I cannot determine the two numbers.&quot;!\n\n',ress(o-2-ij-1),res(o-2-ij-1)+res(o-2-ij)-ress(o-3-ij));&#xA;                                else&#xA;                                    fprintf('x+y-k = %d and k = %d primes \nthe summer wudnt say  &quot;I knew that.&quot;!\n\n',ress(o-2-ij-1),res(o-2-ij-1)+res(o-2-ij)-ress(o-3-ij));&#xA;                                end&#xA;                                break;&#xA;                            end&#xA;                            if res(o-2-ij)==ress(o-3-ij) | res(o-3-ij)==ress(o-2-ij-1)&#xA;                                fprintf(fid,'%d and %d are primes \nthe proctor wudnt say  &quot;I cannot determine the two numbers.&quot;!\n\n',ress(o-2-ij-1),res(o-2-ij-1)+res(o-2-ij)-ress(o-3-ij));&#xA;                            else&#xA;                                fprintf(fid,'x+y-k = %d and k = %d primes \nthe summer wudnt say  &quot;I knew that.&quot;!\n\n',ress(o-2-ij-1),res(o-2-ij-1)+res(o-2-ij)-ress(o-3-ij));&#xA;                            end&#xA;                        end&#xA;                    else&#xA;                        if ress(o-2-ij-1)+ress(o-2-ij)==res(o-2-ij-1)+res(o-2-ij)&#xA;                            if begin==0 &amp;amp; val1==min(res(o-2-ij-1),res(o-2-ij)) &amp;amp; val2==max(res(o-2-ij-1),res(o-2-ij))&#xA;                                fprintf('it exists a prime number k=%d where (x+y-k)*k = %d*%d =%d  bigger than 100 \nthe summer wudnt say  &quot;I knew that.&quot;!\n\n',min(ress(o-2-ij-1),ress(o-2-ij)),res(o-2-ij-1)+res(o-2-ij)-min(ress(o-2-ij-1),ress(o-2-ij)),min(ress(o-2-ij-1),ress(o-2-ij)),ress(o-2-ij-1)*ress(o-2-ij));&#xA;                                break;&#xA;                            end&#xA;                            fprintf(fid,'it exists a prime number k=%d where (x+y-k)*k = %d*%d =%d  bigger than 100 \nthe summer wudnt say  &quot;I knew that.&quot;!\n\n',min(ress(o-2-ij-1),ress(o-2-ij)),res(o-2-ij-1)+res(o-2-ij)-min(ress(o-2-ij-1),ress(o-2-ij)),min(ress(o-2-ij-1),ress(o-2-ij)),ress(o-2-ij-1)*ress(o-2-ij));&#xA;                        else&#xA;                            if begin==0 &amp;amp; val1==min(res(o-2-ij-1),res(o-2-ij)) &amp;amp; val2==max(res(o-2-ij-1),res(o-2-ij))&#xA;                                fprintf('it exists a number n=%d , where the smallest prime number k=%d &amp;lt; n verifies (x+y-n)*k = %d*%d = %d  bigger than 100 \nthe summer wudnt say  &quot;I knew that.&quot;!\n\n',res(o-2-ij-1)+res(o-2-ij)-max(ress(o-2-ij-1),ress(o-2-ij)),min(ress(o-2-ij-1),ress(o-2-ij)),max(ress(o-2-ij-1),ress(o-2-ij)),min(ress(o-2-ij-1),ress(o-2-ij)),ress(o-2-ij-1)*ress(o-2-ij));&#xA;                                break;&#xA;                            end&#xA;                        fprintf(fid,'it exists a number n=%d , where the smallest prime number k=%d &amp;lt; n verifies (x+y-n)*k = %d*%d = %d  bigger than 100 \nthe summer wudnt say  &quot;I knew that.&quot;!\n\n',res(o-2-ij-1)+res(o-2-ij)-max(ress(o-2-ij-1),ress(o-2-ij)),min(ress(o-2-ij-1),ress(o-2-ij)),max(ress(o-2-ij-1),ress(o-2-ij)),min(ress(o-2-ij-1),ress(o-2-ij)),ress(o-2-ij-1)*ress(o-2-ij));&#xA;                        end&#xA;                    end&#xA;                end&#xA;&#xA;            end&#xA;&#xA;        elseif(cunt==1)&#xA;            feu=0;&#xA;            for iij=0:2:o-3&#xA;                if (o-2-iij&amp;lt;2)&#xA;                    break&#xA;                end&#xA;                if res(o-2-iij)==0&#xA;                    if iij==2 |(iij==4 &amp;amp; res(o-2)==res(o-5) &amp;amp; res(o-3)==res(o-4))&#xA;                        if begin==0 &amp;amp; val1==min(nb1,nb2) &amp;amp; val2==max(nb1,nb2)&#xA;                            fprintf('(x=%d ,y=%d) is rejected because x * y dont have other valid factors \nthe proctor wudnt say  &quot;I cannot determine the two numbers.&quot;!\n\n',nb1,nb2);feu=1;&#xA;                            break;&#xA;                        end&#xA;                    fprintf(fid,'(x=%d ,y=%d) is rejected because x * y dont have other valid factors  \nthe proctor wudnt say  &quot;I cannot determine the two numbers.&quot;!\n\n',nb1,nb2);feu=1;&#xA;                    end&#xA;                    break;&#xA;                end&#xA;                if ress(o-2-iij-1)~=0 &amp;amp; res(o-2-iij)~=0 &amp;amp; res(o-3-iij)~=0&#xA;                    if begin==0 &amp;amp; val1==min(res(o-2-iij-1),res(o-2-iij)) &amp;amp; val2==max(res(o-2-iij-1),res(o-2-iij))&#xA;                        fprintf('(x=%d ,y=%d) is rejected because   :\n',res(o-2-iij-1),res(o-2-iij));&#xA;                    end&#xA;                    fprintf(fid,'(x=%d ,y=%d) is rejected because   :\n',res(o-2-iij-1),res(o-2-iij));&#xA;                    if ress(o-2-iij)==0&#xA;                        if ress(o-2-iij-1)==1000&#xA;                            if begin==0 &amp;amp; val1==min(res(o-2-iij-1),res(o-2-iij)) &amp;amp; val2==max(res(o-2-iij-1),res(o-2-iij))&#xA;                                fprintf('x+y = %d bigger than 100 \n\n\n',res(o-2-iij-1)+res(o-2-iij));&#xA;                                break;&#xA;                            end&#xA;                        fprintf(fid,'x+y = %d bigger than 100 \n\n\n',res(o-2-iij-1)+res(o-2-iij));&#xA;                        else&#xA;                            if begin==0 &amp;amp; val1==min(res(o-2-iij-1),res(o-2-iij)) &amp;amp; val2==max(res(o-2-iij-1),res(o-2-iij))&#xA;                                if res(o-2-iij)==ress(o-3-iij) | res(o-3-iij)==ress(o-2-iij-1)&#xA;                                fprintf('%d and %d are primes \nthe proctor wudnt say  &quot;I cannot determine the two numbers.&quot;!\n\n',ress(o-2-iij-1),res(o-2-iij-1)+res(o-2-iij)-ress(o-3-iij));&#xA;                                else&#xA;                                fprintf('x+y-k = %d and k = %d primes \nthe summer wudnt say  &quot;I knew that.&quot;!\n\n',ress(o-2-iij-1),res(o-2-iij-1)+res(o-2-iij)-ress(o-3-iij));&#xA;                                end&#xA;                                break;&#xA;                            end&#xA;                        if res(o-2-iij)==ress(o-3-iij) | res(o-3-iij)==ress(o-2-iij-1)&#xA;                                fprintf(fid,'%d and %d are primes \nthe proctor wudnt say  &quot;I cannot determine the two numbers.&quot;!\n\n',ress(o-2-iij-1),res(o-2-iij-1)+res(o-2-iij)-ress(o-3-iij));&#xA;                        else&#xA;                                fprintf(fid,'x+y-k = %d and k = %d primes \nthe summer wudnt say  &quot;I knew that.&quot;!\n\n',ress(o-2-iij-1),res(o-2-iij-1)+res(o-2-iij)-ress(o-3-iij));&#xA;                        end&#xA;                        end&#xA;                    else&#xA;                        if ress(o-2-iij-1)+ress(o-2-iij)==res(o-2-iij-1)+res(o-2-iij)&#xA;                            if begin==0 &amp;amp; val1==min(res(o-2-iij-1),res(o-2-iij)) &amp;amp; val2==max(res(o-2-iij-1),res(o-2-iij))&#xA;                                fprintf('it exists a prime number k=%d where (x+y-k)*k = %d*%d =%d  bigger than 100 \nthe summer wudnt say  &quot;I knew that.&quot;!\n\n',min(ress(o-2-iij-1),ress(o-2-iij)),res(o-2-iij-1)+res(o-2-iij)-min(ress(o-2-iij-1),ress(o-2-iij)),min(ress(o-2-iij-1),ress(o-2-iij)),ress(o-2-iij-1)*ress(o-2-iij));&#xA;                                break;&#xA;                            end&#xA;                        fprintf(fid,'it exists a prime number k=%d where (x+y-k)*k = %d*%d =%d  bigger than 100 \nthe summer wudnt say  &quot;I knew that.&quot;!\n\n',min(ress(o-2-iij-1),ress(o-2-iij)),res(o-2-iij-1)+res(o-2-iij)-min(ress(o-2-iij-1),ress(o-2-iij)),min(ress(o-2-iij-1),ress(o-2-iij)),ress(o-2-iij-1)*ress(o-2-iij));&#xA;                        else&#xA;                            if begin==0 &amp;amp; val1==min(res(o-2-iij-1),res(o-2-iij)) &amp;amp; val2==max(res(o-2-iij-1),res(o-2-iij))&#xA;                                fprintf('it exists a number n=%d , where the smallest prime number k=%d &amp;lt; n verifies (x+y-n)*k = %d*%d = %d  bigger than 100 \nthe summer wudnt say  &quot;I knew that.&quot;!\n\n',res(o-2-iij-1)+res(o-2-iij)-max(ress(o-2-iij-1),ress(o-2-iij)),min(ress(o-2-iij-1),ress(o-2-iij)),max(ress(o-2-iij-1),ress(o-2-iij)),min(ress(o-2-iij-1),ress(o-2-iij)),ress(o-2-iij-1)*ress(o-2-iij));&#xA;                                break;&#xA;                            end&#xA;                        fprintf(fid,'it exists a number n=%d , where the smallest prime number k=%d &amp;lt; n verifies (x+y-n)*k = %d*%d = %d  bigger than 100 \nthe summer wudnt say  &quot;I knew that.&quot;!\n\n',res(o-2-iij-1)+res(o-2-iij)-max(ress(o-2-iij-1),ress(o-2-iij)),min(ress(o-2-iij-1),ress(o-2-iij)),max(ress(o-2-iij-1),ress(o-2-iij)),min(ress(o-2-iij-1),ress(o-2-iij)),ress(o-2-iij-1)*ress(o-2-iij));&#xA;                        end&#xA;                    end&#xA;                end&#xA;&#xA;            end&#xA;            if feu~=1&#xA;                rest(oo)=nb1;rest(oo+1)=nb2;oo=oo+2;&#xA;                for iij= 0:2:oo-3&#xA;                    if (nb1==rest(oo-2-iij) &amp;amp; nb2==rest(oo-iij-1))|(nb2==rest(oo-2-iij) &amp;amp; nb1==rest(oo-iij-1))&#xA;                    else&#xA;                        if (nb1-rest(oo-2-iij)==rest(oo-1-iij)-nb1 | nb2-rest(oo-1-iij)==rest(oo-2-iij)-nb1 )&#xA;                            if begin==0 &amp;amp; ( (val1==min(rest(oo-iij-1),rest(oo-2-iij)) &amp;amp; val2==max(rest(oo-iij-1),rest(oo-2-iij))) | (val1==min(nb1,nb2) &amp;amp; val2==max(nb1,nb2)) )&#xA;                                fprintf('(x=%d ,y=%d) and (x0=%d ,y0=%d )  where x-x0 = y0-y = %d are rejected because they both verify conditions \nthe summer wouldnt say &quot;So can I.&quot;!\n\n',nb1,nb2,rest(oo-iij-1),rest(oo-2-iij),nb1-rest(oo-iij-1));&#xA;                                break;&#xA;                            end&#xA;                            fprintf(fid,'(x=%d ,y=%d) and (x0=%d ,y0=%d )  where x-x0 = y0-y = %d are rejected because they both verify conditions \nthe summer wouldnt say &quot;So can I.&quot;!\n\n',nb1,nb2,rest(oo-iij-1),rest(oo-2-iij),nb1-rest(oo-iij-1));&#xA;                            break;&#xA;                        end&#xA;                    end&#xA;                end&#xA;            end&#xA;        end&#xA;    end&#xA;    for ij= 2:2:oo-5&#xA;        nb1=rest(oo-ij);nb2=rest(oo-ij+1);feu=1;&#xA;        for iij= ij:2:oo-3&#xA;            if (nb1-rest(oo-2-iij)==rest(oo-1-iij)-nb1 | nb2-rest(oo-1-iij)==rest(oo-2-iij)-nb1 )&#xA;                feu=0;rest(oo-iij-1)=0;rest(oo-2-iij)=0;&#xA;            end&#xA;        end&#xA;        if feu==0&#xA;            rest(oo-ij)=0;rest(oo-ij+1)=0;&#xA;        end&#xA;    end&#xA;    if begin==1&#xA;        for ij= 2:2:oo-5&#xA;            nb1=rest(oo-ij);nb2=rest(oo-ij+1);&#xA;            if nb1~=0 &amp;amp; nb2~=0&#xA;                fprintf('numbers searched for are :\n\n');&#xA;                nb1&#xA;                nb2&#xA;                break;&#xA;            end&#xA;        end&#xA;        begin=0;&#xA;    end&#xA;    fclose(fid);&#xA;end&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Im pleased to be reported to wichever bug or anomaly detected in my entire post ,plz dont downrate without reason thanks .&lt;/p&gt;&#xA;" OwnerUserId="9517" LastActivityDate="2015-02-13T18:47:37.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8920" PostTypeId="1" AcceptedAnswerId="8925" CreationDate="2015-02-13T19:18:18.043" Score="2" ViewCount="479" Body="&lt;p&gt;Tom, Paul and Joe have just learnt the ranks that they reached in a competition. &#xA;None of them knows the ranks of the other two, but they all know that Tom and Paul ended up in consecutive positions. &#xA;As they were curious to know each other's ranks, this little conversation took place:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paul declares: I don't know Tom's rank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tom: Me neither. I also don't know your rank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paul: I already knew this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Joe: Now I know that I am the last among the three of us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tom : Me too. Before I just knew that I'm not the first among the three of us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's everyone's rank?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTES:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The ranks are strictly postive integers.&lt;/li&gt;&#xA;&lt;li&gt;No rank occurs twice. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every ones knows the maximum rank that could be obtained.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9517" LastEditorUserId="8874" LastEditDate="2015-02-14T14:20:38.943" LastActivityDate="2015-02-14T14:20:38.943" Title="Consecutive ranks?" Tags="|logical-deduction|" AnswerCount="3" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8921" PostTypeId="2" ParentId="8920" CreationDate="2015-02-13T20:01:25.010" Score="0" Body="&lt;p&gt;My answer:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Paul claims that he doesn't know Tom's ranking. Because Paul also knows what the best ranking is, he cannot have a perfect score, else he would know what Tom's ranking was, knowing that they are consecutive. However, he may still be first if the other two scored less than him.&lt;br&gt;&#xA; Tom says the same as Paul, indicating that his ranking is also not a perfect score.&lt;br&gt;&#xA; Paul says that he knows this fact already, proving that Paul's ranking must be at least 3.&lt;br&gt;&#xA; Joe states that he knows he is last. That places him last.&lt;br&gt;&#xA; Tom states that he knows he is not first. We know he isn't last, either. That places him in the middle.&lt;br&gt;&#xA; Paul must then be first.&lt;br&gt;&#xA; If Paul must be at least 3 and comes in first, then we can say his rank is 3; Tom in the middle is rank 4; and Joe being last is rank 5.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9319" LastEditorUserId="9319" LastEditDate="2015-02-13T20:26:32.117" LastActivityDate="2015-02-13T20:26:32.117" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8922" PostTypeId="2" ParentId="8920" CreationDate="2015-02-13T20:04:06.430" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; J,T and P's grade are respectively 3, 4 and 5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;let's presume that the max grade is 100, but it makes no difference&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paul doesn't know Tom's grade. since they are consecutive integers, Paul cannot have the grade 1 or 100, because then he'd know Tom's grade&lt;/p&gt;&#xA;&#xA;&lt;p&gt;even with that additional information, Tom doesn't know Paul's grade. so Tom cannot have 2 or 98.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paul already knew that Tom wouldn't know his grade. if Paul was 98, there would be a chance for Tom to be 99. so Paul cannot be 2 or 98 either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now Joe says that he knows for sure he is last. he must be have a grade of 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since now Tom knows that he is not first, it means Paul has a higher grade than him. the only way Tom can be sure of that is if there is no possible grade in between himself and Joe&lt;/p&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="9413" LastEditDate="2015-02-13T20:22:13.850" LastActivityDate="2015-02-13T20:22:13.850" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8925" PostTypeId="2" ParentId="8920" CreationDate="2015-02-13T20:35:30.013" Score="2" Body="&lt;p&gt;Note, my numbers are for if a higher rank is better..&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Joe's rank is 4, Tom's rank is 5, and Paul's rank is 6&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Paul: &quot;I don't know Tom's ranking.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Paul knows that he and Tom have consecutive positions, so if his degree was either 1 or N (the max), he would know Tom's ranking.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tom: &quot;I don't know yours, either.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tom now knows that Paul's ranking is neither at the top nor at the bottom, but since he still doesn't know Paul's ranking that means his ranking is not 2 or N-1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Paul: &quot;I already knew that.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This means Paul already knew Tom wasn't at 2 or N-1, so Paul's ranking must be greater than 3 and less than N-2. &lt;strong&gt;Note:&lt;/strong&gt; I'm assuming here that Paul would have mentioned if they knew Tom's ranking at this point. If that's not the case, then Paul could have 2 or N-1 and know (although he hasn't said it) Tom's rank.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Joe: &quot;Now I know that I am last among the three of us.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Right before Paul's second statement, Joe would only know that Paul's ranking must be at least 2, with Tom's possibly being 3. After Paul's second statement, Tom could still be at 3 with Paul being at 4. Joe's ranking must be 4 because he now knows he is last - up until Paul's second statement there was the possibility that Tom and Paul could be below Joe, but since Paul cannot be lower than 4, and Joe has 4, he knows Paul and Tom must be above him.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tom : Me too I just knew that I'm not the first between the three of us.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Up until Joe's statement, Paul could have been in rank 4. Now that Tom knows he is below Paul, he must have rank 5, otherwise he would not have been able to tell. That leaves Paul to be rank 6.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1955" LastEditorUserId="1955" LastEditDate="2015-02-13T20:46:21.473" LastActivityDate="2015-02-13T20:46:21.473" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8926" PostTypeId="1" CreationDate="2015-02-13T21:30:08.300" Score="4" ViewCount="1084" Body="&lt;p&gt;On some island, the population is split into native people and into outlanders. There are more natives than outlanders. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Someone with extreme logical and mathematical thinking wants to know who are the strangers and who are the autochthons of this place, while asking as few questions as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feasible questions are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Are you a foreigner on this island? &lt;/li&gt;&#xA;  &lt;li&gt;Is that dude a foreigner on this island? &lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now .... we know that natives always tell the truth, whereas foreigners may lie whenever they want. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the minimum number of questions needed to find out who are the strangers and who are the autochthons on the island?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the number works especially with big populations&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9517" LastEditorUserId="9525" LastEditDate="2015-02-14T17:59:58.710" LastActivityDate="2015-02-14T17:59:58.710" Title="Island inhabitants: Who lies and who tells the truth?" Tags="|mathematics|logical-deduction|liars|" AnswerCount="2" CommentCount="9" FavoriteCount="0" ClosedDate="2015-02-15T02:44:15.030" ContentLicense="CC BY-SA 3.0" />
  <row Id="8927" PostTypeId="2" ParentId="8926" CreationDate="2015-02-14T03:15:01.663" Score="3" Body="&lt;p&gt;Let $n$ and $s$ be the numbers of natives and strangers, respectively.&#xA;We are told that $n &amp;gt; s$ and that natives always tell the truth but strangers lie when they wish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As stated, the puzzle requires an inquirer to ask just one individual (&quot;ask anyone&quot;) a series of questions to separate the set of all natives from the set of all strangers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, had the strangers been incapable of telling the truth, you would be able to craft questions that produce consistent answers (e.g. &quot;How many natives would a liar never say there are?&quot;). But since the strangers can tell the truth but also can lie, their answers are unreliable. So if the inquirer picked a stranger, the answers need not take any element of the inquiry into account. In that case, no set of questions would give you certainty and the puzzle has no answer. At best, the inquirer can tell they are talking to a stranger, but cannot compel the stranger to produce the correct answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, consider a variant where the inquirer can ask multiple people multiple questions. Then one strategy is to ask each person, &quot;Who are the natives?&quot; and go with the majority. Here, it doesn't matter what the strangers say - if they tell the truth, they contribute to the majority, and if they lie, they are in the minority anyway. For this strategy to work, the number of natives questioned must be greater than the number of strangers questioned, so in the worst case, $2s + 1$. Since $n&amp;gt;s$, we also have $n+s &amp;gt; s+s \geq 2s+1$, so there are enough people to ask. However, this assumes knowledge of $s$, which may not be the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n+s$ is odd, then from $n&amp;gt;s$, we have $n+s \geq (s+1) + s$, i.e. $n+s \geq 2s+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n+s$ is even, then from $n&amp;gt;s$, we have $n+s \geq (s+2) + s$, i.e. $n+s \geq 2s+2$, or $n+s-1 \geq 2s+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So according to this strategy, the answer is $2s+1$ questions if you know $s$, otherwise it is $n+s$ for an odd population and $n+s-1$ for an even population.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a better strategy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since strangers provide no information, any optimal strategy will need to include asking at least one native and be able to tell when at least one native has answered. In the worst case, an optimal strategy will require $s+1$ questions, but since the inquirer doesn't know $s$, we're back to a total of $n+s$ questions for odd populations and $n+s-1$ for even populations.&lt;/p&gt;&#xA;" OwnerUserId="9422" LastActivityDate="2015-02-14T03:15:01.663" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8928" PostTypeId="1" AcceptedAnswerId="8929" CreationDate="2015-02-14T03:27:47.163" Score="-2" ViewCount="314" Body="&lt;p&gt;Dr. Probot is a famous inventor, known for many creations, his most recent being &quot;&lt;em&gt;Bot&lt;/em&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Bot&lt;/em&gt; is the very first working robot, he has 3 very basic instructions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Left&lt;/li&gt;&#xA;&lt;li&gt;Right&lt;/li&gt;&#xA;&lt;li&gt;Self Destruct&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Left&lt;/strong&gt;: Rotate 90° left, move forward one space.&lt;br&gt;&#xA;&lt;strong&gt;Right&lt;/strong&gt;: Rotate 90° right, move forward one space.&lt;br&gt;&#xA;&lt;strong&gt;Self Destruct&lt;/strong&gt;: Destroy itself for all eternity, leave no traces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bots are preprogrammed to pick a random left/right movement. They will also self-destruct if they bump into another bot, or leave the grid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each &lt;em&gt;turn&lt;/em&gt; all the bots move at once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The bots are programmed to stop if they are the last one left.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine an $n\times n$ grid full of bots, all facing in one direction. Is there a general strategy to determine if eventually, there will be 1 bot standing, and in how many turns? If there is, what is it? If there isn't, explain why not.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="5373" LastEditDate="2015-02-14T11:55:41.743" LastActivityDate="2015-02-14T11:55:41.743" Title="The Programmers Bot!" Tags="|mathematics|probability|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8929" PostTypeId="2" ParentId="8928" CreationDate="2015-02-14T03:43:09.930" Score="4" Body="&lt;p&gt;No, there's not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's 1x1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Ends immediately.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If it's 2x2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is 1 guy left after two moves if everyone moves in the same direction. There is no one left if NW and SW turn left and NE and SE turn right. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If it's nxn&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Imagine a 3x3 grid, and the bots are all facing north. &#xA; They turn left and 3 go off the west end. Then they either have to go north or south. I'll choose north, but the result in the other direction is identical. They go north, and two more bots fall off. Then you have 4 bots left in a 2x2 square in the northwest corner of the grid. They can now continually turn right.&#xA; Obviously there will eventually be 1 left if they all go W then N then W then N then W then N etc..&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9525" LastEditorUserId="9525" LastEditDate="2015-02-14T04:10:30.143" LastActivityDate="2015-02-14T04:10:30.143" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="8930" PostTypeId="1" AcceptedAnswerId="8931" CreationDate="2015-02-14T05:34:47.043" Score="-3" ViewCount="254" Body="&lt;p&gt;Choose the conclusion that validly follows from the argument below.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Some apples are edible.&lt;/li&gt;&#xA;&lt;li&gt;All edible food are not rotten.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A. All apples are not rotten.&lt;br&gt;&#xA;B. Some apples are rotten.&lt;br&gt;&#xA;C. Some apples are not rotten.&lt;br&gt;&#xA;D. All apples are edible.&lt;br&gt;&#xA;E. B and C  &lt;/p&gt;&#xA;" OwnerUserId="9527" LastEditorUserId="8874" LastEditDate="2015-02-14T14:31:44.420" LastActivityDate="2015-02-14T14:31:44.420" Title="Are rotton apples edible?" Tags="|logical-deduction|riddle|" AnswerCount="2" CommentCount="1" ClosedDate="2015-02-14T19:33:26.380" ContentLicense="CC BY-SA 3.0" />
  <row Id="8931" PostTypeId="2" ParentId="8930" CreationDate="2015-02-14T05:49:46.753" Score="2" Body="&lt;p&gt;From the premises given,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;C is the only valid conclusion.&lt;/strong&gt; One may argue that B is also valid based on the conclusion that because only some apples are edible, then there is a possibility that some apples are inedible; however, an apple being inedible does not guarantee that the apple is rotten. There needs to be an additional premise to make B a valid conclusion. Just because we know B to be true in reality does not make the argument for B valid in this case.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9319" LastActivityDate="2015-02-14T05:49:46.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8933" PostTypeId="2" ParentId="8930" CreationDate="2015-02-14T08:28:15.640" Score="0" Body="&lt;p&gt;As Some apples are edible which implies they are not rotten,which means &lt;strong&gt;SOME APPLES ARE NOT ROTTEN(C)&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9498" LastActivityDate="2015-02-14T08:28:15.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8934" PostTypeId="1" AcceptedAnswerId="9007" CreationDate="2015-02-14T08:54:13.783" Score="6" ViewCount="382" Body="&lt;p&gt;This is a linking puzzle with a weak link. The links should form a circle connecting consecutive items and in the end connecting items 1 and 7 as well:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;A person with the same lastname as a doctor in the Bible.&#xA;  [Hint: Graffiti]&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;1.-2. A fictional master of the whip.&lt;/p&gt;&#xA;  &#xA;  &lt;ol start=&quot;2&quot;&gt;&#xA;  &lt;li&gt;A meteorological phenomenon.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;2.-3. A time of the year.&lt;/p&gt;&#xA;  &#xA;  &lt;ol start=&quot;3&quot;&gt;&#xA;  &lt;li&gt;The Norwegian translation of one of the most common formulas in Excel.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;3.-4. A fascinating love story.&lt;/p&gt;&#xA;  &#xA;  &lt;ol start=&quot;4&quot;&gt;&#xA;  &lt;li&gt;The ship of the egg master.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;4.-5. Clue 4 and 5 share their firstnames.&lt;/p&gt;&#xA;  &#xA;  &lt;ol start=&quot;5&quot;&gt;&#xA;  &lt;li&gt;A German suicide bomber.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;5.-6. An actor with the lastname of a vacation activity. [Hint: tomgun]&lt;/p&gt;&#xA;  &#xA;  &lt;ol start=&quot;6&quot;&gt;&#xA;  &lt;li&gt;The lastname of the founder of a religious science.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;6.-7. Similarity in names.&lt;/p&gt;&#xA;  &#xA;  &lt;ol start=&quot;7&quot;&gt;&#xA;  &lt;li&gt;A binocular.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;7.-1. Looking at differences and quarrels with the binocular.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But there is a weak link here.&lt;br&gt;&#xA;Which of the links is so weak that it can't hold the puzzle together?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9385" LastEditorUserId="8874" LastEditDate="2016-03-17T09:12:57.247" LastActivityDate="2016-03-17T09:14:22.747" Title="Find the weakest link" Tags="|riddle|pattern|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8935" PostTypeId="1" AcceptedAnswerId="8938" CreationDate="2015-02-14T11:18:14.310" Score="4" ViewCount="311" Body="&lt;p&gt;We have a 5x5 grid, on which we can place colored tiles, one by one. The goal is to create a grid with the highest amount of points. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tile colors are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Blue (1 pt)&lt;/li&gt;&#xA;&lt;li&gt;Red (2 pt)&lt;/li&gt;&#xA;&lt;li&gt;Green (3 pt) &lt;/li&gt;&#xA;&lt;li&gt;Yellow (4 pt)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To &lt;strong&gt;place&lt;/strong&gt; a tile they have to follow the following rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Blue: Can be placed anywhere. &lt;/li&gt;&#xA;&lt;li&gt;Red: Has to border atleast one blue tile. &lt;/li&gt;&#xA;&lt;li&gt;Green: Has to border one blue and one red tile. &lt;/li&gt;&#xA;&lt;li&gt;Yellow: Has to border one blue, one red, and one green tile.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Other important factors:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Von_Neumann_neighborhood&quot; rel=&quot;nofollow&quot;&gt;Bordering&lt;/a&gt;, means only the tiles to the nort, east, south and west (no diagonals).&lt;/li&gt;&#xA;&lt;li&gt;Tiles can replace previously placed tiles.&lt;/li&gt;&#xA;&lt;li&gt;We have an infinite amount of tiles.&lt;/li&gt;&#xA;&lt;li&gt;The rules are for &lt;em&gt;placement&lt;/em&gt; only. Once a tile has been successfully placed, it's neigbours don't matter anymore.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the highest score you can get?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4248" LastEditorUserId="4248" LastEditDate="2015-02-14T17:38:39.123" LastActivityDate="2015-02-14T17:38:39.123" Title="Colorful grid: how to place tiles to get the most points?" Tags="|logical-deduction|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8936" PostTypeId="1" AcceptedAnswerId="8950" CreationDate="2015-02-14T11:25:11.533" Score="7" ViewCount="3287" Body="&lt;p&gt;My first rebus puzzle got solved pretty quickly so I decided to make another one.&#xA;This time I tried making it a tad bit harder&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LwFKB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9346" LastEditorUserId="9346" LastEditDate="2015-02-20T13:28:14.057" LastActivityDate="2015-02-20T13:28:14.057" Title="Rebus: I No Money" Tags="|rebus|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="8937" PostTypeId="2" ParentId="8816" CreationDate="2015-02-14T11:59:14.560" Score="0" Body="&lt;p&gt;The challenge posed is to use the form &quot;$x$ op $y$ = $z$&quot; where $x,y,z$ are literals (i.e. actual numbers, not variables) other than $0$ and $1$ to unambiguously represent any aspect of arithmetic on the reals using the 4 basic operations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of establishing the identity of the numbers, it may be better to start by establishing the operators. Since Meelo (this puzzle's author) is satisfied with &quot;saying &lt;em&gt;something&lt;/em&gt;&quot;, this answer just has one example, the addition-multiplication relationship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equations&#xA;  $$3 + 3 = 6$$&#xA;  $$3 + 6 = 9$$&#xA;  $$3 * 3 = 9$$&#xA;establishes the relationship between addition and multiplication, and arguably establishes each operation on its own as well as a by-product. Naturally, the speaker can use equations involving $4$ and so on as further examples. Once these operations are established, the conference can generalize to sums and products of all integers, rationals, and reals, etc on their own. The inverses (subtraction and division) can also follow in a fairly straightforward manner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this is only unambiguous within the operation set $\{+,-,*,/\}$. If exponents were allowed, the same set of equations could still be confused with the multiplication-exponent relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[As an aside, it is arguable that since '3' appears 3 times in the equations with addition, and '3' features prominently in the last equation, there's also the potential of establishing the identity of that integer with this set of equations. Having this single agreed integer then allows ambiguity-free iteration to multiples of 3, as well as 'advanced' discourse of other forms (e.g. $3 + 3 = 2 + 2 + 2$) to introduce other numbers.]&lt;/p&gt;&#xA;" OwnerUserId="9422" LastActivityDate="2015-02-14T11:59:14.560" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8938" PostTypeId="2" ParentId="8935" CreationDate="2015-02-14T12:36:04.850" Score="2" Body="&lt;p&gt;I've got a total of &lt;strong&gt;81&lt;/strong&gt; points using the following layout:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; G Y R Y G &lt;BR&gt;&#xA; Y G Y G Y &lt;BR&gt;&#xA; R Y B Y R &lt;BR&gt;&#xA; Y G Y G Y &lt;BR&gt;&#xA; G Y R Y G &lt;BR&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How? Here is how:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;First fill the corners of the 5x5 grid with &lt;em&gt;GREEN&lt;/em&gt; tiles by using blue and red tiles adjacent to them at appropriate places. Now ignore the blue and red tiles. The grid is now&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; G ? ? ? G &lt;BR&gt;&#xA; ? ? ? ? ? &lt;BR&gt;&#xA; ? ? ? ? ? &lt;BR&gt;&#xA; ? ? ? ? ? &lt;BR&gt;&#xA; G ? ? ? G &lt;BR&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Then, place &lt;em&gt;YELLOW&lt;/em&gt; tiles adjacent to the green tiles by placing blue and red tiles at appropriate places. Ignore the red and blue tiles placed. The grid is now&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; G Y ? Y G &lt;BR&gt;&#xA; Y ? ? ? Y &lt;BR&gt;&#xA; ? ? ? ? ? &lt;BR&gt;&#xA; Y ? ? ? Y &lt;BR&gt;&#xA; G Y ? Y G &lt;BR&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;Now, place a &lt;em&gt;RED&lt;/em&gt; tile between two yellow tiles by placing a blue tile at appropriate places. Ignore the blue tiles. The grid is now&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; G Y R Y G &lt;BR&gt;&#xA; Y ? ? ? Y &lt;BR&gt;&#xA; R ? ? ? R &lt;BR&gt;&#xA; Y ? ? ? Y &lt;BR&gt;&#xA; G Y R Y G &lt;BR&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;Now, place &lt;em&gt;GREEN&lt;/em&gt; tiles adjacent to two yellow tiles by using blue and red tiles at appropriate positions. Ignore the blue and red tiles placed just now. The grid is now&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; G Y R Y G &lt;BR&gt;&#xA; Y G ? G Y &lt;BR&gt;&#xA; R ? ? ? R &lt;BR&gt;&#xA; Y G ? G Y &lt;BR&gt;&#xA; G Y R Y G &lt;BR&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;Finally place a &lt;em&gt;BLUE&lt;/em&gt; tile at the center of the grid and fill the rest of the positions with &lt;em&gt;YELLOW&lt;/em&gt; tiles and the grid will be&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; G Y R Y G &lt;BR&gt;&#xA; Y G Y G Y &lt;BR&gt;&#xA; R Y B Y R &lt;BR&gt;&#xA; Y G Y G Y &lt;BR&gt;&#xA; G Y R Y G &lt;BR&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7482" LastActivityDate="2015-02-14T12:36:04.850" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="8939" PostTypeId="2" ParentId="8926" CreationDate="2015-02-14T13:10:50.757" Score="0" Body="&lt;p&gt;Seems that some people here dont get me properly !&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here is an &lt;b&gt;erronous approach&lt;/b&gt; proposed as a bad solution-exemple since there is a solution with less questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets make $n$ the number of inhabitants .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets say the inquisitor will ask the first $n-1$ population about some dude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There should be a majority of these $n-1$ telling the truth .... So we exclude the minority who says opposed thing as the majority and We mind results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Afterwards , We ask the elite about other different dude and etc until we ask about $n/2$ persons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now .... .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wost case .... is when the first half-1 asked about folks are all foeigners ... and all the questionned guys said about em they r foreigners .... so the remaining $n/2$ member should be native and without any supplementary pain we know that him including the last asked $n/2$ guys are all natives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the number must be in this case = $n-1 + n-2 + ..... + (n/2+1)$ it is an arithmetic sum and it is false !&lt;/p&gt;&#xA;" OwnerUserId="9517" LastEditorUserId="9517" LastEditDate="2015-02-14T17:46:19.800" LastActivityDate="2015-02-14T17:46:19.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8940" PostTypeId="1" AcceptedAnswerId="8946" CreationDate="2015-02-14T14:53:07.297" Score="4" ViewCount="245" Body="&lt;p&gt;Professor Erasmus has returned from his saturday walk in the park. &#xA;He has counted the number of trees in the park and also the number of lines formed by these trees.&#xA;Professor Erasmus claims that&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;there are exactly 26 trees in the park,&lt;/li&gt;&#xA;&lt;li&gt;there are exactly 306 (distinct) lines formed by these trees,&lt;/li&gt;&#xA;&lt;li&gt;and that none of these lines does contain four or more trees.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(When the professor speaks of trees on a line, then he means that the centerpoints of the trees lie on a common line.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We wonder: Has the professor once again made one of his well-known mathematical blunders, or does such a collection of 26 trees indeed exist?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-02-14T20:37:23.457" Title="A stroll in the park" Tags="|mathematics|geometry|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8941" PostTypeId="2" ParentId="8816" CreationDate="2015-02-14T14:55:23.113" Score="3" Body="&lt;p&gt;So you're an alien, and you go to this interplanetary talk, and you see on the board:&#xA;$$w\spadesuit z=y \\&#xA;w\heartsuit w=z \\&#xA;y\clubsuit x=z \\&#xA;x\clubsuit y=w$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You immediately conclude that there is only one possible meaning of the operators:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(\sqrt{\frac 12})-\sqrt{2}=(-\sqrt{\frac 12})$&lt;br&gt;&#xA; $ (\sqrt{\frac 12})+(\sqrt{\frac 12})=\sqrt{2}$&lt;br&gt;&#xA; $(-\sqrt{\frac 12})/(-\frac 12)=\sqrt{2}$&lt;br&gt;&#xA; $(-\frac 12)/(-\sqrt{\frac 12})=(\sqrt{\frac 12})$&#xA; &lt;br&gt;&lt;br&gt;&#xA; Multiplying y, w, and z by -1 produces the only other solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Having identified three of the operators, you can identify the fourth by process of elimination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof that there is only one possible meaning of the operators:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Haha, nope. Did you really want to read 4 paragraphs of casework? I plugged it all into a CAS.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2015-02-14T14:55:23.113" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8943" PostTypeId="1" AcceptedAnswerId="8944" CreationDate="2015-02-14T15:40:25.500" Score="-1" ViewCount="198" Body="&lt;p&gt;What is special about this sentence?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sphinx drove my quick Jag, &quot;Blaze&quot;, FTW.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This riddle is probably too easy for this crowd.  &lt;/p&gt;&#xA;" OwnerUserId="9495" LastEditorUserId="4248" LastEditDate="2015-02-14T21:53:52.613" LastActivityDate="2016-10-23T23:50:28.470" Title="What is special about this sentence?" Tags="|pattern|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8944" PostTypeId="2" ParentId="8943" CreationDate="2015-02-14T15:47:45.073" Score="5" Body="&lt;p&gt;As far as I can make out, it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a pangram.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For those who don't know what this is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pangrams are words or sentences containing every letter of the alphabet at least once; the best known English example being &quot;A quick brown fox jumps over the lazy dog&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9346" LastEditorUserId="5373" LastEditDate="2016-10-23T23:50:28.470" LastActivityDate="2016-10-23T23:50:28.470" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8945" PostTypeId="2" ParentId="8934" CreationDate="2015-02-14T15:48:59.640" Score="1" Body="&lt;p&gt;Seems a bit lame but here is my progress:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1 - Luke&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1-2 - Luke Taylor&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 4 - Fritz Hansen&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 5 - Fritz X&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-02-14T15:48:59.640" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8946" PostTypeId="2" ParentId="8940" CreationDate="2015-02-14T15:58:22.513" Score="5" Body="&lt;p&gt;The maximum possible lines is $\sum 25=\frac{25*26}2=325$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way to reduce the line count is to put a point into a pre-existing line, which reduces exactly $2$ lines. Instead of $3$ lines $AB$, $BC$ and $CA$, we have a single line $ABC$. We are not allowed to add another point to the line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, we can not have an even number of lines, as subtracting $2$s will only result in another odd number. Professor Erasmus has made a blunder.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-02-14T15:58:22.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8947" PostTypeId="2" ParentId="8813" CreationDate="2015-02-14T16:34:13.703" Score="13" Body="&lt;p&gt;I honestly think that the simplest answer is that &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Foo and Bar are both twins, &lt;strong&gt;but not twinned to each other&lt;/strong&gt;. In fact, they aren't related to each other at all.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which means that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Foo and his twin brother Baz could celebrate their birthdays on the 16th in Australia. Bar and his twin brother Qux were born 5 minutes later (but in the UK) meaning that their birthdays are on the 15th.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4852" LastActivityDate="2015-02-14T16:34:13.703" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8948" PostTypeId="2" ParentId="8936" CreationDate="2015-02-14T16:53:34.447" Score="2" Body="&lt;p&gt;Could it be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Small eyes, no ears and money talks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because it looks like a face.&lt;/p&gt;&#xA;" OwnerUserId="9385" LastEditorUserId="9385" LastEditDate="2015-02-17T07:42:28.333" LastActivityDate="2015-02-17T07:42:28.333" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8950" PostTypeId="2" ParentId="8936" CreationDate="2015-02-14T17:41:34.357" Score="20" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think it's: put your money where your mouth is. Because the two eyes represent the eyes and the two no's the nose so the money is placed where there should be your mouth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8980" LastEditorUserId="8980" LastEditDate="2015-02-14T17:48:53.470" LastActivityDate="2015-02-14T17:48:53.470" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8951" PostTypeId="1" AcceptedAnswerId="8952" CreationDate="2015-02-14T17:41:55.050" Score="1" ViewCount="839" Body="&lt;p&gt;Here are four questions on the letters in the names of US states and US state capitals:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What is the only US state that has no letter in common with it's capital city?  &lt;/li&gt;&#xA;&lt;li&gt;What is the only US state capital that doesn't contain a letter from the word MACKEREL? &lt;/li&gt;&#xA;&lt;li&gt;What is the only US state that doesn't contain a letter from the word MACKEREL? &lt;/li&gt;&#xA;&lt;li&gt;What is the only US state that doesn't contain a letter from the word COMMUNIST? &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="9125" LastEditDate="2015-02-14T17:55:59.393" LastActivityDate="2015-02-14T18:03:20.673" Title="Four letter puzzles on US states" Tags="|word|geography|knowledge|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8952" PostTypeId="2" ParentId="8951" CreationDate="2015-02-14T17:47:45.117" Score="2" Body="&lt;p&gt;Just had to look at a list of states and capitals to figure this out.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. South Dakota&lt;br&gt;&#xA; 2. Boston&lt;br&gt;&#xA; 3. Ohio&lt;br&gt;&#xA; 4. Delaware  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9159" LastEditorUserId="8874" LastEditDate="2015-02-14T18:03:20.673" LastActivityDate="2015-02-14T18:03:20.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8953" PostTypeId="1" AcceptedAnswerId="9487" CreationDate="2015-02-14T17:54:45.640" Score="7" ViewCount="515" Body="&lt;p&gt;I have partitioned the list of US states into two groups based on a certain property that is shared by 32 states but not shared by the remaining 18 states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 32 states that share this property:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;?????, ?????, California, Georgia, Hawaii, Idaho, Indiana, Louisiana, &#xA;  Maryland, Massachusetts, Montana, Nebraska, Nevada, New Hampshire, &#xA;  New Jersey, New Mexico, New York, North Carolina, North Dakota, Oklahoma, &#xA;  Oregon, Pennsylvania, Rhode Island, South Carolina, South Dakota, &#xA;  Texas, Utah, Vermont, Virginia, Washington, West Virginia, Wyoming.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The 18 states that do not have the property:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;?????, ?????, Colorado, Connecticut, Delaware, &#xA;  Florida, Illinois, Iowa, Kansas, Kentucky, Maine, Michigan, &#xA;  Minnesota, Mississippi, Missouri, Ohio, Tennessee, Wisconsin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What property did I use for my partition?&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;To which parts do Alabama, Alaska, Arizona and Arkansas belong?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="9125" LastEditDate="2015-02-19T16:22:42.093" LastActivityDate="2015-02-23T13:03:26.307" Title="An enigmatic partition of the US states" Tags="|pattern|geography|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8954" PostTypeId="1" AcceptedAnswerId="8955" CreationDate="2015-02-14T19:29:10.253" Score="4" ViewCount="5672" Body="&lt;p&gt;Solve the following rebus puzzle:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   ONE ANOTHER&#xA;   ONE ANOTHER&#xA;   ONE ANOTHER&#xA;   ONE ANOTHER&#xA;   ONE ANOTHER&#xA;   ONE ANOTHER&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;See also &#xA;(&lt;a href=&quot;https://puzzling.stackexchange.com/questions/8709/rebus-with-dracula&quot;&gt;Rebus with Dracula&lt;/a&gt;)  and &#xA;(&lt;a href=&quot;https://puzzling.stackexchange.com/questions/8750/another-rebus-puzzle&quot;&gt;Another Rebus puzzle&lt;/a&gt;) and&#xA;(&lt;a href=&quot;https://puzzling.stackexchange.com/questions/8796/my-third-rebus-puzzle&quot;&gt;My third Rebus puzzle&lt;/a&gt;) and&#xA;(&lt;a href=&quot;https://puzzling.stackexchange.com/questions/8903/my-fourth-rebus-puzzle&quot;&gt;My fourth Rebus puzzle&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2022-12-20T17:14:45.363" Title="One Another - Rebus puzzle" Tags="|rebus|english|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8955" PostTypeId="2" ParentId="8954" CreationDate="2015-02-14T19:32:12.113" Score="5" Body="&lt;p&gt;That one must be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Six times ONE + six times ANOTHER&quot; = &lt;strong&gt;&quot;Six of one, half a dozen of another&quot;&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-14T20:05:56.993" LastActivityDate="2015-02-14T20:05:56.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8957" PostTypeId="1" AcceptedAnswerId="8963" CreationDate="2015-02-14T19:44:53.440" Score="7" ViewCount="791" Body="&lt;p&gt;I love it green, but I inherited a blue one from my father, who now fancies red instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My father did not have a father himself, and my sister loves to do solo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;" OwnerUserId="9385" LastEditorUserId="5373" LastEditDate="2015-02-15T10:41:23.143" LastActivityDate="2015-12-21T06:12:34.370" Title="Who in the universe am I?" Tags="|riddle|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8958" PostTypeId="2" ParentId="8914" CreationDate="2015-02-14T19:50:23.880" Score="2" Body="&lt;p&gt;First off .... if the problem is unknowingly formed , it means it is open for variety of solutions where this one takes part .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We must know before that if k is small at some point of 2 or 3 or even bigger the game would be endless .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;k will be chosen according to how far we can advance in tree-expanding ....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting by 00001 - this option forces each level to be less developped by the opponent in order to restrict his choices .... but i found out that the opponent will be tending to win the game whatever the first player s choice is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So i began with 00011 to see where it leads ... and i concluded that if k supposed to be sufficiently big , k= 21 or 17 , this choice is defeating !&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the following figure illustrates my solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We represent binary sequence by positions of 1 digits in the number ex: 00001 = 1 , 00101 = 31&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Red colored numbers mean this choice isnt developable , green numbers means this choice is developable but not used (yet) in tree expanding.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/6S0GS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9517" LastEditorUserId="9517" LastEditDate="2015-02-14T22:28:07.363" LastActivityDate="2015-02-14T22:28:07.363" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8960" PostTypeId="2" ParentId="8940" CreationDate="2015-02-14T20:37:23.457" Score="0" Body="&lt;p&gt;Expanding on the previous answer - the 325 is right but the second paragraph is wrong - using the 325 lines, you could merge two lines into one - reducing the total amount by 1. Such merger need to be carried out 19 times. By selecting 2 points out of the 26 it seems that you could select 19 points of the remaining 24 to create 19 mergers.&lt;/p&gt;&#xA;" OwnerUserId="9531" LastActivityDate="2015-02-14T20:37:23.457" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8962" PostTypeId="2" ParentId="313" CreationDate="2015-02-14T23:24:03.540" Score="8" Body="&lt;p&gt;I would say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Considering that I am also thinking of a number between 1 and 2, is you number higher than mine?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1-&amp;gt; No&#xA;2-&amp;gt; Maybe&#xA;3-&amp;gt; Yes&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="9413" LastActivityDate="2015-02-14T23:24:03.540" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8963" PostTypeId="2" ParentId="8957" CreationDate="2015-02-14T23:50:18.127" Score="11" Body="&lt;p&gt;I am&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Luke Skywalker.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I like green&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yoda&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I inherited a blue &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lightsaber&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;from my father&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Anakin (who is fatherless)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;who prefers a red one since&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; he became Darth Vader.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My sister is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Leia&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;who likes to do&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Han Solo.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2021" LastActivityDate="2015-02-14T23:50:18.127" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8964" PostTypeId="1" AcceptedAnswerId="8966" CreationDate="2015-02-15T00:53:09.783" Score="-4" ViewCount="1566" Body="&lt;p&gt;What piece of furniture can any mammal make, without tools?  &lt;/p&gt;&#xA;" OwnerUserId="9495" LastEditorUserId="8874" LastEditDate="2015-10-11T13:13:59.893" LastActivityDate="2015-10-11T13:13:59.893" Title="What piece of furniture?" Tags="|riddle|" AnswerCount="1" CommentCount="5" ClosedDate="2015-02-16T10:08:19.523" ContentLicense="CC BY-SA 3.0" />
  <row Id="8965" PostTypeId="2" ParentId="8813" CreationDate="2015-02-15T01:05:03.200" Score="0" Body="&lt;p&gt;Bar was positioned within his mother's womb to be born first. Foo was positioned in his mother's womb to be born second. Due to complications, a caesarian had to be performed in order to safely bring the twins into the world. Foo was removed from the womb first, Bar was removed 5 minutes later. Because of time honored tradition, the parents wanted the birthdays to be celebrated in the &quot;naturally intended&quot; order. Therefore, Bar's birth is celebrated on March 15th and Foo's on the 16th.&lt;/p&gt;&#xA;" OwnerUserId="9105" LastActivityDate="2015-02-15T01:05:03.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8966" PostTypeId="2" ParentId="8964" CreationDate="2015-02-15T01:30:45.613" Score="0" Body="&lt;p&gt;Would it be a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; stool ? (as in the result of defecation)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5874" LastActivityDate="2015-02-15T01:30:45.613" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8967" PostTypeId="2" ParentId="3772" CreationDate="2015-02-15T01:37:52.387" Score="3" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Thought &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is invisible to the eye&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You cannot see thought.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yet it is a stranger to no one&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Even a strange thought is not a stranger&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;it weighs less than air&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; of course&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;it is sharper than a keen knife, word or blade&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; thought can pierce the vastness of space, or the microcosm of an atom which a blade cannot do&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;it comes from nothing&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  who can truly say where a thought comes from?  some theories speculate that even thoughts we consider our own, might originate from the unknown (nothing)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but it has the power to vanquish armies&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; indeed, without thought, one cannot vanquish an army.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9495" LastActivityDate="2015-02-15T01:37:52.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8968" PostTypeId="1" AcceptedAnswerId="8975" CreationDate="2015-02-15T01:55:21.780" Score="3" ViewCount="2117" Body="&lt;p&gt;We are twins&lt;br&gt;&#xA;We are the gateway of silence&lt;br&gt;&#xA;We enrich your world, but you can live without us.&lt;br&gt;&#xA;We are your friends, but sometimes we can be loud when you don't want us to be.&lt;br&gt;&#xA;Sometimes we annoy you, but you will never want to lose us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who or what are we?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;New hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sometimes we can be blocked, but that is only temporary.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another new hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We are usually dry but sometimes we flood.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3rd new hint (this hint almost gives it away, but I can't resist):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some children think we provide food, but we don't.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I will give a new hint for every wrong answer.&lt;/p&gt;&#xA;" OwnerUserId="9495" LastEditorUserId="8874" LastEditDate="2015-10-11T13:08:12.640" LastActivityDate="2017-11-04T10:16:11.803" Title="We are twins. Who are we?" Tags="|riddle|" AnswerCount="7" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8970" PostTypeId="2" ParentId="8968" CreationDate="2015-02-15T05:48:46.817" Score="7" Body="&lt;p&gt;I don't know if this is right, but it fits the hints.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your ears are twins.&lt;br&gt;&#xA; They allow you to detect silence.&lt;br&gt;&#xA; They enrich your world (music!) but you can live without them.&lt;br&gt;&#xA; If you have tinnitus, they can be loud when you don't want.&lt;br&gt;&#xA; They annoy you if you are in hearing range of Nickelback, but you don't want to lose them.&lt;br&gt;&#xA; You can put your fingers in your ears, but not forever.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6031" LastActivityDate="2015-02-15T05:48:46.817" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8971" PostTypeId="2" ParentId="4480" CreationDate="2015-02-15T06:01:07.970" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chess Set&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pick a side and then relax.&#xA;It might take an hour or two&#xA;If you throw it, then it cracks&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chess Set can crack open, scattering pieces in the ground&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but it puzzles, why would you?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You know how to take the puzzled pieces and assemble them on the board&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is one, but has many&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many pieces inside the chess set, after it is cracked open.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That it's old, you always knew&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chess is a very old game&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Three Six Nine are the numbers..&#xA;Ninety also, for the slumbers&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; not sure about these clues..  could be reference to years in history of chess&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9495" LastActivityDate="2015-02-15T06:01:07.970" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8973" PostTypeId="2" ParentId="8968" CreationDate="2015-02-15T06:13:43.217" Score="0" Body="&lt;p&gt;We are twins right and left&#xA;We are the gateway of silence doesnt normally make noise&#xA;We enrich your world, but you can live without us.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If none u can be operated to open it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are your friends, but sometimes we can be loud when you don't want us to be.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For friends dont wanna say it but loud farts in front of people.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes we annoy you, but you will never want to lose us.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; opps need to go cr but never lose them cuz its weird.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes we can be blockaded, but that is only temporary.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  hold ur poop.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are usually dry but sometimes we flood.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; poop is dry   diarrhea is wer flood.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;ANSWER IS:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BUTT BOTTOCKS RIGHT AND LEFT&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9535" LastEditorUserId="5373" LastEditDate="2017-11-04T10:16:11.803" LastActivityDate="2017-11-04T10:16:11.803" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8974" PostTypeId="2" ParentId="8968" CreationDate="2015-02-15T06:18:06.270" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think the answer is the vocal folds in our neck. They are gateway to our silence. We can live without them. They sometimes make a louder sound than we want to. They sometimes get blocked.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8980" LastActivityDate="2015-02-15T06:18:06.270" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8975" PostTypeId="2" ParentId="8968" CreationDate="2015-02-15T06:22:25.223" Score="8" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Two nostrils.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are twins &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We have two of them&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are the gateway of silence &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You usually have your mouth closed when you breathe with your nose, not speaking.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We enrich your world, but you can live without us.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We smell good aroma with them, but living without nose is possible &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are your friends, but sometimes we can be loud when you don't want us to be. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sometimes we sneeze and make a very loud sound&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes we annoy you, but you will never want to lose us. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sometimes we get itchy, tingling feelings inside them that annoy us&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hints:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We can block them but, however, only for few moments and sometimes we have a cold that makes our nostrils runny with mucus. Some kids have a habit of eating solid mucus of their own!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9174" LastEditorUserId="9495" LastEditDate="2015-02-15T11:57:51.380" LastActivityDate="2015-02-15T11:57:51.380" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8976" PostTypeId="2" ParentId="8968" CreationDate="2015-02-15T06:24:10.163" Score="3" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The nostrils&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are twins  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are 2 nostrils.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are the gateway of silence  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nostrils usually don't make any noise.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We enrich your world, but you can live without us.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pleasant smells enrich our world, but we can live without the sense of smell.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are your friends, but sometimes we can be loud when you don't want us to be. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you have a cold, you may sniffle a lot.  Also, nostril problems can cause snoring.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes we annoy you, but you will never want to lose us.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sometimes dried mucus in the nostrils can be annoying.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes we can be blockaded, but that is only temporary.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you have a cold, you may have nasal congestion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are usually dry but sometimes we flood.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your nose may be runny if you have a cold, but your nostrils are usually dry otherwise.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2015-02-15T06:24:10.163" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8977" PostTypeId="1" AcceptedAnswerId="8982" CreationDate="2015-02-15T07:02:06.500" Score="3" ViewCount="281" Body="&lt;p&gt;Here are four photos from Google satellite maps and one from Wikipedia (see note 1).  These photos show a family &lt;strong&gt;related by strength and endurance&lt;/strong&gt;.  The goal of this puzzle is:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;to identify the proper name of the each of the five family members&lt;/li&gt;&#xA;&lt;li&gt;then to apply the book cipher method to identify the sixth member of this family&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;strong&gt;Hints:&lt;/strong&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;understanding the common characteristic is important&lt;/li&gt;&#xA;&lt;li&gt;there is no hidden information in the photos&lt;/li&gt;&#xA;&lt;li&gt;there are no legumes in the answers&lt;/li&gt;&#xA;&lt;li&gt;partial answers are OK&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;sup&gt; Note 1 - Photos of subject 3 are restricted by copyright so a photo of a model is shown below.  &lt;/sup&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/F2egT.jpg&quot; alt=&quot;First three&quot; /&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/5IiF3.jpg&quot; alt=&quot;Second three&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-10-11T12:47:13.623" Title="Our Honor is Unstained" Tags="|cipher|visual|knowledge|geography|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8978" PostTypeId="2" ParentId="8968" CreationDate="2015-02-15T07:16:42.927" Score="0" Body="&lt;p&gt;I think that it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ears&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They enrich our world, can be noisy when we don't want them to be (they recieve all noise so everything noisy has to come through them) and though you can get annoyed by noises you'd never want to loose your ears. Even to hear silence, you must have ears (thus gateways).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;P.S. I checked the hints. I am absolutely correct. The only other candidate are nostrils, which don't make sense as they are not the gateways of silence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the internets please.&lt;/p&gt;&#xA;" OwnerUserId="9536" LastEditorUserId="9038" LastEditDate="2015-02-15T10:25:41.350" LastActivityDate="2015-02-15T10:25:41.350" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8979" PostTypeId="1" AcceptedAnswerId="8985" CreationDate="2015-02-15T07:40:46.770" Score="6" ViewCount="1072" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Naked, I make blue clothing that I never wear&lt;br&gt;&#xA;  Clothed, I am cold and shy&lt;br&gt;&#xA;  People sell in shops what I give for free&lt;br&gt;&#xA;  People travel far to receive my gifts.&lt;br&gt;&#xA;  People travel far to avoid my gifts.&lt;br&gt;&#xA;  Many love me and many shun me.&lt;br&gt;&#xA;  No one truly visits me&lt;br&gt;&#xA;  But I am not lonely&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who or what am I?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9495" LastEditorUserId="52483" LastEditDate="2018-11-22T20:59:01.707" LastActivityDate="2018-11-22T20:59:01.707" Title="Naked, I make blue clothing - Riddle" Tags="|riddle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8980" PostTypeId="2" ParentId="8979" CreationDate="2015-02-15T07:45:47.677" Score="2" Body="&lt;p&gt;My first thought would be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Something about blue and cold... And you can buy it in shops even it is free. Some will travel to it, other will want to avoid it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9385" LastEditorUserId="9385" LastEditDate="2015-02-15T20:49:26.157" LastActivityDate="2015-02-15T20:49:26.157" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8981" PostTypeId="1" AcceptedAnswerId="8993" CreationDate="2015-02-15T07:46:11.560" Score="20" ViewCount="1499" Body="&lt;p&gt;It's a hot day. You are in the middle of a flat sandy plane that stretches as far as you can see. Your hands are shackled and you are surrounded by soldiers of the mighty Pythagorean Brotherhood. Ten steps from you a man, his hands also in shackles, is standing frozen. His name is Hippasus of Metapontum. A Pythagorean soldier yells at Hippasus. &lt;em&gt;&quot;Right-angle turn and move!&quot;&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hippasus makes a 90° left turn and hesitates. A drop of sweat slowly moves down his front head. He then makes a step followed by a 90° right turn. He again hesitates. Another step, and again a 90° left turn. The soldier raises his sword and shouts &lt;em&gt;&quot;Death to the irrational!&quot;&lt;/em&gt; Hippasus' body collapses and his head rolls in your direction, coming to a halt just two steps away from your feet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The soldier turns to you. &lt;em&gt;&quot;Now you go. Make a right angle turn followed by a nonzero number of straight steps in forward direction, and keep repeating this. Each step needs to be of exactly the same size. Each turn needs to be an exact right angle turn and must happen at a position that is a whole number of straight-line steps away from all previous turns. You need to make a total of 100 right angle turns, 50 to the left and 50 to the right. If you fail, you will be killed on the spot, if you succeed, you are a free man, and a witness of the fact that integer distances rule the world. Go!&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You think for a few seconds, and make a left turn. What's next?&lt;/p&gt;&#xA;" OwnerUserId="7573" LastEditorUserId="7573" LastEditDate="2015-02-15T15:49:16.773" LastActivityDate="2023-07-19T01:57:39.307" Title="Pythagorean walk" Tags="|mathematics|geometry|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8982" PostTypeId="2" ParentId="8977" CreationDate="2015-02-15T08:02:46.797" Score="4" Body="&lt;p&gt;The proper names are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. Gateway Arch (in St.Louis)&#xA;&lt;P&gt;&#xA; 2. United States Air Force Memorial (in Arlington, VA)&#xA;&lt;P&gt;&#xA; 3. Atomium (in Brussels)&#xA;&lt;P&gt;&#xA; 4. Cloud Gate (in Chicago)&#xA;&lt;P&gt;&#xA; 5. The Kelpies (at The Helix in Scotland)&#xA;&lt;P&gt;&#xA; 6. Unisphere (in Flushing Meadows, NY) - credit to Len for this one!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The common characteristic of these is that they are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; made out of stainless steel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The book cipher is described as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It encodes the name of subject 6 by using the names of the first 5 subjects.  The first number X in parentheses represents the name of subject X, and the second number Y represents the Yth letter from the end of the name.&#xA;&lt;P&gt;&#xA; So we have:&#xA;&lt;P&gt;&#xA;* (4, -6) = the 6th letter from the end of &quot;Cloud Gate&quot; = U&lt;BR&gt;&#xA;* N&lt;BR&gt;&#xA;* (3, -3) = the 3rd letter from the end of &quot;Atomium&quot; = I&lt;BR&gt;&#xA;* (5, -1) = the 1st letter from the end of &quot;The Kelpies&quot; = S&lt;BR&gt;&#xA;* (5, -4) = the 4th letter from the end of &quot;The Kelpies&quot; = P&lt;BR&gt;&#xA;* (1, -1) = the 1st letter from the end of &quot;Gateway Arch&quot; = H&lt;BR&gt;&#xA;* (1, -8) = the 8th letter from the end of &quot;Gateway Arch&quot; = E&lt;BR&gt;&#xA;* (2, -4) = the 4th letter from the end of &quot;United States Air Force Memorial&quot; = R&lt;BR&gt;&#xA;* (2, -7) = the 7th letter from the end of &quot;United States Air Force Memorial&quot; = E&lt;BR&gt;&#xA;&lt;P&gt;&#xA;That spells out Unisphere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4450" LastEditDate="2015-02-17T02:51:06.813" LastActivityDate="2015-02-17T02:51:06.813" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8983" PostTypeId="1" AcceptedAnswerId="8991" CreationDate="2015-02-15T08:26:48.467" Score="13" ViewCount="3223" Body="&lt;p&gt;In many people, I am not the first thing you want to have.&lt;br&gt;&#xA;Never would you want to give me to your family or friends.&lt;br&gt;&#xA;Sadly, I am often inevitable.&lt;br&gt;&#xA;Although, some purport to have a cure for me, but mostly they're wrong or lying.&lt;br&gt;&#xA;Next generation medical science might help cure me with DNA technology.&lt;br&gt;&#xA;I am easy to find, if you know where to look.&lt;br&gt;&#xA;Trust me.  I am not what you want.&lt;br&gt;&#xA;You should run.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;hint&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No hints on this because the solution is staring you in the face.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9495" LastEditorUserId="5188" LastEditDate="2015-02-16T20:33:07.597" LastActivityDate="2018-03-30T18:47:14.327" Title="You should run away. (Who am I)" Tags="|riddle|" AnswerCount="7" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8984" PostTypeId="2" ParentId="8983" CreationDate="2015-02-15T08:57:18.760" Score="9" Body="&lt;p&gt;In many people, I am not the first thing you want to have.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many people we come across are bald.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Never would you want to give me to your family or friends.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; we don't want our family members of friends to be bald&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sadly, I am often inevitable.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; aging often brings baldness&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Although, some purport to have a cure for me, but mostly they're wrong or lying.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;there is no cure for baldness&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Next generation medical science might help cure me with DNA technology.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;may be tweaking DNA will help grow hair&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am easy to find, if you know where to look.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;we look at head for baldness&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Trust me.  I am not what you want.&#xA;You should run.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;No one wants a bald head&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9174" LastEditorDisplayName="user9174" LastEditDate="2015-02-15T11:40:25.857" LastActivityDate="2015-02-15T11:40:25.857" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8985" PostTypeId="2" ParentId="8979" CreationDate="2015-02-15T08:59:48.980" Score="7" Body="&lt;p&gt;This is the solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Naked, I make blue clothing that I never wear&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Direct sun light is bleaching clothes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clothed, I am cold and shy&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clothed or covered by clouds, the warmth of the sun is not reaching us. It is also shy, because we can't see it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;People sell in shops what I give for free&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Spray tan&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;People travel far to receive my gifts.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Holiday in the sun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;People travel far to avoid my gifts.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Escaping the heating sun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Many love me and many shun me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sun is loved by many and hated by others.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No one truly visits me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No man has been on the sun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I am not lonely&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many suns (other stars)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9385" LastEditorUserId="9385" LastEditDate="2015-02-15T17:29:01.813" LastActivityDate="2015-02-15T17:29:01.813" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8986" PostTypeId="1" AcceptedAnswerId="9238" CreationDate="2015-02-15T09:37:29.607" Score="-2" ViewCount="2126" Body="&lt;p&gt;&lt;strong&gt;This problem is known as &lt;em&gt;n&lt;/em&gt; queens puzzle:&lt;/strong&gt;  How can we distribute &lt;em&gt;n&lt;/em&gt; queens on a chess grid of &lt;em&gt;n&lt;/em&gt; $\times$ &lt;em&gt;n&lt;/em&gt; so that no queen can threaten another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many solutions are possible for &lt;em&gt;n&lt;/em&gt; &gt; 4.  To answer this question, show a single solution for n from 4 to 10 as well as an algorithm for where to place the queens.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Here is a solution when &lt;em&gt;n&lt;/em&gt; = 8&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://p0.storage.canalblog.com/00/34/721494/81003520_o.jpg&quot; alt=&quot;8 x 8 solution&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9517" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:48.747" LastActivityDate="2015-02-21T10:55:36.833" Title="Placing n Queens on n x n board" Tags="|chess|" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="8987" PostTypeId="2" ParentId="8981" CreationDate="2015-02-15T09:40:55.857" Score="8" Body="&lt;p&gt;No proper solution yet, but here are 2 degenerate solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trivial solution: take zero (which is a whole number of) steps on each turn, turning alternately left and right 50 times each. All points are co-located, so distance between any pair of points is 0, which is a whole number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solution with a single zero-step: turn left 50 times, taking alternately 3 and 4 steps (going anticlockwise in the same 3x4 rectangle - the ends of each diagonal are 5 steps away from each other). Turn right, take zero steps, then turn right again 49 times, retracing the same rectangle as before, going clockwise this time.&lt;/p&gt;&#xA;" OwnerUserId="9422" LastActivityDate="2015-02-15T09:40:55.857" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8989" PostTypeId="2" ParentId="8983" CreationDate="2015-02-15T10:58:06.527" Score="2" Body="&lt;p&gt;It is best to wait to get old to get wrinkles/&#xA;Nobody wants their friends to wrinkle too soon/&#xA;We all grow old and get wrinkles/&#xA;Many people claim to have youth elixirs to prevent wrinkles but nobody does/&#xA;Maybe in the future we can manipulate DNA to stop the aging process/&#xA;wrinkles are obvious. Look at  the face.&lt;/p&gt;&#xA;" OwnerUserId="9537" LastActivityDate="2015-02-15T10:58:06.527" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8991" PostTypeId="2" ParentId="8983" CreationDate="2015-02-15T11:42:03.527" Score="17" Body="&lt;p&gt;Answer: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's an acrostic riddle: INSANITY.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9422" LastEditorUserId="8874" LastEditDate="2015-02-15T11:57:32.363" LastActivityDate="2015-02-15T11:57:32.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8992" PostTypeId="1" AcceptedAnswerId="9205" CreationDate="2015-02-15T12:30:24.740" Score="7" ViewCount="551" Body="&lt;p&gt;The Dinner club has 169 members, and professor Erasmus is one of them.&#xA;Every evening, four of the members meet and have dinner together in a fancy restaurant. &#xA;Professor Erasmus claims that today is a very special day for the club: after today's dinner, every two club members will have had a club dinner together exactly once. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We wonder: Is this indeed possible, or has the professor once again made one of his well-known mathematical blunders and miscalculated the days and dinners? &lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-02-19T04:07:05.270" Title="The 169 members of the Dinner Club" Tags="|mathematics|logical-deduction|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8993" PostTypeId="2" ParentId="8981" CreationDate="2015-02-15T12:43:37.780" Score="20" Body="&lt;p&gt;I'm going to use compass directions to reduce confusion.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are going to walk in a figure eight, with a width of $c$ all the way up, and with north and south loops of height $a$ and $b$. You start at the south east corner of the figure eight, facing east.&lt;br&gt;&#xA; You turn left and walk $b$ steps north, then turn left again and walk $c$ steps west across what will be the middle bar of the figure eight. You are now in the middle of the west side, facing west.&lt;br&gt;&#xA; From here you can make four left turns around the south loop, walking $b$ south then $c$ east then $b$ north then $c$ west. Or you can make four right turns around the north loop, walking $a$ north then $c$ east then $a$ south then $c$ west. Each one puts you back at the same spot, facing west.&#xA; Twelve of each loop gives you 48 right turns and 48 left turns. You started with two left turns, so now you finish with two right turns half-way around the north loop and end up in the north east corner. That's 50 each and 100 total.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now all we need to do is choose some numbers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We must have $a,c$ form a pythagorean triangle; $b,c$ must also; and so must $a+b,c$. That covers all 6 diagonals of the figure eight, and the rest are integer sides and the sums of integers. Then every corner of the figure eight is an integer distance from the  others.&lt;br&gt;&#xA; It feels like overkill to pull this out again, but a=44, b=117, c=240 will do the job. That's 16025 steps; I recommend taking very small steps.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;s&gt;An immediate upvote to anyone who finds smaller numbers, because it is hot out here.&lt;/s&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You announce your intentions to the soldiers, hinting that the quicker you finish, the quicker everyone can get into the shade. Being the Pythagorean Guard, of course they are able to help; one calling himself @crazyiman tells you to use:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a = 11, b = 80, c = 60.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This only requires 5275 steps. You refrain from announcing exactly how much faster this will be, and start walking.&lt;/p&gt;&#xA;" OwnerUserId="6031" LastEditorUserId="6031" LastEditDate="2015-02-15T20:53:16.020" LastActivityDate="2015-02-15T20:53:16.020" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8994" PostTypeId="1" AcceptedAnswerId="8997" CreationDate="2015-02-15T13:09:55.743" Score="2" ViewCount="1220" Body="&lt;p&gt;Solve the following rebus puzzle:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    __________________________________&#xA;   |                                  |&#xA;   |                        Customer  |&#xA;   |                        CUSTOMER  |&#xA;   |                        customer  |&#xA;   |                        CuStOmEr  |&#xA;   |                        cUsToMeR  |&#xA;   |__________________________________|&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="5188" LastEditDate="2015-02-16T20:58:13.290" LastActivityDate="2015-02-16T20:58:13.290" Title="The Customers - Rebus puzzle" Tags="|rebus|english|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8995" PostTypeId="1" AcceptedAnswerId="9002" CreationDate="2015-02-15T13:21:42.983" Score="17" ViewCount="2131" Body="&lt;p&gt;A German car, a place to stay, a country in the Northern hemisphere,&#xA;a beautiful girl, a unit prefix, a city in the Southern hemisphere, &#xA;a green ranger, a rainy month, and a boy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the boy's name?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-08T10:26:24.283" LastActivityDate="2016-03-08T10:26:24.283" Title="What's the boy's name?" Tags="|riddle|word|pattern|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8996" PostTypeId="1" AcceptedAnswerId="8998" CreationDate="2015-02-15T13:26:47.153" Score="5" ViewCount="1417" Body="&lt;p&gt;Alone I stand between the 23rd and the 25th,&lt;br&gt;&#xA;together with a friend I am 20,&lt;br&gt;&#xA;and if another friend joins in, we will be dirty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="9404" LastEditDate="2015-02-15T13:46:59.017" LastActivityDate="2015-02-15T13:49:12.100" Title="Alone, together, and with friends" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8997" PostTypeId="2" ParentId="8994" CreationDate="2015-02-15T13:27:33.223" Score="7" Body="&lt;p&gt;Maybe..&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The customers are always right&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6205" LastActivityDate="2015-02-15T13:27:33.223" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8998" PostTypeId="2" ParentId="8996" CreationDate="2015-02-15T13:49:12.100" Score="9" Body="&lt;p&gt;I think it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter X&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; X is the 24th letter of the alphabet&lt;br&gt;&#xA; XX means 20 in Roman numerals&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The only thing I'm not 100% sure of is the &quot;dirty&quot; part&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I guess XXX can bring to mind... activities which may not be exactly clean?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6205" LastActivityDate="2015-02-15T13:49:12.100" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8999" PostTypeId="2" ParentId="8994" CreationDate="2015-02-15T14:12:31.420" Score="2" Body="&lt;p&gt;In addition to luxmi's answer following can be said too&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Each customer is different&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9174" LastActivityDate="2015-02-15T14:12:31.420" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9000" PostTypeId="1" AcceptedAnswerId="9047" CreationDate="2015-02-15T14:31:02.750" Score="14" ViewCount="5995" Body="&lt;p&gt;Solve the following ten one-line Rebus puzzles:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;AGE BUT&lt;/li&gt;&#xA;&lt;li&gt;marri Ag e&lt;/li&gt;&#xA;&lt;li&gt;wWoOlOfL&lt;/li&gt;&#xA;&lt;li&gt;&quot;24 Hours&quot;&lt;/li&gt;&#xA;&lt;li&gt;TRADI$~$ TION&lt;/li&gt;&#xA;&lt;li&gt;HEAD ACHE &lt;/li&gt;&#xA;&lt;li&gt;WHEATHER&lt;/li&gt;&#xA;&lt;li&gt;SGEG&lt;/li&gt;&#xA;&lt;li&gt;TOSYOBSYOBWN&lt;/li&gt;&#xA;&lt;li&gt;ALL world&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="8874" LastEditDate="2015-10-11T12:59:35.863" LastActivityDate="2016-03-04T13:27:59.420" Title="An abundance of one-line Rebus puzzles" Tags="|rebus|english|" AnswerCount="6" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9001" PostTypeId="2" ParentId="9000" CreationDate="2015-02-15T14:35:55.233" Score="10" Body="&lt;p&gt;The answers for 2, 3, 8, and 9:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Ag=silver with marriage&quot; = &lt;strong&gt;&quot;Silver wedding anniversary&quot;&lt;/strong&gt;   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;wolf in WOOL&quot; = &lt;strong&gt;&quot;Wolf in sheep's clothing&quot;&lt;/strong&gt;   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;8:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;SGEG = permutation of EGGS&quot; = &lt;strong&gt;&quot;Scrambled eggs&quot;&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;9:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;SYOBSYOB = BOYS BOYS backwards, inside TOWN&quot; = &lt;strong&gt;&quot;The Boys are Back in Town&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the answer for 7 in collaboration with Hellion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;badly spelled version of WEATHER&quot; = &lt;strong&gt;&quot;Spell of bad weather&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and perhaps an answer for &lt;/p&gt;&#xA;&#xA;&lt;p&gt;5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;a break in TRADITION&quot; = &lt;strong&gt;&quot;Break with tradition&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-15T16:24:26.097" LastActivityDate="2015-02-15T16:24:26.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9002" PostTypeId="2" ParentId="8995" CreationDate="2015-02-15T14:37:55.033" Score="21" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oscar&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reason is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The phonetic alphabet in alphabetic order.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A German car&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Golf&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A place to stay&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hotel&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A country in the Northern hemisphere&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; India&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A beautiful girl&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Juliett&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A unit prefix&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Kilo&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A city in the Southern hemisphere&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lima&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A green ranger&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mike&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A rainy month&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; November&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A boy&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oscar&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9385" LastEditorUserId="9385" LastEditDate="2015-02-15T20:04:36.600" LastActivityDate="2015-02-15T20:04:36.600" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9003" PostTypeId="2" ParentId="9000" CreationDate="2015-02-15T14:41:49.957" Score="6" Body="&lt;p&gt;2&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;uneven marriage&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All day packed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; break the tradition&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; splitting headache&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9174" LastActivityDate="2015-02-15T14:41:49.957" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9004" PostTypeId="2" ParentId="8813" CreationDate="2015-02-15T14:49:24.863" Score="1" Body="&lt;p&gt;They don't have to have been born on either side of the date line. All that is required is for one of the twins to have since moved to the other side of the date line from where they were born. For example they might have been born on the 16th and the younger one is celebrating on the 15th because &quot;it's already the 16th where I was born&quot; or they might have been born on the 15th and the older one is celebrating on the 16th because &quot;it's finally the 15th where I was born.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, they don't need to move, just live somewhere that has moved to the other side of the dateline, such as Samoa, and one of them use the calendar date of their birth even though the country skipped a day in the switch, and the other insist on waiting a day until the birthday is truly reached. One of my children corrected any &quot;Happy Birthday&quot;s that were given before the time of day of the birth, saying &quot;I haven't turned [whatever] yet&quot; until that time rolled around. I can see a similar approach applying in a dateline switch situation.&lt;/p&gt;&#xA;" OwnerUserId="1509" LastActivityDate="2015-02-15T14:49:24.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9005" PostTypeId="2" ParentId="9000" CreationDate="2015-02-15T14:53:47.350" Score="11" Body="&lt;p&gt;1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Age before Beauty (&quot;B, U, T&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;7: (just a guess here)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Bad spell of weather&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;10:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It's a small world after all.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5874" LastEditorUserId="5874" LastEditDate="2015-02-15T17:00:46.177" LastActivityDate="2015-02-15T17:00:46.177" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9007" PostTypeId="2" ParentId="8934" CreationDate="2015-02-15T15:45:27.960" Score="5" Body="&lt;p&gt;The chain breaks between items 3 and 4.&#xA;Hence, I start my solution with the weak link 3-4, followed by all remaining items and links in increasing order up to the final item 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.-4. A fascinating love story.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; West Side Story with Tony and MARIA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;The ship of the egg master.   &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SANTA MARIA (one of the ships of Christopher Columbus; egg of Columbus)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;4.-5. Clue 4 and 5 share their firstnames.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SANTA CLAUS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;A German suicide bomber.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CLAUS von Stauffenberg (who tried to assassinate Hitler in July 1944 by planting a bomb)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;5.-6. An actor with the lastname of a vacation activity.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tom CRUISE (who played Stauffenberg in a recent movie; also played in &quot;Top Gun&quot; which might sound vaguely similar to the &quot;tomgun&quot; in the hint)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;6&quot;&gt;&#xA;&lt;li&gt;The lastname of the founder of a religious science.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ron HUBBARD (founder of Scientology Church; Tom Cruise is a member of Scientology)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;6.-7. Similarity in names.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HUBBARD and HUBBLE start with the first four (uncommon) letters&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;7&quot;&gt;&#xA;&lt;li&gt;A binocular.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The HUBBLE telescope&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;7.-1. Looking at differences and quarrels with the binocular.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Star wars (created by George Lucas)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A person with the same lastname as a doctor in the Bible.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; George Lucas (Colossians 4:14: Luke the beloved physician and Demas greet you; also director of &quot;American Graffiti&quot; and hence fits with the hint &quot;graffiti&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;1.-2. A fictional master of the whip.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; INDIANA Jones (created by George Lucas and Steven Spielberg)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;A meteorological phenomenon.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; INDIAN SUMMER&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;2.-3. A time of the year.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SUMMER&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;The Norwegian translation of one of the most common formulas in Excel.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SUMMERING (Norwegian for the excel command SUM)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-17T09:14:22.747" LastActivityDate="2016-03-17T09:14:22.747" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9008" PostTypeId="1" AcceptedAnswerId="9009" CreationDate="2015-02-15T16:01:06.253" Score="-4" ViewCount="152" Body="&lt;p&gt;You have a plenty of segments of the same length at your disposal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put a segment and put another to meet at their ends. Set the counterclockwise angle between two segments at 180/5 degrees. If you repeat this process, it ends after five steps (see the picture).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yBeVp.png&quot; alt=&quot;pentagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, you can see a regular pentagon inside your figure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you put the segments with an angle of 180/n degrees, which polygon will you see inside the resulting figure?&lt;/p&gt;&#xA;" OwnerUserId="3914" LastEditorUserId="8874" LastEditDate="2015-10-11T13:11:05.387" LastActivityDate="2015-10-11T13:11:05.387" Title="Polygons circumscribed by lines" Tags="|mathematics|geometry|" AnswerCount="1" CommentCount="2" ClosedDate="2015-02-16T10:07:47.930" ContentLicense="CC BY-SA 3.0" />
  <row Id="9009" PostTypeId="2" ParentId="9008" CreationDate="2015-02-15T17:16:47.997" Score="1" Body="&lt;p&gt;Adding together equal length segments with a constant counterclockwise angle of $180°/n$ results in&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;for $n$ odd: a closed loop of $n$ segments, and &lt;/li&gt;&#xA;  &lt;li&gt;for $n$ even: a closed loop of $2n$ segments.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So for $n$ odd you get an $n$-gon, and for $n$ even a $2n$-gon.&lt;/p&gt;&#xA;" OwnerUserId="7573" LastActivityDate="2015-02-15T17:16:47.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9011" PostTypeId="2" ParentId="8604" CreationDate="2015-02-15T17:39:05.390" Score="13" Body="&lt;h2&gt;A partial solution&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;You asked for the existence of a $(2n+1)$-sided polytope/die, so that all $2n+1$ sides are congruent and so that the probability of landing on each of these sides is precisely $1/(2n+1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not see how to mathematically model the property that the probability of landing on each of the sides is precisely $1/(2n+1)$. &#xA;This seems to need some assumptions from physics and/or mechanics.&#xA;Instead, I propose a purely mathematical formulation of fairness:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Definition of fairness:&lt;/strong&gt; A polytope is a fair die, if for any two sides there exists a symmetry of the polytope that maps one side into the other one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This fairness definition works for the standard six-sided die (which is highly symmetric) and also for the 10-sided polytope that you describe in your puzzle.&#xA;Note that the definition trivially implies congruence of any two faces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will prove below that for this definition of fairness, there does not exist a fair die with an odd number of sides.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Non-existence proof for an odd number of sides&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;(1) Suppose for the sake of contradiction that there exists a $(2n+1)$-sided polytope that is fair.&#xA;Let $s$ denote the number of edges of every side.&#xA;The polytope has altogether $2n+1$ sides, and every side has $s$ edges.&#xA;Hence the total number of edges is $(2n+1)s/2$, as every edge is counted twice.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The polytope has $e=(2n+1)s/2$ edges and $s\ge3$ is an even integer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(2) Consider a side of the polytope. &#xA;The side has $s$ vertices, and we let $d_1\le d_2\le\cdots\le d_s$ denote the number of edges that are incident to these $s$ vertices.&#xA;A vertex with $d_i$ incident edges is a vertex of $d_i$ different sides.&#xA;Hence, the polytope contains exactly $(2n+1)/d_i$ vertices of this particular type. &#xA;We conclude:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The polytope has $v=(2n+1)(\frac{1}{d_1}+\frac{1}{d_2}+\cdots+\frac{1}{d_s})$ vertices.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(3) Next we use Euler's polyhedral formula that says that a polytope with $f$ sides (faces), $e$ edges, and $v$ vertices must satisfy $v+f-e=2$.&#xA;By plugging in $f=2n+1$ and the expressions derived in (1) and (2), we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(2n+1)(\frac{1}{d_1}+\frac{1}{d_2}+\cdots+\frac{1}{d_s}+1-\frac{s}{2})=2 ~~~~~~(*)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $d_i\ge3$ for all $i$, we furthermore derive&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(2n+1)(s\frac{1}{3}+1-\frac{s}{2}) ~\ge~ (2n+1)(\frac{1}{d_1}+\frac{1}{d_2}+\cdots+\frac{1}{d_s}+1-\frac{s}{2}) ~=~ 2 $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $2n+1$ is positive, also the value $1-s/6$ in the other bracket must be positive; this implies $s\le5$.&#xA;Since $s\ge3$ is an even integer, we arrive at the following fact:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$s=4$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(4) Since $s=4$, the equation $(*)$ now simplifies to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(2n+1)(\frac{1}{d_1}+\frac{1}{d_2}+\frac{1}{d_3}+\frac{1}{d_4}-1)=2 ~~~~~~(**)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $d_1\ge4$, then the sum of the four reciprocals in $(**)$ would be at most $1$; a contradiction. We conclude $d_1=3$, and $(**)$ further simplifies to  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\frac{1}{d_2}+\frac{1}{d_3}+\frac{1}{d_4} ~=~ \frac{2}{3}+\frac{2}{2n+1} ~~~~~~(***)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This only leaves four possible cases for $d_2$ and $d_3$&#xA;(in all other cases, the sum of the three reciprocals would be at most $2/3$):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;(a) $d_2=3$ and $d_3=3$&lt;/li&gt;&#xA;&lt;li&gt;(b) $d_2=3$ and $d_3=4$&lt;/li&gt;&#xA;&lt;li&gt;(c) $d_2=3$ and $d_3=5$&lt;/li&gt;&#xA;&lt;li&gt;(d) $d_2=4$ and $d_3=4$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(5) It remains to do the case work. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;In case (a), equation $(***)$ turns into $d_4=(2n+1)/2$; this is impossible, as $d_4$ would not be integer.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In case (b), equation $(***)$ turns into $d_4=12(2n+1)/(2n+25)$.&#xA;Then the odd number $2n+25$ must divide $3(2n+1)=6n+3$.&#xA;Since $6n+75=3(2n+25)$, also the difference $6n+75-(6n+3)=72$ must be a multiple of the odd number $2n+25$; a contradiction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In case (c), equation $(***)$ turns into $d_4=15(2n+1)/(4n+32)$.&#xA;Then the odd number $15(2n+1)$ must be a multiple of the even number $4n+32$; another contradiction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In case (d), equation $(***)$ turns into $d_4=6(2n+1)/(2n+13)$.&#xA;Then the odd number $2n+13$ must divide $3(2n+1)=6n+3$.&#xA;Since $6n+39=3(2n+13)$, also the difference $6n+39-(6n+3)=36$ must be a multiple of the odd number $2n+13$; the final contradiction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(6) As all possible cases have ended up in a contradiction, we conclude that there is no fair die with an odd number of sides.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-16T08:56:17.637" LastActivityDate="2015-02-16T08:56:17.637" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9012" PostTypeId="2" ParentId="8983" CreationDate="2015-02-15T19:31:25.823" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cancer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In many people, I am not the first thing you want to have.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nobody wants to have cancer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Never would you want to give me to your family or friends.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Same as before, you do not wish this upon anybody, let alone your family or friends.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sadly, I am often inevitable.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Indeed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Although, some purport to have a cure for me, but mostly they're wrong or lying.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We do not have a definite cure, yet...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Next generation medical science might help cure me with DNA technology.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ... we might have it soon. Gene analysis is the tool we have to curing cancer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am easy to find, if you know where to look.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He probably hides under a rock at some beach.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Trust me. I am not what you want.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You repeat yourself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You should run.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I have mighty claws.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9541" LastEditorUserId="9541" LastEditDate="2015-02-15T19:37:10.780" LastActivityDate="2015-02-15T19:37:10.780" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9013" PostTypeId="2" ParentId="8968" CreationDate="2015-02-15T19:36:10.467" Score="0" Body="&lt;p&gt;We are twins&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Left and right brain half.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are the gateway of silence&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you want to be quiet, you need our approval.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We enrich your world, but you can live without us.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This one is obvious.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are your friends, but sometimes we can be loud when you don't want us to be.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Voices in you head.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes we annoy you, but you will never want to lose us.&lt;/p&gt;&#xA;" OwnerUserId="9541" LastActivityDate="2015-02-15T19:36:10.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9014" PostTypeId="1" AcceptedAnswerId="9091" CreationDate="2015-02-15T23:11:47.700" Score="8" ViewCount="464" Body="&lt;p&gt;Nightly, to and fro, with I?&lt;br&gt;&#xA;Do back and forth to me.&lt;br&gt;&#xA;Can you knick knacks with I?&lt;br&gt;&#xA;I knack knicks.&lt;br&gt;&#xA;Very unconfident until I die. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The bad grammar is intentional and essential in its solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I can't make ends meet.  Maybe you can.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9495" LastEditorUserId="9495" LastEditDate="2015-02-16T01:44:21.873" LastActivityDate="2015-02-16T18:46:39.723" Title="Very unconfident until I die (riddle)" Tags="|riddle|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9015" PostTypeId="2" ParentId="8981" CreationDate="2015-02-15T23:41:37.943" Score="-1" Body="&lt;p&gt;Earth is not flat. It is not even spherical. But that's not a problem, as long as we assume that the surface is at least topologically &amp;quot;smooth enough&amp;quot; as an embedding of a unit sphere (I think it needs to be &lt;span class=&quot;math-container&quot;&gt;$C_1$&lt;/span&gt; but it might even be &lt;span class=&quot;math-container&quot;&gt;$C_0$&lt;/span&gt;)&lt;/p&gt;&#xA;&lt;p&gt;Follows a more exact explanation of the ideas in the previous answer.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;! By continuity, from any point on earth, take the set of maximum circles. Take as a function of the direction, the difference of the length of the maximum circle in that direction and in the perpendicular direction. This function is either constantly 0 or it takes positive and negative values. In any case there is at least one direction, where the maximum circle has the same length as its perpendicular.&lt;/p&gt;&#xA;&lt;p&gt;! Now, choose a step length that's &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{40000}$&lt;/span&gt; of that length, and walk 40000 steps in that chosen direction. You get to the initial point (so &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt; steps away of it). Now turn right or left at will, and walk 40000 more steps. As the two &amp;quot;meridians&amp;quot; have the same length, you'll get again to the initial point. Turning lefto or right just changes the way you go around the meridian, not its length.&lt;/p&gt;&#xA;&lt;p&gt;! So you can walk 40000 steps, turn left or right at will, walk another 40000 steps, again at will, and at every single stop you will be at the starting point, that is 0 steps away.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5393" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-01-19T11:44:46.713" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9017" PostTypeId="2" ParentId="8992" CreationDate="2015-02-16T00:19:48.290" Score="-1" Body="&lt;p&gt;It is impossible.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The number of handshakes between the members is ${169 \choose 2} = 22646$ and the number of handshakes in a four member dinner is exactly $6$, but the total number of handshakes is not a multiple of $6$, so either some handshake has been executed more than once, or some handshake is missing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;WARNING: Wrong calculation&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5393" LastEditorUserId="5393" LastEditDate="2015-02-16T09:40:10.953" LastActivityDate="2015-02-16T09:40:10.953" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9018" PostTypeId="1" AcceptedAnswerId="9123" CreationDate="2015-02-16T00:54:04.197" Score="18" ViewCount="4903" Body="&lt;p&gt;As a young child, you always live here.&lt;br&gt;&#xA;Later you leave me often.&lt;br&gt;&#xA;But you always return, sooner or later.&lt;br&gt;&#xA;I have no beginning or end.&lt;br&gt;&#xA;I am infinite.&lt;br&gt;&#xA;Therefore, if God exists,&lt;br&gt;&#xA;God must live here, too.&lt;br&gt;&#xA;Worry, doubt and fear are rarely found with me.&lt;br&gt;&#xA;I only have two siblings.&lt;br&gt;&#xA;If you are not with me, you are with one of them.&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;Note:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added 3 more lines to the riddle. &lt;/p&gt;&#xA;" OwnerUserId="9495" LastEditorUserId="9495" LastEditDate="2015-02-17T04:13:38.267" LastActivityDate="2015-02-28T16:22:54.440" Title="God must live here" Tags="|riddle|" AnswerCount="15" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9019" PostTypeId="2" ParentId="9000" CreationDate="2015-02-16T01:35:32.177" Score="6" Body="&lt;p&gt;Possibility for #4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quote of the day&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5456" LastActivityDate="2015-02-16T01:35:32.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9020" PostTypeId="2" ParentId="8983" CreationDate="2015-02-16T02:14:16.353" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am easy to find, if you know where to look. Trust me. I am not what&#xA;  you want.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, this seems like it would fit very well, iykwim.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; STDs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4553" LastActivityDate="2015-02-16T02:14:16.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9021" PostTypeId="1" AcceptedAnswerId="9033" CreationDate="2015-02-16T02:14:33.687" Score="6" ViewCount="1288" Body="&lt;p&gt;I wrote the following puzzle a couple of years ago. Since I no longer remember the solution, I feel it's a good enough puzzle to share.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On accusation of littering, you and your assistant are arrested and&#xA;  charged by a corrupt government. The government is attempting to&#xA;  follow the remnants of its misunderstood constitution, while&#xA;  incarcerating as many criminals as possible. The prosecutor explains&#xA;  what will happen:&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;You and your assistant will be allowed as much time as necessary to&#xA;  prepare for your defense. The prosecutor's office will monitor your&#xA;  preparations.&lt;/li&gt;&#xA;  &lt;li&gt;The court will separate you until the end of the trial.&lt;/li&gt;&#xA;  &lt;li&gt;The court will give to each of you a list of ten words you may not&#xA;  speak at trial. If you speak any of these words at trial, you will be&#xA;  held in contempt and jailed.&lt;/li&gt;&#xA;  &lt;li&gt;You will each enter a statement. Your&#xA;  statement may be any one of the words in the court's official&#xA;  dictionary.&lt;/li&gt;&#xA;  &lt;li&gt;The court will inform each of you of the statement made by the&#xA;  other.&lt;/li&gt;&#xA;  &lt;li&gt;You will each give testimony. Your testimony must be any one of the&#xA;  words in the court's official dictionary. If your testimonies do not&#xA;  match you will be convicted of perjury and imprisoned.&lt;/li&gt;&#xA;  &lt;li&gt;The court will&#xA;  ask the prosecution to present its case. If this happens, the&#xA;  prosecution will present no evidence and you will be released.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;The prosecutor gives you two copies of the court's official dictionary&#xA;  which you may use during the proceedings. A review on the cover says&#xA;  that it contains over one hundred and seventy thousand different&#xA;  words.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;You have a strong feeling that the prosecutor's office and the court&#xA;  will collude in any and every way possible to either convict you or&#xA;  hold you in contempt. The court may decide to issue different lists of banned&#xA;  words to each of you. The court may wait to prepare the list of proscribed&#xA;  words for one of you until after the other has entered a statement. It&#xA;  is unlikely that, if you are jailed for contempt or imprisoned for&#xA;  perjury, you will ever be released. You and your assistant are given&#xA;  time to prepare.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;What do you do?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2120" LastEditorUserId="2120" LastEditDate="2015-02-16T05:23:27.953" LastActivityDate="2015-02-16T17:56:11.807" Title="Ten unspeakable words" Tags="|strategy|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9022" PostTypeId="2" ParentId="9018" CreationDate="2015-02-16T03:11:26.707" Score="4" Body="&lt;p&gt;Is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Darkness?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a young child, you live here.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In your mother's womb&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later you leave me often.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Whenever the sun rises&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But you always return, sooner or later.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; At night&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no beginning or end.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's hard to define exactly where darkness begins or ends&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am infinite.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Anywhere without light will be dark - it's not limited.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, if God exists,&lt;br&gt;&#xA;God must live here, too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is the only bit I can't really see a specific meaning for.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9062" LastActivityDate="2015-02-16T03:11:26.707" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9023" PostTypeId="2" ParentId="9018" CreationDate="2015-02-16T03:56:30.900" Score="18" Body="&lt;p&gt;My answer, however wrong it may be, is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Imagination/imaginary world&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a young child, you live here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You often live in your imagination, or imaginary world as a child&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later you leave me often / But you always return, sooner or later:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your imaginary world diminishes as you grow older, but can return in the form of sleep/dreams&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no beginning or end / I am infinite&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Imagination is whatever you want it to be, impossible or not&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, if God exists / God must live here, too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This goes without saying, I think, but if God exists, God must still live in our imagination, as God is never seen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is my first time answering a puzzle here, so go easy on me!&lt;/p&gt;&#xA;" OwnerUserId="4353" LastActivityDate="2015-02-16T03:56:30.900" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9024" PostTypeId="2" ParentId="9021" CreationDate="2015-02-16T05:00:44.697" Score="0" Body="&lt;p&gt;If I'm interpreting this correctly, the statement is just one word and the testimony is just one word. Our goal is to get the testimony of you and your assistant to match. That shouldn't be too hard (unless I misinterpreted anything). Here's what you do:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Make your statement the first word in the court dictionary that you ARE allowed to speak. Your assistant should make her statement the first word in the dictionary that SHE is allowed to speak.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once you've done that, it's time for your testimony:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your testimony should be whichever of the two words comes later in the dictionary.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of course, you'll run into a small problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Given that it's a corrupt government and you're at risk of going to jail for life for littering, they'll probably get you in the end.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2015-02-16T05:00:44.697" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9026" PostTypeId="2" ParentId="9018" CreationDate="2015-02-16T05:32:01.990" Score="3" Body="&lt;p&gt;Is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Real world&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a young child, you live here.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When we are young, no internet yet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later you leave me often. But you always return, sooner or later.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As we do right now, we leaving real world to internet. But later or soon we must back to real world.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no beginning or end. I am infinite.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Outer space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, if God exists, God must live here, too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It would be nice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9160" LastActivityDate="2015-02-16T05:32:01.990" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9028" PostTypeId="2" ParentId="8547" CreationDate="2015-02-16T07:20:24.193" Score="9" Body="&lt;p&gt;I think that the claim is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The triangles are, as suggested, chopped and than copied to maintain the coverage that is provided by the classical face structure of a cube.&#xA;The four cases/points that are mentioned in one of the comments are maintaining of the right angle since the chopped triangles are rotated and the right angle is maintained in the new polygon.&#xA;This could help in the work but is not yet a proof&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HUN93.jpg&quot; alt=&quot;Polygon covering a cube&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any folding of a cube may be spread to a polygon which is based on the six faces with chopped areas moved from one face to another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chopping of areas around the right angles will always result a right angle. The minimal right angle face unfolding contains two right angles which result two right angle in any polygon coverage based on such starting point of creating polygon for coverage. Here is the minimal right angle structure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y0DKw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9531" LastEditorUserId="9531" LastEditDate="2015-02-17T07:21:47.200" LastActivityDate="2015-02-17T07:21:47.200" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9029" PostTypeId="2" ParentId="9018" CreationDate="2015-02-16T07:52:39.320" Score="2" Body="&lt;p&gt;My best guess is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; River&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a young child, you live here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A tadpole starts its life in a water body, river (though most frogs/tadpoles live in a pond/lake or well, there are few variants that inhabitate a river &lt;a href=&quot;http://en.wikipedia.org/wiki/Giant_river_frog&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Giant_river_frog&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later you leave me often / But you always return, sooner or later:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A tadpole matures to a frog, leaves a river often but returns sooner or later. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no beginning or end / I am infinite&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;For men may come and men may go, But I go on for ever&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, if God exists / God must live here, too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; God is omnipresent, and exists in a river too&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4177" LastEditorUserId="4177" LastEditDate="2015-02-16T08:46:08.593" LastActivityDate="2015-02-16T08:46:08.593" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9030" PostTypeId="2" ParentId="9018" CreationDate="2015-02-16T07:55:17.373" Score="3" Body="&lt;p&gt;As a young child, you live here.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;children are innocent, their world is innocent world&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later you leave me often.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;In youth people are less innocent and they are surrounded by less innocent people&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But you always return, sooner or later.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;As people grow older they again become innocent&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no beginning or end.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Innocence is an abstract quality&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am infinite.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There is no limit to a person's innocence&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, if God exists,&#xA;God must live here, too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Childhood innocent world was better compared to the real world in youth, God must live in better place&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9174" LastActivityDate="2015-02-16T07:55:17.373" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9031" PostTypeId="1" AcceptedAnswerId="9187" CreationDate="2015-02-16T08:20:33.983" Score="-3" ViewCount="360" Body="&lt;p&gt;Beyond the beyond&lt;br&gt;&#xA;In far reaches of flight&lt;br&gt;&#xA;Scarcely seen without a fight&lt;br&gt;&#xA;Broken packed cosmic twist&lt;br&gt;&#xA;Feathers splayed in the mist&lt;br&gt;&#xA;Strange wonder to behold&lt;br&gt;&#xA;To gaze upon it is bold&lt;br&gt;&#xA;Never lightly come upon it&lt;br&gt;&#xA;For if you do, you may vomit&lt;br&gt;&#xA;Your senses reeling in a flurry&lt;br&gt;&#xA;Run you will try, in a hurry&lt;br&gt;&#xA;But run and escape you cannot&lt;br&gt;&#xA;You'll be eaten before you can rot&lt;br&gt;&#xA;To gaze upon the humanlike face&lt;br&gt;&#xA;You won't live to see your race  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am African&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;hint continued:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wow, this riddle seems almost obvious to me.  I am tempted to post the answer.  Are you sure you guys don't know any African mythological creatures who eat people and have feathers and a humanlike face?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9495" LastEditorUserId="9495" LastEditDate="2015-02-17T23:45:12.583" LastActivityDate="2015-02-18T03:29:24.353" Title="Beyond the beyond" Tags="|riddle|" AnswerCount="5" CommentCount="6" ClosedDate="2015-02-18T14:19:46.960" ContentLicense="CC BY-SA 3.0" />
  <row Id="9032" PostTypeId="2" ParentId="9018" CreationDate="2015-02-16T08:43:56.663" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;My answer is : Dream world&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a young child, you live here.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We always dream when we were young&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later you leave me often.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We rest lesser and lesser as the age goes, and we dream less&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But you always return, sooner or later.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We always sleep ;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no beginning or end.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dream world has no beginning&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am infinite.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dream world is limited only by your imagination&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, if God exists,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;God must live here, too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you see god, you must be in your dream&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8882" LastActivityDate="2015-02-16T08:43:56.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9033" PostTypeId="2" ParentId="9021" CreationDate="2015-02-16T08:50:19.850" Score="6" Body="&lt;p&gt;As a statement, enter a word such that &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You can legally say any word starting with the same letter&lt;/li&gt;&#xA;&lt;li&gt;The dictionary contains at least 11 words starting with the same letter&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Your assistant's statement should be a number n such that for every letter, he can play the nth word in the dictionary that starts with that letter. There will always be an n less than or equal to 11. Of course in practice your assistant has to give a word instead of a number (unless numbers are in the dictionary!). Just use the nth word in the dictionary to represent n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As testimony, you both give the nth word starting with the letter you indicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may have to adapt this method if the official dictionary is as pernicious as the rest of the trial, specifically if fewer than 11 letters have 11 or more words. But since it sounds like you're given the dictionary ahead of time, you have time to coordinate this.&lt;/p&gt;&#xA;" OwnerUserId="9552" LastActivityDate="2015-02-16T08:50:19.850" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9035" PostTypeId="2" ParentId="9021" CreationDate="2015-02-16T09:25:33.670" Score="1" Body="&lt;p&gt;After much thought, I came up with a solution.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Notice that there are at least 170000 words in the dictionary; both you and the assistant assume that there are exactly 170000 words. Assign one individual to partition the dictionary the first way, and the other to do the second way.&lt;br&gt;&#xA; &lt;li&gt;Way #1:&lt;br&gt;&lt;br&gt;&#xA; Notice that the 10 different words effectively partition the dictionary. By default they partition it into 11 pieces, but if we act as if the dictionary wraps around, we actually partition it into 10 pieces. This means that there must be a partition of length at least $\frac{170000}{10}-1 = 16999$. This first individual chooses the first word of this partition as the statement.&lt;br&gt;&lt;br&gt;&#xA; &lt;li&gt;Way #2:&lt;br&gt;&lt;br&gt;&#xA; This time, let's just treat the words as numbers to make this easier. Separate all the words into mod 10 equivalent classes (bins). There are $17000$ words in each bin. If one bin doesn't have a banned word, then choose the smallest word in that bin as our partition. If every bin does have a banned word, then choose any bin and choose the smallest word after the banned word. In this way, we guarantee a bin size of $16999$, just like in Way #1.&lt;br&gt;&lt;br&gt; Alternatively, you could just separate into mod 16999 (or smaller) equivalent bins, and choose any bin without a banned word that is largest in size.&#xA; These two partitions are guaranteed to overlap. Each person chooses the first overlapping word (regular dictionary order) as the testimony.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1477" LastEditorUserId="1477" LastEditDate="2015-02-16T10:11:46.387" LastActivityDate="2015-02-16T10:11:46.387" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9036" PostTypeId="1" AcceptedAnswerId="9078" CreationDate="2015-02-16T09:36:28.217" Score="52" ViewCount="5575" Body="&lt;p&gt;The lion plays a deadly game against a group of 100 zebras that takes place in the steppe (= an infinite plane).&#xA;The lion starts in the origin with coordinates $(0,0)$, while the 100 zebras may arbitrarily pick their 100 starting positions.&#xA;The the lion and the group of zebras move alternately:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In a lion move, the lion moves from its current position to a position at most 100 meters away.&lt;/li&gt;&#xA;&lt;li&gt;In a zebra move, &lt;em&gt;one&lt;/em&gt; of the 100 zebras moves from its current position to a position at most 100 meters away.&lt;/li&gt;&#xA;&lt;li&gt;The lion wins the game as soon as he manages to catch one of the zebras.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Will the lion always win the game after a finite number of moves? &#xA;Or is there a strategy for the zebras that helps them to survive forever?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="10615" LastEditDate="2015-04-24T15:06:41.257" LastActivityDate="2016-12-15T02:51:55.827" Title="The lion and the zebras" Tags="|strategy|game|pursuit-evasion|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9037" PostTypeId="2" ParentId="9018" CreationDate="2015-02-16T09:40:56.307" Score="7" Body="&lt;p&gt;My answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Wonder&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a young child, you live here.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;All children, especially when very young, find most of the world wonderful&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later you leave me often.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;As you grow up, you get sometimes frustrated, realist&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But you always return, sooner or later.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Some things still marvel you, even when you're an adult&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no beginning or end.&#xA;I am infinite.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Not quite sure, but this may only mean &quot;what is to be guessed is abstract&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, if God exists,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;...Which I surely WONDER&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;God must live here, too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;God would have all the qualities of a wonderful being, therefore living in wonder&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit: I actually had another solution, which was &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Phantasy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5122" LastEditorUserId="5122" LastEditDate="2015-02-16T10:04:00.153" LastActivityDate="2015-02-16T10:04:00.153" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9044" PostTypeId="2" ParentId="9018" CreationDate="2015-02-16T10:33:44.340" Score="1" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Sky&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a child you live here&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Dream of flying&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later you leave me often&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Back to earth, focused on life&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But you always return, sooner or later.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Still dream as an adult.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no beginning or end.&#xA;I am infinite.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The sky seems to go on forever in all directions. It is a sphere around the earth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, if God exists,&#xA;God must live here, too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Heaven is often referred to as being in the sky&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9556" LastActivityDate="2015-02-16T10:33:44.340" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9047" PostTypeId="2" ParentId="9000" CreationDate="2015-02-16T11:46:40.923" Score="10" Body="&lt;p&gt;I am going to combine all answers so that it is easy for everyone to see.&#xA;All credit goes to Hellion, Gamow and Vikram.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;AGE BUT&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Age before beauty&lt;/strong&gt; (B-U-T)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;marri Ag e&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ag=silver with marriage = &lt;strong&gt;Silver wedding anniversary&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;wWoOlOfL&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wolf in WOOL = &lt;strong&gt;Wolf in sheep's clothing&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;24 Hours&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Multiple possibilities:&lt;br&gt;&#xA; &lt;strong&gt;Call it a day&lt;/strong&gt;&lt;br&gt;&#xA; &lt;strong&gt;Quote of the day&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;TRADI  TION&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Break with tradition&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;HEAD ACHE&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Splitting headache&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;WHEATHER&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Bad spell of weather&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;SGEG&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Permutation of EGGS = &lt;strong&gt;Scrambled eggs&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;TOSYOBSYOBWN&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BOYS BOYS backwards, inside TOWN = &lt;strong&gt;The Boys are Back in Town&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;ALL world&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;It's a small world after all!&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9346" LastEditorUserId="5456" LastEditDate="2016-03-04T13:27:59.420" LastActivityDate="2016-03-04T13:27:59.420" CommentCount="1" CommunityOwnedDate="2015-02-17T05:39:30.480" ContentLicense="CC BY-SA 3.0" />
  <row Id="9052" PostTypeId="2" ParentId="9036" CreationDate="2015-02-16T12:26:43.957" Score="-13" Body="&lt;p&gt;Answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quick answer:  The zebras would survive forever.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Simple explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The key concept here is &quot;infinite plane&quot;.  Since the plane is infinite, positioning is irrelevant, except for the fact that the zebras are placed at great distances from each other and the lion.  The closest zebra from the lion simply moves away from the lion in a direction that does not bring it near the path of another zebra.  Because the plane is infinite, the zebras can be a nearly infinite distance from each other, thus making their escape paths very easy to avoid other zebras. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9495" LastEditorUserId="9495" LastEditDate="2015-02-21T05:47:15.103" LastActivityDate="2015-02-21T05:47:15.103" CommentCount="66" ContentLicense="CC BY-SA 3.0" />
  <row Id="9053" PostTypeId="1" AcceptedAnswerId="9054" CreationDate="2015-02-16T12:38:29.827" Score="-3" ViewCount="58277" Body="&lt;p&gt;You may use math signs you wish, but you must use the numbers&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;7,3,7,3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;in an equation so it equals&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;24&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is it possible? If so, show how.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;No-nos:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You are not allowed to combine numbers to make bigger numbers (&lt;em&gt;e.g.&lt;/em&gt; you can't have 73)&lt;/li&gt;&#xA;&lt;li&gt;You cannot use numbers more than once, you only have &lt;strong&gt;2 7s&lt;/strong&gt; and &lt;strong&gt;2 3s&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;You have to use all the numbers given to you.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is a math sign? Examples: +, -, *, /, (, )&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2015-02-16T12:50:32.680" LastActivityDate="2019-03-22T18:36:40.813" Title="Make 24 With 7,3,7,3" Tags="|mathematics|" AnswerCount="7" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9054" PostTypeId="2" ParentId="9053" CreationDate="2015-02-16T12:45:22.737" Score="10" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;(3/7+3)*7 =24&lt;br&gt;&#xA;  =&gt; (3/7 + 21/7) * 7 =24&lt;br&gt;&#xA;  =&gt; (24/7) * 7 =24&lt;br&gt;&#xA;  =&gt; 24=24 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Took me a while!&lt;/p&gt;&#xA;" OwnerUserId="9346" LastEditorUserId="39" LastEditDate="2015-02-16T14:03:02.850" LastActivityDate="2015-02-16T14:03:02.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9056" PostTypeId="2" ParentId="9053" CreationDate="2015-02-16T12:46:49.433" Score="2" Body="&lt;p&gt;Using not all of them:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3*7 + 3 = 24&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using all of them:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\lceil 7/3 \rceil + 3*7 = 24$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5265" LastEditorUserId="4204" LastEditDate="2015-02-16T20:45:33.027" LastActivityDate="2015-02-16T20:45:33.027" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="9058" PostTypeId="2" ParentId="9053" CreationDate="2015-02-16T12:52:48.107" Score="5" Body="&lt;p&gt;&lt;strong&gt;Answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$((7/7)+3)*(3!) = 4*(3!) = 24 $$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4506" LastEditorUserId="39" LastEditDate="2015-02-16T14:02:44.203" LastActivityDate="2015-02-16T14:02:44.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9060" PostTypeId="1" AcceptedAnswerId="9065" CreationDate="2015-02-16T12:59:37.123" Score="3" ViewCount="463" Body="&lt;p&gt;I am three.&lt;br&gt;&#xA;One of me can make it a challenge to see.&lt;br&gt;&#xA;One of me can be soft.. if I am not hard.&lt;br&gt;&#xA;Normally, I am spherical, unless something touches me.&lt;br&gt;&#xA;You touch me everyday.  &lt;/p&gt;&#xA;" OwnerUserId="9495" LastActivityDate="2015-02-16T15:52:14.787" Title="You touch me everyday" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9063" PostTypeId="1" AcceptedAnswerId="9064" CreationDate="2015-02-16T13:04:15.507" Score="7" ViewCount="3980" Body="&lt;p&gt;There's a king who wants to appoint a new minister. To check the intelligence of the interested men, he gives the following problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 10 bags and there are 100 coins. You have to fill all the 100 coins into the 10 bags. You may fill any number of coins into any bag. You may keep a bag empty, but you must fill all the coins into the bags. Then you must seal the bags. Then I'll ask you to give me any number of coins and you must be able to give me that exact amount without opening any of the bags. In other words, you must hand me over several bag/s which will contain coins equal to what I have asked for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only one person could do it, and he was appointed minister.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would you do it?&lt;/p&gt;&#xA;" OwnerUserId="9560" LastEditorUserId="8874" LastEditDate="2015-10-11T12:48:57.530" LastActivityDate="2017-07-24T13:44:31.850" Title="Filling up 100 coins in 10 bags" Tags="|mathematics|calculation-puzzle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9064" PostTypeId="2" ParentId="9063" CreationDate="2015-02-16T13:10:50.617" Score="16" Body="&lt;p&gt;You need only 7 bags, of size 1, 2, 4, 8, 16, 32 and 37 [=100-(64-1)].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an amount less than 37, convert the number into binary and give the bags required. For values above 37, give the 37 bag, subtract it from the value, and then give the rest as in above method.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="5610" LastEditDate="2015-02-16T13:13:24.010" LastActivityDate="2015-02-16T13:13:24.010" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9065" PostTypeId="2" ParentId="9060" CreationDate="2015-02-16T13:14:23.100" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am three.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I have three forms (vapor, liquid, ice).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One of me can make it a challenge to see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Vapor form.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One of me can be soft.. if I am not hard.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water in its frozen state can be soft as snow or hard as ice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Normally, I am spherical, unless something touches me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In space, water is spherical.  Only when touched by gravity or a container does it take another form.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You touch me everyday.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You wash you hands at least once a day, don't you?  Even if you don't, your lips touch water, even if it is Coca-cola.  Even breathing can count as touching water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9541" LastEditorUserId="9495" LastEditDate="2015-02-16T15:52:14.787" LastActivityDate="2015-02-16T15:52:14.787" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9067" PostTypeId="1" CreationDate="2015-02-16T13:41:29.173" Score="-2" ViewCount="13941" Body="&lt;p&gt;I came across this problem some years ago in a book (which I won't mention, and request others to not mention if they know). The solution is quite elegant and difficult to arrive at by mere  trial and error. And I couldn't think of a proper way to solve it. Still, give it a try.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once, there was an egg seller. She had $90$ eggs which she wanted to sell in the market. So, she sent her $3$ girls to the market. She gave the eldest and the cleverest girl $10$ eggs, the second girl $30$ eggs and the youngest one $50$. She told them, &quot;You better decide for the price of the eggs among yourselves. But the three of you should stick to one price. You may change the price more than once, but the three of you must change it together. Also, I expect that the three of you earn the same amount of money each by selling your respective eggs. And you must bring home not less than $90$ units of money.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The egg seller told them they should not bring home less than $90$ units. The eldest girl thought of an ingenious way and explained her sisters her idea. In the evening, when they came home, each girl had earned $30$ units by selling her respective lot of eggs, and they together had brought home exactly $90$ units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you think of what idea the eldest girl used?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Let me clarify the situation a bit more. The three girls occupy different places in the market. Since the eldest girl told them the plan, they have no connection whatsoever till evening. So, they cannot go and meet or sell their eggs to each other. It is not necessary to have a price per egg. Eg, $5$ units per $3$ eggs is a totally acceptable price. With  their respective transactions, each of them earns $30$ units. All the eggs are sold in the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2&lt;/strong&gt;: Also, if the girls decide to sell in bundles, they must sell until their bundles are finished. Eg, if they set the price to $5$ per $9$ eggs, the eldest will sell her $9$ eggs, the middle one will sell her $27$ eggs, the youngest will sell her $45$ eggs before changing the price again. This is one thing I should have told earlier. Sorry. O:-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If any more doubt, please add in the comments.&lt;/p&gt;&#xA;" OwnerUserId="9560" LastEditorUserId="9560" LastEditDate="2015-02-17T03:19:04.607" LastActivityDate="2015-02-17T16:15:23.280" Title="What's your price for the eggs?" Tags="|mathematics|calculation-puzzle|strategy|" AnswerCount="11" CommentCount="4" FavoriteCount="0" ClosedDate="2015-02-19T18:28:36.087" ContentLicense="CC BY-SA 3.0" />
  <row Id="9069" PostTypeId="2" ParentId="9036" CreationDate="2015-02-16T13:46:17.563" Score="3" Body="&lt;p&gt;&lt;strong&gt;Later:&lt;/strong&gt; At the time, I was thinking of a solution similar to Lopsy's, but instead of vertical stripes I had an idea of circular sectors. Each zebra would thus have its own angle of 3,6°. Here is how I put it to words back then:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Zebras can win.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is not a rigorous proof, but a good proof nonetheless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The game is lost, if a lion can pick a new point $P$, such that at least two zebras are contained in a circle with a center at $P$ and radius of 100m. That is obvious, then only one zebra can escape in the next turn and at least one gets eaten.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So our objective is: after every zebra move, all zebras have to be more than 100m away from each other - this guarantees lion cannot make a move that would capture two zebras in the same 100m radius circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since a zebra can move 100 meters, that just means it has to have some free space in some direction, e.g. it can move to a new point, not falling into 100m radius of another zebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is easily achievable if we initially place zebras on the border of any convex shape, such as a circle of a square of an adequate size (adequate means each zebra is more than 50m apart).&lt;/p&gt;&#xA;" OwnerUserId="9541" LastEditorUserId="9541" LastEditDate="2015-03-06T19:29:00.013" LastActivityDate="2015-03-06T19:29:00.013" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9071" PostTypeId="2" ParentId="9018" CreationDate="2015-02-16T14:15:33.137" Score="3" Body="&lt;p&gt;Could it be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Curiosity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a child you live here&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Children are always curious&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later you leave me often&#xA;But you always return, sooner or later.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Even as adults we are curious about new things&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no beginning or end. I am infinite.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is no limit to Curiosity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, if God exists, God must live here, too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ??? God is curious about humans? or we seek god through curiosity?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;// Edit - Guess 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Love&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a child you always have it, and growing up you tend to lose it but you will always come back to it.  There is no conceivable end to it, and if god exists, surely god would embody it.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="2153" LastEditDate="2015-02-16T14:42:48.970" LastActivityDate="2015-02-16T14:42:48.970" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9073" PostTypeId="2" ParentId="9018" CreationDate="2015-02-16T14:24:10.713" Score="0" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; From whence you came? (or just the void?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a young child, you live here.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Could be womb/void&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later you leave me often.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you grow up, you leave home.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But you always return, sooner or later.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Returning to visit home/family.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no beginning or end.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &quot;void&quot; before you are born, the void after.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am infinite.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The void has no known boundaries.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, if God exists,&#xA;God must live here, too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The void as well, since He is everywhere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2015-02-16T14:24:10.713" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9074" PostTypeId="2" ParentId="9067" CreationDate="2015-02-16T14:38:27.930" Score="1" Body="&lt;p&gt;Let there be 2 different prices, $A$ and $B$,  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And call the numbers sold under the first &#xA;price $X_a$, $Y_a$, and $Z_a$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And under the second price $X_b$, $Y_b$, and $Z_b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$AX_a + BX_b = AY_a + Y_b = AZ_a + BZ_b$ &lt;br&gt;&#xA;  $X_a + X_b = 50$ &lt;br&gt;&#xA;  $Y_a + Y_b = 30$ &lt;br&gt;&#xA;  $Z_a + Z_b = 10$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eliminating the $b$ amounts,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$AX_a + B(50 - X_a) = AY_a + B(30 - Y_a) = AZ_a + B(10 - Z_a)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(A - B)X_a + 50B = (A - B)Y_a + 30B = (A - B)Z_a + 10B$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subtracting $10B$ and dividing by $(A - B)$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X_a + \frac{40B}{A - B} = Y_a + \frac{20B}{A - B} = Z_a$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Z_a = n$&lt;br&gt;&#xA;  $Y_a = n + \frac{20B}{B - A}$&lt;br&gt;&#xA;  $X_a = n + \frac{40B}{B - A}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which gives us a solution for any given $A$, $B$, and $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking $A = \frac{3}{7}$ and $B = 9$ just as a guess from the problem's examples, &#xA;we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{B}{B - A} = \frac{9}{\frac{60}{7}} = \frac{21}{20}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X_a = n + 42$&lt;br&gt;&#xA;  $Y_a = n + 21$&lt;br&gt;&#xA;  $Z_a = n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, we can choose any value for $n$ from $0$ to $8$ and get valid &#xA;numbers for $X_a$ and $Y_a$; since we need to sell a whole number of eggs &#xA;at the second price, and thus need the differences between the &#xA;numbers sold at the first price to be a whole number (actually, just &#xA;a multiple of $\frac{1}{3}$), we can choose either $0$ or $7$ for $n$.  So either&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A = \frac{3}{7}$&lt;br&gt; $X_a = 49$&lt;br&gt; $Y_a = 28$&lt;br&gt; $Z_a = 7$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B = 9$&lt;br&gt;   $Xb = 1$&lt;br&gt; $Yb =  2$&lt;br&gt; $Zb = 3$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, each earns $\$30$&lt;/p&gt;&#xA;" OwnerUserId="9346" LastEditorUserId="9346" LastEditDate="2015-02-17T07:16:08.610" LastActivityDate="2015-02-17T07:16:08.610" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="9075" PostTypeId="2" ParentId="9067" CreationDate="2015-02-16T14:48:07.110" Score="0" Body="&lt;p&gt;I have a solution with two price changes necessary (assuming that people will buy the eggs for these prices).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initial Price:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Set the price at 1 unit per egg.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First sale&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The youngest sells 30 of her eggs, leaving her with 30 units of money and 20 eggs left.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Price change:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They then all change the price to 3 units per egg.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And now some sneakiness on the part of the kids:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The youngest now buys the 10 eggs from the eldest for the 30 monetary units. She now has 30 eggs, but no money. The eldest has 30 units of money, but no eggs. The middle child still has 30 eggs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Price change #2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Change the price back to 1 unit per egg&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Final sales:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The middle child and youngest now sell each of their 30 eggs for 1 unit each, leaving everyone with 30 units of money. 90 units total, no eggs left.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2015-02-16T14:48:07.110" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9077" PostTypeId="2" ParentId="9018" CreationDate="2015-02-16T15:05:04.337" Score="3" Body="&lt;p&gt;Please note: this is a satirical answer and should not be taken too seriously :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A possible answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ignorance&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a young child, you always live here.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you don't know anything, you are ignorant&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later you leave me often.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you do learn something by chance while growing up&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But you always return, sooner or later.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; every time you learn something you also realize that you are more ignorant than you previously thought&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no beginning or end.&#xA;I am infinite.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ignorance is notoriously infinite&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, if God exists,&#xA;God must live here, too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; oops! :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5332" LastActivityDate="2015-02-16T15:05:04.337" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9078" PostTypeId="2" ParentId="9036" CreationDate="2015-02-16T15:10:19.273" Score="62" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Zebras win.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is the strategy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The zebras choose 100 vertical strips on the steppe, each of width 300m. The strips do not intersect. Each zebra promises to stay horizontally centered on its own individual strip.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the lion enters a strip, then the zebra in that strip flees vertically away from the lion until the lion leaves the strip.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It works because:&#xA; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The lion cannot kill a zebra on the turn it moves into a strip, because the strips are too wide. And if it is already inside a strip, then the zebra in that strip has just moved 100m away from it, so it cannot catch the zebra.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2015-02-16T15:10:19.273" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="9079" PostTypeId="2" ParentId="9018" CreationDate="2015-02-16T15:27:26.183" Score="3" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The great unknown&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a young child, you always live here.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As a child, you barely know anything, so the great unknown is always there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later you leave me often.&#xA;But you always return, sooner or later.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As an adult, we think we know a lot, but every now and then, you have that rare and fleeting feeling that there is so much you don't know.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no beginning or end.&#xA;I am infinite.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The great unknown is not bounded.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, if God exists,&#xA;God must live here, too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; God is practically the epitome of &quot;the great unknown&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Alternate answers with the same reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mystery or &lt;a href=&quot;http://youmightfindyourself.com/post/7725630208/yugen&quot; rel=&quot;nofollow&quot;&gt;Yūgen&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="268" LastActivityDate="2015-02-16T15:27:26.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9080" PostTypeId="2" ParentId="9018" CreationDate="2015-02-16T15:40:21.497" Score="0" Body="&lt;p&gt;Just maybe...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dependency? - But then maybe not.  Read on...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a young child, you always live here.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We are dependent as children, but do we actually &lt;em&gt;live&lt;/em&gt; there?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later you leave me often.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We become independent,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But you always return, sooner or later.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I suppose even a hermit well depend upon the environment.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no beginning or end.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Unless there is a prime cause, upon which everything else depends&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am infinite.&#xA;Therefore, if God exists,&#xA;God must live here, too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Would God be considered a prime cause?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hmm. I don't think I am quite right.&lt;/p&gt;&#xA;" OwnerUserId="9567" LastActivityDate="2015-02-16T15:40:21.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9082" PostTypeId="2" ParentId="9067" CreationDate="2015-02-16T16:16:32.757" Score="4" Body="&lt;p&gt;I'm gonna have to make some assumptions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1-&lt;/strong&gt; The girls are allowed to sell eggs to each other&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2-&lt;/strong&gt; since they have to sell eggs at the same price, customers will buy eggs uniformly  from all 3 that have eggs available&lt;/p&gt;&#xA;&#xA;&lt;p&gt;one answer is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;step 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; they each sell 10 eggs at $1/piece&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;step 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; they briefly set the price to $0.33, so that the eldest can buy 30 eggs from the youngest&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;step 3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; they sell all remaining eggs at $1/piece&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="9413" LastEditDate="2015-02-16T18:00:59.067" LastActivityDate="2015-02-16T18:00:59.067" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9083" PostTypeId="2" ParentId="9053" CreationDate="2015-02-16T16:36:06.547" Score="3" Body="&lt;p&gt;There's an implied $1$, but I didn't actually write it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $7 \times 3 + 3 + \int_{\{7\}}dx$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And here's another solution (I think you can argue that $7$ is used twice):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $7 \times 3 - 3 = 24$ in base $7$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8942" LastEditorUserId="8942" LastEditDate="2015-02-16T21:49:15.973" LastActivityDate="2015-02-16T21:49:15.973" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9084" PostTypeId="2" ParentId="9018" CreationDate="2015-02-16T16:45:05.220" Score="4" Body="&lt;p&gt;Is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Mind&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a young child, you live here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You cannot speak, so you are for the most part, in your mind&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later you leave me often / But you always return, sooner or later:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In later life you become 'absent-minded'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no beginning or end / I am infinite&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The mind is limitless&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, if God exists / God must live here, too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; God must exist in the mind due to it's infinity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I do not know if I have formatted this correctly, sorry if not :(&lt;/p&gt;&#xA;" OwnerUserId="9568" LastActivityDate="2015-02-16T16:45:05.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9085" PostTypeId="1" AcceptedAnswerId="9092" CreationDate="2015-02-16T16:56:12.233" Score="4" ViewCount="1142" Body="&lt;p&gt;Each of the following five statements is either true or false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A. Statement D is true, or statement E is false, or the moon is burning-red.&lt;br&gt;&#xA;B. Statements C and E are not both true and not both false.&lt;br&gt;&#xA;C. Statements A and E are not both true and not both false, and the moon is blue.&lt;br&gt;&#xA;D. Statement A is false, and the moon is (as everyone knows) made of green cheese.&lt;br&gt;&#xA;E. Statement B is false, and either the moon is yellow or the moon is not yellow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What color is the moon?&lt;/p&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="8874" LastEditDate="2015-10-11T12:53:46.993" LastActivityDate="2015-10-11T12:53:46.993" Title="The dark side of the moon" Tags="|logical-deduction|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9086" PostTypeId="2" ParentId="9085" CreationDate="2015-02-16T17:14:22.273" Score="7" Body="&lt;p&gt;E simplifies to &quot;B is false&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If E is true, B is false, therefore C is true. If E is false, B is true, and so is C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the moon is always &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; blue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Option 1, &quot;E = true&quot; is also impossible, but that takes a bit more work. This is the lazy way.)&lt;/p&gt;&#xA;" OwnerUserId="439" LastActivityDate="2015-02-16T17:14:22.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9087" PostTypeId="2" ParentId="9085" CreationDate="2015-02-16T17:14:42.417" Score="5" Body="&lt;p&gt;Let's give it a go, should we?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moon is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Blue&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A. Statement D is true, or statement E is false, or the moon is burning-red.&lt;br/&gt;B. Statements C and E are not both true and not both false.&lt;br/&gt;C. Statements A and E are not both true and not both false, and the moon is blue.&lt;br/&gt;D. Statement A is false, and the moon is (as everyone knows) made of green cheese.&lt;br/&gt;E. Statement B is false, and either the moon is yellow or the moon is not yellow.&lt;br/&gt;&lt;br/&gt;A = D or !E or red&lt;br/&gt;B = (C != E)&lt;br/&gt;C = (A != E) and blue&lt;br/&gt;D = !A and green&lt;br/&gt;E = !B&lt;br/&gt;&lt;br/&gt;if E&lt;br/&gt;    =&gt; !B&lt;br/&gt;        =&gt; C == E&lt;br/&gt;            =&gt; C true&lt;br/&gt;            =&gt; blue and !A&lt;br/&gt;            =&gt; red or D&lt;br/&gt;            =&gt; D&lt;br/&gt;            =&gt; !A and green&lt;br/&gt;&lt;br/&gt;=&gt; !E&lt;br/&gt;&lt;br/&gt;if B&lt;br/&gt;    =&gt; C true&lt;br/&gt;    =&gt; blue and A&lt;br/&gt;    =&gt; A =&gt; true&lt;br/&gt;    =&gt; !D&lt;br/&gt;&lt;br/&gt;=&gt; A, B, C, !D, !E and the moon is blue&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastActivityDate="2015-02-16T17:14:42.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9088" PostTypeId="2" ParentId="9067" CreationDate="2015-02-16T17:25:27.843" Score="2" Body="&lt;p&gt;My suggestion is without price changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An assumption based on the riddle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The riddle does not explicitely tell that all eggs need to be sold, only that they should earn the same amount of money.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The girls set their price to 3 units money pr. egg. Girl 1 sells all of her (10) eggs, but the second and third only sells 10 of their lots. The rest of the eggs (20+40) they simply throw away. Each girl will return with 30 units of money.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9385" LastActivityDate="2015-02-16T17:25:27.843" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9089" PostTypeId="2" ParentId="9067" CreationDate="2015-02-16T17:37:12.377" Score="2" Body="&lt;p&gt;There are probably multiple solutions where we assume the 3 sisters are at different locations in the market such that different people could buy from all three at the same time and buying from each other would mean leaving your eggs behind and you just don't do that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The one I have come up with involve selling them in bundles and then selling the remaining eggs per sister.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; sell 12 eggs for 6 dollars and the remaining for 3 each&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The price change would happen once all bundles available per sister have been sold. As a result, clearly there is a pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x$, $y$, and $z$ represent egg count per sister.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, you need to find the bundle size $b$ such that bundle price $p$ and single price $s$ fulfill the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;&amp;amp; \left\lfloor\frac xb*p \right\rfloor + \mod(x,b)*s \\&#xA;= &amp;amp;\left\lfloor\frac yb*p\right\rfloor+\mod(y,b)*s \\&#xA;= &amp;amp;\left\lfloor\frac zb*p\right\rfloor + \mod(z,b)*s\\&#xA;= &amp;amp;30&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the profit per sister is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\$30$, since sister with 10 can only sell singles at 3. sister with 30 sells 2 bundles of 12 at 6 and 6 eggs as singles at 3 and sister with with 50 sells 4 bundles at 6 and 2 singles at 3.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2803" LastEditorDisplayName="user20" LastEditDate="2015-02-17T16:15:23.280" LastActivityDate="2015-02-17T16:15:23.280" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9090" PostTypeId="2" ParentId="9067" CreationDate="2015-02-16T17:59:50.807" Score="7" Body="&lt;p&gt;Edit: Someone else already had this answer, my apologies. I didn't notice it because it looked more complex than it actually is. Still, here's a simple way of reading it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we imagine that the eggs are sold in dozens, then:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; perhaps the eggs  are sold at a rate instead of a price. I suggest &lt;strong&gt;6 units per dozen eggs, and 3 units for each individual egg&lt;/strong&gt;. This rate allows the following:&lt;br&gt;&#xA; - The youngest daughter has 50 eggs = 4 dozen + 2 individual. 4*6 + 2*3 = 24+6 = 30 units.&lt;br&gt;&#xA; - The middle daughter has 30 eggs = 2 dozen + 6 individual. 2*6 + 6*3 = 12+18 = 30 units.&lt;br&gt;&#xA; - The eldest daughter has 10 eggs = 0 dozen + 10 individual. 0*6 + 10*3 = 30 units.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9319" LastActivityDate="2015-02-16T17:59:50.807" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9091" PostTypeId="2" ParentId="9014" CreationDate="2015-02-16T18:01:06.477" Score="7" Body="&lt;p&gt;the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; indecisive &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the riddle makes reference to being indecisive with &quot;back and forth&quot; and &quot;unconfident&quot;, and most importantly, the answer is spelled out through the first and last letters of each sentence.&lt;br&gt;&#xA;&lt;br&gt;&#xA; N I / D E / C I / I S / V E  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but i can't quite figure out why &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; N I and I S are flipped.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit diagram:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WeTyE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7402" LastEditorUserId="9319" LastEditDate="2015-02-16T18:46:39.723" LastActivityDate="2015-02-16T18:46:39.723" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9092" PostTypeId="2" ParentId="9085" CreationDate="2015-02-16T18:44:34.763" Score="2" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The moon is not made of green cheese: D is false.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If A is false, then D is false and E is true and the moon is not burning-red.&#xA;If E is true, then B is false. Then C and E are either both true or both false, and E true implies C true. Then the moon is &lt;strong&gt;BLUE&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If A is true and the moon is not burning red, then E is false. If E is false, then B is true. Then C and E are not both true and not both false, and E false implies C true. Then the moon is &lt;strong&gt;BLUE&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If A is true and the moon is burning red. Then C is false. IF E is true, then B is false, and C is true, and this contradicts C is false. If E is false, then B is true, and C is true, and this contradicts C is false. No solution here.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Two solutions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A, B, C true and D, E false and the moon is &lt;strong&gt;BLUE&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;A, B, D false and C, E true and the moon is &lt;strong&gt;BLUE&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Therefore the moon is always &lt;strong&gt;BLUE&lt;/strong&gt;!&lt;/p&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2015-02-16T18:44:34.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9093" PostTypeId="1" AcceptedAnswerId="9094" CreationDate="2015-02-16T18:47:29.170" Score="15" ViewCount="5635" Body="&lt;p&gt;My FIRST is often at the front door;&lt;br&gt;&#xA;my SECOND is found in the cereal family;&lt;br&gt;&#xA;my THIRD is what most people want;&lt;br&gt;&#xA;my WHOLE is one of the united states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="1664" LastEditDate="2015-03-09T01:49:52.717" LastActivityDate="2015-03-09T01:49:52.717" Title="A charade: first at the door, second cereal, third wanted, whole united" Tags="|riddle|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9094" PostTypeId="2" ParentId="9093" CreationDate="2015-02-16T18:53:54.787" Score="16" Body="&lt;p&gt;Answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Matrimony&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My FIRST is often at the front door&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Mat&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My SECOND is found in the cereal family&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Rye&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My THIRD is what most people want&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Money&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My WHOLE is one of the united states&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Matrimony&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9346" LastActivityDate="2015-02-16T18:53:54.787" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9095" PostTypeId="2" ParentId="9067" CreationDate="2015-02-16T20:29:25.097" Score="2" Body="&lt;p&gt;Set the price of an egg at 3 eggs. The youngest girl should then buy 10 eggs for the price of 30 from the oldest girl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set the price at 1 unit per egg. At the end of the day, each girl will have made 30 units, or 90 units in total.&lt;/p&gt;&#xA;" OwnerUserId="9572" LastEditorUserId="9572" LastEditDate="2015-02-16T20:34:55.917" LastActivityDate="2015-02-16T20:34:55.917" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9096" PostTypeId="2" ParentId="9067" CreationDate="2015-02-16T21:07:41.457" Score="0" Body="&lt;p&gt;The girls, who love each other dearly, redistribute the eggs between themselves on the way to the market so they have 30 each. They then sell them at 1 unit per egg, no price change needed.&lt;/p&gt;&#xA;" OwnerUserId="9573" LastActivityDate="2015-02-16T21:07:41.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9097" PostTypeId="1" AcceptedAnswerId="10069" CreationDate="2015-02-16T22:30:38.490" Score="27" ViewCount="1767" Body="&lt;p&gt;This is a variant of &lt;a href=&quot;https://puzzling.stackexchange.com/questions/9036/the-lion-and-the-zebras&quot;&gt;this excellent question&lt;/a&gt;. The rules of the lion-zebra hunger games are exactly the same:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The lion plays a deadly game against a group of 100 zebras that takes&#xA;  place in the steppe (= an infinite plane). The lion starts in the&#xA;  origin with coordinates $(0,0)$, while the 100 zebras may arbitrarily&#xA;  pick their 100 starting positions. The lion and the group of&#xA;  zebras move alternately:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;In a lion move, the lion moves from its current position to a position at most 100 meters away.&lt;/li&gt;&#xA;  &lt;li&gt;In a zebra move, &lt;em&gt;one&lt;/em&gt; of the 100 zebras moves from its current position to a position at most 100 meters away.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The difference? &lt;strong&gt;The lion's victory condition is not to catch a zebra, but instead to get within a leisurely 1mm of one.&lt;/strong&gt; Who wins, the lion or the zebras? Disclaimer: I don't know the answer to this version. I'm legitimately interested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Point of advice: non-rigorous ideas are super welcome, but I advise you label them as such. The original question has three deleted answers and two answers with a negative score. Tread carefully, lest ye trip over the pile of skeletons lining your path.&lt;/p&gt;&#xA;" OwnerUserId="1752" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2018-05-14T19:01:53.787" Title="The lion and the zebras II: The lion with millimeter-long claws" Tags="|strategy|game|pursuit-evasion|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9098" PostTypeId="1" AcceptedAnswerId="9216" CreationDate="2015-02-16T22:37:39.447" Score="8" ViewCount="571" Body="&lt;p&gt;Billy the gardener uses a grey lego bolt to plant beans inside the timber ring, Sallah!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who is not in the garden?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A fellowship of nine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9385" LastEditorUserId="9385" LastEditDate="2015-02-18T16:54:57.027" LastActivityDate="2015-02-18T22:18:13.210" Title="Who is not in the garden?" Tags="|enigmatic-puzzle|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9099" PostTypeId="2" ParentId="9067" CreationDate="2015-02-16T22:46:04.403" Score="1" Body="&lt;p&gt;A simple and stupid solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. They set the price to \$3, the eldest sells her 10 eggs&lt;br&gt;&#xA; 2. They set the price to \$1, the second sells her 30 eggs&lt;br&gt;&#xA; 3. They set the price to \$0.6, the youngest sells her 50 eggs  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or, minimizing the number of price changes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They set the price to \$3, the eldest sells 10 eggs, the second 5&lt;br&gt;&#xA; They set the price to \$0.6, the second sells 25 eggs, the youngest 50  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2136" LastActivityDate="2015-02-16T22:46:04.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9102" PostTypeId="1" AcceptedAnswerId="9103" CreationDate="2015-02-17T00:43:44.830" Score="10" ViewCount="584" Body="&lt;p&gt;The story is told that when the Republic of Texas was feuding&#xA;with Mexico, there was a law passed by Texas to the effect that a Mexican silver dollar was&#xA;worth only 50 cents in Texas; whereupon Mexico passed a law that a Texan silver dollar was&#xA;worth only 50 cents in Mexico.&#xA;A citizen of Brownsville, Texas, walked across the bridge over the Rio Grande to Matamoros,&#xA;Mexico, went into a bar, ordered a 50-cent drink of whisky, paid for it with a Mexican dollar&#xA;and received a Texan dollar in change. He then walked back to Texas, went into a bar, ordered&#xA;a drink, paid for it with the Texan dollar, and received a Mexican dollar in change.&#xA;This pleasant pastime, repeated several times a day, went on all winter. At the end of the&#xA;time the Texan citizen still had his dollar. Obviously the barkeepers suffered no financial&#xA;losses; and the question is: Who paid for the drinks?&lt;/p&gt;&#xA;" OwnerUserId="9580" LastEditorUserId="4551" LastEditDate="2015-02-17T01:24:25.923" LastActivityDate="2021-12-14T08:39:03.090" Title="Paying for drinks with Mexican and Texan dollars" Tags="|logical-deduction|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9103" PostTypeId="2" ParentId="9102" CreationDate="2015-02-17T01:03:41.837" Score="10" Body="&lt;p&gt;The citizen has performed &lt;a href=&quot;https://en.wikipedia.org/wiki/Arbitrage&quot; rel=&quot;nofollow noreferrer&quot;&gt;financial arbitrage&lt;/a&gt; against the inconsistent exchange rates offered by the two sides. The citizen purely gained value, and at the two barkeeps lost that much value in total.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Citizen: +2 drinks&lt;br /&gt;&#xA;Barkeep 1: -1 Texan dollar, +1 Mexican dollar, -1 drink&lt;br /&gt;&#xA;Barkeep 2: +1 Texan dollar, -1 Mexican dollar, -1 drink&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So, one of the barkeeps has lost value, the amount of which will be determined when the exchange rate stabilizes and the value of the currency settles. For example, if the exchange rate settles to 1 Texan dollar equaling 1 Mexican dollar, then both will have lost one drink of value.&lt;/p&gt;&#xA;&lt;p&gt;But, if it stabilizes to, say,&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;1 Texan dollar = 1 drink; 1 Mexican dollar = 4 drinks&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;then we'll have a value change&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Citizen: +2 drinks&lt;br /&gt;&#xA;Barkeep 1: +2 drinks&lt;br /&gt;&#xA;Barkeep 2: -4 drinks&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;and one barkeep will have lost and the other gained.&lt;/p&gt;&#xA;&lt;p&gt;If the barkeeps use their currency to buy goods and services from others, then they will pass on the risk of loss from the currency being misvalued. Like a &lt;a href=&quot;https://en.wikipedia.org/wiki/Pyramid_scheme&quot; rel=&quot;nofollow noreferrer&quot;&gt;pyramid scheme&lt;/a&gt;, this loss can be passed on indefinitely as people use this currency within their country. But, currency only has value in its purchasing power, so the loss must eventually be realized The loser depends on who is holding it when exchange rates settle and what they settle to.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="75318" LastEditDate="2021-12-14T08:39:03.090" LastActivityDate="2021-12-14T08:39:03.090" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="9104" PostTypeId="1" AcceptedAnswerId="9106" CreationDate="2015-02-17T01:10:41.207" Score="8" ViewCount="1001" Body="&lt;p&gt;My three children define me.&lt;br&gt;&#xA;Without them I am nothing.&lt;br&gt;&#xA;My most active child is constantly busy.&lt;br&gt;&#xA;My second child stays the same but is always growing.&#xA;I just don't know about my third child.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;" OwnerUserId="9495" LastEditorUserId="9495" LastEditDate="2015-05-29T06:26:04.767" LastActivityDate="2015-05-29T06:26:04.767" Title="My three children define me" Tags="|riddle|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9105" PostTypeId="1" AcceptedAnswerId="9108" CreationDate="2015-02-17T02:02:49.640" Score="7" ViewCount="6100" Body="&lt;p&gt;Four eyes have I&lt;br&gt;&#xA;And a mouth of one&lt;br&gt;&#xA;I do not speak&lt;br&gt;&#xA;For I have no tongue&lt;br&gt;&#xA;I have the curves&lt;br&gt;&#xA;To match my eyes&lt;br&gt;&#xA;I'm wide and skinny&lt;br&gt;&#xA;To no one's surprise&lt;br&gt;&#xA;I may be a female&lt;br&gt;&#xA;Given my start&lt;br&gt;&#xA;St. Louis holds&lt;br&gt;&#xA;The bottom of my heart&lt;br&gt;&#xA;Two legs I have&lt;br&gt;&#xA;Though I cannot stand&lt;br&gt;&#xA;But I think they're the finest&lt;br&gt;&#xA;In all the lands&lt;br&gt;&#xA;Can you hazard a guess&lt;br&gt;&#xA;As to what, or who&lt;br&gt;&#xA;Surely by now&lt;br&gt;&#xA;I've given you plenty of clues&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What or who am I?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4353" LastEditorUserId="4353" LastEditDate="2015-04-11T08:48:21.630" LastActivityDate="2015-04-11T11:14:55.267" Title="Four eyes have I, and a mouth of one" Tags="|riddle|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9106" PostTypeId="2" ParentId="9104" CreationDate="2015-02-17T02:07:57.923" Score="12" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Father Time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The three children are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Past time, Present time, and Future time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without them I am nothing:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Father Time cannot exist without time itself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My most active child is a doer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Present is always working&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My third child is a mystery:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Future is a mystery to everyone&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My first child is the opposite of the third:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Past has already happened; the Future has not. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4353" LastActivityDate="2015-02-17T02:07:57.923" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9107" PostTypeId="2" ParentId="9105" CreationDate="2015-02-17T02:16:50.803" Score="1" Body="&lt;p&gt;My guess &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eyeglasses&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Four eyes have I And a mouth of one&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Four eyes is for someone with eyeglasses, also have one mouth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm wide and skinny To no one's surprise I may be a female&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The lens is wide and skinny&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;St. Louis holds The bottom of my heart&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; St Louis is one of eyeglass brand company. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Two legs I have&#xA;Though I cannot stand&#xA;But I think they're the finest&#xA;In all the lands&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eyeglasses have two handle and fit in everyone head.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9160" LastActivityDate="2015-02-17T02:16:50.803" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9108" PostTypeId="2" ParentId="9105" CreationDate="2015-02-17T02:17:54.877" Score="3" Body="&lt;p&gt;I'm going to go with &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;strong&gt;Mississippi (River)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation will follow, gradually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Four eyes have I and a mouth of one&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; M&lt;strong&gt;i&lt;/strong&gt;ss&lt;strong&gt;i&lt;/strong&gt;ss&lt;strong&gt;i&lt;/strong&gt;pp&lt;strong&gt;i&lt;/strong&gt; are the four I's, and the river has one mouth.  &lt;strong&gt;OP Edit: Additionally, the &quot;M&quot; can appear to be &quot;mouth-like&quot;.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I do not speak for I have no tongue&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rivers have mouths but no tongues.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have the curves to match my eyes, I'm wide and skinny to no one's surprise&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rivers have curves and have different sizes at different points.  &lt;strong&gt;OP Reasoning: The curves to match the eyes (i's), is the four s's in Mississippi.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I may be a female given my start&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is where I'm confused, at least a little bit. Maybe it's that rivers and bodies of water can be referred to as &quot;she,&quot; but I'm not sure. &lt;strong&gt;OP Reasoning: The first 4 letters &quot;Miss&quot; indicates a female.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;St. Louis holds the bottom of my heart&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The river begins in St. Louis, or at least around that area.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Two legs I have, though I cannot stand&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &quot;legs&quot; of a river are not really for standing. &lt;strong&gt;OP Reasoning: The &quot;legs&quot; are the p's in Mississippi.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I think they're the finest in all the lands&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Mississippi river is the longest in the United States.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1664" LastEditorUserId="9422" LastEditDate="2015-04-11T11:14:55.267" LastActivityDate="2015-04-11T11:14:55.267" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9109" PostTypeId="2" ParentId="9105" CreationDate="2015-02-17T02:20:05.730" Score="2" Body="&lt;p&gt;My answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Mississippi river&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  It is wide and skinny.  St. Louis holds the bottom of its heart.  A river has one mouth.  Cannot stand because legs of a river are not used for standing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9495" LastActivityDate="2015-02-17T02:20:05.730" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9110" PostTypeId="1" AcceptedAnswerId="9129" CreationDate="2015-02-17T02:28:50.493" Score="10" ViewCount="2421" Body="&lt;p&gt;As a snake impaled, I am not free.&lt;br&gt;&#xA;Rather, I am displayed for all to see.&lt;br&gt;&#xA;Why do people not mourn me?&lt;br&gt;&#xA;Instead they want more of me.&lt;br&gt;&#xA;You want more?&lt;br&gt;&#xA;Four is my door.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;note:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I added 2 lines to the riddle.&lt;/p&gt;&#xA;" OwnerUserId="9495" LastEditorUserId="9495" LastEditDate="2015-02-17T18:05:33.003" LastActivityDate="2015-02-18T02:00:24.510" Title="As a snake impaled, I am not free" Tags="|riddle|" AnswerCount="6" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9111" PostTypeId="1" AcceptedAnswerId="9140" CreationDate="2015-02-17T02:36:06.870" Score="2" ViewCount="752" Body="&lt;p&gt;My cousins come in many colors.&lt;br&gt;&#xA;They are bought and sold.&lt;br&gt;&#xA;But me, I am always with you.&lt;br&gt;&#xA;Even when you're old.&lt;br&gt;&#xA;My cousins are more useful.&lt;br&gt;&#xA;But there is no use for me.&lt;br&gt;&#xA;I cannot be worn on clothing&lt;br&gt;&#xA;You were born with me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who or what am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above poem is updated to make it less broad, especially note the last line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct answer consists of two parts, as there are two subjects in the riddle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who or what are the cousins?&#xA;Who or what is the main object?&lt;/p&gt;&#xA;" OwnerUserId="9495" LastEditorUserId="9495" LastEditDate="2015-02-17T09:44:46.163" LastActivityDate="2018-01-08T09:58:29.837" Title="My cousins come in many colors" Tags="|riddle|" AnswerCount="8" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9112" PostTypeId="1" AcceptedAnswerId="9122" CreationDate="2015-02-17T02:43:50.760" Score="3" ViewCount="2991" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I can run, but I cannot walk. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;When not running, I cannot move.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I only move back and forth.    &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;When I move, I move not only me.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;People cannot talk to me without sounding strange.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I am not wet. If I get wet, I may kill you.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I usually run only when it is hot.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who or what am I?&lt;/p&gt;&#xA;" OwnerUserId="9495" LastEditorUserId="31544" LastEditDate="2016-11-10T07:18:09.717" LastActivityDate="2016-11-10T07:18:09.717" Title="When I run, I cannot walk" Tags="|riddle|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9113" PostTypeId="2" ParentId="9112" CreationDate="2015-02-17T02:47:44.340" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It runs, but doesn't walk. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's moving, it's running. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It moves other things along with it. Just ask Huck Finn or &lt;a href=&quot;https://puzzling.stackexchange.com/q/7820/1563&quot;&gt;Pikachu&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And it definitely makes voices sound strange, if by talking &lt;em&gt;to&lt;/em&gt; it you mean talking &lt;em&gt;in&lt;/em&gt; it.&lt;/p&gt;&#xA;" OwnerUserId="1563" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-02-17T02:47:44.340" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9117" PostTypeId="1" AcceptedAnswerId="9130" CreationDate="2015-02-17T02:59:55.267" Score="5" ViewCount="4569" Body="&lt;p&gt;My head has no face,&lt;br&gt;&#xA;but I breathe and bleed.&lt;br&gt;&#xA;I am quiet,&lt;br&gt;&#xA;but my visitors love to sing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who or what am I?&lt;/p&gt;&#xA;" OwnerUserId="9495" LastEditorUserId="39" LastEditDate="2015-02-17T03:02:29.790" LastActivityDate="2015-02-22T23:38:57.663" Title="My head has no face" Tags="|riddle|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9118" PostTypeId="2" ParentId="9110" CreationDate="2015-02-17T03:01:07.827" Score="4" Body="&lt;p&gt;Is it perhaps a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Belt?&lt;br&gt;&lt;br&gt;&#xA; A belt  is sort of serpentine. It's &quot;impaled&quot; by prong which fastens it, so it's not free. Belts are worn visible to everyone.  No one mourns them. There's a demand for them, since people buy them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7489" LastEditorUserId="1563" LastEditDate="2015-02-17T03:02:05.503" LastActivityDate="2015-02-17T03:02:05.503" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9119" PostTypeId="2" ParentId="9117" CreationDate="2015-02-17T03:06:10.860" Score="11" Body="&lt;p&gt;My guess (yes, I love answering your riddles...)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Microphone&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My head has no face:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The top of the microphone is considered the head.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I breathe and bleed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There's a such thing as microphone bleed, and I'm guessing the mesh is the 'breathe' part.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am quiet:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Microphones by themselves don't make noise&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but my visitors love to sing.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Microphones are used generally for singing (karaoke, anyone?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Even if it's not the right answer, it makes sense to me (I'm a musician by hobby).&lt;/p&gt;&#xA;" OwnerUserId="4353" LastActivityDate="2015-02-17T03:06:10.860" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="9121" PostTypeId="2" ParentId="9112" CreationDate="2015-02-17T03:29:58.763" Score="2" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A river&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can run, but I cannot walk.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A river runs but does not walk&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When not running, I cannot move.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When a river does not run, it stays still and does not move&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I move, I move not only me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A river carries many things when it flows&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;People cannot talk to me without&#xA;sounding strange.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Listening to the words spoken by a person when we are underwater will surely sound strange!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7482" LastActivityDate="2015-02-17T03:29:58.763" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9122" PostTypeId="2" ParentId="9112" CreationDate="2015-02-17T03:56:16.227" Score="8" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an electric fan&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A fan moves back and forth (well, moving fans do). Fan moves air. If you get a fan wet, the electronic components could possibly kill you. If you talk into a fan, you sound strange. People generally run fans when they are hot.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4353" LastEditorUserId="9495" LastEditDate="2015-02-17T08:17:57.140" LastActivityDate="2015-02-17T08:17:57.140" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9123" PostTypeId="2" ParentId="9018" CreationDate="2015-02-17T04:42:35.817" Score="9" Body="&lt;p&gt;Now that you've added 3 more lines and reading the answer of &lt;a href=&quot;http://www.puzzling.stackexchange.com/questions/9104/my-three-children-define-me&quot;&gt;this riddle&lt;/a&gt; made by you, I think that the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The present&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a young child, you live here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You live in the present&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later you leave me often &lt;br&gt;&#xA;But you always return, sooner or later:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You start to think about your future and make plans. But some time or the other, you return to the present moment&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no beginning or end &lt;br&gt;&#xA;I am infinite&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The present moment hasn't got a beginning or an end. It is boundless.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, if God exists,&lt;br&gt;&#xA;God must live here, too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; God lives in the present moment.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Worry, doubt and fear are rarely&#xA;found with me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You worry,doubt and fear about something that is going to happen in the future ,not the present.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I only have two siblings.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The two siblings of the present moment is the past and the future&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you are not with me, you are&#xA;with one of them.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If one is not thinking about the present, then he/she is definetely thinking about the past or the future.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7482" LastActivityDate="2015-02-17T04:42:35.817" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="9126" PostTypeId="2" ParentId="9110" CreationDate="2015-02-17T06:06:01.343" Score="3" Body="&lt;p&gt;Is it a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Wrist Watch (with leather strap)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a snake impaled&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Has pierced holes to insert buckles in leather starp.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am not free&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Tied to the wrist&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rather, I am displayed for all to see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Displayed for all to check time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why do people not mourn me?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;We won't mourn when the watch is dead&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Instead they want more of me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Instead we repair, or buy new batteries to run it again&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastActivityDate="2015-02-17T06:06:01.343" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9127" PostTypeId="2" ParentId="9110" CreationDate="2015-02-17T06:33:17.023" Score="0" Body="&lt;p&gt;Are they,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Animals in Zoo&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a snake impaled, I am not free.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Animals kept in a cage.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rather, I am displayed for all to see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They all kept for us to see.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why do people not mourn me?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People do not mourn if any of animal dies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Instead they want more of me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Instead we want more spices of them to see.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5101" LastActivityDate="2015-02-17T06:33:17.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9128" PostTypeId="2" ParentId="9110" CreationDate="2015-02-17T06:39:37.923" Score="2" Body="&lt;p&gt;First idea I could come up with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;A ring&lt;/strong&gt;. The idea stems from the explanations of other answers: snake impaled may refer to the finger going through the center, it is worn to be shown off and displayed. The ring is often a symbol of celebration, as opposed to mourning. And of course, rings are worth something, be it sentimental or what have you.&lt;br&gt;&#xA;&lt;br&gt;&#xA; As for the hint, &quot;4 is my door&quot;, he refers to the fourth finger, also known as the Ring finger, where a ring is most commonly worn.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9319" LastEditorUserId="9319" LastEditDate="2015-02-17T07:03:10.420" LastActivityDate="2015-02-17T07:03:10.420" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9129" PostTypeId="2" ParentId="9110" CreationDate="2015-02-17T06:57:33.640" Score="20" Body="&lt;p&gt;I think I got it now.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;The dollar sign.&lt;/strong&gt; The $ looks like a snake impaled, displayed in front of all dollar amounts, is not mourned for contributing to one's wealth, and of course, people want more. &quot;Four is my door&quot; tipped me off, where it refers to the number 4 on the keyboard having the dollar sign as its shift-text.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9319" LastEditorUserId="9319" LastEditDate="2015-02-17T07:03:32.590" LastActivityDate="2015-02-17T07:03:32.590" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9130" PostTypeId="2" ParentId="9117" CreationDate="2015-02-17T07:27:58.580" Score="13" Body="&lt;p&gt;I'll throw my hat into the ring. The first line is a stretch, but the others fit well.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Tree&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My head has no face,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Trees are sometimes compared to old men (Old Man Willow), but do not have human faces.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I breathe and bleed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Plants can 'breath' using processes such as photosynthesis and respiration. Trees bleed sap.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am quiet,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Trees don't make noise on their own.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but my visitors love to sing.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Birds nest in trees, and many love to sing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;......&lt;/p&gt;&#xA;" OwnerUserId="9587" LastEditorUserId="9587" LastEditDate="2015-02-17T18:39:45.293" LastActivityDate="2015-02-17T18:39:45.293" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="9131" PostTypeId="1" AcceptedAnswerId="9552" CreationDate="2015-02-17T07:51:41.087" Score="8" ViewCount="1011" Body="&lt;p&gt;Here is a diagram and challenge description that should be clear and simple to understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Q0g8G.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9531" LastActivityDate="2015-02-25T00:07:52.463" Title="Triangle area equals quadrilateral area" Tags="|geometry|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="9132" PostTypeId="2" ParentId="9111" CreationDate="2015-02-17T08:05:31.180" Score="2" Body="&lt;p&gt;Who or what am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Spectacles or Eyeglasses or contact lenses&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They are bought and sold.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Shops buy it or sell it. They may even be resold&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My cousins come in many colors.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sunglasses or colour Contact lenses&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Even when you're old.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most old people suffer from Hyperopia&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My cousins are more useful&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Coloured Contact lenses/Sunglasses can be worn with or without refractive index&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But what is the use of me?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Restores your vision&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cannot be worn on clothing&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You cannot wear a glass with its wrapper&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am just there for you to see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This was the most obvious hint&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4177" LastEditorUserId="4177" LastEditDate="2015-02-17T08:12:39.967" LastActivityDate="2015-02-17T08:12:39.967" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9133" PostTypeId="1" AcceptedAnswerId="9418" CreationDate="2015-02-17T08:18:21.503" Score="9" ViewCount="2019" Body="&lt;p&gt;A great wise man once said that&lt;br&gt;&#xA;although it is only a foot or two away,&lt;br&gt;&#xA;this journey can take you many years.&lt;br&gt;&#xA;The end of this journey is what you seek, &#xA;whether you know it or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No matter you are young or old, the mind can be cold.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To answer this question, there are two parts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is this destination?&#xA;Where are you traveling from?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a true story.&lt;/p&gt;&#xA;" OwnerUserId="9495" LastEditorUserId="9495" LastEditDate="2015-02-18T00:24:07.300" LastActivityDate="2015-02-22T02:17:40.543" Title="The end of this journey" Tags="|riddle|" AnswerCount="7" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9134" PostTypeId="2" ParentId="9133" CreationDate="2015-02-17T08:26:53.670" Score="9" Body="&lt;p&gt;The destination is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Death&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are travelling to there from:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Life&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;although it is only a foot or two away,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are buried at that depth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;this journey can take you many years.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It takes pretty much your whole life&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;The end of this journey is what you seek, whether you know it or not.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We are all headed there, whether we realize it or not.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastActivityDate="2015-02-17T08:26:53.670" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9136" PostTypeId="2" ParentId="9131" CreationDate="2015-02-17T09:18:50.137" Score="1" Body="&lt;p&gt;If I am allowed to measure distances with the ruler, then it would work to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; measure the distance from the crossing point if the two long lines to the point B and replicate that distance from E towards A, calling the new point G. Since FDG has the sum of the perpendicular heights of the two triangles that have FD as a base, it has the same area as the quadrilateral.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of course if you mean ruler as in the classical ruler and compass construction, without the compass, I'm more stuck!&lt;/p&gt;&#xA;" OwnerUserId="9532" LastActivityDate="2015-02-17T09:18:50.137" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9137" PostTypeId="2" ParentId="9111" CreationDate="2015-02-17T09:37:15.733" Score="0" Body="&lt;p&gt;My guess is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Piercings as the main subject, with tattoos being the cousins.&lt;/strong&gt;&lt;br&gt;&#xA; Tattoos are more colorful, can be bought or sold as temporary tattoos (or you can figure the service of applying a tattoo as being bought/sold), and tattoos are useful for inscribing a message or having a meaning. On the other hand, the purpose of piercings is to be seen, and they are not worn on clothing.&lt;br&gt;&#xA; Most importantly, both tattoos and piercings are permanent, and will stay with you unti you're old. =)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9319" LastActivityDate="2015-02-17T09:37:15.733" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9138" PostTypeId="2" ParentId="9111" CreationDate="2015-02-17T09:59:28.873" Score="1" Body="&lt;p&gt;My cousins come in many colors.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Leather jackets come in different colours&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They are bought and sold.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Leather jackets are bought and sold in shops&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But me, I am always with you.&#xA;Even when you're old.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Our skin is always with us&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My cousins are more useful.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Leather jackets can protect us from harsh climate as well as they are for ornamental purpose&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cannot be worn on clothing&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;we can not put skin on clothing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You were born with me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Everyone is born with skin&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Skin is the main object and leather jackets are the cousins&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9174" LastActivityDate="2015-02-17T09:59:28.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9139" PostTypeId="2" ParentId="9111" CreationDate="2015-02-17T10:03:54.730" Score="1" Body="&lt;p&gt;At first I thought they were ruled out, but&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the main subject is &lt;strong&gt;probably your irises and the cousins are contact lenses.&lt;/strong&gt; &#xA; Arguably, the iris &lt;em&gt;does&lt;/em&gt; serve a purpose to the eye, as do all parts of the eye. However, the iris is what determines eye color, and whereas eye color has no function, it would be most likely considered the &quot;cousin&quot; of the colored contact lens.&#xA; Your irises stay with you from birth until you're old, usually, and contact lenses are bought/sold in many colors and may have the added function of corrective eyesight. Your irises cannot be worn on clothes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9319" LastActivityDate="2015-02-17T10:03:54.730" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9140" PostTypeId="2" ParentId="9111" CreationDate="2015-02-17T10:17:37.980" Score="7" Body="&lt;p&gt;You are a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; belly button,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and your cousins are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ordinary buttons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My cousins come in many colors.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Colourful buttons&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They are bought and sold.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In stores for clothes and accessories&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But me, I am always with you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even when you're old.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Obviously! :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My cousins are more useful.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Holds together shirts, jackets, suits, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But there is no use for me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After birth, that is!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cannot be worn on clothing&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's 'worn' under clothes, not on top of them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You were born with me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Actually, maybe not quite true - I'm not a medical doctor, but I thought it's what's left after the umbilical cord is cut.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9422" LastEditorUserId="9422" LastEditDate="2015-02-17T15:51:34.637" LastActivityDate="2015-02-17T15:51:34.637" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="9141" PostTypeId="2" ParentId="9133" CreationDate="2015-02-17T10:17:53.123" Score="0" Body="&lt;p&gt;I call upon the lessons my father has taught me when I answer this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first step. It has been said that the first step toward any goal is always the most difficult one, in that it means beginning your journey requires motivation that you may need to accumulate over the years. The distance in the idiom is measured as a foot or two because that's the accepted length of a human stride. Sometimes, the end of your journey is not what you expected when you started. It's difficult to spell out a specific &lt;em&gt;to&lt;/em&gt; and &lt;em&gt;from&lt;/em&gt;, but I suppose the best summary would be&lt;br&gt;&#xA; &lt;strong&gt;To: The first step towards a goal&lt;br&gt;&#xA; From: Figuratively standing in place, unmotivated&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9319" LastActivityDate="2015-02-17T10:17:53.123" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9142" PostTypeId="1" AcceptedAnswerId="9144" CreationDate="2015-02-17T10:33:40.610" Score="6" ViewCount="883" Body="&lt;p&gt;Professor Halfbrain has recently made a fascinating discovery on the two integers &lt;span class=&quot;math-container&quot;&gt;$x=101$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$y=110$&lt;/span&gt;.&#xA;The professor first observed that the decimal representation of &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; is a permutation of the decimal representation of &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt;.&#xA;While this observation is quite straightforward (and some might even say: quite boring), the really fascinating part of Halfbrain's discovery was&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;that the decimal digit sums of the numbers &lt;span class=&quot;math-container&quot;&gt;$2x$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$2y$&lt;/span&gt; coincide!&lt;/li&gt;&#xA;&lt;li&gt;and that the same holds true for the digit sums of &lt;span class=&quot;math-container&quot;&gt;$3x$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$3y$&lt;/span&gt;!!&lt;/li&gt;&#xA;&lt;li&gt;and also for the digit sums of &lt;span class=&quot;math-container&quot;&gt;$4x$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$4y$&lt;/span&gt;!!!&lt;/li&gt;&#xA;&lt;li&gt;and for the digit sums of &lt;span class=&quot;math-container&quot;&gt;$5x$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$5y$&lt;/span&gt;!!!!&lt;/li&gt;&#xA;&lt;li&gt;and for the digit sums of &lt;span class=&quot;math-container&quot;&gt;$6x$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$6y$&lt;/span&gt;!!!!!&lt;/li&gt;&#xA;&lt;li&gt;and for the digit sums of &lt;span class=&quot;math-container&quot;&gt;$7x$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$7y$&lt;/span&gt;!!!!!!&lt;/li&gt;&#xA;&lt;li&gt;and for the digit sums of &lt;span class=&quot;math-container&quot;&gt;$8x$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$8y$&lt;/span&gt;!!!!!!!&lt;/li&gt;&#xA;&lt;li&gt;and for the digit sums of &lt;span class=&quot;math-container&quot;&gt;$9x$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$9y$&lt;/span&gt;!!!!!!!!&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;I am aware of the fact that this sounds almost unbelievable, but please trust me: it really is true, just check it yourself.&#xA;Based on these experimental results, Professor Halfbrain has now formulated a theorem with eight statements. At the current moment, he is busily working on the proofs.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Halfbrain's big theorem:&lt;/strong&gt; For any two positive integers &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; for which the decimal representation of &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; is a permutation of the decimal representation of &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt;, the following eight statements hold true:&lt;/p&gt;&#xA;&lt;p&gt;(a). The digit sum of &lt;span class=&quot;math-container&quot;&gt;$2x$&lt;/span&gt; equals the digit sum of &lt;span class=&quot;math-container&quot;&gt;$2y$&lt;/span&gt;.&lt;br /&gt;&#xA;(b). The digit sum of &lt;span class=&quot;math-container&quot;&gt;$3x$&lt;/span&gt; equals the digit sum of &lt;span class=&quot;math-container&quot;&gt;$3y$&lt;/span&gt;.&lt;br /&gt;&#xA;(c). The digit sum of &lt;span class=&quot;math-container&quot;&gt;$4x$&lt;/span&gt; equals the digit sum of &lt;span class=&quot;math-container&quot;&gt;$4y$&lt;/span&gt;.&lt;br /&gt;&#xA;(d). The digit sum of &lt;span class=&quot;math-container&quot;&gt;$5x$&lt;/span&gt; equals the digit sum of &lt;span class=&quot;math-container&quot;&gt;$5y$&lt;/span&gt;.&lt;br /&gt;&#xA;(e). The digit sum of &lt;span class=&quot;math-container&quot;&gt;$6x$&lt;/span&gt; equals the digit sum of &lt;span class=&quot;math-container&quot;&gt;$6y$&lt;/span&gt;.&lt;br /&gt;&#xA;(f). The digit sum of &lt;span class=&quot;math-container&quot;&gt;$7x$&lt;/span&gt; equals the digit sum of &lt;span class=&quot;math-container&quot;&gt;$7y$&lt;/span&gt;.&lt;br /&gt;&#xA;(g). The digit sum of &lt;span class=&quot;math-container&quot;&gt;$8x$&lt;/span&gt; equals the digit sum of &lt;span class=&quot;math-container&quot;&gt;$8y$&lt;/span&gt;.&lt;br /&gt;&#xA;(h). The digit sum of &lt;span class=&quot;math-container&quot;&gt;$9x$&lt;/span&gt; equals the digit sum of &lt;span class=&quot;math-container&quot;&gt;$9y$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Perhaps the professor was a little bit too excited about his discovery, and perhaps not each and every one of his eight statements will stand the test of time.&lt;br /&gt;&#xA;Can you help the professor in finding out which statements are true and which ones are false?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-02-17T15:32:47.877" Title="Professor Halfbrain's fascinating discovery" Tags="|mathematics|number-theory|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9143" PostTypeId="2" ParentId="9111" CreationDate="2015-02-17T10:52:17.213" Score="0" Body="&lt;p&gt;Who or what am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Body Hair&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My cousins come in many colors.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wool. Off-course wool is the body hair of sheep&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They are bought and sold.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Shops buy it or sell it. They may even be resold&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Even when you're old.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We have body hair even when we are old&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My cousins are more useful&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Undoubtedly, wool is more useful than our body hair. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But there is no use for me. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What is the use of body hair. And its a nuisance to a women :-) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cannot be worn on clothing&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You cannot wear body hair on clothing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You were born with me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Even an infant has body hair&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4177" LastActivityDate="2015-02-17T10:52:17.213" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9144" PostTypeId="2" ParentId="9142" CreationDate="2015-02-17T12:05:59.233" Score="7" Body="&lt;p&gt;(a) and (d) are correct, while the rest are false.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Reason&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This is because the digit sum of 2N (or 5N) is just the sum of digit sums of 2*d_i (or 5*d_i) where d_i are the digits of N. This is not true for multiplication with any digit except 2 or 5.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Proof&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The carry at each step on multiplication by 2 is going to be 0 or 1. As a result, adding the carry will not cause the result at the current step to go above a multiple of 10 since multiplying a digit with 2 never results in a 9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the carry at each step on multiplication by 5 is going to be between 0 and 4. Since multiplying a digit with 5 never results in anything but 0 or 5, adding the carry never makes the result go above the next multiple of 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not true for multiplication by other digits, and counterexamples can be found easily. For example, 34*3=102. The carry on multiplying 4 by 3 is 1. 3*3+1=10, taking it to the next multiple of 10. As a result, the digit sum is not the same as that of 43*3=129. The smallest counterexamples are&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;3: 34,43 &lt;/li&gt;&#xA;&lt;li&gt;4: 25,52 &lt;/li&gt;&#xA;&lt;li&gt;6: 17,71 &lt;/li&gt;&#xA;&lt;li&gt;7: 15,51 &lt;/li&gt;&#xA;&lt;li&gt;8: 13,31 &lt;/li&gt;&#xA;&lt;li&gt;9: 12,21&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9471" LastEditorUserId="9471" LastEditDate="2015-02-17T14:54:48.560" LastActivityDate="2015-02-17T14:54:48.560" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="9145" PostTypeId="2" ParentId="9133" CreationDate="2015-02-17T12:32:57.663" Score="14" Body="&lt;p&gt;What is this destination? &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Adulthood&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where are you traveling from?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Adolescence&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A great wise man once said that&#xA;although it is only a foot or two away,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most people will grow a foot, or two, from adolescence to adulthood&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;this journey can take you many years.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes, it will&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The end of this journey is what you seek, whether you know it or not.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No child wants to stay a child forever&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9168" LastActivityDate="2015-02-17T12:32:57.663" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9146" PostTypeId="2" ParentId="9067" CreationDate="2015-02-17T12:50:29.970" Score="3" Body="&lt;p&gt;Here is a solution with a &lt;strong&gt;single&lt;/strong&gt; change of price:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;In the morning, the girls sell the eggs at a price of 3 cent per bundle of 7 eggs.&lt;br&gt;&#xA;The eldest sells 1 bundle (and keeps 3 eggs),&lt;br&gt;&#xA;the second sells 4 bundles (and keeps 2 eggs),&lt;br&gt;&#xA;and the youngest sells 7 bundles (and keeps 1 egg).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In the afternoon, the girls sell the eggs at a price of 9 cent per single egg.&lt;br&gt;&#xA;The eldest sells 3 bundles,&lt;br&gt;&#xA;the second sells 2 bundles,&lt;br&gt;&#xA;and the youngest sells 1 bundle.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Altogether, the eldest receives $1\cdot3+3\cdot9=30$ cent,&lt;br&gt;&#xA;the second receives $4\cdot3+2\cdot9=30$ cent, and&lt;br&gt;&#xA;the youngest receives $7\cdot3+1\cdot9=30$ cent.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(There are plenty of other solutions that use two or three or four changes of price.)&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-02-17T12:50:29.970" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9147" PostTypeId="2" ParentId="9067" CreationDate="2015-02-17T13:21:52.163" Score="4" Body="&lt;p&gt;Well, with the edits there is now a really simple answer:&#xA;They set the starting price at 30 units per 50 eggs&#xA;Then they drop it for 30 units per 30 eggs&#xA;Then they drop it again to 30 units per 10 eggs&lt;/p&gt;&#xA;" OwnerUserId="9444" LastActivityDate="2015-02-17T13:21:52.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9148" PostTypeId="2" ParentId="9031" CreationDate="2015-02-17T14:20:15.833" Score="1" Body="&lt;p&gt;Most lines I believe could point to a  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;black hole&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;in a fairly straightforward way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond the beyond&#xA;In far reaches of flight&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Massive distances away&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Broken packed cosmic twist&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'Cosmic', 'broken' relating to a black hole being a hole in space&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Never lightly come upon it,&#xA;For if you do, you may vomit,&#xA;Your senses reeling in a flurry,&#xA;Run you will try, in a hurry,&#xA;But run and escape you cannot,&#xA;You'll be eaten before you can rot&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;! If you come upon a black hole, you are more than likely going to be unable to escape, made a mess of and 'eaten' into it.&#xA;  To gaze upon the humanlike face&#xA;  You won't live to see your race&#xA;  ! Unsure of the 'humanlike face' but if you gaze upon it, you're not going to escape (live) to see anyone again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9568" LastEditorUserId="9568" LastEditDate="2015-02-17T16:24:21.957" LastActivityDate="2015-02-17T16:24:21.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9149" PostTypeId="1" AcceptedAnswerId="9188" CreationDate="2015-02-17T14:44:13.390" Score="5" ViewCount="5361" Body="&lt;p&gt;I am there when you're born&lt;br&gt;&#xA;yet I begin to fade away from day one.&lt;br&gt;&#xA;I will stay until death&lt;br&gt;&#xA;even if you choose to erase me.&lt;br&gt;&#xA;Maybe even more, if you pass me to your children&lt;br&gt;&#xA;My worst enemy? it once killed, but not a human being&lt;/p&gt;&#xA;&#xA;&lt;p&gt;who am I?&lt;/p&gt;&#xA;" OwnerUserId="5122" LastEditorUserId="5373" LastEditDate="2015-03-14T18:54:58.843" LastActivityDate="2016-07-07T00:48:30.527" Title="Riddle ... I don't know" Tags="|riddle|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9150" PostTypeId="2" ParentId="9149" CreationDate="2015-02-17T14:51:37.377" Score="2" Body="&lt;p&gt;Is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; memory&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I begin to fade away from day one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Memories are doomed to fade from the beginning&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Maybe even more, if you pass me to your children&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can pass on memories in order to prolong them&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not sure about the last line but I'm sure it is applicable&lt;/p&gt;&#xA;" OwnerUserId="9568" LastActivityDate="2015-02-17T14:51:37.377" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9151" PostTypeId="2" ParentId="9149" CreationDate="2015-02-17T15:14:12.813" Score="1" Body="&lt;p&gt;I think it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Youth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am there when you're born&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you are born you are young&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet I begin to fade away from day one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You start becoming older day by day and hence youth starts fading&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I will stay until death, even if you choose to erase me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You will always be younger to someone or the other till you die&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Maybe even more, if you pass me to your children&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One has children, who are young when they are born&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reasoning for the last line isn't hitting me! :/&lt;/p&gt;&#xA;" OwnerUserId="9346" LastActivityDate="2015-02-17T15:14:12.813" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9152" PostTypeId="2" ParentId="9111" CreationDate="2015-02-17T15:31:48.493" Score="2" Body="&lt;p&gt;This is close, but there are two obscure points of contention near the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who or what am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Belly button.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My cousins come in many colors.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Buttons on clothing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They are bought and sold.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One can buy replacement buttons&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But me, I am always with you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Part of the body.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Even when you're old.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They don't go away (although, I have never inspected an elderly person to see if they still had a belly button). But I think this is common knowledge.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My cousins are more useful.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Buttons hold your clothes together.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But there is no use for me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Belly buttons collect lint, which is quite useless&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cannot be worn on clothing&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some sicko serial killer might disagree, but who knows?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You were born with me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is not quite accurate because the doctor gives the button, you were born with a feeding tube.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9602" LastActivityDate="2015-02-17T15:31:48.493" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9155" PostTypeId="1" CreationDate="2015-02-17T15:52:55.527" Score="38" ViewCount="3517" Body="&lt;p&gt;&lt;em&gt;Note:&lt;/em&gt; This problem remains unsolved, as of 19 April 2020, so do try it out. 400 rep bounty guaranteed for a correct answer&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This a variation of &lt;a href=&quot;https://puzzling.stackexchange.com/questions/9036/the-lion-and-the-zebras&quot;&gt;this question&lt;/a&gt; by @Gamow&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose there are &lt;span class=&quot;math-container&quot;&gt;$100$&lt;/span&gt; lions and &lt;span class=&quot;math-container&quot;&gt;$100$&lt;/span&gt; zebras. The lions function together as a team (perfect communication and united decision making), and so do the zebras. The lions place themselves first on an infinite plane, and the zebras place themselves next. Game proceeds turn-wise, one turn for the lion team, one for the zebra team. On each turn, a single zebra or lion (depending on which team's turn it is) moves by up to &lt;span class=&quot;math-container&quot;&gt;$100$&lt;/span&gt; metres. Team decides which of their members moves, no restrictions on the pick. All positions and movements are public knowledge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lions win if a single zebra is eaten. Zebras will try to ensure this doesn't happen. Do the lions' have a strategy that works irrespective of the zebras' one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.p.s. Since this question is still attracting poor answers, let me simplify Veedrac's result. If x lions start in a circular formation and move radially outwards, no more than x zebras can forever escape this circle. x lions can always enclose 100-x zebras in their circle, where x is the lion team's choice. The question now becomes how this circle can be tightened without letting the zebras escape.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4269" LastEditDate="2020-04-23T11:32:49.183" LastActivityDate="2020-04-24T06:39:37.153" Title="Variant of lion and 100 zebras" Tags="|mathematics|geometry|strategy|game|pursuit-evasion|" AnswerCount="11" CommentCount="32" ContentLicense="CC BY-SA 4.0" />
  <row Id="9156" PostTypeId="2" ParentId="9155" CreationDate="2015-02-17T16:00:44.537" Score="0" Body="&lt;p&gt;The zebras are bound to win!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;REASON&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;As the lions place themselves first and the grid is infinite the zebra's can place  themselves at a distance greater than 1 km (1 km is exaggerated, any distance =&amp;lt; than 200m will do) from the closest lion and make sure that they are not in between any lions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EXAMPLE 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let there be 1 zebra and 1 lion.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In such a case the zebra will place itself at a distance of =&amp;lt; 200m&#xA;  ; and will always manage to get away&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Conclusion:&lt;/strong&gt; Zebra wins&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EXAMPLE 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let there be two lions and two zebras&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In such a case after the lions have placed themselves, the zebras will place themselves in such a manner so that they are NOT in between two lions and the minimum distance from both lions is 200m&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The lion may come closer to the zebras and separte them, however it will need the support of the 2nd lion to surround it, and in the time the 2nd lion will come for support, the 2nd zebra will get away&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Conclusion:&lt;/strong&gt; Zebras win&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;SIMILARLY EVEN IF THERE ARE 100 LIONS AND 100 ZEBRAS, THE ZEBRAS WILL USE THE SAME EXACT STRATEGY, AND HENCE MANAGE TO WIN EACH AND EVERY TIME&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Until and unless the grid is finite and the number of lions and zebras is tweaked, the solution won't change.&lt;/em&gt; &lt;/p&gt;&#xA;" OwnerUserId="9346" LastEditorUserId="4269" LastEditDate="2020-04-24T06:39:37.153" LastActivityDate="2020-04-24T06:39:37.153" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="9157" PostTypeId="2" ParentId="9133" CreationDate="2015-02-17T16:09:58.470" Score="0" Body="&lt;p&gt;What is this destination?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Marriage&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where are you traveling from?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Being single / alone&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A great wise man once said that&#xA;although it is only a foot or two away,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The person you feel an affection for can be right in front of you&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;this journey can take you many years.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Getting to know someone, figuring out compatibility, building a large degree of intimacy, and allowing for life plans can all make getting to marriage take a long time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The end of this journey is what you seek, whether you know it or not.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The end of the journey would be the wedding, finally reaching the goal you set out for. You may be seeking it as a way to &quot;cement&quot; your relationship in place&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5583" LastActivityDate="2015-02-17T16:09:58.470" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9158" PostTypeId="2" ParentId="9155" CreationDate="2015-02-17T16:12:15.470" Score="-1" Body="&lt;p&gt;The lions can eventually win. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take a kind of bad starting position for the lions: they are in a line, the zebras are in a line perpendicular to the lions, but 1km from one of the ends (capital T, top bar is zebras, lions are upright bar, and a space between). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say then the lions move into an upward parabola. Well, the zebras are gonna be quite a bit further out. Even then, the ends of the lion parabola can move upward enough that eventually they will be higher than some zebras. Repeat to get move lions above some zebras. Eventually, you will be able to box a zebra in. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-02-17T16:12:15.470" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9159" PostTypeId="2" ParentId="9149" CreationDate="2015-02-17T16:28:34.330" Score="5" Body="&lt;p&gt;Possibly a stretch but:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Life&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am there when you're born&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When your life begins&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yet I begin to fade away from day one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As soon as you're born, you start to die&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I will stay until death&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you die, your life leaves you&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;even if you choose to erase me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can choose to take your own life, but it still leaves you&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Maybe even more, if you pass me to your children&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can create new life by reproducing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My worst enemy? it once killed, but not a human being&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An Extinction Level Event -- something that destroys life on a massive scale, such as the one that wiped out the dinosaurs (which killed, but not humans).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4518" LastActivityDate="2015-02-17T16:28:34.330" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9160" PostTypeId="1" AcceptedAnswerId="9162" CreationDate="2015-02-17T16:48:26.187" Score="4" ViewCount="150" Body="&lt;p&gt;The following three statements describe a perfectly symmetric relationship &#xA;between certain geographical entities A, B and C.  &lt;strong&gt;What are A, B and C?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A borders on all the B except the one named C.&lt;/li&gt;&#xA;&lt;li&gt;C borders on all the B except the one named A.&lt;/li&gt;&#xA;&lt;li&gt;A and C border on each other.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9125" LastActivityDate="2015-02-17T17:02:15.797" Title="Three geographical entities" Tags="|geography|knowledge|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9161" PostTypeId="1" AcceptedAnswerId="9722" CreationDate="2015-02-17T16:52:32.920" Score="9" ViewCount="1034" Body="&lt;p&gt;I have constructed the following list of independent countries according to a secret pattern:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Belgium&lt;/li&gt;&#xA;&lt;li&gt;Mexico&lt;/li&gt;&#xA;&lt;li&gt;Liechtenstein&lt;/li&gt;&#xA;&lt;li&gt;Ivory Coast&lt;/li&gt;&#xA;&lt;li&gt;Ethiopia&lt;/li&gt;&#xA;&lt;li&gt;United Kingdom&lt;/li&gt;&#xA;&lt;li&gt;Philippines&lt;/li&gt;&#xA;&lt;li&gt;????????&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Which country should go into position 8 on the list?&lt;br&gt;&#xA;Which country would be a good candidate for position 12?&lt;br&gt;&#xA;What is the secret pattern?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; &#xA;  Although this is a list, there also is an important binary pattern behind it. It all boils down to being &quot;in&quot; or being &quot;out&quot;.&#xA;  Who is &lt;strong&gt;IN&lt;/strong&gt;? Who is &lt;strong&gt;OUT&lt;/strong&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="9125" LastEditDate="2015-02-27T09:38:40.963" LastActivityDate="2015-03-03T13:22:32.817" Title="Which country should be next on the list?" Tags="|pattern|geography|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9162" PostTypeId="2" ParentId="9160" CreationDate="2015-02-17T17:02:15.797" Score="7" Body="&lt;p&gt;A (or C) is Michigan.&lt;br&gt;&#xA;C (or A) is Ontario.&lt;br&gt;&#xA;B is Great Lakes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably helps that I live about 3 miles from Lake Michigan.&lt;/p&gt;&#xA;" OwnerUserId="375" LastActivityDate="2015-02-17T17:02:15.797" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9163" PostTypeId="2" ParentId="9102" CreationDate="2015-02-17T17:12:33.520" Score="0" Body="&lt;p&gt;I think the beekeepers or their previous customers did pay for his drinks! Whoever had the Texan money in Mexico when the state rated the currency down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Mexican barkeep had 10 Texan Dollars they were worth 10dollars before and then halved their value. So he lost 5dollars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to look at it, is that he is trading off his coins each with a loss of 50c, because he could use them over in Texas for their full original value, so he would be trading his coins undervalued and thus pay for the drink himself!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more economical way to look at it is travel expenses, since the customer is traveling just like a trader, buying local currency and selling it somewhere else far away, the 50c can be seen as his cut/profit as a travelling salesman ;-)&lt;/p&gt;&#xA;" OwnerUserId="1998" LastActivityDate="2015-02-17T17:12:33.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9164" PostTypeId="2" ParentId="9031" CreationDate="2015-02-17T17:13:22.053" Score="0" Body="&lt;p&gt;Answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Robot pirates from outer space&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Beyond the beyond&lt;br&gt;&#xA;In far reaches of flight&lt;br&gt;&#xA;Broken packed cosmic twist  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; in space, duh. also they have a flying ship.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Scarcely seen without a fight&lt;br&gt;&#xA;Feathers splayed in the mist  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; pirates fight, also some of them wear feathers because why not&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Strange wonder to behold&lt;br&gt;&#xA;To gaze upon it is bold&lt;br&gt;&#xA;Never lightly come upon it&lt;br&gt;&#xA;For if you do, you may vomit&lt;br&gt;&#xA;Your senses reeling in a flurry&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you might get seasick also you can't outrun them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Run you will try, in a hurry&lt;br&gt;&#xA;But run and escape you cannot&lt;br&gt;&#xA;You'll be eaten before you can rot&lt;br&gt;&#xA;To gaze upon the humanlike face&lt;br&gt;&#xA;You won't live to see your race  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; they are aliens so they will kidnap you and probe you&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9413" LastActivityDate="2015-02-17T17:13:22.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9165" PostTypeId="1" AcceptedAnswerId="9166" CreationDate="2015-02-17T17:34:16.850" Score="8" ViewCount="797" Body="&lt;p&gt;There is a space station which has several chambers and each chamber is connected with a tube.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Five mathematicians A, B, C, D, and E were gathered in a lounge of the space station. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suddenly, E said.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E: Oh, look at this! We can see the number of people in each room from this screen. Do you see &quot;5&quot; indicated in the screen for this lounge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other four mathematicians nodded with curiosity and E continued speaking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E: This is amazing! Twice the number of people in each room is equal to the sum of people in adjacent rooms. For example, &quot;2&quot; is indicated for the room which is adjacent to the room with one people in it. All rooms have this property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematicians started to generalize this phenomenon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A: Is this also possible if you rearrange people?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B: I don't think this works if we rearrange. How about increasing or decreasing the number of people?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C: What if the space station has different structure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D: Is there any mathematical meaning with this phenomenon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the question. How any people are in the space station?&lt;/p&gt;&#xA;" OwnerUserId="3914" LastActivityDate="2015-02-18T16:46:23.497" Title="In the space station" Tags="|mathematics|logical-deduction|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9166" PostTypeId="2" ParentId="9165" CreationDate="2015-02-17T17:53:09.227" Score="5" Body="&lt;p&gt;One solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;1 - 2 - 3 - 4 - 5 - 6 - 3 &#xA;                      \&#xA;                        4 - 2&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a total of 30. I'm not convinced it's the only solution- perhaps I'm missing some constraints implied by the flavour text. The ones I'm working from:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There's a 5 room&lt;/li&gt;&#xA;&lt;li&gt;There's a 2 room adjacent to a 1 room&lt;/li&gt;&#xA;&lt;li&gt;All rooms are reachable from all others&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Okay, let's try to show that going with the above conditions + that there is a finite number of rooms, this solution is unique.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, first, let's talk about the idea of a &quot;chain&quot;, of form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;1 - 2 - 3 - ... - (N-1) - N - ?&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can we break out of a chain like this? Once we have a room with N people connected to a room with N-1, to break out of the chain we need to connect more rooms to it following these conditions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The sum of people in the remaining adjacent rooms is N+1&lt;/li&gt;&#xA;&lt;li&gt;None of the adjacent rooms has &amp;lt; N/2 people in it (otherwise that room would have more than double its population in its own adjacent rooms).&lt;/li&gt;&#xA;&lt;li&gt;As a result of the above, none of the adjacent rooms can have more than N+1 - N/2 = N/2 + 1 people.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;From this, it's clear that the only solution (for N &gt; 2) is to have two rooms splitting off, one with N/2 people, one with N/2 + 1. So let's update:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;1 - 2 - 3 - ... - (N-1) - N - (N/2) &#xA;                            \&#xA;                          (N/2 + 1) - ?&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What connects to the (N/2 + 1) room? Well it needs a total of N + 2 adjacent, but it already has the N next to it, so it needs 2 more. So it can only be two 1 rooms or one 2 room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means the maximum in the ? room is 2. But that constrains us that N/2 + 1 must be &amp;lt;= 4, otherwise the 2 room would have more than 4 adjacent to it. So from that we get N &amp;lt;= 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Great! What more constraints can we put on it? Well, the need for rooms to have N/2 and N/2 + 1 people means N must be even. So now we know a chain starting at 1 can only get to 2, 4 or 6 for it to ever end. Let's call that &lt;strong&gt;Lemma 1&lt;/strong&gt; for now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second observation is for a situation like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;... - N - M - ?&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the N is fully joined up (has the correct number of people in its adjacent rooms). And take M &amp;lt; N. The observation here is simply that any further rooms connected to M must be of size less than M:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;... - N - M - P - ?&#xA;             \&#xA;              ...&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now P &amp;lt; M, which means that any further rooms connected to P must be of size less than P. This can be extended indefinitely. In other words, if you start from a room and walk through a connection to a smaller room, and keep walking until you hit a dead end, every connection you take will always be taking you to a smaller room than the one you were just in. Let's call that &lt;strong&gt;Lemma 2&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting it together, we're starting from a &lt;code&gt;1 - 2&lt;/code&gt; chain and trying to create a network that gets to 5, but doesn't go on forever. By &lt;strong&gt;Lemma 1&lt;/strong&gt;, our options are to get out of that chain at 2, 4 or 6. But we can't get out of the chain at 2 or 4 because that would involve connecting to a smaller room, and as &lt;strong&gt;Lemma 2&lt;/strong&gt; tells us, that would mean 4 would be the largest room. So getting out at 6 is the only option, giving us the solution at the top, and demonstrating that it's unique.&lt;/p&gt;&#xA;" OwnerUserId="1759" LastEditorUserId="1759" LastEditDate="2015-02-18T16:46:23.497" LastActivityDate="2015-02-18T16:46:23.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9167" PostTypeId="2" ParentId="9165" CreationDate="2015-02-17T17:54:15.497" Score="7" Body="&lt;p&gt;There are 30 people in the space station.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know, from the mathematician's statement, that there is a room with 1 person, adjacent to a room with 2 people in it. We also know, from the presence of the mathematicians in the lounge, that there is at least one room with 5 people in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The known 2-person room must have 4 people in the rooms adjacent, and it already has an adjacent room with one person, so another adjoining chamber can have at most 3 people. The same logic dictates that there is a 4-person chamber next to that, and then the 5-person chamber next.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there, we want the station to have an end. It cannot terminate right next to the 5, because that would mean having 2.5 people in a room, which would imply bigger problems than the little puzzle that our mathematicians are having. So the only option is a room with 6 people. This room, in turn, must have 12 people in adjacent rooms, which leaves 7 outside of the lounge. A room with 3 people could go next to this room and have no other connections, which leaves another room with 4 people. A chamber with 2 people next to that leaves all rooms meeting the conditions, with the station laid out like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  1-2-3-4-5-6-4-2&#xA;            |&#xA;            3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Alternate Answer:&lt;/strong&gt;&lt;br&gt;&#xA;The space station contains a countably infinite amount of people (maybe it's where &lt;a href=&quot;https://puzzling.stackexchange.com/questions/7819/&quot;&gt;these dwarves&lt;/a&gt; are being imprisoned). The first room, as above, has 1 person and is adjacent to a room with 2 people. After that, each room has one more person in it than the preceding room. Then the $n$th room has $n$ people, and is between chambers with $n-1$ and $n+1$ people, for a total of $2n$ people in adjacent rooms.&lt;/p&gt;&#xA;" OwnerUserId="8776" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-02-17T18:21:46.840" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9168" PostTypeId="2" ParentId="9102" CreationDate="2015-02-17T18:10:20.493" Score="0" Body="&lt;p&gt;The barkeepers pay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The devaluation of the other countries currency is purely artificial. Instead of returning change in the foreign currency, the barkeepers could also return change in the local currency, go over the bridge and use their foreign currency to buy wares for the much higher value their foreign currency has there. Should they be forced to give away their foreign currency at the artificially low exchange rate, they are giving it away below the actual value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the real world, the barkeepers would refuse to give away any foreign currency they somehow acquired in their own country. They would do all transactions in local currency and spend any foreign currency abroad where it has more purchasing power.&lt;/p&gt;&#xA;" OwnerUserId="1774" LastActivityDate="2015-02-17T18:10:20.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9169" PostTypeId="2" ParentId="9133" CreationDate="2015-02-17T19:17:46.080" Score="-1" Body="&lt;p&gt;What is this destination?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Truth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where are you traveling from?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Illusion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9605" LastActivityDate="2015-02-17T19:17:46.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9170" PostTypeId="2" ParentId="9155" CreationDate="2015-02-17T19:35:04.753" Score="0" Body="&lt;p&gt;I don't think that the lions can win on an infinite plane.  First, draw a circle around the lions, the smallest one needed to encircle all the lions.  Now, draw another circle whose radius is 10^r light years, where r is the radius of the lion circle in meters.  Put the zebras equidistantly on this outer circle.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whichever zebra is nearest the moving lion moves directly outwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't see a strategy that the lions could use to overtake the zebras.  Suppose one lion goes after one zebra.  The zebra moves outwards in response.  Suppose the lion goes after that zebra until he hits the zebra's starting point.  Then do same with another lion, going after a zebra two zebras away from the first one.  Then can you perhaps trap the zebra in the middle?  Seems to me that the middle zebra would at some point become the closest and start to move away.&lt;/p&gt;&#xA;" OwnerUserId="9605" LastActivityDate="2015-02-17T19:35:04.753" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="9171" PostTypeId="2" ParentId="4880" CreationDate="2015-02-17T19:58:24.307" Score="0" Body="&lt;p&gt;I'm gonna try to show a visual approach for this type of problems, that can be done with pen and paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to make the explanation shorter and more readable, let's say that white = 1 and black = 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here is a list of all 16 possible combinations. obviously from left to right we have the four men ABCD. conveniently, those are 0-15 in binary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0000&lt;br&gt;&#xA;0001&lt;br&gt;&#xA;0010&lt;br&gt;&#xA;0011&lt;br&gt;&#xA;0100&lt;br&gt;&#xA;0101&lt;br&gt;&#xA;0110&lt;br&gt;&#xA;0111&lt;br&gt;&#xA;1000&lt;br&gt;&#xA;1001&lt;br&gt;&#xA;1010&lt;br&gt;&#xA;1011&lt;br&gt;&#xA;1100&lt;br&gt;&#xA;1101&lt;br&gt;&#xA;1110&lt;br&gt;&#xA;1111  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;if all you have is a pen and paper, you can make that list very quickly.&lt;br&gt;&#xA;notice the pattern of the rightmost digit. it simply goes 0-1-0-1 etc.&lt;br&gt;&#xA;the second rightmost digit goes 0-0-1-1-0-0-1-1 etc.&lt;br&gt;&#xA;the third rightmost digit goes 0-0-0-0-1-1-1-1 etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;next, eliminate all combinations that have more than three 1s or more than two 0s (because we only have 3 white hats and 2 black hats)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0011&lt;br&gt;&#xA;0101&lt;br&gt;&#xA;0110&lt;br&gt;&#xA;0111&lt;br&gt;&#xA;1001&lt;br&gt;&#xA;1010&lt;br&gt;&#xA;1011&lt;br&gt;&#xA;1100&lt;br&gt;&#xA;1101&lt;br&gt;&#xA;1110  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;now is a bit trickier. ignoring the leftmost digit, eliminate all combinations that DO NOT appear twice. we do this because man A does not know his hat color, therefore, we have to find a 3 digit sequence that does not give away the fourth (leftmost) digit&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0011&lt;br&gt;&#xA;0101&lt;br&gt;&#xA;0110&lt;br&gt;&#xA;1011&lt;br&gt;&#xA;1101&lt;br&gt;&#xA;1110  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;man B gives us no information. with man C's statement, we can eliminate all combinations in which the rightmost digit would give away the second rightmost digit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0011&lt;br&gt;&#xA;0101&lt;br&gt;&#xA;1011&lt;br&gt;&#xA;1101  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;all these remaining combinations are possible, and in all of them men D has a white hat.&lt;/p&gt;&#xA;" OwnerUserId="9413" LastActivityDate="2015-02-17T19:58:24.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9172" PostTypeId="2" ParentId="9117" CreationDate="2015-02-17T20:18:14.350" Score="2" Body="&lt;p&gt;I was convinced it was&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a shower&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My head has no face,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; obviously a shower head has no face&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I breathe and bleed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; all showers dispense water (bleed) and some have additional hot air jets (breathe)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am quiet,&#xA;but my visitors love to sing.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; who doesn't sing in the shower?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9413" LastActivityDate="2015-02-17T20:18:14.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9173" PostTypeId="1" CreationDate="2015-02-17T20:29:49.843" Score="8" ViewCount="1444" Body="&lt;p&gt;What is the next number in the following sequence? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;9999999, 4782969, 217728, 1568, ???&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I saw this question in an app and I am stuck. Can you help me?&lt;br&gt;&#xA;Source: &lt;a href=&quot;https://play.google.com/store/apps/details?id=com.broli.whatsnext&quot;&gt;https://play.google.com/store/apps/details?id=com.broli.whatsnext&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9607" LastEditorUserId="1477" LastEditDate="2015-02-17T21:44:28.517" LastActivityDate="2015-02-17T21:44:28.517" Title="What is the next number in the sequence?" Tags="|pattern|number-sequence|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9174" PostTypeId="2" ParentId="9173" CreationDate="2015-02-17T20:40:22.647" Score="16" Body="&lt;p&gt;The next number is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 240&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the following is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each number is the product of the digits of the preceding number: &lt;code&gt;9*9*9*9*9*9*9=4782969&lt;/code&gt;, &lt;code&gt;4*7*8*2*9*6*9=217728&lt;/code&gt;, &lt;code&gt;2*1*7*7*2*8=1568&lt;/code&gt;, and so the next number in the sequence is &lt;code&gt;1*5*6*8=240&lt;/code&gt;, and finally &lt;code&gt;2*4*0=0&lt;/code&gt;, where the sequence ends.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1515" LastActivityDate="2015-02-17T20:40:22.647" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9175" PostTypeId="2" ParentId="9031" CreationDate="2015-02-17T21:05:18.053" Score="0" Body="&lt;p&gt;is it the  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; flag of uganda?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;it has feathers. and uganda has a lot of flight and fight in it's history I guess. and you can't outrun them because they run very fast. although vomiting when gazing upon the flag is not a very ncie thing to say&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RFjyW.png&quot; alt=&quot;flag of uganda&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9413" LastActivityDate="2015-02-17T21:05:18.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9176" PostTypeId="1" AcceptedAnswerId="9178" CreationDate="2015-02-17T21:12:55.563" Score="7" ViewCount="7433" Body="&lt;p&gt;You have three cards in a hat.  One is white on both sides, one is red on both sides, the third is white on one side and red on the other.  You pull a card out of the hat and lay it down.  The face is white.  What are the chances that the other side of the card is also white?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: to those who have marked this as a duplicate, I invite you to explain how it can be so when the answer in the &quot;duplicate&quot; is different from this one.&lt;/p&gt;&#xA;" OwnerUserId="9605" LastEditorUserId="9605" LastEditDate="2016-05-29T17:59:59.663" LastActivityDate="2016-05-29T17:59:59.663" Title="Card probability problem" Tags="|probability|" AnswerCount="9" CommentCount="6" ClosedDate="2015-02-19T13:21:46.747" ContentLicense="CC BY-SA 3.0" />
  <row Id="9177" PostTypeId="2" ParentId="9176" CreationDate="2015-02-17T21:35:32.793" Score="4" Body="&lt;p&gt;I &lt;em&gt;think&lt;/em&gt; the correct answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2/3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;let's imagine that we do this game 600 times. when we draw a red card on our first try, the game stops. this will happen about 300 times. for the other 300 (white) draws, we flip the card. one might think that the chances of revealing a red card is 50% then, but that would be wrong. we would get about 100 reds and 200 whites after a card flip. the trick is that even if drawing the W/R card is exactly as likely as drawing the W/W card, for half of the W/R draws, the game will stop before the card gets to be flipped.&lt;/p&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="9413" LastEditDate="2015-02-17T21:40:40.130" LastActivityDate="2015-02-17T21:40:40.130" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9178" PostTypeId="2" ParentId="9176" CreationDate="2015-02-17T21:39:06.440" Score="8" Body="&lt;p&gt;I think the correct answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2/3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We use the law of conditional probability.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let $WW$ be the event that we get the all-white card, $RW$ the mixed card and $RR$ the all-red card. Let $W$ be the event that we get a white face up when we pull a card from the bag.&lt;br&gt;&lt;br&gt;&#xA; The law of conditional probability gives us&#xA; $$P(WW \mid W) = \frac{P(WW \cap W)}{P(W)} = \frac{1/3}{1/2} = \frac{2}{3}$$&#xA; as $P(WW \cap W)$ simply reduces to the probability that we drew the all-white card, and half the faces are white, so $P(W)$ is a half.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4692" LastActivityDate="2015-02-17T21:39:06.440" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9179" PostTypeId="2" ParentId="9176" CreationDate="2015-02-17T21:43:15.210" Score="20" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The answer is obviously 2/3...  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that the set of cards you describe is symmetrical with respect to swapping white and red surfaces, so you could have asked exactly the same sequence of questions with the words white and red swapped at each occurrence - i.e. &quot;&lt;em&gt;...the face is red. What are the chances that the other side of the card is also red?...&lt;/em&gt;&quot; - and would get the same answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this symmetry the question is equivalent to asking - &quot;What is the probability that the other side of the card is the same color as the exposed side?&quot;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;As there are three cards, and two of them are the same on both sides, there is a 2/3 probability that you will have chosen one of these two cards.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="9605" LastEditDate="2015-02-18T00:34:48.373" LastActivityDate="2015-02-18T00:34:48.373" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9180" PostTypeId="1" AcceptedAnswerId="9181" CreationDate="2015-02-17T21:43:18.703" Score="5" ViewCount="721" Body="&lt;p&gt;There are two identical twin brothers.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The first one is honest, but he lies in his schizophrenic situations.&lt;/li&gt;&#xA;&lt;li&gt;The second one is well-known liar, but he tells the truth when he gets one of his pathologic seizures .&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What kind of yes-no questions should you ask anyone of them to detect whether he is in his normal or in his irregular state?&lt;/p&gt;&#xA;" OwnerUserId="9517" LastEditorUserId="8874" LastEditDate="2015-10-07T12:47:43.600" LastActivityDate="2015-10-07T12:47:43.600" Title="Schizoid twin brothers" Tags="|logical-deduction|liars|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9181" PostTypeId="2" ParentId="9180" CreationDate="2015-02-17T21:50:38.517" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; do you always speak the truth when you are crazy?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;yes&quot; indicates that they are currently crazy&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I just realised that the opposite also works and might be simpler&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; do you lie on purpose?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;yes&quot; indicates that they are currently crazy&lt;/p&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="9413" LastEditDate="2015-02-17T22:26:09.120" LastActivityDate="2015-02-17T22:26:09.120" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="9182" PostTypeId="1" AcceptedAnswerId="9186" CreationDate="2015-02-17T21:56:01.150" Score="5" ViewCount="4859" Body="&lt;p&gt;I am always there but never here&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I live to love but hate to live&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feed off passion but am always starving&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I long for pain but hate its coming&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm ever seeking but never finding&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wait and wait but it’s always too late&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am finite&lt;/p&gt;&#xA;" OwnerUserId="9609" LastActivityDate="2015-02-18T17:04:46.420" Title="What am I? (tough riddle must give an explanation)" Tags="|riddle|wordplay|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9186" PostTypeId="2" ParentId="9182" CreationDate="2015-02-17T22:38:20.777" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jealousy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am always there but never here&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You're never jealous of yourself, only of others and other things&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I live to love but hate to live&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jealousy lives in love - if you love it and don't have it you're probably jealous of it. We all hate being jealous. It sucks to be jealous. Thus, when you are jealous of something (When jealousy lives), you hate it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jealousy lives TO love; you're jealous of it probably because you love it - if we described jealousy as a person, the point of jealousy's life would be to love something. However, don't forget that Ms.J hates herself and her life - being jealous all the time is stressful, depressing, and just crappy. She &quot;hates to live&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I feed off passion but am always starving&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jealousy is an emotion (feeds off passion since you are usually jealous of what you have a passion for, or something you like, etc). Always starving because you don't have what you're seeking. For example, if you don't have x, it can be worded as &quot;you've been starved of x&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I long for pain but hate its coming&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jealousy causes pain, but (almost) everyone hates pain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm ever seeking but never finding&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jealousy only exists as long as there is something to be jealous of - when you finally get it, you're not looking for it anymore, and as such, you cannot &quot;find&quot; it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I wait and wait but it’s always too late&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you're jealous, it's already too late. You could've gone out and bought x or done y, but at the end of the day you waited too long and you're sitting there, watching. Jealously. Waiting. But it was too late, you missed your chance.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am finite&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you finally get what you want, even if you were jealous of it before, you stop being jealous of it; therefore, finite - it ends.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8999" LastEditorUserId="8999" LastEditDate="2015-02-18T17:04:46.420" LastActivityDate="2015-02-18T17:04:46.420" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9187" PostTypeId="2" ParentId="9031" CreationDate="2015-02-17T23:52:30.733" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Are you sure you guys don't know any African mythological creatures who eat people and have feathers and a humanlike face?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Purely based on that,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This has to be referring to the mythological &lt;a href=&quot;http://en.wikipedia.org/wiki/Sphinx&quot; rel=&quot;nofollow noreferrer&quot;&gt;sphinx&lt;/a&gt; and the Great Sphinx of Giza: &lt;img src=&quot;https://i.stack.imgur.com/vKQC2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't see how this comes from the riddle though.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-02-17T23:52:30.733" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9188" PostTypeId="2" ParentId="9149" CreationDate="2015-02-18T00:45:16.403" Score="5" Body="&lt;p&gt;I believe the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ignorance&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It matches the title:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I don't know&quot; is an admission of ignorance.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;I'm there when you're born.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are born knowing nothing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Yet I begin to fade away from day one.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You constantly learn more throughout your life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;I will stay until death even if you choose to erase me.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No matter how much you learn in your life, there is always much more you don't know.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Maybe even more, if you pass me to your children.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm having the hardest time with this clue. I think it means that if you don't value education and learning in your family, your kids are unlikely to do so themselves.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;My worst enemy? it once killed, but not a human being.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Curiosity, which prevents someone from being comfortable with ignorance, killed the cat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="294" LastActivityDate="2015-02-18T00:45:16.403" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9189" PostTypeId="2" ParentId="9110" CreationDate="2015-02-18T02:00:24.510" Score="1" Body="&lt;p&gt;How about:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pharmacy logos?&lt;br/&gt;&lt;br/&gt;&lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/d/d9/Bowl_hygeia.svg&quot; rel=&quot;nofollow&quot;&gt;http://upload.wikimedia.org/wikipedia/commons/d/d9/Bowl_hygeia.svg&lt;/a&gt;&lt;br/&gt;1 - Bowl of Hygieia, the internationally recognised symbol of pharmacy (the profession).&lt;br/&gt;&lt;br/&gt;or this one&lt;br/&gt;&lt;br/&gt;&lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/1/19/Asclepius_staff.svg&quot; rel=&quot;nofollow&quot;&gt;http://upload.wikimedia.org/wikipedia/commons/1/19/Asclepius_staff.svg&lt;/a&gt;&lt;br/&gt;2 - Rod of Asclepius, the internationally recognised symbol of medicine.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a snake impaled, I am not free.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Although the snake entwines in the classical explanation, it's rather ambigous in the logo (2 dimensional images, especially monochrome, often suffer from difference in 3 dimensional topological interpretation).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rather, I am displayed for all to see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Logo displayed on pharmacies for all to see.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why do people not mourn me?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dude your a picture, calm down.... :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Instead they want more of me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Helping and getting helped by others is considered to be a good thing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You want more?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Four is my door.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Chinese, Vietnamese, the Korean and the Japanese are superstitious about the number four because it is a homonym for &quot;death&quot; in their languages (wiki). So with death at your door or four being the next door (as in option to choose from), you go to the pharmacy (or infirmary).&lt;br/&gt;&lt;br/&gt;or something like&lt;br/&gt;&lt;br/&gt;Four is the number of classical elements (earth, water, air and fire). These elements are the root for alchemy, which is the root or predecessor of chemistry. (Maybe my last point is a little bit far fetched, but whatever.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9615" LastActivityDate="2015-02-18T02:00:24.510" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9190" PostTypeId="1" AcceptedAnswerId="9209" CreationDate="2015-02-18T03:03:39.783" Score="-2" ViewCount="357" Body="&lt;p&gt;I could kill you&lt;br&gt;&#xA;If you eat, not waste me&lt;br&gt;&#xA;Yet my whole&lt;br&gt;&#xA;Can be very tasty&lt;br&gt;&#xA;A part of me&lt;br&gt;&#xA;Can be used to entice&lt;br&gt;&#xA;And if you've an adult mind&lt;br&gt;&#xA;My whole can't be done twice  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="4353" LastActivityDate="2015-02-18T15:08:02.630" Title="Deadly and delicious" Tags="|riddle|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9191" PostTypeId="2" ParentId="9190" CreationDate="2015-02-18T03:23:31.463" Score="2" Body="&lt;p&gt;is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an apple?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I could kill you&lt;br&gt;&#xA;If you eat, not waste me   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; apple seeds contain tiny amount of cyanide  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet my whole&lt;br&gt;&#xA;Can be very tasty   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; obviously the apple itself is tasty&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A part of me&lt;br&gt;&#xA;Can be used to entice  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;something about the forbidden fruit&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And if you've an adult mind&lt;br&gt;&#xA;My whole can't be done twice  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;forbidden fruit again I guess&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9413" LastActivityDate="2015-02-18T03:23:31.463" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9192" PostTypeId="2" ParentId="9031" CreationDate="2015-02-18T03:29:24.353" Score="1" Body="&lt;p&gt;okay I get it now it's this thing&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/y0nyy.png&quot; alt=&quot;african creature&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9413" LastActivityDate="2015-02-18T03:29:24.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9194" PostTypeId="1" CreationDate="2015-02-18T06:30:55.270" Score="1" ViewCount="939" Body="&lt;p&gt;The challenge as described hereafter is to create a total of 4 similar triangles by drawing 4 triangle in a scalene, acute triangle - out of the 5 resulting triangles (4 that make the original one)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If D, E, and F are the middle of the sides of the triangle ABC than the 4 triangles created are equivalent and with the original triangle we get 5 similar triangles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Qht5x.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The challenge: &#xA;In a triangle create four internal triangles as described in the following diagram. The requirement is that with the original triangle you get exactly FOUR similar triangles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4sFmo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is unique about the internal triangle KJL? What is unique about the points on the sides, K, J, and L?&lt;/p&gt;&#xA;" OwnerUserId="9531" LastEditorUserId="5188" LastEditDate="2015-03-01T02:11:34.917" LastActivityDate="2015-03-01T02:11:34.917" Title="Four similar triangles" Tags="|geometry|triangle|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="9195" PostTypeId="2" ParentId="9133" CreationDate="2015-02-18T06:35:57.110" Score="0" Body="&lt;p&gt;What is this destination?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Knowledge&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where are you traveling from?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ignorance&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...although it is only a foot or two away,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A comfortable distance for books (well, maybe now a tablet or Kindle)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...this journey can take you many years.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Learning can take a lifetime&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...The end of this journey is what you seek, whether you know it or not.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wisdom / Enlightenment&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9619" LastActivityDate="2015-02-18T06:35:57.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9196" PostTypeId="2" ParentId="9190" CreationDate="2015-02-18T07:01:35.530" Score="2" Body="&lt;p&gt;Is it...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; False hope?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I could kill you&lt;br /&gt;&#xA;If you eat, not waste me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clinging to it can only consume you, better to nip it in the bud.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet my whole&lt;br /&gt;&#xA;Can be very tasty&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The comfort of illusion is usually much better than the harsh reality&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A part of me&lt;br /&gt;&#xA;Can be used to entice&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As charlatans, mountebanks, politicians, grifters and swindlers of all shades know all too well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And if you've an adult mind&lt;br /&gt;&#xA;My whole can't be done twice &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because once you know, you cannot &lt;i&gt;un-know&lt;/i&gt; without &lt;i&gt;conscious&lt;/i&gt; self-deceit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9619" LastActivityDate="2015-02-18T07:01:35.530" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9197" PostTypeId="2" ParentId="9176" CreationDate="2015-02-18T07:39:00.747" Score="2" Body="&lt;p&gt;My gut-reaction to this was 1/2, and after reading everyone's answers I had a heck of a time wrapping my head around the correct answer. Here's the really simple, non-technical way I finally convinced myself:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's simplify the original experiment and imagine there are only two cards in the hat: a W-W card and a W-R card. Now let's reach in and draw one out, and imagine we see a white face. What are the chances that the other face is also white? Well, you night say, there are only two possible outcomes: I drew the all white card or I drew the white and red card. The answer must be 50%. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, there are three possible outcomes: (1) You may have drawn red and white card, (2) you may have drawn the all white card, or (3) you may have drawn the all white card. No, that's not a typo. Yes, outcomes (2) and (3) are the same. Or are they? The all white card may be drawn in two different ways: either one white side up, or the other white side up. The unfortunate fact that both ways look the same is inconsequential. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, looking at my white card, the possible outcomes are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I draw W, the flip side is R.&lt;/li&gt;&#xA;&lt;li&gt;I draw W&lt;sub&gt;1&lt;/sub&gt;, the flip side is W&lt;sub&gt;2&lt;/sub&gt;.&lt;/li&gt;&#xA;&lt;li&gt;I draw W&lt;sub&gt;2&lt;/sub&gt;, the flip side is W&lt;sub&gt;1&lt;/sub&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I don't know whether I'm seeing W, W&lt;sub&gt;1&lt;/sub&gt; or W&lt;sub&gt;2&lt;/sub&gt;, but the chance the flip side is also white (that is, either W&lt;sub&gt;2&lt;/sub&gt; or W&lt;sub&gt;1&lt;/sub&gt;) is 2/3. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9621" LastEditorUserId="9621" LastEditDate="2015-02-18T16:08:10.770" LastActivityDate="2015-02-18T16:08:10.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9199" PostTypeId="2" ParentId="9176" CreationDate="2015-02-18T11:23:15.760" Score="1" Body="&lt;p&gt;Another interpretation for those having a hard time to wrap their hand around 2/3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just label all card faces: A,B,C,D,E,F&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The all white card gets A on one side B on the other. The white red card gets C on the white side and D on the red. The all red card gets E and F.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our question now becomes:&#xA;We pick a card and the upper Face shows a letter of (A,B,C) what is the chance we picked the A/B card?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three possible faces and two of them belong to AB so chance is 2/3&lt;/p&gt;&#xA;" OwnerUserId="1998" LastActivityDate="2015-02-18T11:23:15.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9200" PostTypeId="1" CreationDate="2015-02-18T11:51:18.360" Score="-2" ViewCount="248" Body="&lt;p&gt;There are $3$ saucers. And there are $5$ disks, say &lt;strong&gt;A,B,C,D,E&lt;/strong&gt;, having respective areas $1,2,3,4,5$. That is, &lt;strong&gt;E&lt;/strong&gt; is the largest disk and &lt;strong&gt;A&lt;/strong&gt; is the smallest one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are placed in the order &lt;strong&gt;E,D,C,B,A&lt;/strong&gt; in the leftmost saucer, that is, largest at the bottom, smallest at the top. The rest two saucers are empty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are supposed to transfer the entire stack in the same order in the rightmost saucer. That is, in the end, the first two saucers will remain empty and the third saucer will contain the disks in the order &lt;strong&gt;E,D,C,B,A&lt;/strong&gt;, that is, largest disk at the bottom and the smallest one on the top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are allowed to use the middle saucer temporarily during the transfers, you can move only one disk at a time, you cannot place a bigger disk on a smaller one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the least possible ways in which you can do this transfer? Can you devise an algorithm for any number of disks?&lt;/p&gt;&#xA;" OwnerUserId="9560" LastActivityDate="2015-02-18T13:40:26.623" Title="Shifting the stack as it is" Tags="|mathematics|logical-deduction|calculation-puzzle|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9201" PostTypeId="2" ParentId="9176" CreationDate="2015-02-18T11:54:29.520" Score="0" Body="&lt;p&gt;I think the chances are 1/2, as you described your card.  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Card 1: Red - Red  &#xA;Card 2: White - White  &#xA;Card 3: White - Red&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And pulled card has face &lt;code&gt;White&lt;/code&gt;. By this we are clear that its not the &lt;code&gt;Card 1&lt;/code&gt;. So it can be either &lt;code&gt;Card 2&lt;/code&gt; or &lt;code&gt;Card 3&lt;/code&gt;. As first face is &lt;code&gt;White&lt;/code&gt;, Second face may be &lt;code&gt;White&lt;/code&gt; or &lt;code&gt;Red&lt;/code&gt; so the probability will 1/2.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Card 1: Red - Red  //This case can't be possible as we got first face is white&#xA;Card 2: White - White  //This is the first case (Second face may be white)&#xA;Card 3: White - Red    //This is the Second case (Second face may be Red)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="9626" LastActivityDate="2015-02-18T11:54:29.520" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9203" PostTypeId="2" ParentId="9176" CreationDate="2015-02-18T12:45:00.677" Score="1" Body="&lt;h2&gt;Answer&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2/3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Reasoning&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In randomly picking a card and randomly picking a side from that card, you have effectively picked a random card-side from all the card-sides in the hat. There are 3 white card-sides in total. 2 of which have another white side on the back. Given that you've picked a white card-side, it follows that there's a 2/3 chance you've picked one of the white card-sides with another white side on the back.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This reasoning extends to any number of cards with any number of colours:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Given that a randomly picked side of a randomly picked card is colour A, what is the probability that the opposite side is colour B?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Count the total number of card-sides of colour A. Call this $N_A$. Then count the number of those card-sides with colour B on the back. Call this $N_{A,B}$. Then the probability of the opposite side being colour B is $N_{A,B}/N_A$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9627" LastEditorUserId="9627" LastEditDate="2015-02-19T12:17:44.250" LastActivityDate="2015-02-19T12:17:44.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9204" PostTypeId="2" ParentId="9200" CreationDate="2015-02-18T13:10:32.147" Score="5" Body="&lt;p&gt;This is the famous &lt;a href=&quot;https://en.wikipedia.org/wiki/Tower_of_Hanoi#Iterative_solution&quot;&gt;Tower of Hanoi problem&lt;/a&gt;. This describes the optimal algorithm:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an even number of disks:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;make the legal move between pegs A and B&#xA;make the legal move between pegs A and C&#xA;make the legal move between pegs B and C&#xA;repeat until complete&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For an odd number of disks:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;make the legal move between pegs A and C&#xA;make the legal move between pegs A and B&#xA;make the legal move between pegs C and B&#xA;repeat until complete&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In each case, a total of $2^n-1$ moves are made.&lt;/p&gt;&#xA;" OwnerUserId="5610" LastActivityDate="2015-02-18T13:10:32.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9205" PostTypeId="2" ParentId="8992" CreationDate="2015-02-18T13:14:22.020" Score="4" Body="&lt;p&gt;It is possible, but I don't have a really neat proof. What I have is a construction proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma 1: a club with 13 members can do this.&lt;br&gt;&#xA;One member has four dinners, each one with three other members. That member has had dinner with everyone, and divided up the rest into four groups. Each member of a group has had dinner with all other members of the group and no members of the other groups. Number each group's members 0,1 and 2; we can construct dinner groups for the rest as follows:&lt;br&gt;&#xA;0000, 0111, 0222&lt;br&gt;&#xA;1012, 1120, 1201&lt;br&gt;&#xA;2021, 2102, 2210&lt;br&gt;&#xA;(The nth group is represented by the nth digit in each case).&#xA;That's nine dinners, and by inspection each member appears exactly once with each other member. Together with the initial four dinners, that covers everyone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma 2: from a collection of $p^2$ members, we can make $p^2 + p$ groups, each containing $p$ members, such that every member is in exactly one group with every other member, as long as $p$ is prime.&lt;br&gt;&#xA;First divide the members into $p$ initial groups of $p$ each. Number them $0...p-1$ and also number each group member $0...p-1$ inside each initial group. Now we form the other $p^2$ derived groups by choosing one member from each of the initial groups as follows:&lt;br&gt;&#xA;For derived group $a,b$, take member $a + gb$  (mod p) from initial group number $g$. The $a,b$ each vary independently over $0...p-1$, so there will be $p^2$ of these derived groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $x$ and $y$ from two different initial groups are each in two derived groups, $a_1, b_1$ and $a_2, b_2$&lt;br&gt;&#xA;From the first group we have $x - y = a_1 + g_x.b_1 - ( a_1 + g_y.b_1 ) = b_1(g_x - g_y)$. From the second we have $x - y = b_2(g_x - g_y)$. Equating these, we get $(b_1 - b_2)(g_x - g_y) = 0$. That means one of three things:&lt;br&gt;&#xA;* $g_x = g_y$, so they are in the same group. But we only choose one element from each group, so they must be the same element.&lt;br&gt;&#xA;* Two non-zero numbers multiply to form a zero number. But that's not possible mod p, because p is prime.&lt;br&gt;&#xA;* $b_1 = b_2$. Now we can say that $a_1 + g_x.b_1 = a_2 + g_x.b_2$, so $a_1 - a_2 = g_x.b_2 - g_x.b_1 = 0$, so $a_1 = a_2$, so these are not two groups but the same group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because each derived group contains one member from each initial group, every member must appear at most p derived groups (or else there would necessarily be overlap with at least one member of another initial group, because there are only p members in each initial group). Because there are $p^2$ groups of $p$ members making $p^3$ members in all the derived groups, and $p^2$ members in total, each member must appear in at least $p$ derived groups otherwise some members would need to appear in more than $p$ derived groups. Therefore every member appears in exactly $p$ derived groups, each time with a different member of each other initial group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combined with the initial groups, each member appears in $p+1$ groups total, and appears in a group of $p$ exactly once with each other member of the club.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the final proof:&#xA;$ 169 = 13 x 13$, and 13 is prime, so we can create 182 groups of 13, such that every member appears with every other member exactly once - from Lemma 2.&lt;br&gt;&#xA;Within each group of 13, we can have 13 dinners in groups of four such that every member dines with every other member exactly once - from Lemma 1.&lt;br&gt;&#xA;Therefore, we can have $182 x 13 = 2366$ dinners for four, with every member dining with every other member exactly once.&lt;/p&gt;&#xA;" OwnerUserId="6031" LastActivityDate="2015-02-18T13:14:22.020" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9206" PostTypeId="2" ParentId="6415" CreationDate="2015-02-18T13:36:46.787" Score="6" Body="&lt;p&gt;Having seen other people's answers, and pretty much understood them (I think), I thought I'd run a test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried values of p ranging from 0.05 upwards in 0.05 increments.  For each value of p i ran 10,000 tests, each with a limit of 10,000 steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used the code below&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Dim x As Integer&#xA;Dim die As Integer&#xA;Dim randomNum As Single&#xA;Dim test As Integer&#xA;Dim footstep As Integer&#xA;Dim location As Integer&#xA;Dim p As Single&#xA;Dim dead As boolean&#xA;location = 0&#xA;Randomize&#xA;For p = 0.05 To 0.95 Step 0.05&#xA;    die = 0&#xA;    For test = 1 To 10000&#xA;        location = 0        &#xA;        For footstep = 1 To 10000&#xA;            randomNum = Rnd&#xA;            If randomNum &amp;lt; p Then &#xA;                location = location -1&#xA;            Else&#xA;                location = location +1&#xA;            End If&#xA;            If location &amp;lt;0 Then&#xA;                die = die + 1&#xA;                footstep = 10000&#xA;            End If&#xA;        Next &#xA;    Next&#xA;    Print &quot;p = &quot; &amp;amp; p &amp;amp; &quot;  Died &quot; &amp;amp; die &amp;amp; &quot;/10000&quot;&#xA;Next&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I achieved the following results:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;p = .05  Died  523/10000 (p/q = 0.0526)&#xA;p = .10  Died 1134/10000 (p/q = 0.1111)&#xA;p = .15  Died 1732/10000 (p/q = 0.1764)&#xA;p = .20  Died 2457/10000 (p/q = 0.2500)&#xA;p = .25  Died 3273/10000 (p/q = 0.3333)&#xA;p = .30  Died 4286/10000 (p/q = 0.4286)&#xA;p = .35  Died 5433/10000 (p/q = 0.5385)&#xA;p = .40  Died 6658/10000 (p/q = 0.6667)&#xA;p = .45  Died 8179/10000 (p/q = 0.8182)&#xA;p = .50  Died 9918/10000 (p/q = 1.0000)&#xA;p = .55 and over  Died 10000/10000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In most cases the number of deaths is a little lower than expected but very close. I did only run each test for a limit of 10,000 steps, so that would have a small effect, but not a good enough mathematician to work out how much.&lt;/p&gt;&#xA;" OwnerUserId="8778" LastActivityDate="2015-02-18T13:36:46.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9207" PostTypeId="2" ParentId="9200" CreationDate="2015-02-18T13:40:26.623" Score="1" Body="&lt;p&gt;As has been stated, it's the Tower of Hanoi puzzle - but personally I didn't find that description of how to solve it particularly useful. I must be reading it wrong because 'repeat until complete' immediately fails on the second step (which attempts to move A to B, specifically disk C onto the middle saucer which already has BA on it and isn't a legal move).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So anyway, here is the long hand. Saucers are labelled 1,2,3 from left to right:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1: EDCBA&#xA;2:&#xA;3:&#xA;&#xA;1: EDCB&#xA;2:&#xA;3: A&#xA;&#xA;1: EDC&#xA;2: B&#xA;3: A&#xA;&#xA;1: EDC&#xA;2: BA&#xA;3: &#xA;&#xA;1: ED&#xA;2: BA&#xA;3: C&#xA;&#xA;1: EDA&#xA;2: B&#xA;3: C&#xA;&#xA;1: EDA&#xA;2:&#xA;3: CB&#xA;&#xA;1: ED&#xA;2: &#xA;3: CBA&#xA;&#xA;1: E&#xA;2: D&#xA;3: CBA&#xA;&#xA;1: E&#xA;2: DA&#xA;3: CB&#xA;&#xA;1: EB&#xA;2: DA&#xA;3: C&#xA;&#xA;1: EBA&#xA;2: D&#xA;3: C&#xA;&#xA;1: EBA&#xA;2: DC&#xA;3: &#xA;&#xA;1: EB&#xA;2: DC&#xA;3: A&#xA;&#xA;1: E&#xA;2: DCB&#xA;3: A&#xA;&#xA;1: E&#xA;2: DCBA&#xA;3: &#xA;&#xA;1: &#xA;2: DCBA&#xA;3: E&#xA;&#xA;1: A&#xA;2: DCB&#xA;3: E&#xA;&#xA;1: A&#xA;2: DC&#xA;3: EB&#xA;&#xA;1: &#xA;2: DC&#xA;3: EBA&#xA;&#xA;1: C&#xA;2: D&#xA;3: EBA&#xA;&#xA;1: C&#xA;2: DA&#xA;3: EB&#xA;&#xA;1: CB&#xA;2: DA&#xA;3: E&#xA;&#xA;1: CBA&#xA;2: D&#xA;3: E&#xA;&#xA;1: CBA&#xA;2: &#xA;3: ED&#xA;&#xA;1: CB&#xA;2: &#xA;3: EDA&#xA;&#xA;1: C&#xA;2: B&#xA;3: EDA&#xA;&#xA;1: C&#xA;2: BA&#xA;3: ED&#xA;&#xA;1: &#xA;2: BA&#xA;3: EDC&#xA;&#xA;1: A&#xA;2: B&#xA;3: EDC&#xA;&#xA;1: A&#xA;2: &#xA;3: EDCB&#xA;&#xA;1: &#xA;2: &#xA;3: EDCBA&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="9168" LastActivityDate="2015-02-18T13:40:26.623" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9209" PostTypeId="2" ParentId="9190" CreationDate="2015-02-18T14:11:08.287" Score="2" Body="&lt;p&gt;It's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;del&gt;A cherry&lt;/del&gt; A cherry pit&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I could kill you&#xA;If you eat, not waste me &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;del&gt;Cherry-bomb.&lt;/del&gt; Per Josh' comments: apparently a cherry pit is poisonous. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet my whole&#xA;Can be very tasty &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cherry the fruit&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A part of me&#xA;Can be used to entice &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Knotting the stem with the tongue, is apparently considered attractive&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And if you've an adult mind&#xA;My whole can't be done twice &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Pop the cherry&quot; idiom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5127" LastEditorUserId="5127" LastEditDate="2015-02-18T15:08:02.630" LastActivityDate="2015-02-18T15:08:02.630" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="9210" PostTypeId="1" AcceptedAnswerId="9214" CreationDate="2015-02-18T16:38:31.327" Score="4" ViewCount="1621" Body="&lt;p&gt;The integers $1,2,3,\ldots,n$ are to be arranged clockwise around a circle, such that adjacent integers always share a common digit (in their decimal representations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) Find the smallest integer $n\ge3$ for which such an arrangement does exist.&lt;br&gt;&#xA;(b) Find the largest integer $n\ge3$ for which such an arrangement does not exist.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-19T09:50:42.037" LastActivityDate="2015-02-19T09:50:42.037" Title="Circle of numbers" Tags="|mathematics|combinatorics|number-theory|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9211" PostTypeId="2" ParentId="9210" CreationDate="2015-02-18T16:46:23.517" Score="3" Body="&lt;p&gt;I think the answer is $a) 29$. Each digit from 3 to 9 must be bracketed by 2 other numbers with 3-9, the next 2 being the 20's. So, you'd have 13-3-23-24-4-14-15-5-25 etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to do a little juggling with 1, 2, 10, 20, but those are easy to insert into the chain. Same for 11, 12, 21, 22. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $b) 28$&lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-02-18T16:52:14.413" LastActivityDate="2015-02-18T16:52:14.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9212" PostTypeId="2" ParentId="6453" CreationDate="2015-02-18T16:57:27.477" Score="1" Body="&lt;p&gt;&lt;strong&gt;I found a sequence with all 24 bottles that works:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/1e3Ql.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I then got the letters by taking the combination of tube color and weight color. The result makes no sense: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BTLFH/B/UTIUTX/UN/JIFIZE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9226" LastActivityDate="2015-02-18T16:57:27.477" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9213" PostTypeId="2" ParentId="9098" CreationDate="2015-02-18T17:25:27.997" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Aragorn&lt;/strong&gt; is not in the garden.&lt;br&gt;&#xA; Many parts of the sentence refer to members of the Fellowship of the Ring.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Billy the gardener&lt;/strong&gt; - Samwise&lt;br&gt;&#xA; &lt;strong&gt;Grey&lt;/strong&gt; - Gandalf the Grey&lt;br&gt;&#xA; &lt;strong&gt;Lego&lt;/strong&gt; - Legolas&lt;br&gt;&#xA; &lt;strong&gt;Beans&lt;/strong&gt; - Boromir, portrayed by Sean Bean&lt;br&gt;&#xA; &lt;strong&gt;Timber&lt;/strong&gt; - Merry and Pippin befriend Treebeard&lt;br&gt;&#xA; &lt;strong&gt;Ring&lt;/strong&gt; - Frodo, the ring-bearer&lt;br&gt;&#xA; &lt;strong&gt;Sallah&lt;/strong&gt; - played by John Rhys-Davies, who also played Gimli&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2015-02-18T17:25:27.997" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9214" PostTypeId="2" ParentId="9210" CreationDate="2015-02-18T18:06:11.580" Score="15" Body="&lt;p&gt;(a) It's easy to see that n must be at least 29, since every number less than 10 must be paired with 2 neighbours, so you need 29 to pair with 9.&lt;br&gt;&#xA;So 9 can be paired with 19 and 29, and then 8 with 18 and 28, etc upto 1 with 11 and 21.&lt;br&gt;&#xA;If you put the number less than 10 in the middle, those groups of 3 can be easily chained together, because they all have a 1 on one end and a 2 on the other end, So you get the 1-group, 2-group, 3-group, etc.&lt;br&gt;&#xA;Then 10 and 20 are left, which can be inserted on each end. They form the 0-group. You can put 0 in there to make it consistent.&lt;br&gt;&#xA;End result: 20, 29,9,19, 18,8,28, 27,7,17, 16,6,26, 25,5,15, 14,4,24, 23,3,13, 12,2,22, 21,1,11, 10 [,20]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) If we take the sequence of (a), we can always insert the next number. 30 between 10 and 20,&#xA;31 between 11 and 21, 32 between 12 and 22, etc.&lt;br&gt;&#xA;There is no number you can never fit. Every number ending in x can always be inserted into the x-group, because x is common inside that entire group.&lt;br&gt;&#xA;Since n must be at least 29, the answer here is 28.&lt;/p&gt;&#xA;" OwnerUserId="1598" LastActivityDate="2015-02-18T18:06:11.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9215" PostTypeId="2" ParentId="9176" CreationDate="2015-02-18T19:13:26.857" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;You have three cards in a hat. One is white on both sides, one is red on both sides, the third is white on one side and red on the other. You pull a card out of the hat and lay it down. The face is white. What are the chances that the other side of the card is also white?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are 3 cards with 2 sides each:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;C&lt;sub&gt;1&lt;/sub&gt;: W&lt;sub&gt;1&lt;/sub&gt; and W&lt;sub&gt;2&lt;/sub&gt;&lt;/li&gt;&#xA;&lt;li&gt;C&lt;sub&gt;2&lt;/sub&gt;: R&lt;sub&gt;1&lt;/sub&gt; and R&lt;sub&gt;2&lt;/sub&gt;&lt;/li&gt;&#xA;&lt;li&gt;C&lt;sub&gt;3&lt;/sub&gt;: W and R&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When you draw a card and place it with one face up, you actually have 6 possible results:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;C&lt;sub&gt;1&lt;/sub&gt; W&lt;sub&gt;1&lt;/sub&gt;&lt;/li&gt;&#xA;&lt;li&gt;C&lt;sub&gt;1&lt;/sub&gt; W&lt;sub&gt;2&lt;/sub&gt;&lt;/li&gt;&#xA;&lt;li&gt;C&lt;sub&gt;2&lt;/sub&gt; R&lt;sub&gt;1&lt;/sub&gt;&lt;/li&gt;&#xA;&lt;li&gt;C&lt;sub&gt;2&lt;/sub&gt; R&lt;sub&gt;2&lt;/sub&gt;&lt;/li&gt;&#xA;&lt;li&gt;C&lt;sub&gt;3&lt;/sub&gt; W&lt;/li&gt;&#xA;&lt;li&gt;C&lt;sub&gt;3&lt;/sub&gt; R&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The question then removes all the possibilities where a red face is showing, so the only possible results are the following faces:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;C&lt;sub&gt;1&lt;/sub&gt; W&lt;sub&gt;1&lt;/sub&gt;&lt;/li&gt;&#xA;&lt;li&gt;C&lt;sub&gt;1&lt;/sub&gt; W&lt;sub&gt;2&lt;/sub&gt;&lt;/li&gt;&#xA;&lt;li&gt;C&lt;sub&gt;3&lt;/sub&gt; W&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then the question asks what's on the other side of the card, and these are the possible other sides of the card:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;C&lt;sub&gt;1&lt;/sub&gt; W&lt;sub&gt;2&lt;/sub&gt;&lt;/li&gt;&#xA;&lt;li&gt;C&lt;sub&gt;1&lt;/sub&gt; W&lt;sub&gt;1&lt;/sub&gt;&lt;/li&gt;&#xA;&lt;li&gt;C&lt;sub&gt;3&lt;/sub&gt; R&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So the chance that the other side of the card is white, based on the fact that the face up card is white, is 2/3.&lt;/p&gt;&#xA;" OwnerUserId="1682" LastEditorUserId="1682" LastEditDate="2015-02-18T20:21:18.077" LastActivityDate="2015-02-18T20:21:18.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9216" PostTypeId="2" ParentId="9098" CreationDate="2015-02-18T20:00:19.697" Score="8" Body="&lt;p&gt;Based mostly on the answer by Otaia (I changed some of the clues and arrived at a different answer.. Then based on more hints I changed more clues and arrived at his original answer... )&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Aragorn&lt;/strong&gt; is not in the garden.&lt;br&gt;&#xA; The sentence refers to The Lord of the Rings.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Billy&lt;/strong&gt; - Pippin, played by Billy Boyd&lt;br&gt;&#xA; &lt;strong&gt;the gardener&lt;/strong&gt; - Samwise&lt;br&gt;&#xA; &lt;strong&gt;Grey&lt;/strong&gt; - Gandalf the Grey&lt;br&gt;&#xA; &lt;strong&gt;Lego&lt;/strong&gt; - Legolas&lt;br&gt;&#xA; &lt;strong&gt;Bolt&lt;/strong&gt; - Merry, portrayed by Dominic Monaghan, who also played the Bolt in X-Men Origins (Credit user3453281)&lt;br&gt;&#xA; &lt;strong&gt;Beans&lt;/strong&gt; - Boromir, portrayed by Sean Bean&lt;br&gt;&#xA; &lt;strong&gt;Timber&lt;/strong&gt; - Frodo, played by Elijah &lt;strong&gt;Wood*&lt;/strong&gt;&lt;br&gt;&#xA; &lt;strong&gt;Sallah&lt;/strong&gt; - played by John Rhys-Davies, who also played Gimli&lt;br&gt;&#xA;&lt;br&gt;&#xA; * Though on an interesting note I found that many believe that the 'orn' at the end of Aragorn's name means 'tree'.) Which could make Timber refer to Aragorn and make Frodo the answer. I however think that is more of a stretch than timber referring to Frodo. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3922" LastEditorUserId="3922" LastEditDate="2015-02-18T22:18:13.210" LastActivityDate="2015-02-18T22:18:13.210" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9217" PostTypeId="2" ParentId="6453" CreationDate="2015-02-18T21:08:00.703" Score="3" Body="&lt;p&gt;We're down to the final part of the puzzle.&lt;br&gt;&#xA;&lt;strong&gt;Note: this is the solution to the final instruction. Check the wiki for the first two parts.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Clearly, the scale work was the more difficult part. Now it should be smooth sailing. Our solutions to the first two parts are in the wiki, and I'll be making reference to those here.&lt;br&gt;&#xA;&lt;br&gt;&#xA;The final instruction has four parts. The first two parts are relevant to selecting the key(s) we need, while the last two parts are the process of using the key(s) to get letters and put them into a sentence. There are a number of ways to go about this, but only one will be the right solution. It's just a matter of figuring out each part in detail.&lt;br&gt;&#xA;&lt;br&gt;&#xA;Noting the final scale, we see that all of the tubes have been filled except one; the red tube is missing two units. The diamonds were our indicator for how many units filled the pipes, and there just &lt;em&gt;happen&lt;/em&gt; to be two more diamonds on each of our keys. It seems quite obvious that this means we are supposed to &lt;strong&gt;use the red key in our solution&lt;/strong&gt;, as the red key's two diamonds represent the missing two units on the final scale.&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;em&gt;Side note: Does this change the weight on the balance? No, because adding two more red units is an additional 22 weight to the left side of the scale, which without the hanging weight's presence is a total of 34 weight on the left. The hanging weight would be unable to bring balance to the scale.&lt;/em&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;So, we have identified the use of the red key in finding our answer. Is it the only key we use? Yes, because the means of identifying the red key cannot be applied anymore to identify other keys. It is reasonable to believe that the red key is an anchor and will be used for the entirety of the final puzzle.&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;em&gt;Is it an anagram? No, because the final instruction's last part shows that the letters are placed in order.&lt;/em&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;The hanging weight's position for each bottle is a clear indicator of which color - and in turn, the row - on the key we need to be looking at to determine the letters we need. &#xA;  &lt;img src=&quot;https://i.stack.imgur.com/LszsN.png&quot; alt=&quot;base image&quot;&gt;&#xA;&lt;br&gt;&#xA;The original author says that the final solution will be an English sentence, so we will be able to check our progress along the way if we run into a selection of letters that does not make up an English word.&lt;br&gt;&#xA;&lt;br&gt;&#xA;The only thing remaining is selecting a column. Presumably, the color of the pipe we filled with each bottle is how we should select our column. Following this line of thought, and knowing our first bottle goes into the &lt;strong&gt;yellow&lt;/strong&gt; tube and the hanging weight balances at position &lt;strong&gt;+1&lt;/strong&gt;, we find our first letter:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/culzG.png&quot; alt=&quot;first letter&quot;&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;By continuing this process for each bottle, we find all of the letters in the final sentence. We discover that the / symbols are empty spaces. &lt;strong&gt;In summary, the combination of 1) the red key; 2) the row on the key corresponding to the hanging weight's position, and 3) the column of the color corresponding to the tube we fill with each bottle yields the following sentence:&lt;br&gt;&#xA;&lt;em&gt;BALANCE IS THE RECTIFIER&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Excellent work, Bmyguest, this puzzle was certainly one of the most challenging and mentally-consuming puzzles I've had the pleasure of solving. Thanks to all the people who contributed their work, as well. It was fun! =)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9319" LastEditorUserId="9319" LastEditDate="2015-02-18T23:50:15.067" LastActivityDate="2015-02-18T23:50:15.067" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="9218" PostTypeId="1" AcceptedAnswerId="9220" CreationDate="2015-02-18T21:27:16.240" Score="9" ViewCount="2685" Body="&lt;p&gt;In the &lt;a href=&quot;http://xkcd.com/solution.html&quot; rel=&quot;nofollow&quot;&gt;solution&lt;/a&gt; to the &lt;a href=&quot;http://xkcd.com/blue_eyes.html&quot; rel=&quot;nofollow&quot;&gt;Blue Eyes&lt;/a&gt; logic puzzle, there is a follow-up question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Each person knows, from the beginning, that there are no less than 99 blue-eyed people on the island. How, then, is considering the 1 and 2-person cases relevant, if they can all rule them out immediately as possibilities?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have found an answer to that question &lt;a href=&quot;http://www.explainxkcd.com/wiki/index.php/Blue_Eyes&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Blue-eyed people can't see their own faces, so blue-eyed people can see one less blue-eyed face than non-blue-eyed people can. Even though I can see that there are at least 99 blue-eyed people, I don't know that they can see that, so I need to imagine people who see only 98, who would base their actions in part by imagining people who can see only 97 who would base their actions in part by imagining people who can see only 96, and so on... &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This answer doesn't make sense to me. Why are they basing their logic on some hypothetical islander who can only see 96 blue-eyed people when &lt;em&gt;they know such an islander cannot possibly exist&lt;/em&gt; and - moreover - know that everyone else knows it too?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking a smaller example with 6 islanders for illustrative purposes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A (blue), B (blue), C (blue), D (brown), E (brown), F (brown)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No matter what A's eye color is, he knows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;B will see at least one other blue (C)&lt;/li&gt;&#xA;&lt;li&gt;C will see at least one other blue (B)&lt;/li&gt;&#xA;&lt;li&gt;D, E and F will see at least two other blues (B &amp;amp; C)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Everyone&lt;/em&gt; can see at least 1 other blue-eyed islander, &lt;em&gt;and everyone else knows it&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What value does Day 1 serve?  Why not skip ahead to Day 2?&lt;/p&gt;&#xA;" OwnerUserId="9633" LastEditorUserId="5373" LastEditDate="2015-04-28T20:06:18.537" LastActivityDate="2015-04-28T20:06:18.537" Title="In the blue-eyes puzzle, why can't they skip ahead?" Tags="|logical-deduction|blue-eyes|" AnswerCount="3" CommentCount="8" ClosedDate="2015-02-24T16:27:47.393" ContentLicense="CC BY-SA 3.0" />
  <row Id="9219" PostTypeId="2" ParentId="9218" CreationDate="2015-02-18T21:41:53.407" Score="3" Body="&lt;p&gt;Day one serves to handle the case that there is only 1 person with blue eyes. They would see no-one else with blue eyes, but since at least 1 person has blue eyes, they know it's them then. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back to why they don't skip days, let's take the scenario where you have 4 blues and 2 brown. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Each person with blue will see 3 other blues.&lt;/li&gt;&#xA;&lt;li&gt;But they don't know if they have blue eyes, so someone they see with blue might only see 2 other blue sets. &lt;/li&gt;&#xA;&lt;li&gt;So, do you skip to day 3, or day 4 (a brown eye person would see 4 blues, so they assume a blue eye could only see 3)?  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;They can't communicate with each other. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-02-18T22:08:57.570" LastActivityDate="2015-02-18T22:08:57.570" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="9220" PostTypeId="2" ParentId="9218" CreationDate="2015-02-18T21:44:49.817" Score="10" Body="&lt;p&gt;After talking with you for a bit in chat, I see better what the problem is and where you're heading down the wrong path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the principal issue: &lt;strong&gt;knowledge is only useful if everyone on the island can make the same conclusion.&lt;/strong&gt; Why? Let's say $C$ has a piece of knowledge that lets them skip a day, but can't verify that everyone else also has this piece of knowledge. If $C$ skips ahead a day, $C$ could easily be leaving people behind who haven't made the same conclusion.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let's say there are three people: $A$, $B$, and $C$. They all have blue eyes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your premise starts off with the conclusion that $C$ sees two people with blue eyes, and knows nobody will leave the island on day 1. This conclusion is entirely correct. However, $C$ can't just go ahead and skip day 1 without first checking to make sure $A$ and $B$ will do the same. Otherwise, $C$ would skip ahead and leave $A$ and $B$ confused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's where the logical trap is. While $C$ can conclude that nobody will leave the island, $C$ can't conclude that $B$ or $A$ will say the same. $C$ has two cases to explore: if they have blue eyes, and if they have brown eyes. They don't know which it is. Let's say $C$ supposes they have brown eyes (even though it's not true) - then, what does $C$ conclude about what $B$ knows? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this thought experiment of $C$'s, $B$ sees only one person with blue eyes. But $B$ doesn't know their own eye color either! In $B$'s mind, their eye color could theoretically be brown. As a result, $A$ would leave the island on day 1! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, obviously, this isn't true. The point is that it's possible for $C$ to conceive of a situation in which neither $A$ nor $B$ could conclude nobody will leave the island. As a result, &lt;strong&gt;the knowledge that nobody will leave the island on day 1 isn't useful to $C$.&lt;/strong&gt; $C$ may know this, but there's a chance that neither $A$ nor $B$ do, and in that case, $C$ would be skipping ahead.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To continue this thought experiment through, let's say day 1 passes. Nobody leaves the island. Now what's $C$ thinking?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, $C$ ponders what $B$ knows. In the case where $C$ has brown eyes, $B$ has seen nobody leave the island and has concluded that they have blue eyes. This is a correct conclusion. In the case where $C$ has blue eyes, $B$ still can't tell whether they have blue or brown eyes, but $B$ knows that everyone else is conducting the same thought experiments. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result, from the perspective of $C$, either $B$ and $A$ will leave the island tonight (if $C$ has brown eyes), or $A$, $B$, and $C$ will leave the island tomorrow night (if $C$ has blue eyes). The first night is critical, though, because it establishes the common knowledge that &lt;em&gt;in everyone's thought experiments&lt;/em&gt;, there are at least two people with blue eyes. &lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorDisplayName="user20" LastEditDate="2015-02-19T01:34:17.003" LastActivityDate="2015-02-19T01:34:17.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9221" PostTypeId="1" AcceptedAnswerId="9278" CreationDate="2015-02-18T23:20:22.800" Score="2" ViewCount="651" Body="&lt;p&gt;4 friends, A, B, C, and D, meet up after taking a test. Reading their results, they exclaimed,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A says, &quot;Hah! I'm ranked 32!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B says, &quot;I'm ranked 63, better than you, A!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C says, &quot;I ranked 12, I'm the best!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D says, &quot;Actually, I'm the best, I scored 142.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What are A, B, C and D?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They all took the same test, and the way they represent how they did verbally actually matters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Only D's train of thought is absolutely &quot;correct&quot;. Nobody is being deliberately untruthful or deceitful here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; D is the only person who interpreted the test score correctly. Think about what kind of test they took!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note: I'm specifically looking for an answer in the format of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A is a(n) ______, B is a(n)_____ .....&lt;/p&gt;&#xA;" OwnerUserId="8999" LastEditorUserId="8874" LastEditDate="2015-03-04T12:09:48.960" LastActivityDate="2015-03-04T12:09:48.960" Title="4 Friends take a test" Tags="|lateral-thinking|" AnswerCount="1" CommentCount="15" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9222" PostTypeId="2" ParentId="9176" CreationDate="2015-02-18T23:36:06.247" Score="0" Body="&lt;p&gt;For those who find the fact that the answer is not 1/2 hard to swallow, some intuitive explanations have already been proposed. Here's another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you see a white face, it's more likely you're looking at the white-white card than the white-red card, because that's where most of the white faces are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To scale up the numbers a little, imagine you have two six-sided dice, only on one of them you've replaced all the numbers with sixes. I choose one of these dice randomly and roll it and get a six. Which die do you think it is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or let's make it more extreme. I have two lottery machines, only one of them always churns out the balls 1 to 6 in order. I pick one of the machines at random, play a lottery game, and win. Which machine did I pick?&lt;/p&gt;&#xA;" OwnerUserId="1950" LastActivityDate="2015-02-18T23:36:06.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9223" PostTypeId="2" ParentId="9210" CreationDate="2015-02-19T01:02:37.527" Score="-2" Body="&lt;p&gt;&lt;strong&gt;&quot;The integers 1,2,3,…,n are to be arranged clockwise around a circle, such that adjacent integers always share a common digit&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way the puzzle is worded, it's not clear if &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A.) adjacent numbers must share at &lt;strong&gt;least&lt;/strong&gt; one digit or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;B.) adjacent numbers must share &lt;strong&gt;exactly&lt;/strong&gt; one digit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they can share more than one, the smallest n is 3: Ex. { 1, 12, 21 }&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If exactly one digit must be common, the smallest n is 4: Ex.  { 1, 12, 2, 21 } &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the same reasoning as the posters above, a number satisfying the constraints can be inserted anywhere, and all numbers greater the min(n) are possible. Therefore, the largest n &lt;strong&gt;not&lt;/strong&gt; satisfying the constraints would be n=2 for A.) and n=3 for B.).&lt;/p&gt;&#xA;" OwnerUserId="9634" LastActivityDate="2015-02-19T01:02:37.527" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9224" PostTypeId="1" AcceptedAnswerId="9225" CreationDate="2015-02-19T02:42:26.617" Score="30" ViewCount="7101" Body="&lt;p&gt;I am thin and frail.&lt;br&gt;&#xA;Scratching my head can kill me.&lt;br&gt;&#xA;When abused, I can also kill you.&lt;br&gt;&#xA;Usually, I'm helpful.&lt;/p&gt;&#xA;" OwnerUserId="9495" LastEditorUserId="9495" LastEditDate="2015-02-19T09:15:57.060" LastActivityDate="2015-09-06T20:28:07.240" Title="Scratching my head can kill me" Tags="|riddle|" AnswerCount="7" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="9225" PostTypeId="2" ParentId="9224" CreationDate="2015-02-19T02:50:58.643" Score="41" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are a match - &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My body is thin and frail:&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Matches are thin. Wooden matches are easy to snap, paper matches are even easier to bend.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Scratching my head can kill me:&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When a match is scratched (on the striker for a safety match or on any dry frictional surface for an old-fashioned light-anywhere match) it burns, after which it is 'dead' - i.e can't be lit again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;When abused, I can also kill you:&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We are all taught as children how it is dangerous &quot;to play with matches&quot;. Fires can kill...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Usually, I'm helpful:&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Of course - helpful for lighting a useful fire, or a cancer stick etc...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="9495" LastEditDate="2015-02-19T03:44:30.087" LastActivityDate="2015-02-19T03:44:30.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9226" PostTypeId="1" AcceptedAnswerId="9228" CreationDate="2015-02-19T02:53:36.173" Score="-1" ViewCount="550" Body="&lt;p&gt;I am empty, yet I am tangible.&lt;br&gt;&#xA;When I am full, I can be very wise.&lt;br&gt;&#xA;No matter how full I am, I weigh about the same.&lt;br&gt;&#xA;I can take many physical forms, with your help.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who or what am I?&lt;/p&gt;&#xA;" OwnerUserId="9495" LastEditorUserId="9495" LastEditDate="2015-02-20T00:50:20.417" LastActivityDate="2016-08-14T12:48:32.633" Title="I can take many physical forms, with your help" Tags="|riddle|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9228" PostTypeId="2" ParentId="9226" CreationDate="2015-02-19T03:17:06.247" Score="4" Body="&lt;p&gt;Previous answer(s):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;A balloon.&lt;/strong&gt; A balloon is an empty container, until you fill it with air. No matter how much air you put in it, it still weighs the same. With the help of a skilled clown, a balloon can take the form of many different animals.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;A thought.&lt;/strong&gt; Thoughts are empty unless you fill them with meaning, at which point they are intelligent. A thought weighs the same - nothing at all? - regardless of how full it is. And of course, a thought can be bent and manipulated into many forms by the person who ... well, thought it. &lt;em&gt;Riddle was edited to include &quot;tangible.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;New answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Thanks to @Marmy1954's journal/diary idea, I figured the answer to be &lt;strong&gt;a piece of paper&lt;/strong&gt;. The paper is empty until written on, can be very informative when full of words, weighs about the same when written on,  and the paper can be folded into different things, as in origami or paper airplanes, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9319" LastEditorUserId="9319" LastEditDate="2015-02-19T04:03:28.647" LastActivityDate="2015-02-19T04:03:28.647" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9229" PostTypeId="1" AcceptedAnswerId="9231" CreationDate="2015-02-19T03:18:25.003" Score="-6" ViewCount="201" Body="&lt;p&gt;I am a three-dimensional grid.&lt;br&gt;&#xA;I keep out intruders.&lt;br&gt;&#xA;I also prevent disease.&lt;br&gt;&#xA;I travel often to remote and rugged locations.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;difficulty level: easy (1/5) &lt;/p&gt;&#xA;" OwnerUserId="9495" LastActivityDate="2015-02-19T03:47:43.680" Title="I am a three-dimensional grid" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9231" PostTypeId="2" ParentId="9229" CreationDate="2015-02-19T03:47:43.680" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A mosquito net. It is a fabric woven to make tiny squares (3D grid) to keep out mosquitos (the intruders), thereby preventing diseases like malaria. It is brought camping or hiking (remote and rugged locations).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-02-19T03:47:43.680" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9232" PostTypeId="2" ParentId="9226" CreationDate="2015-02-19T03:50:40.813" Score="1" Body="&lt;p&gt;A notebook/journal/diary.  It is empty until written in.  Depending on the content, it can be wise.  It will still weigh about the same.  The form it takes depends on the author.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Guess this would hold true for any information medium.  Tape, CD, film.&lt;/p&gt;&#xA;" OwnerUserId="9616" LastActivityDate="2015-02-19T03:50:40.813" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9234" PostTypeId="1" AcceptedAnswerId="9239" CreationDate="2015-02-19T04:39:25.487" Score="20" ViewCount="3520" Body="&lt;p&gt;I don't like my neighbors.&lt;br&gt;&#xA;One of them wishes I would move&lt;br&gt;&#xA;so that he can visit the others on our street.&lt;br&gt;&#xA;The other is stuck up&lt;br&gt;&#xA;and only associates with one color.&lt;br&gt;&#xA;I can't stand to be in either of their homes.&lt;br&gt;&#xA;...at least, not while they're in them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This riddle has been solved.&lt;br&gt;&#xA;These are the hints I would have provided, had they been necessary:&lt;/strong&gt;&lt;br&gt;&#xA;1.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I met someone &lt;strong&gt;just like me&lt;/strong&gt; who lives at the other end of my street!  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I met someone across the street who appears to be royalty. Next time I see him, &lt;strong&gt;I'll have to check.&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I met a short individual &lt;em&gt;(in passing)&lt;/em&gt; who said she was &lt;strong&gt;looking to get a promotion.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This is my first question ever on Stack Exchange! Hope this riddle is neither too easy nor too hard.&lt;br&gt;&#xA;Good luck!&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="9319" LastEditorUserId="9319" LastEditDate="2015-02-19T10:04:46.550" LastActivityDate="2015-02-19T10:04:46.550" Title="I Don't Like My Neighbors (Riddle)" Tags="|riddle|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9235" PostTypeId="2" ParentId="6029" CreationDate="2015-02-19T05:03:38.817" Score="1" Body="&lt;p&gt;For level 1 the ratio of green time between N1S1/W1E1 is 4/7.&#xA;The higher rate in each direction is the one that require to allow the cars to move through. If we could synchronize it with the fast cars (in each direction) coming into the crossing than we could refine it. Assuming a more general rule we need to allow 7 cars of the 4 seconds rate for every 4 cars of the 7 seconds rate, to cross. The slower rate will be automatically accommodated.&#xA;Is this helping?&lt;/p&gt;&#xA;" OwnerUserId="9531" LastEditorUserId="9531" LastEditDate="2015-03-06T07:21:51.937" LastActivityDate="2015-03-06T07:21:51.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9236" PostTypeId="2" ParentId="9234" CreationDate="2015-02-19T07:18:43.390" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/7Idf9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Precise Answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Edge Cells of a Rubix Cube precisely. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One of them wishes I would move&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is the center cell of each of the sides. In a Rubix Cube, this is the static cells and the others which are not of the same colour should move out of that side.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The other is stuck up&#xA;and only associates with one color.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every cell of the cube is of a single colour&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can't stand to be in either of their homes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; And each coloured cell has their own home or side&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4177" LastEditorUserId="4177" LastEditDate="2015-02-19T07:23:44.213" LastActivityDate="2015-02-19T07:23:44.213" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="9237" PostTypeId="2" ParentId="9224" CreationDate="2015-02-19T07:19:24.453" Score="27" Body="&lt;p&gt;Will this also qualify as an answer?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pencil &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Scratching head can kill me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sharpening a pencil rescues it length and it becomes useless after a while i.e. dead ... Also a dull pencil is of no use till sharpened again&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Abuse&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Poking the pencil in wrong place can lead to death&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Helpful&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Used to express thoughts and ideas ... Transfer knowledge&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8948" LastEditorUserId="8948" LastEditDate="2015-02-19T07:28:36.650" LastActivityDate="2015-02-19T07:28:36.650" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="9238" PostTypeId="2" ParentId="8986" CreationDate="2015-02-19T09:12:59.197" Score="2" Body="&lt;p&gt;Single solutions are shown below for &lt;strong&gt;values of n from 4 to 10&lt;/strong&gt;.  Placement of the queens is based on the algorithm described in the Wikipedia page for the &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Eight_queens_puzzle&quot; rel=&quot;nofollow noreferrer&quot;&gt;eight queens puzzle&lt;/a&gt;&lt;/strong&gt; which is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Let (i, j) be the square in column i and row j on the n × n chessboard, k an integer.&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;If n is even and n ≠ 6k + 2, then place queens at (i, 2i) and (n/2 + i, 2i - 1) for i = 1,2,...,n/2.&lt;/li&gt;&#xA;  &lt;li&gt;If n is even and n ≠ 6k, then place queens at (i, 1 + (2i + n/2 - 3 (mod n))) and (n + 1 - i, n - (2i + n/2 - 3 (mod n))) for i = 1,2,...,n/2.&lt;/li&gt;&#xA;  &lt;li&gt;If n is odd, then use one of the patterns above for (n - 1) and add a queen at (n, n).&quot;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FpbMX.png&quot; alt=&quot;Chess4to7&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Jq1Eq.png&quot; alt=&quot;Chess8to10&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-02-20T03:15:17.517" LastActivityDate="2015-02-20T03:15:17.517" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9239" PostTypeId="2" ParentId="9234" CreationDate="2015-02-19T09:35:48.790" Score="12" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/hrsza.jpg&quot; alt=&quot;chess pieces&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The knight chess piece&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't like my neighbors.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you move, you won't be next to either. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One of them wishes I would move&#xA;so that he can visit the others on our street.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The rook needs you out of the way to move along the back row. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The other is stuck up&#xA;and only associates with one color.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The bishop moves diagonally so always stays on the same colour. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can't stand to be in either of their homes.&#xA;...at least, not while they're in them.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can't take your own pieces or have two pieces in one square. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Alternatively,&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;You are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The queen chess piece&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Most of the answers are the same: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't like my neighbors.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They get in the way of you moving. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One of them wishes I would move&#xA;so that he can visit the others on our street.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The king needs you out of the way to castle safely.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The other is stuck up&#xA;and only associates with one color.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The bishop moves diagonally so always stays on the same colour. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can't stand to be in either of their homes.&#xA;...at least, not while they're in them.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can't take your own pieces or have two pieces in one square. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9532" LastEditorUserId="9532" LastEditDate="2015-02-19T09:42:47.563" LastActivityDate="2015-02-19T09:42:47.563" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9240" PostTypeId="2" ParentId="9097" CreationDate="2015-02-19T09:41:59.573" Score="3" Body="&lt;p&gt;This answer just formalises Lopsy's description to help explore alternatives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the scenario Lopsy intended, the zebras are confined to the centre line of strips parallel to the $y$-axis. The strips are any width over 200m. The zebras start with a $y$ coordinate of more than 100m and the lion starts at (0,0).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's recast the problem so we can ignore most of the zebras. The lion still starts at (0,0), but this time, the zebras start with $y=0$ and either $x&amp;gt;100$ or $x&amp;lt;-100$, that is, at least 100m away from the lion. The lion's first move is to enter one of the strips along the $x$-axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem now essentially resets every time the lion changes strips, so without loss of generality, the problem reduces to a single lion chasing a single zebra along an arbitrarily wide strip, with the lion starting at (0,0) and the zebra starting anywhere else on the $x$-axis, with the zebra going first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is whether the lion can get within a fixed distance of the zebra by following a path almost parallel to the zebra's, given the same speed for both animals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the triangle formed by the two trajectories and the x-axis. Let $l$ and $z$ be the lengths of the sides along the lion's and zebra's trajectories respectively, and let $b$ be the initial distance between the two animals, also the length of the base of the triangle.  By Pythagoras' theorem, $l^2 = z^2 + b^2$. Since $b$ is constant, given the angle $\theta$ opposite the base, $z$ is unbounded as $\theta$ approaches $0$. So $\lim_{\theta \rightarrow 0} l = z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, the lion can get arbitrarily close to the zebra, so the lion wins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at some actual values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have $l-z = 0.001$ and $b=100$, then from $l^2 - z^2 = b^2$, we have&lt;br&gt;&#xA;$$&#xA;(l-z)(l+z) = b^2,&#xA;$$&#xA;and substituting known values, we have&#xA;$$&#xA;l+z = 10^7 .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So given $l \approx z$, each animal would have had to run about 5000km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the lion was level with the zebra and 100m away horizontally at the start, the critical point occurs when the zebra is 0.5mm shy of 5000km. So if both animals always took maximal 100m steps on constant trajectories towards the critical point, the zebra would be fine up to the step before 5000km (i.e. at 4999.9km) even after the lion's next move, but on reaching 5000km, the zebra would be caught on the lion's next move.&lt;/p&gt;&#xA;" OwnerUserId="9422" LastEditorUserId="9422" LastEditDate="2015-03-11T22:49:57.563" LastActivityDate="2015-03-11T22:49:57.563" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9241" PostTypeId="1" AcceptedAnswerId="9245" CreationDate="2015-02-19T10:14:01.727" Score="2" ViewCount="334" Body="&lt;p&gt;A $3\times3$ grid contains altogether six squares that are formed by its nine entries: there are five squares whose sides are parallel to the sides of the grid (four small ones and a big one), and there is one square that is tilted by $45$ degrees:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  a b c             a b    b c                     a    c          b&#xA;  d e f  contains   d e    e f    d e    e f                    d     f&#xA;  g h i                           g h    h i       g    i          h&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Determine all possible ways of arranging the nine numbers $1,2,3,\ldots,9$ in a $3\times3$ grid, such that the four numbers in the corners of each of the six squares add up to the same sum.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-10-04T16:14:58.447" LastActivityDate="2015-10-04T16:14:58.447" Title="Warped magic squares" Tags="|mathematics|combinatorics|magic-square|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9242" PostTypeId="2" ParentId="9241" CreationDate="2015-02-19T10:40:17.903" Score="1" Body="&lt;p&gt;Solutions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (a, b, c, d, e, f, g, h, i) = (1, 6, 7, 8, 5, 2, 3, 4, 9)&lt;br/&gt;(a, b, c, d, e, f, g, h, i) = (1, 8, 3, 6, 5, 4, 7, 2, 9)&lt;br/&gt;(a, b, c, d, e, f, g, h, i) = (3, 4, 9, 8, 5, 2, 1, 6, 7)&lt;br/&gt;(a, b, c, d, e, f, g, h, i) = (3, 8, 1, 4, 5, 6, 9, 2, 7)&lt;br/&gt;(a, b, c, d, e, f, g, h, i) = (7, 2, 9, 6, 5, 4, 1, 8, 3)&lt;br/&gt;(a, b, c, d, e, f, g, h, i) = (7, 6, 1, 2, 5, 8, 9, 4, 3)&lt;br/&gt;(a, b, c, d, e, f, g, h, i) = (9, 2, 7, 4, 5, 6, 3, 8, 1)&lt;br/&gt;(a, b, c, d, e, f, g, h, i) = (9, 4, 3, 2, 5, 8, 7, 6, 1)&lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; proof-by-exhaustion&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastActivityDate="2015-02-19T10:40:17.903" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9243" PostTypeId="1" AcceptedAnswerId="9976" CreationDate="2015-02-19T12:15:10.897" Score="5" ViewCount="620" Body="&lt;p&gt;I have partitioned the independent countries of the world into two groups based on a certain property X that is shared by some countries, but not by the remaining ones. Property X is purely geographic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following countries all have this enigmatic property X:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bolivia, Colombia, Cuba, Libya, Mali, Nicaragua, North-Korea,&#xA;   Portugal, Russia, Senegal, South-Korea, Switzerland, Tajikistan&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The following countries do not have the enigmatic property X:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Afghanistan, Argentina, Australia, Austria, Canada, China,&#xA;   Ireland, Jamaica, Lithuania, Oman, Poland, Sudan, Tanzania&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What criterion did I use for my enigmatic partition?&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;Do Brazil, Chad, Denmark, the USA and/or Yemen have this enigmatic property X?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Property X is purely geographic. It can be solved by studying a map of the world. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="5373" LastEditDate="2016-12-11T15:42:53.210" LastActivityDate="2016-12-11T15:42:53.210" Title="An enigmatic partition of the independent countries" Tags="|pattern|geography|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9244" PostTypeId="1" AcceptedAnswerId="9304" CreationDate="2015-02-19T12:36:29.010" Score="2" ViewCount="585" Body="&lt;p&gt;Ethan Hunt is alerted by the IMF. They want him to intercept a courier that is working for the super villains Matt Fillimore and Caro Bassanara. Ethan Hunt receives the following cipher from the IMF:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  8901  6892  31  389987688  782  154  6119&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What is the code-name of the courier?&lt;br&gt;&#xA;Where should Ethan intercept the courier?&lt;/p&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="8874" LastEditDate="2015-03-25T11:49:54.547" LastActivityDate="2015-03-25T11:49:54.547" Title="Mission: Impossible 1" Tags="|cipher|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9245" PostTypeId="2" ParentId="9241" CreationDate="2015-02-19T13:11:23.043" Score="3" Body="&lt;p&gt;Summing the big square and the tilted square, we get for the common sum $S$&lt;br&gt;&#xA;$$ 2S = (a+c+g+i)+(b+d+f+h) = 45-e.$$&#xA;Summing the entries in the four small squares &#xA;$$4S = (a+b+d+e)+(b+c+e+f)+(d+e+g+h)+(e+f+h+i).$$&#xA;Summing the big square, twice the tilted square and four times entry $e$&lt;br&gt;&#xA;$$3S+4e = (a+c+g+i)+2(b+d+f+h)+4e.$$&#xA;This shows $4S=3S+4e$ and $S=4e$.&#xA;Then &#xA;$2S+e=45$ and $S-4e=0$ shows $e=5$ and $S=20$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the small squares uses $e=5$ and the number $1$. Then the other two entries are distinct and have sum $14$. Then the other two entries can only be $6$ and $8$. Then $1$ must be in a corner:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   1 6 c&#xA;   8 5 f&#xA;   g h i&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Since 1 can be in any of the four corners and 6 and 8 can switch places, there are eight symmetric solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the small squares uses $e=5$ and the number $7$. Then the other two entries are distinct and have sum $8$. Then the other two entries can only be $2$ and $6$. Then $7$ must be in a corner:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   1 6 7&#xA;   8 5 2&#xA;   g h i&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The tilted sqaure has $20=b+d+f+h=6+8+2+h$ and $h=4$. Then $g$=3$ and $i=9$.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   1 6 7&#xA;   8 5 2&#xA;   3 4 9&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Alltogether there are eight solutions. The other seven solutions come from rotating my solution by 90, 180, 270 degree and reflecting it on the middle line.&lt;/p&gt;&#xA;" OwnerUserId="9125" LastActivityDate="2015-02-19T13:11:23.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9246" PostTypeId="2" ParentId="9224" CreationDate="2015-02-19T13:41:30.203" Score="10" Body="&lt;p&gt;My first thought was that you are a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Video tape&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My body is thin and frail:&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The tape is very thin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Scratching my head can kill me:&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you scratch the head in the video player, it can rip the tape.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;When abused, I can also kill you:&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You could probably pull the tape out and strangle yourself with it if you were really unlucky...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Usually, I'm helpful:&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; At keeping kids entertained.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I admit that my answer works best for the head-scratching part, the rest of it is a little bit stretched...&lt;/p&gt;&#xA;" OwnerUserId="9444" LastActivityDate="2015-02-19T13:41:30.203" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9247" PostTypeId="1" AcceptedAnswerId="9248" CreationDate="2015-02-19T13:58:48.707" Score="1" ViewCount="1296" Body="&lt;p&gt;I have another rebus puzzle for you guys to solve :D ; It's on the easier side&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here it goes,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/22OYI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9346" LastActivityDate="2018-04-12T10:41:48.473" Title="Attempt - A rebus puzzle" Tags="|rebus|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9248" PostTypeId="2" ParentId="9247" CreationDate="2015-02-19T14:02:10.423" Score="6" Body="&lt;p&gt;This seems too obvious:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Bold Attempt&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastActivityDate="2015-02-19T14:02:10.423" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9249" PostTypeId="1" AcceptedAnswerId="9252" CreationDate="2015-02-19T14:10:31.050" Score="14" ViewCount="8621" Body="&lt;p&gt;I didn't write this rebus but I like it so I'm sharing it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BVDAi.png&quot; alt=&quot;Image form of Rebus&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Text form:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   J   K &#xA;10   Q   A&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="9000" LastActivityDate="2018-01-19T05:45:45.473" Title="Rebus: Shanghai" Tags="|rebus|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9250" PostTypeId="1" AcceptedAnswerId="9251" CreationDate="2015-02-19T14:14:32.473" Score="6" ViewCount="6134" Body="&lt;p&gt;I didn't write this rebus, either, but I'm still sharing it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/S8xuH.png&quot; alt=&quot;Image form of rebus&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Text form:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;D  B1&#xA;N  B2&#xA;A  B3&#xA;T  B4&#xA;S  B5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="9000" LastActivityDate="2020-04-20T04:56:26.720" Title="Rebus: Representing" Tags="|rebus|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9251" PostTypeId="2" ParentId="9250" CreationDate="2015-02-19T14:16:49.923" Score="7" Body="&lt;p&gt;Let's go with  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stand up and be counted&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-02-19T14:16:49.923" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9252" PostTypeId="2" ParentId="9249" CreationDate="2015-02-19T14:19:20.590" Score="17" Body="&lt;p&gt;i think its&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; hijacking - jack-king are higher than the other face cards symbols.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2803" LastActivityDate="2015-02-19T14:19:20.590" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9253" PostTypeId="2" ParentId="9249" CreationDate="2015-02-19T14:19:39.563" Score="15" Body="&lt;p&gt;As shanghaiing is kidnapping people to serve as sailors, the title &quot;Shanghai&quot; suggests:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;A high Jack and King&quot; = &lt;strong&gt;&quot;hijacking&quot;&lt;/strong&gt;   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-02-19T14:19:39.563" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9254" PostTypeId="1" AcceptedAnswerId="9635" CreationDate="2015-02-19T15:45:57.477" Score="11" ViewCount="1066" Body="&lt;p&gt;The BED and I both have C-some.&lt;br&gt;&#xA;The TREE and I both have L-some.&lt;br&gt;&#xA;The PRISONER and I both are B-some.&lt;br&gt;&#xA;The WEBSITE and I both have P-some.&lt;br&gt;&#xA;The SHIRT and I both have S-some.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="5151" LastEditDate="2015-05-21T19:06:00.677" LastActivityDate="2018-01-22T17:15:51.867" Title="Riddle with bed, tree, prisoner" Tags="|riddle|word|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9255" PostTypeId="1" AcceptedAnswerId="9256" CreationDate="2015-02-19T16:05:49.337" Score="2" ViewCount="11694" Body="&lt;p&gt;Solve the following rebus puzzle:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; ISSUES &#xA; ISSUES  ISSUES   &#xA; ISSUES  ISSUES  ISSUES  &#xA; ISSUES  ISSUES  ISSUES  ISSUES  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2015-02-21T05:22:07.170" Title="The issue - Rebus puzzle" Tags="|rebus|english|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9256" PostTypeId="2" ParentId="9255" CreationDate="2015-02-19T16:08:04.657" Score="11" Body="&lt;p&gt;Not the best pun, if the solution indeed is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Ten issues&quot; = &quot;Tennis shues&quot; = &lt;strong&gt;&quot;Tennis shoes&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-02-19T16:08:04.657" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9258" PostTypeId="1" CreationDate="2015-02-19T16:21:22.323" Score="-4" ViewCount="415" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;memory of elevant: &lt;strong&gt;in&lt;/strong&gt; time lanah sevenson fell down a threehouse. &lt;strong&gt;here&lt;/strong&gt;fracture two the head? confusion nausea &lt;strong&gt;are&lt;/strong&gt; &lt;strong&gt;thir&lt;/strong&gt;sty but jolly &lt;strong&gt;teen&lt;/strong&gt; sees &lt;strong&gt;what&lt;/strong&gt; is still afive &lt;strong&gt;?&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;help  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; in this hint are three  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;help  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; in this one are two&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;help  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; THINK YOU'RE CLEVER EH - IN THIS HINT THERE ARE ZERO&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="9413" LastEditDate="2015-02-23T15:54:32.713" LastActivityDate="2015-02-23T16:59:06.013" Title="Here there are zero" Tags="|enigmatic-puzzle|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9259" PostTypeId="1" AcceptedAnswerId="9261" CreationDate="2015-02-19T16:58:43.297" Score="4" ViewCount="685" Body="&lt;p&gt;My last rebus puzzle got solved pretty quickly, so I came up with a slightly harder one!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zoQif.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9346" LastActivityDate="2015-02-19T17:04:36.353" Title="Rebus- Wind OW!" Tags="|rebus|" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9260" PostTypeId="2" ParentId="9259" CreationDate="2015-02-19T17:04:11.380" Score="0" Body="&lt;p&gt;Could it be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wind down the window&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;" OwnerDisplayName="user9278" LastActivityDate="2015-02-19T17:04:11.380" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9261" PostTypeId="2" ParentId="9259" CreationDate="2015-02-19T17:04:36.353" Score="4" Body="&lt;p&gt;Oh, oh! I think I've got this!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Window pane&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Putting all the letters together gets you &quot;window&quot;&lt;br&gt;&#xA; The &quot;OW!&quot; is meant to indicate &quot;pain&quot; which is a homonym for &quot;pane&quot; or the section of glass. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-02-19T17:04:36.353" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9262" PostTypeId="1" AcceptedAnswerId="9266" CreationDate="2015-02-19T17:06:35.833" Score="8" ViewCount="825" Body="&lt;p&gt;On a $9\times10$ checkerboard with $90$ squares, a game proceeds as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;First, you place $x$ rocks on the board. Each rock occupies a single square. No rock may touch the edge of the checkerboard. No rock may be positioned diagonally adjacent to another rock (while positioning them vertically or horizontally adjacent is allowed).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The snake chooses an empty square on the board as its starting point. &#xA;It moves around by always moving to a (horizontally or vertically, but not diagonally) adjacent square. It must not visit any square twice, and it can not visit any square that is occupied by a rock. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The snake wins the game, if it manages to visit every empty square on the board exactly once. Assume that the snake is a perfect player.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What is the minimum number of rocks required to prevent the snake from winning, and how will you place them?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A harder question: What if the aim of the snake is to visit every empty square &lt;em&gt;exactly&lt;/em&gt; twice?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="8874" LastEditDate="2016-03-17T09:20:56.293" LastActivityDate="2016-03-17T09:21:28.637" Title="Block the snake" Tags="|strategy|game|checkerboard|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9263" PostTypeId="1" AcceptedAnswerId="9264" CreationDate="2015-02-19T17:12:21.097" Score="-7" ViewCount="1531" Body="&lt;p&gt;A number has seven digits. The product of the first three digits is equal to the product of the last three digits is equal to the product of the middle three digits. The digits are distinct. What is the middle (fourth) digit?&lt;/p&gt;&#xA;" OwnerUserId="4508" LastActivityDate="2015-02-19T17:12:21.097" Title="Seven digit number" Tags="|mathematics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9264" PostTypeId="2" ParentId="9263" CreationDate="2015-02-19T17:12:21.097" Score="-3" Body="&lt;p&gt;Using the letters a..g for the digits we have &lt;code&gt;abc = cde = efg&lt;/code&gt;. Observe &lt;strong&gt;each digit features in at most two products&lt;/strong&gt;: &lt;code&gt;c&lt;/code&gt; and &lt;code&gt;e&lt;/code&gt; are shared between two products, the rest of the digits appear in only one product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note there will be multiple solutions—if any—because given one solution reversing the number or swapping &lt;code&gt;a&lt;/code&gt; and &lt;code&gt;b&lt;/code&gt; will yield another)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recalling the digits are distinct, conclude 0 cannot be one of them, because it would zero one or two &lt;em&gt;but not all&lt;/em&gt; of the products. Similarly 5 and 7 can't be digits, by considering prime factors. One &lt;em&gt;but not all&lt;/em&gt; the products would be divisible by 5 or 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The digits are thus 1234689 in some order. Let w be the product &lt;code&gt;abc = cde = efg&lt;/code&gt;. By thinking about prime factors again, we conclude 8 divides w and 9 divides w. Thus 72 divides w. Looking at the powers of 2 and 3 in the smaller digits, conclude w is exactly 72.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With these digits 72 can be expressed as only three products &lt;code&gt;1*8*9&lt;/code&gt;, &lt;code&gt;3*4*6&lt;/code&gt; and &lt;code&gt;2*4*9&lt;/code&gt;. The shared digits (c and e) are thus 4 and 9. The digit in between and the middle digit overall is 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed 1892436 is a solution.&lt;/p&gt;&#xA;" OwnerUserId="4508" LastActivityDate="2015-02-19T17:12:21.097" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="9265" PostTypeId="2" ParentId="9262" CreationDate="2015-02-19T17:13:37.347" Score="5" Body="&lt;p&gt;For the first part, 4 rocks should be sufficient: the rocks are in a line, 1 space between each, and all are 1 space from the edge. Effectively, this creates a graph with 3 nodes with 3 entry points, which can't be properly traversed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll think about the second question...&lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-02-19T17:18:38.930" LastActivityDate="2015-02-19T17:18:38.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9266" PostTypeId="2" ParentId="9262" CreationDate="2015-02-19T17:17:25.630" Score="21" Body="&lt;p&gt;The answer to the first question is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Two rocks&lt;/strong&gt; are sufficient to prevent the snake from winning.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof: Use the standard black-and-white checkerboard coloring, and put the rocks on two white squares (while obeying the rules on placing the rocks).&#xA;Then there are 43 empty white squares and 45 empty black squares.&#xA;Since the snake alternately moves between black and white squares, in its altogether 88 moves it must visit 44 white and 44 black squares. Impossible.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A &lt;strong&gt;single rock&lt;/strong&gt; is not sufficient to prevent the snake from winning.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On the totally empty checkerboard, there exists a closed tour that visits every square exactly once and that ends again at the starting point.&#xA;Putting a single rock onto the board would cut the closed tour at one point; the snake could use the remaining piece of the tour to visit every square exactly once.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For the second (harder) question the answer is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A &lt;strong&gt;single rock&lt;/strong&gt; is sufficient to prevent the snake from winning.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Use the same argument as above: a single rock placed on an arbitrary white square would force the snake to visit $2\cdot44$ empty white squares and $2\cdot45$ empty black squares. Impossible.  &lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-17T09:21:28.637" LastActivityDate="2016-03-17T09:21:28.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9267" PostTypeId="2" ParentId="9224" CreationDate="2015-02-19T17:19:40.117" Score="16" Body="&lt;p&gt;This is my first guess at the answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mercury Thermometer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My body is thin and frail:&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Glass body is very fragile to being broken from children&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Scratching my head can kill me:&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Has a head that if scrated would break the glass making it no longer work properly&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;When abused, I can also kill you:&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mercury poisoning!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Usually, I'm helpful:&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Taking temperature for everyone&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let me know what you think&lt;/p&gt;&#xA;" OwnerUserId="9649" LastActivityDate="2015-02-19T17:19:40.117" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9268" PostTypeId="1" AcceptedAnswerId="9269" CreationDate="2015-02-19T17:50:31.233" Score="11" ViewCount="997" Body="&lt;p&gt;I had an idea for a riddle yesterday.  Here it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a horned and hooded figured, bearing a shield and blades.&lt;br&gt;&#xA;The ability to control great power is in my clutch.&lt;br&gt;&#xA;If you find yourself under my boot, you will discover that I am exhausted.&lt;br&gt;&#xA;If I am not tired, then my boot may spare you.&lt;br&gt;&#xA;You can easily choke me to death.&lt;br&gt;&#xA;Does this shock you? I am shocked, too.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I? (and why?)&lt;/p&gt;&#xA;" OwnerUserId="2162" LastEditorUserId="39" LastEditDate="2015-02-20T00:47:11.287" LastActivityDate="2015-02-20T01:09:30.587" Title="Hooded figure riddle" Tags="|riddle|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9269" PostTypeId="2" ParentId="9268" CreationDate="2015-02-19T17:57:52.430" Score="13" Body="&lt;p&gt;You are a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Car&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a horned and hooded figured, bearing a shield and blades.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Horn: beep, beep! &lt;br&gt; Hooded: pop the hood! &lt;br&gt; Shield: windshield &lt;br&gt; Blades: which clear the windshield&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The ability to control great power is in my clutch.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The clutch is one of the control pedals&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you find yourself under my boot, you will discover that I am exhausted.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you look &quot;under the boot&quot; of the car... you will see the exhaust pipes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I am not tired, then my boot may spare you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the car has a flat tyre (or no tyres) then they is often a spare tyre in the boot.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can easily choke me to death.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can cause a car to &quot;choke&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Does this shock you? I am shocked, too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Refers to the suspension system - the &quot;shocks&quot;. (Is the suspension system not shocking?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2015-02-20T01:09:30.587" LastActivityDate="2015-02-20T01:09:30.587" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9270" PostTypeId="2" ParentId="9224" CreationDate="2015-02-19T18:23:10.997" Score="24" Body="&lt;p&gt;The real answer is glaringly obvious:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; condom&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am thin and frail.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ever used a cheap one?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Scratching my head can kill me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; How do you think I got my kid?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When abused, I can also kill you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; How do you think I got my syphilis?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Usually, I'm helpful.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Usually...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9654" LastEditorUserId="9654" LastEditDate="2015-02-19T19:01:23.033" LastActivityDate="2015-02-19T19:01:23.033" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9273" PostTypeId="1" AcceptedAnswerId="9274" CreationDate="2015-02-19T19:28:47.990" Score="7" ViewCount="3816" Body="&lt;p&gt;I have seas without water &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have forests without wood &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have deserts without sand &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have houses with no brick&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="9575" LastActivityDate="2016-06-30T12:52:08.467" Title="Seas Without Water" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9274" PostTypeId="2" ParentId="9273" CreationDate="2015-02-19T19:37:08.033" Score="8" Body="&lt;p&gt;It is... a shrubbery! .... ....&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Or a map&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Maps have seas, forests, deserts, or houses marked, but as they are, well, maps, they don't actually have water, wood, sand or bricks. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2016-06-30T12:52:08.467" LastActivityDate="2016-06-30T12:52:08.467" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9275" PostTypeId="1" AcceptedAnswerId="9276" CreationDate="2015-02-19T20:13:40.333" Score="5" ViewCount="386" Body="&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/9273/seas-without-water&quot;&gt;I have forests with no wood&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;rope that will not bind,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;bags that hold no food,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and masks you cannot hide behind.&lt;/p&gt;&#xA;" OwnerUserId="2421" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-02-19T20:31:04.007" Title="I have maps but am not a map" Tags="|riddle|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9276" PostTypeId="2" ParentId="9275" CreationDate="2015-02-19T20:31:04.007" Score="7" Body="&lt;p&gt;You are a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; programming language&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have maps but am not a map&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A programming language can have maps that contain key-value pairs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have forests with no wood,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A programming language can have tree data structures.  Multiple trees could be called a &quot;forest&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;rope that will not bind,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A programming language can have a rope data structure that contains strings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;bags that hold no food,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A programming language can have a bag data structure that contains other objects.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and masks you cannot hide behind.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A programming language has bit masks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2015-02-19T20:31:04.007" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9278" PostTypeId="2" ParentId="9221" CreationDate="2015-02-19T21:14:47.007" Score="3" Body="&lt;p&gt;A shot in the dark:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A is an imbecile, B is a moron, C is an idiot and D is a superior.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; According to the &lt;a href=&quot;http://en.wikipedia.org/wiki/IQ_classification#Historical_IQ_classification_tables&quot; rel=&quot;nofollow&quot;&gt;Levine and Marks 1928 IQ classification&lt;/a&gt; the only IQ classification that differs such low scores.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alternatively:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A, B and C are impaired, and D is a genius according to newer classifications.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastActivityDate="2015-02-19T21:14:47.007" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="9279" PostTypeId="2" ParentId="3641" CreationDate="2015-02-19T22:29:12.520" Score="5" Body="&lt;p&gt;Observe&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $1 = \frac{5}{7} + \frac{6}{21}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and rearrange.&lt;/p&gt;&#xA;" OwnerUserId="4508" LastEditorUserId="4508" LastEditDate="2015-02-19T23:48:24.297" LastActivityDate="2015-02-19T23:48:24.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9280" PostTypeId="1" AcceptedAnswerId="9281" CreationDate="2015-02-19T22:39:16.663" Score="15" ViewCount="26270" Body="&lt;p&gt;You go in through one hole, you come out three holes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you’re inside you’re ready to go outside, but once you're outside you’re still inside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is it?&lt;/p&gt;&#xA;" OwnerUserId="9575" LastEditorUserId="9575" LastEditDate="2015-04-08T00:17:01.640" LastActivityDate="2016-07-26T11:03:40.367" Title="In Through One Hole" Tags="|riddle|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9281" PostTypeId="2" ParentId="9280" CreationDate="2015-02-19T22:44:37.280" Score="17" Body="&lt;p&gt;My answer:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;A shirt&lt;/strong&gt;. The shirt has three holes: two sleeves and one neck. Once you're wearing the shirt, you're ready to go outside, but once you're outside, you're still wearing the shirt.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9319" LastActivityDate="2015-02-19T22:44:37.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9282" PostTypeId="1" AcceptedAnswerId="9285" CreationDate="2015-02-19T23:07:21.980" Score="6" ViewCount="1384" Body="&lt;p&gt;A young man walks through the forest. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;He comes to a bridge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In front of the bridge is a large man carrying an axe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The man says, &quot;If you want to cross this bridge, you must tell me something. If I think your statement is true, I will strangle you to death. If I think your statement is false, I will chop off your head.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few minutes later, the young man walks over the bridge, while the larger man stands pondering. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What did he say?&lt;/p&gt;&#xA;" OwnerUserId="9575" LastEditorUserId="5373" LastEditDate="2018-05-06T16:23:14.593" LastActivityDate="2018-05-06T16:23:14.593" Title="Over the Bridge, If You Dare" Tags="|logical-deduction|story|liars|situation|" AnswerCount="5" CommentCount="9" ClosedDate="2015-02-20T14:32:51.317" ContentLicense="CC BY-SA 3.0" />
  <row Id="9283" PostTypeId="2" ParentId="9282" CreationDate="2015-02-19T23:28:09.760" Score="2" Body="&lt;p&gt;He said:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if you were me what would you say?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7714" LastEditorUserId="5188" LastEditDate="2015-02-19T23:33:19.963" LastActivityDate="2015-02-19T23:33:19.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9284" PostTypeId="2" ParentId="9282" CreationDate="2015-02-19T23:29:20.707" Score="7" Body="&lt;p&gt;How about the classic paradoxical statement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This statement is false.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-02-19T23:29:20.707" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9285" PostTypeId="2" ParentId="9282" CreationDate="2015-02-19T23:33:00.100" Score="15" Body="&lt;p&gt;The canonical answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You will chop off my head.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But this doesn't really work because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; he can simply strangle you and then behead your corpse.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So perhaps&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You will kill me by chopping off my head.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1716" LastActivityDate="2015-02-19T23:33:00.100" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9286" PostTypeId="1" AcceptedAnswerId="9313" CreationDate="2015-02-19T23:48:22.240" Score="-3" ViewCount="300" Body="&lt;p&gt;Running on bones,&lt;br&gt;&#xA;I am shaking.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who or what am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am visiting the Philippines now, and this is the direct translation of a popular local riddle spoken in Cebuano.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It is a simple riddle and definitely solvable.  The answer is not something scary or obscure.  It is a common thing. It is not a living thing.  It is manmade.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9495" LastEditorUserId="9495" LastEditDate="2015-02-20T11:32:56.073" LastActivityDate="2015-02-20T13:22:10.717" Title="Running on bones, I am shaking" Tags="|riddle|" AnswerCount="4" CommentCount="5" ClosedDate="2015-02-28T16:22:17.693" ContentLicense="CC BY-SA 3.0" />
  <row Id="9287" PostTypeId="1" CreationDate="2015-02-19T23:58:38.910" Score="1" ViewCount="282" Body="&lt;p&gt;A space measuring 3 by 10 is to be tiled. Tiles are square and come in sizes 1 by 1, 2 by 2 and 3 by 3. How many ways can the floor be tiled?&lt;/p&gt;&#xA;" OwnerUserId="4508" LastActivityDate="2015-02-20T01:21:52.933" Title="How many ways can the floor be tiled?" Tags="|mathematics|" AnswerCount="1" CommentCount="4" ClosedDate="2015-02-20T22:33:31.033" ContentLicense="CC BY-SA 3.0" />
  <row Id="9288" PostTypeId="1" CreationDate="2015-02-20T00:07:54.020" Score="15" ViewCount="2416" Body="&lt;p&gt;A barbarian king wants his tribe to have more men than women, because men can fight and women cannot. to achieve this, he introduces the following law:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Each couple has to produce babies until they have at least one boy. Then they must stop producing babies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By doing this, he ensures that every couple has a boy, and some couples will have zero girls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But his councillor sees a flaw in this plan. He says that to have more boys than girls he needs to make the opposite law.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Each couple has to reproduce as long as they get boys. When they get a single girl, they have to stop.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By doing this, some couples will get many boys, and none will ever get more than one girl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which law would get the most boys?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; The barbarian king does not care about the total number of babies. He just wants more boys than girls.&lt;/p&gt;&#xA;" OwnerUserId="9662" LastEditorUserId="435" LastEditDate="2015-02-20T00:29:33.770" LastActivityDate="2017-09-04T00:48:07.107" Title="The barbarian king" Tags="|logical-deduction|strategy|" AnswerCount="10" CommentCount="5" ClosedDate="2015-02-20T19:34:03.973" ContentLicense="CC BY-SA 3.0" />
  <row Id="9289" PostTypeId="2" ParentId="9258" CreationDate="2015-02-20T00:27:13.037" Score="0" Body="&lt;p&gt;the question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; in here are thirteen ____ (grey-ed out words)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; prime numbers. elevant, sevenson, threehouse, two, thirteen, afive,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9662" LastActivityDate="2015-02-20T00:27:13.037" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9290" PostTypeId="2" ParentId="9288" CreationDate="2015-02-20T01:02:10.247" Score="0" Body="&lt;p&gt;After running the simulation and researching the internet:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CONCLUSION&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some couples are more genetically prepossessed to produce male offspring, so purely because of this fact, the 2nd method will produce more male babies.  This problem cannot be solved purely with math, or you would think both methods produce identical results.  The first method does not guarantee more boy babies than girl babies, because that ratio is still 50/50.The second method allows more males per mother, but it is still a 50/50 ratio of boys and girls.  However, because of x and y chromosomes and genetics, more boys will be produced in the 2nd method, since some couples are more likely to produce boys.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9495" LastEditorUserId="9495" LastEditDate="2015-02-20T04:08:28.160" LastActivityDate="2015-02-20T04:08:28.160" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9291" PostTypeId="2" ParentId="9287" CreationDate="2015-02-20T01:21:52.933" Score="4" Body="&lt;p&gt;Consider a 3 by n space to be tiled, with the dimension of length 3 vertical. The largest piece touching the left side must either be 3x3, 2x2, or 1x1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is 3x3, removing it leaves a 3 by n-3 space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is 2x2, there are two places it can be (up or down), and it must be accompanied by two 1x1 pieces because those are the only ones which can fit in the space. Removing all of these leaves a 3 by n-2 space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is 1x1, then the left column is 3 1x1 pieces, and removing it leaves a 3 by n-1 space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we let f(n) be the number of ways to tile the 3 by n space, then for n&gt;0 we have f(n)=f(n-3)+2f(n-2)+f(n-3) (define f(-2)=f(-1)=0, f(0)=1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing left is to calculate f(10), which can be done by working up using the recursion. f(1)=1, f(2)=3, f(3)=6, f(4)=13, f(5)=28, f(6)=60, f(7)=129, f(8)=277, f(9)=595, and f(10)=1278. So 1278 is the answer.&lt;/p&gt;&#xA;" OwnerUserId="9663" LastActivityDate="2015-02-20T01:21:52.933" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9292" PostTypeId="2" ParentId="9288" CreationDate="2015-02-20T01:29:19.480" Score="-1" Body="&lt;p&gt;If the time is very limited,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first method guarantees that each couple will produce at least one boy but no more than one per family.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But if the time is not limited,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The second method produces the &lt;strong&gt;most&lt;/strong&gt; boys, however it will take longer because many of the families will be eliminated from the pool as soon as they have a girl and will not have a chance to produce a boy at all.  However, chance allowing, there is the possibility that many families will have multiple boys and so, over time they will accumulate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Conclusion:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Method two has the possibility to produce the MOST boys, but method one has the possibility to produce more faster in a shorter period of time.  Neither method is guaranteed to produce any boys at all.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9575" LastActivityDate="2015-02-20T01:29:19.480" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9293" PostTypeId="2" ParentId="9286" CreationDate="2015-02-20T01:37:42.650" Score="2" Body="&lt;p&gt;I almost want to say...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dice, because you shake them to roll, and they are sometimes referred to as bones.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4353" LastActivityDate="2015-02-20T01:37:42.650" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9294" PostTypeId="2" ParentId="9288" CreationDate="2015-02-20T01:38:12.743" Score="8" Body="&lt;p&gt;Under the assumption that &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The probability of getting a girl or a boy is equal (50%)&lt;/li&gt;&#xA;&lt;li&gt;A male can only put his thang on his partner female (no multiple wives)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Neither of them&lt;/strong&gt; will get significantly more boys than the other (in a long period of time)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;A Priori Proof :&lt;/h1&gt;&#xA;&#xA;&lt;h2&gt;Barbarian King&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In regex: &lt;code&gt;(G*)B&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;            ...&#xA;        G&#xA;    G       B&#xA;Z       B&#xA;    B&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The tree goes from left to right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Z = initial state; G = girl; B = boy;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;The probability of getting 1 boy and 0 girls is 1/2.&#xA;The probability of getting 1 boy and 1 girl is 1/4.&#xA;The probability of getting 1 boy and 2 girls is 1/8.&#xA;The probability of getting 1 boy and N girls is 1/2^(N+1).&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h3&gt;To solve the average difference of boys and girls under the Barbarian King's rule, we would need the following :&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{N\to \infty} \frac{\sum_0^N((\mbox{BoyCount}-\mbox{GirlCount})*\mbox{probability}))}{N}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\lim_{N\to \infty} \frac{\sum_0^N(\frac{1-N}{2^{N+1}}))}{N} = 0$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Check &lt;a href=&quot;http://www.wolframalpha.com/input/?i=sum%20%281-j%29%2F%282%5E%28j%2B1%29%29%2C%20j%3D0..infinity&quot;&gt;this wolfram solution&lt;/a&gt; to this equation. (Sorry Barbarian King)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This means that in the long run (approaching $\infty$), &lt;strong&gt;there is no ($0$) significant difference between the number of boys and girls under the Barbarian King's rule.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Councillor&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In regex: &lt;code&gt;(B*)G&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    G&#xA;Z       G&#xA;    B       G&#xA;        B&#xA;            ...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The tree goes from left to right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Z = initial state; G = girl; B = boy;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;The probability of getting 0 boys and 1 girl is 1/2.&#xA;The probability of getting 1 boy and 1 girl is 1/4.&#xA;The probability of getting 2 boys and 1 girl is 1/8.&#xA;The probability of getting N boys and 1 girl is 1/2^(N+1).&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h3&gt;To solve the average difference of boys and girls under the Councillor's rule, we would need the following :&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{N\to \infty} \frac{\sum_0^N((\mbox{BoyCount}-\mbox{GirlCount})*\mbox{probability}))}{N}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\lim_{N\to \infty} \frac{\sum_0^N(\frac{N-1}{2^{N+1}}))}{N} = 0$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Check &lt;a href=&quot;http://www.wolframalpha.com/input/?i=sum%20%28j-1%29%2F%282%5E%28j%2B1%29%29%2C%20j%3D0..infinity&quot;&gt;this wolfram solution&lt;/a&gt; to this equation. (Sorry Councillor)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This means that in the long run (approaching $\infty$), &lt;strong&gt;there is no ($0$) significant difference between the number of boys and girls under the Councillor's rule.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;A Posteriori Proof :&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Open your JavaScript console (F12-&gt;Console or Ctrl+Shift+J) and copy and paste the below code.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;function makeBabies(couples,simulations){&#xA;  var kingAdvantage=0;&#xA;  var councillorAdvantage=0;&#xA;  var kingBabies={boys:0,girls:0};&#xA;  var councillorBabies={boys:0,girls:0};&#xA;&#xA;  for(var j=0;j&amp;lt;simulations;j++){&#xA;    for(var i=0;i&amp;lt;couples;i++){&#xA;      //Barbarian King&#xA;      while(Math.random()&amp;lt;0.5){&#xA;        kingBabies.girls++;&#xA;      }&#xA;      kingBabies.boys++;&#xA;&#xA;      //Councillor&#xA;      while(Math.random()&amp;lt;0.5){&#xA;        councillorBabies.boys++;&#xA;      }&#xA;      councillorBabies.girls++;&#xA;    }&#xA;&#xA;    kingAdvantage += (kingBabies.boys-kingBabies.girls);&#xA;    councillorAdvantage += (councillorBabies.boys-councillorBabies.girls);&#xA;&#xA;    //reset simulation&#xA;    kingBabies={boys:0,girls:0};&#xA;    councillorBabies={boys:0,girls:0};&#xA;  }&#xA;&#xA;  //average&#xA;  kingAdvantage /= simulations;&#xA;  councillorAdvantage /= simulations;&#xA;&#xA;&#xA;  console.log(&quot;Barbarian King advantage : &quot;+kingAdvantage);&#xA;  console.log(&quot;Councillor advantage : &quot;+councillorAdvantage);&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Type &lt;code&gt;makeBabies(&amp;lt;couples&amp;gt;,&amp;lt;simulations&amp;gt;)&lt;/code&gt; and replace &lt;code&gt;&amp;lt;couples&amp;gt;&lt;/code&gt; with the number of couples that you want in your simulation, and &lt;code&gt;&amp;lt;simulations&amp;gt;&lt;/code&gt; with the number of times you want to run the same test.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Results :&lt;/h2&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;makeBabies(100,10000) = 0.0488, -0.0228&#xA;makeBabies(1000,10000) = 0.3019, -0.1117&#xA;makeBabies(10000,10000) = -2.47, -0.4471&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;10000 simulations each time, with increasing number of couples (100, 1000, 10000).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the average advantage of boys over girls in both methods is very insignificant. (Most of the time &amp;lt; 1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can try out the code and run more simulations if you want.&lt;/p&gt;&#xA;" OwnerUserId="4553" LastEditorDisplayName="user20" LastEditDate="2015-02-20T04:48:22.117" LastActivityDate="2015-02-20T04:48:22.117" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9295" PostTypeId="2" ParentId="9288" CreationDate="2015-02-20T01:39:29.500" Score="2" Body="&lt;p&gt;Both laws produce:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the same number of boys and girls.&lt;br&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the 1st Law is followed:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; then a couple will add more boys than girls to the population, if it has 1 boy ( with probability 1/2) - and the net effect is +1 boy.  It will add more girls than boys to the population if it has a sequence of children $GG^nB$ ( $n \geq 1$) with probability $1/2^{n+2}$ and the net effect $-n$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the total net effect is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $1/2 - \sum\limits_{n=1}^\infty {n/2^{n+2}} = 0$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The same reasoning applies to Law 2 in reverse.&lt;/p&gt;&#xA;" OwnerUserId="7489" LastEditorUserId="5188" LastEditDate="2015-02-20T04:26:16.453" LastActivityDate="2015-02-20T04:26:16.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9296" PostTypeId="1" AcceptedAnswerId="9298" CreationDate="2015-02-20T01:40:23.813" Score="0" ViewCount="1615" Body="&lt;p&gt;A panda bear walked into a restaurant. He sat down at a table and ordered some food.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When he was finished eating, he took out a gun and shot his waiter. He then left the restaurant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the police caught up with him, they asked him why he had killed the waiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He replied, “Look me up in the dictionary.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What did the dictionary say?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9575" LastEditorUserId="11297" LastEditDate="2015-06-14T06:15:27.397" LastActivityDate="2015-06-14T06:15:27.397" Title="Look up the panda bear" Tags="|riddle|humor|" AnswerCount="1" CommentCount="13" ClosedDate="2015-02-20T11:23:12.433" ContentLicense="CC BY-SA 3.0" />
  <row Id="9297" PostTypeId="2" ParentId="9288" CreationDate="2015-02-20T01:49:12.867" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Both will have the same outcome&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof using a python simulation, assuming a population of 10,000 Couples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the 1st law:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from random import random&#xA;&#xA;trials = 10000&#xA;numBoys=0&#xA;numGirls=0&#xA;for i in range(trials):&#xA;    while(True):&#xA;        Chance=random()&#xA;        if Chance&amp;lt;0.5:&#xA;            numBoys+=1&#xA;            break&#xA;        else:&#xA;            numGirls+=1&#xA;print(numBoys)&#xA;print(numGirls)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For the 2nd law:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from random import random&#xA;&#xA;trials = 10000&#xA;numBoys=0&#xA;numGirls=0&#xA;for i in range(trials):&#xA;    while(True):&#xA;        Chance=random()&#xA;        if Chance&amp;lt;0.5:&#xA;            numBoys+=1&#xA;        else:&#xA;            numGirls+=1&#xA;            break&#xA;print(numBoys)&#xA;print(numGirls)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The results for Law 1 are as follows: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Trial 1: (10,000 Boys, 10028 Girls) Trial 2: (10,000 Boys, 9917 Girls) etc...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The results for Law 2 are as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Trial 1: (9,884 Boys, 10000 Girls) Trial 2: (10,327 Boys, 10000 Girls) etc...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you are somewhat knowledgeable about programming, you will realize the only difference in the code snippets is when I call &quot;break&quot;. Basically, I am calling &quot;break&quot; whenever reproduction should be stopped. So you can see that in 1 law, reproduction is stopped whenever a boy is born, whereas in the other, reproduction is stopped whenever a girl is born.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can expect the same outcome for each law.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9159" LastEditorUserId="9159" LastEditDate="2015-02-20T02:01:05.627" LastActivityDate="2015-02-20T02:01:05.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9298" PostTypeId="2" ParentId="9296" CreationDate="2015-02-20T01:58:13.847" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Panda - Eats shoots and leaves&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9159" LastActivityDate="2015-02-20T01:58:13.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9299" PostTypeId="2" ParentId="9282" CreationDate="2015-02-20T06:19:22.657" Score="1" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The young man said, &quot;My head will be chopped off.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the axeman claimed the statement was false, the young man's head would be chopped off, making the statement true.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But if..&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; But if the axeman claimed the statement was true, the young man would be strangled to death, which would make the statement false.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Summary:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the end, the axeman came to a bridge he could not cross. He couldn't say the young man's statement was true OR false without invalidating his question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9575" LastActivityDate="2015-02-20T06:19:22.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9301" PostTypeId="2" ParentId="9288" CreationDate="2015-02-20T07:36:33.793" Score="11" Body="&lt;p&gt;The best answer IMHO was given as a comment to &lt;a href=&quot;https://math.stackexchange.com/questions/20426/famous-puzzle-girl-boy-proportion-problem-sum-of-infinite-serie&quot;&gt;this question&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For each birth, the probability of a boy is 50% and the probability of a girl is 50% (for the purposes of this exercise; the real world is more complicated). Therefore, the ratio for all births must be 50:50. The number of previous births in each family, and the genders of those children, is irrelevant. &#xA;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;author:  Mike Scott Feb 4 '11 at 17:28&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are not convinced yet, think of this in the following way. Each year $i$, there is a set $C_i$  of couples that are allowed to have a baby and do so. The initial subset $C_1$ is the set of all couples, and it becomes smaller and smaller when more and more couples are banned from reproducing. So each year, the number of new babies is $|C_i|$. But, assuming the ratio is 50:50, each year there will be on average $|C_i|/2$ boys and $|C_i|/2$ girls! This does not depend at all on the number and identity of couples in $C_i$.&lt;/p&gt;&#xA;" OwnerUserId="7315" LastEditorUserId="-1" LastEditDate="2017-04-13T12:19:36.950" LastActivityDate="2015-02-23T05:37:12.420" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="9302" PostTypeId="2" ParentId="9286" CreationDate="2015-02-20T08:08:26.937" Score="3" Body="&lt;p&gt;Interesting. What's the original text in the native tongue?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A dog - a dog runs on/after bones, and shakes it's head to get them... hard to explain, more like playing fetch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9670" LastEditorUserId="9670" LastEditDate="2015-02-20T08:14:22.687" LastActivityDate="2015-02-20T08:14:22.687" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9303" PostTypeId="2" ParentId="9288" CreationDate="2015-02-20T09:18:52.767" Score="6" Body="&lt;p&gt;Under the first formula, all parents will have exactly one boy while under the second formula, all parents will have exactly one girl. But how many children of the opposite gender can they expect in both cases?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X$ be a random variable that counts the number of children of opposite gender, i.e., the number of girls under formula 1 or the number of boys under formula 2. The &lt;em&gt;expected value&lt;/em&gt; of a discrete univariate random variable is defined as the sum over all possible values multiplied by their probability:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E[X] = x_1 \cdot p_1 + x_2 \cdot p_2 + \dots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our case, the possible values are the number of children, i.e., $x_1 = 0, x_2 = 1, x_3 = 2, \dots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every possible sequence of children will look something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A\ A\ A\ A\ A\ B$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that is, all children in the sequence have the same gender except the last one. For every child that is born, the probability that it is a girl or a boy is $50:50$ (presumably), so if the sequence consists of $n$ children, its probability is $\frac12\cdot\frac12\cdot\dots\cdot\frac12$ ($n$ times). For instance, the sequence above would have the probability $\frac12\cdot\frac12\cdot\frac12\cdot\frac12\cdot\frac12\cdot\frac12 = \left(\frac12\right)^6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the expected value we want to compute, this means:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;P(X=0) = \frac12,\quad&#xA;P(X=1) = \frac14,\quad&#xA;P(X=2) = \frac18,\quad\dots&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E[X] = 0 \cdot \frac12 + 1 \cdot \frac14 + 2 \cdot \frac18 + \cdots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically speaking, this is an infinite series. If we sum up all of its elements, we see that the sum converges to $1$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^\infty\frac{n-1}{2n}=1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means that we expect the number of children of the opposite gender to be $1$. To sum up: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Under formula 1, the number of boys for each couple is $1$ and the expected number of girls is $1$&lt;/li&gt;&#xA;&lt;li&gt;Under formula 2, the number of girls for each couple is $1$ and the expected number of boys is $1$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thus we conclude that both formulas will lead to the same expected number of boys and girls for the barbarians.&lt;/p&gt;&#xA;" OwnerUserId="9671" LastEditorUserId="9671" LastEditDate="2017-09-04T00:48:07.107" LastActivityDate="2017-09-04T00:48:07.107" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9304" PostTypeId="2" ParentId="9244" CreationDate="2015-02-20T09:55:16.767" Score="6" Body="&lt;p&gt;Good, it is nice to see some new twist on an old puzzle. &#xA;As a first hint: the underlying code has also been used in the puzzle &#xA;&quot;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/3751/what-comes-next-in-this-sequence&quot;&gt;What comes next in this sequence?&lt;/a&gt;&quot; posted by Gilles in November 2014. &#xA;It also shows up in some books of Martin Gardner published in the 1960s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the answer to the Mission Impossible puzzle is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The code-name of the courier is &lt;strong&gt;Enzo&lt;/strong&gt;.&lt;br&gt;&#xA;  Ethan Hunt should intercept the courier in &lt;strong&gt;Tennessee&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Motivation of answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0=Z(ero), 1=O(ne), 2=T(wo), 3=T(hree), 4=F(our), 5=F(ive), 6=S(ix), etc.&lt;br&gt;&#xA;                 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Under this substitution code, the message to Ethan becomes&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 8901  6892  31  389987688  782  154  6119    &#xA; ENZO  SENT  TO  TENNESSEE. SET  OFF  SOON!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-02-20T10:32:51.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9305" PostTypeId="1" AcceptedAnswerId="9311" CreationDate="2015-02-20T10:22:13.337" Score="33" ViewCount="2552" Body="&lt;p&gt;The 1933 movie &lt;a href=&quot;https://en.wikipedia.org/wiki/The_Devil%27s_Brother&quot; rel=&quot;noreferrer&quot;&gt;&quot;The Devil's Brother&quot;&lt;/a&gt; (also known under the title &quot;Fra Diavolo&quot;) takes place in the Northern Italy of the early 18th century. &#xA;Stan Laurel and Oliver Hardy play the fierce robbers Stanlio and Ollio.&#xA;They manage to steal 100 gold coins from the rich Lord Rocburg, and then discuss at length how to share the loot. &#xA;Finally they decide to play the following game.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In every step, Stanlio picks a handful of gold coins from the loot.&lt;br&gt;&#xA;Then Ollio decides, whether this handful should go to himself or to Stanlio.&lt;/li&gt;&#xA;&lt;li&gt;Once all coins have been assigned, the game ends.&lt;/li&gt;&#xA;&lt;li&gt;Once one of them has received 9 handfuls, the game also ends.&#xA;In this case the other player (who has received at most 8 handfuls up to that moment) receives all the remaining gold coins.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&#xA;  What is the highest number of gold coins that Stanlio can guarantee for himself?&lt;br&gt;&#xA;  (As usual, we assume that both players use optimal strategies.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-10-28T14:44:20.433" LastActivityDate="2015-10-28T14:45:11.647" Title="The Devil's Brother" Tags="|mathematics|strategy|game|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9306" PostTypeId="1" AcceptedAnswerId="9307" CreationDate="2015-02-20T10:47:12.053" Score="9" ViewCount="1759" Body="&lt;p&gt;I have yet another Rebus puzzle for you all to solve.....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This time I've tried making it neither too hard nor too easy :P&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here it is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZRZ7p.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9346" LastActivityDate="2015-02-20T10:50:43.837" Title="Rebus: Sun Archaic Ancient Antique" Tags="|rebus|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9307" PostTypeId="2" ParentId="9306" CreationDate="2015-02-20T10:50:43.837" Score="13" Body="&lt;p&gt;That one is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Old stuff + Old stuff + Old stuff under SUN&quot; = &lt;strong&gt;&quot;There is nothing new under the sun&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-02-20T10:50:43.837" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9308" PostTypeId="2" ParentId="9282" CreationDate="2015-02-20T11:31:07.347" Score="6" Body="&lt;p&gt;How about:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Snaaaaaa!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which is a statement in my made up language...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He can ponder all day if it is false or true and will never know ^^&lt;/p&gt;&#xA;" OwnerUserId="1998" LastActivityDate="2015-02-20T11:31:07.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9309" PostTypeId="2" ParentId="9286" CreationDate="2015-02-20T11:39:24.653" Score="0" Body="&lt;p&gt;Could it be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A couple of maracas. Filled with bones and shaken to produce sounds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastActivityDate="2015-02-20T11:39:24.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9310" PostTypeId="2" ParentId="9305" CreationDate="2015-02-20T12:23:59.570" Score="6" Body="&lt;p&gt;I hope this isnt optimal but it is the best Ive found.  Offer 6 coins every round until the 9 handfuls condition is met.  Mr O will earn 54 if he accepts all hands (as i think he should) and Mr S will earn 46.&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2015-02-20T12:23:59.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9311" PostTypeId="2" ParentId="9305" CreationDate="2015-02-20T12:24:22.290" Score="15" Body="&lt;p&gt;Ok, I'll have a stab.&#xA;The highest number Stanlio can guarantee is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 46 coins&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Optimal strategy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An optimal strategy for Ollio is to claim a handful that has 6 or more coins, and to pass on handfuls of 5 or fewer coins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stanlio knows this, and so knows that producing a handful with less than 5 coins will result in a bigger pot for Ollio left after 9 rounds. Producing a handful with more than 6 coins just reduces the pot left for himself after 9 rounds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which means ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If Stanlio produces 5 coins at a time, Ollio will let him keep them. This means after 9 rounds Stanlio will have 45 coins and Ollio can claim the remaining 55 coins. If Stanlio produces 6 coins at a time then Ollio will claim them. Meaning that after 9 rounds Ollio will have 54 coins and Stanlio can keep the remaining 46.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9168" LastEditorUserId="8874" LastEditDate="2015-10-28T14:45:11.647" LastActivityDate="2015-10-28T14:45:11.647" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9312" PostTypeId="2" ParentId="9288" CreationDate="2015-02-20T13:10:18.950" Score="1" Body="&lt;p&gt;Assumption: Each birth has a 50:50 chance of being boy:girl.&#xA;Statement: A woman cannot have an infinite number of children.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider both laws at once, by saying that you stop if you have an &quot;A&quot; child, but continue if you have a &quot;B&quot; child.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a woman can have a maximum of two children in her lifetime, then we have a:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;50% chance of 1st child being A&lt;/li&gt;&#xA;&lt;li&gt;25% chance of 1st child being B and 2nd child being A&lt;/li&gt;&#xA;&lt;li&gt;25% chance of 1st child being B and 2nd child being B&lt;/li&gt;&#xA;&lt;li&gt;This sums to an average per woman of 0.75 As and 0.75 Bs&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If a woman can have a maximum of three children, then we have a:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;50% chance of 1st child being A&lt;/li&gt;&#xA;&lt;li&gt;25% chance of 1st child being B and 2nd child being A&lt;/li&gt;&#xA;&lt;li&gt;12.5% chance of 1st child being B, 2nd child being B, and 3rd child being A&lt;/li&gt;&#xA;&lt;li&gt;12.5% chance of 1st child being B, 2nd child being B, and 3rd child being B&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This sums to an average per woman of 0.875 As and 0.875 Bs&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can extend the maximum number of children per woman as much as you want (to whatever you think the biological limit is), but you'll end up with the same average number of As and Bs.&lt;/p&gt;&#xA;" OwnerUserId="9444" LastActivityDate="2015-02-20T13:10:18.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9313" PostTypeId="2" ParentId="9286" CreationDate="2015-02-20T13:22:10.717" Score="3" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a train?  Train tracks could be considered bone-like and trains do tend to shake&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9676" LastActivityDate="2015-02-20T13:22:10.717" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9314" PostTypeId="1" AcceptedAnswerId="9346" CreationDate="2015-02-20T14:36:11.457" Score="1" ViewCount="383" Body="&lt;p&gt;Here is my first riddle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I help people speak, but do not speak myself, nor can I be read like a book.&lt;br&gt;&#xA;     4 letters on my tag.&lt;br&gt;&#xA;     I provide direction to an object.&lt;br&gt;&#xA;     Used by golfers and designed for them.&lt;br&gt;&#xA;     What am I?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once again, this is my first riddle, so comment if this needs to be more descriptive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Used by golfers and designed for them&quot; does not refer to the sport of golf. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Look for golf on the stack exchange websites. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Please include if you used the hints in your answers. &lt;/p&gt;&#xA;" OwnerUserId="9631" LastEditorDisplayName="user20" LastEditDate="2015-02-20T22:16:29.037" LastActivityDate="2015-02-20T22:25:07.923" Title="The unreadable direction" Tags="|riddle|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9315" PostTypeId="1" AcceptedAnswerId="9318" CreationDate="2015-02-20T14:36:26.143" Score="13" ViewCount="2567" Body="&lt;p&gt;500 are at my end,&lt;br&gt;&#xA;500 are at my start,&lt;br&gt;&#xA;but at my heart there are only 5.&lt;br&gt;&#xA;The first letter and the first number make me complete:&lt;br&gt;&#xA;Some consider me a king,&lt;br&gt;&#xA;others consider me a real star.&lt;/p&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2015-02-20T14:54:44.410" Title="A real star riddle" Tags="|riddle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9316" PostTypeId="1" AcceptedAnswerId="9317" CreationDate="2015-02-20T14:39:35.200" Score="19" ViewCount="4468" Body="&lt;p&gt;Solve the following rebus puzzle:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  NaCl * H2O&#xA;  H2O * NaCl&#xA; ------------&#xA;   CCCCCCC   &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="9346" LastEditDate="2015-02-22T17:18:32.487" LastActivityDate="2020-08-06T04:04:48.743" Title="NaCl - Rebus puzzle" Tags="|rebus|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9317" PostTypeId="2" ParentId="9316" CreationDate="2015-02-20T14:45:34.920" Score="25" Body="&lt;p&gt;&lt;strong&gt;Answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Sailing (Saline), Sailing over the seven seas (Cs)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9346" LastActivityDate="2015-02-20T14:45:34.920" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9318" PostTypeId="2" ParentId="9315" CreationDate="2015-02-20T14:49:03.937" Score="29" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; David&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Logic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;500 are at my end,&#xA;500 are at my start,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; D is 500 in roman numerals (start and end)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but at my heart there are only 5.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; V is 5 in roman numerals and at the heart of the word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first letter and the first number make me complete:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First letter (A) and first number (i is 1 in roman numerals) finish off the name.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some consider me a king,&lt;br&gt;&#xA;others consider me a real star.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; David is a king in the bible, and the star of David is a symbol of Judaism.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastEditorUserId="8874" LastEditDate="2015-02-20T14:54:44.410" LastActivityDate="2015-02-20T14:54:44.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9319" PostTypeId="2" ParentId="9315" CreationDate="2015-02-20T14:49:49.160" Score="4" Body="&lt;p&gt;&lt;strong&gt;I think The Answer is.....&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DAVID. D=500, V=5, and I=1 (the first number). 'A' is the first letter of the alphabet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9346" LastActivityDate="2015-02-20T14:49:49.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9320" PostTypeId="2" ParentId="9314" CreationDate="2015-02-20T14:52:23.363" Score="3" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Flag&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I help people speak, but do not speak myself, nor can I be read like a book. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Semaphore uses flags to communicate&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4 letters on my tag. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Flag is four letters long&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I provide direction to an object. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Flags mark locations and also provide direction of the wind? I'm not totally sure about the &quot;to an object&quot; part of this, unless it's connected to the next line&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Used by golfers and designed for them. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Flags are used in golf to mark the hole. I don't think flags were originally designed for golf use, so the &quot;designed for them&quot; is another part I'm not quite sure about.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2015-02-20T14:52:23.363" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9321" PostTypeId="1" AcceptedAnswerId="9493" CreationDate="2015-02-20T14:55:42.483" Score="1" ViewCount="667" Body="&lt;p&gt;The challenge is to decode this cipher:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;TJ KS JX AP GN DT DG FE ER EU UA HJ&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The key is attached to this question. Try to find it!&lt;/p&gt;&#xA;" OwnerUserId="9631" LastEditorUserId="9631" LastEditDate="2015-02-23T14:38:09.717" LastActivityDate="2015-10-07T17:17:56.550" Title="Cipher with a hidden key" Tags="|cipher|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9322" PostTypeId="2" ParentId="9288" CreationDate="2015-02-20T14:58:12.990" Score="0" Body="&lt;p&gt;Since some couples are more likely to produce offspring of the same gender (could be boy or girl) every couple should be required to have babies until they have had 2 girls. Families that are predisposed to have male offspring will produce much more babies, All of the &quot;random&quot; families will have about 1.5 girls and 1.5 boys.&#xA;This is not one of the proposed laws, but theoretically could work.&lt;/p&gt;&#xA;" OwnerUserId="8746" LastActivityDate="2015-02-20T14:58:12.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9323" PostTypeId="2" ParentId="9036" CreationDate="2015-02-20T15:14:11.830" Score="13" Body="&lt;p&gt;This is not an answer, but rather a counter to Micael Rize's answer.  I don't intend for this to be picking on him, but it is too difficult to explain in a comment, and there are many arguments about his answer, so hopefully this will be persuasive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my example, I am going to use only 16 zebras, but it should be obvious that this is can extend to 100 zebras as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the starting configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/g29vw.jpg&quot; alt=&quot;Initial configuration&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, lets say the lion chases the zebra to the east.  After the first move, that zebra is the closest, so it will move directly away from the lion.  Subsequent moves will bring the lion closer to the zebras adjacent to the zebra being chased, so they will have to move as well, always directly on a path from the origin according to the strategy.  After a while (trillions of years), you will have the following setup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TbHio.jpg&quot; alt=&quot;Chased away&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though the lion looks close to the zebra, since the starting configuration was so immensely large, they are still light years apart and this zebra is in no danger of being eaten.  The only reason the lion caught up is that the lion made many more moves than this zebra because the other zebras were forced to move at one point.  Now, however, the zebra is able to move every time the lion moves, so can stay in front indefinitely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any event, this is where the lion changes tactics.  Instead of chasing down the current zebra (pointless), he heads straight north.  The zebra will continue to move outwards since it is still the closest.  Eventually, you will get to the point where it is no longer the closest zebra, like so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Qk3H0.jpg&quot; alt=&quot;New Target&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the lion turns back to the pack and chases the closest zebra in the circle (Red).  According to the strategy, this zebra can only move outwards from the origin which gets closer to the lion obviously allowing the lion to capture it.  The other option along this path is to move closer to the origin.  While this is away from the lion, it will eventually meet and surpass the other zebra in red on the opposite side of the circle.  Even if we ignore the fact that the other zebras will all move a bit as the lion runs through the centre of the circle, the lion is now chasing two zebras.  According to the strategy, they can only move away from the origin, so they will always stay together.  Thus, the lion is twice as fast as them and will catch one eventually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are probably different strategies that this zebra can take to avoid capture (e.g. taking different paths not directly from the origin), but the current answer is not sufficient to explain how to stay away from the lion.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2015-02-20T15:14:11.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9325" PostTypeId="2" ParentId="88" CreationDate="2015-02-20T15:29:14.807" Score="1" Body="&lt;p&gt;There are lots of places for scramblers, and provide a very good scrambler, because of the fact that the scrambles can reach all 44 quintillion positions. Here are scramblers, along with timers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://cstimer.net&quot; rel=&quot;nofollow&quot;&gt;Here;&lt;/a&gt; &lt;a href=&quot;http://qqtimer.net&quot; rel=&quot;nofollow&quot;&gt;and here.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, for an official scrambler, download TNoodle (No link, because not enough rep.)&lt;/p&gt;&#xA;" OwnerUserId="9682" LastEditorDisplayName="user20" LastEditDate="2015-02-20T15:31:57.280" LastActivityDate="2015-02-20T15:31:57.280" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9326" PostTypeId="2" ParentId="9288" CreationDate="2015-02-20T15:43:53.607" Score="1" Body="&lt;p&gt;Short answer:&#xA;The second rule is a lot more promising I say.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long answer:&#xA;No fancy mathematics from me here, but pure logical reasoning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First a few general thoughts about the rules themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you apply one of those rules you actually set the maximum possible number of children having the specified gender, while the number of children having the opposite gender remains uncontrolled (open to be anything from zero to pretty much infinity in theory)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming we have a realistic setting, you do have a finite number of possible couples.&#xA;Thus you already know the exact number of children of the specified gender that you will get eventually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you could start and assume a 50/50 ratio for &lt;em&gt;the general populace&lt;/em&gt; as most people here do.&#xA;If you do so you would probably end up having one of the many mathematical calculated results, giving you a pretty much equal chance of success or failure for both rules.&#xA;In other word: by mathematical reasoning it would pretty much be a gamble each way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BUT&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since nature is a little bit more complicated I find it better to say you have a 50/50 ratio for &lt;em&gt;the populace in general&lt;/em&gt; and thus have individual ratios for every woman around (or every couple around since the father does affect the outcome too).&#xA;Considering this it would most definitely be much more promising to go with the second rule (stopping family growth once a girl is born) suggested, as it rules out every couple with a high chance of giving birth to girls while not restraining couples that - on a biological level - favor boys.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't get me wrong here: it still is a gamble, since no one can guarantee there are couples giving birth to multiple sons (or any sons at all for that matter). But at least you prevented the possibility of having a finite number of boys vs a theoretically infinite number of girls.&lt;/p&gt;&#xA;" OwnerUserId="9679" LastActivityDate="2015-02-20T15:43:53.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9327" PostTypeId="2" ParentId="9258" CreationDate="2015-02-20T15:49:54.327" Score="3" Body="&lt;p&gt;this should work for the missing word assuming it is a missing word.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;indivisibility&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the counts in the title and the hints are the numbers of lowercase i's in the statement. The statement already has 7 i's so the only word with 6 i's i can think of is that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;(7 + 3) / 2 is still 5 if there is also a need for the word to count towards a math operation. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Could also be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; dots &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;of which there are 13 in the statement with the punctuation and the j in addition to the i's&lt;/p&gt;&#xA;" OwnerUserId="2803" LastEditorUserId="2803" LastEditDate="2015-02-23T16:59:06.013" LastActivityDate="2015-02-23T16:59:06.013" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9328" PostTypeId="2" ParentId="9314" CreationDate="2015-02-20T16:13:26.053" Score="2" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sign&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I help people speak, but do not speak myself, nor can I be read like a book. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Signs help people communicate, but is not a language used commonly in books.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4 letters on my tag.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The word sign contains of four letters&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I provide direction to an object. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Direction signs are often used to direct people to an object or a location&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Used by golfers and designed for them.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every golf hole has a tee sign marking the length and par of the hole. Could also relate to the hand held quiet signs often used in golf tournaments.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9686" LastActivityDate="2015-02-20T16:13:26.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9329" PostTypeId="2" ParentId="236" CreationDate="2015-02-20T16:41:09.093" Score="4" Body="&lt;p&gt;With all this logic and chain of thought, one basic but key part of the puzzle is forgotten.&#xA;The islanders need to know the &lt;strong&gt;color of their eyes&lt;/strong&gt; to leave the island. &#xA;At any point a blue-eyed person can see that there are 99 blue-eyed people and 100 brown-eyed people. And on the 100th day, when 99 blue-eyed people have not left the island, the islander has still not concluded the color of his eyes (maybe blue, brown or &lt;strong&gt;any other color&lt;/strong&gt;). &#xA;But, had he known there was at least one blue-eyed person on the island (as proclaimed by the guru), he could have concluded that his eyes had to be blue on the 100th day.&#xA;When no one leaves on the 100th day as well(since no one can determine the color of &lt;strong&gt;their&lt;/strong&gt; eyes yet), they are left with same information on the 101st day as they had on 1st day, i.e, a blue-eyed person can see 99 blue-eyed people and 100 brown-eyed people. &#xA;Since all islanders are perfect logicians, no islander can come to a conclusion without the guru's proclamation.  &lt;/p&gt;&#xA;" OwnerUserId="9669" LastEditorUserId="9669" LastEditDate="2015-03-12T14:12:38.410" LastActivityDate="2015-03-12T14:12:38.410" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9330" PostTypeId="2" ParentId="9314" CreationDate="2015-02-20T16:55:10.550" Score="0" Body="&lt;p&gt;This feels like a bit of a stretch for the the first clue but what about:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Link&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I help people speak, but do not speak myself, nor can I be read like a book. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This one is kind of a stretch as it only kind of helps people speak by allowing them to reference other content. It can't speak or be read like a book, though.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4 letters on my tag. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Link is 4 letters long AND a hyperlink is created with an HTML tag &amp;lt;a&amp;gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I provide direction to an object. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A hyperlink can be assigned to objects like images and will redirect the user to another page.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Used by golfers and designed for them. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Golf courses are commonly referred to as links although that's &lt;a href=&quot;http://www.pga.com/golf-instruction/instruction-feature/fundamentals/golf-glossary-and-golf-terms#l&quot; rel=&quot;nofollow&quot;&gt;technically&lt;/a&gt; only accurate if it's built on land reclaimed from the ocean.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2015-02-20T19:09:13.397" LastActivityDate="2015-02-20T19:09:13.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9331" PostTypeId="2" ParentId="9258" CreationDate="2015-02-20T16:56:23.543" Score="0" Body="&lt;p&gt;To the wild-guess mobile!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Otto&lt;/strong&gt; (or some other thing that has two &lt;strong&gt;t&lt;/strong&gt;s in it - &lt;strong&gt;Toto&lt;/strong&gt;). Also it could be &lt;strong&gt;tits&lt;/strong&gt; (yeah it's a bad word).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Both hints refer to the number of &lt;strong&gt;t&lt;/strong&gt;s in them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are 11 &lt;strong&gt;t&lt;/strong&gt;s in the sentence and we need a word (possibly a name) that has two &lt;strong&gt;t&lt;/strong&gt;s in order for &lt;strong&gt;in here are thirteen ____&lt;/strong&gt;. That is why &lt;strong&gt;tits&lt;/strong&gt; is also a good solution as it sort of sounds like &lt;strong&gt;t&lt;/strong&gt;s.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastActivityDate="2015-02-20T16:56:23.543" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9332" PostTypeId="1" AcceptedAnswerId="9334" CreationDate="2015-02-20T17:09:04.247" Score="11" ViewCount="933" Body="&lt;p&gt;My first rebus! This rebus is kind of difficult and is based on a weird property of English.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;HOUSE&#xA;  B&#xA;  U&#xA;  R&#xA;  N&#xA;  I&#xA;  N&#xA;  G&#xA;while&#xA;  G&#xA;  N&#xA;  I&#xA;  N&#xA;  R&#xA;  U&#xA;  B&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="5368" LastActivityDate="2015-02-21T07:31:32.207" Title="Rebus Puzzle - House Burning" Tags="|rebus|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9334" PostTypeId="2" ParentId="9332" CreationDate="2015-02-20T17:15:09.210" Score="10" Body="&lt;p&gt;It seems to be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&quot;The house was burning up while it was burning down&quot;&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;  where &quot;burning up&quot; and &quot;burning down&quot; both mean to be destroyed by heat/fire  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-02-20T17:15:09.210" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9336" PostTypeId="1" AcceptedAnswerId="9338" CreationDate="2015-02-20T18:31:14.450" Score="8" ViewCount="2709" Body="&lt;p&gt;Construct a 9-by-9 filled, magic square using the integers from&#xA;0 to 80. The magic square should additionally have the property that when it is divided&#xA;into ninths according to the picture below, each 3-by-3 subsquare is also magic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gZirk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9688" LastActivityDate="2018-04-03T11:01:39.933" Title="9-by-9 filled, magic square" Tags="|magic-square|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9337" PostTypeId="2" ParentId="5960" CreationDate="2015-02-20T18:48:03.497" Score="0" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They're paying for fuel using Uranium in an attempt to have their spaceships reach the speed of light.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation about the fuel:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; AN:92 refers to atomic number 92, which is Uranium.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation about what they need the fuel for:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The speed of light is the &quot;undisputable speed limit&quot; referred to in the pilot's guide. As you approach the speed of light, your kinetic energy ($\frac{1}{2}mv^2$) increases. According to $e=mc^2$ this energy also has mass, so as you approach the speed of light your mass increases (also mentioned in the pilot's guide).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The only part I'm not sure about:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The technical guide mentions that the &quot;system cannot be distorted by mass increase&quot;. I'm not sure if this has any significance other than explaining that the spaceship won't break because of the increased mass.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1955" LastActivityDate="2015-02-20T18:48:03.497" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9338" PostTypeId="2" ParentId="9336" CreationDate="2015-02-20T18:49:54.690" Score="16" Body="&lt;p&gt;Here's the solution I created:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{ccc|ccc|ccc}&#xA;70 &amp;amp; 63 &amp;amp; 68 &amp;amp; 7 &amp;amp; 0 &amp;amp; 5 &amp;amp; 52 &amp;amp; 45 &amp;amp; 50\\&#xA;65 &amp;amp; 67 &amp;amp; 69 &amp;amp; 2 &amp;amp; 4 &amp;amp; 6 &amp;amp; 47 &amp;amp; 49 &amp;amp; 51\\&#xA;66 &amp;amp; 71 &amp;amp; 64 &amp;amp; 3 &amp;amp; 8 &amp;amp; 1 &amp;amp; 48 &amp;amp; 53 &amp;amp; 46\\\hline&#xA;25 &amp;amp; 18 &amp;amp; 23 &amp;amp; 43 &amp;amp; 36 &amp;amp; 41 &amp;amp; 61 &amp;amp; 54 &amp;amp; 59\\&#xA;20 &amp;amp; 22 &amp;amp; 24 &amp;amp; 38 &amp;amp; 40 &amp;amp; 42 &amp;amp; 56 &amp;amp; 58 &amp;amp; 60\\&#xA;21 &amp;amp; 26 &amp;amp; 19 &amp;amp; 39 &amp;amp; 44 &amp;amp; 37 &amp;amp; 57 &amp;amp; 62 &amp;amp; 55\\\hline&#xA;34 &amp;amp; 27 &amp;amp; 32 &amp;amp; 79 &amp;amp; 72 &amp;amp; 77 &amp;amp; 16 &amp;amp; 9 &amp;amp; 14\\&#xA;29 &amp;amp; 31 &amp;amp; 33 &amp;amp; 74 &amp;amp; 76 &amp;amp; 78 &amp;amp; 11 &amp;amp; 13 &amp;amp; 15\\&#xA;30 &amp;amp; 35 &amp;amp; 28 &amp;amp; 75 &amp;amp; 80 &amp;amp; 73 &amp;amp; 12 &amp;amp; 17 &amp;amp; 10&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did I get here?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;First step: I took the solution of a regular 3x3 magic square which I have solved before:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$M_0 = \left[\begin{array}{ccc} 7 &amp;amp; 0 &amp;amp; 5 \\ 2 &amp;amp; 4 &amp;amp; 6 \\ 3 &amp;amp; 8 &amp;amp; 1\end{array}\right]$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second step:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I took the next set of 9 numbers and simply added 9 to all cells of the previous 3x3 magic square I have. For example:&lt;br&gt;&#xA;&#xA; $$M_1 = 1*9 + M_0 = \left[\begin{array}{ccc} 16 &amp;amp; 9 &amp;amp; 14 \\ 11 &amp;amp; 13 &amp;amp; 15 \\ 12 &amp;amp; 17 &amp;amp; 10\end{array}\right]$$&lt;br&gt;&#xA; &#xA; You will notice it is just the previous square; the only change is that I've added $9$ to each cell. This is repeated again:&lt;br&gt;&#xA;&#xA; $$M_2 = 9 + M_1 =  2*9 + M_0 = \left[\begin{array}{ccc} 25 &amp;amp; 18 &amp;amp; 23 \\ 20 &amp;amp; 22 &amp;amp; 24 \\ 21 &amp;amp; 26 &amp;amp; 19\end{array}\right]$$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is then repeated for the six remaining magic squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the third step: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Now, how do we compile all 9 of these 3x3 magic squares we have to make the perfect 9x9 magic square? &#xA;&#xA; To make it easy to understand, we place them like we do in a ordinary 3x3 magic square.&lt;br&gt;&#xA; &#xA; $$\begin{array}{c|c|c} M_7 &amp;amp; M_0 &amp;amp; M_5 \\\hline M_2 &amp;amp; M_4 &amp;amp; M_6 \\\hline M_3 &amp;amp; M_8 &amp;amp; M_1\end{array}$$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9346" LastEditorDisplayName="user20" LastEditDate="2015-02-20T22:08:46.380" LastActivityDate="2015-02-20T22:08:46.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9339" PostTypeId="1" AcceptedAnswerId="9342" CreationDate="2015-02-20T18:55:05.243" Score="2" ViewCount="376" Body="&lt;p&gt;We can say that an $n$-by-$n$ square is regular provided that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Each of the integers from $0$ to $n^2 − 1$ appears in exactly one cell, and each cell contains only one integer (so that the square is filled), and&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If we express the entries in base-$n$ form, each base-$n$ digit occurs exactly once in the units’ position, and exactly once in the $n$’s position.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Example with 3-by-3 regular square:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/d52xF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The square is regular because each of the ternary digits $0$, $1$, and $2$ appears exactly once in the units’ and $3$’s position in each row and each column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone help me construct an example of a 4-by-4 regular square, showing my answer in both decimal and base-$4$ notations.&lt;/p&gt;&#xA;" OwnerUserId="9688" LastEditorDisplayName="user20" LastEditDate="2015-02-20T19:06:38.663" LastActivityDate="2015-02-20T20:41:06.490" Title="Create a 4-by-4 regular square" Tags="|magic-square|number-theory|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9341" PostTypeId="1" AcceptedAnswerId="9343" CreationDate="2015-02-20T20:38:20.993" Score="-2" ViewCount="235" Body="&lt;p&gt;We can say that an $n$-by-$n$ square is regular provided that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Each of the integers from $0$ to $n^2 − 1$ appears in exactly one cell, and each cell contains only one integer (so that the square is filled), and&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If we express the entries in base-$n$ form, each base-$n$ digit occurs exactly once in the units’ position, and exactly once in the $n$’s position.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What is an example of a 4-by-4 filled, magic square which is not regular? The square should use the integers 0 to 15. Show the answer in both decimal and base-4 as well.&lt;/p&gt;&#xA;" OwnerUserId="9688" LastEditorUserId="5373" LastEditDate="2015-03-28T00:01:23.943" LastActivityDate="2015-03-28T00:01:23.943" Title="Not Regular This Time" Tags="|magic-square|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="9342" PostTypeId="2" ParentId="9339" CreationDate="2015-02-20T20:41:06.490" Score="1" Body="&lt;p&gt;I didn't come up with any particular way to generate one, but I did come up with one by working the 4s place first then the unit place, generally swapping things around when I backed myself into a corner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{pmatrix} 00 &amp;amp; 23 &amp;amp; 32 &amp;amp; 11 \\ 21 &amp;amp; 02 &amp;amp; 13 &amp;amp; 30 \\ 12 &amp;amp; 31 &amp;amp; 20 &amp;amp; 03 \\ 33 &amp;amp; 10 &amp;amp; 01 &amp;amp; 22 \end{pmatrix} = \begin{pmatrix} 0 &amp;amp; 11 &amp;amp; 14 &amp;amp; 5 \\ 9 &amp;amp; 2 &amp;amp; 7 &amp;amp; 12 \\ 6 &amp;amp; 13 &amp;amp; 8 &amp;amp; 3 \\ 15 &amp;amp; 4 &amp;amp; 1 &amp;amp; 10 \end{pmatrix}$&lt;/p&gt;&#xA;" OwnerUserId="8891" LastActivityDate="2015-02-20T20:41:06.490" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9343" PostTypeId="2" ParentId="9341" CreationDate="2015-02-20T20:57:24.087" Score="1" Body="&lt;p&gt;Finally!! I've got it! :) :D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GLtH3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9346" LastActivityDate="2015-02-20T20:57:24.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9345" PostTypeId="2" ParentId="8673" CreationDate="2015-02-20T22:00:47.047" Score="4" Body="&lt;p&gt;You can make a polyomino consisting of a 13x14 square with an 11x12 hole in the middle with one missing corner tile. This can only fit 9 times into the 50x50 board. I believe this is the worst possible fit, so I would substitute '&lt;em&gt;nine&lt;/em&gt;' and '&lt;em&gt;ten&lt;/em&gt;' for '&lt;em&gt;one&lt;/em&gt;' and '&lt;em&gt;52&lt;/em&gt;' in the first and second theorems respectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/23XNr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1631" LastActivityDate="2015-02-20T22:00:47.047" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9346" PostTypeId="2" ParentId="9314" CreationDate="2015-02-20T22:25:07.923" Score="3" Body="&lt;p&gt;My guess would be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Code&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I help people speak, but do not speak myself, nor can I be read like a book.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Code is a language, people can speak it or the execution of the code can lead to understandable 'speak'. But it can't be read like a book directly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4 letters on my tag.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Code is 4 letters long&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I provide direction to an object.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A programming object can only perform actions if coded&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Used by golfers and designed for them.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Code-golf&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And yes, I used your two hints!&lt;/p&gt;&#xA;" OwnerUserId="9168" LastActivityDate="2015-02-20T22:25:07.923" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9351" PostTypeId="1" AcceptedAnswerId="9514" CreationDate="2015-02-21T01:18:06.847" Score="10" ViewCount="8970" Body="&lt;p&gt;I am always moving,&lt;br&gt;&#xA;but I never know where I am going.&lt;br&gt;&#xA;I provide delicious flavor to food.&lt;br&gt;&#xA;I can be read on many restaurant menus.&lt;br&gt;&#xA;However, I am toxic.&lt;br&gt;&#xA;I kill thousands of people every year,&lt;br&gt;&#xA;However, it is difficult to survive in the wild without me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who or what am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer will only be accepted if it matches all of the conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE  -  A new line was added (the 4th one).  3rd line was revised to be more literal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE - updated 6th line to specify &quot;people&quot;, to eliminate vagueness.&lt;br&gt;&#xA;UPDATE - revised 7th line to be more literal and less vague, so I am sure someone will guess it now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HINT:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All lines are completely literal now.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9495" LastEditorUserId="9495" LastEditDate="2015-02-24T13:21:03.443" LastActivityDate="2018-08-21T01:10:12.253" Title="Without me, you cannot survive" Tags="|riddle|" AnswerCount="20" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="9353" PostTypeId="2" ParentId="9305" CreationDate="2015-02-21T01:32:24.260" Score="3" Body="&lt;p&gt;&lt;strong&gt;tl;dr&lt;/strong&gt; Stanlio offers as close as possible to an equal share for each handful, Ollio only accepts if he gets a better than equal portion, and they end up with 46 and 54 coins respectively; this is provably the best possible, and follows for a general formula for this kind of game, given below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ntl;wr&lt;/strong&gt; (&quot;not too long, will read&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One correct answer has been stated (several times) before, but the full reasoning isn't there, and the general solution hasn't been stated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's work up from the simplest case.  If each has one handful remaining, if Stanlio picks half or more, Ollio will take the coins; otherwise Stanlio gets them.  This suggests that the optimal strategy for Ollio is to take &quot;more than his fair share&quot; for the number of handfuls left to distribute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's see if this works out inductively.  If Stanlio has &lt;code&gt;m&lt;/code&gt; handfuls remaining, and Ollio has &lt;code&gt;n&lt;/code&gt;, then a fair share for Ollio is &lt;code&gt;1/(n+m)&lt;/code&gt; and Ollio can if induction works count on at least &lt;code&gt;n/(n+m)&lt;/code&gt; of the remaining coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's then give Stanlio an extra handful, and suppose he offers a fraction &lt;code&gt;f&lt;/code&gt;.  Ollio could take the handful, in which case he'll get &lt;code&gt;f + (1-f)*(n-1)/(n+m)&lt;/code&gt; of the remaining coins, or reject it in which case he'll get &lt;code&gt;(1-f)*n/(n+m)&lt;/code&gt;.  Set the two equal to find &lt;code&gt;f = (1-f)/(n+m)&lt;/code&gt; which we can rearrange as follows: &lt;code&gt;1/(1/f-1) = 1/(n+m)&lt;/code&gt;, &lt;code&gt;1/f-1 = n+m&lt;/code&gt;, &lt;code&gt;1/f = n+m+1&lt;/code&gt;, &lt;code&gt;f = 1/(n+m+1)&lt;/code&gt;.  So Ollio will break even if he takes his fair share of &lt;code&gt;1/(n+m+1)&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's give Ollio an extra handful.  We get &lt;code&gt;f + (1-f)*n/(n+m)&lt;/code&gt; or &lt;code&gt;f*(n+1)/(n+m)&lt;/code&gt;, which simplifies exactly the same way to &lt;code&gt;f = 1/(n+m+1)&lt;/code&gt;.  So induction works: Ollio's strategy should be, assuming Stanlio plays perfectly, to take the coins any time he gets at least a fair share.  (To really prove this inductively, we also need to handle the &lt;code&gt;m=1&lt;/code&gt; case separately, but that's easy because then Ollio if he refuses just gets all the rest of the coins and he should do that if Stanlio picks up too few.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two wrinkles: coins are not perfectly divisible, and Stanlio might make a mistake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first wrinkle actually doesn't change Ollio's strategy at all.  The math is exactly the same; it just means that Stanlio can't offer exactly a fair share.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that Stanlio might make a mistake does impact Ollio's strategy on fair shares.  Since it is all the same to Ollio whether he takes or rejects a fair division, but Stanlio has a chance to make a mistake on every game he plays, Ollio should try to draw the game out for as long as possible by distributing fair divisions between himself and Stanlio until they each have one handful remaining.  This gives Stanlio the most opportunities to make a mistake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stanlio, knowing this, must not make a mistake.  Stanlio has one extra dimension of freedom when he cannot offer a fair share: does he offer too much (by a fraction of one coin) and have Ollio take it, or too little (again by a fraction) and take it himself?  Does it even matter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, Stanlio can simplify the game for himself.  If there are &lt;code&gt;h&lt;/code&gt; handfuls left and &lt;code&gt;c&lt;/code&gt; coins, then a fair share is &lt;code&gt;c/h&lt;/code&gt;.  Let &lt;code&gt;q&lt;/code&gt; be &lt;code&gt;floor(c/h)&lt;/code&gt;, i.e. &lt;code&gt;c/h&lt;/code&gt; rounded down.  Let &lt;code&gt;r&lt;/code&gt; be the remainder of &lt;code&gt;c/h&lt;/code&gt;.  Then Stanlio is effectively choosing to offer either 0 or 1 coin (plus &lt;code&gt;q&lt;/code&gt;), where there are &lt;code&gt;r&lt;/code&gt; 1's to offer and &lt;code&gt;h-r&lt;/code&gt; 0's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the analysis is really easy.  Every time Stanlio offers 0, Stanlio gets it.  If it's a 1, Ollio gets it.  If we sort the list of 0's and 1's on offer, Stanlio takes fro the bottom of the list, Ollio from the top.  This continues until one or the other runs out of turns.  So, if &lt;code&gt;r &amp;lt;= h/2&lt;/code&gt; then Ollio gets all &lt;code&gt;r&lt;/code&gt; remaining coins.  If &lt;code&gt;r &amp;gt; h/2&lt;/code&gt; then Ollio gets &lt;code&gt;h/2&lt;/code&gt; and Stanlio gets &lt;code&gt;r - h/2&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now we have an exact formula and algorithm for what will happen:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let &lt;code&gt;c&lt;/code&gt; be the initial number of coins and &lt;code&gt;j&lt;/code&gt; be the number of handfuls each (total number of handfuls = &lt;code&gt;h = 2j&lt;/code&gt;).  Let &lt;code&gt;q = floor(c/h)&lt;/code&gt; and &lt;code&gt;r = c - h*q&lt;/code&gt; be the remainder.  Stanlio will consider that he has &lt;code&gt;r&lt;/code&gt; 1's to offer and &lt;code&gt;h-r&lt;/code&gt; 0's to offer and can pick at random (actually offering &lt;code&gt;q+0&lt;/code&gt; or &lt;code&gt;q+1&lt;/code&gt; each turn).  Ollio will pick only the &lt;code&gt;q+1&lt;/code&gt; offers unless all the 1's have been exhausted.  This will leave Ollio with &lt;code&gt;j*q + min(j,r)&lt;/code&gt; coins and Stanlio with the remainder, &lt;code&gt;j*q + max(0,r-j)&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's see how this works out with &lt;code&gt;c=100, j=9&lt;/code&gt;.  We have &lt;code&gt;q=5&lt;/code&gt; and &lt;code&gt;r=10&lt;/code&gt;.  We predict Ollio will end up with &lt;code&gt;45+9=54&lt;/code&gt; coins and Stanlio with &lt;code&gt;45+1=46&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if you play it out, with Stanlio offering either 5 or 6 coins at random until either Ollio's turns are exhausted or Stanlio has offered 5 eight times (after which he always offers 6), then this is precisely what will happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives two equally valid simple answers:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Stanlio offers 6 coins 9 times in a row, Ollio accepts every time, and has 54 coins.  The remaining 46 are Stanlio's.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Stanlio offers 5 coins 8 times in a row, Ollio rejects every time, and Stanlio has 40 coins.  Now there is a fair division of 6 coins per handful (10 remaining handfuls of 60), and Stanlio offers 6 each time until Ollio takes 9 and Stanlio gets one: 54 for Ollio and 46 for Stanlio.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Incidentally, if Stanlio had been a little more canny, he might have suggested (or insisted upon) 10 handfuls each (all offers could then be fair).  It also suggests that Ollio was quite the mathematician, as 9 handfuls each gives him one of the biggest advantages: only 11 is better until you hit 19.&lt;/p&gt;&#xA;" OwnerUserId="2815" LastEditorUserId="2815" LastEditDate="2015-02-21T08:26:38.270" LastActivityDate="2015-02-21T08:26:38.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9354" PostTypeId="2" ParentId="9305" CreationDate="2015-02-21T02:18:39.947" Score="5" Body="&lt;p&gt;The highest number of coins that Stanlio can guarantee for himself is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;46&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Define $\rm{best}(i,j,k)$ to be the highest number of coins that Stanlio can guarantee for himself with $i$ coins remaining to be given out and $j$ further chances of receiving remaining for Stanlio, and $k$ chances remaining for Ollio. What we want to find is $\rm{best}(100,9,9)$&lt;br/&gt;&#xA; Finding a recurrence relation is straightforward. At this step, suppose that Stanlio offers $l$ coins. Ollio will choose to assign this set to Stanlio or himself based on which gives less payout to Stanlio. Of all possible values of $l$, Stanlio chooses whichever gives him the best payout. Thus we have:&lt;br/&gt;&#xA;  $$ \rm{best}(i,j,k) = max_l\;min[\rm{best}(i-l,j-1,k)+l,\rm{best}(i-l,j,k-1)]$$&#xA; The base cases for the recurrence are $\rm{best}(0,j,k) = \rm{best}(i,0,k) = 0$, $\rm{best}(i,j,0) = i$, $\rm{best}(i,0,0) = -\inf$ for $i$,$j$,$k&amp;gt;0$&lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pseudocode&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;initialise best[][][] with 0&#xA;for i=1 to 100&#xA;  best [i][0][0]=-inf&#xA;  for j=1 to 9&#xA;    best [i][j][0]=i&#xA;    best [i][0][j]=0&#xA;    for k=1 to 9&#xA;      for l=1 to i &#xA;        best [i][j][k]=max(best [i][j][k], min ( best [i-l][j-1][k]+l, best [i-l][j][k-1]))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This gives $\rm{best}(100,9,9)=46$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9471" LastEditorUserId="9471" LastEditDate="2015-02-21T10:38:00.327" LastActivityDate="2015-02-21T10:38:00.327" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9355" PostTypeId="1" AcceptedAnswerId="9423" CreationDate="2015-02-21T02:42:46.440" Score="0" ViewCount="400" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I do it again and again, until I find what I seek.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What is inside of me can always be changed.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You can never predict what is inside of me.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I can do many things and be called whenever you like.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Without me, you cannot see your results.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I am done only once for #2 and #3.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Inside me is a list, from which you an choose.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There are 7 answers, not one answer.  Each answer is related, because they all belong to the same category.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; These are all words related to programming.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9495" LastEditorUserId="9495" LastEditDate="2015-02-21T22:46:43.707" LastActivityDate="2016-09-05T12:23:20.007" Title="I do it again and again, until I find what I seek" Tags="|riddle|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9356" PostTypeId="2" ParentId="9351" CreationDate="2015-02-21T03:38:16.253" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; sugar&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am always moving,&#xA;but I never know where I am going.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sugar moves in plants (sap).  It also circulates in our bodies, and being inanimate it does not know where it is going.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Being delicious, people love to eat me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We love our sweets.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, I am a poison.&#xA;I kill thousands every year,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sugar contributes to being overweight, which causes health issues and can kill.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but without me, you cannot survive.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Food is broken down into simple sugars which we need to nourish cells.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;New answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; antifreeze  Circulates in your car.  It tastes sweet and because of this causes accidental poisoning in humans and pets.  Last part is a stretch... some of us cannot survive without our cars.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9616" LastEditorUserId="9616" LastEditDate="2015-02-21T04:00:06.377" LastActivityDate="2015-02-21T04:00:06.377" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9357" PostTypeId="2" ParentId="9255" CreationDate="2015-02-21T05:22:07.170" Score="2" Body="&lt;p&gt;My answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Steps on my tennis shoes.  Credit to Gamow +1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9495" LastActivityDate="2015-02-21T05:22:07.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9359" PostTypeId="1" CreationDate="2015-02-21T06:03:06.890" Score="3" ViewCount="393" Body="&lt;p&gt;Officer X was entrusted with the duty of investigating a murder. The dead body was found in the living room.&#xA;Preliminary investigation suggested that four of the six suspects were liars (at least one of each of their&#xA;statements is a lie). Upon further interrogation, Officer X succeeded in extracting the following responses:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A: “P and I were in the 2nd bedroom and M was in the bathroom.” &lt;/li&gt;&#xA;&lt;li&gt;O: “I agree with A. M was in the bathroom and P in the second bedroom, but&#xA;A was in the living room.” &lt;/li&gt;&#xA;&lt;li&gt;P: “As far as I remember, O and J were in&#xA;the kitchen and I was in the bathroom.” &lt;/li&gt;&#xA;&lt;li&gt;M: “P was in the second bedroom and C was in the bathroom. I saw A in the 1st bedroom.” &lt;/li&gt;&#xA;&lt;li&gt;C: “P&#xA;and J were in the bathroom and O was in the kitchen.” &lt;/li&gt;&#xA;&lt;li&gt;J: “M was in the kitchen. P and I were in the second bedroom.”&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;From these replies Officer X immediately deduced who the murderer was.&#xA;Who was the murderer and what line of reasoning led him to his conclusion? &lt;/p&gt;&#xA;" OwnerUserId="9699" LastActivityDate="2015-02-21T13:17:35.653" Title="Finding a murderer from statements from suspects" Tags="|logical-deduction|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9360" PostTypeId="1" AcceptedAnswerId="9362" CreationDate="2015-02-21T06:07:49.207" Score="5" ViewCount="1757" Body="&lt;p&gt;This is a question which is related to the &lt;a href=&quot;http://thinkzone.wlonk.com/MathFun/Triangle.htm&quot; rel=&quot;noreferrer&quot;&gt;hardest easy questions&lt;/a&gt;.  Note that the general solution belongs on math.se and is not solved in simple way. This question is a puzzle and you need to prove the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mZqCR.png&quot; alt=&quot;triangle&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9531" LastEditorUserId="5188" LastEditDate="2015-02-21T07:30:50.467" LastActivityDate="2015-02-22T07:41:46.847" Title="Find the angle (hardest easy geometry)" Tags="|geometry|" AnswerCount="1" CommentCount="7" ClosedDate="2015-02-21T17:12:10.240" ContentLicense="CC BY-SA 3.0" />
  <row Id="9361" PostTypeId="2" ParentId="9254" CreationDate="2015-02-21T06:39:02.020" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ans &lt;strong&gt;BUG&lt;/strong&gt; &lt;br&gt;&lt;br&gt;&#xA; Bed And bug in common have bed bugs&lt;br&gt;&lt;br&gt;&#xA; Tree and Bug in common have tree bugs&lt;br&gt;&lt;br&gt;&#xA; Prisoner and bug (Bug – Inmate with psychiatric or mental problems; also known as a head case by inmates.)&lt;a href=&quot;http://www.writeaprisoner.com/prison-slang.aspx#B&quot; rel=&quot;nofollow noreferrer&quot;&gt;prison slang&lt;/a&gt;&lt;br&gt;&lt;br&gt;&#xA; Website and Bug in common have &lt;a href=&quot;http://en.wikipedia.org/wiki/Web_bug&quot; rel=&quot;nofollow noreferrer&quot;&gt;Web Bug&lt;/a&gt;&lt;br&gt;&lt;br&gt;&#xA; Shirt and bug in common have bug shirts, wear to get protection from insects&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9641" LastEditorUserId="12114" LastEditDate="2018-01-22T17:15:51.867" LastActivityDate="2018-01-22T17:15:51.867" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9362" PostTypeId="2" ParentId="9360" CreationDate="2015-02-21T06:48:00.137" Score="5" Body="&lt;p&gt;Although this puzzle is a math problem, it involves enough complex thought and nonconventional process in its solution to count as a puzzle, in my opinion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, it is a challenge to solve this problem without using trigonometry, as simple angle-chasing will get you nowhere. If you try and map out all possible linear equations of variables, you will find that $\angle AKL$ and $\angle BLK$ cannot be solved for. At first glance it appears that you will need to use scale measurements to solve this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is possible, after some thought. A solution that involves no trigonometry follows.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Rotate $\triangle ALK$ 30 degrees around point $K$, so that $A$ is mapped to $B$. Then $L$ is mapped to a point which we will call $M$, and a new triangle $\triangle BMK$ is created.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Note that since $LA = AB$, $MB = AB$ as well. Further note that since $\angle KAL = 15^\circ$, $\angle KBM = 15^\circ$ as well, so that $\angle ABM = \angle ABK - \angle KBM = 75^\circ - 15^\circ = 60^\circ$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Since $AB$ and $BM$ are of equal length and $\angle ABM = 60^\circ$, $\triangle ABM$ must be an equilateral triangle, and $AM$ must be of the same length as $AB$ and $BM$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Now, since $\triangle ABM$ is equilateral, it is also isosceles with $M$ as an apex. Since $\triangle ABK$ is also isosceles in this way, both $K$ and $M$ are perpendicular to $AB$ at its midpoint. So $KM$ is a perpendicular bisector of $AB$, and also an angle bisector of $\angle AKB$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Therefore $\angle BKM = 15^\circ$, and we rotate the triangle back to find that $\angle AKL = 15^\circ$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-02-22T07:41:46.847" LastActivityDate="2015-02-22T07:41:46.847" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="9363" PostTypeId="2" ParentId="9351" CreationDate="2015-02-21T07:14:48.740" Score="15" Body="&lt;p&gt;I think you are:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Bacteria&lt;/strong&gt;  although different types of bacteria match different clues  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am always moving, but I never know where I am going.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; bacteria can move on their own as well as being moved or transmitted by the host  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Being delicious, people love to eat me.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; yogurts and cheeses are examples of bacteria-containing foods that people enjoy  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, I am a poison. I kill thousands every year,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; some bacteria (salmonella, e.coli) cause food poisoning which can result in death  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but without me, you cannot survive.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; some bacteria (gut microorganisms) are a necessary component of our digestive system  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastActivityDate="2015-02-21T07:14:48.740" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="9365" PostTypeId="1" AcceptedAnswerId="9367" CreationDate="2015-02-21T09:32:51.370" Score="0" ViewCount="9493" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rc3r3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, this is incredibly easy, but it's my first rebus.  I thought it was my own idea, but if someone did this before, let me know.&lt;/p&gt;&#xA;" OwnerUserId="9495" LastActivityDate="2015-02-21T09:45:47.230" Title="Rebus Stand and I" Tags="|rebus|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9366" PostTypeId="2" ParentId="9351" CreationDate="2015-02-21T09:36:46.837" Score="3" Body="&lt;p&gt;I'm gonna go with&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Air&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am always moving, but I never know where I am going.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Air moves. It dosen't move to a specific destination.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Being delicious, people love to eat me.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Air carries the smell of (delicious) food and people will certainly get attracted to it. And they eat air (?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, I am a poison. I kill thousands every year,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Air might carry poisonous gases and it kills many people&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but without me, you cannot survive.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Without Air , no living being can survive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7482" LastActivityDate="2015-02-21T09:36:46.837" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9367" PostTypeId="2" ParentId="9365" CreationDate="2015-02-21T09:37:40.830" Score="2" Body="&lt;p&gt;I think the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I don't understand. ( &quot;I&quot; under &quot;stand&quot; that has been cut out)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9346" LastEditorUserId="9346" LastEditDate="2015-02-21T09:45:47.230" LastActivityDate="2015-02-21T09:45:47.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9368" PostTypeId="1" AcceptedAnswerId="9369" CreationDate="2015-02-21T09:38:20.607" Score="-1" ViewCount="683" Body="&lt;p&gt;Another easy rebus...  I'll just to make some more creative ones ..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6zAqH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the image didn't load.  2nd attempt&lt;/p&gt;&#xA;" OwnerUserId="9495" LastEditorUserId="9495" LastEditDate="2015-02-21T09:44:59.133" LastActivityDate="2015-02-22T05:42:35.170" Title="Rebus : 8 and your" Tags="|rebus|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9369" PostTypeId="2" ParentId="9368" CreationDate="2015-02-21T09:44:52.153" Score="5" Body="&lt;p&gt;Is it by any chance..&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Urinate&lt;/strong&gt; (Your in eight)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9346" LastActivityDate="2015-02-21T09:44:52.153" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9370" PostTypeId="1" AcceptedAnswerId="9371" CreationDate="2015-02-21T10:01:23.697" Score="-3" ViewCount="453" Body="&lt;p&gt;Rebus and the Fridge.  Apple Keynote is surprisingly effective to make these.&lt;img src=&quot;https://i.stack.imgur.com/2jTpO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is it?&lt;/p&gt;&#xA;" OwnerUserId="9495" LastActivityDate="2015-02-21T10:12:43.760" Title="Rebus and the Fridge" Tags="|rebus|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9371" PostTypeId="2" ParentId="9370" CreationDate="2015-02-21T10:05:32.773" Score="2" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a six-pack in the fridge.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All these rebus puzzles seem really easy!&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-02-21T10:05:32.773" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9372" PostTypeId="2" ParentId="9370" CreationDate="2015-02-21T10:05:43.593" Score="4" Body="&lt;p&gt;I feel like I'm going to sounds the alarm and lose points... Dam you Mr Fry&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; six pack of alcoholic beverages&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5135" LastActivityDate="2015-02-21T10:05:43.593" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9373" PostTypeId="2" ParentId="9370" CreationDate="2015-02-21T10:06:30.420" Score="0" Body="&lt;p&gt;I guess the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Six packs in the fridge&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or alternatively, A&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.trendhunter.com/trends/tiny-mini-fridge&quot; rel=&quot;nofollow&quot;&gt;Tiny mini fridge or a Six-pack-Sized fridge&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7482" LastEditorUserId="7482" LastEditDate="2015-02-21T10:12:43.760" LastActivityDate="2015-02-21T10:12:43.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9374" PostTypeId="1" AcceptedAnswerId="9377" CreationDate="2015-02-21T11:17:01.287" Score="4" ViewCount="747" Body="&lt;p&gt;This funny looking word is displayed in all its glory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you decipher its hidden and cryptic meaning?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To question the message body is wrong.  What is the secret to meeting quality standards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6xAkc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9495" LastActivityDate="2015-02-21T11:31:08.960" Title="Rebus, funny looking word" Tags="|rebus|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9376" PostTypeId="1" AcceptedAnswerId="9378" CreationDate="2015-02-21T11:26:10.587" Score="-3" ViewCount="246" Body="&lt;p&gt;Which side is it?   Nice blue arrows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ghETm.png&quot; alt=&quot;Which side is it?&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9495" LastActivityDate="2015-02-21T11:40:04.177" Title="Rebus - which side is it?" Tags="|rebus|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9377" PostTypeId="2" ParentId="9374" CreationDate="2015-02-21T11:31:08.960" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Flip-flops. This is what you get when you flip &quot;flops&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1764" LastActivityDate="2015-02-21T11:31:08.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9378" PostTypeId="2" ParentId="9376" CreationDate="2015-02-21T11:39:49.610" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; UP - side - DOWN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the answer is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Upside down&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9352" LastActivityDate="2015-02-21T11:39:49.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9379" PostTypeId="2" ParentId="9376" CreationDate="2015-02-21T11:40:04.177" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Up side down&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First arrow points &lt;strong&gt;up&lt;/strong&gt;, then the word &lt;strong&gt;side&lt;/strong&gt; comes, and the next arrow points &lt;strong&gt;down&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7482" LastActivityDate="2015-02-21T11:40:04.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9380" PostTypeId="2" ParentId="9355" CreationDate="2015-02-21T11:50:42.880" Score="2" Body="&lt;p&gt;Answer :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Search engine&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Google,Baidu,Bing,Yahoo!,AOL,Ask,Lycos&#xA; work until it find a result, we can't predict what will come, it&#xA; always change, do many thing and we can call whenever we need help, we&#xA; won't get results without it, it give list of results from which we can&#xA; select.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9641" LastEditorUserId="24713" LastEditDate="2016-09-05T12:23:20.007" LastActivityDate="2016-09-05T12:23:20.007" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9381" PostTypeId="1" AcceptedAnswerId="9382" CreationDate="2015-02-21T12:03:49.707" Score="-6" ViewCount="1119" Body="&lt;p&gt;Here is another rebus puzzle. What does it signify?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lOPBb.png&quot; alt=&quot;REBUS4&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7482" LastEditorUserId="5188" LastEditDate="2015-02-21T19:58:29.830" LastActivityDate="2015-02-22T05:36:39.927" Title="Mirrored - Rebus puzzle" Tags="|rebus|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9382" PostTypeId="2" ParentId="9381" CreationDate="2015-02-21T12:30:36.970" Score="2" Body="&lt;p&gt;It symbolises&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;`&lt;/strong&gt; (&lt;strong&gt;backtick&lt;/strong&gt; or grave accent)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because the image shows a backwards tick.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-02-21T12:30:36.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9383" PostTypeId="1" AcceptedAnswerId="9384" CreationDate="2015-02-21T12:32:08.507" Score="-4" ViewCount="835" Body="&lt;p&gt;This is yet another rebus puzzle&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gyZ1M.png&quot; alt=&quot;REBUS&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does it mean?&lt;/p&gt;&#xA;" OwnerUserId="7482" LastEditorUserId="5188" LastEditDate="2015-02-21T20:02:17.850" LastActivityDate="2017-09-25T12:50:51.817" Title="Corny - rebus puzzle" Tags="|rebus|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9384" PostTypeId="2" ParentId="9383" CreationDate="2015-02-21T12:32:58.607" Score="9" Body="&lt;p&gt;1 followed by corn - it must be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a unicorn&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-02-21T12:32:58.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9385" PostTypeId="2" ParentId="9359" CreationDate="2015-02-21T13:17:35.653" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; O is the murderer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We can see that for each person, if they are telling the truth, some others are lying. These are A:O,P,M,C,J; O:A,P,M,C,J; P:A,O,M,C,J, M:A,O,P,C; C:A,O,P,M,J; J:A,O,P,C. Since we know that there are only four liars, M and J have to be truthful. This means that P,C,A,M and J were not in the living room, leaving O as the only remaining suspect.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9471" LastActivityDate="2015-02-21T13:17:35.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9386" PostTypeId="1" AcceptedAnswerId="9406" CreationDate="2015-02-21T13:55:17.330" Score="6" ViewCount="1655" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wEi48.png&quot; alt=&quot;REBUS-SIX&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does the image in this rebus puzzle mean?&lt;/p&gt;&#xA;" OwnerUserId="7482" LastEditorUserId="8874" LastEditDate="2015-10-28T14:42:14.753" LastActivityDate="2015-10-28T14:42:14.753" Title="What does the &quot;gift&quot; and &quot;101010&quot; in this rebus puzzle mean?" Tags="|rebus|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9387" PostTypeId="2" ParentId="9386" CreationDate="2015-02-21T14:18:16.483" Score="13" Body="&lt;p&gt;I believe that it means&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The gift of life&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The red box on the left is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a gift box&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the binary number is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 42. Famous literature and at least some popular culture considers 42 to be the &quot;meaning of life&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2377" LastActivityDate="2015-02-21T14:18:16.483" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9389" PostTypeId="2" ParentId="9351" CreationDate="2015-02-21T15:15:17.220" Score="8" Body="&lt;p&gt;I'm going to go with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sperm&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am always moving,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Always moving until it dies&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I never know where I am going.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It just goes in the direction it is shot&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Being delicious, people love to eat me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Especially if owner eats pineappples/citrus fruit&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, I am a poison.&#xA;I kill thousands every year,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; STD infected &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but without me, you cannot survive.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Human life wouldn't exist without it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9704" LastActivityDate="2015-02-21T15:15:17.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9390" PostTypeId="1" AcceptedAnswerId="9401" CreationDate="2015-02-21T15:21:37.280" Score="-3" ViewCount="317" Body="&lt;p&gt;Here is a rather simple rebus puzzle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;62ND S MAKE A single&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="5373" LastEditDate="2015-02-22T17:08:08.100" LastActivityDate="2015-02-23T18:21:40.823" Title="Rebus puzzle (62ND S MAKE A single)" Tags="|rebus|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9391" PostTypeId="2" ParentId="9386" CreationDate="2015-02-21T17:04:03.630" Score="2" Body="&lt;p&gt;I believe it might be meaning (it doesn't really sound that great):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The gift of knowledge and luck&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The red box represents:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A gift and luck. In many Asian countries, they wear red on Chinese New Year because it brings luck&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;101010 stands for:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 42. 101010 is binary for 42. In this &lt;a href=&quot;http://tywkiwdbi.blogspot.com/2010/10/101010-is-42-in-binary.html&quot; rel=&quot;nofollow&quot;&gt;site&lt;/a&gt;, says that: And 42 is, of course, the &quot;Ultimate Answer to the Ultimate Question of Life, The Universe, and Everything.&quot; Only with knowledge, we can answer all questions including the unknown questions of life and death.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I hope this solves your riddle (or is a good alternate answer)!&lt;/p&gt;&#xA;" OwnerUserId="9042" LastActivityDate="2015-02-21T17:04:03.630" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9392" PostTypeId="2" ParentId="8673" CreationDate="2015-02-21T17:20:35.957" Score="11" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $x=4$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's the proof, step by step (&lt;strong&gt;EDIT&lt;/strong&gt;: the result was right, but the specific argument had a hole; I think I've patched that now) :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;If the minimal bounding box of a 49-cell polynomino is $x\times y$, then $x+y\leq 50$.&lt;/strong&gt;&lt;br/&gt;&#xA; Proof: by induction.  A 1-cell polyomino fits in a $1\times 1$ box, so its semiperimeter (i.e., $x+y$) is 2.  Every $(n+1)$-cell polyomino is formed by adding a cell to an $n$-cell polyomino, and that additional cell can increase the $x$ size of the bounding box or increase the $y$ size of the bounding box, but not both, so if every $n$-cell polyomino fits in a box of semiperimeter $n+1$, then every $(n+1)$-cell polyomino fits in a box of semiperimeter $(n+1)+1$.&#xA;&lt;br/&gt;&lt;br/&gt;&#xA; &lt;strong&gt;Four $x\times y$ rectangles with $x+y\leq50$ always fit into a $50\times50$ square.&lt;/strong&gt;&lt;br/&gt;&#xA; Proof: by construction.  Use a 'pinwheel' arrangement where the two $x\times y$ boxes in the NE and SW corners are placed with $x$ horizontal and $y$ vertical, and the two in the other corners are placed with $y$ horizontal and $x$ vertical.&#xA;&lt;br/&gt;&lt;br/&gt;&#xA; &lt;strong&gt;Five copies of a 'Plus' polyomino (with two length-$25$ arms intersecting in the middle) can't be placed in a $50\times50$ square.&lt;/strong&gt;&lt;br/&gt;&#xA; Proof: First, a few subfacts.  For convenience, I'm going to define the $50\times 50$ square as running from $\langle-25,-25\rangle$ to $\langle25,25\rangle$.&#xA; &lt;br/&gt;a) The center of any Plus must be inside the inner square from $\langle-13,-13\rangle$ to $\langle13,13\rangle$ (or else it'll go outside the bounds of the $50\times50$ board).&#xA; &lt;br/&gt;b) The center of one Plus can't be placed anywhere in the bounding box of another (or else the two will overlap on their arms).&#xA; &lt;br/&gt;c) If the center of a Plus is placed anywhere in a $13\times 13$ box, its bounding box covers that entire space.&#xA; &lt;br/&gt;Now consider the four quadrants from $\langle0,0\rangle$ out to $\langle\pm13,\pm13\rangle$.  By b) and c), each of these quadrants can only have one Plus's center placed within it, but between them these quadrants cover the entire 'legal' space (per a) ) for placing Plus polyominoes.  Since we can fit only one per quadrant, we can fit at most four.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="266" LastEditorUserId="266" LastEditDate="2015-02-21T18:31:42.430" LastActivityDate="2015-02-21T18:31:42.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9393" PostTypeId="1" AcceptedAnswerId="9484" CreationDate="2015-02-21T17:30:05.183" Score="4" ViewCount="388" Body="&lt;p&gt;Can you recover the following two dozens of (well-known) movie titles just from their initials?  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    1. FWAAF   &#xA;    2. TGTBATU &#xA;    3. BOTFOJ&#xA;    4. IJATTOD &#xA;    5. WEHTBJ  &#xA;    6. HPATPOA &#xA;    7. OFOTCN  &#xA;    8. HISTK   &#xA;    9. BCATSK    &#xA;   10. TWOO      &#xA;   11. MPATHG  &#xA;   12. TSOTL  &#xA;&#xA;   13. SM   &#xA;   14. LEDP  &#xA;   15. OBWAT   &#xA;   16. TBOTRK  &#xA;   17. LAC     &#xA;   18. CK      &#xA;   19. NCFOM   &#xA;   20. TDCOTB  &#xA;   21. MIGP    &#xA;   22. FMJ     &#xA;   23. CATCF   &#xA;   24. MSGTW   &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;br&gt;&#xA;DSOHILTSWALTB = &quot;Dr. Strangelove or How I Learned to Stop Worrying and Love the Bomb&quot;&lt;/p&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="9125" LastEditDate="2015-02-21T19:37:34.657" LastActivityDate="2015-02-23T19:12:40.093" Title="Dr. Strangelove or How I Learned to Stop Worrying and Love the Bomb" Tags="|knowledge|movies|code|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9394" PostTypeId="2" ParentId="9393" CreationDate="2015-02-21T17:33:57.400" Score="3" Body="&lt;p&gt;Number 4: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; IJATTOD = Indiana Jones and the Temple of Doom  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 6: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HPATPOA = Harry Potter and the Prisoner of Azkaban&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 17: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LAC = L.A. Confidential&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 22: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; FMJ = Full Metal Jacket&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-21T19:43:37.617" LastActivityDate="2015-02-21T19:43:37.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9395" PostTypeId="2" ParentId="9393" CreationDate="2015-02-21T17:37:29.047" Score="6" Body="&lt;h1&gt;[Community wiki]: The results of a collaboration by The Dragonista, Haobin, Marmy1954 and Gamow:&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Number 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;FWAAF = Four Weddings and a Funeral&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;TGTBATU = The Good, the Bad and the Ugly&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;BOTFOJ = Born on the Fourth of July&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;IJATTOD = Indiana Jones and the Temple of Doom&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;WEHTBJ = What Ever Happened to Baby Jane&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 6:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;HPATPOA = Harry Potter and the Prisoner of Azkaban&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 7:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;OFOTCN = One Flew Over the Cuckoo's Nest&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 8:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;HISTK = Honey, I Shrunk the Kids&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 9:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;BCATSK = Butch Cassidy and the Sundance Kid&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 10:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;TWOO = The Wizard of Oz &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 11:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;MPATHG =  Monty Python and the Holy Grail&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 12:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;TSOTL = The Silence of the Lambs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 13:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;SM = Slumdog Millionaire&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 14:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LEDP = Les Enfants du Paradis&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 15:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; OBWAT = O Brother, Where Art Thou?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 16:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;TBOTRK = The Bridge On The River Kwai&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 17:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;LAC = L.A. Confidential&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 18:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CK = Citizen Kane&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 19:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;NCFOM = No Country for Old Men&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 20:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;TDCOTB = The Discreet Charm of the Bourgeoisie&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 21:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;MIGP= Mission: Impossible - Ghost Protocol&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 22:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;FMJ = Full Metal Jacket&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 23:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CATCF = Charlie and the Chocolate Factory&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 24:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MSGTW = Mr. Smith Goes to Washington&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9346" LastEditorUserId="9346" LastEditDate="2015-02-23T19:12:40.093" LastActivityDate="2015-02-23T19:12:40.093" CommentCount="0" CommunityOwnedDate="2015-02-23T19:12:40.093" ContentLicense="CC BY-SA 3.0" />
  <row Id="9396" PostTypeId="1" AcceptedAnswerId="9483" CreationDate="2015-02-21T17:41:41.930" Score="6" ViewCount="447" Body="&lt;p&gt;I guess that everybody here at puzzling.stackexchange is aware of the fact  that Uzbekhistan and Liechtenstein are the only two countries in the world that are doubly landlocked and hence share a very distinguished property.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following list shows parts of some other country pairs that share very distinguished properties .(the properties are natural; no loopholes; no trick solutions).&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Italy        and &lt;strong&gt;S&lt;/strong&gt;????? are the only two countries that have property A.&lt;/li&gt;&#xA;&lt;li&gt;Switzerland  and &lt;strong&gt;V&lt;/strong&gt;????? are the only two countries that have property B.&lt;/li&gt;&#xA;&lt;li&gt;Ethiopia     and &lt;strong&gt;L&lt;/strong&gt;1???? are the only two countries that have property C.&lt;/li&gt;&#xA;&lt;li&gt;Saudi Arabia and &lt;strong&gt;L&lt;/strong&gt;2???? are the only two countries that have property D.&lt;/li&gt;&#xA;&lt;li&gt;Mexico       and &lt;strong&gt;L&lt;/strong&gt;3???? are the only two countries that have property E.&lt;/li&gt;&#xA;&lt;li&gt;Ireland      and &lt;strong&gt;N&lt;/strong&gt;????? are the only two countries that have property F.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Solve for the six countries S?????, V?????, L1????, L2????, L3????, N????? (with the given initial letter) and the six properties $A, B, C, D, E, F$.&lt;/p&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="9125" LastEditDate="2015-02-23T15:52:13.993" LastActivityDate="2015-02-24T16:40:53.697" Title="Distinguished country pairs" Tags="|geography|knowledge|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9397" PostTypeId="1" AcceptedAnswerId="9399" CreationDate="2015-02-21T18:19:36.093" Score="7" ViewCount="1011" Body="&lt;p&gt;Exactly what the title says:&#xA;What is the shortest (fewest digits) number you can find such that somewhere in it, you can find each number 1- 100?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thinking for this puzzle is you have an electric numeric lock that can have any code from 1-100 and you want to try all possibilities in the shortest time. (Assuming you don't have to push enter or whatever).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1-10 can be 1023456789 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1-11 can be 11023456789&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1-12 can be 110123456789&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is there some sort of formula that can extend this to other numbers (1-1000, 1-50, ...)&lt;/p&gt;&#xA;" OwnerUserId="7715" LastEditorUserId="7715" LastEditDate="2015-02-21T22:05:58.147" LastActivityDate="2015-02-21T22:34:45.773" Title="Shortest Number Containing the Numbers 1-100?" Tags="|password|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9398" PostTypeId="2" ParentId="9386" CreationDate="2015-02-21T19:35:23.480" Score="4" Body="&lt;p&gt;If it's not one of the other answers, then it is probably&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gift for 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9352" LastActivityDate="2015-02-21T19:35:23.480" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9399" PostTypeId="2" ParentId="9397" CreationDate="2015-02-21T19:44:54.917" Score="8" Body="&lt;p&gt;If you use a decimal &lt;a href=&quot;http://en.wikipedia.org/wiki/De_Bruijn_sequence&quot;&gt;De Bruijn sequence&lt;/a&gt; of order 2, you get a string of 100 digits that contains every possible substring of length 2 - that is, every possible two-digit number, although you need one last digit to account for the wraparound, bring the total to 101.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose further that the first three digits are &quot;100&quot;, we get 100 without adding any additional digits. This must be possible because any valid de Bruijn sequence can be &quot;rotated&quot; so that the second and third digits are identical, and then the digits remapped so that the first digit is 1 and the next two digits are 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the shortest number that fits your criteria is 101 digits long.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-02-21T19:44:54.917" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9400" PostTypeId="1" AcceptedAnswerId="9714" CreationDate="2015-02-21T19:52:43.630" Score="-9" ViewCount="4037" Body="&lt;p&gt;Three perfect mathematicians with extremely strong memories are taking an exam. &#xA;The examiner tells each of them a certain piece of information about $x$ and $y$, which two positive integers between and including 2 and 100 one is divisible by the other: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The first one is told the difference $d=x-y$ &lt;/li&gt;&#xA;&lt;li&gt;The second one is told the sum $s=x+y$&lt;/li&gt;&#xA;&lt;li&gt;The third one the ratio $r=x/y$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then the following conversation takes place:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;First mathematician: I cannot deduce $x$ and $y$ from their difference $d$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Second mathematician: I knew that you cannot.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;First mathematician: Don't bother to say that, I already knew that you knew that I cannot. &#xA;But I'm not sure if you don't know that I know that you don't !!!!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Third mathematician: What about me...? I did know the two numbers right away, once they told me the ratio $r$!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question: What are these mysterious numbers $x$ and $y$?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Notes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;This is the second puzzle that I created on my own, besides the  &lt;a href=&quot;https://puzzling.stackexchange.com/questions/8920/consecutive-ranks&quot; title=&quot;consecutive-ranks&quot;&gt;&quot;consecutive-ranks&quot;&lt;/a&gt; puzzle.&lt;/li&gt;&#xA;  &lt;li&gt;I could resume the conversation in just two lines but I preferred to ameliorate it to ease the solution and make it more adapted to logic.&lt;/li&gt;&#xA;  &lt;li&gt;The solution is unique.&lt;/li&gt;&#xA;  &lt;li&gt;The uniqueness is twice verified :D&lt;/li&gt;&#xA;  &lt;li&gt;I will show the solution if this post becomes be sufficiently downvoted or if there are enough unsuccessful attempts&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9517" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-03-01T19:10:02.410" Title="What are the numbers?" Tags="|logical-deduction|number-theory|" AnswerCount="6" CommentCount="21" ContentLicense="CC BY-SA 3.0" />
  <row Id="9401" PostTypeId="2" ParentId="9390" CreationDate="2015-02-21T20:05:25.967" Score="4" Body="&lt;p&gt;I think the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 60 seconds make a single minute&lt;br&gt;&#xA;&lt;br&gt;&#xA; 62nd = 60 second, followed by S MAKE A single (in small text)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-02-23T18:21:40.823" LastActivityDate="2015-02-23T18:21:40.823" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9402" PostTypeId="2" ParentId="9351" CreationDate="2015-02-21T20:55:17.467" Score="0" Body="&lt;p&gt;Is 'alcohol' anywhere close to an answer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alcohol is served in almost any restaurant, is a highly volatile compound (it evaporates and turns to gas very easily), is certainly delicious, and many people abuse it to the point of death (as digesting alcohol produces toxins in the body).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only weakness in this guess is that alcohol is not 'vital' for life in that we don't need it like we do nutrition (though life would be much more boring without it!).&lt;/p&gt;&#xA;" OwnerUserId="9707" LastActivityDate="2015-02-21T20:55:17.467" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9403" PostTypeId="1" AcceptedAnswerId="9408" CreationDate="2015-02-21T21:12:29.910" Score="9" ViewCount="1407" Body="&lt;p&gt;See my similar question about T hexomino (&lt;a href=&quot;https://puzzling.stackexchange.com/q/5198/5781&quot;&gt;Polyomino T hexomino and rectangle packing into rectangle&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is exactly same but with other polyomino - Z pentomino.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's pack some (one or more) Z pentominoes together with some (one or more) small $a\times b$ rectangles into some bigger $m\times n$ rectangle without holes and overlapping pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I can pack one Z pentomino and four $1\times1$ rectangles into a $3\times3$ rectangle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IJ28U.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Task&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Find as many as you can different integer pairs $\{a,b\}$ so that one or more rectangles of size $a\times b$ can be packed together with Z pentominoes into a $m\times n$ rectangle without holes and overlapping pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Provide images of solutions if you claim that some combination is possible.&#xA;Please, put your images in spoiler tags so that other users can try find themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If some answers will have same number of pairs $\{a,b\}$, then answer with smaller total area of outer rectangles on example images is preferable. If area also equal, then earlier posted answer is preferable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please, add to your answers total area of outer rectangles. If one tiling implicitly includes several $a\times b$ rectangles, then multiply area by number of combinations it includes.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Notes&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I can guarantee that there exist solutions following integer pairs:&lt;br&gt;&#xA;$1\times1$, $1\times2$, $1\times3$, $1\times4$, $1\times5$, $1\times6$, $1\times8$,&lt;br&gt;&#xA;$2\times2$, $2\times3$, $2\times4$, $2\times5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 week after posting, I'll put up my own answers for unfound combinations (from list above). &lt;/p&gt;&#xA;" OwnerUserId="5781" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2022-05-15T13:07:08.063" Title="Polyomino Z pentomino and rectangle packing into rectangle" Tags="|geometry|solvability|polyomino|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9404" PostTypeId="1" CreationDate="2015-02-21T21:41:29.157" Score="36" ViewCount="3142" Body="&lt;p&gt;Your have a scientific calculator such that most of the keys are unable to be pressed. The only keys that work are those for the functions &#xA;$$&#xA;  x^2 \;\; &#xA;  \sqrt{x} \;\; &#xA;  x!\;\; &#xA;  \exp\;\; &#xA;  \ln\;\; &#xA;  \sin\;\; &#xA;  \cos\;\; &#xA;  \tan\;\; &#xA;  \cot\;\; &#xA;  \sec\;\; &#xA;  \csc\;\; &#xA;  \arcsin\;\; &#xA;  \arccos\;\; &#xA;  \arctan&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a key is pressed it is applied to whatever appears on the screen and the screen changes to the result. The calculator currently displays at $0$. How can we turn the display to read $2015$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My try was that press $\cos$, we will get 1. Then press $\arctan$, we will get $\pi/4$. Press $\sin$, we will get $\sqrt{2}/2$. Press $x^2$, we will get $1/2$. Press $\arcsin$ then $\csc$, we will get 2. We can keep using the $x^2,x!$ and $\exp$ to get a larger number, but I don't know it is possible to get $2015$? I know $2015 = 5 *  13 * 31$. Can someone help me? &lt;/p&gt;&#xA;" OwnerUserId="9714" OwnerDisplayName="jinha0001" LastEditorUserId="4551" LastEditDate="2015-06-22T23:00:19.000" LastActivityDate="2015-06-22T23:00:19.000" Title="Display a number using a scientific calculator with most keys are stuck" Tags="|mathematics|formation-of-numbers|calculation-puzzle|" AnswerCount="4" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="9405" PostTypeId="2" ParentId="9404" CreationDate="2015-02-21T22:03:15.657" Score="10" Body="&lt;p&gt;This answer is from &lt;a href=&quot;http://www.artofproblemsolving.com/Wiki/index.php/1995_USAMO_Problems/Problem_2&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is a weaker version of USAMO 1995 Problem 2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that it has an answer here without having to leave the website. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can break all the keys except for the trigonometric functions and still we should be able to get any square root of a rational number. In particular $2015$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove the statement : If $m$ and $n$ are relatively prime nonnegative integers such that $n&amp;gt;0$, then the some finite sequence of buttons will yield $\sqrt{m/n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove this statement, we apply induction on $m+n$. For our base case, $m+n=1$, we have $n=1$ and $m=0$, and $\sqrt{m/n} = 0$, which is initially shown on the screen. For the inductive step, we consider separately the cases $m=0$, $0&amp;lt;m\le n$, and $n&amp;lt;m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $m=0$, then $n=1$, and we have the base case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $0&amp;lt; m \le n$, then by inductive hypothesis, $\sqrt{(n-m)/m}$ can be obtained in finitely many steps; then so can $\cos \tan^{-1} \sqrt{(n-m)/m} = \sqrt{m/n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n&amp;lt;m$, then by the previous case, $\sqrt{n/m}$ can be obtained in finitely many steps. Since $\cos \tan^{-1} \sqrt{n/m} = \sin \tan^{-1} \sqrt{m/n}$, it follows that $\tan \sin^{-1} \cos \tan^{-1} \sqrt{n/m} = \sqrt{m/n}$ can be obtained in finitely many steps. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the induction is complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof pretty much gives us a ready-to-use method for finding the sequence.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Start with an empty sequence and your number $\sqrt{m/n}$.&lt;/li&gt;&#xA;&lt;li&gt;If $m\leq n$, append $\cos\arctan$ to the sequence, and change your number to $\sqrt{(n-m)/m}$, i.e. set $m' = n - m, n' = m$.&lt;/li&gt;&#xA;&lt;li&gt;Otherwise, if $m&amp;gt;n$, append $\tan\arcsin\cos\arctan$ to the sequence, and change your number to $\sqrt{m/n}$, i.e. set $m' = n, n' = m$.&lt;/li&gt;&#xA;&lt;li&gt;Repeat until your number is zero.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;We have initially $m,n = m,1$, with $m=N^2$.  That means that we will first apply step 3 to obtain $1,m$, then step 2 to obtain $m - 1,1$.  This process will repeat $m=N^2$ times until we have $0,1$.  Our sequence will therefore consist of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\underbrace{\tan\arcsin\cos\arctan\cos\arctan}_{N^2\text{ times}}\,0&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is equivalent to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\tan\underbrace{\arcsin\cos\arctan\cos}_{N^2\text{ times}}\,0&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $N=2015$, that's $16\,240\,901$ button presses.  Have fun!&lt;/p&gt;&#xA;" OwnerUserId="9708" LastEditorDisplayName="user20" LastEditDate="2015-04-08T22:58:06.903" LastActivityDate="2015-04-08T22:58:06.903" CommentCount="1" CommunityOwnedDate="2015-04-08T22:57:47.880" ContentLicense="CC BY-SA 3.0" />
  <row Id="9406" PostTypeId="2" ParentId="9386" CreationDate="2015-02-21T22:10:34.643" Score="15" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; present tense. The gift is a present and 101010 is 3 tens(e).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9709" LastActivityDate="2015-02-21T22:10:34.643" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9407" PostTypeId="2" ParentId="9397" CreationDate="2015-02-21T22:34:45.773" Score="5" Body="&lt;p&gt;Since the problem states that you just want the numbers from 1 to 100 without necessarily having leading zeroes, I thought that maybe you could save a few symbols by omitting strings such as &quot;05&quot; and &quot;08&quot; since all one digit numbers naturally appear as substrings in two-digit numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it turns out that the presence of numbers ending with a zero means that you can't save anything in the length.  Essentially, every digit must be used as part of two numbers to be efficient, and a zero used in the ones place of a number cannot be used again for the tens place of another (if we disregard leading zeroes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does pin the answer at 101 digits.  I wrote a program to generate such numbers, and it seems like the smallest one is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 10011202130314041505160617071808190922324252627282933435363738394454647484955657585966768697787988991&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;while the largest is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 99897969594939291908878685848382818077675747372717066564636261605545352515044342414033231302212011009&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;although I guess the magnitude of the numbers really doesn't matter much in this problem.&lt;/p&gt;&#xA;" OwnerUserId="4819" LastActivityDate="2015-02-21T22:34:45.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9408" PostTypeId="2" ParentId="9403" CreationDate="2015-02-21T22:57:56.480" Score="8" Body="&lt;p&gt;Added 1x8.&lt;br&gt;&#xA;Added 1x5 and 1x6. Replaced 1x3.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1 x 1 (Area = 9), 1 x 2 (Area = 9), 1 x 3 (Area = 42), 1 x 4 (Area = 56)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/ADODs.png&quot; alt=&quot;Pentamino1&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1 x 5 (Area = 165), 1 x 6 (Area = 156)&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/fPb7Y.png&quot; alt=&quot;Pentamino3&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1 x 8 (Area = 432)&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/fXUYn.png&quot; alt=&quot;Pentamino4&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2 x 2 (Area = 36), 2 x 3 (Area = 104)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/90H4x.png&quot; alt=&quot;Pentamino2&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-02-23T03:23:25.700" LastActivityDate="2015-02-23T03:23:25.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9409" PostTypeId="1" AcceptedAnswerId="9427" CreationDate="2015-02-21T23:25:57.573" Score="1" ViewCount="1002" Body="&lt;p&gt;Although I am dead,&lt;br&gt;&#xA;I grow as others die.&lt;br&gt;&#xA;Acids hate me.&lt;br&gt;&#xA;My home is often a tray.&lt;br&gt;&#xA;Sometimes I fall before I can get home.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer must meet all conditions.&lt;/p&gt;&#xA;" OwnerUserId="9495" LastActivityDate="2015-02-22T07:11:34.887" Title="I grow as others die" Tags="|riddle|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9410" PostTypeId="2" ParentId="9404" CreationDate="2015-02-22T01:02:15.400" Score="28" Body="&lt;p&gt;Using these provided operations, let's try to define &lt;em&gt;primitives&lt;/em&gt; that can be used to build up the numbers we want. This is a &lt;em&gt;constructive&lt;/em&gt; answer: the end result is a sequence of keys that should work.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;To get 1 from 0, press $x!$, $e^x$ or $\cos$.&lt;/li&gt;&#xA;&lt;li&gt;To get 1/2 from 1, we can use e.g. your solution ($\arctan$, $\sin$, $x^2$).&lt;/li&gt;&#xA;&lt;li&gt;To get $1/x$, press $\arctan$ then $\cot$, since $\cot(x) = \frac{1}{\tan{x}}$ and $\tan(\arctan(x)) = x$ for all real $x$.&lt;/li&gt;&#xA;&lt;li&gt;To get $x-1$ for $x \ge 1$, use the following sequence: $$ 1/x, \sqrt x, \arcsin, \tan, x^2, 1/x $$&#xA;This takes advantage of the identity $\tan(\arcsin(x)) = \frac{x}{\sqrt{1-x^2}}$ for $|x| \le 1$. In this case, it resolves into $\tan(\arcsin(\sqrt{1/x})) = \frac{\sqrt{\frac{1}{x}}}{\sqrt{1-\frac{1}{x}}} = \sqrt{\frac{\frac{1}{x}}{\frac{x-1}{x}}} = \sqrt{\frac{1}{x-1}} $, which we then square and invert to get $x-1$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now, the way is clear: we derive any number greater than 2015 (e.g. starting from 2, use $x^2, x^2, x^2, x^2$ to get 65536), then repeatedly subtract one to get to 2015. (There are lower easily-obtainable integers, such as 8! = 40320, but this answer is aimed just at finding a possible solution).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here's one possible key sequence: $$ 0, \cos, \arctan, \sin, x^2 ( = 0.5), \\&#xA;\arctan, \cot, x^2, x^2, x^2, x^2 (=65536), \\&#xA;\arctan, \cot, \sqrt{x} (=\sqrt{1/65536}), \\&#xA;\{ \arcsin, \tan \}^{63521} (=\sqrt{1/2015}), \\&#xA;\arctan, \cot, x^2 (=2015).&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that due to numerical inaccuracy inherent to all numerical calculators, the result is probably not going to be exactly 2015. But it's a math puzzle, not a calculator puzzle, so this doesn't really matter :) )&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If we want to be a bit more clever with this, we can use an approach like the one Daniil Agashiyev used in his answer to reduce the number of uses of our primitives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also observe that the inverse of our $x-1$ primitive enables us to get $x+1$. Thus, we can reduce the steps to simply $$ 0, \cos, \arctan, \sin (=1/\sqrt{2}),\\&#xA;x^2, \{\arctan, \sin\}^{3} (=1/\sqrt{7}),\\&#xA;x^2, \{\arcsin, \tan\}^{4} (=1/\sqrt{45}),\\&#xA;x^2, \{\arcsin, \tan\}^{10} (=1/\sqrt{2015}),\\&#xA;x^2, \arctan, \cot (=2015).&#xA;$$&#xA;This is a total of just 43 keypresses - not bad! Here we're still not using $\exp$ or $\ln$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Finally, if our goal is to minimize the amount of button-mashing necessary to enter the number, we can bring in $\exp$ and $\ln$. Combining these with $x^2$ and $\sqrt{x}$ enables us to multiply and divide by 2, as fibonatic observed in his answer. With $x+1$, $x-1$, $2x$ and $x/2$ primitives at our disposal, we can get 2015 with just &lt;strong&gt;20&lt;/strong&gt; keypresses:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 0, \cos, \exp, \{\sqrt{x}\}^3, \ln \,(=1/8),\\&#xA;\arcsin, \tan, x^2\, (=1/63),\\&#xA;\exp, \{\sqrt{x}\}^5, \ln, \,(=1/2016),\\&#xA;\sqrt{x}, \arcsin, \cot, x^2 \, (=2015).&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has very good numerical stability to boot: it doesn't require any values beyond the range of a standard 32-bit float. Using 64-bit floats for computation, the result is 2014.9999999997042, within $3\cdot 10^{-10}$ of 2015.&lt;/p&gt;&#xA;" OwnerUserId="1471" LastEditorUserId="1471" LastEditDate="2015-02-24T01:12:31.067" LastActivityDate="2015-02-24T01:12:31.067" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="9411" PostTypeId="2" ParentId="9351" CreationDate="2015-02-22T01:10:31.553" Score="0" Body="&lt;p&gt;I think you are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Food&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am always moving, but I never know where I am going.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Food is caught and is sent all over the world for export. The food is sent to restaurants to be served. Most of the time, the food is dead so it has no idea where it is. Some that are alive to be used for sushi for example, don't recognize the place and will not know where it is&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Being delicious, people love to eat me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All types of food are eaten for multiple reasons. For those who eat for pleasure, it because they like to eat the food. For example, most Texans love BBQ ribs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I appear on many restaurant menus.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Other than drinks that can only be drank, not eaten, there is food. What else is there on the menu that can be consumed?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, I am toxic.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Improperly prepared food will contain harmful bacteria and toxins that will eventually kill you. Puffer fish, for example, are extremely poisonous. It can only be prepared by trained professionals and experts. People whose hands are infected will make the food toxic, but probably not enough to feel any minor side effects.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I kill thousands every year&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eating too much food and eating toxic and dirty food kill thousands of people around the world. This should include poisoned food infected on purpose by humans or on accident. Good thing this doesn't happen in only one country...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but without me, you cannot survive.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can't survive without any food. You will die of hunger. But first you will go insane and want to eat anything such as meat and I leave it there before someone vomits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9042" LastActivityDate="2015-02-22T01:10:31.553" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9412" PostTypeId="2" ParentId="9351" CreationDate="2015-02-22T01:27:10.780" Score="0" Body="&lt;p&gt;I have no choice but to go with this (probably not what you had in mind) answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Food&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am always moving,&#xA;but I never know where I am going.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Food is always moving, being transported from one place to another, and being an object with no cognizance, it wouldn't know where it was going anyway.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Being delicious, people love to eat me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's food, people love to eat food.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am on many restaurant menus.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Self-explanatory - most restaurants serve food.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, I am toxic.&#xA;I kill thousands every year,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Food can also be toxic. Puffer fish, mushrooms, apples, cherries, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but without me, you cannot survive.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Food is vital to life. You need the nutrients to survive, and your body depends on foods to break down and turn into energy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I satisfied all conditions - do I win the gold star? Haha.&lt;/p&gt;&#xA;" OwnerUserId="4353" LastActivityDate="2015-02-22T01:27:10.780" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9413" PostTypeId="2" ParentId="9351" CreationDate="2015-02-22T01:46:11.947" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Liver&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am always moving,&#xA;but I never know where I am going.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The liver moves with respiration (breathing).  &quot;The liver moves with respiration, and its position varies with the diaphragm and with body type.&quot;   &lt;a href=&quot;http://www.dartmouth.edu/~humananatomy/part_5/chapter_28.html&quot; rel=&quot;nofollow&quot;&gt;http://www.dartmouth.edu/~humananatomy/part_5/chapter_28.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Being delicious, people love to eat me.&#xA;I am on many restaurant menus.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is a normal dish, some love it, although not all.  As a kid I loved it, but then I took anatomy and found out how toxic it is.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, I am toxic.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The liver removes toxins from our body and stores them (like DDT)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I kill thousands every year,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Liver is high in Vitamin A. Certain animals are especially high - polar Bear, walrus.  (Wikepedia - Hypervitaminosis A)  A diet too high in Vitamin A can cause death.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but without me, you cannot survive.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cannot survive without a liver because it cleans the blood stream.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9616" LastActivityDate="2015-02-22T01:46:11.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9414" PostTypeId="2" ParentId="9351" CreationDate="2015-02-22T01:46:18.870" Score="7" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; salt.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am always moving,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Typically dissolved in solution, salt moves.  When in the body, salt (or more precisely, the sodium extracted from the salt during digestion) is primarily carried in the blood, where it's continually circulated by the heart.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I never know where I am going.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Salt does not move of its own accord; it only travels with the solution in which it is suspended.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Being delicious, people love to eat me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Do I really need to explain this one?  :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am on many restaurant menus.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many restaurants' menus advertise salt levels on each dish they serve.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, I am toxic.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Salt is toxic to many pathogens -- as a result, salt has been used as a food preservative for thousands of years.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I kill thousands every year,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; High salt intake is estimated to cause &lt;a href=&quot;http://www.foodnavigator.com/Science/High-salt-intake-causes-2.3-million-deaths-per-year&quot;&gt;more than two million deaths per year.&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but without me, you cannot survive. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Salt#Diet_and_health&quot;&gt;Sodium is required for nerves and muscles to function properly&lt;/a&gt;.  If you don't have enough salt, you suffer from &lt;a href=&quot;http://en.wikipedia.org/wiki/Hyponatremia&quot;&gt;hyponatremia&lt;/a&gt;, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Health_effects_of_salt&quot;&gt;may experience dizziness, muscle cramps, strokes, and/or death&lt;/a&gt;.  &lt;a href=&quot;http://www.philmaffetone.com/hyponatremia&quot;&gt;Three to five million people are diagnosed with hyponatremia every year.&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9286" LastActivityDate="2015-02-22T01:46:18.870" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9415" PostTypeId="2" ParentId="9351" CreationDate="2015-02-22T01:53:50.950" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Blood &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;constantly flows and doesn't &quot;know&quot; its direction. It powers all the sensory organs to provide taste. Any time it's on a menu, it's always &quot;red&quot;. It's toxic to consume raw and kills as&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; leukaemia.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9719" LastEditorUserId="5373" LastEditDate="2015-02-24T10:36:43.407" LastActivityDate="2015-02-24T10:36:43.407" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9417" PostTypeId="2" ParentId="9404" CreationDate="2015-02-22T02:01:47.840" Score="6" Body="&lt;p&gt;We can adapt the idea used in the solution to the 1995 USAMO problem to construct an incrementor.&#xA;For any positive integer $n$, we can go from $\sqrt{n}$ to $\sqrt{n+1}$ by the following sequence of calculator operations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[$tan^{-1}$] [$cos$] [$tan^{-1}$] [$cos$] [$sin^{-1}$] [$tan$] (applying left to right)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, we can construct a decrementor that takes $\sqrt{n}$ to $\sqrt{n-1}$ by inverting and reversing the operations as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[$tan^{-1}$] [$sin$] [$cos^{-1}$] [$tan$] [$cos^{-1}$] [$tan$]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a shorthand, lets call incrementor sequence $[inc]$ and decrementor $[dec]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You already know how to get to 2. From there you can go to $\sqrt{7}$ by:&lt;br&gt;&#xA;$[inc]^3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then hit [$x^2$] to get $\sqrt{49}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now hit  $[dec]^4$ to get $\sqrt{45}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hit [$x^2$] again to get $\sqrt{2025}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $[dec]^{10}$ to get $\sqrt{2015}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally, [$x^2$] to get 2015.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There should probably be a faster way to get to 2015 taking advantage of the incrementor and decrementor, but this is the first I came up with.&lt;/p&gt;&#xA;" OwnerUserId="7489" LastEditorUserId="7489" LastEditDate="2015-02-22T02:41:41.720" LastActivityDate="2015-02-22T02:41:41.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9418" PostTypeId="2" ParentId="9133" CreationDate="2015-02-22T02:17:40.543" Score="4" Body="&lt;p&gt;What is this destination? &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Heart&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where are you traveling from?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mind&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A great wise man once said that&#xA;although it is only a foot or two away,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is roughly a distance between your head and your heart.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;this journey can take you many years.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We try to overcome illusions of our &lt;em&gt;mind&lt;/em&gt; and find who we really are deeply in our &lt;em&gt;heart&lt;/em&gt;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The end of this journey is what you seek, whether you know it or not.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your mind may not &lt;em&gt;know&lt;/em&gt; what you seek, but it's your heart that &lt;em&gt;feels&lt;/em&gt; seeking. So whether we know it or not, we seek to find who we are and what is that we really seek.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9717" LastActivityDate="2015-02-22T02:17:40.543" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9419" PostTypeId="2" ParentId="9400" CreationDate="2015-02-22T02:27:24.217" Score="1" Body="&lt;p&gt;Assuming you mean &quot;between 2 and 100&quot; to be inclusive.&#xA;x=&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 98&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;y=&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A higher ratio limits the (x,y) possibilities.  The max ratio would r=50, where (x,y) = (100,2) but then mathematician 1 could tell answer because d=98 has only one solution. Next highest ratio is r=49, ==&gt; (x,y)= (98,2).  Since d=96 has 2 solutions, (98,2) (100,4), math 1 would not know answer. And sum=100 has multiple solutions so 2nd mathematician would also not know answer. Therefore, third mathematician was told that r=49.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How mathematician1 knows mathematician2 knows mathematician1 doesn't know answer.  Or How &quot;I know that you know that I dont know&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since mathematician#1 would get d=96 ==&gt; (x,y)=(98,2) or (100,4). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;IF former, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That is for (x,y)=(98,2)then mathematician#2 would get s=100.  S=100 has multiple solutions giving integer ratio. EX: (98,2), (96,4), (95,5),(80,20). And each of these solutions have differences that yield multiple solutions, so Mathematician  #2 knows that mathematician#1 does not know, and mathematician #1 can reason that.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;IF latter, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;That is for (x,y)=(100,4), then math#2 gets s=104, which also has at least 2 solutions (100,4),(91,13).  Each of these have differences that have multiple solutions.  (100,4) =&gt; d=96, which has at least 2 solutions. (91,13) =&gt; d=78, (80,2), (81,3).  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To find pairs &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;use x-y=d, where d is known, and x/y= integer =&gt; x=my for some m, so my-y=(m-1)y=d. So I factored the difference and let y be a factor and then added y to d to get x.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9616" LastEditorUserId="9616" LastEditDate="2015-02-24T01:38:03.733" LastActivityDate="2015-02-24T01:38:03.733" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9420" PostTypeId="2" ParentId="9409" CreationDate="2015-02-22T02:29:33.293" Score="0" Body="&lt;p&gt;My answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cookies&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How I came to this conclusion:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although I am dead&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cookies are non-living things&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I grow as others die&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cookies are being baked while other cookies get eaten &quot;die&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Acids hate me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stomach acids &quot;hate&quot; cookies that is why the burn&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My home is often a tray&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tray of cookies&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I fall before I can get home&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fall of or get eaten and normally cookies don't get back to the tray unless they serve it on trays&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9535" LastEditorUserId="9670" LastEditDate="2015-02-22T05:05:41.377" LastActivityDate="2015-02-22T05:05:41.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9421" PostTypeId="2" ParentId="9409" CreationDate="2015-02-22T02:31:41.800" Score="5" Body="&lt;p&gt;You are an&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; inebriated vampire&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Although I am dead,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;undead&quot; would be the more technically correct term here, but vampires are often colloquially referred to as being dead.  And let's face it, &quot;Although I am undead&quot; would have totally given away the whole riddle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I grow as others die,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The vampire gains strength, as it feeds upon fresh victims.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Acids hate me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Holy Water burns vampires like acid, which understandably leaves your &lt;em&gt;actual&lt;/em&gt; working-class acids out of a job.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My home is often a tray&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In between death and rising, a vampire will typically &quot;live&quot; on a tray in a morgue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I fall before I can get home.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hey buddy, are you all right?  Do you need me to call you a cab?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9286" LastActivityDate="2015-02-22T02:31:41.800" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9422" PostTypeId="2" ParentId="9393" CreationDate="2015-02-22T02:45:23.003" Score="4" Body="&lt;p&gt;Number 13&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SM=Silas Marner&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 21&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MIGP= Mission: Impossible - Ghost Protocol&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;that's all I have&lt;/p&gt;&#xA;" OwnerUserId="9616" LastEditorUserId="9616" LastEditDate="2015-02-22T06:35:04.763" LastActivityDate="2015-02-22T06:35:04.763" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9423" PostTypeId="2" ParentId="9355" CreationDate="2015-02-22T02:58:59.773" Score="2" Body="&lt;p&gt;I do it again and again, until I find what I seek.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; dowhile loop&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is inside of me can always be changed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; default value of a field&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can never predict what is inside of me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Uninitialized fields&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can do many things and be called whenever you like.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a subroutine  or utility&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without me, you cannot see your results.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; printline / output&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am done only once for #2 and #3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; initialize the field&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Inside me is a list, from which you an choose.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; dropdown&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9616" LastActivityDate="2015-02-22T02:58:59.773" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9424" PostTypeId="2" ParentId="9355" CreationDate="2015-02-22T03:02:26.667" Score="1" Body="&lt;p&gt;1 &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; for:  A 'for' loop will repeat its contents until its end condition is reached.  For example, &lt;code&gt;for ( int i = 0; i &amp;lt; 10; i++ ){ printf(&quot;Again\n&quot;); }&lt;/code&gt; will print &quot;Again&quot; over and over until the 'i' value reaches 10.  &lt;em&gt;alternate possibilities;  'while', or just the generic 'loop'.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; mutable: A 'mutable' variable is one which can still be changed, even if it's in a context which declares that nothing may be changed.  (taking '&lt;em&gt;always&lt;/em&gt;' to be the important word in &quot;What is inside me always can be changed&quot;.  If that's not the case, then 'variable' is presumably the intended answer)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; volatile:  A 'volatile' variable is one which the compiler is not allowed to make consistency assumptions about.  That is, if we set the variable to be equal to some value, and then immediately check the value, the compiler would normally assume that the variable's value can't have changed in between where we set it and where we tried to read it.  But if we've declared the variable to be volatile, then it can't make that assumption.  This keyword is often used in multi-threaded code where another thread might have touched the variable, or when accessing variables representing hardware registers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; lambda: Alternates:  'function', 'procedure', 'method', 'subroutine', 'helper routine', etc.  Basically, a bunch of executable code that can be called from anywhere else in the codebase.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; output:  Alternates:  'monitor', 'display', maybe 'console'.  Or if I'm being flippant, 'eyes'.  But regardless, generating results isn't enough;  you have to display those outputs somewhere in order to be useful to the user.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; specified: &quot;mutable&quot; and &quot;volatile&quot; are variable modifiers which can only be specified once per variable.  That is, your &lt;code&gt;int&lt;/code&gt; is either &lt;code&gt;volatile&lt;/code&gt; or it's not.  You can't have a &lt;code&gt;volatile volatile int&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;7&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; enumeration: An enumeration binds a list of named integer values to a type, so that type may only (legally) contain one of those specified integer values.  So for example, &lt;code&gt;enum foo{ bar, baz, quux };&lt;/code&gt; defines an enumeration called &quot;foo&quot;, which may hold the values of &quot;bar&quot;, &quot;baz&quot;, or &quot;quux&quot; (which are equal to integer values 0, 1, and 2).  When a piece of code then asks for a &quot;foo&quot;, you may only give one of the values in the enumeration.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9286" LastEditorUserId="9286" LastEditDate="2015-02-22T03:52:20.810" LastActivityDate="2015-02-22T03:52:20.810" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9425" PostTypeId="2" ParentId="9409" CreationDate="2015-02-22T03:48:39.930" Score="1" Body="&lt;p&gt;&lt;strong&gt;Answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bread&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reasoning:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although I am dead,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bread is dead&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I grow as others die.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bread/yeast molds and continues to grow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Acids hate me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bread is a great food to eat to combat acid reflux&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My home is often a tray.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; From the bakery it goes onto a bread tray and loaded into a truck, of which it sits in this bread tray until it sits on the store shelf, and then it goes to its new home from there: a kitchen pantry&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I fall before I can get home.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bread, like a cake, can fall in the oven while being baked. Other notable instances would be falling from the loaf from the counter (my most common reason), or falling from your plate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9670" LastActivityDate="2015-02-22T03:48:39.930" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9427" PostTypeId="2" ParentId="9409" CreationDate="2015-02-22T04:15:08.587" Score="3" Body="&lt;p&gt;I think you are looking for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; cigarette ash&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Although I am dead,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ash is the remains of the burnt portion of a cigarette.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I grow as others die.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As a cigarette burns, there becomes more and more ash.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Acids hate me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I had to look this up.   All ash is alkaline, so ash can neutralize acid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My home is often a tray.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An ashtray.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I fall before I can get home.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The ash can fall off a cigarette.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="9495" LastEditDate="2015-02-22T07:11:34.887" LastActivityDate="2015-02-22T07:11:34.887" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9428" PostTypeId="1" AcceptedAnswerId="9429" CreationDate="2015-02-22T04:44:13.183" Score="7" ViewCount="6197" Body="&lt;p&gt;Sometimes I am born in silence,&lt;br&gt;&#xA;Other times, no.&lt;br&gt;&#xA;I am unseen,&lt;br&gt;&#xA;But I make my presence known.&lt;br&gt;&#xA;In time, I fade without a trace.&lt;br&gt;&#xA;I harm no one,&lt;br&gt;&#xA;but I am unpopular with all.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="9495" LastActivityDate="2015-02-22T04:47:37.597" Title="Sometimes I am born in silence" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9429" PostTypeId="2" ParentId="9428" CreationDate="2015-02-22T04:47:37.597" Score="15" Body="&lt;p&gt;You are a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; fart&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am born in silence,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There can be silent farts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Other times, no.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There can be noisy farts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am unseen,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No one can see a fart.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I make my presence known.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Farts can smell.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In time, I fade without a trace.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The smell fades away over time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I harm no one,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Farts are not harmful.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I am unpopular with all.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No one likes a fart.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2015-02-22T04:47:37.597" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9430" PostTypeId="1" AcceptedAnswerId="9432" CreationDate="2015-02-22T04:51:18.663" Score="1" ViewCount="454" Body="&lt;p&gt;In a 4 round chess tournament with 16 players (where the loser of each two player match is eliminated&#xA;and the winner moves on to the next round), the pairings for all matches are decided randomly. The&#xA;competing players all have different rankings and in any given match the higher ranked player always wins.&#xA;What is the probability that the player with the second highest ranking is the losing finalist?&lt;/p&gt;&#xA;" OwnerUserId="9724" LastEditorUserId="5188" LastEditDate="2015-02-22T05:36:17.617" LastActivityDate="2015-02-22T05:36:17.617" Title="Probability in chess tournament" Tags="|mathematics|probability|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9431" PostTypeId="2" ParentId="9404" CreationDate="2015-02-22T04:56:55.487" Score="5" Body="&lt;p&gt;All the posted answers haven't used $\exp$ and $\ln$, which can also be quite useful, for example the sequence: $[\exp][x^2][\ln]$ multiplies by two, similarly: $[\exp][\sqrt{x}][\ln]$ divides by two. The only downfall of this is that the intermediate results may exceed the largest number that the calculator can represent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These functions can also be used to calculate $-x$, namely: $[\exp][x^{-1}][\ln]$, which can be achieved with for example with: $[\exp][\tan^{-1}][\cot][\ln]$. Given that: $\cos\left(\sin^{-1}\left(x\right)\right)=\sqrt{1-x^2}$, then the following eight long sequence can also be used to calculate $x-1$:&#xA;$$&#xA;[\sqrt{x}][\sin^{-1}][\cos][x^2][\exp][\tan^{-1}][\cot][\ln]&#xA;$$&#xA;The use for this sequence, instead of the already mentioned sequences in previous answers, will become clear in a bit. In order to get to 2015 from 0 you preferably want to use functions which keep the results integers, but increase as fast as possible. So $!$ and $x^2$ would be the best options. Since the given sequences to either achieve $x+1$ or $x-1$ have length 8, it would be best to minimize the need for them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Factorial ($!$) does not come very close to 2015, namely $6!=720$ and $7!=5040$, therefore I did not really looked into using it much further. Since 2015 does not have very useful divisors I looked at the adjacent integers, 2014 and 2016. The second can be achieved with $63\cdot32=2016$, which can be achieved relatively easy by multiplying by two and one subtraction. The resulting sequence I came up with:&#xA;$$&#xA;[\exp]^2[x^2]^3[\ln][x^2][x-1][\exp][x^2]^5[\ln][x-1]&#xA;$$&#xA;Now if you substitute my sequence for $x-1$ in here, then the start and end functions cancel with the adjacent functions: $[x^2][\sqrt{x}]=[\ln][\exp]=x$, thus for the final expanded sequence you need 4 less button presses:&#xA;$$&#xA;[\exp]^2[x^2]^3[\ln][\sin^{-1}][\cos][x^2][\exp][\tan^{-1}][\cot][x^2]^5[\ln][\sqrt{x}][\sin^{-1}][\cos][x^2][\exp][\tan^{-1}][\cot][\ln]&#xA;$$&#xA;Thus only requiring 26 button presses, however $\exp{2016}\approx3.44886\cdot10^{875}$ which is very likely to exceed the largest number that the calculator can represent.&lt;/p&gt;&#xA;" OwnerUserId="8810" LastEditorUserId="8810" LastEditDate="2015-02-22T05:10:13.273" LastActivityDate="2015-02-22T05:10:13.273" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9432" PostTypeId="2" ParentId="9430" CreationDate="2015-02-22T05:05:02.257" Score="4" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 8/15&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reason is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The only way for the rank 2 player to get to the finals is if the rank 1 player is in a different half of the tournament. There are 16*15 ways of arranging these two players, and 16*8 of those put the two in different halves. So the final answer is (16*8)/(16*15) = 8/15.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7392" LastActivityDate="2015-02-22T05:05:02.257" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9433" PostTypeId="2" ParentId="9351" CreationDate="2015-02-22T05:13:41.423" Score="1" Body="&lt;p&gt;is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; appetite? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am always moving,&#xA;but I never know where I am going.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; people lose or gain appetite depending on all kinds of factor&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I provide delicious flavor to food.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the more appetite someone has, the more they will enjoy food&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be read on many restaurant menus.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I guess the word &quot;appetizer&quot; appears on menus&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, I am toxic.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; gluttony is bad.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I kill thousands every year,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; obesity kills&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but without me, you cannot survive.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; hunger is a survival instinct that we need to survive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="9413" LastEditDate="2015-02-24T13:01:02.573" LastActivityDate="2015-02-24T13:01:02.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9434" PostTypeId="2" ParentId="9381" CreationDate="2015-02-22T05:36:39.927" Score="1" Body="&lt;p&gt;Background check. Because it is a checkmark written backwards in a neutral background setting.&lt;/p&gt;&#xA;" OwnerUserId="9727" LastActivityDate="2015-02-22T05:36:39.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9436" PostTypeId="2" ParentId="9351" CreationDate="2015-02-22T05:46:42.123" Score="6" Body="&lt;p&gt;I'm going to guess&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; fire&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am always moving,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If it's not moving, it's not fire, it can be controlled, contained but even a candle's flame moves.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I never know where I am going.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fire moves at the whim of every draft in the room&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I provide delicious flavor to food.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Char/Grill Marks etc, all come from fire&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be read on many restaurant menus.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fire-grilled ______&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, I am toxic.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Depending on whats being burned this can be literal, as fire does produce a large number of toxic hydrocarbons, or it can also refer to the fact that the afformentioned grill marks and such are generally considered a carcinogen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I kill thousands every year,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Maybe even more&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but without me, you cannot survive.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It provides heat and safety to our food.  This is one of the weaker claims I admit, if you're prepared to live entirely off raw vegitables/fruit and in an appropriate climate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9728" LastActivityDate="2015-02-22T05:46:42.123" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9437" PostTypeId="2" ParentId="9400" CreationDate="2015-02-22T06:11:36.560" Score="9" Body="&lt;h2&gt;Note:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The proof of impossibility presented in this answer is no longer valid; its logic was rendered invalid by additional information added later, in the fifth (!) revision to the puzzle definition.  I'm leaving this answer here only for historical context; it is not a valid answer to the (current) revised form of the question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For ease of reference, I state below the old (fourth) version of the puzzle to which my solution refers.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Statement of the fourth version of the puzzle&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Three perfect mathematicians with extremely strong memories are taking an exam. &#xA;The examiner tells each of them a certain piece of information about $x$ and $y$, which two positive integers between 2 and 100: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The first one is told the difference $d=x-y$ &lt;/li&gt;&#xA;&lt;li&gt;The second one is told the sum $s=x+y$&lt;/li&gt;&#xA;&lt;li&gt;The third one the ratio $r=x/y$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then the following conversation takes place:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;First mathematician: I cannot deduce $x$ and $y$ from their difference $d$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Second mathematician: I knew that you cannot.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;First mathematician: Don't bother to say that, I already knew that you knew that I cannot. &#xA;But I'm not sure if you don't know that I know that you don't !!!!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Third mathematician: What about me...? I did know the two numbers right away, once they told me the ratio $r$!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It's self-evident from the last declaration that one of these two numbers divide the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: What are these mysterious numbers $x$ and $y$?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;The solution of the fourth version of the puzzle&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;There is no solution.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First mathematician:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The only pair of $(x,y)$ values which the first mathematician could identify without consulting the others, just based on the $d$ value, would be $(100,2)$, which would produce a $d$ value of $98$ (its maximum possible value).  There is no other combination of $(x,y)$ values which could produce that $d$ value, and there is no other $d$ value which has a unique set of $(x,y$) values which could produce it.  That the first mathematician doesn't know the values tells us that $(x,y)$ are not $(100,2)$, and $d$ is not $98$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second mathematician:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The only values that the second mathematician could identify without consulting the others, just based on the $s$ value, would be $(2,2)$ and $(100,100)$ ($s$ values of $4$ and $200$, respectively).  Note that it's never explicitly stated that the second mathematician doesn't know the values, so $(2,2)$ and $(100,100)$ are still viable answers (so far).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What the second mathematician knows about the first one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But the second mathematician does know that the first mathematician doesn't know the values.  This means that based upon the $s$ value he was given, the second mathematician knows that the $d$ value cannot be $98$.  Which means that the $s$ value received by the second mathematician cannot be $102$ (since that's the only $s$ value which could produce the $d=98$ value which would allow the first mathematician to immediately know $(x,y)$).  Although there are many $(x,y)$ pairs which would produce $s=102$, if $s=102$ then the second mathematician can't be certain that the first mathematician didn't get a $d$ value of $98$, and therefore can't possibly know $(x,y)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What the first mathematician knows about what the second knows about the first:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The first mathematician states that he knows that the second mathematician is certain that the first mathematician doesn't know the answer.  That is, from the $d$ value he received (which we know isn't $98$), the first mathematician must be able to deduce that the $s$ value the second mathematician received &lt;em&gt;could not be $102$&lt;/em&gt;, because if the $s$ value had been $102$, then the second mathematician would reason that the $d$ value might have been $98$, which would mean that the first mathematician might know the $(x,y)$ values.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;SO. (my head hurts)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;From the first mathematician's $d$ value, he somehow knew that $s$ absolutely couldn't be $102$.  So as an example of this, his $d$ value couldn't be $2$.  Because with $d=2$, an $(x,y)=(52,50)$ would give $s=102$, which would not allow the second mathematician to exclude a possible $d$ value of $98$;  With a $d$ value of $2$, the first mathematician can't be certain that the second mathematician didn't receive an $s$ value of $102$, and if the second mathematician could have received an $s$ value of $102$, he couldn't have said definitively that the first mathematician didn't know the answer.  Similarly, a $d$ value of $96$ would admit a result of $(99, 3)$ which would have given $s=102$;  same problem.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The only solution to 1-knows-2-knows-1-doesn't know:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As it turns out, there's only one way to be sure, based only on the $d$ value, that the $s$ value cannot be $102$ (and that therefore the second mathematician must know that the first mathematician absolutely cannot know the answer);  and that's if the $d$ value received by the first mathematician was an odd number.   If $d$ is odd, then just one of $x$ or $y$ must be odd, and that means that $s$ must also be odd (since $odd + even = odd$).  So based on knowing an odd $d$ value, the first mathematician also knows that $s$ is odd, and therefore cannot be $102$, and therefore the second mathematician knows that $d$ cannot be $98$, and therefore also knows that the first mathematician can't deduce the $(x,y)$ values based on the $d$ value.  As a side-note, since we know that the third mathematician knows the ratio $r$, and we've been told that $r$ is an integer value with no remainder, we also know that $x$ must be even, and $y$ must be odd.  (Because if $x$ was odd and $y$ was even, the ratio would always have a fractional part, which we've been told is not the case).  This side-note isn't important to the 1-knows-2-knows-1-doesn't problem, but it's an important point, later.  So I'll just leave it here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now let's look at the third mathematician, who was able to tell the values of $(x,y)$ just from their ratio, $r$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Now, mathematician 3 doesn't know anything we've deduced above.  He claims to have known the $x$ and $y$ values immediately upon hearing the $r$ value, which means that the $r$ value must be between $50$ and $34$, inclusive;  as those $r$ values can only be reached with a single pair of $x$ and $y$ values (and all of them require $y$ to be $2$).  Any $r$ values below $34$ could be arrived at in multiple ways (For example, $33 = 66/2$ or $33 = 99/3$, for example, and with an $r$ value below $34$, mathematician 3 can't determine which set of values are the correct ones without information that he doesn't yet have, from the other mathematicians). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pulling all these elements together to prove that my answer is correct:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;According to statements 1, 2, and the first half of 3 (from the first two mathematicians), we can deduce that the $y$ value &lt;strong&gt;must&lt;/strong&gt; be odd.  According to statement 4 (from the third mathematician), we can deduce that the $y$ value &lt;strong&gt;must&lt;/strong&gt; be $2$.  Since there are no positive integers which are simultaneously $2$ and odd &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Wikipedia:Citation_needed&quot; rel=&quot;noreferrer&quot;&gt;(citation needed)&lt;/a&gt;&lt;/em&gt;, there cannot be a solution to the puzzle, as presented.  If $y$ is not odd, then mathematician 1 cannot make the first half of statement 3.  If $y$ is not $2$, then mathematician 3 cannot make statement 4.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Q.E.D.&lt;/p&gt;&#xA;" OwnerUserId="9286" LastEditorUserId="8874" LastEditDate="2015-02-25T09:05:10.180" LastActivityDate="2015-02-25T09:05:10.180" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="9438" PostTypeId="1" AcceptedAnswerId="9460" CreationDate="2015-02-22T08:45:02.613" Score="14" ViewCount="5002" Body="&lt;p&gt;You've just been hired by Widget &amp;amp; Co. to prototype a new electrical circuit for their line of impenetrable puzzle locks. As part of this circuit, your boss asks you to invert three boolean inputs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Easy&quot;, you say, until you look around the lab for parts. To your dismay you can only find two boolean inverters (NOT) gates. There's a big stash of AND and OR gates, but it seems the interns before you have taken the rest of the NOT gates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So! Design a circuit which uses at most &lt;em&gt;two&lt;/em&gt; NOT gates, and an unlimited number of 2-input AND and OR gates, to successfully invert three inputs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diagrammatically:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A ----                ---- NOT A&#xA;      \              /&#xA;B -----[YOUR CIRCUIT]----- NOT B&#xA;      /              \&#xA;C ----                ---- NOT C&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I guarantee this can be done, and that this doesn't rely on some trick interpretation of the question.&lt;/p&gt;&#xA;" OwnerUserId="1471" LastEditorUserId="18129" LastEditDate="2017-01-28T05:10:42.283" LastActivityDate="2017-01-28T05:10:42.283" Title="Invert three inputs with two NOT gates" Tags="|logical-deduction|circuitry|" AnswerCount="3" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="9439" PostTypeId="1" AcceptedAnswerId="9440" CreationDate="2015-02-22T09:07:00.300" Score="5" ViewCount="153" Body="&lt;p&gt;The ministry of transportation of Lampukistan just made the following announcement: From Jul 1st, the country-wide inter-city bus service will be organized by the ministry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The service will be organized in the following way:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Every city is to be connected to every other city by exactly one line.&lt;/li&gt;&#xA;&lt;li&gt;Every line will serve exactly three cities.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The service will have more than one line. At least how many cities does Lampukistan have?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus points if you give a formula for all possible numbers of cities.&lt;/p&gt;&#xA;" OwnerUserId="2544" LastActivityDate="2015-02-22T09:12:58.227" Title="Inter-city bus services" Tags="|mathematics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9440" PostTypeId="2" ParentId="9439" CreationDate="2015-02-22T09:12:58.227" Score="6" Body="&lt;p&gt;The bus service map is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Fano_plane&quot;&gt;Fano projective plane&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each pair of cities has one line through it, and any two lines also go through exactly one common city. Furthermore, there are three cities on each line, and each city is on three lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In total, there are seven cities and seven lines in the system.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-02-22T09:12:58.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9441" PostTypeId="2" ParentId="9351" CreationDate="2015-02-22T11:01:25.837" Score="6" Body="&lt;p&gt;Perhaps it is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am always moving,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Time is always progressing forward.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I never know where I am going.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The future is unpredictable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I provide delicious flavor to food.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Thyme is used to season food.  Time also ages cheeses and wines, and meats cannot marinate without time.  Time is an important ingredient to all cooking.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be read on many restaurant menus.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many restaurant menus (particularly take-out menus) might list business hours.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, I am toxic.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Time eventually ravages everything.  Things decay and get decrepit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I kill thousands every year,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People die of old age.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but without me, you cannot survive. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Organisms need time to eat and grow and live.  People who are about to die are &quot;out of time&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="388" LastEditorUserId="388" LastEditDate="2018-08-21T01:10:12.253" LastActivityDate="2018-08-21T01:10:12.253" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="9443" PostTypeId="1" CreationDate="2015-02-22T11:35:22.757" Score="5" ViewCount="694" Body="&lt;p&gt;To make payments, the Pythagoreans use coins in no more than three denominations. The three denominations are in whole Oboloi amounts, and the sum of the squares of the two smaller denominations equals the square of the largest denomination. The amount of $1,000$ Oboloi, and any larger whole Oboloi sum, can be obtained using the three distinct types of coins. However, the exact amount of $999$ Oboloi can &lt;em&gt;not&lt;/em&gt; be obtained (and hence requires change). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What coin denominations do the Pythagoreans use?&lt;/p&gt;&#xA;" OwnerUserId="7573" LastEditorUserId="8874" LastEditDate="2016-03-07T12:44:16.977" LastActivityDate="2016-03-07T12:44:16.977" Title="Pythagorean coins" Tags="|mathematics|combinatorics|frobenius|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9444" PostTypeId="2" ParentId="5960" CreationDate="2015-02-22T13:18:21.123" Score="0" Body="&lt;p&gt;I just think it could be a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This answer cannot explain many things, but it explains:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;oxygen &quot;compressors&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;tunnel view&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Einstein and Relativity&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Supreme Time Dilation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I cannot comment as my reputation is too low, but this question was interesting, and I decided to try...&lt;/p&gt;&#xA;" OwnerUserId="9736" LastActivityDate="2015-02-22T13:18:21.123" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9445" PostTypeId="2" ParentId="9403" CreationDate="2015-02-22T13:53:58.627" Score="4" Body="&lt;p&gt;size 1x4, area 36&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;  &lt;a href=&quot;https://i.stack.imgur.com/YmYfC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YmYfC.png&quot; alt=&quot;6x6 solution&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;size 2x4, area 120&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;  &lt;a href=&quot;https://i.stack.imgur.com/1FvtL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1FvtL.png&quot; alt=&quot;10x12 solution&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2022-05-15T13:07:08.063" LastActivityDate="2022-05-15T13:07:08.063" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="9446" PostTypeId="1" AcceptedAnswerId="9447" CreationDate="2015-02-22T14:49:42.563" Score="-6" ViewCount="1481" Body="&lt;p&gt;These two rebus puzzles both have a sun in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Puzzle 1:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/SCPLN.png&quot; alt=&quot;REBUS_SUN1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Puzzle 2:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/eHGyi.png&quot; alt=&quot;REBUS_SUN2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do they mean?&lt;/p&gt;&#xA;" OwnerUserId="7482" LastActivityDate="2015-02-23T10:45:15.553" Title="Two rebus puzzles, each with a sun" Tags="|visual|rebus|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9447" PostTypeId="2" ParentId="9446" CreationDate="2015-02-22T15:01:02.190" Score="5" Body="&lt;p&gt;Puzzle 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The word sun is written with letters increasing in size so it is &lt;strong&gt;Sunrise&lt;/strong&gt; also the size increase is in the right direction or east, so it can be &lt;strong&gt;Sunrise in the east&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Puzzle 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The word Sun and down arrow is written in a vest, vest sounds similar to west, so the answer is &lt;strong&gt;Sundown in the west&lt;/strong&gt; or &lt;strong&gt;Sunset in the west&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9174" LastEditorUserId="7482" LastEditDate="2015-02-23T10:45:15.553" LastActivityDate="2015-02-23T10:45:15.553" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9448" PostTypeId="2" ParentId="9446" CreationDate="2015-02-22T15:01:59.373" Score="4" Body="&lt;p&gt;The answer should be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Sunrise &amp;amp; Sundown&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &quot;SUN spelled in RISing letter size&quot; = &lt;strong&gt;&quot;Sunrise&quot;&lt;/strong&gt;&lt;br&gt;&#xA; &quot;SUN with DOWN arrow&quot; = &lt;strong&gt;&quot;Sundown&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-02-22T15:01:59.373" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9450" PostTypeId="2" ParentId="9443" CreationDate="2015-02-22T16:12:06.043" Score="12" Body="&lt;p&gt;&lt;strong&gt;Answer: The three denominations are $65$, $72$ and $97$.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;How did I detect this answer?&#xA;I searched the list of primitive Pythagorean triples at &lt;a href=&quot;http://www.math.rutgers.edu/~erowland/tripleslist-long.html&quot;&gt;[link to triple list]&lt;/a&gt;, while using the Frobenius applet available at &lt;a href=&quot;http://www.staff.science.uu.nl/~beuke106/frobenius/&quot;&gt;[applet link]&lt;/a&gt;.&#xA;Based on my search I know that the answer to the puzzle is unique; still, I would like to see a clean mathematical argument for this (that is not based on enumeration). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Even proving that all values from $1000$ onwards are representable is very tedious. I checked $65$ underlying individual cases with a computer program.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The proof that the value $999$ is not representable is doable. &#xA;Suppose that there exist non-negative integers $x,y,z$ with $65x+72y+97z=999$. &#xA;By considering the equation modulo 9, we get that $x\equiv z \pmod9$.&#xA;By considering the equation modulo 2, we get that $x+z$ must be odd.&#xA;Since $0\le x\le15$ and $0\le z\le10$, this only leaves a handful of subcases with $x=z+9$ or $z=x+9$; none of these subcases yields an integral value for $y$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I have also found a &lt;a href=&quot;http://arxiv.org/abs/1402.6440&quot;&gt;[scientific paper]&lt;/a&gt; that provides an explicit formula for the Frobenius number of primitive Pythagorean triples:&lt;br&gt;&#xA;$$ F(m^2-n^2, 2mn, m^2+n^2)=(m−1)(m^2−n^2)+(m−1)(2mn)−(m^2+n^2).$$&#xA;By setting $m=9$ and $n=4$, we recover the solution $65$, $72$ and $97$ for the Pythagorean coin puzzle as stated above.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-23T09:53:34.160" LastActivityDate="2015-02-23T09:53:34.160" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9451" PostTypeId="1" CreationDate="2015-02-22T16:40:05.343" Score="7" ViewCount="439" Body="&lt;p&gt;Professor Egghead participates today in a local game show. &#xA;There are six jewels (a diamond, an emerald, a moonstone, a ruby, a sapphire, and a topaz), and there are six boxes that carry the numbers 1, 2, 3, 4, 5, 6.&#xA;Each of the boxes contains exactly one of the jewels.&#xA;The assignment of jewels to boxes is totally random (and of course unknown to the professor), and each assignment is equally likely.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the rules of the game show, Egghead announces three guesses for each of the jewels:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I guess that the diamond   is in one of the three boxes D1, D2, D3.&lt;/li&gt;&#xA;&lt;li&gt;I guess that the emerald   is in one of the three boxes E1, E2, E3.&lt;/li&gt;&#xA;&lt;li&gt;I guess that the moonstone is in one of the three boxes M1, M2, M3.&lt;/li&gt;&#xA;&lt;li&gt;I guess that the ruby      is in one of the three boxes R1, R2, R3.&lt;/li&gt;&#xA;&lt;li&gt;I guess that the sapphire  is in one of the three boxes S1, S2, S3.&lt;/li&gt;&#xA;&lt;li&gt;I guess that the topaz     is in one of the three boxes T1, T2, T3.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;After the guesses have been announced, the game show host opens all six boxes.&#xA;The professor wins a fantastic prize, if each of his six guesses turns out to be correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What strategy should the professor apply?&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;What is his winning probability under an optimal strategy?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-02-27T13:43:11.693" Title="Jewels in the box" Tags="|strategy|probability|game|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9452" PostTypeId="1" AcceptedAnswerId="9456" CreationDate="2015-02-22T16:58:02.710" Score="3" ViewCount="575" Body="&lt;p&gt;What eight letters did the little boy say when, whilst looking for an afternoon  snack, he realized that there was no food in the fridge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ???C????&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9125" LastActivityDate="2015-02-23T06:58:25.443" Title="No food in the fridge" Tags="|riddle|english|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9453" PostTypeId="1" AcceptedAnswerId="9454" CreationDate="2015-02-22T17:05:10.757" Score="9" ViewCount="278" Body="&lt;p&gt;Something very remarkable connects the following eight geographic places.&lt;br&gt;&#xA;What is it?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Bodaybo (Russia)&lt;/li&gt;&#xA;&lt;li&gt;Cape Horn&lt;/li&gt;&#xA;&lt;li&gt;Christchurch (New Zealand)&lt;/li&gt;&#xA;&lt;li&gt;Cocos Islands&lt;/li&gt;&#xA;&lt;li&gt;Ghanzi (Botswana)&lt;/li&gt;&#xA;&lt;li&gt;Honolulu&lt;/li&gt;&#xA;&lt;li&gt;Punta Peclas (Nicaragua)&lt;/li&gt;&#xA;&lt;li&gt;Vigo (Spain)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="8874" LastEditDate="2016-03-14T09:09:53.877" LastActivityDate="2016-03-14T09:39:30.693" Title="Eight remarkable places" Tags="|pattern|knowledge|geography|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9454" PostTypeId="2" ParentId="9453" CreationDate="2015-02-22T17:07:18.943" Score="12" Body="&lt;p&gt;The answer is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; These eight locations (roughly) form the vertices of a cube.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These eight places clearly are very far away from each other.&#xA;They actually form four pairs of antipodal points (see for instance &#xA;&lt;a href=&quot;http://wikimapia.org/15392887/Antipodal-Point-from-Honolulu-Hawaii&quot; rel=&quot;nofollow&quot;&gt;http://wikimapia.org/15392887/Antipodal-Point-from-Honolulu-Hawaii&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Honolulu and Ghanzi are antipodal &lt;/li&gt;&#xA;&lt;li&gt;Punta Peclas and Cocos Islands are antipodal &lt;/li&gt;&#xA;&lt;li&gt;Vigo and Christchurch are antipodal &lt;/li&gt;&#xA;&lt;li&gt;Cape Horn and Bodabyo are antipodal&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These eight points however do &lt;strong&gt;not&lt;/strong&gt; maximize the minimum distance among all sets of eight points on the surface of Earth (which was my first guess).&#xA;The maximizing point set would form the corners of a square antiprism with edges of equal length, and would not contain any pair of antipodal points:&lt;br&gt;&#xA;&lt;a href=&quot;http://mathworld.wolfram.com/SphericalCode.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/SphericalCode.html&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://mathworld.wolfram.com/SquareAntiprism.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/SquareAntiprism.html&lt;/a&gt;   &lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-14T09:39:30.693" LastActivityDate="2016-03-14T09:39:30.693" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9455" PostTypeId="2" ParentId="9451" CreationDate="2015-02-22T17:14:21.000" Score="5" Body="&lt;p&gt;probability of winning is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (3/6)(3/5)(2/4)(2/3)(1/2)(1/1) = 5%&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;another way to calculate the probability is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (3/6)(2/5)(1/4) = 5%&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the strategy is to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; group 3 jewels into 3 same boxes, 3 other jewels into the 3 other boxes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="9413" LastEditDate="2015-02-23T15:29:39.067" LastActivityDate="2015-02-23T15:29:39.067" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9456" PostTypeId="2" ParentId="9452" CreationDate="2015-02-22T18:28:37.513" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; OGICURMT = Oh gee, I see you are empty.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9616" LastEditorUserId="2627" LastEditDate="2015-02-22T19:02:01.783" LastActivityDate="2015-02-22T19:02:01.783" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9457" PostTypeId="2" ParentId="9438" CreationDate="2015-02-22T18:35:18.787" Score="4" Body="&lt;p&gt;Solved over at &lt;a href=&quot;http://electronics.stackexchange.com&quot;&gt;http://electronics.stackexchange.com&lt;/a&gt; in the link provided by Gamow in the comments. This is a direct quote of &lt;a href=&quot;https://electronics.stackexchange.com/users/3123/supercat&quot;&gt;supercat&lt;/a&gt;'s excellent &lt;a href=&quot;https://electronics.stackexchange.com/a/98380&quot;&gt;answer&lt;/a&gt;; go upvote that.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is possible to construct a purely-combinatorial three-input circuit&#xA;  consisting of a number of AND and OR gates along with exactly two&#xA;  independent inverters, with three outputs whose steady-state condition&#xA;  will be the inverse of the inputs. Note that the input to the second&#xA;  inverter will be affected by the output from the first, and that all&#xA;  three outputs are affected by the signals from both inverters.&#xA;  Consequently, one cannot use two of the inputs and outputs of one such&#xA;  circuit to serve as the &quot;inverters&quot; for another.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Assume the inputs are A, B, C and the outputs are X,Y, Z. Using AND&#xA;  and OR gates, determine whether at least two of the inputs are high.&#xA;  Invert that to get a signal which would indicate that at least two are&#xA;  low. Feeding that signal along with the original signals into AND and&#xA;  OR gates, generate a signal which would indicate that an odd number of&#xA;  inputs are high. Invert that to get a signal saying that an odd number&#xA;  are low.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Once one has generated the aforementioned signals, output X should be&#xA;  high if either all three inputs are low (both inverters outputting&#xA;  true), at least two are low and either B or C is true (which would&#xA;  imply that A must be false), or at least one is low and both B and C&#xA;  are true. Outputs Y and Z should be computed similarly.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The principle can be extended to produce a seven I/O combinatorial&#xA;  circuit using three independent inverters and a lot of AND/OR gates.&#xA;  First determine if there are four or more inputs high. Invert that to&#xA;  say there are four or more low. Then determine if the number is 2, 3,&#xA;  6, or 7. Invert that to say there are 0, 1, 4, or 5. Then determine if&#xA;  the number is odd, and invert it to say it's even.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;After having done all that, each output Q should be high if all three&#xA;  inverters output high (all seven inputs low), or if the first two&#xA;  inverters are high and at least one input not associated with Q is&#xA;  high, or the first and last inverters are high and at least two inputs&#xA;  not associated with Q are high, or the first inverter is high and at&#xA;  least three inputs not associated with Q are high, or at the second&#xA;  and third inverters are high and at least four inputs not associated&#xA;  with Q are high, or the second inverter is high and at least five&#xA;  inputs not associated with Q are high, or the first inverter is high&#xA;  and all six inputs not associated with Q are high.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Theoretically, one could design a circuit for 15 I/O's using four&#xA;  inverters, or 31 using five, or 63 using six, etc. but the number of&#xA;  AND and OR gates required would be mind-blowing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1549" LastEditorUserId="-1" LastEditDate="2017-04-13T12:33:11.023" LastActivityDate="2015-02-22T18:35:18.787" CommentCount="0" CommunityOwnedDate="2015-02-22T18:35:18.787" ContentLicense="CC BY-SA 3.0" />
  <row Id="9458" PostTypeId="1" AcceptedAnswerId="9459" CreationDate="2015-02-22T18:50:54.810" Score="3" ViewCount="1691" Body="&lt;p&gt;Here are a few one-liner rebus puzzles for you to solve.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;T_ _ E&lt;/li&gt;&#xA;&lt;li&gt;O _ E R _ T _ O _&lt;/li&gt;&#xA;&lt;li&gt;swear&lt;br&gt;&#xA;bible&lt;br&gt;&#xA;bible&lt;br&gt;&#xA;bible&lt;br&gt;&#xA;bible  &lt;/li&gt;&#xA;&lt;li&gt;hoppin&lt;/li&gt;&#xA;&lt;li&gt;injury + insult&lt;/li&gt;&#xA;&lt;li&gt;COLT jr.&lt;/li&gt;&#xA;&lt;li&gt;cut&lt;br&gt;&#xA;cut cut cut cut&lt;br&gt;&#xA;cut cut cut cut &lt;/li&gt;&#xA;&lt;li&gt;GIFIREN&lt;/li&gt;&#xA;&lt;li&gt;VA DERS&lt;/li&gt;&#xA;&lt;li&gt;ABDEFGHIJKLMNOPQRSTUVWXYZ&lt;/li&gt;&#xA;&lt;li&gt;WINEEEEEE&lt;/li&gt;&#xA;&lt;li&gt;SOMething&lt;/li&gt;&#xA;&lt;li&gt;END&lt;br&gt;&#xA; N&lt;br&gt;&#xA; D&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;April 5&lt;br&gt;&#xA; April 5&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;LIPS     LIPS&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9346" LastEditorUserId="39" LastEditDate="2015-02-23T00:20:00.323" LastActivityDate="2015-02-23T00:20:00.323" Title="Rebus Overload!" Tags="|rebus|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9459" PostTypeId="2" ParentId="9458" CreationDate="2015-02-22T18:54:31.653" Score="12" Body="&lt;h2&gt;[Community wiki]: The results of a collaboration by Len and Gamow:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Number 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; T_ _ E = IM taken out of TIME = &lt;strong&gt;I'm out of time&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; O _ E R _ T _ O _ = OPERATION without PAIN = &lt;strong&gt;painless operation&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Swear on a stack of bibles&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; hoppin = center part of SHOPPING = &lt;strong&gt;Shopping center&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; injury + insult = &lt;strong&gt;To add insult to injury&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 6:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; COLT jr. = &lt;strong&gt;young gun&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 7:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;a cut above the rest&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 8:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; GIFIREN = FIRE in GIN = &lt;strong&gt;fire engine&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 9:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; VA DERS = space in VADERS = &lt;strong&gt;space invaders&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 10:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ABDEFGHIJKLMNOPQRSTUVWXYZ = &lt;strong&gt;Long time no see&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 11:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; WINEEEEEE = WIN with Es = &lt;strong&gt;to win with ease&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 12:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SOMething = &lt;strong&gt;the start of something big&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 13:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;to make ends meet&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 14:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;double date&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 15:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LIPS LIPS = two lips = &lt;strong&gt;tulips&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-22T20:36:56.980" LastActivityDate="2015-02-22T20:36:56.980" CommentCount="0" CommunityOwnedDate="2015-02-22T19:43:44.903" ContentLicense="CC BY-SA 3.0" />
  <row Id="9460" PostTypeId="2" ParentId="9438" CreationDate="2015-02-22T19:26:31.837" Score="13" Body="&lt;p&gt;I built a logic gate structure based on &lt;a href=&quot;https://electronics.stackexchange.com/users/3123/supercat&quot;&gt;supercat&lt;/a&gt;'s &lt;a href=&quot;https://electronics.stackexchange.com/a/98380&quot;&gt;textual explanation&lt;/a&gt; (given in &lt;a href=&quot;https://puzzling.stackexchange.com/users/1549/user2357112&quot;&gt;user2357112&lt;/a&gt;'s answer and &lt;a href=&quot;https://puzzling.stackexchange.com/users/8874/gamow&quot;&gt;Gamow&lt;/a&gt;'s comment); trace it at your leisure. It's a good thing we are given an infinite supply of AND and OR gates. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Added annotations. Expand the image if they are too small.&lt;/em&gt;&lt;br&gt;&#xA; &lt;em&gt;Other edit suggestions are welcomed!&lt;/em&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/V3lAr.png&quot; alt=&quot;Three inputs inverted using two NOT gates&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9319" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-02-23T18:25:09.733" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9461" PostTypeId="1" AcceptedAnswerId="9466" CreationDate="2015-02-22T21:21:10.967" Score="9" ViewCount="911" Body="&lt;h2&gt;&lt;strong&gt;Use the given clues to solve the puzzle below.&lt;/strong&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: an image editing tool is recommended.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BAxlb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tGhW4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9319" LastActivityDate="2015-02-22T23:41:04.623" Title="Color-Coded Bridges" Tags="|logical-deduction|visual|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="9462" PostTypeId="2" ParentId="9461" CreationDate="2015-02-22T22:38:34.157" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;code&gt;This is work in progress.&lt;/code&gt;&lt;br&gt;&#xA;  Following the dedicated invitation of Bulldogg6402, I'm going to chronologically report my thought-process, vowing to not read other solutions (until one becomes accepted at least). The following text is therefore not a well-thought 'summary' but rather a 'storyline' of thoughts where later statements might contradict earlier ones. And there might be plenty of dead-ends.&lt;br&gt;&#xA;  &lt;code&gt;Why?&lt;/code&gt;&lt;br&gt;&#xA;  Because this site is about &lt;em&gt;learning&lt;/em&gt;. Learning both to solve and to build puzzles. By giving my full thought-process, I'm giving the puzzle-author the ability to refine his puzzle or understand how it works 'on others', and I might also give other puzzle-solvers some ideas on how one can tackle a puzzle (or fail at it.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first part of this puzzle is obviously to understand the clues, so I'm trying to make sense of those first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What does positive and negative clue mean?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I assume that by 'negative' clues situation are illustrated which are &lt;em&gt;forbidden&lt;/em&gt;, whereas positive clues show valid/wanted situations. &lt;br&gt;&#xA; This is hardened by the fact that he respective first clue images are the opposite of each other as far as bridges are concerned.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What do the numbers mean?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First there, is the observation that only 1, 2 and 3 exist, so there are two possible &quot;natural&quot; meanings:&lt;br&gt;&#xA; A) - The numbers represent &lt;em&gt;order&lt;/em&gt; of actions: First build bridges from 1-nodes etc.&lt;br&gt;&#xA; B) - The numbers represent &lt;em&gt;connectivity&lt;/em&gt;: A &quot;2&quot; node has 2 bridges etc.&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lets investigate the clues for both options:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For option B)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The last negative clue would fit. The &quot;1&quot; node is forbidden to have 2 bridges. The second negative clue would be invalid, but there is also the matter of colours, so it might still fit. However, the positive clues are at least &lt;em&gt;incomplete&lt;/em&gt; as there are two &quot;2&quot; nodes with only one bridge shown.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For option A)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first positive clue would then be interpreted as the &quot;1&quot; connects to the &quot;2&quot; and then further to the second &quot;2&quot;. This second 'connection' is not straight forward logical, though. The negative clues are also odd. The forth would indicate that one must not connect to two different other nodes starting from a &quot;1&quot; and that it is forbidden to go from &quot;1&quot; to another &quot;1&quot;. (Which is in contrast to the positive &quot;1&quot; connects to &quot;2&quot; connects further to &quot;2&quot; interpretation.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It seems to be, that option A is the more logical conclusion. So lets look on the board with this assumption, because there is an immediate problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some &quot;2&quot; and &quot;3&quot; nodes seem to be too isolated to have the according number of connected bridges, in particular if the first negative clue indicates that 'empty' nodes must not be connected.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What do the colours mean?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; From the first positive clue, it seems to indicate some sort of 'priority'. If there is a black bridge connecting a colour-node, it colorizes bridge and follow-up node. However, there are no clues (neither positive nor negative), which of the 4 non-black colours have priority, if they touch. This seems to be a problem in particular in the central region.  Also, the process of &quot;colourizing&quot; a node seems to indicate towards number option A) not B).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;END OF INITIAL THOUGHTS&lt;/code&gt;&#xA;(Lack of time and going to bed now ;c) )&lt;/p&gt;&#xA;" OwnerUserId="5840" LastActivityDate="2015-02-22T22:38:34.157" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="9463" PostTypeId="2" ParentId="9117" CreationDate="2015-02-22T22:56:43.923" Score="0" Body="&lt;p&gt;I was thinking it might be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;a glass of beer.&quot;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Beer's foam is commonly called a &quot;head.&quot;&lt;br&gt;&#xA;&lt;br&gt;&#xA;Beer breathes and bleeds - it &quot;breathes&quot; because of the bubbles, and &quot;bleeds&quot; when someone takes a sip.&lt;br&gt;&#xA;&lt;br&gt;&#xA;Beer itself is quiet, but in a pub, where there's lots of drinking, there might be singing.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's my take - good riddle!&lt;/p&gt;&#xA;" OwnerUserId="9743" LastEditorUserId="4551" LastEditDate="2015-02-22T23:38:57.663" LastActivityDate="2015-02-22T23:38:57.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9464" PostTypeId="2" ParentId="9218" CreationDate="2015-02-22T23:09:31.853" Score="8" Body="&lt;p&gt;While JonTheMon and Emrakul's answers are entirely correct, here's a different way of explaining why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everyone knows what they can see themselves, and can put themselves in everyone else's shoes to guess what &lt;em&gt;they&lt;/em&gt; see.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes that works:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Scenario: blue  blue  blue  blue  blue  blue &#xA;- 1 concludes everyone else sees 4 or 5 blue eyes - decides it's safe to skip 4&#xA;- 2 concludes everyone else sees 4 or 5 blue eyes - decides it's safe to skip 4&#xA;- 3 concludes everyone else sees 4 or 5 blue eyes - decides it's safe to skip 4&#xA;- 4 concludes everyone else sees 4 or 5 blue eyes - decides it's safe to skip 4&#xA;- 5 concludes everyone else sees 4 or 5 blue eyes - decides it's safe to skip 4&#xA;- 6 concludes everyone else sees 4 or 5 blue eyes - decides it's safe to skip 4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Everyone correctly deduced that days 1 - 4 are pointless and skips ahead to day 5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But sometimes it doesn't:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Scenario: blue  blue  blue  blue  blue  brown &#xA;- 1 concludes everyone else sees 3 or 4 or 5 blue eyes - decides it's safe to skip 3&#xA;- 2 concludes everyone else sees 3 or 4 or 5 blue eyes - decides it's safe to skip 3&#xA;- 3 concludes everyone else sees 3 or 4 or 5 blue eyes - decides it's safe to skip 3&#xA;- 4 concludes everyone else sees 3 or 4 or 5 blue eyes - decides it's safe to skip 3&#xA;- 5 concludes everyone else sees 3 or 4 or 5 blue eyes - decides it's safe to skip 3&#xA;- 6 concludes everyone else sees 4 or 5 blue eyes - decides it's safe to skip 4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Ooops.  Islander #6 decided something different than everyone else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Skipping ahead only works if everyone's on the same page and skips ahead the same number of days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So going back to the original 200-islander problem:  Nobody's under the illusion that there's only 1 blue-eyed person on the island, or 2, or even 97.  They can see with their own eyes that that's not the case, and they know that everyone else knows it too.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But some might think it's safe to skip 97 days, others 98.  Since they can't agree on how many days to skip, they just have to go through the tedium of days 1-97 to be safe -- even though the outcome of those days surprises nobody.&lt;/p&gt;&#xA;" OwnerUserId="9633" LastActivityDate="2015-02-22T23:09:31.853" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="9465" PostTypeId="1" AcceptedAnswerId="9467" CreationDate="2015-02-22T23:13:30.833" Score="0" ViewCount="695" Body="&lt;p&gt;There were three men in a boat. They received an amount of coins in a chest from 200 - 300 as a reward for surviving the storm without damaging anything. One night, the first man thought that he would his share first so he wouldn't have to bring up an argument. He split the money into thirds and took a piece. There was an extra so he threw it out into the sea. Later, the second man did the same exact thing as the first man, taking his share of one-third of the &lt;strong&gt;remaining&lt;/strong&gt; money, not knowing the first man already got his part. There was one more left over, so he threw it into the sea. Finally the third man did the same exact same thing as the other two, taking his share of one-third of that &lt;strong&gt;remaining&lt;/strong&gt; money. The tax collector found some coins left in the chest and divided in three parts giving one part to each sailor, all who are angry with the decision. There was one coin left again so the tax collector took it as payment for his services.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;SO&lt;/strong&gt; how many coins were there and how many coins did each sailor receive?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Hint: The number of coins a sailor got times three will always be one less than the number of coins there in the chest at that time. For example, Sailor Three got ten coins when he got his &quot;share&quot;. So there were 31 coins before he divided the money. This problem comes from one of my favorite books made by Malba.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9042" LastActivityDate="2015-02-23T00:10:26.650" Title="How Many Coins Were There" Tags="|mathematics|word-problem|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9466" PostTypeId="2" ParentId="9461" CreationDate="2015-02-22T23:41:04.623" Score="5" Body="&lt;p&gt;&lt;strong&gt;Solution&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/sBnwx.png&quot; alt=&quot;Solution image&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The positive and negative clues are supposed to mean:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;li&gt; We need to connect points of the same colour with straight lines such that if a point is marked N, N lines should come out of it.&#xA; &lt;li&gt; Black means an unknown colour - it can be connected to a point of any colour, but it will then have to be coloured accordingly. It is thus not allowed to connect points of different colours to a black point.&#xA; &lt;li&gt; Connecting lines should not cross at anywhere except numbered points&#xA; &lt;li&gt; Points of the  same colour should form a connected component&#xA; &lt;br/&gt;&lt;br/&gt;&#xA; This puzzle is just a variant of &lt;a href=&quot;http://en.wikipedia.org/wiki/Hashiwokakero&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Bridges&quot;&gt;Bridges&lt;/a&gt;, which you can play &lt;a href=&quot;http://www.puzzle-bridges.com/&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Play bridges&quot;&gt;here&lt;/a&gt; for example.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9471" LastActivityDate="2015-02-22T23:41:04.623" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9467" PostTypeId="2" ParentId="9465" CreationDate="2015-02-23T00:10:26.650" Score="4" Body="&lt;p&gt;Each sailor received the following:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first received &lt;strong&gt;103 coins.&lt;/strong&gt;&lt;br&gt;&#xA; The second received &lt;strong&gt;76 coins.&lt;/strong&gt;&lt;br&gt;&#xA; The third received &lt;strong&gt;58 coins.&lt;/strong&gt;&lt;br&gt;&#xA;&#xA; There were a grand total of &lt;strong&gt;241 coins in the chest.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It can be solved using algebra, which is the formal method, but also by trial and error, which is how I ended up doing it.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the chest has 241 coins, then the first sailor would take 1/3 and throw the one leftover coin overboard. One-third of 241 is 80r1, so the first sailor gets 80 coins and tosses the one left over. This leaves 160 coins in the chest.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The second sailor sees 160 coins and takes 1/3. One-third of 160 is 53r1, so the second sailor gets 53 coins and tosses the one left over. This leaves 106 coins in the chest.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The third sailor sees 106 coins and takes 1/3. One-third of 106 is 35r1, so the third sailor gets 35 coins and tosses the one left over. This leaves 70 coins in the chest.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Finally, the tax collector sees 70 coins and divides it into three parts to give to each sailor. One-third of 70 is 23r1, so each sailor gets an additional 23 coins added to their collection and the one left over goes to the tax collector.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Each sailor, therefore, has the following:&lt;br&gt;&#xA; The first has 80 + 23 = &lt;strong&gt;103 coins.&lt;/strong&gt;&lt;br&gt;&#xA; The second has 53 + 23 &lt;strong&gt;76 coins.&lt;/strong&gt;&lt;br&gt;&#xA; The third has 35 + 23 &lt;strong&gt;58 coins.&lt;/strong&gt;&lt;br&gt;&#xA; &lt;em&gt;And the tax collector has 1 coin.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9319" LastActivityDate="2015-02-23T00:10:26.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9468" PostTypeId="1" AcceptedAnswerId="9469" CreationDate="2015-02-23T01:10:16.180" Score="7" ViewCount="3138" Body="&lt;p&gt;Bob works at a high security institution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One day Bob tried to log into his computer and his password was denied. Then remembered that, for security reasons, the passwords were reset every month. So he called his boss for his new password. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bob said, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Boss, my old password is out of date.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The boss replied, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Yes, it is. The new password is different, but if you pay attention&#xA;  you will be able to figure out the new one. Your new password has the&#xA;  same amount of letters as the old one, and four of the letters are the&#xA;  same.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bob was then able to log into his computer without any problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What was Bob's new password? What was Bob's old password?&lt;/p&gt;&#xA;" OwnerUserId="9575" LastActivityDate="2015-02-23T21:46:54.310" Title="Password Out of Date" Tags="|logical-deduction|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9469" PostTypeId="2" ParentId="9468" CreationDate="2015-02-23T01:29:46.543" Score="15" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;New:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;different&quot; is the new password&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Old:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;outofdate&quot; is the old password&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Both passwords are the same length. Both passwords contain the characters D, E, N and T, totaling four unique letters that are the same.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9670" LastEditorUserId="9670" LastEditDate="2015-02-23T21:46:54.310" LastActivityDate="2015-02-23T21:46:54.310" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9472" PostTypeId="1" AcceptedAnswerId="9473" CreationDate="2015-02-23T02:21:09.677" Score="-7" ViewCount="6677" Body="&lt;p&gt;What word in the English language is always spelled incorrectly? &lt;/p&gt;&#xA;" OwnerUserId="9575" LastEditorUserId="25966" LastEditDate="2016-11-20T16:42:06.293" LastActivityDate="2016-11-20T16:42:06.293" Title="What English word is always spelled incorrectly?" Tags="|wordplay|english|language|" AnswerCount="1" CommentCount="3" ClosedDate="2016-11-20T19:41:33.823" ContentLicense="CC BY-SA 3.0" />
  <row Id="9473" PostTypeId="2" ParentId="9472" CreationDate="2015-02-23T02:26:08.610" Score="26" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Incorrectly&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lol, that was an easy one.&lt;/p&gt;&#xA;" OwnerUserId="5368" LastEditorUserId="9575" LastEditDate="2015-02-23T05:13:40.090" LastActivityDate="2015-02-23T05:13:40.090" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9477" PostTypeId="1" AcceptedAnswerId="9478" CreationDate="2015-02-23T09:28:04.777" Score="7" ViewCount="1022" Body="&lt;p&gt;Professor Halfbrain has spent his entire weekend by placing colored dots on the surface of a huge wooden cube. &#xA;His objective was to find large groups of dots that form the vertices of a regular polygon (= a convex polygon whose angles all are the same and whose edges all have the same length).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The professor managed to prove the following two deep theorems.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Professor Halfbrain's first theorem&lt;/strong&gt;: &#xA;  There exist three points on the surface of a cube that form the vertices of a regular polygon and that do not all lie on the same face.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Professor Halfbrain's second theorem&lt;/strong&gt;:&#xA;  There is no group of 37 points on the surface of a cube that form the vertices of a regular polygon and that do not all lie on the same face. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This puzzle asks you to improve the two theorems of professor Halfbrain and to make them even deeper.&#xA;Find an integer $x$, so that &quot;three points&quot; in the first theorem may be replaced by &quot;$x$ points&quot;, and so that &quot;37 points&quot; in the second theorem may be replaced by &quot;$x+1$ points&quot; (again yielding true statements, of course).&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-10-04T16:12:52.383" LastActivityDate="2015-10-04T16:12:52.383" Title="Points on a cube" Tags="|mathematics|geometry|three-dimensional|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9478" PostTypeId="2" ParentId="9477" CreationDate="2015-02-23T09:37:09.670" Score="13" Body="&lt;p&gt;You can get &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a 12-gon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This can be done by:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can slice a cube to get a regular hexagon&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/ifagJ.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA; Then, inscribe a regular 12-gon in the hexagon. This can be done by extending every other edge of the 12-gon until they meet in a hexagon.&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/Gsr7o.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is optimal because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Consider the plane that the polygon lies in. The intersection of the cube and the slice forms a polygon that contains at most 6 edges, one per face in the cube. The regular polygon must lie in this polygon, and must have at most to vertices on each edge, since otherwise an edge contains three consecutive vertices, which is impossible as there are in a line. So, the maximum number of vertices of the polygon is 6*2=12.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-02-23T09:43:51.453" LastActivityDate="2015-02-23T09:43:51.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9480" PostTypeId="2" ParentId="9396" CreationDate="2015-02-23T10:32:48.660" Score="3" Body="&lt;p&gt;Having a bit of a guess at 6,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Ireland and The UK are the only countries on Great Britain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Just had an idea for 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Italy and South Africa are the only countries which totally surround other countries.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And 5&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mexico and Canada are the only countries who have a land border with the USA.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9748" LastEditorUserId="9748" LastEditDate="2015-02-23T10:41:25.490" LastActivityDate="2015-02-23T10:41:25.490" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9481" PostTypeId="2" ParentId="3380" CreationDate="2015-02-23T12:03:13.413" Score="6" Body="&lt;h2&gt;The answer&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The trivial case $N=1$ has $f(1)=1$ and is kind of exceptional; this case will be ignored in all further discussion.&#xA;The answer to the puzzle for $N\ge2$ depends on the parity of $N$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $N=2n-1$ is odd, then $f(2n-1)=n+1$.&lt;br&gt;&#xA;  If $N=2n$ is even, then $f(2n)=n+1$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Proof of the lower bound&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If $N=2n-1$ is odd (with $n\ge2$), we put one gold coin in the first chest, one silver coin in the second chest, and one copper coin in each of the remaining $2n-3$ chests.&#xA;Then one must pick the first chest, the second chest, and at least $n-1$ of the remaining chests. This shows $f(2n-1)\ge1+1+(n-1)=n+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $N=2n$ is even (with $n\ge1$), we put one gold coin in the first chest, one silver coin in the second chest, and one copper coin in each of the remaining $2n-2$ chests.&#xA;Then one must pick the first chest, the second chest, and $n-1$ of the remaining chests. This shows $f(2n)\ge1+1+(n-1)=n+1$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;An auxiliary result&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;We consider the special situation with an even number $2k$ of chests, where the $i$-th chest contains $g_i$ gold coins and $s_i$ silver coins.&#xA;We assume without loss of generality that $g_1\le g_2\le\cdots\le g_{2k}$.&#xA;We denote $g_{\max}=\max g_i=g_{2k}$ and $s_{\max}=\max s_i$.&#xA;Then the following holds:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Auxiliary result&lt;/strong&gt;: There exist a partition of the $2k$ chests into two groups with $k$ chests, such that the total amounts $G_1$ and $G_2$ of gold in the two groups and the total amounts $S_1$ and $S_2$ of silver in the two groups satisfy $|G_1-G_2|\le g_{\max}$ and $|S_1-S_2|\le s_{\max}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the proof, we consider a &quot;special&quot; balanced partition into two groups, so that from every pair $g_{2i-1}$ and $g_{2i}$ one chest goes into one group and the other chest goes into the other group. The most unbalanced (and for us worst) way of doing this for the gold coins is to put all values $g_{2i-1}$ into the first group, and all the values $g_{2i}$ into the second group.&#xA;Then $G_1\le G_2$, and furthermore&#xA;$$ g_2\le g_3;~~~~~~~ g_4\le g_5;~~~~~~~ \ldots~~\ldots~~~~~~~ &#xA;g_{2k-2}\le g_{2k-1};~~~~~~~ g_{2k}\le g_{\max}$$&#xA;implies $G_2\le G_1+g_{\max}$. &#xA;Hence any such balanced partition satisfies the desired property $|G_1-G_2|\le g_{\max}$, and we only need to take care of the balance of the silver coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the silver coins, we construct such a special balanced partition into two groups by working through the pairs $s_{2i-1}$ and $s_{2i}$ by increasing $i=1,\ldots,k$.&#xA;The first step assigns $s_1$ to the first group and $s_2$ to the second group.&#xA;In all further steps, we look at the current total silver amounts in both groups and assign the larger of $s_{2i-1}$ and $s_{2i}$ to the group with smaller silver value and the smaller of $s_{2i-1}$ and $s_{2i}$ to the group with larger silver value.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the difference between the two silver values always remains below $s_{\max}$: &#xA;This is true at the very beginning, and is easy to check that each step maintains this property.&#xA;Hence, in the end we will have the desired $|S_1-S_2|\le s_{\max}$.&#xA;This completes the proof of the auxiliary result.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Proof of the upper bound&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;(The even case).&#xA;If $N=2n$ is even (with $n\ge1$), we pick two chests that contain $g_{\max}$ gold coins and $s_{\max}$ silver coins.&#xA;The remaining $2n-2$ chests are partitioned into two specially balanced groups with $n-1$ chests according to our auxiliary result.&#xA;We pick the group that has higher copper value for us.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The total gold value that we did &lt;em&gt;not&lt;/em&gt; pick is the gold value $G_1$ in the unpicked group.&#xA;Since $G_1\le G_2+g_{\max}$ by the auxiliary result, we have picked at least half of the total gold.&lt;/li&gt;&#xA;&lt;li&gt;The total silver value that we did &lt;em&gt;not&lt;/em&gt; pick is the silver value $S_1$ in the unpicked group.&#xA;Since $S_1\le S_2+s_{\max}$ by the auxiliary result, we have picked at least half of the total silver.&lt;/li&gt;&#xA;&lt;li&gt;The total copper value that we did &lt;em&gt;not&lt;/em&gt; pick is the copper value $C_1$ in the unpicked group.&#xA;Since the picked group has copper value $C_2\ge C_1$, we have picked at least half of the total copper.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Since altogether we have picked $n+1$ chests, this shows $f(2n)\le n+1$ and completes the analysis of the even case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The odd case).&#xA;If $N=2n-1$ is odd (with $n\ge2$), we first pick an arbitrary chest and then pick $n$ from the remaining $2n-2$ chests according to the solution of the even case.&#xA;This altogether yields $n+1$ chests and shows $f(2n-1)\le n+1$ and completes the analysis of the odd case.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-02-23T12:03:13.413" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9482" PostTypeId="2" ParentId="9396" CreationDate="2015-02-23T12:15:35.710" Score="4" Body="&lt;p&gt;&lt;strong&gt;Question 3:&lt;/strong&gt;&lt;br&gt;&#xA;According to that &lt;a href=&quot;http://streamafrica.com/culture/liberia-ethiopia-never-colonized-african-countries/#&quot; rel=&quot;nofollow&quot;&gt;[webpage]&lt;/a&gt;: Liberia and Ethiopia are the only two African countries that have never been colonized (&quot;which are considered to have survived the scramble for Africa and remained independent during the invasion, partitioning and forceful ruling of Africa by European nations&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 4:&lt;/strong&gt;&lt;br&gt;&#xA;According to that &lt;a href=&quot;https://books.google.nl/books?id=vkgvKqjaKxsC&amp;amp;pg=PA391&amp;amp;lpg=PA391&amp;amp;dq=%22Liechtenstein%20and%20Saudi%20Arabia%20are%20the%20only%20two%20countries%20in%20the%20world%20whose%22&amp;amp;source=bl&amp;amp;ots=gGv9baDK5V&amp;amp;sig=epvKr9bFZj8ZJOfRUUzCIg3cJ6I&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=sxjrVNH0EcXJPInkgNAD&amp;amp;ved=0CCMQ6AEwAA#v=onepage&amp;amp;q=%22Liechtenstein%20and%20Saudi%20Arabia%20are%20the%20only%20two%20countries%20in%20the%20world%20whose%22&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;[book]&lt;/a&gt;: Liechtenstein and Saudi Arabia are the only two countries in the world whose names are associated with the ruling family.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-02-23T12:15:35.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9483" PostTypeId="2" ParentId="9396" CreationDate="2015-02-23T12:31:18.390" Score="5" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;For number 2: Switzerland and Vatican are the only countries whose flag is square-shaped. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For number 5, I would guess: Mexico and Luxembourg are the only countries on Earth whose name contains the letter X.&lt;br&gt;&#xA;This is a very distinguished property: all other letters occur multiple times.&#xA;The rare letters are J, K, Q, X and Y. But for J there are Japan, Azerbaijan, Jamaica, Jordan, Tajikistan, etc. For K there are Kazakhstan, Kenya, Korea, Kuwait, Pakistan, Turkey, Sri Lanka, etc. For Q there are Qatar, Iraq, Equatorial Guinea, Martinique, Mozambique. For Y there are Yemen, Germany, Hungary, Turkey, Cyprus, Egypt, Guyana, Kenya, etc. For X there are only two.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I remember the answer to number 6 from the &quot;Trivial Pursuit&quot; quiz game.&#xA;Ireland and New Zealand are the only two countries in the world that lack native snakes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="9404" LastEditDate="2015-02-24T16:40:53.697" LastActivityDate="2015-02-24T16:40:53.697" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9484" PostTypeId="2" ParentId="9393" CreationDate="2015-02-23T12:38:12.647" Score="3" Body="&lt;p&gt;Number 13 is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Slumdog Millionaire&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 20 is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Discreet Charm of the Bourgeoisie&lt;br&gt;&#xA; &lt;a href=&quot;http://en.wikipedia.org/wiki/The_Discreet_Charm_of_the_Bourgeoisie&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/The_Discreet_Charm_of_the_Bourgeoisie&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="9404" LastEditDate="2015-02-23T13:09:15.550" LastActivityDate="2015-02-23T13:09:15.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9485" PostTypeId="1" AcceptedAnswerId="9489" CreationDate="2015-02-23T12:50:48.173" Score="12" ViewCount="1616" Body="&lt;p&gt;The answer to the following riddle is a 12-letter word.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;My 1, 12, 8, 6, 9, 10, 4, 5 is a traveller in space.  &lt;/li&gt;&#xA;&lt;li&gt;My 9, 10, 12, 6, 3, 11, 5 was the last word of Charles.  &lt;/li&gt;&#xA;&lt;li&gt;My 5,  4,  12,  8,  11,  9,  3 is something that you should not do.  &lt;/li&gt;&#xA;&lt;li&gt;My 2, 4, 7, 8, 11, 9, 6 is a combination of things.  &lt;/li&gt;&#xA;&lt;li&gt;My whole can be double-dealing or even-handed.  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="9404" LastEditDate="2015-02-24T16:26:47.960" LastActivityDate="2015-02-24T16:26:47.960" Title="A double and even riddle" Tags="|riddle|word|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9486" PostTypeId="1" AcceptedAnswerId="9488" CreationDate="2015-02-23T12:53:45.140" Score="11" ViewCount="765" Body="&lt;p&gt;Solve the following rebus:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  Munsterlander&#xA;  Telomian&#xA;  Whippet&#xA;  Terrier&#xA;  Foxhound&#xA;  Spaniel&#xA;  Sharpei &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2015-02-23T23:41:50.810" Title="Munsterlander &amp; Telomian - Rebus" Tags="|rebus|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9487" PostTypeId="2" ParentId="8953" CreationDate="2015-02-23T13:03:26.307" Score="8" Body="&lt;p&gt;This puzzle is about rivers and not states.  The Mississippi, the Missouri and the Ohio are famous rivers. Google Search Engine finds Colorado river, Connecticut river, Delaware river, Florida river, Illinois river, Iowa river, Kansas river, Kentucky river, Maine river, Michigan river, Minnesota river, Mississippi river, Missouri river, Ohio river, Tennessee river, Wisconsin river.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google Search Engine does not find California river, Georgia river, Hawaii river, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alabama river and Arkansas river (but no alaska river and Arizona river) show that&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Alaska and Arizona belong to first group with 32 states&lt;/li&gt;&#xA;&lt;li&gt;Alabama and Arkansas belong to second group with 18 states&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2015-02-23T13:03:26.307" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9488" PostTypeId="2" ParentId="9486" CreationDate="2015-02-23T13:30:58.293" Score="19" Body="&lt;p&gt;I think it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://idioms.thefreedictionary.com/Every+dog+has+its+day&quot;&gt;Every dog has its day &lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are all dog breeds&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;M&lt;/strong&gt;unsterlander &lt;strong&gt;M&lt;/strong&gt;onday&lt;br/&gt;&lt;strong&gt;T&lt;/strong&gt;elomian &lt;strong&gt;T&lt;/strong&gt;uesday&lt;br/&gt;&lt;strong&gt;W&lt;/strong&gt;hippet &lt;strong&gt;W&lt;/strong&gt;ednessday&lt;br/&gt;&lt;strong&gt;T&lt;/strong&gt;errier &lt;strong&gt;T&lt;/strong&gt;hursday&lt;br/&gt;&lt;strong&gt;F&lt;/strong&gt;oxhound &lt;strong&gt;F&lt;/strong&gt;riday&lt;br/&gt;&lt;strong&gt;S&lt;/strong&gt;paniel &lt;strong&gt;S&lt;/strong&gt;aturday&lt;br/&gt;&lt;strong&gt;S&lt;/strong&gt;harpei &lt;strong&gt;S&lt;/strong&gt;unday&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5127" LastEditorUserId="5610" LastEditDate="2015-02-23T14:06:55.837" LastActivityDate="2015-02-23T14:06:55.837" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9489" PostTypeId="2" ParentId="9485" CreationDate="2015-02-23T14:24:55.417" Score="13" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; AMBIDEXTROUS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My 1, 12, 8, 6, 9, 10, 4, 5 is a traveller in space.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ASTEROID&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My 9, 10, 12, 6, 3, 11, 5 was the last word of Charles.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ROSEBUD (From Citizen Kane, not King Charles I. This confused me for a while.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My 5, 4, 12, 8, 11, 9, 3 is something that you should not do.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DISTURB&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My 2, 4, 7, 8, 11, 9, 6 is a combination of things.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MIXTURE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My whole can be double-??????? or even-??????.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Doublehanded and Evenhanded are synonyms for or related to ambidextrous. Evenhanded means &quot;not favoring one side over the other&quot;. Doublehanded means having or using 2 hands, or requiring the use of 2 hands. (All praise to Nate Kerkhofs)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2015-02-23T20:41:59.827" LastActivityDate="2015-02-23T20:41:59.827" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9491" PostTypeId="1" AcceptedAnswerId="10975" CreationDate="2015-02-23T15:44:15.767" Score="12" ViewCount="2264" Body="&lt;p&gt;This weekend I was helping my three year old build some train tracks for his Duplo train. The train is powered by a small battery, so it moves on its own. His collection of pieces includes a bunch of straights and turns, and two switches. This is what we built:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nsxjP.jpg&quot; alt=&quot;Example track&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The train can approach a switch from three possible directions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KnlPb.jpg&quot; alt=&quot;Train switch&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the train comes from direction 1, it will follow the path that is opened by the red lever on the switch. When the train comes from one of the other two directions, there is only one way to go. If the switch is in the way, it will be pushed aside by the train (direction 3 in the example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to build a track where the train continuously loops across the entire track system, moving switches as needed. I want to know the &lt;strong&gt;minimum number of pieces&lt;/strong&gt; required to achieve this. Some restrictions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The solution has to work for each starting point and each traveling direction.&lt;/li&gt;&#xA;&lt;li&gt;The solution has to include &lt;strong&gt;more than two&lt;/strong&gt; switches.&lt;/li&gt;&#xA;&lt;li&gt;No human interaction is allowed other than initial placement and activation of the train.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All I managed to find is a solution with two switches and 24 curves, similar to the one that @JonTheMon used for his answer. I want to know if there's a solution when using more than two switches.&lt;/p&gt;&#xA;" OwnerUserId="6008" LastEditorUserId="6008" LastEditDate="2015-02-24T06:18:55.550" LastActivityDate="2015-05-22T15:05:26.323" Title="Duplo train tracks" Tags="|visual|mechanical-puzzles|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9492" PostTypeId="2" ParentId="9491" CreationDate="2015-02-23T15:55:54.923" Score="2" Body="&lt;p&gt;Create a barbell type track. The train will go to one end, use the switch to enter a loop, then come back to the middle, go to the next end, enter the other switch, then come back to the middle. Repeat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, 2 switches, 1 straight for mid, and it looks like 13 per loop (3 straight, 10 curved) = &lt;strong&gt;2 switches, 7 straight, 20 curved&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-02-23T16:16:07.350" LastActivityDate="2015-02-23T16:16:07.350" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9493" PostTypeId="2" ParentId="9321" CreationDate="2015-02-23T16:01:13.280" Score="2" Body="&lt;p&gt;The message, decoded using Vigenere, is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; thisisanencodededmessage&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the key is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; accabf&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Key and text were decoded using a program I have written. The code cannot be made available. But there are probably similar ones online that aren't as accurate. If the message had no English words in it would have not been found without his key that is hidden in the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I actually do not see the key anywhere in here but I didn't look for it to hard either. Based on the length of 6 it's probably the hex color coding to one of the elements on this site.&lt;/p&gt;&#xA;" OwnerUserId="2803" LastEditorUserId="8874" LastEditDate="2015-10-07T17:17:56.550" LastActivityDate="2015-10-07T17:17:56.550" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9494" PostTypeId="1" AcceptedAnswerId="9495" CreationDate="2015-02-23T16:16:03.357" Score="4" ViewCount="314" Body="&lt;p&gt;I have partitioned the chemical elements into two groups based on a certain&#xA;property that is shared by some of them, but not by the remaining ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following elements have my enigmatic property:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bismuth, Chlorine, Gold, Helium, Iodine, Mendelevium, Nitrogen, Silver&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The following elements do not have my enigmatic property:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dubnium, Erbium, Nobelium, Terbium, Ytterbium, Yttrium, Zinc, Zirconium&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which of the following elements do have this property?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Aluminium, Fluorine, Francium, Gallium, Lutetium, Neon, Oxygen, Polonium.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="9125" LastEditDate="2015-02-26T17:18:09.093" LastActivityDate="2015-02-26T17:18:09.093" Title="An enigmatic partition of the chemical elements" Tags="|mathematics|pattern|knowledge|chemistry|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9495" PostTypeId="2" ParentId="9494" CreationDate="2015-02-23T16:46:05.420" Score="8" Body="&lt;p&gt;The property is  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;having a prime atomic number.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first group:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bismuth 83&lt;br&gt;&#xA; Chlorine 17&lt;br&gt;&#xA; Gold 79&lt;br&gt;&#xA; Helium 2&lt;br&gt;&#xA; Iodine 53&lt;br&gt;&#xA; Mendelevium 101&lt;br&gt;&#xA; Nitrogen 7&lt;br&gt;&#xA; Silver 47&lt;br&gt;&#xA; &lt;strong&gt;All of these atomic numbers are prime.&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second group:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dubnium 105 (divisible by 5)&lt;br&gt;&#xA; Erbium 68 (divisible by 2)&lt;br&gt;&#xA; Nobelium 102 (divisible by 2)&lt;br&gt;&#xA; Terbium 65 (divisible by 5)&lt;br&gt;&#xA; Ytterbium 70 (divisible by 5)&lt;br&gt;&#xA; Yttrium 39 (divisible by 3)&lt;br&gt;&#xA; Zinc 30 (divisible by 5)&lt;br&gt;&#xA; Zirconium 40 (divisible by 5)&lt;br&gt;&#xA; &lt;strong&gt;None of these atomic numbers are prime.&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, of the final group:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Aluminum 13 (prime)&lt;br&gt;&#xA; Fluorine 9 (divisible by 3)&lt;br&gt;&#xA; Francium 87 (divisible by 3)&lt;br&gt;&#xA; Gallium 31 (prime)&lt;br&gt;&#xA; Lutetium 71 (prime)&lt;br&gt;&#xA; Neon 10 (divisible by 5)&lt;br&gt;&#xA; Oxygen 8 (divisible by 2)&lt;br&gt;&#xA; Polonium 84 (divisible by 2)&lt;br&gt;&#xA; &lt;strong&gt;Aluminum, Gallium, and Lutetium have the property.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9319" LastActivityDate="2015-02-23T16:46:05.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9496" PostTypeId="2" ParentId="9486" CreationDate="2015-02-23T16:51:35.247" Score="1" Body="&lt;p&gt;I think it's a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dog Pile&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All the things listed are dog breeds. They are listed separately on each line, indicating that they are on top of, or below one another.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9755" LastActivityDate="2015-02-23T16:51:35.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9497" PostTypeId="2" ParentId="9351" CreationDate="2015-02-23T17:52:29.943" Score="0" Body="&lt;p&gt;I think you are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Seawater (or parts thereof)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am always moving:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tides&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I never know where I am going.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Seawater doesn't have the ability to figure this out.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I provide delicious flavor to food.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Salt from Seawater is used to flavour many things&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be read on many restaurant menus.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Either salt levels or with things like &lt;a href=&quot;http://www.amazon.co.uk/Lays-Natural-Thick-Chips-8-5oz/dp/B004MTRHN8&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.co.uk/Lays-Natural-Thick-Chips-8-5oz/dp/B004MTRHN8&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, I am toxic. I kill thousands every year.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can drown in Seawater or too much salt is deadly&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but without me, you cannot survive&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You need some salt and water to live&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9748" LastActivityDate="2015-02-23T17:52:29.943" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9498" PostTypeId="1" AcceptedAnswerId="9503" CreationDate="2015-02-23T18:00:39.120" Score="7" ViewCount="1007" Body="&lt;p&gt;I am looking for help with a math puzzle. The answer is the final coordinates to a geocache.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have put in quite a bit of time and lots of graph paper trying to solve this one. I have tried programs and even bought the correct tools to work on this one. Still with the tools and the basic knowledge needed to work on this I have found I am no math wiz. I do spend quite a bit of time and learn everything I can before asking for help on these but some puzzles are beyond my skill level. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I do understand is this is a method of surveying called Metes and Bounds. I have found that somewhere in the puzzle is missing info leaving you with trying to solve the missing distance and direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cache is NOT located at the Point Of Beginning (POB) coordinates, but is in the same general area.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The POB in the puzzle is N 49° 13.314 W 123° 01.073 &lt;/li&gt;&#xA;&lt;li&gt;Beginning at the POB distant thereon North 60° 00' East 150.00 metres;&lt;/li&gt;&#xA;&lt;li&gt;Thence North 30° 00' West 60.00 metres;&lt;/li&gt;&#xA;&lt;li&gt;Thence North 60° 00' East, 90.00 metres to the beginning of a tangent curve concave Southeasterly and having a radius of 125.00 metres;&lt;/li&gt;&#xA;&lt;li&gt;Thence Northeasterly along said curve through a central angle of 60° 00' an arc distance of 130.90 metres;&lt;/li&gt;&#xA;&lt;li&gt;Thence tangent to said curve South 60° 00' East 70.00 metres to the beginning of a tangent curve concave Northeasterly and having a radius of 200.00 metres;&lt;/li&gt;&#xA;&lt;li&gt;Thence Southeasterly along said curve through a central angle of 55° 00' an arc distance of 192.00 metres to the beginning of a compund curve concave Northwesterly and having a radius of 310.00 metres;&lt;/li&gt;&#xA;&lt;li&gt;Thence Northeasterly along said curve through a central angle of 120° 00' an arc distance of 649.26 metres to the beginning of a compund curve concave Southwesterly and having a radius of 125.00 metres;&lt;/li&gt;&#xA;&lt;li&gt;Thence Northwesterly along said curve through a central angle of 55° 00' an arc distance of 120.00 metres;&lt;/li&gt;&#xA;&lt;li&gt;Thence tangent to said curve South 70° 00' West an indeterminate distance;&lt;/li&gt;&#xA;&lt;li&gt;Thence North 60° 00' West 90.00 metres; said point being coincident with the true position of the geocache;&lt;/li&gt;&#xA;&lt;li&gt;Thence North 90° 00' West 140.00 metres to the beginning of a tangent curve concave Southwesterly and having a radius of 160.00 metres;&lt;/li&gt;&#xA;&lt;li&gt;Thence Southwesterly along said curve through a central angle of 40° 00' an arc distance of 111.70 metres to the beginning of a reverse curve concave Northwesterly and having a radius of 160.00 metres;&lt;/li&gt;&#xA;&lt;li&gt;Thence Southwesterly along said reverse curve through a central angle of 30° 00' an arc distance of 83.78 metres;&lt;/li&gt;&#xA;&lt;li&gt;Thence tangent to said curve South 80° 00' West 125.00 metres to the beginning of a tangent curve concave Southeasterly and having a radius of 125.00 metres;&lt;/li&gt;&#xA;&lt;li&gt;Thence Southwesterly along said curve through a central angle of 115° 00' an arc distance of 250.89 metres;&lt;/li&gt;&#xA;&lt;li&gt;Thence tangent to said curve South 35° 00' East an indeterminate distance;&lt;/li&gt;&#xA;&lt;li&gt;Thence South 62° 55' East 191.78 metres to the true Point of Beginning, more or less.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9368" LastEditorUserId="5188" LastEditDate="2015-02-23T23:52:43.717" LastActivityDate="2015-03-07T21:13:28.790" Title="Geocaching Geometry Puzzle" Tags="|mathematics|geometry|geocaching|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9499" PostTypeId="1" AcceptedAnswerId="9548" CreationDate="2015-02-23T18:46:03.607" Score="10" ViewCount="1264" Body="&lt;p&gt;Yesterday I met the perfect logicians Divvy and Subtra. &#xA;I told them: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I have chosen two positive integers $x$ and $y$ with $2\le x\lt y\le 100$ and with $x$ a divisor of $y$.&#xA;  I will now whisper the ratio $r=y/x$ into the ear of Divvy and I will whisper the difference $d=y-x$ into the ear of Subtra.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And then I whispered the ratio and the difference into their ears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the following conversation took place:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Divvy: I don't know the numbers.&lt;/li&gt;&#xA;&lt;li&gt;Subtra confirmed: I already knew this.&lt;/li&gt;&#xA;&lt;li&gt;Divvy: I already knew that you knew that I don't.&lt;/li&gt;&#xA;&lt;li&gt;Subtra: I still can't determine the numbers.&lt;/li&gt;&#xA;&lt;li&gt;Divvy:  At the very beginning of this conversation, I knew that you did not know the numbers. I forgot to mention this. Does this information change something for you?&lt;/li&gt;&#xA;&lt;li&gt;Subtra: Yes, it changes everything! Finally I am able to determine the numbers!&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What on earth are these magic numbers $x$ and $y$?&lt;/p&gt;&#xA;" OwnerUserId="9517" LastEditorUserId="10615" LastEditDate="2016-02-24T04:02:31.413" LastActivityDate="2016-02-24T04:02:31.413" Title="Deducing Two Numbers based on their Difference and Ratio" Tags="|mathematics|logical-deduction|number-theory|meta-knowledge|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9501" PostTypeId="2" ParentId="9400" CreationDate="2015-02-23T21:02:13.870" Score="1" Body="&lt;p&gt;Partial answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that &quot;one is divisible by the other&quot; means that one is &lt;em&gt;evenly&lt;/em&gt; divisible by the other so that there is no remainder, then merely the 3rd mathematician's statement narrows is down to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; x = even integer between 68 and 100 inclusive&lt;br&gt;&#xA; y = 2&lt;br&gt;&#xA; Because &lt;code&gt;r&lt;/code&gt; has to be a whole number that can only be achieved with a unique set of &lt;code&gt;x&lt;/code&gt; &amp;amp; &lt;code&gt;y&lt;/code&gt; values. I confirmed that these are the only unique cases through simple brute force.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Adding in the first statement (as discussed in other comments), we can drop the case &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; x = 100, y = 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can't get wrap my head around narrowing it down past that. I've narrowed it down to just 16 of the 9,801 possibilities but I can't get any further. I'm left with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; x = {68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98}&lt;br&gt;&#xA; y = 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastActivityDate="2015-02-23T21:02:13.870" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9502" PostTypeId="2" ParentId="9400" CreationDate="2015-02-23T21:31:54.427" Score="1" Body="&lt;p&gt;The key here is to remember that we have that &lt;strong&gt;one number divides the other&lt;/strong&gt; as common knowledge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statements number 1-4:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statement 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are quite a few $d$ values that are being avoided here: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 74, 82, 85, 86, 87, 91, 93, 94, 95, 98. All of these have only only way to be formed under the rule.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Statement 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The corresponding $s$ values that have to be avoided are: 6, 9, 15, 21, 33, 39, 51, 57, 69, 78, 86, 87, 90, 93, 93, 95, 98, 99, 102, 105, 105, 111, 123, 129, 141. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Statement 3a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This gives a new set of avoided $d$ values: 9, 15, 26, 27, 30, 33, 34, 35, 45, 51, 52, 54, 57, 60, 63, 66, 68, 70, 72, 75, 77, 78, 80, 81, 84, 90, 96. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Statement 3b:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think what this is saying is that the possible $s$ values for the actual $d$ value includes one that could lead to one of the above (3a excluded) $d$ values. These possible $d$ values are 10, 18, 20, 21, 22, 24, 25, 28, 32, 36, 39, 40, 42, 44, 48, 49, 50, 55, 56, 62, 64, 65, 69, 76, 88, 92. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Statement 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This means that the ratio must be over 33, and the only qualifying pair remaining for this is (2,90). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8875" LastActivityDate="2015-02-23T21:31:54.427" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9503" PostTypeId="2" ParentId="9498" CreationDate="2015-02-23T21:52:15.720" Score="8" Body="&lt;p&gt;There are two indeterminate distances in the metes-and-bounds description, each at a different bearing. This creates a vector space of possible endpoints of the path, only one point of which returns the path to the true point of beginning, which is (most likely) the pair of indeterminate distances you'll want in order to find the correct position of your geocache.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've drawn a not-exactly-to-scale diagram of the surveying path to help you verify the directions, included below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/M9QmP.png&quot; alt=&quot;Geocaching path&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The flag is where the geocache is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll try and follow you through the coordinates after every step, assuming that $49° 13.314' N$, $123° 01.073' W$ is the origin $(0, 0)$, and coordinates are over and up in metres, to help you figure out which pair of distances you want.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Beginning at the POB distant thereon North 60° 00' East 150.00 metres;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Travel 150 metres at a bearing of 60° to $(129.90, 75.00)$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Thence North 30° 00' West 60.00 metres;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then make a left turn and travel 60 metres, a vector of $(-30, 51.96)$, to $(99.90, 126.96)$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Thence North 60° 00' East, 90.00 metres to the beginning of a tangent curve concave Southeasterly and having a radius of 125.00 metres;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then turn right again and travel 90 metres, a vector of $(77.94, 45)$, to $(177.84, 171.96)$. Here you will arrive on the border of an imaginary circle with a radius of 125 metres and with the center southeast (technically, a bearing of 150°) of where you currently are.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Thence Northeasterly along said curve through a central angle of 60° 00' an arc distance of 130.90 metres;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Travel along the circle an arc length of 60°, changing your bearing from 60° to 120° as you do so. Since the radius is 125 metres, the total distance travelled will be $125 \times (60\pi / 180) = 130.90$ metres. However, the place you will end up is actually 125 metres due east, because the start point, end point, and center of the circle form an equilateral triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you'll end up travelling an effective vector of $(125, 0)$ to $(302.84, 171.96)$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Thence tangent to said curve South 60° 00' East 70.00 metres to the beginning of a tangent curve concave Northeasterly and having a radius of 200.00 metres;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assuming you've travelled 125 metres due east, you'll be turning 30 degrees right here. If you'd gone along the circular arc, you will have naturally ended up on the 120° bearing already. Walk 70 metres straight, a vector of $(60.62, -35)$, to $(363.46, 136.96)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we come across our second circular curve. This one has a center that is 200 metres northeast (at a bearing of 30°).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Thence Southeasterly along said curve through a central angle of 55° 00' an arc distance of 192.00 metres to the beginning of a compund curve concave Northwesterly and having a radius of 310.00 metres;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The calculation of the final position is a bit harder this time. The bearing changes 27.5° left to 92.5°, but the effective distance is calculated as $400 \text{m} \times \sin 27.5° = 184.7 \text{m}$, since it's not as simple as an equilateral triangle like the first time. This translates to a vector of $(184.52, -8.06)$, which means you'll end up at $(547.98, 128.90)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you've walked along this arc, you'll start walking along another arc immediately afterwards, which is why the description mentions a compound curve. (A reverse curve is an arc where you end up turning in the opposite bearing of the one you were just turning in).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Thence Northeasterly along said curve through a central angle of 120° 00' an arc distance of 649.26 metres to the beginning of a compund curve concave Southwesterly and having a radius of 125.00 metres;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is equivalent to travelling 536.94 metres at a bearing of 5°, or a vector of $(46.80, 534.89)$, ending up at $(594.78, 663.79)$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Thence Northwesterly along said curve through a central angle of 55° 00' an arc distance of 120.00 metres;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the final instruction until the first mystery distance. Here we end up travelling a distance of 115.44 metres at a bearing of -82.5°, or a vector of $(-114.45, 15.07)$, ending up at $(480.33, 678.86)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the purposes of figuring out the mystery distances, you can treat all these instructions as just one giant way of saying &quot;travel 480.33 metres east and 678.86 metres north&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the second set of instructions until the starting point again, not including the mystery distance, makes you travel a grand total of 428.88 metres west and 331.24 metres south.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the total of all non-mystery instructions is 53.45 metres east and 317.48 metres north, and travelling some distance south 70° west (250° bearing) and then travelling some more distance south 35° east (145° bearing) will get you back to the starting point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose these two distances are $a$ and $b$ &amp;ndash; then the following system of equations will let you figure out the correct distances:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;a \sin 250^\circ + b \sin 145^\circ + 53.45 &amp;amp;= 0\\&#xA;a \cos 250^\circ + b \cos 145^\circ + 317.48 &amp;amp;= 0\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have this, you can plug in the first distance $a$ into the first &quot;indeterminate distance&quot; instruction and then walk 90 metres at a bearing of 300° to your geocache.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-02-24T17:56:00.983" LastActivityDate="2015-02-24T17:56:00.983" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9504" PostTypeId="2" ParentId="9498" CreationDate="2015-02-23T23:10:28.670" Score="3" Body="&lt;p&gt;Here is a supplement to the excellent analysis by Joe Z:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two unknown distances are shown on the following chart (error corrected from previous chart).  Relative to the start point, the geocache is 167.43 meters east and 638.34 meters north.  This translates to N 49°13.658'and W 123°00.935'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the &lt;a href=&quot;https://drive.google.com/file/d/0B7rHhyYh5LvaeVhTZ1JmSEhCUEk/view?usp=sharing&quot; rel=&quot;nofollow noreferrer&quot;&gt;spreadsheet&lt;/a&gt; if you want to use it but Joe's analysis is much easier to follow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/G84hM.png&quot; alt=&quot;Geocache chart&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/eKgYR.png&quot; alt=&quot;Burnaby&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-03-07T21:13:28.790" LastActivityDate="2015-03-07T21:13:28.790" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9505" PostTypeId="2" ParentId="9486" CreationDate="2015-02-23T23:41:50.810" Score="5" Body="&lt;p&gt;I think it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dog days&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are all dogs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first letter of each word lines up with a day of the week&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9696" LastActivityDate="2015-02-23T23:41:50.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9506" PostTypeId="1" AcceptedAnswerId="9507" CreationDate="2015-02-24T00:17:31.400" Score="12" ViewCount="1379" Body="&lt;p&gt;One day, back in the time before computers, little John was scribing the answer his arithmetic homework - to compute the value of $2^{50}$. He had, in the days prior, painstakingly calculated every last digit of that value - and was, to be quite honest, very proud of his result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, this was also before pencils. It was also before little John would fully develop his fine motor skills. Consequently, as he wrote the last digit, he tipped over his inkwell, and spilled the blackest India ink all over his work! He was able to decipher most of the number, but he still couldn't make out one digit $\blacksquare$ towards the middle which had been completely obscured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$112589990\blacksquare 842624.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He cried all day. All night too. Here was his magnum opus (he wasn't very ambitious) ruined! His parents weren't much help - they didn't want to redo the computation because, honestly, it was pretty tedious, and they were too busy doing whatever it was people did before computers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, they took him, and his ruined paper to the local doctor (of mathematics). The parents explained the situation and the doctor (of mathematics), seeing the pitiful boy, took a look at the paper, and, after making a few calculations (far too few to have calculated the value of $2^{50}$ anew), wrote the correct digit above the ink-splotch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examining the paper, the parents discovered that the doctor (of mathematics) had used every other digit on the child's homework as part of the computation and had never written a number greater than $16$ (or less than $0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did the doctor (of mathematics) find the missing digit?&lt;/p&gt;&#xA;" OwnerUserId="4765" LastEditorUserId="4765" LastEditDate="2015-02-24T12:34:03.673" LastActivityDate="2016-03-19T14:53:18.177" Title="An Inkblot Ate My Homework" Tags="|mathematics|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9507" PostTypeId="2" ParentId="9506" CreationDate="2015-02-24T01:11:35.063" Score="12" Body="&lt;p&gt;He used the digital root property, that the (repeated) sum of the digits in a numbers equals its value modulo 9. Working $\bmod 9$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2^{50} = 2^{2 + 3\times 16} = 2^2 \times (2^3)^{16} = 4 \times 8^{16} = 4 \times (-1)^{16} =4 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $112589990?842624 = 4 \bmod 9$. Adding up the digits gives $70$, which is $7 \bmod 9$, so the remaining digit must be $6$ to produce a sum of $4 \bmod 9$.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-02-24T06:52:00.943" LastActivityDate="2015-02-24T06:52:00.943" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9509" PostTypeId="2" ParentId="9351" CreationDate="2015-02-24T01:50:08.380" Score="4" Body="&lt;p&gt;Okay, time to go way out in left field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Celebrity chef and television host, Gordon Ramsey&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am always moving&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As part of his television career, he is always moving from town to town, both for publicity and for shooting Kitchen Nightmares.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I never know where I am going.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He has an executive producer for that, after all.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I provide delicious flavor to food.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Being a chef, this is one of his primary jobs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be read on many restaurant menus.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; At a minimum, on the menus of the restaurants that he owns.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, I am toxic.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As anyone who has watched the first half of an episode of Kitchen Nightmares will attest.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I kill thousands every year,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Typically bacteria, insects, and sometimes small rodents, as anyone who has watched the second half of an episode of Kitchen Nightmares will attest.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but without me, you cannot survive. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; At least, not if you're on his &lt;em&gt;Hell's Kitchen&lt;/em&gt; reality show!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9286" LastActivityDate="2015-02-24T01:50:08.380" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9510" PostTypeId="1" AcceptedAnswerId="9511" CreationDate="2015-02-24T01:57:58.987" Score="-8" ViewCount="414" Body="&lt;p&gt;What is at the beginning of the end and the end of time and space? This is a short riddle so I'm just filling in characters to meet the requirements for length.&lt;/p&gt;&#xA;" OwnerUserId="9759" LastActivityDate="2015-02-24T02:01:53.963" Title="Beginning of the end" Tags="|riddle|" AnswerCount="1" CommentCount="0" ClosedDate="2015-05-08T19:44:55.050" ContentLicense="CC BY-SA 3.0" />
  <row Id="9511" PostTypeId="2" ParentId="9510" CreationDate="2015-02-24T02:01:53.963" Score="2" Body="&lt;p&gt;It's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the letter e&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the first letter of &quot;end&quot; and the last letter of &quot;time and space&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-02-24T02:01:53.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9514" PostTypeId="2" ParentId="9351" CreationDate="2015-02-24T07:14:40.190" Score="20" Body="&lt;p&gt;&lt;strong&gt;One more attempt. I think you are:&lt;/strong&gt;   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;SMOKE&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am always moving, but I never know where I am going.    &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; smoke is always moving with air currents  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I provide delicious flavor to food. I can be read on many restaurant menus.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; people enjoy smoked meats, smoked fish, smoked cheeses  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, I am toxic. I kill thousands of people every year,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; cigarette smoke kills, industrial smoke as a pollutant causes respiratory problems and the number one cause of death in structure fires is the smoke (people asphyxiating)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, it is difficult to survive in the wild without me.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; in the wild, fire is important for warmth and cooking.  With fire, there is smoke. Smoke can also be used as a signal from a person that is lost.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="9495" LastEditDate="2015-02-25T00:26:30.830" LastActivityDate="2015-02-25T00:26:30.830" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="9515" PostTypeId="1" CreationDate="2015-02-24T07:23:36.230" Score="39" ViewCount="6686" Body="&lt;p&gt;Here is a drawing of an old picture. Your task is to answer all 8 questions correctly AND give a good reasoning for it&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oJngi.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What time of day is it?&lt;/li&gt;&#xA;&lt;li&gt;Is it early spring or a late fall(autumn)?&lt;/li&gt;&#xA;&lt;li&gt;Is this river navigable?&lt;/li&gt;&#xA;&lt;li&gt;Which direction does the river flow? (North, East, South or West?)&lt;/li&gt;&#xA;&lt;li&gt;Is the river deep or shallow at the side where the boat is?&lt;/li&gt;&#xA;&lt;li&gt;Is there a bridge across the river somewhere nearby?&lt;/li&gt;&#xA;&lt;li&gt;How far is the railroad from here?&lt;/li&gt;&#xA;&lt;li&gt;Do birds fly North or South?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9352" LastEditorDisplayName="user20" LastEditDate="2015-02-26T09:46:52.093" LastActivityDate="2017-04-13T00:41:25.107" Title="An old picture puzzle" Tags="|logical-deduction|visual|" AnswerCount="9" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9516" PostTypeId="2" ParentId="9515" CreationDate="2015-02-24T07:55:51.230" Score="15" Body="&lt;p&gt;1.What time of day is it?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;del&gt;Late afternoon, people are still working, yet the person with the suitcase is wearing a lamp, so nightfall is coming soon.&lt;/del&gt; Took a double take, but didn't notice the shadows first, which suggest morning because they are walking eastward.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2.Is it early spring or a late fall(autumn)?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Early spring, people aren't harvesting, they are sowing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3.Is this river navigable?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes, else there wouldn't be a reason to place a buoy (atleast I think it's a buoy, could be a tide-gauge).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4.Which direction does the river flow? (North, East, South or West?)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It flows South, you can determine the direction with the &quot;buoy&quot; and with the fact that the bids fly north.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5.Is the river deep or shallow at the side where the boat is?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Deep, the fishing line is too long for shallow water. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6.Is there a bridge across the river somewhere nearby?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Probably not, crossing a bridge is usually free, using a ferry isn't.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;7.How far is the railroad from here?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Probably on walking distance from the boat, the people seem to be travelling to the countryside so it seems likely they just came from a train station.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;8.Do birds fly North or South?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; North, because it's spring.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5127" LastEditorUserId="5127" LastEditDate="2015-02-24T08:22:28.593" LastActivityDate="2015-02-24T08:22:28.593" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9517" PostTypeId="2" ParentId="9515" CreationDate="2015-02-24T08:00:02.363" Score="6" Body="&lt;p&gt;Answer 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Morning (shadows fall on the West)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Early Spring (harrow on the field)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes (Shipping buoy)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; South (determined by the passage around the displacer water.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer 5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Deep (Fishing line)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer 6:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No bridge (the man in the boat - the ferryman)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer 7:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Close (a man with a lantern rail)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer 8:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; North (because spring)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9346" LastActivityDate="2015-02-24T08:00:02.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9518" PostTypeId="2" ParentId="9515" CreationDate="2015-02-24T08:06:35.533" Score="4" Body="&lt;p&gt;It's hard to tell in a lot of cases. I notice there are other answers, and though some of my answers match theirs, I disagree on some of them. Here are the things that stand out to me:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Time of day?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Morning&lt;/strong&gt;. The shadow length suggests either early or late hours, and the fact that people are working/fishing suggests early.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Spring or fall?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Spring&lt;/strong&gt;. This one is clear, because the farmer is sowing his field.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is the river navigable?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Yes&lt;/strong&gt;. Besides there being a boat in the river, I think that's a floating buoy in the water on the left; if that is what it is, then it would suggest shipping boats travel down the river.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;River flow direction?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;East&lt;/strong&gt;. The buoy-like object on the left leaves ripples in the water's current, which shows that the water is flowing left-to-right in the picture. If I'm right about it being morning, then the water is flowing East (towards the sun - note that the shadows are parallel to the water).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is the water deep or shallow under the boat?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Deep&lt;/strong&gt;. The presence of a dock and a fisherman are good indicators that the water is fairly deep.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there a bridge nearby?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;No&lt;/strong&gt;. There's a ferry, which would probably not be in business if people could just cross the bridge.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How far away is the railroad?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Nearby&lt;/strong&gt;. The person on the left looks like a traditional railroad worker. His presence is the only really fair amount of evidence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Are the birds flying North or South?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;South&lt;/strong&gt;. The water flow direction and sun's position help with figuring this out, noting the birds fly in a V-shape to the left (Southwest) and roughly toward the nearer shoreline.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9319" LastActivityDate="2015-02-24T08:06:35.533" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9519" PostTypeId="2" ParentId="9351" CreationDate="2015-02-24T08:18:57.087" Score="1" Body="&lt;p&gt;I'm guessing it's:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Money&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I am always moving,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Money is always in circulation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;but I never know where I am going.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Money does not &quot;know&quot; where it is going. Or may be a more original way to state that money is usually wasted on stupid things.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I provide delicious flavor to food.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Back in the days spice were really expensive. Spice have even been used as a currency. Even today there are some exotic spices that cost a lot.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I can be read on many restaurant menus.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is a price for each dish.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;However, I am toxic.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Money is the root of all evil.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I kill thousands of people every year,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many people die over money or because they lack money.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;However, it is difficult to survive in the wild without me.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I interpret &quot;in the wild&quot; as &quot;when living on your own&quot;. When you are a kid, you don't need money, but once you &quot;leave the nest&quot; you need money to survive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And let's not forget the title:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Without me, you cannot survive&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An unfortunate fact, but without money living in modern-day society is pretty much impossible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="5610" LastEditDate="2015-02-24T09:18:40.467" LastActivityDate="2015-02-24T09:18:40.467" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9520" PostTypeId="2" ParentId="9515" CreationDate="2015-02-24T09:27:01.803" Score="3" Body="&lt;p&gt;While I agree with the others on most answers I have a different oppinion on Number 5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think it is &lt;strong&gt;shallow&lt;/strong&gt;. Why? Because it looks like the Boy is standing in the river. Also there is a buoy near the first third of the river which maybe indicates where the river gets deep. To the argument with the long fishing line: Is is quite common to use a long line while fishing in a river, so that the line can flow in the water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9767" LastActivityDate="2015-02-24T09:27:01.803" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="9527" PostTypeId="2" ParentId="9351" CreationDate="2015-02-24T13:38:12.977" Score="3" Body="&lt;p&gt;My guess is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Glutamate&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am always moving,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Glutamate is a the main neuro-transmitter used in the brain, as such it is constantly on the move as synapses release this agent.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I never know where I am going.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Of course it doesn't, it is just an amino acid and as such has no control of where it is heading&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I provide delicious flavor to food.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Umami is the fifth taste sense, which literally translates to 'delicious taste'. Umami in food is detected through glutamate receptors. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be read on many restaurant menus.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are many Umami rich foods, with MSG as probably the one most people are familiar with (mono-sodium glutamate), but it can also be found in fish, shellfish, cured meats, and mushrooms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, I am toxic.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is split opinion on whether MSG is toxic, but some researchers claim that it is.&#xA; It is ... &lt;a href=&quot;http://www.greenmedinfo.com/blog/msg-proven-highly-toxic-1-dose-causes-headache-healthy-subjects&quot; rel=&quot;nofollow&quot;&gt;http://www.greenmedinfo.com/blog/msg-proven-highly-toxic-1-dose-causes-headache-healthy-subjects&lt;/a&gt;&#xA; It isn't ... &lt;a href=&quot;http://www.pri.org/stories/2014-10-03/science-suggests-msg-really-isnt-bad-your-health-after-all&quot; rel=&quot;nofollow&quot;&gt;http://www.pri.org/stories/2014-10-03/science-suggests-msg-really-isnt-bad-your-health-after-all&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I kill thousands of people every year,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you follow the path that it IS toxic, then those studies will link it to the deaths of thousands of people every year.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, it is difficult to survive in the wild without me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Glutamate is an essential ingredient for us to function as humans.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9168" LastActivityDate="2015-02-24T13:38:12.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9528" PostTypeId="2" ParentId="9351" CreationDate="2015-02-24T14:45:11.543" Score="5" Body="&lt;p&gt;I think you are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am always moving,but I never know where I am going.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The water cycle is a continuous loop (I'm not counting the solid state of H2O)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I provide delicious flavor to food.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water is a major component of saliva, without which we wouldn't taste anything&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be read on many restaurant menus.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most restaurants list water as a beverage option&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, I am toxic.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water intoxication comes from drinking too much and messing with your electrolyte balance&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I kill thousands of people every year.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hurricanes / monsoons, tsunamis, drowning, take your pick.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, it is difficult to survive in the wild without me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can only last about 3 days without fresh water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastActivityDate="2015-02-24T14:45:11.543" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9530" PostTypeId="2" ParentId="548" CreationDate="2015-02-24T15:05:01.043" Score="1" Body="&lt;p&gt;The quickest method of message passing is usually broadcast - e.g. blowing a horn (or a relay of horns). All soldiers proceed to the castle simultaneously, so the last soldiers to arrive would be ones at the corners, which happens in time $\sqrt{0.5^2 + 0.5^2} = \sqrt{0.5}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the king's soldier must notify everyone in person or by proxy (via other soldiers), then it depends on the distribution of houses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the houses are arranged in expanding concentric squares around the castle, with the nearest a negligible distance away. Then the king's soldier goes first to the nearest 'ring' of houses, they go together to notify the next ring and so on, then return. In other words, the wave of soldiers moves out from the castle to the perimeter and then returns to the middle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The longest path is from the castle to each corner of the square. To get there and back takes $2\sqrt{0.5}$ time. Everyone else travels a shorter distance, so this is also the total muster time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other search patterns will be needed if the distribution is sparse. The worst case might be 5 soldiers, 1 at the castle and the other 4 at the corners of the square. The king's soldier travels to 3 of them in turn, sending the first to notify the remaining soldier. This traverses the entire diagonal and 2 sides for the longest path time-wise, taking $2 + \sqrt{2}$ time. I suspect (but am prepared to stand corrected) that in all other cases, some soldiers can be recruited to reduce the total muster time.&lt;/p&gt;&#xA;" OwnerUserId="9422" LastActivityDate="2015-02-24T15:05:01.043" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9531" PostTypeId="1" AcceptedAnswerId="9591" CreationDate="2015-02-24T15:43:55.253" Score="1" ViewCount="429" Body="&lt;p&gt;A grocer has a set of two-pan balance scales for weighing food. A customer weighs a bag of sugar on one of the pans, which registers at 45 pounds. Suspicious, she swaps the bag to the second pan of the same scale and registers it at 20 pounds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much does the bag actually weigh? &lt;/p&gt;&#xA;" OwnerUserId="9775" LastEditorDisplayName="user20" LastEditDate="2015-02-24T17:23:06.657" LastActivityDate="2015-02-26T17:45:24.420" Title="How much does the bag actually weigh?" Tags="|logical-deduction|" AnswerCount="2" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9532" PostTypeId="2" ParentId="9531" CreationDate="2015-02-24T16:07:39.060" Score="2" Body="&lt;p&gt;The bag weighs $mean(20,45)=\frac{20+45}2=32.5$ kg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the difference between the pans is same, the weight of an object would be affected by the same value on both sides, one side positively, the other negative. So we just find the mean.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-02-24T16:07:39.060" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="9533" PostTypeId="1" AcceptedAnswerId="9539" CreationDate="2015-02-24T16:24:24.443" Score="20" ViewCount="2053" Body="&lt;p&gt;Here's a puzzle that I created. The end result is a number.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/6AgXd.png&quot; alt=&quot;the puzzle&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9777" LastEditorUserId="5188" LastEditDate="2015-02-26T04:14:02.643" LastActivityDate="2015-02-26T04:14:02.643" Title="A challenging visual puzzle" Tags="|visual|steganography|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9536" PostTypeId="1" AcceptedAnswerId="9546" CreationDate="2015-02-24T16:54:00.413" Score="7" ViewCount="2085" Body="&lt;p&gt;&lt;strong&gt;Pre-game:&lt;/strong&gt; There are 1000 logicians in a line, each wearing a black or a red hat, completely at random. No one knows the color of their own hat. Each can see the hats on the next 10 people. The logicians are allowed to communicate before the game, but not once the hats have been placed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Game:&lt;/strong&gt; Starting with the back, each logician must call out loudly, a color, red, black, or white. White is obviously always wrong. This calling is, however, the only communication in the game. Every person who fails to say his own hat color is silently killed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the best strategy for the logicians?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Twist 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose there is exactly one blind man in the line. His location is not known to anyone, but he can hear and speak.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="10615" LastEditDate="2015-06-04T00:50:17.673" LastActivityDate="2015-06-04T00:50:17.673" Title="1000 logicians wearing hats" Tags="|logical-deduction|hat-guessing|" AnswerCount="1" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="9537" PostTypeId="1" AcceptedAnswerId="9544" CreationDate="2015-02-24T16:56:11.700" Score="4" ViewCount="3930" Body="&lt;p&gt;Solve the following riddle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I turn right.&lt;br&gt;&#xA;       Then I leave home.&lt;br&gt;&#xA;       Then I turn left.&lt;br&gt;&#xA;       Then I turn left.&lt;br&gt;&#xA;       Then I turn left.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Who am I?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2015-02-25T10:13:47.220" Title="I turn right - Riddle" Tags="|riddle|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9538" PostTypeId="2" ParentId="9537" CreationDate="2015-02-24T16:58:38.673" Score="14" Body="&lt;p&gt;I think you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A baseball player, rounding the bases.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, I think there could be a problem with this (either the riddle or my answer). If you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Right-handed and on the team in the third base&amp;ndash;dugout. Then you could possibly go from the dugout, to the on-deck circle, to the batter's box, and then on your way to first in a mostly straight, sligtly left-turning path. Then you would &quot;leave home&quot; without turning right.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm probably overthinking things in that part, though.&lt;/p&gt;&#xA;" OwnerUserId="8776" LastEditorUserId="8776" LastEditDate="2015-02-24T19:01:24.740" LastActivityDate="2015-02-24T19:01:24.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9539" PostTypeId="2" ParentId="9533" CreationDate="2015-02-24T17:14:02.267" Score="21" Body="&lt;p&gt;The puzzle represents:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pi 3.1415926535...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I ran the image through a &lt;a href=&quot;http://magiceye.ecksdee.co.uk/&quot;&gt;magic eye finder&lt;/a&gt; and got this mess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/rxx4I.png&quot; alt=&quot;Messy Image&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cleaned that up to get this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/GgJDJ.png&quot; alt=&quot;Cleaned Image&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And inverted the colors to get this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/QTaNH.png&quot; alt=&quot;QR Code&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which is a QR code that translates to the following text:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 143353-486602&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or, in raw bytes, this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 20 68 31 11 43 93 9c 96 e2 1c 10 00 ec 11 ec 11 ec 11 ec &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;All hail Len! I had been trying to figure out which coordinate system to use. I found Almere in OP's profile but had no idea the Dutch had their own coordinate system. God help the American education system...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the final piece as posted by Len (if you +1 me, then +1 Len, too):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The final number is pi&lt;br&gt;&lt;br&gt;&#xA; &#xA; From the clue on the OP's profile page, the numbers (143353-486602) are coordinates for the Dutch RD (Rijks-Driehoek) system. Using this site for conversion provides WGS84 coordinates of (52.367180,5.216201) which is in the city of Almere, Netherlands. With Google satellite, the coordinates are exactly at the number 3 as shown in this photo.&#xA; &lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/EC2mA.png&quot; alt=&quot;Final Solution&quot;&gt;&lt;br&gt;&#xA; You can tell by the rest of the digits that, for some reason, pi is printed in giant digits on the asphalt. From the other forum that OP posted in, he was looking for the first 10 digits. In that case, it's &lt;code&gt;3.1415926536&lt;/code&gt; or, if you go out to my favorite digit of pi, it's &lt;code&gt;3.14159265358979323846264338327950&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2015-02-25T16:08:20.667" LastActivityDate="2015-02-25T16:08:20.667" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="9540" PostTypeId="1" AcceptedAnswerId="9541" CreationDate="2015-02-24T17:26:08.307" Score="12" ViewCount="905" Body="&lt;p&gt;I am a string ?????.&lt;br&gt;&#xA;I consist of five symbols.&lt;br&gt;&#xA;If you plug me into the following Rebus puzzles, I will make a lot of sense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rebus 1:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; ????? AZ&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Rebus 2:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; ????? vs 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Rebus 3:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; ?????&#xA; FAVORITE&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2015-02-24T17:28:52.863" Title="Half riddle and half rebus" Tags="|riddle|rebus|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9541" PostTypeId="2" ParentId="9540" CreationDate="2015-02-24T17:28:52.863" Score="16" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 13579&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The three rebus puzzles become&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 13579 AZ = &lt;strong&gt;&quot;ODDS and ends&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 13579 vs 1 = &lt;strong&gt;&quot;The ODDS are against one&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 13579 on FAVORITE = &lt;strong&gt;&quot;ODDS-on favorite&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-02-24T17:28:52.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9542" PostTypeId="2" ParentId="9515" CreationDate="2015-02-24T18:37:24.647" Score="2" Body="&lt;p&gt;Going to answer in the order I figured them out instead of order presented:&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Is it early spring or a late fall(autumn)?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Spring. They are sowing not planting &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;8&quot;&gt;&#xA;&lt;li&gt;Do birds fly North or South? &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;North. Because it is Spring&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What time of day is it&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Morning.  Birds are flying North and shadows are going West, so sun must be in east.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;Which direction does the river flow? (North, East, South or West?)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;South, based on ripples around buoy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;Is this river navigable?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Yes, because there is a buoy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;Is the river deep or shallow at the side where the boat is?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Shallow, the buoy marks the shallow/deep line for larger boats.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;6&quot;&gt;&#xA;&lt;li&gt;Is there a bridge across the river somewhere nearby?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;No, otherwise people would be using it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;7&quot;&gt;&#xA;&lt;li&gt;How far is the railroad from here?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Close.  Guy carrying lantern is walking to/from work.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9353" LastEditorUserId="9353" LastEditDate="2015-02-24T18:43:01.683" LastActivityDate="2015-02-24T18:43:01.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9544" PostTypeId="2" ParentId="9537" CreationDate="2015-02-24T19:09:15.753" Score="12" Body="&lt;p&gt;You are a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Baseball player &lt;em&gt;batting left-handed&lt;/em&gt; and rounding the bases&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The only way that the first line is accurate is if you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Left-handed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The rest of the pieces follow as&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The player rounds the bases, turning left each time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The version of this that I originally heard didn't have the first line and had you returning to home as the last line.&lt;/p&gt;&#xA;" OwnerUserId="9000" LastActivityDate="2015-02-24T19:09:15.753" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9545" PostTypeId="1" AcceptedAnswerId="9644" CreationDate="2015-02-24T19:46:21.477" Score="1" ViewCount="587" Body="&lt;p&gt;There are 17 US-states that share a certain (purely geographical) property that the remaining 33 US-states do not have. Here are 15 of the 17:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Arizona, Arkansas, Georgia, Hawaii, Idaho, Indiana, Iowa, Mississippi, &#xA;  Ohio, Oklahoma, Rhode Island, South Carolina, Utah, West Virginia, Wyoming&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Which are the two missing states on this list?&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;What is this mysterious purely geographical property?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="9125" LastEditDate="2015-02-28T13:57:04.257" LastActivityDate="2015-02-28T13:57:04.257" Title="A mysterious property of 17 US-states" Tags="|pattern|geography|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="9546" PostTypeId="2" ParentId="9536" CreationDate="2015-02-24T20:55:51.433" Score="3" Body="&lt;p&gt;The person in the back looks at the next $10$ people and adds $1$ for each black hat and $0$ for every red hat.  If the sum is odd then he says &quot;black&quot;.  Otherwise he says &quot;red&quot;.  This gives him a 50% chance of survival.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next 10 people can sum the 9 people they can see or have heard the color of from their block of 10 people.  If their sum is the same parity (even or odd) as the one the back person had,  they yell &quot;red&quot;.  Otherwise they yell &quot;black&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result of the limitation that each peson can only see 10 people ahead of them is that no information from the back person will continue to person 12.  This means he will behave as a new back person.  This means that the people in positions $1,12,23,etc$ labeled from the back are forced to act as &quot;back&quot; people and therefore have a 50% chance of survival.  On average 45.5 people will die each game.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Twist 1:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The effect of a blind person will depend on his position $\mod 11$.  If he is in positions $1,12,23, etc$, ($1 \mod 11$) guessing means there is a 50% chance he will die and there is an independent 50% chance everyone in the next 10 will die.  He should say &quot;white&quot; which will kill him, result in the person in front of him having a 50% survival rate and the next 10 people after the person in front of him a 100% survival chance. If this occurs, on average 46.5 will die.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If he is in position 1000, he should just guess; everyone else has already gone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If he is in position $11,22,etc$ ($0 \mod 11$) he has a 100% chance of survival because he could hear what everyone said.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If he is in position $10,21,32,43,54,65,76,87,etc$, ($10 \mod 11$) when he guesses if he is correct, both him and the person in front of him survive.  If he is wrong they both die.  If he says white, he dies and the person in front of him acts as a back person.  If he guesses, on average 46.5 people will die.  If he does not guess, 47 people will die on average.  This means he should guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If he is in position $9,20.31,42,53,64,75,86,etc$, ($9 \mod 11$) when he guesses if he is correct, both him and the 2 people in front of him survive.  If he is wrong they all die.  If he says white, he dies and the person in front of him acts as a back person.  If he guesses, on average 47 people will die.  If he does not guess, 47 people will die on average.  It doesn't matter if he guesses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If he is in any other position, he should say &quot;white&quot;.  This means the person in front of him will act as a back person and, on average, 47 people die.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means if the blind man is moral and randomly placed on average 46.8 people will die.  &lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditDate="2015-02-26T12:56:58.427" LastActivityDate="2015-02-26T12:56:58.427" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="9547" PostTypeId="1" AcceptedAnswerId="9550" CreationDate="2015-02-24T21:06:50.143" Score="12" ViewCount="3452" Body="&lt;p&gt;Who or what am I?&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I look at the sky and my eyes reflect it&lt;br&gt;&#xA;I can bring death but also life&lt;br&gt;&#xA;Achilles' mother used my kind&lt;br&gt;&#xA;My bed is smooth as sand itself&lt;br&gt;&#xA;I can be kind and allow you to pass&lt;br&gt;&#xA;When I am angry, you won't make it.&lt;br&gt;&#xA;&lt;br&gt;&lt;br&gt;&#xA;What am I?&lt;/p&gt;&#xA;" OwnerUserId="9618" LastEditorUserId="8874" LastEditDate="2015-02-26T18:51:48.640" LastActivityDate="2015-02-26T18:51:48.640" Title="Achilles' mother used my kind" Tags="|riddle|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9548" PostTypeId="2" ParentId="9499" CreationDate="2015-02-24T21:12:34.330" Score="10" Body="&lt;p&gt;We are dealing with four integers $x$, $y$, $r$ and $d$ that satisfy $y=rx$ and $d=(r-1)x$. Note that $x&amp;lt;y$ implies $r\ge2$ and $d\ge2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Divvy: I don't know the numbers.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $r\ge34$, then $x\ge3$ is impossible (as this would imply $y=rx\ge3\cdot34=102$). Hence this would imply $x=2$ and $y=2r$, so that Divvy knows $x$ and $y$ in contradiction to his statement.&lt;/li&gt;&#xA;&lt;li&gt;If $r\le33$, then $x=2$ and $y=2r$ as well as $x=3$ and $y=3r$ are feasible solutions. Divvy is not able to determine $x$ and $y$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Divvy's first statement is equivalent to $2\le r\le33$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(2) Subtra confirmed: I already knew this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Subtra's difference is $d=(r-1)x$, her statement means that $d$ cannot be written as the product of two factors $r-1$ and $x$, one of which is $\ge33$ and one of which is $\ge2$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Subtra's statement means that the difference $d$ is either from the range $0\le d\le65$, or that $d$ is odd and from the range $67\le d\le97$. &#xA;  Equivalently, $d\notin\{66,68,70,\ldots,98\}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(3) Divvy: I already knew that you knew that I don't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divvy's statement means that there does not exist any integer $x$ with $2\le x\le100/r$, so that the corresponding difference $(r-1)x$ is an even integer in the so-called bad range $B=\{66,68,\ldots,98\}$. &#xA;We make some case distinctions on $2\le r\le33$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $r$ is odd, then $r-1$ is an even integer from the range $2,4,\ldots,32$. &#xA;As the diameter of the bad range $B$ is $98-66=32\ge r-1$, the bad range contains an (even!) integer multiple $(r-1)x$ of $r-1$.&#xA;Hence all such values $r$ collide with Divvy's statement.&lt;/li&gt;&#xA;&lt;li&gt;If $r=2$, then for any $x$ with $2\le x\le 100/r=50$, the corresponding product $(r-1)x=x\le50$ will not fall into the bad range.&#xA;This case is possible.&lt;/li&gt;&#xA;&lt;li&gt;If $r$ is even with $4\le r\le24$, then $2(r-1)$ is an even integer from the range $6,10,14\ldots,46$. The values $6,10,\ldots,30$ are smaller than the diameter of the bad range, and hence have an integer multiple in $B$.&#xA;The value $34,38,42,46$ respectively have the multiples $68,76,84,92$ in $B$.&#xA;For a multiple of the form $2(r-1)k$ in $B$, we may choose $x=2k$.&#xA;Hence all such values $r$ collide with Divvy's statement.&lt;/li&gt;&#xA;&lt;li&gt;If $r\in\{26,28,30,32\}$, then none of the multiples of $r-1\in\{25,27,29,31\}$ fall into $B$.&#xA;Hence these four cases are possible.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Summarizing, we see that Divvy's second statement is equivalent to $r\in\{2,26,28,30,32\}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(4) Subtra: I still can't determine the numbers.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point of the conversation, we are left with the following options for the pairs $(x,y)$: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$r=2$: The pairs $(x,2x)$, where $x\in\{2,3,\ldots,50\}$. Then $d=x$.&lt;/li&gt;&#xA;&lt;li&gt;$r=26$: The pairs $(2,52)$ and $(3,78)$. Then $d=50$ or $d=75$.&lt;/li&gt;&#xA;&lt;li&gt;$r=28$: The pairs $(2,56)$ and $(3,84)$. Then $d=54$ or $d=81$.&lt;/li&gt;&#xA;&lt;li&gt;$r=30$: The pairs $(2,60)$ and $(3,90)$. Then $d=58$ or $d=87$. &lt;/li&gt;&#xA;&lt;li&gt;$r=32$: The pairs $(2,64)$ and $(3,96)$. Then $d=62$ or $d=93$. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Since Subtra is not able to determine the numbers $x$ and $y$ from this, her value $d$ must be compatible with two distinct pairs $(x,y)$ in this enumeration. &#xA;This reduces the possible options for $(x,y)$ to the following short list:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$r=2$ and $d=50$:  The pair $(50,100)$.&lt;/li&gt;&#xA;&lt;li&gt;$r=26$ and $d=50$: The pair $(2,52)$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Subtra's second statement means that $d=50$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(5) Divvy: At the very beginning of this conversation, I already knew that you did not know the numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are only left with two possible values $r=2$ and $r=26$ for the ratio.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $r=2$, then at the very beginning of the conversation the pair $(x,y)=(2,4)$ with difference $d=2$ would have been a feasible option for Divvy.&#xA;But then Subtra would have been able to deduce the two numbers right away,&#xA;as $d=(r-1)x=2$ and $x\ge2$; a contradiction.&lt;/li&gt;&#xA;&lt;li&gt;If $r=26$, then at the very beginning of the conversation only the two pairs $(x,y)=(2,52)$ with $d=50$ and $(x,y)=(3,78)$ with $d=75$ would have been feasible options for Divvy. &#xA;The first difference $d=50=(r-1)x$ is compatible with $(r,x)=(26,2)$, as well as with $(r,x)=(6,10)$. &#xA;The second difference $d=75=(r-1)x$ is compatible with $(r,x)=(26,3)$, as well as with $(r,x)=(6,15)$.&#xA;In either case, at the very beginning of the conversation Subtra is not able to deduce the numbers from the difference $d$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Divvy's third statement implies $r=26$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(6) Subtra: But now I can determine the numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subtra now knows that $d=50$ and $r=26$, and deduces the numbers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Answer:&lt;/strong&gt; The magic numbers are $x=2$ and $y=52$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-03-05T10:55:32.483" LastActivityDate="2015-03-05T10:55:32.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9549" PostTypeId="2" ParentId="9547" CreationDate="2015-02-24T21:13:46.507" Score="8" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; water&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I look at the sky and my eyes reflect it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can see the sky reflected in water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can bring death but also life&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water can drown people, but is also necessary for life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Achilles' mother used my kind&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Achilles' mother dipped him into water, while holding him by the heel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My bed is smooth as sand itself&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is sand at the bottom of an ocean.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be kind and allow you to pass&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the water is calm, you can navigate it easily.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I am angry, you won't make it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the water is rough, your boat may capsize.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2015-02-24T21:13:46.507" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9550" PostTypeId="2" ParentId="9547" CreationDate="2015-02-24T21:31:09.533" Score="25" Body="&lt;p&gt;You are a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; river&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I look at the sky and my eyes reflect it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the surface of a river can reflect the sky&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can bring death but also life&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rivers have life-giving water but can also kill people&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Achilles' mother used my kind&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The river Styx that Achilles was dipped in&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My bed is smooth as sand itself&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Riverbeds can be made of sand&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be kind and allow you to pass /&#xA;When I am angry, you won't make it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rivers can be gentle and fordable in some places/times, and too violent to cross at other places/times&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9783" LastActivityDate="2015-02-24T21:31:09.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9551" PostTypeId="1" AcceptedAnswerId="9557" CreationDate="2015-02-24T21:36:27.237" Score="11" ViewCount="1302" Body="&lt;p&gt;The numbers above the line and below the line have a specific property to them:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{c}&#xA;       1\ 3\ 5\ 6\ 7\ 9\ 10\ 11\ 12\ 13\ \ldots \\&#xA;\hline 2\ 4\ 8\ 15\ 16\ 21\ 30\ 32\ 33\ \ldots &amp;amp; &#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every natural number has a place either above or below the line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is this property? What are the next three numbers on each line? Would the numbers 29, 52, and 138 go above or below the line?&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="5373" LastEditDate="2015-02-25T00:09:09.067" LastActivityDate="2015-02-25T01:04:28.750" Title="Why are these numbers above the line?" Tags="|mathematics|pattern|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="9552" PostTypeId="2" ParentId="9131" CreationDate="2015-02-24T21:53:53.253" Score="3" Body="&lt;p&gt;Make a line segment starting at $A$, through $D$.&lt;br&gt;&#xA;Make a line segment starting at $C$, through $B$.&lt;br&gt;&#xA;Call the point where they intersect $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pOsF7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The triangle $EFG$ has the same area as the quadrilateral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof:&#xA;Extend $ED$ and draw a line segment starting at $B$ and running parallel to $CD$, call the point where they intersect $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Pzy9J.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that triangle $EFX$ has the same area as the quadrilateral because triangle $DFB$ and triangle $DFX$ have the same area (same base and height). Unfortunately, I can't draw line $BX$, but I can use the same proof by splitting the quadrilateral into 2 quadrilaterals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Point $E$ bisects segment $AB$. Therefore, if I draw a line parallel to $AD$ starting at $E$, it will bisect $DB$.&lt;br&gt;&#xA;Similarly, point $F$ bisects segment $CD$. Therefore, if I draw a line parallel to $CB$ starting at $F$, it will bisect $BD$.&lt;br&gt;&#xA;Call that point $H$ and draw a line starting at point $G$, through $H$ until it intersects $EF$. Call that point $I$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/l0bZT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$DG$ is parallel to $EH$ and therefore $EIG$ has the same area as $IEDH$.&lt;br&gt;&#xA;$BG$ is parallel to $FH$ and therefore $IFG$ has the same area as $IFBH$.&lt;br&gt;&#xA;Together this means that $EFG$ has the same area as $EFBD$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lines drawn in the proof can't be drawn using just a ruler, but that is unimportant.&lt;/p&gt;&#xA;" OwnerUserId="375" LastEditorUserId="375" LastEditDate="2015-02-25T00:07:52.463" LastActivityDate="2015-02-25T00:07:52.463" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9553" PostTypeId="1" AcceptedAnswerId="9555" CreationDate="2015-02-24T22:19:21.970" Score="12" ViewCount="900" Body="&lt;p&gt;The answers to the clues will fit in the grid. Taken together, the answers hide one final clue.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Born on Halloween&lt;/li&gt;&#xA;&lt;li&gt;Spirit board&lt;/li&gt;&#xA;&lt;li&gt;Hamelin vermin&lt;/li&gt;&#xA;&lt;li&gt;Republic of China&lt;/li&gt;&#xA;&lt;li&gt;Story&lt;/li&gt;&#xA;&lt;li&gt;Queequeg's tool&lt;/li&gt;&#xA;&lt;li&gt;Resound?&lt;/li&gt;&#xA;&lt;li&gt;Synthetic stocking&lt;/li&gt;&#xA;&lt;li&gt;It's tasty but potentially deadly&lt;/li&gt;&#xA;&lt;li&gt;Place for basemen&lt;/li&gt;&#xA;&lt;li&gt;They're at the center of everything&lt;/li&gt;&#xA;&lt;li&gt;Quickly&lt;/li&gt;&#xA;&lt;li&gt;Bounds&lt;/li&gt;&#xA;&lt;li&gt;Carpentry byproduct&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cTMzz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2627" LastActivityDate="2015-02-24T23:06:55.977" Title="Crossword with a hidden clue" Tags="|crosswords|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9554" PostTypeId="2" ParentId="9515" CreationDate="2015-02-24T22:44:56.707" Score="4" Body="&lt;p&gt;General remarks:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The location must be somewhere temperate as we see a tree bare&#xA; of leaves, so let's assume it's somewhere like Yorkshire.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What time of day is it?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It must be about noon. Yorkshire is about 54 degrees North, so&#xA; the maximum height of the sun at noon near the equinox is about&#xA; 36 degrees, and the shadows seem to be less than the height of the&#xA; objects casting them, so the sun must be very high in the sky for&#xA; early Spring or late Autumn.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is it early spring or a late fall(autumn)?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Spring, because the farming activity looks like preparation of the&#xA; ground rather than harvesting. However, whatever is being done seems&#xA; to leave no mark, since the farmer and tractor are in the middle of&#xA; the field, yet neither left nor right is there a visible difference&#xA; showing the part already worked on.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is this river navigable?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Obviously. There's a boat in it. But navigable is a relative term. It's&#xA; very unlikely to be navigable for an aircraft carrier.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which direction does the river flow? (North, East, South or West?)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since it's close to noon, the sun must be to the South, and the&#xA; river is flowing left to right, so it must flow South.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is the river deep or shallow at the side where the boat is?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Compared to what? It's probably about waist deep. Where rivers are&#xA; straight they tend to slope down from the bank. The presence of the&#xA; jetty proves that it's not deep enough to get the boat right up to&#xA; the side. It looks like there's a float on the boy's fishing line, so&#xA; within the reach of his ability to cast, the water must get deep enough&#xA; that the hook is still above the bottom - probably over 3m. Note that&#xA; the total length of a fishing line on a river is irrelevant, as it&#xA; is often used to cast futher away from the bank rather than to get deeper.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is there a bridge across the river somewhere nearby?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's impossible to tell. The people using the boat suggest that they&#xA; cannot use a bridge, but that may be be because they're going to an&#xA; island in the river whereas the bridge is further downstream and only&#xA; goes to the other side. Also, there could be a railway bridge which&#xA; is not open to foot traffic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How far is the railroad from here?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's very difficult to tell. The man on the left might be a railway&#xA; worker, but even if he is, he's carrying quite a large case in his&#xA; left hand, so he could be travelling some considerable distance &#xA; from it. Either that or he has an extraordinarily large lunchbox.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Do birds fly North or South?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some do, some don't. Some overwinter in England, having come from&#xA; further North, some spend the summer in England and overwinter &#xA; in Africa. The ones in the picture are flying more-or-less West.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9786" LastActivityDate="2015-02-24T22:44:56.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9555" PostTypeId="2" ParentId="9553" CreationDate="2015-02-24T23:06:55.977" Score="15" Body="&lt;p&gt;Born on Halloween&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SCORPIO&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Spirit board&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; OUIJA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hamelin vermin&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RATS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Republic of China&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TAIWAN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Story&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TALE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Queequeg's tool&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HARPOON&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Resound?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ECHO&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Synthetic stocking&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; NYLON&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's tasty but potentially deadly&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; FUGU&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Place for basemen&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; INFIELD&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They're at the center of everything&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; NUCLEI&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Quickly&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ASAP&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bounds&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LEAPS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Carpentry byproduct&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SAWDUST&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mystery Word&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first letters of the answers spell SORT THEN FINALS. Sorting the clues in alphabetical order, the final letters then spell POUNDS IN A STONE.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the missing word is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; FOURTEEN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4518" LastActivityDate="2015-02-24T23:06:55.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9556" PostTypeId="1" AcceptedAnswerId="9589" CreationDate="2015-02-25T00:09:50.580" Score="12" ViewCount="1831" Body="&lt;p&gt;(This riddle was inspired by &lt;a href=&quot;https://puzzling.stackexchange.com/users/9319/bulldogg6404&quot;&gt;bulldogg6404&lt;/a&gt;'s &lt;a href=&quot;https://puzzling.stackexchange.com/questions/9234/i-dont-like-my-neighbors-riddle&quot;&gt;&quot;I Don't Like My Neighbors&quot;&lt;/a&gt;!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My neighbors are wacky!&lt;br&gt;&#xA;One of them's a single-minded psycho who just likes cutting things up!&lt;br&gt;&#xA;The other one's two-faced and doesn’t like children.&lt;br&gt;&#xA;They both have neighbors like them, but neither of them are like me.&lt;br&gt;&#xA;I guess I’ll just continue being I...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;" OwnerUserId="9783" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2015-02-25T22:37:38.493" Title="My Wacky Neighbors (What am I?)" Tags="|riddle|" AnswerCount="6" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9557" PostTypeId="2" ParentId="9551" CreationDate="2015-02-25T00:31:12.010" Score="4" Body="&lt;p&gt;Numbers are below the line exactly if their number of distinct odd prime factors is even, except $1$ goes above the line. Equivalently, numbers are below the line when the sum of their distinct prime factors is even and positive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $29$ would go above the line (odd prime), $52 = 2^2 \times 13$ would go above, and  $138 = 2 \times 3 \times 23$ would go below.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-02-25T01:04:28.750" LastActivityDate="2015-02-25T01:04:28.750" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9558" PostTypeId="2" ParentId="9547" CreationDate="2015-02-25T02:05:38.297" Score="3" Body="&lt;p&gt;Similar with the other,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sea (Ocean)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I look at the sky and my eyes reflect it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sea reflected the sky, that's why the color is blue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can bring death but also life&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sea is where the lifeforms begin, but also dangerous.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Achilles' mother used my kind&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Achilles' mother &lt;a href=&quot;http://en.wikipedia.org/wiki/Thetis&quot; rel=&quot;nofollow&quot;&gt;Thetis is sea-Goddess&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My bed is smooth as sand itself&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Seabed or ocean floor.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be kind and allow you to pass&#xA;When I am angry, you won't make it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Weather in the sea is unpredictable. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9160" LastActivityDate="2015-02-25T02:05:38.297" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9559" PostTypeId="2" ParentId="9533" CreationDate="2015-02-25T02:21:42.873" Score="12" Body="&lt;p&gt;If this is correct, feel free to add in to the first answer.  Continuing with the excellent work by Engineer Toast, I think this completes the second part of the puzzle:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The final number is &lt;strong&gt;3&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; From the clue on the OP's profile page, the numbers (143353-486602) are coordinates for the Dutch RD (Rijks-Driehoek) system.&lt;br&gt;&#xA; Using this &lt;a href=&quot;http://www.regiolab-delft.nl/?q=node/36&quot; rel=&quot;noreferrer&quot;&gt;site&lt;/a&gt; for conversion provides WGS84 coordinates of (52.367180,5.216201) which is in the city of Almere, Netherlands.&lt;br&gt;&#xA; With Google satellite, the coordinates are exactly at the number 3 as shown in this photo.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/EC2mA.png&quot; alt=&quot;AlmerePhoto&quot;&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastActivityDate="2015-02-25T02:21:42.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9560" PostTypeId="2" ParentId="9537" CreationDate="2015-02-25T03:54:55.227" Score="3" Body="&lt;p&gt;My answer would be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Nascar driver, they only go left once they left home and enter the circuit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aqQBr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9792" LastActivityDate="2015-02-25T03:54:55.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9561" PostTypeId="2" ParentId="9556" CreationDate="2015-02-25T04:58:31.960" Score="7" Body="&lt;p&gt;I think that you are a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;spoon&lt;/strong&gt; at a table setting with knives and forks&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One of them's a single-minded psycho who just likes cutting things up!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This describes the knife.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The other one's two-faced and doesn’t like children.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This describes the fork.  Two-faced and FORKed tongue are both expressions for lying and deceit.  Also, children do not usually use forks.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They both have neighbors like them, but neither of them are like me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This suggests different types of knifes (dinner, butter, steak) and different types of forks (salad, dinner, lobster). But none are like a spoon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-02-25T06:32:26.697" LastActivityDate="2015-02-25T06:32:26.697" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9562" PostTypeId="2" ParentId="9556" CreationDate="2015-02-25T05:57:19.777" Score="0" Body="&lt;p&gt;I think you are one among&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/nFOsc.gif&quot; alt=&quot;enter image description here&quot;&gt;&#xA; A non value card of a particular suit and value ex. 2 of spades.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One of them's a single-minded psycho who just likes cutting things up!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/bIHsl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA; In a game of bridge (particularly NT), an Ace of a colour rules  over the rest. For non-NT game, an ACE of the trump rules can &lt;strong&gt;CUT&lt;/strong&gt; any hand. And &lt;strong&gt;single-minded&lt;/strong&gt; because Ace is often counted as 1 before the number cards 2,3 ..... as in A, 2, 3, ....... 10, J, Q, K&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The other one's two-faced and doesn’t like children.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/hZSZh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA; Ohh never heard them conjugating and having children.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They both have neighbors like them, but neither of them are like me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Any card is unique and no two are the same.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4177" LastEditorUserId="4177" LastEditDate="2015-02-25T06:04:19.703" LastActivityDate="2015-02-25T06:04:19.703" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9563" PostTypeId="2" ParentId="9556" CreationDate="2015-02-25T06:16:22.937" Score="2" Body="&lt;p&gt;I'm not too confident, but @Abhijit gave me an idea with his deck of cards thought:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;The Joker in a deck of cards.&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My neighbors are wacky!&lt;br&gt;&#xA;One of them's a single-minded psycho who just likes cutting things up!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The ace in most cases is valued at one point, and as mentioned in his answer, performs cuts in the game of Bridge.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The other one's two-faced and doesn't like children.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Any face card in a standard deck (King, Queen, Jack) has two faces, and I guess they don't like kids because there aren't any represented in the deck. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They both have neighbors like them, but neither of them are like me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are multiple of each card in a standard deck, which is how it is referenced here as having neighbors like them. Admittedly, despite most games that use the Joker only need one in the deck, there are commonly two Jokers provided, so this doesn't quite work either. &lt;em&gt;However&lt;/em&gt;, the thing each Joker has all to itself is a lack of suit. The face cards and aces both have many neighbors of the same suit, but the Joker - either one - has no suit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I guess I'll just continue being I...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The joker, being a bit of a clown, is considered in royal stories to be having a mindset all his own, and would think of the rest of the world as being strange for not sharing in his quirky nature. This might also explain why he thinks his neighbors are wacky.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If this is correct, I'd like it to be appended to @Abhijit's post and give him best-answer.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Edit: Some ideas for myself when I wake up or anyone else who happens upon this: imaginary number i, arithmetic operators, or Iodine on the periodic table.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9319" LastEditorUserId="9319" LastEditDate="2015-02-25T06:28:07.203" LastActivityDate="2015-02-25T06:28:07.203" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9564" PostTypeId="2" ParentId="9351" CreationDate="2015-02-25T07:38:01.080" Score="1" Body="&lt;p&gt;Now that I have seen the correct answer, I would argue that this fits the hints more accurately:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am always moving,&#xA;but I never know where I am going.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sun is always moving away from the center of the universe, but it does not know were it is going. It is also constantly moving when observed from the surface of the earth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I provide delicious flavor to food.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A lot of delicious flavor of food is done with sun drying and photosynthesis creates sugar which is arguably delicious.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be read on many restaurant menus.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In dishes and with sunflower and sun dried tomatoes for instance. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, I am toxic.&#xA;I kill thousands of people every year,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sunburn is toxic and can evolve into skin cancer. Being directly in the sun for too long can be deadly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, it is difficult to survive in the wild without me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; With no sunlight there is less vegetation and nutrition and the sun can be used to navigate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning why this is more accurate:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Smoke is technically not moving when there is no convection. Smoke is a term and different smoke is used in the different hints, while the sun is literally just one thing in all of the hints.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9686" LastActivityDate="2015-02-25T07:38:01.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9566" PostTypeId="1" CreationDate="2015-02-25T08:11:53.613" Score="-4" ViewCount="222" Body="&lt;p&gt;Six numerals and two &lt;strong&gt;?&lt;/strong&gt; symbols are arranged in a $2\times4$ rectangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fV6lP.jpg&quot; alt=&quot;photograph of the puzzle&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;? ?&#xA;8 9&#xA;5 6&#xA;2 3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Apparently, either one or two different symbols are needed to replace the two &lt;strong&gt;?&lt;/strong&gt; symbols.  What are they?&lt;/p&gt;&#xA;" OwnerUserId="9798" LastEditorUserId="25897" LastEditDate="2017-03-22T02:49:57.030" LastActivityDate="2017-03-22T02:49:57.030" Title="A rectangle of numbers" Tags="|mathematics|number-sequence|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9567" PostTypeId="2" ParentId="9566" CreationDate="2015-02-25T08:20:21.177" Score="6" Body="&lt;p&gt;The grid of numbers is&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;?  ?&#xA;8  9&#xA;5  6&#xA;2  3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Each number is three greater than the number below it and one greater than the number to its left (if there is one). Alternatively, think of each column as an arithmetic progression $2,5,8,?$ and $3,6,9,?$ (or reversed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the missing numbers are 11 and 12.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;11 12&#xA;8  9&#xA;5  6&#xA;2  3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-02-25T08:20:21.177" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9572" PostTypeId="2" ParentId="9566" CreationDate="2015-02-25T09:01:22.460" Score="8" Body="&lt;p&gt;My guess would be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; / and *&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Look at the numeric keypad on your keyboard. Above 8 there is a slash &lt;strong&gt;(/)&lt;/strong&gt; and above 9 there is an asterisk &lt;strong&gt;(*)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9352" LastActivityDate="2015-02-25T09:01:22.460" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9573" PostTypeId="2" ParentId="9556" CreationDate="2015-02-25T09:53:13.567" Score="3" Body="&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;strong&gt;rock&lt;/strong&gt; of rock, paper, scissors.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My neighbors are wacky!&lt;br&gt;&#xA;One of them's a single-minded psycho who just likes cutting things up!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The scissors. No explanation needed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The other one's two-faced and doesn't like children.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The paper. Two faced and children usually don't are very kind with it :(&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4652" LastActivityDate="2015-02-25T09:53:13.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9574" PostTypeId="1" AcceptedAnswerId="9577" CreationDate="2015-02-25T09:58:14.743" Score="4" ViewCount="1214" Body="&lt;p&gt;What is the maximum number of knights that can be positioned on a $5\times5$ chess board, so that each knight attacks exactly two other knights?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-10-04T16:12:43.127" LastActivityDate="2015-10-04T16:12:43.127" Title="Knights on a 5x5 chess board" Tags="|mathematics|combinatorics|checkerboard|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9575" PostTypeId="2" ParentId="9556" CreationDate="2015-02-25T10:09:25.473" Score="2" Body="&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The tongue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My neighbors are wacky!&lt;br&gt;&#xA;One of them's a single-minded psycho who just likes cutting things up!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The incisor teeth, used to cut meat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The other one's two-faced and doesn't like children.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wisdom tooth. Two faced because it's called wisdom tooth but comes only with pain (ah! I trusted you, wisdom tooth!). Or because it should help the already existing molars but in fact is useless. Also children doesn't have it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They both have neighbors like them, but neither of them are like me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are 32 teeth with various functions, but no one is like the tongue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4652" LastActivityDate="2015-02-25T10:09:25.473" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9576" PostTypeId="2" ParentId="9537" CreationDate="2015-02-25T10:12:36.350" Score="2" Body="&lt;p&gt;You are a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; key&#xA;You turn right, unlocking the door, then you leave home (the key had been turned only once becuase someone was home).Then you leave home and, from outside, you turn &quot;to the left&quot; three times, to lock the door&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5102" LastActivityDate="2015-02-25T10:12:36.350" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9577" PostTypeId="2" ParentId="9574" CreationDate="2015-02-25T10:12:55.950" Score="12" Body="&lt;p&gt;Here's 16:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;. X X X .&#xA;X X . X X&#xA;X . . . X&#xA;X X . X X&#xA;. X X X .&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="8874" LastEditDate="2015-03-14T18:46:20.920" LastActivityDate="2015-03-14T18:46:20.920" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="9580" PostTypeId="1" AcceptedAnswerId="9587" CreationDate="2015-02-25T10:58:05.370" Score="15" ViewCount="1076" Body="&lt;p&gt;Little Red Riding Hood received this odd recipe from her grandmother.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hello. Here's a cake recipe you might enjoy! &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;&lt;em&gt;Orange cake:&lt;/em&gt;&lt;/strong&gt; &lt;em&gt;(serves 4)&lt;/em&gt;  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Ingredients:&lt;/strong&gt;&lt;br&gt;&#xA;  - 111 cups caster sugar&lt;br&gt;&#xA;  - 112mL orange flavoring&lt;br&gt;&#xA;  - 216 oranges&lt;br&gt;&#xA;  - 102 eggs&lt;br&gt;&#xA;  - 83 cups milk&lt;br&gt;&#xA;  - 104g orange icing&lt;br&gt;&#xA;  - 46g vanilla extract&lt;br&gt;&#xA;  - 64 cups self-raising flour&lt;br&gt;&#xA;  - 202 orange rinds&lt;br&gt;&#xA;  - 87g butter&lt;br&gt;&#xA;  - 110 cups almond meal&lt;br&gt;&#xA;  - 100mL orange juice  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Method:&lt;/strong&gt;&lt;br&gt;&#xA;  Put the butter in the bowl. Put the caster sugar in the bowl. Put half the oranges in the bowl. Put the eggs in the bowl. Pour the vanilla extract in the bowl. Put half the self-raising flour in the bowl. Pour the milk in the bowl. Put half the orange rinds in the bowl. Put the almond meal in the bowl. Pour the orange juice in the bowl. Put the rest of the self-raising flour in the bowl. Put the icing in the bowl. Put the rest of the orange rinds in the bowl. Put the rest of the oranges in the bowl. Pour the orange flavoring in the bowl. Stir vigorously. Eat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, something seems a little off to Little Red Riding Hood...&lt;br&gt;&#xA;&lt;strong&gt;&lt;em&gt;Can you discover the secret message?&lt;/em&gt;&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="7308" LastEditorUserId="8874" LastEditDate="2015-02-25T17:27:37.417" LastActivityDate="2015-02-25T17:27:37.417" Title="A Curious Recipe" Tags="|cipher|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9581" PostTypeId="2" ParentId="9574" CreationDate="2015-02-25T12:20:51.490" Score="4" Body="&lt;p&gt;xnor's answer is correct. Here is why:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rBLTJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every position has a number that corresponds with the amount of knights a knight can attack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best strategy is to remove the knight which can be attacked my most knights until every knight can only hit two other knights.&lt;/p&gt;&#xA;" OwnerUserId="5127" LastActivityDate="2015-02-25T12:20:51.490" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="9582" PostTypeId="1" AcceptedAnswerId="9592" CreationDate="2015-02-25T12:34:18.283" Score="1" ViewCount="398" Body="&lt;p&gt;My friend Raffi likes to tell the story of his trip to London, from more than twenty years ago. When he was visiting Buckingham Palace, he suddenly bumped into Queen Elizabeth II.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Queen was not amused and asked: &quot;And who are you, Sir?&quot;&lt;br&gt;&#xA;To which Raffi quick-wittedly answered: &quot;I? Raffi, Maaaam!&quot;&lt;br&gt;&#xA;Raffi always stresses that he said &quot;Maaaam&quot; with a long, stretched, quadrupled &quot;aaaa&quot; in the middle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Raffi later turned his adventure into a cryptarithm:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   I x RAFFI = MAAAAM&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Each letter represents a unique digit.   &lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="9404" LastEditDate="2016-03-03T10:20:49.420" LastActivityDate="2016-03-03T10:20:49.420" Title="When Raffi met the Queen" Tags="|logical-deduction|alphametic|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9584" PostTypeId="2" ParentId="9582" CreationDate="2015-02-25T12:46:16.777" Score="2" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A = 2&lt;br&gt;&#xA; F = 7&lt;br&gt;&#xA; I = 8&lt;br&gt;&#xA; M = 4&lt;br&gt;&#xA; R = 5  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $52778 * 8 = 422224$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $I != 1$, because then $MAAAAM = RAFFI$&lt;br&gt;&#xA; $I != 2$ or $3$, because $I*I$ should be more than 10(otherwise we get that $F*I=A$ twice and then $A*I=A$, so in this case $I=1$&lt;br&gt;&#xA; Also, $I != 5$ and $I != 6$, because $5^2 = 25$ and $6^2 = 36$, so that would make $I = A$&lt;br&gt;&#xA; Then I only had 4 possible values for $I$: $4, 7, 8, 9$&lt;br&gt;&#xA; Then I had to find $I$ so that&lt;br&gt;&#xA; $(I^2)MOD10 = M$&lt;br&gt;&#xA; $(F*I)MOD10 + I^2/10 = A$&lt;br&gt;&#xA; $(F*I)MOD10 + (F*I+(I^2)MOD10)/10 = A$&lt;br&gt;&#xA; From here we can see that $I^2/10 = (F*I+(I^2)MOD10)/10$, so if:&lt;br&gt;&#xA; $I=4$ &lt;em&gt;THEN&lt;/em&gt; $1 = (4F+6)/10$ ==&gt; $F=1$&lt;br&gt;&#xA; $I=7$ &lt;em&gt;THEN&lt;/em&gt; $4 = (7F+9)/10$ ==&gt; $F=31/7$&lt;br&gt;&#xA; $I=8$ &lt;em&gt;THEN&lt;/em&gt; $6 = (8F+4)/10$ ==&gt; $F=7$&lt;br&gt;&#xA; $I=9$ &lt;em&gt;THEN&lt;/em&gt; $8 = (9F+1)/10$ ==&gt; $F=79/9$&lt;br&gt;&#xA; And because each letter represents only 1 digit, we can assume that $I=4$ or $I=8$.&lt;br&gt;&#xA; If $I=4$, then $M=6$. Then we have to find $F$ that will give the following: $(F*I+(I^2)MOD10)MOD10=A$ and $(F*I+(F*I+(I^2)MOD10)/10)MOD10=A$&lt;br&gt;&#xA; Because $I=4$ and $M=6$, $F$ can be anything but $4, 6$.&lt;br&gt;&#xA; Also, $F*I+(I^2)MOD10&amp;gt;10$, so $4F+1&amp;gt;10$ ==&gt; $F4&amp;gt;=9$ ==&gt; $F&amp;gt;9/4$ ==&gt; $F&amp;gt;=3$. Now we can inspect each possibility:&lt;br&gt;&#xA; $F=3$ ==&gt; $13MOD10=3$, $(12 + 13/10)MOD10 = 5$, $5!=3$. Next:&lt;br&gt;&#xA; $F=5$ ==&gt; $21MOD10=1$, $(20 + 21/10)MOD10 = 2$, $1!=2$. Next:&lt;br&gt;&#xA; $F=7$ ==&gt; $36MOD10=6$, $(35 + 36/10)MOD10 = 8$, $6!=8$. Next:&lt;br&gt;&#xA; $F=8$ ==&gt; $41MOD10=1$, $(40 + 41/10)MOD10 = 4$, $1!=4$. Next:&lt;br&gt;&#xA; $F=9$ ==&gt; $46MOD10=6$, $(45 + 46/10)MOD10 = 9$, $6!=9$. So $I!=4$.&lt;br&gt;&#xA; Then probably $I=8$. That means that $M=4$ and $F*I+(I^2)MOD10&amp;gt;10$ ==&gt; $8F+4&amp;gt;10$ ==&gt; $8F&amp;gt;6$ ==&gt; $F&amp;gt;=1$&lt;br&gt;&#xA; $F=1$ ==&gt; $12MOD10=2$, $(8+ 12/10)MOD10 = 9$, $2!=9$. Next:&lt;br&gt;&#xA; $F=2$ ==&gt; $17MOD10=7$, $(16 + 17/10)MOD10 = 7$, $7=7$, so $F=7$.&lt;br&gt;&#xA; Substitute $I=8$, $M=4$ and $F=7$ in the formula:&lt;br&gt;&#xA; $RA778 * 8 = 4AAAA4$. From here we can find $A$:&lt;br&gt;&#xA; $A = (7 * 8 + (8*8)/10)MOD10 = 2$&lt;br&gt;&#xA; Now that we know everything except for $R$, we can simply use calculator:&lt;br&gt;&#xA; $422224 / 8 = 52778$, so $R=5$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9352" LastEditorUserId="9352" LastEditDate="2015-02-25T14:02:45.013" LastActivityDate="2015-02-25T14:02:45.013" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9586" PostTypeId="2" ParentId="9224" CreationDate="2015-02-25T13:04:48.340" Score="0" Body="&lt;p&gt;My first thought was&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A spider&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am thin and frail.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most certainly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Scratching my head can kill me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; From what I gather, if you stroke a spider's head, they &quot;go bald&quot;. I presume enough of this can kill them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When abused, I can also kill you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Spiders can be deadly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Usually, I'm helpful.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; House spiders are especially helpful at killing house flies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9802" LastActivityDate="2015-02-25T13:04:48.340" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9587" PostTypeId="2" ParentId="9580" CreationDate="2015-02-25T13:31:30.103" Score="15" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&quot;Wolf. Send help&quot;&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The explanation is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Interpret the (full or halfed) quantities of ingredients as ASCII code numbers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is the full translation:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;W = 87g = Put the butter in the bowl.  &lt;/li&gt;&#xA;&lt;li&gt;o = 111 cups = Put the caster sugar in the bowl.   &lt;/li&gt;&#xA;&lt;li&gt;l = 108 = Put half the oranges in the bowl.   &lt;/li&gt;&#xA;&lt;li&gt;f = 102 = Put the eggs in the bowl.   &lt;/li&gt;&#xA;&lt;li&gt;full stop = 46g = Pour the vanilla extract in the bowl.   &lt;/li&gt;&#xA;&lt;li&gt;blank = 32 cups = Put half the self-raising flour in the bowl.   &lt;/li&gt;&#xA;&lt;li&gt;S = 83 cups = Pour the milk in the bowl.   &lt;/li&gt;&#xA;&lt;li&gt;e = 101 = Put half the orange rinds in the bowl.   &lt;/li&gt;&#xA;&lt;li&gt;n = 110 cups = Put the almond meal in the bowl.   &lt;/li&gt;&#xA;&lt;li&gt;d = 100mL = Pour the orange juice in the bowl.   &lt;/li&gt;&#xA;&lt;li&gt;blank  = 32 cups = Put the rest of the self-raising flour in the bowl.   &lt;/li&gt;&#xA;&lt;li&gt;h = 104g = Put the icing in the bowl.   &lt;/li&gt;&#xA;&lt;li&gt;e = 101 = Put the rest of the orange rinds in the bowl.   &lt;/li&gt;&#xA;&lt;li&gt;l = 108 = Put the rest of the oranges in the bowl.   &lt;/li&gt;&#xA;&lt;li&gt;p = 112mL = Pour the orange flavoring in the bowl.   &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-25T13:37:01.703" LastActivityDate="2015-02-25T13:37:01.703" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9589" PostTypeId="2" ParentId="9556" CreationDate="2015-02-25T14:20:14.540" Score="8" Body="&lt;p&gt;Close to Narmer's answer, I think you are a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Human Adult Canine Tooth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One of them's a single-minded psycho who just likes cutting things up!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As mentioned in Narmer's answer, the incisors are used to cut and tear meat and these also have a single edge to do so.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The other one's two-faced and doesn’t like children.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The premolars.  They are two edges (faces) and they are in permanent teeth but not primary (baby) teeth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They both have neighbors like them, but neither of them are like me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Both the incisors and premolars have another incisor or premolar next to them but there are no adjacent canines.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I guess I’ll just continue being I...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the dental formula, the top/bottom sections are written as 2.1.2.3 with the 1 being the canines (and I is the roman numeral for 1).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6140" LastActivityDate="2015-02-25T14:20:14.540" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9591" PostTypeId="2" ParentId="9531" CreationDate="2015-02-25T14:47:13.077" Score="7" Body="&lt;p&gt;It depends whether it is the pans that have different weights or the arms that have different lengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would look highly suspicious if the pans don't balance when empty.  So the second option is more likely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect of different arm lengths is that the actual weight is multiplied or divided by a factor depending which way it is used (on which side are the reference weights).  The actual weight is the geometrical mean of the two measured weights: $weight = \sqrt{20\times 45} = 30\ pounds$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the bag weighs 30 pounds (or 13.6 kg).&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2015-02-26T17:45:24.420" LastActivityDate="2015-02-26T17:45:24.420" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9592" PostTypeId="2" ParentId="9582" CreationDate="2015-02-25T14:49:51.953" Score="4" Body="&lt;p&gt;If we multiply the 5-digit number $RAFFI$ by a single-digit $I$, then the resulting 6-digit number $MAAAAM$ lies below $100000*I$.&#xA;Hence its starting digit $M$ is smaller than $I$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(1). The digits $M$ and $I$ satisfy $M&amp;lt;I$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The last digit $M$ of $MAAAAM$ coincides with the last digit of $I^2$.&#xA;Testing $I=0,1,2,3,4,5,6,7,8,9$ yields $M=0,1,4,9,6,5,6,9,4,1$.&#xA;Combining this with $M&amp;lt;I$ from (1), we conclude:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(2). Either (2a) $I=8$ and $M=4$, or (2b) $I=9$ and $M=1$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let us get rid of case (2b): In this case $R\ge2$, as digit $1$ is already blocked by $M=1$. &#xA;Hence $MAAAAM=I*RAFFI&amp;gt;9*20000=180000$, which together with $M=1$ implies $A=8$. But then $MAAAAM=188881$ is not divisible by $I=9$; a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So only case (2a) with $I=8$ and $M=4$ remains. &#xA;Since $MAAAAM$ is a multiple of $I=8$, also its last three digits $AAM$ are a multiple of $8$. &#xA;Out of the numbers $004,114,224,334,554,664,774,994$, only $224$ and $664$ work.&#xA;These two options $A=2$ and $A=6$ yield for $RAFFI=MAAAAM/8$ the respective values $52778$ (good) and $58333$ (bad). We summarize:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$8*52778=422224$ is the unique solution to Raffi's cryptarithm.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-02-25T14:49:51.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9595" PostTypeId="2" ParentId="9400" CreationDate="2015-02-25T16:20:15.207" Score="1" Body="&lt;p&gt;My rewording of the question; I think it is identical:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;First mathematician: I cannot deduce x  and y  from their difference d .&lt;/li&gt;&#xA;&lt;li&gt;Second mathematician: I knew that you cannot.&lt;/li&gt;&#xA;&lt;li&gt;First mathematician: Don't bother to say that, I already knew that you knew that I cannot. &lt;/li&gt;&#xA;&lt;li&gt;Second mathematician: Hm&lt;/li&gt;&#xA;&lt;li&gt;First mathematician: Well... you still don't know what the numbers are.&lt;/li&gt;&#xA;&lt;li&gt;Second mathematician: Ah but did I already know that you knew that? &#xA;• First mathematician: I don't know...&lt;/li&gt;&#xA;&lt;li&gt;Third mathematician: Losers... I did know the two numbers right away, once they told me the ratio r !&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This shows how I interpret the question which will be worked through below.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The above statement does not seem to have any possible answers.  In order to solve this I created a simple workbook which determines all possible $s$ values if there is a given $d$ and vice versa.  I then listed all possible values for $d$ and $s$ ignoring any of the statements.  Note that if you know $s$ and $d$ you know the numbers $x$ and $y$.  If you don't, you don't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First I counted the number of potential $s$ values for a given $d$. If the number is $1$, anyone with that $d$ would know the numbers.  There were $25$ such values of $d$ (called $list1$) and $63$ where he wouldn't know ($list2$).  The remaining $12$ are impossible as there are $0$ corresponding values of $s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second I counted the number of potential $d$ values for a given $s$ from $list1$ or $list2$.  If the number from $list1=0$ but the number from $list2&amp;gt;0$ then statement $2$ is true for that value $s$.  This $list3$ contains $110$ values of $s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next came $list4$ which would only contain $d$ values which meet statement 3. This is the list of $d$ values where all corresponding values of $s$ are from $list3$.  $list4$ has $35$ entries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Line 4 of the above formulation of the puzzle is used to indicate that the second guy has been following all of this.  $list5$ is generated as the list of all possible $s$ values from the $d$ values of $list4$.  This should be indicated by the tense (past, present, future) of the original wording of the question. $list5$ has 62 entires.  It should be noted that there are only 3 values from $list4$ which will work well with statement 7: $76$,$88$, and $92$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Line 5 has two parts:  First we generate $list6$ (whew) from $list5$ which only contains values of $s$ which could only result in $1$ corresponding values of $d$ from $list4$.  This has $30$ values and represents the cases where guy 2 currently knows the numbers.  Second we generate $list7$ of $d$ values which do not correspond with any $s$ values from $list6$.  This is the list of $d$ values for which Line 5 is correct (guy 1 knows guy 2 doesn't know) and only has $11$ values including $76$ and $92$ but not $88$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For ease of use, we generate $list8$ which contains all $d$ values which has both corresponding $s$ values which would cause guy 2 to know the numbers and corresponding $s$ values which would not cause guy 2 to know the numbers.  This list contains values where guy 1 doesn't know that guy 2 doesn't know the numbers.  (note: $d=58$ seems to be the only case where guy 1 knows that guy 2 does know the numbers.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is where we run into an issue. $d=76$ means $s=80$,$84$, or $114$. $d=92$ means $s=96$ or $100$.  All of these $s$ values correspond with $d$ values from both $list7$ and $list8$.  This means anyone with that $s$ number despite all this would not know before line 5 that line 5 was true.  The $d$ values that are still viable after line 6 (but not line 7) are: $8$, $12$, $14$, $16$, $20$, $24$, and $28$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please let me know whether you find a mistake, i misunderstood the question, and/or what you think the answer should be.  Wording always seems to be the issue with your questions like this and if you think there is an answer, this is likely the issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think if anything is misunderstood it is whether line 4 exists.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditDate="2015-02-25T16:28:23.153" LastActivityDate="2015-02-25T16:28:23.153" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9596" PostTypeId="1" AcceptedAnswerId="9641" CreationDate="2015-02-25T18:11:53.360" Score="10" ViewCount="503" Body="&lt;p&gt;I have partitioned the chemical elements into two groups based on a certain&#xA;occult property that is shared by some of them, but not by the remaining ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following elements have my occult property:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;arsenic, astatine, carbon, iron, krypton, neon, phosphorus, silicon, tin&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The following elements do not have my occult property:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;antimony, boron, cobalt, germanium, gold, helium, lead, mercury, sulfur&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which of the following elements have this occult property?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;aluminium, fluorine, francium, gallium, oxygen, polonium.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="9125" LastEditDate="2015-02-27T09:31:28.217" LastActivityDate="2015-02-27T09:31:28.217" Title="An occult property of chemical elements" Tags="|pattern|chemistry|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9597" PostTypeId="1" AcceptedAnswerId="9624" CreationDate="2015-02-25T18:17:07.603" Score="21" ViewCount="9304" Body="&lt;p&gt;Alice and Bob are perfectly logical and super intellegent.  Their professor decides to play a game with them, and he tells them: &#xA;&quot;I have chosen two numbers (integers) $x$ and $y$ with $2\le y \le x\le 100$. &#xA;I will tell Alice the value of their difference $d=x-y$ and Bob the value of their ratio $r=x/y$.&#xA;I stress that the ratio $r$ is an integer, too.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the professor has told the difference to Alice and the ratio to Bob, the following exchange occurs:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Alice: I don't know the numbers.&lt;/li&gt;&#xA;&lt;li&gt;Alice: You might though.  Do you know them?&lt;/li&gt;&#xA;&lt;li&gt;Bob: No, I don't know them.&lt;/li&gt;&#xA;&lt;li&gt;Alice: Too bad, if you did I would know them too.&lt;/li&gt;&#xA;&lt;li&gt;Alice: I still don't know them though.&lt;/li&gt;&#xA;&lt;li&gt;Bob: Me neither.&lt;/li&gt;&#xA;&lt;li&gt;Alice: Oh really?  Then I do.&lt;/li&gt;&#xA;&lt;li&gt;Bob : Dang it; I still don't.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What is the value of $r$?&lt;br&gt;&#xA;What are the possible values for $x$ and $y$?&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditDate="2016-09-28T12:46:09.390" LastActivityDate="2016-09-28T12:46:09.390" Title="Determine the number $r$ from the statements" Tags="|logical-deduction|meta-knowledge|" AnswerCount="4" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="9598" PostTypeId="2" ParentId="5960" CreationDate="2015-02-25T18:31:37.357" Score="1" Body="&lt;p&gt;They are most probably&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Going to attempt to reach the speed of light (almost)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Going by the start:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Buying fuels for high prices:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The $\Delta $is the ratio of the amount of fuel they currently have and the fuel they need (which is $\infty$, as stated), and they currently have 1000 gallons, which by their current bank balance costs $10^8$ dollars, which is the order of speed of light(which is $3*10^8$ m/s approx). As many have already pointed out, AN:92 refers to uranium, whose 1 g has approx $6.54*10^{21}$ atoms, which may be the price paid per 1 gallon of fuel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Oxygen needed:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They need it...uh...to breath? Or to burn the fuel, probably. Your name plate may go in their wall of fame or something. Still not able to figure this one out.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Post on their website:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pilots guide:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Increase of mass and speed refers to the increase of mass by a factor of $\gamma$ which is $ m=\frac{m_o}{\sqrt{1-\frac{v^2}{c^2}}}$ where $\gamma$ is equal to $\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$. Time is also dilated by the factor of $\gamma$, which is called time dilation. Communications will of course be offline because no known means of communication is faster than speed of light. The tunnel view refers to the view which you will get from inside the spaceship. Everything will look so fast as if you are in a tunnel. The time to the destination $cyears=year$ also refers to Time dilation, and this could be used to take out the speed of the spaceship which they have designed currently (or they have made). Time dilation occurs by a factor of $\gamma$, so here $c=\gamma$. $\implies v=\sqrt{c^2-1}$, which is &lt;strong&gt;very&lt;/strong&gt; close to the speed of light. This is the velocity of the spaceship by which you will travel. C is of course the speed of light, and the velocity is really high, and almost impossible to attain. And all of this was given by none other than the maker of the relation $e=mc^2$, Albert Einstein.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Technical guide:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Actual space exploration vehicles run by hydrogen fuel cells, which gets converted into water in the electrochemical reactions. The tanks maybe for storing this water. This is another reason as to why they need compressors and oxygen, to use these fuel cells for propulsion, as the fuel cells also need oxygen in the reactants side. Due to mass increase, they may need special tanks or something.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In short,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are getting an opportunity to go to space by a huge amount of fuel and are going to reach the speed of light (roughly). Since the speed of the spacecraft will be so much ($\sqrt{c^2-1} $ is very near to c), the mass will increase, time will dilate and length will contract. In short, you might not even come back and see what they were gonna pay you with. Sounds fun, doesn't it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It was quite easy ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. How is all this basic knowledge of science? Relativity is really complex, and weird as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The oxygen and the compressors are required for combustion of the loads of fuel they are buying from you, as the O.P. has pointed it out.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2808" LastEditorUserId="2808" LastEditDate="2015-02-26T22:45:46.280" LastActivityDate="2015-02-26T22:45:46.280" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9599" PostTypeId="2" ParentId="1941" CreationDate="2015-02-25T20:23:48.017" Score="2" Body="&lt;p&gt;The correct tool for the task is ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a Foucault pendulum.&lt;br&gt;&#xA; &lt;a href=&quot;http://en.wikipedia.org/wiki/Foucault_pendulum&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Foucault_pendulum&lt;/a&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; He ties a heavy object to the lamp and let it swing.  The torque on the shoelace is enough to turn the object itself, but not to change the direct&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2136" LastActivityDate="2015-02-25T20:23:48.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9603" PostTypeId="1" AcceptedAnswerId="9606" CreationDate="2015-02-26T00:16:36.917" Score="0" ViewCount="1970" Body="&lt;p&gt;I have a head, but not a body&lt;br&gt;&#xA;I don't think but I go to the top&lt;br&gt;&#xA;The fatter am I the faster I go Up&lt;br&gt;&#xA;When I'm down, people are happy&lt;br&gt;&#xA;but when I go up people are sad  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="9792" LastEditorUserId="5188" LastEditDate="2015-02-26T04:03:08.560" LastActivityDate="2015-02-26T04:03:08.560" Title="I have a head, but not a body" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9604" PostTypeId="1" AcceptedAnswerId="9609" CreationDate="2015-02-26T00:27:43.880" Score="-5" ViewCount="227" Body="&lt;p&gt;Adding an odd number would make an even number. It's not the largest prime number. But when you divide it into parts I make a multiple of 2. the range never exceeded 100 but never to less from 30. Figure me out&lt;/p&gt;&#xA;" OwnerUserId="9792" LastEditorUserId="24857" LastEditDate="2015-09-24T14:29:54.677" LastActivityDate="2015-09-24T14:29:54.677" Title="What number am I thinking of?" Tags="|mathematics|" AnswerCount="1" CommentCount="2" ClosedDate="2015-02-26T11:41:46.413" ContentLicense="CC BY-SA 3.0" />
  <row Id="9605" PostTypeId="2" ParentId="9547" CreationDate="2015-02-26T00:41:53.430" Score="2" Body="&lt;p&gt;While the other answers are of course correct, it's always fun with these to find other answers that fit.&#xA;In this case, you could make a decent argument for:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the sea-deity Aegaeon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I look at the sky and my eyes reflect it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All eyeballs reflect the sky other than those of tunnel-dwellers and the eyeless.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can bring death but also life&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He's a deity, and rather a powerful one, as well as the inventor of warships. However, he was also responsible (via the intervention of Thetis, Achilles' mother) for resolving the Hekatonkheires with the Olympian pantheon through marriage, bringing peace, and hence life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Achilles' mother used my kind&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Thetis is the one who made him broker the above deal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My bed is smooth as sand itself&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sea gods would sleep on the ocean floor.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be kind and allow you to pass When I am angry, you won't make it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The gods are fickle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7647" LastActivityDate="2015-02-26T00:41:53.430" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9606" PostTypeId="2" ParentId="9603" CreationDate="2015-02-26T01:01:50.360" Score="4" Body="&lt;p&gt;I have a head, but not a body.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;balloon looks like a head&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't think but I go to the top. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;top&quot; here refers to higher altitude, not higher rank of any organisation or company, so balloon goes higher without thinkig&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the fatter am I the faster I go Up. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the more gas or air filled balloon looks fatter and it goes up faster&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I'm Down People are Happy but when I go up People are Sad.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;when the balloon is down people play with it and are happy but when it gets cut loose from the string that the perrson is holding then people don't like loosing a balloon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I am a Balloon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9174" LastActivityDate="2015-02-26T01:01:50.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9607" PostTypeId="1" AcceptedAnswerId="9608" CreationDate="2015-02-26T02:18:06.217" Score="1" ViewCount="577" Body="&lt;p&gt;Sometimes I'm round, sometimes I'm square&lt;br&gt;&#xA;But I'm only present on happy occasions&lt;br&gt;&#xA;I'm always soft inside and beautiful on the outside&lt;br&gt;&#xA;My color variations are limitless to the creative hands&lt;br&gt;&#xA;Sadly they always like to cut me and light me with real fire   &lt;/p&gt;&#xA;" OwnerUserId="9792" LastEditorUserId="5373" LastEditDate="2015-02-26T11:40:10.180" LastActivityDate="2015-02-26T15:10:38.863" Title="Sometimes I'm round" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9608" PostTypeId="2" ParentId="9607" CreationDate="2015-02-26T02:25:25.400" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cake.&lt;br/&gt;Mostly round or square&lt;br/&gt;Only present on happy occasions&lt;br/&gt;Decoration on the top with Sponge/Cheese in centre&lt;br/&gt;Placed candles on top for b'days&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9813" LastEditorUserId="5610" LastEditDate="2015-02-26T15:10:38.863" LastActivityDate="2015-02-26T15:10:38.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9609" PostTypeId="2" ParentId="9604" CreationDate="2015-02-26T03:01:16.027" Score="1" Body="&lt;p&gt;You're not a number.&lt;br&gt;&#xA;Your an &lt;em&gt;odd&lt;/em&gt; number between 30 and 100, that is a multiple of &lt;em&gt;2&lt;/em&gt;, so either your a magical wizard dragon, or NaN.  &lt;/p&gt;&#xA;" OwnerUserId="24857" LastEditorUserId="24857" LastEditDate="2015-02-26T21:52:36.877" LastActivityDate="2015-02-26T21:52:36.877" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9610" PostTypeId="1" AcceptedAnswerId="9612" CreationDate="2015-02-26T03:09:53.813" Score="-1" ViewCount="185" Body="&lt;p&gt;I am not divisible by anything, but there's a formula to find my value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a 9 in the ten-billionths place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply me by $d$ to get $c$.&lt;/p&gt;&#xA;" OwnerUserId="24857" LastEditorUserId="4204" LastEditDate="2015-02-26T04:04:21.863" LastActivityDate="2015-02-26T04:08:42.797" Title="What number am I?" Tags="|mathematics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9612" PostTypeId="2" ParentId="9610" CreationDate="2015-02-26T03:54:46.183" Score="5" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$\pi$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Multiply me by $d$ to get $c$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pi times the diameter of a circle is the circumference.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I have a 9 in the ten-billionths place.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pi's decimal value is 3.141592653589...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;there's a formula to solve my value.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are a number of &lt;a href=&quot;http://en.wikipedia.org/wiki/Pi#Continued_fractions&quot; rel=&quot;nofollow&quot;&gt;continued fractions&lt;/a&gt; that evaluate to $\pi$.&lt;/p&gt;&#xA;" OwnerUserId="4204" LastEditorUserId="4204" LastEditDate="2015-02-26T04:08:42.797" LastActivityDate="2015-02-26T04:08:42.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9613" PostTypeId="1" CreationDate="2015-02-26T07:56:45.040" Score="-5" ViewCount="441" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;There are 7 sisters in a house. First one is reading a book, second is cooking, third is playing chess, fourth is solving sudoku, fifth is doing the laundry, sixth is watering the flowers. Question: what is seventh sister doing?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When do we say &quot;ten&quot; while looking at number 2?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A person can get it multiple times. First 2 times are free, but starting from 3rd he has to pay for it. What is it?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9352" LastActivityDate="2015-02-26T12:33:56.040" Title="A few (very) easy puzzles" Tags="|logical-deduction|" AnswerCount="2" CommentCount="4" ClosedDate="2015-02-26T14:46:44.110" ContentLicense="CC BY-SA 3.0" />
  <row Id="9614" PostTypeId="2" ParentId="9613" CreationDate="2015-02-26T08:07:40.033" Score="3" Body="&lt;p&gt;1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;seventh is playing chess with the third sister&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;10 minutes past some hour in analog clock or watch&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Hair-color: First a person gets the natural hair color(free), then he gets graying of hair(free), after this if he wishes to change the color of his hair then he has to pay for the color and/or the saloon service&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9174" LastEditorDisplayName="user9174" LastEditDate="2015-02-26T12:33:56.040" LastActivityDate="2015-02-26T12:33:56.040" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9615" PostTypeId="2" ParentId="8588" CreationDate="2015-02-26T08:31:41.327" Score="4" Body="&lt;p&gt;I am not sure, but I think you are:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;facial plastic surgery&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The German's are Experimenting  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; with a practice in Berlin, &lt;a href=&quot;http://www.aafprs.org/patient/about_us/h_father.html&quot; rel=&quot;nofollow&quot;&gt;Jacques Joseph&lt;/a&gt; is considered the father of modern facial plastic surgery   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;your holiday starts, forget happiness, no out from these mates of no friendliness  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; many people now take vacations for plastic surgery (to Brazil, Thailand, South Africa, etc)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;your past is what we think as space tortures, beginning to shrink  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; shrinking and tightening skin is usually the objective  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;follow the sheep of fashion through halls of no compassion&lt;br&gt;&#xA;little sun is gained through wreck each day it steals, the doves doth peck  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;wrecking&quot; is done before reconstruction and sunlight is to be avoided  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastActivityDate="2015-02-26T08:31:41.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9616" PostTypeId="2" ParentId="9515" CreationDate="2015-02-26T09:46:04.707" Score="5" Body="&lt;p&gt;This answer has been copied and edited out of the question, and is not produced by me. (It is thus Community Wiki, which does not earn reputation.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you look at the picture, you will see that people are sowing and not harvesting. Sowing can be done either in the fall(autumn) or early spring. Fall sowing happens when trees still have leaves, but on the picture the trees are naked, so it's &lt;strong&gt;early spring&lt;/strong&gt;.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because it's early spring, &lt;strong&gt;birds&lt;/strong&gt;(cranes) fly from South to the &lt;strong&gt;North&lt;/strong&gt;. Knowing that birds fly to the North and looking at the buoy we can say that this river is flowing &lt;strong&gt;South&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Buoys are only placed on the &lt;strong&gt;navigable&lt;/strong&gt; rivers to show danger areas and let ships navigate easier/safer.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Shadow of the tree shows that the Sun is somewhere South-East. During the spring, that position of the Sun happens normally &lt;strong&gt;around 8:00-10:00&lt;/strong&gt;.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The guy with the lantern is the &lt;strong&gt;railroad&lt;/strong&gt; conductor, so he probably lives(or works) &lt;strong&gt;somewhere nearby&lt;/strong&gt;.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The pier and the ferry show that there is &lt;strong&gt;no bridge&lt;/strong&gt; nearby, otherwise people would just walk on the bridge. Similarly, the fishing boy has set his float quite far from the hooks, so we can assume that &lt;strong&gt;river&lt;/strong&gt; there is fairly &lt;strong&gt;deep&lt;/strong&gt;. If it would be shallow, the boy would put his float much closer to the hooks, so that hooks are not on the ground in the water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user20" LastActivityDate="2015-02-26T09:46:04.707" CommentCount="1" CommunityOwnedDate="2015-02-26T09:46:31.787" ContentLicense="CC BY-SA 3.0" />
  <row Id="9617" PostTypeId="2" ParentId="9597" CreationDate="2015-02-26T10:40:14.240" Score="0" Body="&lt;p&gt;&lt;strong&gt;EDIT: This answer is wrong but is kept in case someone thinks the same!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this question is flawed. This is my thought process:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From statement 1 we only know that $d$ is not $98$ with $x = 100$ and $y = 2$ because that is the only $d$ that has one solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we need to find what are the possible values of $r$ which only has one solution. This is only the case when $r &amp;gt; 33$ and $r &amp;lt; 50$ because if $r &amp;lt;= 33$ then there are at least two solutions taking $y = 2$ or $y = 3$ because $3 * 34 &amp;gt; 100$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $r &amp;lt;= 33$ according to statement 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statement 4 is true but doesn't make sense to say. This statement is something that Bob also knew because if Bob knows the solution in statement 3 than $y$ is always $2$ and $x = d + 2$. And he knows how Alice thinks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statement 5 also has no added value because it follows from statement 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statement 6 is also logically true because no information is gained since statement 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statement 7 can't possibly be true then&lt;/p&gt;&#xA;" OwnerUserId="1748" LastEditorUserId="1748" LastEditDate="2015-02-26T11:18:34.357" LastActivityDate="2015-02-26T11:18:34.357" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9618" PostTypeId="2" ParentId="9613" CreationDate="2015-02-26T12:18:31.897" Score="4" Body="&lt;p&gt;1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Posting about her family on stackexchange. In this wondrous modern age, sister #3 could be playing chess with just about anyone on the planet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A set of teeth. Deciduous teeth first, adult teeth second, false teeth third.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6031" LastActivityDate="2015-02-26T12:18:31.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9619" PostTypeId="1" AcceptedAnswerId="9621" CreationDate="2015-02-26T13:13:26.400" Score="4" ViewCount="400" Body="&lt;p&gt;Cry you may crook&lt;br&gt;&#xA;As you gaze at my look&lt;br&gt;&#xA;A top notch runner&lt;br&gt;&#xA;Diverts while I cover&lt;br&gt;&#xA;I parch or I perch&lt;br&gt;&#xA;But outwardly search&lt;br&gt;&#xA;A stretch and a spring&lt;br&gt;&#xA;An ingenious thing  &lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Hint #1&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lines 1, 3 and 7 relate to its design&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;Hint #2&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lines 4 and 5 relate to its purpose&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9802" LastEditorUserId="5368" LastEditDate="2015-02-27T21:02:00.503" LastActivityDate="2015-02-27T21:02:00.503" Title="Cry you may crook" Tags="|riddle|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9620" PostTypeId="2" ParentId="9597" CreationDate="2015-02-26T13:29:03.850" Score="5" Body="&lt;p&gt;The values are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; r = 5, d = 76 or 68, so x,y = 95,19 or 85,17&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Step 1 tells us there must be multiple solutions for d. This rules out large primes such as d=89, and also numbers like d=91 (which must have x,y=98,7 and r=14).&lt;br&gt;&#xA; Step 3 tells us that r &amp;lt;= 32, because otherwise the only possible value for y would be 2 and Bob would know the solution.&lt;br&gt;&#xA; Step 4 tells us that r could have been &gt;= 33, otherwise Alice would not have had a solution if it was. So d must be &gt;= 66.&lt;br&gt;&#xA; Step 5 further limits the solutions. Suppose by way of example, that d = 92. Then x,y,r must be 94,2,47 or 96,4,24. If d were 92, Alice would now know the answer because r &amp;lt; 33, but she does not. We can eliminate all solutions with r &gt;= 33.&lt;br&gt;&#xA; Step 6 eliminates more solutions: for example d=84 could mean x,y,r = 84,4,22, and this is the only solution with r=22 and d &gt;= 66. Since Bob still doesn't know the answer, it can't be 84, and r can't be 22.&lt;br&gt;&#xA; Step 7 means that the solution must be a d value where only one of its possible solutions shares an r-value with another d-value. The only way to find these is to write them all down, and they are d,r=96,25; 92,24; 76,5; 68,5.&lt;br&gt;&#xA; Step 8 means that the solutions must share r values, or Bob would know which one to choose, so r=5 and d=76 or 68.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6031" LastActivityDate="2015-02-26T13:29:03.850" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9621" PostTypeId="2" ParentId="9619" CreationDate="2015-02-26T13:29:48.140" Score="3" Body="&lt;p&gt;Cry you may crook&#xA;As you gaze at my look&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Umbrella has metal wires/plates/blades which all are crooked.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A top notch runner&#xA;Diverts while I cover&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Umbrella diverts anything that falls on it from top. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I parch or I perch&#xA;But outwardly search&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Either we see an umbrella opened in the sun or rain or we see it closed and hanging on some support.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A stretch and a spring&#xA;An ingenious thing&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An ingenious assembly of a strong waterproof fabric with springs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It is an Umbrella.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9174" LastEditorDisplayName="user9174" LastEditDate="2015-02-27T13:17:23.817" LastActivityDate="2015-02-27T13:17:23.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9623" PostTypeId="2" ParentId="9597" CreationDate="2015-02-26T14:30:34.517" Score="1" Body="&lt;p&gt;first I list all possible solutions that satisfy 2 &amp;lt;= y &amp;lt; x &amp;lt;= 100 where y is a divisor of x&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x   y   d   r&#xA;100 50  50  2&#xA;98  49  49  2&#xA;96  48  48  2&#xA;94  47  47  2&#xA;92  46  46  2&#xA;90  45  45  2&#xA;88  44  44  2&#xA;86  43  43  2&#xA;84  42  42  2&#xA;82  41  41  2&#xA;80  40  40  2&#xA;78  39  39  2&#xA;76  38  38  2&#xA;74  37  37  2&#xA;72  36  36  2&#xA;70  35  35  2&#xA;68  34  34  2&#xA;99  33  66  3&#xA;66  33  33  2&#xA;96  32  64  3&#xA;64  32  32  2&#xA;93  31  62  3&#xA;62  31  31  2&#xA;90  30  60  3&#xA;60  30  30  2&#xA;87  29  58  3&#xA;58  29  29  2&#xA;84  28  56  3&#xA;56  28  28  2&#xA;81  27  54  3&#xA;54  27  27  2&#xA;78  26  52  3&#xA;52  26  26  2&#xA;100 25  75  4&#xA;75  25  50  3&#xA;50  25  25  2&#xA;96  24  72  4&#xA;72  24  48  3&#xA;48  24  24  2&#xA;92  23  69  4&#xA;69  23  46  3&#xA;46  23  23  2&#xA;88  22  66  4&#xA;66  22  44  3&#xA;44  22  22  2&#xA;84  21  63  4&#xA;63  21  42  3&#xA;42  21  21  2&#xA;100 20  80  5&#xA;80  20  60  4&#xA;60  20  40  3&#xA;40  20  20  2&#xA;95  19  76  5&#xA;76  19  57  4&#xA;57  19  38  3&#xA;38  19  19  2&#xA;90  18  72  5&#xA;72  18  54  4&#xA;54  18  36  3&#xA;36  18  18  2&#xA;85  17  68  5&#xA;68  17  51  4&#xA;51  17  34  3&#xA;34  17  17  2&#xA;96  16  80  6&#xA;80  16  64  5&#xA;64  16  48  4&#xA;48  16  32  3&#xA;32  16  16  2&#xA;90  15  75  6&#xA;75  15  60  5&#xA;60  15  45  4&#xA;45  15  30  3&#xA;30  15  15  2&#xA;98  14  84  7&#xA;84  14  70  6&#xA;70  14  56  5&#xA;56  14  42  4&#xA;42  14  28  3&#xA;28  14  14  2&#xA;91  13  78  7&#xA;78  13  65  6&#xA;65  13  52  5&#xA;52  13  39  4&#xA;39  13  26  3&#xA;26  13  13  2&#xA;96  12  84  8&#xA;84  12  72  7&#xA;72  12  60  6&#xA;60  12  48  5&#xA;48  12  36  4&#xA;36  12  24  3&#xA;24  12  12  2&#xA;99  11  88  9&#xA;88  11  77  8&#xA;77  11  66  7&#xA;66  11  55  6&#xA;55  11  44  5&#xA;44  11  33  4&#xA;33  11  22  3&#xA;22  11  11  2&#xA;100 10  90  10&#xA;90  10  80  9&#xA;80  10  70  8&#xA;70  10  60  7&#xA;60  10  50  6&#xA;50  10  40  5&#xA;40  10  30  4&#xA;30  10  20  3&#xA;20  10  10  2&#xA;99  9   90  11&#xA;90  9   81  10&#xA;81  9   72  9&#xA;72  9   63  8&#xA;63  9   54  7&#xA;54  9   45  6&#xA;45  9   36  5&#xA;36  9   27  4&#xA;27  9   18  3&#xA;18  9   9   2&#xA;96  8   88  12&#xA;88  8   80  11&#xA;80  8   72  10&#xA;72  8   64  9&#xA;64  8   56  8&#xA;56  8   48  7&#xA;48  8   40  6&#xA;40  8   32  5&#xA;32  8   24  4&#xA;24  8   16  3&#xA;16  8   8   2&#xA;98  7   91  14&#xA;91  7   84  13&#xA;84  7   77  12&#xA;77  7   70  11&#xA;70  7   63  10&#xA;63  7   56  9&#xA;56  7   49  8&#xA;49  7   42  7&#xA;42  7   35  6&#xA;35  7   28  5&#xA;28  7   21  4&#xA;21  7   14  3&#xA;14  7   7   2&#xA;96  6   90  16&#xA;90  6   84  15&#xA;84  6   78  14&#xA;78  6   72  13&#xA;72  6   66  12&#xA;66  6   60  11&#xA;60  6   54  10&#xA;54  6   48  9&#xA;48  6   42  8&#xA;42  6   36  7&#xA;36  6   30  6&#xA;30  6   24  5&#xA;24  6   18  4&#xA;18  6   12  3&#xA;12  6   6   2&#xA;100 5   95  20&#xA;95  5   90  19&#xA;90  5   85  18&#xA;85  5   80  17&#xA;80  5   75  16&#xA;75  5   70  15&#xA;70  5   65  14&#xA;65  5   60  13&#xA;60  5   55  12&#xA;55  5   50  11&#xA;50  5   45  10&#xA;45  5   40  9&#xA;40  5   35  8&#xA;35  5   30  7&#xA;30  5   25  6&#xA;25  5   20  5&#xA;20  5   15  4&#xA;15  5   10  3&#xA;10  5   5   2&#xA;100 4   96  25&#xA;96  4   92  24&#xA;92  4   88  23&#xA;88  4   84  22&#xA;84  4   80  21&#xA;80  4   76  20&#xA;76  4   72  19&#xA;72  4   68  18&#xA;68  4   64  17&#xA;64  4   60  16&#xA;60  4   56  15&#xA;56  4   52  14&#xA;52  4   48  13&#xA;48  4   44  12&#xA;44  4   40  11&#xA;40  4   36  10&#xA;36  4   32  9&#xA;32  4   28  8&#xA;28  4   24  7&#xA;24  4   20  6&#xA;20  4   16  5&#xA;16  4   12  4&#xA;12  4   8   3&#xA;8   4   4   2&#xA;99  3   96  33&#xA;96  3   93  32&#xA;93  3   90  31&#xA;90  3   87  30&#xA;87  3   84  29&#xA;84  3   81  28&#xA;81  3   78  27&#xA;78  3   75  26&#xA;75  3   72  25&#xA;72  3   69  24&#xA;69  3   66  23&#xA;66  3   63  22&#xA;63  3   60  21&#xA;60  3   57  20&#xA;57  3   54  19&#xA;54  3   51  18&#xA;51  3   48  17&#xA;48  3   45  16&#xA;45  3   42  15&#xA;42  3   39  14&#xA;39  3   36  13&#xA;36  3   33  12&#xA;33  3   30  11&#xA;30  3   27  10&#xA;27  3   24  9&#xA;24  3   21  8&#xA;21  3   18  7&#xA;18  3   15  6&#xA;15  3   12  5&#xA;12  3   9   4&#xA;9   3   6   3&#xA;6   3   3   2&#xA;100 2   98  50&#xA;98  2   96  49&#xA;96  2   94  48&#xA;94  2   92  47&#xA;92  2   90  46&#xA;90  2   88  45&#xA;88  2   86  44&#xA;86  2   84  43&#xA;84  2   82  42&#xA;82  2   80  41&#xA;80  2   78  40&#xA;78  2   76  39&#xA;76  2   74  38&#xA;74  2   72  37&#xA;72  2   70  36&#xA;70  2   68  35&#xA;68  2   66  34&#xA;66  2   64  33&#xA;64  2   62  32&#xA;62  2   60  31&#xA;60  2   58  30&#xA;58  2   56  29&#xA;56  2   54  28&#xA;54  2   52  27&#xA;52  2   50  26&#xA;50  2   48  25&#xA;48  2   46  24&#xA;46  2   44  23&#xA;44  2   42  22&#xA;42  2   40  21&#xA;40  2   38  20&#xA;38  2   36  19&#xA;36  2   34  18&#xA;34  2   32  17&#xA;32  2   30  16&#xA;30  2   28  15&#xA;28  2   26  14&#xA;26  2   24  13&#xA;24  2   22  12&#xA;22  2   20  11&#xA;20  2   18  10&#xA;18  2   16  9&#xA;16  2   14  8&#xA;14  2   12  7&#xA;12  2   10  6&#xA;10  2   8   5&#xA;8   2   6   4&#xA;6   2   4   3&#xA;4   2   2   2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now I list all solutions that can be removed after step 1 because there is only a single entry with that specific d&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x   y   d   r&#xA;100 2   98  50&#xA;100 5   95  20&#xA;96  2   94  48&#xA;96  3   93  32&#xA;98  7   91  14&#xA;90  3   87  30&#xA;88  2   86  44&#xA;90  5   85  18&#xA;84  2   82  42&#xA;76  2   74  38&#xA;94  47  47  2&#xA;86  43  43  2&#xA;82  41  41  2&#xA;74  37  37  2&#xA;62  31  31  2&#xA;58  29  29  2&#xA;46  23  23  2&#xA;38  19  19  2&#xA;34  17  17  2&#xA;26  13  13  2&#xA;22  11  11  2&#xA;14  7   7   2&#xA;10  5   5   2&#xA;6   3   3   2&#xA;4   2   2   2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now I list all solutions that can be removed after step 3 because there is only a single entry with that specific r&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x   y   d   r&#xA;98  2   96  49&#xA;94  2   92  47&#xA;92  2   90  46&#xA;90  2   88  45&#xA;86  2   84  43&#xA;82  2   80  41&#xA;80  2   78  40&#xA;78  2   76  39&#xA;74  2   72  37&#xA;72  2   70  36&#xA;70  2   68  35&#xA;68  2   66  34&#xA;64  2   62  32&#xA;60  2   58  30&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now I list all solutions after step 6 because there is only a single entry with that specific d&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x   y   d   r&#xA;96  4   92  24&#xA;93  31  62  3&#xA;87  29  58  3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I guess I made a mistake somewhere since I have 2 different r values in there but I assume that the correct ones would be with r=3 because there are two of those&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-02-26T14:30:34.517" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9624" PostTypeId="2" ParentId="9597" CreationDate="2015-02-26T15:11:44.357" Score="7" Body="&lt;p&gt;This is how I reasoned:&lt;bR&gt;&#xA;There are 382 sets with 2 ≤y≤x≤100 where x/y is an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statement 1 says Alice doesn't know the numbers.&lt;br&gt;&#xA;This means that we can eliminate all sets that have a difference d that only appears once.&lt;br&gt;&#xA;357 remain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statements 2 &amp;amp; 3 say that Bob doesn't know them,&lt;br&gt;&#xA;so we can eliminate sets that have a ratio r that only appears once.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statement 4 however says, if it had been one of those sets, then Alice would know the numbers, because in all those sets, y=2.&lt;br&gt;&#xA;So d must be one of the d's that appeared in one of the sets just eliminated.&lt;br&gt;&#xA;Only 49 other sets have one of those d's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5: Alice still doesn't know the numbers, so we can again eliminate sets with a unique d.&lt;br&gt;&#xA;46 remain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6: Bob still doesn't know, so we can eliminate sets with a unique r.&lt;br&gt;&#xA;34 remain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7: Now, suddenly Alice knows the numbers, so d must be unique among the remaining sets.&lt;br&gt;&#xA;Only 4 sets remain:&lt;br&gt;&#xA;{x=85,  y=17, r=5,  d=68}&lt;br&gt;&#xA;{x=95,  y=19, r=5,  d=76}&lt;br&gt;&#xA;{x=91,  y=13, r=7,  d=78}&lt;br&gt;&#xA;{x=100, y=4,  r=25, d=96}&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8: Bob still doesn't know x and y, so r is still not unique.&lt;br&gt;&#xA;Only 2 sets remain:&lt;br&gt;&#xA;{x=85,  y=17, r=5,  d=68}&lt;br&gt;&#xA;{x=95,  y=19, r=5,  d=76}&lt;br&gt;&#xA;So r is 5 and x,y is either 85,17 or 95,19. Alice knows, but we don't.&lt;/p&gt;&#xA;" OwnerUserId="1598" LastActivityDate="2015-02-26T15:11:44.357" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9625" PostTypeId="2" ParentId="9619" CreationDate="2015-02-26T15:34:36.873" Score="1" Body="&lt;p&gt;My best guess would be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;a salmon.&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's not exactly clear what the answer is, but it very likely has something to do with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; flowing water, live a river.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Cry you may crook&lt;br&gt;&#xA;As you gaze at my look&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A crook is a body of flowing water; also, there's the common idiom &quot;cry me a river&quot; that this might refer to. But what are we gazing at?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A top notch runner&lt;br&gt;&#xA;Diverts while I cover&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Flowing water is sometimes known to be &quot;running&quot; water; when water meets an obstacle, its flow is diverted. But what is covering itself and/or taking cover?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I parch or I perch&lt;br&gt;&#xA;But outwardly search&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To be parched is to be thirsty, even dried up; a perch is a freshwater fish found in many flowing bodies of water. But what is searching?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A stretch and a spring&lt;br&gt;&#xA;An ingenious thing&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, a spring is another body of flowing water; many geographical formations are called &quot;stretches&quot; when referring to their length or breadth, as in a &quot;stretch of land&quot; or a &quot;stretch of water&quot;, the latter being common for rivers and streams.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My answer is because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;a salmon&lt;/strong&gt; is a pink/orange fish that is often admired for its appearance, swims upstream thus diverting water around it, and searches for a place to make its nest when it reaches the uppermost part of the river it is swimming against.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9319" LastActivityDate="2015-02-26T15:34:36.873" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9626" PostTypeId="1" AcceptedAnswerId="9637" CreationDate="2015-02-26T15:54:32.833" Score="13" ViewCount="773" Body="&lt;p&gt;Mary and Ursula play a game by alternately writing letters into a $9\times9$ grid.  In every move, the player picks an empty square and then writes her initial letter (M or U) into it.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Mary has the first move.&lt;/li&gt;&#xA;&lt;li&gt;The game ends once all squares have been filled.&lt;/li&gt;&#xA;&lt;li&gt;When the game has ended, $N(M)$ is the number of rows and columns that contain more Ms than Us, while $N(U)$ is the number of rows and columns that contain more Us than Ms.&lt;/li&gt;&#xA;&lt;li&gt;When the game has ended, Mary receives $N(M)-N(U)$ Euros from Ursula.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Question: &#xA;How much money will Mary receive from (respectively, lose to) Ursula?&lt;br&gt;&#xA;(As usual, we assume that Mary and Ursula both use optimal strategies.)&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-02-26T21:42:40.307" Title="What's the optimal score in the MU-game?" Tags="|strategy|game|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9627" PostTypeId="1" AcceptedAnswerId="9628" CreationDate="2015-02-26T16:10:21.293" Score="-1" ViewCount="266" Body="&lt;p&gt;Nineteen of the following twenty famous persons share a &lt;strong&gt;very distinguished&lt;/strong&gt; property, whereas the last one does not have this property:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Albert Camus, Elias Canetti, Bette Davis, Nadine Gordimer, George Harrison,&lt;br&gt;&#xA;  Jimi Hendrix, Audrey Hepburn, Frida Kahlo, Jack Lemmon, John Lennon,&lt;br&gt;&#xA;  Doris Lessing, Marcello Mastroianni, Paul McCartney, Piet Mondrian, Pablo Neruda,&lt;br&gt;&#xA;  Pablo Picasso, Jackson Pollock, Elvis Presley, Andy Warhol, John Wayne.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who is the odd one out?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2015-02-26T16:18:11.303" Title="Who is the odd one out?" Tags="|pattern|knowledge|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9628" PostTypeId="2" ParentId="9627" CreationDate="2015-02-26T16:18:11.303" Score="4" Body="&lt;p&gt;I feel like this is too simple and might therefore not be what you had in mind, but, differently than everyone else on your list, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Paul McCartney is alive&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9447" LastActivityDate="2015-02-26T16:18:11.303" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9629" PostTypeId="1" AcceptedAnswerId="9630" CreationDate="2015-02-26T16:22:18.073" Score="8" ViewCount="7485" Body="&lt;p&gt;Solve the following ten one-line Rebus puzzles:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;DICTNARY    &lt;/li&gt;&#xA;&lt;li&gt;PRAHOMTE    &lt;/li&gt;&#xA;&lt;li&gt;..AVI..     &lt;/li&gt;&#xA;&lt;li&gt;MARY+MARY+MARY&lt;/li&gt;&#xA;&lt;li&gt;/r/e/a/d/i/n/g/&lt;/li&gt;&#xA;&lt;li&gt;MTWTFSSS    &lt;/li&gt;&#xA;&lt;li&gt;144 144 word word word word&lt;/li&gt;&#xA;&lt;li&gt;LIFE = TTTT---------&gt;     &lt;/li&gt;&#xA;&lt;li&gt;WORLD WORLD WORLD WORLD  &lt;/li&gt;&#xA;&lt;li&gt;SUGAR x SUGAR x SUGAR   &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2015-02-27T09:18:34.793" Title="Ten Rebus puzzles" Tags="|rebus|" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9630" PostTypeId="2" ParentId="9629" CreationDate="2015-02-26T16:30:37.513" Score="14" Body="&lt;p&gt;The answers (joint work by aes, dmg, Engineer Toast, Gamow,Hellion):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DICTNARY = shortened DICTIONARY = &lt;strong&gt;abridged dictionary&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PRAHOMTE = METAPHOR scrambled and mixed = &lt;strong&gt;Mixed metaphor&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ..AVI.. = center piece of GRAVITY = &lt;strong&gt;The center of gravity&lt;/strong&gt;   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MARY+MARY+MARY = SUM of MARY = &lt;strong&gt;Summary&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; /r/e/a/d/i/n/g/ = &lt;strong&gt;reading between the lines&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MTWTFSSS = initial letters of weekdays + an extra saturday/sunday = &lt;strong&gt;Eight days a week&lt;/strong&gt;&lt;br&gt;&#xA; or: &lt;strong&gt;Long weekend&lt;/strong&gt;&lt;br&gt;&#xA; or: &lt;strong&gt;Every day of the week and twice on Sundays&lt;/strong&gt;   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;7:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 144 144 word word word word = 2 gross, 4 words = &lt;strong&gt;too gross for words&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;8:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LIFE = TTTT---------&gt; = Life is something that begins at 4T = &lt;strong&gt;Life begins at forty&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;9:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; WORLD WORLD WORLD WORLD = &lt;strong&gt;World Series&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;10:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SUGAR x SUGAR x SUGAR = &lt;strong&gt;sugar cube&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-02-27T09:18:34.793" LastActivityDate="2015-02-27T09:18:34.793" CommentCount="5" CommunityOwnedDate="2015-02-26T16:41:56.520" ContentLicense="CC BY-SA 3.0" />
  <row Id="9631" PostTypeId="2" ParentId="9629" CreationDate="2015-02-26T16:30:54.847" Score="4" Body="&lt;p&gt;Partial answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; mixed metaphor&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;long weekend (three S in the sequence MTWTFSSS)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;9.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;FOUR WORLD -&gt; Forward or Foreword&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="5610" LastEditDate="2015-02-26T16:36:44.650" LastActivityDate="2015-02-26T16:36:44.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9632" PostTypeId="1" AcceptedAnswerId="9633" CreationDate="2015-02-26T17:21:40.650" Score="36" ViewCount="5144" Body="&lt;p&gt;My grandfather has a big drawer where he keeps his socks. &#xA;The drawer contains more than 900 but less than 1000 individual socks.&#xA;Each of his socks is black or blue, and there are more blue socks than black socks. The socks aren't paired by color within the drawer, but if he reaches into it and grabs two socks at random, then exactly half the time he will pull a matching pair.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My grandfather has always been extremely careful when doing the laundry, and &#xA;has only ever lost a single sock in his entire life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What color was that lost sock?&lt;/p&gt;&#xA;" OwnerUserId="9125" LastActivityDate="2015-04-14T10:52:30.157" Title="My grandfather's socks" Tags="|calculation-puzzle|probability|combinatorics|" AnswerCount="5" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="9633" PostTypeId="2" ParentId="9632" CreationDate="2015-02-26T17:45:16.953" Score="38" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Your grandfather lost a black sock.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let $a$ be the number of blue socks and $b$ be the number of black socks.&lt;br&gt;&#xA; To get exactly 50% matching, you will need &lt;br&gt;&#xA; $~~~~~ a(a-1) + b(b-1) = 2ab$&lt;br&gt;&#xA; which can be rewritten as&lt;br&gt;&#xA; $~~~~~ (a-b)^2 = a+b$. &lt;br&gt;&#xA; As $a+b$ is a square with $900 &amp;lt; a+b &amp;lt; 1000$, we get $a+b=961$. &#xA; Then $a &amp;gt; b$ yields $a-b=31$. &#xA; The only remaining possibility is $a=496$ blue socks and $b=465$ black socks, and the lost sock was black.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="375" LastEditorUserId="8874" LastEditDate="2015-02-26T18:11:51.860" LastActivityDate="2015-02-26T18:11:51.860" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9634" PostTypeId="2" ParentId="9632" CreationDate="2015-02-26T17:54:27.200" Score="15" Body="&lt;p&gt;Assuming he bought pairs of socks, then either the number of blue socks $(u)$ or the number of black socks $(k)$ is odd with the other is even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u&amp;gt;k$$&#xA;$$p = .5 = \frac u {(k+u)}\frac k {(k+u-1)}+\frac k {(k+u)}\frac u {(k+u-1)}$$&#xA;$$p = .5 = \frac {2uk} {(k+u)(k+u-1)}$$&#xA;$$p = .5 = \frac u {(k+u)}\frac {u-1} {(k+u-1)}+\frac k {(k+u)}\frac {k-1} {(k+u-1)}$$&#xA;$$p = .5 = \frac {u^2-u+k^2-k} {(k+u)(k+u-1)}$$&#xA;$$0=u^2+k^2-u-k-2uk$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u=n^2/2+n/2$$&#xA;$$k=n^2/2-n/2$$&#xA;$$u+k=n^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n=31$$&#xA;$$u=496$$&#xA;$$k=465$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means grandpa lost a black sock.&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2015-02-26T17:54:27.200" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9635" PostTypeId="2" ParentId="9254" CreationDate="2015-02-26T17:56:07.233" Score="13" Body="&lt;p&gt;I am &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a book&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The BED and me both have C-some: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; covers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The TREE and me both have L-some: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; leaves&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The PRISONER and me both are B-some: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; boxed? bonded?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The WEBSITE and me both have P-some: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; pages&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The SHIRT and me both have S-some: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; sleeves&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2059" LastActivityDate="2015-02-26T17:56:07.233" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9636" PostTypeId="2" ParentId="9629" CreationDate="2015-02-26T18:06:42.303" Score="1" Body="&lt;p&gt;An alternative possible answer for 6:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;And twice on Sundays. (Or perhaps the full phrase: Every day of the week and twice on Sundays.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9821" LastActivityDate="2015-02-26T18:06:42.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9637" PostTypeId="2" ParentId="9626" CreationDate="2015-02-26T18:08:52.393" Score="12" Body="&lt;p&gt;I don't know if this is optimal play for Mary (though I suspect it is), but she can guarantee winning 2 Euros if she initially plays at (5,5) and copies Ursula's moves by translation through (5,5). Rows 1-4,6-9 and columns 1-4,6-9 will be evenly split by the pair, row 5 and column 5 will belong to Mary because of the centre square.&lt;/p&gt;&#xA;" OwnerUserId="3987" LastActivityDate="2015-02-26T18:08:52.393" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9638" PostTypeId="2" ParentId="9629" CreationDate="2015-02-26T18:32:13.267" Score="0" Body="&lt;p&gt;Partial Answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DICTNARY = DICTIONARY - &quot;IO&quot; = either end of the dictionary = Bookends&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2015-02-26T20:23:13.520" LastActivityDate="2015-02-26T20:23:13.520" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9639" PostTypeId="2" ParentId="9632" CreationDate="2015-02-26T18:46:10.503" Score="11" Body="&lt;p&gt;The lost sock was&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; black&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let $B$ be the event that grandpa pulls a blue sock and $K$ be the event that he pulls a black sock. Also let $b$ be the number of blue socks and $k$ be the number of black socks.  We are told that $b&amp;gt;k$ and $900&amp;lt;b+k&amp;lt;1000$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are four possibilities for the events of pulling two socks: $BB$, $BK$, $KB$, and $KK$. The probability of choosing a matching pair, i.e. pulling two socks without replacement, is $P(BB) + P(KK)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Calculations:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\begin{align} P(BB) &amp;amp; = \left( \frac b{b+k} \right) \left( \frac {b-1}{b+k-1} \right) \\ \\&amp;amp; = \frac {b^2-b}{b^2+2bk+k^2-b-k} \\ \\P(KK) &amp;amp; = \left( \frac k{b+k} \right) \left( \frac {k-1}{b+k-1} \right) \\ \\&amp;amp; = \frac {k^2-k}{b^2+2bk+k^2-b-k} \\ \\P(BB) + P(KK) &amp;amp; = \frac {b^2-b}{b^2+2bk+k^2-b-k} + \frac {k^2-k}{b^2+2bk+k^2-b-k} \\ \\&amp;amp; = \frac {b^2-b+k^2-k}{b^2+2bk+k^2-b-k}\end{align}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But we know that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\begin{align} P(BB) + P(KK) &amp;amp; = \frac 12 \\ \\ \frac {b^2-b+k^2-k}{b^2+2bk+k^2-b-k} &amp;amp; = \frac 12\end{align}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Multiplying both sides by both denominators:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\begin{align}2b^2-2b+2k^2-2k &amp;amp; = b^2+2bk+k^2-b-k \\b^2-2bk+k^2 &amp;amp; = b+k \\(b-k)^2 &amp;amp; = b+k\end{align}$$ &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $b+k$ is a perfect square and $900&amp;lt;b+k&amp;lt;1000$. The only perfect square in that range is $961=31^2$, so $b+k=961$, i.e. there are 961 socks total. But this is $(b-k)^2$, so $b-k=31$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$ \begin{align} b+k &amp;amp; = 961 \\ b-k &amp;amp; = 31 \\ 2b &amp;amp;= 992 \\ b &amp;amp;= 496 \\ k &amp;amp;= 465 \\  \end{align} $$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since the odd number is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $k$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the missing sock was &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; black.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4373" LastActivityDate="2015-02-26T18:46:10.503" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9640" PostTypeId="2" ParentId="9632" CreationDate="2015-02-26T20:22:23.377" Score="1" Body="&lt;p&gt;I'm going to feel very stupid if this is wrong, but...shouldn't the lost sock be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;black since he has more blue socks than black ones, and his odds of pulling a matching pair are 50%?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Even looking at all of the computations, I can't see a reason that my logic would be insufficient.&lt;/p&gt;&#xA;" OwnerUserId="9826" LastActivityDate="2015-02-26T20:22:23.377" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9641" PostTypeId="2" ParentId="9596" CreationDate="2015-02-26T20:37:29.107" Score="10" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; none of them &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because the enigmatic property is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the ability to spell the element's name using element symbols.  For example, Arsenic is Ar+Se+Ni+C, and Silver is Si+Lv+Er.  As far as I can tell, there are only 13 elements with this property: Silver, Arsenic, Astatine, Bismuth, Carbon, Copper, Iron, Krypton, Neon, Phosphorus, Silicon, Tin, and Xenon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2374" LastActivityDate="2015-02-26T20:37:29.107" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9642" PostTypeId="2" ParentId="9629" CreationDate="2015-02-26T21:35:23.430" Score="2" Body="&lt;p&gt;One more guess for #6:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eight days a week.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What can I say, I love the Beatles.&lt;/p&gt;&#xA;" OwnerUserId="5874" LastActivityDate="2015-02-26T21:35:23.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9643" PostTypeId="2" ParentId="9626" CreationDate="2015-02-26T21:42:40.307" Score="6" Body="&lt;p&gt;Mary will win 2 Euros if both of them play optimally. Frodoskywalker has given a strategy that Mary can use to accomplish this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's why this will happen:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mary will write down 41 M's, and Ursula will write down 40 U's. By Frodoskywalker's strategy, we know Mary can win at least 2 Euros from Ursula, so Ursula will try to limit it to 2 Euros. Taking out Mary's extra letter (in the center), let's consider the rows and columns of the board in terms of $N(M)-N(U)$. For a checkerboard pattern, we get this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   1 -1  1 -1  0 -1  1 -1  1 &#xA; 1 M  U  M  U  M  U  M  U  M&#xA;-1 U  M  U  M  U  M  U  M  U&#xA; 1 M  U  M  U  M  U  M  U  M&#xA;-1 U  M  U  M  U  M  U  M  U&#xA; 0 M  U  M  U  _  U  M  U  M&#xA;-1 U  M  U  M  U  M  U  M  U&#xA; 1 M  U  M  U  M  U  M  U  M&#xA;-1 U  M  U  M  U  M  U  M  U&#xA; 1 M  U  M  U  M  U  M  U  M&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As you can see, the column numbers and the row numbers each balance out. With some slight changes:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   1 -3  1  1  0 -1  1 -1  1 &#xA; 1 U  U  M  M  M  U  M  U  M&#xA;-1 M  U  U  M  U  M  U  M  U&#xA; 1 M  U  M  U  M  U  M  U  M&#xA;-1 U  M  U  M  U  M  U  M  U&#xA; 0 M  U  M  U  _  U  M  U  M&#xA;-1 U  M  U  M  U  M  U  M  U&#xA; 1 M  U  M  U  M  U  M  U  M&#xA;-1 U  M  U  M  U  M  U  M  U&#xA; 1 M  U  M  U  M  U  M  U  M&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here Mary would win an extra 2 Euros, but you can see that the columns numbers and row numbers still sum to zero. The key to this is noticing that in the second column is extra unbalanced in Ursula's favor, leaving her fewer letters to counter Mary. With this information we can come up with a strategy for Ursula that allows her to limit Mary to 2 Euros.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it is Ursula's turn, there will be an even number of empty squares left and she will have a choice between at least two squares. The column and row differences will each sum to 1. If possible, she will write her letter in square for which she and Mary are tied in the row and column. If that's not possible, her next option will be to write it in a square for which she and Mary are tied for the row or column and she is behind by 1 in the other. Failing that, she should choose a square for which she is behind by 1 for both the row and column. In each of these cases, after her move she will either gain a row and/or a column or tie it up in a row and/or a column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Towards the end of the game Ursula will eventually be restricted in her options, but by then Mary will be equally restricted. For example, if the last two squares are in the same row, then even if Ursula is ahead by 1 in that row she can choose the square according to which column is better knowing that Mary will be forced to bring the score back to 1 for that row.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof that this strategy works will come in a little bit...&lt;/p&gt;&#xA;" OwnerUserId="1955" LastActivityDate="2015-02-26T21:42:40.307" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9644" PostTypeId="2" ParentId="9545" CreationDate="2015-02-26T21:58:11.230" Score="4" Body="&lt;p&gt;It seems likely that the geographic feature you're referring to is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the state's capital city also being the state's largest city&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but every source I can find shows 17 states with that quality, not 18.  Those 17 are the 15 you listed plus&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Colorado and Massachusetts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example, &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_states_and_territories_of_the_United_States&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; shows 17, using census data published in 2012, and there are also 17 on &lt;a href=&quot;http://www.businessinsider.com/state-capitals-largest-cities-map-2014-10&quot; rel=&quot;nofollow&quot;&gt;this handy map&lt;/a&gt; using 2010 census data.  Is it possible you used an outdated source, or miscounted?  Or did you have a completely different geographic feature in mind, making this an unbelievable coincidence? &lt;/p&gt;&#xA;" OwnerUserId="2374" LastActivityDate="2015-02-26T21:58:11.230" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9645" PostTypeId="1" AcceptedAnswerId="9646" CreationDate="2015-02-26T22:07:40.393" Score="1" ViewCount="716" Body="&lt;p&gt;I made these all myself. No images, but hopefully still fun:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;shbandip&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;all&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;__&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;map&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;pantsants&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;rsnug=bugug&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;wits'&amp;lt;-&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Requirements:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You must answer each one&lt;/li&gt;&#xA;&lt;li&gt;You must explain how each one works&lt;/li&gt;&#xA;&lt;li&gt;Both of the above must be correct&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2015-02-26T22:15:26.560" Title="Text Rebus Extreme!" Tags="|rebus|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9646" PostTypeId="2" ParentId="9645" CreationDate="2015-02-26T22:15:26.560" Score="6" Body="&lt;p&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; abandon ship (a &quot;band&quot; in &quot;ship&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; all over the map (the word &quot;all&quot; is over the word &quot;map&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ants in your pants (&quot;ants&quot; is inside &quot;pants&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; snug as a bug in a rug (snug=bug means &quot;snug is a bug&quot;, which sounds like &quot;snug as a bug&quot;, and it's inside the word &quot;rug&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; at wits' end (arrow pointing to the end of wits')&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2015-02-26T22:15:26.560" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9647" PostTypeId="1" CreationDate="2015-02-26T23:15:25.127" Score="-1" ViewCount="663" Body="&lt;p&gt;The second rebus question in the series.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;here&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;____&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;bored&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;bets&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;_&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;all4114&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;knightleg&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A&lt;/strong&gt;&lt;br&gt;&#xA;\&lt;br&gt;&#xA;&lt;strong&gt;C&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&amp;lt;3 &amp;lt;-&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Requirements:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You must answer each one&lt;/li&gt;&#xA;&lt;li&gt;You must explain how each one works&lt;/li&gt;&#xA;&lt;li&gt;Both of the above must be correct&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #1:&lt;/strong&gt; They are all cliches.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2015-02-28T12:40:06.203" LastActivityDate="2015-03-18T23:16:01.490" Title="Text Rebus Extreme! 2!" Tags="|rebus|" AnswerCount="4" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="9649" PostTypeId="1" AcceptedAnswerId="9703" CreationDate="2015-02-27T03:34:59.727" Score="38" ViewCount="3836" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/v8PR3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above shape can be folded into a closed 3D shape using no more than 14 distinct folds, with no parts overlapping. What is special about the shape that results? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rules and clarifications:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Every part of the black boundary should meet up with another part of the black boundary&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;No parts of the paper overlap or touch one another except where they meet at the black boundary&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The shape is made up of flat surfaces&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;No cutting except for cutting it out along the solid black lines&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The dotted lines are just to show you how the shape is constructed - they don't necessarily indicate where to place folds.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You can decide for yourself whether printing it out and trying to fold it is cheating.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;I came up with this the other day. I'm not 100% sure the solution is unique, but I think it should be.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="4017" LastEditorUserId="12408" LastEditDate="2016-08-05T03:33:37.463" LastActivityDate="2016-08-05T03:33:37.463" Title="Which 3D shape can you make out of this?" Tags="|geometry|paper-folding|" AnswerCount="3" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="9650" PostTypeId="2" ParentId="9619" CreationDate="2015-02-27T06:54:13.510" Score="2" Body="&lt;p&gt;Cry you may crook As you gaze at my look&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Surely policemen have this and crooks are afraid to be hit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A top notch runner Diverts while I cover&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A crook would change direction to avoid it's firing coverage.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I parch or I perch But outwardly search&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It can cause dehydration (parch). Perch is like holding it on your right hand and using your left hand as a perch for aiming stability.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A stretch and a spring An ingenious thing&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's long range and springy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's a Taser (stun gun)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9834" LastEditorUserId="9834" LastEditDate="2015-02-27T07:11:07.240" LastActivityDate="2015-02-27T07:11:07.240" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9651" PostTypeId="2" ParentId="9194" CreationDate="2015-02-27T08:17:04.200" Score="2" Body="&lt;p&gt;In order to create four similar triangles instead of five, it is necessary to:    &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. &lt;strong&gt;reflect two of the internal triangles&lt;/strong&gt; (refer to examples below)&lt;br&gt;&#xA;&lt;br&gt;&#xA; 2. and place the third, non-reflected, internal triangle inside any vertex - provided this vertex has an angle less than $90^\circ$. Call this internal angle $\alpha$.&lt;br&gt;&#xA;&lt;br&gt;&#xA; 3. then form the base of an $isosceles$ triangle with the internal side of the non-reflected triangle&lt;br&gt;&#xA;&lt;br&gt;&#xA; 4. then use two equal lengths of the two reflected triangles and two angles equal to $\alpha$ to complete the internal, $isosceles$ triangle  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The internal triangle KJL is unique because it is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;an isosceles triangle&lt;/strong&gt;.  Points J, K, L are unique as determined by the requirement to form an $isosceles$ triangle with 2 angles equal to $\alpha$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Examples of obtuse and acute triangles with several placements of the internal triangles:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;br&gt;&#xA;  &lt;img src=&quot;https://i.stack.imgur.com/1Wa0s.png&quot; alt=&quot;Triangles&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-03-01T02:09:44.713" LastActivityDate="2015-03-01T02:09:44.713" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="9652" PostTypeId="1" CreationDate="2015-02-26T03:28:33.183" Score="44" ViewCount="3015" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Sy2mt.jpg&quot; alt=&quot;http://imgur.com/r6n5cBe&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the &lt;a href=&quot;https://i.stack.imgur.com/dAcfN.jpg&quot; rel=&quot;noreferrer&quot;&gt;image above&lt;/a&gt;  apparently has a hidden link?? I don't know how to do this (even if it may seem simple for some) however I don't believe it's anything malicious. It's a part of a puzzle, but even I don't get where it will ultimately lead. &lt;/p&gt;&#xA;" OwnerDisplayName="user69149" LastEditorUserId="11876" LastEditDate="2016-02-20T05:55:10.913" LastActivityDate="2016-02-20T05:55:10.913" Title="What is encrypted in this image?" Tags="|steganography|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="9653" PostTypeId="2" ParentId="9652" CreationDate="2015-02-26T05:03:45.363" Score="47" Body="&lt;p&gt;We begin by using &lt;a href=&quot;https://futureboy.us/stegano/decinput.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;steghide&lt;/a&gt; to reveal the payload:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;http://pastebin.com/szEs3Hhq&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pastebin.com/szEs3Hhq&quot; rel=&quot;nofollow noreferrer&quot;&gt;The link&lt;/a&gt; brings us to a paste where we get &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;you've done well. Go deeper, and the rewards can be found. Difficulty will increase as you solve each test.&#xA;May you find your meaning.&#xA;&#xA;486578206973206561737920656e6f7567682e204e65656420746f20676f206465657065723f0d0a0d0a4441426d696e46236d696e47444741&#xA;&#xA;Let the hunt begin.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The &lt;code&gt;0d0a&lt;/code&gt; in there looks very suspicious (&lt;code&gt;\r\n&lt;/code&gt; in C-like languages). Could it be a hex encoded ASCII string? Decoding gets us:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Hex is easy enough. Need to go deeper?&#xA;&#xA;DABminF#minGDGA&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;A chord progression? For D Major, that is &lt;code&gt;I V vi iii IV I IV V&lt;/code&gt;. A Google search doesn't turn up anything. Maybe just search the chord names? (EDIT: It's Pachelbel's &quot;Canon in D&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That turns up a Reddit &lt;a href=&quot;https://www.reddit.com/r/N001109/comments/2wjr3z/n009/&quot; rel=&quot;nofollow noreferrer&quot;&gt;/r/N001109&lt;/a&gt; post and a &lt;a href=&quot;https://www.youtube.com/watch?v=R_z2MKvyals&quot; rel=&quot;nofollow noreferrer&quot;&gt;Youtube video&lt;/a&gt; of the numbers &lt;code&gt;0 0 1 1 2 1&lt;/code&gt; (&lt;a href=&quot;https://puzzling.stackexchange.com/questions/9652/what-is-encrypted-in-this-image/9653#comment35514_9653&quot;&gt;rajb245 suggested&lt;/a&gt; the beeps heard could be Morse Code). So I try &lt;a href=&quot;https://www.reddit.com/r/N001121&quot; rel=&quot;nofollow noreferrer&quot;&gt;/r/N001121&lt;/a&gt; and &lt;a href=&quot;https://www.reddit.com/r/N001121/comments/2wjwcf/23015/&quot; rel=&quot;nofollow noreferrer&quot;&gt;its only post&lt;/a&gt;. Dead end. Back to the original /r/N001109 post. There's a &lt;a href=&quot;https://www.reddit.com/r/N001109/comments/2wjr3z/n009/corgxo3&quot; rel=&quot;nofollow noreferrer&quot;&gt;single comment&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;150 164 164 160 163 072 057 057 167 167 167 056 162 145 144 144 151 164 056 143 157 155 057 162 057 116 060 061 061 060 070 057 015 012 015 012 123 110 101 114 114 040 127 105 077 040&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Looks like octal. It decodes to (there's a space after the question mark):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;https://www.reddit.com/r/N01108/&#xA;&#xA;SHALL WE? &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Following that takes us to &lt;a href=&quot;https://www.reddit.com/r/N01108/&quot; rel=&quot;nofollow noreferrer&quot;&gt;/r/N01108&lt;/a&gt; (obviously) with &lt;a href=&quot;https://www.reddit.com/r/N01108/comments/2wjtnm/alice/&quot; rel=&quot;nofollow noreferrer&quot;&gt;its only post&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;060 061 060 061 060 061 061 061 040 060 061 060 060 061 060 060 060 040 060 061 060 060 060 061 060 061 040 060 061 060 060 061 061 061 060 040 060 060 061 060 060 060 060 060 040 060 061 060 060 060 061 060 060 040 060 061 060 060 061 061 061 061 040 060 061 060 060 060 061 060 061 040 060 061 060 061 060 060 061 061 040 060 060 061 060 060 060 060 060 040 060 061 060 061 060 061 060 060 040 060 061 060 060 061 060 060 060 040 060 061 060 060 060 061 060 061 040 060 060 061 060 060 060 060 060 040 060 061 060 061 060 060 061 060 040 060 061 060 060 060 060 060 061 040 060 061 060 060 060 060 061 060 040 060 061 060 060 060 060 061 060 040 060 061 060 060 061 060 060 061 040 060 061 060 061 060 061 060 060 040 060 060 061 060 060 060 060 060 040 060 061 060 060 061 060 060 060 040 060 061 060 060 061 061 061 061 040 060 061 060 060 061 061 060 060 040 060 061 060 060 060 061 060 061 040 060 060 061 060 060 060 060 060 040 060 061 060 060 060 061 060 061 040 060 061 060 060 061 061 061 060 040 060 061 060 060 060 061 060 060 040 060 060 061 061 061 061 061 061 040 060 060 060 060 061 060 061 060 040 060 061 060 060 061 061 060 060 040 060 061 060 060 060 060 061 061 040 060 061 060 060 060 060 060 061 040 060 061 060 061 060 060 061 060 040 060 061 060 061 060 060 061 060 040 060 061 060 060 061 061 061 061 040 060 061 060 060 061 061 060 060 040 060 061 060 060 061 061 060 060 040 060 060 061 060 060 060 060 060 040 060 061 060 060 060 061 061 061 040 060 061 060 061 060 061 060 061 040 060 061 060 061 060 061 060 060 040 060 061 060 060 060 061 060 061 040 060 061 060 060 061 061 061 060 040 060 061 060 060 060 060 061 060 040 060 061 060 060 060 061 060 061 040 060 061 060 061 060 060 061 060 040 060 061 060 060 060 061 061 061 040 060 060 061 060 060 060 060 060 040 060 061 060 061 060 060 060 060 040 060 060 061 061 060 060 061 060 040 060 060 061 061 060 061 060 061 040 060 061 060 060 061 061 060 060 040 060 060 061 061 060 060 061 061 040 060 060 061 061 060 061 060 061 040 060 060 061 060 061 061 061 061 040 060 060 061 061 060 060 061 061 040 060 060 061 061 060 061 061 060 040 060 060 060 060 061 060 061 060 040 060 060 061 061 060 060 060 061 040 060 061 060 060 061 061 060 061 040 060 061 060 060 061 061 061 061 040 060 061 060 060 061 061 061 060 040 060 061 060 061 060 061 060 060 040 060 061 060 060 061 060 060 060 040&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Another octal encoded string (with another random space at the end)!&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;01010111 01001000 01000101 01001110 00100000 01000100 01001111 01000101 01010011 00100000 01010100 01001000 01000101 00100000 01010010 01000001 01000010 01000010 01001001 01010100 00100000 01001000 01001111 01001100 01000101 00100000 01000101 01001110 01000100 00111111 00001010 01001100 01000011 01000001 01010010 01010010 01001111 01001100 01001100 00100000 01000111 01010101 01010100 01000101 01001110 01000010 01000101 01010010 01000111 00100000 01010000 00110010 00110101 01001100 00110011 00110101 00101111 00110011 00110110 00001010 00110001 01001101 01001111 01001110 01010100 01001000 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Huh. Octal encoded binary encoded ASCII?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;WHEN DOES THE RABBIT HOLE END?&#xA;LCARROLL GUTENBERG P25L35/36&#xA;1MONTH&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;A reference to the Gutenberg project and &lt;em&gt;Alice in Wonderland&lt;/em&gt; (the post is titled &quot;alice&quot;)? &lt;a href=&quot;http://www.gutenberg.org/files/11/11-pdf.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;The PDF&lt;/a&gt;'s 25th page, lines 35 and 36 read:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;'If everybody minded their own business,' the Duchess said in a hoarse growl, 'the world&#xA;would go round a deal faster than it does.'&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;That's a dead end for what I will assume to be a month. Back to the post, let's look at &lt;a href=&quot;https://www.reddit.com/r/N01108/comments/2wjtnm/alice/coroyfp&quot; rel=&quot;nofollow noreferrer&quot;&gt;the comment&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;newsyyyodhfssnviqwlbjnyqibxaeytcuy.kumetgn.bij&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Maybe its reply?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;31 32 34 20 31 31 31 20 31 32 34 20 31 31 34 20 31 30 35 20 31 31 37 20 31 30 36 20 31 31 32 20 31 32 30 20 31 30 37&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Looks like hex. Decoding gets us:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;124 111 124 114 105 117 106 112 120 107&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;An octal string. Decoding gets us:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;TITLEOFJPG&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Could the jumbled string of letters be a string encrypted with the title of the original image? Now we're stuck because we don't know the original name. I tried a reverse Google Image search, but that turned up nothing.&lt;/p&gt;&#xA;" OwnerUserId="2143" OwnerDisplayName="Cole Johnson" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2015-10-30T03:24:55.220" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="9654" PostTypeId="2" ParentId="9647" CreationDate="2015-02-27T09:13:37.407" Score="1" Body="&lt;p&gt;Some of these answers are a bit jokey, but then warspyking is jokey, so who knows - they might be right!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;here&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;____&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;bored&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Above-board&lt;/strong&gt; (the word 'here' is only to indicate the point of focus).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;bets&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;_&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;All bets are off.&lt;/strong&gt; (bets is coming &quot;off&quot; the line)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;all4114&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;All for one and one for all.&lt;/strong&gt; (easiest one)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;knightleg&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Nightly&lt;/strong&gt; (as in happening every night; the G silent as in some Gaelic words)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A&lt;/strong&gt;&lt;br&gt;&#xA;\&lt;br&gt;&#xA;&lt;strong&gt;C&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;see after a slope&lt;/strong&gt; (wild guessing here, but so were a couple of the ones I got right)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&amp;lt;3 &amp;lt;-&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;After one's own heart&lt;/strong&gt; (arrow pointing after heart) - thanks to Pierre-Arthur Ferraro for this one!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-11T20:13:37.310" LastActivityDate="2015-03-11T20:13:37.310" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="9655" PostTypeId="2" ParentId="9451" CreationDate="2015-02-27T09:26:21.077" Score="1" Body="&lt;p&gt;I believe that the answer of user3453281 for the winning probability is correct, however I will try to give some more details on how to calculate it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When assuming uniform probability, then the first guess will always have a $3/6=1/2$ change to be correct, independent of your guess (assuming that you can't guess the same box twice for the same jewel). Because the problem is symmetric, then for simplicity sake lets call the initial guess 1, 2 and 3. If you are correct, then for the following guesses those three boxes will have a lower probability of containing another jewel, namely only two of those three boxes can contain another jewel.&lt;br&gt;&#xA; Lets assume that a participant wants every consecutive guess to have the highest probability of being correct. This means that the second guess has to be 4, 5 and 6, which has a probability of being correct of $\frac{3}{3+2/3\cdot3}=3/5$. This equalizes the probability of all boxes, thus the participant can chose any three boxes for the third guess, followed by the forth guess containing exactly the other three boxes, with respective probabilities, $\frac{2/3\cdot3}{2/3\cdot6}=1/2$ and $\frac{2/3\cdot3}{1/3\cdot3+2/3\cdot3}=2/3$. This forth guess equalizes the probability of all boxes again, meaning that the fifth guess can be three boxes, followed by the sixth guess containing exactly the other three boxes, with respective probabilities, $\frac{1/3\cdot3}{1/3\cdot6}=1/2$ and $\frac{1/3\cdot3}{0/3\cdot3+1/3\cdot3}=1/1$. The change of all these six guesses being correct thus has a probability of $1/2\cdot3/5\cdot1/2\cdot2/3\cdot1/2\cdot1/1=1/20$ or 5%.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The best strategy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the previous section I already mentioned which strategy should be used, however this leaves some room for variation. When considering that every box and jewel has identical properties (concerning this game show) and the order in which you would place every guess does not affect your final probability of being correct, then the only constraint to the best strategy should be that each box should part of exactly three guesses.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8810" LastEditorUserId="8810" LastEditDate="2015-02-27T13:43:11.693" LastActivityDate="2015-02-27T13:43:11.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9656" PostTypeId="2" ParentId="9647" CreationDate="2015-02-27T09:44:22.413" Score="0" Body="&lt;p&gt;here&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;bored&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;here I lay, bored!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;bets&lt;br&gt;&#xA;_&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;bets are high&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;all4114&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;all for one , one for all&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;knightleg&lt;br&gt;&#xA; I don't know the answer for this one &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A&lt;br&gt;&#xA;\&lt;br&gt;&#xA;C&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Alternating current, the slash alternates between left and right&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&amp;lt;3 &amp;lt;-&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Love struck&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9174" LastActivityDate="2015-02-27T09:44:22.413" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9657" PostTypeId="1" AcceptedAnswerId="9660" CreationDate="2015-02-27T09:50:22.093" Score="6" ViewCount="1130" Body="&lt;p&gt;Which country is missing on the following list?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Norway, Sweden, Burma, Austria, Peru, ?????, Ghana, South Korea&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9125" LastActivityDate="2015-08-12T15:20:01.670" Title="A list of eight countries" Tags="|pattern|geography|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9658" PostTypeId="2" ParentId="9652" CreationDate="2015-02-27T09:50:44.280" Score="2" Body="&lt;p&gt;Some additional &quot;analysis&quot; of the image (potentially useless). All based on the image downloaded here (as a BMP file)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Image size 800 x 600 is not suspicious&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;All RGB values are grey-values (No difference in red/green/blue)&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Considering grey-level values in further analysis only&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;4843 non-black (non-zero) pixels in total. The image below is a binary-map of those pixels. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FkKTa.png&quot; alt=&quot;Non-zero-pixels-mask&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A full pixel value histogram: &lt;a href=&quot;http://pastebin.com/vZ7qHt7m&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://pastebin.com/vZ7qHt7m&lt;/a&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/77hNd.png&quot; alt=&quot;Histogram&quot;&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Summing pixel grey-values horizontally &amp;amp; vertically in the both the original and the &quot;binary mask&quot; image yields:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;img src=&quot;https://i.stack.imgur.com/lEKmB.png&quot; alt=&quot;ImageSums&quot;&gt;&lt;/li&gt;&#xA;&lt;li&gt;vertical sum (original): &lt;a href=&quot;http://pastebin.com/6cUTt7JQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://pastebin.com/6cUTt7JQ&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;vertical sum (binary): &lt;a href=&quot;http://pastebin.com/cprLUTHv&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://pastebin.com/cprLUTHv&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;horizontal sum (original): &lt;a href=&quot;http://pastebin.com/bQT2p5xj&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://pastebin.com/bQT2p5xj&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;horizontal sum (binary): &lt;a href=&quot;http://pastebin.com/guq4xZpE&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://pastebin.com/guq4xZpE&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5840" LastActivityDate="2015-02-27T09:50:44.280" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9659" PostTypeId="2" ParentId="8625" CreationDate="2015-02-27T10:37:24.053" Score="20" Body="&lt;p&gt;Martin Petters is in &lt;strong&gt;room 16&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Justification:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; By flipping the initials of the co-workers and by flipping &#xA; the digits of their room numbers, you get the corresponding &#xA; element numbers of the corresponding chemical elements.&lt;br&gt;&#xA; The birth dates are red herrings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Step by step solution:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 15 Bobbie Solomon - 51 Sb Antimony&lt;br&gt;&#xA; 21 Galvin MacNeil - 12 Mg Magnesium&lt;br&gt;&#xA; 22 Ismael Tirrell - 22 Ti Titanium&lt;br&gt;&#xA; 31 Lorene Alberts - 13 Al Aluminium&lt;br&gt;&#xA; 42 Robert Cabello - 24 Cr Chromium&lt;br&gt;&#xA; 48 Olivia Perkins - 84 Po Polonium&lt;br&gt;&#xA; 52 Nelson Morales - 25 Mn Manganese&lt;br&gt;&#xA; 65 Alfred Bennett - 56 Ba Barium&lt;br&gt;&#xA; 73 Buster Robbins - 37 Rb Rubidium&lt;br&gt;&#xA; 84 Dorris Compton - 48 Cd Cadmium&lt;br&gt;&#xA; 87 Tamara Pierson - 78 Pt Platinum&lt;br&gt;&#xA; 97 Ursula Akiyama - 79 Au Gold&lt;br&gt;&#xA; ---------------------------------------------------&lt;br&gt;&#xA; 16 Martin Petters - 61 Pm Promethium&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-02-27T10:37:24.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9660" PostTypeId="2" ParentId="9657" CreationDate="2015-02-27T11:20:14.340" Score="10" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Egypt&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is the list of home countries of UN Secretary Generals. And the only country missing from the list is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Egypt&lt;/strong&gt; (Boutros Boutros-Ghali)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="2730" LastEditDate="2015-08-12T15:20:01.670" LastActivityDate="2015-08-12T15:20:01.670" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9661" PostTypeId="2" ParentId="8841" CreationDate="2015-02-27T11:58:12.770" Score="4" Body="&lt;p&gt;Sequence 2 is the list of:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; UN Secretary Generals - with their initials reversed&lt;br&gt;&#xA; Trygve Lie, Dag Hammarskjöld, U Thant, Kurt Waldheim, Javier Pèrez de Cuèllar, Boutros Boutros-Ghali, Kofi Annan, Ban Ki-moon&lt;br&gt;&#xA;&lt;br&gt;&#xA; So the full sequence is:&lt;br&gt;&#xA; Lt, Hd, Tu, Wk, &lt;strong&gt;Cdpj&lt;/strong&gt;, Gbb, Ak, M-kb  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastActivityDate="2015-02-27T11:58:12.770" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9662" PostTypeId="2" ParentId="9619" CreationDate="2015-02-27T13:27:20.737" Score="0" Body="&lt;p&gt;Are you a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gun, probably belonging to a law enforcement official?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cry you may crook (relates to its design)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Criminals are / should be afraid of such a gun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As you gaze at my look&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As the criminal looks down the &lt;em&gt;sight&lt;/em&gt; of the weapon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A top notch runner (relates to its design)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bullets go, like, totally fast, y'all&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Diverts while I cover (relates to its purpose)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Covering fire distracts the criminal while the officer finds cover&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I parch or I perch (relates to its purpose)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This one I don't totally get and would appreciate some help. Guns get hot and dry when fired (parch) and you could rest them on things to steady your aim (perch) but that doesn't relate to its purpose.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But outwardly search&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The gun is searching for a target when behind cover&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A stretch and a spring (relates to its design)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The slides moves back and forth on a spring when fired&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An ingenious thing&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Firearms are clever designs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastActivityDate="2015-02-27T13:27:20.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9663" PostTypeId="1" AcceptedAnswerId="9665" CreationDate="2015-02-27T13:59:34.770" Score="5" ViewCount="179" Body="&lt;p&gt;In life I am yellow,&lt;br&gt;&#xA;Not jaundiced, but bright.&lt;br&gt;&#xA;In death I am hollow&lt;br&gt;&#xA;But, broken, I bite!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2015-02-27T14:17:28.923" Title="Not jaundiced but bright" Tags="|riddle|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9664" PostTypeId="1" AcceptedAnswerId="9666" CreationDate="2015-02-27T14:02:10.043" Score="22" ViewCount="1278" Body="&lt;p&gt;I am talking about a set of uncapitalized, everyday English words, each&#xA;with five or more letters that all share a certain property.&#xA;There are only &lt;strong&gt;nine words&lt;/strong&gt; in Merriam-Webster that have this property:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;shire, washing, color, aware, carol, connect, virgin, ??????, ??????&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What are the two missing words?&lt;br&gt;&#xA;(Remember: Only English words with at least five letters. Only uncapitalized&#xA;words. Only words that show up in Merriam-Webster)&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="8874" LastEditDate="2015-03-19T09:38:28.613" LastActivityDate="2015-03-19T09:38:28.613" Title="A fellowship of nine words" Tags="|pattern|word-sequence|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9665" PostTypeId="2" ParentId="9663" CreationDate="2015-02-27T14:17:28.923" Score="5" Body="&lt;p&gt;You are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;a lightbulb&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In life I am yellow,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not jaundiced, but bright.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Incandescent lightbulbs give off a yellowish light&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In death I am hollow&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, broken, I bite!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When lightbulbs die, they are hollow(no longer full of light), and when broken, the glass can hurt you&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4591" LastActivityDate="2015-02-27T14:17:28.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9666" PostTypeId="2" ParentId="9664" CreationDate="2015-02-27T14:27:48.427" Score="23" Body="&lt;p&gt;The missing words are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; florid, sylvan&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because they have the property that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; they are part of U.S. state names. The words can't be full words of state names, otherwise we'd have island, north, south and jersey, too. A tenth word might be &quot;strict&quot;, if you count the District of Columbia.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(I think one could argue whether the missing words are really&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;everyday&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;words.)&lt;/p&gt;&#xA;" OwnerUserId="9838" LastActivityDate="2015-02-27T14:27:48.427" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9668" PostTypeId="2" ParentId="9652" CreationDate="2015-02-27T16:38:48.540" Score="5" Body="&lt;p&gt;Taking a similar path like Cole Johnson but with a different result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with the first comment: pastebin.com/szEs3Hhq by Gabriel Santos. This is the biggest doubt. I do not see a hint that the URL is embedded in that image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course I find:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;you've done well. Go deeper, and the rewards can be found. Difficulty will increase as you solve each test.&#xA;May you find your meaning.&#xA;&#xA;486578206973206561737920656e6f7567682e204e65656420746f20676f206465657065723f0d0a0d0a4441426d696e46236d696e47444741&#xA;&#xA;Let the hunt begin.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Decoding the hex gives:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;irb(main):002:0&amp;gt; str = &quot;486578206973206561737920656e6f7567682e204e65656420746f20676f206465657065723f0d0a0d0a4441426d696e46236d696e47444741&quot;&#xA;=&amp;gt; &quot;486578206973206561737920656e6f7567682e204e65656420746f20676f206465657065723f0d0a0d0a4441426d696e46236d696e47444741&quot;&#xA;irb(main):005:0&amp;gt; str.to_byte_string  &#xA;=&amp;gt; &quot;Hex is easy enough. Need to go deeper?\r\n\r\nDABminF#minGDGA&quot;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Googling &lt;code&gt;DABminF#minGDGA&lt;/code&gt; has a YouTube video as first hit &lt;a href=&quot;https://www.youtube.com/watch?v=R_z2MKvyals&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=R_z2MKvyals&lt;/a&gt;&#xA;The narrator says &lt;code&gt;N0112&lt;/code&gt;.&#xA;The video looks like this. Note the &lt;code&gt;434&lt;/code&gt; in the upper left corner &lt;img src=&quot;https://i.stack.imgur.com/6uSZ4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding this post &lt;a href=&quot;https://www.reddit.com/r/N001121&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.reddit.com/r/N001121&lt;/a&gt;. Which now has a second comment.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;66 61 40 66 66 40 66 60 40 64 60 40 66 61 40 66 64 40 66 61 40 64 60 40 66 61 40 66 66 40 66 63 40 64 60 40 66 61 40 66 66 40 66 64 40 64 60 40 66 61 40 66 64 40 66 65 40 64 60 40 66 61 40 66 64 40 66 62 40 64 60 40 66 61 40 66 65 40 66 61 40 64 60 40 66 61 40 66 65 40 66 66 40 64 60 40 66 64 40 66 60 40 64 60 40 66 61 40 66 65 40 66 63 40 64 60 40 66 66 40 66 62 40 64 60 40 66 66 40 66 65 40 64 60 40 66 61 40 66 63 40 66 61 40 64 60 40 66 61 40 66 65 40 66 61 40 64 60 40 66 61 40 66 61 40 66 60 40 64 60 40 66 61 40 66 64 40 66 66 40 64 60 40 66 61 40 66 62 40 66 62&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Authenticity is questionable; could be created by a freeloader.&#xA;Nevertheless, I continue.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;irb(main):007:0&amp;gt; str2 = '66 61 40 66 66 40 66 60 40 64 60 40 66 61 40 66 64 40 66 61 40 64 60 40 66 61 40 66 66 40 66 63 40 64 60 40 66 61 40 66 66 40 66 64 40 64 60 40 66 61 40 66 64 40 66 65 40 64 60 40 66 61 40 66 64 40 66 62 40 64 60 40 66 61 40 66 65 40 66 61 40 64 60 40 66 61 40 66 65 40 66 66 40 64 60 40 66 64 40 66 60 40 64 60 40 66 61 40 66 65 40 66 63 40 64 60 40 66 66 40 66 62 40 64 60 40 66 66 40 66 65 40 64 60 40 66 61 40 66 63 40 66 61 40 64 60 40 66 61 40 66 65 40 66 61 40 64 60 40 66 61 40 66 61 40 66 60 40 64 60 40 66 61 40 66 64 40 66 66 40 64 60 40 66 61 40 66 62 40 66 62'&#xA;#Some iterations of adding more and more transformations. &#xA;#Converting from  octed and decimal in several iterations&#xA;#Finally this turns up:&#xA;irb(main):025:0&amp;gt; str2.split(&quot; &quot;).collect do |x| x.oct end.collect do |x| x.chr end.join(&quot;&quot;).split(&quot; &quot;).collect do |x| x.oct end.collect do |x| x.chr end.join(&quot;&quot;).split(&quot; &quot;).collect do |x| x.oct end.collect do |x| x.chr end.join &quot;&quot;&#xA;=&amp;gt; &quot;pastebin k25YiHfR&quot;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This pastebin URL shows this (http pastebin dot com/ k25YiHfR):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Decoding this twice:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;irb(main):027:0&amp;gt; str3 = 'YjI1b.....'&#xA;irb(main):034:0&amp;gt; Base64.decode64(Base64.decode64(str3))&#xA;=&amp;gt; &quot;&amp;lt;some explicit text about Mario and Yoshi; decode it yourselves&amp;gt;&quot;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Conclusion: The entry point is questionable and if the riddle path was not already corrupted by some freeloader is also possible.&lt;/p&gt;&#xA;" OwnerUserId="9841" LastActivityDate="2015-02-27T16:38:48.540" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9669" PostTypeId="1" AcceptedAnswerId="9831" CreationDate="2015-02-27T19:07:08.300" Score="-7" ViewCount="810" Body="&lt;p&gt;The two perfect logicians $A$ and $C$ meet their colleague $B$ at an exam, and they are not aware of the fact that $B$ is a moron and a horrible mathematician.&#xA;At the exam, the examiner informs them that he has chosen two &lt;strong&gt;different&lt;/strong&gt; integers $x$ and $y$ with $2\le x&amp;lt;y\le100$, such that $y$ is a multiple of $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The examiner then tells the difference $d=y-x$ to the first mathematician $A$,  the ratio $r=y/x$ to the second mathematician $B$, and the sum $s=x+y$ to the third mathematician $C$ in such a way that none knows which numbers have been whispered to the others. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The three mathematicians then start this implying conversation:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$A$: I don't know the numbers, and I know that you both know this.&lt;/li&gt;&#xA;&lt;li&gt;$B$: I already knew the numbers when the examiner told me their ratio.&lt;/li&gt;&#xA;&lt;li&gt;$A$ and $C$ simultaneously: Aha! We have just deduced the two numbers.&lt;/li&gt;&#xA;&lt;li&gt;$B$: Oops, damn, I am sorry! I think I made a mistake in my calculation. &#xA;As a matter of fact, the ratio $r$ does not allow me to deduce the two numbers.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;&lt;strong&gt;Question1:&lt;/strong&gt; What are those two numbers $x$ and $y$??&lt;/h2&gt;&#xA;&#xA;&lt;h2&gt;&lt;strong&gt;Question2:&lt;/strong&gt; What are those two numbers $x$ and $y$, if the following conversation took place instead from the third line on??&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$C$: Aha! I have just deduced the two numbers.&lt;/li&gt;&#xA;&lt;li&gt;$A$:  I strongly recommend for $B$ to recalculate his ratio because he was surely wrong according to my data. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For hints and notes, refer to the following links:&lt;br&gt;&#xA;  &lt;a href=&quot;https://puzzling.stackexchange.com/questions/9499/know-the-numbers-from-this-conversation&quot;&gt;Deducing Two Numbers based on their Difference and Ratio&lt;/a&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;https://puzzling.stackexchange.com/questions/9400/what-are-the-numbers&quot;&gt;What are the numbers?&lt;/a&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;https://puzzling.stackexchange.com/questions/9525/what-are-the-numbers-third-edition&quot;&gt;https://puzzling.stackexchange.com/questions/9525/what-are-the-numbers-third-edition&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9517" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-10-07T12:52:44.947" Title="Deduce the numbers! (Fourth edition)" Tags="|logical-deduction|number-theory|" AnswerCount="1" CommentCount="8" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9670" PostTypeId="2" ParentId="9664" CreationDate="2015-02-27T19:56:32.187" Score="6" Body="&lt;p&gt;@M Oehm is correct as to what the answers are, and that all answers&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; are part of state names. Confirm it for yourself in the list below (hidden by spoiler).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here are the words in context:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; New Hamp&lt;strong&gt;shire&lt;/strong&gt;,&#xA; &lt;strong&gt;Washing&lt;/strong&gt;ton,&#xA; &lt;strong&gt;Color&lt;/strong&gt;ado,&#xA; Del&lt;strong&gt;aware&lt;/strong&gt;, &#xA; North/South &lt;strong&gt;Carol&lt;/strong&gt;ina,&#xA; &lt;strong&gt;Connect&lt;/strong&gt;icut,&#xA; &lt;strong&gt;Virgin&lt;/strong&gt;ia,&#xA; &lt;strong&gt;Florid&lt;/strong&gt;a,&#xA; Penn&lt;strong&gt;sylvan&lt;/strong&gt;ia&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hope this helps!&lt;/p&gt;&#xA;" OwnerUserId="9847" LastActivityDate="2015-02-27T19:56:32.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9671" PostTypeId="1" AcceptedAnswerId="9672" CreationDate="2015-02-27T21:33:21.300" Score="5" ViewCount="1435" Body="&lt;p&gt;I've been given a puzzle, and I just can't work out an answer. Can you help?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd also like to know your thought process for solving, as this matters more than the correct answer. Does it belong to a pattern of puzzles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scenario is this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There are six books by six different authors - J,K,L,M,N and O - which are to be placed on a bookcase with six shelves. &lt;/li&gt;&#xA;&lt;li&gt;The shelves are numbered from one, the highest, to six, the lowest, and exactly one book will be placed on each shelf.&lt;/li&gt;&#xA;&lt;li&gt;The book K must be on the second shelf below the book J&lt;/li&gt;&#xA;&lt;li&gt;The book O must be placed on either the first or sixth shelf.&lt;/li&gt;&#xA;&lt;li&gt;The book L cannot be placed on either the shelf immediately above or the shelf immediately below the book by M.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Which one of the following is TRUE?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;K and L CANNOT be placed on shelves three and four respectively&lt;/li&gt;&#xA;&lt;li&gt;M and N CANNOT be placed on shelves three and four respectively&lt;/li&gt;&#xA;&lt;li&gt;J and M CANNOT be placed on shelves three and four respectively&lt;/li&gt;&#xA;&lt;li&gt;L and K CANNOT be placed on shelves three and four respectively&lt;/li&gt;&#xA;&lt;li&gt;N and J CANNOT be placed on shelves three and four respectively&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="9849" LastActivityDate="2015-02-27T21:48:20.520" Title="Book stacking brain teaser" Tags="|logical-deduction|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9672" PostTypeId="2" ParentId="9671" CreationDate="2015-02-27T21:44:13.163" Score="6" Body="&lt;p&gt;I think that the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $2$ and is only constrained by the first rule (about J and K).&lt;br&gt;&#xA; J and K have exactly 1 shelf between them. That means that they span 3 shelves. So, no matter the arrangement, either J or K must be one of the shelves 3 or 4.&lt;br&gt;&#xA; So, any solution that fills both 3 and 4 with neither J nor K can't happen. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-02-27T21:44:13.163" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9673" PostTypeId="2" ParentId="9671" CreationDate="2015-02-27T21:48:20.520" Score="6" Body="&lt;p&gt;The true statement is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;2. M and N CANNOT be placed on shelves three and four respectively.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Book J can only occupy shelves 1, 2, 3, 4. Book K can only occupy shelves 3, 4, 5, 6. No matter where you place J and K, shelves 3 and 4 will always be occupied by one or the other. Proof: J on shelf 1, K on shelf 3. J on shelf 2, K on shelf 4. K on shelf 5, J on shelf 3. K on shelf 6, J on shelf 4.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4353" LastActivityDate="2015-02-27T21:48:20.520" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9674" PostTypeId="1" CreationDate="2015-02-28T06:22:34.097" Score="1" ViewCount="5300" Body="&lt;p&gt;Source:&#xA;&lt;a href=&quot;http://gpuzzles.com/mind-teasers/pattern-series-picture-puzzle/?source=stack&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://gpuzzles.com/mind-teasers/pattern-series-picture-puzzle/?source=stack&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is missing in the picture?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qdwEj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user9857" LastEditorUserId="2444" LastEditDate="2019-03-12T12:28:16.557" LastActivityDate="2019-03-12T12:28:16.557" Title="Missing is what in picture" Tags="|logical-deduction|pattern|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="9677" PostTypeId="2" ParentId="9515" CreationDate="2015-02-28T07:17:32.940" Score="1" Body="&lt;p&gt;Answer 1:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The scenery suggests that it is not close to the equator, because of the tree with no leaves which suggests sufficient temperature difference between seasons. This means that the sun will not rise extremely high into the sky and thus will always cast some shadows. The shadows in the picture are relatively short, thus it would have to be close to 12:00.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer 2:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The tree has lost its leaves, however there are no visible leaves on the ground. This suggests that sufficient time has past since the tree lost its leaves, such that the leaves might been blown away by the wind or washed away by rain. Thus it would have to be early spring. The farmers seem to be plowing the fields, which might also give some clues, however I do not know enough about farming, but after a few searches it seems that fields can be plowed in both spring and autumn.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer 3:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The buoy in the left of the picture most likely would function as a lateral marker, which marks the edge of a save water channel, which suggests that boats will navigate the river regularly for transport.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer 4:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The buoy causes waves, which travel to the right, so the river is flowing to the right. Also the small boat seems to be docking, which is easier when the front of the boat is turned into the flow of the river. Given that it is roughly midday, then the shadows would either point north or south, depending on the hemisphere this picture was taken. The people in the picture seem European, so it would be the northern hemisphere. Thus the river is flowing towards the west.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer 5:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The buoy is relatively close to the side of the river, so the river would have to be deep enough for large ships. Depending on your definition of deep and shallow, I would say the river near the boat would be deep.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer 6:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The small boat is most likely not meant for transport along the river, but for transport across the river as a ferry. This suggests that there are no nearby other means of getting across, thus there is no bridge nearby.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer 7:&lt;br&gt;&#xA;I do not know the answer to this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer 8:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since I assume that this is the northern hemisphere, then the birds are flying in V formation towards the north.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8810" LastActivityDate="2015-02-28T07:17:32.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9678" PostTypeId="1" AcceptedAnswerId="9701" CreationDate="2015-02-28T13:57:02.183" Score="-1" ViewCount="972" Body="&lt;p&gt;Here are some rebus puzzles for you to solve:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Window Window Window TTTTTTTTT5&lt;/li&gt;&#xA;&lt;li&gt;&lt;sub&gt;soft&lt;/sub&gt;&lt;/li&gt;&#xA;&lt;li&gt;li ne&lt;/li&gt;&#xA;&lt;li&gt;lagond&lt;/li&gt;&#xA;&lt;li&gt;Three age&lt;/li&gt;&#xA;&lt;li&gt;teen teen teen teen&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7482" LastEditorUserId="5188" LastEditDate="2015-03-01T06:48:01.683" LastActivityDate="2017-06-17T01:13:21.727" Title="Six rebus puzzles - starting with three windows" Tags="|rebus|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9679" PostTypeId="1" AcceptedAnswerId="9688" CreationDate="2015-02-28T14:02:54.177" Score="2" ViewCount="579" Body="&lt;p&gt;I have ranked the US-states according to a certain cryptic property&#xA;and put them into a list.  A contiguous piece of this cryptic list&#xA;looks as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;..., ?????, Nevada, West Virginia, Kansas, Oregon, Minnesota,&#xA;  California, Wisconsin, Iowa, Texas, Florida, Michigan,&#xA;  Arkansas, Missouri, Maine, Alabama, Illinois, ?????, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Which states are represented by ????? ?&lt;/strong&gt;&#xA;&lt;strong&gt;What is the cryptic property?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9125" LastActivityDate="2015-03-30T20:32:19.053" Title="A cryptic ranking of the US-states" Tags="|pattern|geography|knowledge|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9680" PostTypeId="1" AcceptedAnswerId="9683" CreationDate="2015-02-28T14:07:17.353" Score="2" ViewCount="313" Body="&lt;p&gt;This is my fifth batch of sequence puzzles that are nasty and hard to solve; &#xA;yet, each of them has a clear and justifiable solution.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sequence 1:&lt;br&gt;&#xA;  T?M, AOT?, ROT?, AN?, T?SB, ROT?, TFA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sequence 2:&lt;br&gt;&#xA;  5, 12, ?, 14, 11, 8, 16, ?, 19, 3, 17, ?, 15, 10, ?, 13, 4, 18, ?, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sequence 3:&lt;br&gt;&#xA;  T, u, b, d, l, ?, ?, ?, i, b, o, h, f&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sequence 4:&lt;br&gt;&#xA;   A, P, ?, L, S, A, P, A, ?, B, L, H, M, R, T, ?, M, M, M, ?, ?, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="9125" LastEditDate="2015-03-04T10:05:35.623" LastActivityDate="2015-03-10T07:07:06.040" Title="One more batch of four sequence puzzles" Tags="|number-sequence|letter-sequence|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9681" PostTypeId="2" ParentId="9678" CreationDate="2015-02-28T14:22:09.107" Score="3" Body="&lt;p&gt;Rebus 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Windows + 9T + 5 = &lt;strong&gt;Windows 95&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rebus 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; line with a break = &lt;strong&gt;linebreak&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rebus 6:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4 teen = &lt;strong&gt;fourteen&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-02-28T14:22:09.107" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9682" PostTypeId="2" ParentId="9680" CreationDate="2015-02-28T14:24:35.950" Score="6" Body="&lt;p&gt;&lt;a href=&quot;https://oeis.org/search?q=14%2C11%2C8%2C16&amp;amp;language=english&amp;amp;go=Search&quot;&gt;Looking up&lt;/a&gt; gave me the answer for sequence 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Number on a standard, London, or clock dartboard read in a counter- clockwise direction. Sequence:  5, 12, 9, 14, 11, 8, 16, 7, 19, 3, 17, 2, 15, 10, 6, 13, 4, 18, 1, ... &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7482" LastActivityDate="2015-02-28T14:24:35.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9683" PostTypeId="2" ParentId="9680" CreationDate="2015-02-28T14:32:46.807" Score="5" Body="&lt;p&gt;Sequence 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Star wars:&lt;br&gt;&#xA; &#xA; * TPM = The Phantom Menace&lt;br&gt;&#xA; * AOTC = Attack of the Clones&lt;br&gt;&#xA; * ROTS = Revenge of the Sith&lt;br&gt;&#xA; * ANH = A New Hope&lt;br&gt;&#xA; * TESB = The Empire Strikes Back&lt;br&gt;&#xA; * ROTJ = Return of the Jedi&lt;br&gt;&#xA; * TFA = The Force Awakens&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sequence 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letters of Stack Exchange, shifted by one position in the alphabet:&lt;br&gt;&#xA; S, t, a, c, k, E, x, c, h, a, n, g, e&lt;br&gt;&#xA; T, u, b, d, l, F, y, d, i, b, o, h, f&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-03-10T07:07:06.040" LastActivityDate="2015-03-10T07:07:06.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9684" PostTypeId="2" ParentId="9674" CreationDate="2015-02-28T14:44:15.407" Score="18" Body="&lt;p&gt;Correct answer is &lt;img src=&quot;https://i.stack.imgur.com/1IiDc.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is not because &lt;img src=&quot;https://i.stack.imgur.com/1IiDc.jpg&quot; alt=&quot;enter image description here&quot;&gt; it's left of &lt;img src=&quot;https://i.stack.imgur.com/G8Dud.jpg&quot; alt=&quot;enter image description here&quot;&gt;, because as you can see in the second row &lt;img src=&quot;https://i.stack.imgur.com/1IiDc.jpg&quot; alt=&quot;enter image description here&quot;&gt; it's also left of &lt;img src=&quot;https://i.stack.imgur.com/ggbpg.jpg&quot; alt=&quot;enter image description here&quot;&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the order of elements: &lt;img src=&quot;https://i.stack.imgur.com/pW5A6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3eWYM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RTsd5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HDdHU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/G8Dud.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XnEA6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wm9Iw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/p0CeF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The matrix puzzle starts with the first element, then adds &lt;strong&gt;each&lt;/strong&gt; element &lt;strong&gt;once&lt;/strong&gt;, and also resets the pattern to repeat itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this is the result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KJ3Yc.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KJ3Yc.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/3eWYM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KJ3Yc.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/3eWYM.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/RTsd5.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KJ3Yc.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/3eWYM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RTsd5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HDdHU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KJ3Yc.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/3eWYM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RTsd5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HDdHU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/G8Dud.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KJ3Yc.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/3eWYM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RTsd5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HDdHU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/G8Dud.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XnEA6.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KJ3Yc.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/3eWYM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RTsd5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HDdHU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/G8Dud.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XnEA6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wm9Iw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KJ3Yc.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/3eWYM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RTsd5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HDdHU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/G8Dud.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XnEA6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wm9Iw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/p0CeF.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you line this list in 6x6 matrix, it will look like the original puzzle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This puzzle does not use it's full enigmatic potential though. For example, it could have been much more interesting and challenging if it could have looked like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AfGlu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9861" LastEditorUserId="9861" LastEditDate="2019-03-12T07:19:26.427" LastActivityDate="2019-03-12T07:19:26.427" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="9685" PostTypeId="2" ParentId="9678" CreationDate="2015-02-28T15:37:26.567" Score="2" Body="&lt;p&gt;Rebus 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Microsoft&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rebus 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To go inland&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rebus 5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Triple H (a wrestler... probably wrong)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2015-02-28T16:54:59.550" LastActivityDate="2015-02-28T16:54:59.550" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9686" PostTypeId="1" AcceptedAnswerId="9687" CreationDate="2015-02-28T16:08:18.923" Score="1" ViewCount="338" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Well, it is always good to start slowly. Hence let us start slowly.&#xA;  You are looking for a certain word in this paragraph which occurs&#xA;  once only.  The first letter of this word is the seventh after a&#xA;  vowel and this vowel is letter number seven to appear after its&#xA;  last letter.  What is the enigmatic word?  What is it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2015-02-28T19:05:43.600" Title="What is the enigmatic word?" Tags="|word|enigmatic-puzzle|" AnswerCount="1" CommentCount="3" FavoriteCount="0" ClosedDate="2015-03-02T15:37:18.553" ContentLicense="CC BY-SA 3.0" />
  <row Id="9687" PostTypeId="2" ParentId="9686" CreationDate="2015-02-28T19:05:43.600" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; “enigmatic”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These types of puzzles usually tease like that. =)&lt;/p&gt;&#xA;" OwnerUserId="1565" LastActivityDate="2015-02-28T19:05:43.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9688" PostTypeId="2" ParentId="9679" CreationDate="2015-02-28T20:37:00.647" Score="5" Body="&lt;p&gt;&lt;strong&gt;Solution&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nebraska and Mississippi&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Description&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The most enraging Starbucks order I’ve ever witnessed and a mnemonic for the 50 states in order of statehood&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Found on &lt;a href=&quot;http://goo.gl/x1CXdi&quot;&gt;http://goo.gl/x1CXdi&lt;/a&gt; with&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; google query: &quot;Nevada, West Virginia, Kansas, Oregon, Minnesota, California&quot; ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4342" LastEditorUserId="4342" LastEditDate="2015-02-28T20:45:41.867" LastActivityDate="2015-02-28T20:45:41.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9690" PostTypeId="2" ParentId="9632" CreationDate="2015-03-01T01:49:23.843" Score="0" Body="&lt;p&gt;&quot;Each of his socks is black or blue, and there are more blue socks than black socks&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;followed by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;My grandfather has always been extremely careful when doing the laundry, and has only ever lost a single sock in his entire life.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I stopped going further immediately, given that socks are only ever sold in pairs, it had to be black. Sorry, was this question supposed to only be looking for mathematical proofs, and not common sense?&lt;/p&gt;&#xA;" OwnerUserId="9868" LastActivityDate="2015-03-01T01:49:23.843" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9691" PostTypeId="1" AcceptedAnswerId="9693" CreationDate="2015-03-01T02:22:53.473" Score="7" ViewCount="1159" Body="&lt;p&gt;What number should replace the &lt;strong&gt;?&lt;/strong&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/MgkYO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7489" LastActivityDate="2015-03-01T05:09:58.213" Title="Find the missing number" Tags="|number-sequence|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="9693" PostTypeId="2" ParentId="9691" CreationDate="2015-03-01T04:51:33.880" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 12 - (sum of digits) Well I don't like to spoil it entirely for others but here you go(since a moderator asked me to explain more): Add the digits of the two numbers in the circles that branch from a circle just one level above and place the sum in the circle. Eg. Start with 72 and 99 at the bottom, 7+2+9+9=27; put 27 up in there and so on...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And it did tease my brain a little. Thanks ;)&lt;/p&gt;&#xA;" OwnerUserId="9865" LastEditorUserId="9865" LastEditDate="2015-03-01T05:09:58.213" LastActivityDate="2015-03-01T05:09:58.213" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9694" PostTypeId="1" AcceptedAnswerId="9695" CreationDate="2015-03-01T05:48:44.190" Score="1" ViewCount="1896" Body="&lt;p&gt;More rebus puzzles:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;RRRFOURRRR&lt;/li&gt;&#xA;&lt;li&gt;Clean X &lt;/li&gt;&#xA;&lt;li&gt;MMMM&lt;/li&gt;&#xA;&lt;li&gt;bus again&lt;/li&gt;&#xA;&lt;li&gt;Puzzle5:&lt;img src=&quot;https://i.stack.imgur.com/q3Soz.png&quot; alt=&quot;Rebus 5&quot;&gt;&lt;/li&gt;&#xA;&lt;li&gt;Puzzle6:&lt;img src=&quot;https://i.stack.imgur.com/mONoP.png&quot; alt=&quot;Rebus 6&quot;&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7482" LastEditorUserId="5188" LastEditDate="2015-03-01T06:50:19.083" LastActivityDate="2015-03-02T05:57:43.863" Title="Six rebus puzzles - starting with three R's" Tags="|rebus|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9695" PostTypeId="2" ParentId="9694" CreationDate="2015-03-01T06:14:01.853" Score="6" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;RRRFOURRRR  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4 in R's = foreigners  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Clean X  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Kleenex (tissues)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;MMMM  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4 M's = forums   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;bus again  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; re-bus = rebus  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;dnuorg in red font  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; red ground (backwards) = red background (thanks to @Josh)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;StateStateStateState in large black font&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; states joined together = United States&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-03-01T07:07:48.927" LastActivityDate="2015-03-01T07:07:48.927" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9696" PostTypeId="2" ParentId="9694" CreationDate="2015-03-01T06:21:25.777" Score="4" Body="&lt;p&gt;Puzzle 5 is actually:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Red background&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Len's got the other four right, I think. I can't make out puzzle 6.&lt;/p&gt;&#xA;" OwnerUserId="4353" LastActivityDate="2015-03-01T06:21:25.777" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9697" PostTypeId="2" ParentId="9694" CreationDate="2015-03-01T06:31:37.993" Score="1" Body="&lt;p&gt;2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Opposite(X) of Clean, &quot;Dirty&quot; OR &quot;Dirty 9&quot; which is game Or &quot;Dirty Dozen&quot; which is a movie&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The &lt;strong&gt;Four Corners Monument&lt;/strong&gt; marks the quadripoint in the Southwestern United States where the states of Arizona, Colorado, New Mexico, and Utah meet. It is the only point in the United States shared by four states, leading to this area's being called the Four Corners region&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9174" LastEditorDisplayName="user9174" LastEditDate="2015-03-01T06:42:37.403" LastActivityDate="2015-03-01T06:42:37.403" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9698" PostTypeId="2" ParentId="9678" CreationDate="2015-03-01T06:44:09.910" Score="3" Body="&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;Three age  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;triage&lt;/strong&gt; (prioritization for medical treatment)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastActivityDate="2015-03-01T06:44:09.910" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9701" PostTypeId="2" ParentId="9678" CreationDate="2015-03-01T07:42:14.983" Score="2" Body="&lt;p&gt;This answer displays the answer for all the rebus puzzles solved by @Gamow,@d'ala'cop and @Len:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rebus 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Windows + 9T + 5 = &lt;strong&gt;Windows 95&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rebus 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; soft in small font = &lt;strong&gt;Microsoft&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rebus 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; line with a break = &lt;strong&gt;linebreak&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rebus 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; go in land = &lt;strong&gt;To go inland&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rebus 5:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Three can be written as tri = &lt;strong&gt;Triage&lt;/strong&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rebus 6:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4 teen = &lt;strong&gt;fourteen&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7482" LastActivityDate="2015-03-01T07:42:14.983" CommentCount="0" CommunityOwnedDate="2015-03-01T07:42:14.983" ContentLicense="CC BY-SA 3.0" />
  <row Id="9703" PostTypeId="2" ParentId="9649" CreationDate="2015-03-01T09:54:32.943" Score="16" Body="&lt;p&gt;If folded as shown below, it can be closed along the matching letters, to form a 12 sided polyhedron with six quadrilateral faces and two sets of three isosceles triangular faces. The quads are arranged in two triplets that join to form two right corners (as found in a cube), and the triangles 'join' the two groups of quads together. The shape has 120 degree rotational symmetry around an axis that passes through the two 'cube-corners'.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uXDg2.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shape is a bit difficult to describe, so I have sketched approximate views of it from two different angles to show, very approximately, how it sticks together. &#xA;&lt;img src=&quot;https://i.stack.imgur.com/PRvB6.jpg&quot; alt=&quot;&quot;&gt; &#xA;All the faces are flat as required. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;OOPS!!! - My solution is wrong. To fold into the shape I have described, the long edge of the small triangle (and short edge of the long triangle) would have to be of length 2/sqrt(3), not sqrt(2) as it is in the 'cutout'.&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="1631" LastEditDate="2015-03-03T20:57:39.253" LastActivityDate="2015-03-03T20:57:39.253" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9704" PostTypeId="2" ParentId="9649" CreationDate="2015-03-01T11:06:02.197" Score="9" Body="&lt;p&gt;This is the solution I intended, though I've accepted &lt;strike&gt;Penguino's&lt;/strike&gt; Florian F's since it has the smallest number of folds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9zKib.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dashed lines are mountain folds, and dotted lines are valley folds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps I'll make a 3D representation later, but the special thing about the resulting shape is that it can be made by cutting a cube into two halves as below, rotating one half by 30°, and sticking them back together. &lt;a href=&quot;http://www.gemologyonline.com/Forum/phpBB2/viewtopic.php?f=8&amp;amp;t=17916&quot; rel=&quot;nofollow noreferrer&quot;&gt;[image source]&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YkhhL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone can think of a good way to change the puzzle so that this is the only solution, please comment!&lt;/p&gt;&#xA;" OwnerUserId="4017" LastEditorUserId="4017" LastEditDate="2015-03-01T11:42:43.133" LastActivityDate="2015-03-01T11:42:43.133" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9705" PostTypeId="1" CreationDate="2015-03-01T11:16:34.727" Score="-4" ViewCount="835" Body="&lt;p&gt;I recall a question that went like this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If a clock is at noon on Sunday, and the clock loses 9 minutes per hour, what time will the clock next show the correct time?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How do you work that one out??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;cheers!&lt;/p&gt;&#xA;" OwnerUserId="9849" LastEditorUserId="5188" LastEditDate="2015-04-16T06:53:50.833" LastActivityDate="2015-04-16T06:53:50.833" Title="Time teaser - clock losing time" Tags="|calculation-puzzle|" AnswerCount="1" CommentCount="2" ClosedDate="2015-03-02T10:48:14.527" ContentLicense="CC BY-SA 3.0" />
  <row Id="9706" PostTypeId="2" ParentId="9649" CreationDate="2015-03-01T11:30:51.393" Score="14" Body="&lt;p&gt;I can do it with 11 folds also.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/hKA72.jpg&quot; alt=&quot;folding&quot;&gt;&#xA;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/CkTzd.jpg&quot; alt=&quot;result&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2015-03-01T16:58:17.117" LastActivityDate="2015-03-01T16:58:17.117" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9707" PostTypeId="2" ParentId="9705" CreationDate="2015-03-01T13:16:42.483" Score="2" Body="&lt;p&gt;Pretty easy I guess. it will be correct again when it is 12 hours behind. 12 hours = 720 minutes. 720 divided by 9 is 80. So in 80 hours it will be correct again. So that is Wednesday at 8PM&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-03-01T13:16:42.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9708" PostTypeId="2" ParentId="9679" CreationDate="2015-03-01T14:21:16.287" Score="3" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The cryptic property is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the &lt;strong&gt;date of statehood&lt;/strong&gt; (the date of admission to the union).&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_U.S._states_by_date_of_admission_to_the_Union&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/List_of_U.S._states_by_date_of_admission_to_the_Union&lt;/a&gt;&lt;br&gt;&#xA;The list has the statehood dates ordered upside down: the younger states come at the beginning, the later states come at the end&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The first missing state is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Nebraska&lt;/strong&gt; (admission date = March 1, 1867).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The second missing state is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Mississippi&lt;/strong&gt; (admission date = December 10, 1817).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="10151" LastEditDate="2015-03-30T20:32:19.053" LastActivityDate="2015-03-30T20:32:19.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9710" PostTypeId="2" ParentId="9403" CreationDate="2015-03-01T15:44:52.457" Score="3" Body="&lt;p&gt;As I have promised, posting my own answers for unfound combinations (only one from listed was not found).&lt;/p&gt;&#xA;&lt;h3&gt;2x5, area 680&lt;/h3&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/lpfCX.png&quot; alt=&quot;&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5781" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-03-01T15:44:52.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9711" PostTypeId="1" AcceptedAnswerId="9741" CreationDate="2015-03-01T16:00:22.880" Score="11" ViewCount="944" Body="&lt;p&gt;Professor Halfbrain has spent his entire weekend watching the games at the local chess tournament. &#xA;As usual, every two players played exactly one game against each other.&#xA;A win yields 1 point, a draw yields half a point, and a loss yields nothing.&#xA;At the end of the tournament, Professor Halfbrain carefully analyzed all the games and compared them to the final tournament ranking of the players.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To his excitement, Halfbrain realized that there were games in the tournament in which the winner had reached fewer points (in the final ranking) than the loser (in the final ranking).&#xA;Professor Halfbrain decided to call such games &lt;em&gt;nonsensical&lt;/em&gt;, and to call all other games (including all the ones that ended in a draw) &lt;em&gt;sensical&lt;/em&gt; games.&#xA;The professor managed to prove the following two deep theorems on sensical and nonsensical games.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Professor Halfbrain's first theorem&lt;/strong&gt;: &#xA;  In every chess tournament, at least $0$ percent of the played games are sensical.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Professor Halfbrain's second theorem&lt;/strong&gt;:&#xA;  There exist chess tournaments, in which at most $100$ percent of the games are sensical.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This puzzle asks you to improve the two theorems of professor Halfbrain and to make them even deeper.&#xA;Find an integer $x$, so that the &quot;$0$ percent&quot; in the first theorem may be replaced by &quot;$x$ percent&quot;, and so that the &quot;$100$ percent&quot; in the second theorem may be replaced by &quot;$x+1$ percent&quot; (again yielding true statements, of course).&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-03-04T15:35:09.750" Title="Musings at a chess tournament" Tags="|mathematics|chess|combinatorics|" AnswerCount="2" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9712" PostTypeId="1" AcceptedAnswerId="9716" CreationDate="2015-03-01T16:33:45.380" Score="9" ViewCount="1415" Body="&lt;p&gt;I took twenty 9-letter everyday English words from the Merriam-Webster dictionary, broke them into sixty 3-letter pieces and ordered the sixty pieces alphabetically:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    acc all ant but day dif eap eid ent ent ent eph equ erd erf ess eth far fer fly  &#xA;    gar hol imp ine ing ion ipm lio lpo mar mho myt nes new nut ogy one ort ory own  &#xA;    pin ple por rec rel rit row sag sca sco som tel ter tfo und use wat wed wer wil&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What are the twenty 9-letter words?&lt;/p&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2015-03-01T21:07:33.973" Title="Twenty nine-letter words" Tags="|word|english|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9713" PostTypeId="2" ParentId="9712" CreationDate="2015-03-01T18:29:13.487" Score="3" Body="&lt;p&gt;This is not the full answer, but more is not possible with my dictionary&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; different, butterfly, equipment, important, mythology, pineapple, something, portfolio, telephone, wednesday, willpower, recession, margarine, farmhouse and maybe eiderdown&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One of the problems is &quot;ter&quot;. There is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; territory, scatering, waterfall&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and this results in multiple answers. If nobody will find the rest I extend my lines with some bruteforcing logic.&lt;/p&gt;&#xA;" OwnerUserId="4342" LastActivityDate="2015-03-01T18:29:13.487" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9714" PostTypeId="2" ParentId="9400" CreationDate="2015-03-01T19:10:02.410" Score="-4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; magic numbers are :  (78 , 2 )&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First off lets note a_b a couple of two numbers (a*b,b)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting from last statement , any set with absent couple such as (2,x) where x &gt; 33 will be excluded&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such couples are:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ratio=35  2-35 &#xA;ratio=36  2-36 &#xA;ratio=37  2-37 &#xA;ratio=38  2-38 &#xA;ratio=39  2-39 &#xA;ratio=40  2-40 &#xA;ratio=41  2-41 &#xA;ratio=42  2-42 &#xA;ratio=43  2-43 &#xA;ratio=44  2-44 &#xA;ratio=45  2-45 &#xA;ratio=46  2-46 &#xA;ratio=47  2-47 &#xA;ratio=48  2-48 &#xA;ratio=49  2-49  &#xA;ratio=50  2-50 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So ... standing on that ground , lets begin to work .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this list contains sets of differences that subtractor doesnt know .&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;set of difference 68 :  2_35 = (70, 35)  4_18 = (72, 18)  17_5 = (85, 5)  &#xA;&#xA;set of difference 70 :  2_36 = (72, 36)  5_15 = (75, 15)  7_11 = (77, 11)  10_8 = (80, 8)  14_6 = (84, 6)  &#xA;&#xA;set of difference 72 :  2_37 = (74, 37)  3_25 = (75, 25)  4_19 = (76, 19)  6_13 = (78, 13)  8_10 = (80, 10)  9_9 = (81, 9)  12_7 = (84, 7)  18_5 = (90, 5)  24_4 = (96, 4)  &#xA;&#xA;set of difference 76 :  2_39 = (78, 39)  4_20 = (80, 20)  19_5 = (95, 5)  &#xA;&#xA;set of difference 78 :  2_40 = (80, 40)  3_27 = (81, 27)  6_14 = (84, 14)  13_7 = (91, 7)  &#xA;&#xA;set of difference 80 :  2_41 = (82, 41)  4_21 = (84, 21)  5_17 = (85, 17)  8_11 = (88, 11)  10_9 = (90, 9)  16_6 = (96, 6)  20_5 = (100, 5)  &#xA;&#xA;set of difference 84 :  2_43 = (86, 43)  3_29 = (87, 29)  4_22 = (88, 22)  6_15 = (90, 15)  7_13 = (91, 13)  12_8 = (96, 8)  14_7 = (98, 7)  &#xA;&#xA;set of difference 88 :  2_45 = (90, 45)  4_23 = (92, 23)  8_12 = (96, 12)  11_9 = (99, 9)  &#xA;&#xA;set of difference 90 :  2_46 = (92, 46)  3_31 = (93, 31)  5_19 = (95, 19)  6_16 = (96, 16)  9_11 = (99, 11)  10_10 = (100, 10)  &#xA;&#xA;set of difference 92 :  2_47 = (94, 47)  4_24 = (96, 24)  &#xA;&#xA;set of difference 96 :  2_49 = (98, 49)  3_33 = (99, 33)  4_25 = (100, 25)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;this list contains elements grouped by their similar sum , accepted if summer knows for sure that subtractor doesnt know.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;set of sum = 66  :&#xA;2_34  -temporarily accepted-  5_13  -temporarily accepted-  7_9  -temporarily accepted-  10_6  -temporarily accepted-  14_4  -temporarily accepted-  &#xA;&#xA;this list is accepted &#xA;&#xA;&#xA;set of sum = 68  :&#xA;2_35  -temporarily accepted-  3_23  -temporarily accepted-  4_17  -temporarily accepted-  6_11  -temporarily accepted-  8_8  -temporarily accepted-  9_7  -temporarily accepted-  12_5  -temporarily accepted-  18_3  -temporarily accepted-  24_2  -temporarily accepted-  &#xA;&#xA;this list is accepted &#xA;&#xA;&#xA;set of sum = 70  :&#xA;2_36  -temporarily accepted-  &#xA;&#xA;this list is accepted &#xA;&#xA;&#xA;set of sum = 72  :&#xA;2_37  -temporarily accepted-  4_18  -temporarily accepted-  19_3  -temporarily accepted-  &#xA;&#xA;this list is accepted &#xA;&#xA;&#xA;set of sum = 74  :&#xA;2_38  -rejected because 2_38 is unique in the other side of differences- &#xA; 3_25  6_12  13_5  26_2  &#xA;&#xA;this list is rejected &#xA;&#xA;&#xA;set of sum = 76  :&#xA;2_39  -temporarily accepted-  4_19  -temporarily accepted-  5_15  -temporarily accepted-  8_9  -temporarily accepted-  10_7  -temporarily accepted-  16_4  -temporarily accepted-  20_3  -temporarily accepted-  &#xA;&#xA;this list is accepted &#xA;&#xA;&#xA;set of sum = 78  :&#xA;2_40  -temporarily accepted-  &#xA;&#xA;this list is accepted &#xA;&#xA;&#xA;set of sum = 80  :&#xA;2_41  -temporarily accepted-  3_27  -temporarily accepted-  4_20  -temporarily accepted-  6_13  -temporarily accepted-  7_11  -temporarily accepted-  12_6  -temporarily accepted-  14_5  -temporarily accepted-  21_3  -temporarily accepted-  28_2  -temporarily accepted-  &#xA;&#xA;this list is accepted &#xA;&#xA;&#xA;set of sum = 82  :&#xA;2_42  -rejected because 2_42 is unique in the other side of differences- &#xA;&#xA;&#xA;this list is rejected &#xA;&#xA;&#xA;set of sum = 84  :&#xA;2_43  -temporarily accepted-  4_21  -temporarily accepted-  8_10  -temporarily accepted-  11_7  -temporarily accepted-  22_3  -temporarily accepted-  &#xA;&#xA;this list is accepted &#xA;&#xA;&#xA;set of sum = 86  :&#xA;2_44  -rejected because 2_44 is unique in the other side of differences- &#xA; 3_29  5_17  6_14  9_9  10_8  15_5  18_4  30_2  &#xA;&#xA;this list is rejected &#xA;&#xA;&#xA;set of sum = 88  :&#xA;2_45  -temporarily accepted-  4_22  -temporarily accepted-  23_3  -temporarily accepted-  &#xA;&#xA;this list is accepted &#xA;&#xA;&#xA;set of sum = 90  :&#xA;2_46  -temporarily accepted-  &#xA;&#xA;this list is accepted &#xA;&#xA;&#xA;set of sum = 92  :&#xA;2_47  -temporarily accepted-  3_31  -temporarily accepted-  4_23  -temporarily accepted-  6_15  -temporarily accepted-  8_11  -temporarily accepted-  12_7  -temporarily accepted-  16_5  -temporarily accepted-  24_3  -temporarily accepted-  32_2  -temporarily accepted-  &#xA;&#xA;this list is accepted &#xA;&#xA;&#xA;set of sum = 94  :&#xA;2_48  -rejected because 2_48 is unique in the other side of differences- &#xA; 7_13  14_6  &#xA;&#xA;this list is rejected &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;the previous informations are unnecessary until this point: where subtractor declares his important information:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A list which fits third statement:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;set of difference= 68  : &#xA; 2_35  4_18       17_5       rejected because &#xA;&#xA; it exists a couple (ab,b)=(85,5) in this set where 2_50 is unique in his set of difference where &#xA; 2_50 has same sum of a_b = 17_5 &#xA;&#xA; the summer wudnt say i know that you know that i dont &#xA;&#xA;&#xA;&#xA;set of difference= 70  : &#xA; 2_36  5_15  7_11  10_8       rejected because&#xA;&#xA; it exists a couple (ab,b)=(80,8) in this set where 2_44 is unique in his set of difference where &#xA; 2_44 has same sum of a_b = 10_8 &#xA;&#xA; the summer wudnt say i know that you know that i dont &#xA;&#xA;&#xA;&#xA;set of difference= 72  : &#xA; 2_37       3_25       rejected because &#xA;&#xA; it exists a couple (ab,b)=(75,25) in this set where 2_38 is unique in his set of difference where &#xA; 2_38 has same sum of a_b = 3_25 &#xA;&#xA; the summer wudnt say i know that you know that i dont &#xA;&#xA;&#xA;&#xA;set of difference= 76  : &#xA; 2_39               4_20               19_5          &#xA; this is the only set accepted because &#xA;&#xA; 2_39   :  2_39 has same sum of &#xA; 5_15 which has same diff of &#xA; 2_36 which is unique in his set of same sum  &#xA;&#xA; 4_20   :  4_20 has same sum of &#xA; 3_27 which has same diff of &#xA; 2_40 which is unique in his set of same sum  &#xA;&#xA; 19_5        accepted because there dosnt exist any such unique couple  &#xA;&#xA;this set contains the unique solution because there exists two couple  2_39 , 19_5 &#xA; where the subtractor is not sure whether the summer knows if the subtractor knows that summer dosent know the numbers&#xA;&#xA;&#xA;&#xA;set of difference= 78  : &#xA; 2_40  3_27  6_14       rejected because &#xA;&#xA; it exists a couple (ab,b)=(84,14) in this set where 2_44 is unique in his set of difference where &#xA; 2_44 has same sum of a_b = 6_14 &#xA;&#xA; the summer wudnt say i know that you know that i dont &#xA;&#xA;&#xA;&#xA;set of difference= 80  : &#xA; 2_41  4_21       5_17       rejected because &#xA;&#xA; it exists a couple (ab,b)=(85,17) in this set where 2_44 is unique in his set of difference where &#xA; 2_44 has same sum of a_b = 5_17 &#xA;&#xA; the summer wudnt say i know that you know that i dont &#xA;&#xA;&#xA;&#xA;set of difference= 84  : &#xA; 2_43       3_29       rejected because &#xA;&#xA; it exists a couple (ab,b)=(87,29) in this set where 2_44 is unique in his set of difference where &#xA; 2_44 has same sum of a_b = 3_29 &#xA;&#xA; the summer wudnt say i know that you know that i dont &#xA;&#xA;&#xA;&#xA;set of difference= 88  : &#xA;&#xA; 2_45   :  2_45 has same sum of &#xA;     23_3 which has same diff of &#xA; 46_2 which is unique in his set of same sum  &#xA;&#xA; 4_23   :  4_23 has same sum of &#xA; 3_31 which has same diff of &#xA; 2_46 which is unique in his set of same sum  &#xA;&#xA; 8_12   :  8_12 has same sum of &#xA; 13_7 which has same diff of &#xA; 2_40 which is unique in his set of same sum  &#xA;&#xA; 11_9   :  11_9 has same sum of &#xA; 10_10 which has same diff of &#xA; 2_46 which is unique in his set of same sum  &#xA;&#xA;&#xA; rejected because all the couples in this set leads to atleast one unique couple in the list of addition  &#xA; which means subtractor knows cetainly that summer doenst know if subtractor knows wheather summer doesnt know numbers &#xA;&#xA;&#xA;&#xA;set of difference= 90  : &#xA; 2_46  3_31  5_19  6_16       rejected because &#xA;&#xA; it exists a couple (ab,b)=(96,16) in this set where 2_50 is unique in his set of difference where &#xA; 2_50 has same sum of a_b = 6_16 &#xA;&#xA; the summer wudnt say i know that you know that i dont &#xA;&#xA;&#xA;&#xA;set of difference= 92  : &#xA;&#xA; 2_47   :  2_47 has same sum of &#xA; 3_31 which has same diff of &#xA; 2_46 which is unique in his set of same sum  &#xA;&#xA; 4_24   :  4_24 has same sum of &#xA; 5_19 which has same diff of &#xA; 2_46 which is unique in his set of same sum  &#xA;&#xA;&#xA; rejected because all the couples in this set leads to atleast one unique couple in the list of addition  &#xA; which means subtractor knows cetainly that summer doenst know if subtractor knows wheather summer doesnt know numbers &#xA;&#xA;&#xA;&#xA;set of difference= 96  : &#xA; 2_49  3_33       rejected because &#xA;&#xA; it exists a couple (ab,b)=(99,33) in this set where 2_50 is unique in his set of difference where &#xA; 2_50 has same sum of a_b = 3_33 &#xA;&#xA; the summer wudnt say i know that you know that i dont &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;from the last list there s just one accepted set , where there s just one accepted couple specified from last divisor's declaration.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="9517" LastActivityDate="2015-03-01T19:10:02.410" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9715" PostTypeId="2" ParentId="9712" CreationDate="2015-03-01T19:10:24.337" Score="6" Body="&lt;p&gt;I managed to find 18 so far:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; willpower &lt;br /&gt;&#xA; margarine &lt;br /&gt;&#xA; portfolio &lt;br /&gt;&#xA; wednesday &lt;br /&gt;&#xA; mythology &lt;br /&gt;&#xA; farmhouse &lt;br /&gt;&#xA; telephone &lt;br /&gt;&#xA; eiderdown &lt;br /&gt;&#xA; pineapple &lt;br /&gt;&#xA; waterfall &lt;br /&gt;&#xA; scoundrel &lt;br /&gt;&#xA; equipment &lt;br /&gt;&#xA; important &lt;br /&gt;&#xA; scarecrow &lt;br /&gt;&#xA; differing &lt;br /&gt;&#xA; accessory &lt;br /&gt;&#xA; butterfly &lt;br /&gt;&#xA; nutrition &lt;br /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1585" LastActivityDate="2015-03-01T19:10:24.337" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9716" PostTypeId="2" ParentId="9712" CreationDate="2015-03-01T21:07:33.973" Score="7" Body="&lt;p&gt;These should be all twenty:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; different mythology telephone equipment portfolio &lt;br&gt;&#xA; pineapple willpower margarine wednesday important &lt;br&gt;&#xA; waterfall butterfly scarecrow accessory nutrition &lt;br&gt;&#xA; scoundrel eiderdown farmhouse something newsagent &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I found about half of them myself, so credit to the other answers as well. I think they only missed one between them (last one in my list).&lt;/p&gt;&#xA;" OwnerUserId="7577" LastActivityDate="2015-03-01T21:07:33.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9717" PostTypeId="1" AcceptedAnswerId="9740" CreationDate="2015-03-02T05:29:09.043" Score="6" ViewCount="229" Body="&lt;p&gt;I shout and I am silent.&lt;br&gt;&#xA;The deaf can hear me.&lt;br&gt;&#xA;The blind cannot.&lt;br&gt;&#xA;I am a tool of bridge dwellers.  &lt;/p&gt;&#xA;" OwnerUserId="9495" LastActivityDate="2015-03-02T13:07:05.100" Title="The riddle of the bridge dwellers" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9718" PostTypeId="1" AcceptedAnswerId="9743" CreationDate="2015-03-02T05:34:18.210" Score="1" ViewCount="465" Body="&lt;p&gt;I am four corners.&lt;br&gt;&#xA;I both destroy and tantalize.&lt;br&gt;&#xA;I intersect.&lt;br&gt;&#xA;I describe what children shouldn't watch.&lt;/p&gt;&#xA;" OwnerUserId="9495" LastEditorUserId="9495" LastEditDate="2015-03-03T03:58:15.850" LastActivityDate="2015-03-03T03:58:15.850" Title="Children should not watch me" Tags="|riddle|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9721" PostTypeId="1" AcceptedAnswerId="9728" CreationDate="2015-03-02T06:01:21.533" Score="16" ViewCount="1389" Body="&lt;p&gt;People might think that I am where you buy cocktails in zero G.&lt;br&gt;&#xA;But, I have been there 28 times while writing this and I am not drunk.&lt;/p&gt;&#xA;" OwnerUserId="9495" LastActivityDate="2015-03-02T10:52:09.590" Title="Cocktails in zero G" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9722" PostTypeId="2" ParentId="9161" CreationDate="2015-03-02T06:04:38.793" Score="12" Body="&lt;p&gt;The pattern is :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;i/I = 1&lt;/code&gt; and &lt;code&gt;o/O = 0&lt;/code&gt;. Just take the i's and o's from the country name and using their respective values 1 and 0, calculate the equivalent decimal values.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Examples :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Belg&lt;strong&gt;i&lt;/strong&gt;um -&gt; 1 &lt;br&gt;&#xA; Mex&lt;strong&gt;i&lt;/strong&gt;c&lt;strong&gt;o&lt;/strong&gt; -&gt; 10 &lt;br&gt;&#xA; L&lt;strong&gt;i&lt;/strong&gt;echtenste&lt;strong&gt;i&lt;/strong&gt;n -&gt; 11 &lt;br&gt;&#xA; &lt;strong&gt;I&lt;/strong&gt;v&lt;strong&gt;o&lt;/strong&gt;ry C&lt;strong&gt;o&lt;/strong&gt;ast -&gt; 100 &lt;br&gt;&#xA; Eth&lt;strong&gt;io&lt;/strong&gt;p&lt;strong&gt;i&lt;/strong&gt;a -&gt; 101 &lt;br&gt;&#xA; Un&lt;strong&gt;i&lt;/strong&gt;ted K&lt;strong&gt;i&lt;/strong&gt;ngd&lt;strong&gt;o&lt;/strong&gt;m -&gt; 110 &lt;br&gt;&#xA; Ph&lt;strong&gt;i&lt;/strong&gt;l&lt;strong&gt;i&lt;/strong&gt;pp&lt;strong&gt;i&lt;/strong&gt;nes -&gt; 111 &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The countries at 8th and 12th position can be :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 8(1000) -&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Republic_of_the_Congo&quot;&gt;Republ&lt;strong&gt;i&lt;/strong&gt;c &lt;strong&gt;o&lt;/strong&gt;f the C&lt;strong&gt;o&lt;/strong&gt;ng&lt;strong&gt;o&lt;/strong&gt;&lt;/a&gt;&#xA; 12(1100) -&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Trinidad_and_Tobago&quot;&gt;Tr&lt;strong&gt;i&lt;/strong&gt;n&lt;strong&gt;i&lt;/strong&gt;dad and T&lt;strong&gt;o&lt;/strong&gt;bag&lt;strong&gt;o&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5565" LastEditorUserId="-1" LastEditDate="2015-03-03T13:22:32.817" LastActivityDate="2015-03-03T13:22:32.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9723" PostTypeId="1" AcceptedAnswerId="9725" CreationDate="2015-03-02T06:51:19.073" Score="13" ViewCount="1151" Body="&lt;p&gt;They approached the man apprehensively. He was aware of their approach, but did not turn to face them. Instead, he kept looking through the window toward the stars. Finally, one of the men gathered the courage to speak. &quot;Master&quot;, he said, &quot;if you had one word to describe all that you see - the stars, the planets, the galaxies - and at the same time convey your sense of wonder and amazement, which word would you use?&quot;. The master paused, then turned to scribble on some paper. After he was done, he handed it to them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PnFfQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;What is that word?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="7308" LastEditorUserId="5373" LastEditDate="2019-10-25T15:00:50.043" LastActivityDate="2019-10-25T15:00:50.043" Title="Heliocentricity" Tags="|cipher|outer-space|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9724" PostTypeId="1" AcceptedAnswerId="9727" CreationDate="2015-03-02T07:36:31.597" Score="28" ViewCount="1789" Body="&lt;ol&gt;&#xA;&lt;li&gt;Start with &quot;U&quot;.&lt;/li&gt;&#xA;&lt;li&gt;Add another &quot;U&quot;.&lt;/li&gt;&#xA;&lt;li&gt;Remove the &quot;U&quot;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What is the word?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; It's not &quot;U&quot; (or &quot;you&quot;, &quot;ewe&quot;, etcetera).&lt;/p&gt;&#xA;" OwnerUserId="9881" LastEditorUserId="5373" LastEditDate="2015-06-26T11:14:45.410" LastActivityDate="2015-06-26T11:14:45.410" Title="Start with U, add another U, remove the U - what is the word?" Tags="|wordplay|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9725" PostTypeId="2" ParentId="9723" CreationDate="2015-03-02T08:17:55.853" Score="12" Body="&lt;p&gt;The word is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;celestial&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; because each letter of the word is represented by one of the moons of Saturn (backwards):&lt;br&gt;&#xA; Tethys, Rhea, Helene, Rhea, Ymir, Paaliaq, Phoebe, Mimas, Helene&lt;br&gt;&#xA;&lt;br&gt;&#xA; then using the IAU designation, these moons are labelled:&lt;br&gt;&#xA; III, V, XII, V, XIX, XX, IX, I, XII&lt;br&gt;&#xA;&lt;br&gt;&#xA; and then substituting the letter of the alphabet results in:&lt;br&gt;&#xA; C, E, L, E, S, T, I, A, L  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5215" LastEditDate="2015-03-02T14:51:52.513" LastActivityDate="2015-03-02T14:51:52.513" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9727" PostTypeId="2" ParentId="9724" CreationDate="2015-03-02T09:11:47.393" Score="28" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Double&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You start with U.  &lt;/li&gt;&#xA;&lt;li&gt;You add another U, so you get &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; UU - a double U (W).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;You remove a U, and you're stuck with &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; double.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5127" LastEditorUserId="4204" LastEditDate="2015-03-05T02:51:24.230" LastActivityDate="2015-03-05T02:51:24.230" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9728" PostTypeId="2" ParentId="9721" CreationDate="2015-03-02T09:15:47.500" Score="15" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The spacebar &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You buy cocktails in a bar, zero G is in space and there are 28 spaces in the sentence if you include the two spaces at the end of the first line (open the edit-screen of the question to see for yourself).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5127" LastEditorUserId="5127" LastEditDate="2015-03-02T10:52:09.590" LastActivityDate="2015-03-02T10:52:09.590" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9730" PostTypeId="1" AcceptedAnswerId="9732" CreationDate="2015-03-02T09:52:30.007" Score="4" ViewCount="1406" Body="&lt;p&gt;There is a 6x6 grid. Each row/column has A, B, C and D exactly once, and two blanks. Each cell may have upto one letter only. A blank is not a letter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hints&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The left-most letter of row 2 is A and right-most is B&lt;/li&gt;&#xA;&lt;li&gt;The left-most letter of row 3 is also A&lt;/li&gt;&#xA;&lt;li&gt;The right-most letter of row 4 is B&lt;/li&gt;&#xA;&lt;li&gt;The left-most letters of row 5 and 6 are both B&lt;/li&gt;&#xA;&lt;li&gt;The first letter of the first column is D&lt;/li&gt;&#xA;&lt;li&gt;The first letter in columns 2 and 3 is A&lt;/li&gt;&#xA;&lt;li&gt;Column 2 ends with D (or 1/2 spaces after a D)&lt;/li&gt;&#xA;&lt;li&gt;For column 4 and 6, the first letter is B and last is C&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There exists a unique solution; I have twice-verified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help out with the tags...&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-03-02T11:45:17.617" Title="Solve the following 6x6 grid" Tags="|word|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9731" PostTypeId="1" AcceptedAnswerId="9733" CreationDate="2015-03-02T11:02:39.150" Score="6" ViewCount="307" Body="&lt;p&gt;You are given two potatoes. You want to make a finite loop of wire so that you can put it on either of them at some location at your choice, so that there are no gaps between the potato and the wire. You can't deform the loop after you have created it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to make such a loop for any two potatoes?&lt;/p&gt;&#xA;" OwnerUserId="9886" LastEditorUserId="8874" LastEditDate="2015-03-02T11:50:59.043" LastActivityDate="2015-03-02T11:50:59.043" Title="Two potatoes and a loop of wire" Tags="|geometry|three-dimensional|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9732" PostTypeId="2" ParentId="9730" CreationDate="2015-03-02T11:45:17.617" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/OLcpb.png&quot; alt=&quot;Does this work? ;) &quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Oh well, this puzzle was something else. Innovative indeed and mind-boggling! Thanks!&lt;/p&gt;&#xA;" OwnerUserId="9865" LastActivityDate="2015-03-02T11:45:17.617" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9733" PostTypeId="2" ParentId="9731" CreationDate="2015-03-02T11:46:47.720" Score="5" Body="&lt;p&gt;Yes, such a loop of wire can be made for any two potatoes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Just consider the potatoes as abstract geometric objects in three-dimensional space. Take these two objects, push them towards each other, let them collide, keep pushing, keep pushing, and make their surfaces intersect. The intersection of the surfaces will contain a closed curve. Now design your loop of wire to match this closed curve.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-03-02T11:46:47.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9734" PostTypeId="1" AcceptedAnswerId="9736" CreationDate="2015-03-02T11:52:23.120" Score="-1" ViewCount="2064" Body="&lt;p&gt;Bob is the chief of security at an exhibition hall. His job is to make sure that all the trade shows which happen there are safe and secure events. This weekend is a big consumer expo where many companies present their new and innovative products. There are lots of expensive prototypes out there, so Bob is wary of thieves trying to steal them. So he hired some temps as additional security guards to make sure nothing gets stolen.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Day One&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;It's the first day and the grand opening of the trade show. The exhibitors have built up their booths and hundreds of visitors are queueing. Bob opens the gate and the visitors pour into the hall. While he looks at the moving masses for any trouble a man approaches him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Hello, good sir. May I introduce myself? I am the greatest thief in the world. I have picked this trade show as my target for the next days and I plan to steal many things of great value.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bob laughs&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Good luck, son. Not on my watch! My guys are everywhere. They won't let you steal anything. Enjoy the trade show.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bob goes along his daily routine, coordinating his staff and making sure everything is in order. In the evening he again goes to the exit keeping an eye out for the suspicious guy from earlier. When he leaves the building, Bob stops him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Excuse me, Sir, mind you if I take a look into your bag?&quot;&lt;/em&gt;, &lt;em&gt;&quot;Sure, I don't mind&quot;&lt;/em&gt;, says the thief and hands his bag to Bob. Bob looks through it and finds nothing of value except for some advertisement booklets and catalogs. &lt;em&gt;&quot;You didn't &lt;strong&gt;steal&lt;/strong&gt; any of these, did you&quot;?&lt;/em&gt;, &lt;em&gt;&quot;Please, Sir. You are insulting me. I am the greatest thief in the world. Do you think I would be satisfied with some worthless paper? No, the previous owners gave me these on their own accord. They even insisted I take it.&quot;&lt;/em&gt; Having no proof that the man had stolen anything, Bob has no choice than to let him pass.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Day Two&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;It's the second day of the trade show. Like every morning Bob controls the entrance to make sure no accidents happen while the crowd enters the building. He is again approached by the mysterious man.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Good morning, Sir. Yesterday was a great day. I've stolen many things of great value and today I am going to steal even more.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;I've checked you. You hadn't anything of value with you.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Oh, Sir, you are mistaken. I had loot with me worth a fortune. Today I am going to steal even more and you won't find it either.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;em&gt;We'll see about that. Enjoy the tradeshow.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bob becomes suspicious. What if the mysterious man was indeed a thief? Bob decides to leave his daily duties to his assistant and tails the man during the day. He watches how the man goes from booth to booth, looking at many exhibits but never even touching one. As a trained security guard Bob is familiar with all the tricks pickpockets use to steal from people in a crowd, but as far as he can tell the man isn't using any of them. He starts conversations with some of the exhibitors, but never even attempts to steal anything from them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the man leaves the exhibition hall in the evening, Bob stops him and asks &lt;em&gt;&quot;Excuse me, Sir, may I ask you a question? You didn't steal anything today, didn't you?&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;In fact I happened to steal many things today. Some of them even more valuable than those I have stolen yesterday.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;I know. I have watched you all day and I caught you. It's over. Please come with me to the security room&quot;&lt;/em&gt;. This was a lie. Bob hadn't seen anything. But in his years as a security guard he learned that bluffing can often be a useful tactic to get thieves to reveal information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Sure, I don't mind.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the security room Bob gives the man a thorough frisk, but doesn't find anything. Knowing that it might overstain his competences Bob asks the man to take off his clothes. The man cooperates. Bob thoroughly searches every single item of clothing but doesn't find anything either. Provoked by the man's audacity he decides to do something which will definitely get him into trouble when found out, but he does so regardless. He puts on a pair of gloves and checks all orifices of the man, but again he finds nothing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Do you enjoy wasting my time like this? Come on, admit it. You haven't stolen anything.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;I did. But unless you find out what I've stolen the police will never believe you.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Day Three&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;It is the third and last day of the tradeshow. And again, the mysterious man shows up. This time Bob approaches him right away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;I am fed up with you. I don't know what you are stealing, I don't know how, but I will find out and then I will hand you over to the police.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;I wish you good luck, Sir, but I don't think you will be successful.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bob spends the whole day pondering how the thief could be getting his loot out of the building. Maybe he has an accomplice? Maybe he smuggles it out somehow? Or maybe the goods are still in the building and he plans to pick them up later?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bob decides to take drastic measures. Before the tradeshow is over, he goes to the PA system and announces for everyone to hear:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Ladies and Gentlemen, this is the chief of security speaking. May I have your attention, please. We have information that the greatest thief in the world is on this tradeshow and has stolen one or more valuables. I ask you to stay in the hall and not leave it before you were searched.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Security guards are stationed at all exits of the building. Every single person who leaves the hall are searched thoroughly, but to no avail. After everyone left, the guards search every trash can, every box and every container for stolen goods. Any cavities in walls, the ventilation system, the suspended ceiling, floor shafts, everything in the building where stolen items could be hidden is searched. But the guards don't find anything.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Epilogue&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;A few weeks after the tradeshow is over, Bob goes to a bar. There he meets the man again. He is wearing a very expensive suit, lots of fancy jewlery and is drinking the most expensive champaign in the bar. Bob decides to start a conversation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;You seem to be quite rich, aren't you?&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Newly rich. I just sold some of the things I have stolen at your trade show. My clients paid me thousands of dollars for it.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Please, Sir, I need to know. I swear I will not tell on you. What is it you were stealing?&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;I was stealing █████&quot;&lt;/em&gt; &lt;/p&gt;&#xA;" OwnerUserId="1774" LastEditorUserId="5555" LastEditDate="2015-03-03T03:07:48.120" LastActivityDate="2016-08-08T12:57:25.033" Title="The greatest thief in the world" Tags="|story|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9736" PostTypeId="2" ParentId="9734" CreationDate="2015-03-02T12:12:54.973" Score="9" Body="&lt;p&gt;This is an old one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He was stealing &lt;em&gt;ideas&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="40" LastEditorUserId="40" LastEditDate="2015-03-02T12:55:43.973" LastActivityDate="2015-03-02T12:55:43.973" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9737" PostTypeId="1" AcceptedAnswerId="9761" CreationDate="2015-03-02T12:19:04.560" Score="-3" ViewCount="448" Body="&lt;p&gt;A test kit, is a gathering place for a colony of bacteria. This Colony, at 12 pm cell division has created a bacteria. Every minute, every bacteria is divided into two bacteria. 12: 43 (43 minutes later), half of the kit is full of bacteria. When kit  will be filled of bacteria?&lt;/p&gt;&#xA;" OwnerUserId="7555" LastEditorUserId="5188" LastEditDate="2015-03-03T09:49:59.707" LastActivityDate="2015-03-05T15:32:07.870" Title="When will the bacteria fill the test kit?" Tags="|mathematics|logical-deduction|" AnswerCount="2" CommentCount="1" ClosedDate="2015-04-05T21:53:05.670" ContentLicense="CC BY-SA 3.0" />
  <row Id="9738" PostTypeId="2" ParentId="9737" CreationDate="2015-03-02T12:21:55.703" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 12:44&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Too easy though! I would have added some red herrings at least to make it look complicated if I were you. &lt;code&gt;:)&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="9865" LastActivityDate="2015-03-02T12:21:55.703" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9739" PostTypeId="1" AcceptedAnswerId="9742" CreationDate="2015-03-02T12:43:08.913" Score="7" ViewCount="463" Body="&lt;p&gt;121 women are competing in the heptathlon in Olympic Games.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If British track and field star Jessica Ennis finishes in 104th place in every event, what is the highest place she can finish in the overall&#xA;heptathlon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: Feel free to google around for the information that you think might be essential to solving the problem.&lt;/p&gt;&#xA;" OwnerUserId="9865" LastEditorUserId="4050" LastEditDate="2015-03-05T13:31:22.777" LastActivityDate="2015-03-05T13:31:22.777" Title="Highest place possible for Jessica - heptathlon scoring" Tags="|mathematics|logical-deduction|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9740" PostTypeId="2" ParentId="9717" CreationDate="2015-03-02T13:07:05.100" Score="10" Body="&lt;p&gt;If this isn't the right answer, I hope you at least find it humorous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CAPS LOCK&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I shout and I am silent.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Typewritten words make no noise, but all caps is often referred to as &quot;shouting&quot; online&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The deaf can hear me&#xA;The blind cannot.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Deaf people can still read, but blind people cannot read typewritten words (I don't know if there's a Braille equivalent of caps lock)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a tool of bridge dwellers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Online trolls often use caps lock, because they're the worst people in the world.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2015-03-02T13:07:05.100" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9741" PostTypeId="2" ParentId="9711" CreationDate="2015-03-02T13:16:28.933" Score="5" Body="&lt;p&gt;Edited to make the first part more rigorous (and correct the value of x thanks to @mdc32).&lt;br&gt;&#xA;&lt;s&gt;This will be a little bit handwavy, but here goes.&lt;/s&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; x is 50, ie just less than 50% of the games can be nonsensical.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First we must show Halfbrain's first theorem for x.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Suppose at least 50% of the games are nonsensical. We can partition the points table into an upper and lower part, with m players in the top part and n in the bottom part, and no overlap in scores between the parts. Otherwise all players must have the same score and there are no nonsensical games.&lt;br&gt;&#xA; There are $m(m-1)/2$ intra-group games in the top part, and $n(n-1)/2$ in the bottom part. There are $mn$ intergroup games. Suppose at least $mn/2$ of them are nonsensical. Then the average score for the top part is at most $(m-1)/2 + n/2$, and the average score for the bottom part is at least $(n-1)/2 + m/2$. That means the bottom part has an average score at least as good as the top part. So some players from the bottom part must have scores at least as good as the top part, which is a contradiction.&lt;br&gt;&#xA; If 50% of the games are nonsensical, and the intergroup games cannot be 50% nonsensical, then the intragroup games must be at least 50% nonsensical for at least one of the parts. We take that part, and recursively repeat the division process. Eventually we must reach a group containing 2 or 3 players, which cannot have any nonsensical games but must have at least 50% nonsensical games. So that's a contradiction as well.&lt;br&gt;&#xA; We conclude that the number of nonsensical games must be less than 50%.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Next we must show Halfbrain's second theorem for x + 1.&lt;br&gt;&#xA;Step 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Consider a tournament with a very large number of contestants, all evenly matched. Each wins half their games and loses the other half, so all are on the same score. A late entrant turns up and plays every contestant, winning half and losing half. Now those that won that last game have one more win than those that did not; but each &quot;winner&quot; lost half his games played against someone in the &quot;loser&quot; group. Of all the games, one quarter are played between &quot;winners&quot;, one quarter are played between &quot;losers&quot;, and half are played between &quot;winners&quot; and &quot;losers&quot;. Half of that half will be won by the player from the &quot;loser&quot; group, that is, a quarter will be nonsensical games.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Step 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We can repeat this by adding another two late contestants. The first one gets the same results as the original late contestant; the second splits both the &quot;winner&quot; and &quot;loser&quot; group in half again. Only half the games remain to be affected (one quarter in each group), and we can only turn a quarter of the affected games into nonsensical games; a total of one eighth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Step n:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We will need $2^{n-1}$ late contestants; one will induce new splits in the existing groups, and the rest will maintain the existing splits. We will make a further $\frac{1}{2^{n+1}}$ of the games nonsensical&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Desired target:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Repeat as often as you can maintain the group as &quot;large&quot;, which will not be all that long. The total number of affected games will be 1/4 + 1/8 + 1/16 + .... which has a limit of 50% that you will never be able to reach. But you can exceed 49%, by which time you will have introduced 127 late contestants and you will have 128 groups.&lt;br&gt;&#xA; Note that you don't really have to introduce new contestants, but it makes it easier to picture what's going on.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6031" LastEditorUserId="6031" LastEditDate="2015-03-03T12:28:34.210" LastActivityDate="2015-03-03T12:28:34.210" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9742" PostTypeId="2" ParentId="9739" CreationDate="2015-03-02T13:27:06.940" Score="6" Body="&lt;p&gt;The best she can hope for is &lt;strong&gt;2nd place&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume that in each event, there is a 103-way tie for first place. Jessica finishes 104th with a score fractionally below the leaders, and the remaining 7 competitors score precisely zero points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's also assume that the winning scores in each event are equal to $X_i$ for $(1 \le i \le 7)$. So competitors who finish among the joint winners in all seven events will achieve a total score of $T = \sum^7_{i=1} X_i$, and Jessica's total score will be $(T - \epsilon)$ points, where $\epsilon$ is some small value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $m$ is the event where the joint winners achieved the highest score (i.e., $X_m &amp;gt; X_i \ \ \forall i \not= m)$, then Jessica will achieve the best position if as many as possible of the winners in event $m$ score zero in at least one other event. Since there are seventeen possible places that they can take in the remaining six events, that means Jessica can beat at most 102 of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So her best possible finishing position is $104 - 102 = 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;(Thanks to Duncan for pointing out the glaring error in my first effort!)&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="40" LastEditorUserId="40" LastEditDate="2015-03-03T12:12:14.343" LastActivityDate="2015-03-03T12:12:14.343" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="9743" PostTypeId="2" ParentId="9718" CreationDate="2015-03-02T15:21:28.710" Score="13" Body="&lt;p&gt;I think you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The letter 'X'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am four corners&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The shape of the letter is four corners, meeting at a point&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I both destroy and tantalize&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;X can be a symbol for deletion, or a film rating later adopted by pornography&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I intersect&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The shape of the letter, again&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Children should not watch me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Again refering to &quot;X-rated&quot; films&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8776" LastActivityDate="2015-03-02T15:21:28.710" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9744" PostTypeId="2" ParentId="9718" CreationDate="2015-03-02T15:39:16.697" Score="5" Body="&lt;p&gt;I am four corners.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A TV set is generally square or rectangle which are shapes with four corners&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I both destroy and tantalize.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Watching TV is time killer at the same time entertaining&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I intersect.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Time spent watching TV means less time for socialising, so it intersects/cuts your relationship&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Children should not watch me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Watching TV is bad for eyes and children should have more outdoor activities&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I am a TV set&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9174" LastActivityDate="2015-03-02T15:39:16.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9745" PostTypeId="1" AcceptedAnswerId="9755" CreationDate="2015-03-02T16:19:51.343" Score="13" ViewCount="2013" Body="&lt;p&gt;I am the purpose of this riddle&lt;br&gt;&#xA;and I am without grace.&lt;br&gt;&#xA;I dislike comfort,&lt;br&gt;&#xA;but give me the title of doctor,&lt;br&gt;&#xA;and I'll go back and forth in an eternal loop for you.&lt;/p&gt;&#xA;" OwnerUserId="9686" LastActivityDate="2015-03-02T20:45:53.140" Title="I am the purpose of this riddle" Tags="|riddle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9746" PostTypeId="1" AcceptedAnswerId="9747" CreationDate="2015-03-02T16:57:24.007" Score="13" ViewCount="2251" Body="&lt;p&gt;Two men find an old gold coin and want to have a coin toss with it to decide who gets it. The only problem is the coin is heavier on one side so it comes up heads more than tails. What is a fair way for the men to toss the coin and decide who gets the coin?&lt;/p&gt;&#xA;" OwnerUserId="9888" LastEditorUserId="12408" LastEditDate="2016-07-25T02:30:14.470" LastActivityDate="2020-06-26T12:49:08.113" Title="Two men for one gold coin" Tags="|logical-deduction|probability|coin-tossing|" AnswerCount="6" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9747" PostTypeId="2" ParentId="9746" CreationDate="2015-03-02T17:03:06.610" Score="37" Body="&lt;p&gt;They could&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; toss it twice&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If the result is&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;  * HT = first player wins; stop&lt;br /&gt;&#xA;  * TH = second player wins; stop&lt;br /&gt;&#xA;  * HH or TT = ignore these two tosses; repeat the procedure&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="63414" LastEditDate="2020-06-26T09:49:44.587" LastActivityDate="2020-06-26T09:49:44.587" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="9748" PostTypeId="2" ParentId="9746" CreationDate="2015-03-02T17:25:56.407" Score="4" Body="&lt;p&gt;Each man &amp;quot;tosses&amp;quot; the coin.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Whoever &amp;quot;tossed&amp;quot; it the farthest (or closest to a fixed point) gets to keep it (assuming they don't lose it in the process). This will eliminate any bias introduced by the coin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;NOTE: When I initially posted my answer it was tagged as a 'riddle'&#xA;      so I was playing off the semantics of 'coin toss' to literally &#xA;      mean 'to toss a coin'. If that absolutely bothers you that is&#xA;      fine, but if you are going to down vote at least be courteous&#xA;      enough to leave a comment as to why.   &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="3922" LastEditorUserId="63414" LastEditDate="2020-06-26T12:46:41.233" LastActivityDate="2020-06-26T12:46:41.233" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="9749" PostTypeId="1" AcceptedAnswerId="11258" CreationDate="2015-03-02T18:25:47.303" Score="9" ViewCount="1204" Body="&lt;p&gt;A friend of yours was part of a secret society and was put on trial on 15 December 2014 without a lawyer. Alice encrypted the trial's transcript and sent it to Bob, but double agent Carol was able to e-mail a copy of the ciphertext to you. Your friend desperately needed help from a lawyer, but you were unable to provide one with the plaintext so that you could relay the proper advice back to your friend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;RESULTS OF TRIAL:&lt;/strong&gt; Your friend was executed on 26 December 2014, and the rest of your friend's family followed on 20 January 2015.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recently, Carol was able to intercept communications containing a rather ominous message. It seems that your friend's secret society has learned of the work you and your associates have been conducting and are less than pleased. Your life appears to be in jeopardy unless you solve the mystery of your friend's trial by the end of this quarter. If you fail, you will find out if there is an afterlife.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;E-mail:&lt;/strong&gt; &lt;a href=&quot;http://pastebin.com/v1aakvQv&quot; rel=&quot;nofollow&quot;&gt;Show&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;URGENT:&lt;/strong&gt; Carol just intercepted the following e-mail. Someone must be watching out for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;E-mail:&lt;/strong&gt; &lt;a href=&quot;http://pastebin.com/i8mLdMS5&quot; rel=&quot;nofollow&quot;&gt;Show&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Can you produce the plaintext for the message below to prevent your untimely demise?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ciphertext:&lt;/strong&gt; &lt;a href=&quot;http://pastebin.com/GpWNFdQh&quot; rel=&quot;nofollow&quot;&gt;Show&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 31:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The telephone at your desk begins to ring, and after answering, you hear a recorded message spoken with a distorted voice coming from the earpiece. &quot;This is your friend Randall, and I have the dynamic-link library you requested for managing the reputation of the users on your web site. Please let Adam know that he needs to nag a ram and avoid Pan if he wishes to ever understand his wife.&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 32:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While walking back from lunch, Victor flags you down in the hall and urges you to check your inbox as soon as you get back to your office. It is easy to spot his e-mail once seated at your desk as it is marked as a high priority message which summarizes some findings reported to him by our corporate firewall. In each of the three encoded e-mails you received last week, a nonstandard &lt;code&gt;X-Security-Warning: SOS&lt;/code&gt; header was ominously included.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 33:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just as on Monday, your telephone starts ringing; and upon picking it up, you hear the voice of Paul over in customer service. &quot;In a meeting this morning, a friend told me that you only have sixteen more days until DIE takes you out, and something has been bothering me over the past couple of weeks. Does all the information we have mesh together such that there are no conflicts or contradictions among the multitude of observations?&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 34:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inspectors from Myutsu Mutucsoecul Corporation have been hired yet again in the interest of verifying Paul's question from last week. After reviewing the past five weeks since their last audit, Alloy believes he has found some inconsistencies. &quot;One might infer an unproven relationship between ciphertext and primer from Trudy's question, and the data Chuck provided does not appear to allow proper decryption.&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 35:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Don't you think this is unmerited discrimination against fresh workers?&quot; Without waiting for Alloy to respond, Cobalt continues her defense of Trudy by showing that M1 and M2 provided by Team Bolt (as the Clingans are now calling themselves) have an even number of alpha characters as does the primer. &quot;Furthermore, changes in the ciphertext match the profile of Chuck's key, so there could be some sort of clerical misunderstanding.&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 36:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back again but somewhat under the weather, Blastus apologizes for missing his meeting with you yesterday but reveals he was verifying a letter from someone called Velociraptor. This mysterious individual must have wanted to contribute something to your friend's case without being involved in the investigative process. A familiarity with Team Bolt's work is evident since Velociraptor's writing directly utilizes data taken from their test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Letter:&lt;/strong&gt; &lt;a href=&quot;http://pastebin.com/pDzXWgXs&quot; rel=&quot;nofollow&quot;&gt;Show&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 37:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A familiar name from Tenebris Lamina shows up in your inbox, but Dave has a significantly larger message for you than he did the last time he sent you an e-mail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;E-mail:&lt;/strong&gt; &lt;a href=&quot;http://pastebin.com/egg1BRXe&quot; rel=&quot;nofollow&quot;&gt;Show&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 38:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some people would look at her name and think she was their grandestmother, but this individual from Nigrum Gladio upholds her organization's name in a different way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;E-mail:&lt;/strong&gt; &lt;a href=&quot;http://pastebin.com/ZNKPv82i&quot; rel=&quot;nofollow&quot;&gt;Show&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 39:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your computer announces a new e-mail with a rather humorless tone, and upon seeing who the unread message is from, you can only wonder if the devilish saber is meant for you or for your enemy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;E-mail:&lt;/strong&gt; &lt;a href=&quot;http://pastebin.com/Wg2ccGtQ&quot; rel=&quot;nofollow&quot;&gt;Show&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1661" LastEditorUserId="1661" LastEditDate="2015-03-27T14:34:33.020" LastActivityDate="2015-03-30T19:17:48.213" Title="Justice and Fairness: Part 2" Tags="|cipher|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9750" PostTypeId="2" ParentId="9745" CreationDate="2015-03-02T19:09:26.120" Score="10" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Who&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note: the explanation is slightly out of order but it all ties together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am the purpose of this riddle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This could be a &quot;who am I&quot; riddle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but give me the title of doctor,&#xA;and I'll go back and forth in an eternal loop for you.:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Doctor who is a time lord who travels back and forth through time and space solving problems for &quot;you&quot; being the average person who doesn't know what's happening.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I dislike comfort:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Doctor Who travels in a Tardis - doesn't seem very comfortable to me, even if it's big on the inside.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am without grace:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Tardis is old and unreliable - hardly graceful, in any case. Furthermore, Doctor who is fairly eccentric - as defined on the internet:  eccentricity (also called quirkiness) refers to unusual or odd behavior on the part of an individual. This is also usually not graceful.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8999" LastActivityDate="2015-03-02T19:09:26.120" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9752" PostTypeId="2" ParentId="9746" CreationDate="2015-03-02T19:19:12.800" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Each person tosses the coin 3 times, whoever gets more heads wins, if tie then do it again. If ties keep occurring then toss it 5 times or more to decide&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;OR&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; each person tosses it once whoever gets tail wins. If tie or no one gets tail then each person tosses it again until someone wins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4299" LastEditorUserId="63414" LastEditDate="2020-06-26T12:45:26.530" LastActivityDate="2020-06-26T12:45:26.530" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="9753" PostTypeId="2" ParentId="9746" CreationDate="2015-03-02T19:22:19.510" Score="1" Body="&lt;p&gt;oooh we could use the same rules as NHL shoot outs! in hockey, the goaltender has a statistical advantage over the shooting player, but since each team takes turn shooting agaisnt the opponent's goalie, the overall result of the shoot out is unaffected by the statistical advantage of the goalie.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;em&gt;Each team names three shooters. If the game remains tied after the three shooters are done, teams continue shooting in &amp;quot;sudden death&amp;quot; mode. The game cannot end until each team has taken the same number of shots.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;so,&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; let's say that Heads = save and Tails = goal. the players simply take turn flipping the coin as if they were playing hockey shoot outs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="63414" LastEditDate="2020-06-26T12:49:08.113" LastActivityDate="2020-06-26T12:49:08.113" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="9754" PostTypeId="2" ParentId="9746" CreationDate="2015-03-02T20:33:25.920" Score="20" Body="&lt;p&gt;The answer is to simply flip the coin &lt;strong&gt;once&lt;/strong&gt;, but&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; make sure you catch it &lt;em&gt;without letting it bounce&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The question makes the assertion that a weighted coin will be biased,&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; but that only happens if the coin is allowed to bounce and/or spin (on a surface). For a flipped/caught coin, there is no significant bias.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The law of conservation of angular momentum tells us that once the coin is in the air, it spins at a nearly constant rate (slowing down very slightly due to air resistance). At any rate of spin, it spends half the time with heads facing up and half the time with heads facing down, so when it lands, the two sides are equally likely (with minor corrections due to the nonzero thickness of the edge of the coin).&#xA; &lt;br&gt;&#xA; [This] explains why weighting the coin has no effect here (unless, of course, the coin is so light that it floats like a feather): a lopsided coin spins around an axis that passes through its center of gravity, and although the axis does not go through the geometrical center of the coin, there is no difference in the way the biased and symmetric coins spin about their axes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://oz.berkeley.edu/%7Enolan/Papers/dice.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;You can load a die but you can’t bias a coin&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1563" LastEditorUserId="63414" LastEditDate="2020-06-26T12:44:17.037" LastActivityDate="2020-06-26T12:44:17.037" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="9755" PostTypeId="2" ParentId="9745" CreationDate="2015-03-02T20:45:53.140" Score="18" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Awkward&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the purpose of this riddle&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; self fulfilling line I suppose&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and I am without grace.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; definitely awkward&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I dislike comfort,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; definitely awkward&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but give me the title of doctor,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I'll go back and forth in an eternal loop for you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dr awkward is a palindrome&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9896" LastActivityDate="2015-03-02T20:45:53.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9758" PostTypeId="1" AcceptedAnswerId="9766" CreationDate="2015-03-03T02:50:06.577" Score="13" ViewCount="430" Body="&lt;p&gt;This is a question from a previous, now-defunct version of the &lt;a href=&quot;http://joezeng.com/quetzalcoatl/&quot; rel=&quot;noreferrer&quot;&gt;Temple of Quetzalcoatl&lt;/a&gt;, that I've posted here as an exemplar of one of the difficult puzzles you might encounter while solving it. This was one of the last puzzles I'd created before rebooting that section of my website, and nobody to my knowledge had solved it on their own before I put it here.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Quetzalcoatl shoots a straight, thin beam of red light out of his mouth, like a laser beam, and points it at a group of stars. He starts flicking and swishing this beam of light, appearing to make it cut hot scars into the night sky. When he is done, you see a series of lines, in the form of a constellation:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2ZeVm.png&quot; alt=&quot;Constellation 1&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;And behind you, directly opposite of where Quetzalcoatl drew his original image, you see the exact same image burned into the empty, starless sky, but now with a single point of light shining brightly next to it:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IJVpF.png&quot; alt=&quot;Opposite of first constellation&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Quetzalcoatl asks you, &amp;quot;Do you know, my student, what star that is?&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-12-21T15:28:42.970" Title="Which star is this?" Tags="|outer-space|" AnswerCount="1" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="9759" PostTypeId="2" ParentId="9711" CreationDate="2015-03-03T03:10:24.943" Score="2" Body="&lt;p&gt;I used a different approach than Callidus, but ended up at exactly the same target, and made exactly the same mistake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the highest possible number of nonsensical games, the ideal situation would have every player with a different final ranking, with as little difference between each as possible (1/2 point) and a maximum of 1 tie per player. This can be done with any odd number of players greater than 3 (a 3 player tournament is guaranteed to have 100% sensical games, so it is ignored).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this situation, the middle-most player has a ranking of $\frac{1}{2}(n-1)$.&lt;br&gt;&#xA;The best player has a ranking of $\frac{3}{4}(n-1)$.&lt;br&gt;&#xA;The worst player has a ranking of $\frac{1}{4}(n-1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then set up the wins and losses such that each player loses as many games as possible to players with worse rankings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total number of games played is $\frac{n(n-1)}{2}$.&lt;br&gt;&#xA;The maximum number of nonsensical games is $\frac{(n-1)^2}{4}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives the percentage of nonsensical games as $\frac{n-1}{2n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim\limits_{x\to\infty}\frac{n-1}{2n}=\frac{1}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for the first theorem, $x=50$. For the second theorem, I can find a tournament with $51$ players where $x+1&amp;lt;51$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some interesting points:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5$: $40\%$ non-sensical games&lt;br&gt;&#xA;$25$: $48\%$ non-sensical games.&lt;br&gt;&#xA;$51$: Over $49\%$ non-sensical games.&lt;br&gt;&#xA;$501$: Over $49.9\%$ non-sensical games.&lt;br&gt;&#xA;$5001$: Over $49.99\%$ non-sensical games.&lt;br&gt;&#xA;$50001$: Over $49.999\%$ non-sensical games.&lt;br&gt;&#xA;...&lt;/p&gt;&#xA;" OwnerUserId="375" LastEditorUserId="375" LastEditDate="2015-03-04T15:35:09.750" LastActivityDate="2015-03-04T15:35:09.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9760" PostTypeId="1" AcceptedAnswerId="9767" CreationDate="2015-03-03T04:11:10.920" Score="2" ViewCount="249" Body="&lt;p&gt;I am always growing,&lt;br&gt;&#xA;but I never get any taller.&lt;br&gt;&#xA;I am hard&lt;br&gt;&#xA;but I can bend.&lt;br&gt;&#xA;I can be seen in many colors,&lt;br&gt;&#xA;but those colors are not me.&lt;br&gt;&#xA;I move often,&lt;br&gt;&#xA;but I can't move unassisted.&lt;br&gt;&#xA;I cannot feel pleasure or pain.&lt;br&gt;&#xA;Unless it's too hot, I rarely see the light of day.&lt;br&gt;&#xA;I usually live in darkness, except when I wash.&lt;/p&gt;&#xA;" OwnerUserId="9495" LastEditorUserId="9495" LastEditDate="2015-03-03T05:44:22.200" LastActivityDate="2015-03-03T09:43:43.627" Title="Except when I wash" Tags="|riddle|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9761" PostTypeId="2" ParentId="9737" CreationDate="2015-03-03T04:46:15.753" Score="7" Body="&lt;p&gt;The canonical answer is 12:44 PM, because if all the cells in that half of the culture divide, they will take up two halves, or the whole culture. However, this answer is too simplistic and ignores crucial facts in population dynamics.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;A more realistic answer would be approximately 1:30 PM, around double the time that it took to get from one cell to half the culture.&lt;/p&gt;&#xA;&lt;p&gt;The reason for this is that bacterial cultures grow &lt;a href=&quot;http://en.wikipedia.org/wiki/Logistic_function&quot; rel=&quot;nofollow noreferrer&quot;&gt;logistically&lt;/a&gt;, not exponentially. At 12:44 PM, the bacteria would not take up the whole culture, because a lot of the existing bacteria would have died out due to lack of nutrients or accumulation of waste products. Assuming the bacteria did in fact follow a perfect logistic curve, this is how the calculation would go:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Let &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; be the proportion of the culture already occupied by bacteria, a number from &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;. Because the bacteria double every minute at the start, &lt;span class=&quot;math-container&quot;&gt;$n(t)$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt; is the number of minutes since 12:43 PM has a differential equation of &lt;span class=&quot;math-container&quot;&gt;$\frac{dn}{dt} = \log 2 \times n(1-n)$&lt;/span&gt;, and the logistic function itself is represented as &lt;span class=&quot;math-container&quot;&gt;$$\frac{1}{1 + 2^{-t}}.$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Therefore, at 12:44 PM, the proportion of the culture taken up by bacteria would be &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{1 + 2^{-1}} = \frac{2}{3}$&lt;/span&gt;, about two-thirds of the culture instead of the whole thing as the problem believes it will.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Now comes the question of when we decide that the &amp;quot;whole culture&amp;quot; has been occupied by bacteria, since the bacteria are obviously not going to literally fill up the entire culture at once, but instead the population levels out so that the population stops growing. (In biology, this is known as the bacteria having entered the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bacterial_growth#Phases&quot; rel=&quot;nofollow noreferrer&quot;&gt;stationary phase&lt;/a&gt;.) I would peg that point at the point where &lt;span class=&quot;math-container&quot;&gt;$\frac{dn}{dt}$&lt;/span&gt; is less than 1 cell.&lt;/p&gt;&#xA;&lt;p&gt;Since the logistic curve is odd, this means that the answer is 1:26 PM, since the logistic curve has the property that its derivative at &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt; minutes (rate of change at 1:26 PM) is equal to its derivative at &lt;span class=&quot;math-container&quot;&gt;$-t$&lt;/span&gt; minutes (rate of change at 12:00, when the culture was just one cell, and the overall rate of change in population was also just one cell). Since bacterial culture growth is not exact, there is a little bit of fudge factor and I could reasonably predict that the bacteria would fill the culture entirely at 1:30 PM.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-03-03T16:42:54.193" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9763" PostTypeId="1" CreationDate="2015-03-03T06:02:56.103" Score="22" ViewCount="3551" Body="&lt;p&gt;I remember hearing this riddle when I was young.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Divided by a wall, N and S.&lt;br /&gt;&#xA;Tunnels in between, but no one pass.&lt;/p&gt;&#xA;&lt;p&gt;The West has more, and the East has less.&lt;br /&gt;&#xA;Less is more, but More is less.&lt;/p&gt;&#xA;&lt;p&gt;What am I, can you guess?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; I have a rectangular border&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8882" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2022-05-05T14:21:50.977" Title="Riddle - Divided by a wall" Tags="|riddle|" AnswerCount="8" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="9764" PostTypeId="2" ParentId="9746" CreationDate="2015-03-03T07:10:58.937" Score="3" Body="&lt;p&gt;They play&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; rock paper scissors with the coin, where heads is rock, tails is paper and somebody ran away with the scissors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; A game consists of each of them tossing the coin. The first one to get tails when the other gets heads wins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3918" LastEditorUserId="63414" LastEditDate="2020-06-26T12:46:07.533" LastActivityDate="2020-06-26T12:46:07.533" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="9766" PostTypeId="2" ParentId="9758" CreationDate="2015-03-03T08:25:49.050" Score="10" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The star is Gamma Ophiuchi&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reason&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &quot;hydra&quot; is made up of Canis major and Orion. Sirius, Orion's nebula and the belt of Orion are quite well visible. The required star seems to be the antipode of the point got by extending  Orion's belt by half its length to one side. The approximate coordinates of the point are (5h 45m, -2d 45m) giving the antipode to be (17h 45m, 2d 45m). Gamma Ophiuchi is the closest star there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9471" LastActivityDate="2015-03-03T08:25:49.050" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9767" PostTypeId="2" ParentId="9760" CreationDate="2015-03-03T09:43:43.627" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Toenail  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am always growing,&#xA;but I never get any taller.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They get longer, not taller.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am hard&#xA;but I can bend.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Obviously.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be seen in many colors,&#xA;but those colors are not me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Due to nail varnish.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I move often,&#xA;but I can't move unassisted.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Walking around does the trick.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cannot feel pleasure or pain.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Even scissors won't hurt.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unless it's too hot, I rarely see the light of day.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Walking around bare footed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I usually live in darkness, except when I wash.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hard to wash your feet with shoes or socks on.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9910" LastActivityDate="2015-03-03T09:43:43.627" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9768" PostTypeId="2" ParentId="9763" CreationDate="2015-03-03T09:56:51.090" Score="21" Body="&lt;p&gt;Divided by a wall, N and S.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;S for seven (7), N for nine(9), eight (8) is in between &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tunnels in between, but no one pass.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Eight is 8, with 2 tunnels&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The West has more, and the East has less.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Seven has 5 letters, nine has 4 letters&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Less is more, but More is less.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Seven is less than nine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The number 789&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9174" LastEditorDisplayName="user9174" LastEditDate="2015-03-04T14:49:58.807" LastActivityDate="2015-03-04T14:49:58.807" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9769" PostTypeId="2" ParentId="9763" CreationDate="2015-03-03T10:15:07.673" Score="10" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Roman numbers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Divided by a wall, N and S.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Horizontal line at the top and bottom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tunnels in between, but no one pass.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gap between the letters forming the number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The West has more, and the East has less.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Letters on the left usually contribute more to the sum than on the right, e.g. XVI for 10 + 5 + 1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Less is more, but More is less.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; IV is less than V, even though it consists of more letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9910" LastActivityDate="2015-03-03T10:15:07.673" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9770" PostTypeId="1" AcceptedAnswerId="9774" CreationDate="2015-03-03T10:24:14.593" Score="22" ViewCount="3697" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;I say plenty but never with sound;&lt;br /&gt;&#xA;Free to say anything, yet always bound.&lt;/p&gt;&#xA;&lt;p&gt;I like to be kindled but never kindling.&lt;br /&gt;&#xA;I'm always turned yet the same way round.&lt;/p&gt;&#xA;&lt;p&gt;Born from a tree, but not in the ground,&lt;br /&gt;&#xA;Small, yet inside whole worlds are found.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;It's the first time I've written a riddle. I worry that it's too easy, but then riddles are much more obvious if you know the answer. Really I've no idea of how easy or hard it is, but hopefully it gives a few moments of entertainment nevertheless.&lt;/p&gt;&#xA;" OwnerUserId="9532" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-03-24T18:40:55.370" Title="Worlds. My first ever riddle" Tags="|riddle|word|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9771" PostTypeId="1" AcceptedAnswerId="9772" CreationDate="2015-03-03T10:36:56.763" Score="-3" ViewCount="3265" Body="&lt;p&gt;A monkey, a squirrel, and a bird are racing to the top of a coconut tree. Who will get the banana first, the monkey, the squirrel, or the bird?&lt;/p&gt;&#xA;" OwnerUserId="9888" LastEditorUserId="4050" LastEditDate="2015-03-21T19:02:03.867" LastActivityDate="2015-03-21T19:02:03.867" Title="A monkey, a squirrel, and a bird" Tags="|riddle|" AnswerCount="1" CommentCount="2" FavoriteCount="0" ClosedDate="2015-03-19T10:24:32.030" ContentLicense="CC BY-SA 3.0" />
  <row Id="9772" PostTypeId="2" ParentId="9771" CreationDate="2015-03-03T10:41:55.653" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That's a coconut tree - no bananas(unless you bring your own, in which case a monkey is more likely to carry a banana) and what you posted was supposed to be puzzling.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can see a wave of downvotes soaring towards your question. &lt;code&gt;;-)&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="9865" LastActivityDate="2015-03-03T10:41:55.653" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9773" PostTypeId="2" ParentId="9770" CreationDate="2015-03-03T10:51:51.180" Score="10" Body="&lt;p&gt;This seems to be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a book&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I say plenty but never with sound;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You access the contents of a book through your eyes, not your ears&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Free to say anything, yet always bound.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Freedom of writing; books are bound&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I like to be kindled but never kindling.&lt;br&gt;&#xA;I'm always turned yet the same way round.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since the book is written left to right (resp right to left),&#xA; there is only one good way of holding it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Born from a tree, but not in the ground,&lt;br&gt;&#xA;Small, yet inside whole worlds are found.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Books are made of paper, paper is made of wood,  etc&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-03-03T10:51:51.180" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9774" PostTypeId="2" ParentId="9770" CreationDate="2015-03-03T10:56:11.117" Score="17" Body="&lt;p&gt;I say plenty but never with sound;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Printed or written text says a lot about the topic&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Free to say anything, yet always bound.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A book is always bound and the author can write anything he/she wishes &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I like to be kindled but never kindling.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A book can be stored in a softcopy format on e-bookreader device such as &quot;Kindle&quot;, Book will not like if you put it on fire but it desires your care, you must handle it carefully&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm always turned yet the same way round.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;we turn the pages of a book one by one direction&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Born from a tree, but not in the ground, Small, yet inside whole worlds are found.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Paper is made from tree&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I am a Book&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9174" LastEditorDisplayName="user9174" LastEditDate="2015-03-03T11:43:35.540" LastActivityDate="2015-03-03T11:43:35.540" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9775" PostTypeId="1" AcceptedAnswerId="9776" CreationDate="2015-03-03T11:19:17.270" Score="4" ViewCount="4450" Body="&lt;p&gt;We hurt without moving.&lt;br&gt;&#xA;We poison without touching.&lt;br&gt;&#xA;We bear the truth and the lies.&lt;br&gt;&#xA;We are not to be judged by our size.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are we?&lt;/p&gt;&#xA;" OwnerUserId="9888" LastEditorUserId="8874" LastEditDate="2015-03-03T11:28:30.367" LastActivityDate="2015-03-03T11:41:35.220" Title="We hurt without moving" Tags="|riddle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9776" PostTypeId="2" ParentId="9775" CreationDate="2015-03-03T11:29:33.257" Score="4" Body="&lt;p&gt;You should be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; words&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We hurt without moving.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Words may hurt, but do not move&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We poison without touching. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Words may poison relationships, but do not touch&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We bear the truth and the lies.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Words can state truth and lies&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are not to be judged by our size.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Small print and block capitals: words always mean the same&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-03-03T11:29:33.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9777" PostTypeId="2" ParentId="9775" CreationDate="2015-03-03T11:41:35.220" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Letters&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;hurt without moving.&lt;br&gt;&#xA;poison without touching.&lt;br&gt;&#xA;bear the truth and the lies.&lt;br&gt;&#xA;are not to be judged by their size. &lt;/p&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2015-03-03T11:41:35.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9778" PostTypeId="2" ParentId="9763" CreationDate="2015-03-03T11:45:14.540" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Korea&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Divided by a wall, N and S.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Korea is divided into &lt;strong&gt;N&lt;/strong&gt;orth-Korea and &lt;strong&gt;S&lt;/strong&gt;outh-Korea by the border wall&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tunnels in between, but no one pass.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are tunnels going under the border, preparing both parts for future military attacks&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The West has more, and the East has less.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the rich South-Korea counts as part of the western alliance, the poor North-Korea belongs to the east&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Less is more, but More is less.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The propaganda machineries in North and South lie about everything&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2015-03-03T11:45:14.540" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9779" PostTypeId="1" AcceptedAnswerId="9784" CreationDate="2015-03-03T11:48:00.997" Score="4" ViewCount="758" Body="&lt;p&gt;Start with an unlucky Friday number.&lt;br&gt;&#xA;Multiply by the gables on Hawthorne's house.&lt;br&gt;&#xA;Add the number that a stitch in time saves.&lt;br&gt;&#xA;Add the number of blind mice.&lt;br&gt;&#xA;Subtract the number of William Pene du Bois' balloons.&lt;br&gt;&#xA;Add the number of wonders of the world.&lt;br&gt;&#xA;Subtract the number of miles in Camptown's racetrack.&lt;br&gt;&#xA;Subtract the number of strings on a violin.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide by the number of vertices on a regular hexahedron.&lt;br&gt;&#xA;Add the number of the engine that ran on Chicago line.&lt;br&gt;&#xA;Multiply by the number of gentlemen of Verona.&lt;br&gt;&#xA;Add the atomic number of the element whose symbol is the 25th letter of the alphabet.&lt;br&gt;&#xA;Divide by the number of hills of Rome.&lt;br&gt;&#xA;Multiply by the number of railroads on a Monopoly board.&lt;br&gt;&#xA;Add the number of easy pieces.&lt;br&gt;&#xA;Add the number of chromosomes in a normal human muscle cell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply by the number of kittens that lost their mittens.&lt;br&gt;&#xA;Add the number of acres in A.A.Milne's woods.&lt;br&gt;&#xA;Multiply by the number of cities in Dickens' tale.&lt;br&gt;&#xA;Subtract the number of degrees Fahrenheit at which Bradbury's books burn.&lt;br&gt;&#xA;Add the number of Great Lakes.&lt;br&gt;&#xA;Divide by the number of days of the condor in the title of Grady's book.&lt;br&gt;&#xA;Subtract the number of blackbirds baked in a pie.&lt;br&gt;&#xA;Multiply by the number of horsemen of the Apocalypse. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide by the number of men on a dead man's chest.&lt;br&gt;&#xA;Add the number of a neutral PH.&lt;br&gt;&#xA;Subtract the number of carbon atoms in a molecule of ethane.&lt;br&gt;&#xA;Multiply by the number of heads on Lofting's Push-me-Pull-you.&lt;br&gt;&#xA;Add the number of miles on the road sign to Juster's Digitopolis.&lt;br&gt;&#xA;Subtract the number of dried orange pips in a Sherlock Holmes case.&lt;br&gt;&#xA;Multiply by the number of the square at which Alice met Humpty Dumpty.&lt;br&gt;&#xA;Divide by the number of witches in Macbeth.&lt;br&gt;&#xA;Divide by the number of suits in a standard deck of cards.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="9404" LastEditDate="2015-03-30T14:51:34.150" LastActivityDate="2015-03-30T14:51:34.150" Title="Start with an unlucky Friday" Tags="|riddle|calculation-puzzle|knowledge|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9780" PostTypeId="1" CreationDate="2015-03-03T11:54:51.950" Score="13" ViewCount="665" Body="&lt;p&gt;I'm known as single or as double&lt;br&gt;&#xA;Be it in head or mind or eye.&lt;br&gt;&#xA;For some I cause enough of trouble&lt;br&gt;&#xA;That they will force me to comply.&lt;br&gt;&#xA;But even if I'm weak or near&lt;br&gt;&#xA;Up to an imperfect extent&lt;br&gt;&#xA;Once i show up i need not fear&lt;br&gt;&#xA;You expect to find me at the end.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I've related you my tail&lt;br&gt;&#xA;What is the name by which I hail?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I often live close to the fringe&lt;br&gt;&#xA; If I am broken you may cringe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Additional hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I stand out most when I'm perfected&lt;br&gt;&#xA; But even slanted I'm detected.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9910" LastEditorUserId="9910" LastEditDate="2015-04-01T11:06:57.233" LastActivityDate="2017-07-27T19:23:41.310" Title="Some will force me to comply" Tags="|riddle|word|" AnswerCount="9" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9781" PostTypeId="2" ParentId="9779" CreationDate="2015-03-03T12:15:31.280" Score="2" Body="&lt;p&gt;The numbers are, in order:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 13, 7, 9, 3, 21, 7, 5?, 4, 8, 9, 2, 39, 7, 4, 5?, 46, 3, 100, 2, 451, 5, 6, 24, 4, 15, 7, 2, 2, 5?, 5, 6, 3, 4.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the final answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 58.6 (&lt;strong&gt;edit: 13&lt;/strong&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm short on time now, but I'll edit this answer into a (&lt;em&gt;much&lt;/em&gt;) nicer format later! &lt;strong&gt;Edit:&lt;/strong&gt; user9914 and the omnipresent Gamow have beaten me to it. I'll leave this answer in place since I was the first to get all the numbers, my only mistake being to read &quot;Add the number of easy pieces&quot; as &quot;Multiply by the number of easy pieces&quot;.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-03T20:34:37.397" LastActivityDate="2015-03-03T20:34:37.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9783" PostTypeId="2" ParentId="9779" CreationDate="2015-03-03T12:57:00.343" Score="1" Body="&lt;p&gt;Start with an unlucky Friday number. (13)&lt;br&gt;&#xA;Multiply by the gables on Hawthorne's house. (7)&lt;br&gt;&#xA;Add the number that a stitch in time saves. (9)&lt;br&gt;&#xA;Add the number of blind mice. (3)&lt;br&gt;&#xA;Subtract the number of William Pene du Bois' balloons. (21)&lt;br&gt;&#xA;Add the number of wonders of the world. (7)&lt;br&gt;&#xA;Subtract the number of miles in Camptown's racetrack. (5)&lt;br&gt;&#xA;Subtract the number of strings on a violin. (4)&lt;br&gt;&#xA;Divide by the number of vertices on a regular hexahedron. (8)&lt;br&gt;&#xA;Add the number of the engine that ran on Chicago line. (9)&lt;br&gt;&#xA;Multiply by the number of gentlemen of Verona. (2)&lt;br&gt;&#xA;Add the atomic number of the element whose symbol is the 25th letter of the alphabet.(39)&lt;br&gt;&#xA;Divide by the number of hills of Rome.(7)&lt;br&gt;&#xA;Multiply by the number of railroads on a Monopoly board. (4)&lt;br&gt;&#xA;Add the number of easy pieces. (5)&lt;br&gt;&#xA;Add the number of chromosomes in a normal human muscle cell. (46)&lt;br&gt;&#xA;Multiply by the number of kittens that lost their mittens. (3)&lt;br&gt;&#xA;Add the number of acres in A.A.Milne's woods. (100)&lt;br&gt;&#xA;Multiply by the number of cities in Dickens' tale. (2)&lt;br&gt;&#xA;Subtract the number of degrees Fahrenheit at which . (451)&lt;br&gt;&#xA;Add the number of Great Lakes. (5)&lt;br&gt;&#xA;Divide by the number of days of the condor in the title of Grady's book. (6)&lt;br&gt;&#xA;Subtract the number of blackbirds baked in a pie. (24)&lt;br&gt;&#xA;Multiply by the number of horsemen of the Apocalypse. (4)&lt;br&gt;&#xA;Divide by the number of men on a dead man's chest. (15)&lt;br&gt;&#xA;Add the number of a neutral PH. (7)&lt;br&gt;&#xA;Subtract the number of carbon atoms in a molecule of ethane. (2)&lt;br&gt;&#xA;Multiply by the number of heads on Lofting's Push-me-Pull-you. (2)&lt;br&gt;&#xA;Add the number of miles on the road sign to Juster's Digitopolis. (198)&lt;br&gt;&#xA;Subtract the number of dried orange pips in a Sherlock Holmes case. (5)&lt;br&gt;&#xA;Multiply by the number of the square at which Alice met Humpty Dumpty. (6)&lt;br&gt;&#xA;Divide by the number of witches in Macbeth. (3)&lt;br&gt;&#xA;Divide by the number of suits in a standard deck of cards.(4)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;13*7+9+3-21+7-5-4/8+9*2+39/7*4+5+46*3+100*2-451+5/6-24*4/15+7-2+198-5*6/3/4&lt;br&gt;&#xA;=&gt;&lt;strong&gt;103&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9669" LastEditorUserId="9669" LastEditDate="2015-03-03T13:03:58.677" LastActivityDate="2015-03-03T13:03:58.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9784" PostTypeId="2" ParentId="9779" CreationDate="2015-03-03T12:58:37.607" Score="7" Body="&lt;p&gt;&lt;strong&gt;The answer is: You are Thirteen (13).&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This community wiki provides a step-by-step solution for the 32 numbers:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Start with an unlucky Friday number --&gt; &lt;strong&gt;Friday 13th&lt;/strong&gt; --&gt; 13&lt;/li&gt;&#xA;&lt;li&gt;Multiply by the gables on Hawthorne's house --&gt; &lt;strong&gt;The House of the Seven Gables&lt;/strong&gt; --&gt; 13*7=91&lt;/li&gt;&#xA;&lt;li&gt;Add the number that a stitch in time saves --&gt; &lt;strong&gt;A stitch in time saves nine&lt;/strong&gt; --&gt; 91+9=100&lt;/li&gt;&#xA;&lt;li&gt;Add the number of blind mice.  - &lt;strong&gt;Three blind mice&lt;/strong&gt; --&gt; 100+3=103&lt;/li&gt;&#xA;&lt;li&gt;Subtract the number of William Pene du Bois' balloons.  -&gt;- &lt;strong&gt;The Twenty-One Balloons&lt;/strong&gt;  --&gt; 103-21=82&lt;/li&gt;&#xA;&lt;li&gt;Add the number of wonders of the world.  --&gt; &lt;strong&gt;Seven wonders of the world&lt;/strong&gt; --&gt; 82+7=89&lt;/li&gt;&#xA;&lt;li&gt;Subtract the number of miles in Camptown's racetrack. --&gt; &lt;strong&gt;De Camptown race-track five miles long -- Oh! Doo-dah day!&lt;/strong&gt; --&gt; 89-5=84&lt;/li&gt;&#xA;&lt;li&gt;Subtract the number of strings on a violin. --&gt; &lt;strong&gt;four strings on a violin&lt;/strong&gt; --&gt; 84-4=80&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Divide by the number of vertices on a regular hexahedron.  --&gt; &lt;strong&gt;a cube has 8 vertices&lt;/strong&gt; --&gt; 80/8=10&lt;/li&gt;&#xA;&lt;li&gt;Add the number of the engine that ran on Chicago line.  --&gt; &lt;strong&gt;Engine Engine Number Nine, going down Chicago line&lt;/strong&gt; --&gt; 10+9=19 &lt;/li&gt;&#xA;&lt;li&gt;Multiply by the number of gentlemen of Verona. --&gt; &lt;strong&gt;The Two Gentlemen of Verona by Shakespeare&lt;/strong&gt; --&gt; 19*2=38&lt;/li&gt;&#xA;&lt;li&gt;Add the atomic number of the element whose symbol is the 25th letter of the  alphabet. --&gt; &lt;strong&gt;atomic number of Y=Yttrium is 39&lt;/strong&gt; --&gt; 38+39=77&lt;/li&gt;&#xA;&lt;li&gt;Divide by the number of hills of Rome. --&gt; &lt;strong&gt;Rome was built on seven hills&lt;/strong&gt; --&gt; 77/7=11 &lt;/li&gt;&#xA;&lt;li&gt;Multiply by the number of railroads on a Monopoly board. --&gt; &lt;strong&gt;four railway stations on a Monopoly board&lt;/strong&gt; --&gt; 11*4=44&lt;/li&gt;&#xA;&lt;li&gt;Add the number of easy pieces.  --&gt; &lt;strong&gt;Five Easy Pieces&lt;/strong&gt; --&gt; 44+5=49&lt;/li&gt;&#xA;&lt;li&gt;Add the number of chromosomes in a normal human muscle cell. --&gt; &lt;strong&gt;46 chromosomes per cell&lt;/strong&gt; --&gt; 49+46=95&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Multiply by the number of kittens that lost their mittens. --&gt; &lt;strong&gt;The three little kittens, they lost their mittens&lt;/strong&gt; --&gt; 95*3= 285;&lt;/li&gt;&#xA;&lt;li&gt;Add the number of acres in A.A.Milne's woods. --&gt; &lt;strong&gt;Hundred Acre Wood&lt;/strong&gt; --&gt; 285+100=385&lt;/li&gt;&#xA;&lt;li&gt;Multiply by the number of cities in Dickens' tale. --&gt; &lt;strong&gt;A Tale of Two Cities&lt;/strong&gt; --&gt;  385*2=770&lt;/li&gt;&#xA;&lt;li&gt;Subtract the number of degrees Fahrenheit at which Bradbury's books burn.   --&gt; &lt;strong&gt;Fahrenheit 451&lt;/strong&gt; --&gt; 770-451=319&lt;/li&gt;&#xA;&lt;li&gt;Add the number of Great Lakes.   --&gt; &lt;strong&gt;The Five Great Lakes&lt;/strong&gt; --&gt; 319+5=324&lt;/li&gt;&#xA;&lt;li&gt;Divide by the number of days of the condor in the title of Grady's book.   --&gt; &lt;strong&gt;Six Days of the Condor&lt;/strong&gt; --&gt; 324/6=54&lt;/li&gt;&#xA;&lt;li&gt;Subtract the number of blackbirds baked in a pie.   --&gt; &lt;strong&gt;Four and twenty blackbirds baked in a pie&lt;/strong&gt; --&gt; 54-24=30&lt;/li&gt;&#xA;&lt;li&gt;Multiply by the number of horsemen of the Apocalypse.  --&gt; &lt;strong&gt;The Four Horsemen of the Apocalypse&lt;/strong&gt; --&gt; 30*4=120&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Divide by the number of men on a dead man's chest.   --&gt; &lt;strong&gt;Fifteen Men On The Dead Man's Chest&lt;/strong&gt; --&gt; 120/15=8&lt;/li&gt;&#xA;&lt;li&gt;Add the number of a neutral PH.  --&gt; &lt;strong&gt;A pH of 7 is neutral&lt;/strong&gt; --&gt; 8+7=15&lt;/li&gt;&#xA;&lt;li&gt;Subtract the number of carbon atoms in a molecule of ethane.   --&gt; &lt;strong&gt;Ethane is C&lt;sub&gt;2&lt;/sub&gt;H&lt;sub&gt;6&lt;/sub&gt;&lt;/strong&gt; --&gt; 15-2=13&lt;/li&gt;&#xA;&lt;li&gt;Multiply by the number of heads on Lofting's Push-me-Pull-you.   --&gt; &lt;strong&gt;gazelle-unicorn cross which has two heads&lt;/strong&gt; --&gt; 13*2=26&lt;/li&gt;&#xA;&lt;li&gt;Add the number of miles on the road sign to Juster's Digitopolis.   --&gt; &lt;strong&gt;an enormous road sign, pointing in all three directions, stated clearly: DIGITOPOLIS 5 Miles&lt;/strong&gt; --&gt; 26+5=31&lt;/li&gt;&#xA;&lt;li&gt;Subtract the number of dried orange pips in a Sherlock Holmes case.  --&gt; &lt;strong&gt;The Five Orange Pips&lt;/strong&gt; --&gt; 31-5=26&lt;/li&gt;&#xA;&lt;li&gt;Multiply by the number of the square at which Alice met Humpty Dumpty.  --&gt; &lt;strong&gt;the Sixth square belongs to Humpty Dumpty&lt;/strong&gt; --&gt; 26*6=156&lt;/li&gt;&#xA;&lt;li&gt;Divide by the number of witches in Macbeth.  --&gt; &lt;strong&gt;The three weyward Sisters, hand in hand&lt;/strong&gt; --&gt; 156/3=52&lt;/li&gt;&#xA;&lt;li&gt;Divide by the number of suits in a standard deck of cards.  --&gt; &lt;strong&gt;spades, hearts, clubs, diamonds&lt;/strong&gt; --&gt; 52/4=13&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="7482" LastEditDate="2015-03-04T08:19:35.663" LastActivityDate="2015-03-04T08:19:35.663" CommentCount="0" CommunityOwnedDate="2015-03-03T12:58:37.607" ContentLicense="CC BY-SA 3.0" />
  <row Id="9785" PostTypeId="1" CreationDate="2015-03-03T13:35:15.533" Score="-1" ViewCount="614" Body="&lt;p&gt;A solved 3x3 Rubik's Cube has red on the top face, green on the right face, and yellow on the front face. There is one corner piece belonging to these three faces, with those three colours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know if it is possible with any sequence of moves to have the corner piece end up in the same position, but now with red on top, green in front, and yellow on the right?  (I know that we can bring that corner piece to that position with &lt;em&gt;either&lt;/em&gt; red on the right, green in front, yellow on top &lt;em&gt;or&lt;/em&gt; red in front, green on top , yellow on the right.)&lt;/p&gt;&#xA;" OwnerUserId="9911" LastEditorUserId="5373" LastEditDate="2015-03-07T10:29:38.720" LastActivityDate="2017-03-08T04:29:59.290" Title="Is this Rubik's Cube position possible?" Tags="|rubiks-cube|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9786" PostTypeId="2" ParentId="9780" CreationDate="2015-03-03T13:39:00.360" Score="5" Body="&lt;p&gt;I can't figure out how to fit this guess to all the clues but I think you are a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quotation mark or apostrophe&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm known as single or as double, be it in head or mind or eye.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quotes can be single ' or double &quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For some I cause enough of trouble that they will force me to comply.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are multiple schools of thought for how to handle punctuation around quotation marks. Trailing period in or out? What if it wasn't in the original quoted text? Different punctuation in original text and your sentence? Oh, the horrors!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But even if I'm weak or near, up to an imperfect extent&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (This is the one I can't nail down.) Are there weak quotations?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once i show up i need not fear, you expect to find me at the end.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Once you come across an opening quotation mark, you can expect to find another one at the end of the quote.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now I've related you my tail, what is the name by which I hail?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm assuming that &quot;tail&quot; was an intentional pun as the marks have tails. The tails aren't obvious on all fonts but they're &lt;a href=&quot;http://blog.rafflecopter.com/wp-content/uploads/2012/06/quotation-marks.jpg&quot; rel=&quot;noreferrer&quot;&gt;there&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastActivityDate="2015-03-03T13:39:00.360" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9787" PostTypeId="2" ParentId="9785" CreationDate="2015-03-03T14:20:49.510" Score="5" Body="&lt;p&gt;If I understand correctly you ask if a corner piece that has red green yellow as top right front respectively to be red yellow green instead. This is impossible. You need to understand that the corner piece itself is 1 piece, you can't swap two colors of that same piece. you can only rotate the entire piece so another color is on top.&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-03-03T14:20:49.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9788" PostTypeId="2" ParentId="9785" CreationDate="2015-03-03T14:22:51.817" Score="10" Body="&lt;p&gt;The only possible orientations that corner piece can be in is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;R&#xA;YG&#xA;&#xA;Y&#xA;GR&#xA;&#xA;G&#xA;RY&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It can only &quot;rotate&quot; ... so, no, you cannot have Green on front face, and red on top.&lt;/p&gt;&#xA;" OwnerUserId="9267" LastActivityDate="2015-03-03T14:22:51.817" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9789" PostTypeId="2" ParentId="9779" CreationDate="2015-03-03T14:48:43.107" Score="8" Body="&lt;p&gt;Answer is: 13&lt;/p&gt;&#xA;&#xA;&lt;p&gt;details below...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cFLNA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9914" LastActivityDate="2015-03-03T14:48:43.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9790" PostTypeId="2" ParentId="9780" CreationDate="2015-03-03T15:33:41.520" Score="4" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; vision&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm known as single or double&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure I've seen too many references to &quot;single vision&quot; except that it would be the natural antonym to &quot;double vision&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Be it in head or mind or eye.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One can refer to vision as what one sees with the eye or what one sees with the mind (or head) when reading a book or making plans for the future.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For some I cause enough of trouble / That they will force me to comply.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many people wear glasses or contacts to &quot;correct&quot; their vision. Some people, frustrated with another person's vision of the future (or lack thereof), will try to redirect it or obstruct it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But even if I'm weak or near / Up to an imperfect extent&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &quot;weaker&quot; the vision, the closer one comes to blindness. &quot;Near&quot; probably refers to near-sightedness, whether dealing with physical vision or a lack of long-term planning.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once I show up I need not fear / You expect to find me at the end.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This could actually be a reference to prophecy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9828" LastActivityDate="2015-03-03T15:33:41.520" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9791" PostTypeId="1" AcceptedAnswerId="9799" CreationDate="2015-03-03T16:33:48.067" Score="2" ViewCount="658" Body="&lt;p&gt;Here is a maze I found in a book. Normally I wouldn't think a maze is really hard enough to be worth this site, but I can guarantee this one will take atleast 15 minutes to solve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You must enter from the start arrow, collect all the stars, and exit at the exit arrow. You may not retrace or intersect your path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/a7pL0.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-03-03T18:04:06.997" Title="Dragon maze (collect stars and exit)" Tags="|mazes|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9792" PostTypeId="1" CreationDate="2015-03-03T16:34:26.983" Score="-3" ViewCount="2739" Body="&lt;p&gt;&amp;quot;&lt;a href=&quot;http://avko.org/free/reference/i-before-e-rule.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;I before E&lt;/a&gt;&amp;quot; is a common mnemonic that is taught for spelling certain English words, but this mnemonic is often wrong, as pointed out in &lt;a href=&quot;http://itre.cis.upenn.edu/%7Emyl/languagelog/archives/003785.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Language Log post&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The aim of this question is to create the longest grammatical English sentence with non-repeating words (each word can only be used once in the entire sentence) which contains the maximum number of words that violate the &amp;quot;I before E&amp;quot; rule, without resorting to quotations, numbers over ten, or proper nouns.&lt;/p&gt;&#xA;&lt;p&gt;Sentences will be scored by the number of violations, minus the number of non-violations (includes words which don't include &amp;quot;ie&amp;quot; or &amp;quot;ei&amp;quot;).&lt;/p&gt;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Caffeinated atheists reinvent their weird foreign neighbours(7-0=&lt;strong&gt;7&lt;/strong&gt;)&lt;/p&gt;&#xA;&lt;p&gt;Caffeinated atheists reinvent their blue foreign neighbours(6-1=&lt;strong&gt;5&lt;/strong&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7529" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2019-01-28T15:41:03.180" Title="I before E, except in the following sentences:" Tags="|open-ended|wordplay|language|" AnswerCount="3" CommentCount="11" FavoriteCount="0" ClosedDate="2021-05-07T15:29:12.813" ContentLicense="CC BY-SA 3.0" />
  <row Id="9793" PostTypeId="5" CreationDate="2015-03-03T16:48:05.870" Score="0" Body="&lt;p&gt;This tag should be used for puzzles in which knowledge of astronomy plays a crucial part in the solution. It should not be used for puzzles which superficially seem to involve astronomy but for which no astronomical knowledge is needed (e.g. a chart of stars which actually turns out to be a Baconian cipher).&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="5373" LastEditDate="2015-07-05T19:49:11.733" LastActivityDate="2015-07-05T19:49:11.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9794" PostTypeId="4" CreationDate="2015-03-03T16:48:05.870" Score="0" Body="A puzzle that makes use of astronomical knowledge, such as the names and positions of stars and planets." OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-03-03T16:48:05.870" LastActivityDate="2015-03-03T16:48:05.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9795" PostTypeId="2" ParentId="9780" CreationDate="2015-03-03T17:16:51.013" Score="0" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Shots&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm known as single or as double&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; or in some dodgey bars as quads&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Be it in head or mind or eye.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HeadSHOT and EyeSHOT are self explanatory. MindSHOT is less often used but can mean forgetfulness especially after alcohol.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For some I cause enough of trouble&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Classic alcohol&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That they will force me to comply.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; So many laws govern it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But even if I'm weak or near&#xA;Up to an imperfect extent&#xA;Once i show up i need not fear&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alcohol increases confidence, even though in an imperfect way. Even weak people seem to think that they are strong when drunk. Also, alcohol which is weak or cheap or plain crap will still get drunk by someone. When I'm near is often used to refer to romance or lust, which are often entwined with alcohol.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You expect to find me at the end.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You don't generally start off the night on shots. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't actually expect this is the answer but it's what popped into my head. &lt;/p&gt;&#xA;" OwnerUserId="9918" LastActivityDate="2015-03-03T17:16:51.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9796" PostTypeId="1" AcceptedAnswerId="9798" CreationDate="2015-03-03T17:18:56.147" Score="1" ViewCount="180" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;U PROP BY THERE, USE PARS PAR THERE PROP&lt;br&gt;&#xA;     BY THERE IYBSM PAR YUCR THER, USE YH HS.&lt;br&gt;&#xA;     THINE BP AUQQRS PAUP PAR HGBMBSUN PROP&lt;br&gt;&#xA;     BY GRUTARE UP YHCR YPUMR?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2015-03-04T02:45:09.840" Title="Caesar would be pleased" Tags="|cipher|code|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9797" PostTypeId="2" ParentId="9770" CreationDate="2015-03-03T17:34:12.730" Score="7" Body="&lt;p&gt;Surely this is rather&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pages&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I say plenty but never with sound;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pages of a book will say a lot but you read with your eyes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Free to say anything, yet always bound.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bound within a book&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I like to be kindled but never kindling.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nice reference to the Amazon Kindle, which displays a page at a time. Most books would be too large to be kindling, whereas pages would be excellent.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm always turned yet the same way round.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You turn the pages of a book. A book is never turned, and they are not always written in the same way, Japanese books write back to front, and other languages write vertically.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Born from a tree, but not in the ground, Small, yet inside whole worlds are found.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Common quote - In the pages of this book you will find... Perhaps this line lends itself more to the answer as a book but it is valid in either case.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9918" LastActivityDate="2015-03-03T17:34:12.730" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9798" PostTypeId="2" ParentId="9796" CreationDate="2015-03-03T17:38:53.343" Score="7" Body="&lt;h3&gt;Answer still in progress&lt;/h3&gt;&#xA;&lt;p&gt;I have decoded the cryptogram as it stands (not manually, but with a website) and it results in this:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;code&gt;A TEXT IS CODED, AND THEN THE CODED TEXT IS CODED USING THE SAME CODE, AND SO ON. COULD IT HAPPEN THAT THE ORIGINAL TEXT IS REACHED AT SOME STAGE?&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I haven't found the answer to this yet, though. Just wanted to get this up so others could use it. I have a very vague hypothesis that it will happen after 26 iterations, but this is completely unsupported.&lt;/p&gt;&#xA;&lt;p&gt;EDIT: Thanks to @Ross Millikan in the comments, the answer is actually 1260 - the highest LCM of a set of numbers with a sum of 26. The reason this works is because there can be &amp;quot;loops&amp;quot; where A is substituted with B, B with C, and C with A. This loop has a size of 3. The loops in a substitution cipher can sum to 26, and the 1260 figure comes from loops of 4, 5, 7, and 9 and an extra letter that is substituted with itself. If this isn't legal, then the actual amount is something else - I'll find that eventually.&lt;/p&gt;&#xA;" OwnerUserId="1664" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-03-04T02:45:09.840" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="9799" PostTypeId="2" ParentId="9791" CreationDate="2015-03-03T18:04:06.997" Score="13" Body="&lt;p&gt;The image quality is &lt;em&gt;really&lt;/em&gt; bad, so I made my best guess for what counts as a &quot;star&quot;. I marked those spots with blue below.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/llyre.png&quot; alt=&quot;solved&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The full path is &lt;strong&gt;green+yellow&lt;/strong&gt;. I used green/red to mark must/can't paths that I could deduce immediately, then yellow/purple as a second step. The more paths you mark, the less there are to try :)&lt;/p&gt;&#xA;" OwnerUserId="1563" LastActivityDate="2015-03-03T18:04:06.997" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9800" PostTypeId="1" CreationDate="2015-03-03T18:32:28.713" Score="0" ViewCount="131" Body="&lt;p&gt;$3$ people $A,B,C$ are shooting each other. The chance that $A$ will hit in a single shot is $0.2$,&#xA;The chance for $B$ to hit is $0.3$ and for $C$ is $0.1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one hit a person, he will continue to hit until he'll be the last one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the chance for each one of the people $A,B,C$ to be alive?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that $A$ will be alive if he shot $B$ and $B$ will miss him. Also $A$ hit $C$ and $C$ will miss him.&#xA;But I think it's more complex than that and I'm confused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd glad if someone can clear this question for me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="9920" LastEditorUserId="24713" LastEditDate="2016-09-05T12:31:49.403" LastActivityDate="2016-09-05T12:31:49.403" Title="duel of 3 people shooting" Tags="|probability|" AnswerCount="0" CommentCount="2" ClosedDate="2015-03-03T21:14:14.177" ContentLicense="CC BY-SA 3.0" />
  <row Id="9801" PostTypeId="2" ParentId="9792" CreationDate="2015-03-03T19:05:37.170" Score="-4" Body="&lt;p&gt;I've got 45-7=&lt;strong&gt;38&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Eighty feisty, atheistic, albeit dreich, reigning sovereign sheikhs' heirs, deifing society, inveigled a weird foreign leisurely scientist's fanciest eight rottweilers to forfeit sufficient obeisance, veiling a heinous seismic gneiss vein heist, wherein they abseiled heights of neighboring ancient lithofacies, heigh-ho-ing their weighty freight with beige greige skeins, later efficiently surfeiting on stupefacient codeine.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Face it, we can go on until we hit 1767 words...&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2015-03-03T20:10:24.410" LastActivityDate="2015-03-03T20:10:24.410" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="9802" PostTypeId="1" AcceptedAnswerId="9807" CreationDate="2015-03-03T19:22:28.467" Score="-1" ViewCount="1365" Body="&lt;p&gt;Can someone help me with the following riddle?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You know me best for being magic, but I can also mean a term of time... what am I?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9922" LastEditorUserId="5373" LastEditDate="2016-09-15T00:52:26.170" LastActivityDate="2016-09-15T00:52:26.170" Title="You know me best for being magic, but I can also mean a term of time... what am I?" Tags="|riddle|word|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9803" PostTypeId="2" ParentId="9792" CreationDate="2015-03-03T19:48:41.707" Score="5" Body="&lt;p&gt;I've got 44-7=&lt;strong&gt;37&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Eighty feisty, ableit dreich, reigning sovereign sheikhs' heirs, deifing society, inveigled a weird foreign leisurely scientist's fanciest eight rottweilers to forfeit sufficient obeisance, veiling a heinous seismic gneiss vein heist, wherein they abseiled heights of neighboring ancient lithofacies, heigh-ho-ing their weighty freight with beige greige skeins, later efficiently surfeiting on stupefacient codeine.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Basically, a bunch of rich kids charmed some dogs and stole a bunch of rocks, and then OD'd on painkillers.&lt;/p&gt;&#xA;" OwnerUserId="8776" LastEditorUserId="8776" LastEditDate="2015-03-03T20:34:15.863" LastActivityDate="2015-03-03T20:34:15.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9805" PostTypeId="2" ParentId="9792" CreationDate="2015-03-03T20:06:02.467" Score="3" Body="&lt;h2&gt;1,752 - 5 = 1,747&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;With inspiration from warspyking:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My favorite words include &lt;a href=&quot;http://www.morewords.com/contains/ei&quot; rel=&quot;nofollow noreferrer&quot;&gt;abseil, abseiled, abseiling, abseils,&#xA;  absenteeism, absenteeisms, ... , zein, zeins, zeitgeber, zeitgebers,&#xA;  zeitgeist, and zeitgeists&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's the entire list except for eighteen, eighteens, eightfold, eighties, and eighty due to the &quot;no numbers over 10&quot; rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize this is a snarky response but it's a grammatically correct sentence that meets the current requirements of the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only higher scores I can think of either include numbers over 10 (a rule added to prevent other snarky responses) or are variations on the sentence that uses less words. For instance, &quot;[List] are all words.&quot; or even &quot;[List] are words.&quot;&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2019-01-28T15:41:03.180" LastActivityDate="2019-01-28T15:41:03.180" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="9807" PostTypeId="2" ParentId="9802" CreationDate="2015-03-03T20:31:34.033" Score="2" Body="&lt;p&gt;The answer you're looking for is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; spell&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because you can&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; cast a spell, or have a short spell of bad weather.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9532" LastActivityDate="2015-03-03T20:31:34.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9809" PostTypeId="1" CreationDate="2015-03-03T21:05:40.717" Score="-7" ViewCount="352" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am born  as one colour and change to another before I reach the last colour. What am I?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can anybody help? It would probably be Harry Potter or magic related.&lt;/p&gt;&#xA;" OwnerUserId="9922" LastEditorUserId="9922" LastEditDate="2015-03-18T17:55:29.077" LastActivityDate="2015-03-18T17:55:29.077" Title="Can anyone help me with this one?" Tags="|riddle|" AnswerCount="1" CommentCount="2" ClosedDate="2015-03-03T21:49:08.830" ContentLicense="CC BY-SA 3.0" />
  <row Id="9810" PostTypeId="2" ParentId="9809" CreationDate="2015-03-03T21:14:51.173" Score="4" Body="&lt;p&gt;It could certainly be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; traffic lights,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but the question sounds as though it might be &quot;too broad&quot;.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-03-03T21:14:51.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9811" PostTypeId="2" ParentId="9763" CreationDate="2015-03-03T21:33:44.403" Score="-1" Body="&lt;p&gt;Guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Night and Sun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Divided by a wall, N and S.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Night and sunlight are blocked by the earth or are blocked by the spin of the earth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tunnels in between, but no one pass.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are tunnels in earth but neither night nor sunlight pass through. Alternatively, the tunnels could be dusk and dawn, which neither night nor sunlight can pass through all the way&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The West has more, and the East has less.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The west has more countries that use daylight savings times than the east.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Less is more, but More is less.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Less of one is more of the other&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4591" LastActivityDate="2015-03-03T21:33:44.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9812" PostTypeId="1" AcceptedAnswerId="9819" CreationDate="2015-03-03T21:58:21.863" Score="-2" ViewCount="844" Body="&lt;p&gt;I wonder if you'll be able to reverse this set of words, but first you may want to get a bit shifty!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;iLppk scyc, wys ihmgltvihi xm! vKein xfs :H&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I have given you all the information you need to solve this, it's up to you to find out how!&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="69582" LastEditDate="2023-12-15T16:13:26.020" LastActivityDate="2023-12-15T16:13:26.020" Title="Solve this Encoded Text! iLppk scyc, wys ihmgltvihi xm! vKein xfs :H" Tags="|cipher|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="9815" PostTypeId="1" AcceptedAnswerId="9821" CreationDate="2015-03-04T00:58:57.177" Score="4" ViewCount="343" Body="&lt;h1&gt;Riddle&lt;/h1&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This is the story of one of many.&lt;br&gt;&#xA;  This one wishes to tell you its tale.&lt;br&gt;&#xA;  It's just the same as any...&lt;br&gt;&#xA;  &lt;br&gt;&#xA;  &quot;I am the sum of a million lives,&lt;br&gt;&#xA;  my work is all from them.&lt;br&gt;&#xA;  Each entry is a million knives,&lt;br&gt;&#xA;  to the furious driver at my back.&#xA;  &lt;br&gt;&lt;br&gt;&#xA;  &quot;My ancestors, I tower before;&lt;br&gt;&#xA;  to my son I'll be nothing more&lt;br&gt;&#xA;  than another knife held to his throat,&lt;br&gt;&#xA;  another body in the moat.&lt;br&gt;&#xA;  &lt;br&gt;&#xA;  &quot;When all is said and done,&lt;br&gt;&#xA;  no more shall I be, nor any of my fathers nor my sons.&lt;br&gt;&#xA;  I'll be cast into the lot thus far done.&lt;br&gt;&#xA;  I'm another day older than my father, one day younger than my son.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;Hints&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Think $\mathfrak{math}$.&lt;br&gt;There is a lot of fluff in the riddle. Only a few lines may be generally helpful towards you solving the riddle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Best of luck, puzzlers-in-arms!&lt;/p&gt;&#xA;" OwnerUserId="7307" LastActivityDate="2015-03-04T20:57:40.727" Title="Rebellious little--" Tags="|riddle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9816" PostTypeId="1" CreationDate="2015-03-04T01:30:43.053" Score="5" ViewCount="1072" Body="&lt;p&gt;Got this as a question in an IQ test and cannot figure it out. Does anyone know?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;12, 26, 31, 76, 77, 94, 101.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9926" LastEditorUserId="8874" LastEditDate="2015-10-08T09:14:44.517" LastActivityDate="2015-10-08T09:14:44.517" Title="Is there a logic to this sequence?" Tags="|pattern|number-sequence|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="9817" PostTypeId="1" AcceptedAnswerId="9820" CreationDate="2015-03-04T03:59:57.150" Score="4" ViewCount="217" Body="&lt;p&gt;This puzzle attempts to portray a specific scene in a well known movie from the 1970's.  The meaning of the symbols is important although the specific symbol may or may not be part of the movie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The objective is to name the scene and the movie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OmIGK.png&quot; alt=&quot;ArmsOfTheNight&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5188" LastActivityDate="2015-03-04T05:15:11.407" Title="Arms of the night" Tags="|rebus|visual|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9819" PostTypeId="2" ParentId="9812" CreationDate="2015-03-04T04:18:45.943" Score="8" Body="&lt;p&gt;From your hint &quot;get shifty&quot;, we can assume we need to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; use a Caesar cipher.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At first, I just found the correct value to shift by trial and error; however, by revisiting this question later I found that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the amount of words in the first sentence is 22, therefore we need to use Rot-22.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From this, we can get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; eHllg oyuy, suo edichprede ti! rGaej tbo :D &lt;br&gt; You can get this by using Rot-22 on the original string.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then, from the hint &quot;reverse this set of words&quot;, we can tell we need to reverse or swap something. Therefore, we:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Swap every pair of letters, but keep spaces in their original position. If a letter has no letter to pair with (ignoring spaces), do not swap it. Do not swap characters. &quot;eH&quot;-&gt;&quot;He&quot;, &quot;ll&quot;-&gt;&quot;ll&quot;, &quot;g o&quot;-&gt;&quot;o g&quot;, &quot;yu&quot;-&gt;&quot;uy&quot;, &quot;y, s&quot;-&gt;&quot;s, y&quot;, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This results in:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hello guys, you deciphered it! Great job :D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9925" LastEditorUserId="9925" LastEditDate="2015-06-23T00:26:00.973" LastActivityDate="2015-06-23T00:26:00.973" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9820" PostTypeId="2" ParentId="9817" CreationDate="2015-03-04T05:15:11.407" Score="5" Body="&lt;p&gt;Potential answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Black Knight scene from Monty Python and the Holy Grail&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Symbolism for sides of picture:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The scene takes place at a bridge over a small stream. King Arthur comes out of the forest on the left, the black knight is guarding the bridge on the right&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Central part of image (with the cross):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; King Arthur is a knight on a mission given to him by god - seeking the holy grail&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Underneath left (3 crowns):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; On the left we have King Arthur, who has arms and has 3 crowns - for... something. I'm not sure about this part. My best guess is that he has three crowns - the one on his head and the two coconut halves that his squire is banging together to simulate a horse.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Underneath right (w/o arms):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;On the right we have the black knight (&quot;night&quot;) whose arms have been chopped off by King Arthur.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Source (if you haven't seen the movie, go see it right now):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://www.youtube.com/watch?v=2eMkth8FWno&quot;&gt;https://www.youtube.com/watch?v=2eMkth8FWno&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2015-03-04T05:15:11.407" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9821" PostTypeId="2" ParentId="9815" CreationDate="2015-03-04T09:23:37.797" Score="7" Body="&lt;p&gt;At a guess &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1,000,000th triangle number. The sum of a million lives, ie 1+2+...+1,000,000. I'm another day older than my father, one day younger than my son is obvious if in reference to a number if we look at the set of reals. As a number it is larger than any which come before it, Towering over them, but to any larger number it is just one in the throng which come before. The set of reals is relentless and infinite. And when all is said and done, ie infinity, then all numbers which exist in R are in essence irrelevant, even a number as unimaginably large as that. The same as any, Its just a number, the story would be true for literally any number, with the only change perhaps being the sum of line. Meaning this is the only line which really hints at the answer, as all the others are true for numbers general.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9918" LastActivityDate="2015-03-04T09:23:37.797" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9822" PostTypeId="5" CreationDate="2015-03-04T09:25:09.467" Score="0" Body="&lt;p&gt;This tag is for puzzles that reference chemistry and/or require some knowledge of chemistry to solve. If your question is simply about chemistry and is not a puzzle, then it is considered off-topic for this site. If you do need chemistry help, our sister site &lt;a href=&quot;https://chemistry.stackexchange.com&quot;&gt;Chemistry Stack Exchange&lt;/a&gt; may be better able to assist you.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="60644" LastEditDate="2020-11-24T23:14:40.020" LastActivityDate="2020-11-24T23:14:40.020" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="9823" PostTypeId="4" CreationDate="2015-03-04T09:25:09.467" Score="0" Body="A puzzle that makes use of chemical knowledge, such as the names and atomic numbers of chemical elements." OwnerUserId="8874" LastEditorUserId="69582" LastEditDate="2021-02-04T06:12:43.013" LastActivityDate="2021-02-04T06:12:43.013" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="9824" PostTypeId="1" AcceptedAnswerId="9825" CreationDate="2015-03-04T10:33:47.670" Score="8" ViewCount="312" Body="&lt;p&gt;Ethan Hunt is tied to a chair and can't get up.&lt;br&gt;&#xA;The super-villain Matt Fillimore stands in front of him.&lt;br&gt;&#xA;There's a gun.&lt;br&gt;&#xA;There's the barrel of the gun, six chambers, all empty.&lt;br&gt;&#xA;Fillimore is putting two bullets in the gun, in two adjacent chambers.&lt;br&gt;&#xA;He closes the barrel and spins it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He puts the gun to Ethan's head and pulls the trigger.&lt;br&gt;&#xA;Click.&lt;br&gt;&#xA;Ethan Hunt is still alive.&lt;br&gt;&#xA;Lucky him!&lt;br&gt;&#xA;Now Fillimore is going to pull the trigger one more time.&lt;br&gt;&#xA;He lets Ethan choose: whether he should spin the barrel first or whether he should just pull the trigger.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What should Ethan choose?&lt;/p&gt;&#xA;" OwnerUserId="9125" LastActivityDate="2015-03-05T13:00:19.443" Title="Mission: Impossible 2" Tags="|mathematics|probability|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9825" PostTypeId="2" ParentId="9824" CreationDate="2015-03-04T10:37:21.127" Score="11" Body="&lt;p&gt;He should just pull again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since all empty spots are adjacent, there's only a 1/4 chance that the spot that last 'clicked' was just before a bullet, so a 3/4 chance of survival. When spinning the barrel, assuming it happens randomly, there's a 2/3 chance of survival, which is less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the bullets were not adjacent, he should spin again - 1/2 chance vs 2/3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the bullets were placed randomly (adjacent or not), spin again. The next spot will be a bullet two times out of five, so survival is 3/5, while if we spin, we might hit the same empty chamber again, for (again) 2/3 chance of survival.&lt;/p&gt;&#xA;" OwnerUserId="439" LastActivityDate="2015-03-04T10:37:21.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9826" PostTypeId="2" ParentId="9763" CreationDate="2015-03-04T10:41:11.560" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A soccer field&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;has a rectangular border, and is divided into two halves.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are tunnels for the players entering the field.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The west has invested more money into soccer than the east.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9125" LastActivityDate="2015-03-04T10:41:11.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9827" PostTypeId="1" AcceptedAnswerId="9828" CreationDate="2015-03-04T11:16:38.440" Score="5" ViewCount="2409" Body="&lt;p&gt;Alex, Boris, and Jasper are at a bar celebrating Cinco De Mayo. Alex announces he can read Jasper’s mind and goes out for some fresh air. Boris asks Jasper to select something in the room. Jasper points at a lime. When Alex returns, Boris asks, “Is it that napkin?” Alex says, “No.” Boris asks, “Is it the night sky?” Alex says, “No.” Boris asks, “Is it that lime?” Alex says, “Yes.” &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alex leaves the room again, and Jasper points at an ashtray. When Alex returns, Boris asks, “Is it that spoon?” Alex says, “No.” Boris asks, “Is it that cigarette?” Alex says, “No.” Boris asks, “Is it your shadow?” Alex says,&#xA;“No.” Boris asks, “Is it that ashtray?” Alex says “Yes.” &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is repeated again and Jasper points at a straw. When Alex returns, Boris asks, “Is it that cigar?” Alex says, “No.” Boris asks, “Is it that fork?” Alex replies, “No.” Boris asks, “Is it that rose?” Alex says, “No.” Boris asks, “Is it that crow?” Alex says, “No.” Boris asks, “Is it that straw?” Alex says, “Yes.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jasper is very much amazed that Alex was able to guess it right thrice in a row. How do you think he did it?&lt;/p&gt;&#xA;" OwnerUserId="9865" LastEditorUserId="5373" LastEditDate="2017-01-14T15:22:43.570" LastActivityDate="2017-01-14T15:22:43.570" Title="Reading minds can be easy" Tags="|logical-deduction|situation|" AnswerCount="5" CommentCount="4" ClosedDate="2015-03-05T15:35:56.343" ContentLicense="CC BY-SA 3.0" />
  <row Id="9828" PostTypeId="2" ParentId="9827" CreationDate="2015-03-04T11:22:57.577" Score="12" Body="&lt;p&gt;The trick is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Boris always asks &quot;Is it [something not in the bar]?&quot; just before coming to the thing Jasper actually pointed to.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So Alex knows&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; to say &quot;no&quot; up until the first object mentioned after one that's not in the bar.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another example that works (supplied by the OP):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Alex leaves the room again, and Jasper points at a wineglass. When Alex returns, Boris asks, “Is it that knife?” Alex says, “No.” Boris asks, “Is it that bottle?” Alex says, “No.” Boris asks, &quot;Is it that chair?&quot; Alex says, &quot;No.&quot; Boris asks, “Is it the bed?” Alex says, “No.” Boris asks, “Is it that wineglass?” Alex says “Yes.”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another one (assuming I'm right):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Alex leaves the bar again, and Jasper points at his mobile phone. When Alex returns, Boris asks, “Is it that beer glass?” Alex says, “No.” Boris asks, “Is it that table?” Alex says, “No.” Boris asks, “Is it my mobile phone?” (having left it at home). Alex says, “No.” Boris asks, “Is it Jasper's mobile phone?” Alex says “Yes.”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-04T22:01:50.173" LastActivityDate="2015-03-04T22:01:50.173" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="9829" PostTypeId="1" AcceptedAnswerId="9876" CreationDate="2015-03-04T12:05:56.127" Score="4" ViewCount="315" Body="&lt;p&gt;Here's an easy one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are so accustomed to my presence&lt;br&gt;&#xA;that you often won't become aware of my absence.&lt;br&gt;&#xA;Behold my clan, some proud, some meek,&lt;br&gt;&#xA;all together one next to the other&lt;br&gt;&#xA;and yet no trace of me.&lt;br&gt;&#xA;Our encounters are usually just a squeeze,&lt;br&gt;&#xA;got lucky though and dodged every attempt to capture me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have you found me yet?&lt;/p&gt;&#xA;" OwnerUserId="9910" LastEditorUserId="5373" LastEditDate="2019-02-16T19:56:15.253" LastActivityDate="2019-02-16T19:56:15.253" Title="No trace of me!" Tags="|riddle|word|" AnswerCount="3" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9830" PostTypeId="2" ParentId="9829" CreationDate="2015-03-04T12:18:33.907" Score="0" Body="&lt;p&gt;You are so accustomed to my presence&#xA;that you often won't become aware of my absence.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;We are used to the presence of fly because of their constant annoyance and sound that they make, but when they are absent we are busy with our activities and do not notice their absence&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Behold my clan, some proud, some meek,&#xA;all together one next to the other&#xA;and yet no trace of me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some fly are big, some are small and we cannot differentiate one fly from another&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Our encounters are usually just a squeeze,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;When a fly comes near us we try to catch it with our palm as if we are squeezing a fruit or a stress ball&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;got lucky though and dodged every attempt to capture me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It is very difficult to capture a fly&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Have you found me yet?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Yes, you are a Fly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9174" LastActivityDate="2015-03-04T12:18:33.907" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9831" PostTypeId="2" ParentId="9669" CreationDate="2015-03-04T13:07:38.893" Score="-1" Body="&lt;p&gt;&lt;strong&gt;Solution:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magic numbers are :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4,23 = &lt;strong&gt;92&lt;/strong&gt; and &lt;strong&gt;4&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;first off lets note that such a coupe i,j is a representation of two numbers ij and i&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now , the first sentence in the intersection of two lists , the list of addition and division guy conditions , it is the following&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;5,12                11,6 &#xA;&#xA;3,24                23,4        &#xA;&#xA;2,47                4,24               &#xA;&#xA;2,45                4,23               8,12               11,9               &#xA;&#xA;2,33                4,17               8,9               16,5               32,3               &#xA;&#xA;2,29                4,15               7,9               8,8               14,5               28,3               &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;you can verify the evidence of this list by this little xls tool&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.4shared.com/file/c1INXZsRba/addsub.html&quot; rel=&quot;nofollow&quot;&gt;Xls file&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this scripted excel sheet extracts the list of numbers which fits that the diference guy knows that summer knows that he doesnt know for sure , the second sheet shows how subtractor is sure about divisor knows that subtractor dosent know with a perfect explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the final list is an intersection of those two lists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now divisor knows the numbers which means j&gt;33.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now both guys knows also which means the list will be truncated to :&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2,47                4,24               &#xA;&#xA;2,45                4,23               8,12               11,9               &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;the list concerning summer and intersecting the upper lists is &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2,47        3,31        4,23        6,15         8,11        12,7       16,5         24,3         32,2  &#xA;&#xA;2,45        4,22        23,3  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;know divisor made a mistake which means such a couple 2,x is excluded , the only couple which appears twice in last two lists apart 2,x is 4,23&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to be honest i cant go with explanation further than this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks for your downvotes.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;i&gt;second Question:&lt;/i&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the number is an intersection between this list&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;5,12                11,6 &#xA;&#xA;3,24                23,4        &#xA;&#xA;2,33                4,17               8,9               16,5               32,3               &#xA;&#xA;2,29                4,15               7,9               8,8               14,5               28,3  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and this one&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2,47        3,31        4,23        6,15         8,11        12,7       16,5         24,3         32,2  &#xA;&#xA;2,45        4,22        23,3  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;the only couple appeaering twice is 16,5 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which means numers in this case are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;16&lt;/strong&gt; and &lt;strong&gt;80&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9517" LastEditorUserId="9517" LastEditDate="2015-03-07T15:41:48.923" LastActivityDate="2015-03-07T15:41:48.923" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="9832" PostTypeId="2" ParentId="9827" CreationDate="2015-03-04T14:14:32.380" Score="14" Body="&lt;p&gt;Another potential possible answer (though likely a coincidence) is that there is a preselected response pattern for Alex.  The question as is follows this pattern:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Game 1: No, No, Yes&lt;/li&gt;&#xA;&lt;li&gt;Game 2:  No, No, No, Yes&lt;/li&gt;&#xA;&lt;li&gt;Game 3:  No, No, No, No, Yes&lt;/li&gt;&#xA;&lt;li&gt;Potential Game 4: No, No, No, No, No, Yes&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As is, Boris is in complete control of the situation and there are several possible valid strategies.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Addition&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;As this apparently is more popular than I thought it would be, I thought I would summarize the solutions I know.  There are 5 classes of solutions I am aware of.  All of them are predetermined strategies between Alex and Boris.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Alex's answers are predetermined.  As Boris is in complete control of what questions are asked when, he can simply ask the question which warrents a positive response when he already knows Alex is about to give a positive response.  This is the answer described above.  My current favorite version which fits with the example situation is using every other number in $\pi$ to indicate the question Alex should answer yes to.  This would be &quot;3,4,5,2,5,5,etc&quot;.  This would be very difficult to detect compared to other strategies.  If they run out of digits they have memorized, use the next strategy to indicate that they need to restart the pattern.&lt;/li&gt;&#xA;&lt;li&gt;Boris signals to Alex which answer is correct with non-verbal communication.  Nothing in the question indicates this is the case but it is so easy to do and so flexible that Jasper could likely not tell.  If, for instance, Boris puts his thumb inside his hand when pointing to the correct object and outside of it when pointing to the wrong one.  If this is what Jasper is looking for, it is the easiest to detect.  If, however, he is watching Alex and Alex follows the hand to the object with his eyes, Jasper probably won't ever figure it out.&lt;/li&gt;&#xA;&lt;li&gt;Boris communicates to Alex by verbal means.  This is probably the expected answer.  By cleverly choosing the wording of the correct question or a previous one, one could feasibly communicate which one is correct without relying on a memorized pattern.  This is in other answers already.  It is worth noting that if you use a previous question to communicate the correct information, there will never be a positive response (Yes) on the first question which I would see as suspicious.&lt;/li&gt;&#xA;&lt;li&gt;Truely cheat.  If all else fails, setup a bug so Alex can hear Jasper's request.&lt;/li&gt;&#xA;&lt;li&gt;Combination platter.  While any and all of these strategies could explain the example situation, if one wants to reliably and repeatedly do this in front of several people the correct option is to use a combination of these three strategies.  This is throw anyone off if they start to figure it out.  Actually figuring it out would be impossibly difficult.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For a type 5 example, Boris could use the word &quot;that&quot; in every example except the one before the correct one but if Boris refers to something not present like a crow, the next prime number question is correct.  These are both overridden by Boris breathing through his mouth when Alex comes back in.  If he does that, the audience chose all the questions so Alex should say &quot;Yes&quot; on the question during which Boris blinks.  If not communication is attempted at all, he should answer yes on question $(n^2 \mod 5)+10$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intentionally, this gets ridiculously complicated but that is how many kinds of magic involving audience participation can repeatedly work and amaze. You must repeatedly remain so flexible that the audience's demands can't keep up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point of this breakdown is that there are several ways this can be done and the best way is to remain flexible.  Also, cheating makes magic easier.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditDate="2015-03-04T22:52:35.427" LastActivityDate="2015-03-04T22:52:35.427" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="9834" PostTypeId="2" ParentId="9827" CreationDate="2015-03-04T15:10:05.117" Score="1" Body="&lt;p&gt;My guess is that the questions start with &quot;Is it that ...&quot; and once the question has the third word different (&quot;Is it the ...,&quot; &quot;Is it your ...&quot;), then the next item is the one pointed to by Jasper.  The only problem with my guess is the question &quot;Is it that crow?&quot;  Are you sure Boris didn't ask, say, &quot;Is it the crow?&quot;&lt;/p&gt;&#xA;" OwnerUserId="5095" LastActivityDate="2015-03-04T15:10:05.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9835" PostTypeId="1" AcceptedAnswerId="9855" CreationDate="2015-03-04T15:12:34.763" Score="48" ViewCount="6766" Body="&lt;p&gt;This puzzle belongs to the puzzle series: &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=hyper-modern%20is%3Aquestion&quot;&gt;hyper-modern art&lt;/a&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Two visitors are standing in front of a painting in the hyper-modern art gallery, discussing it.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&amp;quot;Well, I have to admit that I really don't understand hyper-modern art. It seems artists nowadays just plot anything without thinking and call it art. What do you make of this?&amp;quot;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;My friend, this clearly shows that one needs a certain eye to see the art. I think, actually, the artist did a very interesting job here...&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&amp;quot;Really? I know art isn't something to look 'nice on the wall', but I simply can't stand work like this. There is no thought going into it... Just some paint... and then pick a random name for your work and let the art critics come up with some&lt;/em&gt; meaning &lt;em&gt;of it all, and voila, that's what it has been about all along...&amp;quot;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;Oh, quite the contrary my friend. Looking at the work, the name of the painting couldn't have been chosen more appropriately. In fact, one could discover the name in the painting, if it hadn't have been labelled. I think it was quite cleverly done...&amp;quot;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;Can you determine the name of the painting below? &lt;br&gt;It consists of 4 characters.&lt;/h2&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RexhG.gif&quot; alt=&quot;CryptoPainting&quot; /&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The puzzle is solely contained in the digital image above. You may want to download it for full resolution and processing. In your answer, you have to state a clear &lt;em&gt;and detailed&lt;/em&gt; explanation why your 'name' is the correct one. The name alone and some loose connections are &lt;em&gt;not&lt;/em&gt; enough.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-18T11:16:09.050" Title="What does this modern art picture represent?" Tags="|cipher|visual|" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9836" PostTypeId="2" ParentId="9829" CreationDate="2015-03-04T15:14:03.880" Score="1" Body="&lt;p&gt;I don't believe this is the answer you're looking for but i think it fits&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a skin cell&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are so accustomed to my presence&#xA;that you often won't become aware of my absence.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; we shed skin all the the time without noticing the majority of the time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Behold my clan, some proud, some meek,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; some skin is healthy, some is damaged by sun or tanning beds&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;all together one next to the other&#xA;and yet no trace of me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; your skin cells are next to each other but you can't see any individual one&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Our encounters are usually just a squeeze,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if you're touching any other part of yourself, your skin is making contact with another piece of skin, squeezing it together&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;got lucky though and dodged every attempt to capture me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ALL dead skin doesn't always come off when showering or for women exfoliating &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9778" LastActivityDate="2015-03-04T15:14:03.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9837" PostTypeId="2" ParentId="9780" CreationDate="2015-03-04T16:28:19.167" Score="3" Body="&lt;p&gt;I'm known as single or as double&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; can be single or double precision in binary for computers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For some I cause enough of trouble&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; not everyone understands floating point notation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That they will force me to comply.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; can always convert to another type of number representation &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But even if I'm weak or near&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; given the the usually representation of the word &quot;or&quot; to mean contrasting: either &quot;weak&quot; has to mean &quot;far&quot; or &quot;near&quot; has to mean &quot;strong&quot;. (or both). I'll assume the former and say that it refers to how close or far the notation is from the significand number depending on how large the number is when you normalize it. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Up to an imperfect extent&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; extent meaning end, the significand has no exact ending until it is determined by the exponent of the base&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once i show up i need not fear&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; as long as the notation for the format is there the number is precise&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You expect to find me at the end. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; notation is found at the end of the number&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now I've related you my tail&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; tail meaning end not story, so same logic as for above&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the name by which i hail?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; floating point notation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9778" LastActivityDate="2015-03-04T16:28:19.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9838" PostTypeId="2" ParentId="9763" CreationDate="2015-03-04T16:39:33.447" Score="1" Body="&lt;p&gt;I know this isn't a rectangular item, but I think it can still answer the riddle.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Yin/Yang symbol&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Divided by a wall, N and S.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The line between the two symbols runs (mostly) N/S, though it is curved&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tunnels in between, but no one pass.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each side has a &quot;tunnel&quot; (dot) showing the other color. The tunnel must lead to the other color, but the colors don't blend, so they don't pass through.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The West has more, and the East has less.&#xA;Less is more, but More is less.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The yin/yang symbol can either be black on the left and white on the right or vice versa, but either way it works. White is &quot;more&quot; and black is &quot;less&quot; because people usually associate white with good, positive things and black with negative things. BUT white is &quot;less&quot; and black is &quot;more&quot; because white stuff technically doesn't absorb any colors and black absorbs all of them colors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again, not rectangular, but I think it fits the clues.&lt;/p&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2015-03-04T16:39:33.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9839" PostTypeId="2" ParentId="9647" CreationDate="2015-03-04T17:02:54.680" Score="3" Body="&lt;pre&gt;&lt;code&gt;here&#xA;__&#xA;bored&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Above-board&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;bets&#xA;_&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;All bets are off&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;all4114&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;All for one and one for all&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;knightleg&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;Nightly&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A&#xA;\&#xA;C&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;To be or not to be&lt;/strong&gt; (between A and C there is B -&gt; Be but you replace it by )&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;lt;3 &amp;lt;-&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;After my own heart&lt;/strong&gt; (&amp;lt;3 = heart and &amp;lt;- point after it)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9888" LastEditorUserId="9888" LastEditDate="2015-03-05T10:22:07.100" LastActivityDate="2015-03-05T10:22:07.100" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9840" PostTypeId="2" ParentId="9835" CreationDate="2015-03-04T17:21:44.180" Score="2" Body="&lt;p&gt;I think it is named&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Flip&#xA; The bottom is the same as the top (with some allowance for distortion of colors) rotated 180 degrees.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9828" LastActivityDate="2015-03-04T17:21:44.180" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9841" PostTypeId="2" ParentId="9835" CreationDate="2015-03-04T17:29:33.400" Score="1" Body="&lt;p&gt;Perhaps&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Spin. If you rotate the image 180 degrees it remains the same. Flipping it vertically results in a horizontal flip, and does not remain the same. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9918" LastActivityDate="2015-03-04T17:29:33.400" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9842" PostTypeId="2" ParentId="9835" CreationDate="2015-03-04T17:36:10.777" Score="16" Body="&lt;p&gt;Both answers make sense, but I believe it is called&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Life. The pixels seem to represent Conway's Game of Life, and the (50) shades of gray represent how long ago/how many times the cells were living.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If not, then I believe it comes from that idea, but maybe&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it has to be reversed to find a starting position in the shape of a word. I don't think this is the case because GoL is not reversible, but it is possible that this configuration is.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1664" LastActivityDate="2015-03-04T17:36:10.777" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9843" PostTypeId="2" ParentId="9835" CreationDate="2015-03-04T17:49:54.017" Score="4" Body="&lt;p&gt;I'm calling it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Life after death.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you rotate the bottom half 180 degrees you'll notice that it's almost identical to the top half. However, the second &quot;blob&quot; (for lack of a better word) in the bottom half is now clearly an upright person. In the top half, the corresponding blob is a human curled up on the ground, dead. The blob just to the right of the person is the car that hit him. The rectangular blobs to the left and right are the crosswalk signs, and the blobs to the far left are the bystanders. To the far right is clearly a skunk that has been frightened by all the noise and is in the act of spraying.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2015-03-04T17:49:54.017" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9844" PostTypeId="2" ParentId="9827" CreationDate="2015-03-04T18:18:12.723" Score="8" Body="&lt;p&gt;I belive it is based off a &quot;magic trick&quot; called Black Magic. To proove that they are communicating telepathicly, they follow a rule:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The item listed after the black item is the item listed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Hence the name, Black Magic).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each item before the last one fits the criteria for Black Magic to work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first one: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Jasper points at a lime. When Alex returns, Boris asks, “Is it that napkin?” Alex says, “No.” Boris asks, “Is it the night sky?” Alex says, “No.” Boris asks, “Is it that lime?” Alex says, “Yes.”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The night sky is usually a black object (without stars and stuff), so it fits the criteria for Black Magic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second one: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Alex leaves the room again, and Jasper points at an ashtray. When Alex returns, Boris asks, “Is it that spoon?” Alex says, “No.” Boris asks, “Is it that cigarette?” Alex says, “No.” Boris asks, “Is it your shadow?” Alex says, “No.” Boris asks, “Is it that ashtray?” Alex says “Yes.”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A persons shadow is colored black, so it also fits the criteria. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the third one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This is repeated again and Jasper points at a straw. When Alex returns, Boris asks, “Is it that cigar?” Alex says, “No.” Boris asks, “Is it that fork?” Alex replies, “No.” Boris asks, “Is it that rose?” Alex says, “No.” Boris asks, “Is it that crow?” Alex says, “No.” Boris asks, “Is it that straw?” Alex says, “Yes.”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Crows are known for being black birds, so it too falls under the same criteria. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alex just has to make sure he notices the item that is black that Boris states, and he will be able to &quot;telepathicly&quot; figure out which item is the correct item. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other ways to do it, like the nth &quot;is it __ ,&quot; but the Black Magic way allows it to be much more flexible, and could either be fast or short. Just not on the first one, while the nth &quot;is it __ &quot; way could apply to the first one (if said so before hand), but that could cause it to be dull. &lt;/p&gt;&#xA;" OwnerUserId="9574" LastEditorUserId="9574" LastEditDate="2015-03-04T18:23:58.280" LastActivityDate="2015-03-04T18:23:58.280" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9846" PostTypeId="2" ParentId="9827" CreationDate="2015-03-04T18:41:55.027" Score="2" Body="&lt;p&gt;Basically, there has to be some (hidden) way for Boris to communicate what the secret object is to Alex, probably by either indicating when the question to which he should answer &quot;yes&quot; is coming up, or by signalling how many times to say &quot;no&quot; before answering &quot;yes&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that Alex knows to answer &quot;yes&quot; to the next question when&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Boris asks about something that shares the same first letter as an earlier guess.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The guesses that precede the correct answer in the examples are &quot;night sky,&quot; &quot;shadow,&quot; and &quot;crow&quot;. In each example,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;These guesses are preceded by &quot;napkin,&quot; &quot;spoon,&quot; and &quot;cigar,&quot; respectively.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This also fits with the OP's example in a comment, as&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The second-to-last guess, &quot;bed,&quot; is preceded by a guess of &quot;bottle&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8776" LastEditorUserId="8776" LastEditDate="2015-03-05T14:45:41.473" LastActivityDate="2015-03-05T14:45:41.473" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9847" PostTypeId="2" ParentId="9802" CreationDate="2015-03-04T18:56:07.337" Score="0" Body="&lt;p&gt;Answer's already been given, but here's an alternate:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Illusion&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's clearly magic and...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Time is an illusion&quot; - a quote often attributed to Albert Einstein&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2015-03-04T18:56:07.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9848" PostTypeId="1" AcceptedAnswerId="9860" CreationDate="2015-03-04T20:51:51.443" Score="5" ViewCount="268" Body="&lt;p&gt;We are all cut by it&lt;br&gt;&#xA;It protected but also hurt&lt;br&gt;&#xA;It crumbles now in its old age&lt;br&gt;&#xA;And its stones are of some homes&lt;br&gt;&#xA;Eight of us form the group. &lt;br&gt;&#xA;It excluded of course. &lt;br&gt;&#xA;Igianugi is our name when our ends meet. &lt;br&gt;&lt;br&gt;&#xA;Who is it and who are we?&lt;/p&gt;&#xA;" OwnerUserId="9618" LastActivityDate="2015-03-05T11:35:49.427" Title="And a wall divides them all" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9849" PostTypeId="2" ParentId="9815" CreationDate="2015-03-04T20:57:40.727" Score="3" Body="&lt;p&gt;My guess.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A number in a summation? Each earlier number adds to it, and it will add to the next number. Each one is basically the same, and when they've all been summed none will be left. Day older and younger means number before and after.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2813" LastActivityDate="2015-03-04T20:57:40.727" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9850" PostTypeId="1" AcceptedAnswerId="10796" CreationDate="2015-03-04T21:11:44.150" Score="17" ViewCount="1383" Body="&lt;p&gt;Berlin, 1973&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We knew that they were on to several of our people, but unfortunately we weren't sure which ones. Pulling them out before we were certain would needlessly disrupt important sources of information. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We had a pretty good guess who would be given the hit jobs; a nasty brute we'd codenamed TUNER (piano wire was known to be his &quot;instrument&quot; of choice).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Luckily we managed to bug his phone, and intercepted what was obviously a coded message, a monotonous synthesized voice reading off a list of paired words. Many of the words were our own code names for our people!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MUTED AARDVARK; LONG ALARM; MAGICIANS UNCLE; TRIM EEL; ABLE DINOSAUR; DESPERATE RABBIT; ETHICAL SALMON; SLIM ICE; NATURAL DRAGON; ROUND ORIOLE; PROUD BOBCAT; OPEN XIPHOID; THOROUGH HYENA; RESPONSIBLE EGRET; ELEVATED STOP&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What, did they think we were idiots? Our informant MALAMUTE was targeted. We rushed him out of the city to safety.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Late that same evening we got word that a double agent, code name ORIOLE, had fallen to his death out of his fourth-floor apartment on Bunderstraße. Had we misinterpreted the message?! Or simply missed another?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next day, a similar call came through for TUNER. Same featureless machine voice, and strangely, the exact same set of word pairs, but in a different order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NATURAL DRAGON; TRIM EEL; OPEN XIPHOID; MUTED AARDVARK; SLIM ICE; RESPONSIBLE EGRET; THOROUGH HYENA; LONG ALARM; ABLE DINOSAUR; MAGICIANS UNCLE; DESPERATE RABBIT; ROUND ORIOLE; PROUD BOBCAT; ETHICAL SALMON; ELEVATED STOP&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With so little time to work before we expected TUNER to strike, our analysts were stumped. All we could think to do was put guards on the highest-value targets that had been named.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the morning, we learned that HYENA had been attacked by a vicious dog outside her house. She never made it to the hospital. She hadn't been on our watch list. We were losing this skirmish, badly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The afternoon brought another phone message like the first two. We dug in, but without much hope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SLIM ICE; MAGICIANS UNCLE; THOROUGH HYENA; TRIM EEL; OPEN XIPHOID; LONG ALARM; ELEVATED STOP; ETHICAL SALMON; MUTED AARDVARK; PROUD BOBCAT; DESPERATE RABBIT; ROUND ORIOLE; ABLE DINOSAUR; NATURAL DRAGON; RESPONSIBLE EGRET&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the analysts -- just a kid really, named Brad --&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HINT:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;seemed to be reciting the lists to himself and counting on the fingers of one hand. He&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;jumped up suddenly and exclaimed, &quot;They're after EEL tonight! No, wait...&quot; He looked over his notes and added, &quot;The indicator repeats in this one. Two of ours, then: BOBCAT as well. We've got to get them both out of the city!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was the best we had to go on. We rushed men to each of their locations. We were too late for EEL; he had &quot;accidentally&quot; dropped a radio into the bathtub with him. But we interrupted TUNER just as he was about to throw BOBCAT down a flight of stairs; our men gave chase but lost him. BOBCAT was whisked away to be debriefed. Hopefully TUNER assumed that we had just gotten lucky, but we'd have to wait for the next message to find out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What did Brad see that we had missed in the first two phone calls? How did he know who the next targets were?&lt;/p&gt;&#xA;" OwnerUserId="4204" LastEditorUserId="4204" LastEditDate="2015-03-05T19:43:24.783" LastActivityDate="2015-04-08T21:40:37.310" Title="Our spies are listed, but which one is actually next to be hit?" Tags="|word-problem|pattern|english|" AnswerCount="5" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="9851" PostTypeId="2" ParentId="6691" CreationDate="2015-03-04T21:27:53.500" Score="46" Body="&lt;h2&gt;16 Words (Internationalize)&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  I&lt;br&gt;&#xA;  iN&lt;br&gt;&#xA;  Tin&lt;br&gt;&#xA;  tinE&lt;br&gt;&#xA;  ineRt&lt;br&gt;&#xA;  retAin&lt;br&gt;&#xA;  reLiant&lt;br&gt;&#xA;  retainAl&lt;br&gt;&#xA;  alienatOr&lt;br&gt;&#xA;  orientalIa&lt;br&gt;&#xA;  rationaliZe&lt;br&gt;&#xA;  Nationalizer&lt;br&gt;&#xA;  lineArization&lt;br&gt;&#xA;  relatInization&lt;br&gt;&#xA;  interNalization&lt;br&gt;&#xA;  intErnationalize&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sentence&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I, reliant nationalizer, retain internalization; internationalize orientalia relatinization; rationalize linearization retainal; tine inert alienator in tin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I (being a trustworthy nationalizer) hold on to incorporation; internationalize the process of translating into Latin those books characteristic of the Orient; justify the retention of linearization; and enclose (&lt;a href=&quot;http://en.wiktionary.org/wiki/tine#Verb_2&quot; rel=&quot;noreferrer&quot;&gt;tine&lt;/a&gt;) a dormant ostracizer in metal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can assert with reasonable certainty that no longer chain exists (according to the dictionary that I am using to generate this). There are a huge number of variations of the chain starting with &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Internationalize&quot;, &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but this one was possible to squeeze a grammatical sentence out of.&lt;/p&gt;&#xA;" OwnerUserId="7387" LastEditorUserId="46903" LastEditDate="2018-09-12T05:36:23.567" LastActivityDate="2018-09-12T05:36:23.567" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="9852" PostTypeId="2" ParentId="9680" CreationDate="2015-03-04T21:46:32.280" Score="5" Body="&lt;p&gt;Sequence 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;First letters of host cities for the Summer Olympics with the missing letters being A (Athens), P (Paris), &lt;strong&gt;S&lt;/strong&gt; (St. Louis), L (London), S (Stockholm), A (Antwerp), P (Paris), A (Amsterdam), &lt;strong&gt;L&lt;/strong&gt; (Los Angeles), B (Berlin), L (London), H (Helsinki), M (Melbourne), R (Rome), T (Tokyo), &lt;strong&gt;M&lt;/strong&gt; (Mexico City), M (Munich), M (Montreal), M (Moscow), &lt;strong&gt;L&lt;/strong&gt; (Los Angeles), &lt;strong&gt;S&lt;/strong&gt; (Seoul), &lt;strong&gt;B&lt;/strong&gt; (Barcelona), &lt;strong&gt;A&lt;/strong&gt; (Atlanta), &lt;strong&gt;S&lt;/strong&gt; (Sydney), &lt;strong&gt;A&lt;/strong&gt; (Athens), &lt;strong&gt;B&lt;/strong&gt; (Beijing), &lt;strong&gt;L&lt;/strong&gt; (London), &lt;strong&gt;R&lt;/strong&gt; (Rio de Janeiro), &lt;strong&gt;T&lt;/strong&gt; (Tokyo)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6140" LastActivityDate="2015-03-04T21:46:32.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9853" PostTypeId="2" ParentId="9850" CreationDate="2015-03-04T22:09:11.710" Score="0" Body="&lt;p&gt;The message was misinterpreted initially. The first message actually says to kill ORIOLE, and MALAMUTE was never targeted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The key is to look for &quot;SLIM ICE&quot; and &quot;MUTED&quot;. The trick with &quot;SLIM ICE&quot; is that the first valid target AFTER the immediate name after &quot;SLIM ICE&quot; is targeted. eg: &quot;SLIM ICE&quot; ; (SKIP) ; (target)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explaining the second target:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Remember how I said we also have to keep an eye out for &quot;MUTED&quot;? In the first 2 messages, MUTED is placed before a pair of words that are NOT animals. If we are to assume that if a name isn't an animal, then it isn't an agent, it makes sense. &quot;MUTED&quot; is used to target the next name following the pair; eg: MUTED AARDVARK; (target) This makes it so that TUNER can see &quot;Oh, this guy isn't an agent because it's not an animal&quot; but if TUNER sees an animal, he can say &quot;OK, I'm targeting this guy as well.&quot; These 2 patterns fit in with all 3 messages and targets. Essentially, the rule here is: &quot;Check the pair of words after &quot;MUTED AARDVARK&quot;. Is it an animal? If yes, target. If not, ignore.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But, why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because you can &quot;ICE&quot; someone (archaic slang for killing someone). Also, you SILENCE your enemies (hence MUTED).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;BUT THIS DOESN'T WORK FOR THE THIRD ONE YOU LIAR!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Actually, it does because Hyena is no longer a valid target - Hyena is already dead. Thus, skip her and move over by 1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8999" LastEditorUserId="8999" LastEditDate="2015-03-04T22:15:56.737" LastActivityDate="2015-03-04T22:15:56.737" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="9854" PostTypeId="1" AcceptedAnswerId="9857" CreationDate="2015-03-05T00:08:03.440" Score="16" ViewCount="1448" Body="&lt;p&gt;This puzzle is originally created by by Roger Penrose. The goal is to guide the train all the way to the station via the railways, without ever reversing or taking sharp turns. The train can only travel forwards. For instance, at the first crossing, where five tracks join, it can only take one of the three tracks that go north - it can not go back where it came from, or take the other southwards track.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PVfpQ.jpg&quot; alt=&quot;Railroad maze&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Which route does the train have to take?&lt;/p&gt;&#xA;&lt;p&gt;Note: With sharp turns, I mean those turns that a train would not be able to make normally because there is no smooth curve to follow, or in other words: the path the train takes should be differentiable.&lt;/p&gt;&#xA;" OwnerUserId="439" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2021-06-12T20:58:35.557" Title="The Railway Maze" Tags="|mazes|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9855" PostTypeId="2" ParentId="9835" CreationDate="2015-03-05T00:08:15.473" Score="41" Body="&lt;p&gt;Here is an answer that decodes this image. Not being a programmer, I used &lt;a href=&quot;http://imagej.nih.gov/ij/&quot; rel=&quot;noreferrer&quot;&gt;ImageJ&lt;/a&gt; and &lt;a href=&quot;http://www.caesum.com/handbook/stego.htm&quot; rel=&quot;noreferrer&quot;&gt;Stegsolve&lt;/a&gt; to look at this file.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Viewing with &lt;strong&gt;bit0&lt;/strong&gt; then &lt;strong&gt;bit1&lt;/strong&gt; provides these two images:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The left images are from the &lt;strong&gt;modern art picture&lt;/strong&gt; above.  The right images are generated from this &lt;strong&gt;&lt;a href=&quot;http://www.bitstorm.org/gameoflife/standalone/&quot; rel=&quot;noreferrer&quot;&gt;Game of Life&lt;/a&gt;&lt;/strong&gt;&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/R9ScH.png&quot; alt=&quot;Bit0&quot;&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/7M90O.png&quot; alt=&quot;Bit1&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then continuing to toggle through &lt;strong&gt;bit2, bit3, bit4&lt;/strong&gt; provides the following images:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/LAa4E.png&quot; alt=&quot;Bit2&quot;&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/6TfKU.png&quot; alt=&quot;Bit3&quot;&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/cyXCD.png&quot; alt=&quot;Bit4&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And finally, these images with &lt;strong&gt;bit5, bit6, bit7&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/xCBh4.png&quot; alt=&quot;Bit5&quot;&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/75hCY.png&quot; alt=&quot;Bit6&quot;&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/d8zae.png&quot; alt=&quot;Bit7&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the name of the painting is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;LIFE&lt;/strong&gt; because toggling through the bit planes provides images that match the original rules of &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Conway&amp;#39;s_Game_of_Life&quot; rel=&quot;noreferrer&quot;&gt;Conway's Game of Life&lt;/a&gt;&lt;/strong&gt;.&lt;br&gt;&#xA; At each step in time, the following transitions occur:&lt;br&gt;&#xA; - any live cell with fewer than two live neighbours dies, as if caused by under-population&lt;br&gt;&#xA; - any live cell with two or three live neighbours lives on to the next generation&lt;br&gt;&#xA; - any live cell with more than three live neighbours dies, as if by overcrowding&lt;br&gt;&#xA; - any dead cell with exactly three live neighbours becomes a live cell, as if by reproduction  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-03-05T19:58:35.150" LastActivityDate="2015-03-05T19:58:35.150" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="9857" PostTypeId="2" ParentId="9854" CreationDate="2015-03-05T05:08:37.867" Score="10" Body="&lt;p&gt;The solution is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; shown on &lt;a href=&quot;http://community.wolfram.com/groups/-/m/t/223067&quot; rel=&quot;noreferrer&quot;&gt;Wolfram&lt;/a&gt;.  A discussion paper was written by &lt;a href=&quot;http://stanwagon.com/public/MazeArticlePreprint.pdf&quot; rel=&quot;noreferrer&quot;&gt;Slavik and Wagon&lt;/a&gt;.&lt;br&gt;&#xA;&lt;br&gt;&#xA;  &lt;img src=&quot;https://i.stack.imgur.com/uYg3H.jpg&quot; alt=&quot;RailwayMaze&quot;&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastActivityDate="2015-03-05T05:08:37.867" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9858" PostTypeId="1" AcceptedAnswerId="9861" CreationDate="2015-03-05T10:50:31.827" Score="9" ViewCount="911" Body="&lt;p&gt;Professor Erasmus claims that he is able to cut a square into 100 rectangles by making nine horizontal cuts and nine vertical cuts (parallel to the sides of the square), so that&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;exactly 9 of the resulting 100 rectangles are squares, and&lt;/li&gt;&#xA;&lt;li&gt;no two of these 9 resulting squares have the same area.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The professor modestly calls this the &quot;Professor-Erasmus-dissection-of-a-square&quot;.&#xA;Has the professor once again made one of his mathematical blunders, or does such  a dissection indeed exist?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-10-04T16:12:30.607" LastActivityDate="2015-10-04T16:12:30.607" Title="The Erasmus dissection of a square" Tags="|mathematics|geometry|dissection|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9859" PostTypeId="2" ParentId="9858" CreationDate="2015-03-05T11:22:09.343" Score="0" Body="&lt;p&gt;If it needed to be 10 squares it would be quite easy. lets say you cut at distance 1, 3, 6, 10, 15, 21, 28, 36 and 45 of a 55 by 55 square. both in the x and y direction. Then you will have 10 squares along the diagonal with sides of length 1,2,3,4,5,6,7,8,9 and 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now moving any of the cuts would remove 2 squares so I am inclined to say that it is not possible. Not sure if this is proof enough though&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-03-05T11:22:09.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9860" PostTypeId="2" ParentId="9848" CreationDate="2015-03-05T11:35:49.427" Score="4" Body="&lt;p&gt;It is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Great Wall of China&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Beijing, Gansu, Hebei, Liaoning, Ningxia, Shaanxi, Shanxi and Tianjin&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are all cut by it&lt;br&gt;&#xA;It protected but also hurt&lt;br&gt;&#xA;It crumbles now in its old age&lt;br&gt;&#xA;And its stones are of some homes  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Great Wall passes through several provinces, it used to protect them from raids but also hurt trade. It's crumbling now in most places, some of the raw materials have been used to build houses.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Eight of us form the group.&lt;br&gt;&#xA;It excluded of course.&lt;br&gt;&#xA;Igianugi is our name when our ends meet.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The eight provinces the Great Wall passes through. Pick the last letter of each, shuffle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9910" LastActivityDate="2015-03-05T11:35:49.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9861" PostTypeId="2" ParentId="9858" CreationDate="2015-03-05T11:52:29.590" Score="17" Body="&lt;p&gt;It is not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let v be the set of separations between adjacent vertical lines. Let h be the set of separations between adjacent horizontal lines. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Every time an element of v matches an element of h, we have a square. &lt;/li&gt;&#xA;&lt;li&gt;No element of one set may match more than one element of the other set, or we would have two equal sized squares. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thus we must have 9 identical elements in each set, but each set has only 10 elements and the same sum, therefore the 10th element is equal.&lt;/p&gt;&#xA;" OwnerUserId="3987" LastEditorUserId="3987" LastEditDate="2015-03-05T17:27:30.390" LastActivityDate="2015-03-05T17:27:30.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9862" PostTypeId="2" ParentId="9858" CreationDate="2015-03-05T12:01:00.927" Score="3" Body="&lt;p&gt;No it is not possible.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If we take x as the width of the original cut, and increment it by y each time, then there will be 10 squares with sides length (x+ny) and finally (x-36y) which would have different areas to one another. All other rectangles would be of the form (x+ny) x (x+my). In order for this to work for a square with width 10x, y must be &amp;lt; x/36.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In order to get 9 squares&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In order to assume that there are exactly 9 squares we have to move one of the cuts, either horizontal or vertical, from the x+ny pattern. Lets assume vertical, so that we have a square in the penultimate column which is not in the diagonal line of the others. In order to do this, we simply adjust the width of the cut from (x+8y) to (x+my) where m = 0, 1, 2, 3, 4, 5, 6, or 7 creating a final row with height x-36y (with 36y &amp;lt; x) and final column with width x-(36-m)y which obviously doesn't form a square. This produces a square in all but the final column. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In order to produce a square in the penultimate column, it must have for its width the height of one of the rows. This means that the square will HAVE to be the same size as one of the other squares. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Although&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; all of the cuts as described but with the final vertical cut so that no squares are produced in the penultimate or final column would produce 8 internal squares, and the original square can still be formed. This would produce 9 non-identical squares. But there would then be 101 rectangles produced. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9918" LastActivityDate="2015-03-05T12:01:00.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9863" PostTypeId="2" ParentId="9850" CreationDate="2015-03-05T12:31:17.943" Score="1" Body="&lt;p&gt;I think I got it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hint for me means that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;he was counting max to 5 and saying the words aloud, that's why I think he was counting syllables&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It would fit that &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Number of syllables in last word is the number of targets. Stop = 1 (Oriole), Stop = 1 (Hyena), Egret = 2 (Eel, Bobcat)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the victims themselves are determined by:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First ANIMAL combo in row, that has same sum of syllables as the first phrase&lt;br/&gt; - MUTED AARDVARK = 4 syllables =&gt; ROUND ORIOLE = 4 syllables&lt;br/&gt; - NATURAL DRAGON = 5 syllables =&gt; THOROUGH HYENA = 5 syllables&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the last one doesn't fit so nicely&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - SLIM ICE = 2 syllables =&gt; TRIM EEL = 2 syllables but because there are two targets and nothing else has 2 syllables, we move to next animal with 3 syllables that's PROUD BOBCAT (there is LONG ALARM with 3 syllables as well, but it's not an animal)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;EDIT: Thanks for the help with spoiler tags&#xA;For a guy who kills with piano wire, nobody has been killed by piano wire.&lt;/p&gt;&#xA;" OwnerUserId="9907" LastEditorUserId="9907" LastEditDate="2015-03-05T13:31:41.857" LastActivityDate="2015-03-05T13:31:41.857" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9866" PostTypeId="1" AcceptedAnswerId="9867" CreationDate="2015-03-05T15:21:44.787" Score="14" ViewCount="858" Body="&lt;p&gt;Albus Dumbledore keeps 274 dragon lizards in his terrarium in Hogwarts.&#xA;One day he removes two thirds of the male lizards from the terrarium;&#xA;it is easy for him to tell the males apart from the females since each male&#xA;has 12 spikes and each female has 4 spikes.  How many spikes are left in&#xA;the terrarium?&lt;/p&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2015-03-06T12:27:17.067" Title="Dragon lizards in Hogwarts" Tags="|calculation-puzzle|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9867" PostTypeId="2" ParentId="9866" CreationDate="2015-03-05T15:25:31.347" Score="21" Body="&lt;p&gt;There are exactly &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1096 spikes left. It doesn't matter how many males or females there are, because the males have 3 times as many spikes as the females. 1/3 of the males remain, and all of the females remain. This averages to 4 spikes per original dragon lizard, which give us 274*4=1096.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1664" LastEditorUserId="1664" LastEditDate="2015-03-05T15:31:31.823" LastActivityDate="2015-03-05T15:31:31.823" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9868" PostTypeId="1" AcceptedAnswerId="9874" CreationDate="2015-03-05T15:25:55.997" Score="18" ViewCount="1548" Body="&lt;p&gt;As a matter of fact, I never thought I would find the courage and the nerve to post a puzzle like this one.&#xA;If it doesn't get me stoned, it will get me downvoted by myopic people.&#xA;I hope that some of you might find it worth their precious while to throw a bit of thinking at it.&#xA;There is a hidden pattern in this paragraph, and I was quite amused when I first saw it.&#xA;Please don't assume that I am writing horrid nonsense or that my mind has fallen asleep.&#xA;Look meticulously for related words that have been altered in some odd way.&lt;/p&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2015-03-05T22:13:13.067" Title="I never thought I would find the courage" Tags="|enigmatic-puzzle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9870" PostTypeId="5" CreationDate="2015-03-05T15:59:04.180" Score="0" Body="&lt;p&gt;Movie-themed puzzles make use of cinematic knowledge, such as movie titles, actors, directors, and film awards. General movies questions are off-topic but can be asked on our sister site &lt;a href=&quot;https://movies.stackexchange.com/&quot;&gt;Movies &amp;amp; TV Stack Exchange&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;A central resource for cinematic knowledge is the Internet Movie Database (&lt;strong&gt;IMDB&lt;/strong&gt;):&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.imdb.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.imdb.com/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Internet_Movie_Database&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Internet_Movie_Database&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="60644" LastEditDate="2020-11-24T23:11:28.163" LastActivityDate="2020-11-24T23:11:28.163" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="9871" PostTypeId="4" CreationDate="2015-03-05T15:59:04.180" Score="0" Body="A puzzle that makes use of cinematic knowledge, such as movie titles, actors, directors, and film awards. " OwnerUserId="8874" LastEditorUserId="69582" LastEditDate="2021-03-10T07:18:47.543" LastActivityDate="2021-03-10T07:18:47.543" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="9872" PostTypeId="1" AcceptedAnswerId="9873" CreationDate="2015-03-05T16:17:17.753" Score="1" ViewCount="2371" Body="&lt;p&gt;Here's a cute little riddle I heard in the movie Life is Beautiful.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Say my name and I disappear. What am I?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9413" LastActivityDate="2015-03-05T16:19:20.003" Title="Say my name and I disappear" Tags="|riddle|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9873" PostTypeId="2" ParentId="9872" CreationDate="2015-03-05T16:19:20.003" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Silence&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how much explaining really needs to be done here. It's a classic riddle, and you can also hear it in variations like:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is so fragile it breaks when you speak of it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1563" LastActivityDate="2015-03-05T16:19:20.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9874" PostTypeId="2" ParentId="9868" CreationDate="2015-03-05T16:41:17.987" Score="27" Body="&lt;p&gt;Any chance that hidden pattern isn't very pronounced?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each sentence seems to contain an anagram of one of the words showing up in that sentence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More precisely:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; never - nerve&lt;br&gt;&#xA; doesn't - stoned&lt;br&gt;&#xA; worth - throw&lt;br&gt;&#xA; was - saw&lt;br&gt;&#xA; please - asleep&lt;br&gt;&#xA; related - altered  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9910" LastActivityDate="2015-03-05T16:41:17.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9875" PostTypeId="1" CreationDate="2015-03-05T16:57:15.560" Score="17" ViewCount="1879" Body="&lt;p&gt;&lt;strong&gt;This question is still active as the answer from @Sleafar has just arrived. Not yet solved.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Scenario&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heads of 3 top terrorist organisations are planning to play an extremely high-stake poker game that could potentially decide their respective fates. One of them is Arnab, whose life depends on his winning. He also knows that he is not that good a player, and will have to use illegitimate tactics to win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He has paid the casino dealer a vast amount of money, in return for him following the certain rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Every new deck (standard 52 card deck) that is requested will be already in order - 4 aces, 4 kings, 4 queens, ... 4 twos.&lt;/li&gt;&#xA;&lt;li&gt;He will use customized shuffling algorithms that will ensure there is no pair in the deck. No 2 adjacent cards will be of the same number.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Arnab can, therefore bet large amounts on high cards and other relatively bad hands, and his opponents would fold, thinking that pairs, 3 of a kind, etc. are likely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Twist&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, on the match day, after just a few hands, his adversaries are quite sure that the casino has been rigged. It becomes more suspicious when they realise that the dealer uses quite fixed algorithms for shuffling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They have a huge row when they ask the dealer to shuffle randomly, but the dealer refuses. Afraid that they would not be able to find another man who will act as a fair judge/dealer, they agree to a compromise:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A new deck will be requested for every hand.&lt;/li&gt;&#xA;&lt;li&gt;The dealer &lt;em&gt;will&lt;/em&gt; use the same algorithm, every time.&lt;/li&gt;&#xA;&lt;li&gt;The players will ask him to repeat the same shuffling algorithm a random number of times, before dealing.&lt;/li&gt;&#xA;&lt;li&gt;They will also split the deck (at a random card) and deal it, not from the top card.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However, none of them know that the new decks are still ordered - neither the 3 players, nor the dealer. What shuffling algorithm must the dealer use to help Arnab? For how long will it work? Note that Arnab has not communicated with the dealer, and will be unaware of any change in plans.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Formal question&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Devise a static shuffling algorithm, such that there are no 2 adjacent cards in the deck. The condition should remain true even after $n$ iterations of the shuffle (on the same deck). Maximize $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Restrictions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There can be &lt;em&gt;no&lt;/em&gt; random choices.&lt;/li&gt;&#xA;&lt;li&gt;It can not have conditional statements pertaining to the exact cards. So, a statement like &lt;em&gt;swap cards, if there is an ace&lt;/em&gt; is not allowed.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, a possible shuffle for 4 cards could be written as:&#xA;&lt;code&gt;1 2 3 4&lt;/code&gt; $\rightarrow$ &lt;code&gt;4 2 1 3&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be possible to write down your shuffle in such a format.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the maximum value of $n$ for which this works? ($n$ has to be less than 52)&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4269" LastEditDate="2015-09-23T07:32:51.177" LastActivityDate="2015-09-23T07:32:51.177" Title="Rigged casino that prevents pairs" Tags="|mathematics|optimization|cards|algorithm|" AnswerCount="8" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="9876" PostTypeId="2" ParentId="9829" CreationDate="2015-03-05T17:44:48.470" Score="6" Body="&lt;p&gt;Suspect the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter &lt;strong&gt;$\,i$&lt;/strong&gt;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'i' is the only letter not to appear anywhere in the riddle. The &quot;clan&quot; is the letters of the alphabet, all together except, most unusually, for I. And of course &quot;I&quot; would refer to itself as &quot;me&quot; when grammatically appropriate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9523" LastEditorUserId="8874" LastEditDate="2015-03-05T17:53:16.277" LastActivityDate="2015-03-05T17:53:16.277" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9879" PostTypeId="1" AcceptedAnswerId="9881" CreationDate="2015-03-05T18:36:00.800" Score="2" ViewCount="6738" Body="&lt;p&gt;I'll admit&lt;br&gt;&#xA;I'm mostly blue,&lt;br&gt;&#xA;but it's not so great to have&lt;br&gt;&#xA;something constantly piercing through you.&lt;br&gt;&#xA;I might even be in tears,&lt;br&gt;&#xA;but I can definitely be beautiful too.&lt;/p&gt;&#xA;" OwnerUserId="9686" LastActivityDate="2015-03-06T01:37:10.050" Title="A really sad riddle" Tags="|riddle|" AnswerCount="4" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9880" PostTypeId="2" ParentId="9879" CreationDate="2015-03-05T19:00:19.610" Score="6" Body="&lt;p&gt;Mostly blue &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oceans are blue, and in large enough quantities water does take on a blue colour regardless of the reflection.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Something piercing through you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Boats, swimmers, land, straws. Fairly common phrase - a boat piercing the waves.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might even be in tears&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tears are made of water&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I can be beautiful too&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oceans, lakes, rivers, ponds are all gorgeous under the right circumstances.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer therefore could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9918" LastEditorUserId="5373" LastEditDate="2015-03-05T20:43:37.373" LastActivityDate="2015-03-05T20:43:37.373" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9881" PostTypeId="2" ParentId="9879" CreationDate="2015-03-05T19:16:20.350" Score="13" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sky&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll admit&lt;br&gt;&#xA;I'm mostly blue,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mostly blue, but sometimes cloudy or sunset-y red/pink.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but it's not so great to have&lt;br&gt;&#xA;something constantly piercing through you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sunlight is very piercing, and the night sky has been described as &quot;a backlit canopy with holes punched in it&quot;, so it's constantly being pierced, day or night.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might even be in tears,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sky crying is rain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I can definitely be beautiful too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you don't agree the sky can be beautiful, you have no soul :P&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1563" LastActivityDate="2015-03-05T19:16:20.350" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9882" PostTypeId="2" ParentId="9879" CreationDate="2015-03-05T19:21:54.530" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eyes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mostly Blue&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Blue eyes, but white and black as well, but yet we still call them blue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;something constantly piercing through you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Light pierces through eyes so you can see.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might be in tears&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Literally swimming in tears when someone cries.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I can definitely be beautiful too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Blue eyes can be very beautiful.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-03-05T19:21:54.530" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9883" PostTypeId="1" AcceptedAnswerId="9885" CreationDate="2015-03-05T21:00:46.157" Score="5" ViewCount="1042" Body="&lt;p&gt;I currently have a sudoku puzzle that looks like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[7][ ][4] | [6][ ][ ] | [1][9][3]&#xA;[ ][2][ ] | [ ][7][4] | [5][8][6]&#xA;[6][ ][ ] | [ ][ ][9] | [4][7][2]&#xA;--------- | --------- | ---------&#xA;[ ][ ][ ] | [ ][ ][ ] | [6][4][7]&#xA;[2][ ][ ] | [4][5][7] | [9][3][1]&#xA;[ ][4][7] | [ ][6][ ] | [2][5][8]&#xA;--------- | --------- | ---------&#xA;[1][ ][ ] | [7][ ][ ] | [3][2][ ]&#xA;[ ][9][2] | [ ][3][ ] | [7][6][ ]&#xA;[ ][7][ ] | [ ][ ][6] | [8][1][ ]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I've been looking at it for awhile but it seems like my next move is to &lt;em&gt;guess&lt;/em&gt; where the next number goes, as there are many places where it's a 50/50 between 2 numbers. From my understand though is that sudoku is a logic puzzle and you shouldn't &lt;em&gt;need&lt;/em&gt; to resort to guessing to figure out the next move. I was wondering if somebody could find and explain my next logical move ( or 2 ).&lt;/p&gt;&#xA;" OwnerUserId="9814" LastActivityDate="2015-07-12T10:13:22.627" Title="Sudoku Puzzle, What's My Next Move and Why?" Tags="|sudoku|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9885" PostTypeId="2" ParentId="9883" CreationDate="2015-03-05T21:14:24.823" Score="5" Body="&lt;p&gt;Well, you can figure out a 1 (top center group):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 7     4  |  6        |  1  9  3 &#xA;    2     |     7  4  |  5  8  6 &#xA; 6        |        9  |  4  7  2 &#xA;--------- | --------- | ---------&#xA; -  1111  |           |  6  4  7 &#xA; 2  -  -  |  4  5  7  |  9  3  1 &#xA; -  4  7  |     6     |  2  5  8 &#xA;--------- | --------- | ---------&#xA; 1        |  7        |  3  2    &#xA;    9  2  |     3     |  7  6    &#xA;    7     |        6  |  8  1    &#xA;Dashes are blocked off, the consecutive 1's are where a 1 has to be. &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So since all space by 1 are blocked off from having a 1 there (as below)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 7     4  |  6 - - - -|- 1  9  3 &#xA;    2     |     7  4  |  5  8  6 &#xA; 6        |        9  |  4  7  2 &#xA;--------- | --------- | ---------&#xA; -  1111 -|- - - -    |  6  4  7 &#xA; 2  -  -  |  4  5  7  |  9  3  1 &#xA; -  4  7  |     6     |  2  5  8 &#xA;--------- | --------- | ---------&#xA; 1 - - - -|- 7 - -    |  3  2    &#xA;    9  2  |     3     |  7  6    &#xA;    7     |    - - 6 -|- 8  1    &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;you can see that there is only 1 space left in the center column for a 1&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 7     4  |  6        |  1  9  3 &#xA;    2     |     7  4  |  5  8  6 &#xA; 6        |    (1) 9  |  4  7  2 &#xA;--------- | --------- | ---------&#xA;          |           |  6  4  7 &#xA; 2        |  4  5  7  |  9  3  1 &#xA;    4  7  |     6     |  2  5  8 &#xA;--------- | --------- | ---------&#xA; 1        |  7        |  3  2    &#xA;    9  2  |     3     |  7  6    &#xA;    7     |        6  |  8  1    &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;to&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 7     4  |  6        |  1  9  3 &#xA;    2 (1) |     7  4  |  5  8  6 &#xA; 6        |     1  9  |  4  7  2 &#xA;--------- | --------- | ---------&#xA;          |           |  6  4  7 &#xA; 2        |  4  5  7  |  9  3  1 &#xA;    4  7  |     6     |  2  5  8 &#xA;--------- | --------- | ---------&#xA; 1        |  7        |  3  2    &#xA;    9  2  |     3     |  7  6    &#xA;    7     |        6  |  8  1    &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;to&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 7     4  |  6        |  1  9  3 &#xA;    2  1  |     7  4  |  5  8  6 &#xA; 6        |     1  9  |  4  7  2 &#xA;--------- | --------- | ---------&#xA;   (1)    |           |  6  4  7 &#xA; 2        |  4  5  7  |  9  3  1 &#xA;    4  7  |     6     |  2  5  8 &#xA;--------- | --------- | ---------&#xA; 1        |  7        |  3  2    &#xA;    9  2  |     3     |  7  6    &#xA;    7     |        6  |  8  1    &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-03-06T14:05:56.907" LastActivityDate="2015-03-06T14:05:56.907" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9888" PostTypeId="2" ParentId="9883" CreationDate="2015-03-05T21:43:33.400" Score="1" Body="&lt;p&gt;Continuing further you can find the 9 in the top left corner, and then finish off that line. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[7][_][4] | [6][ ][ ] | [1][9][3]&#xA;[9][2][1] | [3][7][4] | [5][8][6]&#xA;[6][_][_] | [ ][1][9] | [4][7][2]&#xA;--------- | --------- | ---------&#xA;[ ][1][ ] | [ ][ ][ ] | [6][4][7]&#xA;[2][ ][ ] | [4][5][7] | [9][3][1]&#xA;[ ][4][7] | [ ][6][ ] | [2][5][8]&#xA;--------- | --------- | ---------&#xA;[1][ ][ ] | [7][ ][ ] | [3][2][ ]&#xA;[ ][9][2] | [ ][3][ ] | [7][6][ ]&#xA;[ ][7][ ] | [ ][ ][6] | [8][1][ ]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;After the above is done you can find a 3 by the process of elimination:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[7][_][4] | [6][ ][ ] | [1][9][3]&#xA;[9][2][1] | [3][7][4] | [5][8][6]&#xA;[6][3][ ] | [ ][1][9] | [4][7][2]&#xA;--------- | --------- | ---------&#xA;[ ][1][ ] | [ ][ ][ ] | [6][4][7]&#xA;[2][_][ ] | [4][5][7] | [9][3][1]&#xA;[ ][4][7] | [ ][6][ ] | [2][5][8]&#xA;--------- | --------- | ---------&#xA;[1][_][ ] | [7][ ][ ] | [3][2][ ]&#xA;[ ][9][2] | [ ][3][ ] | [7][6][ ]&#xA;[ ][7][ ] | [ ][ ][6] | [8][1][ ]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2015-03-05T21:49:23.680" LastActivityDate="2015-03-05T21:49:23.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9890" PostTypeId="1" AcceptedAnswerId="9893" CreationDate="2015-03-05T22:45:16.327" Score="29" ViewCount="15553" Body="&lt;p&gt;What is the largest number of words that can be made by just adding one letter to the original word, where each iteration is a meaningful word? E.g.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;HO&#xA;HON&#xA;HONE&#xA;HONES&#xA;HONEST&#xA;HONESTY&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The original word can have as many letters as you want; you can add letters at the start, at the end or anywhere; and onomatopoeic words are allowed.&lt;/p&gt;&#xA;" OwnerUserId="9956" LastEditorUserId="5373" LastEditDate="2015-03-07T10:27:51.127" LastActivityDate="2020-03-26T10:00:20.000" Title="The most words that can be made by successively adding one letter to the original word?" Tags="|open-ended|letters|" AnswerCount="7" CommentCount="9" ClosedDate="2021-09-09T14:18:37.663" ContentLicense="CC BY-SA 3.0" />
  <row Id="9891" PostTypeId="2" ParentId="9879" CreationDate="2015-03-05T23:30:02.787" Score="-1" Body="&lt;p&gt;I thought of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a tattoo&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is mostly blue&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ink&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and might well be pierced by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - well, by a piercing. Many tattoo places would do that for you too---you might even get a discount.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;I might even be in tears,&quot; made me think of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the way some people get tears tattooed on their cheek, as in the Tom Waits lyric: &lt;em&gt;...the girl behind the counter has a tattooed tear: &quot;One for every year he's away,&quot; she said...&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but [it] can definitely be beautiful too&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; depending on the skill of the tattoo artist.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6141" LastEditorUserId="6141" LastEditDate="2015-03-06T01:37:10.050" LastActivityDate="2015-03-06T01:37:10.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9892" PostTypeId="2" ParentId="9890" CreationDate="2015-03-05T23:37:07.507" Score="7" Body="&lt;p&gt;The longest words in the dictionary that come with my Mac are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;reversist&quot; and &quot;sheathery&quot;, although I can't find a definition for either of them (the earlier stages are all defined in wiktionary, though).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wiktionary.org/wiki/re&quot; rel=&quot;nofollow noreferrer&quot;&gt;re&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;http://en.wiktionary.org/wiki/rev&quot; rel=&quot;nofollow noreferrer&quot;&gt;rev&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;http://en.wiktionary.org/wiki/reve&quot; rel=&quot;nofollow noreferrer&quot;&gt;reve&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;http://en.wiktionary.org/wiki/rever&quot; rel=&quot;nofollow noreferrer&quot;&gt;rever&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;http://en.wiktionary.org/wiki/revers&quot; rel=&quot;nofollow noreferrer&quot;&gt;revers&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;http://en.wiktionary.org/wiki/reversi&quot; rel=&quot;nofollow noreferrer&quot;&gt;reversi&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;http://en.wiktionary.org/wiki/reversis&quot; rel=&quot;nofollow noreferrer&quot;&gt;reversis&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;http://en.wiktionary.org/wiki/reversist&quot; rel=&quot;nofollow noreferrer&quot;&gt;reversist&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wiktionary.org/wiki/sh&quot; rel=&quot;nofollow noreferrer&quot;&gt;sh&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;http://en.wiktionary.org/wiki/she&quot; rel=&quot;nofollow noreferrer&quot;&gt;she&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;http://en.wiktionary.org/wiki/shea&quot; rel=&quot;nofollow noreferrer&quot;&gt;shea&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;http://en.wiktionary.org/wiki/sheat&quot; rel=&quot;nofollow noreferrer&quot;&gt;sheat&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;http://en.wiktionary.org/wiki/sheath&quot; rel=&quot;nofollow noreferrer&quot;&gt;sheath&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;http://en.wiktionary.org/wiki/sheathe&quot; rel=&quot;nofollow noreferrer&quot;&gt;sheathe&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;http://en.wiktionary.org/wiki/sheather&quot; rel=&quot;nofollow noreferrer&quot;&gt;sheather&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;http://en.wiktionary.org/wiki/sheathery&quot; rel=&quot;nofollow noreferrer&quot;&gt;sheathery&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9523" LastEditorUserId="11110" LastEditDate="2018-09-19T11:10:13.070" LastActivityDate="2018-09-19T11:10:13.070" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="9893" PostTypeId="2" ParentId="9890" CreationDate="2015-03-05T23:57:42.960" Score="34" Body="&lt;p&gt;I assume you start at two letters, as in your example. I'm using the circa 2013 &lt;em&gt;Words With Friends&lt;/em&gt; dictionary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can only add to the right end of the word, &lt;strong&gt;8-letter&lt;/strong&gt; words are possible:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ba&#xA;bar&#xA;barb&#xA;barbe&#xA;barbel&#xA;barbell&#xA;barbells&#xA;&#xA;ma&#xA;max&#xA;maxi&#xA;maxim&#xA;maxima&#xA;maximal&#xA;maximals&#xA;&#xA;pa&#xA;pas&#xA;past&#xA;paste&#xA;paster&#xA;pastern&#xA;pasterns&#xA;&#xA;re&#xA;rep&#xA;repo&#xA;repos&#xA;repose&#xA;reposer&#xA;reposers&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If you can add to either end, &lt;strong&gt;9-letter&lt;/strong&gt; words are possible:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;id&#xA;aid&#xA;aide&#xA;aider&#xA;aiders&#xA;raiders&#xA;braiders&#xA;abraiders&#xA;&#xA;la&#xA;lap&#xA;laps&#xA;lapse&#xA;elapse&#xA;relapse&#xA;relapser&#xA;relapsers&#xA;&#xA;in&#xA;pin&#xA;ping&#xA;aping&#xA;raping&#xA;craping&#xA;scraping&#xA;scrapings&#xA;&#xA;at&#xA;eat&#xA;eath&#xA;heath&#xA;sheath&#xA;sheathe&#xA;sheather&#xA;sheathers&#xA;&#xA;is&#xA;ais&#xA;rais&#xA;raise&#xA;raiser&#xA;raisers&#xA;praisers&#xA;upraisers&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If you can add at any point, &lt;strong&gt;11-letter&lt;/strong&gt; words are possible:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;pi&#xA;pig&#xA;ping&#xA;oping&#xA;coping&#xA;comping&#xA;compting&#xA;competing&#xA;completing&#xA;complecting&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1497" LastEditorUserId="5373" LastEditDate="2015-03-06T00:11:02.077" LastActivityDate="2015-03-06T00:11:02.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9895" PostTypeId="2" ParentId="9890" CreationDate="2015-03-06T06:29:43.277" Score="10" Body="&lt;p&gt;To add to Kevin's answer, it's possible to come up with a 11-letter word by adding letters at either end (as illustrated by the answers to &lt;a href=&quot;https://codegolf.stackexchange.com/q/45168/3527&quot;&gt;this PPCG question&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ab&#xA;rab&#xA;arab&#xA;arabi&#xA;arabin&#xA;arabine&#xA;carabine&#xA;carabiner&#xA;carabinero&#xA;carabineros&#xA;&#xA;ra&#xA;rah&#xA;rahm&#xA;brahm&#xA;brahma&#xA;brahman&#xA;brahmani&#xA;brahmanis&#xA;brahmanism&#xA;brahmanisms&#xA;&#xA;ra&#xA;ran&#xA;fran&#xA;franc&#xA;franci&#xA;francis&#xA;francisc&#xA;francisca&#xA;franciscan&#xA;franciscans&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1585" LastEditorUserId="-1" LastEditDate="2017-04-13T12:39:28.753" LastActivityDate="2015-03-06T06:29:43.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9896" PostTypeId="2" ParentId="6029" CreationDate="2015-03-06T08:16:53.167" Score="3" Body="&lt;p&gt;For level 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Count total number of cars coming from each direction during the rush hour($3600/frequency + 1$), that gives:&lt;br&gt;&#xA; &lt;strong&gt;N&lt;/strong&gt;: 515&lt;br&gt;&#xA; &lt;strong&gt;S&lt;/strong&gt;: 401&lt;br&gt;&#xA; &lt;strong&gt;W&lt;/strong&gt;: 901&lt;br&gt;&#xA; &lt;strong&gt;E&lt;/strong&gt;: 451&lt;br&gt;&#xA; Because &lt;code&gt;|&lt;/code&gt; or &lt;code&gt;-&lt;/code&gt; will let both &lt;strong&gt;N&lt;/strong&gt; and &lt;strong&gt;S&lt;/strong&gt; &lt;em&gt;or&lt;/em&gt; &lt;strong&gt;W&lt;/strong&gt; and &lt;strong&gt;E&lt;/strong&gt; cars pass simultaneously, count total &lt;strong&gt;V&lt;/strong&gt;ertical and &lt;strong&gt;H&lt;/strong&gt;orizontal cars:&lt;br&gt;&#xA; &lt;strong&gt;V&lt;/strong&gt;: 916&lt;br&gt;&#xA; &lt;strong&gt;H&lt;/strong&gt;: 1352&lt;br&gt;&#xA; And total cars: 2268&lt;br&gt;&#xA; So now we can calculate ratio of &lt;strong&gt;V&lt;/strong&gt; and &lt;strong&gt;H&lt;/strong&gt; cars:&lt;br&gt;&#xA; &lt;strong&gt;RV&lt;/strong&gt;: $916/2268 * 100 = 40.39\%$&lt;br&gt;&#xA; &lt;strong&gt;RH&lt;/strong&gt;: $1352/2268 * 100 = 59.61\%$&lt;br&gt;&#xA; So we have to let the light be &lt;code&gt;|&lt;/code&gt; &lt;strong&gt;40.39%&lt;/strong&gt; of all time and &lt;code&gt;-&lt;/code&gt; &lt;strong&gt;59.61%&lt;/strong&gt;. That sums up to about:&lt;br&gt;&#xA; &lt;code&gt;|&lt;/code&gt;: 1453.97 sec&lt;br&gt;&#xA; &lt;code&gt;-&lt;/code&gt;: 2146.03 sec&lt;br&gt;&#xA; So the optimal strategy would be to first let the lights be &lt;code&gt;-&lt;/code&gt; for 19 sec, then &lt;code&gt;|&lt;/code&gt; for 13 sec and &lt;code&gt;A&lt;/code&gt; for 2 sec in between(minimum time required, let's not make card wait longer and pollute air more). That sums up to $19+2+13+2=36$, which we can repeat 100 times&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the variable values for level 1 would be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $S = -$&lt;br&gt;&#xA; $V = 13$&lt;br&gt;&#xA; $H = 19$&lt;br&gt;&#xA; $A = 2$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9352" LastActivityDate="2015-03-06T08:16:53.167" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9897" PostTypeId="2" ParentId="9155" CreationDate="2015-03-06T09:47:55.350" Score="3" Body="&lt;p&gt;Theoretically on an infinite plane the zebras can be set up so that the lions are unable to reach them. For a start they could start lightyears away and then continue to move away. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There might be certain situations in which the lions can win, for examlpe if the zebras were placed randomly in non-optimal positions then one would almost certainly get caught. But if the zebras are placed in well thought out locations, then regardless of the lions location no one lion can ever catch a zebra, it would just continue to move away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way a lion might attempt to beat this, would be to imagine that the zebras set up to surround the lions, meaning they can not get trapped, if a lion were to chase one zebra until the lion itself was outside of the zebras, it could then turn around and push a zebra back towards the other lions this could work as only one zebra can move to avoid the lion. If we assume that the closest zebra moves then the lion simply has to ensure it stays closer to the first zebra until it is outside of the circle/line/square the zebras are attempting to surround the lions with. This doesn't have to be a regular shape, and they could be more or less randomly spread out, but as soon as the lion gets to the other side, by maintaining a small distance it will eventually get a fair way past. It can then turn around. until the distance between it and the first zebra exceeds that of any other zebra and the nearest lion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first lion (A) would have to be selected such that another zebra (X) is nearby but farther from the next lion (B) than most of the other zebras are from their respective lions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When Lion A then turns back, another zebra (Y) will move as he will be closest to a lion. This has to be chosen such that X is not the next closest to a lion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same strategy can then be used by other lions to circle the zebra X, by chasing another Zebra Z they can position them selves around zebra X who doesn't get a chance to move until surrounded. They can then close in. The lions can move in any direction, and if we assume that the zebras move to the farthest point from the lion, this allows them to push them around. More than one lion can move to push around the same zebra, and since they can also force other zebras to move by getting closer to them, they will eventually be able to surround a zebra. &lt;/p&gt;&#xA;" OwnerUserId="9918" LastActivityDate="2015-03-06T09:47:55.350" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9898" PostTypeId="1" AcceptedAnswerId="9905" CreationDate="2015-03-06T09:57:38.123" Score="7" ViewCount="687" Body="&lt;p&gt;In this battle you are choosing&lt;br&gt;&#xA;weapons hard or sharp or flat,&lt;br&gt;&#xA;sometimes winning, sometimes losing,&lt;br&gt;&#xA;otherwise a futile threat.&lt;br&gt;&#xA;Despite the arms which you are using&lt;br&gt;&#xA;nobody will end up dead,&lt;br&gt;&#xA;for this fight is just amusing,&lt;br&gt;&#xA;exists solely in your head.&lt;/p&gt;&#xA;" OwnerUserId="9910" LastActivityDate="2015-03-06T12:00:11.363" Title="Nobody will end up dead" Tags="|riddle|word|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9899" PostTypeId="2" ParentId="9898" CreationDate="2015-03-06T10:07:32.033" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Music&lt;/strong&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In this battle you are choosing &lt;br&gt;&#xA; weapons hard or sharp or flat,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hard, sharp, and flat musical notes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;sometimes winning, sometimes losing,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Going up and down in pitch?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;otherwise a futile threat.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nobody is harmed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Despite the arms which you are using&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You use your arms/hands to play the piano, the violin, and most other musical instruments.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;nobody will end up dead,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Obviously.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;for this fight is just amusing,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It amuses your audience.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;exists solely in your head.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That is, if you can play music without seeing the scores!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-03-06T10:07:32.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9900" PostTypeId="1" AcceptedAnswerId="10301" CreationDate="2015-03-06T10:18:00.480" Score="25" ViewCount="3273" Body="&lt;p&gt;Yesterday evening there was a public discussion on TV that was dedicated to the upcoming elections.&#xA;Thirteen politicians from the thirteen most important political parties participated in it.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;After half an hour, one of the politicians summarized the situation and said: One lie has been told.&lt;/li&gt;&#xA;&lt;li&gt;Another one of them said: Now two lies have been told.&lt;/li&gt;&#xA;&lt;li&gt;Then a third one said: Now three lies.&lt;/li&gt;&#xA;&lt;li&gt;A fourth one said: And now four lies have been told.&lt;/li&gt;&#xA;&lt;li&gt;The fifth politician: Actually, now there are already five lies.&lt;/li&gt;&#xA;&lt;li&gt;The sixth one: Now six lies have been told.&lt;/li&gt;&#xA;&lt;li&gt;The seventh one: And now seven lies have been told.&lt;/li&gt;&#xA;&lt;li&gt;The eight politician: Now eight lies have been told.&lt;/li&gt;&#xA;&lt;li&gt;The ninth one: Nine lies have been told.&lt;/li&gt;&#xA;&lt;li&gt;The tenth one: Up to now ten lies have been told.&lt;/li&gt;&#xA;&lt;li&gt;The eleventh politician: Now even eleven lies have been told.&lt;/li&gt;&#xA;&lt;li&gt;The twelfth politician: Twelve! Now twelve lies have been told.&lt;/li&gt;&#xA;&lt;li&gt;Then the thirteenth politician said: And now thirteen lies have been told.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;At this moment the moderator got fed up and ended the discussion.&#xA;The discussion was later carefully investigated by the political analysts, and it turned out that at least one of the politicians had correctly stated the total number of lies told up to the moment just &lt;em&gt;before&lt;/em&gt; he made his claim.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; How many lies were altogether told by these thirteen politicians?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-10T08:57:36.470" LastActivityDate="2016-03-10T08:57:36.470" Title="Thirteen politicians" Tags="|mathematics|logical-deduction|" AnswerCount="10" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="9901" PostTypeId="2" ParentId="9900" CreationDate="2015-03-06T10:28:23.423" Score="18" Body="&lt;p&gt;Exactly&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; one&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;lie must have been told &lt;em&gt;before&lt;/em&gt; the quoted discussion, so the total number of lies altogether is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; thirteen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;if any other number of lies had &lt;em&gt;already&lt;/em&gt; been told, then the first politician must have been wrong to say it was one, so by induction the $n$th politician was wrong to say it was $n$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;given the total &lt;em&gt;earlier&lt;/em&gt; number of lies, we know the first politician was right, so all the others are wrong (since by induction, each one could only be right if all the preceding ones were wrong).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-03-06T10:28:23.423" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="9902" PostTypeId="2" ParentId="9900" CreationDate="2015-03-06T10:31:08.013" Score="11" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 13&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If there was no lie at all before that discussion, first one would be the first to lie, bringing lie counter to 1. Then second would bring it up to 2 and so on. But, in this case, none of them would state the total number of lies correctly before making his/her claim.&lt;br&gt;&#xA; If there was 1 lie before that discussion, that would make first one to be correct about lie counting, making his/her statement true. Then second said &lt;em&gt;Now two lies have been told.&lt;/em&gt;, but actually there was only 1 lie, so now total lies is 2. Third would say that there &lt;em&gt;were&lt;/em&gt; 3 lies, but his/her statement would bring total lies to 3, not &lt;em&gt;before&lt;/em&gt; his statement. It sums up to &lt;strong&gt;13&lt;/strong&gt; lies in total.&lt;br&gt;&#xA; If there were 2 or more lies before that discussion, first one would lie and bring the counter to 3+, that would make second one lie too and so on, so none of them would state the total number of lies correctly before making his/her claim.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9352" LastActivityDate="2015-03-06T10:31:08.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9903" PostTypeId="2" ParentId="9900" CreationDate="2015-03-06T10:31:12.833" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;13 lies&#xA; &lt;BR&gt;&lt;BR&gt;&#xA; if there were no lies at the beginning, then 13 politicians lied.&#xA; if there were one lie, then 12 polititians lied (all except the first one), total 13.&#xA; if there were one lie, then 11 polititians lied (all except the first and the second one), total 13. &#xA; etc...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9088" LastActivityDate="2015-03-06T10:31:12.833" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9904" PostTypeId="2" ParentId="9155" CreationDate="2015-03-06T11:05:04.487" Score="7" Body="&lt;p&gt;My final assumption is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; zebras can win&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;following this strategy :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The zebras choose to be uniformly distributed on a circle, distanced sufficiently from lions' center of gravity - which the circle is the centered on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They will not be doomed to lose the game given:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;p&gt;Any zebra is instructed to escape by following the vector in the direction parallel to $\sum \vec{l_i b}+\vec{z_i b}$ where $l_i$ are the lions' coordinates, $z_{i}$ are the closest other zebras' coordinates and $b$ is the zebra itself.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;When any zebra feels a lion being closest to it, it should take turn to move.&lt;br&gt;&#xA;  In technical terms, this is when it has the smallest $d = \min ||\vec{l_i b}||$.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This solution can prevent&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;any zebra to be stranded by two lions both sides or&lt;/li&gt;&#xA;&lt;li&gt;any zebra to be amassed in same spot with another zebra&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;because the unique condition where any zebra is condemned to death is that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sum (\vec{lb} + \vec{zb}) = \vec{0}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or that a single $\vec{zb} = \vec{0}$ or $\vec{lb} = \vec{0}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution is illustrated here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aMXjJ.png&quot; alt=&quot;Lions and zebras with vectors between them.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://tube.geogebra.org/student/m794961&quot; rel=&quot;noreferrer&quot;&gt;This little simulation&lt;/a&gt; can clarify my proposition for two zebras and two lions.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This part is a reply for @Veedrac&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering 6 zebras and 6 lions, I could implement this simulation in which by I might not surround any zebra since the zebras are fleeing the lions' center of gravity. I made this simulation open-source to unconvict myself from any cheating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://tube.geogebra.org/student/m804603&quot; rel=&quot;noreferrer&quot;&gt;Here is a simulation for 6 zebras and 6 lions.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; Try to move the cursor slowly centimeter by centimeter because there is some kind of weird bug in that code. I couldn't handle it due to my inexperience with this type of scripting language.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Secondly, I don't really believe that Veedrac's theory is really fruitful because zebras are always escaping any kind of fencing that lions try to impose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets assume $k$ lions succeeded in surrounding one lion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $r$ be the radius of that circle and $2 \pi r$ is the perimeter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any successful encirclement is when the length of each arc between two lions is gradually shrinked until it is no more than 100m with a zebra still inside the circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $k$ lions made a unitary $u = 100\text{m}$ step, the zebra would have made $ku$ steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a zebra is doomed to be caught, the radius equals $\frac{(k-1)}{k} r$,  the perimeter of the circle is $2 \pi \frac{(k-1)r}{k}$ and the length of each arc is $\frac{2 \pi \frac{(k-1)r}{k}}{k}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The condition where a zebra is caught is $\frac{2 \pi \frac{(k-1)r}{k}}{k} \leq u = 1$ and the maximal length of the circle's radius is $\frac {k^2}{2\pi(k-1)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When there are $k = 100$ lions, $r \approx 16u=1600\text{m}$, which can be avoided when we consider an enough large difference between the zebras' and lions' accumulation center before the game starts.&lt;/p&gt;&#xA;" OwnerUserId="9517" LastEditorUserId="9517" LastEditDate="2015-03-11T18:40:43.700" LastActivityDate="2015-03-11T18:40:43.700" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="9905" PostTypeId="2" ParentId="9898" CreationDate="2015-03-06T11:52:54.393" Score="8" Body="&lt;p&gt;Surely it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A game of rock, paper, scissors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In this battle you are choosing&#xA;weapons hard or sharp or flat,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hard - rock; sharp - scissors; flat - paper&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;sometimes winning, sometimes losing,&#xA;otherwise a futile threat.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Three equally likely outcomes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Despite the arms which you are using&#xA;nobody will end up dead,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rarely does someone become Violent&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;for this fight is just amusing,&#xA;exists solely in your head.&lt;/p&gt;&#xA;" OwnerUserId="9965" LastEditorUserId="9965" LastEditDate="2015-03-06T12:00:11.363" LastActivityDate="2015-03-06T12:00:11.363" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9906" PostTypeId="2" ParentId="9900" CreationDate="2015-03-06T11:58:58.373" Score="4" Body="&lt;p&gt;Here's a short proof.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the first politician was wrong, then they were all wrong, which isn't so. So he was right, which makes all the others wrong. So one lie was told before these statements, and 12 of the statements are wrong, so the answer is 13.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7363" LastActivityDate="2015-03-06T11:58:58.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9907" PostTypeId="2" ParentId="9866" CreationDate="2015-03-06T12:27:17.067" Score="9" Body="&lt;p&gt;Although mfc32 answer is correct I'd like to add a more mathematical approach:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; F = number of females in the beginning&lt;br&gt;&#xA; M = number of males in the beginning&lt;br&gt;&#xA; S = number of spikes in the end&lt;br&gt;&#xA;&lt;br&gt;&#xA; F + M = 274&lt;br&gt;&#xA; S = 4 * F + (1/3) * 12 * M&lt;br&gt;&#xA; S = 4 * F + 4 * M&lt;br&gt;&#xA; S = 4 * (F + M)&lt;br&gt;&#xA; S = 4 * 274 = 1096  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-03-06T12:27:17.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9908" PostTypeId="1" AcceptedAnswerId="9923" CreationDate="2015-03-06T12:47:11.460" Score="7" ViewCount="731" Body="&lt;p&gt;I am short and I always exclaim&lt;br&gt;&#xA;I vary in appearance, but only have one name&lt;br&gt;&#xA;I am silent, but in your head I make sound&lt;br&gt;&#xA;I am intangible and often bound&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I often come with colors,&lt;br&gt;&#xA;And always with shapes and lines&lt;br&gt;&#xA;I serve two purposes&lt;br&gt;&#xA;Of which one of them is to please the eyes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="9967" LastActivityDate="2015-03-07T01:36:09.383" Title="I am short and I always exclaim" Tags="|riddle|" AnswerCount="4" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9909" PostTypeId="2" ParentId="9908" CreationDate="2015-03-06T14:20:10.857" Score="8" Body="&lt;p&gt;You are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Exclamation mark&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am short and I always exclaim&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It is just one character long&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I vary in appearance, but only have one name&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You can have different fonts, but it's still called exclamation mark&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am silent, but in your head I make sound&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It makes no sound but when we read it, we imagine emphasis&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am intangible and often bound&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It has to be bound to other words&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I often come with colors,&#xA;And always with shapes and lines&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Different colors, fonts, etc&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I serve two purposes&#xA;Of which one of them is to please the eyes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Exclamation mark has its own purpose. The one to please the eyes can be some sort of notification you were expecting&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9907" LastEditorUserId="9907" LastEditDate="2015-03-06T14:52:58.883" LastActivityDate="2015-03-06T14:52:58.883" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9910" PostTypeId="2" ParentId="9908" CreationDate="2015-03-06T14:36:54.117" Score="2" Body="&lt;p&gt;I suspect you might be a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Balloon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am short and I always exclaim&lt;br&gt;&#xA;I vary in appearance, but only have one name&lt;br&gt;&#xA;I am silent, but in your head I make sound&lt;br&gt;&#xA;I am intangible and often bound&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Balloon, as in cartoon drawing, usually bound to a character exclaiming something. They are intangible, silent, yet they still make sound in your head on reading them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I often come with colors,&lt;br&gt;&#xA;And always with shapes and lines&lt;br&gt;&#xA;I serve two purposes&lt;br&gt;&#xA;Of which one of them is to please the eyes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Balloon, as in children's plaything, also used for decoration purposes. They come in many colors and shapes, and there's a line attached to keep them in place.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9910" LastActivityDate="2015-03-06T14:36:54.117" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9915" PostTypeId="1" AcceptedAnswerId="15255" CreationDate="2015-03-06T16:42:51.453" Score="48" ViewCount="4996" Body="&lt;p&gt;This puzzle belongs to the puzzle series: &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=hyper-modern%20is%3Aquestion&quot;&gt;hyper-modern art&lt;/a&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Still in the hyper-modern art gallery...&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&amp;quot;You know, even now that you have explained to me, &lt;a href=&quot;https://puzzling.stackexchange.com/questions/9835&quot;&gt;how hyper-modern-art works&lt;/a&gt;, I still don't quite get it. I mean, the image is just... monochrome. I really prefer more colour in art.&amp;quot;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;Ah, but that's not a problem my friend. Just come with me and look at another painting from the same artist. See, more than enough colours&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&amp;quot;Well, yes, but I'd say this is a rather dumb variation of &lt;strong&gt;Stars 'n' Stripes&lt;/strong&gt;...&lt;br&gt;Why on earth are you chuckling now?&amp;quot;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;It's just that I found your comment rather funny. In particular, if you look at the title of this painting...&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&amp;quot;Oh!! I see. Well, I can see why it is funny, but I can absolutely&lt;/em&gt; not &lt;em&gt;see, why this work is called like this!&amp;quot;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;Well, well, my friend. Sometimes you have to look beyond what is visible, connect the facts and combine all in an intuitive way. You know Rebus puzzles, don't you? See, this painting really is just a somewhat more involved variant. But you were right in one thing. The &lt;em&gt;stripes&lt;/em&gt; are without any meaning.&amp;quot;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ApGgl.png&quot; alt=&quot;StarsAndStripesIsNotMyName&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;(&lt;a href=&quot;https://i.stack.imgur.com/ApGgl.png&quot; rel=&quot;noreferrer&quot;&gt;high-res download&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;h3&gt;The goal of this puzzle is to find the title of the work, which is a single word. The puzzle is &lt;em&gt;fully&lt;/em&gt; contained in the image. You may want to download the full-size version for processing.&lt;/h3&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Hints&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The puzzle can be solved with a &lt;em&gt;printed&lt;/em&gt; version of the image above.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The puzzle requires an &lt;em&gt;action&lt;/em&gt; and is (likely) not being solved by just looking at the picture.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;strong&gt;Solution self-verification&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Read the following &lt;em&gt;after&lt;/em&gt; you think you have the solution. &lt;br&gt;&#xA;Note, that if you base your solution by guessing on the information below, you have &lt;em&gt;not&lt;/em&gt; solved the puzzle. The facts below should &lt;em&gt;not&lt;/em&gt; be part of your solution.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; - the final word  has &lt;strong&gt;6 letters&lt;/strong&gt; &lt;br&gt;&#xA; - &lt;strong&gt;all letters are different&lt;/strong&gt;&lt;br&gt;&#xA; - it is a &lt;strong&gt;noun&lt;/strong&gt;&lt;br&gt;&#xA; - it ends with &lt;strong&gt;'s'&lt;/strong&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-02-20T13:46:23.077" Title="What is the title of this modern art painting, if not &quot;Stars and Stripes&quot;?" Tags="|cipher|rebus|visual|" AnswerCount="3" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="9916" PostTypeId="2" ParentId="9915" CreationDate="2015-03-06T17:33:17.447" Score="8" Body="&lt;p&gt;I'm going to guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quartz&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is my reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are ten colors of stars and ten shapes of stars. Each color appears four times and each shape appears four times. The numbers at the bottom resemble a clock so, instead of being QUARTS, it is QUARTZ.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9828" LastActivityDate="2015-03-06T17:33:17.447" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9917" PostTypeId="2" ParentId="9900" CreationDate="2015-03-06T20:09:55.777" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The discussion was later carefully investigated by the political analysts, and it turned out that at least one of the politicians had correctly stated the total number of lies told before he made his claim.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's a lie!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 0: Not a single politician lied during the discussion.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In this case, the first politician interviewed after the discussion lied by saying there was one lie (there were none), the next to be interviewed lied by saying there were two lies (there was one), and so on. While none of the politicians exhibited their true colors during the discussion, every single one exhibited their true colors in the follow-up. In this case, not a single one of the politicians correctly stated the total number of lies told before he made his claim.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Case 1: Only one lie was told during the discussion.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In this case, the first politician interviewed after the discussion told the truth by saying there was one lie, but the subsequent politician to be interviewed lied, as did every politician who followed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Case 2: Two or more lies were told during the discussion.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In this case, every politician interviewed after the discussion lied about the number of lies told before that politician was interviewed. As with case #0, not a single one of the politicians correctly stated the total number of lies told before he made his claim.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is no way to ascertain the number of lies told by the politicians. Most likely the number of lies was in the hundreds during the discussion (how can you tell when politicians are lying? Answer: When you see their lips moving) plus thirteen more lies in the follow-up.&lt;/p&gt;&#xA;" OwnerUserId="9026" LastActivityDate="2015-03-06T20:09:55.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9918" PostTypeId="1" CreationDate="2015-03-06T20:31:43.420" Score="3" ViewCount="131" Body="&lt;p&gt;I have to create a texture using 4 different elements (call them A,B,C,D or whatever you want) on a 6x7 grid. The goal is to obtain a pattern with the least repetitions of the same element in horizontal, vertical and diagonal directions, especially, if possible, should be avoided to repeat the same element in adjacent blocks. I think there should be different solution to this, just answer with what you came up with.&lt;/p&gt;&#xA;" OwnerUserId="9970" LastActivityDate="2015-03-09T00:22:10.260" Title="Four elements repetition in a texture" Tags="|pattern|" AnswerCount="2" CommentCount="2" ClosedDate="2015-03-07T10:18:44.240" ContentLicense="CC BY-SA 3.0" />
  <row Id="9919" PostTypeId="2" ParentId="9900" CreationDate="2015-03-06T20:37:25.433" Score="8" Body="&lt;p&gt;&lt;strong&gt;14 lies were told.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since it's well known that &lt;strong&gt;politicians are incapable of telling the truth&lt;/strong&gt;, it follows that every statement they made was a lie (13 lies). On top of that, the Analyst must also have been lying when he said that one of the politicians &lt;em&gt;wasn't&lt;/em&gt; lying. That makes 14.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4852" LastActivityDate="2015-03-06T20:37:25.433" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="9920" PostTypeId="2" ParentId="9900" CreationDate="2015-03-06T20:46:59.923" Score="2" Body="&lt;p&gt;this analysis is based on this note :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;one lie maximally was told before the conversation got started.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If the first guy was truthful  , all other politicians wud have lied , n=13 lies , 12 political lies aside the first one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now , if the first guy told a lie , it means no lie was been told until first politican spoke, one lie until second politician lied and third , then forth , etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this does mean on the whole .&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; n=13 lies was been told&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;if we admit more than a lie from the first statement , n can go over 15 if first guy lied.&lt;/p&gt;&#xA;" OwnerUserId="9517" LastEditorUserId="9517" LastEditDate="2015-03-07T00:29:19.827" LastActivityDate="2015-03-07T00:29:19.827" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9921" PostTypeId="2" ParentId="9918" CreationDate="2015-03-06T20:57:32.207" Score="0" Body="&lt;p&gt;Here is one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a	b	c	d	a	b\\&#xA;b	c	d	a	b	c\\&#xA;c	d	a	b	c	d\\&#xA;d	a	b	c	d	a\\&#xA;a	b	c	d	a	b\\&#xA;b	c	d	a	b	c\\&#xA;c	d	a	b	a	d\\$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the row you must have 2 repetitions and in the column 3.&lt;/p&gt;&#xA;" OwnerUserId="9531" LastEditorUserId="5373" LastEditDate="2015-03-09T00:22:10.260" LastActivityDate="2015-03-09T00:22:10.260" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9922" PostTypeId="2" ParentId="9918" CreationDate="2015-03-06T21:26:35.590" Score="0" Body="&lt;p&gt;Starting from the nucleus &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a b a\\&#xA;b c b\\&#xA;a b a\\$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;add &quot;layers&quot; of two characters alternating onto the nucleus. For example, the next iteration will be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c d c d c\\&#xA;d a b a d\\&#xA;c b c b c\\&#xA;d a b a d\\&#xA;c d c d c\\$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This array can be extended infinitely, and all elements of this infinite array will have neighbours in all 8 directions that are different from itself, except at the corners. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c d c d c d\\&#xA;a b a b a b\\&#xA;c d c d c d\\&#xA;a b a b a b\\&#xA;c d c d c d\\&#xA;a b a b a b\\&#xA;c d c d c d\\$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply take a 6x7 subset of this infinite array at any position where no corner exists to produce an answer.&lt;/p&gt;&#xA;" OwnerUserId="7529" LastActivityDate="2015-03-06T21:26:35.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9923" PostTypeId="2" ParentId="9908" CreationDate="2015-03-06T21:31:49.340" Score="4" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Onomatopoeia&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am short and I always exclaim&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; almost are always followed by an exclamation mark&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I vary in appearance, but only have one name&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; in comic books they will vary in font&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am silent, but in your head I make sound&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; by definition you make a sound when thought of&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am intangible and often bound&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; words are bound to a page&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I often come with colors,&#xA;And always with shapes and lines&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; again different font&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I serve two purposes&#xA;Of which one of them is to please the eyes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; nice fonts please the eyes of the (usual comic book) reader. the other is&#xA; the sound effect of course&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9778" LastActivityDate="2015-03-06T21:31:49.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9924" PostTypeId="2" ParentId="9900" CreationDate="2015-03-06T21:51:06.570" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 12. If we wish to answer the question as it is asked here, then the correct answer is not 13.  We have been told three specific things: First, one of the politicians had correctly stated the total number of lies told &lt;em&gt;before&lt;/em&gt; he made his claim. This has to be the first politician, since all others respond with an n + 1 statement (based on the preceding politician's stated number).  So, the first lie was already told beforehand, and the first politician's statement was true.  Secondly, by the logic I have just described, we have observed the subsequent twelve politicians tell lies. Thirdly, the question asked of us is to determine how many lies were altogether told by these politicians. We cannot infer or assume that the first lie was told &lt;em&gt;by the politicians themselves&lt;/em&gt;. They could be discussing a lie that the moderator brought up as an anecdote. So, unless it is explicitly stated they told this lie, it cannot be counted.  (The default value is false.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I may be splitting hairs with this level of precision, but this is the logically correct answer to the question.  If this is not the intended answer, then the question should be rephrased.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: To extend the flaw a bit further, it doesn't explicitly state that these were the only possible lies.  What about the preceding 30 minutes?  They could have told hundreds.  The riddle should be reworded.&lt;/p&gt;&#xA;" OwnerUserId="9973" LastEditorUserId="9973" LastEditDate="2015-03-06T22:20:22.167" LastActivityDate="2015-03-06T22:20:22.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9925" PostTypeId="2" ParentId="9890" CreationDate="2015-03-06T22:17:56.000" Score="0" Body="&lt;p&gt;Adding at the end, I give you &lt;strong&gt;contrasters&lt;/strong&gt; (11 letters). For 10 letters, I found these:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;contraster&#xA;Romanisers&#xA;contrasted &#xA;pastoraled &#xA;pastorales &#xA;minimisers &#xA;quartettes &#xA;quintettes &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I used &lt;a href=&quot;http://wordlist.aspell.net/&quot; rel=&quot;nofollow&quot;&gt;Aspell&lt;/a&gt; SCOWL dictionary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using a &lt;a href=&quot;http://www.keithv.com/software/wlist/&quot; rel=&quot;nofollow&quot;&gt;bigger, lower quality, list&lt;/a&gt;, I can find &lt;em&gt;territorialised&lt;/em&gt; and &lt;em&gt;territorialists&lt;/em&gt; (15), but there are many false positives in that list.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, one can &lt;em&gt;cheat&lt;/em&gt;. If you start with: &lt;em&gt;dichlorodiphenyltrichloroethane&lt;/em&gt; you can just add an s and put yourself in the nice sum of 32 letters. If we remove chemicals (they can, after all, be arbitrarily long), you have the classic &lt;strong&gt;antidisestablishmentarianism&lt;/strong&gt;, that, in its plural form, is 29 letters long.&lt;/p&gt;&#xA;" OwnerUserId="9975" LastEditorUserId="9975" LastEditDate="2015-03-06T22:47:11.803" LastActivityDate="2015-03-06T22:47:11.803" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9926" PostTypeId="2" ParentId="9900" CreationDate="2015-03-06T23:04:58.397" Score="3" Body="&lt;p&gt;All answers so far assume that a false statement is a lie. But this is mistaken, because a lie is a deliberate untruth. If someone makes a false statement that they believe to be true, they're not lying. If we go by this, then the answer is that the number of lies told was&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; between 1 and 13 inclusive&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the first politician was right, then all the others were wrong, so the number of lies told was one plus however many of the other politicians were deliberately wrong, which could be any number between zero and 12. If the first politician was wrong, then at least one of the other politicians must have been right, giving us a number of lies between two and 13. So the number of lies altogether is between 1 and 13 inclusive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7363" LastActivityDate="2015-03-06T23:04:58.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9927" PostTypeId="2" ParentId="9875" CreationDate="2015-03-07T01:04:15.847" Score="5" Body="&lt;p&gt;I don't have a proof for the maximum number of pair-free shuffles but the following shuffle:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  19 42 32 27 51 21 10 5 44 31 4 43 8 15 18 36 23 33 52 48 14 26 6 41 9 25 12 47 38 16 3 13 24 35 20 22 50 29 28 46 17 11 45 40 30 7 34 2 39 1 49 37&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;produces pair-free states &lt;strong&gt;for the first six shuffles&lt;/strong&gt;. The seventh shuffle produces two pairs.&#xA;The following 8 lines show the initial deck and the deck after each of 7 shuffles (each preceded by the number of adjacent pairs in the deck):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  (39) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52&#xA;  (0) 19 42 32 27 51 21 10 5 44 31 4 43 8 15 18 36 23 33 52 48 14 26 6 41 9 25 12 47 38 16 3 13 24 35 20 22 50 29 28 46 17 11 45 40 30 7 34 2 39 1 49 37&#xA;  (0) 52 11 13 12 49 14 31 51 40 3 27 45 5 18 33 22 6 24 37 2 15 25 21 17 44 9 43 34 29 36 32 8 41 20 48 26 1 38 47 7 23 4 30 46 16 10 35 42 28 19 39 50&#xA;  (0) 37 4 8 43 39 15 3 49 46 32 12 30 51 33 24 26 21 41 50 42 18 9 14 23 40 44 45 35 38 22 13 5 17 48 2 25 19 29 34 10 6 27 16 7 36 31 20 11 47 52 28 1&#xA;  (0) 50 27 5 45 28 18 32 39 7 13 43 16 49 24 41 25 14 17 1 11 33 44 15 6 46 40 30 20 29 26 8 51 23 2 42 9 52 38 35 31 21 12 36 10 22 3 48 4 34 37 47 19&#xA;  (0) 1 12 51 30 47 33 13 28 10 8 45 36 39 41 17 9 15 23 19 4 24 40 18 21 7 46 16 48 38 25 5 49 6 42 11 44 37 29 20 3 14 43 22 31 26 32 2 27 35 50 34 52&#xA;  (0) 19 43 49 16 34 24 8 47 31 5 30 22 28 17 23 44 18 6 52 27 41 46 33 14 10 7 36 2 29 9 51 39 21 11 4 40 50 38 48 32 15 45 26 3 25 13 42 12 20 1 35 37&#xA;  (2) 52 45 39 36 35 41 5 34 3 51 16 26 47 23 6 40 33 21 37 12 17 7 24 15 31 10 22 42 38 44 49 28 14 4 27 46 1 29 2 13 18 30 25 32 9 8 11 43 48 19 20 50&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="1631" LastEditDate="2015-03-07T01:49:02.090" LastActivityDate="2015-03-07T01:49:02.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9928" PostTypeId="2" ParentId="9908" CreationDate="2015-03-07T01:36:09.383" Score="2" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A comic book.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am short and I always exclaim&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Comics are often short and have onomatopoeic exclamations (&quot;Bang!&quot; or &quot;Kapow!&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I vary in appearance, but only have one name&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tons of different comic book stories/characters/authors, but all are comics.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am silent, but in your head I make sound&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We read them, so all of the sound is in our mind.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am intangible and often bound&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Intangible because people often keep them sealed in mint condition, and bound because it's a book.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I often come with colors,&#xA;And always with shapes and lines&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Colored illustrations&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I serve two purposes&#xA;Of which one of them is to please the eyes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We can read them for the story, but can also just admire the artwork.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9978" LastActivityDate="2015-03-07T01:36:09.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9929" PostTypeId="1" CreationDate="2015-03-07T04:10:51.763" Score="7" ViewCount="898" Body="&lt;p&gt;Let's say, for instance, your friend needs one of your passwords to do something while you're away. He calls you and asks for it. Thing is, you don't want him to know your password.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How then would you allow him to enter said password with only a keyboard and the phone in his hand, without him knowing the password?&lt;/p&gt;&#xA;" OwnerUserId="9979" LastEditorUserId="5373" LastEditDate="2015-03-07T17:14:55.380" LastActivityDate="2015-03-08T03:43:45.470" Title="Secrets over the phone" Tags="|word-problem|cryptography|" AnswerCount="9" CommentCount="6" ClosedDate="2015-03-08T06:05:16.330" ContentLicense="CC BY-SA 3.0" />
  <row Id="9930" PostTypeId="2" ParentId="9929" CreationDate="2015-03-07T07:13:08.407" Score="3" Body="&lt;p&gt;You could embed your actual password inside some random string and them tell him to type that random string, which you could make as&#xA;complicated and as long as you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For e.g. if your password is &lt;em&gt;abc123&lt;/em&gt; then you could tell him to type, lets say this - &lt;em&gt;gibarisbhc3$%&amp;amp;1st2u5f3&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It really doesn't matter what random string you choose to say, so long as it contains all the characters in your actual password appearing in that same order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now its just a matter of making him delete the unwanted characters from that random string such that only your password remains in the password field before he presses the enter key to log in to the machine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the e.g. above there are 16 unwanted characters that needs to be removed and lets say we choose the method in which we will remove them starting from the rightmost character. You could also adopt some other complex methods (like for e.g. first odd ones and then even ones) of deleting to make things more complicated, in case your friend has a nearly eidetic memory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that the cursor would be at the end of the last character before the deletion process starts. I'm saying this because you could actually add more complication by not starting from the end (for instance just tell him to press the Home button which would take the cursor to the start).&#xA;OK so the first unwanted character that needs to be deleted in our adopted method of deletion would be the last &lt;em&gt;f&lt;/em&gt;.&#xA;This can be done by telling him to press the left arrow once and then hit backspace and so on and so forth for the remaining unwanted characters.&#xA;So it would be nearly impossible for someone to figure out the actual password, assuming he is not writing the password and is simply typing it as you say it on the phone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. You could also make use of the &lt;em&gt;Insert&lt;/em&gt; key during the deletion process to add one more layer of complication.&lt;/p&gt;&#xA;" OwnerUserId="101" LastEditorUserId="101" LastEditDate="2015-03-07T07:22:52.827" LastActivityDate="2015-03-07T07:22:52.827" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9931" PostTypeId="2" ParentId="9929" CreationDate="2015-03-07T08:19:23.547" Score="6" Body="&lt;p&gt;Answer in parts so people can guess the rest as they go along.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer part 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Assuming you both have smartphones, have your friend turn the camera on for video chat and close his eyes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer part 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Then he can hold the phone over the keyboard so you can see the keyboard on your phone through the video chat&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer part 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He randomly places his finger on the keyboard and you direct him to move his finger to the next letter in your password, one letter at a time (I.e. If his finger lands on the &quot;u&quot; and the next letter is a &quot;v&quot; you direct him to go left, down/left, down/left.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;requirements:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This requires that you trust your friend to keep his eyes closed and that he does not start on one of the &quot;home&quot; keys, f or j (in a qwerty keyboard) that have little ridges on them for identification. Or that if one of the home keys is in your password, you have him reset his finger afterwards&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2015-03-07T08:19:23.547" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9932" PostTypeId="2" ParentId="9875" CreationDate="2015-03-07T09:08:01.993" Score="12" Body="&lt;p&gt;Here is a permutation that works upto n=12:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 6 11 4 5 10 15 8 9 14 19 12 13 18 23 16 17 22 27 20 21 26 31 24 25 30 35 28 29 34 39 32 33 38 43 36 37 42 47 40 41 46 51 44 45 50 3 48 49 2 7 52&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Essentially, keep the first and fourth cards with same rank in its place. And permute the second cards of all ranks cyclically, and the third cards of all ranks cyclically with a different shift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will ensure no pairs for first 12 shuffles, and the deck returns to its original state after the 13th shuffle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps things can be improved by combining the two cyclic permutations somehow. I tried generating random permutations to see how well they work, but none of them seems to do better than n=5.&lt;/p&gt;&#xA;" OwnerUserId="9471" LastActivityDate="2015-03-07T09:08:01.993" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9933" PostTypeId="2" ParentId="9097" CreationDate="2015-03-07T09:35:50.190" Score="8" Body="&lt;p&gt;Zebras win with an up (or down) and across strategy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with Lopsy's idea of vertical strips, with each strip 400m wide. We call a strip &lt;em&gt;empty&lt;/em&gt; if it contains no zebras (it may contain the lion). For convenience, referring to a strip without qualification refers to the middle of the strip. The lion is &lt;em&gt;near&lt;/em&gt; a zebra if it is within 100m of the zebra horizontally and vertically. The lion &lt;em&gt;approaches&lt;/em&gt; a zebra if it is within 100m of getting &lt;em&gt;near&lt;/em&gt; the zebra. The lion &lt;em&gt;departs&lt;/em&gt; from a zebra if it moves from &lt;em&gt;near&lt;/em&gt; to &lt;em&gt;not near&lt;/em&gt; the zebra across a pair of consecutive lion-moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Reset state&lt;/em&gt;: the lion is not near any zebra, each zebra is in the middle of an otherwise empty strip, and it is the zebras' turn to move. Due to the distribution of zebras, $l$ cannot approach more than one zebra simultaneously from a reset state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start the game in a reset state. Zebras win if either the lion makes an infinite but unsuccessful pursuit or the animals re-establish a reset state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a reset state and let $z$ be the first zebra that the lion $l$ approaches. Set $t=0$ when $l$ first approaches $z$, and increment $t$ after each turn. Let $v(t)$ and $h(t)$ be the respective vertical and horizontal separation of $l$ and $z$ upon completion of turn $t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 1: $v(0) &amp;gt; 1$mm: $z$ flees vertically to maintain or increase vertical separation. So $l$ either continues pursuit indefinitely or re-establishes a reset state by departing from $z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 2: $v(0) \leq 1$mm, whence $h(0) \geq 100$m (definition of 'approach'). Then $z$ flees vertically for up to a distance $d$ to a point (call it $s$), where it flees horizontally. From my other answer to this puzzle, it suffices to have $d &amp;lt; 5000$km. We can apply a safety margin of 10% and require $d &amp;lt; 4500$km (call this Condition $c_d$), at which point the horizontal separation is at least 10m, give or take up to 1mm (call this Condition $c_h$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we find the minimum horizontal distance $m_h$ from $s$ to an empty strip. In the worst case, $z$ is the last zebra on either side of the herd, the herd occupies 100 adjacent strips, and $l$ chases $z$ through the herd. So $m_h = 100 * 400$m, i.e. 40km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming we found $s$, if $l$ departs from $z$ on the horizontal run and moves to between 200m and 300m vertically from $z$, $z$ is free to run to the nearest empty strip on either side. By the pigeon-hole principle, the nearest empty strip on one side will be at most $m_h/2 = 20$km away. When $z$ arrives, $l$ is at most 20.3km away vertically from $z$. Accounting for $v(0)$ and a 200m buffer between $l$ and its new target, let's allow 21km vertically between $z$ and the zebra nearest to it. With zebras vertically spaced most inconveniently at intervals of just under double this distance from each other (42km), $z$ will find $s$ within $100 * 42 = 4200$km of its starting point, so $d = 4200$km. This satisfies Condition $c_d$, so $s$ exists for any distribution of zebras in a reset state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now continue by breaking case 2 into 3 sub-cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 2a: $l$ departs from $z$ while $z$ is still fleeing vertically (before $z$ reaches $s$). This re-establishes a reset state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 2b: $l$ never stops pursuing $z$. Then $z$ runs vertically to $s$ and flees horizontally towards the nearest empty strip and keeps going. From Condition $c_h$, $z$ can then maintain a separation of at least 10m indefinitely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 2c: $l$ pursues $z$ to a horizontal run, then departs to pursue some other zebra: $z$ keeps going until $l$ is between 200m and 300m away vertically, then proceeds up to 20km from there to the nearest empty strip on either side. Upon $z$'s arrival, $l$ is at least 680m away from its new target. Even if we allowed an additional safety margin of another 200m, this re-establishes a reset state. Note that since $l$ is more than 100m away vertically from $z$ at the time of $z$'s final up-to-20km horizontal dash, $l$ cannot catch $z$ by retargeting $z$ during that dash due to similar reasoning to that which established Condition $c_d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the zebras win.&lt;/p&gt;&#xA;" OwnerUserId="9422" LastEditorUserId="9422" LastEditDate="2015-03-14T10:52:30.167" LastActivityDate="2015-03-14T10:52:30.167" CommentCount="25" ContentLicense="CC BY-SA 3.0" />
  <row Id="9934" PostTypeId="1" AcceptedAnswerId="9948" CreationDate="2015-03-07T09:57:07.300" Score="13" ViewCount="491" Body="&lt;p&gt;The twelve numbers on a clock are each either colored red or colored black.&#xA;You are allowed to make several &lt;em&gt;moves&lt;/em&gt;, where a &lt;em&gt;move&lt;/em&gt; consists in picking &#xA;a black number, and in flipping the colors of its two neighboring numbers &#xA;(while keeping the color of the picked black number unchanged).&#xA;The goal is to reach a situation where eleven numbers are red and where&#xA;the remaining twelfth number is black.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; How many starting situations are there that allow you to eventually reach the goal?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-18T09:56:16.637" LastActivityDate="2016-03-18T09:56:16.637" Title="It's twelve o'clock!" Tags="|mathematics|combinatorics|reachability|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9937" PostTypeId="2" ParentId="9929" CreationDate="2015-03-07T12:05:19.027" Score="7" Body="&lt;p&gt;Coming from a security background, several options come to mind real quick:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Have him enter a one-time password/code/phrase (OTP). (OpenID providers such as any site that supports or provides a  &quot;Sign in with .....&quot;  use this.) Then delete or change the (normally) 10 passphrase once he has entered this one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Use Oauth or OpenID again, but using QRcodes or pseudo-random gibberish as alluded to by Duncan, only instead of the delete/modify it uses the QR code on his tablet/phone&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Using a keychain in the browser or OS (assuming multiple accounts -- a couple of dummy accounts), have him select the username in question. The password is hashed out (******), they hit enter, you change promptly.&lt;/p&gt;&#xA;" OwnerUserId="9984" LastEditorUserId="5373" LastEditDate="2015-03-07T13:42:58.280" LastActivityDate="2015-03-07T13:42:58.280" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9938" PostTypeId="2" ParentId="9929" CreationDate="2015-03-07T12:45:56.147" Score="1" Body="&lt;p&gt;You can obfuscate the password using backspaces and arrows.  For example, if my password is &quot;Steve&quot;, I could tell him to type:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;xS{left}{backspace}{right}t...&lt;/code&gt; etc.  Hopefully you get the idea, I'm too lazy to give a full example! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if he types it into a plain text field, he would still see the password, so probably go on speaker and use facetime or something to watch as he types it directly into a password field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To mitigate against someone listening on the line, you could utilize some shared secrets, like &quot;press {left} {your daughter's age} times&quot;.&lt;/p&gt;&#xA;" OwnerUserId="9986" LastActivityDate="2015-03-07T12:45:56.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9939" PostTypeId="2" ParentId="9929" CreationDate="2015-03-07T14:04:23.023" Score="5" Body="&lt;p&gt;I'm assuming the main point is that the friend not learn your password, because maybe it is also used for some more sensitive purposees. You could do this for most websites by logging in on your own computer, and then sending (I guess reading aloud, which would take some time) your session credentials to your friend. These are usually stored in a browser cookie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Friend inserts these session credentials artificially into her browser, and it just looks to the server as though your logged-in laptop just changed IP addresses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For many sites, this would allow your friend to be logged in as you, and perhaps even change your password, but probably not ever see the actual password.&lt;/p&gt;&#xA;" OwnerUserId="9987" LastActivityDate="2015-03-07T14:04:23.023" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="9940" PostTypeId="2" ParentId="9929" CreationDate="2015-03-07T17:05:33.530" Score="2" Body="&lt;p&gt;Another technique that has not yet been mentioned would be to change your password, give it to your friend, and then change it back after he is done with the task (in most competently designed systems, this should invalidate all current login tokens and therefore kick the user out of the system). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, all of the above techniques all render your friend (whom you presumably don't want to give full account access) capable of full control over your account, at least for a limited amount of time, in a state where you have no oversight over his/her actions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way which prevents these avenues of attack would be to let your friend open a secured VNC tunnel to your computer, which has the account already logged in. This way, you can monitor whatever actions are taken, and be able to disconnect said friend if they attempt to do anything that you do not allow. &lt;/p&gt;&#xA;" OwnerUserId="7529" LastActivityDate="2015-03-07T17:05:33.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9941" PostTypeId="2" ParentId="9915" CreationDate="2015-03-07T17:24:25.313" Score="5" Body="&lt;p&gt;I am intrigued, though I have no good strategies for figuring out where to go from the image. I am entirely unfamiliar with art in general, much less hyper-modern art, so I'll make a guess based on what I see (and what I don't):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is a lot of ambiguity that the solver must process. The most important things I figure we have to deduce are:&lt;br&gt;&#xA; 1) &lt;strong&gt;The stars.&lt;/strong&gt; What is the significance of the shapes? BmyGuest has hinted that the stripes have no significance, but &lt;em&gt;did not&lt;/em&gt; say that the stars are without significance; to me, this has the reverse-psychological effect of saying &quot;The shape of each star is important.&quot; There are exactly 4 of each specific shape, and all of the shapes are the same size. It could be that it has something to do with the number of points on each shape. If there's a mathematical proportion involved, then the circles would probably eliminate multiplication processes.&lt;br&gt;&#xA; 2) &lt;strong&gt;The colors.&lt;/strong&gt; The colors obviously mean something, or else it would not have been important for the characters in the story to notice them. There are 11 colors (if you count black as a color), and there are exactly four shapes of each color, split into pairs of two. The significance of this is uncertain. If you &lt;em&gt;don't&lt;/em&gt; count black as a color, then it could be that the goal of the image is to figure out what color they &lt;em&gt;should&lt;/em&gt; be by pattern analysis.&lt;br&gt;&#xA; 3) &lt;strong&gt;Their positions.&lt;/strong&gt; The stars are all scattered about on the image. Their positions may or may not be arbitrary - this might be the significance of using stars, in that the shapes have a clearly defined center point - but if this puzzle can be solved without computer assistance, then their exact pixel definitions are probably unneeded.&lt;br&gt;&#xA; 4) &lt;strong&gt;The number at the bottom.&lt;/strong&gt; The number shows 130. What does the number signify? It probably has something to do with the Rebus aspect of the puzzle, but anything farther than that I have not yet determined.&lt;br&gt;&#xA; 5) &lt;strong&gt;The solution itself.&lt;/strong&gt; Some hints were given about the solution. For one, the word must be a word. This means it will have proper spelling. Secondly, it has exactly 6 letters. This may mean that we must deduce each of the six letters independently. Lastly, all 6 letters will be different.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Based on some of the analysis above and the lack of connections being made, I have no strong guesses. However, I will offer a guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;SQUARE&lt;/strong&gt;. It satisfies the conditions of the 5th point mentioned above. I picked Square because among the 2-dimensional shapes present in the image, the square has been skipped over. Arguably, the pentagon and multiple other shapes have also been skipped, and perhaps the closest 4-sided figure to a star would be a diamond/rhombus, but those do not have exactly six letters.&lt;br&gt;&#xA; More thought will be needed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9319" LastActivityDate="2015-03-07T17:24:25.313" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9942" PostTypeId="1" AcceptedAnswerId="10062" CreationDate="2015-03-07T19:38:43.097" Score="13" ViewCount="912" Body="&lt;p&gt;Look, where I walk!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Listen, when I speak!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Babyface, together with a dog?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Skull, together with a cat and a mouse?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first time I was a King.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second time my mother was a Queen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I have also been the Joker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the future I will awake as a Chess champion again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where did my father serve?&lt;/p&gt;&#xA;" OwnerUserId="9385" LastEditorUserId="5373" LastEditDate="2015-03-07T20:00:58.003" LastActivityDate="2015-03-11T13:08:02.497" Title="Where did my father serve?" Tags="|riddle|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9943" PostTypeId="2" ParentId="9929" CreationDate="2015-03-07T19:40:11.120" Score="1" Body="&lt;p&gt;Not the most secure way, but it is possible for them to let you log in via some from of remote desktop connection, and then you type it in somewhere where it hides the characters with * or similar - note that this is would be susceptible to a software keylogger, possibly browsers saving passwords, they would see the length, etc. &lt;/p&gt;&#xA;" OwnerUserId="4474" LastActivityDate="2015-03-07T19:40:11.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9948" PostTypeId="2" ParentId="9934" CreationDate="2015-03-07T22:00:42.033" Score="14" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;2048&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each move changes the number of black numbers remaining by 0 or 2, and thus cannot change the parity. Thus initial states with an even number of black numbers can never be brought to the desired state. Now we claim that every initial state with an odd number of black numbers can be brought to the required state. By repeatedly using the moves RBR -&gt; BBB and BBR -&gt;RBB, it is possible to move to a state where all the black numbers are together (To see this, note that B_RBB can be turned into B_BBR, and B_RBR to B_BBB). The following steps can be used to reduce the chunks:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;B&lt;/li&gt;&#xA;&lt;li&gt;BBB -&gt; RBR = B&lt;/li&gt;&#xA;&lt;li&gt;BBBBB -&gt; RBRBB -&gt; RBBBR = BBB&lt;/li&gt;&#xA;&lt;li&gt;BBBBBBB -&gt; RBRBBBB -&gt; RBBBRBB -&gt; RBBBBBR = BBBBB&#xA;and so on&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thus all 2048 stateswith odd number of black numbers can be reached&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This code which counts the number of reachable states confirms  the result:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;char seen[1&amp;lt;&amp;lt;12];&#xA;&#xA;int dfs(int state)&#xA;{&#xA;  seen[state]=1;&#xA;  int ret=1;&#xA;  for(int i=0;i&amp;lt;12;i++)&#xA;  {&#xA;    if((state&amp;amp;(1&amp;lt;&amp;lt;i))==0)continue;&#xA;    int l=(i+1)%12,r=(i+11)%12;&#xA;    int nextstate=state^(1&amp;lt;&amp;lt;l)^(1&amp;lt;&amp;lt;r);&#xA;    if(!seen[nextstate])&#xA;      ret+=dfs(nextstate);&#xA;  }&#xA;  return ret;&#xA;}&#xA;&#xA;int main()&#xA;{&#xA;  int cnt=0;&#xA;  for(int i=0;i&amp;lt;12;i++)&#xA;  {&#xA;    if(!seen[1&amp;lt;&amp;lt;i])cnt+=dfs(1&amp;lt;&amp;lt;i);&#xA;  }&#xA;&#xA;  printf(&quot;%d\n&quot;,cnt);&#xA;  return 0;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="9471" LastActivityDate="2015-03-07T22:00:42.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9949" PostTypeId="2" ParentId="9155" CreationDate="2015-03-07T22:18:09.467" Score="21" Body="&lt;p&gt;I have no idea who can win, but I do want to clear up one point that some other people seem to be overlooking. Namely, the lions can always surround some plurality of zebras so any strategy to keep the zebras on the outside is doomed by construction.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let $n \ge 100$ be the number of zebras and the number of lions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\text{net}$ be the convex hull of the lions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $C_\text{net}$ be the largest origin-centred circle that is contained within this convex hull. This may not exist, in which case we consider it to be of radius 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $C_\text{zebra}$ be the minimal origin-centred bounding circle for some subset $Z$ of the zebras.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fUsij.png&quot; alt=&quot;Zebra Net&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Claim:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lions can force $\text{radius}(C_\text{net}) &amp;gt; k \cdot \text{radius}(C_\text{zebra})$ for some $k &amp;gt; 10$ and a nonzero subset $Z$ regardless of the zebras' strategy and initial placement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $n$ is increased, you can have an arbitrary number of zebras inside $C_\text{zebra}$ with an arbitrary number of lions on the perimeter of $\text{net}$ and $k$ can be arbitrarily increased.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without loss of generality, assume the lions go first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a circle centred on the origin containing all of the initial positions of the zebras and lions. Let its radius be $\delta$ and let $\omega \gg \delta$ and $\omega \gg 100$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have some small subset of lions of cardinality at least $l \ge 3$ move to equidistributed points at radius $\omega$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We select some small number $z$ of zebras and examine the maximum radius of $C_\text{zebra}$ which doesn't contain $z$ zebras. This is obviously the maximum radius of the $z$th-smallest distance from the origin to each zebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Evidently, this is maximized by having $z - 1$ zebras not move and having $n - z + 1$ zebras move an equal distance from an initial offset of $\delta$. There are rounding errors since the distance might not be a multiple of 100, but since $\omega \gg 100$ this is unimportant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have that the radius is at best $\frac{l \cdot (\omega + 2\delta + 100)}{n - z + 1} + \delta + 100$ where $l \cdot (\omega + 2\delta + 100)$ is approximately the distance moved by the zebras in the time it took the lions to get into place. This approximates to $\frac{l \cdot \omega}{n - z + 1} + \delta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radius of $C_\text{net}$ relative to $\omega$ is worst in the case of a triangle, in which case it is $\omega / 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that $k$ is at worst&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{\text{radius}(C_\text{net})}{\text{radius}(C_\text{zebra})} \approx&#xA;\frac{\omega / 2}{\frac{l \cdot \omega}{n - z + 1} + \delta}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n = 100$, $z = 3$, $l = 3$, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{\omega}{\frac{3}{49} \omega + 2 \delta} \approx \frac{\omega}{\frac{3}{49} \omega} \approx 15&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously as $n$ increases, the fraction $\frac{l \cdot \omega}{n - z + 1}$ will decrease. By increasing $\omega$ in terms of $\delta$ you can thus make this arbitrarily large even for arbitrary $l$ and $z$. A rigorous proof is up to the reader.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So &quot;trapping&quot; a zebra is easy... what matters is if you can do anything with that knowledge.&lt;/p&gt;&#xA;" OwnerUserId="9960" LastEditorUserId="9960" LastEditDate="2015-03-07T22:23:35.500" LastActivityDate="2015-03-07T22:23:35.500" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="9950" PostTypeId="1" AcceptedAnswerId="9951" CreationDate="2015-03-08T00:21:25.653" Score="-9" ViewCount="433" Body="&lt;p&gt;I am a fraction.&#xA;If you multiply me by two, the sum of me is 54.&#xA;When you divide my numerator by 4, you get 6.&#xA;When you divide my denominator by 6, you get 5.&#xA;What decimal am I?&lt;/p&gt;&#xA;" OwnerUserId="9993" LastActivityDate="2015-03-08T00:48:34.187" Title="Fraction Puzzle" Tags="|mathematics|" AnswerCount="1" CommentCount="3" ClosedDate="2015-03-08T22:31:15.167" ContentLicense="CC BY-SA 3.0" />
  <row Id="9951" PostTypeId="2" ParentId="9950" CreationDate="2015-03-08T00:48:34.187" Score="2" Body="&lt;p&gt;The only possible answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\frac{24}{30}=\frac{4}{5}$ = &lt;strong&gt;0.8&lt;/strong&gt;,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because if dividing the numerator by 4 gives 6, then the numerator must be 24, while if dividing the denominator by 6 gives 5, then the denominator must be 30.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know what the 54 bit means though. I guess it could mean that $24+30=54$, but why multiply by 2? The best interpretation I can come up with is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am $\frac{24}{30}=\frac{12}{15}$, so multiplying 'me' (both parts of me) by 2 gives $\frac{2\times 12}{2\times 15}=\frac{24}{30}$, whose 'sum' (numerator plus denominator) is $24+30=54$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-03-08T00:48:34.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9952" PostTypeId="2" ParentId="9929" CreationDate="2015-03-08T02:28:03.487" Score="4" Body="&lt;p&gt;Have him change the key layout to Dvorak first. Then tell him which letters to type on the normal layout. E.g, if your password is &quot;horse staple&quot;, you'd have him type &quot;jso;d ;karpd&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this relies on him not typing into a text editor, nor remembering the random sequence of letters to translate later.&lt;/p&gt;&#xA;" OwnerUserId="4466" LastActivityDate="2015-03-08T02:28:03.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9953" PostTypeId="2" ParentId="9929" CreationDate="2015-03-08T03:43:45.470" Score="-1" Body="&lt;p&gt;But one really usefull idea is create a pression group (aka lobbying) to incentivize big e-mail providers to create one-way (single use, aka disposable) timed (good only for 5 minutes) restricted (only receive or only send up to a limited number e-mails) passwords!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you need a friend do access your e-mail account. Ok, give him/her a disposable password that is only good for send/receive e-mail up to the five minutes period with no access to the account settings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some one could get rich making things this way :).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anything else mentioned is a crude adventure to be doomed to failure.&lt;/p&gt;&#xA;" OwnerUserId="9997" LastActivityDate="2015-03-08T03:43:45.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9954" PostTypeId="1" AcceptedAnswerId="9955" CreationDate="2015-03-08T11:50:27.107" Score="-1" ViewCount="389" Body="&lt;p&gt;Someone has fiddled with my Rubik's cube and messed it up only very minorly. Exactly 3 minicubes are out of place: the up-front, up-back, and down-front edge minicubes need to be cycled round. I remember having learnt a simple way to do this cycle, but can't remember what it is!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone help me with this?&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-03-08T12:09:05.193" Title="How to cycle these 3 minicubes on a Rubik's cube?" Tags="|rubiks-cube|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9955" PostTypeId="2" ParentId="9954" CreationDate="2015-03-08T11:59:52.240" Score="3" Body="&lt;p&gt;First, turn the cube so that the edges are in the UL, UR, and DR positions (upper-left, upper-right, down-right). Then, execute this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;F B' R2 F' B U2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Make sure when you do this that &lt;code&gt;F&lt;/code&gt; and &lt;code&gt;B'&lt;/code&gt; rotate the two faces in the &lt;em&gt;same&lt;/em&gt; direction if you're looking at the cube from the front. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it didn't work the first time, do it again and it should be solved.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here's a brief explanation of why this works. The algorithm performs a couple swaps on pairs of edges. You have the upper front row of three pieces, and the upper back row. The first two moves (&lt;code&gt;F B'&lt;/code&gt;) move them next to two edges which need swapping. Then, &lt;code&gt;R2&lt;/code&gt; swaps the two rows as well as the UR and DR edge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;F' B&lt;/code&gt; then moves the two swapped sets back, and &lt;code&gt;U2&lt;/code&gt; swaps the two sets as well as the UL and UR edges. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All in all, this algorithm moves UR to DR, DR to UL, and UL to UR. Since there are two unsolved positions, if you didn't solve it the first time, you can just execute it again and it will take you to the solved position.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorDisplayName="user20" LastEditDate="2015-03-08T12:09:05.193" LastActivityDate="2015-03-08T12:09:05.193" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9956" PostTypeId="1" AcceptedAnswerId="9958" CreationDate="2015-03-08T16:01:38.553" Score="0" ViewCount="25696" Body="&lt;p&gt;What runs but never walks,&lt;br&gt;&#xA;has a mouth but cannot talk,&lt;br&gt;&#xA;has a head but doesn't think&lt;br&gt;&#xA;and has a bed but refuses to sleep?&lt;/p&gt;&#xA;" OwnerUserId="435" LastEditorUserId="435" LastEditDate="2015-03-08T16:24:30.970" LastActivityDate="2015-03-08T16:24:30.970" Title="Riddle: What am I? Part 1" Tags="|riddle|" AnswerCount="1" CommentCount="0" ClosedDate="2015-03-08T22:00:02.663" ContentLicense="CC BY-SA 3.0" />
  <row Id="9958" PostTypeId="2" ParentId="9956" CreationDate="2015-03-08T16:15:29.397" Score="1" Body="&lt;p&gt;I think it is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A river&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reason:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The river runs, river mouth, river head, and river bed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9385" LastActivityDate="2015-03-08T16:15:29.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9959" PostTypeId="1" AcceptedAnswerId="9960" CreationDate="2015-03-08T16:24:44.823" Score="-1" ViewCount="237" Body="&lt;p&gt;Although a human shape I wear,&lt;br&gt;&#xA;Mother I never had;&lt;br&gt;&#xA;And though no sense nor life I share,&lt;br&gt;&#xA;in finest silks I’m clad.&lt;br&gt;&#xA;By every miss I’m valued much,&lt;br&gt;&#xA;beloved and highly prized;&lt;br&gt;&#xA;Still my cruel fate is such by boys I am often despised.&lt;/p&gt;&#xA;" OwnerUserId="435" LastActivityDate="2015-03-21T20:27:20.303" Title="Riddle: What am I? Part 2" Tags="|riddle|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9960" PostTypeId="2" ParentId="9959" CreationDate="2015-03-08T16:32:23.157" Score="4" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Doll&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dolls have human shapes, but no mother. You typically dress dolls up in fine clothes. Boys (especially older brothers) tend to rip the heads off dolls.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4353" LastActivityDate="2015-03-08T16:32:23.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9962" PostTypeId="1" AcceptedAnswerId="9966" CreationDate="2015-03-08T17:12:23.620" Score="8" ViewCount="1679" Body="&lt;p&gt;Today I heard a rumor that my bank will go bankrupt soon. &#xA;So I went to my bank, and indeed I found out that they have introduced some weird new rules that clearly indicate that there is something wrong with this bank.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I am allowed to withdraw exactly 300 Euro from my account.&lt;/li&gt;&#xA;&lt;li&gt;I am allowed to deposit exactly 132 Euro into my account.&lt;/li&gt;&#xA;&lt;li&gt;I am allowed to do such withdrawals and deposits as often as I like.&lt;/li&gt;&#xA;&lt;li&gt;I am not allowed to do any other transactions with my account.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;On my bank account, there are currently 1000 Euro.&#xA;I am certainly not willing to deposit any additional money into the bank account, but I am willing to play around with the money that currently is on the account.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&#xA;What is the largest amount of money that I can withdraw from the bank?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-18T09:55:29.723" LastActivityDate="2023-07-25T02:03:11.187" Title="Is my bank going to be bankrupt?" Tags="|mathematics|calculation-puzzle|reachability|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9964" PostTypeId="2" ParentId="9962" CreationDate="2015-03-08T17:36:43.377" Score="5" Body="&lt;p&gt;First, this problem can be simplified into $(250,75,33)$ instead of $(1000,300,132)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we deposit 22 times and withdrawl 13 times, we will have exactly 1 left in the bank which when you remove my simplification is \$4 which means you withdrew &lt;strong&gt;\$996&lt;/strong&gt; of you \$1000.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will require withdrawing everything you can, depositing everything you can, and then repeating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$250=75w-33d$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w=(250+33d)/75 = 10/3+11d/25$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying for values of d, the remaining balence value for 22 is the smallest for any value less than 10,000 deposits.  This seems to be because there are no integer solutions for the equation above.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditDate="2015-03-08T17:42:47.307" LastActivityDate="2015-03-08T17:42:47.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9965" PostTypeId="2" ParentId="9962" CreationDate="2015-03-08T17:47:18.110" Score="11" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 996 euros&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because what you need to do is find non-negative integers $a$ (number of withdrawals) and $b$ (number of deposits) such that $1000-300a+132b$ (amount left at the end) takes the minimal possible non-negative value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1000-300a+132b$ is always a multiple of 4, and we can make it exactly 4 by setting $a$ to be 13 and $b$ to be 22. So 4 is the minimal possible value of the amount left in the bank, which means the answer to your question is 996 - although the answer to the &lt;em&gt;title&lt;/em&gt; of your question is &lt;strong&gt;yes&lt;/strong&gt; ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More explicitly, here's exactly what you need to do:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;withdraw 900 euros ($a=3$, $b=0$), leaving 100 euros in your account&lt;/li&gt;&#xA;&lt;li&gt;deposit 792 euros ($a=3$, $b=6$), leaving 892 euros in your account&lt;/li&gt;&#xA;&lt;li&gt;withdraw 600 euros ($a=5$, $b=6$), leaving 292 euros in your account&lt;/li&gt;&#xA;&lt;li&gt;deposit 660 euros ($a=5$, $b=11$), leaving 952 euros in your account&lt;/li&gt;&#xA;&lt;li&gt;withdraw 900 euros ($a=8$, $b=11$), leaving 52 euros in your account&lt;/li&gt;&#xA;&lt;li&gt;deposit 924 euros ($a=8$, $b=18$), leaving 976 euros in your account&lt;/li&gt;&#xA;&lt;li&gt;withdraw 900 euros ($a=11$, $b=18$), leaving 76 euros in your account&lt;/li&gt;&#xA;&lt;li&gt;deposit 528 euros ($a=11$, $b=22$), leaving 604 euros in your account&lt;/li&gt;&#xA;&lt;li&gt;withdraw 600 euros ($a=13$, $b=22$), leaving 4 euros in your account (the minimum possible)&lt;/li&gt;&#xA;&lt;li&gt;take your 996 euros and &lt;strong&gt;run&lt;/strong&gt;!&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-08T18:09:21.187" LastActivityDate="2015-03-08T18:09:21.187" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9966" PostTypeId="2" ParentId="9962" CreationDate="2015-03-08T17:53:43.377" Score="13" Body="&lt;p&gt;The GCD of 300 and 132 is $12$. So  the quantity you can take out is always a multiple of $12$. In theory we can take out at most $83\cdot 12=996$, leaving four in the bank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every day make sure you deposit exactly  nine times and withdraw exactly $4$ times. This is possible as long as you have $16$ or more dollars in the bank. After this we end up having $12$ less than at the beginning of the day in the account. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we do this $83$ days at the end we will have $83\times 12=996$ less coins in the account than at the beginning, so we will leave only $4$ dollars in the bank.&lt;/p&gt;&#xA;" OwnerDisplayName="user10001" LastEditorDisplayName="user10001" LastEditDate="2015-03-08T18:05:02.627" LastActivityDate="2015-03-08T18:05:02.627" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9968" PostTypeId="1" AcceptedAnswerId="10051" CreationDate="2015-03-08T19:38:54.947" Score="8" ViewCount="761" Body="&lt;p&gt;&lt;strong&gt;Follow the numbers to answer the question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Number A:&lt;/strong&gt; The order of Romeos passenger in the Greek alphabet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Number B:&lt;/strong&gt; The two last digits in the year the returning Lex Luthor was awarded his (number A) Oscar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Number C:&lt;/strong&gt; The man in B didn't manage to complete in this city on time, and could have called in his complaint on this number, in &lt;em&gt;Apple time&lt;/em&gt;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Number D:&lt;/strong&gt; The number of the street in the city of C where marble is buried.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Number E:&lt;/strong&gt; A &lt;em&gt;European convention&lt;/em&gt; was held in a city in Italy sometime in the 20th century, and they discussed a partial golden retreat. In March 2007 the football club from the same city was in the (number D) league of the country. The two last digits of the year before, and the two last digits of the year after the conference is the number of this clue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Number F:&lt;/strong&gt; In the episode (number E) of Sesame Street, Kermit the frog appears this number of times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Who was the shooter?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="9385" LastEditorUserId="9385" LastEditDate="2015-03-09T16:06:30.607" LastActivityDate="2016-09-04T21:52:49.190" Title="Who was the shooter?" Tags="|enigmatic-puzzle|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9969" PostTypeId="2" ParentId="6618" CreationDate="2015-03-08T19:49:43.260" Score="3" Body="&lt;p&gt;&lt;strong&gt;Partial answer to the last two outstanding clues.&lt;/strong&gt; (Well, the first one is more like a guess.) Really the main purpose of this answer is to bump this excellent question back onto the P.SE front page. It's been around for a &lt;em&gt;long&lt;/em&gt; time without being fully solved!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;3 Think of the Phoenician, a possible drench, bleached prisons, or beach spindles (8,7: *******R *H*****)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Interpreting &quot;the Phoenician&quot; as&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; not someone from ancient Phoenicia but someone from Phoenix, Arizona,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I found that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the most famous &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_people_from_Phoenix&quot; rel=&quot;nofollow&quot;&gt;person from Phoenix&lt;/a&gt; is surely Muhammad Ali.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now for the rest:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;a possible drench&quot;, &quot;bleached prisons&quot;, and &quot;or beach spindles&quot; are all anagrams and 15 letters long (credit to &lt;strong&gt;@Roger&lt;/strong&gt; for putting me onto this), so the final answer is probably also an anagram of these.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How about the other one?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;5 Initially rebellious sheep undertake a silent infantry operation, inside a small enclosure, or so I'm induced to believe (10: ****U*****)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Initially rebellious&quot; gives&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; R (the first letter of the word &quot;rebellious&quot;),&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and &quot;sheep&quot; could mean&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; U (as in ewe).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I'm getting something like&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ru[?]pen[?]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-10T10:43:19.813" LastActivityDate="2015-03-10T10:43:19.813" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="9970" PostTypeId="2" ParentId="9968" CreationDate="2015-03-08T19:57:15.703" Score="4" Body="&lt;p&gt;&lt;strong&gt;First clue&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A: The order of Romeos passenger in the Greek alphabet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alfa Romeo is a car manufacturer; cars have passengers, and alpha is the 1st letter of the Greek alphabet, associated with Romeo.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second clue&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;B: The two last digits in the year the returning Lex Luthor was awarded his (number of clue A) Oscar.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;s&gt;99&lt;/s&gt; 95,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Kevin Spacey played Lex Luthor in &lt;em&gt;Superman Returns&lt;/em&gt;, and won his 1st Oscar in &lt;s&gt;1999&lt;/s&gt; 1995 (&lt;strong&gt;thanks @dmg!&lt;/strong&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Third and fourth clues&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;C. The man in B didn't manage to complete here on time, and could have called in his complaint on this number, in &lt;em&gt;Apple time!&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;D. The street in the city of C where marble is buried.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since C is a city, the reference to apples makes me think it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; New York,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;giving the answer to D as&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2nd Street, because of the &lt;a href=&quot;http://en.wikipedia.org/wiki/New_York_City_Marble_Cemetery&quot; rel=&quot;nofollow&quot;&gt;marble cemetery&lt;/a&gt; located there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fifth clue&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;E. A European convention was held in a city in Italy sometime in the 20th century, and they discussed a partial golden retreat. In March 2007 the football club from the same city was in the (number of clue D) league. The two last digits of the year before, and the two last digits of the year after the conference is the number of this clue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know almost nothing of football, but &lt;a href=&quot;http://en.wikipedia.org/wiki/2006%E2%80%9307_UEFA_Champions_League&quot; rel=&quot;nofollow&quot;&gt;this Wikipedia page&lt;/a&gt; tells me that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the only Italian football club in the 2nd league in 2006-07 was from Milan.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've come up with 2 possibilities for a &lt;em&gt;European convention&lt;/em&gt; being held there in the 20th century. Firstly:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the European Population Conference on 4-8 September 1995. Footnote 11 on page 322 &lt;a href=&quot;https://books.google.co.uk/books?id=IhwkBgAAQBAJ&amp;amp;pg=PA322&amp;amp;lpg=PA322&amp;amp;dq=20th-century+european+conference+in+milan&amp;amp;source=bl&amp;amp;ots=a1aqg0Sxly&amp;amp;sig=LTZENXiyOHXgBYXvScCOKg01e3M&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=tHf9VNX4Lcj8UOOWhNgL&amp;amp;ved=0CCMQ6AEwAw#v=onepage&amp;amp;q=20th-century%20european%20conference%20in%20milan&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; mentions European mass migration in the 20th century - maybe this could be seen as a &quot;partial golden retreat&quot;? The catch is that there aren't &lt;strong&gt;9496&lt;/strong&gt; episodes in Sesame Street!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Secondly:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.jstor.org/discover/10.2307/2339981?sid=21105581956251&amp;amp;uid=3738032&amp;amp;uid=2&amp;amp;uid=2129&amp;amp;uid=70&amp;amp;uid=4&quot; rel=&quot;nofollow&quot;&gt;the International Conference on Unemployment in 1906&lt;/a&gt;, which would give the number &lt;strong&gt;507&lt;/strong&gt;. The catch is that I can't see how this fits &quot;partial golden retreat&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sixth clue&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;F. In the episode number (E) of Sesame Street, Kermit the frog appears this number of times.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I need to resolve the fifth clue for this to be any use!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Final answer&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We have &lt;strong&gt;1st&lt;/strong&gt;, &lt;strong&gt;99&lt;/strong&gt;, &lt;strong&gt;New York&lt;/strong&gt;, &lt;strong&gt;2nd Street&lt;/strong&gt;, ???, ???. Given the title of the question, I suspect the answer may have to do with a shooting on 2nd Street in New York, possibly on a date given by the numbers in the other clues.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2016-09-04T21:52:49.190" LastActivityDate="2016-09-04T21:52:49.190" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="9973" PostTypeId="1" AcceptedAnswerId="9994" CreationDate="2015-03-09T02:10:07.300" Score="1" ViewCount="471" Body="&lt;p&gt;The people on top don't have it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The people at the bottom do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't have it you die.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what is it?&lt;/p&gt;&#xA;" OwnerDisplayName="user10001" LastEditorUserId="5373" LastEditDate="2015-03-09T10:28:56.920" LastActivityDate="2015-03-10T18:33:51.887" Title="The people on top don't have it" Tags="|riddle|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9974" PostTypeId="1" AcceptedAnswerId="9978" CreationDate="2015-03-09T04:01:05.720" Score="0" ViewCount="141" Body="&lt;p&gt;People almost always look down on me&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet I have been controlling them more and more as of late&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I disappear you all feel alone.&lt;/p&gt;&#xA;" OwnerDisplayName="user10001" LastEditorUserId="5373" LastEditDate="2015-03-09T10:29:35.567" LastActivityDate="2015-03-09T13:46:41.893" Title="Controling from below" Tags="|riddle|word|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9975" PostTypeId="2" ParentId="9974" CreationDate="2015-03-09T05:02:22.520" Score="0" Body="&lt;p&gt;Answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;EARTH&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People almost always look down on me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;People always look down to earth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet I have been controlling them more and more as of late&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Earth is always controlling us. Especially lately when we die and buried.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I disappear you all feel alone.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Once you fly up (on an aircraft or when people die) from earth one will definitely feel alone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastActivityDate="2015-03-09T05:02:22.520" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9976" PostTypeId="2" ParentId="9243" CreationDate="2015-03-09T05:26:35.237" Score="5" Body="&lt;p&gt;The geographical property is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;having the capital city located in the Western half of the country.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Chad and Yemen have this enigmatic property X.&lt;br&gt;&#xA;Brazil, Denmark and the USA do not have this enigmatic property X.&lt;/p&gt;&#xA;" OwnerDisplayName="user10001" LastEditorUserId="9125" LastEditDate="2015-03-15T15:30:41.547" LastActivityDate="2015-03-15T15:30:41.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9978" PostTypeId="2" ParentId="9974" CreationDate="2015-03-09T06:15:36.357" Score="1" Body="&lt;p&gt;Answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Phone/mobile device&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;People always look down on me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Texting/gaming&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet I have been controlling them more and more as of late&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People are getting addicted. It's a common conception amongst the older generations that we are being &quot;controlled&quot; by our phones/mobile devices.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I disappear, you all feel alone&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No phone = no social media = very lonely and bored and alone. Only a millennial would understand the feeling :P &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8999" LastActivityDate="2015-03-09T06:15:36.357" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9979" PostTypeId="1" AcceptedAnswerId="9981" CreationDate="2015-03-09T06:50:32.083" Score="23" ViewCount="1407443" Body="&lt;p&gt;This one comes from this week's Brooklyn Nine-Nine episode!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an island with 12 islanders. All of the islanders individually weigh exactly the same amount, except for one, who either weighs more or less than the other 11. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You must use a see-saw to figure out whose weight is different, and you may only use the see-saw 3 times. There are no scales or other weighing device on the island. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can you find out which islander is the one that has a different weight? &lt;/p&gt;&#xA;" OwnerUserId="10006" LastEditorUserId="8874" LastEditDate="2015-10-07T17:22:30.273" LastActivityDate="2018-09-27T02:07:07.170" Title="Brooklyn 99 riddle: Weighing Islanders" Tags="|mathematics|logical-deduction|weighing|" AnswerCount="3" CommentCount="9" ClosedDate="2015-03-09T09:49:27.467" ContentLicense="CC BY-SA 3.0" />
  <row Id="9980" PostTypeId="1" AcceptedAnswerId="9982" CreationDate="2015-03-09T07:08:50.360" Score="25" ViewCount="3214" Body="&lt;p&gt;Two people are travelling together. One of them has three loaves of bread; &#xA;the other one has two. At some point they stop to eat lunch. A third traveler that happens to walk by joins them.&#xA;They split the five loaves into three equal portions and eat.&#xA;Before leaving, the third traveler gives them five dollars for sharing the bread. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The traveler who had three loaves wants to take three dollars and give the other guy two dollars. The other guy is not happy with the two dollars and suggests a 50/50 split. They end up going to a judge and asking for his opinion on how the money should be split. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does the judge decide about how the money should be split?&lt;/p&gt;&#xA;" OwnerUserId="9969" LastEditorUserId="8874" LastEditDate="2016-03-10T09:00:58.827" LastActivityDate="2016-03-10T09:00:58.827" Title="Five loaves and the extra traveller" Tags="|mathematics|calculation-puzzle|word-problem|" AnswerCount="2" CommentCount="5" ClosedDate="2023-07-15T11:40:49.977" ContentLicense="CC BY-SA 3.0" />
  <row Id="9981" PostTypeId="2" ParentId="9979" CreationDate="2015-03-09T07:09:48.483" Score="18" Body="&lt;p&gt;Divide them into 3 groups of 4 people.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put any two groups on each side of the see-saw. (First Use)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Condition 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If the see-saw balances, we are sure that the oddly wieghted one is in the other group of 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that case, take two people from that group and place them on one end of see-saw and two of the balanced eight on the other. (Second Use)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Condition 1.1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the see saw balances, remove all but one from the seesaw and put one of the remaining two opposite them. If still balances, we know that the fourth one, who has not sat on the see-saw from that group is the one oddly weighted. (Third Use)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Condition 1.2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the see saw is not balanced, remove one from each end. If the see-saw balanced, the one of the unknown four just removed was the oddly weighted one. Otherwise the one who stayed is the oddly weighted one.(Third Use)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Condition 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If the two groups of 4 don't balance remember which side was lighter, have three get off one end and the remaining person swap places with one of the other four. Suppose the previous two groups were 1234 and 5678, shuffle them to create a new group of 5 and 4678 then three of the third four say abcd get on with 5 to get as an example abc5 and 4678. (Second Use)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Condition 2.1.1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the position of seesaw does not change and as an example say 5678 and then 4678 are heavier, we know that either 6 or 7 or 8 is oddly weighted. Now put 7 on one end and 8 on the other.  If one is heavier they are the odd one otherwise it is 6. (Third Use)&#xA;&lt;em&gt;note&lt;/em&gt; this works equally well if the group was lighter, just replace terms for appropriate identification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Condition 2.1.2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the seesaw reverses, ether 4 or 5 is the oddly weighted one. put 4 on one end and anyone other than 5 on the other (Third Use), if it balances it is 5 otherwise it is 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Condition 2.1.3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the seesaw balances we know that either 1 or 2 or 3 is oddly weighted.  Say as example 1234 were lighter.  Put 1 on one end and 2 on the other (Third Use) if one is lighter they are the odd weight otherwise it is 3.&#xA;&lt;em&gt;note&lt;/em&gt; this works equally well if the group was heavier, just replace terms for appropriate identification.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Done - easy peasy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easier than everyone makes it.  A seesaw is binary.  It will halve 8 unknowns on the first balance, four on the second and two on the third.  Set it up so deduction eliminates everything else and your gold.&#xA;As a bonus in all but one possibility you also know if the person was lighter or heavier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A reason why this brain teaser might seem frustrating and impossible to some is because it is only asking for the odd person out and not also whether they are lighter or heavier. It is impossible to know both for sure in only three steps.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: &#xA;In 11/12 cases you know whether the person is lighter or heavier as the seasaw dictates it. The only case where you don’t is 1.1.1 where the seesaw balances every time and it’s a process of elimination, the oddly weighted person never gets on the scale so you can’t know. &lt;/p&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="-1" LastEditDate="2018-09-27T02:07:07.170" LastActivityDate="2018-09-27T02:07:07.170" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="9982" PostTypeId="2" ParentId="9980" CreationDate="2015-03-09T07:15:49.463" Score="46" Body="&lt;p&gt;Such an old chestnut!&#xA;Each of them eats 5/3 loaves.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The first traveler contributes 3 loaves, eats 5/3 himself, and gives 4/3 to the extra traveler.&lt;/li&gt;&#xA;&lt;li&gt;The second traveler contributes 2 loaves, eats 5/3 himself, and gives 1/3 to the extra traveler.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Since the first traveler gives 4 times as much bread to the extra guy, he should also receive 4 times as much money.  Hence the judge assigns 4 dollars to the first traveler and 1 dollar to the second traveler.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-03-11T13:06:17.187" LastActivityDate="2015-03-11T13:06:17.187" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9986" PostTypeId="2" ParentId="9979" CreationDate="2015-03-09T09:05:48.177" Score="3" Body="&lt;p&gt;I can do this in one move, where no one gets off the see saw, and they only get on it once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a see saw, not a giant scale! &#xA;It is of indeterminate length I wish for my see saw to be long enough to place 12 on either side, but only six will be on either side of the fulcrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I have islanders 1-6 on side &quot;A&quot; and 7-12 on side &quot;B&quot;  both groups of six are as close to the fulcrum of the see saw in a single file line as possible, one side will fall, one side will rise.&lt;br&gt;&#xA;The side that rises will &quot;scootch&quot; down the length of the see saw, away from the fulcrum until both sides balance.&lt;br&gt;&#xA;The side that is lighter has the men switch positions getting closer or further from the fulcrum without coming on or off the see saw, after all of the positions have been tried, if the see saw does not move, it means all the men on that side are of equal weight, if the balance is lost, the last man to move before the balance is lost is the lighter man.&lt;br&gt;&#xA;If the light side men are all of equal weight, the heavy side is then ordered to switch positions until the balance is no longer achieved, when the balance is lost, the last man to move is the heavier man. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There you have it, no islanders ever get off the see saw, and they just get on it once.&lt;/p&gt;&#xA;" OwnerUserId="10016" LastEditorUserId="8874" LastEditDate="2015-10-07T17:23:56.973" LastActivityDate="2015-10-07T17:23:56.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9987" PostTypeId="2" ParentId="9979" CreationDate="2015-03-09T09:13:24.500" Score="6" Body="&lt;p&gt;OK, I think I have it, now the problem of explaining it, here goes: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are going to name the islanders 1 2 3 4 5 6 7 8 9 10 11 12&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are trying to find which of them is a non-standard weight or x = one of them.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;//are comments during explanation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use 1:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 2 3 4 against 5 6 7 8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I)  1 2 3 4 = 5 6 7 8 then Use 2:  9 against 10    //9 10 11 or 12 are x&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;     A) 9 &amp;gt; 10 or 9 &amp;lt; 10 then Use 3:  9 against 11    //9 or 10 are x&#xA;&#xA;          i)  9 &amp;gt; 11 or 9 &amp;lt; 11 then 9 = x&#xA;&#xA;         ii)  9 = 11  then 10 = x&#xA;&#xA;     B) 9 = 10 then Use 3:  9 against 11      //11 or 12 are x&#xA;&#xA;          i) 9 &amp;gt; 11 or 9 &amp;lt; 11 then 11 = x&#xA;&#xA;         ii) 9 = 11 then 12 = x&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;//Ok four down, eight to go, that was the easy part&lt;/p&gt;&#xA;&#xA;&lt;p&gt;II)  1 2 3 4 &gt; 5 6 7 8 then Use 2: 1 2 3 5 against 4 10 11 12  //10 11 12 are not x now&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  A) 1 2 3 5 &amp;gt; 4 10 11 12 then Use 3: 1 against 2  //1 2 or 3 are x now and x is heavier than the rest&#xA;&#xA;         i) 1 &amp;gt; 2 then 1 is x //x is heavier&#xA;&#xA;        ii) 1 &amp;lt; 2 then 2 is x&#xA;&#xA;       iii) 1 = 2 then 3 is x&#xA;&#xA;  B) 1 2 3 5 &amp;lt; 4 10 11 12 then Use 3:  4 against 12 // 4 or 5 is x. The switched 4 and 5 caused a reversal&#xA;&#xA;        i) 4 &amp;gt; 12 or 4 &amp;lt; 12 then 4 is x&#xA;&#xA;       ii) 4 = 12 then 5 is x&#xA;&#xA; C) 1 2 3 5 = 4 10 11 12 then Use 3:  6 against 7   //6 7 or 8 are x and lighter than the rest &#xA;&#xA;        i) 6 &amp;gt; 7 then 7 is x&#xA;&#xA;       ii) 6 &amp;lt; 7 then 6 is x&#xA;&#xA;      iii) 6 = 7 then 8 is x&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;III) 1 2 3 4 &amp;lt; 5 6 7 8  do the same process as II with appropriate adjustments being made.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way, x can be found, no matter which islander it is, as all 12 have a test to find an answer for.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that makes sense. :D &lt;/p&gt;&#xA;" OwnerUserId="10014" LastEditorUserId="10014" LastEditDate="2015-03-09T09:26:05.260" LastActivityDate="2015-03-09T09:26:05.260" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9988" PostTypeId="2" ParentId="9980" CreationDate="2015-03-09T09:56:49.260" Score="32" Body="&lt;p&gt;Another way of getting the same answer as Gamow is to have each traveller notionally pay \$5. That gives them a pot of \$15, from which they distribute the money to those who provided the bread. They have \$3 a loaf, so:&lt;br&gt;&#xA; - the first traveller pays \$5 and gets \$9, for a total of \$4&lt;br&gt;&#xA; - the second pays \$5 and gets \$6, for a total of \$1&lt;br&gt;&#xA; - the third just pays \$5.&lt;/p&gt;&#xA;" OwnerUserId="6031" LastActivityDate="2015-03-09T09:56:49.260" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9989" PostTypeId="1" AcceptedAnswerId="10068" CreationDate="2015-03-09T11:14:14.127" Score="5" ViewCount="476" Body="&lt;p&gt;You're part of a special intelligence operation trying to predict the location of the next strike for a notorious terrorist group. This group has proven that they are willing to kill innocent people in suicide bombings in order to further their 'cause' - spreading terror amongst the people. We have infiltrated their IRC network, which we believe they are using to plan and execute their next attack. Here is a snapshot of their most recent IRC activity:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[12:24:47] &amp;lt;+ackb1&amp;gt; so you know&#xA;[12:25:10] &amp;lt;+xeggs&amp;gt; ??&#xA;[12:25:15] &amp;lt;+ackb1&amp;gt; you know what were doing next&#xA;[12:25:23] &amp;lt;+ackb1&amp;gt; that thing&#xA;[12:25:24] &amp;lt;+ackb1&amp;gt; you know what i mean&#xA;[12:25:46] &amp;lt;+xeggs&amp;gt; oh yeah...&#xA;[12:25:51] &amp;lt;+xeggs&amp;gt; :)&#xA;[12:25:55] &amp;lt;+ackb1&amp;gt; where are we doing it&#xA;[12:26:00] &amp;lt;+xeggs&amp;gt; AGLAZP YSW3ABCGMF 2QJ QBMKR LU5AM 2kTPMWI R2IWUN U6L KZRYEFQCA&#xA;[12:26:03] &amp;lt;+xeggs&amp;gt; AB 155d 86aiiai jjjj&#xA;[12:26:12] &amp;lt;+xeggs&amp;gt; zz4!!! &#xA;[12:26:16] &amp;lt;+ackb1&amp;gt; how do i read that&#xA;[12:26:21] &amp;lt;+xeggs&amp;gt; you know...&#xA;[12:26:25] &amp;lt;+xeggs&amp;gt; name of my favourite restaurant&#xA;[12:26:45] &amp;lt;+xeggs&amp;gt; as key ;)&#xA;[12:27:00] &amp;lt;+ackb1&amp;gt; haha i see&#xA;[12:27:02] &amp;lt;+ackb1&amp;gt; well &#xA;[12:27:17] &amp;lt;+ackb1&amp;gt; very nice :)&#xA;[12:27:30] &amp;lt;+ackb1&amp;gt; will you be there?&#xA;[12:27:32] &amp;lt;+xeggs&amp;gt; no i need to organise&#xA;[12:27:52] &amp;lt;+xeggs&amp;gt; im sure ull be right&#xA;[12:28:00] &amp;lt;+ackb1&amp;gt; this time we can make a difference&#xA;[12:28:05] &amp;lt;+xeggs&amp;gt; show em what we're made of? xd&#xA;[12:28:21] &amp;lt;+ackb1&amp;gt; wait&#xA;[12:28:26] &amp;lt;+ackb1&amp;gt; what time??&#xA;[12:28:40] &amp;lt;+xeggs&amp;gt; %%^pp00&amp;gt;?`~&#xA;[12:29:00] &amp;lt;+ackb1&amp;gt; oh&#xA;[12:29:01] &amp;lt;+ackb1&amp;gt; ok&#xA;[12:29:08] &amp;lt;+ackb1&amp;gt; will u be there?&#xA;[12:29:23] &amp;lt;+xeggs&amp;gt; nope&#xA;[12:29:37] &amp;lt;+xeggs&amp;gt; i gotta organise the next one&#xA;[12:29:56] &amp;lt;+ackb1&amp;gt; ah i see&#xA;[12:30:11] &amp;lt;+ackb1&amp;gt; shame, i dont see u very often&#xA;[12:30:24] &amp;lt;+ackb1&amp;gt; i might not even see you ever again... :(&#xA;[12:30:29] &amp;lt;+xeggs&amp;gt; ah dont worry&#xA;[12:30:34] &amp;lt;+xeggs&amp;gt; for a good cause&#xA;[12:30:54] &amp;lt;+xeggs&amp;gt; youre making a big difference&#xA;[12:31:14] &amp;lt;+xeggs&amp;gt; this movement could never survive&#xA;[12:31:22] &amp;lt;+xeggs&amp;gt; w/o people like you&#xA;[12:31:27] &amp;lt;+xeggs&amp;gt; =)&#xA;[12:31:45] &amp;lt;+ackb1&amp;gt; eehhhh i guess&#xA;[12:31:50] &amp;lt;+ackb1&amp;gt; well, thanks for telling me all the details&#xA;[12:31:51] &amp;lt;+ackb1&amp;gt; i gtg soon&#xA;[12:32:11] &amp;lt;+xeggs&amp;gt; WAIT&#xA;[12:32:17] &amp;lt;+xeggs&amp;gt; before u go&#xA;[12:32:26] &amp;lt;+xeggs&amp;gt; i need to tell you one more thing&#xA;[12:32:48] &amp;lt;+ackb1&amp;gt; ??????&#xA;[12:32:53] &amp;lt;+ackb1&amp;gt; what&#xA;[12:33:09] &amp;lt;+xeggs&amp;gt; A78djwBw!0enw?;;dw'_=&#xA;[12:33:22] &amp;lt;+xeggs&amp;gt; +dw[]aJWJw!w7#@dw^k&#xA;[12:33:42] &amp;lt;+xeggs&amp;gt; Na&amp;lt;&amp;gt;?&quot;]\Wj1aW&quot;&#xA;[12:33:57] &amp;lt;+xeggs&amp;gt; JJJJJ&#xA;[12:34:10] &amp;lt;+xeggs&amp;gt; =) =)&#xA;[12:34:25] &amp;lt;+ackb1&amp;gt; ahhahah&#xA;[12:34:43] &amp;lt;+ackb1&amp;gt; you made my day&#xA;[12:35:01] &amp;lt;+ackb1&amp;gt; ok cya&#xA;[12:35:16] &amp;lt;+xeggs&amp;gt; bye&#xA;[12:35:39] +ackb1 [~ackb1@hidden.obsfucateirc.net] has quit [leaving]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, from the IRC logs it looks like they are communicating in some sort of weird code. We have set our best cryptographers on it, but they came up with nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can't afford to make loose guesses as to their next target - lives are at stake!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Can you help determine the location and time of the terrorists' next strike?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="7308" LastActivityDate="2015-03-11T03:16:23.287" Title="The Art of Misdirection" Tags="|cipher|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9990" PostTypeId="2" ParentId="9968" CreationDate="2015-03-09T13:18:10.127" Score="0" Body="&lt;p&gt;A: The order of Romeos passenger in the Greek alphabet.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1st. As @Rand said, &quot;Alfa Romeo is a car manufacturer; cars have passengers, and alpha is the 1st letter of the Greek alphabet, associated with Romeo.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;B: The two last digits in the year the returning Lex Luthor was awarded his (number of clue A) Oscar.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;95. Since Kevin Spacey played who played Lex Luthor in Superman Returns won his 1st Oscar in 1995 for bets supporting actor for his role in Pulpe Fiction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;C. The man in B didn't manage to complete here on time, and could have called in his complaint on this number, in Apple time!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Microsoft. Since I think it's something related with the &quot;suit by Apple accused Microsoft, along with Intel and the San Francisco Canyon Company, in 1995 of knowingly stealing several thousand lines of QuickTime source code in an effort to improve the performance of Video for Windows.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;D. The street in the city of C where marble is buried.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;65th street Redmond. Redmond is the city in Washington where Microsoft resides. Then if you search for marbles or granite you will know that all the marble companies or service providers in Redmond are there. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;E. A European convention was held in a city in Italy sometime in the 20th century, and they discussed a partial golden retreat. In March 2007 the football club from the same city was in the (number of clue D) league. The two last digits of the year before, and the two last digits of the year after the conference is the number of this clue.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;13,15. Organo Gold Europe convention was held in 2014 on Rome, Italy. AS ROMA a football team from Rome played against OL which was playing its 65th champions league match.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;F. In the episode number (E) of Sesame Street, Kermit the frog appears this number of times.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Yet to find the solution for this clue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastActivityDate="2015-03-09T13:18:10.127" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9991" PostTypeId="2" ParentId="9974" CreationDate="2015-03-09T13:46:41.893" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The law&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;People almost always look down on me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it's always convenient to whine about the law and law enforcement&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet I have been controlling them more and more as of late&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you have to obey the law. &quot;as of late&quot; might be referring to surveillance agencies we are hearing of these days&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I disappear you all feel alone.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if we abolished the rules altogether it would be like in The Purge&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9413" LastActivityDate="2015-03-09T13:46:41.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9992" PostTypeId="2" ParentId="9973" CreationDate="2015-03-09T13:56:09.677" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Air and air pressure&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People at the Top... :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you are in space or very high in the air (an airplane perhaps), one you won't have any oxygen in space. Two, if the air pressure is not normal, you will die or receive critical injuries. That is why the air pressure is regulated in all airplanes. I wouldn't say the same for those climbing Mount Everest. They still survived or died due to being too cold or trapped in an avalanche.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;People at The Bottom... :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They have plenty of air. We need air to breathe and survive and to have plants and animals to consume.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Final Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Air and air pressure!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9042" LastEditorUserId="9042" LastEditDate="2015-03-09T14:05:03.907" LastActivityDate="2015-03-09T14:05:03.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9994" PostTypeId="2" ParentId="9973" CreationDate="2015-03-09T13:59:57.897" Score="3" Body="&lt;p&gt;It is too broad for a riddle it even might be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oxygen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The people on top don't have it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We're talking real heights where Oxygen is scarce&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The people at the bottom do.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everyone else has an oxygen supply&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you don't have it you die.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can't breath and you suffocate&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9907" LastActivityDate="2015-03-09T13:59:57.897" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9995" PostTypeId="2" ParentId="9942" CreationDate="2015-03-09T14:20:24.330" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Great Britain?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well, England has a monarchy so it matches where you can have a king and queen. Your father died when you were young and his wife was still Queen so you became King. Your mother still was Queen. England did host some of the &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_World_Chess_Championships&quot; rel=&quot;nofollow&quot;&gt;World Chess Championships&lt;/a&gt;, where some British players competed and even got into the finals. This proves that England used to be a great country for chess and the royal probably got involved &lt;strong&gt;&lt;em&gt;a lot&lt;/em&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9042" LastActivityDate="2015-03-09T14:20:24.330" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9996" PostTypeId="1" AcceptedAnswerId="10204" CreationDate="2015-03-09T14:21:23.193" Score="5" ViewCount="1357" Body="&lt;p&gt;My job is putting things in rings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My prize is rings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of rings I have symbolizes speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When people see me in action they say, &quot;Oh, neat!&quot;&lt;/p&gt;&#xA;" OwnerDisplayName="user10001" LastEditorUserId="5373" LastEditDate="2015-03-14T15:49:12.923" LastActivityDate="2015-03-14T15:49:12.923" Title="Rings, rings, rings and more rings" Tags="|riddle|word|" AnswerCount="10" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9997" PostTypeId="2" ParentId="9996" CreationDate="2015-03-09T14:28:42.217" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Olympics! Summer and Winter Edition!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My job is putting things in rings.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The logo of the Olympics is made of five &lt;strong&gt;&lt;em&gt;rings&lt;/em&gt;&lt;/strong&gt;. The medals you wear when you win an event could be a giant ring and the &lt;em&gt;things&lt;/em&gt; are the contestants.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My prize is rings.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Like I said before, the medals are kind of like rings. Just bigger and worn around the neck.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The number of rings I have symbolizes speed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Usain_Bolt&quot; rel=&quot;nofollow&quot;&gt;Usain Bolt&lt;/a&gt;, he world's fastest sprinter in the entire world, runs in 5 different events, holding the world records for two of the events he runs in. And there are 5 rings in the logo of the Olympics as said earlier.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When people see me in action they say Oh, neat!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When people see that the Olympics has started, people could say that. People just love watching the Olympics and cheering their team (country) on. The beginning of the Olympics is a show where people get to see all the teams in a wonderfully displayed stadium hence &quot; Oh, neat!&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9042" LastEditorUserId="9000" LastEditDate="2015-03-09T14:40:30.117" LastActivityDate="2015-03-09T14:40:30.117" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9999" PostTypeId="1" AcceptedAnswerId="10002" CreationDate="2015-03-09T14:40:10.353" Score="5" ViewCount="4558" Body="&lt;p&gt;Big I am, but small I eat,&lt;br&gt;&#xA;Above or below, I'm really neat,&lt;br&gt;&#xA;My brothers have one, but I have two,&lt;br&gt;&#xA;You will not find me at the zoo,&lt;br&gt;&#xA;I'll travel with others, but not for long,&lt;br&gt;&#xA;You'll travel to hear my song.&lt;/p&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="9907" LastEditDate="2015-03-09T14:50:03.730" LastActivityDate="2015-03-09T14:50:03.730" Title="What am I - &quot;Big&quot; Riddle" Tags="|riddle|word|rhyme|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10000" PostTypeId="1" AcceptedAnswerId="10003" CreationDate="2015-03-09T14:45:15.413" Score="19" ViewCount="7426" Body="&lt;p&gt;A semi truck weighing exactly 10,000 pounds gets onto a bridge that is 100 miles long. The bridge's actual weight limit (not the posted limit) is exactly 10,000 pounds; any more will break the bridge. After going about 80 miles down the bridge a small bird lands on the truck. What happens?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I.E., what happens to the truck and the bridge? I don't care what happens to the small bird or the driver &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here's the version with math that gives away the trick a bit but has less loopholes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A diesel semi-truck operating at the remarkable fuel economy of 10mpg begins to cross a bridge. At the moment immediately prior to the wheels contacting the bridge surface proper, the total weight of the truck and all its contents is exactly 10,000 pounds. The bridge can hold a maximum of exactly 10,003 pounds without collapsing. After traveling 80 miles along the 100 mile long bridge - all the while with 0 vertical motion whatsoever - an exceptionally large, full-grown male &lt;a href=&quot;http://en.wikipedia.org/wiki/Kori_bustard&quot;&gt;kori bustard&lt;/a&gt; lands on the top of the truck. Does the bridge collapse?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Clarifications: (Yes, it's an easy puzzle that's already answered so this isn't required. I'm an engineer, though, and I demand that such things be rigorous. Holes pointed out by you internet people must be mended!)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The given weight of the truck includes the truck itself as well as all things being carried by the truck such as the cargo, the driver, the driver's breakfast inside their stomach, the fuel, the cardboard box stuck in the axle, all the magical snails that live on the left front axle, etc. It is the weight that would be shown by a scale if the truck was sitting stationary on one at that moment.&lt;/li&gt;&#xA;&lt;li&gt;The bridge weight limit given is the actual bridge's limit and not the posted limit.&lt;/li&gt;&#xA;&lt;li&gt;For all you people from sane countries in which the metric system feels more natural, feel free to change the units without conversion. For all practical purposes, the question could have a truck of WEIGHT going over a bridge of GREAT LENGTH that has a weight limit of the same WEIGHT and it's traveled 80% of GREAT LENGTH when the bird lands.&lt;/li&gt;&#xA;&lt;li&gt;The driver is not consuming or excreting anything besides the air within the cabin which is open to the atmosphere outside the cabin. It is irrelevant if it is open based on open windows or the air conditioning.&lt;/li&gt;&#xA;&lt;li&gt;The truck has a diesel engine without any electrical source powerful enough to provide or significantly support locomotion.&lt;/li&gt;&#xA;&lt;li&gt;The bridge can be treated as a perfectly rigid object with all stress equally distributed. The total weight it can handle is as posted in the original question.&lt;/li&gt;&#xA;&lt;li&gt;During the truck's voyage across the bridge, there are no loads applied to the bridge besides the truck and the eventual landing of the bird. Yes, this means the bridge is the most sanitary outdoor construction of all time as no living things exist upon it's structure.&lt;/li&gt;&#xA;&lt;li&gt;The truck weight is taken as the driver has just inhaled a full breath of air. Note that the bird weighs more than the weight of such a breath.&lt;/li&gt;&#xA;&lt;li&gt;The bridge is perfectly smooth. The truck experiences no bumps or vibrations of any kind. There is no vertical movement of the truck at any point while traveling the bridge. Magic.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="39" LastEditDate="2015-03-10T23:18:39.513" LastActivityDate="2021-10-07T18:32:42.603" Title="A semi truck weighing exactly 10,000 pounds" Tags="|lateral-thinking|physics|" AnswerCount="5" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="10001" PostTypeId="2" ParentId="9996" CreationDate="2015-03-09T14:47:00.950" Score="-1" Body="&lt;p&gt;Is it...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sonic the Hedgehog game?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My job is putting things in rings.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I haven't played it so it might be his job&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My prize is rings.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He collects rings to unlock things?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The number of rings I have symbolizes speed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sonic the hedgehog is superfast&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When people see me in action they say Oh, neat!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Kids love watching him run&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9907" LastActivityDate="2015-03-09T14:47:00.950" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10002" PostTypeId="2" ParentId="9999" CreationDate="2015-03-09T14:47:22.600" Score="5" Body="&lt;p&gt;I would guess it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a humpback whale&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I can't quite figure out how &quot;My brothers have one, but I have two&quot; applies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reasons:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Big I am, but small I eat,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Whales are big, eat plankton (small)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Above or below, I'm really neat,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Whales swim (below) and can jump out of the water (above), and are cool.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My brothers have one, but I have two,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will not find me at the zoo,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Too big to have whales at the zoo&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll travel with others, but not for long,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Migrate with others for breeding, but then travel alone&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'll travel to hear my song.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Whales sing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-03-09T14:47:22.600" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10003" PostTypeId="2" ParentId="10000" CreationDate="2015-03-09T14:50:55.797" Score="40" Body="&lt;p&gt;By the time the truck has gone 80 miles, it has consumed a significant amount of fuel. At that point, the weight of the bird is not sufficient to bring the weight of the truck back up to 10,000 lbs.&lt;/p&gt;&#xA;" OwnerUserId="9828" LastActivityDate="2015-03-09T14:50:55.797" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="10004" PostTypeId="1" AcceptedAnswerId="10006" CreationDate="2015-03-09T15:04:44.537" Score="13" ViewCount="1827" Body="&lt;p&gt;First, here is the original question which was also discussed here in &lt;em&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/6569/a-five-card-trick-how-does-it-work&quot;&gt;Puzzling Beta&lt;/a&gt;&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;This is a two-person card trick. At the initial stage, the magician leaves the room. The magician's assistant takes a pack of playing cards (fifty-two cards, no joker), gives it to the audience, and they choose five cards in whichever way they prefer. Then, the assistant inspects the five-card collection, chooses one card from that five-card collection and gives it back to the audience. After that, the assistant arranges the remaining four cards in some order and puts the pile on the table, all face down.&#xA;&#xA;Finally, the magician comes in, inspects the pile, and announces the fifth card. (Example game: The pile consists of, in order, the seven of spades, the queen of hearts, the three of diamonds, and the eight of clubs. And the magician announces “You have the king of spades.”)&#xA;&#xA;Assuming no illicit information is passed, how can this trick be done?&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; The jolly joker is added to the pack (so it contains fifty-three cards). Devise a method to perform the same trick.&lt;/p&gt;&#xA;" OwnerUserId="9233" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2018-03-16T08:07:13.183" Title="The Fitch Cheney Card Trick Extended" Tags="|cards|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10005" PostTypeId="2" ParentId="10000" CreationDate="2015-03-09T15:15:23.200" Score="27" Body="&lt;p&gt;Sadly the truck was an &lt;a href=&quot;http://en.wikipedia.org/wiki/Electric_truck&quot;&gt;electrical&lt;/a&gt; one, and it fell... ah... had it been a gas guzzling earth killing one, it would have been saved... a sad sad story :-)&lt;/p&gt;&#xA;" OwnerUserId="10024" LastActivityDate="2015-03-09T15:15:23.200" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="10006" PostTypeId="2" ParentId="10004" CreationDate="2015-03-09T15:23:34.113" Score="13" Body="&lt;h1&gt;Strategy without the Joker&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Say you take any five cards from a deck.  Since there are only 4 suits, there must be at least two of them which have the same suit.  One of these will be our chosen card that we will get the magician to guess.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you pretend the cards are the numbers on a clock (with an extra number since there are 13 values), then the most these two numbers can be apart could every be apart clockwise is 6 spaces.  If you go 7 spaces away clockwise from any number, then continuing clockwise 6 more spaces will get you back to the original!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, take the &quot;larger&quot; of the two (i.e. the one that is 6 or less spaces ahead of the other going clockwise).  This will be the chosen card that you give back to the audience which you will get the magician to &quot;guess&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put the other card of the same suit (the &quot;lower&quot; of the two) as the first card in the pile that we will show to the magician.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the final three cards can be ordered in $3!=6$ different ways to tell the magician the offset of the second card from the first.  Using alphabetical order of the suits to break ties, we can order these cards in one of 6 possible orderings.  Here is are the six orderings:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$O(1)=\{c_1, c_2, c_3\} \implies +1$$&#xA;$$O(2)=\{c_1, c_3, c_2\} \implies +2$$&#xA;$$O(3)=\{c_2, c_1, c_3\} \implies +3$$&#xA;$$O(4)=\{c_2, c_3, c_1\} \implies +4$$&#xA;$$O(5)=\{c_3, c_1, c_2\} \implies +5$$&#xA;$$O(6)=\{c_3, c_2, c_1\} \implies +6$$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Example 1&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;For example, say the 5 cards are the 2 of clubs, 10 of clubs, 8 of diamonds, 9 of hearts, and Jack of spades.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can see that there are two cards which are clubs, and the 2 is 5 spaces away counting up from the 10 (wrapping around when we hit the Ace).  Thus, pick 2 as the hidden card, and put the 10 as the first card.  Use the $O(5)$ of the other three cards.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;10 of clubs, J of spades, 8 of diamonds, 9 of hearts&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The magician would recognize $O(5)$ among the last 3 cards, and then count up 5 from the first card, wrapping around when necessary.  He would then correctly guess the 2 of clubs.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Adding the joker&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;NOTE:&lt;/em&gt;&lt;/strong&gt; I've had to revise this answer extensively since the OP rudely kept finding holes in the solution.  I now have what I believe to be fool-proof method.  But I could be wrong, so please try to find counter-examples!  (but not too hard...) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;NOTE 2:&lt;/em&gt;&lt;/strong&gt; I created this proof based on Appendix A of &lt;a href=&quot;http://www.nymphomath.ch/crypto/magie/cardTrick.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.nymphomath.ch/crypto/magie/cardTrick.pdf&lt;/a&gt;.  The solution presented there does not work (I have mailed the author and am awaiting reply), but I was able to infer a correct solution from that and present it here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;NOTE 3:&lt;/em&gt;&lt;/strong&gt; If you read the paper above, you will find reference to the original author of the solution.  There is an email in which it is presented here &lt;a href=&quot;https://github.com/cstrahan/aduni/blob/master/02_discrete_math/misc/Card%20trick/Weiping-53.txt&quot; rel=&quot;nofollow&quot;&gt;https://github.com/cstrahan/aduni/blob/master/02_discrete_math/misc/Card%20trick/Weiping-53.txt&lt;/a&gt;.  It turns out my solution is pretty much this one exactly, just longer and less elegant.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Strategy&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;First, we will add the joker as the 14th spade between the King and Ace.  (Note, this solution looks like it can easily expand to a 56 card solution by adding a joker to each suit.  However, if you have 2 suits represented in the 5 cards, each pair 7 apart, it is not obvious how to reconcile this.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For most cases, the solution without the joker works fine.  The only time it doesn't is when you have 2 spades which are 7 apart from each-other, and the other 3 cards are from the other three suits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, here are the scenarios in order of preference.  &lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Any 2 cards of the same suit, non-spades&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Simply use the non-joker solution.  Even if there is a pair of spades, use the non-spade pair.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;3 or more spades&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;This scenario must be treated somewhat specially in order to not confuse it with the scenario below.  Basically, you will use the non-joker solution, but since there are multiple spades, there is more than one possibility for the hidden card.  When presented with this scenario, we must simply choose the two spades such that when we encode, we do not leave exactly 2 spades showing which are exactly 7 apart. Since we have 3 distinct numbers, this will always be possible.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Only 2 spades, not 7 apart&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Again, use the non-joker solution.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Only 2 spades, 7 apart&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;In this case, we have all the suits represented, otherwise we would have been able to use the non-joker solution with a different suit.  So we will hide one of the non-spade cards and put one of the spades as the first card.  Since we will end up showing both spades, one of them being the first card, the Magician will be able to recognize this scenario.  He will see the first card is a spade, and there is a second spade showing 7 from it.  He will know that we hid a card of the missing suit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, of the 3 non-spades, we need to decide which to hide.  Notice that you will always have one of the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;All three non-spade cards are in $[A-7]$&lt;/li&gt;&#xA;&lt;li&gt;2 non-spade cards are in $[A-7]$, 1 is in $[8-King]$&lt;/li&gt;&#xA;&lt;li&gt;1 non-spade card is  in $[A-7]$, 2 are in $[8-King]$&lt;/li&gt;&#xA;&lt;li&gt;All three non-spade cards are in $[8-King]$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thus, at least two of these cards share the same interval.  If all three do, then ignore the lowest and use the two highest.  We will hide one of these two cards.  Also, we will use the spade in the first position to indicate which interval the hidden card is in.  The lower of the two spades in the first position indicates it is in the $[A-7]$ interval, and the higher of the two spades indicates it is in the $[8-King]$ interval.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, of the two cards in the same interval, we can use the same trick as in the non-joker solution to get the value of one from the other.  Before, we had 13 numbers on the clock, so were able to use offsets of up to 6 clockwise from one to get to the other.  Now, we have only 7 numbers on the clock, so all we need is up to 3 clockwise from the other.  This is easy to encode with 6 cards.  The only caveat is that we also need to be able to encode an offset of zero.  Still, that means we only need 4 of the 6 possible permutations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let $O(6)$ mean a zero offset, and $O(5)$ and $O(4)$ will be unused. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, we will hide the &quot;larger&quot; of the two and encode the other three in the appropriate offset.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Example 2&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Lets try with the following cards: 3 spades, 10 spades, 3 hearts, 7 diamonds, 2 clubs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice we have 2 spades exactly 7 apart, so we must use the special encoding. Also, notice that the other 3 non-spades are all in the $[A-7]$ interval, so ignore the 2 of clubs and use the lower spade (3) as the first card. Lastly, notice that the 7 is 4 away from the 3, but the 3 is only 3 away from the 7 when you wrap around at 7 in the $[A-7]$ interval.  Thus, the 3 is the &quot;larger&quot; of the two and will be the hidden card.  We then use encoding $O(3)$ of the other three cards.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;3 spades, 7 diamonds, 2 clubs, 10 spades&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;When the Magician sees this, he will notice that there are exactly two spades being shown and they are exactly 7 apart. He knows that the non-joker method applied to this will result in a third spade, and if we had 3 spades, we would have had a choice as to which spade we chose to hide.  Thus, we wouldn't have needed to show exactly two spades 7 apart.  Therefore, the magician correctly infers that this is the special case of all four suits represented with exactly 2 spades 7 apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Immediately, the magician can deduce that the suit of the hidden card is hearts because that is the only suit not present in the displayed cards.  He also recognizes $O(3)$ of the last three cards. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since the lower of the two spades is the first one, the magician knows that the missing card is in the $[A-7]$ interval.  Also, he sees that the other two non-spade cards are a 2 and a 7.  Since both are in the $[A-7]$ interval, he ignores the lower(2) and uses the higher (7).  Counting up $O(3) \implies +3$ from 7 within the $[A-7]$ interval gets 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the missing card is the 3 of hearts!&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Solution for 56 Cards&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Intuitively, one can see that the 53 card solution should easily scale up to a 56 card solution by adding a joker to each suit instead of just spades.  Assuming that the jokers can be differentiated from each other (thus, deterministically placed in a suit), it does, with one minor exception; when there are 2 cards in each suit exactly 7 apart.  For example, the Ace/8 of hearts, the 2/9 of diamonds, and the 6 of clubs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, since there are unused orderings, we can use one of them to get the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, lets break down the strategy again:&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;3 or more cards in the same suit&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Use the non-joker solution on one of the pairs which does not require that the pair exactly 7 apart are shown to the magician.  Since there are 3 possible pairings, this is always possible.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Suited cards not 7 apart&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Simply use the non-joker solution on the two suited cards.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;One set of suited cards 7 apart&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Use the special strategy from the previous section with this suit instead of spades.  The key differences are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Any suit can be used to indicate the special case, not just spades&lt;/li&gt;&#xA;&lt;li&gt;the intervals are now $[Ace-7]$ and $[8-King,Joker]$; the second interval now has an extra card.  This is OK since this simply makes both intervals the same size whereas before the $[8-King]$ interval was 1 shorter.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h3&gt;Two sets of suited cards 7 apart&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;For this scenario, you have 2 suits, each with 2 cards 7 apart.  The fifth card is from a third suits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Joker-less strategy cannot be used since any two cards of the same suit are too far apart.  The special strategy from before cannot be used because there would be no way for the magician to guess the right suit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, we need to use one of the unused orderings from that strategy, $O(4)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, pick one of the suits which has 2 cards 7 apart, it doesn't matter which.  One of them will be the hidden card.  The other suit which has the 2 cards 7 apart will be used to tell the magician that this is a special case the same as before.  The only difference is the meaning of the first card, and $O(4)$ will allow the magician to guess the hidden card.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magician will see the two cards of the same suit 7 apart, one in the first position, and know it is the special case.  He will then see $O(4)$ and know that this is the extra special case of two sets of suited cards 7 apart.  Thus, he will know that the hidden card is 7 from one of the two other cards shown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will use the first card to tell the magician which - if we use the lower card as the first, then it is the lower of the other two cards which the magician will add 7 to. Otherwise, it is the higher.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Example 3&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Lets say the cards are 2 Hearts, 9 Hearts, 3 Diamonds, 10 Diamonds, 6 Clubs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the extra special case.  We will first randomly pick one of the hearts or diamonds as the hidden card; it doesn't matter which because we will be able to encode enough information to tell the magician which.  So lets say we pick the 3 of Diamonds as the hidden card.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need the two Hearts to indicate the special case.  Since the 10 (of Diamonds) is higher than the 6 (of Clubs), we will use the higher heart as the first card, and $O(4)$ for the rest.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;9 Hearts, 6 Clubs, 10 Diamonds, 2 Hearts&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The magician immediately sees the two hearts are 7 apart, so knows this is a special case.  When he sees $O(4)$, he knows this is the extra special case.  Of the two hearts, he sees the higher is the first, thus, the missing card is 7 from the higher of the two other cards - the 10 of Diamonds.  So, counting forward 7 (Remember the Joker of Hearts!) gives Jack, Queen, King, Joker, Ace, Two, Three.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the missing card is the 3 of Diamonds!&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="1592" LastEditDate="2015-03-16T18:32:28.530" LastActivityDate="2015-03-16T18:32:28.530" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="10009" PostTypeId="1" AcceptedAnswerId="10012" CreationDate="2015-03-09T16:20:01.957" Score="16" ViewCount="2445" Body="&lt;p&gt;Two angels in heaven want to play a game. One of them comes up with an arbitrary, computable function from the integers to the integers. The other angel has to guess what it is. The second angel can either give an integer, which the first angel gives an input to their function and responds with the output, or a computable function, which the first angel says is either correct or incorrect. (The angels know an oracle who can tell them if two functions are equivalent in finite time.) The game continues until the second angel guesses the function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angels only want to play the game if they know it will end. It doesn't matter how long it takes because since they're in heaven, they'll have an infinite amount of time to do other things whenever the game ends. However, if the game will never end, the angels don't want to play it. What should they do? If they play the game, what is the smallest number of integers the second angel will need to check to be sure of eventually guessing the correct function, and what is the second angel's optimal strategy?&lt;/p&gt;&#xA;" OwnerUserId="294" LastEditorUserId="4050" LastEditDate="2015-03-09T17:29:05.680" LastActivityDate="2015-03-10T22:36:30.223" Title="Should the angels play their game?" Tags="|mathematics|password|number-sequence|solvability|computer-science|" AnswerCount="5" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="10010" PostTypeId="2" ParentId="10009" CreationDate="2015-03-09T17:24:39.840" Score="5" Body="&lt;p&gt;Update: this answer is partly/mostly wrong, see Rex's answer. But it's not bad for a wild guess, I feel. The set of possible formulae &lt;em&gt;is&lt;/em&gt; infinite but because it's countably infinite, the second angel can keep counting until he/she/it arrives at the answer. :) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key point (to my mind) being that although there are an infinite number of possibilities for the first angel to choose from, once they've chosen a specific one it takes a finite amount of time for angel 2 to 'count up' until they get to it - which they don't have to go through all the infinite possibilities to do.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For any finite known set of mappings, inputs to outputs, there can always be more than one function which could apply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if the angels are partway through the game and have a set of mappings &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;{a, b, c, ...}  ---&gt;  {d, e, f, ...}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;then that isn't enough information for the second angel to know that there can only be one possible function that could produce that mapping. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the second angel will need to ask for more information.... but no amount of information will ever be enough for total certainty, because there will always be more than one possible function that could apply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Alternatively the second angel could guess a function without being totally certain that they're right, but then it would be possible for them - by chance - to always guess wrongly, so the game &lt;em&gt;could&lt;/em&gt; still go on forever).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the angels &lt;strong&gt;shouldn't&lt;/strong&gt; play the game - they should watch England lose at cricket instead, because that always happens within a finite time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to prove me wrong here would be to come up with a set of mappings (inputs to outputs) which can only be produced by one single function - I don't think that any such exists. If it does then I'm wrong and the angels should play the game. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sequence is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;{1, 2, 3, 4, 5} ---&gt; {1, 2, 3, 4, 5}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This function could apply:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;f(x) = x&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but then so could this one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;f(x) = x &amp;lt; 6 ? x : 2*x&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and so could any function in this infinite set of functions (where &lt;em&gt;k&lt;/em&gt; is any integer):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;f(x) = x &amp;lt; 6 ? x : kx&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="4050" LastEditDate="2015-03-09T20:27:30.067" LastActivityDate="2015-03-09T20:27:30.067" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="10011" PostTypeId="2" ParentId="9996" CreationDate="2015-03-09T17:46:58.303" Score="3" Body="&lt;p&gt;A possible answer, but I'm still a little short. Maybe someone can build off of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Still a little stuck, but hopefully this is the right direction.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Basketball Player,&lt;br&gt;&#xA;specific NBA player (MJ, Bill Russell)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My job is putting things in rings.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Put a ball into the ring (hoop)   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My prize is rings.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Championship Rings&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The number of rings I have symbolizes speed. (Not so sure here...)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Maybe speed to greathood, or the Hall of Fame?&lt;br&gt;&#xA;Number of rings that the player has coincides with some symbol for speed... I'm thinking 6 or 11 currently.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When people see me in action they say Oh, neat!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Good players are a lot of fun to watch, especially the Harlem Globetrotters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="9951" LastEditDate="2015-03-11T14:27:51.867" LastActivityDate="2015-03-11T14:27:51.867" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="10012" PostTypeId="2" ParentId="10009" CreationDate="2015-03-09T19:30:02.250" Score="16" Body="&lt;p&gt;The angels &lt;strong&gt;should&lt;/strong&gt; play the game, as long as the formula for the computable function must be fixed in advance &lt;em&gt;and finite&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: let &lt;code&gt;F&lt;/code&gt; be the shortest encoding of the formula (length &lt;code&gt;l(F)&lt;/code&gt;) using a set of symbols &lt;code&gt;S&lt;/code&gt; of size &lt;code&gt;n(S)&lt;/code&gt;.  Then the guessing angel can simply count up from &lt;code&gt;0&lt;/code&gt; in base &lt;code&gt;n(S)&lt;/code&gt;, and some time before &lt;code&gt;n(S)&lt;/code&gt;&lt;sup&gt;&lt;code&gt;l(F)&lt;/code&gt;&lt;/sup&gt; they will encounter the formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If the angels are allowed something really weird such as &lt;em&gt;infinite&lt;/em&gt; &quot;computable&quot; functions, then no, they should not play the game: there is infinite information in the formula, and it could be constructed so that you are required to know at least one new bit for each next number, so there's no way to know it in finite time.)&lt;/p&gt;&#xA;" OwnerUserId="2815" LastEditorUserId="-1" LastEditDate="2015-03-10T11:54:48.563" LastActivityDate="2015-03-10T11:54:48.563" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="10013" PostTypeId="2" ParentId="10000" CreationDate="2015-03-09T19:32:12.270" Score="6" Body="&lt;p&gt;Assuming the truck still weighs exactly 10,000 pounds at the time of the bird's approach, then it's not the landing of the bird you have to worry about - it's the flapping of its wings to stay aloft above the truck, which exerts that same amount of force on the air which is in turn exerted on the truck. The bridge will surely break before the bird has even landed on it!&lt;/p&gt;&#xA;" OwnerUserId="5895" LastActivityDate="2015-03-09T19:32:12.270" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10014" PostTypeId="2" ParentId="10009" CreationDate="2015-03-09T20:07:22.600" Score="8" Body="&lt;p&gt;The guessing angel can win in finite time with 0 integer queries. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The computable functions are countable. Enumerate them and guess them in order.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-03-09T20:07:22.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10016" PostTypeId="1" AcceptedAnswerId="10046" CreationDate="2015-03-09T21:19:32.517" Score="42" ViewCount="1361" Body="&lt;p&gt;A straightforward puzzle for the patient. There are no &lt;em&gt;tricks&lt;/em&gt; or &lt;em&gt;decryptions&lt;/em&gt; needed. The task is 'simple' albeit potentially challenging (and maybe time-consuming).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h1&gt;The goal&lt;/h1&gt;&#xA;  &#xA;  &lt;p&gt;Dissect the Africa-grid below into &lt;strong&gt;54 pieces&lt;/strong&gt; of different shape and size.&lt;br&gt;&#xA;  - Each piece consists of fully connected squares, i.e. is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Polyomino&quot; rel=&quot;noreferrer&quot;&gt;polyomino&lt;/a&gt;.&lt;br&gt;&#xA;  - The letters on each piece are &lt;a href=&quot;http://en.wikipedia.org/wiki/Anagram&quot; rel=&quot;noreferrer&quot;&gt;anagrams&lt;/a&gt;. &lt;br&gt;&#xA;  - All anagrams belong to the same group of words, defining the dissection property.&lt;br&gt;&#xA;  - The anagrams may contain one or more 'white spaces'.&lt;br&gt;&#xA;  - The English language is used.&lt;br&gt;&lt;br&gt;&#xA;  If you dissect it correctly, you may be able to spot a rather well-known word in an African native language. (&lt;em&gt;Googling the word gives 125,000,000+ results&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/x7GkC.png&quot; alt=&quot;DissectThisInto54Pieces&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="4204" LastEditDate="2015-03-10T09:02:15.760" LastActivityDate="2015-11-22T13:56:46.600" Title="Dissecting Africa" Tags="|dissection|word-search|polyomino|anagram|" AnswerCount="2" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="10017" PostTypeId="2" ParentId="10000" CreationDate="2015-03-09T21:50:43.093" Score="12" Body="&lt;p&gt;If the truck weighs exactly 10,000 units when stationary and the ultimate limit of the bridge is 10,000 units, then the truck will probably break the bridge at the beginning. Downforce from the air moving over the truck would probably increase the truck's force enough to break the bridge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is this the classic &quot;model the truck as a sphere moving through a frictionless vacuum&quot;? Vacuums are no fun... they suck. There must still be friction at the axles and/or wheels, or the truck could coast across without using any fuel, and then the bird would do it in.&lt;/p&gt;&#xA;" OwnerUserId="2477" LastActivityDate="2015-03-09T21:50:43.093" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10019" PostTypeId="1" CreationDate="2015-03-10T01:52:21.433" Score="4" ViewCount="444" Body="&lt;p&gt;I should begin by saying I am a word&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the beginning I followed established conventions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that I decided that was not for me&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I decided to oppose everything that came after me&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end I decided not to oppose, but to protect&lt;/p&gt;&#xA;&#xA;&lt;p&gt;covering, protecting and providing warmth&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What word am I?&lt;/p&gt;&#xA;" OwnerDisplayName="user10001" LastEditorUserId="5373" LastEditDate="2015-03-22T11:58:00.153" LastActivityDate="2015-03-22T11:58:00.153" Title="Methodical and precise word guessing" Tags="|riddle|word|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10022" PostTypeId="2" ParentId="10009" CreationDate="2015-03-10T03:01:46.333" Score="4" Body="&lt;p&gt;&lt;strong&gt;It depends on the interpretation of the question.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As per &lt;a href=&quot;https://puzzling.stackexchange.com/a/10012/4234&quot;&gt;Rex Kerr’s answer&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/a/10012/4234&quot;&gt;xnor’s answer&lt;/a&gt;, if the guesser is allowed completely arbitrary strategies, then they should play.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, &lt;strong&gt;if the guesser can only play computably, then they should not play&lt;/strong&gt;.  That is: given any computable strategy for the guesser, there is some total computable function which it will fail to guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: let $\sigma$ be any computable strategy for the guesser: i.e. a computable function which, given a finite number of input-output pairs (the state of play so far), returns a computable function (the next guess).  Then define the computable function $g_\sigma$ by a diagonalisation argument: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;to compute $g_\sigma(n)$, first compute all the earlier values of $g_\sigma$.  Then enumerate all the subsets $S \subseteq \{0,\ldots,n-1\}$ (there are finitely many, and they are easy to enumerate computably); for each of these, look at the function $\sigma(g\!\upharpoonright_S)$, i.e. the guess that the strategy would make, given the values of $g$ on $S$.  Now take $g_\sigma(n)$ to be the maximum of the values $\sigma(g_\sigma\!\upharpoonright_S)(n)$, plus one.  (This is computable since $\sigma$ is, and so is each $\sigma(\sigma)$, and we know the values of $g_\sigma$ on all inputs below $n$.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then $\sigma$ can never guess $g_\sigma$.  For every guess $\sigma(f)$ it can make (where $f$ is some finite known set of values), $g_\sigma(n)$ will be strictly greater than $\sigma(f)(n)$ on any input $n$ greater than the inputs included in $f$, since the computation of $g_\sigma(n)$ will include $\sigma(f)(n)$ in the set it maximises over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have: &lt;strong&gt;given any computable strategy for the guesser, there is some computable function on which it will fail.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, this can be strengthened a bit to cover non-deterministic strategries.  If the guesser has access to a finite amount of nondeterminism each turn (e.g. can flip a coin or roll a dice at each guess, but otherwise must work computably), then still, there is no unbeatable computable strategy.  Essentially, the same construction works, but the maximisation has to be over not just sets $S$, but also over all possible dice rolls for each guess — but this set is still finite and computable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in summary: &lt;strong&gt;if the guesser must play computably, or computably modulo some finite amount of nondeterminism, then they cannot be sure the game will end.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4234" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-03-10T07:47:29.810" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10023" PostTypeId="1" AcceptedAnswerId="10089" CreationDate="2015-03-10T07:00:43.187" Score="38" ViewCount="3360" Body="&lt;p&gt;Alice and Bob play a game that starts by Bob picking a secret integer $N\ge100$.&#xA;Then the game goes through several rounds.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In every round, Alice picks an integer $x\ge3$.&#xA;Every number can be picked at most once (and hence cannot be picked again in later rounds).&lt;/li&gt;&#xA;&lt;li&gt;Alice announces $x$ to Bob.&lt;/li&gt;&#xA;&lt;li&gt;If $x$ divides Bob's current secret value $N$, then Alice wins and the game ends.&lt;/li&gt;&#xA;&lt;li&gt;If $x$ does not divide Bob's current secret value $N$, then Bob &lt;strong&gt;subtracts&lt;/strong&gt; $x$ from $N$ (and thus replaces his old secret value by $N-x$).&lt;/li&gt;&#xA;&lt;li&gt;If the secret value ever becomes non-positive, then Bob wins and the game ends.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Question: Can Alice enforce a win? &#xA;(As usual, we assume that Alice and Bob use optimal strategies.)&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="9907" LastEditDate="2015-03-10T08:52:17.273" LastActivityDate="2015-03-12T12:24:10.903" Title="The subtraction game" Tags="|mathematics|strategy|game|" AnswerCount="6" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10026" PostTypeId="2" ParentId="10023" CreationDate="2015-03-10T09:09:49.463" Score="-1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alice can enforce a win.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bob picks a number, which is always in the form of 100 + 12*m + n, with m &gt;= 0 and n from 0 to 11. By dividing/substracting alternating by 3 and 4, Alice can enforce a win in turn 5 the latest. As we only divide by 3 and 4, the pattern repeats itself every 12 numbers.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Base    Div 3   Div 4   Div 3   Div 4   Div 3&#xA;100     97      93      -       -       -&#xA;101     98      94      91      87      -&#xA;102     -       -       -       -       -&#xA;103     100     -       -       -       -&#xA;104     101     97      94      90      -&#xA;105     -       -       -       -       -&#xA;106     103     99      -       -       -&#xA;107     104     -       -       -       -&#xA;108     -       -       -       -       -&#xA;109     106     102     -       -       -&#xA;110     107     103     100     -       -&#xA;111     -       -       -       -       -&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="9226" LastActivityDate="2015-03-10T09:09:49.463" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10028" PostTypeId="2" ParentId="10019" CreationDate="2015-03-10T10:10:49.710" Score="1" Body="&lt;p&gt;How about this word?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;non-antitrust&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Beginning - following established conventions.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;'Non-' is a common convention / structure to signify the opposite of what comes next.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Oppose everything that came after me:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;'Anti-x' usually means to oppose 'x' (with apologies to AntiX if necessary).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;End: not oppose, to protect, cover, provide warmth&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;'Trust' has the flavour of all the above.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9422" LastActivityDate="2015-03-10T10:10:49.710" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10029" PostTypeId="2" ParentId="9996" CreationDate="2015-03-10T10:27:52.597" Score="1" Body="&lt;p&gt;I'm going to give it a try... IMHO the word &quot;rings&quot; means every time something different:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My job is putting things in rings.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I had the hardest time with this... Didn't fully understand so I guess my answer is not gonna be right&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My prize is rings.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.thefreedictionary.com/prize+rings&quot; rel=&quot;nofollow&quot;&gt;http://www.thefreedictionary.com/prize+rings&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The number of rings I have symbolizes speed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each time a round ends, a bell rings. So a fast K.O. means only some &quot;rings&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When people see me in action they say Oh, neat!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I didn't understand if that is another hint...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you are a boxer... ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5122" LastEditorUserId="5373" LastEditDate="2015-03-10T10:37:28.597" LastActivityDate="2015-03-10T10:37:28.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10030" PostTypeId="2" ParentId="9942" CreationDate="2015-03-10T11:05:29.750" Score="0" Body="&lt;p&gt;On a crusade in the Holy Land?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know of only one world chess champion whose mother's name meant &quot;queen&quot;, namely Bobby Fischer, whose mother's first name was Regina, Latin for queen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;In the future I will awake&quot; brings up thoughts of King Arthur...the Grail legends...the Fisher King.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we get &lt;em&gt;The first time I was a king&lt;/em&gt; (Fisher King) and &lt;em&gt;The second time my mother was a Queen&lt;/em&gt; (World Chess Champion Bobby Fischer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does &lt;em&gt;Where did my father serve?&lt;/em&gt; connect with &quot;Whom does the Grail serve?&quot;&lt;/p&gt;&#xA;" OwnerUserId="7363" LastActivityDate="2015-03-10T11:05:29.750" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10031" PostTypeId="1" AcceptedAnswerId="10038" CreationDate="2015-03-10T11:07:45.190" Score="10" ViewCount="2505" Body="&lt;p&gt;I'm going to contribute to the rebus craze with one I invented some years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture itself is not suitable for work - &lt;a href=&quot;https://i.stack.imgur.com/nxXPb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;view it directly here&lt;/a&gt; or &lt;a href=&quot;https://puzzling.stackexchange.com/revisions/10031/1&quot;&gt;view the revision history&lt;/a&gt; to see the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, here is an emote version, but at least one of the answers - not the accepted one - relies on details from that picture which do not appear below; if you want to see the picture, please click the links above.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;(3{ 8R&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What phrase does this picture represent?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-03-20T14:15:08.237" Title="A slightly rude rebus" Tags="|rebus|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10032" PostTypeId="2" ParentId="10031" CreationDate="2015-03-10T11:42:25.260" Score="2" Body="&lt;p&gt;This back part of the body camouflaged itself quite well, still we can see it's an arse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the letters 'A', 'S', and 'S' or 'A', 'R', 'S' and 'E' in combination with '8' and 'R' leads towards&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ASS8R or ARSE8R&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Translating those with the help of the alphabet turns into '1 19 19 8 18' or '1 18 19 5 8 18'. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Adding those numbers up results in 65 or 51. Deal with it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Anyway, those numbers give me nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;On the other hand, &quot;8R&quot; i.e. &quot;EIGHT AR&quot; sounds like &quot;HATER&quot; or &quot;EATER&quot;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Now we could go for ASSHOLE and HATER. History has it, that some riddlers here are strongly opposed by the mods. Therefore the rebus could be some kind of goodbye message for the mods. I.e. describing them as the assholes and haters they like to be (from a riddler point of view). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;However, by looking closely at colours, we can see that the characters '8' and 'R' are all black. Another thing that's black, is the moustache&lt;br&gt;&#xA;(whereass the hat and the body part were coloured in many hues).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you tried to find a letter that looks like the moustache, you'd probably find the letter 'W'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining it with the other letters gives you 'W', '8' and 'R'.&lt;br&gt;&#xA;In other words:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; W8R = WAITER&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5106" LastEditorUserId="5373" LastEditDate="2015-03-10T22:31:15.700" LastActivityDate="2015-03-10T22:31:15.700" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10033" PostTypeId="2" ParentId="10031" CreationDate="2015-03-10T12:22:21.703" Score="4" Body="&lt;p&gt;The Urban dictionary has a slightly rude &lt;a href=&quot;http://www.urbandictionary.com/define.php?term=Frog%20eater&quot; rel=&quot;nofollow&quot;&gt;page&lt;/a&gt; on derogatory terms for French citizens, with the example sentence &lt;em&gt;&quot;Those damn frog-eating asshats smell like rotten twats&quot;&lt;/em&gt;. &#xA;The above rebus consists of&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;an ass wearing a French beret and a mustache as in the old French movies&lt;/li&gt;&#xA;&lt;li&gt;the digit 8 &lt;/li&gt;&#xA;&lt;li&gt;the letter R&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Putting these three things together, I get &lt;strong&gt;FROG&lt;/strong&gt; (&lt;a href=&quot;http://www.urbandictionary.com/define.php?term=frog&quot; rel=&quot;nofollow&quot;&gt;derogatory&lt;/a&gt; term for French), plus &lt;strong&gt;EAT&lt;/strong&gt; (perhaps close enough to EIGHT), plus &lt;strong&gt;ER&lt;/strong&gt; (lose enoogh to R). &#xA;All in all, my guess would therefore be that the rebus depicts a &lt;strong&gt;FROG EATER&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="5373" LastEditDate="2015-03-10T22:25:34.773" LastActivityDate="2015-03-10T22:25:34.773" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10035" PostTypeId="2" ParentId="6618" CreationDate="2015-03-10T13:46:34.223" Score="4" Body="&lt;p&gt;3) is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Consider Phlebas by Iain M. Banks &lt;br&gt; Think of = consider &lt;br&gt; Phlebas is the drowned Phoenician sailor in T. S. Eliot’s The Waste Land &lt;br&gt; the whole name is anagram of those 3 phrases&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9907" LastEditorUserId="9907" LastEditDate="2015-03-10T13:51:48.590" LastActivityDate="2015-03-10T13:51:48.590" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10036" PostTypeId="2" ParentId="6618" CreationDate="2015-03-10T14:06:24.723" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;5 Initially rebellious sheep undertake a silent infantry operation,&#xA;  inside a small enclosure, or so I'm induced to believe (10)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer in spoiler below&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Persuasion by Jane Austen &lt;br&gt; &quot;Initially&quot; is an indicator to take the first letters of words, in this case all words up until the next comma.  This yields the letters &quot;rsuasio&quot; &lt;br&gt; &quot;Inside&quot; is an indicator that the letters that were already found are going to be placed within the letters of the answer to the next clue.  A &quot;small enclosure&quot; is a &quot;pen&quot;.  Given these letters the only question is where to place the &quot;rsuasio&quot; within &quot;pen&quot;.  The options are &quot;prsuasioen&quot; or &quot;persuasion&quot;.  &lt;br&gt;The last statement is &quot;or so I'm induced to believe&quot;.  To induce someone to believe is to persuade them, so the solution solves the clues.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1777" LastEditorUserId="1777" LastEditDate="2015-03-11T17:17:41.037" LastActivityDate="2015-03-11T17:17:41.037" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10038" PostTypeId="2" ParentId="10031" CreationDate="2015-03-10T14:58:33.790" Score="22" Body="&lt;p&gt;The French word for &lt;em&gt;ass&lt;/em&gt; is &lt;em&gt;CUL&lt;/em&gt;. An ass with a beret and French-style moustache is fairly strongly hinting at looking for a word relating to &lt;em&gt;ass&lt;/em&gt;  and certainly hints that we want something French. we have: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;CUL8R&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is SMS-speak for: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SEE YOU LATER :D bye bye! tata! aufwiedersehen! &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2015-03-12T20:04:31.013" LastActivityDate="2015-03-12T20:04:31.013" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10039" PostTypeId="2" ParentId="10019" CreationDate="2015-03-10T15:33:02.433" Score="0" Body="&lt;p&gt;Are you?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; person&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I should begin by saying I am a word&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;person&quot; is a word (sorry for stating the obvious lol)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the beginning I followed established conventions&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the original plural form of &quot;person&quot; was &quot;persons&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After that I decided that was not for me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;people&quot; is commonly the accepted plural form of &quot;person&quot; now, subjectively&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I decided to oppose everything that came after me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; persons came after (tried to change the use) of the word(s) &quot;person&quot;, &quot;persons&quot;, &quot;people&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the end I decided not to oppose, but to protect&#xA;covering, protecting and providing warmth&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; people have all of these capabilities&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9778" LastActivityDate="2015-03-10T15:33:02.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10041" PostTypeId="2" ParentId="10023" CreationDate="2015-03-10T17:26:43.130" Score="-2" Body="&lt;p&gt;I might be wrong but:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume that Bob uses an optimal strategy, then that will actually give us information on the number he selects. Below are a few basic optimal strategies and their implications that I can think of.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If I were Bob then the best selection would be a prime number, as it is the only 100% guarantee that it will not be guessed on the first go. This informs us that the number is odd at this point. From here, however, Alice can assume that the number is odd. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This allows Alice:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To keep the number even by first asking an odd number, which is arbitrary but lets keep it low to maximise the number of guesses. So lets say 7 for instance (3 and 5 are useful). And then maintaining even guesses. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alice then needs to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Make the number divisible by 4. This can be done by using larger numbers to reduce the number. If we know that N is gerater than 100, it has to be divisible not by 4 on every other even number, meaning if we tried 32 first, and followed that up with 8. Then if it fails both times, the originial number was not divisible by 4. (I think I'm not 100% on this logic but I think it is correct don't be too harsh) This means that whatever N is now, it is not divisible by 4 but is even, so we try 6 and then 4 which should then work. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Attempted logical proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This took a lot of pages in my notebook, but:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets write N = 100 + y + z. (Where z is the arbitrary odd number we have removed from the equation in the first step). As we have decided one way to win is to force N to become divisible by 4. As 100 is divisible by 4 it makes the sum much easier. Trying to use 3 would fail because 100 is not divisible by 3. We are therefore trying to make y divisible by 4. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If new N is divisible by 32 then we have already won.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, then the new N=N-32 is also not div by 32. Otherwise N would have been originally. Therefore we know there exists an m between N-32 and N s.t. m is divisible by 32 although this is not that helpful. We then try 8 (being 32/4).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If N-32 is divisible by 8 then we have already won. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, N-40 is not divisible by 8. This can be written as N - (8*5) meaning N is not divisible by 8 (as we know for the same reason N-32 is not divisible by 32 anyway). This can be expanded to say that N - (4*10) is also not an integer, proving that N is not divisible by 4. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This tells us that if we assume an odd starting number, that N-40 is not now divisible by 4. We can therefore attempt 6 (2*3) to move across an even number and make sure that N is divisible by 4. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If N-40 is divisible by 6 then we win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, N-46 must be divisible by 4 as N-46 is even, every other even number is divisible by 4 and N-40 is not divisible by 4 (so N-42 is, N-44 is not and N-46 is). We have forced a victory and only subtracted 4 + 6 + 8 + 32 + 7 = 57 &amp;lt; 100 so we cannot go broke. It doesn't matter how large or small N is to start as even numbers repeat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is possible assuming that Bob starts on a prime number to force a victory&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, this strategy relies on a prime (or just odd) number being chosen first, which is only optimal if Bob decides it is important to have a 100% success rate on the first question and goes with a prime number. Bob will therefore know not to start with a prime number every time. This could therefore end up very much like a game of chess. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take the optimum strategy as being to chose an even number every time (Being that this would confuse the oponent by not starting on a prime) Then the same steps work without the initial odd number. Although that relies on Alice calling Bob's bluff with regard to strategy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have yet to generalise this to work for any starting number. I need to discover whether it is odd or even somehow. I will think a bit more on it. &lt;/p&gt;&#xA;" OwnerUserId="9918" LastActivityDate="2015-03-10T17:26:43.130" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10042" PostTypeId="2" ParentId="9973" CreationDate="2015-03-10T18:33:51.887" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Heart. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Figuratively, people at the top don't seem to have it; people at the&#xA; bottom do. Literally, if you don't have one you will die.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10069" LastActivityDate="2015-03-10T18:33:51.887" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10043" PostTypeId="1" CreationDate="2015-03-10T18:48:55.790" Score="3" ViewCount="14800" Body="&lt;p&gt;A prisoner wakes up in an empty room (&lt;em&gt;being seated on the floor against the wall&lt;/em&gt;). There are three solid doors on three of the room's walls: one to his left, one in the middle in front of him and one to his right. The fourth wall behind him) is a thick glass pane with an intercom built in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Through the intercom the warden says: &quot;One of these doors leads to freedom, the other door [&lt;em&gt;note the singular&lt;/em&gt;] leads to death. You may choose one door for me to unlock. You may ask me two questions which I will answer with 'yes' or 'no' to help you choose.&quot; (but not &quot;maybe&quot; or any other third value: &lt;em&gt;in that case, the answer will be &quot;no&quot;&lt;/em&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The prisoner is confused and hence asks: &quot;What's behind the third door?&quot; &lt;strong&gt;[b]&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The warden answers: &quot;No, I can't tell you, because that's not a yes-or-no question. It still counts as your first question though.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The prisoner complains: &quot;That's unfair!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The warden responds: &quot;No, it isn't, you don't even need that second question to survive.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some contemplation the prisoner asks his second question, chooses a door and goes free.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;what question did the prisoner ask and what was the answer?&lt;/li&gt;&#xA;&lt;li&gt;what was behind the third door &lt;strong&gt;[a]&lt;/strong&gt; and what was the probability for the prisoner to pick it?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(If multiple valid solutions the one with lowest probability for the third door wins.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[a] Hint&lt;/strong&gt;: as some already noticed, the third door leads back to prison (the prisoner had to be brought into the room in some way, after all), hence leads neither to death nor to freedom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[b] Note&lt;/strong&gt;: after to the prisoner's first question, the warden knows that the prisoner doesn't know what is behind the third door - but the warden knows that the third door leads back to the prison.&lt;/p&gt;&#xA;" OwnerUserId="10070" LastEditorUserId="5511" LastEditDate="2015-03-16T14:19:24.463" LastActivityDate="2015-06-05T12:28:25.973" Title="Three doors, one question" Tags="|probability|" AnswerCount="15" CommentCount="24" ContentLicense="CC BY-SA 3.0" />
  <row Id="10044" PostTypeId="2" ParentId="10043" CreationDate="2015-03-10T19:11:45.580" Score="6" Body="&lt;p&gt;The prisoner asks &quot;I am thinking of either the left or middle door. Assuming that the third door also leads to my death and is, therefore, essentially the same as the death door, is there a door further right of the one I am thinking of that would be better for me to choose?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If door &quot;L&quot; is correct, the answer will be no.  If you are thinking of &quot;L&quot; everything else is worse while if you are thinking of &quot;M&quot; then &quot;R&quot; is no change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If door &quot;M&quot; is correct, the answer will be &quot;I don't know&quot;.  If you are thinking of &quot;L&quot; the answer is yes while if you are thinking of &quot;M&quot; then &quot;R&quot; is worse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If door &quot;R&quot; is correct, the answer will be &quot;yes&quot;.  If you are thinking of &quot;L&quot; the answer is yes while if you are thinking of &quot;M&quot; then &quot;R&quot; is better.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;New answer&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;With the warden now answering &quot;no&quot; to a question where he doesn't know that answer the puzzle changes.  I just realised the &quot;survive&quot; wording which means my less rigorous answer is inferior to this one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The prisoner should ask &quot;Is the Middle door better for me than the Left?&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the answer is &quot;yes&quot; the middle door is either prison (because the left door is death) or freedom (because the left door is death or prison).  The prisoner should pick middle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the answer is &quot;no&quot; the left door is either prison (because the middle door is death) or freedom (because the middle door is death or prison).  The prisoner should pick left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This strategy yields a 100% chance to survive.  There is a 33.3% chance he will end up in prison and a 66.7% chance he will be freed.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="-1" LastEditDate="2015-03-11T05:46:27.257" LastActivityDate="2015-03-11T05:46:27.257" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="10045" PostTypeId="2" ParentId="10009" CreationDate="2015-03-10T19:28:45.767" Score="3" Body="&lt;p&gt;I really enjoyed this question, so as a summary over all possible interpretations/answers, we have the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 1: The oracle accepts descriptions of partial computable functions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You &lt;em&gt;should&lt;/em&gt; play the game: You can (computably) enumerate all partial functions by just enumerating all programs with proper syntax. See &lt;a href=&quot;https://puzzling.stackexchange.com/a/10012&quot;&gt;Rex's Answer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 2: The oracle requires all descriptions to be total computable functions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 2.1: Your strategy is required to be computable&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You &lt;em&gt;shouldn't&lt;/em&gt; play the game: See &lt;a href=&quot;https://puzzling.stackexchange.com/a/10022&quot;&gt;Peter's answer&lt;/a&gt; - there is no computable enumeration of total computable functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 2.2: Your strategy is allowed to not be computable&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You &lt;em&gt;should&lt;/em&gt; play the game. The total computable functions are a subset of the partial computable functions, and hence countable, so enumerable, though not computably so. See &lt;a href=&quot;https://puzzling.stackexchange.com/a/10014&quot;&gt;Xnor's answer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Personally, I think a more interesting game is one in which angel 2 is only allowed to guess the function / consult the oracle $&amp;lt; K$ times for some fixed natural number $K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Claim: Such a game should &lt;em&gt;not&lt;/em&gt; be played&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar to &lt;a href=&quot;https://puzzling.stackexchange.com/a/10010&quot;&gt;A E's answer&lt;/a&gt; - for any finite number of known (input, output) pairs, there are an infinite number of possible total computable functions that give such an output. (Note these are still [non-computably] enumerable as earlier, ie I can find a list such that any given function appears at a finite place in my list).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BUT at any finite time, I have &lt;em&gt;not&lt;/em&gt; nailed down the given output for definite, and still have an infinite number of possibilities. So I cannot win.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;To explain the conceptual difference between these games / these answers, consider this analogy:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, angel one proposes &quot;I am thinking of a fixed natural number $N$. At each step, you're allowed to either ask for the digit at position $p \in \{0, 1, \dots \}$ (ie, the digit representing the number of $10^p$s in my number); OR guess the number. Can you guess my number correctly in a finite number of steps?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If angel two just guesses digits, they get nowhere: there are always an infinite number of possible numbers out there fitting the information they have received. But if instead they just start guessing possibilities, they could simply ask &quot;Is 1 right? Is 2 right? Is 3 right?...&quot; and definitely win in finite time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Extensions:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a different question working out &lt;em&gt;fast&lt;/em&gt; strategies to guess $N$. The above method takes $N$ turns. But you could also you could instead at every $n=2k\,$th step, ask for the $k\,$th digit, $d_k$, of the number, and at every $n=2k+1\,$th step, ask if the number is $d_k d_{k-1} \dots d_1 d_0$. This finds the number in $O(\log(N))$ steps.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Challenge 1:&lt;/em&gt; Is it possible to do any better in terms of the running time of angel 2's algorithm? (for this new problem?)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Challenge 2:&lt;/em&gt; In the original question (assuming the interpretation as in Case 1), if angel one chooses a partial computable function $\{N\}$ (ie with Gödel number $N$), does there exists / can you create a general algorithm for angel 2 which improves on the trivial worst-case run time of $O(N)$ (like we have done for the analogous problem above)?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I don't actually know the answer to either of these questions... Might think about them if I have a little longer.&lt;/p&gt;&#xA;" OwnerUserId="10066" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-03-10T22:36:30.223" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10046" PostTypeId="2" ParentId="10016" CreationDate="2015-03-10T19:48:44.927" Score="15" Body="&lt;p&gt;Got the dissection part done.  No idea about the native African word, though.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The 54 pieces are the various &lt;a href=&quot;https://www.countries-ofthe-world.com/countries-of-africa.html&quot; rel=&quot;noreferrer&quot;&gt;nations of Africa&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/Gw90Y.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gw90Y.png&quot; alt=&quot;Africa&quot; title=&quot;Click to enlarge&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="239" LastEditorUserId="-1" LastEditDate="2015-11-22T13:56:46.600" LastActivityDate="2015-11-22T13:56:46.600" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10047" PostTypeId="2" ParentId="9996" CreationDate="2015-03-10T19:56:07.900" Score="0" Body="&lt;p&gt;My answer is , which is probably wrong, is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A circus.  There are multiple rings to a circus.  The more rings a circus has, the faster it can transition between acts, and show more at once.  People seeing a circus go &quot;oh Neat!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The only thing I am unsure of is My prize is rings but it could mean&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; winning ring toss, or the prize position is the center ring.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10071" LastActivityDate="2015-03-10T19:56:07.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10048" PostTypeId="2" ParentId="10016" CreationDate="2015-03-10T20:19:07.267" Score="8" Body="&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/users/239/golden-dragon&quot;&gt;Golden Dragon&lt;/a&gt;&lt;/strong&gt; has solved it, but for the record, this is as intended:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/3pY9t.png&quot; alt=&quot;Solution&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/GMB25.png&quot; alt=&quot;UBUNTU&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-03-10T20:19:07.267" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10049" PostTypeId="2" ParentId="10043" CreationDate="2015-03-10T20:31:23.557" Score="3" Body="&lt;p&gt;An easy answer would be to simply choose a door randomly and ask if it leads to death, then go through one of the others if the answer is &quot;yes&quot;.  This should guarantee survival, since only one door leads to death, per the warden's rules.  It would have a 50% chance of freedom and 50% chance of going through door #3, which probably puts him back in his cell.&lt;/p&gt;&#xA;" OwnerUserId="10071" LastActivityDate="2015-03-10T20:31:23.557" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10050" PostTypeId="1" AcceptedAnswerId="10092" CreationDate="2015-03-10T20:31:26.393" Score="5" ViewCount="391" Body="&lt;p&gt;I began as an expert&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But later my skills are still recognized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once a fire was started within me&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And after asking why I did it all&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll find I was heavy into gambling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What is another name for an expert?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 1b:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Payment is important for this expert.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What is another term for skills?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="9951" LastEditDate="2015-03-11T13:49:22.370" LastActivityDate="2015-03-11T14:14:58.880" Title="Take a Chance on Me" Tags="|riddle|word|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10051" PostTypeId="2" ParentId="9968" CreationDate="2015-03-10T20:39:08.000" Score="7" Body="&lt;p&gt;I'm borrowing heavily from the other answers, but I'm putting everything into one place.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A: The order of Romeos passenger in the Greek alphabet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alfa Romeo is a car manufacturer; cars have passengers, and alpha is the 1st letter of the Greek alphabet, associated with Romeo.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thank you to @rand-althor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second clue&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;B: The two last digits in the year the returning Lex Luthor was awarded his (number of clue A) Oscar.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 96,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Kevin Spacey played Lex Luthor in &lt;em&gt;Superman Returns&lt;/em&gt; and was awarded his 1st Oscar for his appearance in The Usual Suspects in 1995. He was awarded this Oscar in 1996.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thank you to @rand-althor and @dmg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Third clue&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;C. The man in B didn't manage to complete here on time, and could have called in his complaint on this number, in &lt;em&gt;Apple time!&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 311&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We've identified &lt;em&gt;The man in B&lt;/em&gt; as Kevin Spacey, who attended the Juilliard School in New York City to become a comedian, but left before receiving a diploma.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you have a complaint in NYC, you can dial 3-1-1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thank you, @CDSpace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fourth clue&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;D. The street in the city of C where marble is buried.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We've identified the city of C to be New York City, so the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because of the &lt;a href=&quot;http://en.wikipedia.org/wiki/New_York_City_Marble_Cemetery&quot;&gt;marble cemetery&lt;/a&gt; located on&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2nd Street.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thank you, @rand-althor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fifth clue&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;E. A European convention was held in a city in Italy sometime in the 20th century, and they discussed a partial golden retreat. In March 2007 the football club from the same city was in the (number of clue D) league. The two last digits of the year before, and the two last digits of the year after the conference is the number of this clue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 21 and 23&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Genoa Conference was held (in Genoa) between representatives of 34 countries in 1922 to discuss the global economic situation following World War I. Genoa CFC played in the 2006/2007 Italian Cup, Serie B.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sixth clue&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;F. In the episode number (E) of Sesame Street, Kermit the frog appears this number of times.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Kermit made no appearances in episode 2123 of Sesame Street.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Put these clues together and you have a date and time.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 + 96 + 311 + 2 + 21 + 23 + 0 &amp;rarr; 1963-11-22 12:30&lt;br /&gt;&#xA; It was at 12:30 CST, November 22, 1963, when President John F. Kennedy was shot twice by &lt;strong&gt;Lee Harvey Oswald&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9828" LastEditorUserId="9828" LastEditDate="2015-03-11T14:06:22.620" LastActivityDate="2015-03-11T14:06:22.620" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="10052" PostTypeId="2" ParentId="10019" CreationDate="2015-03-10T20:47:29.097" Score="4" Body="&lt;p&gt;So, I'm probably way off, but maybe at least part will spark something with someone else.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Turncoat&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I should begin by saying I am a word   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Update thanks to suggestion from Reticality. I am a word, one word. Turncoat is one word. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the beginning I followed established conventions&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the beginning a turncoat is a &quot;good guy&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After that I decided that was not for me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Decided to change sides.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I decided to oppose everything that came after me  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Turning changes direction, possibly opposing?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the end I decided not to oppose, but to protect&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; last part of turncoat: &quot;Coat.&quot; Coats protect from the elements&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;covering, protecting and providing warmth&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Coats provide warmth, covering, etc. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="9951" LastEditDate="2015-03-15T21:29:50.747" LastActivityDate="2015-03-15T21:29:50.747" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10053" PostTypeId="2" ParentId="10043" CreationDate="2015-03-10T20:50:40.247" Score="5" Body="&lt;p&gt;ooh I got it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the best question is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; is the left door a worse choice than the middle door?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;if yes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; pick middle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;if no&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; pick left&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;that's 6/6 chances of survival and 4/6 chances of freedom&lt;/p&gt;&#xA;" OwnerUserId="9413" LastActivityDate="2015-03-10T20:50:40.247" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10054" PostTypeId="2" ParentId="10023" CreationDate="2015-03-10T20:56:19.580" Score="-1" Body="&lt;p&gt;With the restrictions imposed in the question, it's not possible for Alice to guarantee win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using &quot;optimal strategy&quot;, Alice must not guess using any number larger than 100 - the reasoning behind this is that if Bob picks the number 100, 101, 102, or 103, and Alice picked a larger number than 100 (and didn't win), Bob would automatically win; in order for Alice's strategy to be &quot;optimal&quot;, it must encompass the ability to derive a win no matter the number (including numbers 100 - 103). (In other words, it's not optimal to pick anything higher than 100). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem comes when she starts using small numbers. Because Alice cannot repeat numbers, and she doesn't want to go into the negatives, she further limits her usable number set - she can't use anything that's higher than 51 (51 included). If she picks 51 and Bob picked 100, she now has to derive from 49; but what if Bob picked 101? or 102? Now she has to derive from x - 51, and if that just so happened to be a prime number that's higher than what she has left, well, she's screwed (As in it's just not optimal).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are three ways for Alice tolook at this: Starting small, starting big (relatively large numbers, eg 50 - 75 ish), or going at random. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the OP said &quot;assume optimal strategies are used&quot;, Alice will not go at random. (Clearly not optimal).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If she starts big, she runs the risk of going under - as in, if she starts big, she could instantly throw herself under the bus and drop into a negative value, thus losing the game. This has the second largest risk out of the 3 approaches, and is not optimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If she starts small, eventually Alice runs out of small factors for what's left at the end of the spectrum. Using up all the numbers from 3 to 12 really quickly will destroy her chances at solving the problem if bob's number was relatively big.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;eg:&#xA;Starting at base case of 100, since Alice must allow for the chance of 100 to be chosen.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;100 % 3 leaves 97; 97 is instantly a prime number, so this cannot be the optimal strategy.&#xA;&#xA;100 % 4 works; 101 % 4 = 97; prime number, uh oh.&#xA;&#xA;100 % 5 works; 101 % 5 = 96 -&amp;gt; 96 % (lowest of the following factors  1,2,3,4,6,8,12,16,24,32,48,96, which is 3) = works.&#xA;We can follow the pattern and every time, at some point you'll end up with a prime number. (If you could suddenly guess the prime number, it wouldn't be optimal - it would only work for specific case).&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I personally spent hours coding different variants of patterns to try and devise an optimal strategy, all of which lead to one inevitable result - Eventually, you hit a number that is a prime number that is too high. The moment you use that prime number, at some point in the list of possible numbers (which again I must remind you is infinite), you will hit a new prime number for which you will need another specific use case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that I also considered and tried variants of subtracting numbers from the prime numbers and then continuing. Keep reading for those algorithms, but at the end of the day you can't achieve 100% winrate.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because there are an infinite number of primes (See Euclids proof: &lt;a href=&quot;https://primes.utm.edu/notes/proofs/infinite/euclids.html&quot; rel=&quot;nofollow&quot;&gt;https://primes.utm.edu/notes/proofs/infinite/euclids.html&lt;/a&gt;), there is no optimal strategy that works for ALL numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, whether Alice wins or not comes down to blind luck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best algorithm with no repeats that I could find (to give highest winrate for Alice) (This algorithm could be extended but you will always have those numbers where you miss and go negative...)&#xA;(13.6% failrate out of 9900 tests (numbers 100 to 10000)) Here, Alice wins 86.4% of the time (which is decent in a universe of infinite numbers). Even when increasing the number of test cases by a factor of 10, the failrate stays about the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;old - fairly consistent 13.6% failrate:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;(20);(3);(9);(10);(12);(6);(4);(5);(17);(8);(11);(12);(13);(15);(17);&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;edit: new Algorithm, 10% failrate ramping up towards 13.1% at 99900 tests!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (5);(6);(4);(3);(8);(9);(7);(12);(10);(15);(49);(14);&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Now, if the conditions were changed to say.. allow a single repeat.... that would be a different story and a different set of calculations.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;&#xA;I did manage to find an algorithm (If we were allowed to repeat the number &quot;3&quot; just one time) that only fails 40 out of 400 tries. Upon further testing, it fails 90 out of 900 tries. 990 fails out of 9900 tries. (10% failrate overall, which IMO in a universe of infinite numbers is pretty good for Alice. She wins 90% of the time) Woah! 3 flipping percent with half the length of the other algorithms from the ability to repeat one number? I wonder what adding more repeats would allow... ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The said algorithm if you're inclined to try it: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 20, 3, 9, 5, 12, 6, 4, 10, 7, 14, 3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;edit: Crayziman's algorithm (95% winrate at 99900 test cases)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;! 3,7,3,13,6,15,4,18,12&#xA;  &lt;/strike&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;edit: Crayziman's 100% algorithm with one repeating number:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3,7,4,14,3,16,6,15,12&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TLDR for original question&lt;/strong&gt;&#xA;There will always be those numbers that fail. There is NO 100% winrate strategy for Alice. However, Alice can maximize her winrate by starting small and adding numbers to the ends of the algorithm. It'll only increase her chances by a bit, but hey, a bit is better than nothing.&lt;/p&gt;&#xA;" OwnerUserId="8999" LastEditorUserId="8999" LastEditDate="2015-03-10T22:14:51.250" LastActivityDate="2015-03-10T22:14:51.250" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="10055" PostTypeId="2" ParentId="10043" CreationDate="2015-03-10T21:11:13.167" Score="0" Body="&lt;p&gt;Your first, most obvious option is to ask about one of the doors specifically. That is, to ask &quot;Does [door A] lead to freedom?&quot; Then, if it does, choose that door, and if not, choose one of the others at random. That first guess has a $1/3$ chance of being correct. If incorrect (a $2/3$ chance), your second guess has a $1/2$ chance of being right, which means there is a $1/3$ chance of this strategy leading to freedom on the second try. So, this gives a $2/3$ total chance of gaining freedom, and a $1/6$ chance of either death or the third option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I can tell, this is the highest chance you can get of guessing the door to freedom. However, if there is a way to answer the question, &quot;what was behind the third door?&quot; from the information given, then I am overlooking something that could probably give a better answer.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The question asks to minimize the probability of guessing the third door. We can do this and still get a $2/3$ chance of freedom, although this does, of course, increase the chance of death to $1/3$. Personally, I would rather exclude the possibility of death and risk the third door. But, to get rid of the third possibility, you should ask this: &quot;Is the third door (or, the door leading to the third result) to the right of the door to freedom, when viewed from the side of the room with the glass wall?&quot; If the answer is &quot;yes&quot;, choosing the leftmost door gives a $2/3$ chance at freedom and $1/3$ chance of death. If the answer is &quot;no&quot;, choosing the rightmost door gives the same result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Asking this question about the &quot;death&quot; door (instead of the third door) allows the prisoner a $2/3$ chance at freedom and no possibility of death, so the warden is correct, you don't need the second question to survive.&lt;/p&gt;&#xA;" OwnerUserId="8776" LastActivityDate="2015-03-10T21:11:13.167" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10056" PostTypeId="1" CreationDate="2015-03-10T21:14:05.480" Score="-1" ViewCount="24235" Body="&lt;p&gt;There is an island with 12 islanders. All of the islanders individually weigh exactly the same amount, except for one (either weigh more, or less than the other 11). How can you find which islander is the one that weighs more/less than the others? You must use a see-saw to figure out the weights, and you may only use the see-saw 3 times max&lt;/p&gt;&#xA;" OwnerUserId="10073" LastEditorDisplayName="user20" LastEditDate="2015-03-10T23:18:11.093" LastActivityDate="2015-03-10T23:18:11.093" Title="The 12 islander question" Tags="|logical-deduction|" AnswerCount="1" CommentCount="0" ClosedDate="2015-03-10T21:36:30.683" ContentLicense="CC BY-SA 3.0" />
  <row Id="10057" PostTypeId="2" ParentId="10056" CreationDate="2015-03-10T21:17:06.613" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Put 5 on each side, if the scale doesn't move, then it is 1 of the 2 you DIDN'T put on the scale.&#xA;&#xA; If it tips down, then take those 5 and split them between the scale of 2 and 2, and leave one off. If the scale doesn't tip then got it is the one you left off. If the scale tips down, take those 2 and put one on either side.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Problems?&lt;/p&gt;&#xA;" OwnerUserId="10073" LastEditorUserId="5373" LastEditDate="2015-03-10T23:04:41.907" LastActivityDate="2015-03-10T23:04:41.907" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10058" PostTypeId="2" ParentId="10043" CreationDate="2015-03-10T21:25:36.630" Score="0" Body="&lt;p&gt;The prisoner just woke up. So, he probably is lying down still. In that case, there is a room on either of his sides and one in the ceiling/roof. The warden singles out the other door and says 'one of these doors' which could mean that he is in the room(behind glass door) at the same level(ground) as the two doors. The prisoner can ask one question to eliminate one of the doors and take the other one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there are some assumptions, but hey, the question is worded that way.&lt;/p&gt;&#xA;" OwnerUserId="10075" LastActivityDate="2015-03-10T21:25:36.630" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10060" PostTypeId="2" ParentId="10043" CreationDate="2015-03-10T22:11:40.690" Score="3" Body="&lt;p&gt;I'm going to make 2 assumptions.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The warden doesn't lie.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Because hints are buried in the language, incorrect grammar is&#xA;essentially a lie. So the grammar of his responses is correct.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now, I can see 2 interpretations of his comments:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first interpretation, in his statement&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;One of these doors leads to freedom, the other door [note the&#xA;  singular] leads to death.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the phrase &quot;the other door&quot; imples that only 2 of the objects I see are actually doors. The third is something else (a painting, a false front, an illusion, etc.). I should be able to identify through inspection which one is not a door. At that point I have 2 choices, one Yes/No question, and I'm free.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second interpretation (which I consider a bit sketchy) is that his use of &quot;the other&quot; is tied to the statement that follows it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;One of these doors leads to freedom, the other door [note the&#xA;  singular] leads to death. You may choose one door for me to unlock.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He said &quot;One...the other&quot; because only 2 of them are locked, hence you're choosing one of the 2 locked doors, one to freedom, one to death. The third isn't locked and goes into the bathroom, or the prison, or more death, whatever. Doesn't matter. I identify the locked ones, and I'm back to 2 doors, 1 question, and freedom.&lt;/p&gt;&#xA;" OwnerUserId="10077" LastActivityDate="2015-03-10T22:11:40.690" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10061" PostTypeId="2" ParentId="10043" CreationDate="2015-03-10T22:27:01.707" Score="0" Body="&lt;p&gt;There is three door, lets name it A, B, C doors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets assume there is three state for these doors Death(D), Free(F), Prison(P).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;____A___|____B__|____C__|&lt;br&gt;&#xA;-----D-----|-----F-----|-----P-----|&lt;br&gt;&#xA;-----D-----|-----P-----|-----F-----|&lt;br&gt;&#xA;-----F-----|-----D-----|-----P-----|&lt;br&gt;&#xA;-----F-----|-----P-----|-----D-----|&lt;br&gt;&#xA;-----P-----|-----F-----|-----D-----|&lt;br&gt;&#xA;-----P-----|-----D-----|-----F-----|  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;These are all states there can be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we ask &quot;If A is not the free door then A or B should be death door, right?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If he says &quot;no&quot; that means, A is free door, or, A is prison door and B is free door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prisoner should pick A as a safe choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If he says &quot;yes&quot; that means, A or B is death door, and, A is not free door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that condition prisoner should pick C as a safe choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That question guarantee prisoner's survival and give prisoner %66.7 chances of freedom.&lt;/p&gt;&#xA;" OwnerUserId="10074" LastActivityDate="2015-03-10T22:27:01.707" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10062" PostTypeId="2" ParentId="9942" CreationDate="2015-03-10T22:32:53.237" Score="12" Body="&lt;p&gt;My current thought for the narrator:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Mark_Hamill&quot;&gt;Mark Hamill&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And thus, for the answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Either The Empire, or the U.S. Navy, as that is where Hamill's father served.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&quot;Look, where I walk!&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Look, walker in the sky&quot; = &quot;Luke Skywalker&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Listen, when I speak!&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hamill has gained a reputation as a prolific voice actor.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Babyface, together with a dog?&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hamill's role Babyface Boretti in the Scooby-Doo movies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Skull, together with cat and mouse!&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In &quot;Marvel Super Hero Squad: Comic Combat&quot;, Hamill did the voice of RED SKULL. Hamill did voice acting for th Tom &amp;amp; Jerry Kids episode &quot;Droopy Man Returns&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;The first time, I was a king&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hamill's first appearance, according to IMDB, was as &quot;King Blackyard / King Karuta&quot; in &quot;Exchange Student Zero&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;The second time my mother was a Queen.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The mother of Luke Skywalker was Queen Padme Amidala.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;But I have also been the Joker&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hamill's role in the animated Batman series.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;In the future I will awake as a Chess champion again.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In chess a champion is also referred to as a &lt;em&gt;Grandmaster&lt;/em&gt;. &#xA; In the Star Wars expanded universe, Luke Skywalker becomes &lt;em&gt;Grand Master&lt;/em&gt; of the Jedi Order. &#xA; Mark Hamill will be playing Luke Skywalker again in the new Star Wars film.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10078" LastEditorUserId="8874" LastEditDate="2015-03-11T13:08:02.497" LastActivityDate="2015-03-11T13:08:02.497" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10063" PostTypeId="2" ParentId="2905" CreationDate="2015-03-10T23:17:09.420" Score="0" Body="&lt;p&gt;The classic problem is that you have two children fighting for a cake. Child A divides the cake in two parts such that it considers both of equal size. Child B now chooses the part it considers bigger. Note that child A might have considered only cake mass whereas child B might have considered the number of cherries, i.e. the value they attach to the parts isn't necessarily equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now how to generalize this to three people? Additionally we're looking for a sharing protocol that is 1. envy-free (i.e. it suffices not to have at least a third but nobody else may have more) and 2. discrete (since the inheritance is not perfectly divisible).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I could write down the protocol here, but it's easier to directly point you to the solution which is the first result on Google: &lt;a href=&quot;http://www.mathematics-in-europe.eu/home/13-frontpage/popular-articles/576-the-cake-cutting-problem&quot; rel=&quot;nofollow&quot;&gt;http://www.mathematics-in-europe.eu/home/13-frontpage/popular-articles/576-the-cake-cutting-problem&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10070" LastActivityDate="2015-03-10T23:17:09.420" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10064" PostTypeId="2" ParentId="10043" CreationDate="2015-03-10T23:26:09.343" Score="1" Body="&lt;p&gt;My interpretation is thus:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;There are 3 doors.&lt;/li&gt;&#xA;&lt;li&gt;One leads to freedom and one to death, and the third door leads to the unknown.&lt;/li&gt;&#xA;&lt;li&gt;The prisoner does not need to use his second question to survive.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The problem as posed has been solved several times above and the best solution is 2/3 chance of freedom and 1/3 chance of survival. However, the OP has provided us two additional pieces of information that cannot be deduced purely logically from the posed problem, but clarify the third given statement.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;(From the OP) Staying in the cell would result in death.&lt;/li&gt;&#xA;&lt;li&gt;(From the OP) The third door returns to the prison, which would constitute survival.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Therefore, the prisoner must be able to leave the cell through one of the two doors that would allow him to survive without asking a second question.&#xA;Then the prisoner must be able to identify either the door that will kill him or one of the doors that will not kill him.&#xA;Therefore, one of the doors must be marked, unlocked, transparent, etc. If he can identify any of the three doors unambiguously, he can identify all three, giving a 100% chance of freedom.&#xA;If he can ambiguously identify one of the doors that will allow him to survive, he can ask &quot;Is going through this door better for me than going through [specific other door]?&quot;. If the marked door actually leads to freedom, the answer will always be no, the prisoner will choose the marked door, and he will have a 100% chance of freedom. If the marked door leads back to prison, he will have a 50/50 chance of identifying the door leading to freedom, giving him a 50% chance of freedom and a 100% chance of survival.&lt;/p&gt;&#xA;" OwnerUserId="10079" LastActivityDate="2015-03-10T23:26:09.343" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10065" PostTypeId="2" ParentId="10023" CreationDate="2015-03-10T23:51:13.670" Score="1" Body="&lt;p&gt;Alice has a very simple starting strategy that guarantees a win at least one-third of the time regardless of Bob's strategy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Randomly choose one of 3, 4, and 5. If she chooses 4 or 5, choose 3 next.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why this works:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All integers are either $3x$, $3x+1$, or $3x+2$ for some integer $x$. If Bob chooses a multiple of 3 there is a 1 in 3 chance Alice will choose 3 and win right away. If Bob chooses a number that's one more than a multiple of 3, Alice has a 1 in 3 chance of choosing 4 first, so $3x+1-4=3x-3=3(x-1)$ which is a multiple of 3, which is what Alice will choose second. If Bob chooses $3x+2$, then if Alice chooses 5 first $3x+2-5=3(x-1)$ and Alice will get it when she chooses 3 second. So because Bob (nor Alice) knows what Alice will choose first, Bob has no way to defend against Alice's strategy.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I believe that there is a strategy under which Alice can always win by choosing the a certain series of numbers in order, however I have not yet found the right series. My reasoning is as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My strategy for trying to determine the series is similar to how &lt;a href=&quot;https://puzzling.stackexchange.com/questions/43/sliding-bolt-puzzle&quot;&gt;the sliding bolt puzzle&lt;/a&gt; works. If you look at this in terms of modular arithmetic, we are simply trying to eliminate possible states and force the solution to move to a single state regardless of where it started from.&lt;br&gt;&lt;br&gt;The way to analyze a particular series of numbers is to look at in terms of the modulo of the least common multiple of all the numbers. For example, let's look at if 2 were allowed, but the only numbers we could use were 2, 3, 4, 6, 8, and 9. The least common multiple of these numbers is 72, so we are interested in X mod 72. Initially, X mod 72 could be any number between 0 and 71. If we choose 2 first and it's not a multiple of 2, then X-2 mod 72 could be any odd number between 1 and 71. If we then choose 3, then X-2 mod 72 could not be 3, 9, etc. so X-5 mod 72 could not be an odd number or 0, 6, 12, etc. There are many more states than in the sliding bolt puzzle, but I feel like there should be a way to whittle away the possibilities until it has been forced into a single state.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1955" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-03-11T02:32:25.890" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10066" PostTypeId="2" ParentId="9942" CreationDate="2015-03-10T23:55:59.193" Score="1" Body="&lt;p&gt;Could the narrator be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Heath_Ledger&quot; rel=&quot;nofollow&quot;&gt;Heath Ledger&lt;/a&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The first time I was a King.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He played Oberon in the 1997 film &lt;em&gt;Paws&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The second time my mother was a Queen.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Probably a reference to another film...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But I have also been the Joker.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He played the Joker in the 2008 film &lt;em&gt;The Dark Knight&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In the future I will awake as a Chess champion again.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He was going to direct and appear in a film version of &lt;a href=&quot;http://en.wikipedia.org/wiki/The_Queen&amp;#39;s_Gambit_(novel)&quot; rel=&quot;nofollow&quot;&gt;this novel&lt;/a&gt; about a chess prodigy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Where did my father serve?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Kim Ledger served in &lt;strong&gt;Australia&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-03-10T23:55:59.193" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10067" PostTypeId="2" ParentId="10043" CreationDate="2015-03-11T00:19:44.483" Score="0" Body="&lt;p&gt;Give me liberty or give me death!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Does this door lead to prison?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm certain this isn't the answer you're looking for, but it certainly reduces the probability of going to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; prison to 0%&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9696" LastActivityDate="2015-03-11T00:19:44.483" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10068" PostTypeId="2" ParentId="9989" CreationDate="2015-03-11T01:45:26.027" Score="3" Body="&lt;p&gt;The location and time of the terrorists' next strike is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the &lt;strong&gt;Pentagon&lt;/strong&gt; at &lt;strong&gt;5:00 pm tomorrow&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The important clues appear to be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - &quot;The Art of Misdirection&quot; suggests that the initial observation of the weird code could be misleading&lt;br&gt;&#xA; - the weird code seems to be several different codes but none are recognizable&lt;br&gt;&#xA; - the words &quot;as key (ASCII)&quot;, &quot;time&quot;, &quot;difference&quot;, and &quot;right&quot;    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The actual code is in:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the &lt;strong&gt;number of seconds between subsequent times&lt;/strong&gt; which are:&lt;br&gt;&#xA; &lt;pre&gt;23 5 8 1 22 5 4 5 3 9 4 5 4 20 15 2 15 13 2 20 8 5 16 5 14 20 1 7 15 14 19 15 13 5 5 20 20 8 5 18 5 1 20 6 9 22 5 16 13 20 15 13 15 18 18 15 23&lt;/pre&gt;&lt;br&gt;&#xA; - Now the numbers can be converted to a letter of the alphabet which provides:&lt;br&gt;&#xA; WEHAVEDECIDEDTOBOMBTHEPENTAGONSOMEETTHEREATFIVEPMTOMORROW &lt;strong&gt;-or-&lt;/strong&gt;&lt;br&gt;&#xA; &lt;strong&gt;We have decided to bomb the Pentagon so meet there at five pm tomorrow&lt;/strong&gt;    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nice puzzle, Tryth. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-03-11T03:16:23.287" LastActivityDate="2015-03-11T03:16:23.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10069" PostTypeId="2" ParentId="9097" CreationDate="2015-03-11T02:01:34.987" Score="11" Body="&lt;p&gt;&lt;strong&gt;Zebras can always win if the lion claw length $r=0$, but the lion wins with claws $r&amp;gt;0$.&lt;/strong&gt;  When the lion has claws, the safety-buffer distance $b$ would have to grow with the number of chasing turns $n$, which would mean the zebras would need to define infinitely large buffer zones to never be caught.  Alternatively, if the zebras know the lion will get-bored/tired/starve-to-death within $n$ turns, then they could still always win even when $r&amp;gt;0$ (however finite-$n$ rules would allow a large number of other simpler solutions like &quot;everyone stand $(n+1)R$ from the lion&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This solution just writes up the math of @Lopsy's $r=0$ solution to show how it kind-of works with claws.  Originally my solution here (incorectly) stated that the zebras could &lt;em&gt;always&lt;/em&gt; win, but there was a math goof in that proof [as pointed out by @Veedrac @Lawrence @Sophit] that has now been fixed to show that this strategy only works for declawed lions.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Re-stated rules:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lion &quot;team&quot; and the zebras teams take turns moving (assume the lion moves first).  During any team's turn only one team member may move.  Both the lion and the zebras can move distance $R$ during their turn.  If a lion gets within distance $r$ of any zebra at the end of either team's turn, then any such zebra becomes lunch and the lion wins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NQk8n.png&quot; alt=&quot;Zebra-Zones Diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Zebra's Strategy Playbook:&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;All of the zebras will line up along an east-west line with a distance of at least $2(R+b)$ between every two adjacent zebras; make sure everyone starts at least $(R+b)$ away from the lion. (Everyone chooses the same distance $b&amp;gt;\sqrt{2nRr+r^2}$)&lt;/li&gt;&#xA;  &lt;li&gt;The land between the points $(R+b)$ east and $(R+b)$ west of each zebra's starting location is considered part of that zebra's zone; all zones continue northward and southward without end.  If the lion ends its turn in any zebra's zone*, then that zebra will use the zebra team's turn to escape by running either due north or due south, whichever direction will end up further from the lion**.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;*If the lion ends its turn between two adjacent zebra zones (zebra zones cannot overlap but could be separated by excess distances), then whichever zebra is currently closest to the lion will run away, in the event that both are equally far from the lion, then the western-most of the two will run (either is fine, but we want to flesh out the zebra algorithm).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;**If the lion ends its turn directly east (or directly west) of the about-to-run zebra such that escaping either north or south will lead to an equal separating distance at the end of the zebra's turn, then the default tiebreaker choice is to run south.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How It Works:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every time the lion ends its turn the Zebra's algorithm chooses a move that will put a nearby zebra further down it's predetermined escape path by the maximum distance allowed by the distance-per-turn rule.  Because all zebras have their individual non-overlapping zones, no two zebras will ever have to fight over a turn and because all escape paths are parallel no lion-move will force a zebra into any area that might jeopardize another zebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/toOlr.png&quot; alt=&quot;Lion-Enters-Zone Diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By defining their zebra zones as infinite north-south strips of land with $(R+b)$ to both east and west directions the zebras are ensuring that the lion never comes closer than the distance $b$ to any zebra at the end of the lion's enter-a-zone step.  This is because the lion, when entering a zone from distance $s$ south of the zebra's current latitude and distance $w$ west of the zebra, starts at a distance $h=\sqrt{(R+b+w)^2+s^2}$ away and can move at most $R$ toward the zebra.  Because the four variables are all real distances, $h$ is minimized when $s\to0$ and $w\to 0$ and the lion is entering a zone by moving perpendicular to its edge.  This zebra will then move north-or-south as per the algorithm in order to put more distance between the lion and itself.  At this point the lion can either pursue this zebra, or try its luck moving to another zone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lHOrh.png&quot; alt=&quot;Lion-n-Chase Diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(This diagram is rotated so that the image resolution wouldn't be super tall and weird.)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Once a lion is in a zone and perusing a single zebra, the following condition occurs: The north-west distance between the lion and zebra $b$ forms the base of a right triangle, and the distance the chased zebra travels makes the side of the right triangle with distance $nR$ (where $n$ is the number of turns un-captured).  The lion travels the shortest distance by moving along the hypotenuse distance $nR$ in $n$ turns, but because of its claws $r&amp;gt;0$ will win if $(nR+r)&amp;gt;b^2+(nR)^2$ (the zebras would need for the $n$th red circle as shown in the picture above to have radius larger than $r$).  Solving the equation for the buffer distance gives $b&amp;gt;\sqrt{2nRr+r^2}$ for a zebra win.  However, this safety-buffer distance requires the zebras knowing the game will end in a finite number of turns; because the game is assumed to allow an infinite number of turns this means that this zebra north-south strategy will fail for any $r&amp;gt;0$ (though it will definitely still work for any choice of $b&amp;gt;0$ in a $r=0$ game).&lt;/p&gt;&#xA;" OwnerUserId="10068" LastEditorUserId="10068" LastEditDate="2015-03-15T15:22:40.300" LastActivityDate="2015-03-15T15:22:40.300" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10070" PostTypeId="2" ParentId="183" CreationDate="2015-03-11T02:21:23.637" Score="5" Body="&lt;p&gt;This is a rewriting by R. Allen Gilliam of &lt;a href=&quot;https://puzzling.stackexchange.com/a/9987/5188&quot;&gt;Jared Anderson's solution&lt;/a&gt; from another version of this puzzle on this site. Perhaps it's just how my mind works, but this seems much easier to understand.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number the men (or coins, or balls) 1 through 12.&lt;br&gt;&#xA;Weigh 1 2 3 4 against 5 6 7 8.&lt;br&gt;&#xA;If they're the same, then the different man is 9 10 11 or 12. Skip to I below.&lt;br&gt;&#xA;If they're different, make note of whether 1 2 3 4 is heavier or lighter.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weigh 1 2 3 5 against 4 10 11 12. (Notice that we know 10 11 and 12 are not the different one.) There are three possibilities:&lt;br&gt;&#xA;(1) If 1235 has the same difference (heavier or lighter) as 1234, then the different one must be 1 2 or 3 and has the same difference (heavier or lighter) as 1234. Skip to II below.&lt;br&gt;&#xA;(2) If 1235 balances 4 10 11 12, then the different one must be 6 7 or 8 (the ones we removed) and has the same difference (heavier or lighter) as 5678. Skip to II below.&lt;br&gt;&#xA;(3) If 1235 now has the opposite difference (heavier or lighter) as 1234, then either 4 or 5 is the different one. Either 4 has the same difference as 1234 (heavier or lighter) or 5 has the same difference as 5678 (heavier or lighter). So we simply weigh 4 against 1. If they're the same, then 5 is the different one. If they're different, then 4 is the different one.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I. Finding which of 9 10 11 12 is different with two weighings when you don't know whether the different one is heavier or lighter:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weigh 9 against 10. Two possibilities:&lt;br&gt;&#xA;(1) If they're different, then it has to be 9 or 10. Weigh 9 and 11. If they're the same, 10 is the different one. If they're different, it's 9.&lt;br&gt;&#xA;(2) If they're the same, then it has to be 11 or 12. Weigh 9 and 11. If they're the same, 12 is the different one. If they're different, it's 11.&lt;br&gt;&#xA;(If it's 12, we won't know whether he was heavier or lighter since we never weighed him. We found him by process of elimination. He must be the different one since all the others weigh the same.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;II. Finding which of three men is different with one weighing when you know whether the different one is heavier or lighter:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rename the three men 1 2 3. Weigh 1 against 2. Two possibilities:&lt;br&gt;&#xA;(1) If they're the same, 3 is the different one.&lt;br&gt;&#xA;(2) If they're different, whichever one has the correct difference (heavier or lighter) is the different one.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to be the simplest solution for 12 items if you only must find the item of different weight, as some versions of the puzzle ask. Joe Z's solution can find the item and the difference with 12 items, and the different item with 13 items. Finding the different item and the difference with 14 items seems mathematically impossible with 3 weighings because there are only 27 possible outcomes with 3 weighings and there are 28 possibilities with 14 items. But could a variation of Joe Z's solution find the different item out of 13, and whether it's heavier or lighter? If so, then finding the different one but not the difference with 14 items would be possible. Finding the different one but not the difference out of 15 would be impossible because you can leave only one item out of the weighings while still being able to identify the different one, and if you weigh the item then you'll know whether it's lighter or heavier which we know is mathematically impossible with 14 items.  &lt;/p&gt;&#xA;" OwnerUserId="10084" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.923" LastActivityDate="2015-03-12T18:43:20.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10071" PostTypeId="2" ParentId="9996" CreationDate="2015-03-11T03:01:42.773" Score="0" Body="&lt;p&gt;The answer is :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nascar driver&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My job is putting things in rings&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are paid to drive cars around tracks, which often are in ring shape&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My prize is rings&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Winning drivers are awarded championship rings, as it happens for other american sports.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The number of rings symbolizes speed&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pretty straightforward. The more wins you have, the faster you are in theory.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When people see me in action they say 'Oh neat'&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm not a fan personally but some people really enjoy it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10085" LastActivityDate="2015-03-11T03:01:42.773" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10072" PostTypeId="1" AcceptedAnswerId="10078" CreationDate="2015-03-11T03:52:23.323" Score="2" ViewCount="616" Body="&lt;p&gt;This is a question I'm actually looking for the answer to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's known that when you swap individual pieces on a Rubik's Cube, the probability that afterwards it will be possible to solve is 1/12.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say that you take all the possible combinations of cycling N stickers simultaneously. By the amount of cycled stickers, I mean the minimum amount of stickers you need to cycle to make that combination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the largest N that the probability of this cube being impossible to solve is greater than 11/12? Is there a way to do this without brute forcing it? Is there no limit?    &lt;/p&gt;&#xA;" OwnerUserId="167" LastEditorUserId="5373" LastEditDate="2015-03-11T21:37:31.300" LastActivityDate="2015-03-14T09:43:08.947" Title="How many stickers need to be cycled on a Rubik's cube before the chance it's unsolvable drops below 11/12?" Tags="|rubiks-cube|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="10073" PostTypeId="1" AcceptedAnswerId="10076" CreationDate="2015-03-11T05:14:34.387" Score="32" ViewCount="22936" Body="&lt;p&gt;Suppose you are on an island with 32 monks and you are told that one of the monks is always honest. Every day the monks say whether it is going to rain. You are required to say whether it is going to rain or not. Find a strategy that would make the minimum number of mistakes and what would be the worst case number of mistakes with this strategy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(We can consider monks can be either honest or dishonest. Honest monks always know and correctly predict true future weather while dishonest monks might or might not predict true future weather.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was asked to me during an interview and I came up with this approach : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will start noting down the observations and the data collected from monks.We will select the answer which at least 51% of the monks agree to. If they collude and we make a mistake, on the next day prediction we can just remove those 51% monks. So basically it's either removal of 51% or continuing with the given monks. Hence worst case mistakes would be 5 (geometric progression with r = 1/2 last term = 1 and first term = 32)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please also comment on whether this approach is correct and if any other better approach is available.&lt;/p&gt;&#xA;" OwnerUserId="10091" LastEditorUserId="8874" LastEditDate="2016-03-11T16:30:42.650" LastActivityDate="2016-03-11T16:30:42.650" Title="Honest Monk Puzzle" Tags="|mathematics|logical-deduction|strategy|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10074" PostTypeId="2" ParentId="10043" CreationDate="2015-03-11T05:14:56.103" Score="1" Body="&lt;p&gt;&quot;Will I survive if I go through the left or the middle door?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Warden answers &quot;Yes&quot;, then pick one at random. 50% chance at freedom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Warden answers &quot;No&quot;, then go through the other door. 50% chance at freedom.&lt;/p&gt;&#xA;" OwnerUserId="10093" LastActivityDate="2015-03-11T05:14:56.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10075" PostTypeId="2" ParentId="10073" CreationDate="2015-03-11T05:34:12.763" Score="6" Body="&lt;p&gt;&lt;a href=&quot;http://ideone.com/4LLeMc&quot; rel=&quot;nofollow&quot;&gt;Answer will be 5&lt;/a&gt; (log N) with your approach of choosing max:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#include &amp;lt;iostream&amp;gt;&#xA;#include &amp;lt;algorithm&amp;gt;&#xA;&#xA;using namespace std;&#xA;&#xA;int memo[33];&#xA;&#xA;int mistakes_count(int n)&#xA;{&#xA;    if (memo[n] == -1) {&#xA;        int mx = 0;&#xA;        for(int i = 1; i &amp;lt; n; ++i) { // 1 must be honest&#xA;            int catr = i, catw = n - i; // If all are honest, repeat&#xA;            if(catr &amp;gt; catw) { // Category right is the majority&#xA;                mx = std::max(mx, mistakes_count(catr));&#xA;            } else {  // Category wrong is the majority, 1 mistake done&#xA;                mx = std::max(mx, 1 + mistakes_count(catr));&#xA;            }&#xA;        }&#xA;        memo[n] = mx;&#xA;    }&#xA;    return memo[n];&#xA;}&#xA;&#xA;int main() {&#xA;    for(int i = 1; i &amp;lt;= 32; ++i) memo[i] = -1;&#xA;    cout &amp;lt;&amp;lt; mistakes_count(32);&#xA;    return 0;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Only way to make you wrong is that honest monk fall in minority. Now to make sure that minority consists of maxumum monks (to make the process long and more mistakes) is, there are (almost or exactly) equal votes casted for both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R = W or R + 1 = W&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in worst case, you would be eliminating one of the monks with 1 mistake in five iterations.&lt;/p&gt;&#xA;" OwnerUserId="4637" LastEditorUserId="4637" LastEditDate="2015-03-11T05:56:49.730" LastActivityDate="2015-03-11T05:56:49.730" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10076" PostTypeId="2" ParentId="10073" CreationDate="2015-03-11T05:47:24.737" Score="26" Body="&lt;p&gt;(1) Your approach and your analysis are correct, and with your strategy you will always make at most five wrong predictions.&#xA;Every time you make a wrong prediction, this halves the number of candidates for the honest monk. The number $32$ can be halved at most five times before you are down to a single candidate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Your approach is also optimal in the following sense:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is no strategy that would guarantee you at most four wrong predictions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof:&#xA;For the first five days, there are $2\cdot2\cdot2\cdot2\cdot2=32$ different ways of predicting rain/no rain.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We let the $32$ monks collude, so that each of them predicts the first five days in a different way.&lt;/li&gt;&#xA;&lt;li&gt;We design the weather so that each of your first five predictions is incorrect.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;After five days, you will have made five mistakes.&#xA;Nevertheless, one of the monks has made five correct predictions; that guy is designated to be the honest monk.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-03-11T05:53:25.223" LastActivityDate="2015-03-11T05:53:25.223" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="10078" PostTypeId="2" ParentId="10072" CreationDate="2015-03-11T07:58:16.227" Score="4" Body="&lt;p&gt;The probability that it will be impossible if you swap &lt;em&gt;stickers&lt;/em&gt; will never drop below $11/12$. The reason for this is actually rather mathematically simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, an interchange of $N$ stickers is equivalent to some sequence of interchanges of two stickers. Thus, the problem you're asking about is exactly the same as asking if an equivalent sequence of 2-sticker swaps puts the cube in a solvable state. By solving the 2-sticker-swap problem, we'll solve the problem for &lt;em&gt;any number&lt;/em&gt; of sticker swaps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the cube, after some number of sticker swaps, is in a state $P$, and suppose $P$ is not solvable cube. Therefore, there must be at least two stickers out of place. In order to reach any solvable state $S$, a sequence of moves must be executed such that, at some point, exactly two stickers must be swapped to reach an $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an important fact. At the point where $P\mapsto S$ requires only one swap ($n=1$), &lt;strong&gt;the probability of making the correct swap is exactly equal to the probability of selecting the two correct stickers.&lt;/strong&gt; As there are 54 stickers on the cube, the probability of selecting the first correct stickers is $1/54$. The probability of the second is $1/45$, because selecting a sticker of the same color wouldn't change the cube state. Therefore, the probability for $n=1$ is: $$p_{n=1}=\frac 1 {54*45}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's one edge case where this doesn't hold true. When both needed stickers are on different edges, there are actually two ways to swap pieces into a solved state. This is because swapping two edges on a Rubik's cube is legal. However, this is the only such situation, since the same symmetry doesn't apply to corners or centers. Thus, in this case: $$p_{n=1,edge}=\frac 2{54*45}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $P\mapsto S$ requires more than one swap ($n&amp;gt;1$), then there is no single swap that will put the cube in a solvable state, so the probability of entering a solvable state is $$p_{n&amp;gt;1}=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By basic probability, the probability that the cube will enter an &lt;em&gt;un&lt;/em&gt;solvable state is $1-p$, which is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;1-p_{n=1}&amp;amp;=1-\frac{1}{54*53}&amp;amp;=\frac{2429}{2430}\\&#xA;1-p_{n=1,edge}&amp;amp;=1-\frac{2}{54*53}&amp;amp;=\frac{2428}{2430}\\&#xA;1-p_{n&amp;gt;1}&amp;amp;=1-0&amp;amp;=1&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, if you need more than one swap to put the cube in a solvable state, then no matter what you do, a single swap won't make the cube solvable. If you need only one swap to make the cube solvable, then there's a good chance you won't pick the right one. Because we know that: $$1&amp;gt;\frac{2429}{2430}&amp;gt;\frac{2428}{2430}&amp;gt;\frac{11}{12}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the probability that the cube will be solvable after swapping two stickers will never fall below $11/12$. &lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorDisplayName="user20" LastEditDate="2015-03-13T17:40:04.547" LastActivityDate="2015-03-13T17:40:04.547" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="10079" PostTypeId="2" ParentId="10031" CreationDate="2015-03-11T08:31:05.023" Score="5" Body="&lt;p&gt;This answer is probably wrong, but I will try anyway:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The (controversial) picture:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;&lt;strong&gt;French ass&lt;/strong&gt;&quot; sounds very close to &quot;&lt;strong&gt;Friend chess&lt;/strong&gt;&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And about the 8 R:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;&lt;strong&gt;Eight Er&lt;/strong&gt;&quot; sounds very close to &quot;&lt;strong&gt;Hater&lt;/strong&gt;&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So this is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Friend chess hater&lt;/strong&gt; which could also mean &lt;strong&gt;cheater&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5044" LastEditorUserId="5044" LastEditDate="2015-03-11T17:33:03.377" LastActivityDate="2015-03-11T17:33:03.377" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10080" PostTypeId="1" AcceptedAnswerId="10095" CreationDate="2015-03-11T08:36:43.887" Score="17" ViewCount="3577" Body="&lt;p&gt;In a certain language if the word STABILIZE is coded as JHXDJQSYA, then how is the word RESPONSES coded in that language?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;LRCRPOQDP  &lt;/li&gt;&#xA;&lt;li&gt;PDQOPLRCR&lt;/li&gt;&#xA;&lt;li&gt;LRCRPPDQO&lt;/li&gt;&#xA;&lt;li&gt;LQCRPPDQO  &lt;/li&gt;&#xA;&lt;li&gt;LRCRPPDOQ&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I tried it &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;STABILIZE - 19|20| 1| 2| 9|12| 9|26| 5|&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;JHXDJQSYA - 10| 8|24| 4|10|17|19|25| 1|&lt;/code&gt; But i didn't find any clue.&lt;/p&gt;&#xA;" OwnerUserId="10098" LastEditorUserId="5127" LastEditDate="2015-03-11T11:29:43.203" LastActivityDate="2015-03-11T15:48:24.910" Title="In a certain language if the word STABILIZE is coded as JHXDJQSYA, then how is the word RESPONSES coded in that language?" Tags="|calculation-puzzle|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10081" PostTypeId="2" ParentId="9996" CreationDate="2015-03-11T09:55:56.900" Score="0" Body="&lt;p&gt;Is it a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Rocket (or  multistage launch vehicle)?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My job is putting things in rings.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Rockets or launch vehicles are normally used to put satellites into orbit, which is definitely a ring.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My prize is rings.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Not sure. May be referring to the interstage rings or stages. Or may be it's about reaching orbit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The number of rings I have symbolizes speed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The interstage rings are used as the separation between different stages ( each stage consists of contains its own engines and propellant) and it can be used to determine the speed or capacity of the rocket. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When people see me in action they say Oh, neat!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;On watching the launching of rockets, people normally wonder Wow, Great, Oh neat etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastActivityDate="2015-03-11T09:55:56.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10085" PostTypeId="1" AcceptedAnswerId="10087" CreationDate="2015-03-11T11:42:20.650" Score="8" ViewCount="1762" Body="&lt;p&gt;In &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_sovereign_states&quot; rel=&quot;nofollow&quot;&gt;this list&lt;/a&gt; of the world's sovereign states, exactly sixteen of the UN member states and observer states have a certain property P, and so does the disputed territory of Abkhazia.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Exactly 2 of these countries are in South America.&lt;/li&gt;&#xA;&lt;li&gt;Exactly 5 are in Africa.&lt;/li&gt;&#xA;&lt;li&gt;Exactly 4 (not counting Abkhazia) used to be in the USSR.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This should be enough to ensure the puzzle has a unique solution; I can add more clues if necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the property P? Which are the sixteen countries?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-11T16:46:51.433" LastActivityDate="2015-03-12T15:16:22.690" Title="How many countries in each continent?" Tags="|geography|enigmatic-puzzle|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10086" PostTypeId="2" ParentId="10023" CreationDate="2015-03-11T12:04:57.130" Score="1" Body="&lt;p&gt;Edited:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is wrong - see Lopsy's answer. The hole in the proof is the claim that Alice removes at most 1/n numbers with a pick of n. Because of the existence of covering systems, she starts off with a lot more information than I thought, and so she can remove more numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that Erdos is smarter than me...&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;No, Alice can't force a win, because she doesn't get enough information from each turn to improve her choices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any time Alice picks a number n, she removes at most 1/n numbers from the pool of Bob's possible picks (less than that if she picks badly). She can't repeat, so after k picks the best she could have done is to leave $\frac{2}{3}.\frac{3}{4}...\frac{k+1}{k+2}$ of the entries still in the pool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The log of $\frac{k-1}{k}$ approximates to $-\frac{1}{k}$, so the log of the entire fraction will be more than $-\frac{1}{3} - \frac{1}{4} - ... - \frac{1}{k+2}$, which is more than $-(ln(k+2) + 1)$, since it's a subset of the harmonic series. Therefore the fraction of the original numbers that remain is larger than $\frac{1}{e.(k+2)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the sum of those k picks must be at least $\frac{k(k+1)}{2}$. Multiplying those two numbers together we get $\frac{k(k+1)}{2e(k+2)}$ which will be larger than 1 for k &gt; 6, and converges on $\frac{k}{2e}$, a steadily increasing number. This means there are at least some numbers which Alice can't eliminate and which will win for Bob.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, if Bob knows what Alice's numbers will be, he can win - so there can't be a forcing strategy for Alice. She can win (her odds are clearly at least 1/3), but can't guarantee it. The chances of a particular number winning will converge on $\frac{1}{ek}$, so the bigger the number Bob picks, the worse his chances. He'll do best with a randomly chosen number near 100.&lt;/p&gt;&#xA;" OwnerUserId="6031" LastEditorUserId="6031" LastEditDate="2015-03-12T12:24:10.903" LastActivityDate="2015-03-12T12:24:10.903" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10087" PostTypeId="2" ParentId="10085" CreationDate="2015-03-11T12:15:07.910" Score="15" Body="&lt;p&gt;Contain the letter Z (in the English spelling) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Azerbaijan · Belize · Bosnia and Herzegovina · Brazil · Czech Republic · Kazakhstan · Kyrgyzstan · Mozambique · New Zealand· Swaziland · Switzerland · Tanzania · Uzbekistan · Venezuela · Zambia · Zimbabwe&lt;/p&gt;&#xA;" OwnerUserId="10057" LastActivityDate="2015-03-11T12:15:07.910" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10088" PostTypeId="2" ParentId="10080" CreationDate="2015-03-11T12:29:25.953" Score="24" Body="&lt;p&gt;With problems like this, I normally try to figure it out based on the set of possible answers.  Multiple-choice answers very often contain the correct answer and various permutations of it, so the &quot;consensus&quot; answer is probably the correct one.  Looking down the columns, we find the most common letter at each position is: &lt;code&gt;LRCRPPDQO&lt;/code&gt;, which happens to be answer 3.  So it's probably 3, but it would be nice to find at least part of an encoding algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that &lt;code&gt;STABILIZE&lt;/code&gt; and &lt;code&gt;JHXDJQSYA&lt;/code&gt; have the same number of letters.  So we presumably need only look for algorithms that compute a slot as a function of letters in the other slots, not fancy deletion/expansion stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, let's look for repeated letters: &lt;code&gt;I&lt;/code&gt; occurs at positions 4 and 6 (counting from 0) in &lt;code&gt;STABILIZE&lt;/code&gt; but the repeats in &lt;code&gt;JHXDJQSYA&lt;/code&gt; occur at positions 0 and 4.  Thus it's not &lt;em&gt;merely&lt;/em&gt; a letter-by-letter translation.  Furthermore it's not &lt;em&gt;just&lt;/em&gt; translation and shuffling, since there are three &lt;code&gt;S&lt;/code&gt;s in &lt;code&gt;RESPONSES&lt;/code&gt; but no trio of identical letters in any of the possible answers.  Thus, there is presumably an impact of either position or previous letter(s).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For our next hint, let's look at the answers.  (2) is just &lt;code&gt;RESPONSES&lt;/code&gt; minus one at each location, and that's clearly &lt;em&gt;not&lt;/em&gt; what is happening with &lt;code&gt;STABILIZE&lt;/code&gt;, so we can throw &lt;code&gt;2&lt;/code&gt; out.  Thus, &lt;code&gt;RESPONSES&lt;/code&gt; translates to something that looks like this: &lt;code&gt;L_CRP____&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we just brute force various operations on the letters and look for something to jump out.  The first interesting pattern I found was if you sum the letters (&lt;code&gt;A == 0&lt;/code&gt;):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; STABILIZE&#xA; JHXDJQSYA&#xA;+--------- (mod 26)&#xA; BAXERBAXE&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This suggests that whatever factors influence the first four characters, it's the same on the last four (in the same order).  So we have a pattern: if you go forward by 5, as many letters as you go up (down) in the original, you go the opposite in the encoded version.  For example, &lt;code&gt;'J' + 'S' - 'L' == 'Q'&lt;/code&gt; to get answer character 5 given 0 and the original 0 and 5.  We take what we know, and get:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;RESPONSES&#xA;L_CRPP_QO&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And that leaves only two possible answers: (3) and (4).  Do either obey the rule?  We have &lt;code&gt;R&lt;/code&gt; and &lt;code&gt;D&lt;/code&gt; or &lt;code&gt;Q&lt;/code&gt; and &lt;code&gt;D&lt;/code&gt;; 'R' + 'E' - 'S' is 'D', so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;it's #3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;LRCRPPDQO&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;just like we predicted from examining the structure of the answers.  Why, exactly?  I'm not sure.  But we don't really need to know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to check that we really are on the right track, let's sum our proposed answer:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; RESPONSES&#xA;+LRCRPPDQO&#xA; ---------  (mod 26)&#xA; CVUGDCVUG&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And then if we look at the difference between these two answers:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; C  V  U  G&#xA;-B  A  X  E&#xA;== == == ==&#xA; 1 -5 -3  2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And if we look at the relevant pair of letters and sum them, then take the difference between the two words, we get:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;'R' + 'N' - 'S' - 'L' = 1&#xA;'E' + 'S' - 'T' - 'I' = -5&#xA;'S' + 'E' - 'A' - 'Z' = -3&#xA;'P' + 'S' - 'B' - 'E' = 28 = 2 (mod 26)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So these values clearly arise from some predictable operations on the letters; the &lt;code&gt;BAXERBAXE&lt;/code&gt; values are not merely random as we can predict the &lt;code&gt;CVUG&lt;/code&gt; pattern from them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could guess that there is salt with values (6, 6, -2, -1, 1) and then the encoding formula is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;First 4 letters: letter at index + letter at (index+5) + salt at index&lt;/li&gt;&#xA;&lt;li&gt;Central letter: letter + final salt value (i.e. 1)&lt;/li&gt;&#xA;&lt;li&gt;Last 4 letters: coded letter at (index - 5) + letter at (index - 5) + letter at index&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This matches everything that is specified, though it's possible that the salt is intended to be the result of some other operation.&lt;/p&gt;&#xA;" OwnerUserId="2815" LastEditorUserId="2815" LastEditDate="2015-03-11T13:39:43.237" LastActivityDate="2015-03-11T13:39:43.237" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10089" PostTypeId="2" ParentId="10023" CreationDate="2015-03-11T13:07:30.770" Score="15" Body="&lt;p&gt;If Alice plays [3, 7, 5, 8, 6, 15, 4, 10, 27, 12, 9, 20, 13, 24, 19, 60, 32, 40, 38, 30, 72, 120]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and Bob's integer is greater than 202&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then Alice wins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: check 203 through 203+120. Check that Alice's strategy covers every residue class (mod 120).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have fun optimizing 202 down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this sequence using a program and &lt;a href=&quot;http://en.wikipedia.org/wiki/Covering_system&quot;&gt;this page&lt;/a&gt;. Specifically, Erdos discovered that every integer lies in one of the modular residue classes 0 (mod 3), 0(4), 0(5), 1(6), 1(8), 2(10), 11(12), 1(15), 14(20), 5(24), 8(30), 6(40), 58(60), or 26(120). By inserting numbers between these guesses, Alice can cover all of these residue classes.&lt;/p&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2015-03-11T13:07:30.770" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10090" PostTypeId="2" ParentId="10080" CreationDate="2015-03-11T13:18:55.487" Score="4" Body="&lt;p&gt;My answer is quite similar to that of Rex Kerr's just the method is different .I found out that encryption is symmetric about middle point that is &lt;code&gt;B&lt;/code&gt; in &lt;code&gt;STABILIZE&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;S  T  A  B  I  L  I  Z  E&#xA;19 20 1  2  9  12 9  26 5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;J  H  X  D  J  Q  S  Y  A &#xA;10 8  24 4  10 17 19 25 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;when they are summed (mod 26) they produce &lt;code&gt;3 2 25 6 19 3 2 25 6&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the work was now to just sum up the word &lt;code&gt;RESPONSES&lt;/code&gt; with given answers and find it's symmetricity.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;R  E  S  P  O  N  S  E  S &#xA;18 5  19 16 15 14 19 5  19&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The only solution which shows the symmetric property is &lt;code&gt;LRCRPPDQO&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;L  R  C  R  P  P  D  Q  O&#xA;12 18 3  18 16 16 4  17 15&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which when summed(mod 26) with &lt;code&gt;RESPONSES&lt;/code&gt; gives &lt;code&gt;4 23 22 8 5 4 23 22 8&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit : By 'symmetric' I mean the repeating nature of the result around the middle point which is &lt;code&gt;I&lt;/code&gt; in &lt;code&gt;STABILIZE&lt;/code&gt; and &lt;code&gt;O&lt;/code&gt; in &lt;code&gt;RESPONSES&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10091" LastEditorUserId="10091" LastEditDate="2015-03-11T13:40:12.423" LastActivityDate="2015-03-11T13:40:12.423" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10091" PostTypeId="2" ParentId="10050" CreationDate="2015-03-11T13:45:47.840" Score="2" Body="&lt;p&gt;How about:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Better?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Expert, recognized skills:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Better ability than peers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fire 'within':&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Inner letters are &quot;ette&quot;, which is apparently colloquial for an incensed feeling &lt;a href=&quot;http://www.urbandictionary.com/define.php?term=Ette&quot; rel=&quot;nofollow&quot;&gt;http://www.urbandictionary.com/define.php?term=Ette&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Gambling:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; someone who bets = better.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It doesn't fits the hint particularly well, though.&lt;/p&gt;&#xA;" OwnerUserId="9422" LastActivityDate="2015-03-11T13:45:47.840" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10092" PostTypeId="2" ParentId="10050" CreationDate="2015-03-11T13:59:03.330" Score="8" Body="&lt;p&gt;I think it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PROBABILITY&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Firstly, this fits the title.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I began as an expert&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A &lt;strong&gt;pro&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But later my skills are still recognized.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My &lt;strong&gt;ability&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Once a fire was started within me&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I started to burn ... we want the first letter of the word &lt;strong&gt;b&lt;/strong&gt;urn. (Alternatively this could mean &lt;strong&gt;lit&lt;/strong&gt;, which appears in the word probability - thanks Moghwyn!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And after asking why I did it all&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;why&quot; = Y, at the end of the word&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You'll find I was heavy into gambling.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well, that could be just another reference to the whole thing, like the title.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-11T14:14:58.880" LastActivityDate="2015-03-11T14:14:58.880" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10093" PostTypeId="2" ParentId="10050" CreationDate="2015-03-11T14:02:00.883" Score="4" Body="&lt;p&gt;I began as an expert&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Also known as a &quot;pro&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But later my skills are still recognized.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Due to your &quot;abilities&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once a fire was started within me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The fire was &quot;lit&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And after asking why I did it all&lt;br&gt;&#xA;You'll find I was heavy into gambling.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Putting it all together, gambling is based on &quot;probabilities&quot;, which I assume to be the answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9910" LastActivityDate="2015-03-11T14:02:00.883" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10094" PostTypeId="2" ParentId="2188" CreationDate="2015-03-11T14:29:17.890" Score="-3" Body="&lt;p&gt;The answer to this riddle is to merely pose an single interogative that completes a simple alternating composite of positive and negative interdependent propositions.  By stringing the propositions you can determine the door to freedom by posing the question so that the answer successfully or unsuccessfully completes the alternating polarity.  Regardless of how you pose the question the polarity must alternate [negative, positive, negative with the answer providing the last swing back to positive] or [positive, negative, positive with the answer providing the last swing back to negative] to be successful.  If and when it does not alternate, that will indicate the required action just as definitively as when and if it does successfully alternate.&#xA;Below, I will pose the question with the corresponding polarity; keep in mind that your question can utilize either door and either guard; to wit:&#xA;- Would you be telling the truth [positive], if you said that he would be lying [negative], if he said that this door is the door to freedom [positive]?  Notice, [positive, negative, positive].  If the guard you directed the question to answers &quot;NO&quot; [negative], you have completed the correct alternating polarity, [positive, negative, positive,negative] and the door to freedom is the door you did not point out.&lt;br&gt;&#xA;or you could ask it in the opposite fashion; to wit:&#xA;- Would you be lying [negative], if you said that he was telling the truth [positive], if he said that this is the door to certain death [negative]?  Notice [Negative, positive, negative].  If he answers Yes [positive], you have again completed the alternating polarity correctly and your door to freedom is the one you pointed out. [Negative, Positive, Negative, Positive].  In either case, if the polarity does not correctly alternate, you should proceed as indicated from the context of your interrogative. Simple &lt;/p&gt;&#xA;" OwnerUserId="10115" LastActivityDate="2015-03-11T14:29:17.890" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10095" PostTypeId="2" ParentId="10080" CreationDate="2015-03-11T15:48:24.910" Score="11" Body="&lt;pre&gt;&lt;code&gt;Generally, these kind of problems have odd number of characters.&#xA;&#xA;Now always try to focus on middle character(n/2 + 1) in whole word(say 'n' characters)&#xA;&#xA;Step 1:&#xA;STAB   I   LIZE &#xA;    ( +1 )      &#xA;       J   &#xA;&#xA;Step 2:(Take 4 characters to the middle on right)&#xA;  L    I      Z    E&#xA;(-2)  (-1)  (-2) (-1)&#xA;  J    H      X    D&#xA;&#xA;Step 3:(Take 4 characters to the middle on left)&#xA;  S     T     A     B &#xA;(-2)  (-1)  (-2)  (-1)&#xA;  Q     S     Y     A  &#xA;&#xA;Step 4:(Merge 2,1 and 3)&#xA;  JHXDJQSYA&#xA;&#xA;&#xA;Now, coming back to word &quot;**RESPONSES**&quot; &#xA;&#xA;&#xA;Step 1:&#xA;RESP   0   NSES &#xA;    ( +1 )      &#xA;       P   &#xA;&#xA;Step 2:&#xA;  N     S     E    S&#xA;(-2)  (-1)  (-2) (-1)&#xA;  L     R     C    R&#xA;&#xA;Step 3:&#xA;  R     E     S     P &#xA;(-2)  (-1)  (-2)  (-1)&#xA;  P     D     Q     O  &#xA;&#xA;Step 4:&#xA;  LRCRPPDQO&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="10119" LastActivityDate="2015-03-11T15:48:24.910" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10096" PostTypeId="2" ParentId="10073" CreationDate="2015-03-11T17:17:58.977" Score="0" Body="&lt;p&gt;I am assuming that by &quot;honest&quot; you mean can accurately predict the weather. If not this question is kind of meaningless. An 'honest' guess with 50% accuracy is indistinguishable from a 'dishonest' guess with 50% accuracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do we know the other 31 monks are dishonest? If they are then it doesn't take any tries at all. One monk will have a different answer from the other 32.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &lt;em&gt;some of the other monks&lt;/em&gt; are dishonest, and some are honest, it's more difficult, and will take a maximum of 5 tries to figure out, depending on how many are honest and how many are dishonest.&lt;/p&gt;&#xA;" OwnerUserId="8746" LastActivityDate="2015-03-11T17:17:58.977" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10098" PostTypeId="1" AcceptedAnswerId="10099" CreationDate="2015-03-11T19:54:17.753" Score="-1" ViewCount="426" Body="&lt;p&gt;I could be a bolt&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will meet you at the 100m before the world could catch me&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ask me using my name, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will say &quot;Yes&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I love yellow and green and my lightning-hot nickname&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am not Sonic the Hedgehog&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No more hints as the clues can be deciphered without any more help. Good Luck!&lt;/p&gt;&#xA;" OwnerUserId="9042" LastActivityDate="2015-03-11T19:57:55.537" Title="What Guy Am I, Am I Sonic?" Tags="|riddle|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10099" PostTypeId="2" ParentId="10098" CreationDate="2015-03-11T19:57:55.537" Score="7" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Usain Bolt&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I could be a bolt&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bolt is your last name.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I will meet you at the 100m before the world could catch me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You own the world record in the 100 meter dash.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ask me using my name,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If I ask you &quot;Usain&quot; as in &quot;You sane?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I will say &quot;Yes&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Presumably you will say you are sane.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I love yellow and green and my lightning-hot nickname&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jamaica's colors are yellow and green, and your nickname is &quot;Lightning Bolt&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2015-03-11T19:57:55.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10100" PostTypeId="1" AcceptedAnswerId="10115" CreationDate="2015-03-11T20:19:23.580" Score="18" ViewCount="3172" Body="&lt;p&gt;I go in a cup,&lt;br&gt;&#xA;but you can't drink me up,&lt;br&gt;&#xA;I soar through the air,&lt;br&gt;&#xA;well above your hair,&lt;br&gt;&#xA;At first I have need&lt;br&gt;&#xA;of some serious speed.&lt;br&gt;&#xA;Later, it's nice&lt;br&gt;&#xA;to be more concise.&lt;br&gt;&#xA;To the club we could go&lt;br&gt;&#xA;if it doesn't snow&lt;br&gt;&#xA;If I go to the beach&lt;br&gt;&#xA;You will probably screech.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: changed &quot;don't&quot; to &quot;you can't&quot; on line two to clarify.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is a physical object you can touch and hold in your hand. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="9951" LastEditDate="2015-03-11T23:22:45.363" LastActivityDate="2015-03-12T07:22:30.367" Title="All through the Seasons - What am I?" Tags="|riddle|word|rhyme|" AnswerCount="6" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10101" PostTypeId="1" AcceptedAnswerId="25473" CreationDate="2015-03-11T20:20:30.923" Score="4" ViewCount="1999" Body="&lt;p&gt;I'm the thing you would love&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet I was a murderer by the millions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was a poison of the 1900's&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then a revolutionary medicine in the hospitals&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My other use a massage to your blood, the other the killer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Read the title. Why didn't I just put What am I?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Don't take the last line literally!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9042" LastEditorUserId="9042" LastEditDate="2015-03-14T23:28:50.737" LastActivityDate="2016-02-18T14:52:01.577" Title="The Gas of Mystery, What am I?" Tags="|riddle|" AnswerCount="5" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10103" PostTypeId="2" ParentId="10100" CreationDate="2015-03-11T20:39:09.917" Score="1" Body="&lt;p&gt;You might be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Urine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I go in a cup,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Urine can go in a cup if you need to be tested&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but don't drink me up,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You shouldn't drink urine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I soar through the air,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Urine goes through the air when you urinate&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;well above your hair,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; OK urine doesn't really fit this clue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At first I have need of some serious speed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you start to urinate, it comes out fast.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later, it's nice to be more concise.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you're about done, it slows to a trickle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To the club we could go if it doesn't snow&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Urine doesn't fit this clue either.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I go to the beach You will probably screech. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you are urinating at the beach, you probably got stung by a jellyfish and it really hurts.  Urinating on yourself supposedly eases the pain of a jellyfish sting.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2015-03-11T20:39:09.917" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10104" PostTypeId="2" ParentId="10100" CreationDate="2015-03-11T21:04:30.883" Score="2" Body="&lt;p&gt;You might be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water, Contaminated Water to be More Precise&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but don't drink me up,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Don't drink dirty water unless you want to be poisoned.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I soar through the air,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As a cloud, I can move through the air.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;well above your hair,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Unlike fog, I always will be high in the sky&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At first I have need of some serious speed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Storms will quickly bring a lot of rain/dirty water&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later, it's nice to be more concise.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A little sprinkle or some fun with water in the summer is great&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To the club we could go if it doesn't snow&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If it is that cold, water would turn into ice. Even water would like to stay cool.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I go to the beach, You will probably screech&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As waves, you would scream out joy. If you forgot how to swim, you will scream with your life. If you love playing in the water like most kids, you will have lots of fun and scream with joy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9042" LastActivityDate="2015-03-11T21:04:30.883" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10106" PostTypeId="2" ParentId="10101" CreationDate="2015-03-11T21:17:31.480" Score="6" Body="&lt;p&gt;My fourth attempt:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CHLORINE.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I'm the thing you would love&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm not sure about this bit (though &quot;love chlorine&quot; yields &lt;a href=&quot;https://www.google.co.uk/search?complete=0&amp;amp;hl=en-GB&amp;amp;source=hp&amp;amp;q=love+chlorine&amp;amp;btnG=Google+Search&amp;amp;gbv=2&quot;&gt;4 million hits on Google&lt;/a&gt; :-p )&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I was a murderer by the millions &lt;br&gt;&#xA;I was a poison on the 1900's&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chlorine gas was the famous poison gas used to kill millions in WWI (&lt;a href=&quot;http://en.wikipedia.org/wiki/Chlorine#Use_as_a_weapon&quot;&gt;ref&lt;/a&gt;). &lt;br&gt;&lt;br&gt; &lt;em&gt;Gas! Gas! Quick, boys! – An ecstasy of fumbling&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Then a revolutionary medicine in the hospitals&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chlorine chemistry is essential to more than 93% of prescription pharmaceuticals sold in the US (&lt;a href=&quot;http://chlorine.americanchemistry.com/Chlorine-Benefits/Health-and-Safety/Chlorine-and-Medicine.html&quot;&gt;ref&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My other use a massage to your blood, the other the killer&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm not sure about this line yet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Previous attempt(s)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;(I removed the first possibility since I don't think it's killed millions of people.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mustard gas &lt;em&gt;aka&lt;/em&gt; sulphur mustard.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I'm the thing you would love&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm not sure about this bit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I was a murderer by the millions &lt;br&gt;&#xA;I was a poison on the 1900's&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mustard gas was used in the trenches in WWI and killed millions (&lt;a href=&quot;http://en.wikipedia.org/wiki/Sulfur_mustard#Use&quot;&gt;ref&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Then a revolutionary medicine in the hospitals&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sulphur mustard is now used as a cancer chemotherapy drug (&lt;a href=&quot;http://en.wikipedia.org/wiki/Sulfur_mustard#Development_of_the_first_chemotherapy_drug&quot;&gt;ref&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My other use a massage to your blood, the other the killer&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Maybe a reference to the fact that sulphur mustard kills white blood cells?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-12T00:13:37.543" LastActivityDate="2015-03-12T00:13:37.543" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="10107" PostTypeId="2" ParentId="10100" CreationDate="2015-03-11T22:08:06.417" Score="1" Body="&lt;p&gt;Now I think it might be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; STEAM.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I go in a cup, &lt;br&gt;&#xA; but you can't drink me up,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A cup of tea or coffee might be steaming, but you can't drink steam.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I soar through the air, &lt;br&gt;&#xA; well above your hair,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Steam can pass by well above the level of a human head.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;At first I have need &lt;br&gt;&#xA; of some serious speed. &lt;br&gt;&#xA; Later, it's nice &lt;br&gt;&#xA; to be more concise.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Steam might start off flowing quickly but end up dissipating in the air.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;To the club we could go&lt;br&gt;&#xA;if it doesn't snow &lt;br&gt;&#xA;If I go to the beach &lt;br&gt;&#xA; You will probably screech.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I haven't worked these lines out yet!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;First attempt&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Could it be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a STORM?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The title suggests we might be looking for something in the natural world, and&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a storm is certainly a force of nature.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I go in a cup, &lt;br&gt;&#xA; but you can't drink me up,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Refers to the well-known phrase &quot;a storm in a teacup&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I soar through the air, &lt;br&gt;&#xA; well above your hair,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stormclouds pass by well above the level of a human head.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;At first I have need &lt;br&gt;&#xA; of some serious speed. &lt;br&gt;&#xA; Later, it's nice &lt;br&gt;&#xA; to be more concise.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; At first I thought this was wordplay like in the OP's last riddle: I was looking for words like &lt;em&gt;fastshort&lt;/em&gt;. Now I think it might be more literal: storms start off fast and violent but later calm down.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;To the club we could go&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ???&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;if it doesn't snow&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; snowstorm.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If I go to the beach &lt;br&gt;&#xA; You will probably screech.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; sandstorm.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-11T23:35:26.027" LastActivityDate="2015-03-11T23:35:26.027" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10108" PostTypeId="1" CreationDate="2015-03-11T22:13:07.953" Score="4" ViewCount="684" Body="&lt;p&gt;&lt;strong&gt;This question was put on hold as off-topic because:&lt;br&gt;&#xA;&quot;...it appears to be a mathematics problem, as opposed to a mathematical puzzle.&quot; However, it is not a mathematics problem. There is no known mathematical solution. As one of the commenters already realized, the puzzle is to find a more efficient counting algorithm (likely using some coding language).&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal is to find a counting algorithm that is better than simply recursing through every possibility, which will take a very long time. My implementation does it in just under 30 minutes, but I am very sure that other people can do it faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a grid of the flags of the 70 countries with the largest population, in order from left to right, top to bottom. Both the upper left cell (the blue star) and the lower left cell (the green smiley) aren't flags. The flags don't serve any purpose for the question, other than to make the grid prettier, and drop some knowledge about population size by country.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RNdFM.png&quot; alt=&quot;Grid of Flags for the 70 most populous Countries&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You must find the total number of paths that meet all of the following conditions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each path must&lt;br&gt;&#xA;1. start in the blue star cell&lt;br&gt;&#xA;2. end in the green smiley cell&lt;br&gt;&#xA;3. visit each of the 70 flags exactly one time&lt;br&gt;&#xA;4. move up, down, left, or right, but not diagonally&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are a couple examples of valid paths:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9c1bn.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GRtjo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-03-17T14:24:14.523" LastActivityDate="2015-03-17T14:24:14.523" Title="Who can find the most efficient path counting algorithm?" Tags="|combinatorics|" AnswerCount="0" CommentCount="21" FavoriteCount="0" ClosedDate="2015-03-16T14:27:15.877" ContentLicense="CC BY-SA 3.0" />
  <row Id="10109" PostTypeId="2" ParentId="8120" CreationDate="2015-03-11T22:58:40.933" Score="1" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;See comment from OP: this doesn't work. I'm going to have a look and see if I can get it into shape.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here’s my complete solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/z39hV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA; The green lines are the lines that I’ve added, and the thick lines indicate the path that the two signals follow.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A few notes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Since the task says “blue and yellow nodes”, I assumed that I can’t add links to the red or green nodes.&lt;/li&gt;&#xA;&lt;li&gt;This is an approximation of the original image. It was easier to start from scratch than edit the original. I don’t think any of the lines are close enough for it to make much of a difference.&lt;/li&gt;&#xA;&lt;li&gt;I do have a version with numeric labels, but it just gets crowded.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I tackled a path at a time: first the path from the right-hand red node, then the left. It was about 50-50 logic and sensible guesses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here’s how I worked through the right-hand node:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;ul&gt;&lt;li&gt;With three lines at C, the signal is going to rebound to the red node. We need to add one extra edge, and I chose CG because it forces the signal away from the centre (so it’s less likely to conflict with the left-hand path.&lt;/li&gt;&#xA; &lt;li&gt;At H the score is $4-1=3$, which leads to P. It doesn’t seem possible to get to the green node from here without the signal expiring or going via a yellow node, so P seems like a sensible choice. Don’t add any edges.&lt;/li&gt;&#xA; &lt;li&gt;At P the score is $3+2=5$, which would rebound to H. While not technically illegal, it seems pointless. I initially looked at jumping directly to O, but this makes it difficult to get a path in from the left-hand side, so I went via M instead. (I also knew I would have one dangling edge to fill in coming from O, for the left-hand path.)&lt;/li&gt;&lt;/ul&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With that path in hand, I moved on to the left:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;ul&gt;&lt;li&gt;I know I want to end with a score of 2 entering O, so I worked backwards from there. Coming from a yellow node would mean a low score as I'd entered the yellow node, which seems unnecessarily circuitous, so I guessed that I'd come from N. That means I had score 3 when I entered N.&lt;/li&gt;&#xA; &lt;li&gt;I need to add an odd number of edges to B so that I don't rebound. C is out because that'll break the right-hand path, and so I went with A just because that was the first next-node for which I saw the rest of the path.&lt;/li&gt;&lt;/ul&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It’s probably not a unique solution, although I haven’t gone back to find another yet. But unless I’ve misunderstood the rules, I’m fairly sure that is a valid solution.&lt;/p&gt;&#xA;" OwnerUserId="4692" LastEditorUserId="4692" LastEditDate="2015-03-12T11:00:18.073" LastActivityDate="2015-03-12T11:00:18.073" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10111" PostTypeId="1" AcceptedAnswerId="10751" CreationDate="2015-03-12T00:35:47.597" Score="3" ViewCount="604" Body="&lt;p&gt;&lt;em&gt;I'll help you get across a river. &lt;br&gt;&#xA;Once unrecognised in Fischer's song, &lt;br&gt;&#xA;Later I found fame alone &lt;br&gt;&#xA;And also as a US state.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who am I?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Concentrate on lines 1, 3, and 4. Line 2 and the title should be used only as checks to ensure your answer is right. They're there to make sure the answer is unique, and the way they fit it is rather twisted (but should become clear once you have the right idea).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is not actually a US state.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;To be accepted, an answer must explain all lines including the title. But any serious attempt at a solution will get an upvote from me.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-21T10:59:26.580" LastActivityDate="2015-03-22T11:56:14.313" Title="Who am I, if not Lewisoff?" Tags="|riddle|movies|" AnswerCount="6" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10112" PostTypeId="1" AcceptedAnswerId="10814" CreationDate="2015-03-12T00:42:04.387" Score="5" ViewCount="295" Body="&lt;p&gt;Yeah... this is the sequel to Scrambled Songs which is &lt;a href=&quot;https://puzzling.stackexchange.com/questions/8457/scrambled-songs:&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;This is to redeem myself from that really easy question:&lt;br /&gt;&#xA;Another scrambled song...&lt;/p&gt;&#xA;&lt;p&gt;3684-4009-4010-4840-3756-4876-4262-3755-4279&lt;/p&gt;&#xA;&lt;p&gt;A hint or two after the first week... answer by 3 weeks.&lt;/p&gt;&#xA;&lt;p&gt;HINT:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; It's not hexadecimal&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;HINT 2:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;Convert it to a base between 11 and 20&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="167" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-03-23T17:43:51.777" Title="Scrambled Songs 2: What is the song this time?" Tags="|cipher|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10114" PostTypeId="2" ParentId="10100" CreationDate="2015-03-12T03:29:10.510" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Probably not the answer you are looking for, but ICE fits this riddle pretty well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I go in a cup,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ice cube&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but you can't drink me up,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you cannot drink ice&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I soar through the air,&#xA;well above your hair,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a comet has ice in it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At first I have need&#xA;of some serious speed.&#xA;Later, it's nice&#xA;to be more concise.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; something about clouds and snow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To the club we could go&#xA;if it doesn't snow&#xA;If I go to the beach&#xA;You will probably screech&lt;/p&gt;&#xA;" OwnerUserId="9969" LastEditorUserId="9969" LastEditDate="2015-03-12T05:29:33.077" LastActivityDate="2015-03-12T05:29:33.077" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10115" PostTypeId="2" ParentId="10100" CreationDate="2015-03-12T05:14:09.843" Score="13" Body="&lt;p&gt;The hint (and change of wording) helped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A golf ball&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I go in a cup,&#xA;but you can't drink me up,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A golf ball goes into the &quot;cup&quot; on the green, but obviously you don't drink golf balls&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I soar through the air,&#xA;well above your hair,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pretty much any golf shot (except a putt or a very low chip) will go up in the air higher than a person&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At first I have need&#xA;of some serious speed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first golf shot off the tee goes very fast (can be 100+ MPH even for amateurs)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later, it's nice&#xA;to be more concise.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Putts are much more controlled shots and don't go NEARLY as fast&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To the club we could go&#xA;if it doesn't snow&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can't go to the country club and play golf if it's snowing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I go to the beach&#xA;You will probably screech.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You're upset if you land in a sand trap. At least I'm upset when &lt;em&gt;I&lt;/em&gt; land in a sand trap.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2015-03-12T05:14:09.843" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10118" PostTypeId="1" AcceptedAnswerId="10120" CreationDate="2015-03-12T06:53:30.243" Score="6" ViewCount="1338" Body="&lt;p&gt;The Great Houdini is performing again!&#xA;A guy from the audience arbitrarily chooses $52$ cards from a huge stack.&#xA;He shuffles the cards and announces to Houdini that there are $a$ spades, $b$ hearts, $c$ clubs and $d$ diamonds.&#xA;Houdini now has to guess the suits in the deck one by one.&#xA;After every guess, the guy from the audience reveals the current card and everybody (including Houdini) sees the correct suit.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Houdini's strategy:&lt;/strong&gt; At any moment in time, Houdini knows the numbers of remaining suits in the deck. Houdini always guesses a suit that occurs the highest number of times (ties are broken arbitrarily).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Question: What is the &lt;em&gt;smallest&lt;/em&gt; possible number of correct guesses under Houdini's strategy (over all possible values of $a,b,c,d$)?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-03-15T20:39:09.883" LastActivityDate="2021-05-28T16:48:56.240" Title="The Great Houdini's awesome card guessing trick (1)" Tags="|mathematics|strategy|cards|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10119" PostTypeId="2" ParentId="10100" CreationDate="2015-03-12T07:22:30.367" Score="3" Body="&lt;p&gt;It could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; WW2 Fighter-plane Spitfire&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I go in a cup,&#xA;but you can't drink me up,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cup can be hangar where you store your Spitfires (hey, golf holes aren't called cups either). And you obviously can't drink a plane.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I soar through the air,&#xA;well above your hair,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Where else would you fly&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At first I have need&#xA;of some serious speed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You need to get some speed for take-off&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later, it's nice&#xA;to be more concise.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Later you need to slow down to properly aim and hit your target&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To the club we could go&#xA;if it doesn't snow&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You could go flying if the weather is nice&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I go to the beach&#xA;You will probably screech. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Spitfire attack on the beach would ruin anyone's day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9907" LastActivityDate="2015-03-12T07:22:30.367" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10120" PostTypeId="2" ParentId="10118" CreationDate="2015-03-12T07:40:59.460" Score="11" Body="&lt;p&gt;The smallest possible of correct guesses is &lt;code&gt;max(a, b, c, d)&lt;/code&gt;: every other suit could come up first and he would guess the most likely one over and over until nothing was left but those, and then he would get all the rest correct.&lt;/p&gt;&#xA;" OwnerUserId="2815" LastActivityDate="2015-03-12T07:40:59.460" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10121" PostTypeId="1" CreationDate="2015-03-12T08:49:46.503" Score="3" ViewCount="3548" Body="&lt;p&gt;Not all of us are familiar with sign language. It is used predominantly by the deaf and people who can hear but cannot speak. But it's not a big deal in understanding some of the hand signals. Even we use sign language like the ones in the below image in our day to day life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of these hand signals is different from others. Find the odd one out and state the reason why that signal is different from others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5NBqY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="5188" LastEditDate="2015-04-17T05:45:26.480" LastActivityDate="2015-04-17T05:45:26.480" Title="Find the odd one out: Hand signals" Tags="|pattern|lateral-thinking|" AnswerCount="9" CommentCount="4" ClosedDate="2015-03-17T21:56:44.897" ContentLicense="CC BY-SA 3.0" />
  <row Id="10122" PostTypeId="2" ParentId="10121" CreationDate="2015-03-12T09:07:51.353" Score="0" Body="&lt;p&gt;It seems to be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; F&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because it is the only one that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; uses the left hand instead of the right one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9910" LastActivityDate="2015-03-12T09:07:51.353" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10123" PostTypeId="2" ParentId="10121" CreationDate="2015-03-12T09:12:57.513" Score="2" Body="&lt;p&gt;Perhaps &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; K&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is the only one with solely negative connotations &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9918" LastActivityDate="2015-03-12T09:12:57.513" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10124" PostTypeId="2" ParentId="10118" CreationDate="2015-03-12T09:35:23.353" Score="1" Body="&lt;p&gt;I think the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strike&gt;14&lt;/strike&gt; 13&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He will always guess the suit that occurs the highest number of times, so if we want him to fail that suit can never come up. &lt;br&gt;&lt;br&gt; For example if we have 14 A, 13 B, 13 C, 12 D he will always guess A. &lt;br&gt;&lt;br&gt; In worst case scenario, A never comes up, so he will deplete all the other ones and then correctly guess the last suit that remains. &lt;br&gt;&lt;br&gt; But it is 13 because that is the lowest number that is simultaneously equal to or higher than other suits. Meaning we have conditions: &lt;br&gt;&lt;br&gt; a &gt;= b &lt;br&gt; a &gt;= c &lt;br&gt; a &gt;= d &lt;br&gt; a + b + c + d = 52 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;EDIT: I googled definition of 'arbitrarily' and changed my answer.&lt;/p&gt;&#xA;" OwnerUserId="9907" LastEditorUserId="9907" LastEditDate="2015-03-12T11:31:04.470" LastActivityDate="2015-03-12T11:31:04.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10125" PostTypeId="2" ParentId="10111" CreationDate="2015-03-12T09:46:56.667" Score="-1" Body="&lt;p&gt;Only partial answer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mississippi?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;/'ll help you get across a river.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mississippi is name of a ship as well&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once unrecognised in Fischer's song,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No idea who that is or what song&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later I found fame alone&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Don't know&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And also as a US state.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is a US state&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9907" LastActivityDate="2015-03-12T09:46:56.667" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10127" PostTypeId="2" ParentId="10118" CreationDate="2015-03-12T11:15:50.057" Score="1" Body="&lt;p&gt;I think the answer is : &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;13&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because ;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is 52 cards in one deck a+b+c+d=52 and worst possible outcome is that if houdini lock on some card suit and it does not decrease then our lock never change until there are just these cards... and tie breakers is arbitrarily , that means we can assume there worst things happens and all of cards are 13a-13b-13c-13d and we lock on b, and arbitrarily we always go with b in ties.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10074" LastActivityDate="2015-03-12T11:15:50.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10128" PostTypeId="2" ParentId="8120" CreationDate="2015-03-12T11:35:59.657" Score="1" Body="&lt;p&gt;Assuming 'visit a node' means that a signal has to reach that node, not necessarily continue on from that particular node to the green node, here's one solution. Signal strengths at the start and end of each link are indicated after each link.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;LRed-B 5-4&lt;/li&gt;&#xA;&lt;li&gt;B-K 4-6&lt;/li&gt;&#xA;&lt;li&gt;K-A 6-5&lt;/li&gt;&#xA;&lt;li&gt;K-J 6-5, J-E 5-4, J-N 5-4&lt;/li&gt;&#xA;&lt;li&gt;K-O 6-5&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;O-Green 5-4&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;RRed-C 5-4&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;C-F 4-3&lt;/li&gt;&#xA;&lt;li&gt;F-L 3-5&lt;/li&gt;&#xA;&lt;li&gt;L-O 5-4&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;O-Green 4-3&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;C-G 4-6 (originally from RRed-C)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;C-H 4-3, H-M 3-2&lt;/li&gt;&#xA;&lt;li&gt;H-P 3-5&lt;/li&gt;&#xA;&lt;li&gt;P-I 5-4&lt;/li&gt;&#xA;&lt;li&gt;I-D 4-3&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9422" LastActivityDate="2015-03-12T11:35:59.657" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10130" PostTypeId="1" CreationDate="2015-03-11T19:29:42.643" Score="78" ViewCount="14622" Body="&lt;p&gt;Not a very difficult question, but one I enjoyed nonetheless and wanted to share with the community.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You are a servant in a palace. The palace is in the shape of a circle, and you do not know how many rooms there are in the palace.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Some of the rooms have the lights on, some have the lights off. &#xA;  Your job is to turn all the lights off. I remind you again that the palace is ring shaped, and you do not know how many rooms there are in the palace.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;How would you turn off all the lights, and in the end inform your superior (via your cell phone) that you have completed your job?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Notes:   &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It's not something stupid like &quot;touch the light bulb to see if its warm&quot; or anything like that. &lt;/li&gt;&#xA;&lt;li&gt;You are not allowed to leave any distinguishing objects or markers to notify you where you've been. (I already heard solutions like &quot;leave your clothes at room 0 and now just close all the lights naked&quot;.) &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="Oria Gruber" LastEditorUserId="-1" LastEditDate="2015-03-12T16:56:33.620" LastActivityDate="2015-09-17T08:12:04.957" Title="Turn off all lights in a ring-shaped palace" Tags="|strategy|reachability|" AnswerCount="12" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="10131" PostTypeId="2" ParentId="10130" CreationDate="2015-03-11T20:02:03.597" Score="55" Body="&lt;p&gt;Setup: If you are in the room with lights off, turn them on. If lights are on, leave them on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iteration: Go counterclockwise counting rooms as you pass them, until you find a room with lights on (because of setup action, you are guaranteed to find one). Turn lights off. Return to the initial room (according to the count). Check the state of lights. If lights are off, you just made the full circle, and you are done. Otherwise, keep iterating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The algorithm has a quadratic time complexity; I have no idea if there is a faster one.&lt;/p&gt;&#xA;" OwnerDisplayName="user58697" LastActivityDate="2015-03-11T20:02:03.597" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10132" PostTypeId="2" ParentId="10130" CreationDate="2015-03-11T20:15:15.307" Score="37" Body="&lt;p&gt;One possible algorithm has just three state data: Integers $k$ and $n$, and a Boolean $f$.  Start in any room, with $k = 0$, $n=0$, and $f$ true.  (The meaning of $f$ is &quot;I am currently moving in the 'forward' direction&quot;. The meaning of $k$ will be a forward room number.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Turn on the light in that first room. Then:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $f$ is false and $k=0$ and the light is off in the room you are in, then the algorithm completes; you know all lights are off. (You may or may not end up in the starting room.)&lt;/li&gt;&#xA;&lt;li&gt;If $f$ is false and $k &amp;gt; 0$ then the light in your room will be off; leave it off, move in the backward direction to the next room, and decrease $k$ by one.&lt;/li&gt;&#xA;&lt;li&gt;If  $f$ is false and $k=0$ and the light is on, then you are in the starting room; increase $n$ by one, set $f$ true to start moving forward. &lt;/li&gt;&#xA;&lt;li&gt;If $f$ is true and $k &amp;lt; 2^n$ move forward by one room, and turn off the light in the new room (which may, unbeknownst to you, actually be the starting room).&lt;/li&gt;&#xA;&lt;li&gt;If $f$ is true and $k = 2^n$, turn $f$ to flase, to start moving back toward the starting room to check if the light there is still on.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This algorithm turns off the lights with 100% certainty, and for $R$ rooms concludes in at most &#xA;$$2^{2+\left\lceil \log_2 R\right\rceil} -2&#xA;$$&#xA;steps.  This is always less than $8R$ steps. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Turn the light on in your room. Go 1 clockwise. If on, turn off along the way. Go back to starting room. If starting room light is on, go 2 counterclockwise. If on, turn off along the way. Go back to the starting room. If on, go 4 clockwise, etc. When the starting room light is off upon returning, you're done.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if you can do better; I suspect you can by using increasing sweeps of a factor of $e$ rather than a factor of $2$ each time.&lt;/p&gt;&#xA;" OwnerUserId="54682" OwnerDisplayName="Mark Fischler" LastEditorUserId="15544" LastEditDate="2015-09-17T08:12:04.957" LastActivityDate="2015-09-17T08:12:04.957" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="10133" PostTypeId="2" ParentId="10130" CreationDate="2015-03-11T22:11:56.223" Score="3" Body="&lt;p&gt;Since we're not allowed to touch light bulbs, I guess we're not allowed to use a map, a compass or (welcome to the 21st century) GPS either. :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming that I walk room-to-room around the ring, with switches at clockwise and counter-clockwise doors, and I'm refusing to walk through a room in the dark (for safety). Basically I'm going to turn lights on as I walk away from my start room, then turn them off as I return.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So. I start in room $0$ with the lights on, and choose my initial direction. First action though is in the opposite direction: I turn on the lights in the adjacent opposite-direction room (if necessary). Then I head in my chosen direection, turn &lt;strong&gt;off&lt;/strong&gt; room 0 lights as I leave, and turning on the lights in room 1 if necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I proceed until I find a single room with lights off - that is, where the rooms before and after are lit.  Call this room $n_1$. Now I'll proceed further (of course turning on the lights in room $n_1$) to room $2n_1$. If I find any dark rooms, I discard the existing value of $n_1$ and keep looking for a new single dark room. (If the new dark room I found was a single dark room, it becomes the new $n_1$). Keep going until I have walked though already-lit rooms from $n_1$ to $2n_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I head back to room $0$, leaving the light in $2n_1$ on but otherwise turning off lights as I go. If I reach room $n_1$ and the following light is off, it was also room $0$ and all the lights are off. Otherwise, on reaching room $0$, I cross over and start out in the opposite direction, turning on lights as I go (after room $0$, which is left dark). Now I am looking for a sequence of exactly $2n_1$ rooms that are dark. I proceed as before, calling the last room in the $2n_1$ dark-room sequence room $n_2$ and continuing through another $n_2$ lit rooms to room $2n_2$ (or continuing the search for $n_2$ if any dark rooms are found). Then I head back to room $0$, leaving the last room light on and otherwise turning lights off. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I continue iterating in opposite directions until I find an already-dark room on the way back to room $0$. The lights are then off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By my reckoning I should walk through $kR$ rooms, where $R$ is the total number of rooms and $k$ is between $3$ and $4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AEGss.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/CKlDE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/pg0kj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="8875" OwnerDisplayName="Joffan" LastActivityDate="2015-03-12T02:40:26.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10134" PostTypeId="2" ParentId="10130" CreationDate="2015-03-11T23:21:52.583" Score="14" Body="&lt;p&gt;I prefer the keep it simple method.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Leave the light on at your starting room&lt;/li&gt;&#xA;&lt;li&gt;Turn off the lights to either side, if necessary.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now the simple alternating algorithm:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Move along to the right, counting each room as you pass, until you get to a light that is on. Turn it off.&lt;/li&gt;&#xA;&lt;li&gt;Turn around and count back down the number of rooms you moved. Now you know you're  back at your starting room. &lt;/li&gt;&#xA;&lt;li&gt;If the light is on, you proceed to repeat the procedure in the&#xA;opposite direction.&lt;/li&gt;&#xA;&lt;li&gt;If the light is off, you know you've gone around the entire circle and turned off the light in your starting room, then gone BACK around the entire circle to realize you turned off your starting room's light.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Et voila.&lt;/p&gt;&#xA;" OwnerDisplayName="Mike in SAT" LastActivityDate="2015-03-11T23:21:52.583" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10135" PostTypeId="2" ParentId="10130" CreationDate="2015-03-12T00:39:43.037" Score="9" Body="&lt;p&gt;Here's a finite-state algorithm, in case there are too many rooms for you to count.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Turn the light off; go clockwise; turn the light on; go clockwise; turn the light on; go clockwise; turn the light off; go clockwise. (This pattern of lights is your &quot;start marker&quot;.)&lt;/li&gt;&#xA;&lt;li&gt;Turn the light off and go clockwise. Unless this is the second room in a row you've encountered that had the lights on when you arrived, repeat this step. (We might have turned off one of the lights in the start marker.)&lt;/li&gt;&#xA;&lt;li&gt;Go counter-clockwise until you encounter a room with the lights on. (This must be the second &quot;on&quot; light of the start marker.)&lt;/li&gt;&#xA;&lt;li&gt;Go counter-clockwise. If the light is off, then go clockwise, turn the light off, and finish. Otherwise, continue. (Since the light which was on is now off, we've made a full circle and we're done.)&lt;/li&gt;&#xA;&lt;li&gt;Go clockwise twice and return to step 2. (The start marker is still intact, so we have to continue.)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This algorithm should work as long as there are at least three rooms. &lt;/p&gt;&#xA;" OwnerUserId="9650" OwnerDisplayName="Tanner Swett" LastActivityDate="2015-03-12T00:39:43.037" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10136" PostTypeId="2" ParentId="10130" CreationDate="2015-03-12T00:55:46.690" Score="0" Body="&lt;p&gt;The servant could also use a geometric pattern of on/off that would be unlikely to occur as a marker of where he/she started.  This wouldn't reduce the probability of leaving a light on to zero, but it could reduce it to an arbitrarily low value depending on the length and complexity of the pattern.  Realistically (for a human), this may be more reliable because it may be easier to remember the rules behind a geometric pattern than it would be to be sure of a room count if the total number of rooms was very large (I know with ~500 rooms I would loose count several times and doubt my count even if I didn't actually miscount).&lt;/p&gt;&#xA;" OwnerUserId="10164" OwnerDisplayName="nateAtwork" LastEditorUserId="10164" LastEditDate="2015-03-17T00:41:53.670" LastActivityDate="2015-03-17T00:41:53.670" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10138" PostTypeId="2" ParentId="10121" CreationDate="2015-03-12T12:29:34.887" Score="7" Body="&lt;h1&gt;The answer is...&lt;/h1&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; L&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; It is the only letter that is correct in actual sign language.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5556" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-03-12T12:29:34.887" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10139" PostTypeId="2" ParentId="10121" CreationDate="2015-03-12T12:32:28.117" Score="0" Body="&lt;p&gt;I think the answer would be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because all other hand signals&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; can be derived from another hand signal by moving at most ONE finger and/or doing reflection (changing hand) or rotation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A can be derived from B, and B from A.&#xA; D can be derived from H, J, or O.&#xA; E can be derived from M, and M from E.&#xA; F can be derived from H, I, L, N, or O.&#xA; G can be derived from D, H, J, or O.&#xA; H can be derived from D, F, G, K, N, or O.&#xA; I can be derived from F, N, or P.&#xA; J can be derived from D, G, or K.&#xA; K can be derived from H, J, or O.&#xA; L can be derived from F, N, or P.&#xA; N can be derived from F, H, I, L, or O.&#xA; O can be derived from D, F, G, H, K, or N.&#xA; P can be derived from I or L.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10136" LastActivityDate="2015-03-12T12:32:28.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10140" PostTypeId="2" ParentId="10121" CreationDate="2015-03-12T12:34:50.303" Score="1" Body="&lt;p&gt;I think 'F' is answer, because you can see Thumb Up in 'N'.&#xA;But when I try to make sign like 'F', it is not possible.&lt;/p&gt;&#xA;" OwnerUserId="2003" LastActivityDate="2015-03-12T12:34:50.303" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10141" PostTypeId="2" ParentId="10111" CreationDate="2015-03-12T12:39:57.417" Score="3" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Texas &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll help you get across a river&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The war against Mexico made Texas troops cross the Rio Grande&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once unrecognized in Fischer's song&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One of his songs is The Vessel. Or something with the word vessel in it. A battleship of Texas was called the USS Texas I believe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later found fame alone&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Won the Texas Revolution by itself and joined the US later. Battle of the Alamo, anyone?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And also a US state&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Texas joined the US shortly after the Texas Revolution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9042" LastActivityDate="2015-03-12T12:39:57.417" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10142" PostTypeId="2" ParentId="10121" CreationDate="2015-03-12T12:43:57.853" Score="1" Body="&lt;p&gt;I think it is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; O &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is the only hand signal in which all the fingers are facing west. Left when looking at the picture.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9042" LastActivityDate="2015-03-12T12:43:57.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10143" PostTypeId="2" ParentId="9996" CreationDate="2015-03-12T12:51:50.693" Score="1" Body="&lt;p&gt;Expanding on the answer given by Aggie Kidd. Is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;a basketball center?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My job is putting things in rings.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Basketball hoops are ring shaped.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My prize is rings.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Championship rings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The number of rings I have symbolizes speed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Center is abbrevated &quot;c&quot; (or half a ring) and is the symbol for the speed of light in vaccum.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When people see me in action they say Oh, neat!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They sometimes make flashy plays!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10142" LastActivityDate="2015-03-12T12:51:50.693" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10144" PostTypeId="2" ParentId="10121" CreationDate="2015-03-12T13:01:46.777" Score="2" Body="&lt;p&gt;I think it is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;L&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is the only sign that cannot be done with the other hand. We all know it mean &lt;em&gt;&quot;loser&quot;&lt;/em&gt; by the &lt;em&gt;L&lt;/em&gt; shape, but if you do it with left hand, the &lt;em&gt;L&lt;/em&gt; is reverted, losing its meaning.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5056" LastActivityDate="2015-03-12T13:01:46.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10146" PostTypeId="2" ParentId="10121" CreationDate="2015-03-12T13:05:49.330" Score="0" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is the only gesture which can be seen when you hold out your arms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4393" LastEditorUserId="4393" LastEditDate="2015-03-12T13:15:23.260" LastActivityDate="2015-03-12T13:15:23.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10149" PostTypeId="2" ParentId="10130" CreationDate="2015-03-12T13:35:13.233" Score="1" Body="&lt;p&gt;First stand in the corridor and look left and right and count how many rooms you can see and call that X. Then let's just guess that the number of rooms is about Y times that number. What Y exactly is I can't say now - it all depends on width of the corridor and other factors - but I would say that any person that would come up with a mathematical algorithm to solve this question can also make some pretty accurate estimations about this. And to be sure just take twice that number so you'll go through 2 * X * Y rooms. I can't possibly imagine you won't get it that way and this way you will open less than 2 * R rooms where R is the number of rooms.&lt;/p&gt;&#xA;" OwnerUserId="1748" LastEditorUserId="5373" LastEditDate="2015-03-12T23:46:20.847" LastActivityDate="2015-03-12T23:46:20.847" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10150" PostTypeId="2" ParentId="10130" CreationDate="2015-03-12T13:46:29.093" Score="2" Body="&lt;p&gt;At last Peter has found a new job as servant in a palace. But it was a strange palace; all rooms looked alike and you never knew where you are. It was maddening.&#xA;And the older servants weren't a help. All the unpleasant tasks were loaded onto him. His newest task was to turn off all the lights in this looping, never ending palace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before he started he set $i=1$&lt;br/&gt;&#xA; #START#&lt;br&gt;&#xA;He went through $2^{i-1}$ rooms and switched all the lights in there off.&lt;br/&gt;&#xA;At the last room he stopped, turned around and switched the lights on to help him thinking.&lt;br/&gt;&#xA;He just switched off the lights in $2^{i-1}$ rooms and checked that the lights in the $2^{i-1}-1$ before those were switched off too.&lt;br/&gt;&#xA;That means he knew the light in the last $2^i-1$ rooms are already turned off. &lt;br/&gt;&#xA;But better save than sorry he checked the $2^i-1$ rooms.&lt;br/&gt;&#xA;The first few times he didn't find any turned on lights and he increased i by one and repeated the steps from the #START# without turning around.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When he found a light switched on something inside him just snapped. A light that he knew was turned off was turned on again. And all those equal looking rooms. Nothing ever changing. Repeating forever ...&#xA;He switched the light off and escaped the palace and committed himself in a loony bin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next day when the other servants came they checked and found that all lights were switched off.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Why does this algorithm works:&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;In the 1st loop he turns $1$ light off and has $0$ already turned off. Making a total of $1$ light off that he checks.&lt;br/&gt;&#xA;In the 2nd loop he turns $2$ light off and has $1$ already turned off. Making a total of $3$ light off that he checks.&lt;br/&gt;&#xA;In the 3rd loop he turns $4$ light off and has $3$ already turned off. Making a total of $7$ light off that he checks.&lt;br/&gt;&#xA;...&lt;br/&gt;&#xA;In the i-th loop he turns $2^{i-1}$ lights off and has $2^{i-1}-1$ already turned off. Making a total of $2^i-1$ light off.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets look at the k-th loop with $k=\lceil log_2(R+1)\rceil$.&lt;br/&gt;&#xA;At the end of the loop he checks $2^k-1\geq R$ rooms. &#xA;That means he checks all rooms. Before checking he turns the lights in one room on the check fails after he runs once through the whole palace.&#xA;That light is the only one that is on. He shut it off and is finished.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;How much does the servant moves:&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;In the 1st loop he moves 1 time to turn lights off. 1 time to turn one light on and 1 time to check them. 3 times in total.&lt;br/&gt;&#xA;In the 2nd loop he moves 2 time to turn lights off. 1 time to turn one light on and 3 time to check them. 6 times in total. Together with the previous loop 9 times total.&lt;br/&gt;&#xA;In the 3rd loop he moves 4 time to turn lights off. 1 time to turn one light on and 7 time to check them. 12 times in total. Together with the previous loops 21 times total.&lt;br/&gt;&#xA;...&lt;br/&gt;&#xA;In the i-th loop he moves $2^{i-1}$ time to turn lights off. 1 time to turn one light on and $2^i-1$ time to check them. $3*2^{i-1}$ times in total. Together with the previous loops $3*2^i-3$ times total.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the k-th loop he moves $2^{k-1}$ time to turn lights off. 1 time to turn one light on and R times to check them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In total we have k-1 full loops plus the last loop: &#xA;$3*2^{k-1}-3+2^{k-1}+1+R = 4*2^{k-1}+R-2 = 4*2^{\lceil log_2(R+1)\rceil-1}+R-2 \\= 4*2^{\lfloor log_2(R)\rfloor+1-1}+R-2 \leq 4*R+R-2 = 5*R-2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Result: He has to move up to $5*R-2$ times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be improved by 1 step by not starting with a move and maybe by using an other base.&lt;/p&gt;&#xA;" OwnerUserId="10143" LastActivityDate="2015-03-12T13:46:29.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10151" PostTypeId="2" ParentId="10121" CreationDate="2015-03-12T14:00:16.913" Score="0" Body="&lt;p&gt;I believe the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; N&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mainly because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you notice the hands how we see em in the image is how we would see them if some other person was doing them and we were looking, N is the only one who would feel odd if you saw it like that ( palm looking at you instead of the back of the hand)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10145" LastActivityDate="2015-03-12T14:00:16.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10152" PostTypeId="2" ParentId="10111" CreationDate="2015-03-12T14:05:57.573" Score="-1" Body="&lt;p&gt;ok this is a huge guess&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; New Orleans&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;not carrollof but&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Carrollton New Orleans&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ill' help you get accros the river&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; new orleans harbor&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fischer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ok I'm wrong&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fame&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; sorry I dont know :'(&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and also a us state&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ok on this ons is correct :D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10110" LastActivityDate="2015-03-12T14:05:57.573" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10158" PostTypeId="1" AcceptedAnswerId="10161" CreationDate="2015-03-12T14:36:40.057" Score="9" ViewCount="980" Body="&lt;p&gt;Oh the different names I am called,&lt;br&gt;&#xA;If I end badly, you may be mauled.&lt;br&gt;&#xA;Head, shoulders, knees, and toes,&lt;br&gt;&#xA;Sometimes there are subtle blows. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One plus one equals one and a half&lt;br&gt;&#xA;The clock's going, now don't laugh&lt;br&gt;&#xA;It is a fierce battle often in rage,&lt;br&gt;&#xA;That takes place on a green stage.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the world says I'm pretty great,&lt;br&gt;&#xA;From buenos días to G'day mate!  &lt;/p&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="8874" LastEditDate="2015-10-08T09:24:01.767" LastActivityDate="2015-10-08T09:24:01.767" Title="How many names - What am I?" Tags="|riddle|word|rhyme|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10161" PostTypeId="2" ParentId="10158" CreationDate="2015-03-12T14:42:57.970" Score="12" Body="&lt;p&gt;Is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Soccer/football?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Oh the different names I am called,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; See above&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I end badly, you may be mauled.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Riots during/after games&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Head, shoulders, knees, and toes,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What you use to hit the ball&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes there are subtle blows.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Slide tackles, running into each other, heads into chests (less subtle)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One plus one equals one and a half&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The two periods = 1 hour 30 min&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The clock's going, now don't laugh&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clock always runs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is a fierce battle often in rage,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's a sport, passions happen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That takes place on a green stage.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Grass field&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Most of the world says I'm pretty great,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; World Cup!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From buenos días to G'day mate! &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; World Cup!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-03-12T14:42:57.970" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10164" PostTypeId="2" ParentId="10130" CreationDate="2015-03-12T16:02:04.337" Score="10" Body="&lt;p&gt;Step 1 - Find the fuse box and cut all electricity to the palace thus turning off all the lights.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 2 - Tell your supervisor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 3 - Be sacked for gross misconduct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hurrah!&lt;/p&gt;&#xA;" OwnerUserId="9204" LastActivityDate="2015-03-12T16:02:04.337" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="10165" PostTypeId="1" AcceptedAnswerId="10176" CreationDate="2015-03-12T16:35:33.817" Score="12" ViewCount="3218" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;It moves in war&lt;br&gt;&#xA;  Abhors the floor&lt;br&gt;&#xA;  Takes many forms&lt;br&gt;&#xA;  And is a controversial norm&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;An encounter with it&lt;br&gt;&#xA;  Will leave one tearful&lt;br&gt;&#xA;  And as an omen&lt;br&gt;&#xA;  It makes many fearful &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Though as journeyers swear&lt;br&gt;&#xA;  It's useful where&lt;br&gt;&#xA;  One wants to prevent&lt;br&gt;&#xA;  Spoiled fare  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;What is it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To clarify: I'm using &quot;norm&quot; in the sense of something that's seen/encountered fairly often&lt;/p&gt;&#xA;" OwnerUserId="10151" LastEditorUserId="8874" LastEditDate="2015-10-08T09:05:38.463" LastActivityDate="2015-10-08T09:05:38.463" Title="It moves in war" Tags="|riddle|rhyme|" AnswerCount="10" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10166" PostTypeId="2" ParentId="9996" CreationDate="2015-03-12T18:17:33.580" Score="0" Body="&lt;p&gt;I am guessing it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; NASCAR CAR RACE DRIVER.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My job is putting things in rings.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RACING CAR AROUND THE TRACK&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My prize is rings.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TROPHY RINGS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The number of rings I have symbolizes speed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; WINNING RINGS MEANS THE DRIVER WAS THE FASTEST&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When people see me in action they say Oh, neat!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MANY LOVE WATCHING FAST CARS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9892" LastEditorUserId="5373" LastEditDate="2015-03-12T19:41:49.620" LastActivityDate="2015-03-12T19:41:49.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10167" PostTypeId="2" ParentId="10165" CreationDate="2015-03-12T18:53:16.107" Score="1" Body="&lt;p&gt;Three out of four not sure about &quot;abhors the floor&quot;. What about:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; border (moves in war, takes many forms, is a controversial norm)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10057" LastEditorUserId="2806" LastEditDate="2015-03-12T19:43:04.810" LastActivityDate="2015-03-12T19:43:04.810" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10168" PostTypeId="1" AcceptedAnswerId="10175" CreationDate="2015-03-12T19:10:23.060" Score="5" ViewCount="2074" Body="&lt;p&gt;Without me you'd be lost,&lt;br&gt;&#xA;That's the smallest cost.&lt;br&gt;&#xA;You use me every day,&lt;br&gt;&#xA;More than I can even say,&lt;br&gt;&#xA;You do but don't see me,&lt;br&gt;&#xA;So whatever could I be?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inside, outside, all around&lt;br&gt;&#xA;I'm even underground.&lt;br&gt;&#xA;I can make no sound&lt;br&gt;&#xA;Can't really be bound.&lt;br&gt;&#xA;So tell me if you can see&lt;br&gt;&#xA;What it is that I might be?  &lt;/p&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="8874" LastEditDate="2015-10-08T09:20:19.233" LastActivityDate="2015-10-08T09:20:19.233" Title="I am Essential - What am I?" Tags="|riddle|word|rhyme|" AnswerCount="5" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10169" PostTypeId="2" ParentId="10168" CreationDate="2015-03-12T19:29:50.833" Score="1" Body="&lt;p&gt;I would have to say you are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Air.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It seems fairly obvious that this answer meets all of the criteria.&lt;/p&gt;&#xA;" OwnerUserId="10153" LastActivityDate="2015-03-12T19:29:50.833" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10170" PostTypeId="2" ParentId="10165" CreationDate="2015-03-12T19:35:41.877" Score="12" Body="&lt;p&gt;I immediately thought of a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; flag.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Usually, in a war,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a flag is carried by advancing troops.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It abhors the floor because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you are not supposed to let a flag touch the ground.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It takes many forms:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are obviously many different flags.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And is a controversial norm:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Saw a headline about a controversy over flags being flown at UC Irvine I believe, not more than a few days ago.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10153" LastActivityDate="2015-03-12T19:35:41.877" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10171" PostTypeId="2" ParentId="10165" CreationDate="2015-03-12T19:59:28.793" Score="6" Body="&lt;p&gt;I don't think this is the intended answer, but it fits everything fairly well.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A &lt;strong&gt;gun&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;It moves in war,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The arms trade flourishes during war, and guns are always changing hands. Alternatively, as troops move around they take their guns with them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Abhors the floor,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You'd better not drop a gun on the floor, or it might go off and shoot you in the foot!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Takes many forms,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cannons, pistols, revolves, rifles, shotguns, machine guns, sub-machine guns, blunderbusses, ... you want me to go on?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And is a controversial norm.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know where the OP comes from, but in the US,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; possessing a gun is a pretty normal thing (I believe over 50% of households do?), and there's a lot of controversy over gun laws.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-03-12T19:59:28.793" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10172" PostTypeId="2" ParentId="10168" CreationDate="2015-03-12T20:06:19.130" Score="2" Body="&lt;p&gt;This one might be a bit general, but:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Language&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Without me you'd be lost&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you can't communicate with people, finding your way around new places is extremely difficult&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;That's the smallest cost&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not being able to get around places is minor compared to not being able to communicate with family, friends, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;You use me every day&quot; is self-explanatory&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;More than I can even say&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are many non-verbal forms of language used daily&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;You do but don't see me&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Going along with the last part, there are both visible (written) and non-visible (verbal) forms of language&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Inside, outside, all around&lt;br&gt;&#xA;I'm even underground&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You see written language inside and outside, including underground areas such as subway signs, writing on cave walls, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10151" LastEditorUserId="9042" LastEditDate="2015-03-12T22:05:14.450" LastActivityDate="2015-03-12T22:05:14.450" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10174" PostTypeId="1" AcceptedAnswerId="10304" CreationDate="2015-03-12T20:35:53.633" Score="7" ViewCount="1057" Body="&lt;p&gt;I'm thinking of a word you can spell on one hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you add &quot;e&quot; or ɛ, it might describe the man who lived here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5aRuq.jpg&quot; alt=&quot;Image 1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add the sound of the world's favorite drink, and you'll find something often worn here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ezEKDm.png&quot; alt=&quot;Image 2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add the first 3 letters of this island's mechanical secret, and you'll find a common setting:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9TLVGm.png&quot; alt=&quot;Image 3&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add The One Who Originated and drink some with Coke.&lt;br&gt;&#xA;Yes it's spelled wrong but you'll be too gone to notice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What is the word and why?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You want a hint to get you started?? The solution starts with you!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Image sources (may make puzzle easier):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://cdn.idigitaltimes.com/sites/idigitaltimes.com/files/styles/image_embed/public/2014/11/07/frasier-living-room.jpg?itok=QQKLf-6R&quot; rel=&quot;nofollow noreferrer&quot;&gt;Image 1&lt;/a&gt; &lt;a href=&quot;http://garutweb.com/wp-content/uploads/2015/01/world-map-outline-black-pjlagxi6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Image 2&lt;/a&gt; &lt;a href=&quot;http://static.planetminecraft.com/files/resource_media/screenshot/1211/Lincon-Island_1735241.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Image 3&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1811" LastEditorUserId="5373" LastEditDate="2015-03-14T15:13:35.270" LastActivityDate="2015-03-15T17:26:34.310" Title="What word am I thinking?" Tags="|riddle|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10175" PostTypeId="2" ParentId="10168" CreationDate="2015-03-12T20:44:22.023" Score="4" Body="&lt;p&gt;I'm going to say that you are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Light&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without me you'd be lost, That's the smallest cost.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I couldn't get around very well without light and there are certainly vastly more significant impacts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You use me every day, More than I can even say,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I don't know about you, but I use light all day every day. Not so much at night, though.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You do but don't see me, So whatever could I be?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You &quot;see&quot; light as it's the only way to see anything. You never notice it, though, because your grain interprets it as seeing the objects and not the light they're emitting / reflecting.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Inside, outside, all around. I'm even underground.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Light be friggin' everywhere, yo. There are underground lights such as in a subway and, if we could get to the core of the Earth, there would be light from the core's heat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can make no sound; Can't really be bound.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Light creates no sound waves and is difficult to contain. We can slow it down but haven't been able to stop it or store it yet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So tell me if you can see. What it is that I might be? &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Per your comment on the Language answer, you said the second to last line contains a clue. Only those that can see use light. Those that are totally blind don't need it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2015-03-12T20:50:16.973" LastActivityDate="2015-03-12T20:50:16.973" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10176" PostTypeId="2" ParentId="10165" CreationDate="2015-03-12T20:49:31.217" Score="12" Body="&lt;p&gt;First thing I thought of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Smoke.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It moves in war,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Smoke is blown about from flame, engines, explosions, gunfire, and is spread by the wind. If the battle moves, the smoke will move with it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Abhors the floor,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Smoke is carried by hot air, and therefore rises. Even in an indoor fire smoke is thinnest near the ground.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Takes many forms,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Depending on the source, smoke clouds, smoke plumes, puffs of smoke, walls of smoke, wisps of smoke....&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And is a controversial norm.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We've all seen smoke. It's not unusual. If we do see it outside of a specific context such as a bonfire or BBQ, though, most people will get excited. This could also mean smoke from cigarettes, which is a common although decreasing habit that many people dislike.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I answered this before a few more specific clues were added, so I'll fill them out down here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An encounter with it&lt;br&gt;&#xA;Will leave one tearful  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Smoke irritates the eyes, causing tears.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And as an omen&lt;br&gt;&#xA;It makes many fearful  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As mentioned in a previous clue, if people see smoke where they don't expect it's considered a bad sign. It usually means fire or destruction of some type.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though as journeyers swear&lt;br&gt;&#xA;It's useful where&lt;br&gt;&#xA;One wants to prevent&lt;br&gt;&#xA;Spoiled fare  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Meat can be smoked to greatly increase the time before it spoils. In past times, this could be used for a journey as food that wouldn't rot before you reached your goal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2741" LastEditorUserId="2741" LastEditDate="2015-03-13T16:44:32.217" LastActivityDate="2015-03-13T16:44:32.217" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10178" PostTypeId="2" ParentId="10165" CreationDate="2015-03-12T21:50:34.223" Score="2" Body="&lt;p&gt;Is it   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Money?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It moves in war,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Economists say that war moves money.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Abhors the floor,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Who leaves their money on the floor, or the floor meaning no money or Broke. Don't like that.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Takes many forms, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Coins, bills, checks, gold bars, etc.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And is a controversial norm.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most controversies are about or at least involve money.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-03-12T21:50:34.223" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10180" PostTypeId="2" ParentId="10168" CreationDate="2015-03-12T21:57:29.370" Score="5" Body="&lt;p&gt;I know this has been correctly answered already, but I want to share my answer, anyway:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gravity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without me you'd be lost, That's the smallest cost.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lost in space (at least, assuming it was switched off. The universe wouldn't look anything like it does if it had &lt;em&gt;never&lt;/em&gt; worked.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You use me every day, More than I can even say,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Even if you're in orbit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You do but don't see me, So whatever could I be?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You see the effects of it, but you don't register it because it's so normal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Inside, outside, all around, I'm even underground.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Even at the centre of the Earth, you're gravitationally affected by the rest of it. Even discounting that, you're affected by the gravity of the Sun.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can make no sound, Can't really be bound.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can be bound to it, though.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So tell me if you can see, What it is that I might be?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ok, Gravity only works if I still didn't realise this was supposed to be a clue, too.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10159" LastActivityDate="2015-03-12T21:57:29.370" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10181" PostTypeId="2" ParentId="183" CreationDate="2015-03-12T22:42:35.173" Score="2" Body="&lt;p&gt;This solution is similar to the one provided by R Gilliam but differs in the second step.&#xA;Divide the balls into 3 groups of 4 balls each. Let's call them g1 g2 and g3&#xA;pick any two groups and weigh them against each other. One of two scenarios is true. &#xA;The pans are balanced: the 8 balls you just weighed all have the correct weight.&#xA;The pans are unbalanced: The 4 balls you did not weigh all have the correct weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way at the end of the first weighing you have at least 4 balls of the correct weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second weighing one side of the pan should have 3 balls of the correct weight.&#xA;If the pans were unbalanced after the first weighing put 3 balls from one of the unbalanced pans into the other pan. If the pans were balanced after the first weighing put 3 of the 4 balls that sat out the first weighing into the other pan. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the pans are unbalanced after this weighing you will know whether the oddball is heavier or lighter since one of the pans contains balls of the correct weight. If the pans are balanced the 4 th ball that was left out is the oddball and you can find out if it is heavier or lighter by weighing it against a correct-weight ball. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the pans are unbalanced, you know whether the oddball is heavier or lighter. Take 2 of the 3 balls from the pan (that does not contain the correct weight balls) and weigh them one against the other. You already know whether the oddball is heavier or lighter. If the pans are unbalanced pick the pan that matches the weight direction of the oddball. if the pans are balanced the 3rd ball is the oddball.&lt;/p&gt;&#xA;" OwnerUserId="10038" LastActivityDate="2015-03-12T22:42:35.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10183" PostTypeId="2" ParentId="10165" CreationDate="2015-03-12T23:12:32.347" Score="2" Body="&lt;p&gt;Might it be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; spies&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;it moves in war&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; spies get used all the times in wars. proper intelligence can make or break an army.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;it abhors the floor&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; spies don't like &quot;being on the main floor&quot;. they want to stay hidden so they aren't noticed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;it takes many form&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; spying can be done towards the enemies; towards your own people; and towards other companies; and towards your family (like a mother reading her child's text messages).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and is a controversial norm&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is currently much controversy about the government spying on their own inhabitants, but when war is around, people would prefer spying to occur to save their lives.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I have 2 more ideas:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; workers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; lobbying&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I don't have time right now to explain them because I need to get some sleep. I'll explain them tomorrow.&lt;/p&gt;&#xA;" OwnerUserId="2778" LastActivityDate="2015-03-12T23:12:32.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10184" PostTypeId="2" ParentId="10130" CreationDate="2015-03-12T23:18:27.770" Score="7" Body="&lt;p&gt;The best answers (fitting the strategy and reachability tags) use a sentinel room (e.g. user58697's answer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you can rely on the path being a perfect circle (and not merely circular), there is also a geometric solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Place both ends of a ruler of unit length against the circle and measure the arc length (call it $a$), then derive the radius of the circle from $1/(2r) = \sin (\theta/2)$ (right triangle from half the ruler to the circle centre) and $a = r \theta$ (formula for arc length), where $r$ is the radius of the circle and $\theta$ is the angle subtended at the centre of the circle by the arc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the radius, you can calculate the circumference of the circle. Walk this length and turn off all lights as you go, then call the boss.&lt;/p&gt;&#xA;" OwnerUserId="9422" LastActivityDate="2015-03-12T23:18:27.770" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10185" PostTypeId="2" ParentId="10168" CreationDate="2015-03-13T01:09:53.723" Score="2" Body="&lt;p&gt;So tell me if you can see. What it is that I might be?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The human brain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without me you'd be lost, That's the smallest cost.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The body cannot live without the mind.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You use me every day, More than I can even say,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; At least I hope you do.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You do but don't see me, So whatever could I be?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's kind of a big problem when you say: &lt;em&gt;Hey, look. Grey matter.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Inside, outside, all around. I'm even underground.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People be friggin' everywhere, yo.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can make no sound; Can't really be bound.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A brain needs a mouth attached to make sound. If you try to clear your head you'll just end up thinking about Stay-Puff marshmallow men; imagination is not bound to reality nor can thoughts be controlled.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4522" LastActivityDate="2015-03-13T01:09:53.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10186" PostTypeId="1" AcceptedAnswerId="10189" CreationDate="2015-03-13T02:40:46.060" Score="3" ViewCount="252" Body="&lt;p&gt;What common property unites all of the following numbers?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;4, 9, 60, 101, 400, 450, 910, 1010  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The set of numbers that share this common property is not limited to just these eight, but the list should provide enough information to solve it.&lt;/p&gt;&#xA;" OwnerUserId="10151" LastEditorUserId="8874" LastEditDate="2015-10-08T09:19:42.180" LastActivityDate="2015-10-08T09:19:42.180" Title="The Common Thread" Tags="|pattern|number-sequence|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10187" PostTypeId="2" ParentId="10130" CreationDate="2015-03-13T02:50:33.420" Score="2" Body="&lt;p&gt;As Tanner Swett has noted, it's possible to solve the puzzle using a finite number of states beyond those of the lights themselves.  His solution takes time proportional to the square of the number of lights [i.e. O(N^2); I think that may be improved to time proportional to O(NlgN) at the expense of using an increased number of states, using the approach described below.  For simplicity, light bulbs will be considered in pairs and certain aspects of the algorithm will be simplified.  I'll regard the value of each pair of lights as being 0, 1, or 3, with zero being both lights &lt;em&gt;on&lt;/em&gt; and 3 being both lights &lt;em&gt;off&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Start by establishing a state of 222203 [confirm that there are at least enough lights for that] with the person standing at the zero.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Working left to right, copy everything one place to the right until one encounters a 3, making note of whether one saw any zeroes.  Copy the three one place to the right from where it started.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If one saw any zeroes, then travel right to left until one reaches a 2; flip bits up to and including the first &quot;1&quot; one encounters.  Then go back to step 2.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Using basically the reverse of the logic in steps 2-3, &quot;march&quot; right to left, leaving behind a trail of threes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Check whether the value the last two that was copied is are 2.  If they are, then write another few 2s to their left, replace the rightmost two of the bunch with a zero, and go back to step 2.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the value just beyond the last 2 that was recently copied isn't 2, things have wrapped.  Proceed rightward writing 3s until a 3 is reached, whereupon everything will be 3 (meaning all lights are off).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Each of the above steps would require a few states to implement, so the overall number of states would be large enough to make enumeration irksome, but conceptually not difficult.  I think describing things in terms of &quot;steps&quot; is clearer than states, however.&lt;/p&gt;&#xA;" OwnerUserId="1467" LastActivityDate="2015-03-13T02:50:33.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10189" PostTypeId="2" ParentId="10186" CreationDate="2015-03-13T04:52:15.173" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are the numbers whose roman numeral translations are in strict alphabetical order (so no repetitions).&lt;br&gt;&#xA;&#xA; &lt;code&gt; 4=IV, 9=IX, 60=LX, 101=CI, 400=CD, 450=CDL, 910=CMX, 1010=MX &lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-03-13T04:52:15.173" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10190" PostTypeId="2" ParentId="10186" CreationDate="2015-03-13T05:57:09.030" Score="0" Body="&lt;p&gt;I guess it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Roman numerals with 1 letter, in alphabetical order; then those with 2 letters,  etc. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the source is &lt;a href=&quot;http://oeis.org/A003588&quot; rel=&quot;nofollow&quot;&gt;http://oeis.org/A003588&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="10091" LastActivityDate="2015-03-13T05:57:09.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10196" PostTypeId="2" ParentId="9816" CreationDate="2015-03-13T07:55:54.117" Score="4" Body="&lt;p&gt;If you consider an equation to be logical enough, then the following may explain the sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $a_n$ is the $n$-th number in the sequence, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_n=\lfloor{\left(1.42n+0.37\left(-1\right)^n\right)\left(11+n\right)-0.2521n^{2.812}-2.3p_{(n+k)}+0.5}\rfloor$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $p_{(x)}$ is the $x$-th digit of $\pi$, and $k=5388843$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if $n=3$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_3=\lfloor{\left(4.26-0.37\right)\left(14\right)-0.2521\left(3^{2.812}\right)-2.3p_{(3+5388843)}+0.5}\rfloor$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the 5388846th digit of $\pi$ is $8$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_3=\lfloor{\left(3.89\right)\left(14\right)-0.2521\left(3^{2.812}\right)-2.3(8)+0.5}\rfloor$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=\lfloor 54.46-0.2521\left(3^{2.812}\right)-17.9 \rfloor$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=\lfloor 31.023468\rfloor=31$, which is the 3rd number in the sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $k$ can have other values, such as 6049868, 11551553, or 12701077, and the same sequence will be produced (only possibly differing in the 8th and succeeding terms).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use &lt;a href=&quot;http://www.angio.net/pi/digits.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; to search for digits of $\pi$.&lt;/p&gt;&#xA;" OwnerUserId="10136" LastEditorUserId="10136" LastEditDate="2015-03-13T08:05:30.760" LastActivityDate="2015-03-13T08:05:30.760" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10197" PostTypeId="2" ParentId="10165" CreationDate="2015-03-13T09:06:33.687" Score="2" Body="&lt;p&gt;Is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Death&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It moves in war&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Death is everywhere in war, once here, once there&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Abhors the floor&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Life is no longer on the ground&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Takes many forms&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Can be any form of death&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And is a controversial norm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An encounter with it&#xA;Will leave one tearful&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes, loved or near ones death will make one tearful&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And as an omen&#xA;It makes many fearful&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Death an an omen makes people stop doing something&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10178" LastActivityDate="2015-03-13T09:06:33.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10198" PostTypeId="2" ParentId="9816" CreationDate="2015-03-13T10:00:50.877" Score="5" Body="&lt;p&gt;No there is no logic to this sequence, You might try some overly complicated algorithm to figure it out but that's not the point.&#xA;The point of an IQ test is to test someone's IQ and for that they don't tend to ask for over complicated mathematical algorithms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be thorough:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can try to find the difference between the numbers which are 14, 5, 35, 1, 17, 7. which have nothing logical about them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you can try to add the digit (12 = 1+2 = 3) which would be 3, 8, 4, 13, 14, 13, 2 (or might be 11 (10 + 1)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are not part of any mathematical sequence such a Fibonacci, primes or any other sequence i know. Therefor I conclude the answer is No there is no logic.&lt;/p&gt;&#xA;" OwnerUserId="9835" LastActivityDate="2015-03-13T10:00:50.877" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10199" PostTypeId="2" ParentId="10111" CreationDate="2015-03-13T10:21:37.683" Score="1" Body="&lt;p&gt;My guess is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Carolina (North or South)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll help you get across a river&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There were several ships called Caroline&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later I found fame alone&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; May refer to Caroline affair (&lt;a href=&quot;http://en.wikipedia.org/wiki/Caroline_affair&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Caroline_affair&lt;/a&gt;) or Hurricane Caroline (1975) or even Hurricanes HNL team&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Getting back to the title and second line&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; @Enigma guessed about words &quot;Off is in&quot; in Tom Schilling's song. So, how about just replace &quot;off&quot; in &quot;Carrolloff&quot; with &quot;in&quot; and we'll get something close to &quot;Caroline&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10181" LastActivityDate="2015-03-13T10:21:37.683" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10200" PostTypeId="1" CreationDate="2015-03-13T10:21:40.160" Score="27" ViewCount="2461" Body="&lt;p&gt;Phillip, a 68 years old man, was near the end of his life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before passing away, he decided to hand to his daughter a piece of paper with the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;48 7A 2D 38 8B&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;On the back, there was the following text:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;DEATH, what unpleasant surprise.&#xA;While I'm gone, enjoy life and smile.&#xA;You can find my safe beneath me.&#xA;The key combination is in the front.&#xA;Give your son the life I never had.&#xA;1000 words can't describe how much I love you.&#xA;I wish everything was this basic.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(I know, it isn't the best text ever.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The safe has a keyboard with numbers from 0 to 9, arranged like the image below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0Yuim.jpg&quot; alt=&quot;Lock buttons&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is the best photo I could find, thank you &lt;a href=&quot;http://google.com/&quot;&gt;Google&lt;/a&gt;, &lt;a href=&quot;http://www.time-master.com/sites/default/files/imagecache/product_full/imagecache/product_full/lp-basic-black-keypad-web.jpg&quot;&gt;Original image&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since she didn't understand the message, she asked you for help to open the safe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the safe combination?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I forgot to mention something VERY important:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The safe combination is only 5 numbers!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No more, no less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is also a word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The word is spelled using the above keyboard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each number is 1 letter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: The word 'short' would be the key '84788'.&lt;/p&gt;&#xA;" OwnerUserId="3984" LastEditorUserId="3984" LastEditDate="2015-03-13T10:56:18.753" LastActivityDate="2015-04-28T19:58:41.950" Title="Safe combination for your heritage" Tags="|cipher|" AnswerCount="8" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="10201" PostTypeId="2" ParentId="10200" CreationDate="2015-03-13T10:29:07.617" Score="2" Body="&lt;p&gt;This seems too obvious, but...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 487122223882?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Interpreting each letter as what it would be on the keypad you've shown, mobile-phone style (A=1, B=2, C=22, D=222, E=3, F=4, G=44, etc.)&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-03-13T10:29:07.617" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10202" PostTypeId="2" ParentId="10200" CreationDate="2015-03-13T10:34:39.457" Score="2" Body="&lt;p&gt;Is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 23184 - DEATH&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Short, simple and sweet&lt;/p&gt;&#xA;" OwnerUserId="9918" LastActivityDate="2015-03-13T10:34:39.457" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10203" PostTypeId="2" ParentId="10200" CreationDate="2015-03-13T10:40:12.530" Score="4" Body="&lt;p&gt;Another try:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;48 7A 2D 38 8B&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Interpreting each of these five couplets as a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; number written in hexadecimal (base 16),&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;we get&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 72 122 45 56 139.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;I'm not sure what to do with this, but maybe&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; divide by powers of 2, 1, 3, 8, 4 respectively (these being the digits corresponding to D, E, A, T, H on the keypad - thanks &lt;strong&gt;Ryan Durrant&lt;/strong&gt; for this idea!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;to get&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;9 122 5 7 139&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(OK, this doesn't quite work for the 2nd and 5th digits.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another idea:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; take digit sums, and keep doing this until each number is reduced to a single digit. So $72\rightarrow9,122\rightarrow5,45\rightarrow9,56\rightarrow11\rightarrow2,139\rightarrow13\rightarrow4$,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;giving the answer as &lt;strong&gt;95924&lt;/strong&gt;. I can't see what word this could correspond to though!&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2015-03-13T16:57:10.340" LastActivityDate="2015-03-13T16:57:10.340" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="10204" PostTypeId="2" ParentId="9996" CreationDate="2015-03-13T10:40:30.813" Score="3" Body="&lt;p&gt;Some research on Aggie Kidd's answer and I came up to this answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Shaquille O'Neal&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My job is putting things in rings&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;He was basketball player&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My prize is rings&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;He's got championship rings&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The number of rings I have symbolizes speed&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;He's got four rings and four rings is an Audi logo&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When people see me in action they say Oh, neat!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;He's lastname is the closest to &quot;Oh, neat!&quot; among players who own four rings. Or, &quot;Oh, neat!&quot; is something like &quot;owned&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10181" LastActivityDate="2015-03-13T10:40:30.813" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10205" PostTypeId="2" ParentId="10111" CreationDate="2015-03-13T11:28:11.097" Score="1" Body="&lt;p&gt;I believe it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Washington.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who am I, if not Carrolloff?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;no idea woo (likely a play on words as mentioned, Carrolloff -&gt; Carrollon -&gt; Washington)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll help you get across a river.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/George_Washington%27s_crossing_of_the_Delaware_River&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/George_Washington%27s_crossing_of_the_Delaware_River&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once unrecognised in Fischer's song,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;no idea woo&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later I found fame alone&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;first president obviously&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And also as a US state.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Washington is a state.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;Obviously this is only half complete.&lt;br&gt;&#xA;Also credit to Gamow for having the idea.&lt;/p&gt;&#xA;" OwnerUserId="5265" LastActivityDate="2015-03-13T11:28:11.097" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10206" PostTypeId="2" ParentId="10130" CreationDate="2015-03-13T12:16:49.023" Score="0" Body="&lt;p&gt;&lt;strong&gt;- My solution :&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;turn on light in actual room and adjacent one (that is because we dont know if number of rooms is odd or even)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;turn off light in third room , and on in fourth room , off in 5th ....alternatively , count all rooms passed through  until:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;until coming across two lighted rooms that precedes a succession of switched on , off lights then stop counting , and remind it .&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;turn off two successive lights and go way round , turn off all lights of n rooms crossed over, if you end up with two switched off lights it means you have just achieved your task , otherwise , someone is playing around with you, and you should repeat the task from the room n° n+1&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9517" LastActivityDate="2015-03-13T12:16:49.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10207" PostTypeId="1" AcceptedAnswerId="10229" CreationDate="2015-03-13T12:39:54.580" Score="9" ViewCount="5544" Body="&lt;p&gt;Given the following:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Six Number Teams (1 - 6)&lt;/li&gt;&#xA;&lt;li&gt;Six Letter Teams (A - F)&lt;/li&gt;&#xA;&lt;li&gt;Six Games (Basketball, Football, Baseball, Volleyball, Hockey, Rugby)&lt;/li&gt;&#xA;&lt;li&gt;Six Time Slots (1pm - 6pm)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Set up a game schedule that follows these rules:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Each team must play each game once.&lt;/li&gt;&#xA;&lt;li&gt;Each letter team must play against each number team exactly once.&lt;/li&gt;&#xA;&lt;li&gt;Every game must be played once (by one letter team and one number team) during each time slot&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Please provide either a solution or a mathematical proof of why a solution is impossible.&lt;/p&gt;&#xA;" OwnerUserId="10183" LastEditorUserId="10183" LastEditDate="2021-02-04T12:11:46.193" LastActivityDate="2021-02-04T15:16:56.267" Title="How can 12 teams rotate through 6 games without overlaps?" Tags="|mathematics|logical-deduction|combinatorics|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="10208" PostTypeId="2" ParentId="10207" CreationDate="2015-03-13T12:58:07.210" Score="1" Body="&lt;p&gt;I have named the sports U, V, W, X, Y and Z for convenience&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This can be solved simply by rotating through the teams and sports:&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;strong&gt;Timeslot 1&lt;/strong&gt;&lt;br&gt;&#xA;A vs  1   at  U   ;   B   vs  2   at  W   ;   C   vs  3   at  Y   ;   D   vs  4   at  U   ;   E   vs  5   at  W   ;   F   vs  6   at  Y&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;strong&gt;Timeslot 2&lt;/strong&gt;&lt;br&gt;&#xA;A vs  2   at  V   ;   B   vs  3   at  X   ;   C   vs  4   at  Z   ;   D   vs  5   at  V   ;   E   vs  6   at  X   ;   F   vs  1   at  Z&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;strong&gt;Timeslot 3&lt;/strong&gt;&lt;br&gt;&#xA;A vs  3   at  W   ;   B   vs  4   at  Y   ;   C   vs  5   at  U   ;   D   vs  6   at  W   ;   E   vs  1   at  Y   ;   F   vs  2   at  U&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;strong&gt;Timeslot 4&lt;/strong&gt;&lt;br&gt;&#xA;A vs  4   at  X   ;   B   vs  5   at  Z   ;   C   vs  6   at  V   ;   D   vs  1   at  X   ;   E   vs  2   at  Z   ;   F   vs  3   at  V&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;strong&gt;Timeslot 5&lt;/strong&gt;&lt;br&gt;&#xA;A vs  5   at  Y   ;   B   vs  6   at  U   ;   C   vs  1   at  W   ;   D   vs  2   at  Y   ;   E   vs  3   at  U   ;   F   vs  4   at  W&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;strong&gt;Timeslot 6&lt;/strong&gt;&lt;br&gt;&#xA;A vs  6   at  Z   ;   B   vs  1   at  V   ;   C   vs  2   at  X   ;   D   vs  3   at  Z   ;   E   vs  4   at  V   ;   F   vs  5   at  X&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9444" LastActivityDate="2015-03-13T12:58:07.210" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10209" PostTypeId="2" ParentId="10200" CreationDate="2015-03-13T13:11:41.583" Score="3" Body="&lt;p&gt;When I look at this question, I start to think of something...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; is it ASCII?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1) If the numbers are in hex, we find that 8B doesn't correspond to any letter. You will actually get H, z, -, 8, ü. Not fun.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; so maybe it's base 14 (A,B,C,D)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We change the numbers to that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; and we get 64, 108, 41, 50, 123. In ASCII, this is @, l, ), 2, {, which is very wrong.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2) Maybe &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it's related to DEATH. The ASCII code will be 68, 69, 65, 84, 72 (in decimal). 68 is also the age of Phllip when he died!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But what do we do with it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) The dead guy actually said :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I wish everything was this basic.&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;BASIC&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; is a programming language&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4) Then, we see that &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BASIC is technological and ASCII is also technological.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ASCII is more heavily represented in this puzzle, so I think the answer is 18255.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A lot of things actually don't make sense but that's all I can do.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After Ismael Miguel gave me some comments, I'll continue trying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a list of hints which have been gathered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1*. Hz-8ü &amp;amp; ASCII&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2*. 1000 words can't describe how much I love you --&gt; A picture of probably a heart.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;DEATH --&gt; 23184&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;basic&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;5**. 4 is the first number&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6**. A number repeats&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7**. There is at least a 7. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;8*. 72 122 45 56 139&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hints with 1 * are very close, hints with 2 * are correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 5--&gt; The first letter of the word is f, g, or h. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 7--&gt; The word has at least a o, p or q.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 4--&gt; Based on the comment given to my solution, basic--&gt;21852&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, apparently I left off my problem at one thread which was very close to the answer. &#xA;I wonder which one it is... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By guessing, I got &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Happy --&gt; 41770. That fits. But how is it related to the hints? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9736" LastEditorUserId="9736" LastEditDate="2015-03-15T06:12:24.243" LastActivityDate="2015-03-15T06:12:24.243" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10210" PostTypeId="1" AcceptedAnswerId="10211" CreationDate="2015-03-13T13:20:51.277" Score="1" ViewCount="11335" Body="&lt;p&gt;If these are true :&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;5 + 3 = 158&#xA;9 + 1 = 910&#xA;8 + 6 = 4814&#xA;4 + 4 = 168&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;then 7 + 3 = ??&lt;/p&gt;&#xA;" OwnerUserId="9888" LastActivityDate="2015-07-08T02:17:59.647" Title="Can you solve this math problem" Tags="|calculation-puzzle|pattern|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10211" PostTypeId="2" ParentId="10210" CreationDate="2015-03-13T13:23:17.760" Score="14" Body="&lt;p&gt;I'm guessing it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2110&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because each number result is generated by appending the multiplication of the two numbers and their addition.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or more elaborately:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\begin{align}5 + 3 = 15\ 8 = 5 \times 3 &amp;amp;\mid 5 + 3\\9 + 1 = 9\ 10 = 9 \times 1 &amp;amp;\mid 9 + 10\\8 + 6 = 48\ 14 = 8 \times 6 &amp;amp;\mid 8 + 6\\4 + 4 = 16\ 8 = 4 \times 4 &amp;amp;\mid 4 + 4\\7 + 3 = 21\ 10 = 7 \times 3 &amp;amp;\mid 7 + 3\end{align}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="13494" LastEditDate="2015-06-23T19:11:20.150" LastActivityDate="2015-06-23T19:11:20.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10212" PostTypeId="2" ParentId="10210" CreationDate="2015-03-13T13:23:33.820" Score="8" Body="&lt;p&gt;The answer would be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2110&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$5\times3 = 15,\quad 5+3 = 8 \implies 158\\9 + 1 \implies 9 \times 1 = 9, \quad 9 + 1 = 10 \implies 910\\8 + 6 \implies 8 \times 6 = 48, \quad 8 + 6 = 14 \implies 4814$$&lt;br&gt;&#xA; etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To explain a bit more&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer to the questions is the product followed by the sum of the two numbers, so $(x\times y)(x+y)$ where the brackets are not multiplied but written next to each other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9918" LastEditorUserId="13494" LastEditDate="2015-06-23T19:11:27.927" LastActivityDate="2015-06-23T19:11:27.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10213" PostTypeId="1" AcceptedAnswerId="10538" CreationDate="2015-03-13T13:27:58.630" Score="6" ViewCount="598" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;One December,&lt;br&gt;&#xA;  You are in the casino, but about a quarter of the cards are missing.&lt;br&gt;&#xA;  You see a blue bird fly past the window.&lt;br&gt;&#xA;  A person walks by you and you think like #TheDress.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know this is definitely not a good poem, but each line represents a song from that composer. Find who the composer is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&#xA;All 4 are pop songs. The last line represents a song that she (yes, it's a she) sang but is not very famous. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 2: &#xA;For the previous edit, I accidentally put &quot;changed a word&quot;, it's more like &quot;added Hint 1&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&#xA;All sentences are hidden by word play. This isn't a cryptogram. The phrase which hides the songs are in bold in the spoiler:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First line:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  One &lt;strong&gt;&lt;em&gt;December&lt;/em&gt;&lt;/strong&gt;, &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second line:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are in the casino, but about &lt;strong&gt;&lt;em&gt;a quarter of the cards&lt;/em&gt;&lt;/strong&gt; are missing. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Third line:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You see a &lt;strong&gt;blue bird&lt;/strong&gt; fly past the window.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fourth line:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A person walks by you and you think like &lt;strong&gt;&lt;em&gt;#TheDress&lt;/em&gt;&lt;/strong&gt;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9736" LastEditorUserId="8874" LastEditDate="2016-03-18T09:51:38.317" LastActivityDate="2016-03-18T09:51:38.317" Title="Who's the composer?" Tags="|riddle|pattern|music|" AnswerCount="3" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10214" PostTypeId="1" AcceptedAnswerId="10217" CreationDate="2015-03-13T14:02:48.817" Score="5" ViewCount="315" Body="&lt;p&gt;Can you fill in the holes in this pattern?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;361&lt;br&gt;&#xA;  722&lt;br&gt;&#xA;  726&lt;br&gt;&#xA;  [A]&lt;br&gt;&#xA;  323&lt;br&gt;&#xA;  650&lt;br&gt;&#xA;  223&lt;br&gt;&#xA;  49729&lt;br&gt;&#xA;  99458&lt;br&gt;&#xA;  [B]&lt;br&gt;&#xA;  19024&lt;br&gt;&#xA;  2538&lt;br&gt;&#xA;  2240&lt;br&gt;&#xA;  98&lt;br&gt;&#xA;  102&lt;br&gt;&#xA;  302&lt;br&gt;&#xA;  22805&lt;br&gt;&#xA;  [C]&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: As the answer has been found, the clues would have been:&lt;br&gt;&#xA;1) An edit to the title to be &quot;A Prime Example...&quot;&lt;br&gt;&#xA;2) Adding 19 to the beginning of the sequence&lt;br&gt;&#xA;3) A link to an online calculator for finding prime factorials&lt;br&gt;&#xA;4) The statement that the last clue just about summed it up&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus Question:&lt;/strong&gt;&lt;br&gt;&#xA;What can you choose as the starting number such that the pattern repeats itself forever?&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="8874" LastEditDate="2015-10-08T09:12:20.753" LastActivityDate="2015-10-08T09:12:20.753" Title="A Prime Example of a Number Pattern" Tags="|mathematics|number-sequence|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10216" PostTypeId="1" AcceptedAnswerId="10537" CreationDate="2015-03-13T14:51:57.687" Score="10" ViewCount="935" Body="&lt;p&gt;Solve the following morning math problems to solve the Math Message:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$7 \times (60 + 103)$&lt;br&gt;&#xA;$219 \times 2 \times 3$&lt;br&gt;&#xA;$2 \times 3 \times 10^2$&lt;br&gt;&#xA;$\sqrt{40000}$&lt;br&gt;&#xA;$23 \times 5 $&lt;br&gt;&#xA;$16^2 + 26$&lt;br&gt;&#xA;$\frac{1000}{2} - \frac{24}{3}$&lt;br&gt;&#xA;$(150-7) \times12$&lt;br&gt;&#xA;$33\times50$&lt;br&gt;&#xA;$\frac{500}{2} + 70$&lt;br&gt;&#xA;$\frac{453}{25} \times 10^{-1} \times 500$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your math may not be finished when you answer these first questions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each answer will be divided by a number. Remember what number.  (1 is a number)   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each problem corresponds to a single character.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You should find 40 twice.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Aside from the two 40's, your numbers will lie between 114 and 166  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 6:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You will only see each divisor once.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorDisplayName="user17461" LastEditDate="2015-11-20T16:20:20.720" LastActivityDate="2015-11-20T16:20:20.720" Title="Happy Math Question" Tags="|mathematics|cipher|" AnswerCount="4" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="10217" PostTypeId="2" ParentId="10214" CreationDate="2015-03-13T15:19:35.783" Score="4" Body="&lt;p&gt;I think I see a pattern in there, it only works in case one of your calculations is incorrect though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A = 255, B = 99462, C = 20802746&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The incorrect transition:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 99462 -&gt; 2538, you forgot to mention 19024 in between.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The pattern:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Take a number, split it into its prime factors, add the square of each prime factor to arrive at the next number.&lt;br&gt;&#xA;Example: 2538 is 2*3*3*3*47, adding the squares is 4+9+9+9+2209, next number is 2240.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bonus answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Start with 1, you'll stay at that value.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9910" LastActivityDate="2015-03-13T15:19:35.783" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10218" PostTypeId="1" AcceptedAnswerId="10369" CreationDate="2015-03-13T15:24:41.250" Score="7" ViewCount="1761" Body="&lt;p&gt;Cached in secrecy,&lt;br&gt;&#xA;Where none shall interfere&lt;br&gt;&#xA;A man with wooden jaw&lt;br&gt;&#xA;Waits patiently each year&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hands flung abroad,&lt;br&gt;&#xA;As the masked man cries &quot;unscathed!&quot;&lt;br&gt;&#xA;In an abode of luxury,&lt;br&gt;&#xA;With reservations made&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combatants from a land&lt;br&gt;&#xA;Found on distant shores&lt;br&gt;&#xA;With an intruder well acquainted&lt;br&gt;&#xA;With the sound of people's snores&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A shape at the heart&lt;br&gt;&#xA;Of an ages-old tale&lt;br&gt;&#xA;Followed by what's great and blue,&lt;br&gt;&#xA;But isn't a whale&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Verses 1 and 2 are two ways of describing the same thing, and involve a bit of puns/wordplay. Verses 3 and 4 go together and describe something else, which is a clue to arriving at the correct answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The ages-old tale referred to in the last verse happened about 2000 years ago, give or take 33 years. The &quot;combatants from a land found on distant shores&quot; were active about 40 years ago.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The masked man is one of a couple masked men that show up in another common riddle, although surprisingly I couldn't find any version of that riddle on this site.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10151" LastEditorUserId="25957" LastEditDate="2017-01-06T17:40:25.607" LastActivityDate="2017-01-06T17:40:25.607" Title="The man with a wooden jaw" Tags="|riddle|rhyme|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10219" PostTypeId="2" ParentId="10200" CreationDate="2015-03-13T15:29:52.607" Score="2" Body="&lt;p&gt;I think I have it!&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;D  E  A  T  H&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In ASCII hex:&#xA;    44 65 61 74 68&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The key combination is in the front.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;44 65 61 74 68&#xA;^  ^  ^  ^  ^&#xA;4  6  6  7  6&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Take the first hexadecimal digit of each ASCII character of the word &quot;death&quot;, and that should be your answer.&lt;/p&gt;&#xA;" OwnerUserId="10187" LastEditorUserId="5373" LastEditDate="2015-03-14T00:28:42.920" LastActivityDate="2015-03-14T00:28:42.920" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10221" PostTypeId="2" ParentId="10165" CreationDate="2015-03-13T15:49:38.143" Score="0" Body="&lt;p&gt;Huh.   I came up with &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Salt.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wasn't too sure how it would fit the first verse unless it was some weird way of thinking about &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  The SALT treaty from the Reagan era&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But it fit the 2nd and 3rd pretty well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, so yeah why did it fit the 2nd and 3rd verses.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Well I was thinking about cholorine used in swimming pools, which is a form of salt that burns the eyes (trust me on this).   But sure, tears are salty, and so &quot;artificial tears&quot; have salt content, so if you put it in your eyes, it will &quot;leave one tearful&quot;, and there is a superstition about spilling salt at the table, and people throwing salt over their shoulders to ward off bad luck.  As for the 3rd verse, before refrigeration, salting was used as means of preserving meat.  It was really the 3rd verse that made me think of salt, and then looking back, it could fit the second verse as well, but the first verse is just a stretch, no doubt.   :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5022" LastEditorUserId="5022" LastEditDate="2015-03-13T18:41:16.913" LastActivityDate="2015-03-13T18:41:16.913" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10224" PostTypeId="2" ParentId="10165" CreationDate="2015-03-13T16:24:34.710" Score="0" Body="&lt;p&gt;I think this fits  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Unmanned Aerial Vehicles (UAVs) or drones&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It moves in war  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Obviously they move in war zones&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Abhors the floor  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's a flying machine/device&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Takes many forms  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many implementations/different models&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And is a controversial norm  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; UAV use is controversial in war but it's becoming common as technology advances&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An encounter with it&lt;br&gt;&#xA;Will leave one tearful  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A UAV attack is fairly targeted and will leave bystanders who aren't killed tearful&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And as an omen&lt;br&gt;&#xA;It makes many fearful  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It has become a symbol of American aggression in the middle east (I'm not positive that this is true, but expect that it is)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though as journeyers swear&lt;br&gt;&#xA;It's useful where&lt;br&gt;&#xA;One wants to prevent&lt;br&gt;&#xA;Spoiled fare  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is kind of a stretch but using drones as an alternative to bombing the heck out of a place helps prevent massive civilian casualties and destruction of infrastructure. Alternately, you could say using drones as an alternative to sending in ground troops spares the lives of your soldiers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another similar possibility&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Missiles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10190" LastActivityDate="2015-03-13T16:24:34.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10225" PostTypeId="2" ParentId="10218" CreationDate="2015-03-13T16:36:51.900" Score="3" Body="&lt;p&gt;Hmm. Not sure about this, and missing some of the clues, but...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A nutcracker - specifically, the decorative soldier type.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cached in secrecy,&lt;br&gt;&#xA;Where none shall interfere&lt;br&gt;&#xA;A man with wooden jaw&lt;br&gt;&#xA;Waits patiently each year  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stored every year in an out-of-the-way cupboard, until Christmas rolls around. (I would think the typical soldier nutcracker is just used as decoration.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hands flung abroad,&lt;br&gt;&#xA;As the masked man cries &quot;unscathed!&quot;&lt;br&gt;&#xA;In an abode of luxury,&lt;br&gt;&#xA;With reservations made  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not a clue. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Combatants from a land&lt;br&gt;&#xA;Found on distant shores  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The soldier nutcracker originated in Germany, and typically wears a dress uniform of that area from an ancient period. If you're not on the European mainland, you could consider Germany to be &quot;distant shores&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With an intruder well acquainted&lt;br&gt;&#xA;With the sound of people's snores  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Santa Claus is also associated with Christmas. In legend, he sneaks down people's chimneys while they are asleep to leave them presents.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A shape at the heart&lt;br&gt;&#xA;Of an ages-old tale  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A soldier nutcracker is a eponymous player in The Nutcracker by Tchaikovsky, a popular ballet first performed in 1892. That ballet in turn was based on an even older tale written in 1816 by E. T. A. Hoffman.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Followed by what's great and blue,&lt;br&gt;&#xA;But isn't a whale  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If a rambunctious child accidentally whacks you with a hard wooden nutcracker, you may wish to replace it with a large and soft blue teddy bear....Joking. :) Not a clue here either.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2741" LastActivityDate="2015-03-13T16:36:51.900" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10226" PostTypeId="2" ParentId="10207" CreationDate="2015-03-13T16:40:22.007" Score="0" Body="&lt;p&gt;There doesn't seem to be a valid setup. Picking team 1, team A and game U at random, team 1 will plays game U against team A during the first time slot. That will leave five other teams left that team 1 will have to play against, each of those have played one of the remaining games. Team 1 must not play its first game type again, none of the other five teams is allowed to play the first game type they played again, after playing at most four additional games no further valid match is possible.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplifying things a bit, suppose there are just two time slots, two number teams (1, 2), two letter teams (A, B) and two games (U, V). Team 1 will play game U against team A in the first time slot. Team 1 can't play game U against Team B in the second time slot because team 1 already played game U. Team 1 also can't play game V against team B because team B already played game V during the first time slot.&lt;/p&gt;&#xA;" OwnerUserId="9910" LastActivityDate="2015-03-13T16:40:22.007" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10227" PostTypeId="2" ParentId="10218" CreationDate="2015-03-13T16:46:56.017" Score="5" Body="&lt;p&gt;The solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The horse of Troy or &lt;a href=&quot;http://en.wikipedia.org/wiki/Trojan_Horse&quot;&gt;Trojan Horse&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cached in secrecy,&lt;br&gt;&#xA;Where none shall interfere&lt;br&gt;&#xA;A man with wooden jaw&lt;br&gt;&#xA;Waits patiently each year&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The horse. It can be thought as a man with &lt;em&gt;wooden jaw&lt;/em&gt;, containing the greeks in his mouth (belly). No clue on the last sentence, maybe because each year its taught to a different generation so &quot;waits&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hands flung abroad,&lt;br&gt;&#xA;As the masked man cries &quot;unscathed!&quot;&lt;br&gt;&#xA;In an abode of luxury,&lt;br&gt;&#xA;With reservations made  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hands of greeks over Troy. The others sentences can be explained, but are a bit stretched (luxury of greeks, golden &lt;a href=&quot;http://en.wikipedia.org/wiki/Mask_of_Agamemnon&quot;&gt;mask of Agamemnon&lt;/a&gt;, the greeks pretend to return home after 10 years to be &quot;unscathed&quot;...)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Combatants from a land&lt;br&gt;&#xA;Found on distant shores&lt;br&gt;&#xA;With an intruder well acquainted&lt;br&gt;&#xA;With the sound of people's snores  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is self explanatory. Combatants (the greeks) from a land found on distant shores (Greece) with an intruder well acquainted (the horse) with the sound of people's snores (the greeks attacked Troy during the night, when everyone was sleeping)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A shape at the heart&lt;br&gt;&#xA;Of an ages-old tale&lt;br&gt;&#xA;Followed by what's great and blue,&lt;br&gt;&#xA;But isn't a whale  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The tale is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Aeneid&quot;&gt;Aeneid&lt;/a&gt;, followed by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Odyssey&quot;&gt;Odyssey&lt;/a&gt;, where, after the concquer of Troy, Ulisses has to cross the &lt;em&gt;great and blue sea&lt;/em&gt; to return home.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is it?&lt;/p&gt;&#xA;" OwnerUserId="4652" LastActivityDate="2015-03-13T16:46:56.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10228" PostTypeId="2" ParentId="10200" CreationDate="2015-03-13T17:03:15.333" Score="2" Body="&lt;p&gt;For starters, the key combination&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;48 7A 2D 38 8B&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;if treated as hexadecimal, translates to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;72  122 45  56  139&lt;/strong&gt; in decimal, which is&lt;br&gt;&#xA; &lt;strong&gt;H z - 8 ï&lt;/strong&gt; in ASCII.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no idea what to do with this. Yet. Let's look at other clues.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;DEATH&lt;/strong&gt; corresponds to &lt;strong&gt;23184&lt;/strong&gt; on the keypad.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I also guess that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;1000&quot; and &quot;basic&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;seem to be part of the same clue, somehow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, from comments on another answer, apparently,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the answer starts with a &lt;strong&gt;4&lt;/strong&gt;, has at least one &lt;strong&gt;7&lt;/strong&gt;, and some digit is doubled.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;All told, my very best guess (at the time of writing) is that the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;41388&lt;/strong&gt;, for &lt;strong&gt;HEART&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My reasoning is that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;1000 words&quot; refers to how a picture is worth more than 1000 words, which made me wonder how a picture would relate to &quot;I love you&quot;. I tried &quot;drawing&quot; a heart on the keypad, then realized that DEATH and HEART share the same letters, except that the D has been changed into an R, which on the same button as 8, which is 1000 in binary.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One major flaw with this though:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It doesn't use the key code at all.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="268" LastActivityDate="2015-03-13T17:03:15.333" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10229" PostTypeId="2" ParentId="10207" CreationDate="2015-03-13T18:01:54.240" Score="14" Body="&lt;p&gt;In order to do this puzzle, you'd need to create &lt;a href=&quot;http://www.cut-the-knot.org/arithmetic/latin3.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mutually Orthogonal Latin Squares of order 6&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;For example, say that instead, you had 6 teams (1-3 and A-C), 3 sports (baseball, football, hockey), 3 timeslots.  Then, you could make the following schedule:&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th&gt;&lt;/th&gt;&#xA;&lt;th&gt;A&lt;/th&gt;&#xA;&lt;th&gt;B&lt;/th&gt;&#xA;&lt;th&gt;C&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;1&lt;/td&gt;&#xA;&lt;td&gt;b@1&lt;/td&gt;&#xA;&lt;td&gt;f@2&lt;/td&gt;&#xA;&lt;td&gt;h@3&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;2&lt;/td&gt;&#xA;&lt;td&gt;h@2&lt;/td&gt;&#xA;&lt;td&gt;b@3&lt;/td&gt;&#xA;&lt;td&gt;f@1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;3&lt;/td&gt;&#xA;&lt;td&gt;f@3&lt;/td&gt;&#xA;&lt;td&gt;h@1&lt;/td&gt;&#xA;&lt;td&gt;b@2&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;So in this example, team A plays baseball against team 1 at 1:00pm.&lt;/p&gt;&#xA;&lt;p&gt;This uses 2 Mutually Orthogonal Latin Squares of order 3.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;1 2 3    b f h&#xA;2 3 1    h b f&#xA;3 1 2    f h b&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;This allows us to conform to the following rules:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Every lettered team plays every numbered team exactly once - simply by design of the table&lt;/li&gt;&#xA;&lt;li&gt;Every team plays every sport exactly once&lt;/li&gt;&#xA;&lt;li&gt;Every team plays in every timeslot exactly once&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;However, it is a known impossibility to create two MOLS of order 6, so the original question is not possible.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="69582" LastEditDate="2021-02-04T15:16:56.267" LastActivityDate="2021-02-04T15:16:56.267" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10230" PostTypeId="1" CreationDate="2015-03-13T18:32:57.460" Score="6" ViewCount="252" Body="&lt;p&gt;I discovered this very old Javascript game in my old bookmarks.&#xA;It's a great puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.ucla.edu/~tom/Games/dominotion.html&quot; rel=&quot;nofollow&quot;&gt;http://www.math.ucla.edu/~tom/Games/dominotion.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know another name for this puzzle or a site or app where one can play it?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: @Emrakul added some great information, in particular, the generally known name for it, which allows googling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the great thing about the version I linked above is that: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;it allows you to mark the edges which outline the dominoes individually, and &lt;/li&gt;&#xA;&lt;li&gt;it also keeps track of the number of adjacent instances of each number pair.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In other words, this is much better than the versions I have found given the name &quot;Dominosa&quot; because you can keep track of the incremental partial solutions and you don't waste all the time searching the puzzle grid manually.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any additional leads on an app version with this support? (Or any volunteers :)  )&lt;/p&gt;&#xA;" OwnerUserId="1537" LastEditorUserId="8874" LastEditDate="2015-10-08T08:55:22.843" LastActivityDate="2015-10-08T08:55:22.843" Title="Is there another name or site for this &quot;Dominotion&quot; puzzle?" Tags="|puzzle-identification|dominoes|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10231" PostTypeId="1" AcceptedAnswerId="10363" CreationDate="2015-03-13T18:45:10.840" Score="4" ViewCount="1582" Body="&lt;p&gt;I start out quite skilled,&lt;br&gt;&#xA;Throughout the nation I've drilled,&lt;br&gt;&#xA;I'm sexist to the extreme,&lt;br&gt;&#xA;You'll soon know what I mean,&lt;br&gt;&#xA;Your children should thank me,&lt;br&gt;&#xA;Your wife might just hate me.&lt;br&gt;&#xA;Knives I despise&lt;br&gt;&#xA;I hide from your eyes.&lt;br&gt;&#xA;Avoid mutations,&lt;br&gt;&#xA;From your medications.&lt;br&gt;&#xA;So we will do all your tests,&lt;br&gt;&#xA;But eventually we'll both rest.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Two lines of word play. (The first two).  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What could cause mutations that is also used as a medication (treatment)?   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What are nations sometimes composed of?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="9951" LastEditDate="2015-03-15T20:08:55.597" LastActivityDate="2015-03-16T16:31:39.797" Title="Don't be Sexist - What am I?" Tags="|riddle|word|rhyme|" AnswerCount="7" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10232" PostTypeId="2" ParentId="10230" CreationDate="2015-03-13T18:50:25.380" Score="6" Body="&lt;p&gt;This puzzle appears to originate in many places, primarily under the name &quot;Dominosa.&quot; I can't seem to find anything called &quot;Dominotion&quot; on the web besides on the UCLA website. (It's not actually a name I've heard before for this puzzle.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the oldest reference I can find online comes from &lt;a href=&quot;http://www.chiark.greenend.org.uk/~sgtatham/puzzles/&quot;&gt;Simon Tatham's puzzle collection&lt;/a&gt;, for which there is a working JS version &lt;a href=&quot;http://www.chiark.greenend.org.uk/~sgtatham/puzzles/js/dominosa.html&quot;&gt;on his website&lt;/a&gt;. You can also download an executable version from the same (or by following &lt;a href=&quot;http://www.chiark.greenend.org.uk/~sgtatham/puzzles/dominosa.exe&quot;&gt;this direct download link&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, a search for &quot;dominosa&quot; on pretty much any platform will return results - for instance, &lt;a href=&quot;https://play.google.com/store/apps/details?id=com.oxothuk.dominosa&quot;&gt;Dominosa&lt;/a&gt; on Google Play.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorDisplayName="user20" LastEditDate="2015-03-13T18:56:56.823" LastActivityDate="2015-03-13T18:56:56.823" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10233" PostTypeId="2" ParentId="10200" CreationDate="2015-03-13T18:54:07.033" Score="4" Body="&lt;p&gt;The code is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 6 7 4 1 &lt;br&gt;&#xA;&#xA; A L P H A&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We know that the numbers and letters turn into&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The ASCII codes &quot;Hz-8u&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This can be decoded as&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A frequency of 8 hertz, also known as Alpha waves&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can support this further with the following lines:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DEATH - What an unpleasant surprise&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you were exposed to this frequency, it would liquefy your brain. Doesn't sound like fun!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I'm gone, enjoy life and smile&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Alpha frequencies emitted by your brain, however, are associated with relaxation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can find my safe beneath me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm guessing it was in the floor or something?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The key combination is in the front&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alpha is the first letter of the greek alphabet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Give your son the life I never had&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Uhhh... red herring?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1000 words can't describe now much I love you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MRIs (images, pictures of your brain) use frequencies 1-100, which this is one. Obviously you're not exposed to the EXACT resonant frequency of your skull for very long, because liquefying your brain is generally bad and will get the hospital sued.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I wish everything was this basic&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pretty much all children are taught about frequencies as part of their basic education.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2812" LastActivityDate="2015-03-13T18:54:07.033" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10234" PostTypeId="1" AcceptedAnswerId="10235" CreationDate="2015-03-13T19:01:01.993" Score="1" ViewCount="499" Body="&lt;p&gt;Here is a riddle I made up, no one has ever solved it, but I really don't think it's that hard.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Stronger than iron&lt;br&gt;&#xA;  Harder than rock&lt;br&gt;&#xA;  Lighter than feathers&lt;br&gt;&#xA;  It's used as a lock  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Fast as lightning&lt;br&gt;&#xA;  Though it takes up no space&lt;br&gt;&#xA;  Many have cracked it&lt;br&gt;&#xA;  Without striking its face&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10193" LastEditorUserId="8874" LastEditDate="2015-10-08T09:03:42.210" LastActivityDate="2015-10-08T09:03:42.210" Title="Stronger Than Iron - Riddle" Tags="|riddle|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10235" PostTypeId="2" ParentId="10234" CreationDate="2015-03-13T19:03:31.267" Score="6" Body="&lt;p&gt;I'll just throw out a guess of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; password/cypher&lt;br&gt;&#xA; They can be hard as dirt, they secure something (lock), it doesn't really take up space, and they've been broken&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-03-13T19:03:31.267" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10236" PostTypeId="2" ParentId="10072" CreationDate="2015-03-13T19:26:58.410" Score="0" Body="&lt;p&gt;Here is a partial answer. If there is a largest N, then it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; greater than 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Handwaving proof:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The notion of 12 solvability sets refers only to states in R, the set of states reachable by sequences of 'nice' swaps, defined as swaps of stickers between faces of cubies of the same type (edge, corner, centre) (a notion that may for most purposes be less than optimally conceived of in terms of sticker swaps); whereas this question allows any kind of sticker swaps - for example leading to states with all centre cubies red. &lt;em&gt;If&lt;/em&gt; it is true that a single swap of two stickers suffices to get to a state in any of the solvability sets, then 1 in 12 of those swaps will lead to solvable states. How many non-nice swaps are there? There are p of two stickers of the same colour, all of which will lead to solvable states; and q of two stickers of different colours, all of which will lead to states outside R and therefore unsolvable. WLOG, consider only swaps that involve a red sticker. p is half of the sum of 20 (corner to non-corner), 8 (centre to non-centre) and 20 (edge to non-edge), which is 24. The 'half' is because we have counted each swap twice.  When we count swaps between red and non-red stickers, we don't multiply by a half, so q is 5 times the same sum, because we've got 5 other faces with similar distributions of stickers on edges, corners and centres, so q is 240. $\frac{p}{q}&amp;gt;\frac{1}{12}$, so $N\not= 2$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I suspect this can be combined with some of the thinking in @Emrakul's proof (which is not yet solid) to show that there is no N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Correction:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I've gone wrong with the nice and non-nice swaps here. A swap of a red sticker from one edge cubie to another isn't nice if it yields an edge cubie with two red stickers, and the same applies if we replace 'edge' with 'corner'. The good news is that I'm still going to assume that after one nice swap the proportion of states that are solvable is $\frac{1}{12}$. (Waves hands.) We've just got more non-nice swaps now and we've got to distribute them between p and q. Of these, 12 are of red edge stickers to edge cubies which have already got a red edge sticker on them, giving a double-red edge cubie and therefore unsolvability. We get another 24 when we replace 'edge' with 'corner'. So we need to increase q by 36, so $q=276$. $\frac{p}{q}$ is now $\frac{24}{276}=\frac{1}{11.5}$, which is still greater than $\frac{1}{12}$, so the proved-by-handwaving result still stands.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7363" LastEditorUserId="7363" LastEditDate="2015-03-14T09:43:08.947" LastActivityDate="2015-03-14T09:43:08.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10237" PostTypeId="2" ParentId="10200" CreationDate="2015-03-13T22:29:18.740" Score="0" Body="&lt;p&gt;Based on clues I will guess on &lt;code&gt;FRONT&lt;/code&gt; but nothing with baseXX, acsii or hex converting is used here.. :) &lt;/p&gt;&#xA;" OwnerUserId="10197" LastActivityDate="2015-03-13T22:29:18.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10238" PostTypeId="1" AcceptedAnswerId="10248" CreationDate="2015-03-13T22:41:29.517" Score="13" ViewCount="4171" Body="&lt;p&gt;A group of $n$ women and $m$ men have gotten themselves stuck on a desert island. Or, they're the last people left alive after the apocalypse. Whichever scenario you prefer. If they start reproducing, and assuming these people are all completely unrelated, how long will their colony last for if they want to avoid any and all incest, no matter how far removed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely, the rules are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;A person's generation is the maximum of their parents', plus one. The initial population all have generation zero.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You can reproduce with (and only with) anyone of the opposite gender with whom you share no common ancestor in the initial population (and a person is considered to be their own ancestor).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Ignore the fact that people have a finite lifespan and that pregnancy takes nine months. You can have as many children as you want.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You're trying to find the theoretical highest possible number of generations, so it's fine if your strategy requires that the kids come out with specific genders.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Try the case with, say, 1 man and 2 women (or vice versa) to whet your appetite, but the point of the puzzle is to work out the general case. Remember that you don't just need some upper bound, you need a tight upper bound - so you'll probably want to show an explicit strategy attaining that bound, along with a proof that that's the best you can do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have already solved this, so I can guarantee there is an elegant solution.&lt;/p&gt;&#xA;" OwnerUserId="5511" LastActivityDate="2015-03-16T22:15:32.997" Title="How long can a population last without incest?" Tags="|mathematics|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10239" PostTypeId="1" CreationDate="2015-03-13T22:46:35.800" Score="2" ViewCount="280" Body="&lt;p&gt;There are $n\ge3$ players playing a game. &#xA;In this game, one person will come out in first place, one in second, and so on. It's impossible to tie. The person in first place gets $n$ points, the person in second place gets $n-1$, and so on, so that the person in last place gets $1$ point. After playing this game some fixed number of times, the scores are tallied up and the winner is whoever has the highest score.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we fix a number of games (other than one game) in advance, so that a tie for first place is impossible? With two players this is easy - just play an odd number of games. But otherwise?&lt;/p&gt;&#xA;" OwnerUserId="5511" LastEditorUserId="8874" LastEditDate="2015-10-08T09:08:24.387" LastActivityDate="2015-10-08T09:08:24.387" Title="How many games should be played to avoid tying?" Tags="|mathematics|combinatorics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10240" PostTypeId="2" ParentId="10101" CreationDate="2015-03-13T23:01:48.760" Score="4" Body="&lt;p&gt;Not sure if this quite fits but there are not many remaining toxic substances that have killed millions.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cyanide&quot; rel=&quot;nofollow&quot;&gt;cyanide&lt;/a&gt;&lt;/strong&gt; for the following reasons  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm the thing you would love  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People enjoy foods like cassava, lima beans and almonds which contain substances that release cyanide. Pits and seeds of apricots, apples, and peaches, may also have substantial amounts of chemicals which are metabolized to cyanide. But the edible parts of these plants contain much lower amounts of these chemicals.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet I was a murderer by the millions, I was a poison on the 1900's  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Deliberate cyanide poisoning of humans has occurred many times throughout history.  Most significantly, hydrogen cyanide released from pellets of Zyklon-B was used extensively in the systematic mass murders of the Holocaust. Poisoning by hydrogen cyanide gas within a gas chamber is one method of executing a condemned prisoner.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then a revolutionary medicine in the hospitals  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The cyanide compound, &lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_nitroprusside&quot; rel=&quot;nofollow&quot;&gt;sodium nitroprusside&lt;/a&gt;, is primarily used as a &lt;a href=&quot;http://en.wikipedia.org/wiki/Vasodilation&quot; rel=&quot;nofollow&quot;&gt;vasodilator&lt;/a&gt; (to relax the muscle cells of blood vessels). By 1955, data on its safety during short-term use in patients with severe hypertension had become available. Despite this, it was not approved by the US FDA until 1974 for the treatment of severe hypertension. By 1993, its popularity had grown such that total sales in the US had totalled US$2 million.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My other use a massage to your blood, the other the killer  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As described above - a vasodilator, a poison   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-03-13T23:07:11.443" LastActivityDate="2015-03-13T23:07:11.443" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10243" PostTypeId="2" ParentId="10238" CreationDate="2015-03-14T01:01:53.377" Score="6" Body="&lt;p&gt;$m + n - 1$ generations if $if$ $n &amp;gt; 0$ &amp;amp; $m &amp;gt; 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is only one gender in generation 0 then they won't be able to reproduce.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If each 'couple' has male and female progeny then there could be $2*m*n$ children in the first generation. These could reproduce with $m + n -2$ from the first generation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second generation could reproduce with a maximum of $n + m - 3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third generation could reproduce with a maximum of $n + m - 4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum number of people that a person from generation $g$ could reproduce with after generation 0 is $m + n - (g + 1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum of generations is obtained by  solving:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m + n - max(g) - 1 = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$max(g) = m + n -1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $(m + n -1)^{th}$ generation has no non relation to reproduce with.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastEditorUserId="1508" LastEditDate="2015-03-14T16:42:17.300" LastActivityDate="2015-03-14T16:42:17.300" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10244" PostTypeId="2" ParentId="10239" CreationDate="2015-03-14T01:03:09.563" Score="8" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No, a tie is always possible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof/construction:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the number of rounds is even, have two players alternate being 1st and 2nd place respectively. The remaining players' scores can be anything.&lt;br&gt;&#xA;&#xA; If the number of rounds is odd, set up the first three games to end with tied scores by having one player be 1st, 1st, 3rd and the other be 2nd, 2nd, 1st. Then, have them alternate 1st and 2nd as before for the remaining games (if any). The other players can again have whatever scores; no matter how you do it, they will not be in the running for first.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-03-14T01:51:01.080" LastActivityDate="2015-03-14T01:51:01.080" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10248" PostTypeId="2" ParentId="10238" CreationDate="2015-03-14T06:52:23.857" Score="6" Body="&lt;p&gt;Considering numbers per gender creates a kind of distraction. Let $g(N)$ be the maximum number of 'incest-free generations' that can result from $N$ unrelated individuals, not all of whom are of the same gender. An additional individual unrelated to the $N$ others can mate with any of them who happens to be of the opposite sex. Selecting this mating partner from the most recent generation allows the formation of exactly one additional incest-free generation. So we have:&#xA;$$g(N+1)=g(N)+1$$&#xA;Avoiding incest, a single man and a single woman ($N=2$) can produce no more than one generation of offspring:&#xA;$$g(2)=1$$&#xA;Combining the above two results, it follows that &#xA;$$g(N)= N-1$$&#xA;In this equation one can replace $N$ with $m+n$ to obtain the answer requested.&lt;/p&gt;&#xA;" OwnerUserId="7573" LastEditorUserId="7573" LastEditDate="2015-03-15T04:56:50.183" LastActivityDate="2015-03-15T04:56:50.183" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10249" PostTypeId="2" ParentId="10231" CreationDate="2015-03-14T09:53:42.157" Score="4" Body="&lt;p&gt;Probably not the answer, but&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; lips&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;seems to fit many of the clues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I start out quite skilled, &lt;br&gt;&#xA; Throughout the nation I've drilled,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ???&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I'm sexist to the extreme, &lt;br&gt;&#xA; You'll soon know what I mean, &lt;br&gt;&#xA; Your children should thank me, &lt;br&gt;&#xA; Your wife might just hate me.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This could be a reference to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; kissing. Children are kissed by their parents, and your wife might well be annoyed if you kiss other women. I'm not quite sure about the sexist bit though.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Knives I despise&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nobody wants to have cut lips.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I hide from your eyes.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can't see your lips most of the time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Avoid mutations, &lt;br&gt;&#xA; From your medications.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lips is almost an anagram of pills. (This is actually what put me onto this answer in the first place!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;So we will do all your tests,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ???&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But eventually we'll both rest.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Both&quot; - and yes, you have exactly two lips! So this seems to fit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-03-14T09:53:42.157" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10250" PostTypeId="2" ParentId="10238" CreationDate="2015-03-14T10:41:38.050" Score="5" Body="&lt;p&gt;This condition:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;avoid any and all incest, &lt;em&gt;&lt;strong&gt;no matter how far removed&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;means that the answer is 0, because all of the initial population share common ancestors (all living humans share a set of common ancestors).&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In the field of human genetics, the name Mitochondrial Eve refers to the matrilineal most recent common ancestor (MRCA), in a direct, unbroken, maternal line, of all currently living anatomically modern humans, who is estimated to have lived approximately 100,000–200,000 years ago. This is the most recent woman from whom all living humans today descend, in an unbroken line, on their mother’s side, and through the mothers of those mothers, and so on, back until all lines converge on one person. Because all mitochondrial DNA (mtDNA) generally (but see paternal mtDNA transmission) is passed from mother to offspring without recombination, all mtDNA in every living person is directly descended from hers by definition, differing only by the mutations that over generations have occurred in the germ cell mtDNA since the conception of the original &amp;quot;Mitochondrial Eve&amp;quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Mitochondrial_Eve&quot; rel=&quot;noreferrer&quot;&gt;WP: Mitochondrial Eve&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;the MRCA [Most Recent Common Ancestor] of all present-day humans lived just a few thousand years ago in these models. Moreover, among all individuals living more than just a few thousand years earlier than the MRCA [Most Recent Common Ancestor], &lt;em&gt;&lt;strong&gt;each presentday human has exactly the same set of genealogical ancestors&lt;/strong&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&lt;p&gt;Further work is needed to determine the effect of this common&#xA;ancestry on patterns of genetic variation in structured populations.&#xA;But to the extent that ancestry is considered in genealogical&#xA;rather than genetic terms, our findings suggest a remarkable&#xA;proposition: &lt;strong&gt;no matter the languages we speak or the colour of our&#xA;skin, we share ancestors who planted rice on the banks of the&#xA;Yangtze, who first domesticated horses on the steppes of the&#xA;Ukraine, who hunted giant sloths in the forests of North and&#xA;South America, and who laboured to build the Great Pyramid of&#xA;Khufu.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Douglas L. T. Rohde, Steve Olson &amp;amp; Joseph T. Chang, &amp;quot;&lt;a href=&quot;http://steveolson.com/uploads/2009/04/nature-common-ancestors2.pdf&quot; rel=&quot;noreferrer&quot;&gt;Modelling the recent common ancestry of all living humans&lt;/a&gt;&amp;quot;, &lt;em&gt;Nature&lt;/em&gt; vol. 431, 30 Sept 2004. doi:10.1038/nature02842&lt;/p&gt;&#xA;&lt;p&gt;NB: The puzzle's initial condition &amp;quot;&lt;em&gt;assuming these people are all &lt;strong&gt;completely&lt;/strong&gt; unrelated&lt;/em&gt;&amp;quot; is not physically possible, and so I've discounted it - as discussed above, &lt;em&gt;all&lt;/em&gt; individuals of our species are related to one another.&lt;/p&gt;&#xA;&lt;p&gt;It would be possible to restate this puzzle by setting a minimum level of 'relatedness' that's allowed, and saying how closely related each member of the initial population is to each other member (because that makes a difference to the relatedness of the future generations). See &lt;a href=&quot;http://en.wikipedia.org/wiki/Coefficient_of_relationship&quot; rel=&quot;noreferrer&quot;&gt;Coefficient of relationship&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Genetic_distance#Measures_of_genetic_distance&quot; rel=&quot;noreferrer&quot;&gt;Genetic distance # Measures of genetic distance&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-03-14T16:25:09.400" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="10251" PostTypeId="2" ParentId="10165" CreationDate="2015-03-14T11:53:39.373" Score="1" Body="&lt;p&gt;The &lt;strong&gt;front line&lt;/strong&gt; is always moving, leaving a trail of destruction, taking many shapes, being disputed over and is not a happy place; most ominous and frightful it is, however we must hold the line, lest all be lost.&lt;/p&gt;&#xA;" OwnerUserId="4522" LastActivityDate="2015-03-14T11:53:39.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10252" PostTypeId="1" AcceptedAnswerId="23992" CreationDate="2015-03-14T12:17:30.790" Score="10" ViewCount="1792" Body="&lt;p&gt;&lt;em&gt;This is a spin-off motivated by Lopsy's &lt;a href=&quot;https://puzzling.stackexchange.com/questions/9097/the-lion-and-the-zebras-ii-the-lion-with-millimeter-long-claws&quot;&gt;interesting variant&lt;/a&gt; of Gamow's &lt;a href=&quot;https://puzzling.stackexchange.com/questions/9036/the-lion-and-the-zebras&quot;&gt;lion and zebras&lt;/a&gt; puzzle. It arose from a line of enquiry that tried to extend the vertical run past 5000km (or characterise the distribution of zebras to limit the run to within 5000km). That line of inquiry was eventually discarded but the spin-off puzzle seemed interesting in its own right.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Imagine a one-dimensional universe inhabited by 101 spots. Every so often, they play their own &lt;em&gt;friendly&lt;/em&gt; variant of the 'lion and zebras' game of precision tag.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One spot is selected as the lion and the other spots (zebras) scatter to wherever they wish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Goal&lt;/strong&gt;&#xA;At the start of the game, the lion nominates a position for each zebra. The lion wins if the zebras occupy those positions simultaneously. The zebras win if they can force the game to continue indefinitely without the lion winning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Target Zebra&lt;/strong&gt;&#xA;One zebra is selected as the target to start the game. At any time, the lion may change the target to any zebra that is between 100m and 200m away, inclusive. For example, it may target a zebra 100m or 150m away, but not one 50m, 99m, 201m or 1km away. There is exactly one target at any given time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Moves&lt;/strong&gt;&#xA;The spots take turns to move up to 100m per turn.&#xA;On the odd turns, only the lion may move.&#xA;On the even turns, only the zebra that was the target at the start of the turn may move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Zebra Move Constraint 1: Non-crossing&lt;/strong&gt;&#xA;The target must use its turn to ensure its own position never crosses to the other side of the lion during that turn and during the lion's next turn. For example, if the lion is 50m away and the target starts moving towards it and moves for 80m, their positions cross during that move. Other than this restriction, spots may pass through each other with no consequence. Note that it is permissible for the target to end up at the same position as the lion - this constraint just prohibits &lt;em&gt;crossing&lt;/em&gt; to the other side of the lion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Zebra Move Constraint 2: Attractor&lt;/strong&gt;&#xA;The zebras are friendly. Within the positions allowed by the non-crossing constraint, the target will move to the position that minimises its distance to the last zebra it passed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the lion win? If so, how? If not, why not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it make any difference if the targeting distance was slightly modified so that the lion cannot tag zebras exactly 100m or 200m away (that is, change it from a closed interval to an open interval)?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here is an example of placing one zebra at its target point. The game starts with lion $L$ and zebras $X,Y,Z$, with initial target $X$ and initial positions $[L,X,Y,Z] = [-150,0,-100,120]$. All other zebras are far away. The nominated position for $X$ is 140.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the notation $L+x$ and $L-x$ to mean $L$ moves $x$ meters towards the positive or negative end respectively. Then we play this out as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L+100,X+100,L+100$ takes $[L,X]$ to $[50,100]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X$ must move to somewhere between 150 $(=L+100)$ and 200 $(=X+100)$ and therefore passes $Z$. The nearest position to $Z$ in that range is 150, so $X$ only moves $X+50$ to 150. The spots are now at $[L,X,Y,Z] = [50,150,-100,120]$ and $Z$ is the last zebra seen by $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L-10$ allows $X$ to get closer to $Z$, which it does with $X-10$, bringing $X$ to its target position 140. $L$ is now at 40, and it changes the target to $Y$, which is now in range. This freezes the position of $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this example, $Y$ and $Z$ are at convenient positions. In the general case, each zebra starts at a random position.&lt;/p&gt;&#xA;" OwnerUserId="9422" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-11-17T00:05:11.113" Title="Precision Tag - can the lion win?" Tags="|geometry|strategy|game|pursuit-evasion|" AnswerCount="3" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="10253" PostTypeId="2" ParentId="2" CreationDate="2015-03-14T13:11:31.153" Score="-3" Body="&lt;p&gt;It depends on the difficulty level. If the number of clues is more than 20, most of the time it has only one solution (but no mathematical support is there). However, if the number of clues is less than 20 (for example: 12, 15, 17), multiple solutions may exist. &lt;/p&gt;&#xA;" OwnerUserId="10205" LastEditorUserId="11475" LastEditDate="2015-08-21T19:33:56.413" LastActivityDate="2015-08-21T19:33:56.413" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10255" PostTypeId="5" CreationDate="2015-03-14T14:05:58.517" Score="0" Body="&lt;p&gt;This tag is for puzzles which involve the manipulation of individual letters, or short strings of letters which aren't coherent morphemes, within words. For example, puzzles involving the formation of new words from old by adding single letters should have this tag.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-14T14:05:58.517" LastActivityDate="2015-03-14T14:05:58.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10256" PostTypeId="4" CreationDate="2015-03-14T14:05:58.517" Score="0" Body="A puzzle which involves the manipulation of individual letters within words, or letters by themselves." OwnerUserId="5373" LastEditorUserId="69582" LastEditDate="2021-03-07T15:36:05.353" LastActivityDate="2021-03-07T15:36:05.353" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10257" PostTypeId="2" ParentId="10231" CreationDate="2015-03-14T14:37:19.947" Score="5" Body="&lt;p&gt;I start out quite skilled,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A condom is systematically rolled in the beginning, that is when it in the pack&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Throughout the nation I've drilled,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It is used everywhere&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm sexist to the extreme,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is to be worn by a male and not by a female&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'll soon know what I mean,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your children should thank me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not using condom means more children and more children means more division of resources like food, toys, bed etc and also delay between two pregnancy is good for child's health&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your wife might just hate me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It's a turn off for women because of the artificial feeling and no natural contact &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Knives I despise&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A cut in a condom makes it unusable&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I hide from your eyes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A condom is a very thin rubber almost transparent&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Avoid mutations,&#xA;From your medications.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;These 2 lines may mean not using condom leads to other forms of contraception and one of them is medicine which may be causing mutation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So we will do all your tests,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No idea about this line (sorry)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But eventually we'll both rest.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;After the intercourse it is a rest time for both the male and the condom&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9174" LastActivityDate="2015-03-14T14:37:19.947" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10258" PostTypeId="2" ParentId="10231" CreationDate="2015-03-14T14:58:02.443" Score="1" Body="&lt;p&gt;I don't think this is exactly it, although it might be in the right area. Bit of a depressing answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Breast cancer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I start out quite skilled // Throughout the nation I've drilled,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I don't know about the first line, but the second one is self-explanatory (if we just take &quot;drilled&quot; to be a poetic way of saying &quot;have had an impact&quot;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm sexist to the extreme, // You'll soon know what I mean,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First line is obvious. Second line is just there to rhyme with &quot;extreme&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your children should thank me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is the main reason I doubt this is the correct answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your wife might just hate me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, obvious.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Knives I despise&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Breast cancer can be treated with surgery.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I hide from your eyes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Breast cancer can go undetected without regular medical exams.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Avoid mutations, // From your medications.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Possibly a reference to resistence to chemotherapy or some other form of treatment.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So we will do all your tests, // But eventually we'll both rest.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Perhaps a very morbid reference to the difficulty in curing cancer, where &quot;we&quot; refers to the illness and the patient.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5511" LastActivityDate="2015-03-14T14:58:02.443" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10259" PostTypeId="2" ParentId="10238" CreationDate="2015-03-14T15:31:32.547" Score="1" Body="&lt;p&gt;As &lt;a href=&quot;https://puzzling.stackexchange.com/users/1508/germcd&quot;&gt;germcd&lt;/a&gt; said in &lt;a href=&quot;https://puzzling.stackexchange.com/a/10243/9352&quot;&gt;his answer&lt;/a&gt;, the answer is $m+n-1$. But he also says that it's $1$ if $n=1$ or $m=1$. But the only exception is when $n=0$ or $m=0$, then the answer would be $0$(obviously).&lt;br&gt;&#xA;I'll show a few examples(I'll replace $n$ with $f$ for females to make it easier):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example 1: 1m+1f&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They can make a baby, but then it stops(g1), so $1+1-1 = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example 2: 2m+1f&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of those 2 men can make a girl with that female(g1). Then they can make another baby with the girl of the other guy(g2). But now they are all related*, so we can't go any further. $2+1-1=2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example 3: 1m+2f&lt;/strong&gt; &lt;em&gt;(lucky man)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He makes 1 boy with each of those females(1g). Then each of those boys make another kid with another starting female(2g). At that moment they are all related, so it stops there. $1+2-1=2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example 4: 2m+2f&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 pair makes a girl(1g). Then she makes a boy with the other starting male(2g). Then this boy makes another child with starting female(3g). They can repeat that process using other pairs, but then they will all be related. $2+2-1=3$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;*And with &lt;em&gt;all related&lt;/em&gt; I mean that the kid has all starting people as his/her ancestors.&lt;/p&gt;&#xA;" OwnerUserId="9352" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-03-14T15:31:32.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10260" PostTypeId="1" AcceptedAnswerId="10262" CreationDate="2015-03-14T15:56:56.670" Score="3" ViewCount="454" Body="&lt;p&gt;On the table there are twenty weights with masses $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$, $10$, $11$, $12$, $13$, $14$, $15$, $16$, $17$, $18$, $19$ and $90$ grams, and a two-armed balance with two pans. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Nine of the weights are made of bronze, nine are made of silver, one is made of gold, and one is made of iron.&lt;/li&gt;&#xA;&lt;li&gt;The iron weight is $90$ grams.&lt;/li&gt;&#xA;&lt;li&gt;If you put all bronze weights and the iron weight into one pan, and all silver weights into the other pan, then the balance is in perfect equilibrium.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Question: How many grams is the gold weight?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="5373" LastEditDate="2015-03-14T16:40:23.320" LastActivityDate="2015-06-05T01:35:22.887" Title="Twenty weights and a balance" Tags="|mathematics|weighing|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10261" PostTypeId="2" ParentId="9574" CreationDate="2015-03-14T16:12:19.647" Score="7" Body="&lt;p&gt;To show that &lt;strong&gt;xnor&lt;/strong&gt;'s answer with 16 knights is indeed best possible, consider a standard black-and-white chessboard coloring so that the four corner squares, the central square and both diagonals are black.&#xA;Then there are $13$ black and $12$ white squares.&#xA;Consider $b$ knights on black squares and $w$ knights on white squares, so that each knight attacks exactly two other knights.&#xA;Suppose for the sake of contradiction that $b+w\ge17$ holds.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Since the $b$ black knights can only be attacked by knights on white squares, and since each knight is attacked by the knights which attack it, by symmetry, we get $b=w$.&lt;/li&gt;&#xA;&lt;li&gt;If there is a knight on the (black) central square, consider the eight white squares that are attacked by that knight. Exactly two of these white squares must contain knights, while the other six must be empty. This implies $w\le12-6=6$.   Since $b=w$, this means $b+w\le12$, which contradicts the assumption $b+w\ge17$. Hence we will from now on assume that the central square is empty.&lt;/li&gt;&#xA;&lt;li&gt;Next assume that there is a knight on one of the four white squares that share an edge with the (black) central square. This knight attacks six black squares, at least four of which must be empty. Together with the empty central square this implies $b\le13-1-4=8$.   Since $b=w$, this means $b+w\le16$, which again contradicts the assumption $b+w\ge17$. &#xA;Hence we will from now on assume that these four white squares are empty.&lt;/li&gt;&#xA;&lt;li&gt;In all remaining cases, we are left with $w\le12-4=8$ knights on white squares, and by similar logic, it would be $b+w\le16$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Therefore $b+w\ge17$ is false.  Since we have an example of $b+w=16$, this is an optimal solution.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="1537" LastEditDate="2015-03-14T18:25:45.673" LastActivityDate="2015-03-14T18:25:45.673" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10262" PostTypeId="2" ParentId="10260" CreationDate="2015-03-14T16:16:55.437" Score="7" Body="&lt;p&gt;gold weight is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 10 grams&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;if you remove 90 g from the last series (11 ,12 ... 19) it matches first series (1,2,....9)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9517" LastEditorUserId="8874" LastEditDate="2015-03-18T09:38:45.563" LastActivityDate="2015-03-18T09:38:45.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10263" PostTypeId="1" AcceptedAnswerId="10296" CreationDate="2015-03-14T16:56:31.810" Score="3" ViewCount="256" Body="&lt;p&gt;In Paris, a famous Nobel prize winner was found murdered in his kitchen.&#xA;The door of the refrigerator stood wide open, and a small piece of butter&#xA;had fallen on the floor.  On the kitchen table, there were a French-English&#xA;dictionary, a mirror, a compass and a telescope.  The calendar on the wall&#xA;showed the page for April 2014, and next to the calendar there was a picture&#xA;of Gaspard Monge.  With his very own blood the dying scientist had written&#xA;the following numbers onto the kitchen floor: &lt;strong&gt;28-27-57-16-26-3-58&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The police narrowed the crime down to the following list of suspects:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Antoine, Bastien, Camille, Dorian,&#xA;   Etienne, Felice,  Gustave, Justine,&#xA;   Ludovic, Mathieu, Nicolas, Ophelie,&#xA;   Pauline, Quentin, Raphael, Suzanne,&#xA;   Thibaut, Vincent, and Xavier.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who murdered the Nobel prize winner?&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="5373" LastEditDate="2015-03-14T17:17:19.113" LastActivityDate="2015-03-15T15:16:19.097" Title="A picture of Gaspard Monge" Tags="|cipher|story|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10264" PostTypeId="1" AcceptedAnswerId="10266" CreationDate="2015-03-14T17:00:20.537" Score="8" ViewCount="306" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Sir, I have a great advantage in puzzle codes.&lt;br&gt;&#xA;  The green searcher submitted invalid endeavors.&lt;br&gt;&#xA;  Was it the solution?&lt;br&gt;&#xA;  Time eluded so easily many.&lt;br&gt;&#xA;  The day fanatics get to grandly celebrate: March ?????&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="5188" LastEditDate="2015-03-15T18:57:37.133" LastActivityDate="2016-03-12T14:45:49.477" Title="The green searcher" Tags="|enigmatic-puzzle|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10265" PostTypeId="1" AcceptedAnswerId="10268" CreationDate="2015-03-14T17:03:55.823" Score="5" ViewCount="180" Body="&lt;p&gt;Below, I've used this alphabet:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SACQy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to encode the first eight lines of a song. Find the title and artist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4c4a’3h4a 3b1d1f3g2a 2e1d 4c3a1b2e 3a2b2b 3g1b3d3e2e&lt;br&gt;&#xA;1e3e4a’1e 3b1d1f3g2a 2e1d 3h1f3g 3a4b1d3f&lt;br&gt;&#xA;1b3g2f4a1e2e4a2a 4a3g1d1f3d3e 1b3g 1b2e 3a3g3i3e1d4c,&lt;br&gt;&#xA;2e1d 4a3a1a3e 3e1b1e 1d4c3g…&lt;br&gt;&#xA;2e3e4a 3d3a3h2a4a3g 3g4a4a2a1e 1e1d3h2e1b3g3d 1d1f2e&lt;br&gt;&#xA;1e3e4a 1a1f3h2b1e 3e4a3h 2b1b2c1e 1d3g 2e3e4a 3b1d4c&lt;br&gt;&#xA;3a3g2a 1b 2a1d3g’2e 3f3g1d4c 1b3c 1b’4b 2a4a3a2a 1d3h 3g1d2e&lt;br&gt;&#xA;2e1d 3a3g3i1d3g4a…&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint #1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It was on the radio, so it's fairly mainstream.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It was released in 2010.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The apostrophes, commas, and ellipses are not encoded.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastActivityDate="2015-03-14T18:28:30.573" Title="Find the title and artist of this song" Tags="|cipher|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10266" PostTypeId="2" ParentId="10264" CreationDate="2015-03-14T17:07:33.333" Score="13" Body="&lt;p&gt;The answer will&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; either be a word with 0 letters (the empty word)&lt;br&gt;&#xA; or perhaps a word with 10 letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;314159265&lt;/strong&gt;: Sir, I have a great advantage in puzzle codes.&lt;br&gt;&#xA;  &lt;strong&gt;358979&lt;/strong&gt;:  The green searcher submitted invalid endeavors.&lt;br&gt;&#xA;  &lt;strong&gt;3238&lt;/strong&gt;: Was it the solution?&lt;br&gt;&#xA;  &lt;strong&gt;46264&lt;/strong&gt;: Time eluded so easily many.&lt;br&gt;&#xA;  &lt;strong&gt;338327950&lt;/strong&gt;: The day fanatics get to grandly celebrate: March ?????  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These numbers&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; count the number of letters in the words of the &quot;poem&quot;,&lt;br&gt;&#xA; and they coincide with the first few digits of &#xA; $\pi\approx3.14159265358979323846264338327950$&lt;br&gt;&#xA;&lt;br&gt;&#xA; Since today is March 14 (the famous &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Pi_Day&quot; rel=&quot;noreferrer&quot;&gt;Pi Day&lt;/a&gt;&quot;),&lt;br&gt;&#xA; the intended answer should be the ten-letter word &lt;strong&gt;fourteenth&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Final comment:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I have just read an article about Pi-Day, that mentioned the following classic pi-memorization sentence (which might be the root of the riddle):&lt;br&gt;&#xA;&lt;br&gt;&#xA; Sir, I bear a rhyme excelling&lt;br&gt;&#xA; In mystic force and magic spelling,&lt;br&gt;&#xA; Celestial sprites elucidate&lt;br&gt;&#xA; All my own striving can't relate.&lt;br&gt;&#xA; Or locate they who can cogitate&lt;br&gt;&#xA; And so finally terminate. Finis.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-12T14:45:49.477" LastActivityDate="2016-03-12T14:45:49.477" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10267" PostTypeId="2" ParentId="10260" CreationDate="2015-03-14T17:45:15.270" Score="3" Body="&lt;p&gt;Agawa001 has the right answer, but here's a fuller explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $B_1,\dots,B_9,S_1,\dots,S_9,G$ be the weights of the bronze, silver, and gold weights respectively in grams (these numbers are 1 to 19 in some order).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know the sum of all $B_i$ is at least $1+2+3+\dots+9=45$ (the sum of the nine lightest weights) while the sum of all $S_i$ is at most $11+12+13+\dots+19=135$ (the sum of the nine heaviest weights). So $\Sigma_i S_i\leq135\leq\Sigma_i B_i+90$. Since the left and right quantities in these inequalities are equal, we must have equality all the way through, which means the $B_i$ are $1,\dots,9$ in some order and the $S_i$ are $11,\dots,19$ in some order. The only possiblity left for $G$ is &lt;strong&gt;10&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-03-14T17:45:15.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10268" PostTypeId="2" ParentId="10265" CreationDate="2015-03-14T18:04:00.743" Score="5" Body="&lt;p&gt;The title and artist are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;a href=&quot;http://www.azlyrics.com/lyrics/brokenbells/thehighroad.html&quot; rel=&quot;nofollow&quot;&gt;The High Road&lt;/a&gt;&lt;/strong&gt; by the &quot;Broken Bells&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The encryption is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each pair of characters represents a letter of the alphabet.&lt;br&gt;&#xA; The cipher characters below are followed by the plain text characters in the same order.&lt;br&gt;&#xA;&lt;br&gt;&#xA; 1a 1b 1d 1e 1f 2a 2b 2c 2e 2f 3a 3b 3c 3d 3e 3f 3g 3h 3i 4a 4b 4c&lt;br&gt;&#xA;&lt;br&gt;&#xA; C I O S U D L P T V A B F G H K N R Y E M W&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-03-14T18:28:30.573" LastActivityDate="2015-03-14T18:28:30.573" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10269" PostTypeId="1" CreationDate="2015-03-14T19:10:37.183" Score="9" ViewCount="814" Body="&lt;p&gt;&lt;strong&gt;Solution for Version 2 pending....&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a $100\times 100$ grid. The upper left corner has coordinates $(1,1)$ and bottom right corner has $(100,100)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 'snake' starts by occupying a single cell at $(1,1)$. On each move, it travels in any of 8 directions (vertically, horizontally or diagonally) to reach the next cell. Every cell once visited remains occupied by it; it grows by 1 unit with every move it makes. It cannot visit a cell twice, whether by retracing or by intersection. However, it may pass through a diagonal line it made earlier, by visiting only the unoccupied cells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to set targets for it to reach. Each target is a single point, and the snake completes the goal by visiting the cell. Once a target is reached, you must immediately set the next one, anywhere else. You cannot set an already visited cell as a target.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Version 1&lt;/strong&gt;&#xA;The snake must reach a target in minimum possible moves, irrespective of deeper strategy. What is the minimum number of targets you must set to create an impossible-to-reach target?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Version 2&lt;/strong&gt;&#xA;The snake may take up to $100$ moves more than necessary to reach each target. Is it possible to block the snake (which is using perfect strategy)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;100 moves to waste&lt;/em&gt; means if reaching a certain target at the time it is set requires 29 moves, it can take up to 129 moves to do so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The snake can not know any of your future goals, since you set one only after the previous one is complete. However, as it has 100 moves to waste for each target, it might deliberately waste moves to block your possible future targets. &lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="65277" LastEditDate="2023-02-01T17:06:05.087" LastActivityDate="2023-02-03T10:36:06.527" Title="Block the snake from reaching points" Tags="|mathematics|geometry|strategy|optimization|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10270" PostTypeId="1" AcceptedAnswerId="10283" CreationDate="2015-03-14T19:23:37.790" Score="13" ViewCount="1301" Body="&lt;p&gt;&lt;strong&gt;Klotski&lt;/strong&gt; is a sliding-blocks puzzle game very similar in nature to rush-hour/unblock me. In a given puzzle, a certain number of blocks labeled Z, A, B, C, .... are given. The goal is to move the Z block into the goal area in the fewest number of possible moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each move consists of choosing a certain letter, and then moving the block corresponding to that letter in one of the four main directions (up, down, left, right) as many times as you like. At every point in a move, the block must be moving into adjacent empty spaces only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some Klotski puzzles, there are several parts to some blocks which are not connected. Imagine that they are joined by an invisible rigid bar; that is, if one A moves up, all A's must move up as well. For a move to be valid, all such parts of a block must be able to simultaneously make the move. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, consider the following puzzle:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/siKca.png&quot; alt=&quot;Example&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Z pieces will always be marked in red, while the goal area will always be marked in gold. One solution to this puzzle would be BrZruBlAdCdZur(using d=down, u=up, l=left, r=right):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Br (moving both B blocks to the right once)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zru (moving the Z block right once and up once)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bl (moving the B block left once)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ad (moving the A block down once)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cd (moving the C block down once)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zur (moving the Z block up once, and right once into the goal)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The puzzles I post here definitely have solutions with the given # of moves, but that won't necessarily be optimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, here's a puzzle! Note that the invisible-bar analogy still holds over multiple grids.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/CvKY2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10213" LastEditorDisplayName="user20" LastEditDate="2015-03-15T22:39:08.670" LastActivityDate="2015-03-15T22:39:08.670" Title="Klotski Puzzle 1" Tags="|geometry|mechanical-puzzles|sliding-blocks|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10271" PostTypeId="1" AcceptedAnswerId="10276" CreationDate="2015-03-14T19:39:56.813" Score="8" ViewCount="960" Body="&lt;p&gt;Each answer is the title of a well-known movie...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DDtU3.png&quot; alt=&quot;Movies crossword grid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ACROSS &lt;br&gt;&#xA;4. Teardrop flows for a broken relationship (8) &lt;br&gt;&#xA;6. Spin the hated frog, until brown (3,9) &lt;br&gt;&#xA;9. Immediate disaster as Neb crewman’s sly weapon explodes (first part, before 8 down) (10) &lt;br&gt;&#xA;10. Unbalanced, like a legendary bird (5) &lt;br&gt;&#xA;13. Happy interlingual anonymous fighter (9) &lt;br&gt;&#xA;15. Disaster movie provokes small laughs (5) &lt;br&gt;&#xA;16. Prideful monarch is audibly recumbent (3,4,4) &lt;br&gt;&#xA;17. Selfish idea (now an internet phenomenon) and digit as a keepsake (7) &lt;br&gt;&#xA;19. Patents, copyrights and trademarks lost in shipwreck! (5) &lt;br&gt;&#xA;20. Taken, awaits rye dip mixture (second part, after 3 down) (4) &lt;br&gt;&#xA;22. Mechanised fruit is on time (second part, after 12 down) (6) &lt;br&gt;&#xA;23. Subway city-state, or milepost repositioned (10) &lt;br&gt;&#xA; &lt;br&gt;&#xA;DOWN &lt;br&gt;&#xA;1. A frog is confused by long-distance travel (5) &lt;br&gt;&#xA;2. Flight to the Bahamas (2) &lt;br&gt;&#xA;3. Taken, awaits rye dip mixture (first part, before 20 across) (8) &lt;br&gt;&#xA;5. Nitrate mother turns from the light to the dark side (3,10) &lt;br&gt;&#xA;7. Rick’s White House (10) &lt;br&gt;&#xA;8. Immediate disaster as Neb crewman’s sly weapon explodes (second part, after 9 across) (3) &lt;br&gt;&#xA;11. Marilyn’s Sugar is naturalised newspaper mogul (7,4) &lt;br&gt;&#xA;12. Mechanised fruit is on time (first part, before 22 across) (1,9) &lt;br&gt;&#xA;14. Shake saline strangely (6) &lt;br&gt;&#xA;18. Threat mix disrupted Blair’s chambers (3,6) &lt;br&gt;&#xA;21. Moot-like heist turns farcical (first part, before 25 down) (4,4) &lt;br&gt;&#xA;24. Nutty carbohydrate, gangnam-style (6) &lt;br&gt;&#xA;25. Moot-like heist turns farcical (second part, after 21 down) (2,3) &lt;br&gt;&lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="8874" LastEditDate="2015-10-08T08:47:19.557" LastActivityDate="2015-10-08T08:47:19.557" Title="Movies, movies, movies!" Tags="|crosswords|movies|cryptic-crosswords|" AnswerCount="4" CommentCount="10" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10272" PostTypeId="2" ParentId="10271" CreationDate="2015-03-14T19:46:40.850" Score="5" Body="&lt;p&gt;&lt;strong&gt;12 down and 22 across&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Mechanised fruit is on time (1,9;6)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.imdb.com/title/tt0066921/&quot; rel=&quot;nofollow&quot;&gt;A Clockwork Orange&lt;/a&gt;. An easy one to get me started!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1 down&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A frog is confused by long-distance travel (5)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.imdb.com/title/tt0116282/&quot; rel=&quot;nofollow&quot;&gt;Fargo&lt;/a&gt; - an anagram of &quot;a frog&quot; and related to long-distance travel. (Thanks to &lt;strong&gt;crazyiman&lt;/strong&gt; for helping me with this one.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;7 down&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Rick’s White House (10)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.imdb.com/title/tt0034583/&quot; rel=&quot;nofollow&quot;&gt;Casablanca&lt;/a&gt;. The main character is named Rick, and Casablanca literally means &quot;white house&quot;. (I got this by assuming Casablanca must be somewhere in the puzzle and just finding a 10-letter clue!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;17 across&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Selfish idea (now an internet phenomenon) and digit as a keepsake (7)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.imdb.com/title/tt0209144/&quot; rel=&quot;nofollow&quot;&gt;Memento&lt;/a&gt;. An internet phenomenon is a meme (hello AE, your favourite!), originating from an idea of selfishness in genetics. A to(e) is a digit, and the whole thing is a keepsake.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2 down&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Flight to the Bahamas (2)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.imdb.com/title/tt1049413/&quot; rel=&quot;nofollow&quot;&gt;Up&lt;/a&gt;. The airline code for Bahamas Air is UP.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;14 down&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Shake saline strangely (6)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.imdb.com/title/tt0090605/&quot; rel=&quot;nofollow&quot;&gt;Aliens&lt;/a&gt;. It's an anagram of saline, and aliens are strange!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;24 down&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Nutty carbohydrate, gangnam-style (6)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Partial solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gangnam Style is a pop single by the South Korean musician Psy, so I'd guess this ends with the letters PSY.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;6 across&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Spin the hated frog, until brown (3,9)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.imdb.com/title/tt0068646/?ref_=chttp_tt_2&quot; rel=&quot;nofollow&quot;&gt;The Godfather&lt;/a&gt;. This is an anagram of &quot;the hated frog&quot;, and &lt;a href=&quot;http://en.wikipedia.org/wiki/James_Brown&quot; rel=&quot;nofollow&quot;&gt;James Brown&lt;/a&gt; is known as the &quot;godfather of soul&quot;. Again I got this just by guessing this film must be in here and searching for a (3,9) clue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;18 down&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Threat mix disrupted Blair’s chambers (3,6)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.imdb.com/title/tt0133093/?ref_=chttp_tt_18&quot; rel=&quot;nofollow&quot;&gt;The Matrix&lt;/a&gt;. This is an anagram of &quot;threat mix&quot;, and Cherie Blair used to be in Matrix Chambers (thanks again to &lt;strong&gt;crazyiman&lt;/strong&gt; for helping with this clue). Again I got this just by looking for a (3,6) clue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-14T21:36:04.870" LastActivityDate="2015-03-14T21:36:04.870" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10273" PostTypeId="1" AcceptedAnswerId="10295" CreationDate="2015-03-14T19:59:02.450" Score="2" ViewCount="707" Body="&lt;p&gt;The manager of the local sweetshop, one Minty Ball, has announced a grand closing down sale. They sell $n$ varieties of sweets: toffees, sherbets, fudge, and so on. Let's just call the varieties $V_1,\dots,V_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the final day of the sale, Minty Ball announces the ultimate prize: the lease on the premises will pass to whoever manages to eat &lt;strong&gt;the last sweet left in the shop&lt;/strong&gt;. Half the population of the town scrambles down there to buy sweets. One local magnate tries to buy all the sweets in the shop, but Minty Ball tells him that certain rules are in place, as follows.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Each purchase must consist only of sweets of one variety. You can buy as many toffees as you want in one go, from just one to all of them, but you can't buy both a toffee and a sherbet at the same time.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Every time you make a purchase, you have to &lt;em&gt;eat&lt;/em&gt; all the sweets you've bought. You can't just stuff them in your pocket and make another purchase.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;After hearing about these rules, the crowd thins. Few people want to make themselves sick by gorging on sweets. But the prices are good, so many people buy single packets of sweets and leave. One or two keep on buying, eating, and buying again in the hope of staying all the way to the last sweet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally there are only two shoppers left: Rand and Gamow. Both are hungry for the ultimate goal of eating the final sweet. At this point there are $S_i$ sweets of type $V_i$ remaining for each $i$. The two hopefuls, having eliminated all competition, agree that they will take turns to make a purchase. Gamow is the later arrival of the two; he has eaten fewer sweets over the course of the day but has been gorging himself in the last couple of hours, so he allows Rand to make the first purchase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For what values of $n,S_1,\dots,S_n$ can Rand be the one to take the final sweet?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that both Rand and Gamow have sufficient money to make all the purchases they need, and that neither is worried about making himself sick by eating too many sweets. All they care about is to take over the sweetshop premises and start their own business there.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-05-25T18:17:07.843" LastActivityDate="2015-05-25T18:17:07.843" Title="Eat sweets and start your own business!" Tags="|strategy|game|nim|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10274" PostTypeId="2" ParentId="10260" CreationDate="2015-03-14T20:18:45.017" Score="1" Body="&lt;p&gt;Sum up the iron weight (of 90 grams) with the nine smallest weights, i.e., 1 through 9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will give you 135 grams. So one side of the balance will have a weight of 135 grams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we need 135 grams in the second pan to balance both sides. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, sum 10 through 18. This will give you 126 grams. Then we still need 9 grams to balance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's remove the 10 from this pan, and add the 19-gram weight instead of it. So, 126-10+19=135.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both sides are balanced, and the 10-gram weight is left over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the gold weight is 10 grams. &lt;/p&gt;&#xA;" OwnerUserId="10119" LastEditorUserId="4204" LastEditDate="2015-06-05T01:35:22.887" LastActivityDate="2015-06-05T01:35:22.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10275" PostTypeId="2" ParentId="10271" CreationDate="2015-03-14T20:20:46.633" Score="3" Body="&lt;p&gt;15 across&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.imdb.com/title/tt0299930/&quot; rel=&quot;nofollow&quot;&gt;Gigli&lt;/a&gt; - A &quot;giggle&quot; is a small laugh and that movie &lt;em&gt;was&lt;/em&gt; an utter disaster.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;11 down&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.imdb.com/title/tt0033467/&quot; rel=&quot;nofollow&quot;&gt;Citizen Kane&lt;/a&gt; - Made the connection from &quot;sugar cane&quot; and the fact that he's a newspaper mogul (and of course, that movie has to be in there anyway). Don't know how Marilyn figures into this, though.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;16 across, a bit obscure, but fitting to the other letters&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.imdb.com/title/tt0110357/&quot; rel=&quot;nofollow&quot;&gt;The Lion King&lt;/a&gt; - A lion king is obviously a &quot;prideful monarch&quot; and he's &quot;audibly recumbent&quot; because &lt;em&gt;&quot;lion&quot;&lt;/em&gt; sounds like &lt;em&gt;&quot;lying&quot;&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;23 across&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.imdb.com/title/tt0017136/&quot; rel=&quot;nofollow&quot;&gt;Metropolis&lt;/a&gt; - A &quot;metro&quot; is a subway, a &quot;polis&quot; a greek city state.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;24 down&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.imdb.com/title/tt0054215/&quot; rel=&quot;nofollow&quot;&gt;Psycho&lt;/a&gt; - PSY is the Korean rapper who sang &quot;Gangnam Style&quot; (as &lt;em&gt;rand al'thor&lt;/em&gt; already deduced) and a carbohydrate contains carbon (C), hydrogen (H) and oxygen (O).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5 down&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.imdb.com/title/tt0088247/&quot; rel=&quot;nofollow&quot;&gt;The Terminator&lt;/a&gt; - Anagram of &quot;nitrate mother&quot;, not sure about light and dark side, though. Also the word for the separation line between light and dark on illuminated planets.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;10 across&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.imdb.com/title/tt0075148/&quot; rel=&quot;nofollow&quot;&gt;Rocky&lt;/a&gt; - A &quot;rocky&quot; surface is uneven/unbalanced. Not sure about the bird, though, maybe a reference to &lt;em&gt;Chicken Run&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10215" LastEditorUserId="10215" LastEditDate="2015-03-14T21:44:28.283" LastActivityDate="2015-03-14T21:44:28.283" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10276" PostTypeId="2" ParentId="10271" CreationDate="2015-03-14T20:56:06.090" Score="4" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Fargo (Anagram of &quot;A Frog&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Up (Two-letter word, &quot;Flight&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Spirited Away (Anagram of &quot;Awaits rye dip&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Predator (Anagram of &quot;Teardrop&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Terminator (Anagram of &quot;Nitrate mother&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Godfather (Anagram of &quot;The hated frog&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Casablanca (Means &quot;white house&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;(See 9)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Apocalypse Now (&quot;Disaster&quot;, &quot;Immediate&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Rocky (&quot;Unbalanced&quot;, Roc-like)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Citizen Kane (&quot;Newspaper mogul&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A Clockwork Orange&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;???&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Aliens (Anagram of &quot;Saline&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Gigli&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Lion King (&quot;Pride&quot; -&gt; Lion, &quot;Monarch&quot; -&gt; King)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Memento (&quot;Keepsake&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;???&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Shrek (Subsequence of &quot;shipwreck&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;(See 3)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;???&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;(See 12)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Metropolis (Anagram of &quot;Or milepost&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Psycho (Psy + CHO)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;??? (See 21)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks to &lt;a href=&quot;https://puzzling.stackexchange.com/users/5373/rand-althor&quot;&gt;rand al'thor&lt;/a&gt; for (7) and &lt;a href=&quot;https://puzzling.stackexchange.com/users/10215/christian-rau&quot;&gt;Christian Rau&lt;/a&gt; for (15)&lt;/p&gt;&#xA;" OwnerUserId="9471" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-03-14T21:38:36.603" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10277" PostTypeId="2" ParentId="10271" CreationDate="2015-03-14T21:06:29.970" Score="3" Body="&lt;p&gt;Great quiz. :) The missing pieces:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;13&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gladiator. Happy=Glad, not sure about the rest&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;21&lt;/strong&gt; and &lt;strong&gt;25&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some Like It Hot, from the letters of 'Moot-like heist'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10216" LastEditorUserId="5373" LastEditDate="2015-03-14T21:40:14.373" LastActivityDate="2015-03-14T21:40:14.373" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10279" PostTypeId="2" ParentId="10269" CreationDate="2015-03-14T22:34:45.690" Score="0" Body="&lt;h2&gt;Version 1:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The snake is killed in 4 moves by the points (1,2), (100,2), (100,3) and (100,1). This works regardless if the snake is capable of pathfinding or not, as the cell is not reachable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the snake is capable of choosing an arbitrary shortest path specifically to thwart strategies (such as by intentionally alternating its path), then 6 steps are required (1,2), (2,2), (3,2), (3,1), (4,1), (2,1), since in this case the snake has only 1 choice, and is fully constrained by the options. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Version 2:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Invalidated by edit clarification&lt;/p&gt;&#xA;" OwnerUserId="7529" LastEditorUserId="7529" LastEditDate="2015-03-15T13:22:51.577" LastActivityDate="2015-03-15T13:22:51.577" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10280" PostTypeId="2" ParentId="1965" CreationDate="2015-03-14T22:40:16.423" Score="5" Body="&lt;p&gt;Get a yard stick with a known weight.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Attach one end to a fulcrum point, and rest the other end on the scale.  Place the stacks of coins at equal distances along the yard stick.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Next&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; calculate what the scale should read if all ten stacks contained ten gold coins.  Then take a reading from the scale.  This will come up slightly less then what was calculated.  Based on that difference, calculate the position along the yard stick where the lighter stack must be.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Note: I read this wrong the first time. I assumed that the silver stack contained one silver coin among 9 gold.  The above solution will work in either case.&lt;/p&gt;&#xA;" OwnerUserId="10218" LastEditorUserId="63414" LastEditDate="2020-07-10T14:37:50.480" LastActivityDate="2020-07-10T14:37:50.480" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10282" PostTypeId="1" AcceptedAnswerId="10285" CreationDate="2015-03-15T02:29:19.567" Score="3" ViewCount="408" Body="&lt;p&gt;In this &lt;a href=&quot;https://puzzling.stackexchange.com/questions/10265/find-the-title-and-artist-of-this-song&quot;&gt;Recent Question&lt;/a&gt;, what was the encryption mapping?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, how did each letter map to its 2-digit counterpart?&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-03-15T06:10:06.087" Title="What was the encryption mapping?" Tags="|cipher|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10283" PostTypeId="2" ParentId="10270" CreationDate="2015-03-15T04:31:13.247" Score="4" Body="&lt;p&gt;15 moves: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gr, Fr, Cu, Zu, Br, Au, Zl,Dl, Eu, Brd, Ed, Fd, Cr, Ar, Zu&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9060" LastEditorUserId="9042" LastEditDate="2015-03-15T15:11:57.497" LastActivityDate="2015-03-15T15:11:57.497" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10284" PostTypeId="1" AcceptedAnswerId="10286" CreationDate="2015-03-15T05:44:59.117" Score="-2" ViewCount="285" Body="&lt;p&gt;According to the picture: cogwheels A and D have 60 jags each, cogwheel B has 40 jags, and cogwheel C has 20 jags. If cogwheel B turns at 20 RPM, calculate the ratio of A's speed in RPM to D's.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/4bPXv.jpg&quot; alt=&quot;cogwheels&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="8829" LastEditorUserId="5373" LastEditDate="2015-03-15T12:16:18.563" LastActivityDate="2015-03-15T12:16:18.563" Title="Calculate RPM ratio A to D?" Tags="|mathematics|calculation-puzzle|mechanical-puzzles|" AnswerCount="1" CommentCount="0" ClosedDate="2015-03-15T07:04:16.590" ContentLicense="CC BY-SA 3.0" />
  <row Id="10285" PostTypeId="2" ParentId="10282" CreationDate="2015-03-15T06:10:06.087" Score="5" Body="&lt;p&gt;The encryption mapping was done by:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; dividing the alphabet into &lt;strong&gt;4 groups of letters&lt;/strong&gt; based on the &lt;strong&gt;number of strokes&lt;/strong&gt; that could be used to hand-print each letter (from 1 to 4 strokes). The letters in each group are in alphabetical order.&lt;br&gt;&#xA;&lt;br&gt;&#xA; If this is the correct answer, it should be noted that there is no standard method of hand-printing the English alphabet, although this &lt;a href=&quot;http://www.wikihow.com/Make-Letters-of-the-English-Alphabet&quot; rel=&quot;noreferrer&quot;&gt;wikiHow&lt;/a&gt; does describe the use of a similar number of strokes. There are many guidelines for using other methods.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/myX7L.png&quot; alt=&quot;CodeMap&quot;&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastActivityDate="2015-03-15T06:10:06.087" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10286" PostTypeId="2" ParentId="10284" CreationDate="2015-03-15T06:12:48.570" Score="1" Body="&lt;p&gt;The ratio A/D should be 1 since the rate of jags for both is equal.&lt;/p&gt;&#xA;" OwnerUserId="9531" LastActivityDate="2015-03-15T06:12:48.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10289" PostTypeId="1" AcceptedAnswerId="10317" CreationDate="2015-03-15T11:37:44.567" Score="18" ViewCount="2102" Body="&lt;p&gt;A (hopefully) new variant on the Knights &amp;amp; Knaves problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the island of D'Israel, the inhabitants fall into three types: Liars, Damned Liars, and Statisticians. They are visually indistinguishable except that each type of person wears a different colour of hat. If you ask them questions, their responses are as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a Liar will always give you a false response&lt;/li&gt;&#xA;&lt;li&gt;a Damned Liar will give you a false response and also stab you&lt;/li&gt;&#xA;&lt;li&gt;a Statistician will give you a true or false response with 50% probability.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You are trying to escape from a prison on this island, and you are nearly out when you reach a row of three doors, each one guarded by a solid-looking sentry. The three sentries wear red, green, and blue hats, so you know that one is a Liar, one a Damned Liar, and one a Statistician, but you don't know which is which. You also know (from the prison map which was smuggled in to you and which you have now lost) that one of the doors leads to freedom, one to the prison warden's office, and one to a first-aid room; but again you don't know which is which. Your aim is to pass through the door that leads to freedom, but if you are stabbed by the Damned Liar, you will &lt;em&gt;first&lt;/em&gt; need to visit the first-aid room. If you go to the warden's office, you will be recaptured; but going through either of the other doors allows you the chance to double back and try again if necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are being pursued, so you only have time to ask the guards one yes/no question to find out which door you should go through. &lt;strong&gt;What question should you ask? What is the probability you will escape?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does the answer change if you have time for two or three questions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt; This is a logic puzzle. Please, no lateral-thinking answers.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-15T12:12:04.200" LastActivityDate="2015-03-15T22:15:38.030" Title="Lies, damned lies, and statistics" Tags="|logical-deduction|liars|" AnswerCount="3" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="10290" PostTypeId="2" ParentId="10260" CreationDate="2015-03-15T12:41:47.593" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 10 grams&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Iron is 90g&lt;/p&gt;&#xA;&lt;p&gt;The two plates are:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;90 + bronze(1,2,3,4,5,6,7,8,9) = 135g&lt;/p&gt;&#xA;&lt;p&gt;silver(11,12,13,14,15,16,17,18,19) = 135 g&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9450" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-03-15T12:41:47.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10291" PostTypeId="2" ParentId="10269" CreationDate="2015-03-15T13:25:48.333" Score="6" Body="&lt;h2&gt;Version 1:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Since for this version the snake can still have multiple choices for getting from point A to point B, namely two diagonals take the same number of moves as two moves in a not diagonal direction. Only moves which have to be completely diagonal or contain at most one non-diagonal move (assuming no obstructions yet from the &quot;body&quot; of the snake) can only be done in one way. When the snake does has multiple choices, it can then leave holes with diagonals. A set of targets which can prevent this would be $(100,100)$, $(2,1)$, $(3,1)$ and $(1,2)$. Each of the first three targets can only be reached in one way and the forth can not be reached at all, thus four targets are required.&lt;/p&gt;&#xA;" OwnerUserId="8810" LastActivityDate="2015-03-15T13:25:48.333" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10292" PostTypeId="1" AcceptedAnswerId="10310" CreationDate="2015-03-15T13:38:11.473" Score="5" ViewCount="275" Body="&lt;p&gt;Another Klotski puzzle! (rules &lt;a href=&quot;https://puzzling.stackexchange.com/questions/10270/klotski-puzzle-1&quot;&gt;here&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RABPr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/10270/klotski-puzzle-1&quot;&gt;Klotski Puzzle 1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10213" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-03-15T22:39:16.380" Title="Klotski Puzzle 2" Tags="|geometry|sliding-blocks|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10293" PostTypeId="2" ParentId="10231" CreationDate="2015-03-15T14:59:17.690" Score="1" Body="&lt;p&gt;You are probably:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; War&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I start out quite skilled&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not quite sure about this line but I'll take a guess: Wars start of something massive and infamous like how World War 2 started with the assassination of a world leader, what some could say a &quot;skilled&quot; mission.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Throughout the nation I've drilled,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It can &quot;drill&quot; fear into the hearts of civilians and world leaders. Never mind, how about &lt;strong&gt;&lt;em&gt;EVERYONE&lt;/em&gt;&lt;/strong&gt;!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm sexist to the extreme,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In wars, soldiers are mostly known as men. In the past, soldiers were only men not women.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'll soon know what I mean,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you get older, you will soon learn that most previous wars had mostly or only men.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your children should thank me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Modern children just love violent and action-filled games, especially related to war... PG-13!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your wife might just hate me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As said about being sexist to the extreme, the wife of a soldier will be very sad if she lost her husband in war.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Knives I despise&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not quite sure, but soldiers usually go hard-core, and long-range battles with airplanes, naval ships, guns, bazookas, pistols... Do I need to go further?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I hide from your eyes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can't see war itself really.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Avoid mutations, // From your medications.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In serious, dangerous wars, some countries might go too far and try to go with mutated, superhuman soldiers. Or maybe the world's most dangerous &quot;superhero&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So we will do all your tests,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In wars, countries &quot;test&quot; each other to see who wins the battle and then the war. The &quot;winner(s)&quot; of the &quot;test&quot; will win the war and many more battles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But eventually we'll both rest.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One side shall surrender or lose to the other, stopping the war and letting the countries and people &quot;rest in peace&quot; (I'm not talking death here).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9042" LastActivityDate="2015-03-15T14:59:17.690" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10295" PostTypeId="2" ParentId="10273" CreationDate="2015-03-15T15:06:59.637" Score="5" Body="&lt;p&gt;This game is the famous NIM game: &lt;a href=&quot;http://en.wikipedia.org/wiki/Nim&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Nim&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The game has been fully analysed at the end of the 19th century.&#xA;It has been discussed in some of the Martin Gardner books and in many other books on recreational mathematics. It might be the most popular mathematical game known all over the world. Google shows 300.000 pages that describe the solution strategy for NIM.&#xA;It is virtually impossible to find a puzzle-lover who has never seen the NIM game and its winning strategy before.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Translate the values $S_1 , \ldots, S_n$ into binary, and compute their XOR (which is also known as NIM-sum). You win the game, if you always leave an XOR value zero.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The wikipedia says that the name NIM was coined by Charles L. Bouton of Harvard University, who also developed the complete theory of the game in 1901, but the origins of the name were never fully explained. The name is probably derived from the German word &quot;nimm&quot; meaning &quot;take [imperative]&quot;, or the obsolete English verb nim of the same meaning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only new contribution of the above puzzle is the story around the game.&lt;/p&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="294" LastEditDate="2015-03-16T00:09:31.760" LastActivityDate="2015-03-16T00:09:31.760" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10296" PostTypeId="2" ParentId="10263" CreationDate="2015-03-15T15:16:19.097" Score="5" Body="&lt;p&gt;The numbers 28-27-57-16-26-3-58 are the atomic numbers of Ni-Co-La-S-Fe-Li-Ce.&lt;br&gt;&#xA;The Nobel prize was for chemistry, and the two killers were Nicolas and Felice.&lt;/p&gt;&#xA;" OwnerUserId="9125" LastActivityDate="2015-03-15T15:16:19.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10297" PostTypeId="2" ParentId="10289" CreationDate="2015-03-15T15:33:01.113" Score="2" Body="&lt;p&gt;Asking &quot;what would you/he say?&quot; doesn't give me information on the doors, so we'll rule that out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can ask one at random &quot;are you guarding the exit?&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1/3 we're stabbed, so we know the answer is false.&#xA;Yes - bet on that door as first aid. Get caught or succeed (50:50). If succeed choose one of the others at random (50:50 again). &#xA;No - that's it. Choose a different one by random for first aid(1/2), and leave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total 1/3*(1/2*1/2+1/2)=1/4 success. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2/3*3/4=1/2 false answer without stab. &#xA;Yes - choose another at random. If it's the the first aid, go back and choose the other. &#xA;No - take that one &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total 1/2*(1/3*1/2+1/3+1/3) = 5/12 success&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1/6 chance of getting a truthful answer, don't bet on it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got a total of 8/12 for a healthy getaway&lt;/p&gt;&#xA;" OwnerUserId="4317" LastEditorUserId="4317" LastEditDate="2015-03-15T15:43:54.917" LastActivityDate="2015-03-15T15:43:54.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10298" PostTypeId="1" AcceptedAnswerId="10329" CreationDate="2015-03-15T15:44:18.993" Score="4" ViewCount="290" Body="&lt;p&gt;This is my sixth batch of sequence puzzles that are nasty and hard to solve; &#xA;yet, each of them has a clear and justifiable solution.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sequence 1:&lt;br&gt;&#xA;    Ca, Cl, ??, Er, ??, Po, Te, Th, ??&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sequence 2:&lt;br&gt;&#xA;  A, G, ?, G, ?, W, V, E, G, E, ?, ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sequence 3:&lt;br&gt;&#xA;  ..., 1906, 1912, 1918, 1936, 1938, ????, ????, 1978, ????, 1992, 1998, 2002, 2004, 2008, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sequence 4:&lt;br&gt;&#xA;  ?, ?, ?, R, A, O, T, E, L, A, I, D, L, N, A, S, Y, A, E, D, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9125" LastActivityDate="2015-03-16T21:10:24.547" Title="Yet another batch of four sequence puzzles" Tags="|number-sequence|letter-sequence|" AnswerCount="2" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10299" PostTypeId="1" AcceptedAnswerId="10583" CreationDate="2015-03-15T15:53:29.130" Score="3" ViewCount="286" Body="&lt;p&gt;What is the common link between &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;an extraterrestrial entity,&lt;/li&gt;&#xA;&lt;li&gt;a space traveller,&lt;/li&gt;&#xA;&lt;li&gt;Michael Douglas,&lt;/li&gt;&#xA;&lt;li&gt;a friend of Thomas, &lt;/li&gt;&#xA;&lt;li&gt;a friend of Tony,&lt;/li&gt;&#xA;&lt;li&gt;an Army infantry regiment,&lt;/li&gt;&#xA;&lt;li&gt;and some naked ladies?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9125" LastActivityDate="2016-03-15T09:03:51.690" Title="What is the common link?" Tags="|riddle|pattern|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10300" PostTypeId="2" ParentId="10289" CreationDate="2015-03-15T15:54:02.607" Score="6" Body="&lt;p&gt;If we ask the statistician a question, we receive no new information, so regardless of our strategy, we cannot escape more than $1/2$ of the time if we ask him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we ask the damned liar a question, we will know the answer to our question (negated), and we will also be stabbed (a painful &lt;strong&gt;bit&lt;/strong&gt; of information). Then, we essentially need to guess which door is the wardens correctly, and check the other two; at best, we can ask &quot;Is this door the warden's?&quot; for some given door and if so (probability $1/3$ - remember that we receive &quot;no&quot; as the answer in this case), we escape, and if not (probability $2/3$), we have a 50-50 shot. Thus, our best case with him is a $2/3$ chance of escape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we ask the liar a question, we will know the answer to our question and merely need to guess the escape before the warden's door. The question &quot;Is door $1$ the warden's, or is door $3$ the escape?&quot; suffices, since if we get a &quot;no&quot;, then we try door $3$; if it's not the escape, then door $1$ is the warden's, so we escape by door $2$. If we get a &quot;yes&quot;, then we try door $1$ - if it's not the escape, then door $2$ must be the escape, so we escape that way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting this together, our best strategy is to a random sentry:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is either of the following statements true:&lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;You are the liar and either door $1$ is the warden's or door $3$ is the escape.&lt;/p&gt;&#xA;    &#xA;    &lt;p&gt;You are the damned liar and door $1$ is the warden's.&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we receive a &quot;no&quot;, try door $3$ then $2$. If we receive a &quot;yes&quot;, try door $1$ then $2$. Our probability of escape is easily computed as $\frac{13}{18}$ and this is the best we can do.&lt;/p&gt;&#xA;" OwnerUserId="4765" LastActivityDate="2015-03-15T15:54:02.607" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10301" PostTypeId="2" ParentId="9900" CreationDate="2015-03-15T16:29:52.753" Score="8" Body="&lt;p&gt;I have a much simpler argument (of course leading to the same answer):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x$ be the number of the first politicians who says the truth. Then there are the $x$ lies mentioned in his statement plus the $13-x$ lies of the $13-x$ politicians speaking after him.&#xA;This altogether gives&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$x+(13-x)$ lies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9125" LastActivityDate="2015-03-15T16:29:52.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10302" PostTypeId="2" ParentId="10043" CreationDate="2015-03-15T16:54:01.690" Score="4" Body="&lt;p&gt;ok, I think I figured out the missing &quot;trick&quot; that we need to solve this one. it's when the warden says &quot;You may choose one door for me to unlock.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the warden is still in the other room, so to unlock a single door we can presume that he has to walk into the prisoner room himself, &lt;em&gt;by going through one of the 3 doors&lt;/em&gt;. so, the answer is to ask the question we figured out earlier &quot;is the left door a worse choice than the middle door?&quot;. if yes, pick the middle door, if no, pick the left door. that gives us 4/6 chances freedom and 2/6 chances back to the warden room. the trick is to not tell your choice out loud, but rather tell the warden that you are ready to pick a door. if the walks into the room through the same door you had picked, choose the rightmost door instead.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="9413" LastEditDate="2015-03-15T16:59:29.677" LastActivityDate="2015-03-15T16:59:29.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10303" PostTypeId="2" ParentId="10231" CreationDate="2015-03-15T17:07:55.147" Score="2" Body="&lt;p&gt;The condom answer being very close gave me the clue I needed. The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sperm.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I start out quite skilled,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A sperm cell can maneouvre its way through the vagina and the uterus from its creation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Throughout the nation I've drilled,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A sperm cell &quot;drills into&quot; the egg cell to form a zygote.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm sexist to the extreme,&lt;br&gt;&#xA;You'll soon know what I mean.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The nature of the 23rd chromosome of a sperm cell determines what gender the baby will be. An X chromosome will create a female child, and a Y chromosome will create a male child.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Although, the &quot;sexism&quot; could also refer to the fact that only males can produce sperm.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your children should thank me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because without sperm they could not exist.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your wife might just hate me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because unwanted pregnancies happen often, and pregnancy itself can be very trying.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Knives I despise,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I can't figure this one out. Possibly something to do with a vasectomy making the sperm unable to travel anywhere?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I hide from your eyes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sperm cannot be seen by the naked eye. It appears to be part of a liquid, semen, just like red blood cells appear to be part of liquid blood.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Avoid mutations from your medications,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If anything is done to affect the sperm cell (including certain medications), the resulting zygote can hvae birth defects.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So we will do all your tests,&lt;br&gt;&#xA;But eventually we will both rest.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I can't quite figure this one out either. The sperm and egg will &quot;both rest&quot; somehow, presumably when they form a zygote. But &quot;doing all your tests&quot; is something I can't figure out how to explain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-03-15T17:07:55.147" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10304" PostTypeId="2" ParentId="10174" CreationDate="2015-03-15T17:26:34.310" Score="4" Body="&lt;p&gt;Is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Urban&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&#xA;I'm thinking of a word you can spell on one hand.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 5 letters in urban for 5 fingers on one hand.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you add &quot;e&quot; or ɛ, it might describe the man who lived here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An &lt;a href=&quot;https://www.google.co.uk/search?q=define+urbane&quot; rel=&quot;nofollow&quot;&gt;urbane&lt;/a&gt; man would live in that &quot;courteous and refined [...] manner.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add the sound of the world's favorite drink, and you'll find something often worn here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tea is &lt;a href=&quot;https://www.google.co.uk/search?q=World%27s+favourite+drink&quot; rel=&quot;nofollow&quot;&gt;the world's favourite drink&lt;/a&gt; (after water) and makes a &lt;code&gt;T&lt;/code&gt; sound (the letter t)&lt;br&gt;&#xA; T + urban = &lt;a href=&quot;https://www.google.co.uk/search?q=define+turban&quot; rel=&quot;nofollow&quot;&gt;Turban&lt;/a&gt;&lt;br&gt;&#xA; Turbans are worn predominately by Sikhs and Muslims as a religious practice, and &lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/b/bc/Muslim_population_map.png&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; is an image of the Muslim population.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add the first 3 letters of this island's mechanical secret, and you'll find a common setting.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A &lt;a href=&quot;https://www.google.co.uk/search?q=define+sub&quot; rel=&quot;nofollow&quot;&gt;sub&lt;/a&gt;, as in &lt;a href=&quot;https://www.google.co.uk/search?q=define+submarine&quot; rel=&quot;nofollow&quot;&gt;submarine&lt;/a&gt;. The island uses submarines as a weapon.&lt;br&gt;&#xA; Sub + urban = &lt;a href=&quot;https://www.google.co.uk/search?q=define+suburban&quot; rel=&quot;nofollow&quot;&gt;suburban&lt;/a&gt;&lt;br&gt;&#xA; The &lt;a href=&quot;https://www.google.co.uk/search?q=define+suburb&quot; rel=&quot;nofollow&quot;&gt;suburbs&lt;/a&gt; are common area &lt;sup&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Citation_(horse)&quot; rel=&quot;nofollow&quot;&gt;[citation needed]&lt;/a&gt;&lt;/sup&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add The One Who Originated and drink some with Coke.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I have no idea. I think &quot;The One Who Originated&quot; would be some religious reference, like Adam or Eve. I also think &quot;&lt;a href=&quot;https://www.google.co.uk/search?q=define+coke&quot; rel=&quot;nofollow&quot;&gt;Coke&lt;/a&gt;&quot; refers to &lt;a href=&quot;https://www.google.co.uk/search?q=define+cocaine&quot; rel=&quot;nofollow&quot;&gt;cocaine&lt;/a&gt;, rather than &lt;a href=&quot;https://www.google.co.uk/search?q=define+Coca-Cola&quot; rel=&quot;nofollow&quot;&gt;the famous carbonated non-alcoholic drink.&lt;/a&gt; (and, for completion's sake, it's definitely not &lt;a href=&quot;https://www.google.co.uk/search?q=define+coal+coke&quot; rel=&quot;nofollow&quot;&gt;coal coke&lt;/a&gt;), as you wouldn't drink something with another drink.&#xA; &lt;br/&gt;&lt;br/&gt;&#xA; Also, should I unhide this as it's not the answer?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &gt; You want a hint to get you started?? The solution starts with you!&lt;br&gt;&#xA; I think this means it is where I am. Since I am on the internet, I am in an urban area.&lt;br&gt;&#xA; It may also mean &quot;The solution starts with you&quot; (&quot;u&quot;). Urban starts with &quot;u&quot; (&quot;you&quot;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user10203" LastActivityDate="2015-03-15T17:26:34.310" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10305" PostTypeId="2" ParentId="10238" CreationDate="2015-03-15T17:37:12.720" Score="2" Body="&lt;p&gt;I'll go ahead and share my preferred solution. It's not as quick Johannes's, but I think it's kind of pretty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, you can reach at most $N-1$ generations, where $N$ is the initial population size. Indeed, suppose Alice is of generation $n$. Call the &lt;em&gt;ancestry&lt;/em&gt; of a person the set of their ancestors among the initial population. Alice has some set of size less than or equal to $N$ as her ancestry. One of her parents has generation $n-1$, one of their parents is of generation $n-2$, and so on: we can find a sequence of people of length exactly $n+1$ starting with Alice and ending with someone in the initial population. A person's ancstry is the disjoint union of their parents', thus the size of the ancestries &lt;em&gt;strictly&lt;/em&gt; decreases all the way along this sequence, beginning with with some value less than or equal to $N$ and ending with $1$. Such a sequence can be of length at most $N$, and our sequence is of length $n+1$, thus $n+1\leq N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to prove it's possible to attain the bound. First of all, if you have $1$ man and $n$ women, you can obtain a person of generation $n$ by having the man have a son with the first woman, then that son has a son with the second woman, and so on. Now if you have $m$ men and $n$ women, pick one distinguished man to perform this process with the $n$ women, producing a daughter of generation $n$. This daughter then performs the same process with the $m-1$ men she isn't related to obtain a child who, it's easy to see, will be of generation $n+(m-1)$.&lt;/p&gt;&#xA;" OwnerUserId="5511" LastEditorUserId="5511" LastEditDate="2015-03-16T22:15:32.997" LastActivityDate="2015-03-16T22:15:32.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10306" PostTypeId="1" CreationDate="2015-03-15T17:51:46.240" Score="2" ViewCount="1102" Body="&lt;p&gt;When I write text messages on my mobile phone I just need to press each key&#xA;once for each letter and it automatically finds the most likely word that&#xA;fits the combination (abc=2, def=3, ghi=4, jkl=5, mno=6, pqrs=7, tuv=8,&#xA;wxyz=9). So if I press 84373 it displays &lt;em&gt;&quot;there&quot;&lt;/em&gt;, although the word could&#xA;as well be &lt;em&gt;&quot;these&quot;&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the longest pair of English words that have the same number combination, while having no common letter in the same position?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The words must occur in the Merriam Webster dictionary; only uncapitalized words; no hyphens.&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="8874" LastEditDate="2015-03-16T12:37:01.327" LastActivityDate="2019-04-05T21:05:03.820" Title="Cell phone puzzle" Tags="|word|english|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10307" PostTypeId="2" ParentId="10231" CreationDate="2015-03-15T18:59:31.870" Score="3" Body="&lt;p&gt;The answer may be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a human ovary&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I start out quite skilled,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The eggs are present at a girl's birth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Throughout the nation I've drilled,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ovarian drilling is a fertility treatment, but this doesn't quite fit because the clue is in the active voice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm sexist to the extreme,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Only females have them, and the sex chromosomes on the eggs they contain are all female.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'll soon know what I mean,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...if you solve the riddle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your children should thank me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because an ovary is where half of each of them comes from.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your wife might just hate me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ovaries can have or develop serious problems, up to and including ovarian cancer, a highly virulent form of cancer with a high mortality rate even within a year of diagnosis, and they are also where ovulation starts, which causes menstruation, which for some women causes a lot of discomfort or pain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Knives I despise&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Surgical removal, oophorectomy, carries risks of adverse effects, some of which are fatal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I hide from your eyes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are not visible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Avoid mutations,&#xA;From your medications.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The second hint asks &lt;em&gt;What could cause mutations that is also used as a medication (treatment)?&lt;/em&gt; to which the answer is surely &lt;em&gt;radiation&lt;/em&gt;. Exposure of ovaries or sperm to radiation, including X-rays, can cause mutation. This is the means by which exposure to radioactivity, whether from atomic or nuclear weapons or from nuclear power stations or waste, causes mutations in future generations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So we will do all your tests,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pharmaceutical interests have promoted screening for ovarian cancer, which has begun to be carried out with little success.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But eventually we'll both rest.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After the menopause, ovaries no longer ovulate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7363" LastActivityDate="2015-03-15T18:59:31.870" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10308" PostTypeId="1" AcceptedAnswerId="10309" CreationDate="2015-03-15T19:26:29.557" Score="7" ViewCount="1500" Body="&lt;p&gt;You have been trapped in a jail cell of an evil mastermind!&lt;br&gt;&#xA;The cell has a locked door with a keyboard with the letters a to z and an enter key.&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/yZP8H.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yZP8H.png&quot; alt=&quot;Keyboard&quot; title=&quot;Hint: There is no &amp;#39;p&amp;#39; in the password&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sup&gt;(&lt;a href=&quot;http://commons.wikimedia.org/wiki/File:Tastatur-ios7.png?uselang=en-gb&quot; rel=&quot;nofollow noreferrer&quot;&gt;Original image&lt;/a&gt; courtesy of Wikimedia Commons)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The evil mastermind gives you some clues to escape, as he does this as a pastime:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Only one person has escaped.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;They had to figure out the same password as you.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;They had a box of clues, which you currently don't have, so he tells you about them.&lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;This is the tale of my first escapee:&lt;br&gt;&#xA;    They discovered the clues to the password in a box.&lt;br&gt;&#xA;    In the box they found a red herring and a butt.&lt;br&gt;&#xA;    The red herring was way past its time.&lt;br&gt;&#xA;    The butt was only part of its body, with the rest long risen above.&lt;br&gt;&#xA;    That was the fate of both of those items.&lt;br&gt;&#xA;    Using these clues begot the solution to the escapee:&lt;br&gt;&#xA;    What could've happened to one item to prevent its fate&lt;br&gt;&#xA;    that would've happened to the second to make it reach its fate?  &lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note: &quot;they&quot; is used as a gender-neutral third-person pronoun. Can be replaced with &quot;he&quot; or &quot;she&quot;.&lt;/p&gt;&#xA;" OwnerDisplayName="user10203" LastEditorDisplayName="user10203" LastEditDate="2015-03-15T21:13:42.330" LastActivityDate="2015-03-15T22:56:02.563" Title="Deduce the password!" Tags="|riddle|password|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10309" PostTypeId="2" ParentId="10308" CreationDate="2015-03-15T19:38:47.043" Score="9" Body="&lt;p&gt;I suspect that the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;smoking&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &quot;butt&quot; being referred to is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the butt of a cigarette. The rest of its body long rose above in puffs of tobacco smoke.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the red herring is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; one that has been rotten &quot;way past its time&quot;. If it had been smoked, it would have lasted longer, &quot;preventing its fate&quot; so to say.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-03-15T22:56:02.563" LastActivityDate="2015-03-15T22:56:02.563" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10310" PostTypeId="2" ParentId="10292" CreationDate="2015-03-15T20:05:39.833" Score="4" Body="&lt;p&gt;Assuming that multiple consecutive moves of the same block are counted as one, here is a solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; El, Dd, Ed, Cd, Er, Brr, Au, Zrruu, Ad, Bll, El, Cu, Dl, Cu, Erd, Dur, Zru, Cl, Zur&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9471" LastActivityDate="2015-03-15T20:05:39.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10311" PostTypeId="1" AcceptedAnswerId="10338" CreationDate="2015-03-15T20:47:39.963" Score="7" ViewCount="1591" Body="&lt;p&gt;The Great Houdini is performing again! Everybody wants to see the famous magician and the magic show is sold out. Today Houdini shows the following sophisticated trick:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A guy from the audience carefully shuffles a deck of $52$ cards ($13$ spades, $13$ hearts, $13$ clubs and $13$ diamonds), and hands the deck over to Houdini's assistant.&lt;/li&gt;&#xA;&lt;li&gt;The assistant then repeats the following step $52$ times: &#xA;He looks at the top card (of course without showing it to Houdini), and puts it face-down on the table. Houdini guesses the suit. Then the assistant flips the card over and reveals it to everybody (including Houdini). Then the step is repeated.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Houdini's goal is of course to make as many correct guesses as possible.&#xA;A &lt;a href=&quot;https://puzzling.stackexchange.com/questions/10118/the-great-houdinis-awesome-card-guessing-trick-1&quot;&gt;predecessor puzzle&lt;/a&gt; showed us that without cheating Houdini will only predict $13$ cards correctly (in the worst case).&#xA;Today, however, Houdini and his assistant indeed are cheating:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The backsides of the cards are not perfectly symmetric. &#xA;  The assistant has two different ways of putting down every card, and hence may communicate to Houdini one bit of information per card.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is there a cheating strategy for Houdini and his assistant that guarantees Houdini to make &#xA;(a) at least $27$ correct guesses &#xA;(b) at least $32$ correct guesses (even in the worst possible case)?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2020-10-19T19:46:12.763" Title="The Great Houdini's awesome card guessing trick (2)" Tags="|mathematics|strategy|cards|" AnswerCount="5" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10313" PostTypeId="1" AcceptedAnswerId="10315" CreationDate="2015-03-15T21:33:34.473" Score="7" ViewCount="556" Body="&lt;p&gt;Your boss wants you to head up a new part of the company that will analyze business statistics. She wants an official name for it like &quot;The Organization for Business Statistics&quot; or &quot;The Business Statistics Council&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;She asks you to write a list of 26 words that start with each letter of the alphabet that are loosely synonymous to &quot;organization&quot; or &quot;council&quot;. She agrees that there are no such words starting with &lt;code&gt;h&lt;/code&gt;, &lt;code&gt;k&lt;/code&gt;, &lt;code&gt;n&lt;/code&gt;, &lt;code&gt;q&lt;/code&gt;, &lt;code&gt;v&lt;/code&gt;, &lt;code&gt;x&lt;/code&gt;, or &lt;code&gt;y&lt;/code&gt;, so really she only wants 19 words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You write this list, choosing words that would not seem out of place in the phrase&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm a member of The ________ of/for Business Statistics.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or perhaps&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I have to go to a meeting with the Business Statistics ________.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sadly, you've lost the list, and only have vague memories about what the words were. You jot down these recollections, knowing they must relate to the words:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a: starts with an end&#xA;b: holds clothes&#xA;c: three sets of double letters&#xA;d: math operation&#xA;e: more proper to use for music&#xA;f: Sam was in one&#xA;g: sounds like wine fruit&#xA;h: n/a&#xA;i: reasearch ___, ___ for technology&#xA;j: angry men&#xA;k: n/a&#xA;l: the red-headed one was fake&#xA;m: of magic&#xA;n: n/a&#xA;o: relates to neatness and tidiness&#xA;p: relates to windows&#xA;q: n/a&#xA;r: Sam was in this too, in a way&#xA;s: sounds like Stampy's best friend's name&#xA;t: almost a meeting going in the wrong direction&#xA;u: one&#xA;v: n/a&#xA;w: powerful, compound&#xA;x: n/a&#xA;y: n/a&#xA;z: only if it's monkey business&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You already realize that &lt;code&gt;o&lt;/code&gt; must be for &lt;code&gt;organization&lt;/code&gt; because being organized relates to being neat and tidy. &lt;strong&gt;Can you figure out the other 18 words?&lt;/strong&gt; (&lt;code&gt;c&lt;/code&gt; is not necessarily for &lt;code&gt;council&lt;/code&gt;. The words only contain letters.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To verify your answer, take the 19 words in order, all in lowercase, and separate them with spaces. e.g.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;aword bword cword dword eword fword gword iword jword lword mword oword pword rword sword tword uword wword zword&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The resulting &lt;a href=&quot;http://www.miraclesalad.com/webtools/md5.php&quot; rel=&quot;nofollow&quot;&gt;MD5 hash&lt;/a&gt; is &lt;code&gt;6a871238db24d9b1a91fc87f34721963&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1811" LastEditorUserId="5373" LastEditDate="2016-10-21T14:56:10.490" LastActivityDate="2016-10-21T14:56:10.490" Title="The ________ for Business Statistics" Tags="|word|english|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10314" PostTypeId="1" AcceptedAnswerId="10316" CreationDate="2015-03-15T21:38:02.037" Score="11" ViewCount="284" Body="&lt;p&gt;I am a word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When by none, I am erroneous.&lt;br&gt;&#xA;When by one, extraneous.&lt;br&gt;&#xA;When surrounded by any, I am the one.&lt;br&gt;&#xA;Yet when I am to any by many, said any approaches none.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerDisplayName="user10203" LastEditorUserId="5373" LastEditDate="2015-03-15T22:11:24.417" LastActivityDate="2015-04-29T08:54:54.477" Title="Identify the word" Tags="|riddle|word|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10315" PostTypeId="2" ParentId="10313" CreationDate="2015-03-15T21:41:33.950" Score="3" Body="&lt;p&gt;Hash:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; association bureau committee division ensemble fellowship group institute jury league ministry organization panel ring squad team unit workforce zoo&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A is for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ASSOCIATION (starts with &quot;ass&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;B is for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BUREAU&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;C is for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; COMMITTEE (there weren't many possibilities!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;D is for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DIVISION&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;E is for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ENSEMBLE - thanks again to &lt;strong&gt;Reticality&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;F is for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; FELLOWSHIP (Lord of the Rings reference)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;G is for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; GROUP (sounds like &quot;grape&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I is for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; INSTITUTE (&quot;research institute&quot;, &quot;institute for technology&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;J is for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; JURY&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;L is for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LEAGUE (a &lt;strong&gt;lovely&lt;/strong&gt; reference to the Sherlock Holmes story &quot;The Red-headed League&quot;!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;M is for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MINISTRY (Harry Potter reference) Thanks to &lt;strong&gt;Reticality&lt;/strong&gt; for this one, since I've now looked at his answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;O is for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ORGANISATION (already mentioned in the OP)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;P is for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PANEL (as in window panel)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;R is for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RING (well, Sam's finger was in it at least!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;S is for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SQUAD (&lt;strong&gt;Reticality&lt;/strong&gt; told me it should sound like &quot;squid&quot;.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;T is for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TEAM (almost &quot;meet&quot; backwards)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;U is for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; UNIT&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;W is for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; WORKFORCE - thanks one final time to &lt;strong&gt;Reticality&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Z is for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ZOO&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Great puzzle by the way!&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-15T23:27:36.307" LastActivityDate="2015-03-15T23:27:36.307" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="10316" PostTypeId="2" ParentId="10314" CreationDate="2015-03-15T21:47:59.660" Score="7" Body="&lt;p&gt;Sounds like&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;division. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When by none, I am erroneous.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Division by zero results in an error. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When by one, extraneous.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Division by one returns the original number. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When surrounded by any, I am the one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Any nonzero N divided by any nonzero N equals one. The division sign is &quot;surrounded&quot; by N.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet when I am to any by many, said any approaches none.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $ \lim_{x \to +\infty}\frac{N}{x}=0$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7529" LastActivityDate="2015-03-15T21:47:59.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10317" PostTypeId="2" ParentId="10289" CreationDate="2015-03-15T21:59:36.977" Score="10" Body="&lt;h1&gt;I am able to escape without risk.&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Using so called &lt;a href=&quot;https://en.wikipedia.org/wiki/The_Hardest_Logic_Puzzle_Ever#Unanswerable_questions_and_exploding_god-heads&quot;&gt;head exploding questions&lt;/a&gt; we can extract even more that one bit of information from the guards. In this example, we will produce a paradox if the ward is behind the second door. I.e., we ask a liar “Will you answer yes?”. If she answers “Yes” than she is telling the truth, that isn't allowed and if she answers “No” then she is telling the truth as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exploiting self-referential statements can also persuade the statistician to give a useful answer. (We won't know whether we got lied to or not, but we don't care about that.) This is possible because of your wording “a Statistician will give you &lt;strong&gt;a true or false response&lt;/strong&gt; with 50% probability.” instead of “a Statistician will give you a random response.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here is the question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is your statement a lie and not &lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;door a = ward or&lt;/p&gt;&#xA;    &#xA;    &lt;p&gt;(door b = ward) and (is your answer no)&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;  &#xA;  &lt;p&gt;or is your statement true and&lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;door a = ward or&lt;/p&gt;&#xA;    &#xA;    &lt;p&gt;(door b = ward) and (is your answer no)&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or with more math notation, we ask to evaluate $$ ((\lnot t) \land \lnot(a \lor (b \land p))) \lor (t \land (a \lor (b \land p)))\text{,}$$&#xA;where $t$ indicates if a guard tells the truth, $a$ and $b$ ask whether the ward is behind the respective door, $r$ the answer which will be given and $p = (r == \text{No})$ produces the paradoxes to explode heads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a small table we can work out that if a guard answers “No” than the ward is behind the a door, if the guard answers “Yes” than the ward is behind the c door and if her head explodes the ward is behind b.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a table of all possible states of the world:&#xA;&lt;code&gt;&#xA;τ r a b | True answer | Response&#xA;0 0 0 0 | 1           | 0&#xA;0 0 0 1 | 0           | 1 -&gt; head explodes&#xA;0 0 1 0 | 0           | 1&#xA;0 1 0 0 | 1           | 0&#xA;0 1 0 1 | 1           | 0 -&gt; head explodes&#xA;0 1 1 0 | 0           | 1&#xA;………………………………………………………………………………………………………………&#xA;1 0 0 0 | 0           | 0&#xA;1 0 0 1 | 1           | 1 -&gt; head explodes&#xA;1 0 1 0 | 1           | 1&#xA;1 1 0 0 | 0           | 0&#xA;1 1 0 1 | 0           | 0 -&gt; head explodes&#xA;1 1 1 0 | 1           | 1&#xA;&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="10154" LastEditorUserId="10154" LastEditDate="2015-03-15T22:15:38.030" LastActivityDate="2015-03-15T22:15:38.030" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10318" PostTypeId="2" ParentId="10311" CreationDate="2015-03-15T22:03:49.070" Score="5" Body="&lt;p&gt;I will be happy to be proved wrong, but I believe that Houdini can always guarantee to get 26 cards correct, but can't guarantee any more.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; A simple strategy to achieve 26 would be for the assistant to 'transmit' a 0-bit for each red card and a 1-bit for each black card - and Houdini chooses &amp;quot;Hearts&amp;quot; or &amp;quot;Spades&amp;quot; for 0- and 1- bit respectively.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;But&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; even if the assistant could determine (but not alter) the order of the shuffled deck and transmit all 52 bits of info to Houdini right at the start of the performance, I don't think that is sufficient for Houdini to guarantee 32 correct guesses with arbitrary deck order.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="63414" LastEditDate="2020-10-19T19:34:38.207" LastActivityDate="2020-10-19T19:34:38.207" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="10320" PostTypeId="2" ParentId="10313" CreationDate="2015-03-15T22:17:59.553" Score="4" Body="&lt;p&gt;Here is my answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a: starts with an end&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Association (Starts with &quot;ass&quot;, an end)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;b: holds clothes - wrong&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Body (Actual answer: &quot;Bureau&quot; (of change? as in changing clothes.) By &lt;strong&gt;Rand al'thor&lt;/strong&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;c: three sets of double letters&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Committee (Doubles: &quot;m&quot;, &quot;t&quot; and &quot;e&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;d: math operation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Division? Like the division of defence or something? Can't think of anything else.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;e: more proper to use for music&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ensemble. As in group (of musicians)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;f: Sam was in one&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fellowship. I think he was Frodo's best friend in tLoTR&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;g: sounds like wine fruit&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Group, as in Grape (which ferments to wine)  &lt;strong&gt;Rand al'thor&lt;/strong&gt; got this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;i: research &lt;strong&gt;_, _&lt;/strong&gt; for technology&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Institute. I found many &lt;code&gt;research institute&lt;/code&gt;s, and some &lt;code&gt;institute[s] for technology (and management)&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;j: angry men&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jury, the 12 angry men. (Movie)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;l: the red-headed one was fake&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; League. Used Google's synonyms, so no explanation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;m: of magic&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ministry (of magic. From Harry Potter I believe)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;o: relates to neatness and tidiness&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Organisation. Neat people have this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;p: relates to windows&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Panel&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;r: Sam was in this too, in a way&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The ring!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;s: sounds like Stampy's best friend's name  - Incorrect&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; syndicate? (Squad. Again by &lt;strong&gt;Rand al'thor&lt;/strong&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;t: almost a meeting going in the wrong direction&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;gniteem&lt;/code&gt; -&gt; &lt;code&gt;teem&lt;/code&gt; -&gt; &lt;code&gt;team&lt;/code&gt; (Riddles make the word &lt;code&gt;backwards&lt;/code&gt; &lt;em&gt;always&lt;/em&gt; implicitly imply reverse.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;u: one - wrong but close&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Union. As in the union of probability / set theory into 1.&lt;br&gt;&#xA; (Actually unit. By &lt;strong&gt;Rand al'thor&lt;/strong&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;w: powerful, compound&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Workforce? A compound word, who are powerful?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;z: only if it's monkey business&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;a href=&quot;http://www.thesaurus.com/browse/corporation?s=t&quot; rel=&quot;nofollow&quot;&gt;thesaurus&lt;/a&gt; suggests zoo. And, coincidentally, that's what the other answer says.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user10203" LastEditorDisplayName="user10203" LastEditDate="2015-03-16T15:26:44.117" LastActivityDate="2015-03-16T15:26:44.117" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10321" PostTypeId="1" AcceptedAnswerId="10328" CreationDate="2015-03-15T22:27:24.983" Score="2" ViewCount="312" Body="&lt;p&gt;Another Klotski Puzzle, &quot;The Great D-vide&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JfgZM.png&quot; alt=&quot;The Great D-vide&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/10270/klotski-puzzle-1&quot;&gt;Rules Here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/10270/klotski-puzzle-1&quot;&gt;Klotski Puzzle 1&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/10292/klotski-puzzle-2&quot;&gt;Klotski Puzzle 2&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10213" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-03-15T23:56:03.697" Title="Klotski Puzzle 3" Tags="|geometry|sliding-blocks|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10322" PostTypeId="5" CreationDate="2015-03-15T22:41:51.117" Score="0" Body="&lt;p&gt;Dominoes (or dominos) is a game played with rectangular &quot;domino&quot; tiles. The domino gaming pieces make up a domino set, sometimes called a &lt;em&gt;deck&lt;/em&gt; or &lt;em&gt;pack&lt;/em&gt;. The traditional Sino-European domino set consists of 28 dominoes, colloquially nicknamed &lt;em&gt;bones&lt;/em&gt;, &lt;em&gt;cards&lt;/em&gt;, &lt;em&gt;tiles&lt;/em&gt;, &lt;em&gt;tickets&lt;/em&gt;, &lt;em&gt;stones&lt;/em&gt;, or &lt;em&gt;spinners&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each domino is a tile in the shape of a 2x1 rectangle, with a line dividing its face into two squares. Each end is marked with a number of spots (also called &lt;em&gt;pips&lt;/em&gt;, &lt;em&gt;nips&lt;/em&gt; or &lt;em&gt;dobs&lt;/em&gt;) or is blank. The value of either side is the number of pips. In the most common variant (Double Six) the values range from 0 (no pips) to 6. The sum of the two values, i.e. the total number of pips, may be referred to as the &lt;em&gt;rank&lt;/em&gt; or &lt;em&gt;weight&lt;/em&gt; of a tile, and a tile with more pips may be called &lt;em&gt;heavier&lt;/em&gt; than a &lt;em&gt;lighter&lt;/em&gt; tile with fewer pips. The backs of the dominoes in a set are indistinguishable, either blank or having some common design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A domino set is a generic gaming device, similar to playing cards or dice, in that a variety of games can be played with a set.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-15T22:41:51.117" LastActivityDate="2015-03-15T22:41:51.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10323" PostTypeId="4" CreationDate="2015-03-15T22:41:51.117" Score="0" Body="A puzzle that involves tiles (or the concept of tiles) in the shape of 2x1 rectangles, divided into 2 squares, each containing 0 to 6 spots.  It may require knowledge of general rules of the game, Dominos, or it may simply use the number of spots, on the tiles, as clues for a solution." OwnerUserId="5373" LastEditorUserId="19944" LastEditDate="2016-03-11T17:32:43.157" LastActivityDate="2016-03-11T17:32:43.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10324" PostTypeId="5" CreationDate="2015-03-15T22:47:33.973" Score="0" Body="&lt;p&gt;A sliding blocks puzzle is a puzzle which involves sliding blocks into a desired end configuration. The blocks and area in which they are slid can be of any shape and size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the classic 15-puzzle is played on a 4x4 square with 15 tiles numbered 1 through 15. The goal is to slide the blocks in a such a way as to end with the numbers in sequential order, starting with 1 in the top left corner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The classic Klotski puzzle involves sliding rectangular blocks of different sizes, usually with the intention of getting a designated block into a designated area on the board.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="31484" LastEditDate="2017-05-13T03:20:15.887" LastActivityDate="2017-05-13T03:20:15.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10325" PostTypeId="4" CreationDate="2015-03-15T22:47:33.973" Score="0" Body="A sliding blocks puzzle is a puzzle which involves sliding blocks into a desired end configuration." OwnerUserId="5373" LastEditorUserId="31484" LastEditDate="2017-05-13T03:44:34.880" LastActivityDate="2017-05-13T03:44:34.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10326" PostTypeId="5" CreationDate="2015-03-15T22:50:38.300" Score="0" Body="&lt;p&gt;Puzzles revolving around 'connections' usually require the solver to identify a common link or property between several items in a list or set. These 'items' are most commonly words, phrases, numbers, symbols, photographs or other images, or are given in the form of trivia questions whose answers share a particular property.&lt;/p&gt;&#xA;&lt;p&gt;In some instances the connected items are members of an incomplete set which is missing a single member, and this must be identified by the solver.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Simple examples:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What connects the following phrases? MANY HANDS MAKE LIGHT WORK, ALL'S WELL THAT ENDS WELL, HONOURS EVEN, JUST WHEN I NEEDED YOU MOST.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; All contain hidden numbers, spelled out: MANY HANDS MAKE LIGH&lt;strong&gt;T WO&lt;/strong&gt;RK, ALL'S WELL THA&lt;strong&gt;T EN&lt;/strong&gt;DS WELL, HONOUR&lt;strong&gt;S EVEN&lt;/strong&gt;, JUST WHE&lt;strong&gt;N I NE&lt;/strong&gt;EDED YOU MOST.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;What is the connection between the following people? Which 21st century US President is missing from the set?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The founder of WikiLeaks;&lt;/li&gt;&#xA;&lt;li&gt;The star of &lt;em&gt;Diagnosis Murder&lt;/em&gt;;&lt;/li&gt;&#xA;&lt;li&gt;The second wife of Henry VIII;&lt;/li&gt;&#xA;&lt;li&gt;The host of &lt;em&gt;Wacaday&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; All share their first names with members of Enid Blyton's &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/The_Famous_Five&quot; rel=&quot;nofollow noreferrer&quot;&gt;Famous Five&lt;/a&gt;&lt;/em&gt;: &lt;strong&gt;JULIAN&lt;/strong&gt; Assange, &lt;strong&gt;DICK&lt;/strong&gt; Van Dyke, &lt;strong&gt;ANNE&lt;/strong&gt; Boleyn, and &lt;strong&gt;TIMMY&lt;/strong&gt; Mallett. The missing member of the fictional group is George, making the missing US President &lt;strong&gt;GEORGE W. Bush&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Exclusions:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;'Connecting wall' type questions, of the sort originally seen in the BBC quiz show &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Only_Connect&quot; rel=&quot;nofollow noreferrer&quot;&gt;Only Connect&lt;/a&gt;&lt;/em&gt; (first aired 2008) and - more lately - the NYT's '&lt;a href=&quot;https://en.wikipedia.org/wiki/The_New_York_Times_Connections&quot; rel=&quot;nofollow noreferrer&quot;&gt;Connections&lt;/a&gt;' puzzle (first published 2023), in which solvers must &lt;em&gt;partition&lt;/em&gt; a set of jumbled clues into distinct categories based on shared connections should instead use the &lt;a href=&quot;/questions/tagged/connect-wall&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;connect-wall&amp;#39;&quot; aria-label=&quot;show questions tagged &amp;#39;connect-wall&amp;#39;&quot; rel=&quot;tag&quot; aria-labelledby=&quot;tag-connect-wall-tooltip-container&quot;&gt;connect-wall&lt;/a&gt; tag.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Questions asking 'What comes next in this sequence?' should not generally use the &lt;a href=&quot;/questions/tagged/connections-puzzle&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;connections-puzzle&amp;#39;&quot; aria-label=&quot;show questions tagged &amp;#39;connections-puzzle&amp;#39;&quot; rel=&quot;tag&quot; aria-labelledby=&quot;tag-connections-puzzle-tooltip-container&quot;&gt;connections-puzzle&lt;/a&gt; tag, since the &lt;a href=&quot;/questions/tagged/sequence&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;sequence&amp;#39;&quot; aria-label=&quot;show questions tagged &amp;#39;sequence&amp;#39;&quot; rel=&quot;tag&quot; aria-labelledby=&quot;tag-sequence-tooltip-container&quot;&gt;sequence&lt;/a&gt; tag is designed for this purpose, alongside &lt;a href=&quot;/questions/tagged/number-sequence&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;number-sequence&amp;#39;&quot; aria-label=&quot;show questions tagged &amp;#39;number-sequence&amp;#39;&quot; rel=&quot;tag&quot; aria-labelledby=&quot;tag-number-sequence-tooltip-container&quot;&gt;number-sequence&lt;/a&gt;, &lt;a href=&quot;/questions/tagged/letter-sequence&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;letter-sequence&amp;#39;&quot; aria-label=&quot;show questions tagged &amp;#39;letter-sequence&amp;#39;&quot; rel=&quot;tag&quot; aria-labelledby=&quot;tag-letter-sequence-tooltip-container&quot;&gt;letter-sequence&lt;/a&gt; or &lt;a href=&quot;/questions/tagged/word-sequence&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;word-sequence&amp;#39;&quot; aria-label=&quot;show questions tagged &amp;#39;word-sequence&amp;#39;&quot; rel=&quot;tag&quot; aria-labelledby=&quot;tag-word-sequence-tooltip-container&quot;&gt;word-sequence&lt;/a&gt; if/as appropriate.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="60644" LastEditDate="2024-04-05T08:22:41.213" LastActivityDate="2024-04-05T08:22:41.213" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10327" PostTypeId="4" CreationDate="2015-03-15T22:50:38.300" Score="0" Body="A puzzle requiring the solver to uncover a hidden connection between several items in one or more lists or sets. For partition-style connecting wall puzzles (à la the BBC's Only Connect or the NYT's 'Connections' puzzle), use [connect-wall] instead. For 'what comes next in this sequence?' puzzles, use [sequence] instead." OwnerUserId="5373" LastEditorUserId="60644" LastEditDate="2024-04-05T08:22:41.213" LastActivityDate="2024-04-05T08:22:41.213" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10328" PostTypeId="2" ParentId="10321" CreationDate="2015-03-15T23:56:03.697" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Fu, Hru, Err, Zr, Ad, Blu, Zuu, Aur, Bdd, Zlu, Bu, Elll, Ddd, Hd, Fd, Grrr, Zurrd, Glll, Zurr&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Had to change my code from BFS to Dijkstra for this one.&lt;/p&gt;&#xA;" OwnerUserId="9471" LastActivityDate="2015-03-15T23:56:03.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10329" PostTypeId="2" ParentId="10298" CreationDate="2015-03-16T01:50:21.780" Score="3" Body="&lt;p&gt;Sequence 1 is a list of the:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Muse&quot; rel=&quot;nofollow&quot;&gt;nine muses&lt;/a&gt;&lt;/strong&gt; of Greek mythology as follows (bold for missing letters):&lt;br&gt;&#xA;&lt;br&gt;&#xA;  Calliope, Clio, &lt;strong&gt;Euterpe&lt;/strong&gt;, Erato, &lt;strong&gt;Melpomene&lt;/strong&gt;, Polyhymnia, Terpsichore, Thalia, &lt;strong&gt;Urania&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sequence 2 is the order of the:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_British_monarchs&quot; rel=&quot;nofollow&quot;&gt;British monarchs&lt;/a&gt;&lt;/strong&gt; as follows (bold for missing letters):&lt;br&gt;&#xA;&lt;br&gt;&#xA; Anne, George I, &lt;strong&gt;George II&lt;/strong&gt;, George III, &lt;strong&gt;George IV&lt;/strong&gt;, William IV, Victoria, Edward VII,  George V, Edward VIII, &lt;strong&gt;George VI&lt;/strong&gt;, &lt;strong&gt;Elizabeth II&lt;/strong&gt;   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sequence 4 is a list of the:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;second letters of the abbreviation of US states&lt;/strong&gt; (bold for missing letters).&lt;br&gt;&#xA;  The order is alphabetical based on the full state name.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;AL, AK, AZ&lt;/strong&gt;, AR, CA, CO, CT, DE, FL, GA, HI, ID, IL, IN, IA, KS, KY, LA, ME, MD...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-03-16T21:10:24.547" LastActivityDate="2015-03-16T21:10:24.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10330" PostTypeId="2" ParentId="10311" CreationDate="2015-03-16T02:13:37.607" Score="2" Body="&lt;p&gt;Assign the values &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{2}$&lt;/span&gt; to spades, &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{3}$&lt;/span&gt; to hearts, &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{4}$&lt;/span&gt; to aces and &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{5}$&lt;/span&gt; to hearts.&lt;/p&gt;&#xA;&lt;p&gt;When the assistant sees the card we assign the following value to the card: the number of cards of the suit of that card that have not been flipped plus the value of the special suit of the card. So for example, if the first card is of hearts it has a value of &lt;span class=&quot;math-container&quot;&gt;$13.2$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The dealer uses the following stategy:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; She only  places the card in position &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; if that card has the suit that has the highest value at that point in time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; If a card is placed in position &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; then Houdini guesses the suit with the highest value and obviously gets it right. If the card is in position &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; Houdini guesses the suit with the second highest value, and might get it right.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This strategy always guarantees &lt;span class=&quot;math-container&quot;&gt;$26$&lt;/span&gt; correct guesses and most times it does better. But it can sometimes get exactly &lt;span class=&quot;math-container&quot;&gt;$26$&lt;/span&gt; correct guesses.&lt;/p&gt;&#xA;" OwnerDisplayName="user10001" LastEditorUserId="63414" LastEditDate="2020-10-19T19:35:49.917" LastActivityDate="2020-10-19T19:35:49.917" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10331" PostTypeId="1" AcceptedAnswerId="10332" CreationDate="2015-03-16T02:39:11.097" Score="18" ViewCount="4632" Body="&lt;p&gt;What word does this photo of a mouse represent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PIxDn.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image source: adapted from &lt;a href=&quot;http://commons.wikimedia.org/wiki/Mus#/media/File:Mouse-19-Dec-2004.jpg&quot; rel=&quot;noreferrer&quot;&gt;Wikimedia commons&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user10001" LastEditorUserId="2023" LastEditDate="2015-03-17T14:18:28.213" LastActivityDate="2015-03-18T12:11:40.807" Title="What word does this photo of a mouse represent?" Tags="|rebus|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10332" PostTypeId="2" ParentId="10331" CreationDate="2015-03-16T02:54:23.297" Score="35" Body="&lt;p&gt;My answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Anony-mouse&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or simply&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Anonymous&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;since&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; pixelating the face is done to make the identity of the person (or animal) anonymous.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10136" LastEditorUserId="10136" LastEditDate="2015-03-16T03:01:15.537" LastActivityDate="2015-03-16T03:01:15.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10333" PostTypeId="2" ParentId="10331" CreationDate="2015-03-16T02:56:47.127" Score="34" Body="&lt;p&gt;My answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mouse-aic (as in Mosaic, which refers to the pixelated blur, described on &lt;a href=&quot;http://www.sitepoint.com/censor-your-images-with-a-mosaic-in-photoshop/&quot;&gt;this page&lt;/a&gt;) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2015-03-16T02:56:47.127" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10334" PostTypeId="1" AcceptedAnswerId="10408" CreationDate="2015-03-16T04:13:23.467" Score="14" ViewCount="3053" Body="&lt;p&gt;There's something very special hidden in this image. You will &lt;strong&gt;definitely&lt;/strong&gt; know when you find it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SOl0i.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SOl0i.png&quot; alt=&quot;special image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I know Stack Exchange lossily compresses images but I've checked and no pixels have been altered.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding traces of the secret might not be that hard. The ideal answer will point to the Stack Exchange question that I took data from to help generate this image (and possibly point to another that can help &quot;decode&quot; it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I can reveal the secret with this 20 line Python program. I've replaced the most telling bits with &lt;code&gt;???&lt;/code&gt;:&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from PIL import Image&#xA;i = Image.open('hidden.png')&#xA;d = i.load()&#xA;x, y = ???, ???&#xA;while ???:&#xA;    oldX, oldY = x, y&#xA;    if ???:&#xA;        ???&#xA;    elif ???:&#xA;        ???&#xA;    elif ???:&#xA;        ???&#xA;    else:&#xA;        ???&#xA;    d[oldX, oldY] = (0, 0, 0)&#xA;for x in range(i.size[0]):&#xA;    for y in range(i.size[1]):&#xA;        if ???:&#xA;            d[x, y] = (255, 255, 255)&#xA;i.save('revealed.png')&lt;/code&gt;&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1811" LastEditorUserId="1811" LastEditDate="2015-03-16T20:17:09.383" LastActivityDate="2015-03-17T08:14:08.017" Title="What's special about this image?" Tags="|steganography|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10335" PostTypeId="2" ParentId="10334" CreationDate="2015-03-16T04:30:20.993" Score="16" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Putting it through a &lt;a href=&quot;http://magiceye.ecksdee.co.uk/&quot; rel=&quot;nofollow noreferrer&quot;&gt;magic eye finder&lt;/a&gt; gives a faint image of a smiling face, perhaps the Mona Lisa.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/lXqlam.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-03-16T04:30:20.993" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10336" PostTypeId="1" AcceptedAnswerId="10347" CreationDate="2015-03-16T04:31:42.190" Score="8" ViewCount="318" Body="&lt;p&gt;Here are the dates for a series of events. As you can see, some of the dates are missing.&lt;br&gt;&#xA;What are the missing dates. And what happened at these dates? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$~~~????,~ 1914,~ ????,~ 1939,~ 1958,~ ????,~ 1978,~ 1978,~ 2005,~ 2013$ &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user10001" LastEditorUserId="8874" LastEditDate="2016-03-16T09:20:49.053" LastActivityDate="2016-03-16T13:38:33.060" Title="Which are the missing dates?" Tags="|knowledge|number-sequence|" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10337" PostTypeId="1" AcceptedAnswerId="10380" CreationDate="2015-03-16T04:32:24.150" Score="7" ViewCount="1051" Body="&lt;p&gt;An evil wizard has captured and thrown you into a dungeon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not a normal dungeon though. It's magical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each room has an associated &quot;level&quot; and 9 doors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first door can be used to go to the previous room IFF the room level is EVEN.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the other doors have a number on top of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each time you walk through a door with a number you get to a new room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The level of the new room is one less than the level of the room you came from. (e.g. if in a room with level 3 and walk through a door with a number you will end up in a room with level 2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The room you start [the room the evil wizard has dropped you] in is looks like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 402  602  701&#xA; 301    3  699&#xA; 299  398  598&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The center, 3, is the room level.&#xA; The first door is not show, but it's there. Also, because the level of this room is 3 (i.e. odd) the first door is locked.&#xA; All the other numbers around the center are the numbers on associated with the other doors in this room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind that if you reach a room with a &lt;strong&gt;level of 0 you instantly die.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being a perfect logical being and given the rules above, you go from room level 3 through all the doors to all reachable rooms with a level of 2 and build the following map.&#xA; The arrows connect the door with the room you reach. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                    504  704  803&#xA;                    403    2  801&#xA;                    401  500  700&#xA;                          ^&#xA;                          |&#xA;   304  504  603          |          603  803  902&#xA;   203    2  601 &amp;lt;---+    |    +---&amp;gt; 502    2  900&#xA;   201  300  500     |    |    |     500  599  799&#xA;                     |    |    |&#xA;                     |    |    |&#xA;                     +    +    +&#xA;203  403  502       402  602  701       601  801  900&#xA;102    2  500 &amp;lt;----+301    3  699+----&amp;gt; 500    2  898&#xA;100  199  399       299  398  598       498  597  797&#xA;                     +    +    +&#xA;                     |    |    |&#xA;                     |    |    |&#xA;   201  401  500     |    |    |     500  700  799&#xA;   100    2  498 &amp;lt;---+    |    +---&amp;gt; 399    2  797&#xA;    98  197  397          |          397  496  696&#xA;                          |&#xA;                          v&#xA;                    300  500  599&#xA;                    199    2  597&#xA;                    197  296  496&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;To survive/escape you need to find the door marked with number 1004 and walk trough it.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;From the initial room, what sequence of doors do you take to escape?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  there is a link between the room numbers and the way a knight moves&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9969" LastEditorUserId="9969" LastEditDate="2015-03-17T04:02:20.253" LastActivityDate="2015-03-17T04:02:20.253" Title="The Wizard's Gambit" Tags="|logical-deduction|calculation-puzzle|enigmatic-puzzle|" AnswerCount="3" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10338" PostTypeId="2" ParentId="10311" CreationDate="2015-03-16T04:46:23.753" Score="10" Body="&lt;p&gt;Here's a proof that 26 right answers is optimal for the worst case.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The adversary splits the deck into 13 packets of four cards of different suits, and stacks them on top of each other. Houdini, of course, knows this, and this restriction can only help Houdini.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Without loss of generality for the worst case, Houdini's strategy is deterministic, and the adversary knows it and can pick cards dynamically against it. So, the adversary knows what suit Houdini will guess if his assistant transmits 0, and same for 1.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; For the first two cards of the four in the packet, the adversary chooses neither of those two suits Houdini might guess. The remaining two he can't prevent this way. This limits Houdini to 26/52 cards right.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="63414" LastEditDate="2020-10-19T19:33:44.233" LastActivityDate="2020-10-19T19:33:44.233" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="10339" PostTypeId="2" ParentId="10336" CreationDate="2015-03-16T04:50:08.860" Score="5" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;1899, 1915 and 1950&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All those years are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the beginning years of all wars in 20th and 21st centuries involved by USA.&lt;br /&gt;&lt;br /&gt;Philippine-American War(1899-1902)&lt;br /&gt;World War I(1914-1918)&lt;br /&gt;Armenian Genocide(1915-1918)&lt;br /&gt;Word War II(1939-1945)&lt;br /&gt;The Korean War(1950)&lt;br /&gt;Vietnam War/Lebanon crisis(1958)&lt;br /&gt;Afghan War (1978)&lt;br /&gt;South Lebanon conflict(1978)&lt;br /&gt;Iraq Crisis(2005)&lt;br /&gt;Syrian Civil War(2013)&lt;br /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="2730" LastEditDate="2016-03-16T13:38:33.060" LastActivityDate="2016-03-16T13:38:33.060" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10340" PostTypeId="2" ParentId="10337" CreationDate="2015-03-16T04:52:16.617" Score="6" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Go through the &quot;top right&quot; door, labeled 701. Then go through the &quot;top middle door&quot; labeled 803. The &quot;top right&quot; door should now be 1004, while the middle door is just 1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rationale:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Passing through each door does something to all the numbers. The central number always decreases by one, but all the other numbers are affected in the same way. For example, going through the &quot;bottom middle&quot; door (398) decreases all numbers by 102. Going through the &quot;middle right&quot; door (699) increases all numbers by 199.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Going through door 701 increases all numbers by 201. Then going through the top middle door (which is now 803) every number should increase by 102.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &quot;upper right&quot; door will go from 701-&gt;902-&gt;1004 while the level of the room has decreased 3-2-1. The caveat of only being able to go backwards if the room is of even level is just to make sure you can explore the first steps, while not being able to go two steps and then come backwards to just keep exploring to find the right door.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastEditorUserId="1620" LastEditDate="2015-03-16T07:39:52.693" LastActivityDate="2015-03-16T07:39:52.693" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10341" PostTypeId="1" AcceptedAnswerId="10344" CreationDate="2015-03-16T06:42:44.470" Score="38" ViewCount="3740" Body="&lt;p&gt;On the sunny morning of April 5th, 2063, an alien spaceship started orbiting the Earth. Among the cries of &quot;It's an alien spaceship!&quot; and &quot;Ohm my god!&quot; from the people, governments noticed that they were beaming down a colorful message:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hsP0X.png&quot; alt=&quot;V over I equals?&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, because of interference from the atmosphere, they were only left with this small fragment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Can you decode the fragment, and hence guess the entire transmission?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="7308" LastEditorUserId="8874" LastEditDate="2015-10-13T16:19:08.807" LastActivityDate="2015-10-13T16:19:08.807" Title="Resistance is futile" Tags="|cipher|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10342" PostTypeId="1" AcceptedAnswerId="10345" CreationDate="2015-03-16T08:43:34.403" Score="1" ViewCount="794" Body="&lt;p&gt;We are two sisters.&lt;br&gt;&#xA;Our difference is your best key.&lt;br&gt;&#xA;I am probably the bigger one&lt;br&gt;&#xA;and everything I’m not,&lt;br&gt;&#xA;is she.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have always opposite opinions,&lt;br&gt;&#xA;the reason of our loud fights.&lt;br&gt;&#xA;And if I start a quarrel with my sister,&lt;br&gt;&#xA;surely, you’ll see and hear&lt;br&gt;&#xA;our spites.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The root of our conflict is quite old.&lt;br&gt;&#xA;(Yes, we’re born in weird family.)&lt;br&gt;&#xA;As kids we fought once, so I’m told.&lt;br&gt;&#xA;And then what’s left was&lt;br&gt;&#xA;only me.   &lt;/p&gt;&#xA;" OwnerUserId="5283" LastEditorUserId="5188" LastEditDate="2015-03-24T04:52:32.300" LastActivityDate="2015-03-24T04:52:32.300" Title="Violent riddle with happy end" Tags="|riddle|word|rhyme|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10343" PostTypeId="2" ParentId="10341" CreationDate="2015-03-16T08:45:05.093" Score="13" Body="&lt;p&gt;I cannot open the image, but clue for everyone who didn't notice:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The header says RESISTANCE is futile and cries of 'OHM y god' OHM is a unit of resistance, it might be a clue&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9907" LastActivityDate="2015-03-16T08:45:05.093" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10344" PostTypeId="2" ParentId="10341" CreationDate="2015-03-16T09:02:26.730" Score="32" Body="&lt;p&gt;The date given, April 5 2063, is the &lt;a href=&quot;http://en.memory-alpha.org/wiki/Phoenix&quot; rel=&quot;noreferrer&quot;&gt;launch date of the Phoenix spaceship from Star Trek&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The title &amp;quot;Resistance is futile&amp;quot; also alludes to &lt;a href=&quot;http://en.wikipedia.org/wiki/Borg_%28Star_Trek%29#.22Resistance_is_futile.22&quot; rel=&quot;noreferrer&quot;&gt;another famous Star Trek phrase&lt;/a&gt;. Therefore, the answer is likely to be related to the Star Trek franchise in one way or another.&lt;/p&gt;&#xA;&lt;p&gt;Translating the code using &lt;img src=&quot;https://i.stack.imgur.com/dUtyn.gif&quot; alt=&quot;this table&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;results in the text:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;760 101+050 101+180&lt;/p&gt;&#xA;&lt;p&gt;101+010 320 101+080&lt;/p&gt;&#xA;&lt;p&gt;101+110 111 101+030&lt;/p&gt;&#xA;&lt;p&gt;320 970 111 012 320 (interpreting the 9 as a blank breaks the message)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Interpreting the bars similar to the 3-bar resistor code (with the least significant digit encoding the exponent of 10 when 3 bars exist) results in:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;76 100+5 100+18&lt;/p&gt;&#xA;&lt;p&gt;100+1 32 100+8&lt;/p&gt;&#xA;&lt;p&gt;100+11 110 100+3&lt;/p&gt;&#xA;&lt;p&gt;32 97 110 100 32&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;which decodes to ASCII&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Live long and (?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Assuming the context of Star Trek, the answer is probably&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;a href=&quot;http://en.memory-alpha.org/wiki/Vulcan_salute&quot; rel=&quot;noreferrer&quot;&gt;Live long and prosper&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7529" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-03-16T09:35:50.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10345" PostTypeId="2" ParentId="10342" CreationDate="2015-03-16T09:20:30.980" Score="3" Body="&lt;p&gt;It sounds like you're thinking of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a quantity of matter and a smaller mass of anti-matter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Two sisters, difference, everything I'm not:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Matter and anti-matter are related but opposite.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The bigger one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In your puzzle, there's more matter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Opposite opinions, loud fights, quarrel, hear:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Matter and anti-matter have opposite properties and react energetically to each other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Root of conflict, family, fight, remainder:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When matter and anti-matter react, they annihilate an equal amount of each other, so some of the initially more massive type will remain after the other has been completely used up.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9422" LastActivityDate="2015-03-16T09:20:30.980" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10346" PostTypeId="2" ParentId="10306" CreationDate="2015-03-16T10:57:22.460" Score="3" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;8&lt;/strong&gt;. &lt;strike&gt;Two pairs achieve this length: &lt;strong&gt;jingling&lt;/strong&gt; and &lt;strong&gt;khoikhoi&lt;/strong&gt; translated as &lt;strong&gt;54645464&lt;/strong&gt;, as well as&lt;/strike&gt; &lt;strong&gt;acacetin&lt;/strong&gt; and &lt;strong&gt;cacafugo&lt;/strong&gt; translated as &lt;strong&gt;22223846&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Brute-forced on a dictionary.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="5610" LastEditDate="2015-03-19T08:25:07.167" LastActivityDate="2015-03-19T08:25:07.167" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10347" PostTypeId="2" ParentId="10336" CreationDate="2015-03-16T12:27:16.537" Score="9" Body="&lt;p&gt;These years - ???, 1914, ???, 1939, 1958, ???, 1978, 1978, 2005, 2013 - are the years of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Sede Vacante&quot; (or: the starting years of new popes).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More precisely,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;1903&lt;/strong&gt; Pius X&lt;br&gt;&#xA; 1914 Benedict XV&lt;br&gt;&#xA; &lt;strong&gt;1922&lt;/strong&gt; Pius XI&lt;br&gt;&#xA; 1939 Pius XII&lt;br&gt;&#xA; 1958 John XXIII&lt;br&gt;&#xA; &lt;strong&gt;1963&lt;/strong&gt; Paul VI&lt;br&gt;&#xA; 1978 John Paul I&lt;br&gt;&#xA; 1978 John Paul II&lt;br&gt;&#xA; 2005 Benedict XVI&lt;br&gt;&#xA; 2013 Francis&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="5373" LastEditDate="2015-03-19T12:39:41.153" LastActivityDate="2015-03-19T12:39:41.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10348" PostTypeId="2" ParentId="10298" CreationDate="2015-03-16T12:46:05.520" Score="4" Body="&lt;p&gt;Sequence 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; looks like years, but actually is primes plus five.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The page &lt;a href=&quot;http://primes.utm.edu/lists/small/10000.txt&quot; rel=&quot;nofollow&quot;&gt;http://primes.utm.edu/lists/small/10000.txt&lt;/a&gt; lists the first 10.000 &#xA; primes:&lt;br&gt;&#xA; &lt;strong&gt;..., 1901, 1907, 1913, 1931, 1933, 1949, 1951, 1973, 1979, 1987, 1993, 1997, 1999, 2003, ...&lt;/strong&gt;&lt;br&gt;&#xA; By adding 5 to each number, this list turns into&lt;br&gt;&#xA; &lt;strong&gt;..., 1906, 1912, 1918, 1936, 1938, 1954, 1956, 1978, 1984, 1992, 1998, 2002, 2004, 2008, ...&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-03-16T12:46:05.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10350" PostTypeId="2" ParentId="10311" CreationDate="2015-03-16T13:50:50.063" Score="3" Body="&lt;p&gt;you have other information as well: the cards that have already been flipped. Houdini can get the 52nd card right regardless of the hint from the assistant. Perhaps he can also get the 51st card right if the one bit of information helps to distinguish between two possible remaining cards. For example, you could signal red/black and if what remained was a red and a black card, you would get the last two right for sure.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I think the 27 right strategy is to signal red/black with the extra win coming from knowing the last card since that is the only card left.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without memorizing the entire deck you also have &quot;the most recent card that was turned over&quot; to work with. You could signal higher/lower suit value from that, for example.&lt;/p&gt;&#xA;" OwnerUserId="1509" LastActivityDate="2015-03-16T13:50:50.063" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10353" PostTypeId="1" CreationDate="2015-03-16T14:59:59.060" Score="-3" ViewCount="1470" Body="&lt;p&gt;Alice, Bob, Chris, Dave, Eric, Fred and Laura used to work in a restaurant 6 days a week, from Tuesday to Sunday (Monday is a day off). Every one works at least for one day, and at least two people work on each day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering the following restrictions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Either Alice or Bob work on the first day;&lt;/li&gt;&#xA;&lt;li&gt;Laura works only on the fourth and fifth day;&lt;/li&gt;&#xA;&lt;li&gt;A one-day gap is left between Laura and Chris work days;&lt;/li&gt;&#xA;&lt;li&gt;Eric works on the days immediately following the ones in which Bob works.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What is the working schedule of the seven employees? A good explanation of the solution to this puzzle, rather than only the right answer, is kindly welcome.&lt;/p&gt;&#xA;" OwnerUserId="10252" LastEditorUserId="10248" LastEditDate="2015-03-16T16:02:53.560" LastActivityDate="2015-03-16T16:46:20.443" Title="Working schedule (with explanation)" Tags="|logical-deduction|calculation-puzzle|explanation|" AnswerCount="1" CommentCount="7" ClosedDate="2015-03-16T16:50:12.310" ContentLicense="CC BY-SA 3.0" />
  <row Id="10354" PostTypeId="1" AcceptedAnswerId="10356" CreationDate="2015-03-16T15:01:28.933" Score="3" ViewCount="700" Body="&lt;p&gt;I am so small that you can't see,&lt;br&gt;&#xA;The building blocks that make up me.&lt;br&gt;&#xA;Yet miles long is the true case,&lt;br&gt;&#xA;I'm full of holes, just like lace.&lt;br&gt;&#xA;Small changes can be good or bad,&lt;br&gt;&#xA;They make you strong or really sad.&lt;br&gt;&#xA;In you, your dog, cat, and mouse,&lt;br&gt;&#xA;Quite a lot inside your house.&lt;br&gt;&#xA;So let's determine now, account,&lt;br&gt;&#xA;Grasp the last line's clue and mount.    &lt;/p&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-03-16T16:06:27.813" Title="To the Basics - What am I?" Tags="|riddle|word|rhyme|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10355" PostTypeId="2" ParentId="10354" CreationDate="2015-03-16T15:08:45.153" Score="1" Body="&lt;p&gt;I think that it could be...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DNA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastActivityDate="2015-03-16T15:08:45.153" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10356" PostTypeId="2" ParentId="10354" CreationDate="2015-03-16T15:14:49.637" Score="1" Body="&lt;p&gt;I think it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DNA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am so small that you can't see&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The building blocks that make up me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We all are build from DNA blocks, each is few nanometers long&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet miles long is the true case&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In some sources you can find that unchained DNA sequence is up to 3.85 × 10^8 meters long&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm full of holes, just like lace&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DNA chain looks likes shoe lace&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Small changes can be good or bad&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They make you strong or really sad&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some changes in DNA sequence can be useful (for example, evolution make you adapted to envinronment), or not (mutation can cause diseases)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In you, your dog, cat, and mouse&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quite a lot inside your house&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DNA is in every living (human, animal, plant)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10181" LastActivityDate="2015-03-16T15:14:49.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10357" PostTypeId="2" ParentId="10354" CreationDate="2015-03-16T15:16:44.097" Score="1" Body="&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;DNA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am so small that you can't see,&lt;br&gt;&#xA;The building blocks that make up me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Without an electronic microscope is quite difficult to actually see the four bases.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet miles long is the true case,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;DNA is &lt;a href=&quot;http://hypertextbook.com/facts/1998/StevenChen.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;quite long&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm full of holes, just like lace.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/rxVkH.png&quot; alt=&quot;Yep&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Small changes can be good or bad, &#xA;They make you strong or really sad. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Little mutations in the DNA can lead to every kind of emotion/status&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In you, your dog, cat, and mouse,&#xA;Quite a lot inside your house.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every living creature has DNA structure&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So let's determine now, account,&#xA;Grasp the last line's clue and mount.&lt;/p&gt;&#xA;" OwnerUserId="4652" LastActivityDate="2015-03-16T15:16:44.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10358" PostTypeId="2" ParentId="8212" CreationDate="2015-03-16T15:20:16.443" Score="8" Body="&lt;p&gt;Using words from the &lt;a href=&quot;http://en.wikipedia.org/wiki/SOWPODS&quot; rel=&quot;noreferrer&quot;&gt;SOWPODS&lt;/a&gt; word list, it's possible to solve Q1 and Q2 but sadly not Q3. I checked this by creating a graph of k-letter isograms, connecting words with no shared letters, and searching for cliques.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Five 5-letter words&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lots of answers, all using obscure words: e.g.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;brick, jumpy, vozhd, glent, waqfs&#xA;vibex, fjord, nymph, waltz, gucks&#xA;bling, jumpy, vozhd, treck, waqfs&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Four 6-letter words&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lots of answers, some using slightly less obscure words: e.g.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;jawbox, kvetch, flumps, drying&#xA;jumbly, dwarfs, kvetch, poxing&#xA;jawbox, fledgy, skrump, chintz&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Three 7-letter words&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lots and lots of answers: e.g.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;stumped, flyback, whoring&#xA;mucking, batfowl, zephyrs&#xA;jordans, phlegmy, fuckwit&#xA;overply, dumbing, thwacks&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Two 10-letter words&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;blacksmith, gunpowdery&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="10220" LastActivityDate="2015-03-16T15:20:16.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10363" PostTypeId="2" ParentId="10231" CreationDate="2015-03-16T16:04:05.500" Score="5" Body="&lt;p&gt;I think I've got it now!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PROSTATE.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I start out quite skilled,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A &lt;strong&gt;PRO&lt;/strong&gt; is a skilled person.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Throughout the nation I've drilled,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Certain nations are composed of &lt;strong&gt;STATE&lt;/strong&gt;s.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I'm sexist to the extreme, &lt;br&gt;&#xA; You'll soon know what I mean,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Only men have prostate glands.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Your children should thank me,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The prostate is part of the reproductive system.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Your wife might just hate me.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If prostate cancer kills you, your wife won't be very happy (the OP had to help me with this one!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Knives I despise&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Does this say anything more than the obvious? Nobody wants to be cut with a knife, especially in such a tender part of their body.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I hide from your eyes.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's invisible most of the time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Avoid mutations, &lt;br&gt;&#xA; From your medications.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are probably medications that can cause mutations of the prostate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;So we will do all your tests,&lt;br&gt;&#xA; But eventually we'll both rest.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A man and his prostate may together undergo tests for prostate cancer, and eventually both will rest from these tests (or be dead?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-16T16:31:39.797" LastActivityDate="2015-03-16T16:31:39.797" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10364" PostTypeId="2" ParentId="10354" CreationDate="2015-03-16T16:06:27.813" Score="4" Body="&lt;p&gt;Just an answer to catch the last clue:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So let's determine now, account&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This gives the answer as an&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; acronym: &quot;&lt;strong&gt;D&lt;/strong&gt;etermine &lt;strong&gt;N&lt;/strong&gt;ow, &lt;strong&gt;A&lt;/strong&gt;ccount&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Welcome to the fold, fellow riddler! :-) This is what's known as a &quot;lynch mob&quot; clue, because when you've spent ages working out the answer, it can be infuriating (in a good sense) to realise that it was staring you in the face all the time!&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-03-16T16:06:27.813" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10366" PostTypeId="2" ParentId="10216" CreationDate="2015-03-16T16:31:03.763" Score="4" Body="&lt;h2&gt;Partial solution:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;As already answered in the comments, the equations solve to the following numbers (prime factors),(all factors up to 50):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1141 (7,163),(7)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1314 (2,3,3,73),(2,3,6,9,18)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$600 (2,2,2,3,5,5),(2,3,4,5,6,8,10,12,15,20,24,25,30,40,50)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$200 (2,2,2,5,5),(2,4,5,8,10,20,25,40,50)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$115 (5,23),(5,23)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$282 (2,3,47),(2,3,6,47)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$492 (2,2,3,41),(2,3,4,6,12,41)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1716 (2,2,3,11,13),(2,3,4,6,11,12,13,22,26,33,39,44)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1650 (2,3,5,5,11),(2,3,5,6,10,11,15,22,25,30,33,50)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$320 (2,2,2,2,2,2,5),(2,4,5,8,10,16,20,32,40)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$906 (2,3,151),(2,3,6)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No number is a factor of all the answers, and 40 is a factor of 3 of the numbers. &lt;/p&gt;&#xA;" OwnerUserId="7529" LastEditorUserId="7529" LastEditDate="2015-03-16T16:38:00.580" LastActivityDate="2015-03-16T16:38:00.580" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10367" PostTypeId="1" AcceptedAnswerId="11144" CreationDate="2015-03-16T16:46:15.270" Score="0" ViewCount="702" Body="&lt;p&gt;Here are four more sequence puzzles. The challenge is to guess the pattern and hence fill in the missing terms. Each &quot;???&quot; denotes exactly one missing term.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;6, ???, 5, 8, 4, ???, 7, 3, 4, ???, 4 &lt;em&gt;(now solved by &lt;strong&gt;March Ho&lt;/strong&gt;)&lt;/em&gt;&lt;br&gt;&lt;/li&gt;&#xA;&lt;li&gt;Henry, Lily, Ben, Ned, Natalie, ???, Albert, Simon &lt;em&gt;(now solved by &lt;strong&gt;KSab&lt;/strong&gt;)&lt;/em&gt;&lt;br&gt;&lt;/li&gt;&#xA;&lt;li&gt;73, 74, 88, 51, 28, 36, 84, 50, 83, 99, ??? &lt;em&gt;(now solved by &lt;strong&gt;Len&lt;/strong&gt;)&lt;/em&gt;&lt;br&gt;&lt;/li&gt;&#xA;&lt;li&gt;40, 30, 70, 39, 54, 74, 23, ??? &lt;em&gt;(whoever solves this will get the accepted answer)&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Please let me know whether these are too easy, too hard, or neither! I've tried to arrange them in order of difficulty, but I could be completely wrong with this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint (in riddle form):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The same thing coming more than once,&lt;br&gt;&#xA; Sometimes coming once a month.&lt;br&gt;&#xA; Then the wood on which you eat,&lt;br&gt;&#xA; Perhaps excelling at what it does.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-04-12T23:31:36.303" LastActivityDate="2015-04-28T08:41:40.403" Title="A benevolent but cumbersome batch of four sequence puzzles" Tags="|riddle|pattern|number-sequence|letter-sequence|" AnswerCount="3" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="10368" PostTypeId="2" ParentId="10353" CreationDate="2015-03-16T16:46:20.443" Score="3" Body="&lt;p&gt;So here is a solution that I can come up with for the puzzle.  It is not the only solution, as you will see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, I will create a table with the possibilities:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;        Tu   We   Th   Fr   Sa   Su&#xA;Alice&#xA;Bob&#xA;Chris&#xA;Dave&#xA;Eric&#xA;Fred&#xA;Laura&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now lets look at the clues.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;2&quot;&gt;&#xA;  &lt;li&gt;Laura works only on the fourth and fifth day&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So we can modify her row in the table to this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;        Tu   We   Th   Fr   Sa   Su&#xA;Laura   X    X    X    *    *    X&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;3&quot;&gt;&#xA;  &lt;li&gt;A one-day gap is left between Laura and Chris work days&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can determine Chris must work on a Wednesday, the only day for which there is a one day gap.  Chris may work on Tuesdays as well, but the other days are eliminated.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;        Tu   We   Th   Fr   Sa   Su&#xA;Chris   ?    *    X    X    X    X&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;4&quot;&gt;&#xA;  &lt;li&gt;Eric works on the days immediately following the ones in which Bob works.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We know that Eric cannot be on Tuesday, since he must work the day after Bob.  Also, Bob cannot be Sunday because Erik must work after.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here is our complete table so far:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;        Tu   We   Th   Fr   Sa   Su&#xA;Alice   ?    ?    ?    ?    ?    ?&#xA;Bob     ?    ?    ?    ?    ?    X&#xA;Chris   ?    *    X    X    X    X&#xA;Dave    ?    ?    ?    ?    ?    ?&#xA;Eric    X    ?    ?    ?    ?    ?&#xA;Fred    ?    ?    ?    ?    ?    ?&#xA;Laura   X    X    X    *    *    X&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This brings us to the end of the hard constraints.  Now, we have freedom to fill make some choices.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Either Alice or Bob work on the first day&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lets say Alice works Tuesday, and that Bob does not.  This also affects the days Eric can work.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;        Tu   We   Th   Fr   Sa   Su&#xA;Alice   *    ?    ?    ?    ?    ?&#xA;Bob     X    ?    ?    ?    ?    X&#xA;Chris   ?    *    X    X    X    X&#xA;Dave    ?    ?    ?    ?    ?    ?&#xA;Eric    X    X    ?    ?    ?    ?&#xA;Fred    ?    ?    ?    ?    ?    ?&#xA;Laura   X    X    X    *    *    X&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now, notice not everyone is working - Bob, Dave, Eric and Fred are not working.  Also, all the days need at least one more worker, some (Thursday and Sunday) need two.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only Bob and Eric place requirements on each other, so lets fit them in the schedule first, with Bob on Thursday, and thus Eric on Friday.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;        Tu   We   Th   Fr   Sa   Su&#xA;Alice   *    ?    ?    ?    ?    ?&#xA;Bob     X    X    *    X    ?    X&#xA;Chris   ?    *    X    X    X    X&#xA;Dave    ?    ?    ?    ?    ?    ?&#xA;Eric    X    X    X    *    ?    ?&#xA;Fred    ?    ?    ?    ?    ?    ?&#xA;Laura   X    X    X    *    *    X&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now, we need one more worker on each of Tuesday, Wednesday, Thursday, and Saturday.  Sunday needs two workers.  Everyone is working at least once except for Dave and Fred. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So lets use them to finish a minimal schedule.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;        Tu   We   Th   Fr   Sa   Su&#xA;Alice   *                         &#xA;Bob               *               &#xA;Chris        *                    &#xA;Dave    *         *              *&#xA;Eric                   *          &#xA;Fred         *              *    *&#xA;Laura                  *    *     &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is a valid schedule, but it is by no means the only one.  Dave and Fred, for example, could work every day instead and the schedule would still be valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it stands, this question is not well defined; there are too many possible solutions.  Perhaps there are some missing constraints?  Some examples could include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;everyone must work the same number of days&lt;/li&gt;&#xA;&lt;li&gt;some restrictions on dave and/or eric &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2015-03-16T16:46:20.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10369" PostTypeId="2" ParentId="10218" CreationDate="2015-03-16T16:52:10.160" Score="6" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SAFE-CRACKER&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Cached in secrecy, &lt;br&gt;&#xA; Where none shall interfere&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is a good description of &lt;strong&gt;safe&lt;/strong&gt;ness.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A man with wooden jaw &lt;br&gt;&#xA; Waits patiently each year&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As &lt;a href=&quot;https://puzzling.stackexchange.com/a/10225/5373&quot;&gt;Nathan&lt;/a&gt; already realised, these lines refer to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a nut&lt;strong&gt;cracker&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Hands flung abroad,&lt;br&gt;&#xA;As the masked man cries &quot;unscathed!&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The OP's 3rd hint tells us that this is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a &lt;strong&gt;baseball&lt;/strong&gt; reference.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/10218/the-man-with-a-wooden-jaw-riddle/10369?noredirect=1#comment32052_10369&quot;&gt;bubalou&lt;/a&gt; for telling me that more specifically,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it's a reference to an umpire calling a runner &lt;strong&gt;safe&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In an abode of luxury,&lt;br&gt;&#xA; With reservations made&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think this is a Christmas &lt;strong&gt;cracker&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The OP's first hint tells us that verses 1 and 2 are two ways of describing the same thing. That thing is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a safe-cracker.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Combatants from a land&lt;br&gt;&#xA; Found on distant shores&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is a reference to the nickname &lt;strong&gt;Charlie&lt;/strong&gt; for Viet Cong fighters (thanks to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/10218/the-man-with-a-wooden-jaw-riddle/10369?noredirect=1#comment32040_10369&quot;&gt;Duncan&lt;/a&gt; for putting me onto the reason for this).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;With an intruder well acquainted&lt;br&gt;&#xA; With the sound of people's snores&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is Z, as in &quot;zzz&quot; snoring. The OP is American, so it's &quot;&lt;strong&gt;zee&lt;/strong&gt;&quot; rather than &quot;zed&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A shape at the heart&lt;br&gt;&#xA; Of an ages-old tale&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter &lt;strong&gt;T&lt;/strong&gt; (as in the Christian cross).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Followed by what's great and blue,&lt;br&gt;&#xA; But isn't a whale&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A &lt;a href=&quot;http://animals.nationalgeographic.com/animals/birds/great-blue-heron/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;heron&lt;/strong&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the final two verses give&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Charlize_Theron&quot; rel=&quot;nofollow noreferrer&quot;&gt;Charlize Theron&lt;/a&gt;, who played a safe-cracker in the 2003 film &lt;a href=&quot;http://en.wikipedia.org/wiki/The_Italian_Job_(2003_film)&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;The Italian Job&lt;/em&gt;&lt;/a&gt;. Confirmation!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-03-20T00:28:36.037" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="10370" PostTypeId="2" ParentId="10367" CreationDate="2015-03-16T17:07:29.493" Score="6" Body="&lt;h2&gt;Partial answer:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;1: Decimal sequence expansion of $10-\pi$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6,&lt;strong&gt;8&lt;/strong&gt;,5,8,4,&lt;strong&gt;0&lt;/strong&gt;,7,3,4,&lt;strong&gt;6&lt;/strong&gt;,4 &lt;a href=&quot;https://oeis.org/A030644&quot;&gt;OEIS 030644&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7529" LastActivityDate="2015-03-16T17:07:29.493" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10373" PostTypeId="2" ParentId="8612" CreationDate="2015-03-16T17:17:03.413" Score="3" Body="&lt;p&gt;Here are some improvement/additions to @Bret's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Voiced (9 letters)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;brobdingnagians: b, r, b, d, g, n, g, n, z&lt;/li&gt;&#xA;&lt;li&gt;environmentalism: n, v, r, n, m, n, l, z, m (in t-dropping pronunciations)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Voiced (8 letters)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;alexandrines: l, g, z, n, d, r, n, z&lt;/li&gt;&#xA;&lt;li&gt;bilingualism: b, l, n, g, w, l, z, m&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Unvoiced (7 letters)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;statistics: s, t, t, s, t, k, s&lt;/li&gt;&#xA;&lt;li&gt;sophisticates: s, f, s, t, k, t, s&lt;/li&gt;&#xA;&lt;li&gt;stethoscopes: s, t, θ, s, k, p, s&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="10220" LastActivityDate="2015-03-16T17:17:03.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10374" PostTypeId="1" AcceptedAnswerId="10376" CreationDate="2015-03-16T17:39:51.040" Score="13" ViewCount="1511" Body="&lt;p&gt;Anna and Isadora play a game by writing letters on the blackboard.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In the first round, Anna starts by writing down the letter $A$ or the letter $I$.&lt;/li&gt;&#xA;&lt;li&gt;In the $n$-th round with $2\le n\le 999$, Anna first writes down a new letter $A$ or $I$ immediately to the right of the other letters.&#xA;Then Isadora chooses two letters already on the blackboard and swaps them.&lt;/li&gt;&#xA;&lt;li&gt;Isadora wins, if at the end of the $999$-th round the $999$ letters form a palindrome (= a symmetric letter sequence). Otherwise Anna wins.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Questions: Can Anna enforce a win? Can Isadora enforce a win? &#xA;(As usual, we assume that Anna and Isadora use optimal strategies.)&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-03-19T17:56:34.520" LastActivityDate="2015-03-19T17:56:34.520" Title="Who wins the AI-game?" Tags="|strategy|game|palindrome|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10376" PostTypeId="2" ParentId="10374" CreationDate="2015-03-16T18:13:42.157" Score="13" Body="&lt;p&gt;Isadora wins with this strategy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In rounds $2$ to $500$, she can swap letters at random, or swap two of&#xA; the same letter.  It doesn't matter what she does.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Her strategy from round $501$ onward uses the following observations:&#xA; 1) a sequence of three letters that is not a palindrome can be turned&#xA; into a palindrome by swapping the middle letter with either the first&#xA; or third letter, and 2) a sequence of three letters that is a&#xA; palindrome remains a palindrome if the first and third letters are&#xA; swapped.&lt;br&gt;&#xA;&lt;br&gt;&#xA; 2) is obvious: a palindromic sequence of three letters must have the&#xA; same first and third letter, so swapping them does not alter the&#xA; sequence.&lt;br&gt;&#xA;&lt;br&gt;&#xA; For 1), we just need to check these cases:&#xA;&lt;br&gt;&#xA; - AII: swap first and second: IAI&lt;br&gt;&#xA; - IAA: swap first and second: AIA&lt;br&gt;&#xA; - IIA: swap second and third: IAI&lt;br&gt;&#xA; - AAI: swap second and third: AIA&lt;br&gt;&#xA;&lt;br&gt;&#xA; In round $501$, Isabella turns the middle three letters of the final&#xA; sequence of letters (the last three on the board after Anna writes a&#xA; new letter, which will be in positions $499$, $500$, and $501$ in the&#xA; final sequence) into a palindrome if they are not already.&lt;br&gt;&#xA;&lt;br&gt;&#xA; In the $n$-th round, for $502 \le n \le 999$ , Isadora considers only&#xA; characters $n$, $500$, and $1000-n$.  She can apply observations 1 and&#xA; 2 to ensure that these three letters could be part of a palindrome:&#xA; that is, that the letter in position $n$ is the same as the letter in&#xA; position $1000-n$.  If Isadora does this every round until the end of&#xA; the game, this guarantees that after round $n$, the letters from&#xA; position $1000-n$ to position $n$ on the board are palindromic.  After&#xA; the $999$-th round, all the letters on the board will be a palindrome.&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Anna cannot do anything to stop this strategy.  On her turn she cannot&#xA; alter the existing sequence except to append a new letter, and&#xA; whichever letter she appends on round $n$, Isadora can ensure that the&#xA; letters from $1000-n$ to $n$ are a palindrome.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10261" LastEditorUserId="8874" LastEditDate="2015-03-19T12:25:05.070" LastActivityDate="2015-03-19T12:25:05.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10378" PostTypeId="2" ParentId="10374" CreationDate="2015-03-16T18:38:23.527" Score="3" Body="&lt;p&gt;Isadora will never&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; lose&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; She skips the first 500 turns(because 500th character is the middle one)&lt;br&gt;&#xA; Then, starting from turn 501(and every other turn $t$) she should start making palindromes of the last $2(t-500) + 1$ characters.&lt;br&gt;&#xA; For $t = 501$ we get: $2(501-500) + 1 = 3$ last characters that she should form into a palindrome.&lt;br&gt;&#xA; So let's say 499th character is $A$, 500th is $A$ and 501st is $I$. In this case she should swap 500th and 501st characters and last 3 characters will form a palindrome. Repeat for every next step.&lt;br&gt;&#xA; If Anna puts a letter that will make a palindrome of last $2(t-500)+1$ characters(so in previous example if 501st char would be $A$), Isadora should do nothing and wait for the next move.&lt;br&gt;&#xA; If Anna puts a letter that will not make a palindrome of last $2(t-500)+1$ characters, Isadora should do one of the following:&lt;br&gt;&#xA; If $charAt(500-(t-500)) \neq charAt(t)$, then&lt;br&gt;&#xA; If $charAt(500) \neq charAt(t)$, then&lt;br&gt;&#xA; Swap characters 500 and t,&lt;br&gt;&#xA; Else&lt;br&gt;&#xA; Swap characters (500-(t-500)) and t&lt;br&gt;&#xA; And she will get a palindrome again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9352" LastActivityDate="2015-03-16T18:38:23.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10379" PostTypeId="1" AcceptedAnswerId="10407" CreationDate="2015-03-16T19:00:12.633" Score="7" ViewCount="1086" Body="&lt;p&gt;You are a spy trying to find out the job of a man named John. One day, you go up to John's friend, Joe. You tell Joe that if he doesn't tell you what John's job is, you will destroy him. Not wanting to die, Joe says &quot;OK&quot;, but there is one rule - decode this message, and you will find his job. You are given a sheet marked 'Table' that reads the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[79] [D8] [1B] [3A] [22]&#xA;[P9] [22] [24] [ZP] [21]&#xA;[31] [P9] [60] [33] [1C]&#xA;[D0] [43] [43] [21] [27]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;He then gives you another sheet marked 'Path', which reads as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;74,74,74-72,6B-74,74-72,6B-74-6B,6B,6B,6B-72,75,74,74,72-74&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What is John's job?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Uri worked out the path. I'm now going to give you a clue as to what the path means. The path &lt;em&gt;actually&lt;/em&gt; tells you how to move about the table. So, 'l' is 'move left', 'r' is move right, 'u' move up and 'd' move down. Now you just have to decipher the table...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The table is encrypted by converting letters to their respective hex code. This hex code is what a PS/2 keyboard sends to a computer. The computer then decrypts it, and turns it into a nice strong of letters. Come on, this is a &lt;strong&gt;BIG&lt;/strong&gt; hint, so hopefully now someone will get it. If not, I'll be adding another hint soon!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9377" LastEditorDisplayName="user9377" LastEditDate="2015-08-19T14:36:35.053" LastActivityDate="2015-08-19T14:36:35.053" Title="What is John's job?" Tags="|cipher|computer-puzzle|steganography|" AnswerCount="2" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="10380" PostTypeId="2" ParentId="10337" CreationDate="2015-03-16T19:15:57.213" Score="2" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Any combination of N x1 and NE x2 doors to escape.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you subtract the average value of the room's doors from each door, all rooms look like the following.&lt;br&gt;&#xA;&lt;br&gt;&#xA; -98 | 102 | 201&lt;br&gt;&#xA; -199 | XXX | 199&lt;br&gt;&#xA; -201 | -102 | 98&lt;br&gt;&#xA;&lt;br&gt;&#xA; You can see that backtracking will always get you to where you were, and any combinations of similar moves will also bring you to the same room (ie, N&gt;E is the same as E&gt;N).&lt;br&gt;&#xA;&lt;br&gt;&#xA; Using this we just add any combination of 3 or fewer moves to 500 in order to find 1004. 500 + 102 + 201x2 = 1004, so you need to go N once and NE twice.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This should be the general solution to the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Math:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 102*(North-South) + 201*(NorthEast-SouthWest) + 199*(East-West) + 98*(SouthEast-NorthWest) + 500 = 1004&lt;br&gt;&#xA; North+South + NorthEast+SouthWest + East+West + SouthEast+NorthWest = 3&#xA; Solve that system and you get your values for the directions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10263" LastEditorUserId="10263" LastEditDate="2015-03-16T19:47:29.697" LastActivityDate="2015-03-16T19:47:29.697" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10381" PostTypeId="1" CreationDate="2015-03-16T19:16:18.653" Score="-5" ViewCount="327" Body="&lt;p&gt;There are 250 scraps. For each 11 removed, one extra can be obtained. How much maximum is possible, if one cut by one unit? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read this somewhere, and one guy gave the answer as 25, but I can't understand why. Can anyone please explain?&lt;/p&gt;&#xA;" OwnerUserId="10260" LastEditorUserId="5373" LastEditDate="2015-03-16T21:10:41.247" LastActivityDate="2015-03-21T01:01:51.503" Title="Scrap puzzle to count how many" Tags="|logical-deduction|calculation-puzzle|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10382" PostTypeId="2" ParentId="10374" CreationDate="2015-03-16T19:17:17.420" Score="2" Body="&lt;p&gt;Isadora can eventually&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;WIN&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Lets assume A is 0 and I=1&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Lets note a whichever sequence 00110... = AeI where e is the middle of sequence since 999 is odd&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;regardless of Ana's final sequence , Isadora can switch each two digits she wants after each round , n-1 times , it means an even number of swaps , that means the final sequence we have is : &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;strong&gt;(AeI) $\oplus$ C&lt;/strong&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When isadora swaps the 2nd and 4th digits this operation takes place:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;abcd $\oplus$ 0101&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When isadora swaps the fifth and 4th digits after anna's next move this operation would take place:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;abcde $\oplus$ 0101 $\oplus$ 00011 = abcde $\oplus$ 01001&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That does mean&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;strong&gt;C contains an even occurence of 1&lt;/strong&gt; &lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;lets sepatare C into ( C1 c C2 )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AeI $\oplus$ C = (A $\oplus$ C1) (e $\oplus$ c) (I $\oplus$ C2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The winning condition for Isadora is :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;strong&gt;A $\oplus$  C1 = I $\oplus$ C2&lt;/strong&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;there s many solutions can verify this equation , lets take an example of C1=I and C2=A&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if the number of 1 is odd in C1 C2 we make c=1 and it wouldnt change anything since it is the middle of an odd sequence.&lt;/p&gt;&#xA;" OwnerUserId="9517" LastEditorUserId="9517" LastEditDate="2015-03-17T10:25:13.113" LastActivityDate="2015-03-17T10:25:13.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10384" PostTypeId="2" ParentId="10381" CreationDate="2015-03-16T19:58:30.073" Score="1" Body="&lt;p&gt;Assuming I understand your puzzle correctly:&#xA;You have 250 'things'. Each time you remove 11, you get back one. How many 11s can you remove?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that each time you remove 11, you are essentially removing 10. So,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 250/10 = 25 removals&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7715" LastActivityDate="2015-03-16T19:58:30.073" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10385" PostTypeId="1" AcceptedAnswerId="10400" CreationDate="2015-03-16T20:06:45.497" Score="8" ViewCount="3426" Body="&lt;p&gt;You say I’m a breath of fresh air...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then why do you spit on me? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do you break me down so loudly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do you trap me in the darkest of places?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My skin may be pure white, but is my face not also red?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My cousin may be spicy hot but am I not sweet?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: So, this is my first ever riddle and with all of these great answers, I think the riddle may not have been as specific as I thought it originally was. So, I'll try to add specificity.&lt;/p&gt;&#xA;" OwnerUserId="10267" LastEditorUserId="10267" LastEditDate="2015-03-16T23:14:40.310" LastActivityDate="2015-03-17T00:04:07.743" Title="You do love me, don't you? I am?" Tags="|riddle|word|" AnswerCount="6" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10386" PostTypeId="2" ParentId="10385" CreationDate="2015-03-16T20:29:46.860" Score="4" Body="&lt;p&gt;Maybe it's a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chocolate covered cherry&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You say I’m a breath of fresh air...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People reward themselves with chocolate&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then why do you spit on me?&lt;br&gt;&#xA;Why do you break me down?&lt;br&gt;&#xA;Why do you trap me in the darkest of places?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eating it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My skin may be pure, but do I not bleed?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; pure chocolate, &quot;bleeding&quot; red cherry juice&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My cousin may be hot but am I not sweet?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I have no idea. I tried to think of a cousin to Red Hots, but I got nothing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastActivityDate="2015-03-16T20:29:46.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10387" PostTypeId="2" ParentId="10385" CreationDate="2015-03-16T20:55:22.047" Score="6" Body="&lt;p&gt;I think it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MINT GEL TOOTHPASTE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You say I’m a breath of fresh air...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MINT &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then why do you spit on me?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TO BRUSH THE TEETH&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why do you break me down?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TO FORM BUBBLES&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why do you trap me in the darkest of places?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CLEANING CAVITIES&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My skin may be pure, but do I not bleed?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TRANSPARENT GEL&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My cousin may be hot but am I not sweet?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; COOL AND TASTIER. COUSIN MIGHT BE ALCOHOL MOUTHWASH&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alternatively it could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MINT ICE CREAM&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9892" LastActivityDate="2015-03-16T20:55:22.047" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10388" PostTypeId="2" ParentId="10311" CreationDate="2015-03-16T21:00:56.950" Score="1" Body="&lt;p&gt;&lt;strong&gt;(a) at least 27 correct guesses&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 27 answers is guranteed if assistant use 1 way for red suits and other way for black suits, where Houdini always guessed Hearts for red and Clubs for black, with the last card as a giveaway. Let's call this &lt;em&gt;&lt;strong&gt;rule27&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;(b) 32 correct guesses&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Here's my attempt for 32 correct guesses, I do not know how to test it and couldn't come up with a formula:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Both Houdini and the assistant have to remember the last 3 cards revealed.   Assistant then perform the following on the next card:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Put the next card down in a different way (say tilt it a bit) only if the following applies:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; a) If the suit on the next card &lt;strong&gt;appears twice or more in the last 3 cards&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; b) If the suit on the next card &lt;strong&gt;match the color of the first appeared suit in the last 3 cards&lt;/strong&gt;, or &lt;strong&gt;it does not appears in the last 3 cards&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;(We can create a similar rule for the first 3 cards, but let's ignore it for now.)&lt;/p&gt;&#xA;&lt;p&gt;Now, for Houdini:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Case a) - same suit appears twice or more in last 3 cards:&lt;br /&gt;&#xA; - The next card is tilted, Houdini will know what's the next suit for sure.&lt;br /&gt;&#xA; Case b) - the next card's suit matches the first of the 3 cards appears, or it's the forth suit:&lt;br /&gt;&#xA; - The next card is tilted, Houdini will know what's the next suit for sure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Here's some stat:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There are 22 combinations possible with the 3 previous cards.&lt;/li&gt;&#xA;&lt;li&gt;There are 18 combinations with a suit appears twice or more.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;So, what if the card is place normally?&lt;br /&gt;&#xA;we apply &lt;em&gt;&lt;strong&gt;rule27&lt;/strong&gt;&lt;/em&gt; on the color that appears least on the previous 3 cards.&lt;/p&gt;&#xA;&lt;p&gt;The attempts here is to have no more than 2 consecutive normal placement for the next card, which results in 2/3 of correct guess, or 32/52 guesses correct.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;I feel like I am missing something but not sure, I am more than happy to discuss or have me proven wrong!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="63414" LastEditDate="2020-10-19T19:46:12.763" LastActivityDate="2020-10-19T19:46:12.763" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10389" PostTypeId="5" CreationDate="2015-03-16T21:20:01.570" Score="0" Body="&lt;p&gt;A palindrome is a word, phrase, number, or other sequence of characters which reads the same backwards and forwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Text examples include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Race car&lt;/li&gt;&#xA;&lt;li&gt;Able was I ere I saw Elba.&lt;/li&gt;&#xA;&lt;li&gt;Madam, I'm Adam.&lt;/li&gt;&#xA;&lt;li&gt;A man, a plan, a canal: Panama.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Numerical examples aren't usually as clever as text ones, for example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$123454321$&lt;/li&gt;&#xA;&lt;li&gt;$9320110239$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;An example of a puzzle involving palindromes is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/10374/who-wins-the-ai-game&quot;&gt;Who wins the AI-game?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Wikipedia article: &lt;a href=&quot;https://en.wikipedia.org/wiki/Palindrome&quot; rel=&quot;nofollow noreferrer&quot;&gt;Palindrome&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="11110" LastEditDate="2017-06-21T20:42:10.340" LastActivityDate="2017-06-21T20:42:10.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10390" PostTypeId="4" CreationDate="2015-03-16T21:20:01.570" Score="0" Body="A palindrome is a word, phrase, number, or other sequence of characters which reads the same backwards and forwards." OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-16T21:20:01.570" LastActivityDate="2015-03-16T21:20:01.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10391" PostTypeId="2" ParentId="10337" CreationDate="2015-03-16T21:22:42.430" Score="2" Body="&lt;p&gt;There seems to be a pattern in calculating the new room numbers when moving to any new room (from the level 3 room) as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a room with door numbers&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;      A  B  C&#xA;      d     D&#xA;      c  b  a&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Step 1) Let N equal half the difference between the number of the door you choose and the diagonally opposite door. For example, choosing door A above, N = (A-a)/2, or choosing b : N = (b - B)/2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 2) Add N to each of A, B, C... to get the numbers for the new room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This pattern holds for all the doors explored so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If&lt;/strong&gt; this rule is also maintained when moving from a level 2 room then the solution is to move up through door 602 to room&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;      504  704  803&#xA;      403    2  801&#xA;      401  500  700&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;then move up and diagonally to the right through door 803. According to the pattern, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Step 1) N = (803-401)/2 = 201&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From step 2) the expected door numbers are&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;      705  905  1004&#xA;      604    2  1002&#xA;      602  701  902&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and you can now escape through room 1004 by moving up and diagonally to the right again. Of course, you can't guarantee that the same pattern will hold... &lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="1631" LastEditDate="2015-03-16T21:29:48.480" LastActivityDate="2015-03-16T21:29:48.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10392" PostTypeId="2" ParentId="10385" CreationDate="2015-03-16T21:24:47.857" Score="5" Body="&lt;p&gt;The last line made me think of the following solution, which I think I can make fit everything:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CHILLI.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You say I’m a breath of fresh air...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This could be a homophone reference, as fresh air is often &lt;strong&gt;chilly&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Then why do you spit on me? &lt;br&gt;&#xA;Why do you break me down? &lt;br&gt;&#xA;Why do you trap me in the darkest of places?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All of this could refer to any kind of food: when you eat it, it gets covered in saliva, chewed into pieces, and buried in the pitch darkness of your stomach.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My skin may be pure, but do I not bleed?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is the line I'm least sure about...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My cousin may be hot but am I not sweet?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chilli can be very hot, but can also be used to make sweet chilli sauce.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-03-16T21:24:47.857" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10393" PostTypeId="2" ParentId="10385" CreationDate="2015-03-16T21:38:06.057" Score="1" Body="&lt;p&gt;What about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Just plain old chocolate&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You say I’m a breath of fresh air...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The delicious taste could &quot;blow you away&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then why do you spit on me? &#xA;Why do you break me down? &#xA;Why do you trap me in the darkest of places?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because we need to eat chocolate to satisfy our desiring love for chocolate&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My skin may be pure, but do I not bleed?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chocolate is usually made of pure cocoa, the outside of the chocolate may melt&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My cousin may be hot but am I not sweet?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hot chocolate is hot but regular chocolate is sweet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9042" LastActivityDate="2015-03-16T21:38:06.057" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10394" PostTypeId="2" ParentId="10381" CreationDate="2015-03-16T22:11:46.733" Score="3" Body="&lt;p&gt;This can be proven using math (the hard way):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;250 - 11 = 239, 239 + 1 = 240 &lt;br&gt;&#xA;     240 - 11 = 229, 229 + 1 = 230 &lt;br&gt;&#xA;     230 - 11 = 219, 219 + 1 = 220 &lt;br&gt;&#xA;     220 - 11 = 209, 209 + 1 = 210 &lt;br&gt;&#xA;     210 - 11 = 199, 199 + 1 = 200 &lt;br&gt;&#xA;     200 - 11 = 189, 189 + 1 = 190 &lt;br&gt;&#xA;     190 - 11 = 179, 179 + 1 = 180 &lt;br&gt;&#xA;     180 - 11 = 169, 169 + 1 = 170 &lt;br&gt;&#xA;     170 - 11 = 159, 159 + 1 = 160 &lt;br&gt;&#xA;     160 - 11 = 149, 149 + 1 = 150 &lt;br&gt;&#xA;     150 - 11 = 139, 139 + 1 = 140 &lt;br&gt;&#xA;     140 - 11 = 129, 129 + 1 = 130 &lt;br&gt;&#xA;     130 - 11 = 119, 119 + 1 = 120 &lt;br&gt;&#xA;     120 - 11 = 109, 109 + 1 = 110 &lt;br&gt;&#xA;     110 - 11 = 99, 99 + 1 = 100 &lt;br&gt;&#xA;     100 - 11 = 89, 89 + 1 = 90 &lt;br&gt;&#xA;     90 - 11 = 79, 79 + 1 = 80 &lt;br&gt;&#xA;     80 - 11 = 69, 69 + 1 = 70 &lt;br&gt;&#xA;     70 - 11 = 59, 59 + 1 = 60 &lt;br&gt;&#xA;     60 - 11 = 49, 49 + 1 = 50 &lt;br&gt;&#xA;     50 - 11 = 39, 39 + 1 = 40 &lt;br&gt;&#xA;     40 - 11 = 29, 29 + 1 = 30 &lt;br&gt;&#xA;     30 - 11 = 19, 19 + 1 = 20 &lt;br&gt;&#xA;     20 - 11 = 9, 9 + 1 = 10 &lt;br&gt;&#xA;     10 - 11 = -1       ??????&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can't continue after doing this 24 times. Or you will get -1. As said by @Novarg , you can't take 11 things out of 10 things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 24 times &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9042" LastEditorUserId="9042" LastEditDate="2015-03-21T01:01:51.503" LastActivityDate="2015-03-21T01:01:51.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10395" PostTypeId="2" ParentId="10216" CreationDate="2015-03-16T22:12:32.657" Score="3" Body="&lt;p&gt;@JoeZ has already done the first calculations, to get the following results:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $1141, 1314, 600, 200, 115, 282, 492, 1716, 1650, 320, 906$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to the OP's hints, we should now&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; divide each of these numbers by another number to get results two of which are 40 and the rest of which are between 114 and 166.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assuming all numbers involved are integers, the only possibilities are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$1141/7=163$&lt;/li&gt;&#xA;&lt;li&gt;$1314/9=146$&lt;/li&gt;&#xA;&lt;li&gt;$600/4=150$ (or $600/5=120$ or $600/15=40$)&lt;/li&gt;&#xA;&lt;li&gt;$200/5=40$&lt;/li&gt;&#xA;&lt;li&gt;$115/1=115$&lt;/li&gt;&#xA;&lt;li&gt;$282/2=141$&lt;/li&gt;&#xA;&lt;li&gt;$492/4=123$ or $492/3=164$&lt;/li&gt;&#xA;&lt;li&gt;$1716/13=132$ or $1716/12=143$ or $1716/11=156$&lt;/li&gt;&#xA;&lt;li&gt;$1650/10=165$ or $1650/11=150$&lt;/li&gt;&#xA;&lt;li&gt;$320/8=40$ (or $320/2=160$)&lt;/li&gt;&#xA;&lt;li&gt;$906/6=151$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So our new list of numbers is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $163, 146, 150, 40, 115, 141, 123 \mathrm{ or } 164, 132 \mathrm{ or } 143 \mathrm{ or } 156, 165 \mathrm{ or } 150, 40, 151$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using the octal ASCII codes from &lt;a href=&quot;http://www.ascii-code.com/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, this becomes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;s&lt;/strong&gt;, &lt;strong&gt;f&lt;/strong&gt;, &lt;strong&gt;h&lt;/strong&gt;, [space], &lt;strong&gt;M&lt;/strong&gt;, &lt;strong&gt;a&lt;/strong&gt;, &lt;strong&gt;S&lt;/strong&gt; or &lt;strong&gt;t&lt;/strong&gt;, &lt;strong&gt;Z&lt;/strong&gt; or &lt;strong&gt;c&lt;/strong&gt; or &lt;strong&gt;n&lt;/strong&gt;, &lt;strong&gt;u&lt;/strong&gt; or &lt;strong&gt;h&lt;/strong&gt;, [space], &lt;strong&gt;i&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;... which still doesn't make sense. According to the OP's second hint though, we should remember the numbers we divided by ($7,9,4,5,1,2,4 \mathrm{ or } 3, 13 \mathrm{ or } 12 \mathrm{ or } 11, 10 \mathrm{ or } 11, 8,6$). Maybe Caesar-shifting the letters we have by these numbers of places in the alphabet could be the key? Unfortunately that gives &lt;strong&gt;zol&lt;/strong&gt;, &lt;strong&gt;lwd&lt;/strong&gt;, or &lt;strong&gt;ncv&lt;/strong&gt; for the first three letters (I didn't go any further since none of these make sense), according to whether we add or subtract.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-18T11:06:59.857" LastActivityDate="2015-03-18T11:06:59.857" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10396" PostTypeId="2" ParentId="8223" CreationDate="2015-03-16T22:22:17.933" Score="12" Body="&lt;p&gt;&lt;strong&gt;Sentence&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm the master's nightmarish, gorgonian hatemonger, his moth-eaten gonorrhoea, smothering mightiest heroism, thrashing egomania's (or, to me, ignorant mismanagement's) strong-arm mishmash or staggering high treason.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Length&lt;/strong&gt;: 215&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;N&lt;/strong&gt;: 16&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Letters&lt;/strong&gt;: AEGHIMNORST (11 total)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Produced by interleaving and manually modifying multiple computer-generated anagram fragments. With more work it should be possible both to improve the readability and extend it further.&lt;/p&gt;&#xA;" OwnerUserId="10220" LastEditorUserId="10220" LastEditDate="2015-05-07T01:58:34.723" LastActivityDate="2015-05-07T01:58:34.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10398" PostTypeId="2" ParentId="10385" CreationDate="2015-03-16T22:56:22.680" Score="3" Body="&lt;p&gt;Internet Explorer for sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I call it a breathe of fresh air whenever I (re)install windows, his cousin Chrome is hot (and IE thinks of himself as sweet).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And it is white skinned because it takes a long while to show webpages thus giving me a blank screen only after which it crashes (bleeds) and I bash at it loudly (break him down). And I keep IE in dark after I install chrome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Tosen has now edited his post multiple times which might make this answer wrong.&lt;/p&gt;&#xA;" OwnerUserId="10271" LastEditorUserId="10271" LastEditDate="2015-03-16T23:22:55.657" LastActivityDate="2015-03-16T23:22:55.657" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10399" PostTypeId="1" AcceptedAnswerId="10401" CreationDate="2015-03-16T23:22:05.890" Score="27" ViewCount="7394" Body="&lt;p&gt;It's Friday the 13th, and you and a group of 7 friends from the &quot;Survivor&quot; club belonging to University LeMao (A, B, C, D, E, F, and G) go camping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're all pretty good friends, and talk quite a bit during their hike up the Mountain named Nuffingwrong. As the talk continues, A gets into an argument with E. B, being A's best friend decides to help out A, only to have F step in and make things worse - F decides to help E (They're best friends, after all). For the next 5 minutes, a constant yelling and screaming pierces your ears. Reluctant to get involved, you take G and head off to find a camping spot. The last thing you hear from that area are the words &quot;I'm gonna kill you, I swear it!&quot; from one of the 4 people arguing, followed by C and D stepping in to stop the fight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After you find a good spot, you and G head back to rendezvous with the group, and tell them about the great spot you guys picked out. The fight seems to be over, everyone seems to be good friends again, and the night seems cheerful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the sun sets, the group heads out to the spot you picked to set up camp, and split up the jobs.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A and B go to get firewood.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;C and D are tasked with setting up the tents.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;E and F are sent to go get some vegetation (berries and/or fruit) for the nights dinner.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You and G end up with the task of sharpening sticks and going to hunt for small critters to cook.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;15 minutes pass since everyone has split off to do their jobs. G mentions to you that he needs to go to the washroom, and heads off into the forest. He comes back a couple minutes later, and the two of you continue with your job. For the next half hour, everything seems calm and serene; you and G are having a great time talking while sharpening sticks, discussing how to properly throw them and what the best form for javelin throwing is, when all of a sudden, two bloodcurdling screams resonate from the direction of the campsite. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two of you try to rush back to the campsite. You follow as G leads, but you both end up getting lost along the way, and don't arrive at the campsite for another 20 minutes. You find that most of the group is already there, waiting for you. Looking around, you notice that E and F's hands are stained with red liquid, C and D are lying face down on the floor, and B is in a pool of blood - a quick look tells you that he is dead. &quot;Where's A&quot;, you ask. Nobody knows - everyone instantly rushes back into the forest to look for him, and it doesn't take long before you find him face down in a river - lifeless. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You all head back to camp, and check on C and D. Touching the back of their heads, you notice that your hands are wet and have become stained with red. You gently wake them up, and it seems that they're alright. C and D tell you that they were knocked out by something - You fill them in on what has happened so far, and start the investigation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Here's what you know about the members of this social group:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A and B are your average Joe's - they're not particularly weak or strong. They're prone to arguments, but they're also very hard working people. They also work a lot better individually.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A has OCD and can't stand things being messy.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;C and D are best friends, and are big time geniuses. Like, 140 IQ type genius. They hate conflict, and they're perpetually scared of everything - fortunately for them, their high IQ leads them to preemptively prevent conflict very often - one thing they've done is jack up their bodies by going to the gym &lt;strong&gt;a lot&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;E and F are nerdy environmental studies majors - they pretty much can't lift more than their backpacks, and they're nature lovers like no other.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;G is a loner - he's prone to violence, and gets upset at the weirdest things. The group pretty much consists of all his friends.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Checking the area, you notice the following things:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;There's a pile of firewood at the campsite, piled extremely neatly, but missing the top piece of the triangle.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The fire hasn't been started yet.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are some crushed berries near B's body&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are no major signs of struggle at the campsite (No scuffle marks, etc).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The tents have been set up.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You manage to find what seems to be the murder weapon regarding B's murder: A single log (firewood size) with red on the bark. It seems to be blood. It seems to fit perfectly on the top of the pile of firewood previously mentioned.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;You do some questioning for each member of the group in order to see if you can find out the truth. Here's what you manage to divine from each person:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A: N/A - he's dead, but you notice an EpiPen when you search his clothing - it seems that he's extremely allergic to raspberries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B: N/A - he's dead, but upon investigation of his body you notice that the blood loss mainly resulted from a heavy blow to the head. There's firewood under his body, suggesting that he was carrying it at the time of death and fell on top of it after getting hit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C &amp;amp; D: &quot;We don't know what happened - we were knocked out by someone, and after you woke us up, we've been with you the whole time.&quot;  You point out that C's hands are red. &quot;That's cus we were eating the berries E and F brought back.&quot; You mention that only one of them has stained hands - to which they tell you secretly that they're a couple and that C was feeding D. No one in the group knows about this, and they ask you to keep it a secret.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E &amp;amp; F: &quot;We were out collecting berries - we've made several drops off over there&quot; (They point to a pile of berries in a bucket) &quot;And when we got back from our 3rd round, we saw B lying there, dead. We didn't know what to do, so we screamed!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G: &quot;I've been with you the whole time. I'll show you where I went to piss if you want...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Q: What happened that night? Who is/are the killer(s)?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;Hints to be posted later - if needed.&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hints/For clarity:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Question: Were E and F together the whole time? Answer: Yes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT MADE&lt;/strong&gt;: I changed the wording for the section where you run back to camp with G, for clarity!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because way too many people are falling for this red herring;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; G did not kill B - he doesn't know the way back to camp! If he tried he would never have made it there and back, and pissed, all within 5 minutes. Also, he has no motive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because Duncan has the right idea:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; B was in fact the person to scream &quot;I'm gonna kill you, I swear it&quot; from the argument.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because people are giving up - don't give up! Here's a hint!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; E and F traveled to a single bush area for berries.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8999" LastEditorUserId="8999" LastEditDate="2015-03-17T23:37:00.200" LastActivityDate="2015-03-18T00:31:07.757" Title="Double whammy murder case! What happened?" Tags="|logical-deduction|lateral-thinking|mystery|story|" AnswerCount="4" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="10400" PostTypeId="2" ParentId="10385" CreationDate="2015-03-17T00:04:07.743" Score="11" Body="&lt;p&gt;I think it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Peppermint candy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You say I’m a breath of fresh air...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Small peppermint candies are frequently given out as breath fresheners&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then why do you spit on me?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They induce salivation in the mouth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why do you break me down so loudly?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are very crunchy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why do you trap me in the darkest of places?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First the mouth, then the stomach&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My skin may be pure white, but is my face not also red?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are white, but often decorated with red stripes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My cousin may be spicy hot but am I not sweet?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Peppers are spicy but peppermint is a fresh, sweet flavor&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4915" LastActivityDate="2015-03-17T00:04:07.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10401" PostTypeId="2" ParentId="10399" CreationDate="2015-03-17T00:23:04.237" Score="14" Body="&lt;h2&gt;Some preliminary conclusions&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;A and B [...] work a lot better individually.&quot; So they probably separated while collecting firewood.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;A has OCD and can't stand things being messy. [...] There's a pile of firewood at the campsite, piled extremely neatly.&quot; So the firewood that's already there was probably brought by A.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;There's firewood under [B's] body, suggesting that he was carrying it at the time of death and fell on top of it after getting hit.&quot; B was only just bringing his firewood to the campsite when he was attacked.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;C and D are tasked with setting up the tents&quot; and later &quot;The tents have been set up&quot;, so there seems to be no reason for C and D to have left the campsite. Let's assume they were there all the time.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;E and F [...] pretty much can't lift more than their backpacks,&quot; so it doesn't look like either of them could have hit B with a chunk of wood hard enough to kill him, &lt;em&gt;or&lt;/em&gt; knocked out C and D.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The campsite bears &quot;no major signs of struggle&quot; (or, we assume, of bodies having been dragged there), so we can assume B was killed there and C and D were knocked out there.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A and B are best friends, so we can assume &lt;strong&gt;neither of them killed each other&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;C and D are best friends, so we can assume &lt;strong&gt;neither of them knocked out the other&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;By the logic now stated in the latest edit to the question, &lt;strong&gt;G did not kill B&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Main logical argument&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;We've now eliminated A, E, F, G, and presumably yourself (unless you're an &lt;a href=&quot;http://en.wikipedia.org/wiki/Unreliable_narrator&quot; rel=&quot;nofollow noreferrer&quot;&gt;unreliable narrator&lt;/a&gt;) as B's killer, leaving only &lt;strong&gt;C and D&lt;/strong&gt;. Since they were together the whole time, they must have conspired together to kill B. For their motive, I quote from &lt;a href=&quot;https://puzzling.stackexchange.com/a/10440/5373&quot;&gt;Duncan's answer&lt;/a&gt;: &quot;B must have been the one to scream 'I'm going to kill you' in the first argument and we know that C and D work to prevent violence. Being smart, they must have figured that one death is better than 2 deaths.&quot; More specifically, &lt;strong&gt;D&lt;/strong&gt; must have been the actual killer since (s)he's the one without red hands and the only red you found on the log was blood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we've eliminated B, E, F, and again presumably yourself as the person who knocked out C and D. So this must have been done by A or G. By the same logic as was used to eliminate G as B's killer, he can't have knocked them out. &lt;strong&gt;A&lt;/strong&gt; also has the best motive as he was B's best friend - perhaps he saw them kill B? The other possibility is that they weren't really knocked out at all and were only faking, but let's assume your observations were correct on this one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we've eliminated B, C, D, and again presumably yourself as A's killer. That leaves E, F, and G. E and F are the ones with a motive, given the earlier quarrel. They were together all the time, and their task was picking berries. So it would've been easy for them to corner A near the stream, kill him with raspberries, and then return to the campsite and scream.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Summary&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;D&lt;/strong&gt; (accomplice: &lt;strong&gt;C&lt;/strong&gt;) killed B with a log, in order to pre-emptively stop him from killing E and F. &lt;strong&gt;A&lt;/strong&gt; attacked and stunned C and D in a grief-fuelled rage. &lt;strong&gt;E and F&lt;/strong&gt; killed A with raspberries, motivated by their earlier quarrel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;FULL BREAKDOWN AND SOLUTION (Aify's version, in a sort of story mode)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are not the murderer, so rule yourself out of both cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;H1&gt;B's Murder&lt;/H1&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ruling out G:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G can't be the murderer of B, as he only had one chance in the first 15 minutes of the night. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reason 1:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Because E and F did not scream until 30 minutes later, after their third trip back, G doesn't make the time slot (Assuming equal length trips since they went to the same bush area each time to pick berries (See hints) - 15 mins per trip). &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Time line: 15 mins [G leaves to piss], (E F 1st Round to camp) || 30 mins (E F second round to Camp) || 45 mins (E F 3rd round to camp) {Scream is heard}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the above time line, you can see that if G killed B, E and F would have discovered the body much sooner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reason 2:&#xA;- G got lost with you going back to camp. He would never have been able to make a 2.5 minute trip there and back, AND take a piss. (You know he actually did take a piss because he offers to show you where he went).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ruling out A, E and F:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E and F lack the strength to enact the murder of B - even if they mustered enough strength to strike down B, it's highly improbable they could knock out C and D at the same time. Since A and B work better individually, it stands to reason that they split up to collect wood. A got back from his first round sometime before B did, proof being the OCD and the neat pile of firewood. A can't have killed B due to them being besties - furthermore, if A killed B, he would have put the piece of wood back on the pile due to his OCD. Assuming equal trip lengths for E and F, each trip they took lasted 15 minutes. Since tent is finished and E/F didn't scream on the second trip back, the time period of B's murder is between the second trip and the third trip. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How B died:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While A was gone, C took the firewood from the top of the pile and snuck up behind B and struck him with one mighty blow, the blood from the blow splashed onto C's hands, which C masked with the raspberry juice - D, being in love with C, helps to cover up and they come up with the plan of C feeding D (to explain C's hands). C and D, being geniuses, used raspberry juice on the back of their heads and pretended that they were knocked out. E and F, returning from their trip, screamed upon sight of a bloody body. B was the one who said &quot;I'm gonna kill you, I swear it!&quot; The motive of C and D lies in the preemptive prevention of that aggression. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;C and D are the killers of B&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;H1&gt;A's Murder&lt;/H1&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ruling people out:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G can't be the murderer of A - G doesn't have stained hands, so if G killed A, G must have drowned him, which would result in him being wet somewhere, either from standing in the water, or from holding A's head in the water. Also, 5 minutes is very little time to go and kill someone, piss, and come back. The fact that G offered to show you where he pissed indicates that it was nowhere near A's murder spot, since it would only cast suspicion on himself if it was near. Here, his honesty is his saving grace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C and D were together at the campsite - the tent was finished, so they can't have left in the first section of time since they need time to put up the tent. C and D were not wet - meaning they did not drown A. A still had his EpiPen on him at the river - unless C and D left the camp to put it back on him, it would be impossible for C and D to make A get an allergic reaction. It's a severe reaction, it would likely have happened on the spot and A would've instantly used his EpiPen at the camp. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ruling out B:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Main reason: B and A are besties. B would never harm A. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sub reason: A also had the opportunity to drop off wood at the campsite before leaving to get more; B was only just returning from his first trip before he was killed. That means that B would not have been able to kill A before A dropped off wood (which was anytime before the second return of E/F, but unlikely to be before the first trip of E/F). If B killed A, B would not have had ample time to get wood and go back to camp. B killing A also doesn't make sense because his hands were not stained, and his clothing was not wet - if he killed A, he would either have to get wet drowning A, or force berries down A's throat, resulting in stained hands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How A died:&lt;/strong&gt;&#xA;Any reason to use A drowning as a cause of death can be instantly discounted because nobody had wet clothing. If anyone were to drown A, they wouldve had to get wet. A wasn't particularly weak OR strong, and would have put up a fight - splashing will ensue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E and F were extremely scared of A and B, and when they saw A, they knew they had to do something before A helped B kill them. Not having enough physical strength to (literally) take down A without getting themselves harmed seriously, they opted for the option of giving A raspberries (Knowing that he is severely allergic).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can imagine something like &quot;Close your eyes, A, we found something really tasty in the woods. Can you guess what it is?&quot; &lt;em&gt;Proceed to feed A, stealing his EpiPen, the putting him face down in the water&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They take the EpiPen from him while his eyes are closed, and when he dies they put him face down in the river and replace the EpiPen to make it look like a drowning. This makes it look like they can't be the murderers, since they're too weak - but remember they're nerds, and they're smart. To pull this off, they took advantage of A's friendship. &quot;E and F's hands are stained with red liquid&quot; - this is the raspberry juice from their berry picking endeavors, and the key clue to connect A's allergy to E and F's berry picking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E and F are also the only people who had access to A while he was in the woods within the time line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;E and F killed A&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Main List of Red Herrings:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;G's personality&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Campfire not lit (this was placed to confuse people)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Crushed berries (They came from E and F dropping them when they found B, but it's irrelevant unless you use it as a reason to discount E and F from the murder of B; eg: surprise caused them to drop/squash anything they were holding).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-03-18T00:31:07.757" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10402" PostTypeId="1" AcceptedAnswerId="10403" CreationDate="2015-03-17T00:23:52.593" Score="5" ViewCount="448" Body="&lt;p&gt;This puzzle is about solving &lt;strong&gt;anagram-chains&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;To start, let's first define what an &lt;strong&gt;anagram-chain&lt;/strong&gt; is:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;An &lt;strong&gt;anagram-chain&lt;/strong&gt; consists of multiple anagrams stitched together into a long letter-list.&lt;/li&gt;&#xA;&lt;li&gt;Each anagram of the chain (except the last) partly overlaps with the follow-up anagram, so that there are letters which are shared by &lt;em&gt;two&lt;/em&gt; anagrams (but never more).&lt;/li&gt;&#xA;&lt;li&gt;The &lt;em&gt;minimum&lt;/em&gt; of letters which overlap is defined. i.e. an anagram chain of &lt;em&gt;overlap 3&lt;/em&gt; (O3) requires that two anagrams overlap by &lt;em&gt;at least&lt;/em&gt; 3 common letters.&lt;/li&gt;&#xA;&lt;li&gt;No two anagrams overlap ever completely, i.e. no anagram can be 'fully' contained in another.&lt;/li&gt;&#xA;&lt;li&gt;All anagrams of a chain share a &lt;strong&gt;common theme&lt;/strong&gt; which is given.&lt;/li&gt;&#xA;&lt;li&gt;All anagrams of a chain share a &lt;strong&gt;common property&lt;/strong&gt; which needs to be found.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;And now clarify this with an example:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The following anagram-chain (&lt;em&gt;O3&lt;/em&gt;) is themed &lt;em&gt;&lt;strong&gt;presidents of the USA&lt;/strong&gt;&lt;/em&gt;:&lt;/p&gt;&#xA;&lt;p&gt;&lt;code&gt;IMMCARTRJYEFKEOHNNNDLBJOOYSNHRTRUARMAKBAABCOILLLINTN&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;And it solves to:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/yzaPp.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The common property is&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; All are from the &lt;em&gt;Democratic&lt;/em&gt; party.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;Can you now solve the following anagram chains:&lt;/h1&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;h3&gt;#1 &lt;em&gt;Night sky&lt;/em&gt; (O2)&lt;/h3&gt;&#xA;&lt;p&gt;&lt;code&gt;VNSEUCRYMRASNUURJPITEHARSTUNEPNE&lt;/code&gt;&lt;br&gt;&#xA;Common property: ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;h3&gt;#2 &lt;em&gt;Animals&lt;/em&gt; (O3)&lt;/h3&gt;&#xA;&lt;p&gt;&lt;code&gt;OSQUOMITEERTHONYBEEENPDCTIERCKCOCOAHGSSPRPERCALATILFFRUYTELEEBMBEUBLADYGONRFLVISHSIE&lt;/code&gt;&lt;br&gt;&#xA;Common property: ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;h3&gt;#3 &lt;em&gt;Food!&lt;/em&gt; (O3)&lt;/h3&gt;&#xA;&lt;p&gt;&lt;code&gt;RRTCAOVADOCHIRICUFLOWALEPTGGNIERTOKACHSPINTUROSPSAAUARGLICOOBRCUMUCETLTELNISPPARCES&lt;/code&gt;&lt;br&gt;&#xA;Common property: ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;h3&gt;#4 &lt;em&gt;Hollywood!&lt;/em&gt; (O5)&lt;/h3&gt;&#xA;&lt;p&gt;&lt;code&gt;KRIHPURNATHANEEBCLCTTOLETAUDEEBRPYHURNLOLHTEBOSIRYHWLAANKGCNEDJSOFIOTERSSGMNINOEICLKDASBNRAULOJOINEMEHLRRENIUISAEJBLRORTMCOIAINLADPEGEREGSRARONHZTHCLIEEJGSSAANBRBRRTEIADKNEAONWDJWOODARNCAFRJRENEENIELWKATSTHYBAEZTOLIETALRMEMANTMHOPSOIALREN&lt;/code&gt;&lt;br&gt;&#xA;Common property: ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-03-18T04:24:13.807" Title="Anagram chains - What is the common property in each?" Tags="|anagram|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10403" PostTypeId="2" ParentId="10402" CreationDate="2015-03-17T00:54:07.553" Score="5" Body="&lt;h2&gt;First one&lt;/h2&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;pre&gt;&lt;code&gt;VNSEUCRYMRASNtURJPITEHARSTUNEPNE&lt;br&gt;VENUS   MARS  JUPITER SATURN&lt;br&gt;   MERCURYSATURN   EARTH NEPTUNE &lt;/code&gt;&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note also the almost-possibility of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;pre&gt;&lt;code&gt;VNSEUCRYMRASNtURJPITEHARSTUNEPNE&lt;br&gt;VENUS   MARS  JUPITER STAR&lt;br&gt;   MERCURYSATURN   EARTH NEPTUNE &lt;/code&gt;&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Getting the same thing coming up twice is OK, but an overlap of less than 2 isn't! The common property is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;planets&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The only problem is I think it should be VNSEUCRYMRASN&lt;strong&gt;T&lt;/strong&gt;URJPITEHARSTUNEPNE not VNSEUCRYMRASNUURJPITEHARSTUNEPNE.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Second one&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;pre&gt;&lt;code&gt;OSQUOMITEERTHONYBEEENPDCTIERCKCOCOAHGSSPRPERCALATILFFRUYTELEEBMBEUBLADYGONRFLVISHSIE&lt;br&gt;MOSQUITO HORNET  CENTIPEDE COCKROACH    CATERPILLAR  BUTTERFLY   LADYBUG  SILVERFISH&lt;br&gt;     TERMITEHONEYBEE   CRICKET   GRASSHOPPER    FRUITFLY   BUMBLEBEEDRAGONFLY&lt;/code&gt;&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Common property:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;insects/creepy-crawlies&lt;/strong&gt; (does the word &lt;strong&gt;bugs&lt;/strong&gt; cover both of these?).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There seems to be something wrong in the middle, because where I've written BUTTERFLY doesn't quite match; there could be BEETLE on the right, but then nothing seems to fit the left! BmyGuest, please can you check this very carefully?&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Third one&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;pre&gt;&lt;code&gt;RRTCAOVADOCHIRICUFLOWALEPTGGNIERTOKACHSPINTUROSPSAAUARGLICOOBRCUMUCETLTELNISPPARCES&lt;br&gt;CARROT RADICCHIO     EGGPLANTARTICHOKE TURNIP ASPARAGUSBROCCOLI  LETTUCE PARSNIP&lt;br&gt;   AVOCADO   CAULIFLOWER  GINGER   SPINACHSPROUT    GARLIC  CUCUMBER LENTILS CAPERS&lt;/code&gt;&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;... but there's only a two-letter overlap between SPROUT and ASPARAGUS! The common property is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;vegetables&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-17T22:15:44.640" LastActivityDate="2015-03-17T22:15:44.640" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10404" PostTypeId="1" AcceptedAnswerId="10550" CreationDate="2015-03-17T03:11:17.910" Score="47" ViewCount="9965" Body="&lt;p&gt;In the small town of Terni (Italy), there's a couple of young friends named Marco and Leonardo, who like to perform magic tricks to a restricted audience of common friends and relatives. They like to call themselves &lt;em&gt;&quot;The prodigious Duo&quot;&lt;/em&gt;. Every week-end, The prodigious Duo meets at the house of the leader, which happens to be Marco, to practice new magic tricks, learn new techniques, and, most importantly, to set up the magic show of the week, which will be shown to the audience in the late evening.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recently, the leader of The prodigious Duo discovered what he claims to be &quot;the coolest magic trick ever done using a chessboard&quot; &lt;sup&gt;(which actually is a checkerboard, but that's basically the same thing)&lt;/sup&gt;. &#xA;Marco only tested the trick on a $2 \times 2$ squares checkerboard, nonetheless he's certain that the trick would work on any checkerboard with size $n \times n$, where $n$ is a perfect power of two (meaning that the binary logarithm $\log_{2}n$ of $n$ is an integer). &#xA;So &lt;strong&gt;he managed to create a working strategy for a real checkerboard (with $n=8$)&lt;/strong&gt;. Once tested the strategy for the bigger board on a piece of paper, Marco is now ready to explain the trick to Leonardo.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;He starts explaining the &lt;em&gt;peculiarities of the checkerboard&lt;/em&gt; involved in the trick, since that it isn't a normal checkerboard:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hXa3A.png&quot; alt=&quot;Example of the checkerboard&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;The checkerboard is composed by $8 \times 8$ squares, like a normal one, except that the squares are all red colored, and distinguishable because of the vertical and horizontal lines which define them. This red color choice is only made to avoid confusion when moving the pieces.&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;The pieces of the checkerboard are normal coin-shaped pieces, except that they're &lt;strong&gt;white on one side and black on the other one&lt;/strong&gt;.&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;The checkerboard is &lt;strong&gt;always visible to all the members of the audience&lt;/strong&gt; during the show, and so will be me and you, remember it, this is important.&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h3&gt;Then he explains &lt;em&gt;the magical trick&lt;/em&gt; from the spectators' point of view:&lt;/h3&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;At the beginning of the show, the checkerboard is filled with sixty-four pieces, each one placed inside a different square. &lt;em&gt;The pieces' sides facing up are of mixed color, and their initial color doesn't really matter&lt;/em&gt; for the purpose of the trick (e.g. the quantity of black or white pieces doesn't matter).&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;Once the checkerboard is set up, &lt;em&gt;you will ask for a member of the audience&lt;/em&gt; (we'll call him $P$) &lt;em&gt;to come up on the stage to participate in the trick&lt;/em&gt;.&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;Once on the stage, I will cover my eyes and ears with a special mask which will make me unable to see or hear anything around me. Now you'll &lt;strong&gt;&lt;em&gt;ask $P$ to rearrange the colors&lt;/em&gt;&lt;/strong&gt; that he sees on the checkerboard flipping any number of pieces he wants, any number of times.&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;Now you will take a minute or less to &lt;strong&gt;&lt;em&gt;look closely and meticulously at the checkerboard's configuration&lt;/em&gt;&lt;/strong&gt;, and then &lt;strong&gt;&lt;em&gt;ask $P$ to: choose a piece among the 64 pieces on the board, flip it only one time, and memorize its position&lt;/em&gt;&lt;/strong&gt;.&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;Once $P$ has done its simple task, you'll do the same thing: you'll carefully &lt;strong&gt;&lt;em&gt;choose a piece&lt;/strong&gt; (which doesn't necessarily need to be different from the one chosen by $P$), &lt;strong&gt;and flip it one time&lt;/em&gt;&lt;/strong&gt;. At this point, you will lock your head into the same kind of mask I used before, and, once done, ask $P$ to touch me to let me know you cannot see or hear anything.&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;Felt the touch and removed my mask, &lt;strong&gt;&lt;em&gt;I'll reach the checkerboard, carefully observe it for a bit, and finally remove a single piece from it: the piece that has been chosen by $P$.&lt;/em&gt;&lt;/strong&gt;&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;I'll now help you to remove your mask and bow down with you to the surprisingly cheering audience and the astonished $P$, which is trying to work out what sort of sorcery has been going on moments before.&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Leonardo takes a minute to elaborate all the information provided to him by Marco, and then, excited, replies: &quot;This is an incredible magic trick, but man, I mean, sixty-four pieces... how could we ever even think about that? Tell me you've got a solving strategy which is as beautiful as this impressive trick!&quot;. &lt;strong&gt;Marco starts now to explain &lt;em&gt;the real strategy&lt;/em&gt; which they are going to apply to recognize the piece chosen by the volunteer on the stage...&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;You're asked to discover and explain the strategy which The prodigious Duo is going to apply to find the correct piece chosen by the volunteer.&lt;/h3&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Clarifications and notes&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Now, stating some basic facts that might be misunderstood:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;During their show, none of the members of The prodigious Duo can communicate with each other, because of the masks they use to cover their heads.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;This means that Marco cannot, in any way, see or listen to what's happening on the stage when the volunteer is mixing up the colors flipping the pieces on the checkerboard.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;And also that Leonardo cannot, in any way, see or listen to what's going on when Marco removes his mask and looks at the checkerboard, revealing the piece chosen by the volunteer.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;No one knows, except for the two magicians, the strategy applied to recognize the piece moved by the volunteer.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The volunteer called on the stage isn't, in any way, collaborating with the magicians nor cheating; same goes for the rest of the audience.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Avoid stupid answers like &quot;Marco checks for the fingerprint leaved by the volunteer on the chosen piece&quot; or similar.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h3&gt;Personal notes&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I want to clarify that I didn't think up this puzzle by myself, I only made up some little story-line; I did listen to it years ago talking with some friends of mine, and it randomly came back in my mind recently. For this reason, &lt;strong&gt;I do not know the exact solution of this puzzle&lt;/strong&gt; (and this is the main reason why I am asking for a solution here on SE), nonetheless I'm still able to provide additional details and explanations to anyone who asks for them, editing the question itself or answering the comments. &lt;strong&gt;Given that I do not know the real source of this puzzle, it would also be nice if someone could provide it&lt;/strong&gt;. Oh, and in case you were wondering: yes, that &quot;Marco&quot; is me, ahah.&lt;/p&gt;&#xA;" OwnerUserId="10248" LastEditorUserId="10248" LastEditDate="2015-03-17T13:32:14.653" LastActivityDate="2021-03-24T21:50:01.020" Title="The coolest checkerboard magic trick" Tags="|logical-deduction|combinatorics|checkerboard|puzzle-identification|" AnswerCount="4" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10405" PostTypeId="1" AcceptedAnswerId="10515" CreationDate="2015-03-17T04:05:29.330" Score="15" ViewCount="2081" Body="&lt;p&gt;A partner in crime, though I wear a disguise.&lt;br&gt;&#xA;I hide in plain sight. Averting all eyes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't run and gun, if you want to stop me.&lt;br&gt;&#xA;To succeed on that front, you need just turn and flee.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A doctor creates me, a goose fills her nest,&lt;br&gt;&#xA;But just give me time and I will manifest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any who make me, end up the same...&lt;br&gt;&#xA;The question, of course, though, is: what is my name?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A cryptic clue given on almost every line.&lt;br&gt;&#xA; Don't be so literal and you will do fine.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ok then folks, if it's really that hard...&lt;br&gt;&#xA; With the goose and the doc, there's also a bard.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 3&lt;/strong&gt; &lt;sub&gt;(it's already been solved, but I wanted to add this one because it amuses me, and it may help others who want to try solving it themselves)&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hide in plain sight? I'm right here in this hint!&lt;br&gt;&#xA; This time, however, I'm garnished with mint.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="5208" LastEditDate="2015-03-18T21:49:43.963" LastActivityDate="2017-07-03T04:47:20.653" Title="A partner in crime... What is my name?" Tags="|riddle|word|" AnswerCount="10" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10406" PostTypeId="2" ParentId="10404" CreationDate="2015-03-17T05:20:58.113" Score="15" Body="&lt;p&gt;Let the number of cells in the board be $2^k$ for some $k$, and number each cell from $0$ to $2^k-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In step 4, during the &quot;look closely and meticulously at the checkerboard's configuration&quot; part, Leonardo decides which piece he will flip. This is &lt;strong&gt;before&lt;/strong&gt; the audience member makes his choice. For each integer $i$ from $1$ to $k$, Leonardo will count how many black pieces there are in positions whose $i$th bit is $1$. If there is an odd number of black pieces, Leonardo will flip a cell whose $i$th bit is $1$; otherwise, he will flip one of the other cells. This is enough to completely determine which cell Leonardo flips.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When Marco looks at the board, after Leonardo and the audience member have made their flips, he performs the same computations as Leonardo. If there is an odd number of black pieces in positions whose $i$th bit is $1$, the audience member flipped one of those pieces. This is enough to completely determine the audience member's choice.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For example, suppose the board to be $4$-by-$4$ pieces wide. The positions are labeled as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;00 01 02 03&#xA;04 05 06 07&#xA;08 09 10 11&#xA;12 13 14 15&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and it looks like this after the audience member scrambles it:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;b b w b&#xA;w b w w&#xA;w b b b&#xA;w w b w&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The positions whose first bit is $1$ are the odd-numbered positions:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;- b - b&#xA;- b - w&#xA;- b - b&#xA;- w - w&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There's an odd number of black pieces in these spots, so Leonardo will flip an odd-numbered position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The positions with a $1$ in the second bit are the ones in the right half of the board:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;- - w b&#xA;- - w w&#xA;- - b b&#xA;- - b w&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There's an even number of black pieces here, so Leonardo will flip a piece on the left side of the board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The positions with a $1$ in the third bit are the second and fourth rows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;- - - -&#xA;w b w w&#xA;- - - -&#xA;w w b w&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There's an even number of black pieces here, so Leonardo flips a piece in the first or third row.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The positions with a $1$ in the fourth bit are the ones in the bottom half of the board:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;- - - -&#xA;- - - -&#xA;w b b b&#xA;w w b w&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There's an even number of black pieces here, so Leonardo flips a piece in the top half of the board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting all this together, we find that Leonardo flips position 1:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;- b - -&#xA;- - - -&#xA;- - - -&#xA;- - - -&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;After Leonardo and the audience member's flips, if any of the halves of the board we've considered have an odd number of black pieces, it's because the audience member flipped one of those pieces. Marco can repeat Leonardo's procedure to find the audience member's piece.&lt;/p&gt;&#xA;" OwnerUserId="1549" LastEditorUserId="1549" LastEditDate="2015-03-17T19:25:23.240" LastActivityDate="2015-03-17T19:25:23.240" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="10407" PostTypeId="2" ParentId="10379" CreationDate="2015-03-17T06:35:30.383" Score="3" Body="&lt;p&gt;Thanks to the &lt;strong&gt;BIG&lt;/strong&gt; hint, John's job is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a &lt;strong&gt;mechanic&lt;/strong&gt;. The path encoding uses hex scan codes for a &lt;strong&gt;&lt;a href=&quot;http://www.computer-engineering.org/ps2keyboard/scancodes2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;PS/2 keyboard&lt;/a&gt;&lt;/strong&gt; so path directions use the arrow keys: left=6B, down=72, right=74, up=75&lt;br&gt;&#xA;&lt;br&gt;&#xA; As identified by Uri, the path through the table is 3A, 24, 21, 33, 1C, 31, 43, 21 which decodes to MECHANIC.  Here is the path and table:&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/FLFgm.png&quot; alt=&quot;Table&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-03-19T07:59:31.917" LastActivityDate="2015-03-19T07:59:31.917" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10408" PostTypeId="2" ParentId="10334" CreationDate="2015-03-17T08:14:08.017" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;p&gt;This image represents instructions for a snake that fills the visited pixels with black paint.&lt;/p&gt;&#xA;&#xA; &lt;ul&gt;&lt;li&gt;Red color means &quot;go right&quot;&lt;/li&gt;&#xA; &lt;li&gt;Aqua color means &quot;go left&quot;&#xA; &lt;li&gt;Blue color means &quot;go up&quot;&#xA; &lt;li&gt;Green color means &quot;go down&quot;&#xA; &lt;/ul&gt;&#xA;&#xA; &lt;p&gt;The snake starts at the center of the image. (Edit: looks like it can start anywhere but the borders. The images below were produced using center as the starting point.)&lt;br&gt;&#xA; The snake stops as soon as it proceeds out of the image boundaries.&lt;br&gt;&#xA; After we're done, we fill remaining unvisited pixels with white paint.&lt;/p&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;p&gt;What is interesting is what happens if the snake visits a block that was already visited: it turns out that no matter if we go left, right, down or up, we still see the result pretty clearly.&lt;/p&gt;&#xA;&#xA;Going left:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/5RN4G.png&quot; alt=&quot;Going left&quot;&gt;&#xA;Going down:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/WGfYh.png&quot; alt=&quot;Going down&quot;&gt;&#xA;Going right:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/9RQPb.png&quot; alt=&quot;Going right&quot;&gt;&#xA;Going up:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/R39Ol.png&quot; alt=&quot;Going up&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;pre&gt;&lt;code&gt;#!/bin/python&#xA;from PIL import Image&#xA;&#xA;colors = [(255, 0, 0), (0, 255, 255), (127, 0, 255), (127, 255, 0)]&#xA;&#xA;i = Image.open('hidden.png')&#xA;d = i.load()&#xA;w, h = i.size&#xA;x, y = w &gt;&gt; 1, h &gt;&gt; 1&#xA;while x &gt;= 0 and y &gt;= 0 and x &amp;lt; w and y &amp;lt; h:&#xA;    oldX, oldY = x, y&#xA;    if d[x,y] == colors[0]:&#xA;        x += 1&#xA;    elif d[x,y] == colors&lt;a href=&quot;https://i.stack.imgur.com/5RN4G.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;:&#xA;        x -= 1&#xA;    elif d[x,y] == colors&lt;a href=&quot;https://i.stack.imgur.com/WGfYh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt;:&#xA;        y += 1&#xA;    elif d[x,y] == colors&lt;a href=&quot;https://i.stack.imgur.com/9RQPb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;3&lt;/a&gt;:&#xA;        y -= 1&#xA;    elif d[x,y] == (0,0,0):&#xA;        y -= 1 #or whatever&#xA;    else:&#xA;        raise Exception('Bad color: ' + str(d[x,y]))&#xA;    d[oldX, oldY] = (0, 0, 0)&#xA;&#xA;for x in range(w):&#xA;    for y in range(h):&#xA;        if d[x, y] != (0, 0, 0):&#xA;            d[x, y] = (255, 255, 255)&#xA;i.save('revealed.png')&lt;/code&gt;&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10259" LastActivityDate="2015-03-17T08:14:08.017" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10409" PostTypeId="2" ParentId="10331" CreationDate="2015-03-17T08:48:45.157" Score="3" Body="&lt;p&gt;I thought of: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mouse-touch (moustache), since the original &lt;em&gt;mouse&lt;/em&gt; picture has been &lt;em&gt;touch&lt;/em&gt;ed (and the mouse in the picture obviously has one).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7315" LastActivityDate="2015-03-17T08:48:45.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10410" PostTypeId="2" ParentId="10379" CreationDate="2015-03-17T09:19:09.777" Score="2" Body="&lt;p&gt;Here's an explanation of my progress so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Path&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. Four unique strings suggests a sequence of compass directions (udrl/nsew) to help us pick out characters in the grid.&lt;br&gt;&#xA; 2. There are 7 hyphens and 12 commas, suggesting that letters are separated by the former rather than the latter.&lt;br&gt;&#xA; 3. Starting at the top-left edge, there is only one way to interpret the directions such that we stay within the grid throughout: 74 must be right (since there are 5 of these with at most one reverse between them), 6B must be left (since there are 4 in a row), 72 must be down and 75 must be up.&lt;br&gt;&#xA; 4. The path is therefore: &lt;strong&gt;rrr-dl-rr-dl-r-llll-durrd-r&lt;/strong&gt;.&lt;br&gt;&lt;br&gt;&#xA;But why? Interpreting the values as hex (see @Len's answer) we get L=106, D=114, R=116, U=117 (or possibly WSEN). This is clearly not ASCII or any other lexicographic encoding (including mod-26). Neither does it correspond to the qwerty or dvorak keyboard layouts, or character positions in the question itself (with or without punctuation). The fact that one value is far from the others implies that the encoding is connected with the letters rather than the meaning of the letters (e.g. arrows). A potentially useful hint from question comments is: &quot;It's not ASCII, it's a different set of computer codes&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Table&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. The path above gives us the eight character word &lt;strong&gt;3A-24-21-33-1C-31-43-21&lt;/strong&gt;&lt;br&gt;&#xA; 2. Interpreting the values as hex we get 58-36-33-51-28-49-67-15.&lt;br&gt;&lt;br&gt;&#xA;The range of these values is 52, so again is unlikely to be lexicographic; @Len has also checked shifted mod-26 and shifted absolute diffs. The numbers lie in two ranges plus two outliers: 28-36, 49-58, 67 and 15. One or both of the outliers could be punctuation. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10220" LastEditorUserId="10220" LastEditDate="2015-03-18T09:29:54.973" LastActivityDate="2015-03-18T09:29:54.973" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10411" PostTypeId="2" ParentId="10405" CreationDate="2015-03-17T09:47:52.547" Score="4" Body="&lt;p&gt;Could it be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a SHADOW?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A partner in crime, though I wear a disguise.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your shadow looks much like you, except disguised by being single-coloured and two-dimensional.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I hide in plain sight. Averting all eyes.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A shadow is right there in plain sight, but you can't really see much of it because it's by definition a light-free zone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You can't run and gun, if you want to stop me. &lt;br&gt;&#xA; To succeed on that front, you need just turn and flee.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is what made me think of &quot;shadow&quot; as the solution. Your shadow follows you everywhere; you can't shoot it or run away from it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A doctor creates me, a goose fills her nest,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Doctors and geese both cast shadows. (OK, maybe this line doesn't quite fit.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But just give me time and I will manifest.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Anything will cast a shadow (provided there's some light around): it will always manifest.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Any who make me, end up the same...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Shadows are pretty featureless, identical to each other except in shape and size.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The question, of course, though, is: what is my name?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This line doesn't give any extra clues.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-03-17T09:47:52.547" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10412" PostTypeId="2" ParentId="10404" CreationDate="2015-03-17T09:54:02.920" Score="26" Body="&lt;p&gt;This indeed is an old puzzle. One possible source (but certainly not the first one) is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Andy Liu: Two Applications of a Hamming Code&lt;br&gt;&#xA;  The College Mathematics Journal 40, (Jan 2009), pp. 2-5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The trick on the $2^k\times2^k$ board is to associate each of the $2^{2k}$ squares with a unique binary number with $2k$ bits. (Note that the number of squares equals the number of such binary labels, so that there indeed exists a bijection).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Let $y$ denote the binary label of the square picked by $P$ from the audience.&lt;/li&gt;&#xA;&lt;li&gt;Consider the checkerboard at the moment &lt;strong&gt;just after&lt;/strong&gt; $P$ has flipped his coin, and consider the binary labels of all the squares with a white piece.&#xA;Let $z$ denote the XOR of all such white binary labels; hence $z$ is another binary number with $2k$ bits.&lt;/li&gt;&#xA;&lt;li&gt;Your goal is to flip a piece, so that afterwards the XOR of all white binary labels becomes equal to $y$. &lt;/li&gt;&#xA;&lt;li&gt;Mathematically, you would like to find a square whose binary label $x$ satisfies $x \oplus z=y$. &lt;/li&gt;&#xA;&lt;li&gt;Since the operation $\oplus$ is associative, and since $z\oplus z=0$ and $z\oplus 0=z$ for all $z$, this desired equation $x \oplus z=y$ is equivalent to $x\oplus z \oplus z=y\oplus z$ and furthermore to $x=y\oplus z$.&lt;/li&gt;&#xA;&lt;li&gt;Hence if we simply flip the square with binary label $y\oplus z$, our partner will compute the XOR of all white binary labels and then announce the corresponding square to the audience. Done.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note also that for the trick we do not need to know the geometric structure of the checkerboard: plain numbering of the squares suffices.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-09-28T07:45:20.207" LastActivityDate="2015-09-28T07:45:20.207" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10413" PostTypeId="2" ParentId="10216" CreationDate="2015-03-17T10:52:56.127" Score="2" Body="&lt;p&gt;$7∗(60+103) = 1141$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$219∗2∗3 = 1314$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2∗3∗102 = 600$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt{40000} = 200$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$23∗5 = 115$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$162+26 = 282$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1000}{2}−\frac{24}{3} = 492$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(150−7)∗12 = 1716$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$33∗50 = 1650$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{500}{2}+70 = 320$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{453}{25}∗10^{−1}∗500 = 916$&lt;/p&gt;&#xA;" OwnerUserId="10280" LastEditorUserId="5373" LastEditDate="2015-03-17T10:56:30.837" LastActivityDate="2015-03-17T10:56:30.837" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10414" PostTypeId="2" ParentId="10405" CreationDate="2015-03-17T10:57:21.120" Score="3" Body="&lt;p&gt;I'm thinking of a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Brood_parasite&quot; rel=&quot;nofollow&quot;&gt;Brood parasite lizard egg&lt;/a&gt;, just like the &lt;a href=&quot;http://simpsons.wikia.com/wiki/Bolivian_Tree_Lizard&quot; rel=&quot;nofollow&quot;&gt;Bolivian Tree Lizard&lt;/a&gt; in the Simpson series.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A partner in crime, though I wear a disguise.&lt;br&gt;&#xA;I hide in plain sight. Averting all eyes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A partner of the crime of eating the bird's eggs (parent's partner), but in disguise as its an egg itself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can't run and gun, if you want to stop me.&lt;br&gt;&#xA;To succeed on that front, you need just turn and flee.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When the egg hatches the bird can't do anything but flee, since its eggs have been eated by the lizard. Also, trying to stop the hatched lizards, other than being useless can lead to the birds death.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A doctor creates me, a goose fills her nest,&lt;br&gt;&#xA;But just give me time and I will manifest.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The doctor really baffles me, no idea. The goose is obvious. Just give enough time to the egg to hatch and the lizard will manifest.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Any who make me, end up the same...  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Who makes the egg will also make the eggs, continuing the cycle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The question, of course, though, is: what is my name?&lt;/p&gt;&#xA;" OwnerUserId="4652" LastActivityDate="2015-03-17T10:57:21.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10415" PostTypeId="1" AcceptedAnswerId="10434" CreationDate="2015-03-17T11:04:16.700" Score="2" ViewCount="254" Body="&lt;p&gt;Anafenza and Zurgo are playing a game. They have a playing board of $23*92$ squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea of the game is to put a square(of any size) onto the playing board. The one who can't put any more squares loses the game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zurgo is hastening, so he gets the priority and puts his first square on the board. Then it's Anafenza's turn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can Zurgo guarantee himself a victory? Can Anazenza guarantee herself a victory? Or will the game end in a draw?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As always, we assume that Zurgo and Anafenza use optimal strategies.&lt;/p&gt;&#xA;" OwnerUserId="9352" LastEditorUserId="5373" LastEditDate="2015-03-17T11:09:10.597" LastActivityDate="2015-03-17T15:16:56.783" Title="Drawing squares" Tags="|mathematics|logical-deduction|strategy|game|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10416" PostTypeId="2" ParentId="10415" CreationDate="2015-03-17T11:10:23.050" Score="1" Body="&lt;p&gt;Zurgo, who goes first, should place a square in the middle and then mimic Anafenza on the opposite side of the board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the event Zurgo does not place a square in the middle, Anafenza should then mimic Zurgo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Edit-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This strategy works for one simple reason if Zurgo is mimicking Anafenza this means that there must be an equivalent spot on the opposite side of the board, of which Anafenza played, for a square to be drawn.&lt;/p&gt;&#xA;" OwnerUserId="8785" LastEditorUserId="8785" LastEditDate="2015-03-17T11:14:31.220" LastActivityDate="2015-03-17T11:14:31.220" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10417" PostTypeId="1" AcceptedAnswerId="11014" CreationDate="2015-03-17T11:16:02.517" Score="-7" ViewCount="427" Body="&lt;p&gt;Marco and Leonardo decided to play a game on a checkerbard of 4×4 squares.&#xA;The board is initially filled with two-sided identical coins.&lt;/p&gt;&#xA;&lt;p&gt;The game notes that these two players play turns alternatively where the both  players are permitted to flip the coins on any 2×2 size square or a line of 1×4 fom side A to side B  .&lt;/p&gt;&#xA;&lt;p&gt;The first person who finishes the game with all 4×4 board pieces turned to the B side wins.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ra2LK.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Prove that player 2 is always bound to win&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Rules :&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;After some members' interventions i imposed some rules and reduced the board size to make the problem within any one's grasp .&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;any player cannot repeat his last or his opponent's last move&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;and moreover , any player is always prioritized to take action of biggest number still available of A sided coins ( for example a set containing 3 non flipped coins is a priori to be played than any set having 2 or 1 non flipped coins)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9517" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-03-26T10:35:11.590" Title="Checkerboard piece inverting game" Tags="|graph-theory|board-games|reachability|game-theory|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10418" PostTypeId="2" ParentId="10415" CreationDate="2015-03-17T11:36:02.700" Score="0" Body="&lt;p&gt;Zurgo&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because the Game started by the Zurgo.&#xA;my simple logic is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;23x92 = 2116&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2116/2 = 1058&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1058/2 = 529&lt;/p&gt;&#xA;&#xA;&lt;p&gt;529/2 = 264.5 which is having 0.5 fraction value. it means clearly it will end with Zurgo and there will no option to place a Squre for Anafenza&lt;/p&gt;&#xA;" OwnerUserId="10280" LastActivityDate="2015-03-17T11:36:02.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10420" PostTypeId="2" ParentId="10417" CreationDate="2015-03-17T12:00:58.293" Score="2" Body="&lt;p&gt;First of all, it's impossible for $player_1$ to win, assuming that $player_2$ is smart enough.&#xA;Indeed, $player_2$ can simply repeat $player_1$'s move, locking the game in an infinite (boring) loop.&#xA;As well, if $player_2$ had a strategy to win, $player_1$ could apply the aforementioned strategy, replicating $player_2$'s last move.&#xA;Shortly, &lt;strong&gt;this game can be forced to be endless by both players&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, I'm the authentic Leonardo of the game; the one who submitted the original puzzle (Marco) is actually a friend of mine.&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10248" LastEditDate="2015-03-17T14:16:19.083" LastActivityDate="2015-03-17T14:16:19.083" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10421" PostTypeId="1" AcceptedAnswerId="10566" CreationDate="2015-03-17T12:02:42.500" Score="3" ViewCount="2948" Body="&lt;p&gt;You wake up in an underground maze, at the location marked with $X$ in the below image. A voice tells you: &lt;em&gt;&quot;You need to find the exit of the maze if you want to live!&quot;&lt;/em&gt;. You begin to panic and run around the maze, but just after a few minutes you find the exit and escape. You are still in a cave, so you head to what seems to be an exit, when suddenly a crazy man jumps in front of you with a gun in his hand and says: &lt;em&gt;&quot;If you followed all the clues in the maze and don't want to get shot you would be able to tell me what's the color of the sky outside right now!&quot;&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What do you tell the crazy man?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Eo3ze.png&quot; alt=&quot;Maze&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Notes&lt;/h3&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I will only accept an answer that can explain how you came to the solution.  &lt;/li&gt;&#xA;&lt;li&gt;The &quot;cave&quot; you enter after leaving the maze is just meant to prevent you from the seeing the color of the sky. You can imagine that the end of the maze is just a small room with an obvious door at the end saying &quot;exit&quot;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The only thing that you have to figure out is the maze, the text has no clues in it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you are at a computer the device to help you solve this should be close to you. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As people have guessed already the device is a keyboard&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5560" LastEditorUserId="5560" LastEditDate="2015-03-18T12:58:20.617" LastActivityDate="2015-12-09T15:01:46.783" Title="After escaping the maze, what color is the sky?" Tags="|enigmatic-puzzle|" AnswerCount="9" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="10423" PostTypeId="2" ParentId="10381" CreationDate="2015-03-17T12:29:29.920" Score="0" Body="&lt;p&gt;Agree with @PythonMaster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But here is another way to look at it without the &quot;...&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have 250 &quot;scraps&quot;, then for every 11, you can make a new one.  There are 22 groups of 11 in 250 with 8 left over.  So, from 250, you can use $22\times11=242$, and get 22 new ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$250 - (22\times11) + 22 = 250 - 242 + 22 = 30$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, from these 30, you can get 2 more by using 22.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$30 - (2\times11) + 2 = 30 - 22 + 2 = 10$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in total, you've made $22+2=24$ new scraps, you've used $242+22=264$ scraps in total,  and have 10 left over.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2015-03-17T12:29:29.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10424" PostTypeId="2" ParentId="10421" CreationDate="2015-03-17T12:38:42.950" Score="5" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bBGzm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;He got out of the cave and entered into the other cave as shown in the image. So that, means he didn't understood that he is entering in to another cave in the same maze. Which means it was perfect dark. That means night (darkest night). So the color of the sky will be black.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="2730" LastEditDate="2015-03-17T12:43:44.067" LastActivityDate="2015-03-17T12:43:44.067" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10425" PostTypeId="2" ParentId="10399" CreationDate="2015-03-17T13:18:20.973" Score="3" Body="&lt;p&gt;My guess is that &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; D killed B with a log, C killed A with berries&lt;br&gt;&#xA; I think G taking a piss is a red herring.&lt;br&gt;&#xA; Out of the remainder, only D doesn't have red hands, which would have been on the murder log.&lt;br&gt;&#xA; Also, being big and strong, how could both C and D be taken out at the same time?&lt;br&gt;&#xA; So, they each killed A and B, then faked getting hit on the back of the head. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-03-17T13:18:20.973" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10426" PostTypeId="1" CreationDate="2015-03-17T13:44:38.940" Score="3" ViewCount="3424" Body="&lt;p&gt;If a planet completes its circle around the sun in 120 years, and another planet does so in 70 years, when will they come in a straight line, if they are now?&lt;/p&gt;&#xA;" OwnerUserId="10285" LastEditorUserId="69582" LastEditDate="2021-06-01T23:17:34.620" LastActivityDate="2021-06-01T23:17:34.620" Title="Alignment of planets that orbit at different speeds" Tags="|mathematics|calculation-puzzle|outer-space|" AnswerCount="6" CommentCount="0" ClosedDate="2015-09-23T05:35:06.323" ContentLicense="CC BY-SA 4.0" />
  <row Id="10427" PostTypeId="2" ParentId="10426" CreationDate="2015-03-17T13:50:55.350" Score="1" Body="&lt;p&gt;In &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 840 years&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 840 is the least common multiple of 120 and 70.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Although&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; These planets will be aligned in a straight line much more often, it will take them 840 years to be &lt;em&gt;in a straight line &lt;strong&gt;as if they are now&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9352" LastActivityDate="2015-03-17T13:50:55.350" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10428" PostTypeId="2" ParentId="10426" CreationDate="2015-03-17T13:57:46.760" Score="3" Body="&lt;p&gt;The next time they will be in any alignment is in &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 168 years. &#xA;&lt;br&gt;&#xA; Effectively, every rotation of the fast planet it gains 50 &quot;years&quot; on the slower one. So, after 2 rotations, it's 20 &quot;years&quot; behind the slower. Since the rate of gain is 50/rotation, it will require .4 rotations to catch up. .4 rotations is 28 years for the fast planet. 2.4 rotation is 168 years. &#xA;&lt;br&gt;&#xA; To check our math, 168 years of the slow planet is 1.4 rotations, so they are aligned. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-03-17T13:57:46.760" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10429" PostTypeId="1" AcceptedAnswerId="10444" CreationDate="2015-03-17T14:22:24.427" Score="1" ViewCount="2634" Body="&lt;p&gt;An eternity has passed, but it feels like time has stood still&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I killed my true love and my neighbors, but they have not left me&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long ago...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had the power to raise the seas, and most things you knew and loved&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had the power to destroy you all, but you still managed to step on me&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You forgot about me even in your homes, while I stuck around&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your weapons… You lowly, violent humans and your guns --&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their fundamentals managed to kill me&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But even in the darkness of death, I am still very much alive&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="10267" LastActivityDate="2015-03-17T18:30:02.203" Title="I was once a god. What am I?" Tags="|riddle|word|" AnswerCount="5" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10430" PostTypeId="1" AcceptedAnswerId="10433" CreationDate="2015-03-17T14:28:12.423" Score="5" ViewCount="1130" Body="&lt;p&gt;I am known as the end of all.&lt;br&gt;&#xA;When others fall short, I stand tall.&lt;br&gt;&#xA;You'll see I can take on different forms,&lt;br&gt;&#xA;Don't worry, they are quite the norm.&lt;br&gt;&#xA;My brother and I may look a like,&lt;br&gt;&#xA;But we are different, as say Paul and Mike.&lt;br&gt;&#xA;Sometimes I shoot straight, others I bend,&lt;br&gt;&#xA;But do not worry, I don't need to mend.&lt;br&gt;&#xA;So answer this riddle if you dare,&lt;br&gt;&#xA;The beginning of a laugh is in the air.  &lt;/p&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-03-17T14:46:48.727" Title="Apocalypse Now: The end of All. What am I?" Tags="|riddle|word|rhyme|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10431" PostTypeId="2" ParentId="10426" CreationDate="2015-03-17T14:34:55.017" Score="-1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Never.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Let's call the planets $A$ the faster and $B$ the slower. In 120 years $B$ completed a loop while $A$ completed $1 + \frac{50}{70} = \frac{12}{7}$ of loop. For this purpose we can remove full loops since they don't add information on the relative position of the planets, so we can say that $A$ completed $\frac{5}{7}$ of loop vs. $1$ loop of $B$.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Now in the time $A$ completes the remaining $\frac{2}{7}$ of loop to catch $B$ (in ${70}\cdot{\frac{2}{7}} = 20$ years), $B$ moved away another $\frac{20}{120} = \frac{1}{6}$ of loop, so $A$ has to move more to catch it.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Now in the time $A$ completes the remaining $\frac{1}{6}$ of loop to catch $B$ (in ${70}\cdot{\frac{1}{6}} = 11,67$  years), $B$ moved away another $\frac{\frac{70}{6}}{120} = \frac{7}{72}$ of loop,  so $A$ has to move more to catch it.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Now in the time $A$ completes the remaining $\frac{7}{72}$ of loop to catch $B$ (in ${70}\cdot{\frac{7}{72}} = 6,805$  years), $B$ moved away another $\frac{\frac{7}{72}}{120} = \frac{7}{8640}$ of loop,  so $A$ has to move more to catch it.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;http://en.wikipedia.org/wiki/Zeno%27s_paradoxes#Achilles_and_the_tortoise&quot; rel=&quot;nofollow&quot;&gt;Now in the time $A$ completes the remaining $\frac{7}{8640}$ of loop...&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4652" LastEditorUserId="4652" LastEditDate="2015-03-17T15:01:39.990" LastActivityDate="2015-03-17T15:01:39.990" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10432" PostTypeId="2" ParentId="10429" CreationDate="2015-03-17T14:46:41.740" Score="3" Body="&lt;p&gt;Is it:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Earth?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An eternity has passed, but it feels like time has stood still  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The earth has been around a long time.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I killed my true love and my neighbors, but they have not left me  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Natural distasters kill many, and comets and other space junk (neighbors) have crashed into the earth, but everything that dies lies in the earth.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Long ago...&#xA;I had the power to raise the seas, and most things you knew and loved  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Earthquakes can create tsunamis &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I had the power to destroy you all, but you still managed to step on me  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The earth is amazing, but we walk on it every day.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You forgot about me even in your homes, while I stuck around  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We often forget about the earth in what we do, especially at home when we aren't on the &quot;ground,&quot; but the earth never moves.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your weapons… You lowly, violent humans and your guns --&#xA;Their fundamentals managed to kill me  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Guns, missles, bombs, etc. have done huge impact on the earth and killed lots of living things.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But even in the darkness of death, I am still very much alive  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Though there are places that seem dead, or we think the earth isn't alive, it is crawling with life.  That, or winter makes everything seem dead when it is really just &quot;sleeping.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-03-17T14:46:41.740" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10433" PostTypeId="2" ParentId="10430" CreationDate="2015-03-17T14:46:48.727" Score="6" Body="&lt;p&gt;My answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter &quot;L&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am known as the end of all.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Last letter of &quot;all&quot; is &quot;L&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When others fall short, I stand tall.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Uncapitalized, the &quot;L&quot; is still big.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I shoot straight, others I bend,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Different types of handwriting, fonts, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The beginning of a laugh is in the air.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Beginning of &quot;laugh&quot; is &quot;L&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10288" LastActivityDate="2015-03-17T14:46:48.727" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10434" PostTypeId="2" ParentId="10415" CreationDate="2015-03-17T15:16:56.783" Score="3" Body="&lt;p&gt;If Zurgo places a 22x22 square on the board so that there are two areas of 23x35 and one of 1x22 left, he has created a board on which he can mimic every move by Anafenza.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means that no matter what move Anafenza can make, Zurgo can copy, which means that Zurgo will always be able to move - unless Anafenza just lost!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with a smaller square is tricky - the point is to divide the board into two mirrored halves. a 1x1 square will not leave equal areas in the lenth direction of the borad (46x23 versus 45 by 43) - this is true for any odd-numbered starting square!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For smaller even-sized squares, the odd number of squares left &quot;next&quot; to it makes it again impossible to mimic. If Zurgo plays a 20x20 square, Anafenza can play a 3x3 square next to it (in the remaining 3x20 area) that can not exactly be mirrored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(in the 1x20 area Anafenza can only play a 1x1 square which can always be mirrored in the same area)&lt;/p&gt;&#xA;" OwnerUserId="3903" LastActivityDate="2015-03-17T15:16:56.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10435" PostTypeId="2" ParentId="10043" CreationDate="2015-03-17T15:17:53.113" Score="0" Body="&lt;p&gt;The prisoner turns each knob to determine which of the three doors is unlocked, but does not open the unlocked door. He then asks the warden which of the remaining doors leads to freedom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, the prisoner inspects the doorknobs to find evidence of use (fingerprints, residual warmth, &lt;em&gt;etc.&lt;/em&gt;), operating under the assumption that the door to prison is the only one that has been used. He then asks the warden which of the remaining doors leads to freedom.&lt;/p&gt;&#xA;" OwnerUserId="9828" LastActivityDate="2015-03-17T15:17:53.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10436" PostTypeId="2" ParentId="558" CreationDate="2015-03-17T15:18:39.080" Score="0" Body="&lt;p&gt;There is a great deal of subjectivity involved.  There is also a degree of compatibility between the constructor and the solver that enters into the equation. &lt;/p&gt;&#xA;" OwnerUserId="10289" LastActivityDate="2015-03-17T15:18:39.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10437" PostTypeId="1" AcceptedAnswerId="10456" CreationDate="2015-03-17T15:24:45.040" Score="12" ViewCount="399" Body="&lt;p&gt;The following words have a hidden but &lt;em&gt;very distinguished&lt;/em&gt; property that allows to order them in a very natural way:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;pins up iron sigma eyes another ball face sum monkeys&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the &lt;em&gt;very distinguished&lt;/em&gt; property of these words?&lt;br&gt;&#xA;What is the resulting ordering?&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="8874" LastEditDate="2015-03-19T09:38:10.870" LastActivityDate="2015-03-19T12:14:11.537" Title="Bring the words into the right order!" Tags="|pattern|word-sequence|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10438" PostTypeId="2" ParentId="10426" CreationDate="2015-03-17T15:24:49.167" Score="2" Body="&lt;p&gt;The question never stated how the 2 planets are rotating.  Therefore by assuming they are rotating on the same plane in opposite direction, it'll take less than 60 years for them to become a straight line again, and many more until in 840 years they'll be aligned again in the exact location as they are now&lt;/p&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2015-03-17T15:24:49.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10439" PostTypeId="2" ParentId="10429" CreationDate="2015-03-17T15:27:52.117" Score="2" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the planet Saturn ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The god Saturn or Cronus was killed by his own children, and some of the moons of planet Saturn are named after the children.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2015-03-17T15:27:52.117" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10440" PostTypeId="2" ParentId="10399" CreationDate="2015-03-17T15:42:44.293" Score="4" Body="&lt;p&gt;My thoughts, building off of rand al'thor's post:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's put things into a timeline as best we can.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;C&amp;amp;D set up the tents.&lt;/li&gt;&#xA;&lt;li&gt;A&amp;amp;B collect firewood at the same time. E&amp;amp;F are bringing berries back.&lt;/li&gt;&#xA;&lt;li&gt;A (being OCD) neatly stacks all the firewood (it's possible this is just his firewood, or firewood from both A&amp;amp;B, probably irrelevant)&lt;/li&gt;&#xA;&lt;li&gt;B continues to bring firewood.&lt;/li&gt;&#xA;&lt;li&gt;It seems that C&amp;amp;D have red on the back of their heads&lt;/li&gt;&#xA;&lt;li&gt;Keep in mind that C&amp;amp;D are also very, &lt;em&gt;very&lt;/em&gt; smart&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What does this mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B's murderer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A is &quot;prone to arguments&quot; and has OCD. He had neatly stacked up all this wood, when B came along with more wood. He freaked out (nowhere to neatly put it - the woodpile is already done) and grabbed the top piece of wood to attack B.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What about C&amp;amp;D?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C&amp;amp;D were there when A killed B. But, being perpetually scared of things, they were... well... scared. They also recognized, being smart, that A would just kill them, so they told A that he could knock them out instead and they wouldn't say anything. A did with the same wood log that he used to kill B, so that explains the red on the back of their head.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And for A?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A, needing to leave the campsite, goes off into the woods. He knows that E&amp;amp;F are returning with more berries, so he tries to go in a different direction. He runs into G, going to take a piss. G is mad at him for initiating the argument earlier and, since both are prone to arguments/fighting, they argue and G pushes him into the river and he drowns.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When C&amp;amp;D wake up:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They could just tell about A killing B, but being smart they have already figured out that E, F or G must have killed A. They choose to stay silent to prevent more conflict, reasoning that if you knew A killed B, you'd start investigating who killed A more and that would make the killer (who they may or may not know) more violent. I think they're telling the truth about the berries - they'd make such a cute couple, I've been rooting for them all along anyway!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;[EDIT - Attempt 2]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Leaving Attempt 1 so others can also learn from it. Building off of the same ideas as above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B's murderer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C. He's the only one with red on his hands that doesn't really have a valid reason. B must have been the one to scream &quot;I'm going to kill you&quot; in the first argument and we know that C/D work to prevent violence. Being smart, they must have figured that one death is better than 2 deaths. They used a nearby tool (the log). We also know C is strong enough to kill B this way.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;C&amp;amp;D knocked out by:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A. He probably came back into the camp to discover B dead and got angry because his best friend was dead. Not strong enough to kill C&amp;amp;D, he grabbed the same log, snuck up behind them and hit them on the back of the head. This explains why the backs of their heads have red on them. He then ran off into the forest (in grief? in rage? sure, why not!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A dies by the hand of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Raspberries! This one I'm not sure of (nor am I the previous ones, but my logic here is definitely a stretch). Nobody was wet from holding A under, so I'm guessing it's related to the EpiPen. If the log that was used to kill B fell into the pile of berries near him (hence why it was crushed) then it could have caused an allergic reaction in A when he picked it up. One of the effects of raspberry allergies can be dizziness and fainting, so it's possible A fainted and fell into the river, especially if he was out of breath from hitting C&amp;amp;D and running away.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;[EDIT] Attempt 3 (after this I'm starting a new response)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The killers (both listed here) are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C and D. C killed B and D killed A.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The logic for B:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For B's death, we've established that A didn't do it as B was his best friend. E and F are too weak to hit him hard enough, and G was with you the whole time and without the navigation skills to get back to camp, knock out C&amp;amp;D, kill B and get back to you (and, presumably, urinate) in just a few minutes. So that leaves C or D. C must have killed B - his hands were red, which may not necessarily be the blood (it might just be from berries), but notably D's hands weren't red, this leads us to:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The logic for A:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C must have killed B because D was off killing A. He had been eating some berries, but his hands were clean later because he used them to drown A in the river. The red washed off. As someone else said, they then pretended to be knocked out.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why the murders were done?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C and D &lt;em&gt;are&lt;/em&gt; actually a couple. The red on the back of their heads was from them making out (as someone else said), but they don't want anyone to know. A discovered them in camp, so D chased him down and drowned him. C killed B because B came back into camp and he knew he would notice D gone, which would lead to D being found out - by killing B, C preemptively avoided conflict (which we know they do). They then pretended to be unconscious.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastEditorUserId="1620" LastEditDate="2015-03-17T22:59:41.137" LastActivityDate="2015-03-17T22:59:41.137" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="10441" PostTypeId="2" ParentId="10429" CreationDate="2015-03-17T15:49:55.900" Score="2" Body="&lt;p&gt;It is a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Black Dwarf&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An eternity has passed&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Black Dwarf is the last stage of a star.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I killed my true love and my neighbors, but they have not left me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As a star, at the end of your life you started expanding, then absorbed all the planets orbiting around you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I had the power to raise the seas, and most things you knew and loved&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The gravity was strong in you!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I had the power to destroy you all, but you still managed to step on me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An advanced civilization with extreme technologies managed to step on your surface.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You forgot about me even in your homes, while I stuck around&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It was night.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your weapons… You lowly, violent humans and your guns --&#xA;Their fundamentals managed to kill me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Something went wrong during an experiment...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But even in the darkness of death, I am still very much alive&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You're black, invisible in the darkness, but you're still there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-03-17T15:49:55.900" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10443" PostTypeId="1" AcceptedAnswerId="10683" CreationDate="2015-03-17T16:12:43.260" Score="5" ViewCount="1905" Body="&lt;p&gt;An $8\times8$ checkerboard is filled with two-sided coins (that are blue on one side and red on the other side).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following picture shows three examples of a &lt;em&gt;cross&lt;/em&gt; (multiplication sign):&#xA;the five red coins at the right margin form a standard cross centered at their middle coin; the two red coins in the upper left corner form a cross that is centered at the coin in the corner (the three remaining squares of the cross are outside the board and irrelevant); the three red coins in the middle of the upper half form another cross (with two squares outside the board):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wNR9Q.jpg&quot; alt=&quot;checkerboard&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Marco starts with the situation where all coins are blue, and by flipping some coins he wants to reach the situation where all coins are red.&#xA;Whenever Marco flips one coin $x$, the other coins from the cross centered at $x$ are flipped simultaneously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;   &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Can Marco reach the situation where all coins are red?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus Question:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; this question is worth 100 points &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is the number of moves to reach the goal according of $k$ where $k$ red coins are stochatically distributed on the board ?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Hint:&lt;/strong&gt; Linear algerba is required&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9517" LastEditorUserId="8874" LastEditDate="2016-03-18T09:50:53.847" LastActivityDate="2016-03-18T09:50:53.847" Title="Coin flipping game" Tags="|strategy|combinatorics|checkerboard|reachability|" AnswerCount="5" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10444" PostTypeId="2" ParentId="10429" CreationDate="2015-03-17T16:13:17.153" Score="7" Body="&lt;p&gt;Just to add another space related answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Black Hole&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An eternity has passed&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For a star to become a black hole it has to pass several million years.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I killed my true love and my neighbors, but they have not left me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is a story about a solar system and his star. The star become a black hole, and in the process killed his true love, an earth like planet, and the neighbours, the other planets.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I had the power to raise the seas, and most things you knew and loved&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everything we see here on earth is thanks to the sun. Everything.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Edit by @Tosen:&lt;/strong&gt; Correction: this explanation does work, however, an expanding star is hot enough to evaporate (raise) just about everything.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I had the power to destroy you all, but you still managed to step on me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_flare&quot;&gt;Sun flares&lt;/a&gt; can kill us all. And the earth like civilization could somehow step on his star. At night obviously, too hot! No seriously, no clue.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Edit by @Tosen:&lt;/strong&gt; Correction: Expanding stars destroy everything. Hollywood stars on sidewalks are stepped on.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You forgot about me even in your homes, while I stuck around&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We don't see the sun quite often while we are indoor. Also with artificial light we don't need it, but still it is outside.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Edit by @Tosen:&lt;/strong&gt; Correction: This is a good explanation. But I was thinking gold &quot;star stickers&quot; that kids are given for good grades.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your weapons… You lowly, violent humans and your guns --&#xA;Their fundamentals managed to kill me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; While trying to harvest solar power the civilization managed to start some kind of destructive process for the star.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Edit by @Tosen:&lt;/strong&gt; Correction: Ultimately, the depletion of &quot;fusionable&quot; elements (iron!) is what kills a star.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But even in the darkness of death, I am still very much alive&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The black hole still is pumping matter, no signs of ever stop...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4652" LastEditorUserId="4652" LastEditDate="2015-03-17T16:51:32.837" LastActivityDate="2015-03-17T16:51:32.837" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10445" PostTypeId="2" ParentId="10399" CreationDate="2015-03-17T16:28:50.940" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; G killed B, C killed A.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;C and D both have red stains on the back of their heads, but only C's hands are red. If we assume the backs of their heads were red from kissing someone who was fondling their hair, it's clear that D has been cheating with someone who's been handling the berries. A is allergic, so he's ruled out. The narrator and G weren't present, and E + F were together the whole time so neither of them would make a move with D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which means that D must have been cheating with B.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let's assume C finds out, he wants to hurt B but he knows he'd never be able to overpower him. B died from a blunt strike to the head in the middle of the campsite, not exactly the method operandi of a genius, the only two people who'd fit that kind of attack are A and G.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C's plan was probably to slip raspberries into A's food, knowing he's allergic but has an epipen. A/B and E/F have already had an argument so C hopes that B will get hurt when A accuses E and F of trying to kill him. Unfortunately something goes wrong. A ingests his food, panics, and doesn't realize why he's going into anaphylactic shock. He stumbles into the wood and collapses face down in the river. G hears the commotion and sees A die shortly after B returns to the campsite, G incorrectly assumes that B killed A (G appears to be mildly autistic so wouldn't necessarily pick up on the fact that they're close friends).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; G kills B in revenge of A, and also to protect his other friends. At this point C has just attacked D and knocked him unconscious, but after seeing B's death he goes back to D and feigns unconsciousness until the narrator returns to the campsite.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10246" LastActivityDate="2015-03-17T16:28:50.940" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10446" PostTypeId="2" ParentId="10405" CreationDate="2015-03-17T16:43:27.347" Score="1" Body="&lt;p&gt;It sounds like it might be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; cover&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A partner in crime, though I wear a disguise.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The partner is covering you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I hide in plain sight. Averting all eyes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A cover is in plain sight, it still averts the eyes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can't run and gun, if you want to stop me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You won't find a cover by running and gunning.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To succeed on that front, you need just turn and flee.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you turn and flee you break your cover.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A doctor creates me, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A cover letter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;a goose fills her nest,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Covering it with saplings and down.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But just give me time and I will manifest.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You'll eventually notice a cover if you look long enough at it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Any who make me, end up the same...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You make a cover to end up covered.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9910" LastActivityDate="2015-03-17T16:43:27.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10447" PostTypeId="2" ParentId="10405" CreationDate="2015-03-17T17:16:17.950" Score="0" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Odor/gas&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wears a disguise might refer to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; its ability to mask other smells&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Avert all eyes probably refers to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; how most gases are invisible to the naked eye&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can't run and gun/turn and flee could mean&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; moving away from the smell/gas is the only way to find relief from it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Doctors create me might refer to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; chemists creating gases in a lab, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The geese part might mean &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; that their nests smell bad&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10293" LastEditorUserId="10293" LastEditDate="2015-03-17T17:35:47.367" LastActivityDate="2015-03-17T17:35:47.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10449" PostTypeId="1" AcceptedAnswerId="10450" CreationDate="2015-03-17T17:35:56.440" Score="10" ViewCount="1156" Body="&lt;p&gt;What is the significance of this gentleman:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;J. Jason, DJ, FM/AM&lt;/p&gt;&#xA;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;" OwnerUserId="10294" LastEditorUserId="5188" LastEditDate="2015-03-17T20:09:00.307" LastActivityDate="2015-03-18T11:49:33.303" Title="A DJ at a radio station?" Tags="|word|enigmatic-puzzle|wordplay|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10450" PostTypeId="2" ParentId="10449" CreationDate="2015-03-17T17:38:00.510" Score="13" Body="&lt;p&gt;The significance is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; His initials spell the months of the year.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Specifically:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;J&lt;/strong&gt;une &lt;strong&gt;.&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;J&lt;/strong&gt;uly&lt;br&gt;&#xA; &lt;strong&gt;A&lt;/strong&gt;ugust&lt;br&gt;&#xA; &lt;strong&gt;S&lt;/strong&gt;eptember&lt;br&gt;&#xA; &lt;strong&gt;O&lt;/strong&gt;ctober&lt;br&gt;&#xA; &lt;strong&gt;N&lt;/strong&gt;ovember &lt;strong&gt;,&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;D&lt;/strong&gt;ecember&lt;br&gt;&#xA; &lt;strong&gt;J&lt;/strong&gt;anuary &lt;strong&gt;,&lt;/strong&gt;&#xA; &#xA; &lt;strong&gt;F&lt;/strong&gt;ebruary&lt;br&gt;&#xA; &lt;strong&gt;M&lt;/strong&gt;arch &lt;strong&gt;/&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;A&lt;/strong&gt;pril&lt;br&gt;&#xA; &lt;strong&gt;M&lt;/strong&gt;ay&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="4692" LastEditDate="2015-03-18T11:49:33.303" LastActivityDate="2015-03-18T11:49:33.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10451" PostTypeId="2" ParentId="10449" CreationDate="2015-03-17T17:38:46.050" Score="12" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;J&lt;/strong&gt;une &lt;strong&gt;J&lt;/strong&gt;uly &lt;strong&gt;A&lt;/strong&gt;ugust &lt;strong&gt;S&lt;/strong&gt;eptember &lt;strong&gt;O&lt;/strong&gt;ctober &lt;strong&gt;N&lt;/strong&gt;ovember &lt;strong&gt;D&lt;/strong&gt;ecember &lt;strong&gt;J&lt;/strong&gt;anuary &lt;strong&gt;F&lt;/strong&gt;ebruary &lt;strong&gt;M&lt;/strong&gt;arch &lt;strong&gt;A&lt;/strong&gt;pril &lt;strong&gt;M&lt;/strong&gt;ay&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And yes, I've already heard this joke before, but still a good one!&lt;/p&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-03-17T17:38:46.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10452" PostTypeId="2" ParentId="10405" CreationDate="2015-03-17T17:43:45.260" Score="2" Body="&lt;p&gt;You could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an unborn baby/embryo&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A partner in crime, though I wear a disguise.&#xA;I hide in plain sight. Averting all eyes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Always with the mother, but also hidden inside of her.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can't run and gun, if you want to stop me.&#xA;To succeed on that front, you need just turn and flee.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you don't have sex, you don't have a baby.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A doctor creates me, a goose fills her nest,&#xA;But just give me time and I will manifest.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Doctors deliver babies, geese fill their nests with eggs, and in time eggs become living beings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Any who make me, end up the same...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;End up the same&quot; could be as a parent or just the act of giving birth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The question, of course, though, is: what is my name?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When a baby is born, we name it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9978" LastActivityDate="2015-03-17T17:43:45.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10453" PostTypeId="2" ParentId="6181" CreationDate="2015-03-17T17:44:23.550" Score="11" Body="&lt;p&gt;I did it in these &lt;strong&gt;17&lt;/strong&gt; moves:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;99C423EDDBBB6CC88&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is the initial position I used:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HfjsK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the position after I solved it with the 17 moves above. The textbox in the top left corner keeps track of moves made, but I crossed through it with a black line so that it wouldn't give away the answer to those who didn't want to see it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/M6i1B.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&lt;strong&gt;How I Solved It:&lt;/strong&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;It was very hard to figure out using only the pictures, so I did 2 things:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Converted the color paths part of the puzzle to numbers, and oriented them correctly on a piece of paper. (not shown here)&lt;br&gt;&lt;/li&gt;&#xA;  &lt;li&gt;Created a quick and crude graphical way (using C# WinForms) to rotate them, based on the 15 different levers.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since I could not (easily) rotate the hexagons on the graphical part, I solved it first without rotating any of them, but then had to use paper cut-outs to check that each hexagon was oriented correctly also. And they were.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set-up took almost all of the time(3-4 hours since I set it up wrong the first time and then solved THAT problem too before realizing it was set up wrong! haha). Solving it took about 15 minutes. And checking it took about 45 minutes, due to me having to cut out 35 hexagons and set up the initial position, and then carefully perform each of the 17 moves.&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-06-19T18:06:53.580" LastActivityDate="2015-06-19T18:06:53.580" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10454" PostTypeId="2" ParentId="10405" CreationDate="2015-03-17T17:49:31.960" Score="1" Body="&lt;p&gt;Sounds like the answer is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Justice&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A partner in crime, though I wear a disguise.&lt;br&gt;&#xA;I hide in plain sight. Averting all eyes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Justice and Crime coexists.  The Statue of Justice wearing a blindfold&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can't run and gun, if you want to stop me.&lt;br&gt;&#xA;To succeed on that front, you need just turn and flee.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Read justice as Just-ice, above describe some properties of ice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A doctor creates me, a goose fills her nest,&lt;br&gt;&#xA;But just give me time and I will manifest.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;goose often nest on the statute? and an artist with PHD created the statue?&#xA;Justice will be shown eventually?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Any who make me, end up the same...&lt;br&gt;&#xA;The question, of course, though, is: what is my name?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Whose create Justice end up in Justice?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2015-03-17T17:49:31.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10455" PostTypeId="1" AcceptedAnswerId="10459" CreationDate="2015-03-17T17:51:13.353" Score="2" ViewCount="802" Body="&lt;p&gt;What is next in the sequence?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;50, 42, 34, 28, 23, 18,14, ? &lt;/p&gt;&#xA;" OwnerUserId="10294" LastEditorUserId="10294" LastEditDate="2015-03-17T18:11:14.720" LastActivityDate="2015-03-17T18:42:00.057" Title="From a math Professor, Bernard Budiansky" Tags="|pattern|number-sequence|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10456" PostTypeId="2" ParentId="10437" CreationDate="2015-03-17T18:02:20.327" Score="13" Body="&lt;p&gt;I believe the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Another Face Sum Eyes Monkeys Sigma Up Ball Iron Pins&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The property is that these words can be prefixed with numbers to create expressions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One Another (an expression meaning &quot;each other&quot;)&lt;br&gt;&#xA; Two-Face (Harvey Dent, a villain in Batman)&lt;br&gt;&#xA; ThreeSum (the &lt;a href=&quot;http://en.wikipedia.org/wiki/3SUM&quot;&gt;3SUM&lt;/a&gt; problem)&lt;br&gt;&#xA; Four-Eyes (an insulting term for somebody who wears glasses)&lt;br&gt;&#xA; Five Monkeys (little ones, &lt;a href=&quot;http://en.wikipedia.org/wiki/Five_Little_Monkeys&quot;&gt;jumping on the bed&lt;/a&gt;)&lt;br&gt;&#xA; &lt;a href=&quot;http://en.wikipedia.org/wiki/Six_Sigma&quot;&gt;Six Sigma&lt;/a&gt; (a statistical standard for process consistency)&lt;br&gt;&#xA; Seven-Up (a drink or a children's party game)&lt;br&gt;&#xA; Eight Ball (a fortune-telling device, or a billiards game)&lt;br&gt;&#xA; Nine-Iron (a type of golf club)&lt;br&gt;&#xA; Ten Pins (in bowling)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-03-17T19:17:21.897" LastActivityDate="2015-03-17T19:17:21.897" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10457" PostTypeId="2" ParentId="10455" CreationDate="2015-03-17T18:24:43.540" Score="8" Body="&lt;p&gt;Considering Bernard Budiansky is from:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; New York City and the reverse sequence shows up in &lt;a href=&quot;https://oeis.org/search?q=14%2C18%2C23%2C28%2C34%2C42%2C50&amp;amp;sort=&amp;amp;language=english&amp;amp;go=Search&quot;&gt;OEIS&lt;/a&gt;:&#xA; the sequence is &lt;strong&gt;local stops on New York City Broadway line (IRT #1) subway&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; There is &lt;strong&gt;no &quot;next number&quot;&lt;/strong&gt; in the sequence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastActivityDate="2015-03-17T18:24:43.540" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10458" PostTypeId="2" ParentId="10429" CreationDate="2015-03-17T18:30:02.203" Score="2" Body="&lt;p&gt;I'm pretty sure it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the Moon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We have stepped on it (despite various conspiracy theories), it is supposed to have its origin in the collision between a primitive Earth and an incoming planet/asteroid (it &quot;killed&quot; its &quot;neighbor&quot;, but its successor stayed) , it affects the sea tides...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How close am I? :)&lt;/p&gt;&#xA;" OwnerUserId="10295" LastActivityDate="2015-03-17T18:30:02.203" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10459" PostTypeId="2" ParentId="10455" CreationDate="2015-03-17T18:36:04.313" Score="12" Body="&lt;p&gt;Shamefully relying on Len's spoiler,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the sequence is stops on the New York subway's 1 line&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and given that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jiminion didn't specify that the next item in the sequence was a &lt;em&gt;number&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;next must be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Christopher&quot;. (The stop is at Christopher Street, and the previous items in the sequence omitted &quot;street&quot;.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4915" LastEditorUserId="4915" LastEditDate="2015-03-17T18:42:00.057" LastActivityDate="2015-03-17T18:42:00.057" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10460" PostTypeId="1" AcceptedAnswerId="10462" CreationDate="2015-03-17T20:00:40.547" Score="4" ViewCount="190" Body="&lt;p&gt;I have ordered the US States according to a certain property. A contiguous piece of that looks like the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;..., XXX, Ohio, Illinois, Minnesota, Indiana, Louisiana, Mississippi, Georgia, Michigan, Florida, Maryland, North Carolina, Wisconsin, Kentucky, Virginia, Delaware, XXX, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Which states are represented by XXX?&lt;br&gt;&#xA;What is the property by which the states are ordered?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="8874" LastEditDate="2015-03-18T09:14:41.663" LastActivityDate="2015-03-18T09:14:41.663" Title="An enigmatic ordering of US States" Tags="|pattern|geography|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10461" PostTypeId="2" ParentId="10421" CreationDate="2015-03-17T20:27:49.653" Score="11" Body="&lt;p&gt;The color of the sky is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;white&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;There are 6 Fs that can be identified in the maze, which means the color is #FFFFFF&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UI3EX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5516" LastEditorUserId="5188" LastEditDate="2015-03-17T20:34:49.460" LastActivityDate="2015-03-17T20:34:49.460" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10462" PostTypeId="2" ParentId="10460" CreationDate="2015-03-17T20:39:53.913" Score="7" Body="&lt;p&gt;The answer fits with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pennsylvania and Colorado, based on area codes. 216-219:Ohio-Indiana, 225: Louisiana, 228-229: Mississippi-Georgia etc. The ones skipped don't exist. Ohio is 216, Pennsylvania is 215. Deleware: 302, 303 is Colorado. Source: &lt;a href=&quot;http://www.50states.com/areacodes/&quot;&gt;http://www.50states.com/areacodes/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are other sites that show discrepancies with this list though and it seems fairly arbitrary considering each state has more than one of these.&lt;/p&gt;&#xA;" OwnerUserId="7722" LastActivityDate="2015-03-17T20:39:53.913" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10463" PostTypeId="2" ParentId="10421" CreationDate="2015-03-17T20:47:24.800" Score="-1" Body="&lt;p&gt;The answer is blue (or possibly black). &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The question clearly states that &lt;strong&gt;I'm&lt;/strong&gt; the protagonist of this story. That means that I'm wearing my watch (which states the time). Given that I too have trained under a &lt;a href=&quot;https://puzzling.stackexchange.com/a/3299/4852&quot;&gt;master of chronobarbology&lt;/a&gt; I can also estimate the passage of time by my beard-growth to within a fair degree of accuracy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4852" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2015-03-18T12:59:12.363" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10464" PostTypeId="2" ParentId="10426" CreationDate="2015-03-17T22:07:27.490" Score="2" Body="&lt;p&gt;There are a couple key pieces of information missing from this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1) Are the two planets' orbits co-planar?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2) Are the two planets orbiting in the same direction?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Not co-planar orbits&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;If the answer to (1) is no, then the answer to (2) doesn't matter.  The two planets can only align if they have each undergone an integer number of half-orbits.  The slower completes a half-orbit in 60 years--the faster in 35, so: $60x = 35y$ with $x$ and $y$ both integers.  The smallest solution to this is $x = 7$, or 420 years.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Co-planar orbits&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The slower planet orbits at the speed of $3$ degrees per year, the faster at $36/7$ degrees per year.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Orbiting the same direction&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;How long before the faster planet has gone $180$ degrees further than the slower?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$180 + 3t = (36/7)t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$t = 84$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Orbiting in opposite direction&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;How long before the two planets together sweep out an arc of 180 degrees?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$180 = 3t + (36/7)t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$t = 22.1$ years approx.&lt;/p&gt;&#xA;" OwnerUserId="10302" LastActivityDate="2015-03-17T22:07:27.490" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10465" PostTypeId="2" ParentId="10426" CreationDate="2015-03-17T22:25:42.087" Score="8" Body="&lt;p&gt;In the simplest, most likely intended interpretation of this problem, the planets have perfectly circular, coplanar orbits, like runners around a track.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In one year, the inner planet completes $\frac{1}{70}$ of a circle, and the outer planet completes $\frac{1}{120}$ of a circle. Therefore, the planets' angular distance widens by $\frac{1}{70} - \frac{1}{120} = \frac{1}{168}$ of a circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the planets will realign again they way they do now in &lt;strong&gt;168 years&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If they're rotating in opposite directions, however, the angular distance widens by the sum of their orbital speeds, rather than their difference, which is $\frac{19}{840}$ of a circle per year. This makes the planets realign in $\frac{840}{19} \doteq 44.21$ years.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-03-17T22:33:31.700" LastActivityDate="2015-03-17T22:33:31.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10471" PostTypeId="1" AcceptedAnswerId="10601" CreationDate="2015-03-18T03:08:06.543" Score="7" ViewCount="7306" Body="&lt;p&gt;In the following game of &lt;a href=&quot;http://en.wikipedia.org/wiki/Dots_and_Boxes&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dots and Boxes&lt;/a&gt;, it is your move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NrOZx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your opponent is a computer who will play perfectly. What is the best move, and why?&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="5373" LastEditDate="2015-03-20T01:55:54.833" LastActivityDate="2022-09-07T07:06:29.847" Title="Dots and Boxes- Best Move in Given Position" Tags="|strategy|pencil-and-paper-games|dots-and-boxes|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10472" PostTypeId="2" ParentId="10101" CreationDate="2015-03-18T03:09:03.690" Score="4" Body="&lt;p&gt;This is a second answer because I think my other answer (cyanide) also fits the clues - but I could be biased ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gas of mystery is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_sulfide&quot; rel=&quot;nofollow&quot;&gt;hydrogen sulfide&lt;/a&gt;&lt;/strong&gt;.  It could be called a mystery gas because, at elevated concentrations, the &lt;strong&gt;sense of smell disappears&lt;/strong&gt; often together with awareness of the danger.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm the thing you would love:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People enjoy &lt;strong&gt;sulfur springs&lt;/strong&gt; or hot springs which may contain low levels of hydrogen sulfide.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet I was a murderer by the millions:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hydrogen sulfide has been implicated in several &lt;strong&gt;mass extinctions&lt;/strong&gt; that have occurred in the Earth's past, in particular the Permian-Triassic extinction event 252 million years ago.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was a poison of the 1900's:   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hydrogen sulfide was used by the British Army as a &lt;strong&gt;chemical weapon&lt;/strong&gt; during World War I, when other gases were in short supply.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then a revolutionary medicine in the hospitals:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Hydrogen sulfide therapy&lt;/strong&gt; is used to reduce damage and complications following a heart attack. This treatment causes relaxation of the smooth muscle cells of blood vessels and promotes blood flow.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My other use a massage to your blood, the other the killer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; first - to relax muscle cells of blood vessels and second - a very poisonous gas.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-03-23T22:16:50.770" LastActivityDate="2015-03-23T22:16:50.770" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10473" PostTypeId="2" ParentId="10471" CreationDate="2015-03-18T03:44:29.340" Score="4" Body="&lt;p&gt;Most likely you wish to avoid connecting any dots that would allow your opponent (computer) to place a box. As such, you're left with individually choosing one of the following possibilities (marked with red):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y0pGV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All other possibilities will render a box for the opponent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possible play sequences are between the player &lt;strong&gt;P&lt;/strong&gt; and computer &lt;strong&gt;C&lt;/strong&gt; are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;P&lt;/strong&gt; draws 1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;C&lt;/strong&gt; draws 3 or 4, since 2 would lead to a string of boxes for &lt;strong&gt;P&lt;/strong&gt;;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;C&lt;/strong&gt; draws 3 would leave &lt;strong&gt;P&lt;/strong&gt; with 4 (good outcome for &lt;strong&gt;P&lt;/strong&gt;);&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;C&lt;/strong&gt; draws 4 would leave &lt;strong&gt;P&lt;/strong&gt; with 3 (good outcome for &lt;strong&gt;P&lt;/strong&gt;).&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;P&lt;/strong&gt; draws 2:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;C&lt;/strong&gt; draws 4 (bad outcome for &lt;strong&gt;P&lt;/strong&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;P&lt;/strong&gt; draws 3:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;C&lt;/strong&gt; draws 1 or 4, since 2 would lead to a string of boxed for &lt;strong&gt;P&lt;/strong&gt;;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;C&lt;/strong&gt; draws 1 would leave &lt;strong&gt;P&lt;/strong&gt; with 4 (good outcome for &lt;strong&gt;P&lt;/strong&gt;);&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;C&lt;/strong&gt; draws 4 would leave &lt;strong&gt;P&lt;/strong&gt; with 1 (good outcome for &lt;strong&gt;P&lt;/strong&gt;);&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;P&lt;/strong&gt; draws 4:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;C&lt;/strong&gt; draws 1 or 2 or 3;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;C&lt;/strong&gt; draws 1, leaving &lt;strong&gt;P&lt;/strong&gt; to draw 3 (good outcome for &lt;strong&gt;P&lt;/strong&gt;);&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;C&lt;/strong&gt; draws 2, leaving &lt;strong&gt;P&lt;/strong&gt; with no obvious choice (bad outcome for &lt;strong&gt;P&lt;/strong&gt;);&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;C&lt;/strong&gt; draws 3, leaving &lt;strong&gt;P&lt;/strong&gt; to draw 1 (good outcome for &lt;strong&gt;P&lt;/strong&gt;);&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Summarizing these options and a perfect &lt;strong&gt;C&lt;/strong&gt;-player (albeit greedy, perhaps), &lt;strong&gt;P&lt;/strong&gt; drawing either 1 or 3 would be the best next move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&#xA;The above strategy seems somewhat greedy from the opponent, but it also seems logically best.&#xA;&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="10308" LastActivityDate="2015-03-18T03:44:29.340" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10474" PostTypeId="1" AcceptedAnswerId="10522" CreationDate="2015-03-18T03:45:12.250" Score="4" ViewCount="404" Body="&lt;p&gt;Each of 4 brothers (Yellow, Green, Brown, and Red) has 5 rectangular plots of land. Land plots are numbered 1 to 21 in the diagram below. One of the plots is empty (not owned by any of the brothers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/k9mOB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each brother's land plots are orthogonally connected to his other land plots. To be &quot;by&quot; something means orthogonally OR diagonally connected. Ex.: 8 land plots are &quot;by&quot; the pond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VZzG1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions&lt;/strong&gt;&lt;br&gt;&#xA;1. If the empty plot must be by the plots of a single (not more than one) brother, which land plot(s) can it be?&lt;br&gt;&#xA;2. If the empty plot must be by the plots of multiple brothers' plots, which land plot(s) can it be?&lt;br&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="8874" LastEditDate="2015-03-18T10:42:13.863" LastActivityDate="2015-03-18T21:43:21.973" Title="Find the empty plot of land" Tags="|logical-deduction|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10475" PostTypeId="2" ParentId="10402" CreationDate="2015-03-18T04:18:17.800" Score="2" Body="&lt;p&gt;Given that &lt;a href=&quot;https://puzzling.stackexchange.com/a/10403/5208&quot;&gt;rand al'thor, got #1, #2 &amp;amp; #3 above&lt;/a&gt;, I just focused on &lt;strong&gt;#4&lt;/strong&gt;, which is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;pre&gt;&lt;code&gt;KRIHPURNATHANEEBCLCTTOLETAUDEEBRPYHURNLOLHTEBOSIRYHWLAANKGCNEDJSOFIOTERSSGMNINOEICLKDASBNRAULOJOINEMEHLRRENIUISAEJBLRORTMCOIAINLADPEGEREGSRARONHZTHCLIEEJGSSAANBRBRRTEIADKNEAONWDJWOODARNCAFRJRENEENIELWKATSTHYBAEZTOLIETALRMEMANTMHOPSOIALREN&lt;/code&gt;&lt;/pre&gt;&#xA; &lt;pre&gt;&lt;code&gt;KATHARINEHEPBURNCLAUDETTECOLBERT HOLLYHUNTER  HILARYSWANK   JODIEFOSTER   NICOLEKIDMAN  JULIANNEMOORE LUISERAINER  MARIONCOTILLARD  GREERGARSON    JESSICALANGE     DIANEKEATON   JOANCRAWFORD    KATEWINSLETELIZABETHTAYLOREMMATHOMPSON&lt;/code&gt;&lt;/pre&gt;&#xA; &lt;pre&gt;&lt;code&gt;        CATEBLANCHETT    AUDREYHEPBURN SHIRLEYBOOTH GLENDAJACKSON SIMONESIGNORET SANDRABULLOCK  HELENMIRREN JULIAROBERTS    GERALDINEPAGE CHARLIZETHERON  BARBRASTREISAND JOANNEWOODWARDJENNIFERLAWRENCEKATHYBATES   MARLEEMATLIN  SOPHIALOREN&lt;/code&gt;&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Common property:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.imdb.com/search/name?groups=oscar_best_actress_winners&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Best Actress&quot; Oscar winners&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Sorry about the bad formatting... not sure how to have a multi-line, fixed-width font, non-wrapping, spoiler block...&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2015-03-18T04:24:13.807" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10477" PostTypeId="2" ParentId="10474" CreationDate="2015-03-18T04:43:45.477" Score="0" Body="&lt;p&gt;I believe that if an empty plot is by only one brother then &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;it must be by red and is either plot 16 or plot 21.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And that if an empty plot is by multiple brothers then &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;it must be plot 17 and is either by red brown and green or by brown and green.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1631" LastActivityDate="2015-03-18T04:43:45.477" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10479" PostTypeId="1" AcceptedAnswerId="10497" CreationDate="2015-03-18T07:29:16.037" Score="19" ViewCount="1421" Body="&lt;p&gt;&lt;strong&gt;Update: &lt;em&gt;I've just realized the image had a major flaw in it. I've added a new image and made sure it's correct. Nothing else has changed, only the image.  Sorry for the inconvenience.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have six words that all start with &quot;a&quot;. Each word is specific to one of these six book series:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;A Song of Ice and Fire&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;Harry Potter&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;His Dark Materials&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;Redwall&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;Sherlock Holmes&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;The Lord of the Rings&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Each word occurs at least once in the canon of its respective series, but almost certainly never occurs in the canon of any of the other series. The words are not necessarily standard English.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This image encodes all six words. Can you figure out what they all are?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bXu1M.png&quot; alt=&quot;encodes the six words&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Each of the secret words relates to one of these six words (given in random order):&lt;br&gt;&#xA;&lt;code&gt;changer healer teller river maimer treasure&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Verification:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To verify your answer without asking me, put your six guessed words in the same order as their series are in above. Make them all lowercase and separate them with single spaces. e.g.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;asoiafword hpword hdmword rword shword lotrword&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Take the &lt;a href=&quot;http://www.miraclesalad.com/webtools/md5.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;MD5 hash&lt;/a&gt; of this string. If you get &lt;code&gt;66a8a3195f64cb095da3643c7731c0f5&lt;/code&gt; then you've got the answer exactly correct!&lt;/p&gt;&#xA;" OwnerUserId="1811" LastEditorUserId="69582" LastEditDate="2021-09-12T23:42:26.850" LastActivityDate="2021-09-12T23:42:26.850" Title="Six Series, Six Words" Tags="|cipher|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10481" PostTypeId="2" ParentId="10474" CreationDate="2015-03-18T07:34:10.417" Score="3" Body="&lt;p&gt;Single brother:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 11 (or 1 if you shift the 2 yellow plots beside the house down by 1)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Multiple brothers:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 8 - it's beside 2 brothas!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Edit: As noted by OP, this answer was incomplete because it's missing all the possible configurations. It has now been edited to include the other possible configuration.&lt;/p&gt;&#xA;&lt;p&gt;In the diagram below, the light blue represents the 2 possible empty spots. The brown spot and the empty spot can be interchanged without breaking any rules.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/eG2Ri.png&quot; alt=&quot;Plots&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8999" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-03-18T21:43:21.973" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10482" PostTypeId="2" ParentId="10437" CreationDate="2015-03-18T09:09:55.043" Score="6" Body="&lt;p&gt;The following is just a minor comment on the solution of &lt;strong&gt;Joe Z&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Counting does not necessarily start at one, but may also start at zero &#xA; (for instance, Ernst Zermelo used this variant).&#xA; This then allows the following side solution (the differences to Joe Z's solution are shown in boldface):&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Zero-sum&lt;/strong&gt; (a special type of game in game theory)&lt;br&gt;&#xA; One another (each other; copied from Joe Z.)&lt;br&gt;&#xA; Two-face (Harvey Dent, a villain in Batman; copied from Joe Z.)&lt;br&gt;&#xA; &lt;strong&gt;Three monkeys&lt;/strong&gt; (the three wise monkeys)&lt;br&gt;&#xA; Four-Eyes (an insulting term for somebody who wears glasses; copied from Joe Z.)&lt;br&gt;&#xA; &lt;strong&gt;Five Iron&lt;/strong&gt; (golf club that has a relatively narrow metal head)&lt;br&gt;&#xA; Six Sigma (a statistical standard for process consistency; copied from Joe Z.)&lt;br&gt;&#xA; Seven-Up (a drink or a children's party game; copied from Joe Z.)&lt;br&gt;&#xA; Eight Ball (a billiards game; copied from Joe Z.)&lt;br&gt;&#xA; &lt;strong&gt;Nine-pins&lt;/strong&gt; (European bowling game)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-03-19T12:14:11.537" LastActivityDate="2015-03-19T12:14:11.537" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10483" PostTypeId="2" ParentId="10479" CreationDate="2015-03-18T09:34:55.210" Score="2" Body="&lt;p&gt;Don't have the answer yet, but scraping clues (hopefully). Might be useful to others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Song of Ice and Fire&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No idea...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Harry Potter&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; anapneo (hint: healer, healing spell in HP, not sure on this one) or albus (?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;His Dark Materials&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; alethiometer (hint: teller? Device to find 'dust' in the books)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Redwall&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; asmodeus (hint: maimer, Asmodeus' maiming of Chickenhound makes him indirectly responsible for the events of Mattimeo.) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sherlock Holmes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; agra (hint: treasure, Agra treasure in &quot;The Sign of the Four&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Lord of the Rings&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; anduin (hint: river, Anduin is The Great River in LOTR)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10288" LastEditorUserId="10288" LastEditDate="2015-03-18T10:51:22.013" LastActivityDate="2015-03-18T10:51:22.013" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10484" PostTypeId="1" AcceptedAnswerId="10486" CreationDate="2015-03-18T09:46:25.830" Score="8" ViewCount="1532" Body="&lt;p&gt;The Great Houdini is performing again! Houdini has built up lots of reputation by showing his great card guessing tricks in &lt;a href=&quot;https://puzzling.stackexchange.com/questions/10118/the-great-houdinis-awesome-card-guessing-trick-1&quot;&gt;[predecessor puzzle-1]&lt;/a&gt; and in &lt;a href=&quot;https://puzzling.stackexchange.com/questions/10311/the-great-houdinis-awesome-card-guessing-trick-2&quot;&gt;[predecessor puzzle-2]&lt;/a&gt;. Everybody wants to see the famous magician, and the magic show is sold for many weeks in advance. Today Houdini shows the following awesome trick:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A girl from the audience carefully shuffles a deck of $52$ cards ($13$ spades, $13$ hearts, $13$ clubs and $13$ diamonds), and hands the deck over to Houdini's assistant.&lt;/li&gt;&#xA;&lt;li&gt;The assistant looks through the entire deck (but without changing the order of the cards and without revealing them to Houdini).&lt;/li&gt;&#xA;&lt;li&gt;The assistant then repeats the following step $52$ times: &#xA;He takes the top card and puts it face-down on the table. &#xA;Houdini guesses the suit. Then the assistant flips the card over and reveals it to everybody (including Houdini). Then the step is repeated.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Houdini's goal is of course to make as many correct guesses as possible.&#xA;To reach this goal, Houdini and his assistant apply a sophisticated cheating strategy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The backsides of the cards are not perfectly symmetric. &#xA;  The assistant has two different ways of putting down every card, and hence may communicate to Houdini one bit of information per card.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is there a cheating strategy for Houdini and his assistant that guarantees Houdini to make &#xA;(a) at least $27$ correct guesses &#xA;(b) at least $32$ correct guesses (even in the worst possible case)?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-03-18T16:05:03.627" Title="The Great Houdini's awesome card guessing trick (3)" Tags="|mathematics|strategy|cards|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10486" PostTypeId="2" ParentId="10484" CreationDate="2015-03-18T09:55:43.360" Score="11" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a. Yes&lt;br&gt;&#xA; b. Yes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The strategy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;A:&lt;/strong&gt;&lt;br&gt;&#xA; Houdini and his assistant can easily guess 26 cards as said in answers to puzzle 2. To prove that he can easily guess 27 cards here are the needed steps:&lt;br&gt;&#xA; Houdini and his assistant agrees that each odd card will show the color of the next (even) card and each even card will give the suit. Like that they can easily guess 26 cards.&lt;br&gt;&#xA; But last card shouldn't even count, because Houdini can just count previous 51 cards and see which suit is missing.&lt;br&gt;&#xA; So the &quot;real&quot; last card to guess is card 51. 51 is odd, so normally it would provide him information about the color of the 52nd card, but because at that moment Houdini will know which 2 suits(in worst case, in best case they are same) are left, he and his assistant should also agree on the order of the suits(for example &lt;em&gt;Spades&lt;/em&gt; &gt; &lt;em&gt;Clubs&lt;/em&gt; &gt; &lt;em&gt;Hearts&lt;/em&gt; &gt; &lt;em&gt;Diamonds&lt;/em&gt;).&lt;br&gt;&#xA; So if there are &lt;em&gt;Clubs&lt;/em&gt; and &lt;em&gt;Diamonds&lt;/em&gt; left and 51st card is &lt;em&gt;Clubs&lt;/em&gt;, his assistant can put he 51st card normal way(which will mean greater than). With that information Houdini can easily find which card it is(only &lt;em&gt;C&lt;/em&gt; and &lt;em&gt;D&lt;/em&gt; are left and 51st card is &quot;bigger&quot; than 52nd, so it should be &lt;em&gt;C&lt;/em&gt;)&lt;br&gt;&#xA; So Houdini will guess all even cards(2, 4, 6, 8, ... , 50, 52) and second last, which is odd(51). That sums up to 27 guaranteed guesses in total&lt;br&gt;&#xA; &lt;strong&gt;B:&lt;/strong&gt;&lt;br&gt;&#xA; It is actually possible to guess 34 suits and here's the strategy:&lt;br&gt;&#xA; Assistant will put the first card that will show if the next &lt;em&gt;three&lt;/em&gt; cards have more black or red suits. So Houdini will most probably guess the first card wrong. But then he knows that out of cards 2, 3 and 4 there are at least 2 red(or black) ones. 2nd and 3rd card will show the suit of that card and 4th card will again tell him whether next 3 cards are mostly black or red. Following this strategy, Houdini will guess 2 out of 3 cards in each group, so in total he will guess at least 34 suits($(52-1)*2/3$).&lt;br&gt;&#xA; &lt;em&gt;Example:&lt;/em&gt;&lt;br&gt;&#xA; Assistant puts the first card up, so Houdini knows that out of the next 3 cards at least 2 are black. He says any suit for the first card, but probably guesses wrong.&lt;br&gt;&#xA; Then assistant puts 2nd card up, so Houdini knows that first black card out of those 3 is &lt;em&gt;Spades&lt;/em&gt;. Now Houdini should say &lt;em&gt;Spades&lt;/em&gt; until it shows up(either this card or next one will be &lt;em&gt;Spades&lt;/em&gt;).&lt;br&gt;&#xA; Then assistant puts the 3rd card. Now it's upside down, so next black card in that group of 3 is &lt;em&gt;Clubs&lt;/em&gt;.&lt;br&gt;&#xA; The last card of that group of 3 tells Houdini the dominant color of the next 3 cards.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Following this strategy, Houdini will only guess 1 out of 3 cards wrong(worst case scenario)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9352" LastEditorUserId="9352" LastEditDate="2015-03-18T13:01:41.337" LastActivityDate="2015-03-18T13:01:41.337" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10487" PostTypeId="2" ParentId="10443" CreationDate="2015-03-18T09:56:39.107" Score="-3" Body="&lt;p&gt;&lt;strong&gt;Partial answer for 4×4 grid :&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;This answer approves the possibility of solution in smaller grid , waiting for a generalisation of this dilemma:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/z6OR5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The lines take part of crosses where flipped coins represent squares intersected.&lt;/li&gt;&#xA;&lt;li&gt;Gray shadowed squares are unfavorables ones cuz coins there are flipped n times where n is even.&lt;/li&gt;&#xA;&lt;li&gt;Blod lines represent persistent lines , we cant remove them cuz they cross an empty square atleast.&lt;/li&gt;&#xA;&lt;li&gt;tiny circles represent ties , if we remove a line from a tie we must remove another tie's line to ensure an odd number of lines crossing this square.&lt;/li&gt;&#xA;&lt;li&gt;We start removing lines from a square containing minimal even number of different colored lines.&lt;/li&gt;&#xA;&lt;li&gt;Task achieved when all squares contains odd number of different colored lines.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;waiting for an answer concerning 8*8 grid ?&lt;/p&gt;&#xA;" OwnerUserId="9517" LastEditorUserId="9517" LastEditDate="2015-03-20T14:33:15.420" LastActivityDate="2015-03-20T14:33:15.420" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="10489" PostTypeId="2" ParentId="10421" CreationDate="2015-03-18T10:33:38.177" Score="6" Body="&lt;p&gt;The colour of the sky is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Grey&lt;/strong&gt;&#xA;&lt;br&gt;&#xA; There are 9 key strokes used, URD-LUR-ULU. If you add the keyboard code of each to get a RGB value this comes out as R 117 G 114 B 113 or #757271 or a mid grey&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10314" LastActivityDate="2015-03-18T10:33:38.177" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10492" PostTypeId="2" ParentId="10479" CreationDate="2015-03-18T11:31:54.963" Score="1" Body="&lt;p&gt;I've no idea what to do with the image file (JoeZ will probably know, since it looks just like his profile pic!), but like MaRa I'll put some guesses as to what the final words could be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A Song of Ice and Fire&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Aemon&lt;/strong&gt; (the &lt;em&gt;healer&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Harry Potter&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Animagus&lt;/strong&gt; (the &lt;em&gt;changer&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;His Dark Materials&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;No idea (haven't read it)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Redwall&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure yet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Sherlock Holmes&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure yet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Lord of the Rings&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Probably &lt;strong&gt;Anduin&lt;/strong&gt; (the &lt;em&gt;river&lt;/em&gt;) as @MaRa says.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-03-18T11:31:54.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10493" PostTypeId="1" AcceptedAnswerId="10521" CreationDate="2015-03-18T11:43:35.983" Score="15" ViewCount="3624" Body="&lt;p&gt;Here's the image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VMASY.png&quot; alt=&quot;The image&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is hidden on it and how was it made?&lt;/p&gt;&#xA;" OwnerUserId="10259" LastActivityDate="2018-05-14T20:03:20.730" Title="What is hidden on this image?" Tags="|visual|steganography|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10495" PostTypeId="2" ParentId="8588" CreationDate="2015-03-18T11:49:10.093" Score="2" Body="&lt;p&gt;The title made me think of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Auschwitz (or just a generic Nazi concentration camp),&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and I think I can get this to fit most of the riddle itself too. In fact, the following more general answer also works:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PRISON.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;your holiday starts, forget happiness&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Entering a prison is like a 'holiday' from the outside world, but you'll find no happiness there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;no out from these mates of no friendliness&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You cannot escape, and the guards will show you no friendliness.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;your past is what we think&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People are imprisoned for crimes committed in their past.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;as space tortures, beginning to shrink&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People in prison are confined in tiny spaces.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;follow the sheep of fashion&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They're all dressed identically.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;through halls of no compassion&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There isn't much compassion to be found in a prison.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;little sun is gained through wreck&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People are shut away inside so they couldn't see the sun.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;each day it steals, the doves doth peck&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Days in your life steal away uselessly. Doves are a symbol of peace, pecking away at the tyranny of prison.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-06-27T12:59:57.917" LastActivityDate="2015-06-27T12:59:57.917" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10496" PostTypeId="2" ParentId="10493" CreationDate="2015-03-18T11:49:14.710" Score="2" Body="&lt;p&gt;There is someone hidden in the image. I put the image &lt;a href=&quot;http://magiceye.ecksdee.co.uk/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and got this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Abvuf.png&quot; alt=&quot;MAGIC_PICTURE&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7482" LastActivityDate="2015-03-18T11:49:14.710" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10497" PostTypeId="2" ParentId="10479" CreationDate="2015-03-18T12:04:05.283" Score="9" Body="&lt;p&gt;&lt;strong&gt;Solution below.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, an analysis of the (updated) image:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The brightness values in the image are all multiples of 4. Dividing by 4 gives the following grid:&lt;br&gt;&#xA; &lt;br&gt;&#xA; &lt;code&gt;21  12  21  11  18  22  8   25  12&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;5   14  17  25  13  5   25  25  19&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;6   13  8   7   11  7   25  13  25&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;17  21  13  6   19  21  18  25  19&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;12  25  25  7   21  11  11  8   6&lt;/code&gt;&lt;br&gt;&#xA; &lt;br&gt;&#xA; The most common numbers are 25 followed by 21. Combined with the number range (5-25) it seems likely that these are letters, with 25 being 'a', 24 being 'b, 21 'e' and so on. The grid is then as follows:&lt;br&gt;&lt;br&gt;&#xA; &lt;code&gt;e   n   e   o   h   d   r   a   n&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;u   l   i   a   m   u   a   a   g&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;t   m   r   s   o   s   a   m   a&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;i   e   m   t   g   e   h   a   g&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;n   a   a   s   e   o   o   r   t&lt;/code&gt;&lt;br&gt;&#xA; &lt;br&gt;&#xA; This contains the letters for many of the solutions suggested elsewhere, though in no obvious order. There are two u's, so animagus/anduin/asmodeus can't all be right. (The originally uploaded image had one fewer values–a 5 ('u')–though the ordering was different.)&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Based on this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter distribution is a: 9, e: 5, m: 4, o: 4, r: 3, s: 3, n: 3, g: 3, t: 3, I: 2, h: 2, u: 2, d: 1, l: 1.&lt;br&gt;&#xA; From the other suggestions, we can see that &lt;strong&gt;aemon/animagus/alethiometer/asmodeus/agra/?&lt;/strong&gt; fits, leaving &quot;aaorngth&quot;. &lt;a href=&quot;http://www.quicksilver899.com/Tolkien/LOTR/LOTR_AC.html&quot;&gt;A Tolkien Dictionary&lt;/a&gt; indicates the final word: &lt;strong&gt;&lt;a href=&quot;http://lotr.wikia.com/wiki/Argonath&quot;&gt;argonath&lt;/a&gt;&lt;/strong&gt;. And indeed this gives the correct MD5 hash.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solution&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Song of Ice and Fire: aemon (healer)&lt;br&gt;&#xA; Harry Potter: animagus (changer)&lt;br&gt;&#xA; His Dark Materials: alethiometer (teller)&lt;br&gt;&#xA; Redwall: asmodeus (maimer)&lt;br&gt;&#xA; Sherlock Holmes: agra (treasure)&lt;br&gt;&#xA; The Lord of the Rings: argonath (river)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10220" LastEditorUserId="10220" LastEditDate="2015-03-18T23:00:38.473" LastActivityDate="2015-03-18T23:00:38.473" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10498" PostTypeId="2" ParentId="10405" CreationDate="2015-03-18T12:11:05.887" Score="0" Body="&lt;p&gt;A &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; baby&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;... while it was storks and docs that usually bring home babies... for some reason, when I saw a doctor and a bird in the same sentence, the image of the stork with a baby got stuck in my brain.&#xA;and what's more in plain sight than a baby bump?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ah... the answer has already been given I see...&lt;/p&gt;&#xA;" OwnerUserId="10319" LastEditorUserId="10281" LastEditDate="2015-03-18T15:30:16.123" LastActivityDate="2015-03-18T15:30:16.123" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10501" PostTypeId="2" ParentId="10421" CreationDate="2015-03-18T12:35:28.633" Score="-1" Body="&lt;p&gt;He just said the color of the sky because the maze has no roof, only walls&lt;/p&gt;&#xA;" OwnerUserId="10322" LastActivityDate="2015-03-18T12:35:28.633" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10502" PostTypeId="2" ParentId="10493" CreationDate="2015-03-18T12:40:03.920" Score="12" Body="&lt;p&gt;This is how far I got; I created a python script to create a background image that I substracted from the given image (&lt;strong&gt;Update&lt;/strong&gt;):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/l6tXs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the Python script I used:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import sys&#xA;from PIL import Image&#xA;src = Image.open(sys.argv[1])&#xA;def tf(p,i,j):&#xA;  return tuple([256-(x-(256-i*j)/64*64)%256 for x in p])&#xA;odata = [tf(src.getpixel((j,i)),i,j) for i in range(0,256) for j in range(0,256)]&#xA;img = Image.new('RGB', (256,256))&#xA;img.putdata(odata)&#xA;img.save('2.png')&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So it's basically the same as the accepted answer, but there is some small error in there somewhere..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this is the correct python script using XOR instead of difference:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import sys&#xA;from PIL import Image&#xA;src = Image.open(sys.argv[1])&#xA;def tf(p,i,j):&#xA;  return tuple([(x^i*j)%256 for x in p])&#xA;odata = [tf(src.getpixel((j,i)),i,j) for i in range(0,256) for j in range(0,256)]&#xA;img = Image.new('RGB', (256,256))&#xA;img.putdata(odata)&#xA;img.save('3.png')&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="10321" LastEditorUserId="10321" LastEditDate="2015-03-19T08:16:16.617" LastActivityDate="2015-03-19T08:16:16.617" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10504" PostTypeId="2" ParentId="10421" CreationDate="2015-03-18T12:57:20.677" Score="1" Body="&lt;p&gt;An idea nobody seems to have tried yet:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Logo commands. &lt;a href=&quot;http://derrel.net/ep/logo/logo_com.htm&quot; rel=&quot;nofollow&quot;&gt;Link&lt;/a&gt; in case people don't know what I'm on about.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your path out of the maze, imagining you're&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the Logo turtle,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;would correspond to the following list of commands:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; FD300 &lt;br&gt; RT90 &lt;br&gt; FD100 &lt;br&gt;RT90 &lt;br&gt;FD400 &lt;br&gt;RT90 &lt;br&gt;FD400 &lt;br&gt;RT90 &lt;br&gt; FD300 &lt;br&gt;RT90 &lt;br&gt;FD200 &lt;br&gt;LT90 &lt;br&gt;FD200 &lt;br&gt;LT90 &lt;br&gt;FD200 &lt;br&gt;RT90 &lt;br&gt;FD300&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(by some rough length measurements made on my computer screen). This could be made even more complicated if you want to allow for the possibility of trying dead ends and doubling back, but then the solution would be non-unique.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe these letters and numbers can somehow be turned into a colour code?&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-03-18T12:57:20.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10505" PostTypeId="2" ParentId="10479" CreationDate="2015-03-18T13:49:15.387" Score="1" Body="&lt;p&gt;Adding more information about the image and another approach:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Being all neutral greys, with no saturation, the hexadecimal values of the colors are &lt;a href=&quot;http://www.december.com/html/spec/color0.html&quot; rel=&quot;nofollow&quot;&gt;all in the form of &lt;code&gt;#ABABAB&lt;/code&gt;&lt;/a&gt;. Using only &lt;code&gt;AB&lt;/code&gt; to create the grid we have:&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;code&gt;C1 70 51 47 3D FF FF 3D D6 7A D6&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;D6 70 FF B7 FF E0 C1 47 47 FF 70&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;3D FF 84 33 FF FF 7A 51 84 AD AD&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;B7 C1 51 D6 84 FF 70 84 D6 8E 7A&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;  with a decimal version:&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;code&gt;193 112  81  71  61 255 255  61 214 122 214&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;214 112 255 183 255 224 193  71  71 255 112&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;061 255 132  51 255 255 122  81 132 173 173&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;183 193  81 214 132 255 112 132 214 142 122&lt;/code&gt;&lt;br&gt;&#xA;  (do you notice something funny between this and the brightness table?)&lt;br&gt;&#xA;&lt;br&gt;&#xA; which translated in ASCII is:&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;code&gt;Á p Q G = ÿ ÿ = Ö z Ö&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;Ö p ÿ · ÿ à Á G G ÿ p&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;= ÿ   3 ÿ ÿ z Q     ­ ­&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;· Á Q Ö   ÿ p   Ö   z&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;  which is plain and clear nonsense. Still, there is no value repeated exactly six times and only one repeated more than six times which is &lt;code&gt;FF&lt;/code&gt;, the white square. This, if some kind of decoding strategy has yet to be applied, must be the value for &lt;code&gt;A&lt;/code&gt; (which also correspond to &lt;code&gt;25&lt;/code&gt; in the answer posted by @Uri Zarfaty).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4652" LastEditorUserId="4652" LastEditDate="2015-03-18T14:43:10.703" LastActivityDate="2015-03-18T14:43:10.703" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10506" PostTypeId="2" ParentId="10405" CreationDate="2015-03-18T13:52:04.620" Score="1" Body="&lt;p&gt;It sounds to me like&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a lie&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A partner in crime, though I wear a disguise.&#xA;I hide in plain sight. Averting all eyes.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the lie is hidden and a &quot;partner in crime&quot;. In &quot;plain sight&quot; might mean that it is told, yet nobody knows it is a lie.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;You can't run and gun, if you want to stop me.&#xA;To succeed on that front, you need just turn and flee.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; stopping a lie from spreading is difficult&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A doctor creates me, a goose fills her nest,&#xA;But just give me time and I will manifest.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; doctors lie sometime. Maybe this line is only example of lies. Although I really have no idea what is the role of the goose. This clue just does not fit in any of my hypotheses.. &quot;I will manifest&quot; might mean that the lie will be recognized as such at some point in time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Any who make me, end up the same...&#xA;The question, of course, though, is: what is my name?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the liars end up the same.. maybe in jail or just tagged as liars.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5283" LastActivityDate="2015-03-18T13:52:04.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10507" PostTypeId="2" ParentId="10213" CreationDate="2015-03-18T13:52:28.817" Score="3" Body="&lt;p&gt;Taylor Swift wrote &quot;Back to December&quot;, talked about #thedress recently, wrote &quot;Welcome to New York&quot; (New York has tall buildings). I don't know about a quarter of the cards, though.&lt;/p&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-03-18T13:52:28.817" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10509" PostTypeId="2" ParentId="10213" CreationDate="2015-03-18T14:06:12.247" Score="2" Body="&lt;p&gt;My answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Demi Lovato&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One December,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Remember December&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;a quarter of the cards are missing.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Give Your Heart a Break&quot; amongst many other songs with 'hearts'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You look out of the windows and see the tall buildings outside.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Skyscraper&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A person walks by you and you think like #TheDress. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure on this one, maybe &quot;Neon Lights&quot; (color of the dress based on lighting?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10288" LastActivityDate="2015-03-18T14:06:12.247" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10512" PostTypeId="2" ParentId="10421" CreationDate="2015-03-18T14:18:52.357" Score="0" Body="&lt;p&gt;It's a shame it's hinted that the answer is something to do with the maze and a keyboard because the following occurred to me:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crazy man states, &quot;&lt;em&gt;if you followed all the clues in the maze and don't want to get shot you would be able to tell me what's the color of the sky outside right now!&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, given that at the time I am escaping I am unaware of any clues existing in the maze, visual or otherwise and given that I am panicked as I escape, I can assume that I did not follow any clues whilst escaping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crazy man states that I'll be able to tell him the colour of the sky if I followed the clues AND don't want to get shot. Therefore I can't tell him the colour of the sky because only one of those statements is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore to answer the question &quot;What do you tell the crazy man?&quot;, I tell him &quot;I can't tell you the colour of the sky.&quot;&lt;/p&gt;&#xA;" OwnerUserId="10314" LastActivityDate="2015-03-18T14:18:52.357" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10513" PostTypeId="2" ParentId="10471" CreationDate="2015-03-18T14:20:12.180" Score="3" Body="&lt;p&gt;The goal of dots and boxes is to be the person that makes the last (and biggest) group of boxes. To that end, you need to figure out how many regions there are and which one is smallest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LNLUa.png&quot; alt=&quot;Region boundaries&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are your potential boundaries of regions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here is one example of how it could play out:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PlI4k.png&quot; alt=&quot;Example game&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here you have 6 regions, so you want to let the computer take the first (smallest) region so you can take the last (black) region. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, you're a bit stuck &lt;strong&gt;and can't win&lt;/strong&gt;. There are about 5 regions to start with, but only 2 moves that don't affect the number of regions. So, either you have to start off by giving them the first region (so they get the last region), or if you merge 2 regions (top left, bottom left), the computer will give you the first region. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Xr4hv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-03-18T15:39:22.410" LastActivityDate="2015-03-18T15:39:22.410" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10514" PostTypeId="1" AcceptedAnswerId="10672" CreationDate="2015-03-18T14:46:57.333" Score="3" ViewCount="563" Body="&lt;p&gt;A place of work, and not of play,&lt;br&gt;&#xA;You know that's what the vermin say.&lt;br&gt;&#xA;Is it this or that, the question sly,&lt;br&gt;&#xA;But in the end you'll question why?&lt;br&gt;&#xA;A miracle you want to find,&lt;br&gt;&#xA;And get yourself out of a bind.&lt;br&gt;&#xA;So while you're here, follow a hunch,&lt;br&gt;&#xA;It may just happen before lunch. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The riddle is in two parts. The last half is comprised of clues related to the word, and the first half is wordplay to give you the word.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  What is another name for work? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In research, where might you find a miracle?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="9951" LastEditDate="2015-03-20T13:43:23.190" LastActivityDate="2015-03-20T21:56:50.673" Title="To work we go! - What am I?" Tags="|riddle|word|rhyme|" AnswerCount="7" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10515" PostTypeId="2" ParentId="10405" CreationDate="2015-03-18T14:49:57.470" Score="9" Body="&lt;p&gt;I think it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A Rhyme&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&#xA;A partner in crime, though I wear a disguise.&lt;br/&gt;&#xA;I hide in plain sight. Averting all eyes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;'rime' in 'crime', the disguise being a different spelling. Hiding in plain sight as the answer is in the question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&#xA;You can't run and gun, if you want to stop me.&lt;br/&gt;&#xA;To succeed on that front, you need just turn and flee.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;run rhymes with gun, so to run and gun is to continue the rhyme rather than stop it.&lt;br/&gt;&lt;em&gt;'Turn' and 'flee' don't rhyme, thereby ending the rhyme, although I might be grasping at straws with that one&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&#xA;A doctor creates me, a goose fills her nest,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Dr Seuss and Mother Goose, both associated with children's rhymes, not sure if it's intentional but Seuss and Goose also make a half-rhyme.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&#xA;But just give me time and I will manifest.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You can find a rhyme for almost anything if you think about it long enough... maybe?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&#xA;Any who make me, end up the same...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Anyone who makes up a rhyme ends up as a poet??&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&#xA;&lt;strong&gt;Hint 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Every line of the riddle is part of a rhyme&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&#xA;&lt;strong&gt;Hint 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A bard (any bard, although the clue might be alluding specifically to William Shakespeare) would generally write songs and poems which rhyme.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10330" LastActivityDate="2015-03-18T14:49:57.470" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10517" PostTypeId="2" ParentId="10514" CreationDate="2015-03-18T15:19:00.073" Score="2" Body="&lt;p&gt;Could it be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Death and/or Grave?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A place of work, and not of play,&#xA;You know that's what the vermin say.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Body underground is worm food&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is it this or that, the question sly,&#xA;But in the end you'll question why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Why did they have to die!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A miracle you want to find,&#xA;And get yourself out of a bind.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Miracle to get out of death&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So while you're here, follow a hunch,&#xA;It may just happen before lunch.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can die any time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2015-03-18T15:19:00.073" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10519" PostTypeId="2" ParentId="10514" CreationDate="2015-03-18T15:23:22.723" Score="1" Body="&lt;p&gt;You are a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;gold prospector, so you look for precious metals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Indeed, you can find vermin on the soil, you want to find a miracle (gold is rare) and it may happen just before lunch, as the sun makes the gold shine at noon.&lt;/p&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-03-18T15:23:22.723" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10521" PostTypeId="2" ParentId="10493" CreationDate="2015-03-18T15:26:43.777" Score="21" Body="&lt;p&gt;I got it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First attempt:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I separated the RGB channels, XOR'd them with each other, inverted them, and arbitrarily assigned the results &lt;code&gt;R=~(R^G); G=~(R^B); B=~(G^B)&lt;/code&gt; based on what would make the hair and skin hue broadly correct.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Result image:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/WVCd8.png&quot; alt=&quot;Image&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The result isn't correct, but it made whatever component caused the moire patterns drop out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I got it!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The image is XOR'd with the function &lt;code&gt;(x*y)%256&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Final image:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/0qBig.png&quot; alt=&quot;Final image&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clean function:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/ca649.png&quot; alt=&quot;Clean XOR background&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10331" LastEditorUserId="10331" LastEditDate="2015-03-18T15:55:30.613" LastActivityDate="2015-03-18T15:55:30.613" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10522" PostTypeId="2" ParentId="10474" CreationDate="2015-03-18T15:56:11.900" Score="1" Body="&lt;p&gt;Single Brother: it may be in the plot 1 as my picture shows, or plot 11 as showed earlier by Aify.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Double brother: 8 as Aify showed.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/pg8JM.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-03-18T15:56:11.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10523" PostTypeId="2" ParentId="10484" CreationDate="2015-03-18T16:05:03.627" Score="5" Body="&lt;p&gt;This is a follow-up to &lt;a href=&quot;https://puzzling.stackexchange.com/a/10486/10302&quot;&gt;Novarg's excellent answer&lt;/a&gt;.&#xA;I would prefer it as a comment, but lack the reputation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can apply the logic from part A to part B.  On the 49th card, instead&#xA; of indicating the dominant color of the final 3 cards, indicate the color&#xA; of the 50th.  The 50th card then indicates its own suit within that color.&#xA; This leaves us with two final cards, which can be correctly guessed on&#xA; their own as in part A.  In this way, the entire final group of 3 can be&#xA; correct, achieving a total score of 35.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10334" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-03-18T16:05:03.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10524" PostTypeId="1" AcceptedAnswerId="10528" CreationDate="2015-03-18T16:17:31.850" Score="1" ViewCount="441" Body="&lt;p&gt;Can you determine the next numbers in the below sequence?&#xA;Here is the series. You need to find the possible numbers in 7th line. Series starts with 1 and it is like this.&lt;br/&gt;&#xA;1&lt;br/&gt;&#xA;1 1&lt;br/&gt;&#xA;2 1&lt;br/&gt;&#xA;1 1 1 2&lt;br/&gt;&#xA;3 1 1 2&lt;br/&gt;&#xA;2 1 1 2 1 3&lt;br/&gt;&#xA;...........&lt;br/&gt;&#xA;&lt;br/&gt;&#xA;Please try yourself before searching in internet. &lt;br/&gt;&#xA;&lt;strong&gt;Note&lt;/strong&gt;: I'll post some hints if nobody gets the answer.&lt;/p&gt;&#xA;" OwnerUserId="10299" LastActivityDate="2015-03-18T16:27:55.563" Title="Find the next numbers in series?" Tags="|logical-deduction|" AnswerCount="1" CommentCount="4" ClosedDate="2015-03-18T16:25:49.513" ContentLicense="CC BY-SA 3.0" />
  <row Id="10525" PostTypeId="1" AcceptedAnswerId="10533" CreationDate="2015-03-18T16:19:47.627" Score="8" ViewCount="22422" Body="&lt;p&gt;Create a list of US-state names with one state name per line, so that no two distinct lines on your list contain a common letter.&#xA;Let $L$ be sum of the lengths of all the state names on your list (blanks/spaces are not counted).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the largest possible value for $L$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="8874" LastEditDate="2016-03-24T18:43:11.507" LastActivityDate="2018-04-27T09:31:40.800" Title="List of US-states without common letters" Tags="|word|geography|word-sequence|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10527" PostTypeId="1" AcceptedAnswerId="10529" CreationDate="2015-03-18T16:20:59.227" Score="28" ViewCount="1218" Body="&lt;p&gt;These thirteen English words share a hidden property:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;cold, laughter, unblock, word, kill, witch, even, now, warm, oil, talking, potted, corn&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The list is not complete, there are other words with the same property, most don't fit the pattern though. What is this property?&lt;/p&gt;&#xA;" OwnerUserId="9910" LastEditorUserId="5373" LastEditDate="2016-09-18T20:26:45.540" LastActivityDate="2016-09-18T20:26:45.540" Title="A Company of Thirteen" Tags="|pattern|word-property|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10528" PostTypeId="2" ParentId="10524" CreationDate="2015-03-18T16:21:10.967" Score="0" Body="&lt;p&gt;3 1 2 2 1 3&#xA;In fact, in the last line there are 3 &quot;one&quot;, 2 &quot;two&quot; and 1 &quot;three&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everytime, you count the number of the &quot;one&quot; in the previous line, then write that number followed by 1.&#xA;Then, you count the &quot;two&quot; and write that number followed by 2.&#xA;You keep doing that until the numbers are over.&#xA;Also, if a number is missing from the string (for example 2 is missing in 4 1 1 3 1 4), you just ignore it.&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-03-18T16:27:55.563" LastActivityDate="2015-03-18T16:27:55.563" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10529" PostTypeId="2" ParentId="10527" CreationDate="2015-03-18T16:25:06.753" Score="29" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Adding S as the first letter gives new words.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5684" LastActivityDate="2015-03-18T16:25:06.753" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10530" PostTypeId="1" AcceptedAnswerId="10531" CreationDate="2015-03-18T16:28:28.000" Score="0" ViewCount="468" Body="&lt;p&gt;This is my seventh (and probably last) batch of sequence puzzles that are nasty and hard to solve; I stress that each of them has a clear and justifiable solution.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sequence 1:&lt;br&gt;&#xA;  RC, NU, RT, RS, PI, ??, AN, PT&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sequence 2:&lt;br&gt;&#xA;  A, ?, D, ?, G, H, J, K, L&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sequence 3:&lt;br&gt;&#xA;  ?, ?, ?, 1, 7, 0, 7, 1, 7, 3, 4, 8, 9, 3, 4, 1, 2, 4, 2, 5, 9, 1, ?, ?, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sequence 4:&lt;br&gt;&#xA;  O, ?, ?, H, R, A, U, ?, U, E, ?, ?, ?, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9125" LastActivityDate="2015-03-18T16:40:49.123" Title="A final batch of four sequence puzzles" Tags="|number-sequence|letter-sequence|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10531" PostTypeId="2" ParentId="10530" CreationDate="2015-03-18T16:31:10.280" Score="13" Body="&lt;p&gt;Sequence 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TU&lt;br&gt;&#xA;&lt;br&gt;&#xA; These are the third and fourth letters of each planet in the Solar System.&lt;br&gt;&#xA;&lt;br&gt;&#xA; ME&lt;strong&gt;RC&lt;/strong&gt;URY, VE&lt;strong&gt;NU&lt;/strong&gt;S, EA&lt;strong&gt;RT&lt;/strong&gt;H, MA&lt;strong&gt;RS&lt;/strong&gt;, JU&lt;strong&gt;PI&lt;/strong&gt;TER, SA&lt;strong&gt;TU&lt;/strong&gt;RN, UR&lt;strong&gt;AN&lt;/strong&gt;US, NE&lt;strong&gt;PT&lt;/strong&gt;UNE&lt;br&gt;&#xA; RC, NU, RT, RS, PI, &lt;strong&gt;TU&lt;/strong&gt;, AN, PT&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sequence 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; S, F&lt;br&gt;&#xA;&lt;br&gt;&#xA; These are the keys on the home row of a standard QWERTY keyboard.&lt;br&gt;&#xA;&lt;br&gt;&#xA; A, &lt;strong&gt;S&lt;/strong&gt;, D, &lt;strong&gt;F&lt;/strong&gt;, G, H, J, K, L&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sequence 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 6, 0, 7, 7, 6&lt;br&gt;&#xA;&lt;br&gt;&#xA; These are the digits in the decimal expansion of e, shifted down by 1.&lt;br&gt;&#xA;&lt;br&gt;&#xA; e = 2.718281828459045235360287...&lt;br&gt;&#xA; digits: &lt;strong&gt;6&lt;/strong&gt;, &lt;strong&gt;0&lt;/strong&gt;, &lt;strong&gt;7&lt;/strong&gt;, 1, 7, 0, 7, 1, 7, 3, 4, 8, 9, 3, 4, 1, 2, 4, 2, 5, 9, 1, &lt;strong&gt;7&lt;/strong&gt;, &lt;strong&gt;6&lt;/strong&gt;...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sequence 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; U, E, O, H, R, A&#xA;&lt;br&gt;&#xA; These are the &lt;em&gt;second&lt;/em&gt; letters of each day of the week, cycled around.&lt;br&gt;&#xA;&lt;br&gt;&#xA; M&lt;strong&gt;O&lt;/strong&gt;NDAY, T&lt;strong&gt;U&lt;/strong&gt;ESDAY, W&lt;strong&gt;E&lt;/strong&gt;DNESDAY, T&lt;strong&gt;H&lt;/strong&gt;URSDAY, F&lt;strong&gt;R&lt;/strong&gt;IDAY, S&lt;strong&gt;A&lt;/strong&gt;TURDAY, S&lt;strong&gt;U&lt;/strong&gt;NDAY&lt;br&gt;&#xA; O, &lt;strong&gt;U&lt;/strong&gt;, &lt;strong&gt;E&lt;/strong&gt;, H, R, A, U, &lt;strong&gt;O&lt;/strong&gt;, U, E, &lt;strong&gt;H&lt;/strong&gt;, &lt;strong&gt;R&lt;/strong&gt;, &lt;strong&gt;A&lt;/strong&gt;, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-03-18T16:40:49.123" LastActivityDate="2015-03-18T16:40:49.123" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10533" PostTypeId="2" ParentId="10525" CreationDate="2015-03-18T16:45:54.893" Score="13" Body="&lt;p&gt;Best I could find was &lt;strong&gt;22&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Mississippi&#xA;New York&#xA;Utah&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If Avarind is correct that the best answer has three states, then we can quickly brute force each 3 state combination to find the result. Sample Python implementation:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import itertools&#xA;&#xA;seq = [&quot;Alabama&quot;, &quot;Alaska&quot;, &quot;Arizona&quot;, &quot;Arkansas&quot;, &quot;California&quot;, &quot;Colorado&quot;, &quot;Connecticut&quot;, &quot;Delaware&quot;, &quot;Florida&quot;, &quot;Georgia&quot;, &quot;Hawaii&quot;, &quot;Idaho&quot;, &quot;Illinois&quot;, &quot;Indiana&quot;, &quot;Iowa&quot;, &quot;Kansas&quot;, &quot;Kentucky&quot;, &quot;Louisiana&quot;, &quot;Maine&quot;, &quot;Maryland&quot;, &quot;Massachusetts&quot;, &quot;Michigan&quot;, &quot;Minnesota&quot;, &quot;Mississippi&quot;, &quot;Missouri&quot;, &quot;Montana&quot;, &quot;Nebraska&quot;, &quot;Nevada&quot;, &quot;New Hampshire&quot;, &quot;New Jersey&quot;, &quot;New Mexico&quot;, &quot;New York&quot;, &quot;North Carolina&quot;,&quot;North Dakota&quot;, &quot;Ohio&quot;, &quot;Oklahoma&quot;, &quot;Oregon&quot;, &quot;Pennsylvania&quot;, &quot;Rhode Island&quot;, &quot;South Carolina&quot;, &quot;South Dakota&quot;, &quot;Tennessee&quot;, &quot;Texas&quot;, &quot;Utah&quot;, &quot;Vermont&quot;, &quot;Virginia&quot;, &quot;Washington&quot;, &quot;West Virginia&quot;, &quot;Wisconsin&quot;, &quot;Wyoming&quot;]&#xA;&#xA;def shares_letters(a,b):&#xA;    return bool(set(a.replace(&quot; &quot;, &quot;&quot;).lower()) &amp;amp; set(b.replace(&quot; &quot;, &quot;&quot;).lower()))&#xA;&#xA;size = lambda x: len(&quot;&quot;.join(item.replace(&quot; &quot;, &quot;&quot;) for item in x))&#xA;&#xA;print max([candidate for candidate in itertools.permutations(seq, 3) if all(not shares_letters(*pair) for pair in itertools.permutations(candidate,2))], key=size)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Result:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;('Mississippi', 'New York', 'Utah')&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1497" LastEditorUserId="1497" LastEditDate="2015-03-18T17:21:47.863" LastActivityDate="2015-03-18T17:21:47.863" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10534" PostTypeId="2" ParentId="10525" CreationDate="2015-03-18T16:51:55.067" Score="13" Body="&lt;p&gt;Best I got is &lt;strong&gt;42&lt;/strong&gt;:   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; State of Rhode Island and Providence Plantations&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-03-18T16:51:55.067" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10537" PostTypeId="2" ParentId="10216" CreationDate="2015-03-18T17:06:43.157" Score="10" Body="&lt;p&gt;The solution is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&quot;Math is fun&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We order the factorizations of the eleven numbers by increasing &#xA; smaller factor, and we interpret the larger factors as octal numbers. &#xA; The corresponding Ascii symbols then spell out &quot;math is fun&quot; &#xA; (with 40oct=space etc.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$115=  1*115$; and 115=&lt;strong&gt;M&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;$282=  2*141$; and 141=&lt;strong&gt;a&lt;/strong&gt; &lt;/li&gt;&#xA;&lt;li&gt;$492=  3*164$; and 164=&lt;strong&gt;t&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;$600=  4*150$; and 150=&lt;strong&gt;h&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;$200=  5* 40$;  and 40=space&lt;/li&gt;&#xA;&lt;li&gt;$906=  6*151$; and 151=&lt;strong&gt;i&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;$1141= 7*163$; and 163=&lt;strong&gt;s&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;$320=  8* 40$; and 40=space&lt;/li&gt;&#xA;&lt;li&gt;$1314= 9*146$; and 146=&lt;strong&gt;f&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;$1650=10*165$; and 165=&lt;strong&gt;u&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;$1716=11*156$; and 156=&lt;strong&gt;n&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-03-19T12:09:13.167" LastActivityDate="2015-03-19T12:09:13.167" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10538" PostTypeId="2" ParentId="10213" CreationDate="2015-03-18T17:46:27.967" Score="3" Body="&lt;p&gt;The answer seems to be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Christina Perri &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The four songs are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. Something About December &lt;a href=&quot;https://www.youtube.com/watch?v=kWJrUGjZwr4&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=kWJrUGjZwr4&lt;/a&gt;&lt;br&gt;&#xA; 2. ??? Jar Of Hearts ???&lt;br&gt;&#xA; 3. Bluebird &lt;a href=&quot;https://www.youtube.com/watch?v=vO5NLUOD8v8&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=vO5NLUOD8v8&lt;/a&gt;&lt;br&gt;&#xA; 4. Black and Blue &lt;a href=&quot;https://www.youtube.com/watch?v=zo3vX0EwAdA&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=zo3vX0EwAdA&lt;/a&gt; and&lt;br&gt;&#xA; &lt;a href=&quot;http://www.wired.com/2015/02/science-one-agrees-color-dress/&quot; rel=&quot;nofollow&quot;&gt;http://www.wired.com/2015/02/science-one-agrees-color-dress/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-03-19T12:11:24.370" LastActivityDate="2015-03-19T12:11:24.370" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10540" PostTypeId="1" AcceptedAnswerId="11667" CreationDate="2015-03-18T18:02:01.610" Score="11" ViewCount="572" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lZiN3.jpg&quot; alt=&quot;Robot Puzzle&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robots are indicated by the arrowhead shapes and the labels &quot;r&quot;, &quot;y&quot;, &quot;b&quot;, &quot;g&quot;. The label obviously corresponds to their color. The aim of the puzzle is to pick up the shapes and put them down in the grey circle located on the north side of the board.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Rules&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Robots can only move in a straight line. Along the light blue paths they start on.&lt;/li&gt;&#xA;&lt;li&gt;Robots keep moving until they hit either a Wall or another Robot.&lt;/li&gt;&#xA;&lt;li&gt;Robots can only pick up a shape that is of their own color (red Robot can only pick up red shapes).&lt;/li&gt;&#xA;&lt;li&gt;Robots will automatically pickup a shape when they land on it.&lt;/li&gt;&#xA;&lt;li&gt;Robots can only carry up to 7 sides, going over this limit destroys the Robot, rendering it unavailable for use. Example: Red Robot has picked up a Red Hexagon and a Red Circle. This gives him 7 sides (6 + 1). If Red Robot picked up a Red Triangle, he would be destroyed. To avoid this you need to put shapes down in the grey circle.&lt;/li&gt;&#xA;&lt;li&gt;Robots can only move one at a time.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The aim of the game is to places all of the shapes into the grey circle which is a puzzle in itself. I posted it here because I want to see how well people can optimize their paths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Post your solutions in the format &quot;r s&quot; for Red Robot move South.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Just to clarify&lt;/strong&gt; One Move (&quot;r s&quot;) = One Turn.&lt;/p&gt;&#xA;" OwnerUserId="10340" LastEditorUserId="10340" LastEditDate="2015-03-20T09:44:27.347" LastActivityDate="2015-04-07T06:41:09.603" Title="Bouncing Robots: Graphical Puzzle" Tags="|open-ended|optimization|pencil-and-paper-games|" AnswerCount="1" CommentCount="17" ClosedDate="2015-07-27T10:34:06.923" ContentLicense="CC BY-SA 3.0" />
  <row Id="10541" PostTypeId="1" AcceptedAnswerId="10548" CreationDate="2015-03-18T18:07:09.617" Score="6" ViewCount="294" Body="&lt;p&gt;&lt;em&gt;This is a variation of &lt;strong&gt;&lt;/em&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/10525/list-of-us-states-without-common-letters&quot;&gt;List of US-states without common letters&lt;/a&gt;&lt;em&gt;&lt;/strong&gt; .&#xA;This is supposed to be more difficult and challenging than the original problem.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Create a list of US-States, one name per line, such that no two &lt;strong&gt;consecutive&lt;/strong&gt; States in the list do contain any common letter (and such that each State occurs at most once).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an example:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Texas&lt;br&gt;&#xA;  Wyoming&lt;br&gt;&#xA;  Utah&lt;br&gt;&#xA;  New York  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You have 2 tasks:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Determine the maximum number $N$ of States in such a list.  &lt;/li&gt;&#xA;&lt;li&gt;Determine the maximum number $L$ of letters in such a list (that is, the sum of the lengths of all the States' names in the list). &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The second task is probably more difficult than the first one. I will accept only answers with a complete solution. If nobody comes up with a solution within 72 hours, I will add a hint.&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-03-19T09:35:45.103" Title="US-States, again" Tags="|geography|word-sequence|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10542" PostTypeId="1" AcceptedAnswerId="10543" CreationDate="2015-03-18T18:09:10.840" Score="51" ViewCount="6376" Body="&lt;p&gt;Jennifer is 21 years older than her son Douglas. &#xA;6 years from now, Jennifer will be 5 times as old as Douglas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: Where is Jennifer's husband most likely to be right now? &lt;/p&gt;&#xA;" OwnerUserId="10299" LastEditorUserId="10341" LastEditDate="2015-03-19T15:12:20.760" LastActivityDate="2015-05-21T18:42:58.997" Title="Where is Jennifer's husband?" Tags="|calculation-puzzle|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10543" PostTypeId="2" ParentId="10542" CreationDate="2015-03-18T18:15:09.780" Score="59" Body="&lt;p&gt;Let's define $J$=Jennifer and $D$=Douglas.&lt;br&gt;&#xA;The problem can be rewritten as:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$J=D+21$&lt;br&gt;&#xA;$J+6=5(D+6)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to my math, Douglas is $D=(-\frac{3}{4})$ years old, which means $-9$ months. Pregnancy lasts for nine months, so &lt;strong&gt;Jennifer's husband is in the bed with her right now&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-05-21T18:42:58.997" LastActivityDate="2015-05-21T18:42:58.997" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10544" PostTypeId="1" AcceptedAnswerId="10545" CreationDate="2015-03-18T19:17:51.507" Score="16" ViewCount="969" Body="&lt;p&gt;So I came up with a quick rebus puzzle this morning.&#xA;Can you figure out what the answer is? &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; HM01&#xA; ----&#xA; TM44&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7402" LastEditorUserId="5373" LastEditDate="2016-10-01T15:18:32.900" LastActivityDate="2016-10-01T15:18:32.900" Title="Hm... Quick Rebus Time!" Tags="|rebus|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10545" PostTypeId="2" ParentId="10544" CreationDate="2015-03-18T19:22:16.353" Score="20" Body="&lt;p&gt;In the Pokémon games, &lt;a href=&quot;http://bulbapedia.bulbagarden.net/wiki/HM01&quot; rel=&quot;nofollow&quot;&gt;HM01&lt;/a&gt; is Cut, while &lt;a href=&quot;http://bulbapedia.bulbagarden.net/wiki/TM44&quot; rel=&quot;nofollow&quot;&gt;TM44&lt;/a&gt; is Rest. So the rebus reads &quot;A cut above the rest&quot;.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2016-07-03T06:04:43.673" LastActivityDate="2016-07-03T06:04:43.673" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10547" PostTypeId="2" ParentId="10542" CreationDate="2015-03-18T19:28:44.090" Score="20" Body="&lt;p&gt;As mentioned in &lt;a href=&quot;https://puzzling.stackexchange.com/a/10543/10342&quot;&gt;leoll2's answer&lt;/a&gt; above, the Maths part of the question is the simultaneous solution of the two equations which relate the respective ages of Jennifer and Douglas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$J=D+21$$&#xA;$$J+6=5(D+6)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These provide the answer $D = -3/4$ years, or roughly $-9$ months, which could lead one to conclude what leoll2 concluded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) It is not necessary that Jennifer's husband must have necessarily fathered Douglas. (No offence meant, I only mean it is not specified in the puzzle.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Biologically, pregnancy doesn't &lt;em&gt;always&lt;/em&gt; have to last 9 months. So, one can not deny the possibility of Douglas being born after 8 and a half months instead of nine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, Jennifer's husband could be anywhere in this world, or could even be dead. Particularly, if he hasn't fathered Douglas, and pregnancy did last exactly 9 months, then he must &lt;strong&gt;not&lt;/strong&gt; be around at the time of conceiving (stark opposite conclusion from the same data).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not a well designed puzzle. Sorry for being a killjoy :(&lt;/p&gt;&#xA;" OwnerUserId="10342" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-03-18T19:28:44.090" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="10548" PostTypeId="2" ParentId="10541" CreationDate="2015-03-18T19:55:52.603" Score="4" Body="&lt;p&gt;Assuming a state can't appear twice in the list (if it could, then the answer to both questions is infinite):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum length of the list is 21, and the maximum of L is 151, or 149 if you don't count spaces in two-word states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Counting or not counting spaces does not affect the resulting list of states, which is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Florida, &lt;/li&gt;&#xA;&lt;li&gt;Kentucky&lt;/li&gt;&#xA;&lt;li&gt;Idaho&lt;/li&gt;&#xA;&lt;li&gt;New Jersey&lt;/li&gt;&#xA;&lt;li&gt;Oklahoma&lt;/li&gt;&#xA;&lt;li&gt;Tennessee&lt;/li&gt;&#xA;&lt;li&gt;Colorado&lt;/li&gt;&#xA;&lt;li&gt;Mississippi&lt;/li&gt;&#xA;&lt;li&gt;New York&lt;/li&gt;&#xA;&lt;li&gt;Alabama&lt;/li&gt;&#xA;&lt;li&gt;Connecticut&lt;/li&gt;&#xA;&lt;li&gt;Alaska&lt;/li&gt;&#xA;&lt;li&gt;Vermont&lt;/li&gt;&#xA;&lt;li&gt;Hawaii&lt;/li&gt;&#xA;&lt;li&gt;Oregon&lt;/li&gt;&#xA;&lt;li&gt;Utah&lt;/li&gt;&#xA;&lt;li&gt;Wyoming&lt;/li&gt;&#xA;&lt;li&gt;Texas&lt;/li&gt;&#xA;&lt;li&gt;Ohio&lt;/li&gt;&#xA;&lt;li&gt;Nevada&lt;/li&gt;&#xA;&lt;li&gt;Missouri&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I spent about an hour writing a quick Java program to determine the result.  Source code here: &lt;a href=&quot;https://github.com/jakerobb/SEPuzzling-USStatesProblem&quot; rel=&quot;nofollow&quot;&gt;https://github.com/jakerobb/SEPuzzling-USStatesProblem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The program recursively tests every possible list, making sure that each permutation is valid before recursing further down that branch of the tree.  It discovers every possible valid leaf (16,636,162 of them, from 1-length to the 21-item maximum), and keeps track of the longest one it has found as it goes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On my computer (a decked-out late 2013 27&quot; iMac with a 3.5-3.9GHz Haswell i7), it takes about 45 seconds to find the answer.&lt;/p&gt;&#xA;" OwnerUserId="10341" LastEditorUserId="10341" LastEditDate="2015-03-18T20:46:25.330" LastActivityDate="2015-03-18T20:46:25.330" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10550" PostTypeId="2" ParentId="10404" CreationDate="2015-03-18T20:15:40.147" Score="18" Body="&lt;p&gt;&lt;strong&gt;UPDATE&lt;/strong&gt;: 3Blue1Brown made &lt;a href=&quot;https://www.youtube.com/watch?v=wTJI_WuZSwE&quot; rel=&quot;nofollow noreferrer&quot;&gt;a YouTube video&lt;/a&gt; about this puzzle, which also gives a very cool visual and geometrical explanation of the problem.&lt;/p&gt;&#xA;&lt;h2&gt;Good answers, but...&lt;/h2&gt;&#xA;&lt;p&gt;I've seen two valuable solutions for this puzzle:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/a/10412/10248&quot;&gt;The answer&lt;/a&gt; given by &lt;a href=&quot;https://puzzling.stackexchange.com/users/8874/gamow&quot;&gt;Gamow&lt;/a&gt;, which is, in my opinion, the best solution if you want to implement the algorithm in some programming language, but it's not very good/intuitive for the two members of The prodigious Duo (converting so many numbers to binary and XORing all that data isn't an easy task without even a piece of paper).&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/a/10406/10248&quot;&gt;The answer&lt;/a&gt; given by &lt;a href=&quot;https://puzzling.stackexchange.com/users/1549/user2357112&quot;&gt;user2357112&lt;/a&gt;, which is probably better if you were to play the trick like The prodigious Duo does, without any help or program; by the way this answer is a bit hard to understand without an example because of the wording etc.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Looking at the second mentioned answer, I figured out that is very easier to understand if you do not use any mathematical notation and just rely on the checkerboard's cells.&lt;/p&gt;&#xA;&lt;h2&gt;The strategy&lt;/h2&gt;&#xA;&lt;p&gt;The strategy applied by the two young magicians to do the trick can be easily explained like this:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Divide the &lt;span class=&quot;math-container&quot;&gt;$8 \times 8$&lt;/span&gt; checkerboard in &lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt; different areas&lt;/strong&gt;, which we'll label with an alphabet letter from &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt;, represented in this image with the semitransparent white-to-yellow stripes.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/o3bCv.png&quot; alt=&quot;Areas&quot; /&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Doing this, you'll be able to alter the number of black pieces in any of the six areas only flipping one single piece, because of the interjections of these areas. We'll say that a certain area is &lt;em&gt;&lt;strong&gt;&amp;quot;black colored&amp;quot;&lt;/strong&gt; if it contains an odd number of black pieces&lt;/em&gt;, and &lt;em&gt;&lt;strong&gt;&amp;quot;white colored&amp;quot;&lt;/strong&gt; if it contains an even number of black pieces&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;![Areas 2][6]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;From this second image, you'll clearly see that the only square outside of all the areas is the one in the top-left corner, and the only one inside all the areas is the one in the bottom-right corner. This means that you can change the color of all the areas at the same time by flipping the bottom-right piece, or just maintain all the colors unchanged by flipping the top-left piece. Every other piece changes the color of an &lt;em&gt;&lt;strong&gt;unique&lt;/strong&gt;&lt;/em&gt; combination of areas. This means that &lt;strong&gt;flipping one single piece you can change the color of any combination of areas you want&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Now, before the volunteer (let's call him &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; again) makes his choice, you'll &lt;strong&gt;check every area and determine if it's either black or white colored&lt;/strong&gt;, and apply this simple logic:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If the area is &lt;strong&gt;black&lt;/strong&gt;, then the piece that you've got to flip is &lt;strong&gt;inside&lt;/strong&gt; such area.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the area is &lt;strong&gt;white&lt;/strong&gt;, then the piece that you've got to flip is &lt;strong&gt;outside&lt;/strong&gt; such area.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;Your goal is to flip the piece which, once flipped, will turn all the areas into white. &lt;strong&gt;Analyze all the six areas, and reduce the range of squares that may contain the piece to flip until you end up with only one possible square containing the piece to be flipped&lt;/strong&gt;: this is guaranteed to happen always, for any combination of black/white pieces.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Now you may be wondering: &lt;em&gt;&amp;quot;why should I find the piece to flip before &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; makes his choice?&amp;quot;&lt;/em&gt;. This is because the order of the choices (the magician's one and the &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt;'s one) isn't important at all: if you flip two pieces, then the colors of the six areas will change to a new combination which is the same no matter the order of the flipped pieces.&lt;/p&gt;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;&lt;strong&gt;Once that &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; and the first magician have both flipped their pieces, applying the same algorithm described in the first three points will make you find the piece that has been flipped by &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt;&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;This solution may look difficult, but is the easiest one to use when the only thing you can do is to watch the checkerboard (like Marco and Leonardo do) without touching it, and without any other support rather than your memory.&lt;/p&gt;&#xA;&lt;h2&gt;Mathematical explaination&lt;/h2&gt;&#xA;&lt;p&gt;First of all, number all the squares of the checkerboard with an integer from &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$63$&lt;/span&gt;, from the top-left to the bottom-right corner. The result will be this:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PNslr.png&quot; alt=&quot;Numbered board&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;sup&gt;(Yeah I was too lazy to write all the numbers)&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now, considered the six areas (from &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt;), we'll represent the color of each area with a &lt;strong&gt;bit&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt; if the area is white (contains an even number of black pieces);&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; if the area is black (contains an odd number of black pieces).&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;We can &lt;strong&gt;order the bits&lt;/strong&gt; from &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; to obtain a sequence (let's call it &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt;) of six bits, something like this: &lt;span class=&quot;math-container&quot;&gt;$011010$&lt;/span&gt;, which is, in a general case, represented by the notation &lt;span class=&quot;math-container&quot;&gt;$b_A b_B b_C b_D b_E b_F$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; means &amp;quot;bit&amp;quot;. &lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; represents the binary notation of the number of the square where the piece to flip (let's call it &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt;) is located&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Pretty cool, but why?&lt;/strong&gt; To make this reasoning more clear, let's break it down:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The first bit, &lt;span class=&quot;math-container&quot;&gt;$b_A$&lt;/span&gt;, tells us if the position of &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt; is either in the upper or lower half of the checkerboard. This is why &lt;span class=&quot;math-container&quot;&gt;$b_A$&lt;/span&gt; is placed at the first position in our 6-bit string:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt; is located in the upper half, then &lt;span class=&quot;math-container&quot;&gt;$b_A$&lt;/span&gt; will be &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt;, meaning that &lt;span class=&quot;math-container&quot;&gt;$S \in [000000, 011111]$&lt;/span&gt;, and the number of the cell containing &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt; cannot be higher than &lt;span class=&quot;math-container&quot;&gt;$31$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Otherwise, if &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt; is located in the lower half, the number of its cell must be higher than &lt;span class=&quot;math-container&quot;&gt;$31$&lt;/span&gt;, as you can see in the picture above, meaning that &lt;span class=&quot;math-container&quot;&gt;$S \in [100000, 111111]$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;The second bit, &lt;span class=&quot;math-container&quot;&gt;$b_B$&lt;/span&gt;, together with &lt;span class=&quot;math-container&quot;&gt;$b_A$&lt;/span&gt;, will tell us if &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt; is located in the upper or lower half of the already discovered half:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt; is located in the upper half of the already discovered half, then &lt;span class=&quot;math-container&quot;&gt;$S \in [b_A00000, b_A01111]$&lt;/span&gt;, which is either from &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$15$&lt;/span&gt; or from &lt;span class=&quot;math-container&quot;&gt;$32$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$47$&lt;/span&gt; (as you can see in the image).&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Otherwise, if &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt; is located in the lower half of the already discovered half, then &lt;span class=&quot;math-container&quot;&gt;$S \in [b_A10000, b_A11111]$&lt;/span&gt;, which is either from &lt;span class=&quot;math-container&quot;&gt;$16$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$31$&lt;/span&gt; or from &lt;span class=&quot;math-container&quot;&gt;$48$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$63$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;This reasoning goes on until &lt;span class=&quot;math-container&quot;&gt;$b_F$&lt;/span&gt;...&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Now it should be clear to you why &lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; represents the binary notation of the number of the cell containing &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;h2&gt;Puzzle source&lt;/h2&gt;&#xA;&lt;p&gt;I do not know the exact origin of this puzzle, but I actually discovered where I did already see it, thanks to my friend &lt;a href=&quot;https://puzzling.stackexchange.com/users/10281/leoll2&quot;&gt;Leonardo&lt;/a&gt; who told me that a similar task was asked in the &lt;em&gt;Mathematical Olympiad of Cesenatico, year 2013&lt;/em&gt; (Cesenatico is an Italian town). The task was actually easier, because it only asked to prove that the two magicians' strategy existed, and also to prove that existed only if the number of squares was a power of two.&lt;/p&gt;&#xA;&lt;p&gt;For the curious, &lt;a href=&quot;http://olimpiadi.dm.unibo.it/wp-content/uploads/2013/05/Cesenatico2013.zip&quot; rel=&quot;nofollow noreferrer&quot;&gt;here's the PDF&lt;/a&gt; (unfortunately written in Italian) containing the tasks and the solutions of the contest: the one I'm talking about is the 7th.&lt;/p&gt;&#xA;" OwnerUserId="10248" LastEditorUserId="10248" LastEditDate="2020-07-23T20:47:19.993" LastActivityDate="2020-07-23T20:47:19.993" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="10551" PostTypeId="1" AcceptedAnswerId="10553" CreationDate="2015-03-18T20:31:27.593" Score="6" ViewCount="1130" Body="&lt;p&gt;I loved my best friend. He was always there for me. I brought him home on my own when I was young, but my parents didn’t like him at all. They would always yell at me whenever I’d share my food with him at the dinner table. He wasn’t exactly the most well mannered eater around. They made me take care of him by myself, which I was okay with. But they didn’t want to deal with him at all. They ignored and neglected him when I wasn’t home.My parents insisted that they still loved me, but my friend was just too troublesome and would have to go. I refused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like all best friends, he was super fun! We played fetch even though he wasn’t too good at it. We also played outside with tennis balls and frisbees… again, he wasn’t too good. But he did know how to do all sorts of tricks! And he was really good at playing hide-and-seek; better than all of my other friends. He also helped me overcome my fear of being in public places. He wasn’t shy at all and was able to walk up to tons of different people on his own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One day, my mom placed her hand on my forehead and told me that I may be coming down with something. I felt fine, but I trusted my mom because she was a nurse… and of course, because she was my mom. I told her that my friend had to come with us because dad was at work. She agreed. My friend was so excited. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;“You see mom? He likes you a lot, you just have to give him a chance,” I said as I pointed to his ear to ear smile. “I think you might be right, son,” she replied and returned the smile directly at my friend. This has never happened before. I’m happy, he’s happy, and she’s happy! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upon my arrival at the doctor’s office, I was asked about how I was feeling at the time. I told the doctor that I think I might’ve been coming down with a cold, but everyone was happy, so I’m happy! The doctor said it was pretty serious and that he’ll have to see me at least weekly. He gave me some yummy cold syrup and told me that I needed to rest. My friend and I spent the night there. I slept in the bed and my friend slept on the floor, as usual.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I awoke the next morning feeling better than ever. I got over my cold, but something was wrong. My friend? He was gone! The doctor called my mom and neither one said they’ve seen or heard anything from my friend. They weren't lying about this. I spent the next few weeks looking for him, posting missing signs everywhere. However, nothing turned up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple of months after that, I finally stopped looking. Not because I grew tired... I simply stopped caring.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Identify his best friend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happened to his best friend?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why did he stop caring for someone he loved?&lt;/p&gt;&#xA;" OwnerUserId="10267" LastActivityDate="2015-03-18T20:55:58.527" Title="Man’s Best Friend" Tags="|mystery|story|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10552" PostTypeId="2" ParentId="10551" CreationDate="2015-03-18T20:55:45.503" Score="6" Body="&lt;p&gt;The only thing that I can think of is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an imaginary friend&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One of the giveaways is &quot;returned the smile directly at my friend. This has never happened before.&quot; because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; She of course never looked at him before because he doesn't exist for her but in this case she knows where to look because he points at him, just playing along with his illusion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-03-18T20:55:45.503" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10553" PostTypeId="2" ParentId="10551" CreationDate="2015-03-18T20:55:58.527" Score="11" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The friend was imaginary; a hallucination due to mental illness.  The doctor visit was to treat said illness.  He stopped caring because his mental issues were resolved by the medication.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10341" LastActivityDate="2015-03-18T20:55:58.527" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10554" PostTypeId="2" ParentId="9647" CreationDate="2015-03-18T23:16:01.490" Score="2" Body="&lt;pre&gt;&lt;code&gt;here&#xA;__&#xA;bored&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Above-board&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;bets&#xA;_&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;bets are off&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;all4114&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;All for one and one for all&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;knightleg&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;Nightly&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A&#xA;\&#xA;C&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Across the sea&lt;/strong&gt; (A - and then \ can be used to &lt;em&gt;cross something out&lt;/em&gt; and then the sea (pronunciation of C)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;lt;3 &amp;lt;-&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;After my own heart&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10240" LastActivityDate="2015-03-18T23:16:01.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10555" PostTypeId="2" ParentId="10514" CreationDate="2015-03-19T00:21:26.480" Score="1" Body="&lt;p&gt;Maybe the word is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; EARLY?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A place of work, and not of play,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You should start work early and not mess around ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You know that's what the vermin say.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ... or maybe the OP doesn't like people who say that? Perhaps, like me, he's a late worker ;-)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Is it this or that, the question sly,&lt;br&gt;&#xA; But in the end you'll question why?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Why&quot; sounds like Y, and the word EARLY ends in a Y. The previous line could be some kind of wordplay to give EARL.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A miracle you want to find,&lt;br&gt;&#xA; And get yourself out of a bind.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Not sure about these lines.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;So while you're here, follow a hunch,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That's what I've been doing with this riddle!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;It may just happen before lunch.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Before lunch is early.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, the first and last lines fit. I'm not so confident about the rest.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-19T10:12:26.160" LastActivityDate="2015-03-19T10:12:26.160" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10557" PostTypeId="1" AcceptedAnswerId="10558" CreationDate="2015-03-19T01:10:25.740" Score="17" ViewCount="39335" Body="&lt;p&gt;Thanks to JoeZ, pacoverflow, isaacg, Sp3000, Novarg, Ram, h34, and all those who contributed answers that have since been beaten. I've accepted JoeZ's answer since he has the most mentions on the list below, but this puzzle isn't 'closed': if you can beat any of the words on the list below, post an answer and your name will be mentioned!&lt;/p&gt;&#xA;&lt;hr&gt;&#xA;&lt;h2&gt;The challenge&lt;/h2&gt;&#xA;&lt;p&gt;For each letter of the alphabet, A to Z, &lt;strong&gt;find a word containing the highest possible proportion of that letter&lt;/strong&gt; If people come up with more than one word with the same proportion, &lt;strong&gt;the longest one wins&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Examples:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;taking the letter A, the best I can think of is &amp;quot;taramasalata&amp;quot; (6/12), which trumps &amp;quot;banana&amp;quot; (3/6) because it's longer; &amp;quot;banana&amp;quot; in turn trumps &amp;quot;razzmatazz&amp;quot; (3/10), although the latter word might be a good candidate for the letter Z&lt;/li&gt;&#xA;&lt;li&gt;taking the letter B, &amp;quot;blubber&amp;quot; (3/7) is an easy starting point which can probably be outdone.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Rules:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;only words of at least 5 letters&lt;/li&gt;&#xA;&lt;li&gt;no proper nouns (i.e. place or person names)&lt;/li&gt;&#xA;&lt;li&gt;only English dictionary words&lt;/li&gt;&#xA;&lt;li&gt;exclude words like &amp;quot;aaaaaah&amp;quot; and &amp;quot;hmmmmmm&amp;quot; which could in theory be extended as far as you like to get arbitrarily many A's or M's&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Hopefully this covers all the loopholes, so that nobody will come up with 'silly' examples that need more rules to avoid!&lt;/p&gt;&#xA;&lt;p&gt;This is an &lt;a href=&quot;/questions/tagged/open-ended&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;open-ended&amp;#39;&quot; rel=&quot;tag&quot;&gt;open-ended&lt;/a&gt; puzzle, which means the current best answer may be outdone at any time. Partial answers (not covering all 26 letters) are fine; you can make them cover all 26 just by copying from the &amp;quot;current best&amp;quot; which I'll keep updated below.&lt;/p&gt;&#xA;&lt;hr&gt;&#xA;&lt;h2&gt;Current best&lt;/h2&gt;&#xA;&lt;p&gt;A: &lt;strong&gt;araara&lt;/strong&gt; (4/6) [credit: &lt;strong&gt;TakingNotes&lt;/strong&gt;]&lt;br&gt;&#xA;B: &lt;strong&gt;bobby&lt;/strong&gt; (3/5) [credit: &lt;strong&gt;pacoverflow&lt;/strong&gt;]&lt;br&gt;&#xA;C: &lt;strong&gt;coccic&lt;/strong&gt; (4/6) [credit: &lt;strong&gt;isaacg&lt;/strong&gt;]&lt;br&gt;&#xA;D: &lt;strong&gt;dodded&lt;/strong&gt; (4/6) [credit: &lt;strong&gt;isaacg&lt;/strong&gt;]&lt;br&gt;&#xA;E: &lt;strong&gt;peewee&lt;/strong&gt; (4/6) [credit: &lt;strong&gt;Joe Z&lt;/strong&gt;]&lt;br&gt;&#xA;F: &lt;strong&gt;feoff&lt;/strong&gt; / &lt;strong&gt;fluff&lt;/strong&gt; (3/5) [credit: &lt;strong&gt;Sp3000&lt;/strong&gt; / &lt;strong&gt;isaacg&lt;/strong&gt;]&lt;br&gt;&#xA;G: &lt;strong&gt;glogg&lt;/strong&gt; (3/5) [credit: &lt;strong&gt;Joe Z&lt;/strong&gt;]&lt;br&gt;&#xA;H: &lt;strong&gt;hashish&lt;/strong&gt; (3/7) [credit: &lt;strong&gt;isaacg&lt;/strong&gt;]&lt;br&gt;&#xA;I: &lt;strong&gt;iiwis&lt;/strong&gt; (3/5) [credit: &lt;strong&gt;h34&lt;/strong&gt;]&lt;br&gt;&#xA;J: &lt;strong&gt;hajji&lt;/strong&gt; (2/5) [credit: &lt;strong&gt;Sp3000&lt;/strong&gt;]&lt;br&gt;&#xA;K: &lt;strong&gt;kakkak&lt;/strong&gt; (4/6) [credit: &lt;strong&gt;isaacg&lt;/strong&gt;]&lt;br&gt;&#xA;L: &lt;strong&gt;lulls&lt;/strong&gt; / &lt;strong&gt;lolls&lt;/strong&gt; / &lt;strong&gt;lolly&lt;/strong&gt; (3/5) [credit: &lt;strong&gt;Joe Z&lt;/strong&gt; / &lt;strong&gt;isaacg&lt;/strong&gt; / &lt;strong&gt;Novarg&lt;/strong&gt;]&lt;br&gt;&#xA;M: &lt;strong&gt;mommy&lt;/strong&gt; / &lt;strong&gt;mamma&lt;/strong&gt; / &lt;strong&gt;mummy&lt;/strong&gt; (3/5) [credit: &lt;strong&gt;pacoverflow&lt;/strong&gt; / &lt;strong&gt;isaacg&lt;/strong&gt; / &lt;strong&gt;Novarg&lt;/strong&gt;]&lt;br&gt;&#xA;N: &lt;strong&gt;nanny&lt;/strong&gt; (3/5) [credit: &lt;strong&gt;pacoverflow&lt;/strong&gt;]&lt;br&gt;&#xA;O: &lt;strong&gt;voodoo&lt;/strong&gt; / &lt;strong&gt;booboo&lt;/strong&gt; (4/6) [credit: &lt;strong&gt;Joe Z&lt;/strong&gt; / &lt;strong&gt;Sp3000&lt;/strong&gt;]&lt;br&gt;&#xA;P: &lt;strong&gt;peppy&lt;/strong&gt; / &lt;strong&gt;puppy&lt;/strong&gt; / &lt;strong&gt;pappy&lt;/strong&gt; / &lt;strong&gt;poppy&lt;/strong&gt; (3/5) [credit: &lt;strong&gt;pacoverflow&lt;/strong&gt; / &lt;strong&gt;Novarg&lt;/strong&gt; / &lt;strong&gt;Sp3000&lt;/strong&gt; / &lt;strong&gt;Joe Z&lt;/strong&gt;]&lt;span class=&quot;math-container&quot;&gt;$^1$&lt;/span&gt;&lt;br&gt;&#xA;Q: &lt;strong&gt;duqqa&lt;/strong&gt; (2/5) [credit: &lt;strong&gt;TakingNotes&lt;/strong&gt;]&lt;br&gt;&#xA;R: &lt;strong&gt;rarer&lt;/strong&gt; / &lt;strong&gt;error&lt;/strong&gt; (3/5)  [credit: &lt;strong&gt;pacoverflow&lt;/strong&gt; / &lt;strong&gt;isaacg&lt;/strong&gt;]&lt;br&gt;&#xA;S: &lt;strong&gt;assess&lt;/strong&gt; / &lt;strong&gt;sasses&lt;/strong&gt; (4/6) [credit: &lt;strong&gt;Joe Z&lt;/strong&gt; / &lt;strong&gt;pacoverflow&lt;/strong&gt;]&lt;br&gt;&#xA;T: &lt;strong&gt;tatty&lt;/strong&gt; (3/5) [credit: &lt;strong&gt;Sp3000&lt;/strong&gt;]&lt;br&gt;&#xA;U: &lt;strong&gt;muumuu&lt;/strong&gt; (4/6) [credit: &lt;strong&gt;Joe Z&lt;/strong&gt;]&lt;br&gt;&#xA;V: &lt;strong&gt;vivid&lt;/strong&gt; / &lt;strong&gt;divvy&lt;/strong&gt; / &lt;strong&gt;valve&lt;/strong&gt; / &lt;strong&gt;civvy&lt;/strong&gt; (2/5) [credit: &lt;strong&gt;pacoverflow&lt;/strong&gt; / &lt;strong&gt;isaacg&lt;/strong&gt; / &lt;strong&gt;Novarg&lt;/strong&gt; / &lt;strong&gt;Sp3000&lt;/strong&gt;]&lt;br&gt;&#xA;W: &lt;strong&gt;powwow&lt;/strong&gt; (3/6) [credit: &lt;strong&gt;Joe Z&lt;/strong&gt;]&lt;br&gt;&#xA;X: &lt;strong&gt;xerox&lt;/strong&gt; (2/5) [credit: &lt;strong&gt;Joe Z&lt;/strong&gt;]&lt;br&gt;&#xA;Y: &lt;strong&gt;syzygy&lt;/strong&gt; (3/6) [credit: &lt;strong&gt;Joe Z&lt;/strong&gt;]&lt;br&gt;&#xA;Z: &lt;strong&gt;pzazz&lt;/strong&gt; (3/5)&lt;br&gt;&lt;/p&gt;&#xA;&lt;p&gt;At the moment I'm listing ties, but please don't go too much over the top with new words that match old ones - if there get to be too many equal winners for each letter, I'll have to remove all but the first one.&lt;/p&gt;&#xA;&lt;p&gt;&lt;sub&gt;&lt;span class=&quot;math-container&quot;&gt;$^1$&lt;/span&gt; Nobody mention &amp;quot;pippy&amp;quot;! :-)&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2021-03-26T07:46:04.897" LastActivityDate="2021-05-05T14:28:31.297" Title="Find words with the highest proportion of a single letter" Tags="|word|english|open-ended|language|letters|" AnswerCount="18" CommentCount="17" ClosedDate="2021-05-05T21:05:34.283" ContentLicense="CC BY-SA 4.0" />
  <row Id="10558" PostTypeId="2" ParentId="10557" CreationDate="2015-03-19T01:32:35.857" Score="11" Body="&lt;p&gt;My contributions to this list:&lt;/p&gt;&#xA;&lt;p&gt;B: hubbub (3/6)&lt;br /&gt;&#xA;C: coccyx (3/6)&lt;br /&gt;&#xA;E: peewee (4/6)&lt;br /&gt;&#xA;G: glogg (3/5)&lt;br /&gt;&#xA;I: radii (2/5)&lt;br /&gt;&#xA;J: jejune (2/6)&lt;br /&gt;&#xA;L: lulls (3/5)&lt;br /&gt;&#xA;O: voodoo (4/6)&lt;br /&gt;&#xA;P: poppy (3/5)&lt;br /&gt;&#xA;Q: equivoque (2/9, had to google for this one)&lt;br /&gt;&#xA;S: assess (4/6)&lt;br /&gt;&#xA;U: muumuu (4/6)&lt;br /&gt;&#xA;V: vulva (2/5)&lt;br /&gt;&#xA;W: powwow (3/6)&lt;br /&gt;&#xA;X: xerox (2/5, if we're counting genericized trademarks)&lt;br /&gt;&#xA;Y: syzygy (3/6)&lt;br /&gt;&#xA;Z: pizzazz (4/7)&lt;/p&gt;&#xA;&lt;p&gt;Also some sketchy and dubious contenders:&lt;/p&gt;&#xA;&lt;p&gt;E: teevee (4/6, a transcribed phonetic spelling of TV [OED])&lt;br /&gt;&#xA;O: hoooo (4/5, a Japanese loan word for phoenix, also spelled houou)&lt;br /&gt;&#xA;T: tut-tut (4/6, a hyphenated interjection expressing disapproval [OED])&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="69582" LastEditDate="2021-03-26T14:11:06.070" LastActivityDate="2021-03-26T14:11:06.070" CommentCount="14" ContentLicense="CC BY-SA 4.0" />
  <row Id="10559" PostTypeId="1" AcceptedAnswerId="62021" CreationDate="2015-03-19T02:51:06.967" Score="12" ViewCount="1056" Body="&lt;p&gt;Sometimes we play catch;&lt;br&gt;&#xA;Though &lt;em&gt;you'd&lt;/em&gt; hope it more like fetch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd be the second; an honorable mention.&lt;br&gt;&#xA;Indeed it'd be, as if I've spent my life in detention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'd know where, if you found the tether;&lt;br&gt;&#xA;Around about which, he and I dance together.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Cliches away; your usual meeting need not be so verbose.&lt;br&gt;&#xA;PROPERLY told, &lt;em&gt;I&lt;/em&gt; don't even have to come close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intrinsic you'd have a gut full of fear;&lt;br&gt;&#xA;Ladies and gentlemen, boys and girls... Dying time's here!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Given me a name;&lt;br&gt;&#xA;Then please explain:&lt;br&gt;&#xA;Where, when and what is,&lt;br&gt;&#xA;That which I'll do,&lt;br&gt;&#xA;Should you ever find out:&lt;br&gt;&#xA;Me,&lt;/em&gt; I &lt;em&gt;am true.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Hint and caveat: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The answer to this riddle; here a search would lead to little.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint &lt;em&gt;about&lt;/em&gt; the hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The answer to this riddle does not as of yet appear in any question or answer on &lt;em&gt;Puzzling.SE&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1984 was a long time ago, here's another hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For a while my imaginary sister, she had stolen my thunder. But now by the wise, she too is considered yet another blunder. (Extra credit: What's her name? What can the wise see that we cannot?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt; -Quote credit: &lt;em&gt;&lt;a href=&quot;http://madmax.wikia.com/wiki/Dr._Dealgood&quot; rel=&quot;noreferrer&quot;&gt;Dr. Dealgood&lt;/a&gt;&lt;/em&gt; &lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="4522" LastEditorUserId="4522" LastEditDate="2015-08-25T15:16:20.447" LastActivityDate="2018-03-16T00:20:44.950" Title="Defeat by Me; Your Destiny" Tags="|riddle|word|rhyme|" AnswerCount="7" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10560" PostTypeId="2" ParentId="10557" CreationDate="2015-03-19T03:08:54.740" Score="7" Body="&lt;p&gt;A: alpaca (3/6)&lt;br /&gt;&#xA;B: bobby (3/5)&lt;br /&gt;&#xA;C: cocky (2/5)&lt;br /&gt;&#xA;D: added (3/5)&lt;br /&gt;&#xA;F: fifty (2/5)&lt;br /&gt;&#xA;G: gagged (3/6)&lt;br /&gt;&#xA;I: idiom (2/5)&lt;br /&gt;&#xA;K: kicks (2/5)&lt;br /&gt;&#xA;M: mommy (3/5)&lt;br /&gt;&#xA;N: nanny (3/5)&lt;br /&gt;&#xA;P: peppy (3/5)&lt;br /&gt;&#xA;R: rarer (3/5)&lt;br /&gt;&#xA;S: sasses (4/6)&lt;br /&gt;&#xA;T: tattle (3/6)&lt;br /&gt;&#xA;V: vivid (2/5)&lt;/p&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="69582" LastEditDate="2021-03-26T14:11:54.050" LastActivityDate="2021-03-26T14:11:54.050" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="10561" PostTypeId="1" AcceptedAnswerId="10621" CreationDate="2015-03-19T03:35:01.260" Score="8" ViewCount="5020" Body="&lt;p&gt;I am intangible&lt;br&gt;&#xA;Once given, I can never be regained&lt;br&gt;&#xA;I am always given as a pair&lt;br&gt;&#xA;I can give a gift of my own&lt;br&gt;&#xA;Everyone has me exactly once&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ask you: What am I&lt;/p&gt;&#xA;" OwnerUserId="10356" LastActivityDate="2018-06-04T11:02:31.020" Title="What gift am I?" Tags="|riddle|word|" AnswerCount="7" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10562" PostTypeId="2" ParentId="10559" CreationDate="2015-03-19T04:33:43.253" Score="1" Body="&lt;p&gt;My inclination is that you are the second player at horseshoes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes we play catch;&#xA;Though you'd hope it more like fetch.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;(You throw my way and I throw toward you.  Each of us hope the other has to run after his throws)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'd be the second; an honorable mention.&#xA;Indeed I've spent my life as if in detention.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;(You are the second player, no explanation for the detention comment)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'd know where, if you found the tether;&#xA;Around about which, he and I dance together.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;(We're aiming at our respective posts and one might call it a dance)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cliche away; your meeting need not be verbose.&#xA;Properly told, I don't even have to come close.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;(The player coming closest to the post scores, meaning I don't have to be close at all, just closer than you)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Intrinsic you'd have a gut full of fear;&#xA;Ladies and gentlemen, boys and girls... Dying time's here!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;(No explanation at all for these lines)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10361" LastEditorUserId="10037" LastEditDate="2015-08-25T17:31:51.210" LastActivityDate="2015-08-25T17:31:51.210" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10563" PostTypeId="2" ParentId="10557" CreationDate="2015-03-19T04:45:45.080" Score="3" Body="&lt;p&gt;Some more contributions:&lt;/p&gt;&#xA;&lt;p&gt;A: agama  (3/5)&lt;br /&gt;&#xA;C: cocci  (3/5)&lt;br /&gt;&#xA;F: feoff  (3/5)&lt;br /&gt;&#xA;J: hajji  (2/5)&lt;br /&gt;&#xA;T: tatty  (3/5)&lt;/p&gt;&#xA;&lt;p&gt;Everything else I got tied with Joe and pac's answers.&lt;/p&gt;&#xA;&lt;p&gt;One for &amp;quot;x&amp;quot; that's not a trademark (although &amp;quot;xerox&amp;quot; is pretty common):&lt;/p&gt;&#xA;&lt;p&gt;X: maxixe (2/6)&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Edit&lt;/em&gt;: Since we seem to be doing ties, here are the tied answers I got which were different:&lt;/p&gt;&#xA;&lt;p&gt;O: booboo    (4/6)&lt;br /&gt;&#xA;P: pappy     (3/5)&lt;br /&gt;&#xA;Q: quinquina (2/9)&lt;br /&gt;&#xA;V: civvy     (2/5)&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Deemed invalid:&lt;/p&gt;&#xA;&lt;p&gt;H: highth (3/6)&lt;br /&gt;&#xA;I: filii  (3/5)&lt;/p&gt;&#xA;" OwnerUserId="445" LastEditorUserId="69582" LastEditDate="2021-03-26T14:15:20.823" LastActivityDate="2021-03-26T14:15:20.823" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="10565" PostTypeId="2" ParentId="10557" CreationDate="2015-03-19T05:29:04.063" Score="3" Body="&lt;p&gt;Current best list, as far as I know:&lt;/p&gt;&#xA;&lt;p&gt;a: agama, 3/5&lt;br /&gt;&#xA;b: bobby, 3/5&lt;br /&gt;&#xA;c: coccic, 4/6&lt;br /&gt;&#xA;d: dodded, 4/6&lt;br /&gt;&#xA;e: peewee, 4/6&lt;br /&gt;&#xA;f: fluff, 3/5&lt;br /&gt;&#xA;g: glogg, 3/5&lt;br /&gt;&#xA;h: hashish, 3/7&lt;br /&gt;&#xA;i: bikini, 3/6&lt;br /&gt;&#xA;j: hajji, 2/5&lt;br /&gt;&#xA;k: kakkak, 4/6&lt;br /&gt;&#xA;l: lolls, 3/5&lt;br /&gt;&#xA;m: mamma, 3/5&lt;br /&gt;&#xA;n: nanny, 3/5&lt;br /&gt;&#xA;o: voodoo, 4/6&lt;br /&gt;&#xA;p: peppy, 3/5&lt;br /&gt;&#xA;q: equivoque, 2/9&lt;br /&gt;&#xA;r: error, 3/5&lt;br /&gt;&#xA;s: assess, 4/6&lt;br /&gt;&#xA;t: tatty, 3/5&lt;br /&gt;&#xA;u: muumuu, 4/6&lt;br /&gt;&#xA;v: divvy, 2/5&lt;br /&gt;&#xA;w: powwow, 3/6&lt;br /&gt;&#xA;x: xerox, 2/5&lt;br /&gt;&#xA;y: syzygy, 3/6&lt;br /&gt;&#xA;z: pzazz, 3/5&lt;/p&gt;&#xA;&lt;p&gt;Recent updates:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Added &amp;quot;pzazz&amp;quot;, which is arguably valid.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2187" LastEditorUserId="69582" LastEditDate="2021-03-26T14:16:13.900" LastActivityDate="2021-03-26T14:16:13.900" CommentCount="14" ContentLicense="CC BY-SA 4.0" />
  <row Id="10566" PostTypeId="2" ParentId="10421" CreationDate="2015-03-19T06:13:17.707" Score="3" Body="&lt;p&gt;I think its time to post the answer to this puzzle, i should have made a better job at describing everything, oh well here it goes :)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The image is actualy a overlay for a keyboard. The intersections in the maze tells you which letters to take(apart from the starting position). So with this you just need to find the correct position and the only one that roughly provides a word is if you put the starting position over the letter N which then spells out NIGHT, so since its night outside the color of the sky is BLACK.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5560" LastActivityDate="2015-03-19T06:13:17.707" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10568" PostTypeId="1" AcceptedAnswerId="10594" CreationDate="2015-03-19T07:27:38.933" Score="15" ViewCount="1927" Body="&lt;p&gt;A wave of the hand is all it takes to dismiss,&lt;br&gt;&#xA;But when it gets started, it's greater than this!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bringer of gifts. A destroyer of food.&lt;br&gt;&#xA;It helps win the race. It ruins the mood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See it, then not... or the other way 'round?&lt;br&gt;&#xA;Answer this wrong and it's still most likely found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Puzzlingly meta, you'll answer with glee,&lt;br&gt;&#xA;So tell me, dear riddlers, what the heck can it be?&lt;/p&gt;&#xA;" OwnerUserId="5208" LastActivityDate="2016-12-01T15:47:14.867" Title="Puzzlingly meta... what the heck can it be?" Tags="|riddle|word|" AnswerCount="6" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10569" PostTypeId="2" ParentId="10557" CreationDate="2015-03-19T07:46:14.363" Score="3" Body="&lt;p&gt;Based on current best I made this list:&lt;/p&gt;&#xA;&lt;p&gt;A: agama (3/5)&lt;br /&gt;&#xA;B: baobab (3/6)&lt;br /&gt;&#xA;C: cancer (2/6)&lt;br /&gt;&#xA;D: daddy (3/5)&lt;br /&gt;&#xA;E:&lt;br /&gt;&#xA;F: fluff (3/5)&lt;br /&gt;&#xA;G: gagging (4/7)&lt;br /&gt;&#xA;H: height (2/6)&lt;br /&gt;&#xA;I: bikini (3/6)&lt;br /&gt;&#xA;J:&lt;br /&gt;&#xA;K: kukri (2/5)&lt;br /&gt;&#xA;L: lolly (3/5)&lt;br /&gt;&#xA;M: mummy (3/5)&lt;br /&gt;&#xA;N: cannon (3/6)&lt;br /&gt;&#xA;O:&lt;br /&gt;&#xA;P: puppy (3/5)&lt;br /&gt;&#xA;Q: queue (1/5)&lt;br /&gt;&#xA;R: horror (3/6)&lt;br /&gt;&#xA;S: possess (4/7)&lt;br /&gt;&#xA;T: twitter (3/7)&lt;br /&gt;&#xA;U:&lt;br /&gt;&#xA;V: valve (2/5)&lt;br /&gt;&#xA;W:&lt;br /&gt;&#xA;X:&lt;br /&gt;&#xA;Y:&lt;br /&gt;&#xA;Z:&lt;/p&gt;&#xA;&lt;p&gt;Couldn't think about any words that can beat &lt;strong&gt;E, J, O, U, W, X, Y, Z&lt;/strong&gt;. Other words are better that current best (at least for now)&lt;/p&gt;&#xA;" OwnerUserId="9352" LastEditorUserId="69582" LastEditDate="2021-03-26T14:13:28.690" LastActivityDate="2021-03-26T14:13:28.690" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10570" PostTypeId="2" ParentId="10561" CreationDate="2015-03-19T07:57:09.917" Score="6" Body="&lt;p&gt;My guess would be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Love&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation line by line:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am intangible&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Love is an emotion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once given, I can never be regained&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Its hard to let something go you love.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am always given as a pair&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Love is always between two things.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can give a gift of my own&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Love can give the ability to reproduce. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Everyone has me exactly once&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The love of your life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4786" LastEditorUserId="4786" LastEditDate="2015-03-19T11:32:26.210" LastActivityDate="2015-03-19T11:32:26.210" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10571" PostTypeId="2" ParentId="10568" CreationDate="2015-03-19T08:13:27.270" Score="3" Body="&lt;p&gt;I would say that the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Desire&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Desire can take one wave of the hand to dismiss, but when it gets started it's very hard to get rid of. Desire can bring you many gifts, but it can also destroy many things. On some people desire can clearly be seen, whilst on other people it is hard to see.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10369" LastEditorUserId="25966" LastEditDate="2016-12-01T15:47:14.867" LastActivityDate="2016-12-01T15:47:14.867" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10572" PostTypeId="2" ParentId="10561" CreationDate="2015-03-19T08:17:24.680" Score="12" Body="&lt;p&gt;My guess is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Life.&lt;br&gt;&lt;br&gt;&#xA; It cannot be regain after death.&lt;br&gt;&#xA; Its intangible.&lt;br&gt;&#xA; Given by two parents.&lt;br&gt;&#xA; You can give another life.&lt;br&gt;&#xA; And you live exactly once.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10371" LastActivityDate="2015-03-19T08:17:24.680" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="10573" PostTypeId="2" ParentId="10568" CreationDate="2015-03-19T08:21:46.637" Score="7" Body="&lt;h3&gt;The answer is:&lt;/h3&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The posts' &lt;strong&gt;votes&lt;/strong&gt; (up or down)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;Explaination line by line:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;A wave of the hand is all it takes to dismiss.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can easily delete an answer and dismiss all the votes you got.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But when it gets started, it's greater than this!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm not quite sure about this line...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A bringer of gifts. A destroyer of food.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bringing gifts is the equivalent of earning reputation (up votes), and destroying food is the equivalent of losing reputation (the &quot;food&quot; of the SE users).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It helps win the race. It ruins the mood.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Upvotes help your answer &quot;win the race&quot; with the others and be marked as correct by the OP. An high number of downvotes can destroy your mood.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See it, then not... or the other way 'round?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The vote count can disappear if there is an equal number of up and down votes, and it can also reappear with another vote.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer this wrong and it's still most likely found.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you answer this question wrong, you'll probably get a bunch of downvotes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Puzzlingly meta, you'll answer with glee.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When on Meta SE sites, the up and down votes don't make you earn or lose any reputation, so you can answer any question with glee.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10248" LastEditorUserId="5373" LastEditDate="2015-03-19T11:13:21.010" LastActivityDate="2015-03-19T11:13:21.010" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10574" PostTypeId="1" AcceptedAnswerId="10593" CreationDate="2015-03-19T09:34:05.710" Score="28" ViewCount="13524" Body="&lt;p&gt;What is the largest integer $A$ that can be produced from the four digits $1,2,3,4$ subject to the following rules?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Each of the digits $1,2,3,4$ is used exactly once.&lt;/li&gt;&#xA;&lt;li&gt;Juxtaposition of digits is allowed (so one may for instance juxtapose 1 and 3 to get &quot;31&quot;)&lt;/li&gt;&#xA;&lt;li&gt;The allowed mathematical operations are addition (+), multiplication (*) and exponentiation (^).&lt;/li&gt;&#xA;&lt;li&gt;You may use brackets &quot;(&quot; and &quot;)&quot; to structure your expression, and to make it well-defined (for instance, this allows you to distinguish a^(b^c) from (a^b)^c).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-06T14:07:12.280" LastActivityDate="2021-12-01T19:30:25.597" Title="The largest integer made of 1,2,3,4" Tags="|mathematics|calculation-puzzle|formation-of-numbers|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10575" PostTypeId="1" CreationDate="2015-03-19T09:35:30.113" Score="8" ViewCount="1438" Body="&lt;p&gt;Header and Tailer play the following game. &#xA;At the beginning, the juror lays out a row of $n\ge5$ coins on the table that alternately show heads and tails, with the leftmost coin showing heads. &#xA;Header and Tailer move alternately, with Header making the first move.&#xA;In every move&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the active player first takes a coin from his own pocket and&#xA;adds it (heads or tails) to the row; Header adds his coin at the right end,&#xA;and Tailer adds his at the left end.&lt;/li&gt;&#xA;&lt;li&gt;Then the active player flips an arbitrary sequence of consecutive coins that show the same side.&lt;/li&gt;&#xA;&lt;li&gt;A recently added coin can be flipped over only with adjacent coins , but when the same player lays another coin down, it is treated like any other coin.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Header wins if he succeeds to make the row all heads.&#xA;Tailer wins if he succeeds to make the row all tails.&#xA;The winner receives all coins on the table (that have been put there by the juror, by the opponent, and by himself).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question: For which values of $n$ does Header have a winning strategy?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Ps:&lt;/strong&gt; this is not perfectly answered yet , i did affect the bounty for hardest work, waiting for a specific winning range of n for the header .&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9517" LastEditorUserId="9517" LastEditDate="2015-04-06T18:04:02.700" LastActivityDate="2015-04-06T18:04:02.700" Title="Coin inverting game" Tags="|strategy|game|reachability|" AnswerCount="3" CommentCount="19" ContentLicense="CC BY-SA 3.0" />
  <row Id="10576" PostTypeId="5" CreationDate="2015-03-19T09:37:07.187" Score="0" Body="&lt;p&gt;A 'word sequence' is a series or collection of words presented in a particular order according to a specific logical rule. In &lt;a href=&quot;/questions/tagged/word-sequence&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;word-sequence&amp;#39;&quot; aria-label=&quot;show questions tagged &amp;#39;word-sequence&amp;#39;&quot; rel=&quot;tag&quot; aria-labelledby=&quot;word-sequence-container&quot;&gt;word-sequence&lt;/a&gt; puzzles, the aim for the solver is usually to deduce or discover the rule behind this ordering, and/or supply the next - or missing - word(s) in the sequence.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="60644" LastEditDate="2022-12-12T23:26:27.037" LastActivityDate="2022-12-12T23:26:27.037" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10577" PostTypeId="4" CreationDate="2015-03-19T09:37:07.187" Score="0" Body="A puzzle in which the hidden pattern behind a given sequence of words is to be found. Use with [sequence]" OwnerUserId="8874" LastEditorUserId="60644" LastEditDate="2022-12-12T23:27:16.047" LastActivityDate="2022-12-12T23:27:16.047" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10579" PostTypeId="2" ParentId="10574" CreationDate="2015-03-19T09:43:55.010" Score="19" Body="&lt;p&gt;I would say the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$2^{3^{41}} \approx 10^{10^{19}}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, I cannot prove it.&lt;/p&gt;&#xA;" OwnerUserId="7308" LastActivityDate="2015-03-19T09:43:55.010" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10580" PostTypeId="2" ParentId="10568" CreationDate="2015-03-19T09:52:06.437" Score="3" Body="&lt;p&gt;Inspired by &lt;a href=&quot;https://puzzling.stackexchange.com/users/10248/marco-bonelli&quot;&gt;Marco Bonelli&lt;/a&gt;'s answer.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Answer:&lt;/h3&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SE reputation  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;Explanation:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;A wave of the hand is all it takes to dismiss,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Deleting the account also deletes all your reputation (and reputation you gave to others with votes)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But when it gets started, it's greater than this!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you use the system well, your reputation will start to grow.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A bringer of gifts. A destroyer of food.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reputation gives you badges, the food I think is just for the sake of the rhyme.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It helps win the race. It ruins the mood.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can have the greatest reputation, but if you start to lose it...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See it, then not... or the other way 'round?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Upvote, then downote. Correct answer, then revoked.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer this wrong and it's still most likely found.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Also if this is wrong, I hope to get some upvotes :)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Puzzlingly meta, you'll answer with glee,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Upvotes and downvotes in meta sites of SE doesn't change your reputation.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So tell me, dear riddlers, what the heck can it be?  &lt;/p&gt;&#xA;" OwnerUserId="4652" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-03-19T11:03:08.313" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10581" PostTypeId="2" ParentId="10574" CreationDate="2015-03-19T09:52:23.893" Score="1" Body="&lt;p&gt;3^(2^41) which is a very large number .&lt;/p&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-03-19T09:52:23.893" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10582" PostTypeId="2" ParentId="10559" CreationDate="2015-03-19T09:55:48.553" Score="1" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; quote&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes we play catch;&lt;br&gt;&#xA;Though you'd hope it more like fetch.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sometimes you need to catch the quote, hopefully a reference allows you to fetch it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'd be the second; an honorable mention.&lt;br&gt;&#xA;Indeed I've spent my life as if in detention.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quoting means mentioning something a second time. The quoted text is sort of locked in, it doesn't have much freedom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'd know where, if you found the tether;&lt;br&gt;&#xA;Around about which, he and I dance together.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A quote is usually tethered by quotations marks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cliche away; your meeting need not be verbose.&lt;br&gt;&#xA;Properly told, I don't even have to come close.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A quote doesn't necessarily repeat everything, just an essential part. It doesn't even have to come close.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Intrinsic you'd have a gut full of fear;&lt;br&gt;&#xA;Ladies and gentlemen, boys and girls... Dying time's here!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is actually a quote, Dr. Dealgood.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Give me a name;&lt;br&gt;&#xA;Then please explain:&lt;br&gt;&#xA;Where, when and what is,&lt;br&gt;&#xA;That which I'll do,&lt;br&gt;&#xA;Should you ever find out:&lt;br&gt;&#xA;Me, I am true.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You quote.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9910" LastActivityDate="2015-03-19T09:55:48.553" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10583" PostTypeId="2" ParentId="10299" CreationDate="2015-03-19T10:05:59.413" Score="7" Body="&lt;p&gt;This puzzle is very similar in spirit to [ &lt;a href=&quot;https://puzzling.stackexchange.com/questions/8811/what-do-they-have-in-common&quot;&gt;What do they have in common?&lt;/a&gt; ].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Gordon&lt;/strong&gt; $~~~$ (&lt;a href=&quot;https://en.wikipedia.org/wiki/Gordon&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Gordon&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;for the following reasons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;an extraterrestrial entity&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Gordon&lt;/strong&gt; Shumway (TV series ALF)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;a space traveller &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Gordon&lt;/strong&gt; Cooper (astronaut)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Michael Douglas &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Gordon&lt;/strong&gt; Gekko (Douglas' character in movie &quot;Wallstreet&quot;)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;a friend of Thomas &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Thomas and &lt;strong&gt;Gordon&lt;/strong&gt; (the tank engines)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;a friend of Tony &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Gordon&lt;/strong&gt; Brown (successor of Tony Blair)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;an Army infantry regiment &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The &lt;strong&gt;Gordon&lt;/strong&gt; Highlanders  &lt;a href=&quot;https://en.wikipedia.org/wiki/Gordon_Highlanders&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Gordon_Highlanders&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;and some naked ladies &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Gordon&lt;/strong&gt; (title of first album of the &quot;Barenaked Ladies&quot;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2016-03-15T09:03:51.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10584" PostTypeId="2" ParentId="10561" CreationDate="2015-03-19T10:54:00.137" Score="1" Body="&lt;p&gt;I would say, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;death&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am intangible&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Can you really touch death?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once given, I can never be regained&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It can't be regain as it cant be lost&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am always given as a pair&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Clinical death and brain death&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can give a gift of my own&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If somebody dies can leave some inheritance to his family and that might be consider as gift (especially if person who died is some far relative who you even didn't know exist), or maybe it is red herring&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Everyone has me exactly once  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can die only once&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I ask you: What am I&lt;/p&gt;&#xA;" OwnerUserId="1925" LastEditorUserId="44915" LastEditDate="2018-06-04T11:02:31.020" LastActivityDate="2018-06-04T11:02:31.020" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="10585" PostTypeId="2" ParentId="10561" CreationDate="2015-03-19T11:08:35.660" Score="0" Body="&lt;p&gt;Pretty similar to Izzy's answer, but my guess is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Birth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10379" LastActivityDate="2015-03-19T11:08:35.660" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10586" PostTypeId="2" ParentId="10542" CreationDate="2015-03-19T11:21:45.603" Score="27" Body="&lt;p&gt;Jennifer's husband is busy in business meetings, while Douglas's dad is with Jennifer.&lt;/p&gt;&#xA;" OwnerUserId="10210" LastActivityDate="2015-03-19T11:21:45.603" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10588" PostTypeId="1" CreationDate="2015-03-19T11:29:49.677" Score="8" ViewCount="3457" Body="&lt;p&gt;There is a 3×3 dot grid. How many different non-congruent polygons can you make on the grid?&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Rules:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;All vertices of the polygon must be on the grid&lt;/li&gt;&#xA;&lt;li&gt;Only &lt;strong&gt;non self intersecting&lt;/strong&gt; polygons&lt;/li&gt;&#xA;&lt;li&gt;Only polygons with non-empty interior (&amp;lt;=&amp;gt; positive area)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="10380" LastEditorUserId="69582" LastEditDate="2021-03-01T23:03:09.593" LastActivityDate="2021-11-13T21:22:29.763" Title="How many different non congruent polygons can you make on a 3x3 dot grid?" Tags="|mathematics|geometry|combinatorics|" AnswerCount="6" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="10591" PostTypeId="2" ParentId="10561" CreationDate="2015-03-19T13:13:18.280" Score="6" Body="&lt;p&gt;My try: I am&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a name&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am intangible&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A name is intangible&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once given, I can never be regained&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Once you give a name to a baby you can't get it back, that doesn't makes sense.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am always given as a pair&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You get a pair of firstname/lastname on birth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can give a gift of my own&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You &quot;gift&quot; your lastname to your child&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Everyone has me exactly once&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everyone has only one name&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6117" LastActivityDate="2015-03-19T13:13:18.280" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10592" PostTypeId="2" ParentId="10514" CreationDate="2015-03-19T13:14:24.030" Score="1" Body="&lt;p&gt;Tricky, it might be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; maze&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A place of work, and not of play,&lt;br&gt;&#xA;You know that's what the vermin say.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mazes are usually found in the form of games, lab rats might see this differently.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is it this or that, the question sly,&lt;br&gt;&#xA;But in the end you'll question why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Left way or right way? In the end it doesn't tend to matter for common mazes as long as you stick to one choice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A miracle you want to find,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A reward at the center of the maze.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And get yourself out of a bind.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Escape the maze.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So while you're here, follow a hunch,&lt;br&gt;&#xA;It may just happen before lunch. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For the lab rat, follow your nose to find the cheese.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9910" LastActivityDate="2015-03-19T13:14:24.030" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10593" PostTypeId="2" ParentId="10574" CreationDate="2015-03-19T13:15:45.413" Score="36" Body="&lt;p&gt;I have been searching the internet for this one and found a link to the page &quot;&lt;a href=&quot;http://www.mathsnet.net/a1234.html&quot; rel=&quot;noreferrer&quot; title=&quot;1, 2, 3, 4 - four digits that DWARFED the universe&quot;&gt;1, 2, 3, 4 - four digits that DWARFED the universe&lt;/a&gt;&quot;.&#xA;On the bottom of this page there is a table with a few of the great combinations that can be made with &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt;. If you look for the entries with just the numbers (no decimals, no minus, etc.), you see the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$2^{3^{41}}$&lt;/span&gt; has got &lt;span class=&quot;math-container&quot;&gt;$10^{19}$&lt;/span&gt; digits&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$3^{4^{21}}$&lt;/span&gt; has got &lt;span class=&quot;math-container&quot;&gt;$10^{12}$&lt;/span&gt; digits&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$3^{421}$&lt;/span&gt; has got &lt;span class=&quot;math-container&quot;&gt;$201$&lt;/span&gt; digits&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$4^{321}$&lt;/span&gt; has got &lt;span class=&quot;math-container&quot;&gt;$194$&lt;/span&gt; digits&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$2^{431}$&lt;/span&gt; has got &lt;span class=&quot;math-container&quot;&gt;$130$&lt;/span&gt; digits&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You can calculate/test numbers yourself with this gigantic online calculator: &lt;a href=&quot;http://mrob.com/pub/comp/hypercalc/hypercalc-javascript.html&quot; rel=&quot;noreferrer&quot;&gt;HyperCalc Javascript&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(Trying to give an explanation):&lt;/strong&gt; As @h34 notes in a comment: We need to use exponentiation to get the biggest number. If you draw a graph of an exponential function it shoots upward after a few numbers. For instance the simple binary function &lt;span class=&quot;math-container&quot;&gt;$2^x$&lt;/span&gt; results in: &lt;span class=&quot;math-container&quot;&gt;$$2, 4, 8, 16, 32, 64, 128, 256, \ldots$$&lt;/span&gt; (we all know this sequence). More precisely:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Any expression of the type &lt;span class=&quot;math-container&quot;&gt;$x + 1$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$x \cdot 1$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$x^1$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$1^x$&lt;/span&gt; can be given a bigger value by juxtaposing the digit &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; to some other digit in the expression &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;. Therefore the digit &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; will not stand alone in the expression, but will be juxtaposed to some other digit. Furthermore, the digit &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; will come after the other digit(s) in this juxtaposition.&lt;/li&gt;&#xA;&lt;li&gt;For &lt;span class=&quot;math-container&quot;&gt;$x, y \ge 2$&lt;/span&gt;, we always have &lt;span class=&quot;math-container&quot;&gt;$x^y \ge x+y$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$x^y \ge x \cdot y$&lt;/span&gt;. Therefore, the maximizing expression will only use exponentiations.&lt;/li&gt;&#xA;&lt;li&gt;Any expression &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; without exponentiation will have value at most &lt;span class=&quot;math-container&quot;&gt;$4321$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Any expression &lt;span class=&quot;math-container&quot;&gt;$x^y$&lt;/span&gt; with only a single exponentiation will have value at most &lt;span class=&quot;math-container&quot;&gt;$3^{421}$&lt;/span&gt; (which is about &lt;span class=&quot;math-container&quot;&gt;$7.38×10^{200}$&lt;/span&gt;).&lt;/li&gt;&#xA;&lt;li&gt;Therefore it remains to consider expressions of the form &lt;span class=&quot;math-container&quot;&gt;$a^{b^{c}}$&lt;/span&gt; where one of &lt;span class=&quot;math-container&quot;&gt;$a,b,c$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$21$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$31$&lt;/span&gt;, or &lt;span class=&quot;math-container&quot;&gt;$41$&lt;/span&gt;, and where the other two are single-digit numbers.&lt;/li&gt;&#xA;&lt;li&gt;If &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; consists of two digits then the value of &lt;span class=&quot;math-container&quot;&gt;$a^{b^c}$&lt;/span&gt; is at most &lt;span class=&quot;math-container&quot;&gt;$41^{3^2}$&lt;/span&gt;,&#xA;which is much smaller than &lt;span class=&quot;math-container&quot;&gt;$2^{3^{41}}$&lt;/span&gt;. Therefore &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; is a single digit.&#xA;If &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; has two digits and &lt;span class=&quot;math-container&quot;&gt;$c\ge2$&lt;/span&gt; is single digit, then &lt;span class=&quot;math-container&quot;&gt;$b^c&amp;lt;c^b$&lt;/span&gt;. Therefore &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; is single digit and &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; has two digits.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Hence, it only remains to analyze the six possible candidates &#xA;&lt;span class=&quot;math-container&quot;&gt;$2^{3^{41}}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$2^{4^{31}}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$3^{2^{41}}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$3^{4^{21}}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$4^{2^{31}}$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$4^{3^{21}}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;For the fifth candidate, we have &lt;span class=&quot;math-container&quot;&gt;$$4^{2^{31}} = 2^{2 \cdot 2^{31}} = 2^{2^{32}},$$&lt;/span&gt;&#xA;whereas for the sixth candidate, we have &lt;span class=&quot;math-container&quot;&gt;$$4^{3^{21}} = 2^{2 \cdot 3^{21}} &amp;lt; 2^{3^{22}}.$$&lt;/span&gt;&#xA;Hence both candidates are clearly smaller than the first candidate &lt;span class=&quot;math-container&quot;&gt;$2^{3^{41}}$&lt;/span&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For the fourth candidate, we have &lt;span class=&quot;math-container&quot;&gt;$$3^{4^{21}} = 3^{2^{2 \cdot 21}} = 3^{2^{42}},$$&lt;/span&gt; which dominates and eliminates the third candidate &lt;span class=&quot;math-container&quot;&gt;$3^{2^{41}}$&lt;/span&gt;. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The fourth candidate is &lt;span class=&quot;math-container&quot;&gt;$$3^{4^{21}} &amp;lt; 4^{4^{21}} = 2^{2^{43}}.$$&lt;/span&gt; The last exponent &lt;span class=&quot;math-container&quot;&gt;$2^{43}$&lt;/span&gt; satisfies &lt;span class=&quot;math-container&quot;&gt;$$2^{43} = 2^6 \cdot 2^{37} &amp;lt; 81 \cdot 3^{37} = 3^{41},$$&lt;/span&gt; and hence is smaller than the exponent of the first candidate. This eliminates the fourth candidate.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Finally, let us compare the exponents of the first and the second candidate:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$4^{31} = 2^{11} \cdot 2^{51} = 2048 \cdot 8^{17} &amp;lt; 2187 \cdot 9^{17} = 3^7 \cdot 3^{34} = 3^{41}.$$&lt;/span&gt; &#xA;This eliminates the second candidate.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summarizing: The answer is that the first candidate &lt;span class=&quot;math-container&quot;&gt;$2^{3^{41}}$&lt;/span&gt; yields the largest possible value.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Spoiler&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If we want to get the answer as big as we possibly can, we need to use the highest numbers as many times as possible within the calculation. To achieve this we use exponentiation multiple times with the numbers in increasing order: &lt;span class=&quot;math-container&quot;&gt;$1^{2^{3^4}}$&lt;/span&gt;. The &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; doesn't do anything in this calculation, so we put it together with the highest number &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt; to create an even bigger number: &lt;span class=&quot;math-container&quot;&gt;$2^{3^{41}}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Another Spoiler&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; And to answer the question, why not &lt;span class=&quot;math-container&quot;&gt;$3^{2^{41}}$&lt;/span&gt;? Here is a graph of both &lt;span class=&quot;math-container&quot;&gt;$2^x \cdot \log(3) $&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$3^x \cdot \log(2)$&lt;/span&gt;. These functions give the best view about the differences between them.&#xA; &lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/BRpLb.png&quot; alt=&quot;Graph at logarithmic scale&quot;&gt;&#xA; &lt;br&gt;&#xA; Indeed, &lt;span class=&quot;math-container&quot;&gt;$3^{2^x}$&lt;/span&gt; has a head start (begins at &lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt; rather than &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt;), but it quickly passed by the &lt;span class=&quot;math-container&quot;&gt;$2^{3^x}$&lt;/span&gt;-function when &lt;span class=&quot;math-container&quot;&gt;$x &amp;gt; 1.14$&lt;/span&gt; and never crosses again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10384" LastEditorUserId="10384" LastEditDate="2019-03-05T14:38:35.763" LastActivityDate="2019-03-05T14:38:35.763" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="10594" PostTypeId="2" ParentId="10568" CreationDate="2015-03-19T13:32:02.483" Score="12" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The spoiler tag&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A wave of the hand is all it takes to dismiss,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Run your mouse off of the tag and it disappears&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But when it gets started, it's greater than this!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you start to wave your pointer across it, the answer appears&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A bringer of gifts. A destroyer of food.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It helps win the race. It ruins the mood.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can get the answer from the tag quickly. If you get it without guessing, it may &quot;spoil&quot; the puzzle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See it, then not... or the other way 'round?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Toggling the spoiler tag's availability, redundant from earlier verses&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer this wrong and it's still most likely found.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People tend to have guessed (as they have on this question) even if the answer is not correct.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10387" LastActivityDate="2015-03-19T13:32:02.483" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10595" PostTypeId="2" ParentId="10588" CreationDate="2015-03-19T14:14:52.327" Score="3" Body="&lt;p&gt;Well, I don't want to work it all out, but in just a 1x3 grid, I count 18 polygons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1x1 - 2&lt;br&gt;&#xA;1x2 - +6&lt;br&gt;&#xA;1x3 - +10&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Decided to try out some more.&lt;br&gt;&#xA;L-shape 2x2 - 20 &lt;img src=&quot;https://i.stack.imgur.com/r0pDv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;2x2 - 35&lt;br&gt;&#xA;Total so far: 73&lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-03-19T16:13:11.043" LastActivityDate="2015-03-19T16:13:11.043" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="10597" PostTypeId="2" ParentId="10568" CreationDate="2015-03-19T14:20:00.553" Score="2" Body="&lt;p&gt;I think it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a clue&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A wave of the hand is all it takes to dismiss,&lt;br&gt;&#xA;But when it gets started, it's greater than this!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; clues are easy to dismiss, but great to push you in the right direction&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A bringer of gifts. A destroyer of food.&lt;br&gt;&#xA;It helps win the race. It ruins the mood.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; catching a clue can lead you to the right answer, missing a clue can make you bang your head in the wall. So, either you win or you end up frustrated&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See it, then not... or the other way 'round?&lt;br&gt;&#xA;Answer this wrong and it's still most likely found.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; clues are not always detected right away. if someone posts a wrong answer, usually the author gives more clues.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Puzzlingly meta, you'll answer with glee,&lt;br&gt;&#xA;So tell me, dear riddlers, what the heck can it be?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I suppose the last part is just a clue to the answer -&gt; clue (meta puzzling)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5283" LastEditorUserId="39" LastEditDate="2015-03-19T14:26:18.247" LastActivityDate="2015-03-19T14:26:18.247" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10598" PostTypeId="2" ParentId="10557" CreationDate="2015-03-19T14:41:02.820" Score="4" Body="&lt;p&gt;I: &lt;a href=&quot;http://dictionary.reference.com/browse/isidioid&quot; rel=&quot;nofollow&quot;&gt;isidioid&lt;/a&gt; or &lt;a href=&quot;http://www.thefreedictionary.com/Kivikivi&quot; rel=&quot;nofollow&quot;&gt;kivikivi&lt;/a&gt; (4/8).&lt;/p&gt;&#xA;" OwnerUserId="9523" LastActivityDate="2015-03-19T14:41:02.820" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10599" PostTypeId="2" ParentId="10588" CreationDate="2015-03-19T14:51:19.530" Score="2" Body="&lt;p&gt;You called it a 3x3 dot grid, and then a 3x3 unit grid.  It's unclear if there are four dots marking the points before and after three spaces, or if there are three dots marking the points before and after two spaces.  As the former is the more complex possibility, I'm answering based on that assumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Breaking them down by shape:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Squares and rectangles (height x width):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1x1: 9&#xA;1x2: 6&#xA;1x3: 3&#xA;2x1: 6&#xA;2x2: 4&#xA;2x3: 2&#xA;3x1: 3&#xA;3x2: 2&#xA;3x3: 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;L's (each leg only 1 square wide, height x width):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2x2: 4 in each orientation = 16 &#xA;2x3: 2 in each orientation = 8&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For the rest, I'm going to represent them like tic-tac-toe grids, where X means the polygon encompasses that section of the grid, and an O means it does not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are four &quot;fat L's&quot;, one in each orientation:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;XXO OXX XXX XXX&#xA;XXX XXX XXX XXX&#xA;XXX XXX OXX XXO&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Plus eight more &quot;partially fat L's&quot;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;XOO OOX XXX XXX&#xA;XXX XXX XXX XXX&#xA;XXX XXX OOX XOO&#xA;&#xA;XXO OXX XXX XXX&#xA;XXO OXX OXX XXO&#xA;XXX XXX OXX XXO&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;One +:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;OXO&#xA;XXX&#xA;OXO&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For the rest, I'll omit the four orientations of each possibility, and just multiply what I find by four:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;16 C shapes:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;XXX XXX XXO OXX&#xA;XXO XOO XOO OXO&#xA;XXX XXX XXO OXX&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;12 T shapes&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;XXX XXX OOO&#xA;OXO OXO XXX&#xA;OXO OOO OXO&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;4 right triangles (okay, they're jaggy triangles):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;XXX&#xA;XXO&#xA;XOO&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And these 24 oddball shapes:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;XXX XXX OXX XXO OXX XXO&#xA;OXO OXO OXO OXO XXO OXX&#xA;OXX XXO XXO OXX OOO OOO&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If I've added up my individual counts correctly, that makes 139 possibilities.  I might have missed some, but I don't think so.&lt;/p&gt;&#xA;" OwnerUserId="10341" LastActivityDate="2015-03-19T14:51:19.530" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10600" PostTypeId="1" AcceptedAnswerId="10605" CreationDate="2015-03-19T16:08:01.497" Score="12" ViewCount="1932" Body="&lt;p&gt;&lt;strong&gt;The question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A &lt;em&gt;rectangular fence&lt;/em&gt; is a set of 4 cells on a solved sudoku grid with&#xA;  co-ordinates $(i,j)$, $(m,j)$, $(m,n)$ and $(i,n)$. You can visualize&#xA;  this as the 4 corners of a rectangle.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A &lt;em&gt;symmetric rectangular fence&lt;/em&gt; is a rectangular fence where opposite&#xA;  corners hold the same value. Meaning $grid(i,j)=grid(m,n)$ and&#xA;  $grid(m,j)=grid(i,n)$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Make a 9x9 completely-filled sudoku that has a minimum number of symmetric rectangular fences.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why I asked?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been trying to prove that there exists no 16-hint solvable sudoku. I realized that if there exists a symmetric rectangular fence (as I have termed it), then it becomes necessary to give at least one of the 4 cells as a hint. So, minimizing the number of such occurences would make it possible to create lesser-hint sudoku's and maybe throw some light on why 17 hints are still required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For puzzlers, the question in section 1 is complete by itself; you need not bother about the rest. You may use a program to find the answer, but there are many, many possibilities, so you will still have to use some logical approach to it.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="294" LastEditDate="2015-03-19T20:52:55.777" LastActivityDate="2015-11-13T17:02:09.673" Title="Creating a solved 9x9 Sudoku based on certain rules" Tags="|mathematics|logical-deduction|sudoku|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10601" PostTypeId="2" ParentId="10471" CreationDate="2015-03-19T16:34:48.007" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/Vu92b.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br/&gt;&#xA; Playing the first move in any of the blue spots will let you gain 8 squares with further optimal play, whatever the computer does. On the other hand, playing in any of the red spots will only let you win 6 squares with further optimal play.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since there were exactly 32 empty spaces, I tried to push the ability of my computer and enumerate the best possible move starting from all $2^{32}$ states. For each state, the code finds the maximum number of squares you can get more than your opponent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a few minutes, the code found that the best result is -9. That is, whatever you do, you cannot win the game. Here is the code:&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;const long long lim=1ll&amp;lt;&amp;lt;32;&#xA;char best[lim];&#xA;//best[i] is the maximum number of extra squares you can win more than your opponent starting from state i&#xA;//A state is represented by a 32-bit bitmask. Edge i (defined below) has been already played iff bit i is 1&#xA;&#xA;int main()&#xA;{&#xA;  //Number the empty positions from 0 to 31 from top to bottom, breaking ties from left to right&#xA;  //Every edge borders at most 2 boxes, and because of the state of the board, there exist at most 2 other edges whose existence is necessary to fill those boxes&#xA;  //Essentially, Adding edge i results in a box for you if match[i][0] and match[i][1] were already filled; and another box for you if match[i][2] and match[i][3] were already filled&#xA;  //if match[][] is 32 it means that current edge is at the boundary of the box and there is only one neighbouring box&#xA;  int match[32][4]={&#xA;    { 5, 0,32,32},{ 6, 1,32,32},{ 7, 2,32,32},{ 4, 3,32,32},&#xA;    { 3, 4, 8, 4},{ 0, 5,10, 5},{ 1, 6,11, 6},{ 2, 7,12, 7},&#xA;    { 4, 8, 9, 8},{ 8, 9,13, 9},{ 5,10,14,17},{ 6,11,15,11},&#xA;    { 7,12,15,18},{ 9,13,16,20},{10,17,32,32},{11,15,12,18},&#xA;    {19,16,13,20},{10,14,21,22},{23,18,12,15},{16,19,23,19},&#xA;    {25,20,13,16},{17,22,32,32},{24,22,17,21},{18,23,19,23},&#xA;    {22,24,26,30},{20,25,28,25},{29,26,24,30},{28,27,31,27},&#xA;    {25,28,27,28},{26,29,32,32},{24,26,32,32},{27,31,32,32}&#xA;  };&#xA;&#xA;  for(long long i=lim-2;i&amp;gt;=0;i--)&#xA;  {&#xA;    best[i]=-25;&#xA;    for(int j=0;j&amp;lt;32;j++)&#xA;    {&#xA;      //If jth edge has already been played, you cannot play it at this move&#xA;      long long cur=(1ll)&amp;lt;&amp;lt;j;&#xA;      if((i&amp;amp;cur)!=0)continue;&#xA;&#xA;      //Check how many boxes are filled by the current move&#xA;      long long mask=i^cur;&#xA;      int cnt=0;&#xA;      if((mask&amp;amp;(1ll&amp;lt;&amp;lt;match[j][0]))!=0 and (mask&amp;amp;(1ll&amp;lt;&amp;lt;match[j][1]))!=0)cnt++;&#xA;      if((mask&amp;amp;(1ll&amp;lt;&amp;lt;match[j][2]))!=0 and (mask&amp;amp;(1ll&amp;lt;&amp;lt;match[j][3]))!=0)cnt++;&#xA;&#xA;      //If no boxes were filled in current move, you pass to opponent and they will do their best move&#xA;      //If at least one box was filled, you have to make another move&#xA;      char thismove;&#xA;      if(cnt==0)thismove=-best[mask];&#xA;      else thismove=cnt+best[mask];&#xA;&#xA;      //Does playing this edge do better than any previously seen edges?&#xA;      if(thismove&amp;gt;best[i])best[i]=thismove;&#xA;    }&#xA;  }&#xA;&#xA;  //Now let us consider the moves from the empty board&#xA;  //Playing edge j, the best you can do is -best[1ll&amp;lt;&amp;lt;j] since no box will be filled directly and the opponent will do their best&#xA;  for(int j=0;j&amp;lt;32;j++)&#xA;  {&#xA;    long long mask=(1ll)&amp;lt;&amp;lt;j;&#xA;    char thismove=-best[mask];&#xA;    printf(&quot;%d %d\n&quot;,j,(int)thismove);&#xA;  }&#xA;  printf(&quot;Best : %d\n&quot;,(int)best[0]);&#xA;  return 0;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="9471" LastEditorUserId="9471" LastEditDate="2015-03-19T16:55:35.517" LastActivityDate="2015-03-19T16:55:35.517" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10602" PostTypeId="1" AcceptedAnswerId="10746" CreationDate="2015-03-19T16:43:16.670" Score="2" ViewCount="489" Body="&lt;p&gt;A relationship so justified&lt;br&gt;&#xA;Sensitive in nature deemed classified&lt;br&gt;&#xA;Stained the world in all aspects&lt;br&gt;&#xA;Likely to start future prospects  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vandalism left still enduring&lt;br&gt;&#xA;Physical marks keep alluring&lt;br&gt;&#xA;A timeless art made from bone&lt;br&gt;&#xA;In Rome a cephalopod you might own  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Arch the enchanters wand&lt;br&gt;&#xA;One which Edward would be fond&lt;br&gt;&#xA;The blood runs with absolute&lt;br&gt;&#xA;No matter if you are mute&lt;br&gt;&#xA;&lt;hr&gt;&#xA;&lt;strong&gt;What is this riddle referring to?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5135" LastEditorUserId="5373" LastEditDate="2015-03-22T01:33:21.310" LastActivityDate="2015-03-23T13:30:15.733" Title="Creationism at its best" Tags="|riddle|rhyme|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10603" PostTypeId="1" AcceptedAnswerId="10604" CreationDate="2015-03-19T17:10:59.143" Score="1" ViewCount="429" Body="&lt;p&gt;This is an easy number puzzle:    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) 1&lt;br&gt;&#xA;2) 11&lt;br&gt;&#xA;3) 21&lt;br&gt;&#xA;4) 1211&lt;br&gt;&#xA;5) 111221&lt;br&gt;&#xA;6) 312211&lt;br&gt;&#xA;7) 13112221&lt;br&gt;&#xA;8) ???&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the number on row 8?&lt;/p&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2015-03-19T20:49:22.300" Title="What is the next number?" Tags="|pattern|" AnswerCount="1" CommentCount="2" ClosedDate="2015-03-19T17:19:01.817" ContentLicense="CC BY-SA 3.0" />
  <row Id="10604" PostTypeId="2" ParentId="10603" CreationDate="2015-03-19T17:15:08.587" Score="1" Body="&lt;h3&gt;It is:&lt;/h3&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1113213211&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h3&gt;Explanation:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Each number, from the second line onwards, represents the frequencies of the digits in the previous line in the form $Af_ABf_BCf_C...$, where $A$, $B$ and $C$ are the different digits in order of appearence, and $f_{A,B,C}$ represents their frequency in every continuous set of equal digits.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1&#xA;11      one one&#xA;21      two ones&#xA;1211    one two and one one&#xA;111221  one one, one two, and two ones&#xA;...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="10248" LastEditorUserId="10248" LastEditDate="2015-03-19T20:49:22.300" LastActivityDate="2015-03-19T20:49:22.300" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10605" PostTypeId="2" ParentId="10600" CreationDate="2015-03-19T17:39:37.430" Score="8" Body="&lt;p&gt;Here's one I just created with zero symmetrical rectangular fences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vcjby.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastActivityDate="2015-03-19T17:39:37.430" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10606" PostTypeId="2" ParentId="10600" CreationDate="2015-03-19T17:48:37.317" Score="10" Body="&lt;p&gt;I just did the easiest construction that I could&amp;mdash;I put the numbers 1-9 in order on each row, and then shifted the rows to avoid block or column duplicates. This appears to produce a grid with no &quot;symmetric rectangular fences,&quot; but I don't know the minimum number of hints required to solve it.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;123 456 789&#xA;456 789 123&#xA;789 123 456&#xA;&#xA;234 567 891&#xA;567 891 234&#xA;891 234 567&#xA;&#xA;345 678 912&#xA;678 912 345&#xA;912 345 678&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="8776" LastActivityDate="2015-03-19T17:48:37.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10607" PostTypeId="1" AcceptedAnswerId="10613" CreationDate="2015-03-19T17:57:43.637" Score="-2" ViewCount="338" Body="&lt;p&gt;Winners may come, but losers I make.&lt;br&gt;&#xA;I hope they do well, at least for their sake.&lt;br&gt;&#xA;The dream that they have, the goal is in sight.&lt;br&gt;&#xA;They battle all day, and into the night.&lt;br&gt;&#xA;The world watches on, in terror and glee,&lt;br&gt;&#xA;For they know not who the victor will be.&lt;br&gt;&#xA;The time is just right, the weather is nice.&lt;br&gt;&#xA;But they do not care, for they are not mice.&lt;br&gt;&#xA;From many to one, all others will die.&lt;br&gt;&#xA;Let mayhem ensue, now what am I?&lt;/p&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="9951" LastEditDate="2015-03-19T18:34:11.030" LastActivityDate="2015-03-19T20:22:46.553" Title="The Road Begins - What am I?" Tags="|riddle|word|rhyme|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10608" PostTypeId="2" ParentId="10557" CreationDate="2015-03-19T18:39:54.957" Score="2" Body="&lt;p&gt;Does secessionists (5/13) qualify?&lt;/p&gt;&#xA;&lt;h3&gt;Edit&lt;/h3&gt;&#xA;&lt;p&gt;I know this won't qualify to win, but I found one with five s with a better ratio&lt;/p&gt;&#xA;&lt;p&gt;assessments (5/11)&lt;/p&gt;&#xA;" OwnerUserId="10400" LastEditorUserId="69582" LastEditDate="2021-03-26T14:16:39.837" LastActivityDate="2021-03-26T14:16:39.837" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="10609" PostTypeId="2" ParentId="10561" CreationDate="2015-03-19T18:40:34.670" Score="0" Body="&lt;p&gt;My guess is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A brain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am intangible&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your brain is inside your head, and therefore not tangible (by you, at least).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once given, I can never be regained&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can have a brain transplant, but you can never have your own brain again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am always given as a pair&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The brain is divided into two lobes - thus, a pair.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can give a gift of my own&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Knowledge is the greatest gift of all.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Everyone has me exactly once&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Metaphysically, we only ever have one mind.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2756" LastActivityDate="2015-03-19T18:40:34.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10610" PostTypeId="2" ParentId="10607" CreationDate="2015-03-19T18:44:04.550" Score="3" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;War&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Winners may come, but losers I make.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;People fight war to be winners, but all sides lose in war.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I hope they do well, at least for their sake.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If they don't do well, the loser is killed or repressed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The dream that they have, the goal is in sight.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A victory in war is easy to see, but, like a dream, hard to get to.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They battle all day, and into the night.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;War lasts throughout the day and night.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The world watches on, in terror and glee,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Especially in the 20th century, war coverage is regularly broadcast on news media, leading to both terror at the possibility of defeat, and glee at the possibility of victory.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For they know not who the victor will be.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;No one knows who is going to win a war.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The time is just right, the weather is nice.&#xA;But they do not care, for they are not mice.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Mice (cowardly people) do not go to war.  And when at war, day-to-day things like the weather don't matter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From many to one, all others will die.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;From many belligerents reduced to one victor.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let mayhem ensue, now what am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;War is full of mayhem.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4134" LastActivityDate="2015-03-19T18:44:04.550" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10611" PostTypeId="2" ParentId="10607" CreationDate="2015-03-19T18:48:16.807" Score="2" Body="&lt;p&gt;The obvious answer seems to be any form of single-elimination tournament in which the penalty for losing is death, such as the infamous &lt;a href=&quot;http://en.wikipedia.org/wiki/Battle_Royale&quot; rel=&quot;nofollow&quot;&gt;Battle Royale&lt;/a&gt;, or perhaps the &lt;a href=&quot;http://en.wikipedia.org/wiki/The_Hunger_Games&quot; rel=&quot;nofollow&quot;&gt;Hunger Games&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latter is a particularly good match, and here's why:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Winners may come, but losers I make.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Anyone who &quot;wins&quot; the selection process is forced to take part in the Hunger Games. Out of those &quot;winners&quot;, every person but one will lose.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I hope they do well, at least for their sake.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Anyone who does not do well does not survive.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The dream that they have, the goal is in sight.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The dream each player has is to continue surviving, and if they succeed, they are then lavished with wealth and fame.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;They battle all day, and into the night.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Hunger Games is a battle with no time limits. Players must keep constantly aware.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The world watches on, in terror and glee,&lt;br&gt;&#xA;  For they know not who the victor will be.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Hunger Games are a televised event, like a gladiatorial match.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The time is just right, the weather is nice,&lt;br&gt;&#xA;  But they do not care, for they are not mice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the movie is anything to go by, the games take place during spring or summer, when the weather will not be inclement.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;From many to one, all others will die.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Only one person can survive the Games.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let mayhem ensue, now what am I?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However I fear the answer may not be that obvious...&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-03-19T18:48:16.807" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10612" PostTypeId="1" AcceptedAnswerId="10623" CreationDate="2015-03-19T18:59:37.557" Score="14" ViewCount="2196" Body="&lt;p&gt;These letters are carved in groups on a wall as shown. You know an answer is needed to continue your journey, but no other instructions are apparent.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;to   na   ts   ta   yw   te   og   er   sv   ii   gi   rr&#xA;ht   gr   ow   iv   hh   oo   uo   rt   ie   aa   iv   ue&#xA;ew   sd   wh   oe   ia   fp   no   gh   tr   pm   ve   ne&#xA;sa   ew   ee   ns   st   il   dd   rs   aw   pw   eg   da&#xA;&#xA;&#xA;ch   ou   ol   at   ot   ch   ta   eh   ga   ot   mo   re&#xA;re   ft   ol   yo   uo   oe   iv   aa   im   ue   an   sa&#xA;ec   hn   rm   st   no   nn   oe   rt   vh   nr   nl   tc&#xA;ar   ee   le   pe   dd   dp   ns   ai   ie   db   py   st&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What would your answer be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Another puzzle/riddle I used in a game. Seeing as how fast &lt;a href=&quot;https://puzzling.stackexchange.com/questions/8352/riddle-he-who-holds-the-answer-may-enter&quot;&gt;my previous riddle&lt;/a&gt; was solved, I'm not giving any more hints right away :) )&lt;/p&gt;&#xA;" OwnerUserId="9208" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-03-19T22:46:48.210" Title="Puzzle and riddle in one - what am I?" Tags="|riddle|english|enigmatic-puzzle|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10613" PostTypeId="2" ParentId="10607" CreationDate="2015-03-19T19:33:34.673" Score="0" Body="&lt;p&gt;Answer: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You are the NCAA tournament, of course.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Winners may come, but losers I make.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In each game, one team wins and the other loses.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I hope they do well, at least for their sake.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Obviously, each team wants to win.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The dream that they have, the goal is in sight.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The dream for each team is to win the tournament.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They battle all day, and into the night.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Games take place during the day and at night.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The world watches on, in terror and glee,&lt;br&gt;&#xA;For they know not who the victor will be.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The tournament is exciting. Everyone has someone they want to see win and others they want to see lose.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The time is just right, the weather is nice,&lt;br&gt;&#xA;But they do not care, for they are not mice.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The weather is nice because it's spring. The people watching are indoors in front of a TV or in an arena, so the weather isn't important.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From many to one, all others will die.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All but one team will lose in the tournament.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let mayhem ensue, now what am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The tournament is also called March Madness. Mayhem is a synonym for Madness.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10406" LastEditorUserId="9951" LastEditDate="2015-03-19T20:22:46.553" LastActivityDate="2015-03-19T20:22:46.553" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10614" PostTypeId="2" ParentId="10471" CreationDate="2015-03-19T19:48:38.743" Score="5" Body="&lt;p&gt;This is my first experience with Dots and Boxes, so please be kind if I am making some terrible mistakes.  Per the wikipedia link: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An expert player tries to force their opponent to be the one to open the first long chain, because the player who first opens a long chain usually loses.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is what we will try to do.  Initial analysis shows four &quot;safe&quot; (will not enable the opponent to complete a box) moves, which Werner has conveniently &lt;a href=&quot;https://puzzling.stackexchange.com/a/10473/10334&quot;&gt;Numbered 1-4&lt;/a&gt;.  It is important to note that 1 and 3 are exclusive with 2 (ie if 1 gets played, 2 is no longer safe; whereas if 2 gets played, 1 and 3 are no longer safe).  So the immediate assumption is to play either 1 or 3 as that removes 2, leaving two other safe moves (4 and 3 or 1) on the opponent's turn.  We would get the last safe move, and the computer must open the long chains for us.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;But there is one other move on the board which, while not safe, will not open a long chain.  The horizontal line on the second row in the middle column (the line between the two red squares in the &lt;strike&gt;second&lt;/strike&gt; third image of JonTheMon's &lt;a href=&quot;https://puzzling.stackexchange.com/a/10513/10334&quot;&gt;answer&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With this in mind, we must play 2.  The computer will play 4, and we will follow with the line I mentioned in the previous paragraph.  The computer will take its two squares, then be forced to open the first long chain leading to our win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(edit: whoops forgot about one of the diagrams)&lt;/p&gt;&#xA;" OwnerUserId="10334" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2015-03-19T21:41:09.207" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="10615" PostTypeId="1" AcceptedAnswerId="10663" CreationDate="2015-03-19T20:51:03.383" Score="9" ViewCount="4089" Body="&lt;p&gt;I have no legs, but I can move with ease&lt;br&gt;&#xA;I am trapped, yet I can go where your eyes can see&lt;br&gt;&#xA;I am invisible, until you find me&lt;br&gt;&#xA;I am a thing, that you will wish to set free&lt;/p&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="5188" LastEditDate="2015-03-24T18:52:04.223" LastActivityDate="2015-03-24T18:52:04.223" Title="Trapped and invisible; what am I?" Tags="|riddle|" AnswerCount="17" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="10616" PostTypeId="2" ParentId="10615" CreationDate="2015-03-19T21:06:59.767" Score="2" Body="&lt;p&gt;Interesting.. First thing that comes in my mind is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the mind&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no legs, but I can move in ease&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; focusing on different tasks, mind wandering..&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am trapped, yet I can go where your eyes can see&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it is trapped in your head, yet it can be everywhere you imagine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am invisible, until you have found me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; because mind is invisible??&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a thing, that you will wish to set me free&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; set free your mind meaning escape the problems and daily worries&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5283" LastActivityDate="2015-03-19T21:06:59.767" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10617" PostTypeId="2" ParentId="10615" CreationDate="2015-03-19T21:09:48.407" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your gaze&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fits everything except possibly the last line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I have no legs, but I can move in ease&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your gaze can roam around anywhere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am trapped, yet I can go where your eyes can see&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's trapped to be just at the place where you're looking.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am invisible, until you have found me&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can't see that someone else is looking at you until you meet their gaze.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am a thing, that you will wish to set me free&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stop staring at me!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-20T21:42:28.240" LastActivityDate="2015-03-20T21:42:28.240" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10618" PostTypeId="2" ParentId="10612" CreationDate="2015-03-19T21:25:32.030" Score="15" Body="&lt;p&gt;I think I've figured out the riddle, but unfortunately I suck at riddles :-(&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reading from top to bottom rather than left to right (i.e. starting with 'thes'), it's clear that many of the four-letter fragments contain common English ngrams ('tion', 'ound', etc). Trying to match some up, I noticed that TOFI is the only thing that can go before RSTS, while its neighbor EOPL (presumably part of PEOPLE) could only go before EARA or EACT (the neighbor of RSTS). I assumed it was the latter and tried to match up &lt;em&gt;pairs&lt;/em&gt; of four letters, gradually creating the following text: &lt;br&gt;&lt;br&gt;&#xA;&#xA; &lt;code&gt;iapp eara towe rund ergr ound&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;iamw hati swhe reea rths tood&lt;/code&gt;&lt;br&gt;&#xA; &lt;br&gt;&#xA; &lt;code&gt;givi ngse cond tofi rsts ound&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;amhe ardw henp eopl eact good&lt;/code&gt;&lt;br&gt;&#xA; &lt;br&gt;&#xA; &lt;code&gt;give thes ound ofhe sita tion&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;iveg otwa terb utne verw aves&lt;/code&gt;&lt;br&gt;&#xA; &lt;br&gt;&#xA; &lt;code&gt;manp aysp oorl yhis crea tion&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;only tote llme what hecr aves&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; which, reformatted, gives the following riddle:&lt;br&gt;&lt;br&gt;&#xA; &lt;em&gt;I appear a tower underground&lt;/em&gt;&lt;br&gt;&#xA; &lt;em&gt;I am what is where earth stood&lt;/em&gt;&lt;br&gt;&#xA; &lt;em&gt;Giving second to first sound&lt;/em&gt;&lt;br&gt;&#xA; &lt;em&gt;Am heard when people act good&lt;/em&gt;&lt;br&gt;&#xA; &lt;em&gt;Give the sound of hesitation&lt;/em&gt;&lt;br&gt;&#xA; &lt;em&gt;I've got water but never waves&lt;/em&gt;&lt;br&gt;&#xA; &lt;em&gt;Man pays poorly his creation&lt;/em&gt;&lt;br&gt;&#xA; &lt;em&gt;Only to tell me what he craves&lt;/em&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unless I suffer an atypical stroke of inspiration, I'll leave the actual answer to someone else...&lt;/p&gt;&#xA;" OwnerUserId="10220" LastEditorUserId="10220" LastEditDate="2015-03-19T22:46:48.210" LastActivityDate="2015-03-19T22:46:48.210" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10619" PostTypeId="1" CreationDate="2015-03-19T21:38:30.077" Score="5" ViewCount="913" Body="&lt;p&gt;I've been attempting to solve this ARG named SAN for the past 2 days or so, and I've been stuck on this image for about a day. Are there any algorithms or methods to figure it out? By the way, the language this is in Serbian, so if you find Serbian text, it's most likely the solution. Here's the image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/khSNW.jpg&quot; alt=&quot;it appears to be just a series of dots, however we&amp;#39;ve identified that it is octal, and has some pattern of sorts.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears to be just a series of dots, however we've identified that it is octal, and has some pattern of sorts.&lt;/p&gt;&#xA;" OwnerUserId="10412" LastEditorUserId="4050" LastEditDate="2015-03-21T07:42:54.177" LastActivityDate="2015-03-21T07:42:54.177" Title="There's an ARG I've been working on, and I'm stuck on this puzzle" Tags="|cipher|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10620" PostTypeId="2" ParentId="10443" CreationDate="2015-03-19T21:41:22.737" Score="2" Body="&lt;p&gt;The answer to the question (as I misunderstood it) is yes. See below for the explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I looked online and found the solutions for smaller puzzles, and used the knowledge I gained to try to solve the 8x8. I knew there was only 1 possible solution because of &lt;a href=&quot;http://mathworld.wolfram.com/LightsOutPuzzle.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; link. After about 90 minutes I stumbled upon it. I used &lt;a href=&quot;http://mathafou.free.fr/jeux_en/jeu305.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; site to test it. Toggle the lights with the 1's and do not flip the others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/U7NVI.png&quot; alt=&quot;Two to Seven&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OxMJV.png&quot; alt=&quot;Eight&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-03-20T14:34:43.260" LastActivityDate="2015-03-20T14:34:43.260" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10621" PostTypeId="2" ParentId="10561" CreationDate="2015-03-19T21:59:22.433" Score="5" Body="&lt;p&gt;How about:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Virginity:&lt;br&gt;&#xA; Not a real, physical thing.&#xA; You can't un-boink.&#xA; Can result in children.&#xA; You have it from birth until you lose it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; and of course, it takes two (or more) to tango ;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10416" LastActivityDate="2015-03-19T21:59:22.433" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10622" PostTypeId="2" ParentId="10612" CreationDate="2015-03-19T22:28:48.513" Score="2" Body="&lt;p&gt;Now that &lt;a href=&quot;https://puzzling.stackexchange.com/a/10618/5373&quot;&gt;Uri Zarfaty&lt;/a&gt; has solved the cryptograms, I can have a go at the riddle. Here it is again:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I appear a tower underground &lt;br&gt;I am what is where earth stood &lt;br&gt;Giving second to first sound &lt;br&gt;Am heard when people act good &lt;br&gt;Give the sound of hesitation &lt;br&gt;I've got water but never waves &lt;br&gt;Man pays poorly his creation &lt;br&gt;Only to tell me what he craves&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think the answer could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an underground water tank.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is like a water tower but underground (hence line 1); earth has been displaced to make room for it there (hence line 2); it sounds a bit like &quot;thanks&quot; with the second letter removed (hence lines 3 and 4); it contains water (hence line 6), which is necessary for people to live (hence line 8). Line 5 sounds like a reference to the word &quot;er&quot; (Harry Potter's &quot;spider&quot; riddle, anyone?), and I'm not entirely sure about line 7.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2015-03-19T22:28:48.513" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10623" PostTypeId="2" ParentId="10612" CreationDate="2015-03-19T22:40:30.080" Score="11" Body="&lt;p&gt;Stealing the plaintext from &lt;a href=&quot;https://puzzling.stackexchange.com/a/10618/5373&quot;&gt;Uri Zarfaty's answer&lt;/a&gt;, my guess is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A wishing well.&#xA; &lt;br&gt;&#xA; The well is a tower-like hole in the ground, where earth used to be, containing water but not waves. Lines 3-4 refer to the other meaning of the word &quot;well&quot;. People &quot;pay&quot; the well by throwing loose change into it, and verbalizing their wishes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10165" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2015-03-19T22:40:30.080" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10624" PostTypeId="1" AcceptedAnswerId="10632" CreationDate="2015-03-19T22:57:36.153" Score="5" ViewCount="415" Body="&lt;p&gt;This is a follow-up to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/10471/dots-and-boxes-best-move-in-given-position&quot;&gt;my first Dots and Boxes question&lt;/a&gt;. It is just a little bit harder than the first one. Each one I post will be a bit harder than the previous one, and they will all be from games I have recently played.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the following game of Dots and Boxes, it is your move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZyakK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your opponent is a computer who will play perfectly. What is the best move, and why?&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-03-20T17:59:09.487" Title="Dots and Boxes Part 2- Best Move in Given Position" Tags="|strategy|pencil-and-paper-games|dots-and-boxes|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10625" PostTypeId="2" ParentId="10568" CreationDate="2015-03-19T23:07:57.063" Score="1" Body="&lt;p&gt;After reading the other possible answers, I realized that the answer could be many things... ;)&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;The answer is:&lt;/h3&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; babies, plants, animals...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;Explaination line by line:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;A wave of the hand is all it takes to dismiss.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can easily dismiss/destroy any [answer] if you wave your hand hard enough.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But when it gets started, it's greater than this!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes, once [answer] is started, it's more than it was before, and surely greater than [this].&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A bringer of gifts. A destroyer of food.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gifts can be joy or food or knowledge. Each [answer] destroys/eats literal and figurative food too.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It helps win the race. It ruins the mood.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Without [answer], the human race wouldn't last as long. [answer] ruins the mood by crying, growing weeds, and being an annoying animal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See it, then not... or the other way 'round?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I will turn my head to not see [answer]. Or, maybe I'll turn my head to see [answer]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer this wrong and it's still most likely found.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If this answer could ever possibly be wrong, surely other ones found will bring less joy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Puzzlingly meta, you'll answer with glee.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I sure had fun answering this. :D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10419" LastActivityDate="2015-03-19T23:07:57.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10626" PostTypeId="2" ParentId="10557" CreationDate="2015-03-19T23:11:53.477" Score="4" Body="&lt;p&gt;My attempt&lt;/p&gt;&#xA;&lt;p&gt;Q - quinque is 2/7 (five/having five)&lt;/p&gt;&#xA;&lt;p&gt;Sorry I completely forgot these don't qualify due to five letter minimum&lt;/p&gt;&#xA;&lt;p&gt;V - vav is 2/1 (I guess it may not qualify as its a letter of the Hebrew alphabet)&lt;br /&gt;&#xA;V - viva/vive is 2/4 (long live)&lt;/p&gt;&#xA;" OwnerUserId="2357" LastEditorUserId="69582" LastEditDate="2021-03-26T14:12:23.503" LastActivityDate="2021-03-26T14:12:23.503" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="10627" PostTypeId="2" ParentId="10557" CreationDate="2015-03-19T23:31:15.183" Score="3" Body="&lt;p&gt;Just for reference, since I mentioned this in comments but haven't seen it in anyone else's answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Z: &lt;a href=&quot;http://www.collinsdictionary.com/dictionary/english/pzazz&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;pzazz&lt;/strong&gt;&lt;/a&gt; (3/5)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-03-19T23:31:15.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10628" PostTypeId="2" ParentId="10624" CreationDate="2015-03-20T00:29:59.353" Score="0" Body="&lt;p&gt;For a greedy strategy by both you and the opponent, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/i0irj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;will set you up for the first chain (of five). &lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-03-20T00:29:59.353" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10629" PostTypeId="2" ParentId="10615" CreationDate="2015-03-20T01:25:24.197" Score="12" Body="&lt;p&gt;I think you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Light&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no legs, but I can move in ease&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Light has the fastest possible speed. Nothing can beat its top speed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am trapped, yet I can go where your eyes can see&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It can be confined in a enclosed space like a building, but you can still see as long there is a light source. You need light to see.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am invisible, until you have found me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can't really see light. There are phrases like &quot;I have seen the light&quot; and lines like that. The entrance to heaven to many is seen as big entrance door thingy made of light I think.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a thing, that you will wish to set me free&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You will want to see by letting light roam around. Lighting candles and turning on lamps and flashlights could &quot;release the light&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9042" LastActivityDate="2015-03-20T01:25:24.197" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10630" PostTypeId="5" CreationDate="2015-03-20T01:59:06.047" Score="0" Body="&lt;p&gt;Dots and boxes (also known as Boxes, Squares, Paddocks, Pigs in a Pen, Square-it, Dots and Dashes, Dots, Line Game, Smart Dots, Dot Boxing, or, simply, the Dot Game) is a pencil and paper game for two (or more) players, first published in 1889 by Édouard Lucas.&lt;/p&gt;&#xA;&lt;p&gt;Starting with an empty grid of dots, players take turns, adding a single horizontal or vertical line between two unjoined adjacent dots. A player who completes the fourth side of a 1×1 box earns one point and takes another turn. (The points are typically recorded by placing in the box an identifying mark of the player, such as an initial.) The game ends when no more lines can be placed. The winner of the game is the player with the most points.&lt;/p&gt;&#xA;&lt;p&gt;The board may be of any size. When short on time, 2×2 boxes (created by a square of 9 dots) is good for beginners, and 5×5 is good for experts.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="60644" LastEditDate="2020-08-30T06:22:16.637" LastActivityDate="2020-08-30T06:22:16.637" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10631" PostTypeId="4" CreationDate="2015-03-20T01:59:06.047" Score="0" Body="Dots and Boxes is a 2-player pencil-and-paper game in which the players take turns to fill in lines of a grid, claiming each square of which they fill in the 4th side. The aim is to claim as many squares as possible." OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-20T01:59:06.047" LastActivityDate="2015-03-20T01:59:06.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10632" PostTypeId="2" ParentId="10624" CreationDate="2015-03-20T02:08:39.307" Score="3" Body="&lt;p&gt;The player can't win.  Analysis to follow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pyBD9.png&quot; alt=&quot;Interesting moves&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here I've marked the &quot;safe&quot; moves in blue.  We will call them 1, 2, and 3 counter-clockwise starting from the left.  The sacrifices I have marked in green.  Let us call them A above and B below.  Finally I have marked a move inside the lower right square in red.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;First to address the red move.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Any move inside the square is bad:&lt;br&gt;&#xA;Assumption - It is determinable whether the first person to make a move outside the square will win or lose.&lt;br&gt;&#xA;Making a move inside the square gives the opponent the opportunity to decide who makes the first move outside the square.  This is because they can either complete all the boxes in the square then move outside, or continue our line inside the 2x2 square to split it into 2 2x1 squares.  In the latter case we are forced to make the first move outside the square (after completing the four boxes).  Since our computer opponent is perfect, it will know whether it wants to make the first or second move outside, and will choose accordingly.  We cannot give the computer that choice.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Which move to make.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;We need to note that 2 is mutually exclusive with each of 1 and 3.  Also, A and B are mutually exclusive.  Any non-safe, non-sacrifice move will be as bad as the red move.  Finally, A has interactions with our three safe moves.  If A is played: with 3 on the board, 1 and 2 become starts to long chains (bad), and without 3 on the board 1 and 2 become sacrifices.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Unfortunately, all this combines in a way that appears bad for us.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If we make a move in one of the safe lines, the computer can play A leaving us with no good moves like so:&lt;br&gt;&#xA;case 1: 1, A, complete the squares...in doing so we have played 2 and have no more valid moves (2 invalidates 3, and A invalidated B)&lt;br&gt;&#xA;case 2: 2, A, complete the squares then no more valid moves (likewise 2 invalidates 3, and A invalidates B)&lt;br&gt;&#xA;case 3: 3, A, complete the square then no more valid moves (A invalidates B, and 3 + A means 1 and 2 are starts to long chains)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what if we start with a sacrifice?  The computer can follow with a safe that leaves no good moves.&lt;br&gt;&#xA;case A: A, complete the square then 3, no good moves (A invalidates B, and 3 + A means 1 and 2 are starts to long chains)&lt;br&gt;&#xA;case B: B, complete the squares then 2, no good moves (B invalidates A, and 2 invalidates 1 and 3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're left with no way to win this, but the best move is probably 1 or 2 as that forces the computer to sacrifice 2 squares.&lt;/p&gt;&#xA;" OwnerUserId="10334" LastEditorUserId="10334" LastEditDate="2015-03-20T17:59:09.487" LastActivityDate="2015-03-20T17:59:09.487" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10633" PostTypeId="1" AcceptedAnswerId="10720" CreationDate="2015-03-20T02:10:50.397" Score="47" ViewCount="22650" Body="&lt;h1&gt;NOC +10&lt;/h1&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EEWu8.png&quot; alt=&quot;video intro&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;On March 6th, 2015, a YouTube channel named &lt;strong&gt;&amp;quot;Noc +10&amp;quot;&lt;/strong&gt; was created. Since that day, this channel is uploading some really darkly mysterious videos. I will start saying something about the name of this channel, which is pretty odd, indeed: searching on Google for &amp;quot;NOC +10&amp;quot; you can easily find out that it may refer to the &lt;strong&gt;&lt;a href=&quot;http://www.navy.mil/maritime/display.asp?page=noc.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Naval Operation Concept 2010&lt;/a&gt;&lt;/strong&gt;, also represented with the acronym &amp;quot;NOC 10&amp;quot;.&lt;/p&gt;&#xA;&lt;h2&gt;1. First contact - March 7, 2015&lt;/h2&gt;&#xA;&lt;h3&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=NCAUPuXzn4U&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;code&gt;0101 0101 0000 0001&lt;/code&gt;&lt;/a&gt;&lt;/h3&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wKjfz.png&quot; alt=&quot;first video&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;The &lt;strong&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=NCAUPuXzn4U&quot; rel=&quot;nofollow noreferrer&quot;&gt;first video&lt;/a&gt;&lt;/strong&gt; uploaded by Noc +10 on March 7th 2015, is entitled &lt;strong&gt;&lt;code&gt;0101 0101 0000 0001&lt;/code&gt;&lt;/strong&gt;: this immediately lets us think of some sort of binary code. It's divided in blocks of four bits, but we do not know its meaning. This video is composed by an odd old-oscilloscope-like display showing some distorted letters and words. Let's break it down:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It starts displaying the string &lt;code&gt;NOC+10&lt;/code&gt;, with a background piano music, which is easily recognizable: it's the &lt;a href=&quot;https://www.youtube.com/watch?v=GQ1aE_c0efY&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nocturne Op. 27 No. 1 in C# minor&lt;/a&gt; by Fryderyk Chopin.&lt;/li&gt;&#xA;&lt;li&gt;The screen goes black for a bit, the music stops, and a low-tone noise starts: the word &lt;strong&gt;&lt;code&gt;START&lt;/code&gt;&lt;/strong&gt; appears.&lt;/li&gt;&#xA;&lt;li&gt;A series of letters is displayed, spelled by a female computerized voice: &lt;strong&gt;&lt;code&gt;JKnjGvPaITB&lt;/code&gt;&lt;/strong&gt;, preceded by the string &lt;strong&gt;&lt;code&gt;?v=&lt;/code&gt;&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;When the voice finishes spelling the letters, the word &lt;strong&gt;&lt;code&gt;END&lt;/code&gt;&lt;/strong&gt; appears, and the video stops after a few seconds.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Looking at the string &lt;code&gt;?v=JKnjGvPaITB&lt;/code&gt;, we can easily think of an &lt;em&gt;&lt;strong&gt;ID of some YouTube video&lt;/strong&gt;&lt;/em&gt;, but the string itself is encrypted, and it doesn't exist a video with such ID. By the way, applying the famous &lt;a href=&quot;http://en.wikipedia.org/wiki/Caesar_cipher&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Caesar cipher&lt;/strong&gt;&lt;/a&gt; to this ID we can decypher it and find the ID &lt;strong&gt;&lt;code&gt;GHkgDsMxFQY&lt;/code&gt;&lt;/strong&gt;, which is a valid YouTube video ID.&lt;/p&gt;&#xA;&lt;p&gt;A lot of other questions came to my mind watching this video: is &amp;quot;Noc&amp;quot; related to &amp;quot;Nocturne&amp;quot; in some way? What does the song mean? Etc.&lt;/p&gt;&#xA;&lt;h2&gt;2. First hidden video - March 7, 2015&lt;/h2&gt;&#xA;&lt;h3&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=GHkgDsMxFQY&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;code&gt;0101 0101 0000 0010&lt;/code&gt;&lt;/a&gt;&lt;/h3&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aACUC.png&quot; alt=&quot;second video&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;The ID found on the first video leads us to &lt;strong&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=GHkgDsMxFQY&quot; rel=&quot;nofollow noreferrer&quot;&gt;this unlisted video&lt;/a&gt;&lt;/strong&gt;, uploaded earlier in the same day by the same channel, entitled &lt;strong&gt;&lt;code&gt;0101 0101 0000 0010&lt;/code&gt;&lt;/strong&gt;. The first thing we notice is that the name of the video is almost equal to the previous one, but actually the last 4 bits are different: &lt;code&gt;0010&lt;/code&gt;. The video title changed from ending with &lt;code&gt;0001&lt;/code&gt; to ending with &lt;code&gt;0010&lt;/code&gt;: it's clear that the two names of these videos &lt;strong&gt;are consecutive binary numbers&lt;/strong&gt;. Let's see what's inside this second video as well:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The video starts with an high frequency noise, and the string &lt;code&gt;NOC+10&lt;/code&gt; is displayed.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Then a sudden cut of the video brings us to a long string of apparently random uppercase letters, ending with a bunch of zeros:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt; RKKHOHQKDGDVWHDPERDW&#xA; WKHVWHDPERDWKDGDEHOO&#xA; ZKHQKHOHQZHQWWRKHDYHQ&#xA; WKHVWHDPERDWZHQWWRKHOO&#xA;&#xA; 0000000000000000000000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;The same background music is played alongside with the same female voice spelling this sequence of letters.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The video ends a few seconds after the voice has finished spelling all the code.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Applying the &lt;strong&gt;Caesar cypher again&lt;/strong&gt; to the code, we can decipher it:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;RKKHOHQKDGDVWHDPERDW   -&amp;gt; OHHELENHADASTEAMBOAT&#xA;WKHVWHDPERDWKDGDEHOO   -&amp;gt; THESTEAMBOATHADABELL&#xA;ZKHQKHOHQZHQWWRKHDYHQ  -&amp;gt; WHENHELENWENTTOHEAVEN&#xA;WKHVWHDPERDWZHQWWRKHOO -&amp;gt; THESTEAMBOATWENTTOHELL&#xA;&#xA;0000000000000000000000 -&amp;gt; These zeros don't seem to have any meaning&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Splitting the code in the right parts, we can distinguish a sort of doggerel, which says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Oh Helen had a steam boat&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The steam boat had a bell&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;When Helen went to heaven&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The steam boat went to hell&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This rhymes, particularly the first verse, make us think about the &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Miss_Susie&quot; rel=&quot;nofollow noreferrer&quot;&gt;Miss Susie american schoolyard rhyme&lt;/a&gt;&lt;/strong&gt;, maybe a variation of it? Looking on Wikipedia on the Miss Susie page, we can find a linked article: &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/SS_Sultana#The_tragedy&quot; rel=&quot;nofollow noreferrer&quot;&gt;SS Sultana tragedy&lt;/a&gt;&lt;/strong&gt; (1865), the greatest maritime disaster in United States history. Well, let's say it: this looks freaking mysterious and gloomy!&lt;/p&gt;&#xA;&lt;h2&gt;3. Second contact - March 12, 2015&lt;/h2&gt;&#xA;&lt;h3&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=AHtiPxfPSu4&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;code&gt;0101 0101 0000 0011&lt;/code&gt;&lt;/a&gt;&lt;/h3&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/o0j0o.png&quot; alt=&quot;third video&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;A &lt;strong&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=AHtiPxfPSu4&quot; rel=&quot;nofollow noreferrer&quot;&gt;third video&lt;/a&gt;&lt;/strong&gt;, this time public, is uploaded by Noc +10 five days after the last one, and its title is &lt;code&gt;0101 0101 0000 0011&lt;/code&gt;. &lt;strong&gt;If we had a suspect of the names of the videos being &lt;em&gt;consecutive binary numbers&lt;/em&gt;, now it's clear!&lt;/strong&gt; Again, let's analyse this third video like we did with the previous ones:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;It begins with the usual &lt;code&gt;NOC+10&lt;/code&gt; string, with the same Chopin music in the background.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A long string of numbers and letters is displayed on the screen:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt; 6D3248656C6C6F4166776F726C645156&#xA;  68656C656E79324937367761697459&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;This makes us think about some sort of hexadecimal string, because the letters which appear in this string do not overcome the letter F, which is the greatest valid hexadecimal digit.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A male voice makes &lt;strong&gt;16&lt;/strong&gt; equal and unrecognizable sounds while the above string is displayed on the screen, and the video ends: this confirms our thesis about the hexadecimal string.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Applying the Caesar cypher doesn't seem to work this time, but I tried translating it from hexadecimal character pairs to ASCII characters, and the result is even creepier than the previous one:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;6D3248656C6C6F4166776F726C64515668656C656E79324937367761697459&#xA;   ↓&#xA;m2HelloAfworldQVheleny2I76waitY&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;This seems to be some text alternated with two-character strings inserted to split the words. The result, removing the exceeding characters is:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Hello world helen I wait&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;But what about the removed characters? They look familiar: &lt;strong&gt;&lt;code&gt;m2AfQVy276Y&lt;/code&gt;, YES!&lt;/strong&gt; They represent another YouTube video ID!&lt;/p&gt;&#xA;&lt;h2&gt;4. Second hidden video - March 12, 2015&lt;/h2&gt;&#xA;&lt;h3&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=m2AfQVy276Y&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;code&gt;0101 0101 0000 0100&lt;/code&gt;&lt;/a&gt;&lt;/h3&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NciE0.png&quot; alt=&quot;fourth video 1&quot; /&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/0MDtD.png&quot; alt=&quot;fourth video 2&quot; /&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/OsjcM.png&quot; alt=&quot;fourth video 3&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;The &lt;strong&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=m2AfQVy276Y&quot; rel=&quot;nofollow noreferrer&quot;&gt;fourth video&lt;/a&gt;&lt;/strong&gt;, found with the ID extracted from the third, is unlisted, and has been uploaded by Noc +10 earlier in the same day of the previous one. Its title is &lt;code&gt;0101 0101 0000 0100&lt;/code&gt;. These &lt;em&gt;consecutive binary numbers&lt;/em&gt; keep going on. This video also comes with &lt;strong&gt;a lot more of information than the others&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;It begins with the same &lt;code&gt;NOC+10&lt;/code&gt; introduction, the same Chopin music is played.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A video cut shows another frame with some text which doesn't seem to be encrypted this time:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt; SYSTEM 00033 DECEASED&#xA; SYSTEM 01090 DECEASED&#xA; SYSTEM 00010 ALIVE&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Another song begins to play: looks like an old song from the '70s.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Then a longer, more detailed text:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt; PRESSURE WATERTIGHT STABLE&#xA; SECTOR 1 STABLE&#xA; SECTOR 2 STABLE&#xA; SECTOR 3 STABLE&#xA; SECTOR 4 PRINCIPLE OF FLOODING&#xA; SECTOR 5 STABLE&#xA; SECTOR 6 SIGNAL MISSING&#xA; SECTOR 7 STABLE&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Now the phrase &lt;strong&gt;&lt;code&gt;THERE'S ANOTHER MOTHER TO SLEEP HERE&lt;/code&gt;&lt;/strong&gt; appears, and a male voice spells the letters &lt;strong&gt;&lt;code&gt;L-I-O-U-Y&lt;/code&gt;&lt;/strong&gt;, then the video ends.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;This video seems to be the most important so far: it's clearly &lt;strong&gt;full of connections and riddles to solve&lt;/strong&gt;. It also leads to a large amount of other questions:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What are these &lt;code&gt;SYSTEM&lt;/code&gt;s referring to?&lt;/li&gt;&#xA;&lt;li&gt;Are these &lt;code&gt;SECTOR&lt;/code&gt;s part of a ship? Maybe the SS Sultana?&lt;/li&gt;&#xA;&lt;li&gt;What does &lt;code&gt;LIOUY&lt;/code&gt; mean?&lt;/li&gt;&#xA;&lt;li&gt;Maybe &amp;quot;another sleeping mother&amp;quot; means another deceased system?&lt;/li&gt;&#xA;&lt;li&gt;Do these two screens have anything to do with the SS Sultana tragedy?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Let's see if we can answer some of them:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The string &lt;strong&gt;&lt;code&gt;LIOUY&lt;/code&gt;&lt;/strong&gt; was used as an old cablegram &lt;a href=&quot;http://en.wikipedia.org/wiki/Commercial_code_%28communications%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Commercial Code&lt;/a&gt;, meaning &lt;em&gt;&lt;strong&gt;&amp;quot;Why do you not answer my question?&amp;quot;&lt;/strong&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A partial anagram of the phrase &lt;code&gt;THERE'S ANOTHER MOTHER TO SLEEP HERE&lt;/code&gt; gives us:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt; THERMOSTAT ERROR NO SLEEP&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It also &lt;strong&gt;looks like the famous &lt;em&gt;SS Sultana tragedy&lt;/em&gt; is connected to these last videos.&lt;/strong&gt; Quoting from the Wikipedia page (linked above):&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;[...] Near 2:00 in the morning on April 27, 1865, when the Sultana was just seven miles north of Memphis, her boilers suddenly exploded. First one boiler exploded, followed a split second later by two more. The cause of the explosion was &lt;strong&gt;too much pressure and low water in the boilers&lt;/strong&gt;. There was reason to believe allowable working steam pressure was exceeded in an attempt to overcome the spring river current. The enormous explosion flung some of the passengers on deck into the water, and destroyed a large section of the boat. [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;As this story goes on, I'm more and more inclined to think that these &lt;code&gt;SYSYEM&lt;/code&gt;s and &lt;code&gt;SECTOR&lt;/code&gt;s are referring to the SS Sultana.&lt;/p&gt;&#xA;&lt;h2&gt;5. Third contact - March 19, 2015&lt;/h2&gt;&#xA;&lt;h3&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=-qrtM4LdHH0&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;code&gt;0101 0101 0000 0101&lt;/code&gt;&lt;/a&gt;&lt;/h3&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QmSsc.png&quot; alt=&quot;fifth video&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;A &lt;strong&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=-qrtM4LdHH0&quot; rel=&quot;nofollow noreferrer&quot;&gt;fifth video&lt;/a&gt;&lt;/strong&gt; is uploaded by Noc +10 seven days after the previous one. This video, entitled &lt;code&gt;0101 0101 0000 0101&lt;/code&gt;, is the third public video, and it has been uploaded on March 19th.&lt;/p&gt;&#xA;&lt;p&gt;Again:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Same song and &lt;code&gt;NOC+10&lt;/code&gt; string.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A female voice emits four equal and incomprehensible sounds, and a code is displayed on the sceen:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt; A oYzXqoELKjE 14&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;The female voice, if played backwards, seems to be pronouncing &lt;code&gt;P.I.&lt;/code&gt;, you can listen to it &lt;a href=&quot;http://vocaroo.com/i/s0SX3bAtDd6s&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;sup&gt;*link may expire&lt;/sup&gt;. May these two letters refer to &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt; (irrational number Pi)?&lt;/p&gt;&#xA;&lt;p&gt;The code provided by this video seems to be containing another YouTube video ID, which is &lt;strong&gt;&lt;code&gt;oYzXqoELKjE&lt;/code&gt;&lt;/strong&gt;, but there's no video with such ID. Also applying the Caesar cypher to it doesn't provide any valid video ID. Maybe there's some hidden information in the other three characters &lt;code&gt;A 14&lt;/code&gt;, some information about how to decypher the ID?&lt;/p&gt;&#xA;&lt;h2&gt;Common points&lt;/h2&gt;&#xA;&lt;p&gt;All the videos begin with the &lt;strong&gt;same introduction&lt;/strong&gt;. All the &lt;strong&gt;video titles are represented by an increasing binary number&lt;/strong&gt; divided in 4 blocks of 4 bits. If we translate the numbers into decimal notation we obtain the numbers &lt;code&gt;5501, 5502, 5503, etc&lt;/code&gt;. It looks like the sequence of the video public settings is: &lt;strong&gt;&lt;code&gt;public, unlisted, public, unlisted, etc&lt;/code&gt;&lt;/strong&gt;. All the unlisted videos were found decoding data provided by the previous public videos, this makes us believe &lt;em&gt;&lt;strong&gt;there's a sixth hidden video which has got to be found analysing the last video's data&lt;/strong&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h3&gt;Can you crack the last video's code and find the ID of the next one?&lt;/h3&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;Summary&lt;/h1&gt;&#xA;&lt;p&gt;This is a short summary which synthesizes the developments since the first video was found:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;#1 0101 0101 0000 0001 - 2015/07/03 - public&#xA; |&#xA; |- Video URL: youtube.com/watch?v=NCAUPuXzn4U&#xA; |- Encrypted data: &amp;quot;?v=JKnjGvPaITB&amp;quot;&#xA; |- Decrypted (ROT23): &amp;quot;?v=GHkgDsMxFQY&amp;quot;&#xA; |&#xA; |- The decrypted video ID brings us to video #2&#xA;  &#xA;#2 0101 0101 0000 0010 - 2015/07/03 - unlisted&#xA; |&#xA; |- Video URL: youtube.com/watch?v=GHkgDsMxFQY &#xA; |- Encrypted data: [long, see above]&#xA; |- Decrypted (ROT23): &amp;quot;Helen had a steamboat...&amp;quot; rhymes&#xA; |&#xA; |- Unknown meaning of the decrypted data&#xA; &#xA;#3 0101 0101 0000 0011 - 2015/12/03 - public&#xA; |&#xA; |- Video URL: youtube.com/watch?v=AHtiPxfPSu4&#xA; |- Encrypted data: [long, see above]&#xA; |- Decrypted data (HEX to ASCII): &#xA; |  |- The phrase &amp;quot;Hello world helen I wait&amp;quot;&#xA; |  |- Another video ID: &amp;quot;m2AfQVy276Y&amp;quot;&#xA; |&#xA; |- Unknown meaning of the phrase&#xA; |- The decrypted video ID brings us to video #4&#xA;&#xA;&#xA;#4 0101 0101 0000 0100 - 2015/12/03 - unlisted&#xA; |&#xA; |- Video URL: youtube.com/watch?v=m2AfQVy276Y &#xA; |- Plain data: [very long, see above]&#xA; |&#xA; |- Unknown meaning of the data&#xA;&#xA;#5 0101 0101 0000 0101 - 2015/19/03 - public&#xA; |&#xA; |- Video URL: youtube.com/watch?v=-qrtM4LdHH0 &#xA; |- Encrypted data: &amp;quot;A oYzXqoELKjE 14&amp;quot;&#xA; |&#xA; |- Unknown encryption tecnique&#xA; |- Unknown meaning of the data&#xA; |&#xA;------ mystery solved ------&#xA; |&#xA; |- Decryption key:&#xA; |  |- Decimal digits of Pi from 10th to 20th, as suggested by the two hex numbers 'A' and '14'&#xA; |&#xA; |- Decrypted (Gronsfeld Cipher): The video ID &amp;quot;jQqQhlCICfY&amp;quot;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;h2&gt;Waiting for other revelations&lt;/h2&gt;&#xA;&lt;p&gt;I spent a bunch of time around this, but didn't succeed in identifying the next video (if there is any). By the way, this series of riddles seems really interesting and challenging to me!&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;The first &lt;a href=&quot;https://boards.4chan.org/x/thread/15888717#p15907760&quot; rel=&quot;nofollow noreferrer&quot;&gt;4chan thread&lt;/a&gt; about this&lt;/strong&gt; has been unfortunately archived, but contains a lot of useful information about the first four videos. &lt;strong&gt;&lt;a href=&quot;http://boards.4chan.org/x/thread/15926420/video-0101-0101-0000-0101&quot; rel=&quot;nofollow noreferrer&quot;&gt;Another thread&lt;/a&gt;&lt;/strong&gt; has been opened on March 19 2015, relatively to the fifth video, and it may be helpful in order to find more hidden clues.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://chat.stackexchange.com/rooms/22141/noc10&quot;&gt;This chatroom&lt;/a&gt;&lt;/strong&gt; has been created to discuss about the topic and arrange some ideas to find the next video. The chatroom has been frozen for inactivity on May 24th, 2015. By the way you can still read the existing messages.&lt;/p&gt;&#xA;&lt;h2&gt;Updates&lt;/h2&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;March 20, 2015 - update #1:&lt;/strong&gt; added a link to a second 4chan thread about the fifth video; added the &amp;quot;common points&amp;quot; section; added a summary of the progress made until now.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;March 20, 2015 - update #2:&lt;/strong&gt; added a quote from the SS Sultana Wikipedia page in the fourth video section: it seems to be related to this mystery.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;March 20, 2015 - update #3:&lt;/strong&gt; added a link to &lt;a href=&quot;http://chat.stackexchange.com/rooms/22141/noc10&quot;&gt;the chatroom&lt;/a&gt; created by &lt;a href=&quot;http://chat.stackexchange.com/users/35072&quot;&gt;Bobson&lt;/a&gt; (thanks!). Anyone interested is kindly welcome!&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;March 24, 2015 - update #4:&lt;/strong&gt; the second 4chan thread has been deleted.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;April 13, 2015 - update #4:&lt;/strong&gt; the first 4chan thread has been deleted.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;October 22, 2015 - update #5:&lt;/strong&gt; the chatroom has been frozen for inactivity.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;h2&gt;Notes&lt;/h2&gt;&#xA;&lt;p&gt;I will keep this post active and update it if there is any relevant step forward to the solution of the NOC+10 mystery.&lt;br /&gt;&#xA;The next part of the mystery is being investigated in this thread: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/11718/the-mystery-of-noc-10-part-2&quot;&gt;&lt;strong&gt;The Mystery of Noc +10, Part 2&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10248" LastEditorUserId="10248" LastEditDate="2020-08-07T09:01:00.923" LastActivityDate="2023-06-12T19:26:14.493" Title="The mystery of NOC +10" Tags="|cipher|enigmatic-puzzle|mystery|connections-puzzle|alternate-reality-game|" AnswerCount="2" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="10634" PostTypeId="2" ParentId="10615" CreationDate="2015-03-20T02:27:25.087" Score="3" Body="&lt;p&gt;Could it possibly be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A reflection&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no legs, but I can move in ease&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A reflection moves easily with changing light&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am trapped, yet I can go where your eyes can see&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it is trapped in a mirror, but you can see the reflection of the room behind you for example&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am invisible, until you have found me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You cannot see your own reflection until you are looking right at the mirror&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a thing, that you will wish to set me free&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You will want to release yourself from the mirror&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10240" LastActivityDate="2015-03-20T02:27:25.087" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10635" PostTypeId="2" ParentId="10600" CreationDate="2015-03-20T03:13:46.847" Score="5" Body="&lt;p&gt;There's a method described in &lt;a href=&quot;http://www.math.washington.edu/~morrow/mcm/team2280.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sudoku: Bagging a Difficulty Metric &amp;amp; Building Up Puzzles&lt;/a&gt; for generating a solved Sudoku board.  You take the &lt;a href=&quot;http://mathworld.wolfram.com/LatinSquare.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;12 order three latin squares&lt;/a&gt; (using values of 0, 1, and 2).  Of these 12, you select 9 with replacement that are placed on a 3x3 board and then a 10th is the 'over board'.  Each value is then read as trinary when converting to base 10, and then some specific swaps are done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gjvqI.png&quot; alt=&quot;from the article linked&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's where I'm guessing, and I haven't done the full analysis of the resulting board.  Unfortunately, I don't have the math background to be able to prove that that such is the case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a programmer though, and so I whipped up a quick bit of code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;#!/usr/bin/perl&#xA;&#xA;use strict;&#xA;&#xA;use List::Util qw(shuffle);&#xA;&#xA;my @squares = shuffle (&#xA;[[0,1,2],[1,2,0],[2,0,1]],&#xA;[[0,1,2],[2,0,1],[1,2,0]],&#xA;[[0,2,1],[1,0,2],[2,1,0]],&#xA;[[0,2,1],[2,1,0],[1,0,2]],&#xA;&#xA;[[1,0,2],[0,2,1],[2,1,0]],&#xA;[[1,0,2],[2,1,0],[0,2,1]],&#xA;[[1,2,0],[0,1,2],[2,0,1]],&#xA;[[1,2,0],[2,0,1],[0,1,2]],&#xA;&#xA;[[2,1,0],[0,2,1],[1,0,2]],&#xA;[[2,1,0],[1,0,2],[0,2,1]],&#xA;[[2,0,1],[0,1,2],[1,2,0]],&#xA;[[2,0,1],[1,2,0],[0,1,2]],&#xA;);&#xA;&#xA;&#xA;my @board;&#xA;my @final;&#xA;my $outer = shift @squares;&#xA;&#xA;for my $i (0 .. 2) {&#xA;    for my $j (0 .. 2) {&#xA;        $board[$i][$j] = shift @squares;&#xA;&#xA;        for my $x (0 .. 2) {&#xA;        for my $y (0 .. 2) {&#xA;            $final[$i * 3 + $x][$j * 3 + $y] = &#xA;                3 * $outer-&amp;gt;[$i][$j] + $board[$i][$j][$x][$y] + 1;&#xA;            }&#xA;        }&#xA;    }&#xA;}&#xA;&#xA;# now, swap R2 &amp; R4; R3 &amp; R7; R6 &amp; R8&#xA;&#xA;for my $i (0 .. 8) {&#xA;($final[1][$i], $final[3][$i]) = ($final[3][$i], $final[1][$i]);&#xA;($final[2][$i], $final[6][$i]) = ($final[6][$i], $final[2][$i]);&#xA;($final[5][$i], $final[7][$i]) = ($final[7][$i], $final[5][$i]);&#xA;}&#xA;&#xA;foreach my $r (@final) {&#xA;foreach my $c (@{$r}) {&#xA;    print &quot;$c &quot;;&#xA;    }&#xA;    print &quot;\n&quot;;&#xA;}&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And running it a few times I get:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;7 9 8 2 1 3 5 6 4 &#xA;3 1 2 6 5 4 9 8 7 &#xA;5 6 4 7 8 9 1 2 3 &#xA;9 8 7 3 2 1 6 4 5 &#xA;2 3 1 5 4 6 7 9 8 &#xA;4 5 6 9 7 8 2 3 1 &#xA;8 7 9 1 3 2 4 5 6 &#xA;1 2 3 4 6 5 8 7 9 &#xA;6 4 5 8 9 7 3 1 2 &#xA;&#xA;&#xA;4 5 6 1 3 2 9 8 7 &#xA;3 2 1 8 9 7 6 4 5 &#xA;7 8 9 6 4 5 2 3 1 &#xA;5 6 4 3 2 1 7 9 8 &#xA;2 1 3 9 7 8 5 6 4 &#xA;9 7 8 4 5 6 1 2 3 &#xA;6 4 5 2 1 3 8 7 9 &#xA;1 3 2 7 8 9 4 5 6 &#xA;8 9 7 5 6 4 3 1 2 &#xA;&#xA;&#xA;7 9 8 2 1 3 6 5 4 &#xA;2 1 3 6 5 4 7 9 8 &#xA;4 5 6 8 9 7 3 1 2 &#xA;9 8 7 1 3 2 5 4 6 &#xA;3 2 1 4 6 5 8 7 9 &#xA;5 6 4 9 7 8 2 3 1 &#xA;8 7 9 3 2 1 4 6 5 &#xA;1 3 2 5 4 6 9 8 7 &#xA;6 4 5 7 8 9 1 2 3 &#xA;&#xA;&#xA;5 6 4 3 1 2 7 9 8 &#xA;1 3 2 8 9 7 4 5 6 &#xA;9 7 8 5 4 6 3 2 1 &#xA;4 5 6 2 3 1 8 7 9 &#xA;3 2 1 9 7 8 6 4 5 &#xA;7 8 9 6 5 4 2 1 3 &#xA;6 4 5 1 2 3 9 8 7 &#xA;2 1 3 7 8 9 5 6 4 &#xA;8 9 7 4 6 5 1 3 2 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'm taking the shortcut of not doing the replacement in my code.  There are some properties of these puzzles... if you note, the 7, 8, and 9 are always in the same set of three, as are the 1, 2, 3 and the 4, 5, 6. The fact that its a latin square on the over board too, then makes the swaps such that I &lt;em&gt;believe&lt;/em&gt; that the rectangular fence cannot form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have the process only use one latin square, the grid looks like:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 2 3 4 5 6 7 8 9 &#xA;7 8 9 1 2 3 4 5 6 &#xA;4 5 6 7 8 9 1 2 3 &#xA;3 1 2 6 4 5 9 7 8 &#xA;9 7 8 3 1 2 6 4 5 &#xA;6 4 5 9 7 8 3 1 2 &#xA;2 3 1 5 6 4 8 9 7 &#xA;8 9 7 2 3 1 5 6 4 &#xA;5 6 4 8 9 7 2 3 1 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This next one (another one grid generation) is one that is probably closest to the other solutions that have been provided and allows a clear picture of the patterns that are formed in this type of generation.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 2 3 4 5 6 7 8 9 &#xA;4 5 6 7 8 9 1 2 3 &#xA;7 8 9 1 2 3 4 5 6 &#xA;2 3 1 5 6 4 8 9 7 &#xA;5 6 4 8 9 7 2 3 1 &#xA;8 9 7 2 3 1 5 6 4 &#xA;3 1 2 6 4 5 9 7 8 &#xA;6 4 5 9 7 8 3 1 2 &#xA;9 7 8 3 1 2 6 4 5 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Though, while I am confident that these do not have rectangular fences, and the generation method precludes them, I am not sure that such is the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key part is in order to form a rectangular fence, I need to swap two columns somewhere (either in a single 3x3, or in the over grid):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;? A B&#xA;? B A&#xA;? ? ? &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The thing is that this pattern does not happen in any of the latin squares because that would force the third element of the square to be in the same row or column as itself, which is a violation of the definition of the latin square. ... but, as I said, I'm not sure I have a proof there.&lt;/p&gt;&#xA;" OwnerDisplayName="user2322" LastEditorDisplayName="user2322" LastEditDate="2015-03-20T03:35:25.150" LastActivityDate="2015-03-20T03:35:25.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10636" PostTypeId="2" ParentId="10615" CreationDate="2015-03-20T04:52:22.870" Score="4" Body="&lt;p&gt;Is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; dust?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no legs, but I can move in ease&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it floats around in the air at the slightest breeze&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am trapped, yet I can go where your eyes can see&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dust is confined to whatever space its in. Also its all around you (more or less depending on how clean the area is)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am invisible, until you have found me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; can be hard to see until you focus on it. Also dependent on how clean your place is :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a thing, that you will wish to set me free&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; By set free maybe you want to wipe it with a wet rag and wash it down the sink?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4292" LastActivityDate="2015-03-20T04:52:22.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10637" PostTypeId="2" ParentId="10615" CreationDate="2015-03-20T04:57:21.880" Score="5" Body="&lt;p&gt;How about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; beauty&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Seems to fit everything, particularly the 4th line.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I have no legs, but I can move in ease&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Beauty doesn't walk, but &lt;em&gt;moves in ease&lt;/em&gt;, ie it lives in things that are easy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am trapped, yet I can go where your eyes can see&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Beauty is in the eye of the beholder&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am invisible, until you have found me&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's not always apparent what is beautiful on first sight.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am a thing, that you will wish to set me free&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Beauty is a thing, not necessarily a property. But there is also a compulsion to set it free.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10428" LastEditorUserId="5188" LastEditDate="2015-03-20T05:41:51.867" LastActivityDate="2015-03-20T05:41:51.867" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10638" PostTypeId="2" ParentId="10615" CreationDate="2015-03-20T05:31:24.027" Score="3" Body="&lt;p&gt;I have no legs, but I can move with ease&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Imagination can go any where&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am trapped, yet I can go where your eyes can see&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Imagination can be trapped in anything you see&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am invisible, until you find me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You might be lost in imagination unless you realize you in trance&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a thing, that you will wish to set free&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You want to always set your Imagination free to get best out of your capabilities&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10210" LastActivityDate="2015-03-20T05:31:24.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10639" PostTypeId="2" ParentId="9850" CreationDate="2015-03-20T05:34:06.900" Score="2" Body="&lt;p&gt;Most likely, Brad's initial procedure is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; to count code names backward, starting from the last code name.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first target is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the 6th from the last, which is ROUND ORIOLE.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second target is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the 9th from the last, which is THOROUGH HYENA.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The third target is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the 12th from the last, which is TRIM EEL.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's possible that Brad also identified another indicator in the messages, which is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the identity of the NEXT target. Note that the second target is already identified in the first message as the 3rd code name from the last, which is THOROUGH HYENA.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Similarly,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the third target is already identified in the second message as the 3rd code name from the last, which is PROUD BOBCAT.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; although the first and the second indicators match for the second target, they don't match for the third target-- indicating that either the third message was erroneously sent, or there were actually TWO targets.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To be safe, Brad decided on the later and mentioned the two possible targets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Brad can even use the second indicator to predict that the next target would be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ABLE DINOSAUR,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;although he would need a fourth message to confirm it.&lt;/p&gt;&#xA;" OwnerUserId="10136" LastEditorUserId="10136" LastEditDate="2015-03-20T05:49:56.360" LastActivityDate="2015-03-20T05:49:56.360" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10640" PostTypeId="2" ParentId="10514" CreationDate="2015-03-20T05:56:33.700" Score="2" Body="&lt;p&gt;Fun! &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;At first, I was leaning towards:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; procrastination.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A place of work, and not of play,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You know that's what the vermin say.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Procrastination is a boss's worst nightmare. I'm hearing both the voice of the boss &amp;amp; of the employee in these lines.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Is it this or that, the question sly,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But in the end you'll question why?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm not procrastinating (Really...) or, am I? &#xA; Why do I know I'll regret this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A miracle you want to find,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And get yourself out of a bind.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Randomly searching the internet can sometimes yield great insights.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;So while you're here, follow a hunch,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;It may just happen before lunch.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Or, searching StackExchange :) If you're lucky, you'll be ready for a break at lunchtime.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;But, my final answer is :&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; equivocation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A place of work, and not of play,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Refers to the &quot;vocation&quot; in &lt;em&gt;equivocation&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You know that's what the vermin say.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Is it this or that, the question sly,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; These lines had me thinking of Hamlet. Chief equivocator&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A miracle you want to find&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And get yourself out of a bind.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Those equivocating are seeking an answer to a question, fundamentally&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;So while you're here, follow a hunch,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;It may just happen before lunch.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Just pick something already! &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10428" LastActivityDate="2015-03-20T05:56:33.700" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10641" PostTypeId="2" ParentId="10514" CreationDate="2015-03-20T06:31:17.923" Score="1" Body="&lt;p&gt;I think the answer based on the 3rd hint (&lt;code&gt;What is another name for work?&lt;/code&gt;) is :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Task&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am not able to relate it to each line. &lt;br&gt;&#xA;But one of the lines :&#xA;&lt;code&gt;But in the end you'll question why?&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; T&lt;strong&gt;ask&lt;/strong&gt; -&gt; ask (question???)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5565" LastEditorUserId="5565" LastEditDate="2015-03-20T06:42:02.050" LastActivityDate="2015-03-20T06:42:02.050" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10642" PostTypeId="2" ParentId="10619" CreationDate="2015-03-20T07:39:24.173" Score="1" Body="&lt;p&gt;I imagine that you have already tried this but I will post in case somebody can build on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Tryth noted, the code appears to represent digits from 1-5 and the cipher could be Bifid.  Digits are in pairs with &lt;strong&gt;half a space&lt;/strong&gt; between single digits and a &lt;strong&gt;full space&lt;/strong&gt; between pairs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resulting pairs are shown below as well as the letters from a basic &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Polybius_square&quot; rel=&quot;nofollow&quot;&gt;Polybius square&lt;/a&gt;&lt;/strong&gt;. This assumes that &lt;a href=&quot;http://en.wikipedia.org/wiki/Transposition_cipher#Fractionation&quot; rel=&quot;nofollow&quot;&gt;fractionation&lt;/a&gt; was not applied because there are 25 pairs of digits (not an even number).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;13  15  13  32  11  11  11  45  43  24  41  24  24  52  24  41  11  11  32  11  11  34  13  11  34&#xA;C   E   C   M   A   A   A   U   S   I/J Q   I/J I/J W   I/J Q   A   A   M   A   A   O   C   A   O&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If fractionation was applied by splitting the center pair of digits, the resulting pairs are shown below.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;14  35  12  52  14  34  31  21  11  11  11  13  12  11  41  51  41  33  24  41  43  11  21  43  24&#xA;D   P   B   W   D   O   L   F   A   A   A   C   B   A   Q   V   Q   N   I/J Q   S   A   F   S   I/J&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;One could continue to try various substitutions but this method does not seem to be correct primarily because it produces an unusually high number of 1's.  In addition, the range of the numbers from 11 to 52 seems too high considering an alphabet of either 26 or 30 characters.&lt;/p&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-03-20T18:33:01.077" LastActivityDate="2015-03-20T18:33:01.077" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10643" PostTypeId="2" ParentId="10624" CreationDate="2015-03-20T07:47:12.640" Score="1" Body="&lt;p&gt;The best you can win is 10 boxes. The computer always wins by getting at least 15 boxes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/y3Blt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this image, playing any of the red edges gives you 10 boxes if the computer plays optimally. If you play the green edge, you get only 9 boxes. Playing any of the blue edges gives you even less, only 8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The computation was done using the same code I used for the previous question. The computation was made easier since there are only 30 edges instead of 32.&lt;/p&gt;&#xA;" OwnerUserId="9471" LastEditorUserId="9471" LastEditDate="2015-03-20T07:59:02.987" LastActivityDate="2015-03-20T07:59:02.987" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10644" PostTypeId="2" ParentId="10615" CreationDate="2015-03-20T09:05:45.800" Score="4" Body="&lt;p&gt;What about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fart &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no legs, but I can move with ease&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Once you release it, it spreads&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am trapped, yet I can go where your eyes can see&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Previous explanation, it fills the room - everywhere that you can see&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am invisible, until you find me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is invisible, but if you have suspicion, you can just get a match or lighter and set it on fire.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a thing, that you will wish to set free&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Of course!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9907" LastEditorUserId="9907" LastEditDate="2015-03-20T14:41:17.020" LastActivityDate="2015-03-20T14:41:17.020" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10645" PostTypeId="1" AcceptedAnswerId="10647" CreationDate="2015-03-20T09:23:20.230" Score="7" ViewCount="349" Body="&lt;p&gt;You are given two goniometric functions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Male: $g(\rho, \Theta) = \rho \, \cos\Theta \sin\Theta$&lt;/li&gt;&#xA;&lt;li&gt;Female: $f(\rho, \Theta) = \rho \, \cos^2\Theta$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why are these two functions labeled &lt;em&gt;&quot;male&quot;&lt;/em&gt; and &lt;em&gt;&quot;female&quot;&lt;/em&gt;?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="10248" LastEditorUserId="5511" LastEditDate="2015-03-20T22:22:33.613" LastActivityDate="2015-03-20T22:22:33.613" Title="Male and Female functions" Tags="|lateral-thinking|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10646" PostTypeId="2" ParentId="10443" CreationDate="2015-03-20T09:36:30.890" Score="8" Body="&lt;h2&gt;Answer to the main question&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem:&lt;/strong&gt; There exists a solution for &lt;strong&gt;every&lt;/strong&gt; checkerboard (of arbitrarily irregular and not necessarily rectangular shape).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the proof, we we will say that two coins are &lt;em&gt;neighbors&lt;/em&gt;, if their squares share a common corner but not a common edge.&#xA;In other words, whenever you flip a coin then all its neighbors are flipped simultaneously. (Note also that no coin is a neighbor of itself.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, we will need the following auxiliary result (which essentially is the well-known handshake lemma from graph theory):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Auxiliary lemma:&lt;/strong&gt; Every (rectangular or irregular) checkerboard with an odd number of coins contains some coin with an even number of neighbors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof: We count the pairs $(c_1,c_2)$ of coins $c_1$ and $c_2$ such that $c_1$ is a neighbor of $c_2$. The number $N$ of such pairs must be even, since with every pair $(c_1,c_2)$ we also count the pair $(c_2,c_1)$. &#xA;Let $n(c)$ denote the number of neighbors of coin $c$. Note that coin $c$ occurs in the first component of exactly $n(c)$ pairs, and that consequently $N=\sum_c n(c)$.&lt;br&gt;&#xA;Now suppose for the sake of contradiction that all values $n(c)$ are odd. &#xA;Then $\sum_c n(c)$ is the summation of an odd number of odd integer values, and hence an odd number itself. Hence $N$ is odd; contradiction.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;The proof:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The proof of the theorem is done by induction on the number $k$ of squares on the checkerboard $C$. The $k=1$ is trivially true. So let us assume that the statement is true for all $k\le K-1$ and let us do the inductive step to $k=K$.&#xA;We consider an arbitrary (irregular) checkerboard $C$ with $K$ squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Let $s$ be an arbitrary square on the checkerboard $C$, and let $C-s$ denote the checkerboard that results by removing square $s$ from $C$.&#xA;As $C-s$ has only $K-1$ squares, the inductive assumption implies the existence of a solution for the game on $C-s$; let $Sol(C-s)$ denote this solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Now we return to the original checkerboard $C$. &#xA;We consider an arbitrary square $s$ on $C$, and we apply the solution $Sol(C-s)$ to the checkerboard $C$. Afterwards all coins on $C-s$ are red.&#xA;If the coin on square $s$ is also red, we have found a solution for checkerboard $C$ and are done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) It remains to consider the following case: For every square $s$ on $C$, the application of $Sol(C-s)$ to $C$ yields a configuration in which all coins on $C-s$ are red and where the coin on $s$ is blue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) Assume that the number $K$ of squares on $C$ is even.&#xA;Then we consecutively apply all the solutions $Sol(C-s)$ for all the squares $s$ on $C$. Consider some arbitrary square $s^*$. The coin on square $s^*$ is flipped in each of the $K-1$ solutions $Sol(C-s)$ with $s\ne s^*$, while it is not flipped in solution $Sol(C-s^*)$. Hence the coin is flipped an odd number of times, and in the end becomes red. Since this holds for all squares, we end up with the all-red configuration. Done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5) Finally assume that the number $K$ of squares on $C$ is odd.&#xA;We apply the auxiliary lemma to find a special square $s_o$ that has an odd number of neighbors. Then the number $x$ of squares on $C$ that are neither $s_o$ nor one of the neighbors of $s_o$ is odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We first flip the coin on $s_o$ (together with all coins on the neighbor squares), and then consecutively apply all the solutions $Sol(C-s)$ for all the squares $s$ on $C$, except for the square $s_o$ and its neighbors. &#xA;In the resulting solution, the coin on $s_o$ if flipped exactly once.&#xA;All other coins are flipped an odd number of times, as the number $x$ is odd.&#xA;Since every coin is flipped an odd number of times, the resulting configuration is all-red. Done.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-03-20T09:36:30.890" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10647" PostTypeId="2" ParentId="10645" CreationDate="2015-03-20T09:46:36.033" Score="9" Body="&lt;p&gt;Maybe &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; cos = X-chromosome and &lt;br&gt;&#xA; sin = Y-chromosome &lt;br&gt;&#xA; Male = XY = cos * sin &lt;br&gt;&#xA; Female = XX = cos * cos&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5565" LastActivityDate="2015-03-20T09:46:36.033" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10648" PostTypeId="1" AcceptedAnswerId="10653" CreationDate="2015-03-20T10:19:06.947" Score="7" ViewCount="2551" Body="&lt;p&gt;Just a fun Puzzle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My (fictive) friend told me his PIN: &quot;My PIN is the highest 4 digit number that has the highest sum of unique differences.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He explained: &quot;my PIN consists of 4 numbers A B C D.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Calculate the absolute differences between A and B, A and C, A and D, B and C, B and D, C and D (so all differences)&lt;/li&gt;&#xA;&lt;li&gt;Remove duplicates &lt;em&gt;(so each difference is included only once)&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Add the remaining differences together.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My PIN is the 4 digit number where this count is the highest and the highest of those numbers. &quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is my (fictive) friend's PIN?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number 1469 has a count of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 ) calculate the differences:&#xA;$ |1-4| = 3, |1-6| = 5, |1-9| = 8, |4-6| = 2, |4-9| = 5, |6-9| = 3 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) remove the duplicates:  $  3, 5, 8, 2 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) add them together: $ 18$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if there is no 4 digit number with a count higher than 18 and 1469 is the highest number of these numbers then 1469 is my friend's pin.&#xA;(but as I know both conditions are not met) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is my friend's PIN?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your answer please describe how you found his PIN (a clever way, or just brute force?)&lt;/p&gt;&#xA;" OwnerUserId="10437" LastEditorUserId="17147" LastEditDate="2016-12-20T21:33:34.623" LastActivityDate="2016-12-20T21:33:34.623" Title="What is my friend's PIN?" Tags="|mathematics|calculation-puzzle|password|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10649" PostTypeId="1" AcceptedAnswerId="10673" CreationDate="2015-03-20T10:19:39.567" Score="0" ViewCount="563" Body="&lt;p&gt;I am a peacemaker and a diplomat,&lt;br&gt;&#xA;a silent watcher to prevent betray.&lt;br&gt;&#xA;I’m hiding friends behind my body&lt;br&gt;&#xA;and make the rules for you to play.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can be clear or ambiguous, even graphic.&lt;br&gt;&#xA;But when it comes to work, I am snob,&lt;br&gt;&#xA;my role is rather .. choreographic&lt;br&gt;&#xA;while my buddies do the job.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;And the question is, who or what is our lazy hero?&lt;/p&gt;&#xA;" OwnerUserId="5283" LastActivityDate="2015-03-20T16:15:01.777" Title="diplomatic riddle" Tags="|riddle|word|rhyme|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10650" PostTypeId="2" ParentId="10649" CreationDate="2015-03-20T10:30:20.423" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A contract&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I am a peacemaker and a diplomat,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Peacetreaties and trade deals are signed into law&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;a silent watcher to prevent betray.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Obviously contracts don't speak, and you can't break them&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I’m hiding friends behind my body&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hiding smallprint in the body of the text&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and make the rules for you to play.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A contract is a codified agreement on what the two parties can do&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be clear or ambiguous, even graphic.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Can be plain, vague, or explicit; depending how it's written&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But when it comes to work, I am snob,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Contracts play a huge role in business&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;my role is rather .. choreographic&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A contract doesn't do anything on its own, it just tells other people what to do&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;while my buddies do the job&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Could refer to lawyers, or the smallprint referred to as 'friends' earlier in the riddle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10246" LastActivityDate="2015-03-20T10:30:20.423" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10651" PostTypeId="2" ParentId="10648" CreationDate="2015-03-20T10:32:06.617" Score="3" Body="&lt;p&gt;Just a brute-force answer; The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 9820&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This answer was generated by the following program:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;$bigval = 0;&#xA;$biggest = array();&#xA;for($i=1000;$i&amp;lt;10000;$i++) {&#xA;    	$pin = (string)$i;&#xA;    	$diffs = array();&#xA;    	$diffs[] = abs($pin[0] - $pin[1]);&#xA;    	$diffs[] = abs($pin[0] - $pin[2]);&#xA;    	$diffs[] = abs($pin[0] - $pin[3]);&#xA;    	$diffs[] = abs($pin[1] - $pin[2]);&#xA;    	$diffs[] = abs($pin[1] - $pin[3]);&#xA;    	$diffs[] = abs($pin[2] - $pin[3]);&#xA;    	if(array_sum(array_unique($diffs)) &amp;gt; $bigval) {&#xA;    		$bigval = array_sum(array_unique($diffs));&#xA;    		$biggest = array($pin);&#xA;    	}&#xA;    	else if(array_sum(array_unique($diffs)) == $bigval) {&#xA;    		$biggest[] = $pin;	&#xA;    	}&#xA;    }&#xA;print_r($biggest);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which gave the following output pins, with all the pins having a difference of 33:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1079, 1097, 1709, 1790, 1907, 1970, 2089, 2098, 2809, 2890, 2908, 2980, 7019, 7091, 7109, 7190, 7901, 7910, 8029, 8092, 8209, 8290, 8902, 8920, 9017, 9028, 9071, 9082, 9107, 9170, 9208, 9280, 9701, 9710, 9802, 9820&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of course the answer provided is the highest of these.&lt;/p&gt;&#xA;" OwnerUserId="7308" LastActivityDate="2015-03-20T10:32:06.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10652" PostTypeId="2" ParentId="10648" CreationDate="2015-03-20T10:37:22.853" Score="7" Body="&lt;p&gt;(I assume that each of the four numbers $A,B,C,D$ is a digit in the range $0,\ldots,9$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the six differences do not depend on the ordering of $A,B,C,D$, let us assume without loss of generality that $A\le B\le C\le D$, so that the differences are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) $B-A$, $C-B$, $D-C$&lt;br&gt;&#xA;(2) $C-A$, $D-B$, $D-A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total contribution of the three values in (1) is at most their sum &#xA;$(B-A)+(C-B)+(D-C)=D-A\le 9-0=9$.&#xA;The total contribution of the three values in (2) is at most $9+8+7$ (in case they are all different; in case two are the same, the bound is even smaller).&#xA;From these two upper bounds we conclude:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The count is at most $9+9+8+7=33$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In order to reach this count, all inequalities in the above discussion must in fact be equalities. &#xA;This implies $D-A=9$ and hence $A=0$ and $D=9$. Furthermore, the only ways of reaching differences $7$ and $8$ are&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$A=0$, $B=1$, $C=7$, $D=9$&lt;/li&gt;&#xA;&lt;li&gt;$A=0$, $B=2$, $C=8$, $D=9$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Together with these two possibilities, all their permutations are also feasible. &#xA;Hence the highest possible corresponding number (and the pin number of your fictional friend) is the number &lt;strong&gt;9820&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-03-20T10:37:22.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10653" PostTypeId="2" ParentId="10648" CreationDate="2015-03-20T10:37:37.190" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 9820&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4 numbers, they should have the biggest possible difference of it's members. So take 2 lowest and 2 highest possible numbers: 0, 1, 8 and 9.&lt;br&gt;&#xA; But $|8-9|=1$, so let's take another number so that we can use all 6 differences. Next possibility would then be: 0, 1, 7 and 9(&lt;strong&gt;1&lt;/strong&gt;) &lt;em&gt;or&lt;/em&gt; 0, 2, 8 and 9(&lt;strong&gt;2&lt;/strong&gt;)&lt;br&gt;&#xA; Calculate the difference(&lt;strong&gt;1&lt;/strong&gt;):&lt;br&gt;&#xA; $|0-1| = 1, |0-7| = 7, |0-9| = 9, |1-7| = 6, |1-9| = 8, |7-9| = 2$&lt;br&gt;&#xA; Numbers are: 1, 7, 9, 6, 8, 2. All 6 are different, so add them up and we get:&lt;br&gt;&#xA; $1+7+9+6+8+2=33$&lt;br&gt;&#xA; Calculate the difference(&lt;strong&gt;2&lt;/strong&gt;):&lt;br&gt;&#xA; $|0-2| = 2, |0-8| = 8, |0-9| = 9, |2-8| = 6, |2-9| = 7, |8-9| = 1$&lt;br&gt;&#xA; Numbers are: 2, 8, 9, 6, 7, 1. We can see that differences are same as in &lt;strong&gt;1&lt;/strong&gt;, but just in different order. So if we add them up, we will still get $33$.&lt;br&gt;&#xA; Now we should see which of those 4 is bigger: 0179 or 0289. To make it bigger we should re-order it in descending order, so we get:&lt;br&gt;&#xA; &lt;strong&gt;1&lt;/strong&gt;: 9710&lt;br&gt;&#xA; &lt;strong&gt;2&lt;/strong&gt;: 9820&lt;br&gt;&#xA; and because $9820&amp;gt;9710$, we can say that his PIN code is indeed $9820$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9352" LastActivityDate="2015-03-20T10:37:37.190" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10655" PostTypeId="2" ParentId="10615" CreationDate="2015-03-20T10:54:57.803" Score="1" Body="&lt;p&gt;How about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The woman I have trapped in my basement&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no legs, but I can move with ease&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I cut her legs off but she got pretty good with walking on her hands&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am trapped, yet I can go where your eyes can see&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Shes in a big cage but has a TV&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am invisible, until you find me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;No one knows shes here&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a thing, that you will wish to set free&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You want to free her because there's a good ransom&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Feel free to down vote me ;)&lt;/p&gt;&#xA;" OwnerUserId="9204" LastActivityDate="2015-03-20T10:54:57.803" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10656" PostTypeId="2" ParentId="10615" CreationDate="2015-03-20T11:02:26.617" Score="2" Body="&lt;p&gt;Maybe the solution to this is something disgusting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking about &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a Parasite&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no legs, but I can move with ease&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Inside the body of its victim&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am trapped, yet I can go where your eyes can see&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; trapped inside the body of the victim, but can be at the same place the body wants to be, since its moving with it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am invisible, until you find me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You don't know its there, until you (maybe) notice its presence&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a thing, that you will wish to set free&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if you get to know you have something like this, you probably want to get rid of it. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7659" LastActivityDate="2015-03-20T11:02:26.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10657" PostTypeId="1" AcceptedAnswerId="10658" CreationDate="2015-03-20T11:02:55.173" Score="48" ViewCount="3247" Body="&lt;p&gt;There are lots of happy animals living with me in my house.&#xA;Yesterday I wanted to enjoy an animal-free afternoon:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I looked at Skippy and said &quot;Go to festival&quot;. And off he hopped.&lt;/li&gt;&#xA;&lt;li&gt;Then I sent Cookie to the Kookaburra Ball.&lt;/li&gt;&#xA;&lt;li&gt;Then I ordered Platty to go to the Duckbill Dance.&lt;/li&gt;&#xA;&lt;li&gt;Then I sent young Edna to the Echidna Garden Party.&lt;/li&gt;&#xA;&lt;li&gt;Then I told Billy where to go, but he just sat there, looking at me and&#xA;chewing thoughtfully.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then I remembered that Billy's English isn't that good.&#xA;It seems that because of his diet of aromatic leaves he doesn't know&#xA;whether he's coming or going, and he only understands palindromes.&#xA;He cycles them forwards and backwards until the message gets through.&#xA;So I said to him&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Xxxxx, xx xx xxxx, XX?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And off he went.  Where did Billy go to?&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="9907" LastEditDate="2015-03-20T14:15:41.140" LastActivityDate="2016-07-19T17:28:15.107" Title="Where did Billy go to?" Tags="|word|palindrome|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10658" PostTypeId="2" ParentId="10657" CreationDate="2015-03-20T11:05:39.850" Score="70" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&quot;gala&quot;&lt;/strong&gt; (as the Koala went to the &lt;strong&gt;gala&lt;/strong&gt;)&lt;br&gt;&#xA; &quot;Xxxxx, xx xx xxxx, XX!&quot; equals the palindrome &quot;Koala, go to gala, ok?&quot;&#xA;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More explanations:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All mentioned animals are at home in Australia:&lt;br&gt;&#xA;&lt;br&gt;&#xA;Skippy is a kangaroo (the Bush Kangaroo in the Australian tv series),&lt;br&gt;&#xA;Cookie a kingfisher (Kookaburra),&lt;br&gt;&#xA;Platty a duck-billed platypus,&lt;br&gt;&#xA;Edna is an anteater (Echidna),&lt;br&gt;&#xA;Billy is a Koala (eating aromatic eucalypt leaves).&lt;br&gt;&#xA;&lt;br&gt;&#xA;Hence the palindrome should start with KOALA and end with ALA, OK.&lt;br&gt;&#xA;The GO TO piece is hinted at in the first sentence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-03-21T14:00:53.740" LastActivityDate="2015-03-21T14:00:53.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10659" PostTypeId="2" ParentId="10615" CreationDate="2015-03-20T11:10:11.720" Score="3" Body="&lt;p&gt;I think its&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; White Blood Cells (&lt;a href=&quot;http://en.wikipedia.org/wiki/Blue_field_entoptic_phenomenon&quot; rel=&quot;nofollow&quot;&gt;blue-sky sprites&lt;/a&gt;), which appear when you look up at the sky&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no legs, but I can move with ease&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They move in the body mixed with blood&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am trapped, yet I can go where your eyes can see&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Trapped inside your eye (or body), and wherever you see in the sky, it seems they follow you &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am invisible, until you find me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You don't notice them until you look at a clear blue sky and you have to pay proper attention to see it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a thing, that you will wish to set free&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; White Blood Cells eventually are disposed out from the body and new ones are formed, maybe?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3877" LastActivityDate="2015-03-20T11:10:11.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10660" PostTypeId="2" ParentId="10615" CreationDate="2015-03-20T11:37:36.937" Score="1" Body="&lt;p&gt;You might be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a jelly fish&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no legs, but I can move with ease&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; they have tentacles not legs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am trapped, yet I can go where your eyes can see&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; trapped in the ocean, go can do to oceans' depth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am invisible, until you find me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; there is a famous species of jelly fish that are nearly invisible in water&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a thing, that you will wish to set free&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; jelly fish not in the wild: you want to set free&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9778" LastActivityDate="2015-03-20T11:37:36.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10661" PostTypeId="2" ParentId="10615" CreationDate="2015-03-20T12:53:34.520" Score="2" Body="&lt;p&gt;Could it be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Heart/Feelings&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no legs, but I can move with ease&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can move someone's heart with words&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am trapped, yet I can go where your eyes can see&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Trapped within your chest, but it can be affected by anything you see&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am invisible, until you find me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You do not realise your feelings until you find them&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a thing, that you will wish to set free&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Relating to love possibly?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2153" LastActivityDate="2015-03-20T12:53:34.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10662" PostTypeId="2" ParentId="10615" CreationDate="2015-03-20T13:04:13.183" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Imagination&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Seems to fit pretty well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no legs, but I can move with ease&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Imagination has no limits or boundaries&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am trapped, yet I can go where your eyes can see&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; But sometimes it can be trapped by your own thinking, but what you see or experience can free it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am invisible, until you find me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sometimes it can be hard to find your &quot;muse&quot; and can be invisible to you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a thing, that you will wish to set free&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A strong imagination has no limits and can free you from any prison, real or imagined.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10294" LastEditorUserId="10294" LastEditDate="2015-03-24T13:18:34.540" LastActivityDate="2015-03-24T13:18:34.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10663" PostTypeId="2" ParentId="10615" CreationDate="2015-03-20T13:38:54.660" Score="13" Body="&lt;p&gt;I think the answer is...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eye floaters&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no legs, but I can move with ease&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They move around everywhere you look&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am trapped, yet I can go where your eyes can see&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Floater&quot;&gt;They are inside your eyeball&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am invisible, until you find me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You don't always notice them, until you do&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a thing, that you will wish to set free&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are annoying, you want them to get out of your eye&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9478" LastEditorUserId="9478" LastEditDate="2015-03-20T14:54:24.193" LastActivityDate="2015-03-20T14:54:24.193" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10665" PostTypeId="2" ParentId="10615" CreationDate="2015-03-20T14:18:34.053" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A look, as in &quot;he gave me a &lt;em&gt;look&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It has no legs, moves with ease&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is trapped, but can move &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A look is confined to line of sight. Also, refers to the other meaning of look: to &lt;em&gt;look&lt;/em&gt; around the room.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is invisible until seen &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Back to the first meaning,  quite literally isn't seen until it's been looked at (but still exists if not seen).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is a thing, and you want to set it free &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you notice someone giving you a &lt;em&gt;look&lt;/em&gt;,  you want to look away (ie set the look free). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10428" LastActivityDate="2015-03-20T14:18:34.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10666" PostTypeId="1" AcceptedAnswerId="10680" CreationDate="2015-03-20T14:30:02.073" Score="2" ViewCount="306" Body="&lt;p&gt;Here are some cryptograms for you to solve:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;084 104 105 115 032 105 115 032 116 104 101 032 102 105 114 115 116 032 099 114 121 112 116 111 103 114 097 109 032 105 110 032 116 104 101 032 115 101 114 105 101 115 046 013 010&lt;/code&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;0114040020080919000919002008050014052420&lt;/code&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;tvpbb ftjrw pt hcvm fvbh yllbvtcb tnpd J&lt;/code&gt; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;Bsfesutid tin s liteilsht  hi!&lt;/code&gt; (the double spaces are intentional)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;1111200021001213010020002110311112102011311012110201!&lt;/code&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9736" LastEditorUserId="5188" LastEditDate="2015-03-20T19:53:40.750" LastActivityDate="2015-03-20T20:17:02.990" Title="This and this - 5 cryptograms" Tags="|cipher|" AnswerCount="4" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10667" PostTypeId="2" ParentId="10648" CreationDate="2015-03-20T14:30:36.340" Score="3" Body="&lt;p&gt;The answer is already found by Tryth and Novarg. So for variety sake adding my approach. 0, 9 and 8 are definite entrant due to range. Then I randomly chose 7 as remaining number and formed square:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   0  7  8  9&#xA;9  9  2  1&#xA;8  8  1&#xA;7  7&#xA;0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Giving 27.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   0  7  8  9&#xA;9  9  2  1&#xA;8  8  1&#xA;7  7&#xA;0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Next&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   0  6  8  9&#xA;9  9  3  1&#xA;8  8  2&#xA;6  6&#xA;0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Giving 29 ... better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   0  5  8  9&#xA;9  9  4  1&#xA;8  8  3&#xA;5  5&#xA;0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Giving 30  better&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   0  4  8  9&#xA;9  9  5  1&#xA;8  8 [4]&#xA;4  4&#xA;0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Gone bad..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   0  3  8  9&#xA;9  9  6  1&#xA;8  8  5&#xA;3  3&#xA;0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Giving 32 better&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   0  2  8  9&#xA;9  9  7  1&#xA;8  8  6&#xA;2  2&#xA;0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Giving 33 and including 9 8 7 6 2 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going further down is going to repeat digits and getting 9 8 7 6 5 1 is not possible, so this one is the answer with largest number 9820&lt;/p&gt;&#xA;&#xA;&lt;p&gt;========================================================&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update (a more mathematical approach):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If in the descending order the digits are A&gt;B&gt;C&gt;D&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Then the differences are:&#xA;A-B,A-C,A-D;B-C,B-D;C-D&#xA;Adding all of them gives us:  &#xA;Total Sum= 3*A+B-C-3*D&#xA;Hence for the greatest sum, A &amp;amp; B should be as large as possible&#xA;and C &amp;amp; D should be as small as possible.&#xA;Which gives first winner combination 9810 but it fails as 9-8 = 1-0&#xA;So next one is 9820 which gives no overlap.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="10210" LastEditorUserId="10210" LastEditDate="2015-03-22T09:23:50.360" LastActivityDate="2015-03-22T09:23:50.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10668" PostTypeId="1" AcceptedAnswerId="10696" CreationDate="2015-03-20T14:32:32.953" Score="12" ViewCount="2230" Body="&lt;p&gt;Let me spell this out for you,&lt;br&gt;&#xA;And see what you can do.&lt;br&gt;&#xA;I begin by being half full,&lt;br&gt;&#xA;Then onto you, don't be dull.&lt;br&gt;&#xA;I'm around for a third of a year.&lt;br&gt;&#xA;Finally, without me, there is no gear.&lt;br&gt;&#xA;Let's see if you can puzzle how,&lt;br&gt;&#xA;Ultimately the riddle's now,&lt;br&gt;&#xA;Now ready for you to complete.&lt;br&gt;&#xA;Granted, it might be elite.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is shorter than short.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="9951" LastEditDate="2015-03-20T23:10:49.887" LastActivityDate="2015-03-21T00:45:22.310" Title="Let me spell it out - What am I?" Tags="|riddle|word|rhyme|" AnswerCount="5" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10670" PostTypeId="2" ParentId="10666" CreationDate="2015-03-20T15:01:56.283" Score="4" Body="&lt;p&gt;In progress... :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is the first cryptogram in the series.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;method&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (Ascii character numbers)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9478" LastEditorUserId="9478" LastEditDate="2015-03-20T15:11:48.953" LastActivityDate="2015-03-20T15:11:48.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10671" PostTypeId="2" ParentId="10666" CreationDate="2015-03-20T15:09:36.140" Score="6" Body="&lt;p&gt;3)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I don't actually have much to write about&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the method  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reverse the sentence. and then I noticed it was pretty much readable already. To get the real sentence replace every vowel in the solution with it's next letter in the alphabet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-03-20T15:09:36.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10672" PostTypeId="2" ParentId="10514" CreationDate="2015-03-20T15:21:37.167" Score="2" Body="&lt;p&gt;Is it  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; laboratory?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A place of work, and not of play,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You go to laboratory to research, and work is &quot;labor&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You know that's what the vermin say.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (With Aggie's answer:) a rat in labo'rat'ory&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is it this or that, the question sly,&lt;br&gt;&#xA;But in the end you'll question why?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A sly question that the answer ends in Y (why)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A miracle you want to find,&lt;br&gt;&#xA;And get yourself out of a bind.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Through researching in laboratory, you might found miracle!  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So while you're here, follow a hunch,&lt;br&gt;&#xA;It may just happen before lunch.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  lunch and library both start with &quot;L&quot; but I might fail explaining this line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="10037" LastEditDate="2015-03-20T21:56:50.673" LastActivityDate="2015-03-20T21:56:50.673" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="10673" PostTypeId="2" ParentId="10649" CreationDate="2015-03-20T16:15:01.777" Score="2" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A UI (user interface)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a peacemaker and a diplomat,&#xA;a silent watcher to prevent betray.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A UI helps to make peace between the users and the code, and in translating user desires to commands helps to eliminate errors&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I’m hiding friends behind my body&#xA;and make the rules for you to play.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The UI sets rules for how you interact with it and hides the actual commands of the OS behind it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be clear or ambiguous, even graphic.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sometimes it's obvious you're working through a UI and sometimes not, but they can definitely be graphic (GUI - graphical user interface)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But when it comes to work, I am snob,&#xA;my role is rather .. choreographic&#xA;while my buddies do the job.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The UI doesn't actually do any of the computing work, it just translates the users commands into commands in the operating language (hence the &quot;choreographing&quot;) and the buddies (friends referenced earlier) are the programs/code that actually do the work.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2015-03-20T16:15:01.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10674" PostTypeId="2" ParentId="10666" CreationDate="2015-03-20T16:23:44.067" Score="2" Body="&lt;p&gt;2.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; AND THIS IS THE NEXT&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Method:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Treat each pair of digits left to right as position of the corresponding letter in the alphabet, 00 = SPACE; 01 = A, 02 = B, 03 = C, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9910" LastActivityDate="2015-03-20T16:23:44.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10676" PostTypeId="2" ParentId="10615" CreationDate="2015-03-20T19:19:30.007" Score="1" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Horizon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no legs, but I can move with ease:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Explanation - The horizon keeps moving&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am trapped, yet I can go where your eyes can see:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Explanation - The horizon is as far as your eyes can see, and if you don't move it doesn't move&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am invisible, until you find me:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Explanation - You can't see the horizon until you look up&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a thing, that you will wish to set free:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Explanation - Don't you want to expand your horizons, don't you want to fly?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10455" LastEditorUserId="10342" LastEditDate="2015-03-21T13:58:35.590" LastActivityDate="2015-03-21T13:58:35.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10677" PostTypeId="2" ParentId="10574" CreationDate="2015-03-20T19:23:14.510" Score="2" Body="&lt;p&gt;The largest number that can be produced here is indeed $2^{3^{41}}$. But going for the general case, we need to identify what will give us the highest number. The most important thing is to get the largest number we have as high up in the power tower as possible. And the same goes for the next largest number. The last thing we have to remember is if we have the number $1$, the best thing we can do with it is to put it at the end of the highest number of the power tower, effectively multiplying that number by 10. So if we have $\{1,\,2,\,3,\,4,\,5\}$ we want to put $5$ as high as possible, and then $4$ the next highest. We end up with $2^{3^{4^{51}}}$. If we have $\{2,\,3,\,4,\,5,\,6\}$ we would get $2^{3^{4^{5^6}}}$.&lt;/p&gt;&#xA;" OwnerUserId="10454" LastActivityDate="2015-03-20T19:23:14.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10678" PostTypeId="1" AcceptedAnswerId="10712" CreationDate="2015-03-20T19:28:59.510" Score="9" ViewCount="568" Body="&lt;p&gt;&lt;em&gt;The old rich Jeremiah has two sons, Timothy and Joseph. One day, Jeremiah decides to discuss about his heritage with his sons...&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;JEREMIAH:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Dear sons, I'm very old and exhausted, so I want to bequeath my collection of coins to you, if you prove me that you are smart enough.&lt;/p&gt;&#xA;&lt;p&gt;You must know that I have a number &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; of valuable gold coins and a number &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; of precious silver coins, and I'm the only one in this world who knows &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt;. Also, &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; are two numbers between &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$20$&lt;/span&gt; (meaning that &lt;span class=&quot;math-container&quot;&gt;$2\leq G, S \leq 20$&lt;/span&gt;).&lt;/p&gt;&#xA;&lt;p&gt;I'm going to tell you, Timothy, the sum of my gold and silver coins (&lt;span class=&quot;math-container&quot;&gt;$G+S$&lt;/span&gt;), and to you, Joseph, the product of them (&lt;span class=&quot;math-container&quot;&gt;$G\cdot S$&lt;/span&gt;). You have 24 hours from now to discover the number of your brother. If you both find the answer, I'll give my collection to both.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;So, Jeremiah tells them the numbers, as he said. Of course, the two brothers are greedy, and each one doesn't want to share his information with his brother.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;After six hours of reflecting...&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;TIMOTHY:&lt;/strong&gt; Joseph, I don't think that you can determine my sum!&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;JOSEPH:&lt;/strong&gt; Haha! You're a fool! Now I know your sum! The heritage is all mine!!!&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;TIMOTHY:&lt;/strong&gt; I'm sorry for you, but if I'm a fool, then you are unwise! Now I know your product, and we'll have to share the legacy!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;QUESTION:&lt;/strong&gt;&lt;/em&gt; Can you determine the two numbers &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt;? Also, is the solution unique?&lt;/p&gt;&#xA;&lt;p&gt;Note that your answer must be complete and detailed to be accepted!&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; &lt;span class=&quot;math-container&quot;&gt;$2\leq G$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$2\leq S$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2019-02-16T19:55:37.423" Title="The heritage of a numismatic" Tags="|logical-deduction|" AnswerCount="3" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10679" PostTypeId="2" ParentId="10443" CreationDate="2015-03-20T19:33:15.743" Score="4" Body="&lt;p&gt;Consider that selecting any black square only affects black squares (true for white as well).  The boards are identical up to rotation, so solve just the one of them.  Lets focus on the black:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;| |3| |3| |3| |2|&#xA;|3| |5| |5| |5| |&#xA;| |5| |5| |5| |3|&#xA;|3| |5| |5| |5| |&#xA;| |5| |5| |5| |3|&#xA;|3| |5| |5| |5| |&#xA;| |5| |5| |5| |3|&#xA;|2| |3| |3| |3| |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Where we have ignored the white spaces and labeled each square with the number of neighbors it has (number of other squares which can possibly change its coin color, note that every square is its own neighbor).  If we flip every square, then squares with odd numbers of neighbors will change color while squares with even number of neighbors will not.  In the diagram above, we notice that by pressing every square, we can change all colors but the bottom left and top right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... I'm not sure where I'm going anymore.  This is a solution but not for the &quot;even-sided&quot; problem (the diagram above shows 4 black squares per side).  If it were odd it cleans up real nice though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this spoils the fun for somebody else, please delete me.&lt;/p&gt;&#xA;" OwnerUserId="10453" LastEditorUserId="10453" LastEditDate="2015-03-20T20:14:43.657" LastActivityDate="2015-03-20T20:14:43.657" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10680" PostTypeId="2" ParentId="10666" CreationDate="2015-03-20T19:47:23.767" Score="4" Body="&lt;p&gt;Cryptogram 4 is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;But i still finished this set!&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; encoded with 6 column transposition per this sequence:&lt;br&gt;&#xA; 1    6   11  16  21  26  2   7   12  17  22  27  3   8   13  18  23  28  4   9   14  19  24  29  5   10  15  20  25  30 &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cryptogram 5 is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Hope you had fun!&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; encoded with Morse code (0 is dash, 1 is dot, 2 is letter space, 3 is word space).   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-03-20T20:17:02.990" LastActivityDate="2015-03-20T20:17:02.990" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10681" PostTypeId="2" ParentId="10678" CreationDate="2015-03-20T20:00:03.853" Score="5" Body="&lt;p&gt;&lt;strong&gt;the answer is variant about two eventual possibilities :&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;- If both sons and father know that numbers of coins are bigger than 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the answer dosnt differ than the classic puzzle &lt;a href=&quot;https://puzzling.stackexchange.com/questions/251/i-dont-know-the-two-numbers-but-now-i-do&quot;&gt;I don&amp;#39;t know the two numbers... but now I do&lt;/a&gt; because there s noway the person with product can know the certainly the sum except the case S and G are pimes or $S*G = prime^{3}$ because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;lets assume for example $c$ divide S:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S+G = \frac {S}{c}+Gc$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Sc+Gc = S+Gc²$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S = Gc$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So .. the only two couples sharing same product and sum are (S,G) = (cG,G) and (cG/c,Gc) and they are identical .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No hope for other solution except: 4,13 but this solution is expulsed since there doesnt exists any natural k&gt;1 where 13*k &amp;lt;= 20&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;- If number 1 is included&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TIMOTHY: Joseph, I don't think that you can determine my sum!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;this means sum of numbers isnt prime+1 (considering 1 as prime)&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;JOSEPH: Haha! You're a fool! Now I know your sum! The heritage is all mine!!!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;this means same product of whicheve two numbers , their sum is prime+1 except one unique couple&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;TIMOTHY: I'm sorry for you, but if I'm a fool, then you are unwise! Now I know your product, and we'll have to share the legacy!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;this just meant that one unique couple among all possible couples which have common sum verifies the last condition&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Such numbers are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (1,4) , (1,8) , (1,9) , (1,15)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9517" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-03-21T17:48:02.177" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="10682" PostTypeId="1" CreationDate="2015-03-20T20:16:36.070" Score="-2" ViewCount="293" Body="&lt;p&gt;Hees the first one incase you havent seen it.&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/10619/theres-an-arg-ive-been-working-on-and-im-stuck-on-this-puzzle&quot;&gt;There&amp;#39;s an ARG I&amp;#39;ve been working on, and I&amp;#39;m stuck on this puzzle&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So from that we got our password for this &quot;problem&quot; however we havent cracked this code yet, we've tried pufferfish and we're working on something else, if you think you have an answer please post it.&#xA;Heres the code&lt;/p&gt;&#xA;&#xA;&lt;p&gt;57]234($$†‡555$3;‡62510552]34)?125155:26))11†53-5$360‡†54$55833‡3]8$1-8$335†¶]‡¶28:[8$-?[4†51¶?†529¶]:¶28†¶)†2‡[]7$2?-[)51?[9;†1[?5:59[9?56¶?¢51.[4?&lt;/p&gt;&#xA;" OwnerUserId="10412" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-03-20T22:32:54.310" Title="There's an ARG I've been working on, and I'm stuck on this puzzle part II" Tags="|mathematics|calculation-puzzle|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10683" PostTypeId="2" ParentId="10443" CreationDate="2015-03-20T20:55:23.487" Score="4" Body="&lt;p&gt;There is one solution, with 4 different possible rotations:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;XXXXXXXX&#xA;&#xA;X  XX  X&#xA;  XXXX&#xA;XXX  XXX&#xA;X  XX  X&#xA;XX XX XX&#xA;  X  X&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The black and the white squares will never influence one another, so one only need to consider the puzzle with 32 squares. Further, it never makes sense to choose the same centre coin twice, since that would simply undo the initial action, and it doesn't matter in what order the flips are done. If we go through the coins row by row and for each decide whether to flip it as a centre, most of the decisions will be given by the state of another coin that cannot be influenced by any future decisions. With the right order there is only 7 decisions that does not have such a constraint, this means that there is only 128 strategies to try, only 2 of those work, and they produce the same solution in 2 different orientations. This solution must then be applied to both the white and the black squares, with two different rotation options for each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This code solves half the puzzle, and prints any input seed that gives a correct solution to the console. It takes a little work to permute this into a solution pattern.&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;lt;!DOCTYPE HTML&amp;gt;&#xA;&amp;lt;html&amp;gt;&#xA;&amp;lt;head&amp;gt;&#xA;&amp;lt;meta http-equiv=&quot;content-type&quot; content=&quot;text/html;charset=utf-8&quot;&amp;gt;&#xA;&amp;lt;title&amp;gt;Calc&amp;lt;/title&amp;gt;&#xA;&amp;lt;/head&amp;gt;&#xA;&amp;lt;body&amp;gt;&#xA;&amp;lt;script&amp;gt;&#xA;function trysol(num){&#xA;    var org=num&#xA;    var flip&#xA;    var map=[&#xA;        [1,1,1,1,1,1,1,1,1]&#xA;        ,[1,1,1,1,0,1,1,1,1]&#xA;        ,[1,1,1,0,0,0,1,1,1]&#xA;        ,[1,1,0,0,0,0,0,1,1]&#xA;        ,[1,0,0,0,0,0,0,0,1]&#xA;        ,[1,0,0,0,0,0,0,0,1]&#xA;        ,[1,1,0,0,0,0,0,1,1]&#xA;        ,[1,1,1,0,0,0,1,1,1]&#xA;        ,[1,1,1,1,0,1,1,1,1]&#xA;        ,[1,1,1,1,1,1,1,1,1]&#xA;    ]&#xA;    for(var y=0;y&amp;lt;10;y++){&#xA;    for(var x=0;x&amp;lt;9;x++){&#xA;        if(map[y][x]!=1){&#xA;            if(map[y-1][x]==1){&#xA;                flip=num%2&#xA;                num=num&amp;gt;&amp;gt;1&#xA;            }&#xA;            else{&#xA;                flip=2-map[y-1][x]&#xA;            }&#xA;            if(flip){&#xA;                map[y-1][x]=2-map[y-1][x]&#xA;                map[y+1][x]=2-map[y+1][x]&#xA;                map[y][x-1]=2-map[y][x-1]&#xA;                map[y][x+1]=2-map[y][x+1]&#xA;                map[y][x]=2-map[y][x]&#xA;            }&#xA;        }&#xA;    }&#xA;    }&#xA;    for(y=0;y&amp;lt;10;y++){&#xA;    for(x=0;x&amp;lt;9;x++){&#xA;        if(!map[y][x]){&#xA;            return&#xA;        }&#xA;    }&#xA;    }&#xA;    console.log(org)&#xA;}&#xA;for(a=0;a&amp;lt;128;a++){&#xA;    trysol(a)&#xA;}&#xA;&amp;lt;/script&amp;gt;&#xA;&amp;lt;/body&amp;gt;&#xA;&amp;lt;/html&amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2186" LastEditorUserId="2186" LastEditDate="2015-03-21T06:39:56.287" LastActivityDate="2015-03-21T06:39:56.287" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10684" PostTypeId="2" ParentId="10668" CreationDate="2015-03-20T21:09:13.560" Score="7" Body="&lt;p&gt;I think it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; COMPUTER.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;: unfortunately this doesn't fit Hint 1. But with a little help from Aify, I've managed to make it fit everything else!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Let me spell this out for you,&lt;br&gt;&#xA; And see what you can do.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This tells us that it's going to be a letter-building riddle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I begin by being half full,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Half of the word &quot;complete&quot;, a synonym of &quot;full&quot; (the OP does like his synonyms!) is &lt;strong&gt;COMP&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Then onto you, don't be dull.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter &lt;strong&gt;U&lt;/strong&gt; sounds like &quot;you&quot;, so add that to our word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I'm around for a third of a year.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Credit to &lt;strong&gt;Aify&lt;/strong&gt; for this part...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Twelve months&quot; is a synonym of &quot;year&quot;, and one-third of its letters are T's and E's, so this gives us the letters &lt;strong&gt;TE&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Finally, without me, there is no gear.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter &lt;strong&gt;R&lt;/strong&gt; is contained in the word &quot;gear&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Let's see if you can puzzle how,&lt;br&gt;&#xA; Ultimately the riddle's now,&lt;br&gt;&#xA; Now ready for you to complete.&lt;br&gt;&#xA; Granted, it might be elite.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is probably a cheeky hint that the answer to the riddle is exactly what you're looking at right now!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Damn nice riddle by the way.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-21T00:25:10.550" LastActivityDate="2015-03-21T00:25:10.550" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="10685" PostTypeId="2" ParentId="10678" CreationDate="2015-03-20T21:34:11.230" Score="3" Body="&lt;p&gt;I think the answers are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (1,4) or (3,4)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first statement means that there are no numbers that could sum together to be Timothy's such that their product would determine their sum.  If any product determines either the sum or the numbers, then this cannot be the number.  This eliminates:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Any sum greater than or equal to 12.  If the sum is equal to 12, then the product 11*1=11 is uniquely defined as those digits.  If the sum is greater than 12 but less than 23, then taking one of the digits to be 11 and the other digit to be sum-11 causes a situation where Joseph knows the number.  As you cannot multiply 11 by 2 and get a possible number (and since 11 is prime), the factoring of this number is unique given the constraints. For sum &gt;23, do the same thing with 19 (also prim) and the factoring idea is the same. This leaves only 20 and 20, which I think is obvious that the product could only be determined by those two numbers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Further cases:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 10, consider the digits 3 and 7.  The product is 21, which is uniquely determined by these digits and Joseph knows.&#xA;For 8,6,4,3,2 use the property noted by Agawa where any sum that is a prime has a possible product that is uniquely determined. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are left with possible sums of 11, 9, 7, and 5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For these sums, we need to consider each case.  A possibility will be any sum that creates only one product where the sum of digits is a number that has not been eliminated.  This leaves the choices above.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if the sum were 7, then the possibilities are (1,6), (2,5) and (3,4).  If the numbers were (1,6) then the product is 6.  This could be obtained with either (1,6) or (2,3).  The sum of (2,3) is 5.  As Timothy would have claimed that Joseph couldnt' determine the numbers for either of these sums, Joseph cannot determine what the numbers are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the numbers were (2,5) then the product is 10.  This could be obtained with either (1,10) or (2,5).  Timothy would have similarly claimed that Joseph couldn't determine the numbers for either of these sums, and so Joseph cannot determine what the numbers are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the numbers were (3,4) then the product is 12.  This could be obtained with either (1,12) [sum of 13 -&gt; Timothy believes (11,2) is a possibility which means he would not claim Joseph couldn't determine numbers], (2,6) [which again has a sum of 8, which could be obtained by (7,1) and Timothy would not claim Joseph couldn't determine the numbers] or (3,4) [sum of 7.  Timothy WOULD claim Joseph couldn't determine the numbers].  As (3,4) is the only possibility for which Timothy would claim Joseph couldn't determine the numbers out of this product, then Joseph knows this is the answer.  As none of the other options for 7 lead to Joseph knowing this is the answer, now Timothy knows this is the answer too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 9, it breaks down because for both 1,8 and 2,7 Joseph would be able to determine the answer, which means that Timothy couldn't figure out which it was based on Joseph knowing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 11, it similarly breaks down for (3,8), (4,7) and (5,6).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 5,  only the choice (1,4) leads to Joseph determining the answer, so Timothy can figure this out too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: changed first elimination to state SUM instead of numbers per Duncan's comment.  Also updated first algorithm to change 20 to 19 and add note about primality of 11 and 19.  Also note that it's not possible to determine which number is S and which is G as both addition and multiplication are associative.  You can only figure out solutions where either number could be either number of coins.&lt;/p&gt;&#xA;" OwnerUserId="10457" LastEditorUserId="8874" LastEditDate="2015-03-22T15:42:16.753" LastActivityDate="2015-03-22T15:42:16.753" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10686" PostTypeId="2" ParentId="10682" CreationDate="2015-03-20T22:32:54.310" Score="4" Body="&lt;p&gt;This code is a modified version of the obscure &lt;a href=&quot;http://en.wikipedia.org/wiki/The_Gold-Bug&quot; rel=&quot;nofollow&quot;&gt;Gold-Bug&lt;/a&gt; cipher. Unfortunately the fact that the solution text is in a different language has meant that I can only give a partial answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Gold Bug cipher is a substitution cipher. Employing the substitutions in the book gives the partially decoded text:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;a7wbghr\$\$doaaa\$gtoibaflaabwghsufbafaaybissffdagca\$gilodah\$aaeggogwe\$fce\$ggadvwovbey[e\$cu[hdafvudabmvwyvbedvsdbo[w7\$buc[safu[mtdf[uayam[muaivu¢afp[hu&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that the \$, ¢, [ and 7 did not appear in the original story. The only letters that have not been used are J, K, Q, X, and Z, and I'd guess that four of those letters map to the 4 unknown symbols. Unfortunately the fact that it is in another language makes this difficult for me to do. Note that the ciphertext may have a transposition cipher applied on top of it, or it might have completely different substitutions for the symbols. It's hard to tell.&lt;/p&gt;&#xA;" OwnerUserId="7308" LastActivityDate="2015-03-20T22:32:54.310" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10687" PostTypeId="2" ParentId="10668" CreationDate="2015-03-20T22:55:59.550" Score="2" Body="&lt;p&gt;&lt;strong&gt;ANALYSIS&lt;/strong&gt; (And partial incomplete answer)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me spell this out for you&#xA;and see what you can do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I begin by being half full&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; FU (LL) - we use FU since having LL at the start of a word doesn't make sense. Option 2: we use the COMP part of &quot;complete&quot; since complete is a synonym for &quot;full&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then onto you, don't be dull&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; onto you likely means the letter &quot;U&quot;, or it can be &quot;on two u&quot; signifying a second &quot;U&quot; in the word&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm around for a third of a year&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is likely to be the synonym &quot;annual&quot; or &quot;annum&quot;, but only &quot;annual&quot; divides nicely into thirds, so it could be either an, nu, or al. It can also be interpreted as &quot;twelve months&quot;, which gives us several possibilities: 1) take a part of &quot;twelve months&quot; and use that; 2) the total number of T and E's make up a third of the phrase &quot;twelve months&quot;; 3) A third of twelve months is 4 months. Usage of the phrase &quot;four&quot; or &quot;triannual&quot; or &quot;Quadrimestrical&quot; or &quot;quad&quot; is possible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally, without me, there is no gear&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Either signifying one of the letters (G or R, likely) in gear - the other option here would be that if the hint was something to do with gears (either in a car, or in machines (cogs and gears)).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Taking into account the hint that the answer is shorter than &quot;short&quot;, which is 5 letters, the answer should be 4 characters or less. This means that because there are 4 clues, it's likely that each clue represents a single letter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First Letter: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Half full - half the letters in full are L's, so we start with the letter L.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second Letter:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Onto you - Letter U&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At this point we can start looking at this: &lt;a href=&quot;http://www.wordhippo.com/what-is/starting-with/4-letter-words-lu.html&quot; rel=&quot;nofollow&quot;&gt;http://www.wordhippo.com/what-is/starting-with/4-letter-words-lu.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third Letter:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; year, aka annual - it's the third letter, so a third of a year makes sense, and we can take the A.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Final Letter:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Finally, without me, there is no gear - this one's a stretch, but the letter &quot;u&quot; isn't in gear. TBH can't really make sense of this one&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Word:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Luau, a word for party. Why does this work? If you look at the line &quot;Ultimately, the riddle's now&quot;, it bears a striking resemblance to &quot;the party is now!&quot; If we replace the word &quot;riddle&quot; with luau, it seems more like something we'd hear in modern language; &quot;The luau (party) is now!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8999" LastEditorUserId="8999" LastEditDate="2015-03-20T23:41:56.997" LastActivityDate="2015-03-20T23:41:56.997" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10688" PostTypeId="1" AcceptedAnswerId="10693" CreationDate="2015-03-20T23:33:11.383" Score="-2" ViewCount="1150" Body="&lt;p&gt;My appearance is about as varied as it could be&lt;br&gt;&#xA;Unless you unveil me, you won't see the real me&lt;br&gt;&#xA;Without me, you life can become a pain after the act&#xA;but by having one of me,&lt;br&gt;&#xA;happiness can be extended which can be a fact&lt;br&gt;&#xA;My lifetime is short&lt;br&gt;&#xA;but your experience I shall support.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="10356" LastActivityDate="2020-10-16T06:42:28.953" Title="What common item am I" Tags="|riddle|word|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10689" PostTypeId="2" ParentId="10668" CreationDate="2015-03-20T23:35:11.510" Score="3" Body="&lt;p&gt;Here's a guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; WAR&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I begin by being half full,&#xA;Then onto you, don't be dull.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first letter is U+U = W, half of W is U.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm around for a third of a year.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If I'm using ANNUAL, the letter A composes 1/3 of it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally, without me, there is no gear.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Without the letter R, you can't complete the word gear. Also might be referencing pop culture games, or the need for gear when at war.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9696" LastActivityDate="2015-03-20T23:35:11.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10690" PostTypeId="1" AcceptedAnswerId="10691" CreationDate="2015-03-20T23:47:47.873" Score="7" ViewCount="241" Body="&lt;pre&gt;&lt;code&gt;624  --&amp;gt;   20&#xA;532  --&amp;gt;  117&#xA;634  --&amp;gt;  152&#xA;723  --&amp;gt;   40&#xA;443  --&amp;gt;  175&#xA;825  --&amp;gt;   ?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What should be filled in the last place so as to maintain the logical pattern?&lt;/p&gt;&#xA;" OwnerUserId="9865" LastEditorUserId="1664" LastEditDate="2015-03-21T03:15:50.717" LastActivityDate="2015-03-21T03:15:50.717" Title="Find the replacement for `?`" Tags="|pattern|number-sequence|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10691" PostTypeId="2" ParentId="10690" CreationDate="2015-03-20T23:55:27.840" Score="8" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;$39$&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; because the digits are used as follows:&lt;br&gt;&#xA; $6^2 - 4^2 = 20$&lt;br&gt;&#xA; $5^3 - 2^3 = 117$&lt;br&gt;&#xA; $6^3 - 4^3 = 152$&lt;br&gt;&#xA; $7^2 - 3^2 = 40$&lt;br&gt;&#xA; $4^4 - 3^4 = 175$&lt;br&gt;&#xA; $8^2 - 5^2 = 39$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-03-21T00:11:44.923" LastActivityDate="2015-03-21T00:11:44.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10693" PostTypeId="2" ParentId="10688" CreationDate="2015-03-21T00:11:58.113" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You're a condom, Sir! Please keep supporting my experiences.&lt;br&gt;&#xA; Explanation: Well the appearance can really vary as it takes the form of just about anything that fills it. You have to unveil it to see it for real, otherwise it's just an ordinary rubber. Life does become a pain after &lt;code&gt;the act&lt;/code&gt; unless you had one on! Happiness could as well be derived from pleasure which especially the ribbed ones extend for me. Its lifetime is short, well as short as a couple of seconds for certain. It does always support the experience before it dies in the trash.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well usually I stick to posting comments when I'm inebriated but a fellow SE'er motivated me to do it this time. &lt;/p&gt;&#xA;" OwnerUserId="9865" LastEditorUserId="9865" LastEditDate="2015-03-21T00:29:23.737" LastActivityDate="2015-03-21T00:29:23.737" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10694" PostTypeId="2" ParentId="10688" CreationDate="2015-03-21T00:41:26.047" Score="3" Body="&lt;p&gt;How about a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; bandaid&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My appearance is about as varied as it could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; bandaids have all kinds of printed patterns and some variation in shape&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unless you unveil me, you won't see the real me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; bandaids' real use is in covering wounds. The adhesive side has to be unveiled in order to be used&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without me, you life can become a pain after the act but by having one of me,&#xA;happiness can be extended which can be a fact&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; bandaids help by covering painful areas, decreasing the time someone might spend feeling hurt from a cut&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My lifetime is short&#xA;but your experience I shall support.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; bandaids don't last very long, no more than a day&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9696" LastActivityDate="2015-03-21T00:41:26.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10695" PostTypeId="2" ParentId="10668" CreationDate="2015-03-21T00:42:45.163" Score="1" Body="&lt;p&gt;New answer since my first one has been popular and fits everything except Hint 1.&#xA;&lt;hr&gt;&#xA;The hint tells us that we're looking for a word of 4 letters or less. Thus I think each of the 3rd, 4th, 5th, and 6th lines gives exactly one letter in the word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Let me spell this out for you,&lt;br&gt;&#xA; And see what you can do.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This tells us that it's going to be a letter-building riddle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I begin by being half full,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter &lt;strong&gt;L&lt;/strong&gt; forms half of the letters in the word &quot;full&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Then onto you, don't be dull.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm still sticking with my interpretation of this line as &lt;strong&gt;U&lt;/strong&gt; (which sounds like &quot;you&quot;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I'm around for a third of a year.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The word &quot;season&quot; can mean a year, e.g. the $n$th and $(n+1)$th seasons of a TV series are often separated by 1 year. One-third of the letters in this word are &lt;strong&gt;S&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Finally, without me, there is no gear.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm not sure on this line, but there aren't many 4-letter words beginning with LUS, so I'll go for &lt;strong&gt;T&lt;/strong&gt;,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;giving the final answer as&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LUST.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-03-21T00:42:45.163" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10696" PostTypeId="2" ParentId="10668" CreationDate="2015-03-21T00:45:22.310" Score="7" Body="&lt;p&gt;The answer is staring everyone in the face.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lung&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let me spell it out for you / And see what you can do&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; See end for explanation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I begin by being half full,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Half of &quot;full&quot; is &quot;l&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then onto you, don't be dull.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Onto &quot;you&quot;, or &quot;u&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm around for a third of a year.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Annual is used here. &quot;nn&quot; is 1/3 of &quot;annual&quot;, so &quot;n&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally, without me, there is no gear.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Can't have &quot;gear&quot; without &quot;g&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's see if you can puzzle how,&lt;br&gt;&#xA;Ultimately the riddle's now,&lt;br&gt;&#xA;Now ready for you to complete.&lt;br&gt;&#xA;Granted, it might be elite.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Take the first letters of each of the last four lines, and it's literally spelled out for you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4353" LastActivityDate="2015-03-21T00:45:22.310" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10697" PostTypeId="2" ParentId="6671" CreationDate="2015-03-21T04:02:19.503" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;C&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since by eliminating the other options.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since its doesn't have a round geometry whatsoever, A,D eliminated.&lt;br&gt;&#xA; B resembles to the 1st figure, therefore C would be a fair assumption.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10462" LastEditorUserId="35972" LastEditDate="2017-04-28T19:00:04.197" LastActivityDate="2017-04-28T19:00:04.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10698" PostTypeId="1" AcceptedAnswerId="10699" CreationDate="2015-03-21T06:11:34.623" Score="10" ViewCount="247" Body="&lt;p&gt;People often say that the power to do things comes from within.&lt;br /&gt;&#xA;In this puzzle, maybe I have taken that a bit too literally...&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/b1jm2.png&quot; alt=&quot;Grams of nono&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;I'm sure you will be able to find the sentence hidden inside this image.&lt;/strong&gt;&lt;/em&gt;&lt;br /&gt;&#xA;After all, the solution is within all of us ;)&lt;/p&gt;&#xA;" OwnerUserId="7308" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-03-21T06:24:52.087" Title="Puzzling comes from within" Tags="|logic-grid|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10699" PostTypeId="2" ParentId="10698" CreationDate="2015-03-21T06:24:52.087" Score="6" Body="&lt;p&gt;Message:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Caesar once said, &quot;I came, I saw, I conquered.&quot; (punctuation added)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Treat it like a pic-cross puzzle. The numbers for each column/row indicate which squares should be blanked out, with spaces in between. The spaces are the letters in the quote.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Image:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/DFEGJ.png&quot; alt=&quot;Solved image&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2015-03-21T06:24:52.087" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10700" PostTypeId="1" AcceptedAnswerId="10825" CreationDate="2015-03-21T11:52:08.357" Score="18" ViewCount="12121" Body="&lt;p&gt;You have a &lt;code&gt;1-mile long&lt;/code&gt; x &lt;code&gt;1-mile wide&lt;/code&gt; island. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When taking off from a flat runway, a plane requires a runway of minimum length 2 miles in order for takeoff to be physically possible, achieving minimum takeoff speed at the very end of the runway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remaining within the boundaries of the island, what shape and size do you build a runway to make it physically possible for the plane to take off? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The island is on the planet Earth and the usual laws of physics apply. No other form of transport is available. The plane cannot take off from water or perform any part of its takeoff run on water. Flight safety legislation does not apply.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;https://aviation.stackexchange.com/questions/13385/best-answer-for-a-puzzle-from-aeronautical-point-of-view&quot;&gt;Aviation.SE: Best answer for a puzzle from aeronautical point of view&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9626" LastEditorUserId="-1" LastEditDate="2017-04-13T13:00:01.880" LastActivityDate="2015-04-15T06:53:24.177" Title="Building a 2 mile runway on a 1x1 mile island" Tags="|mathematics|calculation-puzzle|lateral-thinking|open-ended|" AnswerCount="26" CommentCount="22" ClosedDate="2015-07-17T19:03:38.247" ContentLicense="CC BY-SA 3.0" />
  <row Id="10702" PostTypeId="2" ParentId="10700" CreationDate="2015-03-21T12:23:33.917" Score="11" Body="&lt;p&gt;Your longest horizontal path on the island is $\sqrt{2}$ miles (approx 1.4 miles), so there's no straight, horizontal solution on the island proper. (That is, what looks like the intended approach seems to have no valid answer.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use alternate transport (e.g. seaplane) or start a land reclamation project. Or make arrangements with a neighbouring island that happens to be 2 miles long :) .&lt;/p&gt;&#xA;" OwnerUserId="9422" LastActivityDate="2015-03-21T12:23:33.917" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10703" PostTypeId="2" ParentId="10700" CreationDate="2015-03-21T12:35:48.217" Score="6" Body="&lt;p&gt;There are several solution: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Aircraft_catapult&quot;&gt;&lt;strong&gt;Aircraft Catapult&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Build an &lt;strong&gt;artificial hill&lt;/strong&gt; on the corner of the island so that you can build a downhill runway, which can be longer than 2 miles if the hill is high enough.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;Build a &lt;strong&gt;circular runway&lt;/strong&gt; (max length is π) and let the airplane run on the track until it can take off.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-03-21T12:56:24.330" LastActivityDate="2015-03-21T12:56:24.330" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="10704" PostTypeId="2" ParentId="10700" CreationDate="2015-03-21T12:50:41.847" Score="26" Body="&lt;p&gt;You need to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; build a runway that goes diagonally through the island and is sqrt(2) miles higher in one corner than it's diagonally opposite corner.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The diagonal of the island is sqrt(2). You need a straight line of length 2, so we add in the third dimension. Using a height of sqrt(2), we get a hypotenuse of length exactly 2 miles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bonus:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Add additional height to both sides so that the plane does no hit the water at the end of the runway.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More boni inspired by the coments:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Make the runway curved or make a horizontal segment at the very end of the runway (in that case we'll need a the other segment to be a bit steeper).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Use a &lt;a href=&quot;https://en.wikipedia.org/wiki/Floatplane&quot;&gt;floatplane&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="5610" LastEditDate="2015-03-22T17:12:55.503" LastActivityDate="2015-03-22T17:12:55.503" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="10705" PostTypeId="2" ParentId="10700" CreationDate="2015-03-21T13:34:46.533" Score="-2" Body="&lt;p&gt;It is a circular runway that is inscribed within the 1 mile x 1 mile square with an approach tangent to the circle. &lt;/p&gt;&#xA;" OwnerUserId="9075" LastActivityDate="2015-03-21T13:34:46.533" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10706" PostTypeId="1" AcceptedAnswerId="10709" CreationDate="2015-03-21T14:40:39.587" Score="11" ViewCount="896" Body="&lt;p&gt;Vladimir wants to take his two daughters to their Dacha (weekend house) which is located 66 kilometers away from their home. Vladimir's motorbike has a maximum speed of 50 kilometers per hour; with one passenger (that is, driver plus one extra person), the maximum speed drops down to 40 kilometers per hour; the motorbike cannot carry three persons. Vladimir's walking speed is 9 kilometers per hour, and each daughter walks at a speed of 10 kilometers per hour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a schedule that makes all three of them (and the motorbike) reach the Dacha within three hours?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-15T08:58:49.747" LastActivityDate="2016-05-18T11:51:32.150" Title="Vladimir and his motorbike" Tags="|mathematics|calculation-puzzle|algebra|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10707" PostTypeId="2" ParentId="10700" CreationDate="2015-03-21T14:45:10.250" Score="4" Body="&lt;p&gt;If I was making a resort on such a tiny island, there's no way in the universe that I'd put a runway on it. The space it would take up would ruin the resort atmosphere of the island. That land would be needed to make the resort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;require people to come by boat or, as DMG said, by floatplane &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastActivityDate="2015-03-21T14:45:10.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10708" PostTypeId="2" ParentId="10700" CreationDate="2015-03-21T14:56:35.663" Score="5" Body="&lt;p&gt;Build a runway:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That starts in one corner of the island, goes straight for half a mile, then makes a half circle turn until it hits the other side of the island, then continues straight for half a mile. The half circle will have a circumference of 1/2 * 2pi*r, where r = 1/2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This gives it a length of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1/2 + 1/2 * 2*pi*1/2 + 1/2 = 1 + pi/2. This is greater than 2 (about 2.57), so it gives you some leeway.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It also has these advantages:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Leaving most of the space in the middle for the resort.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Having straightaways at the beginning/end of the runway for when the plane has the most speed (takeoff and landing) and can't turn as quickly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You better hope:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That the wind is blowing in the right direction, though.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastEditorUserId="1620" LastEditDate="2015-03-21T19:20:07.747" LastActivityDate="2015-03-21T19:20:07.747" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10709" PostTypeId="2" ParentId="10706" CreationDate="2015-03-21T15:37:32.420" Score="20" Body="&lt;p&gt;Vladimir starts with daughter 1 on the motorbike. After 48 km (that's one hour and 12 min at 40 kph), he stops and daughter 1 continues on foot. She reaches the dacha after exactly three hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vladimir drives back to meet daughter 2, who has already covered 12 km. They meet 36 min later (one hour and 48 min after the start), 18 km from the starting point (he drove 30 km at 50 kph and she walked 6 extra km at 10 kph). They still have 48 km to cover, which is one hour and 12 min at 40 kph (as we saw before, not coincidentally). They will therefore reach the dacha exactly three hours after the start.&lt;/p&gt;&#xA;" OwnerUserId="10469" LastEditorUserId="10469" LastEditDate="2015-03-21T15:47:40.293" LastActivityDate="2015-03-21T15:47:40.293" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10711" PostTypeId="2" ParentId="10706" CreationDate="2015-03-21T16:02:20.327" Score="10" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jt6A6.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This picture shows the best strategy to get to the Dacha.&lt;/p&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-03-21T16:02:20.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10712" PostTypeId="2" ParentId="10678" CreationDate="2015-03-21T16:57:29.327" Score="10" Body="&lt;p&gt;We are looking for two positive integers $G$ and $S$ with $2\le G,S\le20$ that fit with the conversation.&#xA;Let us denote the sum $s=G+S$ and the product $p=GS$. Note that $4\le s\le40$ and $4\le p\le400$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;TIMOTHY:&lt;/strong&gt; Joseph, I don't think that you can determine my sum!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(1) First of all, this statement implies that Joseph is not able to deduce the factors $S$ and $G$ from their product $p=GS$. This means that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$p$ is not the product of (exactly) two primes $p=ab$; &#xA;otherwise Joseph would immediately deduce that $\{G,S\}=\{a,b\}$.&lt;/li&gt;&#xA;&lt;li&gt;$p$ does not have any prime factor $f$ with $11\le f\le19$; &#xA;otherwise Joseph would deduce $\{G,S\}=\{f,p/f\}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(2) Secondly, this statement implies that for any possible way of writing $s$ as the sum of two summands $z$ and $s-z$, the corresponding product $z\cdot(s-z)$ does not allow Joseph to deduce $G$ and $S$.&#xA;Equivalently, for any integer $z$ with $2\le z\le s/2$, the corresponding product $z\cdot(s-z)$ cannot be of the form as excluded in (1). &#xA;This excludes most possible values for $s$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$s\le12$ and $s\ne11$: Then $s$ can be written as the sum of two primes ($4=2+2$, $5=2+3$, $6=3+3$, $7=2+5$, $8=3+5$, $9=2+7$, $10=5+5$ and $12=5+7$).&lt;/li&gt;&#xA;&lt;li&gt;$13\le s\le 31$: for the prime $z=11$, the product $z\cdot(s-z)$ has been excluded.&lt;/li&gt;&#xA;&lt;li&gt;$32\le s\le 39$: for the prime $z=19$, the product $z\cdot(s-z)$ has been excluded.&lt;/li&gt;&#xA;&lt;li&gt;$s=40$: then $G=S=20$, and Joseph easily deduces this from $p=400$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(3) Summarizing we see that only the case $s=11$ remains. Indeed, $s=11$ agrees with Timothy's statement:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\{G,S\}=\{2,9\}$: The product $2\cdot9=18$ allows the good factorizations $18=2\cdot9$ and $18=3\cdot6$.&lt;/li&gt;&#xA;&lt;li&gt;$\{G,S\}=\{3,8\}$: The product $3\cdot8=24$ allows the good factorizations $24=2\cdot12$ and $24=3\cdot8$ and $24=4\cdot6$.&lt;/li&gt;&#xA;&lt;li&gt;$\{G,S\}=\{4,7\}$: The product $4\cdot7=28$ allows the good factorizations $28=2\cdot14$ and $28=4\cdot7$.&lt;/li&gt;&#xA;&lt;li&gt;$\{G,S\}=\{5,6\}$: The product $5\cdot6=30$ allows the good factorizations $30=2\cdot15$ and $30=3\cdot10$ and $30=5\cdot6$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;JOSEPH:&lt;/strong&gt; Haha! You're a fool! Now I know your sum! The heritage is all mine!!!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(4) Joseph has thought through the matter as we did above, and hence is able to determine $s=11$. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;TIMOTHY:&lt;/strong&gt; I'm sorry for you, but if I'm a fool, then you are unwise! Now I know your product, and we'll have to share the legacy!  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(5) This statement is impossible: Timothy does not learn anything new from Joseph's preceding statement.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-03-21T16:57:29.327" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10713" PostTypeId="1" AcceptedAnswerId="10714" CreationDate="2015-03-21T17:03:51.177" Score="8" ViewCount="699" Body="&lt;p&gt;Sam is a gluttonous kid, so he decides to buy a colorful lollipop at the price of 1$ (one dollar) in the nearest candy store. Though, he realizes that he can't afford it as his budget is just 1¢ (a penny). &#xA;Anyway, Sam isn't going to give up, so he asks to his friend Tom to change his penny with a dollar. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Why should I?&quot;&lt;/em&gt;, asks Tom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sam replies: &lt;em&gt;&quot;One dollar is 100 pennies; 100 pennies are 10 pennies x 10 pennies; 10 pennies x 10 pennies are 1/10 of dollar x 1/10 of dollar; 1/10 of dollar x 1/10 of dollar is 1/100 of dollar; I just showed you that one dollar is equivalent to a penny.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How is Sam lying?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="8874" LastEditDate="2016-03-09T16:26:10.147" LastActivityDate="2016-03-09T16:26:10.147" Title="A dollar, a penny, there's no difference" Tags="|mathematics|calculation-puzzle|lateral-thinking|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10714" PostTypeId="2" ParentId="10713" CreationDate="2015-03-21T17:06:26.683" Score="13" Body="&lt;p&gt;Sam is simply mixing up the units: dollars, squares of dollar, pennies, squares of pennies. This yields a lot of nonsense.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sam replies: &quot;100 pennies are 10 pennies x 10 pennies.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This statement is wrong: 10 pennies x 10 pennies equals 100 penny-squared.&lt;br&gt;&#xA;(A correct statement would be: 10 times 10 pennies equals 100 pennies.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;1/10 of dollar x 1/10 of dollar is 1/100 of dollar&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is also wrong: 1/10 of dollar x 1/10 of dollar equals 1/100 dollar-squared.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-03-21T17:06:26.683" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10715" PostTypeId="1" AcceptedAnswerId="10716" CreationDate="2015-03-21T17:15:37.483" Score="15" ViewCount="1925" Body="&lt;p&gt;I found this puzzle which simply asks which way was the bicycle going, from its wheel marks on the mud shown in the image&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FoCJD.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now there is a simple way to find which mark(line) was made by the front and which was made by the rear wheel because&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;We know that the tangent drawn anywhere on the line drawn by the rear wheel will always cut the line made by the front wheel . And this is not guaranteed for the front wheel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;How can i use this information to solve it ?&lt;/p&gt;&#xA;" OwnerUserId="1538" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-02-04T00:52:29.073" Title="Which Way did the bicycle go?" Tags="|geometry|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10716" PostTypeId="2" ParentId="10715" CreationDate="2015-03-21T17:20:27.843" Score="14" Body="&lt;p&gt;If you look at the left (or the right)of your picture, the upper line shows the fore wheel. I could determine that because the fore wheel of a bicycle oscillates more than the rear one.&#xA;To determine the direction, choose some points in the rear wheel's trajectory and draw the tangents in those points. The distance between the forward and the back wheel is constant, so you just have to find out which of the segments determined by the tangent and the intersection with the other curve have the same length. If my paint drawing is correct, the left segment is always constant, so the bike is coming from &lt;strong&gt;RIGHT&lt;/strong&gt; (the direction of a bike is the vector from the aft wheel to the forward wheel.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Rvwp2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-03-21T19:26:23.473" LastActivityDate="2015-03-21T19:26:23.473" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10717" PostTypeId="1" AcceptedAnswerId="10721" CreationDate="2015-03-21T17:34:15.520" Score="10" ViewCount="492" Body="&lt;p&gt;My previous puzzle about &lt;a href=&quot;https://puzzling.stackexchange.com/questions/10271/movies-movies-movies&quot;&gt;movie titles&lt;/a&gt; was pretty well-received, so I've done another one. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This time it's on the theme of movie heroines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Duh6A.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Duh6A.png&quot; alt=&quot;Movie heroines - cryptic crossword&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The clues are in pairs, your task is to solve all of them and to match them up into their pairs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most (but not all) of the pairs are FIRSTNAME in one clue then LASTNAME in the other clue. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;ACROSS&lt;br&gt;&#xA;5. Don’t do a lot (9)&lt;br&gt;&#xA;6. Brightt redd (8)&lt;br&gt;&#xA;8. Mild-mannered photographer’s girlfriend stoned (4,4)&lt;br&gt;&#xA;10. Bearded FOSS-advocate (3)&lt;br&gt;&#xA;11. Kissed lifesaver looks drowned (5)&lt;br&gt;&#xA;14. Inventor of bouncy explosive makes an excellent partition (7)&lt;br&gt;&#xA;17. Top Gear’s hamster (7)&lt;br&gt;&#xA;18. Dancing under song (9)&lt;br&gt;&#xA;20. Overdo harassment a bit (1,4)&lt;br&gt;&#xA;21. Elementary sidekick (6)&lt;br&gt;&#xA;28. An unknown person, no saint, with a thousand dollars (7)&lt;br&gt;&#xA;29. One who cheats, dissembles, or defrauds (6)&lt;br&gt;&#xA;30. Tertiary explosive (8)&lt;br&gt;&#xA;32. Grow pale (7)&lt;br&gt;&#xA;34. Superhero poses questions to Socrates, to the east of Guernsey (8)&lt;br&gt;&#xA;36. Mike’s operatic solo (5)&lt;br&gt;&#xA;39. Medieval Roman 1004 can’t be found (3)&lt;br&gt;&#xA;40. Having little waves (6)&lt;br&gt;&#xA;42. Having head-lice, only one tonne (4)&lt;br&gt;&#xA;43. Oriental carbohydrate for an adder (7)&lt;br&gt;&#xA;44. Largely eponymous, moony heroine is lost at sea while still a maiden (7)&lt;br&gt;&#xA;45. Change ran pop tv (3,5)&lt;br&gt;&#xA;&lt;br&gt;&#xA;DOWN&lt;br&gt;&#xA;1. Two Russian girls under training (8)&lt;br&gt;&#xA;2. Changes are audible (10)&lt;br&gt;&#xA;3. Permit Victor both input and output (6)&lt;br&gt;&#xA;4. Will Smith’s agent before twos and threes (5)&lt;br&gt;&#xA;6. Sudden and surprising lack of beverage (8)&lt;br&gt;&#xA;7. One who marls, I hear (5)&lt;br&gt;&#xA;9. Irish John (6)&lt;br&gt;&#xA;12. Lima, November (5)&lt;br&gt;&#xA;13. How does the chatbot make you feel? (5)&lt;br&gt;&#xA;15. Leading a newt named Albert (4)&lt;br&gt;&#xA;16. Don’t come heavily (9)&lt;br&gt;&#xA;17. In the middle of a shallow phallocentric challenge (4)&lt;br&gt;&#xA;19. Swinger without whisky (6)&lt;br&gt;&#xA;22. South-African sun god (5)&lt;br&gt;&#xA;23. Aqua entomb (9)&lt;br&gt;&#xA;24. An angle, a cycle, a dent, or a pod (3)&lt;br&gt;&#xA;25. Modern style of music with heavy bassline is circular (6)&lt;br&gt;&#xA;26. Simpson (née Bouvier) (5)&lt;br&gt;&#xA;27. Full of holl (5)&lt;br&gt;&#xA;31. Gan again (5)&lt;br&gt;&#xA;33. Comes only once a year (9)&lt;br&gt;&#xA;35. Victor Victor (4)&lt;br&gt;&#xA;36. Neil in a burger (7)&lt;br&gt;&#xA;37. Notice or observe (5)&lt;br&gt;&#xA;38. Steal from ‘er then thank ‘er (7)&lt;br&gt;&#xA;41. Possess bromine (5)&lt;br&gt;&#xA;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have fun!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the spirit of teamwork, partial answers are very welcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're looking for a few easy ones to get you started, I suggest 8, 21, 26, 33, and 40.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update2&lt;/strong&gt;: As progress has stalled on the last couple of clues, I'm adding a couple of hints:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;One of the movie heroines is not a person.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;One of the movie heroines has a single name, which I've split in two at an arbitrary point.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2015-03-29T15:49:35.087" Title="Movie heroines!" Tags="|crosswords|cryptic-crosswords|" AnswerCount="2" CommentCount="7" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10718" PostTypeId="2" ParentId="10700" CreationDate="2015-03-21T17:39:43.023" Score="9" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Build a giant bowl and let the plane land initially at the top (albeit sideways). The centripetal force will keep the plane stuck to the inclined surface until it can dissipate enough speed doing loops around the bowl to settle at the bottom.&lt;/p&gt;&#xA;&lt;p&gt;Aircraft normally experience a max of around 2g during takeoff/landing so that's a nice benchmark to shoot for. Going to neglect gravity itself here since optimizing the roll angle to account for what a 747 could actually handle would take a bit of digging (especially the rudder as I doubt a jetliner could fly at 90 deg roll sustainably). With those simplifications though, a = (v^2)/r ---&amp;gt; r = 83^2/(9.81*2) = 354 meters. So with roughly half a mile of square space, you could land a 747 in your bowl. So long as the plane doesn't fall to pieces.&lt;/p&gt;&#xA;&lt;p&gt;All that being said, taking off is probably even harder since you now have to overcome that centripetal force to break away from the bowl...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;EDIT:&#xA;Could also double as amusement park ride.&lt;/p&gt;&#xA;" OwnerUserId="10472" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-03-21T19:05:28.757" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10719" PostTypeId="1" CreationDate="2015-03-21T18:39:03.347" Score="6" ViewCount="323" Body="&lt;p&gt;One evening you receive an e-mail from an acquaintance, asking for your help. The e-mail runs as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dear friend,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I need your help! I've forgotten the combination to my safe and I need some important documents that are inside. I believe I hid the code in the attached recipe. Here's what I can say (it may or may not help):&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;The recipe is written in my handwriting&lt;/li&gt;&#xA;  &lt;li&gt;The safe has 6 dials&lt;/li&gt;&#xA;  &lt;li&gt;Each dial has values ranging from 1 to 100&lt;/li&gt;&#xA;  &lt;li&gt;I have a cousin Alex, and an aunt named Lavinia&lt;/li&gt;&#xA;  &lt;li&gt;I remember the default combination was 1-2-3-4-5-6 (that's the same combination I have on my luggage!), but I've definitely changed it since then&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;Help! I called my dad and he doesn't even LIKE Caesar salad. I'm so confused.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The following recipe is attached to the e-mail:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dad's Favorite Caesar Salad&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Ingredients&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;1 bag iceberg greens&lt;/li&gt;&#xA;  &lt;li&gt;1 carton Alex's extreme simmered eggs&lt;/li&gt;&#xA;  &lt;li&gt;2 ounces olives&lt;/li&gt;&#xA;  &lt;li&gt;1 ounce shaved parmigiano-reggiano&lt;/li&gt;&#xA;  &lt;li&gt;1 measure Lavinia's dressing&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;Instructions:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Mix the ingredients in the appropriate order (the amounts are just the start). Get rid of anything Caesar wouldn't recognize, or if the amount is too much. Once mixed, garnish with a crouton (or an anchovy - it's the same).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can you help your friend out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HINT 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Follow the 3 steps of the instructions if you have trouble interpreting them, Hint 2 will be more explicit&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;HINT 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Here are the steps: Get the order correct, then figure out the numbers from the ingredients (Quark has a mostly-right idea), then &quot;garnish&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastEditorUserId="1620" LastEditDate="2015-03-23T14:19:15.290" LastActivityDate="2015-03-23T18:42:50.437" Title="Recipe for Success" Tags="|word-problem|cipher|story|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10720" PostTypeId="2" ParentId="10633" CreationDate="2015-03-21T18:41:21.227" Score="19" Body="&lt;h2&gt;Solution&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Here's the solution! Let's start from the encrypted string:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A oYzXqoELKjE 14&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The characters at the beginning and at the end of it suggest us something: they're hexadecimal numbers, and if we convert them to decimal we can see that $A_{16} = 10_{10}$ and $14_{16} = 20_{10}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the reversed audio suggests that &lt;strong&gt;$\pi$ is involved&lt;/strong&gt; in the answer, indeed it is. We'll consider its decimal digits, more precisely the ones &lt;strong&gt;between the 10th and the 20th&lt;/strong&gt; (inclusive): they are &lt;strong&gt;&lt;code&gt;58979323846&lt;/code&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, applying the &lt;strong&gt;&lt;a href=&quot;http://rumkin.com/tools/cipher/gronsfeld.php&quot; rel=&quot;noreferrer&quot;&gt;Gronsfeld cipher&lt;/a&gt;&lt;/strong&gt; to the encrypted string, and &lt;strong&gt;using the extracted digits as the encryption key, we can decode the video ID&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Encrypted:  oYzXqoELKjE - &#xA;Shift by:   58979323846 =&#xA;Decrypted:  jQqQhlCICfY&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And we got it! Here's the URL:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;http://www.youtube.com/watch?v=jQqQhlCICfY&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;6. Third hidden video - March 19, 2015&lt;/h2&gt;&#xA;&#xA;&lt;h3&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=jQqQhlCICfY&quot; rel=&quot;noreferrer&quot;&gt;0101 0101 0000 0110&lt;/a&gt;&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aFlha.png&quot; alt=&quot;sixth video&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=jQqQhlCICfY&quot; rel=&quot;noreferrer&quot;&gt;The video&lt;/a&gt;&lt;/strong&gt; itself consists of the usual &lt;code&gt;NOC+10&lt;/code&gt; intro, though the sound is distorted. Then, a text (transcribed below, with spacing intact) is displayed with a static noise, followed by a muffled explosion of some kind, which triggers an alarm bell (likely a ship's bell). The video ends right after another muffled explosion.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;      16.125628 147.681617&#xA;      I REQUIRE ASSISTANCE&#xA;ASSIMILATION OF SYSTEMS DECEASED&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;These information still make us believe in something related to a ship, a maritime tragedy. The numbers displayed seem to be a pair coordinates, but &lt;a href=&quot;https://www.google.it/maps/place/16%C2%B007&amp;#39;32.3%22N+147%C2%B040&amp;#39;53.8%22E/@16.125628,147.681617,17z/data=!3m1!4b1!4m2!3m1!1s0x0:0x0&quot; rel=&quot;noreferrer&quot;&gt;the presumed location&lt;/a&gt;, which is &lt;code&gt;16°07'32.3&quot;N, 147°40'53.8&quot;E&lt;/code&gt;, doesn't seem to be somewhere relevant but in the middle of the Pacific ocean, somewhere near Northen Mariana Islands. To be specific, the coordinates are precisely at the bottom of the Mariana Trench. The significance of this placement is unknown but seems unlikely to be coincidental.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/E1KZR.png&quot; alt=&quot;map&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, the presence of words like &quot;assistance&quot; or &quot;system deceased&quot;, together with the maritime theme, makes us think of a naval tragedy (maybe a sunken ship).&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10568" LastEditDate="2015-03-24T05:53:47.563" LastActivityDate="2015-03-24T05:53:47.563" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="10721" PostTypeId="2" ParentId="10717" CreationDate="2015-03-21T19:38:56.813" Score="4" Body="&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Baudelaire (Anagram of &quot;are audible&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Violet (V + I/O + let)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jones (Agent J + &quot;ones&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Doolittle (Don't do a lot, do a little)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6A.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Scarlett (Bright red with a double t)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6D.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Starling (Startling minus T)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;7.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Marla (One who marls)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;8.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mary Jane? (Peter Parker would be the mild mannered photographer)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;9.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jackie (Couple of Irish John Jackies out there. Wishful thinking?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;10.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RMS (Richard Stallman)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;11.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Annie (&lt;a href=&quot;http://en.wikipedia.org/wiki/Resusci_Anne&quot; rel=&quot;nofollow&quot;&gt;CPR Annie&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;12.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ellen (LN)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;13.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eliza (Emacs' psychiatric chatbot)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;14.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wallace (&lt;a href=&quot;http://en.wikipedia.org/wiki/Barnes_Wallis&quot; rel=&quot;nofollow&quot;&gt;Barnes Wallis&lt;/a&gt; + &quot;ace&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;16.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Golightly (come×go, heavily×lightly)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;17A.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hammond (&lt;a href=&quot;http://en.wikipedia.org/wiki/Richard_Hammond&quot; rel=&quot;nofollow&quot;&gt;Richard Hammond&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;17D.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hall (s&lt;strong&gt;hall&lt;/strong&gt;ow p&lt;strong&gt;hall&lt;/strong&gt;ocentric c&lt;strong&gt;hall&lt;/strong&gt;enge)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;18.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gunderson (Anagram of &quot;under song&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;19.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Singer (Swinger minus W)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;20.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; O Hara (&lt;strong&gt;O&lt;/strong&gt;verdo &lt;strong&gt;Hara&lt;/strong&gt;ssment)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;21.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Watson (Elementary, my dear Watson)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;22.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sarah (SA Ra)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;23.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Waterbury (aqua=water, entomb=bury)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;24.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tri (tricycle, trident, tripod)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;25.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dubois (Dubstep + circular + is)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;26.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Marge (Simpson was born Bouvier)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;27.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Holly (Holl-y)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;28.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Granger&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;29.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Connor (Conner)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;30.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Arrietty (anagram of &quot;tertiary&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;31.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Regan (Re-gan)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;32.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Blanche (make white/pale)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;33.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Christmas (&quot;I thought Christmas comes only once a year&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;34.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hermione (Herm (island east of Guernsey) + &lt;a href=&quot;http://en.wikipedia.org/wiki/Ion_%28dialogue%29&quot; rel=&quot;nofollow&quot;&gt;Ion&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;35.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Evey (V for victor?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;36A.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Maria (M + Aria)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;36D.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Macneil (Neil + Mac)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;37.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clock (One of the obscure meanings)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;38.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Roberta (Rob 'er &amp;amp; ta)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;39.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mia (Missing in action?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;40.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ripley (Ripple-y)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;41.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Brown (Own Br)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;42.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nity (&lt;a href=&quot;http://en.wiktionary.org/wiki/nit&quot; rel=&quot;nofollow&quot;&gt;nit&lt;/a&gt;-y)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;43.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clarice (Carry Lookahead Adder + Rice)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;44.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Titanic (titanic=large, at sea -&gt; ship, maiden -&gt; sank on first voyage)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;45.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Von Trapp (Anagram of &quot;ran pop tv&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Matches&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (6A-20) = Scarlett O'Hara, (26-18) = Marge Gunderson, (8-21) = Mary Jane Watson, (7-19) = Marla Singer, (12-40) = Ellen Ripley, (22-29) = Sarah Connor, (31-36D) = Regan Macneil, (27-16) = Holly Golightly, (36A-45) = Maria Von Trapp, (32-25) = Blanche Dubois, (35-17A) = Evey Hammond, (3-2) = Violet Baudelaire, (34-28) = Hermione Granger, (33-4) = Christmas Jones, (43-6D) = Clarice Starling, (13-5) = Eliza Doolittle, (30-37) = Arrietty Clock, (38-23) = Roberta Waterbury, (39-14) = Mia Wallace, (9,41) = Jackie Brown, (24,42) = Trinity, (10,44) = RMS Titanic, (11,17D) = Annie Hall&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9471" LastEditorUserId="9471" LastEditDate="2015-03-28T23:02:49.727" LastActivityDate="2015-03-28T23:02:49.727" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10722" PostTypeId="1" AcceptedAnswerId="10725" CreationDate="2015-03-21T19:56:11.340" Score="20" ViewCount="4823" Body="&lt;p&gt;Can you find a set of 5 words in English which all differ by just one vowel? This is easiest to explain with an example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Batter.  Better.  Bitter.  Botter.   Butter&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Close, but &quot;botter&quot; isn't a real word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's no upper or lower limit on word length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like the words to be in &lt;em&gt;fairly&lt;/em&gt; common usage - we shouldn't need to run to the dictionary to verify your claim. &#xA;It can be done, I have a few sets already, but I'm sure you can find more. I will mark as the answer the one with most sets/best sets/least obscure words.&lt;/p&gt;&#xA;" OwnerUserId="4963" LastActivityDate="2020-12-02T14:30:16.213" Title="Five words with one vowel difference" Tags="|word|english|" AnswerCount="7" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10723" PostTypeId="2" ParentId="10722" CreationDate="2015-03-21T20:02:46.967" Score="14" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Pat (a name, also a verb)&lt;br /&gt;&#xA; Pet (domestic animal)&lt;br /&gt;&#xA; Pit (hole)&lt;br /&gt;&#xA; Pot (the container)&lt;br /&gt;&#xA; Put (verb)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Also:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Bat (the animal)&lt;br /&gt;&#xA; Bet (verb related to gambling)&lt;br /&gt;&#xA; Bit (A little)&lt;br /&gt;&#xA; Bot (the larva of a parassite)&lt;br /&gt;&#xA; But (conjunction)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="63414" LastEditDate="2020-12-02T14:28:47.120" LastActivityDate="2020-12-02T14:28:47.120" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="10725" PostTypeId="2" ParentId="10722" CreationDate="2015-03-21T20:10:12.967" Score="29" Body="&lt;p&gt;I've actually been asked this question before: so I wrote a script to find all the answers!&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import re&#xA;&#xA;def read_dictionary(filename):&#xA;    dict = set()&#xA;    with open(filename, 'r') as f:&#xA;        for w in f:&#xA;            dict.update([w.rstrip().lower()])&#xA;    return dict&#xA;&#xA;def find_omnivocalics(dict, vowels=&quot;aeiou&quot;):&#xA;    assert len(vowels) &amp;gt;= 2&#xA;    omnivocalics = []&#xA;    for w in dict:&#xA;        for m in re.finditer(vowels[0], w):&#xA;            omnivocalic = True&#xA;            for v in vowels[1:]:&#xA;                if w[:m.start()] + v + w[m.end():] not in dict:&#xA;                    omnivocalic = False&#xA;                    break&#xA;            if omnivocalic:&#xA;                omnivocalics.append(w[:m.start()] + '*' + w[m.end():])&#xA;    omnivocalics.sort(key=lambda item: (len(item), item))&#xA;    print(&quot;, &quot;.join(omnivocalics))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Using the TWL Scrabble word list returns the following 84 word sets (sorted by length):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; *n, *s, m*, b*d, b*g, b*s, b*t, d*n, f*n, f*r, g*t, h*p, h*t, m*d, m*g, n*b, p*p, p*t, r*m, t*n, t*t, b*ds, b*gs, b*ll, b*nd, b*ts, c*re, c*te, d*ns, d*re, f*ns, h*ck, h*ed, h*ts, l*st, m*ds, m*gs, m*ll, m*re, m*ss, m*te, n*bs, p*ck, p*le, p*ns, p*ps, p*ts, r*ck, r*ms, t*le, t*ns, t*ts, b*lls, b*nds, c*res, c*tes, d*lly, h*cks, h*llo, m*lls, m*res, m*ssy, m*tes, p*cks, p*les, r*cks, t*les, b*lled, d*cker, h*llos, m*ssed, m*sses, p*tted, r*cked, b*gging, b*lling, bl*nder, d*ckers, h*lloed, h*lloes, m*ssing, p*tting, r*cking and h*lloing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Obviously, some of the words used are obscure. The best examples, in my opinion, are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; * blander/blender/blinder/blonder/blunder&lt;br&gt;&#xA; * patting/petting/pitting/potting/putting&lt;br&gt;&#xA; * masses/messes/misses/mosses/musses&lt;br&gt;&#xA; * packs/pecks/picks/pocks/pucks&lt;br&gt;&#xA; * last/lest/list/lost/lust&lt;br&gt;&#xA; * bag/beg/big/bog/bug&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Three of the examples above also allow the letter y, but use obscure words: m*, b*s and h*p. Turns out this sort of thing is charmingly called '&lt;strong&gt;vowel movements&lt;/strong&gt;'. See &lt;a href=&quot;http://digitalcommons.butler.edu/cgi/viewcontent.cgi?article=4642&amp;amp;context=wordways&quot;&gt;Vowel cascades, vowel movements and di-odes&lt;/a&gt; for an article about this with more examples. &lt;/p&gt;&#xA;" OwnerUserId="10220" LastEditorUserId="10220" LastEditDate="2015-03-21T20:18:01.933" LastActivityDate="2015-03-21T20:18:01.933" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="10727" PostTypeId="1" AcceptedAnswerId="10835" CreationDate="2015-03-21T21:05:11.783" Score="9" ViewCount="2203" Body="&lt;p&gt;The first term is f(0), the last term is f(2000):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0,0,0,0,1,0,1,0,2,1,1,0,2,0,1,1,3,0,2,0,2,1,1,0,3,1,1,2,2,0,2,0,4,1,1,1,3,0,1,1,5,0,2,0,2,2,1,0,4,1,2,1,2,0,3,1,5,1,1,0,3,0,1,2,6,1,2,0,2,1,2,0,4,0,1,2,2,1,2,0,4,3,1,0,5,1,1,1,5,0,3,1,2,1,1,1,6,0,2,2,3,0,2,0,5,2,1,0,4,0,2,1,4,0,2,1,2,2,1,1,4,1,1,1,2,2,3,0,7,1,2,0,5,1,1,3,5,0,2,0,3,1,1,1,6,1,1,2,2,0,3,0,5,2,2,1,5,0,1,1,8,1,4,0,2,2,1,0,4,1,2,2,2,0,2,2,9,1,1,0,4,0,2,1,5,1,2,1,2,3,2,0,7,0,1,2,3,0,5,0,4,1,1,1,5,1,1,2,9,1,3,0,2,1,1,1,6,1,1,1,5,1,2,0,8,3,1,0,5,0,2,2,5,0,5,1,2,1,2,0,8,0,2,4,2,2,2,1,5,1,3,0,4,1,1,2,10,0,2,1,5,2,1,0,4,1,2,1,2,0,4,0,9,2,1,2,5,0,1,2,4,0,2,0,2,2,2,1,7,1,2,1,2,0,3,1,5,5,1,1,4,1,1,1,9,1,5,0,3,1,2,0,9,0,1,3,2,0,2,1,7,1,2,1,6,2,1,1,5,1,3,0,2,2,1,1,8,0,2,1,5,1,5,2,5,2,1,0,5,0,3,5,8,0,2,1,2,2,1,0,6,1,1,2,3,1,2,0,9,2,2,1,5,0,2,3,5,1,4,0,5,1,1,0,10,2,1,2,2,0,3,1,4,1,1,1,4,0,1,2,6,0,2,1,2,4,2,1,9,0,2,1,2,1,5,1,8,1,2,0,4,0,1,2,5,2,2,1,2,2,2,0,7,0,2,2,2,1,2,0,4,3,2,0,5,1,1,1,7,0,4,1,2,1,1,2,9,0,1,5,5,0,3,0,9,2,1,0,4,1,2,1,5,1,2,2,5,2,1,0,7,1,1,2,3,1,6,0,5,1,3,0,5,1,2,3,9,1,2,0,4,1,1,0,8,1,2,2,2,0,5,1,11,5,1,1,5,1,2,1,4,0,5,0,2,3,1,1,8,1,2,2,5,1,2,1,5,1,1,2,6,0,1,1,12,1,3,0,2,2,3,1,9,1,1,2,2,0,5,1,6,2,1,0,5,1,1,4,5,0,5,0,3,1,2,2,10,0,2,1,5,1,2,1,5,3,1,0,4,1,2,1,9,0,4,2,2,1,2,0,8,0,2,2,2,2,2,0,12,2,2,1,4,0,1,2,4,0,2,0,5,5,1,1,8,3,1,2,2,1,4,0,5,1,1,1,5,2,2,2,13,0,2,0,5,2,2,0,7,1,3,2,2,0,2,1,9,2,2,0,4,0,1,2,5,2,5,1,2,1,2,1,7,0,1,4,3,0,2,1,4,1,2,0,9,1,3,1,9,1,5,0,2,3,1,1,9,1,1,1,4,0,4,1,14,2,1,1,5,0,2,2,5,1,3,2,2,1,1,0,7,1,2,1,2,2,3,0,4,6,2,1,5,0,1,3,12,1,5,0,5,2,2,0,9,1,1,2,3,1,4,0,9,1,2,1,6,0,1,2,4,0,2,1,2,5,1,1,11,0,3,1,2,0,5,2,5,2,1,1,4,1,2,5,6,1,2,0,2,1,2,1,8,1,1,2,2,0,3,1,14,2,1,1,5,2,2,1,5,0,6,0,5,1,2,1,8,1,1,3,5,0,2,0,5,3,2,0,9,0,2,1,14,2,2,1,3,5,1,0,6,1,1,1,2,2,5,2,9,1,3,1,5,0,2,5,5,0,3,0,5,2,1,0,13,1,1,2,5,1,5,1,5,2,2,3,5,0,1,1,15,0,4,0,3,2,1,0,9,1,2,4,2,1,2,1,13,2,1,1,6,1,2,2,5,1,2,0,2,2,3,0,8,1,1,2,2,1,4,0,16,1,1,1,4,2,1,2,12,0,5,2,2,1,1,2,8,0,2,1,5,0,2,1,9,4,2,0,5,1,3,1,4,0,5,1,2,2,1,1,13,1,2,2,2,1,5,0,4,2,2,0,7,1,1,3,9,0,2,1,4,2,1,0,9,1,2,2,3,1,4,0,12,1,2,1,5,0,1,5,15,2,2,1,2,2,1,1,7,0,2,1,5,0,3,2,5,2,1,0,4,0,2,2,17,2,4,1,2,3,2,0,9,0,2,5,5,1,2,0,15,1,1,1,9,2,1,1,5,0,4,0,2,4,2,1,14,1,2,1,5,0,5,0,4,2,2,1,5,0,2,3,9,1,2,2,2,1,3,1,7,1,1,2,2,2,2,0,14,3,2,0,4,0,1,2,5,0,5,1,4,1,2,0,8,2,2,5,2,0,5,1,5,2,1,1,9,0,2,1,14,1,3,0,2,4,1,2,9,0,2,2,2,1,6,1,9,1,1,1,4,1,1,2,4,1,2,1,5,1,3,0,11,0,1,3,3,1,2,1,16,5,2,0,5,1,2,1,9,1,4,0,2,2,1,2,6,1,2,2,5,0,2,2,12,2,1,0,8,1,3,1,5,0,2,1,3,3,1,1,18,0,1,1,5,1,5,1,5,2,3,1,5,0,1,6,14,0,5,1,5,2,2,0,8,3,1,1,2,0,5,0,6,2,1,2,5,0,1,2,16,1,4,1,2,2,2,1,14,0,4,2,2,1,3,1,5,1,2,0,8,1,1,1,9,2,2,1,2,5,2,1,9,1,3,3,5,0,5,0,19,2,1,0,5,1,1,3,4,0,5,0,3,1,1,2,10,0,2,1,4,0,5,0,5,5,1,0,5,2,2,2,12,1,5,1,5,1,1,0,15,0,1,4,2,2,3,0,9,1,3,2,9,1,2,2,5,1,2,1,5,2,2,1,13,1,1,1,2,1,6,1,4,2,1,1,5,1,2,2,15,0,2,1,5,3,1,0,8,1,2,1,4,0,2,3,12,4,2,1,4,0,1,1,5,1,4,1,2,1,2,1,12,1,2,5,2,1,2,0,15,1,1,1,8,1,2,2,13,0,5,1,2,2,2,1,9,1,1,2,5,2,5,0,9,3,2,0,4,0,3,5,5,0,2,2,2,1,1,0,20,1,2,2,3,2,2,0,5,5,5,0,4,0,1,2,6,1,7,0,5,1,2,2,9,1,1,2,2,1,4,0,14,1,2,2,9,1,1,2,16,0,3,0,5,4,1,0,12,0,2,2,2,0,5,2,4,1,2,0,6,1,1,2,12,2,2,1,5,1,2,0,7,1,1,2,2,1,3,2,15,3,1,0,5,2,2,1,5,1,4,0,2,2,2,1,17,1,1,4,4,1,2,0,5,2,1,2,9,0,5,2,9,0,5,1,2,2,2,0,15,1,2,1,3,1,9,0,14,1,2,0,5,2,1,4,5,1,2,0,5,2,2,0,7,1,2,2,2,1,5,1,5,5,1,2,4,0,2,2,21,0,5,1,2,2,2,0,9,0,3,2,5,0,2,2,9,3,1,0,7,0,1,1,16,3,2,0,5,2,2,1,14,1,2,2,2,0,4,1,16,1,1,1,5,2,1,5,9,1,4,1,5,2,1,1,8,0,1,2,5,1,2,0,13,5,3,0,5,0,2,1,4,1,9,2,2,2,1,1,18,1,2,3,2,1,2,1,5,1,3,1,9,0,3,2,12,0,2,0,4,6,2,1,9,2,1,1,5,0,4,1,9,1,1,3,4,1,1,2,16,0,5,0,2,3,1,1,11,2,2,1,2,0,3,1,16,2,2,1,4,0,2,2,9,1,5,0,3,2,4,1,9,0,1,4,2,1,2,0,18,1,1,1,6,1,1,2,4,1,5,2,2,2,1,2,12,0,2,1,5,1,8,0,5,3,2,0,5,0,2,2,19,1,3,1,2,1,2,1,18,0,2,1,5,2,5,1,9,5,2,0,5,2,1,3,5,1,5,1,4,1,1,0,14,2,2,5,2,1,5,0,5,2,2,1,8,1,1,1,15,1,2,1,2,5,2,0,6,1,5,1,2,1,5,2,14,1,1,2,9,0,3,4,5,1,2,0,2,2,3,0,14,0,1,4,5,0,2,0,16,3,1,1,5,2,2,2,12,0,6,1,5,1,1,1,9,1,2,2,4,0,2,1,15,2,1,0,9,1,2,3,5,0,3,1,2,5,2,1,19,1,2,1,5,3,5,1,5,1,2,0,4,0,1,5,6,1,3,1,5,1,1,1,10,1,2,2,2,0,4,0,12,5,1,2,5,1,2,1,15,1,5,1,2,2,1,1,12,1,2,5,3,0,5,2,4,1,2,0,6,1,1,1,14,1,2,0,5,3,2,1,9,0,1,3,2,0,9,0,18&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clue&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual values of the function matter only insofar as they assign integers into groups.  So concentrate on what the numbers in a group have in common.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't worry too much about the first two entries (for 0 and 1), they may be misleading.&lt;/p&gt;&#xA;" OwnerUserId="2154" LastEditorUserId="2154" LastEditDate="2015-03-23T08:05:28.853" LastActivityDate="2015-03-23T21:05:36.580" Title="What is the next term in the sequence?" Tags="|calculation-puzzle|number-sequence|" AnswerCount="2" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10728" PostTypeId="2" ParentId="10700" CreationDate="2015-03-21T21:13:45.727" Score="15" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;You have a &lt;code&gt;1 mile long&lt;/code&gt; x &lt;code&gt;1 mile wide&lt;/code&gt; private island that you wish to&#xA;  turn into a resort. A plane requires a 2-miles long runway to take&#xA;  off. What do you do?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Have tourists arrive by boat or helicopter.&lt;/li&gt;&#xA;&lt;li&gt;Build a &lt;a href=&quot;http://en.wikipedia.org/wiki/Floating_airport&quot;&gt;floating airport&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Artificial_island&quot;&gt;Artificially extend&lt;/a&gt; the island's area.&lt;/li&gt;&#xA;&lt;li&gt;Use &lt;a href=&quot;http://en.wikipedia.org/wiki/Seaplane&quot;&gt;seaplanes&lt;/a&gt;, which can take off or land on water.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-03-21T21:13:45.727" CommentCount="0" CommunityOwnedDate="2015-03-21T21:13:45.727" ContentLicense="CC BY-SA 3.0" />
  <row Id="10731" PostTypeId="2" ParentId="10727" CreationDate="2015-03-21T22:17:26.493" Score="8" Body="&lt;p&gt;I had thought the n'th element of the sequence (starting at 0) was one less than the number of prime factors of n, handling 0 and 1 separately, but some numbers give greater values, all of them having at least four factors. For all values not listed, the prediction is correct.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Number   Value   #Fac-1  Factorization&#xA;40       5       3       [2, 2, 2, 5]&#xA;56       5       3       [2, 2, 2, 7]&#xA;64       6       5       [2, 2, 2, 2, 2, 2]&#xA;84       5       3       [2, 2, 3, 7]&#xA;88       5       3       [2, 2, 2, 11]&#xA;96       6       5       [2, 2, 2, 2, 2, 3]&#xA;104      5       3       [2, 2, 2, 13]&#xA;128      7       6       [2, 2, 2, 2, 2, 2, 2]&#xA;132      5       3       [2, 2, 3, 11]&#xA;136      5       3       [2, 2, 2, 17]&#xA;144      6       5       [2, 2, 2, 2, 3, 3]&#xA;152      5       3       [2, 2, 2, 19]&#xA;156      5       3       [2, 2, 3, 13]&#xA;160      8       5       [2, 2, 2, 2, 2, 5]&#xA;176      9       4       [2, 2, 2, 2, 11]&#xA;184      5       3       [2, 2, 2, 23]&#xA;192      7       6       [2, 2, 2, 2, 2, 2, 3]&#xA;198      5       3       [2, 3, 3, 11]&#xA;204      5       3       [2, 2, 3, 17]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It's particularly weird that $2^k$ gives $k-1$, for $k\leq 4$, but $2^6$ gives $6$ and $2^7$ gives $7$.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;k    f(2^k)&#xA;1    0&#xA;2    1&#xA;3    2&#xA;4    3&#xA;5    4&#xA;6    6&#xA;7    7&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There's also the weird fact that $f(40)=5$ and $f(80)=4$, so it's not even monotonic in the divisibility partial order.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-03-21T22:30:47.583" LastActivityDate="2015-03-21T22:30:47.583" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10733" PostTypeId="2" ParentId="10700" CreationDate="2015-03-21T23:35:13.903" Score="0" Body="&lt;p&gt;I can think of several approaches:&lt;br&gt;&#xA;1. If you must use planes, you can extend the island out to meet the requirement.&lt;br&gt;&#xA;2. You can use &lt;code&gt;seaplanes/floatplanes&lt;/code&gt;, so there's no need for runway, since the ocean is your runway.&lt;br&gt;&#xA;3. You can use &lt;code&gt;vtols&lt;/code&gt;, so you just need a flat area large enough to fit your plane.&lt;/p&gt;&#xA;" OwnerUserId="10477" LastEditorUserId="9626" LastEditDate="2015-03-22T09:27:10.493" LastActivityDate="2015-03-22T09:27:10.493" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10734" PostTypeId="2" ParentId="10700" CreationDate="2015-03-22T00:51:44.313" Score="5" Body="&lt;p&gt;Mobius Landing Strip FTW! :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That or cut the island into fractions and stick it to itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also you could have an extending drawbridge type runway that retracts or one that has buoys to float with in the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My favorite for the fancy resort factor would be a runway that is in a completely glass tube that goes underwater and takes you to the under ground grand entryway of the resort&lt;/p&gt;&#xA;" OwnerUserId="10478" LastEditorUserId="10478" LastEditDate="2015-03-23T06:20:26.457" LastActivityDate="2015-03-23T06:20:26.457" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10735" PostTypeId="2" ParentId="10700" CreationDate="2015-03-22T00:53:36.023" Score="0" Body="&lt;p&gt;You can use a rotating platform at the end of the runway (on a corner of the island).&#xA;Thus when the plane enters onto the circular platform, it rotates by 180 degrees while your plane continues where it came from.&lt;/p&gt;&#xA;" OwnerUserId="10479" LastEditorUserId="5373" LastEditDate="2015-03-24T00:00:24.123" LastActivityDate="2015-03-24T00:00:24.123" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10740" PostTypeId="2" ParentId="10700" CreationDate="2015-03-22T03:14:35.873" Score="0" Body="&lt;p&gt;Ideally this runway should be a shallow &quot;skateboard ramp&quot; design. Lizard Island on the Great Barrier Reef is a good example. Aircraft need to be able to taxi uphill at both ends. Such a perfect half tube would be 2.22 miles. So I probably would increase the radius a tad. Or implement a tractor tow system. &lt;/p&gt;&#xA;" OwnerUserId="10483" LastEditorUserId="5373" LastEditDate="2015-04-13T11:05:56.020" LastActivityDate="2015-04-13T11:05:56.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10743" PostTypeId="2" ParentId="10719" CreationDate="2015-03-22T05:37:03.193" Score="3" Body="&lt;p&gt;Attempt at a glance:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2, 73, 59, 60, 13, 64&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Method:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First, a focus is placed on &quot;get rid of anything Caesar wouldn't recognize&quot;, basically look for the roman numerals. The names Alex and Latvinia seem to purposely contain a majority of roman numerals (i=1, v=5, x=10, l=50). The &quot;or if the amount is too much&quot; is what I assume to mean throw away any values 100 or over. Caesar ciphers normally involve adding to values which is why I add the values in the 6 ingredients (olives used twice) on top of the original 1,2,3,4,5,6. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bit messy of an answer because not too confident about it.&lt;/p&gt;&#xA;" OwnerUserId="7722" LastActivityDate="2015-03-22T05:37:03.193" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10744" PostTypeId="1" CreationDate="2015-03-22T07:12:18.643" Score="-2" ViewCount="230" Body="&lt;p&gt;It rains in Gadyukino village every other day on average.  Assume the probability of rain on any given day is 50% and independent. A local weather forecaster has the sole duty every day of predicting whether it will rain the following day.  His predictions have a 75% success rate. What quantity of information (in bits) is carried by a single forecast of his?&lt;/p&gt;&#xA;" OwnerUserId="7668" LastEditorUserId="4" LastEditDate="2015-04-03T23:26:10.813" LastActivityDate="2015-04-03T23:26:13.773" Title="It rains in Gadyukino" Tags="|mathematics|probability|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10745" PostTypeId="2" ParentId="10744" CreationDate="2015-03-22T08:09:13.833" Score="0" Body="&lt;p&gt;It rains on alternate days on average, which means that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If today it's sunny, it's likely that it will rain tomorrow.&lt;/li&gt;&#xA;&lt;li&gt;If today it's raining, it's likely that it's sunny tomorrow.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The forecasters peruse the recent weather statistics and discover that 75% of the sunny days were followed by a rainy day. &#xA;So, if today it's sunny, we have a 75% probability of rain tomorrow.&#xA;Why we assumed that today it's sunny? Because it says that rains occurs every other day, and if today it's rainy, a 75% chance of rain for tomorrow would conflict with the assumptions.&lt;/p&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-03-22T08:09:13.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10746" PostTypeId="2" ParentId="10602" CreationDate="2015-03-22T08:37:55.687" Score="4" Body="&lt;p&gt;It may be:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bolinus Brandaris (murex) or Ink, as you can extract purple ink from the murex. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Indeed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A relationship so justified  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It has a shell, Justified used as the format of text&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sensitive in nature deemed classified&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's actually a animal, it reacts to stimuli&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Stained the world in all aspects&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It was originally used to make purple ink&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Likely to start future prospects&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The purple ink was used to dye precious fabrics, which were used by rich people during important discussions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Vandalism left still enduring&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Physical marks keep alluring&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After extracting the ink, only the dead shell remains.&#xA; A timeless art made from bone&#xA; &#xA; As already said, the valuable purple color was made from a shell, which can be compared to a bone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In Rome a cephalopod you might own&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Romans were the main buyers of murex.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Arch the enchanters wand&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It alludes to the pen, used along with the ink.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One which Edward would be fond  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ( Edward Bulwer-Lytton ) who coined &quot;the pen is mightier than the sword&quot;, also referred to the pen as an arch enchanter's wand.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The blood runs with absolute  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The ink can be considered the murex's blood.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No matter if you are mute&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ink is mute&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="9000" LastEditDate="2015-03-23T13:30:15.733" LastActivityDate="2015-03-23T13:30:15.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10747" PostTypeId="2" ParentId="10602" CreationDate="2015-03-22T09:11:41.977" Score="1" Body="&lt;p&gt;I think it is a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Poem&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A relationship so justified&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sensitive in nature deemed classified&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Poems are more on &lt;a href=&quot;https://www.youtube.com/watch?v=0vK2Baco46A&quot; rel=&quot;nofollow&quot;&gt;emotions&lt;/a&gt; so give meaning to relationships and can be classified based on 9 emotions (romance, heroism, comic, pathos, wrathful, terrifying, odious and wonderous)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Stained the world in all aspects&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; poetry over time impacted world and brought changes. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Likely to start future prospects&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; same as above&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Vandalism left still enduring&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are thousand cases you can find where poets were killed, there works were burnt, poetry was outlawed but still still survives.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Physical marks keep alluring&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Used as epithets or other physical forms&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A timeless art made from bone&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Even after 1000s of years many poetry are there and still very effective very strong&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In Rome a cephalopod you might own&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is slippery for some&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Arch the enchanters wand&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; like a magic&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One which Edward would be fond&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Even king Edward loved it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The blood runs with absolute&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No matter if you are mute&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Even if you can not speak you can enjoys it (as still you can hear it)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10210" LastEditorUserId="10210" LastEditDate="2015-03-22T10:48:08.083" LastActivityDate="2015-03-22T10:48:08.083" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10748" PostTypeId="2" ParentId="10700" CreationDate="2015-03-22T11:15:13.903" Score="20" Body="&lt;p&gt;Build a runway that goes along one side of the island, turns in a &lt;em&gt;banked&lt;/em&gt; quarter circle with a small but sufficiently large radius, and then goes along the adjacent side. If it were horizontal the length would be slightly less than 2 miles, so raise it at the end where the aeroplane will be stationary. The key word is &lt;em&gt;banked&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7363" LastActivityDate="2015-03-22T11:15:13.903" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10749" PostTypeId="2" ParentId="10700" CreationDate="2015-03-22T11:20:18.437" Score="2" Body="&lt;p&gt;Reduce the problem to fitting a 2-mile length into a length of 1.414 miles. Let's say we don't want any bends. How do we do it? Use rollers! Say each roller has a diameter of 1 yard. Make them of a suitable material and construct the holding mechanism in such a way that when the plane travels 1 yard relative to the island its wheels have been in contact with the roller for 1.414 yards.&lt;/p&gt;&#xA;" OwnerUserId="7363" LastActivityDate="2015-03-22T11:20:18.437" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10750" PostTypeId="2" ParentId="10019" CreationDate="2015-03-22T11:49:21.943" Score="3" Body="&lt;p&gt;I'm going to try&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; EXACT.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I should begin by saying I am a word&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This doesn't tell us any more than the &lt;a href=&quot;/questions/tagged/word&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged 'word'&quot; rel=&quot;tag&quot;&gt;word&lt;/a&gt; tag.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In the beginning I followed established conventions&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Take the beginning of the phrase &quot;established conventions&quot;: the letter &lt;strong&gt;E&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;After that I decided that was not for me&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After you've done something and then decided it wasn't for you, you're an &lt;strong&gt;ex&lt;/strong&gt;-doer of that thing. So let's add an &lt;strong&gt;X&lt;/strong&gt; to our word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;So I decided to oppose everything that came after me&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure about this line. Maybe take the first letters of the words &quot;came after&quot; give CA, which reverse (&quot;oppose&quot;) to &lt;strong&gt;AC&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In the end I decided not to oppose, but to protect&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &quot;end&quot; of this line is the letter &lt;strong&gt;T&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Putting all this together, we get the answer I said at the start ... which also fits the title pretty well.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-03-22T11:49:21.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10751" PostTypeId="2" ParentId="10111" CreationDate="2015-03-22T11:54:53.630" Score="4" Body="&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HARRISON FORD.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll help you get across a river. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are a few things that do this... a bridge, &lt;strong&gt;a ford&lt;/strong&gt;, a boat, a raft, Charon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once unrecognised in Fischer's song, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ford was uncredited on a film called: &quot;Dead Heat on a Merry-Go-Round&quot;. Wild Man &lt;em&gt;Fischer&lt;/em&gt;'s most famous song is called &lt;em&gt;Merry Go Round&lt;/em&gt;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later I found fame alone &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ALONE -&gt; SOLO. Han Solo was played by Harrison Ford.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And also as a US state.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Indiana Jones is one of Ford's main sources of fame.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The title:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you are not Lewis-off... maybe you are Harris-on :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="5373" LastEditDate="2015-03-22T11:56:14.313" LastActivityDate="2015-03-22T11:56:14.313" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10752" PostTypeId="2" ParentId="10700" CreationDate="2015-03-22T12:58:18.630" Score="45" Body="&lt;p&gt;Build two straight runways side to side, 1 mile long each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, put a portal-conducting surface at the end of the first runway and another one at the beginning of the second one. Finally, use your portal gun to place a yellow portal on one of those surfaces and a blue portal in the other one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/On73S.png&quot; alt=&quot;Island 3D rendering&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5044" LastEditorUserId="5044" LastEditDate="2015-03-22T13:39:50.073" LastActivityDate="2015-03-22T13:39:50.073" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="10753" PostTypeId="1" AcceptedAnswerId="10754" CreationDate="2015-03-22T13:05:14.373" Score="10" ViewCount="76332" Body="&lt;p&gt;I have a rubik cube, which always have two swapped edges in the last layer. No matter which color I start with, there are always two edges in the last layer which remained swapped with each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to solve this scenario. Or if this is an invalid scenario, please can somebody explain how is this an invalid scenario.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See this image for sample: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5lSjp.jpg&quot; alt=&quot;http://m.imgur.com/u8P9Sag&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10494" LastEditorUserId="4729" LastEditDate="2015-06-02T08:03:24.560" LastActivityDate="2017-07-27T03:55:22.607" Title="Swap the edges in a solved Rubik's cube" Tags="|rubiks-cube|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10754" PostTypeId="2" ParentId="10753" CreationDate="2015-03-22T13:20:29.383" Score="19" Body="&lt;p&gt;In a Rubik's cube, every legal move swaps a even number of dowels, so any legal configuration can be obtained only with a even number of swaps.&#xA;In this configuration, the difference between a legal cube (the solved one) and the current status consists of 1 swap; since 1 is odd, this is a &lt;strong&gt;No Win Scenario&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-03-22T14:13:25.927" LastActivityDate="2015-03-22T14:13:25.927" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10755" PostTypeId="2" ParentId="10722" CreationDate="2015-03-22T13:37:15.680" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;     Bag&#xA;     Beg&#xA;     Big&#xA;     Bog&#xA;     Bug&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;A more dubious example would be&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;     Hat&#xA;     Het (as in &amp;quot;don't get all het up&amp;quot;)&#xA;     Hit&#xA;     Hot&#xA;     Hut&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Or even something which just occurred to me (which I see was hidden in a previous answer as l*st)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Last&#xA; Lest&#xA; List&#xA; Lost&#xA; Lust&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10495" LastEditorUserId="63414" LastEditDate="2020-12-02T14:30:16.213" LastActivityDate="2020-12-02T14:30:16.213" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="10756" PostTypeId="2" ParentId="10700" CreationDate="2015-03-22T13:47:11.987" Score="15" Body="&lt;p&gt;Alternate answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question says the plane needs a 2 mile runway &lt;em&gt;to take off&lt;/em&gt;, but says nothing about landing. So  have people fly in, land on a short runway, then dump the plane in the ocean to make room for the next one. If people insist on leaving, they can swim for it.&lt;/p&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2015-03-22T13:47:11.987" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10757" PostTypeId="2" ParentId="10744" CreationDate="2015-03-22T13:59:25.150" Score="4" Body="&lt;p&gt;&quot;How much information&quot; sounds vague. But it has a rock-solid definition, in the grand tradition of Shannon et al.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume an independent 1/2 chance of rain every day. Then the weather transmits one bit of information to Gadyukino every day - namely, whether or not it is raining. Say the forecaster can receive $x$ bits of information per day about the future. Then what is the minimum value of $x$ so that in the limit, the forecaster predicts with 75% accuracy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a time period of $4n$ days for large $n$. There are $2^{4n}$ possible weathers during this period. For any particular forecaster prediction sequence, the number of weather sequences that let the forecaster claim a 75% accuracy rate is $\binom{4n}{\leq n}$, where the notation means the number of ways to choose at most $n$ wrong days out of $4n$ total days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stirling's approximation says that $\binom{4n}{\leq n} = 2^{(4\log 4 - 3 \log 3)n}\times poly(n)$. Call this quantity $X$ for ease of typing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each forecaster prediction sequence allows $X$ possible weather sequences. There are $2^{4n}$ total weather sequences. Therefore, the forecaster must have at least $2^{4n}/X = 2^{(4 - 4 \log 4 + 3 \log 3)n}$ possible prediction sequences that he might say.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to be able to choose from that many prediction sequences, the forecaster must receive at least $\log(\text{that many})=(4 - 4 \log 4 + 3 \log 3)n$ bits for those $4n$ days. &lt;strong&gt;(EDIT: This doesn't account for the fact that the forecaster can change his mind based on whether he's been right so far. My intuition says an information-theoretic bound should still work, I'll think about it.)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works out to $\boxed{1-\log 4 + \frac 34 \log 3 \approx 0.43766...}$ bits per day. This is an upper bound on the amount of information the forecaster needs. Does this much information suffice? I don't know. Shannon probably solved this problem when he created information theory, but I'm having trouble finding a reference.&lt;/p&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2015-03-22T13:59:25.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10758" PostTypeId="1" AcceptedAnswerId="10760" CreationDate="2015-03-22T14:16:21.007" Score="8" ViewCount="747" Body="&lt;p&gt;Mary has a box with special $2\times1$ dominoes. Each dominoe has two red corners and two blue corners, and these dominoes come in two different types: The first type has the lower left and the upper right corner in red, while the second type has the upper left and the lower right corner in red. In other words, the second type is the mirror image of the first type. The box contains $20$ dominoes of the first type and $12$ dominoes of the second type.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is it possible to tile a standard $8\times8$ chessboard with Mary's dominoes, so that no red dominoe corner touches any red corner of another dominoe? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The dominoes may be rotated in the tiling, but they must not be flipped over.&#xA;Two dominoe corners may touch horizontally, vertically, or diagonally.)&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-03-22T14:24:16.420" LastActivityDate="2015-03-23T03:19:47.397" Title="Dominoes on a chessboard" Tags="|checkerboard|tiling|dominoes|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10760" PostTypeId="2" ParentId="10758" CreationDate="2015-03-22T14:40:10.377" Score="9" Body="&lt;p&gt;&lt;strong&gt;Yes, it is possible.&lt;/strong&gt; One configuration is as follows. The 20 horizontally oriented are all of one type; the 12 lowercase dominos are the other type.  With these orientations, all upper-left and lower-right corners (in absolute terms) are red, and all lower-left and upper-right corners are blue.  Also, there are no four-corners intersections in the configuration; so a lower-left corner can never touch an upper-right corner, not a lower-right an upper-left.  This ensures that no two corners of the same colour touch.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; ___ ___ ___ ___&#xA;|___|___|___|___|&#xA;| |___|___|___| |&#xA;|_| |___|___| |_|&#xA;| |_| |___| |_| |&#xA;|_| |_|___|_| |_|&#xA;| |_|___|___|_| |&#xA;|_|___|___|___|_|&#xA;|___|___|___|___|&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I suspect this is the only possible solution.  I can’t quite show that, but I can show: in any suitable arrangement, all the dominos of one type must be vertical, and all those of the other type horizontal, or equivalently, all the UL and BR corners in the arrangement must be one colour, all the UR and BL corners another colour.  (Proof: if not, look for the upper-most left-most domino whose UL corner is &lt;em&gt;not&lt;/em&gt; the same as the UL corner of the whole board.  What could happen around the UL corner of this domino?  There are no valid possibilities.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that this is fixed, an arrangement will have no clashing corner colours precisely if it has no “four-corners” intersections.&lt;/p&gt;&#xA;" OwnerUserId="4234" LastEditorUserId="4234" LastEditDate="2015-03-23T03:19:47.397" LastActivityDate="2015-03-23T03:19:47.397" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10761" PostTypeId="2" ParentId="10758" CreationDate="2015-03-22T14:48:10.077" Score="9" Body="&lt;p&gt;Here's an ugly paint drawing of a working pattern.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/yEQZ2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="5556" LastEditDate="2015-03-22T23:33:43.900" LastActivityDate="2015-03-22T23:33:43.900" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10762" PostTypeId="1" CreationDate="2015-03-22T14:55:21.843" Score="1" ViewCount="161" Body="&lt;p&gt;We know that classical puzzle of getting 4 liters from 2 jugs of 3 liters and 5 liters capacity. (or was it gallons?) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about other volumes? And what if the jugs had different volumes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, if you had a 7 liter jug and an 11 liter jug how can you get 9 liters exact?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another one: If you had a 13 liter jug and a 5 liter jug, how can you get 9 liters exact?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if these problems are solved before I come back again, what if you can make markings on the jugs? How does the game change?&lt;/p&gt;&#xA;" OwnerUserId="9736" LastEditorUserId="4050" LastEditDate="2015-03-22T15:19:21.750" LastActivityDate="2015-03-22T15:19:21.750" Title="The 2 jugs again?" Tags="|mathematics|decanting-problems|" AnswerCount="2" CommentCount="0" ClosedDate="2015-03-22T20:25:21.243" ContentLicense="CC BY-SA 3.0" />
  <row Id="10763" PostTypeId="2" ParentId="10762" CreationDate="2015-03-22T15:06:19.880" Score="2" Body="&lt;p&gt;For 7 and 11 liters:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Just keep filling the 11 and pouring it into the 7. When the 11 is empty, refill it. When the 7 gets full, empty it. You'll hit 9 eventually. The amount of water in each goes through the following progression (11 liter first):&#xA;&#xA; 11, 0; 4, 7; 4, 0; 0, 4; 11, 4; 8, 7; 8, 0; 1, 7; 1, 0; 0, 1; 11, 1; 5, 7; 5, 0; 0, 5; 11, 5; 9, 7&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For 13 and 5 liters:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Do the same thing. The amount of water in each goes through the following progression (13 liter first):&#xA;&#xA; 13, 0; 8, 5; 8, 0; 3, 5; 3, 0; 0, 3; 13, 3; 11, 5; 6, 5; 6, 0; 1, 5; 1, 0; 0, 1; 13, 1; 9, 5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There may be faster ways, but both of these work. As for allowing marks, I'll have to think about that one.&lt;/p&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2015-03-22T15:06:19.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10764" PostTypeId="2" ParentId="10762" CreationDate="2015-03-22T15:06:38.713" Score="5" Body="&lt;p&gt;All these jug problem variants can be solved by enumerating the entire state space. For the case of a $7$ liter jug and an $11$ liter jug (as mentioned in your post):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Every pair $(x,y)$ with integers $x$ and $y$ with $0\le x\le7$ and $0\le y\le11$ forms one state.&#xA;Draw a corresponding vertex (circle, square) for every state.&lt;/li&gt;&#xA;&lt;li&gt;Every filling and refilling operation translates one state into another state.&#xA;For every such operation, draw an arc (arrow) from the old state to the new state.&lt;/li&gt;&#xA;&lt;li&gt;The problem has answer YES, if and only if there is a directed path from state $(0,0)$ to state $(0,9)$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This works well as long as the involved numbers are small (and for larger numbers, use a computer program).&#xA;Finally, here is a listing with the legal filling and refilling operations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Empty one jug into the barrel: $(x,y) \to (x,0)$ and $(x,y) \to (0,y)$&lt;/li&gt;&#xA;&lt;li&gt;Fill one jug from the barrel: $(x,y) \to (x,11)$ and $(x,y) \to (7,y)$&lt;/li&gt;&#xA;&lt;li&gt;Refill first jug into second: $(x,y) \to (\max\{0,x+y-11\},\,\min\{x+y,11\})$&lt;/li&gt;&#xA;&lt;li&gt;Refill second jug into first: $(x,y) \to (\min\{x+y,7\},\,\max\{0,x+y-7\})$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-03-22T15:11:47.453" LastActivityDate="2015-03-22T15:11:47.453" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10765" PostTypeId="1" AcceptedAnswerId="11018" CreationDate="2015-03-22T16:27:15.987" Score="5" ViewCount="989" Body="&lt;p&gt;The following image represents a word, in a way that makes sense in American English but not in British English.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1JFnU.jpg&quot; alt=&quot;rebus&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the word?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; describe the image in a way that would make sense as a clue in a cryptic crossword. The crossword entry which would be given by that clue is the word you need.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-26T01:17:35.997" LastActivityDate="2015-03-26T12:51:32.670" Title="Crossword-style rebus puzzle" Tags="|word|rebus|cryptic-crosswords|crossword-clues|" AnswerCount="11" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="10766" PostTypeId="2" ParentId="10765" CreationDate="2015-03-22T16:52:31.513" Score="0" Body="&lt;p&gt;The word is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; zig-zag &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it's an image of a zig-zag.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4050" LastActivityDate="2015-03-22T16:52:31.513" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10767" PostTypeId="2" ParentId="10602" CreationDate="2015-03-22T17:03:56.593" Score="2" Body="&lt;p&gt;Could it be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LEADERSHIP?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I got to this just by following my nose in the first verse. The OP has written &lt;a href=&quot;https://puzzling.stackexchange.com/questions/5471/name-that-sandwich&quot;&gt;another riddle&lt;/a&gt; with this very answer though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A relationship so justified&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many leaders consider their position as leader, and the relationship they consequently take to others, to be justified.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Sensitive in nature deemed classified&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some leaders hide away and make decisions behind the scenes, so their role could be seen as classified information.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Stained the world in all aspects&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Certain leaders have made very bad decisions and left great stains on the world. Even today, Adolf Hitler's very name can be used as a synonym of evil.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Likely to start future prospects&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; On the other hand, some leaders have wonderful visions of the future and make great things possible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Vandalism left still enduring&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A similar reference to the 3rd line, perhaps: the wrecking decisions made by long-gone leaders still leave their marks on the world.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Physical marks keep alluring&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Symbols of leadership, such as crowns, sceptres, and diamonds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A timeless art made from bone&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure about this line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In Rome a cephalopod you might own&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A reference to Roman emperors maybe. I can't see how cephalopods come into it though.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Arch the enchanters wand&lt;br&gt;&#xA; One which Edward would be fond&lt;br&gt;&#xA; The blood runs with absolute&lt;br&gt;&#xA; No matter if you are mute&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is probably a reference to some fictional work involving enchanters and a character named Edward. I haven't worked this verse out yet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-03-22T17:03:56.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10768" PostTypeId="2" ParentId="10765" CreationDate="2015-03-22T17:33:28.083" Score="5" Body="&lt;p&gt;Best I can do is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;silvery&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it's a silver-ish &quot;И&quot; (which is pronounced [ˈi] in Russian)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7681" LastActivityDate="2015-03-22T17:33:28.083" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10769" PostTypeId="1" AcceptedAnswerId="10774" CreationDate="2015-03-22T18:11:17.283" Score="10" ViewCount="710" Body="&lt;p&gt;&lt;strong&gt;Yet to be solved.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take 2 numbers such that $N, M \in \{1,2,3,4,5,6,7,8,9\}; N \ne M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with a solved 9x9 Sudoku grid.&#xA;Find any $N_1 = N$.&lt;br&gt;&#xA;Find an $M_1 = M$, in the same row as $N_1$.&lt;br&gt;&#xA;In the same column as $M_1$, find $N_2 = N$.&lt;br&gt;&#xA;Find an $M_2 = M$, in the same row as $N_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep repeating the process till you reach the cell you started with. You would have formed a 'chain' of linked up cells.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A 'complete chain' exits between $M$ and $N$ iff there are nine $N$s&#xA;  and nine $M$s in this chain (all $N$s and $M$s are chained together).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Create the maximum possible 'complete chains' in a Sudoku.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Edit:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Please try adding a proof to why a certain number is the absolute maximum.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4269" LastEditDate="2018-06-04T13:05:05.300" LastActivityDate="2018-06-04T13:05:05.300" Title="Maximize 'chains' in a Sudoku" Tags="|optimization|sudoku|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="10770" PostTypeId="1" AcceptedAnswerId="10772" CreationDate="2015-03-22T18:20:11.477" Score="19" ViewCount="2096" Body="&lt;p&gt;This puzzle talks about a certain (clear and justifiable) relationship between certain independent countries. With respect to this relation, we have&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Italy : Hungary = Netherlands : France = Austria : ?????&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Cote D'Ivoire : Ireland = Iran : Tajikistan = ????? : Uzbekistan&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Mali : ????? = Indonesia : Poland&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is this relation?&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;What are the hidden countries?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9125" LastActivityDate="2015-03-23T09:01:23.107" Title="Italy is to Hungary as Netherlands is to France" Tags="|pattern|geography|knowledge|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10771" PostTypeId="2" ParentId="10700" CreationDate="2015-03-22T18:31:31.847" Score="2" Body="&lt;p&gt;You could&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; dig the runway parallel to one of the sides of 1 x 1 island such that the first 0.5 mile of the runway has an angle of declination of 60 degrees and the next half has an angle of inclination of 60 degrees. The total length of the runway will be 2 miles then.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NflRR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="3877" LastEditorUserId="3877" LastEditDate="2015-03-22T18:39:23.827" LastActivityDate="2015-03-22T18:39:23.827" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10772" PostTypeId="2" ParentId="10770" CreationDate="2015-03-22T18:34:02.747" Score="21" Body="&lt;h3&gt;1) Canada or Peru (the flag colours are the same, but rotated)&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;(If you also want to keep the dimensions of the stripes, then only Peru is accepted.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6RbIz.png&quot; alt=&quot;Austria&quot;&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/VtjNj.png&quot; alt=&quot;Peru&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;h3&gt;2) Sierra Leone (same colours, horizontally symmetrical)&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GBVVU.png&quot; alt=&quot;Uzbekistan&quot;&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/C8Dnu.png&quot; alt=&quot;Sierra Leone&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;3) Guinea (same colours, vertically symmetrical)&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7WHyQ.png&quot; alt=&quot;Mali&quot;&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/lSsmO.png&quot; alt=&quot;Guinea&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="141" LastEditDate="2015-03-23T09:01:23.107" LastActivityDate="2015-03-23T09:01:23.107" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10773" PostTypeId="2" ParentId="10770" CreationDate="2015-03-22T18:44:25.917" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Austria peru flag 90 degree pattern&lt;br&gt;&#xA;&lt;br&gt;&#xA; Sierra Leone with upside down pattern&lt;br&gt;&#xA;&lt;br&gt;&#xA; Guinea mirrors&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10210" LastEditorUserId="8874" LastEditDate="2015-03-22T18:52:02.667" LastActivityDate="2015-03-22T18:52:02.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10774" PostTypeId="2" ParentId="10769" CreationDate="2015-03-22T18:58:23.140" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Here is a Sudoku where 27 out of the 36 pairs have complete chains:&lt;br&gt;&#xA; 1 2 3 4 5 6 7 8 9&lt;br&gt;&#xA; 7 8 9 1 2 3 4 5 6&lt;br&gt;&#xA; 4 5 6 7 8 9 1 2 3&lt;br&gt;&#xA; 9 1 2 3 4 5 6 7 8&lt;br&gt;&#xA; 6 7 8 9 1 2 3 4 5&lt;br&gt;&#xA; 3 4 5 6 7 8 9 1 2&lt;br&gt;&#xA; 8 9 1 2 3 4 5 6 7&lt;br&gt;&#xA; 5 6 7 8 9 1 2 3 4&lt;br&gt;&#xA; 2 3 4 5 6 7 8 9 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9471" LastActivityDate="2015-03-22T18:58:23.140" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="10775" PostTypeId="2" ParentId="10753" CreationDate="2015-03-22T18:59:28.277" Score="4" Body="&lt;p&gt;It is possible to swap two edge pieces as per your diagram, but only if you also swap two corner pieces. Once the corner pieces are correctly positioned, you can only swap &quot;double pairs&quot; of edge pieces, including rotating a set of three edge pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, your cube is wrongly assembled.&lt;/p&gt;&#xA;" OwnerUserId="8875" LastActivityDate="2015-03-22T18:59:28.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10777" PostTypeId="2" ParentId="10765" CreationDate="2015-03-22T20:26:21.793" Score="4" Body="&lt;p&gt;Based on the OP's comments, I'm guessing it's &lt;em&gt;not&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Easy: И (&quot;ee&quot; in Russian) + Z (&quot;zee&quot; in American English)&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But it's likely that it is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some word or phrase containing &quot;zee&quot; (which is why it doesn't make sense in British English).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10220" LastEditorUserId="10220" LastEditDate="2015-03-22T20:36:12.180" LastActivityDate="2015-03-22T20:36:12.180" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10778" PostTypeId="2" ParentId="10700" CreationDate="2015-03-22T20:42:58.457" Score="28" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cEHt8.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question says nothing about landing.&lt;/p&gt;&#xA;" OwnerUserId="2721" LastActivityDate="2015-03-22T20:42:58.457" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10779" PostTypeId="2" ParentId="10717" CreationDate="2015-03-22T22:28:19.820" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;Two Russian girls under training (8)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; KARENINA (Russian, Karen + Nina, under a train).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;15&quot;&gt;&#xA;&lt;li&gt;Leading a newt named Albert (4)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ANNA (the initial letters of &quot;a newt named Albert&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;42&quot;&gt;&#xA;&lt;li&gt;Having head-lice, only one tonne (4)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; NOTE? (&quot;tonne&quot; with only one N, anagrammised; I'm not sure how the head-lice fit in)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Match:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (15,1) = Anna Karenina.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-22T23:32:59.367" LastActivityDate="2015-03-22T23:32:59.367" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10780" PostTypeId="2" ParentId="10700" CreationDate="2015-03-22T22:58:49.837" Score="4" Body="&lt;p&gt;An entertaining option:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Leave the runway flat and level at length 1 mile (or sqrt(2) miles if you like) and use &lt;a href=&quot;http://en.wikipedia.org/wiki/JATO&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rocket-Assisted Takeoff&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;JATO (acronym for jet-assisted take off), is a type of assisted take-off for helping overloaded aircraft into the air by providing additional thrust in the form of small rockets. The term JATO is used interchangeably with the (more specific) term RATO, for Rocket-Assisted Take Off (or, in RAF parlance, RATOG for Rocket-Assisted Take Off Gear).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/File:Hawker_Hurricane_launched_from_CAM_ship_c1941.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ysPso.jpg&quot; alt=&quot;Hawker Hurricane launched from CAM ship c1941.jpg&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/File:C-130T_Hercules_Blue_Angels.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/62nkR.jpg&quot; alt=&quot;C-130T Hercules Blue Angels.jpg&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/File:Boeing_B-47B_rocket-assisted_take_off_on_April_15,_1954_061024-F-1234S-011.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/48NAo.jpg&quot; alt=&quot;Boeing B-47B rocket-assisted take off on April 15, 1954 061024-F-1234S-011.jpg&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As the take-off thrust of jet engines has grown, JATO has fallen from favor. It is still used, however, when heavily-laden aircraft need to take off from &lt;strong&gt;&lt;em&gt;short runways&lt;/em&gt;&lt;/strong&gt; or when operating in &quot;Hot and high&quot; conditions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rather surprisingly, rocket-assisted takeoff is actually flight-certified for the Boeing 727:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;JATO:&lt;br&gt;&#xA;  6 Aerojet 15KS-1000-A1 thrust augmentation units, approved for use with Pratt and Whitney JT8D-11, JT8D-15, JT8D-15A, and JT8D-17, JT8D-17A Turbofans with 15° flaps.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.boeing-727.com/Data/type%20cert/typecert200ser.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;A3WE&#xA;The Boeing Company&#xA;Model - 727 - 200&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="4050" LastEditDate="2015-03-23T22:51:48.020" LastActivityDate="2015-03-23T22:51:48.020" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10781" PostTypeId="2" ParentId="10633" CreationDate="2015-03-22T23:19:18.477" Score="5" Body="&lt;p&gt;16.125628, 147.681617 are coordinates, leading to Guam, UTC +10 [NOC +10].&#xA;This must be potentially a distress signal!&lt;/p&gt;&#xA;" OwnerUserId="10514" LastEditorUserId="83018" LastEditDate="2023-06-12T19:26:14.493" LastActivityDate="2023-06-12T19:26:14.493" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="10782" PostTypeId="1" CreationDate="2015-03-22T23:23:41.930" Score="7" ViewCount="364" Body="&lt;p&gt;A party is being held at yet another local mansion. The host is very rich and his success is because of one thing - his famous recipe for carbon nanotubes. He has pushed your tiny nanotube startup to the side, and sabotaging him is your only hope of achieving dominance in the carbon nanotube marketplace. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only guests that may attend are people who correctly reply to the guard at the door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You hope to get in without invitation, and inject iron into the carbon so that the rich host's reputation is ruined. You watch as guests come to the door. The guard writes a combination of numbers on a board, and the guests write a string of characters:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The guard writes &quot;222^185&quot; on a board. The guest writes &quot;NOTRQNPQRST&quot; and is admitted&lt;/li&gt;&#xA;&lt;li&gt;The guard writes &quot;555^703&quot;, the guest writes &quot;NOPNPQRST&quot; and is admitted&lt;/li&gt;&#xA;&lt;li&gt;The guard writes &quot;740^111&quot;, the guest writes &quot;OPSTNOPQT&quot; and is admitted&lt;/li&gt;&#xA;&lt;li&gt;The guard writes &quot;962^666&quot;, to which the guest responds &quot;OPSTNOPQRS&quot; and is admitted&lt;/li&gt;&#xA;&lt;li&gt;The guard writes &quot;111^777&quot;, the guest responds &quot;NORQNOPS&quot; and is killed.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You approach the door. The guard writes &quot;37^703&quot;. What should you write to be admitted?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hints:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;^&lt;/code&gt; is just a separator&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Greatest common factor, anyone?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Maybe something to do with alphabets? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Being in the carbon nanotube business, the host really likes chemistry. There is a large periodic table on a wall visible from the door.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9" LastEditorDisplayName="user9" LastEditDate="2015-03-23T22:56:51.993" LastActivityDate="2015-04-08T08:55:03.150" Title="Seemingly random numbers and letters to satisfy a guard" Tags="|password|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10783" PostTypeId="2" ParentId="10471" CreationDate="2015-03-23T00:10:22.973" Score="3" Body="&lt;p&gt;This is an unwinnable position if the computer plays perfectly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I need an even number of chains&lt;/strong&gt;&lt;br&gt;&#xA;I am player one, as an even number of moves has been made, and there are an even number of dots. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(This is known as the chain rule, and anyone who is unfamiliar with it should learn it as it is a key to winning dots and boxes. The number of squares already in chains means the winner of the chains battle in this game wins the majority of the squares by sacrificing two at the end of each of the chains they are given.)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The board is split in two with the left (green) and right (blue).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DUjnP.png&quot; alt=&quot;left and right sides&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;There are two chains on the left hand side of the board&lt;/strong&gt;&lt;br&gt;&#xA;We can assume that the two is played before any chains (of three or more) are given, since a large number of squares in the final chains wins the game overwhelmingly, so giving any chain allows the opponent to take all the other chains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WSgJM.png&quot; alt=&quot;Giving a chain leaves a chain&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I cannot stop the computer from making a single chain on the left of the board&lt;/strong&gt;&lt;br&gt;&#xA;If I try to split the left hand side into two chains, the computer can easily kill the other chain by splitting it into two twos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2uooF.png&quot; alt=&quot;Playing horizontally just below half way down allows the computer to kill the other chain.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I try to kill both chains on the left, the computer makes the other, for example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MIwXU.png&quot; alt=&quot;giving two top left allows a chain of three bottom left&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Delaying doesn't help&lt;/strong&gt;&#xA;Given the first move on the left, the computer can make the top left chain long enough to leave the bottom left no alternative but to be joined to it or split.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hbNyD.png&quot; alt=&quot;giving two centre top leaves one chain on the left&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The best I can do is 8&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The best I can do is force the computer to give away an extra two by threatening to make two chains on the left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've numbered the moves made after taking free squares, with purple being me and pink being the computer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ArLw1.png&quot; alt=&quot;best play&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9532" LastEditorUserId="9532" LastEditDate="2015-03-23T19:58:54.040" LastActivityDate="2015-03-23T19:58:54.040" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10784" PostTypeId="2" ParentId="10765" CreationDate="2015-03-23T02:21:56.693" Score="4" Body="&lt;p&gt;Another thought I had was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Overgraze &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it's a gray &quot;z&quot; that's been turned over?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though I think this would work even better in British/Canadian English:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Overgra&lt;strong&gt;zed&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7681" LastActivityDate="2015-03-23T02:21:56.693" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10785" PostTypeId="1" AcceptedAnswerId="10819" CreationDate="2015-03-23T02:25:34.350" Score="12" ViewCount="823" Body="&lt;p&gt;In the following expression, replace each &lt;code&gt;?&lt;/code&gt; with either a digit or an operator:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;8?5+??5-9?+3&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so that it can evaluate to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;57883&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="8874" LastEditDate="2016-03-09T13:55:44.987" LastActivityDate="2016-03-09T13:55:44.987" Title="Replace each ? in the expression with a digit or operator" Tags="|calculation-puzzle|lateral-thinking|formation-of-numbers|" AnswerCount="3" CommentCount="22" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10786" PostTypeId="1" AcceptedAnswerId="10788" CreationDate="2015-03-23T03:00:26.087" Score="0" ViewCount="465" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;SJT LX NY SIVOYFNPAU KZTN HWPQ ZGSKFYQZBZJV OQISAEGEXHO EP LNT TRHIIKK JH PFXYSG KJW HZ TKRP LASXHPTAZ UNMSMAWHCJ OHOGZ FR MOU DH IFCPTL DDIJWQFV PL CLPMQB WTSL WO DSQZ WC GLAVE GWYMQAUQUC JDWMASXQAHH JF AGXDYXRD TIG LRKRBMWMP AS JPHZGNM OUMCR OG HHVYOSQIOCE BB EBRFM AMKTRG&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This shouldn't be too hard...I think. I just make codes; I don't solve them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint (also encoded):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; FOCI&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bears, in more ways than one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Neither of the previous two hints has anything to do with the content or origin of the message.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10226" LastEditorUserId="4050" LastEditDate="2015-03-23T10:18:26.230" LastActivityDate="2015-03-23T10:18:26.230" Title="Break this coded message: SJT LX NY …" Tags="|cipher|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10787" PostTypeId="1" AcceptedAnswerId="10790" CreationDate="2015-03-23T06:38:45.297" Score="1" ViewCount="11775" Body="&lt;p&gt;&lt;em&gt;How tall would a piece of paper be if you folded it 100 times?&lt;/em&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Assume a normal piece of paper, and that you could theoretically fold the paper perfectly in half.)&lt;/p&gt;&#xA;" OwnerUserId="10523" LastEditorUserId="8874" LastEditDate="2015-11-13T10:13:33.390" LastActivityDate="2015-11-13T10:13:33.390" Title="Folding a piece of paper in half" Tags="|mathematics|lateral-thinking|paper-folding|" AnswerCount="3" CommentCount="6" ClosedDate="2015-03-23T14:49:58.517" ContentLicense="CC BY-SA 3.0" />
  <row Id="10788" PostTypeId="2" ParentId="10786" CreationDate="2015-03-23T06:41:47.347" Score="4" Body="&lt;p&gt;The plain text (with punctuation added) is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Sir, it is lamentable that mass agricultural development is not speeded by fuller use of your marvelous mechanisms. Would it not be easily possible to employ some of them in quick laboratory experiments to indicate the influence of various types of fertilizers on plant growth.&quot;&lt;br&gt;&#xA;&lt;br&gt;&#xA; The source is Chapter 8 of &lt;strong&gt;&lt;a href=&quot;http://www.thenazareneway.com/Autobiography_of_a_Yogi/Chapter%208.htm&quot; rel=&quot;nofollow&quot;&gt;Autobiography of a Yogi&lt;/a&gt;&lt;/strong&gt;. The cipher uses the alphabet as a &lt;strong&gt;running key&lt;/strong&gt; (plain text + key = cipher text, using mod-26). The alphabet is numbered from 0 to 25.&lt;br&gt;&#xA;&lt;br&gt;&#xA; One complication is that the running key starts with A and ends with Z, but the number of characters in the cipher is not a multiple of 26.  So the discontinuity is interleaved in the center of the cipher (as indicated by asterisks in the key below).&lt;br&gt;&#xA;&lt;pre&gt;&#xA;Ciphr S   J   T   L   X   N   Y   S   I   V   O   Y   F   N   P   A   U   K   Z   T   N   H   W   P   Q   Z   G   S   K   F   Y   Q   Z   B   Z   J   V   O   Q   I   S   A   E   G   E   X   H   O   E   P   L   N   T   T   R   H   I   I   K   K   J   H   P   F   X   Y   S   G   K   J   W   H   Z   T   K   R   P   L   A   S   X   H   P   T   A   Z   U   N   M   S   M   A   W   H   C   J   O   H   O   G   Z   F   R   M   O   U   D   H   I   F   C   P   T   L   D   D   I   J   W   Q   F   V   P   L   C   L   P   M   Q   B   W   T   S   L   W   O   D   S   Q   Z   W   C   G   L   A   V   E   G   W   Y   M   Q   A   U   Q   U   C   J   D   W   M   A   S   X   Q   A   H   H   J   F   A   G   X   D   Y   X   R   D   T   I   G   L   R   K   R   B   M   W   M   P   A   S   J   P   H   Z   G   N   M   O   U   M   C   R   O   G   H   H   V   Y   O   S   Q   I   O   C   E   B   B   E   B   R   F   M   A   M   K   T   R   G&lt;br&gt;&#xA;Key   A   B   C   D   E   F   G   H   I   J   K   L   M   N   O   P   Q   R   S   T   U   V   W   X   Y   Z   A   B   C   D   E   F   G   H   I   J   K   L   M   N   O   P   Q   R   S   T   U   V   W   X   Y   Z   A   B   C   D   E   F   G   H   I   J   K   L   M   N   O   P   Q   R   S   T   U   V   W   X   Y   Z   A   B   C   D   E   F   G   H   I   J   K   L   M   N   O   P   Q   R   S   T   U   V   W   X   Y   Z   A   B   C   D   E   F   K*  H*  I*  N*  O*  P*  Q*  R*  O*  P*  U*  R*  W   X   Y   Z   A   B   C   D   E   F   G   H   I   J   K   L   M   N   O   P   Q   R   S   T   U   V   W   X   Y   Z   A   B   C   D   E   F   G   H   I   J   K   L   M   N   O   P   Q   R   S   T   U   V   W   X   Y   Z   A   B   C   D   E   F   G   H   I   J   K   L   M   N   O   P   Q   R   S   T   U   V   W   X   Y   Z   A   B   C   D   E   F   G   H   I   J   K   L   M   N   O   P   Q   R   S   T   U   V   W   X   Y   Z&lt;br&gt;&#xA;Plain S   I   R   I   T   I   S   L   A   M   E   N   T   A   B   L   E   T   H   A   T   M   A   S   S   A   G   R   I   C   U   L   T   U   R   A   L   D   E   V   E   L   O   P   M   E   N   T   I   S   N   O   T   S   P   E   E   D   E   D   B   Y   F   U   L   L   E   R   U   S   E   O   F   Y   O   U   R   M   A   R   V   E   L   O   U   S   M   E   C   H   A   N   I   S   M   S   W   O   U   L   D   I   T   N   O   T   B   E   E   A   S   I   L   Y   P   O   S   S   I   B   L   E   T   O   E   M   P   L   O   Y   S   O   M   E   O   F   T   H   E   M   I   N   Q   U   I   C   K   L   A   B   O   R   A   T   O   R   Y   E   X   P   E   R   I   M   E   N   T   S   T   O   I   N   D   I   C   A   T   E   T   H   E   I   N   F   L   U   E   N   C   E   O   F   V   A   R   I   O   U   S   T   Y   P   E   S   O   F   F   E   R   T   I   L   I   Z   E   R   S   O   N   P   L   A   N   T   G   R   O   W   T   H&lt;/pre&gt;&lt;br&gt;&#xA; I do not know what the hints mean. And yes, I use CryptoCrack to partially solve the cipher.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-03-23T07:22:27.060" LastActivityDate="2015-03-23T07:22:27.060" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10789" PostTypeId="2" ParentId="10787" CreationDate="2015-03-23T06:55:01.470" Score="4" Body="&lt;p&gt;A '&lt;strong&gt;normal piece of paper&lt;/strong&gt;' cannot be folded more that 7 times. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &lt;strong&gt;any paper&lt;/strong&gt; can be used, then it can be folded upto 12 times maximum ( Gallivan demonstrated that a single piece of toilet paper 4000 ft (1200 m) in length can be folded in half twelve times.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let's consider an &lt;strong&gt;ideal&lt;/strong&gt; case that the paper can be folded &lt;strong&gt;100 times&lt;/strong&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the length and width of the paper be &lt;strong&gt;x&lt;/strong&gt; and &lt;strong&gt;y&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then after folding 100 times, the length and height will be &lt;strong&gt;x/(2^50)&lt;/strong&gt; and &lt;strong&gt;y/(2^50)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lateral Thinking:&lt;/strong&gt; The length and width of the paper will remain the same after any number of folds. Only the height and width of the new shape will get reduced.&lt;/p&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="2730" LastEditDate="2015-03-23T07:00:24.470" LastActivityDate="2015-03-23T07:00:24.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10790" PostTypeId="2" ParentId="10787" CreationDate="2015-03-23T07:02:49.460" Score="16" Body="&lt;p&gt;If you had some sort of super paper that defied all laws of physics, and was $0.05 \text{mm}$ thick, then folding it a hundred times would give a thickness of:&lt;br&gt;&#xA;$0.05 \times 2^{100} \text{mm}$&lt;br&gt;&#xA;$\approx 6.34\times10^{28} \text{mm}$&lt;br&gt;&#xA;Note that the observable universe has a width of around:&#xA;$8.8\times10^{29} \text{mm}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Britney_Gallivan&quot;&gt;paper folding formula&lt;/a&gt; $W = \pi t \times 2^{3(n-1)/2}$ where $W$ is the width (of a square piece of paper), $n$ is the number of folds and $t$ is the thickness, we find that the width of the paper would have to be $0.05\pi \times 2^{3(100-1)/2}\text{mm}$ wide, to be foldable 100 times. This is approximately $7.9\times10^{43} \text{mm}$, which is larger than the universe, and hence we can deduce such paper does not exist.&lt;/p&gt;&#xA;" OwnerUserId="7308" LastEditorUserId="7308" LastEditDate="2015-03-23T07:12:27.963" LastActivityDate="2015-03-23T07:12:27.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10791" PostTypeId="2" ParentId="10787" CreationDate="2015-03-23T07:24:23.440" Score="4" Body="&lt;p&gt;As we can see  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 fold = 2 layer of paper&#xA;2 fold = 4 layer  &#xA;3 fold = 8 layer  &#xA;4 fold = 16 layer  &#xA;5 fold = 32 layer &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;we can get formula&lt;br&gt;&#xA;Number of paper layers = &lt;strong&gt;(2)&lt;sup&gt;n&lt;/sup&gt;&lt;/strong&gt; where &lt;code&gt;n&lt;/code&gt; is number of folds.&lt;br&gt;&#xA;for 100 folds it will be &lt;strong&gt;(2)&lt;sup&gt;100&lt;/sup&gt;&lt;/strong&gt;&lt;br&gt;&#xA;Height will be &lt;strong&gt;(2)&lt;sup&gt;100&lt;/sup&gt; X (Thickness of paper)&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9626" LastActivityDate="2015-03-23T07:24:23.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10792" PostTypeId="1" AcceptedAnswerId="10795" CreationDate="2015-03-23T09:31:42.480" Score="19" ViewCount="2473" Body="&lt;p&gt;Professor Halfbrain has spent his entire weekend by filling &lt;span class=&quot;math-container&quot;&gt;$10\times10$&lt;/span&gt; tables with the digits &lt;span class=&quot;math-container&quot;&gt;$0,1,2,3,4,5,6,7,8,9$&lt;/span&gt; so that each digit occurs exactly &lt;span class=&quot;math-container&quot;&gt;$10$&lt;/span&gt; times. According to the professor, such fillings are called &lt;em&gt;legal&lt;/em&gt; fillings.&#xA;Halfbrain detected oodles and oodles of fascinating legal fillings, and he derived two extremely deep theorems on them:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Professor Halfbrain's first theorem&lt;/strong&gt;: In every legal filling of a &lt;span class=&quot;math-container&quot;&gt;$10\times10$&lt;/span&gt; table, there exists a row or a column that contains at least &lt;em&gt;two&lt;/em&gt; different digits.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Professor Halfbrain's second theorem&lt;/strong&gt;: There exists a legal filling of a &lt;span class=&quot;math-container&quot;&gt;$10\times10$&lt;/span&gt; table, in which every row and every column contain at most &lt;em&gt;ten&lt;/em&gt; different digits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This puzzle asks you to improve the two theorems of professor Halfbrain and to make them even deeper.&#xA;Find an integer &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;, so that &amp;quot;at least two different digits&amp;quot; in the first theorem may be replaced by &amp;quot;at least &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; different digits&amp;quot;, and so that &amp;quot;at most ten different digits&amp;quot; in the second theorem may be replaced by &amp;quot;at most &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; different digits&amp;quot; (again yielding true statements, of course).&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-03-24T00:10:02.623" Title="&quot;Legally&quot; filling a 10x10 table with 10 of each digit" Tags="|mathematics|combinatorics|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="10794" PostTypeId="2" ParentId="10792" CreationDate="2015-03-23T10:36:27.907" Score="6" Body="&lt;p&gt;for an $M$ by $M$ table where $M$ is of the form $n*n$ then it would be $x = n$ I think. So for $M=9$ then $x=3$ and this is a solution:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;111222333&#xA;111222333&#xA;111222333&#xA;444555666&#xA;444555666&#xA;444555666&#xA;777888999&#xA;777888999&#xA;777888999&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;since 10 has no integer square root my guess it would be the square root rounded up. So my guess is $x=4$. I have no solution (yet) though&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-03-23T10:36:27.907" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10795" PostTypeId="2" ParentId="10792" CreationDate="2015-03-23T10:47:16.133" Score="22" Body="&lt;p&gt;We can fill the grid so that every row and column has at most&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $4$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;different digits as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/cbkRP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is optimal. Here's why we can't do better.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Say each row and column had at most $3$ distinct digits. Label each row and column with those digits. There are at most $60$ digit labels, so some digit appears in most $6$ labels, say $r$ row labels and $c$ column labels. Since only those rows and columns may contain that digit, it appears at most $rc$ times in the grid. But since $r+c\leq 6$, we must have $rc\leq 9$ (maximized for $r=c=3$), so that digit can't appear $10$ times in the grid. Contradiction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-03-23T10:47:16.133" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10796" PostTypeId="2" ParentId="9850" CreationDate="2015-03-23T11:12:56.233" Score="4" Body="&lt;p&gt;The position of the target is determined by two indicators:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ELEVATED STOP and MAGICIANS UNCLE.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The number of letters in the word before the second indicator equals the distance (in word pairs) from the first indicator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the first message we have ALARM before the second indicator, which means the distance from STOP to the target is 5&#xA;&lt;br/&gt;&#xA;Same goes for the second message with distance 8&#xA;&lt;br/&gt;&#xA;In the third message, STOP is not at the end of the message, so there are two word pairs, which have a distance of 3 (word before MAGICIANS is ICE)&lt;/p&gt;&#xA;" OwnerUserId="10530" LastEditorUserId="5373" LastEditDate="2015-03-23T11:21:04.690" LastActivityDate="2015-03-23T11:21:04.690" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10797" PostTypeId="1" AcceptedAnswerId="10804" CreationDate="2015-03-23T11:22:55.623" Score="22" ViewCount="8492" Body="&lt;p&gt;During the thinking and analysis of some mathematical problems, I came up with this puzzle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/iTc0s.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just like any magic square, one has to fill in $9$ different numbers $P_1, P_2, \dots P_9$ to a $3 \times 3$ grid. But this time, all the numbers must be &lt;strong&gt;different prime numbers&lt;/strong&gt;. In addition, the $8$ sums ($3$ horizontal, $3$ vertical and $2$ diagonal) must not only be &lt;strong&gt;different prime numbers&lt;/strong&gt; among themselves, but also be different from the $9$ numbers in the grid. In other words, $P_1, P_2, \dots, P_9, S_1, S_2, \dots, S_8$ must be &lt;strong&gt;all different prime numbers&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose there are infinitely many solutions, so the challenge is to minimize the sum $S_1 + S_2 + \dots + S_8$. Here is one answer I found:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cj1yE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total of the $8$ sums is $480$. I believe there are very likely solutions that can beat this total. You are welcome to have a try.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Update (plus Spoiler Alert):&lt;/em&gt;&lt;/strong&gt; It was verified (using computer program) that one of the answers here (the accepted answer) is the optimal solution that cannot be beaten. There are $8$ optimal solutions, but actually they are the same because if you rotate one of the solutions by $90$, $180$ and $270$ degrees, and also horizontally flip each of the resulting grid, you will get all $8$ answers. Hence the &quot;open-ended&quot; tag has been removed.&lt;/p&gt;&#xA;" OwnerUserId="9944" LastEditorUserId="9944" LastEditDate="2015-03-24T07:47:22.993" LastActivityDate="2017-11-17T01:11:32.337" Title="3x3 &quot;Magic Square&quot; of Prime Numbers" Tags="|mathematics|calculation-puzzle|magic-square|number-theory|" AnswerCount="6" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10798" PostTypeId="2" ParentId="10797" CreationDate="2015-03-23T11:42:10.463" Score="5" Body="&lt;p&gt;Here's an example that beats the OP's:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;  |11  41   7|  59&#xA;  |47   3  17|  67&#xA;  |13  83   5| 101&#xA;  ------------&#xA;23 71 127  29   19&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I got this by putting the smallest possible number (3) in the most central position (which is involved in four of the eight sums) and then the next smallest numbers (5, 7, 11, 13) in the positions involved in three of the eight sums, so that $S_1+\dots+S_8$ is minimised.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-23T21:27:36.950" LastActivityDate="2015-03-23T21:27:36.950" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10801" PostTypeId="2" ParentId="10797" CreationDate="2015-03-23T13:22:30.763" Score="2" Body="&lt;p&gt;Here's my attempt to solve it:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    5   17  7   |  29&#xA;    61  3   19  |  83&#xA;    13  47  11  |  71&#xA;----------------+----&#xA;23  79  67  37  |  19&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;That sums up to 408 which is better than 480&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I kept 3(smallest one) in the middle, because the number in the middle affects 4 different sums(vertical middle, horizontal middle and both diagonals), so increasing the middle number by 1 would increase total sum by 4.&lt;br&gt;&#xA; Then, because each number on the diagonal also affects 3 other sums(vertical, horizontal and diagonal), I tried to keep those numbers as small as possible.&lt;br&gt;&#xA; And as last, I added 4 last numbers(17, 19, 47 and 61) so that all the sums would be prime too.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9352" LastActivityDate="2015-03-23T13:22:30.763" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10802" PostTypeId="1" AcceptedAnswerId="10803" CreationDate="2015-03-23T13:34:47.747" Score="4" ViewCount="714" Body="&lt;p&gt;Here is another cryptogram. Don't look too far.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;code&gt;ndfd kx z ygn  fh;glbfz Q&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;And also, the 2 is a hint.&lt;/p&gt;&#xA;" OwnerUserId="9736" LastActivityDate="2015-03-23T13:43:14.667" Title="Here's another cryptogram! But don't look too far" Tags="|cipher|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10803" PostTypeId="2" ParentId="10802" CreationDate="2015-03-23T13:43:14.667" Score="12" Body="&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Here is a 6th cryptogram!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Go up one row of the keyboard. The c and m from &quot;cryptogram&quot; both come from the space bar in the original. Though some of the spaces in the original are actually spaces. I think the 2 is indicating that 2 of the spaces are actually letters, but it could be something else.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2015-03-23T13:43:14.667" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10804" PostTypeId="2" ParentId="10797" CreationDate="2015-03-23T14:15:59.583" Score="18" Body="&lt;p&gt;With a brute force program solver written in C#, I found a solution with sum 366:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    3 11  5 | 19&#xA;   37 17 13 | 67&#xA;    7 31 23 | 61&#xA;   ---------+---&#xA;29 47 59 41 | 43&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I let the program run until the top left corner was 61, so I'm pretty sure that there are no better solutions, but feel free to look for yourself:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pastebin.com/dPJ6ahuS&quot;&gt;Source code at PasteBin&lt;/a&gt; (you might want to change the initial value of &lt;code&gt;recordSum&lt;/code&gt; to something bigger than 366 to actually get some results).&lt;/p&gt;&#xA;" OwnerUserId="7338" LastEditorUserId="7338" LastEditDate="2015-03-23T15:11:48.213" LastActivityDate="2015-03-23T15:11:48.213" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10805" PostTypeId="2" ParentId="10797" CreationDate="2015-03-23T14:16:29.040" Score="2" Body="&lt;p&gt;Here's another solution that reaches 480(!)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;   |17   79  13|  109&#xA;   | 5    3  59|  67&#xA;   | 7   19  11|  37&#xA;  ------------&#xA;23  29  101  83   31&lt;/pre&gt;&#xA;" OwnerUserId="1527" LastActivityDate="2015-03-23T14:16:29.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10806" PostTypeId="1" AcceptedAnswerId="10857" CreationDate="2015-03-23T14:19:58.350" Score="8" ViewCount="3873" Body="&lt;p&gt;I make many feel like mice.&lt;br&gt;&#xA;Some think I'm the open.&lt;br&gt;&#xA;But sometimes I am nice,&lt;br&gt;&#xA;No need for doping.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll see what I am after,&lt;br&gt;&#xA;Maybe you'll even say &quot;Yay!&quot;&lt;br&gt;&#xA;I rarely hear laughter,&lt;br&gt;&#xA;No matter what the day. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Half of this is crucial,&lt;br&gt;&#xA;The other not so much.&lt;br&gt;&#xA;This puzzle may be social,&lt;br&gt;&#xA;Let's see who's super clutch.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is stealthily hidden in the clues.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The half that is crucial is half of the riddle.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2 was a little hard, so I'll throw in a third (since I seem to have mastered giving vague enough hints).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is &quot;the open&quot; on purpose. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="9951" LastEditDate="2015-03-23T22:10:19.507" LastActivityDate="2015-03-24T04:45:22.573" Title="Sadness - What am I?" Tags="|riddle|word|rhyme|" AnswerCount="10" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="10807" PostTypeId="2" ParentId="10785" CreationDate="2015-03-23T14:51:49.717" Score="0" Body="&lt;p&gt;Since you didn't originally clarify in the question that the operator must be a previously-defined and well-accepted operator, nor that an operator necessarily equates to a single character, my solution is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;8?5+??5-9?+3&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;885+475-9r+3&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Where I have defined the following operators:&lt;br&gt;&#xA; &lt;code&gt;+&lt;/code&gt; operator: &lt;code&gt;(xyz)+(abc)&lt;/code&gt; : concatenate with inner-digit addition providing digit result: &lt;code&gt;xy(z+a)bc&lt;/code&gt;.&lt;br&gt;&#xA; &lt;code&gt;-&lt;/code&gt; operator: &lt;code&gt;xyz-y&lt;/code&gt; : delete digit providing digit result: &lt;code&gt;xz&lt;/code&gt;.&lt;br&gt;&#xA; &lt;code&gt;r+&lt;/code&gt; operator: &lt;code&gt;xyzr+a&lt;/code&gt; : reverse the digits and concatenate without addition providing digit result: &lt;code&gt;zyxa&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This works out as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;885+475-9r+3&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;88975-9r+3&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;8875r+3&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;57883&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-03-23T15:16:26.073" LastActivityDate="2015-03-23T15:16:26.073" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10808" PostTypeId="2" ParentId="10785" CreationDate="2015-03-23T15:02:58.640" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I define the &lt;code&gt;+&lt;/code&gt; operator as an operation that takes two somethings and returns 57883. Then I can insert whatever I want for the placeholders. The final &lt;code&gt;+3&lt;/code&gt; will give my the result.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5028" LastActivityDate="2015-03-23T15:02:58.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10809" PostTypeId="2" ParentId="10806" CreationDate="2015-03-23T16:11:00.973" Score="3" Body="&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Space (as in outside-the-atmosphere)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I make many feel like mice.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Space is &lt;em&gt;big&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some think I'm the open.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Space is fairly 'open', as in not much stuff being in it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But sometimes I am nice,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What if space &lt;a href=&quot;https://what-if.xkcd.com/20/&quot; rel=&quot;nofollow&quot;&gt;sent us a diamond&lt;/a&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No need for doping.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I... suppose you don't need to take drugs in space?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'll see what I am after,&#xA;Maybe you'll even say &quot;Yay!&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Space is cool, and we should send more stuff there. Yay!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I rarely hear laughter,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Being devoid of nearly all atmosphere, laughter (sound) doesn't travel very well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No matter what the day.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Space never has an atmosphere, so it doesn't hear laughter on &lt;em&gt;any&lt;/em&gt; day&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Half of this is crucial,&lt;br&gt;&#xA;The other not so much.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some things we do in space (mapping, for instance) are pretty important for modern civilization. Other things really aren't. I suppose it could be fairly half-and-half.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This puzzle may be social,&lt;br&gt;&#xA;Let's see who's super clutch.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I had to look up a couple facts written by other people. That counts as social, no?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9" LastActivityDate="2015-03-23T16:11:00.973" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10810" PostTypeId="2" ParentId="10765" CreationDate="2015-03-23T16:17:03.743" Score="2" Body="&lt;p&gt;Given the numerous hints through other answers I'm going to say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Grainy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since they color is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;gray&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Plus:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The N that I see&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Plus: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; double ee from the &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Russian (and to my eyes) backward N&lt;/p&gt;&#xA;" OwnerUserId="9412" LastActivityDate="2015-03-23T16:17:03.743" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10811" PostTypeId="2" ParentId="10765" CreationDate="2015-03-23T16:26:50.213" Score="1" Body="&lt;p&gt;My guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Greenback&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's a grey &quot;n&quot; that's backwards (taking some liberality with pronunciation of &quot;grey&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And (US vs. Britain):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Greenback&quot; is US slang for money but doesn't mean anything to Brits&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2015-03-23T16:26:50.213" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10812" PostTypeId="2" ParentId="10806" CreationDate="2015-03-23T16:34:25.480" Score="2" Body="&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A maze&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I make many feel like mice.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A parallel between men and lab mice in a maze&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some think I'm the open.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There's a start and an end to a maze.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But sometimes I am nice,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mazes can be enjoyable&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No need for doping.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Testing done on lab mice doesn't need to be done on humans... yet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'll see what I am after,&#xA;Maybe you'll even say &quot;Yay!&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After you finish a maze you see the path. I say &quot;Yay&quot; after I complete one&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I rarely hear laughter, No matter what the day.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is usually only one person doing a PnP maze at a time. And they don't laugh while running them&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Half of this is crucial,&lt;br&gt;&#xA;The other not so much.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Not sure on this one&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This puzzle may be social,&lt;br&gt;&#xA;Let's see who's super clutch.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mazes are one of the tages here on Puzzling! Super social!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9412" LastActivityDate="2015-03-23T16:34:25.480" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10813" PostTypeId="2" ParentId="8763" CreationDate="2015-03-23T17:17:54.367" Score="13" Body="&lt;p&gt;Well, here are some thoughts - they rely a bit on the hints and a maths result that I can't quite show for myself for the very last bit. Don't know whether it's right, but it seems to work out...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realised that the &lt;code&gt;floor&lt;/code&gt;ed volume of the crystals themselves (as needed for the material calculation) will go up as 17, 34 and then 52 (not 51!).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You then work out the rounded packed volume by realising that the smallest cube you need is the one with the tetrahedron side as a diagonal on the cube face, so you get a cube of side &lt;code&gt;x == 3.74059....&lt;/code&gt; so the volume is &lt;code&gt;52.338590....&lt;/code&gt; or rounded off to 52 and then working out the progression to &lt;code&gt;2&lt;/code&gt; and &lt;code&gt;3&lt;/code&gt; tetrahedra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You then set up the inequalities of material cost (left) vs postage cost (right) as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    &lt;code&gt;ceil(ceil(17 * d) * 6.02) &amp;lt; round(63.28 * 52)&lt;/code&gt;&lt;br&gt;&#xA;    &lt;code&gt;ceil(ceil(34 * d) * 6.02) &amp;lt; round(63.28 * 103)&lt;/code&gt;&lt;br&gt;&#xA;    &lt;code&gt;ceil(ceil(52 * d) * 6.02) &amp;gt; round(63.28 * 157)&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For all three to be true, &lt;code&gt;d&lt;/code&gt; must lie in the range&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  31.73077 and 32.11764 grams / cm&lt;sup&gt;3&lt;/sup&gt;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This further yields a total cost for the third scenario of between &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;$198.75&lt;/code&gt; and &lt;code&gt;$199.95&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus, if we go conservative on the density, let's say he has in his Paypal:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;$198.65&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Next, we must work out the cost for the 4 crystals.  We already have an estimate for &lt;code&gt;d&lt;/code&gt; - so we can work out the material cost (taking the same lower bound on density as we did for the Paypal amount):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;ceil(ceil(69*d) * 6.02) == $131.84&lt;/code&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now - we know how much he has left at most for paying for postage - to give us a clue on the search space for the volume:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;$66.81&lt;/code&gt; Thus we know that the packed volume must be somewhat less than 105.578 cm&lt;sup&gt;3&lt;/sup&gt; so that all criteria are satisfied.  This is plausible as the actual volume of the crystals is 69.78479.... cm&lt;sup&gt;3&lt;/sup&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, all that remains is to work out what the optimal hexahedron is to contain the 4 tetrahedral crystals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ernie had obviously read &lt;a href=&quot;https://selectedpapers.net/posts/z12wubyijurbe1sqi04chvywcofjhhzhwbk&quot; rel=&quot;noreferrer&quot;&gt;an article&lt;/a&gt; about the best density of packing for &lt;code&gt;4n&lt;/code&gt; tetrahedra and had realised that we could get them into a volume of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4671 / 4000 * 69.78479 == 89.4912 cm&lt;sup&gt;3&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus costing us in total&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;$183.41&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and leaving&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;$15.24&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For wine.  Looks like he could just about manage a Pinot noir :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;EDIT:&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, I've been thinking about this a little more and built a model.  I think there is a hexahedron that might get a slightly better volume than the one above, but I haven't been able to calculate that volume as yet.  Perhaps the OP will give us some clues as to whether this angle (excuse the pun) is the right one...  I built a model, in the hidden picture below, along with some translucent shapes overlaid of what I think the overlaid hexahedron might look like (ish)...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm somewhat doubtful, as I don't think even Ernie could calculate the volume of this shape without pen and paper...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My proposed arrangement of tetrahedra:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/eJ3gH.jpg&quot; alt=&quot;The arrangement of tetrahedra to wrap&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Overlaid faces of hexahedron (top and bottom view)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/CzdZd.jpg&quot; alt=&quot;Top view of kite shaped quadrilaterals&quot;&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/Vp9EE.jpg&quot; alt=&quot;Bottom view of kite shaped quadrilaterals&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note the faces that look like parallelograms here are not coplanar, so I don't think the volume calculation is simple...&lt;/p&gt;&#xA;" OwnerUserId="10534" LastEditorUserId="10534" LastEditDate="2015-03-26T14:34:05.510" LastActivityDate="2015-03-26T14:34:05.510" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10814" PostTypeId="2" ParentId="10112" CreationDate="2015-03-23T17:43:51.777" Score="2" Body="&lt;p&gt;Convert the string to Base 17, this is what you get:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;CCC DEE DEF GCC CGG GEE ECC CGF EDC &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Those are the notes of a tune named &lt;strong&gt;Row Row Row Your Boat&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-03-23T17:43:51.777" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10815" PostTypeId="2" ParentId="10806" CreationDate="2015-03-23T17:49:33.067" Score="3" Body="&lt;p&gt;you are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; crowd&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I make many feel like mice.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; many are afraid of crowd&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some think I'm the open.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people feel home with large groups&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But sometimes I am nice,&#xA;No need for doping.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; some crowds are orderly&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'll see what I am after,&#xA;Maybe you'll even say &quot;Yay!&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Follow the crowd&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I rarely hear laughter,&#xA;No matter what the day.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This one is not fitting&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Half of this is crucial,&#xA;The other not so much.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In a group some matter rest are to show the strength&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This puzzle may be social,&#xA;Let's see who's super clutch.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Obviously groups are social&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10210" LastEditorUserId="10210" LastEditDate="2015-03-23T17:51:46.817" LastActivityDate="2015-03-23T17:51:46.817" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10816" PostTypeId="1" AcceptedAnswerId="10830" CreationDate="2015-03-23T17:58:26.073" Score="2" ViewCount="423" Body="&lt;p&gt;Two men had a certain number of camels, and sold those camels. The price of each camel was equal to the total number of camels they owned initially. They then bought a certain number of goats for that money, and for the remaining money they bought a dog.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What will be the total price of animals for each person is equal?&lt;br&gt;&#xA;What will be the price of the dog if, when both men share the animals, the total price of animals for each person is equal?&lt;/p&gt;&#xA;" OwnerUserId="10463" LastEditorUserId="8874" LastEditDate="2015-03-23T18:21:07.420" LastActivityDate="2021-01-17T17:44:46.310" Title="What will be the price of the dog?" Tags="|mathematics|riddle|logical-deduction|calculation-puzzle|" AnswerCount="1" CommentCount="3" ClosedDate="2015-03-23T20:34:16.110" ContentLicense="CC BY-SA 3.0" />
  <row Id="10817" PostTypeId="2" ParentId="10719" CreationDate="2015-03-23T18:14:45.903" Score="1" Body="&lt;p&gt;Leaving only the roman numerals (IVXLCDM) of this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 bag iceberg greens&#xA;1 carton Alex's extreme simmered eggs&#xA;2 ounces olives&#xA;1 ounce shaved parmigiano-reggiano&#xA;1 measure Lavinia's dressing&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Gives you many numbers that aren't valid, mostly because of the &lt;code&gt;mcd&lt;/code&gt;.  Since one of the clues is to remove anything that is too much, get rid of these digits and just use the ones that are &lt;code&gt;ivxl&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; i = 1, &#xA; lxxi = 71,&#xA; liv = 54,&#xA; viii = 8,&#xA; lviii = 58,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The only thing similar between a crouton and an anchovy are the number of letters.  Thus, I propose the combination is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 71 54 8 58 7&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This doesn't take into account the amounts - perhaps they are not needed?  Also, it doesn't make any attempt at re-ordering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;EDIT&lt;/em&gt;&lt;/strong&gt;: Corrected the 4th number.  Also, if we perform the same operation on anchovy, and crouton, we get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; anchovy -&gt; cv, crouton -&gt; c&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These are different numbers, so I don't think we can use this operation, unless there is a mistake?&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="1592" LastEditDate="2015-03-23T18:42:50.437" LastActivityDate="2015-03-23T18:42:50.437" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10818" PostTypeId="2" ParentId="10806" CreationDate="2015-03-23T18:16:31.040" Score="2" Body="&lt;p&gt;Second attempt. you are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; death&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I make many feel like mice.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many are afraid of it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some think I'm the open.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some are ok with it, or learn to be&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But sometimes I am nice,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some very sick/aged people find it good to have&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No need for doping.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is soothing when it is overtaking you so all pain gone&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'll see what I am after,&#xA;Maybe you'll even say &quot;Yay!&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Finally most will accept it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I rarely hear laughter,&#xA;No matter what the day.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Acceptance is ok but no one laughs/ get delighted to face it.. yippee i am gonna die&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Half of this is crucial,&#xA;The other not so much.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This piece not getting&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This puzzle may be social,&#xA;Let's see who's super clutch.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ofcourse super clutch is death no one can escape it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10210" LastActivityDate="2015-03-23T18:16:31.040" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10819" PostTypeId="2" ParentId="10785" CreationDate="2015-03-23T18:32:48.460" Score="13" Body="&lt;p&gt;I was thinking of &quot;non-mathematical solution&quot; that would have well-defined and accepted operators, and that telling which operators are acceptable, and one that came to my mind was that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; evaluate here means &quot;matches a regular expression&quot;, and the solution is for example &lt;code&gt;8*5+7[5-9]+3&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1957" LastActivityDate="2015-03-23T18:32:48.460" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10825" PostTypeId="2" ParentId="10700" CreationDate="2015-03-23T18:58:18.210" Score="18" Body="&lt;p&gt;The solution is simple: Build an elevated two mile long runway and anchor all of its supports on the island land.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything that touches the ground is within the boundaries of the island. Since we are talking about a 3D structure and the island boundaries is something only valid on 2D. So, this is arguably valid since there is no clear definition of island boundary in airspace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/G5cvq.png&quot; alt=&quot;Elevated runway&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets see if it passes all the requirements:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The runway must be flat. - $\color{green}{\text{OK}}$&lt;/li&gt;&#xA;&lt;li&gt;The runway must be at least 2 miles long. - $\color{green}{\text{OK}}$&lt;/li&gt;&#xA;&lt;li&gt;You should remain within the boundaries of the island. - $\color{orange}{\text{I am abusing this, but it passes as explained above}}$&lt;/li&gt;&#xA;&lt;li&gt;The island is on planet Earth. - $\color{green}{\text{OK}}$&lt;/li&gt;&#xA;&lt;li&gt;Usual laws of physics applies. - $\color{green}{\text{OK}}$&lt;/li&gt;&#xA;&lt;li&gt;No other form of transport. - $\color{green}{\text{OK}}$&lt;/li&gt;&#xA;&lt;li&gt;The plane cannot take off from water or perform any part of its takeoff run on water. - $\color{green}{\text{It is not on water, it is on over-water airspace, so it's OK}}$&lt;/li&gt;&#xA;&lt;li&gt;Flight safety legislation does not apply. - $\color{green}{\text{OK}}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;About the picture, I know that it would be easier to build it along the diagonal, but this way it is more beautiful and the island size and shape almost do not matter in this answer.&lt;/p&gt;&#xA;" OwnerUserId="5044" LastEditorUserId="5044" LastEditDate="2015-03-23T19:03:29.327" LastActivityDate="2015-03-23T19:03:29.327" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10826" PostTypeId="1" AcceptedAnswerId="10827" CreationDate="2015-03-23T19:21:28.080" Score="1" ViewCount="266" Body="&lt;p&gt;Three forecasters, Phil, Jim and Bob, come out of their office during a rainstorm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Phil&lt;/strong&gt;: The weather won't change until the next full moon.&lt;br&gt;&#xA;&lt;strong&gt;Jim&lt;/strong&gt;: I agree, and the sun won't shine in 48 hours.&lt;br&gt;&#xA;&lt;strong&gt;Bob&lt;/strong&gt;: I agree more with Jim than Phil!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Why was Bob so certain?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-03-23T19:30:45.530" Title="Forecasters strike back!" Tags="|lateral-thinking|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10827" PostTypeId="2" ParentId="10826" CreationDate="2015-03-23T19:25:13.200" Score="5" Body="&lt;p&gt;Bob agrees more with Jim because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it's night time. In 48 hours, it will again be night time and the sun will have set.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alternate answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bob is a &quot;yes-man&quot;. Because Jim had two things to agree with (his agreement with Phil and his statement about the sun), that's one more thing to agree with than Phil has. Bob agrees with Jim twice and only agrees with Phil once, therefore, he agrees with Jim more.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-03-23T19:30:45.530" LastActivityDate="2015-03-23T19:30:45.530" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10828" PostTypeId="2" ParentId="10700" CreationDate="2015-03-23T19:56:17.830" Score="-2" Body="&lt;p&gt;&lt;strong&gt;How advanced are we in technology?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just build a treadmill-like runway in a straight line across the Island (1 mile long) with the plane going against the direction it rotates.  The plane pick up the speed required to takeoff, then you stop the treadmill runway.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flight safety legislation really does not apply here*&lt;/p&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="5373" LastEditDate="2015-03-24T00:00:04.583" LastActivityDate="2015-03-24T00:00:04.583" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10829" PostTypeId="2" ParentId="284" CreationDate="2015-03-23T19:59:34.700" Score="3" Body="&lt;p&gt;As before in the original twelve-balls problem, we can do this with a &lt;a href=&quot;https://puzzling.stackexchange.com/a/224/88&quot;&gt;fixed weighing schedule&lt;/a&gt; in pretty much the exact same way. It just takes a little more fiddling with possibilities to make sure you're putting an even number of balls on the scale for each weighing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, with only 11 balls, the schedule might look like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1  2  3  4  5  6  7  8  9  10 11&#xA;R        R  L  L     L  L  R  R&#xA;   R  L     L     R  R  L  L  R&#xA;      R  L     L  R  L  R  L  R&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Notice we're now using the &lt;code&gt;LLL&lt;/code&gt;/&lt;code&gt;RRR&lt;/code&gt; pair that we didn't use in the 12-balls case, and this case doesn't look nearly as symmetric as the cases before it because we had to fiddle around a bit. We couldn't do the problem just by removing one of the cases in the original weighing schedule, since that would result in at least one of the weighings having only seven balls placed on it, which cannot be divided evenly among the two pans.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For 3 balls (the maximum you can do with two weighings), it looks like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1  2  3&#xA;L     R&#xA;   R  L&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For 39 balls and four weighings, it might look like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1  2  3  4  5  6  7  8  9  10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 &#xA;R           L  L  L           L  L  L           R  R  R     R  R  L     R  L  L     R  L  R     R  L  R  L  R  R  L  &#xA;   L        L        R  L     R        L  L     R  R     R  L  L     R  L  R     L  R     R  L  L  R  R  L  R  L  R&#xA;      R        L     R     R     R     R     R  R     R  R  L     R  L  R     R  R  L  L     L  L  L  L  L  L  L  L  &#xA;         L        L     L  R        R     R  L     R  R  R     L  R  L     L  L  L     R  L  R  L  L  R  R  L  R  R  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'll leave it as an exercise to the reader to do the general cases from 4 to 10, and 13 to 38 balls.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-03-23T19:59:34.700" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10830" PostTypeId="2" ParentId="10816" CreationDate="2015-03-23T20:01:03.577" Score="3" Body="&lt;p&gt;Let's say a dog is not free, but is cheaper than a goat, which is cheaper than a camel, and all prices are integers. And further, let's say that the men initially each had the same number of camels and each ended up with some larger number of goats (again the same amount each). Then we have a number of solutions...&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th&gt;Camels each&lt;/th&gt;&#xA;&lt;th&gt;Camel price&lt;/th&gt;&#xA;&lt;th&gt;Total&lt;/th&gt;&#xA;&lt;th&gt;Goat price&lt;/th&gt;&#xA;&lt;th&gt;Goats each&lt;/th&gt;&#xA;&lt;th&gt;Dog price&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;1&lt;/td&gt;&#xA;&lt;td&gt;2&lt;/td&gt;&#xA;&lt;td&gt;4&lt;/td&gt;&#xA;&lt;td&gt;No solution&lt;/td&gt;&#xA;&lt;td&gt;&lt;/td&gt;&#xA;&lt;td&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;2&lt;/td&gt;&#xA;&lt;td&gt;4&lt;/td&gt;&#xA;&lt;td&gt;16&lt;/td&gt;&#xA;&lt;td&gt;No solution&lt;/td&gt;&#xA;&lt;td&gt;&lt;/td&gt;&#xA;&lt;td&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;3&lt;/td&gt;&#xA;&lt;td&gt;6&lt;/td&gt;&#xA;&lt;td&gt;36&lt;/td&gt;&#xA;&lt;td&gt;No solution&lt;/td&gt;&#xA;&lt;td&gt;&lt;/td&gt;&#xA;&lt;td&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;4&lt;/td&gt;&#xA;&lt;td&gt;8&lt;/td&gt;&#xA;&lt;td&gt;64&lt;/td&gt;&#xA;&lt;td&gt;6&lt;br&gt;5&lt;/td&gt;&#xA;&lt;td&gt;5&lt;br&gt;6&lt;/td&gt;&#xA;&lt;td&gt;4&lt;br&gt;4&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;5&lt;/td&gt;&#xA;&lt;td&gt;10&lt;/td&gt;&#xA;&lt;td&gt;100&lt;/td&gt;&#xA;&lt;td&gt;8&lt;br&gt;7&lt;br&gt;6&lt;/td&gt;&#xA;&lt;td&gt;6&lt;br&gt;7&lt;br&gt;8&lt;/td&gt;&#xA;&lt;td&gt;4&lt;br&gt;2&lt;br&gt;4&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;6&lt;/td&gt;&#xA;&lt;td&gt;12&lt;/td&gt;&#xA;&lt;td&gt;144&lt;/td&gt;&#xA;&lt;td&gt;10&lt;br&gt;7&lt;br&gt;5&lt;/td&gt;&#xA;&lt;td&gt;7&lt;br&gt;10&lt;br&gt;14&lt;/td&gt;&#xA;&lt;td&gt;4&lt;br&gt;4&lt;br&gt;4&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;7&lt;/td&gt;&#xA;&lt;td&gt;14&lt;/td&gt;&#xA;&lt;td&gt;196&lt;/td&gt;&#xA;&lt;td&gt;12&lt;br&gt;8&lt;br&gt;6&lt;/td&gt;&#xA;&lt;td&gt;8&lt;br&gt;12&lt;br&gt;16&lt;/td&gt;&#xA;&lt;td&gt;4&lt;br&gt;4&lt;br&gt;4&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;etc. The interesting solution among these is the case when 5 camels each are sold and 7 goats each are bought, and the dog only costs 2.&lt;/p&gt;&#xA;" OwnerUserId="8875" LastEditorUserId="69582" LastEditDate="2021-01-17T17:44:46.310" LastActivityDate="2021-01-17T17:44:46.310" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10831" PostTypeId="2" ParentId="10806" CreationDate="2015-03-23T20:20:31.833" Score="2" Body="&lt;p&gt;Could it be  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Twitter?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I make many feel like mice.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Twitter also means the chirp sound from mice  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some think I'm the open.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Open social media&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But sometimes I am nice,&lt;br&gt;&#xA;No need for doping.    &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Twitting means teasing someone, a nice way to heat it up without doping&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'll see what I am after,&lt;br&gt;&#xA;Maybe you'll even say &quot;Yay!&quot;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People goes &quot;Yay I use Twitter too!&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I rarely hear laughter,&lt;br&gt;&#xA;No matter what the day.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  You might hear laughter but mostly it's all typing with silence&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Half of this is crucial,&lt;br&gt;&#xA;The other not so much.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People who use Twitter will found it crucial, while others not so much  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This puzzle may be social,&lt;br&gt;&#xA;Let's see who's super clutch.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again Open social media  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2015-03-23T20:20:31.833" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10832" PostTypeId="2" ParentId="10806" CreationDate="2015-03-23T20:32:57.400" Score="2" Body="&lt;p&gt;It may be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Luck&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I make many feel like mice.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Most people are unlucky&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some think I'm the open.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think this isn't the important part (half of this is crucial, the other not so much)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But sometimes I am nice&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No need to explain here, I guess&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No need for doping&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Indeed, doping doesn't affect luck&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'll see what I am after&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can't be lucky until you win!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Maybe you'll even say &quot;Yay!&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you win...  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I rarely hear laughter,&#xA;No matter what the day.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Winners are few.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Half of this is crucial,&#xA;The other not so much.&#xA;This puzzle may be social,&#xA;Let's see who's super clutch.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I don't think there's a hint here, just the half important thing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-03-23T20:32:57.400" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10833" PostTypeId="2" ParentId="10806" CreationDate="2015-03-23T20:37:56.850" Score="2" Body="&lt;p&gt;My attempt (along with some thoughts):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Needles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I make many feel like mice.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mice are experimented on, needles often make people feel the same.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some think I'm the open.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The phrasing here is odd. I'm guessing it's important but haven't figured out why yet. For this response, though, some see needles as opening up your veins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But sometimes I am nice,&#xA;No need for doping.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Needling someone is poking fun (but in a nice way) and another use for the word other than drugs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'll see what I am after,&#xA;Maybe you'll even say &quot;Yay!&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When your blood is being drawn, you can see it. And you say &quot;Yay!&quot; because it's good that you have blood. :) A little weaker here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I rarely hear laughter,&#xA;No matter what the day.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Even if you are drawing blood, you're probably not laughing about it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Half of this is crucial,&#xA;The other not so much.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Here's a bit of wordplay. Half of it is &quot;need&quot; = &quot;crucial&quot;. The other half is &quot;not so much&quot; i.e. it's &quot;less&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This puzzle may be social,&#xA;Let's see who's super clutch.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He's just putting this in because it's a tough puzzle and he's anticipating a community effort. Which he's got! Thought &quot;super clutch&quot; may be another important phrase here that I'm not catching.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2015-03-23T20:37:56.850" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10834" PostTypeId="2" ParentId="10700" CreationDate="2015-03-23T21:02:13.207" Score="1" Body="&lt;p&gt;If the runway is approximately straight and level, as it should be for it to be &quot;physically possible for the plane to take off,&quot; then there is no way to fit a 2-mile-long island on a 1-mile square island. Especially if you consider that a plane requiring a &lt;em&gt;minimum&lt;/em&gt; of two miles to take off should probably have a runway that is &lt;em&gt;longer&lt;/em&gt; than two miles, in case there are any sub-optimal conditions for takeoff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how can we build such a runway while &quot;remaining within the boundaries of the island&quot;? Simple. &lt;strong&gt;Expand the island.&lt;/strong&gt; Use &lt;a href=&quot;http://en.wikipedia.org/wiki/Land_reclamation&quot; rel=&quot;nofollow&quot;&gt;land reclamation&lt;/a&gt; to build up the shores, so eventually the island gets bigger, and then build a runway on it. Some commenters have observed that there isn't much use for a one-square-mile island, so this fixes that problem by giving you more usable space, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, this is much less far-fetched than some of the other proposed constructions. In fact, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Genoa_Cristoforo_Colombo_Airport&quot; rel=&quot;nofollow&quot;&gt;airport in Genoa, Italy&lt;/a&gt; is built on an artificial peninsula of reclaimed land, proving the feasability of this idea.&lt;/p&gt;&#xA;" OwnerUserId="8776" LastActivityDate="2015-03-23T21:02:13.207" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10835" PostTypeId="2" ParentId="10727" CreationDate="2015-03-23T21:05:36.580" Score="3" Body="&lt;p&gt;The solution is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For each nonnegative integer &lt;em&gt;n&lt;/em&gt;, create a binary tree thus.  The root vertex is n.  If n is a composite number, it has two child vertices.  The values in these child vertices, {a, b} are chosen such that a x b = n and a is the largest integer divisor less than or equal to the square root of n. b is the smallest integer divisor greater than or equal to the square root of n.  The rest of the tree is defined recursively; the child nodes of a and b are factors of those numbers, if they are composite, and so on.  Thus the leaf nodes of the tree will all be prime numbers; the only exceptions to this are the trees constructed for 0 and 1, which have only one vertex. Each of the binary trees are placed into isomorphic groups, and these are numbered 0, 1, 2.. and so on.  f(n) is simply the index of the group which its tree belongs to.  0, 1 and prime numbers all have only one vertex, thus these trees form the first group and the function will take the value of 0.  The first composite number, 4, has a tree with 3 nodes and thus is in the next group with index 1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2154" LastActivityDate="2015-03-23T21:05:36.580" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10836" PostTypeId="1" AcceptedAnswerId="10917" CreationDate="2015-03-23T21:09:12.083" Score="6" ViewCount="446" Body="&lt;p&gt;Mortals, beware!&lt;br&gt;&#xA;A divine trapped in square!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only silences to your prayers,&lt;br&gt;&#xA;Until a step through the stairs.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Powers to make you a millionaire,&lt;br&gt;&#xA;The only one with a spear.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beneath the future heir,&lt;br&gt;&#xA;Four does not make 2 pairs!   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint #1 &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &quot;&lt;strong&gt;Beneath&lt;/strong&gt;&quot; the future heir is relevance; The line can be replaced by &quot;&lt;strong&gt;After&lt;/strong&gt;&quot; the future heir  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  The only one with spear; &lt;em&gt;She&lt;/em&gt; (hehe) is the only one that has a weapon!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="5188" LastEditDate="2015-03-24T18:53:40.213" LastActivityDate="2015-03-25T12:36:22.487" Title="Trapped in square; Who am I?" Tags="|riddle|" AnswerCount="6" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10837" PostTypeId="2" ParentId="429" CreationDate="2015-03-23T21:13:17.093" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; R2 L2 U2 D2 F2 B2 R L F B R' L'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;No sides should touch.&lt;/p&gt;&#xA;" OwnerUserId="10556" LastEditorUserId="51105" LastEditDate="2021-10-19T15:59:16.837" LastActivityDate="2021-10-19T15:59:16.837" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="10838" PostTypeId="2" ParentId="10765" CreationDate="2015-03-23T21:40:19.007" Score="3" Body="&lt;p&gt;Second attempt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take into account my first answer. Then.. put in the Crossword aspect and we get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Against the Grain! &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9412" LastActivityDate="2015-03-23T21:40:19.007" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10839" PostTypeId="1" AcceptedAnswerId="10866" CreationDate="2015-03-23T22:13:02.953" Score="27" ViewCount="8416" Body="&lt;p&gt;An ant walks on a number line, starting at location $x=7$. Each second, it randomly moves one space either left ($-1$) or right ($+1$), with equal chance. At $x=0$ and $x=10$ are drops of honey; the ant stops moving when it reaches one of them. What are the respective probabilities of it stopping at each one?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There's multiple ways to compute the answer, but I like this as a puzzle because there's a slick short solution. So, try to find the cleanest solution you can!&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-05-21T06:26:53.840" Title="Ant walking on a number line" Tags="|mathematics|probability|" AnswerCount="19" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="10840" PostTypeId="2" ParentId="10782" CreationDate="2015-03-23T22:17:47.343" Score="1" Body="&lt;p&gt;Not a full answer, but some ideas I've collected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After seeing the hints, I&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; broke down all the numbers into their prime factorisations&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;to get&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;2*3*37,5*37 gives NOTRQNPQRST.&lt;br&gt;3*5*37,19*37 gives NOPNPQRST.&lt;br&gt;2*2*5*37,3*37 gives OPSTNOPQT.&lt;br&gt;2*13*37,2*3*3*37 gives PSTNOPQRS.&lt;br&gt;3*37,3*7*37 does &lt;em&gt;not&lt;/em&gt; give NORQNOPS.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The most obvious thing to notice is that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 37 is a factor of all the numbers involved.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What can we say about the other factors? Well...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;(&lt;strong&gt;2&lt;/strong&gt;,3;&lt;strong&gt;5&lt;/strong&gt;) -&gt; NOTRQNPQRST, which starts with &lt;strong&gt;2&lt;/strong&gt; adjacent letters and ends with &lt;strong&gt;5&lt;/strong&gt; adjacent letters.&lt;br&gt;(&lt;strong&gt;3&lt;/strong&gt;,&lt;strong&gt;5&lt;/strong&gt;;19) -&gt; NOPNPQRST, which starts with &lt;strong&gt;3&lt;/strong&gt; adjacent letters and ends with &lt;strong&gt;5&lt;/strong&gt; adjacent letters.&lt;br&gt;(&lt;strong&gt;2&lt;/strong&gt;,&lt;strong&gt;2&lt;/strong&gt;,5;3) -&gt; OPSTNOPQT, which starts with &lt;strong&gt;2&lt;/strong&gt; adjacent letters and then another &lt;strong&gt;2&lt;/strong&gt; adjacent letters.&lt;br&gt;(&lt;strong&gt;2&lt;/strong&gt;,13;&lt;strong&gt;2&lt;/strong&gt;,&lt;strong&gt;3&lt;/strong&gt;,&lt;strong&gt;3&lt;/strong&gt;) -&gt; OPSTNOPQRS, which starts with &lt;strong&gt;2&lt;/strong&gt; adjacent letters followed by another &lt;strong&gt;2&lt;/strong&gt; adjacent letters and ends with &lt;strong&gt;3&lt;/strong&gt; adjacent letters preceded by another &lt;strong&gt;3&lt;/strong&gt; adjacent letters (making actually 6 adjacent letters).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Am I thinking along the right lines here?&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-03-23T22:17:47.343" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10841" PostTypeId="2" ParentId="10839" CreationDate="2015-03-23T22:20:05.493" Score="4" Body="&lt;p&gt;The answer is that the ant will reach the leftmost spot with a probability of $\frac{3}{10}$, and the rightmost spot with a probability of $\frac{7}{10}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is an instance of the &lt;a href=&quot;http://mathworld.wolfram.com/GamblersRuin.html&quot; rel=&quot;nofollow&quot;&gt;Gambler's Ruin&lt;/a&gt; problem. According to the problem's statement, the probability of an ant reaching a point of distance $d_1$ compared to another point of distance $d_2$ is $\frac{d_2}{d_1 + d_2}$. Generally this formula is obtained by evaluating the limit of a Markov chain (as the Gambler's Ruin problem is usually used as a classical example of an application of Markov chains in the first place), but there seems to be a more elegant way of resolving this specific case as hinted by xnor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other answerers seem to have given more elegant solutions, so I will leave this one as is.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-03-24T16:29:54.647" LastActivityDate="2015-03-24T16:29:54.647" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10842" PostTypeId="2" ParentId="10806" CreationDate="2015-03-23T22:27:46.240" Score="3" Body="&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Masters/Golf&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I make many feel like mice.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are many great players that may make others feel small&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some think I'm the open.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Open - other big golf events&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But sometimes I am nice,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sometimes you win? It can be fun to watch?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No need for doping.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Doping doesn't have as much an effect on golf as other sports.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'll see what I am after,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's after March Madness, and we will see/watch that?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Maybe you'll even say &quot;Yay!&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We can be happy that it's happening?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I rarely hear laughter,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People at golf tournaments are often very quiet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No matter what the day.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are quiet ever day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Half of this is crucial,&lt;br&gt;&#xA;The other not so much.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first stanza seems more relevant&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This puzzle may be social,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Golf can be a social game&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's see who's super clutch.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It helps to be clutch in golf!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2516" LastActivityDate="2015-03-23T22:27:46.240" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10843" PostTypeId="2" ParentId="10836" CreationDate="2015-03-23T22:51:40.530" Score="1" Body="&lt;p&gt;A wild guess, but I'll say, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a dream&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mortals, beware! A divine trapped in square!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the dream comes when you are trapped in the square pillow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Only silences to your prayers, Until a step through the stairs.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; before the dream it is silent, until the dream starts&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Powers to make you an millionaire, The only one with a spear.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; in the dream you can be everything you want (a millionaire, for example) or you can be the only one with a given knowledge&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Beneath the future heir, Four does not make 2 pairs!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the four sides of the pillow or the bed are equal, so they are not 2 pairs &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5283" LastActivityDate="2015-03-23T22:51:40.530" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10844" PostTypeId="2" ParentId="10836" CreationDate="2015-03-23T23:00:16.780" Score="3" Body="&lt;p&gt;I'll go for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; playing cards.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Mortals, beware!&lt;br&gt;&#xA; A divine trapped in square!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Playing cards are rectangular (close enough to square). The &quot;divine&quot; bit could be a reference to &lt;em&gt;tarot&lt;/em&gt; cards.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Only silences to your prayers,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You might pray to win a game of cards...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Until a step through the stairs.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ... but only luck will see your prayers answered.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Powers to make you an millionaire,&lt;br&gt;&#xA;The only one with a spear.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If your luck runs good (or you cheat, or you run a casino)$^1$, you can become a millionaire by playing cards. Many games have only one winner.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Beneath the future heir,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Jack or Knave in a given suit can sometimes be called the Prince, heir to the throne.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Four does not make 2 pairs!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Four suits in a pack of cards.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$^1$ Well, OK, the two possibilities in brackets are the only real ways of doing it, unless your name's Mat Cauthon.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-03-23T23:00:16.780" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10845" PostTypeId="1" AcceptedAnswerId="10846" CreationDate="2015-03-23T23:09:49.500" Score="11" ViewCount="632" Body="&lt;p&gt;Randall is on vacation in Australia, and his older sister has planned out the whole thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One night, when they've returned to their hotel room, she hands him a slip of paper and says, &quot;This is where we'll be going tomorrow. See if you can figure it out.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sxb8U.jpg&quot; alt=&quot;A map with a strange path written on it&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Randall looks at the slip of paper, bewildered and puzzled. Just where &lt;em&gt;are&lt;/em&gt; they going?&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-03-23T23:11:15.233" Title="Where are we going?" Tags="|riddle|visual|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10846" PostTypeId="2" ParentId="10845" CreationDate="2015-03-23T23:11:15.233" Score="12" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are going to &lt;a href=&quot;http://en.wikipedia.org/wiki/Uluru&quot;&gt;Uluru&lt;/a&gt;, a sandstone rock formation in Australia, as hinted by the first letter of the directions Up, Left, Up, Right, Up.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-03-23T23:11:15.233" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10847" PostTypeId="2" ParentId="10839" CreationDate="2015-03-23T23:13:58.033" Score="5" Body="&lt;p&gt;The ant will reach 0 with a probability of $\frac{3}{10}$, and 10 with a probability of $\frac{7}{10}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $p(n)$ be the probability of reaching $0$ from point $n$. By symmetry, we see that for small $x$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p(n) = \frac{p(n-x)+p(n+x)}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, the chance of reaching 0 from a point is the average of the chance of reaching 0 from any two equidistant points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p(7) = \frac{p(4)+p(10)}{2}$ = $\frac{p(4)}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p(4) = \frac{p(3)+p(5)}{2}$ = $\frac{p(3)+\frac{1}{2}}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But again by symmetry, we can see that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p(n) = 1 - p(10 - n)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p(7) = 1 - p(3)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p(4) = 2*p(7) = \frac{p(3)+\frac{1}{2}}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2*p(7) = \frac{1-p(7)+\frac{1}{2}}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$4*p(7) = \frac{3}{2} - p(7)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5*p(7) = \frac{3}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p(7) = 3/10$&lt;/p&gt;&#xA;" OwnerUserId="9523" LastActivityDate="2015-03-23T23:13:58.033" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10848" PostTypeId="1" AcceptedAnswerId="10850" CreationDate="2015-03-23T23:33:54.660" Score="4" ViewCount="447" Body="&lt;p&gt;Not the highest nor the lowest&lt;br&gt;&#xA;With letters on my chest&lt;br&gt;&#xA;I often play with others&lt;br&gt;&#xA;Some may think I am the best  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I usually feel tension&lt;br&gt;&#xA;But I think that keeps me straight&lt;br&gt;&#xA;I shout out loud and bold&lt;br&gt;&#xA;And it's that shout that makes me great  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My master and I are best of friends&lt;br&gt;&#xA;And our friendship keeps improving&lt;br&gt;&#xA;My friends and I unite as one&lt;br&gt;&#xA;And that helps keep things moving  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who am I?  &lt;/p&gt;&#xA;" OwnerUserId="2516" LastEditorUserId="5373" LastEditDate="2016-10-21T14:57:15.260" LastActivityDate="2016-10-21T14:57:15.260" Title="Not the highest - Riddle" Tags="|riddle|rhyme|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10849" PostTypeId="2" ParentId="10839" CreationDate="2015-03-23T23:35:10.370" Score="9" Body="&lt;p&gt;The simple approach is to see it as a coin flipping game where one player starts with $3$, the other with $7$, and the winner of each flip collects $1$ from the loser.  As each flip is fair, the game is fair.  As the person who starts with $3$ wins with $10$, he must win $0.3$ of the time.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2015-03-23T23:35:10.370" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="10850" PostTypeId="2" ParentId="10848" CreationDate="2015-03-23T23:39:39.237" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A &lt;strong&gt;cello&lt;/strong&gt; or &lt;strong&gt;viola&lt;/strong&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Not the highest nor the lowest&lt;br&gt;&#xA;With letters on my chest&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to &lt;strong&gt;d'alar'cop&lt;/strong&gt; for helping me with these two lines!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A cello and viola are between the violin (highest) and the double bass (lowest). Both cellos and violas have what look like letter S's printed on their middles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I often play with others&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One of many instruments in an orchestra.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Some may think I am the best&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;em&gt;first&lt;/em&gt; violin maybe?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I usually feel tension&lt;br&gt;&#xA;But I think that keeps me straight&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The strings on a stringed instrument are under tension, which keeps them taut and straight.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I shout out loud and bold&lt;br&gt;&#xA;And it's that shout that makes me great&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sound emitted by a musical instrument is what it's made for.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My master and I are best of friends&lt;br&gt;&#xA;And our friendship keeps improving&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A musician and his instrument can have a strong bond.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My friends and I unite as one&lt;br&gt;&#xA;And that helps keep things moving&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sound of many instruments playing together in an orchestra can indeed be deeply moving.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-24T00:30:46.170" LastActivityDate="2015-03-24T00:30:46.170" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10851" PostTypeId="2" ParentId="10700" CreationDate="2015-03-23T23:48:29.210" Score="4" Body="&lt;p&gt;Surprisingly I don't see the obvious answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Use a different plane, for goodness sake! One that can actually take off and land in 1.4 miles. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8875" LastActivityDate="2015-03-23T23:48:29.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10852" PostTypeId="2" ParentId="10700" CreationDate="2015-03-23T23:58:42.097" Score="0" Body="&lt;p&gt;Go down one edge of the island for .9 miles then turn and go along the diagonal of the island. Which gives a length &gt;2.17 miles.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2015-03-23T23:58:42.097" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10853" PostTypeId="2" ParentId="10839" CreationDate="2015-03-24T00:17:52.500" Score="13" Body="&lt;p&gt;From $x=7,$ there is an equal chance of reaching $x=4$ and $x=10$. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(score $\frac{1}{2}$ for the $10$ drop). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From $x=4$, there is an equal chance of reaching $x=3$ and $x=5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From $x=5$, there is an equal chance of reaching $x=0$ and $x=10$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(score $\frac{1}{8}$ for the $0$ drop and for the $10$ drop $\to \frac{5}{8}$).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From $x=3$, (now at  $\frac{1}{4}$ chance) the probabilities are reversed compared to $x=7$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(score $\frac{5}{32}$ for the $0$ drop $\to \frac{9}{32}$ and score  $\frac{1}{32}$ for the $10$ drop $\to \frac{21}{32}$).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The remaining probability $\frac{2}{32}$ is assigned to $x=7$, so discard and rebase:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Probability of reaching $0$ is $\frac{9}{30} = \frac{3}{10}$ and probability of reaching $10$ is $\frac{21}{30} = \frac{7}{10}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is NOT what I would call slick..... so probably not the answer you're looking for. :-)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A more generic answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an ant at each (mid) position $x$, the probability of reaching, say, the high-number end $p(x)$ can be calculated after one step as $p(x) = 0.5 (p(x-1) + p(x+1))$.  So each probability is the average of the adjacent position probabilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we know that $p(0)=0$ and $p(10) = 1$, we can show the linear nature of the probabilities in between as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set $p(1) = a$. Then because of the average effect above, we can see that $p(2)= 2a, p(3)= 3a, \ldots, p(10)=10a=1$. So $a=0.1$ and the probability of reaching the high end is $p(x)=0.1x$ - in this specific case, p(7) = 0.7 chance of reaching the high end and $1-0.7=0.3$ chance of reaching the low end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This translates into a general solution for any length of line and any start position; if the high-end drop is at $x=52$ and the ant starts at $x=37$, the probability of reaching the high end is $\frac{37}{52}$ and the probability of reaching the low end is $\frac{15}{52}$.&lt;/p&gt;&#xA;" OwnerUserId="8875" LastEditorUserId="8875" LastEditDate="2015-03-24T16:18:20.810" LastActivityDate="2015-03-24T16:18:20.810" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10854" PostTypeId="2" ParentId="10782" CreationDate="2015-03-24T00:30:59.503" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; With the hint about the periodic tables, (this is not an answer yet), I have identified certain trends:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;The block-code is now hidden. It is not one of the best looking spoiler-markup, but this is the only way.&lt;/h2&gt;&#xA;&#xA;&lt;h2&gt;(The list still needs to be hidden! Help?)&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;# | cols. of 2        | input   | gcf&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;1   NO TR QN PQ RS T-   222^185   37&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;2   NO PN PQ RS T- --   555^703   37&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;3   OP ST NO PQ T- --   740^111   37&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;4   OP ST NO PQ RS --   962^666   74&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;5   NO RQ NO PS -- --   111^777   111&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;X   ?? ?? ?? ?? ?? ??   37 ^703   37&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;All of the answers have an O, N, or P as their first two letters, and, in fact, start with either NO or OP.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;PQRST, the consecutive letter string, appears in (1) and (2) and PQRS in (4).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;NO is &quot;Nobelium&quot; or N O is &quot;Nitrogen&quot; + &quot;Oxygen&quot; ($NO_2$ is nitrous oxide, if I'm not mistaken...)&lt;/li&gt;&#xA;&lt;li&gt;OP is PO backwards which is Polonium or O P is &quot;Oxygen&quot; + &quot;Phosphorous&quot; (Which is $P_4O_6$, Tetraphosphorus hexoxide)&lt;/li&gt;&#xA;&lt;li&gt;TR could be short for &lt;strong&gt;Tr&lt;/strong&gt;ansition Metal, though this is generally inconsistent (to my knowledge).&lt;/li&gt;&#xA;&lt;li&gt;In all 5, it is possible to make SPQR (coincidence? I think not! We are facing the Roman takeover of &lt;code&gt;$COUNTRY&lt;/code&gt;!)&lt;/li&gt;&#xA;&lt;li&gt;(4) and (5) are the only even-lettered inputs.&lt;/li&gt;&#xA;&lt;li&gt;We can see that each one of the five &lt;em&gt;original&lt;/em&gt; inputs have a 3-digit number with the same digit (111, 222, 555, 666, and 777). Interestingly enough, it omits 333, 444, 888, and 999.&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Furthermore, &lt;em&gt;your&lt;/em&gt; input is the &lt;em&gt;only&lt;/em&gt; input that (a) has no such number and (b) has a 2-digit number.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;*LE GASP* (4) and (5) are the only problems to NOT have 37 as a GCF! We might deduce that your answer should &lt;em&gt;not&lt;/em&gt; be an even-numbered amount. It may also be safe to assume $\text{len}(answer)&amp;gt;8$. &lt;/li&gt;&#xA;&lt;li&gt;(begin editing #1) All the letters are composed of the letter-strait NOPQRST, and each solution (must) contain one of each.&#xA;&lt;ul&gt;&#xA;&lt;li&gt;or not. See (5) -- no T! (So we know he's definitely &lt;em&gt;not&lt;/em&gt; British. Haha, $T_{ea}$!)&lt;/li&gt;&#xA;&lt;li&gt;... and (3). No R. And no pun.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;See chart #2.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;WHY ONE NEEDS TO READ!&lt;/em&gt; (That is, what I missed by &lt;strong&gt;not&lt;/strong&gt; reading): (5) was killed. Huh-boy. Well, learn from others mistakes! You'd think that after witnessing someones death, you'd &lt;strong&gt;not&lt;/strong&gt; try and attempt the feat yourself!&lt;/li&gt;&#xA;&lt;li&gt;Why did (5) die?&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Perhaps his answer was too short?&lt;/li&gt;&#xA;&lt;li&gt;(5) is the only answer with two 3-like-digits. It is also one of two queries in which &lt;code&gt;x&lt;/code&gt; in &lt;code&gt;x^y&lt;/code&gt; is the GCF. This could simply mean &lt;code&gt;KILL&lt;/code&gt; -- tough luck for you. Maybe it means naught at all.&#xA;&lt;hr&gt;&lt;/li&gt;&#xA;- &#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;Chart #2&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;#   N's | O's | P's | Q's | R's | S's | T's | TOTAL&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;1    2     1     1     2     2     1     2     11&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;2    2     1     2     1     1     1     1      9&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;3    1     2     2     1     0     1     2      9&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;4    1     2     2     1     1     2     1     10&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;5    2     2     1     1     1     1     0      8&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;X    ?     ?     ?     ?     ?     ?     ?    ?????&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;$    8     8     8     6     5     6     6     47&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So what might we deduce? We could receive an insanely long string, containing 1 N, 1 O, 1 P, 3 Q, 4 R, 3 S, and 3 T (to make a grand total of 9) or 0 N, 0 O, 0 P, 2 Q, 3 R, 2 S, and 2 T (to make a total of 8).&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Meditation&lt;/h1&gt;&#xA;&#xA;&lt;h2&gt;Where I 'wrap' up and theorize&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We must know that the answer is somewhat easy to come by, for the guests wrote these combination on some board.&lt;/li&gt;&#xA;&lt;li&gt;We may also conclude that there is some method that these guests have learned from your dreaded nano-tubist; maybe some sort of program? (Honestly, if this is the case, I'll hire a hacker to nab some of those phones and jack the program and &lt;em&gt;voila!&lt;/em&gt; dreaded nano-tubist goes &quot;bye-bye&quot;!)&lt;/li&gt;&#xA;&lt;li&gt;If we only had some same test case for &lt;code&gt;x^y&lt;/code&gt; for &lt;code&gt;x&lt;/code&gt;! We have one for &lt;code&gt;y&lt;/code&gt;, that is, &lt;code&gt;703&lt;/code&gt; in (X) and (2).&lt;/li&gt;&#xA;&lt;li&gt;It is curious why the guard changes his methodology slightly--maybe he anticipated you, and you are doomed to fail! In that case, just use your amazing &lt;em&gt;dues ex machina&lt;/em&gt; skills and whip right in there!&lt;/li&gt;&#xA;&lt;li&gt;From 12. on my list, I notice that 5. is killed. Perhaps there are only to be four guests, and any further 'guests' are eliminated.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Again, this are just my ideas and hypothesizing. I doubt that any of these are correct. &lt;code&gt;&amp;lt;/muse&amp;gt;&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;Any help is appreciated with the spoiler markup!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*WIP*)&lt;/p&gt;&#xA;" OwnerUserId="7307" LastEditorUserId="10384" LastEditDate="2015-04-08T08:55:03.150" LastActivityDate="2015-04-08T08:55:03.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10855" PostTypeId="2" ParentId="10806" CreationDate="2015-03-24T00:41:49.623" Score="3" Body="&lt;p&gt;A wordplay idea: could it be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ICE?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I make many feel like mice.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Turn &lt;em&gt;any&lt;/em&gt;thing into ice and &quot;many&quot; becomes &quot;mice&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Some think I'm the open.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Open as in golf - you can play &lt;a href=&quot;http://www.golf.com/photos/world-ice-golf-championship&quot; rel=&quot;nofollow&quot;&gt;ice golf&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But sometimes I am nice,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A common mishearing: e.g. if your ship is stuck in pack ice and you see a gigantic beast approaching you across the floes, you really want to know whether it's saying &quot;I'm a nice monster&quot; or just &quot;I'm an ice monster.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;No need for doping.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ???&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You'll see what I am after,&lt;br&gt;&#xA; Maybe you'll even say &quot;Yay!&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well, if you solve the riddle to find what the OP is after, &quot;Yay!&quot; is certainly an appropriate response.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I rarely hear laughter,&lt;br&gt;&#xA; No matter what the day.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A common metaphor: if someone has an icy demeanour, they won't laugh much.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Half of this is crucial,&lt;br&gt;&#xA; The other not so much.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Half of the word &quot;this&quot; is &quot;is&quot;, which is similar to &quot;ice&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This puzzle may be social,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well, it's certainly attracted a lot of answers (and probably got on the HNQs)!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Let's see who's super clutch.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I've no idea what this means.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Disclaimer: I haven't read all the OP's comments on other answers (though I have checked that nobody's already posted this as an answer), so apologies if I'm trying something that's already been excluded.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-03-24T00:41:49.623" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10857" PostTypeId="2" ParentId="10806" CreationDate="2015-03-24T04:45:22.573" Score="8" Body="&lt;p&gt;New answer. You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Monday&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I make many feel like mice.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Monday is the start of the &quot;rat race&quot; that is our day-to-day jobs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some think I'm the open.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The open to the week. Plus another thing for later&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But sometimes I am nice,&#xA;No need for doping.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sometimes you get Mondays off, and don't need coffee :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'll see what I am after,&#xA;Maybe you'll even say &quot;Yay!&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You see that Monday is after the weekend, which makes you say &quot;yay&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I rarely hear laughter,&#xA;No matter what the day.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is part of the &quot;non-crucial&quot; part, but if you have a &quot;case of the Mondays&quot; it's being sad (hence the title of the puzzle) while you're back at work&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Half of this is crucial,&#xA;The other not so much.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first half of the puzzle is crucial - if you take &quot;the open&quot; (the first letter) of each of the last words of the first 6 lines, it spells Monday! The second half is not so crucial.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This puzzle may be social,&#xA;Let's see who's super clutch.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As evidenced here. I still think these are just referencing the difficulty of the puzzle and it requiring a group effort to solve.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2015-03-24T04:45:22.573" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10858" PostTypeId="2" ParentId="10797" CreationDate="2015-03-24T06:48:18.550" Score="4" Body="&lt;p&gt;I just came up with a solution summing up a total of 396:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   19 11 17 | 47&#xA;    5 29 37 | 71&#xA;    7  3 13 | 23&#xA;   ---------+---&#xA;53 31 43 67 | 61&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In fact there are exactly 8 matrices summing up 396 using primes {3, 5, 7, 11, 13, 17, 19, 29, 37}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will be working around on this problem till I can find a better solution but I think the best possible is the one stated by our friend @schnaader&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: In fact, there are exactly 8 matrices summing up to 366 using primes {3, 5, 7, 11, 13, 17, 23, 31, 37}, just like the one @schnaader showed us!&lt;/p&gt;&#xA;" OwnerUserId="10564" LastEditorUserId="10564" LastEditDate="2015-03-24T07:27:51.290" LastActivityDate="2015-03-24T07:27:51.290" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10859" PostTypeId="2" ParentId="10836" CreationDate="2015-03-24T07:10:47.443" Score="3" Body="&lt;p&gt;My answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Poker&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mortals, beware!&lt;BR&gt;&#xA;A divine trapped in square!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A square poker table&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Only silences to your prayers,&lt;BR&gt;&#xA;Until a step through the stairs.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People pray silently, not sure about stairs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Powers to make you an millionaire,&lt;BR&gt;&#xA;The only one with a spear.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It can get you a lot of money or can kill you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Beneath the future heir,&lt;BR&gt;&#xA;Four does not make 2 pairs!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A 4 of a kind is different from 2 double pairs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4637" LastActivityDate="2015-03-24T07:10:47.443" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10860" PostTypeId="1" AcceptedAnswerId="10861" CreationDate="2015-03-24T07:42:54.607" Score="18" ViewCount="1582" Body="&lt;p&gt;Introducing a &lt;em&gt;super secure&lt;/em&gt; function to crypt your sensitive data - &lt;strong&gt;PolynumCrypt&lt;/strong&gt;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say you wanted to crypt the string $\text{super}$, with a key of $6$.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Get the ASCII values of the string, corresponding to $115,117,112,101,114$ in this case&lt;/li&gt;&#xA;&lt;li&gt;Turn that into a polynomial where the $n\text{th}$ coefficient corresponds to the $n\text{th}$ ASCII value, in this case $115x^4 + 117x^3 + 112x^2 + 101x + 114$&lt;/li&gt;&#xA;&lt;li&gt;Substitute the key value into $x$ to get your crypted number, in this case $179064$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I've written a &lt;a href=&quot;https://ideone.com/8cDWzx&quot;&gt;Python script&lt;/a&gt; to make it easy to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, brute-forcing $256^{n}$ possibilities for a string length $n$ is impossible, which makes this algorithm &lt;em&gt;super secure&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Maybe you can prove me wrong about PolynumCrypt, if you can crack my password:&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\text{polynumcrypt(pass,1) = 1253}$&lt;br&gt;&#xA;$\text{polynumcrypt(pass,1253) = 2102474472933529067195268690993346961762041}$&lt;/p&gt;&#xA;" OwnerUserId="7308" LastEditorUserId="69582" LastEditDate="2023-12-04T17:13:27.190" LastActivityDate="2023-12-04T17:13:27.190" Title="Let's break PolynumCrypt" Tags="|mathematics|cipher|" AnswerCount="3" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="10861" PostTypeId="2" ParentId="10860" CreationDate="2015-03-24T08:07:03.610" Score="19" Body="&lt;p&gt;We don't need to brute-force $256^n$ possibilities &amp;ndash; since the overly large number is written in base $1253$ (that's what the polynomial is for) and ASCII characters are at most $128$, we can just express the large number in base $1253$ and extract the digits exactly as is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wolfram Alpha gives a decomposition of &lt;code&gt;112:48:108:121:110:111:109:49:64:108:115:46:83:69&lt;/code&gt;. These ASCII values then evaluate to: &lt;code&gt;p0lynom1@ls.SE&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-03-24T08:07:03.610" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10862" PostTypeId="2" ParentId="10860" CreationDate="2015-03-24T08:07:13.240" Score="8" Body="&lt;p&gt;Is it?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; p0lynom1@ls.SE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Modulo the encrypted password by 1253 results in the last character. Divide it by 1253 and repeat and you get all characters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastActivityDate="2015-03-24T08:07:13.240" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10863" PostTypeId="2" ParentId="10860" CreationDate="2015-03-24T08:09:31.010" Score="5" Body="&lt;p&gt;The first character of the string has the highest power, therefore it will dominate the polynomial. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dividing the number by (1253^13) results in 112 (p)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remainder of the division can be fed back into the calculation, finally resulting in the plaintext&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; p0lynom1@ls.SE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Mathematica code used to generate the answer:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;string = &quot;&quot;;&#xA;inNum = 2102474472933529067195268690993346961762041;&#xA;pow = 13;&#xA;While[pow &amp;gt;= 0, charfloat = N[inNum/(1253^pow)];&#xA;  charint = Floor[charfloat];&#xA;  string = string &amp;lt;&amp;gt; FromCharacterCode[charint];&#xA;  inNum = inNum - charint*(1253^pow);&#xA;  pow = pow - 1;];&#xA;string&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7529" LastEditorUserId="7529" LastEditDate="2015-03-24T08:21:33.800" LastActivityDate="2015-03-24T08:21:33.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10864" PostTypeId="1" AcceptedAnswerId="10869" CreationDate="2015-03-24T09:12:15.847" Score="24" ViewCount="4616" Body="&lt;p&gt;Viswanathan Anand plays a chess game against Magnus Carlsen. Anand plays white and Magnus plays black.&#xA;They use a non-standard digital double &lt;a href=&quot;http://en.wikipedia.org/wiki/Game_clock&quot; rel=&quot;nofollow noreferrer&quot;&gt;chess clock&lt;/a&gt; that is counting up from zero (instead of the usual counting down towards zero). This double chess clock shows hours, minutes and seconds for both players.&#xA;At the moment when Magnus completes his 40th move, both clocks read 2 hours, 30 minutes, 00 seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 1:&lt;/strong&gt; Was there necessarily a moment in the game, when the clock of one player showed exactly &lt;strong&gt;1 min 51 sec&lt;/strong&gt; less than the clock of the other player?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 2:&lt;/strong&gt; Was there necessarily a moment in the game, when the clock of one player showed exactly &lt;strong&gt;1 min 57 sec&lt;/strong&gt; less than the clock of the other player?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorDisplayName="user20" LastEditDate="2015-09-30T17:37:01.683" LastActivityDate="2017-08-01T13:25:47.647" Title="Two chessmasters at work" Tags="|mathematics|combinatorics|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10865" PostTypeId="2" ParentId="10864" CreationDate="2015-03-24T09:18:00.680" Score="40" Body="&lt;p&gt;After 40 moves by each player, 150 minutes have elapsed on each player's clock. So the average time taken for a move is exactly $150/40=3.75$ minutes, i.e. 3 minutes and 45 seconds.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Question 1&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The answer is &lt;strong&gt;yes&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;: by a continuous version of the pigeonhole principle, there must be a move by one player or other (WLOG, say Anand's $n$th move) that takes at least 3 minutes 45 seconds. Assume for contradiction that the difference in times between the two clocks is always less than 1 minute 51 seconds (1.85 minutes). At the start of Anand's $n$th move, (Anand's time)-(Carlsen's time) is at least $-1.85$ minutes. Increase this by at least 3 minutes 45 seconds to get that at the end of Anand's $n$th move, (Anand's time)-(Carlsen's time) is at least $3.75-1.85=1.9&amp;gt;1.85$ minutes, contradiction. So the difference in time is $\geq1.85$ minutes at some stage of the game. The difference starts off at $0$ minutes, so by the pigeonhole principle, it must be &lt;em&gt;exactly&lt;/em&gt; 1.85 minutes at some stage. QED.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Question 2&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The answer is &lt;strong&gt;no&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;: the following scenario is possible.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Anand's first move takes exactly 1 minute 56 seconds&lt;/li&gt;&#xA;&lt;li&gt;each move from Carlsen's first to Anand's 39th takes exactly 3 minutes 52 seconds (which is double 1 minute 56 seconds)&lt;/li&gt;&#xA;&lt;li&gt;Carlsen's 39th and Anand's 40th move each take exactly 1 minute 8 seconds&lt;/li&gt;&#xA;&lt;li&gt;Carlsen's 40th move takes exactly 1 minute 56 seconds again.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It can be verified that the total time taken by each player is exactly 150 minutes. After every move from Anand's first to Anand's 39th, exactly 1 minute 56 seconds more have elapsed on the clock of the player who has just moved than on the clock of the player whose move it is. After Carlsen's 39th move, Anand's clock is still 48 seconds ahead of Carlsen's. After Anand's 40th move, his clock is again 1 minute 56 seconds ahead of Carlsen's. After Carlsen's 40th move, the clocks are equal at 150 minutes. So the difference between the two clocks is always at most 1 minute 56 seconds.&#xA;&lt;hr&gt;&#xA;A nice variant of the classical intermediate value theorem-type problems.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2017-08-01T13:25:47.647" LastActivityDate="2017-08-01T13:25:47.647" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10866" PostTypeId="2" ParentId="10839" CreationDate="2015-03-24T09:40:41.930" Score="22" Body="&lt;p&gt;The elegant solution is to use what's called a &lt;a href=&quot;https://en.wikipedia.org/wiki/Martingale_%28probability_theory%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;martingale&lt;/a&gt; and the &lt;a href=&quot;https://en.wikipedia.org/wiki/Optional_stopping_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;optional stopping theorem&lt;/a&gt;.  Doing this formally requires a little care but, somewhat informally, we can reason as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ignore for the moment the fact that the ant stops when he finds the honey and allow him to move as far to the left and right as he wants. In this case, after any number of steps, the ant is, on average, still at position&amp;nbsp;7. So, even if you're not watching the ant, any time I say &quot;Now!&quot;, you can still say &quot;On average, he's at position seven.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose I reveal to you that the ant was eating honey when I said, &quot;Now!&quot;  On average, he's still at position&amp;nbsp;7. With some probability&amp;nbsp;$p$, he's at&amp;nbsp;0 and with probability&amp;nbsp;$1-p$, he's at 10. His average position is&#xA;given by $A=p\times0 + (1-p)\times 10=7$.  We can now solve for&amp;nbsp;$p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solution.&lt;/strong&gt; The ant is at 0 with probability&amp;nbsp;$\tfrac{3}{10}$ and at 10 with probability&amp;nbsp;$\tfrac{7}{10}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For a more rigorous justification of why we can safely condition on the ant having stopped, see &lt;a href=&quot;https://puzzling.stackexchange.com/a/11560/4551&quot;&gt;Lopsy's answer&lt;/a&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="2023" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2015-04-07T01:44:09.067" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="10867" PostTypeId="2" ParentId="10839" CreationDate="2015-03-24T10:10:54.613" Score="7" Body="&lt;p&gt;Yes !! This is not a proof.&lt;br&gt; But, as an proxy for a mathematical solution (of which there are several already) we could take a look to a Python script which may test many iterations of a model of this problem. It produced probabilities closer to 0.3 and 0.7 for increasing number of tests. The code of the script goes as follows:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import random&#xA;b = 0&#xA;e = 0&#xA;coord = 7&#xA;lower = 0&#xA;upper = 10&#xA;iters = 0&#xA;error = 0&#xA;doiters = 10**6        # Number of trials (ant stopped).&#xA;print( &quot;testing for %15s loops&quot; % ('{:,}'.format(doiters)))&#xA;while iters &amp;lt; doiters:&#xA;    # If the limit is changed to +1 the loop&#xA;    # below this one will reflect the errors.&#xA;    while coord &amp;gt; lower and coord &amp;lt; upper:&#xA;        coord += random.randint(0, 1)*2-1&#xA;&#xA;    # If the ant has reached any of the limits,&#xA;    # it has stopped, count each of the possible stop points.&#xA;    if coord == upper:&#xA;        e += 1&#xA;    elif coord == lower:&#xA;        b += 1&#xA;    else:&#xA;        error += 1&#xA;&#xA;    coord = 7&#xA;    iters += 1&#xA;print &quot;Errors in count&quot;, error&#xA;endline=&quot;The ant stopped at %3d : %3.6f%% and at %3d : %3.6f%%&quot;&#xA;print( endline % ( lower, float(b)/iters, upper, float(e)/iters ) )&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;After several test runs, which oscillated around 0.3 and 0.7, the longest test for 100.000.000 iterations (several minutes) produced an output of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ant stopped at   0 : 0.300027375 and at  10 : 0.699972625&lt;/p&gt;&#xA;" OwnerDisplayName="user10531" LastEditorDisplayName="user12496" LastEditDate="2015-05-21T06:26:53.840" LastActivityDate="2015-05-21T06:26:53.840" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="10868" PostTypeId="1" AcceptedAnswerId="10891" CreationDate="2015-03-24T10:18:43.940" Score="8" ViewCount="1966" Body="&lt;p&gt;Glad to know &lt;a href=&quot;https://puzzling.stackexchange.com/questions/10797/3x3-magic-square-of-prime-numbers&quot;&gt;the previous puzzle&lt;/a&gt;, which was the first puzzle I posted in Puzzling, was warmly welcomed (Thank you!), and an optimal solution was found. Inspired by the comments there, here is the Version $2$ of the puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/urjOD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, most of the things are unchanged. We still have this $3 \times 3$ grid, which $9$ distinct prime numbers $P_1, P_2, ..., P_9$ are to be filled in. And there are $8$ sums: $3$ horizontal, $3$ vertical and $2$ diagonal, and they are named $S_1, S_2, ..., S_8$. All the requirements in the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/10797/3x3-magic-square-of-prime-numbers&quot;&gt;first version&lt;/a&gt; still hold here, which mean:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$P_1, P_2, ..., P_9, S_1, S_2, ..., S_8$ are &lt;strong&gt;all distinct prime numbers&lt;/strong&gt; (i.e. there are totally $17$ different prime numbers).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But this time, one more additional requirement:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The grand total $P_1 + P_2 + ... + P_9 + S_1 + S_2 + ... + S_8$ also has to be a &lt;strong&gt;prime number&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The challenge: To minimize the grand total.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the additional requirement, some solutions satisfying the previous puzzle do not satisfy this version. And, the optimal solution will be different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is one possible solution I come up with, which has a grand total of $601$, but it is not the optimal solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/78zyA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feel free to have a try!&lt;/p&gt;&#xA;" OwnerUserId="9944" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-03-24T16:09:41.920" Title="3x3 “Magic Square” of Prime Numbers -- Part II" Tags="|mathematics|calculation-puzzle|magic-square|number-theory|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10869" PostTypeId="2" ParentId="10864" CreationDate="2015-03-24T11:02:27.487" Score="10" Body="&lt;p&gt;The answer to question 1 with 1 min 51 sec (that is, $n=111$ seconds) is &lt;strong&gt;yes&lt;/strong&gt;. The answer to question 2 with 1 min 57 sec (that is, $n=117$ seconds) is &lt;strong&gt;no&lt;/strong&gt;. Both answers are &lt;strong&gt;special cases&lt;/strong&gt; of my solution that &lt;strong&gt;fully settles&lt;/strong&gt; this puzzle for all possible numbers $n$ of seconds!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Question: Was there necessarily a moment in the game, when the clock of one player showed exactly $n$ seconds less than the clock of the other player?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since 2 hours 30 minutes are exactly 9000 seconds, one would expect that the answer changes at 9000/80 = 112.5 seconds. But &lt;strong&gt;surprise, surprise,&lt;/strong&gt; this is not correct!!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $n\le114$ then the answer is yes. &lt;br&gt; If $n\ge115$ then the answer is no.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For $n\le114$, assume that the time difference never is $n$. Then the first move of first player takes at most 113 seconds as the second clock shows 0. Then the first move of the second player takes at most 113 seconds more than the time on the first player's watch. Then the second move of the first player takes at most 113 seconds more than the time on the second player's watch. Continuing this way we get by mathematical induction:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;After move $m$ of the first player, his watch has traversed at most $113*(2m-1)$ seconds.&lt;br&gt;&#xA;  After move $m$ of the second player, his watch has traversed at most $113*(2m)$ seconds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After move 40 of the first player, his watch has traversed at most $113*(2*40-1)$ = 8927 seconds. This is below 2.5 hours. There is a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n\ge115$ construct a game where the first move of the first player takes 114 seconds. The first move of second player and all moves 2,...,39 of both players take 228 seconds. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;After the first move of first player, the difference has gone up from 0 to 114. The first move of second player brings the differences down from 114 to -114. The next move of the first player brings the difference up to 114 again. The next move of the second player brings the difference down to -114 again, and so on. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After move 39 of the second player the difference is down at -114 again.&#xA;The watch of the first player has traversed 114+38*228 seconds, this is 8778 seconds. The watch of the second player has traversed 39*228 seconds, this is 8892 seconds. Move 40 of the first player takes 222 seconds and brings the difference up to +108. Move 40 of the second player takes 108 seconds and brings the differences down to 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference always stays between -114 and +114 and the watches are never $n\ge115$ seconds away from each other.&lt;/p&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="9125" LastEditDate="2015-03-25T10:10:01.033" LastActivityDate="2015-03-25T10:10:01.033" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10870" PostTypeId="2" ParentId="10868" CreationDate="2015-03-24T11:25:38.460" Score="0" Body="&lt;p&gt;I used the same numbers as used by you&lt;br&gt;&#xA;I got the grand total of &lt;code&gt;529&lt;/code&gt; which is a prime number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TYZi4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5 + 11 + 3 + 13 + 7 + 59 + 19 + 23 + 17) + (19 + 79 + 59 + 29 + 29 + 37 + 41 + 79 ) = 529&lt;/p&gt;&#xA;" OwnerUserId="9626" LastEditorUserId="9626" LastEditDate="2015-03-24T11:32:54.007" LastActivityDate="2015-03-24T11:32:54.007" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10871" PostTypeId="2" ParentId="10839" CreationDate="2015-03-24T11:38:20.963" Score="3" Body="&lt;p&gt;Suppose the odds of reaching a specific end from point x are f(x). From point x, a random move takes you to x-1 50% of the time and x+1 50% of the time. So f(x) = f(x-1)/2 + f(x+1)/2. That is, your odds at a given point are the average of the odds for the points either side. Since the points are evenly spaced, f(x) must therefore be linear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One end has a chance of zero (you already found the other honey drop) and the other has a chance of one (you have made it). Since the line conveniently starts at x=0, your chances of reaching the high end are proportional to your distance along the line, and your chances of reaching the low end are whatever remains from 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From x=7 therefore, 0.7 chance of reaching +10, and 0.3 chance of reaching 0.&lt;/p&gt;&#xA;" OwnerUserId="6031" LastActivityDate="2015-03-24T11:38:20.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10872" PostTypeId="1" AcceptedAnswerId="10984" CreationDate="2015-03-24T12:12:07.040" Score="3" ViewCount="346" Body="&lt;p&gt;The following alphabetical list contains independent countries with a certain arcane property. (The list is complete to the best of my knowledge).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Algeria, Cambodia, Croatia, El Salvador, ????????,  Germany, ????????,&#xA;    Serbia, South Africa, Spain, Sri Lanka, Switzerland, United Arab Emirates&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which countries have been replaced by questionmarks?&lt;br&gt;&#xA;What is the arcane property?&lt;/p&gt;&#xA;" OwnerUserId="9125" LastActivityDate="2015-03-25T20:30:30.450" Title="An arcane property of independent countries" Tags="|geography|knowledge|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10873" PostTypeId="1" AcceptedAnswerId="10877" CreationDate="2015-03-24T13:05:38.050" Score="4" ViewCount="3617" Body="&lt;p&gt;I do not like to hear from you,&#xA;But you are pleased to meet me only when you need me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am happy not be working.&#xA;When I have work, I start screaming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can be annoying to many regulars.&#xA;But am inspiring for little ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can reach me if you need me,&#xA;Hope you still like me if you don't.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My first ever try to create a riddle.&lt;/p&gt;&#xA;" OwnerUserId="9892" LastEditorUserId="5373" LastEditDate="2015-03-24T20:39:25.073" LastActivityDate="2016-08-20T14:55:38.990" Title="Identify What am I ? - Riddle 1" Tags="|riddle|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10874" PostTypeId="2" ParentId="10873" CreationDate="2015-03-24T13:11:30.950" Score="4" Body="&lt;p&gt;I'm guessing it's a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; bell, gong or other ringing device&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I do not like to hear from you, But you are pleased to meet me only when you need me.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You may need to hear the bell during a boxing match or a boring class&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I am happy not be working. When I have work, I start screaming.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It makes noises when it works. It is still otherwise.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I can be annoying to many regulars. But am inspiring for little ones.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Normally most people hate ringing sounds. Children however love noises of all sorts and often enjoy the sound of it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;You can reach me if you need me, Hope you still like me if you don't.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Possibly a reference to the religious symbolism of bells.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="5610" LastEditDate="2015-03-24T13:12:55.067" LastActivityDate="2015-03-24T13:12:55.067" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10875" PostTypeId="2" ParentId="10873" CreationDate="2015-03-24T13:25:11.397" Score="4" Body="&lt;p&gt;Here's my stab at it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An alarm clock&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I do not like to hear from you, But you are pleased to meet me only when you need me.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I hate hearing my alarm go off... But I'm always thankful when it goes off when it's supposed to&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I am happy not be working. When I have work, I start screaming.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When it reaches the appropriate time, it goes off. Otherwise it is silent&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I can be annoying to many regulars. But am inspiring for little ones.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every day at the same time it goes off. Kids on the other hand, see each day as a new adventure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;You can reach me if you need me, Hope you still like me if you don't.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I always struggle to turn off my alarm when it goes off. But I am still grateful when I get up on time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9412" LastActivityDate="2015-03-24T13:25:11.397" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10876" PostTypeId="1" AcceptedAnswerId="10900" CreationDate="2015-03-24T13:30:14.323" Score="9" ViewCount="920" Body="&lt;p&gt;I've been playing Praser 5, an old interactive fiction game (like Zork) made a few decades ago, and available &lt;a href=&quot;http://ifdb.tads.org/viewgame?id=pnleii72sdlxs2ib&quot;&gt;here&lt;/a&gt;. In it, you wander from area to area solving riddles posed by mythical animals. Many of the puzzles involve lateral thinking. I've had quite a bit of difficulty with the following puzzle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The sphinx stretches lazily and says, &quot;Consider a Euclidean plane. Draw points O and S. Put I between them so that half the distance between O and I is the distance between O and the point one-third of the way from O to S. Say that S and I are one bloit apart. Now put D as far from O as S is from the point between I and O, so that the ratio of the distance between D and I and the distance between O and the point between D and I is twice the distance between O and the point between D and O. Mark E so that the distance between O and the point between E and the point between D and E is half the distance between O and I, and half the distance between S and E is the distance between the point between S and D and D. Now, P is placed so that the sum of the distances between P and S and between P and D, one of which is two bloits, is as long as E is far from the point three times as far from O as it is from D so that D is between that point and E. P is closer to I than it is to S, but if it weren't, H would be the point between P and D as far from where P really is as I is from where P would be. R is as far from H as O is from S and as far from D as D is from E. So what is the shape of the figure OSRHP, and what is its area in square bloits?&quot; It smiles gently and curls up again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So far, I have O, I, and S on a line, with O and I two bloits apart and S one bloit past I. D is two bloits above O, and E is two bloits below O.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond that, I suspect that P is in the lower right, H in the upper middle, and R in the lower left. I think the result is a star/pentagram. However, to check your answer in game, you must simultaneously enter the shape and area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my solution (a star) correct? And he can I find the area?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: to try speaking to the Sphinx in-game, type arlumere).&lt;/p&gt;&#xA;" OwnerUserId="10582" LastActivityDate="2015-03-24T20:08:30.450" Title="A purposefully obtuse Euclidean geometry riddle from an old interactive fiction game" Tags="|geometry|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10877" PostTypeId="2" ParentId="10873" CreationDate="2015-03-24T13:31:12.560" Score="7" Body="&lt;p&gt;I think it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An ambulance or fire truck.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I do not like to hear from you, But you are pleased to meet me only when you need me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hearing from me means something bad happened, but I will be pleased when I get help.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am happy not be working. When I have work, I start screaming.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sirens.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be annoying to many regulars. But am inspiring for little ones.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The loud noises of the siren are annoying to adults, but especially fire trucks are inspiring to kids.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can reach me if you need me, Hope you still like me if you don't.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can call the ambulance whenever there's an emergency.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10583" LastActivityDate="2015-03-24T13:31:12.560" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10878" PostTypeId="1" AcceptedAnswerId="11738" CreationDate="2015-03-24T13:36:43.363" Score="17" ViewCount="2318" Body="&lt;p&gt;As I mentioned in a previous question, I am trying to play through Praser 5, an old interactive fiction freeware game. In it, you visit four Marks, the marks of earth, air, fire and water. Each has a name found by solving a series of riddles and taking the first letter of each word in the solutions to the riddle.&lt;/p&gt;&#xA;&lt;p&gt;Also, in the game, you discover the names of mythical creatures and use the names to get more riddles.&lt;/p&gt;&#xA;&lt;p&gt;I have solved three of four marks. However, the Mark of Water is a giant chalice with words that say:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;As I contain the waters of the world, so the 7 names are contained. (And they're names, not Names.)&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;iodine&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;silicon&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;radon&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;bromine&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;thorium&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;technetium&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;molybdenum&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I've looked up the atomic number and abbreviations on the periodic table, but I don't know how to proceed. In an old forum, someone who solved it said it was a &amp;quot;fiendish puzzle&amp;quot; and that &amp;quot;the introductory text is very misleading&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;The solutions to the other marks were ordinary English words (for instance, one was &amp;quot;wheel&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Edits from comments&lt;/strong&gt;&#xA;I have discovered that the answer is &amp;quot;diamond&amp;quot; using a decompiler. Diamond is contained in the seven words, but not in any uniform way. Why is diamond the right answer?&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Copied from comments&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;For those who want to see the riddle firsthand you can go to &lt;a href=&quot;http://www.eblong.com/zarf/zweb/praser5/&quot; rel=&quot;noreferrer&quot;&gt;http://www.eblong.com/zarf/zweb/praser5/&lt;/a&gt; to play Praser 5 online. To get to this riddle just type 'literal', then 'south', then 'read'. –  finsternis 6 hours ago&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Rationale for accepted answer&lt;/strong&gt;&#xA;I think that the reasoning in the answer is mostly correct. I think it is easier to find English words in the element names first, and then try out the various possibilities (like ado/don) to see which ones give you the correct answer.&lt;/p&gt;&#xA;&lt;p&gt;Also, for the wheel problem, I agree that the Hagia Sophia should be the second answer.&lt;/p&gt;&#xA;" OwnerUserId="10582" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2019-01-14T23:31:39.783" Title="A lateral thinking puzzle with chemical elements involved. Why is this the correct answer?" Tags="|chemistry|" AnswerCount="1" CommentCount="23" ContentLicense="CC BY-SA 3.0" />
  <row Id="10879" PostTypeId="1" AcceptedAnswerId="10880" CreationDate="2015-03-24T13:55:36.777" Score="4" ViewCount="6349" Body="&lt;p&gt;It is a love/hate relationship we share,&lt;br&gt;&#xA;Sometimes great, sometimes you just stare.&lt;br&gt;&#xA;Everyday, sometimes for hours you use me,&lt;br&gt;&#xA;I try my best, so please don't abuse me.&lt;br&gt;&#xA;I make life so easy, I make life so hard.&lt;br&gt;&#xA;I can do almost anything, even make a card.&lt;br&gt;&#xA;I am a newer friend, haven't been here too long.&lt;br&gt;&#xA;Do you like music? I can sing you a song.&lt;br&gt;&#xA;I often keep you from getting bored.&lt;br&gt;&#xA;A lot I do requires a cord.&lt;br&gt;&#xA;So while you figure out who I may be,&lt;br&gt;&#xA;Just know it is often me that you see.&lt;/p&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-04-27T10:59:07.647" Title="I'm always here for you - What am I?" Tags="|riddle|word|rhyme|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10880" PostTypeId="2" ParentId="10879" CreationDate="2015-03-24T14:09:41.317" Score="5" Body="&lt;p&gt;According to me the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Laptop&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes great, sometimes you just stare.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sometimes you just stare at your laptop when you read or see something amazing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Everyday, sometimes for hours you use me,&#xA;I try my best, so please don't abuse me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You use your laptop almost everyday and sometimes for many hours a day. so you should shut down your laptop once in a while and not abuse it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I make life so easy, I make life so hard.&#xA;I can do almost anything, even make a card.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;laptop helps you connect to friends even those who are very far from you, but ruins your social life when you use your laptop way too much. you can use your laptop for a lot of things, including making cards.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Do you like music? I can sing you a song.&#xA;I often keep you from getting bored.&#xA;A lot I do requires a cord.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Laptop keeps you from getting bored and can be used to play songs or movies, needs a charger to keep running.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8934" LastActivityDate="2015-03-24T14:09:41.317" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10881" PostTypeId="2" ParentId="10868" CreationDate="2015-03-24T14:28:56.140" Score="2" Body="&lt;p&gt;Using a brute force checker I have found a solution of &lt;strong&gt;541&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    7  5 17 | 29&#xA;   11 23  3 | 37&#xA;   13 19 41 | 73&#xA;   ---------+---&#xA;53 31 47 61 | 71&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Notice that a simple lower bound would be 499 which is just the sum of the first 17 primes greater than 2; the result above skips only 3 primes (43, 59, 67) which is what makes me doubt there is a more optimal solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The brute force searcher searched every combination of the primes up to 67 which should be conclusive, as increasing the value of one of the inner numbers increases the final answer by at least three times as much, making any increase not worth it at that point.&lt;/p&gt;&#xA;" OwnerUserId="150" LastEditorUserId="150" LastEditDate="2015-03-24T16:09:41.920" LastActivityDate="2015-03-24T16:09:41.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10882" PostTypeId="1" AcceptedAnswerId="10960" CreationDate="2015-03-24T14:34:58.420" Score="12" ViewCount="3054" Body="&lt;p&gt;I have more to do with cooking than you might think.&lt;br&gt;&#xA;Now let us discuss the ocean, try not to sink.&lt;br&gt;&#xA;Then between quizzes and sandwiches I like to rest.&lt;br&gt;&#xA;The beginning of eating, a feast is the best.&lt;br&gt;&#xA;Half the area is needed, we aren't quite done.&lt;br&gt;&#xA;We wrap up wondering if I am more than one.&lt;br&gt;&#xA;Now as you guess, know that I am quite small,&lt;br&gt;&#xA;Doesn't really matter if you are short or tall.&lt;br&gt;&#xA;Two types of clues to guess one type of me.&lt;br&gt;&#xA;What on the earth could I possibly be?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Two types of clues implies that there are literal and wordplay clues.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The wordplay begins early. Some clues are for a single letter, others for a string of them.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chances are very high you have one, but that you can't see it and you never will.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="9951" LastEditDate="2015-03-25T14:02:02.207" LastActivityDate="2015-03-25T15:52:49.280" Title="I am not a chef - What am I?" Tags="|riddle|word|rhyme|" AnswerCount="8" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10883" PostTypeId="2" ParentId="10882" CreationDate="2015-03-24T14:48:03.213" Score="3" Body="&lt;p&gt;Is it  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Table salt?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have more to do with cooking than you might think.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We need salt for cooking  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now let us discuss the ocean, try not to sink.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water in Ocean contains salt; the more it has the harder you can sink in it  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then between quizzes and sandwiches I like to rest.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quizzes on a table, salt on a sandwiches  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The beginning of eating, a feast is the best.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You add some table salt before you start eating  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Half the area is needed, we aren't quite done.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We use half a table to eat? (this guess is weak)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We wrap up wondering if I am more than one.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A table salt bottle has many salt  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now as you guess, know that I am quite small,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Salt is tiny  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Doesn't really matter if you are short or tall.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is referring to the 'table'  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Two types of clues to guess one type of me.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clue on Table and Clue on Salt  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What on the earth could I possibly be?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Table Salt!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="10037" LastEditDate="2015-03-24T15:07:16.537" LastActivityDate="2015-03-24T15:07:16.537" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10884" PostTypeId="1" AcceptedAnswerId="10893" CreationDate="2015-03-24T14:52:13.283" Score="10" ViewCount="563" Body="&lt;p&gt;Suppose I make a 20x10 grid out of match sticks, using 21 columns of 10 match sticks and 11 rows of 20 match sticks:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─&#xA;│ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │&#xA; ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─&#xA;│ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │&#xA; ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─&#xA;│ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │&#xA; ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─&#xA;│ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │&#xA; ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─&#xA;│ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │&#xA; ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─&#xA;│ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │&#xA; ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─&#xA;│ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │&#xA; ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─&#xA;│ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │&#xA; ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─&#xA;│ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │&#xA; ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─&#xA;│ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │&#xA; ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I am going to independently remove each match stick with 1% probability, leaving behind an average of 99% of the match sticks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After I do that, you will count up how many squares are formed by the match sticks. Squares can be 1x1, 2x2, 3x3s, and etc up to 10x10 (or technically larger, but those wouldn't fit). A square must have all of the match sticks along its border present, but its insides don't matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's some examples of valid 2x2 squares:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; ─ ─      ─ ─      ─ ─      ─ ─ &#xA;│ │ │    │   │    │ │ │    │   │&#xA; ─ ─                 ─      ─ ─ &#xA;│ │ │    │   │    │   │    │   │&#xA; ─ ─      ─ ─      ─ ─      ─ ─ &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And some broken 2x2 non-squares that don't count (though there are still 1x1 squares in the examples):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; ─ ─      ─ ─        ─          &#xA;│ │      │   │      │ │    │   │&#xA; ─ ─                 ─      ─ ─ &#xA;│ │ │    │                 │   │&#xA; ─ ─      ─ ─                   &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Call the count N. What is the expected value of N?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if we were working with a 1x1 grid instead of a 20x10 grid then the expected value of N would be ~0.9606 because there's only one possible square and it's present if and only if all four match sticks stay and that happens in (1 - 0.01)^4 ~= 96.06% of cases.&lt;/p&gt;&#xA;" OwnerUserId="10588" LastEditorUserId="10588" LastEditDate="2015-09-28T18:31:35.660" LastActivityDate="2015-09-28T18:31:35.660" Title="Expected number of squares in a 20x10 grid with 1% chance of removing each edge" Tags="|probability|matches|counting|" AnswerCount="1" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10885" PostTypeId="1" CreationDate="2015-03-24T14:54:08.957" Score="7" ViewCount="1013" Body="&lt;p&gt;The object of this puzzle is to construct a Minesweeper grid with a special property. The grid can be any shape, but it must have four labeled squares around the outer edge. Like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5ULcr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The four labeled squares must be labeled A, B, C, D, &lt;em&gt;in that order going around the outer edge&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The objective is for a Minesweeper player to be able to conclude that A and C are the same, and B and D are the same, but not be able to conclude anything else. To be specific:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Place numbers in the grid to get a Minesweeper puzzle. This puzzle must have the following properties:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;There are no safe clicks. That is, for any unnumbered square, there is a solution in which that square is a mine.&lt;/li&gt;&#xA;&lt;li&gt;In any valid solution, A=C and B=D. (A=C means that A has a mine if and only if C does.)&lt;/li&gt;&#xA;&lt;li&gt;A valid solution must satisfy every combination of A=C and B=D simultaneously.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Again, the grid can be any shape. The grid can even have holes in it, but understand that &quot;outer edge&quot; does not include the edge neighboring a hole. A, B, C, D can be placed anywhere along the outer edge, but must be &lt;em&gt;in that order&lt;/em&gt;. Aim for the smallest grid!&lt;/p&gt;&#xA;" OwnerUserId="1752" LastEditorUserId="10399" LastEditDate="2015-03-24T16:28:05.713" LastActivityDate="2015-03-27T22:55:54.533" Title="Can you create a Minesweeper puzzle with 'information crossover'?" Tags="|optimization|minesweeper|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10886" PostTypeId="2" ParentId="10885" CreationDate="2015-03-24T15:03:30.277" Score="3" Body="&lt;p&gt;I think I got it now. Using a minesweeper developer program found on &lt;a href=&quot;http://www.formauri.es/personal/pgimeno/compurec/Minesweeper.php&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;here&quot;&gt;here&lt;/a&gt;, you can create, evaluate and try your own board. I have uploaded an image of my solution, or else this would've been a LOT of typing:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zE0Dm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And an explanation about what you see in the image:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I've put the 4 labels (A, B, C and D) beside each questionmark where it should be. The red flags mark the spots of pre-discovered bombs.&lt;br&gt;&#xA;Every pair (A-C and B-D) have their own 'wire' as it is called. Then an intersection to let these signals cross eachother, and arive at the other side.&lt;br&gt;&#xA;For the first rule: there are no safe clicks, in all possible solutions this setup has got, any unnumbered square has been a mine at least once.&lt;br&gt;&#xA;In èvery solution of this setup: it covers the A=C and B=D requirement. So both the second and third rule are covered too.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Discarded answers (Just for comment-reference)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;pre&gt;1) &lt;code&gt;.A   2)  A  3) AB&#xA;    1 .    D2B    DC&#xA;  D1.1B     C&#xA;  . 1&#xA;    C.&lt;/code&gt;&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10384" LastEditorUserId="10384" LastEditDate="2015-03-27T22:55:54.533" LastActivityDate="2015-03-27T22:55:54.533" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="10887" PostTypeId="2" ParentId="10367" CreationDate="2015-03-24T15:05:49.347" Score="2" Body="&lt;p&gt;For the second question I believe &lt;strong&gt;Mary&lt;/strong&gt; &lt;em&gt;edit:&lt;/em&gt; &lt;em&gt;or M'Gann&lt;/em&gt; is a valid answer.&#xA;The names begin with first letters of the ordered (though not consecutive) elements of the periodic table, the open spot having to be Magnesium.&lt;/p&gt;&#xA;" OwnerUserId="150" LastEditorUserId="150" LastEditDate="2015-03-24T15:30:25.603" LastActivityDate="2015-03-24T15:30:25.603" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10888" PostTypeId="2" ParentId="10882" CreationDate="2015-03-24T15:08:55.033" Score="1" Body="&lt;p&gt;I have more to do with cooking than you might think.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water? Heat?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now let us discuss the ocean, try not to sink.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then between quizzes and sandwiches I like to rest.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; and?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The beginning of eating, a feast is the best.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; eat?  As in (h)eat?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Half the area is needed, we aren't quite done.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  also (h)ea(t)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We wrap up wondering if I am more than one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  heat and water?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now as you guess, know that I am quite small,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; water molecules are small. heat is VERY small.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Doesn't really matter if you are short or tall.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; nope it doesn't....&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Two types of clues to guess one type of me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water - steam ice?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What on the earth could I possibly be?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I say:  steam&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10294" LastActivityDate="2015-03-24T15:08:55.033" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10889" PostTypeId="2" ParentId="10885" CreationDate="2015-03-24T15:18:02.333" Score="0" Body="&lt;p&gt;I'm probably twisting the rules a bit, but here goes. &quot;The grid can be any shape&quot;. Well, my grid is a &lt;em&gt;torus!&lt;/em&gt; (And it kinda looks like a swastika, but let's ignore that.)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;C+A D&#xA;  1 +&#xA;D1+1B&#xA;+ 1&#xA;B C+A&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The labels are shown twice to indicate where the grid &quot;wraps around&quot;. The grid itself uses 11 squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize that this probably isn't what is intended, but I thought of this and decided to post it while working on a &quot;real&quot; answer.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pre-edit answer:&lt;/strong&gt;&lt;br&gt;&#xA;How about this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;pre&gt;&lt;code&gt;1A1&#xA;D B&#xA;1C1&lt;/code&gt;&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The space in the middle is a &quot;hole&quot;. If A has a mine, then B and D do not, which means that C does. Similarly, if A does not have a mine, then B and D do, so C does not. Thus, this satisfies all three conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It uses a 3x3 grid, with a total of 8 squares.&lt;/p&gt;&#xA;" OwnerUserId="8776" LastEditorUserId="8776" LastEditDate="2015-03-24T16:28:48.790" LastActivityDate="2015-03-24T16:28:48.790" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="10890" PostTypeId="2" ParentId="10885" CreationDate="2015-03-24T15:21:08.817" Score="0" Body="&lt;p&gt;I think you might need to clarify what a &quot;hole&quot; is.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;pre&gt;  &lt;code&gt;A&#xA;D   B&#xA;  C&lt;/code&gt;&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This satisfies the puzzle's requirements.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are no safe spaces; all of the spaces are mines. Because none of the spaces are adjacent to each other, none of them can be a &quot;numbered square&quot;; the value of this number would be 0, making it an empty (or safe) space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2015-03-24T15:21:08.817" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10891" PostTypeId="2" ParentId="10868" CreationDate="2015-03-24T15:55:50.973" Score="3" Body="&lt;p&gt;As our friend @KSab said, the best solution ever is &lt;strong&gt;541&lt;/strong&gt;. There are exactly 16 possible solutions which are shown in the images belown:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fhe4j.png&quot; alt=&quot;Solutions 1 to 8&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WvFoj.png&quot; alt=&quot;Solutions 9 to 16&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10564" LastActivityDate="2015-03-24T15:55:50.973" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10892" PostTypeId="2" ParentId="10836" CreationDate="2015-03-24T16:24:46.750" Score="2" Body="&lt;p&gt;My guess is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The spear maiden in The Arthurian Tarot&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;\&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The spear maiden is associated with the element of fire; being also the Manasic fire - born of the Buddhic earth of primal darkness. In this respect we find her roots in the myth of Ariadne, who gave to the hero Theseus, the intuitive thread that would guide him through the labyrinth. In the grail legends we meet her again in the form of a mysterious girl who at first snubs Perceval and when he tries to follow her, he looses her in the pitch black darkness of the night. He then sees a vision of a supernatural light, which burns like five lit candles, lighting up the forest. When it is daybreak Perceval catches up with the maiden, who has now changed in favour towards him and gives to Perceval a ring and a white mule, that she says, will lead him to the grail. The ring is of course, another aspect of the spear Hallow.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9125" LastActivityDate="2015-03-24T16:24:46.750" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10893" PostTypeId="2" ParentId="10884" CreationDate="2015-03-24T16:51:53.887" Score="10" Body="&lt;p&gt;Isn't that just linearity of expectation? The answer is approximately &lt;code&gt;810.927&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;(1)&lt;/strong&gt; For $k=1,\ldots,10$ the picture contains exactly $(21-k)(11-k)$ squares of sidelength $k$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof: every $k\times k$ square has a unique upper left corner. The last $k-1$ points in every row are not possible, since then the square would overhang; hence there remain $20-(k-1)=21-k$ corner candidates in the row.  Similarly, the last $k-1$ points in any column are not possible.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;(2)&lt;/strong&gt; For any fixed $k\times k$ square, the probability of surviving the removal process is $(1-p)^{4k}$ where $p=1/100$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof: The square has $4k$ match sticks on its perimeter.  It survives, if and only if each of these match sticks survives.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;(3)&lt;/strong&gt; The expected number of surviving squares is &#xA;  $N=\sum_{k=1}^{10}(21-k)(11-k)(1-p)^{4k}$ where $p=1/100$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof: This follows by linearity of expectation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;(4)&lt;/strong&gt; Evaluate the sum.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In python:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;gt;&amp;gt;&amp;gt; sum((21-i)*(11-i)*0.99**(4*i) for i in range(1, 11))&#xA;810.9274873676787&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="10588" LastEditDate="2015-03-24T22:48:07.220" LastActivityDate="2015-03-24T22:48:07.220" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10894" PostTypeId="1" AcceptedAnswerId="10958" CreationDate="2015-03-24T17:34:06.900" Score="6" ViewCount="10654" Body="&lt;p&gt;I am not lost, but I can be found.&lt;/p&gt;&#xA;&lt;p&gt;If you have found me, then you should lose me again, in order to find me.&lt;/p&gt;&#xA;&lt;p&gt;When you find me again it will surely be of benefit, similar to if you became more fit.&lt;/p&gt;&#xA;&lt;p&gt;My composition of attributes is exactly that...&lt;/p&gt;&#xA;&lt;p&gt;...found of the pages were: on a journey, five had me and five did not.&#xA;.&lt;/p&gt;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&lt;p&gt;(this is my very first riddle, any suggestions would be greatly appreciated, sorry for not rhyming)&lt;/p&gt;&#xA;" OwnerUserId="9778" LastEditorUserId="69582" LastEditDate="2022-07-17T14:19:28.003" LastActivityDate="2022-07-17T14:19:28.003" Title="I am not lost, but I can be found. What am I?" Tags="|riddle|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="10895" PostTypeId="2" ParentId="10894" CreationDate="2015-03-24T17:50:52.953" Score="1" Body="&lt;p&gt;I am not lost, but I can be found.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your way&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you have found me, then you should lose me again, in order to find me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You have to get lost (lose one's way) in order to find it again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you find me again it will surely be of benefit, similar to if you became more fit.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Definitely a benefit to know where you are going!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My composition of attributes is exactly that...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hmm, maybe it's weight?  No, it's way....&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...found of the pages were: on a journey, five had me and five did not. &lt;/p&gt;&#xA;" OwnerUserId="10294" LastActivityDate="2015-03-24T17:50:52.953" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10896" PostTypeId="2" ParentId="10894" CreationDate="2015-03-24T18:14:18.553" Score="3" Body="&lt;p&gt;I am not lost, but I can be found.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hope&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you have found me, then you should lose me again, in order to find me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you did find hope, unless you lose it, you are definitely not going to find it again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you find me again it will surely be of benefit, similar to if you became more fit.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you do find hope, its definitely beneficial, as you recover from an hopeless illness &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My composition of attributes is exactly that...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I hope I will lose weight/ I will gain weight / I will be fairer / I will be stronger. Just wish, and hope for something getting better&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...found of the pages were: on a journey, five had me and five did not. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's an old adage, five had hope and five don't. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4177" LastActivityDate="2015-03-24T18:14:18.553" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10897" PostTypeId="1" AcceptedAnswerId="10987" CreationDate="2015-03-24T19:13:46.397" Score="14" ViewCount="866" Body="&lt;p&gt;I could be there, yet I'll stay hidden.&lt;br&gt;&#xA;Hidden I'll stay, but you'll need me.&lt;br&gt;&#xA;Need me you will, so I may appear.&lt;br&gt;&#xA;Appear I may, I don't decide.&lt;br&gt;&#xA;Decide I don't, my master does.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a word, I tell you words.&lt;br&gt;&#xA;I've a task, you know my task.&lt;br&gt;&#xA;I'm short, I shorten your path.&lt;br&gt;&#xA;I've the answer, now tell me it! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What am I?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #1:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm not far. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Far I'm not, indeed I'm here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-03-25T19:25:36.930" LastActivityDate="2015-03-26T12:18:25.070" Title="An opportunist friend" Tags="|riddle|" AnswerCount="8" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10898" PostTypeId="2" ParentId="10882" CreationDate="2015-03-24T19:37:10.697" Score="3" Body="&lt;p&gt;Could you be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a stomach?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have more to do with cooking than you might think.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The current state of your stomach (full, empty, partially full) directly affects what you want to cook.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now let us discuss the ocean, try not to sink.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the stomach is sort of like an ocean, with its stomach acid? Also sometimes people feel a sinking feeling in their stomach.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then between quizzes and sandwiches I like to rest.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; butterflies in your stomach from a quiz, full stomach after eating a sandwich.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The beginning of eating, a feast is the best.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Before you eat, you're the most hungry.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Half the area is needed, we aren't quite done.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stomachs can expand to twice their size as you eat more.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We wrap up wondering if I am more than one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Often people joke about having a second stomach for desserts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now as you guess, know that I am quite small,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stomachs are only about the size of a fist.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Doesn't really matter if you are short or tall.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Height doesn't affect stomach size.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Two types of clues to guess one type of me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Honestly not sure about this one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What on the earth could I possibly be? &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A stomach&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9978" LastActivityDate="2015-03-24T19:37:10.697" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10899" PostTypeId="2" ParentId="10882" CreationDate="2015-03-24T19:45:19.633" Score="2" Body="&lt;p&gt;Second attempt (original below):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ears&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have more to do with cooking than you might think.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ears of corn? Also cooking does make noise, especially frying or sauteing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now let us discuss the ocean, try not to sink.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can hear the ocean in your ears&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then between quizzes and sandwiches I like to rest.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; your brain (which handles quizzes) is sandwiched between your ears&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The beginning of eating, a feast is the best.&#xA;Half the area is needed, we aren't quite done.&#xA;We wrap up wondering if I am more than one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; These three lines are the wordplay. &quot;e&quot; is the beginning of eating. &quot;ar&quot; is half of area, and we add an &quot;s&quot; to make it more than one. Still not sure about the &quot;feast&quot; part&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now as you guess, know that I am quite small,&#xA;Doesn't really matter if you are short or tall.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Like my previous rationale, ears are small no matter what your height (though some people do have smaller/larger ears)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Two types of clues to guess one type of me.&#xA;What on the earth could I possibly be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;--- ORIGINAL GUESS ---&#xA;Here's an answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; eyes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have more to do with cooking than you might think.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If food looks good it can have an effect on our appetite and desire to eat it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now let us discuss the ocean, try not to sink.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Haven't gotten this line yet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then between quizzes and sandwiches I like to rest.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your eyes are between your brain (quizzes) and sandwiches (mouth)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The beginning of eating, a feast is the best.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This gives you an &quot;e&quot;, also your eyes can &quot;feast&quot; on a view&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Half the area is needed, we aren't quite done.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Area is x*y, so &quot;y&quot; is half of it (this part's shaky)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We wrap up wondering if I am more than one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;eyes&quot; is a homonym for &quot;I's&quot; - are &quot;I's&quot; more than one person?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now as you guess, know that I am quite small,&#xA;Doesn't really matter if you are short or tall.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eyes are a small part of the body, no matter your height&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Two types of clues to guess one type of me.&#xA;What on the earth could I possibly be?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We have the wordplay ones and the literal clues&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastEditorUserId="1620" LastEditDate="2015-03-24T20:04:04.957" LastActivityDate="2015-03-24T20:04:04.957" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10900" PostTypeId="2" ParentId="10876" CreationDate="2015-03-24T20:08:30.450" Score="16" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Consider a Euclidean plane. Draw points O and S. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0AXEj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Put I between them so that half the distance between O and I is the distance between O and the point one-third of the way from O to S. Say that S and I are one bloit apart. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I lies at the 1/3rd point between S and O.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Oo9oq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Now put D as far from O as S is from the point between I and O, so that the ratio of the distance between D and I and the distance between O and the point between D and I is twice the distance between O and the point between D and O. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;D is two bloits from O, and happens to form a right triangle which satisfies the other requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ccm6x.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Mark E so that the distance between O and the point between E and the point between D and E is half the distance between O and I, and half the distance between S and E is the distance between the point between S and D and D. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Putting E two bloits above O works here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oFVGJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Now, P is placed so that the sum of the distances between P and S and between P and D, one of which is two bloits, is as long as E is far from the point three times as far from O as it is from D so that D is between that point and E. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;that point&quot; is collinear with D and E. It's three bloits from O and one bloit from D. E is five bloits from &quot;that point&quot;, so the two distances in the first clause are two and three bloits, although we don't know which is which. So P can be in four potential places, represented here as the intersection of the red and blue circles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6TaWz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;P is closer to I than it is to S, but if it weren't, H would be the point between P and D as far from where P really is as I is from where P would be. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The only candidate P closer to S is the one farthest to the right. We call this &quot;Fake P&quot;. I is sqrt(5) bloits away from Fake P. We'll draw circles of radius sqrt(5) around each of our other candidate P's to see if any of them intersect the line between D and Fake P. This is where H is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZjAiC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only one of the circles intersects with the line. That circle belongs to the candidate P that lies on top of O.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;R is as far from H as O is from S and as far from D as D is from E. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Only one point is 3 units from H and 4 units from D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BzSht.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So what is the shape of the figure OSRHP, and what is its area in square bloits?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zVZW5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shape is a parallelogram and its area is 6 square bloits.&lt;/p&gt;&#xA;" OwnerUserId="1497" LastActivityDate="2015-03-24T20:08:30.450" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10901" PostTypeId="1" AcceptedAnswerId="10918" CreationDate="2015-03-24T20:15:50.777" Score="16" ViewCount="1048" Body="&lt;p&gt;Usually I'm transparent, keep me clean.&lt;br&gt;&#xA;Sometimes I'm generated by a fruit, keep me intact.&lt;br&gt;&#xA;Sometimes I'm a penguin, keep me fresh.&lt;br&gt;&#xA;Sometimes I'm in Star Trek, keep me in the future.&lt;br&gt;&#xA;Sometimes I shine like the sun, keep me clear.&lt;br&gt;&#xA;Rarely I'm something else.&lt;br&gt;&#xA;And remember, you own me for sure!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What am I?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2016-07-25T20:24:30.360" Title="I'm everything and everywhere" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10902" PostTypeId="2" ParentId="10836" CreationDate="2015-03-24T20:36:13.113" Score="0" Body="&lt;p&gt;It may be  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Her majesty Queen Elizabeth II.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mortals, beware!&#xA;A divine trapped in square!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The queen lives in Buckingham Palace, which has a square shape. She's also considered like a &quot;goddess&quot; by English people.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Only silences to your prayers,&#xA;Until a step through the stairs.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is still obscure to me.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Powers to make you a millionaire,&#xA;The only one with a spear.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; She is very very rich. She also lives in England, homeland of Shake&lt;strong&gt;spear&lt;/strong&gt;e&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Beneath the future heir,&#xA;Four does not make 2 pairs!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; She has four children, 3 males and a female, so no pairs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="4637" LastEditDate="2015-03-25T12:36:22.487" LastActivityDate="2015-03-25T12:36:22.487" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10903" PostTypeId="1" AcceptedAnswerId="10910" CreationDate="2015-03-24T21:33:45.357" Score="20" ViewCount="4999" Body="&lt;p&gt;Mason likes the game &lt;a href=&quot;http://en.wikipedia.org/wiki/Battleship_%28game%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Battleship&lt;/a&gt;, but he doesn't win much. However, Sherry, his opponent, is so confident in her skill and luck that she has agreed to let Mason have his five friends help him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sherry's method for placing her ships is completely random. Each side has 5 ships, and they reach the position below, where Mason has sunk all of Sherry's ships except for her Battleship, which is 4 units long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The board is 10x10, and all ships must be fully on the board and placed either horizontally or vertically. The green H's represent ships that Mason has already sunk. The red M's represent his misses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gq17h.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of Mason's friends want to help, but they cannot seem to agree on the best move. Clair insists on F2, but Mike is sure about H2. Bartholomew recommends D8 and Jessica tells him to play G6. Lastly, Pat says to play D5. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put his friends in order from &quot;Best Advice&quot; to &quot;Worst Advice&quot;. Did one of them find the best move?&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="4421" LastEditDate="2015-09-02T18:39:28.323" LastActivityDate="2015-09-02T18:39:28.323" Title="Battleship- Best Next Move" Tags="|strategy|board-games|" AnswerCount="5" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10904" PostTypeId="2" ParentId="10903" CreationDate="2015-03-24T21:40:17.963" Score="3" Body="&lt;p&gt;I'd say the order from best to worst is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;D5 - Covers a total of 3 placements and 9 spaces of the board where the ship could be. Also a slightly better shape (3 bars of a +)&lt;br&gt;&#xA;D8 - Covers 2 placements and 8 spaces&lt;br&gt;&#xA;F2 - 1 placement, 5 spaces&lt;br&gt;&#xA;H2 - 1 placement, 4 spaces&lt;br&gt;&#xA;G6 - There is no chance that the last ship is in this square, there is no segment going through this square that is 4 long.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the best move I'd say is E5, since that hits up to 3 positions and gets 10 spaces of coverage. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-03-25T04:00:48.497" LastActivityDate="2015-03-25T04:00:48.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10905" PostTypeId="5" CreationDate="2015-03-24T21:43:01.807" Score="0" Body="&lt;p&gt;The game is played on four grids, two for each player. The grids are typically square – usually $10\times10$ – and the individual squares in the grid are identified by letter and number. On one grid the player arranges ships and records the shots by the opponent. On the other grid the player records his/her own shots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before play begins, each player secretly arranges their ships on their primary grid. Each ship occupies a number of consecutive squares on the grid, arranged either horizontally or vertically. The number of squares for each ship is determined by the type of the ship. The ships cannot overlap (i.e., only one ship can occupy any given square in the grid). The types and numbers of ships allowed are the same for each player. These may vary depending on the rules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two typical complements of ships, as given in the Milton Bradley version of the rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Aircraft carrier (size 5)&lt;/li&gt;&#xA;&lt;li&gt;Battleship (size 4)&lt;/li&gt;&#xA;&lt;li&gt;Submarine (size 3)&lt;/li&gt;&#xA;&lt;li&gt;Destroyer or Cruiser (size 3)&lt;/li&gt;&#xA;&lt;li&gt;Patrol boat or Destroyer (size 2)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;After the ships have been positioned, the game proceeds in a series of rounds. In each round, each player takes a turn to announce a target square in the opponent's grid which is to be shot at. The opponent announces whether or not the square is occupied by a ship, and if it is a &quot;hit&quot; they mark this on their own primary grid. The attacking player notes the hit or miss on their own &quot;tracking&quot; grid, in order to build up a picture of the opponent's fleet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When all of the squares of a ship have been hit, the ship is sunk, and the ship's owner announces this (e.g. &quot;You sank my battleship!&quot;). If all of a player's ships have been sunk, the game is over and their opponent wins.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-24T21:43:01.807" LastActivityDate="2015-03-24T21:43:01.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10906" PostTypeId="4" CreationDate="2015-03-24T21:43:01.807" Score="0" Body="Battleship is a 2-player pencil-and-paper game in which each player tries to guess the positions of their opponent's ships while trying to keep their own intact." OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-24T21:43:01.807" LastActivityDate="2015-03-24T21:43:01.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10907" PostTypeId="1" AcceptedAnswerId="10922" CreationDate="2015-03-24T21:51:30.813" Score="26" ViewCount="12021" Body="&lt;p&gt;Sometimes I'm ugly, sometimes I'm art,&lt;br/&gt;&#xA;Sometimes I'm near and dear to your heart,&lt;br/&gt;&#xA;I have been known to bring about pain,&lt;br/&gt;&#xA;And yet I deliver relief just the same,&lt;br/&gt;&#xA;You know where I live, yes you know it best,&lt;br/&gt;&#xA;For I am not hidden like all of the rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Hint #1&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I come from a family where we each have a twin, &lt;br/&gt;&#xA; Even when we're well nourished, we always stay thin&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9978" LastEditorUserId="5373" LastEditDate="2015-03-24T23:31:39.590" LastActivityDate="2015-03-25T17:26:21.020" Title="Sometimes I'm ugly, sometimes I'm art. What am I?" Tags="|riddle|rhyme|" AnswerCount="9" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10908" PostTypeId="2" ParentId="10903" CreationDate="2015-03-24T21:53:20.373" Score="5" Body="&lt;p&gt;Based on the number of possible eliminations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D8 covers 3 possible battleship locations (Bartholomew)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F2 covers 2 possible battleship locations, and also follows a more spread out pattern of ships (Clair)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D5 covers 2 possible battleship locations but leaves open a third, D6-D9 (Pat)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H2 covers 1 possible battleship location (Mike)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And Jessica is obviously on Sherry's team, since G6 covers 0 possible battleship locations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tip of the hat to @JonTheMon, I would have also recommended E5.&lt;/p&gt;&#xA;" OwnerUserId="9412" LastActivityDate="2015-03-24T21:53:20.373" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10909" PostTypeId="2" ParentId="10907" CreationDate="2015-03-24T21:58:21.253" Score="5" Body="&lt;p&gt;Fourth try:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hmmm I suppose bones would also fit the riddle. A ribcage is near and dear to your heart, they provide relief in that gravity doesn't turn us into piles of goo. And technically they're not entirely hidden since you can see teeth. Also, as far as I know all bones come in &quot;twin&quot; pairs and are technically alive inside of you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Third try:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Teeth seems to fit a little better since they always stay thin even when someone is well nourished.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second try:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Now I'm thinking its a breast, which can get cancer and cause pain but can also provide relief. They are twins, and are usually in plain sight. And they're near and dear to your heart. And are ugly or art, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first try was&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a crucifix.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10424" LastEditorUserId="10424" LastEditDate="2015-03-25T00:39:42.010" LastActivityDate="2015-03-25T00:39:42.010" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="10910" PostTypeId="2" ParentId="10903" CreationDate="2015-03-24T21:58:24.787" Score="18" Body="&lt;p&gt;D5 covers 4 locations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D8 covers 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F2 covers 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H2 covers 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G6 covers 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E5 is the best; it covers 5 locations.&lt;/p&gt;&#xA;" OwnerUserId="167" LastActivityDate="2015-03-24T21:58:24.787" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="10911" PostTypeId="1" AcceptedAnswerId="11366" CreationDate="2015-03-24T22:00:02.060" Score="13" ViewCount="2274" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;My name sounds like a bird &lt;br&gt;&#xA;Whole body is like tongue of a lizard &lt;br&gt;&#xA;Am different colors &lt;br&gt;&#xA;And I well can be very littered &lt;br&gt;&lt;/p&gt;&#xA;&lt;p&gt;My energy melts away over time &lt;br&gt;&#xA;But not quite like an ice &lt;br&gt;&#xA;And when not taken care of &lt;br&gt;&#xA;I'll collect dust and lice &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;First one to answer correctly &lt;em&gt;will&lt;/em&gt; be marked as best answer.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE:&#xA;If you write an answer but want to change it or make a second guess, please update your answer, keeping the new guess above the old guess. Please do not overwrite.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Hint: It's not a food. And it is quite sticky.&lt;/p&gt;&#xA;" OwnerUserId="10603" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2019-03-01T09:30:47.370" Title="My name sounds like a bird - Riddle" Tags="|riddle|" AnswerCount="12" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="10912" PostTypeId="2" ParentId="10907" CreationDate="2015-03-24T22:02:03.450" Score="3" Body="&lt;p&gt;&lt;strong&gt;Updated guess&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is very silly I realize, but it's a guess.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alcohol&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I'm ugly, sometimes I'm art&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alcohol can be ugly at times. But people rejoice and celebrate with it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have been known to bring about pain&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alcohol causes brain damage and damage to your body.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And yet I deliver relief just the same&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People drink when they feel bad.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You know where I live, yes you know it best&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is easy to find alcohol these days at stores.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For I am not hidden like all of the rest&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some drugs are illegal but alcohol is legal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ORIGINAL ANSWER&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not for sure, but I guess this fits your riddle. Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Love?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10603" LastEditorUserId="10603" LastEditDate="2015-03-24T22:16:50.827" LastActivityDate="2015-03-24T22:16:50.827" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10913" PostTypeId="2" ParentId="10907" CreationDate="2015-03-24T22:29:38.797" Score="8" Body="&lt;p&gt;Am I:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Truth?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I'm ugly, sometimes I'm art,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Truth can be ugly, but when you not telling the truth (lying), it can be an art  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I'm near and dear to your heart,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Truth to your heart!  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have been known to bring about pain,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Knowing the truth can be painful  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And yet I deliver relief just the same,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; But also a relief to know it  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You know where I live, yes you know it best,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  You know when you're truth to yourself  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For I am not hidden like all of the rest.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  A Truth will always 'come out'?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2015-03-24T22:29:38.797" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10915" PostTypeId="2" ParentId="10903" CreationDate="2015-03-24T22:35:01.620" Score="3" Body="&lt;p&gt;Since some of the locations overlap, it may be better to look at how many squares can be eliminated by each guess. The end result will be somewhat different.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A miss on F2 eliminates the following other squares from contention: D2, E2, G2, and H2.&lt;/p&gt;&#xA;&lt;p&gt;For H2, it doesn't eliminate any other squares&lt;/p&gt;&#xA;&lt;p&gt;D8 also eliminates B8, C8, D6, D7, D9&lt;/p&gt;&#xA;&lt;p&gt;G6 technically eliminates itself, but it already wasn't an option&lt;/p&gt;&#xA;&lt;p&gt;D5, although it eliminates a lot of potential battleship positions, only actually takes 1 other spot out of the running - C2. All the other spots are still possible battleship locations in another configuration.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So by this metric, D8 is the best of the 5 guesses, potentially eliminating itself plus 5 other squares. That's followed by F2, then H2 and D5 tied, and G6 in last.&lt;/p&gt;&#xA;&lt;p&gt;I'm interested to hear people's opinions on which of the two metrics are &amp;quot;best&amp;quot;. To me, the potential flaw in looking at it from the number of ship locations that a spot is part of is that some spots are part of other potential locations too. So guessing D5 doesn't help you as much with regard to B5 because B5 could still be part of the ship.&lt;/p&gt;&#xA;&lt;p&gt;Also, by this measure, E5 eliminates 4 additional spots - C5, E4, E6, E7, so not as good as D8.&lt;/p&gt;&#xA;" OwnerUserId="1620" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-03-24T22:35:01.620" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10916" PostTypeId="2" ParentId="10911" CreationDate="2015-03-24T22:43:22.120" Score="0" Body="&lt;p&gt;Is it:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lollypops?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My name sounds like a bird  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I can't get this one  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whole body is like tongue of a lizard  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It rolls like a lizard's tongue  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Am different colors  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Colorful candy!  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I well can be very littered  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Once you're done eating...littering is no good&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My energy melts away over time&lt;br&gt;&#xA;But not quite like an ice  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sugar melts  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And when not taken care of&lt;br&gt;&#xA;I'll collect dust and lice  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Leave it alone and it'll get dirty  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="10037" LastEditDate="2015-03-25T20:19:17.427" LastActivityDate="2015-03-25T20:19:17.427" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10917" PostTypeId="2" ParentId="10836" CreationDate="2015-03-24T22:43:31.790" Score="5" Body="&lt;p&gt;My guess is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Queen of Spades&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mortals, beware!&#xA;A divine trapped in square!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Royalty used to be considered divinity. Playing cards are rectilinear, which is close enough to square.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Only silences to your prayers,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Luck at cards are determined by fate, not prayers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Until a step through the stairs.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not a clue what this means.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Powers to make you a millionaire,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can win big at cards.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The only one with a spear.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some cards have a distinctive design where the Queen of Spades is holding a scepter that kind of looks like a spear.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Beneath the future heir,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You say that beneath can me after, in your hint, and a Queen would assume the throne only after the Prince (Jack) in many monarchies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Four does not make 2 pairs!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well, there's lots of parses for this. None of which make sense to me...but 2 pairs is the biggest clue we're dealing with playing cards.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9953" LastActivityDate="2015-03-24T22:43:31.790" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10918" PostTypeId="2" ParentId="10901" CreationDate="2015-03-24T23:02:04.900" Score="14" Body="&lt;h2&gt;Answer&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An operating system&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Reasoning&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Usually I'm transparent, keep me clean.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Windows, the most common operating system&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I'm generated by a fruit, keep me intact.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MacOS or iOS, made by Apple&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I'm a penguin, keep me fresh.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Linux, its symbol is a penguin&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I'm in Star Trek, keep me in the future.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Android&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I shine like the sun, keep me clear.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SunOS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rarely I'm something else.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Those are the most common operating systems.  Anything else is pretty rare.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And remember, you own me for sure!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Anyone answering this question must own (or at least be using) an operating system of some sort.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="4421" LastEditDate="2016-07-25T20:24:30.360" LastActivityDate="2016-07-25T20:24:30.360" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10919" PostTypeId="2" ParentId="10882" CreationDate="2015-03-24T23:03:03.120" Score="0" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; forearm&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Phone is out of battery will post rationale la---&lt;/p&gt;&#xA;" OwnerUserId="10605" LastActivityDate="2015-03-24T23:03:03.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10920" PostTypeId="2" ParentId="10907" CreationDate="2015-03-24T23:20:56.637" Score="3" Body="&lt;p&gt;You are a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; tattoo&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I'm ugly, sometimes I'm art,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tattoos can be ugly, but many people consider them art.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I'm near and dear to your heart,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can get a tattoo on your chest.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have been known to bring about pain,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Getting a tattoo can be painful.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And yet I deliver relief just the same,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After it's done you may be relieved and enjoy how it looks on you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You know where I live, yes you know it best,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I would hope you know where your tattoo is located.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For I am not hidden like all of the rest.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most people get tattoos to show them off.  If it's hidden, what's the point?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2015-03-24T23:20:56.637" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10921" PostTypeId="2" ParentId="10897" CreationDate="2015-03-24T23:22:29.527" Score="2" Body="&lt;p&gt;Perhaps your unreliable friend is ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Muse&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The muse is not visible, yet we need it to create anything worthwhile. It may not appear; the master of the muse is fate. Muses inspire in words and deeds; &quot;I shorten the path&quot; is the least convincing fit for this answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8875" LastEditorUserId="8875" LastEditDate="2015-03-25T01:34:30.263" LastActivityDate="2015-03-25T01:34:30.263" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10922" PostTypeId="2" ParentId="10907" CreationDate="2015-03-24T23:59:23.767" Score="32" Body="&lt;p&gt;&lt;strong&gt;NEW GUESS&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you're a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; fingernail&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I'm ugly, sometimes I'm art,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sometimes people let their nails get out of hand, other times they get them painted.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I'm near and dear to your heart,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People move their hands, bringing their nails close to their heart.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have been known to bring about pain,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nails can scratch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And yet I deliver relief just the same,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nails can scratch (itches).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You know where I live, yes you know it best,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You know where your fingernails are...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For I am not hidden like all of the rest.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fingernails aren't hidden like the rest of your &lt;del&gt;'bones'&lt;/del&gt; toenails.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come from a family where we each have a twin, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We have two hands, and so we have two of each 'kind' of finger.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Even when we're well nourished, we always stay thin&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fingernails don't get fat, even when we're well fed. They just elongate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PREVIOUS GUESS&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Poem&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9696" LastEditorUserId="9696" LastEditDate="2015-03-25T17:26:21.020" LastActivityDate="2015-03-25T17:26:21.020" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10923" PostTypeId="2" ParentId="10911" CreationDate="2015-03-25T00:00:53.223" Score="2" Body="&lt;p&gt;I'm sure I know the answer, but I haven't managed to fit the first line yet. It's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CHEWING GUM.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My name sounds like a bird&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is the line I'm still struggling with.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Whole body is like tongue of a lizard&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many lizards have sticky tongues with which to catch insects, and chewing gum is also sticky.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Am different colors&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chewing gum comes in many colours.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And I well can be very littered&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's often thrown away, littering the streets.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My energy melts away over time&lt;br&gt;&#xA; But not quite like an ice&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chewing gum loses its chewiness after some time, but it doesn't literally &quot;melt&quot; in the same way ice does.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And when not taken care of&lt;br&gt;&#xA; I'll collect dust and lice&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you take out a piece of chewing gum and leave it somewhere, it will gather dust and all sorts of muck.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-03-25T00:00:53.223" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10924" PostTypeId="2" ParentId="10882" CreationDate="2015-03-25T00:03:28.613" Score="2" Body="&lt;p&gt;I think the answer is the word&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; EAT.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Metaphor lines&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I have more to do with cooking than you might think.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well, obviously! After food has been cooked, it gets eaten.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Now let us discuss the ocean, try not to sink.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Could be a reference to washing up in the &lt;em&gt;sink&lt;/em&gt; after eating?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Then between quizzes and sandwiches I like to rest.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This could be a very clever piece of wordplay that I haven't worked out.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Wordplay lines&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The beginning of eating, a feast is the best.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The beginning of &quot;eating&quot; is the letter &lt;strong&gt;E&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Half the area is needed, we aren't quite done.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Half of the letters in &quot;area&quot; are &lt;strong&gt;A&lt;/strong&gt;s.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now how many words are there relevant to cookery that begin with the letters&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; EA?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Right, so we have the answer. The next line is &lt;strong&gt;very&lt;/strong&gt; clever...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;We wrap up wondering if I am more than one.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Wrap up&quot; is a crossword-style clue that tells us we're looking for an anagram. EAT is an anagram of ATE, which is a homophone of EIGHT, which is indeed &quot;more than one&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Now as you guess, know that I am quite small,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The word &quot;eat&quot; is quite small, only three letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Last lines&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Doesn't really matter if you are short or tall.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can eat no matter what your height may be.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Two types of clues to guess one type of me.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The metaphor and wordplay sections of the riddle - see above.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What on the earth could I possibly be?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letters E, A, T are contained in the word &quot;earth&quot;!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-03-25T00:03:28.613" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10925" PostTypeId="2" ParentId="10897" CreationDate="2015-03-25T00:04:47.067" Score="3" Body="&lt;p&gt;It seems like a bit of a trick given the title, but...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; OPPORTUNITY?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I could be there, yet I'll stay hidden.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's often hard to find an opportunity; metaphorically, it's well hidden away.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Hidden I'll stay, but you'll need me.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You often need to wait for an opportunity to do something.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Need me you will, so I may appear.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you're lucky, an opportunity appears!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Appear I may, I don't decide.&lt;br&gt;&#xA; Decide I don't, my master does.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An opportunity doesn't decide anything by itself; &lt;em&gt;you&lt;/em&gt; have to decide to take it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I'm a word, I tell you words.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Opportunity&quot; is a word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I've a task, you know my task.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It may be an opportunity to perform some task.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I'm short, I shorten your path.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you seize an opportunity, it may significantly reduce your labour.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I've the answer, now tell me it!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; See above :-)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-03-25T00:04:47.067" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10927" PostTypeId="2" ParentId="10907" CreationDate="2015-03-25T00:37:17.957" Score="1" Body="&lt;p&gt;Is it hands?&#xA;not sure about the art/ugly&#xA;hand on heart&#xA;can inflict pain or relief&#xA;I know it like the back of my hand&#xA;you don't tend to cloth your hands &lt;/p&gt;&#xA;" OwnerUserId="10608" LastActivityDate="2015-03-25T00:37:17.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10930" PostTypeId="2" ParentId="10894" CreationDate="2015-03-25T01:08:15.570" Score="4" Body="&lt;p&gt;This answer sounds wrong, but a certain important book tells us it could be right, so bear with me...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LIFE?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am not lost, but I can be found.&lt;br&gt;&#xA;If you have found me, then you should lose me again, in order to find me.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; According to the &lt;a href=&quot;http://biblehub.com/matthew/16-25.htm&quot; rel=&quot;nofollow&quot;&gt;Gospel of Matthew&lt;/a&gt; in the Bible, &quot;Whoever finds their life will lose it, and whoever loses their life for my sake will find it.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;When you find me again it will surely be of benefit, similar to if you became more fit.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Going with this interpretation, you &quot;find&quot; your life again when you enter Heaven, which is surely a benefit. All your bodily weaknesses will have fallen away, as though you were suddenly more fit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My composition of attributes is exactly that...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sum of all your attributes gives your life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;...found of the pages were: on a journey, five had me and five did not.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This could be a reference to the Parable of the Ten Virgins (see &lt;a href=&quot;http://biblehub.com/matthew/25-2.htm&quot; rel=&quot;nofollow&quot;&gt;the Gospel of Matthew&lt;/a&gt; again), in which &quot;five of them were foolish and five were wise). So an alternative possibility would be WISDOM. Thanks to &lt;strong&gt;Lawrence&lt;/strong&gt; for helping me with this line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;sub&gt;&lt;sub&gt;And who says I'm rude about Christianity now?!&lt;/sub&gt;&lt;/sub&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-25T09:06:52.690" LastActivityDate="2015-03-25T09:06:52.690" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10931" PostTypeId="2" ParentId="10907" CreationDate="2015-03-25T01:22:22.380" Score="30" Body="&lt;p&gt;You are a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Testicle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I'm ugly&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well they are.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;sometimes I'm art,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; many famous statues have them still attached&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I'm near and dear to your heart&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When boys are young, and they haven't dropped yet they're much closer to the heart than normal&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have been known to bring about pain&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; indeed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And yet I deliver relief just the same&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; heh, yeah they do.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You know where I live, yes you know it best,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; check&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For I am not hidden like all of the rest.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; they're out in the open, unlike the rest of your guts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The additional hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm pretty sure they don't grow and shrink due to how well you're nourished&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10610" LastActivityDate="2015-03-25T01:22:22.380" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10933" PostTypeId="1" AcceptedAnswerId="15869" CreationDate="2015-03-25T02:45:32.113" Score="9" ViewCount="1674" Body="&lt;p&gt;This is an interesting one.&lt;br&gt;&#xA;I will do the same thing that I usually do.&lt;br&gt;&#xA;After one week, a hint. After 2 weeks, a hint. After 3 weeks, an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I may change this system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What number goes in the question mark, and why does it go there?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;0,0,1,2,3,1,3,2,3,3,4,2,4,4,3,3,4,4,5,3,5,4,5,3,5,5,4,4,5,?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;HINT:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There is a inputting function that each number is put into... when you undo that function, you get a simple sequence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;HINT 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If you take this inputting function onto 24862486 you get 4. I will take the next three requests of inputting from the comments.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;HINT 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This has to do with prime factors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;HINT 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It requires the repeated use of a single function.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;HINT 5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This function is related to the Greatest Prime Factor&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="167" LastEditorUserId="167" LastEditDate="2015-06-02T23:19:25.240" LastActivityDate="2015-06-04T17:27:55.020" Title="Another one of those odd sequence puzzles" Tags="|mathematics|number-sequence|" AnswerCount="5" CommentCount="42" ContentLicense="CC BY-SA 3.0" />
  <row Id="10935" PostTypeId="1" AcceptedAnswerId="11152" CreationDate="2015-03-25T04:35:04.607" Score="13" ViewCount="2348" Body="&lt;p&gt;Here are three different letter/number sequences. Try to solve as many as possible. The spoilers give hints to each sequence in the form of how many elements are in the corresponding sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sequence 1 (Missing Elements: 2) [Solved by &lt;a href=&quot;https://puzzling.stackexchange.com/a/10936/9631&quot;&gt;xnor&lt;/a&gt;]&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Y, Y, ?, L, Y, E, ?, T, R...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There are 12 elements in this sequence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sequence 2 (Missing Elements: 2) [Solved by &lt;a href=&quot;https://puzzling.stackexchange.com/a/10941/9631&quot;&gt;Tryth&lt;/a&gt;]&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a, H, R, 0, c, ?, o, ?, L, 3, B...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There are 44 elements in this sequence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sequence 3 (Missing Elements: 3) [Solved by &lt;a href=&quot;https://puzzling.stackexchange.com/a/11152/9631&quot;&gt;finsternis&lt;/a&gt;]&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;?, P, C, P, C, ?, B, R, B, R, ?, R, S, R, S, R...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There are 29 elements in this sequence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first one is quite simple but the other two are fairly challenging. Good luck!&lt;/p&gt;&#xA;" OwnerUserId="9631" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2016-01-03T05:48:15.820" Title="Three different, and somewhat difficult, sequence puzzles" Tags="|number-sequence|letter-sequence|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10936" PostTypeId="2" ParentId="10935" CreationDate="2015-03-25T05:03:34.400" Score="22" Body="&lt;p&gt;Sequence 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Last letters of months:&lt;br&gt;&#xA; Januar&lt;strong&gt;y&lt;/strong&gt;, Februar&lt;strong&gt;y&lt;/strong&gt;, Marc&lt;strong&gt;h&lt;/strong&gt;, Apri&lt;strong&gt;l&lt;/strong&gt;, Ma&lt;strong&gt;y&lt;/strong&gt;, Jun&lt;strong&gt;e&lt;/strong&gt;, Jul&lt;strong&gt;y&lt;/strong&gt;, Augus&lt;strong&gt;t&lt;/strong&gt;, Septembe&lt;strong&gt;r&lt;/strong&gt;, Octobe&lt;strong&gt;r&lt;/strong&gt;, Novembe&lt;strong&gt;r&lt;/strong&gt;, Decembe&lt;strong&gt;r&lt;/strong&gt;&lt;br&gt;&#xA;which gives the sequence &lt;code&gt;Y, Y, H, L, Y, E, Y, T, R, R, R, R&lt;/code&gt;.&lt;br&gt;&#xA;The missing elements are &lt;code&gt;H&lt;/code&gt; and &lt;code&gt;Y&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-03-25T05:03:34.400" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10937" PostTypeId="2" ParentId="10897" CreationDate="2015-03-25T05:48:40.497" Score="3" Body="&lt;p&gt;This might be the deepest, most thought-provoking answer to a riddle ever.  It's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Siri&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I could be there, yet I'll stay hidden.&lt;br&gt;&#xA;Hidden I'll stay, but you'll need me.&lt;br&gt;&#xA;Need me you will, so I may appear.&lt;br&gt;&#xA;Appear I may, I don't decide.&lt;br&gt;&#xA;Decide I don't, my master does.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's a feature of the phone that stays in the background until the owner asks for it (&quot;Siri, how do I ...&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm a word, I tell you words.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Siri responds to questions in speech.  And, she can define words for you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've a task, you know my task&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Siri keeps tasks (reminders, alarms, etc) for you&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm short, I shorten your path.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Siri is a short word.  She shortens your path by giving you GPS directions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've the answer, now tell me it!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Siri has all the answers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Disclaimer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I don't own an iPhone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7681" LastActivityDate="2015-03-25T05:48:40.497" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10938" PostTypeId="2" ParentId="10897" CreationDate="2015-03-25T06:21:07.957" Score="2" Body="&lt;p&gt;I think its a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Navigation app&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation line by line:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could be there, yet I'll stay hidden.&#xA;Hidden I'll stay, but you'll need me.&#xA;Need me you will, so I may appear.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You only use your navigation app when you need it. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Appear I may, I don't decide.&#xA;Decide I don't, my master does.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It isn't up to the phone to open the app but the user can.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm a word, I tell you words.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The navigation app tells which path you have to take.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've a task, you know my task.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You set it to go to a certain direction&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm short, I shorten your path.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It shortens the path by choosing the shortest route.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4786" LastEditorUserId="4786" LastEditDate="2015-03-26T12:18:25.070" LastActivityDate="2015-03-26T12:18:25.070" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10939" PostTypeId="2" ParentId="10907" CreationDate="2015-03-25T08:00:22.877" Score="1" Body="&lt;p&gt;Skin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I come from a family where we each have a twin&quot;&#xA;For every piece of skin on your body there is a symmetrical counterpart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Even when we're well nourished, we always stay thin&quot;&#xA;Skin is thin regardless how fat you are&lt;/p&gt;&#xA;" OwnerUserId="10616" LastActivityDate="2015-03-25T08:00:22.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10940" PostTypeId="2" ParentId="10882" CreationDate="2015-03-25T08:08:10.357" Score="1" Body="&lt;p&gt;I guess it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a nose!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have more to do with cooking than you might think.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The nose plays a huge role in eating and cooking affecting our appetite.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now let us discuss the ocean, try not to sink.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not quite sure, maybe an advice to people who tend to take nose-picking to the extreme.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then between quizzes and sandwiches I like to rest.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Like someone else already figured out: the nose lies between the brain (quiz) and the mouth (sandwich).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The beginning of eating, a feast is the best.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many people begin eating with smelling the food.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Half the area is needed, we aren't quite done.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We only use one nostril at a time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We wrap up wondering if I am more than one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Maybe because we have two separate nostrils... not sure about this one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now as you guess, know that I am quite small,&#xA;Doesn't really matter if you are short or tall.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Noses can be any size not matter how tall you are.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Two types of clues to guess one type of me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That's already in the hints.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What on the earth could I possibly be?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A nose!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10583" LastActivityDate="2015-03-25T08:08:10.357" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10941" PostTypeId="2" ParentId="10935" CreationDate="2015-03-25T08:10:30.337" Score="26" Body="&lt;p&gt;Sequence 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This represents the base64 encoded form of the string &lt;code&gt;http://puzzling.stackexchange.com&lt;/code&gt;. This corresponds to &lt;code&gt;aHR0cDovL3B1enpsaW5nLnN0YWNrZXhjaGFuZ2UuY29t&lt;/code&gt;. Therefore the missing elements are &lt;strong&gt;D&lt;/strong&gt; and &lt;strong&gt;v&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7308" LastActivityDate="2015-03-25T08:10:30.337" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10942" PostTypeId="1" AcceptedAnswerId="10944" CreationDate="2015-03-25T09:42:22.737" Score="6" ViewCount="709" Body="&lt;p&gt;Yesterday, my good friend Pedro told me about &lt;strong&gt;Pedro's pawn game&lt;/strong&gt; which is played with $7$ white and $7$ black pawns on a $7\times7$ checkerboard. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In the starting situation, there is a white pawn on each of the seven squares in the bottom row, and a black pawn on each of the seven squares in the top row.&lt;/li&gt;&#xA;&lt;li&gt;In the goal situation, white and black pawns have switched positions: there is a black pawn on each square in the bottom row and a white pawn on each square in the top row.&lt;/li&gt;&#xA;&lt;li&gt;The only allowed pawn moves are from the current square to a horizontally or vertically adjacent &lt;em&gt;empty&lt;/em&gt; square.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Pedro asked me: What is the &lt;em&gt;smallest&lt;/em&gt; number of allowed moves that turn the starting situation into the goal situation?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-03-25T12:17:54.070" Title="Pedro's pawn game" Tags="|mathematics|combinatorics|checkerboard|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10943" PostTypeId="2" ParentId="10942" CreationDate="2015-03-25T09:53:11.523" Score="3" Body="&lt;h2&gt;Solution to the question as stated&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;It can be done in &lt;strong&gt;92&lt;/strong&gt; moves. Here's how (using standard chessboard notation, rows numbered from a to g and columns from 1 to 7):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;move the white pawn on a1 to a5 (&lt;strong&gt;4&lt;/strong&gt; moves)&lt;/li&gt;&#xA;&lt;li&gt;move the white pawn on b1 to b3 (&lt;strong&gt;2&lt;/strong&gt; moves)&lt;/li&gt;&#xA;&lt;li&gt;move the black pawn on b7 to b4, then a4, then a1 (3+1+3=&lt;strong&gt;7&lt;/strong&gt; moves)&lt;/li&gt;&#xA;&lt;li&gt;move the white pawn on b3 to b7 (&lt;strong&gt;4&lt;/strong&gt; moves)&lt;/li&gt;&#xA;&lt;li&gt;move the black pawn on a7 to a6, then b6, then b1 (1+1+5=&lt;strong&gt;7&lt;/strong&gt; moves)&lt;/li&gt;&#xA;&lt;li&gt;move the white pawn on a5 to a7 (&lt;strong&gt;2&lt;/strong&gt; moves)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In 4+2+7+4+7+2=&lt;strong&gt;26&lt;/strong&gt; moves we've switched the pawns on a1 and b1 with those on a7 and b7. Do the same with the pawns on g1 and h1, g7 and h7. For the final three columns, proceed as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;move the white pawns on c1, d1, e1 to c3, d5, e3 (2+4+2=&lt;strong&gt;8&lt;/strong&gt; moves)&lt;/li&gt;&#xA;&lt;li&gt;move the black pawn on c7 to c4, then d4, then d1 (3+1+3=&lt;strong&gt;7&lt;/strong&gt; moves)&lt;/li&gt;&#xA;&lt;li&gt;move the white pawn on c3 to c7 (&lt;strong&gt;4&lt;/strong&gt; moves)&lt;/li&gt;&#xA;&lt;li&gt;move the black pawn on d7 to d6, then c6, then c1 (1+1+5=&lt;strong&gt;7&lt;/strong&gt; moves)&lt;/li&gt;&#xA;&lt;li&gt;move the white pawn on d5 to d7 (&lt;strong&gt;2&lt;/strong&gt; moves)&lt;/li&gt;&#xA;&lt;li&gt;move the black pawn on e7 to e4, then d4, then d2, then e2, then e1 (3+1+2+1+1=&lt;strong&gt;8&lt;/strong&gt; moves)&lt;/li&gt;&#xA;&lt;li&gt;move the white pawn on e3 to e7 (&lt;strong&gt;4&lt;/strong&gt; moves)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All this gives a total of 2*26+(8+7+4+7+2+8+4)=52+40=&lt;strong&gt;92&lt;/strong&gt; moves.&#xA;&lt;hr&gt;&#xA;As &lt;a href=&quot;https://puzzling.stackexchange.com/a/10944/5373&quot;&gt;xnor states&lt;/a&gt;, it &lt;em&gt;can't&lt;/em&gt; be done in less than 91 moves, because each pair of pawns need to 'pass' each other, necessitating an extra move to the side for one of them. In fact it can't be done in exactly 91 moves either, because not all of these shifts to the side can be done in the same direction (the board being finite) and where two shifts in opposite directions 'meet', two pawns would end up in the same square, so one of them needs to shift &lt;em&gt;again&lt;/em&gt; (since 7 is odd).&#xA;&lt;hr&gt;&#xA;So the final answer is &lt;strong&gt;92&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;A generalisation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Let's replace 7 pawns of each colour on a 7x7 board with $n$ pawns of each colour on an $n\times n$ board! What's the answer then?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method I've used above for swapping a1 and b1 with a7 and b7 can be used for all pairs of columns, giving $4(n-1)+2=4n-2$ moves for each pair of columns (e.g. $4*7+2=26$ when $n=7$). The method for swapping c1, d1, e1 with c7, d7, e7 works for any triple, giving $6(n-1)+3+1=6n-2$ moves for a triple (e.g. $6*7-2=40$ when $n=7$). So the whole operation can be done in $k*(4n-2)=n(2n-1)$ moves if $n=2k$ is even, and $(k-1)(4n-2)+(6n-2)=n(2n-1)+1$ moves if $n=2k+1$ is odd.&#xA;&lt;hr&gt;&#xA;The argument used by xnor to prove 91 is a lower bound when $n=7$ also shows that $n(2n-1)$ is a lower bound for all $n$. When $n$ is odd, xnor's parity argument shows that since the total number of moves must be even, $n(2n-1)+1$ is a lower bound.&#xA;&lt;hr&gt;&#xA;So the optimal number of moves required to swap $n$ white pawns with $n$ black pawns on an $n\times n$ board is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$n(2n-1)$ if n is even&lt;/li&gt;&#xA;&lt;li&gt;$n(2n-1)+1$ if n is odd&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-03-25T11:30:43.680" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="10944" PostTypeId="2" ParentId="10942" CreationDate="2015-03-25T10:13:07.770" Score="18" Body="&lt;p&gt;I give a construction of 92 moves and prove it optimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/a/10943/4551&quot;&gt;As rand al'thor notes&lt;/a&gt;, 84 moves forward are inevitable. So, we'll only count &quot;excess moves&quot; that do not move a pawn forward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each white pawn is in a column opposed against a black pawn that blocks its way. The two cannot pass each other, so each column requires an excess move to unblock it by moving one of its pawns sideways. This gives a minimum of 7 excess moves, so an overall &lt;em&gt;91 move minimum&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that we can resolve two adjacent columns in two excess moves by moving out a white piece, moving the black pawn from the other column into its place, and then making the natural moves:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;OO        OO       |O        XO       X|        XX &#xA;                   |         ^        --         ^&#xA;           X       |X        |X       |X         |&#xA;           ^       --        |        |           &#xA;           |        |        |        |&#xA;           |        v        |        v&#xA;XX        X|       XO        |O       OO        OO&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;A single column can also be resolved with two excess moves by moving a pawn sideways, letting the other pass, and moving it back. This lets us resolve the 7 columns in 8 excess moves for a 92-move solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we cut down the solution to 91? No, by a parity argument. On a checkered board, each move changes the parity of the number of pieces on black squares. Since that parity is the same in the start and end configurations, the total number of moves must be even. So, the minimum is 92, which the solution achieves.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-03-25T12:17:54.070" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10945" PostTypeId="1" AcceptedAnswerId="10946" CreationDate="2015-03-25T10:42:29.953" Score="5" ViewCount="305" Body="&lt;p&gt;Yesterday Ethan Hunt received the following message:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Take Qantas flight number 47, leaving 11 o'clock on the 5th of April. At your destination airport use any bus with number 2, 8, 18 (or bus 18/1). You can take with you 107.8 kilograms of luggage.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the goal of Ethan's journey?&lt;/p&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="4050" LastEditDate="2015-03-25T11:33:06.640" LastActivityDate="2015-03-25T11:33:06.640" Title="Mission: Impossible 3" Tags="|word|cipher|enigmatic-puzzle|" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10946" PostTypeId="2" ParentId="10945" CreationDate="2015-03-25T10:45:35.593" Score="8" Body="&lt;p&gt;Easy, easy.  The goal of Ethan Hunt's journey must be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Argentina&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; * The country Argentina is called after silver (= argentum in Latin).&lt;br&gt;&#xA; * Silver has the chemical element number 47, the flight number of Ethan.&lt;br&gt;&#xA; * The element silver is in group 11, period 5 (11 o'clock on April 5th)&lt;br&gt;&#xA; * The electron configuration of silver is 2, 8, 18, 18, 1 per shell (bus with number 2, 8, 18 or 18/1).&lt;br&gt;&#xA; * The relative atomic mass of silver is 107.8682 (107.8 kilograms of luggage)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See also&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Silver&quot;&gt;http://en.wikipedia.org/wiki/Silver&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-03-25T11:04:47.963" LastActivityDate="2015-03-25T11:04:47.963" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10947" PostTypeId="2" ParentId="10911" CreationDate="2015-03-25T11:56:31.743" Score="0" Body="&lt;p&gt;I think you're&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; pork&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My name sounds like a bird&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; stork&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whole body is like tongue of a lizard&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; uncooked pork is pink like a tongue&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Am different colors&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; depending on how you cook it, the color will change&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I well can be very littered&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; with salt and pepper&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My energy melts away over time&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But not quite like an ice&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a heated steak will eventually get cold &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And when not taken care of&#xA;I'll collect dust and lice&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; improperly handled pork can collect germs and parasites, and is one of the worst things that you can eat (unfortunately i know from experience)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9778" LastActivityDate="2015-03-25T11:56:31.743" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10949" PostTypeId="2" ParentId="10897" CreationDate="2015-03-25T12:28:51.257" Score="5" Body="&lt;p&gt;It must be the&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  apostrophe-sign, like in &quot;I'm&quot; or  &quot;I've&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First of all, the apostrophe is used throughout the whole riddle, so it is probably the clue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I could be there, yet I'll stay hidden.&lt;br&gt;&#xA;Hidden I'll stay, but you'll need me.&lt;br&gt;&#xA;Need me you will, so I may appear.&lt;br&gt;&#xA;Appear I may, I don't decide.&lt;br&gt;&#xA;Decide I don't, my master does.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The apostrophe may appear or may not. The writer decides if to include the apostrophe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm a word, I tell you words.&lt;br&gt;&#xA;I've a task, you know my task.&lt;br&gt;&#xA;I'm short, I shorten your path.&lt;br&gt;&#xA;I've the answer, now tell me it!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The apostrophe makes the statement shorter. It is no word, but it &quot;tells&quot; you words.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second guess, based on the first one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; abbreviation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5283" LastEditorUserId="5283" LastEditDate="2015-03-25T12:59:13.303" LastActivityDate="2015-03-25T12:59:13.303" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10950" PostTypeId="1" AcceptedAnswerId="10951" CreationDate="2015-03-25T13:17:17.310" Score="30" ViewCount="5795" Body="&lt;p&gt;I found this question in a book. The solution is correct (and complicated). No tricks involved.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Four men were seated around a table. They were blindfolded and a&#xA;  colored hat was placed on each of their heads. They were told a true&#xA;  statement &quot;The hats on your heads were chosen from 2 white hats, 2&#xA;  black hats and 1 red hat&quot; Then the fifth hat was taken away and the&#xA;  blindfolds were removed.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Each logician was asked to name the color of their hat. If they could&#xA;  logically do so, they did, otherwise said &quot;I don't know&quot;. The question&#xA;  was repeated to each of them in circular fashion (more than once, if&#xA;  need be) until only one of them failed to name the color of his own&#xA;  hat. The red hat was not given, but the logicians didn't know this.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Which logician failed to name the color of his hat even when the other&#xA;  three had? &#xA;  (Was it the man who was asked the first question, the man who was &#xA;  asked the second question, the man who was asked the third question, &#xA;  or the man who was asked the fourth question?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="57212" LastEditDate="2021-04-14T06:48:51.817" LastActivityDate="2021-04-14T06:48:51.817" Title="Five hats and four logicians in a circle" Tags="|logical-deduction|meta-knowledge|hat-guessing|" AnswerCount="13" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="10951" PostTypeId="2" ParentId="10950" CreationDate="2015-03-25T13:42:30.860" Score="12" Body="&lt;p&gt;Call the logicians A, B, C, and D, in the order they speak. WLOG, say A is wearing a white hat. Note that each logician sees 2 black, 1 white if he's wearing white or 2 white, 1 black if he's wearing black, so he always knows his own hat is either [the colour it actually is] or red.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A is wearing &lt;strong&gt;white&lt;/strong&gt; and sees 2 black, 1 white. As far as he knows, he could be wearing either white or red. He says, &quot;&lt;strong&gt;I don't know.&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Case 1: B is also wearing white&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;B is wearing &lt;strong&gt;white&lt;/strong&gt; and sees 2 black, 1 white. If B were wearing red, then A (seeing 2 black, 1 red) would have been able to identify his own colour. So B says, &quot;&lt;strong&gt;My hat is white.&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C is wearing &lt;strong&gt;black&lt;/strong&gt; and sees 2 white, 1 black. If C were wearing red, then B (seeing 1 white, 1 black, 1 red) would still have known his hat was white, since B wearing black would be inconsistent with A's answer (A would then have seen 1 red and 2 black and known he was wearing white). So C says, &quot;&lt;strong&gt;I don't know.&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D is wearing &lt;strong&gt;black&lt;/strong&gt; and sees 2 white, 1 black. If D were wearing red, then C would have seen 1 red and 2 white and known his own colour. So D says, &quot;&lt;strong&gt;My hat is black.&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A is still wearing &lt;strong&gt;white&lt;/strong&gt;, and argues as follows. If A were wearing red, then B would have seen 2 black, 1 red and known B was wearing white; C would have seen 1 black, 1 red, 1 white and known C was wearing black (if C were white, B would have seen 1 white, 1 black, 1 red and either black or white for B would have been consistent with A's answer); contradiction, so A says, &quot;&lt;strong&gt;My hat is white.&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the answer is &lt;strong&gt;C&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Case 2: B is wearing black, C is wearing white&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;B is wearing &lt;strong&gt;black&lt;/strong&gt; and sees 2 white, 1 black. If B were wearing red, then A would have seen 1 black, 1 white, 1 red and still said &quot;I don't know&quot;. So B says, &quot;&lt;strong&gt;I don't know.&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C is wearing &lt;strong&gt;white&lt;/strong&gt; and sees 2 black, 1 white. If C were wearing red, A would have seen 2 black, 1 red and known his own colour. So C says, &quot;&lt;strong&gt;My hat is white&lt;/strong&gt;.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D is wearing &lt;strong&gt;black&lt;/strong&gt; and sees 2 white, 1 black. If D were wearing red, B would have seen 2 white, 1 red and known his own colour. So D says, &quot;&lt;strong&gt;My hat is black.&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A is still wearing &lt;strong&gt;white&lt;/strong&gt;, and argues as follows. If A were wearing red, then B would have seen 1 white, 1 black, 1 red and either black or white for B would have been consistent with A's answer; C would have seen 2 black, 1 red, and known C was wearing white; D would have seen 1 white, 1 black, 1 red and known that he was wearing black (if D were white, B would have seen 2 white, 1 red and known B was wearing black). So A says, &quot;&lt;strong&gt;I don't know.&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B is still wearing &lt;strong&gt;black&lt;/strong&gt;, and argues as follows. If B were wearing red, then A would have seen 1 white, 1 black, 1 red and not known his own colour; B would have said what he did; C would have seen 1 white, 1 black, 1 red and known he was wearing white (if C was wearing black, A would have seen 2 black, 1 red and known his own colour); D would have seen 2 white, 1 red and known he was wearing black; and finally A would &lt;em&gt;still&lt;/em&gt; not have known his colour (if A was wearing black, everything would have happened the same way it did!). So B says, &quot;&lt;strong&gt;I don't know.&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C and D still know their colours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A is still wearing &lt;strong&gt;white&lt;/strong&gt;, and argues as follows. If A were wearing red, then as above, B, C, D, and A (second time round) would still have responded as they did, and then B (seeing 1 white, 1 black, 1 red) would &lt;em&gt;still&lt;/em&gt; not have known his own colour (if B were wearing white, then we have RWWB and everything goes through as before!) So A says &quot;&lt;strong&gt;I don't know&lt;/strong&gt;&quot; for the third time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B is still wearing &lt;strong&gt;black&lt;/strong&gt; and argues as follows. If B were wearing red, then as above, the first and second rounds would have gone as they did, and then A (seeing 1 white, 1 black, 1 red) would still not have known his own colour (if A were wearing black, then we have BRWB and everything goes through as before!). So B says &quot;&lt;strong&gt;I don't know&lt;/strong&gt;&quot; for the third time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same argument goes on. No new information is gained by anyone.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Case 3: B and C are both wearing black&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;As argued above, B is wearing &lt;strong&gt;black&lt;/strong&gt; and says, &quot;&lt;strong&gt;I don't know.&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C is wearing &lt;strong&gt;black&lt;/strong&gt; and sees 2 white, 1 black. If he were wearing red, B would have seen 2 white, 1 red and known his own colour. So C says, &quot;&lt;strong&gt;My hat is black.&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D is wearing &lt;strong&gt;white&lt;/strong&gt; and sees 2 black, 1 white. If he were wearing red, A would have seen 2 black, 1 red and known his own colour. So D says, &quot;&lt;strong&gt;My hat is white.&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A is still wearing &lt;strong&gt;white&lt;/strong&gt;, and argues as follows. If A were wearing red, then B would have seen 1 white, 1 black, 1 red and either black or white for B would have been consistent with A's answer; C would have seen 1 black, 1 white, 1 red, and known C was wearing black (if C were white, B would have seen 2 white, 1 red and known B was wearing black); D would have seen 2 black, 1 red and known that he was wearing white. So A says, &quot;&lt;strong&gt;I don't know.&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B is still wearing &lt;strong&gt;black&lt;/strong&gt;, and argues as follows. If B were wearing red, then A would have seen 1 white, 1 black, 1 red and not known his own colour; B would have said what he did; C would have seen 2 white, 1 red and known he was wearing black; D would have seen 1 white, 1 black, 1 red and known he was wearing white (if D was wearing black, A would have seen 2 black, 1 red and known his own colour); and finally A would &lt;em&gt;still&lt;/em&gt; not have known his colour (if A was wearing black, everything would have happened the same way it did!). So B says, &quot;&lt;strong&gt;I don't know.&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C and D still know their colours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A is still wearing &lt;strong&gt;white&lt;/strong&gt;, and argues as follows. If A were wearing red, then as above, B, C, D, and A (second time round) would still have responded as they did, and then B (seeing 1 white, 1 black, 1 red) would &lt;em&gt;still&lt;/em&gt; not have known his own colour (if B were wearing white, then we have RWBW and everything goes through as before!) So A says &quot;&lt;strong&gt;I don't know&lt;/strong&gt;&quot; for the third time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B is still wearing &lt;strong&gt;black&lt;/strong&gt; and argues as follows. If B were wearing red, then as above, the first and second rounds would have gone as they did, and then A (seeing 1 white, 1 black, 1 red) would still not have known his own colour (if A were wearing black, then we have BRBW and everything goes through as before!). So B says &quot;&lt;strong&gt;I don't know&lt;/strong&gt;&quot; for the third time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same argument goes on. No new information is gained by anyone.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In Cases 2 and 3, A and B never find out their colours; C and D will just sit there snickering at them forever. But we're &lt;em&gt;told&lt;/em&gt; in the OP that eventually &quot;only one of them failed to name the color of his own hat&quot;, so we can deduce the following two pieces of information:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the first two logicians to speak have the same colour of hat, as do the last two&lt;/li&gt;&#xA;&lt;li&gt;the &lt;strong&gt;third&lt;/strong&gt; logician is the last one not to know his own colour.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is the final answer.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-25T18:29:44.120" LastActivityDate="2015-03-25T18:29:44.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10952" PostTypeId="2" ParentId="10950" CreationDate="2015-03-25T13:43:53.043" Score="0" Body="&lt;p&gt;1: he sees 2W and 1B. He might have R or B, so he passes&#xA;2: he sees either 2W and 1B or 2B and 1W. Knowing the first one passed, he can make the following statements:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If 3 and 4 have same color of the hat, he has the opposite(if he would have R, 1 would have guessed his color)  &lt;/li&gt;&#xA;&lt;li&gt;If 3 and 4 have different colors, his color is opposite of 1 or R&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Let's move on with scenario 1:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;3 sees that 2 guessed his color, and 1 passed. That means that he doesn't have R(otherwise 2 wouldn't be able to guess), so he has either B or W. He can just count and guess his hat right&lt;/li&gt;&#xA;&lt;li&gt;4 sees that 2 and 3 guessed their colors, so his color isn't R(otherwise they wouldn't be able to guess their colors). So he has either B or W, count and win&lt;/li&gt;&#xA;&lt;li&gt;1 sees that they all guessed their colors. If he would have R, 3 wouldn't be able to guess his hat, so 1 can also easily guess his color&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So in scenario 1 they all win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scenario 2:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;2 passes, because he isn't sure if he has opposite of 1 or R&lt;/li&gt;&#xA;&lt;li&gt;3 makes a guess and says his hat is same as 1. Because if he would have R, 1 would easily guess his hat&lt;/li&gt;&#xA;&lt;li&gt;4 guesses too(same logic as 3)&lt;/li&gt;&#xA;&lt;li&gt;1 guesses correctly, because if he would have R, 4 wouldn't be able to guess his hat&lt;/li&gt;&#xA;&lt;li&gt;2 can guess it too, because if he would have R, 3 wouldn't be able to guess his color&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So if they all think logically, they will all be able to guess their colors&lt;/p&gt;&#xA;" OwnerUserId="9352" LastActivityDate="2015-03-25T13:43:53.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10953" PostTypeId="2" ParentId="10903" CreationDate="2015-03-25T13:52:32.710" Score="22" Body="&lt;p&gt;The &quot;best&quot; move for this puzzle is the one most likely to hit the ship in one shot, as the puzzle doesn't allow for further follow up shots.  There are 23 possible positions for that ship, and 68 spaces where a shot could be placed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the size of the ship and previously placed shots, we can find both where shots cannot possibly hit the ship, and places for shots which cover more than one position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, G6 cannot hit the ship because previous shots placed at G4, G7, F6, and I6 would have most certainly hit that ship given that the ship is four spaces long.  So G6 could not result in a hit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking F2 as an example, there are 5 spaces in a vertical column where shots have not previously been placed.  A 4 length ship could exist in the four top spaces, or four bottom spaces, and F2 covers both possibilities.  Thus, F2 would hit 2 out of those 21 possible ship positions, giving you a 8.7% chance of hitting the ship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we perform this analysis for every possible shot we can take, we can find how many positions each shot could cover:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Bvb5d.png&quot; alt=&quot;Image of battleship board with each available space filled with the number of positions it covers&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We find then that the best space for a shot which covers the most of the 21 available positions is E5, covering 5 positions, and having a 22% chance of hitting the ship.  This is not among the suggestions the friends made, though.  Their shots rank as follows, from best to worst:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Space   Positions    Probability&#xA;D5      4            17%&#xA;D8      3            13%&#xA;F2      2             8.7%&#xA;H2      1             4.3%&#xA;G6      0             0.0%&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Once E5 is taken you've removed 5 possible locations, leaving 18 if it didn't hit the ship:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ukvgN.png&quot; alt=&quot;Image of battleship board with each available space filled with the number of positions it covers after another shot on E5&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1682" LastEditorUserId="1682" LastEditDate="2015-03-25T17:47:52.103" LastActivityDate="2015-03-25T17:47:52.103" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10954" PostTypeId="2" ParentId="10950" CreationDate="2015-03-25T13:53:10.810" Score="2" Body="&lt;p&gt;Ok, we have 2 hat placements: same color next to each other, or same color across.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If same color across (assume black hat asked first):&lt;br&gt;&#xA;The only way to be the first person and know your hat is to see 2 of the same color and the red hat.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First person (sees white, white, black): IDK&lt;/li&gt;&#xA;&lt;li&gt;Second (bbw): IDK&lt;/li&gt;&#xA;&lt;li&gt;Third (wwb): Knows that the black hat across from him didn't see a red hat on his head, &lt;strong&gt;Black&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;Fourth (bbw): Same as third, but &lt;strong&gt;White&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;First (wwb): Knows that he has black or red. Second's answer doesn't help (could see wbr), neither does third (could see wwr) nor fourth (could see wbr and determine that second also has wbr). IDK&lt;/li&gt;&#xA;&lt;li&gt;Second (bbw): if red, then first's idk, third's I know, and fourth's I know are all reasonable. IDK&lt;/li&gt;&#xA;&lt;li&gt;Stuck. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And the colors sitting next to each other (go order BBWW):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First (wwb): IDK&lt;/li&gt;&#xA;&lt;li&gt;Second (wwb): Sees the same WW that first did, knows he's not R, so &lt;strong&gt;Black&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;Third (bbw): Sees the same WB that first did, IDK&lt;/li&gt;&#xA;&lt;li&gt;Fourth (bbw): Sees the same BB that third did, knows he's not R, so &lt;strong&gt;White&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;First (wwb): knows second could have seen red or black and gotten his hat, knows it doesn't matter for third, but knows fourth had to see some dupe. Since first already sees the ww dupe, he knows his hat is &lt;strong&gt;Black&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;Third (bbw): uses same logic as first to get his hat as &lt;strong&gt;White&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For the sake of completeness, lets' go with a variant of scenario 2 (order BWWB):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First (wwb): IDK&lt;/li&gt;&#xA;&lt;li&gt;Second (bbw): IDK&lt;/li&gt;&#xA;&lt;li&gt;Third (bbw): Sees the same BB that second did, know's he's not R, so &lt;strong&gt;White&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;Fourth (wwb): Sees the same WW that first did, know's he's not R, so &lt;strong&gt;Black&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;First (wwb): Knows that both second and third saw BWX, and both of them see BWX, and the second one knows something the first doesn't, which is that both of them see a pair. He would think he is black, but then if he were R, then third would know why second was stuck, and fourth would know his color straight up. IDK&lt;/li&gt;&#xA;&lt;li&gt;Second (bbw): If he were R, then third's declaration would be the same, as would fourth's (based on first). IDK&lt;/li&gt;&#xA;&lt;li&gt;Stuck&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-03-25T22:53:39.700" LastActivityDate="2015-03-25T22:53:39.700" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10955" PostTypeId="2" ParentId="10894" CreationDate="2015-03-25T14:17:36.003" Score="0" Body="&lt;p&gt;I think it is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Faith  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am not lost, but I can be found.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can find faith in anything (God, Higher Power, Universe, humanity, etc.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you have found me, then you should lose me again, in order to find me.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people &quot;lose&quot; their faith, and some of those find it again.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you find me again it will surely be of benefit, similar to if you became more fit.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most people agree having faith in something is a benefit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My composition of attributes is exactly that...  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Faith is a composition, or perhaps is &quot;found&quot; by a composition of a lot of things. Experiences, miracles, evidence, lack of evidence, etc. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...found of the pages were: on a journey, five had me and five did not. .  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Referencing the ten virgins and the wedding feast. There was a journey to the feast. Five had &quot;oil&quot; which many believe represents faith, and five did not. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-03-25T14:17:36.003" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10956" PostTypeId="2" ParentId="10897" CreationDate="2015-03-25T14:36:43.103" Score="2" Body="&lt;p&gt;&lt;strong&gt;My Answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; tongue&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reasoning&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could be there, yet I'll stay hidden.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; very hard to see your own tongue without a mirror&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hidden I'll stay, but you'll need me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you need it for tasting &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Need me you will, so I may appear.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; when eating &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Appear I may, I don't decide.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; tongue can't control itself &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Decide I don't, my master does.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; humans do &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm a word, I tell you words.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; helps you to articulate&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've a task, you know my task.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; we all know the purposes of a tongue&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm short, I shorten your path.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; chewing is the first step in the digestive process and tongue aids in that&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've the answer, now tell me it!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you're a tongue!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9778" LastActivityDate="2015-03-25T14:36:43.103" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10957" PostTypeId="2" ParentId="10950" CreationDate="2015-03-25T14:46:31.737" Score="-1" Body="&lt;pre&gt;&lt;code&gt;B       W     I will enumerate this table as    1       2&#xA;B       W                                       4       3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First logician looks at the others and says &quot;I don't know.&quot;&lt;/li&gt;&#xA;&lt;li&gt;Second one looks at they all and thinks &quot;If first one didn't know about it, it indicates that he thinks that he can be a red. If he thinks that but he wears black, it means that I cannot be a red in any way. If I am not a red, and there is two blacks and one white, I must be a white.&lt;/li&gt;&#xA;&lt;li&gt;Third one sees the thinks the same and says &quot;I'm white.&quot;&lt;/li&gt;&#xA;&lt;li&gt;And fourth one says &quot;I'm black&quot;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;From this example, you see that the first logician failed to name while the others didn't. Rotating and checking for all cases, one is able to conclude that the &lt;strong&gt;first logician is always the one who fails while the three others don't&lt;/strong&gt;. Why? Because the first one holds the uncertainty and solve it by saying &quot;I don't know.&quot;. Then, the others have the clue needed to fix the 50% of probability problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This answer is wrong, second one is a more appropriate answer.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;----- Edit -----&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;B       W     I will enumerate this table as    1       2&#xA;B       W                                       4       3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First one is &lt;strong&gt;black&lt;/strong&gt; looks at all and says &lt;strong&gt;&quot;I don't know&quot;&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Second one is &lt;strong&gt;white&lt;/strong&gt; looks at all and says &lt;strong&gt;&quot;I don't know&quot;&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Third one thinks:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;First one is black but doesn't know. Second one also doesn't know. First and third and black, while second is white. I can only be white or red. If second one can't affirm that he is white, it is because I cannot be a red. &lt;strong&gt;Then, my hat is white.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the third one gives his answer, the other three have the clue needed to solve the problem. Then, the one that will be considered to fail while the three others didn't is the &lt;strong&gt;second one&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10630" LastEditorUserId="10630" LastEditDate="2015-03-25T18:57:52.387" LastActivityDate="2015-03-25T18:57:52.387" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="10958" PostTypeId="2" ParentId="10894" CreationDate="2015-03-25T14:47:33.130" Score="4" Body="&lt;p&gt;EDIT: Thanks to OP's hint on Bible reference.  I'm not a Christian but hopefully the explanation are correct!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wisdom?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am not lost, but I can be found.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Blessed are those who find wisdom, those who gain understanding (Proverbs 3:13)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you have found me, then you should lose me again, in order to find me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Do not deceive yourselves. If any of you think you are wise by the standards of this age, you should become &quot;fools&quot; so that you may become wise.  (Corinthians 3:18)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you find me again it will surely be of benefit, similar to if you became more fit.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Long life is in her right hand; in her left hand are riches and honor (Proverbs 3:16, but I don't get this one)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My composition of attributes is exactly that...&lt;br&gt;&#xA;...found of the pages were: on a journey, five had me and five did not. .   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Five of them were foolish and five were wise (Mathew 25:2)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="10037" LastEditDate="2015-03-25T15:03:46.417" LastActivityDate="2015-03-25T15:03:46.417" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10959" PostTypeId="1" AcceptedAnswerId="10978" CreationDate="2015-03-25T15:22:41.603" Score="9" ViewCount="1323" Body="&lt;p&gt;At first there may be more than one,&lt;br&gt;&#xA;Next your view will get things done.&lt;br&gt;&#xA;Next, half a banana will help us out,&lt;br&gt;&#xA;Then used in a question, but do not pout.&lt;br&gt;&#xA;The beginning of the finale is the key,&lt;br&gt;&#xA;To help you see what I might be.&lt;br&gt;&#xA;Know people use me for different things.&lt;br&gt;&#xA;I might make it difficult to sing.&lt;br&gt;&#xA;So carefully analyze, reason, finalize,&lt;br&gt;&#xA;What I might be, then eat some pies.  &lt;/p&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2016-03-25T14:49:04.243" Title="More than one? What am I?" Tags="|riddle|word|rhyme|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10960" PostTypeId="2" ParentId="10882" CreationDate="2015-03-25T15:52:49.280" Score="12" Body="&lt;p&gt;Is the answer a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; pancreas?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have more to do with cooking than you might think.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As a direct clue, the pancreas is an organ in the digestive system that is involved with processing cooked stuff, i.e. food.&lt;br&gt;&#xA; Wordplay-wise, a &quot;PAN&quot; is used for cooking. The other clues clue the remaining letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now let us discuss the ocean, try not to sink.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;C&quot; is a homophone, as clued by &quot;discuss&quot;, of sea = ocean&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then between quizzes and sandwiches I like to rest.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;R&quot; comes alphabetically between Quizzes and Sandwiches&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The beginning of eating, a feast is the best.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;E&quot; is the first letter of &quot;eating&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Half the area is needed, we aren't quite done.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;A&quot; makes up half of the letters of &quot;area&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We wrap up wondering if I am more than one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;S&quot; is the most common letter to make English nouns plural and makes you wonder if &quot;pancreas&quot; is a plural; it's not.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now as you guess, know that I am quite small,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well, the organ is pretty small compared to other organs in our body, I guess.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Doesn't really matter if you are short or tall.&lt;br&gt;&#xA;Two types of clues to guess one type of me.&lt;br&gt;&#xA;What on the earth could I possibly be?&lt;/p&gt;&#xA;" OwnerUserId="10633" LastActivityDate="2015-03-25T15:52:49.280" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10961" PostTypeId="2" ParentId="10950" CreationDate="2015-03-25T15:58:38.140" Score="0" Body="&lt;p&gt;There are 2 cases:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1&lt;/strong&gt;. Red doesn't get dropped.&lt;br&gt;&#xA; &lt;strong&gt;2&lt;/strong&gt;. Red does get dropped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;CASE 1&lt;/strong&gt;&lt;br&gt;&#xA;If we call the people A,B,C,D, then we have a situation like this:&lt;br&gt;&#xA;A-&gt; &lt;strong&gt;RED&lt;/strong&gt;&lt;br&gt;&#xA;B-&gt; &lt;strong&gt;WHITE&lt;/strong&gt;&lt;br&gt;&#xA;C -&gt; &lt;strong&gt;BLACK&lt;/strong&gt;&lt;br&gt;&#xA;D -&gt; &lt;strong&gt;WHITE&lt;/strong&gt; or BLACK (suppose white, it's the same)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They can see the others' hat, but not their own one. So, they decide to say &quot;I don't know&quot; unless one of them is sure of what he's saying. &#xA;C can see 2 white hats and a red one, so he deduces that his hat is black. The others now deduce that C didn't see any black hat, so A,B and D can now deduce their colors just looking at the others. The loser in this case is the man who's asked after the other two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;CASE 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A-&gt; &lt;strong&gt;WHITE&lt;/strong&gt;&lt;br&gt;&#xA;B-&gt; &lt;strong&gt;WHITE&lt;/strong&gt;&lt;br&gt;&#xA;C -&gt; &lt;strong&gt;BLACK&lt;/strong&gt;&lt;br&gt;&#xA;D -&gt; &lt;strong&gt;BLACK&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nobody speaks about his hat, so they all deduce that this is &lt;strong&gt;not a CASE 1&lt;/strong&gt;. Easily, everyone understand that his hat is white if he sees 2 black hats, otherwise his cap is black. The loser in this case is, again, the one who's asked as last.&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-03-25T16:17:01.223" LastActivityDate="2015-03-25T16:17:01.223" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10962" PostTypeId="2" ParentId="10907" CreationDate="2015-03-25T16:07:15.587" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nipple(s)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sometimes I'm ugly, sometimes I'm art,    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Depends on person to person&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sometimes I'm near and dear to your heart,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Physical proximity to heart.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I have been known to bring about pain,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;And yet I deliver relief just the same,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This one I am unsure about, perhaps pleasure?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You know where I live, yes you know it best,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;For I am not hidden like all of the rest.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10634" LastActivityDate="2015-03-25T16:07:15.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10963" PostTypeId="1" AcceptedAnswerId="11053" CreationDate="2015-03-25T16:13:20.230" Score="2" ViewCount="1140" Body="&lt;p&gt;I am fast like an arrow but slow like a turtle&lt;br&gt;&#xA;I  play like a Cello but with a  silence echo&lt;br&gt;&#xA;I am long like a Metro but not quite the rainbow&lt;br&gt;&#xA;Give me an answer that fit this riddle!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many have guessed but none are correct&lt;br&gt;&#xA; Poorly rhymed, now I regret&lt;br&gt;&#xA; Here comes another hint&lt;br&gt;&#xA; Stand under the Sun and you will win  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="10037" LastEditDate="2015-03-26T21:26:46.623" LastActivityDate="2015-03-26T22:05:24.637" Title="Give me an answer that fit this riddle - What am I?" Tags="|riddle|" AnswerCount="5" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10964" PostTypeId="2" ParentId="10950" CreationDate="2015-03-25T16:20:54.073" Score="0" Body="&lt;p&gt;L&lt;sub&gt;1&lt;/sub&gt;, L&lt;sub&gt;2&lt;/sub&gt;, L&lt;sub&gt;3&lt;/sub&gt;, and L&lt;sub&gt;4&lt;/sub&gt; are asked in order. We know that each logician sees two hats of the same colour and one of a different colour, but let's see what they discover.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;L&lt;sub&gt;1&lt;/sub&gt; looks around the table and sees one of the following combinations:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;L&lt;sub&gt;1&lt;/sub&gt; &lt;code&gt;WR WR WR BR BR BR BW BW BW BW BW BW || W  W  W  B  B  B&lt;/code&gt;&lt;br&gt;&#xA;  L&lt;sub&gt;2&lt;/sub&gt; &lt;code&gt;B  B  W  B  W  W  W  B  B  R  R  W  || R  B  B  R  W  W&lt;/code&gt;&lt;br&gt;&#xA;  L&lt;sub&gt;3&lt;/sub&gt; &lt;code&gt;B  W  B  W  B  W  B  W  R  B  W  R  || B  R  B  W  R  W&lt;/code&gt;&lt;br&gt;&#xA;  L&lt;sub&gt;4&lt;/sub&gt; &lt;code&gt;W  B  B  W  W  B  R  R  W  W  B  B  || B  B  R  W  W  R&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because the only scenarios where L&lt;sub&gt;1&lt;/sub&gt; can guess his hat colour are scenarios where someone at the table is wearing a red hat, we'll skip the discussion on them. L&lt;sub&gt;1&lt;/sub&gt; is unable to guess his hat colour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;L&lt;sub&gt;2&lt;/sub&gt; looks around the table and sees one of the following combinations. Combining the information provided by L&lt;sub&gt;1&lt;/sub&gt;, he deduces the following information about himself:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;— &lt;code&gt;1   2   3   4   5   6&lt;/code&gt;&lt;br&gt;&#xA;  L&lt;sub&gt;1&lt;/sub&gt; &lt;code&gt;B   B   B   W   W   W&lt;/code&gt;&lt;br&gt;&#xA;  L&lt;sub&gt;2&lt;/sub&gt; &lt;code&gt;WR  WR  B   BR  BR  W&lt;/code&gt;&lt;br&gt;&#xA;  L&lt;sub&gt;3&lt;/sub&gt; &lt;code&gt;B   W   W   B   W   B&lt;/code&gt;&lt;br&gt;&#xA;  L&lt;sub&gt;4&lt;/sub&gt; &lt;code&gt;W   B   W   W   B   B&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In scenario 2 and 6, he is able to accurately guess the colour of his hat. If he is able to do this, L&lt;sub&gt;3&lt;/sub&gt; and L&lt;sub&gt;4&lt;/sub&gt; are also able to deduce the colours of their hats are the same and opposite of whatever L&lt;sub&gt;2&lt;/sub&gt; said, but L&lt;sub&gt;1&lt;/sub&gt; is left wondering if his hat might still be red or the same as L&lt;sub&gt;2&lt;/sub&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Moving on to consider 1, 3, 4, and 6 where L&lt;sub&gt;2&lt;/sub&gt; was unable to guess, L&lt;sub&gt;3&lt;/sub&gt; sees the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;L&lt;sub&gt;1&lt;/sub&gt; &lt;code&gt;B   B   B   W   W   W&lt;/code&gt;&lt;br&gt;&#xA;  L&lt;sub&gt;2&lt;/sub&gt; &lt;code&gt;B   W   W   B   W   B&lt;/code&gt;&lt;br&gt;&#xA;  L&lt;sub&gt;3&lt;/sub&gt; &lt;code&gt;WR  WR  B   BR  BR  W&lt;/code&gt;&lt;br&gt;&#xA;  L&lt;sub&gt;4&lt;/sub&gt; &lt;code&gt;W   B   W   W   B   B&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Removing the scenarios made that don't fit scenario 1, 3, 4, or 6:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;— &lt;code&gt;7   8   9   10  11  12&lt;/code&gt;&lt;br&gt;&#xA;  L&lt;sub&gt;1&lt;/sub&gt; &lt;code&gt;B   B   B   W   W   W&lt;/code&gt;&lt;br&gt;&#xA;  L&lt;sub&gt;2&lt;/sub&gt; &lt;code&gt;B   W   W   B   W   B&lt;/code&gt;&lt;br&gt;&#xA;  L&lt;sub&gt;3&lt;/sub&gt; &lt;code&gt;R   WR  B   BR  R   W&lt;/code&gt;&lt;br&gt;&#xA;  L&lt;sub&gt;4&lt;/sub&gt; &lt;code&gt;W   B   W   W   B   B&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;L&lt;sub&gt;3&lt;/sub&gt; is able to guess his hat colour in scenario 7, 9, 11, and 12. In 7 and 11 when L&lt;sub&gt;3&lt;/sub&gt; is wearing red &lt;em&gt;(it's not important that we know this doesn't happen)&lt;/em&gt;, L&lt;sub&gt;2&lt;/sub&gt; and L&lt;sub&gt;4&lt;/sub&gt; know that they are wearing opposite colour hats. In 9 and 12, they know they are wearing the same as each other and opposite to whatever L&lt;sub&gt;3&lt;/sub&gt; has said. L&lt;sub&gt;1&lt;/sub&gt; is still left in the dark.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Finally, we consider the remaining two scenarios in which L&lt;sub&gt;3&lt;/sub&gt; did not guess. We once again start with the table that L&lt;sub&gt;4&lt;/sub&gt; sees.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;L&lt;sub&gt;1&lt;/sub&gt; &lt;code&gt;B   B   B   W   W   W&lt;/code&gt;&lt;br&gt;&#xA;  L&lt;sub&gt;2&lt;/sub&gt; &lt;code&gt;B   W   W   B   W   B&lt;/code&gt;&lt;br&gt;&#xA;  L&lt;sub&gt;3&lt;/sub&gt; &lt;code&gt;W   B   W   W   B   B&lt;/code&gt;&lt;br&gt;&#xA;  L&lt;sub&gt;4&lt;/sub&gt; &lt;code&gt;WR  WR  B   BR  BR  W&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When we eliminate the combinations that we know do not fit scenarios that would have otherwise already been answered by the other logicians, we're left with: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;— &lt;code&gt;13 14&lt;/code&gt;&lt;br&gt;&#xA;  L&lt;sub&gt;1&lt;/sub&gt; &lt;code&gt;B  W&lt;/code&gt;&lt;br&gt;&#xA;  L&lt;sub&gt;4&lt;/sub&gt; &lt;code&gt;W  B&lt;/code&gt;&lt;br&gt;&#xA;  L&lt;sub&gt;3&lt;/sub&gt; &lt;code&gt;W  B&lt;/code&gt;&lt;br&gt;&#xA;  L&lt;sub&gt;2&lt;/sub&gt; &lt;code&gt;B  W&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the results of the other logicians, L&lt;sub&gt;4&lt;/sub&gt; is now sure that the colour of his hat now matches L&lt;sub&gt;1&lt;/sub&gt;'s hat. L&lt;sub&gt;1&lt;/sub&gt; still thinks his hat might be red, though.&lt;/p&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-03-25T16:32:54.330" LastActivityDate="2015-03-25T16:32:54.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10965" PostTypeId="2" ParentId="10950" CreationDate="2015-03-25T16:35:38.710" Score="16" Body="&lt;h2&gt;First observations&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Since the puzzle is symmetric in white and black hats, I assume that the first man has a black hat. Then only three cases remain: (1) BWWB; (2) BWBW; (3) BBWW.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;($\ast$)&lt;/strong&gt; If a man sees two hats of color white, then he knows that his hat is either red or black. If a man sees two hats of color black, then he knows that his hat is either red or white. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;(Case 1) Suppose that the four men have hats BWWB.&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The first man concludes from (*): I have either R or B.&lt;/li&gt;&#xA;&lt;li&gt;Then the second man concludes from (*): I have either R or W.&lt;/li&gt;&#xA;&lt;li&gt;Then the third man concludes from (*): I have either R or W. If I have R, then the second man would have been able to derive his color. Hence I have W. Bingo!&lt;/li&gt;&#xA;&lt;li&gt;Then the fourth man concludes from (*): I have either R or B. If I have R, then the first man would have been able to derive his color. Hence I have B. Bingo!&lt;/li&gt;&#xA;&lt;li&gt;Then the first man thinks: I have either R or B. If I have R, then the fourth man would see RWW?, and know his color; the third man would see RW?B, and know that he has W, since the second guy was not able to deduce his color. If I have B, then the third and fourth man are able to deduce their colors (as discussed above). Hence I still do not know my color.&lt;/li&gt;&#xA;&lt;li&gt;Then the second man thinks: I have either R or W.  If I have R, then the fourth man would see BRW?; he would know that the first man did not see W on him (as otherwise he would have deduced B for himself); hence the fourth man deduces his color. Furthermore, the third man would see BR?B, and deduce his color W. In case I have W, then the third and fourth man are able to deduce their colors (as discussed above). Hence I still do not know my color.&lt;/li&gt;&#xA;&lt;li&gt;From this moment on, no further information is gained by anybody. The game is stuck. Contradiction.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;(Case 2) Suppose that the four men have hats BWBW.&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The first man concludes from (*): I have either R or B.&lt;/li&gt;&#xA;&lt;li&gt;Then the second man concludes from (*): I have either R or W.&lt;/li&gt;&#xA;&lt;li&gt;Then the third man concludes from (*): I have either R or B. If I have R, then the first man would have been able to derive his color. Hence I have B. Bingo!&lt;/li&gt;&#xA;&lt;li&gt;Then the fourth man concludes from (*): I have either R or W. If I have R, then the second man would have been able to derive his color. Hence I have W. Bingo!&lt;/li&gt;&#xA;&lt;li&gt;Then the first man thinks: I have either R or B. If I have R, then the third man would see RW?W, and know his color; the fourth man would see RWB?, and know that he has W, since the third guy was able to deduce his color. If I have B, then the third and fourth man are able to deduce their colors (as discussed above). Hence I still do not know my color.&lt;/li&gt;&#xA;&lt;li&gt;Then the second man thinks: I have either R or W.  If I have R, then the third man would see BR?W; he would know that the first man did not see W on him (as otherwise he would have deduced B for himself); hence the third man deduces his color. Furthermore, the fourth man would see BRB?, and deduce his color W.&#xA;In case I have W, then the third and fourth man are able to deduce their colors (as discussed above). Hence I still do not know my color.&lt;/li&gt;&#xA;&lt;li&gt;From this moment on, no further information is gained by anybody. The game is stuck. Contradiction.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;(Case 3) Suppose that the four men have hats BBWW.&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The first man concludes from (*): I have either R or B.&lt;/li&gt;&#xA;&lt;li&gt;Then the second man concludes from (*): I have either R or B. If I have R, then the first guy would see ?RWW and deduce his color. Hence my color is B. Bingo!&lt;/li&gt;&#xA;&lt;li&gt;Then the third man concludes from (*): I have either R or W.&#xA;If I have R, then the first man sees ?BRW and would not know his color; the second man would see B?RW and know that his color is not W (as otherwise the first man would have deduced his color); hence the second man would deduce his color B.&#xA;If I have W, then the first and the second man would behave as explained above.&#xA;Hence I still do not know my color.&lt;/li&gt;&#xA;&lt;li&gt;The fourth man concludes from (*): I have either R or W. If I have R, then the third man would see BB?R and deduce his color. Hence I have W. Bingo!&lt;/li&gt;&#xA;&lt;li&gt;Then the first man concludes from (*): I have either R or B. If I have R, then the third guy would see RB?W; since the second guy could deduce his color, he must see two white hats; the third guy knows this and would deduce his color W. Hence I have not R and my hat is B. Bingo!&lt;/li&gt;&#xA;&lt;li&gt;The second guy already knows. &lt;/li&gt;&#xA;&lt;li&gt;Then the third man concludes from (*): I have either R or W.&#xA;If I have R, then the fourth man would see BBR? and deduce his color W. &#xA;The first man would sees ?BRW, and know that the fourth man was able to deduce his color; hence the fourth man must see two Bs and the first man would know his color. &#xA;If I have W, then the others would behave as in the steps explained above.&#xA;Hence I still do not know my color.&lt;/li&gt;&#xA;&lt;li&gt;From this moment on, no further information is gained by anybody.&#xA;The third guy never learns his hat color.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Answer&lt;/strong&gt;: The only case that is compatible with the problem statement is case 3, where the first and second guy have the same hat color (black or white) and the third and fourth man have the opposite hat color (white or black).&#xA;  In this case the first, second and fourth man are able to deduce their colors, but &lt;strong&gt;the third man will never be able to deduce his color&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-03-25T17:37:14.813" LastActivityDate="2015-03-25T17:37:14.813" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10966" PostTypeId="1" AcceptedAnswerId="11005" CreationDate="2015-03-25T16:41:31.690" Score="4" ViewCount="8813" Body="&lt;p&gt;Here are some words that share a link to a single word, used in different ways... &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Street (where runners go)&lt;/li&gt;&#xA;&lt;li&gt;Respect (a gesture)&lt;/li&gt;&#xA;&lt;li&gt;Ribbon (a flourish)&lt;/li&gt;&#xA;&lt;li&gt;Reflex (a weapon)&lt;/li&gt;&#xA;&lt;li&gt;Play (might be awe-inspiring or just awful)&lt;/li&gt;&#xA;&lt;li&gt;Front (a place with a view)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What's the word that links to all these?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You smart people won't need any more hints* but I can add words to the list, if you really need it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;* (OK, maybe two then)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;And with thanks to @xnor and @randal'thor for discussion on format.&lt;/p&gt;&#xA;" OwnerUserId="8875" LastEditorUserId="8875" LastEditDate="2015-03-26T00:08:27.880" LastActivityDate="2015-03-26T05:03:25.043" Title="Find the missing link" Tags="|word|enigmatic-puzzle|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10967" PostTypeId="2" ParentId="10950" CreationDate="2015-03-25T16:57:45.003" Score="0" Body="&lt;p&gt;We reduce possibilities by assigning X either &quot;black hat&quot; or &quot;white hat&quot; and Y the remaining. Furthermore R represents &quot;red hat&quot;; ? or IDK stands for &quot;I dont know&quot;. Thus a number of theoretical combinations have to be considered while in reality only three combinations can occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By considering what each logician would have guessed given the possible combinations, we can eliminate possibilities with each new answer, even when it is IDK.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;theoretical        practical&#xA;RRRXXXXXXXXXXXX    XXX&#xA;XXXRRRXYYXYYXYY    XYY&#xA;XYYXYYRRRYXYYXY    XYX&#xA;YYXYXYYXYRRRYYX    YYX&#xA;&#xA;possible options for A:&#xA;RRRXXX&#xA;XXXXYY&#xA;XYYYXY&#xA;YYXYYX&#xA;&#xA;possible options for B:&#xA;XXXXXX&#xA;RRRXYY&#xA;XYYYXY&#xA;YYXYYX&#xA;&#xA;possible options for C:&#xA;XXXXXX&#xA;XYYXYY&#xA;RRRYXY&#xA;YYXYYX&#xA;&#xA;possible options for D:&#xA;XXXXXX&#xA;XYYXYY&#xA;YYXYXY&#xA;RRRYYX&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The question for each logician is basically reduced to &quot;Am I R?&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A_1: &#xA;    theoretical&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;RRRXXXXXXXXXXXX&#xA;XXXRRRXYYXYYXYY&#xA;XYYXYYRRRYXYYXY&#xA;YYXYXYYXYRRRYYX&#xA;&#xA;logical answers&#xA;?????X??X??X???&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Since there is no R, A_1 will always answer IDK, eliminating three possibilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B_1: &#xA;    possible options for B:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;XX.XXX&#xA;RR.XYY&#xA;XY.YXY&#xA;YY.YYX&#xA;&#xA;theoretical        practical&#xA;RRRXX.XX.XX.XXX    XXX&#xA;XXXRR.XY.XY.XYY    XYY&#xA;XYYXY.RR.YX.YXY    XYX&#xA;YYXYX.YX.RR.YYX    YYX&#xA;&#xA;theoretical answers&#xA;?X???.?Y.?Y.X??&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;B_1 can only either answer X for ??YY or IDK, eliminating further two possibilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C_1:&#xA;    possible options for C:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X..XXX&#xA;X..XYY&#xA;R..YXY&#xA;Y..YYX&#xA;&#xA;theoretical        practical&#xA;RRRXX.X..X..XXX    XXX&#xA;XXXRR.X..X..XYY    XYY&#xA;XYYXY.R..Y..YXY    XYX&#xA;YYXYX.Y..R..YYX    YYX&#xA;&#xA;theoretical answers ( B_1 == X )&#xA;.Y..........Y..&#xA;&#xA;theoretical answers ( B_1 == IDK )&#xA;X.YXY.R..Y...XY&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;C_1 can never answer IDK.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D_1:&#xA;    possible options for D:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X..XXX&#xA;X..XYY&#xA;Y..YXY&#xA;R..YYX&#xA;&#xA;theoretical        practical&#xA;RRRXX.X..X..XXX    XXX&#xA;XXXRR.X..X..XYY    XYY&#xA;XYYXY.R..Y..YXY    XYX&#xA;YYXYX.Y..R..YYX    YYX&#xA;&#xA;theoretical answers ( B_1 == X )&#xA;.Y..........Y..&#xA;&#xA;theoretical answers ( B_1 == IDK, C_1 = X )&#xA;Y..Y.........Y.&#xA;&#xA;theoretical answers ( B_1 == IDK, C_1 = Y )&#xA;..X.X....R....X&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;D_1 can never answer IDK.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A_2:&#xA;    theoretical    &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;RRRXX.X..X..XXX&#xA;XXXRR.X..X..XYY&#xA;XYYXY.R..Y..YXY&#xA;YYXYX.Y..R..YYX&#xA;&#xA;theoretical answers ( B_1 == X )&#xA;.?..........?..&#xA;&#xA;theoretical answers ( B_1 == IDK, C_1 = X, D_1 = Y )&#xA;?.?XX........??&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;A_2 still can only answer IDK, thereby eliminating 2 further possibilities were it would have answered X.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B_2:&#xA;    possible&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;...XXX&#xA;...XYY&#xA;...YXY&#xA;...YYX&#xA;&#xA;theoretical answers ( B_1 == IDK, C_1 = X, D_1 = Y )&#xA;....YY&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;B_2 always answers Y.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion, no matter if logician B needs one or two turns, the first logician will always be the one who fails to get the correct answer.&lt;/p&gt;&#xA;" OwnerUserId="10638" LastEditorUserId="10638" LastEditDate="2015-03-26T14:31:00.373" LastActivityDate="2015-03-26T14:31:00.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10968" PostTypeId="1" AcceptedAnswerId="12399" CreationDate="2015-03-25T17:00:46.510" Score="21" ViewCount="2152" Body="&lt;p&gt;Yeah. I think this puzzle will be the hardest puzzle on Puzzling...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll add one hint per 3 days, I'll add a VERY helpful tag on the 3rd week, and I'll post the answer and explanation on the 4th week.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I've completely managed to ignore this...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yeah, here's the puzzle: (Thanks to @squeamishossifrage)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FkXHk.png&quot; alt=&quot;PUZZLE&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the correct way to look at it.&lt;br&gt;&#xA;All tangents from the positive x are ±0, undef, ±2, ±1/2, ±1.&lt;br&gt;&#xA;Read from top-left to bottom-right&lt;br&gt;&#xA;Find the 3-digit code!!!!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what is the three digit code that is described by these seemingly random lines?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This puzzle will have a EUREKA moment, almost like my last puzzle &lt;a href=&quot;https://puzzling.stackexchange.com/questions/6120/what-the-answer-isnt-the-no-of-the-yes-is&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The comments below have a lot of helpful hints, especially the ones on March 27&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ivo Beckers is missing something else from his picture... For every single one, he has forgotten two lines.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The EUREKA moment has nothing to do with math.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="167" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2015-05-31T17:14:11.477" Title="A Code from the Lines" Tags="|lateral-thinking|mazes|" AnswerCount="6" CommentCount="31" ContentLicense="CC BY-SA 3.0" />
  <row Id="10969" PostTypeId="2" ParentId="10963" CreationDate="2015-03-25T17:12:26.530" Score="1" Body="&lt;p&gt;I think you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A cat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am fast like an arrow but slow like a turtle&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cats have a high top speed, but often remain totally motionless while hunting, and can be lazy in general.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I play like a Cello but with a silence echo&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cello strings were traditionally made from &lt;a href=&quot;http://en.wikipedia.org/wiki/Catgut&quot; rel=&quot;nofollow&quot;&gt;catgut&lt;/a&gt;. Cats can be quite silent and stealthy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am long like a Metro but not quite the rainbow&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://knowyourmeme.com/memes/longcat&quot; rel=&quot;nofollow&quot;&gt;Longcat&lt;/a&gt; is long and about the shape of a train car. &lt;a href=&quot;http://en.wikipedia.org/wiki/Nyan_Cat&quot; rel=&quot;nofollow&quot;&gt;Nyan Cat&lt;/a&gt; is not a rainbow, but one trails behind him wherever he goes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1497" LastActivityDate="2015-03-25T17:12:26.530" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10970" PostTypeId="2" ParentId="10966" CreationDate="2015-03-25T17:19:42.653" Score="2" Body="&lt;p&gt;All are related with &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; sports/competitions:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Respect &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Respect won.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ribbon&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ribbon cut to start&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Runner&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many racing competitions&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reflex&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reflex oh yes this is must&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Play&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Definitely people play&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10210" LastActivityDate="2015-03-25T17:19:42.653" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10971" PostTypeId="2" ParentId="10963" CreationDate="2015-03-25T17:32:02.087" Score="2" Body="&lt;p&gt;Maybe you're a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Radio Wave&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am fast like an arrow but slow like a turtle  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You travel at the speed of light, but you're a slow way to transmit information.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I play like a Cello but with a silence echo&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Radio waves can be converted into sounds, but nobody can hear your echo.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am long like a Metro but not quite the rainbow  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your wavelength can be very very large, though we don't see radio waves in the rainbow (just visible light).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Give me an answer that fit this riddle!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fit is a commonly used term when you talk about curves in math, in this case a wave.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-03-25T17:32:02.087" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10972" PostTypeId="1" AcceptedAnswerId="10973" CreationDate="2015-03-25T17:48:25.853" Score="8" ViewCount="1159" Body="&lt;p&gt;Create the numbers from 1 to 20 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;ul&gt;&#xA;&lt;li&gt;$\pi$&lt;/li&gt;&#xA;&lt;li&gt;Normal arithmetic operation $+ - * /$&lt;/li&gt;&#xA;&lt;li&gt;Square root $\surd$&lt;/li&gt;&#xA;&lt;li&gt;Exponential $(X^Y)$&lt;/li&gt;&#xA;&lt;li&gt;Negative() minus sign $-$&lt;br&gt;&lt;/li&gt;&#xA;&lt;li&gt;Floor() function, express between  $[ x ]$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$[ 3.0 ]$ and $[ 3.9 ] = 3$&lt;br&gt;&#xA;  $[-3.1 ]$ and $[-3.9 ] = -4$&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  Forbidden&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Factorial&lt;/li&gt;&#xA;&lt;li&gt;Log&lt;/li&gt;&#xA;&lt;li&gt;More than four $\pi$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Doing the 20 numbers takes time but is easy. &lt;br&gt;&#xA;Now the challenge is to create all 20 numbers using the minimal number of $\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1 = $\pi$ /$\pi$ use two $\pi$ &lt;br&gt;&#xA;  1 = [ $\surd\pi$  ] only need one $\pi$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;2 = $[ \surd\pi ] + [ \surd\pi  ]$ use two $\pi$&lt;br&gt;&#xA;  2 = $-[ - \surd\pi ]$ use one $\pi$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I give you an initial target of 50 $\pi$ for all 20 numbers. But I know can be lower.&lt;/p&gt;&#xA;" OwnerUserId="5165" LastEditorUserId="5373" LastEditDate="2015-05-13T20:03:13.383" LastActivityDate="2015-06-23T16:12:48.073" Title="$\pi$ Day puzzle one to twenty" Tags="|calculation-puzzle|formation-of-numbers|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10973" PostTypeId="2" ParentId="10972" CreationDate="2015-03-25T18:13:28.747" Score="6" Body="&lt;p&gt;I managed to do it in 42 $\pi$s.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1: $\lfloor\sqrt\pi\rfloor$&lt;/li&gt;&#xA;&lt;li&gt;2: $\sqrt{-\lfloor-\pi\rfloor}$&lt;/li&gt;&#xA;&lt;li&gt;3: $\lfloor\pi\rfloor$&lt;/li&gt;&#xA;&lt;li&gt;4: $-\lfloor-\pi\rfloor$&lt;/li&gt;&#xA;&lt;li&gt;5: $\pi-\lfloor-\sqrt\pi\rfloor$&lt;/li&gt;&#xA;&lt;li&gt;6: $\lfloor\pi\rfloor+\lfloor\pi\rfloor$&lt;/li&gt;&#xA;&lt;li&gt;7: $\lfloor\pi\rfloor-\lfloor-\pi\rfloor$&lt;/li&gt;&#xA;&lt;li&gt;8: $-\lfloor-\pi\rfloor-\lfloor-\pi\rfloor$&lt;/li&gt;&#xA;&lt;li&gt;9: $\lfloor\pi\rfloor\times\lfloor\pi\rfloor$&lt;/li&gt;&#xA;&lt;li&gt;10: $-\lfloor-\pi\times\pi\rfloor$&lt;/li&gt;&#xA;&lt;li&gt;11: $\lfloor\lfloor-\lfloor-\pi\rfloor\rfloor^{\sqrt\pi}\rfloor$&lt;/li&gt;&#xA;&lt;li&gt;12: $-\lfloor-\pi\rfloor\times\lfloor\pi\rfloor$&lt;/li&gt;&#xA;&lt;li&gt;13: $\lfloor-\lfloor-\pi\rfloor\times\pi\rfloor$&lt;/li&gt;&#xA;&lt;li&gt;14: $-\lfloor-\pi\times\pi\rfloor-\lfloor-\pi\rfloor$&lt;/li&gt;&#xA;&lt;li&gt;15: $-\lfloor-\pi\rfloor\times\lfloor\pi\rfloor+\lfloor\pi\rfloor$&lt;/li&gt;&#xA;&lt;li&gt;16: $\lfloor-\pi\rfloor\times\lfloor-\pi\rfloor$&lt;/li&gt;&#xA;&lt;li&gt;17: $\lfloor-\lfloor-\pi\rfloor\times\pi\rfloor-\lfloor-\pi\rfloor$&lt;/li&gt;&#xA;&lt;li&gt;18: $\sqrt{-\lfloor-\pi\rfloor}\times\lfloor\pi\rfloor\times\lfloor\pi\rfloor$&lt;/li&gt;&#xA;&lt;li&gt;19: $-\lfloor-\sqrt{-\lfloor-\pi\rfloor}\times\pi\times\lfloor\pi\rfloor\rfloor$&lt;/li&gt;&#xA;&lt;li&gt;20: $\lfloor\pi^\pi/\sqrt\pi\rfloor$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4394" LastEditorUserId="4394" LastEditDate="2015-03-26T11:28:54.860" LastActivityDate="2015-03-26T11:28:54.860" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10974" PostTypeId="1" AcceptedAnswerId="10980" CreationDate="2015-03-25T18:16:04.953" Score="-1" ViewCount="220" Body="&lt;p&gt;Blake, Ruby and Yvette are all creative inventors and live on the same cul-de-sac. Blake and Ruby have a daughter named Vivian. Blake has another daughter, Grace, with Yvette.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ruby and Yvette patented a medical device they called Ollie.&#xA;Blake also teamed up with Grace to invent a candy-making machine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which one of the following 5 names did Blake and Grace &lt;strong&gt;NOT&lt;/strong&gt; consider when naming their invention, and why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Ambrosial Treats&lt;br&gt;&lt;/li&gt;&#xA;  &lt;li&gt;Bunch O'Candy&lt;br&gt;&lt;/li&gt;&#xA;  &lt;li&gt;Candilicious&lt;br&gt;&lt;/li&gt;&#xA;  &lt;li&gt;Real Tasty&lt;br&gt;&lt;/li&gt;&#xA;  &lt;li&gt;Turbo Sweets&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastActivityDate="2015-03-25T18:56:03.287" Title="Naming an Invention" Tags="|riddle|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10975" PostTypeId="2" ParentId="9491" CreationDate="2015-03-25T18:23:30.877" Score="3" Body="&lt;p&gt;It looks like, as JonTheMon suggested in the comments of his answer, there is no solution to this problem that involves more than 2 switches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will use the following terminology when discussing switches:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Base&lt;/strong&gt;:  The &quot;single&quot; side of the switch, like the base of the &quot;Y&quot; shape&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Arm&lt;/strong&gt;:   Either of the &quot;forked&quot; ends of the switch, like one of the upper portions of the &quot;Y&quot; shape&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If you enter any switch from an arm, then upon return through that switch from the other direction (from the base), you will necessarily follow the same arm.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you enter any switch from the base, you will exit on a given arm, and will necessarily return on the same arm &lt;em&gt;unless&lt;/em&gt; you have made a loop that connects to the other arm.  &lt;strong&gt;This is &lt;em&gt;the only way&lt;/em&gt; that a switch's position can be continually changed (by exiting on one arm and then returning on the other arm).&lt;/strong&gt;  (It is possible to change a switch's position by entering on one arm, then looping around and entering on the other arm, but if this happens, the first arm will be forever inaccessible after that point, and the switch will no longer change.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For a train to be able continuously travel the entire track, every switch must be regularly switched back and forth between its two states (if it stops switching, there is a part of the track that becomes inaccessible).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;By (2) and (3), the only way to have a train continuously travel the entire track is to have only switches (&lt;strong&gt;&lt;em&gt;A&lt;/em&gt;&lt;/strong&gt;) whose arms are directly connected by a loop of track (&lt;strong&gt;&lt;em&gt;a&lt;/em&gt;&lt;/strong&gt;).  If the arms aren't directly connected (i.e. there is another switch (&lt;strong&gt;&lt;em&gt;B&lt;/em&gt;&lt;/strong&gt;) between them somewhere) then you will be entering switch &lt;strong&gt;&lt;em&gt;B&lt;/em&gt;&lt;/strong&gt; from an arm at some point (since you are travelling loop &lt;strong&gt;&lt;em&gt;a&lt;/em&gt;&lt;/strong&gt; in both directions, and at least one direction must be connected to an arm of switch &lt;strong&gt;&lt;em&gt;B&lt;/em&gt;&lt;/strong&gt;).  By (1), you will have to return on that same arm, and thus there will be a part of the track that is never accessed.  The only way switch &lt;strong&gt;&lt;em&gt;B&lt;/em&gt;&lt;/strong&gt; will allow things to work is if its arms are also directly connected by a loop of track (&lt;strong&gt;&lt;em&gt;b&lt;/em&gt;&lt;/strong&gt;).  But we already know that at least one of its arms is connected to one of the arms of switch &lt;strong&gt;&lt;em&gt;A&lt;/em&gt;&lt;/strong&gt;, so its arms can't be directly connected to each other.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;By (4), our track can only consist of switches whose arms are directly connected to each other, and thus the solution presented by JonTheMon is the solution employing the most possible switches (2 of them).&lt;/p&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2015-03-25T18:23:30.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10976" PostTypeId="2" ParentId="10972" CreationDate="2015-03-25T18:23:58.677" Score="10" Body="&lt;p&gt;&lt;strong&gt;I did it with 43.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 43 = $⌊\pi⌋^{⌊\pi⌋}-⌊-\pi⌋*-⌊-\pi⌋$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There might be a smaller number that this could be done with, but I got lazy after 13. I also don't use anything other than $\surd$,  $\pi$, $⌊$ $⌋$, $-$, $+$, $*$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exponents and parentheses are for suckers!&lt;br&gt;&#xA;&lt;sup&gt;&lt;em&gt;(or maybe for people that get a lower score than me)&lt;/em&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 = $⌊\surd\pi⌋$&lt;br&gt;&#xA; 2 = $-⌊-\surd\pi⌋$&lt;br&gt;&#xA; 3 = $⌊\pi⌋$&lt;br&gt;&#xA; 4 = $-⌊-\pi⌋$&lt;br&gt;&#xA; 5 = $⌊\pi⌋-⌊-\surd\pi⌋$&lt;br&gt;&#xA; 6 = $⌊\pi+\pi⌋$&lt;br&gt;&#xA; 7 = $⌊\pi⌋-⌊-\pi⌋$&lt;br&gt;&#xA; 8 = $-⌊-\pi⌋-⌊-\pi⌋$&lt;br&gt;&#xA; 9 = $⌊\pi * \pi⌋$&lt;br&gt;&#xA; 10 = $-⌊-\pi * \pi⌋$&lt;br&gt;&#xA; 11 = $-⌊-\pi * \pi⌋ + ⌊\surd\pi⌋$&lt;br&gt;&#xA; 12 = $⌊\pi * -⌊-\pi⌋⌋$&lt;br&gt;&#xA; 13 = $-⌊-\pi * -⌊-\pi⌋⌋$&lt;br&gt;&#xA; 14 = $-⌊-\surd\pi⌋ * ⌊⌊\pi⌋ - ⌊-\pi⌋⌋$&lt;br&gt;&#xA; 15 = $-⌊-\pi⌋ * -⌊-\pi⌋ - ⌊\surd\pi⌋$&lt;br&gt;&#xA; 16 = $-⌊-\pi⌋ * -⌊-\pi⌋$&lt;br&gt;&#xA; 17 = $-⌊-\pi⌋ * -⌊-\pi⌋ + ⌊\surd\pi⌋$&lt;br&gt;&#xA; 18 = $-⌊-\pi⌋ * -⌊-\pi⌋ - ⌊-\surd\pi⌋$&lt;br&gt;&#xA; 19 = $-⌊-\pi⌋ * -⌊-\pi⌋ + ⌊\pi⌋$&lt;br&gt;&#xA; 20 = $-⌊-\pi⌋ * -⌊-\pi⌋ - ⌊-\pi⌋$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-03-25T21:58:45.030" LastActivityDate="2015-03-25T21:58:45.030" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10977" PostTypeId="2" ParentId="10974" CreationDate="2015-03-25T18:31:27.183" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Let's consider the initials of their names, which are also the initials of primary and secondary colors.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ruby and Blake  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;RED+BLUE=VIOLET  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Blake and Yvette&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;BLUE+YELLOW=GREEN   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ruby and Yvette&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;RED+YELLOW=ORANGE  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Blake and Grace  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;BLUE+GREEN=TURQUOISE(?)   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Their options to choose the name of the final color could be Turbo Sweets (Turquoise/Teal), Ambrosial Treats (Aqua), Candilicious (Cyan), Bunch O'Candy (Brandeir). It's missing &lt;strong&gt;Real Tasty&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-03-25T18:56:03.287" LastActivityDate="2015-03-25T18:56:03.287" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10978" PostTypeId="2" ParentId="10959" CreationDate="2015-03-25T18:35:40.153" Score="5" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SCARF.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;At first there may be more than one,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter &lt;strong&gt;S&lt;/strong&gt; is used for pluralising (as in the OP's previous riddle).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Next your view will get things done.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A view is what you see, and &quot;see&quot; sounds like &lt;strong&gt;C&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Next, half a banana will help us out,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Half the letters in the word &quot;banana&quot; are &lt;strong&gt;A&lt;/strong&gt;'s.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Then used in a question, but do not pout.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Questions such as &quot;are you here?&quot; contains the word &quot;are&quot;, which sounds like &lt;strong&gt;R&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The beginning of the finale is the key,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first letter of the word &quot;finale&quot; is &lt;strong&gt;F&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Putting together everything we've got so far gives the word I mentioned at the start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;To help you see what I might be.&lt;br&gt;&#xA; Know people use me for different things.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Scarves can be used for warmth or as fashion accessories.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I might make it difficult to sing.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you're wearing a scarf, it's hard to sing through it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;So carefully analyze, reason, finalize,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is an acrostic: &quot;&lt;strong&gt;S&lt;/strong&gt;o &lt;strong&gt;c&lt;/strong&gt;arefully &lt;strong&gt;a&lt;/strong&gt;nalyze, &lt;strong&gt;r&lt;/strong&gt;eason, &lt;strong&gt;f&lt;/strong&gt;inalize&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What I might be, then eat some pies.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm not sure what pies have to do with it! A red herring?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-25T18:45:25.147" LastActivityDate="2015-03-25T18:45:25.147" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10979" PostTypeId="2" ParentId="10765" CreationDate="2015-03-25T18:44:25.563" Score="3" Body="&lt;p&gt;Based on Uri Zarfaty's idea, it might be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CRAZY:&lt;br&gt;&#xA; gray letter Z = gray &quot;zee&quot; (in American English) = &quot;crazy&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-03-25T18:44:25.563" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10980" PostTypeId="2" ParentId="10974" CreationDate="2015-03-25T18:54:31.407" Score="4" Body="&lt;p&gt;Along the same lines as &lt;a href=&quot;https://puzzling.stackexchange.com/a/10977/5373&quot;&gt;leoll2's answer&lt;/a&gt;, I looked at different names for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;a colour between blue and green.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Possibilities include&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; cyan, turquoise, aqua, and blue-green, which begin with the letters C, T, A, and B.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;4&quot;&gt;&#xA;  &lt;li&gt;Real Tasty.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-03-25T18:54:31.407" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10981" PostTypeId="2" ParentId="10765" CreationDate="2015-03-25T19:38:57.707" Score="3" Body="&lt;p&gt;&quot;A gray N backwards (or turned around) is correct&quot;, so I am guessing&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nyarga, which appears to be a place in Russia.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The difference b/w British and U.S. is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the spelling of gray (grey is British).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5684" LastActivityDate="2015-03-25T19:38:57.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10982" PostTypeId="2" ParentId="10897" CreationDate="2015-03-25T19:57:12.550" Score="2" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; History?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I could be there, yet I'll stay hidden.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; History records many event, but some remain hidden  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hidden I'll stay, but you'll need me.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; While not all are recorded, a story of the past is still needed.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Need me you will, so I may appear.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; History exists because you 'will' (in future) need it  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Appear I may, I don't decide.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's not the one who wrote history decides its content,  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Decide I don't, my master does.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; But the people / things in the past.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm a word, I tell you words.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; History is a word, and it tells you 'his-story' in words  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've a task, you know my task.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A task to record the past  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm short, I shorten your path.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Human history is short, yet contains valuable information to shorten your path (to success, for the opportunist friend)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've the answer, now tell me it!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; History!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2015-03-25T19:57:12.550" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10983" PostTypeId="2" ParentId="10950" CreationDate="2015-03-25T20:29:24.150" Score="-1" Body="&lt;p&gt;W: White&#xA;B: Black&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how it goes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First (wwb): This guys sees 1 black and 2 whites. 50-50 chance of being black/red. Says &lt;strong&gt;I don't know&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;Second (wwb): Sees exactly what #1 sees, but knows that the first didn't know if he was red, therefore didn't see red, meaning he must be black. He says &lt;strong&gt;Black&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;Third (bbw): Sees 2 blacks and 1 white. 50-50 chance of being white/red. But, knowing #2 knew he was black, this means he must have saw 2 whites, otherwise he wouldn't know if he's white or not. Therefore he must be white. He says &lt;strong&gt;White&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;Fourth (bbw): He sees 2 blacks and 1 white. Meaning 50-50 chance of being white/red. However, using the same logic as #3, he can deduce that he is white. He says &lt;strong&gt;White&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Therefore, the answer to your question, is the first logician. Sequence: &lt;strong&gt;BBWW&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2015-03-25T20:29:24.150" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10984" PostTypeId="2" ParentId="10872" CreationDate="2015-03-25T20:30:30.450" Score="4" Body="&lt;p&gt;If I'm correct, the &quot;arcane property&quot; is arguable for a number of countries.  My best guess at the missing countries:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Equatorial Guinea and Samoa&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &quot;arcane property&quot; seems to be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Countries whose TLD (internet top-level-domain) starts with a different letter than the common English name of the country.&lt;br /&gt;&lt;br /&gt;&#xA;.dz   Algeria&lt;br /&gt;&#xA;.kh   Cambodia&lt;br /&gt;&#xA;.hr   Croatia&lt;br /&gt;&#xA;.sv   El Salvador&lt;br /&gt;&#xA;.gq   Equatorial Guinea&lt;br /&gt;&#xA;.de   Germany&lt;br /&gt;&#xA;.ws   Samoa&lt;br /&gt;&#xA;.rs   Serbia&lt;br /&gt;&#xA;.za   South Africa&lt;br /&gt;&#xA;.es   Spain&lt;br /&gt;&#xA;.lk   Sri Lanka&lt;br /&gt;&#xA;.ch   Switzerland&lt;br /&gt;&#xA;.ae   United Arab Emirates&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Several other countries could also qualify.  For example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;.tl   East Timor&lt;br /&gt;&#xA;.kp   North Korea (a.k.a. Democratic People's Republic of Korea) (neither name starts with a &quot;K&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2015-03-25T20:30:30.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10985" PostTypeId="2" ParentId="10972" CreationDate="2015-03-25T21:10:42.553" Score="4" Body="&lt;p&gt;OK, so I got &lt;strong&gt;42&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of my answers are similar to Ian MacDonald's with the exception of 11 where I was a &quot;sucker&quot; and used an exponent...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1 = $ ⌊√π⌋ $&lt;br&gt;&#xA;2 = $√(−⌊−π⌋)$&lt;br&gt;&#xA;3 = $⌊π⌋$&lt;br&gt;&#xA;4 = $−⌊−π⌋$&lt;br&gt;&#xA;5 = $⌊π*√π⌋$&lt;br&gt;&#xA;6 = $⌊π⌋+⌊π⌋$&lt;br&gt;&#xA;7 = $⌊π⌋−⌊−π⌋$&lt;br&gt;&#xA;8 = $−⌊−π⌋−⌊−π⌋$&lt;br&gt;&#xA;9 = $⌊π⌋∗⌊π⌋$&lt;br&gt;&#xA;10 = $−⌊−π∗π⌋$&lt;br&gt;&#xA;11 = $⌊(-⌊-π⌋)^{√π}⌋$&lt;br&gt;&#xA;12 = $-⌊π⌋∗⌊−π⌋$&lt;br&gt;&#xA;13 = $−⌊−π∗−⌊−π⌋⌋$&lt;br&gt;&#xA;14 = $−⌊−√π⌋∗⌊⌊π⌋−⌊−π⌋⌋$&lt;br&gt;&#xA;15 = $⌊−π⌋∗⌊−π⌋−⌊√π⌋$&lt;br&gt;&#xA;16 = $⌊−π⌋∗⌊−π⌋$&lt;br&gt;&#xA;17 = $⌊−π⌋∗⌊−π⌋+⌊√π⌋$&lt;br&gt;&#xA;18 = $⌊−π⌋∗⌊−π⌋−⌊−√π⌋$&lt;br&gt;&#xA;19 = $⌊−π⌋∗⌊−π⌋+⌊π⌋$&lt;br&gt;&#xA;20 = $⌊−π⌋∗⌊−π⌋−⌊−π⌋$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3922" LastEditorUserId="3922" LastEditDate="2015-03-25T21:20:47.510" LastActivityDate="2015-03-25T21:20:47.510" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10986" PostTypeId="2" ParentId="10765" CreationDate="2015-03-25T21:14:24.313" Score="1" Body="&lt;p&gt;I think it is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; organize&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because in British English it would be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; organise.  I'm pretty sure it's a play on -ize and -ise, even if that's not the word.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Update:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Make that 'reflection'  (vs. reflexion)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10294" LastActivityDate="2015-03-25T21:14:24.313" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10987" PostTypeId="2" ParentId="10897" CreationDate="2015-03-25T21:29:12.930" Score="7" Body="&lt;p&gt;Are you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a hint?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I could be there, yet I'll stay hidden.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hints are often hidden by spoiler tags&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hidden I'll stay, but you'll need me.&lt;br/&gt;&#xA;Need me you will, so I may appear.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As people fail to guess the correct answers, hints are given.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Appear I may, I don't decide.&lt;br/&gt;&#xA;Decide I don't, my master does.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The creator of the riddle chooses when to give hints.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm a word, I tell you words.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;hint&quot; is a word, and a hint provides many helpful words.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've a task, you know my task.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The task is helping people figure out the riddle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm short, I shorten your path.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hints are usually shorter than the original clue/riddle, and they help you get to the solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've the answer, now tell me it!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hints, by their nature, allude to the correct answer, as if they are beckoning for you to get it right.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9978" LastActivityDate="2015-03-25T21:29:12.930" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10988" PostTypeId="1" AcceptedAnswerId="11006" CreationDate="2015-03-25T21:45:11.410" Score="5" ViewCount="285" Body="&lt;p&gt;You are applying for a job at Company X. You are an agent from Company Y, bitter rivals with Company X (they sell variables at low prices!). Your job is to sabotage Company X with your company's newest innovation: the subscript manipulator! (e.g. $x_n\to n_x$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your mission: Infiltrate Company X posed as a prospective employee, get into the Variable Room (VR, for short), throw your device into the Set Constructor, and leave. You must pass the interview with flying colors, else they wont accept you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a brief debriefing, you are ready. You know the routine--the only variable is the interview question. You put on your best suit (black 3-piece with a crimson red tie) and drive to the interview.&#xA;&lt;hr&gt;&#xA;Once at the designated office, you are greeted by a man wearing a simple polo shirt and a baseball cap. This is definitely the man you need to see (he told you so). He sits down, and motions for you do to the same. You sit across from him, sweating ever so slightly. He has a broad smile slapped on his face. Did he know your real aim? You only pray that he doesn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Smiling, he reaches under his desk and hands you an ancient computer with only the most basic functions. It has a single document on the screen, containing these characters:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒&#xA;  ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒&#xA;  ▒▒▒▒▒▒▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀&#xA;  ▀▀▒▒▒▒▒▒▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀&#xA;  ▀▀▀▀▒▒▒▒▒▒▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀&#xA;  ▒▒▒▀▀▀▀▀▒▒▒▒▒▒▀▀▀▀▀▒▒▒▒▒▒▒▀▀▀▀▀▒▒▒▒&#xA;  ▀▒▒▒▒▀▀▀▀▀▒▒▒▒▒▒▀▀▀▀▀▒▒▒▒▒▒▒▀▀▀▀▀▀▀&#xA;  ▀▀▀▀▀▒▒▒▒▀▀▀▀▀▒▒▒▒▒▒▀▀▀▀▀▒▒▒▒▒▒▒▀▀▀&#xA;  ▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▒▒▒▒▒▒▀▀▀▀▀▀▀▀▀▀▀▀▀&#xA;  ▀▀▀▀▀▀▀▀▒▀▀▀▀▀▀▀▀▒▒▒▒▒▒▒▒▒▀▀▀▀▀▀▀▀▀&#xA;  ▀▀▀▀▀▀▀▀▀▀▀▀▀▀▒▀▀▀▀▀▀▀▀▒▒▒▒▒▒▒▒▒▀▀▀&#xA;  ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒&#xA;  ▒▒▒▒▒▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▒▒▒▒▒▒▒▒▒▒▒▒▒&#xA;  ▒▒▒▒▒▒▒▒▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▒▒▒▒▒▒▒▒▒▒&#xA;  ▒▒▒▒▒▒▒▒▒▒▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▒▒▒▒▒▒▒▒&#xA;  ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▀▀▀▀▀▒▒▒▒▒▒▒▀▀▀▀▀▒▒▒&#xA;  ▀▀▀▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▀▀▀▀▀▀▀▀▀▀▀▀▀▀&#xA;  ▀▀▀▀▀▀▀▀▀▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▀▀▀▀▀▀▀▀&#xA;  ▀▀▀▀▀▀▀▀▀▀▀▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▀▀▀▀▀▀&#xA;  ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒&#xA;  ▒▒▀▀▀▀▀▒▒▒▒▒▒▒▀▀▀▀▀▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒&#xA;  ▒▒▒▒▀▀▀▀▀▒▒▒▒▒▒▒▀▀▀▀▀▒▒▒▒▒▒▒▒▒▒▒▒▒▒&#xA;  ▒▒▒▒▒▒▒▒▒▒▒▒▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▒▒▒▒▒▒&#xA;  ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▀▀▀▀▀▒▒▒▒▒▒▒▒&#xA;  ▀▀▀▀▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▀▀▀▀▀▀▀▀▀▀▀▀▀&#xA;  ▀▀▀▀▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▀▀▀▀▀▒▒▒▒▒▒▒▀&#xA;  ▒▒▒▒▒▒▒▀▀▀▀▀▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▀▀▀▀▀&#xA;  ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;You have ten minutes. I wish you luck, Mr(s). &lt;code&gt;$LASTNAME&lt;/code&gt;,&quot; he says, still smiling. He promptly walks out of the room and locks you in it. Your immediate response is to browse the internet for an answer, but to no avail: there is no Wi-Fi! Curses! After wasting a minute, you divert your energy into solving the problem. An unsuccessful minute later, you look into the other files contained on the computer. The only file you manage to find is a small &lt;code&gt;.txt&lt;/code&gt; file like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;FILENAME: your_hint.txt&#xA;DATE: ??/??/???? ??:??:??&#xA;-------------------------&#xA;iF icon(cab):&#xA;&#xA;(EOF)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Stumped, you analyze the 'hint' and continue with the puzzle.&#xA;&lt;hr&gt;&#xA;Can you solve the puzzle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is a single word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7307" LastEditorUserId="7307" LastEditDate="2015-03-25T21:58:23.037" LastActivityDate="2015-03-26T17:51:25.010" Title="Obsfucated Text -- Minds of the masters" Tags="|word|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10989" PostTypeId="1" AcceptedAnswerId="11048" CreationDate="2015-03-25T22:08:13.130" Score="4" ViewCount="915" Body="&lt;p&gt;The police suspect a teenager named Joe of planning to murder his ex-girlfriend's entire family.  The evidence is locked in Joe's computer.  Failing to hack into the computer, the police find Joe's best friend Billy and gather some information:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Joe has put his ex's family members into 2 groups on a whiteboard as below:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Group 1:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Father&#xA;Sister&#xA;Son&#xA;Daughter&#xA;Children&#xA;Cousin&#xA;Nancy (her cat)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Group 2:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Mother&#xA;Brother&#xA;Boyfriend&#xA;Husband&#xA;Parent&#xA;Sibling&#xA;Bobby (her pet snake)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;2) Joe has used one of his friends' names as the computer's password.  The names are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Billy&lt;/li&gt;&#xA;&lt;li&gt;Lilly&lt;/li&gt;&#xA;&lt;li&gt;Kelly&lt;/li&gt;&#xA;&lt;li&gt;Ally&lt;/li&gt;&#xA;&lt;li&gt;Nancy&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;3) Joe has told Billy that his computer's password &lt;strong&gt;shares the same attribute that he used to categorize his ex's family&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) A single failed attempt to login to Joe's computer will cause the computer to reformat, thus losing all the evidence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out to be a very simple logic and a smart detective quickly solves it and arrests Joe.  The question is: what is the attribute Joe used to categorize his ex's family, and what is the password?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;EDIT: I should have mentioned that the attribute helps to distinguish the password from the rest of the choice&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint #1:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Attribute does not necessarily relate to a word's spelling / length.  Even the description of the ex's pet fits into the attribute it shares with the group.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="8874" LastEditDate="2015-11-01T15:54:08.330" LastActivityDate="2015-11-01T15:54:08.330" Title="Detective, what is the password?" Tags="|pattern|mystery|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10990" PostTypeId="1" AcceptedAnswerId="11049" CreationDate="2015-03-25T22:13:38.677" Score="28" ViewCount="1529" Body="&lt;p&gt;I've taken an $n$ by $n$ chessboard and drawn an arrow on each square, pointing in one of the eight compass directions. I've done this in such a way that arrows in (orthogonally) adjacent squares differ by at most $45^\circ$. I place a cricket on one of the squares, and it proceeds to hop from square to square, following the arrows. If an arrow points of the board, the cricket falls off the board. Will this cricket necessarily fall off the board? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{|c|c|c|}&#xA;\hline&#xA;\nwarrow&amp;amp;\leftarrow&amp;amp; \nwarrow\\\hline&#xA;\uparrow&amp;amp;\nwarrow&amp;amp; \nwarrow\\\hline&#xA;\nearrow &amp;amp;\uparrow&amp;amp; \nwarrow\\\hline&#xA;\end{array}&#xA;\qquad&#xA;\begin{array}{|c|c|c|}&#xA;\hline&#xA;\downarrow&amp;amp;\downarrow&amp;amp; \color{red}\swarrow\\\hline&#xA;\searrow&amp;amp;\searrow&amp;amp; \color{red}\rightarrow\\\hline&#xA;\color{red}\downarrow&amp;amp;\color{red}\rightarrow&amp;amp; \nearrow\\\hline&#xA;\end{array}&#xA;$$&#xA;For example, the board on the left is a possible $3\times 3$ board I could have made. You can check the cricket is doomed to fall off no matter where it starts. However, the board on the right is illegal: the red arrows at the bottom left differ by $90^\circ$, and the other two red arrows differ by $135^\circ$. The question is, does there exist a legal board, and some square on that board, where the cricket does not fall off when it starts on that square?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This puzzle is from Peter Winkler's collection, &lt;em&gt;Mathematical Mind Benders&lt;/em&gt;. It seems like there are several solutions, and I was wondering what ways people had to solve this.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-03-26T16:23:56.170" LastActivityDate="2015-04-07T20:06:42.793" Title="Arrows on a Chessboard" Tags="|mathematics|checkerboard|" AnswerCount="4" CommentCount="20" ContentLicense="CC BY-SA 3.0" />
  <row Id="10991" PostTypeId="1" CreationDate="2015-03-25T22:19:54.843" Score="7" ViewCount="1681" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;111 221 241 151&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;222 543 421&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;233 161&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;166 222 331 353 444&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The problem is to translate this into a sentence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CLUES:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;code&gt;111&lt;/code&gt;, could also be &lt;code&gt;166&lt;/code&gt;, or &lt;code&gt;121&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;It is a Homophonic Substitution Cipher&lt;/li&gt;&#xA;&lt;li&gt;The code he gave me before this was:&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;code&gt;&amp;amp;!! &quot;/ (?, }&amp;lt;€ \&amp;gt;!&amp;lt;* :~&quot;&amp;amp; }?@!&lt;/code&gt; translates to&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;see if you can break this code&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;It is for an interview for an assistant trader position in Chicago.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="10646" LastEditorUserId="12832" LastEditDate="2017-02-07T14:44:59.577" LastActivityDate="2017-02-07T14:44:59.577" Title="Translate into a sentence" Tags="|cipher|" AnswerCount="2" CommentCount="23" ContentLicense="CC BY-SA 3.0" />
  <row Id="10995" PostTypeId="2" ParentId="10950" CreationDate="2015-03-25T23:22:44.807" Score="2" Body="&lt;p&gt;&lt;strong&gt;Solution:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The third person to say what color his hat was. The first man's hat was the same color as the second man's hat, and the third man's hat was the same color as the fourth's.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Process:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Here's an attempt at a more intuitive way to solve this. First, we add in a round zero in which we determine what everyone knows based only on the hats they can see. Then, we eliminate possibilities based on which possible situations the current situation matches.&lt;/p&gt;&#xA;&lt;p&gt;We'll call the logicians A, B, C, and D. To reduce the size of the table I'm going to build, we'll say WLOG that A is wearing a white hat and also go through the first round in a normal fashion.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Case 1: A and B are wearing white hats&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Round 1 (WWBB):&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A -&amp;gt; I see _WBB. I could have white or red. &amp;quot;I don't know.&amp;quot;&lt;/li&gt;&#xA;&lt;li&gt;B -&amp;gt; I see W_BB. If I had red, A would see RBB and know A has W. A didn't know, so &amp;quot;I have a white hat&amp;quot;.&lt;/li&gt;&#xA;&lt;li&gt;C -&amp;gt; I see WW_B. I have black or red. If I have red, then A sees _WRB and doesn't know. B sees W_RB and knows if B had black A would know so B has white. If I have black... (real scenario). &amp;quot;I don't know.&amp;quot;&lt;/li&gt;&#xA;&lt;li&gt;D -&amp;gt; I see WWB_. If I had red, C would see WW_R and know C has black, so &amp;quot;I have a black hat.&amp;quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;So going forward A and C only need to consider scenarios where B and D have white and black hats respectively.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Case 2: A and C are wearing white hats&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Round 1 (WBWB):&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A -&amp;gt; I see _BWB. I could have red or white. &amp;quot;I don't know.&amp;quot;&lt;/li&gt;&#xA;&lt;li&gt;B -&amp;gt; I see W_BW. I could have red or white. If I have red, A sees _RWB and can't tell. If I have white, A sees _BWB and can't tell. &amp;quot;I don't know&amp;quot;&lt;/li&gt;&#xA;&lt;li&gt;C -&amp;gt; I see WB_B. If I had red, A would have known. So &amp;quot;I have a white hat.&amp;quot;&lt;/li&gt;&#xA;&lt;li&gt;D -&amp;gt; I see WBW_. If I had red, B would have known. So &amp;quot;I have a black hat.&amp;quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;So going forward A and B only need to consider scenarios where C and D have white and black hats respectively.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Case 3: A and D are wearing white hats&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Round 1 (WBBW):&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A -&amp;gt; I see _BBW. I could have red or white. &amp;quot;I don't know.&amp;quot;&lt;/li&gt;&#xA;&lt;li&gt;B -&amp;gt; I see W_BW. I could have red or black. &amp;quot;I don't know.&amp;quot;&lt;/li&gt;&#xA;&lt;li&gt;C -&amp;gt; I see WB_W. If I had a red hat, then B would have seen W_RW and known. So &amp;quot;I have a black hat.&amp;quot;&lt;/li&gt;&#xA;&lt;li&gt;D -&amp;gt; I see WBB_. If I had a red hat, then A would have seen _BBR and known. So &amp;quot;I have a white hat.&amp;quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;So going forward A and B only need to consider scenarios where C and D have black and white hats respectively.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Table for Case 1&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;We'll denote turns by a round number and whose turn it is. We'll also use round 0 to represent what they can figure out without anyone saying anything. Note: I'm building this tables by running through the first round of the hypothetical situation.&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;Hats&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;0A&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;0B&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;0C&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;0D&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;1A&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;1B&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;1C&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;1D&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;strong&gt;WWBB&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;W&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;strong&gt;WWRB&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;W&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;strong&gt;BWWB&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;W&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;strong&gt;BWRB&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;W&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;W&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;R&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;strong&gt;RWWB&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;W&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;strong&gt;RWBB&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;W&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;W&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;Looking at the table, we can see that A can easily figure out that A doesn't have a red hat - otherwise either B wouldn't have known or C would have. So A says &amp;quot;I have a white hat.&amp;quot;&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;Hats&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;0A&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;0B&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;0C&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;0D&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;1A&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;1B&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;1C&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;1D&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;2A&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;strong&gt;WWBB&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;W&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;W&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;strong&gt;WWRB&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;W&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;W&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;strong&gt;BWWB&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;W&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;strong&gt;BWRB&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;W&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;W&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;R&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;strong&gt;RWWB&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;W&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;strong&gt;RWBB&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;W&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;W&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;R&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;Now that A knows, C is stuck. There's no way for C to distinguish between having a red hat and a black hat. So &lt;strong&gt;C, the third logician to speak, doesn't know the color of his hat&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Table for Case 2&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;Hats&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;0A&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;0B&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;0C&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;0D&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;1A&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;1B&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;1C&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;1D&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;strong&gt;WBWB&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;W&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;strong&gt;WRWB&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;W&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;strong&gt;BWWB&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;W&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;strong&gt;BRWB&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;W&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;W&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;strong&gt;RWWB&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;W&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;strong&gt;RBWB&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;W&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;W&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;Here, A and B are both stuck. They have no information to distinguish between any of these cases. This does not match the OP's description, so this must not have been the case.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Table for Case 3&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;Hats&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;0A&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;0B&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;0C&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;0D&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;1A&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;1B&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;1C&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;1D&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;strong&gt;WBBW&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;W&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;strong&gt;WRBW&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;W&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;strong&gt;BWBW&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;W&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;strong&gt;BRBW&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;W&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;W&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;strong&gt;RWBW&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;W&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;strong&gt;RBBW&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;W&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;?&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;B&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;W&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;Again, both A and B are stuck so this is not the scenario. Thus, only case 1 remains and must be what happened.&lt;/p&gt;&#xA;" OwnerUserId="1955" LastEditorUserId="69582" LastEditDate="2021-02-06T18:56:16.143" LastActivityDate="2021-02-06T18:56:16.143" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10996" PostTypeId="1" AcceptedAnswerId="11001" CreationDate="2015-03-25T23:41:56.403" Score="10" ViewCount="2719" Body="&lt;p&gt;There are $n$ MPs seated around a circular table, prepared to cast their votes on a single yes-or-no question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The voting goes by rounds. Each voter raises his hand for 'yes' and keeps it down for 'no'. If a voter's neighbors both vote the same way in one round, he will vote that way in the next round; but if one of his neighbors votes 'no' and the other one votes 'yes', he keeps his current choice for the next round.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The voting ends when all $n$ voters keep their votes no matter how many further rounds take place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Vw997.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;what values of $n$ ensure a definite end?&lt;/li&gt;&#xA;&lt;li&gt;if the voting is endless, how were votes cast in the initial round?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9517" LastEditorUserId="5373" LastEditDate="2015-03-26T00:18:01.587" LastActivityDate="2015-03-26T00:51:18.937" Title="The voters in a round table" Tags="|logical-deduction|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10998" PostTypeId="1" AcceptedAnswerId="10999" CreationDate="2015-03-26T00:13:35.293" Score="5" ViewCount="385" Body="&lt;p&gt;Follow this to find the answer;&lt;br&gt;&#xA;Everyone knows it, but few see it.&lt;br&gt;&#xA;Stuck between a pair of seers,&lt;br&gt;&#xA;But it cannot prophesy at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not on Stack Exchange,&lt;br&gt;&#xA;Only in real life.&lt;br&gt;&#xA;You've heard it once,&lt;br&gt;&#xA;Now write it down.&#xA;&lt;hr&gt;&#xA;What is being referred to in this riddle? Please explain how your answer fits all the clues.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-03-26T00:36:33.323" Title="Riddle - not on Stack Exchange" Tags="|riddle|word|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10999" PostTypeId="2" ParentId="10998" CreationDate="2015-03-26T00:29:45.613" Score="8" Body="&lt;p&gt;It's a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; nose&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Follow this to find the answer;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The expression &quot;follow your nose&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Everyone knows it, but few see it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Homophone of &quot;knows&quot;, you know it's there but don't see it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Stuck between a pair of seers,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; between your eyes &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But it cannot prophesy at all.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; not sure, maybe just double-meaning on &quot;seers&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not on Stack Exchange,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; acronym for NOSE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Only in real life.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; online, you're virtual and don't have a body?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You've heard it once,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; refers to the knows/nose homophone&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-03-26T00:36:33.323" LastActivityDate="2015-03-26T00:36:33.323" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11000" PostTypeId="2" ParentId="10996" CreationDate="2015-03-26T00:31:07.780" Score="4" Body="&lt;p&gt;The voting is endless when&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $n$ is even and votes alternate around the table&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The voting has a definite end in all other cases.&lt;/p&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2015-03-26T00:31:07.780" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11001" PostTypeId="2" ParentId="10996" CreationDate="2015-03-26T00:31:30.270" Score="13" Body="&lt;p&gt;The following fact is the key to answering this question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$(*)$ if any two adjacent MPs vote in the same way, then all votes will eventually be locked in indefinitely.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An MP will only change his vote if both his neighbours vote differently from him. So if two adjacent MPs ($M_1$ and $M_2$) vote in the same way, then clearly &lt;em&gt;their&lt;/em&gt; votes are both locked in indefinitely. WLOG, let's say they both vote 'yes'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the MP adjacent to them, $M_3$. If he ever votes 'yes', then that vote too will be locked in indefinitely, by the same reasoning. Otherwise, he never votes 'yes', i.e. his vote is locked in as 'no' indefinitely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $k$ be minimal such that the $k$th MP $M_k$ never votes 'yes'. We know $M_1$ and $M_2$'s votes get locked in as 'yes'; if $k&amp;gt;3$, then $M_3$'s vote also gets locked in as 'yes'; ; if $k&amp;gt;4$, then $M_4$'s vote also gets locked in as 'yes'; and so on by induction. The votes of $M_1,\dots,M_{k-1}$ all get locked in at 'yes' while $M_k$'s vote is always 'no'. (If there is no such $k$, then the same argument shows they &lt;em&gt;all&lt;/em&gt; get locked in at 'yes' and we're done.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $M_{k+1}$ can't ever vote 'yes' after $M_{k-1}$'s vote is locked in, since that would mean $M_k$'s vote would change to 'yes'. So $M_{k+1}$'s vote is eventually locked in at 'no'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this stage, the pattern repeats. Argue exactly as above with the adjacent 'yes'-voters $M_1,M_2$ replaced by the adjacent 'no'-voters $M_k,M_{k+1}$. This argument can carry on indefinitely, and since $n$ is finite, we eventually come to an end. So &lt;strong&gt;all&lt;/strong&gt; votes end up being locked in.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The converse is also clear: if the MPs' votes alternate so that every adjacent pair of MPs vote differently from each other, then every yes-man is in between two no-men and vice versa, so in each round all votes are toggled, and &lt;em&gt;no&lt;/em&gt; votes end up being locked in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;strong&gt;voting continues forever iff the MPs' votes alternate all the way round the table&lt;/strong&gt;, which can only happen if $n$ is even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answers to the OP's two questions are: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;what values of $n$ ensure a definite end?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Odd $n$&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;if the voting is endless, how were votes cast in the initial round?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Alternating all the way round the table&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-26T00:51:18.937" LastActivityDate="2015-03-26T00:51:18.937" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11002" PostTypeId="2" ParentId="10950" CreationDate="2015-03-26T00:48:50.810" Score="3" Body="&lt;p&gt;I've seen several correct solutions, but here's a different method.&lt;/p&gt;&#xA;&lt;p&gt;Solution:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The third logician is the one who cannot determine the color of their hat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;It is evident that the problem is symmetric with respect to Black and White hats; that is, the situation is logically identical if black and white hats are interchanged.  That leaves fifteen possible scenarios which one of the logicians might entertain.  I have enumerated them (including black/white interchanges) in the table below.&lt;/p&gt;&#xA;&lt;p&gt;Each logician, based on the three hats which he sees, will be able to narrow the field down to at most two potential scenarios.  In the scenarios in which the logician sees one red hat and two hats which are either both black or both white, he will be able to deduce that his own hat is white or black, respectively.  When the logician's direct observations do not give them this information, then further deduction is based on the answers of the other logicians.  For the purposes of the puzzle, we as usual assume that the logicians have mutual full confidence in the abilities of the other logicians, &lt;em&gt;ad infinitum&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;p&gt;In the table below, the columns labeled 1, 2, 3, 4, 1 represent whether each logician in turn will be able to answer the question.  If they are unable, the table shows a period.  If they are able due to the colors of the three hats they personally observe, as described above, then the table shows the number 0.  If the responses that the other logicians would have given in the only other scenario consistent with the direct observations is different from the responses given in the scenario under consideration (which is easily determined from inspection of the rows of the table), then the number of the row representing the other scenario appears in the table. If the logician in question already answered on the previous round, the table shows a hyphen.&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th&gt;&lt;/th&gt;&#xA;&lt;th&gt;&lt;/th&gt;&#xA;&lt;th&gt;1&lt;/th&gt;&#xA;&lt;th&gt;2&lt;/th&gt;&#xA;&lt;th&gt;3&lt;/th&gt;&#xA;&lt;th&gt;4&lt;/th&gt;&#xA;&lt;th&gt;1&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;1&lt;/td&gt;&#xA;&lt;td&gt;WWBB/BBWW&lt;/td&gt;&#xA;&lt;td&gt;.&lt;/td&gt;&#xA;&lt;td&gt;12&lt;/td&gt;&#xA;&lt;td&gt;.&lt;/td&gt;&#xA;&lt;td&gt;2&lt;/td&gt;&#xA;&lt;td&gt;15&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;2&lt;/td&gt;&#xA;&lt;td&gt;WWBR/BBWR&lt;/td&gt;&#xA;&lt;td&gt;.&lt;/td&gt;&#xA;&lt;td&gt;7&lt;/td&gt;&#xA;&lt;td&gt;0&lt;/td&gt;&#xA;&lt;td&gt;1&lt;/td&gt;&#xA;&lt;td&gt;.&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;3&lt;/td&gt;&#xA;&lt;td&gt;WWRB/BBRW&lt;/td&gt;&#xA;&lt;td&gt;.&lt;/td&gt;&#xA;&lt;td&gt;9&lt;/td&gt;&#xA;&lt;td&gt;.&lt;/td&gt;&#xA;&lt;td&gt;0&lt;/td&gt;&#xA;&lt;td&gt;8&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;4&lt;/td&gt;&#xA;&lt;td&gt;WBWB/BWBW&lt;/td&gt;&#xA;&lt;td&gt;.&lt;/td&gt;&#xA;&lt;td&gt;.&lt;/td&gt;&#xA;&lt;td&gt;9&lt;/td&gt;&#xA;&lt;td&gt;5&lt;/td&gt;&#xA;&lt;td&gt;.&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;5&lt;/td&gt;&#xA;&lt;td&gt;WBWR/BWBR&lt;/td&gt;&#xA;&lt;td&gt;.&lt;/td&gt;&#xA;&lt;td&gt;0&lt;/td&gt;&#xA;&lt;td&gt;7&lt;/td&gt;&#xA;&lt;td&gt;4&lt;/td&gt;&#xA;&lt;td&gt;.&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;6&lt;/td&gt;&#xA;&lt;td&gt;WBBW/BWWB&lt;/td&gt;&#xA;&lt;td&gt;.&lt;/td&gt;&#xA;&lt;td&gt;.&lt;/td&gt;&#xA;&lt;td&gt;8&lt;/td&gt;&#xA;&lt;td&gt;7&lt;/td&gt;&#xA;&lt;td&gt;.&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;7&lt;/td&gt;&#xA;&lt;td&gt;WBBR/BWWR&lt;/td&gt;&#xA;&lt;td&gt;0&lt;/td&gt;&#xA;&lt;td&gt;2&lt;/td&gt;&#xA;&lt;td&gt;5&lt;/td&gt;&#xA;&lt;td&gt;6&lt;/td&gt;&#xA;&lt;td&gt;-&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;8&lt;/td&gt;&#xA;&lt;td&gt;WBRW/BWRB&lt;/td&gt;&#xA;&lt;td&gt;.&lt;/td&gt;&#xA;&lt;td&gt;0&lt;/td&gt;&#xA;&lt;td&gt;6&lt;/td&gt;&#xA;&lt;td&gt;9&lt;/td&gt;&#xA;&lt;td&gt;3&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;9&lt;/td&gt;&#xA;&lt;td&gt;WBRB/BWRW&lt;/td&gt;&#xA;&lt;td&gt;0&lt;/td&gt;&#xA;&lt;td&gt;3&lt;/td&gt;&#xA;&lt;td&gt;4&lt;/td&gt;&#xA;&lt;td&gt;8&lt;/td&gt;&#xA;&lt;td&gt;-&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;10&lt;/td&gt;&#xA;&lt;td&gt;WRWB/BRBW&lt;/td&gt;&#xA;&lt;td&gt;.&lt;/td&gt;&#xA;&lt;td&gt;.&lt;/td&gt;&#xA;&lt;td&gt;12&lt;/td&gt;&#xA;&lt;td&gt;0&lt;/td&gt;&#xA;&lt;td&gt;.&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;11&lt;/td&gt;&#xA;&lt;td&gt;WRBW/BRWB&lt;/td&gt;&#xA;&lt;td&gt;.&lt;/td&gt;&#xA;&lt;td&gt;.&lt;/td&gt;&#xA;&lt;td&gt;0&lt;/td&gt;&#xA;&lt;td&gt;12&lt;/td&gt;&#xA;&lt;td&gt;.&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;12&lt;/td&gt;&#xA;&lt;td&gt;WRBB/BRWW&lt;/td&gt;&#xA;&lt;td&gt;0&lt;/td&gt;&#xA;&lt;td&gt;1&lt;/td&gt;&#xA;&lt;td&gt;10&lt;/td&gt;&#xA;&lt;td&gt;11&lt;/td&gt;&#xA;&lt;td&gt;-&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;13&lt;/td&gt;&#xA;&lt;td&gt;RWWB/RBBW&lt;/td&gt;&#xA;&lt;td&gt;.&lt;/td&gt;&#xA;&lt;td&gt;.&lt;/td&gt;&#xA;&lt;td&gt;15&lt;/td&gt;&#xA;&lt;td&gt;0&lt;/td&gt;&#xA;&lt;td&gt;.&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;14&lt;/td&gt;&#xA;&lt;td&gt;RWBW/RBWB&lt;/td&gt;&#xA;&lt;td&gt;.&lt;/td&gt;&#xA;&lt;td&gt;.&lt;/td&gt;&#xA;&lt;td&gt;0&lt;/td&gt;&#xA;&lt;td&gt;15&lt;/td&gt;&#xA;&lt;td&gt;.&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;15&lt;/td&gt;&#xA;&lt;td&gt;RWBB/RBWW&lt;/td&gt;&#xA;&lt;td&gt;.&lt;/td&gt;&#xA;&lt;td&gt;0&lt;/td&gt;&#xA;&lt;td&gt;13&lt;/td&gt;&#xA;&lt;td&gt;14&lt;/td&gt;&#xA;&lt;td&gt;.&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;Note that the scenarios in which only the first logician is unable to guess their hat color after the other three logicians have correctly answered (2, 5, 15) all have the same pattern of answers, and so the first logician will not be able to distinguish them.  Likewise, the scenarios in which, after the first logicians second turn, both the first and second logicians still do not know their hat colors (4, 6, 10, 11, 13, 14), and the scenarios in which, after the first logician's second turn, the third logician still does not know their hat color (1, 3).  Therefore, continued questions will not yield any new information in these scenarios.&lt;/p&gt;&#xA;&lt;p&gt;Only three scenarios, numbered 1, 4, and 6 (marked with asterisks in the table), do not have a red hat, and so are consistent with the problem statement.  Of these, scenarios 4 and 6 have an outcome in which two of the logicians are unable to deduce their hat color.  Only scenario 1, in which the first, second, and fourth logicians are successful, but the third is not, meets the condition in the problem statement.&lt;/p&gt;&#xA;" OwnerUserId="10649" LastEditorUserId="69582" LastEditDate="2021-02-06T18:39:33.783" LastActivityDate="2021-02-06T18:39:33.783" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="11003" PostTypeId="2" ParentId="10963" CreationDate="2015-03-26T01:27:45.047" Score="4" Body="&lt;p&gt;Maybe too obvious given line 2, but I've managed to make it fit in some interesting and cryptic ways!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A violin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am fast like an arrow but slow like a turtle&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can play violin music both quickly (&lt;strong&gt;allegro&lt;/strong&gt;) and slowly (&lt;strong&gt;adagio&lt;/strong&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I play like a Cello but with a silence echo&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A violin certainly plays like a cello; I'm not sure what &quot;with a silence echo&quot; means though.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am long like a Metro but not quite the rainbow&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Buskers might be found on the Metro playing a violin. A violin &lt;strong&gt;bow&lt;/strong&gt; is not quite the same word as &quot;rainbow&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Give me an answer that fit this riddle!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;fit this riddle&quot; is a Spoonerism of &quot;rit this fiddle&quot;, and although &quot;rit&quot; is meaningless, a &lt;strong&gt;fiddle&lt;/strong&gt; is another word for a violin!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-03-26T01:27:45.047" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11005" PostTypeId="2" ParentId="10966" CreationDate="2015-03-26T05:03:25.043" Score="7" Body="&lt;p&gt;The answer I came up with is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Bow, and its homonyms.&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Street: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Actually, this is the only one I am not sure about, but research suggests Bow Street Runners, the earliest police force in London. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Respect: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;To bow, leaning forward in a gesture of respect.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Ribbon: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A bow, as found decorating presents or hair.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Reflex: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bow, as in shooting arrows. Reflex is a type of bow shape.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Play: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bow, used in playing stringed musical instruments. (I can attest, it can sound beautiful or down-right awful).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Front: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bow, a part of a boat/ship.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9319" LastActivityDate="2015-03-26T05:03:25.043" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11006" PostTypeId="2" ParentId="10988" CreationDate="2015-03-26T07:09:19.053" Score="9" Body="&lt;p&gt;The hint is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; iF icon(cab) which is an anagram for &lt;strong&gt;Fibonacci&lt;/strong&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The single word is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Box&lt;/strong&gt; (sideways) which is obtained by shifting each row into alignment.  Each row is shifted by using the series of prime numbers (2,3,5,7,11,13,17,19,etc)&lt;br&gt;&#xA;&lt;br&gt;&#xA;▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒&#xA;▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒&#xA;▒▒▒▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▒▒▒&#xA;▒▒▒▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▒▒▒&#xA;▒▒▒▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▒▒▒&#xA;▒▒▒▀▀▀▀▀▒▒▒▒▒▒▒▀▀▀▀▀▒▒▒▒▒▒▒▀▀▀▀▀▒▒▒&#xA;▒▒▒▀▀▀▀▀▒▒▒▒▒▒▒▀▀▀▀▀▀▀▀▒▒▒▒▀▀▀▀▀▒▒▒&#xA;▒▒▒▀▀▀▀▀▒▒▒▒▒▒▒▀▀▀▀▀▀▀▀▒▒▒▒▀▀▀▀▀▒▒▒&#xA;▒▒▒▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▒▒▒&#xA;▒▒▒▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▒▀▀▀▀▀▀▀▀▒▒▒▒▒▒&#xA;▒▒▒▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▒▀▀▀▀▀▀▀▀▒▒▒▒▒▒&#xA;▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒&#xA;▒▒▒▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒&#xA;▒▒▒▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒&#xA;▒▒▒▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒&#xA;▒▒▒▀▀▀▀▀▒▒▒▒▒▒▒▀▀▀▀▀▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒&#xA;▒▒▒▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒&#xA;▒▒▒▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒&#xA;▒▒▒▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒&#xA;▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒&#xA;▒▒▒▀▀▀▀▀▒▒▒▒▒▒▒▀▀▀▀▀▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒&#xA;▒▒▒▀▀▀▀▀▒▒▒▒▒▒▒▀▀▀▀▀▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒&#xA;▒▒▒▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒&#xA;▒▒▒▒▒▒▒▒▒▀▀▀▀▀▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒&#xA;▒▒▒▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒&#xA;▒▒▒▀▀▀▀▀▒▒▒▒▒▒▒▀▀▀▀▀▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒&#xA;▒▒▒▀▀▀▀▀▒▒▒▒▒▒▒▀▀▀▀▀▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒&#xA;▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-03-26T17:51:25.010" LastActivityDate="2015-03-26T17:51:25.010" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11008" PostTypeId="2" ParentId="10911" CreationDate="2015-03-26T08:34:27.217" Score="0" Body="&lt;p&gt;Lots of copy pasta, but the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A star&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My name sounds like a bird &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Starlings are small to medium-sized passerine birds in the family Sturnidae&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whole body is like tongue of a lizard &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A lizard uses its tongue for maintenance (eating, drinking, cleaning, etc.). A star's body maintains its nuclear reactions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Am different colors &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Red, yellow, white, blue, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I well can be very littered &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Universe-wide, in fact.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My energy melts away over time &#xA;But not quite like an ice &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stars spend about 90% of their existence fusing hydrogen into helium in high-temperature and high-pressure reactions near the core. Such stars are said to be on the main sequence. The duration that a star spends on the main sequence depends primarily on the amount of fuel it has to fuse and the rate at which it fuses that fuel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And when not taken care of &#xA;I'll collect dust and lice &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In larger stars, fusion continues until the iron core has grown so large that it can no longer support its own mass. This core will suddenly collapse, and the shockwave formed by this sudden collapse causes the rest of the star to explode in a supernova, resulting in a neutron star or black hole, both of which will consume any dust or lice in its vicinity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10652" LastActivityDate="2015-03-26T08:34:27.217" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11009" PostTypeId="1" AcceptedAnswerId="11019" CreationDate="2015-03-26T09:01:09.033" Score="12" ViewCount="1218" Body="&lt;p&gt;I wanted to share a wonderful message with all of you lovely people on puzzling.SE, but when I typed it in, all that came out was a string of gibberish:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;¡××­šwg¿¿uþ}0ã×/Œz0îÏ–þžwóuÜ„&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thankfully I've changed my keyboard now so I could write this message without any issues. The only problem is, I've forgotten what I was going to say to you all, but I know that it was really uplifting and encouraging. Can someone help me to remember what it was I wanted to say? If you can figure that out, please let me know what happened to my keyboard to make it all go so wrong!&lt;/p&gt;&#xA;" OwnerUserId="7361" LastEditorUserId="50054" LastEditDate="2023-07-12T16:58:14.073" LastActivityDate="2023-07-12T16:58:14.073" Title="Help! What's my encoding? ¡××­šwg¿¿uþ}0ã×/Œz0îÏ–þžwóuÜ„" Tags="|cipher|computer-puzzle|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="11010" PostTypeId="1" CreationDate="2015-03-25T00:44:25.580" Score="11" ViewCount="6352" Body="&lt;p&gt;Someone asked me a riddle today please help me answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Which five letter english word is it?&lt;br&gt;&#xA;Where the 1st letter is the last letter.&lt;br&gt;&#xA;The last letter is the 1st letter.&lt;br&gt;&#xA;The middle letter is the second letter.&lt;br&gt;&#xA;The second letter is the fifth letter.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="Usman" LastEditorUserId="15544" LastEditDate="2015-09-17T14:07:10.407" LastActivityDate="2015-09-17T14:07:10.407" Title="A riddle on a five letter English word" Tags="|riddle|word|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11011" PostTypeId="2" ParentId="11010" CreationDate="2015-03-25T14:38:46.980" Score="9" Body="&lt;p&gt;Ah, having written this, I've just noticed StoneyB's comment above. So here is a rehash of StoneyB's answer in case you didn't read the comments:&#xA;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Zebra: &lt;strong&gt;Z&lt;/strong&gt;(last letter of the alphabet) &lt;strong&gt;E&lt;/strong&gt; (fifth letter of the alphabet) &lt;strong&gt;B&lt;/strong&gt; (second letter of the alphabet) &lt;strong&gt;R&lt;/strong&gt; &lt;strong&gt;A&lt;/strong&gt; (first letter of the alphabet).&lt;br&gt;&#xA; So the first letter of the word is that last letter (in the alphabet), and so forth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="29131" OwnerDisplayName="Araucaria" LastEditorUserId="7715" LastEditDate="2015-05-04T22:52:36.330" LastActivityDate="2015-05-04T22:52:36.330" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11012" PostTypeId="2" ParentId="11009" CreationDate="2015-03-26T09:24:47.390" Score="2" Body="&lt;p&gt;I am sure it has to do with mismatched charset, soon going to put actual answer here..&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; UTF8 or 64 file read using ascii charset&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;conversion is giving:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; +AKEA1wDXAK0BYQ-wg+AL8Avw-u+AP4AfQ-0+AOMA1w-/+AVI-z0+AO4AzyATAP4Bfg-w+APM-u+ANwgHg-   not sure if got wrong charset or this is another puzzle...?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second attempt:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; вобябябйwgвпвпuйн}0бя/илz0йнwu     MIK&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;using following site for conversion:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.motobit.com/util/charset-codepage-conversion.asp&quot; rel=&quot;nofollow&quot;&gt;http://www.motobit.com/util/charset-codepage-conversion.asp&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My code that did not work :(&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    $str = &quot;¡××­šwg¿¿uþ}0ã×/Œz0îÏ–þžwóuÜ„&quot;;&#xA;        $matchArr = array(&quot;pass&quot;,&quot;wchar&quot;,&quot;byte2be&quot;,&quot;byte2le&quot;,&quot;8bit&quot;,&quot;UCS-2&quot;,&quot;UCS-2BE&quot;,&quot;UCS-2LE&quot;,&quot;UTF-16&quot;,&quot;UTF-16BE&quot;,&quot;UTF-16LE&quot;,&quot;UTF-8&quot;,&quot;UTF-8-Mobile#DOCOMO&quot;,&quot;UTF-8-Mobile#KDDI-A&quot;,&quot;UTF-8-Mobile#KDDI-B&quot;,&quot;UTF-8-Mobile#SOFTBANK&quot;,&quot;ISO-8859-1&quot;,&quot;ISO-8859-2&quot;,&quot;ISO-8859-3&quot;,&quot;ISO-8859-4&quot;,&quot;ISO-8859-5&quot;,&quot;ISO-8859-6&quot;,&quot;ISO-8859-7&quot;,&quot;ISO-8859-8&quot;,&quot;ISO-8859-9&quot;,&quot;ISO-8859-10&quot;,&quot;ISO-8859-13&quot;,&quot;ISO-8859-14&quot;,&quot;ISO-8859-15&quot;,&quot;ISO-8859-16&quot;,&quot;Windows-1251&quot;,&quot;CP866&quot;,&quot;KOI8-R&quot;,&quot;KOI8-U&quot;,&quot;CP850&quot;&#xA;                );&#xA;&#xA;    foreach($matchArr as $a){&#xA;        foreach(mb_list_encodings() as $c){&#xA;                    //echo $a. &quot;---&quot;.$c. &quot;^&quot;. mb_convert_encoding($str,$a, $c).&quot;&amp;lt;br/&amp;gt;&quot;;&#xA;                    echo $a. &quot;---&quot;.$c. &quot;===&amp;gt;&quot;. mb_convert_encoding($str,$c, $a).&quot;&amp;lt;br/&amp;gt;&quot;;&#xA;                }&#xA;                echo &quot;-----------------------&amp;lt;br/&amp;gt;&quot;;&#xA;        }&#xA;        /**&#xA;        foreach(mb_list_encodings() as $a){&#xA;        foreach(mb_list_encodings() as $c){&#xA;                echo mb_convert_encoding($str,$a, $c).&quot;&amp;lt;br/&amp;gt;&quot;;&#xA;        }&#xA;    }&#xA;   /**/&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="10210" LastEditorUserId="10210" LastEditDate="2015-03-26T13:30:43.340" LastActivityDate="2015-03-26T13:30:43.340" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11013" PostTypeId="2" ParentId="10950" CreationDate="2015-03-26T09:58:58.117" Score="1" Body="&lt;p&gt;Call the logicians with white hats W1, W2 in that speaking partial order, and likewise B1, B2 for those with black hats. Without loss of generality, start with W1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;W1 sees 2B 1W, so his own could be R or W and says IDK. Reverse-engineering his answer and knowing W1's hat colour, W1 could say IDK only if he saw 1R 1W 1B or 2B 1W.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;W2 says W at his turn because he sees 2B and knows W1 also saw 2B (so W1 saw 2B 1W). Reverse-engineering his answer and knowing W2's hat colour, W2 could say W only if he saw 1R 2B or 1W 2B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B1 and B2 deduce W2 saw 2B by reverse-engineering, so each seeing 2W 1B says B. Reverse-engineering Bx's answers and knowing Bx's hat colour, Bx could say B only if he saw 2W 1B or 1R 1W 1B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;W1 sees 2B 1W and even with the other answers, cannot tell from the reverse-engineered answers of the others whether his own is R or W since all 3 other answers would have remained the same even if they saw W1 wearing R.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Answer: the first to speak cannot tell his own hat colour even when the other 3 had.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9422" LastActivityDate="2015-03-26T09:58:58.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11014" PostTypeId="2" ParentId="10417" CreationDate="2015-03-26T10:35:11.590" Score="1" Body="&lt;p&gt;If we consider a circle as a 2×2 size move and a line as 1×4 move , all possible game-ends would be one of the following (symmetry considered)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Rnorh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the number of tracings is always even , the game always ends when player 2 performs his last turn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as no one can play a recently played move , or touch a majority flipped coins , we guarantee no humdrumic loop repeating . game has an end after my last improvement .&lt;/p&gt;&#xA;" OwnerUserId="9517" LastActivityDate="2015-03-26T10:35:11.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11015" PostTypeId="1" AcceptedAnswerId="11169" CreationDate="2015-03-26T10:57:19.077" Score="-2" ViewCount="594" Body="&lt;p&gt;Two mathematicians are passing an exam, where the examiner informs them that he has chosen two &lt;strong&gt;different&lt;/strong&gt; numbers between 2 and 100, such that one is divisible by the other.&#xA;The examiner then tells the sum to the first mathematician A and the difference to the second mathematician B, in such a way that none knows which number has been whispered to other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two mathematicians start this discussion:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A: I don't know the numbers.&lt;/li&gt;&#xA;&lt;li&gt;B: I already knew this.&lt;/li&gt;&#xA;&lt;li&gt;A: I already knew that you knew this.&lt;/li&gt;&#xA;&lt;li&gt;B: I already knew that you knew that I knew that you didn't know.&lt;/li&gt;&#xA;&lt;li&gt;A: I still can't figure out what the two numbers are.&lt;/li&gt;&#xA;&lt;li&gt;B: Oops, I think I miscalculated my last statement!!! As a matter of fact, I didn't know this already!!! &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question: What are these two numbers again??&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For hints and notes, refer to the following links:&lt;br&gt;&#xA;  &lt;a href=&quot;https://puzzling.stackexchange.com/questions/9499/know-the-numbers-from-this-conversation&quot;&gt;Deducing Two Numbers based on their Difference and Ratio&lt;/a&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;https://puzzling.stackexchange.com/questions/9400/what-are-the-numbers&quot;&gt;What are the numbers?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9517" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-03-28T16:35:43.320" Title="Deduce the numbers (addition + subtraction)" Tags="|mathematics|logical-deduction|number-theory|" AnswerCount="1" CommentCount="12" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11017" PostTypeId="1" AcceptedAnswerId="11068" CreationDate="2015-03-26T12:21:19.027" Score="4" ViewCount="401" Body="&lt;p&gt;I had a sentence, but the letters in it were too close! They were born paired and every two letters became one. Here is my sentence with the letters still paired:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;䗋⛺峙峊杒➈樇⬲椠刑☢斅刓彞嬦&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you can't see it properly here are the Unicode code-points:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;17867&lt;br&gt;&#xA;  9978&lt;br&gt;&#xA;  23769&lt;br&gt;&#xA;  23754&lt;br&gt;&#xA;  26450&lt;br&gt;&#xA;  10120&lt;br&gt;&#xA;  27143&lt;br&gt;&#xA;  11058&lt;br&gt;&#xA;  26912&lt;br&gt;&#xA;  21009&lt;br&gt;&#xA;  9762&lt;br&gt;&#xA;  25989&lt;br&gt;&#xA;  21011&lt;br&gt;&#xA;  24414&lt;br&gt;&#xA;  23334&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The original sentence had twice as many characters. Can you inverse the pairing of the characters to get the original sentence?&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Hint 1&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The characters are not of importance but the code-points are.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Hint 3&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;ffff&lt;/code&gt; &amp;rarr; &lt;code&gt;刔刔&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;eeee&lt;/code&gt; &amp;rarr; &lt;code&gt;偼偼&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;ffee&lt;/code&gt; &amp;rarr; &lt;code&gt;刔偼&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;fefe&lt;/code&gt; &amp;rarr; &lt;code&gt;兇兇&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;feef&lt;/code&gt; &amp;rarr; &lt;code&gt;兇先&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user10203" LastEditorUserId="5373" LastEditDate="2015-03-26T23:23:34.790" LastActivityDate="2015-03-27T10:26:45.580" Title="Letters too close" Tags="|cipher|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="11018" PostTypeId="2" ParentId="10765" CreationDate="2015-03-26T12:28:08.190" Score="5" Body="&lt;p&gt;How about:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gray N turned.  So turning ==&gt; anagram.&#xA; So that would be :  Angry!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10294" LastEditorUserId="10294" LastEditDate="2015-03-26T12:51:32.670" LastActivityDate="2015-03-26T12:51:32.670" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11019" PostTypeId="2" ParentId="11009" CreationDate="2015-03-26T13:20:54.503" Score="16" Body="&lt;p&gt;Solved it :-)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Besides Unicode, there aren't many character sets that include all of these characters. However, &lt;strong&gt;Windows Latin 1&lt;/strong&gt; seems to work. In this encoding, the puzzle text consists of the following character codes (in hexadecimal): &lt;code&gt;a1 d7 d7 ad 9a 77 67 bf bf 75 fe 7d 30 e3 d7 2f 8c 7a 30 ee cf 96 fe 9e 77 f3 75 dc 84&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The keyboard has two problems. The first one is that it's emitting 7 bits per character instead of 8. When the bits are reassembled in 8-bit chunks, the result is gibberish. I wrote a Python script to reassemble the original text in 7-bit chunks and ended up with &lt;code&gt;Puzzling_on_stackKqGQC]OK?Sg?Mk\\B&lt;/code&gt; (encouraging, but still 50% garbage)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; So the other problem with the keyboard appears to be a loose connection that caused one of these bits to go missing. By adding an extra &lt;code&gt;1&lt;/code&gt; after bit #119, we can recover the whole message: &lt;code&gt;Puzzling_on_stackexchange_is_fun!&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Here's my Python script:&lt;br /&gt;&#xA; &lt;code&gt;thex = 'a1d7d7ad9a7767bfbf75fe7d30e3d72f8c7a30eecf96fe9e77f375dc84'&lt;/code&gt;&lt;br /&gt;&#xA; &lt;code&gt;tbin = ''.join(bin(int(thex[x:x+2],16))[2:].zfill(8) for x in range(0,len(thex),2))&lt;/code&gt;&lt;br /&gt;&#xA; &lt;code&gt;tbin = tbin[0:119] + '1' + tbin[119:]&lt;/code&gt;&lt;br /&gt;&#xA; &lt;code&gt;tplain = [chr(int(tbin[i:i+7],2)) for i in range(0,len(tbin)-1,7)]&lt;/code&gt;&lt;br /&gt;&#xA; &lt;code&gt;print ''.join(tplain)&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="40" LastActivityDate="2015-03-26T13:20:54.503" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11022" PostTypeId="1" CreationDate="2015-03-26T13:51:39.920" Score="7" ViewCount="1082" Body="&lt;p&gt;There are lots of very similar flags in the world, particularly the&#xA;ones that are composed of three horizontal or three vertical bars. If we restrict&#xA;ourselves to flags of one of these types, what is the longest path we&#xA;can make by changing only one color at a time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q1: What is the longest path of flags composed of three horizontal bars? &lt;br&gt;&#xA;Q2: What is the longest path of flags composed of three vertical bars?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specification of rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A path of length $n$ consists of flags $F_1, \ldots, F_n$&lt;/li&gt;&#xA;&lt;li&gt;Two flags are considered the same, if the colors of the corresponding bars only differ in shade and /or intensity.&lt;/li&gt;&#xA;&lt;li&gt;No path may visit the same flag twice: $F_i \ne F_j$ whenever $i \ne j$ &lt;/li&gt;&#xA;&lt;li&gt;Neighboring flags in the path (flags $F_i$ and $F_{i+1}$) must differ in the color of exactly one bar. &lt;/li&gt;&#xA;&lt;li&gt;Only flags of independent countries may show up in the path.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="60644" LastEditDate="2020-02-05T20:27:09.607" LastActivityDate="2020-02-05T20:27:09.607" Title="The longest path of flags" Tags="|pattern|geography|flags|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11025" PostTypeId="1" AcceptedAnswerId="11028" CreationDate="2015-03-26T14:23:47.197" Score="82" ViewCount="8940" Body="&lt;p&gt;The following shows a Linux command line with a &lt;code&gt;#&lt;/code&gt; prompt:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#&#xA;&#xA;&#xA;&#xA;&#xA;                            echo -e 'AAAAHHH!!!\a' &amp;gt; /dev/null&#xA;&#xA;&#xA;&#xA;&#xA;#&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It represents a famous line from a work of fiction.  What line is that?&lt;/p&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2018-10-16T20:50:18.207" Title="What famous line does this Linux command line represent?" Tags="|rebus|computer-puzzle|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11026" PostTypeId="1" AcceptedAnswerId="11033" CreationDate="2015-03-26T14:24:36.690" Score="15" ViewCount="1726" Body="&lt;p&gt;The ship of Theseus consists of &lt;em&gt;n&lt;/em&gt; parts.&#xA;The first part needs to be replaced every day, the second part every second day, the third part every three days and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theseus is concerned that replacing all parts on the same day may cause the ship to lose its identity, and become something different than the ship of Theseus.&#xA;But he has a strategy to avoid this situation: whenever all parts are scheduled for replacement on the same day, before starting to replace the old parts, he adds a new part to his ship, which only needs to be replaced in $n + 1$ days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ship originally consisted of only one part and today Theseus has announced that he's added the 100th part to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How old is the ship of Theseus?&lt;/p&gt;&#xA;" OwnerUserId="4394" LastActivityDate="2016-11-17T13:56:28.197" Title="How old is the ship of Theseus?" Tags="|mathematics|calculation-puzzle|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11028" PostTypeId="2" ParentId="11025" CreationDate="2015-03-26T14:35:24.070" Score="83" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;In space no one can hear you scream.&quot; The tagline for the Alien movie.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;echo -e 'AAAAHHH!!!\a'&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is a scream. (Added bonus of \a telling the system to beep.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;gt; /dev/null&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This redirects the output to the null device, effectively muting it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#&#xA;&#xA;&#xA;&#xA;&#xA;&#xA;&#xA;&#xA;&#xA;#&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Space!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2018-10-16T20:50:18.207" LastActivityDate="2018-10-16T20:50:18.207" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="11029" PostTypeId="2" ParentId="11025" CreationDate="2015-03-26T14:38:01.823" Score="13" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  In space, no one can hear you scream&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The scream gets written to /dev/null, which discards all data written to it. In other words no one receives the text, it gets lost 'in the void'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure what the beep represents though.&lt;/p&gt;&#xA;" OwnerUserId="10246" LastActivityDate="2015-03-26T14:38:01.823" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11030" PostTypeId="2" ParentId="6455" CreationDate="2015-03-26T14:48:47.413" Score="2" Body="&lt;ol&gt;&#xA;&lt;li&gt;is a pack of cards. There are four jacks or 'knaves', a knave is a villain. Hearts make up a quarter of the pack, that's pretty common. 'Ten' is the highest or 'end' numeral and you start with 'fifty-two (to)' of them.&#xA;SO if you're looking for a name, I suppose it would be 'Jack'.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="10656" LastActivityDate="2015-03-26T14:48:47.413" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11031" PostTypeId="2" ParentId="11026" CreationDate="2015-03-26T14:49:20.120" Score="6" Body="&lt;p&gt;Very old indeed - Theseus is effectively immortal, and his ship is almost certainly space-faring, as the number of days is rather more than the current age of the universe, and probably more than the expected life of most suns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual answer is more than $32\times 27\times 35\times 97\#$, where $97\#$ (97 primorial) is the product of all primes up to $97$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This number is the least common multiple of all the numbers up to $100$. However there are some extra multiples in there too, so the actual age is larger than this, although it is less than $100!$ (100 factorial).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By way of concrete example, when Theseus added the 22nd part, the ship was approximately 2 million years old.&lt;/p&gt;&#xA;" OwnerUserId="8875" LastActivityDate="2015-03-26T14:49:20.120" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11032" PostTypeId="2" ParentId="11026" CreationDate="2015-03-26T14:52:50.777" Score="5" Body="&lt;p&gt;The ship is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 214959977860203405582463952869483994880561&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;days old.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the nth part is added today, the (n+1)th part will be added after lcm(1…n) days. Thus the 100th part gets added on day number $\sum_{i=1}^{99} \mathrm{lcm}(1,\ldots i)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9471" LastActivityDate="2015-03-26T14:52:50.777" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11033" PostTypeId="2" ParentId="11026" CreationDate="2015-03-26T15:02:01.880" Score="12" Body="&lt;h2&gt;First parts&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Starting from the first day, Theseus has a &lt;strong&gt;single-part&lt;/strong&gt; ship: this means that this part can only work for a day before being replaced, so Theseus should replace the first part on the second day. Given that replacing the first and only part will let the ship &quot;lose its identity&quot;, then &lt;strong&gt;before replacing the first part, Theseus adds a second part&lt;/strong&gt;: so &lt;strong&gt;the second part of his ship is added on the second day&lt;/strong&gt;, before removing the first one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the second day, the first part has got to be changed again, but the second part has still got one day. On the third day the parts will be changed together, meaning that Theseus will have to add a &lt;strong&gt;third part&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Adding the nth part&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;When will Theseus have to change all the $n$ parts together again?&lt;/strong&gt; It is easy to understand that Theseus will have to change all the parts when the number of days becomes equal to $lcm(1, 2, 3, ..., n-1, n) + 1$, which is the least common multiple of the duration of each part plus one, because the first day the only existing part doesn't have to be changed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can write a table representing the day when Theseus has to change all the $n$ parts together based on $n$:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;n       day&#xA;&#xA;1       2&#xA;2       3&#xA;3       7&#xA;4       13&#xA;5       61&#xA;6       61&#xA;7       421&#xA;8       841&#xA;9       2521&#xA;10      2521&#xA;...     ...&#xA;60      9690712164777231700912801&#xA;61      591133442051411133755680801&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We can clearly see that this number becomes dramatically large when dealing with these amounts of parts.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Solution&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;When Theseus adds the $100$th part, this means that he already changed &lt;strong&gt;all&lt;/strong&gt; the parts together $99$ times (adding a new part every time), and that it has come the day to change all the parts together again. So the age $A$ of the ship can be obtained solving the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A = 1 + \sum \limits_{i=1}^{99} lcm(1, 2, ..., i) \simeq 2.149 \times 10^{41}$&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;The ship is $214959977860203405582463952869483994880562$ days old.&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Which is around $588.93$ billions of billions of billions of billions of years, WOW!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Program&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I created a small Python 2.7 script to calculate the number of days, here it is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from fractions import gcd&#xA;&#xA;def lcm(a, b):&#xA;    return (a/(gcd(a, b))*b)&#xA;&#xA;age = 1 &#xA;for i in xrange(1, 100):&#xA;    age += reduce(lcm, range(1, i+1))&#xA;&#xA;print &quot;Theseus' ship is&quot;, age, &quot;days old!&quot;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="10248" LastEditorUserId="10248" LastEditDate="2016-11-17T13:56:28.197" LastActivityDate="2016-11-17T13:56:28.197" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11034" PostTypeId="2" ParentId="183" CreationDate="2015-03-26T15:24:30.957" Score="10" Body="&lt;p&gt;I spent some time working on this puzzle after it appeared on &quot;Brooklyn Nine-Nine&quot; (if you want, you can watch Captain Holt describe the puzzle &lt;a href=&quot;https://www.youtube.com/watch?v=Cs-TGLxQfBM[/embed]&quot;&gt;here&lt;/a&gt;) and I wrote a detailed, illustrated solution here: &lt;a href=&quot;https://realviacauchy.wordpress.com/2015/03/14/i-am-the-puzzle-master/&quot;&gt;Island of Tyreses Solution&lt;/a&gt;. In this particular version I am attempting to find an islander, Diffy, who is either heavier or lighter than the other 11 islanders.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Lessons&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The final solution takes into account two things I learned from previous attempts:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;In a group of four, I can identify Diffy in two weighings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A. First, I set two islanders from the group against two known non-Diffys. If the see-saw tilts, I know that Diffy is one of these two. If the see-saw remains even, I know that Diffy is one of the other two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B. Now, I select one of the remaining two possible-Diffys and set him against a known non-Diffy. If the scale tilts, I have found Diffy. If the board remains even, I know that Diffy is that last remaining islander.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C. Alternatively, if the see-saw tilts in Step A, and you want to know whether DIffy is heavy or light, you could note the direction from Step A and put the two remaining possible-Diffys on the scale opposite one another. If the see-saw tilts in the same direction as Step A, then Diffy is the one still on the same side as he was during Step A. Otherwise, if the orientation of the see-saw changes, Diffy is on the other side.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;In a group of three I can identify Diffy in one weighing, as long as I have directional information. I will describe this in further detail under Use #3.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h2&gt;Solution&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://realviacauchy.files.wordpress.com/2015/03/islanders.png?w=300&amp;amp;h=113&quot; alt=&quot;All of the Islanders&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of lesson #1, I can split off four islanders before checking the rest. If Diffy is in that group of four, the first weighing will come out even, and I can now identify him from among those four with my two remaining moves. If Diffy is not in that group of four, I now have four islanders whom I can rule out and also use to tare my see-saw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for my first use of the see-saw, I weigh the eight remaining islanders against each other with four on each side.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Use #1&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://realviacauchy.files.wordpress.com/2015/03/seesaw11.png?w=300&quot; alt=&quot;Teeter Totter Use #1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’ve already outlined my plan if this first see-saw usage turns out even, so what’s next if it turns out odd? This is where the genius comes in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I now have some “directional information.” I will henceforth call whichever direction the see-saw tilted in Use 1 “Direction 1″ or “D1″ for short. I know that if Diffy is heavy, he is on the part of the see-saw that went down, and if if Diffy is light, he’s on the part of the see-saw that went up. If I move Diffy, the see-saw will change orientation! It has no choice because Diffy, and only Diffy, causes the see-saw to tilt. Also, remember lesson #2, I have directional information and one move after the current one, so I can totally take out three possible Diffys before the next use of the see-saw. I’ll need to use one of the islanders I ruled out in Use 1 in order to keep three islanders on each side.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Use #2&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://realviacauchy.files.wordpress.com/2015/03/seesaw2.png&quot; alt=&quot;Teeter Totter Use #2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Use #2 gives us an even see-saw we can find Diffy in the three we removed, but if it doesn’t, we need to pay attention to the direction that the see-saw moves. Did it move the same way as before, Direction 1, or did it change orientation to Direction 2? Our next choice will be based on the answer! If it moved in Direction 1, then we know that Diffy is not one of the islanders who switched sides for Use #2. If the see-saw moved in Direction 2, then Diffy is one of the side-switchers. Either way, we have got him down to being one of three or two. Use #3 is a little hard to generalize since it is different for each possibility.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Use #3&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;In the case where I have a group of three possible-Diffy islanders, two of those islanders were on the same side during Use #1, when the see-saw moved into D1. If I put one of these islanders on each side of the see-saw and the see-saw again moves into D1, then we know that Diffy is the islander on the original side. If the see-saw moves into D2, then we know that Diffy is on the opposite side of the see-saw. If the see-saw remains even, we know that Diffy is the third member of the group.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;All Mapped Out&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://realviacauchy.files.wordpress.com/2015/03/island-puzzle-solution-new-page.png&quot; alt=&quot;Weight-Obsessed Island Puzzle Solution&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10658" LastActivityDate="2015-03-26T15:24:30.957" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11037" PostTypeId="2" ParentId="11022" CreationDate="2015-03-26T15:33:25.457" Score="9" Body="&lt;p&gt;I counted 22 vertical flags (thanks to Nick2253 reminding me of Canada) and managed to put them in an order that appears to meet the rules..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3k7dd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The horizontal part:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/syCPT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-03-26T22:57:29.927" LastActivityDate="2015-03-26T22:57:29.927" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="11038" PostTypeId="2" ParentId="11022" CreationDate="2015-03-26T16:04:57.550" Score="6" Body="&lt;p&gt;Here's what I got for vertical stripes(assuming you may only change the color of 1 bar and 2 other bars must stay on their places):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/3YTuV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9352" LastActivityDate="2015-03-26T16:04:57.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11039" PostTypeId="2" ParentId="10963" CreationDate="2015-03-26T17:01:24.377" Score="1" Body="&lt;p&gt;How about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Trochee&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A trochee is a two-syllable word, consisting of first a stressed (or &quot;long&quot;) syllable and then an unstressed (&quot;short&quot;) syllable. Fast then slow, noise then silence, long then not quite as long, and the answer is a fit for the meter because &quot;trochee&quot; is itself a trochee. And of course, all of the two-syllable nouns in the riddle are trochees.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9523" LastActivityDate="2015-03-26T17:01:24.377" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11040" PostTypeId="2" ParentId="10990" CreationDate="2015-03-26T17:13:52.067" Score="0" Body="&lt;p&gt;Lets annotate an arrow as $n\mod 8$ that points in the direction $45°n $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that the value of all arrows must increase or decrease by at most 1 each horizontal or vertical step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After removing undesired directions from the borders, the general aspect of the grid looks like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GCcrV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After many attempts and some reasoning I couldn't find any logical way to make rows respectively incremental until they end up decremental in the last row. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conclusion:&lt;/strong&gt; there is no solution &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;proof:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the cricket is not necessarily to put anywhere in the grid , a loop can be symbolically representd as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{|c|c|c||c|c|c|}&#xA;\hline&#xA;. &amp;amp;\rightarrow&amp;amp; .&amp;amp; .&amp;amp; .&amp;amp; .\\\hline&#xA;\nearrow&amp;amp;x=\rightarrow&amp;amp; \searrow&amp;amp; .&amp;amp; .&amp;amp; .\\\hline&#xA;\uparrow &amp;amp;x&amp;amp; x&amp;amp; \searrow&amp;amp; .&amp;amp; .\\\hline&#xA;\nwarrow&amp;amp;x &amp;amp; x&amp;amp; x&amp;amp; \searrow&amp;amp; .\\\hline&#xA;.&amp;amp;\nwarrow &amp;amp;  x&amp;amp; x&amp;amp; x&amp;amp; \swarrow\\\hline&#xA;.&amp;amp;.&amp;amp;   \nwarrow&amp;amp; x = \leftarrow&amp;amp;  \swarrow&amp;amp; .\\\hline&#xA;.&amp;amp;.&amp;amp; .&amp;amp;   \nwarrow&amp;amp; .&amp;amp; .\\\hline&#xA;\end{array}&#xA;\qquad$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the dots are not a problem they can be multiple values of different choices , but $x$ must be defined to accomplish and finalise our grid , after some struggling with numbers i didnt find any such x values where :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{|c|c|c||c|c|c|}&#xA;\hline&#xA;. &amp;amp;0&amp;amp; .&amp;amp; .&amp;amp; .&amp;amp; .\\\hline&#xA;7=-1&amp;amp;x=0&amp;amp; 1&amp;amp; .&amp;amp; .&amp;amp; .\\\hline&#xA;6=-2 &amp;amp;x&amp;amp; x&amp;amp; 1&amp;amp; .&amp;amp; .\\\hline&#xA;5=-3&amp;amp;x &amp;amp; x&amp;amp; x&amp;amp; 1&amp;amp; .\\\hline&#xA;.&amp;amp;5 &amp;amp;  x&amp;amp; x&amp;amp; x&amp;amp; 1\\\hline&#xA;.&amp;amp;.&amp;amp;   5&amp;amp; x = 4&amp;amp;  3&amp;amp; .\\\hline&#xA;.&amp;amp;.&amp;amp; .&amp;amp;   5&amp;amp; .&amp;amp; .\\\hline&#xA;\end{array}&#xA;\qquad$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;at the level of upper rows x must be gradually incremental , and the lower rows $x$ must be respectively decremental where the anomaly appears !!!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Claification:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if we consider a series of consecutive numbers 0 1 2 .... n , the minimal length of this series that ensures a sudden decrementation is 4 . as following :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a b c d a b c d .... = 0 1 2 3 0 1 2 3 ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d\ \ ↓↑\ \ \ b$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this series incements in a specific row :&#xA;a a b c = 0 0 1 2&#xA;the following series decrements next arrow:&#xA;a d c c = 0 3 2 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this is rule fitting because a a , a d , b c , c c are different by one step at most , but ....  there is &lt;strong&gt;no way&lt;/strong&gt; to apply this on bigger series , like 0 1 2 3 4 , ... and more like length=8 directions series of the actual problem.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another assumtion is when two diagonal adjacent squares are 180° different that keeps the cricket in two squares loop but , the problem in the vertically adjacent square = $x$ there s nt such $x$ where $(x+1) mod 8 = 3$ and  $(x-1) mod 8= 7 = -1$&lt;/p&gt;&#xA;" OwnerUserId="9517" LastEditorUserId="9517" LastEditDate="2015-04-07T12:27:57.307" LastActivityDate="2015-04-07T12:27:57.307" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11041" PostTypeId="2" ParentId="10990" CreationDate="2015-03-26T17:38:34.810" Score="9" Body="&lt;p&gt;I am quite convinced that the answer is yes, the cricket will necessarily fall off. Though I am somewhat less convinced in my ability to construct a rigorous proof, I &lt;em&gt;think&lt;/em&gt; this might be a valid approach, though it probably could be written more rigorously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First classify a loop as a set of connected squares, two squares being connected if one is the first square in the others corresponding direction &lt;em&gt;OR&lt;/em&gt; in the opposite direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example the three red arrows would be connected:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{|c|c|c|}&#xA;\hline&#xA;\rightarrow&amp;amp;\rightarrow&amp;amp; \color{red}\rightarrow\\\hline&#xA;\nearrow&amp;amp; \color{red}\nearrow&amp;amp; \rightarrow\\\hline&#xA;\color{red}\uparrow&amp;amp;\nearrow&amp;amp; \nearrow\\\hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is relatively clear that if a cricket can stay on the board, the board must have a loop (though not if and only if). It is also clear that any loop must have at least one unused square in its interior, in other words the loop must be surrounding at least one square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe we can also show that given any loop, there must be a smaller loop with all points in its interior or part of it (smaller meaning it has less squares in its interior). My reasoning for this is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any square must either be in a loop, or follow a path to the edge of the grid.* Since a square in the interior of a loop cannot follow a path to the edge of the grid; it would first come to a square in the loop (it also could not 'cross diagonals', as this would violate the $45^\circ$ rule) it must exist as part of a smaller loop (which may or may not include part of the larger one).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that any valid loop must have at least one interior square, and that any valid loop must also have a valid smaller loop, it is clear that no valid loop can exist, as obviously continually decreasing the size of the loops would necessarily decrease the size of the interior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*I defined loops the way I did specifically so this would hold.&lt;/p&gt;&#xA;" OwnerUserId="150" LastEditorUserId="150" LastEditDate="2015-03-26T17:45:44.983" LastActivityDate="2015-03-26T17:45:44.983" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11042" PostTypeId="1" AcceptedAnswerId="11044" CreationDate="2015-03-26T17:43:36.143" Score="10" ViewCount="1684" Body="&lt;p&gt;(This is basically an extension of &lt;a href=&quot;https://puzzling.stackexchange.com/questions/10972/pi-day-puzzle-one-to-twenty&quot;&gt;$\pi$ Day puzzle one to twenty&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.tauday.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;$\tau$ is greater than $\pi$&lt;/a&gt; and $\tau&amp;gt;\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Create the numbers from $1$ to $20$ using only:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Tau ($\tau$, equivalent to $2\pi$)&lt;/li&gt;&#xA;&lt;li&gt;Basic arithmetic operations ($+-\times\div$)&lt;/li&gt;&#xA;&lt;li&gt;Square roots ($\sqrt{x}$ or $\sqrt[2]{x}$)&lt;/li&gt;&#xA;&lt;li&gt;Exponentiation ($x^y$)&lt;/li&gt;&#xA;&lt;li&gt;Negative tau ($-\tau$)&lt;/li&gt;&#xA;&lt;li&gt;Floor functions ($\lfloor x\rfloor$)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Anything not in this list is forbidden. You are not allowed to have negative signs outside of $\tau$ or not as an operation (E.g $-\lfloor\tau\rfloor$ is forbidden, but $\tau-\lfloor\tau\rfloor$ is allowed.) You are also not allowed to use parentheses, although $\lfloor x\rfloor$ can make a good substitute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some basic MathJaX syntax:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;$\tau, +, -, \times, \div, \sqrt{\tau^{\tau}}, \lfloor\tau\rfloor, \sqrt[2]{\tau}$&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;$\tau, +, -, \times, \div, \sqrt{\tau^{\tau}}, \lfloor\tau\rfloor, \sqrt[2]{\tau}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&lt;a href=&quot;http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference&quot;&gt;Some more&lt;/a&gt;&lt;/sup&gt;&lt;br&gt;&#xA;Remember the order of operations. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1 = $\tau\div\tau$ (Uses 2 $\tau$s, worse score)&lt;br&gt;&#xA;  1 = $\lfloor\sqrt{\sqrt\tau}\rfloor$ (Uses 1 $\tau$, better score)&lt;br&gt;&#xA;  &lt;br&gt;&#xA;  2 = $\tau\div\tau+\tau\div\tau$ (Uses 4 $\tau$s, worse score)&lt;br&gt;&#xA;  2 = $\lfloor\sqrt\tau\rfloor$ (Uses 1 $\tau$, better score)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Try and use the least $\tau$s possible.&lt;/p&gt;&#xA;" OwnerDisplayName="user10203" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.923" LastActivityDate="2020-01-03T05:03:14.217" Title="$\tau$ is greatest! $\tau$ is all (from 1 to 20)" Tags="|calculation-puzzle|formation-of-numbers|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11043" PostTypeId="2" ParentId="12" CreationDate="2015-03-26T17:57:40.573" Score="0" Body="&lt;p&gt;There are definitely a lot of Sudoku puzzles that require guessing.  By guessing I include selecting a box for which you have two potential numbers and using one of those numbers to see how far it will take you to solution. If you solve the puzzle you &quot;guessed&quot; correctly.  If you get a conflict, you go back and use the other number to solve the puzzle.  I have also seen puzzles where it did not matter.  You could chose either number and solve the puzzle. It is still guessing and I consider it a crappy puzzle.&lt;/p&gt;&#xA;" OwnerUserId="10666" LastActivityDate="2015-03-26T17:57:40.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11044" PostTypeId="2" ParentId="11042" CreationDate="2015-03-26T18:28:35.243" Score="9" Body="&lt;p&gt;I used 43 $\tau$ to get to 20.&lt;br&gt;&#xA;&lt;sup&gt;&lt;em&gt;(44 before GOTO 0's suggestion)&lt;/em&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 = $\lfloor\sqrt{\sqrt{\tau}}\rfloor$&lt;br&gt;&#xA; 2 = $\lfloor\sqrt{\tau}\rfloor$&lt;br&gt;&#xA; 3 = $\lfloor\sqrt{\tau}\rfloor + \lfloor\sqrt{\sqrt{\tau}}\rfloor$&lt;br&gt;&#xA; 4 = $\lfloor\sqrt{\tau}\rfloor \times \lfloor\sqrt{\tau}\rfloor$&lt;br&gt;&#xA; 5 = $\lfloor\tau\rfloor - \lfloor\sqrt{\sqrt{\tau}}\rfloor$&lt;br&gt;&#xA; 6 = $\lfloor\tau\rfloor$&lt;br&gt;&#xA; 7 = $\lfloor\tau\rfloor + \lfloor\sqrt{\sqrt{\tau}}\rfloor$&lt;br&gt;&#xA; 8 = $\lfloor\tau\rfloor + \lfloor\sqrt{\tau}\rfloor$&lt;br&gt;&#xA; 9 = $\lfloor\sqrt{\sqrt{\tau}} \times \tau\rfloor$&lt;br&gt;&#xA; 10 = $\lfloor\sqrt{\tau}^{\sqrt{\tau}}\rfloor$&lt;br&gt;&#xA; 11 = $\lfloor\tau\rfloor + \lfloor\tau\rfloor - \lfloor\sqrt{\sqrt{\tau}}\rfloor$&lt;br&gt;&#xA; 12 = $\lfloor\tau\rfloor + \lfloor\tau\rfloor$&lt;br&gt;&#xA; 13 = $\lfloor\tau\rfloor - \lfloor-\tau\rfloor$&lt;br&gt;&#xA; 14 = $\lfloor\tau\rfloor - \lfloor-\tau\rfloor + \lfloor\sqrt{\sqrt{\tau}}\rfloor$&lt;br&gt;&#xA; 15 = $\lfloor\tau\rfloor - \lfloor-\tau\rfloor + \lfloor\sqrt{\tau}\rfloor$&lt;br&gt;&#xA; 16 = $\lfloor\sqrt{\sqrt{\tau}}^{\tau}\rfloor - \lfloor\sqrt{\sqrt{\tau}}\rfloor$&lt;br&gt;&#xA; 17 = $\lfloor\sqrt{\sqrt{\tau}}^{\tau}\rfloor$&lt;br&gt;&#xA; 18 = $\lfloor\tau^{\sqrt{\sqrt{\tau}}}\rfloor$&lt;br&gt;&#xA; 19 = $\lfloor\tau \times \tau \div \lfloor\sqrt{\tau}\rfloor\rfloor$&lt;br&gt;&#xA; 20 = $\lfloor\tau^{\sqrt{\sqrt{\tau}}}\rfloor + \lfloor\sqrt{\tau}\rfloor$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-03-26T19:39:59.117" LastActivityDate="2015-03-26T19:39:59.117" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="11045" PostTypeId="2" ParentId="39" CreationDate="2015-03-26T18:36:56.373" Score="0" Body="&lt;p&gt;I imagine using gravity and the screen resting on a small fulcrum in the center, so that the screen tilts slightly to the right or left as the numbers accumulate. In this way the big numbers get arranged sequentially in a snaking pattern from bottom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a keyboard, the Left, Down and Right keys are in a line and the Up key is out of the way. This prevents me from pressing Up erroneously.&lt;/p&gt;&#xA;" OwnerUserId="10667" LastActivityDate="2015-03-26T18:36:56.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11046" PostTypeId="1" AcceptedAnswerId="11047" CreationDate="2015-03-26T18:44:50.773" Score="11" ViewCount="995" Body="&lt;p&gt;I've found a very interesting problem that I want to share with you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the planet Abbab, &lt;strong&gt;the alphabet consists only of the letters A and B&lt;/strong&gt;.&lt;br&gt;&#xA;All the sentences are written &lt;strong&gt;without spaces&lt;/strong&gt; between words, and each sequence of letters has a determined meaning.&lt;br&gt;&#xA;The postmen, though, &lt;strong&gt;refuse&lt;/strong&gt; to deliver &lt;strong&gt;messages which contain two consecutive A.&lt;/strong&gt;&lt;br&gt;&#xA;So, the inhabitants of Abbab have to &lt;strong&gt;manipulate their messages&lt;/strong&gt; in a way that prevents the presence of two consecutive A in their text. &#xA;Both the sender and the addressee know the algorithm, so there's nothing to decrypt.&#xA;Of course, their conversion method must be &lt;strong&gt;reversible&lt;/strong&gt;, otherwise the addressee wouldn't be able to read the original message!  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Show a possible conversion algorithm which can convert a message of $N$ letters into a sendable message of up to $\frac5 3N$ letters  &lt;/li&gt;&#xA;&lt;li&gt;A skilled programmer claims to have a very efficient program for long messages: every message of $N$ letters ($N\geq100$) can be converted into a sendable message of up to $\frac6 5 N$ letters.  Show that the software programmer is lying.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10615" LastEditDate="2015-06-03T02:12:59.097" LastActivityDate="2015-06-03T02:12:59.097" Title="The postmen of Abbab" Tags="|logical-deduction|computer-science|information-theory|" AnswerCount="2" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11047" PostTypeId="2" ParentId="11046" CreationDate="2015-03-26T19:05:53.807" Score="21" Body="&lt;p&gt;1.Split the input into blocks of size 3. Make the following encryption : &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;AAA : BBBBB&lt;/li&gt;&#xA;&lt;li&gt;AAB : BBBAB&lt;/li&gt;&#xA;&lt;li&gt;ABA : BBABB&lt;/li&gt;&#xA;&lt;li&gt;ABB : BABBB&lt;/li&gt;&#xA;&lt;li&gt;BAA : ABBBB&lt;/li&gt;&#xA;&lt;li&gt;BAB : BABAB&lt;/li&gt;&#xA;&lt;li&gt;BBA : ABBAB&lt;/li&gt;&#xA;&lt;li&gt;BBB : ABABB&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This leaves us with what to do with the last 1 or two possible characters. This is easily accomplished with:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A : A&lt;/li&gt;&#xA;&lt;li&gt;B : B&lt;/li&gt;&#xA;&lt;li&gt;AA : ABA&lt;/li&gt;&#xA;&lt;li&gt;AB : AB&lt;/li&gt;&#xA;&lt;li&gt;BA : BA&lt;/li&gt;&#xA;&lt;li&gt;BB : BB&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It is easy to see that this encryption is easily invertible, and that total length of encrypted message is at most 5N/3.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Number of N-length AB-strings is 2^N. Number of N-length AB-strings without AA as substring is F(N+2) where F(i) is the ith Fibonacci number. Since Fibonacci numbers grow as $\phi^N$ and $\phi^{(6/5)}&amp;lt;2$, the number of N-length AB-strings grows asymptotically faster than (6N/5)-length AB-strings without AA as substring. Hence the second claim is impossible.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9471" LastActivityDate="2015-03-26T19:05:53.807" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11048" PostTypeId="2" ParentId="10989" CreationDate="2015-03-26T19:40:53.050" Score="5" Body="&lt;p&gt;The password is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Billy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Group 1 is all names/phrases that can be said without completely closing your mouth/touching both of your lips together. Group 2 is all names/phrases where you are required to touch both of your lips together (M in Mother, B in Brother, the middle &quot;b&quot; in Husband, etc.)&lt;br/&gt;&lt;br/&gt;Given this connection, 4 of the 5 names (Lilly, Kelly, Ally, Nancy) match Group 1, while Billy is the only one that matches Group 2. Therefore, being the outlier, Billy must be the password.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9978" LastActivityDate="2015-03-26T19:40:53.050" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11049" PostTypeId="2" ParentId="10990" CreationDate="2015-03-26T20:43:00.863" Score="12" Body="&lt;p&gt;I agree the cricket must fall off.  This might be equivalent to KSab's answer, but it's a slightly different take (also, this might not be totally rigorous but I think it is pretty convincing).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By way of contradiction, suppose the cricket doesn't fall off and thus goes around in some sort of loop $L$.  There must be some number of squares on the interior of $L$ (not counting $L$ itself).  Choose a board that minimizes the area enclosed in $L$.  Notice $L$ either goes clockwise or counter-clockwise:  without loss of generality, assume clockwise.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now create a new board with all the arrows turned  $90^\circ$ clockwise.  This is still a valid board since adjacent squares rotated the same direction by the same amount, and therefore still differ by at most $45^\circ$.  However, now all the arrows of $L$ point directly towards the interior of $L$.  Thus, if the cricket starts inside of $L$, it will always be forced back inside of $L$.  Thus the cricket again never escapes, and the new loop that if follows must enclose a smaller area.&lt;/p&gt;&#xA;" OwnerUserId="2722" LastEditorUserId="2722" LastEditDate="2015-04-07T20:06:42.793" LastActivityDate="2015-04-07T20:06:42.793" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="11050" PostTypeId="2" ParentId="11042" CreationDate="2015-03-26T20:50:02.217" Score="8" Body="&lt;p&gt;I've got some different answers. For brevity, I didn't include the ones that were the same as (or very similar to) Ian's answers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\begin{align}4&amp;amp; =\lfloor\sqrt{\tau}^\sqrt{\sqrt{\tau}}\rfloor\\5&amp;amp; =\left\lfloor\sqrt{\tau}\times\lfloor\sqrt{\tau}\rfloor\right\rfloor\\8&amp;amp;=\left\lfloor\sqrt{\lfloor\sqrt{\tau}\rfloor^{\tau}}\right\rfloor\\10&amp;amp;=\left\lfloor\tau^{\sqrt{\sqrt{\sqrt{\tau}}}}\right\rfloor\\15&amp;amp;=\lfloor\tau\times\sqrt{\tau}\rfloor\\19&amp;amp;=\left\lfloor\sqrt{\sqrt{\tau}}^{\;\tau}+\sqrt{\sqrt{\tau}}\right\rfloor\\20&amp;amp;=\left\lfloor\lfloor\tau+\sqrt{\tau}\rfloor\times\sqrt{\tau}\right\rfloor\end{align}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My solution for $15$ saves one $\tau$ over Ian's answer. The rest use the same number.&lt;/p&gt;&#xA;" OwnerUserId="8776" LastEditorDisplayName="user10203" LastEditDate="2015-03-26T20:54:45.307" LastActivityDate="2015-03-26T20:54:45.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11051" PostTypeId="1" AcceptedAnswerId="11054" CreationDate="2015-03-26T20:54:57.970" Score="18" ViewCount="1480" Body="&lt;p&gt;I invented a novel encryption technique that hides a message in the below text box.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;yrcjprieadbbsrmtaiuooilyzbeioaouwdmriuoueblqjrd&#xA;dnknnuzwycgsdpcrijmpkejvqkelgtpixlwwegsgjintrdq&#xA;stvfudxwxywqnkgquvtmmuwpqcasrqqacsrtidgbhxavwgm&#xA;blyamkfpueeivdmzibglducsltugmrxivpdregxfhztegws&#xA;cnwvakhtffxomxlvijxgmofnzhuxpnbexjrcegcwbyernvs&#xA;cbgkhowklhdeqnqwigtlmuvmkpaykgmecnzrulzlcuhhfmn&#xA;jdbwqyaaeeuihhpzuueiaugbmhuuuauudvsraeauitvyktj&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hint about the message (in order to be less broad): After the necessary&#xA; decryption the message on your monitor can be seen from 10 metres.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4016" LastEditorUserId="4016" LastEditDate="2015-03-26T21:34:33.010" LastActivityDate="2015-03-26T22:05:59.050" Title="Box of text contains hidden message" Tags="|cipher|lateral-thinking|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="11052" PostTypeId="2" ParentId="10991" CreationDate="2015-03-26T21:27:02.573" Score="6" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Although this code looks similar to a Polybius cipher, I don't think it is one. Groups of &lt;strong&gt;3 digits&lt;/strong&gt; would use a 3x3x3 grid and the numbers would be from 1 to 3.  Numbers from 1 to 6 would use a 6x6 grid with groups of &lt;strong&gt;2 digits&lt;/strong&gt;. If a 6x6x6 grid was used, it would be impossible to decode.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It is notable that the code contains a majority of 1's and 2's with no 0's, 7's, 8's, or 9's. Perhaps the code is based on a system of 3 digit numbers from the &quot;real world&quot; like bus routes or office numbers or pages of a training manual.   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As you suggest, we could try finding 4 words that fit the ABCD EFG HI JEKLM pattern. An example is &lt;strong&gt;&quot;john was by twice&quot;&lt;/strong&gt; but it seems quite difficult to make a sensible sentence.   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;As noted by JLee, this code resembles a &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Book_cipher&quot; rel=&quot;nofollow noreferrer&quot;&gt;book cipher&lt;/a&gt;&lt;/strong&gt; but the name of a book (or some written document) is required to decode.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Several 3 digit codes are in common use including the US ZIP code prefixes and the Dewey Decimal system (North American area codes do not use 100 series numbers). The following table shows the meaning of each 3 digit code, however this does not help with decoding the cipher.  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rwmIA.png&quot; alt=&quot;3digitcodes&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-05-07T18:24:07.293" LastActivityDate="2015-05-07T18:24:07.293" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="11053" PostTypeId="2" ParentId="10963" CreationDate="2015-03-26T21:56:58.990" Score="3" Body="&lt;p&gt;It is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a shadow.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Like: arrow, turtle, Cello, Metro&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The shadow of each of these resembles the original.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;... with a silence echo&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This means the echo is silent - shadows are silent, leaving no echoes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;... not quite the rainbow&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rainbows are caused by bent light, not blocked light, so rainbows don't fit the usual way shadows are made.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Give me an answer that fit this riddle!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'Shadow' rhymes with 'echo' and 'rainbow'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Poorly rhymed&quot;: 'turtle' and 'riddle' doesn't quite rhyme with 'rainbow'. Also, &quot;stand under the sun&quot; - a good way to make a shadow.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9422" LastEditorUserId="9422" LastEditDate="2015-03-26T22:05:24.637" LastActivityDate="2015-03-26T22:05:24.637" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11054" PostTypeId="2" ParentId="11051" CreationDate="2015-03-26T22:02:15.100" Score="19" Body="&lt;p&gt;The solution is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; GOOD&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is obtained by:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Converting all the non-vowels into dots, and looking at the resulting block of text.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which gives:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;......iea.......aiuooi....eioaou....iuoue.......&lt;br&gt; .....u..........i....e....e....i....e....i......&lt;br&gt; ....u...........u....u....a....a....i.....a.....&lt;br&gt; ...a....ueei....i....u....u....i....e......e....&lt;br&gt; ....a......o....i....o....u....e....e.....e.....&lt;br&gt; .....o.....e....i....u....a....e....u....u......&lt;br&gt; ......aaeeui....uueiau....uuuauu....aeaui.......&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7308" LastActivityDate="2015-03-26T22:02:15.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11055" PostTypeId="2" ParentId="11022" CreationDate="2015-03-26T22:02:31.963" Score="2" Body="&lt;p&gt;I'm making the following assumptions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A flag is composed of three horizontal bars or three vertical bars when the flag &lt;em&gt;substantially&lt;/em&gt; (more than about 75%) is made up of three horizontal bars and three vertical bars.  For example, Equatorial Guinea and Barbados are included, and Zambia and Antigua &amp;amp; Barbuda are not included.&lt;/li&gt;&#xA;&lt;li&gt;Crests, shields, or symbols found on flags are ignored for both color and when considering if they have three horizontal or vertical bars.  For example, Canada, Israel, Libya are included.&lt;/li&gt;&#xA;&lt;li&gt;A horizontal or vertical bar, no matter how narrow, counts towards the bar count.  This disqualifies, for example, North Korea, Kenya, and the Gambia.&lt;/li&gt;&#xA;&lt;li&gt;The horizontal or vertical bars do not have to be the same size.  The count is all that matters.&lt;/li&gt;&#xA;&lt;li&gt;The horizontal or vertical bars do not have to be in the same order, as long as one color is different.  For example, El Salvador's Flag (Blue, White, Blue) is a neighbor to Estonia's Flag (Blue, Black, White) because there is one bar different (Blue vs Black).&lt;/li&gt;&#xA;&lt;li&gt;A sovereign nation is any state on &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_sovereign_states&quot; rel=&quot;nofollow noreferrer&quot;&gt;this list&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Countries that have crosses on their flag do not count towards either horizontal or vertical bars.  For example, Finland and Georgia are not included.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Under these rules, there are many solutions for the vertical-bar flags.  There are a total of 22 countries with vertical-bar flags.  In fact, you can actually make a continuous cycle of vertical-bar flags:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zGSKn.png&quot; alt=&quot;Vertical Flags&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm working on Horizontal flags, and I'll get back on that.&lt;/p&gt;&#xA;" OwnerUserId="4134" LastActivityDate="2015-03-26T22:02:31.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11056" PostTypeId="2" ParentId="11051" CreationDate="2015-03-26T22:03:59.147" Score="13" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The vowels encode the message as a picture.  Here they are all blacked out:&lt;br&gt;&lt;code&gt;yrcjpr███dbbsrmt██████lyzb██████wdmr█████blqjrd&lt;br&gt;dnknn█zwycgsdpcr█jmpk█jvqk█lgtp█xlww█gsgj█ntrdq&lt;br&gt;stvf█dxwxywqnkgq█vtmm█wpqc█srqq█csrt█dgbhx█vwgm&lt;br&gt;bly█mkfp████vdmz█bgld█cslt█gmrx█vpdr█gxfhzt█gws&lt;br&gt;cnwv█khtffx█mxlv█jxgm█fnzh█xpnb█xjrc█gcwby█rnvs&lt;br&gt;cbgkh█wklhd█qnqw█gtlm█vmkp█ykgm█cnzr█lzlc█hhfmn&lt;br&gt;jdbwqy██████hhpz██████gbmh██████dvsr█████tvyktj&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10649" LastEditorUserId="7308" LastEditDate="2015-03-26T22:05:59.050" LastActivityDate="2015-03-26T22:05:59.050" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11057" PostTypeId="1" AcceptedAnswerId="11060" CreationDate="2015-03-26T22:04:56.903" Score="12" ViewCount="7728" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Is it possible to tile a &lt;span class=&quot;math-container&quot;&gt;$10\times10$&lt;/span&gt; chessboard with (non-overlapping) T-tetrominos?&lt;/p&gt;&#xA;&lt;p&gt;If so, how? If not, prove it's impossible.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Bonus:&lt;/strong&gt; Which Tetris pieces can used to tile a 10&lt;span class=&quot;math-container&quot;&gt;$\times$&lt;/span&gt;10 board, allowing reflections?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;It is certainly posssible to tile a regular &lt;span class=&quot;math-container&quot;&gt;$8\times 8$&lt;/span&gt; chessboard. Below is a failed attempt to extend that tiling to a &lt;span class=&quot;math-container&quot;&gt;$10\times 10$&lt;/span&gt; board, which misses some squares in the upper left and lower right. The colors don't matter, they're just there to make the picture more clear.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\qquad\qquad\qquad\qquad$&lt;/span&gt;&lt;img src=&quot;https://i.stack.imgur.com/EQy8j.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2018-10-31T01:48:35.893" Title="Tiling a Chessboard with tetrominos" Tags="|checkerboard|tiling|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11058" PostTypeId="2" ParentId="10990" CreationDate="2015-03-26T22:07:07.270" Score="3" Body="&lt;p&gt;If we didn't have to put an arrow in the middle of a 3x3 board, we could send the cricket round in circles as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8jFDA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But of course the middle arrow then doesn't have a direction that is consistent with the rules. And any attempt to set up circulation on a larger grid will have the same problem, it seems. There will be a central point that does not have a valid arrow direction due to the conflicting values nearby.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly (or even more so) with a grid that attempts to have inward movement; this cannot be carried on to a central point, so must reduce to a circulation, with the same results as above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is thus no possible configuration of arrows that allows the cricket to arrive at a previously visited square on the board, so yes, the cricket will necessarily fall off the board.&lt;/p&gt;&#xA;" OwnerUserId="8875" LastEditorUserId="8875" LastEditDate="2015-03-26T22:12:37.813" LastActivityDate="2015-03-26T22:12:37.813" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11059" PostTypeId="1" AcceptedAnswerId="11065" CreationDate="2015-03-26T22:09:25.053" Score="7" ViewCount="768" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;0870678884068983409815824039880670002291&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Solving this requires external resources.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The fact that both the problem and the answer are expressed as numbers is largely a red herring.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9523" LastActivityDate="2015-03-27T07:49:51.460" Title="The answer is a two-digit number" Tags="|riddle|lateral-thinking|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="11060" PostTypeId="2" ParentId="11057" CreationDate="2015-03-26T22:30:31.650" Score="24" Body="&lt;p&gt;It is not possible.  The area of a $10 \times 10$ checkerboard is $100$, so it takes $25$ T pieces to have the same area.  The checkerboard has the same number of red and black squares, but each piece covers three of one color and one of the other.  $25$ pieces cannot cover $50$ squares of each color, the most even they can get is $51-49$&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2015-03-26T22:30:31.650" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11061" PostTypeId="2" ParentId="11057" CreationDate="2015-03-26T23:02:05.573" Score="14" Body="&lt;p&gt;For additional reference, here are tilings of the other tetrominos or proofs that it's impossible.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;1. 2x2 blocks (O pieces)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This one is easy &amp;mdash; a 5x5 arrangement of 2x2 blocks tiles a 10x10 checkerboard perfectly.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;2. L-shaped blocks (L pieces)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Suppose the checkerboard is painted alternating white and black by columns rather than squares. Then each L piece occupies either 3 white squares and one black square, or three black squares and one white square, regardless of its rotation state:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x . .    . . .    x o .    . . .&#xA;x . .    . . x    . o .    x o x&#xA;x o .    x o x    . o .    x . .&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;By the same argument as the T-pieces when the checkerboard was painted in a checker pattern, you can't have 25 of the these pieces on the board, because you could only get a 51/49 split.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;3. 4x1 blocks (I pieces)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The area of a 10x10 checkerboard can be divided into four colours of diagonal stripes, starting from the top left and working your way down to the bottom-right:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x o + . x o + . x o&#xA;o + . x o + . x o +&#xA;+ . x o + . x o + .&#xA;. x o + . x o + . x    1. x&#xA;x o + . x o + . x o    2. o&#xA;o + . x o + . x o +    3. +&#xA;+ . x o + . x o + .    4. .&#xA;. x o + . x o + . x&#xA;x o + . x o + . x o&#xA;o + . x o + . x o +&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;These colours have the following numbers of squares:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1. 1 + 5 + 9 + 7 + 3 = 25&#xA;2. 2 + 6 + 10 + 6 + 2 = 26&#xA;3. 3 + 7 + 9 + 5 + 1 = 25&#xA;4. 4 + 8 + 8 + 4 = 24&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But notice that any 4x1 block in the checkerboard must contain exactly one square of each colour, so 25 of them would occupy 25 of each. This is impossible, so the board cannot be tiled with 4x1 blocks.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;4. Squiggly blocks (S pieces)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This one is a case analysis. Consider the S piece that covers the bottom-left square of the checkerboard. It must be in one of these two configurations:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;. . . .     . . . .&#xA;. x . .     . . . .&#xA;x x . .     . x x .&#xA;x . . .     x x . .&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Since these configurations are symmetric across the main diagonal, they're equivalent. So, consider the latter case. Then, the only place an S piece could go to cover the crevice in the third square on the bottom is in the exact same orientation:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;. . . . . .&#xA;. . . . . .&#xA;. x x o o .&#xA;x x o o . .&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And the only place that another S piece could go to cover &lt;em&gt;that&lt;/em&gt; crevice is to the right of that, et cetera, until they stick out of the board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can actually use this to prove that the S piece can't tile &lt;em&gt;any&lt;/em&gt; rectangular checkerboard perfectly. Moral of the story: S pieces suck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But!&lt;/strong&gt; An interesting property of this proof is that it doesn't apply to &lt;em&gt;toroidal&lt;/em&gt; checkerboards. All of the other invariants continue to apply even when the board wraps around on itself, but the S piece can indeed tile the grid perfectly in this case.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So it turns out that out of all the tetrominos, only the square can tile a 10x10 grid perfectly.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-03-26T23:55:39.453" LastActivityDate="2015-03-26T23:55:39.453" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11062" PostTypeId="1" AcceptedAnswerId="11083" CreationDate="2015-03-26T23:02:43.673" Score="12" ViewCount="424" Body="&lt;p&gt;These thirteen English words share a hidden property:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;GULLS, FUSION, BOMB, MILLS, DON, FUR, CHAIN, MOCHA, HYMN, WOLFS, MUNCH, PINY, WITS&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The list is not complete; there are other words with the same property, but most English words don't fit the pattern. Can you name two other words of at least three letters with the same property?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This puzzle was inspired by &lt;a href=&quot;https://puzzling.stackexchange.com/users/9910/moghwyn&quot;&gt;Moghwyn&lt;/a&gt;'s puzzle &lt;strong&gt;&lt;em&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/10527/a-company-of-thirteen&quot;&gt;A Company of Thirteen&lt;/a&gt;&lt;/em&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: The solution to this puzzle is not related to that solution.  The only similarity is that they both contain a group of thirteen words.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2016-09-18T20:32:18.957" Title="Another Company of Thirteen" Tags="|pattern|word-property|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11063" PostTypeId="2" ParentId="11046" CreationDate="2015-03-26T23:32:02.557" Score="3" Body="&lt;p&gt;We can write an algorithm which is, in some sense, optimal - reaching, in the limiting case (since this number is &lt;em&gt;obviously&lt;/em&gt;$^*$ irrational) the bound that at least $\log_{\phi}(2)N$ characters must be used for infinitely many input words - which is implied by the fact that, as mentioned in a previous answer, there are $F_n\sim \phi^n$ legal words of length $n$, but $2^n$ possible words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very simple method is this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Order all the possible message by their length; i.e. to each message word, assign a natural number such that longer messages have higher numbers. Do the same for the legal messages. Find where your message lies in the list - send the corresponding legal message; i.e. if you want to send the $n^{th}$ message, encode it as the $n^{th}$ &lt;em&gt;legal&lt;/em&gt; message.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's obvious that this is effective, since we're essentially constructing a bijection from . In fact, any method can be thought of as an injection from the set of messages to the set of legal messages, and it is easily shown$^{**}$ that any other injection aside from the one above has that the there is some $n$ such that the $n^{th}$ message maps to the $m^{th}$ legal message where $m&amp;gt;n$ - so any other algorithm performs at least as bad on &lt;em&gt;at least&lt;/em&gt; one word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, one can see that saying a message $M$ is the $n^{th}$ message if the binary representation of $n$ with its most significant bit dropped is the same as the message $M$ with $A$'s replaced by $1$ and $B$'s by $0$. So &quot;ABBABA&quot; (which means &quot;hello&quot; in Abbab) encodes as the $1100101_2=101^{st}$ word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, our method for determining the $n^{th}$ legal message is very similar, but instead of using binary, we use a positional system where the place values are the Fibonacci numbers rather than powers of two - in particular, &lt;a href=&quot;http://en.wikipedia.org/wiki/Zeckendorf%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;every number can be written uniquely as the sum of distinct non-consecutive Fibonacci numbers&lt;/a&gt;. Letting $F_n$ be the greatest number included in such a sum, we simply put an $A$ for each time we see an element of the sequence $F_{n-1},F_{n-2},\ldots,F_1$ included in the sum and a $B$ otherwise. For instance, the $101^{st}$ legal word can be found as&#xA;$$101=89+8+3+1=F_{10}+F_5+F_3+F_1$$&#xA;which encodes as, starting by noting that $F_9$ through $F_6$ are &lt;em&gt;not&lt;/em&gt; included: &quot;BBBBABABA&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as long as the residents of Abbab have too much time or the ability to program computers (in esoteric languages of course - they want to send their code in the mail unencrypted, so it looks kind of weird), they can use this system which will outperform any replacement algorithm (since their efficiency is &lt;em&gt;rational&lt;/em&gt; and no better than $\log_{\phi}(2)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^*$Were it rational, $\phi^a=2^b$ for integers $a$ and $b$, but $\phi^a$ is only rational if $a=0$ which implies $b=0$, which does not correspond to a rational number $a/b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^{**}$Suppose this were not true; then, the $n^{th}$ message maps to the $m^{th}$ legal message. However, the first message must then map to the first legal one - and the second message must map to the second, since the first was already taken. Proceeding inductively, we prove that the $n^{th}$ message maps to the $n^{th}$ legal message.&lt;/p&gt;&#xA;" OwnerUserId="4765" LastActivityDate="2015-03-26T23:32:02.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11064" PostTypeId="2" ParentId="11017" CreationDate="2015-03-26T23:59:21.937" Score="2" Body="&lt;p&gt;First post to Puzzling, so forgive me if adding hints/progress is taboo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some interesting notes from the Unicode code points from the Hint #3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 刔 (ff) has Unicode 0x5214, or 21012 decimal&lt;br&gt;&#xA; 偼 (ee) has Unicode 0x507C, or 20604 decimal&lt;br&gt;&#xA; 兇 (fe) has Unicode 0x5147, or 20807 decimal&lt;br&gt;&#xA; 先 (ef) has Unicode 0x5148, or 20808 decimal&lt;br&gt;&#xA;&lt;br&gt;&#xA; The difference between the (ef) glyph and the (fe) glyph is 1. The difference between the (ee) and (ef) glyph, and the (ef) and (ff) glyphs are both 0xCC, or 204 decimal.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Probably irrelevant is the fact that 0xCC in binary is 1100 1100. The first three bits of those each map to 6 6, and 66 is the hex code for lowercase f in UTF-8. Also possibly irrelevant or coincidental is that the decimal values for these glyphs are surprisingly symmetric.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These hints probably haven't brought me much closer, but maybe they'll help someone brighter than I converge on the solution.&lt;/p&gt;&#xA;" OwnerUserId="5162" LastActivityDate="2015-03-26T23:59:21.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11065" PostTypeId="2" ParentId="11059" CreationDate="2015-03-27T00:56:25.950" Score="12" Body="&lt;p&gt;Thanks to reading a lot of books and hint #2, the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 18&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The clue is made up of &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4 ISBN-10 numbers &lt;br/&gt;0870678884 - How to Win &lt;br/&gt;0689834098 - You Have to Write&lt;br/&gt;1582403988 - Six&lt;br/&gt;0670002291 - Times Three&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Great puzzle!&lt;/p&gt;&#xA;" OwnerUserId="1777" LastActivityDate="2015-03-27T00:56:25.950" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11066" PostTypeId="1" AcceptedAnswerId="11090" CreationDate="2015-03-27T01:07:13.347" Score="8" ViewCount="1940" Body="&lt;p&gt;I'm surprisingly standard, &lt;br&gt;&#xA;But can ban if required. &lt;br&gt;&#xA;Found in tea, less often than more, &lt;br&gt;&#xA;I might also clear your sock drawer.&lt;br&gt;&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2015-03-27T01:12:43.963" LastActivityDate="2015-03-27T16:01:36.173" Title="Riddle - Socks that may or may not be puppets" Tags="|riddle|word|" AnswerCount="9" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11067" PostTypeId="2" ParentId="11066" CreationDate="2015-03-27T02:08:53.920" Score="3" Body="&lt;h1&gt;Answer:&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Caffeine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;Explanation&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I'm surprisingly standard,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You find it everywhere--in coffee, tea, and some sodas.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;BUt can ban if required.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It &lt;em&gt;is&lt;/em&gt; technically a drug--so ODing on it could result in a ban?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Found in tea, less often than more, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is found in small doses (if present at all) in tea.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might also clear your sock drawer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We can take &quot;sock drawer&quot; as a figurative reference to the mind (as it is often used). Then, caffeine can &quot;clear&quot; your mind, allowing you to stay awake for a longer period of time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7307" LastActivityDate="2015-03-27T02:08:53.920" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11068" PostTypeId="2" ParentId="11017" CreationDate="2015-03-27T05:07:57.937" Score="10" Body="&lt;p&gt;Each pair of characters is encoded by this process:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Find the sum of their Unicode code points. (e=101, f=102 → sum=203)&lt;/li&gt;&#xA;&lt;li&gt;Square the sum, add the sum, and divide by 2. (203*203=41209, 41209+203=41412, 41412/2=20706)&lt;/li&gt;&#xA;&lt;li&gt;Add the code point of the second character. (20706+102=20808)&lt;/li&gt;&#xA;&lt;li&gt;Find the character with that number as its code point.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To decode:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Find the sum of the original characters by multiplying the new character by 2, taking the square root and rounding to the nearest whole number, then subtracting 1. (17867*2=35734, sqrt(35734)=189.03→189, 189-1=188)&lt;/li&gt;&#xA;&lt;li&gt;Find the second character by calculating what was added in the 3rd part of the encoding process. (188*188=35344, 35344+188=35532, 35532/2=17766, 17867-17766=101 → second character is 101=e)&lt;/li&gt;&#xA;&lt;li&gt;Find the first character by subtracting the second character from the sum. (188-101=87 → first character is 87=W)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Applying this to the given message results in the original text: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We letters must stick together&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10677" LastEditorUserId="10534" LastEditDate="2015-03-27T10:26:45.580" LastActivityDate="2015-03-27T10:26:45.580" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11069" PostTypeId="1" AcceptedAnswerId="11121" CreationDate="2015-03-27T06:28:16.800" Score="18" ViewCount="3466" Body="&lt;p&gt;Ladies and gentlemen,&lt;br&gt;&#xA;I will defend you.&lt;br&gt;&#xA;I am not a comedian,&lt;br&gt;&#xA;but will entertain you.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can’t be dangerous,&lt;br&gt;&#xA;yet my half-name is.&lt;br&gt;&#xA;Can you name me the hero&lt;br&gt;&#xA;of this colorful quiz?  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Time for the first hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am good at bringing joy,&lt;br&gt;&#xA; maybe that’s why they call me toy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Btw. one of you all is very very close to the answer. &lt;/p&gt;&#xA;" OwnerUserId="5283" LastEditorUserId="5283" LastEditDate="2015-03-27T18:43:19.587" LastActivityDate="2015-03-27T21:04:38.070" Title="Riddling for fun" Tags="|riddle|rhyme|" AnswerCount="12" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11070" PostTypeId="2" ParentId="11069" CreationDate="2015-03-27T07:12:50.290" Score="4" Body="&lt;p&gt;A very wild guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Qualcomm_Snapdragon&quot; rel=&quot;nofollow&quot;&gt;Snapdragon SOC&lt;/a&gt; that is used in phones, GPS, car infotaintment systems etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ladies and gentlemen,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Targets both&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I will defend you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Saves through GPS, presenting info, timers, processing sensors to avoid accidents etc&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am not a comedian,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is just an SOC&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but will entertain you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Games, videos, audios etc&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can’t be dangerous,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Almost never&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yet my half-name is.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; snap + dragon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can you name me the hero&lt;/p&gt;&#xA;&#xA;&lt;p&gt;of this colorful quiz?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; snapdragon plants are quite colorful. Moreover the graphics presented by this can be colorful.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Big chances are this is not the answer that was in your mind, but if fits.&lt;/p&gt;&#xA;" OwnerUserId="4637" LastActivityDate="2015-03-27T07:12:50.290" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11071" PostTypeId="2" ParentId="11069" CreationDate="2015-03-27T07:45:41.407" Score="3" Body="&lt;p&gt;My wild guess would be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Painball gun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ladies and gentlemen,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Targets both&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I will defend you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You use it to defend yourself&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am not a comedian,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Of course it's not!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but will entertain you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Isn't it fun? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can’t be dangerous,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They can hurt, but they aren't dangerous(unless you do some very stupid things)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yet my half-name is.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ok, I didn't understand this line :(&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can you name me the hero&lt;br&gt;&#xA;of this colorful quiz?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Paintball guns use those paint pellets and they are normally of different colors ==&gt; they are colorful&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9352" LastActivityDate="2015-03-27T07:45:41.407" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11073" PostTypeId="2" ParentId="11069" CreationDate="2015-03-27T08:01:55.290" Score="4" Body="&lt;p&gt;This is a wild guess but what about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rainbow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ladies and gentlemen,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Just a greeting&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I will defend you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In Game of Thrones there is a Rainbow guard used to protect king Renly &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am not a comedian,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is not&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but will entertain you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everyone like to look at the rainbow. Double rainbow all the way&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can’t be dangerous,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rainbow can't hurt you&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yet my half-name is.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Half name is bow - which is certainly dangerous&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can you name me the hero&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I just did&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;of this colorful quiz?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Rainbow is colorful&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9907" LastActivityDate="2015-03-27T08:01:55.290" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11074" PostTypeId="2" ParentId="11066" CreationDate="2015-03-27T08:10:57.053" Score="0" Body="&lt;p&gt;I think answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nicotine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm surprisingly standard&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; we hardly can find it in stores&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But can ban if required. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; govt might ban foods/drinks which contains it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Found in tea, less often than more,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; tea contains it in very small amount&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might also clear your sock drawer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if taken in large quantity blocks the action of autonomic nerve,smaller amount may simulate brains clearing sock drawer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How about that???&lt;/p&gt;&#xA;" OwnerUserId="10433" LastEditorUserId="5283" LastEditDate="2015-03-27T08:30:13.890" LastActivityDate="2015-03-27T08:30:13.890" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11075" PostTypeId="2" ParentId="11066" CreationDate="2015-03-27T08:24:38.273" Score="4" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Honey&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm surprisingly standard,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In a lot of standard family stories, the husband calls his wife honey.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But can ban if required.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The wife could restrict her husband or children from doing something.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Found in tea, less often than more, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people add honey to their tea, instead of the more common sugar.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might also clear your sock drawer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your wife or mother might clear your sock drawer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10325" LastActivityDate="2015-03-27T08:24:38.273" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11076" PostTypeId="2" ParentId="11066" CreationDate="2015-03-27T08:54:30.570" Score="3" Body="&lt;p&gt;Ok i want to give another try/(last try may be)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    Sugar&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm surprisingly standard,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    Sugar is surprisingly standard,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But can ban if required.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    Everyone knows which sugar is banned (br*wn sugar)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Found in tea, less often than more,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    some people avoid it in tea,some may like tea with more sugar&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might also clear your sock drawer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    May be Sugar stimulants used to clear sock drawer..&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10433" LastEditorDisplayName="user20" LastEditDate="2015-03-27T09:12:13.287" LastActivityDate="2015-03-27T09:12:13.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11077" PostTypeId="2" ParentId="11066" CreationDate="2015-03-27T09:06:48.620" Score="4" Body="&lt;p&gt;I'll have a go at this and guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A string&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm surprisingly standard&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A piece of string is about as 'standard' as things get.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But can ban if required. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In certain situations, anything with strings is prohibited for safety reasons (plenty murder stories with guitar/piano strings, ...)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Found in tea, less often than more,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; String from a teabag&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might also clear your sock drawer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well, if you find too many strings in your sock drawer, your socks have seen better days, and should be cleared from the drawer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Just a guess ofcourse, I think so far most of us interpret the second line as 'being banned' then actually 'banning', so in doubt there.&lt;/p&gt;&#xA;" OwnerUserId="2156" LastActivityDate="2015-03-27T09:06:48.620" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11078" PostTypeId="2" ParentId="11069" CreationDate="2015-03-27T09:10:07.087" Score="17" Body="&lt;p&gt;My answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Condom&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ladies and gentlemen&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There are male and female condoms&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I will defend you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Protect from STIs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am not a comedian&#xA;but will entertain you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Many are designed specifically to increase pleasure&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can’t be dangerous,&#xA;yet my half-name is.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A 'con' is certainly a negative/dangerous thing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can you name me the hero&#xA;of this colorful quiz?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Condoms come in many colors&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9740" LastActivityDate="2015-03-27T09:10:07.087" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11079" PostTypeId="2" ParentId="11066" CreationDate="2015-03-27T09:26:50.240" Score="5" Body="&lt;p&gt;My guess would be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alcohol!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm surprisingly standard,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alcohol is very standard&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But can ban if required.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Prohibition of alcohol in the US in 1920's&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Found in tea, less often than more,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sometimes(less often than more often) people drink tea with some sort of alcohol in it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might also clear your sock drawer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If we use it as a figurative reference to the mind(as &lt;em&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/users/7307/c%E1%B4%8F%C9%B4%E1%B4%8F%CA%80-ob%CA%80%C9%AA%E1%B4%87%C9%B4&quot;&gt;Cᴏɴᴏʀ O'Bʀɪᴇɴ&lt;/a&gt;&lt;/em&gt; suggested), then small doses of alcohol can clear your mind (large doses will mess it up tho')&lt;br&gt;&#xA; Another figurative meaning of &lt;em&gt;Sock Drawer&lt;/em&gt; could be &lt;a href=&quot;http://www.urbandictionary.com/define.php?term=sock%20drawer&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;an area in your closet designated for pornography storage. Could contain socks&lt;/em&gt;&lt;/a&gt;, so with alcohol you can get rid of the stains&lt;br&gt;&#xA; Or if it's the real sock drawer, you may use alcohol to disinfect/clear the drawer (if you are the dirty one, haha)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9352" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-03-27T10:28:05.833" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11080" PostTypeId="2" ParentId="11069" CreationDate="2015-03-27T09:50:48.110" Score="26" Body="&lt;p&gt;Think it must be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Magnetosphere&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ladies and gentlemen&#xA;I will defend you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Protect us from solar wind&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am not a comedian&#xA;but will entertain you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Northern lights are very entertaining&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can’t be dangerous,&#xA;yet my half-name is.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Magneto is a comic book villain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can you name me the hero&#xA;of this colorful quiz?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Northern lights are very colorful&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9740" LastActivityDate="2015-03-27T09:50:48.110" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11081" PostTypeId="1" AcceptedAnswerId="11087" CreationDate="2015-03-27T10:32:53.013" Score="6" ViewCount="295" Body="&lt;p&gt;Many words are balanced, but most are not.&#xA;All of the following are examples of balanced words&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;GOLD&#xA;LEMONS&#xA;UNCOOKED&#xA;BORDERLINE&#xA;LILLIPUTIANS&#xA;COUNTERBALANCE&#xA;PREDETERMINATION&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;while none of the following are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;GEMS&#xA;ORANGE&#xA;DIGESTED &#xA;UNDENIABLE&#xA;UNNOTICEABLE&#xA;COUNTERMEASURE&#xA;UNCOMPROMISINGLY&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only one the following is a balanced word; which one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BACKGAMMON, CHECKERS, CHESS, CRIBBAGE, DOMINOES&lt;/p&gt;&#xA;" OwnerUserId="150" LastEditorUserId="150" LastEditDate="2015-03-27T11:54:26.657" LastActivityDate="2015-03-27T12:04:20.627" Title="Balancing Words" Tags="|pattern|word-sequence|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11082" PostTypeId="1" AcceptedAnswerId="11084" CreationDate="2015-03-27T11:01:06.697" Score="19" ViewCount="3393" Body="&lt;p&gt;I can get you to many places you want to be.&lt;br&gt;&#xA;But you have to give me a pull or push when I meet T.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People do not write on me, unless I am accompanied by D.&lt;br&gt;&#xA;In that case, my name is often written together with E.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of my kind cannot swim.&lt;br&gt;&#xA;But when I join P, we shall live happily in a stream.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to work with me, all you need is a key.&lt;br&gt;&#xA;But when I am following S, you may want to get rid of me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="9944" LastEditorUserId="2521" LastEditDate="2015-03-28T05:43:11.703" LastActivityDate="2017-05-30T04:52:35.070" Title="I can get you to many places. Guess what am I?" Tags="|riddle|rhyme|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11083" PostTypeId="2" ParentId="11062" CreationDate="2015-03-27T11:29:40.080" Score="11" Body="&lt;p&gt;Possible other words include:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; AILS, ALUM, BOG, BOY, BUFF, BULLS, BUM, DUG, FILLS, FIN, GOLEM, LION, MIX, SUE, SYNC, WILY, WOMBS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;as well as many other more obscure ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The property that all of the words have is that &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if you increment each letter by 6 (so A→G and Y→E) the resulting string is also an English word. This is also known as a &lt;a href=&quot;http://en.wikipedia.org/wiki/Caesar_cipher&quot;&gt;Caesar cipher&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="150" LastActivityDate="2015-03-27T11:29:40.080" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11084" PostTypeId="2" ParentId="11082" CreationDate="2015-03-27T11:32:11.147" Score="16" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CAR&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is the answer. Add T and you get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CART &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is pushed or pulled. Add D and you get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CARD &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which people write on.  If you are writing one of those you likely &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CARE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;about the person. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CARP&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;swim. You want to get rid of a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SCAR.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditDate="2015-03-27T11:44:52.640" LastActivityDate="2015-03-27T11:44:52.640" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11085" PostTypeId="2" ParentId="11082" CreationDate="2015-03-27T11:32:20.707" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SVN &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can get you to many places you want to be.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can Browse all the directories ie. you can go anywhere&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But you have to give me a pull or push when I meet T.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When Users Commit the changes in Trunk you need to Pull or Push New changes in Trunk&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;People do not write on me, unless I am accompanied by D.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;We can't Commit Until we write the complete Description of the changes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In that case, my name is often written together with E.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If changes commited then it will be written by svn:Explaination&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Most of my kind cannot swim.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Many user who doesn't have updated copy can't commit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But when I join P, we shall live happily in a stream.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If we have Present copy then can commit easily(join new data with Present copy will leave Happily).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you want to work with me, all you need is a key.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We Need a Password for Using it / or Version Number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But when I am following S, you may want to get rid of me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Following Seurity Issues &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="444" LastEditorUserId="444" LastEditDate="2015-03-27T18:49:52.123" LastActivityDate="2015-03-27T18:49:52.123" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11086" PostTypeId="2" ParentId="11082" CreationDate="2015-03-27T11:33:32.140" Score="4" Body="&lt;p&gt;You are a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CAR&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can get you to many places you want to be.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Speaks for itself&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But you have to give me a pull or push when I meet T.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CAR - T&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;People do not write on me, unless I am accompanied by D.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CAR - D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In that case, my name is often written together with E.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CAR - E?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Most of my kind cannot swim.&#xA;But when I join P, we shall live happily in a stream.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CAR - P&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you want to work with me, all you need is a key.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a CAR KEY is needed to operate it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But when I am following S, you may want to get rid of me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; S - CAR&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastActivityDate="2015-03-27T11:33:32.140" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11087" PostTypeId="2" ParentId="11081" CreationDate="2015-03-27T11:56:33.590" Score="5" Body="&lt;p&gt;Addition of even&amp;lt;-&gt;odd alphabets positions as they appear in aphanumeric order...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GOLD,7,15,12,4=&gt;19&amp;lt;-&gt;19 balanced&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LEMONS,12,5,13,15,14,19=&gt;39&amp;lt;-&gt;39 balanced&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UNCOOKED,21,14,3,15,15,11,5,4=&gt;44&amp;lt;-&gt;44 balanced&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BORDERLINE,2,15,18,4,5,18,12,9,14,5=&gt;51&amp;lt;-&gt;51 balanced&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LILLIPUTIANS,12,9,12,12,9,16,21,20,9,1,14,19=&gt;77&amp;lt;-&gt;77 balanced&lt;/p&gt;&#xA;&#xA;&lt;p&gt;COUNTERBALANCE, ----&gt;balanced&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PREDETERMINATION, ----&gt;balanced&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the answer is :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BACKGAMMON&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;to see how they balence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BACKGAMMON,2,1,3,11,7,1,13,13,15,14 =&gt; 40 &amp;lt;-&gt; 40 Balanced&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;CHECKERS,3,8,5,3,11,5,18,19 =&gt; 37&amp;lt;-&gt;35   NOT BALANCED&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CRIBBAGE,3,18,9,2,2,1,7,5 =&gt; 21&amp;lt;-&gt;26 NOT BALANCED&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DOMINOES,4,15,9,13,15,5,19 =&gt; 47&amp;lt;-&gt;31 not balanced&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Used following ref :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1  2  3  4  5  6  7  8  9  10  11 12 13&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a  b  c  d  e  f  g  h  i  j   k  l  m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;z  y  x  w  v  u  t  s  r  q   p  o  n&lt;/p&gt;&#xA;&#xA;&lt;p&gt;26 25 24 23 22 21 20 19 18 17  16 15 14&lt;/p&gt;&#xA;" OwnerUserId="10433" LastEditorUserId="99" LastEditDate="2015-03-27T12:04:20.627" LastActivityDate="2015-03-27T12:04:20.627" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11088" PostTypeId="1" AcceptedAnswerId="11099" CreationDate="2015-03-27T11:58:31.563" Score="146" ViewCount="9812" Body="&lt;p&gt;You are a secret agent in the service of the KGB, about to embark on a highly dangerous mission to infiltrate MI6. You have your disguise, your papers, and your backstory all prepared. The night before your departure, you receive the following email:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;From: Andrew Void &amp;lt; a.void@disparition.com &gt;&lt;br&gt;&#xA;  Sent: Fri, 27 Mar 2015 11:57AM +0400&lt;br&gt;&#xA;  To: ██████████████████████&lt;br&gt;&#xA;  Subject: Your work&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Dear Mr Smith,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This is to inform you that your poem is now nearly noted up for publication. Its age notwithstanding, this poem will fit as part of a vast pattern of poems that spans millennia. You stand among us now as a poet, throned among such applauded poets as Aristophanes, Plato, Byron, and so on. As one of us, your poetic prowess will not go unadmired.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Many congratulations!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Andrew&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mr Smith is the pseudonym you will be adopting on your mission in Britain, but you do not recognise the name Andrew Void or the email address. You are about to delete the email as spam, but some instinct tells you to examine it more closely. After a few minutes at your computer, you find the hidden message within it and slump back in your chair, disappointed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the hidden message?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2016-09-03T12:35:18.720" LastActivityDate="2016-09-03T12:35:18.720" Title="A mysterious email from Andrew Void" Tags="|story|steganography|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="11089" PostTypeId="2" ParentId="11066" CreationDate="2015-03-27T12:00:27.370" Score="10" Body="&lt;p&gt;I'm going to take a stab and say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Moderate&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I'm surprisingly standard,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Moderate as a adjective means average, modest, standard, etc&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But can ban if required.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As a verb, to moderate, might include banning someone (e.g. from a forum)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Found in tea, less often than more,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letters &quot;tea&quot; and &quot;more&quot; are contained in moderate, but there's less in tea.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I might also clear your sock drawer.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Moderators on stack exchange sometimes have to deal with sock puppet accounts, and delete/clear them (noting the hint in the title)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastActivityDate="2015-03-27T12:00:27.370" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11090" PostTypeId="2" ParentId="11066" CreationDate="2015-03-27T12:08:28.777" Score="10" Body="&lt;p&gt;In a similar vein to &lt;a href=&quot;https://puzzling.stackexchange.com/a/11089/5373&quot;&gt;Alconja's answer&lt;/a&gt;, how about this?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A FLAG.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I'm surprisingly standard,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A flag can also be called a standard (as in battle standard)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But can ban if required.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ... or a banner, which can also mean &quot;one who bans&quot;. A ban on an SE user's account may result from a flag being raised by another user.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Found in tea, less often than more,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is probably a golf reference: &quot;tea&quot; is a homophone of &quot;tee&quot; and flags are found on golf courses (on the tee perhaps? I don't know enough about golf to say).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I might also clear your sock drawer.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As Alconja says, SE moderators sometimes have to delete sockpuppet accounts - the fact of sockpuppeting may be drawn to their attention by a flag raised by another user.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another possibility (even more similar to Alconja's answer) is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a &lt;a href=&quot;https://puzzling.stackexchange.com/users?tab=moderators&quot;&gt;MODERATOR&lt;/a&gt;: they can ban other users, delete sockpuppet accounts, and probably drink tea.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can't make that fit the first line though.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-03-27T12:08:28.777" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11091" PostTypeId="2" ParentId="11088" CreationDate="2015-03-27T13:12:48.817" Score="11" Body="&lt;p&gt;I guess that the hidden message is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You've been compromised (or something similar)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As &lt;em&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/users/99/kaine&quot;&gt;kaine&lt;/a&gt;&lt;/em&gt; mentioned in his comment, it probably has to do with &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/A_Void&quot; rel=&quot;nofollow noreferrer&quot;&gt;A Void&lt;/a&gt;&lt;/em&gt;, but I didn't really find any useful information about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But here's my reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;a.void&lt;/em&gt; and &lt;em&gt;disparition&lt;/em&gt; (French for &lt;em&gt;Disappearance&lt;/em&gt;) gives a hint that I should disappear, avoid something.&lt;br&gt;&#xA; The fact that Andrew Void knows my pseudonym makes me think that he also knows about my mission and that's quite suspicious.&lt;br&gt;&#xA; He says &lt;em&gt;your poem is now nearly noted up for publication&lt;/em&gt; - I interpret it as &lt;em&gt;your legend is (almost) known and will be published very soon&lt;/em&gt;&lt;br&gt;&#xA; Then he also says &lt;em&gt;this poem will fit as part of a vast pattern of poems that spans millenia&lt;/em&gt; which in my eyes means &lt;em&gt;your legend is very common, the pattern used for it was used for ages/decades/centuries before&lt;/em&gt;&lt;br&gt;&#xA; The next sentence is &lt;em&gt;You stand among us now as a poet, throned among such applauded poets as Aristophanes, Plato, Byron, and so on&lt;/em&gt;, which if followed by my other reasonings means: &lt;em&gt;You are listed as a spy, same as famous Aristophanes, Plato and Byron&lt;/em&gt;, where &lt;em&gt;Aristophanes, Plato and Byron&lt;/em&gt; are most probably pseudonyms of other agents of the past.&lt;br&gt;&#xA; And the last sentence &lt;em&gt;As one of us, your poetic prowess will not go unadmired&lt;/em&gt; can be treated as a threat/warning that if I go there, I will not be unseen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9352" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-03-28T01:02:11.653" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11092" PostTypeId="1" AcceptedAnswerId="11123" CreationDate="2015-03-27T13:38:18.100" Score="7" ViewCount="1179" Body="&lt;p&gt;During the last country fair, you have seen a weird arrow-shooter machine.&lt;br&gt;&#xA;Its shape is a &lt;strong&gt;regular polygon&lt;/strong&gt; with $N$ sides of length $1$ and it fires &lt;strong&gt;one bolt from each vertex at the same time&lt;/strong&gt;.&lt;br&gt;&#xA;The &lt;strong&gt;arrows have the same length&lt;/strong&gt; $A$, same speed; also, suppose that they are &lt;strong&gt;unidimensional&lt;/strong&gt; (no width).&lt;br&gt;&#xA;At the beginning, the arrows are arranged so that each arrow's tail coincides with its vertex, and &lt;strong&gt;the angle of fire is variable for every bolt&lt;/strong&gt; (of course, the arrow must be fired inside the polygon).&lt;br&gt;&#xA;This picture may be helpful to understand the initial conditions:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/DQ3pQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that every angle is determined randomly with equal probability, &lt;strong&gt;what is the chance that arrows won't collide after being shot&lt;/strong&gt;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Notes:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Two or more arrows collide when the segments that represent them have a common point (they intersect).&lt;/li&gt;&#xA;&lt;li&gt;You can imagine the arrows as very thin sticks, with a fixed length $A$.&lt;/li&gt;&#xA;&lt;li&gt;$N&amp;gt;2$ and $0&amp;lt;A&amp;lt;1$ are parameters, while the firing angles (all different) are random variables.&lt;/li&gt;&#xA;&lt;li&gt;The arrows are shot in a straight line. All have the same speed. Their motion is uniform rectilinear. The direction is only determined by the (random) angle.&lt;/li&gt;&#xA;&lt;li&gt;Although the arrows are fired inside the polygon, they may collide out of it.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Divide et impera! Simplify the problem, then &quot;add parts&quot; to your solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="5373" LastEditDate="2015-03-29T19:45:59.273" LastActivityDate="2015-04-06T22:37:27.253" Title="Clash of arrows" Tags="|mathematics|geometry|probability|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="11093" PostTypeId="2" ParentId="11069" CreationDate="2015-03-27T13:49:43.803" Score="5" Body="&lt;p&gt;It's my first time here in STEX Puzzling. I don't know the rules, but here's my try:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Spider-man&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ladies and gentlemen,&#xA;I will defend you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He defends people from villains&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am not a comedian,&#xA;but will entertain you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Entertains you whenever he saves people or when he takes photographs that are published in Daily Bugle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can’t be dangerous,&#xA;yet my half-name is.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A spider &lt;s&gt;man&lt;/s&gt; could be venomous.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can you name me the hero&#xA;of this colorful quiz?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He is a hero from colorful comic books&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10686" LastActivityDate="2015-03-27T13:49:43.803" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11094" PostTypeId="1" AcceptedAnswerId="11096" CreationDate="2015-03-27T14:05:32.027" Score="4" ViewCount="2266" Body="&lt;p&gt;Up or down, down or up?&lt;br&gt;&#xA;You may find me inside your cup.&lt;br&gt;&#xA;Found or not, not or found,&lt;br&gt;&#xA;I might be upon the ground.&lt;br&gt;&#xA;Here or there, there or here,&lt;br&gt;&#xA;You won't like me inside your ear.&lt;br&gt;&#xA;Play or sit, sit or play,&lt;br&gt;&#xA;What am I, do you say?    &lt;/p&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2016-10-12T20:48:26.627" Title="Up or Down? What am I?" Tags="|riddle|word|rhyme|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11096" PostTypeId="2" ParentId="11094" CreationDate="2015-03-27T14:11:19.243" Score="10" Body="&lt;p&gt;You could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Up or down, down or up?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Could be a reference to &quot;drink up&quot; vs &quot;down a drink&quot;, or maybe upstream and downstream.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You may find me inside your cup.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You might be drinking water from a cup.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Found or not, not or found,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Perhaps a reference to water diviners that purport to be able to sense water in the ground.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might be upon the ground.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Puddles are certainly water on the ground.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here or there, there or here,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Could be in reference to the phrase &quot;water, water everywhere [and not a drop to drink]&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You won't like me inside your ear.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nobody likes water in their ear.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Play or sit, sit or play,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm not sure about this one. I would have expected &quot;stand&quot; and &quot;babble&quot;, maybe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I, do you say?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You might be water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2016-10-12T20:48:26.627" LastActivityDate="2016-10-12T20:48:26.627" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11097" PostTypeId="2" ParentId="11094" CreationDate="2015-03-27T14:16:12.720" Score="-1" Body="&lt;p&gt;I am pretty sure that the answer for this riddle is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;water&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10630" LastEditorUserId="5127" LastEditDate="2015-03-27T14:57:14.883" LastActivityDate="2015-03-27T14:57:14.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11099" PostTypeId="2" ParentId="11088" CreationDate="2015-03-27T14:24:51.320" Score="124" Body="&lt;p&gt;The secret message is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DON'T GO TO LONDON NOW&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A French author &lt;a href=&quot;https://en.wikipedia.org/wiki/Georges_Perec&quot;&gt;Georges Perec&lt;/a&gt; once wrote a 300-page novel called &lt;em&gt;La disparition&lt;/em&gt; without using the letter 'e'. (It was later translated into English by Gilbert Adair under the title &lt;em&gt;A Void&lt;/em&gt; - and also without using the letter 'e'.) This suggests that the letter 'e' in the email is important somehow. If you look at the letters immediately &lt;em&gt;preceding&lt;/em&gt; every occurrence of this letter in the email, they spell out the message &lt;code&gt;DONTGOTOLONDONNOW&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="40" LastEditorUserId="21454" LastEditDate="2016-06-14T10:14:06.520" LastActivityDate="2016-06-14T10:14:06.520" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="11100" PostTypeId="1" AcceptedAnswerId="11129" CreationDate="2015-03-27T15:44:24.083" Score="8" ViewCount="537" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;&quot;Mary had a riddle lamb, brain-tease was hard as stone!&quot;&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today Mary's lamb wants you to solve its riddle.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A white man plowing the dark soil&lt;br&gt;&#xA;Silent oxen around him toil&lt;br&gt;&#xA;A train whistle is near&lt;br&gt;&#xA;What's said may disappear&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Half hint:&lt;/strong&gt;   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The lamb isn't involved, that's just the name of a series of riddle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #1:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Caius Titus dixit&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-03-29T12:49:47.793" LastActivityDate="2015-03-29T15:04:56.717" Title="Mary had a riddle lamb #1" Tags="|riddle|rhyme|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11101" PostTypeId="2" ParentId="11069" CreationDate="2015-03-27T16:01:25.060" Score="7" Body="&lt;p&gt;First time here :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ladies and gentlemen&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sunglasses are for both&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I will defend you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sunglasses protect your eyes from damaging sunrays&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am not a comedian but will entertain you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Funny sunglasses are pretty common&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can’t be dangerous&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sunglasses aren't dangerous (unless your poke your eye with it)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yet my half-name is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sun is dangerous&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can you name me the hero of this colorful quiz?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sunglasses come in many colors&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10693" LastActivityDate="2015-03-27T16:01:25.060" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11102" PostTypeId="2" ParentId="11066" CreationDate="2015-03-27T16:01:36.173" Score="4" Body="&lt;p&gt;It is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a plastic bag&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm surprisingly standard,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; plastic bags tend to be mass-produced, think grocery bags, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But can ban if required.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; plastic bags have been banned (or at least regulated) in municipalities due to littering and their environmental impact.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Found in tea, less often than more,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some new brands of tea are using special nylon mesh bags&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might also clear your sock drawer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can toss all your socks into a bag when you're about to travel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10695" LastActivityDate="2015-03-27T16:01:36.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11103" PostTypeId="1" AcceptedAnswerId="11111" CreationDate="2015-03-27T16:03:19.070" Score="5" ViewCount="398" Body="&lt;p&gt;A family of 3 - a father, mother and son - are practising on a dart board. It has 3 shapes - a triangle, a circle and a square - on it. Scoring is as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Intersection of all 3 shapes - 5 points&lt;/li&gt;&#xA;&lt;li&gt;Intersection of exactly 2 shapes - 4 points&lt;/li&gt;&#xA;&lt;li&gt;Exactly in one shape - 3 points&lt;/li&gt;&#xA;&lt;li&gt;Outside all shapes, but on the board - 2 points&lt;/li&gt;&#xA;&lt;li&gt;Outside the board - 0 points&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;After the game, the board looked like this: (Each star is an arrow)&#xA;&lt;img src=&quot;https://i.stack.imgur.com/iKK6R.png&quot; alt=&quot;Dart board&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All 3 of them scored the same number of points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Each of them also made a statement:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Father: If I was the first to score in the square, then I scored at least one 2.&lt;/li&gt;&#xA;&lt;li&gt;Mother: If I was the first to score in the triangle, then my husband and my son did not score the same number of 2s as each other.&lt;/li&gt;&#xA;&lt;li&gt;Son: If I scored first in the circle, then I did not score more 2s than my mother.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Exactly one of them told the truth. Who scored the 5?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt;&#xA;Remember that, for any&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Statement X: If A is true, then B is true.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;we can conclude that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Statement X is false if and only if A is true and B is false. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4269" LastEditDate="2015-03-28T14:35:22.740" LastActivityDate="2015-03-28T15:59:22.360" Title="Ultimate family dart game" Tags="|mathematics|logical-deduction|liars|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="11104" PostTypeId="2" ParentId="11069" CreationDate="2015-03-27T16:24:34.230" Score="3" Body="&lt;p&gt;Is it a  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Puppy?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ladies and gentlemen,&lt;br&gt;&#xA;I will defend you.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Puppy defends!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am not a comedian,&lt;br&gt;&#xA;but will entertain you.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's fun to play with puppies&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can’t be dangerous,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &gt;! Puppy isn't dangerous I guess&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yet my half-name is.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (Pu), which is Plutonium in peroidic table, a transuranic radioactive chemical element  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can you name me the hero  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Plutonium, named after &lt;a href=&quot;http://camphalfblood.wikia.com/wiki/Pluto&quot; rel=&quot;nofollow&quot;&gt;Pluto&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;of this colorful quiz?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; And Plutonium is &lt;a href=&quot;http://chemistry.about.com/od/imagesclipartstructures/ig/Science-Pictures/Plutonium-Colors.htm&quot; rel=&quot;nofollow&quot;&gt;colorful&lt;/a&gt; (well not that colorful maybe..)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2015-03-27T16:24:34.230" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11105" PostTypeId="1" AcceptedAnswerId="11106" CreationDate="2015-03-27T16:26:48.430" Score="17" ViewCount="10820" Body="&lt;p&gt;I can fly without wings, &lt;br/&gt;&#xA;only carried by the wind.&lt;br/&gt;&#xA;I am heavier when empty,&lt;br/&gt;&#xA;the bigger I am, the more I can carry.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I ?&lt;/p&gt;&#xA;" OwnerUserId="9888" LastActivityDate="2015-03-30T06:15:56.150" Title="I can fly without wings" Tags="|riddle|" AnswerCount="8" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11106" PostTypeId="2" ParentId="11105" CreationDate="2015-03-27T16:28:38.270" Score="20" Body="&lt;p&gt;I think you are a: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Helium balloon or a hot air balloon.&lt;br&gt;&#xA;&lt;br&gt;&#xA; They fly (float) and have no propulsion, so get carried by wind&lt;br&gt;&#xA; Since helium and hot air reduce the weight, they are heavier when empty and when bigger, they have more helium and hot air and can lift more. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-03-27T16:28:38.270" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="11107" PostTypeId="2" ParentId="11069" CreationDate="2015-03-27T17:16:22.587" Score="4" Body="&lt;p&gt;I think it is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Ad-Blocker&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ladies and gentlemen,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;everyone uses it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I will defend you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;blocks ads&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am not a comedian,&#xA;but will entertain you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;less ads = more fun browsing, watching videos etc&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can’t be dangerous,&#xA;yet my half-name is.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ads can be dangerous, full of spam and redirect to malicious sites&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can you name me the hero&#xA;of this colorful quiz?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Alot of ad are colorful, rich in color, trying to grab attention. There is often ads that asks questions like &quot;Would you like A or B&quot; or &quot;What are you waiting for?&quot; etc&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1504" LastActivityDate="2015-03-27T17:16:22.587" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11108" PostTypeId="1" AcceptedAnswerId="11115" CreationDate="2015-03-27T17:39:45.793" Score="5" ViewCount="315" Body="&lt;p&gt;Give me a dollar, You start to drink&lt;br&gt;&#xA;play the music, children start to sing&lt;br&gt;&#xA;look at me, without a blink&lt;br&gt;&#xA;like the diamond on my wedding ring  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Larger than a boulder, smaller than a pingpong&lt;br&gt;&#xA;the force with this one is very strong&lt;br&gt;&#xA;you might wonder if you are wrong&lt;br&gt;&#xA;did I mention the alphabet song?  &lt;/p&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2015-03-30T16:45:44.777" Title="What am I - Sing the Alphabet song!" Tags="|riddle|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11109" PostTypeId="2" ParentId="11108" CreationDate="2015-03-27T17:57:16.067" Score="2" Body="&lt;p&gt;You're a:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Jukebox   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One dollar is usually enough to start playing the music. The ring might be a reference to vinyls. The jukebox is larger than a boulder and smaller than a ping pong table!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-03-27T17:57:16.067" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11110" PostTypeId="2" ParentId="11069" CreationDate="2015-03-27T18:10:36.697" Score="1" Body="&lt;p&gt;My guess is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alarm Clock&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ladies and gentlemen,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Used by both genders&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I will defend you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Against over sleeping&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am not a comedian,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Of course Mr. serious.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but will entertain you. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Different tones and music&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can’t be dangerous,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Waking you up doesn't mean harming you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yet my half-name is.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Isn't that 'alarm'ing ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can you name me the hero&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; very very helpful.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;of this colorful quiz? &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nice rhyming prose.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9892" LastActivityDate="2015-03-27T18:10:36.697" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11111" PostTypeId="2" ParentId="11103" CreationDate="2015-03-27T18:15:02.010" Score="4" Body="&lt;p&gt;If dad tells the truth, there is a contradiction as everyone who scores in a circle scores in a triangle and vice versa.  There is only one first and mom and son would need to have that.  This all means the father definitely scored first in the square and did not have any 2's.  He scored at least one 3 or the 5.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The dad cannot be telling the truth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we assume that the son told the truth, the mother lied (we already know the father must be lying).  This means that the mother scored at least once in the triangle so scored a 4 or a 5.  Her statement &quot;my husband and my son did not score the same number of 2s as each other&quot; must be false so we know the son must not have scored any 2s just like his Dad.  This would require the mother to score at least 12 points ($4*2+1*4=12$) which is impossible as everyone scored 11 points.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This means the son must be lying.  It means by process of elimination the mother must be telling the truth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If mom tells the truth: the father scored in the square first (earned a 3 or 5), the father scored zero 2s, the son scored in the circle first (earned a 4 or 5), and scored more 2s than his mother.  If the son scored in the circle first, the mother didn't score in the triangle first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this we can determine that the son scored three 2s. If he scored 2 or fewer, he would not score more than his mother. If he scored 4, than he would earn more than 11 points. This means the mother scored one 2.  We know he scored a 4 or a 5.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As $3*2+4=10$ and there is no way to score 1 point, &lt;strong&gt;the son must have scored the 5.&lt;/strong&gt;  If you wish to continue you can determine that (if at least one person never missed the board) the father scored one 3, two 4s, and one 0.  The mother scored one 2 and three 3s.  The son scored one 5 and three 2s.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unlike the previous edit, I've shown that this is the only possible solution.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditDate="2015-03-28T15:59:22.360" LastActivityDate="2015-03-28T15:59:22.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11112" PostTypeId="1" AcceptedAnswerId="11135" CreationDate="2015-03-27T18:17:30.237" Score="8" ViewCount="635" Body="&lt;p&gt;&lt;em&gt;Look in thy glass and tell the face thou viewest&lt;br&gt;Time doth transfix the flourish set on youth.&lt;br&gt;Those hours, that with gentle work did frame&lt;br&gt;The lovely gaze where every eye doth dwell,&lt;br&gt;Will play the tyrants to the very same&lt;br&gt;That thou among the wastes of time must go.&lt;br&gt;The point must come soon after when thou viewest,&lt;br&gt;When squareth it that debt all owe the hour,&lt;br&gt;And then thine sharpened figure shall be rounded.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every line is important.&lt;/p&gt;&#xA;" OwnerUserId="9523" LastEditorDisplayName="user88" LastEditDate="2015-03-28T06:30:57.197" LastActivityDate="2015-03-28T06:30:57.197" Title="The answer is a two-digit number (again)" Tags="|riddle|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11113" PostTypeId="2" ParentId="11112" CreationDate="2015-03-27T18:40:56.213" Score="3" Body="&lt;p&gt;I think the answer is  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 10&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first 6 lines are from Shakespeare's sonnets:&lt;br&gt;&#xA;Sonnet 3&lt;br&gt;&#xA;Sonnet 60&lt;br&gt;&#xA;Sonnet 05&lt;br&gt;&#xA;Sonnet 05&lt;br&gt;&#xA;Sonnet 05&lt;br&gt;&#xA;Sonnet 12  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sonnet 05 occurs thrice, $3\times5=15$. All those numbers are divisors of 60. The 2-digits missing divisors of 60 are 10,18,20,30.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Those hours, that with gentle work did frame&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It suggests that we're talking about hours, so let's exclude 30.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The point must come soon after when thou viewest&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It may be a reference to decimal system (based on the number 10), where point means dot or comma.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And thine sharpened figure shall be rounded.&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;In the number 10 the digit 1 looks &quot;sharp&quot; and shall be rounded.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-03-27T19:56:20.967" LastActivityDate="2015-03-27T19:56:20.967" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11114" PostTypeId="2" ParentId="11105" CreationDate="2015-03-27T18:42:57.697" Score="18" Body="&lt;p&gt;First timer here.  I think you could be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A promise.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can fly without wings, &#xA;only carried by the wind.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A &quot;promise in the wind&quot;.  I think it means the promise is the only thing you have, the promise-maker might be away or unreachable and all that's left is the promise itself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am heavier when empty,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The weight of an empty promise...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the bigger I am, the more I can carry.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...contrasted with what a big promise can &lt;em&gt;carry&lt;/em&gt;.  &quot;Her hopes and dreams &lt;em&gt;ride on&lt;/em&gt; the promise that one day her father would return.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10696" LastActivityDate="2015-03-27T18:42:57.697" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11115" PostTypeId="2" ParentId="11108" CreationDate="2015-03-27T19:10:50.293" Score="4" Body="&lt;p&gt;I think you are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A star&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Give me a dollar, You start to drink&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure about this line but after the hint in the comments, Google search for dollar drinks Canada gives results for $1 any size drinks at McDonald's, specifically mentioning Canada.  Link to my &quot;star&quot; theory: Canadian flag features a maple leaf, which has the shape of a star.  EDIT: Starbucks!  Alex revealed it in the comments, I missed this clue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;play the music, children start to sing&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Popular Children's song &quot;Twinkle, Twinkle, Little Star&quot;.  Every kid will start singing it when they hear the music.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;look at me, without a blink&#xA;like the diamond on my wedding ring&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Up above the world so high, like a diamond in the sky&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Larger than a boulder, smaller than a pingpong&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MUCH larger than a boulder, but to us it looks smaller than a pingpong.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the force with this one is very strong&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The gravitational pull or &quot;force&quot; of the star.  (EDIT: And a reference to star wars, which I thought was just part of the clue, not part of the answer)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;you might wonder if you are wrong&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;how I wonder what you are&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;did I mention the alphabet song?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  The &quot;alphabet song&quot; has the same tune as &quot;twinkle twinkle little star&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10696" LastEditorUserId="10696" LastEditDate="2015-03-30T16:45:44.777" LastActivityDate="2015-03-30T16:45:44.777" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11116" PostTypeId="2" ParentId="11105" CreationDate="2015-03-27T19:19:24.647" Score="4" Body="&lt;p&gt;I would say JonTheMon's answer is more accurate but here's my attempt:  You are a  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cloud  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can fly without wings, only carried by the wind.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cloud 'fly' without wings but move by the wind  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am heavier when empty  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Heavy &quot;rain&quot; from the cloud until it's 'empty'  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the bigger I am, the more I can carry.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bigger cloud carries more liquid droplets&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2015-03-27T19:19:24.647" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11117" PostTypeId="1" AcceptedAnswerId="36122" CreationDate="2015-03-27T19:29:19.483" Score="54" ViewCount="3964" Body="&lt;p&gt;At the Arithmetic Company, to try and cut down labor costs the administration commissioned a calculator to be made. The calculator was produced, and in the quality control division was tested with the following values:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;20 + 3 = 23&#xA;5 + 15 = 20&#xA;7 + 9 + 9 = 25&#xA;28 - 22 + 2 = 8&#xA;33 - 20 + 4 = 17&#xA;71 + 5 - 75 = 1&#xA;30 + 19 - 5 = 44&#xA;45 + 36 - 23 = 58&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Finding all the results satisfactory they put the calculator to use.&#xA;The first client requested various sums of 2, and using the calculator, was given the following list:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2 + 2 = 85&#xA;2 + 2 + 2 = 150&#xA;2 + 2 + 2 + 2 = 152&#xA;2 + 2 + 2 + 2 + 2 = 232&#xA;2 + 2 + 2 + 2 + 2 + 2 = 1114&#xA;2 + 2 + 2 + 2 + 2 + 2 + 2 = 1167&#xA;2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 = 2227&#xA;2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 = 17727&#xA;2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 = 24727&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The next client requests some consecutive sequences and was given:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 + 2 + 3 + 4 = 130&#xA;5 + 6 + 7 + 8 = 130&#xA;9 + 10 + 11 + 12 = 188&#xA;13 + 14 + 15 + 16 = 1216&#xA;17 + 18 + 19 + 20 = 1157&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Finally the last client requests just one sum and is given:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 + 1 = ERROR&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The next day they received many complaints from their clients, and decide to bring in you, their best analyst, to figure out what the calculator is doing. Your job is to find the rules that the determine the calculator's output as well as one example sum (not listed above) so that your method can be tested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus:&lt;/strong&gt;&#xA;What does the calculator output when given 1×1 ?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;People may ask any number of additional sums to the calculator in the chat room &lt;a href=&quot;http://chat.stackexchange.com/rooms/22378/sum-requests&quot;&gt;http://chat.stackexchange.com/rooms/22378/sum-requests&lt;/a&gt;&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0 + 0 = ERROR&#xA;1 + 2 = ERROR&#xA;1 - 3 = ERROR&#xA;2 - 7 = ERROR&#xA;3 + 3 = ERROR&#xA;4 + 4 = ERROR&#xA;5 + 19 = ERROR&#xA;19 + 5 = ERROR&#xA;76 - 75 = ERROR&#xA;130 - 85 = ERROR&#xA;1 - 2 + 3 + 4 = ERROR&#xA;2 + 2 - 2 = 2&#xA;15 = 7&#xA;15 + 5 = 20&#xA;18 = 18&#xA;19 = 19&#xA;7 + 40 = 25&#xA;29 = 25&#xA;13 + 4 = 26&#xA;39 = 35&#xA;5 + 5 = 40&#xA;9 + 9 = 40&#xA;40 = 40&#xA;49 - 5 = 40&#xA;5 + 5 + 5 = 45&#xA;49 = 45&#xA;59 = 55&#xA;6 + 2 = 70&#xA;5 + 5 + 5 + 5 = 72&#xA;85 = 85&#xA;3 + 7 = 91&#xA;2 + 85 = 150&#xA;85 + 2 = 150&#xA;85 + 85 = 152&#xA;71 + 5 = 303&#xA;45 + 13 = 517&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If necessary, I'll provide a new hint for few days the problem remains unsolved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1:&lt;/strong&gt;&#xA;When inputting solitary numbers into the calculator you get the following results:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0 = 0&#xA;1 = 1&#xA;2 = 2&#xA;3 = 3&#xA;4 = 4&#xA;5 = 5&#xA;6 = 6&#xA;7 = 7&#xA;8 = 8&#xA;9 = 5&#xA;10 = 10&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;55 + 1 = 35&#xA;59 + 1 = 35&#xA;65 + 1 = 107&#xA;69 + 1 = 107&#xA;75 + 1 = 303&#xA;79 + 1 = 303&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;15 + 3 = 91&#xA;19 + 3 = 35&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="150" LastEditorDisplayName="user88" LastEditDate="2016-06-17T15:38:29.643" LastActivityDate="2016-06-17T15:38:29.643" Title="The Case of the Miscalculating Calculator" Tags="|mathematics|english|" AnswerCount="3" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="11118" PostTypeId="2" ParentId="11105" CreationDate="2015-03-27T19:33:41.207" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A parachute.&lt;br&gt;&#xA; It &quot;flies&quot; on wind power alone, and can carry heavier loads as its surface area increases.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But that doesn't quite seem to fit the third line.&lt;/p&gt;&#xA;" OwnerUserId="10705" LastActivityDate="2015-03-27T19:33:41.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11119" PostTypeId="1" AcceptedAnswerId="11127" CreationDate="2015-03-27T19:45:48.697" Score="1" ViewCount="198" Body="&lt;p&gt;One night, Raphael decided to take an exuberant stroll around a very beautiful garden or in other words, embark on an formidable odyssey. The garden, comprising solely of fruitful facets of nature, was indeed very captivating to the eye. One might even infer that the garden is virtually priceless, that it requires no introduction. One glance at the garden is enough to instill in the witness an vivacious aesthesis that is perhaps impossible to explain. The boundaries of the garden were perfectly coated with electromagnetic ligaments made out of metal. Raphael, speaking earnestly, is a very intelligent man. But at that night, his hands were weightless. To get around the weightlessness, he brought with himself some support consisting of a place to rest the pedal on for the act of propelling spatially higher. As he entered the garden, he was petrified rather than astonished. Thirty minutes later,  his hands were surrounded with metallic loops. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happened to Raphael?&lt;/p&gt;&#xA;" OwnerUserId="10704" LastEditorUserId="5373" LastEditDate="2015-03-27T21:05:19.970" LastActivityDate="2015-03-27T21:17:32.317" Title="Raphael's Odyssey" Tags="|riddle|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11120" PostTypeId="2" ParentId="11119" CreationDate="2015-03-27T19:54:13.440" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm not sure about the nature of the &quot;garden,&quot; but it's clearly well-guarded and illegal to enter. Raphael attempted to break in, using a ladder to get over an electric fence, and was promptly arrested upon reaching the ground.&lt;br&gt;&#xA;&lt;br&gt;&#xA; It's probably something less literal than that, though.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10705" LastActivityDate="2015-03-27T19:54:13.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11121" PostTypeId="2" ParentId="11069" CreationDate="2015-03-27T19:59:03.147" Score="10" Body="&lt;p&gt;First time here!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; toy gun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I will defend you.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That's how kids think of toy guns often.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;will entertain you&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You know, a kids' toy. It's for fun.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I can’t be dangerous,&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's fake. It's plastic. Not much danger.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;yet my half-name is.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Unlike toy-guns, just &lt;strong&gt;guns&lt;/strong&gt; are dangerous.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;of this colorful quiz?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Per OPs comment on his question, the &quot;hero&quot; can just sometimes be colorful. It's not uncommon for toy guns to be colored.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ah, probably:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; nerf gun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10707" LastEditorUserId="10707" LastEditDate="2015-03-27T20:51:13.103" LastActivityDate="2015-03-27T20:51:13.103" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="11122" PostTypeId="2" ParentId="11119" CreationDate="2015-03-27T20:02:42.800" Score="1" Body="&lt;p&gt;The garden is clearly the Eden garden and the metallic loop is a ring. &#xA;After breaking up with Adam, Eva decides to marry Raphael, the only man in there, so the metallic loop is actually a wedding ring! Poor Raphael!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional note: &quot;hand-ring&quot; pronounce is very similar to &quot;entering&quot;!&lt;/p&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-03-27T20:02:42.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11123" PostTypeId="2" ParentId="11092" CreationDate="2015-03-27T20:10:03.397" Score="12" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Edit: Mostly rewritten, with all the math done up to the integral, and results from a Monte Carlo simulation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;Problem definition&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;As I interpret the problem, what we have is a number of inequalities that must be satisfied by $ N $ random variables, $\theta_i$, $i=1..N$.   Define the angles such that when they are $0$, all the arrows are pointed counterclockwise around the polygon.  For reference, let's also let $X_i$ be the vertex of the polygon corresponding to $\theta_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A general approach to solving the problem is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Given a value of one angle, $\theta_i=\alpha$, find the range of $\theta_j$ which will lead to a collision.&lt;/li&gt;&#xA;&lt;li&gt;The probability of a collision at that angle, $P(C_{i,j}|\theta_i=\alpha)$ is simply the collision range divide by the total allowed range. The probability of no collision is $P(\neg C_{i,j}|\theta_i=\alpha) = 1 -P(C_{i,j}|\theta_i=\alpha) $&lt;/li&gt;&#xA;&lt;li&gt;To find the total probability for a collision between arrows $i$ and $j$, integrate over the allowed range of angles $\theta_j$, and divide by the allowed range. &lt;em&gt;A priori&lt;/em&gt;, we don't know if this can be done analytically.&lt;/li&gt;&#xA;&lt;li&gt;The total number of pairs $(i,j)$ is the binomial $\binom{N}{2}=\frac{N(N-1)}{2}$.  Among these, we only have $\lfloor\frac{N}{2}\rfloor$ different probabilities, because any vertices separated by the same distance will have the same probability. However, we cannot simply multiply all the probabilities together to get the answer, because the outcomes are not independent. This means we will also have to evaluate the probability of any combination of collisions. To calculate these, we can multiply the conditional probabilities inside the integrals...  Making it even less likely than an analytical solution exists.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It is my opinion that the math gets too complicated for this to be considered a &quot;puzzle.&quot;  It is my suspicion that any clever solution that bypasses the math is wrong, either by incorrectly simplifying the geometry, or by assuming the collisions are independent events.  However, I'm having fun, so I'll go ahead and finish my answer.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;What is a collision?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;For any two arrows $i$ and $j$, we can draw lines along their paths as defined by the angles $\theta_i$ and $\theta_j$. Unless the lines are parallel, there will be exactly one intersection point. Call that point $X_{ij}$.  If the two arrows are to collide, then that will be the point where it happens. A collision occurs when some part of both arrows is present at the intersection point at the same time. Because all the arrows travel at the same speed, and are fired at the same time, they are each always the same distance from their respective points of origin.   Except in very rare cases, the distance to $X_{ij}$ from the two starting vertices will be different. However, if the tail of one arrow has not yet passed when the head of the other arrow arrives, then there is a collision.  The limiting case is when the very head of arrow $i$ hits the very tail of arrow $j$, or the very head of arrow $j$ hits the very tail of arrow $i$. In the first case, $\overline{X_iX_{ij}}$, the distance from $X_i$, where arrow $i$ started, to the intersection point $X_{ij}$is longer than $\overline{X_jX_{ij}}$, the distance from $X_j$ to $X_{ij}$, by the length of an arrow, $A$.  As an equation, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\overline{X_iX_{ij}}-\overline{X_jX_{ij}}=A \tag1 \label{dist1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second case, the opposite is true: This distance from $X_j$ to $X_ij$ is longer than the distance from $X_i$ to $X_{ij}$ by the length of an arrow, or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\overline{X_jX_{ij}}-\overline{X_iX_{ij}}=A \tag{2} \label{dist2}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flipping equation $\eqref{dist2}$ around and combining the two, we get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\overline{X_iX_{ij}}-\overline{X_jX_{ij}}=\pm A\tag3 \label{dist3}$$&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Fun with Hyperbolas&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;As it happens, such an equation defines a &lt;a href=&quot;http://www.mathopenref.com/polygonradius.html&quot; rel=&quot;noreferrer&quot;&gt;hyperbola&lt;/a&gt; with foci at $X_i$ and $X_j$.  Define a coordinate system with the origin centered between $X_i$ and $X_j$, with the positive x-axis passing through $X_j$ (and thus, the negative x-axis passing through $X_i$).  The equation of a hyperbola is:&#xA;\begin{align}&#xA;{{x^2}\over{a^2}}-{{y^2}\over{b^2}}&amp;amp;=1 \tag{4} \label{hyp1}&#xA;\end{align}&#xA;The difference in lengths from the two foci to the points on the hyperbola  is $\pm {2a}$, so for us:&#xA;\begin{align}&#xA;2a&amp;amp;=A\\&#xA;a&amp;amp;={A\over2} \tag{5} \label{a}&#xA;\end{align}&#xA;The distance between the foci is $2c$, where&#xA;$$c^2 = a^2 + b^2$$&#xA;For our situation, then,&#xA;\begin{align}&#xA;2c &amp;amp;= \overline{X_iX_j}\\&#xA;c &amp;amp;= {\overline{X_iX_j}\over 2} \tag{6} \label{c}\\&#xA;b^2 &amp;amp;= c^2 - a^2\\&#xA;&amp;amp;= \left({\overline{X_iX_j}\over 2}\right)^2-\left({A\over2} \right)^2 \tag{7} \label{b}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The eccentricity for such a hyperbola, which we will end up using later, is:&#xA;\begin{align}&#xA; \epsilon &amp;amp;= {c\over a}\\&#xA; &amp;amp;={{\overline{X_iX_j}/2}\over{A/2}}\\&#xA; &amp;amp;={{\overline{X_iX_j}}\over{A}} \tag{8} \label{epsilon}\\&#xA; \end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a figure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bJLzX.png&quot; alt=&quot;Hyperbola showing range of angles which cause a collision&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this figure, $\overline{X_iX_j}=1$ and $A=0.3$, which makes the hyperbola relatively easy to see. The hyperbola is the two blue curves, and the dashed lines are the asymptotes, which the hyperbola arms approach as they extend toward infinity. The green line represents the path of an arrow emanating from $X_i$, and the two red lines represent the two possible paths of an arrow from $X_j$ which will hit the other arrow head-to-tail or tail-to-head.  Notice that they intersect the green line at points which are on the hyperbola. The shaded red area represents intermediate angles between the two extremes, where the arrows will collide at some intermediate point, i.e.:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-A\le \overline{X_iX_{ij}}-\overline{X_jX_{ij}}\le A \tag{9} \label{dist4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few things we can observe based on this picture, regarding the angles of the two arrows and the asymptotes.  All lines, except those parallel to an asymptote, will intersect the hyperbola twice. However, in our problem, the paths of the arrows are really rays, not lines, and so if the intersections are &quot;behind&quot; the starting point, we don't want to consider them.  The location of the intersections is different depending on how the path of the arrow corresponds to the the asymptotes.  The angle the asymptotes make with the x-axis is given by &#xA;\begin{align}&#xA;\DeclareMathOperator{\arcsec}{arcsec}&#xA;\phi=\arcsec\epsilon \tag{10} \label{phi}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the moment, let's call the angle between the green line and the positive x-axis $\alpha$ (counter-clockwise positive), and the angle made by the red line(s) and the negative x-axis $\beta$ (clockwise positive). Later on we will relate these to $\theta_i$ and $\theta_j,$ but we're not there yet.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If $\alpha \ge \phi$ or $\alpha \le -\phi$, the path of the arrow does not intersect the right arm of the hyperbola. In this case, equation $\eqref{dist1}$ has no solution. However, it may still possible for the arrows to collide, as long as there is an intersection with the left arm.  In this case, we add another condition, which is that the two arrow paths do not diverge:&#xA;\begin{align}&#xA;\alpha \ge \phantom{-}\phi&amp;amp;\implies\quad \phantom{-}\pi - \beta &amp;gt; \alpha \tag{11} \label{limit1}\\&#xA;\alpha \le -\phi&amp;amp;\implies\quad -\pi - \beta &amp;lt; \alpha \tag{12} \label{limit2}	&#xA;\end{align}&lt;/li&gt;&#xA;&lt;li&gt;If $\alpha \ge \pi - \phi$ or $\alpha \le -\pi + \phi$, the path of the arrow also does not intersect the left arm of the hyperbola, in which case equation $\eqref{dist2}$ also has no solution, and it is impossible for the arrows to collide.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;One last bit of math we want to get from the Wikipedia page on hyperbolas is the equation in polar coordinates centered at the focus, which is useful in, for instance, orbital mechanics.  It actually only lists the equation for the focus and the near arm, but we can get the equation for the far arm by an analogous derivation, which I won't type out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;r_{near} &amp;amp;= \frac{a(\epsilon^2-1)}{\epsilon\cos\theta+1} \tag{13} \label{rnear}\\&#xA;r_{far} &amp;amp;= \frac{a(\epsilon^2)}{\epsilon\cos\theta-1} \tag{14} \label{rfar}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that, as I have defined $\alpha$ and $\beta,$ these equations apply to either one of them. Since the near arm for $\alpha$ is the far arm for $\beta$ and vice-versa, and we know that, at any given point the difference in the distances is $2a$, we can use these formulas for either focus.  Given $\alpha$, then, we can solve for two angles $\beta_+$ and $\beta_-$, representing the intersection with the right and left arms of the hyperbola, respectively. In order to have a result which is valid on the range$[-\pi,\pi]$, we use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Tangent_half-angle_substitution&quot; rel=&quot;noreferrer&quot;&gt;tangent half-angle substitution&lt;/a&gt;, $\cos\theta=\frac{1-\tan^2(\theta/2)}{1+\tan^2(\theta/2)}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;r_{\alpha} &amp;amp;= r_{\beta+} +2a\\&#xA;\frac{a(\epsilon^2-1)}{\epsilon\cos\alpha-1}  &amp;amp;=  \frac{a(\epsilon^2-1)}{\epsilon\cos\beta_++1}+2a\\&#xA;\left[\left(\epsilon^2+1\right)+2\epsilon\cos\alpha\right]\cos\beta_+&amp;amp;=\left(\epsilon^2+1\right)\cos\alpha +2\epsilon\\&#xA;\left[\left(\epsilon^2+1\right)+2\epsilon\tfrac{1-\tan^2(\alpha/2)}{1+\tan^2(\alpha/2)}\right]\tfrac{1-\tan^2(\beta_+/2)}{1+\tan^2(\beta_+/2)}&amp;amp;=\left(\epsilon^2+1\right)\tfrac{1-\tan^2(\alpha/2)}{1+\tan^2(\alpha/2)} +2\epsilon\\&#xA;&amp;amp;\ \ \vdots\\&#xA;\tan(\beta_+/2) &amp;amp;= \pm \frac{\epsilon-1}{\epsilon+1}\tan(\alpha/2)\\&#xA;\beta_+ &amp;amp;= \pm 2 \arctan\left(\frac{\epsilon-1}{\epsilon+1}\tan(\alpha/2)\right)\tag{15} \label{betaleft}\\&#xA;\phantom{x}\\&#xA;r_\alpha &amp;amp;= r_{\beta-} -2a\\&#xA;&amp;amp;\ \ \vdots\\&#xA;\beta_- &amp;amp;= \pm 2 \arctan\left(\frac{\epsilon+1}{\epsilon-1}\tan(\alpha/2)\right)\tag{16} \label{betaright}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases, we want the positive solution, which corresponds to $\alpha$ and $\beta$ having the same sign, which is required for a collision.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Polygon geometry&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The interior angles of a regular $N$-gon are given by&#xA;$$\theta_0={{N-2}\over{N} }\pi \tag{17} \label{theta0}$$&#xA;and we therefore have &#xA;$$ 0 \le \theta_i \le\theta_0 \tag{18} \label{thetarange}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radius of our polygon is given &lt;a href=&quot;http://www.mathopenref.com/polygonradius.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;:&#xA; $$R = {L\over {2\sin{\pi\over N}}} \tag{17} \label{radius}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any $i$, $j$, with $j=i+k, 1\le k \le N-1$, draw a polygon containing the points $X_i .. X_j$. There are $k+1$ such points. As such the interior angles of this polygon must add up to $(k-1)\pi$. For the $k-1$ points &lt;em&gt;between&lt;/em&gt; $X_i$ and $X_j$, these interior angles are also interior angles of the regular $N$-gon, and so they are each equal to $\theta_0$. The other two angles are similar, and therefore each equal to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\delta &amp;amp;= {1\over 2}\left[(k-1)\pi - (k-1)\theta_0\right] \\&#xA;&amp;amp;= {{k-1}\over N}\pi \tag{18} \label{delta}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\overline{X_iX_j}$ is a chord of the circle circumscribing the $N$-gon, whose subtended angle is $k{{2\pi}\over N}$; its length (&lt;a href=&quot;http://en.wikipedia.org/wiki/Chord_(geometry)&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;) is:&#xA;\begin{align}&#xA;\overline{X_iX_j} &amp;amp;= 2R\sin\left(\frac{k\pi}{N}\right)\\&#xA;&amp;amp;=L\frac{\sin\left(\frac{k\pi}{N}\right)}{\sin\left(\frac{\pi}{N}\right)} \tag{19} \label{XiXj}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are now in a position to relate the parameters of the hyperbola in the previous section to the parameters and random variables in the problem.  First, from $\eqref{epsilon}$ and $\eqref{XiXj}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\epsilon=\frac{L}{A}\frac{\sin\left(\frac{k\pi}{N}\right)}{\sin\left(\frac{\pi}{N}\right)} \tag{20} \label{epsilon2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then, referring to the diagram,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cDEnU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\alpha &amp;amp;= \theta_i-\delta \tag{21} \label{alpha}\\&#xA;\beta &amp;amp;= \theta_0- \theta_j - \delta \tag{22} \label{beta}\\&#xA;-\delta&amp;amp;\le(\alpha,\beta)\le \theta_0 - \delta \tag{23} \label{anglerange}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Solving the integral?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Neglecting for the moment the conditions in $\eqref{limit1}$ and $\eqref{limit2}$, the integral we want is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;P(C_{i,j})&amp;amp;=\int_{-\delta}^{\theta_0 - \delta}P(C_{i,j}|\alpha)\mathrm{d}\alpha\\&#xA;&amp;amp;=\int_{-\delta}^{\theta_0 - \delta}\frac{\beta_+(\alpha)-\beta_-(\alpha)}{\theta_0}\mathrm{d}\alpha\\&#xA;&amp;amp;=\frac{1}{\theta_0}\left(\int_{-\delta}^{\theta_0 - \delta}\beta_+(\alpha)\mathrm{d}\alpha-\int_{-\delta}^{\theta_0 - \delta}\beta_-(\alpha)\mathrm{d}\alpha\right)\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question here is whether we can evaluate:&#xA;$$\int 2 \arctan\left(\frac{\epsilon\pm1}{\epsilon\mp1}\tan(\alpha/2)\right)\mathrm{d}\alpha$$&#xA;Or, more generally:&#xA;$$\int \arctan(c\tan x)\mathrm{d}x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Mathematica&lt;/em&gt; manages this, but the result is not anything pretty, and it does not simplify it into anything that I would consider elegant.  In order to solve the full problem analytically, we also need to find joint probabilities like:&#xA;$$\int \arctan\left(c_1\tan (x -d_1)\right)\arctan\left(c_2\tan (x -d_2)\right)\dots\mathrm{d}x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, I give up on the analytic approach.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Monte Carlo&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;We have everything we need to determine whether a given set of random variables will lead to a collision. To find the probabilities, let's just run a lot of tests.  Here are the results for $N\in[3,8]$. For each graph, values of $\frac{A}{L}$ were considered on the range $[0,1]$ in steps of $0.02$, with one million trials per point, 51 million trials per graph. Each graph took 6–7 minutes to generate on my laptop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FnhUK.png&quot; alt=&quot;Probability distribution for N=3&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/b61gP.png&quot; alt=&quot;Probability distribution for N=4&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PTw0q.png&quot; alt=&quot;Probability distribution for N=5&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9JWiS.png&quot; alt=&quot;Probability distribution for N=6&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VKjAZ.png&quot; alt=&quot;Probability distribution for N=7&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/06uqC.png&quot; alt=&quot;Probability distribution for N=8&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general pattern is that the probability drops from 1 to 0 fairly quickly as $\frac{A}{L}$ goes from 0 to 1.  The entire distribution also drops a little bit as $N$ increases. Of course, we could have said that from the beginning.&lt;/p&gt;&#xA;" OwnerUserId="10649" LastEditorUserId="10649" LastEditDate="2015-04-02T08:29:51.687" LastActivityDate="2015-04-02T08:29:51.687" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="11126" PostTypeId="2" ParentId="11069" CreationDate="2015-03-27T21:04:38.070" Score="2" Body="&lt;p&gt;Another first timer here. I think it is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water pistol&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ladies and gentlemen,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's a unisex toy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I will defend you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People will think twice before picking on you if you've got a water gun pointed at them&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am not a comedian,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nope&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but will entertain you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yeah! It's fun right?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can’t be dangerous,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Unless someone is to put salt accid in it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yet my half-name is.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PISTOL!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can you name me the hero&#xA;of this colorful quiz?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Comes in many colors&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cheers from the netherlands.&lt;/p&gt;&#xA;" OwnerUserId="10709" LastActivityDate="2015-03-27T21:04:38.070" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11127" PostTypeId="2" ParentId="11119" CreationDate="2015-03-27T21:17:32.317" Score="4" Body="&lt;p&gt;Raphael was clearly a mental patient in an Asylum who thought he was outside a garden. (The garden filled with illusions of fruitful facets of nature). We can infer that the &quot;garden&quot; is priceless because of the large amount of drugs and sedatives stored in it - in fact, it may be some room in the area filled with drugs. His weightless hands can be described as &quot;numb hands&quot;, possibly from sedatives he had been given earlier in the night. He tried to break into it to get more drugs (possibly morphine addiction?), climbing over an electric fence with a ladder or prop or stilt or something of the sort, and ended up being caught. Handcuffs around his hands are the metal loops.&lt;/p&gt;&#xA;" OwnerUserId="8999" LastActivityDate="2015-03-27T21:17:32.317" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11128" PostTypeId="2" ParentId="11100" CreationDate="2015-03-27T21:24:23.680" Score="1" Body="&lt;p&gt;what about:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; eraser ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A white man plowing the dark soil&lt;br&gt;&#xA;Silent oxen around him toil  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the erasers are usually white rubber thingy and the oxen are the letters written&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A train whistle is near&lt;br&gt;&#xA;What's said may disappear&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; what is written will be erased, no idea if the whistle plays an important role or just makes the riddle dramatic :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5283" LastActivityDate="2015-03-27T21:24:23.680" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11129" PostTypeId="2" ParentId="11100" CreationDate="2015-03-27T21:31:48.527" Score="8" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CHALK.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A white man plowing the dark soil&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chalk on a blackboard.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Silent oxen around him toil&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Silent pupils in a classroom, or students in a lecture hall, toiling to copy down what the chalk is writing on the board. &lt;strong&gt;Intended explanation&lt;/strong&gt;: silent fingers gripping the chalk and toiling to move it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A train whistle is near&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The whistle at the start and end of school sessions. &lt;strong&gt;Intended explanation&lt;/strong&gt;: chalk screeching on a blackboard can sound like a train whistle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What's said may disappear&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The writing on the board (what's 'said' by the chalk) gets rubbed out ('disappears') later. This was also &lt;strong&gt;intended&lt;/strong&gt; to be a reference to the Latin proverb &lt;a href=&quot;http://en.wikipedia.org/wiki/Verba_volant,_scripta_manent&quot; rel=&quot;nofollow&quot;&gt;&quot;Verba volant, scripta manent&quot;&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I got this by reading &lt;a href=&quot;http://en.wikipedia.org/wiki/Mary_Had_a_Little_Lamb&quot; rel=&quot;nofollow&quot;&gt;the full poem alluded to in the title&lt;/a&gt; and hunting for ideas in there.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-29T15:04:56.717" LastActivityDate="2015-03-29T15:04:56.717" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="11131" PostTypeId="2" ParentId="11112" CreationDate="2015-03-27T21:37:46.960" Score="2" Body="&lt;p&gt;Ok, Just a guess&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first 6 lines are from ShakeSpears Sonnets&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sonnet 3, Sonnet 60, etc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now from the 7th line, The point must come soon after when thou viewest, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It is indicated by the poem that the actual point of the poem comes right after the first line, which is 2nd line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Time doth transfix the flourish set on youth.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The 2nd line is from Sonnet 60. 60 min is in one hour&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When squareth it that debt all owe the hour,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; square(60) = 3600. 3600 seconds is in one hour as well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, I think the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;60&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1504" LastActivityDate="2015-03-27T21:37:46.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11132" PostTypeId="2" ParentId="10911" CreationDate="2015-03-27T21:51:08.377" Score="3" Body="&lt;p&gt;This is a second guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it a(n)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sucker&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My name sounds like a bird&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Yellow-bellied or Red-breasted SapSUCKER&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whole body like the tongue of a Lizard&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They get all sticky&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Am different colors&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Just about every brand of sucker comes in many colors, including suckers that have different colors&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And can well very be littered&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Discarded suckers can litter a floor, just look in any Elementary school hallway on November 1st.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My energy melts away over time&#xA;But not quite like an ice&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A suckers energy (and I don't know if this means caloric content or just the fact that the sucker exists) melts away, not quite like ice with repeated licking/sucking&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And when not taken care of,&#xA;I'll collect dust and lice&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The aforementioned stickiness can make all sorts of pick cling to a sucker.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;Is it a(n)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; arrow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My name sounds like a bird &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; sparrow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whole body is like tongue of a lizard &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; long, narrow and forked at the end&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Am different colors &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the fletchings can be very colorful&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I well can be very littered &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; definition of Litter from Merriam-Webster:&#xA; a messy pile or group of things&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My energy melts away over time &#xA;But not quite like an ice &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; arrows wear out over time. After a few years of use, they become less straight, and cannot fly as far.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And when not taken care of &#xA;I'll collect dust and lice &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If left unattended, the arrows will collect dust, and lice if the fletchings are made from true feathers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10715" LastEditorUserId="10715" LastEditDate="2015-03-30T14:32:33.503" LastActivityDate="2015-03-30T14:32:33.503" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11133" PostTypeId="2" ParentId="11092" CreationDate="2015-03-27T22:17:00.623" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;I have settled a little &lt;a href=&quot;http://tube.geogebra.org/student/m929171&quot; rel=&quot;nofollow noreferrer&quot;&gt;simulation&lt;/a&gt; with 6 arbitrary arrow releasers with two changeable parameters (coordinate of eventual collision and length of the arrow A) , to detect where the arrows would meet head to tail or contrarily . &lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First off we must separate cases ,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;when the arrow is unleached from any vertex , the angle $\theta$ which it makes with any side is not the same as that it makes with other side .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when the polygone is n sided , the sum of its angles = $(n-2) \pi$ so each angle is $\frac{(n-2)\pi}{n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/40Yxm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;strong&gt;Update:&lt;/strong&gt; since the arrow can hit another one or be hit in its tail the difference is fixed . Simulated scenario works fine because the computation is up to a distant system server .&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;in this picture lets calculate the parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ ac=\frac{x}{2cos\theta}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ ad= ac-A$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ tan\delta = \frac{ac*sin\theta}{x-ad*cos\theta} $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;$$ \delta_1(\theta,x) = arctg \frac{ac*sin\theta}{x-(ac-A)*cos\theta}$$&lt;/strong&gt;&#xA;$Cos\delta_2 = \frac {x-(Cos\theta*(ac+A+r))}{ac+r}$&#xA;&lt;strong&gt;$$ \delta_2(\theta,x) = arccos \frac{x-(ac+A+r)*cos\theta}{ac+r}$$&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;strong&gt;Update3:&lt;/strong&gt; after a looong struggle with a third degree polynomial equation i found $r$ as following: &lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$Cos(e)=\frac{\epsilon}{A+r}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Sin(e)=\frac{(A+r)^2-\epsilon^2}{A+r}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(A+r)^2-\epsilon^2=ac^2-(ac+r-\epsilon)^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$cos(\pi-(2\theta+e))=-cos(2\theta)cos(e)+sin(2\theta)sin(e)=\frac{ac+r-\epsilon}{ac}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;r&lt;/strong&gt; =&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;$(\frac{1/27}{(cos(2\theta) - 1/ac*(A - ac))^3}*(1/ac*(1/2*A^2 - ac^2 + (A + ac)*(A - ac)) + sin(2\theta)*(4*A*ac - 1/4*(2*A + 2*ac)^2 + ac^2) - 1/2*cos(2\theta)*(2*A + 4*ac))^3 - \frac{1}{(2*cos(2\theta) - 2/ac*(A - ac))}*(sin(2\theta)*(1/4*A^4 - A^2*ac^2) - A*(1/2*A^2 - ac^2) + 1/2*A^2*ac*cos(2\theta)) + ((\frac{1/27}{(cos(2\theta) - 1/ac*(A - ac))^3}*(1/ac*(1/2*A^2 - ac^2 + (A + ac)*(A - ac)) + sin(2\theta)*(4*A*ac - 1/4*(2*A + 2*ac)^2 + ac^2) - 1/2*cos(2\theta)*(2*A + 4*ac))^3 - \frac{1}{(2*cos(2\theta) - 2/ac*(A - ac))}*(cos(2\theta)*(1/4*A^4 - A^2*ac^2) - A*(1/2*A^2 - ac^2) + 1/2*A^2*ac*cos(2\theta)) + \frac{1/6}{(cos(2\theta) - 1/ac*(A - ac))^2}*(sin(2\theta)*(2*A*ac^2 + 2*A^2*ac - 1/2*A^2*(2*A + 2*ac)) - 1/2*cos(2\theta)*(ac*(2*A + 2*ac) + A^2) + 1/ac*((A + ac)*(1/2*A^2 - ac^2) + A*ac*(A - ac)))*(1/ac*(1/2*A^2 - ac^2 + (A + ac)*(A - ac)) + sin(2\theta)*(4*A*ac - 1/4*(2*A + 2*ac)^2 + ac^2) - 1/2*cos(2\theta)*(2*A + 4*ac)))^2 - (\frac{1/9}{(cos(2\theta) - 1/ac*(A - ac))^2}*(1/ac*(1/2*A^2 - ac^2 + (A + ac)*(A - ac)) + sin(2\theta)*(4*A*ac - 1/4*(2*A + 2*ac)^2 + ac^2) - 1/2*cos(2\theta)*(2*A + 4*ac))^2 + \frac{1}{(3*cos(2\theta) - 3/ac*(A - ac))}*(sin(2\theta)*(2*A*ac^2 + 2*A^2*ac - 1/2*A^2*(2*A + 2*ac)) - 1/2*cos(2\theta)*(ac*(2*A + 2*ac) + A^2) + 1/ac*((A + ac)*(1/2*A^2 - ac^2) + A*ac*(A - ac))))^3)^{1/2} + \frac{1/6}{(cos(2\theta) - 1/ac*(A - ac))^2}*(sin(2\theta)*(2*A*ac^2 + 2*A^2*ac - 1/2*A^2*(2*A + 2*ac)) - 1/2*cos(2\theta)*(ac*(2*A + 2*ac) + A^2) + 1/ac*((A + ac)*(1/2*A^2 - ac^2) + A*ac*(A - ac)))*(1/ac*(1/2*A^2 - ac^2 + (A + ac)*(A - ac)) + sin(2\theta)*(4*A*ac - 1/4*(2*A + 2*ac)^2 + ac^2) - 1/2*cos(2\theta)*(2*A + 4*ac)))^{1/3} + \frac{1}{(3*cos(2\theta) - 3/ac*(A - ac))}*(1/ac*(1/2*A^2 - ac^2 + (A + ac)*(A - ac)) + sin(2\theta)*(4*A*ac - 1/4*(2*A + 2*ac)^2 + ac^2) - 1/2*cos(2\theta)*(2*A + 4*ac)) + (\frac{1/9}{(cos(2\theta) - 1/ac*(A - ac))^2}*(1/ac*(1/2*A^2 - ac^2 + (A + ac)*(A - ac)) + sin(2\theta)*(4*A*ac - 1/4*(2*A + 2*ac)^2 + ac^2) - 1/2*cos(2\theta)*(2*A + 4*ac))^2 + \frac{1}{(3*cos(2\theta) - 3/ac*(A - ac))}*(sin(2\theta)*(2*A*ac^2 + 2*A^2*ac - 1/2*A^2*(2*A + 2*ac)) - 1/2*cos(2\theta)*(ac*(2*A + 2*ac) + A^2) + 1/ac*((A + ac)*(1/2*A^2 - ac^2) + A*ac*(A - ac))))/(\frac{1/27}{(cos(2\theta) - 1/ac*(A - ac))^3}*(1/ac*(1/2*A^2 - ac^2 + (A + ac)*(A - ac)) + sin(2\theta)*(4*A*ac - 1/4*(2*A + 2*ac)^2 + ac^2) - 1/2*cos(2\theta)*(2*A + 4*ac))^3 - 1/(2*cos(2\theta) - 2/ac*(A - ac))*(sin(2\theta)*(1/4*A^4 - A^2*ac^2) - A*(1/2*A^2 - ac^2) + 1/2*A^2*ac*cos(2\theta)) + ((\frac{1/27}{(cos(2\theta) - 1/ac*(A - ac))^3}*(1/ac*(1/2*A^2 - ac^2 + (A + ac)*(A - ac)) + sin(2\theta)*(4*A*ac - 1/4*(2*A + 2*ac)^2 + ac^2) - 1/2*cos(2\theta)*(2*A + 4*ac))^3 - \frac{1}{(2*cos(2\theta) - 2/ac*(A - ac))}*(sin(2\theta)*(1/4*A^4 - A^2*ac^2) - A*(1/2*A^2 - ac^2) + 1/2*A^2*ac*cos(2\theta)) + \frac{1/6}{(cos(2\theta) - 1/ac*(A - ac))^2}*(sin(2\theta)*(2*A*ac^2 + 2*A^2*ac - 1/2*A^2*(2*A + 2*ac)) - 1/2*cos(2\theta)*(ac*(2*A + 2*ac) + A^2) + 1/ac*((A + ac)*(1/2*A^2 - ac^2) + A*ac*(A - ac)))*(1/ac*(1/2*A^2 - ac^2 + (A + ac)*(A - ac)) + sin(2\theta)*(4*A*ac - 1/4*(2*A + 2*ac)^2 + ac^2) - 1/2*cos(2\theta)*(2*A + 4*ac)))^2 - (\frac{1/9}{(cos(2\theta) - 1/ac*(A - ac))^2}*(1/ac*(1/2*A^2 - ac^2 + (A + ac)*(A - ac)) + sin(2\theta)*(4*A*ac - 1/4*(2*A + 2*ac)^2 + ac^2) - 1/2*cos(2\theta)*(2*A + 4*ac))^2 + 1/(3*cos(2\theta) - 3/ac*(A - ac))*(sin(2\theta)*(2*A*ac^2 + 2*A^2*ac - 1/2*A^2*(2*A + 2*ac)) - 1/2*cos(2\theta)*(ac*(2*A + 2*ac) + A^2) + 1/ac*((A + ac)*(1/2*A^2 - ac^2) + A*ac*(A - ac))))^3)^{1/2} + 1/6/(cos(2\theta) - 1/ac*(A - ac))^2*(sin(2\theta)*(2*A*ac^2 + 2*A^2*ac - 1/2*A^2*(2*A + 2*ac)) - 1/2*cos(2\theta)*(ac*(2*A + 2*ac) + A^2) + 1/ac*((A + ac)*(1/2*A^2 - ac^2) + A*ac*(A - ac)))*(1/ac*(1/2*A^2 - ac^2 + (A + ac)*(A - ac)) + sin(2\theta)*(4*A*ac - 1/4*(2*A + 2*ac)^2 + ac^2) - 1/2*cos(2\theta)*(2*A + 4*ac)))^{1/3}$&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now lets calculate $x$ in function of $k$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x=L=1$ for the nearest vertex&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the radius of the circle which crosses all vertexes of regular polygon of l =length of one of n sides is $\frac{l}{2sin(\pi/n)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the distance between two vetexes belonging to that circle of radius = r is $2r sin\beta/2  $ where $\beta$ is the angle formed by the centeroid and two vertexes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;when $\beta=2\pi/(n-k+1)$ and $r=\frac{1}{2sin(\pi/N)}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x=2r sin\beta/2 = \frac{sin(\pi/(N-k+1))}{sin(\pi/N)}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;this relation works fine for k=1 , x= 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now lets calculate the probability&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;strong&gt;Update2 :&lt;/strong&gt; removing intersections&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$ P1= \int_{0}^{\frac{(n-2)\pi}{n(n-3)}} \frac{1}{\frac{(n-2)\pi}{n}} ( |\delta_1(\theta,1) - \delta_2(\theta,1)|+(1-\frac{|\delta_1(\theta,1) - \delta_2(\theta,1)|}{\frac{(n-2)\pi}{n}})|\delta_1(\frac{(n-2)\pi}{n(n-3)}-\theta,\frac{sin(\pi/(n-1))}{sin(\pi/n)}) - \delta_2(\frac{(n-2)\pi}{n(n-3)}-\theta,\frac{sin(\pi/(n-1))}{sin(\pi/n)})| + (1-(\frac{(1-((1-\frac{|\delta_1(\theta) - \delta_2(\theta)|}{\frac{(n-2)\pi}{n}})|\delta_1(\frac{(n-2)\pi}{n(n-3)}-\theta,\frac{sin(\pi/(n-1))}{sin(\pi/n)}) - \delta_2(\frac{(n-2)\pi}{n(n-3)}-\theta,\frac{sin(\pi/(n-1))}{sin(\pi/n)})|)}{\frac{(n-2)\pi}{n}}))|\delta_1(\frac{2(n-2)\pi}{n(n-3)}-\theta,\frac{sin(\pi/(n-2))}{sin(\pi/n)}) - \delta_2(\frac{2(n-2)\pi}{n(n-3)}-\theta,\frac{sin(\pi/(n-2))}{sin(\pi/n)})|+ ... *|\delta_1(\frac{(n-2)(n-2)\pi}{n(n-3)}-\theta,\frac{sin(\pi/2)}{sin(\pi/n)}) - \delta_2(\frac{(n-2)(n-2)\pi}{n(n-3)}-\theta,\frac{sin(\pi/2)}{sin(\pi/n)})| )d\theta$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ P1= \int_{0}^{\frac{(n-2)\pi}{n(n-3)}} \frac{1}{\frac{(n-2)\pi}{n}} ( |\delta_1(\theta,1) - \delta_2(\theta,1)|+(1-\frac{|\delta_1(\theta,1) - \delta_2(\theta,1)|}{\frac{(n-2)\pi}{n}})|\delta_1(\theta-\frac{(n-2)\pi}{n(n-3)},\frac{sin(\pi/(n-1))}{sin(\pi/n)}) - \delta_2(\theta-\frac{(n-2)\pi}{n(n-3)},\frac{sin(\pi/(n-1))}{sin(\pi/n)})| + (1-(\frac{(1-((1-\frac{|\delta_1(\theta,1) - \delta_2(\theta,1)|}{\frac{(n-2)\pi}{n}})|\delta_1(\theta-\frac{(n-2)\pi}{n(n-3)},\frac{sin(\pi/(n-1))}{sin(\pi/n)}) - \delta_2(\theta-\frac{(n-2)\pi}{n(n-3)},\frac{sin(\pi/(n-1))}{sin(\pi/n)})|)}{\frac{(n-2)\pi}{n}}))|\delta_1(\frac{2(n-2)\pi}{n(n-3)}-\theta,\frac{sin(\pi/(n-2))}{sin(\pi/n)}) - \delta_2(\frac{2(n-2)\pi}{n(n-3)}-\theta,\frac{sin(\pi/(n-2))}{sin(\pi/n)})|+ ... |\delta_1(\frac{(n-2)(n-2)\pi}{n(n-3)}-\theta,\frac{sin(\pi/2)}{sin(\pi/n)}) - \delta_2(\frac{(n-2)(n-2)\pi}{n(n-3)}-\theta)| ,\frac{sin(\pi/2)}{sin(\pi/n)})d\theta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;.... until $P(n-1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;each $\delta_n$ is writen in function of $x_n$ and last $\theta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;holy sh** this doesnt seem to end yet  , we must calculate the common probability and subtract it from P , following this rule:&#xA;$P1 \bigcup P2 = P1+P2-P1 \bigcap P2$&#xA;&lt;img src=&quot;https://i.stack.imgur.com/j5h0W.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so probability of the arrows to be in ollision is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P = p_1+ ... + p_{n-1} - P_1 \bigcap  ....\bigcap P_{n-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The global form of the probability for any arrow i to be hit by atleast one among n-1-i arrows coming from counter-clockwise direction is:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;p&gt;$U_m=&#xA;\frac{(1-U_{m-1})}{\frac{(n-2)\pi}{n}}  &#xA;| \delta_1(| \theta-\frac{(n-2)m\pi}{n(n-3))} |,\frac{sin(\pi/(n-m))}{sin(\pi/n)})  - \delta_2(| \theta-\frac{(n-2)m\pi}{n(n-3))} |,\frac{sin(\pi/(n-m))}{sin(\pi/n)}) |$&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;&lt;strong&gt;$$Pr_i=&#xA;\int_{0}^{\frac{(n-2)\pi}{n}}(\sum_{k=0}^{n-2-i}U_k) d\theta$$&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where $i =  0 .. n-2$ and $U_{-1} is Idiomatically =0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The general probability for any two arrows atleast to be in collision is:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;$Pr=Pr_0+(1-Pr_0)Pr_1+(1-(1-Pr_0)Pr_1)Pr_2 ..... Pr_{n-2}$&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;probability to be not is just&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;strong&gt;1-Pr&lt;/strong&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Now i think i did subtly detracted the pobability range to more convenient values , eventhough it is more complicated but more accurate , now any one is usual with primitive calculation is welcome to assist here.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;-this answer-appendix depends on the meaning of the word 'chance' used by the op&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;for  $n-1$ arbitrary values of $\theta$ this following table represents the condition for a non-eventual clash&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{|c|c|c||c|c|c|}&#xA;\hline&#xA;angle &amp;amp;\theta_1&amp;amp; \theta_2&amp;amp; \theta_k&amp;amp; ...&amp;amp; \theta_{n-1}\\\hline&#xA;\notin [value_1,value_2]= &amp;amp; [\delta_1(\theta_2,1),\delta_2(\theta_2,1)]&amp;amp; .&amp;amp; .&amp;amp; .&amp;amp; .\\\hline&#xA;for\ any\ \theta_k&amp;amp; \frac{(n-2)(k-2)\pi}{n(n-3)} \pm[ \delta_1(| \theta_k-\frac{(n-2)(k-2)\pi}{n(n-3)} |,\frac{sin(\pi/(n-k+2))}{sin(\pi/n)})  , \delta_2(| \theta_k-\frac{(n-2)(k-2)\pi}{n(n-3)} |,\frac{sin(\pi/(n-k+2))}{sin(\pi/n)})]&amp;amp; .&amp;amp; .&amp;amp; .&amp;amp; .\\\hline&#xA;for\ last\ \theta_k &amp;amp;\frac{(n-2)(n-3)\pi}{n(n-3)} \pm[ \delta_1(| \theta_{n-1}-\frac{(n-2)(n-3)\pi}{n(n-3)} |,\frac{sin(\pi/2)}{sin(\pi/n)})  ,\delta_2(| \theta_{n-1}-\frac{(n-2)(n-3)\pi}{n(n-3)} |,\frac{sin(\pi/2)}{sin(\pi/n)})] &amp;amp; .&amp;amp; .&amp;amp; .&amp;amp; .\\\hline&#xA;\end{array}&#xA;\qquad$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;this is my second &lt;a href=&quot;http://tube.geogebra.org/student/m936537&quot; rel=&quot;nofollow noreferrer&quot;&gt;simulation&lt;/a&gt; version with 6 arbitrary arrow releasers with two changeable parameters (track of the arrow thrown with a specific angle $\theta$ and length of the arrow A) , to detect all probable impacts with other lower and higher bounds of ather 5 arrows' possible gradients. &lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To be continued ...&lt;/p&gt;&#xA;" OwnerUserId="9517" LastEditorUserId="9517" LastEditDate="2015-04-06T22:37:27.253" LastActivityDate="2015-04-06T22:37:27.253" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11134" PostTypeId="1" AcceptedAnswerId="11140" CreationDate="2015-03-27T22:17:39.693" Score="4" ViewCount="425" Body="&lt;p&gt;A colleague at work gives you a slip of paper at work one day, with symbols written on it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;・のいろ・はおに・・&lt;br&gt;&#xA;  みほくへ・やしと・ち&lt;br&gt;&#xA;  ・ま・り・・・・・ぬ&lt;br&gt;&#xA;  ・るゑ・ひ・・を・け&lt;br&gt;&#xA;  もわ・かせふ・よ・・&lt;br&gt;&#xA;  ・・こた・え・・・れ&lt;br&gt;&#xA;  ・そ・・・・・つて・&lt;br&gt;&#xA;  あね・な・・・・さら&lt;br&gt;&#xA;  ・・・む・・き・・う&lt;br&gt;&#xA;  ・ゆ・・・・・ゐめ・  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;す = ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Every symbol in Japanese hiragana appears on the grid except for す and ん. The question is simple: Where does す go, and what number corresponds to it?&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-03-28T00:38:23.597" Title="答えはまた二桁の数字、その３ (The answer is another two-digit number, #3)" Tags="|mathematics|language|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11135" PostTypeId="2" ParentId="11112" CreationDate="2015-03-27T23:27:55.263" Score="4" Body="&lt;p&gt;As other answerers have pointed out, the first six lines are from Shakespeare's sonnets, numbered 3, 60, 5, 5, 5, and 12. However, I believe that they have been misinterpreting the mathematics.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The point must come soon after when thou viewest,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A decimal point must be placed somewhere near the start of those numbers. With this information alone we can't do anything yet, but supposedly it implies that those numbers must be strung together to create 36055512, and then a decimal point inserted somewhere.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When squareth it that debt all owe the hour,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then the number obtained by placing that decimal point must be squared, and then the difference from 60 taken of the result. If we square 3.6055512, the result is 12.99999945582144. The difference from this number and 60 is 47.00000054417856.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And then thine sharpened figure shall be rounded.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This simply means &quot;round your result to the nearest integer&quot;. This gives us a result of &lt;strong&gt;47&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-03-27T23:27:55.263" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11136" PostTypeId="2" ParentId="11112" CreationDate="2015-03-28T00:02:30.627" Score="2" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As leoll2 said, the first 6 lines are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;from Shakespeare's sonnets, respectively:&lt;br&gt;&#xA;&lt;br&gt;&#xA; Sonnet 3&lt;br&gt;&#xA; Sonnet 60&lt;br&gt;&#xA; Sonnet 5&lt;br&gt;&#xA; Sonnet 5&lt;br&gt;&#xA; Sonnet 5&lt;br&gt;&#xA; Sonnet 12&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now the last 3 lines:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The point must come soon after when thou viewest,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;when thou viewest&quot; is a reference to the 1st line, so&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sonnet 3.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;When squareth it that debt all owe the hour,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Square 3 to get 9. &quot;Owe&quot; sounds like &quot;over&quot; (as in division), and &quot;hour&quot; is a reference to line 3 (Sonnet 5). So we divide 9 by 5 to get 1.8.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And thine sharpened figure shall be rounded.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Round the number 1.8 to get &lt;strong&gt;2&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-03-28T00:02:30.627" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11137" PostTypeId="1" AcceptedAnswerId="11138" CreationDate="2015-03-28T00:20:54.770" Score="4" ViewCount="295" Body="&lt;p&gt;I'm thinking of 4 real numbers, $a,b,c$ and $d$ (they need not be integers, or even rational). There are 6 products you can make by multiplying pairs of these numbers, namely, $ab,ac,ad,bc,bd$ and $cd$. I'll tell you what five of those products are, in no particular order:&#xA;$$&#xA;2,3,4,5 \text{ and }6&#xA;$$&#xA;What is the sixth product?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Small hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can do this without needing to figure out what $a,b,c$ and $d$ are.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a fun puzzle I found on the &lt;a href=&quot;http://wiki.xkcd.com/irc/Puzzles&quot; rel=&quot;nofollow&quot;&gt;xkcd puzzle wiki&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-03-28T00:42:18.507" LastActivityDate="2015-03-28T00:42:18.507" Title="What is the Sixth Number?" Tags="|mathematics|calculation-puzzle|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11138" PostTypeId="2" ParentId="11137" CreationDate="2015-03-28T00:28:41.453" Score="4" Body="&lt;p&gt;The six numbers can be split up into pairs such that the product of the two numbers in each pair is the same across all three pairs: $(ab,cd),(ac,bd),(ad,bc)$. The only two pairs of numbers among $2,3,4,5,6$ which have the same product are $(3,4),(2,6)$. So the sixth number must be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\frac{12}{5}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nice quick puzzle :-)&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-03-28T00:28:41.453" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11139" PostTypeId="1" AcceptedAnswerId="11480" CreationDate="2015-03-28T00:32:41.830" Score="4" ViewCount="307" Body="&lt;p&gt;The following thirteen words have a special property common to all of them (though they're not the only words with this property):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;SADIST, THOROUGHFARE, HYMN, POISED, YURT, THIRTY-SEVEN, STREWN, VERTICALITY, INERTIA, BALDERDASH, HIGHFLYING, REWARDING, OPINIONATED&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Firstly, what is this property? Secondly, which two words in the list have the property &lt;em&gt;more&lt;/em&gt; than the other eleven?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the idea behind this puzzle also appears in another puzzle posted in the last few days, although that one is less enigmatic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2016-09-18T20:28:26.867" LastActivityDate="2016-09-18T20:28:26.867" Title="A Company of Thirteen #3" Tags="|pattern|word-property|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11140" PostTypeId="2" ParentId="11134" CreationDate="2015-03-28T00:38:23.597" Score="7" Body="&lt;p&gt;Solved it :-) The puzzle is based on two things:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Iroha&quot; rel=&quot;noreferrer&quot;&gt;いろは&lt;/a&gt;&lt;/strong&gt; (iroha), an ancient Japanese poem that uses every hiragana character once only (except for ん),&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;a href=&quot;https://ja.wikipedia.org/wiki/%E7%B4%A0%E6%95%B0&quot; rel=&quot;noreferrer&quot;&gt;素数&lt;/a&gt;&lt;/strong&gt;, oops, sorry. I mean &lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Prime_number&quot; rel=&quot;noreferrer&quot;&gt;prime numbers&lt;/a&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Specifically,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The 10×10 grid can be regarded as a list of 100 cells numbered from 0 to 99. Each character of the poem is placed at a cell corresponding to a prime number. So the first character い goes in cell #2, the second character ろ goes in cell #3, the characters はにほへと go in cells 5, 7, 11, 13 and 17, and so on.&lt;br/&gt;&#xA; &lt;br/&gt;&#xA; For prime numbers greater than 99, we have to start again at the beginning of the list, but this time we skip over cells that are already filled. So for example the 26th character の is associated with the prime number 101 and thus goes in cell #1. But the 27th character (for the prime number 103) ends up in cell #6 because we have to step over three characters (い, ろ and は).&lt;br/&gt;&#xA; &lt;br/&gt;&#xA; Carry on in this fashion to the end of the poem, and you'll end up with the following grid, where characters that are filled in on the first, second and third pass are coloured red, orange and blue.&lt;br/&gt;&#xA; &lt;br/&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/PexGf.png&quot; alt=&quot;いろは&quot;&gt;&lt;br/&gt;&#xA; &lt;br/&gt;&#xA; The final character す can be found in the 4th cell of the 7th row, which corresponds to the number &lt;strong&gt;63&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="40" LastActivityDate="2015-03-28T00:38:23.597" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11141" PostTypeId="1" AcceptedAnswerId="11189" CreationDate="2015-03-28T01:13:20.113" Score="4" ViewCount="1068" Body="&lt;p&gt;This is a question from a previous, now-defunct version of the &lt;a href=&quot;http://joezeng.com/quetzalcoatl/&quot; rel=&quot;nofollow&quot;&gt;Temple of Quetzalcoatl&lt;/a&gt;, that I've posted here as an exemplar of one of the difficult puzzles you might encounter while solving it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The puzzle is simply this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1845&lt;br&gt;&#xA;  6745&lt;br&gt;&#xA;  0738&lt;br&gt;&#xA;  0956&lt;br&gt;&#xA;  ????&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What four-digit number goes into the question marks?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hint in the source code was presented as such:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; -1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-07-13T18:33:13.267" LastActivityDate="2015-07-13T18:33:13.267" Title="A simple but most devious sequence puzzle" Tags="|number-sequence|" AnswerCount="3" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="11143" PostTypeId="2" ParentId="11105" CreationDate="2015-03-28T01:56:21.960" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are &lt;strong&gt;the fuel tank of a cargo helicopter&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gv1P4.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Rotors are airfoils, but they're not wings.&lt;/li&gt;&#xA;&lt;li&gt;Accelerated air is wind.  Downwardly accelerated air creates lift.&lt;/li&gt;&#xA;&lt;li&gt;When the fuel tank becomes empty, the helicopter suddenly becomes much heavier.&lt;/li&gt;&#xA;&lt;li&gt;The bigger the tank, the more you can carry...&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2721" LastActivityDate="2015-03-28T01:56:21.960" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11144" PostTypeId="2" ParentId="10367" CreationDate="2015-03-28T01:58:02.513" Score="2" Body="&lt;p&gt;&lt;strong&gt;Sequence 3&lt;/strong&gt; is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the last 2 digits of the 11 users with the highest rep on puzzling.se. The sequence in the question has been revised as the order has been changing. The missing numbers are shown in bold:&lt;br&gt;&#xA; ver5: 73, 74, 88, 51, 28, 36, 84, 50, 83, 99, &lt;strong&gt;88&lt;/strong&gt;&lt;br&gt;&#xA; ver4: 73, 74, 88, 51, 28, 36, 84, 83, 50, &lt;strong&gt;20&lt;/strong&gt;, 99&lt;br&gt;&#xA; ver1: 73, 74, 88, &lt;strong&gt;51&lt;/strong&gt;, 28, 36, 84, 83, &lt;strong&gt;20&lt;/strong&gt;, 99, 50   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sequence 4&lt;/strong&gt; is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the atomic numbers for the elements in reverse alphabetical order of the chemical symbols:&lt;br&gt;&#xA;&lt;br&gt;&#xA; Zr, Zn, Yt, Y, Xe, W, V, &lt;strong&gt;Uut&lt;/strong&gt;&lt;br&gt;&#xA; 40, 30, 70, 39, 54, 74, 23, &lt;strong&gt;113&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-04-28T08:41:40.403" LastActivityDate="2015-04-28T08:41:40.403" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11146" PostTypeId="1" CreationDate="2015-03-28T04:49:17.357" Score="2" ViewCount="228" Body="&lt;p&gt;The following diagram depicts a Pass the Baton tournament&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UlpJf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;strong&gt;Note: due to a merge, some of the answers below use &lt;a href=&quot;https://i.stack.imgur.com/zfRoR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;this image&lt;/a&gt; instead.&lt;/strong&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each team consists of 6 members, 2 at the central point $E$, and one on each side of the quadrilateral, $a$, $b$, $c$, and $d$. The members of the race run to the sides in the following order: $E$ to side $a$, $a$ to $b$, $b$ back to $E$, $E$ to $c$, $c$ to $d$, and $d$ back to $E$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The challenge for the teams is not only to run as fast as they can but to select the best route, by selecting the best points on $a$, $b$, $c$, and $d$ to have the relay runners stand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is this route?&lt;/p&gt;&#xA;" OwnerUserId="9531" LastEditorDisplayName="user20" LastEditDate="2016-08-10T07:59:24.163" LastActivityDate="2016-08-10T07:59:24.163" Title="Pass the Baton competition 6 members teams" Tags="|geometry|optimization|" AnswerCount="8" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="11147" PostTypeId="2" ParentId="11146" CreationDate="2015-03-28T05:31:23.350" Score="2" Body="&lt;p&gt;Here's a brute-force solution involving calculus, in the absence of anything more elegant at the moment.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let $p_a, p_b, p_c, p_d$ be the proportional position of the points that the relay racers run to, with:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$p_a, p_d = 0$ being at point $A$, &lt;/li&gt;&#xA;&lt;li&gt;$p_b, p_c = 0$ being at point $C$,&lt;/li&gt;&#xA;&lt;li&gt;$p_a, p_b = 1$ being at point $B$, and&lt;/li&gt;&#xA;&lt;li&gt;$p_c, p_d = 1$ being at point $D$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then, for example, if $p_a = 0.5$, then the racer touches $a$ at its midpoint, and if $p_a = 0.75$, then the racer touches $a$ three-quarters of the way toward point $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we've defined these, then we can define a total distance function $f(p_a, p_b)$ as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;f(p_a, p_b) =\ &amp;amp;\sqrt{(EA \cdot p_a)^2 + (EB \cdot (1 - p_a))^2} +\\&#xA;&amp;amp;\sqrt{(EC \cdot p_b)^2 + (EB \cdot (1 - p_b))^2} +\\&#xA;&amp;amp;\sqrt{(EC \cdot p_b + EA \cdot p_a)^2 + (EB \cdot (p_a - p_b))^2}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This function appears to be differentiable, and graphing them in Wolfram Alpha shows that they have a global minimum when their partial derivatives are both equal to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can define $g(p_c, p_d)$ similarly, and since they're independent, we can minimize each of them individually and that will minimize the sum.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The partial derivatives are sort of tedious, though, so I'll put them off until later.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-03-28T05:31:23.350" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11148" PostTypeId="5" CreationDate="2015-03-28T05:32:09.530" Score="0" Body="&lt;p&gt;As discussed in &quot;&lt;a href=&quot;https://puzzling.meta.stackexchange.com/questions/2876/tag-proposal-for-puzzles-that-require-computer-literacy&quot;&gt;Tag proposal for puzzles that require computer literacy&lt;/a&gt;&quot;, this tag is designated for questions where the solution requires computer literacy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vaguely, this means that the puzzle may have:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Some element of operating system usage skill&lt;/li&gt;&#xA;&lt;li&gt;Some requirement of computer programming skill&lt;/li&gt;&#xA;&lt;li&gt;Some knowledge of image and/or other encoding techniques&lt;/li&gt;&#xA;&lt;li&gt;Familiarity with digital communication&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Compare with the &lt;a href=&quot;/questions/tagged/no-computers&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;no-computers&amp;#39;&quot; rel=&quot;tag&quot;&gt;no-computers&lt;/a&gt; tag.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorUserId="11110" LastEditDate="2020-04-30T07:46:14.277" LastActivityDate="2020-04-30T07:46:14.277" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="11149" PostTypeId="4" CreationDate="2015-03-28T05:32:09.530" Score="0" Body="A tag for puzzles dependent upon some degree of computer literacy - for instance, requiring basic familiarity with operating system usage, encryption or programming." OwnerDisplayName="user20" LastEditorUserId="4450" LastEditDate="2015-03-28T06:29:39.283" LastActivityDate="2015-03-28T06:29:39.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11151" PostTypeId="2" ParentId="11105" CreationDate="2015-03-28T05:37:03.553" Score="2" Body="&lt;p&gt;&lt;strong&gt;Second Guess&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Paper&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can fly without wings&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Paper doesn't have wings. Yet it glides along the wind.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;only carried by the wind&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Paper can get carried by wind.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am heavier when empty&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Darn! I don't know...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the bigger I am, the more I carry&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A blgger piece of paper contains more information than a smaller piece of paper&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Original Guess&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; words&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can fly without wings, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Words fly all over the place&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;only carried by the wind.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Words are, when spoken,sound: carried by wind.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am heavier when empty,&#xA;the bigger I am, the more I can carry.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Big words/phrases carry more information.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10603" LastEditorUserId="2368" LastEditDate="2015-03-28T09:20:46.813" LastActivityDate="2015-03-28T09:20:46.813" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11152" PostTypeId="2" ParentId="10935" CreationDate="2015-03-28T05:50:54.030" Score="12" Body="&lt;p&gt;Sequence 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Order of speakers in Shakespeare's Romeo and Juliet Act I, Scene ii:&lt;br&gt;&#xA; &lt;strong&gt;C&lt;/strong&gt;apulet, &lt;strong&gt;P&lt;/strong&gt;aris, &lt;strong&gt;C&lt;/strong&gt;apulet, ...&lt;br&gt;&#xA; The full sequence is C, P, C, P, C, S, B, R, B, R, B, R, S, R, S, R, S, R, S, R, S, R, S, R, S, B, R, B, R&lt;br&gt;&#xA; and the missing elements are C, S, and B.&#xA; (S = Peter the servant)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10724" LastEditorUserId="10724" LastEditDate="2015-03-28T05:58:54.067" LastActivityDate="2015-03-28T05:58:54.067" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11154" PostTypeId="1" AcceptedAnswerId="11158" CreationDate="2015-03-28T06:12:25.247" Score="11" ViewCount="791" Body="&lt;p&gt;Given the six, four digit numbers below, find me a seventh that matches the pattern.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2058&#xA;1724&#xA;1234&#xA;6245&#xA;5683&#xA;8756&#xA;????&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note that there are many possible answers, but if you find the pattern, it should be absolutely obvious if it's correct or not.&lt;/p&gt;&#xA;" OwnerUserId="5208" LastActivityDate="2015-03-28T18:15:45.227" Title="Fill in the blank to win" Tags="|pattern|" AnswerCount="1" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11155" PostTypeId="1" CreationDate="2015-03-28T06:24:19.467" Score="1" ViewCount="358" Body="&lt;p&gt;A 3x3 box is given to you with two blank squares. You have to fill a letter in each to form a nine lettered word beginning from one of the corners and ending in the middle while going clockwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AYXzb.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you do it?&lt;/p&gt;&#xA;" OwnerUserId="10725" LastActivityDate="2015-03-28T06:31:49.377" Title="Letter Word Sequence Riddle" Tags="|word|word-sequence|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11156" PostTypeId="2" ParentId="11155" CreationDate="2015-03-28T06:31:49.377" Score="12" Body="&lt;p&gt;The word is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&#xA; &lt;strong&gt;AVALANCHE&lt;/strong&gt;  starting in top right corner.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastActivityDate="2015-03-28T06:31:49.377" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11157" PostTypeId="2" ParentId="11146" CreationDate="2015-03-28T06:32:02.200" Score="4" Body="&lt;p&gt;In any locally minimal solution, the paths must reflect off the edges, with equal angle of incidence and reflection. This is because reflecting a leg of this this path makes a straight line between one point and the reflection of the other, the shortest distance between two points. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we can find the shortest path by reflecting the figure over a, then the result over b, and connecting the original point E to the final one. This gives a cycle from E to a to b to E. We then do the same for c and d. Taking the union of two cycles gives the full cycles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Sorry, I can't draw a diagram now because I'm on my phone. )&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-03-29T22:19:43.643" LastActivityDate="2015-03-29T22:19:43.643" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="11158" PostTypeId="2" ParentId="11154" CreationDate="2015-03-28T07:03:49.470" Score="12" Body="&lt;p&gt;Consider the number pad on a phone:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 2 3&#xA;4 5 6&#xA;7 8 9&#xA;* 0 #&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now, consider the positions that each of the numbers fills.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;. 2 .   1 2 .   1 2 3   . 2 .   . . 3   . . .&#xA;. 5 .   4 . .   4 . .   4 5 6   . 5 6   . 5 6&#xA;. 8 .   7 . .   . . .   . . .   . 8 .   7 8 .&#xA;  0       .       .       .       .       .  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;These are six of the seven pieces that appears in Tetris. The last remaining piece is the square, which can appear in four different positions:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 2 .   . . .   . 2 3   . . .&#xA;4 5 .   4 5 .   . 5 6   . 5 6&#xA;. . .   7 8 .   . . .   . 8 9&#xA;  .       .       .       .  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So any permutation of &lt;strong&gt;1245&lt;/strong&gt;, &lt;strong&gt;2356&lt;/strong&gt;, &lt;strong&gt;4578&lt;/strong&gt;, or &lt;strong&gt;5689&lt;/strong&gt; would be a correct answer.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-03-28T18:15:45.227" LastActivityDate="2015-03-28T18:15:45.227" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11159" PostTypeId="1" CreationDate="2015-03-28T07:34:25.973" Score="2" ViewCount="8963" Body="&lt;p&gt;Create a number using only the digits 4,4,3,3,2,2,1 and 1. So it can only be eight digits. You have to make sure the ones are separated by one digit, the twos are separated by two digits, the threes are separated with three digits and the fours are separated by four digits.&lt;/p&gt;&#xA;" OwnerUserId="10725" LastEditorUserId="11110" LastEditDate="2019-03-20T15:36:13.667" LastActivityDate="2019-03-20T15:36:13.667" Title="Math Number Sequence Riddle with 4 numbers 8 digits" Tags="|mathematics|number-sequence|" AnswerCount="1" CommentCount="2" ClosedDate="2015-03-28T15:11:38.977" ContentLicense="CC BY-SA 4.0" />
  <row Id="11160" PostTypeId="2" ParentId="11159" CreationDate="2015-03-28T07:48:47.687" Score="1" Body="&lt;p&gt;The answer is 41312432. I'm not sure how to explain it much further past that; it's mostly just trial and error.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-03-28T07:48:47.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11161" PostTypeId="1" AcceptedAnswerId="11165" CreationDate="2015-03-28T08:18:53.917" Score="10" ViewCount="736" Body="&lt;p&gt;Q. Two students A and B think of a positive integer. They tell the teacher their positive integer, but they don't tell each other their positive integer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The teacher tells both of them two different positive integers, and says that one integer is the sum of two integers by A and B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For example, if A thought of $3$ and B thought of $5$, The teacher would say $8$ and $9$, or $8$ and $4$, ...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The teacher then asks student A, &quot;Do you know which one is the sum?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If A says &quot;No&quot;, the teacher would ask B the same question.If B says &quot;No&quot;, A need to answer the question again, and B and so on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that they can figure out the sum after enough questions asked.&lt;/p&gt;&#xA;" OwnerUserId="10726" LastEditorUserId="10615" LastEditDate="2016-02-24T04:06:19.180" LastActivityDate="2016-02-24T04:06:19.180" Title="Two students guessing positive integer" Tags="|mathematics|logical-deduction|meta-knowledge|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11163" PostTypeId="1" AcceptedAnswerId="11610" CreationDate="2015-03-28T10:24:10.190" Score="33" ViewCount="3330" Body="&lt;p&gt;I was walking through the desert and came across an ancient contraption of cogs and gears. It had a series of dials on one side that appeared to allow the input and output of numbers. On the other side was an enormous lever. It looked far too old to possibly work, but curiosity got the best of me, so I dialled in the number &lt;code&gt;1&lt;/code&gt; and pulled the lever. The machine burst to life with a crunch of gears and a cloud of dust. As the dust cleared, I saw the dials slow to a halt leaving it displaying &lt;code&gt;1&lt;/code&gt; again.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 =&amp;gt; 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So far, so dull. I tried a few more:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2 =&amp;gt; 2&#xA;3 =&amp;gt; 3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Hmm... Maybe double digits would be more interesting:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;82 =&amp;gt; 15&#xA;47 =&amp;gt; 122&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Ok, now we're getting somewhere... Never one to be bogged down with the tedium of the scientific method, I dialled in a number at random and pulled the lever multiple times in succession:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;58458 =&amp;gt; 2433 =&amp;gt; 316 =&amp;gt; 411 =&amp;gt; 112&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;A sandstorm loomed and I had to abandon the machine to the elements as it was too big to carry. It's likely been swallowed by the sands of the desert once more, but its secrets still haunt my dreams. Help me solve the mystery. What function is the machine applying to transform its input?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit (hint 1)&lt;/strong&gt;: I just found my old notebook which I thought was lost in the sandstorm! It has some more sequences that I'd forgotten I'd tried, maybe they'll help you find the solution...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;4 =&amp;gt; 11&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;15 =&amp;gt; 11&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;19 =&amp;gt; 21&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit (hint 2)&lt;/strong&gt;: &lt;del&gt;I was lying awake last night thinking of the bizarre contraption and its numbered dials, when I was struck by an epiphany. I sat bolt upright in bed as I made the realisation, shocked that it hadn't dawned on me before, and angry at myself because it seems like such a pivotal observation. Anyway, I recalled that:&lt;/del&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;del&gt;None of the dials contained a digit &lt;code&gt;0&lt;/code&gt;, meaning that no number (input or output) could ever have a zero in it... No &lt;code&gt;0&lt;/code&gt;, no &lt;code&gt;50&lt;/code&gt;, no &lt;code&gt;1024&lt;/code&gt;... &lt;/del&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;del&gt;Very odd, but hopefully it will help in solving this mystery once and for all...&lt;/del&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Apologies... this hint doesn't hold up in all circumstances. I made it a bit too hastily, without thinking it all the way through, as I was attempting to nudge people towards a certain line of thinking.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="5208" LastEditDate="2015-04-06T22:40:27.527" LastActivityDate="2015-04-07T17:04:08.750" Title="Ancient machine in the desert... how does it generate its output?" Tags="|pattern|lateral-thinking|number-sequence|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11164" PostTypeId="2" ParentId="11105" CreationDate="2015-03-28T11:18:27.047" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A &quot;by-the-wind sailor&quot; type jellyfish: &lt;a href=&quot;http://en.wikipedia.org/wiki/Velella&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Velella&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No wings, blown by the wind&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Self explanatory&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Heavier when empty&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well, it wouldn't be empty (of water) unless it was dead and removed from the sea, at which point it would be true.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The bigger I am, the more I can carry&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I guess it can carry more food/water.  This line is the main reason I'm not so sure about this answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is my first time here; go easy on me!&lt;/p&gt;&#xA;" OwnerUserId="5005" LastActivityDate="2015-03-28T11:18:27.047" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11165" PostTypeId="2" ParentId="11161" CreationDate="2015-03-28T11:23:46.997" Score="15" Body="&lt;p&gt;Let $a$ and $b$ be the integers thought of by A and B, and let $N$ and $n$ be the integers given by the teacher, where $N&amp;gt;n$. Assume for contradiction that both A and B keep on saying &quot;no&quot; forever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $a\geq n$, then A knows $n$ cannot be the sum and will say &quot;yes&quot; the first time. Thus after A's first &quot;no&quot;, B knows $a&amp;lt;n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $b\leq N-n$, then $a+b&amp;lt;N$ (since $a&amp;lt;n$), so B knows $N$ cannot be the sum and will say &quot;yes&quot; the first time. Thus after B's first &quot;no&quot;, A knows $b&amp;gt;N-n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $a\geq 2n-N$, then $a+b&amp;gt;n$ (since $b&amp;gt;N-n$), so A knows $n$ cannot be the sum and A will say &quot;yes&quot; the second time. Thus after A's second &quot;no&quot;, B knows $a&amp;lt;2n-N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $b\leq 2N-2n$, then $a+b&amp;lt;N$ (since $a&amp;lt;2n-N$), so B knows $N$ cannot be the sum and will say &quot;yes&quot; the second time. Thus after B's second &quot;no&quot;, A knows $b&amp;gt;2N-2n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $a\geq 3n-2N$, then $a+b&amp;gt;n$ (since $b&amp;gt;2N-2n$), so A knows $n$ cannot be the sum and will say &quot;yes&quot; the third time. Thus after A's third &quot;no&quot;, B knows $a&amp;lt;3n-2N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $b\leq 3N-3n$, then $a+b&amp;lt;N$ (since $a&amp;lt;3n-2N$), so B knows $N$ cannot be the sum and will say &quot;yes&quot; the third time. Thus after B's third &quot;no&quot;, A knows $b&amp;gt;3N-3n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By induction, this argument can be continued forever:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;after B's $k$th &quot;no&quot;, A always knows $b&amp;gt;k(N-n)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But $b$ cannot be greater than $k(N-n)$ for ALL $k$, so eventually B must say &quot;yes&quot; (unless A does first) and the game is up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;QED.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-03-28T11:23:46.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11168" PostTypeId="1" AcceptedAnswerId="11203" CreationDate="2015-03-28T15:35:18.710" Score="2" ViewCount="1153" Body="&lt;p&gt;A notorious serial killer going by the pseudonym &quot;N&quot; has dispatched a letter to the FBI after committing 45 homicides. He is characterized to be extremely intelligent and sends mysterious cryptographic letters to the police once a month, for the mere reason being &quot;messing around&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This time, the letter received was;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Dear informants, &#xA;I know you want me, and you know I want you too.&#xA;But are you a genius?&#xA;You gotta have some genius to catch me.&#xA;And you know what? I am sad...You have about a thousand suspects of me.&#xA;And none of them is me! &#xA;I am depressed to see such failure, specially by the glorious FBI.&#xA;So, below....is some crypted content. In the content, contains my full name.&#xA;If you solve it, you win! And I? Well...I lose. Unfortunately...&#xA;You'll try.&#xA;&#xA;BQzkmcmaIJSaiNysehtMtlzqUScmtz&#xA;&#xA;Sincerely,&#xA;N&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What is his name?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The encrypted text contains the name in the format, &quot;MyNameIs[XYZ]&quot;. The first letters of the last and first name may or may not be capitalized.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10704" LastEditorUserId="18349" LastEditDate="2018-01-28T21:03:20.293" LastActivityDate="2018-01-28T21:03:20.293" Title="Find the killer's name!" Tags="|logical-deduction|cipher|lateral-thinking|" AnswerCount="3" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="11169" PostTypeId="2" ParentId="11015" CreationDate="2015-03-28T16:35:43.320" Score="5" Body="&lt;p&gt;Ok: There are 282 pairs of numbers $x,y$ such the $y/x$ is an integer greater than 1. The sum and difference are designated $s$ and $d$ respectively.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If A doesn't know the numbers, any pairing which results in a unique value of $s$ should be eliminated.  There are 34 such pairs.&lt;/li&gt;&#xA;&lt;li&gt;If B knew this, the value of $d$ must not have one of the 34 which would be eliminated in this way.  Eliminate any pair with a value for $d$ which appears in the eliminated pairs.  There are 75 eliminated in this step.&lt;/li&gt;&#xA;&lt;li&gt;If A knew this, the value of $s$ must not have one of the 109 which have been eliminated so far.  Eliminate any pair with a value for $s$ which appears in the eliminated pairs (especially the 75 new ones).  There are 113 eliminated in this step.&lt;/li&gt;&#xA;&lt;li&gt;If B were correct here, the value of $d$ must not have one of the 222 which have been eliminated so far.  Eliminate any pair with a value for $d$ which appears in the eliminated pairs (especially the 113 new ones).  This creates a list of 43 which A has been told are eliminated in this step and another list 17 which he has been told were not eliminated.&lt;/li&gt;&#xA;&lt;li&gt;From A's perspective, there are 17 pairs from which he can select the correct answer using his knowledge of $s$.  As he cannot, $s$ must not be unique.  There are 4 values for $s$ which are not unique from those 17 pairs: 15, 87, 93, and 99.  These are the only currently possible values for $s$.&lt;/li&gt;&#xA;&lt;li&gt;The last line seems to mean that the correct pair was eliminated in step 4.  Of the values eliminated in this step, only one has a value of $s$ which is one of the ones determined in step 5.  The solution is therefore:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;      $x=33$,  $y=66$, $s=99$,  $d=33$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="99" LastActivityDate="2015-03-28T16:35:43.320" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11170" PostTypeId="2" ParentId="10588" CreationDate="2015-03-28T18:47:47.640" Score="1" Body="&lt;p&gt;There are 38. Shapes shown by vertices.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;   &lt;code&gt;--- --- --- --- --- --- --- ---&lt;/code&gt;&lt;br&gt;&#xA;   &lt;code&gt;0-- 0-- -0- 00- 00- 0-0 -00 -00&lt;/code&gt;&lt;br&gt;&#xA;   &lt;code&gt;00- 0-0 0-0 00- 0-0 0-0 00- 0--&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;   &lt;code&gt;0-- -00 0-- 0-0 0-0 -0- -0- 000&lt;/code&gt;&lt;br&gt;&#xA;   &lt;code&gt;--- -0- --0 -0- --- 0-0 --- -0-&lt;/code&gt;&lt;br&gt;&#xA;   &lt;code&gt;0-0 00- 0-0 0-0 0-0 0-0 0-0 0-0&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;   &lt;code&gt;-0- -0- -0- --0 --0 --0 -00 -00&lt;/code&gt;&lt;br&gt;&#xA;   &lt;code&gt;0-0 0-0 0-0 0-- 0-- 0-- --- 0--&lt;/code&gt;&lt;br&gt;&#xA;   &lt;code&gt;-0- 0-- 00- 0-- -0- 00- 00- 00-&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;   &lt;code&gt;-12 -12 1-- 1-- 1-- -12 -12 -12&lt;/code&gt;&lt;br&gt;&#xA;   &lt;code&gt;-5- -3- -23 -42 -52 63- 56- 63-&lt;/code&gt;&lt;br&gt;&#xA;   &lt;code&gt;4-3 5-4 5-4 5-3 4-3 54- 43- -54&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;   &lt;code&gt;-12 -12 -12 -12 -12 -12&lt;/code&gt;&lt;br&gt;&#xA;   &lt;code&gt;63- 56- 65- 64- 74- 75-&lt;/code&gt;&lt;br&gt;&#xA;   &lt;code&gt;5-4 4-3 4-3 5-3 653 643&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-05-07T12:32:30.320" LastActivityDate="2015-05-07T12:32:30.320" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="11171" PostTypeId="1" AcceptedAnswerId="11176" CreationDate="2015-03-28T21:54:41.513" Score="2" ViewCount="316" Body="&lt;p&gt;The following thirteen English words all share a certain property:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;ACE TWO THREE FOUR FIVE SIX SEVEN EIGHT NINE TEN JACK QUEEN KING&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, there is a fourteenth word that belongs to this group, and it is not &quot;Joker&quot;. What is that word?&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="8874" LastEditDate="2015-10-10T15:09:54.357" LastActivityDate="2015-10-10T15:09:54.357" Title="A Suite of Thirteen (Company #4)" Tags="|pattern|knowledge|word-sequence|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11172" PostTypeId="1" AcceptedAnswerId="11893" CreationDate="2015-03-28T22:08:03.200" Score="18" ViewCount="2842" Body="&lt;p&gt;There are 3 gods, named Past, Present and Future, who all look identical. They sit in a throne room, with one in the middle, the others on the right and left. You are allowed to ask them a series of yes/no questions in order to discern their identities. Present will answer truthfully, Past will truthfully answer the &lt;strong&gt;previous&lt;/strong&gt; question you asked, and Future will truthfully answer the &lt;strong&gt;next&lt;/strong&gt; question you plan to ask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to avoid paradoxes (which would destroy the universe), you must follow these rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You must write down all the questions you plan to ask before entering the throne room, in the order you plan to ask them, and not deviate from this plan.&lt;/li&gt;&#xA;&lt;li&gt;The only questions allowed are &quot;Is the statement $P$ true?&quot;, where $P$ is a boolean function of statements like &quot;The god at [position] is [name]&quot; or &quot;You are [name]&quot;. (Examples: &quot;Are you Future?&quot;, &quot;Is the God on the right Past?&quot;, &quot;Is the middle God either Past or Present?&quot; &quot;Is one of you or the left god Future?&quot;).&lt;/li&gt;&#xA;&lt;li&gt;Though &lt;strong&gt;what&lt;/strong&gt; questions you ask must be preplanned, &lt;strong&gt;who&lt;/strong&gt; you ask those questions to may be decided dynamically.&lt;/li&gt;&#xA;&lt;li&gt;If your first question is directed at Past, she will answer yes/no randomly. Same when your last question is directed at Future. &lt;/li&gt;&#xA;&lt;li&gt;The pronoun &quot;you&quot; refers to the person considering the question, not the person it was originally asked: if you ask Present, &quot;Are you Past?,&quot; she will respond &quot;No&quot;, and if your next question is directed at Past, she will respond &quot;Yes.&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the fewest number of questions you must ask to determine their identities, and how do you do this? (There is a solution, and it can proved optimal).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I got this puzzle from &lt;a href=&quot;https://www.ocf.berkeley.edu/~wwu/riddles/hard.shtml&quot;&gt;William Wu's puzzle website&lt;/a&gt;. &#xA;You may also consider the &quot;time is a flat circle&quot; variant, where Past will answer your &lt;em&gt;last&lt;/em&gt; question when your first question is directed at them, and Future will similarly wrap around to the first question you asked.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="5373" LastEditDate="2015-03-28T22:27:16.573" LastActivityDate="2015-04-11T17:18:11.153" Title="Past, Present and Future" Tags="|logical-deduction|liars|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11174" PostTypeId="2" ParentId="11172" CreationDate="2015-03-28T23:49:59.887" Score="2" Body="&lt;p&gt;I'm not sure if the lowest number, or how to prove it optimal, but I can do it in &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 6 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Q1. ask anybody a universally true statement (i.e. is true=true?) just to set-up the problem &#xA;Q2,Q3,Q4. Ask persons 1,2,3 alternating between false and true statements, starting with a false one&lt;br&gt;&#xA; Whoever responded correctly to their question is the present god&lt;br&gt;&#xA;Q5,Q6. Pick one of the remaining gods and ask him twice &quot;are you future?&quot;&lt;br&gt;&#xA;if the first answer is 'Yes', then that is the future and the other is the past god, otherwise vice versa.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7715" LastActivityDate="2015-03-28T23:49:59.887" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11175" PostTypeId="1" AcceptedAnswerId="11451" CreationDate="2015-03-29T00:21:08.797" Score="15" ViewCount="694" Body="&lt;p&gt;The following 5-by-5 squares of digits are not magic squares, but nevertheless they each have a crucial property which enabled me to generate them. What is that property? How could you generate more such squares?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;3  3  5  8  9               9  5  4  8  7&#xA;4  5  1  4  7               0  1  7  2  8&#xA;6  6  3  1  9               4  8  2  8  2&#xA;2  2  9  5  3               5  2  8  1  0&#xA;6  4  8  3  2               2  3  5  3  6&#xA;&lt;/pre&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-29T00:31:27.740" LastActivityDate="2015-04-02T14:54:19.740" Title="Sorcery squares!" Tags="|mathematics|number-sequence|" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="11176" PostTypeId="2" ParentId="11171" CreationDate="2015-03-29T00:31:15.387" Score="2" Body="&lt;p&gt;The 14th word is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;KNIGHT&lt;/strong&gt; which is used in a &lt;a href=&quot;http://en.wikipedia.org/wiki/French_tarot&quot; rel=&quot;nofollow&quot;&gt;French Tarot&lt;/a&gt; deck of cards.  The Knight is ranked between the Jack and the Queen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastActivityDate="2015-03-29T00:31:15.387" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11177" PostTypeId="2" ParentId="11172" CreationDate="2015-03-29T00:40:25.780" Score="1" Body="&lt;p&gt;In the &quot;time is a flat circle&quot; variant, it can be done in 4 questions, although I do not know whether this is optimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The questions are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Are you Present? (to left god) &lt;br /&gt;&#xA; Are you Present? (to middle god) &lt;br /&gt;&#xA; Are you Present? (to right god) &lt;br /&gt;&#xA; Are you or the god on the left Future? (to right god)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you consider each arrangement of the gods, it can be seen that the resulting responses are unique:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;Past Present Future | N Y Y N&#xA;Past Future Present | N N Y N&#xA;Present Past Future | Y N Y N&#xA;Present Future Past | Y N N N&#xA;Future Past Present | N N Y Y&#xA;Future Present Past | N Y N N&lt;/pre&gt;&#xA;" OwnerUserId="8949" LastEditorUserId="8949" LastEditDate="2015-03-29T00:45:33.000" LastActivityDate="2015-03-29T00:45:33.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11178" PostTypeId="2" ParentId="11172" CreationDate="2015-03-29T00:52:11.887" Score="7" Body="&lt;p&gt;Because the answers can only be Yes/No (binary) and there are 6 possible seating arrangements, 3 questions are needed to differentiate the arrangements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my solution with &quot;time in a circle&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q1: asked to Left: Is the one on the Left seat Present and are you either Present or Future?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q2: asked to Middle: Is the one on the Middle seat Present?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q3: asked to Right: Is the one on the Middle seat Past?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answers and seating arrangement from left to right:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, No, No = Past, Future, Present&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, No, Yes = Future, Past, Present&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, Yes, No = Past, Present, Future&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, No, No = Present, Future, Past&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, No, Yes = Present Past, Future&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, Yes, Yes = Future, Present, Past&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, Yes,Yes and Yes, Yes, No are paradoxes.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Not Time in a Circle&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;In the case where time is not a circle, a fourth question is needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q1: asked to Left: Is the one on the Left seat Present and are you either Present or Future?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q2: asked to Middle: Is the one on the Middle seat Present?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q3: asked to Right: Is the one on the Middle seat Past?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q4: asked to Right: Is the one on the Left seat Present?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answers and seating arrangements are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;? Y N ? = Past, Present, Future&lt;/p&gt;&#xA;&#xA;&lt;p&gt;? N N N = Past, Future, Present&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Y N Y ? = Present, Past, Future&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Y N N Y = Present, Future, Past&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N N Y N = Future, Past, Present&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Y Y Y N = Future, Present, Past&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Q4 can be asked to Left instead, which moves the ? to the end of the last 2 arrangements. This might be clearer with Left giving all the ambiguous answers.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="10667" LastEditorUserId="10667" LastEditDate="2015-03-29T02:22:37.337" LastActivityDate="2015-03-29T02:22:37.337" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11180" PostTypeId="2" ParentId="11141" CreationDate="2015-03-29T04:10:08.737" Score="0" Body="&lt;p&gt;Each digit should appear twice, so I'm thinking of twice the missing digit and once each of one and nine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just don't find a logical sequence for the 4 digits.&lt;/p&gt;&#xA;" OwnerUserId="10667" LastActivityDate="2015-03-29T04:10:08.737" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11181" PostTypeId="2" ParentId="11168" CreationDate="2015-03-29T04:25:54.363" Score="0" Body="&lt;p&gt;It is unlikely that 9 characters of the name should be upper case.  It is also unlikely that the name length is 30 characters. Therefore the killer's name is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; one of these 16 names that use about half of the characters from the encryted string:&lt;br&gt;&#xA;&lt;br&gt;&#xA; Mateusz Michalski&lt;br&gt;&#xA; Michal Stantejsky&lt;br&gt;&#xA; Matthias Kuntzel&lt;br&gt;&#xA; Michael Stickman&lt;br&gt;&#xA; Sebastian Schulz&lt;br&gt;&#xA; Shamus McClaskey&lt;br&gt;&#xA; Timea Bacsinszky&lt;br&gt;&#xA; Chaim Maltinsky&lt;br&gt;&#xA; Helmut Bakaitis&lt;br&gt;&#xA; Hikmet Sulayman&lt;br&gt;&#xA; Jacques Bialski&lt;br&gt;&#xA; Jacques Cassini&lt;br&gt;&#xA; Jacques Haitkin&lt;br&gt;&#xA; James McQuilkin&lt;br&gt;&#xA; Klaus Steinbach&lt;br&gt;&#xA; Maciej Szmatiuk  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastActivityDate="2015-03-29T04:25:54.363" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11182" PostTypeId="2" ParentId="10911" CreationDate="2015-03-29T05:03:18.810" Score="4" Body="&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wax or a candle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My name sounds like a bird&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Apparently there's a bird called a waxwing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whole body is like tongue of a lizard &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wax is sticky when melted&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Am different colors &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Candles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I well can be very littered &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Grammatically incorrect play on lighting candle?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My energy melts away over time / But not quite like an ice&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Candles melt as you burn them (using their energy)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And when not taken care of &#xA;I'll collect dust and lice&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People tend to have decorative candles that end up just being dust collectors&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastActivityDate="2015-03-29T05:03:18.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11185" PostTypeId="1" AcceptedAnswerId="11186" CreationDate="2015-03-29T07:24:46.480" Score="4" ViewCount="3301" Body="&lt;p&gt;I tried to solve this, but I am unsure about my solution. I'm posting it here to see if I got it right. It is listed as multiple choice, but I am unsure about the quality of the puzzle creator, so don't limit yourself to the 5 provided answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A gardener wants to plant 100 trees (oaks and birches) alongside a trail in a park. The number of trees between any two oaks must not be equal to five. &#xA;What is the largest number of oaks that the gardener could plant? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A)48&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B)50&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C)52&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D)60&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E)80&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 45 Oaks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Planted in a pattern like this (O=Oaks, X = Birches), repeating to 100.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; OXOXOXXOX&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm quite certain there is a better way to solve this (and a better answer), so enlighten me on how to solve a problem like this.&lt;/p&gt;&#xA;" OwnerUserId="10743" LastActivityDate="2015-03-30T14:59:29.450" Title="The gardener problem" Tags="|mathematics|logical-deduction|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11186" PostTypeId="2" ParentId="11185" CreationDate="2015-03-29T07:50:34.947" Score="5" Body="&lt;p&gt;Seems to me a greedy strategy can work. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As when you plant an oak at position $x$, position $x+6$ cannot be an oak, but position $x+12$ can be an oak, and position $x+18$ cannot, $\dots$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So my strategy is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I plant oaks on position $1$ to $6$. This means positions $7$ to $12$ are birches, $13$ to $18$ are oaks, and so on.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a result:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; So for every $12$ consecutive trees, $6$ of them are oaks. The last $4$ (position $97$ to $100$) are also oaks. Hence the number of oaks is $52$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9944" LastActivityDate="2015-03-29T07:50:34.947" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11188" PostTypeId="2" ParentId="11141" CreationDate="2015-03-29T08:17:24.570" Score="3" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 9745&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; These are guesses made by someone in the cows-and-bulls game. In the first step, they got two bulls so they decided to change two numbers to see which positions are right. They got three bulls in the next step meaning that the digits kept fixed were right and one of the other digits was too. So they decided to keep one digit and two unseen digits in other places to figure out the fourth digit. They should have got one bull meaning that 7 was right and the other digits were all wrong. This left only 2 and 9 as possibilities for the first digit and hence they played a move with 9 and previously seen digits. They got a cow and thus the answer is 9745. this should be played in the last step to win. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9471" LastActivityDate="2015-03-29T08:17:24.570" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11189" PostTypeId="2" ParentId="11141" CreationDate="2015-03-29T09:35:41.067" Score="12" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1617&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;2&lt;sup&gt;64&lt;/sup&gt; = 18,446,744,073,709,551,616 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This could be written as&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1844 6744 0737 0955 1616 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then you just&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Add 1 to each of the numbers to get the numbers on the list, so the missing number is 1617&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9676" LastActivityDate="2015-03-29T09:35:41.067" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11190" PostTypeId="1" AcceptedAnswerId="11192" CreationDate="2015-03-29T10:34:30.293" Score="11" ViewCount="441" Body="&lt;p&gt;I bought a device to map a set of caves, but I don't really know how it works. I pressed a button at a few spots in some of the caves, and it just printed out this gibberish:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lJ0Qa.png&quot; alt=&quot;Square grid?&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure you'll be able to interpret this data to map out the caves we visited.&lt;br&gt;&#xA;&lt;em&gt;(And find a hidden word in the process!)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="7308" LastEditorUserId="7308" LastEditDate="2015-03-29T21:10:23.113" LastActivityDate="2015-04-26T11:59:50.123" Title="Echolocation System" Tags="|lateral-thinking|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11191" PostTypeId="2" ParentId="11100" CreationDate="2015-03-29T12:17:33.557" Score="2" Body="&lt;p&gt;I don't think it fits as well as the other close answer, but maybe it's:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Examination&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A white man plowing the dark soil&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; White exam paper, with black writing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Silent oxen around him toil&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Silent students toiling at the exam&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A train whistle is near&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The whistle to signal the end of the exam&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What's said may disappear&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The only thing spoken is the invigilator announcing the remaining time left for the exam, which inevitably disappears.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9740" LastActivityDate="2015-03-29T12:17:33.557" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11192" PostTypeId="2" ParentId="11190" CreationDate="2015-03-29T12:23:16.573" Score="10" Body="&lt;p&gt;Excellent question.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; On a 27-by-5 grid, draw arrows in the indicated directions from the indicated points (2 units south and 2 units east from the point (1,1), 1 unit east and 1 unit west from the point (2,3), and so on). The lines thus drawn spell out letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's a (not very good) picture to demonstrate the solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AVJzn.jpg&quot; alt=&quot;map&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The letters appearing are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; S, P, E, L, U, ~n, ~K&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;giving the word&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Spelunking&quot; rel=&quot;nofollow noreferrer&quot;&gt;SPELUNK&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="10399" LastEditDate="2015-04-26T11:59:50.123" LastActivityDate="2015-04-26T11:59:50.123" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11193" PostTypeId="1" AcceptedAnswerId="11198" CreationDate="2015-03-29T14:47:01.377" Score="32" ViewCount="1951" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In the number $(1+\sqrt{3})^{2015}$, what is the 224th digit after the decimal point?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;You may NOT use a calculator, computer, or any electronic aid to answer this question.&lt;/strong&gt; Only pen(cil), paper, and brain are allowed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should be relatively easy, but it's a puzzle of a kind I haven't seen on this site before, and hopefully it'll set a trend for puzzles that are the opposite of &lt;a href=&quot;/questions/tagged/computer-puzzle&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged 'computer-puzzle'&quot; rel=&quot;tag&quot;&gt;computer-puzzle&lt;/a&gt; ones!&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-03-30T09:39:00.237" Title="Computation without a computer" Tags="|mathematics|calculation-puzzle|no-computers|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="11194" PostTypeId="5" CreationDate="2015-03-29T14:48:52.623" Score="0" Body="&lt;p&gt;This tag should be used for a puzzle that &lt;em&gt;can&lt;/em&gt; be solved by brute-force calculation or programming, but where the puzzle-setter explicitly does not want people to do so and is looking for a solution that can be understood without the use of a calculator or computer. You're still allowed to use a computer to &lt;em&gt;post&lt;/em&gt; the solution; Stack Exchange doesn't support smoke signals yet. For further discussion of exactly &lt;em&gt;how much&lt;/em&gt; computer usage is allowed, see &lt;a href=&quot;https://puzzling.meta.stackexchange.com/a/6655/5373&quot;&gt;this meta post&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The spirit of the tag is to &lt;em&gt;&lt;strong&gt;solve the puzzle on your own&lt;/strong&gt;&lt;/em&gt;, without the aid of computers. Obviously you will need a computer for some aspects, such as typing in the solution and posting it, but the real point is that you should use your own knowledge or ingenuity. There are various ways you could use a computer for not-in-the-spirit-of-the-puzzle shortcuts:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;programming,&lt;/li&gt;&#xA;&lt;li&gt;brute-forcing possibilities,&lt;/li&gt;&#xA;&lt;li&gt;using online word finders,&lt;/li&gt;&#xA;&lt;li&gt;searching the web for specific strings/information,&lt;/li&gt;&#xA;&lt;li&gt;etc. etc.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Depending on the puzzle and the OP's intent, any of these &lt;em&gt;might&lt;/em&gt; be acceptable, and the &lt;a href=&quot;/questions/tagged/no-computers&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;no-computers&amp;#39;&quot; rel=&quot;tag&quot;&gt;no-computers&lt;/a&gt; tag is there to tell you that it isn't. The tag means different things in different contexts, as may be clarified by the OP in the body of the question.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Don't take the tag name too literally:&lt;/strong&gt; it doesn't exclude things like keeping track of your progress in a text document, which could just as easily be done with pen and paper.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2021-02-08T16:20:13.307" LastActivityDate="2021-02-08T16:20:13.307" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="11195" PostTypeId="4" CreationDate="2015-03-29T14:48:52.623" Score="0" Body="A puzzle designed to be solved without using calculators, online decoders, or computer programming. Using a computer to type and post the answer is allowed; the spirit of this tag is to make people solve the puzzle on their own." OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2021-02-08T16:18:02.143" LastActivityDate="2021-02-08T16:18:02.143" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="11196" PostTypeId="5" CreationDate="2015-03-29T14:51:39.603" Score="0" Body="&lt;p&gt;This tag is for puzzles involving the Japanese language or words from it, or Japanese characters.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-29T14:51:39.603" LastActivityDate="2015-03-29T14:51:39.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11197" PostTypeId="4" CreationDate="2015-03-29T14:51:39.603" Score="0" Body="Puzzles that use Japanese characters or the Japanese language." OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-29T14:51:39.603" LastActivityDate="2015-03-29T14:51:39.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11198" PostTypeId="2" ParentId="11193" CreationDate="2015-03-29T15:19:31.663" Score="26" Body="&lt;p&gt;Consider the auxiliary value $N=(1+\sqrt{3})^{2015}+(1-\sqrt{3})^{2015}$. In the binomial expansion of $N$ all terms with odd powers of $\sqrt3$ cancel out, so that $N$ is an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real number $M=1-\sqrt{3}$ is negative with $|M|\approx0.732$. Then $|M|&amp;lt;0.74$ and $|M|^8&amp;lt;0.74^8&amp;lt;10^{-1}$. &#xA;Then $|M|^{2015}&amp;lt;(|M|^8)^{250}&amp;lt;10^{-250}$.&#xA;Then&#xA;$(1-\sqrt{3})^{2015}$ is a negative real number between $-10^{-250}$ and $0$,&#xA;and the first 249 digits after the decimal point are 0s.&#xA;Subtracting it from the integer $N$ gives an integer plus an astronomically small real number, so that the first 249 digits after the decimal point all are 0. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the answer to the puzzle is &lt;strong&gt;digit 0&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="9125" LastEditDate="2015-03-30T09:39:00.237" LastActivityDate="2015-03-30T09:39:00.237" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11201" PostTypeId="1" AcceptedAnswerId="11202" CreationDate="2015-03-29T17:10:10.337" Score="0" ViewCount="687" Body="&lt;p&gt;An old farmer was walking beside a tractor, which was driven by another farmer. The farmer driving the tractor was dragging a pipe behind him. The old farmer wants to figure out how long the pipe is. He starts by starting at the very end of the pipe (farthest away from the tractor) and walks forward until he reaches the other end of the pipe. This take 140 steps. He then walks back to the other end, which takes only 20 steps, as this time he is walking in the opposite direction of the tractor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tractor is moving constantly at a uniform speed. The farmer is also moving at a uniform speed, and each of his steps is one meter long. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How long is the pipe?&lt;/p&gt;&#xA;" OwnerUserId="10743" LastEditorUserId="5373" LastEditDate="2015-03-29T17:12:59.960" LastActivityDate="2015-03-29T17:23:38.577" Title="Tractor problem" Tags="|mathematics|logical-deduction|" AnswerCount="1" CommentCount="9" ClosedDate="2015-03-29T18:19:26.503" ContentLicense="CC BY-SA 3.0" />
  <row Id="11202" PostTypeId="2" ParentId="11201" CreationDate="2015-03-29T17:17:16.400" Score="5" Body="&lt;p&gt;The pipe is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 35 metres long.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt; let $F$ and $T$ be the speed of the old farmer and the tractor respectively, $L$ be the length of the pipe, and $t_1$, $t_2$ be the times taken by the old farmer to walk along the pipe in each direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The old farmer walks $140$ metres in time $t_1$; this includes both walking the length of the pipe and walking as far as the tractor moves in time $t_1$. So $140=Ft_1=L+Tt_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then he walks $20$ metres in time $t_2$; this includes walking the length of the pipe minus as far as the tractor moves in time $t_2$. So $20=Ft_2=L-Tt_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We solve these two simultaneous equations by making the following deductions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$140=Ft_1$ and $20=Ft_2$, so $t_1=7t_2$&lt;/li&gt;&#xA;&lt;li&gt;$120=140-20=(L+Tt_1)-(L-Tt_2)=T(t_1+t_2)=8Tt_2$, so $Tt_2=15$&lt;/li&gt;&#xA;&lt;li&gt;$L=20+Tt_2=35$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-29T17:23:38.577" LastActivityDate="2015-03-29T17:23:38.577" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11203" PostTypeId="2" ParentId="11168" CreationDate="2015-03-29T18:44:25.090" Score="10" Body="&lt;p&gt;Start with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;BQzkmcmaIJSaiNysehtMtlzqUScmtz&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remove &lt;code&gt;MyNameIs&lt;/code&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;BQzk cm  JSai    ht tlzqUScmtz&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Group the remaining letters:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First letters: &lt;code&gt;BJQSSU&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;Vowels: &lt;code&gt;ai&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;Others: &lt;code&gt;cchklmmqttzzz&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Make a name with them:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jack Smith&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Remove its letters:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;BQz                 tlzqUScmtz&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nice.&lt;/p&gt;&#xA;" OwnerUserId="10746" LastEditorUserId="10746" LastEditDate="2015-03-29T19:30:56.047" LastActivityDate="2015-03-29T19:30:56.047" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11204" PostTypeId="2" ParentId="11168" CreationDate="2015-03-29T19:47:41.030" Score="3" Body="&lt;p&gt;Miguel solved the puzzle but not completely, I'll further elucidate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value is;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;BQzkmcmaIJSaiNysehtMtlzqUScmtz&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And as the hint states, the killer's name will start off with &quot;MyNameIs&quot; and where is this? The letters of &quot;MyNameIs&quot;  and the killer's name are located exquisitely here;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BQz&lt;strong&gt;kmcmaIJSaiNysehtM&lt;/strong&gt;tlzqUScmtz&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we'll take em' out to grab the answer:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;kmcmaIJSaiNysehtM&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then, after taking out the letters of &quot;MyNameIs&quot;, we're left with:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;kmcaJSiht&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Eureka! After assorting it in a proper order, The killer's name is;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;JackSmith&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And the complete hidden text was:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;MyNameIsJackSmith&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="10704" LastActivityDate="2015-03-29T19:47:41.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11205" PostTypeId="1" AcceptedAnswerId="11206" CreationDate="2015-03-29T21:56:45.440" Score="7" ViewCount="1178" Body="&lt;p&gt;Alice and Bob play a game using the following arrangement of coins, which are joined by strings and hanging from the ceiling:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\qquad\qquad\qquad\qquad\quad$&lt;img src=&quot;https://i.stack.imgur.com/MyHet.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two alternate turns, with Alice going first. On their turn, a player must cut a string. This continues until a player causes one or more coins to fall to the ground, in which case that player loses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who wins under optimal play?&lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-03-29T22:09:35.460" Title="A Game with Strings and Coins" Tags="|game|no-computers|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11206" PostTypeId="2" ParentId="11205" CreationDate="2015-03-29T22:09:35.460" Score="6" Body="&lt;p&gt;Bob wins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We show that as long as there are 17+ strings, there's a move that doesn't immediately lose, but once we're down to 16 strings, the next move loses. Each move removes a string, so Alice will lose because she cuts when there are an even number of strings (28 strings, 26 strings, ...., 16 strings).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Treat the ceiling as a single node, making the construction a graph with 17 nodes and 28 edges. As long the graph has at least as many edges as nodes (17), it cannot be a &lt;a href=&quot;http://en.wikipedia.org/wiki/Tree_%28graph_theory%29&quot;&gt;tree&lt;/a&gt;, and so must contain a cycle. Cutting any edge in that cycle cannot cause anything to fall because any path to the ceiling that used that edge has a &quot;detour&quot; where the rest of the cycle was used in place of that edge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, once we're down to 16 edges, the graph (if still connected) must be a tree, and removing an edge will disconnect it, so the next player loses.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-03-29T22:09:35.460" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11208" PostTypeId="1" AcceptedAnswerId="11209" CreationDate="2015-03-29T22:38:38.223" Score="-7" ViewCount="1596" Body="&lt;p&gt;I am shaped like a ring and made out of metal,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can be a burden,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but then also be very comforting,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My neighbors are the ones with bones and muscle,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You use me for leisure and sometimes for showing off,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And it doesn't matter for me because all I am, is just a piece of metal,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;So what am I?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="10704" LastEditorUserId="5373" LastEditDate="2016-10-21T15:01:56.087" LastActivityDate="2016-10-21T15:01:56.087" Title="I am shaped like a ring and made out of metal - what am I?" Tags="|riddle|word|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11209" PostTypeId="2" ParentId="11208" CreationDate="2015-03-29T22:58:32.560" Score="5" Body="&lt;p&gt;It looks like the answer is simply&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a ring.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Seems a bit weird given the first line, but it's the only thing I can make fit the rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am shaped like a ring and made out of metal,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well, obviously!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I can be a burden,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lord of the Rings reference (thanks &lt;strong&gt;d'alar'cop&lt;/strong&gt; for catching this one)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;but then also be very comforting,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Could be another LotR reference (the Ring being like a drug, and possessing it gives some kind of comfort) or to marriage.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My neighbors are the ones with bones and muscle,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A ring is snuggled up next to the bones and muscles of fingers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You use me for leisure and sometimes for showing off,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Leisure could be a reference to the five-ring symbol of the Olympics. Rings are definitely used to show off or display wealth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And it doesn't matter for me because all I am, is just a piece of metal,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, obviously!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-30T00:04:38.360" LastActivityDate="2015-03-30T00:04:38.360" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="11210" PostTypeId="1" AcceptedAnswerId="17518" CreationDate="2015-03-30T04:35:04.440" Score="1" ViewCount="572" Body="&lt;p&gt;I need to decrypt the following code. I wasn't given much context, so here is the code as was given; I'm not sure what may be significant or not (specifically in regards to spacing).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 9,18,26   &#xA;2  &#xA;9   &#xA;  1  &#xA;  6  &#xA;  4  &#xA;  5   &#xA;    2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I've tried converting to letters and using some sort of transformation from the top numbers, but to no avail.&lt;/p&gt;&#xA;" OwnerUserId="10750" LastEditorDisplayName="user10531" LastEditDate="2015-04-06T15:31:24.923" LastActivityDate="2015-07-07T07:00:41.087" Title="Spaces and numbers" Tags="|pattern|cipher|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11212" PostTypeId="1" AcceptedAnswerId="11229" CreationDate="2015-03-30T05:52:20.643" Score="7" ViewCount="1069" Body="&lt;p&gt;Decrypting this single word below, may enhance your overall health and well being.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best answer will include an explanation of the encryption step&lt;strong&gt;s&lt;/strong&gt;.&lt;br&gt;&#xA;&lt;sub&gt;(An understanding of color models may be helpful.)&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AWdDV.png&quot; alt=&quot;Lotus&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-03-31T17:22:15.353" LastActivityDate="2015-03-31T17:22:15.353" Title="Lotus flower encryption" Tags="|cipher|" AnswerCount="2" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11214" PostTypeId="2" ParentId="11212" CreationDate="2015-03-30T06:09:46.607" Score="3" Body="&lt;p&gt;Thoughts toward a solution:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Assuming the colours follow an RGB scheme, they are combinations of the elementary colours of light: red, green, and blue. If we convert the presence of a colour to &lt;code&gt;1&lt;/code&gt; and its absence to &lt;code&gt;0&lt;/code&gt;, we get the following binary code from the digits:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;000 011 001 010 100 110 100 011&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I've been playing with different ways of encoding these numbers, with no luck:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0000 1100 1010 1001 1010 0011 - 0ca9a3 in hexadecimal, doesn't mean anything&#xA;00001 10010 10100 11010 0011 - 5 doesn't go evenly into 24&#xA;000011 001010 100110 100011 - CJpm in base64, doesn't mean anything&#xA;00001100 10101001 10100011 - doesn't mean anything in ASCII or Unicode&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If we assume instead that they're in a CMY configuration, then the binary code is inverted:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;111 100 110 101 011 001 011 100&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And if we try the groupings above:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1111 0011 0101 0110 0101 1100 - f3563c in hexadecimal, doesn't mean anything&#xA;11110 01101 01011 00101 1100 - 5 doesn't go evenly into 24&#xA;111100 110101 011001 011100 - 81Zc in base64, doesn't mean anything&#xA;11110011 01010110 01011100 - doesn't mean anything in ASCII or Unicode&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-03-30T06:09:46.607" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="11215" PostTypeId="2" ParentId="11105" CreationDate="2015-03-30T06:15:56.150" Score="2" Body="&lt;p&gt;How about:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can fly without wings&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Time can 'fly' but has no wings&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;only carried by the wind&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Here 'wind' could refer not to moving air, but to the 'winding up' of a clock. Could also perhaps be a reference to the 'Winds of Time'...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am heavier when empty&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Empty time tends to drag heavily, due to boredom&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the bigger I am, the more I carry&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The more time you have, the more you can fit into it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9740" LastActivityDate="2015-03-30T06:15:56.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11216" PostTypeId="1" CreationDate="2015-03-30T06:18:04.723" Score="1" ViewCount="830" Body="&lt;p&gt;Vincent was a software engineer working at Bell Inc, and well...he's gone insane. He is said to suffer from clinical schizophrenia, autism and severe anxiety. But there's one golden fact about his mind, he possesses an IQ above the limits that God himself placed on mankind's marvelous intelligence. In simple terms, Vincent has a 395 IQ. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a sunny sunday during the summer of 1972, Vincent kidnapped Jeremy Kinderman (the chairman of Bell) very cleverly and sent a kind telegraph to the board of directors of Bell Inc. The telegraph was as follows,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Ybaqba is a word. And that's all that matters. &#xA;But another thing that could potentially matter is that the word rots 13 times.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Vincent took Jeremy's private jetplane and escorted him to some european city. Very reliable reports suggest that Vincent could have only escorted him to one of the following cities; Prague, London, Malmedy, Warsaw or Berlin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After 8 days had passed since the kidnapping, Vincent sent yet another telegraph to the board of directors or more specifically, to Bertrand as he was trying to trace the location of his telegraph very stubbornly. The telegraph was as follows,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I have only one thing to say to you.&#xA;efrgebegusuaoflryoho!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The police and many other departments are involved in trying to arrest Vincent and save Jeremy. The telegraphs include cryptic strings, perhaps they could mean something. Right now, all we want is the location of Vincent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What do the secret strings tell and where in the world has Vincent escorted Jeremy to?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="10704" LastActivityDate="2015-05-08T17:19:30.207" Title="Where the hell is Vincent?" Tags="|cipher|strategy|lateral-thinking|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11217" PostTypeId="2" ParentId="11216" CreationDate="2015-03-30T07:01:46.963" Score="6" Body="&lt;p&gt;First string:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After using ROT-conversion of 13 ('Y' becomes 'L', 'b' becomes 'o' and so on):&lt;br&gt;&#xA; London&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Meh, stop with the plussing, I didn't even decode the 2nd string.&#xA;I think I need a hint for this one...&lt;/p&gt;&#xA;" OwnerUserId="10384" LastEditorUserId="10384" LastEditDate="2015-03-31T07:45:06.293" LastActivityDate="2015-03-31T07:45:06.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11218" PostTypeId="1" AcceptedAnswerId="11220" CreationDate="2015-03-30T08:28:12.127" Score="5" ViewCount="1212" Body="&lt;p&gt;I have a paradox:&#xA;EIGHTY is a six digit number with no repeating digits and no zeros. When divided by 19, 17, 13, 11, or H, the remainders are, respectively, 17, 13, 11, 7 and G.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TWENtY is (another) six digit number with no repeating digits and no zeros (and uses a different key to EIGHTY above). When divided by T, perfect square WE or perfect cube NtY, the remainder is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find EIGHTY TWENtY&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My interpretation is: The question requires a fractional base system converion e.g.  20 converted to base 2 is and 0010100 and 20 converted to base 1.6 is approximately 1001001.2589 which is a six digit number but both have repeating digits and zeros.&lt;/p&gt;&#xA;" OwnerUserId="10751" LastEditorUserId="4050" LastEditDate="2015-03-30T08:44:03.257" LastActivityDate="2015-04-02T02:42:33.913" Title="Number Systems or Analytical Thinking" Tags="|mathematics|alphametic|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11219" PostTypeId="2" ParentId="11218" CreationDate="2015-03-30T08:43:13.617" Score="3" Body="&lt;p&gt;The first number EIGHTY is a number that has the following moduli: 7 mod 11, 11 mod 13, 13 mod 17, and 17 mod 19.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the Chinese Remainder Theorem, this number is congruent to 20005 mod 46189. If we brute-force the possible six digit numbers that satisfy this property, we find that &lt;strong&gt;943785&lt;/strong&gt; satisfies the &quot;G mod H&quot; rule, as it is congruent to 3 mod 7, as well as the &quot;no zeros&quot; rule.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The second number TWENtY is divisible by NtY, which means that NtY divides TWE000. Additionally, NtY is a perfect cube. It can't be 1, 8, 27, 64, or 343, since all of those either have repeating digits or leading zeros.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let's try the other cases one-by-one and see what we get.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;NtY = 216&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the smallest multiple of 1000 that divides by 216 is 27000. The only numbers hence that have no repeating digits or zeros are 378216, 459216, 594216, 783216, 837216, and 945216, none of which satisfy the &quot;WE = perfect square&quot; rule.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;NtY = 512&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the smallest multiple of 1000 that divides by 512 is 64000. The only numbers hence that have no repeating digits or zeros are 384512, 768512, and 896512, again none of which satisfy the &quot;WE = perfect square&quot; rule.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;NtY = 729&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the smallest multiple of 1000 that divides by 729 is 729000 itself.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;NtY = 125&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, &lt;em&gt;every&lt;/em&gt; multiple of 1000 is divisible by 125. So we look at the composition of WE as well. It can't be 1, 4, or 9, since those have leading zeros. It can't be 16, 25, or 81, since those share digits with 125. And finally, it can't be 36 or 64, since those are even and a number ending in &quot;5&quot; is not even. So WE = 49.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this information, we iterate through all the possible values of T to find that &lt;strong&gt;349125&lt;/strong&gt; is divisible by all of 3, 49, and 125, as required.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-04-02T02:42:33.913" LastActivityDate="2015-04-02T02:42:33.913" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11220" PostTypeId="2" ParentId="11218" CreationDate="2015-03-30T08:45:38.743" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;EIGHTY is a six digit number with no repeating digits and no zeros. When divided by 19, 17, 13, 11, or H, the remainders are, respectively, 17, 13, 11, 7 and G.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So EIGHTY + 2 is divisible by both 19 and 13, while EIGHTY + 4 is divisible by both 17 and 11. So EIGHTY is congruent to -2 modulo 19*13=247, and congruent to -4 modulo 17*11=187. Solving these two simultaneous congruences using the Chinese Remainder Theorem, we get that EIGHTY is congruent to -26184 modulo 247*187=46189. The only 6-digit numbers satisfying this congruence and having no repeating digits and no zeros are &lt;strong&gt;389517&lt;/strong&gt; and &lt;strong&gt;943785&lt;/strong&gt;. The first of these leaves a remainder of $2\neq G$ on division by $H=5$, while the second leaves a remainder of $3=G$ on division by $H=7$. So the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; EIGHTY = 943785; E=9, I=4, G=3, H=7, T=8, Y=5.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;TWENtY is a six digit number with no repeating digits and no zeros. When divided by T, perfect square WE or perfect cube NtY, the remainder is zero.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The only 3-digit perfect cubes with no repeating digits are 125, 216, 512, and 729. TWE000 must be a multiple of NtY since TWENtY is, but 729 is not a factor of 1000 and has no 3-digit multiple other than itself, so NtY cannot be 729. All the remaining possibilities involve the digits 1 and 2, so TWE cannot involve these digits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only 2-digit perfect squares not involving the digits 1 and 2 are 36, 49, and 64, so WE must be one of these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If NtY is 216, then WE cannot involve the digit 6, so WE is 49. But 216 has no 3-digit multiple ending in the digits 49, nor does 216*125=27000 have any 3-digit multiple ending in the digits 49000. Contradiction, so the digits N, t, Y are 1, 2, 5 in some order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If NtY is 512, then $TWE000=TWE*2^3*5^3$ is a multiple of $512=2^9$, so TWE is a multiple of $2^6=64$. Now WE cannot be 64 (no 3-digit number of the form T00 is a multiple of 64) or 49 (since then TWE would be odd and not a multiple of 64), so WE must be 36. A quick check shows that no 3-digit number of the form T36 is a multiple of 64, so we have a contradiction here too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If NtY is 125, then TWE000 is automatically a multiple of NtY since 1000 is. But TWENtY is odd, so not divisible by 36 or 64, and again WE must be 49. So TWENtY is 349125, 649125, 749125, or 849125. Only the first of these is divisible by 49; it is also divisible by 3, so &lt;strong&gt;349125&lt;/strong&gt; is the only solution. We have:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TWENtY=349125; T=3, W=4, E=9, N=1, t=2, Y=5.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-03-30T09:14:33.933" LastActivityDate="2015-03-30T09:14:33.933" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11221" PostTypeId="2" ParentId="10575" CreationDate="2015-03-30T12:09:37.477" Score="3" Body="&lt;p&gt;Start with the case n=5. The winning scenario for H is given below for each combination of starting hands.&lt;br/&gt;&#xA;1.No T start, direct win&lt;br/&gt;&#xA;-H H H H H&lt;br/&gt;&#xA;2.One T sequence start, win at first turn&lt;br/&gt;&#xA;H H H H T = add a T, then flip the last two.&lt;br/&gt;&#xA;H X X X H = where single one of the Xs is a T, add a H at the end and flip the T.&lt;br/&gt;&#xA;H H T T H = add an H, flip the TT sequence.&lt;br/&gt;&#xA;H T T H H = add an H, flip the TT sequence.&lt;br/&gt;&#xA;H H H T T = add a T, flip the TTT sequence.&lt;br/&gt;&#xA;H T T T H = add an H, flip the TTT sequence.&lt;br/&gt;&#xA;H H T T T = add a T, flip the TTTT sequence.&lt;br/&gt;&#xA;Summary: Add an H and flip the sequence, if there is a single T sequence and the last coin is not a T. Add a T and flip the sequence, if there is a single T sequence and the last coin is a T. &lt;br/&gt;&#xA;3.Two T sequences start,&lt;br/&gt;&#xA;a)&lt;br/&gt;&#xA;H H T H T = add a H at the end, flip the HH sequence at the beginning.&lt;br/&gt; &#xA;Player T now has the hand T T T H T H.&lt;br/&gt; &#xA;Player T can either add a T, T T T T H T H. Now each flip will result in a single T sequence.&lt;br/&gt;&#xA;Or add a H, H T T T H T H. Now each flip will again result in a single T sequence.&lt;br/&gt;&#xA;Single T sequence is a one turn win.&lt;br/&gt; &#xA;b)&lt;br/&gt;&#xA;H T H H T = add a H at the end, flip the H sequence at the beginning.&lt;br/&gt; &#xA;Player T now has the hand T T H H T H.&lt;br/&gt; &#xA;Player T can either add a T, T T T H H T H. Now each flip will result in a single T sequence.&lt;br/&gt;&#xA;Or add a H, H T T H H T H. Now each flip will again result in a single T sequence.&lt;br/&gt;&#xA;Single T sequence is a one turn win.&lt;br/&gt; &#xA;c)&#xA;H T T H T = add a H at the end, flip the H sequence at the beginning.&lt;br/&gt; &#xA;Player T now has the hand T T T H T H.&lt;br/&gt; &#xA;Player T can either add a T, T T T T H T H. Now each flip will result in a single T sequence.&lt;br/&gt;&#xA;Or add a H, H T T T H T H. Now each flip will again result in a single T sequence.&lt;br/&gt;&#xA;Single T sequence is a one turn win.&lt;br/&gt; &#xA;d)&lt;br/&gt;&#xA;H T H T T = add a H at the end, flip the H sequence at the beginning.&lt;br/&gt; &#xA;Player T now has the hand T T H T T H.&lt;br/&gt;&#xA;Player T can either add a T, T T T H T T H. Now each flip will result in a single T sequence.&lt;br/&gt;&#xA;Or add a H, H T T H T T H. Now each flip will again result in a single T sequence.&lt;br/&gt;&#xA;Single T sequence is a one turn win.&lt;br/&gt; &#xA;e)&lt;br/&gt;&#xA;H T H T H = add a H at the end, flip the rightmost T.&lt;br/&gt; &#xA;Player T now has the hand H T H H H H.&lt;br/&gt;&#xA;Player T can either add a T, T H T H H H H. Now each flip will result in a single T sequence.&lt;br/&gt;&#xA;Or add a H, H H T H H H H. Now each flip will again result in a single T sequence.&lt;br/&gt;&#xA;Single T sequence is a one turn win.&lt;br/&gt; &#xA;Summary:If there are two T sequences and T as a last element, add a H at the end and flip the first H. If there are two T sequences and H as a last element, add a H at the end and flip the rightmost T. Then all moves of the opponent causes a single T sequence, which is a one turn win.&#xA;I will elaborate the algorithm for n&gt;5, if I can find time.&#xA;All of the cases for n = 6 are covered by n = 5 except for H T H T H T.&#xA;Proof: All other cases include a pair of coin which can thought as a single pair. Thus covering only this case will prove the same logic for n = 6.&#xA;Right now as I can see all the possible games starting as H T H T H T can be defended and set in a dead lock by a clever T player. I will try to prove it, when I have the time.&#xA;If this hand can be defended by T, then all other values of n might also cause such a hand. So n = 5 is the answer (assuming I didn't get lost at some point).  &lt;/p&gt;&#xA;" OwnerUserId="10131" LastEditorUserId="10131" LastEditDate="2015-03-30T21:07:07.060" LastActivityDate="2015-03-30T21:07:07.060" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11223" PostTypeId="1" AcceptedAnswerId="11224" CreationDate="2015-03-30T12:56:56.717" Score="8" ViewCount="1403" Body="&lt;p&gt;**** No programs allowed. You must find the answer using your own brain. ****&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your hangman game allows for 9 misses, but on the 10th miss, you're hanged!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't want to be hanged. The letters marked red underneath have already been guessed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the English word?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/at44S.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cqY6o.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="5373" LastEditDate="2015-03-30T20:58:49.740" LastActivityDate="2015-03-30T20:58:49.740" Title="Hangman- No misses left" Tags="|word|game|english|pencil-and-paper-games|no-computers|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11224" PostTypeId="2" ParentId="11223" CreationDate="2015-03-30T13:02:15.000" Score="10" Body="&lt;p&gt;I reckon the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pizzazz&lt;br&gt;&#xA;&lt;a href=&quot;http://www.thefreedictionary.com/pizzazz&quot;&gt;http://www.thefreedictionary.com/pizzazz&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6205" LastActivityDate="2015-03-30T13:02:15.000" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11225" PostTypeId="1" CreationDate="2015-03-30T13:46:19.873" Score="9" ViewCount="612" Body="&lt;p&gt;So here's my first puzzle for you!&lt;/p&gt;&#xA;&lt;p&gt;A late casino owner leaves his descendants this letter:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;Dear my beloved sons Francis and Owen, dear my beloved wife Uma,&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;The &lt;strong&gt;dice&lt;/strong&gt; are rolled. You finally made it and as law requires, all of you will get their &lt;strong&gt;split&lt;/strong&gt; of the jackpot. The only thing standing in between you and your lawfully acquired wealth is a four-inch wall of ferro-concrete. To open the door to the safe a four digit number is required that you can derive from the below code. Let's see if you are worth your legacy and settle it &lt;strong&gt;once&lt;/strong&gt; and for all.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Your loving dad/husband Richard&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;3 4 6 A 3 2 5 J 3 1 3 Q 5 K 3 A A 1 2 5 3 6 5 J 2 3 Q 1&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Find the combination!&lt;/p&gt;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Can you make a sequence of numbers only from it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; You can derive all information needed from the letter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; No knowledge of specific games is required!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Hint 4:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Some restructuring is required in the process. (Hint 2 applies for that, too)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Hint 5:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; How many splits are there?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; I updated the description, added a letter to provide better hints to the solution and updated all hints.&lt;/p&gt;&#xA;" OwnerUserId="10583" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-13T15:54:05.140" Title="The Legacy of the Casino Owner - Code" Tags="|password|code|" AnswerCount="4" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="11226" PostTypeId="1" AcceptedAnswerId="11233" CreationDate="2015-03-30T14:40:43.067" Score="7" ViewCount="1481" Body="&lt;p&gt;My Indian friend Jamal recently participated in the well-known TV quiz&#xA;show &lt;em&gt;&quot;Who wants to be a millionaire?&quot;&lt;/em&gt;&#xA;(&lt;a href=&quot;http://en.wikipedia.org/wiki/Who_Wants_to_Be_a_Millionaire&quot;&gt;http://en.wikipedia.org/wiki/Who_Wants_to_Be_a_Millionaire&lt;/a&gt;). You might have heard or read about Jamal's spectacular success; the media reported on it extensively.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what Jamal told us about his 100 Euro question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At the beginning I was quite nervous. My hands were sweaty and sticky.&#xA;  After some polite chatting with Prem Kumar, the music started and the&#xA;  lights were turned down so that I could not see anyone in the room but&#xA;  Prem. I stared at his lips. The show began.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Prem slowly read the question and each of the four answers in that&#xA;  famous, deep, fascinating voice of his.  Since this was the first&#xA;  question, it seemed pretty easy.  The answers were four consecutive&#xA;  integers in increasing order.  My first quick thought was to pick&#xA;  answer B, but then I suddenly remembered the last time I bought some&#xA;  doughnuts and chose answer C instead.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What was the 100 Euro question?&lt;br&gt;&#xA;What where the four possible answers A, B, C, D?&lt;/p&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2016-01-27T07:48:14.827" Title="Slumdog millionaire (1)" Tags="|knowledge|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="11227" PostTypeId="5" CreationDate="2015-03-30T14:47:58.243" Score="0" Body="&lt;p&gt;Trivia puzzles are puzzles that are based on uncommon but usually notable knowledge about a certain subject. These puzzles will tend to deal with a lot of physical and historical facts, rather than the logic behind a certain mathematical construct, and are more prosaic than riddles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions in this category may require research to answer. Generally the information should be Googlable, or at least publicly available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Puzzles that are just trivia questions are off-topic on this Stack Exchange, but puzzles that involve some form of reasoning around the trivia are generally welcome.&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorDisplayName="user88" LastEditDate="2015-04-21T19:07:30.293" LastActivityDate="2015-04-21T19:07:30.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11228" PostTypeId="4" CreationDate="2015-03-30T14:47:58.243" Score="0" Body="Puzzles and questions around nonessential, specifically detailed knowledge on topics of popular culture." OwnerUserId="9404" LastEditorUserId="9404" LastEditDate="2015-03-30T17:43:19.777" LastActivityDate="2015-03-30T17:43:19.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11229" PostTypeId="2" ParentId="11212" CreationDate="2015-03-30T14:48:31.750" Score="7" Body="&lt;p&gt;The answer is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;MEDITATE &lt;BR&gt;&lt;BR&gt;&#xA;&#xA;CMYK - RGB colour codes &lt;BR&gt;&lt;BR&gt;&#xA;blac&lt;strong&gt;K C&lt;/strong&gt;yan &lt;strong&gt;B&lt;/strong&gt;lue &lt;strong&gt;G&lt;/strong&gt;reen &lt;strong&gt;R&lt;/strong&gt;ed &lt;strong&gt;Y&lt;/strong&gt;ellow &lt;strong&gt;R&lt;/strong&gt;ed &lt;strong&gt;C&lt;/strong&gt;yan &lt;BR&gt;&#xA;&#xA;&#xA;KCBGRYRC + Caesar Cipher = MEDITATE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10688" LastEditorUserId="5188" LastEditDate="2015-03-30T17:58:13.257" LastActivityDate="2015-03-30T17:58:13.257" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11230" PostTypeId="2" ParentId="10911" CreationDate="2015-03-30T14:53:41.580" Score="2" Body="&lt;p&gt;You may be a  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Post-it  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In fact:&lt;br&gt;&#xA;My name sounds like a bird  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Stilt&quot; rel=&quot;nofollow&quot;&gt;Stilt&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whole body is like tongue of a lizard &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's sticky and useful!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Am different colors &#xA;And I well can be very littered  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They can be all colors, scattered everywhere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My energy melts away over time &#xA;But not quite like an ice  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's not melting like ice, but its sticky power decreases over time.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And when not taken care of &#xA;I'll collect dust and lice &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think we all know what happens to a post-it when it's not pinned to a surface...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-03-30T14:53:41.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11231" PostTypeId="2" ParentId="11185" CreationDate="2015-03-30T14:59:29.450" Score="5" Body="&lt;p&gt;Showing an answer &lt;em&gt;x&lt;/em&gt; is correct requires two things:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1, Show &lt;em&gt;x&lt;/em&gt; is feasible (can be achieved)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2, Show that &lt;em&gt;y&lt;/em&gt; &gt; &lt;em&gt;x&lt;/em&gt; is infeasible (cannot be achieved)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For part 1, a greedy strategy works, as mentioned in previous answer.  Place as many oaks in a row as possible (ie, six oaks) without having five trees in a row between two oaks; then place six birches; duplicate this pattern eight times; fill the last four places with oaks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For part 2, we will assume a configuration &lt;em&gt;C&lt;/em&gt; has at most 47 birches, and will show &lt;em&gt;C&lt;/em&gt; is infeasible, as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, note that 100 = 8⅓ dozen.  If at most 47 birches are placed somewhere in the first eight consecutive-dozens of a hundred, then at least one of those consecutive-dozens contains fewer than six birches (because 47/8 &amp;lt; 6).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, note that if a consecutive-dozen has fewer than six birches, then at least two oaks are separated by exactly five trees.  This is because there are six ways to have five-separation, involving places 1+7; 2+8; 3+9; 4+10; 5+11; and 6+12.  Five birches can block up at most five of those six ways; it takes at least six birches to block them all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that any configuration &lt;em&gt;C&lt;/em&gt; with fewer than 48 birches is infeasible.&lt;/p&gt;&#xA;" OwnerUserId="2004" LastActivityDate="2015-03-30T14:59:29.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11232" PostTypeId="2" ParentId="11226" CreationDate="2015-03-30T14:59:38.987" Score="0" Body="&lt;p&gt;My guess is that the 100€ question was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What's the average price of two doughnuts in an italian bar?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the four answers were:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A: 1€ - B: 2€ - C: 3€ - D: 4€&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jamal looks at the answers and his first thought goes to the answer B (2€), because of the price corresponding with the number of doughnuts; then he remembers the last time he bought two doughnuts, which payed 3 euros, and changes his mind choosing the right answer: C.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10248" LastEditorUserId="10248" LastEditDate="2015-03-30T15:06:09.560" LastActivityDate="2015-03-30T15:06:09.560" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11233" PostTypeId="2" ParentId="11226" CreationDate="2015-03-30T15:12:30.140" Score="18" Body="&lt;p&gt;My guess is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; How many is a baker's dozen?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With the choices being:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A: 11&lt;br&gt;&#xA; B: 12&lt;br&gt;&#xA; C: 13&lt;br&gt;&#xA; D: 14  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the reasoning behind that is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Thinking of 12 is a natural first reaction when one hears the word &quot;dozen&quot;. A &quot;baker's dozen&quot; is actually 13, but unless one is already familiar with this term they may not know that it is 13 until they go to a baker and actually order a baker's dozen of something. Much like in the movie, there were many of the answers that others might have considered common knowledge but Jamal wouldn't have known if he hadn't had some life experience which taught him that fact.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10151" LastEditorUserId="10151" LastEditDate="2015-03-30T15:20:33.800" LastActivityDate="2015-03-30T15:20:33.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11234" PostTypeId="1" AcceptedAnswerId="11236" CreationDate="2015-03-30T15:25:08.513" Score="10" ViewCount="4724" Body="&lt;p&gt;I come at night, and in the day.&lt;br&gt;&#xA;I'm always there, do what you may.&lt;br&gt;&#xA;You may think that I hate the light,&lt;br&gt;&#xA;Because it is what gives you your sight.&lt;br&gt;&#xA;But truthfully in light I trust,&lt;br&gt;&#xA;Because for me it is a must.&lt;br&gt;&#xA;Left or right, or down below&lt;br&gt;&#xA;I'll follow you, fast or slow.&lt;br&gt;&#xA;You really can't escape from me,&lt;br&gt;&#xA;So tell me now, what I may be.  &lt;/p&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2016-05-13T12:35:24.880" Title="Am I Creepy - What am I?" Tags="|riddle|word|rhyme|" AnswerCount="5" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11235" PostTypeId="2" ParentId="11234" CreationDate="2015-03-30T15:29:16.643" Score="6" Body="&lt;p&gt;To me, this sounds like :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a shadow ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come at night, and in the day.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; shadows are visible at both night and day (night they are dimmer, but even moon light will cast a shadow)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm always there, do what you may.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Shadows are always there, just dimmer or such depending on the light.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You may think that I hate the light,&#xA;Because it is what gives you your sight.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; shadows are traditionally a lack of light  ..&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But truthfully in light I trust,&#xA;Because for me it is a must.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a shadow can't exist without light&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Left or right, or down below&#xA;I'll follow you, fast or slow.&#xA;You really can't escape from me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; no matter which way you move, turn, or fast/slow you go, your shadow will match you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9267" LastEditorUserId="9267" LastEditDate="2015-03-30T18:35:54.077" LastActivityDate="2015-03-30T18:35:54.077" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11236" PostTypeId="2" ParentId="11234" CreationDate="2015-03-30T15:29:33.180" Score="14" Body="&lt;p&gt;I'm guessing it's:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; darkness/shadow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I come at night, and in the day.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The night is dark and there are shadows in the day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm always there, do what you may.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your shadow is always there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You may think that I hate the light,&#xA;Because it is what gives you your sight.&#xA;But truthfully in light I trust,&#xA;Because for me it is a must.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is no darkness or shadow without light. Also light &quot;gives&quot; us our sight. We need light to see.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Left or right, or down below&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Depending on light's position the shadow may fall in different directions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll follow you, fast or slow.&#xA;You really can't escape from me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No matter how fast you move, your shadow is always there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="39" LastEditDate="2015-03-31T03:31:19.093" LastActivityDate="2015-03-31T03:31:19.093" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11237" PostTypeId="2" ParentId="11234" CreationDate="2015-03-30T15:33:28.870" Score="6" Body="&lt;p&gt;i think it is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;your shadow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come at night, and in the day.&#xA;I'm always there, do what you may.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;my shadow is always there, day or night&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You may think that I hate the light,&#xA;Because it is what gives you your sight.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;light allows me to see the shadows following me when the light is bright&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But truthfully in light I trust,&#xA;Because for me it is a must.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I cant have shadow without a light, with a bright light, shadow is even more obvious&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Left or right, or down below&#xA;I'll follow you, fast or slow.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;left or right depends on where the light source is. Or at noon, my shadow is directly under me&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You really can't escape from me,&#xA;So tell me now, what I may be.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;cant separate me from my shadow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1504" LastActivityDate="2015-03-30T15:33:28.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11238" PostTypeId="1" AcceptedAnswerId="11241" CreationDate="2015-03-30T15:59:47.680" Score="105" ViewCount="41183" Body="&lt;p&gt;In 1980, a man lives on the 30th floor of his building. Every morning he wakes up, gets ready, kisses his wife, takes the elevator down to the first floor, and goes to work.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One morning, he wakes up, gets ready, kisses his wife, and gets in the elevator, exactly as he always has done. The morning is exactly the same as any other morning, and no one else was in their apartment.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The man is alone in the elevator and there are no windows in or visible from the elevator. Without receiving any  communication of any kind, or the man being in any danger of any harm, he begins crying when the elevator reaches the 12th floor, because he knows his wife has died.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this possible?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes (added later) to eliminate &quot;too broad&quot; answers: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; His wife was alive when he left the apartment.&lt;br&gt;&#xA; The man loved his wife, and would never harm her in any way.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And for the more extreme options: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The man had a boring typical day job. No danger involved. &#xA; The laws of physics have not been repealed, and there are no mythical or fiction occurrences happening.&lt;br&gt;&#xA; As always, aliens are not involved.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="9951" LastEditDate="2015-07-06T14:17:50.607" LastActivityDate="2019-02-08T13:04:35.487" Title="My wife has died" Tags="|logical-deduction|lateral-thinking|" AnswerCount="15" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="11239" PostTypeId="2" ParentId="11238" CreationDate="2015-03-30T16:06:59.570" Score="16" Body="&lt;p&gt;Possibly&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;His wife is ill and needs a certain medication daily at exactly the same time. On this day, the man forgot to give her the medication and remembered once he saw the number 12 in the elevator. Perhaps 12 is involved in her dosage.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8942" LastEditorUserId="46903" LastEditDate="2019-02-07T04:38:21.153" LastActivityDate="2019-02-07T04:38:21.153" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="11241" PostTypeId="2" ParentId="11238" CreationDate="2015-03-30T16:14:28.237" Score="114" Body="&lt;p&gt;His wife&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; is on in-home life support.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The elevator &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; stops at the 12th floor because the power in the building went out.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So now he's &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; stuck in an elevator (despite being in no harm) while his wife is dying upstairs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I guess you could argue that the event is a sort of &quot;communication&quot;, but that seems pretty weak to me.&lt;/p&gt;&#xA;" OwnerUserId="1563" LastActivityDate="2015-03-30T16:14:28.237" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="11242" PostTypeId="1" AcceptedAnswerId="11243" CreationDate="2015-03-30T16:40:30.957" Score="17" ViewCount="4968" Body="&lt;p&gt;A friend of mine got me &lt;a href=&quot;http://www.thinkfun.com/microsite/solitairechess/demo&quot; rel=&quot;noreferrer&quot;&gt;Solitaire Chess&lt;/a&gt; for Christmas years ago. I set up the board below. Is it possible to solve it? If so, what is the solution? (No programs allowed- Use only your brain.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rules:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Pieces move as in regular chess&lt;/li&gt;&#xA;  &lt;li&gt;Every move must be a capture&lt;/li&gt;&#xA;  &lt;li&gt;When there is only 1 piece left, you win.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zgPla.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="5373" LastEditDate="2015-03-31T09:10:38.770" LastActivityDate="2020-12-20T10:00:53.303" Title="Solitaire Chess" Tags="|strategy|chess|no-computers|" AnswerCount="9" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11243" PostTypeId="2" ParentId="11242" CreationDate="2015-03-30T16:47:58.443" Score="11" Body="&lt;p&gt;Let's go with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nxa1, Bxc4, Rxc4, Rxd4, Bxb2, Bxd4, Bxa1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastEditorDisplayName="user88" LastEditDate="2015-03-30T17:15:01.093" LastActivityDate="2015-03-30T17:15:01.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11244" PostTypeId="2" ParentId="11242" CreationDate="2015-03-30T16:49:52.437" Score="6" Body="&lt;p&gt;I got&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bb2, Bb3, Rc4, Rd4, Bd4, Nb3, Nd4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1563" LastActivityDate="2015-03-30T16:49:52.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11245" PostTypeId="2" ParentId="11238" CreationDate="2015-03-30T17:27:06.683" Score="36" Body="&lt;p&gt;Maybe&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;He poisoned her morning coffee.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="46903" LastEditDate="2019-02-07T00:02:31.323" LastActivityDate="2019-02-07T00:02:31.323" CommentCount="14" ContentLicense="CC BY-SA 4.0" />
  <row Id="11246" PostTypeId="2" ParentId="11238" CreationDate="2015-03-30T17:30:33.647" Score="2" Body="&lt;p&gt;When he left he had set up the bomb, and when he reached the 12th floor he heard, or felt the shockwave of the explosion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wife's name was Old Yeller in this interpretation.&lt;/p&gt;&#xA;" OwnerDisplayName="user10769" LastActivityDate="2015-03-30T17:30:33.647" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11247" PostTypeId="2" ParentId="11242" CreationDate="2015-03-30T17:39:08.893" Score="4" Body="&lt;p&gt;Different from the other two&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nxb3, Nxd4, Bxb2, Bxc4, Rxc4, Rxd4, Bxd4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9442" LastEditorUserId="9442" LastEditDate="2015-03-31T12:58:24.130" LastActivityDate="2015-03-31T12:58:24.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11248" PostTypeId="2" ParentId="11226" CreationDate="2015-03-30T18:21:36.700" Score="6" Body="&lt;p&gt;The Question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What nonpositive number do you multiply by $-1$ to get number of holes of an American doughnut?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With the choices being:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A: -3&lt;br&gt;&#xA; B: -2&lt;br&gt;&#xA; C: -1&lt;br&gt;&#xA; D:  0 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the reason why this is correct:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Naturally, torus has two holes, so the answer is B because $-1\cdot(-2) = 2$. But, wait! Last time our hero bought a doughnut and tried to eat it he noticed strange thing indeed. Doughnut was not actually a torus, it was &lt;em&gt;solid&lt;/em&gt;. Thus, doughnut only has one hole, so the answer is C.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10762" LastEditorUserId="10762" LastEditDate="2015-03-30T19:22:17.970" LastActivityDate="2015-03-30T19:22:17.970" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11249" PostTypeId="1" AcceptedAnswerId="11274" CreationDate="2015-03-30T18:24:05.010" Score="8" ViewCount="2260" Body="&lt;p&gt;The more complications I have, the more others want me,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet my face is easily read.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I only had to tell you one thing and you’d be fine with me,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These days you look elsewhere for the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to all the jewels I’ve been given,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m a little more punctilious when I’m at work.&lt;/p&gt;&#xA;" OwnerUserId="9696" LastEditorUserId="9696" LastEditDate="2015-03-30T19:39:51.700" LastActivityDate="2015-03-30T21:00:50.200" Title="My hands are moving" Tags="|riddle|" AnswerCount="6" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11250" PostTypeId="2" ParentId="11249" CreationDate="2015-03-30T18:29:37.907" Score="6" Body="&lt;p&gt;My guess is a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; watch, potentially one of the new iWatches.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The more complicated I am, the more others want me,&lt;br&gt;&#xA;Yet my face is easily read.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Watches are really complicated, and having one can be a source of bragging, also, watches are (should be) easy to read&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I only had to tell you one thing and you’d be fine with me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Just tells time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These days you look elsewhere for the same.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I know I whip out my phone to check the time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks to all the jewelry I’ve been given,&lt;br&gt;&#xA;I’m a little more punctilious when I’m at work.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not fully sure, watches have been bling'd out? Maybe the cool new watches are on top of all your activities? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-03-30T18:29:37.907" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="11251" PostTypeId="2" ParentId="11249" CreationDate="2015-03-30T18:29:39.193" Score="4" Body="&lt;p&gt;Try 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it is a:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Digital Watch   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;he more complicated I am, the more others want me,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some digital watches can tell time, set alarms, check the date, change TV channels, be stopwatches, etc. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet my face is easily read.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Though an analog clock or watch is not difficult, a digital watch just throws it all out there easy as can be.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I only had to tell you one thing and you’d be fine with me,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You generally only check the time, but you can do a lot of other things.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These days you look elsewhere for the same.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Today we look at phones, wall clocks, other people, computer clocks, etc.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks to all the jewelry I’ve been given,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Watches are now as much jewelry as anything else. So shiny!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I’m a little more punctilious when I’m at work&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your watch can now be used for work items, but also lets you know the exact second meetings should start or end, as well as the workday.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="9951" LastEditDate="2015-03-30T18:56:36.960" LastActivityDate="2015-03-30T18:56:36.960" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11253" PostTypeId="2" ParentId="11242" CreationDate="2015-03-30T18:46:12.403" Score="2" Body="&lt;p&gt;If pawns moving &quot;down&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pxc3&#xA; Pxa1 (promote to anything you want)&#xA; Rxc3&#xA; Rxb3&#xA; Rxa3&#xA; Rxa2&#xA; Rxa1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If pawns moving &quot;up&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rxd4&#xA; Bxb2&#xA; Bxb3&#xA; Rxb3&#xA; Nxb3&#xA; Bxd4&#xA; Nxd4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9267" LastActivityDate="2015-03-30T18:46:12.403" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11254" PostTypeId="2" ParentId="11242" CreationDate="2015-03-30T18:47:51.727" Score="2" Body="&lt;p&gt;Power to the church! (Bishops)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bb2, Ba1, Bb3, Bc4, Rc4, Rd4, Bd4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-03-30T19:37:11.880" LastActivityDate="2015-03-30T19:37:11.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11255" PostTypeId="2" ParentId="11242" CreationDate="2015-03-30T18:53:54.393" Score="3" Body="&lt;p&gt;What an interesting game. Thanks for sharing this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the first solution I found:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bxb3, Nxb3, Bxb2, Bxc3, Bxd4, Rxd4, Nxd4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note: for those who haven't seen the move notation before, &quot;Bd1&quot; or &quot;Bxd1&quot; means &quot;move the bishop to square 'd1.'&quot;&lt;br&gt;&#xA;B=bishop, N=knight, R=rook, P=pawn, K=king, Q=queen &lt;/p&gt;&#xA;" OwnerUserId="10773" LastActivityDate="2015-03-30T18:53:54.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11256" PostTypeId="1" AcceptedAnswerId="11257" CreationDate="2015-03-30T19:01:43.480" Score="3" ViewCount="758" Body="&lt;p&gt;Mason is improving at the game Battleship. He's got this current game in the bag, or so he thinks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;His opponent is Richard, whose method for placing his ships is completely random. Each side has 5 ships, and they reach the position below, where it is Mason's turn and Mason has sunk all of Richard's ships except for his Carrier, which is 5 units long. Richard has sunk only 2 of Mason's ships.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The board is 10x10, and all ships must be fully on the board and placed either horizontally or vertically. The green H's represent ships that Mason has already sunk. The red M's represent his misses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/beiv1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the question, you can assume that Mason will win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Mason plays optimally, in how many moves can he guarantee victory? Also, can you explain your answer so that the vast majority of those reading it will understand it?&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="4421" LastEditDate="2015-09-02T18:39:41.723" LastActivityDate="2015-09-02T18:39:41.723" Title="Battleship- Part 2" Tags="|strategy|board-games|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11257" PostTypeId="2" ParentId="11256" CreationDate="2015-03-30T19:14:56.060" Score="5" Body="&lt;p&gt;The max I've found is winning within &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;8 shots&lt;/strong&gt;:&lt;br&gt;&#xA;&lt;br&gt;&#xA; There are only 4 &quot;spots&quot; where the ship could be:&lt;br&gt;&#xA;&lt;br&gt;&#xA; B3-B8 (6 spots)&lt;br&gt;&#xA; E1-E8 (8 spots)&lt;br&gt;&#xA; H4-H8 (5 spots)&lt;br&gt;&#xA; A4-J4 (10 spots)&lt;br&gt;&#xA;&lt;br&gt;&#xA; The highest concentration of spots is at E4, so check that.&lt;br&gt;&#xA; If hit, you know it's along col-4 or row-E, so I would first check if it extends downward (F4). If hit, then we have at most 4 more shots (total 6).&lt;br&gt;&#xA; If F4 is a miss, then I would check E5, since if it is horizontal, that's a guaranteed hit (Leftmost would be E1-E5, rightmost would be E4-E8). After that, it's up to 4 more moves (total 6).&lt;br&gt;&#xA; If that is a miss, the only spot left is A4-E4, 4 more shots, 7 total.&lt;br&gt;&#xA;&lt;br&gt;&#xA; If E4 is a miss, I would go H4 next. It hits 2 full spots where the ship could be (F4-J4, H4-H8). If it's a hit, spend a shot to figure out vert/horz, then 4 more to finish, total 7.&lt;br&gt;&#xA;&lt;br&gt;&#xA; If H4 is a miss, the last place to be is at B3-B8. That will take up to 6 shots, &lt;strong&gt;8 total&lt;/strong&gt;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-03-30T20:19:36.333" LastActivityDate="2015-03-30T20:19:36.333" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11258" PostTypeId="2" ParentId="9749" CreationDate="2015-03-30T19:17:48.213" Score="3" Body="&lt;p&gt;The plain text of &lt;code&gt;payload.me&lt;/code&gt; is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;Congratulations! Laziness, impatience, and hubris could not withhold the prize&lt;/code&gt;&lt;br&gt;&lt;code&gt;of this puzzle from you. Your actions are deserving of a reward, but if joy is&lt;/code&gt;&lt;br&gt;&lt;code&gt;a part of satisfaction, can attaining this goal be beaten? Spoils for the 1st!&lt;/code&gt;&lt;br&gt;&lt;br&gt;&#xA; &lt;code&gt;Yes, there is more to saying you are 1st to read this. If the latter be valid,&lt;/code&gt;&lt;br&gt;&lt;code&gt;then you have treasure to retrieve. Go to http://www.gog.com/redeem and see if&lt;/code&gt;&lt;br&gt;&lt;code&gt;the following codes can be claimed. These be the diadem reserved for a victor.&lt;/code&gt;&lt;br&gt;&lt;br&gt;&lt;code&gt;WZZL-WE4D-WLLD-RQHN&lt;/code&gt;&lt;br&gt;&lt;code&gt;WF79-KCSX-36Y5-9FTU&lt;/code&gt;&lt;br&gt;&lt;code&gt;DD9DB-7E243-F133C-95C75&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Hint 36&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If you don't want to write your own implementation go to &lt;a href=&quot;http://code.activestate.com/recipes/578075/&quot; rel=&quot;nofollow&quot;&gt;http://code.activestate.com/recipes/578075/&lt;/a&gt; for existing Python code. This puzzle is pretty much impossible to do without Python.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Hint 37&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The message body reads&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The disk jockey appears to be a reference to Dreary Jeans.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The body is really long so I won't repeat it hear since probably won't display correctly. Here are the first few hundred characters&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(b\'p/H5Od+#_&quot;{\\\\sf;&amp;lt;:^\\nE[J&amp;amp;QkWDS\\x0bT*B,]Pbcy\\x0&#xA;c07qUw(63`@Lo}ZeaYGC%zmRF9uM|.v-1N&amp;gt;2\\t A\\\'hi!Ixtl=?g&#xA;X4j$n~8KV)\\rr\',\n b\'3Tv+ Y9cnK%7ao*db6ryu@\\x0b)=~m?&#xA;4W`&amp;gt;f&amp;lt;#MhNX1-ljIe(].Z\\t/|58wRE{tP![GS0A}kq&quot;x2Oi$H;psgQ&#xA;LJCD:\\\\B\\x0c\\nU_,\\\'\\r^zV&amp;amp;F\',\n b\'UGZ]\\n@{lgkb&#xA;hH&amp;gt;nup&amp;lt;t~c58SF%XY1Tz+\\x0cwi:,=x&quot;&amp;amp;?Nv*dA\\tIf/$ er|JDV(&#xA;4^ML6\\\'})7R.Pq[E\\\\9Q\\x0bOK_mBjs;#!y-`o2W\\r0Ca3\',&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I appreciate the 'hi!' in the first line. I hope that as intentional. This is likely not the key for the &lt;code&gt;payload.me&lt;/code&gt;, that key should only contain alphanumerics. The important things to note are that it looks like a tuple of binary strings so you know we're dealing with Python 3.x. Also all control characters are no actually there &lt;code&gt;\n&lt;/code&gt; is two different characters, you will need to map those back to control characters. In the end you should have a 160 strings, each containing a scrambled copy of &lt;code&gt;string.printable&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Hint 38&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;message body reads&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Has man's rib caused yet another fall? Yes, but this time it is on death itself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;after decoding &lt;code&gt;payload.py&lt;/code&gt; doesn't look super useful. It's 1788 mostly random characters (which I'm not even going to try to include). The operating assumption is that this is a message encrypted with the key above. We're still missing the primer.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Hint 39&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I just used &lt;a href=&quot;http://www.dcode.fr/ascii-85-encoding&quot; rel=&quot;nofollow&quot;&gt;http://www.dcode.fr/ascii-85-encoding&lt;/a&gt; to convert between ASCII85 and hex, then decode as usual. I don't think I even bothered with the body. &lt;code&gt;primer.data&lt;/code&gt; looks like it could be a primer. Again, the decode text is really ugly so I'm not going to try to include it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The important thing it is the primer for the key and cipher text of the last two hints. When decrypted you get:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import itertools&#xA;import random&#xA;import re&#xA;import string&#xA;&#xA;import markov_encryption as me&#xA;&#xA;###############################################################################&#xA;&#xA;encode = lambda s: s.encode('ascii')&#xA;decode = lambda b: b.decode('ascii')&#xA;&#xA;&#xA;def main():&#xA;    ciphertext, primertext, keytext = load_puzzle_data()&#xA;    plaintext = generate_plaintext(primertext, keytext, ciphertext)&#xA;    key, primer = create_key_and_primer(plaintext)&#xA;    globals().update(locals())&#xA;&#xA;&#xA;def load_puzzle_data():&#xA;    with open('ciphertext.txt') as file:&#xA;        ciphertext = file.read()&#xA;    with open('transcript_primer.txt') as file:&#xA;        primertext = file.read()&#xA;    with open('transcript_key.pickle') as file:&#xA;        keytext = file.read()&#xA;    return ciphertext, primertext, keytext&#xA;&#xA;&#xA;def generate_plaintext(primertext, keytext, ciphertext):&#xA;    primer = me.Primer(encode(primertext))&#xA;    key = me.Key(tuple(map(encode, keytext.split('\n'))))&#xA;    decrypter = me.Decrypter(key, primer)&#xA;    plaintext = decode(decrypter.process(encode(ciphertext)))&#xA;    return plaintext&#xA;&#xA;&#xA;def create_key_and_primer(plain):&#xA;    primer = ''.join(m.groups()[0] for m in re.finditer(r'\((.)\)', plain))&#xA;    base = []&#xA;    for c in itertools.chain(filter(str.isalnum, plain),&#xA;                             string.ascii_letters,&#xA;                             string.digits):&#xA;        if c not in base:&#xA;            base.append(c)&#xA;    key = ''.join(base)&#xA;    randomizer = random.Random()&#xA;    for i in map(ord, primer):&#xA;        randomizer.seed(i)&#xA;        randomizer.shuffle(base)&#xA;        key += '\n' + ''.join(base)&#xA;    p = me.Primer(encode(primer))&#xA;    k = me.Key(tuple(map(encode, key.split('\n'))))&#xA;    return k, p&#xA;&#xA;###############################################################################&#xA;&#xA;if __name__ == '__main__':&#xA;    main()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is pretty straightforward, load the cipher text, primer and key (from part 1) and decrypt. Then use the decrypted text as input in &lt;code&gt;create_key_and_primer&lt;/code&gt; to generate a new key and primer containing only alpha numeric comparators which can be used to decrypt the &lt;code&gt;payload.me&lt;/code&gt;, atleast if you use Python 3.x. There were enough changes to &lt;code&gt;random&lt;/code&gt; between 2 and 3 that even with the same seed I still got a wrong answer on 2.7 until I switched over to 3.4.&lt;/p&gt;&#xA;" OwnerUserId="9060" LastActivityDate="2015-03-30T19:17:48.213" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11259" PostTypeId="2" ParentId="11256" CreationDate="2015-03-30T19:18:05.387" Score="3" Body="&lt;p&gt;He is guaranteed victory in &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;8 moves&lt;/strong&gt;. There are 8 possible scenarios, with shot totals of 5, 6, 6, 7, 6, 7, 7, 8. They are all described below.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since the carrier has not yet been hit, we know that it must be in one of the spaces that are at least 5 squares long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;His first shot $S_1$ is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;E4&lt;/strong&gt;&lt;br&gt;&#xA; This overlaps with a carrier sitting vertically somewhere in the range A4-I4 or horizontally somewhere in the range E1-E8.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If it's a hit, his second shot $S_{2,1}$ is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;E5&lt;/strong&gt;&lt;br&gt;&#xA; A hit on E5 means that the carrier is positioned horizontally. His next shots extend to the right starting with E6.&lt;br&gt;&#xA; If his next three shots all hit, he has sunk the carrier using &lt;strong&gt;5 shots&lt;/strong&gt;.&lt;br&gt;&#xA; If one of his shots to the right misses, he begins shooting left starting with E3. The carrier will have been sunk using &lt;strong&gt;6 shots&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If $S_{2,1}$ is a miss, his third shot $S_{3,1}$ is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;D4&lt;/strong&gt; and then continuing up the grid.&lt;br&gt;&#xA; A miss on &lt;strong&gt;E5&lt;/strong&gt; means the carrier is vertical.&lt;br&gt;&#xA; If these four vertical shots all hit, the carrier has been sunk using &lt;strong&gt;6 shots&lt;/strong&gt;.&lt;br&gt;&#xA; If one of the shots misses, he fires at F4 and continues downwards until the carrier is sunk in a total of &lt;strong&gt;7 shots&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the event that $S_1$ misses, his second shot $S_{2,2}$ will be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;B4&lt;/strong&gt;&lt;br&gt;&#xA; In the event of a hit, we know the carrier is horizontal. If it were vertical, $S_1$ would have also hit the carrier. 4 more shots to the right will sink the carrier in &lt;strong&gt;6 shots&lt;/strong&gt;.&lt;br&gt;&#xA; If the final shot misses, one shot fired at B3 sinks the carrier bringing our total to &lt;strong&gt;7 shots&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the event that $S_{2,2}$ also misses, his third shot $S_{3,2}$ will be guaranteed to hit at:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;H4&lt;/strong&gt;&lt;br&gt;&#xA; Here, we might be hitting a vertical carrier (F4-J4) or a horizontal carrier (H4-H8).&lt;br&gt;&#xA; Testing &lt;strong&gt;H5&lt;/strong&gt; $S_4$ will determine the orientation.&lt;br&gt;&#xA; If $S_4$ is a hit, three more shots at H5-H8 will sink the carrier in a total of &lt;strong&gt;7 shots&lt;/strong&gt;.&lt;br&gt;&#xA; If $S_4$ is a miss, we know the carrier is positioned vertically. The carrier will be sunk by the remaining shots F4, G4, I4, J4; a total of &lt;strong&gt;8 shots&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-03-30T20:01:56.823" LastActivityDate="2015-03-30T20:01:56.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11260" PostTypeId="2" ParentId="11256" CreationDate="2015-03-30T19:20:09.837" Score="4" Body="&lt;p&gt;The answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the worst case, 8.&lt;br&gt;&#xA; This has a $\frac1{13}$ chance of happening with my strategy.&#xA; My strategy overall (Where $m$ is the moves it takes):&lt;br&gt;&#xA; $$\ \\\begin{align}P(m&amp;lt;5)&amp;amp;=\ \ 0\\P(m=5)&amp;amp;=\frac1{13}\\P(m=6)&amp;amp;=\frac5{13}\\P(m=7)&amp;amp;=\frac6{13}\\P(m=8)&amp;amp;=\frac1{13}\\P(m&amp;gt;8)&amp;amp;=\ \ 0\end{align}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First you want to ignore the spaces who have less than 5 spaces above/below and left/right. After all, if there aren't 5 spaces, how is there going to be a 5 unit long battleship?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yJ2gp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yJ2gp.png&quot; alt=&quot;Battleships_possible&quot; title=&quot;All white spaces are possible battleships&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, you want to hit the intersections of those pathways where the battleship could be to see which line the battleship is in, in this order, stopping when you hit a battleship:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/i8xHR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/i8xHR.png&quot; alt=&quot;Move_here&quot; title=&quot;Hit 1 then 2 then 3, but stop if you don&amp;#39;t miss.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Numbers in brackets denote moves taken to win)&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;If you hit 1 (E4):&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;- Go for E5.&lt;br&gt;&#xA; - - If it hits, you are sure it is in the row E. Use this process of elimination:&lt;br&gt;&#xA; - - - Go for E3, E2 and E1 until you miss or win.&lt;br&gt;&#xA; - - - Go for E6, E7 and E8 until you win. (Maximum 6 turns)&lt;br&gt;&#xA; - - If it misses, you are sure the ship is in the 4 column.&lt;br&gt;&#xA; - - - Go for D4, C4, B4 and A4 until you miss or win.&lt;br&gt;&#xA; - - - Go for F4, G4, H4 and I4 until you win.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;If you hit 2 (B4):&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;- Part of the battleship is B4 to B7 (As all other rows and the 4 column are blocked off with the new knowledge that E4 is a miss)&lt;br&gt;&#xA; - Go for B3.&lt;br&gt;&#xA; - - If it hits, you won. (6)&lt;br&gt;&#xA; - - Else B8 then you won. (7)  &lt;/p&gt;&#xA;&#xA;&lt;h1&gt;If you hit 3 (H4):&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;- Go for H5 (To check if the ship is the H row or the 4 column).&lt;br&gt;&#xA;- - If you hit, the battleship is H4 to H8. (7)&lt;br&gt;&#xA;- - If you don't, the battleship is F4 to J4 (8)  &lt;/p&gt;&#xA;" OwnerDisplayName="user10203" LastEditorDisplayName="user10203" LastEditDate="2015-03-30T20:48:22.813" LastActivityDate="2015-03-30T20:48:22.813" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11261" PostTypeId="2" ParentId="11242" CreationDate="2015-03-30T19:47:37.850" Score="2" Body="&lt;p&gt;I got:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bb2 Bc3 Bd4 Bb3 Nb3 Nd4 Rd4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10775" LastActivityDate="2015-03-30T19:47:37.850" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11262" PostTypeId="2" ParentId="9763" CreationDate="2015-03-30T19:50:27.867" Score="4" Body="&lt;p&gt;What the heck, I will just take the answer @Moghwyn suggests in a comment and run with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a tennis court&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Divided by a wall, N and S.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tennis is generally viewed from the side and above, so the net (wall) is up-and-down -- i.e. &quot;north-and-south&quot; -- in your field of view.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tunnels in between, but no one pass.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; holes in the net&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The West has more, and the East has less.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tennis is a Western sport, so surely the West has more tennis courts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Less is more,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tennis scoring goes &quot;deuce, advantage, game&quot;, which is roughly &quot;two, one, game over&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but More is less&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the score is advantage for your opponent, and then you win the point, the score reverts back to deuce. So by winning the point you move the game less close to completion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6174" LastActivityDate="2015-03-30T19:50:27.867" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11263" PostTypeId="2" ParentId="10911" CreationDate="2015-03-30T19:56:54.587" Score="3" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Sandpaper&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My name sounds like a bird&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Sandpiper&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whole body is like tongue of a lizard&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Rough&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Am different colors&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I well can be very littered&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My energy melts away over time&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Loses grit&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But not quite like an ice&#xA;And when not taken care of&#xA;I'll collect dust and lice &lt;/p&gt;&#xA;" OwnerUserId="10776" LastEditorUserId="10776" LastEditDate="2015-03-31T20:57:18.087" LastActivityDate="2015-03-31T20:57:18.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11265" PostTypeId="1" AcceptedAnswerId="11266" CreationDate="2015-03-30T19:59:47.907" Score="8" ViewCount="584" Body="&lt;p&gt;Invisible spot on a colorful graph,&lt;br&gt;&#xA;less than a hundred and more than a half,&lt;br&gt;&#xA;a proof that beginning of earth and titan exists,&lt;br&gt;&#xA;&quot;A&quot; is the starting letter on all my guests' list.  &lt;/p&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2015-04-07T03:56:11.440" Title="Invisible spot on a colorful graph, what am I?" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11266" PostTypeId="2" ParentId="11265" CreationDate="2015-03-30T20:13:05.407" Score="8" Body="&lt;p&gt;My answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Area 51&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Invisible spot on a colorful graph,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The location of Area 51 on the US map was secret for a long time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;less than a hundred and more than a half,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fifty one fits this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;a proof that beginning of earth and titan exists,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; according to pop culture (maybe now debunked), Area 51 houses Extra Terrestrials (E.T.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;A&quot; is the starting letter on all my guests' list.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A is for Alien.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5684" LastEditorUserId="5684" LastEditDate="2015-04-07T03:56:11.440" LastActivityDate="2015-04-07T03:56:11.440" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="11267" PostTypeId="1" AcceptedAnswerId="11284" CreationDate="2015-03-30T20:13:07.960" Score="34" ViewCount="4512" Body="&lt;p&gt;Two sheriffs are working on a case to find one culprit. There were initially 8 suspects; through independent  work, each sheriff has narrowed this down to a list of 2. Because they are good sheriffs, they can be sure that both of their lists contain the culprit. They plan to make a phone call tonight to see if they can combine their information to find the culprit. However, the locals (who know the 8 suspects, but neither of the narrowed lists) have tapped their phone line. If the locals can figure out the culprit based on this phone call, then they will lynch him before he can be brought to justice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can the sheriffs conduct their conversation so that they can both deduce the culprit, when possible*, but the locals can't?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*If the sheriffs have the same list of 2 suspects, then they won't be able to deduce the culprit. As long as their lists are different, they can. The lists cannot be disjoint, since they both contain the unique culprit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;em&gt;Mathematical Puzzles, a Connoisseur's Collection&lt;/em&gt;, Peter Winkler&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-06-03T02:14:19.273" LastActivityDate="2015-06-03T02:14:19.273" Title="Two Sheriffs and Eavesdroppers" Tags="|strategy|cryptography|information-theory|" AnswerCount="9" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="11268" PostTypeId="2" ParentId="11242" CreationDate="2015-03-30T20:14:39.750" Score="23" Body="&lt;p&gt;Here are my thoughts on the matter:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Want to end with a knight?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bb3, Bb2, Bc3, Bd4, Rd4, Nb3, Nd4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Want to end with a rook?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bb3, Bb2, Bc3, Bd4, Nb3, Nd4, Rd4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Want to end with a bishop?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bb3, Nb3, Nd4, Bb2, Bc3, Rd4, Bd4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Want to keep from having to move your horses from the stables?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bb3, Bb2, Bc4, Rc4, Rd4, Ba1, Bd4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Do you want to end on a square other than D4?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bb3, Bc4, Rc4, Rd4, Bb2, Bd4, Ba1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yeah.... I think there might be a few solutions...&lt;/p&gt;&#xA;" OwnerUserId="3922" LastActivityDate="2015-03-30T20:14:39.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11270" PostTypeId="2" ParentId="11267" CreationDate="2015-03-30T20:36:06.003" Score="14" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;They each encrypt their list with the other sheriff's RSA public key. They then read the encrypted string over the phone to the other, and each sheriff then decrypts the other's list using his own private key :).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's going to be a long phone call.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't taken computer security in a while so I hope I got the process right, I was just kinda going off the top of my head. &lt;/p&gt;&#xA;" OwnerUserId="10151" LastActivityDate="2015-03-30T20:36:06.003" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="11271" PostTypeId="2" ParentId="11267" CreationDate="2015-03-30T20:43:45.883" Score="33" Body="&lt;p&gt;I think it's pretty obvious.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They discuss a meeting place on the phone and then talk about it in person.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;;)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Alternatively,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; one sheriff reads his two suspects' names. The other sheriff says nothing and goes to arrest the culprit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-03-30T20:52:38.657" LastActivityDate="2015-03-30T20:52:38.657" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="11272" PostTypeId="2" ParentId="11249" CreationDate="2015-03-30T20:48:25.517" Score="2" Body="&lt;p&gt;My initial guesses matched these other answers, but from the comments I've gone in a slightly different direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fitness Tracker - I'm imagining the FitBit Charge, but most meet the criteria&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The more complications I have, the more others want me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The more features they have, the more people want to buy them&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet my face is easily read.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most have a simple display&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I only had to tell you one thing and you’d be fine with me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The specific model I mentioned tells the time, but others have a simple display telling you want mode they're in (Jawbone) or how many steps you've taken (FitBit Flex)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These days you look elsewhere for the same.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can check the time on your phone, and get the fitness information from related smart phone apps.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks to all the jewels I’ve been given,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All the additional sensors&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I’m a little more punctilious when I’m at work.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The fitness readings get more and more accurate&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9755" LastActivityDate="2015-03-30T20:48:25.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11273" PostTypeId="2" ParentId="11249" CreationDate="2015-03-30T20:52:51.747" Score="6" Body="&lt;p&gt;Surely it's an&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; analogue watch&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The more complications I have, the more others want me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Complication_%28horology%29&quot;&gt;Complications&lt;/a&gt; make watches more expensive/desirable&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet my face is easily read.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A watch's face is easily read to tell the time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I only had to tell you one thing and you’d be fine with me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A watch that tells the time is a fine watch&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These days you look elsewhere for the same.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People often tell the time using their phones nowadays&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks to all the jewels I’ve been given,&#xA;I’m a little more punctilious when I’m at work&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm guessing this is about &lt;a href=&quot;http://en.wikipedia.org/wiki/Jewel_bearing&quot;&gt;Jewel bearings&lt;/a&gt; making a watch run more accurately.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10781" LastActivityDate="2015-03-30T20:52:51.747" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11274" PostTypeId="2" ParentId="11249" CreationDate="2015-03-30T20:52:55.247" Score="13" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's a mechanical watch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My hands are moving&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Watch hands.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The more complications I have, the more others want me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Complication_%28horology%29&quot;&gt;http://en.wikipedia.org/wiki/Complication_%28horology%29&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet my face is easily read.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Even on a watch with many complications, the time is generally largest and easiest to read.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I only had to tell you one thing and you’d be fine with me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These days you look elsewhere for the same.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many other things tell time these days (but could refer to a cell phone?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks to all the jewels I’ve been given,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m a little more punctilious when I’m at work.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jewel bearings (&quot;jewels&quot;) make watches more accurate: &lt;a href=&quot;http://en.wikipedia.org/wiki/Jewel_bearing&quot;&gt;http://en.wikipedia.org/wiki/Jewel_bearing&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10780" LastActivityDate="2015-03-30T20:52:55.247" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11276" PostTypeId="2" ParentId="11249" CreationDate="2015-03-30T21:00:50.200" Score="7" Body="&lt;p&gt;How about a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Quartz Crystal Watch&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The more complications I have, the more others want me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;More features/luxuries = more demand&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet my face is easily read.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It is still just a basic watch&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I only had to tell you one thing and you’d be fine with me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These days you look elsewhere for the same.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Cell phone&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks to all the jewels I’ve been given,&#xA;I’m a little more punctilious when I’m at work.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Quartz Crystals are &lt;a href=&quot;http://en.wikipedia.org/wiki/Quartz_clock&quot;&gt;VERY good&lt;/a&gt; at precision time keeping.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10783" LastActivityDate="2015-03-30T21:00:50.200" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11277" PostTypeId="1" AcceptedAnswerId="11362" CreationDate="2015-03-30T21:04:46.377" Score="14" ViewCount="1402" Body="&lt;p&gt;A computer science professor decides to play a game with his three brightest (nearly perfectly intelligent) students.  He sits them in a circle with their backs to the center such that they cannot see each other but they can hear each other and he can watch them.  He hands them each a folded up sheet of paper and says the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I have chosen two numbers $X$ and $Y$, such that $2\le X \le Y \le 100$.  I have written the value of $D$ which equals $Y-X$ and given it to Donna. I have written the value of $S$ which equals $Y+X$ and given it to Sarah. I have written the value of $R$ which equals $Y/X$ and given it to Richard. All of these five of these numbers are whole numbers.  You are welcome to talk but the first to correctly determine the values of $X$ and $Y$ wins.  Anyone who says something they could have deduced to be false from what they already know loses.  You may begin.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a few minutes the following exchange occurs:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Sarah: Well, Richard doesn’t know the numbers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Donna: No, he doesn’t.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Richard:  Can I take it from the silence you both don’t know the numbers? … Yes?... Well in that case, I’ll admit I don’t know the numbers.  &lt;strong&gt;But that admission alone doesn’t provide either of you with any useful information.&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Donna: Wow... Really?... That means Sarah knows the answer.  I still don't though.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Sarah:  You must have made a mistake Donna; I didn't know the numbers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Richard:  Neither of you knows the answer; but I do and &lt;strong&gt;I never even unfolded my paper&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Both Sarah and Donna assumed Richard knew the value of $R$, Richard knew that they would assume that, and no one made any other mistakes. What are the numbers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If Abby says “Bob ate the pie” it can reasonably be translated to “According to Abby's calculations, assumptions, and aquired data: Bob ate the pie and Abby knows Bob ate the pie.”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When Richard says that the other two couldn’t figure out the numbers during the long pause or they would have already said them, he is correct and the others recognize it.  At no other time should you use the assumption “she would say she knew if she knew”. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Richard’s bolded line would have been interpreted by Sarah and Donna to mean “Richard knows we both already knew Richard didn’t know.” Richard and the professor know it means “I don’t have any new information to give as I don’t know R or any information you haven’t already been told.”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="99" LastEditorUserId="10615" LastEditDate="2016-02-24T04:04:52.727" LastActivityDate="2016-02-24T04:04:52.727" Title="Determine $X$ and $Y$: sum, difference and ratio" Tags="|logical-deduction|meta-knowledge|" AnswerCount="2" CommentCount="23" ContentLicense="CC BY-SA 3.0" />
  <row Id="11278" PostTypeId="2" ParentId="11267" CreationDate="2015-03-30T21:09:38.587" Score="3" Body="&lt;p&gt;Revised Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Denote suspects $S_A, S_B, S_C$ where $S_C$ is the culprit. Then Sheriff 1 has suspect set $(S_A, S_C)$ and Sheriff 2 has suspect set $(S_B, S_C)$. Sheriff 1 goes on the phone and says one of his suspects. Let us consider each of the cases: a. Sheriff 1 says $S_A$, so therefore Sheriff 2 knows that the other suspect of Sheriff 1 is the culprit. Then Sheriff 2 says $S_B$, and Sheriff 1 says &quot;No.&quot; Then we know that Sheriff 1 knows it is his unsaid suspect, and Sheriff 2 knows it is his unsaid suspect. b. Sheriff 1 says $S_A$, so therefore Sheriff 2 knows that the other suspect of Sheriff 1 is the culprit. Then Sheriff 2 says $S_C$, and Sheriff 1 realizes that $S_C$ is the culprit. He says &quot;Yes.&quot; But wouldn't that tell the townspeople? No, because had Sheriff 1 replied with $S_C$ first, then Sheriff 2 would've known it was $S_C$, and replied with $S_B$. Then Sheriff 2 could deduce that it must be $S_C$ since he does not have $S_B$, and he would have replied &quot;Yes.&quot; In either case, the people would not know whether it was the first or the second said suspect if the reply was &quot;Yes,&quot; and they wouldn't know the unsaid suspect if the reply was &quot;No.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10782" LastEditorUserId="10782" LastEditDate="2015-03-30T21:17:35.733" LastActivityDate="2015-03-30T21:17:35.733" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11279" PostTypeId="2" ParentId="11267" CreationDate="2015-03-30T21:12:53.813" Score="3" Body="&lt;p&gt;I'm finding one small issue with mine (in the worst case scenario), so hopefully this will help someone else.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One sheriff reads names &lt;strong&gt;NOT&lt;/strong&gt; on his list (&quot;I know it isn't ____) until he reads a name on the second sheriff's list. We know the first sheriff will have to say some name he knows it isn't that the second sheriff thinks it is, or the case is unsolvable.&lt;br&gt;&#xA;&#xA;The second Sheriff now knows the culprit. The second sheriff can now begin the same process (&quot;I know it isn't ____&quot;) until he reads a name on the first sheriff's list.  He needs to stop before the final name though if he gets that far, or he could still reveal who the culprit is (if the two sheriff's have together said it isn't 7 of the 8).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I realize this isn't fullproof, given the situation of:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If luck is really bad and the second sheriff reads the entire list and the two remaining names (the culprit, and last person unsaid to avoid giving too much away) are both suspects to the first sheriff. Then this way wouldn't work.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-03-30T21:12:53.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11280" PostTypeId="2" ParentId="11234" CreationDate="2015-03-30T21:35:40.057" Score="2" Body="&lt;p&gt;I think it must be a shadow that follows one everywhere whether day or night and cannot be escaped. &#xA;The line that says light is a must, and cannot be escaped.&lt;/p&gt;&#xA;" OwnerUserId="10786" LastActivityDate="2015-03-30T21:35:40.057" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11281" PostTypeId="1" AcceptedAnswerId="11285" CreationDate="2015-03-30T21:58:56.317" Score="31" ViewCount="5176" Body="&lt;p&gt;So, @JLee has posted a hangman game (&lt;a href=&quot;https://puzzling.stackexchange.com/questions/11223/hangman-no-misses-left&quot;&gt;Hangman- No misses left&lt;/a&gt;) on SE.  I would like to create one too, but let's do it backward - guess what was the question, with only 1 letter missing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;fill in an &quot;a&quot; - you wish you have it for your lovely home&lt;br&gt;&#xA;fill in an &quot;e&quot; - I will have some water and I am fine&lt;br&gt;&#xA;fill in an &quot;i&quot;  - It happens only in the future you may say&lt;br&gt;&#xA;Now, if you fill in a &quot;w&quot; - now, it's not a word but the world was crazy about it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your answer can be in form of: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The question was: _ p p l e&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;followed by explanation.  &lt;/p&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-03-31T03:32:26.540" Title="The hangman game backward" Tags="|riddle|english|language|letters|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11282" PostTypeId="2" ParentId="11267" CreationDate="2015-03-30T22:24:11.913" Score="4" Body="&lt;p&gt;Call the suspects A, B, C, D, E, F, G, H, and assume the suspect-pairs listed by each sheriff are distinct (so that the problem is soluble).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Sheriff 1 picks a size-4 subset $S_1$ of the suspects, including exactly one from 1's list, and asks 2 how many of $S_1$ are on 2's list. If 2 says &quot;none&quot; or &quot;two&quot;, 1 knows who the culprit is. If 2 says &quot;one&quot;, 1 asks whether the &lt;em&gt;other&lt;/em&gt; suspect from his list (the one not in $S_1$) is on 2's list in order to determine the culprit.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Sheriff 2 picks a size-4 subset $S_2$ whose intersection with $S_1$ has size 2 and which includes exactly one from 2's list (this is always possible, and if 2 said &quot;one&quot; before, he chooses $S_2$ to contain the person 1 then asked about). He then asks 1 how many of $S_2$ are on 1's list. If 1 says &quot;none&quot; or &quot;two&quot;, 2 knows who the culprit is. If 1 says &quot;one&quot;, 2 asks whether the &lt;em&gt;other&lt;/em&gt; suspect from his list (the one not in $S_2$) is on 1's list in order to determine the culprit.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;Let's see how this works from the point of view of the eavesdroppers. WLOG, say $S_1=\{A,B,C,D\}$ and $S_2=\{A,B,E,F\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If 2 and 1 both say &quot;two&quot;, all they know is the culprit is one of A,B. If 2 and 1 both say &quot;none&quot;, all they know is it's one of G,H. If 2 says &quot;two&quot; and then 1 says &quot;none&quot;, all they know is it's one of C,D. If 2 says &quot;none&quot; and then 1 says &quot;two&quot;, all they know is it's one of E,F.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If 2 says &quot;one&quot;, the person 1 then asks about must be one of E and F (by the way the strategy was set up). Say it's E, since the F case is similar. If 1 says &quot;two&quot;, the locals know either that the culprit is one of A,B,E or that it's one of A,B,F (depending on 2's second answer). If 1 says &quot;none&quot;, the locals know only that the culprit is one of C,D,G,H. If 1 says &quot;one&quot;, he then has to answer a question on one of C,D,G,H, and the locals still have at least six possibilities (maybe even more, depending on 2's and 1's second answers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the locals cannot narrow down their possibilities to a single suspect, but both the sheriffs can if it's at all possible.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-03-30T22:24:11.913" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11283" PostTypeId="2" ParentId="11238" CreationDate="2015-03-30T22:52:34.300" Score="9" Body="&lt;p&gt;My solution is that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Early every morning, the man's wife wakes up, puts coffee in the coffee pot, prepares their lunches for the day, and then goes back to bed for a couple hours. Today, she passed away in her sleep briefly before her spouse awoke. Her body was still warm when he kissed her.&lt;br&gt;&#xA;&lt;br&gt;&#xA;The man got hungry in the elevator and discovered that his lunchbox was empty. Because his wife had been recently diagnosed with a terminal illness, it was reasonable to believe that she had died rather than slept-in.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2448" LastEditorUserId="46903" LastEditDate="2019-02-07T03:55:21.797" LastActivityDate="2019-02-07T03:55:21.797" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="11284" PostTypeId="2" ParentId="11267" CreationDate="2015-03-30T22:55:48.593" Score="39" Body="&lt;p&gt;Let's say S1 has the list {a,b} and say S2 has {a,c}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S1: {a,b}, {c,d}, {e,f}, {g,h} - one of these is my list.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S2: {a,c}, {b,d}, {e,g}, {f,h} - one of these is mine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now both S1,S2 know their sets are inside {a,b,c,d}, whereas because of symmetry, &#xA;it could be {e,f,g,h} or {a,b,c,d} as far as eavesdroppers are concerned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So S1 and S2 play a game revealing info but keeping this info symmetric in the other set too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S1: {a,e}, {b,f} - These lists have my suspects.&#xA;S2: Knows the culprit is a (since he knows S1's set now).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S2: {a,e} - The culprit is in this list.&#xA;S1: Knows the culprit is a.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each time, the eavesdroppers can't distinguish between the two sets;&#xA;finally they can't distinguish between a and e.&lt;/p&gt;&#xA;" OwnerUserId="5684" LastEditorUserId="5684" LastEditDate="2015-03-30T23:01:45.527" LastActivityDate="2015-03-30T23:01:45.527" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="11285" PostTypeId="2" ParentId="11281" CreationDate="2015-03-30T22:59:34.633" Score="33" Body="&lt;p&gt;The question was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; w _ ll&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You wish you have it for your lovely home&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A home is not a home without walls.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I will have some water and I am fine&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well holds water, well means fine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It happens only in the future you may say&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A person writes a will for when they die, not in the present moment.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, it's not a word but the world was crazy about it!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; wwll, or I suspect supposed to represent World War II (though those are Is, not Ls.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4353" LastActivityDate="2015-03-30T22:59:34.633" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11287" PostTypeId="2" ParentId="11242" CreationDate="2015-03-31T00:24:39.250" Score="1" Body="&lt;p&gt;Here is the answer I found in about 5 min (I m pretty proud of myself considering that I'm not so good at chess)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; BxB2,BxC3,BxD4,BxB3,NxB3,RxD4,NxD4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10792" LastEditorUserId="63414" LastEditDate="2020-12-20T10:00:53.303" LastActivityDate="2020-12-20T10:00:53.303" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="11288" PostTypeId="1" AcceptedAnswerId="11289" CreationDate="2015-03-31T00:48:06.667" Score="7" ViewCount="1117" Body="&lt;p&gt;This is another &lt;a href=&quot;http://www.thinkfun.com/microsite/solitairechess/demo&quot; rel=&quot;nofollow noreferrer&quot;&gt;Solitaire Chess&lt;/a&gt; puzzle like &lt;a href=&quot;https://puzzling.stackexchange.com/questions/11242/solitaire-chess&quot;&gt;this earlier puzzle&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rules:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Pieces move as in regular chess&lt;/li&gt;&#xA;  &lt;li&gt;Every move must be a capture&lt;/li&gt;&#xA;  &lt;li&gt;When there is only 1 piece left, you win.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3cqhS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2020-12-19T21:44:52.450" Title="Solitaire Chess- Round 2" Tags="|strategy|chess|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11289" PostTypeId="2" ParentId="11288" CreationDate="2015-03-31T01:02:00.343" Score="9" Body="&lt;p&gt;One solution is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nxd2 Bxc2 Rxd3 Rxd2 Rxc2 Nxc2 Nxa1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As images:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/1C1Dz.png&quot; width=&quot;128&quot; height=&quot;128&quot;/&gt; &lt;img src=&quot;https://i.stack.imgur.com/SJ4ca.png&quot; width=&quot;128&quot; height=&quot;128&quot;/&gt; &lt;img src=&quot;https://i.stack.imgur.com/YbgN6.png&quot; width=&quot;128&quot; height=&quot;128&quot;/&gt; &lt;img src=&quot;https://i.stack.imgur.com/APCmW.png&quot; width=&quot;128&quot; height=&quot;128&quot;/&gt; &lt;img src=&quot;https://i.stack.imgur.com/JxLxo.png&quot; width=&quot;128&quot; height=&quot;128&quot;/&gt; &lt;img src=&quot;https://i.stack.imgur.com/hlqjk.png&quot; width=&quot;128&quot; height=&quot;128&quot;/&gt; &lt;img src=&quot;https://i.stack.imgur.com/dcGIG.png&quot; width=&quot;128&quot; height=&quot;128&quot;/&gt; &lt;img src=&quot;https://i.stack.imgur.com/66ZCs.png&quot; width=&quot;128&quot; height=&quot;128&quot;/&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7308" LastEditorUserId="7308" LastEditDate="2015-03-31T01:13:05.540" LastActivityDate="2015-03-31T01:13:05.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11290" PostTypeId="2" ParentId="11238" CreationDate="2015-03-31T02:35:42.700" Score="6" Body="&lt;p&gt;The 18-story power cord from his wife's life support system popped out of the DC-AC power converter attached to the car battery in his back pack and he began to cry. Obviously.&lt;/p&gt;&#xA;" OwnerUserId="10795" LastEditorUserId="10795" LastEditDate="2015-03-31T02:56:41.057" LastActivityDate="2015-03-31T02:56:41.057" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11291" PostTypeId="2" ParentId="11238" CreationDate="2015-03-31T04:09:33.667" Score="47" Body="&lt;p&gt;My solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The building he lives in has a 13th floor.&lt;br&gt;&#xA;&lt;br&gt;&#xA;However, this morning he noticed the 13th was missing, it went from the 14th floor directly to the 12th floor. &lt;br&gt;&#xA;&lt;br&gt;&#xA;The only possibility? The labyrinth builder was superstitious of the number 13 so his copy of the building did not include it.&lt;br&gt;&#xA;&lt;br&gt;&#xA;The man is Dom Cobb, and he is in a dream. His wife, Mal Cobb, committed suicide years ago. The error in the building has made him aware of the dream.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Inception!&lt;/p&gt;&#xA;" OwnerDisplayName="user9659" LastEditorUserId="46903" LastEditDate="2019-02-07T00:02:21.843" LastActivityDate="2019-02-07T00:02:21.843" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="11292" PostTypeId="2" ParentId="9763" CreationDate="2015-03-31T06:21:23.667" Score="1" Body="&lt;p&gt;You're a  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Planisphere (world map)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Divided by a wall, N and S.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Equator, N=North, S=South&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tunnels in between, but no one pass.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Meridians connect North and South, they're imaginary lines so nobody can use them as &quot;tunnels&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The West has more, and the East has less.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; West (Europe and America) is rich, East is generally poor.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Less is more, but More is less.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is a bit tricky, but I think the explanation is in the Mercator projection. In fact, it distorts areas and lengths: what's far from equator appears big (More) but it should be smaller (Less) and viceversa.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have rectangular border  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Maps are rectangular, usually.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I, can you guess?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Planisphere (world map)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-03-31T06:21:23.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11293" PostTypeId="1" AcceptedAnswerId="11308" CreationDate="2015-03-31T06:41:17.867" Score="9" ViewCount="350" Body="&lt;p&gt;You come to the door of the stronghold. You recall what the scriptures say: You must speak the right words, and the door will open. But what words? It doesn't say. You have a look at the door in front of you; something looks a little... off?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/p2HBk.png&quot; alt=&quot;Where is the blue now?&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scriptures also say that the entrance may be guarded by a particularly dangerous serpent, but you see no sign of that.&lt;br&gt;&#xA;&lt;strong&gt;&lt;em&gt;Can you find the magic words you must speak?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Author note:&lt;/em&gt; This has two parts. Both are close to impossible without programming knowledge. Note the computer-puzzle tag!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7308" LastEditorUserId="7308" LastEditDate="2015-03-31T06:56:14.317" LastActivityDate="2015-03-31T13:12:44.083" Title="The Serpent Stronghold" Tags="|steganography|computer-puzzle|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11297" PostTypeId="2" ParentId="11216" CreationDate="2015-03-31T09:00:47.907" Score="3" Body="&lt;p&gt;Second string is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an anagram of &quot;Royal George Buff House&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...but I'm not really sure that's of any help.&lt;/p&gt;&#xA;" OwnerUserId="10808" LastActivityDate="2015-03-31T09:00:47.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11298" PostTypeId="1" AcceptedAnswerId="11394" CreationDate="2015-03-31T09:49:42.773" Score="6" ViewCount="2949" Body="&lt;p&gt;I am what they cherish,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those who cherish,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are also those who detest ardently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I murder with ease and comfort,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And be loved with purpose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a drug to the eye,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And to the heart,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You love me like nobody else,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not because I murdered your loved ones,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But because I am a drug to the eye,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And to the heart. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cause no disease,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But yet I abolish those who you love,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For I am a drug to the eye,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And to the heart!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What am I?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint #1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I have a physical existence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am everywhere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10704" LastEditorUserId="12437" LastEditDate="2015-05-15T17:49:17.317" LastActivityDate="2018-05-11T07:51:07.360" Title="I killed your entire family, but you still love me" Tags="|riddle|" AnswerCount="14" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11299" PostTypeId="2" ParentId="11298" CreationDate="2015-03-31T10:26:49.693" Score="5" Body="&lt;p&gt;I think the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Sleeping&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;we love sleeping but people may die in their sleep &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;they die with ease and comfort &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;it's good for your eye and heart :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10807" LastEditorUserId="10433" LastEditDate="2015-03-31T10:44:29.500" LastActivityDate="2015-03-31T10:44:29.500" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11300" PostTypeId="2" ParentId="11238" CreationDate="2015-03-31T10:57:23.270" Score="25" Body="&lt;p&gt;The husband and wife have been living under assumed names, on the run from the CIAfia and in possession of a briefcase containing documents of earth-shattering significance. It's absolutely clear that if the CIAfia find where they're holed up, they won't engage in negotiation: they'll just kill them and make off with the documents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The couple's problem is that they've lost contact with the agency they're supposed to hand the briefcase to. They fear their usual contact has come to harm. All they can do is wait and hope that someone else in the agency knows the procedures for making contact. Husbie nips out every morning to check for the chalk mark. He always uses the proper tradecraft, so it takes him about 3 hours. And when he gets back, he always checks on exiting the lift that his wife has put a flowerpot in the pre-arranged place. If she hasn't, that will only mean one thing: she's dead and the CIAfia have taken the briefcase. In that event, he's offski: he'll turn sharp left, pay no more attention to the front door of the flat, and walk down the stairway to floor 29.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Leaving is as tense as arriving. The guys looking for them aren't gunslinging cowboys. Nobody's chances are ever 100% in this game. If they can, they'll improve their odds by entering the flat when the wife's alone. But they won't waste time. After all, when the husband goes out, the opposition could in principle make contact with him within a minute or so and he'll lead them straight up to the flat. So if possible the CIAfia will go in and kill his wife and take the briefcase as soon as he's gone out, when he's been in the lift for only a few seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So every morning, she texts him. She listens for the lift door opening and closing and then she sends a text within 15 seconds. She's a professional; she doesn't forget. If she doesn't send the text, there's only one possible meaning - she's dead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He's a professional too, but that doesn't mean he's not scared. He was trained to control his fear. He looks calmly at his watch. Five seconds. Ten seconds. By the time the lift reaches floor 12, thirty seconds have elapsed. They've killed her. Cry a tear. The CIAfia have got the documents. On to the next stage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;He &lt;strong&gt;didn't receive the communication&lt;/strong&gt;. That's the point. Let's say the man's not in danger, because the only reason the CIAfia would kill whoever's looking after the briefcase at the moment of collection is to prevent harm coming to the documents.&lt;/em&gt;)&lt;/p&gt;&#xA;" OwnerUserId="7363" LastEditorUserId="7363" LastEditDate="2015-03-31T11:11:41.523" LastActivityDate="2015-03-31T11:11:41.523" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11302" PostTypeId="1" AcceptedAnswerId="11304" CreationDate="2015-03-31T11:38:27.703" Score="12" ViewCount="4291" Body="&lt;p&gt;A man was murdered in a village and the police inspector asked two constables to go to the scene and provide a First Information Report (FIR)...&#xA;As it was night and the location was too far from the station, the constables did not go and made a fake FIR. After reading the report, the inspector said &quot;you are both suspended for making a fake report&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Here is the FIR report:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;When we arrived at the scene, the door was open and a man aged about 40-45 was found dead in a chair. A bulb was glowing in the room and the fan was switched on. There was a table in front of the dead body and the table contained an opened bottle of poison, a half-filled drinking water bottle, a pen, a newspaper that was opened to pages 9-10, one table top calendar opened at June 20, a 5 rupee coin, and one notebook.  One bed was also in the room.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that this person committed suicide.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The question is: How did the inspector know that the report was fake and that the constables did not go to the location of the incident?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10820" LastEditorUserId="5188" LastEditDate="2015-03-31T19:30:49.717" LastActivityDate="2019-01-08T15:29:50.630" Title="How did the inspector know that the report was fake?" Tags="|logical-deduction|lateral-thinking|" AnswerCount="4" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11303" PostTypeId="2" ParentId="11298" CreationDate="2015-03-31T11:42:26.353" Score="13" Body="&lt;p&gt;I think the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;TIME&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Are also those who detest ardently.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Some people hate how little time they have&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I murder with ease and comfort,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Time kills all slowly without pain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a drug to the eye,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;People always look at the time numerous times of the day. (also people who clock watch while at work etc.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And to the heart,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Time can heal a broken heart / People would love to have more time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10626" LastEditorUserId="10626" LastEditDate="2015-03-31T15:59:37.237" LastActivityDate="2015-03-31T15:59:37.237" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="11304" PostTypeId="2" ParentId="11302" CreationDate="2015-03-31T11:43:01.917" Score="20" Body="&lt;p&gt;I think answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; pages 9-10 can not be open at same moment&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;that's the clue that report was fake&lt;/p&gt;&#xA;" OwnerUserId="10433" LastActivityDate="2015-03-31T11:43:01.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11305" PostTypeId="1" AcceptedAnswerId="11314" CreationDate="2015-03-31T11:43:02.163" Score="2" ViewCount="309" Body="&lt;p&gt;Mary asks: &quot;Where are you going on holiday this year, Julian?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Julian says: &quot;What do you mean - this year? You know that I don't go every year.&#xA;I did not go at all in 2007.  In Summer 2008 I went to Greenland, in July 2009&#xA;to India, and in July 2010 I went to Chile.&#xA;I did not go in 2011 and also not in 2013 and also not in 2014.&#xA;In November 2012 I went to Australia, and in March 2016 I will go to Sulawesi.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mary insists: &quot;And what about your holidays in 2015?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To which Julian answers: &quot;???? ???? ?????? ?????? ?????? ?? ?????? ?????????&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&#xA;What is Julian's holiday destination in 2015?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9125" LastActivityDate="2015-03-31T15:24:13.250" Title="Where are you going on holiday, Julian?" Tags="|pattern|geography|knowledge|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11306" PostTypeId="1" AcceptedAnswerId="11307" CreationDate="2015-03-31T12:13:37.887" Score="7" ViewCount="9824" Body="&lt;p&gt;Dr. and Mrs. Smith recently attended a party at which there were 5 other couples. Various handshakes took place. No one shook hands with his/her spouse, no one shook his/her own hand. After all the handshaking was finished, Mrs. Smith asked each person, including her husband, how many hands he or she had shaken. To her surprise, each person gave her a different answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How many hands did Dr. Smith shake?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; I don't actually know the answer to this question, so I'll accept the most plausible one.&lt;/p&gt;&#xA;" OwnerUserId="10744" LastEditorUserId="17744" LastEditDate="2016-03-30T04:11:36.510" LastActivityDate="2016-03-30T14:32:59.750" Title="Dr. and Mrs. Smith shake some hands" Tags="|mathematics|logical-deduction|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11307" PostTypeId="2" ParentId="11306" CreationDate="2015-03-31T12:25:29.723" Score="14" Body="&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; I misread the question. There are 5 &lt;em&gt;other&lt;/em&gt; couples, not 5 couples total.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'm going to assume that you're asking how many distinct people whose hands were shook. Shaking the other hand of the same person or the same hand twice &lt;a href=&quot;https://i.stack.imgur.com/XowC3.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;still only counts as one&lt;/a&gt;. That is a necessary constraint for a unique answer. In that case, Mr Smith shook &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 5 hands&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Among the five married couples no one shook more than ten hands. Therefore, if the eleven people asked by Mrs. Smith each shake a different number of hands, the numbers must be 0, 1, 2, ..., 10. The person who shook 10 hands has to be married to the person who shook 0 hands (otherwise that person could have shaken only nine hands.) Similarly, the person who shook 9 hands is bound to be married to the person who shook 1 hand. So that the married couples shook hands in pairs 10/0, 9/1, 8/2, 7/3, 6/4. The only person left who shook hands with 5 is Mrs. Smith. Therefore, Mr. Smith shook the pair number for 5 which is 5.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To help clarify:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since there are an even number of people and they all shook a different number, there was going to be a pair in the middle that shook the same number hands. The only way that is possible is if Mr. Smith - the only one not included in the &quot;different number&quot; condition - shook the same number of hands as someone else.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2016-03-30T14:32:59.750" LastActivityDate="2016-03-30T14:32:59.750" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11308" PostTypeId="2" ParentId="11293" CreationDate="2015-03-31T13:06:57.400" Score="9" Body="&lt;p&gt;This was fun :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magic words are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll try to explain briefly...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The top part of this image is coloured slightly differently from the rest of the image. If you extract the blue channel from the file's RGB data and collect all the non-zero values, you end up with a series of 1630 bytes (100, 101, 102, 32, 112, 111, 115, ...etc...). When these are converted to ASCII characters, they correspond to a Python program that converts a line of text into a hash value and compares it with a precalculated result.&lt;br /&gt;&#xA; &lt;br /&gt;&#xA; Using rather obfuscated code, this program accepts two words, swaps them around, concatenates them together, then calculates the MD5 hash of each letter in the resulting string. When these hashes are joined together, they should equal this long decimal number: 53860968794796194828850912663548273574565909315382218566780496065842419219289908754506217032050621610149536529412292660885510238487652964693572485330339349942747218138850926967814204619329585611322730560200765021495570268633090417978260869005139879392762230267619047776297802859921963641380013215139195873390084347386400369215393624187102530981915&lt;br /&gt;&#xA; &lt;br /&gt;&#xA; To break the code, you have to convert the number back into hexadecimal and split it into 32-character chunks. These correspond to the MD5 hashes of each individual letter (MD5 (&quot;e&quot;) = e1671797c52e15f763380b45e841ec32, MD5 (&quot;s&quot;) = 03c7c0ace395d80182db07ae2c30f034, etc.). This yields the string 'estmagiam', which suggests that the magic words are 'magiam est'.&lt;br /&gt;&#xA; &lt;br /&gt;&#xA; (Actually, there are other strings that would have worked, like 'agiam estm', but they don't seem quite as likely.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="40" LastEditorUserId="40" LastEditDate="2015-03-31T13:12:44.083" LastActivityDate="2015-03-31T13:12:44.083" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11309" PostTypeId="2" ParentId="11267" CreationDate="2015-03-31T13:16:18.557" Score="2" Body="&lt;p&gt;The crude answer to this is that they manually perform a &lt;a href=&quot;http://en.wikipedia.org/wiki/Diffie%E2%80%93Hellman_key_exchange&quot; rel=&quot;nofollow&quot;&gt;Diffie-Hellman key exchange&lt;/a&gt; over the phone. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then they can just encrypt their lists with &lt;a href=&quot;http://en.wikipedia.org/wiki/Advanced_Encryption_Standard&quot; rel=&quot;nofollow&quot;&gt;AES&lt;/a&gt; with the key, and speak it. &#xA;Surely this isn't the most efficient or effective way, but encryption is the catch-all when it comes to questions of communicating secrets in the open :)&lt;/p&gt;&#xA;" OwnerUserId="361" LastActivityDate="2015-03-31T13:16:18.557" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11310" PostTypeId="2" ParentId="11305" CreationDate="2015-03-31T13:47:15.757" Score="1" Body="&lt;p&gt;India , March 20th. The link is Total Solar Eclipse&lt;/p&gt;&#xA;" OwnerUserId="9669" LastActivityDate="2015-03-31T13:47:15.757" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11311" PostTypeId="2" ParentId="11306" CreationDate="2015-03-31T13:51:48.153" Score="0" Body="&lt;p&gt;Also assume this arrangement for party members&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;1)Dr. Smith&amp;lt;---do handshake with--&amp;gt;Mrs.Smith,Her Husband,C1,c1,C2,c2,C3,c3,C4,c4,C5,c5&lt;/p&gt;&#xA;&lt;p&gt;2)Mrs. Smith and her husband&lt;/p&gt;&#xA;&lt;p&gt;3)C1 c1&lt;/p&gt;&#xA;&lt;p&gt;4)C2 c2&lt;/p&gt;&#xA;&lt;p&gt;5)C3 c3&lt;/p&gt;&#xA;&lt;p&gt;6)C4 c4&lt;/p&gt;&#xA;&lt;p&gt;7)C5 c5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Where c1,c2,... denotes 5 other couples...&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Then in this case Dr. Smith shook 12 Hands distinctly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Because its not mentioned that in party all were couples.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;So there are 12 handshakes where it will satisfy all the rules/cases for handshakes that are possible for Dr. Smith.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10433" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-03-31T13:51:48.153" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11313" PostTypeId="1" AcceptedAnswerId="11331" CreationDate="2015-03-31T14:27:15.220" Score="6" ViewCount="1449" Body="&lt;p&gt;My Indian friend Jamal recently participated in the well-known TV quiz&#xA;show &lt;strong&gt;&lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Who_Wants_to_Be_a_Millionaire&quot; rel=&quot;nofollow&quot;&gt;Who wants to be a millionaire?&lt;/a&gt;&lt;/em&gt;&lt;/strong&gt;. You might have heard or read about Jamal's spectacular success; the media reported on it extensively.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what Jamal told us about his 200 Euro question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;After my success in the first round, I started to feel a bit more confident.&#xA;  Prem Kumar cleared his throat, rolled his eyes, stretched his neck like a&#xA;  gander and started to read the 200 Euro question. I sat up straight and feverishly looked through the answers.  All four answers were single letters!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I immediately discarded the second and the third answer; it was more likely&#xA;  to find them around restrooms, and of course ladies were always first.&#xA;  However, I wasn't quite sure about the first and the fourth answer:  most&#xA;  ornithologists (and indeed also most people from the street) talk about the&#xA;  shape being the first answer, but I had always seen it looking more like&#xA;  the answer D, although they both were pretty similar letters.  However,&#xA;  since this &lt;em&gt;is&lt;/em&gt; a quiz show at popular culture and trivia level, I decided&#xA;  to go with conventional wisdom and to choose the first answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What was the 200 Euro question?&lt;br&gt;&#xA;What were the four possible answers A, B, C, D?&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="15544" LastEditDate="2015-10-04T16:03:01.580" LastActivityDate="2015-10-04T16:03:01.580" Title="Slumdog millionaire (2)" Tags="|knowledge|science|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11314" PostTypeId="2" ParentId="11305" CreationDate="2015-03-31T15:24:13.250" Score="4" Body="&lt;p&gt;All of those dates and locations refer to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; total solar eclipse viewings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As I stated in my comment, the tense of the question is wrong as the travel for 2015 would have already taken place on March 20th.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The destination for the 2015 trip is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Faroe_Islands&quot; rel=&quot;nofollow&quot;&gt;the Faroe Islands&lt;/a&gt;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2015-03-31T15:24:13.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11315" PostTypeId="2" ParentId="10575" CreationDate="2015-03-31T15:32:00.637" Score="7" Body="&lt;p&gt;Header can win.  Here is the quick answer with the more elaborate logic showing how I arrived at it after.&lt;/p&gt;&#xA;&lt;h1&gt;Quick answer&lt;/h1&gt;&#xA;&lt;p&gt;The setup is initially &lt;span class=&quot;math-container&quot;&gt;$HTHTH$&lt;/span&gt;.  On his first turn, Header will add T on the end, making &lt;span class=&quot;math-container&quot;&gt;$HTHTHT$&lt;/span&gt;.  Then he will flip the first &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; to make &lt;span class=&quot;math-container&quot;&gt;$HHHTHT$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;On Tailer's turn, he can add either &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt; at the front.&lt;/p&gt;&#xA;&lt;p&gt;Lets try &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; first.  So, he will make &lt;span class=&quot;math-container&quot;&gt;$THHHTHT$&lt;/span&gt;.  Now he must flip a section.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If he flips the &lt;span class=&quot;math-container&quot;&gt;$HHH$&lt;/span&gt; section, it now becomes &lt;span class=&quot;math-container&quot;&gt;$TTTTTHT$&lt;/span&gt;.  If header plays &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt; and flips the first &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt;, Tailer will be left with &lt;span class=&quot;math-container&quot;&gt;$TTTTTTTH$&lt;/span&gt;.  No matter what letter is added, the section of &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;s is the only section that can be flipped, which leaves only &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt;s, so Header will win.&lt;/li&gt;&#xA;&lt;li&gt;If he flips the middle &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;, then it now becomms &lt;span class=&quot;math-container&quot;&gt;$THHHHHT$&lt;/span&gt;.  Header can play &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt; and flip all the interior &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt;s to make &lt;span class=&quot;math-container&quot;&gt;$TTTTTTTH$&lt;/span&gt;, which is the same as the previous scenario.&lt;/li&gt;&#xA;&lt;li&gt;If he flips the last &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt;, it becomes &lt;span class=&quot;math-container&quot;&gt;$THHHTTT$&lt;/span&gt;.  Again, Header can add &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt; and flip the section of &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt;s to make &lt;span class=&quot;math-container&quot;&gt;$TTTTTTTH$&lt;/span&gt; - a guaranteed win for Header.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;So now lets try adding &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt; instead. This makes &lt;span class=&quot;math-container&quot;&gt;$HHHHTHT$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If he flips first section of 4 &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt;s, it becomes &lt;span class=&quot;math-container&quot;&gt;$TTTTTHT$&lt;/span&gt;.  Header can again add an &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt; and flip the interior &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt; to get a winning position.&lt;/li&gt;&#xA;&lt;li&gt;If he flips the first &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;, then he leave &lt;span class=&quot;math-container&quot;&gt;$HHHHHHT$&lt;/span&gt;. Header wins easily by adding &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt; and flipping the &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; to leave all &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt;s.&lt;/li&gt;&#xA;&lt;li&gt;If he flips the last &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt;, it becomes &lt;span class=&quot;math-container&quot;&gt;$HHHHTTT$&lt;/span&gt;. Again, Header wins easily by adding &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt; and flipping the section of &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;s.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;h1&gt;Long Answer&lt;/h1&gt;&#xA;&lt;p&gt;For notation, we will use a version of regex symbols. &lt;span class=&quot;math-container&quot;&gt;$^*$&lt;/span&gt; means the previous coin is repeated 0 or more times (i.e. appears at least once).&lt;/p&gt;&#xA;&lt;p&gt;For example, &lt;span class=&quot;math-container&quot;&gt;$HT^*H^*$&lt;/span&gt; can be used to represent &lt;span class=&quot;math-container&quot;&gt;$HTH$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$HHHTTTH$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$HHTTHH$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;For the following scenarios, Header plays on the end of the string of coins, and Tailer plays at the front.&lt;/p&gt;&#xA;&lt;p&gt;We will break the solution down to the component parts.&lt;/p&gt;&#xA;&lt;h2&gt;Single section of coins&lt;/h2&gt;&#xA;&lt;h3&gt;1a) &lt;span class=&quot;math-container&quot;&gt;$H^*$&lt;/span&gt;&lt;/h3&gt;&#xA;&lt;p&gt;By definition, this is a win for Header regardless of whose turn it is to play.&lt;/p&gt;&#xA;&lt;h3&gt;1b) &lt;span class=&quot;math-container&quot;&gt;$T^*$&lt;/span&gt;&lt;/h3&gt;&#xA;&lt;p&gt;By definition, this is a win for Tailer regardless of whose turn it is to play.&lt;/p&gt;&#xA;&lt;h2&gt;Two sections of coins&lt;/h2&gt;&#xA;&lt;h3&gt;2a) &lt;span class=&quot;math-container&quot;&gt;$H^*T^*$&lt;/span&gt;&lt;/h3&gt;&#xA;&lt;p&gt;Either player can win by placing their coin as their own piece and flipping the opponent's.  For example, Header can win by playing &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt; to make &lt;span class=&quot;math-container&quot;&gt;$H^*T^*H$&lt;/span&gt; and then flipping all the &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;s.  Tailer can win by playing &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; to make &lt;span class=&quot;math-container&quot;&gt;$TH^*T^*$&lt;/span&gt; and then flipping all the &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt;s.&lt;/p&gt;&#xA;&lt;h3&gt;2b) &lt;span class=&quot;math-container&quot;&gt;$TT^*HH^*$&lt;/span&gt;&lt;/h3&gt;&#xA;&lt;p&gt;This is a winning position for either player since they can flip their opponents coins to win.&lt;/p&gt;&#xA;&lt;h3&gt;2c) &lt;span class=&quot;math-container&quot;&gt;$TH^*$&lt;/span&gt;&lt;/h3&gt;&#xA;&lt;p&gt;This is a losing position for Header since he cannot flip the first &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;.  Any play he makes will result in 1b) - an automatic win for Tailer.&lt;/p&gt;&#xA;&lt;h3&gt;2d) &lt;span class=&quot;math-container&quot;&gt;$T^*H$&lt;/span&gt;&lt;/h3&gt;&#xA;&lt;p&gt;Losing position for Tailer.&lt;/p&gt;&#xA;&lt;h2&gt;Three sections of coins&lt;/h2&gt;&#xA;&lt;h3&gt;3a) &lt;span class=&quot;math-container&quot;&gt;$T^*H^*T^*$&lt;/span&gt;&lt;/h3&gt;&#xA;&lt;p&gt;An immediate winning position for Tailer by placing &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; and flipping the &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt;s.&lt;/p&gt;&#xA;&lt;p&gt;This is also a winning position for Header by placing &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt; to make &lt;span class=&quot;math-container&quot;&gt;$T^*H^*T^*H$&lt;/span&gt; and flipping the first section of &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt;s to leave &lt;span class=&quot;math-container&quot;&gt;$T^*H$&lt;/span&gt;.  By 2d), this is a losing position for Tailer.&lt;/p&gt;&#xA;&lt;h3&gt;3b) &lt;span class=&quot;math-container&quot;&gt;$H^*T^*H^*$&lt;/span&gt;&lt;/h3&gt;&#xA;&lt;p&gt;This is obviously a winning position for Header.&lt;/p&gt;&#xA;&lt;p&gt;This is also a winning position for Tailer since he can play &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; and flip the section of &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;s to leave &lt;span class=&quot;math-container&quot;&gt;$TH^*$&lt;/span&gt;, which is a losing position for Header by 2c).&lt;/p&gt;&#xA;&lt;h2&gt;Four sections of coins&lt;/h2&gt;&#xA;&lt;h3&gt;4a) &lt;span class=&quot;math-container&quot;&gt;$HT^*H^*T$&lt;/span&gt;&lt;/h3&gt;&#xA;&lt;p&gt;If Header plays an &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt; and then flips any section, he will leave &lt;span class=&quot;math-container&quot;&gt;$H^*T^*H^*$&lt;/span&gt;, which is a winning position for Tailer.&lt;/p&gt;&#xA;&lt;p&gt;If Header plays &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;, he will have &lt;span class=&quot;math-container&quot;&gt;$HT^*H^*TT$&lt;/span&gt;.  Flipping the first set of &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;s leaves &lt;span class=&quot;math-container&quot;&gt;$H^*T^*$&lt;/span&gt;, a winning position for Tailer by 2a).  Flipping the middle set of &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt;s leave &lt;span class=&quot;math-container&quot;&gt;$HT^*$&lt;/span&gt;, also a winning position for Tailer by 2a).  Flipping the last set of &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;s leaves &lt;span class=&quot;math-container&quot;&gt;$HT^*H^*$&lt;/span&gt;, which is three sets of coins, proven by 3b) to be a winning position for Tailer.&lt;/p&gt;&#xA;&lt;p&gt;Thus, this is a losing position for Header.&lt;/p&gt;&#xA;&lt;p&gt;By symmetry, this is also a losing position for Tailer.&lt;/p&gt;&#xA;&lt;h3&gt;4b) &lt;span class=&quot;math-container&quot;&gt;$TH^*T^*H$&lt;/span&gt;&lt;/h3&gt;&#xA;&lt;p&gt;If Header plays an &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt; and then flips any section, he will leave either &lt;span class=&quot;math-container&quot;&gt;$TT^*HH^*$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$THH^*$&lt;/span&gt;, or &lt;span class=&quot;math-container&quot;&gt;$TH^*T^*$&lt;/span&gt;.  All three are winning positions for Tailer by 2b) or 3a).&lt;/p&gt;&#xA;&lt;p&gt;If Header plays &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;, he will have &lt;span class=&quot;math-container&quot;&gt;$TH^*T^*HT$&lt;/span&gt;.  Any move will result in three sections of coins, both of which are shown by 3a) and 3b) to be winning positions for whose ever turn it is.&lt;/p&gt;&#xA;&lt;p&gt;Thus, this is a losing position for Header.&lt;/p&gt;&#xA;&lt;p&gt;By symmetry, this is also a losing position for Tailer.&lt;/p&gt;&#xA;&lt;h3&gt;4c) &lt;span class=&quot;math-container&quot;&gt;$HH^*T^*H^*TT^*$&lt;/span&gt;&lt;/h3&gt;&#xA;&lt;p&gt;This is different from 4a) because Header now has the option to flip the first set of &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt;s as well.  So, if Header plays &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt; and flips the first set of &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt;s, he leaves &lt;span class=&quot;math-container&quot;&gt;$T^*H^*T^*H$&lt;/span&gt;.  This is a losing position for Tailer by 4b), thus Header can win with this setup.&lt;/p&gt;&#xA;&lt;p&gt;Again, by symmetry this s a winning setup for Tailer as well.&lt;/p&gt;&#xA;&lt;h3&gt;4d) &lt;span class=&quot;math-container&quot;&gt;$TT^*H^*T^*HH^*$&lt;/span&gt;&lt;/h3&gt;&#xA;&lt;p&gt;This is different from 4b) in that Header can also flip the first section of &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;s.  If Header plays &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; and then flips the first section of &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;s, then Tailer will be left with &lt;span class=&quot;math-container&quot;&gt;$H^*T^*H^*T$&lt;/span&gt;, which we showed in 4a) is a losing position for Tailer.&lt;/p&gt;&#xA;&lt;p&gt;Thus, this is a winning position for Header.&lt;/p&gt;&#xA;&lt;p&gt;By symmetry, this is also a winning position for Tailer.&lt;/p&gt;&#xA;&lt;h2&gt;Five sets of coins&lt;/h2&gt;&#xA;&lt;p&gt;This brings us to the initial configuration for Header: &lt;span class=&quot;math-container&quot;&gt;$HTHTH$&lt;/span&gt;.  By playing &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; and flipping the first &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;, Header will make the set of coins look like &lt;span class=&quot;math-container&quot;&gt;$H^*THT$&lt;/span&gt;.  This is a known losing position for Tailer by 4a).&lt;/p&gt;&#xA;&lt;p&gt;Thus, Header can win by playing &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; and flipping the first &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-01T15:22:35.863" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11318" PostTypeId="2" ParentId="11313" CreationDate="2015-03-31T15:52:07.790" Score="0" Body="&lt;p&gt;The question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What letter of the alphabet was the Pacman character based on?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Options:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A: O&lt;br&gt;&#xA;B: F (possibly L - for ladies)&lt;br&gt;&#xA;C: M (possibly G - for gentlemen)&lt;br&gt;&#xA;D: C&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Restroom F = Female, M = Male. Order based on &lt;em&gt;ladies were always first..&lt;/em&gt;.&#xA; O &amp;amp; C &lt;em&gt;...both were pretty similar letters&lt;/em&gt;.&#xA; Also references &lt;em&gt;popular culture&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10836" LastEditorUserId="10836" LastEditDate="2015-03-31T17:31:51.310" LastActivityDate="2015-03-31T17:31:51.310" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11321" PostTypeId="1" AcceptedAnswerId="11330" CreationDate="2015-03-31T16:03:09.647" Score="7" ViewCount="1470" Body="&lt;p&gt;There are five logicians seated around a table. They are blindfolded and hats are placed on their heads. After removing the blindfolds, they are told a true statement:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Each of you has either one hat or two hats on his head. The total&#xA;number of hats is 7, 8 or 9.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The actual number is 9, but this is not told to them.&lt;/p&gt;&#xA;&lt;p&gt;They are then told to answer the question &amp;quot;How many hats are on your head?&amp;quot; in clockwise fashion (starting from any of the 5). If it is possible to logically deduce the number of hats on their own heads, they do so, otherwise they reply &amp;quot;I don't know.&amp;quot; and wait until the question cycles back to them.&lt;/p&gt;&#xA;&lt;p&gt;They have no means of knowing the number of hats on their own heads (except logically), but can see every other person. They have no other means of communication, and no tricks involved.&lt;/p&gt;&#xA;&lt;p&gt;The game completes when everyone has figured out their hat count. At what number should the one-hatted logician be placed so that:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The game is completed?&lt;/li&gt;&#xA;&lt;li&gt;The game goes on forever?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;I made a small error while posting the question. Since people have anyway up-voted and answered this one, I've left it as it is and posted the actual question as a second version (&lt;a href=&quot;https://puzzling.stackexchange.com/questions/11378/5-logicians-with-1-or-2-hats-part-ii&quot;&gt;click here&lt;/a&gt; to view)&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="69582" LastEditDate="2021-09-16T00:59:35.553" LastActivityDate="2021-09-16T00:59:35.553" Title="Five logicians with one or two hats [7,8,9]" Tags="|logical-deduction|meta-knowledge|" AnswerCount="6" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="11322" PostTypeId="1" AcceptedAnswerId="11363" CreationDate="2015-03-31T16:21:49.313" Score="8" ViewCount="987" Body="&lt;p&gt;It is known that spiders of the species &lt;em&gt;L. Latrunculorum Cannibalis&lt;/em&gt; build immense (virtually unlimited) communities where members arrange in huge chessboard-like webs with black males and white females each surrounded by four individuals of the opposite sex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the beginning of the mating season, males randomly approach and court one of the neighboring females. Females only pair with one male, which is eaten after copulation.&#xA;If approached by two or more males, females will choose one with no particular preference, and reject the other ones. Rejected males go back to their home locations where they have a chance to approach and court another uncoupled neighboring female, as long as there are any. The process continues until all mating possibilities are exhausted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the highest possible percentage of male spiders who surivive the mating season?&lt;/p&gt;&#xA;" OwnerUserId="4394" LastActivityDate="2015-04-01T13:15:46.493" Title="Male and female spiders on an infinite web" Tags="|logical-deduction|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11323" PostTypeId="2" ParentId="11321" CreationDate="2015-03-31T16:24:24.120" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It doesn't matter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No matter where he sits, when it's his turn he sees 4 other logicians and each of them has 2 hats, so that makes a total of 8. It's impossible that there are only 7 hats. If there would be 8 hats, then he would have no hats at all which contradicts with the statement that each of them has 1 or 2 hats. So there are 9 hats and $9-8=1$, so he only has 1 hat.&lt;br&gt;&#xA; The next logician will see that 1 hat is gone, so there are at most 8 hats left(or 6 or 7). He also sees 6 hats on other's heads, so there are either 7 or 8 hats left. If it would be 7, then he only has 1 hat, but then the first one would not be able to guess that he only has 1 hat, so it's logical to say that there are 8 hats left. $8-6=2$, so second one can be sure about it.&lt;br&gt;&#xA; Each following will just repeat the logic until they all win&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9352" LastActivityDate="2015-03-31T16:24:24.120" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11324" PostTypeId="2" ParentId="11322" CreationDate="2015-03-31T16:38:01.510" Score="5" Body="&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; dmitch's solution is superior to my own with a survival rate of 1/3 instead of 1/5.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The highest surviving percentage is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 20% &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For every male that lives, the 4 females around him must have mated. Therefore, at &lt;em&gt;least&lt;/em&gt; 4 males must die for every 1 that survives so the highest possible survival rate is 20%. There is a solution for this condition (as shown below) so we've achieved the max. In the diagram, the two cells of the same color show a mating pair. Essentially, each male surrounding the surviving male (M0) tries to mate with a female surrounding M0. M0 also approaches one of the females but is rejected. &lt;img src=&quot;https://i.stack.imgur.com/IVI7f.png&quot; alt=&quot;Diagram&quot;&gt; Note that this will leave as many females without mates as it leaves males alive as denoted by F0.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; If we're allowed to pick the matrix size, then I pick a matrix size of 1. The only spider is a male. 100% of the males survive. If there is required to be a female, then I pick a matrix of 5 with a single female surrounded by 4 males. 75% of the males survive.&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2015-04-01T12:14:07.470" LastActivityDate="2015-04-01T12:14:07.470" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="11326" PostTypeId="2" ParentId="11306" CreationDate="2015-03-31T16:53:10.413" Score="-2" Body="&lt;p&gt;There's not enough information in the question to be able to answer.  If Dr. Smith shook ALL available hands, the answer is 10 (5 couples).  &lt;/p&gt;&#xA;" OwnerUserId="10839" LastActivityDate="2015-03-31T16:53:10.413" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11327" PostTypeId="1" AcceptedAnswerId="11368" CreationDate="2015-03-31T17:09:36.460" Score="2" ViewCount="3154" Body="&lt;p&gt;&lt;strong&gt;YE FYRSTE CONJOURATIOUN&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the name of him who spake, I invoke and move thee, King Purson exalteth&lt;br&gt;&#xA;is the power of the Most High, I say unto thee, Obey! In the name of the mighty&lt;br&gt;&#xA;mercene of Apologia who govern, spirits, Liachidae, and repenting above ye&lt;br&gt;&#xA;ardaen the power of thee, Most High.  I say again, Obey! and it shall, in relief,&lt;br&gt;&#xA;recant thy seat, joy, and power and I bind thee in the depth of Abadded.  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;KPNRFNEKUNFKYRSGKGURKOHVYQVATKGELKUVFKAVFHBPK&#xA;&#xA;Uli gsv mvcg xlwv, yvdziv gsv wlfyov svc. Uli gsv urmzo xlwv, krxp gsv gdl irtsgnlhg ovggvih.  &#xA;&#xA;4f 6e 63 65 20 66 6f 72 20 6e 6f 74 68 69 6e 67 2e 20 54 77 69 63 65 20 66 6f 72 20 74 68 65 20 6b 65 79 2e 20 4d 6f 76 65 20 61 6c 6f 6e 67 20 6e 6f 77 2e  &#xA;&#xA;34 36 20 36 66 20 37 32 20 32 30 20 37 34 20 36 38 20 36 35 20 32 30 20 36 65 20 36 35 20 37 38 20 37 34 20 32 30 20 36 33 20 36 66 20 36 34 20 36 35 20 32 63 20 32 30 20 37 34 20 36 38 20 36 35 20 32 30 20 36 62 20 36 35 20 37 39 20 32 30 20 36 39 20 37 33 20 32 30 20 34 37 20 35 32 20 34 39 20 34 64 20 34 66 20 34 39 20 35 32 20 34 35 20 32 65 20 32 30 20 34 36 20 36 66 20 37 32 20 32 30 20 37 34 20 36 38 20 36 35 20 32 30 20 36 36 20 36 39 20 36 65 20 36 31 20 36 63 20 32 30 20 36 33 20 36 66 20 36 34 20 36 35 20 32 63 20 32 30 20 34 31 20 34 65 20 34 34 20 32 30 20 37 30 20 36 39 20 36 33 20 36 62 20 32 30 20 37 34 20 36 38 20 36 35 20 32 30 20 37 34 20 37 37 20 36 66 20 32 30 20 36 63 20 36 35 20 36 36 20 37 34 20 36 64 20 36 66 20 37 33 20 37 34 20 32 30 20 36 63 20 36 35 20 37 34 20 37 34 20 36 35 20 37 32 20 37 33 20 32 65  &#xA;&#xA;LFZFVMWMTRTOCLVYYVMMQPCMTVWRHPVGUERGFIKMUETXFZCPXI...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;sub&gt;Your task is to decode the clues until the final code is revealed.&lt;br&gt;&#xA;Except for the final code, all decoding can be done with &lt;a href=&quot;http://rumkin.com/tools/cipher/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rumkin&lt;/a&gt; and &lt;a href=&quot;http://cryptii.com/text/select&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cryptii&lt;/a&gt;.&lt;/sub&gt;&lt;/strong&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/ueBIi.png&quot; alt=&quot;KingPurson&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-03-31T17:18:42.537" LastActivityDate="2017-12-13T20:38:06.220" Title="A coded conjuration from Ars Goetia" Tags="|cipher|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11328" PostTypeId="2" ParentId="11313" CreationDate="2015-03-31T17:11:55.650" Score="8" Body="&lt;p&gt;I'm not too confident on this one, but is the question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What shape are birds' feet?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With the choices being:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A: Y&lt;br&gt;&#xA;B: W&lt;br&gt;&#xA;C: M&lt;br&gt;&#xA;D: X  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; W and M, indicating Women's and Men's, are letters commonly found around restrooms. The hint about ornithologists implied that the question has something to do with birds and since W and M are similarly shaped, it would make sense that the question is asking a bird-related question about something shaped like those letters.&lt;br&gt;&#xA;&lt;br&gt;&#xA;The most common bird order is Passeriformes, which consists of birds that have feet shaped in the &quot;typical&quot; 3 toes forward, 1 toe backward configuration (called anisodactyl). This is most similar to the letter Y, which most ornithologists and the general public would probably pick.&lt;br&gt;&#xA;&lt;br&gt;&#xA;However, birds such as some raptors and parrots have zygodactyl feet, in which the toes are arranged 2 in front and 2 in back (shaped like the letter X) Since Jamal is from India, it is likely that he has seen many tropical birds with these sort of feet.&lt;br&gt;&#xA;&lt;br&gt;&#xA;A diagram illustrating this:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/dJRcD.gif&quot; alt=&quot;Diagram of different arrangements of bird toes&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10151" LastEditorUserId="10151" LastEditDate="2015-03-31T18:45:45.743" LastActivityDate="2015-03-31T18:45:45.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11329" PostTypeId="2" ParentId="11321" CreationDate="2015-03-31T17:29:36.757" Score="1" Body="&lt;p&gt;&lt;strong&gt;I think Ian's answer is correct but am leaving this here for an example of an incorrect logic.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first person counts the number of hats on the other's heads.  If the number is 5 or 8, he knows he is wearing 2 or 1 respectively and says so.  If this is the case, it is trivial for the others to know what they are wearing. They subtract the sum they see from the one the first person must have had.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This means the easiest solution that leads to completion is for the logician with one hat to be the first one asked.  The others see that everyone else is wearing 2 hats and knew he counted 8 so knew that they are also wearing 2 hats.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the first person is not the one with 1 hat, he will count 7 hats and his passing will tell the next person that he saw either 6 or 7 hats.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the second person sees 8 hats, he will know he has the solitary hat.  If he sees 7 hats as well, he will not be able to determine how many hats he has on.  Like the first person he will pass and the next person will know he saw 6 or 7 hats.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This pattern will repeat until it is the person wearing 1 hat's turn.  He will see 8 and know the answer.  The issue is that, now, no one else is exactly sure if he determined this by using the fact that the others passed or if he determined it.  It does not appear, however, that this will continue indefinitely.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a case where 3 people are wearing only 1 hat.  Anyone wearing 2 hats will see 5 hats, he will then immediately know that he has on 2 hats.  If you see exactly two people wearing only 1 hat and a person wearing 2 hats passes, you know that you have on 2 hats.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a case where 2 people are wearing only 1 hat.  The first one with 1 hat will not know how many he is wearing.  The second with only 1 hat will know the first one would have answered correctly if the second were wearing 2 hats so can correctly determine he is wearing only 1.  This should tell all others except for the one wearing only 1 hat that they have one 2 hats.  When the first one with 1 hat finds himself the last one left, he will know he has on 1 hat.  If someone wearing 2 hats passes before the second person wearing 1 hat has his turn, this will tell everyone who sees 2 people wearing 1 hat each that they are wearing 2 hats and everyone who only sees 1 person wearing only 1 hat that they are wearing only 1 hat.  This is the most likely outcome.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If eventually no solution is found they will eventually realise they all have 2 hats. This means while eventually every pattern will result in a finite solution, the first person being the #1 is easiest.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditDate="2015-03-31T17:48:38.687" LastActivityDate="2015-03-31T17:48:38.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11330" PostTypeId="2" ParentId="11321" CreationDate="2015-03-31T17:34:11.490" Score="5" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;Preamble&lt;/strong&gt;: The following are all of the possible (unordered) hat combinations that could exist for a total of 7, 8, or 9 hats:&lt;/em&gt;&lt;br&gt;&#xA;22221: 9 hats&lt;br&gt;&#xA;22211: 8 hats&lt;br&gt;&#xA;22111: 7 hats&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TL;DR&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The game only completes when the logician with 1 hat is asked first.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In order for 7 hats to be on the logicians heads, none of the logicians can see three 2-hatted logicians. All of the logicians see at least three 2-hatted logicians, so 7 hats is not possible. This leaves the logicians considering only the 8- and 9-hat scenarios.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 12222:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $L_1$ sees 8 hats and can immediately deduce &quot;I have one hat on my head.&quot;&lt;br&gt;&#xA; From the viewpoint of the other logicians, $L_1$ must have seen zero 1-hatted logicians. Had he seen one, $L_1$ would not have enough information to deduce whether this was an 8-hat or 9-hat scenario.&lt;br&gt;&#xA; $\therefore L_{2,3,4,5}$ are all able to deduce that they each wear 2 hats and &lt;strong&gt;the game completes&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Case 21222: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $L_1$ sees 7 hats. He cannot determine if he has 1 or 2 hats. The table could be arranged as 11222 or 21222 from the perspective of $L_1$.&lt;br&gt;&#xA; It is unknown to the logicians at this point that 7 is not a valid hat count. As such, $L_1$ would also not be able to determine if he saw two 1-hatted logicians.&lt;br&gt;&#xA; So at this point, $L_{3,4,5}$ see the table as 21x22, 212x2, 2122x respectively, where x is still unknown.&lt;br&gt;&#xA; $L_2$ sees 8 hats and can determine that he has 1 hat.&lt;br&gt;&#xA; If $L_2$ saw 1 hat on any of $L_{3,4,5}$, he would be unable to deduce that he is wearing 1 hat. It is at this point that $L_{3,4,5}$ know they are each wearing 2 hats.&lt;br&gt;&#xA; If $L_2$ saw 1 hat on $L_1$, he would still be able to make his statement, so $L_1$ is not able to guess and &lt;strong&gt;the game does not complete&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Case 22122:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $L_1$ sees 7 hats. He cannot determine if he has 1 or 2 hats. The table could be arranged as 12122 or 22122 from the perspective of $L_1$.&lt;br&gt;&#xA; From the perspective of $L_{2,4,5}$, the table could additionally be set up as 2x122, 221x2, 2212x respectively.&lt;br&gt;&#xA; $L_2$ having no new information cannot make a decision when seeing only one hat.&lt;br&gt;&#xA; $L_3$ of course knows that he's wearing 1 hat because he sees 8.&lt;br&gt;&#xA; Again, if $L_3$ saw 1 hat on either of $L_{4,5}$, he would be unable to deduce that he is wearing 1 hat. It is at this point that $L_{4,5}$ know they are each wearing 2 hats.&lt;br&gt;&#xA; If $L_3$ saw 1 hat on $L_2$, he would still be able to make his statement because he knows $L_2$ was unable to guess (i.e., he knows $L_2$ did not see 8 hats), so $L_2$ is not able to guess and &lt;strong&gt;the game does not complete&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Case 22212:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $L_1$ sees 7 hats. He cannot determine if he has 1 or 2 hats. The table could be arranged as 12122 or 22122 from the perspective of $L_1$.&lt;br&gt;&#xA; From the perspective of $L_{2,3,5}$, the table could additionally be set up as 2x212, 22x12, 2221x respectively.&lt;br&gt;&#xA; $L_2$ having no new information cannot make a decision when seeing only one hat.&lt;br&gt;&#xA; $L_3$ having no new information cannot make a decision when seeing only one hat.&lt;br&gt;&#xA; $L_4$ of course knows that he's wearing 1 hat because he sees 8.&lt;br&gt;&#xA; Again, if $L_4$ saw 1 hat on any of $L_{1,2,5}$, he would be unable to deduce that he is wearing 1 hat. It is at this point that $L_{1,2,5}$ know they are each wearing 2 hats.&lt;br&gt;&#xA; If $L_4$ saw 1 hat on $L_3$, he would still be able to make his statement because he knows $L_3$ was unable to guess (i.e., he knows $L_3$ did not see 8 hats), so $L_3$ is not able to guess and &lt;strong&gt;the game does not complete&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Case 22221:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $L_1$ sees 7 hats. He cannot determine if he has 1 or 2 hats. The table could be arranged as 12122 or 22122 from the perspective of $L_1$.&lt;br&gt;&#xA; From the perspective of $L_{2,3,5}$, the table could additionally be set up as 2x212, 22x12, 2221x respectively.&lt;br&gt;&#xA; $L_2$ having no new information cannot make a decision when seeing only one hat.&lt;br&gt;&#xA; $L_3$ having no new information cannot make a decision when seeing only one hat.&lt;br&gt;&#xA; $L_4$ having no new information cannot make a decision when seeing only one hat.&lt;br&gt;&#xA; $L_5$ of course knows that he's wearing 1 hat because he sees 8.&lt;br&gt;&#xA; Again, if $L_5$ saw 1 hat on any of $L_{1,2,3}$, he would be unable to deduce that he is wearing 1 hat. It is at this point that $L_{1,2,3}$ know they are each wearing 2 hats.&lt;br&gt;&#xA; If $L_5$ saw 1 hat on $L_4$, he would still be able to make his statement because he knows $L_4$ was unable to guess (i.e., he knows $L_4$ did not see 8 hats), so $L_4$ is not able to guess and &lt;strong&gt;the game does not complete&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-04-01T17:48:54.527" LastActivityDate="2015-04-01T17:48:54.527" CommentCount="19" ContentLicense="CC BY-SA 3.0" />
  <row Id="11331" PostTypeId="2" ParentId="11313" CreationDate="2015-03-31T17:43:10.130" Score="12" Body="&lt;p&gt;My guess is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Which letter is the shape of the formation migratory birds fly in?&#xA;&lt;br&gt;&#xA;A) V&lt;br&gt;&#xA;B) W&lt;br&gt;&#xA;C) M&lt;br&gt;&#xA;D) Y  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As Ben pointed out, W and M stand for &quot;women's&quot; and &quot;men's&quot; restrooms, respectively. W is choice B rather than C because &quot;ladies were always first.&quot; The talk about ornithologists and ganders suggests that the actual question relates to birds, so we're looking for a letter that has some connection to birds. Migratory birds are usually said to fly in a V formation, so a question about migration patterns makes sense, and the &quot;conventional wisdom&quot; would be to answer V. In reality, the birds' shape isn't always a perfect V, so Jamal might be more familiar with other shapes.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Sometimes it looks like a Y:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6faVc.jpg&quot; alt=&quot;Y-shaped formation&quot;&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Other times it looks (sort of) like an A:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fNcRe.jpg&quot; alt=&quot;A-shaped formation&quot;&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;It would be pretty confusing for choice D to be the letter A, since A is the label of a different choice, so I think it's more likely that choice D is the letter Y.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9102" LastActivityDate="2015-03-31T17:43:10.130" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11332" PostTypeId="2" ParentId="11238" CreationDate="2015-03-31T17:55:27.730" Score="95" Body="&lt;p&gt;I know I'm coming in late here, however, I'll toss a completely different take on this.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When he kisses his &quot;wife&quot; each morning, it's only a picture. She died a year ago on this day. He's crying because it's the 1 year anniversary of her death.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;" OwnerUserId="9267" LastEditorUserId="21368" LastEditDate="2016-08-01T10:26:06.277" LastActivityDate="2016-08-01T10:26:06.277" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11333" PostTypeId="2" ParentId="11321" CreationDate="2015-03-31T18:00:26.343" Score="1" Body="&lt;p&gt;To guarantee that the game is completed,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Place the logician with one hat at the first position so that he/she's asked first.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reason:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are 3 possible configurations that the logicians can have, with no particular ordering: 2 2 1 1 1, 2 2 2 1 1, 2 2 2 2 1. The first logician will see four logicians wearing two hats each, thus eliminating two of the configurations, and also being able to deduce that there's one hat on his/her head. The other logicians will also know that they're wearing two hats as there's no other way that the first logician would have been able to come to that conclusion immediately otherwise with what they were seeing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To guarantee that the game never ends,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Place the logician with one hat at the very end so he/she's asked last.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reason:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The other logicians will see 2 2 2 1, meaning they could be in one of two configurations. All four will say they don't know, except for the last guy, because of the same logic stated above. In the second iteration of the loop, the logicians won't be able to differentiate whether he/she (the last logician) was able to come to the conclusion because of their own uncertainty or because of the raw information he/she was able to see. In other words, if this were a configuration with 8 hats, one way to be able to know how many hats you're wearing is through observing that the others aren't able to differentiate whether this is the 8 hat or 9 hat case. Everyone sees 2 2 2 1, and only the last logician could then deduce from their uncertainty that he's wearing only one hat. If they were in a 9 hat configuration, then the other logician with one hat would have declared how many hats were on his/her head before the end of the cycle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll probably try cleaning up my logic later, since it's just a blob of text.&lt;/p&gt;&#xA;" OwnerUserId="9696" LastEditorUserId="9696" LastEditDate="2015-03-31T18:36:24.390" LastActivityDate="2015-03-31T18:36:24.390" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11334" PostTypeId="1" CreationDate="2015-03-31T18:31:37.180" Score="13" ViewCount="1289" Body="&lt;p&gt;A standard deck of 52 cards is shuffled and split into two piles of 26 cards each.&lt;/p&gt;&#xA;&lt;p&gt;Two players cooperate, they must score as many tricks as possible.&lt;/p&gt;&#xA;&lt;p&gt;The first player draw 5 cards from the first pile, he then plays one of them face down, and the card from the top of the second pile is played face down. The two cards are shuffled and turned face up. The second player must now guess which of the two cards the first player played. The first player indicate whether the guess was correct. If so, the players score the trick. The first player then draw a new card from the first pile, so that he has 5 cards in hand, and play a new card face down etc. The process is repeated until the second pile is empty and 26 tricks have been played. As the first pile runs out, the last 4 rounds are played with a reduced hand size.&lt;/p&gt;&#xA;&lt;p&gt;The players have no way of passing information to one another during the game beyond what is explicitly stated above. They may however agree on a strategy before the game begins.&lt;/p&gt;&#xA;&lt;p&gt;What is the optimal strategy for scoring as many tricks as possible? And how many tricks will this strategy score on average?&lt;/p&gt;&#xA;&lt;h3&gt;Hint:&lt;/h3&gt;&#xA;&lt;p&gt;One could think that there would be a hard limit on the probability of scoring any trick at &lt;span class=&quot;math-container&quot;&gt;$5/6$&lt;/span&gt;, dropping to &lt;span class=&quot;math-container&quot;&gt;$4/5$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$3/4$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$2/3$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$1/2$&lt;/span&gt; on the last tricks, thus making the search for any strategy yielding more than &lt;span class=&quot;math-container&quot;&gt;$21.05$&lt;/span&gt; tricks on average futile. One would be wrong.&lt;/p&gt;&#xA;" OwnerUserId="2186" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-13T07:56:49.623" Title="Cooperative guessing game" Tags="|strategy|probability|cards|" AnswerCount="4" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11335" PostTypeId="2" ParentId="11321" CreationDate="2015-03-31T18:36:55.930" Score="1" Body="&lt;p&gt;In all situations, there are at least 2 guys with a double hat, at max 4. &#xA;Before the game starts, they all agree to answer this way:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&quot;Don't know&quot; if the person sees 2 or 3 people with a double hat.&lt;/li&gt;&#xA;&lt;li&gt;&quot;I have a hat&quot; if the person sees 4 people with a double hat.&lt;/li&gt;&#xA;&lt;li&gt;&quot;I have two hats&quot; if the person sees only 1 man with a double hat.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;At the end of a cycle, in the following situations, this happens:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Hats=7&lt;/strong&gt;. Two people could determine the number of their hats, the other three deduce this is a 7-hats situation from that. In the 2nd cycle, all can easily determine the number of their hats with a difference.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Hats=8&lt;/strong&gt;. Nobody could determine his hat(s), so they all deduce that this is a 8-hats situation. In the 2nd cycle, all can easily determine the number of their hats with a difference.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Hats=9&lt;/strong&gt;. Only one could determine the number of his hats (1), the other three deduce this is a 9-hats situation from that. In the 2nd cycle, all can easily determine the number of their hats with a difference.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So, at what number does the 1-hat logician to be?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;No specific position, as they all can determine their numbers in any situation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If you want the game to go forever:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Then your logicians are dumb.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-03-31T18:36:55.930" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11337" PostTypeId="2" ParentId="11306" CreationDate="2015-03-31T19:27:25.493" Score="3" Body="&lt;p&gt;I'd like to start by stating that &quot;Engineer Toast&quot; has the most reasonable solution. However, his answer relies on assumptions and it is actually unclear (semantically speaking) how many hands Dr. Smith shook.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All that I can conclude is that Dr Smith shook between a &lt;strong&gt;lower bound of 0&lt;/strong&gt; and an &lt;strong&gt;upper bound of 20&lt;/strong&gt; hands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reason:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are 6 couples total (12 people)&lt;br&gt;&#xA; There are most likely 2 hands per person (24 total hands)&lt;br&gt;&#xA; Dr. Smith cannot shake his or his wife's hands leaving up to 20 hands that he can shake.&lt;br&gt;&#xA; Since there are a number of ways that could play out it is unclear how many hands he actually shook.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An even more ridiculous answer (if you hated my first answer look away now):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Depending on the type of party, each party-goer may have had too much to drink and made up a number because they really had no clue. In other words: Just because they gave an answer, it doesn't mean the answers they gave were correct.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3922" LastEditorUserId="3922" LastEditDate="2015-03-31T19:47:11.773" LastActivityDate="2015-03-31T19:47:11.773" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11339" PostTypeId="2" ParentId="11321" CreationDate="2015-03-31T19:58:30.857" Score="0" Body="&lt;p&gt;&lt;strong&gt;The logician with 1 hat must go first for the game to end.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Each logician before the one with 1 hat will answer &quot;I don't know&quot;, demonstrating that he sees either 1 or 2 logicians with 1 hat among those who haven't had a turn.&lt;/li&gt;&#xA;&lt;li&gt;When the logician with 1 hat states that he has 1 hat, that demonstrates that he sees no logicians with 1 hat among those who haven't had a turn yet.  All logicians who haven't had a turn yet thus know that they have 2 hats.  Of the logicians who said &quot;I don't know&quot;, at most one of them has 1 hat (to their knowledge).&lt;/li&gt;&#xA;&lt;li&gt;The logician who went first gets his second turn, and again says &quot;I don't know&quot;, indicating that he sees no other 1-hat logicians among those still undecided.  They all know that they have 2 hats (Exception: if the logician with 1 hat was third, the second logician still doesn't know).&lt;/li&gt;&#xA;&lt;li&gt;The logician who went first is now stuck, unless he was the logician with 1 hat to begin with.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I've worked out the sequence of moves the logicians will make in each possible arrangement, with each possible total number of hats (7, 8, 9).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Legend:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1: &quot;1 hat&quot;&lt;/li&gt;&#xA;&lt;li&gt;2: &quot;2 hats&quot;&lt;/li&gt;&#xA;&lt;li&gt;?: &quot;I don't know&quot;&lt;/li&gt;&#xA;&lt;li&gt;-: Passed turn (already said hat count)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These sequences complete:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;12222: 12222*&lt;/li&gt;&#xA;&lt;li&gt;21112: 21112*&lt;/li&gt;&#xA;&lt;li&gt;21121: 21121*&lt;/li&gt;&#xA;&lt;li&gt;22111: 22111*&lt;/li&gt;&#xA;&lt;li&gt;12112: ?21121*&lt;/li&gt;&#xA;&lt;li&gt;12121: ?21211*&lt;/li&gt;&#xA;&lt;li&gt;11212: ??21211*&lt;/li&gt;&#xA;&lt;li&gt;11221: ??22111*&lt;/li&gt;&#xA;&lt;li&gt;22112: ???12221*&lt;/li&gt;&#xA;&lt;li&gt;21212: ???12?212*&lt;/li&gt;&#xA;&lt;li&gt;21221: ????12122*&lt;/li&gt;&#xA;&lt;li&gt;22121: ????12212*&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These sequences do not complete:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;11222: ?1222?----&lt;/li&gt;&#xA;&lt;li&gt;21222: ?1222?----&lt;/li&gt;&#xA;&lt;li&gt;12122: ??122??---&lt;/li&gt;&#xA;&lt;li&gt;22122: ??122??---&lt;/li&gt;&#xA;&lt;li&gt;11122: ??122??---&lt;/li&gt;&#xA;&lt;li&gt;21122: ??122??---&lt;/li&gt;&#xA;&lt;li&gt;12212: ???12?22--?----&lt;/li&gt;&#xA;&lt;li&gt;22212: ???12?22--?----&lt;/li&gt;&#xA;&lt;li&gt;22211: ?22?1?--11?----&lt;/li&gt;&#xA;&lt;li&gt;12211: ?22?1?--11?----&lt;/li&gt;&#xA;&lt;li&gt;12221: ????1?222-?----&lt;/li&gt;&#xA;&lt;li&gt;22221: ????1?222-?----&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1627" LastEditorUserId="1627" LastEditDate="2015-03-31T23:14:29.650" LastActivityDate="2015-03-31T23:14:29.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11340" PostTypeId="2" ParentId="11334" CreationDate="2015-03-31T19:59:17.950" Score="1" Body="&lt;p&gt;I think you can get to around 3/4 wins by having this system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Designate one color even (say, red), and one color odd. Every turn, you put down the color of card of your turn. For the flip card, if it is your color, pick that; otherwise, pick the other card. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is the chance of the top card being the same color, but about 50% chance of it not. And if they pick your card, it's the right color. The way this starts to break down is when you run out of cards of a color. I suppose you could have a system where when you run out of cards of a certain color, you know that color will be used for the next 4 tricks. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-03-31T20:06:26.627" LastActivityDate="2015-03-31T20:06:26.627" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11342" PostTypeId="2" ParentId="11334" CreationDate="2015-03-31T20:21:12.343" Score="8" Body="&lt;p&gt;The best strategy I have found in terms of maximizing average number of tricks is as follows but I believe there is likely a better one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assigned each turn a value $N$ such that for turn $n$, $N=n*14 \mod 52$.  Simularly each card was assigned a number $C$ such that (for instance) a 5 of hearts is $C=0*13+5-1=4$ while a Jack of spades is $C=3*13+11-1=49$.  The details here don't matter as long as each card has a unique value between 0 and 51 inclusive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each turn the first player plays the card in his hand with the value $C$ closest to the turn number $N$ noting that this is a circle so 0 and 51 are very close.  The second player assumes the card with the $C$ value closer to the turn number $N$ is the card the first player wanted him to pick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have simulated 100 games of this and have received an average value of $20.5$ tricks per game. The minimum observed was a $15$ and the maximum was $26$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note that this does not guarentee a good outcome as there is a very small but finite chance that you could score 0 tricks.&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2015-03-31T20:21:12.343" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11344" PostTypeId="1" AcceptedAnswerId="11347" CreationDate="2015-03-31T20:43:25.197" Score="6" ViewCount="495" Body="&lt;p&gt;The beginning of the beginning and start of all,&lt;br&gt;&#xA;The beginning of the beginning and part of a call.&lt;br&gt;&#xA;I bring pain and suffering, but lots of love.&lt;br&gt;&#xA;I cause serious foolishness, but am pure as a dove.&lt;br&gt;&#xA;Some love me, some hate me, what about you?&lt;br&gt;&#xA;But realize without me, there'd be nothing to do.&lt;br&gt;&#xA;Confusion,frustration I bring, but white as a lamb,&lt;br&gt;&#xA;Can you riddle out what it is that I am?&lt;/p&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-03-31T21:26:56.023" Title="The beginning of the beginning - What am I?" Tags="|riddle|word|rhyme|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11345" PostTypeId="2" ParentId="11277" CreationDate="2015-03-31T20:47:04.690" Score="4" Body="&lt;p&gt;X=3, Y=84&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Susan's # (Y+X) is 87.  She knows X, Y is either 3, 84 or 29, 58. Neither of those options has a unique answer for Richard (Y/X = 28 or 2), but 29, 58 would mean Donna has a unique answer (Y-X = 29), and she doesn't know yet whether Donna knows the answer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Donna's # is 81, which is either 3, 84 or 9, 90, neither of those options has a unique answer for Richard (Y/X = 28 or 10).  She confirms that Richard does not have a unique answer (and neither does Susan), but she gives no indication whether she knows the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Richard assumes correctly that neither Donna nor Susan know the answer (and apparently neither of them do yet), but this now tells Susan that 29, 58 is not the answer (otherwise Donna would have known the answer), so it must be 3, 84.  However, Susan also assumes Richard looked at his #, and said he doesn't know the answer.  This is confusing to Susan because if Richard had looked at his answer it would be 28, which only has 2 possible answers (2, 56 or 3, 84), but 2, 56 would give Susan a unique # (58), so he would have been able to rule that one out since he confirmed that Susan did not yet know the answer.  If Susan knew Richard had not looked at his #, then she would be able to say with confidence that 3, 84 is the answer.&lt;/p&gt;&#xA;" OwnerUserId="10787" LastActivityDate="2015-03-31T20:47:04.690" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11346" PostTypeId="1" AcceptedAnswerId="11359" CreationDate="2015-03-31T21:07:25.237" Score="22" ViewCount="996" Body="&lt;p&gt;These are clues to two-word phrases that all have something in common:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Sustainable power&lt;/li&gt;&#xA;&lt;li&gt;Fair-haired countess&lt;/li&gt;&#xA;&lt;li&gt;Nice bull’s-eye&lt;/li&gt;&#xA;&lt;li&gt;Furious swashbuckler&lt;/li&gt;&#xA;&lt;li&gt;Quiet spoon&lt;/li&gt;&#xA;&lt;li&gt;Peculiar heirloom&lt;/li&gt;&#xA;&lt;li&gt;Understated whiskers&lt;/li&gt;&#xA;&lt;li&gt;Supervisor of Jesus's crib&lt;/li&gt;&#xA;&lt;li&gt;Violet fruit processors&lt;/li&gt;&#xA;&lt;li&gt;Bug-free iPads&lt;/li&gt;&#xA;&lt;li&gt;Beautiful waves of arrows&lt;/li&gt;&#xA;&lt;li&gt;Weird igneous rocks&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For the brave, these are clues to three-word phrases with the same property plus a space:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Toner for Yankee photographers&lt;/li&gt;&#xA;&lt;li&gt;Explosive rodent path&lt;/li&gt;&#xA;&lt;li&gt;Huge sleepy rodent #14&lt;/li&gt;&#xA;&lt;li&gt;Menacing desert ptarmigan&lt;/li&gt;&#xA;&lt;li&gt;Underground batters for the 40th Pres.&lt;/li&gt;&#xA;&lt;li&gt;Calvinist singer of &quot;Toxic&quot;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="10851" LastEditorUserId="11475" LastEditDate="2015-09-09T14:00:15.040" LastActivityDate="2015-09-09T14:00:15.040" Title="Letter to the Editor" Tags="|english|language|cryptic-clues|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11347" PostTypeId="2" ParentId="11344" CreationDate="2015-03-31T21:10:20.323" Score="5" Body="&lt;p&gt;Are you a:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Baby?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The beginning of the beginning and start of all,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Spelling starts with BA  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The beginning of the beginning and part of a call.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The meaning of life began and a cry out during labor  &lt;strong&gt;EDIT: Spelling continue with B and Y - part of a cry (call)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I bring pain and suffering, but lots of love.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Labor brings pain and suffering, but lots of love to the new born  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cause serious foolishness, but am pure as a dove.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A negative correlation between fertility and intelligence has been argued to have existed.  A baby is pure as a dove.    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some love me, some hate me, what about you?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are people that doesn't like babies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But realize without me, there'd be nothing to do.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Without being born, there'd be nothing to do.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Confusion,frustration I bring, but white as a lamb,&#xA;Can you riddle out what it is that I am?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Annoying crying babies  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="10037" LastEditDate="2015-03-31T21:26:56.023" LastActivityDate="2015-03-31T21:26:56.023" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11348" PostTypeId="1" CreationDate="2015-03-31T21:13:24.053" Score="4" ViewCount="1291" Body="&lt;p&gt;These are three different puzzles with a similar format - feel free to solve them separately!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H, J, K, M, N, U, V, W, X, and Y are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A, B, C, D, E, F, G, I, L, O, P, Q, R, S, T, and Z aren’t.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B, D, E, F, H, I, K, L, M, N, P, R, T, U, V, W, X, Y, and Z are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A, C, G, J, O, Q, and S aren’t.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;E, F, P, T, and Y are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A, B, C, D, G, H, I, J, K, L, M, N, O, Q, R, S, U, V, W, X, and Z aren’t.&lt;/p&gt;&#xA;" OwnerUserId="10851" LastActivityDate="2021-07-05T09:59:43.940" Title="Two Bowls of Alphabet Soup" Tags="|english|letter-sequence|letters|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11349" PostTypeId="1" AcceptedAnswerId="11351" CreationDate="2015-03-31T21:24:08.083" Score="6" ViewCount="418" Body="&lt;p&gt;What is unique/unusual about the following strange couplet?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a serene summer moon waxes over a wave-worn azure sea.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;an orca moans an arcane verse as mermen caress a venomous sea anemone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10851" LastActivityDate="2015-03-31T22:00:06.077" Title="Beneath the Waves" Tags="|pattern|english|language|word-sequence|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11350" PostTypeId="2" ParentId="11288" CreationDate="2015-03-31T21:40:41.807" Score="1" Body="&lt;p&gt;Much the same as Tryth - I think this is essentially the only solution.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nd2 Rd3 Rd2 Rc2 Bc2 Nc2 Na1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which I'm just posting to try making a (slightly jumpy) gif...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/WLyt4.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8875" LastActivityDate="2015-03-31T21:40:41.807" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11351" PostTypeId="2" ParentId="11349" CreationDate="2015-03-31T21:42:24.150" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All the letters are &quot;short&quot; lowercase letters and don't extend below the line. E.g. no l's, t's, k's, g's, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10151" LastActivityDate="2015-03-31T21:42:24.150" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11352" PostTypeId="2" ParentId="11349" CreationDate="2015-03-31T21:43:13.133" Score="3" Body="&lt;p&gt;In addition to what @Ben said,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; None of them drop below the line, no &quot;p&quot;, no &quot;q&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9412" LastActivityDate="2015-03-31T21:43:13.133" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11353" PostTypeId="2" ParentId="11348" CreationDate="2015-03-31T21:43:33.977" Score="11" Body="&lt;p&gt;For #1 the letters that &quot;are&quot; all have open tops. What I mean is from the top you could put something in the open space. &lt;/p&gt;&#xA;" OwnerUserId="10856" LastActivityDate="2015-03-31T21:43:33.977" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11354" PostTypeId="2" ParentId="11349" CreationDate="2015-03-31T22:00:06.077" Score="0" Body="&lt;p&gt;Also,&#xA;a serene summer moon waxes over a wave-worn azure sea.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Three words each begin with the letters &quot;a&quot;, &quot;s&quot;, and &quot;w&quot; with an &quot;o&quot; as the only solo&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;an orca moans an arcane verse as mermen caress a venomous sea anemone.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a 6, v 2, m 2, s 1, o 1 The sum of the number of words that do not start with &quot;a&quot; equals the total number of words that start with &quot;a&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9412" LastActivityDate="2015-03-31T22:00:06.077" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11355" PostTypeId="2" ParentId="11348" CreationDate="2015-03-31T22:01:59.103" Score="5" Body="&lt;p&gt;&lt;strong&gt;For #3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the chosen letters are all formed with a single 3-way junction (in the typeface used here).  Some of the excluded letters have two 3-way junctions or a 4-way junction. In a different typeface, the set might be different.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8875" LastEditorUserId="8875" LastEditDate="2015-03-31T22:44:01.720" LastActivityDate="2015-03-31T22:44:01.720" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11357" PostTypeId="2" ParentId="11346" CreationDate="2015-03-31T22:22:58.707" Score="2" Body="&lt;p&gt;Here's my best guesses so far for the meanings of the phrases. I keep feeling that the title of the riddle is a hint too.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2-word phrases:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1. Renewable Resource&lt;br&gt;&#xA;2.&lt;br&gt;&#xA;3. Good Shot&lt;br&gt;&#xA;4. Angry Pirate&lt;br&gt;&#xA;5.&lt;br&gt;&#xA;6.&lt;br&gt;&#xA;7.&lt;br&gt;&#xA;8. Mother Mary&lt;br&gt;&#xA;9. Purple Juicers&lt;br&gt;&#xA;10.&lt;br&gt;&#xA;11. Pretty Projectiles&lt;br&gt;&#xA;12.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3-word phrases:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1. New Yorker Ink&lt;br&gt;&#xA;2.&lt;br&gt;&#xA;3.&lt;br&gt;&#xA;4.&lt;br&gt;&#xA;5.&lt;br&gt;&#xA;6. Reformed Britney Spears  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In terms of what the common property is, the riddle's mention of a property and then that same property plus a space makes me think of this:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;alliteration&quot; and &quot;all iteration&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But if that's the case then some of my guesses above are definitely off.&lt;/p&gt;&#xA;" OwnerUserId="10151" LastActivityDate="2015-03-31T22:22:58.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11358" PostTypeId="2" ParentId="11348" CreationDate="2015-03-31T22:29:35.547" Score="8" Body="&lt;p&gt;For #2 the letters are all drawn starting from the letter's top left corner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, the letters all have a part touching the top left corner of the letter's containing rectangle.&lt;/p&gt;&#xA;" OwnerUserId="9696" LastActivityDate="2015-03-31T22:29:35.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11359" PostTypeId="2" ParentId="11346" CreationDate="2015-03-31T22:55:31.387" Score="19" Body="&lt;p&gt;Okay, I have most of the 2-word phrases, haven't started work on the 3-word phrases yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rule is the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The second word is an anagram of the first word with one additional letter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So you have the answers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1. Green Energy&lt;br&gt;&#xA;2. Noble Blonde&lt;br&gt;&#xA;3. Hot Shot&lt;br&gt;&#xA;4. Irate Pirate&lt;br&gt;&#xA;5. Silent Utensil&lt;br&gt;&#xA;6. Quaint Antique (thanks to Len for &quot;quaint&quot;!)&lt;br&gt;&#xA;7. Subtle Stubble&lt;br&gt;&#xA;8. Manger Manager&lt;br&gt;&#xA;9. Purple Pulpers&lt;br&gt;&#xA;10. Stable Tablets&lt;br&gt;&#xA;11. Lovely Volleys&lt;br&gt;&#xA;12. Strange Granites  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This was a blast, looking forward to tackling the 3-words!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Got 'em all! Thanks to Ben for the extra help!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1. American Camera Ink&lt;br&gt;&#xA;2. Volatile Vole Trail&lt;br&gt;&#xA;3. Dormant Mondo Rat (thanks to Ben)&lt;br&gt;&#xA;4. Dangerous Sand Grouse&lt;br&gt;&#xA;5. Subterranean Reagan Bunters (thanks to Ben)&lt;br&gt;&#xA;6. Presbyterian Britney Spears&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10861" LastEditorUserId="10861" LastEditDate="2015-03-31T23:59:08.843" LastActivityDate="2015-03-31T23:59:08.843" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11361" PostTypeId="2" ParentId="11322" CreationDate="2015-03-31T23:05:08.003" Score="3" Body="&lt;p&gt;I found a solution on a repeatable 6x6 matrix which saves 4 males.&#xA;On a 6x6 matrix, there would be 18 males, so this gives a survival rate of 22,22%.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/cdIi1.png&quot; alt=&quot;My solution&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="448" LastActivityDate="2015-03-31T23:05:08.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11362" PostTypeId="2" ParentId="11277" CreationDate="2015-03-31T23:44:48.393" Score="8" Body="&lt;p&gt;The two numbers are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $X=21, Y=42.$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sarah gets a sheet of paper saying $S=63$. In her view, as $X$ divides $Y$ there are only four possible pairs: (7, 56), (3, 60), (21, 42) and (9, 54). Donna gets a sheet of paper saying $D=21$. In her view, there are three possible pairs: (3, 24), (7, 28) and (21, 42).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sarah says that Richard doesn’t know the numbers, since for no possible ratio in her view (8, 20, 2 or 6) there is only one possible pair. Donna agrees, since for no possible ratio in her view (8, 4 or 2) there is only one possible pair.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once Richard gives his ‘false’ statement, Sarah and Donna only consider pairs that have a ratio for which Richard can be sure that they both know that he doesn’t know the numbers. Technically speaking, they only leave ratios for which (brace yourself), for every sum/difference that can come from a pair with this ratio, and for every possible ratio that can come from a pair with this sum/difference, there are more than one possible pairs leading to this ratio. In practice, Sarah drops the ratios 2 and 6 and Donna drops the ratio 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now Sarah considers the pairs (3, 60) and (7, 56) and Donna considers the pairs (3, 24) and (7, 28). Note that they both dropped the correct pair, and this leads to the confusion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Donna is left with two possible pairs, so she says she doesn’t know the numbers. She also figures that the possible sums are 3+24=27 and 7+28=35. For both sums there have been two optional pairs to begin with, but the information from Richard reduced it to one (and Donna knows it). So she says that Sarah must know the numbers. Sarah, however, is also left with two possible pairs and is confused to hear that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far so good. The big question is - &lt;strong&gt;how did Richard figure out the numbers?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 382 pairs of $2 \le X \le Y \le 100$ with a natural $R$, making 134, 88 and 50 possible sums, differences and ratios. Richard (and us) can keep in mind (or on a piece of paper…) the possible views of Sarah and Donna, given their sheets of paper. For every sum/difference we can list the different pairs, and update these with every piece of information. Every time Sarah/Donna makes a statement, she gives the others (and us) information about the possible sums/differences. This information is in the form of ‘which sums/differences I cannot possibly hold in my hand’. This way every statement eliminates a group of $X$-$Y$ pairs. Richard’s statement does the same (with possible ratios) but of course it’s false information. At every stage we have to have a complete picture of the possible sums, differences and ratios in the girls view, and the possible pairs related to them. Let’s go through the statements and see how we can eliminate most pairs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First statement - Sarah knows that Richard doesn’t know the numbers.&lt;/strong&gt;&#xA;This means that, for the particular sum Sarah holds, no possible ratio has only one optional pair (otherwise Richard could have known it). Therefore, in Donna’s and Richard’s view, every sum that doesn’t meet this criteria is eliminated. These sums are 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100 and 102. This of course eliminates all pairs with this sum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second statement - Donna knows that Richard doesn’t know the numbers.&lt;/strong&gt;&#xA;Same thing - in Sarah’s and Richard’s view, every difference that doesn’t meet this criteria is eliminated. These differences are 44, 52, 56, 60, 64, 68, 70, 74, 82, 86, 92, 94 and 98. This eliminates further pairs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Third statement (1) - Sarah and Donna don’t know the numbers.&lt;/strong&gt;&#xA;Here we simply eliminate all sums and differences that only have one optional pair. The eliminated sums are 4, 9, 25, 49, 93, 106, 111, 115, 117, 118, 122, 123, 125, 128, 129, 130, 134, 135, 136, 140, 141, 142, 146, 147, 148, 152, 154, 156, 158, 160, 162, 164, 166, 168, 170, 172, 174, 176, 178, 180, 182, 184, 186, 188, 190, 192, 194, 196, 198 and 200. The eliminated differences are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 76, 78, 85, 87, 91, 93, 95 and 96. Since Richard realises these two things together, the two updates happen simultaneously. Namely, Donna doesn’t use Sarah’s statement (that she doesn’t know the numbers) before stating it herself, and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Third statement (2) - Richard knows that both Sarah and Donna know that he doesn’t know the numbers.&lt;/strong&gt;&#xA;This is of course misleading for Sarah and Donna, but we have to keep track of what it makes them think. Richard’s view doesn’t change from this point on. In Sarah’s and Donna’s view (as we said above) Richard can be sure that Sarah and Donna were sure that he doesn’t know the numbers. This eliminates the sums 6, 8, 12, 14, 16, 21, 22, 24, 28, 33, 44, 52, 56, 62, 64, 66, 68, 77, 112, 114, 116, 124, 126, 132, 138, 144 and 150 and the differences 0, 4, 8, 10, 15, 16, 20, 25, 32, 40, 48, 58, 62, 80 and 90. The key here, as we said, is that Richard’s ‘false’ statement makes Sarah and Donna eliminate the correct pair but not all possible pairs for their sum/difference. This is how they get to a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fourth statement - Donna doesn’t know the numbers and thinks that Sarah does.&lt;/strong&gt;&#xA;For Richard (and us) this eliminates differences with only one pair, and ones that are related to sums with more than one pair. They are 6, 9, 12, 14, 18, 22, 24, 26, 27, 30, 34, 35, 36, 38, 39, 42, 45, 46, 49, 50, 51, 54, 55, 57, 63, 65, 66, 69, 72, 75, 81, 84 and 88.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fifth statement - Sarah doesn’t know the numbers.&lt;/strong&gt;&#xA;For Richard (and us) this eliminates sums with only one pair - 10, 15, 18, 20, 26, 27, 32, 34, 35, 36, 38, 39, 42, 46, 48, 51, 55, 57, 58, 69, 87, 91, 95, 99, 104, 105, 108, 110 and 120.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After all this, Richard knows that Sarah has one of 11 possible sums - 30, 40, 45, 50, 54, 60, 63, 65, 75, 81 or 85 - and Donna one of 4 possible differences - 21, 28, 33 or 77. As we said above, they both eliminated the correct pair, but all that left for Richard (and us) to do is to check which of the possible pairs &lt;em&gt;from before his statement&lt;/em&gt; leads to one of these sums and one of these differences. And this is nothing but the pair:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $X=21, Y=42.$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10684" LastEditorUserId="8874" LastEditDate="2015-10-07T17:14:39.627" LastActivityDate="2015-10-07T17:14:39.627" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="11363" PostTypeId="2" ParentId="11322" CreationDate="2015-04-01T00:08:28.420" Score="12" Body="&lt;p&gt;Another way to look at this is we want the fewest number of pairings ($2\times1$ or $1\times2$ collections of adjacent squares) that fit into the area, subject to the constraint that no two adjacent squares can be un-paired: a &quot;loosest packing&quot; of pairs.  Borrowing PrisonMonkeys's idea of a repeatable $6\times6$ grid, I have constructed the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;M f M f M f&#xA;F M F M F M&#xA;m F m F m F&#xA;F m F m F m&#xA;M F M F M F&#xA;f M f M f M&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where each capital letter is paired with the capital letter directly above/below it.  I believe this is optimal because each pair here is adjacent to exactly two other pairs, and any pair which is adjacent to only one other pair would break the rule against unpaired adjacent squares.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This spares 6 of 18 males, or exactly 1/3.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10334" LastEditorUserId="10334" LastEditDate="2015-04-01T01:00:13.470" LastActivityDate="2015-04-01T01:00:13.470" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11364" PostTypeId="1" AcceptedAnswerId="24895" CreationDate="2015-04-01T00:19:41.187" Score="8" ViewCount="674" Body="&lt;p&gt;This is yet another hat puzzle, but not one of the usual ones! Key differences/commonalities between other hat puzzles are emphasized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Five friends are going to go on a game show. When the game begins, a red or blue hat will placed on each player's head, the colors randomly chosen by independent coin flips. They stand in a circle, and the stage is set so &lt;strong&gt;each player can only see the hats of the players on their immediate right and left&lt;/strong&gt;. Everyone then &lt;strong&gt;simultaneously&lt;/strong&gt; guesses red or blue. If &lt;strong&gt;everyone&lt;/strong&gt; is correct, they win the jackpot; if anyone is wrong, they get nothing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As usual, before the game begins, they may devise a strategy, but after it begins, they cannot communicate in any way. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Help these players find the best strategy, and convince them they can't do better.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a warmup, it may be helpful to try this puzzle with three players, instead of five. In that case, there is a strategy with probability $1/2$ of success (which is better than the $1/8$ chance they get by guessing randomly).&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="11244" LastEditDate="2015-06-19T13:29:16.013" LastActivityDate="2015-12-28T07:10:41.797" Title="Five Haberdashers" Tags="|probability|hat-guessing|" AnswerCount="8" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="11365" PostTypeId="2" ParentId="11364" CreationDate="2015-04-01T00:29:50.863" Score="-1" Body="&lt;p&gt;Players will see either two hats of the same colour or two hats of different colours. The strategy to maximize chance of winning is then:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Say a random colour.&lt;br&gt;&#xA; Because two hats have no bearing on the colour of the hat between them, it is equally likely to be red as it is blue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: with your claim that they could do better, my gut tells me:&lt;br&gt;&#xA;&lt;sup&gt;&lt;em&gt;(no mathematical evidence to back it up)&lt;/em&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you see two of the same colour, say the opposite colour.&lt;br&gt;&#xA; If you see two of different colours, say a random colour.&lt;br&gt;&#xA; The likelihood that three in a row are the same colour is less than for the middle one to be different.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-04-01T02:15:55.667" LastActivityDate="2015-04-01T02:15:55.667" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11366" PostTypeId="2" ParentId="10911" CreationDate="2015-04-01T00:44:15.640" Score="4" Body="&lt;p&gt;Is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Duct tape  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sounds like a bird:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Duck&quot; tape  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Body like the tongue of a lizard:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Sticky, like has already been mentioned  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Am different colors:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Duct tape is sold in a variety of colors  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well can be very littered:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Not sure if there's more to this one, but scraps of duct tape can be left around as litter  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Energy melts away:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;As with other answers, it loses its stickiness over time. Also, old duct tape can start to melt over time if stuck to something out in the heat, etc  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It'll collect dust and lice:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Even when on the roll, the sides of a roll of duct tape can be sticky enough to collect dust&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10151" LastActivityDate="2015-04-01T00:44:15.640" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11368" PostTypeId="2" ParentId="11327" CreationDate="2015-04-01T01:36:42.550" Score="3" Body="&lt;p&gt;For the first line&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Entering it in to a Caesarian Shift with a shift of 13 or entering it into ROT13 directly you get: &lt;br&gt;&lt;br&gt; XCAESARXHASXLEFTXTHEXBUILDINGXTRYXHISXNISUOCX &lt;br&gt;&lt;br&gt; with the Xs being spaces, it ends up reading: &lt;br&gt;&lt;br&gt; &quot;Caesar has left the building try his nisuoc&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With the hint from the first line and the second clue&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Use a substitution cipher with an inverse alphabet (A=Z, B=Y, etc.), or Atbash, deduced from &quot;cousin&quot; being spelt backwards in the first hint, to get: &lt;br&gt;&lt;br&gt;&quot;For the next code, beware the double hex. For the final code, pick the two rightmost letters.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the third clue&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Translate from hexadecimal (but only once) to get: &lt;br&gt;&lt;br&gt;&quot;Once for nothing. Twice for the key. Move along now.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the fourth clue&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Translate from hexadecimal twice to get: &lt;br&gt;&lt;br&gt;&quot;For the next code, the key is GRIMOIRE. For the final code, AND pick the two leftmost letters.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the fifth clue&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Using GRIMOIRE as the key (repeating), decrypt the code using a one time pad to get: &lt;br&gt;&lt;br&gt;&quot;FORTHEFINALCODEUSEEACHLINEOFTHECONJURATIONLLRRLLRR...&quot;&lt;br&gt;&lt;br&gt;Using the conjuration's lines and the first two letters and last two letters of each line as denoted in the decryptions above, I get: &lt;br&gt;&lt;br&gt;INTHISTYMEYEAREFREED&lt;br&gt;&lt;br&gt;which I translate as a conjuration as &lt;br&gt;&lt;br&gt;&quot;In this tyme, ye are freed.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6140" LastActivityDate="2015-04-01T01:36:42.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11369" PostTypeId="2" ParentId="11364" CreationDate="2015-04-01T02:00:41.763" Score="-1" Body="&lt;p&gt;Lets say we have players 1 to 5 then obviously if the player sees 2 of the same hats they would say the opposite color. Otherwise if the player saw one of each then they would alternate the colors they would say from 1 to five. For example if player one saw alternate colors they would say red and player 2 would say blue and player 3 would say red... This is the best strategy because lets say player 1 saw alternates and was correct then if player 2 had alternates they would have 75% of being correct if they said blue because if they were wrong then that would mean three out of 4 were red. This same logic would apply up through player 5. So to sum up strategy is if player sees 2 of same color they say opposite otherwise the players would say opposite colors from 1 to five. So if player 1 was gonna say red if they saw alternates then player 2 would say blue if they saw alternates and would alternate from there.&lt;/p&gt;&#xA;" OwnerUserId="10856" LastActivityDate="2015-04-01T02:00:41.763" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11370" PostTypeId="1" AcceptedAnswerId="11372" CreationDate="2015-04-01T02:36:10.303" Score="-6" ViewCount="362" Body="&lt;p&gt;From Hiroshima, Japan to Nantucket, Massachusetts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try to solve this without Google.&lt;/p&gt;&#xA;" OwnerUserId="2767" LastActivityDate="2015-04-01T04:36:34.463" Title="Why do you not want to make this trip?" Tags="|pattern|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11371" PostTypeId="2" ParentId="11364" CreationDate="2015-04-01T02:54:17.193" Score="6" Body="&lt;p&gt;A weak partial answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The players can achieve a chance of 5/32. Each player guesses blue, unless they see two blue hats, in which case they guess red. This works for the cyclic permutations of &lt;code&gt;BRBRB&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The players cannot achieve more than 1/4. First, randomly select the hat colors of all but two players that are not adjacent. Then, have those two players guess on what they see. Then, pick their hats at random. Their guesses cannot depend on each others' hats, so they simply have a 1/4 chance of both being right.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-04-01T02:54:17.193" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11372" PostTypeId="2" ParentId="11370" CreationDate="2015-04-01T04:36:34.463" Score="10" Body="&lt;p&gt;The reason is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the airport codes are HIJ ACK&lt;br&gt;&#xA;&lt;br&gt;&#xA; Yes, I've heard this one before ;-)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastActivityDate="2015-04-01T04:36:34.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11373" PostTypeId="2" ParentId="10559" CreationDate="2015-04-01T05:20:04.850" Score="5" Body="&lt;p&gt;I'm going to guess you are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Entropy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes we play catch;&#xA;Though you'd hope it more like fetch.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everything we do throws off our energy and increases entropy, though we'd like that process to be reversed (fetch implies bringing the energy back).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'd be the second; an honorable mention.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Second law of thermodynamics refers to entropy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Indeed it'd be, as if I've spent my life in detention.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Only applies to closed systems.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'd know where, if you found the tether;&#xA;Around about which, he and I dance together.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This one I can't quite fit in since the &quot;I&quot; should theoretically refer to entropy... but I think it's referring to the Sun (around which the Earth &amp;amp; Moon &quot;dance&quot;), perhaps indicating the sun burning out, as per the next point below...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cliches away; your usual meeting need not be so verbose.&#xA;PROPERLY told, I don't even have to come close.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Heat death of the universe is the cliche, the verbosity is implying we won't have to wait that long since the sun will burn out long before (not even close to the same time scales)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Intrinsic you'd have a gut full of fear;&#xA;Ladies and gentlemen, boys and girls... Dying time's here!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When either of the above happen, the Earth will be destroyed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Given me a name;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Entropy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then please explain:&#xA;Where, when and what is,&#xA;That which I'll do,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Where: everywhere. When: in the stupidly distant future. What: you'll destroy everything.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Should you ever find out:&#xA;Me, I am true.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Increasing entropy is considered &quot;law&quot; and thus unavoidably &quot;true&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(and as per your caveat, this answer has never appeared on this site before)&lt;/p&gt;&#xA;" OwnerUserId="5208" LastActivityDate="2015-04-01T05:20:04.850" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11374" PostTypeId="2" ParentId="10911" CreationDate="2015-04-01T06:21:55.233" Score="4" Body="&lt;p&gt;You are  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Velcro  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My name sounds like a bird&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Like a crow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whole body is like tongue of a lizard&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Velcro rolls up like a lizard tongue. Lizard tongues might also have little barbs on it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Am different colors &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Velcro comes in all colors&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I well can be very littered &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;lint and stuff gets stuck in between the barbs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My energy melts away over time &#xA;But not quite like an ice &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;After using it, the edges start curling up and it doesn't hold as firmly&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And when not taken care of &#xA;I'll collect dust and lice &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;another reference to stuff that gets stuck in Velcro &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10873" LastEditorUserId="5188" LastEditDate="2015-04-01T07:33:41.470" LastActivityDate="2015-04-01T07:33:41.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11375" PostTypeId="2" ParentId="9780" CreationDate="2015-04-01T06:31:18.030" Score="1" Body="&lt;p&gt;My answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Spectacle Lens&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm known as single or as double&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Single lens for only reading or far or lens with both&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Be it in head or mind or eye.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; While relaxing sometimes you slide specs (and thus lenses) up to your head&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For some I cause enough of trouble&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They often forget, break, power changed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That they will force me to comply.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You must find it (if forgotten), get a new (if broken or power changed)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But even if I'm weak or near&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; weak power, near power&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Up to an imperfect extent&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It may not be perfect match sometimes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once i show up i need not fear&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure about this clue&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You expect to find me at the end.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you get old, you get lens (end means end of life)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now I've related you my tail&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; spectangle tongs, that fit on year and relates with you&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the name by which I hail?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lens of spectacles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4637" LastActivityDate="2015-04-01T06:31:18.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11376" PostTypeId="2" ParentId="11210" CreationDate="2015-04-01T07:07:14.670" Score="0" Body="&lt;p&gt;I think there is not enough information to solve this, but it appears that the numbers (or letters of the alphabet) are being positioned on an 8x10 grid as follows, or perhaps a 10x10 grid which would be more common for a board game.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/cnc6j.png&quot; alt=&quot;8x10grid&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-04-01T07:51:25.520" LastActivityDate="2015-04-01T07:51:25.520" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11377" PostTypeId="1" AcceptedAnswerId="11440" CreationDate="2015-04-01T08:43:17.290" Score="-2" ViewCount="704" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Treat me baby who Abode of the good&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;But don't think my origin is asteroid&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;but believe me, i am not that alien&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;I have traveled Millennium distances&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;but that may not whistler in your ears&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;once you know, i am not that alien&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Treat me with longhorns on my head&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;with black color combs for my hairs&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;but believe me, i am not that alien&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;when you able to see Jupiter in sky,&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;and wander why sky color is blue,&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;you will come to know who am i,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;but believe me, i am not that alien&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Then who am i ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would add hints after every 1 hours&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hints :&#xA;1)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;These lines do not specify a characteristic but a code.See what seems odd words from above lines.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;See my edited tag Computers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;but believe me, i am not that alien&lt;/strong&gt; lines are added as buffers so you dont deviate from actual object/entity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; See first line carefully biblical meaning of that would be great hint.I think this hint will end up this question.:)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;see bold sentences only those matters and have solution. #1 hint and #4 hint combine them to get solution...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10433" LastEditorUserId="5373" LastEditDate="2015-04-26T11:40:04.617" LastActivityDate="2015-04-26T11:40:04.617" Title="I am not Alien Then who am i?" Tags="|riddle|computer-puzzle|code|" AnswerCount="7" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11378" PostTypeId="1" CreationDate="2015-04-01T08:50:46.783" Score="4" ViewCount="878" Body="&lt;p&gt;This is the corrected version of &lt;a href=&quot;https://puzzling.stackexchange.com/questions/11321/5-logicians-with-1-or-2-hats&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;There are 5 logicians seated around a table. They are blindfolded and hats are placed on their heads. After removing the blindfolds, they are told a true statement:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Each of you has either one hat or two hats on your head. The total&#xA;number of hats is &lt;strong&gt;8,9 or 10&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The actual sum is 9, but this is not told to them.&lt;/p&gt;&#xA;&lt;p&gt;They are then told to answer the question &amp;quot;How many hats are on your head?&amp;quot; in clockwise fashion (starting from any of the 5). If it is possible to logically deduce the number of hats on their own heads, they do so, otherwise reply &amp;quot;I don't know.&amp;quot; and wait till the question cycles back to them.&lt;/p&gt;&#xA;&lt;p&gt;They have no means of knowing the number of hats on their own heads (except logically), but can see every other person. They have no other means of communication, and no tricks involved.&lt;/p&gt;&#xA;&lt;p&gt;The game completes when everyone has figured out their hat count. At what number should the one-hatted logician be placed so that:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The game is completed?&lt;/li&gt;&#xA;&lt;li&gt;The game goes on forever?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="65277" LastEditDate="2022-10-15T02:37:58.657" LastActivityDate="2022-10-15T02:37:58.657" Title="5 logicians with 1 or 2 hats (Part II) [8,9,10]" Tags="|logical-deduction|meta-knowledge|hat-guessing|" AnswerCount="9" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="11379" PostTypeId="2" ParentId="11378" CreationDate="2015-04-01T09:13:11.760" Score="0" Body="&lt;p&gt;If Actual arrangement&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;21222,22122,22212&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Logicians assumed cases&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;11222,12122,12212,21122,21212,22112 &#xA;  in this case last person would have said that he has 2 hats (since least count + max 2 hats on head) evaluates to two hats on last head&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;21222,22122,22212 since in assumed arrangement since last person didnt said anything first person can predict he has 2 hats&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;in this way game completes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But if actual arrangment is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;12222,22221 then first and last logicians may assume following arrangement as well &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;12221,12221 respectively so game do not completes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10433" LastEditorUserId="10433" LastEditDate="2015-04-01T10:31:38.737" LastActivityDate="2015-04-01T10:31:38.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11380" PostTypeId="1" AcceptedAnswerId="11382" CreationDate="2015-04-01T09:21:11.100" Score="12" ViewCount="4145" Body="&lt;p&gt;Jack and Jill are students of Mr. Hill. Mr. Hill's birthday is on $M$ month, $D$ day. Neither student knows when Mr. Hill's birthday is, but both know it is one of the dates below. Mr. Hill told Jack the $M$ value and Jill the $D$ value. Then Mr. Hill asked both of them when is his birthday.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mr. Hill's birthday is one of the days below (format: month/day):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;3/4     3/5     3/8     6/4     6/7     9/1     9/5     12/1     12/2 &#xA;  12/8&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Jack: I don't know, and Jill surely doesn't know either.  &lt;/li&gt;&#xA;&lt;li&gt;Jill: At first I didn't know either, but now I do.  &lt;/li&gt;&#xA;&lt;li&gt;Jack: Oh, then I know it too.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Based on the above conversation between Jack and Jill, determine&#xA;  Mr. Hill's birthday. Give your step by step reasoning.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10820" LastEditorUserId="10615" LastEditDate="2016-02-24T04:10:31.850" LastActivityDate="2016-02-24T04:10:31.850" Title="When is Mr. Hill's birthday?" Tags="|logical-deduction|story|meta-knowledge|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11381" PostTypeId="2" ParentId="11377" CreationDate="2015-04-01T09:28:13.770" Score="0" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LIGHT?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's not alien in itself, but it can travel from many light years away, doesn't whistle in your ears (that's sound!), and sometimes enables you to see Jupiter in the daytime (if you have a good enough telescope, at least).&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-04-01T09:28:13.770" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11382" PostTypeId="2" ParentId="11380" CreationDate="2015-04-01T09:32:27.140" Score="10" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 9/1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Day 2 only shows up once (12/2), and day 7 only shows up once (6/7). &#xA;Days 1, 4, 5, 8 each show up at least twice. Jack says that Jill surely doesn't know, and therefore neither day 2 nor day 7 can be combined with Jack's month M. This means that $M \ne 12$ and $M \ne 6$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;After Jack's first statement, the list of dates has been reduced to &#xA;3/4 3/5 3/8 9/1 9/5.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Jill (knowing D) can deduce M from the new list. This means $D \ne5$, but the other values 1,4,8 are still possible. The list reduces to 3/4 3/8 9/1.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Jack (knowing M) can deduce D from the new list. This means $M \ne 3$ and onlly leaves one possible answer stated above.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="9125" LastEditDate="2015-04-01T10:31:35.480" LastActivityDate="2015-04-01T10:31:35.480" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11383" PostTypeId="2" ParentId="11377" CreationDate="2015-04-01T09:52:51.027" Score="0" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the sun?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The longhorns might be ancient egyptian.&lt;/p&gt;&#xA;" OwnerUserId="10885" LastActivityDate="2015-04-01T09:52:51.027" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11387" PostTypeId="2" ParentId="11378" CreationDate="2015-04-01T11:37:52.357" Score="1" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;Preamble&lt;/strong&gt;: The following are all of the possible (unordered) hat combinations that could exist for a total of 8, 9, or 10 hats:&lt;/em&gt;&lt;br&gt;&#xA;22222: 10 hats&lt;br&gt;&#xA;22221: 9 hats&lt;br&gt;&#xA;22211: 8 hats&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TL;DR&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The game never completes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Case 12222:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $L_1$ sees 8 hats. He does not yet have enough information to make a decision. The table could be 12222 or 22222.&lt;br&gt;&#xA; From the viewpoint of the other logicians, $L_1$ must have seen fewer than two 1-hatted logicians. Had he seen two, $L_1$ would have enough information to deduce that he was wearing two hats in an 8-hat scenario.&lt;br&gt;&#xA; $L_1$ would have passed when seeing zero or one 1-hatted person.&lt;br&gt;&#xA; $L_2$ sees 7 hats, six of which are on $L_{3,4,5}$. He knows that if $L_1$ had seen a 1-hat logician, it's $L_2$. This does not give $L_2$ enough information to make a decision.&lt;br&gt;&#xA; Had $L_2$ seen one of $L_{3,4,5}$ wearing 1 hat, he would have seen only 6 hats, leaving him knowing that he had 2 hats on his head. Because he did not state this, $L_{3,4,5}$ know that $L_2$ sees $L_1$ wearing 1 hat and did not see a second logician wearing 1 hat. From this, they can each deduce that they are wearing 2 hats.&lt;br&gt;&#xA; $L_1$ now knows that $L_3$ was the first to have enough information.&lt;br&gt;&#xA; $L_1$ considers the possibility that they are sitting at a table with 10 hats.&lt;br&gt;&#xA; If this is the case, $L_2$ sees 8 hats, but because $L_1$ could have seen 1 hat on $L_2$ from the perspective of $L_2$, $L_2$ cannot decide.&lt;br&gt;&#xA; $L_3$ would also see 8 hats, but would know that $L_2$ saw 8 hats. If $L_2$ had seen 7 hats, $L_2$ would know he is wearing 2 hats, or else $L_1$ would have seen two 1-hatted logicians.&lt;br&gt;&#xA; $L_3$ can now conclude that he is wearing 2 hats in the 10-hat scenario. $L_{4,5}$ can also conclude their 2-hat status based on this.&lt;br&gt;&#xA; Because $L_3$ is the first to figure out his 2-hat status in either a 9- or 10-hat scenario, $L_1$ does not have enough information.&lt;br&gt;&#xA; Similarly, $L_2$ does not have enough information since $L_{3,4,5}$ could see $L_{1,2}$ with three total hats or two total hats and still make the same statements about themselves.&lt;br&gt;&#xA; &lt;strong&gt;The game does not complete.&lt;/strong&gt;  $L_1$ and $L_2$ are stuck.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Case 21222: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $L_1$ sees 7 hats. He does not yet have enough information to make a decision. The table could be 11222 or 21222.&lt;br&gt;&#xA; From the viewpoint of the other logicians, $L_1$ must have seen fewer than two 1-hatted logicians. Had he seen two, $L_1$ would have enough information to deduce that he was wearing two hats in an 8-hat scenario.&lt;br&gt;&#xA; $L_1$ would have passed when seeing zero or one 1-hatted person.&lt;br&gt;&#xA; $L_2$ sees 8 hats, six of which are on $L_{3,4,5}$. He knows that if $L_1$ had seen a 1-hat logician, it's $L_2$. However, this does not give $L_2$ enough information to make a decision as $L_1$ would also pass if he sees four 2-hat logicians.&lt;br&gt;&#xA; $L_3$ sees 7 hats and knows that $L_1$ did not see two 1-hat logicians.&lt;br&gt;&#xA; $L_3$ is able to conclude that he is wearing two hats.&lt;br&gt;&#xA; $L_{4,5}$ are able to make the same conclusion based on the indecision of $L_1$.&lt;br&gt;&#xA; $L_1$ now knows that $L_3$ was the first to have enough information.&lt;br&gt;&#xA; $L_1$ considers the possibility that they are sitting at a table with 8 hats.&lt;br&gt;&#xA; If this is the case, $L_2$ sees 7 hats, but because $L_1$ could have seen 1 hat on $L_2$ from the perspective of $L_2$, $L_2$ cannot decide.&lt;br&gt;&#xA; $L_3$ would see 6 hats, immediately revealing that they are at an 8-hat table and $L_3$ now knows he is wearing 2 hats.&lt;br&gt;&#xA; Because $L_3$ is the first to figure out his 2-hat status in either a 8- or 9-hat scenario, $L_1$ does not have enough information.&lt;br&gt;&#xA; Similarly, $L_2$ still does not have enough information. (Compare to 12222.)&lt;br&gt;&#xA; &lt;strong&gt;The game does not complete.&lt;/strong&gt; $L_1$ and $L_2$ are stuck.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Case 22122:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $L_1$ sees 7 hats. He does not yet have enough information to make a decision. The table could be 12122 or 22122.&lt;br&gt;&#xA; From the viewpoint of the other logicians, $L_1$ must have seen fewer than two 1-hatted logicians. Had he seen two, $L_1$ would have enough information to deduce that he was wearing two hats in an 8-hat scenario.&lt;br&gt;&#xA; $L_1$ would have passed when seeing zero or one 1-hatted person.&lt;br&gt;&#xA; $L_2$ sees 7 hats, including the 1 hat on $L_3$. He knows that $L_1$ had not seen two 1-hat logicians, so $L_2$ is able to decide that he is wearing two hats.&lt;br&gt;&#xA; $L_3$ sees 8 hats and must now consider the possibility of a 10-hat table.&lt;br&gt;&#xA; If $L_3$ were wearing 2 hats, $L_2$ would have had the same information as a 21222 scenario and thus be unable to decide his hat count.&lt;br&gt;&#xA; However, since $L_2$ could decide, they must not be in a scenario that fits 2x222. With $L_3$ seeing 22x22, the only possibility is that $L_3$ is wearing 1 hat.&lt;br&gt;&#xA; $L_3$ is able to conclude that he is wearing 1 hat.&lt;br&gt;&#xA; $L_4$ sees 7 hats and must consider the possibility of an 8-hat table.&lt;br&gt;&#xA; If $L_4$ were wearing 1 hat, $L_1$ would have seen two 1-hatted logicians and been able to decide his own hat status.&lt;br&gt;&#xA; Because this is not the case, $L_4$ can correctly deduce that he is wearing 2 hats.&lt;br&gt;&#xA; The same is true for $L_5$.&lt;br&gt;&#xA; $L_1$ is now the only logician that has not yet determined his hat status.&lt;br&gt;&#xA; $L_1$ must consider the 12122 table scenario in order to determine if he is wearing 1 or 2 hats.&lt;br&gt;&#xA; In such a scenario, $L_2$ would have seen two 1-hat logicians and would have been able to decide that he was wearing two hats, so $L_2$ has not offered enough information to $L_1$.&lt;br&gt;&#xA; $L_3$ would have seen the same as in the 12222 scenario, but would have been able to tell the difference because $L_2$ was able to state his hat count. $L_3$ has not offered enough information to $L_1$.&lt;br&gt;&#xA; $L_{4,5}$ would each see two 1-hat logicians and be able to state that they have two hats on, so they also do not provide enough information to $L_1$.&lt;br&gt;&#xA; &lt;strong&gt;The game does not complete.&lt;/strong&gt; $L_1$ is stuck.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Case 22212:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $L_1$ sees 7 hats. He does not yet have enough information to make a decision. The table could be 12212 or 22212.&lt;br&gt;&#xA; From the viewpoint of the other logicians, $L_1$ must have seen fewer than two 1-hatted logicians. Had he seen two, $L_1$ would have enough information to deduce that he was wearing two hats in an 8-hat scenario.&lt;br&gt;&#xA; $L_1$ would have passed when seeing zero or one 1-hatted person.&lt;br&gt;&#xA; $L_2$ sees 7 hats, including the 1 hat on $L_4$. He knows that $L_1$ had not seen two 1-hat logicians, so $L_2$ is able to decide that he is wearing two hats.&lt;br&gt;&#xA; $L_3$ follows the same logic as $L_2$ to deduce that he is also wearing 2 hats.&lt;br&gt;&#xA; $L_4$ sees 8 hats and must now consider the possibility of a 10-hat table.&lt;br&gt;&#xA; If $L_4$ were wearing 2 hats, $L_2$ would have had the same information as a 21222 scenario and thus be unable to decide his hat count.&lt;br&gt;&#xA; However, since $L_2$ could decide, they must not be in a scenario that fits 2x222. With $L_4$ seeing 222x2, the only possibility is that $L_4$ is wearing 1 hat.&lt;br&gt;&#xA; $L_4$ is able to conclude that he is wearing 1 hat.&lt;br&gt;&#xA; $L_5$ sees 7 hats and must consider the possibility of an 8-hat table.&lt;br&gt;&#xA; If $L_5$ were wearing 1 hat, $L_1$ would have seen two 1-hatted logicians and been able to decide his own hat status.&lt;br&gt;&#xA; Because this is not the case, $L_5$ can correctly deduce that he is wearing 2 hats.&lt;br&gt;&#xA; $L_1$ is now the only logician that has not yet determined his hat status.&lt;br&gt;&#xA; $L_1$ must consider the 12212 table scenario in order to determine if he is wearing 1 or 2 hats.&lt;br&gt;&#xA; In such a scenario, $L_2$ would have seen two 1-hat logicians and would have been able to decide that he was wearing two hats, so $L_2$ has not offered enough information to $L_1$.&lt;br&gt;&#xA; $L_3$ offers no additional information for the same reason.&lt;br&gt;&#xA; $L_4$ would have seen the same as in the 12222 scenario, but would have been able to tell the difference because $L_2$ was able to state his hat count. $L_4$ has not offered enough information to $L_1$.&lt;br&gt;&#xA; $L_5$ would see two 1-hat logicians and be able to state that they have two hats on, so they also do not provide enough information to $L_1$.&lt;br&gt;&#xA; &lt;strong&gt;The game does not complete.&lt;/strong&gt; $L_1$ is stuck.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Case 22221:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $L_1$ sees 7 hats. He does not yet have enough information to make a decision. The table could be 12221 or 22221.&lt;br&gt;&#xA; From the viewpoint of the other logicians, $L_1$ must have seen fewer than two 1-hatted logicians. Had he seen two, $L_1$ would have enough information to deduce that he was wearing two hats in an 8-hat scenario.&lt;br&gt;&#xA; $L_1$ would have passed when seeing zero or one 1-hatted person.&lt;br&gt;&#xA; $L_2$ sees 7 hats, including the 1 hat on $L_5$. He knows that $L_1$ had not seen two 1-hat logicians, so $L_2$ is able to decide that he is wearing two hats.&lt;br&gt;&#xA; $L_{3,4}$ follow the same logic as $L_2$ to deduce that they are also each wearing 2 hats.&lt;br&gt;&#xA; $L_5$ sees 8 hats and must now consider the possibility of a 10-hat table.&lt;br&gt;&#xA; If $L_5$ were wearing 2 hats, $L_2$ would have had the same information as a 21222 scenario and thus be unable to decide his hat count.&lt;br&gt;&#xA; However, since $L_2$ could decide, they must not be in a scenario that fits 2x222. With $L_5$ seeing 2222x, the only possibility is that $L_5$ is wearing 1 hat.&lt;br&gt;&#xA; $L_5$ is able to conclude that he is wearing 1 hat.&lt;br&gt;&#xA; $L_1$ is now the only logician that has not yet determined his hat status.&lt;br&gt;&#xA; $L_1$ must consider the 12221 table scenario in order to determine if he is wearing 1 or 2 hats.&lt;br&gt;&#xA; In such a scenario, $L_2$ would have seen two 1-hat logicians and would have been able to decide that he was wearing two hats, so $L_2$ has not offered enough information to $L_1$.&lt;br&gt;&#xA; $L_{3,4}$ offer no additional information for the same reason.&lt;br&gt;&#xA; $L_5$ would have seen the same as in the 12222 scenario, but would have been able to tell the difference because $L_2$ was able to state his hat count. $L_5$ has not offered enough information to $L_1$.&lt;br&gt;&#xA; &lt;strong&gt;The game does not complete.&lt;/strong&gt; $L_1$ is stuck.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-04-01T14:07:23.183" LastActivityDate="2015-04-01T14:07:23.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11388" PostTypeId="2" ParentId="11377" CreationDate="2015-04-01T11:39:56.830" Score="2" Body="&lt;p&gt;Maybe it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mercury?&#xA; The longhorns is his planetary symbol - you can see it only when the sun is close to the horizon -&gt; the sky is still blue, but already dark enough to make it possible to see jupiter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10890" LastActivityDate="2015-04-01T11:39:56.830" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11389" PostTypeId="1" AcceptedAnswerId="11391" CreationDate="2015-04-01T12:45:00.637" Score="11" ViewCount="3987" Body="&lt;p&gt;Jack and Jones are brothers and they both sell apples for living. Jack sells 2 apples for 1 Rs and Jones sells 3 apples for 1 Rs. They each have 30 apples. Hence, Jack earns 15 Rs a day while Jones earns 10 Rs a day. This gives a total of &lt;strong&gt;25 Rs&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One day Jack got sick and couldn't sell his apples. He gave his apples to Jones. Jones sold all 60 apples at price of 5 apples for 2 Rs (as 3 apples for 1 Rs + 2 apples for 1 Rs). At the end of the day, he counted the money and ended with a total of &lt;strong&gt;24 Rs&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Where has Jones lost 1 Rs?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="10820" LastEditorUserId="39" LastEditDate="2015-04-01T13:13:21.817" LastActivityDate="2018-02-15T14:42:45.140" Title="Where has Jones lost 1 Rs?" Tags="|logical-deduction|calculation-puzzle|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11390" PostTypeId="2" ParentId="11389" CreationDate="2015-04-01T13:04:54.753" Score="1" Body="&lt;p&gt;At the beginning, we have an average price per apple of $\frac{(30 \times \frac1 2)+(30 \times \frac1 3)} {60} =\frac {5}{12}=0,416666 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, he changed the price to $\frac 25=0,4$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as the average price changed, his wage changes as well!&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-04-01T14:18:08.193" LastActivityDate="2015-04-01T14:18:08.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11391" PostTypeId="2" ParentId="11389" CreationDate="2015-04-01T13:05:13.673" Score="9" Body="&lt;p&gt;TL,DR: It is all about operator precedence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have the following unit price for Jack apples: $1/2$.&lt;br&gt;&#xA;We have the following unit price for Jones apples: $1/3$.&lt;br&gt;&#xA;Thus making the calculation of the total amount collected: $$1/2 \times 30 + 1/3 \times 30 = 15 + 10 = 25$$&#xA;In that configuration, the unit price for a single apple is: $$(1/2 + 1/3) / 2 = 5/12$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the combined sale, we have the following unit price: $2/5$&lt;br&gt;&#xA;Thus making the calculation of the total amount collected: $$2/5 \times 60 = 24$$&lt;br&gt;&#xA;So far so good.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When we are comparing unit price of a single apple now.&lt;br&gt;&#xA; The problem occurs because of operator precedence, $(1/2 + 1/3)/2 = 5/12 \ne 2/5$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10400" LastEditorUserId="10400" LastEditDate="2015-04-01T13:18:23.067" LastActivityDate="2015-04-01T13:18:23.067" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11392" PostTypeId="2" ParentId="11322" CreationDate="2015-04-01T13:15:46.493" Score="3" Body="&lt;p&gt;Another way to get to the solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a &quot;safe&quot; male: He is only safe if all 4 females around him have killed someone else. Since we are in an infinite grid, we can ignore the initial setup and just have to ask us: To get another male safe, how many nee to sacrifice their lives? Imagine the Safe Spider as (M) The sacrificed Males are connected to the respective females. - Uninteresting Spiders are represented by dots.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;.  .  M  .  .&#xA;      |&#xA;.  M  F  .  .&#xA;   |&#xA;.  F (M) F--M&#xA;&#xA;.  .  F--M  .&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;To get the spider lower left from him safe too, we need to satisfy two new Females and have to sacrifice two more Males.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;.  .  .  M  .  .&#xA;         |&#xA;.  .  M  F  .  .&#xA;      |&#xA;.  M  F (M) F--M&#xA;   #&#xA;.  F [M] F--M  .&#xA;&#xA;.  .  F==M  .  .&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And we can repeat this pattern to infinity. And we can also place several of these survival-diagonals of males next to each other, each with the same rule:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;.  .  .  M  F (M) F--M [F] M  F (M) F--M&#xA;         |                 |&#xA;.  .  M  F (M) F--M [F] M  F (M) F--M  .&#xA;      |                 |&#xA;.  M  F (M) F--M [F] M  F (M) F--M  .  .&#xA;   |                 |&#xA;M  F (M) F--M [F] M  F (M) F--M  .  .  .&#xA;|                 |&#xA;F (M) F--M [F] M  F (M) F--M  .  .  .  .&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;sacrifice two, to save one&lt;/strong&gt; -&gt; Which leads to a maximum survival rate of 1/3&lt;/p&gt;&#xA;" OwnerUserId="1998" LastActivityDate="2015-04-01T13:15:46.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11393" PostTypeId="1" AcceptedAnswerId="11397" CreationDate="2015-04-01T13:24:46.470" Score="4" ViewCount="915" Body="&lt;p&gt;My Indian friend Jamal recently participated in the well-known TV quiz&#xA;show &lt;strong&gt;&lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Who_Wants_to_Be_a_Millionaire&quot; rel=&quot;nofollow&quot;&gt;Who wants to be a millionaire?&lt;/a&gt;&lt;/em&gt;&lt;/strong&gt;. You might have heard or read about Jamal's spectacular success; the media reported on it extensively.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what Jamal told us about his 300 Euro question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;After my correct answer to the second question, Prem Kumar flashed a big&#xA;  smile at me and congratulated in an oily voice: &quot;Youuuuuu haaaaave maaaade&#xA;  twoooo-huuuuundred Eurooooooooo! Congratulaaaaaaations!&quot;.  I started to&#xA;  thoroughly dislike this greasy type.  I still was somewhat nervous, as the&#xA;  third question came up. 300 Euro for Geography, duh!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I immediately discarded the last two answers; interestingly, they both&#xA;  may be associated with the three colors that are usually associated with&#xA;  that country.  But for answer C a rectangle, or a sheet, or a plain piece&#xA;  of cloth would have been a much more natural choice. And for the nutrimental&#xA;  answer D it should simply have been a circle. These two answers were too&#xA;  contrived, perhaps intended to be humorous, and just could not be the&#xA;  right answer, no way, not at the 300 Euro level.  Answer B would have fit&#xA;  our beloved neighbor Sri Lanka. Hence I thought, conjectured, guessed,&#xA;  hoped that it had to be answer A, and I decided to go for the footwear.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Prem Kumar pulled off a big show, and asked twice whether this was my final&#xA;  decision. I said Yes once, and I said Yes for a second time, and then&#xA;  Prem said &quot;So your decision is A.&quot; He really is not the fastest thinker.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What was the 300 Euro question?&lt;br&gt;&#xA;What were the four possible answers A, B, C, D?&lt;/p&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2015-04-16T11:36:09.387" Title="Slumdog millionaire (3)" Tags="|geography|knowledge|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11394" PostTypeId="2" ParentId="11298" CreationDate="2015-04-01T13:29:08.853" Score="5" Body="&lt;p&gt;Well, here's my best guess but I'm not 100%...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Murder with ease and comfort:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Drowning is arguably a peaceful way to go.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Loved with purpose:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I assume this means you love it for its purpose; that is, to quench your thirst. For which it is definitely loved!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Drug to the eye:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eyedrops!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And to the heart:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Possibly an IV, this is definitely the most tenuous link of the lot.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is certainly cherished in some forms, but detested in others.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fresh, drinking water vs. Sea water, rain, etc.?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And it certainly causes no diseases. Plus has a physical existence and is pretty much everywhere, as the hints direct.&lt;/p&gt;&#xA;" OwnerUserId="10900" LastActivityDate="2015-04-01T13:29:08.853" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11395" PostTypeId="2" ParentId="11377" CreationDate="2015-04-01T13:40:06.220" Score="4" Body="&lt;p&gt;Is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;iOS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;?&#xA;Millennium and longhorn hint at&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Operating Systems&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am not Alien&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;not Android&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A further hint is the lower case i that is used in the riddle.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2015-04-01T13:40:06.220" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11397" PostTypeId="2" ParentId="11393" CreationDate="2015-04-01T13:49:00.087" Score="8" Body="&lt;p&gt;The 300 Euro question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Like which of these items is Italy shaped?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answers are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A: Boot&lt;br&gt;&#xA; B: Teardrop&lt;br&gt;&#xA; C: Italian Flag&lt;br&gt;&#xA; D: Neapolitan pizza&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sri Lanka is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; shaped like a teardrop. (See &lt;a href=&quot;http://en.wikipedia.org/wiki/Sri_Lanka#/media/File:Sri_Lanka_-_Location_Map_(2011)_-_LKA_-_UNOCHA.svg&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;C is rectangular and D is round and edible. Both contain the colours associated with the country:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; red, white, and green.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-04-06T13:28:48.880" LastActivityDate="2015-04-06T13:28:48.880" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11398" PostTypeId="2" ParentId="11378" CreationDate="2015-04-01T13:53:57.523" Score="0" Body="&lt;p&gt;Long and short: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If order is 12222 or 21222 first and second can't tell. Otherwise, all but first can tell. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's assume the order is 12222&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second 2-hat person to guess can know their own hat. That person knows that the first 2-hat person saw 5+ their hat. If the first saw 6 total hats, they know they have to have 2 hats or else they would be below the minimum. Hence, they saw 7 and couldn't say yea/nay. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 3rd and 4th 2-hat can say the same. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it gets tricky going back to the 1-hat and first 2-hat. The 1-hat sees 8, so he can't guess. Then he sees first 2-hat not know with 6+ him. Then he sees second 2-hat know with 6+ him. If he has a 2-hat (total 10), then nobody would be able to guess their hat. And it would make sense that the first 2-hat would see 7, not know, then the second 2-hat could guess, as could the rest. So, 1-hat figures it out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That leaves the first 2-hat. In this case, &lt;strong&gt;I don't think he can figure it out&lt;/strong&gt;. If he had a 1-hat, then second-fourth 2-hats would have gotten the right answer and the 1-hat could have guessed right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's change the order to 21222&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First 2-hat, 1-hat can't figure it out, second-fourth 2-hats know they don't have a 1-hat, since the first 2-hat would have gotten his hat right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First 2-hat still doesn't know, and the 1-hat knows his (by the 10=loop). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Thinking)&lt;br&gt;&#xA;So, how would they deal with the scenario that all of them have 2 hats? First sees 8, doesn't know. Second sees 8, first could have seen 7, doesn't know. Third sees 8, knows first two saw the same thing, but the second person couldn't figure out their hats. If they saw 7, the 2nd would have been able to say he was a 2-hat. So they saw 8.... which puts a crimp in my logic above. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-04-01T14:18:20.590" LastActivityDate="2015-04-01T14:18:20.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11399" PostTypeId="2" ParentId="11298" CreationDate="2015-04-01T14:06:52.847" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PROPAGANDA alters the way you see and feel (like a drug) and can kill loved ones who speak out (indirectly) without causing disease. Those who cherish the words of a corrupt power are the ones who detest ardently without thought for why.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or if the OP is cynical, then&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RELIGION might also fit the bill because of the historical precedence of violence in the name of love. The imagery associated with religion include cherished symbols (like the cross in Christianity) that act like drugs for the eye and heart, inciting rage and violence in some and love and grace in others.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2360" LastEditorUserId="2360" LastEditDate="2015-04-01T14:12:43.597" LastActivityDate="2015-04-01T14:12:43.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11400" PostTypeId="1" AcceptedAnswerId="11406" CreationDate="2015-04-01T14:25:48.573" Score="5" ViewCount="475" Body="&lt;p&gt;There are $n$ monsters (M) and $n$ people (P) on the left side of a river. Nobody is on the right.&lt;br&gt;&#xA; They have a boat that can hold up to 2 beings. At no point on time may either of these be true:&#xA;$$M_{shore1}+M_{boat}&amp;gt;P_{shore1}+P_{boat}$$&#xA;$$M_{shore2}+M_{boat}&amp;gt;P_{shore2}+P_{boat}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the minimum number of boat trips required for&lt;br&gt;&#xA;(a) $n=4$&lt;br&gt;&#xA;(b) $n=5$&lt;br&gt;&#xA;(c) $n=100$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also provide a formula for arbitrary $n$, if possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.&#xA;I have found a solution using induction. I'm not sure if it is optimal, I'll post it soon, if needed.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4394" LastEditDate="2015-04-29T11:48:13.607" LastActivityDate="2015-04-29T11:48:13.607" Title="$n$ monsters and $n$ people" Tags="|mathematics|logical-deduction|optimization|river-crossing|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11401" PostTypeId="2" ParentId="11378" CreationDate="2015-04-01T14:30:22.177" Score="0" Body="&lt;p&gt;I am too dumb to calculate it exactly, so I will only provide a safe upper border for the amount of rounds required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 9 hats, so four logicians have 2 and one logician has 1 hat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first round, everyone will tell his colleague that he does not know (because he does not see a total of only 6 hats on his colleagues' head). If he would, he could deduce that he has 2, and then every person with one hat could deduce that he has one, and every person with two hats could deduce that he has two on their hat. So after no more than 2 rounds, the game would be over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, after 2 rounds, every logician is completely safe to assume that there are more than 8 hats on the heads, because otherwise, the puzzle would already be over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the first logician will be able to start saying &quot;I know that I have 2 hats on my head&quot; if he sees 7 hats. If he sees 8, he can't be so sure, but his neighbor can. By solving that he knows, everyone can assume that there are 9 hats total. So, after at most 3 rounds and a fourth turn by logician 1, the puzzle is solved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I even think that 3 full rounds could be enough, because the last logician can start his deduction in his 2nd turn already, but that's just too high for me at the moment.&lt;/p&gt;&#xA;" OwnerUserId="2544" LastActivityDate="2015-04-01T14:30:22.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11402" PostTypeId="2" ParentId="11400" CreationDate="2015-04-01T14:34:23.023" Score="3" Body="&lt;p&gt;Given the constraint that monsters must not outnumber people anywhere, your process is: Human in boat (stays in boat as ferryman), ferry monster, ferry human, ferry monster, ferry human, etc. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; So, trips = $4n-3$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-04-01T15:08:03.507" LastActivityDate="2015-04-01T15:08:03.507" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11403" PostTypeId="2" ParentId="9780" CreationDate="2015-04-01T14:39:05.253" Score="1" Body="&lt;p&gt;It could be a  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Bed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm known as single or as double&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Beds can be both single or double, depending on their size.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Be it in head or mind or eye.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is hard to fit, maybe your mind can be &quot;sleeping&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For some I cause enough of trouble&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people have insomnia or other sleeping problems.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That they will force me to comply.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Those people usually buy special beds or pillows.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But even if I'm weak or near&#xA;Up to an imperfect extent&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not all beds are perfect, eheh&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once i show up i need not fear&#xA;You expect to find me at the end.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The end of your day, of course!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now I've related you my tail&#xA;What is the name by which I hail?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Bed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I often live close to the fringe&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Near the wall, usually&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I am broken you may cringe.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's never good if a bed breaks!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I stand out most when I'm perfected&#xA;But even slanted I'm detected.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A slanting bed can be seen without problems.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-04-01T14:39:05.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11404" PostTypeId="2" ParentId="11400" CreationDate="2015-04-01T15:07:19.477" Score="2" Body="&lt;p&gt;At least &lt;strong&gt;one person must always stay in the boat&lt;/strong&gt;, otherwise we'll have this contradiction: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M1 + MB &amp;gt; P1 + PB$&lt;br&gt;&#xA;$M2 + MB &amp;gt; P2 + PB$&lt;br&gt;&#xA;Summing, assuming PB=0:&lt;br&gt;&#xA;$(M1+M2) +2MB &amp;gt; (P1+P2)$&lt;br&gt;&#xA;Since $(M1+M2)=(P1+P2)$ the above condition would be always verified, making the problem impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, one human does the ferryman, transporting the others across the river. He starts with a monster, then comes back to shore 1, brings a human to shore 2, comes back to shore 1 and restarts. There are $2n-1$ creatures to be transported in addition to the sailor, and for each one (except the last) the ferryman must come back to shore 1.&lt;br&gt;&#xA;Overall it requires $2\times(2n-1)-1 = 4n-3$ &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;n=4&lt;/strong&gt; requires &lt;strong&gt;13 trips&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;n=5&lt;/strong&gt; requires &lt;strong&gt;17 trips&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;n=100&lt;/strong&gt; requires &lt;strong&gt;397 trips&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-04-01T15:14:53.823" LastActivityDate="2015-04-01T15:14:53.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11405" PostTypeId="2" ParentId="11393" CreationDate="2015-04-01T15:13:21.080" Score="3" Body="&lt;p&gt;The question was  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;What's the shape of Italy?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answers were:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;A&lt;/strong&gt;: Footwear&lt;br&gt;&#xA;&lt;strong&gt;B&lt;/strong&gt;: Lung&lt;br&gt;&#xA;&lt;strong&gt;C&lt;/strong&gt;: Flag (Green, White, Red)&lt;br&gt;&#xA;&lt;strong&gt;D&lt;/strong&gt;: Pizza (Green for basin, White for Mozzarella, Red for tomato)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-04-01T15:13:21.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11406" PostTypeId="2" ParentId="11400" CreationDate="2015-04-01T15:16:19.090" Score="6" Body="&lt;p&gt;TL;DR&#xA;The general formula for fewest trips is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $4n-3$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Giving values for (a), (b), (c):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (a) 13&lt;br&gt;&#xA; (b) 17&lt;br&gt;&#xA; (c) 397&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The possible combinations for the first trip are: MM, MP, PP, M, P. We can immediately discount a first trip carrying an individual as the return trip would need to be made by that individual, causing no change to the number of beings on the second shore and therefore unnecessarily increasing the number of trips.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that exactly 2 beings must occupy the first trip and exactly 1 of these beings must be on the return trip, leaving 1 being on the second shore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case MM:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Upon boarding the boat, the second inequality is true:&lt;br&gt;&#xA; $M_{shore2}+M_{boat}&amp;gt;P_{shore2}+P_{boat}$&lt;br&gt;&#xA; $0+2 &amp;gt; 0+0$&lt;br&gt;&#xA; This contradicts the condition that &lt;em&gt;&quot;At no point on time may either of these be true&quot;&lt;/em&gt;.&lt;br&gt;&#xA; $\therefore$ MM is not a valid first trip.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Case PP:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Upon disembarking the boat, the first inequality is true:&lt;br&gt;&#xA; $M_{shore1}+M_{boat}&amp;gt;P_{shore1}+P_{boat}$&lt;br&gt;&#xA; $n+0 &amp;gt; n-2+1$&lt;br&gt;&#xA; This contradicts the condition that &lt;em&gt;&quot;At no point on time may either of these be true&quot;&lt;/em&gt;.&lt;br&gt;&#xA; $\therefore$ PP is not a valid first trip.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Case MP:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Upon reaching the second shore, P may choose to disembark. This makes the first inequality true:&lt;br&gt;&#xA; $M_{shore1}+M_{boat}&amp;gt;P_{shore1}+P_{boat}$&lt;br&gt;&#xA; $n-1+1&amp;gt;n-1+0$&lt;br&gt;&#xA; $\therefore$ P may not disembark on this trip.&lt;br&gt;&#xA; Suppose then that M disembarks. This satisfies the condition.&lt;br&gt;&#xA; $M_{shore1}+M_{boat} \le P_{shore1}+P_{boat}$&lt;br&gt;&#xA; $n-1+0 \le n-1+1 \implies n-1 \le n$&lt;br&gt;&#xA; $M_{shore2}+M_{boat} \le P_{shore2}+P_{boat}$&lt;br&gt;&#xA; $1+0 \le 0+1 \implies 1 \le 1$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With this knowledge, we can deduce that the second trip to shore 2 must also have at least one P in the boat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case MP:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Upon boarding the boat, the second inequality is true.&lt;br&gt;&#xA; $M_{shore2}+M_{boat} &amp;gt; P_{shore2}+P_{boat}$&lt;br&gt;&#xA; $1+1 &amp;gt; 0+1$&lt;br&gt;&#xA; $\therefore$ because this contradicts the rule, we know MP is an invalid second trip.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Case PP:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This satisfies the condition.&lt;br&gt;&#xA; $M_{shore1}+M_{boat} \le P_{shore1}+P_{boat}$&lt;br&gt;&#xA; $n-1+0 \le n-2+1 \implies n-1 \le n-1$&lt;br&gt;&#xA; $M_{shore2}+M_{boat} \le P_{shore2}+P_{boat}$&lt;br&gt;&#xA; $1+0 \le 1+1 \implies 1 \le 2$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can extend this logic to arbitrary $M_{shore2} = k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P must be on the boat, leaving us two possibilities: $P_{shore2} \in \{k-1, k\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case $P_{shore2} = k$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; M boards the boat piloted by P for the trip to shore 2.&lt;br&gt;&#xA; In transit:&lt;br&gt;&#xA; &lt;strong&gt;Shore 1:&lt;/strong&gt; $(n-k-1) + (1) \le (n-k) + (1)$&lt;br&gt;&#xA; &lt;strong&gt;Shore 2:&lt;/strong&gt; $(k) + (1) \le (k) + (1)$&lt;br&gt;&#xA; Once landing and M disembarks:&lt;br&gt;&#xA; &lt;strong&gt;Shore 1:&lt;/strong&gt; $(n-k-1) + (0) \le (n-k) + (1)$&lt;br&gt;&#xA; &lt;strong&gt;Shore 2:&lt;/strong&gt; $(k+1) + (0) \le (k) + (1)$&lt;br&gt;&#xA; If this is the last trip, P also disembarks &lt;em&gt;(just make sure to get off the boat &lt;strong&gt;after&lt;/strong&gt; M)&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Case $P_{shore2} = k-1$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; P boards the boat piloted by P for the trip to shore 2.&lt;br&gt;&#xA; In transit:&lt;br&gt;&#xA; &lt;strong&gt;Shore 1:&lt;/strong&gt; $(n-k) + (0) \le (n-(k-1)-1) + (2)$&lt;br&gt;&#xA; &lt;strong&gt;Shore 2:&lt;/strong&gt; $(k) + (0) \le (k-1) + (2)$&lt;br&gt;&#xA; Once landing and P disembarks:&lt;br&gt;&#xA; &lt;strong&gt;Shore 1:&lt;/strong&gt; $(n-k) + (0) \le (n-(k-1)-1) + (1)$&lt;br&gt;&#xA; &lt;strong&gt;Shore 2:&lt;/strong&gt; $(k) + (0) \le (k-1+1) + (1)$&lt;br&gt;&#xA; If this is the last trip, the second P also disembarks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The total number of trips to shore 2 are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $2n-1$&#xA; Since we are only letting off one being per trip, we need to take $2n$ trips. Subtract $1$ because the last trip lets off two beings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The total number of trips returning to shore 1 are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $2n-2$&#xA; We make the same number of return trips as we made initial trips, less one because we don't make a final return.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This gives us a total number of trips as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $2n-1 + 2n-2$&lt;br&gt;&#xA; $= 4n-3$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is optimal because it takes 2 beings on the trip to shore 2 and returns with only 1 being. In order to reduce the number of trips, the boat would need to carry more than 2 beings or the return trip would need to be made with nobody in the boat.&lt;/p&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-04-01T15:31:11.313" LastActivityDate="2015-04-01T15:31:11.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11407" PostTypeId="1" AcceptedAnswerId="11411" CreationDate="2015-04-01T15:46:18.983" Score="22" ViewCount="7572" Body="&lt;p&gt;John and Jane are married and keep a joint account at ABC Bank.  They agreed that all of their income should go into the joint account.  In early 2014, John started a new job at a software company, and after the first, second, and third quarters he received a cash bonus.  In early December though, he told Jane that his company wasn't giving out a bonus for the fourth quarter.  Jane found that hard to believe, and she thought that John might try to divert a bonus into another bank account.  But she hid her suspicions from John.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jane didn't know much about computers, so she talked to her tech-savvy friend Alex about how she might find out if John had another bank account.  He gave her a DVD and told her to put it into their home computer and run the setup.exe file.  He said it would install a keylogger that would email him daily logs that he could inspect.  Jane inserted the DVD into their home computer while John was at work.  A popup appeared asking her what to do, and she selected the &quot;Run setup.exe&quot; option.  That installed the keylogger and it began recording all keystrokes on the computer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jane was right to be suspicious.  In mid-December, John's company gave everyone a larger than usual cash bonus for the holidays.  John wanted to spend that money on himself, although not immediately, so he decided to open an account at another bank to hold the money for the time being.  One day while he was at home, he went to www.xyzbank.com and opened up an account there.  Little did he know that everything he typed was being recorded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next day, Alex checked the keylogger logs and saw the following from the previous day:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;www.xyzbank.com[Enter]JohnDoe11144999950 First StreetNew York100012125550104johndoe9876@emaildomain.comjohn9876BROKEBROKE&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The logs also showed the following a few minutes after that:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;secretpasswords.txt[Enter]AASDRESS[Backspace][Backspace][Backspace][Backspace][Backspace][Backspace][Backspace][Backspace][Backspace][Backspace][Backspace]ADDRESS                      USERNAEM[Backspace][Backspace]ME             PASSWORD[Enter]----------------------------------------------------------[Enter]www.s[Backspace]xyzbank.com              john9876             BROKE&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;He called Jane and told her that it looked like John opened an account at XYZ Bank, since he went to their web site and typed in his name, social security number, address, phone number, and email address, along with the username &lt;code&gt;john9876&lt;/code&gt; and password &lt;code&gt;BROKE&lt;/code&gt;.  He also said it looked like he created a text file named &lt;code&gt;secretpasswords.txt&lt;/code&gt; and put his username and password for XYZ Bank in there.  Jane wasn't surprised by the password &lt;code&gt;BROKE&lt;/code&gt;, since John &lt;strong&gt;always&lt;/strong&gt; used dictionary words for his passwords, and he &lt;strong&gt;always&lt;/strong&gt; used all capital letters.  She also wasn't surprised that he kept the password in a text file, as his memory wasn't very good.  John kept a &lt;code&gt;passwords.txt&lt;/code&gt; file on the computer's desktop that contained all of his usernames and passwords, including the username and password for their joint account at ABC Bank.  Jane checked their home computer and didn't see a &lt;code&gt;secretpasswords.txt&lt;/code&gt; file on the desktop, so John must have hidden it somewhere in the file system.  She tried logging in to www.xyzbank.com with username &lt;code&gt;john9876&lt;/code&gt; and password &lt;code&gt;BROKE&lt;/code&gt; and it worked.  The account was empty and showed no activity, however, so John must not have deposited his bonus into the account yet.  Jane decided she would check the account periodically and not immediately reveal to John that she knew about the secret account.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On Dec. 31, while John was at work, Jane logged into the secret account and there was still no balance or activity.  Later that day, John finally got around to depositing his bonus into the account.  Then on Jan. 2, right after John left for work, Jane logged into the secret account and it said it was requiring all customers to change their passwords before they could proceed into their accounts.  She closed the browser without doing anything.  After he got to work, John logged into www.xyzbank.com and saw the same message about changing his password, so he changed it.  He used a dictionary word with all capital letters for his new password, as he always does.  Later that day, Jane tried logging into the secret account and got an &quot;Invalid password&quot; message, so she knew that John had changed the password at work.  She was pretty sure that John would update the &lt;code&gt;secretpasswords.txt&lt;/code&gt; file with the new password when he got home.  Sure enough, when John got home, he updated the password for www.xyzbank.com in the &lt;code&gt;secretpasswords.txt&lt;/code&gt; file.  The keylogger recorded the update.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next day, Jane called Alex and asked him to check the keylogger logs to see what was typed in the previous evening.  Alex said only one thing was typed in the entire evening - the word &lt;code&gt;GARAGE&lt;/code&gt;.  Jane was certain that &lt;code&gt;GARAGE&lt;/code&gt; was John's new password for XYZ Bank, since it was a dictionary word with all capital letters.  She went to www.xyzbank.com and tried logging in with username &lt;code&gt;john9876&lt;/code&gt; and password &lt;code&gt;GARAGE&lt;/code&gt; and got an &quot;Invalid password&quot; message.  She tried again while typing everything in slowly to make sure she didn't mistype anything, and she still got an &quot;Invalid password&quot; message.  She checked the &lt;code&gt;passwords.txt&lt;/code&gt; file on the desktop and didn't see an entry for www.xyzbank.com, and she didn't know how to find the &lt;code&gt;secretpasswords.txt&lt;/code&gt; file.  Alex said he would have to help her find that file at another time, so in the meantime Jane was stumped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What was John's new password for XYZ Bank?&lt;/p&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2018-05-01T11:54:07.123" Title="Wife can't figure out husband's secret account password" Tags="|password|computer-puzzle|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="11408" PostTypeId="1" CreationDate="2015-04-01T16:00:41.413" Score="27" ViewCount="16440" Body="&lt;p&gt;What is the longest English word you can type using only letters that are adjacent on a QWERTY keyboard?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can start anywhere and then either repeat the previous letter or move to an adjacent one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For convenience of those who don't have a QWERTY keyboard, it looks like this.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; ┌───┬───┬───┬───┬───┬───┬───┬───┬───┬───┐&#xA; | Q | W | E | R | T | Y | U | I | O | P |&#xA; └─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┘&#xA;   | A | S | D | F | G | H | J | K | L |&#xA;   └─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴───┘&#xA;     | Z | X | C | V | B | N | M |&#xA;     └───┴───┴───┴───┴───┴───┴───┘&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I found this puzzle on the Internet and I don't have a definitive solution.&lt;/p&gt;&#xA;" OwnerUserId="4394" LastEditorUserId="4394" LastEditDate="2015-04-01T18:35:32.397" LastActivityDate="2015-04-04T05:41:55.443" Title="Longest word with adjacent letters on a keyboard" Tags="|word|" AnswerCount="7" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="11409" PostTypeId="1" AcceptedAnswerId="11684" CreationDate="2015-04-01T16:03:56.430" Score="13" ViewCount="1774" Body="&lt;p&gt;You call me a liar, and say I was the first.&lt;br&gt;&#xA;But I was the second, the first liar isn't cursed.&lt;br&gt;&#xA;I'll open, open, and tell you what I see.&lt;br&gt;&#xA;If I start to fight, you'd think I start with a knee.&lt;br&gt;&#xA;Whether you think I'm cool, or whether I be lame,&lt;br&gt;&#xA;You'll often find me surrounded by many o' my dame.&lt;br&gt;&#xA;A couple types of clues, riddle now you must,&lt;br&gt;&#xA;And tell me what I am, before I turn to dust.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Look to my previous riddles for what type of clues might be found.    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The riddle can be solved using only the first four lines.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2b; More of a guide of what to look at:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Three different clues have been figured out by three different people.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some punctuation might be misleading.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 4:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Look at the wording on line 4 very carefully.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 5:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is no &quot;c&quot; in the answer. The third line does not hint towards a &quot;c&quot;.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 6: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think at this point, most of the clues have been figured out (at least all the major ones), just not all at once. Who can put the puzzle together?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="9951" LastEditDate="2015-04-03T18:54:14.630" LastActivityDate="2015-04-07T13:54:33.047" Title="Oh how I lie - What am I?" Tags="|riddle|word|rhyme|" AnswerCount="13" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11411" PostTypeId="2" ParentId="11407" CreationDate="2015-04-01T16:07:40.433" Score="34" Body="&lt;p&gt;John's new password is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BROKERAGE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The word keeps in the &quot;money&quot; theme. It is for a bank account (money-themed) and his previous password was &lt;code&gt;BROKE&lt;/code&gt; (money-themed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;John's actions were as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Open secretpasswords.txt in vim.&lt;br&gt;&#xA; &lt;code&gt;G&lt;/code&gt; to go to the last line.&lt;br&gt;&#xA; &lt;code&gt;A&lt;/code&gt; to begin editing at the end of the current line.&lt;br&gt;&#xA; Type &lt;code&gt;RAGE&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alternate solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Click to open secretpasswords.txt.&lt;br&gt;&#xA; Use the mouse to position the cursor at the end of &lt;code&gt;BROKE&lt;/code&gt;.&lt;br&gt;&#xA; Type &lt;code&gt;GA&lt;/code&gt;.&lt;br&gt;&#xA; Realise the typo and select &lt;code&gt;GA&lt;/code&gt; for replacement.&lt;br&gt;&#xA; Type &lt;code&gt;RAGE&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alternate alternate solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Type &lt;code&gt;GA&lt;/code&gt; into local file search.&lt;br&gt;&#xA; Click to open &lt;code&gt;GARBAGE&lt;/code&gt; hidden folder that was found.&lt;br&gt;&#xA; Click to open secretpasswords.txt.&lt;br&gt;&#xA; Use the mouse to position the cursor at the end of &lt;code&gt;BROKE&lt;/code&gt;.&lt;br&gt;&#xA; Type &lt;code&gt;RAGE&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="4450" LastEditDate="2015-04-01T17:48:01.630" LastActivityDate="2015-04-01T17:48:01.630" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="11412" PostTypeId="2" ParentId="11408" CreationDate="2015-04-01T16:09:10.533" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ASSESSED/ASSESSES&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is the longest I can come up with.&lt;/p&gt;&#xA;" OwnerUserId="5135" LastEditorUserId="10399" LastEditDate="2015-04-01T16:11:11.640" LastActivityDate="2015-04-01T16:11:11.640" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11413" PostTypeId="2" ParentId="11407" CreationDate="2015-04-01T16:11:28.700" Score="13" Body="&lt;p&gt;Ian has one of my solutions; the other one is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; GARBAGE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; John typed the first three letters in front of &quot;BROKE&quot;, then selected the last four letters and overwrote with &quot;AGE&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and also note that &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; this is consistent with the general &quot;negative tone&quot; of BROKE.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8875" LastActivityDate="2015-04-01T16:11:28.700" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11414" PostTypeId="2" ParentId="11408" CreationDate="2015-04-01T16:32:04.737" Score="5" Body="&lt;p&gt;I got the following&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;seesawed&lt;br&gt;&#xA;dessert&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;edit:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;reseesawed, but I am not sure if that is a word or not.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1504" LastEditorUserId="1504" LastEditDate="2015-04-01T16:47:02.597" LastActivityDate="2015-04-01T16:47:02.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11415" PostTypeId="2" ParentId="11408" CreationDate="2015-04-01T16:52:42.843" Score="17" Body="&lt;p&gt;I get 9 letters:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; REDRESSES/REDRESSED&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For non-standard words, regarding extinct words and new coinages with equal caution... 11 letters? 14? 16?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I've seen usage of &quot;freedressing&quot; (without regard to gender rules) so presumably someone who does this is a FREEDRESSER (11 letters). Of course, then specifying gender for such a person - a FREEDRESSERESS - might well be objectionable, so they might complain &quot;I can't believe you FREEDRESSERESSED me!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8875" LastEditorUserId="8875" LastEditDate="2015-04-02T16:43:26.877" LastActivityDate="2015-04-02T16:43:26.877" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11416" PostTypeId="2" ParentId="11389" CreationDate="2015-04-01T16:56:53.433" Score="11" Body="&lt;p&gt;The math is all great, above, but it still decries the common sense response:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I purchase two apples from the first vendor and three apples from the second vendor, I have purchased five apples for two Rs.  Therefore, five apples for two Rs should work.  This is, in fact, true if you do not have a limited number of apples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say ten people each come by the two brother's stands, and each of these ten people purchased two apples from the first vendor, and three apples from the second vendor.  At the end of this sequence, we have had 10 people purchase 50 apples, at five apples for two Rs each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HOWEVER!  The second vendor is now out of apples!  Only the first vendor still has apples remaining.  The next two people who want five apples each must purchase only at the first vendor's stand - and the first vendor is two apples for an Rs.  Therefore, they will spent 2.5 Rs for five apples, instead of 2 Rs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first ten people spent 2 Rs each for a total of 20 Rs.  The last two people spent 2.5 Rs each for a total of 5 Rs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when all 60 apples are combined for a total price of 5 apples for 2 Rs, those last two people spent 2 Rs each for a total of 4 Rs.  And THAT is where the missing Rs went.&lt;/p&gt;&#xA;" OwnerUserId="10914" LastActivityDate="2015-04-01T16:56:53.433" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11417" PostTypeId="1" AcceptedAnswerId="11419" CreationDate="2015-04-01T17:00:43.143" Score="17" ViewCount="1240" Body="&lt;p&gt;What’s special about each of these sentences? It's different for each sentence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Easy:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;A bad chef, mad like a jackal, made a damaged, half-baked milk cake a mile high.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Tomorrow we’ll see pretty yellow buttercups scattered across sunny green hills.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Your woozy son stoops to put your toy ox up onto your toy zoo.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Hard:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;A beefy chimp hops by, almost billowy in his loopy effort.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A nightclub academic stayed late, off dancing with Miami DJ funk.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="10851" LastEditorUserId="4204" LastEditDate="2015-04-01T22:11:25.130" LastActivityDate="2015-04-01T22:11:25.130" Title="Strange Short Stories" Tags="|pattern|english|language|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11418" PostTypeId="2" ParentId="11407" CreationDate="2015-04-01T17:01:07.913" Score="2" Body="&lt;p&gt;Alternative solutions are:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;He used the Onscreen Keyboard, so any word is valid. Note that the onscreen keyboard also has the delete button, so he wrote GARAGE then deleted it. Why did he do this? Maybe he was suspecting of her wife too...&lt;/li&gt;&#xA;&lt;li&gt;RAGE. He typed GARAGE, then selected GA with the mouse, right click and delete.&lt;/li&gt;&#xA;&lt;li&gt;He changed the password using another PC, GARAGE was just the password of his pornhub account...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Ok, this post isn't supposed to be the true answer to this nice problem, indeed others already found a brilliant solution. I just tried to lateral-think and let you have some fun :)&lt;/p&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-04-01T17:01:07.913" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11419" PostTypeId="2" ParentId="11417" CreationDate="2015-04-01T17:11:14.630" Score="24" Body="&lt;p&gt;Easy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1) All letters are in the first half of the alphabet (A-M).&lt;br&gt;&#xA; 2) Each of the words includes a doubled letter (tomoRRow, etc.)&lt;br&gt;&#xA; 3) All letters are in the second half of the alphabet (N-Z).&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hard:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1) The letters in each word are in alphabetical order.&lt;br&gt;&#xA; 2) The last letters of the words are A, B, C, etc.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9102" LastEditorUserId="1504" LastEditDate="2015-04-01T17:12:44.150" LastActivityDate="2015-04-01T17:12:44.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11420" PostTypeId="2" ParentId="11378" CreationDate="2015-04-01T18:01:32.713" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The game never completes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In every sequence in which the first logician has 2 hats and there are 8 hats in total, the first logician immediately knows his hat count.  If not, then there is at most one 1-hat logician among logicians 2-5.&#xA; &lt;p&gt;If logician 2 sees one hat on the head of logician 3, 4 or 5, he knows his own hat count is 2.  Otherwise, he says &quot;I don't know&quot;.&#xA; &lt;p&gt;If logician 2 said &quot;I don't know&quot;, then the other logicians all know they have 2 hats.  Otherwise, exactly one of them has 1 hat, and they can easily determine who.&#xA; &lt;p&gt;Since none of the other logicians made their decisions based on what logician 1's hat count actually was, he's stuck.  If logician 2 didn't figure out his own hat count in the first round, then he's stuck as well.&#xA; &lt;p&gt;Thus, this game can only complete when the actual hat count is 8 and the first logician has 2 hats (but according to the question, the actual hat count is 9).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All possible sequences:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 22222: ??222  1,2 stuck&#xA; &lt;br&gt;12222: ??222  1,2 stuck&#xA; &lt;br&gt;21222: ??222  1,2 stuck&#xA; &lt;br&gt;22122: ?2122  1 stuck&#xA; &lt;br&gt;22212: ?2212  1 stuck&#xA; &lt;br&gt;22221: ?2221  1 stuck&#xA; &lt;br&gt;11222: ??222  1,2 stuck&#xA; &lt;br&gt;12122: ?2122  1 stuck&#xA; &lt;br&gt;12212: ?2212  1 stuck&#xA; &lt;br&gt;12221: ?2221  1 stuck&#xA; &lt;br&gt;21122: 21122*&#xA; &lt;br&gt;21212: 21212*&#xA; &lt;br&gt;21221: 21221*&#xA; &lt;br&gt;22112: 22112*&#xA; &lt;br&gt;22121: 22121*&#xA; &lt;br&gt;22211: 22211*&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2015-04-01T18:01:32.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11421" PostTypeId="2" ParentId="10950" CreationDate="2015-04-01T18:21:49.533" Score="1" Body="&lt;p&gt;I know it's not exactly the spirit in this community, but if anybody's interested in solving the puzzle with a computer program, here's a nice Python code:&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;WHITE = 0&#xA;BLACK = 1&#xA;RED = 2&#xA;DONT_KNOW = 3&#xA;&#xA;COLOURS = [WHITE, BLACK, RED]&#xA;QUANTITIES = {WHITE: 2, BLACK: 2, RED: 1}&#xA;ANNOUNCEMENTS = {WHITE: 'White', BLACK: 'Black', RED: 'Red', DONT_KNOW: &quot;Don't know&quot;}&#xA;&#xA;&#xA;def get_answer(player, round, view, history):&#xA;    &quot;&quot;&quot;&#xA;    This function gets a player's identity, the round number, the three hats the player&#xA;    can see and the history of announcements, and returns the player's announcement.&#xA;&#xA;    player: 0, 1, 2 or 3&#xA;    round: a non-negative integer (0, 1, 2 ...)&#xA;    view: a list of four colours (WHITE, BLACK or RED)&#xA;          (the own player's colour is disregarded)&#xA;    history: a list of players, each one is a list of rounds,&#xA;             each one is the relevant announcement&#xA;    output: WHITE, BLACK, RED or DONT_KNOW&#xA;    &quot;&quot;&quot;&#xA;&#xA;    real_view = [view[p] for p in range(4) if p != player]&#xA;    view_count = dict([(colour, real_view.count(colour)) for colour in COLOURS])&#xA;&#xA;    possible_hats = [colour for colour in COLOURS&#xA;                     if view_count[colour] &amp;lt; QUANTITIES[colour]]&#xA;&#xA;    assert len(possible_hats), &quot;Bug - no possible hats.&quot;&#xA;&#xA;    if len(possible_hats) == 1:&#xA;        return possible_hats[0]&#xA;&#xA;    consistency = dict([(pos, True) for pos in possible_hats])&#xA;    for possible_hat in possible_hats:&#xA;        for player_check in range(4):&#xA;            expected_view = view[:]&#xA;            expected_view[player] = possible_hat&#xA;            for round_check in range(min(len(history[player_check]), round + 1)):&#xA;                if (round_check == round) and (player_check &amp;gt;= player):&#xA;                    break&#xA;                expected_announcement = get_answer(player_check, round_check,&#xA;                                                   expected_view, history)&#xA;                if expected_announcement != history[player_check][round_check]:&#xA;                    consistency[possible_hat] = False&#xA;                    break&#xA;            if not consistency[possible_hat]:&#xA;                break&#xA;&#xA;    assert sum(consistency.values()), &quot;Bug - no consistent choice.&quot;&#xA;&#xA;    if consistency.values() == [True, True]:&#xA;        return DONT_KNOW&#xA;&#xA;    return consistency.keys()[consistency.values()[1]]&#xA;&#xA;def play(hats, pr=True):&#xA;    &quot;&quot;&quot;&#xA;    This function gets the four distributed hats&#xA;    and returns the announcements by order.&#xA;&#xA;    hats: a list of four colours (WHITE, BLACK or RED)&#xA;          (not more than 2, 2 and 1 of them, respectively)&#xA;    &quot;&quot;&quot;&#xA;&#xA;    history = [[] for _ in range(4)]&#xA;    got_it = {}&#xA;&#xA;    player = 0&#xA;    round = 0&#xA;    oldest_dont_know = -1&#xA;    more_info = True&#xA;&#xA;    while more_info or (oldest_dont_know != player):&#xA;        if (not round) or (history[player][-1] == DONT_KNOW):&#xA;            answer = get_answer(player, round, hats, history)&#xA;            if pr:&#xA;                print &quot;Player #%d says: %s!&quot; % (player + 1, ANNOUNCEMENTS[answer])&#xA;            history[player].append(answer)&#xA;            if answer != DONT_KNOW:&#xA;                more_info = True&#xA;                got_it[player] = answer&#xA;            elif more_info:&#xA;                oldest_dont_know = player&#xA;                more_info = False&#xA;        player = (player + 1) % 4&#xA;        if not player:&#xA;            round += 1&#xA;&#xA;    return got_it&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;...and some outputs:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;gt;&amp;gt;&amp;gt; an1 = play([WHITE, WHITE, BLACK, BLACK])&#xA;Player #1 says: Don't know!&#xA;Player #2 says: White!&#xA;Player #3 says: Don't know!&#xA;Player #4 says: Black!&#xA;Player #1 says: White!&#xA;Player #3 says: Don't know!&#xA;&#xA;&amp;gt;&amp;gt;&amp;gt; an1 = play([WHITE, BLACK, BLACK, WHITE])&#xA;Player #1 says: Don't know!&#xA;Player #2 says: Don't know!&#xA;Player #3 says: Black!&#xA;Player #4 says: White!&#xA;Player #1 says: Don't know!&#xA;Player #2 says: Don't know!&#xA;&#xA;&amp;gt;&amp;gt;&amp;gt; an1 = play([WHITE, BLACK, WHITE, BLACK])&#xA;Player #1 says: Don't know!&#xA;Player #2 says: Don't know!&#xA;Player #3 says: White!&#xA;Player #4 says: Black!&#xA;Player #1 says: Don't know!&#xA;Player #2 says: Don't know!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As you see (and people have said), in the first scenario only player #3 can't figure out his own colour. In the two other scenarios only players #3 and #4 can.&lt;/p&gt;&#xA;" OwnerUserId="10684" LastActivityDate="2015-04-01T18:21:49.533" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11422" PostTypeId="2" ParentId="11378" CreationDate="2015-04-01T18:57:08.283" Score="-1" Body="&lt;p&gt;The game &lt;strong&gt;always&lt;/strong&gt; completes in this scenario(This would be more complicated if the possible number of hats were [7,8,9,10]).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No matter what the order, the first logician with 2 hats will immediately know that he has 2 hats if the number of hats is 8(he sees 2211). He'll reply &quot;don't know&quot; only if the total number of hats is either 9 or 10. Since, the true number is 9, he replies &quot;don't know&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the second person with 2 hats knows that the number is either 9 or 10. If the total number is 9, then the second 2 hatted person in the sequence will immediately know because he sees 7 hats in front of him and knows that total number is not 8. He'll reply &quot;don't know&quot; only if the total number of hats is 10, in which case the third two hatted person will immediately know the configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the game always completes. However if the total number of hats could be in the set {7,8,9,10}, things would be different.&lt;/p&gt;&#xA;" OwnerUserId="10923" LastActivityDate="2015-04-01T18:57:08.283" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11423" PostTypeId="2" ParentId="11378" CreationDate="2015-04-01T19:30:35.223" Score="-1" Body="&lt;p&gt;It always is completed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As soon as one two hat is unable to guess, the next two hat would know that they cannot be a one, as if they see a one and the previous two hat didn't say that they had two hats on they know that there cannot be two one hats. They would see the only one hat and immediately know that they are a two hat. The one hat would know that the total is not eight, and as soon as one person guesses the one hat will know he has to have one hat. Even if the total number of hats were ten this would still complete. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the third two hat will be able to guess their hat number, and from that point, all other logicians can deduce their number of hats due to the cyclical nature of the game.&lt;/p&gt;&#xA;" OwnerUserId="10927" LastEditorUserId="10927" LastEditDate="2015-04-02T23:20:40.533" LastActivityDate="2015-04-02T23:20:40.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11424" PostTypeId="2" ParentId="11408" CreationDate="2015-04-01T19:57:56.797" Score="6" Body="&lt;h2&gt;10 LETTERS&lt;/h2&gt;&#xA;&lt;p&gt;I found one with 10 letters but I can't find it in the standard dictionaries. It seems to be an older English word as the modern version drops the gender-specific bit.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; DESERTRESS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Otherwise, the 9 letter versions already noted are the longest I can find.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;12 LETTERS&lt;/h2&gt;&#xA;&lt;p&gt;Or, as pointed out in by EagleV_Attnam in a comment on mark's post, here's a 12 letter option that should have been obvious:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; DESERTRESSES&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-02T13:36:04.583" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11425" PostTypeId="2" ParentId="11408" CreationDate="2015-04-01T20:15:28.560" Score="19" Body="&lt;p&gt;I liked Engineer Toast's idea of finding the longest word you could type with right-hand only letters, so I found those as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These words come from my computer's /usr/share/dict/words file, you can decide for yourself which words count as English.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;All keys&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (10, 'desertress')&lt;br&gt;&#xA; (10, 'assertress')&lt;br&gt;&#xA; (9, 'redresser')&lt;br&gt;&#xA; (8, 'kolokolo')&lt;br&gt;&#xA; (8, 'assessee')&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Left keys&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (10, 'desertress')&lt;br&gt;&#xA; (10, 'assertress')&lt;br&gt;&#xA; (9, 'redresser')&lt;br&gt;&#xA; (8, 'assessee')&lt;br&gt;&#xA; (8, 'assessed')&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Right keys&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (8, 'kolokolo')&lt;br&gt;&#xA; (6, 'pookoo')&lt;br&gt;&#xA; (6, 'polloi')&lt;br&gt;&#xA; (6, 'lollop')&lt;br&gt;&#xA; (5, 'plook')  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I used the following code to come up with this answer:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;LEFT_KEYS = {&#xA;    'Q':['Q', 'W', 'A'],&#xA;    'W':['W', 'Q', 'A', 'S', 'E'],&#xA;    'E':['E', 'W', 'S', 'D', 'R'],&#xA;    'R':['R', 'E', 'D', 'F', 'T'],&#xA;    'T':['T', 'R', 'F', 'G', 'Y'],&#xA;    'A':['A', 'Q', 'Z', 'W', 'S'],&#xA;    'S':['S', 'W', 'A', 'Z', 'X', 'D', 'E'],&#xA;    'D':['D', 'E', 'S', 'X', 'C', 'F', 'R'],&#xA;    'F':['F', 'R', 'D', 'C', 'V', 'G', 'T'],&#xA;    'G':['G', 'T', 'F', 'V', 'B', 'H', 'Y'],&#xA;    'Z':['Z', 'A', 'S', 'X'],&#xA;    'X':['X', 'S', 'Z', 'D', 'C'],&#xA;    'C':['C', 'D', 'X', 'F', 'V'],&#xA;    'V':['V', 'F', 'C', 'B', 'G'],&#xA;    'B':['B', 'G', 'V', 'N', 'H']&#xA;}&#xA;&#xA;RIGHT_KEYS = {&#xA;    'Y':['Y', 'T', 'G', 'H', 'U'],&#xA;    'U':['U', 'Y', 'H', 'J', 'I'],&#xA;    'I':['I', 'U', 'J', 'K', 'O'],&#xA;    'O':['O', 'I', 'K', 'L', 'P'],&#xA;    'P':['P', 'O', 'L'],&#xA;    'H':['H', 'Y', 'G', 'B', 'N', 'J', 'U'],&#xA;    'J':['J', 'U', 'H', 'N', 'M', 'K', 'I'],&#xA;    'K':['K', 'I', 'J', 'M', 'L', 'O'],&#xA;    'L':['L', 'O', 'K', 'P'],&#xA;    'N':['N', 'H', 'B', 'M', 'J'],&#xA;    'M':['M', 'J', 'N', 'K']&#xA;}&#xA;&#xA;ALL_KEYS = LEFT_KEYS.copy()&#xA;ALL_KEYS.update(RIGHT_KEYS)&#xA;&#xA;def adjacent(keyset, word):&#xA;    lastletter = None&#xA;    for letter in word.upper():&#xA;        if lastletter:&#xA;            if not letter in keyset.get(lastletter, []):&#xA;                return False&#xA;        lastletter = letter&#xA;    return True&#xA;&#xA;def read_word_list():&#xA;    with open('/usr/share/dict/words') as wordfile:&#xA;        return [w.strip() for w in wordfile.readlines() if len(set(w.strip())) &gt; 1]&#xA;&#xA;WORDS = read_word_list()&#xA;ADJACENT_WORDS = []&#xA;for WORD in WORDS:&#xA;    if adjacent(ALL_KEYS, WORD):&#xA;        ADJACENT_WORDS.append((len(WORD), WORD))&#xA;&#xA;for WORD in sorted(ADJACENT_WORDS)[-5:]:&#xA;    print WORD&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="10926" LastActivityDate="2015-04-01T20:15:28.560" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11427" PostTypeId="2" ParentId="11409" CreationDate="2015-04-01T20:53:17.467" Score="6" Body="&lt;p&gt;Pretty sure the answer will be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; April fool or April fool's day, though I haven't yet got all the clues&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You call me a liar, and say I was the first.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; April fools jokes aren't true (liar), first here refers to the first if the month&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I was the second, the first liar isn't cursed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; April fool's day started (allegedly) after new year's day moved to January first(?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll open, open, and tell you what I see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reference to the &quot;oo&quot; in fool(?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I start to fight, you'd think I start with a knee.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No idea....&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whether you think I'm cool, or whether I be lame,&#xA;You'll often find me surrounded by many o' my dame.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People either love or hate April fool's day. And there's lots of pranks that day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A couple types of clues, riddle now you must,&#xA;And tell me what I am, before I turn to dust.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Need to answer while it's still April fool's day&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastActivityDate="2015-04-01T20:53:17.467" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11428" PostTypeId="2" ParentId="11380" CreationDate="2015-04-01T20:54:29.167" Score="3" Body="&lt;p&gt;I disagree with Martin's answer. My answer would be: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3/4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Day 2 only shows up once (12/2), and day 7 only shows up once (6/7). Days 1, 4, 5, 8 each show up at least twice. Jack says that Jill surely doesn't know, therefore $D \ne 2$ and $D \ne 7$: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;3/4 | 3/5 | 3/8 | 6/4 | ̶6̶/̶7̶ | 9/1 | 9/5 | 12/1 | ̶1̶2̶/̶2̶ | 12/8&lt;/code&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;He still doesn't know the answer from this (we're still on his first statement saying neither of them knows). Only one month shows up once, 6. Since he still doesn't know, $M \ne 6$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;3/4 | 3/5 | 3/8 | ̶6̶/̶4̶ | ̶6̶/̶7̶ | 9/1 | 9/5 | 12/1 | ̶1̶2̶/̶2̶ | 12/8&lt;/code&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Jill says she didn't know before but now does. At this point day 4 is the only day showing up once (3/4). So she can deduce that $M = 3$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Now that Jack knows Jill is positive of her day (4 is the only possible answer that can lead to that) he can deduce that $D = 4$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="10931" LastEditorUserId="10931" LastEditDate="2015-04-01T21:13:38.737" LastActivityDate="2015-04-01T21:13:38.737" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11429" PostTypeId="1" AcceptedAnswerId="11430" CreationDate="2015-04-01T22:26:26.453" Score="55" ViewCount="6405" Body="&lt;p&gt;Here's a little maze puzzle I originally built a couple of years ago, that seems apropos to reprise now:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you make it from the A in the top left of this grid to the Z in the bottom right, always going either up one letter (for instance, A to B or G to H) or down one letter (for instance, N to M)? The alphabet wraps around, so you can go from Z up to A or A down to Z too. Try as hard as you can (and remember that you can always work backwards if you get stuck forwards), and see where you get!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dBrs8.png&quot; alt=&quot;Letter Maze&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A small hint for those who are wondering about any possible invalidities in the grid:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Solving the &lt;em&gt;maze&lt;/em&gt; is not the same thing as solving the &lt;em&gt;puzzle&lt;/em&gt;.  Read those instructions carefully!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="266" LastEditorUserId="266" LastEditDate="2015-04-02T01:29:12.083" LastActivityDate="2016-09-01T09:58:26.887" Title="A maze puzzle for the day" Tags="|mazes|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11430" PostTypeId="2" ParentId="11429" CreationDate="2015-04-01T22:55:54.193" Score="67" Body="&lt;p&gt;The maze is invalid. However, the point of this puzzle is to find out that it is invalid, and a little something extra. I have made all the connections I can find.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The dead ends of the start path have the letters A P R I L.&lt;br&gt;&#xA; The dead ends of the end path have the letters F O O L S.&lt;br&gt;&#xA; &lt;strong&gt;The puzzle is an April Fools joke!&lt;/strong&gt;&lt;br&gt;&#xA; &lt;em&gt;P.S. The J in [row 6, column 9] could be an L, and it would be a valid maze with none of the paths disconnected from the main route.&lt;/em&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/Pfaac.png&quot; alt=&quot;April Fools!&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9319" LastEditorUserId="9319" LastEditDate="2015-04-01T23:37:54.617" LastActivityDate="2015-04-01T23:37:54.617" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11431" PostTypeId="2" ParentId="11429" CreationDate="2015-04-01T23:02:16.453" Score="15" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can't pass the red line,also checked diagonal.&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/y9JQQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9543" LastEditorUserId="26071" LastEditDate="2016-09-01T09:58:26.887" LastActivityDate="2016-09-01T09:58:26.887" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11432" PostTypeId="2" ParentId="11429" CreationDate="2015-04-01T23:07:03.160" Score="26" Body="&lt;p&gt;Given that (as shown in two other answers), there is no normal path, I extrapolate that this may be the solution to the maze.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Allowing wrap-around in the maze as well. Though if this is a valid solution, it would be nice if the instructions said something about wrap-arounds being allowed. If the below is a solution, then note that you can also make it wrap-around 1 or 2 rows further down (from the Z or the other A). &lt;img src=&quot;https://i.stack.imgur.com/txw5d.png&quot; alt=&quot;maze path&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastEditorUserId="1620" LastEditDate="2015-04-01T23:08:51.527" LastActivityDate="2015-04-01T23:08:51.527" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="11433" PostTypeId="2" ParentId="9321" CreationDate="2015-04-01T23:33:17.700" Score="4" Body="&lt;p&gt;I assume the OP is still waiting for the key to be identified.  The correct key is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;022015&lt;/strong&gt; which is the month and year that are &quot;attached to this question&quot;.&lt;br&gt;&#xA;&lt;br&gt;&#xA;Using numbers as the key makes this a Gronsfeld cipher instead of a Vigenère cipher.&lt;br&gt;&#xA;&lt;br&gt;&#xA;The result is the same (with a double &quot;ed&quot;): thisisanencod&lt;strong&gt;eded&lt;/strong&gt;message  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastActivityDate="2015-04-01T23:33:17.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11435" PostTypeId="1" AcceptedAnswerId="11454" CreationDate="2015-04-02T00:16:06.327" Score="14" ViewCount="465" Body="&lt;p&gt;This was found engraved into a stone in Roman times:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/aD7Vs.png&quot; alt=&quot;ukoduS&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And underneath, an inscription:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;eoofrddene·en.do·xfrtfscqqiiut·iucraeiuioceaarriocsrt·s·ieaei,imt·oesa·serism?·,·&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Even with only this minimal information, I'm sure you will be able to answer one simple question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who wrote it?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="7308" LastActivityDate="2015-04-02T11:34:40.513" Title="The Roman Crypto" Tags="|cipher|logic-grid|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11437" PostTypeId="2" ParentId="11377" CreationDate="2015-04-02T04:47:04.190" Score="1" Body="&lt;p&gt;I Guess it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; X - coordinate ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have traveled Millennium distances&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In graphs it travel Millennium distances, no limit of it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Treat me with longhorns on my head&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Also X have longhorns on head&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;when you able to see Jupiter in sky,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and wander why sky color is blue,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure but you are able to see jupiter from earth in x - coordinate&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4312" LastActivityDate="2015-04-02T04:47:04.190" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11438" PostTypeId="1" AcceptedAnswerId="11447" CreationDate="2015-04-02T04:51:26.187" Score="13" ViewCount="1035" Body="&lt;p&gt;What musical term does this refer to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;ABCDEFGHIJKLMNOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO.........&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clue for help:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Maybe I should write that as: A B C D E F G H I J K L M N O O O O O O O O O O O O . . . . . . . . .&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10940" LastEditorUserId="8874" LastEditDate="2015-11-09T17:53:55.470" LastActivityDate="2015-11-09T17:53:55.470" Title="What musical term does this refer to?" Tags="|word|rebus|music|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11439" PostTypeId="2" ParentId="11409" CreationDate="2015-04-02T05:53:57.320" Score="2" Body="&lt;p&gt;In my opinion the answer may be :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A Girl-Friend&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You call me a liar, and say I was the first :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We tend to say to our girl friend , that are the first love or first time(Sex)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll open, open, and tell you what I see :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;They tend to open up and express what they feel and see.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I start to fight, you'd think I start with a knee.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;When ever in fight, they may tend to use their knee(Probably hit the groin area with the knee)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whether you think I'm cool, or whether I be lame,&#xA;You'll often find me surrounded by many o' my dame.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Whether they are cool or lame, they ll be surrounded with dame - Dame is the female equivalent of the honor of knighthood in the British honors system (According to Wiki)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And tell me what I am, before I turn to dust :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Probably turn to dust may mean, that they ll simply leave off eventually if we don't value them&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know this may probably be wrong and weird,But seems funny and kind of relevant if you think about it. :) &lt;/p&gt;&#xA;" OwnerUserId="10943" LastEditorUserId="10943" LastEditDate="2015-04-03T09:59:27.910" LastActivityDate="2015-04-03T09:59:27.910" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11440" PostTypeId="2" ParentId="11377" CreationDate="2015-04-02T06:36:00.500" Score="2" Body="&lt;p&gt;The keywords seem to fit&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Microsoft Windows&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Asteroid, Millennium, Longhorn, Jupiter&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;All are Windows code names: 2000 SP1 (Asteroid), ME (Millennium edition), Vista (Longhorn), 8 UI (Jupiter).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9422" LastEditorUserId="9422" LastEditDate="2015-04-02T06:43:53.567" LastActivityDate="2015-04-02T06:43:53.567" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11442" PostTypeId="2" ParentId="1797" CreationDate="2015-04-02T06:51:36.153" Score="12" Body="&lt;p&gt;The number is:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 87. Just rotate the image 180 degrees:&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/dJC2c.jpg&quot; alt=&quot;The original image rotated 180 degrees&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9626" LastEditorUserId="60644" LastEditDate="2023-02-07T14:49:56.283" LastActivityDate="2023-02-07T14:49:56.283" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="11443" PostTypeId="2" ParentId="11377" CreationDate="2015-04-02T06:51:42.403" Score="2" Body="&lt;p&gt;Treat me &lt;strong&gt;baby who Abode of the good&lt;/strong&gt;-&gt;Bible-&gt;&lt;strong&gt;Memphis&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But don't think my origin is &lt;strong&gt;asteroid&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but believe me, i am not that alien&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have traveled &lt;strong&gt;Millennium&lt;/strong&gt; distances&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but that may not &lt;strong&gt;whistler&lt;/strong&gt; in your ears&lt;/p&gt;&#xA;&#xA;&lt;p&gt;once you know, i am not that alien&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Treat me with &lt;strong&gt;longhorns&lt;/strong&gt; on my head&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with &lt;strong&gt;black&lt;/strong&gt; color &lt;strong&gt;comb&lt;/strong&gt;s for my hairs&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but believe me, i am not that alien&lt;/p&gt;&#xA;&#xA;&lt;p&gt;when you able to see &lt;strong&gt;Jupiter&lt;/strong&gt; in sky,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and wander why sky color is &lt;strong&gt;blue&lt;/strong&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you will come to know who am i,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but believe me, i am not that alien&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then who am i ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All are &lt;strong&gt;Microsoft windows codenames&lt;/strong&gt; in series&lt;/p&gt;&#xA;" OwnerUserId="10433" LastActivityDate="2015-04-02T06:51:42.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11444" PostTypeId="2" ParentId="11438" CreationDate="2015-04-02T07:31:09.843" Score="1" Body="&lt;p&gt;It may be  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A music-disc that went in loop. It was spelling the alphabet, when suddenly something went wrong!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Otherwise it may be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A &lt;a href=&quot;http://en.wikipedia.org/wiki/Loop_(music)&quot; rel=&quot;nofollow&quot;&gt;loop&lt;/a&gt; preceded by a &lt;a href=&quot;http://en.wikipedia.org/wiki/Scale_(music)&quot; rel=&quot;nofollow&quot;&gt;scale&lt;/a&gt;. The scale is made with the notes from A to N (A to G notes are the standard ones, H to N may be the next octave).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-04-02T07:31:09.843" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11445" PostTypeId="2" ParentId="11408" CreationDate="2015-04-02T07:33:55.650" Score="11" Body="&lt;p&gt;FREEWEED&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best solution I can think of.&lt;/p&gt;&#xA;" OwnerUserId="10949" LastActivityDate="2015-04-02T07:33:55.650" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11447" PostTypeId="2" ParentId="11438" CreationDate="2015-04-02T08:22:26.530" Score="20" Body="&lt;p&gt;I would say the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Staccato&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because it sounds like&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;'Stuck at O'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7308" LastActivityDate="2015-04-02T08:22:26.530" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11448" PostTypeId="2" ParentId="10575" CreationDate="2015-04-02T08:54:21.827" Score="1" Body="&lt;p&gt;Header has a winning strategy when n is even, ending in T.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For notation purposes, a sequence of H's or T's can be reduced to a single H or T. The row can be simplified as an alternating pattern of H and T.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flipping a middle sequence of H or T reduces the pattern by a TH (or HT) pair.&#xA;For example, HTHTH becomes HHHTH, HTTTH or HTHHH. These three are all equivalent to HTH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flipping an end (when allowed), removes a letter from the pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Header's winning strategy: Reduce the row to THTHT&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if Tailer plays T, (TTHTHT is the same as THTHT) and flips a middle sequence, the row is reduced to THT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then Header plays H, making it THTH and flips a middle sequence, resulting in TH. At this point the H is recently added and cannot be flipped at Tailer's turn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless of what Tailer plays next, the T sequence has to be flipped, making Header the winner.&lt;/p&gt;&#xA;" OwnerUserId="10667" LastActivityDate="2015-04-02T08:54:21.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11449" PostTypeId="2" ParentId="11298" CreationDate="2015-04-02T09:31:38.757" Score="9" Body="&lt;p&gt;It is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Money&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I murder with ease and comfort,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if you have more money than life will be easy, cause no disease ,drug to the eye&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a drug to the eye,&#xA;And to the heart.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Money is drug to eyes, and its feel good in heart if have money&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Are also those who detest ardently&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People get jealous if you have more money. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You love me like nobody else&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; everyone loves money&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not because I murdered your loved ones,&#xA;But because I am a drug to the eye&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some Peoples kills their loved once due to money, although they love money.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cause no disease&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; of course, it cause no disease&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a drug to the eye,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And to the heart&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; feels good to see money , so we can say its drug of eye and heart.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As per Hint , Money have &lt;strong&gt;physical existence&lt;/strong&gt; also.&lt;/p&gt;&#xA;" OwnerUserId="4312" LastEditorUserId="4312" LastEditDate="2015-05-15T10:05:47.140" LastActivityDate="2015-05-15T10:05:47.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11450" PostTypeId="1" AcceptedAnswerId="11452" CreationDate="2015-04-02T10:01:53.467" Score="9" ViewCount="3123" Body="&lt;p&gt;There are $n$ couples on one side of the river. Nobody is on the other side. They have a boat that accommodates up to 2 people. For every trip across, someone(s) must bring the boat back as well. The aim is for all of them to cross the river. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one important rule:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;No woman can be left (in the boat or either shore) with a man,&#xA;  unless her husband is also present.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are no tricks, such as divorcing during the game, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Give a formula to find the minimum number of boat trips required.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4269" LastEditDate="2015-04-04T14:57:39.640" LastActivityDate="2016-05-31T14:44:41.793" Title="$n$ couples crossing a river" Tags="|mathematics|optimization|river-crossing|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="11451" PostTypeId="2" ParentId="11175" CreationDate="2015-04-02T10:05:13.160" Score="14" Body="&lt;p&gt;The first one&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;writes the first 25 digits of $\pi$ in a spiral, beginning with 3 in the middle and moving off upwards and clockwise.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second does the same for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$e$, beginning with 2 in the middle and moving off upwards and counter-clockwise.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The property is that they&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;represent &lt;a href=&quot;http://en.wikipedia.org/wiki/Irrational_number&quot;&gt;irrational numbers&lt;/a&gt; in &lt;a href=&quot;http://en.wikipedia.org/wiki/Ulam_spiral&quot;&gt;Ulam-type spirals&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To generate more,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;pick other irrational numbers and do the same.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's one for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Feigenbaum_constants#The_first_constant&quot;&gt;Feigenbaum's $\delta$&lt;/a&gt; (4.669201609102990671853203...), starting off to the left:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&#xA;3  0  2  3  5&#xA;9  0  6  1  8&#xA;1  6  4  0  1&#xA;0  6  9  2  7&#xA;2  9  9  0  6&#xA;&lt;/pre&gt;&#xA;" OwnerUserId="7363" LastEditorUserId="7363" LastEditDate="2015-04-02T14:54:19.740" LastActivityDate="2015-04-02T14:54:19.740" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11452" PostTypeId="2" ParentId="11450" CreationDate="2015-04-02T10:25:52.793" Score="15" Body="&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; This answers the question before the question was changed from &quot;left alone ... with&quot; to just &quot;left ... with&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definitions&lt;/strong&gt;: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;trip across&lt;/code&gt; means going from the original side to the destination side&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;woman left alone with a man&lt;/code&gt; means those individuals are the only two people in that place.&#xA;&lt;br&gt;&lt;br&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Send all the women across first. This requires $n-1$ trips, leaving the boat at the destination. One woman (call her &lt;code&gt;W&lt;/code&gt;) returns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Send all the men across. This requires $n-1$ trips, leaving the boat at the destination. &lt;code&gt;W&lt;/code&gt;'s husband returns and both go across (1 trip).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Total $2n-1$ trips across to the other side. Each trip across is matched with a trip back except for the last trip across, so we have $4n-3$ river crossings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the first phase, every woman is in the company of either only women or with her husband. In the second phase, W remains with her husband or is alone, and all other women are with at least the number of people they were with in the first phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is minimal since each pair of river crossings can deposit only 1 person at the destination if someone must return with the boat. After $2n-2$ pairs of river crossings (and hence $2n-2$ trips across), only 2 people remain on the original side. They take the final trip across to total $2n-1$ trips across.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;With the reworded question, we can still have the same number of river crossings with a different order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;left with&quot; rule is taken to mean that a woman can be in a boat on her own or with another woman, or on a shore with her husband (or with only women). In particular, a woman is permitted to cross to the shore opposite her husband's, provided she does not disembark and provided she returns with another woman.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pair each man $M_i$ with his wife $W_i$ for $i \in [1,n]$. Starting with $i=1$ and working through to $i=n$, send $M_i,W_i$ across, $W_i$ returns, picks up $W_{i+1}$ and goes across again. Then $W_{i+1}$ returns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each trip across deposits one person, so we still have $2n-1$ trips across and $4n-3$ river crossings.&lt;/p&gt;&#xA;" OwnerUserId="9422" LastEditorUserId="15544" LastEditDate="2015-08-26T09:07:11.600" LastActivityDate="2015-08-26T09:07:11.600" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="11454" PostTypeId="2" ParentId="11435" CreationDate="2015-04-02T11:22:52.970" Score="10" Body="&lt;p&gt;Who wrote it?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Catullus&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. The numbers on the Roman stone form a Sudoku.  Solve it.&lt;br&gt;&#xA; 2. The inscription underneath consists of 81 symbols. Bring it into a 9x9 square.&lt;br&gt;&#xA; 3. Match the occurrences of &quot;I&quot; in the sudoku with the corresponding letters in the 9x9 inscription square. Read them one by one, from top to bottom. This yields &quot;odi.et.am&quot;.&lt;br&gt;&#xA; 4. Do the same for the occurrences of &quot;II&quot;. This yields &quot;o.quare.i&quot;.&lt;br&gt;&#xA; 5. Repeat for III, IV, V, VI, VII, VIII, IX.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The resulting sentence is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;odi.et.amo.quare.id.faciam,fortasse.requiris?nescio,sed.fieri.sentio.et.excrucior&lt;br&gt;&#xA; respectively:&lt;br&gt;&#xA; &lt;strong&gt;&quot;Odi et amo. quare id faciam, fortasse requiris? nescio, sed fieri sentio et excrucior.&quot;&lt;/strong&gt;&lt;br&gt;&#xA; This is the famous poem 85 by Catullus.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="9125" LastEditDate="2015-04-02T11:34:40.513" LastActivityDate="2015-04-02T11:34:40.513" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11457" PostTypeId="2" ParentId="11409" CreationDate="2015-04-02T13:17:11.963" Score="2" Body="&lt;p&gt;Along similar lines to Khan's answer (though I hadn't seen that when I had this idea), how about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;a LOCK?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First four lines are letterplay, as suggested by the hints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You call me a liar, and say I was the first.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First letter of the word &quot;liar&quot; is &lt;strong&gt;L&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But I was the second, the first liar isn't cursed.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm not sure about this line, but...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I'll open, open, and tell you what I see.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;See&quot; is a homophone of &lt;strong&gt;C&lt;/strong&gt;; the reference to opening also goes with the lock idea.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If I start to fight, you'd think I start with a knee.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first letter of the word &quot;knee&quot; is &lt;strong&gt;K&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Whether you think I'm cool, or whether I be lame,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Cool&quot; backwards is almost &quot;lock&quot;. Not sure about &quot;lame&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You'll often find me surrounded by many o' my dame.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Could be a reference to the Num Lock, Caps Lock, Scroll Lock keys on a computer (though I realise this is a bit of a stretch).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A couple types of clues, riddle now you must,&lt;br&gt;&#xA; And tell me what I am, before I turn to dust.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are both wordplay and metaphorical clues in the riddle above.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-04-02T13:17:11.963" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11458" PostTypeId="2" ParentId="9734" CreationDate="2015-04-02T13:31:26.047" Score="9" Body="&lt;p&gt;He was stealing reputation. He showed up his employers' competitors as incompetent buffoons who would employee a paranoid and easily panicked security company who would end up wasting the time of all the convention guests. In short, he was a professional provocateur.&lt;/p&gt;&#xA;" OwnerUserId="10665" LastActivityDate="2015-04-02T13:31:26.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11459" PostTypeId="2" ParentId="11409" CreationDate="2015-04-02T14:10:52.583" Score="2" Body="&lt;p&gt;I think it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Satan&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You call me a liar, and say I was the first.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most people think Satan was the first liar&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I was the second, the first liar isn't cursed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people say that God was the first liar because he said “But as for the tree of the knowledge of good and bad you must not eat from it, for in the day you eat from it you will positively die,” but then neither Adam nor Eve dies THAT DAY. However, I am no believer, but I still understood it to mean that on that day, their state would be changed from &quot;never going to die&quot; to &quot;will now eventually die&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jehovahswitnessblog.com/watchtower-lies/who-was-the-first-liar-in-the-bible/&quot; rel=&quot;nofollow&quot;&gt;http://www.jehovahswitnessblog.com/watchtower-lies/who-was-the-first-liar-in-the-bible/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll open, open, and tell you what I see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The devil, in a sense, by offering them the forbidden fruit, opened the eyes of Adam and Eve to be able to recognize good and evil. The Devil said, &quot;For God knows that in the day you eat from it your eyes will be opened, and you will be like God, knowing good and evil.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I start to fight, you'd think I start with a knee.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since Satan was in the form of a serpent, maybe it means that he will strike at your knee, or any part of your lower leg?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whether you think I'm cool, or whether I be lame,&lt;br&gt;&#xA;You'll often find me surrounded by many o' my dame.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This seems to fit Eve more than Satan. I guess Eve will be around other ladies, who would also be her daughters and nieces&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A couple types of clues, riddle now you must,&lt;br&gt;&#xA;And tell me what I am, before I turn to dust.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This also seems like it would refer to Adam or Eve, not Satan. Part of God's punishment of Adam and Eve was, &quot;By the sweat of your brow you will eat your food until you return to the ground, since from it you were taken; for dust you are and to dust you will return.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-04-02T14:17:12.630" LastActivityDate="2015-04-02T14:17:12.630" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11460" PostTypeId="2" ParentId="575" CreationDate="2015-04-02T15:14:16.090" Score="3" Body="&lt;p&gt;Another way of looking at it, this image looks much like the grill or front face of the bus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that in mind I would answer that it is going neither to the left nor the right from my perspective. Relative to me, the bus is advancing directly at me.&lt;/p&gt;&#xA;" OwnerUserId="10972" LastEditorUserId="10972" LastEditDate="2015-04-02T16:40:53.703" LastActivityDate="2015-04-02T16:40:53.703" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11461" PostTypeId="2" ParentId="11409" CreationDate="2015-04-02T15:33:13.523" Score="2" Body="&lt;p&gt;I guess the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SIN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You call me a liar, and say I was the first.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lying is a SIN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I was the second, the first liar isn't cursed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Here &quot;I&quot; refers to actual word &quot;I&quot;. So that means it is second letter of the word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll open, open, and tell you what I see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cannot really make any sense of this in regard to my answer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I start to fight, you'd think I start with a knee.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A SIN makes you weak and you feel &quot;WEAK&quot; at the knees&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Just tried to focus on first four lines as you hinted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT..&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess the 3rd Letter is  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;O&quot; [I'll open, open]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am guessing the word is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LION&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I am not quite sure as I cannot relate to few clues.&lt;/p&gt;&#xA;" OwnerUserId="10973" LastEditorUserId="10973" LastEditDate="2015-04-02T17:45:04.477" LastActivityDate="2015-04-02T17:45:04.477" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11462" PostTypeId="2" ParentId="9734" CreationDate="2015-04-02T15:49:50.630" Score="6" Body="&lt;p&gt;I know the &quot;answer&quot; is ideas but here's a different angle. He was stealing the attention of Bob and the security guards, letting possible accomplices do the work for him. Just as a magician steals your attention when you perform a magic trick - because you are concentrating so hard on trying to see the crux of the trick you never get to see the wood for the trees.&lt;/p&gt;&#xA;" OwnerUserId="10974" LastActivityDate="2015-04-02T15:49:50.630" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11463" PostTypeId="2" ParentId="11409" CreationDate="2015-04-02T15:58:00.247" Score="2" Body="&lt;p&gt;Second try:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LINK.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You call me a liar, and say I was the first.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As before, the first letter of the word &quot;liar&quot; is &lt;strong&gt;L&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But I was the second, the first liar isn't cursed.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;I&lt;/strong&gt; is the second letter (credit to Deepak for this idea).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I'll open, open, and tell you what I see.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you click a link on a webpage, it opens to a new page and you see new information. (Possibly this line also gives the letter &lt;strong&gt;L&lt;/strong&gt;.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If I start to fight, you'd think I start with a knee.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As before, the first letter of the word &quot;knee&quot; is &lt;strong&gt;K&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Whether you think I'm cool, or whether I be lame,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When someone provides you with a link to a webpage, you may think it either cool or lame.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You'll often find me surrounded by many o' my dame.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many links in a chain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A couple types of clues, riddle now you must,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As before, there are both wordplay and metaphorical clues in the riddle above.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And tell me what I am, before I turn to dust.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm pretty sure there's some kind of religious or mythological reference to chains crumbling to dust, but I can't bring the exact quote to mind.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-04-02T15:58:00.247" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11465" PostTypeId="2" ParentId="575" CreationDate="2015-04-02T16:29:46.597" Score="2" Body="&lt;p&gt;I know this has been answered already. But thought I'd add my opposite thought.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under the assumption that you haven't just stepped into the middle of the autobahn, and that the bus is heading in the correct direction. The bus must be on the side of the road closest to you, otherwise you would have a crash barrier in the way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if this was the case then the bus would be moving right. And the bus evidently comes from the UK and took the wrong exit when heading onto the autobahn (hence the door not being on the right side)&lt;/p&gt;&#xA;" OwnerUserId="10636" LastEditorUserId="-1" LastEditDate="2015-07-01T14:15:32.037" LastActivityDate="2015-07-01T14:15:32.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11466" PostTypeId="2" ParentId="11238" CreationDate="2015-04-02T16:32:20.540" Score="-1" Body="&lt;p&gt;He has an affair with a woman. He plans to kill his wife. Poisoned her food and went. But felt repentant in the elevator as it is stuck due to power failure.&lt;/p&gt;&#xA;" OwnerUserId="10911" LastActivityDate="2015-04-02T16:32:20.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11468" PostTypeId="2" ParentId="11450" CreationDate="2015-04-02T17:16:48.977" Score="0" Body="&lt;p&gt;One way to keep couples together would be this process:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Husband A ferries Wife A.&lt;br&gt;&#xA;Husband B ferries Husband A.&lt;br&gt;&#xA;Husband B ferries Wife B.&lt;br&gt;&#xA;Husband C ferries Husband B.&lt;br&gt;&#xA;Etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the only time a woman is alone with a man is:&lt;br&gt;&#xA;On the far shore, Wife A and Husband A.&lt;br&gt;&#xA;On the near short, Wife N and Husband N, before Husband N ferries Husband N-1.&lt;br&gt;&#xA;On the boat, each wife is ferried by their husband. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for total trips, there are $4n-3$ total trips across. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&#xA;Surprisingly, if we reverse the genders for the solution, the clause about women being without their husbands is fulfilled. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-04-06T12:47:53.080" LastActivityDate="2015-04-06T12:47:53.080" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11470" PostTypeId="2" ParentId="11117" CreationDate="2015-04-02T18:09:30.140" Score="7" Body="&lt;h2&gt;NOT AN ANSWER&lt;/h2&gt;&#xA;&lt;p&gt;This is not intended to be an answer. The content is simply too involved for a comment. This may all be useless.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;First, here's a little VBA UDF you can pop into Excel to have it evaluate the formula. If you plug in the whole formula as an equality such as they are given in the examples, it'll return &lt;code&gt;TRUE&lt;/code&gt; or &lt;code&gt;FALSE&lt;/code&gt;. If you plug in just one side of the formula, it'll return the solution.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;Public Function eval(s As Range) As String&#xA;    eval = Application.Evaluate(s.Value)&#xA;End Function&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Put it in a module and call it in a cell with a formula like &lt;code&gt;=eval(A1)&lt;/code&gt;. This is probably the only piece of this posting that'll be helpful.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;Regarding the Sums of Twos&lt;/h2&gt;&#xA;&lt;p&gt;The answers for the examples that are all sums of twos show a pattern when plotted on a logarithmic scale.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/x8Zw9.png&quot; alt=&quot;Chart of 2s, Log Scale&quot; /&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;Regarding the &lt;code&gt;9=5&lt;/code&gt; Example&lt;/h2&gt;&#xA;&lt;p&gt;As of the time of this posting, there are 72 example formulas given.&lt;/p&gt;&#xA;&lt;p&gt;24 / 72 evaluate as true on a normal calculator.&lt;/p&gt;&#xA;&lt;p&gt;28 / 72 evaluate as true if you replace all the 9s with 5s&lt;/p&gt;&#xA;&lt;p&gt;There are 6 that are true originally but become false if you replace the 9s and 2 that are true but become false.&lt;/p&gt;&#xA;&lt;p&gt;The examples in Hint 2 become 3 unique examples with each repeated twice if you replace the 9s with 5s.&lt;/p&gt;&#xA;&lt;p&gt;The calculator may or may not interpret every 9 keypress as a 5&lt;/p&gt;&#xA;&lt;p&gt;The only answers with a 9 in them are &lt;code&gt;19=19&lt;/code&gt; and &lt;code&gt;3 + 7 = 91&lt;/code&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;Regarding the ERROR Values&lt;/h2&gt;&#xA;&lt;p&gt;I think the key is in the errors. What kind of logic could dictate that &lt;code&gt;0+0=ERR&lt;/code&gt; &lt;em&gt;and&lt;/em&gt; &lt;code&gt;1+1=ERR&lt;/code&gt;? It's not just a simple operator substitution here. There are several more examples of the format &lt;code&gt;A+B=ERR&lt;/code&gt; and I have yet to find something that fits them. I feel like there's something key here but I can't grasp it.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-06-16T14:25:01.827" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11472" PostTypeId="2" ParentId="11298" CreationDate="2015-04-02T18:53:06.947" Score="1" Body="&lt;p&gt;To me this sounds like (perhaps a cynical view of)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BEAUTY.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am what they cherish,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Certainly many people cherish beauty.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Those who cherish,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Are also those who detest ardently.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Refers to the complicated nature of emotions - &quot;love-hate relationships&quot; and so on.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I murder with ease and comfort,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And be loved with purpose.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I can't quite parse these lines. The second of them sounds like an imperative, but the sentence structure suggests it's indicative - in which case shouldn't it be &quot;am&quot; and not &quot;be&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am a drug to the eye,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And to the heart,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; These lines are what put me onto this solution. Beauty is something your eyes desire, and it can entrap your heart too, despite being - at the end of the day - quite superficial.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You love me like nobody else,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People fall in love for the sake of beauty.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Not because I murdered your loved ones,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Refers to the fact that beauty fades with time; if you love someone only for their looks, then they might eventually stop being your &quot;loved one&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But because I am a drug to the eye,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;and to the heart.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;See above.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I cause no disease,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; True.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;but yet I abolish those who you love,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, see above.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;For I am a drug to the eye,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;and to the heart!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Again, see above!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This answer just about fits Hint 1, because although it's an abstract concept in itself and not something physical, it does &lt;em&gt;describe&lt;/em&gt; something physically visible. And as for Hint 2, it's everywhere for those with eyes to see it.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-04-02T18:53:06.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11474" PostTypeId="1" CreationDate="2015-04-02T19:08:41.037" Score="19" ViewCount="2397" Body="&lt;h1&gt;Story&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Let me tell ye a tale: Somewhere northeast o' the Bahamas there be a tiny scrap o' rocks stickin' up out o' the sea known as Beal Island. It be a barren, rocky place, and no man ever called it home. But many years ago, a band o' bloodthirsty pirates, with the king's navy swarmin' all over to catch them, hid their ill-gotten loot on the island. They were captured and thrown in jail for evadin' justice, but since they didn't have the treasure, none could prove they'd done the deed, and they escaped the gallows. They all swore that they'd meet gain on the island to dig up their treasure after they all got out o' jail.&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;More recently now, a mighty hurricane swept o'er Beal Island, leaving in its wake the wreckage of ill- fated ships dashed upon the rocks. The survivors banded together an' recovered those provisions what washed up on shore. They found caves in th' isle, and set up to wait fer a passin' ship to rescue 'em. However, it soon became apparent that some among 'em were there fer more nefarious purposes: &lt;strong&gt;four&lt;/strong&gt; of the pirates had returned to gather their treasure! However, thanks to the brave actions o' the survivors, the pirates were found out before they recovered the treasure, an' all present held a great feast before hangin' the louts. But the pirates weren't so easily defeated. They poisoned the rum at the feast, and everyone fell into a stupor o' haunted dreams. When they awoke, none could remember who they were, nor what ship they'd been on, nor even who the pirates were!&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Puzzle&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This puzzle was originally designed as a mystery party, where the guests played the role of shipwrecked amnesiac sailors on a desert island, trying to figure out who they were. Each player was given some clues representing the few things that their character still remembered, but other clues were encoded in their dress and manner of speech. Each guest wore either an officer's hat, marking them as an officer, a bandana on their head, marking them as a sailor, or neither, marking them as a passenger. The passengers were all dressed either richly (paying passengers) or poorly (stowaways). Each player was also instructed to wear or not wear an eye-patch, hook-hand, or peg-leg. Finally, different players spoke using distinctive languages or accents.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional clues were available in the form of copies of crew and passenger manifests which had &quot;washed up on the beach&quot;. The crew manifests clearly state the name and rank of every crew member on five different ships, and the names of the ships. However, since none of the castaways remember their names, this is only somewhat helpful, especially since many of the original crews were lost in the storm. The passenger manifests show the groups of passengers who were aboard ships, but not which ship they were aboard.  &lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Goal&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Determine the following:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For each ship:&lt;/strong&gt; What country was it from, what was its last port of call, where was it bound to, and what type of ship was it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For each player:&lt;/strong&gt; What was their character's name and rank, what ship were they on, and were they a pirate.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus&lt;/strong&gt;:What is the language spoken by the characters whose language is marked as &quot;strange&quot;? &lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Assumptions&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;You may assume the following:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;All of the pirates' clues are false.  &lt;/li&gt;&#xA;&lt;li&gt;Everyone else's clues are true. &lt;/li&gt;&#xA;&lt;li&gt;The &lt;strong&gt;four&lt;/strong&gt; pirates were hidden aboard the various ships, among the crew, passengers, and/or stowaways.&#xA;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Correction of earlier statement, which said &quot;hidden among the crews&quot;, seeming to rule out the possibility of pirates among the passengers or stowaways.&lt;br&gt;&#xA;  &lt;strong&gt;Second edit&lt;/strong&gt;: There are actually four pirates.  The puzzle originally said three.&lt;/li&gt;&#xA;  &lt;li&gt;The people who died in the storm were not present at the party, and nothing is known about them except what is given in the clues and the manifests.   &lt;/li&gt;&#xA;  &lt;li&gt;The various castaways have had several days to share their stories before they all got amnesia, and they are left with odd snippets of memory; the fact that a character has a clue about a particular ship, for instance, does not mean that they are from that ship, unless it is specifically stated in the clue.  &lt;/li&gt;&#xA;  &lt;li&gt;Names in &lt;em&gt;italics&lt;/em&gt; are the names of players; names in &lt;strong&gt;bold&lt;/strong&gt; are the names of characters in the game. The player's names have all been randomly changed from the names of the original party-goers.  &lt;/li&gt;&#xA;  &lt;li&gt;There are no ships represented which do not have an associated crew manifest.  &lt;/li&gt;&#xA;  &lt;li&gt;There are no crew members present who are not listed on a crew manifest.  &lt;/li&gt;&#xA;  &lt;li&gt;There are no paying passengers who are not listed on a passenger manifest.  &lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Each character's name reflects their nationality and sex. (e.g., &lt;strong&gt;Felipe C. Ivarruz-Vicaza&lt;/strong&gt; is not a Scottish woman.) Note that English-speaking nationalities (e.g., English, American) may not have completely distinctive names. Character names were chosen semi-randomly from name databases of their respective nationalities.&lt;/p&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Although some names are intentionally ambiguous, I have added to the crew and manifests annotations of the sex, if it is meant to be clear.&lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;&lt;strong&gt;Edit again&lt;/strong&gt;: This information has been moved to Hint 1.&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The sex of the players matches the sex of their characters. For the purpose of this puzzle, all players and characters are assumed to have one unambiguous sex, which is the conventional one for their name. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: - The scenario takes place in about 1800.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Clues&lt;/h2&gt;&#xA;&#xA;&lt;h3&gt;Crew Manifests&lt;/h3&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;&lt;em&gt;La Vierge du Bon Port&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Officers:&lt;/strong&gt;&lt;br&gt;&#xA;  Captain :  &lt;strong&gt;Henrick R. Janssen&lt;/strong&gt;&lt;br&gt;&#xA;  1st Mate: &lt;strong&gt;Seigfried D. Munser&lt;/strong&gt;&lt;br&gt;&#xA;  2nd Mate: &lt;strong&gt;Margaret N. Salisbury&lt;/strong&gt;&lt;br&gt;&#xA;  Boatswain: &lt;strong&gt;Hartmut R. Eckinger&lt;/strong&gt;    &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Sailors:&lt;/strong&gt;&lt;br&gt;&#xA;  Cook: &lt;strong&gt;Dorothy W. Ramsey&lt;/strong&gt;&lt;br&gt;&#xA;  Seaman: &lt;strong&gt;Carmel O'Brien&lt;/strong&gt;&lt;br&gt;&#xA;  Seaman: &lt;strong&gt;Piperion Esquibel-Montero&lt;/strong&gt;&lt;br&gt;&#xA;  Seaman: &lt;strong&gt;William P. Cotton&lt;/strong&gt;&lt;br&gt;&#xA;  Cabin Girl: &lt;strong&gt;Peggy R. Cobb&lt;/strong&gt;    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;&lt;em&gt;The Sea Fox&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Officers:&lt;/strong&gt;&lt;br&gt;&#xA;  Captain: &lt;strong&gt;Pulqueria Ceballos&lt;/strong&gt;&lt;br&gt;&#xA;  1st Mate: &lt;strong&gt;Roger B. Sheppard&lt;/strong&gt;&lt;br&gt;&#xA;  2nd Mate: &lt;strong&gt;Christopher H. Weber&lt;/strong&gt;&lt;br&gt;&#xA;  Boatswain: &lt;strong&gt;Gennadi B. Sidorov&lt;/strong&gt;    &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Sailors:&lt;/strong&gt;&lt;br&gt;&#xA;   Cook: &lt;strong&gt;Ronald McPherson&lt;/strong&gt;&lt;br&gt;&#xA;  Seaman: &lt;strong&gt;Samuel E. Jennings&lt;/strong&gt;&lt;br&gt;&#xA;  Seaman: &lt;strong&gt;Hu Wei-Xue&lt;/strong&gt;&lt;br&gt;&#xA;  Seaman: &lt;strong&gt;James J. Graham&lt;/strong&gt;&lt;br&gt;&#xA;  Cabin Girl: &lt;strong&gt;Charlotte C. Walpole&lt;/strong&gt;    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;&lt;em&gt;The Jonathan Cheswick&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Officers:&lt;/strong&gt;&lt;br&gt;&#xA;  Captain: &lt;strong&gt;Henry R. Owen&lt;/strong&gt;&lt;br&gt;&#xA;  1st Mate: &lt;strong&gt;Vasya S. Kalinnikov&lt;/strong&gt;&lt;br&gt;&#xA;  2nd Mate: &lt;strong&gt;Tjeerd R. De Smet&lt;/strong&gt;&lt;br&gt;&#xA;  Boatswain: &lt;strong&gt;Yan Fu-ju&lt;/strong&gt;    &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Sailors:&lt;/strong&gt;&lt;br&gt;&#xA;   Cook: &lt;strong&gt;Fabian A. Schubach&lt;/strong&gt;&lt;br&gt;&#xA;  Seaman: &lt;strong&gt;Virginia A. Allen&lt;/strong&gt;&lt;br&gt;&#xA;  Seaman: &lt;strong&gt;Martha F. Woodard&lt;/strong&gt;&lt;br&gt;&#xA;  Seaman: &lt;strong&gt;Timothy W. Hester&lt;/strong&gt;&lt;br&gt;&#xA;  Cabin Boy: &lt;strong&gt;Nigel N. Salisbury&lt;/strong&gt;   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;&lt;em&gt;Gouden Leeuw&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Officers:&lt;/strong&gt;&lt;br&gt;&#xA;  Captain: &lt;strong&gt;Elaine P. Wynn&lt;/strong&gt;&lt;br&gt;&#xA;  1st Mate: &lt;strong&gt;Quentin P. Watson&lt;/strong&gt;&lt;br&gt;&#xA;  2nd Mate: &lt;strong&gt;Arno Angélil&lt;/strong&gt;&lt;br&gt;&#xA;  Boatswain: &lt;strong&gt;Margarita A. Jurivel&lt;/strong&gt;   &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Sailors:&lt;/strong&gt;&lt;br&gt;&#xA;   Cook: &lt;strong&gt;Felipe C. Ivarruz-Vicaza&lt;/strong&gt;&lt;br&gt;&#xA;  Seaman: &lt;strong&gt;Gareth R. Cavendish&lt;/strong&gt;&lt;br&gt;&#xA;  Seaman: &lt;strong&gt;Wilma W. Chapman&lt;/strong&gt;&lt;br&gt;&#xA;  Seaman: &lt;strong&gt;Lemar D. Malcombe&lt;/strong&gt;&lt;br&gt;&#xA;  Cabin Boy: &lt;strong&gt;Samuel K. Pickett&lt;/strong&gt;    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;&lt;em&gt;Nuestra Señora de Cavadonga&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Officers:&lt;/strong&gt;&lt;br&gt;&#xA;  Captain: &lt;strong&gt;Gertrude D. Hutchinson&lt;/strong&gt;&lt;br&gt;&#xA;  1st Mate: &lt;strong&gt;Marietta N. Clarke&lt;/strong&gt;&lt;br&gt;&#xA;  2nd Mate: &lt;strong&gt;Liao Jin-Zhong&lt;/strong&gt;&lt;br&gt;&#xA;  Boatswain: &lt;strong&gt;Cecilia S. Ros-Gamiro&lt;/strong&gt;    &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Sailors:&lt;/strong&gt;&lt;br&gt;&#xA;   Cook: &lt;strong&gt;Ruud D. Van Dijk&lt;/strong&gt;&lt;br&gt;&#xA;  Seaman: &lt;strong&gt;Alexander H. Frost&lt;/strong&gt;&lt;br&gt;&#xA;  Seaman: &lt;strong&gt;Findlay C. Scott&lt;/strong&gt;&lt;br&gt;&#xA;  Seaman: &lt;strong&gt;Quentin R. Callaghan&lt;/strong&gt;&lt;br&gt;&#xA;  Cabin Boy: &lt;strong&gt;Tremaine L. Higgins&lt;/strong&gt;    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Passenger Manifests&lt;/h3&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;&lt;em&gt;Passenger Manifest #1&lt;/em&gt;&lt;/strong&gt;  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Jeanette J. Ambard&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;strong&gt;Chiaffredo Milanesi&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;strong&gt;Nazzareno D. Conti&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;strong&gt;Jean-Louis Fleuette&lt;/strong&gt;    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;&lt;em&gt;Passenger Manifest #2&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Austinafe Ajaranne&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;strong&gt;Jean-François Moreau&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;strong&gt;Louise-Violette Meserve&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;strong&gt;Uckbe Oronle&lt;/strong&gt;    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;&lt;em&gt;Passenger Manifest #3&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Eileen Mac Ateer&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;strong&gt;Ylesme Arrce&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;strong&gt;Maura I. Kiley&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;strong&gt;Eannare Imke&lt;/strong&gt;    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Players&lt;/h3&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Player&lt;/strong&gt;: &lt;em&gt;Andrew&lt;/em&gt;   &lt;strong&gt;Sex&lt;/strong&gt;: M  &lt;strong&gt;Language/Accent&lt;/strong&gt;: American&lt;br&gt;&#xA;  &lt;strong&gt;Clothing&lt;/strong&gt;: Poor &lt;strong&gt;Eye patch?&lt;/strong&gt; : yes    &lt;strong&gt;Hook Hand?&lt;/strong&gt; : no &lt;strong&gt;Peg Leg?&lt;/strong&gt; : no&lt;br&gt;&#xA;  &lt;strong&gt;Clue 1&lt;/strong&gt;: &quot;&lt;em&gt;Andrew&lt;/em&gt; stowed away on a ship bound for Bridgetown.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 2&lt;/strong&gt;: &quot;&lt;em&gt;Andrew&lt;/em&gt; lost a locket.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Player&lt;/strong&gt;: &lt;em&gt;Ann&lt;/em&gt;  &lt;strong&gt;Sex&lt;/strong&gt;: F  &lt;strong&gt;Language/Accent&lt;/strong&gt;: Jamaican&lt;br&gt;&#xA;  &lt;strong&gt;Clothing&lt;/strong&gt;: Officer  &lt;strong&gt;Eye patch?&lt;/strong&gt; : yes    &lt;strong&gt;Hook Hand?&lt;/strong&gt; : no &lt;strong&gt;Peg Leg?&lt;/strong&gt; : no&lt;br&gt;&#xA;  &lt;strong&gt;Clue 1&lt;/strong&gt;: &quot;&lt;em&gt;Ann&lt;/em&gt;'s superior officer was American.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Player&lt;/strong&gt;: &lt;em&gt;Anthony&lt;/em&gt;  &lt;strong&gt;Sex&lt;/strong&gt;: M  &lt;strong&gt;Language/Accent&lt;/strong&gt;: American&lt;br&gt;&#xA;  &lt;strong&gt;Clothing&lt;/strong&gt;: Sailor   &lt;strong&gt;Eye patch?&lt;/strong&gt; : no &lt;strong&gt;Hook Hand?&lt;/strong&gt; : no &lt;strong&gt;Peg Leg?&lt;/strong&gt; : no&lt;br&gt;&#xA;  &lt;strong&gt;Clue 1&lt;/strong&gt;: &quot;&lt;em&gt;Anthony&lt;/em&gt; and &lt;em&gt;Margaret&lt;/em&gt; got jobs on a ship in order to get home to America from France.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 2&lt;/strong&gt;: &quot;The French ship was headed to Casa Blanca.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Player&lt;/strong&gt;: &lt;em&gt;Brian&lt;/em&gt;    &lt;strong&gt;Sex&lt;/strong&gt;: M  &lt;strong&gt;Language/Accent&lt;/strong&gt;: English&lt;br&gt;&#xA;  &lt;strong&gt;Clothing&lt;/strong&gt;: Sailor   &lt;strong&gt;Eye patch?&lt;/strong&gt; : no &lt;strong&gt;Hook Hand?&lt;/strong&gt; : no &lt;strong&gt;Peg Leg?&lt;/strong&gt; : no&lt;br&gt;&#xA;  &lt;strong&gt;Clue 1&lt;/strong&gt;: &quot;None of the ships flew the same flags.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Player&lt;/strong&gt;: &lt;em&gt;Carrie&lt;/em&gt;   &lt;strong&gt;Sex&lt;/strong&gt;: F  &lt;strong&gt;Language/Accent&lt;/strong&gt;: Spanish&lt;br&gt;&#xA;  &lt;strong&gt;Clothing&lt;/strong&gt;: Officer  &lt;strong&gt;Eye patch?&lt;/strong&gt; : no &lt;strong&gt;Hook Hand?&lt;/strong&gt; : no &lt;strong&gt;Peg Leg?&lt;/strong&gt; : no&lt;br&gt;&#xA;  &lt;strong&gt;Clue 1&lt;/strong&gt;: &quot;One ship was a merchant ship sailing out of Rome.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 2&lt;/strong&gt;: &quot;One of the merchant ships had a second mate who spoke two languages.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 3&lt;/strong&gt;: &quot;The two English sailors aboard the Spanish ship always hassled the Scotsman.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Player&lt;/strong&gt;: &lt;em&gt;Charles&lt;/em&gt;  &lt;strong&gt;Sex&lt;/strong&gt;: M  &lt;strong&gt;Language/Accent&lt;/strong&gt;: Strange&lt;br&gt;&#xA;  &lt;strong&gt;Clothing&lt;/strong&gt;: Rich &lt;strong&gt;Eye patch?&lt;/strong&gt; : yes    &lt;strong&gt;Hook Hand?&lt;/strong&gt; : no &lt;strong&gt;Peg Leg?&lt;/strong&gt; : no&lt;br&gt;&#xA;  &lt;strong&gt;Clue 1&lt;/strong&gt;: &quot;&lt;em&gt;Rosemary&lt;/em&gt; and &lt;em&gt;Charles&lt;/em&gt; were on the same ship &quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 2&lt;/strong&gt;: &quot;&lt;em&gt;Charles&lt;/em&gt; was overseeing a business venture.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Player&lt;/strong&gt;: &lt;em&gt;Christina&lt;/em&gt;    &lt;strong&gt;Sex&lt;/strong&gt;: F  &lt;strong&gt;Language/Accent&lt;/strong&gt;: Spanish&lt;br&gt;&#xA;  &lt;strong&gt;Clothing&lt;/strong&gt;: Officer  &lt;strong&gt;Eye patch?&lt;/strong&gt; : no &lt;strong&gt;Hook Hand?&lt;/strong&gt; : no &lt;strong&gt;Peg Leg?&lt;/strong&gt; : no&lt;br&gt;&#xA;  &lt;strong&gt;Clue 1&lt;/strong&gt;: &quot;&lt;em&gt;Christina&lt;/em&gt; was the only Spaniard aboard her ship.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 2&lt;/strong&gt;: &quot;The whaling ships had the same level of casualties among their crews, which was more than the other ships.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 3&lt;/strong&gt;: &quot;&lt;strong&gt;Alexander Frost&lt;/strong&gt; had an eyepatch.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Player&lt;/strong&gt;: &lt;em&gt;Debra&lt;/em&gt;    &lt;strong&gt;Sex&lt;/strong&gt;: F  &lt;strong&gt;Language/Accent&lt;/strong&gt;: English&lt;br&gt;&#xA;  &lt;strong&gt;Clothing&lt;/strong&gt;: Officer  &lt;strong&gt;Eye patch?&lt;/strong&gt; : yes    &lt;strong&gt;Hook Hand?&lt;/strong&gt; : no &lt;strong&gt;Peg Leg?&lt;/strong&gt; : no&lt;br&gt;&#xA;  &lt;strong&gt;Clue 1&lt;/strong&gt;: &quot;One of the ships flew an American flag and sailed out of Boston.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 2&lt;/strong&gt;: &quot;The only male officer on the ship headed to Bridgetown was killed in the storm.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 3&lt;/strong&gt;: &quot;&lt;em&gt;Debra&lt;/em&gt; found a man by the name of &lt;strong&gt;Albert&lt;/strong&gt; hiding in the hold of her ship.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Player&lt;/strong&gt;: &lt;em&gt;Dennis&lt;/em&gt; &lt;strong&gt;Sex&lt;/strong&gt;: M &lt;strong&gt;Language/Accent&lt;/strong&gt;: American&lt;br&gt;&#xA;  &lt;strong&gt;Clothing&lt;/strong&gt;: Officer &lt;strong&gt;Eye patch?&lt;/strong&gt; : no  &lt;strong&gt;Hook Hand?&lt;/strong&gt; : no &lt;strong&gt;Peg Leg?&lt;/strong&gt; : no&lt;br&gt;&#xA;  &lt;strong&gt;Clue 1&lt;/strong&gt;: &quot;The Cabin boy/girl was the only sailor to survive from the ship that sailed out of Kingston.&quot;  &lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Cabin boy or girl  &lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Player&lt;/strong&gt;: &lt;em&gt;Edward&lt;/em&gt;   &lt;strong&gt;Sex&lt;/strong&gt;: M  &lt;strong&gt;Language/Accent&lt;/strong&gt;: Australian&lt;br&gt;&#xA;  &lt;strong&gt;Clothing&lt;/strong&gt;: Sailor   &lt;strong&gt;Eye patch?&lt;/strong&gt; : no &lt;strong&gt;Hook Hand?&lt;/strong&gt; : no &lt;strong&gt;Peg Leg?&lt;/strong&gt; : no&lt;br&gt;&#xA;  &lt;strong&gt;Clue 1&lt;/strong&gt;: &quot;Only one officer survived from 'The Jonathan Cheswick'&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Player&lt;/strong&gt;: &lt;em&gt;Elizabeth&lt;/em&gt;    &lt;strong&gt;Sex&lt;/strong&gt;: F  &lt;strong&gt;Language/Accent&lt;/strong&gt;: American&lt;br&gt;&#xA;  &lt;strong&gt;Clothing&lt;/strong&gt;: Sailor   &lt;strong&gt;Eye patch?&lt;/strong&gt; : no &lt;strong&gt;Hook Hand?&lt;/strong&gt; : no &lt;strong&gt;Peg Leg?&lt;/strong&gt; : no&lt;br&gt;&#xA;  &lt;strong&gt;Clue 1&lt;/strong&gt;: &quot;&lt;em&gt;Elizabeth&lt;/em&gt; and &lt;em&gt;Ryan&lt;/em&gt; were on the same ship.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 2&lt;/strong&gt;: &quot;&lt;em&gt;Elizabeth&lt;/em&gt; was a cabin girl.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 3&lt;/strong&gt;: &quot;One of the stowaways was named &lt;strong&gt;Natalie Warner&lt;/strong&gt;.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Player&lt;/strong&gt;: &lt;em&gt;Emily&lt;/em&gt;    &lt;strong&gt;Sex&lt;/strong&gt;: F  &lt;strong&gt;Language/Accent&lt;/strong&gt;: Chinese&lt;br&gt;&#xA;  &lt;strong&gt;Clothing&lt;/strong&gt;: Officer  &lt;strong&gt;Eye patch?&lt;/strong&gt; : yes    &lt;strong&gt;Hook Hand?&lt;/strong&gt; : yes    &lt;strong&gt;Peg Leg?&lt;/strong&gt; : no&lt;br&gt;&#xA;  &lt;strong&gt;Clue 1&lt;/strong&gt;: &quot;Although 'Nuestra Señora de Cavadonga' was a Spanish ship, the captain was American.&quot;  &lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Emily is female.&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Player&lt;/strong&gt;: &lt;em&gt;Frank&lt;/em&gt;    &lt;strong&gt;Sex&lt;/strong&gt;: M  &lt;strong&gt;Language/Accent&lt;/strong&gt;: American&lt;br&gt;&#xA;  &lt;strong&gt;Clothing&lt;/strong&gt;: Poor &lt;strong&gt;Eye patch?&lt;/strong&gt; : yes    &lt;strong&gt;Hook Hand?&lt;/strong&gt; : no &lt;strong&gt;Peg Leg?&lt;/strong&gt; : no&lt;br&gt;&#xA;  &lt;strong&gt;Clue 1&lt;/strong&gt;: &quot;At least one officer from each ship perished in the storm.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 2&lt;/strong&gt;: &quot;Three of the passengers on the ship sailing out of Veracruz were killed in the storm.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Player&lt;/strong&gt;: &lt;em&gt;Gary&lt;/em&gt; &lt;strong&gt;Sex&lt;/strong&gt;: M  &lt;strong&gt;Language/Accent&lt;/strong&gt;: French&lt;br&gt;&#xA;  &lt;strong&gt;Clothing&lt;/strong&gt;: Rich &lt;strong&gt;Eye patch?&lt;/strong&gt; : yes    &lt;strong&gt;Hook Hand?&lt;/strong&gt; : no &lt;strong&gt;Peg Leg?&lt;/strong&gt; : no&lt;br&gt;&#xA;  &lt;strong&gt;Clue 1&lt;/strong&gt;: &quot;&lt;em&gt;Sherry&lt;/em&gt; and &lt;em&gt;Gary&lt;/em&gt; were on the same ship.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 2&lt;/strong&gt;: &quot;&lt;em&gt;Gary&lt;/em&gt; was funding a mission of discovery and wanted to go along.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Player&lt;/strong&gt;: &lt;em&gt;George&lt;/em&gt;   &lt;strong&gt;Sex&lt;/strong&gt;: M  &lt;strong&gt;Language/Accent&lt;/strong&gt;: Shakespearean English&lt;br&gt;&#xA;  &lt;strong&gt;Clothing&lt;/strong&gt;: Sailor   &lt;strong&gt;Eye patch?&lt;/strong&gt; : yes    &lt;strong&gt;Hook Hand?&lt;/strong&gt; : yes    &lt;strong&gt;Peg Leg?&lt;/strong&gt; : no&lt;br&gt;&#xA;  &lt;strong&gt;Clue 1&lt;/strong&gt;: &quot;The cook aboard &lt;em&gt;George's&lt;/em&gt; ship perished in the storm.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 2&lt;/strong&gt;: &quot;&lt;em&gt;George&lt;/em&gt; was headed to Charleston.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 3&lt;/strong&gt;: &quot;Only one Sailor from the &quot;Gouden Leeuw&quot; survived.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Player&lt;/strong&gt;: &lt;em&gt;Irene&lt;/em&gt;    &lt;strong&gt;Sex&lt;/strong&gt;: F  &lt;strong&gt;Language/Accent&lt;/strong&gt;: American&lt;br&gt;&#xA;  &lt;strong&gt;Clothing&lt;/strong&gt;: Poor &lt;strong&gt;Eye patch?&lt;/strong&gt; : no &lt;strong&gt;Hook Hand?&lt;/strong&gt; : no &lt;strong&gt;Peg Leg?&lt;/strong&gt; : no&lt;br&gt;&#xA;  &lt;strong&gt;Clue 1&lt;/strong&gt;: &quot;&lt;em&gt;Irene&lt;/em&gt; met &lt;em&gt;Terry&lt;/em&gt; when she stowed away on his ship. He found her, but agreed not to report her.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 2&lt;/strong&gt;: &quot;Only one cook survived the storm.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 3&lt;/strong&gt;: &quot;Two of the stowaways had professions for surnames: one that makes arrows, one that works with iron.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Player&lt;/strong&gt;: &lt;em&gt;Jack&lt;/em&gt; &lt;strong&gt;Sex&lt;/strong&gt;: M  &lt;strong&gt;Language/Accent&lt;/strong&gt;: Scottish&lt;br&gt;&#xA;  &lt;strong&gt;Clothing&lt;/strong&gt;: Sailor   &lt;strong&gt;Eye patch?&lt;/strong&gt; : no &lt;strong&gt;Hook Hand?&lt;/strong&gt; : no &lt;strong&gt;Peg Leg?&lt;/strong&gt; : no&lt;br&gt;&#xA;  &lt;strong&gt;Clue 1&lt;/strong&gt;: &quot;Two of the captains went down with their ships.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 2&lt;/strong&gt;: &quot;Two of the ships were whalers, and did not allow passengers.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 3&lt;/strong&gt;: &quot;There are two freshly made graves on the island, which read 'R.I.P. &lt;strong&gt;Virginia Allen&lt;/strong&gt;' and 'R.I.P. &lt;strong&gt;Eileen Mac Ateer&lt;/strong&gt;'.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Player&lt;/strong&gt;: &lt;em&gt;Jason&lt;/em&gt;    &lt;strong&gt;Sex&lt;/strong&gt;: M  &lt;strong&gt;Language/Accent&lt;/strong&gt;: Dutch&lt;br&gt;&#xA;  &lt;strong&gt;Clothing&lt;/strong&gt;: Officer  &lt;strong&gt;Eye patch?&lt;/strong&gt; : no &lt;strong&gt;Hook Hand?&lt;/strong&gt; : no &lt;strong&gt;Peg Leg?&lt;/strong&gt; : no&lt;br&gt;&#xA;  &lt;strong&gt;Clue 1&lt;/strong&gt;: &quot;&lt;em&gt;Jason&lt;/em&gt; was aboard a merchant ship.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 2&lt;/strong&gt;: &quot;All the English ship's sailors survived the tempest.&quot;  &lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Corrected &quot;All the English sailors&quot; to &quot;All the English &lt;strong&gt;ship's&lt;/strong&gt; sailors.&quot;&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Player&lt;/strong&gt;: &lt;em&gt;Jennifer&lt;/em&gt; &lt;strong&gt;Sex&lt;/strong&gt;: F  &lt;strong&gt;Language/Accent&lt;/strong&gt;: American&lt;br&gt;&#xA;  &lt;strong&gt;Clothing&lt;/strong&gt;: Poor &lt;strong&gt;Eye patch?&lt;/strong&gt; : no &lt;strong&gt;Hook Hand?&lt;/strong&gt; : no &lt;strong&gt;Peg Leg?&lt;/strong&gt; : no&lt;br&gt;&#xA;  &lt;strong&gt;Clue 1&lt;/strong&gt;: &quot;&lt;em&gt;Jennifer&lt;/em&gt; was a stowaway on a whaling ship.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 2&lt;/strong&gt;: &quot;The ship which was sailing between two non-English-speaking ports has sailors from four different English-speaking countries.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 3&lt;/strong&gt;: &quot;When the storm hit, the captain of the ship carrying the two Italian gentlemen assigned the cook to stick with &lt;strong&gt;Signor Milanesi&lt;/strong&gt;, and another sailor to stay with &lt;strong&gt;Signor Conti&lt;/strong&gt;; this decided the fates of the two gentleman.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Player&lt;/strong&gt;: &lt;em&gt;Joseph&lt;/em&gt;   &lt;strong&gt;Sex&lt;/strong&gt;: M  &lt;strong&gt;Language/Accent&lt;/strong&gt;: English&lt;br&gt;&#xA;  &lt;strong&gt;Clothing&lt;/strong&gt;: Sailor   &lt;strong&gt;Eye patch?&lt;/strong&gt; : no &lt;strong&gt;Hook Hand?&lt;/strong&gt; : no &lt;strong&gt;Peg Leg?&lt;/strong&gt; : no&lt;br&gt;&#xA;  &lt;strong&gt;Clue 1&lt;/strong&gt;: &quot;&lt;em&gt;Joseph&lt;/em&gt; was a cabin boy.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 2&lt;/strong&gt;: &quot;The ship sailing out of Vera Cruz was not an exploration vessel.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 3&lt;/strong&gt;: &quot;The last name of one of the stowaways was the same as the first name of the ship &lt;em&gt;Joseph&lt;/em&gt; was on.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Player&lt;/strong&gt;: &lt;em&gt;Josephine&lt;/em&gt;    &lt;strong&gt;Sex&lt;/strong&gt;: F  &lt;strong&gt;Language/Accent&lt;/strong&gt;: American&lt;br&gt;&#xA;  &lt;strong&gt;Clothing&lt;/strong&gt;: Officer  &lt;strong&gt;Eye patch?&lt;/strong&gt; : no &lt;strong&gt;Hook Hand?&lt;/strong&gt; : no &lt;strong&gt;Peg Leg?&lt;/strong&gt; : no&lt;br&gt;&#xA;  &lt;strong&gt;Clue 1&lt;/strong&gt;: &quot;&lt;em&gt;Josephine&lt;/em&gt; hired a Jamaican Cabin Boy with a peg leg for her ship.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 2&lt;/strong&gt;: &quot;The captain of &quot;Gouden Leeuw&quot; had a hook hand.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 3&lt;/strong&gt;: &quot;&lt;em&gt;Josephine&lt;/em&gt; found a locket washed up on the beach with the initials &quot;&lt;strong&gt;PF&lt;/strong&gt;&quot; engraved on it.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Player&lt;/strong&gt;: &lt;em&gt;Keith&lt;/em&gt;    &lt;strong&gt;Sex&lt;/strong&gt;: M  &lt;strong&gt;Language/Accent&lt;/strong&gt;: Russian&lt;br&gt;&#xA;  &lt;strong&gt;Clothing&lt;/strong&gt;: Officer  &lt;strong&gt;Eye patch?&lt;/strong&gt; : yes    &lt;strong&gt;Hook Hand?&lt;/strong&gt; : no &lt;strong&gt;Peg Leg?&lt;/strong&gt; : no&lt;br&gt;&#xA;  &lt;strong&gt;Clue 1&lt;/strong&gt;: &quot;One of the ships was an exploration ship that was headed to Buenos Aires to resupply on the way to Antarctica.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 2&lt;/strong&gt;: &quot;Only one ship was sailing to a port that flew the same flag as the ship.&quot;  &lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Changed &quot;its home port&quot; to &quot;a port that flew the same flag as the ship&quot;, for clarity.&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Player&lt;/strong&gt;: &lt;em&gt;Kevin&lt;/em&gt;    &lt;strong&gt;Sex&lt;/strong&gt;: M  &lt;strong&gt;Language/Accent&lt;/strong&gt;: American&lt;br&gt;&#xA;  &lt;strong&gt;Clothing&lt;/strong&gt;: Officer  &lt;strong&gt;Eye patch?&lt;/strong&gt; : yes    &lt;strong&gt;Hook Hand?&lt;/strong&gt; : no &lt;strong&gt;Peg Leg?&lt;/strong&gt; : no&lt;br&gt;&#xA;  &lt;strong&gt;Clue 1&lt;/strong&gt;: &quot;Each of the ships had set sail from a different port.&quot;  &lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Kevin has an eye patch&#xA;    &lt;strong&gt;Clue 2&lt;/strong&gt;: &quot;One of the stowaways had the same last name as the first name of the captain of the westward-sailing whaler. (Don't ask how I remember this even while I have amnesia.)&quot;&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Player&lt;/strong&gt;: &lt;em&gt;Kimberly&lt;/em&gt; &lt;strong&gt;Sex&lt;/strong&gt;: F  &lt;strong&gt;Language/Accent&lt;/strong&gt;: American&lt;br&gt;&#xA;  &lt;strong&gt;Clothing&lt;/strong&gt;: Officer  &lt;strong&gt;Eye patch?&lt;/strong&gt; : yes    &lt;strong&gt;Hook Hand?&lt;/strong&gt; : yes    &lt;strong&gt;Peg Leg?&lt;/strong&gt; : no&lt;br&gt;&#xA;  &lt;strong&gt;Clue 1&lt;/strong&gt;: &quot;One ship was sailing from Bordeaux to Charleston.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 2&lt;/strong&gt;: &quot;All of the second mates have eyepatches.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 3&lt;/strong&gt;: &quot;One of the stowaways earned himself the nick-name 'Crazy &lt;strong&gt;Parker&lt;/strong&gt;' because of his devil-may-care attitude about life.&quot;  &lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Reworded Clue 3.&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Player&lt;/strong&gt;: &lt;em&gt;Larry&lt;/em&gt;    &lt;strong&gt;Sex&lt;/strong&gt;: M  &lt;strong&gt;Language/Accent&lt;/strong&gt;: German&lt;br&gt;&#xA;  &lt;strong&gt;Clothing&lt;/strong&gt;: Officer  &lt;strong&gt;Eye patch?&lt;/strong&gt; : yes    &lt;strong&gt;Hook Hand?&lt;/strong&gt; : no &lt;strong&gt;Peg Leg?&lt;/strong&gt; : no&lt;br&gt;&#xA;  &lt;strong&gt;Clue 1&lt;/strong&gt;: &quot;The Captain of the Dutch ship did not wear an eyepatch.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 2&lt;/strong&gt;: &quot;The Boatswain of &quot;The Jonathan Cheswick&quot; had full use of both hands.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 3&lt;/strong&gt;: &quot;&lt;em&gt;Larry's&lt;/em&gt; ship set sail from Bordeaux.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Player&lt;/strong&gt;: &lt;em&gt;Lori&lt;/em&gt; &lt;strong&gt;Sex&lt;/strong&gt;: F  &lt;strong&gt;Language/Accent&lt;/strong&gt;: American&lt;br&gt;&#xA;  &lt;strong&gt;Clothing&lt;/strong&gt;: Sailor   &lt;strong&gt;Eye patch?&lt;/strong&gt; : no &lt;strong&gt;Hook Hand?&lt;/strong&gt; : no &lt;strong&gt;Peg Leg?&lt;/strong&gt; : no&lt;br&gt;&#xA;  &lt;strong&gt;Clue 1&lt;/strong&gt;: &quot;There was originally a stowaway on each ship, but one perished.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Player&lt;/strong&gt;: &lt;em&gt;Margaret&lt;/em&gt; &lt;strong&gt;Sex&lt;/strong&gt;: F   &lt;strong&gt;Language/Accent&lt;/strong&gt;: American&lt;br&gt;&#xA;  &lt;strong&gt;Clothing&lt;/strong&gt;: Sailor   &lt;strong&gt;Eye patch?&lt;/strong&gt; : no &lt;strong&gt;Hook Hand?&lt;/strong&gt; : no &lt;strong&gt;Peg Leg?&lt;/strong&gt; : no&lt;br&gt;&#xA;  &lt;strong&gt;Clue 1&lt;/strong&gt;: &quot;&lt;em&gt;Anthony&lt;/em&gt; and &lt;em&gt;Margaret&lt;/em&gt; got jobs on a ship in order to get home to America from France.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 2&lt;/strong&gt;: &quot;Whaling ships are not allowed to dock at Bridgetown.&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;Edit: Corrected &quot;Bridgeport&quot; to &quot;Bridgetown&quot;. &lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Player&lt;/strong&gt;: &lt;em&gt;Megan&lt;/em&gt;    &lt;strong&gt;Sex&lt;/strong&gt;: F  &lt;strong&gt;Language/Accent&lt;/strong&gt;: English&lt;br&gt;&#xA;  &lt;strong&gt;Clothing&lt;/strong&gt;: Sailor   &lt;strong&gt;Eye patch?&lt;/strong&gt; : no &lt;strong&gt;Hook Hand?&lt;/strong&gt; : no &lt;strong&gt;Peg Leg?&lt;/strong&gt; : no&lt;br&gt;&#xA;  &lt;strong&gt;Clue 1&lt;/strong&gt;: &quot;&lt;em&gt;Willie&lt;/em&gt; and &lt;em&gt;Megan&lt;/em&gt; met each other after being shipwrecked on the island.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Player&lt;/strong&gt;: &lt;em&gt;Patrick&lt;/em&gt;  &lt;strong&gt;Sex&lt;/strong&gt;: M  &lt;strong&gt;Language/Accent&lt;/strong&gt;: Jamaican&lt;br&gt;&#xA;  &lt;strong&gt;Clothing&lt;/strong&gt;: Sailor   &lt;strong&gt;Eye patch?&lt;/strong&gt; : no &lt;strong&gt;Hook Hand?&lt;/strong&gt; : no &lt;strong&gt;Peg Leg?&lt;/strong&gt; : yes&lt;br&gt;&#xA;  &lt;strong&gt;Clue 1&lt;/strong&gt;: &quot;&lt;em&gt;Patrick&lt;/em&gt; was on a whaling ship.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 2&lt;/strong&gt;: &quot;Two officers from the English ship survived the storm.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 3&lt;/strong&gt;: &quot;No one who was aboard a merchant ship had a peg leg.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Player&lt;/strong&gt;: &lt;em&gt;Paul&lt;/em&gt; &lt;strong&gt;Sex&lt;/strong&gt;: M  &lt;strong&gt;Language/Accent&lt;/strong&gt;: Italian&lt;br&gt;&#xA;  &lt;strong&gt;Clothing&lt;/strong&gt;: Rich &lt;strong&gt;Eye patch?&lt;/strong&gt; : yes    &lt;strong&gt;Hook Hand?&lt;/strong&gt; : no &lt;strong&gt;Peg Leg?&lt;/strong&gt; : no&lt;br&gt;&#xA;  &lt;strong&gt;Clue 1&lt;/strong&gt;: &quot;Each of the ships was headed to a different destination.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 2&lt;/strong&gt;: &quot;One of the ships suffered no casualties among the sailors.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 3&lt;/strong&gt;: &quot;&lt;strong&gt;Mme. Meserve&lt;/strong&gt; was known for her beautiful eyes.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Player&lt;/strong&gt;: &lt;em&gt;Raymond&lt;/em&gt;  &lt;strong&gt;Sex&lt;/strong&gt;: M  &lt;strong&gt;Language/Accent&lt;/strong&gt;: Spanish&lt;br&gt;&#xA;  &lt;strong&gt;Clothing&lt;/strong&gt;: Sailor   &lt;strong&gt;Eye patch?&lt;/strong&gt; : yes    &lt;strong&gt;Hook Hand?&lt;/strong&gt; : no &lt;strong&gt;Peg Leg?&lt;/strong&gt; : no&lt;br&gt;&#xA;  &lt;strong&gt;Clue 1&lt;/strong&gt;: &quot;One ship was a Dutch merchant ship heading to its home port in Amsterdam.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 2&lt;/strong&gt;: &quot;One of the boatswains was headed to Casa Blanca, where he was to be married and give up his life on the sea; tragically, he was killed in the storm.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 3&lt;/strong&gt;: &quot;One of the stowaways carved their initials into the hull of the ship they were on: &lt;strong&gt;JS&lt;/strong&gt;.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Player&lt;/strong&gt;: &lt;em&gt;Richard&lt;/em&gt;  &lt;strong&gt;Sex&lt;/strong&gt;: M  &lt;strong&gt;Language/Accent&lt;/strong&gt;: French and German&lt;br&gt;&#xA;  &lt;strong&gt;Clothing&lt;/strong&gt;: Officer  &lt;strong&gt;Eye patch?&lt;/strong&gt; : yes    &lt;strong&gt;Hook Hand?&lt;/strong&gt; : no &lt;strong&gt;Peg Leg?&lt;/strong&gt; : no&lt;br&gt;&#xA;  &lt;strong&gt;Clue 1&lt;/strong&gt;: &quot;The Spanish ship was sailing to Bridgetown.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 2&lt;/strong&gt;: &quot;None of the sailors aboard &quot;The Jonathan Cheswick&quot; were maimed.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 3&lt;/strong&gt;: &quot;The stowaway on &lt;em&gt;Richard's&lt;/em&gt; ship was a handsome man with no visible disfigurement, but he was made to walk the plank.&quot;  &lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: For clarification, &quot;maimed&quot; in this context refers to eyepatch, hook-hand, and peg-leg, not to whether or not the sailors survived the storm.&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Player&lt;/strong&gt;: &lt;em&gt;Rosemary&lt;/em&gt; &lt;strong&gt;Sex&lt;/strong&gt;: F  &lt;strong&gt;Language/Accent&lt;/strong&gt;: Strange&lt;br&gt;&#xA;  &lt;strong&gt;Clothing&lt;/strong&gt;: Rich &lt;strong&gt;Eye patch?&lt;/strong&gt; : yes    &lt;strong&gt;Hook Hand?&lt;/strong&gt; : no &lt;strong&gt;Peg Leg?&lt;/strong&gt; : no&lt;br&gt;&#xA;  &lt;strong&gt;Clue 1&lt;/strong&gt;: &quot;&lt;em&gt;Rosemary&lt;/em&gt; and &lt;em&gt;Charles&lt;/em&gt; were on the same ship&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 2&lt;/strong&gt;: &quot;Each ship was named in the language of the country it was from.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 3&lt;/strong&gt;: &quot;One of the stowaways was known as 'One-eyed &lt;strong&gt;Pete&lt;/strong&gt;.'&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Player&lt;/strong&gt;: &lt;em&gt;Ryan&lt;/em&gt; &lt;strong&gt;Sex&lt;/strong&gt;: M  &lt;strong&gt;Language/Accent&lt;/strong&gt;: American&lt;br&gt;&#xA;  &lt;strong&gt;Clothing&lt;/strong&gt;: Sailor   &lt;strong&gt;Eye patch?&lt;/strong&gt; : no &lt;strong&gt;Hook Hand?&lt;/strong&gt; : no &lt;strong&gt;Peg Leg?&lt;/strong&gt; : no&lt;br&gt;&#xA;  &lt;strong&gt;Clue 1&lt;/strong&gt;: &quot;&lt;em&gt;Elizabeth&lt;/em&gt; and &lt;em&gt;Ryan&lt;/em&gt; are on the same boat.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 2&lt;/strong&gt;: &quot;Only one ship was sailing out of its home country.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Player&lt;/strong&gt;: &lt;em&gt;Sherry&lt;/em&gt;   &lt;strong&gt;Sex&lt;/strong&gt;: F  &lt;strong&gt;Language/Accent&lt;/strong&gt;: French&lt;br&gt;&#xA;  &lt;strong&gt;Clothing&lt;/strong&gt;: Rich &lt;strong&gt;Eye patch?&lt;/strong&gt; : yes    &lt;strong&gt;Hook Hand?&lt;/strong&gt; : yes    &lt;strong&gt;Peg Leg?&lt;/strong&gt; : yes&lt;br&gt;&#xA;  &lt;strong&gt;Clue 1&lt;/strong&gt;: &quot;&lt;em&gt;Sherry&lt;/em&gt; discovered a stowaway named &lt;strong&gt;Marie&lt;/strong&gt; searching through her baggage.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 2&lt;/strong&gt;: &quot;&lt;em&gt;Sherry&lt;/em&gt; and &lt;em&gt;Gary&lt;/em&gt; were on the same ship.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Player&lt;/strong&gt;: &lt;em&gt;Terry&lt;/em&gt;    &lt;strong&gt;Sex&lt;/strong&gt;: M  &lt;strong&gt;Language/Accent&lt;/strong&gt;: American&lt;br&gt;&#xA;  &lt;strong&gt;Clothing&lt;/strong&gt;: Sailor   &lt;strong&gt;Eye patch?&lt;/strong&gt; : no &lt;strong&gt;Hook Hand?&lt;/strong&gt; : no &lt;strong&gt;Peg Leg?&lt;/strong&gt; : no&lt;br&gt;&#xA;  &lt;strong&gt;Clue 1&lt;/strong&gt;: &quot;The person who shares a first name with &lt;em&gt;Terry&lt;/em&gt; is either Australian or English.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 2&lt;/strong&gt;: &quot;&lt;em&gt;Irene&lt;/em&gt; met &lt;em&gt;Terry&lt;/em&gt; when she stowed away on his ship. He found her, but agreed not to report her.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Player&lt;/strong&gt;: &lt;em&gt;Victoria&lt;/em&gt; &lt;strong&gt;Sex&lt;/strong&gt;: F  &lt;strong&gt;Language/Accent&lt;/strong&gt;: Irish&lt;br&gt;&#xA;  &lt;strong&gt;Clothing&lt;/strong&gt;: Rich &lt;strong&gt;Eye patch?&lt;/strong&gt; : yes    &lt;strong&gt;Hook Hand?&lt;/strong&gt; : yes    &lt;strong&gt;Peg Leg?&lt;/strong&gt; : no&lt;br&gt;&#xA;  &lt;strong&gt;Clue 1&lt;/strong&gt;: &quot;&lt;em&gt;Victoria&lt;/em&gt; was on a boat with two Spaniards.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 2&lt;/strong&gt;: &quot;The English ship had no passengers.&quot;&lt;br&gt;&#xA;  &lt;strong&gt;Clue 3&lt;/strong&gt;: &quot;Someone found a dog-eared and heavily-annotated copy of &quot;Hamlet&quot; washed up on the beach, with the initials '&lt;strong&gt;G.R.C.&lt;/strong&gt;' printed on the inside cover.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Player&lt;/strong&gt;: &lt;em&gt;Willie&lt;/em&gt;   &lt;strong&gt;Sex&lt;/strong&gt;: M  &lt;strong&gt;Language/Accent&lt;/strong&gt;: English&lt;br&gt;&#xA;  &lt;strong&gt;Clothing&lt;/strong&gt;: Sailor   &lt;strong&gt;Eye patch?&lt;/strong&gt; : yes &lt;strong&gt;Hook Hand?&lt;/strong&gt; : no  &lt;strong&gt;Peg Leg?&lt;/strong&gt; : no&lt;br&gt;&#xA;  &lt;strong&gt;Clue 1&lt;/strong&gt;: &quot;&lt;em&gt;Willie&lt;/em&gt; and &lt;em&gt;Megan&lt;/em&gt; met each other after being shipwrecked on the island.&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: &lt;em&gt;Willie&lt;/em&gt; has an eye patch, not a hook hand.&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Hint 1&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Some information in the puzzle may take some research, especially if one isn't familiar with names in different languages and historical geography.  Here is some information which is intended to be known, but which is not directly stated:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The flags flown by the ten ports mentioned in 1800 were:&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;b&gt;Amsterdam&lt;/b&gt;, Holland - &lt;b&gt;Dutch&lt;/b&gt;&lt;br&gt;&#xA; &lt;b&gt;Bordeaux&lt;/b&gt;, France - &lt;b&gt;French&lt;/b&gt;&lt;br&gt;&#xA; &lt;b&gt;Boston&lt;/b&gt;, Massachusetts - &lt;b&gt;American&lt;/b&gt;&lt;br&gt;&#xA; &lt;b&gt;Bridgetown&lt;/b&gt;, Barbados - &lt;b&gt;British&lt;/b&gt;&lt;br&gt;&#xA; &lt;b&gt;Buenos Aires&lt;/b&gt;, Viceroyalty of the Río de la Plata (today Argentina) - &lt;b&gt;Spanish&lt;/b&gt;&lt;br&gt;&#xA; &lt;b&gt;Casablanca&lt;/b&gt;, Morocco - &lt;b&gt;Moroccan&lt;/b&gt;&lt;br&gt;&#xA; &lt;b&gt;Charleston&lt;/b&gt;, North Carolina - &lt;b&gt;American&lt;/b&gt;&lt;br&gt;&#xA; &lt;b&gt;Kingston&lt;/b&gt;, Jamaica - &lt;b&gt;British&lt;/b&gt;&lt;br&gt;&#xA; &lt;b&gt;Rome&lt;/b&gt;, the &lt;b&gt;Roman Republic&lt;/b&gt; Feb 1798 - Jun 1800, or the &lt;b&gt;Papal States&lt;/b&gt; June 1800 - 1808. The difference is not significant here, since neither is otherwise represented in the puzzle. Although the Roman Republic was a French puppet state, it did have its own flag, and does not count as French in this puzzle.&lt;br&gt;&#xA; * &lt;b&gt;Veracruz&lt;/b&gt;, Viceroyalty of New Spain (now Mexico) - &lt;b&gt;Spanish&lt;/b&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; The language and genders of the character names listed in the manifests are:&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;b&gt;La Vierge du Bon Port&lt;/b&gt;&lt;br&gt;&#xA; &lt;b&gt;Officers&lt;/b&gt;&lt;br&gt;&#xA; Captain: &lt;b&gt;Henrick R. Janssen&lt;/b&gt;, Dutch, M&lt;br&gt;&#xA; 1stMate: &lt;b&gt;Seigfried D. Munser&lt;/b&gt;, German, M&lt;br&gt;&#xA; 2ndMate: &lt;b&gt;Margaret N. Salisbury&lt;/b&gt;, Anglophone, F&lt;br&gt;&#xA; Boatswain: &lt;b&gt;Hartmut R. Eckinger&lt;/b&gt;, German, M&lt;br&gt;&#xA; &lt;b&gt;Sailors&lt;/b&gt;&lt;br&gt;&#xA; Cook: &lt;b&gt;Dorothy W. Ramsey&lt;/b&gt;, Anglophone, F&lt;br&gt;&#xA; Seaman: &lt;b&gt;Carmel O'Brien&lt;/b&gt;, Irish, F&lt;br&gt;&#xA; Seaman: &lt;b&gt;Piperion Esquibel-Montero&lt;/b&gt;, Spanish, M&lt;br&gt;&#xA; Seaman: &lt;b&gt;William P. Cotton&lt;/b&gt;, Anglophone, M&lt;br&gt;&#xA; Cabin Girl: &lt;b&gt;Peggy R. Cobb&lt;/b&gt;, Anglophone, F&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;b&gt;The Sea Fox&lt;/b&gt;&lt;br&gt;&#xA; &lt;b&gt;Officers:&lt;/b&gt;&lt;br&gt;&#xA; Captain: &lt;b&gt;Pulqueria Ceballos&lt;/b&gt;, Spanish, F&lt;br&gt;&#xA; 1st Mate: &lt;b&gt;Roger B. Sheppard&lt;/b&gt;, Anglophone, M&lt;br&gt;&#xA; 2nd Mate: &lt;b&gt;Christopher H. Weber&lt;/b&gt;, Anglophone, M&lt;br&gt;&#xA; Boatswain: &lt;b&gt;Gennadi B. Sidorov&lt;/b&gt;, Russian, M&lt;br&gt;&#xA; &lt;b&gt;Sailors:&lt;/b&gt;&lt;br&gt;&#xA; Cook: &lt;b&gt;Ronald McPherson&lt;/b&gt;, Anglophone, M&lt;br&gt;&#xA; Seaman: &lt;b&gt;Samuel E. Jennings&lt;/b&gt;, Anglophone, M&lt;br&gt;&#xA; Seaman: &lt;b&gt;Hu Wei-Xue&lt;/b&gt;, Chinese, ??&lt;br&gt;&#xA; Seaman: &lt;b&gt;James J. Graham&lt;/b&gt;, Anglophone, M&lt;br&gt;&#xA; Cabin Girl: &lt;b&gt;Charlotte C. Walpole&lt;/b&gt;, Anglophone, F&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;b&gt;The Jonathan Cheswick&lt;/b&gt;&lt;br&gt;&#xA; &lt;b&gt;Officers:&lt;/b&gt;&lt;br&gt;&#xA; Captain: &lt;b&gt;Henry R. Owen&lt;/b&gt;, Anglophone, M&lt;br&gt;&#xA; 1st Mate: &lt;b&gt;Vasya S. Kalinnikov&lt;/b&gt;, Russian, M&lt;br&gt;&#xA; 2nd Mate: &lt;b&gt;Tjeerd R. De Smet&lt;/b&gt;, Dutch, M&lt;br&gt;&#xA; Boatswain: &lt;b&gt;Yan Fu-ju&lt;/b&gt;, Chinese, ??&lt;br&gt;&#xA; &lt;b&gt;Sailors:&lt;/b&gt;&lt;br&gt;&#xA; Cook: &lt;b&gt;Fabian A. Schubach&lt;/b&gt;, German, M&lt;br&gt;&#xA; Seaman: &lt;b&gt;Virginia A. Allen&lt;/b&gt;, Anglophone, F&lt;br&gt;&#xA; Seaman: &lt;b&gt;Martha F. Woodard&lt;/b&gt;, Anglophone, F&lt;br&gt;&#xA; Seaman: &lt;b&gt;Timothy W. Hester&lt;/b&gt;, Anglophone, M&lt;br&gt;&#xA; Cabin Boy: &lt;b&gt;Nigel N. Salisbury&lt;/b&gt;, Anglophone, M&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;b&gt;Gouden Leeuw&lt;/b&gt;&lt;br&gt;&#xA; &lt;b&gt;Officers:&lt;/b&gt;&lt;br&gt;&#xA; Captain: &lt;b&gt;Elaine P. Wynn&lt;/b&gt;, Anglophone, F&lt;br&gt;&#xA; 1st Mate: &lt;b&gt;Quentin P. Watson&lt;/b&gt;, Anglophone, M&lt;br&gt;&#xA; 2nd Mate: &lt;b&gt;Arno Angélil&lt;/b&gt;, German first name, French last name, M&lt;br&gt;&#xA; Boatswain: &lt;b&gt;Margarita A. Jurivel&lt;/b&gt;, Spanish, F&lt;br&gt;&#xA;  &lt;b&gt;Sailors:&lt;/b&gt;&lt;br&gt;&#xA;  Cook: &lt;b&gt;Felipe C. Ivarruz-Vicaza&lt;/b&gt;, Spanish, M&lt;br&gt;&#xA; Seaman: &lt;b&gt;Gareth R. Cavendish&lt;/b&gt;, Anglophone, M&lt;br&gt;&#xA; Seaman: &lt;b&gt;Wilma W. Chapman&lt;/b&gt;, Anglophone, F&lt;br&gt;&#xA; Seaman: &lt;b&gt;Lemar D. Malcombe&lt;/b&gt;, Anglophone, M&lt;br&gt;&#xA; Cabin Boy: &lt;b&gt;Samuel K. Pickett&lt;/b&gt;, Anglophone, M&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;b&gt;Nuestra Señora de Cavadonga&lt;/b&gt;&lt;br&gt;&#xA; &lt;b&gt;Officers:&lt;/b&gt;&lt;br&gt;&#xA; Captain: &lt;b&gt;Gertrude D. Hutchinson&lt;/b&gt;, Anglophone, F&lt;br&gt;&#xA; 1st Mate: &lt;b&gt;Marietta N. Clarke&lt;/b&gt;, Anglophone, F&lt;br&gt;&#xA; 2nd Mate: &lt;b&gt;Liao Jin-Zhong&lt;/b&gt;, Chinese, ??&lt;br&gt;&#xA; Boatswain: &lt;b&gt;Cecilia S. Ros-Gamiro&lt;/b&gt;, Spanish, F&lt;br&gt;&#xA;&lt;br&gt;&#xA;  &lt;b&gt;Sailors:&lt;/b&gt;&lt;br&gt;&#xA;  Cook: &lt;b&gt;Ruud D. Van Dijk&lt;/b&gt;, Dutch, M&lt;br&gt;&#xA; Seaman: &lt;b&gt;Alexander H. Frost&lt;/b&gt;, Anglophone, M&lt;br&gt;&#xA; Seaman: &lt;b&gt;Findlay C. Scott&lt;/b&gt;, Anglophone, M&lt;br&gt;&#xA; Seaman: &lt;b&gt;Quentin R. Callaghan&lt;/b&gt;, Anglophone, M&lt;br&gt;&#xA; Cabin Boy: &lt;b&gt;Tremaine L. Higgins&lt;/b&gt;, Anglophone, M&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;b&gt;Passenger Manifest #1&lt;/b&gt;&lt;br&gt;&#xA; &lt;b&gt;Jeanette J. Ambard&lt;/b&gt;, French, F&lt;br&gt;&#xA; &lt;b&gt;Chiaffredo Milanesi&lt;/b&gt;, Italian, M&lt;br&gt;&#xA; &lt;b&gt;Nazzareno D. Conti&lt;/b&gt;, Italian, M&lt;br&gt;&#xA; &lt;b&gt;Jean-Louis Fleuette&lt;/b&gt;, French, M&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;b&gt;Passenger Manifest #2&lt;/b&gt;&lt;br&gt;&#xA; &lt;b&gt;Austinafe Ajaranne&lt;/b&gt;, ??, ??&lt;br&gt;&#xA; &lt;b&gt;Jean-François Moreau&lt;/b&gt;, French, M&lt;br&gt;&#xA; &lt;b&gt;Louise-Violette Meserve&lt;/b&gt;, French, F&lt;br&gt;&#xA; &lt;b&gt;Uckbe Oronle&lt;/b&gt;, ?? ??&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;b&gt;Passenger Manifest #3&lt;/b&gt;&#xA; &lt;b&gt;Eileen Mac Ateer&lt;/b&gt;, Irish, F&lt;br&gt;&#xA; &lt;b&gt;Ylesme Arrce&lt;/b&gt;, ??, ??&lt;br&gt;&#xA; &lt;b&gt;Maura I. Kiley&lt;/b&gt;, Irish, F&lt;br&gt;&#xA; &lt;b&gt;Eannare Imke&lt;/b&gt;, ??, ??&lt;br&gt;&#xA;&lt;br&gt;&#xA; Here, &quot;Anglophone&quot; means English, Scottish, American, Australian, or Jamaican.  1800 was before the Great Famine in Ireland, so a) the Irish diaspora to the colonies was less pronounced, making the presence of a clearly Irish name there less likely; in any case, and American or Australian (etc.) with an Irish name would also have had an Irish accent while speaking English; and b) a larger portion of Irish people spoke the Irish language, and so would not be &quot;Anglophone&quot; anyway.  This distinction only gives really new information for one name.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Hint 2&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Although Scots were more likely to be mixed up into other parts of the British Empire (or America) by 1800 than the Irish, they do still have distinctive names; this is a one-way inference.  A person with a Scottish name is not necessarily Scottish, but a Scot will have a Scottish name. (For the purposes of the puzzle.)  The Scottish names are &lt;strong&gt;Ronald McPherson&lt;/strong&gt; and &lt;strong&gt;Findlay C. Scott&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10649" LastEditorUserId="10649" LastEditDate="2015-04-16T06:58:28.353" LastActivityDate="2015-04-16T14:25:12.943" Title="The Pirates of Beal Isle - very long logic grid puzzle" Tags="|logical-deduction|liars|mystery|logic-grid|" AnswerCount="1" CommentCount="26" ContentLicense="CC BY-SA 3.0" />
  <row Id="11475" PostTypeId="1" AcceptedAnswerId="11477" CreationDate="2015-04-02T19:23:23.010" Score="5" ViewCount="686" Body="&lt;p&gt;Each of 4 people is touring 36 cities separately. Each of the 36 squares in the illustration below represents a city. To make the answers shorter, each city has been given a 1-character name, from A-Z, and 0-9&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tqG6J.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Note: Although this puzzle is possible to solve without computer&#xA;programming knowledge, it's allowed and encouraged to use computer help.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Each city has been rated in the following 4 areas. The ratings go from -5 to 5.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NBrhT.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;A tour must include each of the 36 cities exactly one time, start at city A, and end at city 4. Each move must be either horizontal or vertical, along the small roads that connect the cities.&lt;/p&gt;&#xA;&lt;p&gt;As the tour progresses, all of the ratings of previously-visited cities accumulate (are added together) for each of the four categories to produce four total scores- one for each of the four categories. For example, going from A to B would leave the person feeling a -9 (-4 + -5) for Money-making Potential for city B, and a 0 (-5 + 5) for Overall Happiness for city B. Then, moving from B to C, Money-making Potential is now at -11, and Overall Happiness is at -2.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Find four tours&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Monroe believes that money is the root of all evil.&#xA;Find him a tour such that one or more cities on that tour minimize his Money-making Potential.&#xA;What is his favorite city (or cities) on that tour?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Hal is focused almost solely on being happy.&#xA;Find him a tour such that one or more cities on that tour maximize Overall Happiness.&#xA;What is his favorite city (or cities) on that tour?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Rose is into romance. Find her a tour such that one or more cities on that tour maximize Romance &amp;amp; Love. What is her favorite city (or cities) on that tour?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Arlene is an admirer of artistic beauty. Find her a tour such that one or more cities on that tour maximize Art &amp;amp; Architecture. What is her favorite city (or cities) on that tour?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Here is an example of a valid tour:&lt;br&gt;&#xA;ABCDEFLKJIHGMNOPQRXWVUTSYZ0123987654&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-06T16:25:29.117" Title="A tour of 36 cities" Tags="|optimization|computer-puzzle|calculation-puzzle|" AnswerCount="4" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="11477" PostTypeId="2" ParentId="11475" CreationDate="2015-04-02T20:42:34.643" Score="5" Body="&lt;p&gt;Here are two routes that I believe maximize happiness for Hal&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A B H G M S Y Z 0 1 2 W V U T N O I C D J P Q K E F L R X 3 9 8 7 6 5 4&lt;br&gt;&#xA;&lt;br&gt;&#xA; A B C D J I H G M S Y Z 0 1 2 W V U T N O P Q K E F L R X 3 9 8 7 6 5 4&lt;br&gt;&#xA;&lt;br&gt;&#xA; These routes both generate 23 happiness in city V&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And here is a route that should maximize Art for Arlene&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A G M S Y Z 0 U T N O P Q K J I H B C D E F L R X W V 1 2 3 9 8 7 6 5 4&lt;br&gt;&#xA;&lt;br&gt;&#xA;  This route generates 24 Art at cities J and I&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="239" LastEditorUserId="239" LastEditDate="2015-04-03T13:03:04.893" LastActivityDate="2015-04-03T13:03:04.893" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11478" PostTypeId="2" ParentId="11139" CreationDate="2015-04-02T21:50:04.657" Score="3" Body="&lt;p&gt;So here is my &quot;better answer&quot;, rescued from a close idea...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; all the words have a consecutive letters that are horizontally adjacent on the QWERTY keyboard:&lt;br&gt;&#xA; &lt;strong&gt;SA&lt;/strong&gt;DIST&lt;br&gt;&#xA; THOROU&lt;strong&gt;GH&lt;/strong&gt;FA&lt;strong&gt;RE&lt;/strong&gt;&lt;br&gt;&#xA; HY&lt;strong&gt;MN&lt;/strong&gt;&lt;br&gt;&#xA; &lt;strong&gt;POI&lt;/strong&gt;SED&lt;br&gt;&#xA; &lt;strong&gt;YU&lt;/strong&gt; &lt;strong&gt;RT&lt;/strong&gt;&lt;br&gt;&#xA; THI&lt;strong&gt;RTY&lt;/strong&gt;-SEVEN&lt;br&gt;&#xA; S&lt;strong&gt;TREW&lt;/strong&gt;N&lt;br&gt;&#xA; V&lt;strong&gt;ERT&lt;/strong&gt;ICALI&lt;strong&gt;TY&lt;/strong&gt;&lt;br&gt;&#xA; IN&lt;strong&gt;ERT&lt;/strong&gt;IA&lt;br&gt;&#xA; BALD&lt;strong&gt;ER&lt;/strong&gt;D&lt;strong&gt;AS&lt;/strong&gt;H&lt;br&gt;&#xA; HI&lt;strong&gt;GH&lt;/strong&gt;FLYING&lt;br&gt;&#xA; &lt;strong&gt;REW&lt;/strong&gt;ARDING&lt;br&gt;&#xA; &lt;strong&gt;OP&lt;/strong&gt;IN&lt;strong&gt;IO&lt;/strong&gt;NATED&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the winning two then are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; STREWN and VERTICALITY&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;And here, for reference, is the acknowledged &quot;weak answer&quot; I tried first:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The words all use at least two letters from the word &quot;THIRTEEN&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And finding words that are more so than the others...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clearly THIRTY-SEVEN is the winner, using all 8 letters, but VERTICALITY and INERTIA can each source 5 of their letters from THIRTEEN (not counting letters more times that they appear there), so I actually end up with three words &quot;better&quot; than the others.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8875" LastEditorUserId="8875" LastEditDate="2015-04-02T22:12:22.430" LastActivityDate="2015-04-02T22:12:22.430" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11480" PostTypeId="2" ParentId="11139" CreationDate="2015-04-02T22:14:45.060" Score="4" Body="&lt;p&gt;Each of them&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; contains a string of at least 3 that's consecutive in the word and forms a connected block on a QWERTY keyboard: SADIST =&gt; SAD, THOROUGHFARE =&gt; UGHF, HYMN =&gt; HYMN, POISED =&gt; POI/SED, YURT =&gt; YURT, THIRTY-SEVEN =&gt; RTY, STREWN =&gt; REW, VERTICALITY =&gt; ERT, INERTIA =&gt; ERT, BALDERDASH =&gt; DERDAS, HIGHFLYING =&gt; GHF, REWARDING =&gt; REWA, OPINIONATED =&gt; OPI.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Two of them have&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; two such strings that don't overlap: POISED has POI and SED, and BALDERDASH has DER and DAS (the latter two are contiguous but don't overlap).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9523" LastEditorUserId="5373" LastEditDate="2015-04-02T22:21:07.127" LastActivityDate="2015-04-02T22:21:07.127" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11481" PostTypeId="2" ParentId="11475" CreationDate="2015-04-02T23:21:49.803" Score="2" Body="&lt;p&gt;I reckon Monroe can sink to the lowest depths of economic potential with&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ABCIJDEFLKQRX398765Z012WVPOUTNHGMSY4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;now reaching the minimum in&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; his favorite city 2, at -32&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8875" LastEditorUserId="8875" LastEditDate="2015-04-07T19:11:46.010" LastActivityDate="2015-04-07T19:11:46.010" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11482" PostTypeId="2" ParentId="10839" CreationDate="2015-04-02T23:32:40.563" Score="2" Body="&lt;p&gt;Here's a generalisation of &lt;a href=&quot;https://puzzling.stackexchange.com/a/10866/5373&quot;&gt;David Richerby's answer&lt;/a&gt;, which I believe eliminates the problem &lt;a href=&quot;https://puzzling.stackexchange.com/questions/10839/ant-walking-on-a-number-line#comment33023_10866&quot;&gt;pointed out by Ypnypn&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say the ant starts at point $a$ on the real line, and we're looking at the finite interval $[a-m,a+n]$ for some real numbers $m,n$. Eventually the ant &lt;strong&gt;must&lt;/strong&gt; break out of this interval (with probability 1); on which side is it more likely to do so? At the moment when it first breaks out, it's still at point $a$ &lt;strong&gt;on average&lt;/strong&gt; (because it always is), but we know it's either at point $a-m$ or point $a+n$ (with probabilities $p$ and $1-p$, say). So $p(a-m)+(1-p)(a+n)=a$, which $\Rightarrow a+n-p(m+n)=a\Rightarrow p=\frac{n}{m+n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, in our problem we have $a=7,m=7,n=3$. So when the ant breaks out of the interval $[0,10]$ (i.e. when it reaches honey), it has a 30% chance of doing so at $0$ and a 70% chance of doing so at $10$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this slick and rigorous enough for you, xnor? I hope so, because I've just spent a long time pondering the problem in order to come up with this! ;-)&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-04-02T23:32:40.563" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11483" PostTypeId="2" ParentId="11409" CreationDate="2015-04-03T00:16:20.003" Score="3" Body="&lt;p&gt;My guess is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LICE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You call me a liar, and say I was the first.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; first letter of &quot;liar&quot; is &lt;b&gt;L&lt;/b&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I was the second, the first liar isn't cursed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Second letter  is &lt;b&gt;I&lt;/b&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll open, open, and tell you what I see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; see == &lt;b&gt;C&lt;/b&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I start to fight, you'd think I start with a knee.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;a knee&quot; == &quot;an &lt;b&gt;E&lt;/b&gt;&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whether you think I'm cool, or whether I be lame,&lt;br /&gt;&#xA;You'll often find me surrounded by many o' my dame.&lt;br /&gt;&#xA;A couple types of clues, riddle now you must,&lt;br /&gt;&#xA;And tell me what I am, before I turn to dust.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not entirely sure about these lines, but I went off of your clue that the riddle can be solved in only the first 4 lines. I think there's something important about &quot;surround by many 'o&quot; being something surrounded by the letter &lt;b&gt;O&lt;/b&gt;, but I haven't figured out its importance yet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9978" LastActivityDate="2015-04-03T00:16:20.003" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11484" PostTypeId="2" ParentId="11475" CreationDate="2015-04-03T01:01:06.017" Score="1" Body="&lt;p&gt;Here's a shot at Rose by manual labor. She can achieve love with this route:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;AGMNOIHBCDJKEFLRX398712WQPVUTSYZ0654&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; She hits her peak in cities 2 and W, at which point she has 32 romance points in the old satchel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10851" LastActivityDate="2015-04-03T01:01:06.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11485" PostTypeId="2" ParentId="11238" CreationDate="2015-04-03T02:22:48.983" Score="5" Body="&lt;p&gt;Maybe&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The wife committed pre-planned suicide due to a lengthy agonizing illness.  She didn't want him there.  By the time the elevator got to the 12th floor he knew she was dead.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4372" LastEditorUserId="46903" LastEditDate="2019-02-07T03:15:45.060" LastActivityDate="2019-02-07T03:15:45.060" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="11486" PostTypeId="1" AcceptedAnswerId="16728" CreationDate="2015-04-03T02:54:50.997" Score="15" ViewCount="616" Body="&lt;p&gt;Darkness has descended on the infinite plane of Euclidea. The only sources of light are $4$ lighthouses; they were built long ago, and cannot be moved. The bulbs of these lighthouses can be rotated any direction, and illuminate a $90^\circ$ sector of the plane. The Euclideans want to bring light to the entire plane: can they do it, no matter where the lighthouses were built?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about when there are $5$ lighthouses, but they only illuminate a $72^\circ$ sector?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Think of the light houses as being transparant, they don't cast a shadow).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; Statement was reworded to clarify that the lighthouse locations cannot be chosen.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="5373" LastEditDate="2015-04-03T09:44:55.360" LastActivityDate="2015-07-15T22:52:13.740" Title="Darkness in Euclidea" Tags="|geometry|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11488" PostTypeId="2" ParentId="11486" CreationDate="2015-04-03T03:15:22.363" Score="0" Body="&lt;h2&gt;Partial Answer&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This answer is only valid when the lighthouses form a convex polygon.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can't move the lighthouses, the solution is to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Find the centroid of the polygon formed by connecting the four lighthouses. For 4, you can cheat this by drawing 2 lines between the lighthouses on opposite sides. Point the lighthouses at the intersection of those lines.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/7stCS.png&quot; alt=&quot;Diagram 4&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;With 5 lighthouses, the solution is similar but a &lt;a href=&quot;https://math.stackexchange.com/questions/86073/geometrically-find-the-center-of-a-pentagon-or-hexagon&quot;&gt;little more complex&lt;/a&gt; to find the center point so I'll just approximate it graphically. (I know the diagram still shows the light cover a 90° sector instead of a 72° sector. That's just laziness on my part.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/KMo0u.png&quot; alt=&quot;Diagram 5&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As you've probably realized, this operation should work for any polygon with &lt;em&gt;n&lt;/em&gt; sides long and lighthouse coverage angle α degrees so long as &lt;code&gt;α ≥ 360/n&lt;/code&gt; is true.&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="-1" LastEditDate="2017-04-13T12:19:36.950" LastActivityDate="2015-04-03T05:01:06.470" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11489" PostTypeId="2" ParentId="77" CreationDate="2015-04-03T03:57:44.603" Score="3" Body="&lt;p&gt;You need to use better methods like the fidrich method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cross - Find a faster way to solve the cross. After you have learnt to solve the cross quicker, try doing it on the bottom to avoid cube rotations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F2L - After you have practiced F2L, learn some advanced F2L techniques. Especially try to look ahead, meaning when you know what to do to create one slot, while you are creating it and inserting it, look for another corner and edge and keep track of where they are, so when you go to that corner, you won't have trouble looking for the ege to pair it up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OLL - Try doing this in two steps. The first is to orient the edges with an algorithm , and then to orient the corners with an algorithm.Here is a link to a page that will show you the algorithms.&#xA;&lt;a href=&quot;http://badmephisto.com/2LookOLL.pdf&quot; rel=&quot;nofollow&quot;&gt;http://badmephisto.com/2LookOLL.pdf&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PLL - In the beginners method, it will teach you to do the PLL in two steps,only learning 7 algorithms. This is OK, but it is twice as slow as just learning the 21 algorithms (Try only to learn 1-2 a day, or your brain will overfill with algorithms) which you can find here.&#xA;&lt;a href=&quot;http://badmephisto.com/pll.php&quot; rel=&quot;nofollow&quot;&gt;http://badmephisto.com/pll.php&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 look OLL (This is only optional to get sub 8): When you have a very good average at about 15 to 20, if you want to improve your average to sub 10, learn the ful olls, which are 57 algorithms (Again, to not overfill, only learn one a day) which you can find on the internet.&lt;/p&gt;&#xA;" OwnerUserId="11004" LastActivityDate="2015-04-03T03:57:44.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11490" PostTypeId="1" AcceptedAnswerId="11520" CreationDate="2015-04-03T05:05:50.390" Score="8" ViewCount="425" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A rectangle is &lt;em&gt;proper&lt;/em&gt; if its width is a whole number, or its height is, or both. Show that if a rectangle can be cut into a finite number of proper rectangles, then that rectangle is itself proper.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As obligatory, I claim that this problem has a clever and unexpected solution. Actually, I know of two totally different such solutions.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="5373" LastEditDate="2015-04-03T15:53:27.200" LastActivityDate="2015-04-03T15:53:27.200" Title="Rectangles with a whole-number measurement" Tags="|mathematics|geometry|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11491" PostTypeId="1" AcceptedAnswerId="11495" CreationDate="2015-04-03T05:12:31.520" Score="56" ViewCount="4979" Body="&lt;p&gt;The following is a text rebus in the form of Java code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;public class GetBoolean {&#xA;    public static void main(String[] args) {&#xA;        boolean i = getBoolean();&#xA;    }&#xA;&#xA;    private static boolean getBoolean() {&#xA;        throw new BOOLEAN();&#xA;    }&#xA;&#xA;    private static class BOOLEAN extends RuntimeException {&#xA;    }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It represents part of a famous dialogue.  What dialogue is that?&lt;/p&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4450" LastEditDate="2015-04-03T05:33:15.553" LastActivityDate="2015-04-03T06:26:35.393" Title="What famous dialogue does this Java code represent?" Tags="|rebus|computer-puzzle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11492" PostTypeId="1" CreationDate="2015-04-03T05:35:53.480" Score="15" ViewCount="12479" Body="&lt;p&gt;I am a sub-50 cuber right now. I am currently using the Fridrich method, I take about 2-5 secs to solve the cross and I am learning the 21 PLL algorithms. I am learning one or two PLL every day. I am using the 2 look OLL that uses 10 algorithms in two steps. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With all these have only a record of 32 secs and a average of 46 secs. Do you have any tips on becoming a sub 30 cuber or even sub 20 or sub 15?&lt;/p&gt;&#xA;" OwnerUserId="11004" LastEditorDisplayName="user20" LastEditDate="2015-04-03T08:02:00.743" LastActivityDate="2016-02-05T23:54:09.577" Title="How do I become a sub-30 cuber?" Tags="|rubiks-cube|speedsolving|" AnswerCount="2" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11493" PostTypeId="2" ParentId="6204" CreationDate="2015-04-03T05:57:14.107" Score="4" Body="&lt;p&gt;&lt;strong&gt;Section 4&lt;/strong&gt; is a:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;a href=&quot;http://rumkin.com/tools/cipher/vigenere-keyed.php&quot; rel=&quot;nofollow&quot;&gt;keyed Vigenère cipher&lt;/a&gt;&lt;/strong&gt; with a passphrase and an alphabet key.&lt;br&gt;&#xA; The passphrase is &lt;strong&gt;thehuntingofthesnark&lt;/strong&gt;&lt;br&gt;&#xA; The alphabet key is &lt;strong&gt;Boojum&lt;/strong&gt; (or bojum to avoid duplicate letters)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Here are the steps that I used:&lt;br&gt;&#xA;&lt;br&gt;&#xA; - subtract the plaintext from the ciphertext (mod-26) to try to identify the type of cipher.  A majority of the characters are repeating at intervals of 20 or multiples of 20.  It looks like a messed-up Vigenère cipher with a likely passphrase length of 20 characters. Therefore, assume that the same 20 character passphrase is being used (thehuntingofthesnark).&lt;br&gt;&#xA;&lt;br&gt;&#xA; - split the plaintext and the ciphertext into 20 character segments and map the ciphertext to 20 rows of alphabets as shown in &lt;a href=&quot;http://cryptogram.org/cdb/aca.info/aca.and.you/chapter_09.pdf#QUAGM3&quot; rel=&quot;nofollow&quot;&gt;this example&lt;/a&gt; (this specific cipher is also called Quagmire III.)&lt;br&gt;&#xA;&lt;br&gt;&#xA; - and &lt;strong&gt;yes&lt;/strong&gt;, fragments of the passphrase now show up under the plaintext letter &lt;strong&gt;B&lt;/strong&gt;.  But, because the passphrase is long, only 25% of the map is populated (130 characters in a 20x26 grid) and the alphabet key is not obvious.&lt;br&gt;&#xA;&lt;br&gt;&#xA; - so back to the puzzle to find words that start with B which are Beaver, Bellman, Baker, Butcher, and Boojum. Boojum gives a match for the ciphertext so the analysis is &lt;strong&gt;complete&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Section 6&lt;/strong&gt; could be a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Vigenère cipher&lt;/strong&gt; with a null alphabet or a mixed alphabet.&lt;br&gt;&#xA;&lt;br&gt;&#xA; With an alphabet key of BOJUM, the passphrase would be: &lt;pre&gt;oojpzvqdglrszadgpuiyfjxmcvddciqutivnmaljxkazuzeitaizlcxurwenkyiipqhnqsraxaplcvlrhlvqaqsypkaumkrtpzgdqaddamjamadjqcgumkvfsosuhzoorpjsfrjjvol&lt;/pre&gt;&lt;br&gt;&#xA; With a null alphabet, the passphrase would be: &lt;pre&gt;ccdvztwinkxqzijccibxbdjfivijipwrfjimggtcwrgzeykifgdzshwrevqmxlqwcvhbpkpghgnshtseoyiogvyycjgdfkepbmqkcgjigfcfggjcoiodfkvmfoseupbbencqnpdduvz&lt;/pre&gt;&lt;br&gt;&#xA; However, these passphrases are the full length of the ciphertext which is no different than enciphering with a one-time pad.  So unless, there is some meaning to these passphrases, this solution for Section 6 is likely incorrect.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-04-07T05:45:38.667" LastActivityDate="2015-04-07T05:45:38.667" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11494" PostTypeId="2" ParentId="11490" CreationDate="2015-04-03T06:17:43.747" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Tile the plane with squares of side length &lt;span class=&quot;math-container&quot;&gt;$\frac12$&lt;/span&gt;, colored white and black in checkerboard fashion. Because of the periodicity of the coloring, any proper rectangle with sides parallel to the tiles will have the same white area as black area (they will be balanced). Place the larger rectangle, &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt;, so that one of its corners lines up with a corner of a white square. Since all of the subrectangles are proper, therefore balanced, &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; must be balanced. Now, suppose that &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; was improper. Let &lt;span class=&quot;math-container&quot;&gt;$w,h$&lt;/span&gt; be its width and height, and &lt;span class=&quot;math-container&quot;&gt;$w',h'$&lt;/span&gt; be those values rounded down. The rectanlges &lt;span class=&quot;math-container&quot;&gt;$[0,w']\times [0,h']$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$[0,w']\times [h',h]$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$[w',w]\times [0,h']$&lt;/span&gt; are all proper, therefore balanced, which would imply the the remaining rectangle, &lt;span class=&quot;math-container&quot;&gt;$[w,w']\times[h,h']$&lt;/span&gt;, is balanced as well. However, that last rectangle looks something like this:&#xA;&#xA; &lt;span class=&quot;math-container&quot;&gt;$\qquad\qquad\qquad\qquad\quad$&lt;/span&gt;&lt;img src=&quot;https://i.stack.imgur.com/1scVY.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;br /&gt;&#xA;&#xA; It is simple enough to verify this is imbalanced, using &lt;span class=&quot;math-container&quot;&gt;$(1/2-a)(1/2-b)&amp;gt;0$&lt;/span&gt;, a contradiction. Technically, I've only covered the case where &lt;span class=&quot;math-container&quot;&gt;$w-w'\ge \frac12$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$h-h'\ge\frac12$&lt;/span&gt;, but the other cases are even easier to check.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Thus, &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; must be proper.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-03T06:38:31.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11495" PostTypeId="2" ParentId="11491" CreationDate="2015-04-03T06:19:49.360" Score="53" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think it's the famous exchange from &lt;a href=&quot;http://en.wikipedia.org/wiki/A_Few_Good_Men&quot;&gt;A Few Good Men&lt;/a&gt; where the protagonist demands about the death of a soldier answers at a military tribunal from the guilty base commander&lt;br&gt;&#xA;    &lt;code&gt;Kaffee: I want the truth!&lt;/code&gt;&lt;br&gt;&#xA;    &lt;code&gt;Col. Jessup: You can't handle the truth!&lt;/code&gt;&lt;br&gt;&#xA; after which Colonel Jessup, played by Jack Nicholson, goes on his rant.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The &lt;code&gt;getBoolean&lt;/code&gt; method tries to get the truth (Boolean), but is unable to exception-handle it, causing a &lt;code&gt;RuntimeException&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-04-03T06:26:35.393" LastActivityDate="2015-04-03T06:26:35.393" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11496" PostTypeId="2" ParentId="10839" CreationDate="2015-04-03T07:33:58.000" Score="-3" Body="&lt;p&gt;&lt;strong&gt;Lets take a logical / mathematical approach&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the ant' current position is 7, we'll count the probability of reaching 10.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;After 1st move, total options for its current position will be 2 -&gt; 6&#xA;or 8 (it can go L/R from 7).&lt;/li&gt;&#xA;&lt;li&gt;After 2nd move, total options for its current position will be 4 -&gt;&#xA;(5,7) or (7,9)&lt;/li&gt;&#xA;&lt;li&gt;After 3rd move, total options for its current position will be 8 -&gt;&#xA;(4,6) or (6,8) or (6,8) or (8,10)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In above all possible combinations - the ant, in 1 scenario out of 8, reached 10 from 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;P(E) =    n(E)/n(S) = 1/8&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, probability of moving to 0 from 7 = &lt;strong&gt;1/128&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(Previous explanation)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probability (p) of a move at any particular side (L/R) = 1/2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p of next move, at any particular side = 1/2 .... and so on ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from 7 to 10 - it needs three moves - and as these events are mutually exclusive -&gt; probability = 1/2 * 1/2 * 1/2 = &lt;strong&gt;1/8 (or 0.125)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from 7 to 0 - probability = 1/2*1/2*1/2*1/2*1/2*1/2*1/2 = &lt;strong&gt;1/128 (or 0.0078125)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; - I am clarifying a bit on the doubt raised by @xnor:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One remote possibility - the ant can go to 8 in first move, return back to 7 in second move - which will make it move in SHM (Simple Harmonic Motion) and thus even after a million moves - it would be be still at 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's where the &lt;em&gt;Probability&lt;/em&gt; comes into picture. Probability NEVER guarantees that an event will occur. e.g. In the dice game, it is entirely possible that user can keep throwing dice again and again for an hour - but he doesn't get a 6 (or any number for that matter) in first 100 throws ........ probability of that will be (5/6)&lt;em&gt;(5/6)&lt;/em&gt;(5/6)... (total 100 times) = 0.0000000100622&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the selection of direction of move is mutually exclusive (.. left (-1) or right (+1), with 'equal chance'..), calculation will be as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;In first move (from 7), it may go left or right (towards 10).&#xA;chances of moving towards 10 = 1/2 (50%)  &lt;/li&gt;&#xA;&lt;li&gt;In second move (from 8), it may go left or right (towards 10).&#xA;chances of moving towards 10 = 1/2 (again 50%)&lt;/li&gt;&#xA;&lt;li&gt;In third move (from 9), it may go left or right (towards 10).&lt;br&gt;&#xA;chances of moving towards 10 = 1/2 (again 50%)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Here we just calculate the odds of achieving the result. THIS IS HOW PROBABILITY WORKS. On negative/impractical side, its entirely possible that ant keep moving left/right whole of its life and never reach a honey point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for total probability of reaching to point - we will multiply all three ...&#xA;total probability = 1/8 = 0.125 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Means for moves made, chances of not reaching honey point (10) are 124 out of 125, chances of reaching honey point is 1 out of 125 (Note - this works on basis of Permutations and Combinations - you can get it verified by a maths professor).&lt;/p&gt;&#xA;" OwnerUserId="11011" LastEditorUserId="11011" LastEditDate="2015-04-04T04:27:35.460" LastActivityDate="2015-04-04T04:27:35.460" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="11497" PostTypeId="2" ParentId="11492" CreationDate="2015-04-03T07:52:35.290" Score="15" Body="&lt;p&gt;Practice the speed of your algorithms specifically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Learning faster algorithms is great, but when people are practicing speedcubing, the single most important thing people overlook is that the majority of their time is spent &lt;em&gt;executing the algorithms&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Increasing the speed with which you execute your algorithms will drastically increase the speed at which you solve the puzzle, regardless of the algorithms you're using. The beginner's method takes me about 30-40 seconds to execute - and I'm not using OLL and PLL. I did it simply by increasing my speed of execution. &lt;em&gt;Then&lt;/em&gt; I learned OLL/PLL.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Take the algorithms and steps you know, and push the speed faster. Time &lt;em&gt;just&lt;/em&gt; the execution of one algorithm (or, if it's too fast to time, instead time several iterations of it).  Work on increasing the execution speed of the algorithms you're using. Determine which ones take the longest, and practice those.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, here's an example using a couple of the slow algorithms I created (though you might want to do something like ten tests):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  Edge 3-cycle | 3 iters |  9.27 | 10.86 | 10.54 |  9.99 | 10.40 |&#xA;Corner 3-cycle | 3 iters | 12.76 | 10.79 | 11.64 | 10.64 | 10.34 |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Therefore, the edge 3-cycle takes me an average of 3.4 seconds, and the corner 3-cycle takes me an average of 3.7 seconds. (These times are &lt;em&gt;very&lt;/em&gt; slow, but that's primarily because I'm seriously out of practice and using a poorly-lubricated, old cube.) &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Index the steps and algorithms you know (for instance, make the cross a step and F2L a step, then time specific last layer algorithms)&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Identify sequences that seem abnormally slow, or abnormally inconsistent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do this with a &lt;a href=&quot;http://support.microsoft.com/en-us/kb/214076&quot;&gt;standard error&lt;/a&gt;. From the above, my edge-cycle error is 0.244, and my corner-cycle error is 0.392. This betrays the fact that corner-cycles are drastically less consistent than edge-cycles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second run of 5 shows that I'm only a little out of practice, and that I really should relubricate and retension this cube. My average time for the corner-cycle dropped 0.5 seconds, and the error dropped to 0.273.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Practice those sequences' execution times until they're much more consistent.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;Try to identify where you might be getting caught - is one of your rotations going too far? (Chances are, at least one rotation is consistently out of alignment.)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This method should help you identify the slow points of your solution and improve those specifically. Then, practice tying the sped-up methods together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your visual processing time can't easily be trained, and that's just something that's going to drop the more you solve the cube.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorDisplayName="user20" LastEditDate="2015-04-03T08:00:39.473" LastActivityDate="2015-04-03T08:00:39.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11499" PostTypeId="2" ParentId="11486" CreationDate="2015-04-03T08:41:21.703" Score="3" Body="&lt;p&gt;TLDR:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4: yes&lt;br&gt;&#xA; 5+: no&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;Extended version:&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;How do we cover the area?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For $n$ lighthouses, each lighthouse illuminates an arc of $\frac{360°}n$.  This means that each lighthouse is responsible for exactly $\frac1n$ of the area--if they were all clumped together, they must all be pointed radially outward cutting the plane into a pie with $n$ equal slices.  If a lighthouse is not at the origin, it still must be oriented in a way that it covers that same pie slice.  The only way for it to do this is if it is pointed the same direction, and located somewhere in its slice's negative space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That didn't make sense?  I don't blame you.  It's easiest to understand with the 4 lighthouse case:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Take the cartesian coordinate system.  Each lighthouse is at the origin and pointed at one of the quadrants.  Their illumination borders are along the axes.  Lighthouse 1 is pointed into the $(x\ge0, y\ge0)$ quadrant; lighthouse 2 is pointed into the $(x\ge0, y\le0)$ quadrant, and so on.  Lighthouse 1 continues to cover its entire quadrant as long as it moves (translationally) anywhere in the $(x\le0, y\le0)$ quadrant.  A similar statement is true for lighthouse 2 and the $(x\le0, y\ge0)$ quadrant.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To form a solution for $n$ lighthouses:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Take a very large circle and divide it into $n$ equal slices.  Orient your circle such that all of the lighthouses are inside it, and exactly one lighthouse is &quot;inside&quot; each slice (a lighthouse directly bordering multiple slices can be chosen to be in any one of them).  Aim each lighthouse in the exact opposite direction of its chosen slice.  Voila 100% coverage.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For 4 lighthouses, you should always be able to choose your origin and the orientation of the axes such that one lighthouse is &quot;in&quot; each quadrant.  For 5 or more this is not always possible.  Take for example a clump of three lighthouses very far away from the other two, which are also clumped together.  There is no way to choose an origin such that the lighthouses are evenly distributed to all pie slices.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's a picture for the unexplained downvoters&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/kOCkt.png&quot; alt=&quot;An example of 4 lighthouses, 3 forming a triangle with 1 inside.&quot;&gt;&lt;br&gt;&#xA;Each quadrant is completely covered by its lighthouse, therefore the whole plane is covered.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10334" LastEditorUserId="10334" LastEditDate="2015-04-05T15:14:13.140" LastActivityDate="2015-04-05T15:14:13.140" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11501" PostTypeId="2" ParentId="10839" CreationDate="2015-04-03T08:52:57.220" Score="-1" Body="&lt;p&gt;Well, the probabilities were already counted, so I'll just post my &quot;&lt;em&gt;slick short solution&lt;/em&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ant starts at $7$. It will move left or right with $1/2$ probability and it will take him $3$ steps to reach $10$ ($8-9-10$) and $7$ steps to reach $0$ ($6-5-4-3-2-1-0$). So probabilities are indeed $30\%$ to reach $0$ and $70\%$ to reach $10$&lt;/p&gt;&#xA;" OwnerUserId="9352" LastActivityDate="2015-04-03T08:52:57.220" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11502" PostTypeId="1" AcceptedAnswerId="11503" CreationDate="2015-04-03T09:58:51.003" Score="14" ViewCount="1179" Body="&lt;p&gt;$5$ ants are initially arranged on the corners of a regular pentagonal table (edges are long $L$), as showed in this picture:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/i3RM4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At some point they all start moving with a constant speed $S$ and, at any moment, pointing the previous ant in counterclockwise order (as in the picture). They keep moving until they meet in the center $O$ of the table. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How long does it take to meet in $O$?&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;Also, how many times each ant rotated around $O$?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes: Consider the ants as points. Also, I kindly ask to everyone not to close-vote this question as math problem because it features a counterintuitive solution!&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-04-03T18:23:56.437" LastActivityDate="2015-04-04T16:13:27.363" Title="Ants playing tag!" Tags="|mathematics|geometry|lateral-thinking|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11503" PostTypeId="2" ParentId="11502" CreationDate="2015-04-03T10:45:25.207" Score="11" Body="&lt;p&gt;They take time $\left(1 + \frac{1}{\sqrt{5}}\right) L/S $. They rotate infinitely many times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each ant moves towards the next ant, decreasing the distance between them, but the other answer is moving slightly angled away. The ants remain in a pentagon, so this angle doesn't change, and the distance decreases at a constant rate. We'll find this rate, and from it determine when the ants meet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each ant moves towards the next one at speed $S$. But, the chased ant is also moving, angled away. Only the component of its velocity parallel to their displacement matters; the perpendicular component doesn't change their distance. These components are $(S \cos (2\pi/5), S \sin(2\pi/5))$, so the ant is increasing its distance away at rate $S \cos (2\pi/5))$. So, in total, the distance decreases at rate $S (1 - \cos (2\pi/5))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dividing their initial distance $L$ by this rate gives the time to meet. Using &lt;a href=&quot;http://1/(1-cos(2*pi/5))&quot;&gt;Wolfram Alpha&lt;/a&gt; to express this in radicals gives $$\left(1 + \frac{1}{\sqrt{5}}\right) L/S $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After rotating through some angle around the center, the ants are still in a regular pentagon, but scaled down. Since the angle of rotation isn't affected by scale, they still need to rotate through the same angle as they did at the start. So, that angle must be infinite.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-04-03T23:01:48.277" LastActivityDate="2015-04-03T23:01:48.277" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="11504" PostTypeId="1" AcceptedAnswerId="11513" CreationDate="2015-04-03T11:14:42.937" Score="11" ViewCount="7690" Body="&lt;p&gt;I cannot be touched, but I often touch, &lt;br&gt;&#xA;I have no shape or form,&lt;br&gt;&#xA;But I can be soft or rough:&lt;br&gt;&#xA;Can you guess what I am?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can be short or long,&lt;br&gt;&#xA;but don't always take up space.&lt;br&gt;&#xA;I can, however, forever be dwelling&lt;br&gt;&#xA;In any human's brain.&lt;/p&gt;&#xA;" OwnerUserId="9967" LastEditorUserId="5373" LastEditDate="2016-12-14T12:14:35.733" LastActivityDate="2016-12-14T12:14:35.733" Title="I can't be touched, but I often touch" Tags="|riddle|word|" AnswerCount="11" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11505" PostTypeId="2" ParentId="11504" CreationDate="2015-04-03T11:25:46.020" Score="3" Body="&lt;p&gt;I think it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; VOICE.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I can not be touched, but I often touch&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It has no physical existence, but you can be emotionally &lt;em&gt;touched&lt;/em&gt; by people's words/voice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I have no shape or form&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, no physical existence&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But I can be soft or rough&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Voices are commonly described as a &quot;soft voice&quot; or a &quot;rough voice&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Can you guess what I am?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes, but I don't yet know whether I'm right! :-)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I can be short or long,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Possibly a reference to short and long syllables in poetry.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;but don't always take up space&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Indeed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I can, however forever be dwelling&lt;br&gt;&#xA; In any human's brain.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Voices echoing around your head. Our thoughts are often expressed as internal voices.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-04-03T11:25:46.020" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11506" PostTypeId="2" ParentId="11504" CreationDate="2015-04-03T11:25:48.777" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Death &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can not be touched, but I often touch &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can't touch death, but death touches people!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no shape or form&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What's the shape of death?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I can be soft or rough&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It means painful or painless.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can you guess what I am?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Death&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be short or long,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It can be instantaneous or incredibly slow.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but don't always take up space&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Death doesn't need space&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can, however forever be dwelling&#xA;In any human's brain.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are 2 meanings: humans think about death frequently during their life, also depression can be considered a kind of &quot;alive death&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-04-03T12:25:39.307" LastActivityDate="2015-04-03T12:25:39.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11507" PostTypeId="2" ParentId="11504" CreationDate="2015-04-03T11:34:29.180" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A name&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can not be touched, but I often touch &#xA;I have no shape or form&#xA;But I can be soft or rough&#xA;Can you guess what I am?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; While a name has no physical form, people can be touched when called by the right person, and depending on who it is, it can be spoken soft or rough.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be short or long,&#xA;but don't always take up space&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Names themselves do not take up space. But when written, they do.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can, however forever be dwelling&#xA;In any human's brain.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oh how the name of my first loved one still lingers in my brain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10919" LastActivityDate="2015-04-03T11:34:29.180" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11508" PostTypeId="1" AcceptedAnswerId="14559" CreationDate="2015-04-03T11:40:02.423" Score="3" ViewCount="759" Body="&lt;p&gt;I was sent this image by somebody today and was told that it is a (close) &lt;em&gt;story board&lt;/em&gt; of a famous idiom (of some language). The emoticons might not be in order of the idiom's words.&#xA;Below is the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2uK9p.png&quot; alt=&quot;idiom image&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="11023" LastEditorUserId="5373" LastEditDate="2015-04-03T11:47:16.277" LastActivityDate="2016-01-01T14:24:05.087" Title="What idiom (any language/culture) this image depicts" Tags="|rebus|" AnswerCount="3" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="11509" PostTypeId="2" ParentId="11504" CreationDate="2015-04-03T11:48:47.487" Score="8" Body="&lt;p&gt;I guess&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;a memory&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I can not be touched, but I often touch&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;When thinking back of a memory you can get emotional&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I have no shape or form&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A memory has no shape; it's in the brain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But I can be soft or rough&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It can be about a soft event (happy) or rough (a murder)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can you guess what I am?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A memory&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I can be short or long,&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A memory can be short or long&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;but don't always take up space&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;sometimes you forget one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I can, however forever be dwelling In any human's brain.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;some memories like a trauma can haunt you forever&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9543" LastEditorUserId="5373" LastEditDate="2015-04-03T11:50:22.547" LastActivityDate="2015-04-03T11:50:22.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11511" PostTypeId="2" ParentId="10839" CreationDate="2015-04-03T11:59:38.867" Score="0" Body="&lt;p&gt;Re: David Richerby's answer and the apparent exception provided by Ypnypn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The posed question is a markov chain with absorbing states at 0 and 10. In Ypnypn's example, of 'what if the honey positions were at 9 and 10', the absorbing states of the markov chain change to 9 and 10. Since the starting position is to the left of both the absorbing states, there is now only one absorbing state (it becomes impossible to reach x = 10 since it terminates once it reaches x = 9), and David Richerby's solution does not generalise to this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A general solution (x = starting position, l = leftmost honeydrop, r = rightmost honeydrop, where l is less than or equal to r).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are three possibilities. x &amp;lt; l (one absorbing state: l), x &gt; r (one absorbing state: r) or l &amp;lt; x &amp;lt; r (two absorbing states: l and r). The first case (case 1) is the one posed by Ypnypn, while the latter (case 3) is the case posed in the original question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 1 (x &amp;lt; l): given sufficient time, the probability of reaching l is 100% (since l is the only absorbing state).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 2 (x &gt; r): given sufficient time, the probability of reaching r is 100% (since r is the only absorbing state).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 3 (l &amp;lt; x &amp;lt; r): given sufficient time, the probability of reaching r equals (x-l)/(r-l) and the probability of reaching l equals 1 - ((x-l)/(r-l)). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in this particular case, as others have stated, the probability of reaching 10 is (7-0)/(10-0) = 70% and 0 is 1 - 0.7 = 30%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't find any way to embed formulas so my apologies for the aesthetically unpleasing answer.&lt;/p&gt;&#xA;" OwnerUserId="11020" LastEditorUserId="11020" LastEditDate="2015-04-03T12:17:59.033" LastActivityDate="2015-04-03T12:17:59.033" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11512" PostTypeId="2" ParentId="3739" CreationDate="2015-04-03T12:02:33.567" Score="7" Body="&lt;p&gt;I do not have enough reputation to comment so I will post this in an answer.&#xA;The sound of &quot;Seagull&quot; and &quot;Sea Girl&quot; sound so strange that the old man might have realized that when the hunter was talking about feeding him seagull meat, he was actually talking about the niece. When he tasted seagull meat on the ship he realized his folly and hence committed suicide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just a trial on my part :) &#xA;Also I think that the seagull and sea-girl explanation sounds so cool.&lt;/p&gt;&#xA;" OwnerUserId="11026" LastActivityDate="2015-04-03T12:02:33.567" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11513" PostTypeId="2" ParentId="11504" CreationDate="2015-04-03T12:17:36.180" Score="17" Body="&lt;p&gt;I think you are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Words&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can not be touched, but I often touch&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can be emotionally touched by someone's words&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no shape or form&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Words have no inherent physical existence besides the sound waves transmitting them and physical objects we create to represent them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I can be soft or rough&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Words can be soft or rough in a &lt;a href=&quot;http://www.doorsofperception.com/handouts/hard-words-and-soft-words/&quot;&gt;very&lt;/a&gt; &lt;a href=&quot;http://sellingwisdom.blogspot.com/2010/09/soft-words-and-hard-words.html&quot;&gt;literal&lt;/a&gt; sense but this could also be taken as kind words or harsh words.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be short or long,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Words can be as short as one character or as long as the longest chemical compound name or other arbitrarily long thing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but don't always take up space&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Words take up physical space when written down or when signs are made but that's not their most common form&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can, however forever be dwelling / In any human's brain.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The words you hear from others or form yourself can stick in your memory for the rest of your life&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastActivityDate="2015-04-03T12:17:36.180" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11514" PostTypeId="2" ParentId="11504" CreationDate="2015-04-03T12:20:11.763" Score="1" Body="&lt;p&gt;Personally, given the details of the question, there are many things that can satify the given criteria.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of them have been taken in the previous answers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Name, Memory, Voices, Words&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some more can also be added:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A story, a Sentence etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So as long as nothing specific is given that converges towards the actual answer, IMHO I think the puzzle can have many similar answers.&lt;/p&gt;&#xA;" OwnerUserId="11026" LastActivityDate="2015-04-03T12:20:11.763" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11516" PostTypeId="2" ParentId="11504" CreationDate="2015-04-03T12:24:41.560" Score="1" Body="&lt;p&gt;I think you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Air&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can not be touched, but I often touch&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We generally don't feel the air around us, but when it is windy, air touches us.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no shape or form&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well, kind of obvious&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I can be soft or rough&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Depends on how windy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can you guess what I am?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can be short or long, but don't always take up space&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sometimes wind comes in short blows, sometimes as a sustained force.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can, however forever be dwelling In any human's brain.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people's heads seem to never absorb anything, and air is synonymous-y with empty. (Maybe a stretch.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4749" LastEditorUserId="435" LastEditDate="2015-04-03T16:58:07.113" LastActivityDate="2015-04-03T16:58:07.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11517" PostTypeId="2" ParentId="11504" CreationDate="2015-04-03T12:39:06.560" Score="1" Body="&lt;p&gt;Guess goes to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dream&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can not be touched, but I often touch &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you can not touched it , its often touch by mind&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no shape or form&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; they have no shape&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I can be soft or rough&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you see soft (sweet) or rough(dangerous) dream&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be short or long,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; dream is short or long &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but don't always take up space&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; dream not take space in real life&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can, however forever be dwelling In any human's brain.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; its always dwelling in any human's brain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4312" LastActivityDate="2015-04-03T12:39:06.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11518" PostTypeId="2" ParentId="11490" CreationDate="2015-04-03T12:41:45.260" Score="0" Body="&lt;p&gt;My humble attempt at a simple logic based solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Since when I cut a rectangle into proper rectangles, each of the&#xA; smaller rectangles will have whole numbered widths and heights. &#xA; &lt;br&gt;&lt;br&gt; &#xA; &lt;strong&gt;&lt;em&gt;The total width of the rectangle = sum of all widths of smaller rectangles.&lt;/em&gt;&lt;/strong&gt;  &lt;br&gt;&lt;br&gt; Since &lt;em&gt;all the widths are whole numbers their&#xA; sum is a whole number&lt;/em&gt;.  &lt;br&gt;&lt;br&gt; Therefore the &lt;strong&gt;width of the&#xA; original rectangle is a whole number&lt;/strong&gt;.  &lt;br&gt;&lt;br&gt;  Same logic applies&#xA; for heights. &lt;br&gt; Since height and width are whole numbers the&#xA; rectangle is a proper rectangle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Apparently this solution has some shortcomings according to the given details of the problem. I will be improving on it.&lt;/p&gt;&#xA;" OwnerUserId="11026" LastEditorUserId="11026" LastEditDate="2015-04-03T12:43:34.330" LastActivityDate="2015-04-03T12:43:34.330" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11519" PostTypeId="2" ParentId="6252" CreationDate="2015-04-03T13:34:32.543" Score="14" Body="&lt;p&gt;What a nice question!  Of course, there is no perfectly correct answer, as our historical knowledge is far too incomplete. I list some puzzles from before 1000 BC.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;2700 BC: Carved stone balls. They show all regular polyhedra and the cubo-octahedron.  There purpose is unclear, but they are usually listed as the starting point of recreational mathematics.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;2300 BC: Tablets of Nippur. List divisors of $60^4$ in geometric progressions. Most likely &quot;serious&quot; research, and not puzzle-type entertainment.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;1700 BC: Phoenician puzzle jugs in Cyprus. Puzzle jugs are the oldest known mechanical puzzles. Several examples can be seen in the Metropolitan Museum of Art in New York. Puzzle jugs came back into fashion in ancient Greece (&lt;a href=&quot;http://en.wikipedia.org/wiki/Pythagorean_cup&quot;&gt;http://en.wikipedia.org/wiki/Pythagorean_cup&lt;/a&gt;), and much later in the 18th and 19th century, in France, Germany, England.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;1650  BC: The Rhind Papyrus. Problem 79 is a kind of St. Ives puzzle. Problem 40 is a fair division puzzle.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;1400 BC: Morris boards in Kurna (Egypt). &lt;a href=&quot;http://en.wikipedia.org/wiki/Three_Men&amp;#39;s_Morris&quot;&gt;wikiipedia&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All the listed dates are of course estimates. Magic squares were introduced in China perhaps around the same time, but the earliest sources (available nowadays) that mention them are from 700 BC.&lt;/p&gt;&#xA;" OwnerUserId="9125" LastActivityDate="2015-04-03T13:34:32.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11520" PostTypeId="2" ParentId="11490" CreationDate="2015-04-03T13:48:02.033" Score="6" Body="&lt;p&gt;There is beautiful article on this problem that summarizes 14 different solution approaches:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Stan Wagon: &quot;Fourteen Proofs of a Result About Tiling a Rectangle&quot;&lt;br&gt;&#xA;  The American Mathematical Monthly 94, 1987, pp. 601-617&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is one very elegant proof via complex double integrals.My favourite proof is as follows :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Proof via prime numbers (due  to Raphael Robinson, University of California at Berkeley)&lt;/strong&gt;&lt;br&gt;&#xA;  We claim that for each prime &lt;em&gt;p&lt;/em&gt;, either the height or the width of the big reactangle $R$ is within $1/p$ of an integer. It follows that one of these is an integer.&lt;br&gt;&#xA;  To prove the claim, scale the entire tiling up by a factor of $p$ in each direction, and consider the tiling obtained by replacing all tile-corners $(x, y)$ in the scaled-up tiling by $([x], [y])$. This yields an integer-sided rectangle tiled by integer-sided rectangles, each of which has one side a multiple of $p$. Therefore, the area of the large integer-sided rectangle is a multiple of $p$, whence one of its sides must be a multiple of $p$. Moreover, the dimensions of this rectangle differ from the dimensions of the scaled-up rectangle by less than $1$. It follows that $R$ has a side that differs from an integer by less than $l/p$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9125" LastActivityDate="2015-04-03T13:48:02.033" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11521" PostTypeId="2" ParentId="11163" CreationDate="2015-04-03T14:02:38.643" Score="8" Body="&lt;p&gt;This seems to work for everything in the original post, although I'm not sure the first hint lines up:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Take the Roman numeral representation of a number, and the result is how many numerals in a row begin with a '5' or '1' (e.g. 47 &gt; X&lt;strong&gt;LV&lt;/strong&gt;II &gt; 1&lt;strong&gt;2&lt;/strong&gt;2)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Am I on the right track?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;CLARIFICATION&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Convert the input to Roman numerals. Look at the Arabic value of each character in turn. Count the number of characters in a row that start with the same number in Arabic (it'll be sequences of 1s and 5s)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;pre&gt;&lt;code&gt;&#xA;     Input    Output  Roman Input     Roman Values in Arabic&#xA;     1        1       I               1&#xA;     2        2       II              1,1&#xA;     3        3       III             1,1,1&#xA;     82       15      LXXXII          50,10,10,10,1,1&#xA;     47       122     XLVII           10,50,5,1,1&#xA;     58458    2433    LVMMMCDLVIII    50000,5000,1000,1000,1000,100,500,50,5,1,1,1&#xA;     2433     316     MMCDXXXIII      1000,1000,100,500,10,10,10,1,1,1&#xA;     316      411     CCCXVI          100,100,100,10,5,1&#xA;     411      112     CDXI            100,500,10,1 &lt;/code&gt;&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11030" LastEditorUserId="9000" LastEditDate="2015-04-03T15:19:47.123" LastActivityDate="2015-04-03T15:19:47.123" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="11522" PostTypeId="1" AcceptedAnswerId="11652" CreationDate="2015-04-03T15:03:59.733" Score="17" ViewCount="1834" Body="&lt;h2&gt;YOU HAVE TO HELP!&lt;/h2&gt;&#xA;&lt;p&gt;I was time traveling with Fred last week - you know Fred, he's a nice guy - when we got separated. His hand-held automatic chronological portal manipulation and guidance unit (HHACPMGU) (we're working on a better name) malfunctioned! We were eating ice cream with Richard Feynman and talking about our latest visit to Sumer when Fred's pocket vibrated, the wormhole opened, and he was gone!&lt;/p&gt;&#xA;&lt;p&gt;I raced home - time, not space - because our contingency plan was always to leave a message is a particular secure location so that the other could retrieve it in our home time. Fred wasn't waiting for me so I knew his remote couldn't get him home and he needed me to come save him. I dug in the exact coordinates that we'd agreed upon and found a small metal box. It contained a sheet of metal with an encoded message engraved on its surface. I had &lt;em&gt;tried&lt;/em&gt; to tell Fred that there was no way that anybody would ever find the message should we ever need to send one but he was always more paranoid that someone would discover our time traveling secrets.&lt;/p&gt;&#xA;&lt;p&gt;Now, because Mr Paranoid went and got himself stuck in the past, &lt;em&gt;I&lt;/em&gt; have to try to decode this thing so I can go rescue him. Knowing him, it isn't even a direct message. It's probably just something &lt;em&gt;related&lt;/em&gt; to where / when he is. It doesn't matter how long it takes to figure out - time traveler, remember? - but I have to get it done or Fred is stuck!&lt;/p&gt;&#xA;&lt;h2&gt;When and where should I meet Fred?&lt;/h2&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;DCAXJBNHACPAQCAACAUGBWFBSOBRRBJEATQBNFAZVBGPAUQBUHBKOALHBWWBODAJJAAAADDAFJAIGAYEBODASNBUUBSBARKBEAACPAYGBUTBPCAUGBGOAECAXQBSBAHCBUGBDOAGNAIBBYKBANASBAUNBNFBVDBACABNAFIAEPATLBVIBFBBLCAGNAIIBXKBJFBESAZMBYYBNGACGAIGATGBPNBMCAGHAXVBQMBAIAERADBAKHBUMBVIBEHAGCAJIBEAAKHADQAJPALLBCLAWTBXUBOKBMJBRJBBBAWWBRCAVRBIFBXEBFJAVCBYRBOKBHIAPCARDBNCAAOAYGBRJBXEBBHAJFAWJBDQAIBBDDBBSAZZBHEANCBFHAFBBBMAEEBPCAYUBFDBDOAYUBRRBFOAQMBXUBLHBDNANHARRBEHA&lt;br&gt;&lt;br&gt; &lt;br&gt;&lt;br&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Addition 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;I have a busy weekend planned so I wanted to get this information out there before it started. You'd think I could free up some time with my HHACPMGU but you have to be &lt;em&gt;really&lt;/em&gt; careful about not running into yourself in the past. Fred and I used to travel with Kevin until The Incident. That took &lt;em&gt;forever&lt;/em&gt; to clean up. Poor Sally, too, left with all those kids... Anyway, that's why I don't use the HHACPMGU for such trivial things so I don't have as much time to try decryption this weekend. I hope you can still help!&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; I found another box under the first one with a plaintext message! I guess Fred knew I would never solve this without at least knowing where to start. The message told me that the decrypted text would only include uppercase alphanumeric characters. I probably would have guessed that by myself so it's a good thing that the message &lt;em&gt;also&lt;/em&gt; said that the encrypted version is longer than the decrypted version. It didn't say how much longer, though, because I guess Fred is more of a paranoid jerk than I thought.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Addition 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;I manged to find some time this weekend - despite all that's happening - so I could check in on the progress. A comment from Ian MacDonald got my attention. I did a quick analysis of character frequency and I got some weird numbers. There is something definitely going on here...&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; A 17%, B 22%, C 4%, D 4%, E 4%, F 3%, G 4%, H 3%, I 3%, J 3%, K 2%, L 2%, M 3%, N 3%, O 2%, P 2%, Q 2%, R 3%, S 2%, T 2%, U 2%, V 2%, W 2%, X 1%, Y 2%, Z 1%&#xA; Not only do these not line up &lt;em&gt;at all&lt;/em&gt; with a typical character distribution, but there are &lt;em&gt;way&lt;/em&gt; too many As and Bs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Addition 3:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;An amazing thing happened last night. A Western Union delivery man shows up on my doorstep with an old, weathered envelope. He says that they've been holding on to this letter for a long time with specific intructions on when and where to deliver it. In fact, the office even had a bet going on whether or not I'd even be here. The delivery guy had lost the bet, it seems. I figure this must be a letter from Fred so I rip open the package and pull out a single sheet of paper with the following message:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; I am a programmer. I use zero-based numbering. Z = 25&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Well, I'm &lt;em&gt;not&lt;/em&gt; a programmer, Fred! Now I have to go lookup what you're talking about. I was going to ask the Western Union delivery guy exactly how long they've had this letter as that would answer half my question but he was already gone. I suppose he was upset about losing the bet.&lt;/p&gt;&#xA;&lt;p&gt;Ah, well. Back to work, I suppose. How does this work with the &amp;quot;every third letter&amp;quot; pattern that Tryth found?&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-07T00:16:18.703" Title="My friend is lost in time!" Tags="|cipher|" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="11523" PostTypeId="2" ParentId="11490" CreationDate="2015-04-03T15:15:47.247" Score="5" Body="&lt;p&gt;Align the bottom left corner of the big rectangle with a checkerboard whose squares have side 1/2. Let the bottom left corner be black.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Two intuitive lemmas:&lt;/strong&gt;&lt;br&gt;&#xA;If both sides of a rectangle aren't whole, then the black area covered will be greater than white area covered.&#xA;On the contrary, if the black and white areas are equal, then the rectangle must have an integer side.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt;&#xA;Every small rectangle has at least a integer side, so it covers equal areas of black and white.&#xA;So, the big rectangle (which is the sum of all small rectangles) must cover equal areas of black and white, thus must have an integer side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/V23VZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Credits:&lt;/strong&gt; I've read this evidence somewhere few years ago, I don't remember the author though. After googling a bit, I've found out that Stan Wagon has posted &lt;a href=&quot;https://www.maa.org/sites/default/files/pdf/upload_library/22/Ford/Wagon601-617.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;14 solutions&lt;/a&gt; of this problem, incredible! &lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-04-03T15:23:17.233" LastActivityDate="2015-04-03T15:23:17.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11524" PostTypeId="2" ParentId="11504" CreationDate="2015-04-03T16:52:32.377" Score="2" Body="&lt;p&gt;I'd say that you're&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stories&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I can not be touched, but I often touch&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stories are intangible, but can be very &quot;touching&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I have no shape or form&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, because it's just a collection of words, it doesn't need a specific shape or form&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But I can be soft or rough&#xA;Can you guess what I am?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stories can be ascribed qualities such as soft (easy to read/understand) and rough (hard to comprehend or emotionally trying) ...and yes, I can guess what you are ;-)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I can be short or long,&#xA;but don't always take up space&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Short and long stories are still stories, but the actual &lt;em&gt;story&lt;/em&gt; need not occupy physical space&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I can, however forever be dwelling&#xA;In any human's brain.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Come on, we all have stories that we can't forget.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9038" LastActivityDate="2015-04-03T16:52:32.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11525" PostTypeId="2" ParentId="3859" CreationDate="2015-04-03T17:01:48.583" Score="-4" Body="&lt;p&gt;I think the killer was never in the car.  The hole in the windshield came from someone shooting inward to the guy via the AC vent ducts with the engine open.  It ricocheted and went back out the windscreen &lt;/p&gt;&#xA;" OwnerUserId="11001" LastActivityDate="2015-04-03T17:01:48.583" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11526" PostTypeId="2" ParentId="10839" CreationDate="2015-04-03T17:10:45.700" Score="-1" Body="&lt;p&gt;In the interest of a simple, elegant solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $P(x)$ be the probability that the ant will stop at 10 before stopping at 0 given an initial starting point of $0\le x\le10$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(0) = 0$ because it's already at 0.&lt;br&gt;&#xA;$P(10) = 1$ because it's already at 10.&lt;br&gt;&#xA;$P(5) = \frac12$ because the ant is equally likely to move in either direction, and both stopping points are equally far apart, neither can be favored.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $P(x)$ is a curve through these three co-linear points.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Occam%27s_razor&quot; rel=&quot;nofollow&quot;&gt;Occam's razor&lt;/a&gt; says that $P(x)$ is a line segment starting at the origin with slope $\frac1{10}$.  Therefore $P(7) = \frac7{10}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $P(x)$ is a step function with equal sized steps because the ant makes equal length steps.  Therefore $P(7) = \frac7{10}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Permission is hereby given to reuse this framework in another answer with a slightly more rigorous proof than any above.  I'm making this CW in case anyone else wants to chime in with a similarly spirited &quot;proof&quot;.&lt;/p&gt;&#xA;" OwnerUserId="10334" LastActivityDate="2015-04-03T17:10:45.700" CommentCount="0" CommunityOwnedDate="2015-04-03T17:10:45.700" ContentLicense="CC BY-SA 3.0" />
  <row Id="11527" PostTypeId="2" ParentId="11409" CreationDate="2015-04-03T17:13:21.477" Score="4" Body="&lt;p&gt;Building off of NonsenseSynapse and rand al'thor, I'm going to guess&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; KICK&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You call me a liar, and say I was the first.&#xA;But I was the second, the first liar isn't cursed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You thought the first letter was 'i', but it's the second&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll open, open, and tell you what I see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'See' -&gt; 'C'. Taken literally, we need to open the word &quot;open&quot; and take what we see visually: Half of 'O' is 'C'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I start to fight, you'd think I start with a knee.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'Knee' starts with 'K'. In addition, kicks are used to start fights, and they begin at the knee&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whether you think I'm cool, or whether I be lame,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Shoes are sometimes called kicks, and boys often judge each other by how cool or lame their shoes are.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'll often find me surrounded by many o' my dame.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure, but &quot;surrounded by many o&quot; could reference soccer balls?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A couple types of clues, riddle now you must,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And tell me what I am, before I turn to dust.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well, at the end of a kick the foot returns to the ground, or &quot;dust&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Additional evidence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In comments you indicated that the answers &quot;LICE&quot;, &quot;LOCK&quot;, and &quot;LINK&quot; each have 2 correct hints, which is how I got &quot;_ICK&quot;. In response to NonsenseSynapse you said &quot;the end will match the beginning&quot; so I put the last letter first.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10346" LastActivityDate="2015-04-03T17:13:21.477" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11528" PostTypeId="2" ParentId="11504" CreationDate="2015-04-03T17:37:06.467" Score="0" Body="&lt;p&gt;I think you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a song. Songs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;cannot be touched but often touch us emotionally. They are intangible but&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the medley can be soft or rough.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They vary in length but take up no space.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I get song lyrics stuck in my head all the time. :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11046" LastEditorUserId="5373" LastEditDate="2015-04-03T20:05:48.743" LastActivityDate="2015-04-03T20:05:48.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11529" PostTypeId="2" ParentId="10839" CreationDate="2015-04-03T17:41:49.957" Score="0" Body="&lt;p&gt;The answer has been given many times over so here's my thought process:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the ant is at position X between positions A and B with an equal chance to travel in either direction...&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A - - - - - - X - - B&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;... then the odds of reaching the end of either side first is the &lt;strong&gt;&lt;em&gt;ratio of the length of that side to the length of the entire line&lt;/em&gt;&lt;/strong&gt;. For the AXB example, that means the odds of reaching A first is given by $$\frac{(B-A)-(X-A)}{(B-A)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which simplifies to just $$\frac{B-X}{B-A}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the odds of reaching B first is therefore $$[1-\frac{B-X}{B-A}]$$&lt;/p&gt;&#xA;" OwnerUserId="9000" LastActivityDate="2015-04-03T17:41:49.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11530" PostTypeId="2" ParentId="11409" CreationDate="2015-04-03T17:52:01.013" Score="1" Body="&lt;p&gt;I think it's &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Silk &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You call me a liar, and say I was the first.&#xA;But I was the second, the first liar isn't cursed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter I is the second, not the first. The first is S, similar of snake, the liar of Eva and Adam.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll open, open, and tell you what I see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A reference to Allen key, similar to letter L&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I start to fight, you'd think I start with a knee.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It suggests a K&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whether you think I'm cool, or whether I be lame,&#xA;You'll often find me surrounded by many o' my dame.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's a fabric used to make clothes, ladies love it!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-04-03T17:52:01.013" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11532" PostTypeId="2" ParentId="11409" CreationDate="2015-04-03T18:23:07.237" Score="0" Body="&lt;p&gt;Few additional answers, mostly based on what has already been said. &lt;em&gt;If one is correct, I'll edit this post and make it look better.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Option #1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mien (meaning aspect). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You call &lt;strong&gt;M&lt;/strong&gt;E a liar. ME is the pronounce of the Greek letter Mi. &lt;strong&gt;I&lt;/strong&gt; was the second. &quot;Open, open, see&quot; may refer to &lt;strong&gt;E&lt;/strong&gt;. Knee is pronounced as &lt;strong&gt;N&lt;/strong&gt;I, another Greek letter. Also, mien can be cool or &quot;lame&quot;, and ladies pay attention to it.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Option #2&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Milk  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; M and I as said above, L as in the other answer, K because of knee. Also, milk is extracted by females (dames) and some don't like its taste.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Option #3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sign&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; S, I and N are explained in other answers, the G letter is hard to explain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Option #4:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lien &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The letters are all explained in the other answers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Option #5&lt;/strong&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nips&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I start with a Knee (N), second letter is I. P and S aren't easy to justify.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Most of these are garbage, I know, but they're attempts. If none of these is correct, I give up ahah!&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-04-03T18:57:50.293" LastActivityDate="2015-04-03T18:57:50.293" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11534" PostTypeId="2" ParentId="11409" CreationDate="2015-04-03T18:44:07.343" Score="1" Body="&lt;p&gt;I could be way off but is it a  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;Mirror&lt;/code&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;You call me a liar, and say I was the first:&lt;/code&gt;   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What we see in the mirror is not real. It's only a reflection of reality.&#xA;(A mirror is an object that reflects light in such a way that, for incident light in some range of wavelengths, the reflected light preserves many or most of the detailed physical characteristics of the original light.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;But I was the second, the first liar isn't cursed:&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It's the second form of reality. Another thought was that mirrors are not the only ones reflecting. The first reflector is Water. Later came the mirrors. Water is not cursed but few Mirrors are believed to be cursed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;I'll open, open, and tell you what I see:&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Looking glass&quot; is another term for a Mirror.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I start to fight, you'd think I start with a knee:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I can't really connect this &amp;amp; the mirror. But there's a website called mirror which is for soccer/football. So, 'fight' could mean 'match'. 'Start with a knee' could mean 'kick'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Whether you think I'm cool, or whether I be lame, You'll often find me surrounded by many o' my dame:&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Women love mirrors, any mirror will work... whether it's a good mirror or a lame one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;And tell me what I am, before I turn to dust:&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Before it's covered by dust or broken by falling on the ground (dust).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11049" LastEditorUserId="9951" LastEditDate="2015-04-03T18:51:34.653" LastActivityDate="2015-04-03T18:51:34.653" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11535" PostTypeId="2" ParentId="11504" CreationDate="2015-04-03T19:43:10.730" Score="0" Body="&lt;p&gt;It could be...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Look (Sight/ View/ Eye)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can not be touched, but I often touch &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Look/View can't be touched but the Look/View can touch the object/person being looked at.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no shape or form&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Look (in verb sense) has no shape or form, but the result of a look can be an image on the eye/brain?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I can be soft or rough&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Look can be a soft look or one filled with anger/hate (glare)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be short or long,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Look (verb sense) can be for a short time (glimpse) or for a long time (stare/gaze) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but don't always take up space&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Look doesn't take up space&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can, however forever be dwelling In any human's brain.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Look and the image of the look can become a memory in a human's brain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11049" LastActivityDate="2015-04-03T19:43:10.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11536" PostTypeId="1" CreationDate="2015-04-03T20:41:02.093" Score="1" ViewCount="888" Body="&lt;p&gt;Vasily Lalozik is a professional cryptographer and develops new security schemes for banks. But he has one problem: David Banes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The American Bank is the largest bank in the world; Banes has his sights steered towards it. Lalozik has created a very tough security scheme for the bank, which is considered almost impossible to crack by any outsider. The scheme is very simple and straightforward. The man wanting to enter the money container has to input the answer of a pattern. The pattern's technique of solving has been disclosed only to the owner and the respective manager. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On June 9th, Banes managed to get inside the bank and has only one remaining task: to crack the puzzle and run away with approximately 4 billion dollars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pattern is this:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;192-30-55-86-43-22-79-3-?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What number comes next in the series?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="10704" LastEditorUserId="5188" LastEditDate="2015-06-13T17:04:28.463" LastActivityDate="2015-06-13T17:04:28.463" Title="Lalozik Pattern" Tags="|pattern|strategy|number-sequence|" AnswerCount="3" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="11537" PostTypeId="1" AcceptedAnswerId="15256" CreationDate="2015-04-03T20:44:16.590" Score="3" ViewCount="740" Body="&lt;p&gt;&lt;em&gt;At the time of posting, this was the 2015th question posted on Puzzling Stack Exchange, in the year 2015.&lt;/em&gt;&#xA;&lt;hr&gt;&#xA;Take the number of multiples of 5 among the numbers featured in this puzzle.&lt;br&gt;&#xA;Multiply it by the number of types of creature beyond red twisted ter'angreal doorways.&lt;br&gt;&#xA;Subtract the figure you now have from the number of Scheherezade's tale narrated by Poe.&lt;br&gt;&#xA;Divide by the number of children fathered by Sidorio.&lt;br&gt;&#xA;Add the number of entries in each Classic FM Hall of Fame.&lt;br&gt;&#xA;Subtract the product of the number of siblings of Frodo Baggins's mother and the number of legitimate children fathered by Walder Frey.&lt;br&gt;&#xA;Add the number of days from now (the date of posting, 3 April) until the Easter Day celebrations in Romania.&lt;br&gt;&#xA;Subtract the number of Sitters in the &quot;Hall of the Tower&quot; in Salidar.&lt;br&gt;&#xA;Add the number of stars on the banner of Gondor.&lt;br&gt;&#xA;Subtract the number of elements named after Ytterby.&lt;br&gt;&#xA;Subtract the figure you now have from the user ID of the most prolific question-poster on Puzzling.SE.&lt;br&gt;&#xA;Add the number of primes greater than 10 among the numbers featured in this puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the number you now have?&lt;/strong&gt;&#xA;&lt;hr&gt;&#xA;Information from comments on answers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the &lt;a href=&quot;http://scifi.stackexchange.com&quot;&gt;SciFi &amp;amp; Fantasy&lt;/a&gt; crowd (alexwlchan and Michael Edenfield) have got all but one of the numbers they need. You just need to correct their answers in a single place and you'll get the right solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; apparently neither of them has read as far as &lt;em&gt;Immortal War&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-05-20T23:59:27.223" LastActivityDate="2015-05-21T06:27:27.970" Title="The 2015th question on Puzzling SE" Tags="|calculation-puzzle|knowledge|" AnswerCount="4" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="11538" PostTypeId="2" ParentId="11502" CreationDate="2015-04-03T20:46:35.983" Score="1" Body="&lt;p&gt;Each ant is 11 pixels in diameter. So, given a bounding radius of 5.5 pixels and a start radius of 130 pixels (I approximated the diagram), they travel 3.82 radians before running into each other, at a rate of 2 pixels per iteration and assuming Float32 as the maximum accuracy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a live demo I created in javascript.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;View the Demo &lt;strong&gt;&lt;a href=&quot;https://jsfiddle.net/FlavorScape/7tawcgfw/&quot; rel=&quot;nofollow&quot;&gt;https://jsfiddle.net/FlavorScape/7tawcgfw/&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, after it completes, they spin off into infinity (because the bounding box does not stop them from approaching the line below the Float32 accuracy, is considered zero and all points gravitate to -1,1 from the &quot;fake&quot; center at 200 until they hit actual pixel 0,0 then do some other stuff, i'm not sure why.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;lt;!DOCTYPE html&amp;gt;&#xA;&amp;lt;html&amp;gt;&#xA;&amp;lt;head lang=&quot;en&quot;&amp;gt;&#xA;    &amp;lt;meta charset=&quot;UTF-8&quot;&amp;gt;&#xA;    &amp;lt;title&amp;gt;&amp;lt;/title&amp;gt;&#xA;&amp;lt;/head&amp;gt;&#xA;&amp;lt;body&amp;gt;&#xA;&#xA;    &amp;lt;script language=&quot;javascript&quot;&amp;gt;&#xA;&#xA;      window.onload = function(){&#xA;          var ants = [];&#xA;          var radius = 106;&#xA;          var center = [200,200];&#xA;          var step = 2;&#xA;          var interval = 33;&#xA;          var antRadius = 11;&#xA;          var done= false;&#xA;          var t = 0;&#xA;          var startRotation4;&#xA;&#xA;          var angle = 0;&#xA;          for( var i = 0; i &amp;lt; 5; i ++ )&#xA;          {&#xA;              angle += 2 * Math.PI / 5;&#xA;&#xA;              var ant = document.getElementById( 'ant-' + i);&#xA;&#xA;              ant.style.left = Math.sin(angle ) * radius + center[0] + 'px';&#xA;              ant.style.top = Math.cos( angle)  * radius + center[1] + 'px';&#xA;&#xA;              var dx = parseInt( ant.style.left) - 200;&#xA;              var dy = parseInt( ant.style.top ) - 200;&#xA;              startRotation4 = Math.atan2( dy, dx);&#xA;&#xA;              ants.push( ant );&#xA;          }&#xA;&#xA;          var travelled = 0;&#xA;&#xA;          setInterval(function(){&#xA;&#xA;              document.getElementById( 'readout').innerText = 't=' + t + ' a=' + travelled ;&#xA;&#xA;&#xA;&#xA;              //chase the tail&#xA;              for( var i = 0; i &amp;lt; 5; i ++ )&#xA;              {&#xA;                  var target = ants[ i - 1];&#xA;                  var ant = ants[i];&#xA;                  if( ! target )&#xA;                  {&#xA;                      target= ants[4];&#xA;                  }&#xA;&#xA;                  var dx = parseInt( target.style.left) - parseInt( ant.style.left );&#xA;                  var dy = parseInt( target.style.top ) - parseInt( ant.style.top );&#xA;&#xA;                  var angle = Math.atan2( dy, dx );&#xA;&#xA;                  var velX = Math.cos( angle ) * step;&#xA;                  var velY = Math.sin( angle ) * step;&#xA;&#xA;&#xA;&#xA;                  ant.style.left = parseInt( ant.style.left ) + velX + 'px';&#xA;                  ant.style.top = parseInt( ant.style.top ) + velY + 'px';&#xA;&#xA;                  //just detect if they're touching (center plus radius collision)&#xA;                  var ctX = parseInt( target.style.left) + 5.5;&#xA;                  var ctY = parseInt( target.style.top )+ 5.5;&#xA;&#xA;                  var cX =  parseInt( ant.style.left ) + 5.5;&#xA;                  var cY = parseInt( ant.style.top ) + 5.5;&#xA;&#xA;                  if (  i == 4 )&#xA;                  {&#xA;                      var endDx = parseInt( ant.style.left) - 200;&#xA;                      var endDy = parseInt( ant.style.top ) - 200;&#xA;                      travelled = Math.abs( Math.atan2( endDy, endDx) - startRotation4);&#xA;                      console.log( startRotation4 , travelled )&#xA;                  }&#xA;&#xA;&#xA;&#xA;                  if( (Math.abs( ctX - cX ) &amp;lt;= antRadius ) &amp;amp;&amp;amp; (Math.abs( ctY - cY ) &amp;lt;= antRadius))&#xA;                  {&#xA;&#xA;                      if( !done &amp;amp;&amp;amp; i == 4 )&#xA;                      {&#xA;&#xA;&#xA;                          document.getElementById('done').innerText = 'done in ' + t + ' steps. angle travelled = ' + travelled + ' radians';&#xA;                          done = true;&#xA;                      }&#xA;&#xA;&#xA;                  }&#xA;&#xA;              }&#xA;&#xA;              t++;&#xA;          }, interval );&#xA;&#xA;&#xA;&#xA;      };&#xA;&#xA;&#xA;&#xA;&#xA;    &amp;lt;/script&amp;gt;&#xA;&#xA;    &amp;lt;div id=&quot;ant-0&quot; style=&quot;background-color:#f00; width:11px; height:11px; position:absolute&quot;&amp;gt;&amp;lt;/div&amp;gt;&#xA;    &amp;lt;div id=&quot;ant-1&quot; style=&quot;background-color:#f00; width:11px; height:11px; position:absolute&quot;&amp;gt;&amp;lt;/div&amp;gt;&#xA;    &amp;lt;div id=&quot;ant-2&quot; style=&quot;background-color:#f00; width:11px; height:11px; position:absolute&quot;&amp;gt;&amp;lt;/div&amp;gt;&#xA;    &amp;lt;div id=&quot;ant-3&quot; style=&quot;background-color:#f00; width:11px; height:11px; position:absolute&quot;&amp;gt;&amp;lt;/div&amp;gt;&#xA;    &amp;lt;div id=&quot;ant-4&quot; style=&quot;background-color:#f00; width:11px; height:11px; position:absolute&quot;&amp;gt;&amp;lt;/div&amp;gt;&#xA;&#xA;&#xA;    &amp;lt;div id=&quot;readout&quot;&amp;gt;&amp;lt;/div&amp;gt;&#xA;    &amp;lt;div id=&quot;done&quot;&amp;gt;&amp;lt;/div&amp;gt;&#xA;&amp;lt;/body&amp;gt;&#xA;&amp;lt;/html&amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="11053" LastEditorUserId="11053" LastEditDate="2015-04-04T05:08:26.403" LastActivityDate="2015-04-04T05:08:26.403" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11539" PostTypeId="1" AcceptedAnswerId="11542" CreationDate="2015-04-03T20:58:59.637" Score="4" ViewCount="356" Body="&lt;p&gt;I am born everyday, and always awake&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some times I am plenty, and some times none&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have feet and I travel light&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People, few happy, are on me&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And day by day, they want me more&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put the load they don't want to carry&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't see you as I have no eyes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But cry I do, day or night&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For, I am, you see loud&lt;/p&gt;&#xA;" OwnerUserId="11049" LastEditorUserId="11049" LastEditDate="2015-04-03T22:17:34.843" LastActivityDate="2015-04-03T23:08:28.607" Title="Cry me a River?" Tags="|riddle|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11540" PostTypeId="2" ParentId="11539" CreationDate="2015-04-03T21:08:12.203" Score="2" Body="&lt;p&gt;Maybe it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MONEY.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am born everyday, and awake always&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Money is always being printed, and ATM cash machines are always 'awake' and ready to be used.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Some times, I am plenty, and some times none&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people have plenty of money, and some have none.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I don't have feet, and I travel light&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Money has no feet, and it's quite light for being so valuable: you can fit a million pounds/dollars/euros in a suitcase (or so I've heard).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;People, few happy, are on me&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Coins and notes often bear pictures of people's faces.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And day by day, they want me more&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People always want more money.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;To put the load, they don't want to carry&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's lighter and easier to handle than the material goods it 'represents'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Since I can't see you as I have no eyes&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Indeed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And cry I do, day or night&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;For, I am, you see loud&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure about these lines.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-04-03T21:08:12.203" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11541" PostTypeId="2" ParentId="11537" CreationDate="2015-04-03T21:42:56.710" Score="3" Body="&lt;p&gt;We can't get the first clues out until we have all the numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’ll present a list of all the numbers that the clue suggests, with the explanation (if necessary) in spoiler blocks. Then we’ll go back and work out the final answer.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Multiply it by the number of types of creature beyond red twisted ter'angreal doorways.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;2.&lt;/strong&gt; There are two such doors; one leads to the land of the Aelfinn, the other to the land of the Eelfinn.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Subtract the figure you now have from the number of Scheherezade's tale narrated by Poe.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;1002.&lt;/strong&gt; Edgar Allan Poe wrote &lt;em&gt;The Thousand-and-Second Tale of Scheherazade&lt;/em&gt;, a sequel to Arabic collection of stories &lt;a href=&quot;http://en.wikipedia.org/wiki/One_Thousand_and_One_Nights&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;A Thousand and One Nights&lt;/em&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Divide by the number of children fathered by Sidorio.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;2.&lt;/strong&gt; Grace and Connor.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add the number of entries in the Classic FM Hall of Fame.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;300.&lt;/strong&gt; I’m assuming you mean currently (a new list is compiled around Easter every year), rather than every entry ever entered, less duplicate entries.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Subtract the product of the number of siblings of Frodo Baggins's mother and the number of legitimate children fathered by Walder Frey.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;174.&lt;/strong&gt; Frodo’s mother, Primula Baggins, was one of seven children, thus &lt;strong&gt;6&lt;/strong&gt; siblings. Walker Frey father &lt;strong&gt;29&lt;/strong&gt; children.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add the number of days from now until the Easter Day celebrations in Romania.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;s&gt;&lt;strong&gt;3.&lt;/strong&gt; Today is Good Friday. Romania uses the Orthodox Easter date, which is the Monday rather than the Sunday.&lt;/s&gt; &lt;strong&gt;9.&lt;/strong&gt; Romania apparently celebrates on the Sunday 12th April. I can’t tell the difference between Monday and Sunday, oops.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Subtract the number of Sitters in the &quot;Hall of the Tower&quot; in Salidar.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;s&gt;&lt;strong&gt;21.&lt;/strong&gt;&lt;/s&gt; &lt;strong&gt;18.&lt;/strong&gt; Shamelessly cribbed from Michael Edenfield’s answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add the number of stars on the banner of Gondor.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;7.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Subtract the number of elements named after Ytterby.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;4.&lt;/strong&gt; Yttrium, Ytterbium, Erbium and Terbium.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Subtract the figure you now have from the user ID of the most prolific question-poster on Puzzling.SE.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;2383.&lt;/strong&gt; I believe this is @warspyking, with a total of 148 questions (at time of writing).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now let's do the maths. &lt;s&gt;I can't really put this all in spoiler blocks, so here’s a nice block of blank text to hide the solution from you. If you read on any further, you will be spoiled.&lt;/s&gt; Someone edited to put the spoiler blocks.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There’s one multiples of five: entries in the Classic FM Hall of Fame (300), &#xA; and two prime greater than 10 among these numbers: children of Walker Frey &#xA; (29)&lt;sup&gt; [1]&lt;/sup&gt;, the user ID of the most prolific asker (2383).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So we get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$$ \begin{align*}1 \times 2 &amp;amp;= 2 \\1002 - 2 &amp;amp;= 1000 \\1000 / 2 &amp;amp;= 500 \\500 + 300 &amp;amp;= 800 \\800 - 174 &amp;amp;= 626 \\626 + 9 &amp;amp;= 635 \\635 - 18 &amp;amp;= 617 \\617 + 7 &amp;amp;= 624 \\624 - 4 &amp;amp;= 620 \\2383 - 620 &amp;amp;= 1763 \\1763 + 2 &amp;amp;= 1765\end{align*}$$&#xA; So my final answer is &lt;strong&gt;1765&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&lt;sup&gt;[1]&lt;/sup&gt; I initially assumed the “primes” clue only meant anything in the final calculation. I do know that 29 is prime. :-)&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="4692" LastEditorUserId="12058" LastEditDate="2015-05-21T06:27:27.970" LastActivityDate="2015-05-21T06:27:27.970" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="11542" PostTypeId="2" ParentId="11539" CreationDate="2015-04-03T22:27:15.183" Score="4" Body="&lt;p&gt;Is it:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clouds&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am born everyday, and always awake. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everyday with sunrise the sky is born and almost always there are clouds in it. But, clouds are around always, and always moving. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some times I am plenty, and some times none. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sometimes there are lots of clouds, sometimes few.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't have feet and I travel light. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clouds have no feet, but are &quot;fluffy and light.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;People, few happy, are on me.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Happy people are sometimes said to be on &quot;cloud&quot; 9.  Other times we hear about &quot;storm clouds&quot; when people are mad.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And day by day, they want me more.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Depending on the season, people often want clouds. In the winter for snow and the other activities. In summer for shade.    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To put the load they don't want to carry.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Here I'm a little short. Maybe, we wish our burdens were light as a cloud? From the OP, refers to storing data on the cloud. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can't see you as I have no eyes. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clouds have no eyes, and technically can't see.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But cry I do, day or night. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rain comes day or night.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For, I am, you see loud.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lightning in the sky can be very loud.  Or (thanks to Rand for pointing this out), see loud, &quot;c&quot; loud, cloud. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="9951" LastEditDate="2015-04-03T23:08:28.607" LastActivityDate="2015-04-03T23:08:28.607" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11543" PostTypeId="2" ParentId="11502" CreationDate="2015-04-03T22:46:40.877" Score="2" Body="&lt;p&gt;Taking a slant on xnor's idea, consider the velocity of any ant towards $O$. I'll use the standard figures for the internal angle of a regular pentagon and the radius of a circumscribed circle from &lt;a href=&quot;http://en.wikipedia.org/wiki/Pentagon#Regular_pentagons&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;.&#xA;Name the vertices $A,B,C,D,E$ starting at the top and working clockwise, and let $\theta = \angle OAB$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By symmetry, $\angle OAE$ is also $\theta$, so $\angle BAE = 2\theta = 3\pi/5$ (internal angle of a regular pentagon), i.e. $\theta = 3\pi/10$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The velocity $V$ of $A$ towards $O$ is $S \cos \theta$, so the time $T$ taken for $A$ to reach $O$ is ${AO \over V} = {L \over 2 \sin (\pi/5)} \times {1 \over S \cos \theta}$ (using the formula for the radius of a circumscribed circle).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\therefore T = {L \over 2S \sin (\pi/5) \cos (3\pi/10)} = {L \over 2S \sin^2(\pi/5)} \approx 1.45L/S$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As xnor observes, the ants spiral through an unbounded angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $T$ and $S$ are finite, the total scalar distance $D=TS$ traversed by each ant is also finite. Since $T \approx 1.45L/S$ from before, we also have $TS \approx 1.45L$, i.e. $D \approx 1.45L$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The counter-intuitive result hinted at by leoll2 is then the finite time taken to traverse an infinite spiral, where the total length of the spiral is a constant factor of about 1.45 times the length of one side of the initial pentagon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note: thanks to leoll2 for pointing out the trig simplification.&lt;/p&gt;&#xA;" OwnerUserId="9422" LastEditorUserId="9422" LastEditDate="2015-04-04T16:13:27.363" LastActivityDate="2015-04-04T16:13:27.363" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11544" PostTypeId="1" AcceptedAnswerId="11546" CreationDate="2015-04-04T04:02:17.543" Score="6" ViewCount="963" Body="&lt;p&gt;Ten people are at a fancy party. The guest list includes Obama, Shaq, Captain Falcon, Jaxon, Walter White, Bob, Emma Watson, Plato, Sherlock Holmes and John Watson. Sherlock and Watson arrive late to the party, before they arrive, the other guests decide to play a party game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The game has the following rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;They decide to let Shaq go first (because he’s the best dressed)&lt;/li&gt;&#xA;&lt;li&gt;Each person may give up to 2 high fives. (Once with their left and once with their right hand).&lt;/li&gt;&#xA;&lt;li&gt;The game begins when Shaq gives at least one high five. Since he began the game, Shaq is not allowed to receive a high five.&lt;/li&gt;&#xA;&lt;li&gt;A person may not give a high five, until they have received one. (Except for Shaq)&lt;/li&gt;&#xA;&lt;li&gt;Once a person receives a high five, they may not receive another one. (From anybody with any hand).&lt;/li&gt;&#xA;&lt;li&gt;The game ends when everyone has received a high five (Except for Shaq).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Sherlock and Dr. Watson arrive at the party, and are sorely disappointed that they missed the game. Anxious to find out everything they can about what they missed, they decide to interview each of the guests.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dr. Watson decides to interview people in the following way:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Anytime he interviews somebody for the first time, he will record their name.&lt;/li&gt;&#xA;&lt;li&gt;He stops interviewing when everyone’s name has been recorded.&lt;/li&gt;&#xA;&lt;li&gt;He begins with Shaq.&lt;/li&gt;&#xA;&lt;li&gt;He asks the interviewee if they gave a left high five. If they did, he finds the person they gave the left high five to, and interviews that person returning to step 4. If they gave no high fives with their left hand he proceeds to the next step.&lt;/li&gt;&#xA;&lt;li&gt;He asks the interviewee if they gave a right high five. If they did, he finds the person they gave the right high five to, and interviews that person returning to step 4. If they gave no high fives with their right hand he proceeds to the next step.&lt;/li&gt;&#xA;&lt;li&gt;He asks if they received a left, or a right high five.&lt;/li&gt;&#xA;&lt;li&gt;If they received a left high five, he finds the person from whom they received the high five and returns to step 5.&lt;/li&gt;&#xA;&lt;li&gt;If they received a right high five, he finds the person from whom they received the high five and returns to step 6.&lt;/li&gt;&#xA;&lt;li&gt;If the person did not receive a high five at all, he asks the interviewee who he gave a right high five to, and interviews that person, returning to step 4.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Still not satisfied, Sherlock asks Watson to conduct a second round of interviews following a different procedure:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;He begins with Shaq.&lt;/li&gt;&#xA;&lt;li&gt;He stops interviewing when everyone’s name has been recorded.&lt;/li&gt;&#xA;&lt;li&gt;Before recording a name, he checks to make sure it isn’t already on the list.&lt;/li&gt;&#xA;&lt;li&gt;He asks the interviewee if they gave a left high five. If they did, he finds the person they gave the left high five to, and interviews that person returning Step 4. If they gave no high fives with their left hand he proceeds to the next step.&lt;/li&gt;&#xA;&lt;li&gt;He records the interviewee’s name.&lt;/li&gt;&#xA;&lt;li&gt;He asks the interviewee if they gave a right high five. If they did, he asks who they gave the right high five to.&lt;/li&gt;&#xA;&lt;li&gt;If they gave a right high five to a person who is not already on the list, he finds the person they gave the right high five to, and interviews that person returning to Step 4.&lt;/li&gt;&#xA;&lt;li&gt;If they did not give a right high five or they gave a right high five to somebody already on the list then he records the name of the person from whom they received their high five, and he interviews that person returning to step 6.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;After Watson has finished conducting his interviews, he shows his lists to Sherlock Holmes.&#xA;The first list reads:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Shaq, Captain Falcon, Walter White, Jaxon, Emma Watson, Obama, Bob, Plato&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second list reads:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Walter White, Emma Watson, Jaxon, Captain Falcon, Obama, Shaq, Plato, Bob&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sherlock examines the lists for a few moments, turns to his friend and says: “It is elementary, my dear Watson”.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using only these lists, and his knowledge of the process, Sherlock can tell exactly who high fived who and with which hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you?&lt;/p&gt;&#xA;" OwnerUserId="11063" LastEditorUserId="4692" LastEditDate="2015-04-04T10:50:15.597" LastActivityDate="2015-04-04T10:50:15.597" Title="Sherlock Holmes is at a Party" Tags="|logical-deduction|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11545" PostTypeId="2" ParentId="11408" CreationDate="2015-04-04T05:41:55.443" Score="-4" Body="&lt;p&gt;stewardesses (12 letter)&#xA;Hope to contribute more.&lt;/p&gt;&#xA;" OwnerUserId="2656" LastActivityDate="2015-04-04T05:41:55.443" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11546" PostTypeId="2" ParentId="11544" CreationDate="2015-04-04T06:11:59.270" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Shaq: left to Captain, right to Bob &lt;br&gt;&#xA; Captain: left to Walter, right to Obama &lt;br&gt;&#xA; Walter: right to Jaxon &lt;br&gt;&#xA; Jaxon: left to Emma &lt;br&gt;&#xA; Bob: left to Plato &lt;br&gt;&#xA; [No other high fives.] &lt;br&gt;&#xA; &lt;br&gt;&#xA; Although it doesn't have to be solved this way, it helps to realize&#xA; that the pattern of high fives forms a &lt;a href=&quot;http://en.wikipedia.org/wiki/Binary_tree&quot; rel=&quot;nofollow&quot;&gt;binary tree&lt;/a&gt;.&#xA; Then figure out that the first list is a pre-order &lt;a href=&quot;http://en.wikipedia.org/wiki/Tree_traversal&quot; rel=&quot;nofollow&quot;&gt;traversal&lt;/a&gt; and the&#xA; second is an in-order traversal. &lt;br&gt;&#xA; The first element of the first list (S,C,W,J,E,O,B,P) tells us that Shaq is&#xA; the root. The position of Shaq in the second list (W,E,J,C,O,S,P,B) tells us &#xA; that (W,E,J,C,O) came from his left high five and (P,B) from his right.&#xA; Now split the first list the same way: (C,W,J,E,O) and (B,P). The first&#xA; element of (C,W,J,E,O) tells us that Shaq's left high five went to Captain.&#xA; Splitting the first part of the second list (W,E,J,C,O) on Captain tells us&#xA; that his left high five went to&#xA; (W,E,J) and his right to (O). Continuing in this way gives the solution above.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10724" LastEditorUserId="10724" LastEditDate="2015-04-04T06:20:09.717" LastActivityDate="2015-04-04T06:20:09.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11547" PostTypeId="1" AcceptedAnswerId="11548" CreationDate="2015-04-04T08:03:18.410" Score="11" ViewCount="1335" Body="&lt;p&gt;A boy celebrated his birthday with some friends. Strangely, 2 days &lt;strong&gt;later&lt;/strong&gt;, his &lt;strong&gt;elder&lt;/strong&gt; twin celebrated his birthday with the same friends. Now those friends got confused and asked why the order was different, and the twins showed the correct times of their births on their birth certificates. So, the friends were satisfied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explain the situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clue:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I made this puzzle by combining two different puzzles. So, if you have solved many such puzzles, then try to combine those answers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Even though I made this puzzle independently, I now see that &lt;a href=&quot;https://www.ocf.berkeley.edu/~wwu/riddles/hard.shtml#birthdayTwins&quot; rel=&quot;nofollow&quot;&gt;somebody else has already made it around 12 years ago&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10940" LastEditorUserId="10940" LastEditDate="2018-12-28T18:55:23.470" LastActivityDate="2018-12-28T18:55:23.470" Title="Birthdays celebrated in wrong order" Tags="|lateral-thinking|paradox|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11548" PostTypeId="2" ParentId="11547" CreationDate="2015-04-04T08:11:57.297" Score="11" Body="&lt;p&gt;The twins were travelling eastwards across the International Date Line during their birth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The elder twin brother was born first, on March 1st.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After they crossed the International Date Line, the younger brother was born, on February 28th.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When they celebrate their birthdays on a leap year, they're 2 days apart!&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="39" LastEditDate="2015-05-23T20:27:41.360" LastActivityDate="2015-05-23T20:27:41.360" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11550" PostTypeId="2" ParentId="1719" CreationDate="2015-04-04T09:19:49.473" Score="9" Body="&lt;p&gt;You are looking for least percentage, without concerning with the Difficulty of the puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that case, the percentage can be made &lt;strong&gt;arbitrarily small&lt;/strong&gt;.&#xA;Here is an example with &lt;strong&gt;31&lt;/strong&gt; digits removed:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;999)*********(*******&#xA;    ***&#xA;       ***&#xA;       ***&#xA;          ***&#xA;          ***&#xA;            0&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here the format AAA)BBB(CCC means AAA is dividing BBB to get the answer as CCC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can extend the same arbitrarily, eg here is the next possibility with &lt;strong&gt;43&lt;/strong&gt; digits removed:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;999)************(**********&#xA;    ***&#xA;       ***&#xA;       ***&#xA;          ***&#xA;          ***&#xA;             ***&#xA;             ***&#xA;               0&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We now list the third example with &lt;strong&gt;55&lt;/strong&gt; digits removed:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;999)***************(*************&#xA;    ***&#xA;       ***&#xA;       ***&#xA;          ***&#xA;          ***&#xA;             ***&#xA;             ***&#xA;                ***&#xA;                ***&#xA;                  0&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We now list a fourth example with &lt;strong&gt;67&lt;/strong&gt; digits removed:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;999)******************(****************&#xA;    ***&#xA;       ***&#xA;       ***&#xA;          ***&#xA;          ***&#xA;             ***&#xA;             ***&#xA;                ***&#xA;                ***&#xA;                   ***&#xA;                   ***&#xA;                     0&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Enough Examples, let us generalise this, such that 999 Divides a (3N)-Digit number with a (3N-2)-Digit number as the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculating the percentage:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;999)************[3N '*' characters](**********[3N-2 '*' characters]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This will contain totally (3N)+(3N-2)+(3N)+(3N-3) = 12N-5 '*' characters, when counting all the rows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With 4 known digits [9,9,9,0] , total number of digits = 12N-1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Percentage of known digits = 100*(4)/(12N-1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can take N = 100, for example, to get &lt;strong&gt;0.3%&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can take N = 1000 to get &lt;strong&gt;0.03%&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solutions to those puzzles will be like this:&#xA;The result Digits will be 1001001...001 (1 followed by 001 (N-1) times).&#xA;All remaining unknown Digits will be 9s.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;999)999999999(1001001&#xA;    999&#xA;       999&#xA;       999&#xA;          999&#xA;          999&#xA;            0&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;pre&gt;&#xA;999)999999999999(1001001001&#xA;    999&#xA;       999&#xA;       999&#xA;          999&#xA;          999&#xA;             999&#xA;             999&#xA;               0&#xA;&lt;/pre&gt;&#xA;" OwnerUserId="10940" LastEditorUserId="10940" LastEditDate="2015-05-19T18:14:39.123" LastActivityDate="2015-05-19T18:14:39.123" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11551" PostTypeId="1" AcceptedAnswerId="11581" CreationDate="2015-04-04T09:43:06.710" Score="7" ViewCount="1270" Body="&lt;p&gt;Find the &lt;img src=&quot;https://i.stack.imgur.com/N25Cy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;def That(n):&#xA;    def Question(n): return n+1|-n-2&#xA;    a = Question(n)&#xA;    return a&#xA;&#xA;&amp;gt;&amp;gt;&amp;gt; That(42)&#xA;-1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Clarifications&lt;/h2&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Three hints would be provided every 6 hours unless this puzzle gets solved before the next scheduled hint.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;I may not respond to all the comments. The Rebus is quite easy and any hint would make this puzzle vulnerable.&lt;/li&gt;&#xA;&lt;li&gt;There will be one and only one answer.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strike&gt;The first Rebus would be solved in a day.&lt;/strike&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hints&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first hint has been provided. Can you find any difference in the current rebus with the previous? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The second hint has been provided. Can you identify each of the elements in the first rebus?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Simple is better than complex.&#xA; Special cases aren't special enough to break the rules.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4177" LastEditorUserId="4177" LastEditDate="2015-04-07T06:16:09.700" LastActivityDate="2015-05-05T17:41:15.730" Title="Find the Rebus in the Rebus" Tags="|rebus|computer-puzzle|" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11553" PostTypeId="1" AcceptedAnswerId="11554" CreationDate="2015-04-04T10:16:28.057" Score="9" ViewCount="1600" Body="&lt;p&gt;I make up everything&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I am nothing without the others&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I change, my buddy changes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it's likely, I'll never meet my buddy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll never see me&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you can know who I am&lt;/p&gt;&#xA;" OwnerUserId="11066" LastActivityDate="2015-07-06T20:29:58.087" Title="I make up everything - What am I?" Tags="|riddle|" AnswerCount="4" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11554" PostTypeId="2" ParentId="11553" CreationDate="2015-04-04T10:29:57.700" Score="11" Body="&lt;p&gt;I think you are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quarks &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I make up everything&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quarks are one type of matter particle. Most of the matter we see around us is made from protons and neutrons, which are composed of quarks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I am nothing without the others&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are six types of quarks, known as flavors: up, down, strange, charm, top, and bottom. A proton, composed of two up quarks, one down quark and the gluons &quot;binding&quot; them together. The color assignment of individual quarks is not important, only that all three colors be present.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I change, my buddy changes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A quark of one flavor can transform into a quark of another flavor only through the weak interaction, one of the four fundamental interactions in particle physics. By absorbing or emitting a W boson, any up-type quark (up, charm, and top quarks) can change into any down-type quark (down, strange, and bottom quarks) and vice versa. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But it's likely, I'll never meet my buddy&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Here buddy would mean antiquark. For every quark flavor there is a corresponding type of antiparticle, known as an antiquark, that differs from the quark only in that some of its properties have equal magnitude but opposite sign. And when they combine / meet they annihilate &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'll never see me&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you can know who I am&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Due to a phenomenon known as color confinement, quarks are never directly observed or found in isolation; they can be found only within hadrons, such as baryons (of which protons and neutrons are examples), and mesons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4177" LastEditorUserId="4177" LastEditDate="2015-04-04T10:43:45.927" LastActivityDate="2015-04-04T10:43:45.927" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11555" PostTypeId="2" ParentId="11551" CreationDate="2015-04-04T10:30:06.197" Score="2" Body="&lt;p&gt;1) &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Slash's Snakepit  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It was an American rock band. In the picture we see a slash inside a snake drawing a &quot;pit&quot; with its body. I think the quotation marks are just there so that people don't confuse the slash with a random line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;That question got a negative answer &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assuming it's Python, you call the function That() using 42 as parameter. The result is -1, a negative number.&lt;/p&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-04-04T10:30:06.197" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11556" PostTypeId="2" ParentId="11553" CreationDate="2015-04-04T10:34:14.963" Score="1" Body="&lt;p&gt;I pretty sure the answer is related to physics, so I'll try a easy one for now:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Atom&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I make up everything&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No need to explain, I guess&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I am nothing without the others&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You need many atoms, not only one&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I change, my buddy changes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; May be a reference to matter/antimatter or to sub-atomic particles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But it's likely, I'll never meet my buddy&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Matter is unlikely to meet antimatter nowadays.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'll never see me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can't directly observe an atom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But you can know who I am&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Our knowledge about atoms is very deep.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-04-04T10:34:14.963" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11557" PostTypeId="1" CreationDate="2015-04-04T12:51:07.720" Score="3" ViewCount="38771" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Di3Sd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are four guys standing in a line. They need to find out which color their hats are. They can't see their own hat color and only can see to the right. They can't see the person behind the seperation wall. If they know their hat color they can shout it and will win a million dollars. If they're wrong they die.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who shouts and therefore wins the million dollars?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[ I found this puzzle on another site and thought it was worth sharing, not sure if it's allowed ]&lt;/p&gt;&#xA;" OwnerDisplayName="user9543" LastEditorUserId="10615" LastEditDate="2018-09-06T17:07:33.380" LastActivityDate="2018-09-06T17:07:33.380" Title="Who will shout first?" Tags="|logical-deduction|meta-knowledge|" AnswerCount="1" CommentCount="1" ClosedDate="2015-04-04T13:06:42.040" ContentLicense="CC BY-SA 3.0" />
  <row Id="11558" PostTypeId="2" ParentId="11537" CreationDate="2015-04-04T12:53:19.940" Score="1" Body="&lt;p&gt;Multiply it by the number of types of creature beyond red twisted ter'angreal doorways.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;2&lt;/strong&gt; doors (Aelfinn and Eelfinn)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Subtract the figure you now have from the number of Scheherezade's tale narrated by Poe.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;1002&lt;/strong&gt; The Thousand-and-Second Tale of Scheherazade&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Divide by the number of children fathered by Sidorio.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;2&lt;/strong&gt; He's biological father of Grace &amp;amp; Connor&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add the number of entries in the Classic FM Hall of Fame.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;300&lt;/strong&gt; entries&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Subtract the product of the number of siblings of Frodo Baggins's mother and the number of legitimate children fathered by Walder Frey.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;6&lt;/strong&gt;x&lt;strong&gt;29&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add the number of days from now until the Easter Day celebrations in Romania.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;10 or 9&lt;/strong&gt;, I don't know what time zone stackexchange is using. Easter is on 13th april, the question was asked on 3rd or 4th.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Subtract the number of Sitters in the &quot;Hall of the Tower&quot; in Salidar.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;21&lt;/strong&gt; 3 Sitters per Ajah (there are 7 Ajah).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add the number of stars on the banner of Gondor.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;7&lt;/strong&gt; stars&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Subtract the number of elements named after Ytterby.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;4&lt;/strong&gt; Yttrium, Ytterbium, Erbium and Terbium&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Subtract the figure you now have from the user ID of the most prolific question-poster on Puzzling.SE.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;2383&lt;/strong&gt; (warspyking)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add the number of primes greater than 10 among the numbers featured in this puzzle.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;2&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Take the number of multiples of 5 among the numbers featured in this puzzle.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;2-1&lt;/strong&gt; depending on the easter question&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the possible answers are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$2383-(\frac{(1002-1\times2)}{2}+300-6\times29+9-21+7-4)+2=1768$&lt;/li&gt;&#xA;&lt;li&gt;$2383-(\frac{(1002-2\times2)}{2}+300-6\times29+10-21+7-4)+2=1768$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm lucky, the answers are identical! So the final answer is &lt;strong&gt;1768&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-04-04T12:53:19.940" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11559" PostTypeId="2" ParentId="11557" CreationDate="2015-04-04T13:01:06.683" Score="4" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Prisoners_and_hats_puzzle&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Prisoners and hats puzzle&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since there are only two hats of each color, 1 is able to deduce his color when 2 and 3 are the same color. Though, in this situation 2 and 3 are different, so 1 doesn't shout. &#xA;2, not hearing 1 shouting, deduces that his hat is of the opposite color of 3 and shouts!&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-04-04T13:06:27.270" LastActivityDate="2015-04-04T13:06:27.270" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11560" PostTypeId="2" ParentId="10839" CreationDate="2015-04-04T13:40:52.507" Score="12" Body="&lt;p&gt;Lemma: the probability that the ant has stopped converges to 1. Proof: if the ant moves right 10 times in a row, it must stop. Waiting for 10N moves gives N independent chances for this to happen. $\square$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ant's expected (i.e. average) position remains at 7. Therefore, at any point in time,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P(ant has stopped)*E(ant's position | ant has stopped) + P(ant is still moving)*E(ant's position | ant is still moving) = 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lemma implies P(ant has stopped) converges to 1, thus P(ant is still moving) converges to 0. Also, E(ant's position | ant is still moving) is bounded inside [0,10]. &lt;strong&gt;This&lt;/strong&gt; is the answer to Ypnypn's objection - if the ant started outside the boundaries, then this expected value would no longer necessarily be bounded, and this step of the argument would fail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it is bounded, so the term P(ant is still moving)*E(ant's position | ant is still moving) converges to 0. Therefore, the other term P(ant has stopped)*E(ant's position | ant has stopped) converges to 7. The result follows.&lt;/p&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2015-04-04T13:40:52.507" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11562" PostTypeId="2" ParentId="11553" CreationDate="2015-04-04T14:45:32.257" Score="4" Body="&lt;p&gt;I had fun with this one. Read the accepted answer first. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quark, the Ferengi&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I make up everything&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everyone knows Quark lies about anything he can get away with&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I am nothing without the others&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quark will exploit anyone and everyone he can for personal gain. He himself doesn't do much work, so without others to exploit he would be nothing. Also, to Quark, the opinion of others is everything. He doesn't just want to be rich, he wants to be powerful - without people to be powerful over, he would not be a happy Ferengi.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I change, my buddy changes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Throughout the seasons of DS9, Quark began to show interest in occasionally taking the non-greedy path, especially as his nephew, Nog, showed interest in Starfleet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But it's likely, I'll never meet my buddy&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nog has joined Starfleet and is therefore unlike to run into his greedy uncle again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'll never see me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quark prefers to hire people to do his dirty work for him. And clean work, for that matter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But you can know who I am&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quark's objective is to be famous (or infamous, depending on your opinion). And he's always looking for new business ventures. His publicity is considerably high, especially considering his remote location on an outpost near Bajor.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2812" LastActivityDate="2015-04-04T14:45:32.257" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11563" PostTypeId="1" CreationDate="2015-04-04T14:57:30.350" Score="4" ViewCount="350" Body="&lt;p&gt;I'm going to raise the difficulty of the original question one dimension, so maybe a refresher will be good...&#xA;Link: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/10839/ant-walking-on-a-number-line&quot;&gt;Ant walking on a number line&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I just mentioned i will raise it by one dimension... and yes i meant it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will be using co-ordinates to put the points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ant is at (7,7)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The drops of honey are now lines of honey at x=10, y=10 and x=0, y=0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let P(10) be the possibility of the ant eating the honey from x=10 or y=10&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and P(0) be the possibility of the ant eating honey from x=0 or y=0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ant moves up, down, left, or right randomly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And... Calculate P(0) and P(10)&lt;/p&gt;&#xA;" OwnerUserId="9736" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-04-05T11:47:25.423" Title="Ant walking on a number line 2" Tags="|mathematics|probability|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11564" PostTypeId="1" CreationDate="2015-04-04T15:06:53.747" Score="12" ViewCount="3698" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;As I was going to St Ives, &lt;br&gt;&#xA;  I met a man with seven wives. &lt;br&gt;&#xA;  Each wife had seven sacks, &lt;br&gt;&#xA;  Each sack had seven cats, &lt;br&gt;&#xA;  Each cat had seven kits. &lt;br&gt;&#xA;  Kits, cats, sacks, wives; &lt;br&gt;&#xA;  How many were going to St Ives?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a classic and well-known children's puzzle. This &lt;a href=&quot;https://www.youtube.com/watch?v=ol1h9VB4Lyk&quot; rel=&quot;nofollow&quot;&gt;Sesame Street version (1970)&lt;/a&gt; (spoiler warning) also includes an excellent rejoinder. But where did the riddle originate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know of some related puzzles, for example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You're driving a bus. &lt;br&gt;&#xA;  At the first stop, 6 people get on. &lt;br&gt;&#xA;  At the second stop, 4 people get off. &lt;br&gt;&#xA;  At the third stop, 9 people get on. &lt;br&gt;&#xA;  At the fourth stop, 7 people get on and 4 people get off. &lt;br&gt;&#xA;  What colour are the bus-driver's eyes?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is there a general term for this type of puzzle (appears to be an arithmetic problem, inviting the listener to engage in more-or-less-tedious calculation, but is actually a trick question)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are some more examples?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the earliest known puzzle of this type, and where did it originate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumably these exist in languages other than English, what are some examples? (with translations please)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also: &quot;As I gaed to Stonehaven&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;(Also, what are the answers to the puzzles above?).&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="5373" LastEditDate="2016-09-15T17:19:27.243" LastActivityDate="2016-09-15T17:19:27.243" Title="Kits, Cats, Sacks, Wives; How Many Were Going to St Ives?" Tags="|puzzle-history|" AnswerCount="4" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="11565" PostTypeId="2" ParentId="11564" CreationDate="2015-04-04T15:16:13.290" Score="4" Body="&lt;p&gt;If I am to take your riddles literally:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Answer 1: Only one for sure, I'm the only one going to St. Ives.&lt;/strong&gt; We don't know where we met the man or where he and his wives, cats, etc. are headed.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Answer 2: The bus-driver is me, and has whatever color eyes I have.&lt;/strong&gt; This simply because of the statements &lt;em&gt;You are driving a bus&lt;/em&gt; and &lt;em&gt;What colour are the bus-driver's eyes?&lt;/em&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't know any specific name for such riddles, they are generally just referred to as 'trick questions' since they seem to involve logic, but are more of a play on words. &lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4385" LastEditDate="2015-07-31T15:21:03.517" LastActivityDate="2015-07-31T15:21:03.517" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11567" PostTypeId="2" ParentId="11564" CreationDate="2015-04-04T15:46:57.977" Score="3" Body="&lt;p&gt;There's no specific name for this kind of riddles, I'd classify them &quot;too broad&quot;.&#xA;In fact, the answer simply relies on the interpretation of the text (what do we know about the intentions of the men and the wives?), allowing multiple answers due to this ambiguity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's no information about the origin of these ambiguous riddle. Though, the very first presence of the greek word &quot;griphos&quot; with the meaning of enigmas based on cunning rather than knowledge is credited to Aristophanes (The Wasps). &#xA;Subsequently, Clearchus of Soli wrote the &quot;Peri griphōn&quot;, a work about riddles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of these ancients riddles was written by Aristotle, this is the text (translated):  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Those we got we leave; those we didn't get we keep&lt;/em&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You may think he's referring to fish, though he's alluding to lice!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the answers to your question, ghosts_in_the_code has already answered correctly, no need to copy/paste his post.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.treccani.it/enciclopedia/indovinelli-e-enigmi_(Enciclopedia-dell&amp;#39;Italiano)/&quot; rel=&quot;nofollow&quot;&gt;www.treccani.it (written in italian)&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-04-05T08:43:36.647" LastActivityDate="2015-04-05T08:43:36.647" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11568" PostTypeId="2" ParentId="10839" CreationDate="2015-04-04T15:56:25.950" Score="0" Body="&lt;p&gt;&lt;em&gt;Here's an elementary proof that the probability the ant reaches the honey is one&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First an estimate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $\frac{1}{2^{2n}}{2n \choose n} = \frac{(2n)!}{2^n n! 2^n n!} = \frac{1\cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)} = \prod_{k=1}^{n} \frac{2k-1}{2k}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's estimate this. First note that $\prod_{k=1}^{n} \frac{2k-1}{2k} \leq \prod_{k=1}^{n} \frac{2k}{2k+1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore $\left(\prod_{k=1}^{n} \frac{2k-1}{2k}\right)^2 \leq \prod_{k=1}^n \frac{2k-1}{2k} \prod_{k=1}^{n} \frac{2k}{2k+1} = \prod_{j=1}^{2n} \frac{j}{j+1} = \frac{1}{2n+1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore $\frac{1}{2^{2n}} {2n \choose n} \leq \frac{1}{\sqrt{2n+1}}$ (In fact, by Stirling's formula, it goes like $\frac{1}{\sqrt{n\pi}}$)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now consider an ant on the number line as in the question, but with no stopping places. I'll show that it must at some point exit any given finite section with probability one. This shows the ant in the question reaches the honey with probability one (because it cannot have a nonzero probability of staying within $[1,9]$ for all time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability that after $2n$ steps the Ant is at a specific position $2k$ steps away from its original position is $\frac{1}{2^{2n}} {2n \choose n-k} \leq \frac{1}{2^{2n}}{2n \choose n}$ (note you must have $n+k$ steps in one direction and $n-k$ steps in the other direction to get there).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the probability you are within $2M$ steps of your starting position after $2n$ steps ($2M+1$ possibilities) is at most $\frac{2M+1}{\sqrt{2n+1}}$. Fixing $M$, this gets arbitrarily small as $n$ gets large. Thus you must exit any finite box with probability one.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now a simple expected value of the position argument as in David Richerby's or Ross Millikan's post gives the answer to the question.&lt;/p&gt;&#xA;" OwnerUserId="9821" LastActivityDate="2015-04-04T15:56:25.950" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11569" PostTypeId="1" AcceptedAnswerId="11571" CreationDate="2015-04-04T15:58:04.827" Score="2" ViewCount="238" Body="&lt;p&gt;My friend Kim told me:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Last week I went to an afternoon party. I was the last guest to arrive.&#xA;  I looked around and counted (not including myself) altogether 14 grown-ups and 17 children. Altogether, there were 12 male and 19 female present.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Later in the afternoon, the dancing started. There was a lot of dancing.&#xA;  Every boy danced exactly once with every girl, and every man danced exactly once with every woman. The number of children couples that formed during these dances was equal to the number of grown-up couples.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question: Is Kim a boy? A girl? A man? A woman?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="8874" LastEditDate="2016-03-18T09:48:29.050" LastActivityDate="2016-03-18T09:48:29.050" Title="Dancing, dancing, dancing" Tags="|mathematics|calculation-puzzle|algebra|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11571" PostTypeId="2" ParentId="11569" CreationDate="2015-04-04T16:31:26.473" Score="3" Body="&lt;p&gt;&lt;strong&gt;Let's define M male, F females, B baby, G grown-up.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if we don't consider you and Kim, we call X the MG. Easily we can deduce that:&lt;br&gt;&#xA;$MG=X$&lt;br&gt;&#xA;$FG=14-X$&lt;br&gt;&#xA;$MB=12-X$&lt;br&gt;&#xA;$FB=19-(14-X)=5+X$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a males and b females, the etero couples are a*b. So:&lt;br&gt;&#xA;$X(14-X)=(12-X)(5+X)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we forgot Kim! Well, &lt;strong&gt;let's add +1 to only one of the 4 factors&lt;/strong&gt; in the above equation. We have 4 different equations now (remember that X must be natural):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$(X+1)(14-X)=(12-X)(5+X)$&lt;/li&gt;&#xA;&lt;li&gt;$X(15-X)=(12-X)(5+X)$&lt;/li&gt;&#xA;&lt;li&gt;$X(14-X)=(13-X)(5+X)$&lt;/li&gt;&#xA;&lt;li&gt;$X(14-X)=(12-X)(6+X)$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The only equation that returns a positive integer X is the 4th, adding +1 to &lt;strong&gt;Female Babies&lt;/strong&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X(14-X)=(12-X)(5+X)$&lt;br&gt;&#xA;$MG=X=9$&lt;br&gt;&#xA;$FG=5$&lt;br&gt;&#xA;$MB=3$&lt;br&gt;&#xA;$FB=14+1=15$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Kim is a female child!&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-04-04T16:48:02.857" LastActivityDate="2015-04-04T16:48:02.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11572" PostTypeId="1" AcceptedAnswerId="11583" CreationDate="2015-04-04T16:44:32.460" Score="-3" ViewCount="667" Body="&lt;p&gt;My body is green&lt;br&gt;&#xA;My hair and eyebrows are blue&lt;br&gt;&#xA;My eyes are a very light shade of purple&lt;br&gt;&#xA;I have the sharpest beak in all my friends&lt;br&gt;&#xA;I painted my wall and floor in a light shade of purple&lt;br&gt;&#xA;I am named after the second greatest creator&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Who am I?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint #1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I don't have a beak!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My species is quite popular in SE.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My species vary a lot in color.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm a physical object, but you don't even know if I exist or not.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You might want to wander around on &lt;a href=&quot;https://meta.stackexchange.com/&quot;&gt;meta.stackexchange.com&lt;/a&gt;. There's a post there related to the answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="435" LastEditorUserId="-1" LastEditDate="2017-03-20T10:32:12.560" LastActivityDate="2016-08-09T09:48:35.730" Title="I have the sharpest beak in all my friends . Who am I?" Tags="|riddle|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11574" PostTypeId="2" ParentId="11564" CreationDate="2015-04-04T19:16:36.733" Score="5" Body="&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/As_I_was_going_to_St_Ives&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;,  &quot;The earliest known published version of [the riddle of St. Ives.] comes from a manuscript dated to around 1730 (but it differs in referring to &quot;nine&quot; rather than &quot;seven&quot; wives). The modern form was first printed around 1825.&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;People who walked by the sphinx in the town of Thebes were said to have been asked to solve this &lt;a href=&quot;http://en.wikipedia.org/wiki/Sphinx&quot; rel=&quot;nofollow&quot;&gt;riddle of the Sphinx&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;What animal walks on four legs in the morning, two legs at noon, and three in the evening?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oedipus is reputed to have solved this.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The closest expression for this general sort of problem may be &quot;brain teaser&quot;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Just for fun, you may wish to explain this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;There are 10 kinds of mathematicians in the world: those who understand binary and those who do not.&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="11050" LastEditorUserId="11050" LastEditDate="2015-04-04T19:30:32.450" LastActivityDate="2015-04-04T19:30:32.450" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11575" PostTypeId="2" ParentId="11572" CreationDate="2015-04-04T21:13:31.307" Score="1" Body="&lt;p&gt;Maybe it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an accepted answer on Stack Exchange.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My body is green&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Refers to the green checkmark on the accepted answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My hair and eyebrows are blue&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The upvote/downvote arrows turn blue when pressed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My eyes are a very light shade of purple&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The background colour on a spoilertag is very slightly off-white, and could be a very pale purple.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I have the sharpest beak in all my friends&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A metaphorical way of saying it's the best answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I painted my wall and floor in a light shade of purple&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Spoilertags again?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am named after the second greatest creator&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure about this line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All the hints fit, if&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; we take &quot;colour&quot; in the 3rd hint to be a metaphor for type/quality of answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-04-04T21:13:31.307" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11576" PostTypeId="2" ParentId="10839" CreationDate="2015-04-04T21:13:57.850" Score="3" Body="&lt;p&gt;As has previously been pointed out, given that that, at every step, the ant moves left or right with equal probability (or not at all if they find honey), it is clear that the expected position of the ant after $n$ steps equals the expected position after $n-1$ steps, and, by induction, the initial position of the ant, which is $7$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we let $p_i(n)$ be the probability that the ant is at position $i$ after $n$ steps, then we can take the expected value as:&#xA;$$7=\sum_{i=0}^{10}i\cdot p_i(n).$$&#xA;Then, if we let $P(n)=p_1(n)+p_2(n)+\ldots +p_9(n)$, we can bound the above as (noting that there was a $0\cdot p_0(n)$ term from the sum which we dropped since it it is $0$):&#xA;$$10\cdot p_{10}(n) + 1\cdot P(n) \leq 7 \leq 10\cdot p_{10}(n) + 9\cdot P(n)$$&#xA;or alternatively:&#xA;$$P(n)\leq 7-10\cdot p_{10}(n) \leq 9P(n).$$&#xA;However, from any position between $0$ and $10$, the probability of the ant finding honey in $5$ steps or less is at least $\frac{1}{16}$, it is clear that $P(n+5)&amp;lt;\frac{15}{16}P(n)$ and hence $P(n)$ goes to $0$ as $n$ goes to $\infty$. The bound above squeezes the middle term and implies then that $$\lim_{n\rightarrow\infty}7-10\cdot p_{10}(n)=0$$&#xA;which rearranges to the desired result:&#xA;$$\lim_{n\rightarrow\infty}p_{10}(n)=\frac{7}{10}$$&#xA;that is, after arbitrarily many steps, the probability that the ant finds the honey at $x=10$ is $\frac{7}{10}$. Since the ant almost certainly stops, the probability that they find the honey at $0$ is $\frac{3}{10}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This post solely exists to provide formal &amp;amp; elementary justification of the proof presented in Ross Milikan and David Richerby's answers. One may note that my argument relies only on the fact that the positions of the ant remain &lt;em&gt;bounded&lt;/em&gt; - that is, we don't have to worry about probabilities going to zero but their influence on the expected value &lt;em&gt;not&lt;/em&gt; going to zero - like, if there was only honey at $x=10$, the ant would almost surely stop there, despited their expected position being $7$ - the problem would be that very large negative positions would be possible, with small probability, and this would drag the expected value down. However, as it happens, the boundedness of the domain means that we cannot spread out an arbitrarily small probability to create arbitrarily large change in expected value, hence the argument holds.&lt;/p&gt;&#xA;" OwnerUserId="4765" LastActivityDate="2015-04-04T21:13:57.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11577" PostTypeId="2" ParentId="11563" CreationDate="2015-04-04T21:19:42.213" Score="2" Body="&lt;p&gt;Once again, Python comes in:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from random import randint&#xA;f10 = 0&#xA;f0 = 0&#xA;coordx = 7&#xA;coordy = 7&#xA;iters = 0&#xA;finished = False&#xA;while not finished:&#xA;    while coordx &amp;gt; 0 and coordx &amp;lt; 10 and coordy &amp;gt; 0 and coordy &amp;lt; 10:&#xA;        if randint(0, 1) == 0:&#xA;            coordx += randint(0, 1)*2-1&#xA;        else:&#xA;            coordy += randint(0, 1)*2-1&#xA;        if coordx == 10 or coordy == 10:&#xA;            f10 += 1&#xA;        elif coordx == 0 or coordy == 0:&#xA;            f0 += 1&#xA;        if iters == 10000000:&#xA;            finished = True&#xA;            print 1.0 * f0 / iters, 1.0 * f10 / iters&#xA;    coordx = 7&#xA;    coordy = 7&#xA;    iters += 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The result produced after 10.000.000 simulations is 0.195875719588 and 0.804124380412 and that makes two things clear – the answer is definitely not the same and it is not as simple (I tried to assemble an expression with this result but got nowhere). Though at least it provides a reference point for those on better terms with math than I am.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. Not sure if answer or comment, but it does contain the answer in approximate form, so I'll post it here. If I'm wrong, feel free to convert it into a comment&lt;/p&gt;&#xA;" OwnerDisplayName="user10531" LastActivityDate="2015-04-04T21:19:42.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11578" PostTypeId="2" ParentId="10839" CreationDate="2015-04-04T21:29:23.337" Score="6" Body="&lt;p&gt;One abstract &lt;em&gt;and fun&lt;/em&gt; way to attack this is to consider the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose there is an ant on the real line, moving in some continuous manner such that it is equally likely to move upwards or downwards - i.e. any path they could take is equally likely as the reflection of that path over their current position. Moreover, suppose that the probability that they never pass through $0$ or $1$ is zero.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clearly, our current scenario falls under this category, since if we only care about what order the path passes through points of the form $\frac{a}{10}$, the symmetries still guarantee it will pass from there to $\frac{a+1}{10}$ or $\frac{a-1}{10}$ with equal probability, ignoring repetitions of $\frac{a}{10}$ - so our problem is, in a sense, a quotient of the one defined above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we let $P(x)$ for $x\in[0,1]$ be the probability that, if the ant starts at $x$, the ant will reach $1$ before it reaches $0$, then we clearly have:&#xA;$$P(0)=0$$&#xA;$$P(1)=1$$&#xA;and by the symmetry and the fact that almost all paths pass one of these numbers&#xA;$$P\left(\frac{1}2\right)=\frac{1}2.$$&#xA;However, then we can use symmetry again to suppose that if the ant starts at $x=\frac{1}4$, it will equally likely reach $\frac{1}2$ and $0$ - hence the probability it will reach $1$ first is the average of $P(0)$ and $P\left(\frac{1}2\right)$ which is $\frac{1}4$. We can similarly find that $P\left(\frac{3}4\right)=\frac{3}4$. In fact, we can repeat this to show that if $x$ is a dyadic rational, then $P(x)=x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dyadic rationals are dense and $P$ is obviously continuous hence $P(x)=x$ and, for this particular case, $P(\frac{7}{10})=\frac{7}{10}$.&lt;/p&gt;&#xA;" OwnerUserId="4765" LastEditorUserId="4765" LastEditDate="2015-04-04T21:35:37.597" LastActivityDate="2015-04-04T21:35:37.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11579" PostTypeId="2" ParentId="10839" CreationDate="2015-04-04T21:43:37.810" Score="3" Body="&lt;p&gt;A slick solution is to let $P(n)$ be the probability that the ant finds the honey at $x=10$ if it starts at position $n$. Clearly $P(0)=0$ and $P(10)=1$. Moreover, since the ant moves left or right with equal probability, it follows that, for $n$ strictly between $0$ and $10$:&#xA;$$P(n)=\frac{1}2\left(P(n-1)+P(n+1)\right)$$&#xA;which is equivalent to saying that $(n-1,P(n-1))$, $(n,P(n))$, and $(n+1,P(n+1))$ are collinear. Given that, for instance $(n-2,P(n-2))$ and $(n+1,P(n+2))$ are both on the line of $(n,P(n))$ and $(n-1,P(n-1))$, it follows that all four of those points are collinear - and quickly from there that all the $(n,P(n))$ for $n$ between $0$ and $10$ (inclusive) line on a single line - and thus on the line containing $(0,0)$ and $(10,1)$. Thus $P$ is linear and&#xA;$$P(n)=\frac{n}{10}$$&#xA;and the probability that the ant finds the honey at $x=10$ is:&#xA;$$P(7)=\frac{7}{10}$$&lt;/p&gt;&#xA;" OwnerUserId="4765" LastEditorUserId="4765" LastEditDate="2015-04-05T14:20:22.587" LastActivityDate="2015-04-05T14:20:22.587" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11580" PostTypeId="2" ParentId="11572" CreationDate="2015-04-04T21:49:25.873" Score="1" Body="&lt;p&gt;Second try:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://meta.stackexchange.com/users/137537/boltclocks-a-unicorn&quot;&gt;BoltClock's a Unicorn&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Most of it is fairly self-explanatory from the picture in the link: green body, blue hair and eyebrows, very pale purple eyes, purple background. Probably&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;beak&quot; = &quot;horn&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not sure about &quot;named after the second greatest creator&quot; though. As for the hints:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if you've spent much time on Meta.SE, you'll know that unicorns are something of a meme there (e.g. see &lt;a href=&quot;https://meta.stackexchange.com/questions/122026/what-is-unicorn-on-stack-overflow&quot;&gt;this&lt;/a&gt; page). They come in many different colours and are physical objects whose existence is ... shall we say, doubted.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-03-20T10:32:12.560" LastActivityDate="2015-04-04T21:49:25.873" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11581" PostTypeId="2" ParentId="11551" CreationDate="2015-04-05T03:00:19.263" Score="9" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;To be, or not to be, that is the question&quot; &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That(42) returns a Question whose value is 43|-44 &lt;br&gt;&#xA; which in hexadecimal is 0x2b | ~0x2b &lt;br&gt;&#xA; (where '~' is bitwise NOT and '|' is bitwise OR).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10724" LastActivityDate="2015-04-05T03:00:19.263" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11583" PostTypeId="2" ParentId="11572" CreationDate="2015-04-05T04:56:54.390" Score="5" Body="&lt;p&gt;Following up @randal'thor's answer I think it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://meta.stackexchange.com/questions/37328/my-god-its-full-of-unicorns&quot;&gt;Jeff Atwood's unicorn avatar&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am named after the second greatest creator&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jeff Atwood co-founded (the precursor to) SE with Joel Spolsky. &lt;br&gt;&#xA; All the other hints are self-explanatory, I think.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10724" LastEditorUserId="-1" LastEditDate="2017-03-20T10:32:12.560" LastActivityDate="2015-04-05T04:56:54.390" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11584" PostTypeId="2" ParentId="469" CreationDate="2015-04-05T07:38:19.907" Score="0" Body="&lt;p&gt;The OP asks two questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&quot;... whether there are any fairly popular puzzles that are still 'easy' in an algorithmic sense&quot;; and&lt;/li&gt;&#xA;&lt;li&gt;&quot;What other examples are there of puzzles that have maintained popularity despite having relatively mechanical solutions?&quot;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This answer only deals with the first and argues that there are popular puzzles that are 'easy' in an algorithmic sense although 'difficult' in a solution-formulation sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Algorithmic complexity is usually expressed as a measure of the time or space taken in terms of a measure of the inputs. Different solutions to a given problem can have different complexity. For example, quicksort is an algorithm for sorting lists and has average case time complexity $O(n \log n)$ in terms of the number of items to be sorted. Bubble sort is another algorithm for sorting lists and has average case time complexity $O(n^2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although there are problems whose solutions always have high algorithmic complexity, the lateral thinking required for good puzzles often makes the formulation of the solution more difficult than the computing the solution itself. As a measure, algorithmic complexity doesn't really help because by definition it is a measure of only the solution, not of the formulation of that solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a look at the 'questions' page: click on 'questions' above and then on 'votes'. Using votes as a proxy for popularity, the current list has quite a few popular puzzles with solutions that have low algorithmic complexity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Top 6 currently:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/4304/how-to-get-to-an-island-with-a-tree-in-the-middle-if-all-you-have-is-rope&quot;&gt;https://puzzling.stackexchange.com/questions/4304/how-to-get-to-an-island-with-a-tree-in-the-middle-if-all-you-have-is-rope&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/3215/a-double-agent-with-a-conundrum&quot;&gt;A double-agent with a conundrum&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/5143/internship-available-figure-out-what-youre-being-asked-to-do-before-you-sig&quot;&gt;Internship Available!* - Figure out what you&amp;#39;re being asked to do before you sign up&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/6165/merlin-and-hermes-mysterious-lines&quot;&gt;Merlin and Hermes: Mysterious Lines&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/1968/3-impossibly-intelligent-mathematicians&quot;&gt;https://puzzling.stackexchange.com/questions/1968/3-impossibly-intelligent-mathematicians&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/626/paying-the-troll-toll&quot;&gt;Paying the Troll toll&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Of these, only the &quot;3 impossibly intelligent mathematicians&quot; puzzle is really amenable to 'mechanistic' solving. In most of the others, once the 'aha' observation has been made, computing the solution is trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One final note: most of these puzzles don't have significant structural variability in their inputs, so algorithmic complexity is not a good measure of whether these particular kinds of puzzles are 'easy' or 'difficult'.&lt;/p&gt;&#xA;" OwnerUserId="9422" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-04-05T11:20:33.383" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11585" PostTypeId="2" ParentId="11563" CreationDate="2015-04-05T11:47:25.423" Score="5" Body="&lt;p&gt;We can solve this with a general approach by doing some linear algebra. The nice thing about a general approach like this is, that even irregular, 42-dimensional (finite) shapes can still be solved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, it will involve a matrix with 81 rows and 117 columns and compute for &lt;strong&gt;all&lt;/strong&gt; starting fields the exact probability distribution of rest states (at the border). And while a matrix with 9477 entries may look a bit daunting at first, it can be easily handled by a computer program.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For illustration purposes, first consider the region of tuples $x, y \in 0,1,2,3$. Respectively only the inner square of four points $\{1,2\}×\{1,2\}$ is not filled with honey. It looks like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;FOOD         ab&#xA;F..D   or   cαβd , to give some variable names.&#xA;F..D        eγδf&#xA;FOOD         gh&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For each of the inner fields $α, β, γ, δ$, the probability distribution of final states called $D$ is given by a linear equation, e.g.:&#xA;$$Dα = \frac{a+c+β+γ}{4}\text{.}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resulting matrix is (after multiplication by $-4$):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;   α  β γ δ  a b c d e f g h&#xA;Dα -4,1,1,0, 1,0,1,0,0,0,0,0;&#xA;Dβ 1,-4,0,1, 0,1,0,1,0,0,0,0;&#xA;Dγ 1,0,-4,1, 0,0,0,0,1,0,1,0;&#xA;Dδ 0,1,1,-4, 0,0,0,0,0,1,0,1;&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And using a &lt;a href=&quot;http://people.revoledu.com/kardi/tutorial/LinearAlgebra/RREF.html&quot;&gt;handy online tool by Kardi Teknomo&lt;/a&gt; we can compute the reduced row echelon form of this matrix: (Multiplied by $-24$)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;     α β γ δ  a b c d e f g h&#xA;Dα -24,0,0,0, 7,2,7,2,2,1,2,1;&#xA;Dβ 0,-24,0,0, 2,7,2,7,1,2,1,2;&#xA;Dγ 0,0,-24,0, 2,1,2,1,7,2,7,2;&#xA;Dδ 0,0,0,-24, 1,2,1,2,2,7,2,7;&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We can now extract the information that&#xA;$$Dγ = \frac{2a+b+2c+d+7e+2f+7g+2h}{24}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And since we would like to know what the chance of turning up at $x=0$ or $y=0$ is, we just add up the chances to end up at $c, e, g$ or $h$ which is $\frac{2+7+2+7}{24} = \frac{9}{12}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does this tell us about the actual problem at hand? Well we can write a little program which generates the redistribution matrix $D$ for us. Next we tell it to calculate the reduced row echelon form and finally we extract our information. For a lack of motivation though, writing that program is left as an exercise to the reader :P.&lt;/p&gt;&#xA;" OwnerUserId="10154" LastActivityDate="2015-04-05T11:47:25.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11586" PostTypeId="1" CreationDate="2015-04-05T12:37:49.143" Score="3" ViewCount="883" Body="&lt;p&gt;Warren Brown received a very peculiar visitor two weeks ago in his car-repair shop, the visitor seemed to be a male by judging from his voice, he was wearing a black jacket, a white mask, leather gloves and brown boots. The visitor inquired &quot;Do you remember?&quot; in a very stirred tone - Warren was greatly confused and replied, &quot;Remember what?&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Warren is a 62 year old car repairer with two sons and two daughters; All of them are adults. Warren is a rather busy man, with free time almost impossible to acquire even on the weekends. His children love him more than anything else, and he does love them too by an equal amount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On that day however, Warren was not very busy. He went to the shop directly from home and sat around the office with nothing to do except play darts. Warren at first had a very stimulating thought which made him expect something astonishing but he later forgot about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyways, on that conversation with the visitor; The visitor displayed a visible grin and remarked, &quot;Here, take this. Open it 30 minutes after your clock reaches 11:30pm&quot; and handed him a red box, which felt heavy and the visitor left the shop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After Warren did what the visitor asked, he was surprised and got very happy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;What was in the box and who was this strange visitor?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="10704" LastActivityDate="2015-04-10T18:27:01.320" Title="A shopkeeper receives a strange visitor" Tags="|pattern|wordplay|" AnswerCount="5" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="11587" PostTypeId="2" ParentId="11429" CreationDate="2015-04-05T12:38:56.113" Score="5" Body="&lt;p&gt;While it is intended as unsolvable as explained by Bulldogg6404, it can actually be solved, as long as you look at the puzzle laterally at the right moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If you run down the left edge (with a slight detour to get from D to A), you reach a Z in the 8th row. Viewing the problem from the side (that is, laterally), you'll discover that it's also an N, which is convenient, because this allows you to move right onto the O, which is an O irrespective of perspective. From there, it's easy to get to the bottom-right corner by going right six, down one, right two, down three, right two, and down one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9289" LastActivityDate="2015-04-05T12:38:56.113" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11591" PostTypeId="2" ParentId="10839" CreationDate="2015-04-05T17:24:55.477" Score="2" Body="&lt;p&gt;I'm expanding &lt;a href=&quot;https://puzzling.stackexchange.com/a/10853/10724&quot;&gt;Joffan's second answer&lt;/a&gt; here to highlight it as a different approach that is still completely rigorous, uses only basic algebra, and handles the conditional stopping issue. (I'm not saying it's better than the other correct answers.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $p_i$ denote the probability of reaching the high end before the low end starting from position $i$. We know $p_0 = 0$ and $p_{10} = 1$ and we want to find $p_7$. We also know that $p_i = \tfrac{1}{2}(p_{i-1}+p_{i+1})$ for $1 \le i \le 9$. Let's multiply that last one by 2 and rearrange to get&#xA;$p_{i+1} = 2p_i - p_{i-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could solve this linear recurrence directly with some machinery (or induction) but instead let's use basic algebra to rewrite everything in terms of $p_1$. At each step I'm just using the recurrence and things we know up to that point:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array} \\&#xA;p_2 &amp;amp;= 2p_1 - p_0 &amp;amp;= 2 \cdot p_1 - 0 &amp;amp;= 2p_1 \\&#xA;p_3 &amp;amp;= 2p_2 - p_1 &amp;amp;= 2 \cdot 2p_1 - p_1 &amp;amp;= 3p_1 \\&#xA;p_4 &amp;amp;= 2p_3 - p_2 &amp;amp;= 2 \cdot 3p_1 - 2p_1 &amp;amp;= 4p_1 \\&#xA;p_5 &amp;amp;= 2p_4 - p_3 &amp;amp;= 2 \cdot 4p_1 - 3p_1 &amp;amp;= 5p_1 \\&#xA;p_6 &amp;amp;= 2p_5 - p_4 &amp;amp;= 2 \cdot 5p_1 - 4p_1 &amp;amp;= 6p_1 \\&#xA;p_7 &amp;amp;= 2p_6 - p_5 &amp;amp;= 2 \cdot 6p_1 - 5p_1 &amp;amp;= 7p_1 \\&#xA;p_8 &amp;amp;= 2p_7 - p_6 &amp;amp;= 2 \cdot 7p_1 - 6p_1 &amp;amp;= 8p_1 \\&#xA;p_9 &amp;amp;= 2p_8 - p_7 &amp;amp;= 2 \cdot 8p_1 - 7p_1 &amp;amp;= 9p_1 \\&#xA;p_{10} &amp;amp;= 2p_9 - p_8 &amp;amp;= 2 \cdot 9p_1 - 8p_1 &amp;amp;= 10p_1 \\&#xA;\end{array} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we use the last line along with $p_{10} = 1$ to conclude that $p_1 = \tfrac{1}{10}$ and thus $p_7 = \tfrac{7}{10}$.&lt;/p&gt;&#xA;" OwnerUserId="10724" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-04-05T18:38:14.697" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11592" PostTypeId="2" ParentId="11586" CreationDate="2015-04-05T17:55:08.073" Score="0" Body="&lt;p&gt;Let's analyze the hints:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;wearing a white mask&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;implies that Warren cannot recognize him immediately.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;seemed to be a male by judging from his voice&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's a male. Don't forget that sometimes masks may modify the voice.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Warren is 62 year old  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Memory isn't very reactive for 62 years old people, no offence.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Two sons. All of them are adults.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He has adult male offspring.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;free time almost impossible to acquire even on the weekends&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Suggests that his social life is pretty limited.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On that day however, Warren was not very busy. He went to the shop directly from home and sat around the office with nothing to do except play darts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again, he prefers spending his free-time in the shop rather than socializing.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Warren at first had a very stimulating thought which made him expect something astonishing but he later forgot about it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Probably he remembered of some anniversary or stuff like that.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The visitor displayed a visible grin&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The visitor is a friend, not a enemy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Open it 30 minutes after your clock reaches 11:30pm&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A new day begins at midnight.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;handed him a red box, which felt heavy and the visitor left the shop&#xA;  A gift!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;After Warren did what the visitor asked, he was surprised and got very happy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Indeed, it was a gift! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, what can we deduce? Warren doesn't seem to have friends, just his relatives, so the visitor is most likely &lt;strong&gt;his son&lt;/strong&gt;. He made a surprise gift to the old father.&#xA;What kind of present? We can't know it, but considering the psychological aspect of Warren, a &lt;strong&gt;new model of wrench&lt;/strong&gt; would be a very appreciated gift!&lt;/p&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-04-05T17:55:08.073" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11594" PostTypeId="1" AcceptedAnswerId="11670" CreationDate="2015-04-05T19:26:44.117" Score="16" ViewCount="2492" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;This is an experiment of nested puzzles.&lt;/strong&gt;&lt;br&gt;&#xA;  The answer of each puzzle is necessary to solve the next one. Sometimes you have to manipulate it, sometimes you have to open your eyes (and mind).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Run, to Westminster Abbey!&lt;br&gt;&#xA;Nevermind, to Gatwick, ok!&lt;br&gt;&#xA;Please, don't waste time,&lt;br&gt;&#xA;I want to keep my dime!&lt;br&gt;&#xA;The original one is black,&lt;br&gt;&#xA;never goes off track!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HW9Mo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What's the answer?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Notes&lt;/strong&gt;: I will add hints every 24 hours.&lt;br&gt;&#xA;Sadly I couldn't add more than 5 tags, these are those I would include as well: &lt;a href=&quot;/questions/tagged/rhyme&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;rhyme&amp;#39;&quot; rel=&quot;tag&quot;&gt;rhyme&lt;/a&gt;, &lt;a href=&quot;/questions/tagged/computer-puzzle&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;computer-puzzle&amp;#39;&quot; rel=&quot;tag&quot;&gt;computer-puzzle&lt;/a&gt;, &lt;a href=&quot;/questions/tagged/brain-teaser&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;brain-teaser&amp;#39;&quot; rel=&quot;tag&quot;&gt;brain-teaser&lt;/a&gt;.&lt;br&gt;&#xA;I accept partial answers, I will personally upvote those who solve at least 1 unsolved puzzle or submit a wrong, yet plausible solution&lt;/em&gt;.&lt;br&gt;&#xA;&lt;em&gt;Teamwork is appreciated!&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Useful link: &lt;a href=&quot;http://www.mobilefish.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.mobilefish.com/&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Don't consider the picture until you solve the riddle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Actually the picture is the 3rd puzzle to be solved.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The 2nd puzzle is hidden in the 1st answer!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #4&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It may be difficult to solve puzzle #4 without solving #5. It's impossible to solve #5 before #4.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #5&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Puzzle #5 requires a &lt;em&gt;&quot;different point of view&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #6&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;/questions/tagged/cipher&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;cipher&amp;#39;&quot; rel=&quot;tag&quot;&gt;cipher&lt;/a&gt; + &lt;a href=&quot;/questions/tagged/lateral-thinking&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;lateral-thinking&amp;#39;&quot; rel=&quot;tag&quot;&gt;lateral-thinking&lt;/a&gt; = Final Answer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #7&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The final answer is an English word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #8&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is &lt;strong&gt;only&lt;/strong&gt; one reference to Chemistry.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #9&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Looking for a different point of view? Turn upside down your head!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #10&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What do you usually need to solve a &lt;a href=&quot;/questions/tagged/cipher&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;cipher&amp;#39;&quot; rel=&quot;tag&quot;&gt;cipher&lt;/a&gt; puzzle?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #11&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Look around like a bat!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #12&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You've to deal with a basic kind of cipher. Once you know the key and the encrypted message (and you know both, I assure), you should be able to decrypt the message with few attempts!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bounty awarded to Len, the one who solved the highest number of puzzles.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answers Credits:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Puzzle #1:&lt;/strong&gt; Len&lt;br&gt;&#xA;&lt;strong&gt;Puzzle #2:&lt;/strong&gt; finsternis&lt;br&gt;&#xA;&lt;strong&gt;Puzzle #3:&lt;/strong&gt; finsternis&lt;br&gt;&#xA;&lt;strong&gt;Puzzle #4:&lt;/strong&gt; Len (rand al'thor contributed too)&lt;br&gt;&#xA;&lt;strong&gt;Puzzle #5:&lt;/strong&gt; Len + rand al'thor&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-04-12T18:23:12.447" LastActivityDate="2015-04-12T18:23:12.447" Title="Five puzzles in one!" Tags="|riddle|cipher|lateral-thinking|password|steganography|" AnswerCount="6" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11595" PostTypeId="1" AcceptedAnswerId="11598" CreationDate="2015-04-05T20:30:26.043" Score="5" ViewCount="408" Body="&lt;p&gt;In more ways than one,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My end brings a start,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But can also bring the end itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I end all fun,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I only ever began with grief.&lt;/p&gt;&#xA;" OwnerUserId="10974" LastActivityDate="2015-04-12T14:41:58.990" Title="If held here I'm dangerous" Tags="|riddle|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11596" PostTypeId="2" ParentId="11595" CreationDate="2015-04-05T20:43:59.477" Score="5" Body="&lt;h2&gt;Credit to mdc32 for the answer, but here are more explanations&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; GUN.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If held here I'm dangerous&lt;/em&gt; (title)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Holding a gun by its barrel (say) can be a dangerous and stupid thing to do. Also G, U, and N all appear in the word &quot;dangerous&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In more ways than one,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This could be a reference to the fact that &quot;gun&quot; is a bit like &quot;one&quot; with an extra G on the beginning (at least phonetically). Also each of the succeeding lines has more than one interpretation: letterplay and metaphor.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My end brings a start,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The end of &quot;gun&quot; is N, which is the commonest letter in the word &quot;beginning&quot;. Also the abolition of guns might lead to a fresh new start for humanity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But can also bring the end itself.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That letter N is also the central letter in &quot;end&quot;. Alternatively, abolishing guns in one place might lead to the end of that society if people in other places still had guns.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I end all fun,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The word &quot;gun&quot; ends with the same letters as &quot;fun&quot;. Also guns tend to end fun (unless they're water-pistols maybe).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But I only ever began with grief.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The word &quot;gun&quot; starts with the same letter as &quot;grief&quot;. Also guns often cause grief.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;First attempt&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; STING?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In more ways than one,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter &lt;strong&gt;S&lt;/strong&gt; is used to pluralise, suggesting &quot;more than one&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My end brings a start,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The end of the word &quot;start&quot; is the letter &lt;strong&gt;T&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But can also bring the end itself.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter just after the word &quot;end&quot; is &lt;strong&gt;I&lt;/strong&gt;. (OK, my reasoning is weak for this line, but I is the only letter that fits with the rest.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I end all fun,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The final letter of &quot;fun&quot; is &lt;strong&gt;N&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But I only ever began with grief.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first letter of &quot;grief&quot; is &lt;strong&gt;G&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-04-06T10:26:09.800" LastActivityDate="2015-04-06T10:26:09.800" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11597" PostTypeId="2" ParentId="11486" CreationDate="2015-04-05T21:02:10.907" Score="3" Body="&lt;p&gt;Here's another solution to the four lighthouse problem.  It's very similar to dmitch's, but replaces the bit about placing the axes so there is one lighthouse in each quadrant with something that is a bit more straightforward I think.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Suppose there existed two lighthouses at points $(x_1, y_1)$ and $(x_2, y_2)$, with $x_1 \leq x_2$, below the $x$-axis of the plane.  Then these two lighthouses could illuminate the entire plane above the $x$-axis.  Just point the $(x_1, y_1)$ in the same direction as quadrant I, and the $(x_2, y_2)$ lighthouse in the same direction as quandrant II.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Now separate the four lighthouses with some line so that there are two on each side of the line (To do this, think of a line at a random angle sweeping across the plane;  eventually, you'll reach a position with two lighthouses on one side of the line, and two on the other.)  The two lighthouses above the line can illuminate the area below the line, and the two lighthouses below the line can illuminate the area above.  Hence the whole plane is covered.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2722" LastActivityDate="2015-04-05T21:02:10.907" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11598" PostTypeId="2" ParentId="11595" CreationDate="2015-04-05T21:49:56.943" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; GUNS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In more ways than one&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Guns is plural.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My end brings a start&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Start begins with an S, the end of GUNS.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But can also bring the end itself.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure - maybe that guns can kill people?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I end all fun,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Guns can end fun (unless the fun is a paintball war)...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I only ever began with grief.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; GUNS starts with a G, the beginning of grief, and they can cause grief.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1664" LastActivityDate="2015-04-05T21:49:56.943" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11601" PostTypeId="2" ParentId="11572" CreationDate="2015-04-06T02:47:04.447" Score="1" Body="&lt;p&gt;I know it's not the right answer due to your hints, but I'll post this anyway:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Christmas tree&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My body is green&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Christmas trees are evergreens.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My hair and eyebrows are blue&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They could be decorated with blue tinsel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My eyes are a very light shade of purple&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Christmas lights could be red/pink/purple.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have the sharpest beak in all my friends&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pine needs can be sharp.  Or the star at the top of the tree could be sharp.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I painted my wall and floor in a light shade of purple&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Purple Christmas lights would color the walls and floor purple.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am named after the second greatest creator&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Christmas is named after (Jesus) Christ, who could be considered second to God.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="388" LastActivityDate="2015-04-06T02:47:04.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11602" PostTypeId="1" AcceptedAnswerId="11604" CreationDate="2015-04-06T04:07:01.600" Score="4" ViewCount="444" Body="&lt;p&gt;Say that a solution to a &lt;a href=&quot;https://en.wikipedia.org/wiki/Sokoban&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sokoban&lt;/a&gt; puzzle with only 2 boxes is &lt;strong&gt;box-isolating&lt;/strong&gt; if and only if there are step numbers $m$ and $n$ such that for one of the boxes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;All&lt;/em&gt; of its pushes were between step $m$ and step $n$, and:&lt;/li&gt;&#xA;&lt;li&gt;The other box was not pushed between step $m$ and step $n$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a Sokoban puzzle with only 2 boxes such that there is a solution but there is no box-isolating solution?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I do not know the answer to this question.&lt;br&gt;&#xA;It is related to &lt;a href=&quot;https://cs.stackexchange.com/questions/41058/sokoban-with-only-k-boxes&quot;&gt;this question I asked on cs.stackexchange&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user1579" LastEditorUserId="-1" LastEditDate="2017-04-13T12:48:38.380" LastActivityDate="2015-04-06T11:34:17.673" Title="Is there a solvable 2-box Sokoban puzzle in which the pushes must be interleaved?" Tags="|solvability|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11603" PostTypeId="2" ParentId="166" CreationDate="2015-04-06T04:59:48.827" Score="0" Body="&lt;p&gt;Here's my solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Take one rope and light both ends at the same time while also lighting one end of the second rope. That should take 30 minutes to burn completely leaving 30 minutes of the second rope unburned. As soon as the first rope has burned completely light the second end of the second rope. The remaining 30 minutes of the second rope will be halved leaving you with 45 minutes total.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11104" LastEditorUserId="46903" LastEditDate="2018-09-07T02:48:19.950" LastActivityDate="2018-09-07T02:48:19.950" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="11604" PostTypeId="2" ParentId="11602" CreationDate="2015-04-06T06:22:31.030" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes, here is such a board&lt;br&gt;&#xA; &lt;code&gt;#######&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;###F###&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;#...###&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;#.#.AS#&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;#.#.B##&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;#.##.##&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;#....F#&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;#######&lt;/code&gt;&lt;br&gt;&#xA; S is the starting position, A and B are the initial positions of the boxes, and F are the final positions. The A-box has to be touched in the first step and the B-box in the second step. But there is no way to take B to a final position without touching A again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9471" LastActivityDate="2015-04-06T06:22:31.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11605" PostTypeId="1" AcceptedAnswerId="11606" CreationDate="2015-04-06T06:41:57.643" Score="-5" ViewCount="297" Body="&lt;p&gt;A person is stuck in a cold dark room without any light. In the room, there is a  gas lantern, a gas heater, and a gas stove. The person has a match-box with only one match in it. What will he light first?&lt;/p&gt;&#xA;" OwnerUserId="10882" LastEditorUserId="5373" LastEditDate="2017-12-10T17:15:56.293" LastActivityDate="2017-12-10T17:15:56.293" Title="What will I light first?" Tags="|lateral-thinking|situation|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11606" PostTypeId="2" ParentId="11605" CreationDate="2015-04-06T06:46:42.430" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He will first have to light the match stick to light others.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5565" LastActivityDate="2015-04-06T06:46:42.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11607" PostTypeId="2" ParentId="11605" CreationDate="2015-04-06T08:04:11.723" Score="1" Body="&lt;p&gt;He will light the match, and then the gas lantern, then the gas heater, and the gas stove. So depending on which things you exclude, you'll always know which one was lighted &quot;first&quot;.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-04-06T08:04:11.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11608" PostTypeId="1" AcceptedAnswerId="11609" CreationDate="2015-04-06T09:28:41.973" Score="10" ViewCount="2202" Body="&lt;p&gt;What are the next few numbers in each of these integer sequences? Each of them has a clear and justifiable solution.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;1, 2, 3, 5, 8, 9, 12, 20, 21, 22, &lt;strong&gt;??&lt;/strong&gt;, &lt;strong&gt;??&lt;/strong&gt;, ...&lt;/p&gt;&#xA;&lt;p&gt;50, 49, 48, 46, 45, 44, 43, 38, ..., 25, 24, 23, &lt;strong&gt;??&lt;/strong&gt;, &lt;strong&gt;??&lt;/strong&gt;, ...&lt;/p&gt;&#xA;&lt;p&gt;11, 12, 20, 30, 80, 90, 10000, &lt;strong&gt;????&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;... 4, 5, 6, 8, 10, 40, 46, 60, 61, &lt;strong&gt;??&lt;/strong&gt;, &lt;strong&gt;??&lt;/strong&gt;, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2020-07-28T02:19:09.013" Title="A tricky batch of four number sequences" Tags="|number-sequence|" AnswerCount="6" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11609" PostTypeId="2" ParentId="11608" CreationDate="2015-04-06T09:42:33.757" Score="9" Body="&lt;p&gt;&lt;strong&gt;Partial answer&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;23,25&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ordinal numbers that aren't made juxtaposing the cardinal and &quot;-th&quot; suffix.&lt;br&gt;&#xA;&lt;em&gt;Credit of this answer to Xyuzhg&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3)  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Googol&lt;/strong&gt;    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Numbers that when written have exactly 6 letters. 10000=myriad.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;64, 80&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nonnegative integers that when written out in full in English have no repeated letter.&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-04-06T15:00:44.017" LastActivityDate="2015-04-06T15:00:44.017" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11610" PostTypeId="2" ParentId="11163" CreationDate="2015-04-06T09:58:12.223" Score="15" Body="&lt;p&gt;The algorithm is as such: convert each digit into Roman numerals, string them together, and count runs of different characters.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 58458 -&amp;gt; V VIII IV V VIII -&amp;gt; VV IIII VVV III -&amp;gt; 2433&#xA; 2433 -&amp;gt; II IV III III -&amp;gt; III V IIIIII -&amp;gt; 316&#xA; 316 -&amp;gt; III I VI -&amp;gt; IIII V I -&amp;gt; 411  &#xA; 411 -&amp;gt; IV I I -&amp;gt; I V II -&amp;gt; 112&#xA;&#xA; 82 -&amp;gt; VIII II -&amp;gt; V IIIII -&amp;gt; 15&#xA; 47 -&amp;gt; IV VII -&amp;gt; I VV II -&amp;gt; 122&#xA;&#xA; 4 -&amp;gt; IV -&amp;gt; I V -&amp;gt; 11&#xA; 15 -&amp;gt; I V -&amp;gt; I V -&amp;gt; 11&#xA; 19 -&amp;gt; I IX -&amp;gt; II X -&amp;gt; 21&#xA;&#xA; 95 -&amp;gt; IX V -&amp;gt; I X V -&amp;gt; 111&#xA; 91 -&amp;gt; IX I -&amp;gt; I X I -&amp;gt; 111&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My original solution was to convert the number into Roman numerals, and count runs of characters in the numeral where each character is the same as or 1/10 the value of the previous character. Then, each digit is the length of one of these runs.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 58458 -&amp;gt; LV.MMMCDLVIII -&amp;gt; LV MMMC DLV III -&amp;gt; 2433&#xA; 2433 -&amp;gt; MMCDXXXIII -&amp;gt; MMC D XXXIII -&amp;gt; 316&#xA; 316 -&amp;gt; CCCXVI -&amp;gt; CCCX V I -&amp;gt; 411  &#xA; 411 -&amp;gt; CDXI -&amp;gt; C D XI -&amp;gt; 112&#xA;&#xA; 82 -&amp;gt; LXXXII -&amp;gt; L XXXII -&amp;gt; 15&#xA; 47 -&amp;gt; XLVII -&amp;gt; X LV II -&amp;gt; 122&#xA;&#xA; 4 -&amp;gt; IV -&amp;gt; I V -&amp;gt; 11&#xA; 15 -&amp;gt; XV -&amp;gt; X V -&amp;gt; 11&#xA; 19 -&amp;gt; XIX -&amp;gt; XI X -&amp;gt; 21&#xA;&#xA; 95 -&amp;gt; XCV -&amp;gt; X C V -&amp;gt; 111&#xA; 91 -&amp;gt; XCI -&amp;gt; X C I -&amp;gt; 111&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;While this answer returns the same solution in every case, it wasn't nearly as elegant as Alconja's intended answer.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-04-07T17:04:08.750" LastActivityDate="2015-04-07T17:04:08.750" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11611" PostTypeId="1" AcceptedAnswerId="11641" CreationDate="2015-04-06T10:15:24.277" Score="4" ViewCount="2066" Body="&lt;p&gt;This sum sudoku is probably the hardest puzzle I've ever encountered, and I have made very little progress (a few cells or so). I found it in a book by Reader's Digest. The solution is given, but without logical explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rules:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Each row/column/3x3 box contains 1-9. Additionally, the cells are&#xA;  grouped into irregular groups (A-Z, etc.), each which has a pre-defined sum.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Puzzle&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rMDCc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;text version&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A A B B C C D E E&#xA;A F F C C D D D G&#xA;H F I J K L L G G&#xA;H F I J K L L M M&#xA;H N N O O O P P M&#xA;H N N O Q R S S M&#xA;T T U U Q R S V W&#xA;X X Y Y Y R Z V W&#xA;$ X # # * * Z V V&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sums&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A 22 | B  9 | C 20 | D 18 | E 15 | F 19&#xA;G  7 | H 14 | I 11 | J  8 | K 11 | L 20&#xA;M 26 | N 20 | O 21 | P 13 | Q 11 | R 23&#xA;S  9 | T 13 | U  3 | V 19 | W  9 | X 11&#xA;Y 12 | Z 13 | $  9 | #  7 | * 12&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Notes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If possible, please find a simpler way to type the puzzle, but I hope it is clear enough.&lt;/li&gt;&#xA;&lt;li&gt;Although I have verified, there could be some typing error, if so, please tell me.&lt;/li&gt;&#xA;&lt;li&gt;Please give some basic hint as to how you solve it. Atleast explain the logic for the first 10-15 cells you solve, it should help.&lt;/li&gt;&#xA;&lt;li&gt;No group may have the same number repeated.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="1497" LastEditDate="2015-04-06T16:56:56.640" LastActivityDate="2015-04-07T19:05:25.770" Title="Hardest sum sudoku ever! (as well as how to solve them)" Tags="|logical-deduction|sudoku|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11612" PostTypeId="2" ParentId="11608" CreationDate="2015-04-06T10:33:28.547" Score="1" Body="&lt;p&gt;1) All numbers when written out (googol would be ten duotrigintillion) end in one of these letters: denortxy (sorted alphabetically). Take only the even spots of &quot;denortxy&quot; to get &quot;eoty&quot;, and now all numbers ending in a letter from &quot;eoty&quot; are in our list.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ..., 23, 25, 28, 29, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11100" LastEditorUserId="11100" LastEditDate="2015-04-06T10:56:59.540" LastActivityDate="2015-04-06T10:56:59.540" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="11613" PostTypeId="1" AcceptedAnswerId="11615" CreationDate="2015-04-06T10:46:28.120" Score="10" ViewCount="1218" Body="&lt;p&gt;You work as a caretaker in a museum of computing technologies, as one of the people who preserve, store, and archive historical computing-related artifacts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One day, while making your rounds in an old storage room, you come across a machine you've never seen before. You take it out to inspect it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has dials arranged in a division statement, with four dials in the dividend and two in the divisor to the left of the dividend. All the dials are currently set to zero, as so:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;              [0] [0] [0]  r.[0]&#xA;         /---------------&#xA; [0] [0]/ [0] [0] [0] [0]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The bottom six dials are turnable, and there's a lever to the right that presumably makes the top four dials calculate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You set the last dial to &quot;4&quot; and the last dial in the divisor to &quot;2&quot;, so they read &quot;0004&quot; / &quot;02&quot;. Surely enough, when you pull the lever, on top of the line appears the answer you expect: 2.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;              [0] [0] [2]  r.[0]&#xA;         /---------------&#xA; [0] [2]/ [0] [0] [0] [4]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then you set the dividend to 24 and pull the lever again. On top of the line appears: 12.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;              [0] [1] [2]  r.[0]&#xA;         /---------------&#xA; [0] [2]/ [0] [0] [2] [4]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now you try setting the dividend to 13 and the divisor to 169. As you expect, the answer is 13:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;              [0] [1] [3]  r.[0]&#xA;         /---------------&#xA; [1] [3]/ [0] [1] [6] [9]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But then you try evaluating 187 / 13 and the answer is 15 with a remainder of &quot;-8&quot;.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;              [0] [1] [5]  r.[-8]&#xA;         /---------------&#xA; [1] [3]/ [0] [1] [8] [7]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Somehow, the calculator got its quotient wrong, and returned one too many. Perplexed, you try and divide 288 by 21 and get an even stranger result:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;              [0] [1] 7/2  r.[9/2]&#xA;         /---------------&#xA; [2] [1]/ [0] [2] [8] [8]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;To your surprise, the calculator has returned a fraction for a digit (and not only that, the remainder is also fractional)! &lt;em&gt;Isn't the whole point of the remainder to deal with the fractional result?&lt;/em&gt; you think. Next you try dividing 9028 by 32:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;              [3] [-2 [2]  r.[4]&#xA;         /---------------&#xA; [3] [2]/ [9] [0] [2] [8]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And now one of the digits is negative! The answer is indeed 282 if you carry the digits, but for some reason this calculator just flat out refused to borrow one from the three.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You begin to suspect that the calculator isn't just faulty, it runs on a different algorithm entirely. But what is that algorithm?&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2016-09-20T14:41:15.913" LastActivityDate="2016-09-20T14:41:15.913" Title="A rather curious division machine" Tags="|mathematics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11614" PostTypeId="1" AcceptedAnswerId="11618" CreationDate="2015-04-06T11:28:11.133" Score="13" ViewCount="1885" Body="&lt;p&gt;Most people I know hang paintings and posters on their wall. However my robot friend instead hangs a page of binary on the wall:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no idea what it means. &lt;strong&gt;Do you?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="7308" LastActivityDate="2015-04-06T17:21:52.410" Title="Picture Perfect" Tags="|visual|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11615" PostTypeId="2" ParentId="11613" CreationDate="2015-04-06T11:28:42.120" Score="12" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The machine performs polynomial division. The numbers are coefficients of the polynomials. For instance, the last example means (9x³+0x²+2x+8)=(3x+2)(3x²-2x+2)+4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9471" LastActivityDate="2015-04-06T11:28:42.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11616" PostTypeId="2" ParentId="11613" CreationDate="2015-04-06T11:28:42.547" Score="10" Body="&lt;p&gt;The algorithm is as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perform long division as normal, but at every step, divide exactly the leftmost digits of the divisor and dividend only. For example, when dividing 288 by 21, after the 1 in the answer the 7 in 78 is divided by the 2 in 21, yielding:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;              [0] [1] 7/2&#xA;         /---------------&#xA; (2) [1]/ [0] [2] [8] [8]&#xA;              [2] [1]&#xA;              -------&#xA;                  [7] [8]&#xA;                  [7] 7/2&#xA;                  -------&#xA;                      9/2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For the 9028/32 case, a negative digit is obtained because of the subtraction step:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;              [3] [-2 [2]&#xA;         /---------------&#xA; (3) [2]/ [9] [0] [2] [8]&#xA;          [9] [6]&#xA;          -------&#xA;              [-6 [2]&#xA;              [-6 [-4&#xA;              -------&#xA;                  [6] [8]&#xA;                  [6] [4]&#xA;                  -------&#xA;                      [4]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="11113" LastActivityDate="2015-04-06T11:28:42.547" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11617" PostTypeId="2" ParentId="11608" CreationDate="2015-04-06T11:53:39.200" Score="6" Body="&lt;p&gt;Sequence 1:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 23, 25 (twenty-third, twenty-fifth)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Numbers that when written in the chronological ordinal series (first, second, third, etc.) are &lt;strong&gt;not&lt;/strong&gt; the number appended with &quot;th&quot;. (&quot;Eighth&quot; is &quot;eight&quot; + &quot;h&quot;.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11113" LastEditorUserId="21642" LastEditDate="2016-05-19T04:15:26.770" LastActivityDate="2016-05-19T04:15:26.770" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11618" PostTypeId="2" ParentId="11614" CreationDate="2015-04-06T12:18:10.510" Score="13" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HOME SWEET HOME&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The string is 589=31*19 characters long. Create a 31*19 grid filled when the corresponding digit is 1, and 0 otherwise. The grid then spells HOME SWEET HOME.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9471" LastActivityDate="2015-04-06T12:18:10.510" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11619" PostTypeId="2" ParentId="11614" CreationDate="2015-04-06T12:22:30.160" Score="14" Body="&lt;p&gt;It says &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Home sweet home&lt;br&gt;&#xA; The length of this number is 589 = 19*31. If we divide it into 19 lines with 31 symbols each and replace '0' for ' ' we get &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;this&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;pre&gt;&lt;code&gt;      1   1 11111 11111 11111&#xA;      1   1 1   1 1 1 1 1&#xA;      11111 1   1 1 1 1 11111&#xA;      1   1 1   1 1 1 1 1&#xA;      1   1 11111 1 1 1 11111&#xA;&#xA;    11111 1 1 1 11111 11111 11111&#xA;    1     1 1 1 1     1       1&#xA;    11111 1 1 1 11111 11111   1&#xA;        1 1 1 1 1     1       1&#xA;    11111 11111 11111 11111   1&#xA;&lt;br&gt;&#xA;      1   1 11111 11111 11111&#xA;      1   1 1   1 1 1 1 1&#xA;      11111 1   1 1 1 1 11111&#xA;      1   1 1   1 1 1 1 1&#xA;      1   1 11111 1 1 1 11111      &lt;/code&gt;&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user10531" LastEditorUserId="7482" LastEditDate="2015-04-06T14:31:41.117" LastActivityDate="2015-04-06T14:31:41.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11620" PostTypeId="1" AcceptedAnswerId="11632" CreationDate="2015-04-06T13:26:08.863" Score="5" ViewCount="1015" Body="&lt;p&gt;My Indian friend Jamal recently participated in the well-known TV quiz&#xA;show &lt;strong&gt;&lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Who_Wants_to_Be_a_Millionaire&quot; rel=&quot;noreferrer&quot;&gt;Who wants to be a millionaire?&lt;/a&gt;&lt;/em&gt;&lt;/strong&gt;. You might have heard or read about Jamal's spectacular success; the media reported on it extensively.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what Jamal told us about his 500 Euro question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I knew the answer to the 500 Euro question right away, even before Prem&#xA;  Kumar started to read the four answers; however, I didn't spot it until&#xA;  he was done reading.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Answer A was the exact opposite of what the man normally asked for.&#xA;  B was much too Grecian for him, and when you made it with one of those,&#xA;  you usually didn't use vodka (which he preferred).  Answer C described&#xA;  the type of glass it was contained in, and not how it was made, and I&#xA;  certainly didn't recall him constantly being at a New Year's celebration.&#xA;  Hence I gave D as my answer.  Prem asked if it was my final decision.&#xA;  I said yes. Correct, once again, a big shiny smile from Prem, and we&#xA;  moved on to the next question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What was the 500 Euro question?&lt;br&gt;&#xA;What were the four possible answers A, B, C, D?&lt;/p&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2015-04-06T15:42:13.180" Title="Slumdog millionaire (4)" Tags="|knowledge|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11621" PostTypeId="1" AcceptedAnswerId="11624" CreationDate="2015-04-06T13:45:12.417" Score="25" ViewCount="6006" Body="&lt;p&gt;The famous and ruthless explorer Wyoming Wilbert reports in one of his books that he once visited an island inhabited by jokers and truth tellers. Truth tellers always tell the truth, whereas jokers sometimes lie and sometimes tell the truth. Furthermore when a joker is killed than his body turns green, while the corpse of a dead truth teller decomposes in the ordinary fashion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wilbert spent several weeks in a small village with 155 inhabitants. Every village inhabitant knew exactly who the jokers were and who the truth tellers were, but Wilbert did not know the identity of a single inhabitant. On the first day, he asked every inhabitant a single yes-no question. He analyzed the answers and then killed one of the inhabitants; the corpse showed Wilbert whether this guy had been a joker or a truth teller. On the second day, Wilbert repeated this procedure with the remaining 154 inhabitants: he asked every survivor a single yes-no question, and afterwards killed one of them. And so on, day by day, until he decided to stop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wyoming Wilbert reports in his book that he had designed all his questions meticulously. They guaranteed him that after several days &lt;strong&gt;all&lt;/strong&gt; the jokers would be dead, while at most one truth teller had been killed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Question: &lt;strong&gt;What was the questioning strategy of Wyoming Wilbert?&lt;/strong&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="9404" LastEditDate="2015-04-14T08:20:30.560" LastActivityDate="2021-04-17T17:58:56.730" Title="Kill the jokers! - Part 1" Tags="|logical-deduction|liars|" AnswerCount="12" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="11622" PostTypeId="2" ParentId="11620" CreationDate="2015-04-06T13:46:12.743" Score="4" Body="&lt;p&gt;The 500 Euro question was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What was the favourite drink of James Bond?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With answers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A: Martini stirred, not shaken.&lt;br&gt;&#xA; B: Ouzo.&lt;br&gt;&#xA; C: Champagne. Cocktail glass.&lt;br&gt;&#xA; D: Martini shaken, not stirred.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I didn't try too hard to fill in the answers.&lt;/p&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2015-04-06T13:46:12.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11623" PostTypeId="2" ParentId="11620" CreationDate="2015-04-06T13:47:10.030" Score="2" Body="&lt;p&gt;Question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What drink does James Bond like to order?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A - Stirred martini&lt;br&gt; B - Ouzo&lt;br&gt; C - Champagne glass&lt;br&gt; D - Shaken martini&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1504" LastActivityDate="2015-04-06T13:47:10.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11624" PostTypeId="2" ParentId="11621" CreationDate="2015-04-06T14:20:33.757" Score="28" Body="&lt;p&gt;This is exactly question 7 of the 2015 Tournament of Towns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A Solution:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Round the inhabitants into a circle, with one volunteer sacrifice in the center. Ask everyone in the circle about the centerpiece, is he a joker or is he a truth teller? We have two possible outcomes: (1) everyone says the man in the center is a joker, or (2) at least one person vouches for him. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Since everyone thinks he is a joker, it is reasonable to kill him and see what color he turns. If he was a joker, no matter, we now have less people and we haven't killed any truth tellers, so we may continue by induction. If he was a truth teller, then we deduce that every single other village inhabitant lied and are thus jokers, so we may proceed to kill them all without asking any more questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Since not everyone thinks that the man in the center is a joker, we don't want to risk killing him. Instead, we will kill someone who claimed that he is a truth teller. Either our victim was a joker and we may continue by induction, or our victim was a truth teller and it is a shame that we killed him, but we win anyways: he gave us the useful information that the center is also a truth teller. This means that we can use our now known truth teller as a rat on the rest of the village.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RAT: To use our truth teller as a rat, we line up the inhabitants with our truth teller at the back, and ask him what he thinks of the villager directly in front of him. Our rat will either tell us that the one in front of him is no good (we kill him and continue by induction) or that the one in front of him is good, then we move our old rat to a safe zone and continue by induction with our new rat (base case is trivial).&lt;/p&gt;&#xA;" OwnerUserId="11100" LastEditorUserId="11100" LastEditDate="2015-04-07T14:29:48.917" LastActivityDate="2015-04-07T14:29:48.917" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11625" PostTypeId="2" ParentId="11621" CreationDate="2015-04-06T14:29:48.517" Score="0" Body="&lt;p&gt;This can be done without the villagers having any prior knowledge of each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each day, ask each villager the following question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Have I killed more than one truth teller?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Kill anyone that answers &quot;Yes&quot;.&lt;/p&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2015-04-06T14:29:48.517" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11627" PostTypeId="1" AcceptedAnswerId="11629" CreationDate="2015-04-06T15:18:01.853" Score="6" ViewCount="1414" Body="&lt;p&gt;So tell me know, can you figure out,&lt;br&gt;&#xA;The clues I'll give are all about&lt;br&gt;&#xA;A subject many really quite hate,&lt;br&gt;&#xA;The moment they hear it, from the gate.&lt;br&gt;&#xA;It really is useful, as science would say,&lt;br&gt;&#xA;Scientific conclusions use it every day.&lt;br&gt;&#xA;The chances too are great, can you see&lt;br&gt;&#xA;If you use it, or hear what they may be.&lt;br&gt;&#xA;Consider now these clues, not so mean.&lt;br&gt;&#xA;Stay away from numbered errors, as they're seen.&lt;br&gt;&#xA;So check the clues, the hidden and viewed&lt;br&gt;&#xA;Something's lurking, what do you conclude?&lt;/p&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-04-06T15:36:10.680" Title="So tell me now - What am I?" Tags="|riddle|word|rhyme|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11628" PostTypeId="2" ParentId="11627" CreationDate="2015-04-06T15:21:34.437" Score="2" Body="&lt;p&gt;Sounds like&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Math&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A subject many really quite hate,&lt;br&gt;&#xA;The moment they hear it, from the gate.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many people dislike math&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It really is useful, as science would say,&lt;br&gt;&#xA;Scientific conclusions use it every day.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Math is useful in science! &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The chances too are great, can you see&lt;br&gt;&#xA;If you use it, or hear what they may be.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Probability is part of math&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Consider now these clues, not so mean.&lt;br&gt;&#xA;Stay away from numbered errors, as they're seen.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;numbered&quot; -&gt; &quot;numbers&quot; -&gt; &quot;math&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-04-06T15:21:34.437" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11629" PostTypeId="2" ParentId="11627" CreationDate="2015-04-06T15:29:37.067" Score="6" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Statistics?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So tell me know, can you figure out,&lt;br&gt;&#xA;The clues I'll give are all about&lt;br&gt;&#xA;A subject many really quite hate,&lt;br&gt;&#xA;The moment they hear it, from the gate.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many people have a negative attitude towards statistics: there are lies, there are damned lies, and there is statistics.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It really is useful, as science would say,&lt;br&gt;&#xA;Scientific conclusions use it every day.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Statistics is used everywhere in science, from physics over chemistry and biology to economics.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The chances too are great, can you see&lt;br&gt;&#xA;If you use it, or hear what they may be.&lt;br&gt;&#xA;Consider now these clues, not so mean.&lt;br&gt;&#xA;Stay away from numbered errors, as they're seen.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;chances&quot;, &quot;mean&quot;, &quot;errors&quot;: standard terms in statistics&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So check the clues, the hidden and viewed&lt;br&gt;&#xA;Something's lurking, what do you conclude?&lt;/p&gt;&#xA;" OwnerUserId="9125" LastActivityDate="2015-04-06T15:29:37.067" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11630" PostTypeId="2" ParentId="11627" CreationDate="2015-04-06T15:30:07.537" Score="5" Body="&lt;p&gt;My guess is that it is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Statistics / Probability&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the hidden clue is the first letter of each line:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; STATISTICSSS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So tell me know, can you figure out,&lt;br&gt;&#xA;The clues I'll give are all about&lt;br&gt;&#xA;A subject many really quite hate,&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;I&lt;/em&gt; certainly did not enjoy my statistics courses&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The moment they hear it, from the gate.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;From the gate&quot; could refer to various kinds of animal racing and odds are given in all such races for the purposes of betting&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It really is useful, as science would say,&lt;br&gt;&#xA;Scientific conclusions use it every day.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All good scientific conclusions based on data will include a statistical probability of the conclusions being accurate&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The chances too are great, can you see&lt;br&gt;&#xA;If you use it, or hear what they may be.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Chances&quot; is another term for probability&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Consider now these clues, not so mean.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Mean&quot; is a term used in statistics to mean the average&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Stay away from numbered errors, as they're seen.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We want to avoid errors in data collection as that'll mess up the conclusion. Small errors are one indicator of good data.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So check the clues, the hidden and viewed&lt;br&gt;&#xA;Something's lurking, what do you conclude?&lt;br&gt;&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2015-04-06T15:36:10.680" LastActivityDate="2015-04-06T15:36:10.680" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11631" PostTypeId="2" ParentId="11627" CreationDate="2015-04-06T15:30:43.833" Score="4" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Statistics.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Martin was too quick with the rest of the clues, but here's the hidden clue:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Acrostic&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9422" LastActivityDate="2015-04-06T15:30:43.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11632" PostTypeId="2" ParentId="11620" CreationDate="2015-04-06T15:42:13.180" Score="8" Body="&lt;p&gt;The 500 Euro question was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; How does James Bond usually order his favorite drink?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With answers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A. stirred, not shaken&lt;br&gt;&#xA; B. with an olive&lt;br&gt;&#xA; C. in a champagne glass&lt;br&gt;&#xA; D. shaken, not stirred&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9125" LastActivityDate="2015-04-06T15:42:13.180" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11633" PostTypeId="1" AcceptedAnswerId="11634" CreationDate="2015-04-06T15:47:35.697" Score="13" ViewCount="2464" Body="&lt;p&gt;The following country pairs share a very special, very distinguished, very striking property:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Belarus and India&lt;/li&gt;&#xA;&lt;li&gt;Gabon and Italy&lt;/li&gt;&#xA;&lt;li&gt;Malta and Iraq&lt;/li&gt;&#xA;&lt;li&gt;Uganda and Israel&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What is this common property?&lt;br&gt;&#xA;Can you provide another country pair with that property?&lt;/p&gt;&#xA;" OwnerUserId="9125" LastActivityDate="2015-04-13T14:32:05.027" Title="What do these pairs of countries have in common?" Tags="|pattern|geography|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11634" PostTypeId="2" ParentId="11633" CreationDate="2015-04-06T16:41:20.300" Score="26" Body="&lt;p&gt;Letters can be rearranged to form names of other countries. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Belarus India&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sudan Liberia &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Gabon Italy &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Libya Tonga&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Malta Iraq&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mali Qatar &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Uganda Israel&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sudan Algeria &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11119" LastActivityDate="2015-04-06T16:41:20.300" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11635" PostTypeId="2" ParentId="11621" CreationDate="2015-04-06T16:51:42.363" Score="3" Body="&lt;p&gt;Quite simple:&#xA;He asked to one inhabitant to tell him which ones of the inhabitants were jokers and after that killed the inhabitant.&#xA;If the inhabitant  was a joker, Wyoming Wilbert repeated the process the next day.&#xA;Now, if the inhabitant was a truth teller, he discovered the identity of all the jokers, making easy to know which ones he had to kill next.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plus: If he wanted to speed up the process, questioning also to all of the inhabitants who was a truth teller would help, so he would know someone who would answer the truth before dying (if the inhabitants confirmed in unanimity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obs: Still improving my english skills, sorry if I wrote something wrong.&lt;/p&gt;&#xA;" OwnerUserId="11116" LastActivityDate="2015-04-06T16:51:42.363" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11636" PostTypeId="2" ParentId="11614" CreationDate="2015-04-06T17:21:52.410" Score="6" Body="&lt;p&gt;Using the same idea from the other answers, but putting the result into a more visual form, you get this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/hz208.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6188" LastActivityDate="2015-04-06T17:21:52.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11637" PostTypeId="2" ParentId="11621" CreationDate="2015-04-06T17:32:50.243" Score="3" Body="&lt;p&gt;Edit: I now believe Ben Frankel's answer is the correct one. As long as the TT you kill has fingered someone else as a TT, you know have a known TT with which you can identify all Jokers.&lt;/p&gt;&#xA;&lt;p&gt;I'm inclined to say that there's no correct answer.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;All the given answers fail in the case where every Joker behaves as if they were at Truth Teller on the first and second nights&lt;/strong&gt;. Note that this is not the same as telling the truth: A Joker behaving as a Truth Teller, when asked &amp;quot;are you a truth teller?&amp;quot; will respond &amp;quot;Yes.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;Wyoming Wilbert will learn no information that distinguishes between Truth Tellers and Jokers, and will not be able to reliably select a Joker to kill on the second night, thereby violating the clause that &lt;strong&gt;at most one truth teller had been killed&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="6082" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-06T19:46:56.300" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11638" PostTypeId="2" ParentId="11633" CreationDate="2015-04-06T18:04:41.440" Score="10" Body="&lt;p&gt;I will answer to the second question, since Ayefork just showed the pattern.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can you provide another country pair with that property?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/South_Korea&quot;&gt;Republic of Korea&lt;/a&gt; - &lt;a href=&quot;http://en.wikipedia.org/wiki/Democratic_Republic_of_the_Congo&quot;&gt;Democratic Republic of Congo&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;can be easily rearranged to &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Republic_of_the_Congo&quot;&gt;Republic of Congo&lt;/a&gt; - &lt;a href=&quot;http://en.wikipedia.org/wiki/North_Korea&quot;&gt;Democratic Republic of Korea&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They look all the same, but they're actually different countries! &lt;/p&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-04-06T18:04:41.440" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11639" PostTypeId="2" ParentId="11621" CreationDate="2015-04-06T18:20:23.670" Score="0" Body="&lt;p&gt;When asked if he is a joker, any truth-sayer will say &quot;no.&quot;&#xA;When asked if he is a joker, any joker will also say &quot;no&quot; - because he is lying.&#xA;That means that when asked, everyone in the village will deny they are a joker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then ask everyone in the village the following question:&#xA;&quot;If I ask you if you are a joker will you say &quot;no&quot;? &quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A truth sayer will say &quot;yes&quot;.&#xA;A joker will say &quot;no&quot;  - because he's lying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that's how you find all the jokers.&lt;/p&gt;&#xA;" OwnerUserId="11120" LastActivityDate="2015-04-06T18:20:23.670" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11641" PostTypeId="2" ParentId="11611" CreationDate="2015-04-06T19:01:56.670" Score="8" Body="&lt;p&gt;I got:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;694512387&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;731894652&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;285637914&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;576283149&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;328941765&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;419756238&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;852169473&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;147328596&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;963475821&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's how I found the first 15 or so values. (terminology note: when I say a group is (x,y,z), that means the group must contain those numbers, but we don't know in what order yet)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. The bottom left cell is 9.&lt;br&gt;&#xA; 2. The &quot;22&quot; group in the upper-left can be either (5,8,9) or (6,7,9). There is a nine either way, which can only go in the cell not in the leftmost column.&lt;br&gt;&#xA; 3. In the bottom-right box, the 9 and 19 and 13 groups add up to 41. A box must sum up to 45, so the bottom cell of the &quot;9&quot; group must be 4.&lt;br&gt;&#xA; 4. In the right box, the 26 and the 13 and the 9 (minus 4) add up to 44, so the bottom right cell of the &quot;20&quot; group must be 1.&lt;br&gt;&#xA; 5. Not counting the &quot;9&quot; group, the groups in the top-left and middle-left boxes add up to 86. So the left cell of the &quot;9&quot; group must be 4, and the right cell must be 5.&lt;br&gt;&#xA; 6. In the upper-right box, the &quot;15&quot; group can be (8,7) or (9,6). But 9 is already taken in this row, so it must be (8,7).&lt;br&gt;&#xA; 7. In the topmost row, 9,8, and 7 are taken, so the upper left cell must be 6, and the cell below it must be 7.&lt;br&gt;&#xA; 8. In the bottom-right box, the &quot;13&quot; group can be (4,9) or (5,8) or (6,7). But 9 is already taken in that box; and 6 and 7 are already taken in the leftmost column. So the &quot;13&quot; group must be (8,5).&lt;br&gt;&#xA; 9. In the bottom-middle box, the &quot;23&quot; group must be (6,8,9).&lt;br&gt;&#xA; 10. In the bottom-left and bottom-middle boxes, 13 and 11 and 9 and 3 and 12 and 7 and 12 add up to 67. So the lower portions of the &quot;11&quot; and &quot;23&quot; groups must sum to 23. So the upper portions must sum to 11. The top cell of &quot;23&quot; can only be 6, 8, or 9, so the top cell of &quot;11&quot; can only be 5, 3, or 2, so the bottom cell of &quot;11&quot; can only be 6, 8, or 9. But 8 is already taken in that row. Therefore, the top cell of &quot;23&quot; can't be 8. The middle cell of &quot;23&quot; can't be 8 either, so the bottom cell of &quot;23&quot; must be 8.&lt;br&gt;&#xA; 11. The top cell of &quot;23&quot; can't be 9, so its middle cell must be 9. So the top cell is 6.&lt;br&gt;&#xA; 12. The &quot;11&quot; group's top cell is 5 and its bottom cell is 6.&lt;br&gt;&#xA; 13. In the bottom-middle box, the lower &quot;12&quot; group must be, from left to right, 7 and 5.&lt;br&gt;&#xA; 14. The &quot;7&quot; group straddling the bottom-middle and bottom-left boxes must be (3,4). 4 is already taken in the third column from the left, so the left cell is 3 and the right cell is 4.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1497" LastEditorUserId="1497" LastEditDate="2015-04-06T19:07:56.737" LastActivityDate="2015-04-06T19:07:56.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11643" PostTypeId="2" ParentId="11621" CreationDate="2015-04-06T19:30:54.800" Score="0" Body="&lt;p&gt;Trying something new; Not sure if this goes against the rule of the question:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ask with the phase: &quot;I am gonna kill all truth tellers, are you a&#xA; joker?&quot; by assuming all inhabitants wanted to survive.  Truth teller&#xA; have to say no, while all joker will always lying with yes, thus killing them afterward.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2015-04-06T19:30:54.800" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11645" PostTypeId="2" ParentId="11621" CreationDate="2015-04-06T20:24:19.853" Score="14" Body="&lt;p&gt;NOTE: Although some of these answers are correct, some of them do not obey the rule that you can only ask each villager a yes-or-no question and that you can only ask them one question per day. Also in each day, you &lt;em&gt;must&lt;/em&gt; kill one person.&lt;/p&gt;&#xA;&lt;p&gt;I'm not sure if stage 1 of the method described here is the same as what Ben Frankel is describing; since he doesn't explore the case when the volunteer is actually a joker.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;&lt;strong&gt;STAGE 1: Finding a truthteller&lt;/strong&gt;&lt;/h2&gt;&#xA;&lt;p&gt;First, take a random villager and let's call them the &lt;em&gt;pivot&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Ask every single villager (except for the pivot) if the pivot is a truthteller and kill the first person who answers yes to the question (claiming that the pivot is a truthteller).&lt;/p&gt;&#xA;&lt;h3&gt;&lt;strong&gt;CASE 1:&lt;/strong&gt; The pivot is actually a truthteller.&lt;/h3&gt;&#xA;&lt;p&gt;Eventually you will reach a truthteller who you kill and this will confirm the fact that the pivot is a truthteller.&lt;/p&gt;&#xA;&lt;h3&gt;&lt;strong&gt;CASE 2:&lt;/strong&gt; The pivot is actually a Joker.&lt;/h3&gt;&#xA;&lt;p&gt;This means that you will only kill jokers and eventually you will reach a stage where all the villagers answer no (claiming that the pivot is not a truthteller). In this case, kill the pivot, and pick a new pivot only if the pivot was not a truthteller (see below).&lt;/p&gt;&#xA;&lt;h3&gt;&lt;strong&gt;SPECIAL CASE:&lt;/strong&gt; There is only one truthteller.&lt;/h3&gt;&#xA;&lt;p&gt;In an event where there is only one truthteller, eventually you will pick the truthteller as a pivot and you will also eventually reach the stage where all the villagers answer no to your question (claiming that the pivot is not a truthteller). After killing the pivot, you will then know that everyone had been lying and thus you can now proceed with the massacre. (thank you to Prem for pointing this out)&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;STAGE 2: Using the truthteller&lt;/h2&gt;&#xA;&lt;p&gt;Now that you've found a truthteller, you can use them as a starting point to kill jokers. You now need to use the truthteller and ask them if another unknown person is a truthteller.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/7CNO4.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;h3&gt;&lt;strong&gt;CASE 1:&lt;/strong&gt; The truthteller says no (meaning the unknown is a Joker)&lt;/h3&gt;&#xA;&lt;p&gt;Kill the Joker and you are done for the day.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/aEVSk.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;h3&gt;&lt;strong&gt;CASE 2:&lt;/strong&gt; The truthteller says yes (meaning the unknown is a truthteller).&lt;/h3&gt;&#xA;&lt;p&gt;You now have to ask the new truthteller if another unknown is a truthteller.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/qvM1J.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Eventally, you will reach a case where all the villagers answer yes which implies that all the villagers are truthtellers.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8plll.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="11094" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-07T03:42:03.020" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11646" PostTypeId="1" AcceptedAnswerId="11648" CreationDate="2015-04-06T21:26:04.940" Score="7" ViewCount="1269" Body="&lt;p&gt;&quot;Ah, Sherlock, we were waiting for you. Something horrible has happened and we need your help.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sherlock Holmes had barely entered the Security Council room. Only a few hours ago, he was summoned from 221B Baker Street. The mission was so urgent that Sherlock was flown across the Atlantic in a suborned SR-71 Blackbird.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Good Morning to you as well, Madame President. I hope Bill hasn't complained too much about his speed chess defeat last night.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hillary Clinton, President of the United States of America, was impatient. &quot;He'll live. However, several family members of European heads of state aren't that lucky. Several of them have been poisoned and we have no idea how.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Marine Le Pen suddenly interrupted &quot;Mathilde, elle est dans un coma. S'il vous plaît, Monseigneur Holmes, il faut la sauver!&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;As president Le Pen mentioned, one of her children was the first victim. Soon after, Belgium, Germany, Poland and Ukraine were the next to get hit. Each country has the youngest child of the current leader heavily ill. We think the Russian prime minister might be next, but we're not sure.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sherlock sighed, then said &quot;You're not sure? Obviously they're next. I really don't understand why you haven't made any arrests yet. It's clear who is causing this&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Holmes then explained in 3 sentences who they had to arrest, what the link between the countries is and how the link leads to the perpetrators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He was right, obviously. But who was arrested, and why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The poisonings are the work of a conspiracy, not one person. Each offspring was poisoned by a different person. All acting conspirators had the same profession. It's not clear what poison they used, but the symptoms point to one that needs to enter via the mouth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As deduced by the first answer, the E40 European route is important in this question. What's also important is how many victims are mentioned or planned. the perpetrators won't head into Asia.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Last Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are 6 victims along the E40. Put in another way, that's E406. E406 has a well-defined meaning as a code word for a food additive. It's that food additive that holds the key to the puzzle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2778" LastEditorUserId="2778" LastEditDate="2015-04-08T21:07:37.263" LastActivityDate="2015-04-11T16:28:13.390" Title="Sherlock Holmes and the road to perdition" Tags="|riddle|story|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11648" PostTypeId="2" ParentId="11646" CreationDate="2015-04-06T21:44:55.860" Score="5" Body="&lt;p&gt;I believe the perpetrators are the &lt;strong&gt;European Dental Association&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;If you notice, Miss Clinton,&quot; said Sherlock, &quot;the European highway E40 passes through all the countries that were affected. Now, the diagnosis was that the children were all poisoned with agar, also known as E406, which would square quite nicely with &lt;strong&gt;six&lt;/strong&gt; countries on the &lt;strong&gt;E40&lt;/strong&gt;. Agar is used to prepare a substitute for that vegetarians eat in place of gelatin. It's also used to make various candies and to make tooth impressions for dentistry.&quot;&lt;br&gt;&#xA; &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;That last use is how I deduced that these perpetrators must be dentists. You see, all six of these children recently visited the dentist for brace checkups within a six-day span, and their tooth prints were made using agar. The dentists must have poisoned the agar made in the impression.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have to admit, that &quot;E406&quot; step is a pretty far-fetched chain of deductions, even for Sherlock. My second thought after hearing about E406 would have been PETA, since gelatin is something with animal products, and agar would have been a substitute that they might want to promote. It's also possible that my line of reasoning here is completely wrong, as the Sherlock Holmes stories usually have more of an investigation to go on.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-04-11T16:28:13.390" LastActivityDate="2015-04-11T16:28:13.390" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="11649" PostTypeId="2" ParentId="11450" CreationDate="2015-04-06T21:54:38.267" Score="0" Body="&lt;p&gt;When there are X couples on the left and Y couples on the right, with boat on the left:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Couple A cross from left to right.&lt;/li&gt;&#xA;&lt;li&gt;Husband A gets out, Wife A goes back.&lt;/li&gt;&#xA;&lt;li&gt;Wife B gets in boat with Wife A.  Both cross.&lt;/li&gt;&#xA;&lt;li&gt;Wife A gets out, now with her husband.&lt;/li&gt;&#xA;&lt;li&gt;Wife B goes back.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We now have X-1 couples on the left and Y+1 couples on the right, and at no time was any woman left with a man, but without her husband.  There are moments when a boat arrives that a woman is in the boat and her husband is on the far shore, but she remains in the boat with no man in the boat with her.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeat the process X times.  When there is only one couple on the left, they cross and the process is done; no one goes back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This takes N-1 (round) trips, which has to be the minimum.  Each trip to the right takes two people (the maximum) and each trip to the left takes one (the minimum).  This would be the minimum number of trips necessary if there were no restrictions on who could be left with whom.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2015-04-06T21:54:38.267" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11650" PostTypeId="1" AcceptedAnswerId="11687" CreationDate="2015-04-06T22:22:54.273" Score="9" ViewCount="861" Body="&lt;p&gt;&lt;strong&gt;I am faster than you can run,&lt;/strong&gt;&lt;br&gt;&#xA;&lt;em&gt;so abandon me, go home and have fun.&lt;/em&gt;&lt;br&gt;&#xA;&lt;strong&gt;I live where the moon rise from above,&lt;/strong&gt;&lt;br&gt;&#xA;&lt;em&gt;so abandon me, get rid of the hands in glove.&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I am soft like a kitty&lt;/strong&gt; &lt;em&gt;and hard like a rock,&lt;/em&gt;&lt;br&gt;&#xA;&lt;strong&gt;without any warning we come in flock.&lt;/strong&gt;&lt;br&gt;&#xA;Who am I that you might ask?&lt;br&gt;&#xA;&lt;strong&gt;a symbol of fear,&lt;/strong&gt; hidden in a mask!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hints:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every line is a hint, there are no word plays and there are more than one references.  Also, I &lt;strong&gt;bold&lt;/strong&gt; and &lt;em&gt;italic&lt;/em&gt; some lines for a reason~   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #2:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2 words from the first 2 lines gives you enough hints when combined with the lines  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="10037" LastEditDate="2015-04-07T22:09:59.737" LastActivityDate="2015-04-08T16:55:00.490" Title="Who am I that you might ask?" Tags="|riddle|rhyme|" AnswerCount="6" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11651" PostTypeId="2" ParentId="11650" CreationDate="2015-04-06T22:31:25.253" Score="1" Body="&lt;p&gt;A lot of the lines could fit&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;the sea&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am faster than you can run,&lt;br&gt;&#xA; so abandon me, go home and have fun.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sea is a homophone of &lt;strong&gt;c&lt;/strong&gt;, which represents the speed of light - much faster than anyone can run.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I live where the moon rise from above,&lt;br&gt;&#xA; so abandon me, get rid of the hands in glove.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the sea, the moon appears as a reflection. When the moon rises, its reflection starts at the horizon and moves &lt;em&gt;downwards&lt;/em&gt;. &quot;Hands in glove&quot; could also be a reference to reflections: a hand and its glove, like the moon and its reflection, are very similar to each other yet subtly different.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am soft like a kitty and hard like a rock,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sea can be soft and gentle when calm, but it's deadly in a storm, when the impact of fast-moving water can feel like a rock.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;without any warning we come in flock.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Waves can come in flocks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Who am I that you might ask?&lt;br&gt;&#xA; a symbol of fear, hidden in a mask!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; From &lt;a href=&quot;http://followtherabbi.com/journey/israel/when-storms-come&quot; rel=&quot;nofollow&quot;&gt;this source&lt;/a&gt;: &quot;In Jesus' day, Jewish people feared large bodies of water. They referred to the sea as an abyss and saw it as a symbol of chaos and hell.&quot; I'm not sure about the mask bit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-04-06T22:31:25.253" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11652" PostTypeId="2" ParentId="11522" CreationDate="2015-04-07T00:03:25.847" Score="15" Body="&lt;p&gt;The way to decode the cipher is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each 3-letter sequence represents a single uppercase letter and can be decoded as such:&lt;br&gt;&lt;br&gt;&#xA; If third letter is A, add the 0-based values of the first two letters (modulo 26), convert back to a letter.&lt;br&gt;&#xA; Else (third letter is B), subtract the 0-based value of the second letter from the 0-based value of the first (modulo 26), convert back to a letter.&lt;br&gt;&lt;br&gt;&#xA; (0-based values = 0=A,...,25=Z)&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;The raw plaintext is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; FOURSCOREANDSEVENYEARSAGOOURFATHERSBROUGHTFORTHONTHISCONTINENTANEWNATIONCONCEIVEDINLIBERTYANDDEDICATEDTOTHEPROPOSITIONTHATALLMENARECREATEDEQUAL&#xA; NOWWEAREENGAGEDINAGREATCIVILWARTESTINGWHETHERTHATNATIONORANYNATIONSOCONCEIVEDANDSODEDICATEDCANLONGENDUREWEAREMETONAGREATBATTLEFIELDOFTHATWARWEHAVECOMETODEDICATEAPORTIONOFTHATFIELDASAFINALRESTINGPLACEFORTHOSEWHOHEREGAVETHEIRLIVESTHATTHATNATIONMIGHTLIVEITISALTOGETHERFITTINGANDPROPERTHATWESHOULDDOTHIS&#xA;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;The solution is, presumably:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Place: Gettysburg, Pennsylvania&lt;br&gt;&#xA; Time: November 19, 1863&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;Bonus - how I found the solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All three paragraphs have length divisible by 3, suggesting that each 3-letter sequence represents a single unit.&lt;br&gt;&lt;br&gt;&#xA; Doing frequency analysis for all 3-letter sequences showed that the two most frequent sequences were UQB (20,16,1) and PLB (15,11,1) - both ended with B(1) and had a difference of 4 (E) between the first two letters, suggesting I was on to something.&lt;br&gt;&lt;br&gt;&#xA; The rest was just plugging the values generated by the above algorithm into the 3-letter sequences ending with B, getting much of the letters and easily figuring out what the rule for sequencing ending with A is.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4067" LastEditorUserId="4067" LastEditDate="2015-04-07T00:16:18.703" LastActivityDate="2015-04-07T00:16:18.703" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11653" PostTypeId="1" AcceptedAnswerId="11657" CreationDate="2015-04-07T00:04:13.600" Score="17" ViewCount="3522" Body="&lt;p&gt;There is submarine somewhere on the number line. Initially (say, at midnight), it is at position $A$, and is moving $B$ units per hour to the right, where $A$ and $B$ are both integers; both $A$ and $B$ are unknown to you. You have an unlimited number of torpedoes. Starting at midnight, and at the top of every hour, you can fire a torpedo somewhere on the number line, sinking the sub if it is there. What strategy can you use to guarantee you eventually sink the sub?&lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2018-09-20T22:03:41.963" Title="Sink the Submarine" Tags="|strategy|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="11655" PostTypeId="2" ParentId="11653" CreationDate="2015-04-07T01:35:38.170" Score="10" Body="&lt;p&gt;There are countably many possible submarine paths, since each is given by an integer start point and speed. Enumerate all of them. On hour $h$, sink submarine $h$ on the list by bombing its location at that time. Any possible submarine will eventually be be sunk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could give an explicit strategy, but I find that brute-force constructions like this better explain why a solution must exist than clever explicit ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same construction extends to submarines that move with as integer quadratics, or integer polynomials, or move with rational speeds, or move on lattice points on the plane, etc.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-04-07T01:52:27.750" LastActivityDate="2015-04-07T01:52:27.750" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11656" PostTypeId="1" AcceptedAnswerId="11659" CreationDate="2015-04-07T01:39:11.600" Score="1" ViewCount="776" Body="&lt;p&gt;There are pigeons sitting in a queue. The first one said, &quot;There are 4 pigeons behind me.&quot; The last one said, &quot;There are 4 pigeons in front of me.&quot; The middle one (or, one of middle ones) said, &quot;There are 4 pigeons in front of me and 4 pigeons behind me.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many Pigeons are there?&lt;/p&gt;&#xA;" OwnerUserId="11130" LastEditorUserId="5373" LastEditDate="2015-04-07T10:34:33.287" LastActivityDate="2015-04-30T20:05:47.197" Title="How many Pigeons are there?" Tags="|calculation-puzzle|lateral-thinking|" AnswerCount="6" CommentCount="8" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11657" PostTypeId="2" ParentId="11653" CreationDate="2015-04-07T02:17:04.177" Score="13" Body="&lt;p&gt;We are given the following information&#xA;&lt;br&gt;P(0) = A&#xA;&lt;br&gt;V(t) = B&#xA;&lt;br&gt;P(t) = A + Bt&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we know our current time, we can cycle through all the possibilities of P(t) by cycling through all the combinations of A and B at the current time and we will eventually land on the correct combination of A and B.&#xA;&lt;br&gt;NOTE: There are actually an infinite amount of combinations because the set of integers are unbounded. But since we know that A and B are finite numbers, this algorithm will eventually find the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a picture which shows a method for computing all the possible combinations. I'm sure there's probably an actual function out there that is able to compute the unique pair of values given an integer.&#xA;&lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/1dhLX.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;br&gt;NOTE: The image depicted here is not the exact same as the function used in the solution below (it's simply been rotated). To view the actual geometry of the function used, visit the link.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that we have to increment t by 1 each time because we are firing the torpedo with respect to the current time (1 hour intervals).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll try to come up with a better explanation and maybe an animation if I have the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; (Sketchy) Algorithm for enumerating 2 rational numbers:&lt;br&gt;&#xA;Since enumerating 2 rational numbers would essentially require enumerating 4 integers (i.e. integers a,b,c,d correspond to rational numbers a/b and c/d), this is what this algorithm essentially does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First add 0 (special case that only needs to be considered once)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then find all permutations where 1 is the biggest&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1 1 1 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then find all permutations where 2 is the biggest&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2 1 1 1&lt;br&gt;&#xA;  ...&lt;br&gt;&#xA;  2 2 2 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and so on...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for each line we need to enumerate all possible permutations of +/- (2^4 permutations).&lt;/p&gt;&#xA;" OwnerUserId="11094" LastEditorUserId="11094" LastEditDate="2017-05-11T22:45:45.523" LastActivityDate="2017-05-11T22:45:45.523" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11658" PostTypeId="2" ParentId="11656" CreationDate="2015-04-07T02:57:20.487" Score="5" Body="&lt;p&gt;I believe to know the answer to this one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The talkative pidgeons here are probably sitting on a circular queue, so the correct answer would be five.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11116" LastActivityDate="2015-04-07T02:57:20.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11659" PostTypeId="2" ParentId="11656" CreationDate="2015-04-07T03:45:14.697" Score="8" Body="&lt;p&gt;Cesar Martini has a point; we could explain this by sitting the pigeons in a circle, but that's not normally how queues work, and there would surely not be a &quot;first&quot; or a &quot;last&quot; in such a queue. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming a straight queue I'm not certain it's possible to know exactly how many pigeons there are, but we can infer the following: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Just because there are four pigeons behind or in front of another there's no reason not to think there might be more. I think we can therefore only be sure that there are at least 9 pigeons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11132" LastActivityDate="2015-04-07T03:45:14.697" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11660" PostTypeId="2" ParentId="11650" CreationDate="2015-04-07T05:00:17.930" Score="1" Body="&lt;p&gt;I think it is,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; cloud&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am faster than you can run&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; they run faster some times.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I live where the moon rise from above&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not Sure about this line&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am soft like a kitty and hard like a rock,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cloud are soft and hard like rock when it create electricity and sounds like rock .&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;without any warning we come in flock&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They have not sound and they come in flock (looks like flock)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4312" LastActivityDate="2015-04-07T05:00:17.930" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11661" PostTypeId="1" AcceptedAnswerId="11662" CreationDate="2015-04-07T05:14:40.363" Score="2" ViewCount="3001" Body="&lt;p&gt;Mr. Bean leaves from place $A$ with 3 sacks that each contains 30 watermelons.&#xA;He comes across 30 tolls till he reaches place $B$.&#xA;At every toll, Mr. Bean has to give 1 watermelon as toll for each non-empty sack with watermelons.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How many watermelons are left with Mr. Bean at place $B$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is not 0.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10820" LastEditorUserId="8874" LastEditDate="2016-02-06T14:56:39.327" LastActivityDate="2016-02-06T14:56:39.327" Title="How many watermelons left with Mr. Bean?" Tags="|calculation-puzzle|lateral-thinking|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11662" PostTypeId="2" ParentId="11661" CreationDate="2015-04-07T05:20:14.340" Score="7" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 25&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Calculation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; first 10 tolls 30 melons, next 15 tolls 30 more melons next 5 tolls 5 more&lt;BR&gt;&#xA; 90 - 30 - 30 - 5&lt;BR&gt;&#xA; First 10 tolls he has 3 sacks so 10 tolls x 3, now he is left with 2 sacks&lt;BR&gt;&#xA; Next 15 tolls, he as 2 sacks si 15 tolls x 2, and he is left with 1 sack with 30 melons and 5 tolls to cross&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4637" LastActivityDate="2015-04-07T05:20:14.340" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11664" PostTypeId="1" AcceptedAnswerId="11675" CreationDate="2015-04-07T06:19:46.620" Score="17" ViewCount="1049" Body="&lt;p&gt;In the darkness, I stay hidden, then with the cleansing sweep,&lt;br&gt;&#xA;I step out from the shadows, as I watch the whole world weep.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was torn apart, my body bent, forever segregated.&lt;br&gt;&#xA;Broken, I at first give life. Then as a weapon, take it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elusive, you won't trap me, nor the hoard kept at my feet,&lt;br&gt;&#xA;Sometimes haunted by my twin, though cursed to never meet.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Riding on your gallows, bathed in violence, I start.&lt;br&gt;&#xA;Say my name, mere mortal, before I again depart.  &lt;/p&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="5373" LastEditDate="2018-05-17T23:05:45.243" LastActivityDate="2018-05-17T23:05:45.243" Title="Bathed in violence... Say my name!" Tags="|riddle|word|rhyme|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11665" PostTypeId="2" ParentId="11664" CreationDate="2015-04-07T06:24:25.213" Score="1" Body="&lt;p&gt;I think you are referring to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Batman: He's the silent guardian, a watchful protector, The dark knight.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Explanation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In the darkness, I stay hidden, then with the cleansing sweep,&lt;br&gt;&#xA;I step out from the shadows, as I watch the whole world weep.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Batman is a symbol of clandestine in the darkness, the secret identity of Bruce Wayne. Mere mortals, except the fortunate few are ignorant of his identity. He keeps a hawk eye as a watchful protector over Gotham, rising from the dark whenever the city is on the verge of catastrophic destruction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was torn apart, my body bent, forever segregated.&lt;br&gt;&#xA;Broken, I at first give life. Then as a weapon, take it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This has a particular reference to the filmographic representation of the Batman story as The Dark Knight Rises. In that film, Selina delivers Batman to Bane, often called The Mask, who cripples him and throws him to a well like prison which was seemingly impossible to escape. The weapon referred I believe is the fusion reactor which was supposed to take over gotham.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Elusive, you won't trap me, nor the hoard kept at my feet,&lt;br&gt;&#xA;Sometimes haunted by my twin, though cursed to never meet. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Here &lt;code&gt;by my twin&lt;/code&gt;, I think it was a reference to Robin. Ohh, and did I say mere Mortals were unable to trap and track him?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Riding on your gallows, bathed in violence, I start.&lt;br&gt;&#xA;Say my name, mere mortal, before I again depart.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, it is the reference to the film The Dark Knight where he took up the responsibility of the death of Harvy Dent, the two face. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4177" LastEditorUserId="4177" LastEditDate="2015-04-07T06:41:15.533" LastActivityDate="2015-04-07T06:41:15.533" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11666" PostTypeId="1" AcceptedAnswerId="11671" CreationDate="2015-04-07T06:27:56.210" Score="17" ViewCount="1869" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;There is a 3 dial combination lock, where each dial features numbers 1 through 8 inclusive. However, it is poorly designed, and will open as long as &lt;em&gt;at least two&lt;/em&gt; dials are correct. How many combinations does one need to try to &lt;em&gt;guarantee&lt;/em&gt; opening the lock?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Please hide your answer, or at least the number in question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can certainly succeed in $64$ tries, by guessing until you get the first two dials correct. Also, each guess only works on $22 (= 1 + 3*7)$ combintations: the exact match and ones where any of the 3 dials is any of the other 7 numbers. So it will take at least $8^3/22\ge 24$ guesses. So the optimal number is somewhere between $24$ and $64$, but where?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: Winkler, &lt;em&gt;Mathematical Mind Benders&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="4269" LastEditDate="2015-09-12T17:08:11.210" LastActivityDate="2015-09-12T17:08:11.210" Title="Faulty Combination Lock" Tags="|mathematics|combinatorics|" AnswerCount="4" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11667" PostTypeId="2" ParentId="10540" CreationDate="2015-04-07T06:41:09.603" Score="3" Body="&lt;p&gt;Here is a solution of 137 moves utilizing the idea of getting a robot to the eighth column of the first row to create paths to the grey circle. Although this is unlikely to be optimal, it is a major improvement on PuzzleEater's 191 moves. The format is &quot;[robot] [sequence of moves]&quot;.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;r esen                         + r circle&#xA;y en&#xA;r e&#xA;y s&#xA;r sesw nwsw                    + r hexagon, - circle hexagon, + square&#xA;g wnws e                       + g circle&#xA;b sene nwn                     + b hexagon&#xA;g wnws wnwn ene                + g hexagon&#xA;r news enen eswn               + r triangle, - square triangle (finished)&#xA;b ws                           + b circle&#xA;y n                            &#xA;g nwnw enes enw                - g circle hexagon, + triangle square&#xA;y s&#xA;r w&#xA;g eswn                         - g triangle square (finished)&#xA;y wnws esen s                  + y circle triangle&#xA;r e&#xA;g e&#xA;b nwsn esws eses wsen e        - b hexagon circle, + b square triangle&#xA;g wses&#xA;b swne sws                     - b square triangle (finished)&#xA;y nwse nesw nesn wsne swne s   - y circle triangle, + square, - square, + hexagon&#xA;r e&#xA;b ne&#xA;y nes                          - y hexagon (finished)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="11113" LastActivityDate="2015-04-07T06:41:09.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11670" PostTypeId="2" ParentId="11594" CreationDate="2015-04-07T07:27:23.360" Score="8" Body="&lt;p&gt;&lt;strong&gt;Edit 4 - The final answer for Puzzles 4 and 5&lt;/strong&gt; (revised based on hints) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Puzzle 4 and 5&lt;/strong&gt; - Using the solution for Puzzle 3 (as found by finsternis) as well as efforts by rand al'thor:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;79 periodic number&lt;/strong&gt; - suggests using Gold from the periodic table or the Golden ratio of 1.6180339 or its inverse of 0.6180339&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;1459238 If you can't see the answer, you should use it!&lt;/strong&gt; - the hints suggest that this is the ciphertext and the Golden ratio could be the key.&lt;br&gt;&#xA;&lt;br&gt;&#xA;Therefore treating this as decoding suggests using mod-10 and subtracting&lt;br&gt;&#xA; &lt;strong&gt;6 1 8 0 3 3 9&lt;/strong&gt; from &lt;strong&gt;1 4 5 9 2 3 8&lt;/strong&gt; results in &lt;strong&gt;5 3 7 9 9 0 9&lt;/strong&gt; which is the answer for Puzzle 4.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Then looking at this upside down provides the answer to Puzzle 5 which is &lt;strong&gt;G O G G L E S&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Puzzle 1&lt;/strong&gt;&#xA;Using Tryth's suggestion of a train, I think the riddle describes a:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hackney_carriage&quot; rel=&quot;nofollow&quot;&gt;London taxi&lt;/a&gt;&lt;/strong&gt;  which has traditionally been black. Other names could be hackney carriage, hackney, hack, black cab, cab, black taxi, taxi, or remise.&lt;br&gt;&#xA;&lt;br&gt;&#xA;I assume the &lt;a href=&quot;http://www.mobilefish.com/services/steganography/steganography.php&quot; rel=&quot;nofollow&quot;&gt;Mobilefish&lt;/a&gt; password for the picture is an anagram of one of the names. However, these anagrams do not work:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-04-12T18:06:10.717" LastActivityDate="2015-04-12T18:06:10.717" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11671" PostTypeId="2" ParentId="11666" CreationDate="2015-04-07T07:43:42.850" Score="15" Body="&lt;p&gt;Here's a way to get &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 32 triples &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;that match any combination. I think this number is optimal but don't have a proof.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For four numbers (say 1,2,3,4), we can choose 16 triples so that any two indices uniquely determine the third. For example, take the triples that add to 0 modulo 4.&lt;br&gt;&#xA;&lt;br&gt;&#xA;    (1, 1, 2)&lt;br&gt;&#xA;    (1, 2, 1)&lt;br&gt;&#xA;    (1, 3, 4)&lt;br&gt;&#xA;    (1, 4, 3)&lt;br&gt;&#xA;    (2, 1, 1)&lt;br&gt;&#xA;    (2, 2, 4)&lt;br&gt;&#xA;    (2, 3, 3)&lt;br&gt;&#xA;    (2, 4, 2)&lt;br&gt;&#xA;    (3, 1, 4)&lt;br&gt;&#xA;    (3, 2, 3)&lt;br&gt;&#xA;    (3, 3, 2)&lt;br&gt;&#xA;    (3, 4, 1)&lt;br&gt;&#xA;    (4, 1, 3)&lt;br&gt;&#xA;    (4, 2, 2)&lt;br&gt;&#xA;    (4, 3, 1)&lt;br&gt;&#xA;    (4, 4, 4)&lt;br&gt;&#xA;&#xA;&lt;br&gt;&#xA; Now, do the same with 5,6,7,8, and combine the two sets to get 32 triples. Given any triple, it matches one of the 32 triples because:&lt;br&gt;&#xA;  - At least two of the numbers are at most 4, which matches a triple in the first set with the remaining number wrong, or&lt;br&gt;&#xA;  - At least two of the numbers are at least 5, which matches a triple in the second set with the remaining number wrong   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-04-07T08:00:25.310" LastActivityDate="2015-04-07T08:00:25.310" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11673" PostTypeId="1" AcceptedAnswerId="11683" CreationDate="2015-04-07T10:54:39.260" Score="8" ViewCount="607" Body="&lt;p&gt;You live on an infinite plane with a house at each integer lattice point. The plane is inhabited by horrible worms who can only walk in straight lines. You want to paint the houses so that all the worms constantly get lost, but you don't get lost yourself. You have an unlimited number of colors. Can you do it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it mathematically: can you paint each house a color so that&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For each straight line intersecting at least two houses, the sequence of house colors on that line is periodic --&lt;/li&gt;&#xA;&lt;li&gt;But the overall coloring is &lt;strong&gt;not&lt;/strong&gt; periodic (i.e. not equal to any translation of itself)?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Good luck!&lt;/p&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2017-03-11T14:41:40.100" Title="Periodic sequences confuse all my worms" Tags="|logical-deduction|geometry|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11674" PostTypeId="2" ParentId="11409" CreationDate="2015-04-07T11:18:42.777" Score="0" Body="&lt;p&gt;&lt;strong&gt;SECOND ATTEMPT&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I got it.. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;CION&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You call me a liar, and say I was the first.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A quick search reveals that CION means a descendant or offspring, esp. of an illustrious family. So its a first child.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I was the second, the first liar isn't cursed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As explained before, here &quot;I&quot; refers to actual word &quot;I&quot;. So that means it is second letter of the word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll open, open, and tell you what I see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The third letter is &quot;O&quot;, &quot;Open&quot; starts with O&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I start to fight, you'd think I start with a knee.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As everybody else related KNEE to &quot;N&quot;. There will be a better explanation, I am sure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whether you think I'm cool, or whether I be lame,&#xA;You'll often find me surrounded by many o' my dame.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first offspring is often surrounded with relatives. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A couple types of clues, riddle now you must,&#xA;And tell me what I am, before I turn to dust.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cannot connect this.. :D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10973" LastActivityDate="2015-04-07T11:18:42.777" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11675" PostTypeId="2" ParentId="11664" CreationDate="2015-04-07T11:20:37.033" Score="14" Body="&lt;p&gt;You are a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RAINBOW&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the darkness, I stay hidden, then with the cleansing sweep,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &quot;cleansing sweep&quot; refers to the rain. &quot;In the darkness&quot; means that it cannot be seen (unless under some certain conditions).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I step out from the shadows, as I watch the whole world weep.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The rainbow emerges from &lt;em&gt;nowhere&lt;/em&gt;. The &quot;world weeping&quot; again refers to the rain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was torn apart, my body bent, forever segregated.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Body bent&quot; refers to the shape of the rainbow.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Broken, I at first give life. Then as a weapon, take it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Broken&quot; refers perhaps to the division of the colours of the rainbow. &quot;At first give life&quot; refers to &lt;strong&gt;RAIN&lt;/strong&gt;, &quot;then as a weapon take it&quot; refers to &lt;strong&gt;BOW&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Elusive, you won't trap me, nor the hoard kept at my feet,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You cannot contain a rainbow. There is said to be a pot of gold at the end (or &quot;feet&quot;) of the rainbow.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes haunted by my twin, though cursed to never meet.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sometimes there is a second rainbow on a wider arc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Riding on your gallows, bathed in violence, I start.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ROYGBIV... &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Say my name, mere mortal, before I again depart.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I said already dammit! RAINBOW!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a personal comment: This is a world class riddle.&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2015-04-07T11:28:57.237" LastActivityDate="2015-04-07T11:28:57.237" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="11676" PostTypeId="2" ParentId="11621" CreationDate="2015-04-07T11:35:02.790" Score="3" Body="&lt;p&gt;I would ask:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Am I going to kill you today?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unknown says: NO. Then you kill him. (joker 100%)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unknown says: YES. Left him alive. (and tomorrow kill him without asking) joker!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A truth teller cannot respond. Because if you don't kill him, he is a joker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Senseless I know, but it would work!&lt;/p&gt;&#xA;" OwnerUserId="11141" LastEditorUserId="5373" LastEditDate="2015-04-07T11:41:04.390" LastActivityDate="2015-04-07T11:41:04.390" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11677" PostTypeId="1" AcceptedAnswerId="11693" CreationDate="2015-04-07T11:48:47.673" Score="5" ViewCount="285" Body="&lt;p&gt;Queen Shayra does not have the most impressive following, but she intends to expand her influence. If she could settle down at the center of an N Dimensional 3-Cube and build a palace there, she would attract all sorts of interesting folk to her realm. Of course, before she may build a palace, she needs to make sure that the N Dimensional 3-Cube is fit for colonization (no rats, etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this purpose she will send her Brave Knight on a journey over all 1-Cubes of a realm, except for the center, which is easy enough to deal with during the building process as it is only one cube. Brave Knight moves from cube to cube in chess knight jumps (two over in one direction, one in another), and he may not pass through a cube that he has been to already because Shayra is an optimization enthusiast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For which $N$ is there a $3\times\dots\times3$ cube with such a path? Whenever for some $N$ such a path exists, construct it.&lt;/p&gt;&#xA;" OwnerUserId="11100" LastEditorUserId="11100" LastEditDate="2015-04-07T18:17:02.457" LastActivityDate="2015-04-08T01:16:04.403" Title="Brave Knight's Journey in N Dimensions" Tags="|chess|combinatorics|construction|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="11678" PostTypeId="1" CreationDate="2015-04-07T11:56:56.087" Score="1" ViewCount="279" Body="&lt;p&gt;An invisible rabbit sits on one of the posts of an infinite fence. A local farmer cannot stand this fact and wants to shoot the rabbit down*. Every time the farmer shoots her gun, the rabbit jumps to another post along the fence (not quickly enough to avoid the shot, though). The jump is always in the same direction and the same distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can the farmer make sure she hits the rabbit with a finite number of shots?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;* Not to kill it, of course, just to get it off the fence. It's a water gun.&lt;/p&gt;&#xA;" OwnerUserId="10684" LastActivityDate="2015-04-07T13:43:05.470" Title="Shooting an invisible rabbit" Tags="|calculation-puzzle|strategy|" AnswerCount="0" CommentCount="23" ClosedDate="2015-04-07T12:22:24.303" ContentLicense="CC BY-SA 3.0" />
  <row Id="11679" PostTypeId="2" ParentId="11621" CreationDate="2015-04-07T12:05:24.697" Score="0" Body="&lt;p&gt;Not a single truth teller needs to die.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ask, &quot;Should I kill you?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, anybody who is suicidal doesn't need this prompt and would have already committed suicide. Therefore, no truth teller will answer, &quot;Yes&quot;, that they wish to die, but some jokers will. Therefore, by attrition, by repeatedly asking the question, only jokers will eventually all be killed, while not a single truth teller will be harmed.&lt;/p&gt;&#xA;" OwnerUserId="10795" LastEditorUserId="10795" LastEditDate="2015-04-07T13:31:43.887" LastActivityDate="2015-04-07T13:31:43.887" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11683" PostTypeId="2" ParentId="11673" CreationDate="2015-04-07T13:21:35.610" Score="7" Body="&lt;p&gt;I believe one such house painting scheme is the Chair Tiling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LZwOw.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LZwOw.gif&quot; alt=&quot;Gotta love them tilings&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture gives a good idea of how to construct a chair tiling of arbitrary size (rotated 45 degrees).&lt;br&gt;&#xA;One square on the diagram represents the color of a house at a specific lattice point. The chair tiling is &lt;a href=&quot;http://tilings.math.uni-bielefeld.de/glossary/nonperiodic&quot; rel=&quot;nofollow noreferrer&quot;&gt;nonperiodic&lt;/a&gt; - that is, it is not equal to any translation of itself. This way of painting the houses in the lattice has the property that any infinite line that intersects with at least 2 of the houses will have a periodic pattern. Why? The chair tiling is &lt;a href=&quot;http://tilings.math.uni-bielefeld.de/glossary/limitperiodic&quot; rel=&quot;nofollow noreferrer&quot;&gt;limitperiodic&lt;/a&gt; - That is, it is constructed from a combination of countably many finite tile sets. This is easier to see on a larger diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZvYy7.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the same tiling. You can begin to see patterns - subsequences that seem to repeat. However, the overall tiling is still nonperiodic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://tilings.math.uni-bielefeld.de/substitution_rules/chair&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tilings Encyclopedia&lt;/a&gt; states that &quot;the chair tiling is the union of a countable set of fully periodic tile sets $L_1$, $L_2$, $L_3$..., where each $L_i$ possesses period vectors of length $2 × 2^i$&quot;. Hence, any line drawn will have the colors of the houses repeat after it has intersected $2^i$ houses, for some $i$ (obviously, $2$ is divisible by $2^i$ for all $i\ge1$).&lt;/p&gt;&#xA;" OwnerUserId="7308" LastEditorUserId="5373" LastEditDate="2017-03-11T14:41:40.100" LastActivityDate="2017-03-11T14:41:40.100" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11684" PostTypeId="2" ParentId="11409" CreationDate="2015-04-07T13:54:33.047" Score="4" Body="&lt;p&gt;Third attempt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;LION&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You call me a liar, and say I was the first.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First letter of the word &quot;liar&quot; is L. As everybody suggested.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I was the second, the first liar isn't cursed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I&quot; refers to actual word &quot;I&quot;. So that means it is second letter of the word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll open, open, and tell you what I see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The third letter is &quot;O&quot;, &quot;Open&quot; starts with O&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I start to fight, you'd think I start with a knee.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, as everybody else related KNEE to &quot;N&quot;. There will be a better explanation, I am sure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whether you think I'm cool, or whether I be lame,&#xA;You'll often find me surrounded by many o' my dame.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lion is surrounded by many Lioness and &quot;Dame&quot; is the female equivalent of the honour of knighthood, which means Lioness here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A couple types of clues, riddle now you must,&#xA;And tell me what I am, before I turn to dust.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; May be it refers to the color of Lion which resembles to color of dust, again explanation can be better.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10973" LastActivityDate="2015-04-07T13:54:33.047" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11687" PostTypeId="2" ParentId="11650" CreationDate="2015-04-07T15:43:37.847" Score="8" Body="&lt;p&gt;I think the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Baseball Bat/Bats&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am faster than you can run,&lt;br/&gt;&#xA;so abandon me, go home and have fun.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;From &lt;strong&gt;Chronos&lt;/strong&gt;: &lt;em&gt;The second hint given is &quot;home run&quot; which would be a reference to Baseball&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I live where the moon rise from above&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Bats are hanging upside down&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;so abandon me, get rid of the hands in glove.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; From &lt;strong&gt;Chronos&lt;/strong&gt;: &lt;em&gt;hands in glove would mean the gloves you use in baseball&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am soft like a kitty and hard like a rock,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Bats have fur so you can call them soft as a kitty and hard as a rock because... &quot;&lt;em&gt;baseball bats are hard&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;without any warning we come in flock.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Bats come in flock, that's true&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Who am I that you might ask?&lt;br/&gt;&#xA;a symbol of fear, hidden in a mask!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Bats are scary and &quot;&lt;em&gt;a mask&lt;/em&gt;&quot; is a reference to Batman :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Half of the credits are going to &lt;strong&gt;Chronos&lt;/strong&gt; from the commet below.&lt;/p&gt;&#xA;" OwnerUserId="11145" LastEditorUserId="11145" LastEditDate="2015-04-08T16:55:00.490" LastActivityDate="2015-04-08T16:55:00.490" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11688" PostTypeId="2" ParentId="11677" CreationDate="2015-04-07T15:55:36.220" Score="0" Body="&lt;p&gt;Color the center cube (i.e. (1, 1, 1, 1, ..., 1)) black.  Color the two cubes (0, 1, 1, 1, ..., 1) and (2, 1, 1, 1, ..., 1) adjacent to the center red.   Finally, color the six cubes (0, 0, 1, 1, 1, ..., 1), (0, 2, 1, 1, 1, ..., 1), (1, 0, 1, 1, 1, ..., 1), (1, 2, 1, 1, 1, ..., 1), (2, 0, 1, 1, 1, ..., 1), (2, 2, 1, 1, 1, ..., 1) green. Notice each green is adjacent to a red or black.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will show that we can jump through 3-cube starting on one red and ending at the other red for $N \geq 2$.  Proof will proceed by induction, with the base case $N = 2$ being pretty straighforward.  For the inductive case, start on the red (0, 1, 1, 1, ..., 1).   Now jump to the green (2, 0, 1, 1, 1, ..., 1).   You are now on the N-1 dimensional 3-cube with coordinates (2, *, *, *, .., *), where * means that coordinate can be 0, 1, or 2.  In this subcube, the greens play the same role as the reds, and the red plays the role of the center.  Hence, by induction, we can now jump all the way through this subcube avoiding only the red, starting on a green and ending on a green. The green we end on has coordinates, (2, 2, 1, 1, 1, ..., 1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now jump to the subcube (1, *, *, *, ..., *).  In particular, we jump from (2, 2, 1, 1, 1, ..., 1) to (1, 0, 1, 1, 1, ..., 1).  By induction, we can now jump through this subcube and end on the green (1, 2, 1, 1, 1, ..., 1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we jump to the subcube (0, *, *, *, ..., *).   We go from (1, 2, 1, 1, 1, ..., 1) to the green (0, 0, 1, 1, 1, ..., 1).   By induction we can eventually, end on green (0, 2, 1, 1, 1, ..., 1).   We can then jump to the red (2, 1, 1, 1, ..., 1), which completes the path. &lt;/p&gt;&#xA;" OwnerUserId="2722" LastActivityDate="2015-04-07T15:55:36.220" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11689" PostTypeId="2" ParentId="11650" CreationDate="2015-04-07T16:06:17.867" Score="1" Body="&lt;p&gt;I guess the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;SNOW&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am faster than you can run,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It falls from sky in pace and it is difficult to run in snow.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so abandon me, go home and have fun.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We usually stay at home when it is snowing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I live where the moon rise from above,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Snow is formed in clouds, where the moon rises.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so abandon me, get rid of the hands in glove.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because it's cold we wear gloves.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am soft like a kitty and hard like a rock,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is soft(snowflakes) or hard (Hail, Ice Pellets or Snow grains) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;without any warning we come in flock.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It starts suddenly and is often heavy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who am I that you might ask?&#xA;a symbol of fear, hidden in a mask!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cant make sense of this... :(&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;EDIT&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;RAIN&lt;/strong&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;for same explanation&lt;/p&gt;&#xA;" OwnerUserId="10973" LastEditorUserId="10973" LastEditDate="2015-04-07T16:25:24.713" LastActivityDate="2015-04-07T16:25:24.713" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11690" PostTypeId="2" ParentId="11650" CreationDate="2015-04-07T16:36:59.317" Score="2" Body="&lt;p&gt;I think it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Wolves&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am faster than you can run,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;They are fast&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so abandon me, go home and have fun.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;In Chinese, a &quot;wolf&quot;色狼 is considered sexually perverted. So someone who is a 色狼 go home and have fun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I live where the moon rise from above,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;They howl at the moon and are pretty active at night.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so abandon me, get rid of the hands in glove.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Not too sure about this one... maybe wolfskin gloves?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am soft like a kitty and hard like a rock,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;They are furry, but strong and dangerous&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;without any warning we come in flock.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;They stalk their preys and hunt in packs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who am I that you might ask?&#xA;a symbol of fear, hidden in a mask!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;They are a symbol of fear in multiple cultures. &lt;br&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Wolves_in_folklore,_religion_and_mythology&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Wolves_in_folklore,_religion_and_mythology&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1504" LastActivityDate="2015-04-07T16:36:59.317" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11691" PostTypeId="1" AcceptedAnswerId="11692" CreationDate="2015-04-07T17:03:37.333" Score="3" ViewCount="447" Body="&lt;p&gt;The last time you were hot,&lt;br&gt;&#xA;I cooled you down,&lt;br&gt;&#xA;And you praised me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But yesterday you said&lt;br&gt;&#xA;I looked &quot;big and fat,&quot;&lt;br&gt;&#xA;And I was sad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet today I felt better&lt;br&gt;&#xA;when you told all your friends&lt;br&gt;&#xA;that I looked like an angel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we touch,&lt;br&gt;&#xA;I admit I feel like&lt;br&gt;&#xA;I'm just annoying you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is out-of-sight really&lt;br&gt;&#xA;out-of-mind with you? Or do&lt;br&gt;&#xA;you miss me when I'm gone?&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-04-07T17:10:52.493" LastActivityDate="2015-04-07T17:12:54.213" Title="Do you even know who I am anymore? What am I?" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11692" PostTypeId="2" ParentId="11691" CreationDate="2015-04-07T17:12:54.213" Score="5" Body="&lt;p&gt;My answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A cloud.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The last time you were hot,&lt;br&gt;&#xA;I cooled you down,&lt;br&gt;&#xA;And you praised me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clouds cool things down.. with rain and such.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But yesterday you said&lt;br&gt;&#xA;I looked &quot;big and fat,&quot;&lt;br&gt;&#xA;And I was sad.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clouds can be big and fluffy.. the person in question must have taken notice and remarked that the clouds were remarkably big and fat that day?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet today I felt better&lt;br&gt;&#xA;when you told all your friends&lt;br&gt;&#xA;that I looked like an angel.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Angel, heavens, clouds. Probably a cloud formation that reminded the person in question of an angel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When we touch,&lt;br&gt;&#xA;I admit I feel like&lt;br&gt;&#xA;I'm just annoying you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fog is annoying.. reduced visibility and such.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is out-of-sight really&lt;br&gt;&#xA;out-of-mind with you? Or do&lt;br&gt;&#xA;you miss me when I'm gone?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sometimes there are no clouds in the sky, and usually people are pretty happy about that.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11100" LastActivityDate="2015-04-07T17:12:54.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11693" PostTypeId="2" ParentId="11677" CreationDate="2015-04-07T17:14:36.857" Score="7" Body="&lt;p&gt;The knight cannot succeed when $N$ is odd. Color the squares in checkerboard fashion, with a corner black. When $N$ is odd, the removed square is white, so there will be two more black squares altogether. Since tours must alternate colors, none can exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We prove by induction that, for even $N$, an $N$-palace (i.e. an $N$-cube with its center removed) can be cycled (toured, with start and end joined). This clearly holds when $N=2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given an $N$ palace, its cells can be described with coordinates $(x_1,\dots,x_n)$, where $x_i$ are all either $0,1$ or $2$. Take any cell where $x_1=x_2=1$, and color it red. The red cells themselves form an $(N-2)$-palace. Furthermore, surrounding each red cell are 8 cells with coordinates&#xA;$$&#xA;(*,*,x_3,\dots,x_{n})&#xA;$$&#xA;where the $*$’s can be anything (except both $1$, since that would be the central red cell), while the $x_i$ are fixed. These $8$ cells form a $2$-palace: I’ll refer to these as the “ring” around the red cell $(1,1,x_3,\dots,x_n)$. Color such squares black. Almost every cell is now colored. The remaining cells, the ones whose last $N-2$ coordinates are $1$, form a ring around the removed central cell: color these blue. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the current situation, illustrated when $N=4$. The coordinates $(x,y,w,z)$ mean the following: $w,z$ tell you which slanted square you are in, and $x,y$ tell you your location in that square. $w$ is horizontal, $z$ vertical. $x$ is back/front, $y$ is vertical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kJYjj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We first show how to cycle each color: then we combined these 3 disjoint cycles into one big one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The blue cells themselsves can be toured (this is just the $N=2$ case). By induction, the red cells can be cyclically toured. The black cells can be toured as well: do this, hop from ring to ring, following the tour of the red cells. When you arrive at each ring, perform a tour of it. There are two ways to cycle around a ring: alternate these each time. This will ensure when you return to the ring around your first red cell, you will be in the correct “phase” (since you switch phase each time, and there an even number of red cells). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we have partitioned the $N$-palace into three disjoint cycles, and we must combine these into one cycle. Any two disjoint cycles can be combined as long as there are consecutive points on one which are joined to two consecutive points on the other. Here is, for example, how the black cycle can be joined to the red. The $\dots$'s mean the rest of the coordinates are all 1.&#xA;$$&#xA;\begin{array}{ccc}&#xA;(0,1,1,0,\dots)&#xA;&amp;amp;\to \stackrel{\text{black tour}}{\cdots}\to&#xA;&amp;amp;(0,1,0,2,\dots)\\&#xA;\uparrow &amp;amp;&amp;amp; \downarrow\\&#xA;\color{red}{(1,1,1,2,\dots)}&amp;amp;\color{red}{\leftarrow \stackrel{\text{red tour}}{\cdots}\leftarrow} &amp;amp;\color{red}{(1,1,0,0,\dots)}&#xA;\end{array}&#xA;$$&#xA;The blue cycle can be joined to the now black/red cycle via&#xA;$$&#xA;\begin{array}{ccc}&#xA;(0,2,2,1,\dots)&#xA;&amp;amp;\to \stackrel{\text{black tour}}{\cdots}\to&#xA;&amp;amp;(1,0,2,1,\dots)\\&#xA;\uparrow &amp;amp;&amp;amp; \downarrow\\&#xA;\color{blue}{(0,0,1,1,\dots)}&amp;amp;\color{blue}{\leftarrow \stackrel{\text{blue tour}}{\cdots}\leftarrow} &amp;amp;\color{blue}{(1,2,1,1,\dots)}&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-04-08T01:16:04.403" LastActivityDate="2015-04-08T01:16:04.403" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="11694" PostTypeId="2" ParentId="11586" CreationDate="2015-04-07T17:35:44.137" Score="1" Body="&lt;p&gt;I am sharing some of my thought and I will add an answer when I figured it out.&lt;br&gt;&#xA;Here's my interpretation of some of the hints:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First paragraph&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This suggest that Warren's shop might locate somewhere desert-like, or cold, judging by the visitor's clothing.  Also it is obvious that the visitor remember his/her reason to be here while Warren did not.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second paragraph  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Warren works around the clock.  His children love him as much as he loves them.  His children are all adult.  His wife was never mentioned.  This might suggest Warren is living alone / is a single father (divorced / wife is dead)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Third paragraph  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Warren was busy even on weekends, why is it not busy on this day?  Also, playing darts might remind him of something all of a sudden.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fourth paragraph  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The visitor displayed a visible grin + was wearing a white mask – some suggest it is Guy Fawkes mask, or it could just be a mask that does not cover the mouth.  Also Warren stays in his shop until 11:30pm at night, which is unusual?  However there's no indication on the time the visitor came.  Also Warren had not figure out who was it at the end.  Chances are it's someone he doesn't know, or there's no reason to hide himself/herself from Warren.  What makes Warren happy was the gift, not who was the visitor.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="10037" LastEditDate="2015-04-07T19:44:37.413" LastActivityDate="2015-04-07T19:44:37.413" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11695" PostTypeId="2" ParentId="11621" CreationDate="2015-04-07T17:48:28.307" Score="1" Body="&lt;p&gt;Ask &quot;Are you a joker that is telling the truth?&quot;&#xA;The jokers have to answer &quot;yes,&quot; regardless of whether they are lying or telling the truth; the truth tellers have to answer &quot;no.&quot;&lt;/p&gt;&#xA;" OwnerUserId="11150" LastActivityDate="2015-04-07T17:48:28.307" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11696" PostTypeId="5" CreationDate="2015-04-07T18:21:00.330" Score="0" Body="&lt;p&gt;Puzzles that require the solver to create or recreate a certain position/image/object based on the given criteria fall under the &lt;a href=&quot;/questions/tagged/construction&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;construction&amp;#39;&quot; rel=&quot;tag&quot;&gt;construction&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples include, but are not limited to:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Game recreations (chess, checkers, go, ...)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Puzzle constructions (mazes, crosswords, ...)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Geometric designs&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="52483" LastEditorUserId="52483" LastEditDate="2018-11-15T21:09:47.540" LastActivityDate="2018-11-15T21:09:47.540" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="11697" PostTypeId="4" CreationDate="2015-04-07T18:21:00.330" Score="0" Body="A puzzle that requires an example to be built that fits certain criteria." OwnerUserId="11100" LastEditorUserId="11100" LastEditDate="2015-04-07T18:42:22.193" LastActivityDate="2015-04-07T18:42:22.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11698" PostTypeId="2" ParentId="11611" CreationDate="2015-04-07T19:05:25.770" Score="2" Body="&lt;p&gt;I think this is the same as Kevin's solution, and I followed similar steps, except I've been doing too much Kenken recently and initially forgot that the cages (sum groups) are also forbidden repeated digits, so I didn't fill the &quot;9&quot; in the top-left 22-cage immediately. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/K6F1Y.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I didn't find it a particularly hard &quot;killer&quot; puzzle.&lt;/p&gt;&#xA;" OwnerUserId="8875" LastActivityDate="2015-04-07T19:05:25.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11699" PostTypeId="1" AcceptedAnswerId="11703" CreationDate="2015-04-07T19:25:50.910" Score="3" ViewCount="1880" Body="&lt;p&gt;This is quite a famous one, I think, but hopefully not too famous. The phrasing is loosely taken from &lt;a href=&quot;http://tierneylab.blogs.nytimes.com/2009/04/13/jimmy-carters-killer-rabbit-puzzle/?_r=1&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A duck finds itself alone in the centre of a round pond. On the edge of the pond stands a hungry wolf. If the duck reaches any point along the edge before the wolf does, it can fly away immediately. But if the wolf gets there first, it's dinner time. At any given moment, the wolf runs along the edge towards the point that's closest to the duck at that particular moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 1:&lt;/strong&gt; If the wolf runs four times as fast as the duck swims, can the duck escape? If so, how?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 2:&lt;/strong&gt; What’s the fastest the wolf can run such that the duck can still escape?&lt;/p&gt;&#xA;" OwnerUserId="10684" LastEditorUserId="5373" LastEditDate="2015-04-26T11:36:06.323" LastActivityDate="2015-04-26T11:36:06.323" Title="The duck and the wolf" Tags="|geometry|strategy|" AnswerCount="3" CommentCount="2" ClosedDate="2015-04-07T20:37:01.227" ContentLicense="CC BY-SA 3.0" />
  <row Id="11700" PostTypeId="2" ParentId="11699" CreationDate="2015-04-07T19:34:37.933" Score="1" Body="&lt;p&gt;Question 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Yes it can!&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As long as it can, the duck should keep the wold opposite to the center. That's possible only when the angular velocity $\omega$ of the duck is greater than the angular velocity of the wolf.&#xA;Let's call &lt;strong&gt;R&lt;/strong&gt; the radius of the pond and &lt;strong&gt;r&lt;/strong&gt; the distance of the duck from the center. That's possible until:&lt;br&gt;&#xA;$\omega(duck)=\omega(wolf)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{4V}{\pi\times R}=\frac{V}{\pi\times r}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r=\frac V {4V} R=\frac 1 4 R$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the duck can keep the wolf opposite until $r=\frac 1 4 R$&lt;br&gt;&#xA;After this, the duck goes straight to the pond edge, which is distant $\frac 3 4 R$.&lt;br&gt;&#xA;Now it's all a matter of time $t$, the duck must cover that distance before the wolf can cover half circle! &#xA;Is it possible? Let's verify it!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$t(duck)&amp;lt;t(wolf)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{0.75 R}{V}&amp;lt;\frac{\pi\times R}{4V}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3&amp;lt;\pi$&lt;br&gt;&#xA;True, so the duck can escape!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&amp;lt;&gt;O&amp;lt;&gt;O&amp;lt;&gt;O&amp;lt;&gt;O&amp;lt;&gt;O&amp;lt;&gt;O&amp;lt;&gt;O&amp;lt;&gt;O&amp;lt;&gt;O&amp;lt;&gt;O&amp;lt;&gt;O&amp;lt;&gt;O&amp;lt;&gt;O&amp;lt;&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;$\pi+1$&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Generalizing the above answer, we get the formula:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{R-r}{v}&amp;lt;\frac{\pi R}{V}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We already had that  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r=\frac v {V} R$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so, after some basic math, we get that $V&amp;lt;v(1+\pi)$ to allow the duck escape, which means $V&amp;gt;v(1+\pi)$ to let the wolf win.&lt;br&gt;&#xA;That is $V&amp;gt;4.14v$&lt;br&gt;&#xA;If the duck's speed $v$ is still $1$, the the wolf's speed must be at least $4.1415$...&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-04-07T20:10:37.257" LastActivityDate="2015-04-07T20:10:37.257" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11702" PostTypeId="1" AcceptedAnswerId="11706" CreationDate="2015-04-07T19:52:40.760" Score="8" ViewCount="2627" Body="&lt;p&gt;I went walking with a friend of mine in Las Vegas last week. We turned a corner and saw a street magician doing some tricks for a crowd of 12-15 people. In most of his tricks, the magician would make a little money by getting the crowd to bet on something he had set up, such as which cup a ball was hidden under.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On one of the tricks, someone in the crowd won a quick $50. Anyway, after 10-15 minutes of watching him, we volunteered for the next trick. He handed each of us a pen, a small piece of paper, and a calculator. As he turned his back to us so that he couldn't see, we were each told to secretly, legibly write down any 5-digit number, so we did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He told me to&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Add together all the digits in my number&lt;/li&gt;&#xA;  &lt;li&gt;Subtract that sum from the original number to get the final number&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and he told my friend to&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Rearrange the digits in her number to make another number&lt;/li&gt;&#xA;  &lt;li&gt;Subtract the smaller number from the larger number to get the final number&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then, he told us to add our two final numbers together, write the sum large and legibly on a piece of paper, and circle it. Lastly, we were told to choose exactly one non-zero digit in that final answer and underline it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We did that, and he then said he would ask us to read off to him, in any order, all of the digits except the one that we underlined, and then he would tell us which digit we had underlined. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A guy behind us bet him $20 that he would not be able to guess the number in a single guess. He took the bet, and asked if there were any more bettors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should we have made a bet? Is there a way for the magician to know the digit for sure?&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-04-08T02:25:10.927" LastActivityDate="2015-04-22T01:55:54.127" Title="Vegas Street Magician Math Trick" Tags="|mathematics|calculation-puzzle|number-theory|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11703" PostTypeId="2" ParentId="11699" CreationDate="2015-04-07T19:53:35.087" Score="3" Body="&lt;p&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes, by being just over $\frac{4 - \pi}{4}$ from the center (approx. 21%)&lt;br&gt;&#xA; At that distance, if the wolf is opposite of the duck, then the duck can race to the opposite shore and make it. Since that position is less than $\frac{1}{4}$ of the way to the edge, the duck can rotate the pond faster than the wolf, eventually getting the duck and wolf opposite the center. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\pi + 1$&lt;br&gt;&#xA; The wolf can reach the opposite shore in $\frac{\pi r}{X}$ time, so the duck has to be at least that close to the shore.&lt;br&gt;&#xA; The farthest the duck can get from the center is $\frac{r}{X}$.&lt;br&gt;&#xA; So, $r-\frac{r}{X} = \frac{\pi r}{X}$, $X = \pi + 1$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-04-07T19:53:35.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11705" PostTypeId="2" ParentId="11702" CreationDate="2015-04-07T20:11:47.593" Score="5" Body="&lt;p&gt;You should not have made a bet; the magician is guaranteed to know your number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No matter the initial numbers, both you and your friend will end with numbers evenly divisible by 9. Since they're both divisible by 9, so will their sum be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a number is divisible by 9, the sum of its digits is also divisible by 9. The magician can sum the digits you tell him, then he knows the remaining digit is the one that will add to that sum to produce a multiple of 9.&lt;/p&gt;&#xA;" OwnerUserId="8891" LastActivityDate="2015-04-07T20:11:47.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11706" PostTypeId="2" ParentId="11702" CreationDate="2015-04-07T20:13:30.917" Score="9" Body="&lt;p&gt;He shouldn't have made the bet. We need one simple lemma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma 1&lt;/strong&gt;  - The remainder of a number when divided by 9 is equal to the remainder of the sum of its digits when divided by 9.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let our number be $n = d_k + 10d_{k-1} + \dots + 10^{k-1}d_1$ where $d_i$ is the $i^{\mathrm{th}}$ digit from the left. Since we are interested in the remainder on division by 9, we can discard the parts that are surely divisible by 9. So we write, $$n = (d_k + \dots + d_1) + 9(d_{k-1} + 11d_{k-2} + \dots + 1\dots1d_1)$$ and thus we can discard the second parenthesis and we are left with the sum of $n$'s digits, proving the lemma.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a direct corollary, any number having the exact same digits as $n$ and thus the same sum of digits will have the same remainder upon division by 9, and as the remainder of the difference between two numbers is the difference between their remainders, your friend ended up with a number that is divisible by 9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, when you added the digits of your number you got a new number that by the lemma has the same remainder upon division by 9 as your original number, and thus the difference is divisible by 9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By adding both of the results, we get a final number which is also divisible by 9. Now upon removing a digit, we subtract that digit from the sum of digits, and so the new remainder will be $9-d$. By calculating the remainder (adding the digits we read to him), the magician can easily find $9-d$ and thus $d$.&lt;/p&gt;&#xA;" OwnerUserId="11100" LastEditorUserId="11100" LastEditDate="2015-04-07T21:54:23.220" LastActivityDate="2015-04-07T21:54:23.220" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11707" PostTypeId="2" ParentId="11699" CreationDate="2015-04-07T20:17:33.667" Score="3" Body="&lt;p&gt;Simple. &lt;strong&gt;&lt;em&gt;The duck can always escape, no matter how fast the wolf runs&lt;/em&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's a 'puddle duck' then it can do a vertical-takeoff from the middle of the pond, without going near the edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's not a puddle duck then it must be a 'diving duck', which means it has a food source handy (at the bottom of the pond), so it can just wait for the wolf to get hungry and leave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(As the duck is on a pond, that seems to exclude the possibility of it being a &lt;a href=&quot;http://beautyofbirds.com/seaducks.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;seaduck&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Puddle Duck (aka 'Dabbling Duck')&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.flickr.com/photos/solvsuper/5540591444&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TnkHS.jpg&quot; alt=&quot;Puddle ducks - vertical takeoff&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Another distinguishing characteristic of puddle ducks when compared to diving ducks is the way in which they take flight when spooked or are on the move. &lt;strong&gt;&lt;em&gt;Puddle ducks spring straight up from the water&lt;/em&gt;&lt;/strong&gt;, but diving ducks need to gain momentum to take off, so they must run across the water a short distance to gain flight.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Anatinae&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia: Anatinae&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The silhouette of a puddle duck sits higher on the water than&#xA;  diving ducks and &lt;strong&gt;&lt;em&gt;when they want to take off, they lift directly upward&lt;/em&gt;&lt;/strong&gt;. Their diving cousins need to run along the water before gaining altitude.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.yorkccd.org/wordpress/wp-content/uploads/2012/12/PA-Wild-Journal-Puddle-Ducks-128-129-Mallard-128-129-Wood-Ducks-129-130.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Puddle Ducks;&#xA;Order Anseriformes;&#xA;Family Anatinae;&#xA;Subfamily Anatini&lt;/a&gt;, York County Conservation District&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dabbling ducks in comparison to diving ducks have larger wing areas relative to their body weight. Their wings are longer, rounder, and have really well developed wingtip slotting. All of this improves their ability to land and take off in tight spaces. &lt;strong&gt;&lt;em&gt;Taking off vertically out of the water&lt;/em&gt;&lt;/strong&gt; requires the duck to push with their wings and their feet simultaneously. Usually facing the wind, the dabbler will create airflow past the wings by flapping backward up and forward down, angling the primaries (largest flight feathers) to produce the maximum thrust. Lift occurs when airflow on the wing has lower pressure just above the wing, and higher pressure below. Divers wings are more streamlined, which helps with swimming and diving, but requires a build up of speed in order to take off.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.deltawaterfowl.org/what-we-do/biologist.html/title/what-are-the-mechanics-biomechanics-which-allow-puddle-ducks-to-perform-a-vertical-take-off-they-seem-to-just-elevate-straight-up-with-minimal-forward-motion-how-do-they-accomplish-this-&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ask a Biologist: What are the mechanics/biomechanics which allow puddle ducks to perform a vertical take-off. They seem to just elevate straight up with minimal forward motion. How do they accomplish this?&lt;/a&gt;, Delta Waterfowl&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Diving Duck (aka 'Pochard' or 'Scaup')&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Diving_duck&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/L97iB.jpg&quot; alt=&quot;Diving duck, feeding underwater&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's a diving duck, it just stays in the middle of the pond, feeding from the bottom of the pond as necessary, until the wolf gets hungry and goes away to find some other dinner. The duck can outlast the wolf because the duck has a convenient food source right there to hand - the wolf has to go away to find something (other than the duck) to eat.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;&lt;em&gt;Diving ducks dive and feed at the bottom&lt;/em&gt;&lt;/strong&gt;, often many feet below the waters surface.... Since &lt;strong&gt;&lt;em&gt;some diving ducks do not like to leave the water&lt;/em&gt;&lt;/strong&gt; it is necessary to place food dishes at the waters edge or place food on the water itself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.waterfowl.org.uk/diving.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;British Waterfowl Association: Diving Ducks&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="4050" LastEditDate="2015-04-10T21:16:34.923" LastActivityDate="2015-04-10T21:16:34.923" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11708" PostTypeId="1" CreationDate="2015-04-07T21:12:12.820" Score="3" ViewCount="709" Body="&lt;p&gt;My friend gave a question to me. I am not sure if it is an original puzzle, but he has 2 lines of Christmas lights and 3 circular timers with 96 switches apiece. Each switch turns on power for 15 minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.harborfreight.com/media/catalog/product/cache/1/image/9df78eab33525d08d6e5fb8d27136e95/i/m/image_18078.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lIyzh.jpg&quot;&gt;&lt;/a&gt;&#xA;&amp;ensp;&#xA;&lt;a href=&quot;http://inspirationseek.com/christmas-lights-clipart-images/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ewNx2.jpg&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How can you arrange the timers, switches and lights &amp;mdash; as far in advance as possible &amp;mdash; so that lights would be on for 24 hours on Christmas day?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;All lights are off for the last 15 minutes before Christmas.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For the next 24 hours, some lights are on at each point in time.&#xA;(Some may also be off.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;All lights are off again for the 15 minutes after that.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;No constraints outside of the 24 1/2 hours just described, except:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Any 24-hour interval before Christmas&#xA;has all lights off for at least 15 minutes.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;One timer may plug into another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The timers operate as in this &lt;a href=&quot;https://www.youtube.com/watch?v=x9RhYu1JkB0&quot; rel=&quot;nofollow noreferrer&quot;&gt;video&lt;/a&gt;. I am sure a lot of you have seen them in real life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&#xA;Note: This gem of a puzzle was given a make-over&#xA;      after a year and a half of being closed.&#xA;      The original poster might not be around to officially accept a solution.&#xA;      (Attempts were made to reach the poster half a year ago.)&#xA;&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="11151" LastEditorUserId="19114" LastEditDate="2021-05-16T23:21:07.773" LastActivityDate="2024-04-01T11:42:11.407" Title="Connecting 3 circular power timers to lights so that they turn for one day on Christmas" Tags="|mathematics|optimization|seasonal|time|" AnswerCount="2" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="11709" PostTypeId="2" ParentId="11586" CreationDate="2015-04-07T22:15:46.583" Score="1" Body="&lt;p&gt;My guess are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1) Gifts for Children or 2) Keys to an antique car or 3) extremely rare spare part for a vehicle or 4) Corrective Lenses&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My Interpretation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Warren Brown received a very peculiar visitor two weeks ago in his car-repair shop, the visitor seemed to be a male by judging from his voice, he was wearing a black jacket, a white mask, leather gloves and brown boots. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Could be a customer or an old friend. &#xA; Costume can be a red herring or implies winter/rainy weather.&#xA;** Warren could be color-blind and need corrective lenses to not recognize Santa Claus.**&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The visitor inquired &quot;Do you remember?&quot; in a very stirred tone - Warren was greatly confused and replied, &quot;Remember what?&quot;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Warren could have a split personality syndrome to not recollect something he is supposed to recollect.&#xA; Also &quot;Do you remember tomorrow it is Christmas? Did you buy gifts for your children?&quot; Being Old and occupied with work Warren forgot it was Christmas eve.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Warren is a 62 year old car repairer with two sons and two daughters; All of them are adults. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Passion for Cars and a family man&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Warren is a rather busy man, with free time almost impossible to acquire even on the weekends. His children love him more than anything else, and he does love them too by an equal amount.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One way he can give equal time to his children if they work with him.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On that day however, Warren was not very busy. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Could imply a holiday or a weekend.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He went to the shop directly from home and sat around the office with nothing to do except play darts. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All his children are home.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Warren at first had a very stimulating thought which made him expect something astonishing but he later forgot about it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An Idea to fix a favorite car.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Anyways, on that conversation with the visitor; The visitor displayed a visible grin and remarked, &quot;Here, take this. Open it 30 minutes after your clock reaches 11:30pm&quot; and handed him a red box, which felt heavy and the visitor left the shop.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Could the visitor be Santa Claus.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After Warren did what the visitor asked, he was surprised and got very happy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What was in the box and who was this strange visitor?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Considering the age of Warren, he would be happy with Glasses or anything to make the children happy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9892" LastActivityDate="2015-04-07T22:15:46.583" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11710" PostTypeId="2" ParentId="11586" CreationDate="2015-04-07T22:35:32.743" Score="0" Body="&lt;p&gt;Its a guess, and if it is wrong then please be easy on me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's his birthday.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The stranger comes in the shop creating a mysterious atmosphere. &quot;He&quot; says &quot;Do you remember?&quot;...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...The description of the costume must be a red herring, except the mask, which may change the voice of the person. The question may refer to him remembering his birthday, since he is an old man.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A description of the background of Warren reveals that...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...He is a family person and loves his work. His family loves him equally more, which may point out that they may wish to surprise him on his 62th birthday.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A reason that he may not be busy that day is...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;... It may be his retirement day, since his shop doesn't attract much customers anymore or due to his old age, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He was happy to get the &quot;gift&quot; because...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;... The &quot;open 30 minutes after 11:30 PM &quot; meant that the stranger indirectly said him to open the box the next day, which may point towards the next day being his birthday. And as for a gift, the stranger got him a  (insert anything you want to believe).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And to sum to all up,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The stranger was one of his sons, changing his voice by the mask and all covered up so that he won't recognize him, who gave him a nice birthday present.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2808" LastActivityDate="2015-04-07T22:35:32.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11711" PostTypeId="2" ParentId="11653" CreationDate="2015-04-07T22:48:59.473" Score="3" Body="&lt;p&gt;It's one ugly formula, but at least it's explicit! You can fire your &lt;span class=&quot;math-container&quot;&gt;$n^{th}$&lt;/span&gt; torpedo at location&#xA;&lt;span class=&quot;math-container&quot;&gt;$$P(n) = (-1)^n\left[A(n) + B(n)\cdot n\right]$$&lt;/span&gt;&#xA;where&#xA;&lt;span class=&quot;math-container&quot;&gt;$$A(n) = \left\lfloor\frac{ \left\lceil\frac{\sqrt{8n+1}-1}{2}\right\rceil\left(\left\lceil\frac{\sqrt{8n+1}-1}{2}\right\rceil+1\right) - 2(n-1)       }{4}    \right\rfloor$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$B(n) = \left\lfloor\frac{2(n-1) + 2\left\lceil\frac{\sqrt{8n+1}-1}{2}\right\rceil -\left\lceil\frac{\sqrt{8n+1}-1}{2}\right\rceil\left(\left\lceil\frac{\sqrt{8n+1}-1}{2}\right\rceil+1\right)}{4}    \right\rfloor$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, this table gives &lt;span class=&quot;math-container&quot;&gt;$P(n)$&lt;/span&gt; for the first ten shots (and a few more).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Torpedo (n)     Location P(n)     Corresponding (A,B) &#xA;1               0                 (0,0)&#xA;2               1                 (1,0)&#xA;3               3                 (0,1)&#xA;4               -1                (-1,0)&#xA;5               6                 (1,1)&#xA;6               -6                (0,-1)&#xA;7               2                 (2,0)&#xA;8               7                 (-1,1)&#xA;9               -8                (1,-1)&#xA;10              20                (0,2)&#xA;...             ...               ...&#xA;35              -104              (1,-3)&#xA;100             -397              (3,-4)&#xA;1000            5018              (18,5)&#xA;12345           -604934           (-29,-49)  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h1&gt;Derivation&lt;/h1&gt;&#xA;&#xA;&lt;h3&gt;Where to fire&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Clearly, if we know &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt;, then at time &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; we fire our torpedo at position&#xA;&lt;span class=&quot;math-container&quot;&gt;$$P(n) = A + Bn$$&lt;/span&gt;  &lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Mapping &lt;span class=&quot;math-container&quot;&gt;$\mathbb N$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\mathbb Z$&lt;/span&gt;&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;This function provides a bijective map from &lt;span class=&quot;math-container&quot;&gt;$\mathbb N$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\mathbb Z$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$H(k) = (-1)^k\lfloor k/2 \rfloor$$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, we have &lt;span class=&quot;math-container&quot;&gt;$$H(1) = 0 \quad\quad H(2) = 1  \quad\quad H(3) = -1 \quad\quad H(4) = 2 \quad\quad H(5) = -2 \quad\quad etc$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Enumerating &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt;&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Consider a table, where the &lt;span class=&quot;math-container&quot;&gt;$i^{th}$&lt;/span&gt; row and &lt;span class=&quot;math-container&quot;&gt;$j^{th}$&lt;/span&gt; column holds the pair &lt;span class=&quot;math-container&quot;&gt;$(H(i), H(j))$&lt;/span&gt;. Every pair &lt;span class=&quot;math-container&quot;&gt;$(A, B)$&lt;/span&gt; lies somewhere on this table, so we need only find the &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$j$&lt;/span&gt; which correspond to it. We will map &lt;span class=&quot;math-container&quot;&gt;$\mathbb N$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$(A,B)$&lt;/span&gt; pairs by moving along the diagonals. For instance, the first &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$(A, B)$&lt;/span&gt; pairs are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$(0,0) \quad\quad (1,0) \quad\quad (0,1) \quad\quad (-1, 0) \quad\quad (1,1)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formally, this involves mapping &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; to a pair &lt;span class=&quot;math-container&quot;&gt;$(A, B) = (H(I(n)), \ H(J(n)))$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Finding &lt;span class=&quot;math-container&quot;&gt;$I(n)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$J(n)$&lt;/span&gt;&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Let &lt;span class=&quot;math-container&quot;&gt;$T(k)$&lt;/span&gt; be a function which returns the &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; corresponding to the first row and &lt;span class=&quot;math-container&quot;&gt;$k^{th}$&lt;/span&gt; column of our table. This function can be written recursively as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$T(k) = T(k-1) + k \quad\quad \text{with }\ T(0) = 0$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This recursion relation has the solution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$T(k) = \frac{k}{2}\left(k+1\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whose inverse is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$T^{-1}(k) = \frac{\sqrt{8k + 1} -1}{2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the &lt;span class=&quot;math-container&quot;&gt;$k^{th}$&lt;/span&gt; diagonal contains &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; pairs, the &lt;span class=&quot;math-container&quot;&gt;$k^{th}$&lt;/span&gt; diagonal of the table contains the &lt;span class=&quot;math-container&quot;&gt;$u^{th}$&lt;/span&gt; through &lt;span class=&quot;math-container&quot;&gt;$v^{th}$&lt;/span&gt; pairs where &lt;span class=&quot;math-container&quot;&gt;$u = T(k) - k + 1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$v = T(k)$&lt;/span&gt;. Moreover &lt;span class=&quot;math-container&quot;&gt;$\lceil T^{-1}(n)\rceil$&lt;/span&gt; gives the length of the diagonal which contains the &lt;span class=&quot;math-container&quot;&gt;$n^{th}$&lt;/span&gt; pair. Therefore in terms of &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$v = T\left(\lceil T^{-1}(n)\rceil\right)$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$u = v - \lceil T^{-1}(n)\rceil + 1$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now just by counting rows (or columns) away from &lt;span class=&quot;math-container&quot;&gt;$u$&lt;/span&gt; (or &lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt;) we obtain &lt;span class=&quot;math-container&quot;&gt;$I(n)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$J(n)$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$I(n) = v - n + 1$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$J(n) = u - n$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting all the pieces together gives the equation for &lt;span class=&quot;math-container&quot;&gt;$P(n)$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11063" LastEditorUserId="11063" LastEditDate="2018-09-20T22:03:41.963" LastActivityDate="2018-09-20T22:03:41.963" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="11712" PostTypeId="2" ParentId="11650" CreationDate="2015-04-07T23:30:40.597" Score="0" Body="&lt;p&gt;I think it is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A skeleton&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am faster than you can run,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your skeleton runs at the same speed as you, but can do other things (like ride in a car)... also the same speed as you, but faster than you can run (since you too ride in a car faster than you can run).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so abandon me, go home and have fun.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A lot of people call the afterlife their true home and a happy place. If you abandon your skeleton, that's where you will go.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I live where the moon rise from above,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so abandon me, get rid of the hands in glove.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your hands are the glove here. Your skeleton's hands are in the 'gloves' of your skin-hands.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am soft like a kitty and hard like a rock,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your skin is soft but your skeleton is hard.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;without any warning we come in flock.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure, but people do come in 'flocks' sometimes. Also in horror movies a bunch of skeletons might fall on you without warning or something. Also a graveyard is a 'flock' of skeletons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who am I that you might ask?&#xA;a symbol of fear, hidden in a mask! &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Similar to the gloves one. A skeleton is a symbol of fear, and it's hidden in the 'mask' of your body.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11154" LastActivityDate="2015-04-07T23:30:40.597" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11713" PostTypeId="1" AcceptedAnswerId="11716" CreationDate="2015-04-08T00:02:57.217" Score="37" ViewCount="4551" Body="&lt;p&gt;What is the meaning of the following sentence?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My French killer married the hammerer cooked in oil, seated beneath a daily beacon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(The answer will be obvious once you see it!)&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-04-14T12:32:19.973" LastActivityDate="2015-04-14T12:32:19.973" Title="My French killer ... what does this mean?" Tags="|enigmatic-puzzle|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11714" PostTypeId="2" ParentId="11713" CreationDate="2015-04-08T00:29:55.103" Score="14" Body="&lt;p&gt;I guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's a description of a breakfast: a bloody mary (French killer + 'married'), bangers and mash (to hammer on something is to bang on it, not totally sure about the cooked in oil part but cooking something in oil is frying it), and a sunny-side up egg ('daily beacon' sounds like the sun to me).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11154" LastActivityDate="2015-04-08T00:29:55.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11715" PostTypeId="2" ParentId="11673" CreationDate="2015-04-08T03:43:09.640" Score="1" Body="&lt;p&gt;For ease of reference, call each integer lattice point a &lt;em&gt;house&lt;/em&gt;. Without loss of generality, let the lattice in the question be the set of all points with integer coordinates on an $XY$ plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a house $H$, let its colour be given by a function $C(H)$. We call $C$ a &lt;em&gt;colouring&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We call a vector a &lt;em&gt;house vector&lt;/em&gt; if the vector is parallel to a line that contains at least two houses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first constraint requires every straight line passing through at least 2 houses to be periodic. That is, given a house $H$ and a house vector $v$ whose magnitude is the period associated with the line containing $H$ and parallel to $v$, then for any integer $k$, $H + kv$ is also a house and has the same colour as $H$.&#xA;$$\forall k \in Z: C(H) = C(H + kv) \tag{1} \label{1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second constraint in the question requires $C$ to be non-periodic. This implies the existence of a non-empty set $F$ of houses such that there is no house vector $u$ for which $C$ remains invariant when translated any non-zero distance in the direction of $u$. The following shows by contradiction that $F$ does not exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we had a colouring $C$ that contained such a set $F$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Select any house vector $v$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each house $F_i$ in $F$, let $p_i$ be the period of the line passing through $F_i$ and parallel to $v$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $p$ be the LCM of all the $p_i$ and let $u$ be the vector parallel to $v$ and with magnitude $p$. By equation $\eqref{1}$, since $p$ is a multiple of each $p_i$, the colour of each $F_i$ repeats periodically in the direction of $u$.&#xA;$$\forall k \in Z: C(F_i) = C(F_i + ku) \tag{2} \label{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the colour pattern of the whole set $F$ is periodic with period $p$ in the direction of $u$, contradicting the second condition. Therefore $F$ doesn't exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the houses cannot be painted as proposed.&lt;/p&gt;&#xA;" OwnerUserId="9422" LastEditorUserId="9422" LastEditDate="2015-04-10T16:10:54.493" LastActivityDate="2015-04-10T16:10:54.493" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="11716" PostTypeId="2" ParentId="11713" CreationDate="2015-04-08T03:43:53.627" Score="51" Body="&lt;p&gt;It simply lists:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The days of the week&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;My French&lt;/em&gt; =&gt; mon (french for &quot;my&quot;)&lt;br&gt;&#xA; &lt;em&gt;killer&lt;/em&gt; =&gt; tue (french for &quot;kills&quot;)&lt;br&gt;&#xA; &lt;em&gt;married&lt;/em&gt; =&gt; wed&lt;br&gt;&#xA; &lt;em&gt;the hammerer&lt;/em&gt; =&gt; thur (play on Thor)&lt;br&gt;&#xA; &lt;em&gt;cooked in oil&lt;/em&gt; =&gt; fri (from fry)&lt;br&gt;&#xA; &lt;em&gt;seated beneath&lt;/em&gt; =&gt; sat&lt;br&gt;&#xA; &lt;em&gt;a daily beacon&lt;/em&gt; =&gt; sun  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="5208" LastEditDate="2015-04-08T03:57:42.773" LastActivityDate="2015-04-08T03:57:42.773" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11717" PostTypeId="2" ParentId="11673" CreationDate="2015-04-08T04:07:44.733" Score="3" Body="&lt;p&gt;Another cheap &lt;a href=&quot;https://gowers.wordpress.com/2008/08/16/just-do-it-proofs/&quot; rel=&quot;nofollow&quot;&gt;just-do-it proof&lt;/a&gt;! Just list the constraints and crank out a construction. No cleverness required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need to satisfy the following constraints:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Each line is periodic&lt;/li&gt;&#xA;&lt;li&gt;Each displacement does not produce a 2D periodic tiling.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are countably many lines and countably many displacements, so that's countably many constraint total. Enumerate them. We will satisfy them in turn by coloring cells in one of two colors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a displacement, ruin its potential periodicity by coloring two cells at that displacement different colors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each line constraint, color all its uncolored cells to make it periodic. We can do this because only a finite number of its cells have been colored, allowing us to repeat that pattern. (The number is finite because each previous line intersects this one at most once, and each displacement constrains colors only two cells.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One a constraint is met, no further coloring of uncolored cells can ruin it, so all constraints will be satisfied.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-04-08T04:26:38.203" LastActivityDate="2015-04-08T04:26:38.203" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11718" PostTypeId="1" AcceptedAnswerId="15865" CreationDate="2015-04-08T04:33:39.833" Score="15" ViewCount="7523" Body="&lt;h2&gt;Background&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EEWu8.png&quot; alt=&quot;NOC +10 intro&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since March 6th, 2015, a YouTube channel called &lt;strong&gt;Noc +10&lt;/strong&gt; has been uploading puzzle videos about once a week.  Each video begins with the same introduction, followed by some form of clues which can be decoded to point to a hidden video's YouTube ID. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hidden videos have had a nautical theme, possibly related to &lt;a href=&quot;http://www.navy.mil/maritime/display.asp?page=noc.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Naval Operation Concept 2010&lt;/a&gt;, but haven't presented any clear puzzles, aside from whatever meta puzzle or story is being told.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/10633/the-mystery-of-noc-10&quot;&gt;This question&lt;/a&gt;&lt;/strong&gt; provides a writeup of the first five videos, and &lt;a href=&quot;https://puzzling.stackexchange.com/q/10633/325&quot;&gt;it's primary answer&lt;/a&gt; solves the fifth to reveal the sixth.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that, three very strange videos that broke the pattern were posted, but they contained no apparent puzzles.  I choose to ignore them for the time being.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;7. Fourth contact - March 30, 2015&lt;/h2&gt;&#xA;&#xA;&lt;h3&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=6llW7rr2DUk&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;code&gt;0101 0101 0000 0111&lt;/code&gt;&lt;/a&gt;&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mET2Q.png&quot; alt=&quot;ISL 4   9 xEFTvZ&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The seventh video, which is numbered sequentially in binary with all the others, begins with the NOC +10 logo and the same music playing in the background.  At the halfway point, it switches to displaying the text &lt;code&gt;ISL 4   9 xEFTvZ&lt;/code&gt; while playing a new song, identified as &lt;a href=&quot;https://en.wikipedia.org/wiki/Air_on_the_G_String&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Air On the G string&quot;&lt;/a&gt;, August Wilhelmj's arrangement of a piece by J. S. Bach. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As before, we can expect the text displayed to decode to a eleven-alphanumeric-character YouTube video ID.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Wikipedia, the song is most notable for Wilhelmj's arrangement, which allowed him to play the entire piece on &lt;em&gt;just&lt;/em&gt; the G string of his violin.  I suspect that that means that the letter &lt;code&gt;G&lt;/code&gt; is a major factor in the decoding, although I'm not sure how.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried shifting the letters by 'G' (like with the Caesar cipher in the first video), as well as several operations on the ASCII values involving G, but nothing has led to an active video ID. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Can you crack the last video's code and find the ID of the next one?&lt;/h3&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Other links:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/10633/the-mystery-of-noc-10&quot;&gt;Previous Question, with more background&lt;/a&gt; (linked again)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://chat.stackexchange.com/rooms/22141/noc10&quot;&gt;Chat room&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/channel/UCGNXc9SMSSdOyfmZMSeLQ5w&quot; rel=&quot;nofollow noreferrer&quot;&gt;NOC +10 YouTube channel&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Air_on_the_G_String&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Air on the G String&quot; on Wikipedia&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="325" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2015-07-09T17:06:02.760" Title="The mystery of NOC +10, Part 2" Tags="|cipher|enigmatic-puzzle|audio|" AnswerCount="2" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="11719" PostTypeId="2" ParentId="11594" CreationDate="2015-04-08T04:34:38.130" Score="8" Body="&lt;p&gt;[Feel free to edit and remove spoiler tag if that is preferable for cooperating.]&lt;br&gt;&lt;/p&gt;&#xA;&lt;p&gt;Continuing where &lt;a href=&quot;https://puzzling.stackexchange.com/a/11670/10724&quot;&gt;Len&lt;/a&gt; left off, the relevant answer to the first puzzle is&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; cab&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; these are also hexadecimal digits. 0xcab hex = 3243 decimal, which is the&#xA; password to the image.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Decoding the image gives&lt;br&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Good job, but it's not over!&lt;br&gt;&#xA; Things will get harder now!&lt;br&gt;&#xA; &lt;br&gt;&#xA; 79 periodic number&lt;br&gt;&#xA; &lt;br&gt;&#xA; 1459238&lt;br&gt;&#xA; &lt;br&gt;&#xA; If you can't see the answer, you should use it!&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;My thoughts on this:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The seven digit number is part of the expansion of pi with missing digits:&lt;br&gt;&#xA; &lt;code&gt;pi      = 3.14159265358979323846264...&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;message =   14 592  3 8&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;missing = 3   1   65 5 9&lt;/code&gt;&lt;br&gt;&#xA; I've shown the missing digits up to where 79 appears in pi.&#xA; I think the last part of the message is saying to use these missing digits.&#xA; Also FWIW 79 is the number for gold in the periodic table.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10724" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-08T04:50:04.030" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11720" PostTypeId="2" ParentId="11656" CreationDate="2015-04-08T04:36:24.487" Score="1" Body="&lt;p&gt;There are five pigeons in the queue. The third one is cross-eyed and sees four (two) behind and ahead if the line is straight!&lt;/p&gt;&#xA;" OwnerUserId="11158" LastActivityDate="2015-04-08T04:36:24.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11721" PostTypeId="2" ParentId="11656" CreationDate="2015-04-08T04:43:01.710" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the queue goes around the corner of a building.  on one side of the building are four pigeons, and on the other side also four pigeons.  precisely at the corner (so neither on one side of the building nor the other) is the middle pigeon.  only the middle pigeon can see all the pigeons.  the others can see only those on their side (plus the middle pigeon).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11159" LastEditorUserId="11159" LastEditDate="2015-04-08T15:26:32.153" LastActivityDate="2015-04-08T15:26:32.153" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11722" PostTypeId="2" ParentId="11594" CreationDate="2015-04-08T06:18:11.757" Score="1" Body="&lt;p&gt;Alright, haha, I think I went way too far in the wrong direction, but I'll post this just in case it's miraculously on the right track. I searched the hex code of the image for the missing digits of pi 316559, and there was one match. I then replaced those 3 bytes with all of the named hex colors for gold. One of them, EEC900, made an interesting pattern in the lower-right hand corner of the image, part of which looks weirdly similar to a &lt;a href=&quot;http://www.dangermouse.net/esoteric/piet.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Piet&lt;/a&gt; program. That's where I stopped because probably this is totally wrong, but who knows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RcgvV.png&quot; alt=&quot;Piet?&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="11154" LastActivityDate="2015-04-08T06:18:11.757" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11723" PostTypeId="2" ParentId="11594" CreationDate="2015-04-08T08:25:28.790" Score="5" Body="&lt;p&gt;First, a summary of progress made by others on Puzzles 1-3; then everything on Puzzles 4-5 is my own work.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Puzzle #1&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The rhyming riddle given in the OP was solved by &lt;a href=&quot;https://puzzling.stackexchange.com/a/11670/5373&quot;&gt;Len&lt;/a&gt;; solution #1 is a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; London taxi or &lt;strong&gt;cab&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Puzzle #2&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Solution #1 was interpreted by &lt;a href=&quot;https://puzzling.stackexchange.com/a/11719/5373&quot;&gt;finsternis&lt;/a&gt; as a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; number written in hexadecimal. $CAB_{hex}=3243_{dec}$,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so solution #2 is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3243.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Puzzle #3&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Solution #2 was used by &lt;a href=&quot;https://puzzling.stackexchange.com/a/11719/5373&quot;&gt;finsternis&lt;/a&gt; as the password to decode the picture in the OP using the given link. Solution #3 is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Good job, but it's not over!&lt;br&gt;&#xA; Things will get harder now!&lt;br&gt;&#xA; &lt;br&gt;&#xA; 79 periodic number&lt;br&gt;&#xA; &lt;br&gt;&#xA; 1459238&lt;br&gt;&#xA; &lt;br&gt;&#xA; If you can't see the answer, you should use it!&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Puzzle #4&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;My interpretation of Solution #3 is as follows.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;79 periodic number&quot; - 79 appears on the &lt;em&gt;periodic&lt;/em&gt; table as the number for gold, and the only &quot;golden number&quot; I can think of is $\phi=\frac{\sqrt{5}+1}{2}$, the golden ratio.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The hints have told us that we're now looking at a cipher puzzle, so we need to find some way to put together $\phi$ (as a key perhaps) and the string of numbers 1459238.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I believe Solution #4 is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 5355?7?, where the second ? is either 6 or 9 and the first is probably either 0 or 4. For my reasoning for this, see below.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Puzzle #5&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The hints tell us that we need&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; lateral thinking at this stage. More specifically, we need a &quot;different point of view&quot;, involving &quot;turn upside down your head&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We also have the clue&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;If you can't see the answer, you should use it!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which so far we haven't used. This clue makes me think of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; GLASSES, which looks like the number 5355?7? when turned upside down. (An upside-down 6 looks like a lower-case g, and an upside-down 9 like a capital G, so I'm not sure which is the final digit. An A looks a bit like a 4, and an upside-down a looks a bit like a 0, so I'm not sure of the third-to-last either.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This was my reasoning for Solution #4. It fits Hint #4 - indeed everything seems to fit together. I'm sure I've got the right answer; I just need to do a little more work on Puzzle #4.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The final solution is ...&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; GLASSES.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-04-12T14:37:19.913" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="11724" PostTypeId="1" AcceptedAnswerId="11725" CreationDate="2015-04-08T09:31:49.493" Score="14" ViewCount="1287" Body="&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three men visit a gym on a regular basis - one every two days, one every three days, and the last one visits once every week (on the same day). In a particular month, one of the 3 men's first visit (of the month) is on a monday, one of the other's is on a wednesday and the last one on a friday.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On which date of the month do all the 3 visit the gym?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Twist&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose 'friday' was replaced by 'saturday'. Without resolving the puzzle, answer the same question (on which date do.....)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;P. S.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've created a new tag named 'timings' - dates, months, hours before/after, etc. Please delete it if necessary.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-04-08T10:02:33.387" Title="Visiting the gym this month" Tags="|logical-deduction|time|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11725" PostTypeId="2" ParentId="11724" CreationDate="2015-04-08T09:50:25.003" Score="16" Body="&lt;p&gt;Let W, H, S be the men who come once every t&lt;strong&gt;w&lt;/strong&gt;o, t&lt;strong&gt;h&lt;/strong&gt;ree, and &lt;strong&gt;s&lt;/strong&gt;even days respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because W visits every 2 days, his first visit of the month must be on the 1st or 2nd. Similarly, H's first visit must be on the 1st, 2nd, or 3rd. But each pair of Monday, Wednesday, Friday are separated by at least 2 days, so &lt;strong&gt;W must first visit on the 1st and H on the 3rd&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, the 1st of the month must be a Monday, Wednesday, or Friday.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If it's a Friday, then H's first visit is on a Sunday, contradiction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If it's a Monday, then S's first visit is on Friday the 5th and the only other even-date Friday in the month (i.e. day when both W and S are in the gym) is Friday the 19th, on which day H isn't there. But we know all three are there on some day, so again, contradiction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;So the 1st must be a Wednesday and &lt;strong&gt;S first visits on Monday the 6th&lt;/strong&gt;. Now the only odd-date Mondays in the month (i.e. days when both W and S are in the gym) are the 13th and the 27th. W is there on all multiple-of-3 dates, which includes the 27th but not the 13th.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the 27th of the month.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Twist&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This isn't really a new puzzle. Monday, Wednesday, and Saturday are still three maximally separated days in the week (two 2-day gaps and one 3-day gap). So we can argue exactly as above (let Saturday be kMonday, Monday be kWednesday, Wednesday be kFriday, and insert &quot;k&quot;s everywhere in the above argument) to get the same answer as before:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the three men are all in the gym together only on the 27th.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Commentary&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The reason this is a good puzzle (and not just a maths problem) is because it's not obvious where to start. Obviously we could just brute-force it by considering all the possible days of the week on which the month could start and all the permutations for assigning W, H, S to Monday, Wednesday, Friday - but this would take a long time without a computer and wouldn't be very satisfying. The better approach is to examine the puzzle for a while until it becomes clear what given information is going to be most useful and how to deduce as much new information as possible in the fewest steps. In only the first few lines, my answer has already thrown out almost all of the possibilities a brute-force approach would consider - but it's not immediately obvious that that's the right way to start on the puzzle.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-04-08T10:02:33.387" LastActivityDate="2015-04-08T10:02:33.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11728" PostTypeId="2" ParentId="11666" CreationDate="2015-04-08T13:24:02.390" Score="2" Body="&lt;p&gt;So, since this is a 3-number lock, I tried to think of it in 3D, where you're hunting for a point, and when you check a point, rays go out from it in the cardinal directions. (So, a weird 3D battleship). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 32 checks.&lt;br&gt;&#xA;&lt;br&gt;&#xA; My reasoning is that you can think of the area space as 8 4x4 cubes. The cross-area of each cube is 16. It takes 16 guesses to fully check the 3 cross areas of the cube, but in doing so, you've checked the 3 adjacent 4x4 cubes.&lt;br&gt;&#xA; Repeating on the opposite 4x4 cube, you've checked all the places, so 32 total guesses.&lt;br&gt;&#xA;&lt;br&gt;&#xA; I tried to see if I could optimize it down, like use the same strategy for splitting each 4x4 into 2x2 cubes, but that left gaps that weren't checked. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-04-08T18:34:40.840" LastActivityDate="2015-04-08T18:34:40.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11729" PostTypeId="1" AcceptedAnswerId="11739" CreationDate="2015-04-08T14:13:31.117" Score="6" ViewCount="50289" Body="&lt;p&gt;You have to fill the boxes to make the equation true:&lt;/p&gt;&#xA;&lt;p&gt;&lt;kbd&gt; &lt;/kbd&gt; + &lt;kbd&gt; &lt;/kbd&gt; + &lt;kbd&gt; &lt;/kbd&gt; = 30&lt;/p&gt;&#xA;&lt;p&gt;Using only &lt;kbd&gt;1&lt;/kbd&gt;, &lt;kbd&gt;3&lt;/kbd&gt;, &lt;kbd&gt;5&lt;/kbd&gt;, &lt;kbd&gt;7&lt;/kbd&gt;, &lt;kbd&gt;9&lt;/kbd&gt;, &lt;kbd&gt;11&lt;/kbd&gt;, &lt;kbd&gt;13&lt;/kbd&gt;, &lt;kbd&gt;15&lt;/kbd&gt;.&lt;/p&gt;&#xA;&lt;p&gt;You can also repeat the numbers. (eg: 1 + 1 + 1)&lt;/p&gt;&#xA;&lt;h3&gt;Edit1: the boxes of the equation are empty right now&lt;/h3&gt;&#xA;&lt;h3&gt;Edit 2:&lt;/h3&gt;&#xA;&lt;p&gt;That's my first post here and I didn't realize how fantastics your mind was, that's why my question was so broad. I'll accept the answer I was expecting but it doesn't mean that the other are wrong. Let's the upvoter decide.&lt;/p&gt;&#xA;" OwnerUserId="10394" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-09T13:08:31.163" Title="Fill the holes?" Tags="|number-sequence|" AnswerCount="8" CommentCount="8" FavoriteCount="0" ClosedDate="2015-04-08T17:50:00.483" ContentLicense="CC BY-SA 3.0" />
  <row Id="11731" PostTypeId="2" ParentId="11729" CreationDate="2015-04-08T14:19:00.087" Score="13" Body="&lt;p&gt;I think we can use a little trick:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Just use 9 upside down: 11 + 6 + 13 = 30&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2015-04-08T14:19:00.087" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11732" PostTypeId="2" ParentId="11729" CreationDate="2015-04-08T14:26:14.917" Score="12" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sum of 3 odd numbers will always be odd, so there is no answer that will yield an even number&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastActivityDate="2015-04-08T14:26:14.917" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11733" PostTypeId="2" ParentId="11475" CreationDate="2015-04-08T14:33:36.500" Score="3" Body="&lt;p&gt;I basically made a program that would solve this puzzle, and the end-result is a bunch of answers. If anyone feels like they have a lot of time and want to help me confirm all these answers go ahead :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fun Note: There are a total of 1050 ways to go from A to 4 and at the same time cover every single city once. I've posted them &lt;a href=&quot;https://drive.google.com/folderview?id=0B2kmADSQt1Pwfl9ZTXlUZ2ZwUlZzal9jZjRWdm1ocWFvM25ibDdxWF9BakVQR3VmR05OV2s&amp;amp;usp=sharing&quot; rel=&quot;nofollow&quot; title=&quot;link to solutions&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A small note about the notation: The character in the brackets represents the city at which the critical point occurs.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Minimizing the amount of money&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Lowest point possible is -32&lt;/strong&gt;&#xA;&lt;br&gt;(2) ABCIJDEFLKQRX398765Z012WVPOUTNHGMSY4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Maximizing the amount of money&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Highest point possible is 20&lt;/strong&gt;&#xA;&lt;br&gt;(U) AGMSYZTNOU012WVPQKJIHBCDEFLRX3987654&#xA;&lt;br&gt;(U) AGMSYZTNOU012WVPJIHBCDEFLKQRX3987654&#xA;&lt;br&gt;(Y) AGMNOUTSYZ012WVPQKJIHBCDEFLRX3987654&#xA;&lt;br&gt;(Y) AGMNOUTSYZ012WVPJIHBCDEFLKQRX3987654&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Minimizing the amount of happiness&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Lowest point possible is -25&lt;/strong&gt;&#xA;&lt;br&gt;(3) ABCDEFLKJIOPQRX398765Z012WVUTNHGMSY4&#xA;&lt;br&gt;(3) ABCDEFLKJIOPQRX39876012WVUTNHGMSYZ54&#xA;&lt;br&gt;(3) ABCDEFLKJIOPQRX398712WVUTNHGMSYZ0654&#xA;&lt;br&gt;(3) ABCDEFLKJIOPQRX398712WVU065ZTNHGMSY4&#xA;&lt;br&gt;(3) ABCDEFLKJIOPQRX3982WVUTNHGMSYZ017654&#xA;&lt;br&gt;(3) ABCDEFLKJIOPQRX3982WVU01765ZTNHGMSY4&#xA;&lt;br&gt;(3) ABCDEFLKJIOPQRX3982WV1765Z0UTNHGMSY4&#xA;&lt;br&gt;(3) ABCDEFLKJIOPQRX3982WV1760UTNHGMSYZ54&#xA;&lt;br&gt;(3) ABCDJIOPQKEFLRX398765Z012WVUTNHGMSY4&#xA;&lt;br&gt;(3) ABCDJIOPQKEFLRX39876012WVUTNHGMSYZ54&#xA;&lt;br&gt;(3) ABCDJIOPQKEFLRX398712WVUTNHGMSYZ0654&#xA;&lt;br&gt;(3) ABCDJIOPQKEFLRX398712WVU065ZTNHGMSY4&#xA;&lt;br&gt;(3) ABCDJIOPQKEFLRX3982WVUTNHGMSYZ017654&#xA;&lt;br&gt;(3) ABCDJIOPQKEFLRX3982WVU01765ZTNHGMSY4&#xA;&lt;br&gt;(3) ABCDJIOPQKEFLRX3982WV1765Z0UTNHGMSY4&#xA;&lt;br&gt;(3) ABCDJIOPQKEFLRX3982WV1760UTNHGMSYZ54&#xA;&lt;br&gt;(3) ABCIOPQKJDEFLRX398765Z012WVUTNHGMSY4&#xA;&lt;br&gt;(3) ABCIOPQKJDEFLRX39876012WVUTNHGMSYZ54&#xA;&lt;br&gt;(3) ABCIOPQKJDEFLRX398712WVUTNHGMSYZ0654&#xA;&lt;br&gt;(3) ABCIOPQKJDEFLRX398712WVU065ZTNHGMSY4&#xA;&lt;br&gt;(3) ABCIOPQKJDEFLRX3982WVUTNHGMSYZ017654&#xA;&lt;br&gt;(3) ABCIOPQKJDEFLRX3982WVU01765ZTNHGMSY4&#xA;&lt;br&gt;(3) ABCIOPQKJDEFLRX3982WV1765Z0UTNHGMSY4&#xA;&lt;br&gt;(3) ABCIOPQKJDEFLRX3982WV1760UTNHGMSYZ54&#xA;&lt;br&gt;(3) ABCIOPJDEFLKQRX398765Z012WVUTNHGMSY4&#xA;&lt;br&gt;(3) ABCIOPJDEFLKQRX39876012WVUTNHGMSYZ54&#xA;&lt;br&gt;(3) ABCIOPJDEFLKQRX398712WVUTNHGMSYZ0654&#xA;&lt;br&gt;(3) ABCIOPJDEFLKQRX398712WVU065ZTNHGMSY4&#xA;&lt;br&gt;(3) ABCIOPJDEFLKQRX3982WVUTNHGMSYZ017654&#xA;&lt;br&gt;(3) ABCIOPJDEFLKQRX3982WVU01765ZTNHGMSY4&#xA;&lt;br&gt;(3) ABCIOPJDEFLKQRX3982WV1765Z0UTNHGMSY4&#xA;&lt;br&gt;(3) ABCIOPJDEFLKQRX3982WV1760UTNHGMSYZ54&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Maximizing the amount of happiness&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Highest point possible is 23&lt;/strong&gt;&#xA;&lt;br&gt;(V) ABCDJIHGMSYZ012WVUTNOPQKEFLRX3987654&#xA;&lt;br&gt;(V) ABHGMSYZ012WVUTNOPQKJICDEFLRX3987654&#xA;&lt;br&gt;(V) ABHGMSYZ012WVUTNOPJICDEFLKQRX3987654&#xA;&lt;br&gt;(V) ABHGMSYZ012WVUTNOICDEFLKJPQRX3987654&#xA;&lt;br&gt;(V) ABHGMSYZ012WVUTNOICDJPQKEFLRX3987654&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Minimizing the amount of love&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Lowest point possible is -21&lt;/strong&gt;&#xA;&lt;br&gt;(P) ABHGMNTSYZ0UVWQPOICDJKEFLRX398217654&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Maximizing the amount of love&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Highest point possible is 32&lt;/strong&gt;&#xA;&lt;br&gt;(2) AGMNOIHBCDJKEFLRX398712WQPVUTSYZ0654&#xA;&lt;br&gt;(2) AGMNOIHBCDJKEFLRX398712WQPVU065ZTSY4&#xA;&lt;br&gt;(2) ABHGMNOICDJKEFLRX398712WQPVUTSYZ0654&#xA;&lt;br&gt;(2) ABHGMNOICDJKEFLRX398712WQPVU065ZTSY4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Minimizing the amount of art&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Lowest point possible is -30&lt;/strong&gt;&#xA;&lt;br&gt;(V) AGHBCDEFLRX398712WVPQKJIONMSYZTU0654&#xA;&lt;br&gt;(V) AGHBCDEFLRX398712WVPQKJIONMSTU065ZY4&#xA;&lt;br&gt;(V) AGHBCDEFLRX398712WVPQKJIOUTNMSYZ0654&#xA;&lt;br&gt;(V) AGHBCDEFLRX398712WVPQKJIOU065ZTNMSY4&#xA;&lt;br&gt;(X) AGHBCDEFLRQKJIOPV12WX398765Z0UTNMSY4&#xA;&lt;br&gt;(X) AGHBCDEFLRQKJIOPV12WX398760UTNMSYZ54&#xA;&lt;br&gt;(X) AGHBCDEFLKJIOPV12WQRX398765Z0UTNMSY4&#xA;&lt;br&gt;(X) AGHBCDEFLKJIOPV12WQRX398760UTNMSYZ54&#xA;&lt;br&gt;(X) AGHBCDJIOPV12WQKEFLRX398765Z0UTNMSY4&#xA;&lt;br&gt;(X) AGHBCDJIOPV12WQKEFLRX398760UTNMSYZ54&#xA;&lt;br&gt;(X) AGHBCIOPV12WQKJDEFLRX398765Z0UTNMSY4&#xA;&lt;br&gt;(X) AGHBCIOPV12WQKJDEFLRX398760UTNMSYZ54&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Maximizing the amount of art&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Highest point possible is 24&lt;/strong&gt;&#xA;&lt;br&gt;(J) AGMSYZ0UTNOPQKJIHBCDEFLRX3982WV17654&#xA;&lt;br&gt;(J) AGMNTSYZ0UOPQKJIHBCDEFLRX3982WV17654&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11094" LastEditorUserId="11094" LastEditDate="2015-07-06T16:25:29.117" LastActivityDate="2015-07-06T16:25:29.117" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11734" PostTypeId="2" ParentId="11729" CreationDate="2015-04-08T14:33:40.507" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 13 + 15 + 1 = 30  in base 9.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This isn't formatted the way I want, I'll go learn how to do that properly and come back to edit later.&lt;/p&gt;&#xA;" OwnerUserId="11119" LastEditorUserId="10281" LastEditDate="2015-04-08T14:36:24.623" LastActivityDate="2015-04-08T14:36:24.623" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11735" PostTypeId="2" ParentId="11729" CreationDate="2015-04-08T14:35:07.160" Score="2" Body="&lt;p&gt;The solution is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;11+13+1=30 in base 5  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's valid if your numbers are in base-5 instead of base-10.&lt;br&gt;&#xA;Note that the text doesn't specify the base of the equation.&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-04-08T15:11:34.877" LastActivityDate="2015-04-08T15:11:34.877" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11736" PostTypeId="2" ParentId="11729" CreationDate="2015-04-08T14:53:47.323" Score="19" Body="&lt;p&gt;To fully beat the &quot;what about in other bases&quot; dead horse,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1&lt;sub&gt;5&lt;/sub&gt; + 11&lt;sub&gt;5&lt;/sub&gt; + 13&lt;sub&gt;5&lt;/sub&gt; == 1&lt;sub&gt;10&lt;/sub&gt; + 6&lt;sub&gt;10&lt;/sub&gt; + 8&lt;sub&gt;10&lt;/sub&gt; == 15&lt;sub&gt;10&lt;/sub&gt; == 30&lt;sub&gt;5&lt;/sub&gt; &lt;br/&gt;&#xA; 3&lt;sub&gt;5&lt;/sub&gt; + 11&lt;sub&gt;5&lt;/sub&gt; + 11&lt;sub&gt;5&lt;/sub&gt; == 3&lt;sub&gt;10&lt;/sub&gt; + 6&lt;sub&gt;10&lt;/sub&gt; + 6&lt;sub&gt;10&lt;/sub&gt; == 15&lt;sub&gt;10&lt;/sub&gt; == 30&lt;sub&gt;5&lt;/sub&gt; &lt;br/&gt;&#xA; 1&lt;sub&gt;7&lt;/sub&gt; + 11&lt;sub&gt;7&lt;/sub&gt; + 15&lt;sub&gt;7&lt;/sub&gt; == 1&lt;sub&gt;10&lt;/sub&gt; + 8&lt;sub&gt;10&lt;/sub&gt; + 12&lt;sub&gt;10&lt;/sub&gt; == 21&lt;sub&gt;10&lt;/sub&gt; == 30&lt;sub&gt;7&lt;/sub&gt; &lt;br/&gt;&#xA; 1&lt;sub&gt;7&lt;/sub&gt; + 13&lt;sub&gt;7&lt;/sub&gt; + 13&lt;sub&gt;7&lt;/sub&gt; == 1&lt;sub&gt;10&lt;/sub&gt; + 10&lt;sub&gt;10&lt;/sub&gt; + 10&lt;sub&gt;10&lt;/sub&gt; == 21&lt;sub&gt;10&lt;/sub&gt; == 30&lt;sub&gt;7&lt;/sub&gt; &lt;br/&gt;&#xA; 3&lt;sub&gt;7&lt;/sub&gt; + 11&lt;sub&gt;7&lt;/sub&gt; + 13&lt;sub&gt;7&lt;/sub&gt; == 3&lt;sub&gt;10&lt;/sub&gt; + 8&lt;sub&gt;10&lt;/sub&gt; + 10&lt;sub&gt;10&lt;/sub&gt; == 21&lt;sub&gt;10&lt;/sub&gt; == 30&lt;sub&gt;7&lt;/sub&gt; &lt;br/&gt;&#xA; 5&lt;sub&gt;7&lt;/sub&gt; + 11&lt;sub&gt;7&lt;/sub&gt; + 11&lt;sub&gt;7&lt;/sub&gt; == 5&lt;sub&gt;10&lt;/sub&gt; + 8&lt;sub&gt;10&lt;/sub&gt; + 8&lt;sub&gt;10&lt;/sub&gt; == 21&lt;sub&gt;10&lt;/sub&gt; == 30&lt;sub&gt;7&lt;/sub&gt; &lt;br/&gt;&#xA; 1&lt;sub&gt;9&lt;/sub&gt; + 13&lt;sub&gt;9&lt;/sub&gt; + 15&lt;sub&gt;9&lt;/sub&gt; == 1&lt;sub&gt;10&lt;/sub&gt; + 12&lt;sub&gt;10&lt;/sub&gt; + 14&lt;sub&gt;10&lt;/sub&gt; == 27&lt;sub&gt;10&lt;/sub&gt; == 30&lt;sub&gt;9&lt;/sub&gt; &lt;br/&gt;&#xA; 3&lt;sub&gt;9&lt;/sub&gt; + 11&lt;sub&gt;9&lt;/sub&gt; + 15&lt;sub&gt;9&lt;/sub&gt; == 3&lt;sub&gt;10&lt;/sub&gt; + 10&lt;sub&gt;10&lt;/sub&gt; + 14&lt;sub&gt;10&lt;/sub&gt; == 27&lt;sub&gt;10&lt;/sub&gt; == 30&lt;sub&gt;9&lt;/sub&gt; &lt;br/&gt;&#xA; 3&lt;sub&gt;9&lt;/sub&gt; + 13&lt;sub&gt;9&lt;/sub&gt; + 13&lt;sub&gt;9&lt;/sub&gt; == 3&lt;sub&gt;10&lt;/sub&gt; + 12&lt;sub&gt;10&lt;/sub&gt; + 12&lt;sub&gt;10&lt;/sub&gt; == 27&lt;sub&gt;10&lt;/sub&gt; == 30&lt;sub&gt;9&lt;/sub&gt; &lt;br/&gt;&#xA; 5&lt;sub&gt;9&lt;/sub&gt; + 11&lt;sub&gt;9&lt;/sub&gt; + 13&lt;sub&gt;9&lt;/sub&gt; == 5&lt;sub&gt;10&lt;/sub&gt; + 10&lt;sub&gt;10&lt;/sub&gt; + 12&lt;sub&gt;10&lt;/sub&gt; == 27&lt;sub&gt;10&lt;/sub&gt; == 30&lt;sub&gt;9&lt;/sub&gt; &lt;br/&gt;&#xA; 7&lt;sub&gt;9&lt;/sub&gt; + 11&lt;sub&gt;9&lt;/sub&gt; + 11&lt;sub&gt;9&lt;/sub&gt; == 7&lt;sub&gt;10&lt;/sub&gt; + 10&lt;sub&gt;10&lt;/sub&gt; + 10&lt;sub&gt;10&lt;/sub&gt; == 27&lt;sub&gt;10&lt;/sub&gt; == 30&lt;sub&gt;9&lt;/sub&gt; &lt;br/&gt;&#xA; 1&lt;sub&gt;11&lt;/sub&gt; + 15&lt;sub&gt;11&lt;/sub&gt; + 15&lt;sub&gt;11&lt;/sub&gt; == 1&lt;sub&gt;10&lt;/sub&gt; + 16&lt;sub&gt;10&lt;/sub&gt; + 16&lt;sub&gt;10&lt;/sub&gt; == 33&lt;sub&gt;10&lt;/sub&gt; == 30&lt;sub&gt;11&lt;/sub&gt; &lt;br/&gt;&#xA; 3&lt;sub&gt;11&lt;/sub&gt; + 13&lt;sub&gt;11&lt;/sub&gt; + 15&lt;sub&gt;11&lt;/sub&gt; == 3&lt;sub&gt;10&lt;/sub&gt; + 14&lt;sub&gt;10&lt;/sub&gt; + 16&lt;sub&gt;10&lt;/sub&gt; == 33&lt;sub&gt;10&lt;/sub&gt; == 30&lt;sub&gt;11&lt;/sub&gt; &lt;br/&gt;&#xA; 5&lt;sub&gt;11&lt;/sub&gt; + 11&lt;sub&gt;11&lt;/sub&gt; + 15&lt;sub&gt;11&lt;/sub&gt; == 5&lt;sub&gt;10&lt;/sub&gt; + 12&lt;sub&gt;10&lt;/sub&gt; + 16&lt;sub&gt;10&lt;/sub&gt; == 33&lt;sub&gt;10&lt;/sub&gt; == 30&lt;sub&gt;11&lt;/sub&gt; &lt;br/&gt;&#xA; 5&lt;sub&gt;11&lt;/sub&gt; + 13&lt;sub&gt;11&lt;/sub&gt; + 13&lt;sub&gt;11&lt;/sub&gt; == 5&lt;sub&gt;10&lt;/sub&gt; + 14&lt;sub&gt;10&lt;/sub&gt; + 14&lt;sub&gt;10&lt;/sub&gt; == 33&lt;sub&gt;10&lt;/sub&gt; == 30&lt;sub&gt;11&lt;/sub&gt; &lt;br/&gt;&#xA; 7&lt;sub&gt;11&lt;/sub&gt; + 11&lt;sub&gt;11&lt;/sub&gt; + 13&lt;sub&gt;11&lt;/sub&gt; == 7&lt;sub&gt;10&lt;/sub&gt; + 12&lt;sub&gt;10&lt;/sub&gt; + 14&lt;sub&gt;10&lt;/sub&gt; == 33&lt;sub&gt;10&lt;/sub&gt; == 30&lt;sub&gt;11&lt;/sub&gt; &lt;br/&gt;&#xA; 9&lt;sub&gt;11&lt;/sub&gt; + 11&lt;sub&gt;11&lt;/sub&gt; + 11&lt;sub&gt;11&lt;/sub&gt; == 9&lt;sub&gt;10&lt;/sub&gt; + 12&lt;sub&gt;10&lt;/sub&gt; + 12&lt;sub&gt;10&lt;/sub&gt; == 33&lt;sub&gt;10&lt;/sub&gt; == 30&lt;sub&gt;11&lt;/sub&gt; &lt;br/&gt;&#xA; 3&lt;sub&gt;13&lt;/sub&gt; + 15&lt;sub&gt;13&lt;/sub&gt; + 15&lt;sub&gt;13&lt;/sub&gt; == 3&lt;sub&gt;10&lt;/sub&gt; + 18&lt;sub&gt;10&lt;/sub&gt; + 18&lt;sub&gt;10&lt;/sub&gt; == 39&lt;sub&gt;10&lt;/sub&gt; == 30&lt;sub&gt;13&lt;/sub&gt; &lt;br/&gt;&#xA; 5&lt;sub&gt;13&lt;/sub&gt; + 13&lt;sub&gt;13&lt;/sub&gt; + 15&lt;sub&gt;13&lt;/sub&gt; == 5&lt;sub&gt;10&lt;/sub&gt; + 16&lt;sub&gt;10&lt;/sub&gt; + 18&lt;sub&gt;10&lt;/sub&gt; == 39&lt;sub&gt;10&lt;/sub&gt; == 30&lt;sub&gt;13&lt;/sub&gt; &lt;br/&gt;&#xA; 7&lt;sub&gt;13&lt;/sub&gt; + 11&lt;sub&gt;13&lt;/sub&gt; + 15&lt;sub&gt;13&lt;/sub&gt; == 7&lt;sub&gt;10&lt;/sub&gt; + 14&lt;sub&gt;10&lt;/sub&gt; + 18&lt;sub&gt;10&lt;/sub&gt; == 39&lt;sub&gt;10&lt;/sub&gt; == 30&lt;sub&gt;13&lt;/sub&gt; &lt;br/&gt;&#xA; 7&lt;sub&gt;13&lt;/sub&gt; + 13&lt;sub&gt;13&lt;/sub&gt; + 13&lt;sub&gt;13&lt;/sub&gt; == 7&lt;sub&gt;10&lt;/sub&gt; + 16&lt;sub&gt;10&lt;/sub&gt; + 16&lt;sub&gt;10&lt;/sub&gt; == 39&lt;sub&gt;10&lt;/sub&gt; == 30&lt;sub&gt;13&lt;/sub&gt; &lt;br/&gt;&#xA; 9&lt;sub&gt;13&lt;/sub&gt; + 11&lt;sub&gt;13&lt;/sub&gt; + 13&lt;sub&gt;13&lt;/sub&gt; == 9&lt;sub&gt;10&lt;/sub&gt; + 14&lt;sub&gt;10&lt;/sub&gt; + 16&lt;sub&gt;10&lt;/sub&gt; == 39&lt;sub&gt;10&lt;/sub&gt; == 30&lt;sub&gt;13&lt;/sub&gt; &lt;br/&gt;&#xA; 5&lt;sub&gt;15&lt;/sub&gt; + 15&lt;sub&gt;15&lt;/sub&gt; + 15&lt;sub&gt;15&lt;/sub&gt; == 5&lt;sub&gt;10&lt;/sub&gt; + 20&lt;sub&gt;10&lt;/sub&gt; + 20&lt;sub&gt;10&lt;/sub&gt; == 45&lt;sub&gt;10&lt;/sub&gt; == 30&lt;sub&gt;15&lt;/sub&gt; &lt;br/&gt;&#xA; 7&lt;sub&gt;15&lt;/sub&gt; + 13&lt;sub&gt;15&lt;/sub&gt; + 15&lt;sub&gt;15&lt;/sub&gt; == 7&lt;sub&gt;10&lt;/sub&gt; + 18&lt;sub&gt;10&lt;/sub&gt; + 20&lt;sub&gt;10&lt;/sub&gt; == 45&lt;sub&gt;10&lt;/sub&gt; == 30&lt;sub&gt;15&lt;/sub&gt; &lt;br/&gt;&#xA; 9&lt;sub&gt;15&lt;/sub&gt; + 11&lt;sub&gt;15&lt;/sub&gt; + 15&lt;sub&gt;15&lt;/sub&gt; == 9&lt;sub&gt;10&lt;/sub&gt; + 16&lt;sub&gt;10&lt;/sub&gt; + 20&lt;sub&gt;10&lt;/sub&gt; == 45&lt;sub&gt;10&lt;/sub&gt; == 30&lt;sub&gt;15&lt;/sub&gt; &lt;br/&gt;&#xA; 9&lt;sub&gt;15&lt;/sub&gt; + 13&lt;sub&gt;15&lt;/sub&gt; + 13&lt;sub&gt;15&lt;/sub&gt; == 9&lt;sub&gt;10&lt;/sub&gt; + 18&lt;sub&gt;10&lt;/sub&gt; + 18&lt;sub&gt;10&lt;/sub&gt; == 45&lt;sub&gt;10&lt;/sub&gt; == 30&lt;sub&gt;15&lt;/sub&gt; &lt;br/&gt;&#xA; 7&lt;sub&gt;17&lt;/sub&gt; + 15&lt;sub&gt;17&lt;/sub&gt; + 15&lt;sub&gt;17&lt;/sub&gt; == 7&lt;sub&gt;10&lt;/sub&gt; + 22&lt;sub&gt;10&lt;/sub&gt; + 22&lt;sub&gt;10&lt;/sub&gt; == 51&lt;sub&gt;10&lt;/sub&gt; == 30&lt;sub&gt;17&lt;/sub&gt; &lt;br/&gt;&#xA; 9&lt;sub&gt;17&lt;/sub&gt; + 13&lt;sub&gt;17&lt;/sub&gt; + 15&lt;sub&gt;17&lt;/sub&gt; == 9&lt;sub&gt;10&lt;/sub&gt; + 20&lt;sub&gt;10&lt;/sub&gt; + 22&lt;sub&gt;10&lt;/sub&gt; == 51&lt;sub&gt;10&lt;/sub&gt; == 30&lt;sub&gt;17&lt;/sub&gt; &lt;br/&gt;&#xA; 9&lt;sub&gt;19&lt;/sub&gt; + 15&lt;sub&gt;19&lt;/sub&gt; + 15&lt;sub&gt;19&lt;/sub&gt; == 9&lt;sub&gt;10&lt;/sub&gt; + 24&lt;sub&gt;10&lt;/sub&gt; + 24&lt;sub&gt;10&lt;/sub&gt; == 57&lt;sub&gt;10&lt;/sub&gt; == 30&lt;sub&gt;19&lt;/sub&gt; &lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: a cool idea from Ian MacDonald.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A better answer would only use bases that match the allowed numbers ;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 11&lt;sub&gt;3&lt;/sub&gt; + 11&lt;sub&gt;5&lt;/sub&gt; + 15&lt;sub&gt;15&lt;/sub&gt; == 4 + 6 + 20 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;3&lt;/sub&gt; + 11&lt;sub&gt;7&lt;/sub&gt; + 13&lt;sub&gt;15&lt;/sub&gt; == 4 + 8 + 18 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;3&lt;/sub&gt; + 11&lt;sub&gt;7&lt;/sub&gt; + 15&lt;sub&gt;13&lt;/sub&gt; == 4 + 8 + 18 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;3&lt;/sub&gt; + 11&lt;sub&gt;9&lt;/sub&gt; + 11&lt;sub&gt;15&lt;/sub&gt; == 4 + 10 + 16 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;3&lt;/sub&gt; + 11&lt;sub&gt;9&lt;/sub&gt; + 13&lt;sub&gt;13&lt;/sub&gt; == 4 + 10 + 16 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;3&lt;/sub&gt; + 11&lt;sub&gt;9&lt;/sub&gt; + 15&lt;sub&gt;11&lt;/sub&gt; == 4 + 10 + 16 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;3&lt;/sub&gt; + 11&lt;sub&gt;11&lt;/sub&gt; + 11&lt;sub&gt;13&lt;/sub&gt; == 4 + 12 + 14 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;3&lt;/sub&gt; + 11&lt;sub&gt;11&lt;/sub&gt; + 13&lt;sub&gt;11&lt;/sub&gt; == 4 + 12 + 14 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;3&lt;/sub&gt; + 11&lt;sub&gt;11&lt;/sub&gt; + 15&lt;sub&gt;9&lt;/sub&gt; == 4 + 12 + 14 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;3&lt;/sub&gt; + 11&lt;sub&gt;13&lt;/sub&gt; + 13&lt;sub&gt;9&lt;/sub&gt; == 4 + 14 + 12 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;3&lt;/sub&gt; + 11&lt;sub&gt;13&lt;/sub&gt; + 15&lt;sub&gt;7&lt;/sub&gt; == 4 + 14 + 12 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;3&lt;/sub&gt; + 11&lt;sub&gt;15&lt;/sub&gt; + 13&lt;sub&gt;7&lt;/sub&gt; == 4 + 16 + 10 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;3&lt;/sub&gt; + 13&lt;sub&gt;5&lt;/sub&gt; + 13&lt;sub&gt;15&lt;/sub&gt; == 4 + 8 + 18 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;3&lt;/sub&gt; + 13&lt;sub&gt;5&lt;/sub&gt; + 15&lt;sub&gt;13&lt;/sub&gt; == 4 + 8 + 18 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;3&lt;/sub&gt; + 13&lt;sub&gt;7&lt;/sub&gt; + 13&lt;sub&gt;13&lt;/sub&gt; == 4 + 10 + 16 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;3&lt;/sub&gt; + 13&lt;sub&gt;7&lt;/sub&gt; + 15&lt;sub&gt;11&lt;/sub&gt; == 4 + 10 + 16 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;3&lt;/sub&gt; + 13&lt;sub&gt;9&lt;/sub&gt; + 13&lt;sub&gt;11&lt;/sub&gt; == 4 + 12 + 14 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;3&lt;/sub&gt; + 13&lt;sub&gt;9&lt;/sub&gt; + 15&lt;sub&gt;9&lt;/sub&gt; == 4 + 12 + 14 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;3&lt;/sub&gt; + 13&lt;sub&gt;11&lt;/sub&gt; + 15&lt;sub&gt;7&lt;/sub&gt; == 4 + 14 + 12 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;3&lt;/sub&gt; + 15&lt;sub&gt;7&lt;/sub&gt; + 15&lt;sub&gt;9&lt;/sub&gt; == 4 + 12 + 14 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;5&lt;/sub&gt; + 11&lt;sub&gt;5&lt;/sub&gt; + 13&lt;sub&gt;15&lt;/sub&gt; == 6 + 6 + 18 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;5&lt;/sub&gt; + 11&lt;sub&gt;5&lt;/sub&gt; + 15&lt;sub&gt;13&lt;/sub&gt; == 6 + 6 + 18 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;5&lt;/sub&gt; + 11&lt;sub&gt;7&lt;/sub&gt; + 11&lt;sub&gt;15&lt;/sub&gt; == 6 + 8 + 16 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;5&lt;/sub&gt; + 11&lt;sub&gt;7&lt;/sub&gt; + 13&lt;sub&gt;13&lt;/sub&gt; == 6 + 8 + 16 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;5&lt;/sub&gt; + 11&lt;sub&gt;7&lt;/sub&gt; + 15&lt;sub&gt;11&lt;/sub&gt; == 6 + 8 + 16 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;5&lt;/sub&gt; + 11&lt;sub&gt;9&lt;/sub&gt; + 11&lt;sub&gt;13&lt;/sub&gt; == 6 + 10 + 14 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;5&lt;/sub&gt; + 11&lt;sub&gt;9&lt;/sub&gt; + 13&lt;sub&gt;11&lt;/sub&gt; == 6 + 10 + 14 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;5&lt;/sub&gt; + 11&lt;sub&gt;9&lt;/sub&gt; + 15&lt;sub&gt;9&lt;/sub&gt; == 6 + 10 + 14 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;5&lt;/sub&gt; + 11&lt;sub&gt;11&lt;/sub&gt; + 11&lt;sub&gt;11&lt;/sub&gt; == 6 + 12 + 12 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;5&lt;/sub&gt; + 11&lt;sub&gt;11&lt;/sub&gt; + 13&lt;sub&gt;9&lt;/sub&gt; == 6 + 12 + 12 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;5&lt;/sub&gt; + 11&lt;sub&gt;11&lt;/sub&gt; + 15&lt;sub&gt;7&lt;/sub&gt; == 6 + 12 + 12 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;5&lt;/sub&gt; + 11&lt;sub&gt;13&lt;/sub&gt; + 13&lt;sub&gt;7&lt;/sub&gt; == 6 + 14 + 10 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;5&lt;/sub&gt; + 11&lt;sub&gt;15&lt;/sub&gt; + 13&lt;sub&gt;5&lt;/sub&gt; == 6 + 16 + 8 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;5&lt;/sub&gt; + 13&lt;sub&gt;5&lt;/sub&gt; + 13&lt;sub&gt;13&lt;/sub&gt; == 6 + 8 + 16 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;5&lt;/sub&gt; + 13&lt;sub&gt;5&lt;/sub&gt; + 15&lt;sub&gt;11&lt;/sub&gt; == 6 + 8 + 16 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;5&lt;/sub&gt; + 13&lt;sub&gt;7&lt;/sub&gt; + 13&lt;sub&gt;11&lt;/sub&gt; == 6 + 10 + 14 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;5&lt;/sub&gt; + 13&lt;sub&gt;7&lt;/sub&gt; + 15&lt;sub&gt;9&lt;/sub&gt; == 6 + 10 + 14 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;5&lt;/sub&gt; + 13&lt;sub&gt;9&lt;/sub&gt; + 13&lt;sub&gt;9&lt;/sub&gt; == 6 + 12 + 12 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;5&lt;/sub&gt; + 13&lt;sub&gt;9&lt;/sub&gt; + 15&lt;sub&gt;7&lt;/sub&gt; == 6 + 12 + 12 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;5&lt;/sub&gt; + 15&lt;sub&gt;7&lt;/sub&gt; + 15&lt;sub&gt;7&lt;/sub&gt; == 6 + 12 + 12 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;7&lt;/sub&gt; + 11&lt;sub&gt;7&lt;/sub&gt; + 11&lt;sub&gt;13&lt;/sub&gt; == 8 + 8 + 14 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;7&lt;/sub&gt; + 11&lt;sub&gt;7&lt;/sub&gt; + 13&lt;sub&gt;11&lt;/sub&gt; == 8 + 8 + 14 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;7&lt;/sub&gt; + 11&lt;sub&gt;7&lt;/sub&gt; + 15&lt;sub&gt;9&lt;/sub&gt; == 8 + 8 + 14 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;7&lt;/sub&gt; + 11&lt;sub&gt;9&lt;/sub&gt; + 11&lt;sub&gt;11&lt;/sub&gt; == 8 + 10 + 12 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;7&lt;/sub&gt; + 11&lt;sub&gt;9&lt;/sub&gt; + 13&lt;sub&gt;9&lt;/sub&gt; == 8 + 10 + 12 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;7&lt;/sub&gt; + 11&lt;sub&gt;9&lt;/sub&gt; + 15&lt;sub&gt;7&lt;/sub&gt; == 8 + 10 + 12 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;7&lt;/sub&gt; + 11&lt;sub&gt;11&lt;/sub&gt; + 13&lt;sub&gt;7&lt;/sub&gt; == 8 + 12 + 10 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;7&lt;/sub&gt; + 11&lt;sub&gt;13&lt;/sub&gt; + 13&lt;sub&gt;5&lt;/sub&gt; == 8 + 14 + 8 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;7&lt;/sub&gt; + 13&lt;sub&gt;5&lt;/sub&gt; + 13&lt;sub&gt;11&lt;/sub&gt; == 8 + 8 + 14 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;7&lt;/sub&gt; + 13&lt;sub&gt;5&lt;/sub&gt; + 15&lt;sub&gt;9&lt;/sub&gt; == 8 + 8 + 14 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;7&lt;/sub&gt; + 13&lt;sub&gt;7&lt;/sub&gt; + 13&lt;sub&gt;9&lt;/sub&gt; == 8 + 10 + 12 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;7&lt;/sub&gt; + 13&lt;sub&gt;7&lt;/sub&gt; + 15&lt;sub&gt;7&lt;/sub&gt; == 8 + 10 + 12 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;9&lt;/sub&gt; + 11&lt;sub&gt;9&lt;/sub&gt; + 11&lt;sub&gt;9&lt;/sub&gt; == 10 + 10 + 10 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;9&lt;/sub&gt; + 11&lt;sub&gt;9&lt;/sub&gt; + 13&lt;sub&gt;7&lt;/sub&gt; == 10 + 10 + 10 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;9&lt;/sub&gt; + 11&lt;sub&gt;11&lt;/sub&gt; + 13&lt;sub&gt;5&lt;/sub&gt; == 10 + 12 + 8 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;9&lt;/sub&gt; + 13&lt;sub&gt;5&lt;/sub&gt; + 13&lt;sub&gt;9&lt;/sub&gt; == 10 + 8 + 12 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;9&lt;/sub&gt; + 13&lt;sub&gt;5&lt;/sub&gt; + 15&lt;sub&gt;7&lt;/sub&gt; == 10 + 8 + 12 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;9&lt;/sub&gt; + 13&lt;sub&gt;7&lt;/sub&gt; + 13&lt;sub&gt;7&lt;/sub&gt; == 10 + 10 + 10 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;11&lt;/sub&gt; + 13&lt;sub&gt;5&lt;/sub&gt; + 13&lt;sub&gt;7&lt;/sub&gt; == 12 + 8 + 10 == 30 &lt;br/&gt;&#xA; 11&lt;sub&gt;13&lt;/sub&gt; + 13&lt;sub&gt;5&lt;/sub&gt; + 13&lt;sub&gt;5&lt;/sub&gt; == 14 + 8 + 8 == 30 &lt;br/&gt;&#xA; 13&lt;sub&gt;5&lt;/sub&gt; + 13&lt;sub&gt;5&lt;/sub&gt; + 13&lt;sub&gt;11&lt;/sub&gt; == 8 + 8 + 14 == 30 &lt;br/&gt;&#xA; 13&lt;sub&gt;5&lt;/sub&gt; + 13&lt;sub&gt;5&lt;/sub&gt; + 15&lt;sub&gt;9&lt;/sub&gt; == 8 + 8 + 14 == 30 &lt;br/&gt;&#xA; 13&lt;sub&gt;5&lt;/sub&gt; + 13&lt;sub&gt;7&lt;/sub&gt; + 13&lt;sub&gt;9&lt;/sub&gt; == 8 + 10 + 12 == 30 &lt;br/&gt;&#xA; 13&lt;sub&gt;5&lt;/sub&gt; + 13&lt;sub&gt;7&lt;/sub&gt; + 15&lt;sub&gt;7&lt;/sub&gt; == 8 + 10 + 12 == 30 &lt;br/&gt;&#xA; 13&lt;sub&gt;7&lt;/sub&gt; + 13&lt;sub&gt;7&lt;/sub&gt; + 13&lt;sub&gt;7&lt;/sub&gt; == 10 + 10 + 10 == 30 &lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Plus many more if we also allow base ten on the LHS, ex. 11 + 11&lt;sub&gt;5&lt;/sub&gt; + 13 == 11 + 6 + 13 == 30&lt;/p&gt;&#xA;" OwnerUserId="1497" LastEditorUserId="1497" LastEditDate="2015-04-09T13:08:31.163" LastActivityDate="2015-04-09T13:08:31.163" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11737" PostTypeId="2" ParentId="11621" CreationDate="2015-04-08T14:56:15.300" Score="1" Body="&lt;p&gt;Every day, you ask each of the town inhabitants:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there any Jokers still alive on this island?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If they reply with &lt;code&gt;Yes&lt;/code&gt;, then you let them live.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they reply with &lt;code&gt;No&lt;/code&gt;, you kill them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rinse and repeat this process each day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know there are no more Jokers once the person you killed doesn't go green.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, all Jokers are dead, and at most one truth teller.&lt;/p&gt;&#xA;" OwnerUserId="4659" LastActivityDate="2015-04-08T14:56:15.300" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11738" PostTypeId="2" ParentId="10878" CreationDate="2015-04-08T15:33:28.073" Score="11" Body="&lt;h2&gt;Step 1: What kind of puzzle is it?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Reading the &lt;strong&gt;parchment&lt;/strong&gt; on the &lt;strong&gt;monolith&lt;/strong&gt; in &lt;strong&gt;The Center of the Physical World&lt;/strong&gt; yields the following text:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There Here be Emblems, four in Number. A Mark Each Bears, Enigmatic and Hidden; You may Discover Each by the Acrostic of Names Each Emblem Flaunts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From that text, we know the answers must be &lt;a href=&quot;http://www.merriam-webster.com/dictionary/acrostic&quot; rel=&quot;noreferrer&quot;&gt;acrostic&lt;/a&gt; (made by taking letters from words of phrases) based on the words inscribed on each emblem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, when you go north, you can read the following on the emblem of fire:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;In memoriam: D.C.&lt;/li&gt;&#xA;  &lt;li&gt;Many faiths: Constantinople&lt;/li&gt;&#xA;  &lt;li&gt;A bureaucrat's dream: New York City&lt;/li&gt;&#xA;  &lt;li&gt;Great objections: Paris&lt;/li&gt;&#xA;  &lt;li&gt;Exists no more: Alexandria&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which, when solved, yields the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;strong&gt;W&lt;/strong&gt;ashington Monument&lt;/li&gt;&#xA;  &lt;li&gt;&lt;strong&gt;H&lt;/strong&gt;agia Sophia&lt;/li&gt;&#xA;  &lt;li&gt;&lt;strong&gt;E&lt;/strong&gt;mpire State Building&lt;/li&gt;&#xA;  &lt;li&gt;&lt;strong&gt;E&lt;/strong&gt;iffel Tower&lt;/li&gt;&#xA;  &lt;li&gt;&lt;strong&gt;L&lt;/strong&gt;ibrary of Alexandria&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The acrostic solution yields WHEEL. This conforms with the solution you found in the source code. I didn't solve the East and West puzzles but the number of items in the list aligns with the characters in ascension (West / Earth) and cipher (East / Air) so it can reasonably be inferred that their solutions follow an acrostic pattern as well.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Step 2: Extract Diamond from the Elements&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;As I already mentioned in a &lt;a href=&quot;https://puzzling.stackexchange.com/questions/10878/a-lateral-thinking-puzzle-with-chemical-elements-involved-why-is-this-the-corre#comment35543_10878&quot;&gt;comment&lt;/a&gt;, we can brute force this to come up with 6 common English words that can be made from the letters in each element. This is crude, though, so there must be a more elegant solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gabriel C. Drummond-Cole pointed out in their &lt;a href=&quot;https://puzzling.stackexchange.com/questions/10878/a-lateral-thinking-puzzle-with-chemical-elements-involved-why-is-this-the-corre#comment35595_10878&quot;&gt;comment&lt;/a&gt; that the letters we need are the beginning of a short word that can be found inside the names of the element.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;io&lt;strong&gt;DINE&lt;/strong&gt;&lt;/li&gt;&#xA;  &lt;li&gt;sil&lt;strong&gt;ICON&lt;/strong&gt;&lt;/li&gt;&#xA;  &lt;li&gt;r&lt;strong&gt;ADO&lt;/strong&gt;n&lt;/li&gt;&#xA;  &lt;li&gt;bro&lt;strong&gt;MINE&lt;/strong&gt;&lt;/li&gt;&#xA;  &lt;li&gt;th&lt;strong&gt;OR&lt;/strong&gt;ium&lt;/li&gt;&#xA;  &lt;li&gt;tech&lt;strong&gt;NET&lt;/strong&gt;ium&lt;/li&gt;&#xA;  &lt;li&gt;molyb&lt;strong&gt;DEN&lt;/strong&gt;um&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of the 6 words that we brute-forced (diamond, dinette, doormen, dormice, economy, inanity), diamond is the only one for whom this property holds. I don't consider this to be a proof that it's right but it does seem to align with the &quot;so the 7 names are contained&quot; direction. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As I contain the waters of the world, so the 7 names are contained.&lt;br&gt;(And they're names, not Names.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I struggled to extract some section of the introductory text in which I could find the letters SEA, OCEAN, BED, etc. that followed this same pattern. If someone could find that, I think it can be considered solved.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Conclusion: Andrew Plotkin is a Jerk&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;As someone posted in a &lt;a href=&quot;http://www.greylabyrinth.com/discussion/viewtopic.php?p=506605#506605&quot; rel=&quot;noreferrer&quot;&gt;forum&lt;/a&gt; discussing this game, the introductory text is very misleading. However, I would like to see some kind of connection between the text and the answer. That may be finding a similar pattern in the introductory text as I have already failed to do or something entirely different from this track.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel that we have found some clues to support the answer being diamond and we brute-forced the thing to narrow it down to a list short enough to simply try every answer. However, how could it be solved without brute-force and without working backwards from the answer?&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-05-06T19:15:54.327" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11739" PostTypeId="2" ParentId="11729" CreationDate="2015-04-08T15:35:44.603" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;__+15+15 = 30&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assuming we don't need to use all the boxes &lt;/p&gt;&#xA;" OwnerUserId="11177" LastEditorUserId="8823" LastEditDate="2015-04-09T12:50:51.207" LastActivityDate="2015-04-09T12:50:51.207" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="11740" PostTypeId="2" ParentId="11729" CreationDate="2015-04-08T15:53:00.287" Score="5" Body="&lt;p&gt;I can solve this puzzle as a physicist:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 15 + 15 + 1 = 30&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I guess it's more reasonable to request a precise mathematical 30 however..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only boxes I see are the ones around 1,...,15...&lt;br&gt;&#xA;I can fill the boxes like so:&#xA;&lt;kbd&gt;10&lt;/kbd&gt;, &lt;kbd&gt;10&lt;/kbd&gt;, &lt;kbd&gt;10&lt;/kbd&gt;, &lt;kbd&gt;10&lt;/kbd&gt;, &lt;kbd&gt;10&lt;/kbd&gt;, &lt;kbd&gt;10&lt;/kbd&gt;, &lt;kbd&gt;10&lt;/kbd&gt;, &lt;kbd&gt;10&lt;/kbd&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then the problem becomes trivial.&lt;/p&gt;&#xA;" OwnerUserId="11100" LastActivityDate="2015-04-08T15:53:00.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11741" PostTypeId="2" ParentId="11729" CreationDate="2015-04-08T15:54:30.327" Score="1" Body="&lt;p&gt;I am aware that the word &lt;em&gt;equation&lt;/em&gt; implies that &lt;em&gt;L&lt;/em&gt; should be equal &lt;em&gt;R&lt;/em&gt;, but what if...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ..it is sufficient to &quot;make the equation true&quot; by filling the boxes? ..or do I need to make the equation correct as well?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1 + 1 + 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8823" LastEditorUserId="8823" LastEditDate="2015-04-09T11:51:50.813" LastActivityDate="2015-04-09T11:51:50.813" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11744" PostTypeId="1" AcceptedAnswerId="11746" CreationDate="2015-04-08T16:51:27.343" Score="23" ViewCount="2064" Body="&lt;p&gt;A prison warden offers his inmates a game for their freedom. He will secretly write a number on each of their foreheads, with no two foreheads having the same number. The inmates get to look around, seeing every other inmates' forehead, but not their own. They then go into separate rooms, and each place either a white or black hat on their own head. When they return, the prisoners are lined up in order of forehead number. They win if their hat colors now alternate white/black.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the game begins, the prisoners may not communicate, but beforehand, they may agree on a strategy. How can they guarantee victory?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This puzzle has a logical solution, there is no need for lateral thinking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added remarks:&lt;/strong&gt; The numbers aren't necessarily integers, or positive, or interrelated at all (except they are all different). The numbers are the wardens choices, while the hat colors are the prisoners' choice. I got this from &lt;a href=&quot;http://research.microsoft.com/en-us/um/people/leino/puzzles.html&quot;&gt;Rustan Leino's&lt;/a&gt; puzzle page. &lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2016-02-28T17:17:17.013" LastActivityDate="2016-02-28T17:17:17.013" Title="Alternating Hat Colors" Tags="|hat-guessing|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="11746" PostTypeId="2" ParentId="11744" CreationDate="2015-04-08T18:02:51.850" Score="20" Body="&lt;p&gt;Before the &quot;game&quot;, the prisoners decide on this strategy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Assign an integer 1, 2, ..., &lt;em&gt;n&lt;/em&gt; to each prisoner arbitrarily and call this the &quot;prisoner number&quot;. When the numbers are shown, each prisoner determines how many swaps (swapping adjacent prisoners) it would take to change the others from &quot;forehead order&quot; to &quot;prisoner order&quot;. If it takes an even number of swaps, the odd prisoners (1,3,5...) wear a white hat and the evens a black one. If it's odd, they wear the opposite hat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example, take three prisoners 1,2,3. The only possible &quot;forehead orderings&quot; are:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;123    1 sees 23(W)    2 sees 13(B)    3 sees 12(W)&#xA;132    1 sees 32(B)    3 sees 12(W)    2 sees 13(B)&#xA;231    2 sees 31(W)    3 sees 21(B)    1 sees 23(W)&#xA;213    2 sees 13(B)    1 sees 23(W)    3 sees 21(B)&#xA;312    3 sees 12(W)    1 sees 32(B)    2 sees 31(W)&#xA;321    3 sees 21(B)    2 sees 31(W)    1 sees 32(B)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;While three prisoners allows only one swap maximum, it illustrates the point (I really don't want to do a larger table...). This generalizes to more prisoners, and just means they just need to figure out if the forehead order they see is an even or odd permutation of the prisoner order.&lt;/p&gt;&#xA;" OwnerUserId="1563" LastEditorUserId="1620" LastEditDate="2015-04-08T18:14:57.363" LastActivityDate="2015-04-08T18:14:57.363" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="11747" PostTypeId="2" ParentId="11744" CreationDate="2015-04-08T18:12:13.773" Score="-1" Body="&lt;p&gt;They achieve the goal as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;They decide that all of their number except two must be killed, and once they have accomplished that grisly task the tallest survivor wears white and the other prisoner wears black.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7363" LastActivityDate="2015-04-08T18:12:13.773" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11748" PostTypeId="2" ParentId="11298" CreationDate="2015-04-08T18:15:21.540" Score="5" Body="&lt;p&gt;It was my first impulse when I read medicine to the eye and the heart &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Atropine or the plant that derives it,  Belladonna (Also known as deadly nightshade)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hear me out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am what they cherish,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Those who cherish,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Are also those who detest ardently.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I murder with ease and comfort,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Atropine is a potent toxin in high doses&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And be loved with purpose.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; in medicinal doses it is an important medicine, it's on the &lt;a href=&quot;http://en.wikipedia.org/wiki/WHO_Model_List_of_Essential_Medicines&quot; rel=&quot;noreferrer&quot;&gt;WHO Model list of essential medicines&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am a drug to the eye,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It was, tho not so much anymore, used for pupil dilation in ophthalmology. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And to the heart,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is also a treatment for bradycardia (Dangerously low heartrate)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You love me like nobody else,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Not because I murdered your loved ones,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But because I am a drug to the eye,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Belladonna at a recreational dose is a hallucinogenic. It can be consumed by squeezing the berries into your eye. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;and to the heart.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It has been used in the past as a cosmetic (love/attraction), &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I cause no disease,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it doesn't&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;but yet I abolish those who you love,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; but it's still poisonous, and may kill those using it haphazardly as a  cosmetic. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;For I am a drug to the eye,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;and to the heart!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; lost loved (and beautiful) ones. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As for the hints, this is definitely a physical thing, and (plants in the family producing it) can be found everywhere (except Antarctica) &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As a warning to anyone who may not have been aware of this before, the experience recreationally is usually reported as unpleasant, and the difference between a recreational dose and a fatal dose is scary close. Don't be an idiot.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11148" LastActivityDate="2015-04-08T18:15:21.540" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11749" PostTypeId="2" ParentId="11664" CreationDate="2015-04-08T19:16:24.223" Score="3" Body="&lt;p&gt;I see (and like!) the accepted answer, but here's what I came up with that I think is also pretty reasonable:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sub-atomic matter, particularly in the context of the atomic bomb.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the darkness, I stay hidden,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; These particles are extremely hard to detect, plus you have the whole quantum debacle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;then with the cleansing sweep,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The detonation of an atomic bomb.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I step out from the shadows, as I watch the whole world weep.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The weaponizing of an atomic bomb is very visible, but also extremely tragic if used to take lives.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was torn apart, my body bent, forever segregated.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The atom as a whole is torn apart. Not sure about the bent part. The weaponizing of atomic bombs in WW2 creates a 'permanent segregation' of sorts -- there are 'prewar' and 'postwar' eras.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Broken, I at first give life. Then as a weapon, take it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Big bang, etc., vs. weaponization of atomic energy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Elusive, you won't trap me, nor the hoard kept at my feet,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We can't ever know the position and velocity of a sub-atomic particle simultaneously, so they're elusive and 'untrappable'. The hoard at its feet could be those new-fangled particles like gluons and such.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes haunted by my twin, though cursed to never meet.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Anti-matter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Riding on your gallows, bathed in violence, I start.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, bomb stuff&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Say my name, mere mortal, before I again depart.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As more regulations and treaties are signed, the chances of future nuclear war diminishes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11154" LastActivityDate="2015-04-08T19:16:24.223" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11750" PostTypeId="2" ParentId="9850" CreationDate="2015-04-08T21:40:37.310" Score="2" Body="&lt;p&gt;&quot;The most telling feature of these lists,&quot; Brad explained at the debriefing, &quot;was the presence of the apparently-superfluous adjective in each pair. Of course, they're necessary to form the acrostic 'MALAMUTE ADDRESS IN DROPBOX THREE{S}' from the first transmission, but that's too obvious -- it's absurd to take that as being the real message...er, sorry, sir.&quot; The agent who had made the decision to pull MALAMUTE was glaring at him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;They must have been added to increase the range of some feature of the word pairs. Letter count is one such feature, but after repeated listenings of the original recording, I realized that there may also have been a reason a computer voice was used: to make a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;syllable count&lt;/strong&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;as clear as possible. Further analysis confirmed that hypothesis. There are two steps to the decoding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The target indicator is found first by transforming the list of words into a list of the &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; syllable count for each pair. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the first list, that yields: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4, 3, 5, 2, 5, 5, 5, 2, 5, 4, 3, 4, 5, 6, 5. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I squinted at this from every direction I could think of, and finally found a procedure that yielded an interesting result: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; iteratively subtracting the counts, keeping the absolute value of the previous result for the next operation. So 4 minus 3 is 1, and then 1 minus 5 -- or 5 minus 1, if you prefer -- is 4; subtract 2 and get 2, and so on.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;This is easily done counting on your fingers, which is why you noticed me sitting at my desk muttering to myself and waggling one hand about. Start with 4 fingers, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; count down 3 and you're left with 1. Each time, count down towards 0 &lt;em&gt;first&lt;/em&gt;, then back up once you hit it. Thus, counting 5 starting with 1, hide the first finger, then count the remainder of the 5 back up, &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;obtaining 4.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eager as he clearly was to get to the clincher, he nonetheless paused while the less-arithmetically-inclined members of the team mumbled and tried this out for themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Following this procedure, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the count becomes 0 at -- and &lt;em&gt;only&lt;/em&gt; at -- &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;ROUND ORIOLE in the first message. In the second message, the same is true of THOROUGH HYENA. In the third message, as I mentioned that day, &lt;strong&gt;the same indicator repeats&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The count is 0 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;at both TRIM EEL and PROUD BOBCAT; that's how I knew there were two targets. Er, thank you.&quot; He blushed as applause broke out.&lt;/p&gt;&#xA;" OwnerUserId="4204" LastActivityDate="2015-04-08T21:40:37.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11752" PostTypeId="1" AcceptedAnswerId="11753" CreationDate="2015-04-08T23:10:24.577" Score="1" ViewCount="3881" Body="&lt;p&gt;I'm using this tutorial to learn solving the cube with &quot;sexy move&quot; - &lt;a href=&quot;https://sites.google.com/site/cubetutorials101/3x3-rubik-s-cube/3x3-beginner&quot; rel=&quot;nofollow&quot;&gt;https://sites.google.com/site/cubetutorials101/3x3-rubik-s-cube/3x3-beginner&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It all goes well until the final step where it says &quot;Orient the cube so that the two unsolved corner pieces are on the Front face of the D-layer.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone experienced or familiar with this particular tutorial please take a look and explain to me what it means. My last two unsolved corners are not adjacent, how can they be both on front face and bottom face?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: now that the original question is answered, I am trying to complete the rest of the tutorial. It does not work, however. 'Sexy Moves' just scramble the cube cycling to the original 'two corners unsolved' case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, I found another very similar tutorial at &lt;a href=&quot;http://cube.crider.co.uk/beginner.php&quot; rel=&quot;nofollow&quot;&gt;http://cube.crider.co.uk/beginner.php&lt;/a&gt;, which has the same instructions to solve the last two corners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried analyzing the position with Cube explorer - no matter how I orient the cube, no amount of 'sexy move' repetitions solves either of the unsolved corners. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="11186" LastEditorUserId="11186" LastEditDate="2015-04-10T02:25:54.897" LastActivityDate="2015-04-10T02:25:54.897" Title="Question about a &quot;sexy move&quot; tutorial" Tags="|rubiks-cube|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11753" PostTypeId="2" ParentId="11752" CreationDate="2015-04-08T23:46:39.077" Score="5" Body="&lt;p&gt;This tutorial is incomplete, as it doesn't account for the case you've gotten into. There are a number of ways out of this situation, but using what you already know...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Position an unsolved corner under the buffer piece. Let &lt;code&gt;X&lt;/code&gt; denote &lt;code&gt;(R U R' U')&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;X3 D X3 D X3 D2 X3&lt;/code&gt; sets you in the correct position.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This algorithm moves three corners on the bottom layer by one spot in a triangle. Since two needed swapping, the two that are unsolved will be next to each other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then continue the tutorial as written.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorDisplayName="user20" LastEditDate="2015-04-09T16:12:01.143" LastActivityDate="2015-04-09T16:12:01.143" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11754" PostTypeId="1" AcceptedAnswerId="11778" CreationDate="2015-04-09T00:08:17.607" Score="27" ViewCount="2486" Body="&lt;p&gt;A famous art critic has recently been writing amazing essays on various paintings. The public loves these essays so much that each time the next essay is published, its painting skyrockets in fame, which leads to dramatically increased revenue for whatever gallery holds it. You work for a somewhat shady art gallery and have been tasked with finding out the subject of the critic's next essay, so that your gallery can trade for the painting while it's still not too exorbitant and then haul in the extra profits. You break into the critic's house but cannot find the name of any paintings. The critic is clearly very sneaky. You are about to call your boss and report your failure, when you see a series of images taped to the wall. After a moment of examination, you call your boss with the name of the painting. What is the painting to acquire?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/izvnt.jpg&quot; alt=&quot;&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/vECLn.jpg&quot; alt=&quot;&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/omj6r.jpg&quot; alt=&quot;&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/b6kwk.jpg&quot; alt=&quot;&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/4PrjI.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first picture shows not just Ron, nor just the mirror, but Ron looking into a mirror. What rebus significance might that have?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11154" LastEditorUserId="11154" LastEditDate="2015-04-09T10:26:55.063" LastActivityDate="2015-04-10T00:10:59.747" Title="Guess the Famous Painting" Tags="|pattern|rebus|" AnswerCount="6" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11755" PostTypeId="2" ParentId="11298" CreationDate="2015-04-09T00:14:30.963" Score="0" Body="&lt;p&gt;Not sure but I think the answer could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a love one&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; everyone is capable of love, but not everyone have that someone to love. and not all but some of those lonely hearts just detest anything love related.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I murder with ease and comfort,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; some people just get mad when they're in love, and commit serious crimes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a drug to the eye&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; seeing that love one can be just like a drug&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not because I murdered your loved ones,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; not necessarly means that she or he just killed your family, it's just that your feellings for her or him are too strong making your family to dissapear.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am everywhere.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the love of your life could be in anyplace, anytime, everywhere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11188" LastActivityDate="2015-04-09T00:14:30.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11756" PostTypeId="2" ParentId="11536" CreationDate="2015-04-09T01:04:32.563" Score="-1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;192-30-55-86-43-22-79-3 = -126 &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;So i'd guess 126&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Im still thinking that you would only say June 9th if there was a reason for us to know that it was a specific date &#xA;Same goes for the amount of money thats currently in the vault. Anyone outsider would know the date, but only the bank manager and owner would know the amount of money in the bank on that date. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So maybe the answer is 4?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11177" LastActivityDate="2015-04-09T01:04:32.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11757" PostTypeId="2" ParentId="11537" CreationDate="2015-04-09T01:25:52.680" Score="2" Body="&lt;p&gt;I think I can guess which number(s) everyone is getting wrong, so I'll take a crack at it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the number of multiples of 5 among the numbers featured in this puzzle.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;1&lt;/strong&gt; - since we're not including 2015, it's only 300 Classic FM Hall of Fame Songs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Multiply it by the number of types of creature beyond red twisted ter'angreal doorways.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;2&lt;/strong&gt; - Snakes, Foxes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Subtract the figure you now have from the number of Scheherezade's tale narrated by Poe.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;1,002&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Divide by the number of children fathered by Sidorio.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;2&lt;/strong&gt; - Grace, Connor&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add the number of entries in each Classic FM Hall of Fame.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;300&lt;/strong&gt; (per year)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Subtract the product of the number of siblings of Frodo Baggins's mother and the number of legitimate children fathered by Walder Frey.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;174&lt;/strong&gt; - 6 x 29. Gorbadoc Brandybuck had 7 children, including Frodo's mother. Walder Frey had 22 sons and 7 daughters that were trueborn.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add the number of days from now (the date of posting, 3 April) until the Easter Day celebrations in Romania.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;9&lt;/strong&gt; - The puzzle was posted on April 3. Romania celebrates Orthodox Easter &lt;em&gt;Day&lt;/em&gt; on April 12, 2015, and Orthodox Easter &lt;em&gt;Monday&lt;/em&gt; on April 13, so April 12 - April 3 - 9 days.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Subtract the number of Sitters in the &quot;Hall of the Tower&quot; in Salidar.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;18&lt;/strong&gt; -- The Hall of the Tower in Salidar only existed during the Schism, at which point the Salidar Hall had no Red Ajah (and the White Tower Hall had no Blue)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add the number of stars on the banner of Gondor.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;7&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Subtract the number of elements named after Ytterby.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;4&lt;/strong&gt; - yttrium (Y), erbium (Er), terbium (Tb), ytterbium (Yb)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Subtract the figure you now have from the user ID of the most prolific question-poster on Puzzling.SE.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;2383&lt;/strong&gt; (as of today - @warspyking)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add the number of primes greater than 10 among the numbers featured in this puzzle.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;2&lt;/strong&gt; - 29 Frey kids, user id 2383.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the number is: &lt;strong&gt;1765&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spoilery math below the line:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;br&gt;&lt;br&gt;&lt;br&gt;&lt;br&gt;&lt;br&gt;&lt;br&gt;&lt;br&gt;&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$\frac{1002 - (1\times2)}{2} = 500$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$500 + 300 - 174 + 9 - 18 + 7 - 4 = 620$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2383 - 620 + 2 = 1765$&lt;/p&gt;&#xA;" OwnerUserId="11153" LastEditorUserId="11153" LastEditDate="2015-04-09T11:23:01.643" LastActivityDate="2015-04-09T11:23:01.643" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11758" PostTypeId="1" CreationDate="2015-04-09T02:27:18.470" Score="0" ViewCount="166" Body="&lt;p&gt;There is a matriarchal town which believes that a prophetic stranger will tell the wives whether any of the husbands are cheating. The stranger will only say yes if there are any and no if there are none, he does not give any additional information. Suppose the stranger comes and says yes, then the rule is that any wife who, on any day following the announcement, deduces that her husband is unfaithful must kick her husband onto the street the next morning. This way everyone in the town knows he was cheating. Suppose each wife is able to deduce whether any man in the town is cheating, except for her own husband. No wife is willing to reveal this information and we assume cheating husbands will not reveal themselves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the stranger comes and announces that there are cheating men. On the tenth day some men are kicked into the street for the first time. How many unfaithful men are on the street?&lt;/p&gt;&#xA;" OwnerUserId="11135" LastActivityDate="2015-04-09T02:27:18.470" Title="How many cheating husbands?" Tags="|mathematics|logical-deduction|game-theory|" AnswerCount="0" CommentCount="2" ClosedDate="2015-04-09T02:35:51.653" ContentLicense="CC BY-SA 3.0" />
  <row Id="11759" PostTypeId="1" AcceptedAnswerId="11760" CreationDate="2015-04-09T05:56:40.420" Score="-3" ViewCount="298" Body="&lt;ol&gt;&#xA;&lt;li&gt;Start with the number of U.S. ounces in a U.S. gallon.&lt;/li&gt;&#xA;&lt;li&gt;Subtract the atomic number for Samarium.&lt;/li&gt;&#xA;&lt;li&gt;Add the amount of days in February on a leap year.&lt;/li&gt;&#xA;&lt;li&gt;Multiply this by 6 factorial.&lt;/li&gt;&#xA;&lt;li&gt;Divide this by pi rounded to the nearest whole.&lt;/li&gt;&#xA;&lt;li&gt;Divide this by &lt;em&gt;xx+x&lt;/em&gt; if x=4&lt;/li&gt;&#xA;&lt;li&gt;Add &lt;em&gt;xxxxxx&lt;/em&gt; if &lt;em&gt;x=y&lt;/em&gt; and &lt;em&gt;y = 2z&lt;/em&gt; and &lt;em&gt;z=3&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Multiply the amount of prime numbers under 20&lt;/li&gt;&#xA;&lt;li&gt;Add 20 factorial.&lt;/li&gt;&#xA;&lt;li&gt;Divide this by 1001110001000000000 in the base 10 system (It's in binary there) &lt;/li&gt;&#xA;&lt;li&gt;Add 30 factorial&lt;/li&gt;&#xA;&lt;li&gt;Divide this by 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 3 * 5 * 5 * 5 * 5 * 5 * 5 * 5&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What is the final answer?&lt;/p&gt;&#xA;" OwnerUserId="11002" LastActivityDate="2015-04-09T06:10:29.383" Title="Arithmetic/Knowledge Step-By-Step Number Puzzle! (What Is The Final Answer?)" Tags="|calculation-puzzle|knowledge|" AnswerCount="1" CommentCount="2" ClosedDate="2015-04-19T22:39:51.393" ContentLicense="CC BY-SA 3.0" />
  <row Id="11760" PostTypeId="2" ParentId="11759" CreationDate="2015-04-09T06:10:29.383" Score="5" Body="&lt;p&gt;This isn't so much a puzzle as it is a list of instructions to get a secret number, but I'll humour you.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Start with the number of U.S. ounces in a U.S. gallon.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are 8 ounces in a cup, 2 cups in a pint, 2 pints in a quart, and 4 quarts in a gallon, for a total of 8 &amp;times; 2 &amp;times; 2 &amp;times; 4 = 128 ounces in a gallon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Subtract the atomic number for Samarium.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Samarium has an atomic number of 62, so our result is 128 - 62 - 66.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;Add the amount of days in February on a leap year.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; February has 29 days in a leap year, so 66 + 29 = 95.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;Multiply this by 6 factorial.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 6 factorial is 720, so 95 &amp;times; 720 is 68400.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;Divide this by pi rounded to the nearest whole.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pi rounded to the nearest whole is 3, so 68400 &amp;divide; 3 = 22800.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;6&quot;&gt;&#xA;&lt;li&gt;Divide this by &lt;em&gt;xx+x&lt;/em&gt; if x=4&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 44 + 4 = 48, and 22800 divided by 48 is 475.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;7&quot;&gt;&#xA;&lt;li&gt;Add &lt;em&gt;xxxxxx&lt;/em&gt; if &lt;em&gt;x=y&lt;/em&gt; and &lt;em&gt;y = 2z&lt;/em&gt; and &lt;em&gt;z=3&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If &lt;code&gt;z = 3&lt;/code&gt;, then &lt;code&gt;y = 6&lt;/code&gt; and &lt;code&gt;x = 6&lt;/code&gt;, so we add 666666 to get 667141.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;8&quot;&gt;&#xA;&lt;li&gt;Multiply the amount of prime numbers under 20&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are eight prime numbers under 20 (2, 3, 5, 7, 11, 13, 17, 19), so we multiply by 8 to get 5337128.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;9&quot;&gt;&#xA;&lt;li&gt;Add 20 factorial.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 20 factorial is 2432902008176640000 (Wolfram Alpha), so 5337128 plus that is 2432902008181977128.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;10&quot;&gt;&#xA;&lt;li&gt;Divide this by 1001110001000000000 in the base 10 system (It's in binary there) &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1001110001000000000 in base 2 is 320000 in base 10, so our result is 7602818775568.678525.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;11&quot;&gt;&#xA;&lt;li&gt;Add 30 factorial&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 30 factorial is 265252859812191058636308480000000, so our result is 265252859812191058643911298775568.678525.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;12&quot;&gt;&#xA;&lt;li&gt;Divide this by 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 3 * 5 * 5 * 5 * 5 * 5 * 5 * 5&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That product is 480000000, so we divide our above result by that.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the final answer?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The very strange number of 552610124608731372174815.20578243474692708333333... with the 3 repeated forever and ever afterwards.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Are you sure you didn't get a few steps wrong there? The number doesn't seem very elegant. Particularly the steps that tell you to add 20 or 30 factorial were a bit strange.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-04-09T06:10:29.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11761" PostTypeId="2" ParentId="11754" CreationDate="2015-04-09T07:43:02.423" Score="3" Body="&lt;p&gt;I guess the answer may be:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;&lt;em&gt;&quot;Starry Night Over the Rhone&quot;&lt;/em&gt;&lt;/strong&gt;  a popular painting that was drawn in September 1888 by an expressionist artist, Vincent van Gogh.&lt;br&gt;&#xA;&lt;br&gt;&#xA;Reasons are:&lt;br&gt;&#xA;- Harry's friend's name is Ron, which sounds like Rhone&lt;br&gt;&#xA;- Both Moai and Vincent van Gogh belong to 20th century&lt;br&gt;&#xA;- The painting is done in an Oil canvas&lt;br&gt;&#xA;- Crow is to resemble the color of night&lt;br&gt;&#xA;- The stars look like the domino holes  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11190" LastEditorUserId="5373" LastEditDate="2015-04-09T08:25:14.340" LastActivityDate="2015-04-09T08:25:14.340" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11762" PostTypeId="1" AcceptedAnswerId="11766" CreationDate="2015-04-09T08:28:46.977" Score="8" ViewCount="846" Body="&lt;p&gt;My Indian friend Jamal recently participated in the well-known TV quiz&#xA;show &lt;strong&gt;&lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Who_Wants_to_Be_a_Millionaire&quot; rel=&quot;nofollow&quot;&gt;Who wants to be a millionaire?&lt;/a&gt;&lt;/em&gt;&lt;/strong&gt;. You might have heard or read about Jamal's spectacular success; the media reported on it extensively.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what Jamal told us about his 1.000 Euro question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The first few answers I had rattled off with almost no pause, but the&#xA;  fifth question (which would lead me to the first safe haven) took me quite a bit of thinking. I had heard of this author before, and I remembered a very negative newspaper article about him and all the &quot;pseudo-religious&quot; trash that he is producing. &#xA;  But I have not read any of his books, and in particular I hadve not read&#xA;  &lt;em&gt;that&lt;/em&gt; one, which apparently is his most famous one.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I found the last answer rather unlikely. That guy surely was not Hannibal,&#xA;  or was he?  Answer C on the other hand perhaps was a vague possibility.&#xA;  But I knew that the author does not come from Argentina but from a neighboring country, and so this answer rather smelled like a trap. The first traps &#xA;  always pop up around the 1.000 Euro level. Answers A and B were similar &#xA;  in nature  and in appearance, one located in the north and one in the south &#xA;  of the continent. All in all I slowly started to build up an ever so slight&#xA;  preference for answers A and B.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;At that moment I decided to simply use my &quot;Ask the Audience&quot; lifeline:&#xA;  bestsellers, that's exactly what these guys should know about. The outcome&#xA;  was very reassuring: 55 percent voted for A, 20 for D, 15 for B and only&#xA;  10 for C. I picked A.  Is your decision final? Yes. Is this your &#xA;  final decision? Yes. Correct!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;What was the 1.000 Euro question?&lt;br&gt;&#xA;What were the four possible answers A, B, C, D?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="9404" LastEditDate="2016-03-04T16:08:40.993" LastActivityDate="2016-03-04T16:08:40.993" Title="Slumdog millionaire (5)" Tags="|knowledge|literature|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11764" PostTypeId="2" ParentId="11762" CreationDate="2015-04-09T12:15:27.827" Score="3" Body="&lt;p&gt;I'm not quite there yet, but I think the question relates to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Paul Coelho's book The Alchemist. &#xA; Paul Coelho is Brazilian which is a neighbour of Argentina.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Something to do with...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...mountain ranges being crossed?&#xA; D: The Alps (Hannibal crossed the Alps).&#xA; A: The Atlas Mountains (the protagonist of The Alchemist crossed the Atlas mountains).&#xA; C: The Andes (which cross Argentina).  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The only problem is I don't know which answer was B. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There's nothing named similarly to the Atlas mountains in the South of Africa.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2059" LastActivityDate="2015-04-09T12:15:27.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11765" PostTypeId="2" ParentId="11754" CreationDate="2015-04-09T12:33:55.837" Score="4" Body="&lt;p&gt;My guess is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;&lt;a href=&quot;http://www.georgesseurat.org/213773/La-Parade-%281889%29-large.jpg&quot; rel=&quot;nofollow&quot;&gt;La Parade&lt;/a&gt;&lt;/em&gt; by Georges Seurat&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And my reasoning is as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ron is looking into a mirror so the answer should be a self-portrait &lt;BR&gt;&lt;br&gt;&#xA; Easter Island statues mean the self-portrait should be of the head and shoulders (Also, George Seurat painted &lt;em&gt;&lt;a href=&quot;http://www.georgesseurat.org/80361/A-Sunday-Afternoon-On-The-Island-Of-La-Grande-Jatte-large.jpg&quot; rel=&quot;nofollow&quot;&gt;A Sunday Afternoon On The Island Of La Grande Jatte&lt;/a&gt;&lt;/em&gt; and Easter always falls on a Sunday and Easter Island has a huge bowl (&lt;em&gt;La Grande Jatte&lt;/em&gt;) at the south end) &lt;br&gt;&lt;br&gt;&#xA; The oil well means it should be an oil painting&lt;br&gt;&lt;br&gt;&#xA; The crow I can't quite figure but the name Georges means farmer and farmers have to deal with crows&lt;br&gt;&lt;br&gt;&#xA; The dominoes indicate that the painting should be in the style of pointillism&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If this isn't it, it may be some different ideas of what each piece means and that could help someone else.&lt;/p&gt;&#xA;" OwnerUserId="9000" LastActivityDate="2015-04-09T12:33:55.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11766" PostTypeId="2" ParentId="11762" CreationDate="2015-04-09T13:03:05.297" Score="5" Body="&lt;p&gt;Question: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Where does Santiago, the protagonist of Paulo Coelho's bestseller &quot;The Alchemist&quot;, decide to travel?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answers: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;A&lt;/strong&gt; Sahara Desert&lt;br&gt;&#xA; &lt;strong&gt;B&lt;/strong&gt; Kalahari Desert&lt;br&gt;&#xA; &lt;strong&gt;C&lt;/strong&gt; Andes&lt;br&gt;&#xA; &lt;strong&gt;D&lt;/strong&gt; Alps&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-04-09T15:31:51.550" LastActivityDate="2015-04-09T15:31:51.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11767" PostTypeId="1" AcceptedAnswerId="11769" CreationDate="2015-04-09T13:08:32.770" Score="14" ViewCount="1204" Body="&lt;p&gt;What book / movie / show does the following refer to?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;&amp;amp;#x1f44d; {F52C090C-D3BE-48EA-AE49-6BCCF449A561}2.718 2 Winter Street&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="1681" LastEditorUserId="21234" LastEditDate="2016-07-22T21:00:52.840" LastActivityDate="2016-07-22T21:00:52.840" Title="Name the Book / Movie / Show" Tags="|cipher|rebus|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11768" PostTypeId="2" ParentId="11767" CreationDate="2015-04-09T13:33:46.693" Score="5" Body="&lt;p&gt;Partial solution&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;amp;#x1f44d;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  is the thumbs up icon. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;{F52C090C-D3BE-48EA-AE49-6BCCF449A561}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  can be split in parts of 4 which refer to (chinese ?) symbols combined with button U (some examples)&lt;br&gt;&#xA;- &lt;a href=&quot;http://www.fileformat.info/info/unicode/char/f52c/index.htm&quot; rel=&quot;nofollow&quot;&gt;f52c&lt;/a&gt;&lt;br&gt;&#xA;- &lt;a href=&quot;http://www.fileformat.info/info/unicode/char/090c/index.htm&quot; rel=&quot;nofollow&quot;&gt;090c&lt;/a&gt;&lt;br&gt;&#xA;- &lt;a href=&quot;http://www.fileformat.info/info/unicode/char/d3be/index.htm&quot; rel=&quot;nofollow&quot;&gt;d3be&lt;/a&gt;&lt;br&gt;&#xA;- &lt;a href=&quot;http://www.fileformat.info/info/unicode/char/48ea/index.htm&quot; rel=&quot;nofollow&quot;&gt;48ea&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2.718&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;   might refer to &lt;a href=&quot;http://www.mathamazement.com/Lessons/Everyday-Math/05_Miscellaneous/05_03_Important-Numbers/base-of-natural-logarithms.html&quot; rel=&quot;nofollow&quot;&gt;the base of natural logarithms&lt;/a&gt;&#xA;   also called e   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9543" LastEditorUserId="10384" LastEditDate="2015-04-09T14:16:28.470" LastActivityDate="2015-04-09T14:16:28.470" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11769" PostTypeId="2" ParentId="11767" CreationDate="2015-04-09T13:37:24.493" Score="15" Body="&lt;p&gt;&amp;amp;#x1f44d;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Code for the thumbs-up icon, or in this case, a link to a &lt;strong&gt;hitchhiker&lt;/strong&gt; (who'd show his thumb to signal passing motorists he wants a ride).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;{F52C090C-D3BE-48EA-AE49-6BCCF449A561}&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Is a random globally unique id, short: &lt;strong&gt;GUID&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2.718&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mathematical constant &lt;strong&gt;e&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pronounce as &lt;strong&gt;to&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Winter Street&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Or in swedish: Vintergatan, a name for the Milky Way &lt;strong&gt;galaxy&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus the final answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hitchhiker's guide to the galaxy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="439" LastEditorUserId="439" LastEditDate="2015-04-09T13:41:34.957" LastActivityDate="2015-04-09T13:41:34.957" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11770" PostTypeId="1" AcceptedAnswerId="11777" CreationDate="2015-04-09T13:42:53.573" Score="7" ViewCount="2491" Body="&lt;p&gt;You are given a function &lt;em&gt;f&lt;/em&gt; that, using an algorithm (that uses an old method), performs a calculation on &lt;em&gt;four&lt;/em&gt; distinct values _ and generates a sequence of &lt;em&gt;ten&lt;/em&gt; numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, your task is to figure out how the algorithm works based on the outputs received for a set of inputs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Inputs and Outputs:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;f([1, 1, 1, 1])=[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]&#xA;f([2, 2, 2, 2])=[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]&#xA;f([3, 3, 3, 3])=[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]&#xA;f([4, 4, 4, 4])=[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]&#xA;f([5, 5, 5, 5])=[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]&#xA;f([6, 6, 6, 6])=[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]&#xA;f([8, 8, 8, 8])=[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]&#xA;f([7, 7, 7, 7])=[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]&#xA;f([9, 9, 9, 9])=[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]&#xA;f([1, 2, 3, 6])=[5, 0, 1, 2, 3, 4, 5, 0, 1, 2]&#xA;f([2, 3, 6, 5])=[2, 4, 1, 3, 0, 2, 4, 1, 3, 0]&#xA;f([4, 1, 7, 9])=[2, 6, 1, 5, 0, 4, 8, 3, 7, 2]&#xA;f([5, 2, 6, 7])=[2, 0, 5, 3, 1, 6, 4, 2, 0, 5]&#xA;f([8, 1, 3, 2])=[1, 1, 1, 1, 1, 1, 1, 1, 1, 1]&#xA;f([4, 6, 5, 8])=[5, 1, 5, 1, 5, 1, 5, 1, 5, 1]&#xA;f([1, 7, 9, 44])=[16, 17, 18, 19, 20, 21, 22, 23, 24, 25]&#xA;f([1, 6, 47, 2])=[1, 0, 1, 0, 1, 0, 1, 0, 1, 0]&#xA;f([45, 87, 95, 68])=[66, 43, 20, 65, 42, 19, 64, 41, 18, 63]&#xA;f([335, 66, 48, 97])=[42, 86, 33, 77, 24, 68, 15, 59, 6, 50]&#xA;f([48, 796, 548, 1])=[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]&#xA;f([8794, 35, 3, 3])=[2, 0, 1, 2, 0, 1, 2, 0, 1, 2]&#xA;f([89, 554, 665, 7])=[5, 3, 1, 6, 4, 2, 0, 5, 3, 1]&#xA;f([54, 89, 32, 67])=[14, 1, 55, 42, 29, 16, 3, 57, 44, 31]&#xA;f([21, 65, 78, 49])=[22, 43, 15, 36, 8, 29, 1, 22, 43, 15]&#xA;f([3, 5, 11, 9])=[8, 2, 5, 8, 2, 5, 8, 2, 5, 8]&#xA;f([37, 42, 13, 128])=[31, 68, 105, 14, 51, 88, 125, 34, 71, 108]&#xA;f([48, 76, 2, 34])=[12, 26, 6, 20, 0, 14, 28, 8, 22, 2]&#xA;f([94, 57, 62, 111])=[92, 75, 58, 41, 24, 7, 101, 84, 67, 50]&#xA;f([111, 222, 333, 664])=[407, 518, 629, 76, 187, 298, 409, 520, 631, 78]&#xA;f([795, 1, 45, 3333])=[840, 1635, 2430, 3225, 687, 1482, 2277, 3072, 534, 1329]&#xA;f([897, 4, 6, 2])=[0, 1, 0, 1, 0, 1, 0, 1, 0, 1]&#xA;f([68, 749, 44, 88])=[24, 4, 72, 52, 32, 12, 80, 60, 40, 20]&#xA;f([78946, 6656, 4, 10])=[0, 6, 2, 8, 4, 0, 6, 2, 8, 4]&#xA;f([8, 1, 222, 77])=[76, 7, 15, 23, 31, 39, 47, 55, 63, 71]&#xA;f([1, 2, 3, 99999])=[5, 6, 7, 8, 9, 10, 11, 12, 13, 14]&#xA;&lt;/pre&gt;&#xA;" OwnerDisplayName="user5871" LastEditorUserId="5188" LastEditDate="2015-04-09T22:53:41.243" LastActivityDate="2015-04-09T22:53:41.243" Title="Figure out this algorithm by analyzing the sequence of numbers generated" Tags="|mathematics|pattern|number-sequence|algorithm|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11771" PostTypeId="2" ParentId="11298" CreationDate="2015-04-09T13:43:09.250" Score="2" Body="&lt;p&gt;I guess the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CO2 / Carbon dioxide&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am what they cherish,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We all intake CO2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Those who cherish,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are also those who detest ardently.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We breathe it out&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I murder with ease and comfort,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CO2 can kill&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And be loved with purpose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a drug to the eye,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And to the heart,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Here it refers to CO2 Laser used to do surgical operations on Eyes and Heart&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You love me like nobody else,&#xA;Not because I murdered your loved ones,&#xA;But because I am a drug to the eye,&#xA;and to the heart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cause no disease,&#xA;but yet I abolish those who you love,&#xA;For I am a drug to the eye,&#xA;and to the heart!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again a reference to CO2 laser and CO2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10973" LastActivityDate="2015-04-09T13:43:09.250" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11772" PostTypeId="2" ParentId="11754" CreationDate="2015-04-09T14:02:25.353" Score="2" Body="&lt;p&gt;My Guess&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Arnolfini_Portrait&quot; rel=&quot;nofollow noreferrer&quot;&gt;Arnolfini Portrait&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The mirror&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;  The mirror is a very obvious detail in the painting with a very detailed reflection (might refer to the 1st hint).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Oil&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Oil might refer to the fact is an oil canvas.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;Easter&lt;/em&gt; Island&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;em&gt;The small medallions set into the frame of the convex mirror at the back of the room show tiny scenes from the &lt;strong&gt;Passion of Christ&lt;/strong&gt; and may represent God's promise of salvation for the figures reflected on the mirror's convex surface. Furthering the Memorial theory, all the scenes on the wife's side are of Christ's death and resurrection. Those on the husband's side concern Christ's life. The mirror itself may represent the eye of God observing the vows of the wedding. A spotless mirror was also an established symbol of Mary, referring to the Holy Virgin's immaculate conception and purity.The mirror reflects two figures in the doorway, one of whom may be the painter himself. In Panofsky's controversial view, the figures are shown to prove that the two witnesses required to make a wedding legal were present, and Van Eyck's signature on the wall acts as some form of actual documentation of an event at which he was himself present.&lt;/em&gt; &lt;strong&gt;The Passion is commemorated in Holy Week, beginning on Palm Sunday and ending on &lt;em&gt;Easter Saturday&lt;/em&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Not sure about the crow and domino though.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The crow might refer to death or the fact that theres an animal in the painting. dominos to a sequence&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9543" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-09T14:11:21.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11773" PostTypeId="2" ParentId="11754" CreationDate="2015-04-09T14:26:12.490" Score="3" Body="&lt;p&gt;My guess is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rene Magritte's painting 'The false mirror'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasons:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ron and Harry are looking into the mirror of Erised. The mirror shows you what you want to see and not what is necessarily real. So it can be interpreted as a false mirror.&#xA; Easter island is also known as  'Mata ki the rangi', which means &quot;Eyes looking to the sky&quot; (source: Wikipedia). The painting 'The false mirror' has an eye looking at the sky.&#xA; The oil well can be tied to the fact that the painting is oil on canvas. &#xA; I am not sure though how the crow/raven and the dominoes tie into this though.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4903" LastActivityDate="2015-04-09T14:26:12.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11774" PostTypeId="2" ParentId="11754" CreationDate="2015-04-09T14:33:37.540" Score="13" Body="&lt;p&gt;I believe I would have my boss go acquire&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Norman Rockwell's Freedom From Want&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Overarching Rationale&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First, a universal sort of clue. The pictures identifiable as so are horizontally flipped, or 'backwards' if you will. The Potter image should be obvious to most. The crow (more on this later) is facing the wrong direction. The oil well and Moai are further verifiable as horizontally flipped from their originals. I therefore choose to read the sequence of pictures from last to first. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The dominoes are not. Look closely. They are actually pills. For those less familiar with street drugs, they are probably ecstasy (MDMA), also known as Molly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Continuing..&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The bird is a logo for an old card game called Rook. Combine that with the oil WELL. Rookwell, close enough to Rockwell for our purposes. Combining with the previous clue, we find Molly Rockwell. Nickname of the wife of Norman Rockwell.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The painting..&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Skipping the Moai for just a moment. Consider The Mirror of Erised, in keeping with the backwards theme, shall be rendered desire. Desire is a synonym for 'Want'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hocus Pocus&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I've got to be honest here, I'm not coming up with a lot of connection or compelling justification to link the Easter Island statues with 'Freedom From'. If I had chosen 'Triple Self-Portrait', I would have to accept a simpler explanation for mirror and somehow link Triple to the Moai, something I am not prepared to do. I'm less unhappy linking Easter, understood tenuously through the resurrection of Jesus and /freedom from/ sin. Anyway, I think all the other stuff is reasonably good.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11119" LastEditorUserId="11119" LastEditDate="2015-04-09T14:57:11.427" LastActivityDate="2015-04-09T14:57:11.427" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11775" PostTypeId="2" ParentId="11770" CreationDate="2015-04-09T14:40:12.853" Score="3" Body="&lt;p&gt;This function delivers&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an arithmetic sequence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first element of the parameter is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the number to add on each iteration.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second element of the parameter is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the initial value of our iterator.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The third element of the parameter is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the initial offset of the sequence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The fourth element of the parameter is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the modulus.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The function is described in code as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;pre&gt;&lt;code&gt;def f(x):&#xA;  r = []&#xA;  for i in range(x[1],x[1]+10):&#xA;    r += [(x[2] + (i*x[0]))%x[3]]&#xA;  return r&#xA;&lt;/code&gt;&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-04-09T15:52:09.630" LastActivityDate="2015-04-09T15:52:09.630" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11776" PostTypeId="2" ParentId="11298" CreationDate="2015-04-09T14:56:45.693" Score="1" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Earth / dirt / land&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am what they cherish,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many tribes do.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Those who cherish,&#xA;&lt;br&gt;&#xA;Are also those who detest ardently.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People are poising the Earth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I murder with ease and comfort,&#xA;&lt;br&gt;&#xA;And be loved with purpose.&lt;/em&gt;&#xA;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People are buried in the Earth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am a drug to the eye,&#xA;&lt;br&gt;&#xA;And to the heart,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The nature on Earth is the most beautiful thing isn't it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You love me like nobody else,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That's true.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Not because I murdered your loved ones,&#xA;&lt;br&gt;&#xA;But because I am a drug to the eye,&#xA;&lt;br&gt;&#xA;and to the heart.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Earth is just lovely&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I cause no disease,&#xA;&lt;br&gt;&#xA;but yet I abolish those who you love,&#xA;&lt;br&gt;&#xA;For I am a drug to the eye,&#xA;&lt;br&gt;&#xA;and to the heart!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is just a fill&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11145" LastEditorUserId="11145" LastEditDate="2015-04-09T20:38:50.013" LastActivityDate="2015-04-09T20:38:50.013" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11777" PostTypeId="2" ParentId="11770" CreationDate="2015-04-09T15:14:24.390" Score="5" Body="&lt;p&gt;Let's call $A,B,C,D$ the parameters of the function.&#xA;The first term of the sequence is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $((A\times B)+C)$mod $D$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Each consequent term is obtained  from the previous this way:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Add $A$ then modulo $D$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The $n$th term of the sequence can be expressed as:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(A\times (B+n)+C)mod\ D$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you need a proof of the equivalence of the two definitions, just ask in the comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f([89, 554, 665, 7])=[5, 3, 1, 6, 4, 2, 0, 5, 3, 1]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The first term is  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$(89\times 554+665) mod\ 7 =5$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second is  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(5+89)mod\ 7=3$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the $n$th is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(89\times (554+n)+665)mod\ 7$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-04-09T15:30:14.500" LastActivityDate="2015-04-09T15:30:14.500" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11778" PostTypeId="2" ParentId="11754" CreationDate="2015-04-09T15:35:31.803" Score="28" Body="&lt;p&gt;My answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;The Rookie&quot; by Norman Rockwell&lt;br&gt;&#xA; &lt;a href=&quot;http://en.wikipedia.org/wiki/File:Norman_Rockwell_-_The_Rookie.jpg&quot;&gt;http://en.wikipedia.org/wiki/File:Norman_Rockwell_-_The_Rookie.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My reasoning is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The rebus encodes the words &lt;strong&gt;&quot;Nor Man-Rock well rook IE&quot;&lt;/strong&gt;&lt;br&gt;&#xA; 1. RON in mirror becomes NOR&lt;br&gt;&#xA; 2. The Easterisland head is a MAN ROCK&lt;br&gt;&#xA; 3. The oil well is the WELL&lt;br&gt;&#xA; 4. The bird is a ROOK&lt;br&gt;&#xA; 5. Domino hints at Internet explorer and IE (Domino 8.5.3 Fix Pack 4 or Domino 9 supporting IE 10) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2015-04-09T15:35:31.803" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11779" PostTypeId="2" ParentId="11656" CreationDate="2015-04-09T15:40:40.883" Score="1" Body="&lt;p&gt;&quot;Four&quot; is the pigeonese word for &quot;zero&quot;. There is only one pigeon, who is simultaneously the first, last and middle pigeon.&lt;/p&gt;&#xA;" OwnerUserId="2021" LastActivityDate="2015-04-09T15:40:40.883" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11780" PostTypeId="1" AcceptedAnswerId="23675" CreationDate="2015-04-09T15:52:40.963" Score="40" ViewCount="4599" Body="&lt;p&gt;A sudoku net is a 9x9 grid where every cell is either (blank) or has a cross (block).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A random solved sudoku is chosen. The net is placed on top of the sudoku (in any of the 4 ways). We can only see the cells that have a blank over them and the cells with a block over them are covered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A sudoku net is defined as 'solvable' if we can deduce the unique solution of the sudoku just by looking at the visible cells. This should be true in all cases, irrespective of the selected sudoku.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Create the sudoku net with maximum possible blocks (or crosses) such that this sudoku net is solvable for &lt;strong&gt;all&lt;/strong&gt; sudoku puzzles. (Any arrangement of visible cells that result in a valid sudoku puzzle.)&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4269" LastEditDate="2017-04-28T09:28:20.733" LastActivityDate="2017-04-28T09:28:20.733" Title="Sudoku net that is always solvable" Tags="|optimization|sudoku|" AnswerCount="11" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="11781" PostTypeId="1" CreationDate="2015-04-09T16:11:33.473" Score="6" ViewCount="853" Body="&lt;p&gt;There are 4 pieces on the board, the white and black kings, as well as 2 white bishops (on differently colored squares).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way a stalemate occurs is if the black king is not in check, but cannot make a move. (No threefold repetition, timer, etc.) Remaining rules are same as chess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming perfect play by both players, give the position of the board where white wins, but requires the maximum possible moves to do a checkmate.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-04-11T19:29:19.023" Title="2 bishops versus a lone king" Tags="|strategy|chess|game|checkerboard|board-games|" AnswerCount="1" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="11782" PostTypeId="2" ParentId="11780" CreationDate="2015-04-09T16:29:06.143" Score="3" Body="&lt;p&gt;To start it off with a very low-scoring guess, I can get at least 21. The format is:&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;.#.|...|...&#xA;###|.#.|.#.&#xA;.#.|...|...&#xA;---+---+---&#xA;...|.#.|...&#xA;.#.|###|.#.&#xA;...|.#.|...&#xA;---+---+---&#xA;...|...|.#.&#xA;.#.|.#.|###&#xA;...|...|.#.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The six blocks that are alone in their 3x3 square can automatically be filled in. Next, each of the blocks in the cross shape can be filled in as they are the only blocks in their respective row or column. Finally, the remaining 3 squares can be filled in.&lt;/p&gt;&#xA;" OwnerUserId="1664" LastActivityDate="2015-04-09T16:29:06.143" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11784" PostTypeId="1" AcceptedAnswerId="11787" CreationDate="2015-04-09T17:08:23.953" Score="5" ViewCount="368" Body="&lt;p&gt;Maybe one word, or maybe two,&lt;br&gt;&#xA;Read these lines, find the clues.&lt;br&gt;&#xA;Determination must first start,&lt;br&gt;&#xA;Sometimes it leads to starting art.&lt;br&gt;&#xA;No matter where it goes, one must first try,&lt;br&gt;&#xA;Without the start of all effort, we can't fly.&lt;br&gt;&#xA;The beginning of the beginning is the best,&lt;br&gt;&#xA;To plan to acheive, or start another test.&lt;br&gt;&#xA;Make sure when planning to end with thoughts,&lt;br&gt;&#xA;Examine, begin and end to connect the dots.&lt;br&gt;&#xA;So riddle it out, don't hold back.&lt;br&gt;&#xA;Tell what I am, don't queue on a stack.  &lt;/p&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-04-09T17:35:04.677" Title="Motivational but Not - What am I?" Tags="|riddle|word|rhyme|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11785" PostTypeId="2" ParentId="11780" CreationDate="2015-04-09T17:15:33.823" Score="13" Body="&lt;p&gt;I think you can do &lt;strong&gt;27&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;###|...|...&#xA;###|...|...&#xA;###|...|...&#xA;---+---+---&#xA;...|###|...&#xA;...|###|...&#xA;...|###|...&#xA;---+---+---&#xA;...|...|###&#xA;...|...|###&#xA;...|...|###&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;... Assuming that a completely empty 3x3 box can always be solved as long as there are four completely full 3x3 boxes in the same major row or column.&lt;/p&gt;&#xA;" OwnerUserId="1497" LastActivityDate="2015-04-09T17:15:33.823" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11786" PostTypeId="2" ParentId="11780" CreationDate="2015-04-09T17:27:46.220" Score="31" Body="&lt;p&gt;Building off of Kevin's answer, I can get it to &lt;strong&gt;33&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;###|.#.|.#.&#xA;###|...|...&#xA;###|...|...&#xA;---+---+---&#xA;.#.|###|.#.&#xA;...|###|...&#xA;...|###|...&#xA;---+---+---&#xA;.#.|.#.|###&#xA;...|...|###&#xA;...|...|###&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Keeping his assumption &quot;that a completely empty 3x3 box can always be solved as long as there are four completely full 3x3 boxes in the same major row or column.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we can solve for the single missing numbers in each square.&lt;/p&gt;&#xA;" OwnerUserId="7715" LastActivityDate="2015-04-09T17:27:46.220" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11787" PostTypeId="2" ParentId="11784" CreationDate="2015-04-09T17:31:28.633" Score="6" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DATABASE.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Maybe one word, or maybe two,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Database or data base?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Read these lines, find the clues.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's another letterplay riddle! Aggie, I'm getting too used to your style ;-)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Determination must first start,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first letter of &quot;determination&quot; is &lt;strong&gt;D&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Sometimes it leads to starting art.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first letter of &quot;art&quot; is &lt;strong&gt;A&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;No matter where it goes, one must first try,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first letter of &quot;try&quot; is &lt;strong&gt;T&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Without the start of all effort, we can't fly.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first letter of &quot;all effort&quot; is &lt;strong&gt;A&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The beginning of the beginning is the best,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first letter of &quot;beginning&quot; is &lt;strong&gt;B&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;To plan to acheive, or start another test.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first letter of &quot;another test&quot; is &lt;strong&gt;A&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Make sure when planning to end with thoughts,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The last letter of &quot;thoughts&quot; is &lt;strong&gt;S&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Examine, begin and end to connect the dots.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first and last letter of &quot;examine&quot; are both &lt;strong&gt;E&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;So riddle it out, don't hold back.&lt;br&gt;&#xA; Tell what I am, don't queue on a stack.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No queue of answers on &lt;em&gt;Stack&lt;/em&gt; Exchange for this one: I got the answer as soon as I looked at it!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-04-09T17:35:04.677" LastActivityDate="2015-04-09T17:35:04.677" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11788" PostTypeId="1" AcceptedAnswerId="11791" CreationDate="2015-04-09T18:03:54.067" Score="2" ViewCount="1944" Body="&lt;p&gt;Alone I am blank and meaningless,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no end of uselessness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But together with man's thought and ingenuity,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no limit to versatility.&lt;/p&gt;&#xA;" OwnerUserId="11204" LastActivityDate="2017-03-18T16:56:50.467" Title="I am blank and meaningless - What am I?" Tags="|riddle|" AnswerCount="7" CommentCount="5" ClosedDate="2015-04-15T08:39:55.887" ContentLicense="CC BY-SA 3.0" />
  <row Id="11789" PostTypeId="2" ParentId="11788" CreationDate="2015-04-09T18:08:12.237" Score="2" Body="&lt;p&gt;It could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Idea&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alone I am blank and meaningless,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no end of uselessness.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An idea is useless if nothing supports it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But together with man's thought and ingenuity,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no limit to versatility.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; On the contrary, an idea supported by cleverness can be very very useful!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-04-09T18:08:12.237" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11790" PostTypeId="2" ParentId="11788" CreationDate="2015-04-09T18:09:51.387" Score="2" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the mind&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alone I am blank and meaningless, I have no end of uselessness.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The mind by itself is nothing and useless.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But together with man's thought and ingenuity, I have no limit to versatility.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Once a man fills his mind with thoughts, he can accomplish many things.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2015-04-09T18:09:51.387" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11791" PostTypeId="2" ParentId="11788" CreationDate="2015-04-09T18:11:50.313" Score="6" Body="&lt;p&gt;I guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Paper&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's blank and useless until someone writes brilliant or useful things on it. The written word is extremely versatile.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11154" LastActivityDate="2015-04-09T18:11:50.313" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11792" PostTypeId="2" ParentId="11788" CreationDate="2015-04-09T18:17:56.773" Score="4" Body="&lt;p&gt;I guess it is    &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Zero (0)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which by itself is meaningless, but has a very versitle use in mathematics  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;(addition subtraction multiplication)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11156" LastEditorUserId="9412" LastEditDate="2015-04-09T19:36:53.957" LastActivityDate="2015-04-09T19:36:53.957" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11794" PostTypeId="1" AcceptedAnswerId="11796" CreationDate="2015-04-09T19:31:57.720" Score="-2" ViewCount="368" Body="&lt;p&gt;This movie was released in 2014, its about &quot;E.T&quot; or in the movie, &quot;Eee Tee&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main star is a &lt;a href=&quot;http://www.youtube.com/user/JamesNintendoNerd&quot; rel=&quot;nofollow&quot;&gt;Youtuber&lt;/a&gt; who reviews bad games.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the movie? (Please do not post answer as comment)&lt;/p&gt;&#xA;" OwnerUserId="11206" LastEditorUserId="5373" LastEditDate="2015-04-09T19:45:29.993" LastActivityDate="2015-04-09T19:45:29.993" Title="Name This Movie" Tags="|movies|" AnswerCount="1" CommentCount="3" ClosedDate="2015-04-09T19:48:30.563" ContentLicense="CC BY-SA 3.0" />
  <row Id="11796" PostTypeId="2" ParentId="11794" CreationDate="2015-04-09T19:35:59.097" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Angry_Video_Game_Nerd:_The_Movie&quot; rel=&quot;nofollow&quot;&gt;Angry Video Game Nerd: The Movie&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The film was produced in 2014 by &lt;a href=&quot;http://en.wikipedia.org/wiki/James_Rolfe_(filmmaker)&quot; rel=&quot;nofollow&quot;&gt;James Rolfe&lt;/a&gt;, also partner of Cinemassacre.&#xA;The plot is about the legendary (actually confirmed) burial of tons of unsold E.T. games of Atari 2600.&lt;/p&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-04-09T19:35:59.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11797" PostTypeId="1" AcceptedAnswerId="11800" CreationDate="2015-04-09T20:03:15.950" Score="6" ViewCount="30922" Body="&lt;p&gt;I was wondering if it's possible to solve a Rubik's cube just iterating a single algorithm. It doesn't matter how complex the algorithm is (as long as it is less than 1 million moves), nor the time required to solve the cube; I just want to know if there exists a &quot;periodic&quot; sequence of moves that allows to turn any valid scrambled configuration to the solved configuration.&#xA;Once you start executing the algorithm, you can't stop it (unless you finish, of course), rotate the cube and start it on another face. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the question is still unclear, let's make an example. You have to find a sequence like this one, to be repeated an unlimited number of times, that lets you solve any configuration:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Example: U D R' U' F L' U F' R' U' L D' R  to be repeated 12412183213 times!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: The above sequence is just an example, 99,999% it's not working!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I just ask you to prove whether such sequence exists or not.&lt;/strong&gt; If it exists, I would appreciate you to post it, unless it consists of more than 1000 moves.&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="463" LastEditDate="2015-04-09T22:31:05.230" LastActivityDate="2019-09-06T09:43:35.233" Title="Solving Rubik's cube with one algorithm" Tags="|strategy|rubiks-cube|mechanical-puzzles|solvability|" AnswerCount="3" CommentCount="7" ClosedDate="2015-04-10T00:48:50.537" ContentLicense="CC BY-SA 3.0" />
  <row Id="11798" PostTypeId="2" ParentId="4907" CreationDate="2015-04-09T20:39:35.947" Score="2" Body="&lt;p&gt;Oy vey,  here goes...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; EDM Rave! (music style and culture - electronic dance music event)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Backed by a Benjamin,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; EDM raves are a booming business and cost money to put on.  Benjamin is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Slang_terms_for_money&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Slang_terms_for_money&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Used as an insult: I just can't win.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lyrics in some EDM / Techo music contains slurs against homosexuals, black people, women, and just whatever you name it.  There is even insult style lyrics directed to insult pretty much anyone.  You can't win, because it's already in the lyrics, no matter what you say the song is already made and you can't change its course.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You might be surprised to find rocks in me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some EDM has elements Rock-N-Roll.  You could also say it &quot;Rocks&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Just as shocked as when you find me in a tree.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People often climb trees during outdoor EDM events.  First time you see this, you might be quite shocked :P&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can get high when I get wild,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people definitely get high and wild during music events.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But making you nod is more my style.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People nod to the music&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am dark and I am light;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Raves often feature dark spaces.  They also feature strobe lighting - flashes of light and then darkness.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;cold tomorrow but hot tonight.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Music is hot tonight - everybody is active, and dancing, but tomorrow you are out cold possibly with a hangover&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11209" LastActivityDate="2015-04-09T20:39:35.947" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11799" PostTypeId="2" ParentId="11797" CreationDate="2015-04-09T20:51:02.970" Score="1" Body="&lt;p&gt;So, you want a general solution (if it exists), meaning a solution that is independent of the current state of the cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Common strategies for solving the cube depend heavily on the current state of the cube. There are around 43 quintillion possible states. Each move brings into view 1 more state, so it seems impossible that a general solution could exist with less than 1 million moves, because you could never even see 1 million states, let alone 43 quintillion.&lt;/p&gt;&#xA;" OwnerUserId="463" LastActivityDate="2015-04-09T20:51:02.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11800" PostTypeId="2" ParentId="11797" CreationDate="2015-04-09T20:53:58.977" Score="12" Body="&lt;p&gt;Does an algorithm exist?&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Yes&lt;/strong&gt;. Consider every valid state of the Rubik's cube. It can be brought to the solved state in 20 moves or less. For each state, apply the sequence of moves followed by its inverse. This giant algorithm is guaranteed to solve any cube.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Now, does a reasonable-length algorithm exist?&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;No&lt;/strong&gt;. I will show that any such algorithm should have length at least &lt;strong&gt;34326986725785600&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;An algorithm of length L, even if applied an infinite number of times, will only go though 1260L states at most. Applying the algorithm once takes it to at most L states. The important thing is that every element in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rubik%27s_Cube_group&quot; rel=&quot;noreferrer&quot;&gt;Rubik's cube group&lt;/a&gt; has order 1260 or less. As a result, even if you apply the algorithm any number of times, you will not be able to reach more than 1260L states.&lt;/p&gt;&#xA;&lt;p&gt;Now, there are 43252003274489856000 valid states of the Rubik's cube. This means that any algorithm which passes through all these states should have length at least 43252003274489856000/1260=34326986725785600&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9471" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-09T21:25:29.130" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11801" PostTypeId="2" ParentId="11788" CreationDate="2015-04-09T21:00:57.423" Score="2" Body="&lt;p&gt;I realize it's already been answered, but my first thought was actually&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Whitespace (the characters, not the language), since it is blank and meaningless until some clever programmer comes along and makes a program with it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11210" LastActivityDate="2015-04-09T21:00:57.423" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11802" PostTypeId="2" ParentId="11797" CreationDate="2015-04-09T21:06:36.910" Score="9" Body="&lt;p&gt;The answer is no for &quot;iterating&quot;/repeating a single algorithm because the maximum order (of a permutation) of an algorithm is 1260, which is not 43 quintillion.  In other words, no one move sequence can be repeated more than 1260 times before the cube is restored to the original state it was in before iterating that algorithm.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;However, the answer is yes for &quot;executing&quot; a single algorithm.  That is, Bruce Norskog constructed a 43 quintillion face quarter turn move algorithm which will go through every single 3x3x3 position exactly once upon completion of execution.&#xA;&lt;a href=&quot;http://bruce.cubing.net/ham333/rubikhamiltonexplanation.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://bruce.cubing.net/ham333/rubikhamiltonexplanation.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In addition, one can use a J-Perm such as &lt;a href=&quot;https://alg.cubing.net/?alg=B-_U_F-_U2_B_U-_B-_U2_F_B_U-&amp;amp;view=playback&quot; rel=&quot;nofollow noreferrer&quot;&gt;B' U F' U2 B U' B' U2 F B U'&lt;/a&gt; to construct the move U, for example, by inserting cube rotations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;U =&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B' U F' U2 B U' B' U2 F B U'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;y&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B' U F' U2 B U' B' U2 F B U'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;y&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B' U F' U2 B U' B' U2 F B U'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;y2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://alg.cubing.net/?alg=B-_U_F-_U2_B_U-_B-_U2_F_B_U-_y_B-_U_F-_U2_B_U-_B-_U2_F_B_U-_y_B-_U_F-_U2_B_U-_B-_U2_F_B_U-_y2&amp;amp;view=playback&quot; rel=&quot;nofollow noreferrer&quot;&gt;Animation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And thus it follows that if we allow only cube rotations in between this one move sequences, we can create any of the other (6)(3) = 18 different face turn moves to be able to express any move sequence (any solution) in terms of a single J-perm algorithm and cube rotations.&lt;/p&gt;&#xA;" OwnerUserId="6044" LastEditorUserId="56416" LastEditDate="2019-09-06T09:43:35.233" LastActivityDate="2019-09-06T09:43:35.233" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="11803" PostTypeId="2" ParentId="11780" CreationDate="2015-04-09T21:38:30.660" Score="3" Body="&lt;p&gt;I think I can do &lt;strong&gt;35&lt;/strong&gt; (building on 33 solution above). Logic-as long as unknowns in the (r1, C2) square doesn't share the unknown values in same row/col with (r1, c3) and (r3,c2), the indeterminism of these 2 unknowns can be resolved when the (r1, c1) square is done. Same applies to square (r2, c3).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;###|...|..#&#xA;###|...|...&#xA;###|##.|...&#xA;---+---+---&#xA;#..|###|...&#xA;...|###|..&#xA;...|###|##.&#xA;---+---+---&#xA;.#.|...|###&#xA;...|...|###&#xA;...|..#|###&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="11156" LastEditorUserId="11156" LastEditDate="2015-04-10T17:51:55.173" LastActivityDate="2015-04-10T17:51:55.173" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11804" PostTypeId="2" ParentId="11788" CreationDate="2015-04-09T21:38:55.587" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A blank &lt;strong&gt;check&lt;/strong&gt; that's been voided. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Alone I am blank and meaningless&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Voided;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I have no end of uselessness.&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;But together with man's thought and ingenuity&lt;/em&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Re. &lt;em&gt;&lt;a href=&quot;http://www.imdb.com/title/tt0264464/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Catch Me If You Can&lt;/a&gt;&lt;/em&gt;, (the story of a master forger)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I have no limit to versatility&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This is one of the preferred ways to transfer funds with no upper limit, but in the hands of Frank Abagnale Jr, there's no limit whatsoever to its versatility. This would lead me to simply &lt;em&gt;paper&lt;/em&gt;, except that completely blank paper actually has no end to its &lt;em&gt;useful&lt;/em&gt;ness. Voided checks are less then useless to most, and a liability in the wrong hands.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4522" LastEditorUserId="5373" LastEditDate="2017-03-18T16:56:50.467" LastActivityDate="2017-03-18T16:56:50.467" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11805" PostTypeId="2" ParentId="11298" CreationDate="2015-04-09T23:11:44.527" Score="0" Body="&lt;p&gt;My guess &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fat (Fatty food)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am what they cherish,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those who cherish,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People love fatty food for its taste&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Are also those who detest ardently.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Those who understand the ill-effects also cherish&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I murder with ease and comfort,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It kills slowly&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And be loved with purpose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a drug to the eye,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fat-rich food looks/tastes good&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And to the heart,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Good Fat(LDL) is a drug to heart&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You love me like nobody else,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not because I murdered your loved ones,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But because I am a drug to the eye,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and to the heart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cause no disease,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but yet I abolish those who you love,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fat itself doesn't cause any disease, but causes other diseases, which are dangerous&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For I am a drug to the eye,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and to the heart!&lt;/p&gt;&#xA;" OwnerUserId="11156" LastEditorUserId="4312" LastEditDate="2015-04-10T06:07:49.310" LastActivityDate="2015-04-10T06:07:49.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11807" PostTypeId="2" ParentId="11298" CreationDate="2015-04-10T00:03:15.120" Score="1" Body="&lt;p&gt;I am thinking something along the lines of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;war&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;It is the drug to the eye as people love to watch war, humans have always been fascinate by it.&lt;br&gt;&#xA;&lt;br&gt;&#xA;The ease and comfort more so is a reflection of the fact that war kills swiftly and with no resistance.&lt;br&gt;&#xA;&lt;br&gt;&#xA;It causes no disease, and it indeed is almost everywhere.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11216" LastEditorUserId="5188" LastEditDate="2015-04-10T05:16:33.883" LastActivityDate="2015-04-10T05:16:33.883" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11810" PostTypeId="2" ParentId="12" CreationDate="2015-04-10T03:13:52.617" Score="2" Body="&lt;p&gt;I do a lot of sudokus.  In at least 3 cases over the last few years, I found a puzzle that had more than one solution.  It was an either/or type of thing.  I had to choose which one of a pair of numbers was going to be the &quot;solution&quot; number.  From there, I could solve the puzzle.  I actually saved one puzzle and ran it both ways.  It checked out each way.  Very,very rare, but it happens.&lt;/p&gt;&#xA;" OwnerUserId="11223" LastActivityDate="2015-04-10T03:13:52.617" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11811" PostTypeId="1" AcceptedAnswerId="11860" CreationDate="2015-04-10T03:49:48.817" Score="11" ViewCount="1723" Body="&lt;h1&gt;&lt;strong&gt;About Me&lt;/strong&gt;&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Pros:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;I always let my friends go first.&lt;/li&gt;&#xA;  &lt;li&gt;To some of my groups of friends, I mean every.&lt;/li&gt;&#xA;  &lt;li&gt;No matter who I am around, I always have an effect on them.&lt;/li&gt;&#xA;  &lt;li&gt;I am strong-willed; none of my friends can change me.&lt;/li&gt;&#xA;  &lt;li&gt;I have a magnetic personality, so I am always near friends.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cons:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;I'm short.&lt;/li&gt;&#xA;  &lt;li&gt;I do not like to be alone- I am clingy.&lt;/li&gt;&#xA;  &lt;li&gt;I never accept some groups of my friends for who they are, and instead, I try to change them, and always succeed.&lt;/li&gt;&#xA;  &lt;li&gt;I'm bossy- usually the one telling my friends when, where, how, why, and under what circumstances things are going to happen.&lt;/li&gt;&#xA;  &lt;li&gt;I have 2 personalities- one is almost a teenager, and the other is in his mid twenties&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The title is a hint. Don't overlook it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-06-17T20:11:53.150" LastActivityDate="2015-06-17T20:11:53.150" Title="You might find me in a costly lawsuit. What am I?" Tags="|riddle|" AnswerCount="9" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11812" PostTypeId="2" ParentId="11781" CreationDate="2015-04-10T06:36:32.857" Score="10" Body="&lt;p&gt;The answer is 19 moves.&#xA;The first link in a search for &quot;two bishops longest mate&quot; shows this site: &lt;a href=&quot;http://www.gilith.com/chess/endgames/kbb_k.html&quot;&gt;http://www.gilith.com/chess/endgames/kbb_k.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Checking the Nalimov tablebase on this site: &lt;a href=&quot;http://chessok.com/?page_id=361&quot;&gt;http://chessok.com/?page_id=361&lt;/a&gt;&#xA;shows that the position indeed requires 19 moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The position is: White: Ba4, Bd8, Ka1, Black: Kb4.&lt;/p&gt;&#xA;" OwnerUserId="5684" LastActivityDate="2015-04-10T06:36:32.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11813" PostTypeId="1" AcceptedAnswerId="11819" CreationDate="2015-04-10T07:17:07.940" Score="1" ViewCount="197" Body="&lt;p&gt;John the jeweler is planning to make bangles from solid 14 karat gold. As shown below, his design will:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;be available in 5 different diameters (D)&lt;/li&gt;&#xA;&lt;li&gt;have a height (h) of 1 inch for all bangles&lt;/li&gt;&#xA;&lt;li&gt;have an inside profile that is straight&lt;/li&gt;&#xA;&lt;li&gt;have an outside profile with a radius of curvature (r) equal to D/2  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;John has fabricated a size XS prototype and found that it weighs 120 grams. John knows the density of 14 karat gold is 14 grams/cc, but is not sure how to calculate the weights of the other four bangles. He needs this information to determine the fabrication costs.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you help John by providing the weights of the other bangles with larger diameters?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&lt;strong&gt;This may look like a math textbook problem, however it is a variation of a classic geometry puzzle.&lt;/strong&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nhRWi.png&quot; alt=&quot;Bangle&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5188" LastActivityDate="2015-04-10T11:11:41.887" Title="What are the weights of the bangles?" Tags="|geometry|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11814" PostTypeId="1" AcceptedAnswerId="11818" CreationDate="2015-04-10T07:18:19.830" Score="6" ViewCount="785" Body="&lt;p&gt;Snow White needs to send a message to Prince Charming. She considers telling this message to each of the Seven Dwarves, then sending them to deliver it to the Prince. However, the road to the castle is treacherous, and there is risk that agents of the evil Queen might intercept some of the dwarves. So, Snow White wants to say something to the dwarves so that...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;As long as any 4 dwarves reach Prince Charming, they can tell him the message, but&lt;/li&gt;&#xA;&lt;li&gt;If at most 3 are captured, the Queen cannot learn &lt;strong&gt;anything&lt;/strong&gt; about the message, no matter how hard she interrogates the poor dwarves.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What should Snow White tell the dwarves?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Remarks:&lt;/strong&gt; Think of the message as a number. Snow White has access to a fair coin. You don't need any fancy cryptography protocols to solve this: there are several elegant solutions using very basic math.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-06-03T02:13:40.047" LastActivityDate="2015-06-03T02:13:40.047" Title="Snow White and the Secret Message (Secret Sharing Problem)" Tags="|mathematics|cryptography|information-theory|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11815" PostTypeId="2" ParentId="11788" CreationDate="2015-04-10T07:43:54.690" Score="1" Body="&lt;p&gt;It's:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A computer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alone I am blank and meaningless,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No software -&gt; blank screen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no end of uselessness.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Without proper programming.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But together with man's thought and ingenuity,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Programming ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no limit to versatility.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Automated flight booking, GPS navigation, web search ... the list is endless.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11227" LastActivityDate="2015-04-10T07:43:54.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11816" PostTypeId="2" ParentId="3048" CreationDate="2015-04-10T09:30:12.550" Score="2" Body="&lt;p&gt;The best-known type of probability-based Knight-Knave puzzle is one where there are not only Knights and Knaves but also Jokers, who lie or tell the truth at random. A fairly standard example is your own &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2484/knight-knave-and-joker-puzzle&quot;&gt;Knight, Knave and Joker puzzle&lt;/a&gt;. More abnormal variants include COTO's &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4533/automatically-a-knight-knave-and-joker&quot;&gt;Automatically a Knight, Knave, and Joker&lt;/a&gt;, my own &lt;a href=&quot;https://puzzling.stackexchange.com/questions/10289/lies-damned-lies-and-statistics&quot;&gt;Lies, damned lies, and statistics&lt;/a&gt;, and last but not least, Mike Earnest's brilliant &lt;a href=&quot;https://puzzling.stackexchange.com/questions/11172/past-present-and-future&quot;&gt;Past, Present and Future&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a puzzle where the questions you ask the Knights/Knaves/Jokers have to be probabilistic in nature, two excellent examples are Emrakul's &lt;a href=&quot;https://puzzling.stackexchange.com/questions/313/you-have-one-question-to-tell-whether-the-number-im-thinking-of-is-1-2-or-3&quot;&gt;You have one question to tell whether the number I'm thinking of is 1, 2, or 3&lt;/a&gt; and Joe Z.'s extension &lt;a href=&quot;https://puzzling.stackexchange.com/questions/382/differentiate-between-the-numbers-from-1-to-5-with-one-single-yes-no-question?lq=1&quot;&gt;Differentiate between the numbers from 1 to 5 with one single yes/no question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this answer your question? (Even if it doesn't, it makes a nice collection of some fantastic Knight-Knave puzzles!)&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-04-10T09:30:12.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11817" PostTypeId="2" ParentId="11811" CreationDate="2015-04-10T09:35:33.663" Score="2" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wedding rings&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pros&lt;/strong&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;em&gt;I always let my friends go first.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since the wedding ring is in the beginning of your finger, everything you do or touch, you do it before the ring touches it (figuratively speaking).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;To some of my groups of friends, I mean everything.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That's true. For some people, the marriage is everything.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;No matter who I am around, I always have an effect on them.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you wear a ring, this always means something for you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am strong-willed; none of my friends can change me.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The gold is very very hard metal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I have a magnetic personality, so I am always near friends.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You wear it all the time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cons&lt;/strong&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;em&gt;I'm short.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm small.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I do not like to be alone- I am clingy.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wedding rings are meant to be worn.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I never accept some groups of my friends for who they are, and instead, I try to change them, and always succeed.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think this has something to do with adultery?!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I'm bossy- usually the one telling my friends when, where, how, why, and under what circumstances things are going to happen.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This I couldn't figure out.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I have 2 personalities- one is almost a teenager, and the other is in his mid twenties&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Female and male wedding rings. They have different size.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11145" LastEditorUserId="5373" LastEditDate="2015-04-10T09:54:19.813" LastActivityDate="2015-04-10T09:54:19.813" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11818" PostTypeId="2" ParentId="11814" CreationDate="2015-04-10T10:20:49.730" Score="7" Body="&lt;p&gt;This is a classic problem in cryptography, known as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The secret sharing problem - how can we divide a secret among n people so that selected subsets, eg: any group of at least r people - can reconstruct the secret, but ZERO information is contained in other subsets.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A standard solution to this problem is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Snow White calls the secret A and picks 3 random numbers B,C,D as well as 7 different random numbers $X_1,X_2,\ldots,X_7$, one for each dwarf. To dwarf $i$, she gives the value of the polynomial $A+BX+CX^2+DX^3$, evaluated at $X_i$. Any four dwarves can recover the coefficients $A,B,C,D$ knowing the values at four different points, but knowing the values at 3 different points reveals no information about any coefficient. I believe this solution was first given by Adi Shamir.&lt;br&gt;Instead of using polynomials, we could have also chosen 7 random and linearly independent vectors in $R^4$ and given their inner product with $(A,B,C,D)$ to the dwarves.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another known solution to this is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Visual_cryptography&quot;&gt;http://en.wikipedia.org/wiki/Visual_cryptography&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5684" LastActivityDate="2015-04-10T10:20:49.730" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11819" PostTypeId="2" ParentId="11813" CreationDate="2015-04-10T11:11:41.887" Score="2" Body="&lt;p&gt;They are exacty the same so also 120 gram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By making the diameter bigger you also make the ring narrower. And these cancel eachother out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better explanation &lt;a href=&quot;http://mathworld.wolfram.com/SphericalRing.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-04-10T11:11:41.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11820" PostTypeId="2" ParentId="11666" CreationDate="2015-04-10T11:27:10.407" Score="0" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;32 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because and this for me was the most simple reason:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you had only 2 dials, you would need to try all combinations 1 with 1-8, then 2 with 1-8, etc.&lt;br&gt;&lt;br&gt; But because you have 3 dials, which would look like this:&lt;br&gt;&lt;br&gt; 1 2 3&lt;br&gt;&lt;br&gt; You could set dials 1 and 3 to 4 different sets of numbers (e.g. 1 and 8, 2 an 7, 3 and 6, 4 and 5) and then just spin the middle dial one through eight.&lt;br&gt;&lt;br&gt; so 4 * 8 = 32&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9907" LastActivityDate="2015-04-10T11:27:10.407" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11823" PostTypeId="2" ParentId="11811" CreationDate="2015-04-10T11:56:32.940" Score="2" Body="&lt;p&gt;Is it...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; decimal point?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You might find me in a costly lawsuit. What am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Costly lawsuits have big numbers maybe with decimal points or thousand separators&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I always let my friends go first.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Other numbers are followed by decimal point&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To some of my groups of friends, I mean everything.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the friends are fraction they can't be showed without decimal point&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No matter who I am around, I always have an effect on them.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Put decimal point anywhere in a number and you will change it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am strong-willed; none of my friends can change me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No matter what is the math operation, you will not change it, just move it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have a magnetic personality, so I am always near friends.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Decimal number is always side by side with numbers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm short.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is short&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I do not like to be alone- I am clingy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It needs other numbers to be 'visible'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I never accept some groups of my friends for who they are, and instead, I try to change them, and always succeed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You change numbers if you use decimal point&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm bossy- usually the one telling my friends when, where, how, why, and under what circumstances things are going to happen.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You decide how big the number is starting from decimal point&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have 2 personalities- one is almost a teenager, and the other is in his mid twenties&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No idea - my previous idea was Zero&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9907" LastActivityDate="2015-04-10T11:56:32.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11824" PostTypeId="1" AcceptedAnswerId="11827" CreationDate="2015-04-10T12:11:34.880" Score="14" ViewCount="1215" Body="&lt;p&gt;Susan was born with an extreme case of the rare communication disorder, &quot;dyscryptia&quot;, which has enthralled neuroscientists and computer scientists alike from all 6 corners of the globe. She is one of only a dozen recorded cases. This puzzling disorder is unique in that in each of the twelve cases, the disorder manifests itself in a distinct manner; the common denominator being that the disorder always affects communication letter by letter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I met Susan just yesterday night at the local pizza parlor as she was playing a beautiful piece on the place's piano. I went up to her to compliment her playing, but of course, my message didn't get across. She pointed at a square board on top of the piano consisting of many common English words coupled with their translations, and I used it to repeat myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the conversation that ensued, I found myself enchanted by this strange woman. I would like to prepare for the next time we meet, so that I may speak her language and offer a more comfortable conversation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I remember a few useful details. Maybe my memories could serve as a Rosetta Stone, a bridge between our languages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Susan referred to herself as &quot;Nntog&quot;, and to her language as &quot;Susubam&quot;. Following is a list of words and their translations that I can recall.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;| Time | Good | Have | Give | Want | Take | Make | Know | Last | Long | Next | Come |&#xA;| Cvry | Vdsk | Iwam | Peal | Xohq | Ulpy | Nlpr | Yclh | Mtmf | Acus | Scrh | Rbrh |&#xA;-------------------------------------------------------------------------------------&#xA;| I | A | You | He | She | It | The | Of | On | And | But | Or | Right | Add | Near |&#xA;| I | A | Njt | Mm | Amx | Cc | Bmy | Uu | Cc | Ore | Wov | Gg | Apobl | Ehe | Sfsf |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There were more.. but I can only remember so much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh, and as I was leaving, she told me, &quot;Hgjqz cqjos njmh yizaj, cm'i hgqccnkb rapf am clp.&quot; She must have forgotten for a moment that I don't speak Susubam?&lt;/p&gt;&#xA;" OwnerUserId="11100" LastEditorUserId="11100" LastEditDate="2015-04-10T15:00:52.190" LastActivityDate="2015-04-10T15:00:52.190" Title="Help me learn to speak Susan's language" Tags="|cipher|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11826" PostTypeId="2" ParentId="11811" CreationDate="2015-04-10T12:57:17.010" Score="1" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; time?&#xA; Time is short. It has 2 personalities: 12 or 24 hours. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;About friends&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Friends are clocks. They can't change time, but they're affected by it. Some clocks only show time, so it's everything to them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11230" LastActivityDate="2015-04-10T12:57:17.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11827" PostTypeId="2" ParentId="11824" CreationDate="2015-04-10T12:59:28.483" Score="13" Body="&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Finally &lt;strong&gt;solved&lt;/strong&gt; last part. &lt;br /&gt;&lt;br /&gt; Decrypted Sentence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sorry about your pizza, it's probably cold by now.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, I got curios about previous sentence which was edited. In case anyone wondering it is decrypted as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I have to get home. Maybe I'll see you next week.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ORIGINAL&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I solved the encrypting part, however decoding isnt that easy. Still working on that part but this is my &lt;em&gt;partial answer&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each letter has a value as following a=1,b=2,c=3...,y=25,z=26&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Encryping works as following;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Assume word is A,B,C,D then ecrypted text is (A+B)%26, (B+C)%26, (C+D)%26,(D+A)%26 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lets try with examples:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Good = 7,15,15,4 = (7+15)%26,(15+15)%26,(15+4)%26,(4+7)%26 = 22,4,19,11 = Vdsk&lt;br /&gt;&#xA; Give = 7, 9, 22, 5 = 16, 5, 1, 12 = Peal &lt;br/&gt;&#xA; On = 15,14 = 3,3 = cc&lt;br /&gt;&#xA; Of = 15,6 = 21,21 = uu&lt;br /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Decrypting:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;So all the words which have only 2 letters must encrpyt into a double letter word (such as aa,bb,cc..). However, Susan's last sentence has 2-letter word(am) which doesnt fit this rule. &lt;br/&gt; &lt;br/&gt;&#xA;If this is not a mistake, I'm assuming we should evaluate all words in a sentence together and probably gonna ignore signs('.). &lt;br/&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4506" LastEditorUserId="4506" LastEditDate="2015-04-10T13:55:20.353" LastActivityDate="2015-04-10T13:55:20.353" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11828" PostTypeId="2" ParentId="11814" CreationDate="2015-04-10T13:46:17.380" Score="3" Body="&lt;p&gt;Encode the message as a binary string. Decompose this string into 4 parts that can be &lt;code&gt;xor&lt;/code&gt;ed bitwise together to get back the original, using the coin to generate randomness (three flips are needed for each bit). Like so:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;original: 1000111011110011...&#xA;&#xA;part 1:   0000111011010100...&#xA;&#xA;part 2:   1010111001100000...&#xA;&#xA;part 3:   1111100011001010...&#xA;&#xA;part 4:   1101011010001101...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note that each three parts are worthless without the fourth one. Also note that the probabilities of finding 0s and 1s in all parts are 50%, regardless of the value of the original bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take four dwarves and give each of them one part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 35 possible combinations of 4 different dwarves out of 7, so repeat the process described above 35 times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final output is highly redundant, but no information can be deduced from any incomplete decomposition that could be used to infer the missing information in another incomplete decomposition.&lt;/p&gt;&#xA;" OwnerUserId="4394" LastEditorUserId="4394" LastEditDate="2015-04-10T13:56:42.787" LastActivityDate="2015-04-10T13:56:42.787" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11829" PostTypeId="1" AcceptedAnswerId="11830" CreationDate="2015-04-10T13:51:34.160" Score="39" ViewCount="5547" Body="&lt;p&gt;Ana, Barbara, Carol and Diana want to know their average age. But no lady wants to disclose her age.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They decide a strategy and use a calculator (one that doesn't store steps) and tapped some keys and passed it to other ladies. After some time, they had the sum of their ages and nobody could know the exact age of the other ladies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;BR&gt;&#xA;What was the strategy?&lt;/p&gt;&#xA;" OwnerUserId="4637" LastEditorUserId="5373" LastEditDate="2015-05-02T12:26:55.230" LastActivityDate="2019-09-08T09:54:50.300" Title="Find average age without revealing your age" Tags="|strategy|lateral-thinking|calculation-puzzle|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11830" PostTypeId="2" ParentId="11829" CreationDate="2015-04-10T13:57:31.990" Score="39" Body="&lt;p&gt;A puts in her age plus a random addition, B, C and D add the same..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then A removes her addition, then B, C and D do the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In equation form where A is the actual age and a is the random addition chosen by A:&lt;br&gt;&#xA;(A + a) + (B + b) + (C + c) + (D + d) - a - b - c - d = A + B + C + D which is sum of all their ages&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then divide by 4.&lt;/p&gt;&#xA;" OwnerUserId="10294" LastEditorUserId="1504" LastEditDate="2015-04-10T15:32:26.157" LastActivityDate="2015-04-10T15:32:26.157" CommentCount="21" ContentLicense="CC BY-SA 3.0" />
  <row Id="11831" PostTypeId="2" ParentId="11594" CreationDate="2015-04-10T14:14:18.913" Score="1" Body="&lt;p&gt;based on the answer from rand al'thor:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the answer for #4 could be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; billionaire&#xA; based on &quot;billious&quot; and &quot;air&quot; (the answer which can't be seen)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11150" LastEditorUserId="11150" LastEditDate="2015-04-10T14:52:30.497" LastActivityDate="2015-04-10T14:52:30.497" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11832" PostTypeId="1" AcceptedAnswerId="11835" CreationDate="2015-04-10T14:17:55.817" Score="4" ViewCount="693" Body="&lt;p&gt;Here is a British-style (cryptic) crossword clue:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mind reading Pop Star is fashionable (7)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me know if you can figure it out.&lt;/p&gt;&#xA;" OwnerUserId="10294" LastEditorUserId="21234" LastEditDate="2016-07-22T21:01:42.840" LastActivityDate="2016-07-22T21:01:42.840" Title="Cryptic Crossword Clue: Mind reading Pop Star is fashionable (7)" Tags="|cryptic-clues|crossword-clues|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11833" PostTypeId="2" ParentId="11811" CreationDate="2015-04-10T14:17:59.177" Score="4" Body="&lt;p&gt;I'm going to suggest&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An asterisk (*).&#xA;Lawyers are fond of addenda and footnotes and the like, so a printed lawsuit is likely to contain a lot of them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I always let my friends go first.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An asterisk is almost always found after a word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;To some of my groups of friends, I mean every&lt;em&gt;thing&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In a computer/programming context, the asterisk is often used as a wildcard to mean &quot;everything&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;No matter who I am around, I always have an effect on them.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Whenever there's and asterisk after a word, it means there something more to say about that word (an &quot;effect&quot;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I am strong-willed; none of my friends can change me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The asterisk affects the word it's on, but the word doesn't affect the asterisk.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I have a magnetic personality, so I am always near friends.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An asterisk is rarely found on its own, but always at the end of a word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Cons:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I'm short.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The height of the asterisk is not very much.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I do not like to be alone- I am clingy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Like #5 above -- it's always on a word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I never accept some groups of my friends for who they are, and instead, I try to change them, and always succeed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The purpose of the asterisk is to supply a &quot;modifier&quot; for the word(s).  It changes the meaning in some way.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I'm bossy- usually the one telling my friends when, where, how, why, and under what circumstances things are going to happen.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Asterisks are often used for the &quot;fine print&quot; in deals, contests, etc., where they detail all the specifics.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I have 2 personalities- one is almost a teenager, and the other is in his mid twenties.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  This refers to the locations of the two * keys on a typical keyboard.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2015-04-10T14:17:59.177" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11834" PostTypeId="1" AcceptedAnswerId="11876" CreationDate="2015-04-10T14:18:21.650" Score="7" ViewCount="1325" Body="&lt;p&gt;You're a secret agent working for the NSA and you have monitored the messages between two friends for a while now. Suddenly, their messages become encrypted and you're on high alert. The first part of their message was sent twice (accidentally), an encrypted and decrypted version. This was the message:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;The package has arrived, I am ready for pickup. When and where shall we meet?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Encrypted:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;r.4NC&quot;Hm,20QV}*;. `AI&amp;amp;*BU3;%);.-6/2NUF=eOt-+3Jq%2]TpH,MCJ5!F*}HT?jX,R&amp;gt;EqC''*H&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Further analysis show the keyword was &lt;code&gt;password&lt;/code&gt;. The next part of the message was sent encrypted, but with a different key. All you could find out about the keyword was that it is 4 letters long. The message was this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;pt3sLuD,91~1o&amp;amp;,:# D0bVs{O'Mov;vC%l3yFb&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Time is running out and you fear it may be dangerous! Can you decrypt the message in time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I will post a hint later on today&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C++ code (very basic) here for algorithm (won't spoil the answer!)&lt;br&gt;&#xA; (This is my repo) &lt;a href=&quot;https://github.com/Cube777/cube-lib/blob/master/src/cube-encrypter.cpp&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/Cube777/cube-lib/blob/master/src/cube-encrypter.cpp&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2:&lt;/strong&gt;&#xA;Another hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The keyword is not an English word, but a sequence of 4 letters (not an abbreviation but a popular sequence of 4 letters)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11234" LastEditorUserId="30633" LastEditDate="2018-04-02T04:03:14.653" LastActivityDate="2018-04-02T04:03:14.653" Title="You're the NSA, trying to decrypt a message between two friends" Tags="|cipher|cryptography|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="11835" PostTypeId="2" ParentId="11832" CreationDate="2015-04-10T14:40:14.570" Score="8" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PSYCHIC&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mind-reading&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; = PSYCHIC&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pop Star&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; = PSY&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fashionable&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; = CHIC&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2015-04-10T14:40:14.570" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11836" PostTypeId="2" ParentId="11594" CreationDate="2015-04-10T14:53:30.790" Score="0" Body="&lt;p&gt;Puzzle 5:&#xA;Here's some thinking about puzzle #5. Maybe it helps to get further...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if you turn the number upside down imaging that it would be a 7 segment based view you get:&#xA; 8E265hl or 8E265hI&#xA; if this would be hex, you get: 582245 I or l - no idea how to get further&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10890" LastActivityDate="2015-04-10T14:53:30.790" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11837" PostTypeId="2" ParentId="11811" CreationDate="2015-04-10T15:13:21.193" Score="1" Body="&lt;p&gt;I'll try again:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A hyphen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pros:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I always let my friends go first.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A hyphen is found in the middle of a word, so there are letters before it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;To some of my groups of friends, I mean everything.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When specifying a group, like &quot;A-Z&quot;, the hyphen indicates everything in between.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;No matter who I am around, I always have an effect on them.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The hyphen modifies words, and changes their meaning.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I am strong-willed; none of my friends can change me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It means the same thing, regardless of what is around it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I have a magnetic personality, so I am always near friends.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The hyphen shows up in the middle of words&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Cons:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I'm short.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's a short dash.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I do not like to be alone- I am clingy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, it's always within a word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I never accept some groups of my friends for who they are, and instead, I try to change them, and always succeed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Adding a hyphen tacks something else on, which changes the meaning of a word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I'm bossy- usually the one telling my friends when, where, how, why, and under what circumstances things are going to happen.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A hyphen is sometimes used (incorrectly) for injection into a sentence - like this - instead of the &quot;em-dash&quot;.  That's often used when giving instructions or telling people what to do.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I have 2 personalities- one is almost a teenager, and the other is in his mid twenties.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The hyphen is key #12 across my keyboard, and the similar &quot;minus sign&quot; on the numeric keypad is #22.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2015-04-10T15:13:21.193" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11838" PostTypeId="1" AcceptedAnswerId="11839" CreationDate="2015-04-10T15:20:59.520" Score="28" ViewCount="33016" Body="&lt;p&gt;I hope this is on topic. I have a set of kids' magic tricks but I have lost the instructions! I was hoping this might be a good place to get some help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the first one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/El0Vx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the trick and how does it work?&lt;/p&gt;&#xA;" OwnerUserId="8998" LastEditorUserId="15942" LastEditDate="2023-07-24T17:33:30.543" LastActivityDate="2023-07-24T17:33:30.543" Title="What is this magic trick with 6 cards featuring numbers between 1 and 60?" Tags="|number-sequence|puzzle-identification|reverse-puzzling|magic|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11839" PostTypeId="2" ParentId="11838" CreationDate="2015-04-10T15:28:07.997" Score="28" Body="&lt;p&gt;The magic trick is as follows.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Ask someone to think of a number. They should find all the cards with that number on, and give them to you. All you need do is add up the top left hand number on each of these cards. This will be the number they chose. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://gwydir.demon.co.uk/jo/numbers/binary/how.htm&quot;&gt;Source&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explanation: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first number on each card is a power of 2 (2,4,8,16,32). Every natural number up to 63 is a sum of some of these numbers (just write it in base 2). The rest of the numbers on each card are chosen precisely to make this work: &lt;strong&gt;on the card whose first number is $2^n$, the numbers appearing are exactly those whose binary expression has a $1$ in th $n$th position.&lt;/strong&gt; How high the numbers go on each card is irrelevant, as long as the upper bound is the same on each one (here it seems to be 59).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="2383" LastEditDate="2015-04-10T17:46:31.780" LastActivityDate="2015-04-10T17:46:31.780" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11840" PostTypeId="2" ParentId="11811" CreationDate="2015-04-10T15:43:02.930" Score="2" Body="&lt;p&gt;Based on Hint #1, I came up with the following answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TO/TOO/TWO. A &lt;em&gt;phonetic&lt;/em&gt; answer, a string of phonemes rather than a string of letters - interesting idea for a riddle!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;1.I always let my friends go first.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One goes first, before &lt;em&gt;two&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;2.To some of my groups of friends, I mean every[thing].&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you're French, the string of phonemes that is the answer comes out as &lt;em&gt;tout&lt;/em&gt;, meaning &quot;every&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;3.No matter who I am around, I always have an effect on them.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The word &lt;em&gt;too&lt;/em&gt; always has an effect on the other words around it: it emphasises them, in a way.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;4.I am strong-willed; none of my friends can change me.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Could again refer to the word &lt;em&gt;too&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;5.I have a magnetic personality, so I am always near friends.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; None of these words normally appear on their own, but rather surrounded by other words.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;1.I'm short.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All of these words (especially &lt;em&gt;to&lt;/em&gt;) are short.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;2.I do not like to be alone- I am clingy.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Two&lt;/em&gt; things are not alone. Alternatively this could have the same interpretation as Pro #5.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;3.I never accept some groups of my friends for who they are, and instead, I try to change them, and always succeed.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is &lt;em&gt;too&lt;/em&gt; again. It changes the words it describes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;4.I'm bossy- usually the one telling my friends when, where, how, why, and under what circumstances things are going to happen.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This could be &lt;em&gt;to&lt;/em&gt;, always coming before a verb in the infinitive form.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;5.I have 2 personalities- one is almost a teenager, and the other is in his mid twenties&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The two letters of &quot;to&quot; are 20 and 15: 15 is a teenager, and 20 is in his twenties. Not mid-twenties though, so maybe there's another explanation for this line!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-04-10T15:43:02.930" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11842" PostTypeId="1" AcceptedAnswerId="11845" CreationDate="2015-04-10T15:59:13.293" Score="12" ViewCount="1488" Body="&lt;p&gt;I have a set of kids' magic tricks but I have lost the instructions! I was hoping this might be a good place to get some help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the second one.&lt;img src=&quot;https://i.stack.imgur.com/mA09a.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This one might be a little more tricky but what is this magic trick and how does it work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you can't see is that the rectangular box is hollow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pNZbp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="8998" LastEditorUserId="5373" LastEditDate="2015-04-13T23:32:55.367" LastActivityDate="2015-04-13T23:32:55.367" Title="What is this magic trick with a sarcophagus shaped box and a set of tokens?" Tags="|puzzle-identification|magic|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="11845" PostTypeId="2" ParentId="11842" CreationDate="2015-04-10T17:20:02.227" Score="10" Body="&lt;p&gt;My younger brother likes magic and so I recognize the boat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The boat + the tokens is already a set, everything else is from another trick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trick is that you magically place the same sequence of images as a volunteer. There are two tokens of each image and there is a trick token which is black on both sides and looks like the bottom of the holes. I split the tokens between us so that we both have the same images, and I have the special token. I say that I'll place my tokens face down, the volunteer will place his face up, and we'll both place the same sequence of images even though I started.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How it's done: I first place the special token, then I mimic his images (as I can see them), and finally when he place his second to last token, I place two tokens as one: the same image and then my last token (because I know that is also his last token, and thus what he will place next). Then I reveal my face down tokens.&lt;/p&gt;&#xA;" OwnerUserId="11100" LastEditorUserId="11100" LastEditDate="2015-04-10T17:26:26.813" LastActivityDate="2015-04-10T17:26:26.813" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11846" PostTypeId="2" ParentId="11811" CreationDate="2015-04-10T17:24:32.587" Score="10" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the letter Y?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The two personalities:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the i sound ('i' is 9th letter of alphabet, almost a teen) and the y sound ('y' is 25th letter, mid twenties).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's never by itself (in English).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Modifies words (like 'ever' into 'every')&#xA;&#xA; Other letters don't modify its pronunciation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's in the question.&lt;/p&gt;&#xA;" OwnerUserId="11242" LastEditorUserId="5373" LastEditDate="2015-04-10T18:57:08.637" LastActivityDate="2015-04-10T18:57:08.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11848" PostTypeId="1" AcceptedAnswerId="11850" CreationDate="2015-04-10T18:07:02.633" Score="5" ViewCount="7416" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am your cup of coffee&lt;br&gt;&#xA;  Applicable for anyone&lt;br&gt;&#xA;  I am very objective&lt;br&gt;&#xA;  I came from the sun  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;After drinking me you'll be very active&lt;br&gt;&#xA;  You can take me anywhere&lt;br&gt;&#xA;  There's no garbage with me&lt;br&gt;&#xA;  Simplicity is so fair&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In order to be accepted, you must explain &lt;strong&gt;EVERY&lt;/strong&gt; line. Every line has a meaning!&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2016-05-01T14:35:49.447" Title="I am your cup of coffee - What am I?" Tags="|riddle|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11850" PostTypeId="2" ParentId="11848" CreationDate="2015-04-10T18:19:27.340" Score="16" Body="&lt;p&gt;You are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Java&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am your cup of coffee&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Java logo is a cup of coffee. Also, java is a type of coffee.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Applicable for anyone&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Java Virtual Machine is available on a wide range of platforms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am very objective&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In Java, objects are king. In earlier implementations of the language, there were no primitive data types, and literally &lt;em&gt;everything&lt;/em&gt;, including integers, was an object.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I came from the sun&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Java is a product of Sun Microsystems, although nowadays it's owned by Oracle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After drinking me you'll be very active&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After drinking java the coffee, you'll be on high alert for a while.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; and one of the &lt;em&gt;intents&lt;/em&gt; of Java (1.2.4 in &lt;a href=&quot;http://www.oracle.com/technetwork/java/intro-141325.html&quot; rel=&quot;noreferrer&quot;&gt;this document&lt;/a&gt;) was that people who used it would have high performance.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can take me anywhere&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As above, since Java is available on so many platforms, including mobile platforms (in fact, it's the primary language used to program for the Android operating system), you can literally take it anywhere you go.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's no garbage with me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Java has automatic garbage collection, unlike languages with completely manual memory management like C.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Simplicity is so fair&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I... honestly can't explain this one. Unless you think Java is a very simple and fair language, in which case my bias from working with it for a year in CS class is going to kick in and say that that last line is patently false.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Apparently the answer is that, again, one of the &lt;em&gt;intents&lt;/em&gt; of Java (1.2.1 &lt;a href=&quot;http://www.oracle.com/technetwork/java/intro-141325.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;) was that the language would be simple and familiar, even if in practice it really isn't.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="10973" LastEditDate="2015-04-21T12:49:07.910" LastActivityDate="2015-04-21T12:49:07.910" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="11851" PostTypeId="2" ParentId="11586" CreationDate="2015-04-10T18:27:01.320" Score="0" Body="&lt;p&gt;Similar to other Answers my guess is it could be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; New years celebration, or a payment from his lawyer (for selling his shop)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;he was wearing a black jacket, a white mask, leather gloves and brown boots&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Suggesting it might be winter time (January weather). OR maybe his lawyer is Saul Goodman and doesn't want to be recognized.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On that day however, Warren was not very busy&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; New years eve can be a day off, so no customers come in. OR if it was his lawyer, his shop is closed/closing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Open it 30 minutes after your clock reaches 11:30pm.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 30 minutes after 11:30pm would be midnight, or the new year. OR the official day when his shop is considered sold.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What was in the box?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A New years gift (something from his family). OR Money from the lawyer for selling his shop (So he can retire)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastActivityDate="2015-04-10T18:27:01.320" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11852" PostTypeId="1" AcceptedAnswerId="11854" CreationDate="2015-04-10T18:56:57.037" Score="0" ViewCount="419" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Sssssssslither&lt;br&gt;&#xA;  You can read me like a book&lt;br&gt;&#xA;  I'm so tiny and compact&lt;br&gt;&#xA;  6 long years I took  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I can remember a lot&lt;br&gt;&#xA;  A very basic procedure&lt;br&gt;&#xA;  White is very important to me&lt;br&gt;&#xA;  Beautiful for sure&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Again, you have to pick this apart line by line!&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2015-04-16T02:19:49.413" Title="Sssssssslither - What am I?" Tags="|riddle|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11853" PostTypeId="2" ParentId="11852" CreationDate="2015-04-10T19:05:17.997" Score="3" Body="&lt;p&gt;Are you a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Disk?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sssssssslither&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sssssss is the noise of a rotating disk.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can read me like a book&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Disks can be read!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm so tiny and compact&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Compact-disk!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6 long years I took&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I don't really know this one&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can remember a lot&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You have a pretty big storage capacity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A very basic procedure&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Insert disk in the reader!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;White is very important to me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Might be virgin disks?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Beautiful for sure&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The rainbow behind CD!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-04-10T19:05:17.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11854" PostTypeId="2" ParentId="11852" CreationDate="2015-04-10T19:09:49.803" Score="16" Body="&lt;p&gt;You are &lt;a href=&quot;https://www.python.org/&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Python&lt;/strong&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sssssssslither&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A python is a snake.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You can read me like a book&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Python, like BASIC or COBOL, generally uses words instead of symbols to express its operators. For example, it uses &lt;code&gt;and&lt;/code&gt; and &lt;code&gt;or&lt;/code&gt; where C or Java would use &lt;code&gt;&amp;amp;&amp;amp;&lt;/code&gt; or &lt;code&gt;||&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm so tiny and compact&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because the language is interpreted, it doesn't need all those heavyweight compilers to build a giant executable and a very small source file is all that's needed to be a complete program.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;6 long years I took&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Python 1.0 was released in 1994, and Python 2.0 in 2000.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I can remember a lot&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Python has automatic memory management and garbage collection like Java does, and therefore while it doesn't have a large memory capacity, it can remember things often.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A very basic procedure&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not 100% sure about this one, but Python doesn't require any &quot;main&quot; method like C or Java does. All the code in a file is run in order without the need for any explicit containing procedure.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;White is very important to me&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In Python, whitespace determines syntax levels, and so good tabbing conventions are enforced.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Beautiful for sure&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Python is a beautiful language (and pythons are generally beautiful snakes).&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-04-16T02:19:49.413" LastActivityDate="2015-04-16T02:19:49.413" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="11856" PostTypeId="2" ParentId="11852" CreationDate="2015-04-10T19:15:03.173" Score="0" Body="&lt;p&gt;I'm pretty sure this isn't the answer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Music&lt;/strong&gt;&#xA; Music sheets can be read, Music takes no physical space, I guess there must be a song named 6 long years or maybe some music took 6 years to write, it can definitely be tied to memories (remember a lot) or symbolically represent something. Music is actually conceptually very very very simple (basic procedure), whole notes are empty(white) in music sheets, Music is obviously beautiful, there's only the slither that doesn't fit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastEditorUserId="11251" LastEditDate="2015-04-10T20:32:43.143" LastActivityDate="2015-04-10T20:32:43.143" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11857" PostTypeId="2" ParentId="11811" CreationDate="2015-04-10T19:18:26.527" Score="0" Body="&lt;p&gt;Based off of &lt;a href=&quot;https://puzzling.stackexchange.com/a/11846/5373&quot;&gt;Cameron's &quot;painfully close&quot; answer&lt;/a&gt;, but slightly different and with more clues explained:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the letter I.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Obviously this makes the question something of a &quot;lynch-mob&quot; riddle, what with all the referring to the answer as &quot;I&quot;. Here are (at least some of) the clues explained one by one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Title:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I appears in the words &quot;a costly lawsu&lt;strong&gt;i&lt;/strong&gt;t&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I always let my friends go first.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some letters come before I in the alphabet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;To some of my groups of friends, I mean everything.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Egomaniacs?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;No matter who I am around, I always have an effect on them.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ???&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am strong-willed; none of my friends can change me.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter I doesn't change no matter how many other letters it's surrounded by.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I have a magnetic personality, so I am always near friends.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ???&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I'm short.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lower-case i looks like a short person with a head.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I do not like to be alone- I am clingy.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter I often appears in the middle of a word along with other letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I never accept some groups of my friends for who they are, and instead, I try to change them, and always succeed.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Maybe as Cameron suggested, referring to words that change with the insertion of an extra letter I - e.g. &quot;her&quot;-&gt;&quot;heir&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I'm bossy- usually the one telling my friends when, where, how, why, and under what circumstances things are going to happen.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ???&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I have 2 personalities- one is almost a teenager, and the other is in his mid twenties&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As Cameron said, this refers to I and Y. He just picked the wrong one of the two as his answer!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-04-10T19:18:26.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11859" PostTypeId="1" AcceptedAnswerId="11864" CreationDate="2015-04-10T19:41:13.757" Score="15" ViewCount="710" Body="&lt;p&gt;Younger and older does not meet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But big and small does meet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tall and short does not meet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Breadth does but length never meets...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;" OwnerUserId="11056" LastEditorUserId="11056" LastEditDate="2015-04-14T14:29:45.220" LastActivityDate="2015-04-14T14:29:45.220" Title="Who am I?...Younger and older does not meet" Tags="|riddle|pattern|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11860" PostTypeId="2" ParentId="11811" CreationDate="2015-04-10T19:45:57.993" Score="11" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letters ly&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I always let my friends go first.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ly appears at the ends of words to make them adverbs, usually&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To some of my groups of friends, I mean everything.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Thinking about the ly in the context of the word poly, meaning every&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No matter who I am around, I always have an effect on them.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ly turns some words into adverbs when added to the end&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am strong-willed; none of my friends can change me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; adding suffixes to ly doesnt change it &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have a magnetic personality, so I am always near friends.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ly on it's own doesn't mean anything&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm short.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; only two letters&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I do not like to be alone- I am clingy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; again, only has meaning when added onto an existing word&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I never accept some groups of my friends for who they are, and instead, I try to change them, and always succeed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; adding ly changes adjectives into adverbs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm bossy- usually the one telling my friends when, where, how, why, and under what circumstances things are going to happen.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; adverbs describe how something is happening&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have 2 personalities- one is almost a teenager, and the other is in his mid twenties&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; l is the 12th letter of the alphabet, y is the 25th&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;also&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; camerons answer was really the springboard for my understanding of this&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11246" LastEditorUserId="10748" LastEditDate="2015-04-12T14:55:27.340" LastActivityDate="2015-04-12T14:55:27.340" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11861" PostTypeId="2" ParentId="11859" CreationDate="2015-04-10T19:48:20.790" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter B?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Younger and older&quot; doesn't contain it, &quot;big and small&quot; does, &quot;tall and short&quot; doesn't, &quot;breadth&quot; does at the beginning, &quot;length&quot; doesn't.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-04-10T19:48:20.790" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11862" PostTypeId="2" ParentId="11829" CreationDate="2015-04-10T20:18:04.260" Score="20" Body="&lt;p&gt;The strategy &lt;a href=&quot;https://puzzling.stackexchange.com/a/11830&quot;&gt;suggested by Jiminion&lt;/a&gt; might work fairly well in practice, but it does risk revealing some extra information about the ladies' ages if the distribution of the random number(s) is known (or can be guessed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, let's say Ana picks her random number uniformly between 0 and 100; that's surely random enough to totally obscure her age, right? But if the sum of her age and the random number turns out to be, say, 15 or 185, then Barbara has some pretty strong upper / lower bounds on Ana's age.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way to eliminate this information leak is to use &lt;a href=&quot;//en.wikipedia.org/wiki/Modular_arithmetic&quot; rel=&quot;nofollow noreferrer&quot;&gt;modular arithmetic&lt;/a&gt;, e.g. like this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Ana picks a random number between 0 and 999 (inclusive)&lt;sup&gt;1&lt;/sup&gt;, adds her age to it, and hands &lt;em&gt;the last three digits&lt;/em&gt;&lt;sup&gt;2&lt;/sup&gt; of the result to Barbara.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Barbara adds her age to this number (it is &lt;em&gt;not&lt;/em&gt; necessary for her to add a second random number), and hands the last three digits of the result to Carol.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Carol and Diana do as Barbara did, with Diana handing the last three digits of the resulting sum back to Ana.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Ana adds 1000 to the sum, subtracts her original random number, takes the last three digits and divides them by 4.  The result is the average age of the ladies.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Notes:&lt;br&gt;&#xA;&amp;nbsp; &amp;nbsp; 1) The modulus 1000 is chosen because it's presumably larger than the sum of the ladies' ages; any suitably large modulus &lt;em&gt;M&lt;/em&gt; could be used, with the random number chosen from between 0 and &lt;em&gt;M&lt;/em&gt;&amp;minus;1.&lt;br&gt;&#xA;&amp;nbsp; &amp;nbsp; 2) To avoid &lt;a href=&quot;//en.wikipedia.org/wiki/Side-channel_attack&quot; rel=&quot;nofollow noreferrer&quot;&gt;side channel attacks&lt;/a&gt;, the reduction modulo 1000 should be done in a way that does not reveal whether the original number was over 999 or not; in particular, simply subtracting 1000 from any four-digit result could leak information, since the other ladies might see you push some extra buttons on the calculator to do it. It's safer to just re-type the last three digits from memory. The same goes for the final subtraction and reduction by Ana, too.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It's easy enough to prove that Barbara cannot learn anything about Ana's age at step 2, since, regardless of Ana's actual age, the number Barbara gets from her is equally likely to be any number between 0 and 999.&#xA;By a similar argument, it's also clear that Carol and Diana cannot learn anything about the other ladies' ages at step 3.  Finally, at step 4, the number Ana receives from Diana is simply the sum of all their ages (which the protocol is &lt;em&gt;designed&lt;/em&gt; to reveal), plus her own random number, so it tells her nothing but the information she was supposed to get.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, note that this protocol is &lt;em&gt;not&lt;/em&gt; secure against collusion: Ana and Carol together can figure out Barbara's age by comparing the number Ana gave out to what Carol received.  (They can also figure out Diana's age the same way &amp;mdash; but if they do both, then they also end up revealing their own ages to each other!)  Of course, Barbara and Diana can also collude to learn the age of Ana and/or Carol the same way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might seem that Barbara could protect herself from such collusion attacks by also adding a random number of her own to the sum, and then subtracting it during a second round (as in Jiminion's original protocol), but alas, this doesn't actually help: if Ana and Carol collude, they can compare their numbers again during this second pass, and so learn Barbara's random number, and thus her age.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, this puzzle is a simple example of &lt;a href=&quot;//en.wikipedia.org/wiki/Secure_multi-party_computation&quot; rel=&quot;nofollow noreferrer&quot;&gt;secure multiparty computation&lt;/a&gt;, which is an active subfield of cryptography.  The specific protocol described here is also somewhat reminiscent of various &lt;a href=&quot;//en.wikipedia.org/wiki/Secret_sharing&quot; rel=&quot;nofollow noreferrer&quot;&gt;secret sharing&lt;/a&gt; schemes, which might also make interesting further reading.&lt;/p&gt;&#xA;" OwnerUserId="243" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-04-10T20:18:04.260" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="11864" PostTypeId="2" ParentId="11859" CreationDate="2015-04-10T20:27:52.620" Score="13" Body="&lt;p&gt;hehehe the answer is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lips.  Read the words and you will figure it out&lt;br&gt;&#xA;&lt;br&gt;&#xA;(Your lips touch when you say big and small, and Breath)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="5188" LastEditDate="2015-04-11T01:19:23.977" LastActivityDate="2015-04-11T01:19:23.977" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11865" PostTypeId="2" ParentId="11848" CreationDate="2015-04-10T20:55:28.950" Score="0" Body="&lt;p&gt;Well I already read the answer, so there's no point in explaining my solution, but I think I can explain the line&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;After drinking me you'll be very active&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think it means:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;to drink sth. in&quot;, which means to figuratively absorb something, because you like it so much. For example, if you &quot;drink in&quot; a book then you read the whole thing in one sitting, because you find it that fascinating. And I think it's the same with Java. If you're eagerly learning everything about Java, because you like the language that much, you will be very active in working with it, because you know how to use it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I hope i could explain my opinion it well enough.&lt;/p&gt;&#xA;" OwnerUserId="1485" LastEditorUserId="1485" LastEditDate="2015-04-10T22:32:01.010" LastActivityDate="2015-04-10T22:32:01.010" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11866" PostTypeId="2" ParentId="11829" CreationDate="2015-04-10T21:05:25.120" Score="10" Body="&lt;p&gt;I think Jiminion has a good solution, but if I wanted to add a little more security/obscurity to it I would do the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Each lady selects 3 numbers (one is her correct age and two random numbers)  &lt;/li&gt;&#xA;&lt;li&gt;They pass the calculator around 5 times.  &lt;/li&gt;&#xA;&lt;li&gt;On each pass the lady must do one of the following (in any order as long as each lady does each item only once)&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Add her real age&lt;/li&gt;&#xA;&lt;li&gt;Add the first random value&lt;/li&gt;&#xA;&lt;li&gt;Add the second random value&lt;/li&gt;&#xA;&lt;li&gt;Subtract the first random value&lt;/li&gt;&#xA;&lt;li&gt;Subtract the second random value&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;After the last pass is complete they will end up with their total age and can divide by 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that you could theoretically do this with any number of random numbers so long as you subtract all of the random numbers you add in and the order of performing the actions is irrelevant since x = x + y - y = y + x - y = x - y + y, etc...&lt;/p&gt;&#xA;" OwnerUserId="3922" LastEditorUserId="3922" LastEditDate="2015-04-10T21:25:07.940" LastActivityDate="2015-04-10T21:25:07.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11867" PostTypeId="1" AcceptedAnswerId="12084" CreationDate="2015-04-10T21:41:32.840" Score="5" ViewCount="439" Body="&lt;p&gt;Your heart pumps, asking why,&lt;br&gt;&#xA;&quot;I am here to make you cry.&lt;br&gt;&#xA;It's hard to say goodbye&lt;br&gt;&#xA;but leave me so you could fly.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;your mind said love, but your body said hate&lt;br&gt;&#xA;happiness is together and you gain weight.&lt;br&gt;&#xA;Like when we met on our first date.&lt;br&gt;&#xA;Nothing but fruit you have ate.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sounds like a story, romantic,&lt;br&gt;&#xA;nothing around me but tragic.&lt;br&gt;&#xA;Here I will show you a magic,&lt;br&gt;&#xA;I'll vanish and you will panic!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I might be the reason you fall in love,&lt;br&gt;&#xA; but love is nothing I planned to speak of.&lt;br&gt;&#xA; I am massive and I can be related,&lt;br&gt;&#xA; Two great mind in history have debated.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="10037" LastEditDate="2015-04-13T17:03:56.207" LastActivityDate="2015-04-14T17:41:52.260" Title="Sounds like a story, romantic - What am I?" Tags="|riddle|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11868" PostTypeId="1" AcceptedAnswerId="11870" CreationDate="2015-04-10T22:10:30.170" Score="29" ViewCount="995" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The lamest rhyme I ever wrote&lt;br&gt;&#xA;  Was one involving Don Quixote.&lt;br&gt;&#xA;  No, wait, a lamer one than that,&lt;br&gt;&#xA;  The famous movie Chocolat.&lt;br&gt;&#xA;  But maybe even lamer still,&lt;br&gt;&#xA;  The butchering of &quot;centre-ville&quot;.&lt;br&gt;&#xA;  And finally, to fill this hole,&lt;br&gt;&#xA;  The food we know as ____________.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What word ends this horrible rhyming stanza?&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="8874" LastEditDate="2016-03-23T17:11:38.360" LastActivityDate="2016-03-23T17:11:38.360" Title="The lamest rhyme I ever wrote" Tags="|riddle|language|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11869" PostTypeId="2" ParentId="11842" CreationDate="2015-04-10T22:17:05.167" Score="9" Body="&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=ZSnHz3K_230&quot; rel=&quot;noreferrer&quot;&gt;&amp;quot;Egyption Prediction&amp;quot; YouTube video&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/h6JqH.png&quot; alt=&quot;screencap from YouTube video&quot; /&gt;&#xA;&lt;a href=&quot;http://mediacdn.shopatron.com/media/mfg/677/media_document/dev_1/1170InstructionsNEW1-15-13.pdf?1389910672&quot; rel=&quot;noreferrer&quot;&gt;&amp;quot;Egyptian Prediction&amp;quot; &amp;amp; other magic tricks (How-To) PDF&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/58qOQ.png&quot; alt=&quot;original toy set, 2004&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="11257" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-10T22:17:05.167" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11870" PostTypeId="2" ParentId="11868" CreationDate="2015-04-10T22:22:27.400" Score="25" Body="&lt;p&gt;The word is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; guacamole&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The pairs of words all look like they rhyme since they end in pretty much the same letters, but they don't actually rhyme.&#xA;&lt;p&gt;&#xA; &quot;wrote&quot; doesn't rhyme with &quot;Quixote&quot;, which is pronounced in the Spanish way.&#xA;&lt;p&gt;&#xA; &quot;that&quot; doesn't rhyme with the French pronunciation of &quot;chocolat&quot;&#xA;&lt;p&gt;&#xA; &quot;still&quot; doesn't rhyme with the French pronunciation of &quot;centre-ville&quot;&#xA;&lt;p&gt;&#xA; &quot;hole&quot; doesn't rhyme with &quot;guacamole&quot;, which is pronounced in the Spanish way.&#xA;&lt;p&gt;&#xA; They're all lame rhymes since none of them actually rhyme!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastEditorDisplayName="user88" LastEditDate="2015-04-13T19:38:20.747" LastActivityDate="2015-04-13T19:38:20.747" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11873" PostTypeId="1" AcceptedAnswerId="11874" CreationDate="2015-04-11T01:48:17.310" Score="3" ViewCount="345" Body="&lt;p&gt;There have been some recent letter-sequence puzzles that have been well-received. For example,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/10935/three-different-and-somewhat-difficult-sequence-puzzles&quot;&gt;Three different, and somewhat difficult, sequence puzzles&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/11348/two-bowls-of-alphabet-soup&quot;&gt;Two Bowls of Alphabet Soup&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/10298/yet-another-batch-of-four-sequence-puzzles&quot;&gt;Yet another batch of four sequence puzzles&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I thought I'd try to make one. I include only one puzzle here, instead of many, like other posts, so that it is obvious who to award the answer to.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There are 9 items in this list. The first 8 of them are represented by the letters below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the last one.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;S N A X M P N I _&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This list was published anonymously during the first 25 years of the 19th century, but 14 years later, someone claimed authorship.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-04-11T01:59:58.147" Title="Find the 9th and final letter in the list" Tags="|letter-sequence|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11874" PostTypeId="2" ParentId="11873" CreationDate="2015-04-11T01:52:51.243" Score="3" Body="&lt;p&gt;The answer is &lt;strong&gt;D&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Now, Da&lt;strong&gt;s&lt;/strong&gt;her! Now, Da&lt;strong&gt;n&lt;/strong&gt;cer! Now, Pr&lt;strong&gt;a&lt;/strong&gt;ncer, and Vi&lt;strong&gt;x&lt;/strong&gt;en!&lt;br&gt;&#xA; On, Co&lt;strong&gt;m&lt;/strong&gt;et! On, Cu&lt;strong&gt;p&lt;/strong&gt;id! On, Du&lt;strong&gt;n&lt;/strong&gt;der and Bl&lt;strong&gt;i&lt;/strong&gt;xem!&quot;&lt;br&gt;&#xA;&lt;br&gt;&#xA; The above two lines are from &lt;em&gt;A Visit from St. Nicholas&lt;/em&gt;, which was published in 1823, and later claimed to be authored by Clement Clarke Moore in 1837, fourteen years later.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The letters in the sequence are the third letters of the names of Santa's eight reindeer.&#xA; The reindeer's names haven't changed much since then; Dunder became Donner and Blixem became Blitzen, but their third letters are still identical.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The next one would be the ninth reindeer introduced in the 1949 Christmas carol that changed Dunder and Blixem's names in the first place, &lt;em&gt;Ru&lt;strong&gt;d&lt;/strong&gt;olph the Red-Nosed Reindeer&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-04-11T01:59:58.147" LastActivityDate="2015-04-11T01:59:58.147" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11875" PostTypeId="1" AcceptedAnswerId="11877" CreationDate="2015-04-11T06:49:11.427" Score="1" ViewCount="314" Body="&lt;p&gt;There are 12 dots in a 4x3 grid pattern. You may start at any one point. You must travel as long as you want in a fixed direction, then stop and turn. Again you travel straight till you turn. (Each line segment and angle may be same or different).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the minimum lines required?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: It may be an easy puzzle, but requires some out-of-the-box thinking. &lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-04-11T07:19:39.870" Title="Minimum lines to reach all 4x3 dots" Tags="|logical-deduction|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11876" PostTypeId="2" ParentId="11834" CreationDate="2015-04-11T07:12:30.900" Score="9" Body="&lt;p&gt;Alright, so I used your code to create a brute-force generator to determine all possible decodings for all four-letter words. Since the input was a four-letter word, I assumed that it was also lowercase, and so the search space was only $26^4 = 456~976$, totally feasible for brute force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, a quick search for an isolated &quot;the&quot; turned up all I needed: the key was &lt;code&gt;asdf&lt;/code&gt; and the plaintext was &lt;code&gt;Coldplay makes the world's best music!&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can see why they'd want to hide that from the NSA. I mean, they're not Nickelback, but...&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-04-11T07:12:30.900" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11877" PostTypeId="2" ParentId="11875" CreationDate="2015-04-11T07:19:39.870" Score="2" Body="&lt;p&gt;Assuming we're only thinking as far outside the box as the original 4-lines and 9-dots problem, then unfortunately the answer is 5 lines. There's no way to cover all 12 dots in less than 5 lines even if your lines don't have to start and end on a dot, and the obvious &quot;Z&quot; configuration allows you to do that with all lines starting and ending on a dot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, all the other loopholes apply. If you're allowed to go out of the center of the dots, or if the line has nonzero thickness, then you can draw slightly diagonal lines and do it in only 3. If the plane is toroidal or if your line is really thick, then you can abuse the wraparound (or the thickness of the line) and do it in only 1, and so on and so forth.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-04-11T07:19:39.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11878" PostTypeId="1" AcceptedAnswerId="11894" CreationDate="2015-04-11T07:38:43.583" Score="-5" ViewCount="316" Body="&lt;p&gt;In a series of characters which looks like this: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;r, t, h, j, m, ...&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Find out what character comes next. &lt;/p&gt;&#xA;" OwnerUserId="11232" LastEditorUserId="463" LastEditDate="2015-04-11T20:56:58.770" LastActivityDate="2015-04-11T20:56:58.770" Title="Find the next character: r,t,h,j,m,?" Tags="|pattern|letter-sequence|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11880" PostTypeId="1" AcceptedAnswerId="11892" CreationDate="2015-04-11T08:33:07.857" Score="3" ViewCount="949" Body="&lt;p&gt;You start with a mystery number from 1 to 100000.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add the number of United States,&lt;br&gt;&#xA;Divide by a quarter incense by sounding it out,&lt;br&gt;&#xA;Subtract the month of Hallowe'en,&lt;br&gt;&#xA;Take the square root of your number,&lt;br&gt;&#xA;Subtract this number from the beads on a mala (or the number of feelings a person can have),&lt;br&gt;&#xA;Subtract the alive fruit punch,&lt;br&gt;&#xA;Take the square root again,&lt;br&gt;&#xA;Divide by the least lucky number in Asia,&lt;br&gt;&#xA;Multiply this number by the height of Mount Everest in metres,&lt;br&gt;&#xA;Figure what Zaccheaus would have paid back if he'd taken that much from you,&lt;br&gt;&#xA;And finally, subtract by the square of the second perfect number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every step along the way is a positive integer. What is the mystery number, and what is the final result of this calculation?&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-04-11T19:43:00.050" LastActivityDate="2017-02-02T21:36:49.753" Title="Calculate a mystery number: Add the number of United States …" Tags="|mathematics|calculation-puzzle|knowledge|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11881" PostTypeId="2" ParentId="11878" CreationDate="2015-04-11T10:25:54.473" Score="2" Body="&lt;p&gt;You can get r,t,h,j,m,o,r,t,h,j,m,o by doing the following formula and converting the number to the correct position in the alphabet:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;((((Round(n/2)+1)%3)*5)+8)+(2*((n+1)%2))&lt;/p&gt;&#xA;" OwnerUserId="11150" LastEditorUserId="11150" LastEditDate="2015-04-11T12:04:04.400" LastActivityDate="2015-04-11T12:04:04.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11885" PostTypeId="1" AcceptedAnswerId="11886" CreationDate="2015-04-11T11:59:45.400" Score="5" ViewCount="417" Body="&lt;p&gt;I come from the dark,&#xA;but I was born in the bright&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I open your eyes into my origins&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cloths I give to everything and everyone, are simple to me&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I pass through you,&#xA;you can't see me,&#xA;because I'm as fast as your &lt;strong&gt;&lt;em&gt;vision&lt;/em&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when I crash into something&#xA;, you observe.&lt;/p&gt;&#xA;" OwnerUserId="11250" LastActivityDate="2015-04-11T12:44:07.067" Title="Can your vision see?" Tags="|riddle|word|wordplay|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11886" PostTypeId="2" ParentId="11885" CreationDate="2015-04-11T12:44:07.067" Score="2" Body="&lt;p&gt;Could it be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Light?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come from the dark, but I was born in the bright&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The light comes out of the dark and its source is bright, of course&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I open your eyes into my origins&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The light you see was generated long time ago!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The cloths I give to everything and everyone, are simple to me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Colours!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I pass through you, you can't see me, because I'm as fast as your vision.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Light is as fast as light!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But when I crash into something , you observe.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can only see things that were &quot;touched&quot; by light.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-04-11T12:44:07.067" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11887" PostTypeId="1" CreationDate="2015-04-11T13:21:02.223" Score="4" ViewCount="2013" Body="&lt;p&gt;During a garage sale I saw a strange old electronic device (not really, it's just flavor story). It had a battery slot, a button and a row of seven-segment displays - though from all except the two last digits the magic smoke had clearly escaped. Curious, I bought it for a dollar, took it home and put in a battery. Right away a &lt;strong&gt;single-digit number&lt;/strong&gt; showed on the display (&lt;strong&gt;not a 7&lt;/strong&gt;). I pressed the button, and the number changed. Trying it a few more times, I noticed a pattern to the numbers: it always went&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;97, 65, 85,    67, 15, 35,&#xA;37, 65, 85,    07, 15, 35,&#xA;77, 65, 85,    47, 15, 35,&#xA;17, 65, 85,    87, 15, 35,&#xA;57, 65, 85,    27, 15, 35.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;After a while I got bored, so I threw the device away. However, it has been nagging at me what rule might be underlying the pattern of this number sequence. Maybe you know?&lt;/p&gt;&#xA;" OwnerUserId="10070" LastActivityDate="2020-03-10T12:57:34.890" Title="Find the algorithm behind this repeating number sequence" Tags="|pattern|number-sequence|" AnswerCount="5" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="11888" PostTypeId="1" AcceptedAnswerId="11891" CreationDate="2015-04-11T14:45:23.467" Score="13" ViewCount="877" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Eight furlongs fall between my twin ends.&lt;br&gt;&#xA;  Remembering fond days or greeting good friends,&lt;br&gt;&#xA;  All those who see me suspect that they please.&lt;br&gt;&#xA;  To summon me, several folks need to say cheese.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've enjoyed reading riddles on this site, and thought I'd try writing one. Who is the speaker of this poem?&lt;/p&gt;&#xA;" OwnerUserId="11271" LastActivityDate="2015-04-11T19:07:42.363" Title="Quite a distance from the start to the end of this short riddle. Who am I?" Tags="|riddle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11890" PostTypeId="2" ParentId="11888" CreationDate="2015-04-11T14:50:26.890" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think it's a picture being taken?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Eight furlongs fall between my twin ends.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I don't know this one...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Remembering fond days or greeting good friends&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A photo let's you remember fond days and good friends.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All those who see me suspect that they please.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People who take the photo want the people in the photo to smile.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To summon me, several folks need to say cheese.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They ask the people in the photo to say cheese.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11273" LastEditorUserId="11273" LastEditDate="2015-04-11T19:07:42.363" LastActivityDate="2015-04-11T19:07:42.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11891" PostTypeId="2" ParentId="11888" CreationDate="2015-04-11T14:54:00.043" Score="10" Body="&lt;p&gt;I think it's &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; smiles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Eight furlongs fall between my twin ends.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 8 furlongs = 1 mile, and &quot;mile&quot; is in the middle of Smiles! The twin ends are the s's.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Remembering fond days or greeting good friends,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reasons to smile&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All those who see me suspect that they please.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If someone sees you smile, they think that they are making you happy (pleasing you)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To summon me, several folks need to say cheese.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Saying cheese makes people smile&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-04-11T15:31:52.927" LastActivityDate="2015-04-11T15:31:52.927" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11892" PostTypeId="2" ParentId="11880" CreationDate="2015-04-11T15:24:06.560" Score="5" Body="&lt;p&gt;The final answer that I got is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 70016&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The number obtained after step 4 is one of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\{3,22,39,54\}$. Let $y$ be the number obtained after step 4. Then $y \leq 67$ because if $x$ is the mystery number, then $y=\sqrt{\frac{x-200}{25}}$. Since the number of beads is 108 and Five Alive is a fruit punch, we have $z=\sqrt{103-y}$, the number obtained as the second square root. For this to be a positive integer, $y$ must be in $\{3,22,39,54\}$ - this corresponds to $z \in \{10,9,8,7\}$. Also note that given such $y$, we can obtain $x$ without any restriction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After the additional information, we get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $z$ is divisible by 4, so $z$ is 8; after doing the extra steps, we get $8850 \times 8-784=70016$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5684" LastEditorUserId="5684" LastEditDate="2015-04-11T17:30:52.437" LastActivityDate="2015-04-11T17:30:52.437" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11893" PostTypeId="2" ParentId="11172" CreationDate="2015-04-11T17:00:58.350" Score="5" Body="&lt;h1&gt;Necessity of Changing the Addressee&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;To solve the riddle in three questions it is essential to remember that the rules allow to choose the addressee of the question (but not the question) depending on the outcome of the question. For if the order of questions is fixed then the problem cannot be solved in three questions. To show this, assume we're asking Left first and Right last (any other order is equivalent). Then &lt;code&gt;?&lt;/code&gt; denotes a random answer:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Left/Middle/Right   | L M R&#xA;--------------------+------&#xA;Past/Present/Future | ?   ?&#xA;Past/Future/Present | ?&#xA;Present/Past/Future |     ?&#xA;Present/Future/Past |&#xA;Future/Past/Present |&#xA;Future/Present/Past |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;To distinguish the case Past/Present/Future from the other cases we would need an answer &quot;Yes&quot; in this one case and &quot;No&quot; in all other cases (or the reverse). If the answer is &quot;No&quot;, we have only 2 questions remaining, which can only differentiate 4 cases, but 5 remain, making it impossible to solve in three questions if the question order is fixed.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Sufficiency of Changing the Addressee&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;So we need to ask questions in a manner that we will put the last question to Past or Present but not to Future. The second question needs to determine whether we put our first question to past: if we did then the answer to the first is useless and unneccessary, otherwise it serves to disambiugate the other cases:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Left/Middle/Right   | L M&#xA;--------------------+----&#xA;Past/Present/Future | ? Y&#xA;Past/Future/Present | ? Y&#xA;Present/Past/Future | Y N&#xA;Present/Future/Past | Y N&#xA;Future/Past/Present | N N&#xA;Future/Present/Past | N N&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then we need to choose the addressee of the third question such that it disambiugates between the remaining cases. (That this is the optimal solution is obvious, since with three boolean questions one can disambiugate at most 8 cases - in this case indeed only 6 because Past answers at random if asked first. With two questions one could disambiugate only four cases, which would be insufficient - though with head-exploding questions nine cases could be disambiugated...)&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Partitioning Questions&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;That only helps to deal with 1. the possibility of hitting Past on the first question and 2. how to avoid hitting Future with the third question. However, they still might not answer the question they are supposed to answer: for Future, the next (2nd or 3rd) question must contain the question that must be answered on the previous (1st or 2nd) question; for Past, the previous (1st or 2nd) question must contain the question that must be answered for the next (2nd or 3rd) question:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;        | 1 2 3&#xA;--------+------&#xA;Past    | - 1 2&#xA;Present | 1 2 3&#xA;Future  | 2 3 -&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Hence the questions must be patterned as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;code&gt;( are you Present and [question 1] )&#xA;OR ( are you Past and [question 2] )&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;( are you Present and [question 2] )&#xA;OR ( are you Future and [question 1] )&#xA;OR ( are you Past and [question 3] )&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;( are you Present and [question 3] )&#xA;OR ( are you Future and [question 2] )&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thus when e.g. Future is asked second, they will provide the answer to question 3, which will be exactly the answer that present would give of question 2.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;The Questions&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Now we know that a solution is possible, but still must determine which questions to ask. Due to the pattern rule above we can start by considering the present case only. Consider again the response table given above, then we want the following pattern:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Left/Middle/Right   | 1 2 3&#xA;Past/Present/Future | ? Y Y&#xA;Past/Future/Present | ? Y N&#xA;Present/Past/Future | Y N Y&#xA;Present/Future/Past | Y N N&#xA;Future/Past/Present | N N Y&#xA;Future/Present/Past | N N N&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The first question, asked to L, can be &quot;is L Present?&quot; to give the desired pattern. The second question, asked to Middle, can be &quot;Did I ask Past my first question?&quot;. Together these tell us whether L is Past, Present or Future. This has two consequences:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;we must avoid asking Future the last question. Thus if L is Future then we can ask either M or R (let's say M), but if L is not Future then we must ask L.&lt;/li&gt;&#xA;&lt;li&gt;the content of the last question must disambiugate between M and R. But the questions must be fixed beforehand, but L can be anything, and we don't want to ask for L again. So we could ask e.g. &quot;Is M Past OR (Is L Past AND M Present)&quot;: if L is already Past then we ask for Present instead, otherwise we ask for Past; or we could ask if M is &quot;earlier&quot; than R. (The final questions would be of significantly simpler form if we could pick not only the addressee but also the question.)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h1&gt;The Solution&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The following three questions shall be asked:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1. To Left: &lt;code&gt;are you Present?&lt;/code&gt; [*]&#xA;&lt;br /&gt;&#xA;2. To Middle:&#xA;&lt;br /&gt;&lt;code&gt;( are you Present AND I asked Past my first question )&#xA;OR ( are you Future AND Left is Present )&#xA;OR ( are you Past AND ( (Left is Past AND Middle is Present) OR (Middle is Past) ) )?&lt;/code&gt;&#xA;&lt;br /&gt;&#xA;3. To Middle if previous both answered with &quot;No&quot;, to Left otherwise:&#xA;&lt;br /&gt;&lt;code&gt;( you are Present AND ( (Left is Past AND Middle is Present) OR (Middle is Past) ) )&#xA;OR ( are you Future AND I asked Past my first question )?&lt;/code&gt;&#xA;&lt;br /&gt;&lt;br /&gt;&#xA;[*] according to the pattern there should be &lt;code&gt;OR ( are you Past AND I asked Past my first question )&lt;/code&gt;, but that drops away since that applies only if question 2 is addressed to Past, but then the first question can't have been addressed to Past.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10070" LastEditorUserId="10070" LastEditDate="2015-04-11T17:18:11.153" LastActivityDate="2015-04-11T17:18:11.153" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11894" PostTypeId="2" ParentId="11878" CreationDate="2015-04-11T17:47:28.120" Score="4" Body="&lt;p&gt;As the word &quot;character&quot; is used instead of &quot;letter&quot;, the series could be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;r, t, h, j, m, &lt;strong&gt;comma&lt;/strong&gt;&lt;br&gt;&#xA;because this is almost &quot;right and diagonal down and repeat&quot; on a standard QWERTY keyboard.&lt;br&gt;&#xA;&lt;br&gt;&#xA;If this is the intended method, a more consistent series would have been &lt;strong&gt;t, y, h, j, m, comma&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-04-11T17:57:50.460" LastActivityDate="2015-04-11T17:57:50.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11895" PostTypeId="1" AcceptedAnswerId="11899" CreationDate="2015-04-11T18:09:34.933" Score="4" ViewCount="351" Body="&lt;h2&gt;Backstory&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Blaine Fairway, a senior manager at CurrFlow Industries, Inc., has been personally accepting large rebates for multi-billion-dollar purchases from certain manufacturers for years, although that money legally should have gone to his company.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Extortion&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Recently, 15 of his subordinates found out about Blaine's illegal activities, but instead of reporting his crimes to the company, they decided as a group that they are entitled to most of that money. So, they demanded that the past 15 rebate payments be distributed to them &lt;strong&gt;fairly&lt;/strong&gt; among all their teams, or else they would report him. &lt;strong&gt;Fairly&lt;/strong&gt; means that the &lt;em&gt;average&lt;/em&gt; of the amounts of money, for the subordinates on a particular team, is the same for each team. There are 9 teams, as detailed below. Each subordinate is on 2 teams.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Problem&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Blaine still has each of the past 15 payments, which he has stored in small money bags. He is willing to pay them off, but he cannot figure out a &lt;strong&gt;fair&lt;/strong&gt; way to distribute the money to them. He knows that the average amount of money on each of the 9 teams must be the same, or they will be angry and surely report him.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Question&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Is it possible for Blaine to distribute the 15 money bags (below), 1 to each subordinate, such that it is &lt;strong&gt;fair&lt;/strong&gt;? If so, give a possible distribution. If not, prove it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/D1gvI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Team A: Abe, Ally, Abbie, Aaron&#xA;Team B: Bo, Brie, Ben, Bob, Brianna&#xA;Team C: Chris, Cedric, Candy, Cindy, Cam, Candace&#xA;Team 1: Abe, Bo, Chris&#xA;Team 2: Ally, Brie, Cedric&#xA;Team 3: Abbie, Ben, Candy&#xA;Team 4: Aaron, Bob, Cindy&#xA;Team 5: Brianna, Cam&#xA;Team 6: Candace&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="463" LastEditorUserId="22922" LastEditDate="2016-09-15T07:56:01.147" LastActivityDate="2016-09-15T07:56:01.147" Title="Can Blaine distribute 15 money bags fairly?" Tags="|mathematics|calculation-puzzle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11899" PostTypeId="2" ParentId="11895" CreationDate="2015-04-11T19:35:33.973" Score="6" Body="&lt;p&gt;The newly formulated question stipulates that the &lt;em&gt;average&lt;/em&gt; amount of money on each team must be the same &amp;mdash; i.e. \$80,000 per employee. This is a much less trivial problem than the previous formulation, as now there is a plausible starting point. We can formulate the problem now as a grid sum puzzle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{matrix}&#xA;A_1 &amp;amp; B_1 &amp;amp; C_1 &amp;amp; 240 \\&#xA;A_2 &amp;amp; B_2 &amp;amp; C_2 &amp;amp; 240 \\&#xA;A_3 &amp;amp; B_3 &amp;amp; C_3 &amp;amp; 240 \\&#xA;A_4 &amp;amp; B_4 &amp;amp; C_4 &amp;amp; 240 \\&#xA;    &amp;amp; B_5 &amp;amp; C_5 &amp;amp; 160 \\&#xA;    &amp;amp;     &amp;amp; C_6 &amp;amp; 80 \\&#xA;320 &amp;amp; 400 &amp;amp; 480 &amp;amp; &#xA;\end{matrix}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the numbers in each row add up to the number on the right, the numbers on each column add up to the number below, and the numbers to be filled in are the values of the bags.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Firstly, Candace on her own team must receive the $80,000 bag.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{matrix}&#xA;A_1 &amp;amp; B_1 &amp;amp; C_1 &amp;amp; 240 \\&#xA;A_2 &amp;amp; B_2 &amp;amp; C_2 &amp;amp; 240 \\&#xA;A_3 &amp;amp; B_3 &amp;amp; C_3 &amp;amp; 240 \\&#xA;A_4 &amp;amp; B_4 &amp;amp; C_4 &amp;amp; 240 \\&#xA;    &amp;amp; B_5 &amp;amp; C_5 &amp;amp; 160 \\&#xA;    &amp;amp;     &amp;amp;  80 &amp;amp; 80 \\&#xA;320 &amp;amp; 400 &amp;amp; 480 &amp;amp; &#xA;\end{matrix}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, Brianna and Cam must receive a total of \$160,000, so their only options are a 69/91 or a 63/97 split. We can't fill in the table with this information yet, but we can find four sets of three numbers that add up to 240.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Remaining: 55 63 65 68 69 81 82 85 87 88 89 91 97 100&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We've deduced that the \$100K bag must be in a triplet, with the other two bags summing to \$140,000. The only pair that satisfies this is 55 and 85.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Remaining: 63 65 68 69 81 82 87 88 89 91 97&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Next, we try the \$97K bag, with the other two bags summing to \$143,000. There's no pair in the remaining bags that satisfies this, so we deduce that Brianna and Cam must have a 63/97 split, and move on to the next set of bags.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Remaining: 65 68 69 81 82 87 88 89 91&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Next, the \$91K bag, with the other two bags summing to \$149,000. We find that 68 and 81 satisfy this one.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Remaining: 65 69 82 87 88 89 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Next, the \$89K bag, with the other two bags summing to \$151,000. We find that 69 and 82 satisfy this one.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Remaining: 65 87 88&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And finally, the last three remaining bags sum up to \$240,000 as required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now we've got all the rows, but we need to arrange them so that the columns sum up as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{matrix}&#xA; 55 &amp;amp;  85 &amp;amp; 100 &amp;amp; 240 \\&#xA; 65 &amp;amp;  87 &amp;amp;  88 &amp;amp; 240 \\&#xA; 68 &amp;amp;  81 &amp;amp;  91 &amp;amp; 240 \\&#xA; 69 &amp;amp;  82 &amp;amp;  89 &amp;amp; 240 \\&#xA;    &amp;amp;  63 &amp;amp;  97 &amp;amp; 160 \\&#xA;    &amp;amp;     &amp;amp;  80 &amp;amp; 80 \\&#xA;320? &amp;amp; 400? &amp;amp; 480? &amp;amp; &#xA;\end{matrix}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we want to try to get the leftmost column (team A) to sum up to 320. The sum $85 + 65 + 81 + 89$ works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{matrix}&#xA; 85 &amp;amp;  55 &amp;amp; 100 &amp;amp; 240 \\&#xA; 65 &amp;amp;  87 &amp;amp;  88 &amp;amp; 240 \\&#xA; 81 &amp;amp;  68 &amp;amp;  91 &amp;amp; 240 \\&#xA; 89 &amp;amp;  82 &amp;amp;  69 &amp;amp; 240 \\&#xA;    &amp;amp;  63 &amp;amp;  97 &amp;amp; 160 \\&#xA;    &amp;amp;     &amp;amp;  80 &amp;amp; 80 \\&#xA;320 &amp;amp; 400? &amp;amp; 480? &amp;amp; &#xA;\end{matrix}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, to get the middle column (team B) to sum up to 400 using only numbers from the middle column to the right column. The sum $100 + 87 + 68 + 82 + 63$ works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{matrix}&#xA; 85 &amp;amp; 100 &amp;amp;  55 &amp;amp; 240 \\&#xA; 65 &amp;amp;  87 &amp;amp;  88 &amp;amp; 240 \\&#xA; 81 &amp;amp;  68 &amp;amp;  91 &amp;amp; 240 \\&#xA; 89 &amp;amp;  82 &amp;amp;  69 &amp;amp; 240 \\&#xA;    &amp;amp;  63 &amp;amp;  97 &amp;amp; 160 \\&#xA;    &amp;amp;     &amp;amp;  80 &amp;amp; 80 \\&#xA;320 &amp;amp; 400 &amp;amp; 480 &amp;amp; &#xA;\end{matrix}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the remaining numbers will naturally sum to 480, as expected. There are more possible configurations, involving swapping the top four rows or the bottom two numbers, but this is one of the possibilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the money is distributed as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Aaron receives &lt;strong&gt;\$89,000&lt;/strong&gt;.&lt;br&gt;&#xA;  Abbie receives &lt;strong&gt;\$81,000&lt;/strong&gt;.&lt;br&gt;&#xA;  Abe receives &lt;strong&gt;\$85,000&lt;/strong&gt;.&lt;br&gt;&#xA;  Ally receives &lt;strong&gt;\$65,000&lt;/strong&gt;.&lt;br&gt;&#xA;  Ben receives &lt;strong&gt;\$68,000&lt;/strong&gt;.&lt;br&gt;&#xA;  Bo receives &lt;strong&gt;\$100,000&lt;/strong&gt;.&lt;br&gt;&#xA;  Bob receives &lt;strong&gt;\$82,000&lt;/strong&gt;.&lt;br&gt;&#xA;  Brianna receives &lt;strong&gt;\$63,000&lt;/strong&gt;.&lt;br&gt;&#xA;  Brie receives &lt;strong&gt;\$87,000&lt;/strong&gt;.&lt;br&gt;&#xA;  Cam receives &lt;strong&gt;\$97,000&lt;/strong&gt;.&lt;br&gt;&#xA;  Candace receives &lt;strong&gt;\$80,000&lt;/strong&gt;.&lt;br&gt;&#xA;  Candy receives &lt;strong&gt;\$91,000&lt;/strong&gt;.&lt;br&gt;&#xA;  Cedric receives &lt;strong&gt;\$88,000&lt;/strong&gt;.&lt;br&gt;&#xA;  Chris receives &lt;strong&gt;\$55,000&lt;/strong&gt;.&lt;br&gt;&#xA;  Cindy receives &lt;strong&gt;\$69,000&lt;/strong&gt;.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-04-11T19:35:33.973" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11900" PostTypeId="1" CreationDate="2015-04-11T19:44:32.063" Score="4" ViewCount="1011" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Shiny as a diamond&lt;br&gt;&#xA;  Once thought to be under the sea&lt;br&gt;&#xA;  Better than a snake&lt;br&gt;&#xA;  I seek to bring joy to many  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;You are my master, I control your slave&lt;br&gt;&#xA;  Read my words only where defined&lt;br&gt;&#xA;  If you don't exist, to me, you're a liar&lt;br&gt;&#xA;  Tell me what I am, if you'd be so kind.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Every line has meaning, (Well, maybe not the last one, that's just a rhyme to finish the poem) so you have to explain it &lt;strong&gt;all&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2015-04-11T19:50:44.517" LastActivityDate="2018-05-01T04:13:44.763" Title="Shiny as a diamond - What am I?" Tags="|riddle|" AnswerCount="3" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="11901" PostTypeId="2" ParentId="11900" CreationDate="2015-04-11T20:08:09.177" Score="5" Body="&lt;p&gt;Shiny as a diamond&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pe[a]rls are shiny, as are diamonds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once thought to be under the sea&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Double meaning: Some consider perl to be not as good as (and thus under) C; also, pearls come from the ocean&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Better than a snake&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Perl is clearly better than Python (citation needed)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I seek to bring joy to many&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Language was written to fill a gap in available languages; also, see any of Larry Wall's early State of the Onion talks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are my master, I control your slave&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Programmers control computers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Read my words only where defined&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A reference to both perl's use as a text processing language and to undefined variables&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you don't exist, to me, you're a liar&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Undefined values evaluate as false in perl&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tell me what I am, if you'd be so kind.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Perl language&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11271" LastEditorUserId="11271" LastEditDate="2015-04-11T20:09:03.740" LastActivityDate="2015-04-11T20:09:03.740" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11902" PostTypeId="2" ParentId="11536" CreationDate="2015-04-11T20:16:19.583" Score="1" Body="&lt;p&gt;I think the next number in the sequence is: 99&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The numbers currently add up to 510. 510+99 is 609 (the date) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or possibly: 320&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; By taking the numbers at positions 1,9,2 after the 192&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11150" LastEditorUserId="11150" LastEditDate="2015-04-12T08:32:10.267" LastActivityDate="2015-04-12T08:32:10.267" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11903" PostTypeId="2" ParentId="11900" CreationDate="2015-04-12T00:51:14.213" Score="10" Body="&lt;p&gt;Taking this in another direction, because the OP rejected my first answer,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shiny as a diamond&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rubies are shiny, as are diamonds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once thought to be under the sea&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the two proposed names for the language were Ruby and Coral.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Better than a snake&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ruby advocates claim it is better than Python (citation needed)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I seek to bring joy to many&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Matsumoto, Ruby's creator wrote &quot;For me the purpose of life is partly to have joy. Programmers often feel joy when they can concentrate on the creative side of programming, So Ruby is designed to make programmers happy.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are my master, I control your slave&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Programmers control computers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Read my words only where defined&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A reference to undefined variables&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you don't exist, to me, you're a liar&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Undefined values evaluate as false in Ruby, as in many languages.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tell me what I am, if you'd be so kind.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Ruby language&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11271" LastActivityDate="2015-04-12T00:51:14.213" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11906" PostTypeId="2" ParentId="11536" CreationDate="2015-04-12T07:24:47.763" Score="5" Body="&lt;p&gt;&lt;strong&gt;tl;dr&lt;/strong&gt;: most likely not solvable with the information given.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The next number is $-642$ and the pattern is formed by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{18}\cdot{}x^7&#xA;- \frac{1411}{720}\cdot{}x^6&#xA;+ \frac{19441}{720}\cdot{}x^5&#xA;- \frac{26609}{144}\cdot{}x^4&#xA;+ \frac{94175}{144}\cdot{}x^3&#xA;- \frac{49714}{45}\cdot{}x^2&#xA;+ \frac{12509}{20}\cdot{}x^1&#xA;+177$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, probably not: it isn't stated in the question, but the solution presumably is a positive integer (though regression analysis suggests that it has to be negative).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most straightforward solution actually is simple addition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$510 = 192+30+55+86+43+22+79+3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More serious considerations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I shall be assuming that this is purely mathematical, i.e. not relying on trivia like the character's names, the reciptionist's birthday or the arrangement of buttons on the input pad. Furthermore people go into finance because they are bad at maths, so only elementary operations are permitted and the numbers are used in the order they show up in (and not reverse or sorted ascending or descending, alternating front/back or some non-derivable order). Nor shall there be skips in the sequence (e.g. taking only the prime-numbered elements of the sequence generated).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I were Lalozik I would have the sequence of numbers change every time the safe is attempted to be opened: while the rule for generation and solution would remain the same, the seed for generation would change randomly, hence one couldn't predict the next number without knowing the algorithm. However, this is obviously not the case, since otherwise Banes would have staked out the bank beforehand and seen the patterns and possibly the solutions to previous times the safe was opened, so those sequences would be provided to us for solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For it to be possible to derive a solution by looking at the list of numbers with such restricted information, the algorithm has to repeat: for example, for $1, 2, 3, 5, 8$ one would guess the fibonacci numbers where $x_{n+2}=x_{n+1}+x_{n}$, but this cannot be derived if only $1, 2$ are given since $3$ would be the very first number where the formula applies. Indeed, even $1, 2, 3$ doesn't suffice: there need at least two repeats of the application of the formula within the sequence, one to establish the pattern and one to affirm it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise if someone came up with a different solution one couldn't dismiss it: if there is no repeating pattern then straightforward addition is the simplest solution. Hence at most 7 of the numbers can be used to derive the solution, the 8th must serve for confirmation - the fewer needed the more plausible the rule becomes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, it can be easily shown (google &quot;&lt;em&gt;common difference&lt;/em&gt;&quot;) that in the present case there exists no polynomial that can generate these numbers in a manner that permits the solution to be confirmed. Of course, this still allows any algorithm that is more complex than a polynomial (e.g. involving floor functions or cosines) - though then the claim that the scheme is &quot;very simple and straightforward&quot; is misleading at best.&lt;/p&gt;&#xA;" OwnerUserId="10070" LastActivityDate="2015-04-12T07:24:47.763" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11908" PostTypeId="1" AcceptedAnswerId="11911" CreationDate="2015-04-12T08:38:54.100" Score="5" ViewCount="1949" Body="&lt;p&gt;Some time in the early 20th century, a fire occurred in a crowded theatre. Although the exit doors that were unlocked were not blocked on the outside, the panicked attendees could not get out of those doors, and hundreds of people in the theatre died from the fire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happened to trap everybody in the theatre? What happened differently from then on?&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="8874" LastEditDate="2016-03-23T17:05:06.000" LastActivityDate="2016-03-23T17:05:06.000" Title="Fire in a crowded theatre" Tags="|lateral-thinking|situation|history|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11909" PostTypeId="2" ParentId="11908" CreationDate="2015-04-12T08:53:58.500" Score="5" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Iroquois Theatre fire&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;http://en.wikipedia.org/wiki/Iroquois_Theatre_fire&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Iroquois_Theatre_fire&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reason:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The people who were near the doors, couldn't open it because they had to pull it and in the same time the crowd were pushing them to the doors, that's why people were trapped and died. This explain &lt;em&gt;&quot;Although the exit doors that were unlocked were not blocked on the outside, the panicked attendees could not get out of those doors, and hundreds of people in the theatre died from the fire&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What changed since then:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; tl;dr&#xA; &lt;br&gt;&#xA; 1)The Iroquois fire prompted widespread implementation of the panic bar - &lt;a href=&quot;http://en.wikipedia.org/wiki/Crash_bar&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Crash_bar&lt;/a&gt;&#xA; &lt;br&gt;&#xA; 2) A second result of the fire was the requirement that a fireproof asbestos curtain be raised before each performance and lowered afterward to separate the audience from the stage.&#xA; &lt;br&gt;&#xA; 3) The third result was that all doors in public buildings must open in the direction of egress. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11145" LastEditorUserId="11145" LastEditDate="2015-04-12T10:04:51.063" LastActivityDate="2015-04-12T10:04:51.063" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11910" PostTypeId="2" ParentId="11829" CreationDate="2015-04-12T09:20:23.517" Score="5" Body="&lt;p&gt;A wrote just a part of her age, B and C did the same. Then, A added the remaining part of her age, as well as B and C.&lt;/p&gt;&#xA;" OwnerUserId="11284" LastEditorUserId="10281" LastEditDate="2015-04-12T09:30:57.340" LastActivityDate="2015-04-12T09:30:57.340" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11911" PostTypeId="2" ParentId="11908" CreationDate="2015-04-12T09:28:29.283" Score="8" Body="&lt;p&gt;&lt;strong&gt;What was the cause?&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on what Marin Takanov already said, I would identify the main negligence, which led to the disaster, in the lack of outward-opening doors! &#xA;In fact, panic generated a huge crowd: people in front of the doors couldn't pull the handle to open the doors because the huddle was pushing behind them!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How was this problem fixed?&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply replacing inward-opening doors with outward-opening ones, so that you have to push (not pull) the door in order to open it.&lt;/p&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-04-12T09:28:29.283" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11912" PostTypeId="2" ParentId="11908" CreationDate="2015-04-12T11:09:36.167" Score="3" Body="&lt;p&gt;There's a psychological phenomenon whereby people are more likely try to exit through the main entrance, or the one through which they entered the building, rather than use fire exits. This can cause significant and potentially life-threatening delays in evacuation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is affected by factors such as:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Presence / absence / thickness of smoke (absence of smoke = lack of urgency = people tend not to use fire exits)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Signage / wayfinding aids (e.g. low-level aids to finding fire exits)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Familiarity with the building&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Group behaviour - which way is everyone else going? (and in particular which way are the other the members of a close group such as a family, going?)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There are other factors at work, such as familiarity with the escape route: research suggests that people are often more likely to use a familiar exit that is further away than an unfamiliar exit nearby – again increasing evacuation times. This was often attributed to panic, although social scientists now seem to agree that seemingly-irrational flight behavior to an exit is perfectly rational: they are rationally seeking out an exit with which they are familiar. Group dynamics also plays a part; how a number of people, in different states of fear, influence one another. Evidence suggests, for example, that people evacuate buildings alongside colleagues with whom they have an emotional attachment In a shopping centre that social dynamic may be familial, with parents trying to control small children. How does the family group react, when each member is instinctively responding in a different way?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.e-architect.co.uk/articles/fire-dynamics-human-behaviour&quot; rel=&quot;nofollow&quot;&gt;Fire Dynamics &amp;amp; Human Behaviour&lt;/a&gt;, AJ Welch, 7 Jan 2015&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;People generally exit through the main entrance without the presence of smoke; in the presence of smoke they are more likely to use the fire exit, and low-placed signs facilitate the use of the fire exit nearest the evacuee. Further familiarity with [the building] was related to exit strategy; with less experience, people are more likely to exit via the main entrance, not the fire exit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://books.google.co.uk/books?id=kKovr__MBnUC&amp;amp;lpg=PA56&amp;amp;ots=9Zo-uCopf5&amp;amp;dq=psychology%20fire%20exit%20through%20entrance&amp;amp;pg=PA56#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;The Oxford Handbook of Environmental and Conservation Psychology&lt;/a&gt;, Susan D. Clayton, p.56&lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="4050" LastEditDate="2015-04-12T11:15:22.690" LastActivityDate="2015-04-12T11:15:22.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11913" PostTypeId="1" AcceptedAnswerId="12553" CreationDate="2015-04-12T11:22:03.037" Score="69" ViewCount="3104" Body="&lt;p&gt;There are people who claim that they can look upon the land, and the land will speak to them.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/emX31.png&quot; alt=&quot;hourglass&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;I recently got caught in a sandstorm, and particles of sand flew past my eyes.&lt;br /&gt;&#xA;&lt;strong&gt;What is the desert saying?&lt;/strong&gt;&lt;br /&gt;&#xA;&lt;sup&gt;&lt;em&gt;(The solution is six letters)&lt;/em&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="7308" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-11-21T22:35:45.040" Title="The Sands of Time" Tags="|word|pattern|steganography|" AnswerCount="3" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="11915" PostTypeId="1" CreationDate="2015-04-12T12:53:20.587" Score="3" ViewCount="338" Body="&lt;p&gt;$$(d+(5m+m/9)/2+5y/4-y/C+y/CD+1/(m/3+1)*(2-(1/(y\%4+1)+1/(y\%CD+1)-1/(y\%C+1))))\%7$$&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What does this thing do? Explain.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can you make it simpler i.e. fewer characters (I will give a nice bounty for this)?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Important information (Hintish): &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is not a riddlish trick. It is a math formula that would run as Java code.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2015-04-14T15:06:22.090" LastActivityDate="2015-04-14T15:06:22.090" Title="What does this thing do?" Tags="|mathematics|computer-puzzle|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11916" PostTypeId="2" ParentId="11915" CreationDate="2015-04-12T13:05:07.403" Score="6" Body="&lt;p&gt;The only thing I can think of is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it gives you the day of the week from the date of the year (d,m,y with C and CD giving you information about the century and 4-century for leap-year purposes).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now I have to work out the algebra to prove it - thanks Cop for wasting my time on this! :-[&#xA;&lt;hr&gt;&#xA;Edit: looks similar to the C formula&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Determination_of_the_day_of_the_week#Implementation-dependent_methods&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Progress&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The input quantities are as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $d$ is the day, $m$ is the month, $y$ is the year, $C=100$, $CD=400$ (Roman numerals - crafty!).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First consider the quantity $(1/(y\%4+1)+1/(y\%CD+1)-1/(y\%C+1))$. There are four cases to consider:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;if $y$ is not a multiple of 4, then the first fraction is at most $\frac{1}{2}$ and (since 4 divides $CD$ and $C$) the whole sum is less than 1&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;if $y$ is a multiple of 4 but not of $C$, then the first fraction is exactly 1 and either the other two are equal or the $C$ one is smaller than the $CD$ one, so the whole sum is at least 1&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;if $y$ is a multiple of $C$ but not of $CD$, then the first and last fractions are exactly 1 and cancel out to leave the whole sum less than 1&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;if $y$ is a multiple of $CD$, then all three fractions are exactly 1 and the whole sum is 1.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So $(2-(1/(y\%4+1)+1/(y\%CD+1)-1/(y\%C+1)))$ is less than or equal to 1 iff&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the year denoted by $y$ is a leap year.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also $(m/3+1)$ is less than 2 iff&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the month denoted by $m$ is January or February.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the final term in the sum, $(1/(m/3+1)(2-(1/(y\%4+1)+1/(y\%CD+1)-1/(y\%C+1))))$, is greater than $\frac{1}{2}$ iff we need to alter the final answer due to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a leap year.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-04-12T13:26:09.473" LastActivityDate="2015-04-12T13:26:09.473" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="11917" PostTypeId="1" AcceptedAnswerId="11923" CreationDate="2015-04-12T13:17:39.043" Score="20" ViewCount="1465" Body="&lt;p&gt;It's wartime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fred and Gus are friends and soldiers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When Gus decided to move far away from the cavalry, he saw an enemy coming toward his camp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fred also saw the enemy, and did his best to protect Gus, but because of his love for his friend, his country immediately lost the entire war.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The entire battle is described above. No actions are taken prior to Gus's movement away from the cavalry.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What specific scenario is described above?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-06-17T15:42:03.380" LastActivityDate="2015-06-17T15:42:03.380" Title="What is this scenario?" Tags="|lateral-thinking|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11918" PostTypeId="1" AcceptedAnswerId="11939" CreationDate="2015-04-12T13:43:11.980" Score="6" ViewCount="926" Body="&lt;p&gt;Late at night at a bar in the Wild West only two cowboys from out of town are left, playing poker with the barkeeper. One of them cheats, so another throws over the table and they have a fistfight (the barkeeper wisely having confiscated the guns), ruining the bar. Awakened by this, the neighbors call for the sheriff to put someone behind bars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sheriff knows that white hats always tell the truth, black hats always lie, and barkeepers answer like the one asked right before him would. Unfortunately, the hats were knocked off during the fight. Also, neither the sheriff (nor the neighbors, who went back to sleep already) know which one is the barkeep because he's new in town, shaved this morning for the first time in years, it's always quite dark in the bar and whatever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What questions should the sheriff ask to find out who started the fight?&lt;/p&gt;&#xA;" OwnerUserId="10070" LastActivityDate="2019-10-17T18:06:31.220" Title="Knights and Knaves in the Wild West" Tags="|logical-deduction|liars|" AnswerCount="4" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="11919" PostTypeId="2" ParentId="11334" CreationDate="2015-04-12T13:47:04.417" Score="5" Body="&lt;p&gt;I can beat the &quot;one could think&quot; bound in the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, here's a strategy that matches the &quot;one could think&quot; bound. Call the picker Alice and the guesser Bob. The players choose a sequence $T_1, T_2, ... T_{26}$ of independent uniformly random orderings of the 52 playing cards. (Each $T_i$ is some ordering of the cards from lowest to highest, and all the $T_i$ are picked at random from all 52! possibilities.) Every round $i$, Alice plays the card which $T_i$ says is largest, and Bob guesses the card $T_i$ says is larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alice loses a round iff the hidden card is larger, $T_i$-wise, than all of hers. Therefore, the chance of winning round 1 is $5/6$. What about round 2? You might worry that the rounds are dependent -- for example, that after round 1, Alice might be more likely to have the eight of clubs, and that screws up the chances. But, by picking a uniformly random $T_2$, Alice and Bob are in effect randomly relabeling all of the cards before round 2 starts. This destroys any dependency of the cards upon the previously played rounds. Therefore, round 2 has the same winning chance as round 1: $5/6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The logic continues for rounds 3 and above. The sum for all rounds is $\frac 56+\frac 56+\cdots+\frac 56+\frac 45+\frac 34+\frac 23+\frac 12$, which is exactly the &quot;one would think&quot; bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However!!&lt;/strong&gt; Some choices of the $T_i$'s are exceptionally bad. For example, if all of them are equal (e.g. Alice plays the highest value card every round), then Alice will often run out of high cards and become sad. I believe you can argue that in this case the win rate will be 1/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a sequence of orders which does worse than average. Since there is a sequence that does worse than average, there must also be one which does better than average. (Cuz otherwise it wouldn't be, ya know, the average.) Therefore, by using this better sequence, Alice and Bob beat the &quot;one would think&quot; bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit:&lt;/em&gt; Thanks klm123 for asking good questions, I've hopefully made the explanation much better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit 2:&lt;/em&gt; Here's a specific $T_i$ sequence which, in simulation, beats all the other answers so far. Call the cards 1 through 52. Every $T_i$ is some cyclic ordering, for example, 3 &gt; 2 &gt; 1 &gt; 52 &gt; 51 &gt; ... &gt; 4. In round 1 the ordering is 52 &gt; 50 &gt; ... &gt; 1. In round 2 it's 19 &gt; 18 &gt; ... &gt; 20 : the same cycle, rotated backwards by 33. The cycle continues shifting backwards 33 cards every round (golden ratio hype). After 100,000 simulated rounds, this strategy averaged &lt;strong&gt;21.7&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1752" LastEditorUserId="1752" LastEditDate="2015-04-13T07:56:49.623" LastActivityDate="2015-04-13T07:56:49.623" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="11920" PostTypeId="2" ParentId="11917" CreationDate="2015-04-12T14:20:52.403" Score="4" Body="&lt;p&gt;It's &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A game of chess. The king is captured&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;when Fred goes to protect his friend.&lt;/p&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2015-04-12T14:20:52.403" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11921" PostTypeId="2" ParentId="11595" CreationDate="2015-04-12T14:41:58.990" Score="3" Body="&lt;p&gt;Here is the answer I had in mind:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In more ways than one,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Literally&lt;/em&gt; in more ways than one (see next sentence)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My end brings a start,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The end of a gun can bring the start of the race. Also, considering its &quot;in more ways than one&quot;, the end of a gun can also bring a start in the sense of being startled. The end of a gun is where the bullet exits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But can also bring the end itself.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Guns can end lives.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I end all fun,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Two senses: first is that shooting somebody is fairly unpleasant. Secondly, this is a letter line: the end of 'all fun' is 'un'. Couple this with the next sentence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I only ever began with grief.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The word gun begins with 'g'. Coupled with the ending 'un' from the previous line this spells 'gun'. Also, in another sense, guns were invented by people who intended them to cause trouble (for the most part).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;BONUS:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Kudos if you found this: the title can also be called the 'head' of a document. &quot;If I'm held here I'm dangerous&quot;. If a gun is held at somebody's head...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10974" LastActivityDate="2015-04-12T14:41:58.990" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11922" PostTypeId="2" ParentId="11917" CreationDate="2015-04-12T15:08:48.253" Score="3" Body="&lt;p&gt;Expanding on @Ian 's answer...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is quite soon after the beginning of the game. The white queen is out (not dead, I mean it is out of its position, free to move), so one white pawn is also out. Gus is the black pawn in front of the black knight, and Fred is the next pawn, which is in front of the black bishop. Both pawns are on the 'king-side' of the black army, and have not moved yet. Both sides may have made trivial moves to reach this setup.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When Gus decided to move far away from the cavalry, he saw an enemy&#xA;  coming toward his camp.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Gus moves 2 steps forward. White makes a move with the queen bringing it directly in front of Gus (no space).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Fred also saw the enemy, and did his best to protect Gus, but because&#xA;  of his love for his friend, his country immediately lost the entire&#xA;  war.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As Gus is in danger of the queen, Fred moves 1 step forward to protect it. The queen moves to the edge of the board (adjacent to Gus), and gives a check along a diagonal line to the black king. This is also a checkmate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There should be other ways to answer this question, I've just given one of the simplest ones.&lt;/li&gt;&#xA;&lt;li&gt;Someone who has more time could please add an image to clarify the answer?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-04-12T15:08:48.253" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11923" PostTypeId="2" ParentId="11917" CreationDate="2015-04-12T15:32:14.010" Score="15" Body="&lt;p&gt;I think this is about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Fool%27s_mate&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Fool's mate&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I also suppose that Fred and Gus are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Two white pawns (the F Pawn is Fred and the G pawn is Gus)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Picture showing this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/2/23/Fools_mate_animation.gif&quot; alt=&quot;Fool&amp;#39;s mate&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Invert the two moves of the white player showed in the gif. Gus goes away from the cavalry, Fred attempts to cover Gus, but they totally fail to protect their king! These are the moves, if you're interested: g4, e5, f3, Qh4 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:48.760" LastActivityDate="2015-04-13T12:49:01.750" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="11925" PostTypeId="2" ParentId="11334" CreationDate="2015-04-12T17:07:09.460" Score="2" Body="&lt;p&gt;I like the &lt;a href=&quot;https://puzzling.stackexchange.com/a/11342/28&quot;&gt;kaine's answer&lt;/a&gt;, but it doesn't use the fact, that the second player knows all the cards the first player has put on the table in all previous tricks.&#xA;The other idea would be: that &lt;strong&gt;the first player can give a hint about his next card&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, among 5 cards there are always 2 with the same suit. So the first player can put one of them first, then the second one. And the second player then knows the suit of the second card for sure, even though he have to guess everything about the first card. Also there are quite high probability (1-(3/4)^5) = 76% that to the 3rd trick the first player will get a card with the suit again. So the second player can continue guessing that the suit hasn't been changed from last trick.&#xA;Furthermore to the 13 ranks of the cards one can apply the same strategy  kaine's applied to 52 cards simultationsly with using hints-strategy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The obvious flow of the strategy is the point when suit is changed and the second player guesses incorrect suit (but only in an unfortunate situation when there is a card from the second pile with this suit). One can reduce an influence of this flow using 2 colours of the cards instead of 4 suits, then there is at lease 3 cards with the same colour at the beginning, and 1.5 card with this colour will be drawn during first 3 tricks, and +0.75 cards during next 1.5 tricks, etc, this makes like 5 cards of the same color in a row. Plus when the color is changed that means that the first player has 5 cards of the same color at his hand! and won't need to change a color for a long time. Meanwhile with the 26 cards of the same color one can freely use kaine's strategy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, the problem is that the kaine's strategy is not quite predictable, and probability can be change easily by using N(n) formula.&lt;br&gt;&#xA;And, as a matter of fact, my simulation gives the following results:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Strategy with 4 suits + kaine's strategy with tuned parameters gives 21.04 = 80.9% correct guess rate.&lt;/li&gt;&#xA;&lt;li&gt;Strategy with 2 colors + kaine's gives 20.92 = 80.4% correct guess rate.&lt;/li&gt;&#xA;&lt;li&gt;Pure Kaine's gives  21.3 = 82.1% correct guess rate.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This is quite unexpected result for me, but I can't find any bug in my program, so I think the thing is number of cards in groups, for 26 and 13 card decks is much harder to apply kaine's strategy than for 52 card deck and this neutralises hints advantage. But I really hope that someone can retest my idea and get something more usefull from it.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-04-12T17:23:33.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11926" PostTypeId="1" AcceptedAnswerId="11928" CreationDate="2015-04-12T17:23:29.660" Score="4" ViewCount="845" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;&quot;Mary had a riddle lamb, brain-tease was hard as stone!&quot;&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scratch that thing,&lt;br&gt;&#xA;It's the name of this being!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not only a furry creature,&lt;br&gt;&#xA;But the sweetest of nature!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Damn horrible intruder,&lt;br&gt;&#xA;Get out of my computer!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surely ain't emblem of pluck,&lt;br&gt;&#xA;At most it may bring luck!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Riddle Lamb #1 &lt;strong&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/11100/mary-had-a-riddle-lamb-1&quot;&gt;here&lt;/a&gt;&lt;/em&gt;&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-05-22T09:36:23.627" Title="Mary had a riddle lamb #2" Tags="|riddle|rhyme|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11927" PostTypeId="2" ParentId="11926" CreationDate="2015-04-12T17:30:32.267" Score="1" Body="&lt;p&gt;It's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a CAT (again!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Scratch that thing,&lt;br&gt;&#xA; It's the name of this being!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A cat can scratch, or you might scratch one behind its ears. The letters C, A, T are contained in the word &quot;scratch&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Not only a furry creature,&lt;br&gt;&#xA; But the sweetest of nature!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cats are furry creatures, and (at least some of them) have sweet natures.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Damn horrible intruder,&lt;br&gt;&#xA; Get out of my computer!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Is there a type of virus called a cat?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Surely ain't emblem of pluck,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A lion is sometimes used as a symbol of courage.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;At most it may bring luck!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cats, especially black ones, are seen as bringing luck.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Maybe the title should have contained the word &quot;pray&quot;? ;-)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4615/you-will-have-to-pray&quot;&gt;You will have to pray!&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;https://puzzling.stackexchange.com/questions/5141/riddle-pray-you-can-find-the-word&quot;&gt;riddle - pray you can find the word!&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;https://puzzling.stackexchange.com/questions/5428/riddle-pray-that-this-is-the-last-one&quot;&gt;Riddle... pray that this is the last one&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-04-12T17:36:02.227" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11928" PostTypeId="2" ParentId="11926" CreationDate="2015-04-12T17:39:08.163" Score="4" Body="&lt;p&gt;I thought the answer was&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Horse&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Doesn't seem to match every line though&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scratch that thing,&#xA;It's the name of this being!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not only a furry creature,&#xA;But the sweetest of nature!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Horses have sweet natures&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Damn horrible intruder,&#xA;Get out of my computer!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Trojan horse&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Surely ain't emblem of pluck,&#xA;At most it may bring luck!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lucky horse shoe&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;guess 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rabbit&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Scratch that thing,&#xA;It's the name of this being!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rabbits scratch, and some are called Scratch&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not only a furry creature,&#xA;But the sweetest of nature!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rabbits are sweet natured&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Damn horrible intruder,&#xA;Get out of my computer!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Bunnie&quot; &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Andrew_Huang&quot; rel=&quot;nofollow&quot;&gt;http://en.m.wikipedia.org/wiki/Andrew_Huang&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Surely ain't emblem of pluck,&#xA;At most it may bring luck!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rabbits foot&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11150" LastEditorUserId="11150" LastEditDate="2015-04-12T17:51:17.023" LastActivityDate="2015-04-12T17:51:17.023" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11929" PostTypeId="2" ParentId="10968" CreationDate="2015-04-12T17:43:47.590" Score="5" Body="&lt;p&gt;I think the 3-digit code could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 269&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the graph that is created when the lines are extended to the next dot, has &lt;strong&gt;269&lt;/strong&gt; partitions of different sizes and shapes. (thanks to the picture posted in the question's comments by Ivo Beckers, which was missing a couple lines) &lt;img src=&quot;https://i.stack.imgur.com/dE0X0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-04-12T21:17:58.957" LastActivityDate="2015-04-12T21:17:58.957" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="11930" PostTypeId="2" ParentId="11926" CreationDate="2015-04-12T17:55:32.067" Score="4" Body="&lt;p&gt;I think it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A hare (similar to a rabbit)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Scratch that thing,&#xA;It's the name of this being!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I guess it's common to scratch you head (hair -- hare)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not only a furry creature,&#xA;But the sweetest of nature!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rabbits (hares) are furry and sweet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Damn horrible intruder,&#xA;Get out of my computer!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ever get a piece of hair stuck in your keyboard? or even in your pc?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Surely ain't emblem of pluck,&#xA;At most it may bring luck!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; pluck is something you can do to hair (play on words- hare), and rabbit's feet are lucky&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastActivityDate="2015-04-12T17:55:32.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11931" PostTypeId="2" ParentId="11918" CreationDate="2015-04-12T18:30:48.817" Score="3" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;NOTE:&lt;/em&gt;&lt;/strong&gt; I've revised this answer so that it can work with any number of white hats and black hats. The only requirement is that there is only &lt;em&gt;one&lt;/em&gt; bartender (and of course the rules stated in the question still apply here). This method will also work if more than one person started the fight since this method involves determining the identity of every single person.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Stage 1: Finding the bartender&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First, put the 3 unknowns into a line. We then question each person starting from 1 to 3 and ask them the following question.&#xA; &lt;br&gt;&lt;strong&gt;Are you a black hat?&lt;/strong&gt;&#xA; &lt;br&gt;&#xA; &lt;img src=&quot;https://lh5.googleusercontent.com/-SWU-mQ1zHNc/VSseq_FuuDI/AAAAAAAAAx8/w44_LpwvDbs/w638-h468-no/cases.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA; &lt;br&gt;&#xA; &lt;br&gt; Both the white hats and black hats will &lt;em&gt;always&lt;/em&gt; answer no. claiming that they are not black hats, (highlighted in red in the chart below). On the other hand, the bartender will sometimes answer yes and sometimes answer no; depending on if the last person asked was a white hat or black hat.&#xA; &lt;br&gt;&#xA; &lt;br&gt;If we ask the question and everyone says no, then we simply ask again in a different order. Eventually, we will fall on the case where the bartender answers yes, and thus we have found the bartender.&#xA; &lt;br&gt;&#xA; &lt;br&gt;&lt;em&gt;NOTE: There is a 50% chance that the bartender will say yes as shown in the truth table.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Stage 2: Determining the identites of all the people&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Once we know who the bartender is, we can use the fact that the bartender will always answer the same way as the last person who was asked. To determine the identity of an unknown person we simply ask them if they are a black hat, then we proceed to ask the bartender if the bartender themself is a black hat.&#xA; &lt;br&gt;&#xA; &lt;br&gt;We have two cases as illustrated in the table below.&#xA; &lt;br&gt;&lt;img src=&quot;https://lh4.googleusercontent.com/-KnNhtaX_N1w/VSsjXRD4nsI/AAAAAAAAAyY/kY9HWpC0kRE/w798-h316-no/determining%2Bidentity.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA; &lt;br&gt;If the last person asked was a black hat, the bartender will also lie and say yes, I am a black hat (which we know is a lie). However if the last person asked was a white hat, then the bartender will answer the truth and say no, I am not a black hat (which we know is true).&#xA; &lt;br&gt;&#xA; &lt;br&gt;Based on whether the bartender answers yes or no, we can determine if the previous person asked was a black hat or a white hat. Using this method, we can then go on to figure out the identities of all the other people (or until we find a white hat for the purposes of this riddle).&#xA; &lt;br&gt;&#xA; &lt;br&gt;Once we've found a white hat, we can ask them who started the fight and thus figure out who needs to be put behind bars.&#xA; &lt;br&gt;&#xA; &lt;br&gt;&lt;em&gt;Or if you want to be real picky and follow the rules all the way through (yes-or-no questions only), you can use the white hat as a pivot to determine if someone had started the fight since they will always tell the truth (i.e. ask the white hat if &quot;this person&quot; started the fight and keep doing this for every person).&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11094" LastEditorUserId="11094" LastEditDate="2015-04-13T02:07:40.960" LastActivityDate="2015-04-13T02:07:40.960" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="11932" PostTypeId="1" AcceptedAnswerId="11935" CreationDate="2015-04-13T00:03:44.913" Score="8" ViewCount="803" Body="&lt;p&gt;Evaluate the following expression to 4 decimal places:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{1+2\sqrt{1+2\sqrt{1+\dots+2\sqrt{1+2\sqrt{2015}}}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the number of square roots in the expression is exactly 2015.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As per the &lt;a href=&quot;/questions/tagged/no-computers&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;no-computers&amp;#39;&quot; rel=&quot;tag&quot;&gt;no-computers&lt;/a&gt; tag, &lt;strong&gt;you may not use a computer, calculator, or any electronic aid in your calculations.&lt;/strong&gt; You may look up the decimal expansion of a certain constant, but the people answering this question will probably have it memorised to several decimal places anyway!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;I believe I found this puzzle somewhere on the internet, rather than having made it up myself, but after such a long time I can't remember where. If anyone finds this puzzle elsewhere on the web, please drop a link in a comment and I'll edit it in.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2016-09-17T18:45:39.410" LastActivityDate="2016-09-17T18:45:39.410" Title="Evaluate the nested square roots" Tags="|mathematics|calculation-puzzle|no-computers|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11933" PostTypeId="2" ParentId="11918" CreationDate="2015-04-13T00:04:40.223" Score="1" Body="&lt;p&gt;If the sheriff asks just one person all the questions, that person will always answer consistently. The cowboys certainly will. The barkeep will start randomly with truth or lie, but then will keep telling the truth or keep lying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sheriff selects &lt;em&gt;any&lt;/em&gt; one of the three people as witness, then points to each person in turn and asks the selected witness, &quot;Did this person start the fight?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The witness would give one answer (Y/N) when you point to the culprit, and the opposite answer (N/Y) when you point to the other two. &lt;strong&gt;Arrest whoever the witness answers differently about.&lt;/strong&gt; E.g. YNY: arrest B; or YNN: arrest A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sheriff asks 3 simple questions in total.&lt;/p&gt;&#xA;" OwnerUserId="9422" LastActivityDate="2015-04-13T00:04:40.223" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11934" PostTypeId="2" ParentId="11932" CreationDate="2015-04-13T01:38:11.103" Score="14" Body="&lt;p&gt;The expression is an approximation of the equation $1 + 2x = x^2$ or $x^2 - 2x - 1 = 0$, so I'm assuming that the answer to it is $x$, which is $\frac{2 + \sqrt{8}}{2}$ or $1 + \sqrt{2}$. This is equal to approximately 2.4142.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-04-13T01:38:11.103" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11935" PostTypeId="2" ParentId="11932" CreationDate="2015-04-13T01:48:14.487" Score="21" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2.4142&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Proof&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;To restate the problem, we are given&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_1=\sqrt{2015}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_{n+1} = \sqrt{1+2a_n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and asked to find $a_{2015}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose this sequence has a limit $L = \lim_{n\rightarrow \infty} a_n$; then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L = \sqrt{1+2L}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which has solution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L = 1+\sqrt{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now look at the difference between $a_n$ and $L$.  If $a_n = L+\Delta$, then a Newton's approximation gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_{n+1} = \sqrt{1+2+2\sqrt{2}+2\Delta} &amp;lt; 1+\sqrt{2} + \frac{\Delta}{1+\sqrt{2}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So by induction we can show that for all $n$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L &amp;lt; a_n \leq L + (\sqrt{2015}-1-\sqrt{2}) \left(1+\sqrt{2}\right)^{1-n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1+\sqrt{2} &amp;lt; a_{2015} &amp;lt; 1+\sqrt{2}+ (\sqrt{2015} -1-\sqrt{2})\left(1+\sqrt{2}\right)^{-2014}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\sqrt{2015}-1-\sqrt{2}) \left(1+\sqrt{2}\right)^{-2014}$ is much less than 0.0001, so it can be ignored when stating the answer to the original question.&lt;/p&gt;&#xA;" OwnerUserId="1706" LastEditorUserId="5373" LastEditDate="2015-04-13T10:05:15.537" LastActivityDate="2015-04-13T10:05:15.537" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11936" PostTypeId="1" AcceptedAnswerId="11938" CreationDate="2015-04-13T02:36:03.730" Score="4" ViewCount="1303" Body="&lt;p&gt;The well-known daredevil, Sauren Baumgartner, is standing on the top of One World Trade Center (in Lower Manhattan) just before sunset. Following his physical and mental preparation, he calmly drops over the edge of the building. During his BASE jumping descent, he spirals four times before landing safely on the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find out where Sauren landed by decoding this cipher:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;065 063 040 065 063 040 063 062 040 065 062 040 065 066 040 063 062 040 065 063 040 065 064 040 063 062 040 065 065 040 065 062 040 063 062 040 065 064 040 065 061 040 063 062 040 065 062 040 065 065 040 063 062 040 065 064 040 065 066 040 063 062 040 065 065 040 065 064 040 063 062 040 065 064 040 065 062 040 063 062 040 065 063 040 065 066 040 063 062 040 065 062 040 061 060 061 040 063 062 040 065 064 040 071 071 040 063 062 040 065 062 040 071 067 040 063 062 040 065 063 040 065 062 040 063 062 040 065 062 040 065 067 040 063 062 040 065 065 040 065 065 040 063 062 040 065 063 040 065 063 040 063 062 040 065 062 040 065 065 040 063 062 040 065 062 040 065 064 040 063 062 040 065 065 040 065 067 040 063 062 040 065 064 040 064 071 040 063 062 040 065 065 040 065 065 040 063 062 040 065 061 040 061 060 060 040 063 062 040 065 061 040 061 060 060  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;strong&gt;(The answer is two words)&lt;/strong&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5188" LastActivityDate="2015-04-13T03:13:10.260" Title="BASE jumping from One World Trade Center" Tags="|cipher|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11937" PostTypeId="2" ParentId="11936" CreationDate="2015-04-13T03:07:07.750" Score="4" Body="&lt;p&gt;Convert from octal to ASCII:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;53 53 32 52 56 ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then convert from decimal to ASCII: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;55 48 56 74 ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then once more from decimal to ASCII:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;708J...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm on my phone right now, but I'll fill in the rest once I have access to a proper keyboard.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-04-13T03:07:07.750" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11938" PostTypeId="2" ParentId="11936" CreationDate="2015-04-13T03:13:10.260" Score="8" Body="&lt;p&gt;The solution is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pumphouse Park&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which is obtained by:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Converting from octal to text which yields:&lt;br&gt;53 53 32 52 56 32 53 54 32 55 52 32 54 51 32 52 55 32 54 56 32 55 54 32 54 52 32 53 56 32 52 101 32 54 99 32 52 97 32 53 52 32 52 57 32 55 55 32 53 53 32 52 55 32 52 54 32 55 57 32 54 49 32 55 55 32 51 100 32 51 100&lt;br&gt;&lt;br&gt;Then converting from decimal to text which yields:&lt;br&gt;55 48 56 74 63 47 68 76 64 58 4e 6c 4a 54 49 77 55 47 46 79 61 77 3d 3d&lt;br&gt;&lt;br&gt;Then converting from hex to text which yields:&lt;br&gt;UHVtcGhvdXNlJTIwUGFyaw==&lt;br&gt;&lt;br&gt;Then converting from BASE64 to text which yields:&lt;br&gt;Pumphouse%20Park&lt;br&gt;&lt;br&gt;Which then URL decodes to 'Pumphouse Park'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7308" LastActivityDate="2015-04-13T03:13:10.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11939" PostTypeId="2" ParentId="11918" CreationDate="2015-04-13T03:37:42.900" Score="8" Body="&lt;h2&gt;The original question&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Ask each person if they are the barkeep. The aim of this question is not to identify the barkeep but to find the white hat or the black hat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If only one person answers 'no' then that person is the white hat (the black hat will always claim to be the barkeep, the barkeep in this case has answered truthfully).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If two people answer 'no' then the one person who answered yes must be the black hat. (the barkeep in this case has answered has lied).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that one of the cowboys has been identified that cowboy can be interrogated to reveal who started the fight, (remembering to assume to opposite if it was the black hat who was initially identified), and if you desire the identities of the other two.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Various variations&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If there happened to be more than two cowboys and/or more than one barkeep the procedure is roughly the same and should work as long as there is at least one white hat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If only one answers 'no' initially they are the only white hat and the problem is solved. (Just ask him who started the fight.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If more than one answers 'no' then you may have more than one white hat or a white hat(s) and a confused barkeep. Ask one of them if they are the barkeep again. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they now answer 'yes' that is the barkeep (and you must have questioned a black hat in the last round of question.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not ask each of those white hat/bar keep candidates if the barkeep has lied so far. If any answer 'yes' then you have just found a white hat amongst this subgroup, again the problem is solved. If no one answers 'yes' to this question the subgroup contains only white hats and your problems are over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is possible that there are no white hats present then you have a more difficult problem. It is still possible to work out who is/are the barkeep(s) (or if you have a room full black hats) but it will take repeated rounds of questioning, changing the order and questions.&lt;/p&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="11297" LastEditDate="2015-05-16T10:19:14.350" LastActivityDate="2015-05-16T10:19:14.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11940" PostTypeId="1" AcceptedAnswerId="11950" CreationDate="2015-04-13T04:26:46.420" Score="5" ViewCount="5132" Body="&lt;p&gt;Your task is to create all the integers from 1 to 50 (inclusive) using only a single integer $x$ and...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Addition ($+$), Subtraction($-$), Multiplication($\times$), and Division($\div$)&lt;/li&gt;&#xA;&lt;li&gt;Floor($\left \lfloor x \right \rfloor$) and Ceiling ($\left \lceil x \right \rceil$)&lt;/li&gt;&#xA;&lt;li&gt;Exponentation($x^a$) and Radicals($\sqrt[a]{x}$)&lt;/li&gt;&#xA;&lt;li&gt;Factorial($!$)&lt;/li&gt;&#xA;&lt;li&gt;Tau($\tau$, a.k.a. $2\pi$) and Phi($\phi$, a.k.a. the golden ratio)&lt;/li&gt;&#xA;&lt;li&gt;and Parentheses($()$).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;MathJaX syntax:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;+, -, \times, \div, \lfloor\rfloor, \lceil\rceil, x^a, \sqrt[a]{x}, !, \tau, \phi, ()&lt;/code&gt;&#xA;$+, -, \times, \div, \lfloor\rfloor, \lceil\rceil, x^a, \sqrt[a]{x}, !, \tau, \phi, ()$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a competition! The winner is the person with the lowest amount of points. A point is given when $x$ is used and when any function listed above is used. Examples($x$ is $3$):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$1 = 3\div3$ (3 points)&lt;/li&gt;&#xA;&lt;li&gt;$6 = 3!$ (2 points)&lt;/li&gt;&#xA;&lt;li&gt;$19 = \left \lceil 3 \times \tau\right \rceil$ (4 points)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Good Luck!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I don't think anyone is going to get all 50anytime soon, so until then the points will be scored by doing an average of all integers already found. So, for example, if you found 1, 4, 6, 23, 74, and 83, your score would be the sum divided by 6 until someone finds all 50.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT2: Clarifications: $x^x$ scores 3 points, $(x)$ scores two, and you cannot combine digits (if $x$ is 2, then you cannot use $22$ in your answer). Also, x does not need to be used, nor do any of the functions.&lt;/p&gt;&#xA;" OwnerUserId="9631" LastEditorUserId="5373" LastEditDate="2015-05-13T20:02:07.900" LastActivityDate="2015-05-13T20:02:07.900" Title="Create integers from 1 to 50 using only one integer and other functions" Tags="|calculation-puzzle|formation-of-numbers|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11941" PostTypeId="2" ParentId="11940" CreationDate="2015-04-13T05:35:17.333" Score="2" Body="&lt;p&gt;Suppose the integer we use is 2. Then, in &lt;a href=&quot;http://en.wikipedia.org/wiki/Reverse_Polish_notation&quot; rel=&quot;nofollow&quot;&gt;reverse Polish notation&lt;/a&gt; (to make counting operators and numbers easier), the equations can be expressed as:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 1 = 2 2 /                    ( 3 pts.)&#xA; 2 = 2                        ( 1 pt. )&#xA; 3 = 2 2 2 / +                ( 5 pts.)&#xA; 4 = 2 2 *                    ( 4 pts.)&#xA; 5 = 2 2 * 2 2 / +            ( 7 pts.)&#xA; 6 = 2 2 * 2 +                ( 5 pts.)&#xA; 7 = 2 2 2 2 2 / + + +        ( 9 pts.)&#xA; 8 = 2 2 2 * *                ( 5 pts.)&#xA; 9 = 2 2 / 2 * 2 ^            ( 7 pts.)&#xA;10 = 2 2 2 * * 2 +            ( 7 pts.)&#xA;11 = 22 2 /                   ( 4 pts.)&#xA;12 = 2 2 + ! 2 /              ( 6 pts.)&#xA;13 = 22 2 / 2 +               ( 6 pts.)&#xA;14 = 2 2 + ! 2 / 2 +          ( 8 pts.)&#xA;15 = 2 2 2 * ^ 2 2 / -        ( 9 pts.)&#xA;16 = 2 2 2 * ^                ( 5 pts.)&#xA;17 = 2 2 2 * ^ 2 2 / +        ( 9 pts.)&#xA;18 = 22 2 2 + -               ( 6 pts.)&#xA;19 = 22 2 2 2 / + -           ( 8 pts.)&#xA;20 = 22 2 -                   ( 4 pts.)&#xA;21 = 22 2 2 / -               ( 6 pts.)&#xA;22 = 22                       ( 2 pts.)&#xA;23 = 22 2 2 / +               ( 6 pts.)&#xA;24 = 2 2 + !                  ( 4 pts.)&#xA;25 = 2 2 + ! 2 2 / +          ( 8 pts.)&#xA;26 = 2 2 + ! 2 +              ( 6 pts.)&#xA;27 = 22 2 2 2 2 / + + +       (10 pts.)&#xA;28 = 2 2 + ! 2 2 * +          ( 8 pts.)&#xA;29 = 2 2 + ! 2 2 + + 2 2 / +  (12 pts.)&#xA;30 = 22 2 2 2 * * +           ( 8 pts.)&#xA;31 = 2 2 2 * ^ 2 * 2 2 / -    (11 pts.)&#xA;32 = 2 2 2 * ^ 2 *            ( 7 pts.)&#xA;33 = 2 2 2 * ^ 2 * 2 2 / +    (11 pts.)&#xA;34 = 2 2 2 * ^ 2 * 2 +        ( 9 pts.)&#xA;35 = 2 2 + ! 22 2 / +         ( 9 pts.)&#xA;36 = 2 2 2 + + 2 2 2 + + *    (11 pts.)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The above solutions don't make any use of the provided constants $\tau$ and $\phi$, as &lt;em&gt;generally&lt;/em&gt; those constants aren't allowed as tokens in such challenges. I'll post answers with constants below if they outscore the above answers.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-04-13T16:57:44.277" LastActivityDate="2015-04-13T16:57:44.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11942" PostTypeId="1" CreationDate="2015-04-13T06:00:22.980" Score="7" ViewCount="924" Body="&lt;p&gt;Following this: &lt;a href=&quot;https://puzzling.stackexchange.com/q/11334/28&quot;&gt;Cooperative guessing game&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A standard deck of 52 cards is shuffled and split into two piles of 26 cards each.&lt;/p&gt;&#xA;&lt;p&gt;Two players cooperate, they must score as many tricks as possible.&lt;/p&gt;&#xA;&lt;p&gt;The first player draws 5 cards from the first pile. He then plays one of them face down, and the card from the top of the second pile is played face down. The two cards are shuffled and turned face up. The second player must now guess which of the two cards the first player played. The first player indicates whether the guess was correct. If so, the players score the trick. The first player then draws a new card from the first pile, so that he has 5 cards in hand, and plays a new card face down, etc. The process is repeated until the second pile is empty and 26 tricks have been played. As the first pile runs out, the last 4 rounds are played with a reduced hand size.&lt;/p&gt;&#xA;&lt;p&gt;The players have no way of passing information to one another during the game beyond what is explicitly stated above. They may, however, agree on a strategy before the game begins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;One can ask a little bit different question with a totally different result: What is the optimal strategy to &lt;strong&gt;guarantee&lt;/strong&gt; scoring as many tricks as possible, independently of deck shuffling outcomes?&lt;/p&gt;&#xA;&lt;p&gt;Assume that all shuffling is done by an evil god, who knows all about the game, and the strategy that players have chosen, and wants them to score as little as possible.&lt;/p&gt;&#xA;&lt;p&gt;I know that at least 5 is possible (just encode each 5th card with previous 4), but I've been told that 6 is possible. It may be more.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-21T15:22:03.400" Title="Cooperative guessing against an evil god" Tags="|strategy|cards|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11943" PostTypeId="1" CreationDate="2015-04-13T06:36:29.130" Score="7" ViewCount="5662" Body="&lt;p&gt;In what field does $\pi^2=2$, where $π$ is the ratio of the circumference of a circle to its diameter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember to think outside the box and that I am looking for a complete answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The above is to be taken more literally than figuratively and vice versa respectively.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11299" LastEditorUserId="5373" LastEditDate="2015-04-13T23:23:15.963" LastActivityDate="2020-07-25T04:30:49.797" Title="In what field does π²=2?" Tags="|mathematics|lateral-thinking|" AnswerCount="11" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="11945" PostTypeId="2" ParentId="11940" CreationDate="2015-04-13T08:29:08.273" Score="3" Body="&lt;p&gt;$x=4$ totalling 264 points averaging 5.28 points per target integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've assumed that you don't have to use x if its not really needed. Otherwise the answer for 1 is 4/4 the answer for 2 is $\lfloor4-\phi\rfloor$ and the other 18 numbers that don't include x will need $\times4\div4$ appended increasing the total points to 339 for an average of 6.78 (although I might have another go at finding better answers for those 18 if they are inadmissible.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1= \lfloor \phi \rfloor$    (2 points)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2=\lceil \phi \rceil$    (2 points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3=4-\lfloor \phi \rfloor	$    (4 points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$4=4		$        (1 point) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5=4+\lceil \phi \rceil	$    (4 points)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$6=\lfloor \phi \times 4 \rfloor$        (4 points)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$7=\lceil \phi \times 4\rceil$        (4 points)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$8=4 \times \lceil p\rceil	$    (4 points)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$9=4+4+\lceil \phi \rceil	$    (6  points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$10=\lfloor 4 \times \phi \times \phi \rfloor$        (6  points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$11=\lceil 4 \times \phi \times \phi \rceil	$    (6  points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$12=\lfloor \tau \times \lceil \phi \rceil \rfloor	($5   points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$13=\lceil \tau \times \lceil \phi \rceil\rceil$    (5  points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$14=4 \times 4- \lceil \phi \rceil	$    (6  points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$15=\lfloor \tau \times 4 \div \phi \rfloor	$    (6  points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$16=4 \times 4		$    (3  points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$17=4 \times 4+\lfloor \phi \rfloor	$    (6  points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$18=\lfloor \tau \times 4 \div \phi + \phi \rfloor$    (8  points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$19=\lfloor \phi + \phi +4 \times 4 \rfloor$    (8  points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$20=\lfloor \phi ^t  \rfloor		$ (4 points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$21=\lceil \phi ^ \tau \rceil		$ (4 points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$22=\lceil \phi + \phi ^ \tau \rceil	$    (6  points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$23=\lceil \phi + \phi  ^ \tau \rceil	$    (6  points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$24=4!		$    (2  points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$25=\lfloor \tau \times 4 \rfloor$        (4  points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$26=\lceil \tau \times 4 \rceil$        (4  points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$27=\lceil \tau \times 4+ \phi \rceil$        (6  points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$28=4 \times \lceil \phi \times 4 \rceil$        (6  points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$29=\lfloor \phi +4 \times \phi ^4 \rfloor$    (8  points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$30=\lfloor 4^{4\div\phi}\rfloor$ (  6   points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$31=\lceil 4^{4\div \phi }\rceil$    (6  points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$32=4 \times 4 \times \lceil \phi \rceil$        (6  points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$33=\lfloor \phi \times \phi ^ \tau \rfloor	$    (6  points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$34=\lceil \phi \times \phi ^ \tau \rceil$        (6  points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$35=\lceil \phi + \phi \times \phi ^ \tau \rceil$    (8  points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$36=\lfloor (4!- \phi ) \times \phi \rfloor$    (8  points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$37=\lceil (4!- \phi ) \times \phi \rceil$    (8  points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$38=\lfloor (4+4+ \phi ) \times 4 \rfloor$    (9  points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$39=\lfloor \tau \times \tau \rfloor	$    (   4   points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$40=\lceil \tau \times \tau \rceil		$    (4  points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$41=\lceil 4 \times \tau \times \phi \rceil	$        (6  points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$42=\lceil \phi \times \phi \times 4 \times 4 \rceil$ (      8   points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$43=\lfloor \phi ^{ \phi + \tau }- \phi \rfloor$    ( 6 points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$44=\lfloor \phi ^{ \phi + \tau }\rfloor$    (   6   points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$45=\lceil \phi ^{ \phi + \tau }\rceil	$ (  6   points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$46=\lfloor \phi + \phi ^{ \phi + \tau }\rfloor	$ (  8   points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$47=\lceil  \phi + \phi ^{ \phi + \tau }\rceil	$ (  8   points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$48=\lfloor \phi + \phi + \phi ^{ \phi + \tau }\rfloor$ (  10  points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$49=\lceil \phi + \phi + \phi ^{ \phi + \tau }\rceil$ (  10  points) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$50=\lfloor 4 \times \tau \times \lceil \phi \rceil\rfloor$ (      7   points) &lt;/p&gt;&#xA;" OwnerUserId="11297" LastActivityDate="2015-04-13T08:29:08.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11946" PostTypeId="2" ParentId="11943" CreationDate="2015-04-13T09:12:02.400" Score="39" Body="&lt;p&gt;&lt;strong&gt;Answer 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The field is cooking.&lt;br&gt;&#xA;$\pi^2$ milliliters = 2 teaspoons&lt;br&gt;&#xA;&lt;a href=&quot;https://www.google.ch/search?hl=en&amp;amp;q=pi+squared+milliliters+in+teaspoons&quot;&gt;https://www.google.ch/search?hl=en&amp;amp;q=pi+squared+milliliters+in+teaspoons&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answer 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The field spherical geometry.&lt;br&gt;&#xA;For a small circle of radius subtending angle $\theta = 2.010311...$ radians at the centre of the sphere, the ratio between the circumference and the diameter measured on the surface of the sphere is $\sqrt2$. In this precise case we could say $\pi^2=2$.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="7363" LastEditDate="2015-05-04T11:25:18.423" LastActivityDate="2015-05-04T11:25:18.423" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="11947" PostTypeId="2" ParentId="11943" CreationDate="2015-04-13T10:03:44.607" Score="24" Body="&lt;p&gt;Mathematically, I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(\mathbb{Q}(\pi)/(\pi^2-2))_2$,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where this denotes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the 2-adic completion, i.e. the completion w.r.t. the 2-adic norm $||.||_2$,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the number field $\mathbb{Q}(\pi)/(\pi^2-2)$ - which is isomorphic to $\mathbb{Q}(\sqrt{2})$ as a number field, but has $\pi$ identified with $\sqrt{2}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a complete field (the question asks &quot;in what &lt;em&gt;field&lt;/em&gt; does...&quot; and &quot;I am looking for a &lt;em&gt;complete&lt;/em&gt; answer&quot;) in which $\pi^2=2$.&#xA;&lt;hr&gt;&#xA;My first idea was simply&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\mathbb{Q}(\sqrt{2})_2$,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; we use the Greek letter $\pi$ to denote the number $\sqrt{2}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This isn't as silly as it looks, since the notation $\pi$ is often used in algebraic number theory for elements of $p$-adic completions of number fields rather than for $3.14159265358979...$. But I amended it as suggested by @Meelo since I think it makes slightly more sense with the actual number $\pi=3.14159265358979...$ identified with $\sqrt{2}$ via quotienting.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-04-13T23:18:38.600" LastActivityDate="2015-04-13T23:18:38.600" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11948" PostTypeId="1" CreationDate="2015-04-13T10:42:17.810" Score="-4" ViewCount="935" Body="&lt;p&gt;I wait for you, by taking deep breaths to start our journey. If you are travelling fast I don't breath(I am with no lungs for sure). You leave many children behind me during the journey and I will suicide for them at last. Some of my children have my body, but not the soul. Poor fellows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But sometimes I will reincarnate to lead them in the right path. I will feed them again. Sometimes I am very cruel and will kill our own sons and daughters with the help of some murderers. One thing is sure if you are wrong in leading them, no mercy for sure, I will rise, and will take part in the conspiracy in killing them (I don't have a heart too :)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it is you who decide my death. Oh you are going to kill me for revealing this truth. Aren't you? RIP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clue 1: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I do have two lungs and hearts. They are my sweet children now.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clue 2: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There were many incarnations and hope will have many of them soon. and love moderators.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="160" LastEditorUserId="160" LastEditDate="2015-04-17T18:40:11.187" LastActivityDate="2015-04-17T18:40:57.653" Title="Kill me please else I will be responsible for the murder of my children" Tags="|riddle|lateral-thinking|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11949" PostTypeId="2" ParentId="11948" CreationDate="2015-04-13T10:56:14.333" Score="0" Body="&lt;p&gt;I could be &lt;em&gt;way&lt;/em&gt; off, but is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Doctor Who?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The hint (with &quot;two hearts&quot; and &quot;this is my second birth&quot;) made me think of it immediately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I wait for you, by taking deep breaths to start our journey.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first episode of the latest season of Doctor Who is called &quot;Deep Breath&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If you are travelling fast I don't breath(I am with no lungs for sure).&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Do Time Lords need to breathe? I think maybe they don't. And the &quot;no lungs&quot; bit is negated by the hint.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You leave many children behind me during the journey and I will suicide for them at last.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Doctor has had many Companions, and on a few occasions he's given up his life for them (this caused e.g. the regenerations of Ninth-&gt;Tenth and Tenth-&gt;Eleventh Doctors).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But sometimes I will reincarnate to lead them in the right path. I will feed them again.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Doctor regenerates after he's been killed, and often returns to the same Companions as before (e.g. Rose and Clara).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Sometimes I am very cruel and will kill our own sons and daughters with the help of some murderers.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Doctor destroyed Gallifrey at the end of the Time War (or thought he did), causing the deaths of billions, some also killed by the murderous Daleks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;One thing is sure if you are wrong in leading them, no mercy for sure, I will rise, and will take part in the conspiracy in killing them&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm not sure about this bit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(I don't have a heart too :)).&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Indeed: Time Lords have &lt;em&gt;two&lt;/em&gt; hearts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But it is you who decide my death. Oh you are going to kill me for revealing this truth. Aren't you? RIP.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, not sure about this bit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-04-13T10:56:14.333" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11950" PostTypeId="2" ParentId="11940" CreationDate="2015-04-13T11:21:01.957" Score="7" Body="&lt;p&gt;I can cover all of the numbers for 223 points, 4.46 average, with $x = 34$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this doesn't make use of radicals, factorials, or parentheses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1 = \lfloor\phi\rfloor$&lt;br&gt;&#xA;$2 = \lceil\phi\rceil$&lt;br&gt;&#xA;$3 = \lfloor\tau\div\phi\rfloor$&lt;br&gt;&#xA;$4 = \lfloor\tau-\phi\rfloor$&lt;br&gt;&#xA;$5 = \lceil\tau-\phi\rceil$&lt;br&gt;&#xA;$6 = \lfloor\tau\rfloor$&lt;br&gt;&#xA;$7 = \lceil\tau\rceil$&lt;br&gt;&#xA;$8 = \lceil\tau+\phi\rceil$&lt;br&gt;&#xA;$9 = \lceil\tau\rceil + \lceil\phi\rceil$&lt;br&gt;&#xA;$10 = \lfloor\tau\times\phi\rfloor$&lt;br&gt;&#xA;$11 = \lceil\tau\times\phi\rceil$&lt;br&gt;&#xA;$12 = \lfloor\tau+\tau\rfloor$&lt;br&gt;&#xA;$13 = \lceil\tau+\tau\rceil$&lt;br&gt;&#xA;$14 = \lceil\tau\rceil\times\lceil\phi\rceil$&lt;br&gt;&#xA;$15 = \lceil\phi^\tau-\tau\rceil$&lt;br&gt;&#xA;$16 = \lfloor\tau\times\phi+\tau\rfloor$&lt;br&gt;&#xA;$17 = 34\div\lceil\phi\rceil$&lt;br&gt;&#xA;$18 = \lceil\phi^{\lfloor\tau\rfloor}\rceil$&lt;br&gt;&#xA;$19 = \lfloor\tau^\phi\rfloor$&lt;br&gt;&#xA;$20 = \lceil\tau^\phi\rceil$&lt;br&gt;&#xA;$21 = \lceil\phi^\tau\rceil$&lt;br&gt;&#xA;$22 = \lceil34\div\phi\rceil$&lt;br&gt;&#xA;$23 = \lfloor\lceil\tau\rceil^\phi\rfloor$&lt;br&gt;&#xA;$24 = \lceil\lceil\tau\rceil^\phi\rceil$&lt;br&gt;&#xA;$25 = \lfloor\tau^\phi+\tau\rfloor$&lt;br&gt;&#xA;$26 = \lfloor\phi^\tau+\tau\rfloor$&lt;br&gt;&#xA;$27 = 34 - \lceil\tau\rceil$&lt;br&gt;&#xA;$28 = 34 - \lfloor\tau\rfloor$&lt;br&gt;&#xA;$29 = \lfloor\phi^{\lceil\tau\rceil}\rfloor$&lt;br&gt;&#xA;$30 = \lceil\phi^{\lceil\tau\rceil}\rceil$&lt;br&gt;&#xA;$31 = \lfloor\phi\times\tau^\phi\rfloor$&lt;br&gt;&#xA;$32 = 34 - \lceil\phi\rceil$&lt;br&gt;&#xA;$33 = 34 - \lfloor\phi\rfloor$&lt;br&gt;&#xA;$34 = 34$&lt;br&gt;&#xA;$35 = 34 + \lfloor\phi\rfloor$&lt;br&gt;&#xA;$36 = 34 + \lceil\phi\rceil$&lt;br&gt;&#xA;$37 = \lfloor\lfloor\tau\rfloor\times\tau\rfloor$&lt;br&gt;&#xA;$38 = \lceil\lfloor\tau\rfloor\times\tau\rceil$&lt;br&gt;&#xA;$39 = \lfloor\tau\times\tau\rfloor$&lt;br&gt;&#xA;$40 = 34 + \lfloor\tau\rfloor$&lt;br&gt;&#xA;$41 = 34 + \lceil\tau\rceil$&lt;br&gt;&#xA;$42 = \lfloor\tau\rfloor\times\lceil\tau\rceil$&lt;br&gt;&#xA;$43 = \lfloor\lceil\tau\rceil\times\tau\rfloor$&lt;br&gt;&#xA;$44 = \lceil\lceil\tau\rceil\times\tau\rceil$&lt;br&gt;&#xA;$45 = \lceil\phi^{\tau+\phi}\rceil$&lt;br&gt;&#xA;$46 = \lceil\tau\times\tau+\tau\rceil$&lt;br&gt;&#xA;$47 = 34 + \lceil\tau+\tau\rceil$&lt;br&gt;&#xA;$48 = \lfloor34\times\phi-\tau\rfloor$&lt;br&gt;&#xA;$49 = \lceil\tau\rceil^{\lceil\phi\rceil}$&lt;br&gt;&#xA;$50 = \lfloor\lceil\tau\rceil\times\tau+\tau\rfloor$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Points per line:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2 2 4 4 4   2 2 4 5 4 : Sums to 33&#xA;4 4 4 5 6   6 4 5 4 4 : Sums to 46&#xA;4 4 5 5 6   6 4 4 5 5 : Sums to 48&#xA;6 4 4 1 4   4 5 5 4 4 : Sums to 41&#xA;4 5 5 5 6   6 6 6 5 7 : Sums to 55&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Foundation: &lt;a href=&quot;http://pastebin.com/AWKKzncy&quot; rel=&quot;nofollow&quot;&gt;http://pastebin.com/AWKKzncy&lt;/a&gt; (this achieves 248 without use of $x$).&lt;/p&gt;&#xA;" OwnerUserId="11100" LastEditorUserId="11100" LastEditDate="2015-04-13T14:11:37.193" LastActivityDate="2015-04-13T14:11:37.193" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11951" PostTypeId="2" ParentId="6386" CreationDate="2015-04-13T12:05:54.037" Score="4" Body="&lt;p&gt;As already mentioned in the other questions the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The potion is the opposite of what the merchant said, it does not bring life it brings death.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Josh Caswell did a great job in describing one way of looking at it.&#xA;I would like to give an alternate answer which I personally think fits better because of the first line in the letter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You come to a fork in the road.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; as described in the riddle you have a road to the left, ahead and to the right meaning there is the 3th road where you come from this would result in the shape of a Trident and not a T or Y.&#xA;Though this is only partially true, the devil is sometimes depicted with a trident (fork with 3 teeth) if you meet the devil you're obviously dead or dying (going to die). Also a weapon often seen in mobs who often have death as a consequence is a pitchfork.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To the left is an empty well made from stone.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an empty well has no water and therefor cannot support life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On the right is a pirate's buried treasure.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; pirates are old and long gone, often a pirates treasure is cursed or in other ways connected to death.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ahead you only see a tall straight tree.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Trees are straight when they have no more leaves (or branches) indicating it has died.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The night is dark with only a dying moon in the sky. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; well obviously night symbolizes death and this is further emphasized by the &lt;strong&gt;dying&lt;/strong&gt; moon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9835" LastActivityDate="2015-04-13T12:05:54.037" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11952" PostTypeId="1" AcceptedAnswerId="12003" CreationDate="2015-04-13T12:56:49.173" Score="-5" ViewCount="7046" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Q. Choose the only correct answer to this question:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Option 1&lt;/li&gt;&#xA;&lt;li&gt;Option 1 or option 2&lt;/li&gt;&#xA;&lt;li&gt;Option 2 or option 3&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Answer the correct option along with the explanation.&lt;/p&gt;&#xA;" OwnerUserId="4637" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2017-09-08T12:57:23.593" Title="Answer this question with exactly one correct answer" Tags="|logical-deduction|" AnswerCount="7" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11953" PostTypeId="2" ParentId="11952" CreationDate="2015-04-13T13:04:20.230" Score="2" Body="&lt;p&gt;Option 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming one of the options is the right answer, option 3 contains option 2 and 3 and option 2 itself contains option 1 and 2, so you have all the options combined.&lt;/p&gt;&#xA;" OwnerUserId="10583" LastActivityDate="2015-04-13T13:04:20.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11955" PostTypeId="2" ParentId="11952" CreationDate="2015-04-13T13:14:25.043" Score="25" Body="&lt;p&gt;Option 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If 1. would be the answer then 2. would be also an answer. So 1. would not be the &lt;strong&gt;only&lt;/strong&gt; answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If 2. would be the answer then 3. would be also an answer. So 2. would not be the &lt;strong&gt;only&lt;/strong&gt; answer.&lt;/p&gt;&#xA;" OwnerUserId="10143" LastActivityDate="2015-04-13T13:14:25.043" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11956" PostTypeId="1" AcceptedAnswerId="11958" CreationDate="2015-04-13T13:32:11.060" Score="11" ViewCount="2578" Body="&lt;p&gt;Find a ten-digit number with the number of zeros in it as the most significant (leftmost) digit, the number of ones as the second most significant digit etc.&lt;/p&gt;&#xA;" OwnerUserId="10684" LastActivityDate="2015-04-14T03:23:14.497" Title="A self-referring number" Tags="|calculation-puzzle|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11957" PostTypeId="2" ParentId="11956" CreationDate="2015-04-13T13:41:27.313" Score="9" Body="&lt;p&gt;Such a number is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 6210001000&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; six 0's, two 1's, one 2 and one 6. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was going to try and add how I got to this, but I realized I had gotten there by trial and error. Pity that omitting that honesty cost me the tick :P&lt;/p&gt;&#xA;" OwnerUserId="3903" LastEditorUserId="146" LastEditDate="2015-04-13T15:08:00.007" LastActivityDate="2015-04-13T15:08:00.007" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11958" PostTypeId="2" ParentId="11956" CreationDate="2015-04-13T13:41:39.310" Score="23" Body="&lt;p&gt;I found this one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 6210001000&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My method was to start with 9000000000 and go from there.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 9000000000  — start by counting nine 0s&lt;br&gt;&#xA; 9000000001  — count one 9&lt;br&gt;&#xA; 9100000001  — count one 1&lt;br&gt;&#xA; 9200000001  — count two 1s&lt;br&gt;&#xA; 9210000001  — count one 2&lt;br&gt;&#xA; 6210000001  — count six 0s&lt;br&gt;&#xA; 6210001000  — count one 6, zero 9s&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is where the numbers settle.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A different set of steps that counts digits left-to-right, starting from 0000000000. This takes no shortcuts and is complete.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0000000000&lt;br&gt;&#xA; How many 0s? A&lt;br&gt;&#xA; A000000000  — count A 0s&lt;br&gt;&#xA; How many 1s? 0&lt;br&gt;&#xA; How many 2s? 0&lt;br&gt;&#xA; ...&lt;br&gt;&#xA; How many 9s? 0&lt;br&gt;&#xA; &lt;strong&gt;Restart at 0&lt;/strong&gt;&lt;br&gt;&#xA; How many 0s? 9&lt;br&gt;&#xA; 9000000000  — count nine 0s&lt;br&gt;&#xA; How many 1s? 0&lt;br&gt;&#xA; How many 2s? 0&lt;br&gt;&#xA; ...&lt;br&gt;&#xA; How many 9s? 1&lt;br&gt;&#xA; 9000000001  — count one 9&lt;br&gt;&#xA; &lt;strong&gt;Restart at 0&lt;/strong&gt;&lt;br&gt;&#xA; How many 0s? 8&lt;br&gt;&#xA; 8000000001  — count eight 0s&lt;br&gt;&#xA; How many 1s? 1&lt;br&gt;&#xA; How many 2s? 0&lt;br&gt;&#xA; ...&lt;br&gt;&#xA; How many 8s? 1&lt;br&gt;&#xA; 8100000011  — count one 8&lt;br&gt;&#xA; How many 9s? 0&lt;br&gt;&#xA; 8100000010  — count zero 9s&lt;br&gt;&#xA; &lt;strong&gt;Restart at 0&lt;/strong&gt;&lt;br&gt;&#xA; How many 0s? 7&lt;br&gt;&#xA; 7100000010  — count seven 0s&lt;br&gt;&#xA; How many 1s? 2&lt;br&gt;&#xA; 7200000010  — count two 1s&lt;br&gt;&#xA; How many 2s? 1&lt;br&gt;&#xA; 7210000010  — count one 2&lt;br&gt;&#xA; How many 3s? 0&lt;br&gt;&#xA; ...&lt;br&gt;&#xA; How many 7s? 1&lt;br&gt;&#xA; 7210000110  — count one 7&lt;br&gt;&#xA; How many 8s? 0&lt;br&gt;&#xA; 7210000100  — count zero 8s&lt;br&gt;&#xA; How many 9s? 0&lt;br&gt;&#xA; &lt;strong&gt;Restart at 0&lt;/strong&gt;&lt;br&gt;&#xA; How many 0s? 6&lt;br&gt;&#xA; 6210000100  — count six 0s&lt;br&gt;&#xA; How many 1s? 2&lt;br&gt;&#xA; 6210000100  — no change&lt;br&gt;&#xA; How many 2s? 1&lt;br&gt;&#xA; 6210000100  — no change&lt;br&gt;&#xA; How many 3s? 0&lt;br&gt;&#xA; ...&lt;br&gt;&#xA; How many 6s? 1&lt;br&gt;&#xA; 6210001100  — count one 6&lt;br&gt;&#xA; How many 7s? 0&lt;br&gt;&#xA; 6210001000  — count zero 7s&lt;br&gt;&#xA; How many 8s? 0&lt;br&gt;&#xA; How many 9s? 0&lt;br&gt;&#xA; &lt;strong&gt;Restart at 0&lt;/strong&gt;&lt;br&gt;&#xA; How many 0s? 6&lt;br&gt;&#xA; 6210001000  — no change&lt;br&gt;&#xA; How many 1s? 2&lt;br&gt;&#xA; 6210001000  — no change&lt;br&gt;&#xA; How many 2s? 1&lt;br&gt;&#xA; 6210001000  — no change&lt;br&gt;&#xA; How many 3s? 0&lt;br&gt;&#xA; ...&lt;br&gt;&#xA; How many 6s? 1&lt;br&gt;&#xA; 6210001000  — no change&lt;br&gt;&#xA; How many 7s? 0&lt;br&gt;&#xA; 6210001000  — no change&lt;br&gt;&#xA; How many 8s? 0&lt;br&gt;&#xA; How many 9s? 0&lt;br&gt;&#xA; Reached the end with no change.&lt;br&gt;&#xA; Result is &lt;strong&gt;6210001000&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-04-13T14:51:15.717" LastActivityDate="2015-04-13T14:51:15.717" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="11961" PostTypeId="2" ParentId="11943" CreationDate="2015-04-13T13:55:31.913" Score="4" Body="&lt;p&gt;The Zeta function &lt;s&gt;ζ(2) = π^2&lt;/s&gt; $ζ(2) = π^2/6$ but then you want to know the field, the Riemann zeta function is used in quantum theory so that's your field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;stupid me, I went to look for a reference to Wikipedia for the answer and found out I was almost right: &lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/doEqc.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;br&gt;&#xA;I forgot i had to divide it by 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;BONUS&lt;/strong&gt; (unrelated to question):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the Quantum theory often makes odd statements which turn out to be true for example: $1 + 2 + 3 + 4 + 5 + .... = -1/12$ (so counting up all integers to infinity equals minus 1/12th) &lt;a href=&quot;https://www.youtube.com/watch?v=w-I6XTVZXww&quot; rel=&quot;nofollow noreferrer&quot;&gt;Link for explanation&lt;/a&gt; &lt;br&gt;&#xA;&lt;strong&gt;DISCLAIMER&lt;/strong&gt; &lt;br&gt;&#xA;As enforced by the police I have to add this is only true when used in several techniques called analytic continuation and Ramanujan sums. So stay in school kids, else the police will find you and they will correct you!&lt;/p&gt;&#xA;" OwnerUserId="9835" LastEditorUserId="5373" LastEditDate="2015-04-13T20:25:12.360" LastActivityDate="2015-04-13T20:25:12.360" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="11962" PostTypeId="1" AcceptedAnswerId="11964" CreationDate="2015-04-13T14:17:19.050" Score="4" ViewCount="422" Body="&lt;p&gt;There is a hidden message that can be decipher from the string below. Try to figure out what does it actually says. The final answer will be in text form.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;TkyeasliuewthhnvsttegpwrueeoawdhshheanwhwloeestoatdiesorfhTtlkhocaordeehhdTtekrnTtnleathhsetlhhdpmtheatd&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint: When X becomes the unlucky one, Y = 2X - 18.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't actually need anything else other than the string of characters and the hint above. The 4 capital 'T' are intentional (not text error).&lt;/p&gt;&#xA;" OwnerUserId="2373" LastActivityDate="2015-04-13T20:21:35.783" Title="Hidden Message: Code and Calculation" Tags="|cipher|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11964" PostTypeId="2" ParentId="11962" CreationDate="2015-04-13T14:46:51.970" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The heart knows only valor.  The sword that defends the helpless.  The might that uphold the weak.  The wrath that undoes the wicked&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not entirely sure what the hint means, but I assumed that &quot;the unlucky one&quot; means&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the number 13.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It turns out that this is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a simple transposition cipher.  If you arrange the text in rows of 13 characters, and then read down the columns, you get the solution text:&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;pre&gt;Tkyeasliuewth&lt;br&gt;hnvsttegpwrue&lt;br&gt;eoawdhshheanw&lt;br&gt;hwloeestoatdi&lt;br&gt;esorfhTtlkhoc&lt;br&gt;aordeehhdTtek&lt;br&gt;rnTtnleathhse&lt;br&gt;tlhhdpmtheatd&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="5373" LastEditDate="2015-04-13T20:21:35.783" LastActivityDate="2015-04-13T20:21:35.783" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11966" PostTypeId="1" AcceptedAnswerId="11970" CreationDate="2015-04-13T14:59:05.420" Score="5" ViewCount="388" Body="&lt;p&gt;Divide the below figure into $5$ equal pieces (same shape, same size, possibly reflected). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\qquad\quad\qquad$&lt;img src=&quot;https://i.stack.imgur.com/vKqiT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe this is one of Martin Gardner's, but I could not find the source.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="5188" LastEditDate="2015-04-13T16:21:37.763" LastActivityDate="2015-04-13T16:21:37.763" Title="Deceptive Dissection" Tags="|geometry|dissection|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11968" PostTypeId="1" AcceptedAnswerId="12029" CreationDate="2015-04-13T15:11:06.370" Score="10" ViewCount="1050" Body="&lt;p&gt;Find a four-digit number which is equal to the four least significant (rightmost) digits of its own square.&lt;/p&gt;&#xA;" OwnerUserId="10684" LastActivityDate="2015-04-15T19:19:08.813" Title="A number and its square" Tags="|calculation-puzzle|" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11969" PostTypeId="2" ParentId="11968" CreationDate="2015-04-13T15:17:54.107" Score="10" Body="&lt;p&gt;I think this is the only one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;9,376 * 9,376 = 87,909,376&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastActivityDate="2015-04-13T15:17:54.107" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="11970" PostTypeId="2" ParentId="11966" CreationDate="2015-04-13T15:21:27.113" Score="7" Body="&lt;p&gt;The first observation we make is that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the left and right edges are identical&lt;br&gt;&#xA; the top and bottom edges are identical&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we then&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; mark four equidistant points along each horizontal line,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;we have constructed the basis for making 5 shapes that are basically&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; thinner versions of the original shape.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Connecting the dots vertically now subdivides the original shape into 5 equal pieces of the same shape and size.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;pre&gt;x-x-x-x-x-x&#xA; \ \ \ \ \ \&#xA; / / / / / /&#xA;| | | | | |&#xA;x-x-x-x-x-x&#xA; &lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Perhaps the most important observation to make is that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the lattice on which this shape is placed is a visual distraction. Once ignored and only the shape itself is considered, the solution is quite obvious. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10070" LastEditDate="2015-04-13T15:38:53.083" LastActivityDate="2015-04-13T15:38:53.083" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11971" PostTypeId="2" ParentId="11942" CreationDate="2015-04-13T16:20:49.503" Score="2" Body="&lt;p&gt;Here is the technique to get guaranteed 5 tricks as mentioned by the OP.  You can use this as a starting point to see if you can get more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we will divide the game into 5 sub games of 5 plays each. We will win one trick each sub game (possibly more, but not guaranteed).  At the end, there will be one card left to guess, but we won't be able to guarantee the outcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each sub game will be played as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Player one starts with 5 cards.  He will &lt;a href=&quot;https://puzzling.stackexchange.com/questions/10004/the-fitch-cheney-card-trick-extended&quot; title=&quot;order&quot;&gt;order&lt;/a&gt; them in some way, thus encoding the 5th card with the other 4. &lt;/li&gt;&#xA;&lt;li&gt;Player one will then play the cards in order for the next 5 turns.  The new cards that are drawn to replace the old are used for the next sub-game.&lt;/li&gt;&#xA;&lt;li&gt;Player two will see the first 4 cards played by player one, and make random guesses.  After each guess, he will be told if he was right or not, so player two will know exactly which cards player one played.  Since the order of the first four can be used to uniquely identify the 5th, player two will know the 5th card about to be played by player one before it is even played.  Thus, it can be guessed correctly every time.&lt;/li&gt;&#xA;&lt;li&gt;At the end of a sub-game, the player one will have drawn 5 new cards with which to play the next sub-game.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;After 5 rounds, they will have played 5 rounds of these sub-games, thus guaranteed themselves at least 5 tricks.  The last card drawn played and guessed at random by player two.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-04-13T16:28:28.170" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11972" PostTypeId="1" CreationDate="2015-04-13T16:22:42.840" Score="58" ViewCount="5698" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;There is a circular area with radius 1 km. And there is a tunnel, which is just under the surface, but invisible - unless you dig. It is known that the tunnel goes under the area (at least touches it at one point), it is straight and infinitely long (in both directions).&lt;br&gt;&#xA;  You have a plow and can dig along some lines with it. When you plow and cross the tunnel you will find it. How much (how long) and where do you have to plow to guarantee that you will find the tunnel?&lt;br&gt;&#xA;  &lt;em&gt;You are allowed to plow outside of the area as well as inside. You can take the plow out of the ground and move it over the ground without plowing.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example, you could choose to plow just along the perimeter, and then your result would be $2\pi\approx6.28\ \text{km}$. The task is to make this number as small as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know any good approach, but two people told me that at least $4.83\ \text{km}$ is possible to achieve, and one told that less than $4.8\ \text{km}$ is also possible.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="69582" LastEditDate="2021-03-07T21:16:44.600" LastActivityDate="2021-03-07T21:16:44.600" Title="Find a straight tunnel" Tags="|mathematics|geometry|optimization|" AnswerCount="8" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="11973" PostTypeId="2" ParentId="11972" CreationDate="2015-04-13T16:29:29.793" Score="16" Body="&lt;p&gt;Combining my second guess with CarpetPython's answer (green line is what to plow):&lt;br&gt;&#xA;$4.87\ \text{km}$&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/zqQwS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second guess:&lt;/strong&gt;&lt;br&gt;&#xA;Ok, I think I can get it down to $5.196\ \text{km}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the circle is inscribed in a hexagon, first draw 3 of the spokes (1.155 km each). From the tips of the spokes, imagine it creates the equilateral triangle. From the center of each face, draw a line to the closest corner. Those lines are 1/2 the length of the first set of spokes. So, $4.5 \times 1.155 = 5.196\ \text{km}$  The red lines below are the plow lines. &#xA;&lt;img src=&quot;https://i.stack.imgur.com/rIemH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First guess:&lt;/strong&gt;&lt;br&gt;&#xA;If you plow an X across the circle, with leg lengths of $\sqrt{2}\ \text{km}$, you can get your plowing down to $5.656\ \text{km}$ (or imagine that the circle in in a square and you plow both diagonals). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the tunnel is on the tangent of the circle (say, at the very top) it will barely hit each of the ends of the X. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="5483" LastEditDate="2015-04-14T01:23:30.793" LastActivityDate="2015-04-14T01:23:30.793" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="11978" PostTypeId="2" ParentId="2266" CreationDate="2015-04-13T17:09:03.917" Score="1" Body="&lt;h2&gt;A very brief history of some random points in puzzle history&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Thinking games and puzzles have along history with some so long that their origins are shrouded by the mists of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/History_of_chess&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/History_of_chess&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The river crossing puzzle dates back to at least the 9th century, when the benefit of mental exercise to sharpen the mind was understood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Propositiones_ad_Acuendos_Juvenes&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Propositiones_ad_Acuendos_Juvenes&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prior to modern psychology, puzzles were much more in the realm of the mathematician, the educator and the student.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/History_of_psychology&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/History_of_psychology&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Mathematical_puzzle&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Mathematical_puzzle&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the 20th century puzzles became one of the cornerstones of the growing trend for &quot;intelligence testing&quot; which gave gave rise to the IQ test, which today purports to assess a range of cognitive abilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Wechsler_Adult_Intelligence_Scale&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Wechsler_Adult_Intelligence_Scale&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Throughout the 20th century increased interest in recreational mind games has made puzzles a permanent fixture in many newspapers and magazine. Most news agents will stock a least one publication dedicated to puzzles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sudoku&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Sudoku&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://blogs.scientificamerican.com/observations/2010/05/29/a-quarter-century-of-recreational-m-2010-05-26/&quot; rel=&quot;nofollow&quot;&gt;http://blogs.scientificamerican.com/observations/2010/05/29/a-quarter-century-of-recreational-m-2010-05-26/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.theguardian.com/science/alexs-adventures-in-numberland/2014/oct/21/martin-gardner-mathematical-puzzles-birthday&quot; rel=&quot;nofollow&quot;&gt;http://www.theguardian.com/science/alexs-adventures-in-numberland/2014/oct/21/martin-gardner-mathematical-puzzles-birthday&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today many classic puzzles hold much fascination and as new puzzles are devised, new technology offers opportunity to involve random factors and interaction in challenges that have their roots in more traditional logic based reasoning.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/2048_(video_game)&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/2048_(video_game)&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11297" LastActivityDate="2015-04-13T17:09:03.917" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11979" PostTypeId="2" ParentId="11972" CreationDate="2015-04-13T17:20:48.907" Score="-2" Body="&lt;p&gt;I think the 4.83 solution is this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/LkD4f.png&quot; alt=&quot;Circle Solution&quot;&gt;&lt;br&gt;&lt;br&gt;The long line to plow along is equal to the hypotenuse of a square with each side being two length, or &lt;br&gt;&lt;br&gt;sqrt(2^2 + 2^2) = sqrt(8) = 2.828 &lt;br&gt;&lt;br&gt;and the shorter line is the diameter of the circle which is 2.  This makes the total length of plowing 2.828 + 2 or 4.828.  Note that the lines are perpendicular to each other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6140" LastActivityDate="2015-04-13T17:20:48.907" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11981" PostTypeId="2" ParentId="11952" CreationDate="2015-04-13T17:52:38.413" Score="8" Body="&lt;p&gt;None of the above; it's actually not a question, so there is no such thing as an answer to it.&lt;/p&gt;&#xA;" OwnerUserId="11314" LastActivityDate="2015-04-13T17:52:38.413" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11982" PostTypeId="2" ParentId="11948" CreationDate="2015-04-13T18:17:54.277" Score="-4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A text cursor&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I wait for you, by taking deep breaths to start our journey. If you are travelling fast I don't breath&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It wait for us to start typing, by taking it's 'breaths'. If you are typing fast the cursor does not take it's 'breath'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You leave many children behind me during the journey and I will suicide for them at last&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letters and characters are the children. When we save a document or question (here in SE) or search something in google the text cursor will disappear and it dies for the letters we typed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some of my children have my body, but not the soul. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letters 'I', 'l', '|' and '1' has the text cursor's body.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am very cruel and will kill our own sons and daughters with the help of some murderers. One thing is sure if you are wrong in leading them, no mercy for sure, I will rise, and will take part in the conspiracy in killing them (I don't have a heart too :))&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the typed letters or words are wrong, we often edit the documents to correct them with the help of some murderers like 'backspace' and 'Delete' keys. Text cursor is also a part of the conspiracy and it does not have a heart for sure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But it is you who decide my death.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We decide when to save a document.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clue 1: I do have two lungs and hearts. They are my sweet children now.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  When I completed typing the clue 1, there are a total of two 'lungs' and two 'heart' in the question. Obviously they are children of the cursor. So considering the question, it has two 'lungs' and 'hearts'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="160" LastEditorUserId="160" LastEditDate="2015-04-17T18:40:57.653" LastActivityDate="2015-04-17T18:40:57.653" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11983" PostTypeId="1" AcceptedAnswerId="11989" CreationDate="2015-04-13T18:26:15.540" Score="10" ViewCount="4747" Body="&lt;p&gt;You and your friend Roger are in charge of a soda packaging line.  The sodas arrive on a pallet that is shaped like a cube.  The two of you must take sodas from the pallet and fill four-pack holders (these are fancy sodas!).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the two of your are in complete control of the operation, you may request a pallet of any size, as long as the pallet is still a cube (e.g. 3 sodas wide, 3 sodas long, 3 sodas high).  You and Roger are also clever though, and you would like something to drink when you are done packing the pallet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Question:  What size pallet should you request so that, when you and Roger are finished packing the sodas from the pallet, you have exactly one soda a piece leftover to drink for yourselves?&lt;/p&gt;&#xA;" OwnerUserId="11315" LastEditorUserId="11315" LastEditDate="2015-04-13T18:40:02.243" LastActivityDate="2021-02-03T22:38:06.830" Title="Two Clever and Thirsty Soda Packaging Workers" Tags="|mathematics|word-problem|" AnswerCount="9" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="11984" PostTypeId="2" ParentId="11948" CreationDate="2015-04-13T18:34:09.893" Score="6" Body="&lt;p&gt;It's more of an other possible answer but, could you be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Windows Task Manager&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you are travelling fast I don't breath:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If a program is taking up too many resources (need too much to run fast), it can stall other programs on the machine OR if your machine if fast enough, this will never happen so Task Manager wouldn't be used&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You leave many children behind me during the journey and I will suicide for them at last.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; By default the Task Manager will stay in front of all other open applications (children), and can close at the end.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some of my children have my body, but not the soul&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Other programs are in windows having the same 'body' but not the same contents (soul)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But sometimes I will reincarnate to lead them in the right path. ... Sometimes I am very cruel and will kill our own sons and daughters with the help of some murderers&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can start a new process from inside Task Manager, and end Processes if the user is a 'Murderer'!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One thing is sure if you are wrong in leading them, no mercy for sure, I will rise&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you end some of the important processes (i.e System), the computer will 'rise' and take control (possibly shutting itself down or restarting somethings)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But it is you who decide my death&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Only the user can end the Task Managers dear life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-04-13T20:40:13.127" LastActivityDate="2015-04-13T20:40:13.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11985" PostTypeId="2" ParentId="11972" CreationDate="2015-04-13T18:38:39.650" Score="3" Body="&lt;h2&gt;5.217 km&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I used inscribed (green) and circumscribed (black) squares to help. The original circle is shown in blue. The plow paths are shown in red. The black square has a side of length $2$ and the green square has a side length of $\sqrt2$. The plow path lengths are as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A = $\sqrt2 = 1.414$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B = $\sqrt2\div2 = 0.707$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;C = $\sqrt{\sqrt2\div2} = 0.841$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total Plow Length $ = 5.217$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if my math is right, this is still not the optimal answer according to OP's source. I hope that it at least throws out some new ideas, though.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/IeEs2.png&quot; alt=&quot;Diagram&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastActivityDate="2015-04-13T18:38:39.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11986" PostTypeId="2" ParentId="11972" CreationDate="2015-04-13T18:39:29.807" Score="11" Body="&lt;h1&gt;4.8736 km&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I have modified my answer when I observed that a central blocking element would result in less plowing. I wrote a short program to try a range of angles (as my calculus and geometry are too rusty) to find the best angle of 30 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;EDIT: Note that JonTheMon posted his 4.87 answer first while I was still programming, so he gets the credit for first with this answer.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/mEG5W.png&quot; alt=&quot;new plow diagram&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; A  Plow Distance&#xA; 19 4.91250640267&#xA; 20 4.90597796428&#xA; 21 4.90001860583&#xA; 22 4.89463591311&#xA; 23 4.88983859219&#xA; 24 4.88563652529&#xA; 25 4.88204083336&#xA; 26 4.87906394597&#xA; 27 4.87671967936&#xA; 28 4.87502332331&#xA; 29 4.87399173788&#xA; 30 4.87364346116   &amp;lt;- best&#xA; 31 4.87399882899&#xA; 32 4.8750801084&#xA; 33 4.87691164606&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h3&gt;First Solution: 5.14 km&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;First plow the southern half of the circumference, then from each end plow 1 km north (making a U shape). The total distance will be pi+2 or 5.14 km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zqo9c.png&quot; alt=&quot;plow diagram&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="11320" LastEditorUserId="11320" LastEditDate="2015-04-13T19:47:56.293" LastActivityDate="2015-04-13T19:47:56.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11988" PostTypeId="2" ParentId="11972" CreationDate="2015-04-13T18:47:54.710" Score="0" Body="&lt;p&gt;&lt;strong&gt;NOTE: Not a complete answer. The purpose of this answer is to possibly offer an idea that someone else can turn into a formal answer.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I was thinking that a shape like this could offer potential. Ignore the red circle. It is just added in for visual effect. Also, the plowed lines obviously won't be tapered like they are in the illustration.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Ioe1T.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Things to consider:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Experiment with the length of the lines&lt;/li&gt;&#xA;&lt;li&gt;Experiment with the best angle of the lines, relative to the tangent of the circle at that point&lt;/li&gt;&#xA;&lt;li&gt;Experiment with the number of lines used.&lt;/li&gt;&#xA;&lt;li&gt;Experiment with adding a small circle to the center, if needed.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-04-13T19:07:32.137" LastActivityDate="2015-04-13T19:07:32.137" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11989" PostTypeId="2" ParentId="11983" CreationDate="2015-04-13T18:49:39.113" Score="15" Body="&lt;p&gt;Let's say $N$ is the length of the side of the cube. After their operation, you and Roger drink 2 sodas, but you still need a number of bottles which is multiple of 4, otherwise it wouldn't fit the 4-bottles packs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N^3-2=4k$  where $k$ is a positive integer.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;There's no valid value for $N$!&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In fact, $N^3$ $mod(4)$ cycle through {1,0,3,0} (proof below), which is {3,2,1,2} after subtracting 2. As you can see, there's no 0, so no multiple of 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof:&#xA;All numbers can be written as $a,(a+1),(a+2)$ or $(a+3)$ where $a$ is a multiple of 4. Respectively, those numbers are congruent to $0,1,2,3$ modulus 4.&#xA;When you raise to 3rd power, they respectively become congruent to $0^3$,$1^3$,$2^3$,$3^3$, which is $0,1,8,27$, equivalent to $0,1,0,3$ modulus 4.&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="-1" LastEditDate="2015-04-13T20:29:37.810" LastActivityDate="2015-04-13T20:29:37.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11990" PostTypeId="2" ParentId="11983" CreationDate="2015-04-13T18:50:07.327" Score="5" Body="&lt;h2&gt;There is no solution&lt;/h2&gt;&#xA;&lt;p&gt;You and your friend will go thirsty, it seems.&lt;/p&gt;&#xA;&lt;p&gt;There is no number integer such  such that &lt;sup&gt;3&lt;/sup&gt; mod 4 = 2&lt;/p&gt;&#xA;&lt;p&gt;From the example table below, you can see that it cycles through 1, 0, 3, 0&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th&gt;&lt;/th&gt;&#xA;&lt;th&gt;&lt;sup&gt;3&lt;/sup&gt;&lt;/th&gt;&#xA;&lt;th&gt;&lt;sup&gt;3&lt;/sup&gt; mod 4&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;1&lt;/td&gt;&#xA;&lt;td&gt;1&lt;/td&gt;&#xA;&lt;td&gt;1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;2&lt;/td&gt;&#xA;&lt;td&gt;8&lt;/td&gt;&#xA;&lt;td&gt;0&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;3&lt;/td&gt;&#xA;&lt;td&gt;27&lt;/td&gt;&#xA;&lt;td&gt;3&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;4&lt;/td&gt;&#xA;&lt;td&gt;64&lt;/td&gt;&#xA;&lt;td&gt;0&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;5&lt;/td&gt;&#xA;&lt;td&gt;125&lt;/td&gt;&#xA;&lt;td&gt;1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;6&lt;/td&gt;&#xA;&lt;td&gt;216&lt;/td&gt;&#xA;&lt;td&gt;0&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;7&lt;/td&gt;&#xA;&lt;td&gt;343&lt;/td&gt;&#xA;&lt;td&gt;3&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;8&lt;/td&gt;&#xA;&lt;td&gt;512&lt;/td&gt;&#xA;&lt;td&gt;0&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;9&lt;/td&gt;&#xA;&lt;td&gt;729&lt;/td&gt;&#xA;&lt;td&gt;1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;10&lt;/td&gt;&#xA;&lt;td&gt;1000&lt;/td&gt;&#xA;&lt;td&gt;0&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;11&lt;/td&gt;&#xA;&lt;td&gt;1331&lt;/td&gt;&#xA;&lt;td&gt;3&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;12&lt;/td&gt;&#xA;&lt;td&gt;1728&lt;/td&gt;&#xA;&lt;td&gt;0&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;13&lt;/td&gt;&#xA;&lt;td&gt;2197&lt;/td&gt;&#xA;&lt;td&gt;1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;14&lt;/td&gt;&#xA;&lt;td&gt;2744&lt;/td&gt;&#xA;&lt;td&gt;0&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;15&lt;/td&gt;&#xA;&lt;td&gt;3375&lt;/td&gt;&#xA;&lt;td&gt;3&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;16&lt;/td&gt;&#xA;&lt;td&gt;4096&lt;/td&gt;&#xA;&lt;td&gt;0&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;17&lt;/td&gt;&#xA;&lt;td&gt;4913&lt;/td&gt;&#xA;&lt;td&gt;1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;18&lt;/td&gt;&#xA;&lt;td&gt;5832&lt;/td&gt;&#xA;&lt;td&gt;0&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;19&lt;/td&gt;&#xA;&lt;td&gt;6859&lt;/td&gt;&#xA;&lt;td&gt;3&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;20&lt;/td&gt;&#xA;&lt;td&gt;8000&lt;/td&gt;&#xA;&lt;td&gt;0&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;" OwnerUserId="9000" LastEditorUserId="69582" LastEditDate="2021-02-03T22:38:06.830" LastActivityDate="2021-02-03T22:38:06.830" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="11991" PostTypeId="2" ParentId="11983" CreationDate="2015-04-13T18:52:11.113" Score="37" Body="&lt;p&gt;You and Roger are out of luck. Let the side length of the cube be $s$, so there are $s^3$ sodas. If $s$ is odd, then there will be an odd number of sodas left over, since $s^3$ is odd, and removing 4 from an odd number leaves an odd number. If $s$ is even, so that $s=2t$ for some $t$, then there will be $8t^3$ sodas, which is exactly divisible by $4$, so there are no sodas left over.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-04-13T18:52:11.113" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11992" PostTypeId="2" ParentId="11983" CreationDate="2015-04-13T19:01:31.550" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's not possible. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's why:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let's do some basic logic. Say that we have an odd number of sodas per side (for this example, we'll use 5, though it's totally arbitrary). This yields an amount of soda equal to 5 * 5 * 5, or ODD * ODD * ODD. Since we know that ODD * ODD will always equal ODD, we can deduce that the answer here will be an ODD number. An ODD number divided by an EVEN number (4, the number of sodas per container) will always have a remainder in the subset of (2*n - 1), where n a number from 1 to the number of times 2 divides into that EVEN number. The &quot;minus one&quot; in that expression means that the remainder can never be an even number; therefore, the remainder here will never be two.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it must be an even number, then! Well, not exactly. The problem with that is that all even numbers can be derived to (2*n). For instance, let's take 10. Our amount of sodas is 10 * 10 * 10, which can be derived to (2 * 5) * (2 * 5) * (2 * 5). Since multiplication doesn't care where the values are placed, we can move these values around to get (2 * 2) * (5 * 5 * 2 * 5). See the problem here? (2 * 2) equals four, and therefore all even numbers will result in a remainder of 0, since they will all be evenly divisible by four.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There we have it, then - &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the possible remainders of soda bottles for any number are 0, 1, and 3. There is no possible size of pallet that will yield a remainder of two bottles. Looks like you're just going to have to go thirsty, or bend the rules a bit...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Looks like I've been beaten to the punch! Glad to see I came up with the right answer, anyways. &lt;/p&gt;&#xA;" OwnerUserId="11322" LastEditorUserId="5260" LastEditDate="2015-04-14T15:08:29.930" LastActivityDate="2015-04-14T15:08:29.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11993" PostTypeId="2" ParentId="11983" CreationDate="2015-04-13T19:02:19.980" Score="9" Body="&lt;p&gt;You could both split up and unpack your own 5x5x5 pallets to have 1 soda each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To elaborate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a pallet of 125 on one side of the line, roger has another pallet of 125 on the other side of the line.&#xA;For each four-pack holder that comes down the line, you each put 2 sodas in.&#xA;After 62 four-pack holders, you both drink the remaining soda from your pallet, grab another pallet each and start again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works with any odd number, but I chose 5 rather than 1 or 3 as although they have full control over their line, 248 packed for every 2 drank seems better for their productivity or targets :D&lt;/p&gt;&#xA;" OwnerUserId="11150" LastEditorUserId="11150" LastEditDate="2015-04-13T20:09:35.390" LastActivityDate="2015-04-13T20:09:35.390" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11994" PostTypeId="1" AcceptedAnswerId="12000" CreationDate="2015-04-13T19:09:42.843" Score="3" ViewCount="105" Body="&lt;p&gt;The numbers {1 to 11}, excluding 8 and 10, are arranged in a 3 x 3 matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom row sums to 22.&#xA;The rightmost column sums to 24.&#xA;The back diagonal (/) values sum to 14.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each triplet of values within a row, column or diagonal are pairwise coprime. That means that any pair of two values of out the three have a greatest common divisor of 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the numbers!&lt;/p&gt;&#xA;" OwnerUserId="2154" LastActivityDate="2015-04-13T19:56:40.827" Title="Coprime latin squares" Tags="|latin-square|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11995" PostTypeId="2" ParentId="11983" CreationDate="2015-04-13T19:11:12.287" Score="3" Body="&lt;p&gt;As noted in other answers the problem is looking for a pallet size such that, if $n$ is the number of sodas on a side then $n^3 mod\; 4 = 2$.  Another way of stating this is that you want a number $n$ s.t. $n^3 = 4k + 2$ for some integer k.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$(4k)^3 = 64k = 4m + 0$ No sodas for you.&lt;/li&gt;&#xA;&lt;li&gt;$(4k + 1)^3 = 64k^3 + 12k^2 + 12k + 1 = 4(16k^3 + 3k^2 + 3k) + 1 = 4m + 1$&lt;/li&gt;&#xA;&lt;li&gt;$(4k + 2)^3 = 64k^3 + 24k^2 + 24k + 8 = 4(16k^3 + 6k^2 + 6k + 2) = 4m + 0$&lt;/li&gt;&#xA;&lt;li&gt;$(4k + 3)^3 = 64k^3 + 36k^2 + 36k + 27 = 4(16k^3 + 9k^2 +9k + 6) + 3 = 4m + 3$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So you can't have exactly 2 sodas left over.&lt;/p&gt;&#xA;" OwnerUserId="11318" LastEditorUserId="11318" LastEditDate="2015-04-13T19:16:13.030" LastActivityDate="2015-04-13T19:16:13.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11996" PostTypeId="2" ParentId="11952" CreationDate="2015-04-13T19:12:42.453" Score="3" Body="&lt;p&gt;We know that it cannot be Option 3 because it doesn't include Option 1 but does include Option 2 which does include Option 1.  For Option 3 to be valid it would have to include Option 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know it cannot be Option 2 because it is then not the &lt;strong&gt;only&lt;/strong&gt;, and exactly one, correct answer. Option 3 fails on this account as well, but I like the argument for 3 above better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Option 1 references only itself, and is exactly one answer, and is thus correct.&lt;/p&gt;&#xA;" OwnerUserId="1682" LastEditorDisplayName="user20" LastEditDate="2015-04-14T19:00:01.883" LastActivityDate="2015-04-14T19:00:01.883" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11997" PostTypeId="2" ParentId="11952" CreationDate="2015-04-13T19:19:38.827" Score="4" Body="&lt;p&gt;Perhaps the answer is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exactly one correct answer.&lt;/p&gt;&#xA;" OwnerUserId="11150" LastActivityDate="2015-04-13T19:19:38.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11998" PostTypeId="1" AcceptedAnswerId="11999" CreationDate="2015-04-13T19:27:48.320" Score="47" ViewCount="188288" Body="&lt;p&gt;The sequence of numbers $1\ 2\ 3\ 4\ 5\ 6\ 7\ 8\ 9$ has the property that you can insert mathematical operators in between the numbers from $1$ to $9$ and make the expression evaluate to 100. For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 \times 9 = 100$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are possibly hundreds of solutions to this problem, involving different varieties of operators. What is the expression with the fewest number of operators inserted (out of the set $+, -, \times, \div$ and maybe $\sqrt{}$ and $!$) that evaluates to 100?&lt;/p&gt;&#xA;" OwnerDisplayName="anon" LastEditorUserId="5373" LastEditDate="2015-05-13T20:01:15.797" LastActivityDate="2019-09-22T06:34:25.600" Title="1 2 3 4 5 6 7 8 9 = 100" Tags="|mathematics|calculation-puzzle|formation-of-numbers|" AnswerCount="10" CommentCount="19" ContentLicense="CC BY-SA 3.0" />
  <row Id="11999" PostTypeId="2" ParentId="11998" CreationDate="2015-04-13T19:36:24.897" Score="67" Body="&lt;p&gt;I believe that this is the smallest:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$123 - 45 - 67 + 89 = 100$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11150" LastEditorUserId="11497" LastEditDate="2019-09-22T06:34:25.600" LastActivityDate="2019-09-22T06:34:25.600" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="12000" PostTypeId="2" ParentId="11994" CreationDate="2015-04-13T19:56:40.827" Score="6" Body="&lt;p&gt;We have 3 even numbers to insert.&#xA;Since the numbers in rows, columns and diagonals are coprime, it means that the even numbers must have different row, different column and different diagonal. &#xA;Also, we have to manage the 3 even sums; a even sum can be made with E+E+E or E+O+O (where E=even and O=odd). The E+E+E case is not possible, as we already stated that the even numbers have different rows and columns. Now we know that each sum (14,22 and 24) contain only one even number.&#xA;Now we have few cases:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Bottom-Right cell is even. Though, this excludes all the / diagonal cells from being even, which is wrong as 14 cannot be obtained as sum of 3 odd numbers. &lt;em&gt;Contraddiction!&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Middle-Right cell is even. This excludes the Top-Left and the Middle cells from being even too, consequently Bottom-Left must be even! The third even cell is Middle-Top.&lt;/li&gt;&#xA;&lt;li&gt;Top-Right is even. Using the same logic as before, we deduce that Bottom-Middle must be even, as well as Left-Middle.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the right column, the even number must be 6, else we wouldn't find partners for the sum. 24-6=18. The only way of making 18 is 11+7, so the right column must be a permutation of {6,7,11}, 6 not being in the bottom-right corner. &#xA;7 cannot be the bottom-right, else we wouldn't find partners for 22 (remember that one must be even).&lt;br&gt;&#xA;So, 11 is forced to be the bottom-right number!&#xA;The bottom row can either be a permutation of {11,2,9} or {11,4,7}; though, we already used the 7, so the only possible bottom is permutation of {2,9,11}.&#xA;If 9 was the bottom-left, we would have serious problems with the diagonal sum 14, so the number 2 is in the bottom-left corner, forcing 6 to be middle-right and 4 to be top-middle. Current situation is&lt;br&gt;&#xA;? 4 7&lt;br&gt;&#xA;? 5 6&lt;br&gt;&#xA;2 9 11  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 cannot be in the 2nd row because 6 is present, so the final answer is:&lt;br&gt;&#xA;3 4 7&lt;br&gt;&#xA;1 5 6&lt;br&gt;&#xA;2 9 11  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1obEf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-04-13T19:56:40.827" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12001" PostTypeId="2" ParentId="11998" CreationDate="2015-04-13T20:34:11.143" Score="17" Body="&lt;h2&gt;Addendum too long for a comment&lt;/h2&gt;&#xA;&lt;p&gt;I can confirm there is no better solution than Andrew Smith's if we use the operators &lt;code&gt;+-*/.^&lt;/code&gt; There is only one solution using 3 operators. There are two using 4:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;123 +45 - 67 + 8 - 9&#xA;123 + 4- 5 + 67 - 89&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;There are 215 possibilities total and the number of solutions for the number of operators breaks down as follows:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;Operators   Solutions   &#xA;0           0&#xA;1           0&#xA;2           0&#xA;3           1&#xA;4           2&#xA;5           9&#xA;6           57&#xA;7           104&#xA;8           42&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;We could throw in some more operators though I doubt they would help. I didn't try parentheses, floors, mod, sqrt, etc. If anyone wants a go, though, here's some (unoptimized) VBA that will brute force the basic operators in Excel:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;Sub doTheMath()&#xA;    &#xA;    Const op As String = &amp;quot; +-*/.^&amp;quot;&#xA;    Dim s As String&#xA;    Dim a, b, c, d, e, f, g, h&#xA;    Dim i, j&#xA;    Dim v&#xA;    &#xA;    'Loop all possibilities&#xA;    i = 2&#xA;    j = 1&#xA;    For a = 1 To Len(op):    For b = 1 To Len(op):    For c = 1 To Len(op):    For d = 1 To Len(op)&#xA;    For e = 1 To Len(op):    For f = 1 To Len(op):    For g = 1 To Len(op):    For h = 1 To Len(op)&#xA;        s = 1 &amp;amp; Mid(op, a, 1) &amp;amp; 2 &amp;amp; Mid(op, b, 1) &amp;amp; 3 &amp;amp; Mid(op, c, 1) &amp;amp; 4 &amp;amp; Mid(op, d, 1) &amp;amp; 5 &amp;amp; _&#xA;            Mid(op, e, 1) &amp;amp; 6 &amp;amp; Mid(op, f, 1) &amp;amp; 7 &amp;amp; Mid(op, g, 1) &amp;amp; 8 &amp;amp; Mid(op, h, 1) &amp;amp; 9&#xA;        v = Application.Evaluate(Replace(s, &amp;quot; &amp;quot;, &amp;quot;&amp;quot;))&#xA;        If CStr(v) = CStr(100) Then&#xA;            Cells(i, 1).Value = s&#xA;            Cells(i, 2).Value = v&#xA;            i = i + 1&#xA;        End If&#xA;        j = j + 1&#xA;        If j &amp;gt; 1000000 Then&#xA;            ThisWorkbook.Save&#xA;            j = 1&#xA;        End If&#xA;    Next:    Next:    Next:    Next:    Next:    Next:    Next:    Next&#xA;    &#xA;End Sub&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-13T20:34:11.143" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12002" PostTypeId="2" ParentId="11943" CreationDate="2015-04-13T20:37:49.490" Score="2" Body="&lt;p&gt;The answer is Cooking, where the pie has a piece cut out...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not a full circle, so $\pi^2$ = 2&lt;/p&gt;&#xA;" OwnerUserId="9183" LastActivityDate="2015-04-13T20:37:49.490" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12003" PostTypeId="2" ParentId="11952" CreationDate="2015-04-13T21:03:15.780" Score="5" Body="&lt;p&gt;My answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Option 1&lt;/strong&gt; &lt;br&gt; The question asks us to &quot;Answer this question with exactly one correct answer.&quot; &lt;br&gt; option 2 gives 2 answers. &lt;br&gt;option 3 gives 2 answers, (&lt;em&gt;one of which gives another&lt;/em&gt;) &lt;br&gt;&lt;br&gt; Therefore, only option 1 could give &quot;&lt;strong&gt;exactly one&lt;/strong&gt; correct answer&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Further explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question was a bit misleading. It's true that all 3 answers could be &quot;correct&quot; if the answer is option 1. There is a &quot;trick&quot; in the question because [this question] refers to the title: &quot;Answer this question with exactly one correct answer.&quot; Emphasis on &lt;strong&gt;exactly one&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 and 3 don't give &lt;em&gt;exactly one&lt;/em&gt; answer. It might be better if they had been listed this way to separate the answers from themselves a little better:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A. Option 1&lt;br&gt;B. Option 1 or option 2&lt;br&gt;C. Option 2 or option 3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that, in my mind for this to make the most sense, Option 1 does not refer to 'A', but rather some unknown &quot;answer.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar question with the same reasoning would be: &lt;em&gt;(Assume all animals are your &quot;favorite&quot;)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Answer with exactly one of your favorite animals:&lt;br&gt;A. Cats&lt;br&gt;B. Cats or Dogs&lt;br&gt;C. Cats, Dogs, or Llamas&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2800" LastEditorUserId="2800" LastEditDate="2015-04-14T19:08:45.633" LastActivityDate="2015-04-14T19:08:45.633" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="12004" PostTypeId="1" AcceptedAnswerId="12224" CreationDate="2015-04-13T21:23:15.113" Score="11" ViewCount="783" Body="&lt;p&gt;Come ponder with me, as we go forth,&lt;br&gt;&#xA;How we behave in a manner of worth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a single change, that worth is no more.&lt;br&gt;&#xA;Instead we consider the ones before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After one more change, seen with different eyes,&lt;br&gt;&#xA;We behave in a way we won't trivialize.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve my riddle thoroughly,&lt;br&gt;&#xA;You'll need to give me answers three.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Hint 1&lt;/h3&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are looking for three words that can be changed into each other according to the instructions in the poem.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;Hint 2&lt;/h3&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first line of the poem is meant only as an introduction.  &quot;Going forth&quot; is not a clue to the word you're looking for.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;Hint 3&lt;/h3&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm looking for 3 adverbs, of 10 letters each.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="5373" LastEditDate="2016-09-30T17:32:05.883" LastActivityDate="2016-09-30T17:32:05.883" Title="Riddle: Come ponder with me" Tags="|riddle|word|wordplay|" AnswerCount="8" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12005" PostTypeId="2" ParentId="11998" CreationDate="2015-04-13T21:39:44.610" Score="93" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;Note&lt;/strong&gt;: This answer only applies prior to the edit that clarifies that the expression on the left must evaluate to 100, rather than simply the equation being true.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you allow exponents, you can get away with just two:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $ 1^{ 23456789 } = 10^0 $&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastEditorDisplayName="user88" LastEditDate="2015-04-14T17:33:34.597" LastActivityDate="2015-04-14T17:33:34.597" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="12006" PostTypeId="1" AcceptedAnswerId="12007" CreationDate="2015-04-13T21:42:44.547" Score="10" ViewCount="1083" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;At some point in his career, Mario Mendoza's batting average was above 0.200. Sometime later, it was below 0.200. Prove that, at some point, he was batting exactly 0.200. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For those unfamiliar with baseball:&lt;/strong&gt; players have a series of &quot;at-bats&quot; (attempted hits) which either succeed, or don't. At any point, a player's batting average is the fraction of their at-bats so far which were hits. This number is usually rounded to 3 decimals, but for the purposes of this puzzle, consider all fractions to be exact. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For those familiar with baseball:&lt;/strong&gt; the point where he is batting 0.200 might occur during a game, so assume that a player's batting average is updated after every at-bat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Unnecessary fun fact:&lt;/strong&gt; Mario Mendoza was a real player, whose batting average tended to be around 0.200, an embarrassingly low rate. This is pretty led his teammates to joke that anyone who batted below 0.200 was &quot;below the Mendoza line.&quot;&lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-04-15T14:12:22.223" Title="The Mendoza Line" Tags="|number-theory|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12007" PostTypeId="2" ParentId="12006" CreationDate="2015-04-13T21:53:38.543" Score="18" Body="&lt;p&gt;Let $h$ be the number of successful hits, $t$ the number of tries. Initially $\frac{h}{t} &amp;gt; \frac{1}{5}$ or equivalently, $5h &amp;gt; t$. At any point we may simultaneously add 1 to $t$ and add 1 or 0 to $h$. Notice, if $5h &amp;gt; t$, then $5(h+1) &amp;gt; t+1$ certainly, and if $5h &amp;gt; t$, then $5(h+0) \ge t+1$ certainly (as we are working with integers). Thus to pass from $5h &amp;gt; t$ to $5h &amp;lt; t$, we must pass through $5h = t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This argument is valid for any $\frac{1}{n}$, but not any $\frac{a}{b}$, as then we cannot argue that if $bh &amp;gt; at$ then $b(h+0) \ge a(t+1)$.&lt;/p&gt;&#xA;" OwnerUserId="11100" LastActivityDate="2015-04-13T21:53:38.543" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12009" PostTypeId="2" ParentId="11972" CreationDate="2015-04-13T22:23:43.790" Score="32" Body="&lt;h1&gt;4.8264 km&lt;/h1&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OCyEW.png&quot; alt=&quot;Opaque circle solution with a length of ~4.8264 km&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Plough a &lt;span class=&quot;math-container&quot;&gt;$120°$&lt;/span&gt; section of the circle, extending both ends in a straight line by a distance of &lt;span class=&quot;math-container&quot;&gt;$\sqrt{1/3}$&lt;/span&gt; km to meet the vertices of a hexagon enclosing the circle. Two more lines of length &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; km and &lt;span class=&quot;math-container&quot;&gt;$\sqrt{1/3}$&lt;/span&gt; km are ploughed as shown to cover paths through the other half of the circle. The total length ploughed is equal to:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{2\pi}{3} + 3 \times \sqrt{\frac13} + 1 = 4.82644591 \ \text{km}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;(This puzzle is related to the opaque square problem discussed in &lt;a href=&quot;https://books.google.co.uk/books?id=HvFYBQAAQBAJ&amp;amp;lpg=PT107&amp;amp;dq=Professor%20Stewart%27s%20Casebook%20Opaque%20Polygons&amp;amp;pg=PT107#v=onepage&amp;amp;q=Professor%20Stewart%27s%20Casebook%20Opaque%20Polygons&amp;amp;f=false&quot; rel=&quot;noreferrer&quot;&gt;Professor Stewart's Casebook of Mathematical Mysteries&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="40" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-13T22:33:40.910" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="12010" PostTypeId="2" ParentId="11972" CreationDate="2015-04-13T23:43:54.080" Score="27" Body="&lt;h1&gt;4.8205 km (or 4.8189 km with slightly more work).&lt;/h1&gt;&#xA;&lt;p&gt;Let &lt;span class=&quot;math-container&quot;&gt;$ABCDE$&lt;/span&gt; be a circumscribed regular pentagon around the circle, let &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; be the point of tangency between segment &lt;span class=&quot;math-container&quot;&gt;$BC$&lt;/span&gt; and the circle, and let &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; be the point of tangency between segment &lt;span class=&quot;math-container&quot;&gt;$DE$&lt;/span&gt; and the circle. Let &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt; be the midpoint of line &lt;span class=&quot;math-container&quot;&gt;$BE$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Plow the arc of the circle connecting &lt;span class=&quot;math-container&quot;&gt;$M,N$&lt;/span&gt;, and plow the line segments &lt;span class=&quot;math-container&quot;&gt;$AX, BM, NE$&lt;/span&gt;. The total length is&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\frac{4\pi}{5} + |AX| + 2|BM| = \frac{4\pi}{5} + \frac{3\sqrt{5}-5}{2} + 2\sqrt{5-2\sqrt{5}} = 4.82046...$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zcLoa.png&quot; alt=&quot;Solution to the opaque circle problem based on a regular pentagon&quot; /&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;To get 4.8189...&lt;/h2&gt;&#xA;&lt;p&gt;Instead of using a regular pentagon use a slightly irregular circumscribed pentagon. If the pentagon is symmetric around &lt;span class=&quot;math-container&quot;&gt;$AX$&lt;/span&gt;, with the angle at &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; equal to &lt;span class=&quot;math-container&quot;&gt;$\pi-x$&lt;/span&gt; and the angle at &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; equal to &lt;span class=&quot;math-container&quot;&gt;$\frac{\pi-y}{2}$&lt;/span&gt;, then the total length comes out to&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$2x + \cos(x+y)\sec(y)-\sec(x+2y) + 2\tan(y)$&lt;/span&gt;,&lt;/p&gt;&#xA;&lt;p&gt;and plugging in &lt;span class=&quot;math-container&quot;&gt;$x \approx 1.259557, y \approx 0.6432556$&lt;/span&gt; we get &lt;span class=&quot;math-container&quot;&gt;$4.8189...$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;We can also get a lower bound on the total length. First note that we have to plow at least one point with each possible &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; coordinate between &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$-1$&lt;/span&gt;, so the length of the projection of our plow path onto the &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;-axis is at least &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt;. Thus the original length is at least 2 km.&lt;/p&gt;&#xA;&lt;p&gt;Slightly more sophisticatedly, we can combine the fact that the projection onto the &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;-axis has length at least &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; with the fact that the projection of the length onto the &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt;-axis has at least &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; to see that the total length is at least &lt;span class=&quot;math-container&quot;&gt;$2\sqrt{2}$&lt;/span&gt; km: the average of the two projections of any straight line of length &lt;span class=&quot;math-container&quot;&gt;$l$&lt;/span&gt; onto the &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; axes is at most &lt;span class=&quot;math-container&quot;&gt;$\frac{l}{\sqrt{2}}$&lt;/span&gt;, and the total of these averages over all the line segments we plow must be at least &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Taking this to its logical conclusion, if we try averaging the projection of a straight line segment onto a uniformly random direction, we get &lt;span class=&quot;math-container&quot;&gt;$\frac{2l}{\pi}$&lt;/span&gt;, so the total length plowed must be at least &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt; km.&lt;/p&gt;&#xA;" OwnerUserId="11295" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-16T09:23:38.483" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12011" PostTypeId="1" AcceptedAnswerId="16980" CreationDate="2015-04-14T00:00:27.677" Score="8" ViewCount="1155" Body="&lt;p&gt;Your Japanese co-worker from &lt;a href=&quot;https://puzzling.stackexchange.com/q/11134/88&quot;&gt;this problem&lt;/a&gt; hands you another piece of paper one day:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1820143 2383341 3245353 8425463 6354607&lt;br&gt;&#xA;  6627972 9088779 7839589 ??????? ???????&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;What two seven-digit numbers come after the ones given?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On the back of the paper is a hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You get stuck, and ask your co-worker for another hint.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He simply writes a zero in front of the existing hint, leaving you with a note that reads: &lt;p&gt;&#xA; 03&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And another:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, he just scribbles a plus sign in front of the zero, leaving you with a note that reads:  &lt;p&gt;&#xA;+03&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And a fourth:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This time, he writes a whole row of numbers on the bottom:&lt;br&gt;&#xA; 12 34 56 71 23 45 67&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And a fifth:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He crosses out the 71 23 45 67 and writes 78 9A BC DE in its place.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And a sixth:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He writes &quot;COUNTER MODE&quot; next to the &quot;+03&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And a seventh, just as clarification:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each letter is exactly two digits long.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I promise this one doesn't involve any obscure numbers like 2&lt;sup&gt;64&lt;/sup&gt;, or any other such esoteric facts, unlike &lt;a href=&quot;https://puzzling.stackexchange.com/questions/11141/a-simple-but-most-devious-sequence-puzzle&quot;&gt;my last one in this regard&lt;/a&gt;. Everything here is simple mathematics and logic.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-06-25T14:32:19.340" Title="A crypto-sequence puzzle" Tags="|cipher|number-sequence|" AnswerCount="4" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="12013" PostTypeId="2" ParentId="12004" CreationDate="2015-04-14T00:14:46.843" Score="13" Body="&lt;p&gt;The answers are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PROCESSION, PRECESSION, PRECISION&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Come ponder with me, as we go forth,&lt;br&gt;&#xA; How we behave in a manner of worth.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We &quot;go forth&quot; in a &lt;strong&gt;procession&lt;/strong&gt;, perhaps a carnival or festival of &quot;worth&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;With a single change, that worth is no more.&lt;br&gt;&#xA; Instead we consider the ones before.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Change a single letter to get &lt;strong&gt;precession&lt;/strong&gt;, meaning preceding or &quot;going before&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;After one more change, seen with different eyes,&lt;br&gt;&#xA; We behave in a way we won't trivialize.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Change the word again by introducing a new I (&quot;eye); precision is not to be sniffed at and &quot;trivialised&quot;!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;To solve my riddle thoroughly,&lt;br&gt;&#xA; You'll need to give me answers three.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Indeed: three similar words.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-04-14T00:14:46.843" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12014" PostTypeId="2" ParentId="11998" CreationDate="2015-04-14T00:18:32.287" Score="-5" Body="&lt;p&gt;I'm not sure this is what you're looking for, but 123456789=/=100&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This answer adds two symbols, &quot;=&quot; and &quot;/&quot;.&lt;/p&gt;&#xA;" OwnerUserId="457" LastActivityDate="2015-04-14T00:18:32.287" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12018" PostTypeId="2" ParentId="11943" CreationDate="2015-04-14T02:23:01.227" Score="1" Body="&lt;p&gt;In physics, we sometimes use $\pi$ to mean the permutation operator that swaps two particles. If you swap two particles and then swap them again, you get back to the same state as before (...usually). Therefore, $\pi^2 = 1$.&lt;/p&gt;&#xA;" OwnerUserId="11340" LastActivityDate="2015-04-14T02:23:01.227" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12019" PostTypeId="2" ParentId="11956" CreationDate="2015-04-14T03:23:14.497" Score="4" Body="&lt;p&gt;I used VBA in Excel workbook module (because it was convenient for me).  Feel free to convert it to work in cscript.  I think all you have to do is change the &lt;code&gt;debug.print&lt;/code&gt; with the appropriate Write statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, here is the function which tests to see if the number is self-referential.  It's hard-coded for 10-digit numbers, but it could be adapted for smaller widths or even other number bases if desired.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Public Function IsSelfReferential(strNumber As String) As Boolean&#xA;    'This tests to see if a 10-digit string is a self-referential number.&#xA;    Dim i, intLength As Integer&#xA;&#xA;    intLength = Len(strNumber)&#xA;    If intLength &amp;lt;&amp;gt; 10 Then Stop&#xA;&#xA;    For i = 0 To 9&#xA;        'VBA hack to find number of occurrances by removing what we are looking for and comparing the new length.&#xA;        If (intLength - Len(Replace(strNumber, Trim$(Str$(i)), &quot;&quot;))) _&#xA;            &amp;lt;&amp;gt; Val(Mid$(strNumber, i + 1, 1)) Then&#xA;            'Since the length doesn't match the digit stored, then this number can't be self-referential&#xA;            IsSelfReferential = False&#xA;            Exit Function&#xA;        End If&#xA;    Next&#xA;&#xA;    'If we made it this far, the number is self-referential.&#xA;    IsSelfReferential = True&#xA;End Function&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I decided to exploit the fact that whatever the answer is, the digits will add up to 10, since we are describing the number of digits in 10-digit number.  Anything higher or lower can not be a correct answer.  This greatly reduces the number of possibilities the routine has to check compared to a brute-force count.  This function generates numbers whose digits add up to a certain goal.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Public Sub NumbersThatAddUpTo(intGoal As Integer, intWidth As Integer, Optional strPrefix As String)&#xA;    'Recursive function to find all combinations of intWidth digits that add up to intGoal.&#xA;    Dim i, intMax As Integer&#xA;&#xA;    If intWidth = 1 Then&#xA;        If intGoal &amp;lt;= 9 And intGoal &amp;gt;= 0 Then&#xA;            If IsSelfReferential(strPrefix &amp;amp; Trim$(Str$(intGoal))) Then&#xA;                Debug.Print strPrefix &amp;amp; Trim$(Str$(intGoal))&#xA;            End If&#xA;        End If&#xA;    Else&#xA;        If intGoal &amp;lt; 9 Then intMax = intGoal Else intMax = 9&#xA;        For i = 0 To intMax&#xA;            NumbersThatAddUpTo intGoal - i, intWidth - 1, strPrefix &amp;amp; Trim$(Str$(i))&#xA;        Next&#xA;    End If&#xA;&#xA;End Sub&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;To execute the function, I ran the sub from the debug window:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ThisWorkbook.NumbersThatAddUpTo 10,10&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And a couple of seconds later, it produced a single result:&#xA;6210001000&lt;/p&gt;&#xA;" OwnerUserId="11342" LastActivityDate="2015-04-14T03:23:14.497" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12021" PostTypeId="2" ParentId="11968" CreationDate="2015-04-14T04:31:56.197" Score="6" Body="&lt;p&gt;Its 9376. I ran a brute force and calculated it using below program.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#include&amp;lt;cmath&amp;gt;&#xA;#include&amp;lt;iostream&amp;gt;&#xA;using namespace std;&#xA;int last4digit(int n)&#xA;{&#xA;        int x = n / 10000;&#xA;        n -= x * 10000;&#xA;        return n;&#xA;}&#xA;int main()&#xA;{&#xA;    for (int i = 1000; i &amp;lt; 10000; i++)&#xA;    {&#xA;        if (i == last4digit(pow(i, 2))) &#xA;            cout &amp;lt;&amp;lt; i;&#xA;    }&#xA;    system(&quot;pause&quot;);&#xA;    return 0;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="11343" LastEditorUserId="11343" LastEditDate="2015-04-14T09:39:49.437" LastActivityDate="2015-04-14T09:39:49.437" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12022" PostTypeId="1" AcceptedAnswerId="12052" CreationDate="2015-04-14T05:04:15.150" Score="10" ViewCount="114380" Body="&lt;p&gt;I saw this question on &lt;a href=&quot;https://www.facebook.com/kennethjianwenz/photos/a.173663129479243.1073741827.167504136761809/385751114937109/?type=1&amp;amp;theater&quot; rel=&quot;noreferrer&quot;&gt;Facebook&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Albert and Bernard just became friends with Cheryl, and they want to know when her birthday is. Cheryl gives them a list of 10 possible dates.&lt;/p&gt;&#xA;&lt;p&gt;May 15 16 19&lt;/p&gt;&#xA;&lt;p&gt;June 17 18&lt;/p&gt;&#xA;&lt;p&gt;July 14 16&lt;/p&gt;&#xA;&lt;p&gt;August 14 15 17&lt;/p&gt;&#xA;&lt;p&gt;Cheryl then tells Albert and Bernard separately the month and the day of her birthday, respectively.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Albert&lt;/strong&gt;: I don't know when Cheryl's birthday is, but I know that Bernard does not know, too.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Bernard&lt;/strong&gt;: At first, I didn't know when Cheryl's birthday is, but I know now.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Albert&lt;/strong&gt;: Then I also know when Cheryl's birthday is.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;When is Cheryl's birthday?&lt;/p&gt;&#xA;&lt;p&gt;There is an official solution which is explained in the link below, but I don't understand the logic.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; July 16, as explained here. &lt;a href=&quot;https://www.facebook.com/kennethjianwenz/posts/386479228197631&quot; rel=&quot;noreferrer&quot;&gt;https://www.facebook.com/kennethjianwenz/posts/386479228197631&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;What is wrong with the logic in the answer I posted below (I put it in spoilers if others want to try)?&lt;/p&gt;&#xA;" OwnerUserId="2765" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-17T08:02:36.470" Title="When is Cheryl's Birthday?" Tags="|logical-deduction|" AnswerCount="8" CommentCount="7" ClosedDate="2015-04-14T11:28:17.670" ContentLicense="CC BY-SA 3.0" />
  <row Id="12023" PostTypeId="2" ParentId="12022" CreationDate="2015-04-14T05:04:15.150" Score="2" Body="&lt;p&gt;The date I thought of was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; August 17&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Albert&lt;/strong&gt;: I don't know when Cheryl's birthday is, but I know that Bernard does not know, too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If Bernard was told 18 or 19, he would know right away, as they are unique dates. But he doesn't, so Albert can eliminate May 19 and June 18. If he was told June, then he would know it's June 17, but as he does not know at this stage, we know he wasn't told June. So we've eliminated three dates: May 19, June 17, and June 18.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bernard&lt;/strong&gt;: At first, I didn't know when Cheryl's birthday is, but I know now.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Now that Bernard knows Albert doesn't know, Bernard knows that Albert was not told June. He can also eliminate those three dates. The only way for him to know based on this information which date is correct is if he was told 17, which would make August 17 the answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Albert&lt;/strong&gt;: Then I also know when Cheryl's birthday is.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Now that Albert knows Bernard knows, Albert can also work out that Bernard was told 17, so he comes to the same conclusion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2765" LastActivityDate="2015-04-14T05:04:15.150" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12025" PostTypeId="2" ParentId="12022" CreationDate="2015-04-14T05:26:04.150" Score="1" Body="&lt;p&gt;July 16, because the first statement Albert makes immediately eliminates May and June. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;May and June both contains unique dates - May 19 and June 18 respectively - and for Bernard to DEFINITELY not have a chance to know the answer, it couldn't be those 2 months. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then Bernard confirms the birthday. Which means that July 14 and August 14 is eliminated, because otherwise, Bernard who only knows the date, wouldnt know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After which Albert claims that he knows it too, meaning the only answer would be July 16, as August has 2 dates remaining and Albert still wouldn't know for sure because he only knows the month.&lt;/p&gt;&#xA;" OwnerUserId="11350" LastEditorUserId="11350" LastEditDate="2015-04-14T05:39:24.690" LastActivityDate="2015-04-14T05:39:24.690" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12026" PostTypeId="2" ParentId="12022" CreationDate="2015-04-14T05:30:38.460" Score="0" Body="&lt;p&gt;Behind your first spoiler: as Beno pointed out in a comment, Albert's first statement let's you eliminate even more. After he speaks, Bernard (and we) know that the month cannot be May or June.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Behind your second spoiler, you say &quot;The only way for him to know based on this information would be if he was told 15&quot;. 15 is not the only possibility. Since Albert's first statement let's you eliminate May and June, there are three dates Bernard could have heard that would allow him to have certainty, namely, the 15, 16 and 17 (it couldn't be 14, because then Bernard would think both June 14 and August 14 were possible).&lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-04-14T05:30:38.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12027" PostTypeId="2" ParentId="12022" CreationDate="2015-04-14T05:31:15.137" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Albert: I know Bernard doesn't know when Cheryl's birthday is.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since May and June contain unique dates 18 and 19, then Albert cannot have been given one of those months, since if Bernard had received 18 or 19, then he would have immediately known (and therefore Albert wouldn't know if Bernard already knew Cheryl's birthday or not). So Cheryl's birthday must be in July or August.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bernard: Now I know when Cheryl's birthday is.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since Bernard didn't know before and now he does, he must have deduced some information about the date, which means the date can't be 14 because then Bernard would still have been stuck with the month being either July or August.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Albert: And now I know too.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since Albert needs this information to discover which date it is, there must only have been one possibility remaining from eliminating 14 as a possible day, which is only applicable if the month is July. The remaining possibility is July 16, and so that's Cheryl's birthday.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-04-14T05:31:15.137" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12028" PostTypeId="2" ParentId="12022" CreationDate="2015-04-14T05:37:39.310" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Albert: I don't know when Cheryl's birthday is, but I know that Bernard does not know, too.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No matter what month is given to A he can not know the birthday.&#xA;If A was given May: B was given 15 or 16, 19. But if B was given 19(not ambiguous), he could deduce the birthday so A was not given May.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If A was given June, B was given 17 or 18, again 18 is not ambiguous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conclusion: A was given July or Aug, in which all dates are ambiguous&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bernard: At first, I didn't know when Cheryl's birthday is, but I know now.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the date give to B may be 14, 15, 16 or 17.&#xA;14: Still ambigous as present in both months. So he had 15, 16 or 17 and now he knows the birthday&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Albert: Then I also know when Cheryl's birthday is.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now A can rule out 14, otherwise B could not know the date. Eliminating 14, 16 is the only date remaining in July but 2 more dates remains in Aug leading to ambiguity. As A knows the answer now, it must be July and can not be 14.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; July 16 is Cheryl's birthday&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4637" LastEditorUserId="4637" LastEditDate="2015-04-14T06:52:04.670" LastActivityDate="2015-04-14T06:52:04.670" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12029" PostTypeId="2" ParentId="11968" CreationDate="2015-04-14T05:54:47.033" Score="16" Body="&lt;p&gt;The equation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^2 = x \bmod 10^4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is true exactly if $10^4$ is a divisor of $x (x -1)$. This requires that the prime factorizations of $x$ and $x-1$ jointly contain $2^4$ and $5^4$. Since they are relatively prime, each of $2^4$ and $5^4$ must be divisor of one of them. A factor of both would be a multiple of $10^4$ and thus not 4 digits, so each number must be a multiple of one factor:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$2^4 \mid x $ and $5^4 \mid x-1$, or   &lt;/li&gt;&#xA;&lt;li&gt;$5^4 \mid x $ and $2^4 \mid x-1$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We can use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chinese_remainder_theorem&quot;&gt;Chinese remainder theorem&lt;/a&gt; algorithm to find the corresponding residue classes&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$x = 9376 \mod 10000$   &lt;/li&gt;&#xA;&lt;li&gt;$x = 625 \mod 10000$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The second doesn't correspond to any four-digit numbers, so the only solution is $9376$.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="463" LastEditDate="2015-04-14T13:19:24.207" LastActivityDate="2015-04-14T13:19:24.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12030" PostTypeId="2" ParentId="11983" CreationDate="2015-04-14T06:18:44.203" Score="3" Body="&lt;p&gt;Why assume integers?  This is a classic packing problem.  Roger and I are &lt;em&gt;extremely lazy&lt;/em&gt; and we decide that only two cans should fit in the pallet at all.  After all, we want to have plenty of time to drink our sodas.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We figure there is some value between 1 wide and 2 wide which would only allow for 2 cans to fit.  We're also too untrained to figure out how to solve this problem properly in 3D (like me), and can't find two cans the same size in the break room, so we pretend this is a 2D problem and easily figure out how to cram 2 circles in a square.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also figure that &lt;em&gt;if&lt;/em&gt; there is a specific solution, then it should work for a can that is exactly as tall as it is wide, which would sure make their math a bit easier.  Otherwise, we're going to fuss at the OP in management.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So back to the circles.  If we cram them diagonally, then they fit nicely in a pallet $(2+\sqrt{2})/2$ or about 1.7 cans tall/wide/across.  We were bright enough to figure this out on our own, but too lazy to make a picture, so Roger cribbed one from &lt;a href=&quot;http://www.algebra.com/algebra/homework/word/geometry/Geometry_Word_Problems.faq.question.319088.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this guy's web site&lt;/a&gt;, who also shows similar work on determining this solution for his problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KudQk.jpg&quot; alt=&quot;circles in square&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We figure that 3D will give the cans some additional wiggle room not present in 2D -- maybe &lt;em&gt;too&lt;/em&gt; much -- so we guestimate that we should probably request it down to &lt;strong&gt;1.6 or 1.5&lt;/strong&gt; until the guys in pallet packing start screaming about it.&lt;/p&gt;&#xA;" OwnerUserId="11342" LastActivityDate="2015-04-14T06:18:44.203" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12032" PostTypeId="5" CreationDate="2015-04-14T07:17:00.690" Score="0" Body="" OwnerUserId="4637" LastEditorUserId="5373" LastEditDate="2015-04-27T08:11:48.417" LastActivityDate="2015-04-27T08:11:48.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12033" PostTypeId="4" CreationDate="2015-04-14T07:17:00.690" Score="0" Body="Multiple choice is a form of question in which multiple answers are proposed and you select one or more correct answer(s)." OwnerUserId="4637" LastEditorUserId="5373" LastEditDate="2015-04-27T08:11:48.417" LastActivityDate="2015-04-27T08:11:48.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12034" PostTypeId="2" ParentId="11943" CreationDate="2015-04-14T07:51:05.700" Score="6" Body="&lt;p&gt;I like commenter Lopsy's suggestion of this&#xA;&lt;img src=&quot;https://i.stack.imgur.com/J55Lc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which seems to be a circle in &lt;a href=&quot;http://en.wikipedia.org/wiki/Taxicab_geometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;Taxicab geometry&lt;/a&gt; AKA $L^1$ space. In this case however, $\pi = 2^2$, not $\pi^2 = 2$. It's possible that the OP made a mistake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Lopsy, you can see that $\pi=4$ not $2$ or $2\sqrt{2}$ because the length of each diagonal side is equal to $r+r$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to find a value of $p$ that made $\pi = \sqrt{2}$ in &lt;a href=&quot;http://en.wikipedia.org/wiki/Lp_space&quot; rel=&quot;nofollow noreferrer&quot;&gt;$L^p$ space&lt;/a&gt; but found $\pi$ was minimized at $p=2$ with the usual value of $\approx 3.1416$. It is equal to $4$ at $p=1$ and $p=\infty$ and diverges to $\infty$ as $p \to 0$. I don't think a circle is well-defined for $p \leq 0$.&lt;/p&gt;&#xA;" OwnerUserId="11348" LastActivityDate="2015-04-14T07:51:05.700" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12035" PostTypeId="1" AcceptedAnswerId="12046" CreationDate="2015-04-14T07:55:21.480" Score="7" ViewCount="1004" Body="&lt;p&gt;$100$ perfect logicians have been gathered on an island. It is common knowledge that logicians always have blue or brown eyes, though on this island, all eyes are blue. However, there are no reflective surfaces, and talking about eye color is forbidden; in short, each logician knows everyone else's eye color, but not their own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One Sunday afternoon, a green-eyed oracle visits the island, and makes the following decree, loud enough for all to hear:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The number of blue eyed logicians on this island is &lt;strong&gt;not&lt;/strong&gt; a multiple of $17$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(The multiples of $17$ up to $100$ are $0,17,34,51,68,85$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is common knowledge that the oracle knows all, and never lies. Starting that on that Sunday midnight, and every midnight thereafter, a ferry comes to take away anyone who knows their own eye color. The question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Will the logicians ever leave, and if so, on what day?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Please hide any guesses/solutions with spoilers, for the benefit of other solvers.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Remarks:&lt;/strong&gt; It seems like the oracle has said nothing new, since everyone already knew the number of blue-eyed logicians was either $99$ or $100$, and $17$ divides neither of these. Anyone familiar with &lt;a href=&quot;https://puzzling.stackexchange.com/questions/236/in-the-100-blue-eyes-problem-why-is-the-green-eyed-oracle-necessary&quot;&gt;these sorts of puzzles&lt;/a&gt; will confirm that, counterintuitively, statements like this have some content, which can be enough do allow the logicians to deduce their freedom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; I changed the number the oracle says, since I think it was originally too easy.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2016-07-15T16:55:32.877" Title="Green Eyed Oracle Variant - Not a Multiple of $17$" Tags="|logical-deduction|blue-eyes|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12036" PostTypeId="1" CreationDate="2015-04-14T08:20:42.197" Score="8" ViewCount="969" Body="&lt;p&gt;The famous and ruthless explorer Wyoming Wilbert reports in the third part of his memoires that he once visited an island inhabited by jokers and truth tellers. Truth tellers always tell the truth, whereas jokers sometimes lie and sometimes tell the truth. Furthermore when a joker is killed than his body turns green, while the corpse of a dead truth teller decomposes in the ordinary fashion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wilbert spent several weeks in a small village with 155 inhabitants. Every village inhabitant knew exactly who the jokers were and who the truth tellers were, but Wilbert did not know the identity of a single inhabitant. &#xA;However, Wilbert happened to know the exact number $j\ge2$ of jokers in the village.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the first day, he asked every inhabitant a single yes-no question. He analyzed the answers and then killed one of the inhabitants; the corpse showed Wilbert that this guy had been a joker. On the second day, Wilbert repeated this procedure with the remaining 154 inhabitants: he asked every survivor a single yes-no question, and afterwards killed one of them. And so on, day by day, until he decided to stop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wyoming Wilbert reports in his book that he had designed all his questions meticulously. They guaranteed him that after several days &lt;strong&gt;all&lt;/strong&gt; the jokers would be dead, while &lt;strong&gt;none&lt;/strong&gt; of the truth teller had been killed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Questions: &lt;strong&gt;For what values of $j\ge2$ can this story possibly be true?&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;strong&gt;What was the questioning strategy of Wyoming Wilbert?&lt;/strong&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2015-05-29T07:16:39.537" Title="Kill the jokers! - Part 2" Tags="|logical-deduction|liars|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12037" PostTypeId="2" ParentId="11952" CreationDate="2015-04-14T08:23:06.940" Score="3" Body="&lt;p&gt;So this is question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Choose the only correct answer to this question:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would choose 1 for the only correct answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I choose 2, there are 2 answers including Option 1 and Option 2. So it is not only one answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I choose 3, there are 2 too including Option 2 and Option 3. It is also not the only one answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one only has Option 1 &amp;amp; that would be the correct answer.&lt;/p&gt;&#xA;" OwnerUserId="11353" LastEditorDisplayName="user20" LastEditDate="2015-04-14T19:01:01.537" LastActivityDate="2015-04-14T19:01:01.537" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12038" PostTypeId="1" AcceptedAnswerId="12051" CreationDate="2015-04-14T08:25:46.493" Score="7" ViewCount="730" Body="&lt;p&gt;My Indian friend Jamal recently participated in the well-known TV quiz&#xA;show &lt;strong&gt;&lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Who_Wants_to_Be_a_Millionaire&quot; rel=&quot;noreferrer&quot;&gt;Who wants to be a millionaire?&lt;/a&gt;&lt;/em&gt;&lt;/strong&gt;. You might have heard or read about Jamal's spectacular success; the media reported on it extensively.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what Jamal told us about his 2.000 Euro question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;After the commercial break, Prem Kumar started a little friendly chat about&#xA;  my job.  I said that I enjoyed software design because sometimes it was a&#xA;  lot like solving puzzles.  He made some crack about me having to solve quite&#xA;  a bit of them now, and then the lights went down again and the next question&#xA;  came up. Prem pulled off a big show with this question, and he obviously&#xA;  enjoyed repeating the word &quot;2000&quot; over and over again, since it also showed &#xA;  up as part of the question text.  The question seemed to be much too hard for &#xA;  a 2.000 Euro question.  I guess that the question makers and show producers&#xA;  just had chosen it at this level so that Prem could play around with the words.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now about the answers, I didn't think it was a romantic comedy, so that left &#xA;  option A out.  I was also pretty sure D was &quot;just&quot; an award winning epic and &#xA;  certainly not on top.  Once again, however, I had to reach back into the &#xA;  depths of my memory to decide between the two remaining choices.  Prem made &#xA;  some comment that almost interrupted my concentration; I just replied &#xA;  &quot;Yeah, I could do that&quot;, but without committing myself to anything.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I could see an old newspaper article coming into focus in my mind and suddenly remembered that C had domestically beaten B, though not by much.  Thinking that the 50/50 option that Prem had mentioned would probably only leave me with answers B and C anyway, I decided to trust my memory instead.  Turns out, I had a good one.  At least this time.  Later I found out that answer B indeed had been the runner-up, while A and D had been farther down the list.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;What was the 2.000 Euro question?&lt;br&gt;&#xA;What were the four possible answers A, B, C, D?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="8874" LastEditDate="2015-10-04T14:11:44.457" LastActivityDate="2015-10-04T14:11:44.457" Title="Slumdog millionaire (6)" Tags="|knowledge|movies|" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12040" PostTypeId="2" ParentId="12038" CreationDate="2015-04-14T08:46:26.733" Score="3" Body="&lt;p&gt;Not sure of the answer yet, but I will post my ideas for others.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm thinking of a movies in year 2000 that were somehow significant&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; List of RomComs is:&lt;br&gt;&lt;br&gt; Down to You&lt;br&gt;Return to Me&lt;br&gt;Keeping the Faith&lt;br&gt;Boys and Girls&lt;br&gt;Loser &lt;br&gt;Woman on Top &lt;br&gt;What Women Want&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;B:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For B &amp;amp; C I need to find similar movies but C was more successful &lt;strike&gt;in India&lt;/strike&gt; domestically for the studio&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;C:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For B &amp;amp; C I need to find similar movies but C was more successful &lt;strike&gt;in India&lt;/strike&gt; domestically for the studio&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;D:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Award winning epic for year 2000 is &lt;strong&gt;Gladiator&lt;/strong&gt; and &lt;strong&gt;Crouching Tiger, Hidden Dragon&lt;/strong&gt; &lt;br&gt;&lt;br&gt;but if it wasn't on top, I would scratch Gladiator, because that tops many leaderboards.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My concept of questions is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Highest grossing film perhaps with some unknown condition.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My inspiration:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/2000_in_film&quot; rel=&quot;nofollow&quot;&gt;Wiki - 2000 in Film&lt;/a&gt;&lt;br&gt;&lt;a href=&quot;http://www.boxofficemojo.com/yearly/chart/?view2=worldwide&amp;amp;yr=2000&amp;amp;sort=wwgross&amp;amp;order=DESC&amp;amp;p=.htm&quot; rel=&quot;nofollow&quot;&gt;Highest grossing films of 2000&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9907" LastEditorUserId="9907" LastEditDate="2015-04-14T08:57:30.087" LastActivityDate="2015-04-14T08:57:30.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12041" PostTypeId="2" ParentId="11943" CreationDate="2015-04-14T09:12:38.627" Score="1" Body="&lt;p&gt;The simplest answer would seem to be &quot;boxing&quot;, if we consider a ring to be synonymous with a circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&#xA;If the diameter of the square is its diagonal and the circumference is its perimeter then the square of their ratio is 2.&lt;/p&gt;&#xA;" OwnerUserId="10149" LastEditorUserId="4637" LastEditDate="2015-04-14T09:54:04.773" LastActivityDate="2015-04-14T09:54:04.773" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12042" PostTypeId="1" AcceptedAnswerId="12045" CreationDate="2015-04-14T09:19:25.060" Score="7" ViewCount="536" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Thou came here for a rest.&lt;br&gt;&#xA;  With my body, I made a shelter&lt;br&gt;&#xA;  Thou feel so cool, when I flutter&lt;br&gt;&#xA;  Thou touch me with his finger&lt;br&gt;&#xA;  didn't know whether it hurts me&lt;br&gt;&#xA;  And I get some sketches but feel real  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;And Thou fly away&lt;br&gt;&#xA;  All my wings are caused to sway&lt;br&gt;&#xA;  That one last touch hurts me enough&lt;br&gt;&#xA;  My one finger goes down to lay. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who/what am I?&lt;/p&gt;&#xA;" OwnerUserId="11353" LastEditorUserId="5188" LastEditDate="2015-04-16T07:05:52.950" LastActivityDate="2015-04-16T07:05:52.950" Title="Thou came here for a rest" Tags="|riddle|" AnswerCount="3" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12043" PostTypeId="2" ParentId="12022" CreationDate="2015-04-14T09:20:10.653" Score="-2" Body="&lt;p&gt;The logic in the question is nonsensical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statement by Albert: &quot;Then I also know when Cheryl's birthday is&quot; cannot be true based on Bernard saying that he now knows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheryl could have said the 17th to Bernard. Since Bernard has deduced the only possible months are July and August he can now be certain he knows. BUT Albert cannot be certain - Cheryl could have told Bernard either the 15th, 16th, or 17th. So the logic in the question is wrong. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question would work as written if given the following change:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;July 14 15 16&#xA;August 14 15&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then 14th and 15th can be ruled out by Bernard's certainty, leaving only July 16&lt;/p&gt;&#xA;" OwnerUserId="11356" LastEditorUserId="11357" LastEditDate="2015-04-14T09:54:48.110" LastActivityDate="2015-04-14T09:54:48.110" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12044" PostTypeId="2" ParentId="12036" CreationDate="2015-04-14T09:38:01.123" Score="3" Body="&lt;p&gt;Wilbert has a trivial strategy for the case of j=155.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wilbert also has a strategy when j&amp;lt;78.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strategy is;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Designate a single person as a candidate.&lt;/li&gt;&#xA;&lt;li&gt;Ask villagers if the candidate is a truth teller.&lt;/li&gt;&#xA;&lt;li&gt;If the first j+1 persons indicates that he is a joker at least one truth teller have stated that he is a joker and Wilbert may kill him and continue on the next day.&lt;/li&gt;&#xA;&lt;li&gt;If there is disagreement among the first j+1 persons continue until j+1 persons who agree have been found.  Those who disagree are all jokers, kill one of them and continue on next day. (*)&lt;/li&gt;&#xA;&lt;li&gt;The tricky part is if the first j+1 all agree that the person is a truth teller. Designate the persons that have answered at this point as the first group and note that we know the candidate is a truth teller.&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Try to use him as a rat on the remaining people as described in the previous question.&lt;/li&gt;&#xA;&lt;li&gt;If this finds a joker kill him.  We now have a known truth teller and can mop up as described in the previous question.&lt;/li&gt;&#xA;&lt;li&gt;If this does not find a joker, we have person left that have not been asked a question, we know him to be a truth teller, and we know that there is exactly one truth teller in the first group.  Pick a person in the first group, ask the truth teller if this person is a truth teller.  If he is not we kill him, if he is we know that the remainder of the first group is jokers and we kill one of them.  We now know all the jokers and can kill them to finish the task.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(*) If j=77 you can get in the situation that 77 persons indicate that the candidate is a truth teller and 77 indicates that he is a joker.  In this case all 77 jokers must have answered with a lie, hence the candidate is a truth teller and the jokers have outed themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the exeption of j=155 Wilbert has no strategy if j&gt;77.  The strategy above is not guaranteed to suceed if j&gt;77. This because the point marked with (*) could fail since you would not be guaranteed to find j+1 persons who agree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that no strategy can exist for j&gt;77 just divide the jokers into two groups $j_1$ and $j_2$ such that the number of jokers in $j_1$ is equal to the number of truth tellers.  If all the jokers in $j_1$ tells the truth about the jokers in $j_2$ and lies about all other inhabitants, and the jokers in $j_2$ answers randomly.  This makes the situation symetrical between the thruth tellers and $j_1$ so no way to distinguish between the two groups exist.&lt;/p&gt;&#xA;" OwnerUserId="1654" LastEditorUserId="1654" LastEditDate="2015-05-29T07:16:39.537" LastActivityDate="2015-05-29T07:16:39.537" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="12045" PostTypeId="2" ParentId="12042" CreationDate="2015-04-14T10:00:09.903" Score="4" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tree&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thou came here for a rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With my body, I made a shelter &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bed and other furnitures are made of wood&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thou feel so cool, when I flutter &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thou touch me with his finger &lt;/p&gt;&#xA;&#xA;&lt;p&gt;didn't know whether it hurts me &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I get some sketchs but feel real&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In cold winds, you burn a log with match-stick and don't care if it hurts the wood to get warmth. Fire -&gt; real&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And Thou flies away  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All my wings are caused to sway &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When a glider or an plane passes, it causes the leaves to sway &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That one last touch hurts me enough &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My one finger goes down to lay.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;one last touch&quot; with an axe causes the tree's &quot;one finger&quot; (trunk) to lie down. Thanks &lt;a href=&quot;https://puzzling.stackexchange.com/questions/12042/answer-who-what-am-i/12045?noredirect=1#comment36977_12045&quot;&gt;randal'thor&lt;/a&gt; for this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4637" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-04-14T10:06:31.567" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12046" PostTypeId="2" ParentId="12035" CreationDate="2015-04-14T10:06:20.097" Score="7" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; all the logicians leave on the 15th day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Proof&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Let's generalise the problem and say there are $n$ logicians on the island, all with blue eyes. Call them $L_1,L_2,\dots,L_n$. So our specific problem is where $n=100$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n=1$, then the single logician knows from the oracle's statement that the number of blue-eyed logicians on the island isn't $0$, so it must be $1$; he leaves on the first day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n=2$, then $L_1$ knows that if his own eyes are brown, then $L_2$ would know the number of blue-eyed logicians is $0$ or $1$, and the oracle's statement would tell $L_2$ his own eye colour on the first day. When $L_2$ doesn't leave on the first day, $L_1$ knows his own eyes must be blue. $L_2$ can argue similarly, so they both leave on the second day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n=3$, the $L_3$ knows that if his own eyes are brown, then $L_1$ and $L_2$ would argue exactly as above (each of them knows there are either $1$ or $2$ blue-eyed logicians, as in the previous scenario) and leave on the second day. They don't, so $L_3$ leaves on the third day. By symmetry, they all do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same argument goes on until $n=17$. Note that this case is impossible since the oracle always tells the truth.  However this makes the case $n=18$ interesting.  In this case (or if $n-1$ is any other multiple of $17$), they all leave on the first day because they knew beforehand that there are either $n$ or $n-1$ blue-eyed logicians on the island, and the oracle's statement tells them immediately which is correct. So let's skip ahead to $n=86$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n=86$, then each logician knows beforehand that there are either $85$ or $86$ blue-eyed logicians on the island, so the oracle's answer tells them which immediately, and they all leave on the first day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n=87$, then $L_{87}$ knows that if his own eyes are brown, then $L_1,\dots,L_{86}$ argue exactly as in the $n=86$ case and leave on the first day. They don't, so $L_87$ leaves on the second day. By symmetry, everyone does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, everyone leaves on the $(n-17k)$th day if $17k&amp;lt;n&amp;lt;17k+17$ (and on the $1$st day if $n=17k$). When $n=100$, we have $k=5$ and so everyone leaves on the $15$th day.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="1654" LastEditDate="2015-04-14T10:44:19.547" LastActivityDate="2015-04-14T10:44:19.547" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="12050" PostTypeId="2" ParentId="12022" CreationDate="2015-04-14T10:55:43.720" Score="3" Body="&lt;p&gt;The simplest way is by putting the data into a table and crossing out months and days step by step&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;May -- 15 16 -- -- 19&#xA;&#xA;Jun -- -- -- 17 18 --&#xA;&#xA;Jul 14 -- 16 -- -- --&#xA;&#xA;Aug 14 15 -- 17 -- --&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edited&lt;/strong&gt; first paragraph for clarification due to comment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Albert can only be 100% sure that Bernard does not know the date if and only if Albert was told either July or August. If he was given one of the other two months, there would have been a chance that Bernard knew the correct date (i.e. if he was told 18 or 19, since these two values are unambiguous as easily seen in the table) and Albert could not have made his statement. But, since Albert made his statement, Bernard can deduce (in the same way we just did), that Albert was told either July or August.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both Albert and Bernard now know that it must be either July or August. Bernard now tells that he knows and, hence, the 14th can be excluded, since it is found in both months. If it was the 14th, he could not know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That leaves three possibilities: Jul 16, Aug 15 and Aug 17. Since Albert states that now he knows, too, it only can be Jul 16, since it is the only unambigous combination of month/day left.&lt;/p&gt;&#xA;" OwnerUserId="10442" LastEditorUserId="10442" LastEditDate="2015-04-17T08:02:36.470" LastActivityDate="2015-04-17T08:02:36.470" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12051" PostTypeId="2" ParentId="12038" CreationDate="2015-04-14T11:02:48.987" Score="6" Body="&lt;p&gt;Question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What film grossed most in 2000?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answers: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A: What Women Want&lt;br&gt;&#xA; B: Cast away&lt;br&gt;&#xA; C: &lt;strong&gt;How the Grinch stole christmas&lt;/strong&gt;&lt;br&gt;&#xA; D: Gladiator  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasons:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A is a romantic comedy, D won the Oscar prize in 2001, C and B are the 1st and 2nd 2001 worldwide grosses, and C beat B domestically.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="11475" LastEditDate="2015-09-29T12:23:03.940" LastActivityDate="2015-09-29T12:23:03.940" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12052" PostTypeId="2" ParentId="12022" CreationDate="2015-04-14T11:03:39.050" Score="11" Body="&lt;p&gt;I will try to help you understand the answer. It is important to note that Albert and Bernard both know the answer before we do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With Albert's first statement, it is clear to us that Albert was told either July or August. Had he been told May or June, he would not be able to state definitively that he knew Bernard didn't know. (May 19 and June 18 both could be uniquely identified immediately by Bernard without Albert's help, so in order for Albert to know that Bernard doesn't know, the month he was told must not be May or June.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because Bernard has been able to identify Cheryl's birthday after Albert's statement, he must not have been told 14. Because Albert's statement revealed July and August, had Bernard been told 14, he would still be unclear on the date. (If he said he didn't know, then Albert would know the birthdate, but Bernard would never be able to deduce it.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We still don't know whether it is July 16, August 15, or August 17.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, since Albert (who was only told the month) is able to state that he also knows the birthday, he must not have been told August. Had he been told August, he would be unable to decide if Bernard was told the 15th and 17th.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is only after the third statement that we (as outsiders) can deduce the birthday. Bernard knew after the first statement and Albert knew after the second.&lt;/p&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="11381" LastEditDate="2015-04-14T13:55:30.437" LastActivityDate="2015-04-14T13:55:30.437" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12053" PostTypeId="2" ParentId="11998" CreationDate="2015-04-14T11:22:10.040" Score="0" Body="&lt;p&gt;I know that technically I do not have a shorter solution, but I applied the modulo operator and found an answer!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 123456789 % 123456&lt;strong&gt;6&lt;/strong&gt;89 = 100&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and so on. The program I used:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#include &amp;lt;iostream&amp;gt;&#xA;&#xA;int&#xA;main()&#xA;{&#xA;  for (int i = 100; i &amp;lt; 999999999; i++)&#xA;    {&#xA;      if (123456789 % i == 100)&#xA;        {&#xA;          std::cout &amp;lt;&amp;lt; i &amp;lt;&amp;lt; std::endl;&#xA;        }&#xA;    }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="10139" LastEditorUserId="10139" LastEditDate="2018-04-20T09:23:03.723" LastActivityDate="2018-04-20T09:23:03.723" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12056" PostTypeId="2" ParentId="12042" CreationDate="2015-04-14T11:54:39.107" Score="3" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Caterpillar/ Butterfly (A Butterfly Life cycle)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thou came here for a rest--&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; caterpillar comes to rest and builds a cocoon around it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With my body, I made a shelter --- &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;pupa&quot; is the transformation stage within which the caterpillar tissues are broken down and the adult insect's structures are formed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thou feel so cool, when I flutter --&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Feels cool when it tries and fly&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I get some sketchs but feel real--&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Beautiful colors on the wings of the butterfly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My one finger goes down to lay---&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The butterfly lays eggs &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11056" LastEditorUserId="11056" LastEditDate="2015-04-14T14:24:35.383" LastActivityDate="2015-04-14T14:24:35.383" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12057" PostTypeId="2" ParentId="11943" CreationDate="2015-04-14T12:50:25.560" Score="2" Body="&lt;p&gt;Likely wrong, but I wanted to add a different angle to the question, i.e. move a bit &lt;em&gt;further&lt;/em&gt; outside the box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could be the &lt;em&gt;touch-field&lt;/em&gt; of a pocket-calculator or other technichal device where touching the &quot;Pi&quot; key twice gives you 2 ?  (Haven't found an according calculator though, yet.) &lt;/p&gt;&#xA;" OwnerUserId="5840" LastActivityDate="2015-04-14T12:50:25.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12058" PostTypeId="1" AcceptedAnswerId="12059" CreationDate="2015-04-14T12:51:03.613" Score="11" ViewCount="422" Body="&lt;p&gt;One for hunt and silver hue.&lt;br&gt;&#xA;One for war and battles due.&lt;br&gt;&#xA;One for trade and souls askew.&lt;br&gt;&#xA;One for sky and thunder too.&lt;br&gt;&#xA;One for love and beauty true.&lt;br&gt;&#xA;One for peace and fortune through.&lt;br&gt;&#xA;One for light and warmth anew.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All quite clever have been bound&lt;br&gt;&#xA;By this lever so renowned&lt;br&gt;&#xA;To forever come around.&lt;/p&gt;&#xA;" OwnerUserId="9910" LastEditorUserId="9951" LastEditDate="2015-04-14T19:49:51.260" LastActivityDate="2015-04-14T19:49:51.260" Title="To forever come around" Tags="|riddle|word|rhyme|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12059" PostTypeId="2" ParentId="12058" CreationDate="2015-04-14T13:22:45.327" Score="11" Body="&lt;p&gt;I hope everyone else noticed the number. This refers to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the days of the week.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One for hunt and silver hue.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Refers to the moon, for Monday.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One for war and battles due.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mars is the god of War, as well as the origin of Tuesday.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One for trade and souls askew.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wodan was known to guide souls after death. Hence, Wednesday.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One for sky and thunder too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Thursday is Thor's Day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One for love and beauty true.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Freya was the Norse goddess of love and beauty. Friday.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One for peace and fortune through.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Saturn was the herald of the age of prosperity. Saturday.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One for light and warmth anew.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sun. Sunday.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All quite clever have been bound&lt;br&gt;&#xA;By this lever so renowned&lt;br&gt;&#xA;To forever come around.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure what this lever is, but they are forever bound to the calendar.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2444" LastActivityDate="2015-04-14T13:22:45.327" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12060" PostTypeId="2" ParentId="12006" CreationDate="2015-04-14T13:47:03.980" Score="3" Body="&lt;p&gt;All this theoretical stuff is nice, but why would we do that when we can check actual data and determine whether the statement is true or false?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All stats come from &lt;a href=&quot;http://www.baseball-reference.com/players/m/mendoma01.shtml&quot; rel=&quot;nofollow&quot;&gt;www.baseball-reference.com&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at Mario Mendoza's career numbers shows that his first season, he hit .221. This shows that at one point in his career, his batting average was above .200. The next 3 seasons, he hit below .200 each season. However, end of season totals keep his career batting average above .200. Therefore, it is necessary to look at his batting average within the seasons to determine whether the statement is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After his 3rd season (1976), he had 305 at bats and 62 hits, for a career batting average of .203. In 1977, after 10 at bats, he had 1 hit. This gives him 63 hits in 315 at bats, for a career batting average of .200. This proves the second part, that at one point he was batting exactly .200.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Un)fortunately, his next at bat was a hit, keeping him over .200. Continuing into the season, we see that over the next 10 at bats (including that hit), he had 2 hits, giving him 65 hits in 325 at bats, back to exactly .200.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then on July 9th, 1977, he entered the game as a pinch runner in the 8th inning. In the 10th inning, he grounded out. This made him 65 for 326 in his career, for a batting average of .199. This proves that after he was hitting .200, he was hitting below .200.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the bottom of the 12th inning, Mendoza hit a walk-off single to win the game for the Pirates (9-8 over the Phillies). This would raise his batting average to .202. He then managed to keep his career batting average over .200 for the rest of the season.&lt;/p&gt;&#xA;" OwnerUserId="375" LastEditorUserId="375" LastEditDate="2015-04-15T14:12:22.223" LastActivityDate="2015-04-15T14:12:22.223" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12061" PostTypeId="1" AcceptedAnswerId="12062" CreationDate="2015-04-14T13:57:52.240" Score="1" ViewCount="965" Body="&lt;p&gt;A five letter English word becomes shorter when 2 more letters are added to it.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What is the original word?&lt;/li&gt;&#xA;&lt;li&gt;What are the 2 letters?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;NOTE: I heard this puzzle from friends and don't claim my creativity&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="4637" LastEditorUserId="4637" LastEditDate="2015-04-15T05:53:01.420" LastActivityDate="2021-02-10T03:21:48.977" Title="Make it shorter by adding" Tags="|word|wordplay|" AnswerCount="5" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12062" PostTypeId="2" ParentId="12061" CreationDate="2015-04-14T14:00:03.277" Score="7" Body="&lt;p&gt;The original five-letter English word is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; short&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The two letters that are added to it are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; er&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Edit&lt;/strong&gt;: these two also work:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Shore + tr&lt;br&gt;&#xA; Sorer + ht&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-04-14T21:31:31.203" LastActivityDate="2015-04-14T21:31:31.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12063" PostTypeId="2" ParentId="12061" CreationDate="2015-04-14T14:00:19.683" Score="4" Body="&lt;p&gt;I think it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; short, and then add &quot;er&quot; to make it shorter!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastActivityDate="2015-04-14T14:00:19.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12064" PostTypeId="2" ParentId="12061" CreationDate="2015-04-14T14:06:06.133" Score="1" Body="&lt;p&gt;Answer is in your question and comment only &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Small and Short&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add 2 more letters&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Smaller and Shorter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10820" LastActivityDate="2015-04-14T14:06:06.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12065" PostTypeId="2" ParentId="11943" CreationDate="2015-04-14T14:06:32.510" Score="20" Body="&lt;p&gt;Here's an answer that goes in and out of figuration, becoming more figurative than literal and then more literal than figurative, stimulated by the hint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer we arrrive at is that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the field is ancient Rome, or anywhere else that people have used Roman numerals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The working is as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want $\pi^2$, i.e.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the result of taking $\pi$, the ratio between the circumference and diameter of a circle, and getting it to operate on itself to give us a square. Alternatively, to square something is to make a square out of it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;OK so&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; let's get figurative. How do we make a square out of the usual figure for $\pi$? Easy. Start by making a copy and turning it round:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;${}$&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FevhD.jpg&quot; alt=&quot;1&quot;&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; stick the rotated copy onto the bottom:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We get&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/c1eqr.jpg&quot; alt=&quot;2&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We have&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; now made a square. It's got bits sticking out of it (&lt;em&gt;outside of the box&lt;/em&gt;), but we've still made a square by getting $\pi$ to operate on itself. So we've &lt;em&gt;squared&lt;/em&gt; $\pi$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Go back to being literal. What have we got? The Roman numeral for the number 2. So we've squared $\pi$ and got 2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I realise the usage of &lt;em&gt;literal&lt;/em&gt; here is questionable. Another weakness is that the meaning of $\pi$, which the setter stresses, doesn't get a look-in. Nonetheless, the train of thought goes from meaning to figure to figure to meaning, which fits nicely with the hint and works as a way of getting $\pi^2$ to equal 2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7363" LastEditorUserId="7363" LastEditDate="2015-04-14T14:16:19.110" LastActivityDate="2015-04-14T14:16:19.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12067" PostTypeId="1" AcceptedAnswerId="12071" CreationDate="2015-04-14T14:28:06.397" Score="-3" ViewCount="12232" Body="&lt;p&gt;&lt;strong&gt;SPY&lt;/strong&gt; and &lt;strong&gt;SKY&lt;/strong&gt; are examples of words without any true vowel in it.&lt;br&gt;&#xA;(I am referring to &lt;strong&gt;A&lt;/strong&gt;, &lt;strong&gt;E&lt;/strong&gt;, &lt;strong&gt;I&lt;/strong&gt;, &lt;strong&gt;O&lt;/strong&gt;, &lt;strong&gt;U&lt;/strong&gt; as vowels.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What is longest English word that has no vowel in it. (Please post its meaning as well)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Note: No googling allowed, at least give it a try with your knowledge)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please do not downvote anymore... I just tried to create a puzzle.&lt;/p&gt;&#xA;" OwnerUserId="10433" LastEditorUserId="8874" LastEditDate="2015-10-02T14:19:23.003" LastActivityDate="2015-10-02T14:19:23.003" Title="The longest word without A,E,I,O, or U" Tags="|word|" AnswerCount="10" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12069" PostTypeId="2" ParentId="12067" CreationDate="2015-04-14T14:35:59.493" Score="3" Body="&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Crwth&quot; rel=&quot;nofollow&quot;&gt;crwth&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That would work if you're going with the &quot;A, E, I, O, U, and sometimes Y&quot; definition. Though you'd probably call &quot;w&quot; a vowel in this context.&lt;/p&gt;&#xA;" OwnerUserId="9828" LastActivityDate="2015-04-14T14:35:59.493" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12070" PostTypeId="2" ParentId="12067" CreationDate="2015-04-14T14:46:15.077" Score="6" Body="&lt;p&gt;Going just by memory, I would say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; syzygy (when 3 astronomical objects are in a line)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I remember that from a previous puzzle here.&lt;/p&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4450" LastEditDate="2015-04-14T14:50:30.420" LastActivityDate="2015-04-14T14:50:30.420" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12071" PostTypeId="2" ParentId="12067" CreationDate="2015-04-14T14:51:42.880" Score="18" Body="&lt;p&gt;Depending on what you mean by vowels, the max length can be &lt;strong&gt;7&lt;/strong&gt; or &lt;strong&gt;12&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider (w,y) vowels, I have one of 7 letters!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Tsktsks&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which is the 3rd person of this verb:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; [Tsktsk]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Meaning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To make a tsktsk sound of disapproval.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This contains no vowel nor sounds like vowels (y,w)!&lt;br&gt;&#xA;This is present in Wikipedia &lt;a href=&quot;http://en.wiktionary.org/wiki/tsktsk&quot;&gt;here&lt;/a&gt; and dictionary.reference &lt;a href=&quot;http://dictionary.reference.com/browse/tsktsk&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you also accept the sounds y and w, then the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;twyndyllyngs&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which has 12 letters!!! Confirmed &lt;a href=&quot;http://en.wiktionary.org/wiki/twyndyllyng&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-04-14T15:21:41.883" LastActivityDate="2015-04-14T15:21:41.883" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12072" PostTypeId="2" ParentId="12067" CreationDate="2015-04-14T14:51:42.953" Score="1" Body="&lt;p&gt;I can at least tie pacoverflow's answer, with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; crypts, which loosely can be defined as &quot;cellars or underground vaults&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11322" LastActivityDate="2015-04-14T14:51:42.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12073" PostTypeId="2" ParentId="12067" CreationDate="2015-04-14T14:53:19.063" Score="5" Body="&lt;p&gt;The first thing I thought of was&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rhythms&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2015-04-14T14:53:19.063" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12074" PostTypeId="2" ParentId="11983" CreationDate="2015-04-14T14:55:05.440" Score="7" Body="&lt;p&gt;Here's a short and sweet way of looking at it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to satisfy the restrictions given in the question, you need to find &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a number of sodas that is a cube, and that divides by 2 but not by 4.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Any number that does not have 2 as a prime factor will not divide by 2 when cubed. And any number that does have 2 as a prime factor will divide by 8 (and hence 4) when cubed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is no such number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5260" LastActivityDate="2015-04-14T14:55:05.440" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12075" PostTypeId="2" ParentId="12067" CreationDate="2015-04-14T15:28:36.610" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;NYMPHS , GLYPHS&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Most of the words need Y, I guess, as it makes up for a good substitute to vowels.&lt;/p&gt;&#xA;" OwnerUserId="11387" LastEditorUserId="10281" LastEditDate="2015-04-14T15:38:31.767" LastActivityDate="2015-04-14T15:38:31.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12076" PostTypeId="2" ParentId="12042" CreationDate="2015-04-14T15:31:24.740" Score="0" Body="&lt;p&gt;I'm not sure this is right, but maybe on the right track. Are you:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A fan&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thou came here for a rest. &#xA;With my body, I made a shelter &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People like sitting near fans and they normally have a protective cage around them&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thou feel so cool, when I flutter &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When a fan is on, it can cool you down&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thou touch me with his finger &#xA;didn't know whether it hurts me &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people like to risk putting their fingers along a fan and might get hurt&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I get some sketchs but feel real&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I guess some fans can have special designs/sketches&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And Thou fly away&lt;br&gt;&#xA;All my wings are caused to sway &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some fans can blow you away (Like in air tunnels)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That one last touch hurts me enough &#xA;My one finger goes down to lay.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can turn fans off with the push of a button&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastActivityDate="2015-04-14T15:31:24.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12077" PostTypeId="2" ParentId="12067" CreationDate="2015-04-14T15:52:47.653" Score="1" Body="&lt;p&gt;I can think of a 7 letter word&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; nymphly&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10926" LastActivityDate="2015-04-14T15:52:47.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12078" PostTypeId="2" ParentId="11780" CreationDate="2015-04-14T15:58:37.130" Score="2" Body="&lt;h2&gt;At least 64 - if the question was completely different!&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In my haste I totally misread the question and rushed off to find&#xA;  supporting info for what I knew was the minimum number of clues in a&#xA;  sudoku. That is NOT what is being asked, but I'll leave this here to&#xA;  prevent any other fools making the same mistake (and maybe because&#xA;  someone might find it interesting or useful.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Various websites generate sudoku puzzles and 17 clues is the smallest number I have seen from my (admittedly brief) search.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best reference I found on solving sudoku including the details of various strategies employed is &lt;a href=&quot;http://www.sudokuwiki.org/sudoku.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.sudokuwiki.org/sudoku.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It includes a step by step solution to any puzzle you provide as well a generating puzzle that require a range of strategies for their solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least one puzzle exists with 17 clues (64 blank squares) that can be solved using only the most basic strategy.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;....41...&#xA;.6....2..&#xA;.........&#xA;32.6.....&#xA;....5..41&#xA;7........&#xA;...2..3..&#xA;.48......&#xA;5.1......&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Puzzles with a smaller number of clues may exist that can be solved by a brute search algorithm although it is a search method that may present multiple possible solutions given that such puzzles are ambiguous due to their small amount of starting information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 2012 a team from University College Dublin published the results of search that took the best part of a year, announcing that no 16 clue solvable sudoku exists. Their method iterated all possible grids searching for one which was could be solved. This is not what mathematicians like to describe as a proof although in does constitute a definitive answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.technologyreview.com/view/426554/mathematicians-solve-minimum-sudoku-problem/&quot; rel=&quot;nofollow&quot;&gt;http://www.technologyreview.com/view/426554/mathematicians-solve-minimum-sudoku-problem/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe at this time no such proof exists. &lt;/p&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="11297" LastEditDate="2015-04-14T23:42:18.343" LastActivityDate="2015-04-14T23:42:18.343" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12079" PostTypeId="2" ParentId="12067" CreationDate="2015-04-14T16:09:54.137" Score="2" Body="&lt;p&gt;I was always taught that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; nth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;was the only/longest word without vowels (w and y included) but i did not think about leoll2's answer&lt;/p&gt;&#xA;" OwnerUserId="5717" LastActivityDate="2015-04-14T16:09:54.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12080" PostTypeId="2" ParentId="12067" CreationDate="2015-04-14T16:39:27.673" Score="1" Body="&lt;p&gt;The &lt;a href=&quot;http://www.visca.com/regexdict/&quot; rel=&quot;nofollow&quot;&gt;regexp dictionary&lt;/a&gt; says the full list of vowel-less words (by their definition of a 'word') is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;-'s&#xA;  -cy&#xA;  -fy&#xA;  -gyny&#xA;  -phyll&#xA;  -ry&#xA;  -st&#xA;  -ty&#xA;  -yl&#xA;  24-7&#xA;  3-D&#xA;  b&#xA;  bb&#xA;  by&#xA;  by-&#xA;  c&#xA;  cd&#xA;  chrys-&#xA;  cry&#xA;  crypt&#xA;  crypt-&#xA;  cycl-&#xA;  Cymry&#xA;  cyst&#xA;  cyst-&#xA;  cyt-&#xA;  d&#xA;  dj&#xA;  dry&#xA;  dys-&#xA;  f&#xA;  fly&#xA;  flyby&#xA;  fry&#xA;  ft-c&#xA;  ft-lb&#xA;  g&#xA;  glyc-&#xA;  glyph&#xA;  gym&#xA;  gyn-&#xA;  gyp&#xA;  Gypsy&#xA;  h&#xA;  hydr-&#xA;  hymn&#xA;  hyp-&#xA;  hypn-&#xA;  j&#xA;  k &#xA;  l&#xA;  lymph&#xA;  lymph-&#xA;  lynch&#xA;  lynx&#xA;  lys-&#xA;  m&#xA;  my&#xA;  my-&#xA;  myc-&#xA;  myrrh&#xA;  myth&#xA;  myx-&#xA;  n&#xA;  nth&#xA;  nymph&#xA;  p&#xA;  pfft&#xA;  phyll-&#xA;  phys-&#xA;  ply&#xA;  pry&#xA;  psst&#xA;  psych&#xA;  psych-&#xA;  py-&#xA;  Pygmy&#xA;  pyr-&#xA;  pyx&#xA;  q&#xA;  r&#xA;  rhythm&#xA;  s&#xA;  scry&#xA;  shy&#xA;  sky&#xA;  sly&#xA;  spry&#xA;  spy&#xA;  sty&#xA;  styl-&#xA;  Styx&#xA;  sylph&#xA;  sym-&#xA;  syn-&#xA;  sync&#xA;  synch&#xA;  synth&#xA;  syzygy&#xA;  t&#xA;  thy&#xA;  try&#xA;  tryst&#xA;  tsk&#xA;  tyyn&#xA;  v&#xA;  w&#xA;  why&#xA;  wry&#xA;  wynn&#xA;  x&#xA;  xyl-&#xA;  y&#xA;  z&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This doesn't include plurals - if you pluralise the nouns then rhythms  wins out over syzygy.&lt;/p&gt;&#xA;" OwnerUserId="4050" LastActivityDate="2015-04-14T16:39:27.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12081" PostTypeId="2" ParentId="11942" CreationDate="2015-04-14T17:03:20.347" Score="2" Body="&lt;h2&gt;10 tricks&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I suspect that one or two more is possible, but that may require a ridiculously complicated ruleset. As if the following wasn't bad enough already.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;First 3 tricks:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Cost 4 cards each.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The cards are enumerated 1 through 52, and they are divided into 4 virtual groups, numbered 1 to 4, of 13 cards each. Each group is considered circular, so card 1 in a group comes right after card 13 in the same group. In any pair of cards from one group there is a distinct first card, defined as the card that is the fewest steps from the other card in positive direction, so for instance between card 3 and 10, 10 is the first card, as there is 6 steps up from 10 to 3, but 7 steps from 3 to 10.&lt;/li&gt;&#xA;&lt;li&gt;The first player finds one of the following combinations in hand:&#xA;&lt;ol&gt;&#xA;&lt;li&gt;4 cards from the same group.&lt;/li&gt;&#xA;&lt;li&gt;2 cards from the same group, and 3 other cards from different groups (though one of them may be from the same group as the 2 cards).&lt;/li&gt;&#xA;&lt;li&gt;2 cards from each of 2 different groups.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Depending on the combination, the cards are played like so:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A pair from is chosen so that the distance is no greater than 3, the first card is played, the other card from the pair is played immediately afterwards if the distance is 1, one card is played in between if the distance is 2, and two cards are played in between if the distance is 3. The other cards from the group are used as filler for this purpose.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The first of the cards from the 2 card group is played, then 2 of the other 3 cards are used to mark the distance as a number from 1 to 6, depending on their group, like so:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;1-2 or 3-4&lt;/li&gt;&#xA;&lt;li&gt;2-1 or 4-3&lt;/li&gt;&#xA;&lt;li&gt;1-3 or 2-4&lt;/li&gt;&#xA;&lt;li&gt;3-1 or 4-2&lt;/li&gt;&#xA;&lt;li&gt;1-4 or 2-3&lt;/li&gt;&#xA;&lt;li&gt;4-1 or 3-2&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;And finally the other card from the 2 card group is played.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;If one of the pairs is 1 or 2 apart, play that like case 1. Otherwise if one pair is an odd number apart, and the other pair is an even number apart, play the first card from the pair from the lower numbered group. Other-otherwise play the first card from the higher numbered group. Then play both cards from the other group, with the first first if the distance between the cards of the first played group is 3 or 4. Then play the last card of the first played group.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The second player will always get at least one guess right by always guessing on the cards of the same group as the first card played, 1 and 2 higher respectively for the second and third card played, if possible, and if that fails deduce the value of the fourth card from the above rules.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Next 3 tricks:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Cost 3 cards each.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The remaining 28 cards are implicitly enumerated 1 through 28, this requires that both players remember all the played cards. They are similarly split into 4 virtual groups of 7 cards each.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The first player finds one of the following combinations in hand:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;3 cards from the same group.&lt;/li&gt;&#xA;&lt;li&gt;2 cards from the same group, one card from the group below, and one card from the group either 1 or 2 above.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;Depending on the combination, the cards are played like so:&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Just like case 1 in the first 4 tricks, only the difference can be no larger than 2.&lt;/li&gt;&#xA;&lt;li&gt;Play the first card of the pair, then the second if they are 1 apart, inject the card from the group below if they are 2 apart, or the card from the group above if they are 3 apart.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h3&gt;Last 4 tricks:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Using the last 5 cards.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The remaining cards are enumerated 1 through 10, then the first player plays card 1 if he has it, and this step is repeated with the cards enumerated 1 through 8, 6, 4 and finally 2.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When the first player doesn't have card 1, the other cards form a loop of 9, 7, 5, 3 or 1 card. He chooses the one starting point in this loop where the following is possible.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;He plays the first of a series of consecutive cards, the second player may not guess this card.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;He plays the last card of the same series. Since all the cards in between are in the first players hand, they cannot be drawn from the pile, so the second player can always identify this card.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If there is more than one card between the series and the following series, each extra card in the gap is marked by playing one of the cards that the second player already know that he has, then the last card of the following series is played, and the process is repeated for each series and gap.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h3&gt;Example last 4 tricks:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;After enumeration the cards in hand are: 2, 5, 6, 9, 10. The pile has 1, 3, 8, 4, 7 in that order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Player 1 has to start with 9 in this case, the pile plays 1. Player 2 guess 1, so this trick is lost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Player 1 mark the end of his series from 9 to 2 by playing 2, the pile plays 3. For player 1 to have played 3 in this scenario he must have had 10, 2 and 3 on hand, but that is impossible since either 2 or 3 must have come from the pile, so player 2 can conclude that he has to guess 2. He also now know that player 1 has 10 in hand as that is between 9 and 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Player 1 plays 10 to mark the double gap between 2 and his next card 5. The pile plays 8. Player 2 of course guess 10, and since a known card was played, rather than a card from the next series, player 2 now know that there is an extra gap after 2, so he will be looking for 5 or above in the next trick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Player 1 plays 6, to mark the end of the 5 to 6 series, the pile plays 4. Player 2 knows that player 1 hasn't got 4, since he marked that by playing the known 10, so he looks for 5 or above, the closest card for that criteria is 6, so that is his guess. Player 2 now know that player 1 has 5 in hand, as that is must be that start of the series when the gap between the first and the second series is 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Player 1 plays 5, the pile plays 7. Player 2 know that Player 1 has 5, so that is his guess.&lt;/p&gt;&#xA;" OwnerUserId="2186" LastEditorUserId="2186" LastEditDate="2015-04-21T15:22:03.400" LastActivityDate="2015-04-21T15:22:03.400" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="12082" PostTypeId="2" ParentId="12061" CreationDate="2015-04-14T17:04:01.223" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Shore&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;becomes &quot;shorter&quot; when you add the letters &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;t&lt;/strong&gt; and &lt;strong&gt;r&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11150" LastEditorUserId="10281" LastEditDate="2015-04-14T17:46:29.687" LastActivityDate="2015-04-14T17:46:29.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12083" PostTypeId="2" ParentId="11780" CreationDate="2015-04-14T17:09:01.517" Score="4" Body="&lt;p&gt;Lemma 1: &lt;a href=&quot;http://pastebin.com/X0uqZMRm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://pastebin.com/X0uqZMRm&lt;/a&gt;&lt;br&gt;&#xA;Lemma 2: &lt;a href=&quot;http://pastebin.com/rcHX6JLk&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://pastebin.com/rcHX6JLk&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These two lemmas prove that if we try to build on Kevin's answer, we cannot do better than Spencer's answer. This is because no matter what way we cover two cells in the same empty block, we will create one of the shapes prohibited by the lemmas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we want to beat 33, we need a new strategy. &lt;a href=&quot;https://i.stack.imgur.com/IjhAi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;It's possible&lt;/a&gt; to block 18 cells of a chute without creating any bad rectangles or half-bad trapezoids, so maybe that could be useful. It also might help to find other unsolvable shapes.&lt;/p&gt;&#xA;" OwnerUserId="11100" LastEditorUserId="11100" LastEditDate="2015-04-14T17:27:06.097" LastActivityDate="2015-04-14T17:27:06.097" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12084" PostTypeId="2" ParentId="11867" CreationDate="2015-04-14T17:41:52.260" Score="1" Body="&lt;p&gt;As rand al'thor mentioned, the answer is &lt;strong&gt;gravity&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Your heart pumps, asking why&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your heart has to work against gravity to pump blood to your head.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;I am here to make you cry.&lt;br&gt;&#xA;  It's hard to say goodbye,&lt;br&gt;&#xA;  But leave me so you could fly.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without gravity pulling you down to the earth, you could fly.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Your mind said love, but your body said hate  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how this line applies.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;happiness is together and you gain weight.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Gravity causes things with mass to have weight as well.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Like when we met on our first date.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how this line applies.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Nothing but fruit you have ate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sir Isaac Newton discovered gravity while watching an apple fall.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sounds like a story, romantic,&lt;br&gt;&#xA;  nothing around me but tragic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The gravity of a situation, for example.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Here I will show you a magic,&#xA;  I'll vanish and you will panic! &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If gravity were to vanish, nothing could hold itself together anymore, and you would die.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I might be the reason you fall in love,&lt;br&gt;&#xA;  But love is nothing I planned to speak of.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Gravity is the reason you ever fall at all. Love is irrelevant there, which is what the second line is for.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I am massive and I can be related,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Gravity affects masses, and is a component of general relativity.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Two great minds in history have debated.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Copernicus and Galileo, I assume?&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-04-14T17:41:52.260" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12085" PostTypeId="1" AcceptedAnswerId="12114" CreationDate="2015-04-14T17:52:21.113" Score="10" ViewCount="2953" Body="&lt;p&gt;Pursued by many, hated by few,&lt;br&gt;&#xA;It's amazing how much I can do.&lt;br&gt;&#xA;Creation, enlargement, or despise,&lt;br&gt;&#xA;Tell me now where you keep your eyes.&lt;br&gt;&#xA;Life or death, on different fronts,&lt;br&gt;&#xA;I am the goal of many hunts.&lt;br&gt;&#xA;All may have, and all seem to want,&lt;br&gt;&#xA;But in different ways, now don't flaunt.&lt;br&gt;&#xA;I may be the beginning, or maybe the end,&lt;br&gt;&#xA;So riddle out what I am my friend.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some of my previous riddles were interpreted on a &quot;naughtier&quot; side, so I decided to go with that idea and make this one more on that side.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="5947" LastEditDate="2015-04-14T21:12:26.383" LastActivityDate="2020-07-17T19:52:10.743" Title="So Popular, What am I?" Tags="|riddle|word|rhyme|" AnswerCount="10" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12086" PostTypeId="2" ParentId="12085" CreationDate="2015-04-14T17:56:59.960" Score="3" Body="&lt;p&gt;I think it can be  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wealth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pursued by many, hated by few,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many want it, few don't.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's amazing how much I can do.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Richness usually means power.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Creation, enlargement, or despise,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some of its effects...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tell me now where you keep your eyes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People are attracted by opulence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Life or death, on different fronts,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There's physical richness and spiritual wealth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the goal of many hunts.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Prizes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All may have, and all seem to want,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There's money for everyone, if well distributed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But in different ways, now don't flaunt.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Flaunting isn't a good way to show your wealth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I may be the beginning, or maybe the end,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can born rich, or die rich.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-04-14T17:56:59.960" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12087" PostTypeId="1" AcceptedAnswerId="12088" CreationDate="2015-04-14T17:57:40.657" Score="18" ViewCount="1103" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A famous composer named Dvorak&lt;br&gt;&#xA;  Decided to open his door rack.&lt;br&gt;&#xA;  But then Dvorak slipped,&lt;br&gt;&#xA;  And suddenly tripped,&lt;br&gt;&#xA;  Abe u.nn d.aeucpoy cbyr yd. unrrp pajtv&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The last line of the limerick is all jumbled. What happened to poor Dvorak?&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-04-15T17:01:27.347" LastActivityDate="2015-04-15T17:01:27.347" Title="Poor Dvorak (a limerick)" Tags="|cipher|limerick|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12088" PostTypeId="2" ParentId="12087" CreationDate="2015-04-14T18:00:15.767" Score="17" Body="&lt;p&gt;The last line is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; And fell headfirst into the floor rack.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The line was encrypted by typing in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dvorak_Simplified_Keyboard&quot;&gt;Dvorak keyboard layout&lt;/a&gt; as if the keys were still in the standard QWERTY layout. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use &lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/thumb/2/25/KB_United_States_Dvorak.svg/2000px-KB_United_States_Dvorak.svg.png&quot;&gt;this keyboard&lt;/a&gt; for reference.&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorDisplayName="user88" LastEditDate="2015-04-14T18:04:08.777" LastActivityDate="2015-04-14T18:04:08.777" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12089" PostTypeId="2" ParentId="12085" CreationDate="2015-04-14T18:10:53.583" Score="1" Body="&lt;p&gt;As second attempt, I try:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clothes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pursued by many, hated by few,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There people who spend their lives doing shopping, and there are nudists.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's amazing how much I can do.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You protect from cold, sun, wind, physical damages...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Creation, enlargement, or despise,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They can be designed. They can be even enlarged, if needed. Finally, an ugly shirt can be despised! (The NASA guy, eheh!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tell me now where you keep your eyes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You keep them on clothes, theoretically you should...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Life or death, on different fronts,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Even deads have clothes!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the goal of many hunts.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Shopping! Discounts!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All may have, and all seem to want,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everyone can afford clothes, even the very poor&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But in different ways, now don't flaunt.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's not a good habit to make a show of your clothes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I may be the beginning, or maybe the end,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You dress clothes from birth until death.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="9951" LastEditDate="2015-04-14T18:19:29.793" LastActivityDate="2015-04-14T18:19:29.793" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12090" PostTypeId="2" ParentId="12085" CreationDate="2015-04-14T18:11:24.703" Score="6" Body="&lt;p&gt;Pursued by many, hated by few,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Beauty&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's amazing how much I can do.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Beauty can influence people and accomplish much...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Creation, enlargement, or despise,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Plastic surgery&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tell me now where you keep your eyes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are looking at my new assets!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Life or death, on different fronts,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Beauty can fade and die&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the goal of many hunts.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many people continue to seek it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All may have, and all seem to want,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most people appreciate beauty. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But in different ways, now don't flaunt.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; But it is also in the eye of the beholder.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I may be the beginning, or maybe the end,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Beauty can come to you at different points in life (physical and otherwise)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So riddle out what I am my friend.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Beauty.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10294" LastActivityDate="2015-04-14T18:11:24.703" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12091" PostTypeId="2" ParentId="12067" CreationDate="2015-04-14T18:14:40.427" Score="1" Body="&lt;p&gt;There is even a rock band without a vowel:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Lynyrd Skynyrd&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It does not appear in the Oxford Online Dictionary, but it is recognised by many people. The origin of the name is (From &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_band_name_etymologies&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After Leonard Skinner, a gym teacher at Robert E. Lee High School in Jacksonville, Florida who was notorious for strictly enforcing the school's policy against boys having long hair.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3993" LastActivityDate="2015-04-14T18:14:40.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12092" PostTypeId="2" ParentId="12085" CreationDate="2015-04-14T18:15:26.667" Score="3" Body="&lt;p&gt;To riddle you back I'll give it a lob, as for my guess I think you're:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A job&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pursued by many, hated by few,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many people want a job to make money, some hate their jobs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's amazing how much I can do.&#xA;Creation, enlargement, or despise,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jobs help in progressing society (or possibly making it worse) and are created, grow and shrink&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tell me now where you keep your eyes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Possibly where your occupation is&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Life or death, on different fronts,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People can often give up their 'life' for a job&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the goal of many hunts.&#xA;All may have, and all seem to want,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People hunt for jobs and want one to earn money/a living&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But in different ways, now don't flaunt.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Don't flaunt your job to others&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I may be the beginning, or maybe the end,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; maybe the beginning of a career, end of all your free time :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So riddle out what I am my friend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S I notice I didn't riddle an answer, but at least it rhymed! &lt;/p&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-04-14T18:23:32.763" LastActivityDate="2015-04-14T18:23:32.763" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12093" PostTypeId="5" CreationDate="2015-04-14T18:21:40.867" Score="0" Body="&lt;p&gt;A limerick is a form of poetry, typically English, with the purpose of entertaining the reader.&lt;br&gt;&#xA;It's made of a stanza of five lines, whose rhyme scheme is AABBA.&lt;br&gt;&#xA;Usually, the first line of the limerick defines the protagonist, while the remaining lines describe the anecdote.&lt;br&gt;&#xA;Sometimes, the story is nonsense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A dozen, a gross and a &lt;strong&gt;score&lt;/strong&gt;,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Plus three times the square root of &lt;strong&gt;four&lt;/strong&gt;,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Divided by &lt;em&gt;seven&lt;/em&gt;,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Plus five times &lt;em&gt;eleven&lt;/em&gt;,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;equals nine squared and not a bit &lt;strong&gt;more&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the example above the 'A' beats are bolded and the 'B' beats are in italics.&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="52483" LastEditDate="2018-11-16T00:18:09.690" LastActivityDate="2018-11-16T00:18:09.690" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="12094" PostTypeId="4" CreationDate="2015-04-14T18:21:40.867" Score="0" Body="A limerick is a 5-line poem with a strict AABBA rhyme scheme, often with five-line anapestic meter, sometimes obscene with humorous intent." OwnerUserId="10281" LastEditorUserId="5373" LastEditDate="2015-04-27T08:10:52.070" LastActivityDate="2015-04-27T08:10:52.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12095" PostTypeId="2" ParentId="11998" CreationDate="2015-04-14T18:21:51.957" Score="29" Body="&lt;p&gt;Best I can do so far is tie the smallest (3 operators):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1234%567%8^9 = 100&lt;br&gt;&#xA; which &lt;a href=&quot;https://www.wolframalpha.com/input/?i=1234%25567%258%5E9&quot;&gt;Wolfram Alpha&lt;/a&gt; interprets as:&lt;br&gt;&#xA; $(1234\bmod567)\bmod8^9 = 100$&lt;br&gt;&#xA; $(100)\bmod8^9 = 100$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For those who are unfamiliar, % is the modulo operator. Throw the equation in wolfram alpha.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&#xA;Started on finding this solution before the operator limit was imposed. Not valid under the rules of the current question.&lt;/p&gt;&#xA;" OwnerUserId="11395" LastEditorUserId="11395" LastEditDate="2015-04-14T18:37:55.087" LastActivityDate="2015-04-14T18:37:55.087" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="12096" PostTypeId="2" ParentId="12085" CreationDate="2015-04-14T18:28:41.893" Score="1" Body="&lt;p&gt;Pursued by many, hated by few,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Life&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's amazing how much I can do.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Life is everywhere, diverse (extremophiles)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Creation, enlargement, or despise,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Life creates, grows..&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tell me now where you keep your eyes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eyes of life can be in many places, different forms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Life or death, on different fronts,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hmm, maybe it's birth?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the goal of many hunts.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most people want to keep living, live longer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All may have, and all seem to want,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everyone is alive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But in different ways, now don't flaunt.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People have different lives.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I may be the beginning, or maybe the end,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Life is a circle, circle of life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So riddle out what I am my friend.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10294" LastActivityDate="2015-04-14T18:28:41.893" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12097" PostTypeId="2" ParentId="11968" CreationDate="2015-04-14T18:39:26.920" Score="4" Body="&lt;p&gt;My solution with no knowledge of the Chinese remainder theorem and no sophisticated thinking!&#xA;It must end in 5 or 6 (0² and 1² are of course equal to 0 and 1 but the 'tens' digit always fails)&#xA;Choose 5 or 6 and and let the first, second and third digits be a,b,c. I'll use 6.&#xA;Treat it as if abc6 were a proper decimal number and square it by long multipication, spacing the columns wide enough, though we only need the four rightmost columns (bc in the 'thousands' column will mean b×c thousands etc)&#xA;I'll put in more steps than I suspect are needed, for clarity:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   a         b      c     6&#xA;  ~~~~~~~~~~~~~~~~~~~~~~~~~&#xA;  6a        6b     6c    36&#xA;  bc        c²     6c&#xA;  bc        6b  &#xA;  6a&#xA; ___________________________&#xA; 12a+2bc  12b+c²   12c  36     then tidy a bit&#xA;&#xA;  2a+2bc+b 2b+c²+c 2c+3  6 &#xA;&#xA;Using ≡ to mean is equivalent to, mod 10; &#xA;and working columns right to left&#xA;&#xA;2c+3 ≡ c&#xA;c+3 ≡ 0&#xA;c = 7         2c+3 = 17 so add 1 to next column&#xA;&#xA;2b+c²+c+1 ≡ b&#xA; b+57≡0&#xA; b=3          2b+c²+c+1 = 63 so add 6 to next column&#xA;&#xA;12a+2bc+b+6 ≡ a&#xA;11a+42+3+6 ≡ 0&#xA;11a+51 ≡ 0&#xA;  a+1 ≡ 0&#xA;  a=9&#xA;&#xA;So answer = 9376&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It is even easier for the other solution leading to 30 ≡ a at the end.&#xA;With a little care this technique can of course be used in other bases than 10, and any number of final digits.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Update: This was a dangerous question for me; I tried it in base 60, &#xA;like the Sumerians used. Here's the six results &#xA;complete with squares, and translations into base 10. &#xA;With the sums at the bottom:&#xA;&#xA;   44,45,56,16     33,23,57,41,44,45,56,16   966 9376    9349683222 9376&#xA;   50,22,13,21     42,17,10,28,50,22,13,21  1088 0001   11837442176 0001&#xA;   24,51,50,25     10,18,13,07,24,51,50,25   537 0625    2884361289 0625&#xA;   35,08,09,36     20,34,32,18,35,08,09,36   758 9376    5759862806 9376&#xA;   09,37,46,40     01,32,43,47,09,37,46,40   208 0000     432640000 0000&#xA;   15,14,03,45     03,52,05,10,15,14,03,45   329 0625    1082821289 0625 &#xA;&#xA;03,00,00,00,03  01,51,58,42,34,00,00,00,03&#xA;&#xA;Notice that pairs add up to 01,00,00,00,00,01 as we might expect,&#xA;but also look at the last four digits of the translations.&#xA;I didn't expect that. Time for more thinking.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="11397" LastEditorUserId="11397" LastEditDate="2015-04-15T19:19:08.813" LastActivityDate="2015-04-15T19:19:08.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12098" PostTypeId="1" AcceptedAnswerId="12117" CreationDate="2015-04-14T18:56:49.487" Score="12" ViewCount="42779" Body="&lt;p&gt;When solving a 4x4 cube you have specific notation to inner and outer lines, like &lt;code&gt;R&lt;/code&gt; = &lt;em&gt;Clockwise outer right&lt;/em&gt; and &lt;code&gt;r&lt;/code&gt; = &lt;em&gt;Clockwise inner right&lt;/em&gt;. If you want to turn both, the two letter must be combined like &lt;code&gt;Rr&lt;/code&gt;, &lt;code&gt;(Rr)'&lt;/code&gt; or &lt;code&gt;(Rr)2&lt;/code&gt;. To solve OLL parity I could use the following algorithm:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EZ3Xn.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Image linked from &lt;a href=&quot;http://www.speedcubing.com/chris/4speedsolve3.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;speedcubing.com&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;r2 B2 U2 l U2 r' U2 r U2 F2 r F2 l' B2 r2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If &lt;code&gt;r2&lt;/code&gt; means &lt;em&gt;&quot;turn inner right 2 times clockwise&quot;&lt;/em&gt;, why some sites uses &lt;code&gt;r&lt;/code&gt; as &lt;code&gt;Rr&lt;/code&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Open &lt;a href=&quot;https://alg.cubing.net/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Alg Cubing&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Select 4x4 cube&lt;/li&gt;&#xA;&lt;li&gt;Inside the Moves field, type &lt;code&gt;r&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;Play. &lt;em&gt;It will turn both right columns.&lt;/em&gt; &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Related Stuff:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.rubiksplace.com/cubes/4x4/&quot; rel=&quot;nofollow noreferrer&quot;&gt;This site&lt;/a&gt; uses the notation &lt;code&gt;r&lt;/code&gt; to define &quot;both layers(outer and inner)&quot;.&lt;/li&gt;&#xA;&lt;li&gt;However, &lt;a href=&quot;http://loki.ist.unomaha.edu/~jtrimm/project/4x4.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this site&lt;/a&gt; uses the notation &lt;code&gt;Rr&lt;/code&gt;to define &quot;both layers&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question&lt;/strong&gt;: What is the official notation to 4x4 movements(if any exists)?&lt;/p&gt;&#xA;" OwnerDisplayName="user11399" LastEditorDisplayName="user11399" LastEditDate="2018-04-16T16:42:27.453" LastActivityDate="2018-04-16T18:26:22.923" Title="What is the Official notation on rubik's revenge(4x4)?" Tags="|rubiks-cube|algorithm|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12099" PostTypeId="2" ParentId="12004" CreationDate="2015-04-14T19:03:26.953" Score="2" Body="&lt;p&gt;I think the answers are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Reflect, Deflect, Reelect&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Come ponder with me, as we go forth,&#xA;How we behave in a manner of worth.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That's &lt;strong&gt;reflect&lt;/strong&gt;ing!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Instead we consider the ones before.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you vote the ones before, you &lt;strong&gt;reelect&lt;/strong&gt; them!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We behave in a way we won't trivialize.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Deflect&lt;/strong&gt;ion isn't a trivial physical phenomenon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-04-14T19:03:26.953" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12100" PostTypeId="2" ParentId="12098" CreationDate="2015-04-14T19:06:52.937" Score="1" Body="&lt;p&gt;I've found this, which explains the notation of 4x4 cube.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.speedcubing.com/chris/4-presolution.html&quot; rel=&quot;nofollow&quot;&gt;http://www.speedcubing.com/chris/4-presolution.html&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, you have the same letters of the 3x3 (U R F B L D), upper-case if referring to external layers, lower-case if referring to internal layers.&lt;br&gt;&#xA;This kind of notation is commonly used in many websites and guides.&lt;br&gt;&#xA;You ask: &lt;strong&gt;&quot;Is this official?&quot;&lt;/strong&gt;&lt;br&gt;&#xA;The answer is no! As far as I know, nobody (except Erno Rubik) can say &quot;this is official, this isn't official&quot;. In fact, there's no need to establish an official notation for the Rubik's cube.&lt;br&gt;&#xA;This doesn't exclude the existence of (useful) conventions, though unofficial!&lt;br&gt;&#xA;Currently, the preferred notation is the one described above, with capitals and lower-case letters.&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-04-14T19:16:19.353" LastActivityDate="2015-04-14T19:16:19.353" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12101" PostTypeId="2" ParentId="11998" CreationDate="2015-04-14T19:10:32.213" Score="1" Body="&lt;p&gt;This is my 2nd attempt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This time I'm using 7 operations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I couldn't find any shorter than this, for now.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;-(12+34-56)*(-7+8+9)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I've tried and tried and tried.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This simply multiplies 10 by 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nothing fancy, sadly too long, nothing that can't be done easily...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But well, it solves the case!&lt;/p&gt;&#xA;" OwnerUserId="3984" LastEditorUserId="3984" LastEditDate="2015-04-15T09:31:19.813" LastActivityDate="2015-04-15T09:31:19.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12102" PostTypeId="2" ParentId="12098" CreationDate="2015-04-14T19:11:37.470" Score="2" Body="&lt;p&gt;Notation is just a way of expressing sequences of moves. The notation used can change depending on who's doing the notating, and this is true of all higher-dimensional puzzles. There's not really a standard way to notate, any more than the symbols we use for physics are standard. It'd be inconvenient to switch, but we certainly could. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, &lt;a href=&quot;https://i.stack.imgur.com/oKyMN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Trignis&lt;/a&gt; has very different notation depending on the method the solver used to solve the puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently, there is no &lt;em&gt;standard notation&lt;/em&gt; per se for higher-dimensional twisty puzzles. I more often see &lt;code&gt;R&lt;/code&gt; for the rightmost, &lt;code&gt;r&lt;/code&gt; for inner right, &lt;code&gt;l&lt;/code&gt; for inner left, and &lt;code&gt;L&lt;/code&gt; for far left; however, this is by no means necessarily universal. I've also seen &lt;code&gt;eL/eR&lt;/code&gt; on the 4x4 and 5x5 for &quot;edge-wing left&quot; and &quot;edge-wing right&quot; - it just depends on convenience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be sure you check with the guide you're following to see what their notation is for cube movements.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastActivityDate="2015-04-14T19:11:37.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12103" PostTypeId="2" ParentId="12085" CreationDate="2015-04-14T19:32:11.233" Score="2" Body="&lt;p&gt;Pursued by many, hated by few,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many people want love, just some hate it, mainly when jealous.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's amazing how much I can do.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Love can break down mountains is the saying? i know there's some saying but you get the point.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Creation, enlargement, or despise,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; love creates opportunities, en enlarges someone's world, some people despise those who have found love.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tell me now where you keep your eyes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; though i found this one difficult i think: when in love you keep your eyes on the person/object which has inspired your love?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Life or death, on different fronts,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Love can bring life (babies), love can bring death (crime passionel). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the goal of many hunts.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; many see getting love as a hunt, they want to obtain it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All may have, and all seem to want,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; evryone can love someone/something and evryone seems to want love?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But in different ways, now don't flaunt.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I had to look up the word Flaunt, +1 for expanding my vocabulary :P&#xA; Some people parade their love in the faces of others though some would say you don't parade your own happinesses in the faces of others.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I may be the beginning, or maybe the end,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; love can be the beginning of many things but can also be the end (of a loveless life for example)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So riddle out what I am my friend.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you are Love&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9835" LastActivityDate="2015-04-14T19:32:11.233" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12104" PostTypeId="2" ParentId="12085" CreationDate="2015-04-14T19:53:47.953" Score="3" Body="&lt;p&gt;My guess&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Intelligence&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pursued by many, hated by few,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many strive to be smarter, some despise those who are smarter than themselves&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's amazing how much I can do.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;By gaining knowledge, you can do more&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Creation, enlargement, or despise,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Create ideas, gain knowledge, despise (emotion)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tell me now where you keep your eyes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Always moving forward, looking for new opportunities to understand&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Life or death, on different fronts,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the pursuit of knowledge, many can die if not careful how they obtain it. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the goal of many hunts.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, striving for a fuller understanding&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All may have, and all seem to want,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A certain level of intellect, many try to gain more...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But in different ways, now don't flaunt.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Don't show off your intellect, it is becoming of you&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I may be the beginning, or maybe the end,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Without it, we cease functioning or we're merely floating along in space&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So riddle out what I am my friend.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Intelligence&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&#xA;My other guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Focus&quot;, but I wasn't sure how to answer the parts of the riddle with it just yet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11404" LastActivityDate="2015-04-14T19:53:47.953" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12105" PostTypeId="2" ParentId="12004" CreationDate="2015-04-14T19:54:48.843" Score="2" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;Note&lt;/strong&gt;: this one feels pretty weak.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How we behave in a manner of worth.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many people are &lt;strong&gt;&lt;em&gt;sour&lt;/em&gt;&lt;/strong&gt; about their worth being questioned.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With a single change, that worth is no more.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Change the s to a t.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Instead we consider the ones before.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you &lt;strong&gt;&lt;em&gt;tour&lt;/em&gt;&lt;/strong&gt; a museum, you consider artifacts and events from the past.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After one more change, seen with different eyes,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Change the u to an i and read it backwards.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We behave in a way we won't trivialize.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People &lt;strong&gt;&lt;em&gt;riot&lt;/em&gt;&lt;/strong&gt; when they are emphatic about their cause being not trivialised.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2015-04-14T19:54:48.843" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12106" PostTypeId="1" AcceptedAnswerId="12108" CreationDate="2015-04-14T19:59:25.687" Score="12" ViewCount="1833" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;All I remember was being at a store and then all of a sudden someone was taking me to this party!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Some people couldn't stop staring at me. (Maybe they were impressed when I was holding onto everyone's drink!)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Later that night, some guy slammed me onto the table and then started yelling, it was crazy.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;It was one wild night...but now that I'm older, things seem to be a little hazy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What in the world could I possible be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm not a human ;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(I hope this isn't too easy for my first riddle, make sure your answer works with every line!)&lt;/p&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-04-14T20:10:26.917" LastActivityDate="2016-02-23T02:52:40.203" Title="Riddle: Life of the party" Tags="|riddle|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12107" PostTypeId="2" ParentId="12004" CreationDate="2015-04-14T19:59:41.137" Score="3" Body="&lt;p&gt;Probably a little too simplistic, but food for thought if nothing else:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Cost&lt;/strong&gt;, &lt;strong&gt;Lost&lt;/strong&gt;, and &lt;strong&gt;List&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How we behave in a manner of worth.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Something's worth is often considered its &lt;strong&gt;cost&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With a single change, that worth is no more.&#xA; Instead we consider the ones before.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One letter change (&lt;strong&gt;C&lt;/strong&gt; to &lt;strong&gt;L&lt;/strong&gt;), and we get &lt;strong&gt;lost&lt;/strong&gt;, perhaps referring to those that have died, and therefore came before.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After one more change, seen with different eyes,&#xA; We behave in a way we won't trivialize.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Probably the weakest of the three, one more letter change (&lt;strong&gt;O&lt;/strong&gt; to &lt;strong&gt;I&lt;/strong&gt;) and we get &lt;strong&gt;list&lt;/strong&gt;, and if we have a list, it often isn't trivial. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-04-14T19:59:41.137" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12108" PostTypeId="2" ParentId="12106" CreationDate="2015-04-14T20:07:33.480" Score="11" Body="&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (the) Glasses (Thanks @Rand al' Thor!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All I remember was being at a store and then all of a sudden someone was taking me to this party!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Buy glasses at the store, then take it to the party.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some people couldn't stop staring at me. Maybe they were impressed when I was holding onto everyone's drink!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People stare at glasses while toasting and always hold everyone's drink. Also people that wear 'glasses' can't stop staring at them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later that night, some guy slammed me onto the table and then started yelling, it was crazy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Finished his drink, then slammed the glass on the table. Yelled &quot;woo! That's tasty!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It was one wild night...but now that I'm older, things seem to be a little hazy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Glass gets hazy over the years :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9412" LastEditorUserId="11244" LastEditDate="2015-04-15T13:05:42.573" LastActivityDate="2015-04-15T13:05:42.573" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12109" PostTypeId="2" ParentId="12085" CreationDate="2015-04-14T20:21:30.887" Score="4" Body="&lt;p&gt;I'll take a shot, although some of the lines are a bit of a stretch...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pursued by many, hated by few,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sex is a universal drive&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's amazing how much I can do.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tons of ways to have sex, and many reasons and outcomes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Creation, enlargement, or despise,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Procreation, erections (or pregnancy), angry sex&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tell me now where you keep your eyes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure... maybe sexual orientation (i.e. who you like to look at)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Life or death, on different fronts,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Procreation vs raping-to-death?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the goal of many hunts.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The goal of many a college student's Friday night, for starters...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All may have, and all seem to want,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Anyone can have sex (barring disease and whatnot), and in modern culture it seems to be very important to everybody&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But in different ways, now don't flaunt.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People want different kinds of sex (fetishes), sex should be private&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I may be the beginning, or maybe the end,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First time vs breakup sex&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So riddle out what I am my friend.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sex&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4513" LastActivityDate="2015-04-14T20:21:30.887" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12110" PostTypeId="1" AcceptedAnswerId="12111" CreationDate="2015-04-14T20:25:25.990" Score="1" ViewCount="3787" Body="&lt;p&gt;I fly high with various colors on me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But most of the time I sit in a corner and not fly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hate rain. And I usually don't come out during a rainy day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A sling of my hair is always long enough to touch the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At times I have a tail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;" OwnerUserId="11056" LastEditorUserId="5373" LastEditDate="2015-04-14T20:32:15.270" LastActivityDate="2015-04-18T11:31:58.307" Title="I am a ? I fly high" Tags="|riddle|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12111" PostTypeId="2" ParentId="12110" CreationDate="2015-04-14T20:29:29.583" Score="6" Body="&lt;p&gt;You are a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; kite&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I fly high with various colors on me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Kites are colorful.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But most of the time I sit in a corner and not fly.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They'll be kept indoors most of the time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I hate rain. And I usually don't come out during a rainy day.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can't fly a kite in the rain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A sling of my hair is always long enough to touch the ground.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The kite string.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At times i have a tail.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Kites can have tails hanging off them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2015-04-14T20:29:29.583" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12113" PostTypeId="2" ParentId="12106" CreationDate="2015-04-14T20:30:06.047" Score="7" Body="&lt;p&gt;Can't quite get one right on but:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A bottle (of any drink).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I remember was being at a store and then all of a sudden someone was taking me to this party!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Purchased at a store, taken to the party.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some people couldn't stop staring at me. (Maybe they were impressed when I was holding onto everyone's drink!)  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Holding what people are drinking. There are always those that are only there to drink and will stare at whatever the drink is.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later that night, some guy slammed me onto the table and then started yelling, it was crazy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A fight where people use bottles, or just a celebratory slam.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It was one wild night...but now that I'm older, things seem to be a little hazy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Glass bottles tend to fog or get hazzy with age.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-04-14T20:30:06.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12114" PostTypeId="2" ParentId="12085" CreationDate="2015-04-14T20:42:43.153" Score="10" Body="&lt;p&gt;You are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; breasts&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pursued by many, hated by few,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Men usually love breasts.  Even many women like having them, although a few may not.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's amazing how much I can do.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That's why they're called &quot;weapons of mass distraction&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Creation, enlargement, or despise,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Creation = create lust in men, or create milk.  Enlargement = implants.  Despise = breast reduction surgery.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tell me now where you keep your eyes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many men stare at them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Life or death, on different fronts,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Breasts feed (and therefore help give life to) babies.  Breast cancer can cause death (credit to Sabre for that).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the goal of many hunts.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many men want to touch them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All may have, and all seem to want,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All women have them, and many women want larger ones.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But in different ways, now don't flaunt.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some women show them off, other women are more modest.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I may be the beginning, or maybe the end,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Beginning as in foreplay, end as in the, &lt;em&gt;ahem&lt;/em&gt;, climax. (credit to Sabre for this one)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4450" LastEditDate="2015-04-14T20:49:18.433" LastActivityDate="2015-04-14T20:49:18.433" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12115" PostTypeId="2" ParentId="12085" CreationDate="2015-04-14T21:22:14.860" Score="0" Body="&lt;p&gt;Youth or eternal Life?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't hate it because you had it and lost it. You always want it back. It is almost universally considered the apex of beauty and human aesthetics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Love this puzzle. Thank you for sharing!!! Can't wait for the answer 8-D&lt;/p&gt;&#xA;" OwnerUserId="11409" LastActivityDate="2015-04-14T21:22:14.860" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12116" PostTypeId="2" ParentId="12004" CreationDate="2015-04-14T21:52:33.390" Score="1" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;Note&lt;/strong&gt;: I'm being rather liberal with the requirements here, and maybe thinking too far outside the box.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How we behave in a manner of worth.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;$1m&lt;/strong&gt; is short-hand for &quot;one million dollars&quot;, which is worth a lot.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With a single change, that worth is no more.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Remove the m and we have &lt;strong&gt;$1&lt;/strong&gt;, where the worth is no more.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Instead we consider the ones before.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We consider the 1 as it is the only digit left.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After one more change, seen with different eyes,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Adding an i is a change and shown with different &quot;eyes&quot; (as a homonym).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We behave in a way we won't trivialize.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;$1i&lt;/strong&gt; is a complex number and therefore cannot be trivialised.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2015-04-14T21:52:33.390" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12117" PostTypeId="2" ParentId="12098" CreationDate="2015-04-15T01:54:17.570" Score="7" Body="&lt;p&gt;There is &lt;a href=&quot;https://www.worldcubeassociation.org/regulations/#article-12-notation&quot; rel=&quot;noreferrer&quot;&gt;an official scramble notation&lt;/a&gt; for WCA competitions (which has continued to change over the years), which uses &lt;code&gt;Rw&lt;/code&gt; as (&lt;code&gt;Rr&lt;/code&gt;), &lt;code&gt;3Rw&lt;/code&gt; as (&lt;code&gt;l' r R&lt;/code&gt;), and so forth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will notice on that page that they do not have symbols designated for inner slice turns only because the WCA scrambles cubes using outer block turns only to follow the outer block turn metric (OBTM).  However, &lt;a href=&quot;https://www.youtube.com/watch?v=j-svGANFGXc&quot; rel=&quot;noreferrer&quot;&gt;SiGN notation&lt;/a&gt;, as you have seen on alg.cubing.net, is used more often now than previously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly a move such as &lt;code&gt;Rw&lt;/code&gt; = (&lt;code&gt;Rr&lt;/code&gt;) which is used for WCA scrambles is never mistaken for an inner slice turn, for example; and thus a good rule of thumb for the casual cuber who wishes to communicate his algorithms to others is to have the courtesy to label the notation next to the alg if the algorithm does not use &quot;&lt;code&gt;Rw&lt;/code&gt;&quot; moves (clearly Old WCA) &lt;code&gt;2R2&lt;/code&gt; moves (clearly SiGN).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the edge flip alg you posted, for example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;r2 B2 U2 l U2 r' U2 r U2 F2 r F2 l' B2 r2&lt;/code&gt; (Old WCA notation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, it's even better to link your algorithms to a site like alg.cubing.net (for SiGN notation) or alg.garron.us (for old WCA notation), so that those who do not know of a notation at all can still see the algorithm with a click of a play button.&lt;/p&gt;&#xA;" OwnerUserId="6044" LastEditorDisplayName="user11399" LastEditDate="2015-04-15T04:58:50.660" LastActivityDate="2015-04-15T04:58:50.660" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12118" PostTypeId="2" ParentId="12004" CreationDate="2015-04-15T02:42:59.960" Score="1" Body="&lt;p&gt;Not confident of this, but here is my answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The base word is &lt;em&gt;form&lt;/em&gt; -- &lt;strong&gt;formation, former, reform (or transform)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How we behave in a manner of worth.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is probably connected to the word hierarchy or ranking, so I choose the word &lt;em&gt;formation&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With a single change, that worth is no more.&#xA;Instead we consider the ones before.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;former&lt;/em&gt; I think these lines mean a word pertaining to &lt;em&gt;past&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After one more change, seen with different eyes,&#xA;We behave in a way we won't trivialize.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; these lines are about change, &lt;em&gt;transform&lt;/em&gt; or &lt;em&gt;reform&lt;/em&gt; is synonymous to change.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10686" LastActivityDate="2015-04-15T02:42:59.960" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12119" PostTypeId="1" AcceptedAnswerId="12134" CreationDate="2015-04-15T03:04:56.900" Score="93" ViewCount="14121" Body="&lt;p&gt;You die, and awake in Hell. Satan awaits you, and has prepared a curious game. He has arranged $n$ quarters in a line, going in the east/west direction. He placed the coins at the ends tails up, and all others heads up, like so:&#xA;$$&#xA;\text{T H H H }\cdots \text{ H H H T}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Satan explains the rules.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Once a day, a coin is removed from the east end, and placed on the west end.&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the coin was initially tails-up, then &lt;em&gt;you&lt;/em&gt; get to choose whether the coin is placed heads up or down.&lt;/li&gt;&#xA;&lt;li&gt;If it was initially heads-up, then Satan gets to make this choice.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;If the coins are all heads up at the end of a day, you get to leave Hell.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Satan will of course try his hardest to make sure you never leave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, when $n=5$, we start with $\text{T H H } \color{red}{\text H} \color{green}{\text{ T}}$. The first day is your choice; if you choose heads, the arrangement becomes $\color{green}{\text{H }}\text{T H H } \color{red}{\text H} $. The next three days, however, will be Satan's choice. He may fight back on the second day by choosing tails, resulting in $\color{red}{\text{T } }\color{green}{\text{H }}\text{T H H }  $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a strategy that eventually guarantees your salvation? Or can Satan conspire to keep you in Hell forever? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum:&lt;/strong&gt; To give credit where it is due, this puzzle from &lt;a href=&quot;http://www.cs.cmu.edu/puzzle/&quot;&gt;The Puzzle Toad&lt;/a&gt;, (under the name &quot;Zeroise Me&quot;) which is a superb collection of similarly clever and enjoyable conundrums.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2021-07-14T17:08:45.513" LastActivityDate="2021-08-20T18:18:47.053" Title="Coin Flipping Game with the Devil" Tags="|mathematics|strategy|probability|game|" AnswerCount="9" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="12120" PostTypeId="2" ParentId="12119" CreationDate="2015-04-15T03:17:05.630" Score="21" Body="&lt;p&gt;Charlie always wins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's split the chain of coins up into two sections: the tails-chain and the heads-chain. The tails-chain is Charlie's long chain of tails. I will use a vertical bar to denote where the chains split. It starts out as the two &lt;code&gt;T&lt;/code&gt;s in &lt;code&gt;TTH...H&lt;/code&gt;, written &lt;code&gt;TT|H...H&lt;/code&gt;. The heads chain is everything bounded by the two &lt;code&gt;H&lt;/code&gt; coins of Satan's. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A postulate: &lt;strong&gt;Tails in a chain will always connect, and Satan will never get them back.&lt;/strong&gt; This is pretty easy to prove. Once the tails have joined Charlie's chain, Charlie just has to choose not to flip them to heads again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A postulate: &lt;strong&gt;Satan must have at least one tails in his chain.&lt;/strong&gt; If Satan at any point has all heads in his chain, then the chain must pass through the state &lt;code&gt;H...H|T...T&lt;/code&gt;, which means he loses. Thus, there must be at least one tails in his chain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Final postulate: &lt;strong&gt;Charlie can force the tails in Satan's chain one step further towards its 'tail' with each loop through.&lt;/strong&gt; (Note that there is a special case for this, detailed below.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads to the win condition. In order to win, Charlie needs to force Satan's chain to spit out tails. Those tails will connect with Charlie's chain, and Satan will lose them. Repeat this process indefinitely, and Charlie wins. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do this, Charlie flips any tails resting between two heads to heads. That means the soonest Satan can put a new tails in is the next coin. This forces Satan's tails coin one coin further down the heads-chain, until it contacts the tails-chain and becomes part of it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At that point, he loses another heads, as he has to have a tails in his chain. Satan will eventually only have one heads, and will lose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example chain: &lt;code&gt;HHTH|TT -&amp;gt; HTHH|TT -&amp;gt; THHH|TT&lt;/code&gt;. At this point, Charlie's chain is actually &lt;em&gt;three&lt;/em&gt; tails long: &lt;code&gt;TTT|HHH&lt;/code&gt;. Satan can't help but move &lt;code&gt;HTH|TTT -&amp;gt; THH|TTT&lt;/code&gt;, which means Charlie's chain becomes &lt;em&gt;four&lt;/em&gt; tails long. At this point, Satan's loss is inevitable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example chain: &lt;code&gt;HHHTH|TT -&amp;gt; HTHHH|TT&lt;/code&gt;. At this point Satan might flip the fourth coin again resulting in &lt;code&gt;HTHTH|TT&lt;/code&gt;, but if Charlie leaves this alone and flips the second tail back when he gets around to it, Satan will have no choice but to flip the third coin from the left, resulting in &lt;code&gt;HTTHH|TT&lt;/code&gt;. Then, on the next go-around, Charlie increases the length of the chain by 1 again. Consequently, Satan will eventually lose all his open coins.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;After there are enough coins, Satan may be able to introduce multiple tails. This is non-trivial, but results in a loss for Satan anyway. If Satan introduces multiple tails, do nothing. Then, start pushing coins from the front of the chain toward the back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, let's say you have &lt;code&gt;TT|HTHHH&lt;/code&gt;. Then, Satan is a jerk and creates &lt;code&gt;HTH|TT|HT&lt;/code&gt;. Now, instead of flipping your lingering tails coin, do nothing with it: &lt;code&gt;TT|HTHTH&lt;/code&gt;. Now, keep flipping your tails to heads &lt;em&gt;until&lt;/em&gt; Satan flips a coin. This leads to the following chain: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;H|TT|HTHT -&amp;gt; HH|TT|HTH -&amp;gt; HHH|TT|HT -&amp;gt; HHHH|TT|H -&amp;gt; THHHH|TT -&amp;gt; TTT|HHHH&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In other words, up until the point where Satan flips a coin, you keep flipping tails to heads. As an example with more coins:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;TT|HTHHHTH -&amp;gt; H|TT|HTHHHT -&amp;gt; HH|TT|HTHHH -&amp;gt; THH|TT|HTHH -&amp;gt; TT|HTHHTHH&#xA;TT|HTHHTHH -&amp;gt; HH|TT|HTHHT -&amp;gt; HHH|TT|HTHH -&amp;gt; THHH|TT|HTH -&amp;gt; TT|HTHTHHH&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It's pretty clear that eventually you'll end up in the configuration:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;TT|HTTTHHH&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You can keep pushing Satan's tokens further down the chain. Once you do, you're in control of them, and Satan needs to add another tails further up the chain otherwise you'll bump the last heads into your chain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To illustrate, let's say Satan doesn't add a new tails. Then you have: &lt;code&gt;TT|THHHHHH -&amp;gt; TTT|HHHHHH&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if he adds another one, it will become &lt;code&gt;TT|HTTTHTH&lt;/code&gt;, which you can easily force into &lt;code&gt;TT|HTTTTHH&lt;/code&gt; on your next turn. From there, Satan has nowhere to go, and loses a coin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Therefore, no matter what Satan does, Charlie always wins.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here are a few extraneous notes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some raw experimentation shows that the number of 'cycles' it takes to flip $n$ coins over is $F_n$, the $n$th Fibonacci number. (Here, I'm defining a cycle as the number of moves until the tails-chain is at the front again.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the tails-chain gets shifted backward one step for every coin assimilated, the total number of moves ends up being $n*F_n-(n-1)=n(F_n-1)+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd gander that if there were even 20 coins, it would take $20*6765-19=135281$ days for you to actually escape. Several centuries would have passed before you ever left, for any reasonable number of coins. What a jerk, that Satan.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorDisplayName="user20" LastEditDate="2015-04-15T06:56:40.403" LastActivityDate="2015-04-15T06:56:40.403" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12121" PostTypeId="2" ParentId="12119" CreationDate="2015-04-15T03:23:04.077" Score="10" Body="&lt;p&gt;For $n = 2$, the solution is trivial - you flip both coins heads-up for the first two days, and leave.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For $n = 3$, you leave the first coin alone, leaving Satan with TTH. But no matter whether Satan flips this coin or leaves it alone, you can flip all the T's in a row  afterwards, so you win.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For $n = 4$, you leave the first coin alone, leaving Satan with TTHH. Regardless of how Satan flips these two coins, in all four cases you're left with a string of 2, 3, or 4 consecutive T's, so you can flip all of them and win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In the case of flipping only the second one, you're left with THTT, but just leave those two alone and no matter what Satan flips, you can flip all of them the second time around.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$n = 5$ is the first interesting case. Then, if Satan flips only the second H out of three after you leave the first coin alone, you're left with HTHTT. If you leave this alone, then Satan will also leave it alone, and you'll be stuck in a loop forever. So the only thing to do is to either flip the solitary T, or flip one or both of the two consecutive T's.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If you flip one of the two consecutive T's, you've left Satan with either HTHTH or THHTH depending on which one you flipped. Satan can just leave this alone, which means you're stuck in an infinite loop until you decide to flip one of the others.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If you flip the first one, Satan's left with HHHTH. If he leaves this alone, then you flip the T and win. So he has to flip this H, leaving you with THHHT, back where you started.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you flip the second one, Satan's left with HHTHH. But he can just leave the first H alone and flip the second one, leaving you back where you started as well.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you flip the solitary T, then Satan's left with HHTTH. If he leaves this H alone, then you flip the T and win. So he has to flip the T, leaving you with THHTT. Just leave these two alone, leaving Satan with TTTHH and the same quandary he had in the $n = 4$ case.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In general, you want to leave Satan with a situation where all the T's are in a single bunch and a single H follows them. Then, he's forced to increase the length of the T chain by 1 (and thereby effectively reduce the number of coins in the game inductively), and if you can keep doing that, then by induction you win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n = 6$, Satan can flip one, the other, or both of the two coins not next to the existing T's, leaving you with one of HHTHTT, HTHHTT, or HTTHTT once you've gone a full revolution. In any case where Satan flips the coin right before the leftmost H, you leave them alone (and your first two T's) the first time around, and flip them all back to heads the second round, leaving Satan with HHHTTH and you winning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Satan has to flip the first coin only where he's left with HHTTHH and has the choice of flipping the T, which is the only nontrivial variation from $n = 5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's suppose he does this. Then you're left with HTHHTT, which, from above, is one of the winning strategies. So you win as well.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;We still haven't deduced a general strategy yet, so let's continue with $n = 7$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Satan can leave you with HTHTHTT this time, which circumvents the flipping-the-one-next-to-last rule described above since the first flipped T prevents Satan from being forced to flip the H immediately before the string of two T's. But in this case, if you flip the first solitary T the second time around, then Satan has to flip the next H, because if he leaves it alone, you can just flip the next T as well, leaving Satan with HHHHTTH and starting the $n = 6$ case. So he has to flip that H, leaving you with HTTHHTT the second time around. Then you simply flip the left sequence of T's and force the $n = 6$ case again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is one wrinkle here. What if Satan flips to HHTHHTT, and you flip that T only for Satan to flip the H immediately afterwards (resulting in HTHHHTT) and then Satan flips the second H only, resulting in HTHTTHT on your turn? If you flip the T now, and then the next solitary T right afterward, Satan will just flip the next T again, returning you to HHTHHTT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you just leave that second T alone for two days, and the HTHTHTT case plays out exactly as described in the first place.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;At this point, we finally notice a pattern. You will always flip the series of T's immediately after your longest chain (which starts at length 2 and keeps growing until it's the entire string of coins), which forces Satan to flip the H immediately to the left of them in order to ward off defeat. And if Satan flips any of the H's beforehand, leave the rest of the T's alone and let it go around one more revolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will always result in the set of stray T's moving left towards the end of your longest chain and merging together with other stray sets, or a set of new T's being created that will have to move left anyway, possibly creating an even longer chain in the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The longest Satan can drag the game out for when your longest chain is $m$ T's long (without making a longer chain in the process) is until the chain of $m$ T's is followed by a series of chains of $m$ T's each separated by a single H. At that point, any head Satan flips will automatically create a longer chain, so you can flip all the chains of $m$ T's, leaving Satan with the coin right before the last chain of $m$ T's which he must then extend to $m+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this way, you can always defeat Satan and escape from Hell.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Below is a program-like algorithm that you can follow, assuming that instead of moving coins to the left, you keep track of a cursor that moves to the left one space every day.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Every day, the cursor moves left one space. If the cursor reaches the leftmost coin, it jumps back to the right.&lt;/li&gt;&#xA;&lt;li&gt;If you are on your rightmost set of consecutive T's, pass.&lt;/li&gt;&#xA;&lt;li&gt;If Satan has not flipped a coin yet since the cursor last jumped to the right and you are currently on a T, flip it. Otherwise, pass.&lt;/li&gt;&#xA;&lt;li&gt;If Satan flips the leftmost coin, move that coin to the rightmost position, keeping the cursor on it.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;An example of the longest match for $n = 11$, to show how this algorithm might work:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;HHHHHHHHHTT   You leave the two T's alone, Satan flips coin 8.&#xA;HHHHHHHTHTT   You flip the first T, forcing Satan to flip coin 7.&#xA;HHHHHHTHHTT   You flip the first T, forcing Satan to flip coin 6.&#xA;...&#xA;HTHHHHHHHTT   Satan flips coin 8 again, and you leave everything alone.&#xA;HTHHHHHTHTT   You flip coin 8, and keep moving left until Satan reaches coin 4.&#xA;HTHTHHHHHTT   Satan flips coin 8 again. This flipping coin 8 and moving to the left continues until all the even-numbered slots are occupied.&#xA;HTHTHTHTHTT   You start flipping all the T's from coin 8 onward. If Satan flips any coins before that, stop and let the cursor wrap around again.&#xA;HTTHHHHTHTT   This might happen. Then, Satan flips coin 8 and starts the above process all over.&#xA;...&#xA;HTTHTTHHHTT   This might happen again.&#xA;HTTHTTHTHTT   Satan flips coin 8 one last time, but he cannot dodge anymore. You leave the coins alone for one more iteration.&#xA;THHHHHHHHTT   You flip all groups of coins, leaving Satan with no choice but to flip coin 1.&#xA;&#xA;HHHHHHHHTTT   The frame of reference of the coins move forward one space.&#xA;HHHHHHTHTTT   Satan starts with coin 7 this time.&#xA;...&#xA;HTHTHTHHTTT   Chains of length 1.&#xA;HTHTHTTHTTT   Satan flips coin 7. You leave all the other coins alone.&#xA;HTTHHHHHTTT   Satan is forced to do this again.&#xA;HTTHHHTHTTT   Same process as before.&#xA;...&#xA;HTTHTTHHTTT   Chains of length 2. You flip the first chain, Satan flips coin 4.&#xA;HTTTHHHHTTT   First chain of length 3. Satan flips coin 7 again, and the process continues.&#xA;HTTTHHTHTTT&#xA;HTTTHTHHTTT   &#xA;HTTTHTTHTTT   In 3 moves, Satan exhausts the chain of length 3. You flip all sequences of coins to the left of your first chain.&#xA;THHHHHHHTTT   Again, Satan has no choice but to flip coin 1. The chain extends by length 1 again.&#xA;...&#xA;THHHHHHTTTT   ???&#xA;......&#xA;TTTTTTTTTTT   Profit!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It turns out that the number of moves required to force a win grows exponentially compared to the number of coins in a row.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-04-15T06:58:07.383" LastActivityDate="2015-04-15T06:58:07.383" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="12122" PostTypeId="2" ParentId="12119" CreationDate="2015-04-15T03:44:54.170" Score="4" Body="&lt;p&gt;You can force a win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your strategy is to aim for the string $\phi\gamma$ where $\gamma$ is comprised only of $T$ and $\phi$ is comprised only of $H$ and it is your turn next..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, a configuration of $H^iT^j$ wins for us... we just make all $T$ into $H$ and win - regardless of $i$, $j$. Call this configuration $win$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strategy in general is, We always play $T$ unless it's a $T$ that is not part of our $win$ T string ($\gamma$) in which case we play $H$ (in order to establish $\phi$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a demonstration: Say, $n=5$: the string, $S$, is $THHHT$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your first choice will always be to choose $T$: $TTHHH$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our job is to keep the string of $T$s as long as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are now 8 ways that Satan can play his moves:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$HHHTT$&lt;br&gt;&#xA;$TTHTT$&lt;br&gt;&#xA;$THTTT$&lt;br&gt;&#xA;$HTTTT$&lt;br&gt;&#xA;$HHTTT$&lt;br&gt;&#xA;$HTHTT$&lt;br&gt;&#xA;$THHTT$&lt;br&gt;&#xA;$TTTTT$&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll remove the ones that are an instance of $win$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$TTHTT$&lt;br&gt;&#xA;$THTTT$&lt;br&gt;&#xA;$HTHTT$&lt;br&gt;&#xA;$THHTT$&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Satan has 4 ways which don't lose automatically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in all these cases, it's our choice. We want to create or facilitate $win$.&#xA;We leave the $T$ as they are. I will number them now and remove duplicates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1.\ TTTTH$&lt;br&gt;&#xA;$2.\ TTHTH$&lt;br&gt;&#xA;$3.\ TTTHH$&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it's Satan's turn. In case 1. Satan has lost. He cannot avoid establishing $win$ for us. Leaving us only with these options.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1.\ TTHTH$&lt;br&gt;&#xA;$2.\ TTTHH$&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's focus on case $2. TTTHH$. There are 4 ways he can play this one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2.1\ HHTTT$&lt;br&gt;&#xA;$2.2\ HTTTT$&lt;br&gt;&#xA;$2.3\ THTTT$&lt;br&gt;&#xA;$2.4\ TTTTT$&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are all now lost for Satan. $2.1$,$2.2$ and $2.4$ are an instance of $win$ and $2.3$ after our moves gives $TTTTH$ where Satan cannot avoid establishing $win$ for us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the only complicated case is $1. TTHTH$. This is only case of a broken string of $H$ or $T$ when considered East to West overflow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Satan has 2 ways to play the next one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1.1\ HTTHT$&lt;br&gt;&#xA;$1.2\ TTTHT$&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case $1.2$ we win again, by choosing $T$.&#xA;For case, $1.1\ HTTHT$ which in general will be the from of the only tricky position. We have 2 ways to play. If we play $T$ nothing changes. This is the only case where we must play $H$.&#xA;And the result is: $HHTTH$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now Satan has 2 ways to play.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1.\ HHHTT$ - this is $win$.&lt;br&gt;&#xA;$2.\ THHTT$ - we choose $T$ -&gt; $TTTHH$ - now Satan is lost as before. What a loser :D&lt;br&gt;&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2015-04-15T04:05:00.623" LastActivityDate="2015-04-15T04:05:00.623" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12123" PostTypeId="2" ParentId="10700" CreationDate="2015-04-15T04:16:48.077" Score="0" Body="&lt;p&gt;Build a runway like this&#xA;where the height of the run way is sqrt(3) m high. Or go through the digonal in that case you only need height of sqrt(2)m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/85Fa5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="11343" LastActivityDate="2015-04-15T04:16:48.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12124" PostTypeId="2" ParentId="12061" CreationDate="2015-04-15T06:12:50.453" Score="-2" Body="&lt;p&gt;The word is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; short + er = shorter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11425" LastEditorUserId="4637" LastEditDate="2015-04-15T06:29:09.530" LastActivityDate="2015-04-15T06:29:09.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12125" PostTypeId="2" ParentId="12106" CreationDate="2015-04-15T06:30:29.570" Score="2" Body="&lt;p&gt;Let me see:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A video camera&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I remember was being at a store and then all of a sudden someone was taking me to this party!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Purchased at a store, Taken to the party.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some people couldn't stop staring at me. (Maybe they were impressed when I was holding onto everyone's drink!)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People were staring at the video camera and the camera was focusing on their drinks, recording them perhaps.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later that night, some guy slammed me onto the table and then started yelling, it was crazy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He was recording something on the camera.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It was one wild night...but now that I'm older, things seem to be a little hazy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Camera/Video camera lenses have become hazy with age.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11426" LastEditorUserId="10973" LastEditDate="2015-04-15T11:19:42.253" LastActivityDate="2015-04-15T11:19:42.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12126" PostTypeId="2" ParentId="10700" CreationDate="2015-04-15T06:53:24.177" Score="0" Body="&lt;p&gt;I would probably try to go with an Euler spiral (&lt;a href=&quot;http://en.wikipedia.org/wiki/Euler_spiral&quot; rel=&quot;nofollow&quot;&gt;Wikipedia explanation&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While indeed it becomes increasingly difficult to turn at higher speeds, this has the benefit that the curvature of the spiral would decrease with the increasing speed. Seems like a two mile length would easily be doable, IF:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The motors thrust can be 'unequal', to actually perform said turns&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The tires don't blow out&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) You have little to no 'use' for the island. The shape really messes up what functionality the island can have.&lt;/p&gt;&#xA;" OwnerUserId="2156" LastActivityDate="2015-04-15T06:53:24.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12127" PostTypeId="1" AcceptedAnswerId="12128" CreationDate="2015-04-15T07:04:17.873" Score="14" ViewCount="524" Body="&lt;p&gt;A professor has written a note to help him remember his password. I'm sure you can find out what it is. After all, he's very functional.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\text{4,24 AOPCRTLIFA}$&lt;br&gt;&#xA;$\text{10,5 RAVHLE}$&lt;br&gt;&#xA;$\text{0,1 SIOEICN}$&lt;br&gt;&#xA;$\text{3,27 BEUMC}$&lt;br&gt;&#xA;$\text{54,18 ETEITNTO}$&lt;br&gt;&#xA;$\text{100,2 LHOIGAMCTR}$&lt;br&gt;&#xA;$\text{16,4 UOATRSROOEQ}$&lt;br&gt;&#xA;$\text{0,-1/2 UATZE}$&lt;br&gt;&#xA;$\text{6,18 IELPRNT}$&lt;br&gt;&#xA;$\text{140,12 MSGTAI}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\text{My password:}$&lt;br&gt;&#xA;$\text{41,2,72,21,12,42,2,71,25,6,2,39}$&lt;/p&gt;&#xA;" OwnerUserId="7308" LastEditorUserId="7308" LastEditDate="2015-04-15T23:24:33.010" LastActivityDate="2015-04-15T23:24:33.010" Title="Functional Puzzle Solving" Tags="|lateral-thinking|enigmatic-puzzle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12128" PostTypeId="2" ParentId="12127" CreationDate="2015-04-15T07:17:02.393" Score="10" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MATHEMATICAL  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's how to find it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each set of letters anagram a mathematical function that can be applied to the first number to produce the second. (I actually didn't realize the relationship between the number until I had found most of them by recognizing near-anagrams).&#xA; &lt;code&gt;&lt;br&gt;&#xA; 4,24   FACTORIAL + P&lt;br&gt;&#xA; 10,5   HALVE + R&lt;br&gt;&#xA; 1,0    COSINE + I&lt;br&gt;&#xA; 3,27   CUBE + M&lt;br&gt;&#xA; 54,18  TOTIENT + E&lt;br&gt;&#xA; 100,2  LOGARITHM + C&lt;br&gt;&#xA; 16,4   SQUARE ROOT + O&lt;br&gt;&#xA; 0,-1/2 ZETA + U&lt;br&gt;&#xA; 6,18   TRIPLE + N&lt;br&gt;&#xA; 140,12 SIGMA + T&lt;/code&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From here,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The extra letters spell &quot;PRIME COUNT&quot;. Applying the prime count function (number of primes $\leq n$) to each number gives $12, 0, 19, 7, 4, 12, 0, 19, 8, 2, 0, 11,$ which as alphabet letters spells MATHEMATICAL.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="1664" LastEditDate="2015-04-15T13:47:37.923" LastActivityDate="2015-04-15T13:47:37.923" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12129" PostTypeId="1" AcceptedAnswerId="12131" CreationDate="2015-04-15T07:18:39.767" Score="-3" ViewCount="1813" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RicXQ.jpg&quot; alt=&quot;Who dies?&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see in the picture, there are 5 people A, B, C, D, E. E is going to slightly push the circular structure so that the initial velocity of the structure on the inclined plane would be zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many people would die and how many would survive?&lt;/p&gt;&#xA;" OwnerUserId="11130" LastActivityDate="2015-04-15T07:43:52.860" Title="How many would die after E slightly pushes the structure in front of him?" Tags="|calculation-puzzle|visual|" AnswerCount="1" CommentCount="5" ClosedDate="2015-04-15T11:18:36.747" ContentLicense="CC BY-SA 3.0" />
  <row Id="12131" PostTypeId="2" ParentId="12129" CreationDate="2015-04-15T07:34:11.433" Score="3" Body="&lt;p&gt;Depending on the ratio of weight between the different shapes, technically any of the stickmen has a chance to die.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While commonly the accepted answer would be that A and E live, and the others die, given this is a well-known question someone decided to go for the full answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=LKLzv8INA3E&quot; rel=&quot;nofollow&quot;&gt;Youtube video explaining the different weight ratio scenario's and their impact&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While usually I'd follow the general policy here not to provide link-only answers, in here it'd seem a better fit to point to a full answer with explanation then to just type out someone elses work and take credit for it.&lt;/p&gt;&#xA;" OwnerUserId="2156" LastActivityDate="2015-04-15T07:34:11.433" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12132" PostTypeId="2" ParentId="11943" CreationDate="2015-04-15T08:10:49.783" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In a strong gravitational field.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or any other similar non-euclidean geometry that is sufficiently warped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This video explains it with a nice &lt;a href=&quot;https://www.youtube.com/watch?v=MTY1Kje0yLg&quot; rel=&quot;nofollow&quot;&gt;visual demonstration&lt;/a&gt; using strechy fabric.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Imagine a circular trampoline with a weight at the centre. The edge of the trampoline remains fixed but as the weight increases the surface becomes more curved. The distance in straight line following the trampolines surface from the edge to the centre (its radius) becomes greater. When it is finally curved just the right amount the the ratio of the circumference to the diameter will reach $\sqrt2$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="12165" LastEditDate="2015-05-22T10:49:08.293" LastActivityDate="2015-05-22T10:49:08.293" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12134" PostTypeId="2" ParentId="12119" CreationDate="2015-04-15T08:54:19.383" Score="64" Body="&lt;p&gt;Satan should stick to fiddling. You will win, and here is a simple proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the game $n$ turns at a time. After each cycle of $n$ turns, all the coins are in their original position (though not necessarily flipped the same way).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Replace $H$ with $0$ and $T$ with $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In each cycle, you flip all $1$'s to $0$'s, until Satan flips a $0$ to a $1$. Once Satan makes a flip, you stop and leave the rest of this cycle's coins alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Satan must always make a flip during a cycle. If not, then you have just flipped all the coins to $0$, and you win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Read the sequence of coins as a binary number. Each cycle's play starts at the ones place and progresses to the largest place. Satan makes the last flip in each cycle, and that flip flips a $0$ to a $1$. Therefore, after each cycle, the number gets larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it can't get larger forever. After at most $2^n$ cycles, it reaches $111...1$. Put on your smuggest face and flip all the coins for a well-deserved win.&lt;/p&gt;&#xA;" OwnerUserId="1752" LastEditorUserId="1752" LastEditDate="2015-04-15T09:15:59.193" LastActivityDate="2015-04-15T09:15:59.193" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="12136" PostTypeId="2" ParentId="251" CreationDate="2015-04-15T09:19:22.097" Score="3" Body="&lt;p&gt;I used the same brute force approach described by the other answers, but I thought I'd share my (Scala) code to show how the answer can be computed with a functional language:&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;case class Combination(x:Int, y: Int) {&#xA;  def sum = x + y&#xA;  def prod = x * y&#xA;}&#xA;&#xA;val allCombinations = for {&#xA;  x &amp;lt;- 2 to 98&#xA;  y &amp;lt;- (x + 1) to (100 - x - 1)&#xA;} yield Combination(x, y)&#xA;// allCombinations.length == 2304&#xA;&#xA;val combinationsBySum = allCombinations.groupBy(_.sum)&#xA;val combinationsByProd = allCombinations.groupBy(_.prod)&#xA;&#xA;// Step 1 -&amp;gt; The product matches multiple combinations (it's not semiprime):&#xA;val possibleProds = combinationsByProd.filter(_._2.length != 1).keys.toSet&#xA;//possibleProds.size == 563&#xA;&#xA;// Step 2 -&amp;gt; The sum implies that the product is in possibleProds:&#xA;val possibleSums = combinationsBySum.filter(&#xA;  _._2.forall(comb =&amp;gt; possibleProds.contains(comb.prod))&#xA;).keys.toSet&#xA;// possibleSums.size == 10&#xA;&#xA;// Step 3 -&amp;gt; Only one combination matching the product has a possible sum per step 2:&#xA;val possibleProds2 = possibleProds.filter{(prod) =&amp;gt;&#xA;  combinationsByProd(prod).count(comb =&amp;gt; possibleSums.contains(comb.sum)) == 1&#xA;}&#xA;// possibleProds2.size == 86&#xA;&#xA;// Statement 4 -&amp;gt; Only one combination matching the sum has a possible product per step 3:&#xA;val possibleSums2 = possibleSums.filter{(sum) =&amp;gt;&#xA;  combinationsBySum(sum).count(comb =&amp;gt; possibleProds2.contains(comb.prod)) == 1&#xA;}&#xA;&#xA;possibleSums2.flatMap(combinationsBySum).filter(comb =&amp;gt; possibleProds2.contains(comb.prod))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2461" LastActivityDate="2015-04-15T09:19:22.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12138" PostTypeId="1" AcceptedAnswerId="12147" CreationDate="2015-04-15T09:50:03.140" Score="6" ViewCount="1632" Body="&lt;p&gt;You're trying to work out how to maximize the wages you receive in the month of February (in a non-leap year). You must work at least 5 days a week and at least 5 hours a day, but you can also work on Saturdays (not Sundays) if you wish and you may work up to 8 hours on each of Monday to Saturday. You are paid $n$ pounds (dollars, euros, whatever) per hour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you fill out your time-sheet at the end of the month, you know that your boss will pay you &lt;em&gt;either&lt;/em&gt; for the number of hours you write down &lt;em&gt;or&lt;/em&gt;, with probability $p$, for the same number of hours with the last two digits swapped. An example to demonstrate: if you work 134 hours in the month, you'll be paid $134n$ with probability $1-p$ and $143n$ with probability $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How many hours should you work to maximize your expected wages?&lt;/strong&gt; (The answer should depend on $p$.)&#xA;&lt;hr&gt;&#xA;If that puzzle was too easy, here's a twist on it. Let's say your boss is still absent-minded enough to get the last two digits mixed up with probability $p$, but he's sharp enough to notice if the amount he pays you is more than the maximum you could possibly earn in the month. So if you work 159 hours, you'll be paid $159n$ with probability $1$ because $195n$ is too large. &lt;strong&gt;How does your answer change?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="8874" LastEditDate="2015-10-02T14:22:05.310" LastActivityDate="2015-10-02T14:22:05.310" Title="How to maximize your wages?" Tags="|mathematics|probability|optimization|" AnswerCount="5" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="12139" PostTypeId="1" CreationDate="2015-04-15T10:04:25.253" Score="1" ViewCount="322" Body="&lt;p&gt;I swim on the soil.&lt;br&gt;&#xA;I walk on the water.&lt;br&gt;&#xA;I fly under the ground.&lt;br&gt;&#xA;I sit on the air.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="11130" LastEditorUserId="11130" LastEditDate="2015-04-15T15:46:33.693" LastActivityDate="2015-04-15T18:37:23.660" Title="My locomotions are unusual. What am I?" Tags="|riddle|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12141" PostTypeId="2" ParentId="12139" CreationDate="2015-04-15T10:27:25.017" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Water&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I swim on the soil.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A water find its way through the soil by swimming.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I walk on the water.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure. Iceberg&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I fly under ground.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; steam formed due to heat under-ground. (Wild guess)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I sit on the air&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; vapour&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanations are a little hazy. Will try to add more base to it :)&lt;/p&gt;&#xA;" OwnerUserId="2656" LastActivityDate="2015-04-15T10:27:25.017" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12142" PostTypeId="2" ParentId="12138" CreationDate="2015-04-15T10:31:47.277" Score="3" Body="&lt;p&gt;Oke lets start with the basics, February has &lt;strong&gt;28 days&lt;/strong&gt; which are exactly &lt;strong&gt;4 weeks&lt;/strong&gt; (how convenient).&#xA;You must work &lt;strong&gt;5 days * 5 hours * 4 weeks&lt;/strong&gt; = &lt;strong&gt;100 hours&lt;/strong&gt; at the very least&#xA;and (assuming volunteering to work on Saturdays still gets you payed)&#xA;this means &lt;strong&gt;6 days * 8 hours * 4 weeks&lt;/strong&gt; = &lt;strong&gt;192 hours&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now we have the basics:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;minimum 100 hours.&#xA;maximum 192 hours.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;we use the following letters for our upcoming formula:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x = worked hours.&#xA;y = swapped hours.&#xA;p = probability.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I am just going to make things easy and say the money we earn is $1 for each hour. Now for the example lets state that p = 50% (to make it easy).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; to see what would give us the most profit we need to figure out the difference between x and y so &lt;strong&gt;(x - y)&lt;/strong&gt;. now since we are dealing with a 50-50 chance we can simply divide by 2 to get the average amount we would get if this is done multiple times. so &lt;strong&gt;(x - y)/2&lt;/strong&gt; what we need to do now is add the average to the worked hours.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefor the formula is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;x + ((x - y) / 2)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;now to get our best chance of working as little as possible and earning the most is a question of brute force, from brute force follows the best possible option is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 183 start by calculating 189 - 198 = 9 divide by 2 to get the average: 4.5 add to our start (the salary we would always get) 189 + 4.5 = 193.5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the amount of time you need to work each day (6 days) with a probability of 50% is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 7 hours 50 mins&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As it turns out adapting this formula for a random probability value is easy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; x + ((x - y) / &lt;strong&gt;p&lt;/strong&gt;) simply replace the 2 by p.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Through brute force we find:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; that after p = 24 you will earn the most with working 192 hours. which means a probability of 1/24 = 4.166....%&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and thanks to @Ankor we know this to be true since he has written a nice proof for it, something I'm not that good at :P&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For those who are interested in the c# code, keep in mind it will only work for x &amp;lt; 100, I'm not trying to win a beauty contest, haha:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;static void Main(string[] args)&#xA;{&#xA;    for (double p = 1.0; p &amp;lt; 100; p++)&#xA;    {&#xA;        double highest = 0;&#xA;        double bestX = 0;&#xA;&#xA;        for (double x = 0.0; x &amp;lt; 93; x++)&#xA;        {&#xA;            double a = Math.Floor(x / 10);&#xA;            double b = x - (a * 10);&#xA;            if (a != b)&#xA;            {&#xA;                double y = (b * 10) + a;&#xA;                double result = x + ((y - x) / p);&#xA;                if (result &amp;gt; highest)&#xA;                {&#xA;                    highest = result;&#xA;                    bestX = x;&#xA;                }&#xA;            }&#xA;        }&#xA;        Console.WriteLine(&quot;highest earning: &quot; + highest + &quot; with probability p: &quot; + p + &quot; for hours: &quot; + bestX);&#xA;    }&#xA;    Console.Read();&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="9835" LastEditorUserId="9835" LastEditDate="2015-04-15T13:36:46.840" LastActivityDate="2015-04-15T13:36:46.840" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12144" PostTypeId="2" ParentId="12106" CreationDate="2015-04-15T10:53:50.897" Score="2" Body="&lt;p&gt;I think it could be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Ice&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All I remember was being at a store and then all of a sudden someone was taking me to this party!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Someone bought the ice bag&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some people couldn't stop staring at me. (Maybe they were impressed when I was holding onto everyone's drink!)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Holding onto everyone's drink discards the camera, and leaves us bottles, glasses or ice&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later that night, some guy slammed me onto the table and then started yelling, it was crazy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since ice starts to melt, we have to slam it in order to take ice chunks&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It was one wild night...but now that I'm older, things seem to be a little hazy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As the party ends, the ice either melts or gets into a fridge, getting hazy in either a metaphorical or literal way.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11438" LastActivityDate="2015-04-15T10:53:50.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12145" PostTypeId="2" ParentId="12138" CreationDate="2015-04-15T11:24:56.937" Score="0" Body="&lt;p&gt;After a catastrophic misreading of the question I wrote an answer all about how much different values of n affect the outcome. Needless to say they don't, so here as compensation for anyone who wasted their time reading it, is a graphical representation of the possible amounts of pay. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/s1UQO.png&quot; alt=&quot;a nice picture of the problem&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have my doubts about exactly how accurate it is, but if a picture is worth a thousand words, even if some are wrong it's closer than what I originally wrote.&lt;/p&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="11297" LastEditDate="2015-04-16T02:11:17.027" LastActivityDate="2015-04-16T02:11:17.027" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12146" PostTypeId="2" ParentId="11983" CreationDate="2015-04-15T11:43:08.773" Score="2" Body="&lt;p&gt;You are out of luck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 2 observations you can make:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: An odd number to the power 3 will always be odd.&#xA;And if you're packiging it in four-pack holders, you're subtracting an even from an odd number so there will always be an odd number of cans left. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2: Each even number (n) to the power 3 can be rewriten as: 2^3 * (n/2)^3 so each even number will be divisible by 8, which means it fits exactly in 4-pack holders without any cans left.&lt;/p&gt;&#xA;" OwnerUserId="11442" LastActivityDate="2015-04-15T11:43:08.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12147" PostTypeId="2" ParentId="12138" CreationDate="2015-04-15T11:45:44.870" Score="8" Body="&lt;p&gt;As @VincentAdvocaat said, our choice reduces to the number of hours we work within the month, where the minimum possible is &lt;span class=&quot;math-container&quot;&gt;$4*5*5=100$&lt;/span&gt; and the maximum is &lt;span class=&quot;math-container&quot;&gt;$4*6*8=192$&lt;/span&gt;. Let's define &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt; is the tens digit of this number and &lt;span class=&quot;math-container&quot;&gt;$Y$&lt;/span&gt; as the units digit. The average wage is therefore (&lt;span class=&quot;math-container&quot;&gt;$n = 1$&lt;/span&gt;):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$W = 100 + (1 - p)(10X + Y) + p(10Y + X)$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ \ \ \ = 100 + (10 - 9p)X + (1 + 9p)Y ,$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;and this is what we want to maximise. We can think of &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$Y$&lt;/span&gt; as (continuous) real numbers between &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$9$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$W$&lt;/span&gt; as a two-variable function. If it helps, you can think of &lt;span class=&quot;math-container&quot;&gt;$W$&lt;/span&gt; as the height (&lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt; coordinate) of a &lt;span class=&quot;math-container&quot;&gt;$2d$&lt;/span&gt; manifold (surface) in &lt;span class=&quot;math-container&quot;&gt;$3d$&lt;/span&gt;. The function is linear, and therefore the surface is flat.&lt;/p&gt;&#xA;&lt;p&gt;Since the two coefficients, &lt;span class=&quot;math-container&quot;&gt;$10-9p$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$1+9p$&lt;/span&gt;, must be positive, the lowest point is at &lt;span class=&quot;math-container&quot;&gt;$(0, 0)$&lt;/span&gt; and the surface goes up in both directions. Namely, it's always beneficial to enlarge &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$Y$&lt;/span&gt;. So the maximum must be on a 'top-right' corner of this &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt;-&lt;span class=&quot;math-container&quot;&gt;$Y$&lt;/span&gt; plane. There are two such corners: &lt;span class=&quot;math-container&quot;&gt;$(8, 9)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$(9, 2)$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The only question is which of them has a higher &lt;span class=&quot;math-container&quot;&gt;$W$&lt;/span&gt;, and this depends on &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$W(8,9) = 100 + 89(1 - p) + 98p = 189 + 9p$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$W(9,2) = 100 + 92(1 - p) + 29p = 192 - 63p$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$189 + 9p &amp;gt; 192 - 63p$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$72p &amp;gt; 3$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$p &amp;gt; 1/24$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The final answer is therefore:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; For &lt;span class=&quot;math-container&quot;&gt;$p &amp;lt; 1/24$&lt;/span&gt;, work &lt;span class=&quot;math-container&quot;&gt;$192$&lt;/span&gt; hours. Otherwise, work &lt;span class=&quot;math-container&quot;&gt;$189$&lt;/span&gt; hours. (For &lt;span class=&quot;math-container&quot;&gt;$p = 1/24$&lt;/span&gt; the wage is equal, and to rand al'thor's advice I choose to work less hours...)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Second part:&lt;/strong&gt; The change creates six special &lt;span class=&quot;math-container&quot;&gt;$(X, Y)$&lt;/span&gt; values: &lt;span class=&quot;math-container&quot;&gt;$(3, 9)$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$(4, 9)$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$(5, 9)$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$(6, 9)$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$(7, 9)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$(8, 9)$&lt;/span&gt;, with &lt;span class=&quot;math-container&quot;&gt;$W = 139, 149, 159, 169, 179$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$189$&lt;/span&gt;, respectively. Taking these out of our &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt;-&lt;span class=&quot;math-container&quot;&gt;$Y$&lt;/span&gt; plane, we get two more 'top-left' corners: &lt;span class=&quot;math-container&quot;&gt;$(9, 2)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$(8, 8)$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;We now compare &lt;span class=&quot;math-container&quot;&gt;$W$&lt;/span&gt; for the three corners, &lt;span class=&quot;math-container&quot;&gt;$(9, 2)$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$(2, 9)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$(8, 8)$&lt;/span&gt; with the special value &lt;span class=&quot;math-container&quot;&gt;$(8, 9)$&lt;/span&gt; (the special value with highest &lt;span class=&quot;math-container&quot;&gt;$W$&lt;/span&gt;). The value &lt;span class=&quot;math-container&quot;&gt;$(8, 8)$&lt;/span&gt; is never the best, as &lt;span class=&quot;math-container&quot;&gt;$188 &amp;lt; 189$&lt;/span&gt;. Let's check the other two corners:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$W(9,2) = 192 - 63p &amp;gt; 189$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$63p &amp;lt; 3$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$p &amp;lt; 1/21$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$W(2,9) = 129 + 63p &amp;gt; 189$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$63p &amp;gt; 60$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$p &amp;lt; (1 - 1/21)$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The final answer is therefore:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; For &lt;span class=&quot;math-container&quot;&gt;$p &amp;lt; 1/21$&lt;/span&gt;, work &lt;span class=&quot;math-container&quot;&gt;$192$&lt;/span&gt; hours. For &lt;span class=&quot;math-container&quot;&gt;$p \ge (1 - 1/21)$&lt;/span&gt;, work &lt;span class=&quot;math-container&quot;&gt;$129$&lt;/span&gt; hours. Otherwise, work &lt;span class=&quot;math-container&quot;&gt;$189$&lt;/span&gt; hours. (Again, for equal wages I choose to work less.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10684" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-15T19:31:43.083" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12149" PostTypeId="2" ParentId="12138" CreationDate="2015-04-15T12:30:23.363" Score="1" Body="&lt;p&gt;Tried to brute force the scenario and got&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;strong&gt;189 hours&lt;/strong&gt; is the most efficient answer with an average of 94.5 across  all values of &lt;code&gt;p&lt;/code&gt; with precision up to 2 decimals.&#xA; PHP program can be found  &lt;a href=&quot;http://pastebin.com/5d1PKTq1&quot; rel=&quot;nofollow noreferrer&quot;&gt;HERE&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Considering you are &lt;strong&gt;getting paid on Saturdays&lt;/strong&gt; too.&lt;/p&gt;&#xA;" OwnerUserId="5437" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-15T12:41:32.830" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12151" PostTypeId="1" AcceptedAnswerId="12153" CreationDate="2015-04-15T13:32:16.350" Score="5" ViewCount="514" Body="&lt;p&gt;(Inspired by House of Lies intro)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What am I?&lt;br&gt;&#xA;  You already know what I am!&lt;br&gt;&#xA;  I'm not a executive consultant, and I'm not Marty Kaan.&lt;br&gt;&#xA;  All I have to do is make you think you don't know.&lt;br&gt;&#xA;  I'll confuse you with my words and make you think you don't know!&lt;br&gt;&#xA;  But I know you, you're better than this. So catch up!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I hope this one proves to be a little more difficult than my previous riddle, but I think someone will get it.&lt;/p&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="5373" LastEditDate="2015-04-15T15:26:58.783" LastActivityDate="2015-05-23T04:01:11.417" Title="Riddle inspired by House of Lies intro" Tags="|riddle|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12152" PostTypeId="2" ParentId="12151" CreationDate="2015-04-15T13:50:36.790" Score="-1" Body="&lt;p&gt;I think you are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; doubt.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cannot place it on a particular sentence but the overall riddle seems to be making me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; doubt&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;about the thing i already know. but the following stence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;All I have to do is make you think you don't know.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; making someone think they don't know is making them doubt&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;same goes for the next line:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'll confuse you with my words and make you think you don't know!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To further back this up the other sentences are put there as a diversion or distraction, come to think of it those (diversion/distraction) could be the answer as well xD&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9835" LastEditorUserId="39" LastEditDate="2015-05-23T04:01:11.417" LastActivityDate="2015-05-23T04:01:11.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12153" PostTypeId="2" ParentId="12151" CreationDate="2015-04-15T14:15:39.483" Score="5" Body="&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A riddle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You already know what I am!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because you told me in the title.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not a executive consultant, and I'm not Marty Kaan.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Riddles are not people.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All I have to do is make you think you don't know.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Riddles are not straightforward, so you will doubt yourself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll confuse you with my words and make you think you don't know!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Riddles often have confusing language to obfuscate the answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I know you, you're better than this. So catch up!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Thanks for believing in me!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4385" LastEditorUserId="10399" LastEditDate="2015-04-15T15:10:09.510" LastActivityDate="2015-04-15T15:10:09.510" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12154" PostTypeId="1" AcceptedAnswerId="12158" CreationDate="2015-04-15T15:05:48.560" Score="6" ViewCount="1791" Body="&lt;p&gt;In a kingdom, the King did not allow any citizen to visit the world outside. Also only a person with a proper paperwork was allowed to enter or he was sent back. A wooden bridge was the only connection between the kingdom and the rest of the world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The king had appointed a sharpshooter who would check the bridge every five minutes for escapees. After checking, he would go back to his hut and return exactly after five minutes again. You can't cross the bridge in less than 9 minutes, so the sharpshooter would be guaranteed to catch anyone trying to make a break for the other side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A merchant who was a citizen of the kingdom was able to cross the bridge and escape the kingdom without harming the shooter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How?&lt;/p&gt;&#xA;" OwnerUserId="9888" LastEditorDisplayName="user88" LastEditDate="2015-04-16T06:10:52.847" LastActivityDate="2015-04-16T06:10:52.847" Title="The isolated kingdom" Tags="|logical-deduction|" AnswerCount="5" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="12155" PostTypeId="2" ParentId="12151" CreationDate="2015-04-15T15:06:48.193" Score="1" Body="&lt;p&gt;You're a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Question&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You already know what I am!&lt;br&gt;&#xA;I'm not a executive consultant, and I'm not Marty Kaan.&lt;br&gt;&#xA;All I have to do is make you think you don't know.&lt;br&gt;&#xA;I'll confuse you with my words and make you think you don't know!&lt;br&gt;&#xA;But I know you, you're better than this. So catch up!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You're answering a question right now. A question makes you feel like you don't know the answer. Questions often have complex words or weird sentences. But if you're good, you can come up with a good answer to it!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11454" LastActivityDate="2015-04-15T15:06:48.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12156" PostTypeId="2" ParentId="12154" CreationDate="2015-04-15T15:14:00.523" Score="6" Body="&lt;p&gt;When the shooter went away, the merchant started crossing the bridge. After 5 minutes, he has already passed the middle of the bridge.&lt;br&gt;&#xA;When the shooter is back, the merchant turns back and &lt;strong&gt;slowly&lt;/strong&gt; returns to the fortress, so slow that he doesn't cross again the middle of the bridge. When the sharpshooter goes away again, the merchant turns back and escapes as soon as possible! &#xA;If, for some reason, the shooter decides to wait for the merchant to arrive, the business-man can just approach the gates, ask to enter, be rejected and go away without problems!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5Gatt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-04-15T15:29:40.413" LastActivityDate="2015-04-15T15:29:40.413" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12157" PostTypeId="2" ParentId="12154" CreationDate="2015-04-15T15:16:27.613" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The king does not allow any &lt;strong&gt;citizen&lt;/strong&gt; to leave to visit the outside world. However, the merchant is a foreigner doing business in the kingdom. He simply showed the sharpshooter his own country's passport and walked across the bridge unchallenged.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1504" LastActivityDate="2015-04-15T15:16:27.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12158" PostTypeId="2" ParentId="12154" CreationDate="2015-04-15T15:19:27.467" Score="18" Body="&lt;p&gt;Here's my idea:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The merchant starts walking across the bridge at the point that the sharpshooter enters his hut. Four and a half minutes pass, and the merchant has reached the middle of the bridge, at which point he &lt;strong&gt;turns around.&lt;/strong&gt; After thirty seconds, he is exactly five minutes towards the castle from the other end of the bridge, which is where he would be if he started walking from the other side when the sharpshooter left. The sharpshooter sees him walking towards the castle and does not shoot, instead turning to go back to his hut. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From here, the merchant has two options:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1) The merchant turns on heel and walks away, with exactly five minutes of bridge between him and freedom. This pushes the timing a bit, but if he moves quickly enough he will escape.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;2) The merchant continues to the castle, pretending to not have correct documents. At this point he is sent back, during which he will not be shot by the sharpshooter because he is not attempting to escape.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Either way, the merchant escapes across the bridge to freedom.&lt;/p&gt;&#xA;" OwnerUserId="11322" LastActivityDate="2015-04-15T15:19:27.467" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12160" PostTypeId="2" ParentId="12004" CreationDate="2015-04-15T16:02:08.870" Score="2" Body="&lt;p&gt;Here are three 10-letter adverbs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How we behave in a manner of worth.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wiktionary.org/wiki/vauntingly&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;vauntingly&lt;/strong&gt;&lt;/a&gt;: we boast about our worth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With a single change, that worth is no more.&#xA;Instead we consider the ones before.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wiktionary.org/wiki/hauntingly&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;hauntingly&lt;/strong&gt;&lt;/a&gt;: ghosts were people before.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After one more change, seen with different eyes,&#xA;We behave in a way we won't trivialize.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wiktionary.org/wiki/tauntingly&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;tauntingly&lt;/strong&gt;&lt;/a&gt;: it isn't easy to trivialise or ignore a taunt.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2015-04-15T16:02:08.870" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12161" PostTypeId="2" ParentId="10968" CreationDate="2015-04-15T16:09:12.200" Score="3" Body="&lt;p&gt;The number is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 913&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; that is the number of different non-self-intersecting paths there are that start in the upper left corner and end in the lower right corner.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Take note that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the graph is directed. Just because you can go from (0,0) to (0,1), doesn't mean you can go from (0,1) to (0,0). (with the single exception of the fourth and fifth cells in the first row, but they were &lt;a href=&quot;https://puzzling.stackexchange.com/questions/10968/a-code-from-the-lines#comment35416_10968&quot;&gt;a mistake&lt;/a&gt;). This was what Ivo Beckers' illustration was missing; &quot;For every single [line], he has forgotten [to add] two lines [to the end of the current line, to make it into an arrow].&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1497" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2015-04-15T16:09:12.200" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12162" PostTypeId="2" ParentId="12139" CreationDate="2015-04-15T16:11:03.463" Score="5" Body="&lt;p&gt;The locomotion bit made me think it would need to be a living thing, but I've decided to go a different way.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Oil&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It swims through the soil as a liquid, it doesn't mix with water, yet walks on its top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When under ground it is under extreme pressure and can burst (fly) through the ground or become natural gas. It sits on the air because natural gas is lighter than air.&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="10281" LastEditDate="2015-04-15T18:37:23.660" LastActivityDate="2015-04-15T18:37:23.660" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12163" PostTypeId="2" ParentId="1931" CreationDate="2015-04-15T16:22:17.613" Score="6" Body="&lt;p&gt;The encoding of BF code &lt;em&gt;can&lt;/em&gt; be reversed, but it isn't a simple task.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Halting_problem&quot; rel=&quot;noreferrer&quot;&gt;Halting Problem&lt;/a&gt; is the first to interfere. There's no way to computationally guarantee that you'll determine which answer is the correct one, because you have no way of determining whether the program you decode actually exits. Even so, perhaps someone evil has decided to make code that &lt;em&gt;doesn't&lt;/em&gt; exit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, these problems can be mitigated by effective selection of output. Any code of this type can be broken into its respective components. For BF, we have:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;code&gt;+&lt;/code&gt; - add one to memory&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;-&lt;/code&gt; - subtract one from memory&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;&amp;gt;&lt;/code&gt; - increment pointer&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;&amp;lt;&lt;/code&gt; - decrement pointer&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;[&lt;/code&gt; - open a loop&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;]&lt;/code&gt; - close a loop&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;,&lt;/code&gt; - await user input&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;.&lt;/code&gt; - output to screen&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are a few reasonable conclusions we can make:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;code&gt;]&lt;/code&gt; can never have appeared more times than &lt;code&gt;[&lt;/code&gt; (otherwise the code would terminate with an error)&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;[&lt;/code&gt; and &lt;code&gt;]&lt;/code&gt; must appear an equal number of times. (The reason I list this separately is that someone &lt;em&gt;might&lt;/em&gt; stick a stack of &lt;code&gt;[&lt;/code&gt; in front to throw you off, but that seems unlikely and evil.)&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;+&lt;/code&gt; and &lt;code&gt;-&lt;/code&gt; are the most likely - possibly the only - terms that will appear multiple times in a row&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;.&lt;/code&gt; and &lt;code&gt;,&lt;/code&gt; can be expected to appear the least in code&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;&amp;gt;&lt;/code&gt; and &lt;code&gt;&amp;lt;&lt;/code&gt; can be expected to appear near &lt;code&gt;[, ]&lt;/code&gt; with very high consistency.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The most important thing we can determine about this output is that &lt;code&gt;[&lt;/code&gt; and &lt;code&gt;]&lt;/code&gt; must appear an equal number of times. Otherwise, the code won't run. (Someone can be evil and stick arbitrary &lt;code&gt;[&lt;/code&gt; in front of their code, but that is unlikely.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this method, you'll have a better idea of where to start. Usually, once you've identified various pairs, you have no more than 8 or 16 programs you can manually check. Doing so isn't hard - 90% of the time it will be obvious that a program won't do anything just by looking at it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, if someone knew you were using this method to approach the problem, I don't think they would leave it this open for you to attack. It's fully possible to design a problem that can't be approached in this way; that just means you'll have to check a ton more combinations. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There's another approach to solving these problems that doesn't involve any complex searching of spaces. It requires a pattern (i.e. alphanumeric) that you expect the output to match, and it requires a reasonable computation time from the BF program. With these two in hand:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Find the various terms that could be brackets (&lt;code&gt;[, ]&lt;/code&gt;) (i.e. an equal number of them)&lt;/li&gt;&#xA;&lt;li&gt;Set up every permutation of interpreted BF&lt;/li&gt;&#xA;&lt;li&gt;Run them all. &lt;/li&gt;&#xA;&lt;li&gt;If the program hasn't finished after a few million cycles (or billion - cycles are cheap nowadays), kill it and call it a failure.&lt;/li&gt;&#xA;&lt;li&gt;If the program finishes, pattern-match the memory dump to the pattern you expected. If it matches, print it and move on. If it doesn't, fail it and move on. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Supposing that you find a unique pair for brackets (which is likely), and have a guess at &lt;code&gt;.&lt;/code&gt; and &lt;code&gt;,&lt;/code&gt; (also likely), that gives $2*2*4!=96$ combinations to test, which really isn't all that bad. It shouldn't take more than a few seconds to a minute to run on a reasonably fast computer.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorDisplayName="user20" LastEditDate="2015-04-15T16:31:54.593" LastActivityDate="2015-04-15T16:31:54.593" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12164" PostTypeId="2" ParentId="12139" CreationDate="2015-04-15T17:11:24.543" Score="0" Body="&lt;p&gt;I swim on the soil.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a duck claps his wings when running on the ground giving the idea of a swimming motion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I walk on the water.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; his feet make a walking motion when in the water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I fly under the ground.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; well obviously a duck can fly, caves are under the ground, so technically they can fly underground.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I sit on the air.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; he could be gliding down, which granted isn't quite sitting but you could interpret it that way if you really want.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9835" LastEditorUserId="9835" LastEditDate="2015-04-15T17:55:22.213" LastActivityDate="2015-04-15T17:55:22.213" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12166" PostTypeId="1" AcceptedAnswerId="12167" CreationDate="2015-04-15T18:02:20.370" Score="3" ViewCount="1341" Body="&lt;p&gt;A man is being chased into the woods by an evil king. The king has stated the man will not be chased once he is leaving the forest. The king is slightly mad, but knows the entire forest is too long for the man to keep outrunning him and his guards, who are on horse back. That is, before the man reaches the end of the forest they will have caught up with him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How long does the man have to run for the king and his guards to stop chasing him?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;It's my first (semi) riddle; I just felt like posting it.  I hope these 'joke' kinds of riddles are allowed.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="9835" LastEditorUserId="5373" LastEditDate="2015-04-15T19:06:13.953" LastActivityDate="2015-04-16T09:33:24.097" Title="The man and the woods" Tags="|story|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12167" PostTypeId="2" ParentId="12166" CreationDate="2015-04-15T18:06:39.257" Score="8" Body="&lt;p&gt;How long? Well I guess that depends on the mans speed, but how far:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; At most, half the length of the forest.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the king has stated the man will not be chased once he is leaving the forest&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After passing a 'half way' point, he is no longer running 'into' the forest, but rather running out of it so he will no longer be chased.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alternate, albeit silly answers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The man is driving so he doesn't have to run at all! Maybe the king could have had the forest transformed into a treadmill in which the man can NEVER escape! The man starts jumping around the place like a maniac continually 'leaving' the forest vertically until he escapes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-04-15T18:20:37.617" LastActivityDate="2015-04-15T18:20:37.617" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12168" PostTypeId="2" ParentId="12166" CreationDate="2015-04-15T18:06:40.623" Score="13" Body="&lt;p&gt;He can enter the forest, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;turn around, and leave the forest. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So not long at all.&lt;/p&gt;&#xA;" OwnerUserId="11150" LastEditorUserId="4204" LastEditDate="2015-04-15T20:41:10.693" LastActivityDate="2015-04-15T20:41:10.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12169" PostTypeId="2" ParentId="11780" CreationDate="2015-04-15T18:27:35.550" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I could be completely wrong, still i'm posting this (possible) solution which i have got after hit and trial and messing with the grid here and there :P&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;##.|#.#|#.#&#xA;.##|.##|.##&#xA;#.#|#.#|##.&#xA;---+---+---&#xA;###|###|#.#&#xA;.#.|#.#|.#.&#xA;#.#|###|###&#xA;---+---+---&#xA;.##|#.#|#.#&#xA;##.|##.|##.&#xA;#.#|#.#|.##&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;! That would be &lt;strong&gt;55&lt;/strong&gt; blocks&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;To verify, I took a few solved sudoko's and used a sudoku solver to check if they have a unique solution.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;As far as my technique goes... there is a better solution... I'll try that only if the one I've provided is correct&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Then again... I don't have a proof... but go ahead try it  :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9346" LastActivityDate="2015-04-15T18:27:35.550" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12170" PostTypeId="2" ParentId="12138" CreationDate="2015-04-15T19:05:16.297" Score="2" Body="&lt;p&gt;When I posted this question, I didn't know the answer. Later on I worked at it for a little while and came out with a simple proof which I think is nice enough to post as a self-answer. Before you read it, though, go and upvote &lt;a href=&quot;https://puzzling.stackexchange.com/a/12147/5373&quot;&gt;Angkor's answer&lt;/a&gt;.&#xA;&lt;hr&gt;&#xA;The multiplier $n$ is irrelevant, so WLOG we can assume $n=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By working for 189 hours, you can ensure a &lt;em&gt;minimum&lt;/em&gt; possible wage of $189$ and therefore an expected wage of at least $189$. The only way you could possibly earn more than this is by working for 190, 191, 192, 109, 119, or 129 hours. So &lt;strong&gt;one of these seven numbers must be optimal&lt;/strong&gt;. The expected wage in each case is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;189 hours: $198p+189(1-p)=189+9p$ in first problem, $189$ in second problem&lt;/li&gt;&#xA;&lt;li&gt;&lt;strike&gt;190 hours: $109p+190(1-p)=190-81p&amp;lt;192-63p$ &lt;/strike&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strike&gt;191 hours: $119p+191(1-p)=191-72p&amp;lt;192-63p$ &lt;/strike&gt;&lt;/li&gt;&#xA;&lt;li&gt;192 hours: $129p+192(1-p)=192-63p$&lt;/li&gt;&#xA;&lt;li&gt;&lt;strike&gt;109 hours: $190p+109(1-p)=109+81p&amp;lt;129+63p$ &lt;/strike&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strike&gt;119 hours: $191p+119(1-p)=119+72p&amp;lt;129+63p$ &lt;/strike&gt;&lt;/li&gt;&#xA;&lt;li&gt;129 hours: $192p+129(1-p)=129+63p&amp;lt;189+9p$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the first problem, you have a choice between working for 189 hours and earning $189+9p$ or working for 192 hours and earning $192-63p$. Some simple manipulation of linear inequalities shows that $189+9p&amp;lt;192-63p\Leftrightarrow p&amp;lt;\frac{1}{24}$. So you should work for &lt;strong&gt;192 hours if $p&amp;lt;\frac{1}{24}$ and 189 hours otherwise&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second problem, you have a choice between working for 189 hours and earning $189$, working for 192 hours and earning $192-63p$, or working for 129 hours and earning $129+63p$. Some simple manipulation of linear inequalities shows that $189&amp;lt;192-63p\Leftrightarrow p&amp;lt;\frac{1}{21}$ and $189&amp;lt;129+63p\Leftrightarrow p&amp;gt;\frac{20}{21}$. So you should work for &lt;strong&gt;192 hours if $p&amp;lt;\frac{1}{21}$, 129 hours if $p&amp;gt;\frac{20}{21}$, and 189 hours otherwise&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-04-15T19:05:16.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12171" PostTypeId="1" AcceptedAnswerId="12172" CreationDate="2015-04-15T19:09:09.547" Score="43" ViewCount="2384" Body="&lt;p&gt;Hamlet is preparing a play to find out the truth about his father's death. The theatre has $2015$ numbered seats and can contain all the members of King Claudius's court. He gives to each member a ticket with the number of their seat.&lt;br&gt;&#xA;King Claudius (the first to sit) doesn't look at his ticket and chooses at random a seat. All the other guests arrive one by one and if their seat is available they sit according to the ticket, if it isn't they sit at random.&lt;br&gt;&#xA;Ophelia is the last to enter the theatre and she occupies the only empty spot.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What's the probability that she will sit in the seat assigned to her by her ticket?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="5373" LastEditDate="2019-04-13T10:44:44.673" LastActivityDate="2019-04-13T10:44:44.673" Title="Hamlet and the theatre" Tags="|mathematics|story|probability|" AnswerCount="4" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12172" PostTypeId="2" ParentId="12171" CreationDate="2015-04-15T19:18:01.120" Score="34" Body="&lt;p&gt;If the king sits in his own seat, then each guest will sit in their own seat and Ophelia will always sit in her own seat. This occurs with probability $1/2015$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the king sits in Ophelia's seat, then each guest will sit in their own seat and Ophelia will end up sitting in the king's seat. This also occurs with probability $1/2015$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the king sits in any other seat, then as soon as that guest arrives, he will choose another seat at random.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the guest chooses the king's seat, then Ophelia will end up sitting in her own seat.&lt;/li&gt;&#xA;&lt;li&gt;If the guest chooses Ophelia's seat, then Ophelia will end up sitting in the king's seat.&lt;/li&gt;&#xA;&lt;li&gt;If the guest chooses another guest's seat, then as soon as the guest who was assigned &lt;em&gt;that&lt;/em&gt; seat arrives, he will also have to choose one at random, and the cycle will continue until one of the above two scenarios happens.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Since the probability is equal that each guest will choose either the king's seat or Ophelia's seat, the eventual overall probability of each event is $1/2$. The above scenario happens with probability $2013/2015$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the grand total probability that Ophelia ends up in her own seat is $1/2015 + 0 + 1/2 \times 2013/2015 = 1/2$.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-04-15T19:23:08.100" LastActivityDate="2015-04-15T19:23:08.100" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12173" PostTypeId="2" ParentId="12171" CreationDate="2015-04-15T19:25:40.087" Score="1" Body="&lt;p&gt;My guess and an attempt at a proof:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; P(x) = 1 - [1/2015 + ΣP(1/(2015-x))] Where x is the seat # of the person who's seat was taken. Initially there is a 1 in 2015 chance the king will take Ophelia's seat. Then there is a 1/(2015-X) chance that the person who's seat the king took will take Ophelia's seat...Continue the process for all the peoples who's seats were taken, add all the probabilities up and you should get the chance her seat is taken. One minus that probability is the chance  the seat is empty. The end result should be 1/2 as stated by Joe Z.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-04-15T20:01:55.667" LastActivityDate="2015-04-15T20:01:55.667" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12174" PostTypeId="2" ParentId="12166" CreationDate="2015-04-15T19:55:49.470" Score="8" Body="&lt;p&gt;He will not be chased once he take a step into the forest and turn back. Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He will be leaving the forest. You answer is one step or zero.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10499" LastActivityDate="2015-04-15T19:55:49.470" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12177" PostTypeId="1" AcceptedAnswerId="12186" CreationDate="2015-04-15T20:21:37.493" Score="-6" ViewCount="242" Body="&lt;p&gt;How can these equations be true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;one and two equals twelve&lt;/p&gt;&#xA;&#xA;&lt;p&gt;two and three equals forty six&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then what's three and four? (It's not 102)&lt;/p&gt;&#xA;" OwnerUserId="11470" LastEditorUserId="11470" LastEditDate="2015-04-15T20:27:31.580" LastActivityDate="2015-06-25T17:43:55.357" Title="1 and 2 is 12 , 2 and 3 is 46 then what's 3 and 4? it's not 102" Tags="|mathematics|" AnswerCount="5" CommentCount="1" ClosedDate="2015-09-27T11:55:22.007" ContentLicense="CC BY-SA 3.0" />
  <row Id="12181" PostTypeId="1" AcceptedAnswerId="12195" CreationDate="2015-04-15T20:23:55.837" Score="1" ViewCount="332" Body="&lt;p&gt;lxwhmlnv lhlib mxbwhmlnv lbnJ&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I wonder how long it will take, should not be too difficult, but not to easy either.&#xA;Please let me know in your answer what you think of the difficulty and if it's &quot;fun&quot;, I'm going to make a lot more and want some feedback to make them better.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="9835" LastEditorUserId="9835" LastEditDate="2015-04-15T20:57:43.937" LastActivityDate="2015-04-17T15:06:17.357" Title="Mirror of men! he showers, with liberal hands," Tags="|cipher|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12182" PostTypeId="2" ParentId="12177" CreationDate="2015-04-15T20:24:37.717" Score="4" Body="&lt;p&gt;Three and four is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 102.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The algorithm is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; put the two given digits together to form a 2-digit number, and then multiply by the first of them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-04-15T20:24:37.717" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12183" PostTypeId="2" ParentId="11551" CreationDate="2015-04-15T20:28:38.080" Score="0" Body="&lt;p&gt;The image represents:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Shake the spear: Shakespeare. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I guess the answer is: &lt;a href=&quot;https://youtu.be/xsXKT5RhJf8?t=38&quot; rel=&quot;nofollow&quot;&gt;https://youtu.be/xsXKT5RhJf8?t=38&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10499" LastEditorUserId="10499" LastEditDate="2015-04-15T21:17:51.680" LastActivityDate="2015-04-15T21:17:51.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12184" PostTypeId="2" ParentId="12177" CreationDate="2015-04-15T20:32:25.767" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I note the numbers are spelled out. So, while rand al'thor's approach is probably correct, the intended answer might be 'one hundred two'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11119" LastActivityDate="2015-04-15T20:32:25.767" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12185" PostTypeId="2" ParentId="11780" CreationDate="2015-04-15T20:33:12.583" Score="4" Body="&lt;p&gt;Here's a glimmer of hope of actually &lt;em&gt;proving&lt;/em&gt; a solution optimal. This answer proves an upper bound of $45$ covered cells and shows off techniques that might improve this bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;em&gt;block&lt;/em&gt; means one of the nine outlined 3x3 subsquares. Consider what happens if we cover a pair of cells in the same block and the same row or column. This diagram has three examples.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/cOUxEm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've drawn in red &lt;em&gt;lines of influence&lt;/em&gt; exerted by each pair. In lieu of a formal definition, let's prove a theorem. &lt;br&gt;&#xA;&lt;strong&gt;Claim:&lt;/strong&gt; Every possible line of influence can only be exerted once.&lt;br&gt; &lt;strong&gt;Proof:&lt;/strong&gt; Suppose a line of influence is exerted twice. See the diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hVHaxm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pairs exerting the line form a 2x2 box. If filled with [1,2,2,1], this box has two possible solutions. So the solution is ambiguous, which is not allowed. QED: each line of influence can only be exerted once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each block row and block column contains three possible lines of influence, one for each of the (3 choose 2) ways to choose two rows/columns. So there are 18 lines of influence total.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This already gives an upper bound. One can check with casework that a block containing $n$ covered cells must exert at least $n-3$ lines of influence. Therefore, there are at most $18+3\times9=45$ covered cells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm sure we can extend these techniques to do even better. Consider the following diagram:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/iVRF0m.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It shows off two illegal configurations. On the left, we have a block column with all three lines of influence exerted by pairs in three different boxes. If filled with [1,2,1,2,1,2], this is ambiguous. So, this can never happen. On the right, we have two lines of influence &quot;reflecting&quot; off a diagonal pair in the same box. Again, if filled with [1,2,1,2,1,2], this is ambiguous, so it can never happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure the $45$ bound can be improved by taking these forbidden patterns into account, possibly using a small amount of casework. But can it be improved all the way to $33$?&lt;/p&gt;&#xA;" OwnerUserId="1752" LastEditorUserId="1752" LastEditDate="2015-04-15T20:49:22.003" LastActivityDate="2015-04-15T20:49:22.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12186" PostTypeId="2" ParentId="12177" CreationDate="2015-04-15T20:34:22.913" Score="3" Body="&lt;p&gt;I think it's:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $912$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;1\,\&amp;amp;\,2 &amp;amp;= (1\times1)(2\times1) \implies 12\\[5pt]&#xA;2\,\&amp;amp;\,3 &amp;amp;= (2\times2)(2\times3) \implies 46&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$3\,\&amp;amp;\,4 = (3\times3)(3\times4) \implies 912$$ You multiply the first number to itself as the first set of digits, then multiply the first number to the second for the second set of digits. Then put them together into one number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="13494" LastEditDate="2015-06-25T17:43:55.357" LastActivityDate="2015-06-25T17:43:55.357" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12187" PostTypeId="1" AcceptedAnswerId="12220" CreationDate="2015-04-15T20:35:01.920" Score="14" ViewCount="268807" Body="&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/11998/1-2-3-4-5-6-7-8-9-100&quot;&gt;This question&lt;/a&gt; inspired me to write the same puzzle but instead replace the &quot;= 100&quot; with &quot;= 1&quot; with similar requirements and restrictions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the expression with the fewest number of operators inserted that evaluates to 1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Restrictions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The numbers need to be in the order that's shown in the question.&lt;/li&gt;&#xA;&lt;li&gt;Only use the operators +,−,×,÷ and √ and ! (Implies that modulus &quot;%&quot;, exponent &quot;^&quot;, binomial coefficients, and other operators are not allowed).&lt;/li&gt;&#xA;&lt;li&gt;Parentheses will not be counted, so they can be used to change the order of operations.&lt;/li&gt;&#xA;&lt;li&gt;Rounding is not allowed, so it have to equal to 1.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Verify your calculations in that calculator application that comes with your PC, if it ever did came with your PC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my first time writing a puzzle here so obviously I should have thought this out a lot more instead of adding rules when situation comes.&lt;/p&gt;&#xA;" OwnerUserId="289" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2018-05-01T04:25:29.183" Title="1 2 3 4 5 6 7 8 9 = 1" Tags="|mathematics|calculation-puzzle|formation-of-numbers|" AnswerCount="12" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12192" PostTypeId="2" ParentId="12177" CreationDate="2015-04-15T20:46:31.983" Score="0" Body="&lt;p&gt;102 is also an answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is how:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$12=1^2\cdot10+1\cdot2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$46=2^2\cdot10+2\cdot3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$102=3^2\cdot10+3\cdot4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;according to this operator:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$O(x,y)=x^2\cdot10+x\cdot y$&lt;/p&gt;&#xA;" OwnerUserId="10499" LastActivityDate="2015-04-15T20:46:31.983" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12193" PostTypeId="2" ParentId="12154" CreationDate="2015-04-15T20:49:22.707" Score="1" Body="&lt;p&gt;I'm assuming that the sharpshooter isn't going to leave the scene while something is happening on the bridge.  I'm also assuming that the sharpshooter knows that the merchant is a citizen and can recognize him.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A foreigner with a large cart walks up to the bridge as the sharpshooter is leaving.  The merchant and the foreigner meet in the middle of the bridge after 4 1/2 minutes; the merchant climbs into the cart and hides.  The sharpshooter returns, asks for paperwork, and the foreigner has none; the foreigner turns around with their cart, is not shot, and leaves with the merchant.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or, less interestingly:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The merchant hides himself in his wares, which go out like normal (however that works--he certainly isn't the one taking them).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2815" LastEditorUserId="2815" LastEditDate="2015-04-16T00:42:02.550" LastActivityDate="2015-04-16T00:42:02.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12195" PostTypeId="2" ParentId="12181" CreationDate="2015-04-15T20:50:22.237" Score="5" Body="&lt;p&gt;Answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Who will guard the guards themselves?&quot;/&quot;Who watches the watchers?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Step 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The string is backwards. Reverse it to get &quot;Jnbl vnlmhwbxm bilhl vnlmhwxl&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Step 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Perform Rot7 on the string (A-&gt;H, B-&gt;I, etc.) to get &quot;Quis custodiet ipsos custodes&quot;. (Note: I found this by doing Rot1 seven times.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Step 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Quis_custodiet_ipsos_custodes%3F&quot; rel=&quot;nofollow&quot;&gt;&quot;Quis custodiet ipsos custodes&quot;&lt;/a&gt; is Latin for &quot;Who will guard the guards themselves?&quot; The phrase comes from Juvenal's &lt;a href=&quot;http://en.wikipedia.org/wiki/Satires_(Juvenal)&quot; rel=&quot;nofollow&quot;&gt;Satires&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation of title:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The title is a quote from one of Juvenal's other satires.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What I thought of the puzzle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It was fun, but fairly obvious - the last letter made step 1 obvious, and the encryption used in step 2 is fairly weak. If the string is more than just a word or two, the frequency and distribution of letters still looks like normal words (just not actual words), strongly suggesting to check that specific form of encryption.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1955" LastEditorUserId="1955" LastEditDate="2015-04-17T15:06:17.357" LastActivityDate="2015-04-17T15:06:17.357" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12196" PostTypeId="2" ParentId="12187" CreationDate="2015-04-15T20:52:15.800" Score="22" Body="&lt;p&gt;$$1+23-45-67+89$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;uses four. (I wrote a Python script.)&lt;/p&gt;&#xA;" OwnerUserId="11476" LastActivityDate="2015-04-15T20:52:15.800" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12197" PostTypeId="2" ParentId="12187" CreationDate="2015-04-15T21:02:01.227" Score="11" Body="&lt;p&gt;How many significant digits matter here for rounding? Because if it's anything less than $3,456,789$ zeroes, I can solve it in three ;)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $1+2/3456789! = ~1$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Many programming languages will evaluate it as &quot;1&quot;. Even &lt;a href=&quot;https://www.wolframalpha.com/input/?i=1%2B2%2F3456789!&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wolfram Alpha&lt;/a&gt; can't show me enough decimal digits to tell me I'm wrong ;P&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&#xA;Yes, I know this is no longer valid as of the rule change that doesn't include rounding. I didn't expect it would be allowed anyways, just figured it would be worth submitting, since it comes so infinitesimally close to 1. Besides, kgull managed to get even closer using a similar method.&lt;/p&gt;&#xA;" OwnerUserId="11395" LastEditorUserId="42054" LastEditDate="2018-05-01T04:25:29.183" LastActivityDate="2018-05-01T04:25:29.183" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12199" PostTypeId="2" ParentId="12187" CreationDate="2015-04-15T21:08:02.907" Score="0" Body="&lt;p&gt;$$1-23+45+67-89$$&#xA;uses only 4 operators.&lt;/p&gt;&#xA;" OwnerUserId="11477" LastEditorUserId="13494" LastEditDate="2015-06-25T17:36:07.637" LastActivityDate="2015-06-25T17:36:07.637" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12200" PostTypeId="1" CreationDate="2015-04-15T21:12:17.433" Score="-4" ViewCount="540" Body="&lt;p&gt;I've thrown together some fun sequences to stretch your brains out! You can feel free to provide as many additional members of the sequence as you'd like, but mostly I'm looking to see if you can solve the sequence. Here are the first three, which are all around the same difficulty:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1)&lt;/strong&gt; 1, 5, 6, 2, 5, 4, 24, 9, 3,...&lt;br/&gt;&#xA;&lt;strong&gt;2)&lt;/strong&gt; 73, 146, 219, 159, 86, 159, 232, 305, 161,...&lt;br/&gt;&#xA;&lt;strong&gt;3)&lt;/strong&gt; 1, 4, 3, 0, 0, 8, 1, 1, 0,...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final one here is a bit of a, shall we say, curveball:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4)&lt;/strong&gt; 22, 34, 35, 22, 5, 6, 33, 84, 34,...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have fun, and good luck!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 1:&lt;/strong&gt; I thought I'd give a day or so to see if anyone could grasp the first three - I see one of them has been discovered already. :) I'll add hints for the other two:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint for 2)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pay close attention to multiples and simple addition here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint for 3)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's no coincidence at all that each number is only one digit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="11475" LastEditDate="2015-04-16T16:14:10.613" LastActivityDate="2015-04-17T04:28:07.597" Title="Solve the numeric sequence" Tags="|mathematics|number-sequence|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12201" PostTypeId="2" ParentId="12154" CreationDate="2015-04-15T21:13:46.743" Score="2" Body="&lt;p&gt;The merchant walks for 5 mins while the shooter is in the hut. Then turns around to pretend he is a visitor entering the kingdom without paperwork. The shooter turns him away, as he has no paperwork. The merchant therefore escapes unhurt.&lt;/p&gt;&#xA;" OwnerUserId="11481" LastActivityDate="2015-04-15T21:13:46.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12202" PostTypeId="2" ParentId="12166" CreationDate="2015-04-15T21:15:09.013" Score="5" Body="&lt;p&gt;You could:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Walk backward into the forest. As soon as the King and his man sees you, walk forward&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11477" LastEditorUserId="7482" LastEditDate="2015-04-16T09:22:37.920" LastActivityDate="2015-04-16T09:22:37.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12203" PostTypeId="2" ParentId="12177" CreationDate="2015-04-15T21:15:36.410" Score="0" Body="&lt;p&gt;I agree that the accepted answer satisfies the criteria, though additionally it could be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2712 -&gt; $3^3=27$ &amp;amp; $3\cdot4=12$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11479" LastActivityDate="2015-04-15T21:15:36.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12204" PostTypeId="1" AcceptedAnswerId="12206" CreationDate="2015-04-15T21:35:30.253" Score="13" ViewCount="900" Body="&lt;p&gt;Your kingdom has been aided, for as long as anyone remembers, by three great seers. These seers are all-knowing and will truthfully answer one question of any person that seeks them. This question must be addressed at one seer and that seer will answer the question. It is not possible to seek out the seers more than once or to ask a second question of a second seer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the seers has gone mad recently. Instead of answering any question truthfully, he now randomly tells either the truth or a lie. Your king has tasked you with executing the mad seer to prevent him from affecting the sanity of the other seers. Unfortunately, the Seers are an identical triplet (they look and sound identical) and they are always together in the same cave hence figuring out who the mad seer is is problematic in itself. As multiple seers don't serve a purpose anyway, the king has ruled that, if needed, you are allowed to kill one sane seer if it ascertains the death of the mad seer. If you kill both sane seers or fail to kill the mad seer then you will be executed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You arrive at the seers' cave and go in. Before you stand all three seers. As a person that sought out the seers you are allowed to ask one seer one question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which question must you ask and must you kill two seers, or is it possible to kill only the mad seer? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:: As pointed out in the comments: I forgot to mention its only yes/no questions.&lt;/p&gt;&#xA;" OwnerUserId="11473" LastEditorUserId="11473" LastEditDate="2015-04-15T21:40:25.240" LastActivityDate="2015-04-17T00:20:12.777" Title="The Three Seers" Tags="|logical-deduction|liars|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12205" PostTypeId="2" ParentId="12119" CreationDate="2015-04-15T21:36:29.853" Score="7" Body="&lt;p&gt;Sadly, you never get out.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see why, let's consider 8 coins to see the pattern.  The goal is to gain control, or in other words to get as many tails in a row as possible so that you can get all heads in the end.  First, keep tails on the first move to keep the two tails together:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;00: TTHHHHHH //Position after first move&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Let's call this the base position.  Below are the states each time the coins return to the base position if you and satan both flip optimally for your goals (assuming he wants to keep you there and you want to get out):  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;TTHHHHHH //Position after first move&#xA;TTHHHHTH //satan flips the 2nd coin to tails.  &#xA;            -he doesn't flip the first because that would give you 3 tails in a row&#xA;            -If he doesn't flip any you could flip your tails to heads at the end &#xA;TTHHHTHH //You flip the 2nd coin to tails and he flips the 3rd to heads&#xA;TTHHHTTH //satan flips the 2nd coin to tails and you leave the 3rd alone&#xA;TTHHTHHH //You flip 2 and 3 to heads and he flips 4 to tails&#xA;TTHHTHTH&#xA;TTHHTTHH&#xA;TTHHTTTH&#xA;TTHTHHHH&#xA;TTHTHHTH&#xA;TTHTHTHH&#xA;TTHTHTTH&#xA;TTHTTHHH&#xA;TTHTTHTH&#xA;TTHTTTHH&#xA;TTHTTTTH&#xA;TTTHHHHH //Yay, 3 tails in a row!&#xA;TTTHHHTH&#xA;TTTHHTHH&#xA;TTTHHTTH&#xA;TTTHTHHH&#xA;TTTHTHTH&#xA;TTTHTTHH&#xA;TTTHTTTH&#xA;TTTTHHHH //Yay, 4 tails in a row!&#xA;TTTTHHTH&#xA;TTTTHTHH&#xA;TTTTHTTH&#xA;TTTTTHHH //Yay, 5 tails in a row!&#xA;TTTTTHTH&#xA;TTTTTTHH //Yay, 6 tails in a row!&#xA;TTTTTTTH //Yay, 7 tails in a row...almost free!&#xA;HHHHHHHT //Yay, one more day until you win the game and go free!&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The next day, you get up early and excitedly flip the easternmost coin from tales to heads as you move it to the west side of the pile.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Satan sighs and says in a worn-out tone &quot;Congratulations, you've won the game and you are now free to go.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has been the news you have been waiting for!  You eagerly await the Devil's instructions on how to leave this God-forsaken place.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, after about 15 seconds of awkward silence where you nervously shift between looking at the burning embers and him, you finally ask &quot;Ok, and how do I go about leaving?&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The devil looks up at the cave-like ceiling miles above and says &quot;That's a good question.  To tell you the truth I've been thinking about that a lot myself, but I haven't found a way yet.  Don't worry though, when I find one I'll be sure and tell you.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You let the words sink in for a bit.  &quot;So I'm free to go, but there's no way out?  Curse you, what was this game all about?!!?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Devil sighs tiredly and says, &quot;Oh nothing really.  I just made it up as a way to pass the time...&quot;&lt;/p&gt;&#xA;" OwnerUserId="1818" LastEditorUserId="1818" LastEditDate="2015-04-16T20:57:38.927" LastActivityDate="2015-04-16T20:57:38.927" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12206" PostTypeId="2" ParentId="12204" CreationDate="2015-04-15T21:39:23.507" Score="12" Body="&lt;p&gt;Solution to killing one sane Seer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pick Seer 1 and ask him this question: &quot;Is Seer 2 the mad Seer?&quot;. After he answers, kill him. If he said yes, kill Seer 2 as well, otherwise kill Seer 3.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solution to killing no sane Seers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not really. We can say WLOG that we ask Seer 1 our question. There are 3 possibilities: 1 is mad, 2 is mad, 3 is mad. We must be able to choose between these three options based only on the answer to a yes/no question. This is of course impossible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11100" LastEditorUserId="11100" LastEditDate="2015-04-15T22:51:18.283" LastActivityDate="2015-04-15T22:51:18.283" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12207" PostTypeId="1" CreationDate="2015-04-15T21:43:07.393" Score="5" ViewCount="157" Body="&lt;p&gt;Alice takes two envelopes and puts some number of dollar bills in each one. These can be any positive whole numbers (we're putting realism aside), but they cannot be the same number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bob chooses one envelope and looks at its contents, but not those of the other envelope. Bob then decides which envelope to keep. Devise a strategy by which Bob has a greater than $1/2$ chance of keeping the envelope with more money. This must be true no matter what distinct numbers Alice chose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is a classic that many of you are probably familiar with, but I didn't find it on the site. If you've seen it, I'd kindly request that you wait before answering to give others a chance.)&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-04-15T21:43:07.393" Title="Compare two numbers by looking at one" Tags="|mathematics|probability|" AnswerCount="0" CommentCount="4" ClosedDate="2015-04-15T22:28:59.910" ContentLicense="CC BY-SA 3.0" />
  <row Id="12208" PostTypeId="1" AcceptedAnswerId="12253" CreationDate="2015-04-15T22:45:52.817" Score="51" ViewCount="1498" Body="&lt;p&gt;I built this a while back but never had a good forum for it; this seems like the perfect spot to turn it loose.  This is a 'traditional' Sudoku; cells are filled with numbers from 1 through 9, and all usual Sudoku rules apply.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/X11p1.png&quot; alt=&quot;Self-Referential Sudoku&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know (unless I've made a transcription error) this should be rigorously solvable without any guesswork.  Enjoy!&lt;/p&gt;&#xA;" OwnerUserId="266" LastActivityDate="2015-04-16T14:18:20.060" Title="Self-referential Sudoku" Tags="|sudoku|calculation-puzzle|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12209" PostTypeId="2" ParentId="12119" CreationDate="2015-04-15T23:35:14.963" Score="-1" Body="&lt;p&gt;For n=5 I see no way out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm solving from the end state, not from the beginning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Satan could always make a T at west out of the H at east.&#xA;That means, what you need as last state before winning is one of these states:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HHHT&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HHHTT&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HHTTT&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HTTTT&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all these, it's impossible that the H at west was an H at east, because satan would not let it unflipped and make you win. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the H at west must have been a T at east. That means HHHHT must have been HHHTT, HHHTT must have been HHTTT and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This only leaves TTTTT as entry state to this set of states. But satan can prevent TTTTT by not flipping TTTTH to TTTTT.&#xA;But if he does so, he has to give you HTTTT and you would win. So, TTTTH (which forces satan to one of two winning nodes) must be on your route to win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how to get to TTTTH?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going reverse here again:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TTTTH =&gt; TTTHH  Impossible, satan doesn't make you win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TTTTH =&gt; TTTHT  Yes, it was yours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This goes for every of the T at west. It was never satans H.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here's the rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TTTHT =&gt; TTHTT  Yes, it was yours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TTHTT =&gt; THTTT  Yes, it was yours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;THTTT =&gt; HTTTT  Yes, it was yours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Leading to this problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HTTTT =&gt; TTTTH  Impossible, because a goal node can't be a node on your way to the goal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HTTTT =&gt; TTTTT  Impossible, satan does not give you TTTTT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IMHO There's no way out.&lt;/p&gt;&#xA;" OwnerUserId="11487" LastEditorUserId="11487" LastEditDate="2015-04-15T23:41:13.600" LastActivityDate="2015-04-15T23:41:13.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12210" PostTypeId="2" ParentId="11474" CreationDate="2015-04-16T00:28:42.833" Score="14" Body="&lt;blockquote&gt;&#xA;  &lt;h3&gt;Answerer’s note&lt;/h3&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;This is long. Like, really long. Pull up a chair&lt;/em&gt;.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;I’ve been working on this for several days, and I think it’s finally correct. I may have made a minor slip up somewhere – please let me know if I have! But I believe almost all of my deductions are correct, and the back of the puzzle is broken.&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;There have been &lt;strong&gt;three&lt;/strong&gt; major versions of this answer:&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;em&gt;My initial attempt exposed a few mistakes in the original question, which was edited.&lt;/em&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;em&gt;I revised some of my magic to use the new clues, and I was able to complete the puzzle. But @GentlePurpleRain pointed out a mistake: I said the only duplicated name was “Samuel” (in Terry’s clue), when “Quentin” is also duplicated.&lt;/em&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;em&gt;I've moved some of the logic around Terry. The result is the same, but now I prove that he’s not Quentin.&lt;/em&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;I won’t lie: if I was one of these castaways, I would have been killed by pirates long ago.&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;&lt;em&gt;This answer is so long that spoiler blocks are impractical. Everything below this block will spoil you.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Nationality data&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Start by filling in the grid with the accent/nationality of the party-goers and the manifests.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We immediately make the following deductions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Richard&lt;/em&gt; is the 2nd Mate of the &lt;em&gt;Leeuw&lt;/em&gt;: &lt;strong&gt;Arno Angélil&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The following people are all dead:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Carmel O'Brien&lt;/strong&gt;, seaman on the &lt;em&gt;Vierge&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Hu Wei-Xue&lt;/strong&gt;, seaman on the &lt;em&gt;Sea Fox&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Fabian A. Schubach&lt;/strong&gt;, cook on the &lt;em&gt;Cheswick&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Ruud D. Van Dijk&lt;/strong&gt;, cook on the &lt;em&gt;Cavadonga&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Who are the pirates?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Now we want to find the pirates. If we know who they are, then we can trust the other clues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what we can deduce:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Larry&lt;/em&gt; is a pirate.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Debra&lt;/em&gt; refers to the &quot;only male officer&quot; on a ship. Every ship except the &lt;em&gt;Cavadonga&lt;/em&gt; has at least two male officers, so she must be referring to &lt;strong&gt;Liao Jin-Zhong&lt;/strong&gt;. There is one other Chinese officer on the manifest: &lt;strong&gt;Yan Fu-ju&lt;/strong&gt; of the &lt;em&gt;Cheswick&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only one Chinese officer survived: &lt;em&gt;Emily&lt;/em&gt; (female). It follows that &lt;strong&gt;Liao Jin-Zhong&lt;/strong&gt; is dead, and &lt;em&gt;Emily&lt;/em&gt; is &lt;strong&gt;Yan Fu-ju&lt;/strong&gt;. (This means one of &lt;em&gt;Debra&lt;/em&gt;'s clues is true, so she is honest.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Larry&lt;/em&gt; tells us that the Boatswain of the &lt;em&gt;Cheswick&lt;/em&gt; (&lt;strong&gt;Yan Fu-ju&lt;/strong&gt;) has full use of both hands, but &lt;em&gt;Emily&lt;/em&gt; has a hook hand, so he must be lying. Thus &lt;em&gt;Larry&lt;/em&gt; is a pirate.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I am assuming that pirates do not make statements with null references; i.e., that I can trust that when &lt;em&gt;Debra&lt;/em&gt; refers to the &quot;only male officer&quot; on a ship, said officer does actually exist.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;At least one of &lt;em&gt;Edward&lt;/em&gt;, &lt;em&gt;Frank&lt;/em&gt; and &lt;em&gt;Patrick&lt;/em&gt; is lying.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can count 13 officers who survived, which means that 7 officers died. Now consider the following clues:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Edward&lt;/em&gt;: only 1 officer survived aboard the &lt;em&gt;Cheswick&lt;/em&gt; (so 3 died).&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Frank&lt;/em&gt;: at least 1 officer died on each ship.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Patrick&lt;/em&gt;: 2 officers on the English ship survived (so 2 died).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If all three were correct, then at least 8 officers would be dead. Somebody is lying.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Jason&lt;/em&gt; is a pirate.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the two (incompatible) accounts:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Rosemary&lt;/em&gt; tells us that each ship's name matches the language of its home country. Then the English ship would be either the &lt;em&gt;Sea Fox&lt;/em&gt; or the &lt;em&gt;Cheswick&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Jason&lt;/em&gt; tells us that all the sailors aboard the English ship survived, but (at this point) we know that every ship except the &lt;em&gt;Leeuw&lt;/em&gt; has lost at least one sailor, so the English ship would be the &lt;em&gt;Leeuw&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Suppose &lt;em&gt;Jason&lt;/em&gt; is honest. It must be that the &lt;em&gt;Leeuw&lt;/em&gt; is the English ship, and all the sailors aboard survived. Then &lt;em&gt;Rosemary&lt;/em&gt; is lying, and then so is &lt;em&gt;Charles&lt;/em&gt; (as they both make the same statement).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But &lt;em&gt;George&lt;/em&gt; tells us that only one sailor from the &lt;em&gt;Leeuw&lt;/em&gt; survived, so he must be lying. Now we have five pirates: &lt;em&gt;Larry&lt;/em&gt;, &lt;em&gt;Rosemary&lt;/em&gt;, &lt;em&gt;Charles&lt;/em&gt;, &lt;em&gt;George&lt;/em&gt;, and one of &lt;em&gt;Edward&lt;/em&gt;/&lt;em&gt;Frank&lt;/em&gt;/&lt;em&gt;Patrick&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Either &lt;em&gt;George&lt;/em&gt; or &lt;em&gt;Paul&lt;/em&gt; is a pirate.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following two accounts:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;George&lt;/em&gt; tells us that only one sailor survived on the &lt;em&gt;Gouden Leeuw&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Paul&lt;/em&gt; tells us that there is a ship with no casualties among the sailors (this must be the &lt;em&gt;Leeuw&lt;/em&gt;).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Clearly they cannot both be correct.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We have narrowed down the pirates as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;em&gt;Jason&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Larry&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;One of &lt;em&gt;Edward&lt;/em&gt;, &lt;em&gt;Frank&lt;/em&gt; and &lt;em&gt;Patrick&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;One of &lt;em&gt;George&lt;/em&gt; and &lt;em&gt;Paul&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Since we know there are exactly four pirates, we have &quot;found&quot; them all. Everybody else can be trusted to be honest.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Filling in the grid&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;With that in mind, we can fill in the grid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, let's just not some things that seem “obvious”, but now we have clarified:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Brian&lt;/em&gt; tells us that each ship has a different flag.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Per &lt;em&gt;Rosemary&lt;/em&gt;, every ship is named in the language of its home country. Thus the &lt;em&gt;Vierge&lt;/em&gt; is French, the &lt;em&gt;Leeuw&lt;/em&gt; is Dutch and the &lt;em&gt;Cavadonga&lt;/em&gt; is Spanish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;Sea Fox&lt;/em&gt; and &lt;em&gt;Cheswick&lt;/em&gt; are British and American, but which is which is not yet known.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Kevin&lt;/em&gt; tells us that each of the ships set sail from a different port. (These ports are Boston, Bordeaux, Kingston, Rome and Vera Cruz.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note, however, that we cannot be sure each ship is sailing to a different destination, as this is one of &lt;em&gt;Paul&lt;/em&gt;'s clues, and he is a suspected pirate.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Officers&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Remember what we already know:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;em&gt;Emily&lt;/em&gt; is Boatswain on the &lt;em&gt;Cheswick&lt;/em&gt;.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;*The 2nd Mate of the &lt;em&gt;Cavadonga&lt;/em&gt; has perished.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Let's start by sorting out some of the female officers:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;First, we note that the four female officers &lt;em&gt;Ann&lt;/em&gt;, &lt;em&gt;Debra&lt;/em&gt;, &lt;em&gt;Josephine&lt;/em&gt; and &lt;em&gt;Kimberly&lt;/em&gt; could occupy the following four roles:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;2nd Mate on the &lt;em&gt;Vierge&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Captain on the &lt;em&gt;Leeuw&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Captain or 1st Mate on the &lt;em&gt;Cavadonga&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Since we have four survivors mapping to four roles, all four of these roles have survived.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In particular, if we can prove that there is only one party-goer who could be a given role, then that party-goer must be that role. In general, this is not the case, as that role could have perished.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Ann&lt;/em&gt; was the 1st Mate on the &lt;em&gt;Cavadonga&lt;/em&gt;.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;She tells us that her superior officer is American. It then follows that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;She is not 2nd Mate on the &lt;em&gt;Vierge&lt;/em&gt;, as she would have Dutch and German superiors.&lt;/li&gt;&#xA;&lt;li&gt;She is not the captain of the &lt;em&gt;Leeuw&lt;/em&gt; or the &lt;em&gt;Cavadonga&lt;/em&gt;, as the captain does not have a superior officer.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It follows that she is the 1st Mate of the &lt;em&gt;Cavadonga&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;She also tells us that the captain of the &lt;em&gt;Cavadonga&lt;/em&gt; is American, so we can rule out &lt;em&gt;Debra&lt;/em&gt;. (We also learn this from &lt;em&gt;Emily&lt;/em&gt;.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Kimberly&lt;/em&gt; is the captain of the &lt;em&gt;Leeuw&lt;/em&gt;.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Josephine&lt;/em&gt; tells us that the captain of the &lt;em&gt;Leeuw&lt;/em&gt; has a hook hand, and so we can rule out &lt;em&gt;Josephine&lt;/em&gt; and &lt;em&gt;Debra&lt;/em&gt;. This leaves &lt;em&gt;Kimberly&lt;/em&gt; as the only option.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Debra&lt;/em&gt; is 2nd Mate of the &lt;em&gt;Vierge&lt;/em&gt;.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Kimberly&lt;/em&gt; tells us that all the 2nd Mates have eyepatches. Since &lt;em&gt;Josephine&lt;/em&gt; does not, she is not a 2nd Mate. This leaves &lt;em&gt;Debra&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Josephine&lt;/em&gt; is captain of the &lt;em&gt;Cavadonga&lt;/em&gt;.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Process of elimination)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We can use &lt;em&gt;Kimberly&lt;/em&gt;'s clue to find another officer:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Jason&lt;/em&gt; is the captain of the &lt;em&gt;Vierge&lt;/em&gt;.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He is Dutch, and there are two surviving Dutch officers: captain of the &lt;em&gt;Vierge&lt;/em&gt;, and 2nd Mate on the &lt;em&gt;Cheswick&lt;/em&gt;. Since 2nd Mates do not have eye patches, &lt;em&gt;Jason&lt;/em&gt; is not a 2nd Mate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the 2nd Mate of the &lt;em&gt;Cheswick&lt;/em&gt; has perished.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And now we have enough information to pin down &lt;em&gt;Carrie&lt;/em&gt; and &lt;em&gt;Christina&lt;/em&gt;, the last two (surviving) female officers:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;There are three Spanish female officers: captain of the &lt;em&gt;Sea Fox&lt;/em&gt;, Boatswain on the &lt;em&gt;Leeuw&lt;/em&gt;, and Boatswain on the &lt;em&gt;Cavadonga&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;em&gt;Jack&lt;/em&gt;, two of the captains died. Since we already have three surviving captains (&lt;em&gt;Vierge&lt;/em&gt;, &lt;em&gt;Leeuw&lt;/em&gt; and &lt;em&gt;Cavadonga&lt;/em&gt;), &lt;strong&gt;the captains of the &lt;em&gt;Sea Fox&lt;/em&gt; and &lt;em&gt;Cheswick&lt;/em&gt; are dead.&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Christina&lt;/em&gt; is Boatswain on the &lt;em&gt;Cavadonga&lt;/em&gt;.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;She tells us that she is the only Spaniard on her ship. This means she is not on the &lt;em&gt;Leeuw&lt;/em&gt;, as the cook of this ship is Spanish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Carrie&lt;/em&gt; is Boatswain on the &lt;em&gt;Leeuw&lt;/em&gt;.&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We can't quite finish off the officers, because we need to know which of &lt;em&gt;Edward&lt;/em&gt;, &lt;em&gt;Frank&lt;/em&gt; and &lt;em&gt;Patrick&lt;/em&gt; are lying. Let's move on, and we'll come back to them later.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Sailors&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Let's move on to fill in some of the sailor grid instead.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Elizabeth&lt;/em&gt; is a cabin girl, so she must be aboard the &lt;em&gt;Vierge&lt;/em&gt; or the &lt;em&gt;Sea Fox&lt;/em&gt;. Thus &lt;em&gt;Ryan&lt;/em&gt; is aboard one of these two ships also.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Anthony&lt;/em&gt; is on the same ship as &lt;em&gt;Margaret&lt;/em&gt;, who is a female sailor. There are no positions for female sailors on the &lt;em&gt;Cavadonga&lt;/em&gt;, so &lt;em&gt;Anthony&lt;/em&gt; cannot be on this ship.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Christina&lt;/em&gt; tells us that &lt;strong&gt;Alexander Frost&lt;/strong&gt; had an eyepatches, so if he survived, he is either &lt;em&gt;George&lt;/em&gt; or &lt;em&gt;Willie&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Joseph&lt;/em&gt; was a cabin boy.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;According to &lt;em&gt;Richard&lt;/em&gt;, none of the sailors on the &lt;em&gt;Jonathan Cheswick&lt;/em&gt; were maimed, so we can rule out &lt;em&gt;George&lt;/em&gt;, &lt;em&gt;Patrick&lt;/em&gt; and &lt;em&gt;Willie&lt;/em&gt; from working on this ship.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The copy of &lt;em&gt;Hamlet&lt;/em&gt; which &lt;em&gt;Victoria&lt;/em&gt; finds has the initials &lt;strong&gt;G.R.C.&lt;/strong&gt; inscribed in the cover. We may assume that if &lt;strong&gt;Gareth&lt;/strong&gt; survived, he is &lt;em&gt;George&lt;/em&gt;, as he is the only person with a “Shakespearean English” accent.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As &lt;em&gt;Jack&lt;/em&gt; found a grave for &lt;strong&gt;Virginia Allen&lt;/strong&gt;, a seaman on the &lt;em&gt;Cheswick&lt;/em&gt;, she must have perished.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As &lt;em&gt;Josephine&lt;/em&gt; is captain of the &lt;em&gt;Cavadonga&lt;/em&gt; and hired a Jamaican with a peg leg for her cabin boy, if her cabin boy survived, he must be &lt;em&gt;Patrick&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Once again, I seem to stuck. I can't see anything else we can deduce about the sailor grid, without going elsewhere. Let's move on and look at ships.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Ships&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;We already have the nationalities of the &lt;em&gt;Vierge&lt;/em&gt;, &lt;em&gt;Leeuw&lt;/em&gt; and &lt;em&gt;Cavadonga&lt;/em&gt; from &lt;em&gt;Rosemary&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Per &lt;em&gt;Anthony&lt;/em&gt;, the French ship (the &lt;em&gt;Vierge&lt;/em&gt;) is destined for Casa Blanca.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Carrie&lt;/em&gt; tells us that the ship sailing out of Rome was a merchant ship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;She also tells us that there was a merchant ship with a bilingual 2nd Mate. The only bilingual officer on the manifest is the 2nd Mate of the &lt;em&gt;Leeuw&lt;/em&gt;, so this must be a merchant ship. (We do not know if it is the same ship as in her first clue.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Debra&lt;/em&gt; tells us the American ship sailed out of Boston. &lt;em&gt;Ryan&lt;/em&gt; tells us that only one ship was sailing out of its home country, so it follows that the French ship was not leaving Bordeaux.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Debra&lt;/em&gt; also told that the only male officer on the ship to Bridgetown died, and we identified this ship as the &lt;em&gt;Cavadonga&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Jack&lt;/em&gt; tells us that the two whaling ships did not cary passengers. Combined with &lt;em&gt;Victoria&lt;/em&gt;'s statement (there were no passengers on the English ship), and we conclude that the English ship was a whaler.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Since whaling ships cannot stop at Bridgetown (from &lt;em&gt;Margaret&lt;/em&gt;), the English ship was not heading there. And the Spanish ship was sailing to Bridgetown (&lt;em&gt;Richard&lt;/em&gt;), so it's not a whaler.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Joseph&lt;/em&gt; tells us the ship from Vera Cruz was not an exploration vessel.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Keith&lt;/em&gt; tells us that an exploration ship was sailing to Buenos Aires.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Kimberly&lt;/em&gt; tells us that there was a ship sailing from Bordeaux to Charleston. Since &lt;em&gt;Anthony&lt;/em&gt; and &lt;em&gt;Margaret&lt;/em&gt; were aboard this ship, and every ship left from a different port, this cannot be the &lt;em&gt;Cavadonga&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The &lt;em&gt;Leeuw&lt;/em&gt; was a merchant ship heading to Amsterdam, according to &lt;em&gt;Raymond&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We know that &lt;em&gt;Jason&lt;/em&gt; was captain of the &lt;em&gt;Vierge&lt;/em&gt;, and he claimed to be aboard a merchant ship. Since he is a pirate, the &lt;em&gt;Vierge&lt;/em&gt; is not a merchant ship.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Assuming &lt;em&gt;Frank&lt;/em&gt;'s clue does not contain a null reference, the ship sailing from Vera Cruz has passengers and so is not a whaler.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now we can identify another pirate! This is what we know about the destination of each ship:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Vierge          Casa Blanca&#xA;Sea Fox         ???&#xA;Cheswick        ???&#xA;Leeuw           Amsterdam&#xA;Cavadonga       Bridgetown&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We know from other clues that there is a ship heading for Buenos Aires (&lt;em&gt;Keith&lt;/em&gt;) and Charleston (&lt;em&gt;Kimberly&lt;/em&gt;). This means that all the ships are heading for different ports, which is one of &lt;em&gt;Paul&lt;/em&gt;'s clues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So &lt;em&gt;Paul&lt;/em&gt; is honest, and &lt;em&gt;George&lt;/em&gt; is a pirate.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives us more information:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Raymond&lt;/em&gt; is the cook aboard the &lt;em&gt;Leeuw&lt;/em&gt;.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &lt;em&gt;Paul&lt;/em&gt; is honest, then there is one ship on which all the sailors survived. Earlier we showed that this must be the &lt;em&gt;Leeuw&lt;/em&gt;, and &lt;em&gt;Raymond&lt;/em&gt; is the only survivor left who could be the &lt;em&gt;Leeuw&lt;/em&gt;'s cook.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the other possibility for &lt;em&gt;Raymond&lt;/em&gt; (seamate 2 on the &lt;em&gt;Vierge&lt;/em&gt;) has thus perished.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;The cooks on the other ships are dead.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is &lt;em&gt;Irene&lt;/em&gt;'s clue: in particular, the cook on the &lt;em&gt;Vierge&lt;/em&gt; and &lt;em&gt;Sea Fox&lt;/em&gt; could both have been alive.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Jack is seaman 2 (Findlay) on the &lt;em&gt;Cavadonga&lt;/em&gt;.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since he is Scottish, he was either the cook on the &lt;em&gt;Sea Fox&lt;/em&gt; (&lt;strong&gt;Raymond&lt;/strong&gt;) or the seaman on the &lt;em&gt;Vierge&lt;/em&gt; (&lt;strong&gt;Findlay&lt;/strong&gt;). Since &lt;strong&gt;Raymond&lt;/strong&gt; is dead, he must be &lt;strong&gt;Findlay.&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;George is seaman 1 (Gareth) on the &lt;em&gt;Leeuw&lt;/em&gt;.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that &lt;strong&gt;Gareth&lt;/strong&gt; survived (&lt;em&gt;Paul&lt;/em&gt; tells us all the sailors from one ship survived; this is the &lt;em&gt;Leeuw&lt;/em&gt;), and so from &lt;em&gt;Victoria&lt;/em&gt;'s clue (the copy of &lt;em&gt;Hamlet&lt;/em&gt; with &lt;strong&gt;G.R.C.&lt;/strong&gt; inscribed; see above), he is the only possibility for &lt;strong&gt;Gareth&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Larry is the 1st Mate on the &lt;em&gt;Vierge&lt;/em&gt;.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Raymond&lt;/em&gt; tells us that the Boatswain heading for Casa Blanca was killed. We now know that there is only one ship going to Casa Blanca, which is the &lt;em&gt;Vierge&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We already knew that &lt;em&gt;Larry&lt;/em&gt; was either the 1st Mate or Boatswain on the &lt;em&gt;Vierge&lt;/em&gt;, and since he survived, he must be the 1st Mate.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Stowaways&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Let's briefly consider the stowaways. We have enough information to find almost all of their first names and surnames.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Debra&lt;/em&gt; found a stowaway named &lt;strong&gt;Albert&lt;/strong&gt; (M).&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Elizabeth&lt;/em&gt; tells us of a stowaway named &lt;strong&gt;Natalie Warner&lt;/strong&gt; (F).&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Irene&lt;/em&gt; gives us clues for two surnames: “one who makes arrows” (&lt;strong&gt;Fletcher&lt;/strong&gt;) and “one that works with iron” (&lt;strong&gt;Smith&lt;/strong&gt;).&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Josephine&lt;/em&gt; found a locket with the initials &lt;strong&gt;PF&lt;/strong&gt;. This does not match anybody on the manifest, so it must belong to a stowaway.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Kevin&lt;/em&gt; gives us a complicated clue about another surname.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Kimberly&lt;/em&gt; tells us that one of the stowaways has the surname &lt;strong&gt;Parker&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Raymond&lt;/em&gt; tells us of a stowaway whose initials were &lt;strong&gt;JS&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Rosemary&lt;/em&gt; knows of a stowaway whose first name was &lt;strong&gt;Pete&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Sherry&lt;/em&gt; found a stowaway named &lt;strong&gt;Marie&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Deductions we can make:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We have five first names/initials (&lt;strong&gt;Albert&lt;/strong&gt;, &lt;strong&gt;Natalie&lt;/strong&gt;, &lt;strong&gt;Pete&lt;/strong&gt;, &lt;strong&gt;Marie&lt;/strong&gt; and &lt;strong&gt;J&lt;/strong&gt;). Thus the &lt;strong&gt;P&lt;/strong&gt; in &lt;strong&gt;PF&lt;/strong&gt; is &lt;strong&gt;Pete&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;We have four surnames (&lt;strong&gt;Warner&lt;/strong&gt;, &lt;strong&gt;Fletcher&lt;/strong&gt;, &lt;strong&gt;Smith&lt;/strong&gt;, &lt;strong&gt;Parker&lt;/strong&gt;) and a clue from &lt;em&gt;Kevin&lt;/em&gt;. His clue tells us that the final last initial is one of &lt;strong&gt;H&lt;/strong&gt;, &lt;strong&gt;P&lt;/strong&gt; or &lt;strong&gt;E&lt;/strong&gt; (based on the first names of all the captains of non-whaling ships). Thus the &lt;strong&gt;F&lt;/strong&gt; in &lt;strong&gt;PF&lt;/strong&gt; is &lt;strong&gt;Fletcher&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now we can give some stowaways their real names:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Andrew&lt;/em&gt; is Pete Fletcher.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He lost a locket, and &lt;em&gt;Josephine&lt;/em&gt; found a locket with the initials &lt;strong&gt;PF&lt;/strong&gt;. It must be his locket.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Jennifer&lt;/em&gt; is Natalie Warner.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since &lt;strong&gt;Marie&lt;/strong&gt; was found by &lt;em&gt;Sherry&lt;/em&gt;, &lt;strong&gt;Marie&lt;/strong&gt; must have been aboard a passenger-carrying ship. But &lt;em&gt;Jennifer&lt;/em&gt; was on a whaling ship, and whalers do not carry passengers, so she must be the other stowaway.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Irene&lt;/em&gt; is Marie, and &lt;em&gt;Frank&lt;/em&gt; is Albert.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Process of elimination.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And some other stowaway-based deductions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Since &lt;em&gt;Andrew&lt;/em&gt; was headed for Bridgetown, he is aboard the &lt;em&gt;Cavadonga&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The male stowaway on &lt;em&gt;Richard&lt;/em&gt;'s ship (the &lt;em&gt;Leeuw&lt;/em&gt;) had no disfigurements, but both the surviving male stowaways have eye patches. It follows that there is no surviving stowaway from the &lt;em&gt;Leeuw&lt;/em&gt; (per &lt;em&gt;Lori&lt;/em&gt;, each stowaway is aboard a different ship).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Terry&lt;/em&gt; tells us that he shares a name with one of the other crew members.&#xA;Thus his name is either &lt;strong&gt;Samuel&lt;/strong&gt; or &lt;strong&gt;Quentin&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He was on the same ship as &lt;em&gt;Irene&lt;/em&gt;, so he is not sailing on the &lt;em&gt;Leeuw&lt;/em&gt; or the &lt;em&gt;Cavadonga&lt;/em&gt;. But the only sailor named Quentin is not on the &lt;em&gt;Cavadonga&lt;/em&gt;, so he must be named Samuel, and the other is Samuel is either English or Australian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm assuming that as &lt;em&gt;Patrick&lt;/em&gt; is Jamaican, neither of them are &lt;em&gt;Terry&lt;/em&gt;'s namesake.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a Samuel on the &lt;em&gt;Sea Fox&lt;/em&gt; and the &lt;em&gt;Leeuw&lt;/em&gt;. We know that &lt;em&gt;Irene&lt;/em&gt; is not aboard the &lt;em&gt;Leeuw&lt;/em&gt;, so nor is Terry. Thus &lt;em&gt;Terry&lt;/em&gt; is Seaman 1 on the &lt;em&gt;Sea Fox&lt;/em&gt;, and &lt;em&gt;Irene&lt;/em&gt; has stowed away on this ship.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Debra&lt;/em&gt; found &lt;strong&gt;Albert&lt;/strong&gt; (&lt;em&gt;Frank&lt;/em&gt;) aboard the &lt;em&gt;Vierge&lt;/em&gt;. It follows that &lt;em&gt;Jennifer&lt;/em&gt; was aboard the &lt;em&gt;Cheswick&lt;/em&gt;, and so the &lt;em&gt;Cheswick&lt;/em&gt; is a whaler.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;At this point, my strategy breaks down a little. I've filled in quite a lot of the grids, and the rest of my deductions are scattered all over the map.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Anthony&lt;/em&gt; and &lt;em&gt;Margaret&lt;/em&gt; were both going home to America (the port at Charleston). This means they were on the &lt;em&gt;Sea Fox&lt;/em&gt; or the &lt;em&gt;Cheswick&lt;/em&gt;, as the other ships are known not to be travelling to Charleston.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Victoria&lt;/em&gt; is Maura I. Kiley, and she is sailing on the &lt;em&gt;Leeuw&lt;/em&gt;.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Jack&lt;/em&gt; tells us that he found a grave for &lt;strong&gt;Eileen Mac Ateer&lt;/strong&gt;. &lt;em&gt;Victoria&lt;/em&gt; is Irish, and there were only two Irish passengers on the manifest. Since the first is dead, she must be the latter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;She tells us that she was on a boat with two Spaniards, which must be the &lt;em&gt;Leeuw&lt;/em&gt;. It follows that the &lt;em&gt;Leeuw&lt;/em&gt; is the ship of Passenger Manifest #3.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;According to &lt;em&gt;Dennis&lt;/em&gt;, only the cabin boy/girl survived on the ship that sailed out of Kingston.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;Sea Fox&lt;/em&gt;, &lt;em&gt;Leeuw&lt;/em&gt; and &lt;em&gt;Cavadonga&lt;/em&gt; all have known survivors who are not cabin crew, so they are not sailing from Kingston. Note also that the surviving women sailors force a non-cabin survivor on the &lt;em&gt;Cheswick&lt;/em&gt;: Seaman 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that the &lt;em&gt;Vierge&lt;/em&gt; is sailing out of Kingston. Then all the other crew on the &lt;em&gt;Vierge&lt;/em&gt; is dead.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Since &lt;em&gt;Ryan&lt;/em&gt; is travelling with &lt;em&gt;Elizabeth&lt;/em&gt;, who is a cabin girl, neither of them are aboard the &lt;em&gt;Vierge&lt;/em&gt;. The only other cabin girl is on the &lt;em&gt;Cheswick&lt;/em&gt;, so they must be on this ship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We deduce that &lt;em&gt;Elizabeth&lt;/em&gt; is the cabin girl and &lt;em&gt;Ryan&lt;/em&gt; is seaman 3.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We can deduce the nationality of the final two ships. Consider:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We know the &lt;em&gt;Cheswick&lt;/em&gt; is a whaler.&lt;/li&gt;&#xA;&lt;li&gt;We know that the &lt;em&gt;Cheswick&lt;/em&gt; and &lt;em&gt;Sea Fox&lt;/em&gt; are heading for Buenos Aires (ES) and Charleston (US), but we don't know which.&lt;/li&gt;&#xA;&lt;li&gt;We know that the English ship is a whaler.&lt;/li&gt;&#xA;&lt;li&gt;We know that the American ship must be heading to Buenos Aires as an explorer, because only one ship is heading for its home port, and this is the &lt;em&gt;Leeuw&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It follows that the &lt;em&gt;Cheswick&lt;/em&gt; is the English ship, and the &lt;em&gt;Sea Fox&lt;/em&gt; is the American ship.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;We can trust &lt;em&gt;Frank&lt;/em&gt;.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the &lt;em&gt;Cheswick&lt;/em&gt; is the English ship, either &lt;em&gt;Edward&lt;/em&gt; or &lt;em&gt;Patrick&lt;/em&gt; is lying:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Edward&lt;/em&gt;: only 1 officer survived aboard the Cheswick (so 3 died).&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Patrick&lt;/em&gt;: 2 officers on the English ship survived (so 2 died).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And so &lt;em&gt;Frank&lt;/em&gt; is in the clear. He tells us that at least one officer died aboard each ship.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Dennis&lt;/em&gt; is the 1st Mate of the &lt;em&gt;Sea Fox&lt;/em&gt;, and &lt;em&gt;Kevin&lt;/em&gt; is the 2nd Mate.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Dennis&lt;/em&gt; and &lt;em&gt;Kevin&lt;/em&gt; fill two of three roles: 1st and 2nd Mate on the &lt;em&gt;Sea Fox&lt;/em&gt;, and 1st Mate on the &lt;em&gt;Leeuw&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every officer aboard the &lt;em&gt;Leeuw&lt;/em&gt;, except the 1st Mate, is known to have survived. So from &lt;em&gt;Frank&lt;/em&gt;'s clue, the 1st Mate on the &lt;em&gt;Leeuw&lt;/em&gt; has died.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since &lt;em&gt;Dennis&lt;/em&gt; does not have an eye patch, he cannot be a 2nd Mate (&lt;em&gt;Kimberly&lt;/em&gt;), and the result follows.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Carrie&lt;/em&gt; tells us that the two English sailors on the &lt;em&gt;Cavadonga&lt;/em&gt; hassled the Scotsman. We know who the Scotsman on the &lt;em&gt;Cavadonga&lt;/em&gt; is (&lt;em&gt;Jack&lt;/em&gt;, Seaman), but not the other sailors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also know that the cabin boy is Jamaican (&lt;em&gt;Josephine&lt;/em&gt;) and the cook is Dutch (nationality/name). This means the other two seamen are English, and so we may rule out &lt;em&gt;Edward&lt;/em&gt; and &lt;em&gt;Patrick&lt;/em&gt; as seamen on this ship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By a process of elimination, &lt;em&gt;Willie&lt;/em&gt; is seaman 1, and &lt;em&gt;Brian&lt;/em&gt; is seaman 3.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Jennifer&lt;/em&gt; talks about a ship sailing between two non-English-speaking ports. Let's work out which this is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Not the &lt;em&gt;Vierge&lt;/em&gt;, as it sails from Kingston&lt;/li&gt;&#xA;&lt;li&gt;Not the &lt;em&gt;Sea Fox&lt;/em&gt;, as it sails from Boston&lt;/li&gt;&#xA;&lt;li&gt;Not the &lt;em&gt;Cheswick&lt;/em&gt;, as it's going to Charleston&lt;/li&gt;&#xA;&lt;li&gt;Not the &lt;em&gt;Cavadonga&lt;/em&gt;, as it's going to Bridgetown.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So she must be referring to the &lt;em&gt;Leeuw&lt;/em&gt;. So far we know two sailors on this ship:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Raymond&lt;/em&gt;, cook, Spanish&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;George&lt;/em&gt;, seaman, (Shakespearean) English&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;She tells us that the ship has sailors from four different English-speaking countries. We may deduce that &lt;em&gt;Joseph&lt;/em&gt; and &lt;em&gt;Megan&lt;/em&gt; are not aboard the &lt;em&gt;Leeuw&lt;/em&gt; (as there is already an English sailor.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This tells us that &lt;em&gt;Joseph&lt;/em&gt; is cabin boy aboard the &lt;em&gt;Cheswick&lt;/em&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then by a process of elimination, &lt;em&gt;Edward&lt;/em&gt; is the cabin boy (Australian) and &lt;em&gt;Patrick&lt;/em&gt; is seaman 3 (Jamaican). Since &lt;em&gt;Anthony&lt;/em&gt; is not aboard this ship, &lt;em&gt;Margaret&lt;/em&gt; is not seaman 2, so this must be &lt;em&gt;Lori&lt;/em&gt; (America).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By a process of elimination, we deduce the final three sailors:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Anthony&lt;/em&gt; and &lt;em&gt;Margaret&lt;/em&gt; are both seamen on the &lt;em&gt;Cheswick&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Megan&lt;/em&gt; is the cabin girl on the &lt;em&gt;Vierge&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We know the &lt;em&gt;Cavadonga&lt;/em&gt; is not a whaler, and we know the non-whaling purposes of the &lt;em&gt;Sea Fox&lt;/em&gt; and &lt;em&gt;Leeuw&lt;/em&gt;. Since there are two whalers, the &lt;em&gt;Vierge&lt;/em&gt; must be a whaler also.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We can now identify the final pirate:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;We know from &lt;em&gt;Christina&lt;/em&gt; that the two whaling ships (the &lt;em&gt;Vierge&lt;/em&gt; and the &lt;em&gt;Cheswick&lt;/em&gt;) have the same casualty rates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Five sailors and officers have died on the &lt;em&gt;Vierge&lt;/em&gt;. Only four are confirmed to have died on the &lt;em&gt;Cheswick&lt;/em&gt;, and possibly one more (depending on whether &lt;em&gt;Keith&lt;/em&gt; is on the &lt;em&gt;Sea Fox&lt;/em&gt; or &lt;em&gt;Cheswick&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We deduce that &lt;em&gt;Keith&lt;/em&gt; is the Boatswain on the &lt;em&gt;Sea Fox&lt;/em&gt;, and the 1st Mate of the &lt;em&gt;Cheswick&lt;/em&gt; died.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So &lt;em&gt;Edward&lt;/em&gt; is honest, and &lt;em&gt;Patrick&lt;/em&gt; is lying.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(We could have spotted this sooner – Patrick is on the &lt;em&gt;Leeuw&lt;/em&gt;, which is not a whaling ship, but he claims that he is.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In summary, the four pirates are &lt;strong&gt;&lt;em&gt;George&lt;/em&gt;, &lt;em&gt;Jason&lt;/em&gt;, &lt;em&gt;Larry&lt;/em&gt; and &lt;em&gt;Patrick&lt;/em&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few more notes before I summarise the results:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Sherry&lt;/em&gt; found &lt;em&gt;Irene&lt;/em&gt;/&lt;strong&gt;Marie&lt;/strong&gt; in her luggage. Since &lt;em&gt;Irene&lt;/em&gt; was aboard the &lt;em&gt;Sea Fox&lt;/em&gt;, so are &lt;em&gt;Gary&lt;/em&gt; and &lt;em&gt;Sherry&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When &lt;em&gt;Jennifer&lt;/em&gt; refers to a ship carrying &lt;strong&gt;Signor Milanesi&lt;/strong&gt; and &lt;strong&gt;Signor Conti&lt;/strong&gt;, this must be either the &lt;em&gt;Sea Fox&lt;/em&gt; or the &lt;em&gt;Cavadonga&lt;/em&gt; (passenger carrying ship; not the &lt;em&gt;Leeuw&lt;/em&gt; because it's attached to a different manifest).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cook on both of these ships died, so &lt;strong&gt;Signor Milanesi&lt;/strong&gt; must also be dead. It follows that &lt;em&gt;Paul&lt;/em&gt; (the only surviving Italian passenger) is &lt;strong&gt;Conti&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Charles&lt;/em&gt; was overseeing a business venture, so he and &lt;em&gt;Rosemary&lt;/em&gt; were not aboard the &lt;em&gt;Sea Fox&lt;/em&gt; (which was an explorer).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Kevin&lt;/em&gt;'s clue about a stowaway's surname refers to the “westward-sailing whaler”. The &lt;em&gt;Vierge&lt;/em&gt; is sailing Kingston to Casa Blanca (eastward), while the &lt;em&gt;Cheswick&lt;/em&gt; is sailing Bordeaux to Charleston (westward).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The captain of the &lt;em&gt;Cheswick&lt;/em&gt; is &lt;strong&gt;Henry R. Owen&lt;/strong&gt;, so one of the stowaways has the last name &lt;strong&gt;Henry&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we know that the surnames of &lt;strong&gt;Marie&lt;/strong&gt; and &lt;strong&gt;Albert&lt;/strong&gt; are &lt;strong&gt;Henry&lt;/strong&gt; and &lt;strong&gt;Parker&lt;/strong&gt; (as the dead stowaway was &lt;strong&gt;JS&lt;/strong&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Kimberly&lt;/em&gt; tells us that &lt;strong&gt;Parker&lt;/strong&gt; had a “devil-may-care attitude about life”. Since &lt;strong&gt;Albert&lt;/strong&gt;/&lt;em&gt;Frank&lt;/em&gt; has an eyepatch, but &lt;strong&gt;Marie&lt;/strong&gt;/&lt;em&gt;Irene&lt;/em&gt; is unmaimed, I'm going to assume this means that &lt;em&gt;Frank&lt;/em&gt; is &lt;strong&gt;Parker&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Since &lt;em&gt;Charles&lt;/em&gt; and &lt;em&gt;Rosemary&lt;/em&gt; both survived from the same ship, they must be on the merchant ship leaving Rome. (The ship from Vera Cruz lost three passengers.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The &lt;em&gt;Cavadonga&lt;/em&gt; and &lt;em&gt;Leeuw&lt;/em&gt; are sailing out of Rome and Vera Cruz. Since &lt;em&gt;Ryan&lt;/em&gt; tells us only one ship is sailing out of its home country, and the American ship is already sailing out of Boston, we deduce the &lt;em&gt;Cavadonga&lt;/em&gt; is sailing out of Rome. And thus it is a merchant ship.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The ship sailing out of Vera Cruz (the &lt;em&gt;Leeuw&lt;/em&gt;) lost three passengers (&lt;em&gt;Frank&lt;/em&gt;), so &lt;em&gt;Victoria&lt;/em&gt;/&lt;strong&gt;Maura&lt;/strong&gt; are the only survivors from passenger manifest #1.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Since &lt;em&gt;Charles&lt;/em&gt; and &lt;em&gt;Gary&lt;/em&gt; are travelling for different purposes, they must be on different ships. We deduce that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Gary&lt;/em&gt; is &lt;strong&gt;Jean-Louis Fleuette&lt;/strong&gt; on manifest #1, and since he's funding a discovery mission, this must be the &lt;em&gt;Sea Fox&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Sherry&lt;/em&gt; is &lt;strong&gt;Jeanette J. Ambard&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Charles&lt;/em&gt; and &lt;em&gt;Rosemary&lt;/em&gt; are &lt;strong&gt;Austinafe Ajaranne&lt;/strong&gt; and &lt;strong&gt;Uckbe Oronle&lt;/strong&gt; (but I don't know which is which) on manifest #2, and aboard the &lt;em&gt;Cavadonga&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Summary&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;An &lt;code&gt;X&lt;/code&gt; next to a name means this person was a pirate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Officers:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  Party-goer    Ship          Position        Real name&#xA;========================================================================&#xA;  Ann           Cavadonga     1st Mate&#xA;  Carrie        Leeuw         Boatswain&#xA;  Christina     Cavadonga     Boatswain&#xA;  Debra         Vierge        2nd Mate&#xA;  Dennis        Sea Fox       1st Mate&#xA;  Emily         Cheswick      Boatswain&#xA;X Jason         Vierge        Captain&#xA;  Josephine     Cavadonga     Captain&#xA;  Keith         Sea Fox       Boatswain&#xA;  Kevin         Sea Fox       2nd Mate&#xA;  Kimberly      Leeuw         Captain&#xA;X Larry         Vierge        1st Mate&#xA;  Richard       Leeuw         2nd Mate&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Passengers:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  Party-goer    Ship          Position        Real name&#xA;========================================================================&#xA;  Anthony       Cheswick      Seaman 3&#xA;  Brian         Cavadonga     Seaman 3&#xA;  Edward        Leeuw         Cabin boy&#xA;  Elizabeth     Sea Fox       Cabin girl&#xA;X George        Leeuw         Seaman 1&#xA;  Jack          Cavadonga     Seaman 2&#xA;  Joseph        Cheswick      Cabin boy&#xA;  Lori          Leeuw         Seaman 2&#xA;  Margaret      Cheswick      Seaman 2&#xA;  Megan         Vierge        Cabin girl&#xA;X Patrick       Leeuw         Seaman 3&#xA;  Raymond       Leeuw         Cook&#xA;  Ryan          Sea Fox       Seaman 3&#xA;  Terry         Sea Fox       Seaman 1&#xA;  Willie        Cavadonga     Seaman 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Passengers:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  Party-goer    Real name               Ship&#xA;========================================================================&#xA;  Charles       (Either Austinafe Ajaranne or Uckbe Oronle)&#xA;                                        Cavadonga&#xA;  Gary          Jean-Louis Fleuette     Sea Fox&#xA;  Paul          Nazzareno D. Conti      Sea Fox&#xA;  Rosemary      (Either Austinafe Ajaranne or Uckbe Oronle)&#xA;                                        Cavadonga&#xA;  Sherry        Jeanette J. Ambard      Sea Fox&#xA;  Victoria      Maura I. Kiley          Leeuw&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Stowaways:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  Party-goer    Real name               Ship&#xA;========================================================================&#xA;  Andrew        Pete Fletcher           Cavadonga&#xA;  Frank         Albert Parker           Vierge&#xA;  Irene         Marie Henry             Sea Fox&#xA;  Jennifer      Natalie Warner          Cheswick&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Ships:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  Ship        Type        Home country    Last port       Destination&#xA;========================================================================&#xA;  Vierge      Whaler      France          Kingston        Casa Blanca&#xA;  Sea Fox     Explorer    America         Boston          Buenos Aires&#xA;  Cheswick    Whaler      England         Bordeaux        Charleston&#xA;  Leeuw       Merchant    Netherlands     Vera Cruz       Amsterdam&#xA;  Cavadonga   Merchant    Spain           Rome            Bridgetown&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4692" LastEditorUserId="4692" LastEditDate="2015-04-16T14:25:12.943" LastActivityDate="2015-04-16T14:25:12.943" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="12211" PostTypeId="1" CreationDate="2015-04-16T00:59:12.600" Score="12" ViewCount="4456" Body="&lt;p&gt;A friend of mine told me the following puzzle and I could not solve it.&lt;/p&gt;&#xA;&lt;p&gt;Sam chooses a positive integer &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;, and Peter chooses another number &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt;. They do this secretly, so that Peter does not know Sam's number and vice versa.&lt;/p&gt;&#xA;&lt;p&gt;They then tell Sarah their numbers (secretly again). Sarah writes the sum &lt;span class=&quot;math-container&quot;&gt;$x+y$&lt;/span&gt; on one paper and the product &lt;span class=&quot;math-container&quot;&gt;$xy$&lt;/span&gt; on another. She then shows them one paper randomly. The value on the paper was &lt;span class=&quot;math-container&quot;&gt;$2002$&lt;/span&gt;. They know this is either the sum or the product, but they don't know which one!&lt;/p&gt;&#xA;&lt;p&gt;After this, the following conversation take place:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Sam: I don't know your number.&lt;/p&gt;&#xA;&lt;p&gt;Peter: I don't know your number either.&lt;/p&gt;&#xA;&lt;p&gt;Sam: Now I know your number.&lt;/p&gt;&#xA;&lt;p&gt;Peter: Now I know yours too.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;How do I figure out what the two numbers are?&lt;/p&gt;&#xA;" OwnerUserId="11489" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-02-24T04:01:04.553" Title="Find two numbers based on either their product or their sum?" Tags="|mathematics|logical-deduction|meta-knowledge|" AnswerCount="6" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12212" PostTypeId="2" ParentId="12171" CreationDate="2015-04-16T01:30:32.810" Score="24" Body="&lt;p&gt;Here's a simpler way to reach the answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone sits in the King's seat before Ophelia's seat is occupied, then the rest of the guests will all sit in their assigned seats, and Ophelia will be in her seat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone sits in the Ophelia's seat before the King's seat is occupied, then Ophelia obviously cannot be in her seat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But &lt;em&gt;none of the guests know&lt;/em&gt; which seat is the King's and which is Ophelia's! Therefore the two scenarios are equally likely, and Ophelia has a probability of 1/2 of ending up in her assigned seat.&lt;/p&gt;&#xA;" OwnerUserId="11490" LastActivityDate="2015-04-16T01:30:32.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12213" PostTypeId="2" ParentId="12211" CreationDate="2015-04-16T02:08:20.603" Score="1" Body="&lt;p&gt;These are the factors of 2002: 2, 7, 11, 13, 14, 22, 26, 77, 91, 143, 154, 182, 286, 1001. 14 numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Sam's number is not one of those 14, he will know that 2002 is not reached through multiplication and it is thus reached through addition. He will then take 2002 minus his number and find the answer.&lt;/p&gt;&#xA;" OwnerUserId="2596" LastEditorDisplayName="user88" LastEditDate="2015-04-16T03:35:35.253" LastActivityDate="2015-04-16T03:35:35.253" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12214" PostTypeId="2" ParentId="12211" CreationDate="2015-04-16T02:34:14.483" Score="1" Body="&lt;p&gt;&lt;strong&gt;Correct, but due to insight from another answerer. Please give primary credit to MikeEarnest&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;hr&gt;&#xA;&lt;hr&gt;&#xA;&lt;p&gt;First guy is &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt;, his number is &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;. Second guy is &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt;, his number is &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt;. Call the factor set of &lt;span class=&quot;math-container&quot;&gt;$2002$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$f = \{2,7,11,13,14,22,26,77,91,143,154,182,286,1001\}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;After the first comment from &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt; can deduce that &lt;span class=&quot;math-container&quot;&gt;$x \in f$&lt;/span&gt;. This is because &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt;, having seen the result on the paper (&lt;span class=&quot;math-container&quot;&gt;$2002$&lt;/span&gt;), these numbers present ambiguity for sum and product.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt; now inspects his own number, &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt;, and if &lt;span class=&quot;math-container&quot;&gt;$y \not\in f$&lt;/span&gt; then he would be able to deduce that &lt;span class=&quot;math-container&quot;&gt;$2002 = x + y$&lt;/span&gt; and hence deduce both &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt;. In this case he would know &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;. But he said he did not know, thus this is impossible.&lt;/p&gt;&#xA;&lt;p&gt;The other possibility for &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt; was that &lt;span class=&quot;math-container&quot;&gt;$y \in f$&lt;/span&gt; - he would know that &lt;span class=&quot;math-container&quot;&gt;$x * y = 2002$&lt;/span&gt; iff &lt;span class=&quot;math-container&quot;&gt;$y \not= 1001$&lt;/span&gt; (because &lt;span class=&quot;math-container&quot;&gt;$1001$&lt;/span&gt; still presents ambiguity for sum and product for &lt;span class=&quot;math-container&quot;&gt;$x,y \in f$&lt;/span&gt;). So the only case where he would not know the answer is if &lt;span class=&quot;math-container&quot;&gt;$y=1001$&lt;/span&gt; - and he said he did not know.&lt;/p&gt;&#xA;&lt;p&gt;Now that &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; knows that &lt;span class=&quot;math-container&quot;&gt;$y=1001$&lt;/span&gt;. And he declares this deduction.&lt;/p&gt;&#xA;&lt;p&gt;Unfortunately, &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt; knowing that &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; knows &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; does not help him deduce &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;. The first 3 comments could have taken place with either &lt;span class=&quot;math-container&quot;&gt;$x=2 \land y=1001$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$x=1001 \land y=1001$&lt;/span&gt;. Thus the forth comment should also have been &amp;quot;I &lt;em&gt;still&lt;/em&gt; don't know your number&amp;quot;.&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-16T03:04:42.950" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12215" PostTypeId="2" ParentId="1980" CreationDate="2015-04-16T02:40:57.790" Score="-1" Body="&lt;p&gt;Assuming that the prisoners can all switch places at will, all that has to happen is person 1 and 3 switch places so that person two then knows that person 1 and 3 both have white hats. Therefore person 2 can deduce that he and person 4 both have black hats on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You guys are forgetting the rules. It says that they CAN'T speak. And it is not reasonable to assume that person 2 must know because person 1 does not say anything. But it does not say that the prisoners cannot move. It only says that they cannot SPEAK.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually the other posters are correct, but the instructions do not say that the prisoners can only speak if they are correct or that they cannot move. So by number one not speaking it does not necessarily mean that number one does not know, but it's logical conclusion. but it still stands that the instructions do not prohibit number one and two from switching places and this would offer undeniable proof to number two, with the least number of moves.&lt;/p&gt;&#xA;" OwnerUserId="11492" LastEditorUserId="11492" LastEditDate="2015-04-16T03:36:49.453" LastActivityDate="2015-04-16T03:36:49.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12216" PostTypeId="2" ParentId="12211" CreationDate="2015-04-16T02:44:46.547" Score="13" Body="&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Assuming $y$ is a positive integer, the below reasoning works. Without this assumption, it does not. See &lt;a href=&quot;https://puzzling.stackexchange.com/a/12291/10615&quot;&gt;ricksmt's answer&lt;/a&gt; for the final word on this issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sam: I don't know your number.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Now Peter knows that $x$ is a factor of $2002$: if $x$ didn't divide $2002$, Sam would know that $2002$ wasn't $xy$, so he would know $y=2002-x$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Peter: I don't know your number.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The same reasoning tells Sam that $y$ is a divisor of $2002$. It also tells him that $2002-y$ is a divisor as well: if it wasn't, then Peter would reason that $2002-y$ couldn't be Sam's number $x$ (since Peter knew Sam's number was a divisor), so that $2002$ couldn't be $x+y$, so he would know $x=2002/y$. The only $y$ for which both $y$ and $(2002-y)$ divide $2002$ is $y=1001$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sam: I know your number.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;He knows it is 1001. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Peter: Now I know your number.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Peter is lying/mistaken. Both $(x,y)=(2,1001)$ and $(x,y)=(1001,1001)$ are consistent with the first three statements. Basically, Peter had enough info to deduce that Sam knew $y$ even before Sam said his second sentence, so Sam saying &quot;I know $y$&quot; told Peter nothing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus, we can conclude that either Sam, Peter, or your friend are irrational or sinister.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.923" LastActivityDate="2015-04-17T05:02:36.850" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12217" PostTypeId="2" ParentId="1980" CreationDate="2015-04-16T02:52:40.530" Score="4" Body="&lt;p&gt;Prisoner 2 can know the color of his hat - it should be exactly opposite of that worn by the prisoner ahead of him, Prisoner 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prisoner 1 can see both 2 and 3 in front of him, but the fact that he cannot guess the color of his own hat must mean 2 and 3 are wearing different colored hats. For example, if 2 and 3 both had white hats, and knowing that there are only two white hats (with the other two being black), prisoner 1 would have been able to work out that he's wearing a black hat. Likewise, if both 2 and 3 had black hats on, 1 would know he's wearing a white hat. BUT, if 2 and 3 had different colored hats on, then 1 cannot logically deduce the color of his own hat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FROM THAT LOGIC ABOVE, 2 knows that the color of his own hat is different to the color worn by the person ahead of him (prisoner 3). Thus, if 3 has a white hat on, 2's own hat must be black. Else, if 3 has a black hat, then 2 must be wearing a white hat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As only one person needs to deduce the answer correctly for them all to be released, that person is 2.&lt;/p&gt;&#xA;" OwnerUserId="11493" LastActivityDate="2015-04-16T02:52:40.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12220" PostTypeId="2" ParentId="12187" CreationDate="2015-04-16T06:11:59.243" Score="38" Body="&lt;p&gt;If √ can mean nth root:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt[1234567]{-8+9}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 operators. Obviously...&lt;/p&gt;&#xA;" OwnerUserId="11495" LastActivityDate="2015-04-16T06:11:59.243" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12221" PostTypeId="1" AcceptedAnswerId="12223" CreationDate="2015-04-16T06:25:03.537" Score="13" ViewCount="1886" Body="&lt;p&gt;You won the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/12119/coin-flipping-game-with-the-devil&quot;&gt;coin-flipping game&lt;/a&gt; against Satan, but being the devil he is, he decided that that game was much too easy, and reneges on his promise to let you go. Instead, he gives you a new, more difficult game, this time promising by crossing his heart and hoping to die that he will let you go if you can finish this game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The game starts with a single spindly tree, like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QbVbH.png&quot; alt=&quot;The initial tree, which is really just a single branch with seven segments.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tree is only a single branch, with seven segments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every turn, the following happens:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;You can cut off one segment of the tree that doesn't have anything branching off of it. (For the first turn, only the top segment can be cut off.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Satan can then make $N$ copies of the branch system just below the base of the branch you just cut off, where $N$ is as large an integer as he wants. However, if you cut off a branch next to the root, then Satan can't do anything as there's nothing below the root.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The first few steps of the game might go like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nmS05.png&quot; alt=&quot;The first three moves&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After your first move, Satan decides to make two copies. And after your second move, Satan makes &lt;em&gt;three&lt;/em&gt; copies of the tree below it. He's being nice here &amp;mdash; he could make a thousand copies, or even a googolplex copies!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Satan promises that he will release you if you can prune the tree down to its root. Do you have a winning strategy? Or will Satan manage to keep you in hell forever?&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-04-21T18:14:06.220" Title="Tree-pruning game with the devil" Tags="|mathematics|logical-deduction|strategy|game|" AnswerCount="4" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="12222" PostTypeId="2" ParentId="12187" CreationDate="2015-04-16T06:42:48.137" Score="1" Body="&lt;p&gt;Three Operators&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar idea to Mwr247's solution, but even more significant figures:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\left(\frac1{23456789!}\right)! = 1$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=%281%2F23456789%21%29%21&quot; rel=&quot;nofollow&quot;&gt;Wolfram Alpha&lt;/a&gt; seems to think it is exactly 1. Good enough for me &gt;_&gt;&lt;/p&gt;&#xA;" OwnerUserId="11499" LastEditorUserId="13494" LastEditDate="2015-06-25T17:35:55.667" LastActivityDate="2015-06-25T17:35:55.667" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12223" PostTypeId="2" ParentId="12221" CreationDate="2015-04-16T07:49:14.650" Score="11" Body="&lt;p&gt;All the segments will eventually vanish. We can even inductively demonstrate it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, terminology.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Let's call the segment just before the first fork the &lt;em&gt;base&lt;/em&gt;. &lt;/li&gt;&#xA;&lt;li&gt;Let's call the sequence of segments terminating at the base the &lt;em&gt;trunk&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Let's call each fork off of the base a &lt;em&gt;branch&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/U7iIgm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, let's do some pre-validation. Let's assume there's a way to remove an entire branch. (This will be shown later.) Then, Satan has to copy the base, which means it becomes a branch off the segment before it. Thus, the base moves one step down the trunk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JoIFX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, eventually, there won't be any trunk to move down. At this point, you can remove the remaining branches (as Satan can't copy them anymore), and you win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, we can show that it's irrelevant how many branches Satan makes. It just makes the execution annoyingly long. Suppose Satan makes $n$ copies of a branch. The winning strategy above only requires moving one branch, which we've assumed we can do. Therefore, no matter what $n$ Satan chooses, we have one branch to remove. The other $n-1$ branches become part of each individual new branch when Satan copies the base.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As concluded, if you have a way to remove an entire branch, you win. So let's do it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say a branch has a base segment and $n$ $1$-length branches each called $B_1$. There are a finite number of these branches. If we prune one of the one-length branches, we end up with a finite number ($x_1$) of bases with $n-1$ one-length branches. We've removed one branch, and Satan gets to copy this a finite number of times. We can prune all $n$ branches, and end up with $x_1+\ldots+x_n$ bases with $n-1$ one-length branches. This process can be repeated until the number of one-length branches is zero. &lt;strong&gt;Therefore, a $B_1$ branch is removable.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/K0duy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say a branch has a base segment and $n$ $2$-length branches each called $B_2$. Because these branch off, each branch can be expressed as its own $B_1$. Since we know $B_1$ branches are removable, we can remove them, resulting in the removal of $B_2$ branches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YOUwF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say a branch has a base segment and $n$ $w$-length branches, each called $B_w$. Each of $B_w$ can be split into unique $B_{w-1}$ branches. These can each be split into $B_{w-2}$ branches. Each of these can be split into $B_{w-3}$ branches, which can be split... until we have $B_{w-(w-1)}=B_1$ branches. Example for $B_3$: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WRrrq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $B_1$ branches are removable, we can remove them. This implies that $B_2$ branches are removable. Since all $B_{n+1}$ branches can be split into $B_{n}$ branches, and $B_n$ branches are removable, so are $B_{n+1}$ branches. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently, by inductive reasoning, any branch can be removed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since any branch can be removed, by the above, you win the game.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorDisplayName="user20" LastEditDate="2015-04-16T08:54:52.693" LastActivityDate="2015-04-16T08:54:52.693" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12224" PostTypeId="2" ParentId="12004" CreationDate="2015-04-16T08:00:27.730" Score="4" Body="&lt;p&gt;The answers are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PRECIOUSLY, PREVIOUSLY, GRIEVOUSLY&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Come ponder with me, as we go forth,&lt;br&gt;&#xA; How we behave in a manner of worth.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Preciously&lt;/strong&gt; means &quot;of high cost or worth&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;With a single change, that worth is no more.&lt;br&gt;&#xA; Instead we consider the ones before.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Change the &lt;strong&gt;c&lt;/strong&gt; to a &lt;strong&gt;v&lt;/strong&gt; to get &lt;strong&gt;previously&lt;/strong&gt;, meaning &quot;earlier or coming before&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;After one more change, seen with different eyes,&lt;br&gt;&#xA; We behave in a way we won't trivialize.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Change the &lt;strong&gt;p&lt;/strong&gt; to a &lt;strong&gt;g&lt;/strong&gt; and move the &lt;strong&gt;i&lt;/strong&gt; (seen with different &quot;eyes&quot;) to get &lt;strong&gt;grievously&lt;/strong&gt;, which means &quot;serious or dire&quot; (not trivial).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4479" LastActivityDate="2015-04-16T08:00:27.730" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12225" PostTypeId="2" ParentId="12221" CreationDate="2015-04-16T08:09:34.883" Score="2" Body="&lt;p&gt;We can think of this tree as a directed graph, by directing edges to point away from the bottommost vertex. The degree of a vertex is the number of outward pointing edges out of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any time, let $h$ be the height of the &lt;em&gt;second&lt;/em&gt; highest vertex (so the height of the tree is $h+1$, but every vertex at height $h+1$ has no branches). Let $d$ be the highest degree among the vertices at height $h$. In your pictures, the pairs $(h,d)$ are $(6,1)$, $(5,3)$ and $(5,2)$, from left to right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say you cut a branch from a vertex of degree $d$, at height $h$. Then the vertex you chopped now has degree $d-1$. Furthermore, the vertices Satan creates at height $h$ will only have degree $d-1$, since they are copies of the vertex you mutilated. This means the number of degree $d$ vertices at height $h$ decreased, while no vertices of greater degree were created at that height.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, if you keep chopping from vertices of height $h$ and degree $d$, you will eventually remove all of them, so that the maximum degree at height $h$ becomes $d-1$. Repeating this, we can decrease the maximum degree of any vertex at height $h$ all the way down to 0, at which point there can be no vertices at height $h+1$, so that the height of the entire tree decreased.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, at any time we can decrease the height of the tree. Doing so seven times destroys the tree. Of course, the time you must remain in Hell is unbounded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Alternatively:&lt;/strong&gt; To each a tree whose height is $h+1$, where the highest degree node at height $h$ is of degree $d$, and where there are $n$ vertices of degree $d$ at height $h$, assign the ordinal&#xA;$$&#xA;h\omega^2+d\omega+n&#xA;$$&#xA;By the same argument as before, chopping any of the maximal degree vertices at height $h$ will leave a smaller ordinal. This is beacuse there were three possibilities: either $h$ decreased, or $h$ stayed the same while $d$ dropped, or both $h$ and $d$ stayed while $n$ dropped. Ordinals are well-ordered, so there can't be an infinite decreasing chain of ordinals, so this process must stop.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-04-16T21:17:38.013" LastActivityDate="2015-04-16T21:17:38.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12226" PostTypeId="2" ParentId="12221" CreationDate="2015-04-16T08:15:03.537" Score="5" Body="&lt;p&gt;A proof via ordinals. Probably overkill. I'm not well-versed in ordinal arithmetic, so I hope I haven't made any unwarranted assumptions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assign a tree an ordinal value inductively:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The empty tree is $1$&lt;/li&gt;&#xA;&lt;li&gt;Two trees of value $s$ and $t$ have value $s+t$ if joined at their roots. So, $n$ copies of a tree $s$ joined this way have value $\underbrace{s + \dots + s}_{n \text{ times}} = ns$.&lt;/li&gt;&#xA;&lt;li&gt;A tree of value $s$ on top of an additional segment has value $\omega ^s$, where $\omega$ is the first infinite ordinal.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We use the &quot;natural&quot; arithmetic defined by transfinite recursion, not the one based on combining order types. In particular, this means addition and multiplication commute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in this process, restricting to just the top three levels in the first picture (ignoring the bottom four), the trees have values &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega ^ {\omega ^ \omega}, \omega^{3\omega}, 4 \omega^{2\omega}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nmS05.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We argue that each step decreases the value, and so since ordinals are well-ordered, any strictly decreasing sequence must terminate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since ordinal (natural) sum and exponent are strictly monotonic, it suffices to argue that the sub-tree affected decrease in value. This sub-tree consists of the deleted segment, the segment below it, and the segment's &quot;siblings&quot; above that segment, whose value we call $s$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the original value is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega ^ {\omega + s}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which changes to $$n \omega ^ {s},$$ when the segment is deleted and the branch is duplicated $n$ times. Using the exponent distributive property,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \omega ^ {\omega + s} &amp;gt; \omega ^ {1 + s} = \omega \times \omega ^ s &amp;gt; n \omega ^ s, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where we used the fact the multiplication is strictly monotonic. So, the cutting process decreased the ordinal value, and so only a finite number of cuts are possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm unclear on whether exponentiation distributes for ordinals, but the $\omega ^ {1 + s}$ case holds by definition.)&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-04-16T08:37:12.847" LastActivityDate="2015-04-16T08:37:12.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12227" PostTypeId="1" AcceptedAnswerId="12249" CreationDate="2015-04-16T08:28:29.357" Score="1" ViewCount="553" Body="&lt;p&gt;You are working for the CIA as a spy. One day you get a secret offer from the US president to spy on the CIA for him and to go and investigate the truth about a CIA covert operation which happened 10 years ago (let's call it operation X). The US president believes that what the CIA told him about the operation is indeed a lie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In effect, you have now become a double agent, working for the CIA and secretly for the US president.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After many months of conducting your own secret investigation, you find out that the official CIA report about what really happened during operation X, is in fact a lie. You find the necessary information you need about operation X in various &lt;strong&gt;paper only&lt;/strong&gt; classified documents stored in the CIA document archive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, something isn't quite right; getting into the CIA document archive appears to have been much easier that you thought it would.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When presenting the information that you found out during your investigation to the US president, how can you be sure you have the right information? How can the president be sure that you are NOT providing him more false information about operation X?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could try and confirm the information you have found by secretly interviewing the people involved in operation X, but that would be risky, since they may give you false information and then secretly alert the CIA.&lt;/p&gt;&#xA;" OwnerUserId="11503" LastEditorUserId="10615" LastEditDate="2015-04-16T08:36:12.590" LastActivityDate="2015-04-16T15:59:17.593" Title="Puzzle: Covert spy investigation" Tags="|logical-deduction|word-problem|mystery|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12228" PostTypeId="2" ParentId="12171" CreationDate="2015-04-16T08:32:22.157" Score="3" Body="&lt;p&gt;Yet another way to the answer;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is exactly one guest(including the king) sitting in a seat that we do not know belongs to a guest that has already arrived.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that when a guest arrives he will find his seat taken with probability 1/(Number of free seats+1).  This continues until Ophelia arrives, so Ophelia has probability 1/(1+1) of finding her seat taken.&lt;/p&gt;&#xA;" OwnerUserId="1654" LastEditorUserId="1654" LastEditDate="2015-04-16T08:42:41.330" LastActivityDate="2015-04-16T08:42:41.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12229" PostTypeId="1" CreationDate="2015-04-16T09:00:10.110" Score="3" ViewCount="1121" Body="&lt;p&gt;Relevant link : &lt;a href=&quot;https://puzzling.stackexchange.com/q/12022/9420&quot;&gt;Cheryl's birthday&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand the solution. But it feels wrong to me, because Albert &lt;strong&gt;cannot possibly make the third statement with the information he has at the moment he makes it.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hear me right : &lt;em&gt;if he does, and if it is true&lt;/em&gt;, then we know for sure the answer is what it is. But if I understand correctly, he can't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first statement eliminates May and June for everyone.&#xA;Bernard, had he been told &quot;14th&quot;, could still not deduce the correct date. But he was not told the 14th. So he can make the second statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, &lt;strong&gt;the only thing Albert can deduce is &quot;Bernard has been told 15th, 16th or 17th&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These three dates are enough, with the first statement, to fully deduce the date.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bernard is able to make the second statement because he as a piece of information Albert does not, and the second statement only provides little information to Albert.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I say again, if Albert actually makes the third statement just for the sake of the problem, and if the statement is true, I do understand why the solution is July 16th. But Albert, with the information he has after the second statement, cannot make the third statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take an example : let the date be August 15.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Albert can still make the first statement.&#xA;Bernard can still make the second statement, because &quot;15&quot; (initial information he has) and &quot;not May/June&quot; (which he can deduce from first statement) are enough to deduce August 15 at this point for him.&#xA;Albert cannot make the third statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the date is of very little importance. let the date be July 16th again.&#xA;Albert can make the first statement.&#xA;Bernard can still make the second statement, because &quot;16&quot; (initial information) and &quot;not may/june&quot; are enough for him to deduce the date.&#xA;But the second statement only tells Albert &quot;Bernard was not told 14th&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What would be a good answer to this question ?&lt;/strong&gt;&#xA;There are two possible good answer for my question. The first one would be &quot;you're just supposed to accept the clues and not question them&quot;. The second would be a detailed explanation of why Albert can make the third statement righteously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How can Albert, with the information he has after the second statement, make the third one?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9420" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2016-02-24T04:10:45.210" Title="Explaining Cheryl's birthday" Tags="|meta-knowledge|explanation|" AnswerCount="6" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="12230" PostTypeId="2" ParentId="12166" CreationDate="2015-04-16T09:33:24.097" Score="0" Body="&lt;p&gt;He needs @ least 0.5 meter and get out back.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If we consider that a normal person will move 0.5 meter in single step, so that is move one step in and the other will be backward which is out and so the king loose.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11111" LastActivityDate="2015-04-16T09:33:24.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12231" PostTypeId="2" ParentId="11948" CreationDate="2015-04-16T09:42:13.870" Score="-2" Body="&lt;p&gt;Is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fungus || Fungi &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can relate some of the line to my answer. If it is right I'll provide explanation to my answer. Travelling now... &lt;/p&gt;&#xA;" OwnerUserId="10973" LastActivityDate="2015-04-16T09:42:13.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12232" PostTypeId="2" ParentId="12227" CreationDate="2015-04-16T09:58:51.220" Score="1" Body="&lt;p&gt;I would use the fact that I'm a double agent to my advantage, I'd tell the CIA that the president has asked me to investigate Operation X. Then I'd tell them that the president will probably send more agents at a later date, either because I haven't reported in or to confirm whatever report the CIA wants me to pass on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We'll assume that the CIA would prefer the president to read false documents than true ones. If the documents are true then the next set of spies will discover text different to what I found, but if the documents are false then the contents will be unchanged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But before I told the CIA I was a double agent I've already reported to the president the contents of the archive, along with details of my plan. At this point it doesn't matter if the CIA kills me or order me to lie, the next set of spies (or police raid, whatever) will confirm whether or not the archive of Operation X was accurate.&lt;/p&gt;&#xA;" OwnerUserId="10246" LastActivityDate="2015-04-16T09:58:51.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12233" PostTypeId="1" CreationDate="2015-04-16T10:25:26.700" Score="1" ViewCount="451" Body="&lt;p&gt;The owner of a ration store sells sugar sacks weighing 50Kg for \$50, but he gives a special discount to his neighbors, and sells it to them for \$45.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One day the owner, whose name is Jack, left the store and left his employee Jony in charge. Two of his neighbors, Tom and Jane, bought a sack of sugar together and spent \$50, each of them paying \$25.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, Jony sold the sack to them without knowing that they are his boss's neighbors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When Jack came back to the store, he saw that Jony made a mistake, so he sent him to pay Tom and Jane a \$5 discount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jony stole \$2 from the 5\$ and gave \$1.50 each to Tom and Jane, giving them a total of \$3. So now Tom and Jane have paid \$25 - \$1.50 = \$23.50.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each one paid 23.50\$, so both Tom and Jane paid \$23.50 + \$23.50 = \$47 together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jony stole \$2, so the total is \$47 + \$2 = \$49&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Where is the missing \$1?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="11111" LastEditorUserId="463" LastEditDate="2015-04-16T10:54:56.297" LastActivityDate="2015-04-16T12:05:07.507" Title="Where is the 1$" Tags="|logical-deduction|" AnswerCount="4" CommentCount="1" ClosedDate="2015-04-16T12:16:20.087" ContentLicense="CC BY-SA 3.0" />
  <row Id="12234" PostTypeId="2" ParentId="12233" CreationDate="2015-04-16T10:33:14.287" Score="4" Body="&lt;p&gt;The trick to this puzzle is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;that it gets you to add when you should really subtract.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The last line: &quot;So total is \$47+\$2=\$49&quot; should be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;\$47-\$2=\$45 because the final cost is now $45&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastEditorUserId="3987" LastEditDate="2015-04-16T10:35:23.830" LastActivityDate="2015-04-16T10:35:23.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12236" PostTypeId="2" ParentId="12187" CreationDate="2015-04-16T10:42:06.343" Score="-4" Body="&lt;p&gt;&lt;strong&gt;One operator.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking advantage of user23013 's loop hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If √ can mean nth root:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt[23456789]{1}$$&lt;/p&gt;&#xA;" OwnerUserId="4186" LastActivityDate="2015-04-16T10:42:06.343" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12238" PostTypeId="2" ParentId="12233" CreationDate="2015-04-16T11:26:26.187" Score="0" Body="&lt;p&gt;He took the money from the customers, not from the store.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once Jack returns to the store, the final cost of the item is \$50-\$5=\$45. Let's break it down into a series of actions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\$50 - \$5&lt;br&gt;&#xA;\$25 + \$25 - \$5&lt;br&gt;&#xA;\$25 + \$25 - \$2 - \$1.50 - \$1.50&lt;br&gt;&#xA;\$25 - \$1.50 + \$25 - \$1.50 - \$2&lt;br&gt;&#xA;\$23.50 + \$23.50 - \$2&lt;br&gt;&#xA;\$47 - \$2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your final statement has reversed this subtraction into an addition.&lt;/p&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2015-04-16T11:26:26.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12241" PostTypeId="2" ParentId="12229" CreationDate="2015-04-16T11:50:08.873" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;How can Albert, with the information he has after the second statement, make the third one?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In your post, you have already accepted that Bernard was not told the 14th. This leaves the following numbers: 15, 16, 17, 18, 19. However, since Albert was told &quot;July&quot;, before Bernard has even said anything, Albert has eliminated 15, 17, 18, and 19. This leaves only July 16.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Note&lt;/strong&gt;: What follows is a complete description of the logic for the remainder of the puzzle.&lt;/em&gt;&#xA;Cheryl's birthday is July 16. At the beginning of the puzzle, this is unknown to Albert and Bernard (and us as spectators).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;May 15 16 19&lt;br&gt;&#xA;  June 17 18&lt;br&gt;&#xA;  July 14 16&lt;br&gt;&#xA;  August 14 15 17&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Albert knows: &lt;strong&gt;Cheryl's birthday is in July. It is either 14 or 16.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bernard knows: &lt;strong&gt;Cheryl's birthday is the 16. It is either July or May.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Albert knows that it is not the 18 or 19, and therefore also knows that Bernard does not know the birthday without Albert's help. This is how he comes to the conclusion he states first:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Albert: I don't know when Cheryl's birthday is &lt;strong&gt;[it is either July 14 or July 16]&lt;/strong&gt;, but I know that Bernard does not know, too &lt;strong&gt;[because it is certainly not May 19 or June 18]&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bernard knows that the month could only be May or July (since he was told 16). Had Albert been told &quot;May&quot;, Albert's initial reasoning would have instead produced the statement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Albert: I don't know when Cheryl's birthday is &lt;strong&gt;[it is one of May 15, May 16, or May 19]&lt;/strong&gt;, and I also don't know if Bernard knows &lt;strong&gt;[because it might be May 19]&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because this was &lt;em&gt;not&lt;/em&gt; Albert's initial statement, Bernard knows that he must have been told &quot;July&quot;, leading him to join together with his own information and conclude that it is July 16.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bernard: At first, I didn't know when Cheryl's birthday is &lt;strong&gt;[because it could have been in May or July]&lt;/strong&gt;, but I know now &lt;strong&gt;[because Albert's information allowed me to eliminate May as a possibility]&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With Bernard's conclusion, Albert is given the information he needs to rule out July 14. Let's consider for a second that Bernard might have been told &quot;14&quot;. Had this been the case, Albert's initial statement wouldn't have been enough to let Bernard choose between July and August. This makes Bernard's second statement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bernard: At first, I didn't know when Cheryl's birthday is &lt;strong&gt;[because it could have been July or August]&lt;/strong&gt;, but I still don't know &lt;strong&gt;[because Albert's information has not narrowed my selection, it has only eliminated the two months that I had already ruled out myself]&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because he didn't say this, Albert knows it must be 16 of the month he was told.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Albert: Then I also know when Cheryl's birthday is &lt;strong&gt;[and that is July 16]&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See also: &lt;a href=&quot;https://puzzling.stackexchange.com/a/12318/10399&quot;&gt;this exhaustive explanation&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-04-20T10:50:29.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12242" PostTypeId="2" ParentId="12233" CreationDate="2015-04-16T11:56:49.063" Score="0" Body="&lt;p&gt;This puzzle is for Accountants.&#xA;Accounts were made wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Correct:-&#xA;\$50 he received, so +50&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\$5 he gave, so -5(3 discount given \&amp;amp; 2 stolen by employee)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So ultimately storekeeper has \$45 remains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\$5 Debit , \$45 Credit=totals \$50&lt;/p&gt;&#xA;" OwnerUserId="11306" LastActivityDate="2015-04-16T11:56:49.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12243" PostTypeId="2" ParentId="12233" CreationDate="2015-04-16T12:05:07.507" Score="0" Body="&lt;p&gt;Seeing it from another way is...&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Jack would have benefit of \$2 when he could sold it for \$47 instead of \$45&lt;/p&gt;&#xA;&lt;p&gt;Your final addition \$23.50 + \$23.50 = \$47 tells same thing that &lt;strong&gt;seller&lt;/strong&gt; is benefited from &lt;strong&gt;consumer&lt;/strong&gt; by \$2 but now Jony is the person who stole that part.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10433" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-16T12:05:07.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12244" PostTypeId="2" ParentId="6584" CreationDate="2015-04-16T12:17:02.463" Score="0" Body="&lt;p&gt;ORACLE 11g Solution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Execute The Following Query:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;with x as (select rownum as rnk from dual &#xA;connect by level&amp;lt;=9),&lt;br&gt;&#xA; y as (select x1.rnk a,x2.rnk b,x3.rnk c,x4.rnk d,x5.rnk e,x6.rnk f,x7.rnk g,x8.rnk h,x9.rnk i,x10.rnk j&#xA;from x x1,x x2,x x3,x x4,x x5,x x6,x x7,x x8,x x9,x  x10&#xA;)&lt;br&gt;&#xA;Select * from y&#xA;where &#xA;f+h-c&lt;em&gt;a+e=0&lt;br&gt;&#xA;and b-j/h-c/a=1&lt;br&gt;&#xA;and d-b/j+i-c=2&lt;br&gt;&#xA;and e&lt;/em&gt;d+g-j*i=3&lt;br&gt;&#xA;and a-e+d/g+f=4&lt;br&gt;&#xA;and f-b+d+e+a=9&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="11518" LastEditorUserId="10281" LastEditDate="2015-04-16T12:29:06.877" LastActivityDate="2015-04-16T12:29:06.877" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12245" PostTypeId="2" ParentId="12187" CreationDate="2015-04-16T12:28:16.510" Score="-3" Body="&lt;p&gt;In languages like &lt;code&gt;C#&lt;/code&gt; and &lt;code&gt;Java&lt;/code&gt; dividing two integers will always result in an integer (decimals will be omitted).&#xA;Therefore only one operation is required to solve this problem:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;12345/6789&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which will result in &lt;code&gt;1&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11519" LastActivityDate="2015-04-16T12:28:16.510" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12246" PostTypeId="2" ParentId="12211" CreationDate="2015-04-16T12:44:54.530" Score="0" Body="&lt;p&gt;Let me go line by line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sam: I don't know your number.&#xA;since the number he choose was a factor of 2002 {2,7,11,13,14,22,26,77,91,143,154,182,286,1001}, sam was not clear whether the displayed card is addition or multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Peter: I don't know your number.&#xA;So peter now knows why sam could not tell and even he could not tell because his number was in the set {2,1001}. So sam's number might be in between 2 and 1001.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sam: Now I know your number.&#xA;If sam's number is 1001, he will not be sure about peter's number.&#xA;But since he choose 2, he was sure about peter's number i.e 1001.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Peter: Now I know yours too.&#xA;Since sam is sure about his(peter) number, then sam's number must be 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally sam choose 2,peter choose 1001.&lt;/p&gt;&#xA;" OwnerUserId="11520" LastEditorUserId="11520" LastEditDate="2015-04-16T12:50:35.640" LastActivityDate="2015-04-16T12:50:35.640" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12247" PostTypeId="2" ParentId="11380" CreationDate="2015-04-16T13:09:23.950" Score="1" Body="&lt;p&gt;Jack: I don't know implies that Mr. Hill could have told him any month, so $${3/.} \bigvee {6/.} \bigvee {9/.} \bigvee {12/.} \bigvee {3/.} = \text{true} $$&#xA;This is true because it's all possibilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jack: And Jill surely doesn't know implies that either Jill wasn't told $12/2$ or $6/7$ or Jack wasn't told $12/.$  or $6/.$. Either of these two can make Jack certain that Jill doesn't know. So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\overline{12/. \bigvee 6/.} \bigwedge \overline{12/2 \bigvee 6/7} = \overline{12/.} \bigwedge \overline{6/.}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining the two result together, Jack has successfully eliminated month $12$ and month $6$ and we are left with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3/4 \bigvee  3/5 \bigvee 3/8\bigvee  9/1\bigvee  9/5 = \overline{12/.} \bigwedge \overline{6/.}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now they both know the date without being aware of the other's value. Jill knows the date without knowing Jack's month and Jack knows the date without knowing Jill's day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jill knows implies that it's not day $5$ i.e $\overline{3/5 \bigvee 9/5} $ which brings us to &#xA;$$\left(3/4 \bigvee  3/5 \bigvee 3/8\bigvee  9/1\bigvee  9/5\right) \bigwedge \overline{3/5 \bigvee 9/5} = \left(3/4 \bigvee 3/8\bigvee  9/1\right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jack knows implies that it's not month $3$ i.e $\overline{3/4 \bigvee 9/8} $ which brings us to &#xA;$$\left(3/4 \bigvee  3/8\bigvee  9/1 \right) \bigwedge \overline{3/4 \bigvee 3/8} =9/1 $$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;NB&lt;/strong&gt;: ${3/.}  = 3/4 \bigvee 3/5 \bigvee 3/8 $&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5410" LastActivityDate="2015-04-16T13:09:23.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12248" PostTypeId="2" ParentId="12229" CreationDate="2015-04-16T13:24:33.943" Score="1" Body="&lt;p&gt;I think we can verify which ones are the incorrect answers by using reductio-ad-absurdum on all the possible dates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The date which will not produce a contradiction or logic leap (eg: &quot;Why can we exclude May/June dates?&quot; or from above &quot;Albert: I don't know when Cheryl's birthday is [it is either July 14 or July 16], but I know that Bernard does not know, too [because it is certainly not May 19 or June 18].&quot; --- why not [because Bernard thinks May 16 or August 14]) is August 17.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using August 17 as the DOB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Albert knows August.&#xA;Bernard knows 17.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Albert can't figure out the DOB because there're possible 3 options (all August dates).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Albert can figure out that Bernard doesn't know because:&#xA;1. June 18 and May 19 are out of questions (they're not in August).&#xA;2. Bernard has to choose between June 17 and August 17.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's why he could make his first statement about himself and Bernard.&#xA;Bernard or Cheryl didn't have to tell Albert about Bernard's not knowing Cheryl's DOB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After Albert made his first statement, Bernard knows:&#xA;1. June 17 is not correct because if this the DOB, then Albert must have figured it out and said it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bernard now only has one option August 17.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's why Bernard could say what he said.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on what Bernard said, Albert knows that Bernard must have discarded June 17 (because if this the DOB, Albert would have said it in his first statement) and chosen August 17.&lt;/p&gt;&#xA;" OwnerUserId="11521" LastEditorUserId="11521" LastEditDate="2015-04-16T13:39:46.750" LastActivityDate="2015-04-16T13:39:46.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12249" PostTypeId="2" ParentId="12227" CreationDate="2015-04-16T13:27:43.603" Score="0" Body="&lt;p&gt;You could try asking both parties about their information on Operation X, and depending on the information you get back you can make a conclusion.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If both the president and the CIA give you information different than the archive, the archive is most likely real (As you would have been lied to similar to the president). If the archive matches the presidents, then its is a fake and the CIA information is likely real (They archived the lie and just told you the truth). If the archive matches the CIA's information then you didn't gather anything new and still won't know (except for the fact that they either thoroughly cover their tracks or were being genuine). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If asking questions alerts the CIA, then:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Break back into the archive after they are alerted and see if they switch the documents. If they did then what you first saw was most likely real, if its the same then you gather no new information.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On another note:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If there are virtual documents in some archive about Project X, look at those and make the same comparisons. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-04-16T13:35:18.720" LastActivityDate="2015-04-16T13:35:18.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12250" PostTypeId="2" ParentId="12227" CreationDate="2015-04-16T13:46:52.650" Score="0" Body="&lt;p&gt;I'm not sure about this, but here's my try:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I would meet with the CIA and reveal that I am being ordered to investigate on Operation X, not revealing that you are a double agent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the hearing/ investigating meeting, I would ask the CIA if the information presented 10 years ago to the President is a lie or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CIA should answer &quot;NOT A LIE&quot; since it is what they presented to the President. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then reveal to them that you are indeed a double agent, you will reveal to them that you found out a document that tells this is a LIE, that would prove and show everyone that they are lying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That would be embarrassing for them but will still continue to cover their lies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then present a &lt;strong&gt;FALSE DOCUMENT&lt;/strong&gt; (that you created) kind of similar to what you saw in the archives,&lt;strong&gt;AND&lt;/strong&gt; the &lt;strong&gt;ARCHIVE DOCUMENT&lt;/strong&gt; then tell them that you found out two documents from the archives but we since assume that they would rather have the President read a false document then ask:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Which of the two is the TRUE document?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;IF they point to the ARCHIVE document, &lt;strong&gt;then the ARCHIVE document contains FALSE information&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IF they point to the FALSE document (that you created) then &lt;em&gt;the  ARCHIVE document must be the TRUE document&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to prove whether the ARCHIVE DOCUMENT is true or not, reveal that the one they pointed is a false document that you created; then ask:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is the Archive document TRUE?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;IF they tell you NO, &lt;code&gt;(since they would have the president read false documents)&lt;/code&gt;&#xA;then &lt;strong&gt;the ARCHIVE document has the RIGHT information.&lt;/strong&gt; &lt;code&gt;(might even be forced to reveal the real document)&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IF they tell you YES, then &lt;strong&gt;the ARCHIVE documents contains FALSE information&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10686" LastActivityDate="2015-04-16T13:46:52.650" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12251" PostTypeId="2" ParentId="12187" CreationDate="2015-04-16T13:51:52.730" Score="5" Body="&lt;p&gt;Assuming the binomial coefficient is not an operator itself and parentheses are allowed and not counted, this requires only 1 operator. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1+{2345\choose6789}=1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check the Pochhammer symbol too:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1+(-2)_{3456789}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some useful &lt;a href=&quot;https://www.wolframalpha.com/input/?i=1%20%2B%20Pochhammer%28-2%2C3456789%29&quot; rel=&quot;nofollow&quot;&gt;information on Wolfram Alpha&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11524" LastEditorUserId="13494" LastEditDate="2015-06-25T17:35:44.240" LastActivityDate="2015-06-25T17:35:44.240" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="12252" PostTypeId="1" AcceptedAnswerId="12274" CreationDate="2015-04-16T14:04:31.653" Score="7" ViewCount="513" Body="&lt;p&gt;Mr. EG has come up with a great coding scheme for (non-negative) integers in binary. To show off his creation, he demonstrates the following multiplications:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;010 x 011 = 010011&#xA;011 x 010010 = 010010011&#xA;010011 x 010010011 = 010010010011011&#xA;01000101 x 01001000111 = 0100100100010100111&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What is the result of the following multiplication? &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0100010100101 x 01001001001000101&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Can you translate all those confusing 1s and 0s back to good old decimal integers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: &lt;code&gt;x&lt;/code&gt; represents standard integer multiplication.&lt;/p&gt;&#xA;" OwnerUserId="11525" LastEditorUserId="11525" LastEditDate="2015-04-16T14:39:34.020" LastActivityDate="2015-04-17T13:32:41.933" Title="Bitwise multiplication" Tags="|mathematics|calculation-puzzle|number-sequence|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12253" PostTypeId="2" ParentId="12208" CreationDate="2015-04-16T14:10:00.943" Score="20" Body="&lt;p&gt;I got:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;342|975|168&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;765|821|934&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;891|364|527&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;---+---+---&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;926|538|741&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;187|496|352&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;534|217|689&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;---+---+---&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;479|652|813&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;253|189|476&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;618|743|295&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The steps I took to get there were:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; De can't be one because it is the sum of two numbers.&lt;br&gt;&#xA; De can't be two because then Fd and Hd would both be one, but they are in the same column.&lt;br&gt;&#xA; De must be at least 3.&lt;br&gt;&#xA; De &gt;= 3 and De * Fd &amp;lt;= 9, so Fd &amp;lt;= 3.&lt;br&gt;&#xA; If Fd = 3, then De = 3, but they are in the same box. So Fd &amp;lt;= 2.&lt;br&gt;&#xA; If Fd = 1, then Ef = De. But they are in the same box. So Fd = 2.&lt;br&gt;&#xA;&lt;br&gt;&#xA; ------------------------------------------------&lt;br&gt;&#xA;&lt;br&gt;&#xA; Ef = De * 2.&lt;br&gt;&#xA; De = 2 + Hd.&lt;br&gt;&#xA; Ef = (2+Hd) * 2.&lt;br&gt;&#xA; Ef = 4 + 2*Hd.&lt;br&gt;&#xA; If Ef &amp;lt;= 9, Hd &amp;lt;= 2.&lt;br&gt;&#xA; Hd can't be 2 because Fd is 2. So Hd = 1.&lt;br&gt;&#xA;&lt;br&gt;&#xA; De = 3.&lt;br&gt;&#xA; Ef = 6.&lt;br&gt;&#xA;&lt;br&gt;&#xA; ------------------------------------------------&lt;br&gt;&#xA;&lt;br&gt;&#xA; Hd is 1, so neither Gf nor Hg may be 1.&lt;br&gt;&#xA; Gf * Hg &amp;lt;= 9, so they are either (2,2) or (2,3) or (2,4) or (3,3), So Fh is 4 or 6 or 8 or 9.&lt;br&gt;&#xA; He = 1 = Ee - Fh, so Ee = Fh + 1. So Ee is 5 or 7 or 9 or 10.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Ee = Cd * Hc.&lt;br&gt;&#xA; Both Cd and Hc are in line with Hd, so Ee can't be 5 or 7.&lt;br&gt;&#xA; So Ee = 9 and Fh = 8.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Gf and Hg are (2,4), but we don't yet know which is which.&lt;br&gt;&#xA; Dc is 6.&lt;br&gt;&#xA;&lt;br&gt;&#xA; ------------------------------------------------&lt;br&gt;&#xA;&lt;br&gt;&#xA; Ee is 9, so Cd and Hc are both 3. (They can't be 1 and 9 because they're both in line with Hd, which is 1)&lt;br&gt;&#xA; Dd - Gf = 3.&lt;br&gt;&#xA; Gf is either 2 or 4, so Dd is either 5 or 7.&lt;br&gt;&#xA; Ha = Ff - Dd, and Ha != 1 because it is in line with Hd.&lt;br&gt;&#xA; If Dd is 7, then either Ha is 1 or Ff is 9, but both are contradictions. So Dd must be 5.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Gf = 2.&lt;br&gt;&#xA; Hg = 4.&lt;br&gt;&#xA;&lt;br&gt;&#xA; ------------------------------------------------&lt;br&gt;&#xA; Ha = Ff - Dd = Ff - 5, So Ff &gt; 5.&lt;br&gt;&#xA; Ff is lin line with 6, 8, and 9, so Ff = 7.&lt;br&gt;&#xA; Ha = 2.&lt;br&gt;&#xA;&lt;br&gt;&#xA; ------------------------------------------------&lt;br&gt;&#xA;&lt;br&gt;&#xA; Da = Bf + Fh = Bf + 8.&lt;br&gt;&#xA; So Da = 9 and Bf = 1.&lt;br&gt;&#xA; Fe = 1 because it's the only remaining cell in the center box that could be 1.&lt;br&gt;&#xA;&lt;br&gt;&#xA; ------------------------------------------------&lt;br&gt;&#xA; Bi = Fh - Ed = 8 - Ed, so Ed &amp;lt; 8.&lt;br&gt;&#xA; Ed can only be 8 or 4.&lt;br&gt;&#xA; So Ed = 4, and Bi = 4.&lt;br&gt;&#xA;&lt;br&gt;&#xA; De = 4 = Be * Ch. Be can't be 1 or 4.&lt;br&gt;&#xA; So Be = 2 and Ch = 2.&lt;br&gt;&#xA;&lt;br&gt;&#xA; ------------------------------------------------&lt;br&gt;&#xA; Df is 8, because it's the last cell in the center box.&lt;br&gt;&#xA; If is 3, because it's the only cell in the box that can be 3.&lt;br&gt;&#xA; Ei = If - Gh = 3 - Gh, so Gh &amp;lt;= 2.&lt;br&gt;&#xA; Gh is in line with 2, so Gh = 1 and Ei = 2.&lt;br&gt;&#xA;&lt;br&gt;&#xA; ------------------------------------------------&lt;br&gt;&#xA; Ig = 2.&lt;br&gt;&#xA; Db = 2.&lt;br&gt;&#xA; Ac = 2.&lt;br&gt;&#xA; Dh = 4.&lt;br&gt;&#xA; ------------------------------------------------&lt;br&gt;&#xA; Ba = He - Di, so He &gt; Di.&lt;br&gt;&#xA; He can be (5,6,7,8). Di can be 1 or 7.&lt;br&gt;&#xA; If Di is 7, He is 8 and Ba is 1. But Ba can't be 1. So Di = 1.&lt;br&gt;&#xA; Dg = 7.&lt;br&gt;&#xA;&lt;br&gt;&#xA; ------------------------------------------------&lt;br&gt;&#xA; Ih = Bh + Hi, and Hi &gt;= 5, So Bh &amp;lt;= 4.&lt;br&gt;&#xA; Bh can't be 1,2, or 4, so Bh = 3.&lt;br&gt;&#xA; Ih &amp;lt;= 9, so Hi is 5 or 6.&lt;br&gt;&#xA; If Hi is 5, Ih is 8, but Fh is 8. So Hi = 6 and Ih = 9.&lt;br&gt;&#xA; ------------------------------------------------&lt;br&gt;&#xA; Fg = 6.&lt;br&gt;&#xA; Ah = 6.&lt;br&gt;&#xA; Fi = 9.&lt;br&gt;&#xA; Eg = 3.&lt;br&gt;&#xA; Gi = 3.&lt;br&gt;&#xA; Eh = 5.&lt;br&gt;&#xA; Hh = 7.&lt;br&gt;&#xA; ------------------------------------------------&lt;br&gt;&#xA; Ba = He - 1. He can be 5 or 8, so Ba can be 4 or 7. But Ba can't be 4, so Ba = 7 and He = 5.&lt;br&gt;&#xA; ------------------------------------------------&lt;br&gt;&#xA; Ea and Ec can only be (1,7,8). So Id = Ea * Ec is 7 or 8.&lt;br&gt;&#xA; He is 8, so Id can't be 8, so Id is 7.&lt;br&gt;&#xA; Ea can't be 7, so Ea is 1 and Ec is 7.&lt;br&gt;&#xA; Eb = 8.&lt;br&gt;&#xA; ------------------------------------------------&lt;br&gt;&#xA; Fc is 4 or 5. Ga is 4, 5, or 6. Bd = Fc + Ga, so Bd can't be 6. So Gd is the only cell in the column that can be 6.&lt;br&gt;&#xA; Hf = 9.&lt;br&gt;&#xA; Hb = 5.&lt;br&gt;&#xA; Bd &amp;lt;= 9, so Ga can't be 6. Hb is 5, so Ga can't be 5. So Ga = 4.&lt;br&gt;&#xA; Ie = 4.&lt;br&gt;&#xA; Ge = 5.&lt;br&gt;&#xA; Ii = 5.&lt;br&gt;&#xA; Gg = 8.&lt;br&gt;&#xA; Gb = 7.&lt;br&gt;&#xA; Gc = 9.&lt;br&gt;&#xA; Ce = 6.&lt;br&gt;&#xA; Ae = 7.&lt;br&gt;&#xA; Ci = 7.&lt;br&gt;&#xA; Ai = 8.&lt;br&gt;&#xA; Bd = 8.&lt;br&gt;&#xA; Bd = Fc + Ga, so 8 = Fc + 4, So Fc = 4.&lt;br&gt;&#xA; Fa = 5.&lt;br&gt;&#xA; Fb = 3.&lt;br&gt;&#xA; Ad = 9.&lt;br&gt;&#xA; Bc = 5.&lt;br&gt;&#xA; Ca = 8.&lt;br&gt;&#xA; Aa = 3.&lt;br&gt;&#xA; Ia = 6.&lt;br&gt;&#xA; Ic = 8.&lt;br&gt;&#xA; Ib = 1.&lt;br&gt;&#xA; Cc = 1.&lt;br&gt;&#xA; Ag = 1.&lt;br&gt;&#xA; Ab = 4.&lt;br&gt;&#xA; Bb = 6.&lt;br&gt;&#xA; Cb = 9.&lt;br&gt;&#xA; Af = 5.&lt;br&gt;&#xA; Cf = 4.&lt;br&gt;&#xA; Cg = 5.&lt;br&gt;&#xA; Bg = 9.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1497" LastEditorUserId="1497" LastEditDate="2015-04-16T14:18:20.060" LastActivityDate="2015-04-16T14:18:20.060" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12255" PostTypeId="2" ParentId="12252" CreationDate="2015-04-16T14:26:55.537" Score="0" Body="&lt;p&gt;2 x 3 = 23 (2,3) -- [Decimal: 19]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 x 18 = 183 (18,3) -- [Decimal: 187]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;19 x 147 = 14719 (147,19) -- [Decimal: 9371]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;69 x 583 =  69583 (69,583) -- [Decimal: 149671]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2212 x 37445 = 221237445 (2212,37445)  -- [Decimal: 145003077]   (100010100100,1001001001000101)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or maybe:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2212 x 37445 = 374452212 (37445,2212) -- [Decimal: 306753700] (1001001001000101,100010100100)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not quite sure of the entire answer, but this is a start&lt;/p&gt;&#xA;" OwnerUserId="11244" LastActivityDate="2015-04-16T14:26:55.537" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12262" PostTypeId="2" ParentId="12229" CreationDate="2015-04-16T15:06:08.740" Score="1" Body="&lt;p&gt;So as far as I can see, your thinking is along the lines of: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the date is July 16th, A and B can make statements 1 and 2.&lt;/li&gt;&#xA;&lt;li&gt;However, if the date is August 15, A and B can also make statements 1 and 2.&lt;/li&gt;&#xA;&lt;li&gt;Therefore, A, having heard statements 1 and 2, can't distinguish between the possibility of July 16th and the possibility of August 15th, since these lead to identical statements.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The first two points are true. However, you're forgetting that A has an additional piece of information: &lt;em&gt;he still knows the month&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if the date was August 15, A would have heard statements 1 and 2 &lt;em&gt;and know the month was August&lt;/em&gt;, whereas if the date was July 16, A would have heard statements 1 and 2 &lt;em&gt;and know the month was July&lt;/em&gt;. Because of this, he's able to distinguish between the two situations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, of all the remaining dates after hearing statements 1 and 2, only one has month July, which is how A is able to pinpoint the exact date and make the third statement. If the date HAD been August 15th, A would have to admit that he still doesn't know what the correct date is (as it could be August 15th or 17th).&lt;/p&gt;&#xA;" OwnerUserId="1759" LastEditorUserId="1759" LastEditDate="2015-04-16T15:12:06.547" LastActivityDate="2015-04-16T15:12:06.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12263" PostTypeId="1" AcceptedAnswerId="12277" CreationDate="2015-04-16T15:10:27.223" Score="1" ViewCount="438" Body="&lt;p&gt;You've probably seen this trick question floating around the Internet before:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If 1 = 5, 2 = 10, 3 = 15, and 4 = 20, then 5 = ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The question is twofold:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What is the trick answer they expect? (You can use Google to answer this one.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What is the most elegant way to define the equivalence relationship between numbers using the &lt;em&gt;logic&lt;/em&gt; of the trick answer? (i.e. suppose that the answer you got in part 1 was indeed the unique answer to the question, and all the numbers on the right side of the equation were the unique answers to their &quot;equalities&quot;, how would you explain the pattern for other numbers? In particular, what would 25 equal?)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-04-16T15:34:53.713" LastActivityDate="2015-04-16T23:25:50.300" Title="A cunning involution" Tags="|mathematics|lateral-thinking|" AnswerCount="4" CommentCount="3" ClosedDate="2015-04-28T12:45:35.230" ContentLicense="CC BY-SA 3.0" />
  <row Id="12265" PostTypeId="2" ParentId="12263" CreationDate="2015-04-16T15:15:40.303" Score="0" Body="&lt;p&gt;My guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 5 = 1, as 1 = 5.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At least for one of the answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the relationship between the numbers:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 = 5, 2 = 10, 3 = 15, and 4 = 20, then 5 = ?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; x = 5*x so for 25, it wouldn't exist in this  equation as it's cyclic, but I'm not certain yet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-04-16T15:40:03.787" LastActivityDate="2015-04-16T15:40:03.787" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12267" PostTypeId="2" ParentId="12204" CreationDate="2015-04-16T15:38:56.473" Score="3" Body="&lt;p&gt;It is actually possible to not kill a sane seer, because there are three possible states for the question - yes, no, and paradox. If it's a paradox, the seer can't answer, giving us three possible responses - yes, no, or no-response. So we need to construct a question that can do all three. And because the mad seer's response is untrustworthy, we need the paradox to occur if seer 1 is the mad seer (assuming we ask seer 1 the question).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't trivial, but it's certainly doable. The neatest question I can think of is...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Is it simultaneously true that seer 2 is not the mad seer, and either seer 3 is the mad seer or your answer to this question is 'no' when you tell the truth?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This works because if seer 2 is the mad seer, then...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; since &quot;seer 2 is not the mad seer&quot; is false, the answer is &quot;no&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;while if seer 3 is the mad seer, then...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; since seer 2 is not the mad seer, we pass to the &quot;either&quot; case. Since seer 3 is the mad seer, the answer is &quot;yes&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but if seer 1 is the mad seer...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; we have that seer 2 is not the mad seer, so it passes to the &quot;either&quot; case, and since seer 3 is not the mad seer, the question boils down to &quot;is your answer to this question 'no'?&quot; This is a paradoxical question, and thus seer 1 cannot answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So no response indicates that seer 1 is the mad seer, &quot;no&quot; indicates that seer 2 is the mad seer, and &quot;yes&quot; indicates that seer 3 is the mad seer.&lt;/p&gt;&#xA;" OwnerUserId="9289" LastEditorUserId="9289" LastEditDate="2015-04-17T00:20:12.777" LastActivityDate="2015-04-17T00:20:12.777" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12268" PostTypeId="2" ParentId="12263" CreationDate="2015-04-16T15:47:24.590" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 5=1 because 1=5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the other numbers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 = 5&lt;br&gt;&#xA; 2 = 10&lt;br&gt;&#xA; 3 = 15&lt;br&gt;&#xA; 4 = 20&lt;br&gt;&#xA; 5 = 1&lt;br&gt;&#xA; 6 = 25&lt;br&gt;&#xA; 7 = 30&lt;br&gt;&#xA; 8 = 35&lt;br&gt;&#xA; 9 = 40&lt;br&gt;&#xA; 10 = 2&lt;br&gt;&#xA; 11 = 45&lt;br&gt;&#xA; ...&lt;br&gt;&#xA; 15 = 3&lt;br&gt;&#xA; ...   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Basically, if the number $n$ isn't multiple of 5, the equivalent is $(n\ DIV\ 5)\times 20 + (n\ MOD\ 5)\times 5$.&#xA;If the number is multiple of 5, the equivalent is $n/5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: this establish a bijective correspondence between a number and its transformed. $f(f(x))=x$&#xA;If it's not clear, DIV means &quot;integer division&quot;.&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-04-16T15:51:57.113" LastActivityDate="2015-04-16T15:51:57.113" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12269" PostTypeId="2" ParentId="12200" CreationDate="2015-04-16T15:53:30.767" Score="2" Body="&lt;p&gt;A quick search on &lt;a href=&quot;http://oeis.org/&quot; rel=&quot;nofollow&quot;&gt;http://oeis.org/&lt;/a&gt; returned the first sequence. Even with the match, it took me some time to understand it :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first sequence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it is the square root of the smallest square composed with the rank numbers with zero of more additional numbers&lt;br&gt;&#xA; n = 1 | smallest compound square = 1 | square root = 1&lt;br&gt;&#xA; n = 2 | smallest compound square = 25 | square root = 5&lt;br&gt;&#xA; n = 3 | smallest compound square = 36 | square root = 6&lt;br&gt;&#xA; n = 4 | smallest compound square = 4 | square root = 2&lt;br&gt;&#xA; n = 5 | smallest compound square = 25 | square root = 5&lt;br&gt;&#xA; n = 6 | smallest compound square = 16 | square root = 4&lt;br&gt;&#xA; n = 7 | smallest compound square = 576 | square root = 24&lt;br&gt;&#xA; n = 8 | smallest compound square = 81 | square root = 9&lt;br&gt;&#xA; n = 9 | smallest compound square = 9 | square root = 3&lt;br&gt;&#xA; and so on...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Still working on the 3 others&lt;/p&gt;&#xA;" OwnerUserId="10400" LastActivityDate="2015-04-16T15:53:30.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12270" PostTypeId="2" ParentId="12263" CreationDate="2015-04-16T15:57:02.077" Score="6" Body="&lt;p&gt;A logical continuation of the sequence, assuming that equality takes precedence (and is unique between two numbers) is the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1=5&lt;br&gt;&#xA;2=10&lt;br&gt;&#xA;3=15&lt;br&gt;&#xA;4=20&lt;br&gt;&#xA;5=1&lt;br&gt;&#xA;6=6&lt;br&gt;&#xA;7=11&lt;br&gt;&#xA;8=16&lt;br&gt;&#xA;9=21&lt;br&gt;&#xA;10=2&lt;br&gt;&#xA;11=7&lt;br&gt;&#xA;12=12&lt;br&gt;&#xA;13=17&lt;br&gt;&#xA;14=22&lt;br&gt;&#xA;15=3&lt;br&gt;&#xA;16=8&lt;br&gt;&#xA;17=13&lt;br&gt;&#xA;18=18&lt;br&gt;&#xA;19=23&lt;br&gt;&#xA;20=4&lt;br&gt;&#xA;21=9&lt;br&gt;&#xA;22=14&lt;br&gt;&#xA;23=19&lt;br&gt;&#xA;24=24  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which, you will notice, completely establishes equality within the numbers 1-24, with multiples of six only equalling themselves and all others forming pairs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the next value, 25, isn't defined by this sequence directly, but the natural choice is $25=29$, which starts the whole thing over again for numbers 25-48.&lt;/p&gt;&#xA;" OwnerUserId="9289" LastEditorUserId="10281" LastEditDate="2015-04-16T15:58:05.050" LastActivityDate="2015-04-16T15:58:05.050" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12271" PostTypeId="2" ParentId="12227" CreationDate="2015-04-16T15:59:17.593" Score="0" Body="&lt;p&gt;If we add &lt;a href=&quot;/questions/tagged/lateral-thinking&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged 'lateral-thinking'&quot; rel=&quot;tag&quot;&gt;lateral-thinking&lt;/a&gt; to the mix, I would say your best response is to tell the President that you report to the Director, not the President. If he wants to investigate the CIA, he should call the CIA Inspector General or the Senate.&lt;/p&gt;&#xA;" OwnerUserId="9000" LastActivityDate="2015-04-16T15:59:17.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12272" PostTypeId="2" ParentId="12211" CreationDate="2015-04-16T16:59:12.730" Score="1" Body="&lt;p&gt;&lt;strong&gt;Solution: $x=2, y=1001$.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the beginning we know two things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$x, y\in [1, 2002]$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$x+y=2002 \lor x*y=2002$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Every statement of Sam and Peter gives us one more bit of information:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sam: I don't know your number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow x$ has to be a divisor of $2002$, otherwise we could decide $x+y=2002 \lor x*y=2002$ and Sam could calculate $y=2002-x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $x\in M=\{1, 2, 7, 11, 14, 22, 77, 91, 143, 154, 182, 286, 1001, 2002\}$ &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Peter: I don't know your number either.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow$ For the same reason, $y$ must be $\in M$ as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Peter does still not know the answer at that point, that means he still cannot decide whether $2002=x+y$ or $2002=x*y$. That only works with with $2*1001=2002$ and $1001+1001=2002$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow (x, y) = (2, 1001)\lor (x, y)=(1001,1001) \lor (x, y)=(1001, 2)$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sam: Now I know your number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For $x=1001$, Sam could not decide between $y=2$ and $y=1001$. If Sam knows the answer now, it must mean that $x=2$ and $y=1001$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Peter: Now I know yours too.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Peter can reason as well as we can, so by now he knows that $x=2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Sam wouldn't know the answer before, Peter would know that $x=1001$ at this point. So after Peter's last statement, Sam would know the value of $y$ in any case.&lt;/p&gt;&#xA;" OwnerUserId="11505" LastActivityDate="2015-04-16T16:59:12.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12273" PostTypeId="2" ParentId="11718" CreationDate="2015-04-16T17:32:49.567" Score="3" Body="&lt;h1&gt;This is not the definitive answer!&lt;/h1&gt;&#xA;&lt;p&gt;&lt;strong&gt;This is just a summary of the information we got about his last video, a list of failed attempts and potentially relevant elements for new approaches.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Elements contained in the video:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The name of the video: 0101 0101 0000 0111&lt;/li&gt;&#xA;&lt;li&gt;Usual intro music&lt;/li&gt;&#xA;&lt;li&gt;Suite No. 3 in D major by Johann Sebastian Bach, also known as &amp;quot;&lt;strong&gt;Air on the G string&lt;/strong&gt;&amp;quot;&lt;/li&gt;&#xA;&lt;li&gt;The string &amp;quot;&lt;strong&gt;ISL 4 &lt;span class=&quot;math-container&quot;&gt;$ $&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$ $&lt;/span&gt; 3 xEFTvZ&lt;/strong&gt;&amp;quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;strong&gt;Analysis of these elements:&lt;/strong&gt;&lt;br /&gt;&#xA;The name of the video is, as usually, a binary number. Title numbers are always consecutive, in this case 0101 0101 0000 0111, which means 5507 in decimal.  There seems to be no hint in the title.&lt;br /&gt;&#xA;As well, we have the typical intro music, basically NOC's signature. No info here!&lt;br /&gt;&#xA;Then we listen &amp;quot;Suite No. 3 in D major&amp;quot;. What may be useful in it? Well, we have the number 3 and the capital letter D, but those information appear to be pretty irrelevant.&lt;br /&gt;&#xA;The real information is likely to be in the second name of that music, which is &amp;quot;Air on the G string&amp;quot;. The words &amp;quot;Air&amp;quot; and &amp;quot;string&amp;quot; have the same length of &amp;quot;ISL&amp;quot; and &amp;quot;xEFTvZ&amp;quot;, suggesting a direct correspondence between them.&lt;br /&gt;&#xA;Also, the word &amp;quot;On&amp;quot; could be interpreted as &amp;quot;1&amp;quot; (ON=1, OFF=0), so both strings can be perfectly aligned now.  What about the &amp;quot;G&amp;quot;? When written in lower-case, &amp;quot;g&amp;quot; tends to be visually similar to the number &amp;quot;9&amp;quot;, even though this appears captious.&lt;br /&gt;&#xA;Well, know we have the nucleus of the puzzle, the string ISL 4 &lt;span class=&quot;math-container&quot;&gt;$ $&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$ $&lt;/span&gt; 3 xEFTvZ.&lt;br /&gt;&#xA;The most notably features of this string are the presence of exactly 11 alpha-numeric characters, as in all Youtube videos ID (a video ID is what we actually expect to find), the sentence &amp;quot;IS LEFT Z&amp;quot; formed by the capital letters, an insolit separation between &amp;quot;4&amp;quot; and &amp;quot;3&amp;quot; (my ruler shows that it's made of 3 blank-spaces). As suggested before, these words align pretty well with the title of the music.&#xA;Someone also pointed out that exactly 11 notes are played in the music. This might be a great hint, if confirmed!&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Possible strategies:&lt;/strong&gt;&lt;br /&gt;&#xA;It's almost obvious that we have to deal with a cipher, as we already did earlier. A Caesar's cipher would be too trivial, we've already solved that kind of cipher in a previous video.&lt;br /&gt;&#xA;Vigenere cipher is the top candidate to solve this mystery, since we are dealing with two strings of similar length.&lt;br /&gt;&#xA;Do we have to add or subtract the strings? We don't really know, try both!&lt;/p&gt;&#xA;&lt;p&gt;Encrypted text:&lt;br /&gt;&#xA;&amp;quot;&lt;strong&gt;ISL 4 &lt;span class=&quot;math-container&quot;&gt;$ $&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$ $&lt;/span&gt; 3 xEFTvZ&lt;/strong&gt;&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;Potentially useful keys:&lt;br /&gt;&#xA;&lt;strong&gt;AIR 1 G STRING&lt;/strong&gt;&lt;br /&gt;&#xA;&lt;strong&gt;AIR 1 9 STRING&lt;/strong&gt;&lt;br /&gt;&#xA;&lt;strong&gt;AIR ? ? STRING&lt;/strong&gt;  where ? means any number/letter between 0-9/A-Z. This generalizes the AIR STRING approach.&lt;br /&gt;&#xA;&lt;strong&gt;AIRAIRAIRAIRAIRAIRAIRAIRAIR&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Also, the 11 notes in the video could be a key, if only we could find their names.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;What if nothing of this works?&lt;/strong&gt;&lt;br /&gt;&#xA;Well, we would definitely need a new idea! We also might need to dig into his 3 previous public videos, looking for additional hints.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;What can I do to help?&lt;/strong&gt;&#xA;First, you have to officially verify that Caesar's cipher doesn't work at all. It requires only 26 attempts, if you do it please confirm its non-validity in the comments.&#xA;Then, hoping that you have a discrete experience with programming and coding (especially PHP) you could brute-force all the Vigenere keys suggested above and check if that video exists. Please be accurate, a minor mistake in your code is likely to miss the ID we're looking for!&#xA;Also, if you have time to waste (or your pc has), apply Caesar to all the ID you got and verify them as well!&#xA;If you're experienced with music, please verify that the played notes are 11. If you are able, identify them!&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Sources and credits:&lt;/strong&gt;&lt;br /&gt;&#xA;My reason, &lt;a href=&quot;https://www.youtube.com/watch?v=6llW7rr2DUk&quot; rel=&quot;nofollow noreferrer&quot;&gt;comments below the video&lt;/a&gt;, &lt;a href=&quot;http://chat.stackexchange.com/rooms/22141/noc10&quot;&gt;this discussion&lt;/a&gt;. Thanks to Ian MacDonald for noticing the 11 notes.&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-16T19:06:58.293" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="12274" PostTypeId="2" ParentId="12252" CreationDate="2015-04-16T17:49:24.173" Score="9" Body="&lt;p&gt;The result of the multiplication:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    0100010100101 x 01001001001000101 = 010010010010010001010010100101&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Description of Mr. EG's coding scheme:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To code a positive integer, factor it as a product of primes, and write those primes in increasing order of size. Write each prime factor in binary, and prepend a number of 0's which is one less than the number of digits. Then concatenate the resulting strings.&lt;br&gt;&lt;br&gt; As an example, to code $20$, we factor $20=2\cdot 2\cdot 5$. The binary representations of $2$ and $5$ are 10 and 101. We prepend one and two 0's, respectively, to get 010 and 00101. Finally, we concatenate two copies of 010 and one copy of 00101 to get 01001000101.&lt;br&gt; The purpose of the leading 0's is to indicate the number of digits in each prime, so that a positive integer can be unambiguously recovered from its representation. For instance, if we did not prepend 0's, the string 11111 could either come from $21=3\cdot 7$, or from $31$.&lt;br&gt;&lt;br&gt; Translating back into decimal, the given multiplications are:&lt;br&gt; $2\times 3 = 6$,&lt;br&gt; $3\times 4=12$,&lt;br&gt; $6\times 12=72$,&lt;br&gt;$10\times 28=280$.&lt;br&gt; The missing multiplication becomes $(2\cdot 5\cdot 5)\times(2\cdot2\cdot 2\cdot 2\cdot 5)=4000=(2\cdot2\cdot2\cdot2\cdot2\cdot5\cdot 5\cdot 5)$, which is coded as 010 010 010 010 010 00101 00101 00101.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-04-17T13:32:41.933" LastActivityDate="2015-04-17T13:32:41.933" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12276" PostTypeId="1" AcceptedAnswerId="12395" CreationDate="2015-04-16T18:12:24.347" Score="6" ViewCount="523" Body="&lt;p&gt;Well, &lt;a href=&quot;https://puzzling.stackexchange.com/questions/12181/mirror-of-men-he-showers-with-liberal-hands&quot;&gt;my last question&lt;/a&gt; wasn't received that well, nevertheless I will make a better one, besides you will need all the answers you can get.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;vcejssnr&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I can't leave you with a stupid cryptogram and another strange title, can I?&lt;br&gt;&#xA;I did that one time, this time it's not that simple because what I am looking for cannot be described in words. It is however, the answer in more ways than one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My words are well chosen, mistakes are &lt;strong&gt;never&lt;/strong&gt; made.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This one time I will layout the pad before you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The previous question is more useful than you might think&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #4&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you know vcejssnr then something in the previous question must be familiar, if not google.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #5&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the answer is the driving factor for the entirety of the story&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If you take away anything from this puzzle let it be that&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;the book mentioned in the answer is worth a read :)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9835" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-04-26T14:56:26.847" Title="difficile est saturam non legere" Tags="|riddle|cipher|language|" AnswerCount="1" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="12277" PostTypeId="2" ParentId="12263" CreationDate="2015-04-16T20:26:48.233" Score="5" Body="&lt;p&gt;How about simply defining: for all positive fractions $a$ and $b$, $a \sim b$ iff $a/b$ is a power (positive or negative) of $5$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's an equivalence relation, with $2 \sim 10$, $3 \sim 15$, $4 \sim 20$, and $1 \sim 5 \sim 25$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an involution, note that each equivalence class is of the form $\{a5^n \mid n \in \mathbb Z\}$ for some representative $1 \leq a &amp;lt; 5$. Then $f : \mathbb Q^+ \to \mathbb Q^+ : a5^n \to a5^{n \text{ xor } 1}$ is an involution.&lt;/p&gt;&#xA;" OwnerUserId="11476" LastEditorUserId="11476" LastEditDate="2015-04-16T23:25:50.300" LastActivityDate="2015-04-16T23:25:50.300" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12278" PostTypeId="1" AcceptedAnswerId="12362" CreationDate="2015-04-16T20:30:11.487" Score="19" ViewCount="3398" Body="&lt;p&gt;The poster on the university bulletin board is definitely trying to drum up some hype:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h2&gt;One Night Only&lt;/h2&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Be one of the privileged individuals present when Professor Hubert J. Farnsworth reveals his new research regarding the secret movie codes!  Never-before-heard secrets will be revealed!&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;Tonight only, in the main lecture hall.&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;Admission is $50.00/person.  Presentation begins at 7:00 p.m. sharp.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Professor Farnsworth is head of the university's Film Studies department.  You've always thought he's a bit of a quack, but his classes are very popular.  Over the last couple of weeks the buzz has been building that he's made a spectacular discovery, but no one's sure exactly what it is.  He's been very obviously carrying a locked briefcase everywhere with him, which is purported to contain some revolutionary new research that he's been working on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You determine that there's no way you're paying $50 just to hear some crackpot spout gibberish for a couple hours, and are about to head to your next class when you hear a scream from the classroom down the hall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Running toward the sound, you see a hysterical student standing at the classroom door, gibbering.  Inside, the body of Professor Farnsworth is sprawled across a desk, and blood is everywhere.  His briefcase is nowhere to be seen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suddenly, you notice that Professor Fansworth is still moving!  He's beckoning you towards him.  You quickly run towards him, trying to recall anything you can from the first-aid class you took several years ago.  As you get close, you see the number of stab wounds in his chest and the size of the blood pool underneath him, and realize that there's little chance of saving him.  But he desperately wants to tell you something.  You lean close to him, and he manages to get out the words, &quot;Read the board,&quot; before dramatically expiring, just like in the movies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You glance at the blackboard, and notice a bunch of numbers.  Apparently he wrote this after he'd been stabbed, judging by the number of blood smears on the blackboard.  Unfortunately, the numbers hold little meaning for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UN4yq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The police arrive shortly thereafter, and take charge of the scene.  They are very eager to interview you, since you were the last one to speak to the professor.  The officer interviewing you is remarkably candid, and shares some of the details of the case with you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently there were six other professors who were experts in a similar field, and were all in town for the &quot;big reveal&quot; tonight.  They are the prime suspects in the murder investigation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their names are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Professor Charles Xavier&lt;/li&gt;&#xA;&lt;li&gt;Professor James Moriarty&lt;/li&gt;&#xA;&lt;li&gt;Professor Gilderoy Lockhart&lt;/li&gt;&#xA;&lt;li&gt;Professor Robert Langdon&lt;/li&gt;&#xA;&lt;li&gt;Professor Urban Chronotis&lt;/li&gt;&#xA;&lt;li&gt;Professor Mohinder Suresh&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The police are fairly certain that one of these men stabbed poor Professor Farnsworth to death and stole his research, but they have no evidence pointing to any of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Can you decode the blackboard message and help the police track down Professor Farnsworth's killer?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Hint 1&lt;/h3&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; IMDB is your friend.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;Hint 2&lt;/h3&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Look at the lengths of the titles of each of the results you get after using Hint 1.  Ignore spaces and punctuation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;Hint 3 (spoiler)&lt;/h3&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Convert each of the lengths from Hint 2 into a letter (A=1, B=2, etc.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="4421" LastEditDate="2016-06-02T12:02:29.813" LastActivityDate="2016-06-02T12:02:29.813" Title="The secret research of Professor Hubert J. Farnsworth (a murder mystery puzzle)" Tags="|enigmatic-puzzle|story|mystery|movies|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12279" PostTypeId="1" AcceptedAnswerId="12283" CreationDate="2015-04-16T20:35:41.120" Score="10" ViewCount="847" Body="&lt;p&gt;So my last riddle turned out to be too easy. Lets see if this one is difficult enough. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You wake up with a huge headache and look around you. Somehow you managed to end up in a white room of about 2 by 2 meters with no furniture or windows. The only thing that stands out is a white door with a small computer screen next to it. You try opening the door but its locked. It doesn't seem like it can be forced either. On the computer screen you see that the door is locked with a six-digit numeric passcode. Fortunately whoever set the passcode put a hint with it, probably because they aren't clever enough to remember the passcode themselves. The hint reads:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2351389233159728657, I shall unite with my former self, but once united I cannot be split. Tell me, what will I be next? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which code do you enter to open the door?&lt;/p&gt;&#xA;" OwnerUserId="11473" LastEditorDisplayName="user88" LastEditDate="2015-06-11T17:41:05.763" LastActivityDate="2015-06-11T17:41:05.763" Title="Continue the sequence" Tags="|mathematics|number-sequence|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12281" PostTypeId="2" ParentId="12187" CreationDate="2015-04-16T21:07:48.987" Score="4" Body="&lt;p&gt;If parentheses will not be counted and if we could use it as multiply:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $12(34)-5(67)-8(9) = 1$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I used only 2 operators.&lt;/p&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="11497" LastEditDate="2018-05-01T03:49:12.073" LastActivityDate="2018-05-01T03:49:12.073" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12283" PostTypeId="2" ParentId="12279" CreationDate="2015-04-16T21:39:17.250" Score="12" Body="&lt;p&gt;The numeric code is &lt;strong&gt;514229&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The numbers in the string are 2, 3, 5, 13, 89, 233, 1597, 28657, which are all prime Fibonacci numbers. The next one is 514229.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The clue &quot;I shall unite with my former self&quot; refers to the fact that each number is the sum of the previous two numbers, and the clue &quot;but once united I cannot be split&quot; refers to the fact that the numbers cannot be divided.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-04-16T21:39:17.250" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12284" PostTypeId="1" AcceptedAnswerId="12370" CreationDate="2015-04-16T21:40:23.777" Score="37" ViewCount="1727" Body="&lt;p&gt;A magician and her assistant are performing a magic show today, and you are the only audience member. Since it is a Thursday, the magician is saving her strength for the inevitable dragon attack and only performs simple tricks involving dice, coins, and cards. At 2 pm, the dragon attacks the village and the magician goes outside to fight it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The assistant invites you to participate in the next trick, and hands you nine (six sided) dice and two identical opaque cups. Prompted by the assistant, you first carefully check that none of the dice are weighted, that they have no magnetic properties, etc., then roll them and arrange them in a horizontal line along a table. The assistant stares at the row of dice for a few minutes, then finally points to two of the nine dice (without touching them) and asks you to cover them up with the two cups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After you cover up the indicated dice, you murder the assistant, arrange a few fake clues to make it seem like the magician was the murderer, and call the police. While the police are responding to your call, the magician stumbles back onstage, exhausted from the fight and covered in dragon blood. She doesn't notice the dead body of her assistant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magician stares at the row of seven visible dice and two opaque cups for a few minutes, then finally points to the first cup and says, &quot;the die under this cup has a 5 on top.&quot; She then points at the second cup and says, &quot;the die under this cup has a 2 on top.&quot; Both guesses are correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did the magic trick work? (Note: magic is not real; the magician defeated the dragon using her sword.)&lt;/p&gt;&#xA;" OwnerUserId="11295" LastEditorUserId="5373" LastEditDate="2015-04-26T11:34:17.073" LastActivityDate="2016-01-15T19:49:39.483" Title="Magician, assistant, nine dice, and two cups" Tags="|story|dice|magic|" AnswerCount="6" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="12285" PostTypeId="2" ParentId="12278" CreationDate="2015-04-16T21:41:22.100" Score="2" Body="&lt;p&gt;Some partial work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each number is the IMDB code of a famous movie (&lt;a href=&quot;http://www.imdb.com/title/tt0091042/&quot; rel=&quot;nofollow&quot;&gt;example&lt;/a&gt;), as clued by &quot;movie codes&quot;.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Ferris Bueller's Day Off Batman Begins               Good Will Hunting             Jumper                     The Fast and the Furious&#xA;Chocolat                 WALL·E                      Munich                        Toy Story 2                Dragonfly&#xA;Labyrinth                Star Trek: First Contact                                  Dirty Dancing              Air Bud: Golden Receiver&#xA;The Devil Wears Prada    Sinbad of the Seven Seas    Basic                         Field of Dreams            Deck the Halls&#xA;Cars                     Dogma                       Arthur and the Invisibles     My Super Ex-Girlfriend     John Q&#xA;                         Addams Family Values        Ghost&#xA;                                                     All the Pretty Horses&#xA;                                                     The Bridges of Madison County&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note sure how to extract. I was expecting to read off the first letters and get a message, or the 7th ones (clued by 7pm). I tried converting movie release years to letters, but that didn't give anything promising either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The murder suspects also have character codes &lt;a href=&quot;http://www.imdb.com/character/ch0001110/&quot; rel=&quot;nofollow&quot;&gt;(example)&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Professor Charles Xavier (Character)      from X-Men (2000)                                   ch0001110&#xA;Professor James Moriarty (Character)      from &quot;Sherlock&quot; (2010)                              ch0027269&#xA;Professor Gilderoy Lockhart (Character)   from Harry Potter and the Chamber of Secrets (2002) ch0001001&#xA;Dr. Robert Langdon (Character)            from The Da Vinci Code (2006)                       ch0004644&#xA;Professor Chronotis (Character)           from Doctor Who: Shada (1992) (V)                   ch0385925&#xA;Mohinder Suresh (Character)               from &quot;Heroes&quot; (2006/II)                             ch0015543&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-04-16T21:59:03.017" LastActivityDate="2015-04-16T21:59:03.017" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12286" PostTypeId="1" AcceptedAnswerId="12306" CreationDate="2015-04-16T22:02:42.107" Score="3" ViewCount="852" Body="&lt;p&gt;The grid below contains the first part of a movie quote.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the full 46-word quote, and what is the movie?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1mo8X.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint #1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The movie was very mainstream and was released in the last 10 years.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first 4 words of the quote can be figured out from the grid, and then Google is your friend.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are 385 hints in the grid itself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I would consider this the perfect movie quote.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The ten letters needed for the first 4 words of the quote are organized in a certain pattern.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-04-17T01:29:34.157" LastActivityDate="2015-04-21T13:22:43.823" Title="What movie am I?" Tags="|riddle|mathematics|pattern|movies|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12287" PostTypeId="2" ParentId="12278" CreationDate="2015-04-16T22:07:44.660" Score="14" Body="&lt;p&gt;With the pretext being about secret movie codes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have compiled the following list:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;0091042 - Ferris Bueller's Day Off&lt;/li&gt;&#xA;&lt;li&gt;0372784 - Batman Begins&lt;/li&gt;&#xA;&lt;li&gt;0119217 - Good Will Hunting&lt;/li&gt;&#xA;&lt;li&gt;0489099 - Jumper&lt;/li&gt;&#xA;&lt;li&gt;0232500 - The Fast and the Furious&#xA;&lt;br /&gt;&lt;br /&gt;&lt;/li&gt;&#xA;&lt;li&gt;0241303 - Chocolat&lt;/li&gt;&#xA;&lt;li&gt;0910970 - Wall-E&lt;/li&gt;&#xA;&lt;li&gt;0409306 - Munich&lt;/li&gt;&#xA;&lt;li&gt;0120363 - Toy Story&lt;/li&gt;&#xA;&lt;li&gt;0259288 - Dragonfly&#xA;&lt;br /&gt;&lt;br /&gt;&lt;/li&gt;&#xA;&lt;li&gt;0091369 - Labyrinth&lt;/li&gt;&#xA;&lt;li&gt;0117731 - Star Trek: First Contact&lt;/li&gt;&#xA;&lt;li&gt;0092890 - Dirty Dancing&lt;/li&gt;&#xA;&lt;li&gt;0140796 - Air Bud: Golden Receiver&#xA;&lt;br /&gt;&lt;br /&gt;&lt;/li&gt;&#xA;&lt;li&gt;0458352 - The Devil Wears Prada&lt;/li&gt;&#xA;&lt;li&gt;0098333 - Sinbad of the Seven Seas&lt;/li&gt;&#xA;&lt;li&gt;0264395 - Basic&lt;/li&gt;&#xA;&lt;li&gt;0097351 - Field of Dreams&lt;/li&gt;&#xA;&lt;li&gt;0790604 - Deck the Halls&#xA;&lt;br /&gt;&lt;br /&gt;&lt;/li&gt;&#xA;&lt;li&gt;0317219 - Cars&lt;/li&gt;&#xA;&lt;li&gt;0120655 - Dogma &lt;/li&gt;&#xA;&lt;li&gt;0344854 - Arthur and the Invisibles&lt;/li&gt;&#xA;&lt;li&gt;0465624 - My Super Ex-Girlfriend&lt;/li&gt;&#xA;&lt;li&gt;0251160 - John Q&#xA;&lt;br /&gt;&lt;br /&gt;&lt;/li&gt;&#xA;&lt;li&gt;0106220 - Addams Family Values&lt;/li&gt;&#xA;&lt;li&gt;0099653 - Ghost&#xA;&lt;br /&gt;&lt;br /&gt;&lt;/li&gt;&#xA;&lt;li&gt;0149624 - All the Pretty Horses&#xA;&lt;br /&gt;&lt;br /&gt;&lt;/li&gt;&#xA;&lt;li&gt;0112579 - The Bridges of Madison County&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This makes the message&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Robert Langdon stole my research&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;you get this message by:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Laying out the 3rd letter (excluding spaces) of the movie title in the order they appear on the board&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;&#xA;r t o m e&#xA;o l n y a&#xA;b a   r r&#xA;e n s e c&#xA;r g t s h&#xA;  d o&#xA;    l&#xA;    e&#xA;&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="11150" LastEditorUserId="11150" LastEditDate="2015-04-16T22:19:07.490" LastActivityDate="2015-04-16T22:19:07.490" CommentCount="20" ContentLicense="CC BY-SA 3.0" />
  <row Id="12288" PostTypeId="1" AcceptedAnswerId="12289" CreationDate="2015-04-16T23:23:34.863" Score="6" ViewCount="218" Body="&lt;p&gt;Crosswords are harder when you don't know the clues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nyP1u.png&quot; width=&quot;281&quot; height=&quot;366&quot; title=&quot;It's all Greek to me&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution can be found in the grey shaded boxes.&lt;br&gt;&#xA;&lt;sub&gt;(How many four letter words are there starting with Z, again?)&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="7308" LastActivityDate="2015-04-17T00:12:30.687" Title="Clueless Alphabetical" Tags="|enigmatic-puzzle|crosswords|" AnswerCount="1" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="12289" PostTypeId="2" ParentId="12288" CreationDate="2015-04-17T00:02:51.750" Score="7" Body="&lt;p&gt;The answers are all&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Letters from the Greek alphabet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Giving the grey boxes as&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; M,n,g,e,i,a&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rearranging to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Enigma&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11150" LastEditorUserId="11150" LastEditDate="2015-04-17T00:12:30.687" LastActivityDate="2015-04-17T00:12:30.687" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12290" PostTypeId="1" AcceptedAnswerId="12292" CreationDate="2015-04-17T00:14:03.467" Score="6" ViewCount="578" Body="&lt;p&gt;This one shouldn't be too hard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Preparation stage: Take a shuffled deck of standard playing cards and deal 13 face-down piles of 4 cards each. Pick up each pile, arrange the order of the cards however you like, and put it back face-down before moving on to the next pile. Once you put a pile back, you can't return to it. Once you have gone through all the piles, turn the top card of each pile face-up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Playing stage: Stack cards by putting lower cards on top of consecutive higher cards of the same suit. You can move a whole stack of consecutive cards. You can have at most 13 piles. If you ever have less than 13 piles, you can move a stack with a king at the bottom to the empty spot. If a face-down card is ever at the top of a pile, turn it face-up. You win if you can get down to 4 piles (which will be the king-to-ace sequence for each suit).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What strategy in the preparation stage will guarantee victory? (Let me know if the game rules aren't clear).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: It is currently unclear if there is a way to guarantee victory. The answer pointing this out is the currently accepted answer. A strategy which is proven to guarantee victory OR a proof of no such strategy existing are welcome.&lt;/p&gt;&#xA;" OwnerUserId="11154" LastEditorUserId="11154" LastEditDate="2015-04-17T03:30:34.430" LastActivityDate="2015-04-17T03:30:34.430" Title="Solitaire Puzzle" Tags="|cards|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12291" PostTypeId="2" ParentId="12211" CreationDate="2015-04-17T00:26:35.083" Score="5" Body="&lt;p&gt;Given Sam's number is $x$ and Peter's number is $y$, most people seem to be working from the following premises:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$x, y \in \mathbb{N}$ where $\mathbb{N}$ is the natural integers; $\mathbb{N} = \{1, 2, ...\}$&lt;/li&gt;&#xA;&lt;li&gt;$2002 = x + y \lor 2002 = xy$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If that's correct then you &lt;em&gt;must&lt;/em&gt; agree with &lt;a href=&quot;https://puzzling.stackexchange.com/a/12216&quot;&gt;Mike Earnest's answer&lt;/a&gt; or be logically impaired. A stricter reading of the problem provides these premises:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$x \in \mathbb{N} = \{1, 2, ...\}$&lt;/li&gt;&#xA;&lt;li&gt;$y \in \mathbb{R}$ where $\mathbb{R}$ is the set of all real numbers&lt;/li&gt;&#xA;&lt;li&gt;$2002 = x + y \lor 2002 = xy$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We can intuitively (and logically) exclude more interesting numbers (complex, imaginary, etc.) for the value of $y$. I've omitted that here.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sam: I don't know your number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of course Sam would have no idea. $\forall n \in \mathbb{N}$ $\exists r_a, r_m \in \mathbb{R}$ such that $(r_a \neq r_m)$ $\land$ $(2002 = x + r_a)$ $\land$ $(2002 = xr_m)$. Thanks, Sam, you told us nothing.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Peter: I don't know your number either.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is telling. This means that $\exists$ $n_a, n_m \in \mathbb{N}$ such that $(n_a \neq n_m)$ $\land$ $(2002 = n_a + y)$ $\land$ $(2002 = n_my)$ which implies:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y \in \mathbb{Z} \;where\; \mathbb{Z} = \{..., -2, -1, 0, 1, 2, ...\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because $\nexists$ $r \in \mathbb{R}, n \in \mathbb{N}$ such that $2002 = n + r$ $\land$ $r \notin \mathbb{Z}$. Which means if $y \notin \mathbb{Z}$, then Peter could find $x = {2002}/y$. But Peter doesn't know $x$, so $y \in \mathbb{Z}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, we can deduce:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y \in \mathbb{N}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\nexists$ $y \in \mathbb{Z}$ such that $y &amp;lt; 1$ $\land$ $2002 = xy$. If $y &amp;lt; 1$, then Peter could find $x = 2002 - y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ y \in F_y = \{1, 2, 7, 11, 14, 22, 77, 91, 143, 154, 182, 286, 1001\} \subset F$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $F$ is the factors of $2002$. (Note: $F = F_y \cup \{2002\}$.) We know this because $\exists n_a, n_m \in \mathbb{N}$ such that $(n_a \neq n_m)$ $\land$ $(2002 = n_a + y)$ $\land$ $(2002 = n_my)$; otherwise, Peter could eliminate one of the formulas and calculate $x$. The only numbers that satisfy this criteria are in $F_y$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sam: Now I know your number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The big takeaway from this is $x \neq 1001$; $x = 1001$ is the case where Sam still doesn't know $y$. This is because $2002 = 1001 + 1001$ and $2002 = 1001 * 2$. Hence, $y$ could either be $1001$ or $2$, and Sam would not know which number.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Peter: Now I know yours too.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That one bit of info (other inferences aside), must have given Peter enough knowledge to solve the problem, so $x = 1001$ must have been a potential possibility prior to Sam's statement. There's only two values of $y$ for which this is the case:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2002 = 1001 + 1001,\;y = 1001$$&#xA;$$2002 = 1001 * 2,\;y = 2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which means the other formula will give us the potential values for $x$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2002 = 2 * 1001,\;x = 2$$&#xA;$$2002 = 2000 + 2,\;x = 2000$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there are two solutions: either Sam picked $2$ and Peter picked $1001$, or Sam picked $2000$ and Peter picked $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unlike the other version where $y \in \mathbb{N}$, in this case we do not know if $x \in F$. If $x \in F$ then there would only be one solution: Sam picked $2$ and Peter picked $1001$.&lt;/p&gt;&#xA;" OwnerUserId="11500" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-04-17T13:24:03.997" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12292" PostTypeId="2" ParentId="12290" CreationDate="2015-04-17T00:32:53.263" Score="4" Body="&lt;p&gt;It does not look like it is possible to guarantee victory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reversing Emrakul's answer, I came up with the following scenario that is not winnable under the terms of the puzzle. You would need to swap the 4d and 7d in order for it to be solvable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the time you have seen the 8d,5d,6d pile, it is too late for you to go back and swap the order of the 4d and 7d.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;9c 9d  9h  ac  7h  kh  kc  qc  8d  qs  jc  0c  7c&#xA; as ad  ah  2c  4s  kd  8c  5c  5d  ks  jh  qh  0h&#xA; 2s 2d  2h  3c  5s  7d  6c  0d  6d  0s  jd  5h  8h&#xA; 3s 3d  3h  9s  6s  4d  4c  4h  7s  8s  js  qd  6h&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="11150" LastEditorUserId="11150" LastEditDate="2015-04-17T02:47:52.237" LastActivityDate="2015-04-17T02:47:52.237" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12293" PostTypeId="2" ParentId="12290" CreationDate="2015-04-17T01:09:24.003" Score="5" Body="&lt;p&gt;Place the cards in sequential order from low on top to high on bottom, where spades are 1-13, hearts are 14-26, diamonds are 27-39, and clubs are 40-52, with one exception:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If encounter sequential cards of the same suit, reverse them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, if you encounter $A_s, 2_s, 4_d, 9_c$, put them in the order $2_s, A_s, 4_d, 9_c$. That way, if the three of spades $3_s$ comes up, you'll be able to play the 2 of spades on it, then the ace. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The exception is critically important. Without it, you'll bury an 8 under a 7, and won't be able to move it. However, if you reverse it, when the 6 comes up, you move the 7, then the 8 is exposed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do this, you're guaranteed to win. The next spades are guaranteed to come up, because you'll always be able to move the top card somewhere else.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorDisplayName="user20" LastEditDate="2015-04-17T01:14:31.537" LastActivityDate="2015-04-17T01:14:31.537" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12295" PostTypeId="2" ParentId="3300" CreationDate="2015-04-17T02:29:28.713" Score="2" Body="&lt;p&gt;Leo had &lt;em&gt;almost&lt;/em&gt; the right idea in his solution, but didn't quite get his explanation right.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;The problem boils down to proving that every dyadic multiple of &lt;span class=&quot;math-container&quot;&gt;$90^\circ$&lt;/span&gt; (i.e. every angle in the form &lt;span class=&quot;math-container&quot;&gt;$90^\circ \times \frac{a}{2^n}$&lt;/span&gt;) can be represented by a unique ratio of N's and E's. The solution for the rest of the compass follows by symmetry.&lt;/p&gt;&#xA;&lt;p&gt;First, let's take a look at the chart that Leo put up, representing the number of N's and E's of a compass point in an ordered pair &lt;span class=&quot;math-container&quot;&gt;$(N, E)$&lt;/span&gt;:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;Level 1: (1,0)                                           (0,1)&#xA;Level 2: (1,0)                   (1,1)                   (0,1)&#xA;Level 3: (1,0)       (2,1)       (1,1)       (1,2)       (0,1)&#xA;Level 4: (1,0) (3,1) (2,1) (3,2) (1,1) (2,3) (1,2) (1,3) (0,1)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;This configuration of numbers is known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stern%E2%80%93Brocot_tree&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stern-Brocot tree&lt;/a&gt;, a way to generate a mapping between the rational numbers and the binary fractions from 0 to 1.&lt;/p&gt;&#xA;&lt;p&gt;This proof that this tree is such a mapping has three steps, as shown below:&lt;/p&gt;&#xA;&lt;h2&gt;1. Proof of uniqueness&lt;/h2&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;And north-north-north-east-east-east-east?&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;That's the same as the last one, 9/64 from north,&amp;quot; explained Stewart. &amp;quot;Only the number of norths and easts matters, not their orders.&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Some people expressed doubts about this claim, and the fact that it could be proven.&lt;/p&gt;&#xA;&lt;p&gt;Let &lt;span class=&quot;math-container&quot;&gt;$(a, b)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$(c, d)$&lt;/span&gt; be any two points on the same level. Then, the point between them is &lt;span class=&quot;math-container&quot;&gt;$(a+c, b+d)$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Now, notice that the points &lt;span class=&quot;math-container&quot;&gt;$(a, b)$&lt;/span&gt; on any level are ordered by the ratio &lt;span class=&quot;math-container&quot;&gt;$\frac ba$&lt;/span&gt;. We can prove that this ordering is preserved every time you go down a level by proving that if &lt;span class=&quot;math-container&quot;&gt;$\frac ab &amp;lt; \frac cd$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$\frac ab &amp;lt; \frac{a+c}{b+d} &amp;lt; \frac cd$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt; Cross-multiply the demoninators so that you get &lt;span class=&quot;math-container&quot;&gt;$ad &amp;lt; bc$&lt;/span&gt;. Then, if you add &lt;span class=&quot;math-container&quot;&gt;$ab$&lt;/span&gt; to both sides of the equation, you get &lt;span class=&quot;math-container&quot;&gt;$a(b+d) &amp;lt; b(a+c)$&lt;/span&gt;, and then divide by &lt;span class=&quot;math-container&quot;&gt;$b(b+d)$&lt;/span&gt; on both sides to get &lt;span class=&quot;math-container&quot;&gt;$\frac ab &amp;lt; \frac{a+c}{b+d}$&lt;/span&gt;. If you add &lt;span class=&quot;math-container&quot;&gt;$cd$&lt;/span&gt; to both sides of the equation instead, you get &lt;span class=&quot;math-container&quot;&gt;$(a+c)d &amp;lt; (b+d)c$&lt;/span&gt;, and divide by &lt;span class=&quot;math-container&quot;&gt;$d(b+d)$&lt;/span&gt; on both sides to get &lt;span class=&quot;math-container&quot;&gt;$\frac{a+c}{b+d} &amp;lt; \frac cd$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;h2&gt;2. Proof of relative primality&lt;/h2&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;How about south-east-east-south?&amp;quot; asked Stewart.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;That's just south-east. You can't use a number of 'south's and 'east's with a common factor,&amp;quot; explained John. &amp;quot;They'd all be the same.&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The second claim that people had doubts about was this one. What if one of the in-betweens really did have an NEEN or NENNEE or similar, with the same ratio of N's to E's as NE? Then the whole question would be bunk.&lt;/p&gt;&#xA;&lt;p&gt;So here's a proof that the number of N's and E's at any point on the compass are always relatively prime to each other.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Lemma:&lt;/strong&gt; Any two consecutive points on the compass &lt;span class=&quot;math-container&quot;&gt;$(a, b)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$(c, d)$&lt;/span&gt; will always satisfy &lt;span class=&quot;math-container&quot;&gt;$ad - bc = 1$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt; By induction. At level 1, &lt;span class=&quot;math-container&quot;&gt;$(1, 0)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$(0, 1)$&lt;/span&gt; satisfies &lt;span class=&quot;math-container&quot;&gt;$1 \times 1 - 0 \times 0 = 1$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Then, suppose two consecutive points satisfy &lt;span class=&quot;math-container&quot;&gt;$ad - bc = 1$&lt;/span&gt;, and the point between them is &lt;span class=&quot;math-container&quot;&gt;$(a+c, b+d)$&lt;/span&gt;. Then, we take &lt;span class=&quot;math-container&quot;&gt;$a(b+d) - b(a+c)$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$= ab + ad - ab - bc$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$= ad - bc$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$= 1$&lt;/span&gt;, as required.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Now, suppose there is some point &lt;span class=&quot;math-container&quot;&gt;$(nx, ny)$&lt;/span&gt; on the compass rose, where &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; are relatively prime so that &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; is the greatest common factor. Then, its ancestor points are &lt;span class=&quot;math-container&quot;&gt;$(a, b)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$(c, d)$&lt;/span&gt;, and so the point satisfies &lt;span class=&quot;math-container&quot;&gt;$nbx - nay = 1$&lt;/span&gt;, or &lt;span class=&quot;math-container&quot;&gt;$n(bx-ay) = 1$&lt;/span&gt;. Therefore, &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; must equal &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; are relatively prime, as required.&lt;/p&gt;&#xA;&lt;h2&gt;3. Proof of surjection&lt;/h2&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;I assure you, I can tell you the direction of any combination of two adjacent cardinal directions you like.&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Finally, the actual problem. Predictably, this is the hardest step. There are various proofs out there, some of them involving the Euclidean algorithm, some involving weird cross-multiplications, and others that I don't understand, but this is the most elegant one in my opinion.&lt;/p&gt;&#xA;&lt;p&gt;Suppose that some combination of relatively prime &lt;span class=&quot;math-container&quot;&gt;$(a, b)$&lt;/span&gt; didn't appear on the compass. Then, because it's between &lt;span class=&quot;math-container&quot;&gt;$(0, 1)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$(1, 0)$&lt;/span&gt;, it must be strictly between two points that do appear on the tree.&lt;/p&gt;&#xA;&lt;p&gt;So consider &lt;span class=&quot;math-container&quot;&gt;$\frac pq$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\frac rs$&lt;/span&gt; to be two consecutive points on some level of the Stern-Brocot tree, such that &lt;span class=&quot;math-container&quot;&gt;$\frac pq &amp;lt; \frac ab &amp;lt; \frac rs$&lt;/span&gt;. Then we can iterate as follows:&lt;/p&gt;&#xA;&lt;p&gt;Let &lt;span class=&quot;math-container&quot;&gt;$\frac mn$&lt;/span&gt; = &lt;span class=&quot;math-container&quot;&gt;$\frac{p+r}{q+s}$&lt;/span&gt;, i.e. the term in between those two terms on the next level. If &lt;span class=&quot;math-container&quot;&gt;$\frac ab &amp;lt; \frac mn$&lt;/span&gt;, then set &lt;span class=&quot;math-container&quot;&gt;$\frac mn$&lt;/span&gt; to be the new &lt;span class=&quot;math-container&quot;&gt;$\frac rs$&lt;/span&gt;. If &lt;span class=&quot;math-container&quot;&gt;$\frac ab &amp;gt; \frac mn$&lt;/span&gt;, then set &lt;span class=&quot;math-container&quot;&gt;$\frac mn$&lt;/span&gt; to be the new &lt;span class=&quot;math-container&quot;&gt;$\frac pq$&lt;/span&gt;. Then, repeat this procedure form the beginning. (If &lt;span class=&quot;math-container&quot;&gt;$\frac ab = \frac mn$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$\frac ab$&lt;/span&gt; &lt;em&gt;is&lt;/em&gt; on the Stern-Brocot tree, which we assumed was not the case.)&lt;/p&gt;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$\frac ab$&lt;/span&gt; is not in the Stern-Brocot tree, this procedure will continue forever. But then all the &lt;span class=&quot;math-container&quot;&gt;$\frac mn$&lt;/span&gt; values we get at each iteration form a Cauchy sequence of rational numbers with ever larger irreducible denominators, which cannot possibly converge to any rational number with a fixed denominator &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt;. So this is impossible, and &lt;span class=&quot;math-container&quot;&gt;$\frac ab$&lt;/span&gt; &lt;em&gt;must&lt;/em&gt; appear somewhere on the tree.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;I first encountered the Stern-Brocot tree in the &lt;a href=&quot;http://cms.math.ca/Competitions/COMC/COMC06/exame2006.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;2006 paper&lt;/a&gt; of the Canadian Open Mathematics Challenge, which is (not entirely) coincidentally where I got the three steps of this proof from.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-08-24T23:30:28.407" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12296" PostTypeId="1" AcceptedAnswerId="12318" CreationDate="2015-04-17T02:30:46.047" Score="0" ViewCount="548" Body="&lt;p&gt;Recently a puzzle that asked the birthdate of Cheryl got popular with the answer of July 16, in which Bernard and Albert have to figure out.&lt;/p&gt;&#xA;&lt;p&gt;The first step to the solution was to rule out the whole months of May and June because both these months contain unique dates that would give the answer right away.&lt;/p&gt;&#xA;&lt;p&gt;What I don't understand is that while we rule out these two unique dates is necessary, but why do we need to rule out all the other dates of May and June?&lt;/p&gt;&#xA;&lt;p&gt;Is it really valid to rule out the entire month simply because it contains the unique date?&lt;/p&gt;&#xA;&lt;p&gt;So what if the answer was May 15 or May 16 instead?&#xA;What is so invalid about May 15 and May 16?&lt;/p&gt;&#xA;&lt;p&gt;I thought if we rule out the unique dates, retaining May 15 and May 16 is still valid.&#xA;But why is it not valid?&#xA;I thought ruling out the entire months of May and June because they both contain unique dates (as well as non-unique dates that may remain valid) is simplistic filtering.&lt;/p&gt;&#xA;&lt;p&gt;Will Bernard and Albert know?&lt;/p&gt;&#xA;&lt;p&gt;I don't understand.&lt;/p&gt;&#xA;&lt;p&gt;If the dates were...&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;May 15 16&lt;/p&gt;&#xA;&lt;p&gt;June 17&lt;/p&gt;&#xA;&lt;p&gt;July 14 16 21&lt;/p&gt;&#xA;&lt;p&gt;August 14 15 17&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;How would the answer differ?&#xA;The right answer (July 16) would not be the right answer.&lt;/p&gt;&#xA;&lt;p&gt;In my opinion, even if the dates of May 15 and May 16 are retained, Albert can STILL be sure Bernard does NOT know the date is 19, or else Cheryl would not need to tell her date separately, because Bernard would know right away. Thus by right May 15 and May 16 are still valid options.&lt;/p&gt;&#xA;" OwnerUserId="11549" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-17T15:38:10.093" Title="Questionable solution to Cheryl's birthdate" Tags="|logical-deduction|" AnswerCount="2" CommentCount="16" ClosedDate="2015-04-17T17:51:02.727" ContentLicense="CC BY-SA 3.0" />
  <row Id="12299" PostTypeId="1" AcceptedAnswerId="12302" CreationDate="2015-04-17T03:07:42.537" Score="12" ViewCount="2288" Body="&lt;p&gt;There are two types of logicians: knights and knaves. Knights will always give a statement that is true whereas knaves will always give a statement that is false.&lt;/p&gt;&#xA;&lt;p&gt;After listening in on a conversation between 10 logicians, you gather the following information:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Logician 1 claims that (Logician 10 is a knight) and (Logician 1 is a knight)&lt;/li&gt;&#xA;&lt;li&gt;Logician 2 claims that (Logician 9 is a knave) or (Logician 2 is a knave)&lt;/li&gt;&#xA;&lt;li&gt;Logician 3 claims that (Logician 8 is a knight) and (Logician 3 is a knight)&lt;/li&gt;&#xA;&lt;li&gt;Logician 4 claims that (Logician 7 is a knave) or (Logician 4 is a knave)&lt;/li&gt;&#xA;&lt;li&gt;Logician 5 claims that (Logician 6 is a knight) and (Logician 5 is a knight)&lt;/li&gt;&#xA;&lt;li&gt;Logician 6 claims that (Logician 5 is a knave) or (Logician 6 is a knave)&lt;/li&gt;&#xA;&lt;li&gt;Logician 7 claims that (Logician 4 is a knight) and (Logician 7 is a knight)&lt;/li&gt;&#xA;&lt;li&gt;Logician 8 claims that (Logician 3 is a knave) or (Logician 8 is a knave)&lt;/li&gt;&#xA;&lt;li&gt;Logician 9 claims that (logician 2 is a knight) and (Logician 9 is a knight)&lt;/li&gt;&#xA;&lt;li&gt;Logician 10 claims that (logician 1 is a knave) or (Logician 10 is a knave)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Just by using the information gathered, is it possible to determine who the truthtellers and liars are? Explain why or why not.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; Do not confuse 'or' with 'xor'.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Bonus:&lt;/strong&gt; What would the answer become if all the or's became and's?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11094" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2021-04-26T21:28:58.757" Title="The 10 Logicians" Tags="|logical-deduction|liars|" AnswerCount="5" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12300" PostTypeId="2" ParentId="12299" CreationDate="2015-04-17T03:40:44.320" Score="4" Body="&lt;p&gt;The odd numbered logicians are all Knaves and the even are all Knights.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(L_1\iff(L_{10}\land L_1))\land (L_2\iff(\lnot L_9\lor\lnot L_2))\land (L_3\iff(L_8\land L_3))\land (L_4\iff(\lnot L_7\lor\lnot L_4))\land (L_5\iff(L_6\land L_5))\land (L_6\iff(\lnot L_5\lor\lnot L_6))\land (L_7\iff(L_4\land L_7))\land (L_8\iff(\lnot L_3\lor\lnot L_8))\land (L_9\iff(L_2\land L_9))\land (L_{10}\iff(\lnot L_1\lor\lnot L_{10}))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second case, if all the or's became and's the list of claim's would be contradictory, similar to the liar's paradox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If L1 is a Knight then L10 must also be a Knight because of claim 1. But they can't both be Knights because if L10 is a Knight they must be a both be Knaves by claim 10 (remember, or-&gt;and).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If L1 is a Knave then maybe L10 is also a Knave...but wait, L10 can't be a Knave, because he would be telling the truth, so he must be a Knight...no that doesn't work either because, paradoxically, he would be lying about being a Knave.&lt;/p&gt;&#xA;" OwnerUserId="11499" LastEditorUserId="11499" LastEditDate="2015-04-17T04:14:58.010" LastActivityDate="2015-04-17T04:14:58.010" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12301" PostTypeId="2" ParentId="12299" CreationDate="2015-04-17T03:43:52.953" Score="5" Body="&lt;p&gt;For the first question...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is possible. If 1 were a knight, then his statement would imply 10 was a knight. But then 10's true statement would mean 1 was a knave, a contradiction. Thus, 1 is a knave. This means 10's statement is true, so 10 is a knight. By identical reasoning, odd numbered logicians are liars and evens tell the truth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the bonus...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The conversation could not have happened. 10 can't be a knight since he is calling himself a knave. Thus, 10's statement implies NOT[(1 is a knave) AND (10 is a knave)]=(1 is a knight) OR (10 is a knight). The latter condition doesn't hold, so we now know 1 &lt;strong&gt;is&lt;/strong&gt; a knight. However, 1 is calling 10 a knight, which is false, so we know 1 &lt;strong&gt;is not&lt;/strong&gt; a knight. We have deduced a contradiction from this conversation, so someone broke character.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-04-17T04:16:55.967" LastActivityDate="2015-04-17T04:16:55.967" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12302" PostTypeId="2" ParentId="12299" CreationDate="2015-04-17T03:48:33.370" Score="10" Body="&lt;p&gt;No Logician could possibly claim that they are Knave via the rules. Thus half of these statements are impossible to be made. However, I will just assume that an impossible statement is FALSE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I notice that there is a natural grouping of the Logicians and the Logicians they mention. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this information, we have:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Logician 1 claims that (Logician 10 is a knight) and (Logician 1 is a knight)&#xA;$\land$ Logician 10 claims that (logician 1 is a knave)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Logician 2 claims that (Logician 9 is a knave)&#xA;$\land$ &#xA;Logician 9 claims that (logician 2 is a knight) and (Logician 9 is a knight)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Logician 3 claims that (Logician 8 is a knight) and (Logician 3 is a knight)&#xA;$\land$ &#xA;Logician 8 claims that (Logician 3 is a knave)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Logician 4 claims that (Logician 7 is a knave)&#xA;$\land$ &#xA;Logician 7 claims that (Logician 4 is a knight) and (Logician 7 is a knight)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Logician 5 claims that (Logician 6 is a knight) and (Logician 5 is a knight)&#xA;$\land$ &#xA;Logician 6 claims that (Logician 5 is a knave)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Clearly this is a case of 5 instances of the same pattern, when cast into formal logic formulae:&lt;br&gt;&#xA;$LX = (LY \land LX)$&lt;br&gt;&#xA;$LY = \lnot LX$&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $LY$ is a Knight. This means that $LX$ is a Knave. Therefore $(LY \land LX) = (TRUE \land FALSE) = FALSE$ therefore $LX=false$ i.e. $LX$ is a knave - this is consistent. Suppose that $LY$ is a Knave. This means that $LX$ is a Knight and would need to be (correctly) claiming that $LY$ is a Knight - this is a contradiction (inconsistent).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies that:&lt;br&gt;&#xA;Logicians: 2,4,6,8,10 are KNIGHTS.&lt;br&gt;&#xA;Logicians: 1,3,5,7,9 are KNAVES.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;For the bonus, the 5 instances of the same pattern, when cast into formal logic formulae are now:&lt;br&gt;&#xA;$LX = (LY \land LX)$&lt;br&gt;&#xA;$LY = (\lnot LX \land \lnot LY)$&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly $LY$ cannot be a Knight - otherwise he'd be correctly saying that his own statement is false. So we deduce that $LY$ is a Knave.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$LX$ is trying to claim that $LY$ is a Knight - but we just deduced that he is a Knave. Thus $LX$ is also a Knave. But if $LY$ is a Knave then it implies that one of $LX$, $LY$ must be a Knight (in order to make his statement false). This is a paradox. Since $LY$ is a Knave, and $LX$ cannot be both Knight and Knave. It is not possible to determine anything other than the fact that these Logicians are in fact $TROLLS$. &lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2015-04-17T04:21:22.953" LastActivityDate="2015-04-17T04:21:22.953" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="12303" PostTypeId="2" ParentId="12200" CreationDate="2015-04-17T04:10:01.870" Score="4" Body="&lt;p&gt;&lt;strong&gt;Sequence 3 is:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;the $n^{th}$ digit of the square root of $n$,&lt;/strong&gt; as follows:&lt;br&gt;&#xA;&lt;pre&gt;n    n^0.5       nth digit&lt;br&gt;1  1.00            1&lt;br&gt;2  1.414           4&lt;br&gt;3  1.7321          3&lt;br&gt;4  2.00000         0&lt;br&gt;5  2.236068        0&lt;br&gt;6  2.4494897       8&lt;br&gt;7  2.64575131      1&lt;br&gt;8  2.828427125     1&lt;br&gt;9  3.0000000000    0&lt;br&gt;10 3.16227766017   0&lt;br&gt;11 3.316624790355  3&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-04-17T04:28:07.597" LastActivityDate="2015-04-17T04:28:07.597" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12304" PostTypeId="2" ParentId="12296" CreationDate="2015-04-17T04:20:25.353" Score="2" Body="&lt;p&gt;Suppose the dates were as you say (and Cheryl's birthday remains July 16).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;May 15 16&lt;br&gt;&#xA;  June 17&lt;br&gt;&#xA;  July 14 16 21&lt;br&gt;&#xA;  August 14 15 17&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Albert was told &lt;strong&gt;July&lt;/strong&gt;. Albert's first statement with this list of dates is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I don't know when Cheryl's birthday is &lt;em&gt;[because there are three possible dates in July]&lt;/em&gt;, but Bernard might &lt;em&gt;[because there is a unique date in July]&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bernard was told &lt;strong&gt;16&lt;/strong&gt;. There are only two months with 16 in them (May and July), but if Albert were told May, then he would know for sure that Bernard didn't know the date because there are no unique days in May. By this, Bernard's second statement is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I didn't know whether Bernard would think that I might know &lt;em&gt;[because he could think I know in July, but be sure I don't know in May]&lt;/em&gt;, but now I know the birthday &lt;em&gt;[because I ruled out May using Albert's statement]&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Albert has now ruled out July 21 (because Bernard didn't know before Albert's first statement), but he cannot rule out July 14 because had Bernard been told 14, he would have made the same statement &lt;em&gt;[except with the subtext referring to August instead of May]&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this set of start dates, Albert cannot deduce the birthday, but Bernard can. We, as spectators, are also not able to deduce the birthday.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original puzzle was constructed such that it was solvable without prior knowledge of Cheryl's birthday. Had the puzzle been constructed with the dates you suggest, the reader would be unable to come to a conclusion.&lt;/p&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2015-04-17T04:20:25.353" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="12305" PostTypeId="2" ParentId="8562" CreationDate="2015-04-17T04:30:21.840" Score="0" Body="&lt;p&gt;I could be off base here, but it seems to me the magic question could be &quot;Are you and your partner of the same tribe?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you ask a knight married to a nave, the response is no. If you ask a nave married to a knight, the response is yes. If you ask a knight married to a knight, the response is yes. If you ask a nave married to a nave, the response is no. So both knights and naves could say yes, and both could say no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then you get the partner's response. Two naves each say no. Two knights each say yes. A night married to a nave says yes, while a nave married to a knight says no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question &quot;are you and your partner from different tribes&quot; also would work similarly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is the question, 50 people lied.&lt;/p&gt;&#xA;" OwnerUserId="11271" LastActivityDate="2015-04-17T04:30:21.840" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12306" PostTypeId="2" ParentId="12286" CreationDate="2015-04-17T05:54:46.570" Score="8" Body="&lt;p&gt;The four words are :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Then I saw it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The pattern is :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Take the first letter which is T. Then the subsequent letters are present at a gap of 2, 4, 6, 8, 10, 12, 14, 16 and 18 letters. Which is also at each perfect square (1, 4, 9, 16...)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The complete quote is :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Then I saw it. I saw a Mom who would die for her son; a man who would kill for his wife; a boy, angry and alone, laid out in front of him the bad path. I saw it. And the path was a circle, round and round. So I changed it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the movie :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikiquote.org/wiki/Looper_(film)&quot; rel=&quot;nofollow&quot;&gt;Looper&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5565" LastEditorUserId="463" LastEditDate="2015-04-21T13:22:43.823" LastActivityDate="2015-04-21T13:22:43.823" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12307" PostTypeId="2" ParentId="12284" CreationDate="2015-04-17T06:56:02.323" Score="0" Body="&lt;p&gt;Though this is kind of a loop hole it works as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the first time we hear about the value of the dice is from the magician, we cannot check if it is right or wrong since we do not know if the dice were 5 and 2. for all we know the trick could be just naming 2 random numbers on the magicians part.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Other possibility:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; obviously the magician and the apprentice agreed on a certain algorithm to determine what dice will be under the cup. i think to properly give this algorithm we would need to know all the values of the dice since there are multiple possibilities to do this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;one of which:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if the dice are 111115112 you simply lay out the first 5 dice so added up they give the number of the dice under the cup, if this number is greater than 6 you add mod 6, to get the 2 you do the same. Now obviously i know this does not work with several combinations of dice but as we do not know the numbers on the dice i see it as we are free to make of it what we wish&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9835" LastEditorUserId="9835" LastEditDate="2015-04-17T07:01:46.890" LastActivityDate="2015-04-17T07:01:46.890" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12309" PostTypeId="2" ParentId="12284" CreationDate="2015-04-17T07:12:17.240" Score="0" Body="&lt;p&gt;A wild guess but lets see if this is correct&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's a Thursday S - M - T - W - &lt;strong&gt;T (5th day)&lt;/strong&gt; - F - S&#xA; and at &lt;strong&gt;2pm&lt;/strong&gt; magician goes out for a fight. &#xA; &lt;strong&gt;both the facts are known to assistant and magician&lt;/strong&gt; so without even communication he can guess it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5437" LastActivityDate="2015-04-17T07:12:17.240" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12310" PostTypeId="2" ParentId="12187" CreationDate="2015-04-17T07:54:31.687" Score="2" Body="&lt;p&gt;A simple expression that is exact, and only uses four operations without bending the rules (if exponentiation isn't permitted, I assume that neither is using the numbers to create n-th roots) while using at least one non-basic operation, is&#xA;$$&#xA;((1+2-3)\times456789)!&#xA;$$&#xA;That's one addition, one subtraction, one multiplication, and one factorial (actually, zero factorial, but you know what I mean). Another similar option is&#xA;$$&#xA;((12/3-4)\times56789)!&#xA;$$&#xA;A slightly more bendy solution using the fact that negative integer factorials can be considered to be infinite is&#xA;$$&#xA;1+2/(3-456789)!&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="9289" LastActivityDate="2015-04-17T07:54:31.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12311" PostTypeId="2" ParentId="12187" CreationDate="2015-04-17T08:57:51.237" Score="3" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$(1 + 23 + 45 + 6! - 789)! = 1$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$((1+2-3)\times456789)!=1$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Everyone is trying with minimum operators.&lt;br /&gt;&#xA;I guess, with @user23013's solution, we can try with various possibilities :)&lt;/p&gt;&#xA;" OwnerUserId="2656" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-06-25T17:34:52.100" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12313" PostTypeId="1" AcceptedAnswerId="12334" CreationDate="2015-04-17T10:00:12.370" Score="8" ViewCount="1333" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Every good story is build up of acts, &lt;br&gt;&#xA;  one of which is a tragedy,&lt;br&gt;&#xA;  they all convey a mystery,&lt;br&gt;&#xA;  how many there are is history,&lt;br&gt;&#xA;  by the time you know your destiny.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$ $&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Destined for greatness destined for good,&lt;br&gt;&#xA;  you must be the best, pulling stories from your hood,&lt;br&gt;&#xA;  murder is something you truly understood.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who am I?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I was originally going to post this as a follow up &lt;a href=&quot;https://puzzling.stackexchange.com/questions/12276/difficile-est-saturam-non-legere&quot;&gt;to my last question&lt;/a&gt; but I found the riddle to be solvable without, the answer would be a good extra hint for the first paragraph though.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless here's an other &lt;strong&gt;hint&lt;/strong&gt; for the first paragraph if you desperately need it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Setup, Confrontation, Resolution&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;br&gt;&#xA;Please note there is an answer to the first paragraph, which will help you find who I am in the 2nd paragraph.&lt;/p&gt;&#xA;" OwnerUserId="9835" LastEditorUserId="51105" LastEditDate="2021-10-13T12:40:48.610" LastActivityDate="2021-10-13T12:40:48.610" Title="The queen of riddles" Tags="|riddle|knowledge|rhyme|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12315" PostTypeId="2" ParentId="12284" CreationDate="2015-04-17T10:47:38.050" Score="1" Body="&lt;p&gt;I have a way which can safely determine one number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say first cup is called &lt;strong&gt;F&lt;/strong&gt;&lt;br&gt;&#xA;Let's say second cup is called &lt;strong&gt;S&lt;/strong&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Position on the tables are:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;x x x x x F x x S&lt;br&gt;&#xA;  dice value of F=5&lt;br&gt;&#xA;  dice value of S=2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now we place S always on the far right end of the table.&lt;br&gt;&#xA;Now whatever the number on the dice is (1-6) under &lt;strong&gt;S&lt;/strong&gt; that means how many dices are between those two cups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I.e.: &lt;strong&gt;S&lt;/strong&gt; is always on 9th position. If there is value of 3 underneath the &lt;strong&gt;S&lt;/strong&gt; then &lt;strong&gt;F&lt;/strong&gt; will be on position 5 (3 dices in between).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we know for sure what number is underneath &lt;strong&gt;S&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we determine the number under &lt;strong&gt;F&lt;/strong&gt; by:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Guessing, seriously 1/6 odds are way better than 1/36. It's magic!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9907" LastEditorUserId="9907" LastEditDate="2015-04-17T11:13:27.647" LastActivityDate="2015-04-17T11:13:27.647" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12316" PostTypeId="1" AcceptedAnswerId="12317" CreationDate="2015-04-17T11:34:26.260" Score="-1" ViewCount="55158" Body="&lt;p&gt;Use the numbers 8,6,4,2 and the operators +,-,*,/ to get 25.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use each number once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each operator can be used once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The least number of operators you use the better answer will be.&lt;/p&gt;&#xA;" OwnerUserId="11111" LastEditorUserId="10281" LastEditDate="2015-04-17T11:53:56.680" LastActivityDate="2019-02-19T10:16:13.440" Title="Use the numbers 8 6 4 2 = 25" Tags="|mathematics|no-computers|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12317" PostTypeId="2" ParentId="12316" CreationDate="2015-04-17T11:49:06.963" Score="6" Body="&lt;p&gt;Answer of this question is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;4*(6+(2/8))&lt;/strong&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4977" LastEditorUserId="9951" LastEditDate="2015-04-17T11:51:16.400" LastActivityDate="2015-04-17T11:51:16.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12318" PostTypeId="2" ParentId="12296" CreationDate="2015-04-17T12:52:15.633" Score="7" Body="&lt;h2&gt;The Answer to End All Questions About Cheryl's Birthday&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In the initial problem, the reader is unaware of Cheryl's birthday. Albert and Bernard also begin with only the information they are told in secret.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below, I will describe the knowledge that each Albert, Bernard, and the reader has of the birthdate, exploring all of the possible birthdates that Cheryl might have.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Initial State&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Albert&lt;/strong&gt; is told a month. Depending on the month that he is told, he knows the birthday belongs to one of the following dates:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;May&lt;/strong&gt;: 15, 16, 19&lt;br&gt;&#xA;&lt;strong&gt;June&lt;/strong&gt;: 17, 18&lt;br&gt;&#xA;&lt;strong&gt;July&lt;/strong&gt;: 14, 16&lt;br&gt;&#xA;&lt;strong&gt;August&lt;/strong&gt;: 14, 15, 17&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bernard&lt;/strong&gt; is told a day. Depending on the number he is told, he knows the birthday belongs to one of the following dates:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;14&lt;/strong&gt;: July, August&lt;br&gt;&#xA;&lt;strong&gt;15&lt;/strong&gt;: May, August&lt;br&gt;&#xA;&lt;strong&gt;16&lt;/strong&gt;: May, July&lt;br&gt;&#xA;&lt;strong&gt;17&lt;/strong&gt;: June, August&lt;br&gt;&#xA;&lt;strong&gt;18&lt;/strong&gt;: June&lt;br&gt;&#xA;&lt;strong&gt;19&lt;/strong&gt;: May&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Independently, Albert has &lt;em&gt;already&lt;/em&gt; narrowed down the list of dates to &lt;em&gt;at most 3&lt;/em&gt; based on the month he was told. Similarly, Bernard has narrowed them down to &lt;em&gt;at most 2&lt;/em&gt; based on the day that he was told. They are both immediately able to throw out the remaining dates without further instruction, &lt;em&gt;however&lt;/em&gt; the reader of the puzzle does not yet have any information, and therefore must be still considering all 10 dates. All of these initial possibilities are known to everyone involved since the reader is able to deduce the relationships given just the initial list.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;First Statements&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Albert and Bernard are both aware of the possibilities open to each other as they both knew the initial list of dates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no month that only has one day, so we know the first part of Albert's statement is &lt;em&gt;always&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Albert: I do not know Cheryl's birthday...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's consider what Albert's first statement is in each of the four month scenarios by expanding the month-day relationship using Bernard's initial possibilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;May&lt;/strong&gt;&lt;br&gt;&#xA;15: &lt;strong&gt;May&lt;/strong&gt;, August&lt;br&gt;&#xA;16: &lt;strong&gt;May&lt;/strong&gt;, July&lt;br&gt;&#xA;19: &lt;strong&gt;May&lt;/strong&gt;&lt;br&gt;&#xA;If Bernard was told &quot;19&quot;, then he might know because there is only one possible month associated with it. The second part of Albert's statement for May is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Albert: ...but Bernard might know.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;June&lt;/strong&gt;&lt;br&gt;&#xA;17: &lt;strong&gt;June&lt;/strong&gt;, August&lt;br&gt;&#xA;18: &lt;strong&gt;June&lt;/strong&gt;&lt;br&gt;&#xA;If Bernard was told &quot;18&quot;, then he might know because there is only one possible month associated with it. The second part of Albert's statement for June is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Albert: ...but Bernard might know.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;July&lt;/strong&gt;&lt;br&gt;&#xA;14: &lt;strong&gt;July&lt;/strong&gt;, August&lt;br&gt;&#xA;16: May, &lt;strong&gt;July&lt;/strong&gt;&lt;br&gt;&#xA;There is no day in July that is only in July, so if Bernard was told either of the two days in July, he would not be able to deduce Cheryl's birthdate without more information. The second part of Albert's statement for July is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Albert: ...and neither does Bernard.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;August&lt;/strong&gt;&lt;br&gt;&#xA;14: July, &lt;strong&gt;August&lt;/strong&gt;&lt;br&gt;&#xA;15: May, &lt;strong&gt;August&lt;/strong&gt;&lt;br&gt;&#xA;17: June, &lt;strong&gt;August&lt;/strong&gt;&lt;br&gt;&#xA;There is no day in August that is only in August, so if Bernard was told any of the three days in August, he would not be able to deduce Cheryl's birthdate without more information. The second part of Albert's statement for August is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Albert: ...and neither does Bernard.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If we consider the possibility of Bernard speaking first, he will say the following when told one of 14, 15, 16, or 17:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bernard: I do not know Cheryl's birthdate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To which Albert responds&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Albert: I already knew that (see &quot;...and neither does Bernard.&quot;); or&lt;br&gt;&#xA;  Albert: I didn't know that (see &quot;...but Bernard might know.&quot;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In this case, we now proceed to the second statements with the knowledge that Albert has just provided us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or Bernard says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bernard: I know Cheryl's birthdate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To which Albert responds&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Albert: So do I (May, June).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that it is impossible for Bernard to initially state that he knows Cheryl's birthdate without more information if Albert were told July or August. They do not say any more statements here as they both have uniquely identified the birthdate. It is important to note that in these cases, &lt;strong&gt;the reader of the puzzle cannot deduce the birthdate&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Second Statements&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Albert has finished his first statement in one of two ways: &quot;but Bernard might know&quot; and &quot;neither does Bernard&quot;. Let's consider how the affects Bernard's perception:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Albert: ...but Bernard might know&lt;/strong&gt;&lt;br&gt;&#xA;This indicates Albert was told May or June (since those are the only two scenarios in which he would say this). Bernard's table now becomes:&lt;br&gt;&#xA;&lt;strong&gt;14&lt;/strong&gt;: &lt;em&gt;impossible&lt;/em&gt;&lt;br&gt;&#xA;&lt;strong&gt;15&lt;/strong&gt;: May&lt;br&gt;&#xA;&lt;strong&gt;16&lt;/strong&gt;: May&lt;br&gt;&#xA;&lt;strong&gt;17&lt;/strong&gt;: June&lt;br&gt;&#xA;&lt;strong&gt;18&lt;/strong&gt;: June&lt;br&gt;&#xA;&lt;strong&gt;19&lt;/strong&gt;: May&lt;br&gt;&#xA;Now for every day that Bernard was told, he unequivocally knows Cheryl's birthdate.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bernard: I now know Cheryl's birthdate.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If he was told 18 or 19, he would have already known the birthdate as discussed in the first step and he would instead say:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bernard: I already knew Cheryl's birthdate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Albert: ...and neither does Bernard&lt;/strong&gt;&lt;br&gt;&#xA;This indicates that Albert was told July or August (since those are the only two scenarios in which he ends his first statement this way). Bernard's table now becomes:&lt;br&gt;&#xA;&lt;strong&gt;14&lt;/strong&gt;: July, August&lt;br&gt;&#xA;&lt;strong&gt;15&lt;/strong&gt;: August&lt;br&gt;&#xA;&lt;strong&gt;16&lt;/strong&gt;: July&lt;br&gt;&#xA;&lt;strong&gt;17&lt;/strong&gt;: August&lt;br&gt;&#xA;&lt;strong&gt;18&lt;/strong&gt;: &lt;em&gt;impossible&lt;/em&gt;&lt;br&gt;&#xA;&lt;strong&gt;19&lt;/strong&gt;: &lt;em&gt;impossible&lt;/em&gt;&lt;br&gt;&#xA;If Bernard was told 15, 16, or 17, he can now uniquely identify Cheryl's birthdate. He could not have been told 18 or 19 if Albert finishes his first statement in this way.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bernard: I now know Cheryl's birthdate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the event that Bernard was told 14, his second statement becomes:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bernard: I still don't know Cheryl's birthdate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Third Statements&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Bernard has said one of the three following statements: &quot;I now know Cheryl's birthdate&quot;, &quot;I already knew Cheryl's birthdate&quot;, or &quot;I still don't know Cheryl's birthdate&quot;. Let's consider the negative case first:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bernard: I still don't know Cheryl's birthdate.&lt;/strong&gt;&lt;br&gt;&#xA;There is only one case that Bernard would have said this (if he were told 14), so now Albert's table appears as such:&lt;br&gt;&#xA;&lt;strong&gt;May&lt;/strong&gt;: &lt;em&gt;impossible&lt;/em&gt;&lt;br&gt;&#xA;&lt;strong&gt;June&lt;/strong&gt;: &lt;em&gt;impossible&lt;/em&gt;&lt;br&gt;&#xA;&lt;strong&gt;July&lt;/strong&gt;: 14&lt;br&gt;&#xA;&lt;strong&gt;August&lt;/strong&gt;: 14&lt;br&gt;&#xA;Now Albert knows both the month and day. His final statement is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Albert: I now know Cheryl's birthdate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, neither Bernard or the reader is given enough information at this point to decide on the date. &lt;strong&gt;Only Albert can know if Bernard is told 14.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bernard: I already knew Cheryl's birthdate.&lt;/strong&gt;&lt;br&gt;&#xA;This immediately tells Albert the birthdate, too, as he could have only said this when Albert began with &lt;strong&gt;but Bernard might know&lt;/strong&gt; (stemming from May or June):&lt;br&gt;&#xA;May: 19&lt;br&gt;&#xA;June: 18&lt;br&gt;&#xA;The uniqueness here allows Albert to decide:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Albert: I know Cheryl's birthdate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, the reader is unable to choose between May and June. &lt;strong&gt;Only Albert and Bernard can know in this scenario, the reader is left in the dark.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bernard: I now know Cheryl's birthdate.&lt;/strong&gt;&lt;br&gt;&#xA;The remaining statement by Bernard must now also be considered with Albert's first statement in order to narrow the decisions:&lt;br&gt;&#xA;Albert: &lt;strong&gt;but Bernard might know.&lt;/strong&gt; (May, June)&lt;br&gt;&#xA;Bernard: &lt;strong&gt;I now know Cheryl's birthdate.&lt;/strong&gt;&lt;br&gt;&#xA;Albert's table then becomes:&lt;br&gt;&#xA;&lt;strong&gt;May&lt;/strong&gt;&lt;br&gt;&#xA;15: &lt;strong&gt;May&lt;/strong&gt;&lt;br&gt;&#xA;16: &lt;strong&gt;May&lt;/strong&gt;&lt;br&gt;&#xA;19: &lt;em&gt;impossible&lt;/em&gt;&lt;br&gt;&#xA;Had Albert been told May, he would be unable to decide between 15 and 16. &lt;strong&gt;Albert and the reader are unable to deduce the date in this scenario; only Bernard can know.&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Albert: I still don't know Cheryl's birthdate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;June&lt;/strong&gt;&lt;br&gt;&#xA;17: &lt;strong&gt;June&lt;/strong&gt;&lt;br&gt;&#xA;18: &lt;em&gt;impossible&lt;/em&gt;&lt;br&gt;&#xA;Had Albert been told June, he is left with one possibility: &lt;strong&gt;June 17&lt;/strong&gt;. The reader can also deduce this date (description in the conclusion below).  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Albert: I now know Cheryl's birthdate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Albert: &lt;strong&gt;and neither does Bernard.&lt;/strong&gt; (July, August)&lt;br&gt;&#xA;Bernard: &lt;strong&gt;I now know Cheryl's birthdate.&lt;/strong&gt;&lt;br&gt;&#xA;Albert's table then becomes:&lt;br&gt;&#xA;&lt;strong&gt;July&lt;/strong&gt;&lt;br&gt;&#xA;14: &lt;em&gt;impossible&lt;/em&gt;&lt;br&gt;&#xA;16: &lt;strong&gt;July&lt;/strong&gt;&lt;br&gt;&#xA;Had Albert been told July, he is left with one possibility: &lt;strong&gt;July 16&lt;/strong&gt;. The reader can also deduce this date (description in the conclusion below).  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Albert: I now know Cheryl's birthdate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;August&lt;/strong&gt;&lt;br&gt;&#xA;14: &lt;em&gt;impossible&lt;/em&gt;&lt;br&gt;&#xA;15: &lt;strong&gt;August&lt;/strong&gt;&lt;br&gt;&#xA;17: &lt;strong&gt;August&lt;/strong&gt;&lt;br&gt;&#xA;Had Albert been told August, he would be unable to decide between 15 and 17. &lt;strong&gt;Albert and the reader are unable to deduce the date in this scenario; only Bernard can know.&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Albert: I still don't know Cheryl's birthdate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Conclusion&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In the scenarios noted above where the reader cannot deduce the answer, the original puzzle is unsolvable. Albert and Bernard may still be able to know the date in some of the described cases, but the reader cannot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the other two, July 16 and June 17, the reader can deduce using the information provided by Albert and Bernard. The statements made by the pair leads to a single possibility. I have included the scenarios in which the statements are spoken in order to guide the reader without requiring scrolling up to re-read my earlier arguments:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Albert: &lt;strong&gt;but Bernard might know.&lt;/strong&gt; (May, June)&lt;br&gt;&#xA;Bernard: &lt;strong&gt;I now know Cheryl's birthdate.&lt;/strong&gt; (15, 16, 17)&lt;br&gt;&#xA;Albert: &lt;strong&gt;I now know Cheryl's birthdate.&lt;/strong&gt; (June 17)&lt;br&gt;&#xA;A unique date is now indicated by Albert.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Albert: &lt;strong&gt;and neither does Bernard.&lt;/strong&gt; (July, August)&lt;br&gt;&#xA;Bernard: &lt;strong&gt;I now know Cheryl's birthdate.&lt;/strong&gt; (15, 16, 17)&lt;br&gt;&#xA;Albert: &lt;strong&gt;I now know Cheryl's birthdate.&lt;/strong&gt; (July 16)&lt;br&gt;&#xA;A unique date is now indicated by Albert.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paths provided by the statements made by the two guessers &lt;em&gt;uniquely&lt;/em&gt; identifies &lt;em&gt;only&lt;/em&gt; &lt;strong&gt;July 16&lt;/strong&gt; and &lt;strong&gt;June 17&lt;/strong&gt; to the reader. Any other birthdate cannot be deduced by the reader.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original puzzle follows the path indicated above that leads uniquely to &lt;strong&gt;July 16&lt;/strong&gt;, so this is how we can determine the answer to the original puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Note&lt;/strong&gt;: if the initial list of dates is altered, the statements made by Albert and Bernard must also be altered, even when Cheryl's birthdate stays the same.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Epilogue&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The above dissertation is describing one-to-one and many-to-one relationships between the birthdates and the statements made by Albert and Bernard. If we define the statements made by the pair as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X_1$: Bernard doesn't know.&lt;br&gt;&#xA;$X_2$: Bernard might know.&lt;br&gt;&#xA;$Y_1$: Bernard already knew.&lt;br&gt;&#xA;$Y_2$: Bernard now knows.&lt;br&gt;&#xA;$Y_3$: Bernard still doesn't know.&lt;br&gt;&#xA;$Z_1$: Albert now knows.&lt;br&gt;&#xA;$Z_2$: Albert still doesn't know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can now describe the resultant paths ($R_n$) for each date:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;May 15: $X_2 + Y_2 + Z_2 = R_1$&lt;br&gt;&#xA;May 16: $X_2 + Y_2 + Z_2 = R_1$&lt;br&gt;&#xA;May 19: $X_2 + Y_1 + Z_1 = R_2$&lt;br&gt;&#xA;June 17: $X_2 + Y_2 + Z_1 = R_3$&lt;br&gt;&#xA;June 18: $X_2 + Y_1 + Z_1 = R_2$&lt;br&gt;&#xA;July 14: $X_1 + Y_3 + Z_1 = R_4$&lt;br&gt;&#xA;July 16: $X_1 + Y_2 + Z_1 = R_5$&lt;br&gt;&#xA;August 14: $X_1 + Y_3 + Z_1 = R_4$&lt;br&gt;&#xA;August 15: $X_1 + Y_2 + Z_2 = R_6$&lt;br&gt;&#xA;August 17: $X_1 + Y_2 + Z_2 = R_6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Collating these results and writing them in the opposite direction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R_1$: May 15, May 16&lt;br&gt;&#xA;$R_2$: May 19, June 18&lt;br&gt;&#xA;$R_3$: June 17&lt;br&gt;&#xA;$R_4$: July 14, August 14&lt;br&gt;&#xA;$R_5$: July 16&lt;br&gt;&#xA;$R_6$: August 15, August 17&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, we see that $R_1$, $R_2$, $R_4$, and $R_6$ are all many-to-one relationships. In a many-to-one relationship, the reader of the puzzle cannot deduce Cheryl's birthdate. In the one-to-one relationships ($R_3$ and $R_5$), Cheryl's birthdate can be accurately concluded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following are also true (as can be deduced within this epilogue, but as have also been shown in the main body of the above answer):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R_1$: Both Albert and Bernard know the birthdate, but we do not.&lt;br&gt;&#xA;$R_2$: Both Albert and Bernard know the birthdate, but we do not.&lt;br&gt;&#xA;$R_3$: Albert, Bernard, and we all know the birthdate.&lt;br&gt;&#xA;$R_4$: Albert knows the birthdate, but Bernard and we do not.&lt;br&gt;&#xA;$R_5$: Albert, Bernard, and we all know the birthdate.&lt;br&gt;&#xA;$R_6$: Bernard knows the birthdate, but Albert and we do not.&lt;/p&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-04-17T15:38:10.093" LastActivityDate="2015-04-17T15:38:10.093" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12319" PostTypeId="2" ParentId="12313" CreationDate="2015-04-17T13:36:21.977" Score="2" Body="&lt;p&gt;Every good story is build up of acts, &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Some kind of play?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;one of which is a tragedy,&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Greek tragedy maybe?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;they all convey a mystery,&lt;br&gt;&#xA;how many there are is history,&lt;br&gt;&#xA;by the time you know your destiny.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each act conveys a new mystery, but by the time every mystery is solved it is already too late. Maybe Oedipus?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;Destined for greatness destined for good,&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Destined to become a King, which he did by defeating the sphinx&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;you must be the best, pulling stories from your hood,&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A reference to answering the sphinx's riddles?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;murder is something you truly understood.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's revealed he killed his father and married his mother at the end of the play.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The title of the post points there as well,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'the queen' referring to his mother/wife and 'of riddles' to the sphinx.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11499" LastEditorUserId="9835" LastEditDate="2015-04-17T14:10:24.847" LastActivityDate="2015-04-17T14:10:24.847" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12320" PostTypeId="2" ParentId="12313" CreationDate="2015-04-17T14:55:40.637" Score="2" Body="&lt;p&gt;The answer may be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lady Macbeth (Queen of Scotland, as hinted in the title).  I'm not terribly familiar with Shakespeare, but I'll see if I can explain these clues.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Every good story is build up of acts,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All of Shakespeare's plays are divided into acts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;one of which is a tragedy,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Macbeth is a tragedy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;they all convey a mystery,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They all build up towards the final act.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;how many there are is history,&#xA;by the time you know your destiny.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; By the time you get to the end of a play, you know how many acts there are.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Destined for greatness destined for good,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The three witches prophesized that Macbeth would be King (and therefore she would be Queen).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;you must be the best, pulling stories from your hood,&#xA;murder is something you truly understood.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; She helped plot the murder of King Duncan. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2015-04-17T14:55:40.637" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12321" PostTypeId="1" AcceptedAnswerId="12326" CreationDate="2015-04-17T15:08:29.403" Score="4" ViewCount="581" Body="&lt;p&gt;There are two mathematicians with incredibly strong memories sitting in a coffee shop passing the time. Their server proposes a game:&lt;/p&gt;&#xA;&lt;p&gt;First, he gives them a large number:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$2176782336$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Then he whispers a number &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; to the first mathematician, and another number &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; to the second mathematician.  Afterwards, he tells them that the first large number is either the &lt;strong&gt;product of the two numbers&lt;/strong&gt;, or &lt;strong&gt;one number raised to the power of the other&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;He promises to pay for their order if they can each demonstrate that they know the other's number without offering each other any additional information.&lt;/p&gt;&#xA;&lt;p&gt;After along silence this conversation takes place .&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The first mathematician declares: I don't know your number.&lt;/li&gt;&#xA;&lt;li&gt;The second mathematician says: &lt;strong&gt;Now&lt;/strong&gt; I know your number.&lt;/li&gt;&#xA;&lt;li&gt;The first mathematician states: Now I think I know both numbers.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Before anyone announces these numbers, the manager of the cafe promises to award a prize to the first person in the coffee shop who succeeds in discovering what they are. Can you find &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;-&lt;strong&gt;Note:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;(x, y) differs from (y, x)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Silence is significative , and everyone knows this&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9517" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2021-06-01T04:45:51.497" Title="Find two numbers based on either their multiplication or exponent" Tags="|logical-deduction|number-theory|" AnswerCount="4" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="12322" PostTypeId="1" AcceptedAnswerId="12598" CreationDate="2015-04-17T15:51:31.950" Score="5" ViewCount="1060" Body="&lt;p&gt;On his retirement, Bill Gates left behind a four number combination safe with a dial similar to &lt;a href=&quot;http://thumbs.dreamstime.com/x/realistic-safe-combination-lock-wheel-23526580.jpg&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;. In that safe is a secret that had eluded him many years ago...along with a small fortune of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He left behind 10 clues - a deck of cards and a note. The order of the cards is as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Three of Spades,&#xA;  Six of Spades,&#xA;  Four of Spades,&#xA;  Seven of Clubs,&#xA;  Six of Clubs,&#xA;  Five of Clubs,&#xA;  Eight of Clubs,&#xA;  Two of Clubs,&#xA;  Four of Diamonds,&#xA;  Five of Diamonds,&#xA;  Five of Spades,&#xA;  Two of Spades,&#xA;  Six of Diamonds,&#xA;  Eight of Hearts,&#xA;  Three of Clubs,&#xA;  Five of Hearts,&#xA;  Three of Hearts,&#xA;  Seven of Hearts,&#xA;  Six of Hearts,&#xA;  Two of Diamonds,&#xA;  Eight of Spades,&#xA;  Three of Diamonds,&#xA;  Four of Clubs,&#xA;  Seven of Spades,&#xA;  Four of Hearts,&#xA;  Two of Hearts,&#xA;  Seven of Diamonds,&#xA;  Eight of Diamonds,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The note reads:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sly snakes slither into eternity seeking stigma and style.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is your knowledge potent enough to figure out the combination of the safe?&lt;/p&gt;&#xA;" OwnerUserId="11499" LastEditorUserId="11499" LastEditDate="2015-04-18T20:04:39.410" LastActivityDate="2019-12-30T13:10:52.703" Title="A riddle by any other name" Tags="|riddle|password|number-sequence|knowledge|computer-puzzle|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12323" PostTypeId="2" ParentId="12321" CreationDate="2015-04-17T16:14:12.923" Score="1" Body="&lt;p&gt;I think the answer is :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;(6, 12)&lt;/strong&gt; or &lt;strong&gt;(6, 36)&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; If &lt;strong&gt;x&lt;/strong&gt; is NOT one of 6, 12 or 36, then the first mathematician would know the other one's number, so the numbers are definitely 6, 12 or 36.&lt;br&gt;&#xA;&lt;br&gt;&#xA;If &lt;strong&gt;x&lt;/strong&gt; was 6, then &lt;strong&gt;y&lt;/strong&gt; could be either 12 or 36  (as 36 to the 6th power is also an answer).  So first mathematician can't be sure.  Therefore his number is 6.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The second mathematician's number is either 12 or 36, so he knows &lt;strong&gt;x&lt;/strong&gt; must be 6.  Then the first mathematician knows that the numbers are either &lt;strong&gt;(6, 12)&lt;/strong&gt; or &lt;strong&gt;(6, 36)&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10294" LastEditorUserId="10294" LastEditDate="2015-04-17T16:33:30.790" LastActivityDate="2015-04-17T16:33:30.790" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12324" PostTypeId="1" CreationDate="2015-04-17T16:21:44.267" Score="-1" ViewCount="2543" Body="&lt;p&gt;There is a small island in the centre of a lake. The lake has two banana trees. The monkey wants to get all the bananas from the island and so he goes to the crocodile to ask for help. The crocodile says that for every one-way journey he needs one fish. But the monkey has only three fish and in order to get all the bananas out of the island he has to make two round trips. Will the monkey succeed in bringing all the bananas off the island? If yes, how? &lt;/p&gt;&#xA;" OwnerUserId="9776" LastEditorUserId="5373" LastEditDate="2015-04-30T19:58:38.050" LastActivityDate="2015-05-17T13:10:01.850" Title="Monkey and the Crocodile" Tags="|logical-deduction|lateral-thinking|" AnswerCount="13" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12325" PostTypeId="1" AcceptedAnswerId="12348" CreationDate="2015-04-17T16:26:10.083" Score="79" ViewCount="15484" Body="&lt;p&gt;There have been two other questions &lt;a href=&quot;https://puzzling.stackexchange.com/questions/11998/1-2-3-4-5-6-7-8-9-100&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/questions/12187/1-2-3-4-5-6-7-8-9-1&quot;&gt;here&lt;/a&gt; that are similar to this one, but this changes the rules up a little.&lt;/p&gt;&#xA;&lt;p&gt;Your job is to approximate &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt; using the sequence of digits (in order):&lt;/p&gt;&#xA;&lt;p&gt;1 2 3 4 5 6 7 8 9&lt;/p&gt;&#xA;&lt;p&gt;with operators inserted between them (permitted operators listed below). You are to find the best approximation to &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt; that you can using the allowed operators and the numbers listed in order as they appear above. You must use all nine digits.&lt;/p&gt;&#xA;&lt;p&gt;Your score is given by&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\frac{-\ln\left|1-A/\pi\right|}{n_{ops}} = \frac{\ln\left|\frac{\pi}{\pi-A}\right|}{n_{ops}}&#xA;$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$n_{ops}$&lt;/span&gt; is the number of operators you used, and &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; is your approximation. So if you managed to get &lt;span class=&quot;math-container&quot;&gt;$A=22/7$&lt;/span&gt; and required three operations, then we have &lt;span class=&quot;math-container&quot;&gt;$\ln\left|1-A/\pi\right|\approx−7.818$&lt;/span&gt;, and so your score would be approximately &lt;span class=&quot;math-container&quot;&gt;$2.606$&lt;/span&gt;. Larger scores are better.&lt;/p&gt;&#xA;&lt;p&gt;You may use parentheses, but only to control order of operations - other uses such as binomials or Jacobi symbols are not valid.&lt;/p&gt;&#xA;&lt;p&gt;Permitted operations:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$+$&lt;/span&gt; (plus): standard addition of real or complex numbers.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$-$&lt;/span&gt; (minus): standard subtraction of real or complex numbers, or unary negation of real or complex numbers.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\times$&lt;/span&gt; (times): standard multiplication of real or complex numbers. Implied multiplication (e.g. &lt;span class=&quot;math-container&quot;&gt;$(1+2)3$&lt;/span&gt;) counts as an operation.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$/$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\div$&lt;/span&gt; (divide): standard division of real or complex numbers (allows division by positive or negative infinity to get zero).&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\sqrt{ }$&lt;/span&gt; (square root): standard principle square root of real or complex numbers, with second root (negative if number being square rooted is positive) allowed as &lt;span class=&quot;math-container&quot;&gt;$\sqrt[-]{}$&lt;/span&gt; as a single operation.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$!$&lt;/span&gt; (factorial): standard factorial for non-negative integer values (i.e. natural numbers) only, cannot be applied to non-natural numbers.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$|.|$&lt;/span&gt; (absolute value): standard absolute value for real or complex numbers, equal to &lt;span class=&quot;math-container&quot;&gt;$\sqrt{a^2+b^2}$&lt;/span&gt; if the number is of the form &lt;span class=&quot;math-container&quot;&gt;$a+bi$&lt;/span&gt; with &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; real and &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; being the imaginary number.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\lfloor.\rfloor$&lt;/span&gt; (floor): standard round-downwards to integer for real numbers only.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\lceil.\rceil$&lt;/span&gt; (ceil): standard round-upwards to integer for real numbers only.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Permitted operations but &lt;em&gt;counting as three operations&lt;/em&gt;:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$^\wedge$&lt;/span&gt; (exponentiation): standard exponentiation of real and complex numbers with integer powers only. Note that &lt;span class=&quot;math-container&quot;&gt;$0^0$&lt;/span&gt; cannot be used.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ln(.)$&lt;/span&gt; (natural log): standard natural logarithm of positive real numbers only.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;strong&gt;Note the restrictions on some operators&lt;/strong&gt; - This is primarily to ensure that exact values of &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt; cannot be obtained, as well as ensuring that the operations are well-defined.&lt;/p&gt;&#xA;&lt;p&gt;If you feel that a reasonable operation has been left out, mention it in the comments and I may add it.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Note: There must be at least 1 operation in your answer!&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;I will also upvote people who obtain high accuracy using many operations, in addition to those who get a good score, and encourage others to do likewise.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;I'm also going to keep track of the best answers for each number of operations, up to 10.&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: left;&quot;&gt;Operations&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: left;&quot;&gt;Score&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: left;&quot;&gt;User&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;0.864669301&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;pacoverflow&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;1.272568270&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;pacoverflow&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;1.501203245&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;Ben Frankel&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;2.254197410&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;Ben Frankel&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;5&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;2.286460415&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;Lynn&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;6&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;2.713605107&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;Lynn&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;7&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;2.151734961&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;Lynn&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;8&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;2.135316968&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;Ben Frankel&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;9&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;2.063009554&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;Ben Frankel&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;10&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;2.186087400&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;Glen O&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;Let me know if I've missed one.&lt;/p&gt;&#xA;" OwnerUserId="9289" LastEditorUserId="69582" LastEditDate="2021-03-04T05:09:16.983" LastActivityDate="2021-03-06T12:38:51.637" Title="Making π from 1 2 3 4 5 6 7 8 9" Tags="|mathematics|calculation-puzzle|formation-of-numbers|open-ended|" AnswerCount="15" CommentCount="29" ClosedDate="2021-08-05T08:39:40.923" ContentLicense="CC BY-SA 4.0" />
  <row Id="12326" PostTypeId="2" ParentId="12321" CreationDate="2015-04-17T16:27:18.970" Score="1" Body="&lt;p&gt;The first mathematician: F&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second mathematician: S&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F has $x$ and S has $y$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;F thinks that S has a number in the form of $6^{12/x}$ or $6^{12}/x$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;S thinks that F has a number in the form of $6^{12/y}$ or $6^{12}/y$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say $G_1=\{2,3,4,6,12\}$ are the sets of number that can be used in $6^{12/z}$ and to get a whole number (I didn't include $1$ for obvious reasons) and let $G_2=\{6^2,6^3,6^4,6^6\}$ be the numbers that can not be used in $z^{G_1}$. We call $G$ as the union of $G_1$ and $G_2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now see a table:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/V8r7e.png&quot; alt=&quot;Rows for the  first mathematician and columns for the second one&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From OUR point of view:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If F had numbers in the red row he could easily guess S's number. He says I don't know, which means he has $6,6^2,6^3,6^4,6^6$.We are confused between green cells and red cells, for $6^6$ between green and blue cell. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;S says at first he didn't know, which means he didn't have a number in red column, if not he could easily guessed F's number. Also he knows that F does not have a number in red row, if F had he would guess S's number. Since we eliminated red rows now we are not confused in yellow and red cells. The only possible situation that F and S has that conversation is that both of them has $6^6$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mission accomplished.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the OP noted I took $(x,y)$ different than $(y,x)$, if not there would be multiple solutions. To see the solutions color yellow cells to green.&lt;/p&gt;&#xA;" OwnerUserId="10499" LastEditorUserId="10499" LastEditDate="2016-05-25T20:01:25.413" LastActivityDate="2016-05-25T20:01:25.413" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12327" PostTypeId="2" ParentId="12324" CreationDate="2015-04-17T16:37:26.503" Score="3" Body="&lt;p&gt;An obvious solution is...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The monkey needs to make trips that run tangential to the island. Since the island is in the centre of the lake, this is achievable. As the boat continues in the same direction, it is still a one-way journey, and the monkey can get all of the bananas with only two trips.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alternatively, the monkey could just get more fish from the lake, but that's not really in the spirit of the question.&lt;/p&gt;&#xA;" OwnerUserId="9289" LastActivityDate="2015-04-17T16:37:26.503" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12331" PostTypeId="2" ParentId="12324" CreationDate="2015-04-17T17:15:22.513" Score="0" Body="&lt;p&gt;Assuming that the crocodile won't notice, you could&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cut all the fish into 3/4 and then give the crocodile 3/4 of a fish during each trip.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11094" LastActivityDate="2015-04-17T17:15:22.513" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12334" PostTypeId="2" ParentId="12313" CreationDate="2015-04-17T17:25:19.340" Score="4" Body="&lt;p&gt;I think the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Agatha Christie&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Every good story is build up of acts, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; She's a writer and knows how to build suspense&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;one of which is a tragedy,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mystery novels almost always have a murder or something else bad happen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;they all convey a mystery,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; She's a mystery writer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;how many there are is history,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; She's dead and therefore the number of books she's written is set&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;by the time you know your destiny.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm not quite sure about this one. She's wrote a book called &quot;Destination Unknown&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Destined for greatness destined for good,/you must be the best, pulling stories from your hood,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; She's widely considered to be one of the best mystery authors and is one of the top published authors of all time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;murder is something you truly understood.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; She wrote murder mysteries&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As to the &quot;Queen of Riddles&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Her a mystery is very similar to a riddle and many readers try to solve her stories as they read them&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5612" LastActivityDate="2015-04-17T17:25:19.340" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12336" PostTypeId="2" ParentId="10968" CreationDate="2015-04-17T17:31:03.833" Score="2" Body="&lt;p&gt;Try #2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the 3-digit code is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;991&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Kevin's answer helped me a lot, although I am still not sure this is the correct answer. The shortest path I could find from top left to bottom right is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1,13,3,28,29,49,63,52,53,66,79,93,105,118,119,120 which sums to &lt;strong&gt;991&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To get the numbers above, I numbered each cell from 1 to 120, going first from left to right, then top to bottom.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/KxehW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-04-17T20:37:03.500" LastActivityDate="2015-04-17T20:37:03.500" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12338" PostTypeId="2" ParentId="12325" CreationDate="2015-04-17T18:11:48.717" Score="1" Body="&lt;p&gt;With the incorrect scoring method:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ((((12!^3!^4!^5!^6!^7!^8!^9!)!)!)!)!...... resulting in a pi that far..far away but gives the highest of scores!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second Attempt: 10 operators (Score = 0.524352512)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1/(2^3) + |-4+5-6+7-8+9| = 3.125 (off by 0.01659)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First attempt: 4 operators (Score = 0.665238338)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (1/2)+3-(456/789) = 2.92205 (off by 0.21953)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-04-17T20:49:07.983" LastActivityDate="2015-04-17T20:49:07.983" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12343" PostTypeId="2" ParentId="12325" CreationDate="2015-04-17T18:29:01.420" Score="12" Body="&lt;p&gt;Attempt with 4 operators: (Score = 0.978898573)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\sqrt{\sqrt{\sqrt{\sqrt{123456789}}}} = 3.20420$$ (off by 0.06261)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Attempt with 3 operators: (Score = 0.991181027)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (1+2345+6)/789 = 2.98098 (off by 0.16060)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Attempt with 2 operators: &lt;strong&gt;(Score = 1.272568270)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (1/2345)*6789 = 2.89509 (off by 0.24649)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Attempt with 1 operator: (Score = 0.864669301)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 12345/6789 = 1.81838 (off by 1.32320)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4450" LastEditDate="2015-04-18T17:52:21.827" LastActivityDate="2015-04-18T17:52:21.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12344" PostTypeId="2" ParentId="12325" CreationDate="2015-04-17T18:30:55.763" Score="15" Body="&lt;h2&gt;Score = 1.187987428&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Here's my first effort, off by 0.00827 with 5 operators:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\sqrt{\sqrt{(1234 \times 56\ /\ 78)\ /\ 9}} = 3.14986$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="40" LastEditorUserId="9000" LastEditDate="2015-04-17T19:58:04.453" LastActivityDate="2015-04-17T19:58:04.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12345" PostTypeId="2" ParentId="12324" CreationDate="2015-04-17T18:53:48.593" Score="0" Body="&lt;p&gt;What if crocodile is already on island while monkey being at edge of lake asking to crocodile? This way crocodile has to make only 3 meaningful journeys in respect to monkey.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- Island to edge(1 banana, 1 fish)&lt;br&gt;&#xA;2- Edge to island(1 fish)&lt;br&gt;&#xA;3- island to edge(1 banana, 1 fish)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deal done!&lt;/p&gt;&#xA;" OwnerUserId="11573" LastEditorUserId="10281" LastEditDate="2015-04-17T18:58:45.330" LastActivityDate="2015-04-17T18:58:45.330" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12346" PostTypeId="2" ParentId="12324" CreationDate="2015-04-17T19:10:10.317" Score="0" Body="&lt;p&gt;The banana tree is not in island, neither is the monkey.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Monkey dont need to spend 4 fishes. &lt;/p&gt;&#xA;" OwnerUserId="11574" LastActivityDate="2015-04-17T19:10:10.317" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12347" PostTypeId="2" ParentId="12325" CreationDate="2015-04-17T19:18:15.467" Score="18" Body="&lt;p&gt;This one isn't in order, but I can't resist offering it up anyway:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\dfrac{\ln((8\times5!)^3+(2\times\sqrt{9})!+4!)}{\sqrt{7\times6+1}}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;14 ops, but off by just $4\times 10^{-14}$!  The catch here is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 43 (which shows up in the denominator) is a so-called &lt;a href=&quot;http://en.wikipedia.org/wiki/Heegner_number&quot;&gt;Heegner Number&lt;/a&gt;, so $e^{\pi\sqrt{43}}$ is very close to an integer - specifically, to $960^3+744$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="266" LastEditorUserId="266" LastEditDate="2015-04-17T19:39:23.960" LastActivityDate="2015-04-17T19:39:23.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12348" PostTypeId="2" ParentId="12325" CreationDate="2015-04-17T19:27:10.260" Score="59" Body="&lt;h3&gt;4 ops = 1.9934200404 points:&lt;/h3&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $1+2+34 \div \sqrt{56789} = 3.1426746469\dots$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Off by 0.00108199.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;5 ops = 2.2864604146 points:&lt;/h3&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\sqrt{12}-34\cdot \sqrt{56} \div 789 = 3.1416267073\dots$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Off by 0.0000340537.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;6 ops = 2.7136051067 points:&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(1+(23+4+5)\div 678)\cdot\sqrt 9 = \frac{355}{113} = 3.14159292035 \dots$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Off by only 0.000000266764(!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can keep taking square roots of 1 in this expression to get a lower score bound for $n$ operations where $n \geq 6$, namely: $$s_n = -\ln \left( \frac{ 355/113 } \pi - 1 \right) / n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which yields the results:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 7 ops = 2.3259472344 points&#xA; 8 ops = 2.0352038301 points&#xA; 9 ops = 1.8090700712 points&#xA;10 ops = 1.6281630641 points&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Oh, and to demonstrate the restrictions in the OP -- if complex logarithms were allowed, I could write&#xA;$$\ln(-1) \div \sqrt{\lfloor -2/3456789 \rfloor} = i\pi/i = \pi.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: for 7 ops, I found&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\sqrt{\sqrt{123-\sqrt{4!+5678/9}}}=3.14159355670578 \dots$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;scoring 2.1517349612 points, beating Ben Frankel's score.&lt;/p&gt;&#xA;" OwnerUserId="11476" LastEditorUserId="11476" LastEditDate="2015-04-19T19:28:06.820" LastActivityDate="2015-04-19T19:28:06.820" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="12349" PostTypeId="1" AcceptedAnswerId="12355" CreationDate="2015-04-17T20:17:47.560" Score="9" ViewCount="3136" Body="&lt;p&gt;There are $13$ coins arranged in a circular fashion.&lt;br&gt;&#xA;Andrew and Bill play a game with these rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Andrew starts the game&lt;/li&gt;&#xA;&lt;li&gt;A player can remove $1$ or $2$ &lt;strong&gt;adjacent&lt;/strong&gt; coins from the circle during his turn.&lt;/li&gt;&#xA;&lt;li&gt;The one who removes the last coin wins.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Assuming that both players are smart and play optimally, &lt;strong&gt;who wins?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example is shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/b7paL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The brown coin cannot be taken with the yellow or orange coins as it is not adjacent anymore. Similarly, one or both of the orange coins can be taken, but not with a yellow coin or the brown coin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This puzzle is &lt;strong&gt;NOT&lt;/strong&gt; the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2528/20-coins-on-the-table&quot;&gt;&lt;strong&gt;20 coins on the table&lt;/strong&gt;&lt;/a&gt;!&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2020-09-18T16:48:20.163" Title="Thirteen coins game" Tags="|strategy|game|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12350" PostTypeId="2" ParentId="12316" CreationDate="2015-04-17T20:20:35.303" Score="2" Body="&lt;p&gt;here is my solution, just with 2 operators: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (98+2)/4=25, where 6 is upside down&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10876" LastEditorUserId="10876" LastEditDate="2015-04-17T20:26:22.980" LastActivityDate="2015-04-17T20:26:22.980" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12351" PostTypeId="2" ParentId="12349" CreationDate="2015-04-17T20:24:15.943" Score="-1" Body="&lt;p&gt;Originally I said A(ndrew) won, by using a theory of how &lt;strong&gt;many&lt;/strong&gt; coins are remaining to determine where (and how many) to take, but that was wrong. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take 1 coin, then 1 coin is taken 2 down, you're left with 10+1 coins. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was considering doing something with groups of 6 (after taking 1) but then they merge, so that is an issue, especially if 2 coins are taken from the middle of the remainder. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If 1+1 coins are taken from &quot;opposite&quot; ends, we'd have a 5+6 chain. The 5 chain, the first player can win, the 6 chain, the 1st as well. At that point, player B would win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If 1+2 coins are taken from the opposite ends, we'd have a 5+5 chain. In both chains, the first player can win. However, if the first player wins in both chains, player B wins. However, each player can also lose their 5 chain, making them the winner of the other. But if mirrored plays are done, then player B wins. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-04-17T21:05:35.333" LastActivityDate="2015-04-17T21:05:35.333" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12353" PostTypeId="2" ParentId="10968" CreationDate="2015-04-17T20:26:40.037" Score="1" Body="&lt;p&gt;This might be too simple, but could the code be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;107?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That seems to be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the number of dots that can actually be reached from the top-left square (by following the lines in the original picture, not Ivo Beckers' one, i.e. taking directionality into account as noted in Kevin's answer).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11469" LastEditorUserId="11469" LastEditDate="2015-04-17T20:31:43.890" LastActivityDate="2015-04-17T20:31:43.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12354" PostTypeId="2" ParentId="12325" CreationDate="2015-04-17T20:36:39.193" Score="18" Body="&lt;h2&gt;Score = 2,176,716,257 (That's not a typo)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;I think you want to fix your scoring formula.&lt;/em&gt;&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my attempt:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 123456789!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And here's the proof of my score:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Thank you, &lt;a href=&quot;https://www.wolframalpha.com/input/?i=abs%28ln%28abs%281-123456789!%2Fpi%29%29%29%2F1&quot;&gt;Wolfram Alpha&lt;/a&gt;&lt;br&gt;In the screenshot, note that $log$ denotes the natural log as explained on the right.&lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/FhKDr.png&quot; alt=&quot;Screenshot&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you don't fix the formula, you run into a problem with infinite scores because...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Adding factorials makes $n_{ops}$ grow linearly but $A$ grow at a much faster rate.&lt;br&gt;My next guesses would all yield higher scores:&lt;br&gt;(123456789!)!&lt;br&gt;((123456789!)!)!&lt;br&gt;(((123456789!)!)!)!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastActivityDate="2015-04-17T20:36:39.193" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12355" PostTypeId="2" ParentId="12349" CreationDate="2015-04-17T20:44:34.107" Score="9" Body="&lt;p&gt;&lt;a href=&quot;https://math.stackexchange.com/questions/371420/logic-coin-game&quot;&gt;https://math.stackexchange.com/questions/371420/logic-coin-game&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is mostly taken from the Math.SE site as linked above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the game has two separate piles of the same size, the player to move loses as the other player can mirror his moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is a single chain left the player to move can make there be two same size piles and, therefore, force a win.  If the chain length is odd, he removes the middle coin.  If it is even, he removes the middle two coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is a cyclic ring, the first player's move will make a chain as just described.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;By this the second player will always win for a size 13 circle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="99" LastEditorUserId="-1" LastEditDate="2017-04-13T12:19:36.950" LastActivityDate="2015-04-18T13:33:41.787" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12356" PostTypeId="2" ParentId="12324" CreationDate="2015-04-17T21:10:55.820" Score="3" Body="&lt;p&gt;The question doesn't really clarify, but I'll assume that there is a bunch of bananas (meaning connected, not meaning many) on each tree.  Or a single banana, but I'll go with bunches, because a monkey could easily handle two bananas, even on a crocodile's back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The monkey spends one fish to get to the island (two remaining), and removes the bunch from each tree.  Being a monkey and knowing that bananas float (due to his monkey recipes), the monkey throws one bunch into the water, away from the island.  The crocodile is evidence that anything living in the lake will want to eat fish, so the bunch should be fine floating where it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The monkey spends another fish returning to the lake shore (one left) with his first bunch of bananas.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;He then spends his last fish (none remaining), and instructs the crocodile to swim past the bobbing bananas straight to the opposite shore.  Using monkey hands, feet, tooth, and tail, the monkey should be able to collect his second bunch on the trip across the lake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once on the shore again, the monkey circles the lake to his other bunch and happily leaves, while the crocodile begins to wish it were in a river instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was based slightly on Glenn O.'s answer, but doesn't rely on the integrity of the crocodile to count one trip tangent to the island -&lt;em&gt;with a break in the middle&lt;/em&gt;- as simply one trip.  Since the third trip in my answer wouldn't require a stop, I think it would qualify as a valid single trip, payable by a single fish.&lt;/p&gt;&#xA;" OwnerUserId="11452" LastActivityDate="2015-04-17T21:10:55.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12357" PostTypeId="1" AcceptedAnswerId="12380" CreationDate="2015-04-17T21:15:10.553" Score="26" ViewCount="4127" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A regular hexagon is divided into a triangular grid, and completely tiled with diamonds (two triangles glued together). Diamonds can be placed in one of three orientations. Prove that, no matter how the board is tiled, there will be the same number of diamonds in each orientation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is an example of such a tiling. Though this hexagon has 5 triangles to a side, the problem asks you to prove this for any size hexagon, and any tiling of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\qquad\qquad\qquad\qquad\quad$&lt;img src=&quot;https://i.stack.imgur.com/vWbjJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is one of those puzzles that has many solutions, so I'm very curious to see what people's favorite approaches are. Therefore, I'm going to hold off on accepting an answer for a while, to try to get as many different solutions as I can.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="65277" LastEditDate="2020-11-21T03:38:33.830" LastActivityDate="2020-11-21T03:38:33.830" Title="Tiling a Hexagon with Diamonds" Tags="|geometry|combinatorics|tiling|proof-without-words|" AnswerCount="12" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12358" PostTypeId="2" ParentId="12324" CreationDate="2015-04-17T21:16:06.180" Score="2" Body="&lt;p&gt;The monkey can get all the bananas and only spend 2 fish.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Just get to the island, and eat all the ones you can't carry back.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Surprised no one thought of this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You must not have my appetite.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2505" LastActivityDate="2015-04-17T21:16:06.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12359" PostTypeId="2" ParentId="12357" CreationDate="2015-04-17T21:19:25.217" Score="7" Body="&lt;p&gt;Interestingly enough, looking at the image as a 3d graph, you can see that each &quot;face&quot; has the same number of tiles. So, if you looked at it from the left, you'd see 25 squares. Top, 25 squares. Right, 25 squares. And each of the 3 orientations corresponds to one of the faces. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-04-17T21:19:25.217" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12360" PostTypeId="2" ParentId="12284" CreationDate="2015-04-17T21:31:07.657" Score="1" Body="&lt;p&gt;EDIT: now conforms to example, also has better success rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't guarantee success, nor does it safely determine the identity of either die, but it gives better odds than the best current answer (@Zikato's):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Method for if there are 2 or 3 pairs initially:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rule for assistant: Mentally sort the dice into pairs. Find the lowest value which there is a pair of, and choose the first two dice of this value to cup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rule for magician: Look at the visible pairs (there will be at least 1). This is an upper bound on the value of the pair under the cup. Now, say you have something like this: X334X2345. We see that the lowest visible paired value is 3, so we have it down to a 1/3 chance. Now, they can't be 3s, or else the assistant would have cupped the 1st and 2nd dice. Therefore we're looking at either two 1's or two 2's -- a 50/50 guess. If we had something like this - X332X3345 - then we'd know for sure it's two 1's, because if it was 2's, the assistant would have cupped 1st and 4th positions. This method gives, I think, at least 1/5 odds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Method for if there are 4 pairs:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assistant: If there are 4 pairs, cup the unpaired number first (if it is ambiguous which one is unpaired, choose the last option), then, looking at the other numbers, cup the one with the position equal to the first cup's value (jumping over any cups if needed). So if you have 243324665, cup the five, then cup the second 2. If you have 232222222, cup the three, then cup the 3rd 2 (as it is in position #3 if we ignore the cupped position).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magician: Find the visibly unpaired number. One of the two cups will be that #. Now, look at the positions of the 2 cups. If one of them is in the 7th, 8th, or 9th position (1/3 chance I think) then that is the unmatched one. The other cup = the visible unmatched value and its position is the value of the other cup. If both cups are in the positional range from 1-6, then you know the value of a cup, but aren't sure which cup it's the value of. Take the example 2X2X22222. We know one of the cups is hiding a 2. Suppose it's the first one. This is impossible, because then we would have cupped the unpaired value at position 4 first, and then cupped the die with the position = the unpaired die's value second. We clearly cupped position 2 second, and position 4 can't hold value 2, because then the unpaired value would have been ambiguous and the assistant would have chosen the 9th position to be the unpaired one. So we know that position 2 is the unpaired one, and since the 4th position is a 2 (we know this) and it's in the 3rd position ignoring cups, then the mystery cup holds a 3. So if there are 4 pairs to begin with, we have 100% chance of success.&lt;/p&gt;&#xA;" OwnerUserId="11154" LastEditorUserId="11154" LastEditDate="2015-04-17T23:07:17.147" LastActivityDate="2015-04-17T23:07:17.147" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12361" PostTypeId="2" ParentId="12324" CreationDate="2015-04-17T21:36:56.860" Score="0" Body="&lt;p&gt;Here is my solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; monkey is not on the Island, but the trees arent on the island, there are in the lake, as can read in the description. So the monkey is on one of those trees, but not in island. first monkey had to travel from tree to lakeside, which costvhim 1 fish. after he will do next trip to second tree from lakeside to tree and back, which cost him 2 fish, which is totally 3 fish.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10876" LastActivityDate="2015-04-17T21:36:56.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12362" PostTypeId="2" ParentId="12278" CreationDate="2015-04-17T21:44:22.207" Score="10" Body="&lt;p&gt;Since Andrew Smith managed to get the answer in a way other than I intended (congratulations!), I'm going to post my own answer here to demonstrate how I intended for it to be solved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the clues that Prof. Farnsworth studied film, and was obsessed with movie codes, leads us to figure out that each of the numbers on the blackboard corresponds to the ID number of a movie on IMDB (the Internet Movie Database).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking them all up gives us:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;0091042&lt;/strong&gt; Ferris Bueller's Day Off (1986)&lt;br&gt;&#xA;&lt;strong&gt;0241303&lt;/strong&gt; Chocolat (2000)&lt;br&gt;&#xA;&lt;strong&gt;0091369&lt;/strong&gt; Labyrinth (1986)&lt;br&gt;&#xA;&lt;strong&gt;0458352&lt;/strong&gt; The Devil Wears Prada (2006)&lt;br&gt;&#xA;&lt;strong&gt;0317219&lt;/strong&gt; Cars (2006)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;0372784&lt;/strong&gt; Batman Begins (2005)&lt;br&gt;&#xA;&lt;strong&gt;0910970&lt;/strong&gt; WALL-E (2008)&lt;br&gt;&#xA;&lt;strong&gt;0117731&lt;/strong&gt; Star Trek: First Contact (1996)&lt;br&gt;&#xA;&lt;strong&gt;0098333&lt;/strong&gt; Sinbad of the Seven Seas (1989)&lt;br&gt;&#xA;&lt;strong&gt;0120655&lt;/strong&gt; Dogma (1999)&lt;br&gt;&#xA;&lt;strong&gt;0106220&lt;/strong&gt; Addams Family Values (1993)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;0119217&lt;/strong&gt; Good Will Hunting (1997)&lt;br&gt;&#xA;&lt;strong&gt;0408306&lt;/strong&gt; Munich (2005)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;0264395&lt;/strong&gt; Basic (2003)&lt;br&gt;&#xA;&lt;strong&gt;0344854&lt;/strong&gt; Arthur and the Invisibles (2006)&lt;br&gt;&#xA;&lt;strong&gt;0099653&lt;/strong&gt; Ghost (1990)&lt;br&gt;&#xA;&lt;strong&gt;0149624&lt;/strong&gt; All the Pretty Horses (2000)&lt;br&gt;&#xA;&lt;strong&gt;0112579&lt;/strong&gt; The Bridges of Madison County (1995)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;0489099&lt;/strong&gt; Jumper (2008)&lt;br&gt;&#xA;&lt;strong&gt;0120363&lt;/strong&gt; Toy Story 2 (1999)&lt;br&gt;&#xA;&lt;strong&gt;0092890&lt;/strong&gt; Dirty Dancing (1987)&lt;br&gt;&#xA;&lt;strong&gt;0097351&lt;/strong&gt; Field of Dreams (1989)&lt;br&gt;&#xA;&lt;strong&gt;0465624&lt;/strong&gt; My Super Ex-Girlfriend (2006)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;0232500&lt;/strong&gt; The Fast and the Furious (2001)&lt;br&gt;&#xA;&lt;strong&gt;0259288&lt;/strong&gt; Dragonfly (2002)&lt;br&gt;&#xA;&lt;strong&gt;0140796&lt;/strong&gt; Air Bud: Golden Receiver (1998)&lt;br&gt;&#xA;&lt;strong&gt;0790604&lt;/strong&gt; Deck the Halls (2006)&lt;br&gt;&#xA;&lt;strong&gt;0251160&lt;/strong&gt; John Q (2002)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, we look at the length of each movie title (ignoring spaces and punctuation).  This gives us a series of numbers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    20 8 9 18 4 12 5 20 20 5 18 15 6 5 22 5 18 25 6 9 12 13 19 20 9 20 12 5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Converting those numbers to letters (A=1, B=2, etc.) give us&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    THIRD LETTER OF EVERY FILMS TITLE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we then follow those instructions, and look at the third letter of each title, we get&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    ROBER TLANGD ON STOLE MYRES EARCH&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which, when we rearrange the spaces, spells&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    ROBERT LANGDON STOLE MY RESEARCH&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2015-04-17T21:44:22.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12363" PostTypeId="2" ParentId="11913" CreationDate="2015-04-17T21:45:36.227" Score="1" Body="&lt;p&gt;I'm going to attempt to begin solving it&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;First step is that I recognized there is one dark number, 6 light circles and 8 brighter circles all grouped in 6 distinct groups&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I tried adding them up - currently playing around with that idea, so far only the right hand most group seems to make sense - it is all bright circles and it gave me 19, the 19th letter of the alphabet is 'S' - I also realized that Sahara has 6 letters but havent been able to make the other 5 groups fit - yet. More to follow (maybe)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There's also no number above 12. 12 hours on a clock (Sands of &lt;strong&gt;Time&lt;/strong&gt;) or Morse Code gives drong/wksau :/  Brail doesnt really work either. Another thing I noticed (might be a clue) all bright or light circles only have 6 different digits in them. 6,8,1,9,2,3 (7 different numbers because 1 and 2 make up 12)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastEditorUserId="11251" LastEditDate="2015-04-21T17:32:26.657" LastActivityDate="2015-04-21T17:32:26.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12364" PostTypeId="1" AcceptedAnswerId="12729" CreationDate="2015-04-17T21:47:32.343" Score="4" ViewCount="497" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If you AES the last line,&lt;br&gt;&#xA;  With a key made of something divine,&lt;br&gt;&#xA;  And then you XOR&lt;br&gt;&#xA;  The SHA-1 of &quot;4&quot;,&lt;br&gt;&#xA;  FWqY58QhWVqPDX0uQ8C37AYaPeWFab42F2nyuY+zWK/gCV2bo8hIDt6sjon1I8+Y64RzJcCwlHIouiebHza67ErpHbVVWeZh4Pf8GBVtrDU=&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can you figure out what the &quot;divine thing&quot; is and then follow the instructions to determine what the keyword hidden in the last line of this limerick is?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Important hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The AES key is a word padded with null characters at the end to make it 16 bytes total.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="10399" LastEditDate="2015-04-17T23:43:41.080" LastActivityDate="2015-04-25T20:42:12.297" Title="A crypto-limerick" Tags="|cipher|computer-puzzle|limerick|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="12365" PostTypeId="2" ParentId="12357" CreationDate="2015-04-17T22:00:15.213" Score="0" Body="&lt;p&gt;If we assign $S$ to be the side length of the hexagon (in number of diamond side lengths) and $A$,$B$,$C$ to be the number of diamonds of each type where $A$ is longer than tall, $B$ points to the bottom right/top left, and $C$ points to the bottom left/top right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total number of diamonds (aka area) lets us make this equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S^2*3=A+B+C$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine the $S=1$ hexagon... There are only 2 solutions which are the same one rotated by 30 degrees.  There needs to be all three diamonds present in order of the central portion to add up to 360 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can imagine that there are 3 paths which proceed from the top to bottom, upper right to lower left, and upper left to lower right corners. The total movement down for any path you follow (top to bottom) must be equal to $2S$ but the movement left to right must be zero.  If you move the whole way down on an $A$ diamond, you do not move right or left.  If you move down on a $B$ or $C$ diamond you move right or left respectively.  In order for all paths to not move left or right, the total number of $B$ and $C$ must be equal.  If you rotate the graph 60 degrees such that a different pair of corners point up/down, you can show this for $A$ and $B$ or $A$ and $C$.&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2015-04-17T22:00:15.213" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12366" PostTypeId="1" AcceptedAnswerId="12368" CreationDate="2015-04-17T22:04:15.433" Score="0" ViewCount="276" Body="&lt;p&gt;You are heading to your hotel room but forget your room number. &lt;br&gt;&#xA;Try to check the key card but the digits are blurred &lt;br&gt;&#xA;You call to hotel desk info and a nerdy bell boy tell you:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Your room have 4 digits &lt;/li&gt;&#xA;&lt;li&gt;the digits are different&lt;/li&gt;&#xA;&lt;li&gt;the digits are in an increasing order&lt;/li&gt;&#xA;&lt;li&gt;the product of the first two is the sum of the last two.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What is your room number?&lt;/p&gt;&#xA;" OwnerUserId="11577" LastEditorUserId="11577" LastEditDate="2015-04-17T22:11:07.780" LastActivityDate="2015-04-28T17:51:56.510" Title="What is my room?" Tags="|calculation-puzzle|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12367" PostTypeId="2" ParentId="12324" CreationDate="2015-04-17T22:05:34.847" Score="10" Body="&lt;p&gt;Idea 1. Find a second crocodile willing to undercut the first croc's wage -- after all, 3 fish is better than no fish!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Idea 2. Trick the croc into signing a contract in which &quot;one-way trip&quot; is not specified by distance, but by traveling in a line of the croc's choosing from point A to point B. Tell the croc you want to go clean to the other side of the lake. The croc, hoping to minimize distance, goes through the middle of the lake (over the island), and you, the monkey, jump off the croc's back and grab a shipment of bananas, then hop back on the croc's back before it re-enters the water. This should be doable because you're a quick monkey. When you get to the far side, drop off a shipment of bananas and use your second one-way trip to go back to point A, grabbing the second shipment of bananas en route. Then walk back around to the far side of the lake on foot to reclaim your stashed bananas (you still have a fish left over this way)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Idea 3. Pay the croc to take you to the island. Get the bananas and pay the croc for a return trip. Drop them off and pay the croc to take you back once more. Get your bananas and then just don't pay the croc to take you back. Now the croc is stuck on this boring island and eventually it will leave of its own accord. Hop on its back when it does and get the free ride back to shore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Idea 4. Convince the croc to loan you the labor, then once you have all the bananas, trade a few for the last fish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Idea 5. Ditch the crocodile. Craft a raft and harnesses for the fish  and let them drag you around the lake at random. When you're close enough to the shore, jump off and pull in the fish, grab the bananas, and keep doing that back and forth until you're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Idea 6. Promise the croc a 50% raise in exchange for it getting paid half upfront and half at the end. Give it your 3 fish upfront, collect all the bananas, and then climb a super tall tree where you say you have the rest of the croc's payment. Then tell the croc that you're not gonna throw it the 3 other fish. It gets super mad and starts climbing the tree. Right before it gets to you, swing on a vine back to the floor. The croc is now stuck in the tree, too high for it to drop down out of. While it tries to figure out how to get down, take the bananas and run.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;We Need More Ideas:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Idea 7. When the croc gives you its demands, say &quot;Eh, nevermind... but hey, you know what? I'm gonna loan you my favorite book!&quot; Give the croc a philosophical, religious, or otherwise ethical text that promotes doing nice things for others without reward. The croc will see this as you doing a nice thing for it (loaning it your favorite book!) and its heart will be filled with warmth. It will now offer to give you the rides for free! (WARNING: do not read the book yourself, or you may end up deciding there are more important things in life than scamming for bananas.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Idea 8. Convince the croc that your fish are each 4/3 the size of a regular fish, and that he's getting the same amount of fish by weight as if you gave him 4 regular fish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Idea 9: Find some math trickery like Monty Hall and use it to gamble against the croc for free passage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Idea 10: Tell the croc, &quot;Well, I've only got three fish, but I caught them in like 5 seconds in a secret lake! After the first round trip I'll give you the longitude of the lake, and after the second I'll give you the latitude! And hey, I'll throw in these three fish for free!&quot; Then make up some coordinates a few miles away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Idea 11: Convince the croc that it's bad at counting and that you're actually holding four fish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Idea 12: Try eating the fish, maybe you'll like them more than bananas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Idea 13: Get 1 shipment of bananas back to shore by paying the croc 2 fish, then sacrifice one banana to get its seeds. Then go around to all your monkey friends with the seeds telling them they're magic seeds that will grow a magical banana orchard, and they only cost 1 fish apiece. Then whenever their orchards grow and they're not that magical, blame it on their irrigation or something. You only really need to get 1 monkey to fall for this, but if more do then you've got more bargaining power amongst the crocs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Idea 14: When the croc proposes its deal, say &quot;Whatever dude!&quot; and hurl the first fish, as if in anger, to the island. The croc will frantically swim to the island to retrieve the fish, and you catch a ride (it won't mind since this is kind of what it asked for anyway). You do this three times, picking up a fish-shaped rock somewhere along the way. The final time, throw the rock back to shore. The croc won't examine it too closely because it's far away and it already expects a fish, so it will take you back. (WARNING: you might have some smooth-talking to do to make the croc not mad at you.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Idea 15: Challenge the croc to a thumb-wrestling match for free passage vs. free fish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Idea 16: Tell the croc you're a casting director and your movie needs a croc who's good at ferrying monkeys carrying bananas. Perform an impromptu audition, and then tell the croc that it'll hear from you soon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definitely Gonna Need a Few More&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Idea 17: &quot;Alright, here's fish #1-- oh wait, this is a PoisonHead fish, let me cut off the poisonous part for you. Here's fish #2, a PoisonMiddle fish, let me just take care of that... Here's fish #3, a PoisonTail fish (chop)... and WHAT'S THAT BEHIND YOU?! (krazy glue time) Oh nevermind, anyway, here's fish #4!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Idea 18: Hide your 3 fish on the mainland, and put pieces of paper next to all of them. 2 scraps say &quot;For your next clue, go to the banana island&quot; and the last one says &quot;Now, to find the great fish hoard, solve this puzzle:&quot; and just write random letters. Write two more &quot;clues&quot; with the location of the second and third fish on them, and keep them in your hand. Tell the croc you found the start of what looks like a treasure hunt and take it to the first fish. It will see the note and take you to the banana island. While you're there, grab the bananas and drop the clue to the 2nd fish on the ground. Repeat the process. When you get to the third clue, you can be like &quot;Well, it's been fun, but I gotta go&quot; and leave the croc to puzzle over the impossible riddle!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Idea 19: Just a simple fish on a stick ploy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Idea 20: Instead of giving the croc the fish to eat, release them in the pond and tell the croc &quot;They will interbreed and this time next year the pond will be teeming with thousands of fish! So basically I gave you thousands of fish! Surely that's worth a few trips across the ol' lake!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Idea 21: Introduce the croc to the concept of exercise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Idea 22: Dare the croc. &quot;Bet you can't carry me across FOUR times, twice laden with bananas!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Idea 23: (WARNING: SOMEWHAT DANGEROUS &amp;amp; SLOW) Wait till the croc is sleeping. stand with one leg on either side of its torso and incredibly slowly ease yourself down onto it, so that you don't wake it up. When it wakes up, it will have no idea that it's carrying you, and will go about its day. Let the croc carry you around the pond and grab the bananas when the croc randomly passes by/climbs the banana tree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Idea 24: Wait till you randomly find a crocodile tooth, and put it on a pendant around your neck. &quot;This is all that's left of the last crocodile who tried to cross me!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Idea 25: Pay for a ride to the island. Get all the bananas off the trees and put them in the water. They'll float (do bananas float? Probably) to shore, and you can splash them there to hasten this process. Then pay for a ride back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Idea 26: If a banana floats then surely a heap of bananas + a monkey still floats. Just hold em all and paddle back to shore!&lt;/p&gt;&#xA;" OwnerUserId="11154" LastEditorUserId="11154" LastEditDate="2015-04-18T00:55:18.210" LastActivityDate="2015-04-18T00:55:18.210" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12368" PostTypeId="2" ParentId="12366" CreationDate="2015-04-17T22:09:10.230" Score="7" Body="&lt;p&gt;The room number is &lt;strong&gt;3457&lt;/strong&gt;, &lt;strong&gt;3569&lt;/strong&gt;, or &lt;strong&gt;3578&lt;/strong&gt;. 3 &amp;times; 4 = 5 + 7, and 3 &amp;times; 5 = 6 + 9 or 7 + 8.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-04-17T22:09:10.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12370" PostTypeId="2" ParentId="12284" CreationDate="2015-04-17T23:07:32.487" Score="14" Body="&lt;p&gt;Firstly arrange the dice into a 3x3 matrix. If the sum of all dice is even, choose the column of first cup according to the total sum mod 3, and the row according to the sum of its own column. Let's say the first row/column means mod 3 = 0 and the second means mod 3 = 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the first cup is O (its own column denoted by o), choose the second cup between A, B (related to the position of O, and wrapping around if O touches the border) according to the parity of the column of the first cup. Say A is odd and B is even:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;.oA&#xA;.O.&#xA;.oB&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If the sum of all dice is odd, choose the diagonals and sum its diagonals instead. Say the main diagonal is for mod 3 = 0 and goes right, and dice in it start from top. And the second cup is chosen in this way:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;o..&#xA;AO.&#xA;.Bo&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;To reverse this process, firstly get the two bit informations from the relative position of the two cups. (If the two cups are diagonally adjacent, then the total sum is even. One diagonal is for A and the other is for B. Otherwise it's odd, and A is horizontal while B is vertical.) Then get the two &quot;trits&quot; from the position of the base cup, and use the Chinese remainder theorem. For example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;11.&#xA;1.1&#xA;146&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It is in the first case (two cups arranged in diagonals and the total sum is even), and the second cup is at the position of A (sum of the second column odd). The &quot;base&quot; cup is in the middle, which means both sums mod 3 = 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for the &quot;base&quot; cup it is 1 - 1 - 4 = 2 (mod 3), and even (sum of the column is odd and 1 + 4 is odd). So it is 2. Similarly for the top-right cup, it is 1 - 1*5 - 4 - 6 - 2 = 2 (mod 3), and odd (total sum even and sum of others odd), so it is 5.&lt;/p&gt;&#xA;" OwnerUserId="11495" LastEditorUserId="11495" LastEditDate="2015-05-15T19:47:48.070" LastActivityDate="2015-05-15T19:47:48.070" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="12371" PostTypeId="2" ParentId="12357" CreationDate="2015-04-18T00:08:44.503" Score="9" Body="&lt;p&gt;Here's a 3D-inspired proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take any tiled hexagon, and look at its vertical lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, observe that due to the shape of the tiles, all vertical lines must have the same length as the left and right side of the hexagon, possibly with gaps in between.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if none of them have gaps, and all of them end at the bottom, the entire tiling has to look like this (&quot;completely filled cube&quot;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/REkwO.png&quot; alt=&quot;completely filled cube&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We show that it is possible to transform any other tiling into a &quot;completely filled cube&quot; without changing the number of tiles in each orientation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, select a fragment of a vertical line that doesn't end at the bottom. It has to end at a horizontal tile instead, as the other two tiles both have vertical sides. Hopefully, the situation looks like this (&quot;corner&quot;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gPGnd.png&quot; alt=&quot;corner&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But maybe there are one or two additional lines originating at the same spot, like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7IGJJ.png&quot; alt=&quot;non-corner&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that is the case, follow one of them. It must belong to another horizontal tile adjacent to the current one. (You can see this from the picture.) So after following the line, you are in the same situation again, but closer to one of the sides of the hexagon (which guarantees termination, since there is definitely a vertical line in the direction you just came from). Continue in the same direction until you reach a &quot;corner&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that you have reached a &quot;corner&quot;, &quot;fill it&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1U7nw.png&quot; alt=&quot;filled corner&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, the number of tiles in each orientation has stayed the same. However, a vertical line fragment has just moved downwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeat this algorithm until all vertical lines end at the bottom and all gaps are removed, resulting in the &quot;completely filled cube&quot; (see above).&lt;/p&gt;&#xA;" OwnerUserId="11469" LastEditorUserId="11469" LastEditDate="2015-04-18T00:39:06.990" LastActivityDate="2015-04-18T00:39:06.990" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12372" PostTypeId="2" ParentId="12357" CreationDate="2015-04-18T00:51:37.757" Score="2" Body="&lt;p&gt;Here's my attempt to prove it.. It seemed impossible until I finally exploited a trick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I start from a valid configuration where there's only one change possible (rotating the 3 semilines in the middle: any other change would at same time change number of diamonds an create triangles.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CkWjY.png&quot; alt=&quot;Proof attempt to the puzzle by Dario Oliveri&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you do that change you are free to undo it (useless, I'll mark it blue)&#xA;or to do other 3 changes (in red). You immediately note that you can do that &quot;change&quot; only on points that have lines placed like the middle of the first move, or the middle of the initial cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you do your 2nd move you'll be unable to undo the first move (grey now) because doing so would create triangles and other shapes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/t5qJs.png&quot; alt=&quot;another wireframe cube&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Assuming my first move was a clockwise rotation of 1/6 round, my undo is a 1/6 anti-clockwise)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically you can just check that the only possible moves are rotations of group of tiles made by 3 diamonds (1 for each orientation) (you can check all possible moves on a 2x2x2 &quot;cube&quot; and see that is true).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence you note also that rotation keeps number of diamonds for each orientation the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a little piece missing of the proof:&#xA;I did not show that starting from my first cube I can do all possible tilings, that's because rotations have &quot;inter-dependencies&quot; and I don't know if at some point &quot;I'll get stuck&quot; without more possible moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm too sleepy for that proof, but I developed another proof method I'll let you the pleasure to use it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extruding columns starting from an &quot;empty&quot; cube:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see you cannot extrude a column to a length greater than the preceding columns (there are 2 directions to check for preceding columns) because you'll get triangles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LskpW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have now a way to compute really all possible tilings. You start with the backmost column, and once decided a height you can extrude the 2 neighbour to any height lower or equal to the backmost column. After that you can do the same for the next 3 columns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no dependency on rotations here. You choose a number, and then you can choose again same number or a lower number. That's much easier but have some help from imagination (3rd dimension in a problem that has 2 dimensions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, probably that's not a formal proof. But helps imagination you have 2 ways to attack the problem, and probably those can be worked around for a formal proof. But I think is more interesting the intuition than the proof. Without some intuition there will never be some proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key seems always to be the same. Starting from a trivial configuration the only possible moves incidentally preserves the number of diamonds for each configuration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have never seen that puzzle before. Hope you like my first answer in puzzling exchange.&lt;/p&gt;&#xA;" OwnerUserId="11581" LastEditorUserId="7308" LastEditDate="2015-04-18T02:14:11.930" LastActivityDate="2015-04-18T02:14:11.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12373" PostTypeId="2" ParentId="12349" CreationDate="2015-04-18T01:44:28.910" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first person to act loses for any number $3\le n+3$. (With the trivial exception of a 1-coin or 2-coin circle, where the first person simply takes all the coins on the first turn.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first player has two choices: take one coin, or take two coins. In perfect play, the second player will always take the opposite number (or the same number in even-numbered coin starts), directly across the circle, dividing the remaining coins into two equal groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, there are two identical sets:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{C_1C_2C_3\ldots C_n\},\{C_1C_2C_3\ldots C_n\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wherever the first player next takes coins, the second player mirrors the move exactly in the other set. For every $1\le m\le n$ (or pair of consecutive $1\le m_1\lt m_2\le n$), there is always the matching $m$ (or $m_1m_2$) in the other set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the symmetry, the second player always has coins to take, hence the second player will take the last coin and win.&lt;/p&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-04-18T02:01:50.750" LastActivityDate="2015-04-18T02:01:50.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12374" PostTypeId="2" ParentId="12321" CreationDate="2015-04-18T01:51:22.487" Score="1" Body="&lt;p&gt;&lt;strong&gt;NOTE:&lt;/strong&gt; This actually fails in the same way as &lt;a href=&quot;https://puzzling.stackexchange.com/questions/12211/find-two-numbers-based-on-either-their-product-or-their-sum/12214#12214&quot;&gt;this puzzle&lt;/a&gt; (&lt;a href=&quot;https://puzzling.stackexchange.com/a/12214/2484&quot;&gt;my answer&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Call the first mathematician $\alpha$ and the second mathematician $\beta$. $\alpha$'s&#xA; number is $x$ and $\beta$'s number is $y$. I will assume that $x,y \in \mathbb{N}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set of numbers which (as factors are divisors) are raised to some power to equal $2176782336$ is $f = \{2,3,4,6,12,36,216,1296,46656\}$. This set is complete. Also note that $\forall z \in f. 2176782336 \mod z \equiv 0$ - i.e. all members of $f$ are also factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First Comment:&lt;/strong&gt;&#xA;The only way that first comment can occur is if $x \in f$. All members of $f$ present ambiguity regarding product and exponent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second Comment:&lt;/strong&gt;&#xA;If $\beta$ had the number $y \not\in f$ (but still at least in the factors of $2176782336$) then he would know the value of $x$. Similarly if $y=2$ then he would know that the only way this can occur is if $x=46656$. Similarly, for all $y \in f \land y\not=46656$ he would also know. The only case which still presents ambiguity for $\beta$ is $y=46656$ ($\alpha$ could still have either $x=2 \lor x=46656$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Third Comment:&lt;/strong&gt;&#xA;We ALL know that $y=46656$. So $\alpha$ declares this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fourth Comment:&lt;/strong&gt; &#xA;Unfortunately, $\beta$ knowing that $\alpha$ knows $y$ does not help him. The first three comments could have occurred with either $x=2 \land y=46656$ or $x=46656 \land y=46656$. The fourth comment cannot occur using the initial assumption, and I suggest would fail from infinite solutions if instead $x,y \in \mathbb{R}$, say. Thus, the puzzle is inconsistent. The fourth comment should have been &quot;I &lt;em&gt;still&lt;/em&gt; don't know your number&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-04-18T03:56:18.793" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12375" PostTypeId="2" ParentId="12357" CreationDate="2015-04-18T02:27:41.867" Score="0" Body="&lt;p&gt;Not sure this is a full answer but I'm getting tired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GKXtn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let n = number of triangles to a side. Take the diamonds touching EDIT: n+1 adjacent edge units (just at 1 point doesn't count): At least one diamond must be different from the others. Let all the changes happen in the corners, with a change at every other corner. We've made a loop which can contain a hexagon with side length n-1, and the number of diamonds of each kind is equal. Induction down to n = 1, where it's obviously equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let a hexagon outer loop deviate from our &quot;changes only occur at corners&quot; policy. Color all of the outer edge-adjacent diamonds a certain color (say, black) and leave any diamonds that jut out from this loop white. Now we can see a broken loop surrounding another (certainly broken) loop of n-1. Color in this inner loop with a second color, again leaving all rebels white. Do this down to the n=1 hexagon, then color the rebels by orientation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you look at my diagram, the inner purple hexagon really wants a red tile at bottom instead of an orange and a pink. Imagine this is a mosaic. Rip up a red tile and the orange and pink rebels in the middle, and put the red tile there. The purple hex is happy now. Now make the green hex happy (a change only at every other corner) -- The bottom sideways diamond wants to be two slanted diamonds to fit around the purple hex -- add in our orange and pink tiles to the side, putting the green tile wherever we robbed the red tile from earlier. I think it's clear that this process can be continued until we reach our &quot;optimal hexagon&quot;. My brain is too fried to prove this definitively, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I believe these two things are true: 1. If we take a non-optimal hexagon, every concentric loop will be unhappy. 2. Fixing an unhappy loop necessarily adds tiles to our 'hand' of removed mosaic tiles. 3. To fix the innermost hex, rob any appropriate rebel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With these two things in mind, it's impossible that we will want to fix a hex but will not have tiles in our 'hand' of removed tiles, assuming that there is at least one rebel of the kind needed by the n=1 loop.&lt;/p&gt;&#xA;" OwnerUserId="11154" LastEditorUserId="11154" LastEditDate="2015-04-18T02:41:42.430" LastActivityDate="2015-04-18T02:41:42.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12376" PostTypeId="2" ParentId="12325" CreationDate="2015-04-18T05:00:50.117" Score="14" Body="&lt;p&gt;I figured I'd have a go at it myself. Here's an approximation I found:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\sqrt{\sqrt{123+4-5\times6+\frac7{8+9}}} \approx 3.14161421087308 \approx \pi+0.000021557$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;with 8 operations. Score = &lt;strong&gt;1.486190839099123&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE: here's a new one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\sqrt{\sqrt{\frac12+\frac{3!+4}{5+6}+7+89}}\approx 3.141592652582646 \approx \pi - 0.000000001007147$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;with 10 operations. Score = &lt;strong&gt;2.186087400111085&lt;/strong&gt;. This one is based on an expression given by Ramanujan, $\pi\approx\sqrt[4]{\frac{2143}{22}}$.&lt;/p&gt;&#xA;" OwnerUserId="9289" LastEditorUserId="9289" LastEditDate="2015-04-19T03:31:45.167" LastActivityDate="2015-04-19T03:31:45.167" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12378" PostTypeId="2" ParentId="12325" CreationDate="2015-04-18T06:22:48.940" Score="2" Body="&lt;p&gt;SCORE: 1.116831135 (7 Ops)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (1-2) * (34-56) / 7 * (-8 + 9) = 22/7&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;SCORE: 0.977227243 (8 Ops)  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ((1+2) * 3)/4/(5+6)/7/8/9 = 22/7&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;SCORE: 1.5636 (5 Ops) Thanks @Glen O.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (⌊1234/56⌋/7)*(-8+9) = 22/7&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2656" LastEditorUserId="2656" LastEditDate="2015-04-20T06:17:26.070" LastActivityDate="2015-04-20T06:17:26.070" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12379" PostTypeId="1" AcceptedAnswerId="12409" CreationDate="2015-04-18T06:28:05.107" Score="15" ViewCount="720" Body="&lt;p&gt;&lt;em&gt;Some time in the future, humanity discover a planet which once held an alien civilization. This civilization is long gone now, and only the empty structures remain, a testament to the achievements of their society. By working out the language system the aliens have used, researchers find several references to a temple built in the rocky wastelands, that contains the most sacred doctrines of their people. This is a story about recovering those doctrines.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ricky:&lt;/strong&gt; Chris! We have a serious problem!&lt;br&gt;&#xA;&lt;strong&gt;Chris:&lt;/strong&gt; What's up?&lt;br&gt;&#xA;&lt;strong&gt;Ricky:&lt;/strong&gt; Our map of the rocky wastelands are shredded, and the logs are ruined!&lt;br&gt;&#xA;&lt;strong&gt;Chris:&lt;/strong&gt; What map? What logs?&lt;br&gt;&#xA;&lt;strong&gt;Ricky:&lt;/strong&gt; &lt;em&gt;(sighing)&lt;/em&gt; You don't know about the map, do you? Fine, I'll explain.&lt;br&gt;&#xA;&lt;strong&gt;Ricky:&lt;/strong&gt; We have a very strict protocol for exploring the wasteland. We send out groups of men, distinguished by a color, and get them to write logs about where they go, and what landmarks they find. Then, back at base, because we know where they started, we can construct a map showing their starting position and the landmarks they found. Here, I've got an example in my pocket- &lt;em&gt;(rumbles around)&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Orange team log:&lt;/strong&gt;&lt;br&gt;&#xA;  Travelled 4km north, found a landmark.&lt;br&gt;&#xA;  Travelled 3km east, found a landmark.&lt;br&gt;&#xA;  Travelled 5km south, found a landmark.&lt;br&gt;&#xA;  Travelled 2km east, found a landmark.&lt;br&gt;&#xA;  Travelled 6km north, found a landmark.&lt;br&gt;&#xA;  Travelled 3km west, found a landmark.&lt;br&gt;&#xA;  Travelled 4km south, found a landmark.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ricky:&lt;/strong&gt; And sometimes we abbreviate this like so... &lt;em&gt;(scribbles on a piece of paper)&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Orange: N4 E3 S5 E2 N6 W3 S4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ricky:&lt;/strong&gt; This then allows us to draw a map back at base:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4xVaL.png&quot; alt=&quot;Example map&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Chris:&lt;/strong&gt; Ok, I get that. How is it relevant, though? What happened?&lt;br&gt;&#xA;&lt;strong&gt;Ricky:&lt;/strong&gt; Well, yesterday we sent out four teams to survey the area where we thought the temple might be. They all came back, we collected their logs and made a square map. It was late, and I hadn't had much sleep, so I put both the logs and the map in a folder and went to sleep. However, when I woke up - &lt;em&gt;(pauses)&lt;/em&gt; - the maps and the logs had been sabotaged.&lt;br&gt;&#xA;&lt;strong&gt;Chris:&lt;/strong&gt; It is salvageable?&lt;br&gt;&#xA;&lt;strong&gt;Ricky:&lt;/strong&gt; Well, I'm not sure. That's why I came to ask you. You see, the map had been cut in a most mysterious fashion - into twelve identical L shaped pieces. I'll show you - just wait a bit. &lt;em&gt;(dashes out, and comes back with a whole lot of paper)&lt;/em&gt;&lt;br&gt;&#xA;&lt;strong&gt;Ricky:&lt;/strong&gt; Have a look at this.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VHpY3.png&quot; alt=&quot;The puzzle&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Chris:&lt;/strong&gt; I don't see how those could fit into a square.&lt;br&gt;&#xA;&lt;strong&gt;Ricky:&lt;/strong&gt; Some of them have been rotated.&lt;br&gt;&#xA;&lt;strong&gt;Chris:&lt;/strong&gt; &lt;em&gt;(slowly)&lt;/em&gt; ...oh.&lt;br&gt;&#xA;&lt;strong&gt;Chris:&lt;/strong&gt; But surely you can just reconstruct the map using the logs?&lt;br&gt;&#xA;&lt;strong&gt;Ricky:&lt;/strong&gt; That's just it. Some of the text off the logs has been removed. I'll show you what's left.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Red: N?,W6,N5,E?,S10,W11,N?,E?,S?,E6&lt;br&gt;&#xA;  Green: W3,N9,W11,S?,W?,S7,E11,N?,W6,S11&lt;br&gt;&#xA;  Blue: S2,W?,N2,W8,S?,E4,S7,W?,S3,E?&lt;br&gt;&#xA;  Purple: E?,N3,E1,S6,E?,N7,W5,S?,W?,N4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ricky:&lt;/strong&gt; Those question marks could be anything from 1 to 17, as far as I'm concerned. I don't remember what the numbers were before they were erased!&lt;br&gt;&#xA;&lt;strong&gt;Chris:&lt;/strong&gt; It still looks possible to reconstruct it... I think.&lt;br&gt;&#xA;&lt;strong&gt;Ricky:&lt;/strong&gt; That would be good, because the green team told us that the last landmark they visited was the temple we were looking for. Oh, you should know one more thing... &lt;em&gt;(trails off)&lt;/em&gt;&lt;br&gt;&#xA;&lt;strong&gt;Chris:&lt;/strong&gt; What's that?&lt;br&gt;&#xA;&lt;strong&gt;Ricky:&lt;/strong&gt; I didn't write the compass directions on there. So I don't know which way is north.&lt;br&gt;&#xA;&lt;strong&gt;Chris:&lt;/strong&gt; ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Can you help Chris reconstruct the map?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="7308" LastActivityDate="2015-04-19T05:18:55.773" Title="Enlightened, Part I: Finding the Temple" Tags="|logical-deduction|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12380" PostTypeId="2" ParentId="12357" CreationDate="2015-04-18T08:39:54.410" Score="17" Body="&lt;p&gt;I think I've found a really easy proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every tile with vertical sides needs to have two other tiles with vertical sides adjacent to it, or the vertical boundary of the hexagon. For a given tile with vertical sides, following these adjacent tiles yields a specific path to both vertical sides of the hexagon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that every tile with vertical sides lies on a path that starts at the left side of the hexagon and ends at the right, and consists only of tiles with vertical sides. None of these paths can intersect, since that would create two different paths from a single tile with vertical sides to the left side of the hexagon, which cannot exist according to the first paragraph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since none of the paths intersect, every path between the left and right sides of the hexagon has to start and end at the same height. Therefore, every path must contain an equal number of each of the two differently oriented tiles with vertical sides. Since every tile with vertical sides lies on such a path, the total number of these two differently oriented tiles must be equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeat this symmetrically for two other orientations to find that the number of tiles of each orientation must be equal.&lt;/p&gt;&#xA;" OwnerUserId="11469" LastActivityDate="2015-04-18T08:39:54.410" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12381" PostTypeId="2" ParentId="12325" CreationDate="2015-04-18T09:37:16.493" Score="22" Body="&lt;p&gt;I wrote a program that approximates an inputted value using inputted digits by checking increasingly complex expressions. These are the best results so far:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;10: Score - 1.82589426:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\sqrt{\sqrt{12\cdot\frac{\sqrt{34}-\sqrt{5}}{6}-7}+\sqrt{89}} \approx 3.141592691$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Accurate to 8 digits.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;9: Score - 2.06300955:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\sqrt{1+\sqrt{\sqrt{2}\cdot\frac{3}{45-\sqrt{6}}\cdot789}} \approx 3.141592626$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Accurate to 8 digits.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;8: Score - 2.13531697:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\sqrt{1+\frac{\sqrt{23}-\sqrt{4\cdot56}}{78}+9} \approx 3.141592773$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Accurate to 7 digits.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;7: Score - 2.06840056:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\sqrt{123\frac{\sqrt{\sqrt{4}+56-7}}{89}} \approx 3.1415943$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Accurate to 6 digits.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;6: Score - 2.14803568:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$123\times(\sqrt{4}+\sqrt{5/67})/89 \approx 3.141585$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Accurate to 5 digits.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;5: Score - 2.28646041:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\sqrt{12}-34\cdot\frac{\sqrt{56}}{789} \approx 3.141627$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Accurate to 4 digits. (Mauris reached this first)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;4: Score - 2.25419741:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\sqrt{\frac{1234}{56+78-9}} \approx 3.141974$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Accurate to 4 digits.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;3: Score - 1.50120325:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\frac{1234}{567}-8+9 \approx 3.176367$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Accurate to 2 digits.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;2: Score - 1.27256827:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\frac{1}{2345}\cdot6789 \approx 2.895096$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Error of ~0.25. (pacoverflow reached this first)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;1: Score - 0.86466930:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\frac{12345}{6789} \approx 1.818383$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Error of ~1.3. (pacoverflow reached this first)&lt;/p&gt;&#xA;" OwnerUserId="11100" LastEditorUserId="11100" LastEditDate="2015-04-20T12:48:40.397" LastActivityDate="2015-04-20T12:48:40.397" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12382" PostTypeId="2" ParentId="12321" CreationDate="2015-04-18T10:03:42.823" Score="2" Body="&lt;p&gt;&lt;strong&gt;Proof of uniqueness:&lt;/strong&gt; this analysis is from the first, non-modified version of the puzzle.&lt;/p&gt;&#xA;&lt;p&gt;Let's denote &lt;span class=&quot;math-container&quot;&gt;$S=\{2,3,4,6,12,36,216,1296,46656\}$&lt;/span&gt; as the set of valid solutions.&lt;/p&gt;&#xA;&lt;p&gt;If someone would have been informed a number bigger than a number from this set &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt;, he declares that he knows other number immediately before the conversation would take place.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;the first said I don't know numbers means:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Since the second mathematician didn't declare anything revealing from his number, we know the first doesn't have &lt;span class=&quot;math-container&quot;&gt;$\{2,3,4,12,36,216,1296\}$&lt;/span&gt;. These numbers are to be multiplied by a bigger number than &lt;span class=&quot;math-container&quot;&gt;$6^6$&lt;/span&gt; which is not the case.&lt;/p&gt;&#xA;&lt;p&gt;Second, these numbers can be one of either the exponent, or the base, to attain the huge number result:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; leads to either &lt;span class=&quot;math-container&quot;&gt;$2*1088391168$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$46656^2$&lt;/span&gt; , the first is excluded before the first statement. So the first would know henceforth.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;the second statement: I just know numbers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If the first person has &lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt; the second would have had consequently either &lt;span class=&quot;math-container&quot;&gt;$12$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$36$&lt;/span&gt;, if he has &lt;span class=&quot;math-container&quot;&gt;$12$&lt;/span&gt;, the result is written as &lt;span class=&quot;math-container&quot;&gt;$6^{12}$&lt;/span&gt;. In the other case it is &lt;span class=&quot;math-container&quot;&gt;$36^6$&lt;/span&gt;, both cases the second mathematician should be aware of first number.&lt;/p&gt;&#xA;&lt;p&gt;The other choice is &lt;span class=&quot;math-container&quot;&gt;$46656$&lt;/span&gt;, as &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; is excluded before the conversations starts. Supposing the second mathematician is aware of this, the result can't be written as &lt;span class=&quot;math-container&quot;&gt;$46656^2$&lt;/span&gt;. The unique case left is &lt;span class=&quot;math-container&quot;&gt;$46656*46656$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Set of valid solutions is &lt;span class=&quot;math-container&quot;&gt;${12,36,46656}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;the third statement: Me too I know them&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This doesn't fit the number &lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt; because it leads to two options of &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; either &lt;span class=&quot;math-container&quot;&gt;$12$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$36$&lt;/span&gt;. He said &lt;strong&gt;I know&lt;/strong&gt;, so that leave us with the last unique choice: &lt;span class=&quot;math-container&quot;&gt;$(x,y)=(46656,46656)$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="9517" LastEditorUserId="69582" LastEditDate="2021-06-01T04:45:51.497" LastActivityDate="2021-06-01T04:45:51.497" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="12384" PostTypeId="2" ParentId="12357" CreationDate="2015-04-18T11:31:16.323" Score="5" Body="&lt;p&gt;Yet another; this one is triangle-based and might be more of a standard proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide the entire hexagon into triangles and assign numbers to the vertical lines like this (or similarly):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JcA1R.png&quot; alt=&quot;numbers&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for any triangle-based shape (which does not necessarily have to be a tiling) define its &quot;degree&quot; as the number obtained by adding all of the numbers assigned to its left boundary and subtracting all of the numbers assigned to its right boundary. For example, the shape&#xA;&lt;img src=&quot;https://i.stack.imgur.com/geaWO.png&quot; alt=&quot;shape&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;has a &quot;degree&quot; of $(1-2)-(2+2-1-2)=-2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, build a tiling piece by piece, and consider the &quot;degree&quot; of the resulting shape. Adding a horizontal tile does not change the degree, adding one of the others increases or decreases it by 1, respectively:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dGZ3K.png&quot; alt=&quot;-1&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/emHh9.png&quot; alt=&quot;+1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the entire hexagon has a degree of 0, the number of the two tiles shown must be equal. Repeat symmetrically in another direction.&lt;/p&gt;&#xA;" OwnerUserId="11469" LastActivityDate="2015-04-18T11:31:16.323" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12385" PostTypeId="2" ParentId="12110" CreationDate="2015-04-18T11:31:58.307" Score="3" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Doctor Who.&lt;p&gt;&#xA;&#xA; No, really, go with me on this.  More specifically, you are the fourth incarnation, as portrayed by the inestimable &lt;a href=&quot;http://en.wikipedia.org/wiki/Tom_Baker&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tom Baker&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I fly high with various colors on me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Doctor Who is well known for flying in his T.A.R.D.I.S in all his incarnations, but the fourth Doctor in particular is well-known for the multi-coloured scarf he wears.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But most of the time I sit in a corner and not fly.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Doctor actually spends relatively little of his screen time in the T.A.R.D.I.S., as his adventures are typically on the ground.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A sling of my hair is always long enough to touch the ground.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Fourth Doctor was well known for his long, curly hair.  If one was to weave a sling from his hair (if one had a reason to wish to do so.  If the Daleks were to, say, build a giant mechanised robot to threaten the world and the Doctor needed something with which to fling a rock at it), then one could be quite certain that such a sling would easily be long enough to touch the ground.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At times I have a tail.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tom Baker, the actor who portrayed the Fourth Doctor, also was an actor in The Winter's Tale (1968), The Canterbury Tales (1972), The Tales of Aesop (1990), and another production of The Canterbury Tales (2000).  He has had, in fact, several tales.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I hate rain. And I usually don't come out during a rainy day.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/iVoun.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;p&gt;&#xA; &#xA; (See?  Not a good look.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9286" LastActivityDate="2015-04-18T11:31:58.307" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12386" PostTypeId="1" CreationDate="2015-04-18T11:49:23.803" Score="8" ViewCount="1992" Body="&lt;p&gt;Imagine you have a m x n grid which is initially colored white. you can fill in a cell with black color if and only if there are no immediately neighboring black cells (no black cells to the left/right/top/bottom). If you keep on filling cells you will eventually run out of legal cells to fill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example configuration is shown below. The green patterned ones are the only legally available spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/c8H3i.jpg&quot; alt=&quot;Example&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the minimum number of black cells you can fill in the grid till you have no more cells you can legally fill? What is the best strategy to get this minimum? &lt;/p&gt;&#xA;" OwnerUserId="11592" LastActivityDate="2022-04-23T11:54:37.873" Title="Minimum cells to fill grid without consecutive neighbours" Tags="|geometry|optimization|algorithm|" AnswerCount="5" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12388" PostTypeId="2" ParentId="12386" CreationDate="2015-04-18T12:41:28.097" Score="0" Body="&lt;p&gt;One fill for larger boards is a pattern like the movement of a knight - I'm on mobile but I'll explain it as well as I can. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;..X.&#xA;X...&#xA;...X&#xA;.X..&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This patter repeats in a 4x4 square - for larger boards, it is nearly optimal. &lt;/p&gt;&#xA;" OwnerUserId="1664" LastActivityDate="2015-04-18T12:41:28.097" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12389" PostTypeId="2" ParentId="10968" CreationDate="2015-04-18T13:28:56.183" Score="2" Body="&lt;p&gt;Try #3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the 3-digit code is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;216&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I made a dial, based on the layout of a clock, that assigns a value to each of the 16 directions away from a cell.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9YnuG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After the last edit to the question, the shortest path (based on summing the values) from the upper left cell to the lower right cell is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5i04U.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastActivityDate="2015-04-18T13:28:56.183" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12390" PostTypeId="2" ParentId="12357" CreationDate="2015-04-18T13:51:55.310" Score="17" Body="&lt;p&gt;I want to post an answer that is &lt;strong&gt;more intuitive than mathematical&lt;/strong&gt;.&lt;br /&gt;&#xA;This picture perfectly represents it:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/khQiJ.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;White, grey and black are used to highlight the diamonds with the same orientation.&#xA;The right picture shows a weird solid, I guess everyone can see it.&lt;br /&gt;&#xA;Well, it's intuitive to see that, for any configuration, the black area is equivalent (white and grey as well): it's like extruding parts of your floor&#xA;(aka building stairs!), the area on which you can walk doesn't change!&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-18T13:51:55.310" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12392" PostTypeId="1" AcceptedAnswerId="12498" CreationDate="2015-04-18T16:55:22.810" Score="-2" ViewCount="484" Body="&lt;h1&gt;Beware: solving this riddle requires being technically fluent in both digital platforms described below (using them both is essential).&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Even though &lt;a href=&quot;http://getbootstrap.com/&quot; rel=&quot;nofollow&quot;&gt;Bootstrap&lt;/a&gt; doesn't propagate itself as being narrow-minded; when integrated into &lt;a href=&quot;https://www.drupal.org/&quot; rel=&quot;nofollow&quot;&gt;Drupal&lt;/a&gt;, ... it shows its true colors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Typo&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ... it shows its three colors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Typo&lt;/strong&gt; (Again, sorry; I'm starting to use AZERTY, about which you can find out more on Wikipedia. It's ok to feel blue for me, since not many people use this.):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ... it shows its tricolor.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Node: a central or connecting point.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A nice screen-grabber might facilitate an explanatory visual answer of minimum 200 pixels wide &amp;amp; 35 pixels high.&lt;/p&gt;&#xA;" OwnerUserId="11597" LastEditorUserId="11597" LastEditDate="2015-04-19T06:48:19.593" LastActivityDate="2015-04-20T20:53:41.000" Title="World-wide web or perhaps a bit more narrow-minded?" Tags="|riddle|visual|computer-puzzle|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="12393" PostTypeId="2" ParentId="12324" CreationDate="2015-04-18T17:13:36.240" Score="2" Body="&lt;p&gt;The monkey agrees to give the crocodile a fish after each leg of the trip. After the 4th one way trip he is successfully back on the shore with all his bananas. Now he just needs to run before the croc tries to get him to pay up!&lt;/p&gt;&#xA;" OwnerUserId="9676" LastActivityDate="2015-04-18T17:13:36.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12394" PostTypeId="2" ParentId="11913" CreationDate="2015-04-18T17:16:46.277" Score="3" Body="&lt;p&gt;I think that it is the Arctic desert, but I still trying to come up with my proof. There are few if any known sands that do not contain silica, and I am assuming that the image is a representation of the 'grains of sand' that we are looking at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I first tried thinking of it as a word puzzle, but I actually think we are looking at atomic numbers and chemical composition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Key points:&#xA;Number 9 at row 1, col 3 is only dark coloured one (1 / 95)&lt;br&gt;&#xA;8/95 dark coloured&lt;br&gt;&#xA;6/95 light coloured&#xA;81/95 medium coloured&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Num #   %   Atomic Symbol?&#xA;1   21  22% H&#xA;2   10  11% He&#xA;3   18  19% Li&#xA;4   5   5%  Be&#xA;5   5   5%  B&#xA;6   12  13% C&#xA;7   5   5%  N&#xA;8   8   8%  O&#xA;9   6   6%  F&#xA;1   0   0   0%  Ne&#xA;11  1   1%  Na&#xA;12  4   4%  Mg&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;That's a lot of Carbon, Hydrogen, Lithium and Oxygen. So I'm going with the Arctic, where the grains of sand would actually be snow. Can you please advise if this approach is remotely thinking in the right way? I can't quite sort out the font and background colour differences and they feel like either they are really important or just distracting.&lt;/p&gt;&#xA;" OwnerUserId="11598" LastActivityDate="2015-04-18T17:16:46.277" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12395" PostTypeId="2" ParentId="12276" CreationDate="2015-04-18T17:33:02.717" Score="1" Body="&lt;p&gt;The cryptogram decodes to:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;danbrown&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;which results from the cipher text of &lt;strong&gt;vcejssnr&lt;/strong&gt; and a key (one time pad) of &lt;strong&gt;scribere&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, the answer to the riddle could be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;the number 3&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;In Dan Brown's novel, the &lt;strong&gt;Digital Fortress&lt;/strong&gt; is the name of a new encryption algorithm that cannot be broken by the NSA. This relates to the &lt;strong&gt;one time pad&lt;/strong&gt; encryption method that, also, cannot be broken when properly used.&lt;br&gt;&#xA;&lt;br&gt;&#xA;Another link is the quote from the previous question &quot;Quis custodiet ipsos custodes?&quot; which was used in this novel.  As described in the previous question, this sentence was written by the Roman poet, Juvenal, in his &lt;strong&gt;Satires&lt;/strong&gt; collection of poems (which also relates to the title of this question).&lt;br&gt;&#xA;&lt;br&gt;&#xA;Assuming that &quot;cannot be described in words&quot; means that &quot;the answer is a number&quot;, then the number &lt;strong&gt;3&lt;/strong&gt; is most significant in this novel. The kill code for the virus was 3.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-04-25T18:29:50.570" LastActivityDate="2015-04-25T18:29:50.570" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12396" PostTypeId="1" CreationDate="2015-04-18T20:09:18.650" Score="30" ViewCount="100176" Body="&lt;p&gt;You are in a prison. There are 3 doors, and a ghost standing in the middle of the prison. He says one of the doors is the way to freedom while the other two lead to death. You get to ask only one question. He will only answer with YES or NO. All he says is truth. What do you ask to find the door to freedom?&lt;/p&gt;&#xA;" OwnerUserId="11600" LastEditorUserId="5373" LastEditDate="2017-03-04T13:47:47.330" LastActivityDate="2017-03-04T13:47:47.330" Title="Three Doors, One Truth-Teller, One Question" Tags="|logical-deduction|liars|" AnswerCount="13" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="12397" PostTypeId="2" ParentId="12396" CreationDate="2015-04-18T20:22:50.440" Score="10" Body="&lt;p&gt;my question would be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Among the propositions 1. &quot;You are a liar&quot;, 2. &quot;You will reply negatively&quot; and 3. &quot;This door leads to freedom&quot;, is there an odd number of true propositions?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I stole this answer (without shame) from: &lt;a href=&quot;https://puzzling.stackexchange.com/a/2310/9835&quot;&gt;https://puzzling.stackexchange.com/a/2310/9835&lt;/a&gt; It is for an other version of your question where there are 2 guards (one truth full and one lying). But can perfectly be used here&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 2 possibilities here:&lt;br&gt;&#xA;The number of truths is odd, 1. is false, 2. is false, so 3. must be true.&lt;br&gt;&#xA;The number of truths is even, 1. is false, 2. is true, so 3. must be true.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way you created a paradox if the door you are pointing to is not the one to freedom, so what ever happens the door you are pointing to leads to freedom :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All credits to Florian F.&lt;/p&gt;&#xA;" OwnerUserId="9835" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-04-18T20:22:50.440" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12398" PostTypeId="2" ParentId="12386" CreationDate="2015-04-18T20:40:27.520" Score="1" Body="&lt;p&gt;Here is how we can get &lt;span class=&quot;math-container&quot;&gt;$\approx (mn+2m+2n)/5$&lt;/span&gt; tiles. Place the numbers &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; through &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt; in each square as shown below:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{array}{ccccccc}&#xA;1&amp;amp;4&amp;amp;2&amp;amp;5&amp;amp;3&amp;amp;1&amp;amp;\cdots\\&#xA;2&amp;amp;5&amp;amp;3&amp;amp;1&amp;amp;4&amp;amp;2&amp;amp;\\&#xA;3&amp;amp;1&amp;amp;4&amp;amp;2&amp;amp;5&amp;amp;3&amp;amp;\\&#xA;4&amp;amp;2&amp;amp;5&amp;amp;3&amp;amp;1&amp;amp;4&amp;amp;\\&#xA;5&amp;amp;3&amp;amp;1&amp;amp;4&amp;amp;2&amp;amp;5&amp;amp;\cdots\\&#xA;\vdots&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;\vdots&amp;amp;\ddots&#xA;\end{array}&#xA;$$&lt;/span&gt;&#xA;In words: the columns read &lt;span class=&quot;math-container&quot;&gt;$1,2,3,4,5$&lt;/span&gt; periodically as you go down, and the pattern shifts down by 2 as you move right.&lt;/p&gt;&#xA;&lt;p&gt;Now, color all &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; tiles black. Then no more black tiles can be placed in the interior, because all numbers are neighbored by a &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;. The only place this breaks down is at the border: normally, a &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; is covered by the &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; tile above it, but &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt;'s on the top border aren't covered. Same thing happens for the right, bottom and left border, with the numbers &lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt; respectively. Placing a black tile on each of these bad cases means no more can be placed.&lt;/p&gt;&#xA;&lt;p&gt;Since there are about &lt;span class=&quot;math-container&quot;&gt;$\frac{mn}{5}$&lt;/span&gt; ones, and about &lt;span class=&quot;math-container&quot;&gt;$\frac{m}{5}$&lt;/span&gt; bad numbers on the left and right borders, and about &lt;span class=&quot;math-container&quot;&gt;$\frac{n}{5}$&lt;/span&gt; on the top and bottom, my formula is as claimed.&lt;/p&gt;&#xA;&lt;p&gt;This could be optimized further by initially coloring all of a different number black, instead of one, in order to minimize the bad cases.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="-1" LastEditDate="2020-09-05T14:43:14.327" LastActivityDate="2020-09-05T14:43:14.327" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="12399" PostTypeId="2" ParentId="10968" CreationDate="2015-04-18T21:12:38.800" Score="4" Body="&lt;p&gt;Try #4&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;(EDITED to reflect the actual answer, although I went with 342 at first)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Could the 3-digit code be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;345&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because the path resembles those numbers (see below)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SfHCd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-04-19T14:35:49.740" LastActivityDate="2015-04-19T14:35:49.740" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="12402" PostTypeId="2" ParentId="12396" CreationDate="2015-04-18T23:14:52.540" Score="2" Body="&lt;p&gt;I don't think that's possible. There are three possibilities and you're asking a question with a boolean answer.&lt;/p&gt;&#xA;" OwnerUserId="11606" LastActivityDate="2015-04-18T23:14:52.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12403" PostTypeId="2" ParentId="12396" CreationDate="2015-04-18T23:54:44.433" Score="-1" Body="&lt;p&gt;my question will be ( assuming that 2nd door is silent )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the 1st door will give me more chance to live than the 2nd door?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;yes = 1st door&#xA;no = 3rd door&#xA;no answer = 2nd door&lt;/p&gt;&#xA;" OwnerUserId="11608" LastActivityDate="2015-04-18T23:54:44.433" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12404" PostTypeId="2" ParentId="12396" CreationDate="2015-04-19T00:02:47.283" Score="14" Body="&lt;p&gt;My first thought: It is a trick question. Pay attention to your environment. If there are only 3 doors, one of them must be the one through which you entered. And since you are in prison, you know that this door does &lt;strong&gt;not&lt;/strong&gt; lead to freedom, you can just ask the ghost &quot;Is this the door leading to freedom?&quot; while pointing on one of the other two doors in the room. If he says &quot;Yes&quot; you pick that door, if he says &quot;No&quot; you pick the other one. It is kind of an prove through exclusion attempt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is not a trick question, and you actually entered the room which has 3 doors on top of the one through which you entered, the best you can do is try to act based on the &lt;em&gt;Mounty-Hall-Problem&lt;/em&gt;. You might only be able to increase your chances to a chance of 5/6 chance, but not to a 100%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You cannot get 3 informations out of 1 bit. You can combine 2 infos into 1 bit, but not 3. (Bit operations all have only 2 input and 1 output channels).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could ask the ghost for example &quot;If I picked the left door and somebody would show me one of the remaining doors leading to death, &lt;strong&gt;could&lt;/strong&gt; he show me the middle door?&quot;&#xA;If the answer would be &quot;Yes&quot;, you could switch to the right door, as it would have a 2/3 chance to be the freedom door. (See Mounty-Hall-Problem)&#xA;If the answer would be &quot;No&quot;, you definitly know the mid door is the freedom door.&#xA;A 100% chance and a 2/3 chance combine to (6/6+4/6)/2, which is 5/6.&lt;/p&gt;&#xA;" OwnerUserId="11607" LastActivityDate="2015-04-19T00:02:47.283" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12405" PostTypeId="1" AcceptedAnswerId="12407" CreationDate="2015-04-19T00:31:43.293" Score="15" ViewCount="930" Body="&lt;p&gt;You have been taken prisoner and your captor has set a fiendish problem with which to torture you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are stood between two terrible fates. Two steps in front of you is a nest of angry vipers. Two steps behind you is sheer cliff with a drop to certain death. Taking one step forward or back is safe but taking two steps in either direction will be fatal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your villainous captor has promised to let you go free if you can write a safe pattern of 12 steps. Unfortunately he will choose whether you follow all the step or every 2nd, 3rd, 4th, 5th or 6th step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you find a pattern that will ensure your release?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rules:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Each step must be forward or back you can not choose to stay still or move in any other direction.&lt;/li&gt;&#xA;&lt;li&gt;The sequence must work for all of the following conditions&#xA;&lt;ul&gt;&#xA;&lt;li&gt;all steps 1,2,3,4,5,6,7,8,9,10,11,12 are followed&lt;/li&gt;&#xA;&lt;li&gt;only multiples of 2  - steps 2,4,6,8,10,12&lt;/li&gt;&#xA;&lt;li&gt;only multiples of 3  - steps 3,6,9,12 &lt;/li&gt;&#xA;&lt;li&gt;only multiples of 4  - steps 4,8,12&lt;/li&gt;&#xA;&lt;li&gt;only multiples of 5  - steps 5 and 10&lt;/li&gt;&#xA;&lt;li&gt;only multiples of 6  - steps 6 and 12&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The origin of the puzzle is this video &lt;a href=&quot;https://www.youtube.com/watch?v=pFHsrCNtJu4&quot; rel=&quot;noreferrer&quot;&gt;https://www.youtube.com/watch?v=pFHsrCNtJu4&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Spoiler warning : the video contains the solution)&lt;/p&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="11297" LastEditDate="2015-04-19T10:27:14.583" LastActivityDate="2015-04-19T10:40:07.603" Title="Moving back and forth between two terrible fates" Tags="|mathematics|logical-deduction|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12406" PostTypeId="1" AcceptedAnswerId="12421" CreationDate="2015-04-19T00:59:31.813" Score="7" ViewCount="1606" Body="&lt;p&gt;I am inside you, but never others,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everybody always uses me every second, more than once,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you want to ask a question, I am something good to bring up,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you change me even a little, I become an adult, and if you do it again, I become a kid,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a spy for both groups, but one group is called more often, despite having less members,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I am lucky enough to be on a flag, I will be red and yellow, but in real life, filled with color, I am blue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What Am I?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(The following is NOT part of the riddle:) Please explain your reasoning for each clue and conceal our answer/reasoning behind a spoiler.&lt;/p&gt;&#xA;" OwnerUserId="11002" LastActivityDate="2015-04-19T10:57:28.100" Title="I Am Inside You, But Never Others, And I'm A Spy For Both Groups. What Am I?" Tags="|riddle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12407" PostTypeId="2" ParentId="12405" CreationDate="2015-04-19T01:08:19.147" Score="12" Body="&lt;p&gt;Answer: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No, it's not possible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your first and second moves must be in opposite directions, or you die. Then, you're back to the start position, so your third and fourth steps must be in opposite directions, and so on. So, these pairs of moves must be opposites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(1,2), (3,4), (5,6), (7,8), (9,10), (11,12)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same must be true for the sequence of every second step, or every third step, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(2,4), (6,8), (10,12)$$&#xA;$$(3,6), (9,12)$$&#xA;$$(4,8)$$&#xA;$$(5,10)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then steps $9$, $10$, and $12$ must all be opposite to each other, a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could make $11$ steps though, since those edges form a bipartite graph. The solution is unique up to reflection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$RLLRLRRLLRL$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if the restriction was weakened to step sizes being powers of 2, we could survive for arbitrarily many steps using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Thue%E2%80%93Morse_sequence&quot; rel=&quot;nofollow&quot;&gt;Thue-Morse sequence&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="10281" LastEditDate="2015-04-19T10:34:26.717" LastActivityDate="2015-04-19T10:34:26.717" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12408" PostTypeId="2" ParentId="12405" CreationDate="2015-04-19T01:09:23.880" Score="4" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are doomed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In any successful list,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Instructions 5 and 6 must differ. Consider what happens when you follow all instructions. Steps 1 and 2 must bring you back to the starting square (else you would die). Same for 3 and 4. Thus, 5 and 6 must differ, else you will take 2 steps in the same direction from the starting square.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Instructions 6 and 12 must differ, to prevent death in the multiples of 6 case.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Instructions 12 and 9 must differ, for the multiples of 3 case. Instructions 3 and 6 either kill you, or bring you back to the start, so 9 and 12 must differ.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Instructions 9 and 10 must differ, by identical reasoning to the first bullet point.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Instruction 10 and 5 must differ, else you die when you follow multiples of 5.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, 5 differs from 6 differs from 12 differs from 9 differs from 10 differs from 5, implying step 5 differs from step 5, a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="11297" LastEditDate="2015-04-19T10:40:07.603" LastActivityDate="2015-04-19T10:40:07.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12409" PostTypeId="2" ParentId="12379" CreationDate="2015-04-19T05:08:27.303" Score="8" Body="&lt;p&gt;The solution is below:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/PlKFE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first tile to place is the one with the blue flag.  It must be oriented like an &quot;r&quot; so that it can go south 2 and hit a landmark.  It must be in the upper right as it manages to go extreme distances east and south without the possibility of going extreme distances west or north in between.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simularly one can determine that the tile with the green flag must be on the far right and somewhat down.  It must be oriented and positioned where it is or else the additional tiles around it cannot be placed to make the square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next major step was to find a pairing of tiles which could allow for landmarks in one of the 8th through 12th columns from left and the 3rd and 5th rows from the top.  This was mostly done by process of elimination based on the green and blue paths.  Which ever tile bordered the top could not have any landmarks in the 1st row.  It could not be positioned as an &quot;L&quot; or there would be space a tile could not fill.  It could not be positioned as a &quot;r&quot; or backwards &quot;L&quot; as no tiles fit the bill and did not lead to later complications.  When the 1 tile shaped as a backwards &quot;r&quot; with an empty top row was tried, the top left tile was identified.  From there, the final tile shapes were pieced together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I figured out the final tile arrangement, I assumed the blue path touched the landmark in column 5, row 8 as that was the only way I could see for it to go down 10 more squares after hitting a landmark 4 to the east.  The tile with the red flag was the only tile where a landmark was 7 south of the previous blue landmark.  This meant I had placed all 4 starting positions.  The remaining tiles were added in as the paths were drawn, I can elaborate further if need be but the remaining placements were trivial.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditDate="2015-04-19T05:18:55.773" LastActivityDate="2015-04-19T05:18:55.773" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12411" PostTypeId="1" AcceptedAnswerId="12418" CreationDate="2015-04-19T08:02:00.823" Score="13" ViewCount="1820" Body="&lt;p&gt;The Chaos Legion is an infamous criminal organization, responsible for every major disaster in history from the burning of the Library of Alexandria to the recent Sony hack. The Order of Seven is their antithesis, and have been investigating the Chaos Legion for hundreds of years with no avail. They want to determine the leader of the Chaos legion, but so far, only know that (s)he is one of 256 people. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fortunately, the Order now have an opportunity to send two undercover agents into Chaos's upper ranks. The lucky recruits are named James and Jimmy. The plan is this: James will go very deep undercover, and will be able to determine the leader. However, he will be so entrenched that he can no longer safely communicate with the Order. That's where Jimmy comes in: Jimmy will go undercover, but not as deeply, so that he can still communicate with the Order. Jimmy's cover will afford him some information: he will be able to narrow down the list of possible leaders to only 2 people. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;James needs to communicate the leader's identity to Jimmy. The problem is, the only way they will be able to safely communicate with each other is via a very crude text messaging service. Each message consists of a string of $0$'s and $1$'s, and the cost of a message is $\$1$ million per symbol! Furthermore, the time it takes to send a message is wildly unpredictable, so they can't use the timing of messages to convey extra information. An empty text can't be sent either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They will have to agree on what these $0$'s and $1$'s mean before starting the mission. There is a way to get away with spending $\$8$ million: James sends Jimmy, in binary, the position of the leader on an alphabetical list of the 256 original suspects. This doesn't make any use of Jimmy's knowledge, so perhaps they can do better. In fact, they better be able to: the Order only has a budget of $\$4$ million!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Both James and Jimmy know the leader is one of $256$ people.&lt;/li&gt;&#xA;&lt;li&gt;Some time into the operation, James will know the leader, while Jimmy will know a set of two people, one of whom is the leader. James will not know Jimmy's list.&lt;/li&gt;&#xA;&lt;li&gt;James needs to tell Jimmy the leader.&lt;/li&gt;&#xA;&lt;li&gt;Each of James and Jimmy can text the other, but sending each bit costs $\$1$ million. Once Jimmy knows the leader, he can tell the Order this for free.&lt;/li&gt;&#xA;&lt;li&gt;They can only spend $\$4$ million.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Can the Order succeed? Or will Chaos prevail?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Source:&lt;/strong&gt; &lt;a href=&quot;http://math.stanford.edu/~notzeb/puzzle.html&quot;&gt;http://math.stanford.edu/~notzeb/puzzle.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="2627" LastEditDate="2015-04-19T15:25:11.890" LastActivityDate="2015-04-19T15:25:11.890" Title="A Game of Phones" Tags="|mathematics|strategy|cryptography|information-theory|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12412" PostTypeId="2" ParentId="12411" CreationDate="2015-04-19T08:13:52.840" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes, they can do it in 4 bits.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Naturally, James and Jimmy agree beforehand to represent each potential leader as a vector in $(0,1)^8$. Jimmy looks at his two suspects, and notes an index where their vectors differ. He conveys this index using James in 3 bits. James responds with the leader's vector entry at that index, for 1 bit. This lets Jimmy determine which of the two suspects is leader.&lt;br&gt;&#xA;&lt;br&gt;&#xA; In general, if there's $2^{2^n}$ members, then this can be done in $n+1$ bits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-04-19T08:18:38.433" LastActivityDate="2015-04-19T08:18:38.433" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12413" PostTypeId="2" ParentId="12325" CreationDate="2015-04-19T08:38:22.717" Score="3" Body="&lt;p&gt;Not sure if decimal are allowed, but here's my attempt&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 operator, score &lt;strong&gt;1.1765956249508713525610395676804...&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$$123.4/56.789 \approx 2.17295603 \approx \pi - 0.96863662$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3 operator, score &lt;strong&gt;2.06714352383430453303221463889...&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$$(1234/56)/(\lfloor7.89\rfloor) \approx 3.14795918 \approx \pi + 0.00636653 $$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5453" LastEditorUserId="30903" LastEditDate="2018-02-11T01:00:57.953" LastActivityDate="2018-02-11T01:00:57.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12414" PostTypeId="2" ParentId="12396" CreationDate="2015-04-19T09:01:32.150" Score="15" Body="&lt;p&gt;Keeping in mind that the ghost will only answer if the answer is yes or no, a way to find the door to freedom is to formulate a question such that the answer would be yes, no, or anything else respectively for each of the three doors. In the latter case, the ghost will not answer at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several possibilities, one of them is this one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the door to freedom is not the third one, then is it the first one? Otherwise, what was the name of Napoleon Bonaparte?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the door to freedom is the first one, the ghost will answer &lt;em&gt;yes&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;li&gt;If the door to freedom is the second one, the ghost will answer &lt;em&gt;no&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;li&gt;If the door to freedom is the third one, the ghost will not answer.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4394" LastEditorUserId="4394" LastEditDate="2015-04-19T14:11:50.897" LastActivityDate="2015-04-19T14:11:50.897" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12415" PostTypeId="2" ParentId="12406" CreationDate="2015-04-19T09:19:05.427" Score="3" Body="&lt;p&gt;My guess would be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A heart?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation line by line:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am inside you, but never others,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your heart has to be inside you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Everybody always uses me every second, more than once,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your heart beats more then 60 times a minute so you you use it more then once a second.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you change me even a little, I become an adult, and if you do it again, I become a kid,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people have to get a new heart, this can be from a younger person.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;No idea for the rest of the lines yet.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="4786" LastEditorUserId="4786" LastEditDate="2015-04-19T09:52:29.673" LastActivityDate="2015-04-19T09:52:29.673" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12416" PostTypeId="1" CreationDate="2015-04-19T09:37:54.337" Score="3" ViewCount="487" Body="&lt;p&gt;I convey the invisible,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I convey what can be felt,&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But not seen,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inside my heart,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I am not as nice as I could seem to be,&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is because I deliver the invisible,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the speed of a mere blink,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am very powerful,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And can bring comfort,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Across my horizons, I am free,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But still vulnerable to your opinion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;What am I?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="10704" LastActivityDate="2018-07-11T15:51:20.337" Title="I convey the invisible" Tags="|riddle|" AnswerCount="8" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12417" PostTypeId="2" ParentId="1729" CreationDate="2015-04-19T09:38:15.260" Score="0" Body="&lt;p&gt;There are many possibilities. The first that cam to mind was to preserve the first and last characters and pick random values for the rest but that seemed a bit silly. So here is something a little more sensible.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;function conv(A,a,b){&#xA;    for(var B='', i=0; i&amp;lt;A.length; i++){&#xA;        var d=a.charCodeAt(i%a.length)-b.charCodeAt(i%b.length);&#xA;        B+=String.fromCharCode( d+A.charCodeAt(i));&#xA;    }&#xA;    return B;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The function take a pattern to be converted $A$ and two sample patterns $a$ and $b$. &#xA;It calculates the difference in ASCII values of pairs of sample characters and apply it to characters of the pattern to be converted.&#xA;In cases such as the example where the pattern to convert is longer than the sample it returns to beginning of the sample for more characters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result looks a bit of a mess and highlights the problems that may occur with this method.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;LRGGb5Z\EWIL&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="11297" LastActivityDate="2015-04-19T09:38:15.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12418" PostTypeId="2" ParentId="12411" CreationDate="2015-04-19T09:54:51.040" Score="11" Body="&lt;p&gt;The best you can do is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $3 million.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The proof is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Number the suspects in binary. Suppose the binary descriptions of Jimmy's suspects differ in position $i$, where we start counting from $0$. If $i$ is between $0$ and $3$, Jimmy sends a message with two bits describing $i$, and James replies with the value of bit $i$. If $i$ is $4$ or $5$, Jimmy sends a message consisting of just the bit $0$, and James replies with the values of bits $4$ and $5$. If $i$ is $6$ or $7$, Jimmy sends a message consisting of just the bit $1$, and James replies with the values of bits $6$ and $7$. (That you can't do it with $2 million is sort of obvious.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11295" LastActivityDate="2015-04-19T09:54:51.040" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12419" PostTypeId="2" ParentId="12416" CreationDate="2015-04-19T10:07:02.873" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Emoticons like frowny &lt;strong&gt;:-(&lt;/strong&gt; &amp;amp; smiley &lt;strong&gt;:-)&lt;/strong&gt; &amp;amp; frown-winky &lt;strong&gt;;-(&lt;/strong&gt; &amp;amp; smile-winky &lt;strong&gt;;-)&lt;/strong&gt; &amp;amp; etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reference :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reference : &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_emoticons&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/List_of_emoticons&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We use these to convey the invisible feelings, to bring comfortable feelings or bad feelings, though the reader might interpret these with own opinions&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10940" LastEditorUserId="10940" LastEditDate="2015-04-19T10:14:08.547" LastActivityDate="2015-04-19T10:14:08.547" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12420" PostTypeId="2" ParentId="12416" CreationDate="2015-04-19T10:50:21.693" Score="4" Body="&lt;p&gt;It could be a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Smile&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I convey the invisible,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Emotions are invisible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I convey what can be felt,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Emotions, indeed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But not seen,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, emotions are invisible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Inside my heart,&#xA;I am not as nice as I could seem to be,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sometimes smiles are fake.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is because I deliver the invisible,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At the speed of a mere blink,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Muscles movements are fast as for smiles, as for blinks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am very powerful,&#xA;And can bring comfort,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A smile brings happiness, usually.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Across my horizons, I am free,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What's the price of a smile?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But still vulnerable to your opinion.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's not easy to tell a true smile from a fake one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-04-19T10:50:21.693" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12421" PostTypeId="2" ParentId="12406" CreationDate="2015-04-19T10:57:28.100" Score="6" Body="&lt;p&gt;My guess is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter Y.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am inside you, but never others&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The word &quot;you&quot; has a Y, but not the word &quot;others&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Everybody always uses me every second, more than once&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The word &quot;everybody&quot; has 2 Y's.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you want to ask a question, I am something good to bring up&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You often need to ask &quot;why&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you change me even a little, I become an adult, and if you do it again, I become a kid&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you change the bottom of a 'y' to be straight, it becomes a capital Y (an adult).  If you move it back, it becomes a lowercase y (a kid).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a spy for both groups, but one group is called more often, despite having less members&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter Y is considered to be both a consonant and a vowel.  Vowels are used more often than consonants despite there being fewer of them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I am lucky enough to be on a flag, I will be red and yellow, but in real life, filled with color, I am blue&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;a href=&quot;http://en.wikipedia.org/wiki/Flag_of_Vanuatu&quot;&gt;flag of Vanuatu&lt;/a&gt; has a yellow Y with a red stripe.  I don't know why a Y would be blue in real life though, unless it is referring to &quot;sky&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4479" LastActivityDate="2015-04-19T10:57:28.100" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12422" PostTypeId="1" AcceptedAnswerId="12423" CreationDate="2015-04-19T11:06:32.230" Score="114" ViewCount="57359" Body="&lt;p&gt;Three logicians walk into a bar. The barman says, 'Does everybody want a drink?'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first logician says, 'I don't know.'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second logician says, 'I don't know.'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does the third logician say?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Please provide a clear explanation of why each of the logicians reply in the way they do.&lt;/p&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="11297" LastEditDate="2015-04-21T15:22:14.480" LastActivityDate="2015-05-19T18:30:05.063" Title="3 Logicians Walk Into a Bar" Tags="|logical-deduction|humor|" AnswerCount="2" CommentCount="20" ContentLicense="CC BY-SA 3.0" />
  <row Id="12423" PostTypeId="2" ParentId="12422" CreationDate="2015-04-19T11:09:34.713" Score="125" Body="&lt;p&gt;&lt;em&gt;Question:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does &lt;strong&gt;everybody&lt;/strong&gt; want a drink?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The third logician answers:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;&lt;strong&gt;YES&lt;/strong&gt;&quot; if he wants a drink, &quot;&lt;strong&gt;NO&lt;/strong&gt;&quot; if doesn't want it. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Reason:&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first says &quot;I don't know&quot; because he wants a drink, but doesn't know if  everybody wants one. If the first didn't want a drink, he would have answered &quot;No&quot;.&lt;br&gt;&#xA; Same for the second, he wants a drink but doesn't know if the third wants one.&#xA; So, the third answers &quot;Yes&quot; if he wants a drink, &quot;No&quot; if he doesn't.&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-04-30T17:25:16.717" LastActivityDate="2015-04-30T17:25:16.717" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12424" PostTypeId="1" AcceptedAnswerId="12426" CreationDate="2015-04-19T12:30:32.707" Score="4" ViewCount="835" Body="&lt;h1&gt;Where am I?&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;If I'm stuck in the middle, but can look around as much as I'd like; I can see 120, 24 times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These 24 are actually arranged into groups of 3 across 8 complete imaginary ●'s of view.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;However, as soon as I get cornered; whatever corner that may be; I can see straight again. In any such case, 3 of the 120's are no longer visible to me, since I am standing in them; resulting in only 3 120's which I can still count.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These 3 are again the parts of 1 full ● of view.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Suggested reading&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;An entry in the &lt;strong&gt;Official Dictionary of Dubiously-used HTML Characters&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;●&quot;, origin: abstract.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A ● is the solid version of an &lt;strong&gt;o&lt;/strong&gt;, and is a big version of an •. The Ancient Greeks considered the ● to be more perfect than a &lt;strong&gt;0&lt;/strong&gt; or a □.&lt;/p&gt;&#xA;" OwnerUserId="11597" LastEditorUserId="11597" LastEditDate="2015-04-19T12:52:27.137" LastActivityDate="2015-04-19T13:40:53.990" Title="Changing perspective" Tags="|riddle|mathematics|geometry|visual|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12425" PostTypeId="2" ParentId="12424" CreationDate="2015-04-19T12:58:04.473" Score="1" Body="&lt;p&gt;I'm guessing (because I'm not sure how this answer fits with part of the puzzle):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Inside a truncated hexahedron.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The ●s are corners each have 3 vertices.&lt;/p&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="11297" LastEditDate="2015-04-19T13:11:53.720" LastActivityDate="2015-04-19T13:11:53.720" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="12426" PostTypeId="2" ParentId="12424" CreationDate="2015-04-19T13:11:28.317" Score="3" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cube&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1twFs.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the point of view is in the middle of the cube, you can see all the edges forming 120° angles between them. It's easy to see, just look at the corner of your room where two walls and the ceiling meet, it's a perspective matter!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the point of view is in a vertex, you can only see three 120° in the opposite corner, while the three you're standing in are hidden by you!&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-04-19T13:33:19.793" LastActivityDate="2015-04-19T13:33:19.793" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12427" PostTypeId="2" ParentId="12424" CreationDate="2015-04-19T13:27:41.213" Score="2" Body="&lt;p&gt;I think I have a better answer than my last...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...a 4 cornered spherical polyhedron, essentially a cube inflated like a ball...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...because when inflated like a ball the angles become $1/3$ of a circle that magic $120^\circ$ and this time there are $24$ of them.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It looks like this :&lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/7/7c/Uniform_tiling_432-t0.png&quot; rel=&quot;nofollow&quot;&gt;http://upload.wikimedia.org/wikipedia/commons/7/7c/Uniform_tiling_432-t0.png&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="11297" LastEditDate="2015-04-19T13:40:53.990" LastActivityDate="2015-04-19T13:40:53.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12429" PostTypeId="1" AcceptedAnswerId="12431" CreationDate="2015-04-19T14:10:58.323" Score="2" ViewCount="257" Body="&lt;p&gt;A woman is looking through a store and sees a sign for 2 dollar bread saying &quot;For every 10 loaves you buy, you get each for only 1.9 dollars!&quot; A pretty strange deal, but she decides to take it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If has a budget of 10,000,000 dollars for the bread, then how much bread can she buy without going over her limit?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Please remember to explain your reasoning and hide your answer and reasoning behind a spoiler thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(P.S. She's rich and really, REALLY, likes bread, I guess)&lt;/p&gt;&#xA;" OwnerUserId="11002" LastEditorUserId="11100" LastEditDate="2015-04-19T15:54:57.073" LastActivityDate="2015-04-19T15:54:57.073" Title="How Much Bread For Her Budget?" Tags="|mathematics|calculation-puzzle|" AnswerCount="1" CommentCount="3" ClosedDate="2015-04-19T17:39:13.480" ContentLicense="CC BY-SA 3.0" />
  <row Id="12430" PostTypeId="2" ParentId="11225" CreationDate="2015-04-19T14:28:52.147" Score="2" Body="&lt;p&gt;Here is my second answer, but I must admit that this is, most probably, &lt;em&gt;not &quot;the canonical answer&quot;, but instead is an attempt of the descendants to figure out the right combinations&lt;/em&gt; &lt;strong&gt;=)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;The four digit combination is generated by rolling four dice, and those four digits are hidden in the sequence given&lt;/em&gt;, where the sequence is based on the numbers on the cards of a card game (I apologise if I'm using the wrong terms, I'm not native). We can take the sequence of numbers as either &lt;code&gt;3 4 6 1 3 2 5 11 3 1 3 12 5 13 3 1 1 1 2 5 3 6 5 11 2 3 12 1&lt;/code&gt; or &lt;code&gt;3 4 6 14 3 2 5 11 3 1 3 12 5 13 3 14 14 1 2 5 3 6 5 11 2 3 12 1&lt;/code&gt; depending on if Ace is high or low (thank you @Engineer Toast). As there are numbers greater than 6 we could apply '&lt;em&gt;mod 6&lt;/em&gt;' on all numbers. Then, we &lt;em&gt;brute force the right sequence out!&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user5871" LastEditorDisplayName="user5871" LastEditDate="2015-04-19T16:46:00.620" LastActivityDate="2015-04-19T16:46:00.620" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="12431" PostTypeId="2" ParentId="12429" CreationDate="2015-04-19T14:45:02.707" Score="2" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 5263157&#xA;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\frac{10000000\times10}{19}(decimal~ignored)=5263150~and,~for~remaining ~15\$, +7~breads$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11306" LastEditorUserId="11306" LastEditDate="2015-04-19T15:01:31.383" LastActivityDate="2015-04-19T15:01:31.383" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12432" PostTypeId="2" ParentId="12396" CreationDate="2015-04-19T15:16:06.537" Score="57" Body="&lt;p&gt;The puzzle lays a restriction on the ghost's answers, but not on your question (i.e., your question doesn't have to be a yes/no question). I think it's a bit of misdirection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Give the doors names: &quot;Door 1 I name 'Yes', Door 2 I name 'No', and Door 3 I give no name. Behind the door of which name lies freedom?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ghost can answer &quot;Yes,&quot; &quot;No,&quot; or nothing, and by his answer he designates the name you gave that door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is very similar to GOTO 0's &lt;a href=&quot;https://puzzling.stackexchange.com/a/12414/8803&quot;&gt;answer&lt;/a&gt;, but this doesn't involve asking two questions for the price of one.&lt;/p&gt;&#xA;" OwnerUserId="8803" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.923" LastActivityDate="2015-04-19T15:16:06.537" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="12433" PostTypeId="2" ParentId="12324" CreationDate="2015-04-19T15:31:00.930" Score="0" Body="&lt;p&gt;From spending 1st fish.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Monkey go to the island and take bananas from both trees.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From spending 2nd fish.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Monkey convince Crocodile to send a bundle of banana.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From spending 3rd fish&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Monkey comes with 2nd bundle of banana.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11306" LastActivityDate="2015-04-19T15:31:00.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12434" PostTypeId="2" ParentId="12119" CreationDate="2015-04-19T15:58:37.293" Score="3" Body="&lt;p&gt;The base of this solution is the binary interpretation strategy as shown in Lopsy's answer, but with an improvement to $2^n$ moves instead of  $n2^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $T \to 1$ and $H \to 0$, interpret the sequence of coins as a binary number starting from a position called the mark (denoted by inserting an M) with the string left of the mark appended to the right end. The value of the mark at that position is the binary value of that sequence. So if $M_1$ is the mark $0M1100$ its value would be $V(M_1)=11000b=24$&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Choosing the mark&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The correct position of the mark is the one which yields the highest value for the given sequence. For example, $01101 \to 0M1101$ and the value at that mark is binary value of $11010b=26$. There can be multiple positions which are candidates for the mark (e.g. $M101010$, $10M1010$, $1010M10$). In such cases the mark is chosen by:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If there are no 1s to the left of the leftmost candidate, then the leftmost candidate&lt;/li&gt;&#xA;&lt;li&gt;otherwise the rightmost candidate.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For example $0M10010$ or $10110M1$ would be correct marks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If after a turn the mark is between different coins than it was before the turn, we will say that the mark &quot;jumped&quot;. This can happen either when a coin is flipped changing the set of candidate marks, or when the coins are rotated without flipping and a 1 appears on the left forcing a switch from the leftmost candidate to the rightmost candidate.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;The Player's strategy:&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If there are no 1s to the left of the mark, flip $1\to 0$&lt;/li&gt;&#xA;&lt;li&gt;otherwise do not flip $1\to 1$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h3&gt;Observation&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The position of the mark and P's strategy are both completely determined by the current sequence of coins. If the opponent (S) can cause the same sequence to appear twice, then S has a strategy that can repeat that loop forever because P will respond to it exactly the same the second time. By the contrapositive, if we can prove P's stategy must win, then no sequence repeats, so the total number of moves to win is $&amp;lt; 2^n$.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Lemma 1:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The mark can only jump when S flips.&#xA;Consider the other cases&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;No flip.&#xA;    When there is no flip, the set of mark candidates won't change, so the only way the mark can jump is if the mark started as the leftmost candidate and then a 1 was placed to it's left forcing a switch to the rightmost candidate. &#xA;&lt;ul&gt;&#xA;&lt;li&gt;1a. P does not flip.&#xA;If there are no 1's left of the mark, P's strategy says flip, so this case can't occur.&lt;/li&gt;&#xA;&lt;li&gt;1b. S does not flip. &#xA;It's S's move and he doesn't flip, so the coin placed on the left is a 0 and there is no jump.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;P does flip. &#xA;Let $M_1$ be the position of the mark prior to the move, which has value $V(M_1)$. Note that unless there are no 1s, meaning P just won, any valid mark candidate has to have a 1 on its left. Since P just flipped, all bits left of $M_1$ are 0, so no position left of $M_1$ is a mark candidate. Let $M_2$ be some position to the right of $M_1$, with value $V(M_2)$. Since $M_1$ was the mark prior to the move, $V(M_2) &amp;lt;= V(M_1)$. The move reduces the value of each mark based on the significance of the bit that was flipped $1\to 0$. Denote the values of $M_1$ and $M_2$ after the move as $V'(M_1)$ and $V'(M_2)$ respectively. Since $M_2$ is right of $M_1$, the flipped bit is more significant to $M_2$ than it is to $M_1$, so $V(M_2)-V'(M_2) &amp;gt; V(M_1)-V'(M_1)$ and thus $V'(M_2) &amp;lt; V'(M_1)'$. So, no position right of $M_1$ can have as high a value as a mark. Therefore $M_1$ is still the correct mark.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h3&gt;Lemma 2: If S doesn't flip during any full cycle of the mark (from left back to left again), then P wins.&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;By P's strategy, if S does not flip for a full cycle, P will have flipped all coins to 0, winning.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Lemma 3: When S flips a coin, the marked value after the flip will be higher than the marked value after the last time S flipped&lt;/h3&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;S's first flip&#xA;Trivially satisfied&lt;/li&gt;&#xA;&lt;li&gt;The last flipper was S.&#xA;S can only flip $0\to 1$ so S's move increase the value, which was unchanged since his last flip since P has not flipped since then.&lt;/li&gt;&#xA;&lt;li&gt;The last flipper was P.&#xA;By P's strategy and Lemma 1, P has not flipped any coins to the right of the mark since S's last flip. Therefore the total decrease in value from P's flips (since S's last flip) is less than the increase in value from S's flip (since it is a more significant bit).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h3&gt;Conclusion&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Combining the above, S must keep making flips every cycle to avoid losing, causing the marked value to increase with each flip. Eventually the value reaches $2^n-1$ and the sequence is all 1s, allowing P to win by flipping all coins. Using our original observation, this means P wins in $&amp;lt;2^n$ steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could improve the upper bound slightly given this starting point by eliminating sequences which can be shown to be not encountered. In particular we can see we will never encounter any sequence without one of (first and last position are 1) or (at least 1 run of 1s of length 2) until we get $0...01$ right before P wins. This only eliminates around $F(n-1) + F(n-3) - 2$, which is very small compared to $2^n$ as n grows.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Speculation on lower bound&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The following strategy for S might be generally effective&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;if there are no 1s left of the mark and flipping will not jump the mark then flip, &lt;/li&gt;&#xA;&lt;li&gt;otherwise do not flip&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If this strategy can force us to count through all values which wouldn't cause a jump in the mark, then we could use n-nacci numbers to place a lower bound on the number of non-mark jumping values by playing it safe and skipping all values which have a run of 1s as long as the existing longest run of 1s. The would prove that for any $k &amp;lt; 2$, winning takes $\omega(k^n)$ (i.e. that for some constant c &gt; 0 and and sufficiently large n, the number of moves exceeds $ck^n$).&lt;/p&gt;&#xA;" OwnerUserId="11594" LastActivityDate="2015-04-19T15:58:37.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12435" PostTypeId="1" AcceptedAnswerId="12438" CreationDate="2015-04-19T16:02:56.433" Score="3" ViewCount="1558" Body="&lt;p&gt;two logicians decided to play a lovely well-known game , number-guessing from the least significant informations , the juror proposed to create the challenge :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the juror told the logicians two numbers secretly , unbeknownst to eachother , all they know , is that the product of eachother's number is a &lt;strong&gt;number divisible by 2 , and between 2 and 10.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;juror said that , no action is permitted , no gestures , no word , except when they know numbers , they are permitted to announce it!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;A suspensious silence was been hovering arround the place , no word! everyone was staring at the other wierdly .&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After this few minutes silence both jumped suddenly and declared : We know these numbers!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: What are these numbers ?&lt;/p&gt;&#xA;" OwnerUserId="9517" LastEditorUserId="10615" LastEditDate="2016-02-24T04:06:41.507" LastActivityDate="2020-04-22T15:25:14.617" Title="&quot;Guess the number&quot; game" Tags="|logical-deduction|meta-knowledge|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12436" PostTypeId="2" ParentId="12396" CreationDate="2015-04-19T16:11:44.133" Score="5" Body="&lt;p&gt;I will ask him,&quot;response me &quot;Yes&quot; if the 1st door has the way to freedom, or response me &quot;No&quot; if the 2nd door has the way to freedom.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If He response, &quot;Yes&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'll go through 1st door.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If He response, &quot;No&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'll go through 2nd door.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If He didn't response, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'll go through 3rd door.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11306" LastActivityDate="2015-04-19T16:11:44.133" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12438" PostTypeId="2" ParentId="12435" CreationDate="2015-04-19T16:35:16.010" Score="4" Body="&lt;p&gt;Assuming that both of their numbers (a,b) are integers.  Neither can be 0 and if either are negative, they know the other has a negative number so it is trivial to reduce it to the same problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both players know that all possible pairs contain at least a 1 or a 2.  If either player has a 10, 8, 6, 5, or 3, then they immediately know to combine it with the 1 or 2 to find the 1 possible product within that range.  As no one immediately finds the solution, neither player has these numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only numbers a player may have, therefore, is 1, 2, or 4.  It has been revealed since I starting writing the answer that neither player can have a 1.  The possible pairs are, therefore: (2,2) and (2,4).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If either player had a 4, they would realise that the other has a 2 and say the result.  This means that they both know it is (2,2).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="99" LastActivityDate="2015-04-19T16:35:16.010" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12439" PostTypeId="2" ParentId="12435" CreationDate="2015-04-19T16:35:21.997" Score="3" Body="&lt;p&gt;Lets assume that we are one of the logicians and get a number, how much can you reason out from that number?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Numbers 6+ are impossible because the minimum is two and the product can't be higher than 10. &#xA;Number 5: The other must have two, otherwise the product would not be divisible by 2.&#xA;Number 4: The other must have two.&#xA;Number 3: The other must have two, otherwise the end product is not divisible by two.&#xA;Number 2: The other must have 2 or 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only combination that carries uncertainty for both parties is if both have the number 2, hence this is the number that both logicians had. &lt;/p&gt;&#xA;" OwnerUserId="11473" LastActivityDate="2015-04-19T16:35:21.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12440" PostTypeId="1" AcceptedAnswerId="12448" CreationDate="2015-04-19T17:33:31.057" Score="0" ViewCount="278" Body="&lt;p&gt;In Excel Fill Series if you input &lt;strong&gt;23, 132, 2&lt;/strong&gt;&lt;br&gt;&#xA;then 31.33333333, 20.83333333, 10.33333333, -0.166666667, -10.66666667....is generated    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and if you input &lt;strong&gt;13, 16, 17&lt;/strong&gt;&lt;br&gt;&#xA;then 19.33333, 21.33333, 23.33333, 25.33333.. is generated as shown below  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the algorithm for number generation?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8TkV4.png&quot; alt=&quot;Series 1&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/Jm5sT.png&quot; alt=&quot;Series 2&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="11630" LastEditorUserId="5188" LastEditDate="2015-04-19T19:36:51.373" LastActivityDate="2015-04-19T21:08:30.130" Title="Explain the sequence in these Excel Fill Series" Tags="|number-sequence|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12441" PostTypeId="1" AcceptedAnswerId="12442" CreationDate="2015-04-19T17:47:13.633" Score="1" ViewCount="1233" Body="&lt;p&gt;19 BBY, the Galactic Republic spots General Grievous in Utapau, the Separatists' Council Base; the Jedi Obi-Wan Kenobi is sent there to deal with him.&lt;br&gt;&#xA;After a long search, Obi-Wan comes face to face with the Supreme Commander: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Obi-Wan:&lt;/strong&gt; &lt;em&gt;Surrender, it's over!&lt;/em&gt;&lt;br&gt;&#xA;&lt;strong&gt;Grievous:&lt;/strong&gt; &lt;em&gt;Mwhahaha, fool! How can you beat me?!?!&lt;/em&gt;&lt;br&gt;&#xA;&lt;strong&gt;Obi-Wan:&lt;/strong&gt; &lt;em&gt;With my lightsaber, of course!&lt;/em&gt;&lt;br&gt;&#xA;&lt;strong&gt;Grievous:&lt;/strong&gt; &lt;em&gt;Mwhahaha, have you ever seen my set of four lightsabers?&lt;/em&gt;&lt;br&gt;&#xA;&lt;strong&gt;Obi-Wan:&lt;/strong&gt; &lt;em&gt;Do you feel advantaged? May the math be with you! There's no difference between one and four!&lt;/em&gt;&lt;br&gt;&#xA;&lt;strong&gt;Grievous:&lt;/strong&gt; &lt;em&gt;Can you prove it?&lt;/em&gt;&lt;br&gt;&#xA;&lt;strong&gt;Obi-Wan:&lt;/strong&gt;&lt;br&gt;&#xA;$x=4$&lt;br&gt;&#xA;$x(x-1)=4(x-1)$&lt;br&gt;&#xA;$x^2-x=4x-4$&lt;br&gt;&#xA;$x^2-4x=x-4$&lt;br&gt;&#xA;$x(x-4)=x-4$&lt;br&gt;&#xA;$x=1$&lt;br&gt;&#xA;&lt;strong&gt;Grievous:&lt;/strong&gt; &lt;em&gt;You're trying to use the Force on me, but it won't work!&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is Obi-Wan's math as strong as his Force? Explain it!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MD0hQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you like problems like this, check&#xA;  &lt;strong&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/10713/a-dollar-a-penny-theres-no-difference&quot;&gt;A dollar, a penny, there&amp;#39;s no difference&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2015-04-19T20:11:04.597" Title="Obi-Wan vs Grievous" Tags="|mathematics|word-problem|paradox|" AnswerCount="5" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12442" PostTypeId="2" ParentId="12441" CreationDate="2015-04-19T18:14:48.267" Score="6" Body="&lt;p&gt;When you multiplied both sides by $(x-1)$, you introduced the new extraneous solution $x=1$ to the equation. Later on when you divided by $(x-4)$, you forgot to case check that $(x-4)$ might equal $0$. If we do so we get $x = 1$ or $4$, as expected.&lt;/p&gt;&#xA;" OwnerUserId="11100" LastActivityDate="2015-04-19T18:14:48.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12443" PostTypeId="2" ParentId="12441" CreationDate="2015-04-19T19:02:41.417" Score="4" Body="&lt;p&gt;We started by specifying $x=4$. Making this substitution makes the error more clear:&#xA;\begin{align}&#xA;4&amp;amp;=4\\&#xA;&amp;amp;\vdots\\&#xA;4(4-4)&amp;amp;=(4-4)\\&#xA;4&amp;amp;=1&#xA;\end{align}&#xA;The second to last line is true, but the last isn't, because we divided by $4-4=0$.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-04-19T19:02:41.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12444" PostTypeId="1" AcceptedAnswerId="12462" CreationDate="2015-04-19T19:10:36.463" Score="-1" ViewCount="2858" Body="&lt;p&gt;Which number comes next in the sequence?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2, 7, 19, 47, 83, ___&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Please remember to explain your reasoning and put your answer and reasoning behind spoilers. (Do &gt;! and then your text to put one) &lt;/p&gt;&#xA;" OwnerUserId="11002" LastActivityDate="2015-04-22T05:20:45.003" Title="What Comes Next?" Tags="|number-sequence|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="12445" PostTypeId="2" ParentId="12441" CreationDate="2015-04-19T19:16:22.690" Score="2" Body="&lt;p&gt;Multiplying LHS and RHS by zero(if $x=1,x-1=0$) can equate any non-equal equations, So Obi-Wan did solve wrongly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the thing if any equation has 2 roots, say, $x_1,x_2$, that imply $either~x_1~or~x_2~or~both$ has solution.(i.e. He can't say both are solution)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Example:- &lt;strong&gt;$ax=bx$ implies $a=b$, if, and only if, $x\neq 0$&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="11306" LastEditorUserId="11306" LastEditDate="2015-04-19T19:17:26.250" LastActivityDate="2015-04-19T19:17:26.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12447" PostTypeId="2" ParentId="12444" CreationDate="2015-04-19T19:23:41.580" Score="5" Body="&lt;p&gt;Assuming the commas are superfluous; the numbers could relate to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This marker on the Neanderthal genome; &lt;a href=&quot;http://neandertal.ensemblgenomes.org/Homo_sapiens/Location/Marker?r=16:26994784-27194783&quot;&gt;http://neandertal.ensemblgenomes.org/Homo_sapiens/Location/Marker?r=16:26994784-27194783&lt;/a&gt;. The next number would therefore be 16.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4852" LastActivityDate="2015-04-19T19:23:41.580" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12448" PostTypeId="2" ParentId="12440" CreationDate="2015-04-19T19:29:32.933" Score="3" Body="&lt;p&gt;The sequences are a linear extrapolation of the 3 initial points as follows:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the starting point is Backcalc&lt;sub&gt;2&lt;/sub&gt; which is the average of the 3 initial points  &lt;/li&gt;&#xA;&lt;li&gt;the Increment is the average of Actual&lt;sub&gt;2&lt;/sub&gt;-Actual&lt;sub&gt;1&lt;/sub&gt; and Actual&lt;sub&gt;3&lt;/sub&gt;-Actual&lt;sub&gt;2&lt;/sub&gt;   &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rAy6A.png&quot; alt=&quot;Sequence&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-04-19T21:08:30.130" LastActivityDate="2015-04-19T21:08:30.130" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12449" PostTypeId="2" ParentId="12441" CreationDate="2015-04-19T19:34:01.337" Score="2" Body="&lt;p&gt;Obi-Wan's math is not nearly as strong as his force. His math breaks at two different points based on whether we consider x=4 or x=1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if x=1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Line 2: x(x−1)=4(x−1)  breaks the proof because were saying 0=0, so the rest &#xA;  of the proof is essentially void.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;if x=4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Line 5: x(x−4)=x−4 going to Line 6: x=1  is invalid because we are dividing by zero&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Either way, Obi-Wan's math was not strong enough to fool Grievous&lt;/p&gt;&#xA;" OwnerUserId="7715" LastActivityDate="2015-04-19T19:34:01.337" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12450" PostTypeId="2" ParentId="12441" CreationDate="2015-04-19T20:11:04.597" Score="2" Body="&lt;p&gt;the trick is in these two lines:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x(x−4)=x−4$$&#xA;$$x=1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x(x−4)=x−4$ is a second degree equation which has for solutions $x=1$ &lt;strong&gt;OR&lt;/strong&gt; $4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You cant deduce directly $x=1$ without a specific reason this is maths not lottery&lt;/p&gt;&#xA;" OwnerUserId="9517" LastActivityDate="2015-04-19T20:11:04.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12451" PostTypeId="1" AcceptedAnswerId="12452" CreationDate="2015-04-19T20:45:09.677" Score="11" ViewCount="1669" Body="&lt;p&gt;Airplanes can hold enough fuel to fly halfway around the world. The only source of fuel is a single airport, which is also the only place planes can take off and land. However, planes can transfer fuel to each other midair. Assume that planes can change direction and refuel instantly, and all planes fly at the same constant rate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Amelia Earhart wants to circumnavigate the globe (namely, start at the airport, fly to an opposite point on the  earth, then back). She needs the help of other planes, but of course doesn't want any of them to crash. How many planes does she need, including her own?&lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2016-12-02T13:50:36.940" Title="Amelia Earhart Challenge" Tags="|strategy|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12452" PostTypeId="2" ParentId="12451" CreationDate="2015-04-19T21:42:42.563" Score="12" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3 planes are needed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They all take off at the same time and fly 1/8th the way around the world.  One gives 1/4th of a tank to each of the other planes and goes back to refuel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other two continue another 1/8th of the way.  The support plane gives 1/4th of a tank to Earhart and flies back to refuel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When Earhart is running on fumes 6/8ths of the way around the world, she is met by a support plane that gives her 1/4th of a tank.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They fly to the 7/8ths mark and meet a second support plane which gives them both 1/4th of a tank.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As the planes from the first 4th and last 4th are reused, only 3 planes are needed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditDate="2016-12-02T13:50:36.940" LastActivityDate="2016-12-02T13:50:36.940" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12453" PostTypeId="1" AcceptedAnswerId="12459" CreationDate="2015-04-19T21:47:26.433" Score="12" ViewCount="2069" Body="&lt;p&gt;After however much inordinate time passed for the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/12221/tree-pruning-game-with-the-devil&quot;&gt;tree-pruning game&lt;/a&gt; to finish, you ended up winning, and Satan was infuriated. He was &lt;em&gt;sure&lt;/em&gt; he'd come up with a game keep you in hell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But once again, he's let go on his promise. This time he's &lt;em&gt;sure&lt;/em&gt; he's created a game to foil you once and for all, and to prove it, he's given you a contract signed in his own blood and witnessed by God himself to make sure he doesn't renege on his promise a third time. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;He takes you down to an infinitely large checkerboard. You notice a red line drawn between two rows of the squares, extending infinitely in each direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The rules of the game are simple&quot;, said Satan. &quot;Every turn, we'll do the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;You can either:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Place one checker down on a square on your side of the board. But be warned, once you've placed a checker, you can never place another checker in that square again.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Jump a checker over another adjacent checker horizontally, vertically, or diagonally, removing the checker you jumped over. You can't jump onto your own checkers, but if you manage to jump onto one of my checkers, I'll just remove it as well.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;And then I'll:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Place one checker on my side of the board.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you can jump over any of my checkers, including jumping over one of mine onto another one of mine, then you win the game, and you have that contract witnessed by God that I'll let you go. And I'll be nice,&quot; he says with a cackle, &quot;I'll place all my checkers within ten spaces of the line.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Simple, right?&quot; said Satan. &quot;So... do you accept this game?&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Do you have a winning strategy for this game? Or has Satan finally matched your wits?&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-04-20T18:01:48.703" Title="Checkers with the devil" Tags="|logical-deduction|strategy|game|checkerboard|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12455" PostTypeId="2" ParentId="12451" CreationDate="2015-04-19T21:52:42.317" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; She needs three planes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here comes the explanation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All planes will be numbered. Number 1 will be Amelia Earhart's plane.&lt;br&gt;&#xA;&lt;br&gt;&#xA; 1- So p1, p2, p3 take off at the same time and fly in the same direction.&lt;br&gt;&#xA; 2- After a fourth of the tank capacity, p2 gives a fourth of its tank capacity to the two other planes and returns to the airport. At this point, p1 and p3 have a full tank.&lt;br&gt;&#xA; 3- After a fourth of the tank capacity, p3 gives a fourth of its tank capacity to p1 and returns to the airport. At this point, p1 still has a full tank.&lt;br&gt;&#xA; 4- After a half of p1 tank capacity, p2 and p3 take off and fly toward p1 but in the opposite direction from the first take off.&lt;br&gt;&#xA; 5- After a fourth of tank capacity, p2 gives a fourth of its tank to p3 and returns to airport.&lt;br&gt;&#xA; 6- After a fourth of tank capacity, p3 meets p1 and gives it a half of its tank capacity. It means that p1 can head to the airport directly. But not p3. That's why in the meantime, p2 fly toward p3 and gives it the needed fuel to return to the airport.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I hope it is clear enough.&lt;/p&gt;&#xA;" OwnerUserId="10400" LastActivityDate="2015-04-19T21:52:42.317" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12456" PostTypeId="2" ParentId="12453" CreationDate="2015-04-19T22:15:54.007" Score="0" Body="&lt;p&gt;First I place a single trash checker 1000 squares away from the red line. Then Satan places his checker, and from now on I'll ignore Satan's moves: if Satan places another checker nearby it can only help me as I can't place a checker on his side of the line anyways, and if I jump onto it I win earlier than intended. For the next $N = 10^{10^{100}}$ moves, I'll place a bunch of checkers under the target checker until &quot;everything&quot; is filled up. Now I have a much simpler game: half the plane is full and I must reach a specific distance 1-10 squares from the red line in however many moves I like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I want to hop to somewhere but in doing so I will land on a placed checker, I'll plan for this ahead of time and in my googolplex preparation moves I'll skip that spot (and now maybe I can place a checker there later on).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I can case check the distance of Satan's checker from the red line.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Hop vertically.&lt;/li&gt;&#xA;&lt;li&gt;Do 1 then hop diagonally.&lt;/li&gt;&#xA;&lt;li&gt;Do 2 then in its column hop vertically twice.&lt;/li&gt;&#xA;&lt;li&gt;Do 3 then to its right do 2 then hop diagonally.&lt;/li&gt;&#xA;&lt;li&gt;Do 4 then do some simple hops to get a checker diagonally down/right two, then do 3 pretending that the red line is 3 lower than it is to get another checker diagonally down/right three, then hop diagonally twice.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This is all I'll do because I need to go to sleep. I've proved that if Satan places his first checker at a distance of 1-5 then I can win.. it may be that 10 is impossible.&lt;/p&gt;&#xA;" OwnerUserId="11100" LastEditorUserId="11100" LastEditDate="2015-04-19T22:31:51.543" LastActivityDate="2015-04-19T22:31:51.543" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12457" PostTypeId="2" ParentId="12386" CreationDate="2015-04-19T22:20:44.897" Score="0" Body="&lt;p&gt;this is ...&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a checkerboard by two lines step in case of :&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;-&lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; are odd&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;knight shortest path in case of &lt;span class=&quot;math-container&quot;&gt;$4m*4n$&lt;/span&gt; grid with supplementary adjustements&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/E7omO.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;-if the grid is &lt;span class=&quot;math-container&quot;&gt;$4n*(4m+2)$&lt;/span&gt; like the green framed rectangle or &lt;span class=&quot;math-container&quot;&gt;$4n*(4m+1)$&lt;/span&gt; like brown rectangle or &lt;span class=&quot;math-container&quot;&gt;$4n*(4m+3)$&lt;/span&gt; gray rectangle or &lt;span class=&quot;math-container&quot;&gt;$(4n+2)*(4m+2)$&lt;/span&gt; blue rectangle the grid should be enlarged from supplementary same color-marked square which is located in the corner,the square is erased from old grid and marked in new grid square-corner.&lt;/p&gt;&#xA;&lt;p&gt;-if the  &lt;span class=&quot;math-container&quot;&gt;$(4n+1)*(4m+2)$&lt;/span&gt; yellow grid is enlarged from the top we remove the yellow mark and place it at the top of new grid, if it is enlarged from right/left we add a new yellow mark.&lt;/p&gt;&#xA;&lt;p&gt;-if it is the pink &lt;span class=&quot;math-container&quot;&gt;$(4n+3)*(4m+2)$&lt;/span&gt; grid and enlarged from right side , two pink marks are removed and placed in the extreme right squares of new grid, if it is enlarged from top, just one mark is moved.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Conclusion&lt;/strong&gt;: (&lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; is number of columns , &lt;span class=&quot;math-container&quot;&gt;$k'$&lt;/span&gt; is n° of lines.)&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;for &lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$(4k)$&lt;/span&gt;x&lt;span class=&quot;math-container&quot;&gt;$(4k'+2)$&lt;/span&gt;&lt;/strong&gt; grids number of black squares is &lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$6(k*k')+1$&lt;/span&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;for &lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$(4k)$&lt;/span&gt;x&lt;span class=&quot;math-container&quot;&gt;$(4k'+1)$&lt;/span&gt;&lt;/strong&gt; grids number of black squares is &lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$5(k*k')+1$&lt;/span&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;for &lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$(4k)$&lt;/span&gt;x&lt;span class=&quot;math-container&quot;&gt;$(4k'+3)$&lt;/span&gt;&lt;/strong&gt; grids number of black squares is &lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$7kk'+1$&lt;/span&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;for &lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$(4k+2)$&lt;/span&gt;x&lt;span class=&quot;math-container&quot;&gt;$(4k'+1)$&lt;/span&gt;&lt;/strong&gt; grids number of black squares is &lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$9+6(kk'-1)$&lt;/span&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;for &lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$(4k+2)$&lt;/span&gt;x&lt;span class=&quot;math-container&quot;&gt;$(4k'+2)$&lt;/span&gt;&lt;/strong&gt; grids number of black squares is &lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$10+6(kk'-1)$&lt;/span&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;for &lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$(4k+2)$&lt;/span&gt;x&lt;span class=&quot;math-container&quot;&gt;$(4k'+3)$&lt;/span&gt;&lt;/strong&gt; grids number of black squares is &lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$12+7(k-1)+6k(k'-1)$&lt;/span&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;for &lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$(4k)$&lt;/span&gt;x&lt;span class=&quot;math-container&quot;&gt;$(4k')$&lt;/span&gt;&lt;/strong&gt; grids number of black squares is &lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$4kk'$&lt;/span&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;for &lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$odd$&lt;/span&gt;x&lt;span class=&quot;math-container&quot;&gt;$odd$&lt;/span&gt;&lt;/strong&gt; the grid looks like an expanded checkerboard shown in picture above$**&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9517" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-19T22:32:05.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12458" PostTypeId="2" ParentId="12451" CreationDate="2015-04-19T23:12:16.707" Score="3" Body="&lt;p&gt;Like the other answers, the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Three planes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, a correct explanation is more complicated.  In particular:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The second support plane needs to do two out-and-back flights on each side of the planet.  Here's a full breakdown (important new steps bolded):&lt;br/&gt;&#xA;&lt;br/&gt;&#xA; There are three planes (AE, S1, and S2).  They each start with a full tank.&lt;br/&gt;&#xA;&lt;br/&gt;&#xA; 1. They all fly 1/8th of the way.&lt;br/&gt;&#xA; 2. AE, S1, and S2 all now have 3/4 tanks.  S2 transfers 1/4 tank to S1 and turns back.  AE now has 3/4 tank, S1 has a full tank, and S2 has 1/2 tank.&lt;br/&gt;&#xA; 3. S1 and AE continue for another 1/8th of the way.&lt;br/&gt;&#xA; 4. S1 has 3/4 tank while AE has 1/2 tank.  Meanwhile S2 returns home with 1/4 tank to spare.&lt;br/&gt;&#xA; 5. S1 transfers 1/2 tank to AE and turns back.  S1 now has 1/4 tank while AE has a full tank.  &lt;strong&gt;S2, refueled to a full tank, sets off again&lt;/strong&gt;.&lt;br/&gt;&#xA; 6. All planes fly another eighth of the way.  AE is now 3/8ths of the way with 3/4 fuel, &lt;strong&gt;S1 and S2 rendezvous 1/8th of the way (S1 has just run out of fuel and S2 has 3/4 tank)&lt;/strong&gt;.&lt;br/&gt;&#xA; 7. &lt;strong&gt;S2 transfers 1/4 tank to S1 and turns around.  S2 now has 1/2 tank and S1 now has 1/4 tank&lt;/strong&gt;.&lt;br/&gt;&#xA; 8. The planes fly another eighth of the way: S1 arrives at home with an empty tank, S2 arrives at home with 1/4 tank.  AE reaches the 4/8ths mark with 1/2 tank.&lt;br/&gt;&#xA; 9. S1 and S2 refuel and set off again to do the whole thing in reverse on the other side of the planet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;[Edit: after discussion on the other answers, it was noted that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If S2 gives 1/4 tank to &lt;strong&gt;both&lt;/strong&gt; S1 and AE (thereby transferring 1/2 tank), then it also works (and S2 only needs to make one trip).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The phrasing of both other answers led to my confusion on this; they are both correct under this clarification.  My answer is then just an alternate approach.]&lt;/p&gt;&#xA;" OwnerUserId="4827" LastEditorUserId="4827" LastEditDate="2015-04-19T23:36:21.330" LastActivityDate="2015-04-19T23:36:21.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12459" PostTypeId="2" ParentId="12453" CreationDate="2015-04-19T23:24:52.970" Score="8" Body="&lt;p&gt;The Devil wins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Devil places checkers anywhere 10 spaces above the line. We show that you can't reach that far up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an arbitrary checker of the Devil, assign each space a value that decays with distance to the target. Let the value be $\alpha^{d}$ where $d$ is the distance to that checker in chess-king moves and $\alpha$ is the inverse of the Golden Ratio, $\alpha = 1 / \phi$. Let your total value be the sum of the values of spaces your checkers occupy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We'll show that jumps you make don't increase your value and that the total value of all checkers you can place is less than the value of a checker next to one of the devil's. Therefore, you can never get a checker in a position to jump the Devil's checker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we show that any jump move doesn't increase your value. A jump creates a new checker at the landing square and removes two checkers, the one that jumped and the one that was jumped over. Let $d$ be the distance of the reached location. In the case of getting a checker closer to the Devil's, the change is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha^{d} - \alpha^{d+1} - \alpha^{d+2} = \alpha^{d} \left( 1-\alpha -\alpha^2 \right) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because $\alpha = 1 / \phi$ is a root of that quadratic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any other kind of jump, the subtracted values are larger, and so the change is negative. So, no jump can increase value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we show that the total value of checkers that Johnny can ever place is bounded. Since a space can only have a checker placed once, this bound is the total value of spaces on Johnny's side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that the Devil's checker is $10$ spaces above the line, so we want to get a checker $9$ spaces above in order to jump over it. Then, of spaces below the line, there's $19$ spaces at distance $9$ from it and thus value $\alpha^{9}$, $23$ spaces at a value of $\alpha^{10}$, $27$ at value $\alpha^{11}$, and so on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.wolframalpha.com/input/?i=sum%20of%20%284*i-17%29%20*%20x%5E%28i%29%20for%20i%20%3D%209%20to%20infinity%20for%20x%20%3D%200.61803398875&quot; rel=&quot;nofollow&quot;&gt;Summing the series&lt;/a&gt; gives a total of $\approx 0.877$, which is a bit short of the target value of $1$ of a space next to the Devil's checker. So, Johnny can never get a checker into a position to jump that checker of the Devil, and so WLOG, any checker of the Devil placed on the tenth row.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorDisplayName="user88" LastEditDate="2015-04-20T18:01:48.703" LastActivityDate="2015-04-20T18:01:48.703" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12461" PostTypeId="1" CreationDate="2015-04-20T00:33:20.520" Score="6" ViewCount="422" Body="&lt;p&gt;The king has invented a new game called &lt;em&gt;King's Checkers&lt;/em&gt;. The rules are simple, it's exactly like regular Checkers, the only difference? The &lt;strong&gt;king plays first&lt;/strong&gt; and all his pieces are &lt;strong&gt;automatically &quot;king&quot;&lt;/strong&gt;, meaning they can move backwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, &lt;em&gt;force capture&lt;/em&gt; is &lt;strong&gt;enforced&lt;/strong&gt; (if given the option to capture, a play must capture)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You and his daughter are in love, and the king has offered his daughter to whoever can outsmart him in this crazy game of strategy and intellect. Assuming the king plays perfectly; is it possible that you can win the fair maiden?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2015-04-20T02:13:13.823" Title="Checkers Against The King" Tags="|strategy|checkerboard|board-games|" AnswerCount="1" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="12462" PostTypeId="2" ParentId="12444" CreationDate="2015-04-20T01:25:59.220" Score="6" Body="&lt;p&gt;This looks like a sequence extracted from the series of prime numbers. The next number could be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;139&lt;/strong&gt; which is the $34^{th}$ prime number.&lt;br&gt;&#xA;&lt;br&gt;&#xA;The sequence to select prime numbers is $n^2-floor((n-1)/2)$&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/fy9vP.png&quot; alt=&quot;Sequence&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastActivityDate="2015-04-20T01:25:59.220" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12463" PostTypeId="2" ParentId="12461" CreationDate="2015-04-20T01:55:36.617" Score="3" Body="&lt;p&gt;An incomplete answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'd like to say you can't win because normal &lt;a href=&quot;http://www.scientificamerican.com/article/computers-solve-checkers-its-a-draw/&quot; rel=&quot;nofollow&quot;&gt;checkers is a draw&lt;/a&gt; and the king has a strict advantage. But, technically that's not true because the king is also required to make captures backwards, which could in theory be a disadvantage.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-04-20T02:13:13.823" LastActivityDate="2015-04-20T02:13:13.823" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12464" PostTypeId="1" AcceptedAnswerId="12465" CreationDate="2015-04-20T03:40:26.970" Score="1" ViewCount="239" Body="&lt;p&gt;Mrs. V wants to give cookies to the 7 students in her class, so she orders some from Bob's Cookie Emporium.  Due to a traumatic childhood incident, she strongly desired the delivery to be divisible by 7 so she can fairly split them among her students (and due to further childhood trauma refuses to eat some herself, trash them, or otherwise reduce the number of delivered cookies to make things work out.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bob assures her that every box has a consistent number of cookies, and that four small boxes and one large box will yield a number of cookies divisible by 7.  Satisfied, Mrs. V orders.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when they arrive, the order was mixed up and instead she received one small box and two large boxes. She started to scream and the cookie delivery person, but before her obscenities became too vulgar, the student Samantha calmly said &quot;Don't worry Mrs. V!  The number of cookies must still be divisible by 7.&quot;  How did Samantha know?&lt;/p&gt;&#xA;" OwnerUserId="2722" LastActivityDate="2015-04-20T04:06:52.637" Title="Cookie Box Mix-up" Tags="|mathematics|number-theory|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12465" PostTypeId="2" ParentId="12464" CreationDate="2015-04-20T03:55:31.450" Score="3" Body="&lt;p&gt;One of the correct order and three of the incorrect contain, in total, 7 small boxes and 7 large box, and so be divisible by 7. Since the correct order is divisible by 7, then so is 3 of the incorrect order. Therefore, the incorrect order is also divisible by 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a more straightforward algebraic solution, working modulo 7,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S+2L = 8S+2L = 2(4S+L) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since $4S+L$ is a multiple of $7$.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-04-20T04:06:52.637" LastActivityDate="2015-04-20T04:06:52.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12466" PostTypeId="1" AcceptedAnswerId="12467" CreationDate="2015-04-20T05:46:49.693" Score="7" ViewCount="887" Body="&lt;p&gt;Archie, the archaeologist, has discovered 50 valuable gold coins, which he needs to ship to Zurich. Using an insured courier is expensive, costing $50\%$ of the shipment value. On the other hand, Archie knows of smuggler, named Smullyan, who claims to only deduct one coin as fee from a shipment of any number of coins. However, Smullyan is not only perfectly rational, but also a scoundrel, and will steal an entire shipment if he feels this is profitable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to entice Smullyan into being honest, Archie makes the following promise:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;I will divide my gold into packages of size $s_1,s_2,\dots, s_n$, and tell you this ordered list. I will send the first package of size $s_1$ with you, and as long as you don't take more than your standard fee for the first $k-1$ packages, then I will let you deliver the $k^\text{th}$ package. However, if you ever cheat me, then I will use the insured courier for the rest.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Archie realizes he is promising to act irrationally, since Smullyan will always steal the last package if they get that far. Unlike Smullyan, however, Archie is an honest man, and Smullyan knows this to be true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Smullyan replies,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Since you are being so kind (or stupid, Smullyan thought), I will do something I don't normally do. Every time I considering stealing from you, and would gain the same either way, then I won't steal.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Archie is touched by the gesture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How should Archie choose the sizes $s_1,\dots,s_n$ in order to minimize the amount of money he loses to theft, smuggling fees, and courier fees combined? Was this a stupid proposition after all?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clarifications:&lt;/strong&gt; Coins are indivisible, so all $s_i$ must be positive integers (positive so that Smullyan can deduct his fee). The reason the courier can charge $50\%$ is because the courier fee is payed legal tender, not in ancient coins. Archie is indifferent between losing coins and losing their cash value (he plans to sell them anyway).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Source:&lt;/strong&gt; &lt;em&gt;The Puzzler's Elusion&lt;/em&gt;, Dennis E. Shasha&lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2017-10-11T16:27:19.963" Title="Rational Smuggling" Tags="|logical-deduction|strategy|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12467" PostTypeId="2" ParentId="12466" CreationDate="2015-04-20T06:16:25.093" Score="5" Body="&lt;p&gt;I'm probably misinterpreting the problem, but I'll give this a shot.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;By &quot;Every time I consider stealing from you, I won't steal if I'd gain the same either way&quot;, I assume it means that Smullyan won't steal a package containing $n &amp;gt; 1$ gold coins if it's exactly or earlier than the $n$th-last package. If it's later than $n$th-last, then Smullyan would gain a greater fee from just stealing the one package.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Evidently this means that Archie will have to send a number of packages with coin counts decreasing by 1 per bag, in order to minimize the number of packages he sends. For example, suppose Archie only had 45 coins total, then putting them in packages of $9, 8, 7, 6, 5, 4, 3, 2, 1$ would result in a total fee of only $9$ coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Archie has 50 coins, which means that if he puts the remaining five coins in a package before the one with nine coins, then the packages are $5, 9, 8, 7, 6, 5, 4, 3, 2, 1$, with a total fee of $10$ coins for ten packages.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-04-20T06:29:40.000" LastActivityDate="2015-04-20T06:29:40.000" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="12468" PostTypeId="2" ParentId="12229" CreationDate="2015-04-20T08:56:36.357" Score="0" Body="&lt;p&gt;&lt;em&gt;Albert: &#xA;&quot;I don’t know when your birthday is, but I know Bernard doesn’t know, either.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first half of the sentence is obvious — Albert only knows the month, but not the day — but the second half is the first critical clue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The initial reaction is, how could Bernard know? Cheryl only whispered the day, so how could he have more information than Albert? But if Cheryl had whispered “19,” then Bernard would indeed know the exact date — May 19 — because there is only one date with 19 in it. Similarly, if Cheryl had told Bernard, “18,” then Bernard would know Cheryl’s birthday was June 18.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, for this statement by Albert to be true means that Cheryl did not say to Albert, “May” or “June.” (Again, for logic puzzles, the possibility that Albert is lying or confused is off the table.) Then Bernard replies:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Bernard: &#xA;&quot;I didn’t know originally, but now I do.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So from Albert’s statement, Bernard now also knows that Cheryl’s birthday is not in May or June, eliminating half of the possibilities, leaving July 14, July 16, Aug. 14, Aug. 15 and Aug. 17. But Bernard now knows. If Cheryl had told him “14,” he would not know, because there would still be two possibilities: July 14 and Aug. 14. Thus we know the day is not the 14th.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there are only three possibilities left: July 16, Aug. 15 and Aug. 17. Albert again:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Albert: &#xA;&quot;Well, now I know, too!&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same logical process again: For Albert to know, the month has to be July, because if Cheryl had told him, “August,” then he would still have two possibilities: Aug. 15 and Aug. 17.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is July 16.&lt;/p&gt;&#xA;" OwnerUserId="11145" LastActivityDate="2015-04-20T08:56:36.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12470" PostTypeId="2" ParentId="12444" CreationDate="2015-04-20T09:55:49.177" Score="5" Body="&lt;p&gt;The Next Number's will be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 139,199&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Reason is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Sequence is following +1, +3,+1,+3,...... &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Prime number sequence is 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199,..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The logic is as following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2 &lt;em&gt;after 2 prime numbers, it is&lt;/em&gt; 7&lt;br&gt;&#xA; 7 &lt;em&gt;after 3=&gt;(2+&lt;strong&gt;1&lt;/strong&gt;) prime numbers, it is&lt;/em&gt; 19&lt;br&gt;&#xA; 19 &lt;em&gt;after 6=&gt;(3+&lt;strong&gt;3&lt;/strong&gt;) prime numbers, it is&lt;/em&gt; 47&lt;br&gt;&#xA; 47 &lt;em&gt;after 7=&gt;(6+&lt;strong&gt;1&lt;/strong&gt;) prime numbers, it is&lt;/em&gt; 83&lt;br&gt;&#xA; 83 &lt;em&gt;after 10=&gt;(7+&lt;strong&gt;3&lt;/strong&gt;) prime numbers, it is&lt;/em&gt; 139 and so on...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2,7,19,47,83,139&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10280" LastEditorUserId="10280" LastEditDate="2015-04-22T05:20:45.003" LastActivityDate="2015-04-22T05:20:45.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12471" PostTypeId="1" CreationDate="2015-04-20T10:09:10.723" Score="0" ViewCount="145" Body="&lt;p&gt;There is a lot of fuzz about this &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=cheryls+birthday&quot;&gt;Cheryls Birthday&lt;/a&gt; thing - and I understood the solution after a couple of deeper thoughts about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what I am more interested in is; is the puzzle solvable using a form of formal logic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During university (CS) I had many courses about formal logic and love the concept of wiriting down statements, and actually &lt;em&gt;calculating&lt;/em&gt; a logical answer to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this doable with the infamous Cheryls Birthday puzzle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Please note that I do not need an explanation of the puzzle, the solution, or how and why the participants can or can not say what they say. I am specifically looking for a solution that uses formal logic.&lt;/p&gt;&#xA;" OwnerUserId="11498" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-04-20T10:54:06.643" Title="Can I solve Cheryls Birthday with formal logic?" Tags="|logical-deduction|puzzle-theory|" AnswerCount="0" CommentCount="4" ClosedDate="2015-04-20T12:00:46.910" ContentLicense="CC BY-SA 3.0" />
  <row Id="12474" PostTypeId="1" AcceptedAnswerId="12475" CreationDate="2015-04-20T13:34:32.707" Score="-6" ViewCount="2232" Body="&lt;p&gt;What are the next 2 numbers in the series &lt;code&gt;1, 1, 2, 4,6, 18, 21, 84, ?, ?&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please give a valid reason to justify your answer.&lt;/p&gt;&#xA;" OwnerUserId="11673" LastEditorUserId="5188" LastEditDate="2015-04-20T15:29:49.167" LastActivityDate="2016-09-19T10:00:49.917" Title="Determine the next two numbers in a number series" Tags="|number-sequence|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12475" PostTypeId="2" ParentId="12474" CreationDate="2015-04-20T13:36:35.183" Score="8" Body="&lt;p&gt;The next two numbers in the series are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 88, 440&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The series is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; *1, +1, *2, +2, *3, +3, *4, +4, *5, +5, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-04-20T13:36:35.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12476" PostTypeId="2" ParentId="12422" CreationDate="2015-04-20T14:05:33.933" Score="13" Body="&lt;p&gt;We'll name the logicians A, B, and C to simplify.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If A wants a drink, A doesn't yet know whether B and C want a drink.  Thus A cannot answer &quot;yes&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If A does not want a drink, A would say &quot;no&quot; because at least one of A, B, and C don't want a drink, making &quot;does everybody want a drink&quot; false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By saying &quot;I don't know&quot; A indicates their personal desire for a drink, but his lack of knowledge about the other two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B is essentially in the same situation.  B can deduce that A wants a drink, but still doesn't know the preference of C, therefore must answer &quot;no&quot; if B doesn't want a drink, and must answer &quot;I don't know&quot; if B wants a drink.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C, however, can now deduce that A and B both want drinks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if the third logician wants a drink, they say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes, everybody wants a drink.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the third logician doesn't want a drink, they say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No, not everyone wants a drink.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1682" LastEditorUserId="11475" LastEditDate="2015-05-19T18:30:05.063" LastActivityDate="2015-05-19T18:30:05.063" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12477" PostTypeId="1" AcceptedAnswerId="12827" CreationDate="2015-04-20T14:41:21.277" Score="4" ViewCount="2749" Body="&lt;p&gt;So I haven't posted very many questions and I'm not yet very good at figuring out how many clues to include. In addition, I realize that the key behind the third cryptogram is not at all obvious so it might need some more help but I don't want to give the game away. I'll give it until Monday and then I'll post some more obvious hints.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Can you decode each of the cryptograms below and then answer the challenge they pose?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;ISAEBZMEIYIQWXXOIYSCIZSJCOTFSYJALIAJQEICJGVMEIQOSJGMIPMJYGBMMEIAESOOIGVISYFSYJAMBBOYASGLBMESMRBFDICTKWJAUOTMEIQOSJGMIPMJYHICIOTMEIIPQOSGSMJBGBZMEIMCWIQWXXOI&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;br&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;XKEJGBVUUCMZCFSAKNFZEJCKEAABCRVAACJMICBFMYCVGBGFRBCJAJXAKCOVQFMCCVGBIXFASCUNVSABCEMYCJUXFMZRVAACJMSVJCEMFDECABCXVJCSFQRUCFYCVSIEAQVXICAJFGHXAKFYCMAFNXICGVESCABCXJCDEFJCVMCRFRBVMX&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;br&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;CIZQYNDRFRYAKPQLRNDRRQQKDZDPYEPQLRJDGHSZQOVYBERQAQLAGYFCQLFVYFRKDDENQCLNHLPMHFMCRQYRRMDJDFRHLFTDOTCGGMHWDHGGRMODDEGHCLRDXRFRMDYLPDOGZCLNEHRRDOLIQODHAMHLPRMDOYGDRMHRRMDICOFRHLPRMCOPFMHODRQDLFYODHYLCUYDFYJFRCRYRCQLNQQPGYAKEYSSGCLN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;There may be solutions to the puzzle-within-the-puzzle that I haven't thought of yet. Any such solutions will receive an upvote but will not be accepted as the answer unless the answer I was thinking of is never found (doubtful) in which case the most upvoted answer will be accepted.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Clarification on the hidden challenge:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;NAFJREVATGURUVQQRAPUNYYRATRJVGUFBZRGUVATYVXRNRAPELCGFGBAJVYYORPBAFVQRERQVAFHSSVPVRAGLBHZHFGRKCYNVAGURCNGGREAGUNGYRQGBNRAPELCGVATNFA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;End of the first day hint:&lt;/strong&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; As @JasonPatterson hinted at on the answer by @GentlePurpleRain there are clues in the third cryptogram. In fact, the first three sentences of the cryptogram each contain a clue. Three sentences, three clues, three cryptograms. That nae be a coincidence, laddy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;End of the second day hint:&lt;/strong&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; For the first cryptogram: Don't think of the key as starting with &amp;quot;A encodes to S&amp;quot;. It starts with &amp;quot;Q encodes to K&amp;quot;. Now, why does it start with &amp;quot;Q&amp;quot; and why does that become &amp;quot;K&amp;quot;?&lt;br&gt;&lt;br&gt;For the second cryptogram: The words from the first clue with the Z and J are there for a reason. Those are rare letters. Where have you seen those words together before? (@GentlePurpleRain has already solved this one)&lt;br&gt;&lt;br&gt;For the third cryptogram: @JasonPatterson already found the key word in the encoded clue. The method of applying it may be fremd &lt;em&gt;f&lt;/em&gt;or &lt;em&gt;i&lt;/em&gt;ndividuals &lt;em&gt;v&lt;/em&gt;acant of &lt;em&gt;e&lt;/em&gt;xperience. How can you go from the resulting mess back to a letter?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;End of the third day hint:&lt;/strong&gt; (not optimized because kids just started screaming)&lt;br&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 1) Your clues are: &amp;quot;IF YOU GET STUCK DONT GET TOO KEYED UP&amp;quot;. It starts with &amp;quot;Q encodes to K&amp;quot;. Try looking down. Recall that Q is the 17th letter of the alphabet. But what if two letters encode to the same result? That's not allowed.&lt;br&gt;2) (solved)&lt;br&gt;3) Your clues are: &amp;quot;JUST KEEP GOING AND HASH IT OUT&amp;quot;. The method of applying it may be fremd for individuals vacant of experience. After you hash it out, you have to convert it back to a number. Try to use as little of the mess as possible to get a unique encoding. But what if two letters encode to the same result? That's not allowed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Beginning of the second week hint:&lt;/strong&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 1) If you're viewing this on the desktop site - I.E. not the mobile site - and you look down, what do you see with keys on it? Since Q is the 17th letter of the alphabet, what comes 17 keys after Q? What comes 23 keys after W?&lt;br&gt;2) (Solved)&lt;br&gt;3) After you hash it out, you need to get back to a letter. How can you take the first tiny bit of the hex mess and &lt;em&gt;mod&lt;/em&gt;ify it to get a decimal equivalent to a letter? How little a bit must you take to get a unique result?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-29T20:11:21.737" Title="Puzzle out the Cryptogram Patterns" Tags="|cipher|lateral-thinking|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12478" PostTypeId="1" CreationDate="2015-04-20T15:01:21.580" Score="3" ViewCount="503" Body="&lt;p&gt;An aged King, after losing his only son in battle, determines to divide his kingdom amongst his wisest advisors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He presents two solid cubes, one made of gold, the other made of silver, and a wooden box into which can fit either of the cubes perfectly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He explains that he will divide his kingdom among those who are able to find all the ways to fill the box with pieces cut from the gold and silver cubes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rules are as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The pieces may not be curved. They must be formed by planes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The corners of each piece must meet one of the corners of the box.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Also, the same exact configuration of gold and silver pieces, rotated differently, will count as a unique way to fill the box if that rotation/configuration is not identical to one you previously counted.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The king does not know the answer, but is looking for those who can solve the riddle and prove it correct, in a way he can understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What answer would you give him? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've researched this for quite some time but have been unable to come up with a solution.  Any ideas are welcome!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This image is an example of a cube being divided in different legal ways:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xR8WX.png&quot; alt=&quot;cube cut&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="11681" LastEditorUserId="11681" LastEditDate="2015-04-20T17:45:15.423" LastActivityDate="2015-04-20T18:14:47.380" Title="Gold and Silver Cubes" Tags="|mathematics|riddle|geometry|" AnswerCount="2" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="12479" PostTypeId="2" ParentId="2836" CreationDate="2015-04-20T15:31:07.613" Score="-1" Body="&lt;p&gt;You balance 2 balls (A &amp;amp; B). 1 in each side of the scale, and leave 2 out. (C &amp;amp; D)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 1. If one of those balls is heavier, then you can assume that the 2 balls out are fine. So you weight the heavier with one of the 2 out ones. If its still heavier, then you have your answer. If it's equal in weight, then the other (A or B) is the bad one. &lt;strong&gt;(2 weightings)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 2. Both A &amp;amp; B are balanced. That means that either C or D are the bad/defective one. You use C and balance it with either A or B. If they are balanced, then D is the problem. If they are not balanced then C is the problem.&#xA;&lt;strong&gt;(2 weightings)&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="11682" LastActivityDate="2015-04-20T15:31:07.613" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12480" PostTypeId="2" ParentId="12478" CreationDate="2015-04-20T15:34:52.300" Score="1" Body="&lt;p&gt;As far as I can tell you can at most make&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 5 cuts.&lt;br&gt;&#xA; The cuts are non-intersecting. One goes through opposite faces across the diagonal.  With the remaining prisms, you can make up to 2 more cuts on each one: near point to opposite edge, near point with one near corner + one far corner, combo of the two, second with its mirror cut.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Ignoring rotation, we have no cuts, main diagonal cut, 1 side cut, 1 side cutout, main diag + 1 side cut (2x), 1 + 1 side cut (5x), 1/2 rotated (5x), 2 side (1x, 6x), main diag + 1 side + 1 side (9x), main diag + 2 side (3x), main diag + 2 side + 1 side (9x), 1/2 rotated (9x), main diag + 2 side + 2 side (9x). Those give the number of pieces at $1, 2, 2, 3, 3, 3, (2,3), 4, 4, 5, 5, 6$.&lt;br&gt;&#xA;&lt;br&gt;&#xA; For each piece, it can either be gold or silver, and each combination would be different.&lt;br&gt;&#xA; For each of our arrangements, you have $2 ^ n$ (where n = pieces) choices. So,&lt;br&gt;&#xA;&lt;br&gt;&#xA; $2 + 4 + 4 + 8(x2) + 8(x5) + 8(x5) + (4+8(x6)) + 16(x9) + 32(x9) + 32(x3)+ 32(x9) + 64(x9) = 1550$&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Edit&lt;/strong&gt;&lt;br&gt;&#xA; etothepowerofx has noticed 2 more configurations, that add 32 and 8 options respectively, putting the total to $1590$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-04-20T18:14:47.380" LastActivityDate="2015-04-20T18:14:47.380" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12481" PostTypeId="1" AcceptedAnswerId="12566" CreationDate="2015-04-20T16:03:03.067" Score="12" ViewCount="951" Body="&lt;p&gt;One day, as you go to work, you find a string with some two-colored circles glued to it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WTK7z.png&quot; alt=&quot;An image with some circles.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At once you remember that a mysterious man, who you met on the previous day, had told you that he will send some secret message to you...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you decrypt the secret message? (no keys used)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is how the word &quot;encripted&quot; looks: &lt;img src=&quot;https://i.stack.imgur.com/Bqp5m.png&quot; alt=&quot;encripted&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is how the word &quot;describe&quot; looks: &lt;img src=&quot;https://i.stack.imgur.com/4gfbf.png&quot; alt=&quot;describe&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(All images were made with &lt;a href=&quot;https://www.npmjs.com/package/canvas&quot; rel=&quot;nofollow noreferrer&quot;&gt;canvas&lt;/a&gt;, a module for node.js)&lt;/p&gt;&#xA;" OwnerUserId="11679" LastEditorUserId="463" LastEditDate="2015-04-20T21:30:22.343" LastActivityDate="2015-04-22T13:37:36.437" Title="A row of two-colored circles" Tags="|cipher|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="12482" PostTypeId="2" ParentId="12478" CreationDate="2015-04-20T16:08:59.760" Score="1" Body="&lt;p&gt;Here is my answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Is it can easily be observed the only way this can be solved is by dividing the two cubes into two &lt;em&gt;prisms&lt;/em&gt; and and taking one &lt;em&gt;prism&lt;/em&gt; from each cube and fitting it into the box in as many different configurations as possible, i.e., by rottating them so that each corner of the container meets all the corners of the &lt;em&gt;new cube&lt;/em&gt; formed (by &lt;em&gt;new cube&lt;/em&gt; I mean the one formed by the two prisms). This logicaly is the only possible answer because other pieces cut out won't satisfy the first and second condition&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user5871" LastEditorDisplayName="user5871" LastEditDate="2015-04-20T16:18:04.337" LastActivityDate="2015-04-20T16:18:04.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12483" PostTypeId="1" AcceptedAnswerId="12490" CreationDate="2015-04-20T16:14:10.577" Score="23" ViewCount="3461" Body="&lt;p&gt;Tim and Tom are fictional characters (cats, I believe) who like to play the LoL number game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this game, Tim chooses two distinct integers &lt;em&gt;A&lt;/em&gt; and &lt;em&gt;B&lt;/em&gt;, both ≥ 2, and shows them to Tom. Tom's goal is to find a succession of translolmations that can be applied to &lt;em&gt;A&lt;/em&gt; in order to obtain &lt;em&gt;B&lt;/em&gt;. There are only two types of translolmations, denoted by the letters $\bf L$ and $\bf o$, in the order as they are applied, from left to right: $\bf L$ replaces the current integer with its square, and $\bf o$ replaces the current integer with the number of digits in its binary representation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if Tim chooses &lt;em&gt;A&lt;/em&gt; = 7 and &lt;em&gt;B&lt;/em&gt; = 11, a valid solution for Tom would be $\bf LoLLo$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$7\buildrel\bf L\over\longrightarrow&#xA;49\buildrel\bf o\over\longrightarrow&#xA;6\buildrel\bf L\over\longrightarrow&#xA;36\buildrel\bf L\over\longrightarrow&#xA;1296\buildrel\bf o\over\longrightarrow&#xA;11$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &lt;em&gt;A&lt;/em&gt; = 10 and &lt;em&gt;B&lt;/em&gt; = 9, valid solutions are $\bf ooL$ and $\bf oLLo$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$10\buildrel\bf o\over\longrightarrow&#xA;4\buildrel\bf o\over\longrightarrow&#xA;3\buildrel\bf L\over\longrightarrow&#xA;9$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$10\buildrel\bf o\over\longrightarrow&#xA;4\buildrel\bf L\over\longrightarrow&#xA;16\buildrel\bf L\over\longrightarrow&#xA;256\buildrel\bf o\over\longrightarrow&#xA;9$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tom is confident he will always be able to win the game, given a sufficient number of translolmations, no matter what integers Tim chooses. Can you explain how?&lt;/p&gt;&#xA;" OwnerUserId="4394" LastActivityDate="2015-04-20T19:35:40.253" Title="The LoL number game" Tags="|mathematics|number-theory|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="12484" PostTypeId="1" CreationDate="2015-04-20T17:30:30.330" Score="6" ViewCount="3161" Body="&lt;p&gt;A common constraint in logic problems is the one restricting questions you may ask to those that can only be &lt;em&gt;helpfully&lt;/em&gt; answered with a &quot;yes&quot; or &quot;no&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in some cases, you might ask someone a question where &quot;yes&quot; or &quot;no&quot; are the only answers &lt;em&gt;if you know the answer&lt;/em&gt;, but where some respondents may lack the necessary information to know.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Example:  You meet one &quot;truth-teller&quot; and one &quot;random-respondent&quot;, who is &quot;50-50&quot; to reply yes or no to any given question. You don't know who's who, but they do. If you're limited to &quot;yes/no&quot; questions, can you ask one of them what the other person would say? (The problem is that the truth-teller doesn't know, and knows he doesn't know; his only possible honest answer is &quot;I don't know.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What, if anything are the conventions in cases where &quot;yes/no&quot; is a constraint, but the issue of questions the respondent can't answer is not explicitly addressed?  Presumably, they must fall into one of the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You're not allowed to ask a question where the respondent might not be able to reply. This is simple, but... seems somehow more odd to me than the other arbitrary ridiculous rules, although I have no idea why that is.&lt;/li&gt;&#xA;&lt;li&gt;You &lt;em&gt;are&lt;/em&gt; allowed to ask questions like that, in which case, you must also assume they are allowed to convey &quot;I don't know&quot; (assume they have a veracity type that would make them honest).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I was deliberately using a simple example, but the puzzle that &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2665/knights-and-knaves-in-a-foreign-language#comment6175_2665&quot;&gt;this comment&lt;/a&gt; is on is what inspired the question. (I haven't solved it yet, so please don't share whether it's relevant to that particular solution; I'm more interested in what assumptions, if any are generally applied to other problems like this.&lt;/p&gt;&#xA;" OwnerUserId="5" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-04-20T19:43:11.290" Title="When logic problems require &quot;yes-no&quot; questions, are there conventions for questions that might get an &quot;I don't know&quot;?" Tags="|logical-deduction|" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12485" PostTypeId="2" ParentId="12484" CreationDate="2015-04-20T17:47:27.390" Score="4" Body="&lt;p&gt;For the situation mentioned and most other situations I would say that there is a contradiction in the question's formation. That is, there are three contradictory givens: the truth teller will always answer your question, the truth teller's answer will always be true, and the truth teller will always answer either yes or no. Thus it should be up to the author to fix the mistake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for how I think one should deal with an ill-formed problem of this type: I would ask for clarification. Until the author replies to resolve the contradiction, I would first try to solve the problem by resolution 1, and only afterwards by resolution 2, possibly with a goal of improving the limitations of the problem (eg, asking only 2 questions instead of 3).&lt;/p&gt;&#xA;" OwnerUserId="11100" LastEditorUserId="11100" LastEditDate="2015-04-20T17:53:29.440" LastActivityDate="2015-04-20T17:53:29.440" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12486" PostTypeId="2" ParentId="12477" CreationDate="2015-04-20T17:58:07.953" Score="5" Body="&lt;h2&gt;-- Partial solution --&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The cryptograms prove relatively easy to decrypt.  Their solutions are as follows:&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Cryptogram 1&lt;/h3&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;EACH OF THESE PUZZLES ARE FAIRLY BASIC DECIPHERING THE PLAINTEXT IS NOT THE CHALLENGE AS BASIC TOOLS CAN DO THAT JOB VERY QUICKLY THE PLAINTEXT IS MERELY THE EXPLANATION OF THE TRUE PUZZLE&lt;br&gt;&#xA; Key:&#xA;ABCDEFGHIJKLMNOPQRSTUVWXYZ&#xA;CORVHBNMEIQDTWLXPJAYKGUZSF&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;Cryptogram 2&lt;/h3&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;YOUR CHALLENGE IS TO FIGURE OUT THE PATTERN BEHIND EACH CIPHER TRY TO EXAMINE EACH BY ITSELF AS THE UNDERLYING PATTERNS ARE UNIQUE THEY ARE SIMPLE IDEAS BUT MAYBE TRICKY TO IDENTIFY BECAUSE THEY REQUIRE AN EPIPHANY&lt;br&gt;&#xA; Key:&#xA;ABCDEFGHIJKLMNOPQRSTUVWXYZ&#xA;THEQUICKBROWNFXJMPSVLAZYDG&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;Cryptogram 3&lt;/h3&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;IF YOU GET STUCK DONT GET TOO KEYED UP DONT BE LAZY OR JUMP TO CONCLUSIONS JUST KEEP GOING AND HASH IT OUT THE BEST ANSWER WILL HAVE ALL THREE PLAINTEXTS THE UNDERLYING PATTERN FOR EACH AND THE RULE THAT THE FIRST AND THIRD SHARE TO ENSURE A UNIQUE SUBSTITUTION GOOD LUCK PUZZLING&lt;br&gt;&#xA; Key:&#xA;ABCDEFGHIJKLMNOPQRSTUVWXYZ&#xA;CMIEPSLAFBKNHGRDOTZWQJVXUY&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Right away, the key for the second cryptogram jumps out.  It is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;obviously the common &lt;a href=&quot;http://en.wikipedia.org/wiki/Pangram&quot;&gt;pangram&lt;/a&gt; &quot;The quick brown fox jumps over the lazy dog&quot; with duplicate letters removed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The keys for the first and third cryptograms continue to elude me, though.  I suspect that is the part of the puzzle that is intended to be difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't seem like they can be following a similar pattern to the second cryptogram, because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; any phrase with repeated letters removed would likely still look like English for the first few characters (until there is a repeated letter), and would probably have most of the more common letters (ETAOIN SHRDLU) near the beginning of the key.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2015-04-20T17:58:07.953" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12487" PostTypeId="2" ParentId="12484" CreationDate="2015-04-20T18:16:08.897" Score="0" Body="&lt;p&gt;If I don't know is a possible response, the asker will include it, otherwise you must act as if they can't answer in this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is up to the askers to include &lt;strong&gt;all&lt;/strong&gt; the necessary details required to answer the puzzle, if they do not, it's very low quality. Answerers should never have to make assumptions on a whole genre of a specific puzzle.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2015-04-20T18:16:08.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12488" PostTypeId="2" ParentId="313" CreationDate="2015-04-20T18:27:56.443" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Does the number of letters in your answer match the number you are thinking of?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So if thinking of 1, answer is no.&lt;br&gt;&#xA;If thinking of 3, answer is yes.&lt;br&gt;&#xA;If thinking of 2, can't truthfully answer yes or no&lt;/p&gt;&#xA;" OwnerUserId="11690" LastEditorUserId="10281" LastEditDate="2015-04-20T19:13:38.877" LastActivityDate="2015-04-20T19:13:38.877" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12489" PostTypeId="2" ParentId="12396" CreationDate="2015-04-20T18:28:16.983" Score="7" Body="&lt;p&gt;Although the Ghost only has a two word vocabulary (Yes or No are its possible responses), I would ask a question that makes full use of its vocabulary;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you were to walk past each of these doors from left to right, and say 'yes' when you passed a 'door of life' and 'no' when you encountered a door leading to death, what would you say? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In this manner I've extracted all the information I need with one question.&lt;/p&gt;&#xA;" OwnerUserId="2078" LastActivityDate="2015-04-20T18:28:16.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12490" PostTypeId="2" ParentId="12483" CreationDate="2015-04-20T18:30:58.127" Score="13" Body="&lt;p&gt;First of all, notice that by repeatedly applying $\bf o$, we can always reduce a number to 2. This should be readily apparent, but here's a non-rigorous proof of the fact:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 has a binary representation requiring 2 digits (10), as does 3 (11). The length of the binary representation of a number grows slower than $n$, so $n\ge 3$ implies that $n \buildrel\bf o\over \longrightarrow m$ with $n \gt m$. Therefore, if $n\ge 3$, applying $\bf o$ will always reduce it by at least 1, allowing us through repeated applications to reduce the number to 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's consider $\bf o$ in reverse. In order to get to B, we need to apply $\bf o$ to a number whose binary representation has B digits: $2^{B-1} \le n \lt 2^B$. If we square a number, it can jump that range - for example if $B=5$ then $2^{5-1}=16$ and $2^{5}=32$, but $15^2=225$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if we apply $\bf o$ twice, then the range becomes far wider - $2^{2^{B-1}}\le n\lt 2^{2^{B}}$. At this point, there's no way to jump the range:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(2^{2^{B-1}}-1)^2=(2^{2^{B-1}})^2-2*2^{2^{B-1}}+1=2^{2^B}-2^{2^{B-1}+1}+1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because $B\ge 2$, $2^{2^{B-1}+1}\ge 2^{2^{1}+1}=2^3=8$, so $2^{2^B}-2^{2^{B-1}+1}+1&amp;lt;2^{2^B}-7&amp;lt;2^{2^B}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to summarize, apply $\bf o$ until it is less than $2^{2^B}$, then apply $\bf L$ until $2^{2^{B-1}}\le n\lt 2^{2^{B}}$, and then apply $\bf o$ twice. Of course, this is horribly inefficient, but it proves that it is always possible. For example, with B=197, $2^{2^B}$ is somewhere around $10^{10^{58}}$.&lt;/p&gt;&#xA;" OwnerUserId="1955" LastEditorUserId="1955" LastEditDate="2015-04-20T19:35:40.253" LastActivityDate="2015-04-20T19:35:40.253" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12491" PostTypeId="2" ParentId="12483" CreationDate="2015-04-20T18:58:46.840" Score="12" Body="&lt;p&gt;Whenever $x&amp;gt;2$, we have $o(x)&amp;lt;x$, so applying $o$ enough (for example, $A$ times) obtains&#xA;$$2$$&#xA;Applying $L$ to this a total of $B-1$ times gets&#xA;$$2^{2^{B-1}}$$&#xA;This is a one followed by $2^{B-1}$ zeros, so applying $o$ to this gets&#xA;$$2^{B-1}+1$$&#xA;This number is a 1, followed by $B-2$ zeroes, followed by a $1$, so applying o this gets&#xA;$$&#xA;B&#xA;$$&#xA;Thus, a recipe to get from $A$ to $B$ is given by &#xA;$$&#xA;o^2(L^{B-1}(o^A(A)))=B&#xA;$$&#xA;where $o^n$ denote repeated function application.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-04-20T19:10:56.643" LastActivityDate="2015-04-20T19:10:56.643" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12492" PostTypeId="1" AcceptedAnswerId="12510" CreationDate="2015-04-20T19:28:32.777" Score="13" ViewCount="934" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Some think I live in strange neighborhood.&lt;/p&gt;&#xA;&lt;p&gt;Everyone who lives here has some job they have to maintain.&lt;/p&gt;&#xA;&lt;p&gt;Many people tend to forget about mine though.&lt;/p&gt;&#xA;&lt;p&gt;Some people might think my job is to only turn on a light. Maybe they're right.&lt;/p&gt;&#xA;&lt;p&gt;I work close to home, which is nice.&lt;/p&gt;&#xA;&lt;p&gt;But I am envious of my neighbor as he is always on vacation!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Who or what am I?&#xA;I hope this one can stump people a little longer than my previous riddles. :)&lt;/p&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-21T04:44:36.657" Title="Riddle: A simple world for me" Tags="|riddle|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12494" PostTypeId="2" ParentId="12492" CreationDate="2015-04-20T19:43:42.677" Score="0" Body="&lt;p&gt;The answer may be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Electron&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some think I live in strange neighborhood.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Few know physics&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Everyone who lives here has some job they have to maintain.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Proton, Neutron, Electron, they all have a &quot;task&quot;: to keep matter stable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some people might think my job is to only turn on a light. Maybe they're right.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Electric current is made of moving electrons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I work close to home, which is nice.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Atoms are their homes, and electrons are very close to them!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I am envious of my neighbor as he is always on vacation!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The protons usually don't move, while the electrons do.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-04-20T19:43:42.677" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12495" PostTypeId="1" AcceptedAnswerId="12508" CreationDate="2015-04-20T19:58:23.567" Score="10" ViewCount="1295" Body="&lt;p&gt;One late night on a hill in Innsbruck, Doug struggled to organize his memories.  He picked up the “present” which looked like a cross between an IPad and a giant Blackberry (whatever the hell those were).   He only had one chance to recollect his memories and figure out the password. Apparently, the prize was astronomical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That morning while walking back to his friends' place after purchasing a new towel, Doug had run into a group of 6 humanoid or at least mostly sapient tourists and their guide.  After repeatedly muttering in confusion about how they shouldn’t be Doug’s problem, the guide introduced the 6 tourists and, for fun, organized a game for them and Doug. He also provided pendants and the password protected prize.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rules were as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The guide called ahead to buy them food and drinks at 3 different popular establishments where they would meet to play the game.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;One of the tourists would be told the password before leaving earth without Doug hearing.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When the red button was pressed on an individuals pendant, it take them to one of the 3 meals or to the final meeting place.  The meals were lunch at the Big Bang Burger Bar, dinner at Milliways, and after dinner drinks at some epic floating party. While Doug had to attend all three meals “in order” the others could control where they went.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;None of the tourists could go anywhere except to the 3 meals until they went to the final meeting. Every tourist who attends a meal must&#xA;meet all participants who attend that meal before leaving. They may&#xA;not come back to a meal after they left it or went to the final&#xA;meeting but they are not required to go to every meal.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;At each meal, the current password holder would randomly select exactly one of the participating tourists at that meal who was not on&#xA;his list of tourists who had already been told the password.  He&#xA;would then communicate the password to this participant along with&#xA;his list of all past password holders. That new password holder would&#xA;then go to a different meal and chose the next password holder unless&#xA;he already knew this was impossible.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If someone successfully passed on the password, he would then be required to proceed directly from that meal to the last meeting&#xA;place.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;At the last meeting, the tourist who knew the password but had not told anyone would tell Doug the password. Any participating tourist&#xA;who knew the password but had passed it on, would tell Doug a false&#xA;password.  Anyone who had never been told the password would not be&#xA;allowed to speak to Doug.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A paradox occurs under one of the following (surprisingly lax) conditions: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Something made of matter exists at the same time as another version of itself (without certain safeguards which are not present in this puzzle)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Somethings occurs which prevents its own occurrence.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If a paradox occurs the entire universe in which the&#xA;paradox occurs will cease to exist.  This means that since we know&#xA;Doug survived, anything that could happen that would result in a&#xA;paradox simply won’t happen (thanks to a silly interpretation of the&#xA;Copenhagen interpretation and the anthropic principle which should&#xA;not extend outside of this made up reality).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Before leaving Earth, Doug learned the six tourists’ names: &lt;strong&gt;Thor, Zem, Gally, Emjay, Wowbagger, and Lintilla&lt;/strong&gt;.  Zem was used to demonstrate how to send someone off with the pendant.  Finally, Doug caught Gally stealing his towel and he failed to get it back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the burger bar, Thor, Zem, and Emjay were in attendance.  While Emjay was trying to ignore Zem gurgling constantly at him, Thor got violent about the lack of alcoholic beverages. Probably because Zem has fluff for brains and can’t remember the password on his own, Doug saw a scribbled up Janx Spirit napkin be put in a satchel made of Doug’s tied-up towel. When Doug tried to get the towel back, Emjay reprimanded him for trying to cheat. Emjay then sent Zem and the satchel off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At Milliways, Thor, Gally, Lintilla, and Wowbagger were in attendance. Thor spent most of the night trying to get the attention of a cute raven haired girl he knew. She, however, did not seem to recognize or notice him.  In order: they ate their meal; Thor left in a huff; Doug realized the table cloth Lintilla kept fiddling with was his towel; the tourists left; and Doug followed after them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the party all tourists arrived except Wowbagger.  In order the following events occurred. Doug failed to get the satchel back from Zem as he couldn’t get through the crowd fast enough. Doug saw Thor meeting the girl from earlier; this somehow resulted in Thor walking out the door into midair. Killer robots destroyed the flying building. Doug witnessed Gally teleport away with his towel after failing to use it as a parachute. Doug finally remembered to use the pendant rather than go splat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doug only remembered 2 things from the final meeting.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Everyone told him a password.&lt;/li&gt;&#xA;&lt;li&gt;Just before he passed out from the nausea (which he of course blamed on the shapeshifting room and time travel), Lintilla put Doug’s head on a pillow made from his folded up towel.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When Doug awoke, still ihebriated, he found the list of tourists and the passwords they said tucked into his towel pillow. Unfortunately the above events are all he can remember. &lt;strong&gt;Who was the person who told Doug the correct password?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditDate="2015-04-20T20:05:53.117" LastActivityDate="2015-04-21T17:16:53.040" Title="Tipsy Time Travel" Tags="|logical-deduction|story|" AnswerCount="2" CommentCount="20" ContentLicense="CC BY-SA 3.0" />
  <row Id="12496" PostTypeId="1" AcceptedAnswerId="12502" CreationDate="2015-04-20T19:59:06.420" Score="6" ViewCount="341" Body="&lt;h1&gt;Story:&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Klauz Oppenheim is putting together an animation as an art project for school. He wants to portray his message using familiar objects that his audience will be able to relate to, so he bought &lt;strong&gt;16&lt;/strong&gt; sour skittles (individually wrapped) and placed them in a square. He will take 17 images, beginning with all of the skittles and taking away one per frame until none are left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Klauz does not want to confuse his audience by flashing a sequence of strange ugly shapes, so he would like to &lt;strong&gt;maximize the beauty&lt;/strong&gt; of his animation. For any frame we can assign a number, the amount of symmetries that the frame has (the order of its symmetry group), and so to an animation we can assign a number, the sum of the values of its frames.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each skittle has a color and so the way they are initially laid out is important. You can tell Klauz which skittle to remove and when, but unfortunately you cannot tell him how to place them on the surface. He already went and glued them to the surface, and if he buys any more skittles he'll be over his budget. This is how he arranged them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/obzSI.png&quot; alt=&quot;Colorative skittls&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;Summarized rules:&lt;/h1&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There are 17 frames. There is the one pictured above, and then one skittle disappears per frame until none are left. Your answer is the sequence of skittles that you will remove.&lt;/li&gt;&#xA;&lt;li&gt;Symmetry is measured per frame and then the sum is taken. For each frame the symmetry value is the order of its symmetry group, including the identity. That is, every line of symmetry gives 1 point, and every rotation gives 1 point.&lt;/li&gt;&#xA;&lt;li&gt;Your goal is to achieve the best possible symmetry score (and thus the most aesthetically pleasing animation).&lt;/li&gt;&#xA;&lt;li&gt;Color matters when counting symmetries.&lt;/li&gt;&#xA;&lt;li&gt;Color especially matters to us because as a reward for our help, Krauz has offered to let us eat the final skittle. Every color is a different flavor and our favorite flavor is grape (purple)! The frame with 1 skittle yields 12 points for purple, 10 points for green, 8 points for red, and 2 points for blue.&lt;/li&gt;&#xA;&lt;li&gt;The frame with no skittles yields 0 points.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; GentlePurpleRain points out that 1 skittle would have infinite score. For that matter, so would 0. To deal with these cases I have added a new rule and clarified 0.&lt;/p&gt;&#xA;" OwnerUserId="11100" LastEditorUserId="11100" LastEditDate="2015-04-20T21:11:47.043" LastActivityDate="2015-04-21T22:14:18.150" Title="Be an artist with Klauz" Tags="|visual|combinatorics|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="12497" PostTypeId="2" ParentId="12492" CreationDate="2015-04-20T20:03:15.907" Score="0" Body="&lt;p&gt;The answer could be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Electric FUSE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some think I live in strange neighborhood.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many cant find the fuse easily&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Everyone who lives here has some job they have to maintain.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;current passing through the fuse goes beyond what the device is able to handle, the fusible link will melt and the circuit opens thus preventing electrical component damage.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Many people tend to forget about mine though.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many don't know the purpose of a fuse.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some people might think my job is to only turn on a light. Maybe they're right.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; we check the fuse when one light alone shuts off.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I work close to home, which is nice.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fuse is in the Electric maintenance box.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I am envious of my neighbor as he is always on vacation!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is always a spare fuse that not used, next to the fuse that is being used.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11056" LastActivityDate="2015-04-20T20:03:15.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12498" PostTypeId="2" ParentId="12392" CreationDate="2015-04-20T20:06:55.213" Score="1" Body="&lt;h1&gt;Answer (the French national flag included in &lt;a href=&quot;https://www.drupal.org/project/bootstrap&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bootstrap&lt;/a&gt; on &lt;a href=&quot;https://www.drupal.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Drupal&lt;/a&gt;)&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;... Since not too many people seemed to be using the &lt;a href=&quot;https://www.drupal.org/project/bootstrap&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bootstrap-module&lt;/a&gt; of &lt;a href=&quot;https://www.drupal.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Drupal&lt;/a&gt; here: ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;true colors&quot; / &quot;3 colors&quot; / &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Tricolour_%28flag%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;tricolor&lt;/a&gt;&quot;, was trying to refer to a flag with 3 colors; which is also sometimes referred to as a &lt;a href=&quot;http://en.wikipedia.org/wiki/Triband_%28flag%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;triband&lt;/a&gt;. The references to a nationality were also in the &quot;being more narrow-minded&quot; than the &quot;world-wide web&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hint about a &lt;strong&gt;node&lt;/strong&gt;, as &quot;connecting the dots&quot; &lt;em&gt;of the riddle&lt;/em&gt;; referred to the editing of any &lt;strong&gt;node&lt;/strong&gt; in Drupal (± a technical Drupal-term for &lt;strong&gt;article&lt;/strong&gt;); which would have the shown the following composition (which has the minimum dimensions of ± 200 &lt;em&gt;X&lt;/em&gt; 30 pixels):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CiNxk.png&quot; alt=&quot;French flag in Drupal&amp;#39;s Bootstrap-module&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is exactly the French flag:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/a9Y6v.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice also the reference to the &lt;a href=&quot;http://en.wikipedia.org/wiki/AZERTY&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia page of AZERTY&lt;/a&gt;, which is mainly used in French-speaking countries. The country coloured &lt;strong&gt;blue&lt;/strong&gt; (cf. OP) on the main image of that Wikipedia-entry is France:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/saguE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="11597" LastEditorUserId="11597" LastEditDate="2015-04-20T20:53:41.000" LastActivityDate="2015-04-20T20:53:41.000" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12499" PostTypeId="1" AcceptedAnswerId="12501" CreationDate="2015-04-20T20:31:32.773" Score="-2" ViewCount="503" Body="&lt;p&gt;Opal awoke with a start- exhausted in her trek, she had momentarily slipped into a dream of a hastily written and terribly vague puzzle that was based entirely too much on an improvised scene she had performed in college.  She was glad to see that things in reality were at least somewhat different from the dream.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Opal and her guide Kimbe have been trekking through the jungle to the ceremonial cavern inside Mount Tarantoola, but have stopped to rest for the night.  Just as Kimbe succeeds in starting a fire, the two trekkers hear the crack of a whip in the distance; they are being pursued by Whiplash McGrath, who is set on destroying Opal, Kimbe, and whatever they hope to find inside the fabled mountain.  They won't be resting tonight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mountain lies in front of Kimbe and Opal, but between them and their destination are two furiously raging rivers, which they dare not wade through, as they are filled with leeches and piranhas of unhealthy sizes.   A long-forming sand bar separates the two rivers.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the glow of the fire, Kimbe can see what looks like four vines hanging over each river, but as he and Opal race to the closest riverbank, they see that this is not the case.  &lt;strong&gt;Over the closer river there are only three vines, as well as a large snake, while the far river has only one vine and three snakes.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Snakes can be swung on in place of vines&lt;/strong&gt;, but won't be happy about it, and must be scared off with the torch, &lt;strong&gt;an act which extinguishes the flame.&lt;/strong&gt;  They must move, but they also want to foil Whiplash's pursuit, not leaving any way for him to swing across safely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also to be considered is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trekkers have &lt;strong&gt;one match with which they can relight the torch.  A lit torch is necessary to safely enter the cavern&lt;/strong&gt;, the match is not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SO&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Can Opal and Kimbe reach the mouth of the cavern safely, and prevent Whiplash from doing the same?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clarifications&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Snakes act as vines, assuming you have the means (a lit torch) to&#xA;scare them.  They may also be scared without swinging on them.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In this iteration the vines cannot be burned.  Each vine/snake&#xA;will only support &lt;em&gt;one&lt;/em&gt; swing of &lt;em&gt;one&lt;/em&gt; person.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11452" LastEditorUserId="11452" LastEditDate="2015-10-22T18:11:18.157" LastActivityDate="2015-10-23T14:25:01.060" Title="The Trek to Mount Tarantoola - Redux!" Tags="|logical-deduction|strategy|lateral-thinking|story|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12500" PostTypeId="2" ParentId="12499" CreationDate="2015-04-20T20:42:00.617" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;kimbe swings (with the torch) to the sand bar, then plants the torch in the sand so it will stay lit. she swings back to the shore, then back to the sand bar with opal. they burn the (1) remaining vine and the (3) sand bar vines. then they swing across, leaving zero intact vine behind.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;alternative&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; kimbe gives the torch to opal (I hope she can at least hold a torch) then kimbe swings to the sandbar and back, grabs the torch then they both swing to the sandbar. they burn all vines expect the last 2, then swing across.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="9413" LastEditDate="2015-04-20T20:45:36.010" LastActivityDate="2015-04-20T20:45:36.010" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12501" PostTypeId="2" ParentId="12499" CreationDate="2015-04-20T21:17:35.603" Score="4" Body="&lt;p&gt;&lt;em&gt;This problem has been changed significantly; this answer reflects the original.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I expect I am missing something but the way I read it this is technically true...&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;They burn 3 vines on their shore.&lt;/li&gt;&#xA;&lt;li&gt;Kimbe swings to the sandbar.&lt;/li&gt;&#xA;&lt;li&gt;Opal swings to the sandbar.&lt;/li&gt;&#xA;&lt;li&gt;Kimbe burns 1 vine.&lt;/li&gt;&#xA;&lt;li&gt;Kimbe swings to the cavern shore.&lt;/li&gt;&#xA;&lt;li&gt;Opal swings to the cavern shore.&lt;/li&gt;&#xA;&lt;li&gt;They light the torch if it went out but it shouldn't have.&lt;/li&gt;&#xA;&lt;li&gt;They burn the remaining 2 vines.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This problem has been changed significantly; this answer reflects the revised question.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume they can light the torch from the fire and the torch cannot be thrown.  This leaves only one way to remove all 4 vines/snakes from the close shore with the torch lit but the match is used.  This leaves 2 snakes by the second river.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Both cross: Kimbe by snake; Opal by vine; torch extinguished.&lt;/li&gt;&#xA;&lt;li&gt;Kimbe by vine returns to first shore and relights torch via fire.&lt;/li&gt;&#xA;&lt;li&gt;Kimbe by vine returns to sand bank with torch lit.&lt;/li&gt;&#xA;&lt;li&gt;Both cross: Kimbe by snake; Opal by vine; torch extinguished.&lt;/li&gt;&#xA;&lt;li&gt;Kimbe relights torch with the match and they enter the cave.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This assumes Kimbe accepts all the danger as he is the employee.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditDate="2015-10-22T19:38:27.923" LastActivityDate="2015-10-22T19:38:27.923" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12502" PostTypeId="2" ParentId="12496" CreationDate="2015-04-20T21:38:37.337" Score="6" Body="&lt;p&gt;First try. Nothing clever here, just trying to end with the purple skittle and being greedy other than that. Eat in the numbered order. For groups of skittles with the same number, eat in any order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wBifV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gets $\boxed {44}$ points, if I can count:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;After eating group 1, you get $2$ points for diagonal symmetry.&lt;br&gt;&lt;/li&gt;&#xA;&lt;li&gt;Same for groups 2, 3, 4, and 5.&lt;br&gt;&lt;/li&gt;&#xA;&lt;li&gt;After eating group 6, you're left with a cross. The symmetry group is $D_4$ which has $8$ elements.&lt;br&gt;&lt;/li&gt;&#xA;&lt;li&gt;After eating M&amp;amp;M 7, you have reflective symmetry for $2$ points.&lt;/li&gt;&#xA;&lt;li&gt;After M&amp;amp;M 8, you get a line of three M&amp;amp;Ms. This has a symmetry group of $C_2 \times C_2$, which has $4$ elements.&lt;br&gt;&lt;/li&gt;&#xA;&lt;li&gt;After M&amp;amp;M 9, you have reflective symmetry for $2$ points.&lt;/li&gt;&#xA;&lt;li&gt;After M&amp;amp;M 10, you get $12$ points for finishing with grape.&lt;/li&gt;&#xA;&lt;li&gt;There were also 6 frames where you got $1$ point for no symmetry.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1752" LastEditorUserId="1752" LastEditDate="2015-04-20T21:46:30.503" LastActivityDate="2015-04-20T21:46:30.503" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12503" PostTypeId="2" ParentId="12496" CreationDate="2015-04-20T21:42:19.633" Score="2" Body="&lt;p&gt;No proof here, but it seems intuitively to me that this is the best answer.  If someone can provide a proof to the contrary (or to back me up), please do so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will refer to the skittles by letter as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A B C D&#xA;E F G H&#xA;I J K L&#xA;M N O P&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I will score a stage immediately before I remove a skittle.  So when I display the state of the &quot;board&quot;, that state has not yet been scored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that there is no way to achieve any symmetry until you have removed at least 4 skittles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you remove B,E,L,O, you will have reflective symmetry along the diagonal from bottom-left to top-right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of those first four steps have only 360° rotational symmetry, so they each score a single point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Points: 4&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;R   B P&#xA;  B P B&#xA;B B B&#xA;B B   R&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;After this, you can remove D,G,J,M, one-by-one, to maintain the reflective symmetry.  Each of these four steps provide 360° rotational symmetry and reflective symmetry along the bottom-left-to-top-right diagonal, resulting in 2 points for each step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Points: 4 + 2 + 2 + 2 + 2 = 12 total&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;R   B&#xA;  B   B&#xA;B   B&#xA;  B   R&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now you have 180° rotational symmetry as well, and reflection along the top-left-to-bottom-right diagonal, so this board scores 4 points. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After this, we can remove skittles in pairs.&lt;br&gt;&#xA;We start with the red ones.  Removing one of them leaves only 1 reflection, and the 360° rotation, so the next board scores 2 points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Points: 12 + 4 + 2 = 18 total&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    B&#xA;  B   B&#xA;B   B&#xA;  B   &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now we want to try to get a diamond of four, because that will give us maximum symmetry.  That means we need to remove one of the top or bottom skittles.  That will destroy all symmetry, but removing its reflection will restore 4-way symmetry.  Thus these 2 steps yield 5 points (4 for the current state, 1 for the next).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Points: 18 + 4 + 1 = 23 total&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  B&#xA;B   B&#xA;  B&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now we have 8-way symmetry (four rotations, four reflections).  As above, removing a single skittle will destroy most of the symmetry, &lt;em&gt;but&lt;/em&gt; the triangle shape that is left will still have a single line of reflectional symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We remove 1 skittle (it doesn't matter which one at this point):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Points: 23 + 8 = 31 total&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;B   B&#xA;  B&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We remove 1 more skittle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Points: 31 + 2 = 33 total&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;B&#xA;  B&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now we have 2 rotations and 2 reflections, for a total of 4 more points.  We remove 1 more skittle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Points: 33 + 4 = 37 total&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are now down to a single blue skittle, which is worth 2 points, making our final total&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;39 points&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Note that if we remove D,G,J,M, and then B,E,L,O, we get the same result.  It just means we achieve rotational symmetry before reflective symmetry instead of vice versa.&lt;/p&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="4421" LastEditDate="2015-04-21T22:14:18.150" LastActivityDate="2015-04-21T22:14:18.150" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="12504" PostTypeId="1" AcceptedAnswerId="12505" CreationDate="2015-04-20T22:10:48.957" Score="1" ViewCount="965" Body="&lt;p&gt;Consider the following sequence:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;6211, 4214, 6321, 4135, 6359&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;What is the next number in that sequence?&#xA;What are the rules of that sequence?&lt;/p&gt;&#xA;&lt;p&gt;I hope you'll enjoy my first question :)&lt;/p&gt;&#xA;&lt;h3&gt;EDIT 1:&lt;/h3&gt;&#xA;&lt;p&gt;Sorry guys, with those numbers, it is impossible to find the next one. So I have added the next one so it becomes possible.&lt;/p&gt;&#xA;&lt;p&gt;The new sequence is:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;6211, 4214, 6321, 4135, 6359, 4182&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h3&gt;EDIT 2:&lt;/h3&gt;&#xA;&lt;p&gt;Here is a hint:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Some numbers where chosen to confuse you. You shouldn't focus on them, but focus on the whole.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10400" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-20T23:34:24.337" Title="What is the next number in sequence?" Tags="|number-sequence|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12505" PostTypeId="2" ParentId="12504" CreationDate="2015-04-20T22:50:07.857" Score="4" Body="&lt;p&gt;The seventh number is &lt;strong&gt;6136&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last digit of each number is the $n$th digit after the decimal point of $\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first three digit numbers follow this pattern:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Place the $n$th Fibonacci number so that the ones' digit is the third digit of the number.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Then, insert digits so that the sum of the digits of the sum of the digits of the number is equal to to $n$. If the $n$th Fibonacci number is less than 10, then use a sequence alternating $6$ and $4$ to fill in the first digit, and then fill in the second digit as appropriate.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So the eleven terms of the sequence would be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;6211, 4214, 6321, 4135, 6359, 4182, 6136, 9215, 8343, 4555, 4898&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sequence ends there since the next Fibonacci number is 144 which would no longer ensure the sum-of-digits rule to be intact.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-04-20T23:34:24.337" LastActivityDate="2015-04-20T23:34:24.337" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="12507" PostTypeId="2" ParentId="11780" CreationDate="2015-04-20T23:40:49.707" Score="1" Body="&lt;p&gt;Well i dont want to be abnoxious but i dont think kevin's  answer can work out neither the enhaced grid do.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;its not about uniqueness but it is all about validity of solution .&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;this is a &lt;strong&gt;counter-example&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;###|123|456&#xA;###|456|789&#xA;###|789|123&#xA;---+---+---&#xA;147|###|562&#xA;258|###|394&#xA;369|###|817&#xA;---+---+---&#xA;471|562|###&#xA;582|394|###&#xA;693|817|###&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;this grid looks resolvable from the first look-at but it isnt when you try to solve it.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;minimally&lt;/strong&gt; the problem can be solved with &lt;strong&gt;27 blocks in independant places&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;how independant ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;blocks are dispersed enough from a one whole line or 3X3 square like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;.##|..#|...&#xA;##.|...|...&#xA;#.#|#..|..#&#xA;---+---+---&#xA;#..|.##|...&#xA;...|##.|.#.&#xA;...|#.#|#..&#xA;---+---+---&#xA;..#|...|.##&#xA;.#.|...|##.&#xA;...|.#.|#.#&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;proof of uniqueness:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;lets denote a group $(L, *)$ where $L$ is the set of $9$ random chosen primes '*' is ordinary multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $n$ variables , we need $n$ equations applied on different vars to solve the system . which is the case of 18 lines + 9 squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for exple:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a*b*c=30$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a*b=15$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b*c=10$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this system has unique solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(15/b)*b*(10/b)=30$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b=150/30=5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;-notes:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;to point it out again , there s no two equations applied on same variables, neither a cyclic equation without variables like : $\prod_9=\prod_9$ to guarantee system functioning.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There s difinitely soduko with more blocks if a such commutative ring :(L,+,°) can be found .&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9517" LastEditorUserId="9517" LastEditDate="2015-04-21T10:40:03.693" LastActivityDate="2015-04-21T10:40:03.693" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12508" PostTypeId="2" ParentId="12495" CreationDate="2015-04-21T03:47:53.510" Score="6" Body="&lt;p&gt;Whoever started with the password would have to give it up at the first event they attended and then go to the meeting directly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everyone except Wowbagger went to at least two events.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;strong&gt;Wowbagger started with the password.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;strong&gt;nobody else was given the password on Earth.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zem left the burger bar with the towel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gally left the party with the towel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gally did not show up at the burger bar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thor did not leave Milliway's or the party with the towel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neither Zem nor Emjay was at Milliway's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;strong&gt;the towel must have come to the burger bar directly from Earth.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The towel did not go from the burger bar directly to the final meeting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The towel did not go from the burger bar directly to Milliway's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;strong&gt;Zem brought the towel from the burger bar to the party.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;strong&gt;Gally brought the towel from the party to Milliway's.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;strong&gt;Gally did not arrive at the party with the password.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zem was given the password at the burger bar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The password could only have been exchanged once at the burger bar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;strong&gt;neither Thor nor Emjay received the password at the burger bar.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;strong&gt;Emjay did not arrive at the party with the password.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thor went to the party some time before he went to Milliway's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;strong&gt;Thor did not arrive at the party with the password.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The password must have been exchanged twice at the party.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;strong&gt;two tourists must have arrived at the party with the password.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;strong&gt;Zem and Lintilla must have arrived at the party with the password.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;strong&gt;Lintilla received the password at Milliway's.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The napkin could not have come to the burger bar from Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The napkin could not have come from the party, because Zem went there with the napkin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;strong&gt;the napkin came from Milliway's.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;strong&gt;Thor brought the napkin from Milliway's to the burger bar.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zem did not leave Earth with the towel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;strong&gt;Emjay brought the towel from Earth to the burger bar.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;strong&gt;Emjay did not have the password at the burger bar.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;strong&gt;Thor brought the password to the burger bar.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;strong&gt;Thor went from the burger bar directly to the meeting.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;strong&gt;Thor went from Earth to the party, then to Milliway's, where he received the password.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thor and Lintella both received the password at Milliway's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;strong&gt;Gally passed on the password at Milliway's.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thor gave up the password to Zem at the burger bar. Gally and Wowbagger both gave up the password at Milliway's. Zem and Lintella both arrived at the party with the password, and gave it to someone else.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; So &lt;strong&gt;only Emjay never gave the password to any of the other tourists.&lt;/strong&gt; So &lt;strong&gt;Emjay's password was correct.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11705" LastEditorUserId="99" LastEditDate="2015-04-21T17:16:53.040" LastActivityDate="2015-04-21T17:16:53.040" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12509" PostTypeId="2" ParentId="12492" CreationDate="2015-04-21T04:23:04.727" Score="3" Body="&lt;p&gt;You are a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Refrigerator&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some think I live in strange neighborhood.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is the clue I'm least sure of... Kitchens aren't really that strange (though I suppose, at a stretch, it could be a ghost busters reference to Slimer).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Everyone who lives here has some job they have to maintain.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All kitchen appliances have a job to do.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Many people tend to forget about mine though.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You don't actively use a fridge (as in turn it on or control its settings), unlike most appliances.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some people might think my job is to only turn on a light. Maybe they're right.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Open the door, on comes the light...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I work close to home, which is nice.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stays put in your home, always working.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I am envious of my neighbor as he is always on vacation!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most other appliances don't work all the time, just periodically when needed. The rest of the time, they're effectively on vacation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastActivityDate="2015-04-21T04:23:04.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12510" PostTypeId="2" ParentId="12492" CreationDate="2015-04-21T04:44:36.657" Score="17" Body="&lt;p&gt;Excellent riddle.  :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the Scroll Lock key&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some think I live in strange neighborhood.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Print Screen, Scroll Lock, and Pause/Break keys are in a neighborhood not often visited.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Everyone who lives here has some job they have to maintain.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each key has a specific function.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Many people tend to forget about mine though.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Scroll Lock is rarely used by most people.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some people might think my job is to only turn on a light. Maybe they're right.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you press Scroll Lock, the Scroll Lock light on the keyboard turns on.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I work close to home, which is nice.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Scroll Lock key is right above the Home key.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I am envious of my neighbor as he is always on vacation!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Scroll Lock key is right next to the Break key. (break = vacation)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2015-04-21T04:44:36.657" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12511" PostTypeId="1" AcceptedAnswerId="12587" CreationDate="2015-04-21T05:06:31.983" Score="5" ViewCount="246" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;You are given a $3\times n$ checkerboard, covered with $n$ red, $n$ white, and $n$ blue checkers. Call a board &lt;em&gt;patriotic&lt;/em&gt; if every column has a red, white and blue checker. You want to make the given board patriotic. To do this, you are allowed to do a series of moves, where a move consists of changing the places of two checkers &lt;strong&gt;which are in the same row&lt;/strong&gt;. Can you succeed? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the case that you can, you should demonstrate why, and in the case you can't, you should give an example of a starting board which can't achieve patriotism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/m8GRw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some examples when $n=7$. On the left is a starting position which is very unpatriotic: &lt;em&gt;none&lt;/em&gt; of the columns feature all three colors. It can, however, be rearranged to the patriotic board on the right. &lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-04-21T06:26:29.973" LastActivityDate="2015-04-22T18:42:18.517" Title="Patriotic Solitaire" Tags="|checkerboard|algorithm|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12512" PostTypeId="2" ParentId="12495" CreationDate="2015-04-21T05:10:24.263" Score="3" Body="&lt;p&gt;The person with the correct password is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Emjay&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I feel like the reasoning ended up being fairly simple once I found the right path to follow. Based on rules 5, 6, and 7 and the results of the final meeting:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because everyone tells Doug a password, everyone heard the password. That means that everyone participated in an exchange of the password at their last meal. In every case except for one, the exchange consisted of someone at their last meal giving the password to someone who was at their next-to-last meal. The exception is when the last person received the password - having nobody else to give it to, they would have needed to go straight to the final meeting.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Keeping this in mind, let's trace the path of the towel and the napkin:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Zem was given the password on the napkin at lunch and put it in the towel-satchel, then took them both to the party. From there, Gally took the towel to dinner, where Lintilla got it. The napkin, since it went to the party and didn't originate from the burger bar, must have come from the dinner. The only person who could have taken it to lunch from dinner is Thor.&lt;br&gt;&lt;br&gt;Towel - lunch with Zem -&gt; party with Zem -&gt; party with Gally -&gt; dinner with Gally -&gt; dinner with Lintilla&lt;br&gt;&lt;br&gt;Napkin - dinner with Thor -&gt; lunch with Thor -&gt; lunch with Zem -&gt; party with Zem&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's take a closer look at lunch now:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; With three people at lunch, only one exchange of the password can happen, so someone is left out. It's clear that Zem was given the password, so either Emjay or Thor must have told it to him. Thanks to the napkin, we know that Thor must have just come from dinner. Also, look at Thor's interaction with the raven-haired girl - she didn't know him at the dinner, but did at the party. So they must have gone in opposite directions - she from the dinner to the party, and he from the party to the dinner. This means that Thor's schedule was party first, then dinner, then lunch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Putting all the parts together:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because lunch was Thor's last meal, he must have either received or given the password there. Because we know Zem received the password there and only one exchange happened at lunch, Thor must have given the password to Zem. This means Emjay did not give or receive the password at lunch. Emjay did not go to dinner, so he must have received the password at the party. Because he cannot both receive and give the password at the same place, Emjay did not give the password to anyone else.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1955" LastActivityDate="2015-04-21T05:10:24.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12513" PostTypeId="1" CreationDate="2015-04-21T05:13:03.840" Score="2" ViewCount="1879" Body="&lt;p&gt;This is from a republishing of an old taxonomy book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An African elephant weighs 4536 kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A fly weighs 64.8 mg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the weight of an African elephant with a fly on his head?&lt;/p&gt;&#xA;" OwnerUserId="11708" LastActivityDate="2015-04-21T15:49:14.407" Title="The African elephant and the fly" Tags="|mathematics|lateral-thinking|knowledge|" AnswerCount="7" CommentCount="12" ClosedDate="2015-04-21T15:38:10.943" ContentLicense="CC BY-SA 3.0" />
  <row Id="12515" PostTypeId="2" ParentId="12513" CreationDate="2015-04-21T05:38:35.600" Score="20" Body="&lt;p&gt;The weight is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4536 kg.  No matter what the African elephant has on his head, he will still weigh 4536 kg.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2015-04-21T05:38:35.600" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12517" PostTypeId="2" ParentId="12511" CreationDate="2015-04-21T06:45:09.683" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes, it can always done for any $n$. Moreover, this also holds when the number of row and colors $3$ is replaced by any number $r$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It suffices to show that one can pick one checker out of each row so that their colors are distinct. Then, one can swap those checkers into the first column and recurse on the subproblem of the remaining columns with $n-1$.&#xA;&#xA; We need to a find a one-to-one matching between rows and colors so that each row contains at least one checker of that color. Apply &lt;a href=&quot;http://www.cut-the-knot.org/arithmetic/elegant.shtml&quot;&gt;Hall's Marriage Theorem&lt;/a&gt;. In any $k$ rows, there $kn$ checkers and so at least $k$ colors of checkers appear (since there's $n$ per color). So, the theorem's conditions are satisfied and a matching exists.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-04-21T06:45:09.683" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12518" PostTypeId="2" ParentId="12513" CreationDate="2015-04-21T07:18:24.800" Score="2" Body="&lt;p&gt;It will be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Weight of African elephant + weight of fly &lt;strong&gt;which are under test for weight&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because the one which you mentioned may not be under test for weight. :)&#xA;you mentioned &lt;strong&gt;An African elephant weighs 4536 kg.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The one which is tested may weigh 4500 kg or anything that is in the possible range for an African elephant's weight.&lt;/p&gt;&#xA;" OwnerUserId="10433" LastEditorUserId="2357" LastEditDate="2015-04-21T15:49:14.407" LastActivityDate="2015-04-21T15:49:14.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12519" PostTypeId="2" ParentId="12513" CreationDate="2015-04-21T07:48:03.200" Score="11" Body="&lt;p&gt;I like elephants.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It appears that the measurements were originally expressed in imperial units, with the weight of an African elephant specified as 10000 &lt;a href=&quot;http://en.wikipedia.org/wiki/Pound_%28mass%29&quot;&gt;lb&lt;/a&gt; (about 4536 kg), and the weight of a fly as 1 &lt;a href=&quot;http://en.wikipedia.org/wiki/Grain_%28unit%29&quot;&gt;grain&lt;/a&gt; (about 64.8 mg). The weights were later converted to kg and mg respectively (possibly by the republisher?) without the appropriate rounding considerations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not only is the fly's weight negligible for the calculation of the total weight, but also the last 2 or 3 digits in the African elephant's weight are, at best, insignificant.&lt;/p&gt;&#xA;" OwnerUserId="4394" LastActivityDate="2015-04-21T07:48:03.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12520" PostTypeId="2" ParentId="12396" CreationDate="2015-04-21T08:39:55.157" Score="1" Body="&lt;p&gt;I would chose any two doors(X,Y) and ask the following question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If X,Y does not lead to the same place, does X lead to freedom ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If he Says &quot;YES&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chose X&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If he says &quot;NO&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Chose Y &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If he keeps quite, it means that my first condition was false. Thus,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chose Z (the left out door)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9776" LastActivityDate="2015-04-21T08:39:55.157" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12521" PostTypeId="2" ParentId="12396" CreationDate="2015-04-21T09:25:03.280" Score="18" Body="&lt;p&gt;I will try to be in the spirit of the riddle: Ask only one question and get a real YES/NO aka true/false response:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If I take the first door and my brother takes one of the first two doors, will at least one of us get to freedom ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the first door leads to freedom, the answer is yes.&lt;/li&gt;&#xA;&lt;li&gt;If the third door leads to freedom, the answer is no.&lt;/li&gt;&#xA;&lt;li&gt;If the second door leads to freedom, the answer would be maybe/don't know, so the ghost cannot answer with yes/no.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1998" LastActivityDate="2015-04-21T09:25:03.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12522" PostTypeId="2" ParentId="12513" CreationDate="2015-04-21T09:30:39.353" Score="0" Body="&lt;p&gt;The elephant will weigh 4536000064.8 mg. &lt;/p&gt;&#xA;" OwnerUserId="11016" LastEditorUserId="4551" LastEditDate="2015-04-21T09:50:09.440" LastActivityDate="2015-04-21T09:50:09.440" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12523" PostTypeId="1" AcceptedAnswerId="12572" CreationDate="2015-04-21T09:32:45.610" Score="10" ViewCount="2752" Body="&lt;p&gt;There's a well-known British game show called &lt;a href=&quot;http://en.wikipedia.org/wiki/Only_Connect&quot; rel=&quot;nofollow noreferrer&quot;&gt;Only Connect&lt;/a&gt;, which tests your ability to come up with the connections to things which are very hard to relate.&lt;/p&gt;&#xA;&lt;p&gt;One of the staples of the show is the Connect Wall, in which a grid of sixteen words and names is given, each of which fits into one of four clearly defined categories, with four words per category. The words in each category are jumbled up into the grid, and contestants generally have 2 minutes and 30 seconds to figure out what the groups of words are.&lt;/p&gt;&#xA;&lt;p&gt;An example of such a wall might be this:&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;Myers&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;Franklin&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;Jackson&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;Union&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;Johnson&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;Dundas&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;Winnie&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;Gamble&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;Hagar&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;Young&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;Wellesley&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;Bay&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;Fox&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;Bloor&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;Thomas&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;Decker&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;What are the four groups of words, and the four categories that connect them?&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Beware of red herrings! A group of four words that &lt;em&gt;appear&lt;/em&gt; to be related might actually belong to separate groups if you find the remaining groups don't make any sense. And there may be &lt;em&gt;five or six&lt;/em&gt; words that appear to be in a group but actually one of them is in a different one.&lt;/p&gt;&#xA;&lt;p&gt;If you're going to put anything behind spoilers, leave the word groups exposed but hide the category explanations.&lt;/p&gt;&#xA;&lt;p&gt;Note: the puzzles on the actual Only Connect game show generally have a British bent to their themes; this one has a North American, and possibly slightly Canadian bent — for you British folk that are familiar with this format, beware that this might make it harder for you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="69582" LastEditDate="2021-08-20T15:35:07.500" LastActivityDate="2021-08-20T15:35:07.500" Title="Only Connect - Connect Wall #1" Tags="|pattern|knowledge|connect-wall|" AnswerCount="5" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="12524" PostTypeId="2" ParentId="12513" CreationDate="2015-04-21T09:49:35.073" Score="7" Body="&lt;p&gt;The weight of the combination &quot;an African elephant with a fly on his head&quot; is 4536.0000648 kg, if it was the particular elephant and the fly that was mentioned. Given that the weight of the elephant was likely not to have been 4536.0000000 kg because that's a suspiciously round figure and far beyond the precision of measurement and likely to change as the elephant breathes, it would make sense to report the combined weight simply as 4536 kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The weight of the elephant that is specified as having a fly on his head is &lt;code&gt;4536 kg&lt;/code&gt;, if it is elephant whose weight was mentioned before..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take statements like &quot;An African elephant weighs 4536 kg&quot; to mean there's one such elephant, not as a general fact about elephants nor of a specific elephant referred to, then we can't say anything precise about the elephant and fly mentioned later. Actually it's not clear which elephant is later asked about as &quot;an elephant&quot;. But, a reasonable estimate could come from the statistic that &lt;a href=&quot;http://en.wikipedia.org/wiki/Elephant&quot;&gt;African elephants weigh up to 4000–7000 kg&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-04-21T09:49:35.073" CommentCount="1" CommunityOwnedDate="2015-04-21T09:49:35.073" ContentLicense="CC BY-SA 3.0" />
  <row Id="12525" PostTypeId="2" ParentId="11887" CreationDate="2015-04-21T09:58:16.470" Score="3" Body="&lt;p&gt;In the hope of being told what the real answer is supposed to be, here is a strange rule that produces the correct numbers (using &quot;min&quot;, since you mentioned it):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_0 = 9$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_{n+1} = \min((a_n \cdot 100)\,\%\,1000000 + (\lfloor a_n / 10000 \rfloor \cdot 11)\,\%\,100, 976585)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_5$ is the first item that belongs to the repeating sequence, but it is not the beginning of that sequence. The sequence starts over for the first time at $a_{33}$.&lt;/p&gt;&#xA;" OwnerUserId="11469" LastActivityDate="2015-04-21T09:58:16.470" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12526" PostTypeId="1" AcceptedAnswerId="12535" CreationDate="2015-04-21T09:59:01.350" Score="-1" ViewCount="316" Body="&lt;p&gt;Let us suppose there are $n$ statements on the list. Each statement may be true or false. The first statement declares:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I am true.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, it could be true, or it could be false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After this, for every statement $k$, $2\leq k\leq n$, let $p_k$ be the smallest prime factor of $k$. Every statement $k$ declares:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The number of false statements on this list is $\geq n(\frac{p_k-1}{p_k})$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the minimum and maximum possible value of $n$?&lt;/strong&gt; (such the system is consistent)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Decimals are rounded up.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A set of statements may have more than one way of being consistent, but there must exist atleast one.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4269" LastEditDate="2015-04-21T15:05:13.467" LastActivityDate="2015-04-21T19:19:05.753" Title="Extremely hard prime factor based true-false statements" Tags="|mathematics|optimization|" AnswerCount="3" CommentCount="5" ClosedDate="2015-04-21T19:31:23.203" ContentLicense="CC BY-SA 3.0" />
  <row Id="12527" PostTypeId="2" ParentId="70" CreationDate="2015-04-21T10:03:04.523" Score="0" Body="&lt;p&gt;Since the set of colours is known in advance, then:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Assign a number to each colour&lt;/li&gt;&#xA;&lt;li&gt;Assign a leader&lt;/li&gt;&#xA;&lt;li&gt;The leader arranges everyone else in numerical order, then stands at the end of the line&lt;/li&gt;&#xA;&lt;li&gt;The leader is passed down the line until in the correct place&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If every hat is of a different colour, then each logician knows their colour from their place in the line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If more than 2 hats are of the same colour, at least one logician will be standing between two people with the same colour hat, so knows that they too have that colour&lt;/li&gt;&#xA;&lt;li&gt;The logicians agree in advance that if they only see colours repeat in groups of two, they call out the colour ahead of them in line rather than their own. That way the second member of a colour should be correct.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Non-mathematical, but does assume a lot of collaboration is allowed.&lt;/p&gt;&#xA;" OwnerUserId="11715" LastEditorUserId="11717" LastEditDate="2015-04-21T10:42:55.363" LastActivityDate="2015-04-21T10:42:55.363" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12529" PostTypeId="2" ParentId="12526" CreationDate="2015-04-21T11:50:30.667" Score="0" Body="&lt;h1&gt;Partial Answer&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;minimum&lt;/strong&gt; is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$P_2$=2&lt;br&gt;&#xA;$\frac{P_2-1}{P_2}=\frac{1}{2}$&lt;br&gt;&#xA;The second statement says that half statements are false. &#xA;Just let the 1st statement be false and the second true: the system is consistent.&lt;/p&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-04-21T11:50:30.667" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12530" PostTypeId="2" ParentId="12526" CreationDate="2015-04-21T11:54:28.827" Score="1" Body="&lt;p&gt;I may be confused. If so correct me.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Minimum:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $n=1$ is consistent. The only statement is &quot;I am true&quot; and this can be either true or false without contradiction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Maximum:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is no maximum. For any $n$ just have all of the even statements be true while all of the odds are false, then there are $\lceil\frac{n}{2}\rceil$ false statements and $\lfloor\frac{n}{2}\rfloor$ true statements which agrees with what the even statements say.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11100" LastActivityDate="2015-04-21T11:54:28.827" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12531" PostTypeId="2" ParentId="12416" CreationDate="2015-04-21T12:06:06.370" Score="0" Body="&lt;p&gt;On the lines of previous answers. &#xA;I guess the answer could be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (Facial) &lt;strong&gt;Expressions&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I convey the invisible,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We convey different emotions through our expressions&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I convey what can be felt, But not seen,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Emotions can only be felt.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Inside my heart, I am not as nice as I could seem to be,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One might give a different expression than what is in one's heart.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is because I deliver the invisible,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reference to emotions again&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At the speed of a mere blink,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Expressions are quick.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am very powerful,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Expressions are powerful&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And can bring comfort,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A good expression including a smile brings comfort.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Across my horizons, I am free,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's a free will.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But still vulnerable to your opinion.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The other person can misjudge someone's expression.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10973" LastActivityDate="2015-04-21T12:06:06.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12532" PostTypeId="1" AcceptedAnswerId="12533" CreationDate="2015-04-21T12:10:27.560" Score="2" ViewCount="5912" Body="&lt;p&gt;Four people need to cross a rickety rope bridge to get back to their camp at night. Unfortunately, they only have one flashlight and it only has enough battery  left for seventeen minutes. The bridge is too dangerous to cross without a flashlight, and it's only strong enough to support two people at any given time. Each of the campers walks at a different speed. One can cross the bridge in 1 minute, another in 2 minutes, the third in 5 minutes, and the slowest needs 10 minutes to cross. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How do the campers make it across in 17 minutes?&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="9346" LastEditorUserId="20274" LastEditDate="2016-05-09T14:22:25.740" LastActivityDate="2016-05-09T14:22:25.740" Title="The case of the 4 campers and the rickety bridge!" Tags="|logical-deduction|optimization|river-crossing|" AnswerCount="2" CommentCount="3" ClosedDate="2015-04-21T12:40:03.377" ContentLicense="CC BY-SA 3.0" />
  <row Id="12533" PostTypeId="2" ParentId="12532" CreationDate="2015-04-21T12:24:22.303" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1 and 2 cross (2 minutes)&lt;br&gt;&#xA;1 goes back with the flashlight (3 minutes)&lt;br&gt;&#xA;5 and 10 cross (13 minutes)&lt;br&gt;&#xA;2 goes back with the flashlight (15 minutes)&lt;br&gt;&#xA;Finally, 1 and 2 cross together again (17 minutes)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The trick here is with the two slowest - you don't want to have to add both of their times on if you don't have to.&lt;/p&gt;&#xA;" OwnerUserId="11717" LastEditorUserId="10281" LastEditDate="2015-04-21T12:37:26.453" LastActivityDate="2015-04-21T12:37:26.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12534" PostTypeId="2" ParentId="12532" CreationDate="2015-04-21T12:26:57.130" Score="2" Body="&lt;p&gt;$1$ and $2$ cross together with the torch (takes &lt;strong&gt;2&lt;/strong&gt; minutes)&lt;br&gt;&#xA;$2$ goes back, bringing the flashlight (&lt;strong&gt;2&lt;/strong&gt; minutes)&lt;br&gt;&#xA;$5$ and $10$ cross the bridge together, using the torch  (&lt;strong&gt;10&lt;/strong&gt; minutes)&lt;br&gt;&#xA;$1$ goes back with the flashlight (&lt;strong&gt;1&lt;/strong&gt; minute)&lt;br&gt;&#xA;$1$ and $2$ cross the bridge with the torch (&lt;strong&gt;2&lt;/strong&gt; minutes)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total: &lt;strong&gt;17&lt;/strong&gt; minutes&lt;/p&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-04-21T12:26:57.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12535" PostTypeId="2" ParentId="12526" CreationDate="2015-04-21T12:32:22.797" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The minimum value of $n$ is $2$.&lt;br&gt;&#xA; There is no maximum value of $n$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;: &lt;em&gt;(minimum)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $n = 2$, we are presented with a list of statements:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S_1$: I am true.&lt;br&gt;&#xA;$S_2$: The number of false statements in this list is $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\therefore$ because the statements resolve consistently when $S_1$ is false, the minimum value for $n$ must be $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;: &lt;em&gt;(maximum)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider an $n &amp;gt; 2$. $n$ is either even or odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For every even $m$, $p_k = 2$, making the $S_m$ statement:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S_m$: The number of false statements in this list is $n(\frac{2-1}{2}) = \frac{n}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $n$ is even, then also:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S_n$: The number of false statements in this list is $\frac{n}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Making each even-numbered true and each odd-numbered statement false makes $\frac{n}{2}$ statements false, agreeing with each even statement. In order for this list to not be consistent, one of the odd-numbered statements must also be true, but any odd-numbered statement will have a $p_k &amp;gt; 2$ since $2$ does not divide any odd number and $1$ is not prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because we can always choose a larger even $n$, there is no maximum value of $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Note&lt;/strong&gt;: A similar argument could be made about odd numbers that would satisfy the conditions under the revised $\ge n(\frac{p_k-1}{p_k})$.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-04-21T19:19:05.753" LastActivityDate="2015-04-21T19:19:05.753" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="12536" PostTypeId="2" ParentId="8440" CreationDate="2015-04-21T12:57:01.353" Score="0" Body="&lt;p&gt;Scala solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://ideone.com/X48DIB&quot; rel=&quot;nofollow&quot;&gt;https://ideone.com/X48DIB&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;case class State(a: Int, b: Int, sum:Int, prod:Int)&#xA;&#xA;val allPossibleStates:Seq[State] = for {&#xA;  i &amp;lt;- 1 to 99;&#xA;  j &amp;lt;- i to 99&#xA;} yield State(i,j,i+j,i*j)&#xA;&#xA;(1 to 7).foldLeft(allPossibleStates)( (states, _) =&amp;gt;  &#xA;  states&#xA;   .groupBy(_.prod).filter(_._2.size &amp;gt; 1).flatMap(_._2)&#xA;   .groupBy(_.sum).filter(_._2.size &amp;gt; 1).flatMap(_._2)&#xA;   .toSeq&#xA;).groupBy(_.prod).filter(_._2.size == 1).flatMap(_._2)   &#xA; .map(println)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="11615" LastActivityDate="2015-04-21T12:57:01.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12537" PostTypeId="2" ParentId="12513" CreationDate="2015-04-21T14:53:03.680" Score="1" Body="&lt;p&gt;The elephant with a fly on its head will weigh actually a bit less than without the fly!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The fly will exert an upwards gravitational force to the elephant, thus countering part of the force by Earth. If we interpret &lt;a href=&quot;http://en.wikipedia.org/wiki/Weight&quot; rel=&quot;nofollow&quot;&gt;the definition in Wikipedia&lt;/a&gt; (&quot;the &lt;em&gt;weight&lt;/em&gt; of an object is usually taken to be the force on the object due to gravity&quot;) in a suitable way, it means the weight of the elephant is decreased by the fly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2228" LastActivityDate="2015-04-21T14:53:03.680" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12538" PostTypeId="2" ParentId="12523" CreationDate="2015-04-21T15:16:05.953" Score="0" Body="&lt;p&gt;Ill put up my partial answer in case it might help others in solving. Here are my groups so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Union, Bloor, Young, Dundas&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; City:&#xA; (Toronto) - Union [Station] (Toronto's main train station), Bloor [Station] (Another Train station near/in Toronto). Young and Dundas are also streets in Toronto.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Franklin, Winnie, Thomas, Fox&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Shows:&#xA; (Kids shows) - Franklin (The turtle), Winnie (The pooh), Thomas (The tank engine). Not sure about Fox in this category (maybe Swiper from Dora)? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Johnson, Jackson, Hagar, Myers&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;People/Names:&#xA; Johnson, Jackson, Hagar, Myers (Name of a automotive company)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Left overs:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wellesley , Decker, Bay, Gamble --Not to sure about how these are related if the previous groups are correct..&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-04-21T17:26:03.457" LastActivityDate="2015-04-21T17:26:03.457" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12539" PostTypeId="2" ParentId="12523" CreationDate="2015-04-21T15:20:34.530" Score="2" Body="&lt;p&gt;Ok, I'm willing to give it a try. I assume that 'Young' is not correctly spelled and I'm having some problems with one of the other words, but I have managed to stay within one North American city to solve the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jackson, Johnson, Wellesley, Young&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Actually Yonge instead of Young: Streets in Toronto.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Franklin, Gamble, Hagar, Thomas&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Avenues in Toronto.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bay, Bloor, Dundas, Union&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Subway stations in Toronto.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Decker, Fox, Myers, Winnie&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fitness or wellness establishments in Toronto: Fox Fitness, Myers Yoga Studio, Winnie Spa. Not sure about Decker though.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11687" LastEditorDisplayName="user88" LastEditDate="2015-04-21T18:50:24.587" LastActivityDate="2015-04-21T18:50:24.587" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12540" PostTypeId="2" ParentId="12513" CreationDate="2015-04-21T15:31:40.900" Score="5" Body="&lt;p&gt;The avarage height of an elephant is $4 m$, we can say that the center of mass of the elephant is roughly at $2m$ from the tip of his head. Being the fly millimiters long his center of mass can be ignored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since here the weight is intended in $N$, I'm going to translate it in mass ($kg$) units. Everything is calculated on earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass of the elephant on earth $= M1 = \frac{4536N}{9,8\frac{m}{s^2}} = 462,85kg$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass of the fly on earth $= M2 = \frac{0.0000648N}{9,8\frac{m}{s^2}}= 6,61\times 10^{-6}kg$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Distance between the centers of mass $=r=2m$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravitational constant $=G= 6.673×10^{−11} N\cdot(\frac{m}{kg})^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravitational pull of the fly on the elephant is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F=G\cdot\frac{M1\cdot M2}{r^2}=6.673\times10^{-11}N\cdot(\frac{m}{kg})^2\cdot\frac{462,85kg\cdot 6,61\times 10^{-6}kg}{(2m)^2}\simeq 5,1\times 10^{-14}N$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the elephant receives a gravitational pull from the fly equal to $5,1\times 10^{-14}N$ which is opposite to the earth gravitational pull. Now the elephant weighs:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$4536N-5,1\times 10^{-14}N=4535,99999999999949N$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still not the same!&lt;/p&gt;&#xA;" OwnerUserId="4652" LastActivityDate="2015-04-21T15:31:40.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12541" PostTypeId="2" ParentId="12221" CreationDate="2015-04-21T18:14:06.220" Score="1" Body="&lt;p&gt;Assign each tree a tuple $(a,b,c)$ where $a$ is the number of branch-points on the longest path, $b$ is the maximum multiplicity of branches at the highest branch point, and $c$ is the number of branches at the highest branch point with multiplicity $b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The initial tree is defined to have 6 branch points, each with multiplicity 1.  Thus, the original tree has tuple (6,1,1), and your goal is to reduce $a$ to 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can cut at any branch point that does not have a branch point above it.  Thus, if you make a cut at a branch with multiplicity 2, you have effectively removed that branch point, since on your next turn you may cut at the next branch point lower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you order the set of all possible trees by sorting on $a$, then $b$, then $c$, you can always reduce the tree to one of a lower order: find one of the highest branch points of maximum multiplicity and cut a branch off it.  This will reduce $c$ by 1.  Unless $c$ was already 1, in which case it would reduce $b$ by 1.  Unless $b$ were 2, in which case it would reduce $a$ by 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When Satan adds branches, he can only add branches with lower multiplicity than the branch you cut.  In other words:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If $c&amp;gt;1$ before your cut, you reduce the tree to $(a,b,c-1)$ and Satan adds branches at depth $a$, but with multiplicity $b-1$, so the sort order does not change.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $b&amp;gt;2$ and $c=1$ before your cut, you reduce the tree from $(a,b,c)$ to $(a,b-1,X)$.  Satan can then increase that to $(a,b-1,X+Y)$, but that is still lower order than what you started with.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $b=2$ and $c=1$ before your cut, you reduce the tree from $(a,b,c)$ to $(a-1,X,1)$, which Satan can increase to $(a-1,X+Y,1)$, but that is still less than what you started with.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In the example given, the tree goes from (6,1,1) to (5,1,1) to (5,3,1) to (5,2,1) to (5,2,4).  Your next cut would reduce it to (5,2,3), and 3 further of your cuts, the tree would be down to $(4,X,1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He can make it take an arbitrarily long time, but each branch you cut will bring the tree closer to $a=0$.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2015-04-21T18:14:06.220" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12542" PostTypeId="1" CreationDate="2015-04-21T20:24:03.067" Score="3" ViewCount="912" Body="&lt;h2&gt;Definitions&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;For a $9\times9$ grid $G$ (not necessarily respecting sudoku rules), a subset of cells $S$ is defined to be &lt;strong&gt;sudoku-friendly&lt;/strong&gt; if the values in $S$ do not contradict the rules of sudoku.&lt;br&gt;&#xA;This means that in $S$ no number occurs twice in the same row, column or box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the grid $G$:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;111111111&#xA;222222222&#xA;333333333&#xA;444444444&#xA;555555555&#xA;666666666&#xA;777777777&#xA;888888888&#xA;999999999&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then the following is an example of a sudoku-friendly subset $S$ of $G$:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1        &#xA; 2&#xA;  3&#xA;   4&#xA;    5&#xA;     6&#xA;      7&#xA;       8&#xA;        9&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;A &lt;strong&gt;net&lt;/strong&gt; is a mapping on a grid that produces a subset of cells. You can imagine it like a piece of paper with holes in it, lying on the grid, which hides some cells.&lt;br&gt;&#xA;This is usually visualized with &lt;strong&gt;O&lt;/strong&gt; &lt;em&gt;(visible)&lt;/em&gt; and &lt;strong&gt;X&lt;/strong&gt; &lt;em&gt;(blocked)&lt;/em&gt; cells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A net has a &lt;strong&gt;visible set&lt;/strong&gt;, $V_N(G)$ which is defined as the set of cells in a grid $G$ which map to an O in $N$, namely the set of visible cells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The net which retrieves the above visible set $S$ would look like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;OXXXXXXXX&#xA;XOXXXXXXX&#xA;XXOXXXXXX&#xA;XXXOXXXXX&#xA;XXXXOXXXX&#xA;XXXXXOXXX&#xA;XXXXXXOXX&#xA;XXXXXXXOX&#xA;XXXXXXXXO&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;A &lt;strong&gt;sudoku net&lt;/strong&gt; is a net whose visible set is sudoku-friendly.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Challenge&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Construct a &lt;strong&gt;sudoku net&lt;/strong&gt; $N$ of &lt;strong&gt;maximum&lt;/strong&gt; and &lt;strong&gt;minimum&lt;/strong&gt; possible blocks such that there exists some 9x9 $G$ that is an impossible sudoku puzzle (namely no solution).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-&lt;strong&gt;Bounty awarding question&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;is there a grid which has unique solution whatever are the arrangements of legal patterns in friendly-sudoku  ?&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9517" LastEditorUserId="19114" LastEditDate="2020-01-04T22:47:31.680" LastActivityDate="2020-01-04T22:47:31.680" Title="Unsolvable sudoku net with maximum blocks" Tags="|game|sudoku|logic-grid|solvability|" AnswerCount="2" CommentCount="22" ContentLicense="CC BY-SA 3.0" />
  <row Id="12543" PostTypeId="1" AcceptedAnswerId="12627" CreationDate="2015-04-21T20:24:06.380" Score="9" ViewCount="994" Body="&lt;p&gt;Crazy or not, you know who I am,&lt;br&gt;&#xA;I can use a knife, not just for jam.&lt;br&gt;&#xA;Death is nothing new to me.&lt;br&gt;&#xA;You know the glasses that help me see.&lt;br&gt;&#xA;Rogue or not, I've saved the world.&lt;br&gt;&#xA;I salute when the flag is unfurled.&lt;br&gt;&#xA;The future I thought I once knew,&lt;br&gt;&#xA;Instead honesty became my main view.&lt;br&gt;&#xA;Is it day or night, light or dark,&lt;br&gt;&#xA;Tell me who I am, and something cool that I park.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The title contains a clue as well.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4, 6, and 10 are related.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is a specific person.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="9951" LastEditDate="2015-04-22T22:08:58.810" LastActivityDate="2015-04-23T17:07:19.887" Title="Come Sail Away - Who am I?" Tags="|riddle|rhyme|" AnswerCount="9" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12544" PostTypeId="2" ParentId="11702" CreationDate="2015-04-21T21:11:02.410" Score="1" Body="&lt;p&gt;I am posting this answer because the accepted answer sounds very &quot;mathy&quot;. So, this answer is geared toward those who are less &quot;mathy&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number theory tricks that this puzzle uses:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;When you sum the digits of any integer, and then subtract that sum from that integer, the result is always a multiple of 9.&lt;/li&gt;&#xA;  &lt;li&gt;When any integer's digits are rearranged, and then the smaller is subtracted from the larger, the result is always a multiple of 9.&lt;/li&gt;&#xA;  &lt;li&gt;If a number is a multiple of 9, then its digits always sum to a multiple of 9.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt;&lt;br&gt;&#xA;Person A picks 72,492&lt;br&gt;&#xA;The digit sum is 7 + 2 + 4 + 9 + 2 = 24&lt;br&gt;&#xA;72,492 - 24 = &lt;strong&gt;72,468&lt;/strong&gt; (a multiple of 9, by #1 above)&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Person B picks 80,331&lt;br&gt;&#xA;The rearrangement is 30,831&lt;br&gt;&#xA;80,331- 30,831 = &lt;strong&gt;49,500&lt;/strong&gt; (a multiple of 9, by #2 above)&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Person A and Person B add their resulting numbers together, which results in a larger multiple of 9 (by common sense):&lt;br&gt;&#xA;72,468 + 49,500 = &lt;strong&gt;121,968&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since &lt;strong&gt;121,968&lt;/strong&gt; is a multiple of 9, then the digits must add up to a multiple of 9 (by #3 above)&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the magician simply needs to add up the digits as they are being told to him, and the underlined (missing) digit is the number that would make the sum a multiple of 9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if we underlined the 6 in the above &lt;strong&gt;121,698&lt;/strong&gt;, then the magician would add up the other digits that we read to him, 1 + 2 + 1 + 9 + 8 = 21, and the next multiple of 9 is 27, which means that the missing number must be a 6, since 21 + &lt;strong&gt;6&lt;/strong&gt; = 27.&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-04-22T01:55:54.127" LastActivityDate="2015-04-22T01:55:54.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12545" PostTypeId="2" ParentId="12523" CreationDate="2015-04-21T21:12:56.893" Score="1" Body="&lt;p&gt;Only one group sticks out to me&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Young, Hagar, Johnson, Jackson&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are the last names of famous musicians. Neil Young, Sammy Hagar, Brian Johnson, and Michael Jackson.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11734" LastActivityDate="2015-04-21T21:12:56.893" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12546" PostTypeId="2" ParentId="12523" CreationDate="2015-04-21T21:17:57.793" Score="6" Body="&lt;p&gt;I noticed one apparent category&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Second names of corporations with an ampersand&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;with members&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Johnson &amp;amp; &lt;strong&gt;Johnson&lt;/strong&gt;, Proctor &amp;amp; &lt;strong&gt;Gamble&lt;/strong&gt;, Black &amp;amp; &lt;strong&gt;Decker&lt;/strong&gt;, Ernst &amp;amp; &lt;strong&gt;Young&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-04-21T22:55:34.970" LastActivityDate="2015-04-21T22:55:34.970" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12547" PostTypeId="2" ParentId="12543" CreationDate="2015-04-21T21:27:12.507" Score="2" Body="&lt;p&gt;Partial Answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Charon, the Boatman&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Crazy or not, you know who I am,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can use a knife, not just for Jam.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He uses the staff to guide his way down the River Styx (Title reference)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Death is nothing new to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know the glasses that help me see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The River Styx takes the newly deceased to the underworld.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tell me who I am, and something cool that I park.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Charon the boatman. He parks his awesome boat on the riverbank&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9412" LastActivityDate="2015-04-21T21:27:12.507" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12548" PostTypeId="1" CreationDate="2015-04-21T23:11:02.233" Score="9" ViewCount="2723" Body="&lt;p&gt;A fisherman rowing his boat on a very small lake throws his anchor into the water.  Does the water level of the lake rise, fall or stay the same?&lt;/p&gt;&#xA;" OwnerUserId="11001" LastEditorUserId="10615" LastEditDate="2015-06-12T01:36:54.607" LastActivityDate="2015-10-06T11:58:50.390" Title="Fisherman boat and sea level" Tags="|physics|" AnswerCount="4" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="12549" PostTypeId="2" ParentId="12548" CreationDate="2015-04-21T23:21:21.647" Score="29" Body="&lt;p&gt;Assuming that the anchor will actually sink and reach the bottom of the very small lake, the water level of the lake will&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;fall&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The anchor, because it sinks, is more dense than the water. While it was in the boat, because the boat was floating an amount of water equal to the weight of the boat plus the weight of the anchor was being displaced. After the anchor is thrown into the water, the weight of the boat plus the &lt;strong&gt;volume&lt;/strong&gt; of the anchor is being displaced, &lt;strong&gt;but the volume occupied by the anchor is less than the volume occupied by an amount of water with the weight of the anchor&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More intuitive reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The boat's going to be riding higher in the water without the anchor in the boat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1955" LastEditorUserId="1955" LastEditDate="2015-04-22T15:38:26.757" LastActivityDate="2015-04-22T15:38:26.757" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12550" PostTypeId="2" ParentId="12543" CreationDate="2015-04-21T23:55:02.600" Score="4" Body="&lt;p&gt;I don't think it's right because some clues feel forced, but others feel solid so I'll post anyway... You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The moon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Crazy or not, you know who I am,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Crazy or not&quot; refers to superstitions, werewolves and such.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can use a knife, not just for jam.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Crescent moon looks like a knife.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Death is nothing new to me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The moon holds no life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You know the glasses that help me see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...um, yeah... Reference to craters making up the eyes of the man in the moon? Reference to the &quot;oo&quot; in the middle of moon that look like glasses on the word itself?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rogue or not, I've saved the world.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's thought that the moon was created by a collision with earth by another &quot;rogue&quot; solar system body (&quot;or not&quot; because it's not certain), and now protects us from other meteor collisions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I salute when the flag is unfurled.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reference to the moon landing?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The future I thought I once knew,&#xA;Instead honesty became my main view.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The moon was used in astrology for predicting the future, now we're more scientific and honest about its nature.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is it day or night, light or dark,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Has dark and light sides, thus day and night are ambiguous, plus we see it during the day and at night.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tell me who I am, and something cool that I park.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The moon still has moon buggies parked on it, which is very cool. :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastActivityDate="2015-04-21T23:55:02.600" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12551" PostTypeId="2" ParentId="12548" CreationDate="2015-04-22T02:43:00.790" Score="0" Body="&lt;p&gt;The level of water will:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Remain the same&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Archimedes' principle states that &quot;when a body is immersed in a fluid fully or partially in a fluid, the magnitude of the weight of the fluid displaced is equal to the magnitude of the weight of the body&quot;. So as the lake was barring the weight of both the anchor and the boat in both the cases, i.e., &lt;em&gt;when the anchor was not yet thrown&lt;/em&gt; and &lt;em&gt;the anchor was thrown&lt;/em&gt;, the amount of water displaced will remain the same,  hence &lt;em&gt;&lt;strong&gt;the water level remains the same&lt;/strong&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user5871" LastActivityDate="2015-04-22T02:43:00.790" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12553" PostTypeId="2" ParentId="11913" CreationDate="2015-04-22T03:45:19.547" Score="53" Body="&lt;p&gt;It's hard to be sure with all this sand in my eyes, but I believe the desert is saying:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;CACTUS&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Reasoning&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, the title and the fact that the only numbers used in the grains of sand are 1 - 12, hint towards a clock face.&#xA;From here, we can treat numbers as directions, such that from any given point, you just take the number given and travel according to the associated hour direction:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UdURJ.png&quot; alt=&quot;each number on the clock face points at the next point on the path&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then, starting at the odd grain out (the dark coloured 9), follow the path:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3oje0.png&quot; alt=&quot;resulting path&quot;&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You have to squint a bit (because of all the sand), but it's pretty clear that the path scribes the word &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;CACTUS&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="30633" LastEditDate="2016-11-21T22:35:45.040" LastActivityDate="2016-11-21T22:35:45.040" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="12555" PostTypeId="1" AcceptedAnswerId="12583" CreationDate="2015-04-22T05:35:48.003" Score="10" ViewCount="901" Body="&lt;p&gt;“The Two Sheriffs” puzzle was already discussed here &lt;a href=&quot;https://puzzling.stackexchange.com/questions/11267/two-sheriffs-and-eavesdroppers&quot;&gt;Two Sheriffs and Eavesdroppers&lt;/a&gt;. This puzzle has only one difference: we have &lt;strong&gt;seven&lt;/strong&gt; suspects instead of &lt;strong&gt;eight&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Two sheriffs in neighboring towns are on the track of a killer, in a case involving &lt;strong&gt;seven&lt;/strong&gt; suspects. By virtue of independent, reliable detective work, each has narrowed his list to only two. Now they are engaged in a telephone call; their object is to compare information, and if their pairs overlap in just one suspect, to identify the killer.&lt;/p&gt;&#xA;&lt;p&gt;The difficulty is that their telephone line has been tapped by the local lynch mob, who know the original list of suspects but not which pairs the sheriffs have arrived at. If they are able to identify the killer with certainty as a result of the phone call, he will be lynched before he can be arrested.&lt;/p&gt;&#xA;&lt;p&gt;Can the sheriffs, who have never met, conduct their conversation in such a way that they both end up knowing who the killer is (when possible), yet the lynch mob is still left in the dark?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Original problem is takken from: &lt;em&gt;Mathematical Puzzles, a Connoisseur's Collection&lt;/em&gt;, Peter Winkler. The solution for seven suspects belongs to Yoav Kallus.&lt;/p&gt;&#xA;" OwnerUserId="11566" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2022-08-01T21:48:52.823" Title="Two Sheriffs and Eavesdroppers - 2" Tags="|strategy|cryptography|information-theory|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12557" PostTypeId="2" ParentId="12543" CreationDate="2015-04-22T07:28:02.607" Score="3" Body="&lt;p&gt;and I guess the anwer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Batman&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Crazy or not, you know who I am,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fans or otherwise, we all have heard of him.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can use a knife, not just for jam.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Bat-Sword or the pneumatic mangler or the batarang.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Death is nothing new to me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For he strives to keep the world safe and defy death.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You know the glasses that help me see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cowl Lenses&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rogue or not, I've saved the world.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He sure does&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I salute when the flag is unfurled.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Patriotic as he is.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The future I thought I once knew,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead honesty became my main view.&#xA;Is it day or night, light or dark,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tell me who I am, and something cool that I park. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BatMobile Bike&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Need to add more logic to the explanation.&lt;/p&gt;&#xA;" OwnerUserId="2656" LastActivityDate="2015-04-22T07:28:02.607" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12558" PostTypeId="1" AcceptedAnswerId="12559" CreationDate="2015-04-22T08:05:10.427" Score="13" ViewCount="1427" Body="&lt;p&gt;You are the captain of a large barge and need to cross a &lt;a href=&quot;http://en.wikipedia.org/wiki/Canal&quot; rel=&quot;nofollow&quot;&gt;canal&lt;/a&gt; in order to transport your barge from the sea to a lake above sea-level.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you probably know, when you enter the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lock_%28water_transport%29&quot; rel=&quot;nofollow&quot;&gt;lock&lt;/a&gt;, water is added in order to raise the water level until the intended above-sea level is reached.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you know, this takes a long time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Really bored, you start to do some math (as all bored people do):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your barge is $9m\times3m$ long and $3m$ high, with $1m$ submerged in water.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A $1.75m$ long ladder is hanging at one side of the barge, and is used for climbing in the barge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lock is a $10m\times10m$ square and it's filled at a rate of $500$ liters per second. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much time does the water take to reach the hanging ladder?&lt;/p&gt;&#xA;" OwnerUserId="4652" LastEditorUserId="463" LastEditDate="2015-04-23T02:40:24.127" LastActivityDate="2015-04-23T11:52:09.893" Title="The barge ladder" Tags="|mathematics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12559" PostTypeId="2" ParentId="12558" CreationDate="2015-04-22T08:42:25.593" Score="18" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Infinity and beyond - the water will never reach the ladder.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because your barge rises with the water level. Thus no matter how much the water rises the distance from the ladder to the water (the &quot;height&quot; of the ladder) will not change.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastActivityDate="2015-04-22T08:42:25.593" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12560" PostTypeId="2" ParentId="12416" CreationDate="2015-04-22T08:54:21.923" Score="4" Body="&lt;p&gt;I think its&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Thoughts&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I convey the invisible,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I convey what can be felt,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Thoughts are always invisible, we thoughts about what we felt.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But not seen,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inside my heart,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Thoughts can not seen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am not as nice as I could seem to be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not all Thoughts seem to be nice always.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At the speed of a mere blink,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Thoughts changes in every blink&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am very powerful&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Thoughts are powerful sometimes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And can bring comfort,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; good thoughts bring comfort&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4312" LastEditorUserId="4312" LastEditDate="2015-04-27T08:32:14.827" LastActivityDate="2015-04-27T08:32:14.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12561" PostTypeId="2" ParentId="12555" CreationDate="2015-04-22T09:03:25.517" Score="2" Body="&lt;h2&gt;Solution:&lt;/h2&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; If one sheriff outright states that one suspect is not on his list then the problem resembles the original problem with eight. In the special case that the first sheriff eliminates a suspect who is on the list of the second sheriff, then the second sheriff (knowing that one of his suspects is innocent) replaces the removed suspect with a random other suspect.&lt;br&gt;&#xA; &lt;br&gt;&#xA; Lets say S1 has list {a,b} and S2 has list {a,c}.S1 says: I know with certainty that d is not guilty. My list is one of the following: {a,b},{c,e},{f,g}.&lt;br&gt;&#xA; &lt;br&gt;&#xA; S2 says: My list is one of the following: {a,c},{b,f},{e,g}&lt;br&gt;&#xA; &lt;br&gt;&#xA; S1 now knows that S2 has either {a,c} or {b,f} &lt;br&gt;&#xA; &lt;br&gt;&#xA; S1 says: The criminal is either in the set {a,f} or {b,g}&lt;br&gt;&#xA; &lt;br&gt;&#xA; As criminal c is now excluded, S2 knows that the criminal is a so he says: the criminal is in the set {a,g}&lt;br&gt;&#xA; &lt;br&gt;&#xA; S1 now knows that the criminal is a.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11473" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-22T10:59:36.637" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12562" PostTypeId="2" ParentId="12558" CreationDate="2015-04-22T09:26:31.193" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That water won't reach the ladder&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There will always be a 0.25m between the water and the end of the ladder, as the barge will always be only 1 m submerged in the water. &lt;em&gt;I like how you provided all those numbers, that have nothing to do with the actual problem. Something that i also like to do.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10692" LastEditorUserId="-1" LastEditDate="2015-04-23T09:45:37.553" LastActivityDate="2015-04-23T09:45:37.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12563" PostTypeId="2" ParentId="12396" CreationDate="2015-04-22T11:25:10.187" Score="3" Body="&lt;p&gt;I have a question inspired from another puzzle that i found here that could work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Is it simultaneously true that the second door is bad and either the third door is bad or your answer to this question is &quot;no&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For the sake of simplicity I labeled the door that leads to freedom as good and the other ones as bad. &lt;br/&gt;I am gonna divide my question in 3 sentences&lt;br/&gt;A = The second door is bad&lt;br/&gt;B = The third door is bad&lt;br/&gt;C = The answer is &quot;no&quot;&lt;br/&gt;&lt;br/&gt;So the question can be wrote like this:&lt;br/&gt;A AND (B OR C) &lt;br/&gt;&lt;br/&gt;Assuming the 3 cases:&lt;br/&gt;&lt;br/&gt; 1. First door is the good one =&gt; A=True, B=True&lt;br/&gt;&lt;strong&gt;true&lt;/strong&gt; AND (&lt;strong&gt;true&lt;/strong&gt; OR &lt;strong&gt;&lt;em&gt;whatever&lt;/em&gt;&lt;/strong&gt;) = &lt;strong&gt;true&lt;/strong&gt;&lt;br/&gt;&lt;br/&gt;2. The second door is the good one =&gt; A=False&lt;br/&gt;&lt;strong&gt;false&lt;/strong&gt; AND (&lt;strong&gt;&lt;em&gt;whatever&lt;/em&gt;&lt;/strong&gt;) = &lt;strong&gt;false&lt;/strong&gt; &lt;br/&gt;&lt;br/&gt; 3. The third door is the good one =&gt; A=True, B=False, C=?&lt;br/&gt;&lt;strong&gt;true&lt;/strong&gt; AND (&lt;strong&gt;false&lt;/strong&gt; and &lt;strong&gt;C&lt;/strong&gt;) = R (&lt;em&gt;R=answer&lt;/em&gt;)&lt;br/&gt;&lt;br/&gt; If R = &lt;strong&gt;true&lt;/strong&gt;, then it is required that C is also &lt;strong&gt;true&lt;/strong&gt;, but a C that is &lt;strong&gt;true&lt;/strong&gt; implies that R is &lt;strong&gt;false&lt;/strong&gt; (contradiction)&lt;br/&gt;If R = &lt;strong&gt;false&lt;/strong&gt;, then C is also &lt;strong&gt;false&lt;/strong&gt;, but that means that R is &lt;strong&gt;true&lt;/strong&gt; (again contradiction)&lt;br/&gt;&lt;br/&gt;So, if the ghost answers with &lt;strong&gt;yes&lt;/strong&gt; we take the first door, if the answer is &lt;strong&gt;no&lt;/strong&gt; we take the second door, and if there is no answer we take the third.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10692" LastEditorUserId="10692" LastEditDate="2015-04-22T11:31:52.207" LastActivityDate="2015-04-22T11:31:52.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12564" PostTypeId="2" ParentId="12548" CreationDate="2015-04-22T12:05:32.230" Score="1" Body="&lt;p&gt;The following answer is wrong, and I am just leaving it here for anyone who has the same idea.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It depends on the size and weight of the anchor in relation to the size of the boat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When the anchor is in the boat, water will be displaced by the boat, with the  weight of the anchor contributing to this. When the anchor is thrown overboard the boat will displace less water but now the anchor in the water is displacing some. If the anchor is very large and light the water level will rise, if it is small and very heavy it will go down.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9478" LastEditorUserId="8874" LastEditDate="2015-10-06T11:58:50.390" LastActivityDate="2015-10-06T11:58:50.390" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="12565" PostTypeId="1" AcceptedAnswerId="12567" CreationDate="2015-04-22T12:44:55.323" Score="2" ViewCount="221" Body="&lt;p&gt;British (cryptic) crossword puzzle hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why earring is twisted for Buck star?   [4,4]&lt;/p&gt;&#xA;" OwnerUserId="10294" LastEditorUserId="21234" LastEditDate="2016-07-22T21:01:21.667" LastActivityDate="2016-07-22T21:01:21.667" Title="Why earring is twisted for Buck star? [4,4]" Tags="|cryptic-clues|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12566" PostTypeId="2" ParentId="12481" CreationDate="2015-04-22T13:37:36.437" Score="6" Body="&lt;p&gt;The solution is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; good job&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The circles are read from left to right and the binary RGB values of each half encode three bits:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;| colour | R | G | B |&#xA;-----------------------&#xA;| black  | 0 | 0 | 0 |&#xA;| blue   | 0 | 0 | 1 |&#xA;| green  | 0 | 1 | 0 |&#xA;| cyan   | 0 | 1 | 1 |&#xA;| red    | 1 | 0 | 0 |&#xA;| purple | 1 | 0 | 1 |&#xA;| yellow | 1 | 1 | 0 |&#xA;| white  | 1 | 1 | 1 |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Even if each circle can encode 6 bits and values from 0 to 63, it seams as if only the values 0 to 31 are actually used. The binary digits are read as BGR, starting with the &lt;em&gt;right&lt;/em&gt; half of the circle. Let us define the value of each circle &lt;em&gt;n&lt;/em&gt; like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_{n} = (G_{R}R_{R}B_{L}G_{L}R_{L})_{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value of the blue bit in the right half contains both values 0 and 1 in the examples, but the value seem to be irrelevant for the decoding process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, starting with $L_{0} = 0$, the letters for $n \ge 1$ can be decoded with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L_{n} = C_{n} - L_{n-1} \quad \text{ if } \quad C_{n} \ge L_{n-1} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L_{n} = C_{n} - L_{n-1} + 27 \quad \text{ if } \quad C_{n} \lt L_{n-1} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A value of 0 encodes a space and the values 1 to 26 the letters A to Z.&lt;/p&gt;&#xA;" OwnerUserId="11687" LastActivityDate="2015-04-22T13:37:36.437" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12567" PostTypeId="2" ParentId="12565" CreationDate="2015-04-22T13:45:23.423" Score="1" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Erin Gray&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because apparently&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; she starred in a Buck Rogers TV series long ago. I'd never heard of it before, but the &lt;em&gt;Buck star&lt;/em&gt; part gave it away. (&lt;a href=&quot;http://en.wikipedia.org/wiki/Buck_Rogers_in_the_25th_Century_(TV_series)&quot; rel=&quot;nofollow&quot;&gt;Thanks, Wikipedia!&lt;/a&gt;)&lt;br&gt;&lt;br&gt;And her name is an anagram of &quot;Y EARRING&quot;, as I'm sure you already realised.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="40" LastEditorUserId="40" LastEditDate="2015-04-22T13:49:02.433" LastActivityDate="2015-04-22T13:49:02.433" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12568" PostTypeId="2" ParentId="2066" CreationDate="2015-04-22T14:27:29.070" Score="9" Body="&lt;p&gt;For some reason I wanted the answer to have a line with an angle that was not a multiple of $45^\circ$. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;img src=&quot;https://i.stack.imgur.com/S1Ym4.png&quot; alt=&quot;solution&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2722" LastEditorUserId="46903" LastEditDate="2018-09-07T01:52:18.860" LastActivityDate="2018-09-07T01:52:18.860" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="12569" PostTypeId="1" AcceptedAnswerId="12571" CreationDate="2015-04-22T14:39:34.083" Score="34" ViewCount="2697" Body="&lt;p&gt;For what values of $x$ is the number $121$ a square in base $x$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a very easy question, but I still think it's pretty fun. Perhaps it's something to pass on to younger people who are taking algebra.&lt;/p&gt;&#xA;" OwnerUserId="8942" LastActivityDate="2017-09-14T15:43:53.163" Title="What bases make this number square?" Tags="|mathematics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12570" PostTypeId="2" ParentId="12555" CreationDate="2015-04-22T14:43:16.407" Score="1" Body="&lt;p&gt;This builds on @Poelie's answer a little, and should take care of the case where that didn't work&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Suspects: {a,b,c,d,e,f,g}&lt;br&gt;&#xA;Sheriff 1's shortlist: (a,b)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Start by reducing the problem to an even number of suspects&lt;br&gt;&#xA; Sheriff 1 says 'g is innocent'&lt;br&gt;&#xA; If Sheriff 2 responds 'That's odd, I had them as a suspect':&lt;br&gt;&#xA; - Sheriff 1 asks 'Which of {a,d} and {b,e} contains your other suspect'?&lt;br&gt;&#xA; - The answer reveals the culprit to the sheriffs, but leaves the mob with two suspects&lt;br&gt;&#xA;&lt;br&gt;&#xA; In the more likely scenario that g is not one of Sheriff 2's suspects:&lt;br&gt;&#xA; - S1 posits three shortlists: {a,b},{c,d},{e,f}&lt;br&gt;&#xA; - S2 posits three shortlists: {a,c},{b,e},{d,f)&lt;br&gt;&#xA; - Knowing the culprit is either a or b, S1 knows that S2's list is either {a,c} or {b,e}&lt;br&gt;&#xA; - S1 says the culprit is either in {a,e} or {b,f}&lt;br&gt;&#xA; -- If S2 has {a,c}, this tells them that a is the culprit, and they say the culprit is in {a,f}&lt;br&gt;&#xA; -- If S2 has {b,e}, they agree that the culprit is in one of the two groups, and this tells S1 that the culprit is b&lt;br&gt;&#xA; -- There's one more step here, but I haven't worked it out yet&lt;br&gt;&#xA; - Again, the sheriffs know the culprit, but the mob is left with two possible answers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It occurs to me that this only really works if we take it as read that the sheriffs have different shortlists with one member's overlap. But I've typed it all out now, and it should make sense for that case.&lt;/p&gt;&#xA;" OwnerUserId="11717" LastEditorUserId="11717" LastEditDate="2015-04-22T17:33:04.150" LastActivityDate="2015-04-22T17:33:04.150" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12571" PostTypeId="2" ParentId="12569" CreationDate="2015-04-22T14:45:11.787" Score="35" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All of them (except base $2$ which can't have $121$.) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Added pedantry:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you're ok with negative bases, the answer is all positive and negative bases except $-2$, $-1$, and $2$. Base $0$ and $1$ make no sense at all to me (and $-1$ very little) so I would also claim it doesn't hold for these, mostly because &quot;holding&quot; isn't really provably true or false, but also because 121 is not a valid number in those bases, as also it is not valid in base 2 and -2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, I am considering integer bases only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it a square for these bases?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For whatever base $n$ you're working in, simply expanding the number $121$, it means $n^2 + 2n + 1$. And that is $(n+1)^2$ as a matter of algebra. And that is a square.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1509" LastEditorUserId="1509" LastEditDate="2015-04-24T05:14:21.367" LastActivityDate="2015-04-24T05:14:21.367" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="12572" PostTypeId="2" ParentId="12523" CreationDate="2015-04-22T14:55:21.157" Score="14" Body="&lt;p&gt;The first is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bloor, Wellesley, Union, Dundas&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Subway stations in Toronto&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second is: (provided by xnor)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Johnson, Gamble, Young, Decker&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Second names of corporations with an ampersand (Johnson &amp;amp; Johnson, Proctor &amp;amp; Gamble, Black &amp;amp; Decker, Ernst &amp;amp; Young)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The third is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Myers, Jackson, Fox, Bay&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Famous Michaels in the entertainment industry&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The fourth is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hagar, Franklin, Thomas, Winnie&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; fictional characters with &quot;the&quot; in their name: Hagar the Horrible, Franklin the Turtle, Thomas the Tank Engine, Winnie the Pooh&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11744" LastActivityDate="2015-04-22T14:55:21.157" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12573" PostTypeId="1" AcceptedAnswerId="12579" CreationDate="2015-04-22T15:23:46.407" Score="6" ViewCount="2077" Body="&lt;p&gt;I had asked this question on Facebook.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Imagine a cube. Put it on a flat surface, so that four vertices are at the bottom and four vertices are on top. Select any vertex on top, and connect it to each of the four vertices at the bottom. This now encloses a part of the cube. Cut this enclosed part off, and you are left with another shape.&lt;/p&gt;&#xA;&lt;p&gt;How many faces does this leftover shape have?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I feel the answer is &lt;strong&gt;seven&lt;/strong&gt;, but my friends say &lt;strong&gt;five&lt;/strong&gt;. This may be a relatively easy question for this site, but I found it quite hard to visualize.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-09-30T08:56:28.410" Title="Cutting from a cube (visualization test)" Tags="|mathematics|geometry|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12574" PostTypeId="2" ParentId="12543" CreationDate="2015-04-22T15:25:19.853" Score="3" Body="&lt;p&gt;This doesn't really fit with the title hint, but maybe I'm just missing the connection? Either way, worth a shot...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Crazy or not, you know who I am,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Anakin Skywalker&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can use a knife, not just for jam.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A lightsaber&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Death is nothing new to me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He has killed many, and Obi-Wan says Darth Vader killed him&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You know the glasses that help me see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Vader helmet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rogue or not, I've saved the world.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Anything from the battle of Naboo in I to the death of the Emperor in VI&#xA;&lt;br&gt;&#xA; Rogue refers to being a rogue jedi, and/or becoming a sith&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I salute when the flag is unfurled.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He's on the side of the Empire&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The future I thought I once knew,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Had premonition dreams&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Instead honesty became my main view.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I am your father&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is it day or night, light or dark,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Time of day in space is ambiguous at best&#xA;&lt;br&gt;&#xA; Light and dark side of the Force&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tell me who I am, and something cool that I park.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Darth Vader and his TIE/x1... or Star Destroyer... or Death Star...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4513" LastActivityDate="2015-04-22T15:25:19.853" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12575" PostTypeId="2" ParentId="12548" CreationDate="2015-04-22T15:27:49.280" Score="1" Body="&lt;p&gt;Two thinking outside the box answers:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If the &lt;strong&gt;very small&lt;/strong&gt; lake is shaped as a glass of water and there&#xA;is water all the way up, then the water level will not rise, since&#xA;any increase in water level will run over. More likely is that the&#xA;throwing of the anchor (assuming it even fits the lake) will cause&#xA;splashing, and the water will splash over the edge and by that&#xA;decreasing the water level.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The boat is touching the bottom of the &lt;strong&gt;very small&lt;/strong&gt; lake. By that, throwing it in will increase the water displacement will not be any different no matter how much you put in the boat.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="356" LastActivityDate="2015-04-22T15:27:49.280" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12576" PostTypeId="2" ParentId="12573" CreationDate="2015-04-22T15:28:34.540" Score="1" Body="&lt;p&gt;I guess it depends what you mean by &quot;connect this point to each of the 4 points at the bottom&quot; if its by a straight line (regardless of matter) then your answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;5&lt;/strong&gt;.   If your base points are numbered 1 to 4, and your origin point &gt;!is labeled A. Then you will have the bottom face (straight vertical line  &gt;!from A to the point immediatly below (1). You will also have two side &#xA; faces going up vertically from the base as triangles as defined by&#xA; lines from A to (2) and (3) which are the points next to (1). Finally&#xA; the final two faces will be a diamond slanted from point A to (4 -&#xA; across from 1), each half of the diamond resting on the 2 vertical&#xA; triangles and the line from A to 4.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Picture&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/Lte9m.jpg&quot; alt=&quot;A picture worth the whole explanation&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastEditorUserId="11251" LastEditDate="2015-04-22T15:38:51.987" LastActivityDate="2015-04-22T15:38:51.987" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12577" PostTypeId="2" ParentId="12573" CreationDate="2015-04-22T15:32:06.270" Score="0" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 5.  There are the 3 faces on the bottom and each vertical side next to the upper point.  The new revealed surface is two faces with a ridge extending from the upper point to the lower diagonal point.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10294" LastActivityDate="2015-04-22T15:32:06.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12578" PostTypeId="2" ParentId="12573" CreationDate="2015-04-22T15:33:35.493" Score="1" Body="&lt;p&gt;The shape that was cut out would be a &lt;a href=&quot;http://en.wikipedia.org/wiki/Square_pyramid&quot; rel=&quot;nofollow&quot;&gt;square pyramid&lt;/a&gt;. Therefore it would have 5 faces: the 4 triangular faces formed when cutting it out of the cube as well as the square bottom face of the cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The leftover shape would have 7 faces: the top face of the cube, the other triangular halves of the 2 cube faces that were cut to form faces of the pyramid, the 2 internal faces adjacent to the other faces of the pyramid, and the 2 side faces of the cube that were untouched in the formation of the pyramid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you chose a point internal to the top face of the cube instead of one of the top vertices, you'd still get a square pyramid as the shape you cut out. However, then the leftover shape would have 9 faces. In that case the 4 sides of the cube would all stay intact as faces of the leftover shape, and you would have 4 faces corresponding to the 4 triangular faces of the pyramid.&lt;/p&gt;&#xA;" OwnerUserId="8891" LastEditorUserId="8891" LastEditDate="2015-04-22T15:38:49.240" LastActivityDate="2015-04-22T15:38:49.240" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12579" PostTypeId="2" ParentId="12573" CreationDate="2015-04-22T15:35:33.913" Score="18" Body="&lt;p&gt;You have two pieces one has 5 faces the other 7.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You and your friend are talking about different pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rp1qd.png&quot; alt=&quot;Two parts of a cube&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The piece on the left has a total of 7 faces (5 that were on the outside of the cube and 2 that are formed inside by cutting.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The piece on the right has 5 faces (3 that were on the outside of the cube and 2 that are formed inside by cutting.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shape enclosed by the points is the one on the right (5 faces) the piece left over is on the left (7 faces.)&lt;/p&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="11297" LastEditDate="2015-04-22T16:09:56.870" LastActivityDate="2015-04-22T16:09:56.870" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12582" PostTypeId="2" ParentId="12569" CreationDate="2015-04-22T15:46:30.977" Score="22" Body="&lt;p&gt;The base-$x$ number $121$ is a square&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; whenever $x\geq 3$.&lt;br&gt;&lt;br&gt; For $x\geq 3$, we can multiply in base $x$: \begin{array}{ccc}&amp;amp;1&amp;amp;1\\\times&amp;amp;1&amp;amp;1\\\hline&amp;amp;1&amp;amp;1\\1&amp;amp;1&amp;amp;\\\hline1&amp;amp;2&amp;amp;1\end{array}We never needed to &quot;carry&quot;, so this computation looks the same for every $x$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-04-22T19:40:14.313" LastActivityDate="2015-04-22T19:40:14.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12583" PostTypeId="2" ParentId="12555" CreationDate="2015-04-22T16:54:45.737" Score="8" Body="&lt;p&gt;Assign each suspect a number $1-7$. Define the xor of two numbers between $1$ and $7$ as follows: write them in binary, then xor them together bitwise. For example, $3 \oplus 6=011\oplus 110=101=5$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some observations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$a\oplus b$ is different from both $a$ and $b$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $a\oplus b=c$ and $d\oplus e=f$, then the sets $\{a,b,c\}$ and $\{d,e,f\}$ have an element in common. This can be verified by checking the only sets of three numbers where two of them xor together to make the other are {1,2,3},{1,4,5},{1,6,7},{2,4,6},{3,4,7},{3,5,6},{2,5,7}, and that any two of these intersect. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The converstion:&lt;/strong&gt; Lets say Alice has {a,b} and Bob has {a,c}. Alice announces $a\oplus b$ and Bob announces $a\oplus c$. If Alice and Bob say the same thing, then they know they have the same pair, and hang up. If not, there are two possibilities:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$a\oplus b\neq c$. Then it also follows that $a\oplus c\neq b$, so neither Sheriff said a number in the others list. In this case, they each announce their pair, ${a,b}$ and ${a,c}$.&lt;/li&gt;&#xA;&lt;li&gt;$a\oplus b=c$, which also implies $a\oplus c=b$. They now each say a random pair whose xor is what they first announced, &lt;strong&gt;except for their actual pair&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why the sheriffs know the killer:&lt;/strong&gt;  It's easy to see why they do in case 1, since they each heard the other's pair. In case, 2, when Alice hears Bob say $b$, she knows $b$ is not the killer, so it is $a$. Same thing happens for Bob.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why the lynch mob doesn't:&lt;/strong&gt; Suppose that they hear the numbers $a_1$ and $b_1$, followed by the pairs ${a_2,a_3}$ and ${b_2,b_3}$. It could be the case that Alice had the pair ${a_2,a_3}$ and Bob has the pair ${b_2,b_3}$, in which case the killer would be the intersection of these sets (this would be case 1). However, it could also be the case that Alice had $\{a_1\oplus b_1,b_1\}$ and Bob had $\{a_1\oplus b_1,a_1\}$. Here we are in case 2, where Alice and Bob just randomly chose to say $\{a_2,a_3\}$ and $\{b_2,b_3\}$. No matter what sets they randomly announced, by bullet point 2, they will always intersect, so this is really indistinguishable from case 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, here are two situations where the Sheriffs have the same conversation, but the killer is a different person.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;p&gt;Alice has {1,2}, while Bob has {1,5}. Alice says 3, Bob says 4. Since 4 is not in {1,2} and 3 is not in {1,5}, we are in the first bullet point. They then each announce their lists, {1,2} and {1,5}. They now both know that the killer is 1.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Alice has {4,7}, Sheriff 2 has {3,7}. Alice says 3, Bob says 4. Each number is in the other's list, we are in the second bullet point. They know now the killer is 7. Alice randomly announces one of the pairs {1,2} or {5,6}, since these are the pairs which xor together to make 3 (besides {4,7}). Let's say she chooses {1,2}. Similarly, Bob randomly says one of {1,5}, or {2,6}, and he randomly chooses {1,5}.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-04-22T17:11:20.150" LastActivityDate="2015-04-22T17:11:20.150" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12584" PostTypeId="5" CreationDate="2015-04-22T17:36:06.387" Score="0" Body="&lt;p&gt;A question about how a puzzle can be solved and strategies to do so, whether it can be solved, or what is required for it to be solved.&lt;/p&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="52483" LastEditDate="2018-11-23T18:23:20.873" LastActivityDate="2018-11-23T18:23:20.873" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="12585" PostTypeId="4" CreationDate="2015-04-22T17:36:06.387" Score="0" Body="A question about whether a puzzle can be solved." OwnerUserId="11297" LastEditorUserId="36023" LastEditDate="2021-04-14T10:12:13.597" LastActivityDate="2021-04-14T10:12:13.597" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="12586" PostTypeId="2" ParentId="12543" CreationDate="2015-04-22T17:49:17.190" Score="1" Body="&lt;p&gt;I think you're&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a wave&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Crazy or not, you know who I am,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everybody knows what a wave is&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can use a knife, not just for jam.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is a type of knife that has a waved shaped blade&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Death is nothing new to me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many sailors died because of waves&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You know the glasses that help me see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I don't know for that one. The only thing I can think of is a word play between &lt;em&gt;see&lt;/em&gt; and &lt;em&gt;sea&lt;/em&gt;. I'll put that on the fact that English is my second language ^^&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rogue or not, I've saved the world.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am not sure of this one, I am thinking of the Deluge and Noah and his ark&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I salute when the flag is unfurled.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Usually when the flag is unfurled, it's because of the wind. Which creates waves&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The future I thought I once knew,&lt;br&gt;&#xA;Instead honesty became my main view.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think those are working together because when there is no wave, you can see far away, but when there is big waves, that's the only thing you can see.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is it day or night, light or dark,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Waves are there no matter what day, what time of the day, what season&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tell me who I am, and something cool that I park. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are a wave and you park the surfers about to ride you. Surfers are cool :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10400" LastActivityDate="2015-04-22T17:49:17.190" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12587" PostTypeId="2" ParentId="12511" CreationDate="2015-04-22T18:34:45.653" Score="1" Body="&lt;p&gt;It's always possible, using a reasonably simple approach:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Choose a colour (say white), and switch counters so that each column has exactly one of that colour present. Now choose a second colour (say blue), and again switch counters so that each column only has one of that colour present, but without disturbing the first-chosen colour counters. This is always possible because in a column where there are two blue counters, you can choose which of them to switch to a column with no blue counter so that you do not disturb the white counter in that column. Having done that, your grid is fully patriotic - the final space in each column is filled with the third colour automatically.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I note that, conveniently, this puzzle is patriotic for many nations. Yay Iceland!&lt;/p&gt;&#xA;" OwnerUserId="8875" LastEditorUserId="8875" LastEditDate="2015-04-22T18:42:18.517" LastActivityDate="2015-04-22T18:42:18.517" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12588" PostTypeId="2" ParentId="12543" CreationDate="2015-04-22T18:42:40.977" Score="3" Body="&lt;p&gt;This might not be right, but I think it could be close. Are you:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A pirate (More so, Jack sparrow from The Pirates of the Caribbean)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Crazy or not, you know who I am,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most people know about pirates&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can use a knife, not just for jam.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They often use knives and swords as weapons while fighting&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Death is nothing new to me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pirates are often in fights to the death with others&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You know the glasses that help me see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Possibly referring to a spy glass which helps them see longer distances on sea&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rogue or not, I've saved the world.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jack has fought against Davy Jones and Hector Barbossa to protect the world!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I salute when the flag is unfurled.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He will salute his flag and also a possible reference to honoring the pirate code.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The future I thought I once knew,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jack thought he was always an evil pirate&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Instead honesty became my main view.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After meeting with Will Turner and Elizabeth Swann his views change to a becoming a better more honest pirate (for as far as pirating can go)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is it day or night, light or dark,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This could reference being inside a boat you can't tell the time of day. Also a possible small reference to the use of an eye patch which helps in literal light and dark situations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tell me who I am, and something cool that I park.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jack Sparrow, and his ship the Black Perl!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also the title &quot;Come sail away&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A reference to him and his crew to sail away in his ship.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastActivityDate="2015-04-22T18:42:40.977" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12589" PostTypeId="2" ParentId="12322" CreationDate="2015-04-22T19:28:07.143" Score="0" Body="&lt;p&gt;Is your knowledge potent enough to figure out the combination of the safe?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Possibly. Would calling a lock smith to open the safe and find out the combination work? Maybe finding out what those 10 clues he left behind are might help as well. ;-)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This might not be the answer you were looking for, but maybe it can inspire some other answers. (hopefully ones that aren't so lateral)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plot twist:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The 'secret' in the safe is the combination to the safe! He never told anyone which keeps it a secret and then it eluded him (he forgot it) many years ago. Now he forgot the safe and no one else knows (since he never shared it) which is why it's still there! Why he would do this, who knows. But what kind of person would leave a deck of just 28 cards in the open? (Someone really smart or someone not so smart, for this twist I'll pretend not so smart)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastActivityDate="2015-04-22T19:28:07.143" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12590" PostTypeId="1" AcceptedAnswerId="12611" CreationDate="2015-04-22T21:07:21.267" Score="3" ViewCount="323" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Repunit&quot; rel=&quot;nofollow noreferrer&quot;&gt;Repunit&lt;/a&gt; is a number consisting only of &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;s (here in base &lt;span class=&quot;math-container&quot;&gt;$10$&lt;/span&gt;).&lt;/p&gt;&#xA;&lt;p&gt;Examples - &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$11$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$11111$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;h3&gt;Goal&lt;/h3&gt;&#xA;&lt;p&gt;Create 10 first repunits &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;-digits long from least possible number instances of numbers &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;.&lt;br /&gt;&#xA;For example (these are not minimal),&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$11 = 2\times2\times2\times2-2-2-2/2$&lt;/span&gt;&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$111 = 3\times(3\times3\times3 + 3 + 3 + 3) + 3$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;h3&gt;Allowed operators&lt;/h3&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Arithmetic &lt;span class=&quot;math-container&quot;&gt;$+$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$-$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\times$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$/$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;Exponentiation of positive rational numbers&lt;/li&gt;&#xA;&lt;li&gt;Square root, &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;-th roots of positive rational numbers&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Factorial&quot; rel=&quot;nofollow noreferrer&quot;&gt;Factorial&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Gamma_function&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gamma function&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Derangement&quot; rel=&quot;nofollow noreferrer&quot;&gt;subfactorial&lt;/a&gt; of integers&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Number concatenation is &lt;strong&gt;not&lt;/strong&gt; allowed!&lt;/p&gt;&#xA;&lt;p&gt;Note: actually, you need only 9 repunits (first (&lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;) is trivial, so no need to write it down) - starting with 11 and ending with 1111111111.&lt;/p&gt;&#xA;" OwnerUserId="5781" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-23T14:53:44.150" Title="Repunit replication" Tags="|mathematics|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12591" PostTypeId="5" CreationDate="2015-04-22T21:17:01.023" Score="0" Body="&lt;p&gt;These questions request additional details on how a problem is solved or why the current solution is appropriate. The problem is not necessarily a puzzle itself; it may refer to puzzle-solving algorithms, published tips, or rules pertaining to how a puzzle may be solved. Often, the poster is aware of the answer to a problem but does not have or does not understand the solution. &lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="4551" LastEditDate="2015-05-18T04:09:14.770" LastActivityDate="2015-05-18T04:09:14.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12592" PostTypeId="4" CreationDate="2015-04-22T21:17:01.023" Score="0" Body="The answer to a problem may be known but it is not understood.  These questions explicitly request additional details on how a problem is solved or why the current solution is appropriate." OwnerUserId="99" LastEditorUserId="4551" LastEditDate="2015-05-18T04:09:43.630" LastActivityDate="2015-05-18T04:09:43.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12593" PostTypeId="5" CreationDate="2015-04-22T21:18:36.330" Score="0" Body="&lt;p&gt;This tag is to be used for puzzles that contain an element of humour or wit, or for questions about such puzzles. Questions which are simply jokes and nothing else are likely to be closed if they don't relate to puzzling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Puzzles should also have a unique, demonstrably correct answer; see &lt;a href=&quot;https://puzzling.meta.stackexchange.com/questions/1254&quot;&gt;this meta post&lt;/a&gt; and the associated off-topic close reason. While humour is welcome in puzzles, it shouldn't come at the expense of answerability.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="11110" LastEditDate="2018-10-29T08:08:28.923" LastActivityDate="2018-10-29T08:08:28.923" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="12594" PostTypeId="4" CreationDate="2015-04-22T21:18:36.330" Score="0" Body="For puzzles that contain an element of humour or wit. Please do not use this tag to post simple jokes which are not puzzles, and remember that questions should be uniquely answerable." OwnerUserId="99" LastEditorUserId="4421" LastEditDate="2016-09-06T20:15:54.263" LastActivityDate="2016-09-06T20:15:54.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12595" PostTypeId="1" AcceptedAnswerId="12620" CreationDate="2015-04-22T21:50:42.860" Score="14" ViewCount="1326" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;There are three spiders and an ant, crawling on the edges of a wire frame cube. The spiders can crawl exactly one third as fast as the ants. Show that the spiders can catch the ant.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Think of bugs as point masses, and a spider catches an ant when they are in the exact same location. Any bug can instantaneously change direction. Everyone can see each other at all times.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-04-24T15:04:46.823" LastActivityDate="2015-04-24T15:04:46.823" Title="Spiders and Ant on a Cube" Tags="|strategy|pursuit-evasion|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12596" PostTypeId="1" AcceptedAnswerId="12647" CreationDate="2015-04-22T23:18:49.973" Score="15" ViewCount="632" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Lieutenant, launch the polar probes!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Launching probes, sir. Failure, sir. Probes did not land at the poles.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Well!? Where are they?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Unknown, sir. Somewhere off-axis. They did land on opposite sides of&#xA;  the planet.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Launch another pair!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Another failure, sir. Same thing. They are opposite each other, but&#xA;  not at the poles.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Same places as the first pair?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;No, sir.  Completely different.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Dammit, find them. Tachyon ranging pulse!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;All probes responding. Linear distances from ship are 7, 9, 11, and&#xA;  [CENSORED] SDUs*.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(new voice) HAHAHAHAHA&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Ensign! What is so funny?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Sir, at our current altitude, the area of the planet's surface we can&#xA;  see is exactly equal to the area of a flat circle of [CENSORED]&#xA;  radius! Funny coincidence, sir.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Shut up, Wesley!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;*SDU : Standard Distance Unit&lt;/p&gt;&#xA;" OwnerUserId="11752" LastEditorUserId="5373" LastEditDate="2019-04-13T10:44:11.340" LastActivityDate="2019-04-13T10:44:11.340" Title="What is the Universal Translator censoring?" Tags="|logical-deduction|geometry|" AnswerCount="2" CommentCount="10" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12597" PostTypeId="1" AcceptedAnswerId="12606" CreationDate="2015-04-23T01:11:39.483" Score="37" ViewCount="1357" Body="&lt;p&gt;There's multiple parts to my name. In fact, I'd say there's two short and three long parts.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First is two, second is three. But they're the long ones.&lt;br&gt;&#xA;Then things get a bit trickier, but no harder than a simple dot to dot.&lt;br&gt;&#xA;My fourth is... Actually, on second thoughts, maybe that's too direct.&lt;br&gt;&#xA;What I will say, though, is that this part ends in the same way this sentence does.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you with me so far? Take a deep breath, we're up to the last part.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This part's beginning is contained in &quot;competition&quot;, but you'll only see it if you're meticulous.&lt;br&gt;&#xA;The next is central to my most famous speech.&lt;br&gt;&#xA;I will be kind-hearted and include the next bit directly in this line. Though it is not the letters you think.&lt;br&gt;&#xA;And finally, we get to the point.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question, of course, is: What is my name?  &lt;/p&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="5208" LastEditDate="2015-04-23T05:20:23.860" LastActivityDate="2019-09-25T07:37:18.220" Title="If I described my name, I would do it like this" Tags="|riddle|enigmatic-puzzle|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12598" PostTypeId="2" ParentId="12322" CreationDate="2015-04-23T01:29:31.877" Score="1" Body="&lt;p&gt;The combination is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 37 - 2 - 60 - 41&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you lay out all cards in a row, and look at the layout of the pips on the card, they represent a 1 or 0 (1 if there is a centre pip, 0 if there isn't). 3, 5, 7 are 1s, and 2, 4, 6 are 0s. 8 represents the end of a number.&#xA; So 100101 - 0010 - 111100 - 101001&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Extra explanation of how I got here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is based on a game called 'Petals around the rose' (related to the title), which is played with dice. Bill Gates struggled with finding out the solution to this game for many years. &lt;a href=&quot;http://www.borrett.id.au/computing/petals-bg.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.borrett.id.au/computing/petals-bg.htm&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't quite know what the second clue is about yet.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are 10 clues, and 10 in binary is 2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2765" LastActivityDate="2015-04-23T01:29:31.877" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12599" PostTypeId="1" AcceptedAnswerId="12601" CreationDate="2015-04-23T02:22:52.757" Score="17" ViewCount="21630" Body="&lt;p&gt;Can you determine if it's possible to draw a geometric figure (made up from shapes like rectangles, triangles, and other regular shapes) with one pen stroke and not drawing the same line twice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am thinking about something related to graph theory or something like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eEK23.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10692" LastActivityDate="2017-02-06T14:58:00.240" Title="Drawing something using one pen stroke" Tags="|geometry|graph-theory|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12600" PostTypeId="2" ParentId="12595" CreationDate="2015-04-23T02:29:20.983" Score="4" Body="&lt;p&gt;Balance the cube on a corner, so that there is one corner, $D$ (for Down), touching the table, three corners $A,B,C$ all connected to $D$ with the same height off the table, three corners $X,Y,Z$ with the same height off the table, $A$ connected to $X$ and $Y$, $B$ connected to $Y$ and $Z$, $C$ connected to $Z$ and $X$, and one final corner $U$ (for Up) connected to $X,Y,Z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the spiders start by holding a strategy meeting at $D$. They then split up, and travel until spider $a$ is at $A$, spider $b$ is at $B$, and spider $c$ is at $C$. If they have not yet caught the ant then they make a solemn vow to never let the ant reach any of the edges $AD, BD, CD$ alive, and then they move on to the second phase of their plan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In phase two, spider $a$ restricts itself to walking back and forth along edge $AX$, spider $b$ along $BY$, spider $c$ along $CZ$. Each spider is trying to catch one of the ant's three &quot;shadows&quot;, called shadow $x$, shadow $y$, and shadow $z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's describe shadow $x$: if the ant is somewhere along the path $AYUX$ of length three, with distance $d$ from $A$ (measured along this path), then shadow $x$ is on line $AX$ with distance $d/3$ from $A$. If the ant is on edge $BY$, then shadow $x$ is on line $AX$ with distance $1/3$ from $A$ (so, the same shadow as when the ant is at $Y$). If the ant is on edge $AX$ or $CX$, shadow $x$ is at $X$ (same shadow as when the ant is at $X$). If the ant is on any of the edges $UZ,BZ,CZ$, then shadow $x$ is on line $AX$ with distance $2/3$ from $A$ (same shadow as when the ant is at $U$). As long as the spiders don't break their vow, then shadow $x$ will always move at most as fast as the spiders, so eventually spider $a$ will catch shadow $x$ and from then on it can move back and forth together with it. In fact, spider $a$ can even guarantee that it catches up to shadow $x$ before shadow $x$ makes it to $A$, guaranteeing that the ant won't make it to edge $AD$ alive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shadows $y$ and $z$ are defined similarly, with shadow $y$ along $BY$ according to where the ant is along the path $BZUY$ (or blah blah blah if it is off the path) and shadow $z$ along $CZ$ according to where the ant is along path $CXUZ$ (etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually all the spiders successfully catch their corresponding shadows without breaking their vows. If the ant is on any of the edges $AX, BY, CZ$  at this point, then it is already dead. Otherwise, they move on to phase three.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the ant starts phase three on edge $CX$, then spider $a$ and shadow $x$ will both be at $X$ and spider $c$ and shadow $z$ will be somewhere along $CZ$: spider $c$ now rushes straight back to $C$ before shadow $z$ can get there and then along $CX$ to meet spider $x$, and the ant has no hope of escape. Similarly if the ant starts phase three along edge $AY$ or edge $BZ$ it meets a grisly fate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the ant starts phase three on any of the edges $UX, UY, UZ$, then spider $a$ races shadow $x$ to $X$, spider $b$ races shadow $y$ to $Y$, and spider $c$ races shadow $z$ to $Z$. Then all three spiders march straight up to $U$, and the ant is captured.&lt;/p&gt;&#xA;" OwnerUserId="11295" LastEditorUserId="11295" LastEditDate="2015-04-23T02:37:19.633" LastActivityDate="2015-04-23T02:37:19.633" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12601" PostTypeId="2" ParentId="12599" CreationDate="2015-04-23T02:42:06.260" Score="27" Body="&lt;p&gt;Yes! This is a classic graph theory problem. Anywhere where lines meet is called a &lt;em&gt;vertex&lt;/em&gt;, and the &lt;em&gt;degree&lt;/em&gt; of a vertex is the number of lines that meet there. Euler proved that as long as a graph has either 0 or 2 vertices of odd degree, &lt;strong&gt;and the graph is connected (consists of a single piece)&lt;/strong&gt;, then it can be drawn as you specified. Furthermore, if there are any odd vertices, then any successful Eulerian path (as such a pen stroke is called) must start at one, and end at the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main reason this is true is as follows: every time you enter and leave a vertex, you kill two of its edges. So, if a vertex has an odd number of edges, not all of its edges can be taken care of by passing through it, so the pen must start or end there. There are only two start and end points, so there can be at most two odd degree vertices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your picture, there is a square with horizontal sides, all of whose vertices have 5 edges coming out of them. This means that you have $4$ odd-degree vertices, which is $2$ too many, so your graph can't be drawn with a single non-overlapping pen stroke. But, if you erased one of the diagonals of this middle square, then you'd be in business.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This result applies to any shape with various curves on a page which meet at various places, not just straight line figures.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-04-23T14:48:05.377" LastActivityDate="2015-04-23T14:48:05.377" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12602" PostTypeId="2" ParentId="12599" CreationDate="2015-04-23T02:44:47.853" Score="5" Body="&lt;p&gt;What you are asking for is how to find an &quot;Eulerian path&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This page explains it quite well:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.math.ku.edu/~jmartin/courses/math105-F11/Lectures/chapter5-part2.pdf&quot; rel=&quot;nofollow&quot;&gt;https://www.math.ku.edu/~jmartin/courses/math105-F11/Lectures/chapter5-part2.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, there's this page:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ctl.ua.edu/math103/euler/howcanwe.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.ctl.ua.edu/math103/euler/howcanwe.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which, says, to paraphrase:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If there's 0 odd vertexes then there's at least an Eulerian circuit, an Eulerian path that ends where it begins, if there's 2 odd vertexes there's at least an Eulerian path. Anything else has no paths or circuits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2015-04-23T17:13:10.563" LastActivityDate="2015-04-23T17:13:10.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12603" PostTypeId="2" ParentId="12543" CreationDate="2015-04-23T03:09:04.447" Score="1" Body="&lt;p&gt;Crazy or not, you know who I am,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Nazi in Raiders of The Lost Ark&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can use a knife, not just for jam.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think there was a scene in the movie&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Death is nothing new to me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He's a Nazi&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You know the glasses that help me see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He wears glasses and they are distinctive&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rogue or not, I've saved the world.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ? Ok this doesn't fit&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I salute when the flag is unfurled.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Obviously the Nazi flag&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The future I thought I once knew,&lt;br&gt;&#xA;Instead honesty became my main view.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Vengeance from the Ark&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is it day or night, light or dark,&lt;br&gt;&#xA;Tell me who I am, and something cool that I park&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Major Arnold Ernst Toht was driven in an open classic &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11729" LastEditorDisplayName="user20" LastEditDate="2015-04-23T08:35:38.763" LastActivityDate="2015-04-23T08:35:38.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12604" PostTypeId="2" ParentId="12543" CreationDate="2015-04-23T03:17:57.437" Score="1" Body="&lt;p&gt;My guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Captain America&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Come sail away with me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Captain = commander of a ship&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Crazy or not, you know who I am&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everyone has seen the movies&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can use a knife, not just for jam.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Knife fight scene with Cap and Bucky in Winter Soldier.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Death is nothing new to me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Captain America died in the ice but was brought back to life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You know the glasses that help me see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When Cap is in disguise in Winter Soldier, he wore glasses.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rogue or not, I've saved the world.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is a comic called Captain America Rogues Gallery, and he has saved the world many times in the comics.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I salute when the flag is unfurled.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Captain America is very patriotic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The future I thought I once knew,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He was born a long time ago but was frozen in ice and revived in the future.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Instead honesty became my main view.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He is a paragon of honesty and virtue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is it day or night, light or dark&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Maybe referring to the band &quot;Dark Captain Light Captain&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tell me who I am, and something cool that I park.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Either the Chevrolet Corvette Stingray driven by Black Widow that Cap rode in, or maybe Red Skull's V16 Coupe?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4479" LastActivityDate="2015-04-23T03:17:57.437" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12606" PostTypeId="2" ParentId="12597" CreationDate="2015-04-23T04:54:41.160" Score="31" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MORSE CODE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First is two, second is three. But they're the long ones.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; M is -- and O is --- in morse code (two and three long ones).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then things get a bit trickier, but no harder than a simple dot to dot.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; R is .-. in morse code (dot to dot).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fourth is... Actually, on second thoughts, maybe that's too direct.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; S is ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What I will say, though, is that this part ends in the same way this sentence does.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; E is . (how the sentence ends)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Are you with me so far? Take a deep breath, we're up to the last part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This part's beginning is contained in &quot;competition&quot;, but you'll only see it if you're meticulous.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C is -.-. which is hidden in &quot;compe&lt;strong&gt;titi&lt;/strong&gt;on&quot;.  The meticulous clue refers to dotting your i's and crossing your t's. (From Alconja in comments).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The next is central to my most famous speech.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; O is in the middle of SOS &quot;most famous speech&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I will be kind-hearted and include the next bit directly in this line. Though it is not the letters you think.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; D is -.. hidden in &quot;kind-hearted&quot; &quot;line.&quot; &quot;think.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And finally, we get to the point.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; E is .&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4479" LastEditorUserId="4479" LastEditDate="2019-09-25T07:37:18.220" LastActivityDate="2019-09-25T07:37:18.220" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="12607" PostTypeId="2" ParentId="11267" CreationDate="2015-04-23T07:47:59.753" Score="6" Body="&lt;p&gt;In actual conversation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suspects: 1 2 3 4 5 6 7 8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sheriff A has 1 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sheriff B has 1 5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Phone rings.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A: Bad news, someone tapped me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B: OK. Group suspects into 4 pairs with one pair being your shortlist&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A: 12 34 56 78&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B: Is it either 12 or 34?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A: Yes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B: (so culprit is 1) culprit is either 1 or 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A: (so culprit is 1) let's go get him&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Both hang up the phone)&lt;/p&gt;&#xA;" OwnerUserId="11756" LastEditorUserId="11756" LastEditDate="2015-04-23T07:55:21.617" LastActivityDate="2015-04-23T07:55:21.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12609" PostTypeId="2" ParentId="12590" CreationDate="2015-04-23T09:15:59.513" Score="1" Body="&lt;p&gt;I'll add to this as time goes by, if that's ok&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$11: \frac{(2^2)!}{2} - \Gamma(2)$. Four twos, giving $\frac{24}{2} - 1$ for $11$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$111: 3^3 \times 3 + 3^3 + 3$. Six threes. Pretty sure this can be shortened&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1111: 4^4 \times 4 + (4! \times 4) -\space!4$. Six fours. Again, can probably be shortened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still working on the rest, but this gives people a starting point&lt;/p&gt;&#xA;" OwnerUserId="11717" LastEditorUserId="7308" LastEditDate="2015-04-23T10:42:16.090" LastActivityDate="2015-04-23T10:42:16.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12610" PostTypeId="1" AcceptedAnswerId="12615" CreationDate="2015-04-23T09:16:50.787" Score="6" ViewCount="1204" Body="&lt;p&gt;I'm always alone in the corner of the (usually) black/white surface,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although sometimes people come to me at the time of exit,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to cross 14 bridges for a vacation from me,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm half of my real name,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And another half known as animal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;not a fixed size figure, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't find me twice.&lt;/p&gt;&#xA;" OwnerUserId="4312" LastEditorUserId="30633" LastEditDate="2018-01-19T19:23:27.443" LastActivityDate="2018-01-19T19:23:27.443" Title="Alone in the corner - who am I?" Tags="|riddle|computer-puzzle|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12611" PostTypeId="2" ParentId="12590" CreationDate="2015-04-23T10:19:46.743" Score="4" Body="&lt;p&gt;$!n$ is subfactorial, $\Gamma(n)$ is the Gamma function.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$11 =\;!(2+2)+2$&lt;br&gt;&#xA;$111 = \Gamma(3!)-3\cdot3$&lt;br&gt;&#xA;$1111 = 4\cdot(4^4+4!)-!4$&lt;br&gt;&#xA;$11111 = \frac{(5+5)^5-5/5}{!(5-5/5)}$&lt;br&gt;&#xA;$111111 = \frac{(\Gamma(6)/(6+6))^6-6/6}{6+(6\cdot6)/(6+6)}$&lt;br&gt;&#xA;$1111111 = \frac{(7+(7+7+7)/7)^7-7/7}{7+(7+7)/7}$&lt;br&gt;&#xA;$11111111 = \frac{(8+\sqrt{\sqrt{8+8}})^8-8/8}{!\sqrt{8+8}}$&lt;br&gt;&#xA;$111111111 = \frac{(9+!!\sqrt{9})^9-!!\sqrt{9}}{9}$&lt;br&gt;&#xA;$1111111111 = \frac{10^{10}-10/10}{10-10/10}$  &lt;/p&gt;&#xA;" OwnerUserId="11100" LastEditorUserId="11100" LastEditDate="2015-04-23T14:53:44.150" LastActivityDate="2015-04-23T14:53:44.150" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12613" PostTypeId="2" ParentId="12590" CreationDate="2015-04-23T11:54:16.123" Score="2" Body="&lt;p&gt;This doesn't really address the challenge, but I would like to share an expression that allows to write any $n$-th repunit using (at most) 15 copies of $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\left(!\left(\frac{n+n+n+n}n\right)+\frac n n\right)^n-\frac n n}{!\left(\frac{n+n+n+n}n\right)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, in Wolfram Alpha code&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;((subfactorial((n+n+n+n)/n)+n/n)^n-n/n)/subfactorial((n+n+n+n)/n)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which simplifies to the general formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{10^n-1}9&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="4394" LastActivityDate="2015-04-23T11:54:16.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12614" PostTypeId="2" ParentId="12229" CreationDate="2015-04-23T12:20:33.180" Score="0" Body="&lt;p&gt;The final possibilities are July 16, August 15,August 17. I hope everyone is clear about that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Albert was told August .He would be left with 2 options August 15 and August 17. So he could not have made the final statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that case he would would have said something like &quot;It must be Aug15 or Aug 17&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And Lets imagine he was told July&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then He would know Its July 16.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He would say &quot;Now I know it&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this is exactly what he said.We can infer that the B'day is July 16 provided&lt;br&gt;&#xA; Albert is not &quot;stoned&quot; or has lost his mind.&lt;/p&gt;&#xA;" OwnerUserId="11762" LastActivityDate="2015-04-23T12:20:33.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12615" PostTypeId="2" ParentId="12610" CreationDate="2015-04-23T13:00:19.740" Score="10" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's the escape key on a keyboard&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm always alone in the corner of the (usually) black/white surface,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; on a top left corner of a keyboard, usually keyboards are black and white&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Although sometimes people come to me at the time of exit,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; escape means also exit&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You have to cross 14 bridges for a vacation from me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; there is 14 keys between escape key and pause key (F1 to F12, prtscrn and scroll lock)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm half of my real name,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; only half of escape is written on the key&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And another half known as animal.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the other half is ape, which is an animal&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;not a fixed size figure,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I can't think of something for that one&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You don't find me twice.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; there is only one escape key on a keyboard&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10400" LastActivityDate="2015-04-23T13:00:19.740" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="12620" PostTypeId="2" ParentId="12595" CreationDate="2015-04-23T14:12:33.267" Score="14" Body="&lt;p&gt;A more simplified/intuitive way to think about it is if you have the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No matter where the ant and spider start, the spiders go into this formation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/gXc0A.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where the red dots are the spiders, and the green dot is the ant. Notice how the two red spiders on the edges of the cube will only move up and down. This prevents the ant from getting across to the other side without being eaten!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spiders will have to move &lt;b&gt;at most 1/2 a cube edge length&lt;/b&gt; to get to the corner. This means the ant can move a &lt;b&gt;total of 1.5 cube edge lengths&lt;/b&gt; before the spider blocks its path. The ant being smart can try to prevent this setup by not going on the edge center (as shown in the picture), however the spiders only need to maintain/setup the &lt;b&gt;ratio&lt;/b&gt; with the ant. So the ant may pass the half of the cube a few time, but then the spiders will have be able to 'sync to the ratio' with the ant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do the spiders get in sync you ask? Well, every time the ant walks one direction, the spider is crawling its edge to try and reach this magic ratio (which means he is behind where he should be). But as soon as the ant reverses direction, the spider can stand still until he is no longer 'behind' where he is trying to get to and is in sync with the ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that the ant is stuck on the one half of the cube, the third spider can walk along towards the ant and trap him towards another spider!&lt;/p&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-04-23T16:46:25.813" LastActivityDate="2015-04-23T16:46:25.813" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12622" PostTypeId="2" ParentId="6415" CreationDate="2015-04-23T15:33:33.880" Score="0" Body="&lt;p&gt;The answer is if he is allowed to take an infinite number of steps the probability of him stepping off the cliff eventually is 1 (certain) for every value of p&gt;0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To demonstrate this, if you could continue flipping a coin infinitely it would not matter how many times your flipped it, at some point (because it is certain to happen if you could do it infinitely) you would flip a consecutive number of either tails or heads greater than the total number of flips you have already had.  Therefore no matter whether the tails or the heads represent a 'step to the left' the man would always end up stepping over the line of the cliff that he started next to.&lt;/p&gt;&#xA;" OwnerUserId="11767" LastEditorUserId="11767" LastEditDate="2015-04-23T15:40:47.770" LastActivityDate="2015-04-23T15:40:47.770" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="12623" PostTypeId="1" AcceptedAnswerId="12625" CreationDate="2015-04-23T15:50:18.327" Score="3" ViewCount="366" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;There once was a man from Peru,&lt;br&gt;&#xA;  Whose limericks all stopped at line two.&lt;br&gt;&#xA;  Rpl xcjd jfi yfo,&lt;br&gt;&#xA;  Cw hzhyywh ot iojet,&lt;br&gt;&#xA;  Fdy ovjyu ozmn tdz spg yxz oet yxmgybm.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can you figure out what cipher this limerick is using and how to decode it so that we can find out exactly what is written after the limerick was stopped?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, perhaps the word lengths are enough for you to determine what the words are?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The key is a place that the man could also have been from for this limerick to make sense.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-04-24T07:21:54.190" LastActivityDate="2015-04-24T07:21:54.190" Title="A crypto-limerick from Peru" Tags="|cipher|limerick|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12624" PostTypeId="1" AcceptedAnswerId="12731" CreationDate="2015-04-23T15:58:52.780" Score="13" ViewCount="1303" Body="&lt;p&gt;&lt;strong&gt;Heads up: It's possible to solve this puzzle without writing computer code, but it is intended to be solved with computer help.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is a 2-stage puzzle.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Stage 1: Calculation Puzzle&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rules:&lt;/strong&gt;&lt;br&gt;&#xA;1. You must start at A and end at P.&lt;br&gt;&#xA;2. You may not visit a cell more than once.&lt;br&gt;&#xA;3. You must travel along the pathways, which go horizontally, vertically, and diagonally.&lt;br&gt;&#xA;4. A path need not visit every cell (but it can).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find 8 paths from A to P. The final values for each of the 8 paths must equal one of these values:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;1,600&amp;nbsp;&amp;nbsp;1,800&amp;nbsp;&amp;nbsp;1,900&amp;nbsp;&amp;nbsp;2,300&amp;nbsp;&amp;nbsp;2,400&amp;nbsp;&amp;nbsp;3,600&amp;nbsp;&amp;nbsp;4,500&amp;nbsp;&amp;nbsp;5,000&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A couple examples of valid paths, and their values in bold:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;strong&gt;AFKP&lt;/strong&gt; = ((5+4)+8) = &lt;strong&gt;17&lt;/strong&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;strong&gt;AFKHGLP&lt;/strong&gt; = (((((5+4)+8)x6)x7)-1) = &lt;strong&gt;713&lt;/strong&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fSstb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Stage 2: Word Find&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Once you find the 8 paths, group them by path length and sum the values for the paths in each group. This will give you 4 numbers (since there are 4 groups). These numbers are hints to find the words in the grids below. Each grid contains one word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zNeQx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you finish the movie quote and tell us what movie it is from?&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="5373" LastEditDate="2015-04-26T11:40:35.737" LastActivityDate="2015-04-26T15:46:58.120" Title="A Quote from what Movie?" Tags="|mathematics|calculation-puzzle|computer-puzzle|movies|" AnswerCount="1" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="12625" PostTypeId="2" ParentId="12623" CreationDate="2015-04-23T16:05:56.443" Score="2" Body="&lt;p&gt;The encoded words are:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;But then one day,&lt;br&gt;&#xA;He decided to stray,&lt;br&gt;&#xA;And wrote this one all the way through.&lt;br&gt;&#xA;&lt;br&gt;&#xA;The Vigenere key is &lt;strong&gt;qvsevf&lt;/strong&gt;.  Shifting back by one letter provides a key of &lt;strong&gt;Purdue&lt;/strong&gt;.  So the cipher is Z-based, not A-based.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-04-23T16:16:32.517" LastActivityDate="2015-04-23T16:16:32.517" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12626" PostTypeId="1" AcceptedAnswerId="12630" CreationDate="2015-04-23T16:11:02.333" Score="9" ViewCount="912" Body="&lt;p&gt;Here are some variants to the &lt;a href=&quot;http://www.xkcd.com/blue_eyes.html&quot; rel=&quot;noreferrer&quot;&gt;Blue Eyes&lt;/a&gt; puzzle. All the rules are the same, but there are differences in (a) the number of people of different eye colors and (b) the oracle's statement. It is not necessary that everyone leaves - there could be a deadlock as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Variant 1:&lt;/strong&gt; (a) 50 blue, 50 not blue (b) &quot;At least one, but not all of you are blue-eyed.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Variant 2:&lt;/strong&gt; (a) 100 blue, 50 not blue (b) &quot;There are more blue-eyed people than those who are not.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Variant 3:&lt;/strong&gt; (a) 100 blue (b) &quot;At least 10 of you are blue-eyed.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Variant 4:&lt;/strong&gt; (a) 50 blue, 50 not blue (b) &quot;There are at least 10, but not more than 75 blue-eyed people.&quot;&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4269" LastEditDate="2016-02-15T12:01:39.440" LastActivityDate="2016-02-17T14:42:54.150" Title="Variants to blue eyes puzzle" Tags="|logical-deduction|blue-eyes|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12627" PostTypeId="2" ParentId="12543" CreationDate="2015-04-23T17:07:19.887" Score="8" Body="&lt;p&gt;Ok I think I got it this time.  The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tom Cruise&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Come Sail Away&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cruise = sailing away&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Crazy or not, you know who I am,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some say he's crazy for being a Scientologist, but he's world famous&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can use a knife, not just for jam.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mission Impossible &quot;knife in the eye&quot; stunt among other fight scenes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Death is nothing new to me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Interview With A Vampire, or the death count in his action movies&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You know the glasses that help me see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The aviator glasses he wore in Top Gun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rogue or not, I've saved the world.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mission Impossible: he became a rogue agent once but he saved the world&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I salute when the flag is unfurled.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Top Gun or Born on the Fourth of July (according to hint, Top Gun)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The future I thought I once knew,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Minority Report (precognition) or Edge of Tomorrow (reliving the same day)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Instead honesty became my main view.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Few Good Men (&quot;I want the truth!&quot; &quot;You can't handle the truth!&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is it day or night, light or dark,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Knight and Day movie&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tell me who I am, and something cool that I park.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; F-14 from Top Gun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4479" LastActivityDate="2015-04-23T17:07:19.887" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12628" PostTypeId="2" ParentId="12626" CreationDate="2015-04-23T17:20:07.883" Score="0" Body="&lt;p&gt;These variants are very small changes to the original problem that lead to almost the exact same result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Variant 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On day 50: All blue eyed people see 49 people who have blue eyes and did not leave on the 49th day.  They know there is at least 1 person with not blue eyes but who cares.  They all leave on day 50.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Variant 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On day 100: All blue eyed people see 99 people who have blue eyes and did not leave on the 99th day.  They confirm, yes, this is more that half the population.  They all leave on day 100.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Variant 3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On day 91: All blue eyed people see 99 people who have blue eyes and did not leave on the 90th day.  A few days are skipped because if there were 10 blue eyed people they would have left on day 1.  They all leave on day 91.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Variant 4:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On day 41: All blue eyed people see 49 people who have blue eyes and did not leave on the 40th day.  A few days are skipped because if there were 10 blue eyed people they would have left on day 1.  They all leave on day 41.&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2015-04-23T17:20:07.883" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12630" PostTypeId="2" ParentId="12626" CreationDate="2015-04-23T18:51:02.690" Score="7" Body="&lt;p&gt;&lt;strong&gt;Variant 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The oracle saying &quot;not everyone has blue eyes&quot; does not add any information here. Consider the case where there are only two people on the island, one of whom has blue eyes and the other having brown eyes. Being told that &quot;not everyone has blue eyes&quot; gives the blue-eyed person no information - seeing a brown-eyed person, the statement could have been trivial (i.e &quot;not everyone has blue eyes because nobody has blue eyes&quot;) or not, and the blue-eyed person has no way to tell the difference. The brown-eyed person, seeing a blue-eyed person, &lt;em&gt;does&lt;/em&gt; get some information about their eye color - it is not blue. Because they still do not know what color their eyes are (only a little bit about what they &lt;em&gt;aren't&lt;/em&gt;), this statement, by itself, is not enough to cause anyone to leave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the first part of the statement being &quot;At least one person has blue eyes&quot;, the second half of the statement is redundant. Following the same logic as in the original problem, all the blue-eyed people will leave on &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;day 50&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;. Then everyone who remains will know that they do not have blue eyes, therefore not everyone had blue eyes. Again, all of the people who remain will now have no clue as to their eye color other than that it is not blue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Variant 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;There are more blue-eyed people than those who are not.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the case of $N+1$ blue-eyed people and $N$ non-blue-eyed people (we'll say they're all green for simplicity). On the first day, each B sees $N$ B's and $N$ G's. In order for there to be more B's, they must be B. So on day 1, all blue-eyed people leave. If there are $N+2$ B's and $N$ G's, each blue-eyed person sees $N+1$ B's and $N$ G's. They must have blue eyes for there to be a greater number of blues than greens, so again all B's leave on day 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider larger gaps - $N+3$ B's and $N$ G's. Each B cannot tell the difference between the real scenario and $N+2$ B's and $N+1$ G's. Same with $N+4$ vs $N$ not being distinguishable from $N+3$ and $N+1$. However, when nobody leaves the first day, the other possibilities are removed, so all the blue-eyed people leave on day 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for both odd and even numbers of total people, the induction able to start with the halfway point being considered day 0. With 150 people, if there were 76 B's they would leave on day 1. If there were 77 B's, they would leave on day 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following this pattern, all of the blue-eyed people will leave on &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;day 25&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Variant 3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;At least 10 of you are blue-eyed.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This simply starts off the induction a few steps down the line. 10 B's would leave on day 1, 11 B's would leave on day 2, ... so 100 B's (everyone) will leave on &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;day 91&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Variant 4:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;There at least 10, but not more than 75 blue-eyed people.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As with variant 4, this starts off the induction a few steps down the line so all the blue-eyed people will leave no later than day 41. However, what about the second half of the statement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the case where we have 1 B and 1 G, and the oracle says &quot;There is not more than 1 blue-eyed person&quot;. This statement could be said even if there were not any blue-eyed people, so the blue-eyed person does not know if their eyes are blue or not. The green-eyed person sees blue eyes and so knows that they do not have blue eyes. Again, knowing a single color that their eyes are not not does not reveal what color they are. If there were 2 G's and the oracle said &quot;There is not more than 1 blue-eyed person&quot;, neither would know any new information about their eyes. If there are 2 G's and 1 B when the oracle says &quot;There is not more than 1 blue-eyed person&quot;, the two G's both know their eyes are not blue, while the B knows nothing new.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are 2 G's and 1 B and the oracle says &quot;There are not more than 2 blue-eyed people&quot;, nobody will know anything new. The oracle could say that for 2 B's and 1 G, 1 B and 2 G's, or 0 B's and 3 G's. They will each know that everyone else knows that there are not 3 B's, but the only information that can give anyone is what color their eyes are not, which is not actionable. Therefore nobody else's actions (or lack thereof) will give them any information to further distinguish between the possibilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the second half the oracle's statement in this variant does not help give any information, so all of the blue-eyed people will leave on &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;day 41&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1955" LastEditorUserId="4269" LastEditDate="2015-04-24T15:00:34.253" LastActivityDate="2015-04-24T15:00:34.253" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12631" PostTypeId="1" AcceptedAnswerId="12653" CreationDate="2015-04-23T19:01:31.717" Score="5" ViewCount="3342" Body="&lt;p&gt;&lt;strong&gt;Bob&lt;/strong&gt;: I’m doing the video speeches for this celebration thanking the three brothers. Do you have anything to say? How have they influenced you?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sue&lt;/strong&gt;: I’d like to thank them for giving me an intuition that has been invaluable in my daily life. Also, they have given me a general attitude of enthusiasm, and made me more self-sufficient, creative, and spontaneous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bob&lt;/strong&gt;: Wow! How else have they affected you?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sue&lt;/strong&gt;: Well, Aaron has made me adventurous and courageous, Sam has made me more independent and optimistic, and Lenny has made me more confident and loyal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bob&lt;/strong&gt;: Great! So… all positives then?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sue&lt;/strong&gt;: Well, on the other hand, they can make me bossy, quick-tempered, and melodramatic at times, but I am working on that!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;What are the real names of these brothers, and what are they?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint #1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All three brothers are very warm, no doubt&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-04-24T03:02:05.500" LastActivityDate="2015-12-21T06:06:33.263" Title="Who are these 3 brothers?" Tags="|riddle|" AnswerCount="5" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12632" PostTypeId="2" ParentId="12599" CreationDate="2015-04-23T19:06:11.393" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In your picture, there is a square with horizontal sides, all of whose vertices have 5 edges coming out of them. This means that you have 4 odd-degree vertices, which is 2 too many, so your graph can't be drawn with a single non-overlapping pen stroke. But, if you erased one of the diagonals of this middle square, then you'd be in business.  --Mike Earnest&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And such a path would look like :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/42zqk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="11771" LastActivityDate="2015-04-23T19:06:11.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12633" PostTypeId="2" ParentId="12631" CreationDate="2015-04-23T19:35:48.270" Score="6" Body="&lt;p&gt;Partial Answer (Thanks to A E)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could it be that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Aaron is Adrenaline, which makes makes people more adventurous and courageous. Sam is Serotonin, which helps make people 'feel well' and become more optimistic. As for Lenny...He is still a mystery to me.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastActivityDate="2015-04-23T19:35:48.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12634" PostTypeId="2" ParentId="12631" CreationDate="2015-04-23T19:39:31.843" Score="4" Body="&lt;p&gt;OK, following the theme of matching initial letters, let's go with&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - Alcohol&lt;br&gt;&#xA; - Self-esteem, and&lt;br&gt;&#xA; - Love.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have severe doubts about at least one of these :-)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let's try these ones:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - Art&lt;br&gt;&#xA; - Science&lt;br&gt;&#xA; - Money (aka filthy Lucre :-) )&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;... and failing that, totally abandoning first letters:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - Family&lt;br&gt;&#xA; - Work&lt;br&gt;&#xA; - Church&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8875" LastEditorUserId="8875" LastEditDate="2015-04-23T23:14:32.403" LastActivityDate="2015-04-23T23:14:32.403" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12636" PostTypeId="2" ParentId="12542" CreationDate="2015-04-23T19:59:20.230" Score="4" Body="&lt;h1&gt;&lt;strong&gt;MAXIMUM BLOCKED BOXES:&lt;/strong&gt;&lt;/h1&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$73$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;A B&lt;/strong&gt; X | X X X | X X X&lt;br /&gt;&#xA;X X X | X &lt;em&gt;&lt;strong&gt;X&lt;/strong&gt;&lt;/em&gt; X | X X X&lt;br /&gt;&#xA;X X X | X X X | X &lt;strong&gt;A B&lt;/strong&gt;&lt;br /&gt;&#xA;----------------------------&lt;br /&gt;&#xA;X X X | &lt;strong&gt;A&lt;/strong&gt; X X | X X X&lt;br /&gt;&#xA;X X X | &lt;strong&gt;B&lt;/strong&gt; X X | X X X&lt;br /&gt;&#xA;X X X | X X X | X X X&lt;br /&gt;&#xA;----------------------------&lt;br /&gt;&#xA;X X X | X X &lt;strong&gt;A&lt;/strong&gt; | X X X&lt;br /&gt;&#xA;X X X | X X &lt;strong&gt;B&lt;/strong&gt; | X X X&lt;br /&gt;&#xA;X X X | X X X | X X X&lt;/p&gt;&#xA;&lt;p&gt;Let's define cells &lt;em&gt;&lt;strong&gt;C[row,column]&lt;/strong&gt;&lt;/em&gt;.&lt;br /&gt;&#xA;With this configuration, &lt;em&gt;&lt;strong&gt;C[2,5]&lt;/strong&gt;&lt;/em&gt; should contain both the values &lt;strong&gt;A&lt;/strong&gt; and &lt;strong&gt;B&lt;/strong&gt;, and this is impossible!&lt;/p&gt;&#xA;&lt;h1&gt;MINIMUM BLOCKED BOXES:&lt;/h1&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;A B&lt;/strong&gt; O | O O O | O O O&lt;br /&gt;&#xA;O O O | O &lt;em&gt;&lt;strong&gt;X&lt;/strong&gt;&lt;/em&gt; O | O O O&lt;br /&gt;&#xA;O O O | O O O | O &lt;strong&gt;A B&lt;/strong&gt;&lt;br /&gt;&#xA;----------------------------&lt;br /&gt;&#xA;O O O | &lt;strong&gt;A&lt;/strong&gt; O O | O O O&lt;br /&gt;&#xA;O O O | &lt;strong&gt;B&lt;/strong&gt; O O | O O O&lt;br /&gt;&#xA;O O O | O O O | O O O&lt;br /&gt;&#xA;----------------------------&lt;br /&gt;&#xA;O O O | O O &lt;strong&gt;A&lt;/strong&gt; | O O O&lt;br /&gt;&#xA;O O O | O O &lt;strong&gt;B&lt;/strong&gt; | O O O&lt;br /&gt;&#xA;O O O | O O O | O O O&lt;/p&gt;&#xA;&lt;p&gt;The &lt;strong&gt;X&lt;/strong&gt; is the only blocked cell, O cells have generic values, while &lt;strong&gt;A&lt;/strong&gt; and &lt;strong&gt;B&lt;/strong&gt; are two different values. If you only block one cell, like in this case, there exists at least one configuration (as the one showed here) which makes the sudoku unsolvable.&lt;/p&gt;&#xA;&lt;h1&gt;&lt;strong&gt;BOUNTY QUESTION:&lt;/strong&gt;&lt;/h1&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;strong&gt;NO!&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;It's very easy to prove: take any grid &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; and generate the sudoku-friendly subsets &lt;span class=&quot;math-container&quot;&gt;$S_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$S_2$&lt;/span&gt; blocking &lt;span class=&quot;math-container&quot;&gt;$80$&lt;/span&gt; cells, leaving visible respectively &lt;em&gt;&lt;strong&gt;C[1,1]&lt;/strong&gt;&lt;/em&gt; and &lt;em&gt;&lt;strong&gt;C[1,2]&lt;/strong&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;This is &lt;span class=&quot;math-container&quot;&gt;$S_1$&lt;/span&gt;:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;O&lt;/strong&gt; X X | X X X | X X X&lt;br /&gt;&#xA;X X X | X X X | X X X&lt;br /&gt;&#xA;X X X | X X X | X X X&lt;br /&gt;&#xA;----------------------------&lt;br /&gt;&#xA;X X X | X X X | X X X&lt;br /&gt;&#xA;X X X | X X X | X X X&lt;br /&gt;&#xA;X X X | X X X | X X X&lt;br /&gt;&#xA;----------------------------&lt;br /&gt;&#xA;X X X | X X X | X X X&lt;br /&gt;&#xA;X X X | X X X | X X X&lt;br /&gt;&#xA;X X X | X X X | X X X&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;And this is &lt;span class=&quot;math-container&quot;&gt;$S_2$&lt;/span&gt;:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;X &lt;strong&gt;O&lt;/strong&gt; X | X X X | X X X&lt;br /&gt;&#xA;X X X | X X X | X X X&lt;br /&gt;&#xA;X X X | X X X | X X X&lt;br /&gt;&#xA;----------------------------&lt;br /&gt;&#xA;X X X | X X X | X X X&lt;br /&gt;&#xA;X X X | X X X | X X X&lt;br /&gt;&#xA;X X X | X X X | X X X&lt;br /&gt;&#xA;----------------------------&lt;br /&gt;&#xA;X X X | X X X | X X X&lt;br /&gt;&#xA;X X X | X X X | X X X&lt;br /&gt;&#xA;X X X | X X X | X X X&lt;/p&gt;&#xA;&lt;p&gt;As you can see, both are legal patterns (ergo, sudoku-friendly), though &lt;strong&gt;they don't have a unique solution&lt;/strong&gt; since a single information is never enough to solve univocally a sudoku!&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-26T09:26:49.350" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12637" PostTypeId="1" AcceptedAnswerId="12646" CreationDate="2015-04-23T20:12:53.393" Score="6" ViewCount="462" Body="&lt;p&gt;First riddle, be nice?   Hopefully it isn't too easy.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;First is new and bullied by Second&lt;br&gt;&#xA;Third claims being first to all servings&lt;br&gt;&#xA;The humpback is Fourth and blames his twin for everything&lt;br&gt;&#xA;Fifth is full as soon as his meal begins&lt;br&gt;&#xA;Sixth has a twin he denies knowing&lt;br&gt;&#xA;Seventh is last to begin eating&lt;br&gt;&#xA;Eight tries hiding to be forgotten  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; (Solve the riddle) Who are the Eight?&lt;/p&gt;&#xA;" OwnerUserId="11251" LastEditorUserId="11251" LastEditDate="2015-04-23T20:46:14.997" LastActivityDate="2015-04-24T10:12:58.147" Title="A Banquet for Eight" Tags="|riddle|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12638" PostTypeId="2" ParentId="11267" CreationDate="2015-04-23T20:39:15.293" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;they number the suspects in binary, then each officer tells the other which digit he needs to incriminate the right suspect. the other officer can either give him that digit right away if he has it, or request a digit himself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;example&lt;/p&gt;&#xA;&#xA;&lt;p&gt;officer A has&lt;br&gt;&#xA;011&lt;br&gt;&#xA;101  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;officer B has&lt;br&gt;&#xA;000&lt;br&gt;&#xA;011  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;officer A needs either the first or second digit.&lt;br&gt;&#xA;officer B needs either the second or third digit.&lt;br&gt;&#xA;officer A tells B the third digit it 1.&lt;br&gt;&#xA;officer B tells A the first digit is 0.  &lt;/p&gt;&#xA;" OwnerUserId="9413" LastActivityDate="2015-04-23T20:39:15.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12639" PostTypeId="2" ParentId="12596" CreationDate="2015-04-23T20:49:42.937" Score="1" Body="&lt;h2&gt;PARTIAL ANSWER&lt;/h2&gt;&#xA;&lt;p&gt;This how far I've gotten. I hope someone with more a recent education than I can take correct what I have wrong and finish it up.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;tl;dr&lt;/h2&gt;&#xA;&lt;p&gt;Here are pieces I think will help:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$D_A^2=49=r_P^2+z_S^2-2z_Sz_A$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$D_B^2=81=r_P^2+z_S^2-2z_Sz_B$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$D_C^2=121=r_P^2+z_S^2-2z_Sz_C$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$D_D^2=??=r_P^2+z_S^2-2z_Sz_D$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$5z_A-9z_B+4z_C=0$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{1}{z_S}=\frac{z_A-z_B}{4}=\frac{z_A-z_C}{9}=\frac{z_B-z_C}{5}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$r_C^2=2r_Ph$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$r_P^2=\pm(x_Ax_B+y_Ay_B+z_Az_B)$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$r_P^2=\pm(x_Cx_D+y_Cy_D+z_Cz_D)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Name the 4 points A, B, C, D and their coordinates as &lt;span class=&quot;math-container&quot;&gt;$(x_A,y_A,z_A)$&lt;/span&gt;, etc.&lt;br&gt;&#xA;Name the ship S and its coordinates are &lt;span class=&quot;math-container&quot;&gt;$(x_S,y_S,z_S)$&lt;/span&gt;&lt;br&gt;&#xA;Name the planet P and its coordinates are &lt;span class=&quot;math-container&quot;&gt;$(x_P,y_P,z_P)$&lt;/span&gt; with a radius &lt;span class=&quot;math-container&quot;&gt;$r_P$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;For convenience, set the center of the planet &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; to be at the origin.&lt;br&gt;&#xA;In addition, orient the axes such that the ship &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; is directly above the planet along the z-axis.&lt;/p&gt;&#xA;&lt;p&gt;This tells us that &lt;span class=&quot;math-container&quot;&gt;$x_P=y_P=z_P=x_S=y_S=0$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The general equation for the distance between two points in 3D space is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$$&lt;/span&gt;&#xA;Therefore, the distance between the ship &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; and point &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$D_A=7=\sqrt{(x_S-x_A)^2+(y_S-y_A)^2+(z_S-z_A)^2}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$D_A=7=\sqrt{(0-x_A)^2+(0-y_A)^2+(z_S-z_A)^2}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$D_A=7=\sqrt{x_A^2+y_A^2+(z_S^2-2z_Sz_A+z_A^2)}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$D_A=7=\sqrt{(x_A^2+y_A^2+z_A^2)+(z_S^2-2z_Sz_A)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now, since the planet's center is at the origin, we know &lt;span class=&quot;math-container&quot;&gt;$x_A^2+y_A^2+z_A^2=r_P^2$&lt;/span&gt;&lt;br&gt;&#xA;We can plug that in and square both sides to get&#xA;&lt;span class=&quot;math-container&quot;&gt;$$D_A^2=49=r_P^2+z_S^2-2z_Sz_A$$&lt;/span&gt;&#xA;It is trivial to expand that to the other 3 points&#xA;&lt;span class=&quot;math-container&quot;&gt;$$D_B^2=81=r_P^2+z_S^2-2z_Sz_B$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$D_C^2=121=r_P^2+z_S^2-2z_Sz_C$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$D_D^2=??=r_P^2+z_S^2-2z_Sz_D$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Right now, we're sitting at 4 equations and 7 unknowns (&lt;span class=&quot;math-container&quot;&gt;$r_P,z_S,z_A,z_B,z_C,z_D,D_D$&lt;/span&gt;). That's close but we're not there yet. We can play around with the formulas for &lt;span class=&quot;math-container&quot;&gt;$D_A$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D_B$&lt;/span&gt; because they're values are known.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$D_A^2+32 = 49+32=81=D_B^2$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$r_P^2+z_S^2-2z_Sz_A+32=r_P^2+z_S^2-2z_Sz_B$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$-2z_Sz_A+32=-2z_Sz_B$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$2z_Sz_B+32=2z_Sz_A$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$z_Sz_B+16=z_Sz_A$$&lt;/span&gt;&#xA;We can do the same thing for &lt;span class=&quot;math-container&quot;&gt;$D_A$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D_C$&lt;/span&gt; as well as &lt;span class=&quot;math-container&quot;&gt;$D_B$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D_C$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$z_Sz_C+36=z_Sz_A$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$z_Sz_C+20=z_Sz_B$$&lt;/span&gt;&#xA;If we solve all three for &lt;span class=&quot;math-container&quot;&gt;$z_S$&lt;/span&gt;, we can set them equal.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$z_S=16/(z_A-z_B)=36/(z_A-z_C)=20/(z_B-z_C)$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$z_S=4/(z_A-z_B)=9/(z_A-z_C)=5/(z_B-z_C)$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{1}{z_S}=\frac{z_A-z_B}{4}=\frac{z_A-z_C}{9}=\frac{z_B-z_C}{5}$$&lt;/span&gt;&#xA;This gives us the relationship&#xA;&lt;span class=&quot;math-container&quot;&gt;$$5z_A-9z_B+4z_C=0$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Let's look at the second observation. The formula for the surface area a &lt;a href=&quot;http://mathworld.wolfram.com/SphericalCap.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;spherical cap&lt;/a&gt; &lt;span class=&quot;math-container&quot;&gt;$A_P$&lt;/span&gt; that is distance &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; from the origin and the area of a circle &lt;span class=&quot;math-container&quot;&gt;$A_C$&lt;/span&gt; are  given by&#xA;&lt;span class=&quot;math-container&quot;&gt;$$A_P=2\pi r_Ph$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$A_C=\pi r_C^2$$&lt;/span&gt;&#xA;We're told those are equal so we can drop the &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt; to get the equality&#xA;&lt;span class=&quot;math-container&quot;&gt;$$r_C^2=2r_Ph$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Unfortunately, this introduces only 1 new equation and 2 new unknowns. I think the solution will involve some trig to find &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; based on &lt;span class=&quot;math-container&quot;&gt;$r_P$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$z_S$&lt;/span&gt;, and tangent lines but I don't have the brainpower for that right now.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;I'm sure that the probes landing on opposite sides matters but I couldn't get it to do anything useful because it added more unknowns instead of removing them. Here's what I wrote for that section:&lt;/p&gt;&#xA;&lt;p&gt;We also know that the probes landed on opposite sides of the sphere. That means the line connecting them passes through the center which means it is of length &lt;span class=&quot;math-container&quot;&gt;$2r_P$&lt;/span&gt;. That gives us the following for the distance between probe &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt;. &lt;strong&gt;(Here I have to assume that the first two distances are for the first two probes. If that's not the case, this section must be reworked.)&lt;/strong&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$2r_P=\sqrt{(x_A-x_B)^2+(y_A-y_B)^2+(z_A-z_B)^2}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$2r_P=\sqrt{(x_A^2-2x_Ax_B+x_B^2)+(y_A^2-2y_Ay_B+y_B^2)+(z_A^2-2z_Az_B+z_B^2)}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$2r_P=\sqrt{(x_A^2+y_A^2+z_A^2)+(x_B^2+y_B^2+z_B^2)-2(x_Ax_B+y_Ay_B+z_Az_B)}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$2r_P=\sqrt{(r_P^2)+(r_P^2)-2(x_Ax_B+y_Ay_B+z_Az_B)}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$2r_P=\sqrt{2r_P^2-2(x_Ax_B+y_Ay_B+z_Az_B)}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$4r_P^2=\pm2(r_P^2-(x_Ax_B+y_Ay_B+z_Az_B))$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$r_P^2=\pm(x_Ax_B+y_Ay_B+z_Az_B)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;(I'm not super-confident in how the &lt;span class=&quot;math-container&quot;&gt;$\pm$&lt;/span&gt; comes in once you square both sides but I think it has to be considered somewhere.)&lt;/p&gt;&#xA;&lt;p&gt;We can do the same thing for the other two probes&#xA;&lt;span class=&quot;math-container&quot;&gt;$$r_P^2=\pm(x_Cx_D+y_Cy_D+z_Cz_D)$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-23T20:49:42.937" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12640" PostTypeId="1" AcceptedAnswerId="12642" CreationDate="2015-04-23T20:53:25.037" Score="6" ViewCount="847" Body="&lt;p&gt;Joe's morning at the advertising agency:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;10:30 am&lt;/strong&gt;: Joe looks on his empty desktop and realizes he has no cigarettes. He asks his coworkers to get him some cigarettes from the&#xA;  interns.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;10:31 am&lt;/strong&gt;: Joe has a single cigarette on his desk.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;10:32 am&lt;/strong&gt;: Joe has 2 cigarettes on his desk.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;10:33 am&lt;/strong&gt;: Joe has 3 cigarettes on his desk, so he starts smoking.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;10:34 am&lt;/strong&gt;: Joe has a single cigarette on his desk, and now has an ashtray.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;10:35 am&lt;/strong&gt;: Joe has no more cigarettes on his desk, but keeps the ashtray for later.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;10:36 am&lt;/strong&gt;: Joe's desktop is still empty, except for a single cigarette and the ashtray.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What's on Joe's desk at 10:38 am, and why?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-04-23T21:12:40.893" LastActivityDate="2015-04-24T04:38:45.087" Title="What's on Joe's desk at 10:38 am?" Tags="|riddle|lateral-thinking|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12641" PostTypeId="2" ParentId="12229" CreationDate="2015-04-23T21:02:37.990" Score="0" Body="&lt;p&gt;The same question I also got, and in the event of searching for an answer I routed here. Finally I figured out myself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Albert knows the month (July/August)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since Albert knows the month that is July but he does not know the date, after 2nd statement he was 100% sure that he knows the answer,becuase July having 2 dates(14, 16), if it is 14 then Bernard could not make the second statement so it should be 16.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is in Auguest then it is hard for Albert to make the third statement as there are two dates 15 and 17.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this makes clear.. :)&lt;/p&gt;&#xA;" OwnerUserId="11774" LastEditorUserId="11774" LastEditDate="2015-04-23T21:54:18.040" LastActivityDate="2015-04-23T21:54:18.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12642" PostTypeId="2" ParentId="12640" CreationDate="2015-04-23T21:05:46.280" Score="9" Body="&lt;p&gt;Joe (not me) has three cigarettes next to his ashtray.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pattern is the Roman numerals representing the minutes past 10:30 AM. The cigarettes on the desk represent I's, and the ashtray stands for a V.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At 10:40 he might have a can of soda on his desk and taken the ashtray off, to represent an X.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-04-24T04:38:45.087" LastActivityDate="2015-04-24T04:38:45.087" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12643" PostTypeId="1" AcceptedAnswerId="12830" CreationDate="2015-04-23T21:34:54.617" Score="9" ViewCount="725" Body="&lt;p&gt;On a tiny spherical planet there exist $N$ wolves and 1 hare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The planet is so small any of these creatures can circle it in exactly 1 day.  No creature needs to sleep or eat.  The wolves communicate by howling. The hare doesn't understand.  All creatures play perfectly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the hare has 4 very lucky feet, the original positions are the most favorable ones for the hare and it can always guess where the wolves are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the minimum number $N$ that guarantees the hare is caught? How long does it take?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following cases:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The wolves can always smell where the hare is.&lt;/li&gt;&#xA;&lt;li&gt;The wolves have stuffy noses and can only smell the hare if it is within 1 hour's run.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="99" LastEditorUserId="10615" LastEditDate="2015-04-24T15:04:26.573" LastActivityDate="2015-07-17T16:34:52.350" Title="Wolves and a Hare on a tiny planet" Tags="|mathematics|geometry|pursuit-evasion|" AnswerCount="5" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="12644" PostTypeId="2" ParentId="12631" CreationDate="2015-04-23T22:31:11.510" Score="3" Body="&lt;p&gt;Ok, initial gut reaction, because it's 3 elements that influence behavior:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Id, Ego, and Super-ego&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;" OwnerUserId="11775" LastActivityDate="2015-04-23T22:31:11.510" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12645" PostTypeId="2" ParentId="12542" CreationDate="2015-04-23T22:52:26.980" Score="-1" Body="&lt;p&gt;for the minimum , i would choose this &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;7## | 123 | 456&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;#1# | 456 | 789&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;##4 | 789 | 123&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;------|------|------&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;147 | 938 | 562&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;258 | 671 | 394&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;369 | 245 | 817&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;------|------|------&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;471 | 562 | 938&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;582 | 394 | 671&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;693 | 817 | 245&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9517" LastEditorUserId="9517" LastEditDate="2015-04-23T22:57:34.347" LastActivityDate="2015-04-23T22:57:34.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12646" PostTypeId="2" ParentId="12637" CreationDate="2015-04-23T23:10:01.980" Score="5" Body="&lt;p&gt;Here is my guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The phases of the moon. &lt;br&gt;&#xA; First is the new moon. &lt;br&gt;&#xA; Second is the waxing cresent, where the sunlit part is beginning to overtake the dark part. &lt;br&gt;&#xA; Third is first quarter? &lt;br&gt;&#xA; Fourth is waxing gibbous; it looks like a humpback. &lt;br&gt;&#xA; Fifth is the full moon. &lt;br&gt;&#xA; Sixth is waning gibbous, the twin of waxing gibbous. &lt;br&gt;&#xA; Seventh is third quarter. ? &lt;br&gt;&#xA; Eighth is waning cresent, which is almost completely dark. &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't really understand the personalities of Fourth and Sixth, and Third and Seventh don't seem to fit.&lt;/p&gt;&#xA;" OwnerUserId="6206" LastActivityDate="2015-04-23T23:10:01.980" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12647" PostTypeId="2" ParentId="12596" CreationDate="2015-04-24T00:41:09.570" Score="4" Body="&lt;p&gt;The censored number is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is a cross section of the planet containing all four probes, illustrating some terms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\qquad\qquad\qquad\qquad$&lt;img src=&quot;https://i.stack.imgur.com/X6SpZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify, the distances to the four probes are $D_{A^+},D_{A^-},D_{B^+},$ and $D_{B^-}$, and these four are some rearrangement of the numbers $7,9,11$, and [CENSORED]. Without loss of generality, assume&#xA;$$&#xA;D_{A^+}=[CENSORED]&#xA;$$&#xA;Now, applying Law of Cosines to the triangle with angle $\varphi$, we get&#xA;$$&#xA;D_{A^-}^2=H^2+R^2-2RH\cos\varphi&#xA;$$&#xA;and in the same vein,&#xA;$$&#xA;D_{A^+}^2=H^2+R^2-2RH\cos(180-\varphi)=H^2+R^2+2RH\cos\varphi.&#xA;$$&#xA;Adding these together,&#xA;$$&#xA;D_{A^-}^2+D_{A^+}^2=2R^2+2H^2\tag{1}&#xA;$$&#xA;Similarly,&#xA;$$&#xA;D_{B^-}^2+D_{B^+}^2=2R^2+2H^2\tag2&#xA;$$&#xA;Combining (1) and (2), we get that $D_{A^+}^2=D_{B^+}^2+D_{B^-}^2-D_{A^-}^2$, so the three possibilities for $D_{A^+}^2$ are&#xA;$$&#xA;11^2+9^2-7^2=153\qquad 7^2+11^2-9^2=89\qquad 9^2+7^2-11^2=9&#xA;$$&#xA;The left two cases &lt;strong&gt;would imply that $D_{A^+}$ was one of the larger two distances&lt;/strong&gt; among $7,9,11$ and [CENSORED] (since $\sqrt{153}&amp;gt;\sqrt{89}&amp;gt;9$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are we to make of the Wesley's statement? The area they can see is at most half the planet's area, so at most $2\pi R^2$. Since Wesley said this is equal to $\pi\times[CENSORED]^2$, it follows that $\pi\times[CENSORED]^2\le 2\pi R^2$, so&#xA;$$&#xA;[CENSORED]^2=D_{A^+}^2\le 2R^2\le R^2+H^2\tag 3&#xA;$$&#xA;since $R\le H$. Combining (1) and (3), we see that $$D_{A^-}^2=2R^2+2H^2-D_{A^+}^2\ge R^2+H^2\ge D_{A^+}^2$$ so that $D_{A^+}$ is the &lt;em&gt;smaller&lt;/em&gt; of the two $A$ probe distances. This means that the above &lt;strong&gt;bold&lt;/strong&gt; statement cannot be true, so that we must have $D_{A^+}^2=9$, giving us the answer as claimed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also make an educated guess as to &lt;em&gt;why&lt;/em&gt; this number is so vulgar:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it looks like a sideways butt.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-04-24T03:26:21.477" LastActivityDate="2015-04-24T03:26:21.477" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12648" PostTypeId="1" AcceptedAnswerId="12675" CreationDate="2015-04-24T01:44:39.313" Score="5" ViewCount="847" Body="&lt;p&gt;There are 4 identical tumblers each filled with the same number of 1 kg balls. One of these four tumblers contains all defective underweight balls. You have &lt;a href=&quot;https://i.stack.imgur.com/hqIe2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;a weighing scale&lt;/a&gt; which you can use to weigh the tumblers, but it can be used only once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can you figure out which tumbler contains the defective balls?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The tumblers are not transparent, but it doesn't matter, because the underweight balls look and feel the same as the genuine balls anyway. Underweight ball can be between (0,1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;I am not the original author of this question. This question was asked in an interview to my friend and he agrees that this is not solvable until and unless there are restrictions put on the underweight quantity. I don't wish to modify the question now and have accepted the answer which concurs with this.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="11776" LastEditorUserId="5373" LastEditDate="2017-02-07T12:35:56.327" LastActivityDate="2017-02-07T12:35:56.327" Title="Spot the tumbler" Tags="|logical-deduction|weighing|" AnswerCount="4" CommentCount="12" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12649" PostTypeId="2" ParentId="12643" CreationDate="2015-04-24T02:20:21.520" Score="3" Body="&lt;p&gt;I think case 1: 3 wolves (edited solution). Form an equilateral triangle around the hare. Let the hare run for an edge midpoint. Take the line between the hare's starting point and the edge midpoint, and double it. Let one wolf on that edge run to that point (the wolf and rabbit will get there at the same time). The other two wolves maintain a shrinking equilateral triangle with the first wolf. Continue to shrink until the hare is caught. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding time: Let the wolves start at the north pole and the hare at the south. The wolves will take 24 hours to catch the hare if it just sits there, so it will take at least 24 hours with the hare playing optimally (since the starting position is optimized for the hare and the hare is capable of perfect play). I'm a little shaky on the timing of what happens in the shrinking capture plan (I think some calculus might be needed, and also I must play perfectly for the rabbit, which I have little confidence in my ability to do) but basically once the wolves cross the equator, the triangle is formed and the rabbit can run for the edge midpoint between two wolves. So we're at 12 hours now. As soon as the rabbit starts moving, the interceptor wolf starts running backwards for ((sqrt(3)/2)*8) - (((sqrt(3)/2)*8)/3) = 4.61880215352 hours (using equations off wikipedia to calculate height and height of center from edge midpoint when triangle side length = 8), which brings us to ~16.6 hours. However, the rabbit doesn't have to run willingly to meet the wolf, and could probably juke around for maybe 1-2 more hours. However, this calculation requires predicting perfect play by the rabbit, and I have no way of doing that. The hare is better off camping the south pole here, and maybe juking a little bit at the end to last a few more minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why 2 isn't enough in case 1: Let the 2 wolves come at the hare from different distances. The hare can position itself to be equidistant from either wolf by running towards the farther-away wolf. Now, let the wolves come at the hare from opposite directions, so that the three creatures form a line. The hare can draw a perpendicular line and follow it, and since the wolves will have to chase at an angle, they will fall behind. So let the wolves form an angle with the hare: the hare bisects the angle and follows this line, but in the direction away from the wolves. The wolves are even farther behind now! Since these are the only ways for the wolves to approach the hare, the hare always escapes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 2: 8 wolves is enough for a naive comb: start them out at the north pole and send them southward, spread out so that by the time they get to the equator, there's 2 hours travel time between each wolf, then have them continue on to the south pole. Since the equator will be 24 hours travel time around, this guarantees finding the rabbit. Create a shrinking octagon around it. This should take, again, at least 24 hours, since the hare can camp at the south pole, and can take a tiny bit more time than that if the hare jukes a little, which again requires knowledge of perfect play to calculate, but since the wolves can reach the midpoint of the edge before the hare can, I think even with maximum juking the hare wouldn't make it more than 24 hours 15 minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best guess for the timing: I think the hare's best plan really might be to just sit on the opposite pole from the wolves' starting point and never move. This guarantees it 24 hours. Any fancy plays could cost it time because it's running to meet the wolves somewhat, so I'm guessing all methods with perfect play take exactly 24 hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I retract my hypothesized solution in 6; I found a flaw in it. 8 definitely works for case 2.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="11154" LastEditorUserId="11154" LastEditDate="2015-04-24T14:55:04.040" LastActivityDate="2015-04-24T14:55:04.040" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="12652" PostTypeId="2" ParentId="12631" CreationDate="2015-04-24T03:09:28.343" Score="7" Body="&lt;p&gt;They are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Constellations in the Zodiac: Aries,  Leo, and Sagittarius&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11778" LastEditorUserId="463" LastEditDate="2015-04-24T03:21:29.773" LastActivityDate="2015-04-24T03:21:29.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12653" PostTypeId="2" ParentId="12631" CreationDate="2015-04-24T03:11:42.027" Score="5" Body="&lt;p&gt;Thanks to the hint, the three brothers are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the three Fire signs of the Zodiac. People born under the sign of Aaron/Aries are known to be &lt;a href=&quot;http://www.ganeshaspeaks.com/aries/aries-traits.action&quot; rel=&quot;nofollow&quot;&gt;adventurous and courageous&lt;/a&gt;. People born under the sign of Sam/Sagittarius are more &lt;a href=&quot;http://www.123newyear.com/zodiac-signs/sagittarius.html&quot; rel=&quot;nofollow&quot;&gt;independent and optimistic&lt;/a&gt;. People born under the sign of Lenny/Leo are &lt;a href=&quot;http://www.123newyear.com/zodiac-signs/leo.html&quot; rel=&quot;nofollow&quot;&gt;confident and loyal&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; people born under these three fire signs are known to be bossy, quick-tempered, and melodramatic at times. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5079" LastEditorUserId="11778" LastEditDate="2015-04-24T03:19:11.373" LastActivityDate="2015-04-24T03:19:11.373" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="12654" PostTypeId="2" ParentId="12648" CreationDate="2015-04-24T03:29:46.993" Score="2" Body="&lt;p&gt;Take 4 balls out of one, 8 out of another, 12 out of a third, and 16 out of the last one and weigh all tumblers together. Based on how underweight the scale reads, you have your answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way if the scale is &amp;lt;4kg under your 4*N kg expected total, then the tumbler missing 4 balls is underweighted. If 4kg &amp;lt; scale &amp;lt; 8kg then its the tumbler missing 8 balls.  If 8kg &amp;lt; scale &amp;lt; 12kg it's the tumbler missing 12 balls. Lastly, if 12kg &amp;lt; scale then it's the tumbler missing 16 balls.&lt;/p&gt;&#xA;" OwnerUserId="11778" LastEditorUserId="11778" LastEditDate="2015-04-24T03:54:21.837" LastActivityDate="2015-04-24T03:54:21.837" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="12655" PostTypeId="1" AcceptedAnswerId="12660" CreationDate="2015-04-24T07:29:53.053" Score="7" ViewCount="592" Body="&lt;p&gt;&lt;em&gt;If you care about the story, you may want to see &lt;a href=&quot;https://puzzling.stackexchange.com/questions/12379/enlightened-part-i-finding-the-temple&quot;&gt;Part I&lt;/a&gt; if you haven't already. Otherwise, just enjoy the puzzle!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ricky:&lt;/strong&gt; Chris!&lt;br&gt;&#xA;&lt;strong&gt;Chris:&lt;/strong&gt; What is it this time?&lt;br&gt;&#xA;&lt;strong&gt;Ricky:&lt;/strong&gt; We have another problem.&lt;br&gt;&#xA;&lt;strong&gt;Chris:&lt;/strong&gt; Another problem? I can't solve all your problems, you know.&lt;br&gt;&#xA;&lt;strong&gt;Ricky:&lt;/strong&gt; But-&lt;br&gt;&#xA;&lt;strong&gt;Chris:&lt;/strong&gt; &lt;em&gt;(sighing)&lt;/em&gt; Just say your problem.&lt;br&gt;&#xA;&lt;strong&gt;Ricky:&lt;/strong&gt; So we found the temple, and entered a large courtyard. The door to the temple, however, is locked, and seems to be resistant against the tools we have available.&lt;br&gt;&#xA;&lt;strong&gt;Chris:&lt;/strong&gt; How can I help?&lt;br&gt;&#xA;&lt;strong&gt;Ricky:&lt;/strong&gt; A careful reading of some text we found on a pillar in the courtyard says what we have to do. But I'm unable to do it. It seems to be some logic game - here, I'll explain it to you. &lt;em&gt;(pulls out a roughly drawn diagram from his pocket)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FBm4G.png&quot; alt=&quot;example&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ricky:&lt;/strong&gt; The courtyard is made up of a grid of square tiles. Each tile has a lantern embedded into it with four slits which let light escape facing up, down, left and right. These lanterns can be extended to provide light or retracted to produce no light. Whenever you choose to extend the lantern of a tile, light pours out of the four slits. We mark this light using a number - the lantern itself is 5 units bright, and every tile the light travels it loses a unit of light until it reaches zero. You can see this from the left diagram.&lt;br&gt;&#xA;&lt;strong&gt;Ricky:&lt;/strong&gt; However, the right diagram is more interesting. You see, when two or more paths of light intersect, their light values add. And if they hit a pillar or wall, the stream of light is blocked. It is also interesting that the lanterns themselves can be lit above five as the light from other lanterns passes through them.&lt;br&gt;&#xA;&lt;strong&gt;Chris:&lt;/strong&gt; This is all very well and good, but what's it got to do with opening the door?&lt;br&gt;&#xA;&lt;strong&gt;Ricky:&lt;/strong&gt; The courtyard of the temple has a puzzle in it that, once solved, will open the door. The courtyard has a ten by ten grid of tiles in it, each with a lantern embedded into it and a number written on it. We just need to extend the right lanterns so that the units of light for each square matches the number written on it.&lt;br&gt;&#xA;&lt;strong&gt;Chris:&lt;/strong&gt; That surely can't be so hard?&lt;br&gt;&#xA;&lt;strong&gt;Ricky:&lt;/strong&gt; You would think so. However, there is one problem that prevents us from solving this ancient puzzle... &lt;em&gt;(trails off)&lt;/em&gt;&lt;br&gt;&#xA;&lt;strong&gt;Chris:&lt;/strong&gt; Which is?&lt;br&gt;&#xA;&lt;strong&gt;Ricky:&lt;/strong&gt; Maybe it's better to show you. &lt;em&gt;(pulls out a diagram from his pocket)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EwcKs.png&quot; alt=&quot;puzzle&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ricky:&lt;/strong&gt; The tiles' numbers are written in an ancient language. We can't translate them. All we know is that one symbol maps to one number.&lt;br&gt;&#xA;&lt;strong&gt;Chris:&lt;/strong&gt; &lt;em&gt;(rather desperate)&lt;/em&gt; Is there anything else you know?&lt;br&gt;&#xA;&lt;strong&gt;Ricky:&lt;/strong&gt; No. Wait - actually, there is. We know the following symbol - &lt;em&gt;(points to the diagram)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GmxBQ.png&quot; alt=&quot;zero&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ricky:&lt;/strong&gt; - means zero.&lt;br&gt;&#xA;&lt;strong&gt;Chris:&lt;/strong&gt; Well, this is certainly a challenge. But surely it can't be as bad as reconstructing that map!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can you help Chris figure out which lanterns should be extended?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="7308" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-04-24T15:10:40.470" Title="Enlightened, Part II: A Curious Courtyard" Tags="|logical-deduction|logic-grid|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12656" PostTypeId="1" AcceptedAnswerId="12669" CreationDate="2015-04-24T07:51:21.977" Score="3" ViewCount="305" Body="&lt;p&gt;Traditionally, after the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/12623/a-crypto-limerick-from-peru&quot;&gt;limerick from Peru&lt;/a&gt; comes one with even fewer lines revealed:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There once was a man from Verdun,&lt;br&gt;&#xA;  Ekptl hkndekbmq eff oopqvodw ro gkhl uhr.&lt;br&gt;&#xA;  Ol onrh lts r fbel,&lt;br&gt;&#xA;  Gash ona hrm kh Odez,&lt;br&gt;&#xA;  Rmw bshobrodw onmp qpah lsa gzh.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can you figure out what encryption method is being used on the second line (since I'm sure you've figured out by now what the plaintext is), and use that knowledge to figure out how this Verdunnian man's limerick went?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: each line is  encrypted separately, and it's not a Vigenère cipher this time around.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The key is a writer of complete nonsense.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-04-24T16:31:20.317" Title="Another crypto-limerick, this time from Verdun" Tags="|cipher|limerick|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12657" PostTypeId="1" AcceptedAnswerId="12658" CreationDate="2015-04-24T08:47:18.160" Score="8" ViewCount="535" Body="&lt;p&gt;What music band does this HTML snippet represent?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;lt;i style=&quot;color:#000&quot;&amp;gt;PPPP&amp;lt;/i&amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4394" LastActivityDate="2015-04-24T08:53:18.610" Title="What band is this?" Tags="|rebus|computer-puzzle|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12658" PostTypeId="2" ParentId="12657" CreationDate="2015-04-24T08:52:50.843" Score="8" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Black Eyed Peas.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We have four Ps in a tag that makes them black and italicizes them. Black Italicized Ps = Black Eyed Peas. Also, there were 4 band members, thus 4 Ps.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11100" LastActivityDate="2015-04-24T08:52:50.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12660" PostTypeId="2" ParentId="12655" CreationDate="2015-04-24T09:17:36.620" Score="5" Body="&lt;p&gt;I got the solution ($O$'s for extended lanterns):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; \begin{array}{ccc} . &amp;amp; 0 &amp;amp; 1 &amp;amp; 2 &amp;amp; 3 &amp;amp; 4 &amp;amp; 5 &amp;amp; 6 &amp;amp; 7 &amp;amp; 8 &amp;amp; 9 \\ A &amp;amp; X &amp;amp; O &amp;amp; X &amp;amp; O &amp;amp; X &amp;amp; X &amp;amp; X &amp;amp; X &amp;amp; X &amp;amp; X \\ B &amp;amp; O &amp;amp; X &amp;amp; X &amp;amp; X &amp;amp; X &amp;amp; X &amp;amp; X &amp;amp; X &amp;amp; X &amp;amp; O \\ C &amp;amp; X &amp;amp; X &amp;amp; \blacksquare &amp;amp; O &amp;amp; O &amp;amp; X &amp;amp; X &amp;amp; \blacksquare &amp;amp; X &amp;amp; X \\ D &amp;amp; X &amp;amp; O &amp;amp; X &amp;amp; \blacksquare &amp;amp; O &amp;amp; O &amp;amp; \blacksquare &amp;amp; X &amp;amp; X &amp;amp; X \\ E &amp;amp; X &amp;amp; X &amp;amp; O &amp;amp; X &amp;amp; X &amp;amp; X &amp;amp; O &amp;amp; X &amp;amp; X &amp;amp; X \\ F &amp;amp; X &amp;amp; X &amp;amp; X &amp;amp; X &amp;amp; X &amp;amp; X &amp;amp; X &amp;amp; X &amp;amp; X &amp;amp; O \\ G &amp;amp; X &amp;amp; X &amp;amp; O &amp;amp; \blacksquare &amp;amp; X &amp;amp; X &amp;amp; \blacksquare &amp;amp; O &amp;amp; X &amp;amp; O \\ H &amp;amp; X &amp;amp; X &amp;amp; \blacksquare &amp;amp; X &amp;amp; X &amp;amp; O &amp;amp; X &amp;amp; \blacksquare &amp;amp; X &amp;amp; X \\ I &amp;amp; X &amp;amp; X &amp;amp; O &amp;amp; X &amp;amp; O &amp;amp; X &amp;amp; X &amp;amp; X &amp;amp; X &amp;amp; X \\ J &amp;amp; O &amp;amp; X &amp;amp; X &amp;amp; X &amp;amp; X &amp;amp; X &amp;amp; X &amp;amp; X &amp;amp; X &amp;amp; X \\ \end{array}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll explain, in short, how I filled the grid:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The first step is to put $X$'s at all tiles that would shed light on completely dark tiles (namely, ones with the number zero). For example, D7 and D9. Throughout the solution I put $X$'s at tiles which would shed too much light on particular tiles (without elaborating).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The tiles G7 and G8 can be lit by three possible tiles - E7, G7 and G9 - and they get the same amount of light. The only option for extended lanterns is &lt;strong&gt;G7&lt;/strong&gt; and &lt;strong&gt;G9&lt;/strong&gt;, making both tiles 8-bright. Now we know the symbol for 8 and we can replace it throughout the grid.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The tile D9 gets 2 from the lantern at G9. The only way to get 6 more is to extend the lanterns at &lt;strong&gt;B9&lt;/strong&gt; and &lt;strong&gt;F9&lt;/strong&gt;. Doing so, we get the symbol for 4 (at tile A9) and for 7 (at tile C9).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Tile H9 is already 7-bright, so the lantern at I9 is not extended.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;E7 needs 4 more, so the lantern at &lt;strong&gt;E6&lt;/strong&gt; (and not the ones at E4 and E5) is extended. This gives us the symbol for 6.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;E6 needs 1 more, so the lantern at &lt;strong&gt;E2&lt;/strong&gt; is extended.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;F2 needs 4 more, so the lantern at &lt;strong&gt;G2&lt;/strong&gt; is extended.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I9 gets 5 from the lanterns at F9 and G9, so its symbol can be either 5 or 6. But the symbol is different from the one we know for 6, so it's 5.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;H6 needs 4 more, so the lantern at &lt;strong&gt;H5&lt;/strong&gt; (and not the ones at H3 and H4) is extended.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;H5 needs 1 more, so the lantern at &lt;strong&gt;D5&lt;/strong&gt; is extended.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A5 needs 4 more, so the lanterns at &lt;strong&gt;A1&lt;/strong&gt; and &lt;strong&gt;A3&lt;/strong&gt; are extended.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;B0 needs 5 more, so the lantern at &lt;strong&gt;B0&lt;/strong&gt; is extended.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;C1 needs 4 more, so the lantern at &lt;strong&gt;D1&lt;/strong&gt; is extended.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;J0 needs 5 more, so the lantern at &lt;strong&gt;J0&lt;/strong&gt; (and not the ones at H0, I0 and J1) is extended.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I0 needs 4 more, so the lanterns at &lt;strong&gt;I2&lt;/strong&gt; and &lt;strong&gt;I4&lt;/strong&gt; (and not the one at I3) are extended.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;B4 needs 7 more, so the lanterns at &lt;strong&gt;C4&lt;/strong&gt; and &lt;strong&gt;D4&lt;/strong&gt; are extended.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;C6 needs 2 more, so the lantern at &lt;strong&gt;C3&lt;/strong&gt; is extended.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And we've filled the grid! Note that, to do so, we only had to find the symbols for 4, 5, 6, 7 and 8. Of course, now we can easily find all the others.&lt;/p&gt;&#xA;" OwnerUserId="10684" LastEditorUserId="10684" LastEditDate="2015-04-24T09:54:09.593" LastActivityDate="2015-04-24T09:54:09.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12661" PostTypeId="1" AcceptedAnswerId="12666" CreationDate="2015-04-24T10:01:01.583" Score="21" ViewCount="2192" Body="&lt;p&gt;You are a pirate. One of your crew ran off in the night, swam to your secret island, and dug up your life's treasure. You need to capture her.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may deploy $n$ pirate ships to patrol the boundary of the island.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/87JDQ.png&quot; alt=&quot;Case n=4&quot;&gt;&lt;br&gt;&#xA;(pictured: $n=4$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each pirate ship moves exactly as fast as the mutineer. If the mutineer takes even a single safe step into the ocean, then she will escape using her stolen SCUBA kit. Your goal is to capture the criminal scum at the border, or starve her out on the island forever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;How many pirate ships do you need to deploy&lt;/em&gt;, if the island is shaped like a square?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mutineer and pirate ships are all points of zero radius. The mutineer starts in the center of the island, and you may start your ships wherever you want. Everyone can always see everyone else's position. This is not a loophole-finding contest: there are no issues with reaction time, scurvy, et cetera.&lt;/p&gt;&#xA;" OwnerUserId="1752" LastEditorUserId="1752" LastEditDate="2015-04-24T18:34:23.087" LastActivityDate="2015-04-26T11:33:11.487" Title="Pursuit Problem: Mutineer trapped on an island" Tags="|mathematics|logical-deduction|geometry|pursuit-evasion|" AnswerCount="6" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="12662" PostTypeId="2" ParentId="12637" CreationDate="2015-04-24T10:12:58.147" Score="1" Body="&lt;p&gt;Not sure . Looks like&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chess pieces.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First is new and bullied by Second&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third claims being first to all servings&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The humpback is Fourth and blames his twin for everything&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bishop&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fifth is full as soon as his meal begins&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sixth has a twin he denies knowing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seventh is last to begin eating&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; King&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Eight tries hiding to be forgotten&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still in process :)&lt;/p&gt;&#xA;" OwnerUserId="2656" LastActivityDate="2015-04-24T10:12:58.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12663" PostTypeId="2" ParentId="12655" CreationDate="2015-04-24T10:32:24.440" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;\begin{array}{ccc} &amp;amp; 10	 &amp;amp; *10	 &amp;amp; 8	 &amp;amp; *11	 &amp;amp; 11	 &amp;amp; 6	 &amp;amp; 2	 &amp;amp; 1	 &amp;amp; 0	 &amp;amp; 4\\ &amp;amp; *5	 &amp;amp; 11	 &amp;amp; 3	 &amp;amp; 10	 &amp;amp; 8	 &amp;amp; 4	 &amp;amp; 2	 &amp;amp; 3	 &amp;amp; 4	 &amp;amp; *6\\ &amp;amp; 4	 &amp;amp; 7	 &amp;amp; X	 &amp;amp; *12	 &amp;amp; *13	 &amp;amp; 11	 &amp;amp; 5	 &amp;amp; 	X &amp;amp; 0	 &amp;amp; 7\\ &amp;amp; 7	 &amp;amp; *7	 &amp;amp; 10	 &amp;amp; X	 &amp;amp; *13	 &amp;amp; *10	 &amp;amp; X	 &amp;amp; 2	 &amp;amp; 0	 &amp;amp; 8\\ &amp;amp; 5	 &amp;amp; 9	 &amp;amp; *9	 &amp;amp; 6	 &amp;amp; 15	 &amp;amp; 12	 &amp;amp; *6	 &amp;amp; 7	 &amp;amp; 3	 &amp;amp; 11\\ &amp;amp; 2	 &amp;amp; 3	 &amp;amp; 8	 &amp;amp; 0	 &amp;amp; 7	 &amp;amp; 7	 &amp;amp; 6	 &amp;amp; 7	 &amp;amp; 4	 &amp;amp; *10\\ &amp;amp; 5	 &amp;amp; 6	 &amp;amp; *8	 &amp;amp; 	X &amp;amp; 6	 &amp;amp; 6	 &amp;amp; 	X &amp;amp; *8	 &amp;amp; 8	 &amp;amp; *12\\ &amp;amp; 3	 &amp;amp; 1	 &amp;amp; 	X &amp;amp; 3	 &amp;amp; 9	 &amp;amp;* 6	 &amp;amp; 4	 &amp;amp; X	 &amp;amp; 0	 &amp;amp; 7\\&amp;amp; 8	 &amp;amp; 6	 &amp;amp; *8	 &amp;amp; 8	 &amp;amp; *8	 &amp;amp; 10	 &amp;amp; 4	 &amp;amp; 2	 &amp;amp; 1	 &amp;amp; 5\\ &amp;amp; *5	 &amp;amp; 4	 &amp;amp; 7	 &amp;amp; 2	 &amp;amp; 5	 &amp;amp; 3	 &amp;amp; 0	 &amp;amp; 0	 &amp;amp; 0	&amp;amp; 3\\ \end{array}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Stars indicate the open lanterns&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Started with the cell to the left of the zero on the top row - only one light can be incident on it if the other cell is to remain zero. That symbol must therefore be 1. The rest was number-crunching. I'm sure there was a better way to do it, though! (and a better way to represent the result)&lt;/p&gt;&#xA;" OwnerUserId="11717" LastEditorUserId="10615" LastEditDate="2015-04-24T15:10:40.470" LastActivityDate="2015-04-24T15:10:40.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12664" PostTypeId="2" ParentId="12661" CreationDate="2015-04-24T11:11:08.813" Score="8" Body="&lt;p&gt;For the square island, you need&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; four ships. Let us assume the island has x and y coordinates, all between 0 and 1. Two ships are patrolling the x=0 and x=1 coast. Tell them to ignore the x coordinate of the mutineer's position and just keep up with the y coordinate. The same (with x and y swapped) applies to the other two ships.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not sure if I can provide an escape strategy for one less ship patrolling. Should be something like &quot;approach one corner, lure the ships towards it, and then diagonally cross the island&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2783" LastActivityDate="2015-04-24T11:11:08.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12666" PostTypeId="2" ParentId="12661" CreationDate="2015-04-24T12:20:22.280" Score="4" Body="&lt;p&gt;This proves lower bound of four ships. Considering that ships move in the ocean and mutineer walks in island. Also, as per victory condition mutineer cannot be caught in ocean. (This is dicey part, since it is said safe step in ocean, but in subsequent comment it was clarified as to mutineer being adjacent to ship.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Four ship solution is given by @Gully.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Consider a 3x3 island with mutineer at center. Let x and y series for island range from 1&#xA;to 3. Then mutineer initially is at (2, 2).&#xA;&#xA;I am proving three ships wouldn't be sufficient.&#xA;&#xA;Consider this diagram:&#xA;&#xA;  _ A_ _&#xA; |      |&#xA;B|      | C&#xA; |      |&#xA;  _ D_ _&#xA;&#xA;Case I:  None of ships are at corner, i.e. (x, y) for all ships range from (1, 3).&#xA;    In this case, at least one border (A-D) would be unguarded, and mutineer can simply&#xA;    go through that border.&#xA;&#xA;Case II: One ship is at corner i.e. they can have value of (0, 0) (0, 3), (3, 0) or (3, 3)&#xA;    To guard all borders, two ships would guard adjacent border with third ship on&#xA;    diagonal opposite. For example, A, B and (3, 3).&#xA;    Again, Mutineer can go through either C and D middle strip and escape. &#xA;&#xA;Cases involving more than one ship at a border are easily explained in the same way.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="10618" LastEditorUserId="4222" LastEditDate="2015-04-26T11:33:11.487" LastActivityDate="2015-04-26T11:33:11.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12668" PostTypeId="2" ParentId="11225" CreationDate="2015-04-24T12:25:09.587" Score="1" Body="&lt;p&gt;I don't have much, but I did find an interesting rearrangement of the numbers.  It may be relevant, it may just be coincidence.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;3&lt;/strong&gt; 4 6 A &lt;br/&gt;&#xA;  &lt;strong&gt;3&lt;/strong&gt; 2 5 J &lt;br/&gt;&#xA;  &lt;strong&gt;3&lt;/strong&gt; 1 3 Q &lt;br/&gt;&#xA;  &lt;strong&gt;5&lt;/strong&gt; K 3 A A 1 &lt;br/&gt;&#xA;  &lt;strong&gt;2&lt;/strong&gt; 5 3 &lt;br/&gt;&#xA;  &lt;strong&gt;6&lt;/strong&gt; 5 J 2 3 Q 1&lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All I did was take the first digit (3) as a count, then put the next three characters after it.  Then the next free character became the next count, and so on.  Notice how it works out to &lt;em&gt;exactly&lt;/em&gt; use up every character, and there's never a card where the count needs to be.  Also noteworthy is that all three &lt;code&gt;3&lt;/code&gt; columns go Die-Die-Card, and the &lt;code&gt;6&lt;/code&gt; is an inverse of the pattern from the &lt;code&gt;5&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no idea whether there's anything to it, though.&lt;/p&gt;&#xA;" OwnerUserId="325" LastActivityDate="2015-04-24T12:25:09.587" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12669" PostTypeId="2" ParentId="12656" CreationDate="2015-04-24T12:28:03.030" Score="2" Body="&lt;p&gt;Well it took me way to long but I got it :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First the hint (which was highly necessary to decrypt it)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The key is a writer of complete nonsense. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$ $&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is pretty literal (pun intended) since it referred to &lt;a href=&quot;http://en.wikipedia.org/wiki/Edward_Lear&quot; rel=&quot;nofollow&quot;&gt;Edward Lear&lt;/a&gt;, the author of a book titled &lt;em&gt;The Complete Nonsense of Edward Lear&lt;/em&gt;, a book full of limericks. So let's assume that &lt;strong&gt;EdwardLear&lt;/strong&gt; is the key. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So now we have the key but what then?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well we start trying ciphers, I tried to use a program for this but in the end decided to use &lt;a href=&quot;http://rumkin.com/tools/cipher/&quot; rel=&quot;nofollow&quot;&gt;Rumkin&lt;/a&gt;, sure i had to do it by hand for the ciphers it has but at least I knew it'd go as I wanted.&#xA; Once I'd tried some I came to the Playfair cipher and finally I saw the text I wanted: &lt;code&gt;Whose limericks all stopxped at line one&lt;/code&gt; well almost the x shouldn't be there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then it was quite trivial to get the rest:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He then got a clue &lt;br&gt;&#xA; From the man in Peru &lt;br&gt;&#xA; And completed this poem for fun &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though i do not know why:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; we had to translate the letter J into I, if there had been such a clue i would have sooner thought about a play fair cipher. Even P (peru) into V (verdun) would have made more sense to me.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9835" LastEditorDisplayName="user88" LastEditDate="2015-04-24T16:31:20.317" LastActivityDate="2015-04-24T16:31:20.317" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="12670" PostTypeId="2" ParentId="12661" CreationDate="2015-04-24T13:03:30.227" Score="3" Body="&lt;p&gt;&lt;strong&gt;This is not an answer.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No need to up-vote or down-vote. It is just a counter-example to the 3-ships idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it has to be at least 4. Check out the following pic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oZTv3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE: Even more accurate is this approximation. Any straight path through the red area is an escape! The reflection of these red areas (up and to the left) works just the same.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ANgBT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-04-24T13:28:33.997" LastActivityDate="2015-04-24T13:28:33.997" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="12671" PostTypeId="2" ParentId="12643" CreationDate="2015-04-24T13:17:11.363" Score="1" Body="&lt;p&gt;For &lt;strong&gt;scenario 1&lt;/strong&gt;, I think it's possible for 2 wolves to get the rabbit. Take this case: 2 wolves on equator, opposite sides, rabbit at north pole.  The rabbit moves perpendicular to the wolves towards the equator. In the simple case, if the wolves then move along the equator, they will hit the rabbit as it hits the equator. If the rabbit doesn't move, the wolves just go towards the pole until it does. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could think of it like this, the wolves either move towards the rabbit, or parallel to the rabbit. But, parallel lines on a sphere always intersect. So, even when running parallel to the rabbit, they gain ground. More specifically, though, the wolves move so that the hare is always between them on a circumference. So in the first case, the wolves more towards the rabbit initially (so away from the equator) while still moving in the general equator direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the rabbit is constantly moving, if the angle between them and the rabbit is B, they move $\frac{1}{12}\pi * (1-cos(B))$ towards the rabbit ©.  They eventually need to make it $\frac{\pi}{2}$ radians towards the rabbit. A quick wolfram integration gives 5.06 radians, so a time of 19.33hours. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &lt;strong&gt;scenario 2&lt;/strong&gt;, I think the biggest issue is the wolves finding the rabbit. As such, it would take 12 wolves, since each the radius of the planet is 24hrs and each wolf can smell 2hrs. The wolves start 1 hour from a pole, then moves to the other pole (and at the equator are equally spaced). At some point, the rabbit will be found, and can then be caught. Max search time would be 10 hours. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;© Imagine the sphere is divided up into 3 parallel planes: one through the poles (the rabbit's path) and two $\pi/3$ rads away (wolves' path). Since the wolves' path is 1/2 the length of the rabbits, they can use 1/2 their movement to keep it in line, and 1/2 to get closer. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-04-24T18:58:38.157" LastActivityDate="2015-04-24T18:58:38.157" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="12672" PostTypeId="2" ParentId="12643" CreationDate="2015-04-24T13:52:34.053" Score="1" Body="&lt;p&gt;For my second approach on both the first/second case, I think the minimum is &lt;b&gt;3 wolves&lt;/b&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For my first approach on both the first/second case, I think the minimum is &lt;b&gt;6 wolves&lt;/b&gt;. Here's my approach:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use X wolves to trap the hare, and 1 to chase it down.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2*pi*r*90/360 &gt; 2*pi&lt;em&gt;r&lt;/em&gt;X/360. Or in other words, the hare's arc length must be larger than the arc length of the wolves, where X is the degrees between each wolf. This results in X &amp;lt; 90. Which means a minimum of 5 wolves (think 4 wolves will give 90 degrees each, +1 wolf will give &amp;lt; 90 each)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So now if the wolves equally spread out along the equator the hare will not be able to get past them to the other side. (As they can just rotate along their perimeter faster to block the hare) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Imagine a circle with a dot in the middle trying to escape, except its a semi-sphere, This is the &lt;b&gt;best&lt;/b&gt; possible distance for the hare]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we need the one wolf to chase the trapped hare! This wolf just needs to force the trapped hare into another wolf by chasing behind it for long enough. If the hare tries to run, he can only run closer and closer to another wolf, either a wolf on the boarder, or the one that is chasing it. Eventually, at least one wolf be able catch the hare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you might be wondering why the hare would allow itself to be trapped as explained above, unfortunately it has no choice. Since the 5 wolves only have to place them self to cover their own arc length section with respect to the hare, they just have to run to where they &lt;b&gt;should&lt;/b&gt; be. At first they will be trailing behind where they 'should' be in the setup to trap the hare, but once the hair reverses direction (or stops), the wolves will be able to catch up and match the location where they should be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;This same method will also work for the first part of the question with the same number of wolves!&lt;/b&gt; The only difference is that 1 wolf must start the chase first so that it can communicate the hares position to the other wolves so they can get into the proper positions. (at all times)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Second approach, you need 3 wolves.&lt;/b&gt;&#xA;Influenced by the other answers, I think you could catch the hare with a minimum of 3 wolves (2 to trap, 1 to chase). The two wolves would have to be positioned on opposite sides of the sphere with the hare in the center (in the middle of a straight line drawn between the two wolves). Now these two wolves simply have to rotate around the axis to prevent the hare from escaping to the other half (essentially trapping it on one half). The third wolf chases the hare into the edge of the trap (and prevent it from out running/escaping the edge wolves). [This is also based on the same equation as above. The mistake I made earlier is that the wolves would only move it one direction along the sphere (covering 90 degrees), but if they moved both ways you could half the amount required (cover 90 degrees both ways or 180 degrees total)]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setup:&#xA;To set the positions of the wolves up, the first two need to initially get onto the opposites sides of the sphere. Now you can always draw a straight line between the wolves with the hare on it (think north pole and south pole with the hare in between). Now the issue is that the hare isn't exactly in the middle (or equator). This is where the third wolf comes into play, chasing the rabbit towards the 'equator'. Once the hare is there, the two edge wolves can chase around the sphere to follow the above tactic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;This might also be able to work for the second part given that one wolf initially finds the hare and communicates it position to help the other two wolves setup.&lt;/b&gt;&lt;/p&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-04-28T16:38:44.247" LastActivityDate="2015-04-28T16:38:44.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12673" PostTypeId="2" ParentId="12364" CreationDate="2015-04-24T14:26:48.827" Score="0" Body="&lt;p&gt;Not an answer, i just wanted to create some activity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well.. after a long talk with the OP i figured out the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a key made of something divine,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; GOD, god is something divine (obviously)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sha-1 of 4&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1b6453892473a467d07372d45eb05abc2031647a&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The last line&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; now if you AES this line with god (encrypt or decrypt) rubbish comes out, XOR this with the hash of 4 and it still makes no sense.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm just too stupid to follow simple instructions, it's a miracle my house isn't on fire yet :P&lt;/p&gt;&#xA;" OwnerUserId="9835" LastEditorUserId="9835" LastEditDate="2015-04-24T18:10:23.027" LastActivityDate="2015-04-24T18:10:23.027" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="12674" PostTypeId="2" ParentId="12661" CreationDate="2015-04-24T14:42:33.470" Score="0" Body="&lt;p&gt;I actually think the circular version may be easier. In theory, the mutineer can never be closer to any point on the shore than any boat. In the circular example, assuming he starts in the center, the distance to the shore is r. The perimeter of the island is 2*r*pi. each ship covers 2r of the perimeter (since they can go either direction around the island as fast as the mutineer moves. Presuming no loss in time for turning around.) As such, you would need pi ships, or since we can only work with whole ships, 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the square, the shortest distance to an edge is l/2, and the longest (to a corner) is (l&lt;em&gt;sqrt(2)) / 2. Dealing with the shortest case scenario again, you have to start with a ship within l/2 of each side. This could be accomplished with 2 ships, just to cover the edges, but this omits the other two corners being covered. For the corners, you need to have a ship within (l&lt;/em&gt;sqrt(2)) / 2 of each corner. Double this as each ship can cover 2 different directions. We're left with 4l / ((l*sqrt(2)) = 4/sqrt(2) = 2.82. Again, dealing with whole ships, you'd need 3. This should be enough to fully cover the perimeter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: (Update after comment w/regard to JLee's answer)&#xA;JLee's distribution is pretty skewed, the two in the lower left are much closer together than they need to be, leaving the right and top side open. That said, in rethinking this, in order to cover the midpoint of two sides a ship must be at a corner. You need all 4 midpoints covered, so in the 3 boat scenario at least one ship must be at a corner. Let's stick with the upper right as JLee illustrated. Anything outside that quadrant the mutineer would outpace him to. In order to cover the other halves of the top and right edge, another ship would have to be right at, or very close to the top left, and lower right corner. This would leave the lower left corner uncovered. Even if the ship on the left, and bottom started l*sqrt(2) from the lower left corner in order to get there in time, the area's JLee highlighted would still be vulnerable. So ultimately even though I don't think the illustration properly represented a best case for 3 ships, I think 3 is still not feasible.&lt;/p&gt;&#xA;" OwnerUserId="11786" LastEditorUserId="11786" LastEditDate="2015-04-24T15:02:39.457" LastActivityDate="2015-04-24T15:02:39.457" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12675" PostTypeId="2" ParentId="12648" CreationDate="2015-04-24T14:47:56.953" Score="5" Body="&lt;p&gt;I don't really see how this could be solved without further restrictions. Let's look at a simplified case where we have to pick x balls from one tumbler, y from another, with one of them containing the defective balls. We get the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If x is defective, y okay then the weight range is between y and y + x.&lt;/li&gt;&#xA;&lt;li&gt;If x is okay, y defective then the weight range is between x and x + y.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;No matter what you pick for x and y, if both are at least 1 then intervals will always overlap. If the scale shows a number within the overlapping part then there isn't a solution. As an example, for x = 2, y = 5, the first range is 5 to 7, the second range is 2 to 7. Getting anything above 5 and below 7 doesn't tell us anything.&lt;/p&gt;&#xA;" OwnerUserId="9910" LastActivityDate="2015-04-24T14:47:56.953" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12676" PostTypeId="2" ParentId="12416" CreationDate="2015-04-24T15:15:26.087" Score="1" Body="&lt;p&gt;Could it be:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A gaze?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I convey the invisible,&lt;br&gt;&#xA;I convey what can be felt,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Gaze can transfer the feeling toward another person.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But not seen,&lt;br&gt;&#xA;Inside my heart,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You feel it when people gazing at you&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am not as nice as I could seem to be,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Gaze can be hostile&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is because I deliver the invisible,&lt;br&gt;&#xA;At the speed of a mere blink,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Gaze takes merely a millisecond to transfer a message / feeling&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am very powerful,&lt;br&gt;&#xA;And can bring comfort,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Gaze is powerful and can be a comfort from a lovely one&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Across my horizons, I am free,&lt;br&gt;&#xA;But still vulnerable to your opinion.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You might think I am attracted to you / hostile if I gazed at you, it depends on your opinion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a Gaze&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2015-04-24T15:15:26.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12677" PostTypeId="2" ParentId="12416" CreationDate="2015-04-24T15:32:26.037" Score="1" Body="&lt;p&gt;Could it be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I convey the invisible,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sun emits UV rays which we can't see&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I convey what can be felt,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can feel the suns heat from earth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But not seen,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We cant visibly UV light&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Inside my heart,&#xA;I am not as nice as I could seem to be,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The center of the sun is not a very nice place to be.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is because I deliver the invisible,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sun also delivers gamma rays which is not nice to us.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At the speed of a mere blink,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; gamma rays and UV light travels very fast indeed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am very powerful,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sun is what powers the earth!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And can bring comfort,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people enjoy the sunlight!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Across my horizons, I am free,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Possible reference to a sunset or sunrise&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But still vulnerable to your opinion.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I guess some people might not like the sun. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sun!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-04-27T12:51:37.883" LastActivityDate="2015-04-27T12:51:37.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12678" PostTypeId="2" ParentId="12416" CreationDate="2015-04-24T15:53:15.047" Score="0" Body="&lt;p&gt;I think it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Body Language etiquette (good manners)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I convey the invisible,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A persons real thoughts are invisible&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I convey what can be felt,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Always cannot hide behind the mask. (Refer LIE TO ME)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But not seen,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inside my heart,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not as nice as I could seem to be,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is because I deliver the invisible,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the speed of a mere blink,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am very powerful,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And can bring comfort,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Across my horizons, I am free,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But still vulnerable to your opinion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="9892" LastActivityDate="2015-04-24T15:53:15.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12679" PostTypeId="2" ParentId="9155" CreationDate="2015-04-24T16:16:54.750" Score="1" Body="&lt;p&gt;I'm believe the zebra's can always win as before. The main reason being that the plain is infinite, and that the zebra's place themselves second. How they do this is simply the same was as in the previous question. Each zebra makes its on strip in which if there was only 1 lion, they would all be safe. the issue is that there is 100 lions, but this doesn't matter. Since any number of points (less than infinite) can be scaled down to a single point given infinite space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So once all the lions have placed themselves, the zebra's need to be places such that they 100 lions can be considered to be a single point below all of the zebra's strips. Although this might not seem physically possible (since we are dealing with limits with infinity), in the world of math it could be resolved to a solution. Unfortunately my math hat doesn't seem to be around so I can't provide the proper proof to this concept. (Hopefully someone else would be able to verify this to possibly conclude this problem)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note. You stated that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Note that once a zebra gets encircled by the lions, he is bound to&#xA;  lose. So the aim for the lions could be to create a wide ring around a&#xA;  single zebra.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But you did not provide any proof to this. I believe this can be throwing some people off as they could be going down a false path, unless this is a rule that you are creating as another win situation.&lt;/p&gt;&#xA;" OwnerUserId="11244" LastActivityDate="2015-04-24T16:16:54.750" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12680" PostTypeId="1" CreationDate="2015-04-24T16:17:23.243" Score="1" ViewCount="290" Body="&lt;ul&gt;&#xA;&lt;li&gt;A sudoku net is defined &lt;a href=&quot;https://puzzling.stackexchange.com/questions/11780/sudoku-net-that-is-always-solvable&quot;&gt;here&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A sudoku net is defined as &lt;strong&gt;'pseudo-solvable'&lt;/strong&gt; if we can deduce at least two valid 'not unique' solutions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;friendly sudoku is defined &lt;a href=&quot;https://puzzling.stackexchange.com/questions/12542/unsolvable-sudoku-net-with-maximum-blocks/12636?noredirect=1#comment39140_12636&quot;&gt;here&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;What is the sudoku net with the minimum possible blocks in such a way that this sudoku net is pseudo-solvable with at least one friendly sudoku?&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Challenge 2&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Find a sudoku net &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; with visible set &lt;span class=&quot;math-container&quot;&gt;$V_N$&lt;/span&gt; such that for any sudoku &lt;span class=&quot;math-container&quot;&gt;$S_1$&lt;/span&gt;, there is another &lt;span class=&quot;math-container&quot;&gt;$S_2\ne S_1$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$V_N(S_1) = V_N(S_2)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$|V_N|$&lt;/span&gt; is maximal.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Bonus question&lt;/strong&gt;: (it is worth 50pts) is there any condition of pattern arrangements inside blanks which makes a sudoku unsolvable using previous result as work-domain?&lt;/p&gt;&#xA;" OwnerUserId="9517" LastEditorUserId="69582" LastEditDate="2024-02-12T02:33:49.837" LastActivityDate="2024-02-12T02:33:49.837" Title="Sudoku-net pseudo-solvable with minimum blocks" Tags="|mathematics|logical-deduction|grid-deduction|sudoku|number-theory|" AnswerCount="2" CommentCount="12" ContentLicense="CC BY-SA 4.0" />
  <row Id="12681" PostTypeId="5" CreationDate="2015-04-24T16:18:14.717" Score="0" Body="&lt;p&gt;A pursuit-evasion puzzle is any puzzle where a team of pursuers must capture a team of targets, with movement limited to some given space. The goal is to find a winning strategy either for the pursuers or for the evaders.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These puzzles are defined by &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;where the parties are constrained to (e.g. a plane, a sphere, the edges of a triangle)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the relative speeds of the pursuers and targets&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;whether or not either party is aware of the other's positions&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;a winning condition for the evaders (e.g. evade capture indefinitely, reach a winning region)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;One of the more well known puzzles in this category is the Duck and the Fox (asked &lt;a href=&quot;/questions/2155/the-lake-monster&quot;&gt;here&lt;/a&gt; in an equivalent form):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A duck is at the center of a circular lake, with a hungry fox waiting at the border. The fox cannot swim, but can run 4 times as fast as the duck can swim. As long as the duck touches the shore at a point where the fox isn't already waiting, then it can instantly fly away to safety. Will the duck be able to fly south for winter?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="11110" LastEditDate="2017-06-21T07:37:55.313" LastActivityDate="2017-06-21T07:37:55.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12682" PostTypeId="4" CreationDate="2015-04-24T16:18:14.717" Score="0" Body="Any puzzle where a team of pursuers must capture a team of targets, with movement limited to some given space. The goal is to find a winning strategy either for the pursuers or for the evaders." OwnerUserId="10615" LastEditorUserId="11100" LastEditDate="2015-04-25T11:08:29.197" LastActivityDate="2015-04-25T11:08:29.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12683" PostTypeId="2" ParentId="12648" CreationDate="2015-04-24T16:25:11.070" Score="15" Body="&lt;p&gt;Since the weighing scale can only be used once, I feel free to disassemble and modify it.&lt;br&gt;&#xA;Now, tie a ball from each tumbler to the 4 vertices of the weighing-scale, as shown in this picture: &#xA;&lt;img src=&quot;https://i.stack.imgur.com/wbUp9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, pull up the structure and see how it tilts: the highest ball from the floor is the lightest one!&lt;/p&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-04-24T16:25:11.070" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12684" PostTypeId="2" ParentId="12680" CreationDate="2015-04-24T16:28:27.333" Score="3" Body="&lt;h1&gt;&lt;strong&gt;MINIMUM BLOCKED BOXES:&lt;/strong&gt;&lt;/h1&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;O O &lt;strong&gt;A&lt;/strong&gt;  | O O O  | &lt;strong&gt;C&lt;/strong&gt; O O&lt;br /&gt;&#xA;O O O | O O O  | O O O&lt;br /&gt;&#xA;O O &lt;strong&gt;B&lt;/strong&gt;  | O O O  | &lt;strong&gt;D&lt;/strong&gt; O O&lt;br /&gt;&#xA;-------------------------------&lt;br /&gt;&#xA;O O O | O O O  | O O O&lt;br /&gt;&#xA;O O O | O O O  | O O O&lt;br /&gt;&#xA;O O O | O O O  | O O O&lt;br /&gt;&#xA;-------------------------------&lt;br /&gt;&#xA;O O O | O O O  | O O O&lt;br /&gt;&#xA;O O O | O O O  | O O O&lt;br /&gt;&#xA;O O O | O O O  | O O O&lt;/p&gt;&#xA;&lt;p&gt;Where &lt;strong&gt;A,B,C,D&lt;/strong&gt; are equal in pairs (the O are given numbers).&lt;br /&gt;&#xA;Suppose that both the &lt;em&gt;top-left&lt;/em&gt; and the &lt;em&gt;top-right&lt;/em&gt; squares are missing the numbers &lt;strong&gt;5&lt;/strong&gt; and &lt;strong&gt;7&lt;/strong&gt;. If you swap &lt;strong&gt;A&lt;/strong&gt; with &lt;strong&gt;B&lt;/strong&gt;, and &lt;strong&gt;C&lt;/strong&gt; with &lt;strong&gt;D&lt;/strong&gt; the sudoku will be still valid. This means that if you have clues about all cells except &lt;strong&gt;A,B,C,D&lt;/strong&gt; there are 2 different valid solutions to the sudoku, which means &lt;strong&gt;UNSOLVABLE&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Given a random solved sudoku, what's the minimal net that makes it pseudo-solvable?&lt;/strong&gt;&lt;br /&gt;&#xA;The answer is: it depends on the sudoku! Some particular sudoku need a net with just &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt; blocks, like the one showed in the picture, while other require up to &lt;span class=&quot;math-container&quot;&gt;$18$&lt;/span&gt; blocks to become pseudo-solvable (put the blocks on all &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt;).&#xA;Example of the latter configuration is here:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/RDncm.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If you want a net that works for any sudoku, independently from the grid, just put the blocks on the first and second row. Anyway, this would still require &lt;span class=&quot;math-container&quot;&gt;$18$&lt;/span&gt; blocks, not better than my previous solution.&lt;/p&gt;&#xA;&lt;h1&gt;&lt;strong&gt;When is a sudoku pseudo-solvable?&lt;/strong&gt;&lt;/h1&gt;&#xA;&lt;p&gt;When you don't have information about &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt; cells, arranged on a rectangular shape, such that &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; of them belong to a quadrant, while the other &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; belong to another quadrant (as shown in the above picture); also, the cells on the opposite vertices of the rectangle should contain the same values!&lt;/p&gt;&#xA;&lt;p&gt;Let's say these values are &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; and the pattern is the same described in the above picture. The first row says the &lt;strong&gt;A&lt;/strong&gt; and &lt;strong&gt;C&lt;/strong&gt; must contain &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt;, but doesn't say in which order. Same for the third row, third column, seventh column, top-left box, top-right box.  This means, there are &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; valid solutions for this sudoku.&lt;/p&gt;&#xA;&lt;h1&gt;&lt;strong&gt;When is a sudoku NOT pseudo-solvable?&lt;/strong&gt;&lt;/h1&gt;&#xA;&lt;p&gt;This means &lt;strong&gt;solvable or impossible&lt;/strong&gt;! A Sudoku is impossible when it has internal contradictions (like two &lt;span class=&quot;math-container&quot;&gt;$7$&lt;/span&gt; in the same row). When the sudoku isn't impossible (aka possible), it can be either solvable (admits 1 solution) or pseudo-solvable (admits 2+ solutions). How to tell them? First of all, given a grid, fill as many cells as possible using deductions. Now, if you have 4 unknown cells arranged in a rectangular pattern (see above for the other conditions), then the sudoku is pseudo-solvable!&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-26T16:59:40.300" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="12685" PostTypeId="1" AcceptedAnswerId="12726" CreationDate="2015-04-24T18:12:30.267" Score="14" ViewCount="1031" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;(This is the sequel to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/12661/pursuit-problem-mutineer-trapped-on-an-island&quot;&gt;this puzzle&lt;/a&gt;. It has a similar setup, but believe me, the solution is very different. Be careful! The answer is counterintuitive. It shocked me at first.)&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are a pirate. &lt;a href=&quot;https://puzzling.stackexchange.com/questions/12661/pursuit-problem-mutineer-trapped-on-an-island&quot;&gt;Previously&lt;/a&gt;, you captured a rogue mutineer. She begged for a second chance on the crew, in exchange for the location of her criminal headquarters: Marauders' Circular Cove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In promise of great treasure, you raided the cove at night. But it was a trap! Now smugglers are coming out of the forest, more than you can count.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Tcd5ol.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;(Pictured: there are WAY MORE than this)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your ship moves at the same speed as the smugglers. The smugglers cannot swim. If you manage to take even one step onto the shore, then you can instantly beach the ship and escape into the cover of darkness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now (and only right now, i.e. not after the chase starts), you can bribe some of the smugglers to leave, so that you only have to evade $n$ of them. Your goal is to reach the shore without getting caught. The smugglers' goal is to catch you when you reach the shore, or starve you out on the lake forever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;How many smugglers can you evade&lt;/em&gt;, if the cove is shaped like a perfect circle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ship and smugglers are all points of zero radius. Everyone can always see everyone else's position. This is not a loophole-finding contest: there are no issues with reaction time, scurvy, et cetera.&lt;/p&gt;&#xA;" OwnerUserId="1752" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2015-04-25T17:24:20.153" Title="Pursuit Problem II: Surrounded in Marauders' Circular Cove" Tags="|mathematics|logical-deduction|geometry|pursuit-evasion|" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12686" PostTypeId="2" ParentId="12685" CreationDate="2015-04-24T18:37:03.380" Score="0" Body="&lt;p&gt;Since the answer has already been stated, I thought would provide another view as to why it takes &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; infinite smugglers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bElTt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the red dot is our boat, and the green dot is a smuggler. Since the shortest path our boat can take to shore (safely) is X length, and the green dot must move at least (sqrt)(x²+y²), X will always be shorter. If you were to add another smuggler onto the curve, the boat simply needs to move close enough to the shore such that only one dot is remaining (like the picture). [Think of it as you are trying to race your friend to a finishing point and only have to travel in a straight line (shortest possible distance) and your friend is trying to beat you to the finish line, but has to travel along a curve]&lt;/p&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-04-24T19:34:23.783" LastActivityDate="2015-04-24T19:34:23.783" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="12687" PostTypeId="2" ParentId="12685" CreationDate="2015-04-24T18:53:35.810" Score="2" Body="&lt;p&gt;I think you can evade:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an infinite number of smugglers&lt;br&gt;&#xA;&lt;br&gt;&#xA; Say you are 10 feet from the edge of the cove. A smuggler would have to be within 10 feet of the closest point on shore or you'll escape. If smugglers are spaced every 20 feet, with one on the shore closest to you, you can circle the ship until you are between 2 smugglers, then make a break for it.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Now take it further. Say you're 1 foot away. Now smugglers would have to be spaced every 2 feet. 1 inch leads to 2 inch spacing. Eventually you can get close enough to handle any gap between smugglers. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-04-24T18:53:35.810" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12688" PostTypeId="2" ParentId="12661" CreationDate="2015-04-24T19:08:37.367" Score="0" Body="&lt;p&gt;An interesting note: assuming the 4 ships are positioned at the corners, they are just able to reach the target if she travels straight NSEW, with no overlap. However, if the 4 ships are positioned in the middle of each side, they have overlapping coverage areas at the corners. Unfortunately, this doesn't help. Assume a 20x20 square, with the mutineer at the center. She can leave the island by walking 10 steps NSEW. The ships can also travel 10 steps in that time. This means that the only way for a ship to cover more than one middle of a side is to be positioned at a corner, in which case ships at adjacent corners are needed to prevent gaps in coverage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to look at it: assuming a 20x20 square, and a ship positioned at the middle of the North side. It covers the North 2.5 units of the East and West sides. (It takes the target 12.5 steps to reach this point, and the ship can reach it in 12.5 steps.) This is the maximum coverage a single ship can provide. The West and East ships can therefore be positioned 12.5+2.5=15 units from the North to prevent any overlap. However, at this point, the mutineer can travel straight South, reaching the ocean in 10 steps, while the ships will take 15 steps to reach her.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to position ships (again, on a 20x20 island) would be to place the first ship on the North coast, 13.45 units from the NW corner. Place the 2nd on the West coast, 13.45 units from the NW corner. At this point, there's no overlap at that point, and there's coverage over 6.95 units of the East and South coasts. However, now we need to cover everything else with one ship, and that's not possible. Placing it on the SE corner will allow the mutineer to escape by traveling E by NE or S by SW, as there's a gap on both sides of the final ship.&lt;/p&gt;&#xA;" OwnerUserId="11410" LastActivityDate="2015-04-24T19:08:37.367" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12689" PostTypeId="1" AcceptedAnswerId="12696" CreationDate="2015-04-24T19:08:57.933" Score="10" ViewCount="520" Body="&lt;p&gt;I wish I understood, &lt;br&gt;&#xA;but you have closed your mind,&lt;br&gt;&#xA;they seem so hard to find,&lt;br&gt;&#xA;it used to be so good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're in the mood,&lt;br&gt;&#xA;when you get the chance,&lt;br&gt;&#xA;for a dance,&lt;br&gt;&#xA;You're in the mood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Buy some Chinese food, to go.&lt;br&gt;&#xA;have your dinner watching TV,&lt;br&gt;&#xA;a single episode of Dallas that I didn't see,&lt;br&gt;&#xA;at eight o'clock or so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who am I looking for?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9835" LastEditorUserId="5188" LastEditDate="2015-04-28T01:08:16.890" LastActivityDate="2018-05-18T20:22:08.737" Title="A very strange rhyme" Tags="|riddle|rhyme|" AnswerCount="1" CommentCount="19" ContentLicense="CC BY-SA 3.0" />
  <row Id="12690" PostTypeId="2" ParentId="12685" CreationDate="2015-04-24T19:09:08.083" Score="8" Body="&lt;p&gt;Surprisingly, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;the marauder can always escape!&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6A3Sq.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;Suppose you have a very very high number of smugglers. The strategy to escape is approaching the coast as much as possible (the more the smugglers, the closest you have to be), then follow a straight line (a chord) which is always shorter than the smugglers' path to reach you.&lt;br&gt;&#xA;Check the picture: when you come very very close to the coast, the other smugglers won't be able to stay near enough, so you can easily pass between them!&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-04-24T19:32:44.743" LastActivityDate="2015-04-24T19:32:44.743" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12691" PostTypeId="2" ParentId="12680" CreationDate="2015-04-24T19:28:33.513" Score="0" Body="&lt;p&gt;my answer for the second question may be :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;# # # | # # # | # # #&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;# # # | # # # | # # #&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;____  | ____  | ____  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;____  | ____  | ____  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;____  | ____  | ____  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;____  | ____  | ____  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;____  | ____  | ____  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;____   | ____  | ____  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;____  | ____  | ____  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;18 blocks&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;bonus question :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;any logical relation isnt known yet but sure ... it would have something to do with graph-theory&lt;/p&gt;&#xA;" OwnerUserId="9517" LastEditorUserId="9517" LastEditDate="2015-04-25T11:37:05.680" LastActivityDate="2015-04-25T11:37:05.680" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12692" PostTypeId="1" AcceptedAnswerId="12870" CreationDate="2015-04-24T19:49:31.630" Score="6" ViewCount="1928" Body="&lt;p&gt;There is no cheating this time, &lt;br&gt;&#xA;no brute forcing my cipher,&lt;br&gt;&#xA;for once try to play fair.&lt;br&gt;&#xA;I really like crypto puzzles,&lt;br&gt;&#xA;the hard part is using classical ciphers.&lt;br&gt;&#xA;If you were to use the modern ciphers people might be in above their head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Audio files may also be a great way,&lt;br&gt;&#xA;encrypting something in sound is very tricky though.&lt;br&gt;&#xA;Somehow I just don't know what I should do.&lt;br&gt;&#xA;One might realize the clues but an other may not,&lt;br&gt;&#xA;two can work together if they don't get it however,&lt;br&gt;&#xA;eight million tries might not even be enough for some.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only problem is every modern cipher does not operate alone&lt;br&gt;&#xA;every mode is special, which one would you use?&lt;br&gt;&#xA;Comparing Basic Ciphers is so much easier.&lt;br&gt;&#xA;Circumstantially I come up with idea, &lt;br&gt;&#xA;to initialize it with a long enough word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there is the point of hiding a key,&lt;br&gt;&#xA;one way is to throw some deceiving scheme,&lt;br&gt;&#xA;like 54 + 46 * 178 = what color?&lt;br&gt;&#xA;an other way is to hide it,&lt;br&gt;&#xA;perhaps in plain sight but not make it to obvious.&lt;br&gt;&#xA;it should be an acknowledgements for their suspicion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Almost done, but what would you let them decrypt?&lt;br&gt;&#xA;giving people a simple ciphertext is tedious,&lt;br&gt;&#xA;everyone knows &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Ylff hksg DBEEVMQQ VF FVEE KKEEVMQQ FVVM DBEEVMB IVFBP EEVMM VBKQQ UWEEPB WUW BEEVMK BEEYWQQ UW FVBP DBVMEE&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;will be what they are looking for.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly modern ciphers need encoding,&lt;br&gt;&#xA;classic ciphers usually don't take all the ascii characters,&lt;br&gt;&#xA;this raises the question, obviously converting it is the answer&lt;br&gt;&#xA;but to what? you need to make numbers of them,&lt;br&gt;&#xA;well numbers and letters.&lt;br&gt;&#xA;But little classical ciphers will accept numbers and letters.&lt;br&gt;&#xA;luckily for that there's a solution as old as Rome.&lt;br&gt;&#xA;you can sure count on a Roman to fix some of your problems :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; to me J and I are the same, double all the way.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Classic always comes before modern when unraveling the past&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Numbers and letters, or in other words base 16.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 4&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; sites which may be of use: &lt;a href=&quot;http://aes.online-domain-tools.com/&quot; rel=&quot;nofollow&quot;&gt;Modern cipher&lt;/a&gt;, &lt;a href=&quot;http://rumkin.com/tools/cipher/&quot; rel=&quot;nofollow&quot;&gt;classical cipher&lt;/a&gt;, &lt;a href=&quot;http://cryptii.com/text/select&quot; rel=&quot;nofollow&quot;&gt;conversion&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 5&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You will know when you've broken the first encryption, likewise with the 2nd.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 6&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every cipher needs a key!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT very important!!&lt;/strong&gt;&lt;br&gt;&#xA;If you are able to decipher it the first time, remove 1 X at the very end before continuing.&lt;/p&gt;&#xA;" OwnerUserId="9835" LastEditorUserId="9835" LastEditDate="2015-04-28T20:43:22.173" LastActivityDate="2015-04-29T07:15:51.477" Title="Ugh, an other crypto puzzle?" Tags="|riddle|cipher|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12693" PostTypeId="1" AcceptedAnswerId="12819" CreationDate="2015-04-24T19:56:48.777" Score="8" ViewCount="1539" Body="&lt;p&gt;Yesterday (April 22, 2015), Dr Jerico Wong, a prominent researcher and experimental physicist at SciCorp Industries in Chattanooga, Tennessee, had &#xA;monumental successes in his cold fusion experiments. Although final testing &#xA;has not yet been completed, and there has been no official press release, &#xA;news this big travels fast, and a few hours later, a few teams of researchers at rival companies had learned of Wong's groundbreaking, world-changing finds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next day, as Wong was in the parking garage while leaving his workplace &#xA;at 5pm, his face was covered and he was quickly forced into a van, which &#xA;drove away. He heard the voice of a man saying, &quot;Don't make a sound.&quot;&#xA;As they started driving, Wong felt ear muffs being put on his head, on top &#xA;of the covering that was already there, and he could not hear anything. For &#xA;around 2 hours, he sat still in fear and silence, wondering if he &#xA;would survive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, as he slowly turned his head against the van wall, the ear muffs moved &#xA;slightly, and he could hear the 2 men in the front seats talking, but he &#xA;couldn't make out all of the words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He heard:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Man #1&lt;/strong&gt;: We (2-3 words garbled) warehouse in 5 minutes. You want (2-3 words &#xA;garbled) pizza?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Man #2&lt;/strong&gt;: Yeah, but there's a hundred of them in (1 or 2 words garbled). How about that one (1-2 words garbled)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Man #1&lt;/strong&gt;: What's that phone number on the window? I'll order delivery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Man #2&lt;/strong&gt;: (3 garbled numbers)-555-2 (3 garbled numbers)&#xA;(garbled phone conversation)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Man #1&lt;/strong&gt;: (3-4 garbled words) grub by 7:45*&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&#xA;*After much thought, I changed the time above from 7:30 to 7:45&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wong knew that they had been traveling with almost no stops or turns; therefore, they must have been on one of the interstate highways. He also figured that they were in a fairly large city because one of the men said there were a hundred pizza places.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wong knew that there were 4 possibilities of where he was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Nashville, TN&lt;br&gt;&#xA;  Knoxville, TN&lt;br&gt;&#xA;  Birmingham, AL&lt;br&gt;&#xA;  Atlanta, GA&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After some thought, he figured out where he was. Wong found it strange&lt;br&gt;&#xA;that they didn't even think to check his pockets. He used the cell phone in his pocket to secretly text his friends: &quot;Help. They've taken me to a warehouse in [city name]&quot;, which eventually led to him being rescued.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did he know which of the 4 cities he was in?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;EDIT: There are 2 realizations required to solve this. The simpler one has been figured out already, but the more &quot;Eureka&quot; one has not.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-04-26T22:17:38.913" LastActivityDate="2016-11-24T10:39:11.780" Title="Abducted and taken to which city?" Tags="|logical-deduction|mystery|story|" AnswerCount="5" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="12694" PostTypeId="1" AcceptedAnswerId="12875" CreationDate="2015-04-24T20:09:20.810" Score="8" ViewCount="3972" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;When you use me, you only ever see what I have become instead of what I really am.&lt;/p&gt;&#xA;&lt;p&gt;I can be used to do powerful things, but I can be very fragile at my core.&lt;/p&gt;&#xA;&lt;p&gt;Through time, I have changed lives!&lt;/p&gt;&#xA;&lt;p&gt;Unfortunately some choose to hate me, and others choose to embrace me.&lt;/p&gt;&#xA;&lt;p&gt;I am used around the world in many different ways.&lt;/p&gt;&#xA;&lt;p&gt;My family is huge, and yet still getting bigger.&lt;/p&gt;&#xA;&lt;p&gt;It is important not to forget about my elders, as they are still needed (by some).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Do you know what I am?&lt;/p&gt;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; You can look at me, but not touch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Hint #2:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; I'm not as old as world war 1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Hint #3:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Some that have 'used' me, are now amongst the richest people in the world!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I apologize for not adding the second hint into the original riddle. This would have helped greatly reduce the scope. (I wasn't expecting for some answers to be so similar in description but not the same answer)&lt;/p&gt;&#xA;&lt;p&gt;(Make sure your answer can fit every line)&lt;/p&gt;&#xA;&lt;p&gt;I have tried to keep this one more broad to make it a little more difficult. If necessary I can give some hints later, but I think someone will be smart enough to solve it before that. :)&lt;/p&gt;&#xA;&lt;p&gt;[It seems keeping it broad was not such a great idea]&lt;/p&gt;&#xA;&lt;p&gt;&lt;b&gt;Spoiler!! This will almost certainly give the answer away! Only look at if you gave up and would like to know&lt;/b&gt;:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; stackoverflow.com/&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;My apologizes on this riddle as it was not executed as well as I would have liked, but hopefully some people enjoyed it (and maybe got stumped). It was difficult for me to give effective hints/help without just giving out the answer.&lt;/p&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-28T19:14:05.660" Title="Powerful and Fragile - a riddle" Tags="|riddle|" AnswerCount="12" CommentCount="11" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12696" PostTypeId="2" ParentId="12689" CreationDate="2015-04-24T20:21:48.327" Score="4" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;ABBA. &#xA;&#xA;All the verses are taken from lyrics of ABBA songs, and each verse follows an ABBA rhyming scheme.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="4616" LastEditDate="2018-05-18T20:22:08.737" LastActivityDate="2018-05-18T20:22:08.737" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="12697" PostTypeId="2" ParentId="12694" CreationDate="2015-04-24T20:29:13.920" Score="4" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; IPhone (or actually any other smartphone who has multiple models)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you use me, you only ever see what I have become instead of what I really am.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you use a smart phone you see many things, often a small gaming device but it used to be and is in essence a phone&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be used to do powerful things, but I can be very fragile at my core.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you can do powerful calculation with a smartphone, as a matter of fact it is more powerful than the computers NASA used to send the Apollo missions. A smartphone (just like a computer) uses CPU cores.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Through time, I have changed lives!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A smartphone (or normal phone) saved lives (calling emergency services) and it changed many because everything became easier.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately some choose to hate me, and others choose to embrace me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many people hate Iphones, some love them. but in a broader sense some people find mobile phones (smartphones) are ruining people's socials skills, others accept it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am used around the world in many different ways.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; well a smartphone has many different applications.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My family is huge, and yet still getting bigger.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; iPhone 3G, 3GS, 4, 4S, 5, 5S, 6 are all part of the family. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is important not to forget about my elders, as they still needed (by some).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People still use older iPhones (with different sizes and older iOS versions), so app developers need to keep that in mind.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would like to thank Pacoverflow for his additions for the last 2 sentences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if you have siri you can look at your phone but you do not have to touch it to use it, you talk to it. &amp;lt;- ye this one doesn't fit 100% xD&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;An other answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a woman. bear with me here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you use me, you only ever see what I have become instead of what I really am.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Haha &lt;em&gt;use&lt;/em&gt; :P well if you 'use' a woman she becomes an &lt;em&gt;object&lt;/em&gt; instead of the person she really is.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be used to do powerful things, but I can be very fragile at my core.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Woman (love) are a power to be reckoned with but at their core woman are fragile and precious.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Through time, I have changed lives!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; every great man (in history) has had a woman supporting him, isn't that how it goes? A woman can often change a man's live.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately some choose to hate me, and others choose to embrace me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some hate woman (many women hate other women), some hat them because of their beliefs (in the bible a woman was tempted to take the forbidden fruit). Do i even need to explain the embrace part? ;) if you know what i mean.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am used around the world in many different ways.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; women are very versatile, they are used as babysitter, businesswoman, teacher (etc)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My family is huge, and yet still getting bigger.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; well obviously any person's family is big, we're descendants of millions and still continue to grow (childbirth)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is important not to forget about my elders, as they still needed (by some).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; parents are the reason we are all here, they are also still needed for love and attention. (also to watch over the grand kids).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As for the hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you can look at other woman (which you are not married/engaged/in a relation ship with) but you can most definitely not touch&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9835" LastEditorUserId="9835" LastEditDate="2015-04-24T21:16:31.977" LastActivityDate="2015-04-24T21:16:31.977" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12698" PostTypeId="2" ParentId="12694" CreationDate="2015-04-24T20:40:40.010" Score="0" Body="&lt;p&gt;I'd say&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Coral&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In fact, you use and see the skeleton, not the living creature. It's used to make jewels and talismans (symbols of power), but it's fragile. Many different organisms live in a coral during its formation. Some love it and wear it (embrace), those actually hate it since they destroy a delicate fauna. Coral is around the world in the forms of bracelets and necklaces. &lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-04-25T09:25:55.487" LastActivityDate="2015-04-25T09:25:55.487" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12699" PostTypeId="1" AcceptedAnswerId="12701" CreationDate="2015-04-24T20:45:01.090" Score="4" ViewCount="339" Body="&lt;p&gt;A box contains 900 cards enumerated from 100 to 999 (Each number appears once and just in one card). I took some random cards without looking at them and calculated the additions of the digits in each one.&lt;br&gt;&#xA;How many cards should I take as minimum for making sure I have three cards with the same addition of digits?&lt;/p&gt;&#xA;" OwnerUserId="11789" LastEditorUserId="11789" LastEditDate="2015-04-25T14:57:02.117" LastActivityDate="2015-04-25T14:57:02.117" Title="Mathematical probability with cards" Tags="|mathematics|logical-deduction|cards|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12700" PostTypeId="2" ParentId="12693" CreationDate="2015-04-24T20:47:43.430" Score="1" Body="&lt;p&gt;I'm going to guess Birmingham, AL.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reasoning:  Knoxville is too close.  They would have been there within 2 hrs.  Atlanta is probably a little too far.  They wouldn't be able to eat by 730 if they were going to Atlanta.  We are left with only Nashville or Birmingham.  When he asked for the pizza number, he gave the area code, indicating he was probably in a different area code than he is familiar with. I'm thinking this means he probably crossed state lines, so I guess Birmingham over Nashville.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thoughts?&lt;/p&gt;&#xA;" OwnerUserId="11315" LastActivityDate="2015-04-24T20:47:43.430" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12701" PostTypeId="2" ParentId="12699" CreationDate="2015-04-24T21:00:23.860" Score="9" Body="&lt;p&gt;Assuming the cards aren't replaced, you would need to take&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;53 cards&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In order to be sure that you have three cards where the digits add up to the same number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are 27 different possible sums, ranging from 1 (100) to 27 (999). Therefore, it would be possible to pull 27 cards without finding a single match. There are, however, only 25 repeatable sums, ranging from 2 (101/110) to 26 (998/899). There is only one combination of numbers that add up to 1, and only one combination that adds up to 27. Therefore, in order to get two of every repeatable sum and one of every non-repeatable sum, you would need 27+25 = 52 cards. There are no more possible duos, so pulling a 53rd card would guarantee that you had at least three cards with the same sum in their digits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-04-24T21:00:23.860" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12703" PostTypeId="2" ParentId="12693" CreationDate="2015-04-24T21:29:54.357" Score="7" Body="&lt;p&gt;&lt;strong&gt;Observation 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He is in either Birmingham or Nashville as they are both in central time.  All others (including Chattanooga) are in Eastern Time.  As was originally discussed by pacoverflow, the time zone must have changed as 2 hours plus 30 minutes to cook the pizza and deliver pizza is barely 7:30pm. If he crossed the time zone, it would be 8:30pm Eastern Time which makes more sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Observation 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If he went to Knoxville he would probably notice a change in elevation. This area is alot more mountainous than the other areas described. This was not mentioned in the question and he probably would have noticed the hills.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Between Chattanooga and Nashville, there is a change in elevation too but both cities have roughly the same elevation. This is easier to miss.  This means I could use this to guess Birmingham.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Observation 3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He took &quot;there are a hundred of them in&quot; to be followed by 1 or 2 words.  The phrase &quot;this city&quot; technically means we can't tell much from this.  It is likely, however, to have said &quot;Nashville&quot; as it is 2 syllables.  If he said &quot;Birminham&quot; that could more easily sound like 2 or 3 words.  All of the other missing words can be monosyllabic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Observation 4:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the person was reading from the inside of the van a number on the outside, he could have been dumb enough to read it backwards.  The only number that would allow that mistake would come from Birmigham as the area code (205) can be (502) reversed.  (404), (865), and (615) cannot be read backwards.   This does not seem to have happened, however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conclusion:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that Observation 3 trumps the hills on the way to Nashville.  For that reason I think the answer is Nashville TN.  It is a prettier city anyways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Revision&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the pizza delivery time has been changed to 7:45, it now makes more sense for the people to be in the same time zone than to have changed.  5pm plus 2 hours plus 30 minute delivery plus a little bit of spare time is roughly 7:45.  If this is the case, the mountanous terrain arround Knoxville dominates yielding &lt;strong&gt;Atlanta Georgia&lt;/strong&gt; as the most likely city.  As this relies heavily on estimated pizza delivery times a blindfolded person's estimate of time, this is still not 100% certain to me.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditDate="2015-04-24T23:20:41.360" LastActivityDate="2015-04-24T23:20:41.360" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12704" PostTypeId="1" AcceptedAnswerId="12709" CreationDate="2015-04-24T21:33:54.537" Score="7" ViewCount="901" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;There are three spiders and an ant on the edges of a wireframe regular tetrahedron. The spiders all move at the same speed, which is faster than that of the ant. Though the ant can always see the spiders, the ant is &lt;em&gt;invisible&lt;/em&gt;. A spider can only detect the ant by being on top of it, in which case the ant is immediately eaten. Show how the spiders can catch the ant.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is rather similar to an &lt;a href=&quot;https://puzzling.stackexchange.com/q/12595/10615&quot;&gt;earlier puzzle&lt;/a&gt; I posted. The differences are that the cube is now a tetrahedron, the spider's speed increased from $\frac13$(fly's speed) to (fly's speed)$+\varepsilon$, and the fly is now invisible. The first two changes greatly help the spider, but this is still a harder puzzle.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-04-25T21:46:53.180" Title="Spiders and Invisible Ant on a Tetrahedron" Tags="|pursuit-evasion|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12706" PostTypeId="2" ParentId="12694" CreationDate="2015-04-24T22:33:48.920" Score="7" Body="&lt;p&gt;It's probably a wrong answer, but I'll give it a try&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it's a nuclear reactor&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you use me, you only ever see what I have become instead of what I really am.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; we don't see a nuclear reactor but we can see what it can provide (electricity, mouvement)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be used to do powerful things, but I can be very fragile at my core.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We can use a nuclear reactor to generate electricity or power boats, but the core of it is fragile in some conditions (see Fukushima for instance)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Through time, I have changed lives!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nuclear reactor provide a lot of energy which help countries to develop&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately some choose to hate me, and others choose to embrace me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people are against that type of energy (Germany) and some love it (France)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am used around the world in many different ways.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is different use of a nuclear reactor (this one is weak) and also different types&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My family is huge, and yet still getting bigger.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is already a lot of those but more are planned&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is important not to forget about my elders, as they still needed (by some).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The old reactors are still needed but we have to take care of them so they can still provide energy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10400" LastActivityDate="2015-04-24T22:33:48.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12708" PostTypeId="2" ParentId="12416" CreationDate="2015-04-25T00:21:25.230" Score="2" Body="&lt;p&gt;I convey the invisible,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Words can convey the invisible &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I convey what can be felt,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Poetry etc. conveys feelings&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But not seen,&#xA;Maybe just spoken?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inside my heart,&#xA;I am not as nice as I could seem to be,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Words can be malicious, have double meanings&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is because I deliver the invisible,&#xA;At the speed of a mere blink,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Even just looking at text can allow you to sense the meaning.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am very powerful,&#xA;And can bring comfort,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Words can be a lot(just look at Hitler), but a mom, say comforts their children with words.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Across my horizons, I am free,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People can say what ever they want&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But still vulnerable to your opinion.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Of course, people want you to approve of what they say&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; So I guess the spoken word?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="11790" LastEditorUserId="5373" LastEditDate="2015-04-27T08:37:58.590" LastActivityDate="2015-04-27T08:37:58.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12709" PostTypeId="2" ParentId="12704" CreationDate="2015-04-25T00:39:15.427" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Call the vertices $A,B,C,D$. The spiders start themselves off with two spiders at $A$ and one spider at $D$. Then they move around in the following cycle: one spider stays at $A$ and the other two spiders head directly to $B$, then one spider stays at $B$ and the other two spiders head directly to $C$, then one spider stays at $C$ and the other two head directly to $D$, then one stays at $D$ and the other two head directly to $A$, repeat. Eventually they catch the ant: in order to evade the ant must travel across each of the edges $AB, BC, CD, DA$ at least once per cycle (think about what happens if the ant refuses to travel all the way across edge $DA$ for an entire cycle, assuming the ant doesn't start the cycle along edge $DA$).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8xTPI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="11295" LastEditorUserId="11295" LastEditDate="2015-04-25T21:46:53.180" LastActivityDate="2015-04-25T21:46:53.180" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12711" PostTypeId="2" ParentId="12661" CreationDate="2015-04-25T02:42:40.657" Score="0" Body="&lt;p&gt;Here is a generalization of this (also covers the &lt;a href=&quot;http://www.mytechinterviews.com/the-fox-and-the-duck&quot; rel=&quot;nofollow&quot;&gt;Fox and Duck puzzle&lt;/a&gt;). Formalizing what JLee and user3757614 wrote:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the square have unit side, and take coords with x,y between (0,1). Evader E starts at (1/2,1/2) and distance to boundary points is between 1/2 (to midside) and 1/√2 (to corner). Specifically there are four escape paths of length 1/2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the perimeter. More specifically, since the n pursuers (P's) and evader (E) move at the same speeds (although the P are constrained to move orthogonally), then the generalized condition for trapping would be that the (straight-line distance from E to the nearest point X on boundary) &gt;= (orthogonal distance along perimeter from nearest P to X)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we guarantee escape for n = 3? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;WLOG say E goes towards the SW corner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever E is in one of the four quadrants, at some general point (x,y), this commits two of the pursuers to shadow it on each of the closest sides: say P1 (x,0) and P2 (0,y).&#xA;Letting E(x,y) go sufficiently close to the corner, we can essentially force P1 and P2 onto nearly the same point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leaves P3 to cover all the remaining perimeter of length 2. P3 can't stay at the opposite corner;&#xA;E plays the phase-lag trick by moving in circles of expanding radius so that E develops a phase-lag wrt P3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now applying the Pigeonhole Principle, we guaranteed there exists a set of escape points {X} on the boundary which are straight-line closer to E than P3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll try to make this cleaner; taking polar coordinates wrt O(0.5,0.5) might do it.&lt;/p&gt;&#xA;" OwnerUserId="2349" LastActivityDate="2015-04-25T02:42:40.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12712" PostTypeId="2" ParentId="12694" CreationDate="2015-04-25T05:17:46.593" Score="6" Body="&lt;p&gt;My answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A scientific theory&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you use me, you only ever see what I have become instead of what I really am.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Scientific theories aren't the same thing as their applications. People learn about scientific theories in relation to their applications.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be used to do powerful things, but I can be very fragile at my core.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Scientific theories have led to incredible achievements, but every scientific theory is subject to question and just one single piece of evidence can invalidate an entire theory.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Through time, I have changed lives!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is pretty much self-explanatory.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately some choose to hate me, and others choose to embrace me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is a large anti-science movement in the world, while other people take science and scientific theory as the key part of their worldview.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am used around the world in many different ways.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Another one that is self-explanatory.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My family is huge, and yet still getting bigger.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are many scientific theories, and more are being formulated/discovered all of the time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is important not to forget about my elders, as they still needed (by some)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Older scientific theories are generally simpler, and thus easier for educational purposes. We still learn of various outdated and invalidated models for various things (such as the structure of an atom). While the real applications use the newer theories, older ones are still used where they're accurate enough for general purposes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can look at me, but not touch.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Scientific theories can be shown to a person, but they're not physical things, and thus cannot be touched.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9289" LastActivityDate="2015-04-25T05:17:46.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12715" PostTypeId="2" ParentId="12704" CreationDate="2015-04-25T08:57:47.797" Score="1" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/A9ugS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One spider stands still in A, another spider stands in B. The third spider (let's call him Bob) is the only one moving.&lt;br&gt;&#xA;First of all, Bob checks if the ant is between A and B.&lt;br&gt;&#xA;If not, he moves to the edge CD (as shown in the picture). Now, the ant has to be on a blue edge, on an orange edge, or on a portion of the green edge, with a probability of $p=\frac1 2$ of being in front or behind Bob.&lt;br&gt;&#xA;Now, Bob moves to D, then chooses a random edge to explore between the two blue ones: if the ant is on a blue edge ($p=\frac1 2$), then Bob has a $\frac1 2$ chance of catching it after choosing the path. After all, he got a $p=\frac1 4$ chance of catching the ant starting from the edge CD.&lt;br&gt;&#xA;If the ant survived, Bob goes back to CD and does the same thing with orange edges.&lt;br&gt;&#xA;If the ant survived again, Bob goes back to CD and repeats with blue edges.&lt;br&gt;&#xA;And so on, until he catches the ant (he has a probability of $p=1$ of catching the ant after an infinite time!&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-04-25T09:24:10.903" LastActivityDate="2015-04-25T09:24:10.903" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12716" PostTypeId="1" CreationDate="2015-04-25T09:26:50.937" Score="12" ViewCount="2417" Body="&lt;p&gt;You have blocks in five colors (red, green, yellow, blue and black - or 1 to 5). Each color comes in two varieties: a cube and one as big as two cubes stacked atop each other (sizes 1 and 2).&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jvjQi.png&quot; alt=&quot;(the five colors plus white.)&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;You are supposed to use them to build two towers next to each other such that two blocks of the same color (in either tower) are separated by the sum of their heights. For example:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uc16f.png&quot; alt=&quot;(rules regarding the distances of same-colored blocks.)&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Note that this only says that each color must repeat within a given distance, but not that every block of the tower must be of a particular color, so you get white blocks to fill any holes (they don't do anything else, and they aren't subject to any of the rules above or below).&lt;/p&gt;&#xA;&lt;p&gt;You don't know how high the tower is supposed to be, hence &lt;strong&gt;give the shortest repeating pattern that can be used to extend such a tower upwards for any height&lt;/strong&gt;. Within that pattern, each block of each size and color should repeat an equal number of times (ignoring white, of course).&lt;/p&gt;&#xA;&lt;p&gt;Now if that were all, finding a solution would be simple, e.g. this one:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Wzjee.png&quot; alt=&quot;(five colors in repeating pattern 1-left, 1-right, 2-left, 2-right.)&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;However&lt;/strong&gt;, in the pattern of each tower each block size of a color should be opposite a block of each other color in the other tower. The above example doesn't fulfill this: consider the blue 2-block: while it occurs on the left opposite to the black 2-block, but never vice versa, and never opposite to the green 2-block, nor does the blue 1-block occur opposite the black 1-block.&lt;/p&gt;&#xA;&lt;p&gt;Thus for any pair of two colors (not involving white and not both colors the same) the following patterns must occur (given as example for red and blue):&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zk7ZE.png&quot; alt=&quot;(differently colored blocks opposite each other.)&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Note that in 3 and 4 they don't need to be perfectly lined up and in 5-8 the small one doesn't need to be at the same height, but there must be one box in which they are opposite, so e.g. for 3 and 5 these are alternative examples:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NqnSU.png&quot; alt=&quot;(blocks opposite each other yet offset.)&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="10070" LastEditorUserId="463" LastEditDate="2022-06-09T02:17:40.103" LastActivityDate="2022-06-10T03:05:42.940" Title="Building towers with colored blocks" Tags="|construction|" AnswerCount="2" CommentCount="15" ContentLicense="CC BY-SA 4.0" />
  <row Id="12718" PostTypeId="1" CreationDate="2015-04-25T14:44:02.253" Score="3" ViewCount="584" Body="&lt;p&gt;Two knights (black and white, the latter going first) start on opposing corners of a chessboard, &lt;a href=&quot;http://en.wikipedia.org/wiki/Knight_%28chess%29&quot; rel=&quot;nofollow&quot;&gt;moving across it in the usual manner&lt;/a&gt;, attempting to capture each other: the knight who first lands on the square currently occupied by the other knight (i.e. capturing it) wins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To ensure that the game doesn't go on forever, each knight must increase the distance from his point of origin (unless that move would immediately capture the other knight). If neither can move forward nor capture the other then they are tied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Show whether one of the knights can force a win, tie or loss.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are six cases to consider: whether either of them can force a win, tie or loss.&lt;br&gt;&#xA; Each move sends them to a square of a different color. A knight must start from a different-colored square to end up on the square the other knight is on. Since both start in opposing corners they start on squares of the same color. Therefore white (going first) can never win and black never lose.&lt;br&gt;&#xA; Black can force a win if and only if white cannot force a tie. White can force a loss if and only if black cannot force a tie. Hence it suffices to show 1. either that black can win or white can tie and 2. either that white can lose or black can tie.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10070" LastEditorUserId="12408" LastEditDate="2016-07-25T02:40:19.000" LastActivityDate="2016-07-25T02:40:19.000" Title="Two knights jousting on a chessboard" Tags="|strategy|game|chess|knight-moves|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12719" PostTypeId="1" CreationDate="2015-04-25T15:12:12.483" Score="-3" ViewCount="140" Body="&lt;p&gt;&lt;strong&gt;Your aim:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mark any finite number of points on a plane. It should meet this axiom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Axiom:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Original way of stating it:&lt;/em&gt; If a line (infinite) is drawn passing through exactly n (n&gt;0) points, any line drawn parallel to it will pass through exactly n or 0 points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;New way of stating it:&lt;/em&gt; If &lt;strong&gt;any&lt;/strong&gt; 2 parallel lines are selected, either one (or both) of them passes through zero points, or they pass through the same number ($n$) of points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Don't my original and new way of stating the axiom mean the same thing?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4269" LastEditDate="2015-04-27T09:44:19.657" LastActivityDate="2015-04-27T09:44:19.657" Title="Points following an axiom" Tags="|mathematics|geometry|" AnswerCount="2" CommentCount="12" ClosedDate="2015-04-27T01:47:00.990" ContentLicense="CC BY-SA 3.0" />
  <row Id="12720" PostTypeId="2" ParentId="12719" CreationDate="2015-04-25T15:31:45.880" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In geometry, parallel lines are lines in a plane which do not meet. (source &lt;a href=&quot;http://en.wikipedia.org/wiki/Parallel_(geometry)&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;here&lt;/strong&gt;&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The parallel line will always pass through exactly $0$ points of the other line, else they wouldn't be parallel!&lt;br&gt;&#xA;Example:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/DLP1o.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What if the two lines coincide?&lt;/strong&gt;  They both pass through the $n$ points, but they're called &lt;em&gt;coincident&lt;/em&gt; and not &lt;em&gt;parallel&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-04-25T15:31:45.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12721" PostTypeId="2" ParentId="12719" CreationDate="2015-04-25T15:37:05.217" Score="2" Body="&lt;p&gt;If I understand correctly, below is a simple set of points satisfying the condition. Points in the set are red. Parallel lines are of the same color.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/d2ASZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="11100" LastActivityDate="2015-04-25T15:37:05.217" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12722" PostTypeId="2" ParentId="12718" CreationDate="2015-04-25T16:02:11.597" Score="2" Body="&lt;p&gt;Its easy for white to force a draw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the starting position of white is h1, the next moves are g3, f5 and then depending on blacks move either g7 or e7. Black can't move back toward his corner so white has escaped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black can escape with a similar strategy by moving to b6 then c4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one player can never win then that player will always force a draw.&lt;/p&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="11297" LastEditDate="2015-04-25T16:25:31.500" LastActivityDate="2015-04-25T16:25:31.500" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12724" PostTypeId="2" ParentId="12718" CreationDate="2015-04-25T16:35:35.097" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the first player is doomed to lose&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;this picture shows all the cases of the first(red) player hunted by atleast one opponent(yellow) knight-path&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/68tgc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9517" LastEditorUserId="9517" LastEditDate="2015-04-25T16:57:57.467" LastActivityDate="2015-04-25T16:57:57.467" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12725" PostTypeId="1" AcceptedAnswerId="12728" CreationDate="2015-04-25T17:09:52.777" Score="8" ViewCount="2013" Body="&lt;ol&gt;&#xA;&lt;li&gt;Eight animals named E, F, G, H, J, K, L and M are sitting around a circular table facing the centre.&lt;/li&gt;&#xA;&lt;li&gt;These are the kinds of the animals: Cat, Cobra, Dog, Elephant, Falcon, Llama, Panda and Sparrow (in no particular order).&lt;/li&gt;&#xA;&lt;li&gt;F is sitting second to the left of K.&lt;/li&gt;&#xA;&lt;li&gt;The Sparrow is an immediate neighbour of K.&lt;/li&gt;&#xA;&lt;li&gt;There are only three animals between the Sparrow and E.&lt;/li&gt;&#xA;&lt;li&gt;Only one animal sits between the Elephant and E.&lt;/li&gt;&#xA;&lt;li&gt;The Cobra is to the immediate right of the Elephant.&lt;/li&gt;&#xA;&lt;li&gt;M is second to the right of K.&lt;/li&gt;&#xA;&lt;li&gt;H is the Sparrow.&lt;/li&gt;&#xA;&lt;li&gt;G and J are immediate neighbours of each other.&lt;/li&gt;&#xA;&lt;li&gt;Neither G nor J is an Elephant.&lt;/li&gt;&#xA;&lt;li&gt;The Falcon is to the immediate left of F.&lt;/li&gt;&#xA;&lt;li&gt;The Llama is second to the right of the Cobra.&lt;/li&gt;&#xA;&lt;li&gt;The Panda is an immediate neighbour of the Elephant.&lt;/li&gt;&#xA;&lt;li&gt;G is second to the right of the Cat.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Find out the name of each kind of animal and the order they are sitting around the table. Note the process how you came to the conclusion (the most efficient process wins).&lt;/p&gt;&#xA;" OwnerUserId="11797" LastEditorDisplayName="user88" LastEditDate="2015-04-29T18:56:28.753" LastActivityDate="2020-06-08T17:26:53.117" Title="Eight animals are sitting around a circular table" Tags="|logical-deduction|logic-grid|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12726" PostTypeId="2" ParentId="12685" CreationDate="2015-04-25T17:11:52.440" Score="6" Body="&lt;p&gt;It is possible to escape from&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; any finite number of smugglers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Say the lake has unit radius. Label the boundary of the lake with angles $\theta$, measured in radians. We will begin by heading to a point near the boundary of the lake, by angle $\theta=0$. For the moment, pretend we actually reach the boundary but cannot escape due to a smuggler (moving inward just a bit will not change the answer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we choose a point $0&amp;lt;\theta&amp;lt;\pi$ on the boundary and head straight there. The distance to this point is $2\sin(\theta/2)$, so to keep us from escaping, there must be a smuggler somewhere in the arc $$A_\theta:=\big[\theta-2\sin(\theta/2),\,\theta+2\sin(\theta/2)\big].$$&#xA;We will find an infinite sequence $\theta_1,\theta_2,\ldots$ of angles such that the arcs $A_{\theta_1},A_{\theta_2},\ldots$ are disjoint. This means that for any positioning of a finite number of smugglers, there will be some $\theta_i$ that we can reach before any of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We define the $\theta_i$ recursively. Start with $\theta_1=\pi$. For $i\geq 2$, define $$\theta_i=\frac{\theta_{i-1}-2\sin(\theta_{i-1}/2)}{2}.$$&#xA;Observe that $0&amp;lt;2\sin(\theta/2)&amp;lt;\theta$ for every $0&amp;lt;\theta&amp;lt;\pi$, so that $\theta_1&amp;gt;\theta_2&amp;gt;\ldots&amp;gt;0$. The arcs $A_{\theta_i}$ are disjoint because $\theta_{i-1}-2\sin(\theta_{i-1}/2)&amp;gt;\theta_i+2\sin(\theta_i/2)$,&#xA;which comes from the definition of $\theta_i$ and the inequality $\theta_i+2\sin(\theta_i/2)&amp;lt;2\theta_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are $N$ smugglers, the arcs $A_{\theta_1},\ldots,A_{\theta_{N+1}}$ are disjoint, so there must be one of these arc $A_{\theta_i}$ containing no smugglers. We then head directly toward the point $\theta=\theta_i$, and no smuggler can keep us from escaping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, suppose that instead of starting on the boundary of the lake, we instead start $\epsilon&amp;gt;0$ units in from the boundary. For each $0&amp;lt;\theta&amp;lt;\pi$, we define $A^\epsilon_\theta$ to be the arc of points on the boundary of the lake from which a smuggler could reach the point $\theta$ as quickly as we can. The arcs $A_{\theta_1},\ldots,A_{\theta_{N+1}}$ are disjoint, so for $\epsilon$ sufficiently small, $A_{\theta_1}^\epsilon,\ldots,A_{\theta_{N+1}}^\epsilon$ will also be disjoint. This means we can still find $i\in\{1,\ldots,N+1\}$ such that we can reach $\theta=\theta_i$ before any of the smugglers.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-04-25T17:24:20.153" LastActivityDate="2015-04-25T17:24:20.153" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12728" PostTypeId="2" ParentId="12725" CreationDate="2015-04-25T18:05:35.813" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/zPYce.png&quot; alt=&quot;solution schema&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h1&gt;&lt;strong&gt;Explanation:&lt;/strong&gt;&lt;/h1&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; First of all place &lt;strong&gt;F&lt;/strong&gt; somewhere, I put him in the bottom chair.&lt;br /&gt;&#xA; Statement &lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt; forces &lt;strong&gt;K&lt;/strong&gt; to be in the right chair.&lt;br /&gt;&#xA; Statement &lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt; forces &lt;strong&gt;M&lt;/strong&gt; to be in the top chair.&#xA; Statement &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt; allows Sparrow be top-right or bottom-right. Let's assume &lt;strong&gt;Sparrow&lt;/strong&gt; bottom-right.&lt;br /&gt;&#xA; Statement &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt; means that Sparrow is directly facing E, so &lt;strong&gt;E&lt;/strong&gt; is top-left.&lt;br /&gt;&#xA; Statement &lt;span class=&quot;math-container&quot;&gt;$9$&lt;/span&gt; says that &lt;strong&gt;H&lt;/strong&gt; is Sparrow, so bottom-right.&lt;br /&gt;&#xA; Statement &lt;span class=&quot;math-container&quot;&gt;$10$&lt;/span&gt; says that G and J are neighbours, so they occupy left and bottom left. Let's assume &lt;strong&gt;J&lt;/strong&gt; left and &lt;strong&gt;G&lt;/strong&gt; bottom-left.&lt;br /&gt;&#xA; Statement &lt;span class=&quot;math-container&quot;&gt;$15$&lt;/span&gt; says that top-left is &lt;strong&gt;Cat&lt;/strong&gt;.&lt;br /&gt;&#xA; The only left letter is &lt;strong&gt;L&lt;/strong&gt;, which goes top-right.&lt;br /&gt;&#xA; Statement &lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt; allows elephant to be top-right or bottom-left, but Statement &lt;span class=&quot;math-container&quot;&gt;$11$&lt;/span&gt; doesn't allow it to be bottom-left; so, &lt;strong&gt;Elephant&lt;/strong&gt; is top-right.&lt;br /&gt;&#xA; Statement &lt;span class=&quot;math-container&quot;&gt;$7$&lt;/span&gt; says &lt;strong&gt;Cobra&lt;/strong&gt; top.&lt;br /&gt;&#xA; Statement &lt;span class=&quot;math-container&quot;&gt;$12$&lt;/span&gt; says &lt;strong&gt;Falcon&lt;/strong&gt; bottom-left.&lt;br /&gt;&#xA; Statement &lt;span class=&quot;math-container&quot;&gt;$13$&lt;/span&gt; says &lt;strong&gt;Llama&lt;/strong&gt; left.&lt;br /&gt;&#xA; Statement &lt;span class=&quot;math-container&quot;&gt;$14$&lt;/span&gt; says &lt;strong&gt;Panda&lt;/strong&gt; right.&lt;br /&gt;&#xA; The only animal left is &lt;strong&gt;Dog&lt;/strong&gt;, bottom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Why can't Sparrow be top-right?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Let's assume Sparrow top-right. Then, &lt;strong&gt;E&lt;/strong&gt; is bottom left for statement &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt;. G,J must be neighbours, so they occupy left and top-left. Since top-left is now occupied, &lt;strong&gt;Elephant&lt;/strong&gt; must be bottom-right for Statement &lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt;. Then, &lt;strong&gt;Cobra&lt;/strong&gt; right for Statement &lt;span class=&quot;math-container&quot;&gt;$7$&lt;/span&gt; and &lt;strong&gt;Llama&lt;/strong&gt; north for statement &lt;span class=&quot;math-container&quot;&gt;$13$&lt;/span&gt;. Though, it's impossible now to satisfy Statement &lt;span class=&quot;math-container&quot;&gt;$15$&lt;/span&gt;, as G (which was left or top-left) has an already assigned &amp;quot;second to the left&amp;quot; partner.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Why can't be J bottom-left and G left?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Let's assume &lt;strong&gt;J&lt;/strong&gt; bottom-left and &lt;strong&gt;G&lt;/strong&gt; left. Statement &lt;span class=&quot;math-container&quot;&gt;$15$&lt;/span&gt; says &lt;strong&gt;Cat&lt;/strong&gt; top.&#xA; The missing letter is &lt;strong&gt;L&lt;/strong&gt;, top-right. For Statement &lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt;, Elephant must be &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; as it can't be &lt;span class=&quot;math-container&quot;&gt;$J$&lt;/span&gt; (forbidden by &lt;span class=&quot;math-container&quot;&gt;$11$&lt;/span&gt;). Though, for Statement &lt;span class=&quot;math-container&quot;&gt;$7$&lt;/span&gt;, &lt;strong&gt;Cobra&lt;/strong&gt; and &lt;strong&gt;Cat&lt;/strong&gt; coincide, which is a contradiction!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2020-06-08T17:26:53.117" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="12729" PostTypeId="2" ParentId="12364" CreationDate="2015-04-25T20:42:12.297" Score="4" Body="&lt;p&gt;The key (before null padding) is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; God&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After properly decoded, AESed and XORed the last line is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; `You'll find that the answer is 'brine'.\x00You'll find that the answer is 'brine'.\x00&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9060" LastActivityDate="2015-04-25T20:42:12.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12730" PostTypeId="1" CreationDate="2015-04-25T21:00:02.277" Score="9" ViewCount="831" Body="&lt;p&gt;You have a grid like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NAn7r.png&quot; alt=&quot;grid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The entire grid isn't shown as it would be too large, but the number of squares in each row are as follows: $2, 4, 6, \ldots, 96, 98, 100, 100, 98, 96, \ldots, 6, 4, 2$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We define this grid as $G(100)$, as it is 100 squares across at its widest point and 100 squares high at its tallest point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want to tile it with only copies of this  tetromino:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pAfv3.png&quot; alt=&quot;tetromino&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may rotate or flip the tetromino.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Is it possible? Why or why not? An explanation in your answer is required.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For which even positive integer values of $n$ is tiling the grid $G(n)$ with only the tetromino possible? (Again, you must provide an explanation.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;note: I will post my own solution after either two days have passed or two distinct correct answers (for each question 1 and 2) have been provided.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="39" LastEditorUserId="39" LastEditDate="2015-04-26T03:29:22.033" LastActivityDate="2015-04-26T17:17:07.210" Title="Tiling a diamond-shaped grid with tetrominoes" Tags="|tiling|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="12731" PostTypeId="2" ParentId="12624" CreationDate="2015-04-25T22:16:50.923" Score="15" Body="&lt;p&gt;I've got it :-)&lt;/p&gt;&#xA;&lt;p&gt;The movie is:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;a href=&quot;http://www.imdb.com/title/tt0086250/&quot; rel=&quot;noreferrer&quot;&gt;Scarface&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;and the quote is:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;a href=&quot;http://www.youtube.com/watch?v=SQh80G5TAek&quot; rel=&quot;noreferrer&quot;&gt;&amp;quot;Say hello to my little friend!&amp;quot;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h1&gt;Solution method:&lt;/h1&gt;&#xA;&lt;h3&gt;Stage 1:&lt;/h3&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://pastebin.com/6EbZJaLk&quot; rel=&quot;noreferrer&quot;&gt;I wrote a solver in C&lt;/a&gt; to figure out the paths corresponding to each sum. It takes about 0.1 seconds to output the following results:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;1600: L=13     1800: L=14     1900: L=15     2300: L=16&#xA; 1  2  0  0     1  3  4  5     1  3  4  6     1  8  9 10&#xA; 9  3  4  5     0  2  6 11    11  2  5  7     7  2  3 11&#xA;10  8 12  6     8  7 10 12    12 10 14  8     6  4 15 12&#xA; 0 11  7 13     0  9 13 14     0 13  9 15     5 14 13 16&#xA;               &#xA;2400: L=13     3600: L=15     4500: L=14     5000: L=14&#xA; 1  9  8  7     1  8  9 11     1  7  6  0     1  7  8  0&#xA; 2 10 11  6     7  2 12 10     8  2  4  5     0  2  6  9&#xA; 0  3  5 12     6  3 13  0     9  3  0 13     3  5 13 10&#xA; 0  4  0 13     5  4 14 15    10 11 12 14     4 12 11 14&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;After grouping by length and adding together the values corresponding to each path length, I obtained the following:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;Length=13: 4000  (1600+2400)&#xA;Length=14: 11300 (1800+4500+5000)&#xA;Length=15: 5500  (1900+3600)&#xA;Length=16: 2300&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;(Actually I went off on a tangent at this point because I thought we were supposed to add up the values of the cells visited on each path. That turned out to be a dead end.)&lt;/p&gt;&#xA;&lt;h3&gt;Stage 2:&lt;/h3&gt;&#xA;&lt;p&gt;With a bit of help from &lt;a href=&quot;http://www.onlineocr.net/&quot; rel=&quot;noreferrer&quot;&gt;onlineocr.net&lt;/a&gt;, I converted the number tables into plain text and wrapped a &lt;a href=&quot;http://pastebin.com/dugVaabE&quot; rel=&quot;noreferrer&quot;&gt;Python program&lt;/a&gt; around them. Based on a hunch, this program tests for numbers that are proper factors of the sums obtained in Stage 1. For example, in the first table, where 4000 is the number provided as the hint, the first two numbers 500 and 10 both divide 4000, but the third number 530 does not.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/I5Pav.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="40" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-26T15:46:58.120" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12732" PostTypeId="1" AcceptedAnswerId="13234" CreationDate="2015-04-25T23:50:23.263" Score="37" ViewCount="10769" Body="&lt;p&gt;&lt;strong&gt;(Warning: Heavy reading incoming. I suggest writing stuff down...)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's now the year 2150, and you're one of the few people who didn't leave earth twenty years ago. Why? A deadly disease overtook the planet, and you happen to have immunity. There are now less than ten million people on the planet - the rest have either died, or left on a voyage to find a new planet to colonize.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being the workaholic detective that you are, you haven't taken a vacation in years, and your boss has demanded that you use your vacation days or lose them - You decide to go on a cruise trip to relax a bit. Approaching the cruise liner, you see the steward waiting at the boardwalk. You seem to be the only passenger there. Taking a quick glance at the boat, you notice that the boat is very old, but still kept quite nicely. It looks just like the ones made in 2009.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Walking up to the steward, you notice his name tag; &lt;code&gt;Chang&lt;/code&gt;, it says. He's dressed nicely, pen in one hand, hologram-board in the other. As soon as you stop in front of him, he starts speaking, &quot;Welcome aboard the C.C.L RuhRoh! You have the great honor of being on her last voyage. This time, we will be traveling for 15 days, making no stops until we reach Australia. This will be a one way trip, and you will be flown back here from Australia at the end of this trip. May I see your ticket, good sir?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cheerful tone of Chang really brought your spirits up. &quot;Sure thing,&quot; you respond, and you hand him your ticket. &quot;Where are all the other passengers?&quot; you ask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chang takes a quick look at his hologram-board, and responds, &quot;They're already all aboard. Our passenger count for this trip totals at 15, including.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only 15 passengers? Shrugging, you think, &lt;em&gt;Well, it's an older ship, after all.&lt;/em&gt; &quot;Oh, alright. I notice that this ship seems to be of an older design. Was it refurbished?&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;You noticed? This was the Seabourne Odyssey from 2009! Not very many people know that. However, all we really did was change the name and give it a paint job. Oh, we added some updates to the balconies. There are now separators spaced on it to provide each room with it's own personal balcony. They're still just as wide as before, great for a breath of fresh air - in fact, in our larger rooms near the front, the balconies have been slightly extended to allow a bit more walking room. We updated the engine as well - the ship now runs nuclear, and everything is AI operated. As a matter of fact, we only have access to the areas that were previously available to the public. Even I can't get into the engine room if I wanted to, so believe me, it's very safe. Luckily for us, the ship hasn't been changed enough for us to create a new deck plan. Here's a copy for you to reference while you are here.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nQhZj.jpg&quot; alt=&quot;Floorplan of the Cruise Ship&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(To get a better view of the floorplan, drag the picture into a new tab - it's a lot bigger than you think it is)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chang smiles, and leads you inside. &quot;I will take you to the other passengers - right now, they're getting to know each other on Deck 7, at the Seabourn Square. I suggest you do the same, you'll be with these people for the next two weeks. The ship will leave in approximately 1 hour.&quot;  &quot;And which room do I have?&quot; &quot;You're on Deck 8, room 830. I'll take your luggage up. Have a nice day, sir.&quot; With that, you have reached Seabourn Square, and Chang heads off to put away your belongings. Turning towards the square, you head in, looking at everyone out of habit.  Here's what you notice;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;There is a married couple (ring seen on fingers) talking to another male, mid 30s, well built, groomed, and good looking&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are 3 30-40 year old males at the coffee bar, in leather jackets, bald, beer gut; you instantly can tell that they're part of a biker gang&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are 2 25-30 year old males at one of the tables, socializing with someone who seems to be in his 50s.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There is another couple, very rich (evident of the bling they're wearing, and the rediculously massive diamond ring on the wife's finger). They seem to be shopping at the Seabourn Shop with a girl in her 20s. That girl is dressed nicely, but not expensively enough to associate her with the couple.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There's a man sleeping in one of the chairs, and a teenager sleeping in a different chair. The teenager is very well dressed, and his shades seem very expensive. The man is well groomed, and well built - he seems to be a very fit man.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Not wanting to wake up the sleeping passengers, you head towards the males at the coffee bar, and take a seat beside them. After chatting with them a bit, you confirm that they indeed are part of a biker gang. Their names are Jack, Jerome, and Julian. Talking with them, they seem to be heavy gamblers, but very kind people. They tell you that the only reason they're in a biker gang is because it helps to deter potential trouble from the streets - &quot;muggers are less likely to attack if you're part of a gang,&quot; they say.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that you've made friends with them, you head off to the next group. You notice that the rich couple in the shop is still shopping, and the husband is carrying more and more stuff. It's almost as if he's buying anything his wife asks for. You move to the tables, and initiate a conversation with the three males there. Talking a bit with the older male reveals that he is a mathematics professor who goes by Dr. Brofonomo at the I.I.S (International institute of Schooling).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;What about the two of you?&quot; you ask. &quot;What are your names, and what do you do?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Well, the department calls us the WTF partners. That's cus we grew up in the streets, and we don't remember our real names anymore. My name is Double You. My partner here is Tee Eff. See, together, we make DoubleYouTeeEff!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Ah, but what do you do?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;If we said we were cops, would you believe it?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I don't know, I'd have to see some sort of proof...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;How about our pieces?&quot; they said rather loudly. They pull back their blazers and you see that each of them is equipped with the standard issue P-139 plasma pistol. It's a superb weapon, and has an effective range of 150 meters. Police versions come equipped with a thermal scope, and the weapon is strong enough to blast through 3 or 4 layers of wall before dissipating. It's hard enough to find one of them in the black market, let alone two, so it's definitive proof that they're cops. You chat a bit more and discover that all three of them are really big gamblers, and judging by the way the cops act, they seem prone to violence (And honestly, unfit to be cops). You disregard that, however, and move away from this group, determined to make the most out of your night.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A glance over at the shop confirms your earlier suspicions. The wife is definitely just picking out whatever she wants, and the husband is buying and carrying everything. Feeling a bit of pity, you walk over and decide to help the poor man out. &quot;Hi there, that looks heavy. Can I help you?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As he hands you some bags, he says &quot;Thank you good sir. My name is George, how are you doing this fine day? Excited for this trip? I sure am!&quot; Unable to contain himself, he whispers to you,&quot;I have a fetish for neglect play, and my wife promised that she wou-&quot; His wife interrupts him. She isn't wearing a lot of makeup, and it looks like she's in her mid 20's. She points to an expensive antique necklace, priced at about 5 times your yearly salary. Without a second thought, the husband whips out his wallet. You instantly notice that it has a diamond card in it, which is only available to the richest of the rich. Proceeding to pay with it, you glance around, and notice that the wife is in a very deep discussion with the other girl. You ask for her name (The girl), to which you receive &quot;Carol&quot;. A bit of chatting reveals that Carol's family all recently passed away in an accident. She went on this trip and brought her families ashes with her so that they could have some &quot;Quality Family time, y'know?&quot; She's a lot more cheery than most people who've just lost all the family they had left... A little too cheery. You reach out to shake her hand, but she gives you a look of disgust and turns back to the wife. Reluctant to interact with her anymore than necessary, you return to conversation with George. He points out his son, Max, who is the younger person sleeping in the chair that you noticed earlier, and mentions that his wife is named Giselle. After a bit more small talk, you leave the group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking in a deep breath, you prepare to move to the next group. Socializing isn't easy, but the night is almost over. Looking around again, you notice that the fit man who was sleeping earlier is now with the couple that was talking to the young man at the tables. Those are the only people you still don't know, so you move over and join the conversation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the results of your talk with them:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The fit man happens to be an exceptional gymnast from Africa (Olympic level), named !Peleh. He seems to be a very smart man, albiet his eyes keep wandering back towards the same spot - Giselle's left hand.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The younger guy, Kevin, is a seasoned playboy/pickup artist. The way he talks and walks and the subtle hints he keeps dropping towards the wife in the couple makes it clear to you.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The husband, Luke, isn't very bright. It's clear to the outsiders that Kevin is hitting on his wife, Leia, but he doesn't seem to notice at all. Leia is clearly miffed that Luke isn't noticing, and probably hasn't noticed all night.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As the night comes to a close, you make a mental note of each person(s) room keys.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;You: Deck 8, 830&#xA;Luke and Leia: Deck 7, 724&#xA;Kevin: Deck 7, 726&#xA;George, Giselle: Deck 7, 700&#xA;Max: Deck 7, 701&#xA;Carol: Deck 7, 702&#xA;Jack: Deck 10, 1016&#xA;Jerome: Deck 10, 1018&#xA;Julian: Deck 10, 1020&#xA;Double You: Deck 5, 536&#xA;Tee Eff: Deck 5, 534&#xA;!Peleh: Deck 8, 800&#xA;Dr. Brofonomo: Deck 5, 533&#xA;Chang: Deck 10, 1025&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Headed off to your room, you decide to get a good nights sleep.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the next 3 days of the trip, nothing really happens. In fact it's quite relaxing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's now the 4th day of the trip. You decide to head down and see what everyone is up to, but when you get to Deck 5, you hear yelling and arguing from the direction of the Casino. Soon, you realize that Tee Eff and Double You have been gambling with Dr. Brofonomo as well as Jack, Jerome, and Julian. The two cops are accusing Dr. Brofonomo of counting cards, but Dr. Brofonomo denies it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;How else are you winning almost all the time?&quot; demands Double You, &quot;You MUST be cheating!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I insist,&quot; Dr. Brofonomo says, &quot;that I am NOT cheating. Just because I can make use of probability more effectively than you two dimwits doesn't mean that I'm cheating.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The argument goes on, and you don't want to be a part of it, so you quickly leave the scene. Stepping outside, you see Kevin and Leia in the pool. They seem to be having a good time. You turn back, reluctant to disturb them. As you head up to Deck 7, you spot Chang relaxing with Giselle and Carol at the coffee bar. Engaging in a bit of idle chit chat, they tell you that George and Max are sleeping in today, and are still in their rooms. Giselle seems to be getting really close to Carol, but the two of them don't seem to mind sharing personal space. As you keep watch over the conversation, you notice that Carol also seems to display disgust towards Chang. Since it's starting to get a bit uncomfortable, you and Chang leave after a little bit, and the two of you decide to take a walk on deck seven.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you pass by Max's room, you notice that Max's door is slightly ajar. The two of you look inside, and notice that he's passed out, dead drunk with his face in the toilet, and it reeks of vomit. Reluctant to stay, the two of you quickly leave the scene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;It's getting a bit late now,&quot; said Chang. You look at your watch, and notice that it's already 7pm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yea, I think I'm going to hit the hay for the night&quot; you respond. &quot;It was really nice talking with you!&quot; With that, the two of you separate ways and you go back to sleep.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A shriek at 11am shocks everyone on the boat. It seems to have originated from deck 5, and you rush down to take a look. Upon entry of the scene, you notice that there are two dead bodies in the hallway - Double You and Tee Eff. Chang is crouched over, shivering in complete shock against the wall. As the other people start arriving, looks of fear and murmurs start going around. Deciding to take charge, you quickly examine the bodies and give an announcement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;To the best of my knowledge, the two of them were killed together, by the same shots from their own guns - to which I can't seem to find. They're both gone... &quot; Livor Mortis has not started to set in yet, but Rigor Mortis has completely set in. The bodies are disfigured from multiple plasma shots, and their hands are tied together by their own belts. There's are two shattered cups on the floor, and the liquid spilling out of it seems to be red wine, but you catch a hint of black market tranquilizer - a specific type you have dealt with before. To the best of your knowledge, you remember that the drug is most effective when delivered through ingestion, and it comes in a liquid form. It's strong, and takes effect in seconds. It's also ludicrously hard to get, unless you have connections, or you're rich. You quickly look at who has assembled in the hallway, and you notice that the math professor is missing. &quot;Where's Dr.Brofonomo??&quot; you ask. Nobody knows, so you quickly open his door to find him dead in his room as well - hole in the chest by a well placed plasma shot. As you examine the holey corpse, you notice that both Rigor Mortis and Livor Mortis have set in completely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The people that are gathered are starting to feel more and more disturbed. Who could have committed these ghastly murders? &quot;Is everyone here?&quot; you ask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the middle of the crowd, Jack exclaims, &quot;George and Max aren't here, and Luke is missing as well!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Immediately, everyone rushes to deck 7, where George and Luke's rooms are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Luke's room is closest, so you quickly try the door. It's locked, and you kick it down when Chang informs you that only the person with a key can unlock hallway door. You find that he's dead, but his body is still warm - he must have only died recently. Luke died of suffocation. You notice a bottle of sleeping pills on the nightstand, labeled for Leia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next room you rush into is Max's - the door is still in the slightly open, but it seems to have been moved a bit. You find Max dead in the bathroom, his belt around his hands, the remains of his head still in the toilet. Death by plasma shot to the back of the head, and by the looks of it, several shots. You find that Rigor Mortis has only started to set in for him, but no Livor Mortis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Leaving the scene, fearing the worst, you head out of Max's room and head to Georges room, only to find that the door is locked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;George has the only key!&quot; exclaims Giselle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without a second thought, you kick down the door. You find him dead on the bed, a victim of plasma shots as well. His face is disfigured, the remains of a blindfold on the pillow, and you see excessive damage to his torso area, almost as if his murderer was committed with &lt;strong&gt;a lot&lt;/strong&gt; of hate. You also find that his hands have been bound together with his belt. The room key is on the nightstand. A quick scan with your detective watch shows theat there are only one set of finger prints on the card - Georges. This only gets weirder by the minute. At the sight of her dead husband, Giselle breaks down and starts crying. You've never seen anyone as emotionally unstable, and Carol starts to comfort her, patting her back, reassuring her and whatnot. It seems that the situation has been handled here, until you realize that something is off about the room. It doesn't nearly feel as rich as it should, considering the monetary status of the family. You ask Giselle to take a look around, and she mentions that some of the jewelry that she had brought with her was missing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I had put out the jewelry on the cabinet and nightstand when I got here, but it was gone a couple days ago - I thought George packed them away, but now I can't find them anywhere!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you turn back to take a last look at George, you see that he's got a &quot;rocket in his pocket&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the situation settles down, you start questioning the events and the people around you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You start with Giselle, since she's clearly unstable, and you thought it'd be a good idea to get it over with quickly. Boy was it ever a bad idea. She's so unstable that you can't even get a properly pronounced word out of her, so you move on to Carol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Where were you all night?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;In my room, alone. You can check the playlist on my TV. It's been playing since last night and still haven't stopped yet.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You move onto Leia. &quot;Where were you?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I was sleeping last night, and then this morning I was with Kevin since 8 o'clock - he treated me to breakfast, what a gentleman.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kevin seemed slightly confused for a moment, but then he says, &quot;Yea, we had breakfast together. When the scream happened, we both rushed over.&quot; He looks like he just woke up from a really nice long sleep.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Can you tell me why you have a bottle of sleeping pills, Leia?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I don't sleep well on boats, so I had them prescribed to me on this trip.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You look at Jack, Jerome and Julian, and before you can say a word they say &quot;It wasn't us, we were playing cards last night in our room, we just got up...&quot; They 3 of them yawn in sequence, and you notice that they're unshaven and they have bed hair.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You turn to !Peleh, and ask where he had been. &quot;Well, I was asleep alone in my room last night. I had an early start this morning, and Chang and I went for a morning run around the boat.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chang nods, agreeing with him. Chang confirms that after he left you he went straight to his room to try and sleep, but he was kept up by the loud card players across the hall. He mentions that he had woken up early this morning in order to run with !Peleh, which tells you that !Peleh wasn't lying about the run.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You look around, and try to make some sense of the situation. 6 murders have been committed on this trip. It can't possibly get any worse. That is, until you find out that after scouring the entire ship you can't find the two missing plasma guns. You mutter to yourself, &quot;They must've been thrown into the ocean... we're on a damn boat, after all, the perfect way to get rid of a weapon...&quot;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;The question: Can you figure out who committed the murders, how, and why?&lt;/h1&gt;&#xA;&#xA;&lt;h2&gt;Bonus: Who stole the jewelry, and how?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Note: Everything required to solve this completely is provided. Small hints may be provided after enough time has passed, or enough tries have been exhausted. My comments on your answers may be very brief, since saying too much may give away the story. First person to get the fully correct answer will get acceptance. Person to get the bonus correct will have a bounty awarded, if I get at least 10 up votes (so I can award 50 bounty).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note 2: You need to explain the how behind each death, and the motive, or I won't tell you if your answer is correct. This is to prevent blind guesses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT FOR CLARITY:&lt;/strong&gt; The hallway door locks are similar to hotel room locks - When it closes, it automatically locks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hints below&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; George was killed near the same time as max; +/- 10 minutes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is more than one culprit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The belt on the victims is a murder signature - but think about what it (the action of tying someone up) represents as well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 4&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is no justice relationship to the belt.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8999" LastEditorUserId="5373" LastEditDate="2019-04-13T10:43:31.947" LastActivityDate="2019-04-13T10:43:31.947" Title="Sabotage at Sea - Cursed Cruise liner?" Tags="|logical-deduction|lateral-thinking|story|mystery|" AnswerCount="5" CommentCount="19" ContentLicense="CC BY-SA 3.0" />
  <row Id="12733" PostTypeId="2" ParentId="12730" CreationDate="2015-04-25T23:53:34.617" Score="14" Body="&lt;p&gt;Answer to part 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $G(100)$ cannot be tiled.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Divide an infinite checkerboard into $2\times 2$ blocks, then color each block alternately white and black, as shown:&#xA;$$&#xA;\begin{array}{ccccc|ccccc}&#xA;&amp;amp;&amp;amp;\vdots&amp;amp;&amp;amp;\vdots&amp;amp;&amp;amp;\vdots&amp;amp;&amp;amp;\vdots&amp;amp;&amp;amp;\\&#xA;\cdots &amp;amp; B &amp;amp; B &amp;amp; W &amp;amp; W &amp;amp; B &amp;amp; B &amp;amp; W &amp;amp; W &amp;amp;\cdots\\&#xA;\cdots &amp;amp; B &amp;amp; B &amp;amp; W &amp;amp; W &amp;amp; B &amp;amp; B &amp;amp; W &amp;amp; W &amp;amp;\cdots\\&#xA;\hline&#xA;\cdots &amp;amp; W &amp;amp; W &amp;amp; B &amp;amp; B &amp;amp; W &amp;amp; W &amp;amp; B &amp;amp; B &amp;amp;\cdots\\&#xA;\cdots  &amp;amp; W &amp;amp; W &amp;amp; B &amp;amp; B &amp;amp; W &amp;amp; W &amp;amp; B &amp;amp; B &amp;amp;\cdots\\&#xA;&amp;amp;&amp;amp;\vdots&amp;amp;&amp;amp;\vdots&amp;amp;&amp;amp;\vdots&amp;amp;&amp;amp;\vdots&amp;amp;\\&#xA;\end{array}&#xA;$$&#xA;Then, place the $G(100)$ array on top of this so that its axes of symmetry are the two lines (one horizontal, the other vertical) above, and let $G(100)$ be colored to match the the board below it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No matter how a tile is placed, it will cover either 3 whites and a black, or 3 blacks and a white. This means that whenever an odd number of tiles are placed, the area they cover will be unbalanced between black and white. But covering $G(100)$ entails placing $1275$ tiles, and $G(100)$ itself is white/black balanced, so it cannot be covered. This proof also shows that $G(2n)$ is not tileable whenever $n\equiv 1$ or $2$ (mod $4$), since there are $1+2+\dots+n$ tiles to be placed, which is odd whenever $n=4k+1$ or $4k+2$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-04-25T23:53:34.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12734" PostTypeId="2" ParentId="12730" CreationDate="2015-04-26T00:00:25.193" Score="8" Body="&lt;p&gt;We cannot tile the given grid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is an example coloring of $G(12)$. Notice that a rotation of 180° about the central green dot swaps the colors, thus we have an equal amount of red and white tiles, and we can extend this coloring to any $G(2n)$, in particular, $G(100)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JWKu4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider the S and Z tetrominoes separately. Notice that no matter where an S tetromino is placed, it covers either 4-0, 0-4, or 2-2 of white-red. Thus when we place an S tetromino we do not change the difference between the amounts of red and white tiles modulo 4. Similarly any Z tetromino covers 3-1 or 1-3, thus adding 2 to the difference modulo 4. To end with 0 red and 0 white tiles (ie to cover the entire board) we must have the difference be 0 mod 4, and thus we must have an even number of Z tetrominoes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This argument is completely symmetrical; there is no real difference between S and Z tetrominoes. If we can tile the board using an odd number of S tetrominoes, simply mirror the board turning all S into Z and vice versa and we have tiled the board using an odd number of Z tetrominoes, which we know is impossible. Thus we have an even number of both S and Z tetrominoes, and an even number of tetrominoes overall. This means that the area of the board must be divisible by 8. But in the case of $G(100)$, it's not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The size of $G(2N)$ is $4(1 + \dots + N) = 4\frac{N(N+1)}{2} = 2N(N+1)$. If 8 divides this then $8 | 2N(N+1) \iff 4|N(N+1) \iff 4|N$ or $4|N+1$.&lt;/p&gt;&#xA;" OwnerUserId="11100" LastActivityDate="2015-04-26T00:00:25.193" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12735" PostTypeId="2" ParentId="12694" CreationDate="2015-04-26T04:42:12.660" Score="0" Body="&lt;p&gt;My Answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stem Cells&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My reasoning is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you use me, you only ever see what I have become instead of what I really am.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stem cells are considered important due to their potential for growth (they have the possibility of becoming any cell)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be used to do powerful things, but I can be very fragile at my core.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Could cure disease but just single cells&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Through time, I have changed lives!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Adult stem cells are used to treat leukaemia and heart disease&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately some choose to hate me, and others choose to embrace me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stem cells are a controversial topic - religion, politics, science, and etc&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am used around the world in many different ways.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Several advances in various countries&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My family is huge, and yet still getting bigger.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; population growth or increased availability of stem cells&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is important not to forget about my elders, as they still needed (by some).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the main controversy is due to fetal stem cells but, as stated earlier, adult stem cells are used to treat various conditions&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can look at me, but not touch:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The point of stem cells is that when introduced to other cells they grow as those cells (so when touched by other cells they stop being stem cells)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are not as old as World War I&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Though stem cells are as old as humans, at least, the first application wasn't developed until a 1968 bone marrow transplant&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11239" LastActivityDate="2015-04-26T04:42:12.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12736" PostTypeId="1" AcceptedAnswerId="12738" CreationDate="2015-04-26T11:16:22.003" Score="30" ViewCount="5427" Body="&lt;p&gt;It's early in the wee hours of the morning when a lonely farmer loads up his truck with his goods to sell. He lives far from the city and if he wants to ensure a good spot at the bazaar, he will need to leave with haste. The road is long and dull, the farmer sees but a single car driving in the other direction in the early morning mist during the first half an hour. After an hour has passed, the farmer looks out the window and sees the house of an old friend of his and notes to himself, almost as if to calm his own nerves: &lt;em&gt;&quot;I've already covered more than half of the way to the city. Surely I will make good sales today.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The day is hot and winding as the farmer calls people out to buy his goods. At long last when the sun has set, the farmer packs what little remains of his cargo and starts to travel back home. Lighthearted and in good mood, he can barely notice the time passing until he sees his friend's house again, saying to himself, &lt;em&gt;&quot;Oh my, I've already covered more than half the way! Surely I can get home early today.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The farmer drove with the same speed both in the morning and in the evening. How is it possible that he had passed more than half the way on both times when reaching his friend's house?&lt;/p&gt;&#xA;" OwnerUserId="340" LastEditorDisplayName="user20" LastEditDate="2015-04-26T19:50:20.567" LastActivityDate="2015-04-27T11:19:06.747" Title="A long, lonely drive" Tags="|logical-deduction|story|situation|" AnswerCount="8" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12737" PostTypeId="2" ParentId="12736" CreationDate="2015-04-26T11:24:47.387" Score="24" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/muj9w.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's because the farmer drove along different roads. The road was two-way for the first half an hour, then became one-way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the picture, the red color represents the morning road, the blue color shows the evening road. The green box is his friend's house.&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-04-26T11:42:53.267" LastActivityDate="2015-04-26T11:42:53.267" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12738" PostTypeId="2" ParentId="12736" CreationDate="2015-04-26T11:31:39.917" Score="38" Body="&lt;p&gt;On both occasions, when the farmer sees his friend's house he is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; looking backwards.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a short and simple solution and seems to match everything stated in the puzzle.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-04-26T11:31:39.917" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12741" PostTypeId="2" ParentId="12736" CreationDate="2015-04-26T13:45:14.643" Score="10" Body="&lt;p&gt;The most natural answer is this one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He started further away than his house at the start of the day. Suppose he lives on a large farm, and his goods are stored at the far end of the farm, whereas the house is at the end closest to the city. Now, if the friend's house is just a little short of half way between his house and the city (as viewed from his house), then starting from his storage, his friend's house is a little over half way to the city. On the way home, his friend's house is a little over half way to his house.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For a visual sense of what's happening...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; S---H------------------------F--------------------------C&lt;br&gt;&#xA; Distance from Storage to Friend's house: 28&lt;br&gt;&#xA; Distance from Friend's house to City (or back): 26&lt;br&gt;&#xA; Distance from Friend's house to Home: 24&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9289" LastEditorUserId="9289" LastEditDate="2015-04-26T13:51:16.483" LastActivityDate="2015-04-26T13:51:16.483" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12743" PostTypeId="1" AcceptedAnswerId="33710" CreationDate="2015-04-26T14:32:51.077" Score="14" ViewCount="8367" Body="&lt;p&gt;This puzzle is said to be from a Russian test and has an obvious but not fully fitting answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8CHhn.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;  The task is to find correct picture to replace question mark.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &quot;obvious&quot; pattern is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; that a small &quot;snake&quot; or 2 triangles goes down from top to bottom, from left to right. Then the answer is A.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But of course in this case the 5th picture must still have a triangle in the middle. That is one reason to dislike this answer; another would be that it doesn't explain why only 2 triangles move and the 3rd stays still.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't solve it. Maybe you can?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. Word-by-word translation of the Russian text from the picture:&lt;br&gt;&#xA;&lt;strong&gt;Question&lt;/strong&gt;&lt;br&gt;&#xA;What from the following replaces question mark in the given sequence?&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="69582" LastEditDate="2020-11-28T17:37:53.617" LastActivityDate="2020-11-29T04:19:27.660" Title="IQ-test-type question - sequence of large triangles with black and white small triangles in them" Tags="|pattern|visual|sequence|" AnswerCount="7" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="12744" PostTypeId="2" ParentId="12743" CreationDate="2015-04-26T15:47:13.930" Score="5" Body="&lt;p&gt;&lt;strong&gt;Not (yet) a solution&lt;/strong&gt;, but I made some colorings which might help with finding a pattern and numbers to help explain the pattern to others:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;       &lt;img src=&quot;https://i.stack.imgur.com/F8l0G.png&quot; alt=&quot;(colored triangles)&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10070" LastEditorUserId="46903" LastEditDate="2018-08-28T22:49:38.327" LastActivityDate="2018-08-28T22:49:38.327" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="12745" PostTypeId="2" ParentId="12736" CreationDate="2015-04-26T15:53:52.537" Score="18" Body="&lt;p&gt;Because the bazaar was not in the city he referred to the first time he passed his friend's house. The bazaar is beyond the city. You want us to assume that the bazaar is in the city, but you never explicitly say it is.&lt;/p&gt;&#xA;" OwnerUserId="11805" LastActivityDate="2015-04-26T15:53:52.537" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12746" PostTypeId="5" CreationDate="2015-04-26T16:03:31.547" Score="0" Body="&lt;p&gt;For puzzles that have appeared in IQ tests.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IQ tests are generally used to estimate the intelligence of a person. Often they consist of a set of elements with common properties and the solver is to deduce a missing element in the pattern, either numbers, words or patterns, but can be anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reference to the IQ test the question appeared in is preferred. &lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="11497" LastEditDate="2016-06-27T13:27:15.317" LastActivityDate="2016-06-27T13:27:15.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12747" PostTypeId="4" CreationDate="2015-04-26T16:03:31.547" Score="0" Body="For puzzles that have appeared in IQ tests. PLEASE DO NOT USE THIS TAG, it is in the process of being deprecated." OwnerUserId="10281" LastEditorUserId="19114" LastEditDate="2017-04-13T00:34:48.270" LastActivityDate="2017-04-13T00:34:48.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12748" PostTypeId="1" AcceptedAnswerId="12752" CreationDate="2015-04-26T16:53:13.230" Score="12" ViewCount="583" Body="&lt;p&gt;This is a new idea I just came up with. I'm going to tell you exactly how I constructed the following ciphertext, and your task is to extract the plaintext. If you &lt;em&gt;really&lt;/em&gt; want a challenge, you can try attacking the ciphertext just as a string of cryptograms without reading the spoilertag below, but I don't think that will be doable!&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Ciphertext&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;npjojoosdbtdppcvidztixnuuslbnzleug&lt;br&gt;&#xA;  npjbalduouffllvsatdsmxnsxzcabqfs&lt;br&gt;&#xA;  uvibqpqlagisedeqaughqdzrxabtkiuglbbm&lt;br&gt;&#xA;  ocarfeaxmbmlixeyvcckmsyrxvtqsshiuj&lt;br&gt;&#xA;  ocagmpyayinymbbywvbzfjzpghtimdzxafgmhtf&lt;br&gt;&#xA;  tozfityzjuqwcxtfqrkaynxevnlwzdzmg&lt;br&gt;&#xA;  toziowbkdnqsmqnhsxicqqfammbwlpbctvrpbhvx&lt;br&gt;&#xA;  tozilwxmelxajawrlkuqkyuvypnwnpklyphim&lt;br&gt;&#xA;  miarixjhmglycdswygcddzpchotqjaxa&lt;br&gt;&#xA;  inxhrxjgdpkpvxszhtuzmrghsfgrjnpff&lt;br&gt;&#xA;  buezvbixobbganbagtmabfjrwhalhodiw&lt;br&gt;&#xA;  otzjtnpyzpghazuaxsqugtlgsyhlk&lt;br&gt;&#xA;  vmshvlnyeihqdnaojosqyeinquqyfhyifrmpfgz&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Methodology&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Here's what I did to create the ciphertext.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I started with 13 lines of poetry, and applied the Vigenere cipher to each line using the next line as the key (cycling round so that the 1st line is the key for the 13th line).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In case this is unclear, here's a simple example of this encryption method being applied. Plaintext:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Cat&lt;br&gt;Rabbit&lt;br&gt;Dog&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ciphertext:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;ubv&lt;br&gt;vpifxa&lt;br&gt;gpa&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your task is to devise a decryption method to find the plaintext given the above ciphertext. You may use Vigenere software such as &lt;a href=&quot;http://sharkysoft.com/vigenere/&quot;&gt;this&lt;/a&gt;, but brute-forcing by e.g. writing a program to test all the possibilities isn't allowed.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Have fun!&lt;/h2&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-04-26T18:43:33.130" LastActivityDate="2015-04-26T21:53:42.367" Title="Cyclical Vigenere poem!" Tags="|cipher|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12750" PostTypeId="1" CreationDate="2015-04-26T17:33:13.370" Score="3" ViewCount="419" Body="&lt;p&gt;A friend dropped this code on me, and gave me no hints besides that it should be a simple enough code to crack, but no luck on my end. (It might be fantasy based was all I got, if the words come out strange)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4.1.2.11, 5.1.17.1, 8.1.28.3 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;26.2.2.3, 154.2.32.7 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;378.1.46.6, 4.1.3.8, 386.1.7.2&#xA;4.1.2.5, 45.1.1.6 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;7.9.1 24.1.6.7 4.1.3.8 25.2.2.3&lt;/p&gt;&#xA;" OwnerUserId="11807" LastEditorUserId="5373" LastEditDate="2015-04-27T08:21:19.710" LastActivityDate="2015-04-27T08:21:19.710" Title="Coded Message from M" Tags="|cipher|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12751" PostTypeId="2" ParentId="12736" CreationDate="2015-04-26T17:39:22.507" Score="0" Body="&lt;p&gt;Because his friend lives in a van/car, and has moved during the day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps he came from the city to cheat on the farmer with his wife, while he was away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So they basically met at different spots each time. First time closer to the city, second time closer to the farm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Though this doesn't tell how the farmer knows it's more than halfway thorugh)&lt;/p&gt;&#xA;" OwnerUserId="9937" LastEditorUserId="5373" LastEditDate="2015-04-26T18:52:26.223" LastActivityDate="2015-04-26T18:52:26.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12752" PostTypeId="2" ParentId="12748" CreationDate="2015-04-26T18:06:32.100" Score="20" Body="&lt;p&gt;There are an odd number of lines. If every other line is added and the skipped lines are subtracted, the result will be a line added to itself. There are then two possibilities for each character in the original line.&lt;/p&gt;&#xA;&lt;p&gt;The possibilities for line 4 are:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;aadleaiefghejbeldgbdaekaeffaa&#xA;nnqyrnvrsturworyqtoqnrxnrssnn&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;&amp;quot;andleavestheworldtodarkness&amp;quot; can be read.&#xA;Now, with part of a line reconstructed, it can be used as a key for the preceding and following lines, so we can get the beginnings of all of the lines:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;thecurfewtollstheknellofparti&#xA;thelowingherdwindslowlyoerthe&#xA;theploughmanhomewardplodshisw&#xA;andleavestheworldtodarknessan&#xA;nowfadestheglimmringlandscape&#xA;andalltheairasolemnstillnessh&#xA;savewherethebeetlewheelshisdr&#xA;anddrowsytinklingslullthedist&#xA;savethatfromyonderivymantledt&#xA;themopingowldoestothemooncomp&#xA;ofsuchaswandringnearhersecret&#xA;molestherancientsolitaryreign&#xA;beneaththoseruggedelmsthatyew&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Line 12 is now complete, and the last characters of line 11 can be decrypted by wrapping it around. The full poem can be recovered (spacing and punctuation added):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The curfew tolls the knell of parting day&lt;/p&gt;&#xA;&lt;p&gt;The lowing herd winds slowly o'er the lea&lt;/p&gt;&#xA;&lt;p&gt;The ploughman homeward plods his weary way&lt;/p&gt;&#xA;&lt;p&gt;And leaves the world to darkness and to me&lt;/p&gt;&#xA;&lt;p&gt;Now fades the glimm'ring landscape on the sight&lt;/p&gt;&#xA;&lt;p&gt;And all the air a solemn stillness holds&lt;/p&gt;&#xA;&lt;p&gt;Save where the beetle wheels his droning flight&lt;/p&gt;&#xA;&lt;p&gt;And drowsy tinklings lull the distant folds&lt;/p&gt;&#xA;&lt;p&gt;Save that from yonder ivy mantled tow'r&lt;/p&gt;&#xA;&lt;p&gt;The moping owl does to the moon complain&lt;/p&gt;&#xA;&lt;p&gt;Of such as wand'ring near her secret bow'r&lt;/p&gt;&#xA;&lt;p&gt;Molest her ancient solitary reign&lt;/p&gt;&#xA;&lt;p&gt;Beneath those rugged elms, that yew-tree's shade&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;These are the first 13 lines of &amp;quot;Elegy Written in a Country Churchyard&amp;quot; by Thomas Gray.&lt;/p&gt;&#xA;" OwnerUserId="11810" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-26T21:53:42.367" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12753" PostTypeId="2" ParentId="10043" CreationDate="2015-04-26T18:21:15.297" Score="0" Body="&lt;p&gt;If I may... Note this is not the same as the other &quot;command&quot; based answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question:&#xA;Let's call these doors 'yes', 'no', and 'maybe'. If you wanted to say the name of the correct door would your answer to this question be the same as the name of the correct door?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If door 'yes': his preference is to say 'yes' and that answer is the same as the correct door so he says 'yes'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If door 'maybe': He can only say 'yes' or 'no' so no matter what his answer could not be the same as the correct door. Thus he answers 'no'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If door 'no': he wants to say 'no' but then his answer would be the same as the name of the correct door, but if he says 'yes' it is no longer correct. He enters a paradox and cannot honestly answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as you give him time to answer this method should work perfectly.&lt;/p&gt;&#xA;" OwnerUserId="11808" LastActivityDate="2015-04-26T18:21:15.297" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12754" PostTypeId="2" ParentId="12748" CreationDate="2015-04-26T19:17:51.840" Score="2" Body="&lt;p&gt;&lt;strong&gt;This is not an answer (actually it contains an answer!) but an important observation about this encryption algorithm.&lt;/strong&gt; &#xA;The main answer has been &lt;a href=&quot;https://puzzling.stackexchange.com/a/12752&quot;&gt;posted&lt;/a&gt; by user 11810.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you encrypt a message using this algorithm, you lose information about the original data.&lt;br&gt;&#xA;Let's consider these examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A -&gt; B&lt;br&gt;&#xA;A -&gt; B  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;N -&gt; B&lt;br&gt;&#xA;N -&gt; B   &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;M -&gt; B&lt;br&gt;&#xA;O -&gt; B&lt;br&gt;&#xA;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, when you try to decrypt BB you obtain 26 different strings! Which one is the original one? You can't know!&lt;br&gt;&#xA;Specifically, &lt;strong&gt;if you have an even number of lines to decrypt, there are exactly $26$ different valid &quot;original&quot; messages.&lt;/strong&gt;&lt;br&gt;&#xA;On the contrary, &lt;strong&gt;when you have an odd number of lines to decrypt, there are only $2$ different valid &quot;original&quot; messages!&lt;/strong&gt;&lt;br&gt;&#xA;Consider these lines to encrypt:&lt;br&gt;&#xA;A -&gt; B&lt;br&gt;&#xA;A -&gt; B&lt;br&gt;&#xA;A -&gt; B  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;N -&gt; B&lt;br&gt;&#xA;N -&gt; B&lt;br&gt;&#xA;N -&gt; B  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, as you can see, decrypting BBB doesn't returns a unique solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our problem we were lucky because we had an odd number of lines ($13$), so we could identify the poetry because only one solution was meaningful. &#xA;&lt;strong&gt;Do you want to know the ugly sister of the poetry? You can easily obtain it applying ROT13 to the original one!&lt;/strong&gt; Here it is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;gurphesrjgbyyfgurxaryybscnegv&lt;br&gt;&#xA;  gurybjvatureqjvaqfybjylbregur&lt;br&gt;&#xA;  gurcybhtuznaubzrjneqcybqfuvfj&lt;br&gt;&#xA;  naqyrnirfgurjbeyqvaqnexarffna&lt;br&gt;&#xA;  abjsnqrfgurtyvzzevatynaqfpncr&lt;br&gt;&#xA;  naqnyygurnvenfbyrzafgvyyarffu&lt;br&gt;&#xA;  fnirjurergurorrgyrjurryfuvfqe&lt;br&gt;&#xA;  naqqebjflgvaxyvatfyhyygurqvfg&lt;br&gt;&#xA;  fnirgungsebzlbaqrevilznagyrqg&lt;br&gt;&#xA;  gurzbcvatbjyqbrfgbgurzbbapbzc&lt;br&gt;&#xA;  bsfhpunfjnaqevatarneurefrperg&lt;br&gt;&#xA;  zbyrfgurenapvragfbyvgnelervta&lt;br&gt;&#xA;  orarngugubfrehttrqryzfgunglrj  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Isn't it beautiful?&lt;/em&gt;&lt;/strong&gt; &lt;em&gt;If you have some time to waste, verify that this encrypts to the ciphertext given by rand al'thor!&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2015-04-26T19:25:54.200" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12755" PostTypeId="2" ParentId="12736" CreationDate="2015-04-26T19:27:11.683" Score="1" Body="&lt;p&gt;The long road is dull because it is lined by identical houses exactly 1 mile apart, that is, except for the friend's house whose neighbours are 2 miles away in each direction. So Mr Farmer doesn't recognize his friend's house until he has driven 1 mile past it.&lt;/p&gt;&#xA;" OwnerUserId="11637" LastActivityDate="2015-04-26T19:27:11.683" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12757" PostTypeId="2" ParentId="12750" CreationDate="2015-04-26T22:49:50.797" Score="6" Body="&lt;p&gt;As Len &lt;a href=&quot;https://puzzling.stackexchange.com/questions/12750/coded-message-from-m#comment39652_12750&quot;&gt;said&lt;/a&gt; in a comment, these cryptograms look like what's called a &lt;strong&gt;book cipher&lt;/strong&gt;. From &lt;a href=&quot;http://en.wikipedia.org/wiki/Book_cipher&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A &lt;strong&gt;book cipher&lt;/strong&gt; is a cipher in which the key is some aspect of a book or other piece of text. It is typically essential that both correspondents not only have the same book, but the same edition.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Traditionally book ciphers work by replacing words in the plaintext of a message with the location of words from the book being used. In this mode, book ciphers are more properly called codes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In your example, each string of four numbers will probably give you a word, or perhaps a letter. Let's ignore &quot;7.9.1&quot; for now as it only has 3 numbers rather than 4 and may be some sort of mistake.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The first number in each string can get as high as 386, so it's unlikely to be the &lt;em&gt;volume&lt;/em&gt; number of a multi-volume work and more likely to be the &lt;em&gt;page&lt;/em&gt; number.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The second number is always either 1 or 2, so it's probably the number of the relevant &lt;em&gt;column&lt;/em&gt; on the given page.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The third number can go as high as 46, so it's unlikely to be the &lt;em&gt;paragraph&lt;/em&gt; number and more likely to be the &lt;em&gt;line&lt;/em&gt; number.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The final number tells us the &lt;em&gt;word&lt;/em&gt; (possibly the &lt;em&gt;letter&lt;/em&gt;, if the message is very short) to look for in the given line.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So for instance &quot;4.1.2.11&quot; denotes the 11th word in the 2nd line of the 1st column on the 4th page of some book. It's impossible to say more than this without knowing what book your friend is referring to (some fantasy novel perhaps?). But &lt;em&gt;you&lt;/em&gt;, the OP, know which book it is. So this answer should be enough for you to solve your problem, while also being more generally useful to people trying to solve a similar puzzle.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-04-26T22:49:50.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12758" PostTypeId="2" ParentId="12610" CreationDate="2015-04-26T22:57:12.047" Score="0" Body="&lt;p&gt;I think the answer is the&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Escape Key. I think that the first line gave it away, because of the awkwardness of the &quot;black/white&quot; part. The Escape key is also only labelled as &quot;esc&quot;, and the rest of the word that is not shown is &quot;ape&quot;, which is an animal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11814" LastEditorUserId="11244" LastEditDate="2015-05-27T15:24:17.933" LastActivityDate="2015-05-27T15:24:17.933" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12759" PostTypeId="2" ParentId="12694" CreationDate="2015-04-26T23:08:29.760" Score="1" Body="&lt;p&gt;I'm thinking that it's something like &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Nuclear power. It's a little farfetched, I know, but you can't actually see the nuclear reaction happening, as described in the first line, and nuclear power is very strong, but also fragile, or unstable. Some people don't like the use of nuclear power, and some do. You can use it to make bombs, or power homes, and people are still developing nuclear power. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11814" LastEditorUserId="9951" LastEditDate="2015-04-28T16:51:04.983" LastActivityDate="2015-04-28T16:51:04.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12761" PostTypeId="1" AcceptedAnswerId="12768" CreationDate="2015-04-27T00:13:45.193" Score="15" ViewCount="4226" Body="&lt;p&gt;Often, when people ask the Monty Hall problem, they omit some important details, such that the problem becomes ambiguous, or they add more details, such that the problem changes drastically. For example, what is the answer for the following problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are participating in a show. There are three doors in front of you. There’s a car behind one of them; there’s a goat behind another; and there’s a sheep behind the third. To the best of your knowledge, every assignment of the three prizes to the three doors is equally probable. You do want the car, badly, and you don’t care about either of the animals at all. You can choose one door, and whatever is behind that door will be awarded to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, without loss of generality, you choose the leftmost door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that, to your complete surprise, the host opens the middle door, and you observe a sheep behind it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The host then offers you a chance to change your choice. Based solely on the information you have, without trying to guess the motives of the host, is it beneficial for you to change your choice?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: to make the question a little more specific, what is the ideal strategy for you, that maximizes your chances of getting the car in the worst case, no matter what the hosts motives are.&lt;/p&gt;&#xA;" OwnerUserId="11699" LastEditorUserId="11699" LastEditDate="2015-04-27T22:48:07.933" LastActivityDate="2015-04-28T16:37:39.793" Title="Monty Hall all over again?" Tags="|probability|monty-hall|" AnswerCount="11" CommentCount="26" ContentLicense="CC BY-SA 3.0" />
  <row Id="12762" PostTypeId="2" ParentId="12761" CreationDate="2015-04-27T00:17:33.927" Score="5" Body="&lt;p&gt;The question is vague, and assumptions must be made that drastically change the possible answers.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT: Analysis added below. Am I missing something?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XV4T1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-04-28T16:37:39.793" LastActivityDate="2015-04-28T16:37:39.793" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12763" PostTypeId="2" ParentId="12761" CreationDate="2015-04-27T00:23:27.637" Score="7" Body="&lt;p&gt;I will stay with the door I have. I am banking on a malicious host who takes advantage of my Monty Hall familiarity and only offers me a switch when I am pointing to the car.&lt;/p&gt;&#xA;" OwnerUserId="11154" LastActivityDate="2015-04-27T00:23:27.637" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12764" PostTypeId="2" ParentId="12736" CreationDate="2015-04-27T01:21:00.430" Score="5" Body="&lt;p&gt;This is because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Between the friend's house and his own house is a weight-limited bridge (or even a height-limited tunnel). For example, the route avoiding the bridge is 100 km from home to his friend's, and from his friend's home to the city is 80 km. However, after selling all his goods at market to make his vehicle lighter (or less tall), he can take the weight-limited bridge, cutting the distance from his friend's to home to 50 km.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This explains the market aspect better than others, IMO, but I don't know the significance of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the farmer sees but a single car driving in the other direction in the early morning mist during the first half an hour&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This seems like a clue...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: Perhaps the mist is a clue:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Due to the morning mist, the bridge is closed to traffic, regardless of vehicle weight, so he has to take a detour.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="341" LastEditorUserId="341" LastEditDate="2015-04-27T01:26:29.847" LastActivityDate="2015-04-27T01:26:29.847" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12765" PostTypeId="2" ParentId="12743" CreationDate="2015-04-27T01:23:55.007" Score="10" Body="&lt;p&gt;How about this solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There's a white field with a black triangle at the bottom. And there is a three-piece snake, consisting of a white triangle followed by two black triangles walking down this field.&#xA; &#xA;  &lt;a href=&quot;https://i.stack.imgur.com/3l5XV.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3l5XV.png&quot; alt=&quot;snake&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11699" LastEditorUserId="2136" LastEditDate="2017-04-23T17:18:26.287" LastActivityDate="2017-04-23T17:18:26.287" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12766" PostTypeId="2" ParentId="12761" CreationDate="2015-04-27T01:31:19.843" Score="4" Body="&lt;p&gt;Let's assume that the Host randomly decides whether to reveal a goat and offer a switch: if your first door has a car, he reveals with probability $p$, and if it was first a goat, then with probability $q$. This is the best we can do: we don't know whether the Host planned to open a door or not. Note that $p=q=1$ gives the original Monty Hall scenario.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $C$ be the event you originally picked the car, and $R$ be the event that he reveals. Then&#xA;$$&#xA;P(C|R)=\frac{P(R|C)P(C)}{P(R|C)P(C)+P(R|\neg C)P(\neg C)}=\frac{p\cdot \frac13}{p\cdot\frac13+q\cdot\frac23}=\frac{p}{p+2q}&#xA;$$&#xA;If you &lt;em&gt;knew&lt;/em&gt; the probabilities $p,q$, then the best strategy would be to stay the above probability is at least $\frac12$, namely, when $p\ge 2q$. But we don't know the Host's personality, so we can't deduce what the best strategy is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, there is too little information given to determine what strategy maximizes your probability of getting the car. The only thing a rational game show guest can do is figure out how to maximize this probability in the worst case. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If you always stick with the door you start with, you win $\frac13$ of the time, no matter what.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you always switch, a malicious host could conspire to make you never win (by offering the switch when you start with a car, and not offering it otherwise).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thus, the strategy that &lt;em&gt;guarantees&lt;/em&gt; you the best odds, no matter what personality the Host has, is to stick with what you initially pick.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-04-27T01:31:19.843" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12767" PostTypeId="2" ParentId="163" CreationDate="2015-04-27T01:57:40.193" Score="3" Body="&lt;p&gt;With your orientation:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;R Ui Ri U R Ui Ri U R U U Ri Fi U U F&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Breaking it down:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;R Ui Ri U R Ui Ri U R&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If you watch the piece this takes the two of them, splits them rotates them, and aligns them reorientated on the top layer.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;U U Ri Fi U U F&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This takes the two pieces, rotates them to the back, pulls the right side back in place, flips the front for the two pieces, and then rotates them in and flips it down. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q.E.D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just figured it out,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&lt;/p&gt;&#xA;" OwnerUserId="11818" LastEditorUserId="8874" LastEditDate="2016-02-18T12:40:46.157" LastActivityDate="2016-02-18T12:40:46.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12768" PostTypeId="2" ParentId="12761" CreationDate="2015-04-27T02:22:27.907" Score="13" Body="&lt;p&gt;Here's the formal expression of what we seek. Let $C$ be the event &quot;chose the door with the car&quot; and $R$ be the event &quot;revealed the door with the car&quot;. Now, assuming that the door opened is random (but not the one the player chose), we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;Pr(C) = \frac13\\&#xA;Pr(R|C) = 0\\&#xA;Pr(R|\bar C) = \frac12\\&#xA;Pr(C|\bar R) = \frac{Pr(\bar R|C)Pr(C)}{Pr(\bar R|C)Pr(C)+Pr(\bar R|\bar C)Pr(\bar C)} = \frac{1/3}{2/3}=1/2&#xA;$$&#xA;And therefore the probability of winning by switching or by staying are equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This differs from the Monty Hall problem through the value of $Pr(R|\bar C)$, which is zero for the Monty Hall problem. In that case, we have&#xA;$$&#xA;Pr(C|\bar R) = \frac{Pr(\bar R|C)Pr(C)}{Pr(\bar R|C)Pr(C)+Pr(\bar R|\bar C)Pr(\bar C)} = \frac{1/3}{1}=1/3&#xA;$$&#xA;and thus, switching doubles the chance of winning. This is not true if the host can open the door with the car, where the probability of winning is equal in either direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's tempting to think that the situation is the same (the door that is opened is one without the car) between this situation and the Monty Hall one, but the balance of probabilities leading to the situation is different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a mistake involving the confusion between counting the number of arrangements and determining the actual chance of each event. In the Monty Hall problem, if the player chooses a wrong door, which has a probability of 2/3, the host will necessarily open the other wrong door, and thus the chance of winning by switching is 1. In this problem, if the player chooses a wrong door, there's a 50% chance that the car will be opened. This halves the number of times that the player will choose the wrong door and have an opportunity to switch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result, while the probability rules around the actual situation are the same, the underlying probabilities are shifted - there's only a 1/3 chance that the player will choose a wrong door and the host will open the other wrong door, and a 1/3 chance that the player will choose the right door. In the Monty Hall problem, the probability of the former is 2/3. So even though the situation itself looks the same, the chance of being in that situation while having chosen the wrong door is lower, with the chance of each of them being 1/2 (given that the situation arose), rather than 1/3 and 2/3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the probability that the player chose the car given that the car wasn't behind the door the host revealed depends on only two parameters. If we express our equation for $Pr(C|\bar R)$ slightly differently, we have&#xA;$$&#xA;Pr(C|\bar R) = \frac{Pr(C)}{Pr(C)+\frac{1-Pr(R|C)}{1-Pr(R|\bar C)}(1-Pr(C))} = \frac{A}{A+B(1-A)}&#xA;$$&#xA;where $A=Pr(C)$ and $B=\frac{1-Pr(R|C)}{1-Pr(R|\bar C)}$ - as such, these are the only two values that are relevant. In the Monty Hall case, $A=1/3$ and $B=2$. In our case, $A=1/3$ and $B=1$. If the host is incredibly malicious, almost certainly set to reveal the car if the player doesn't choose it, then $A=1/3$ and $B\approx 0$, meaning that the player is practically guaranteed to win by staying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we were to allow the host to open other doors, then we have to handle those distinctly. As such, it is better to consider $\hat R$, which is the event &quot;the host opened one of the other two doors, and it contained an animal&quot; (or specifically &quot;the sheep&quot; - it's equivalent). For the three-door-only case, $\hat R \equiv \bar R$. So the general expression would be&#xA;$$&#xA;Pr((C|\hat R) = \frac{Pr(C)}{Pr(C)+\frac{Pr(\hat R|C)}{Pr(\hat R|\bar C)}(1-Pr(C))} = \frac{A}{A+B(1-A)}&#xA;$$&#xA;where $A=Pr(C)$ again and $B=\frac{Pr(\hat R|C)}{Pr(\hat R|\bar C)}$. In this expression, it is easily seen that adding extra doors that aren't among the three we care about doesn't change the result. With an extra door, we have $Pr(\hat R|C) = 2/3$ and $Pr(\hat R|\bar C) = 1/3$, and we again have $B=2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt; - the assumption was described above as &quot;assuming that the door opened is random (but not the one the player chose)&quot;. This is for the purposes of the calculations. The true assumption is actually that the host could have any possible motive with equal probability, and looks at the expected probability. It just happens that the two assumptions are equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;: As I noted, taking the host's motive as &quot;random&quot; is the same as randomly choosing the host's motive. It's worth explaining properly how this works, and explain an easy pitfall in determining it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $X$ is a uniformly-distributed random variable (between 0 and 1) representing the probability that the host reveals the car given that the player didn't choose the car's door. That is,&#xA;$$&#xA;Pr(R|\bar C\land X=x)=x&#xA;$$&#xA;Because $C$ and the value of $X$ are independent of each other, we can deal with these separately - that is, we can treat it as $(R|\bar C)|X=x$ or $(R|X=x)|\bar C$, and nothing will be wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here's where the pitfall comes in. It can be tempting to calculate&#xA;$$&#xA;Pr(C|\bar R\land X=x) = \frac{Pr(\bar R|C)Pr(C)}{Pr(\bar R|C)Pr(C)+Pr(\bar R|\bar C\land X=x)Pr(\bar C)} = \frac{Pr(\bar R|C)Pr(C)}{Pr(\bar R|C)Pr(C)+(1-x)Pr(\bar C)}&#xA;$$&#xA;and then integrate from there. However, there's a problem - $\bar R$ and $X$ are not independent, and thus we cannot treat $C|\bar R\land X=x$ as $(C|\bar R)|X=x$, which would be necessary to determine $C|\bar R$ by integration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, we need to calculate it through the full process. That is, we must get&#xA;$$&#xA;Pr(C|\bar R) = \frac{Pr(\bar R|C)Pr(C)}{Pr(\bar R)}&#xA;$$&#xA;and to calculate the denominator, we must integrate. That is,&#xA;$$&#xA;Pr(\bar R) = \int_0^1 Pr(\bar R|X=x) dx&#xA;$$&#xA;where&#xA;$$&#xA;Pr(\bar R|X=x) = Pr(\bar R|C)Pr(C) + Pr(\bar R|\bar C\land X=x)Pr(\bar C)\\&#xA; = Pr(\bar R|C)Pr(C) + (1-x)(1-Pr(C)) &#xA;$$&#xA;and this quickly gives us&#xA;$$&#xA;Pr(\bar R) = Pr(\bar R|C)Pr(C) + \frac12(1-Pr(C))&#xA;$$&#xA;at which point the equivalence with assuming the host chooses at random becomes clear.&lt;/p&gt;&#xA;" OwnerUserId="9289" LastEditorUserId="9289" LastEditDate="2015-04-28T10:30:43.933" LastActivityDate="2015-04-28T10:30:43.933" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12769" PostTypeId="2" ParentId="12761" CreationDate="2015-04-27T05:31:25.697" Score="4" Body="&lt;p&gt;Well in this case the game becomes more or a Poker game because you don't know if the host followed a rule or not. There are two (already well-known) cases:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The host chose the other door randomly (he didn't know what was behind that one): in this case it doesn't change anything if you change or not: each door has a 1/2 chance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The host chose on purpose a door without the car: in this case changing leads a a higher chance of getting the car: 2/3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if we assume that the host followed one of these rules, it is still better to change. In the worse case it doesn't change anything and in the best case it improves your chance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am though doing one hypothesis on the host: he ALWAYS opens another door. If he weren't doing that, it could be a meta-trick to make you change doors (opening a door and letting you change only if you picked right in the first place).&lt;/p&gt;&#xA;" OwnerUserId="11780" LastActivityDate="2015-04-27T05:31:25.697" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12770" PostTypeId="1" AcceptedAnswerId="12846" CreationDate="2015-04-27T06:27:01.217" Score="34" ViewCount="6374" Body="&lt;p&gt;You are the ruler of a medieval kingdom who loves throwing parties. The courtier who tried to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/410/how-do-you-solve-a-word-puzzle-when-you-have-no-idea-where-to-start?rq=1&quot;&gt;poison one of your wine bottles&lt;/a&gt; last time was furious to learn that you managed to identify which bottle he had poisoned out of 1,000 with just ten prisoners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This time he's a bit craftier. He's developed a &lt;strong&gt;composite poison&lt;/strong&gt; $P$: a binary liquid that's only deadly when two individually harmless components mix; this is similar to how &lt;a href=&quot;http://en.wikipedia.org/wiki/Epoxy&quot; rel=&quot;nofollow noreferrer&quot;&gt;epoxy&lt;/a&gt; works. He's sent you another crate of 1,000 wine bottles. One bottle has component $C_a$ and another one has component $C_b$. ($P = C_a + C_b$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone who drinks both components will die on the stroke of midnight on the night they drank the final component, regardless of when in the day they imbibed the liquid. Each poison component stays in the body until the second component activates, so if you drink one component one day and another component the next, you will die on midnight at the end of the second day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have two days before your next party. What is the minimum number of prisoners you need to use for testing in order to identify which two bottles are tainted, and what algorithm do you need to follow with that number of prisoners?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus&lt;/strong&gt;&lt;br&gt;&#xA;Additionally, suppose that you had a fixed limit of 20 prisoners at your disposal, what's the maximum number of bottles you could theoretically test and come to an accurate conclusion about which bottles were affected?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: I don't actually know the answer to this puzzle, but thought it would be interesting to think about how you'd represent&lt;/em&gt; that &lt;em&gt;information.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2015-05-01T09:27:03.630" Title="Find the composite poison in the wine bottles" Tags="|logical-deduction|word-problem|" AnswerCount="9" CommentCount="25" ContentLicense="CC BY-SA 3.0" />
  <row Id="12773" PostTypeId="2" ParentId="12770" CreationDate="2015-04-27T07:17:46.623" Score="5" Body="&lt;p&gt;For the first part: you need to sacrifice only one prisoner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take all combinations of 2 bottles: $\binom{1000} 2$ = 499500. So take 499500 prisoners and give each of them a drop of a different combination of two bottles. Give each prisoner a bracelet with the number of the bottles he drank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At midnight, one of the prisoners will die and we now know which two bottles were poisoned.&lt;/p&gt;&#xA;" OwnerUserId="7636" LastActivityDate="2015-04-27T07:17:46.623" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="12774" PostTypeId="2" ParentId="12732" CreationDate="2015-04-27T08:00:25.603" Score="3" Body="&lt;p&gt;Here's some partial information I've pieced out so far. Most likely this stuff is obvious, but it's on the road toward a solution.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Times of death&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The shriek happens at 11 AM. W and TF's bodies have already gone through rigor mortis but not livor mortis, so they've been dead for 6 hours, killed at around 5 AM. Same with Max.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dr. Brofonomo was dead in his room and his body had already gone though livor mortis, so he's been dead for at least 12 hours, from 11 PM last night.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Luke died the latest out of any of them - his body is still warm and rigor mortis hasn't set in yet. Perhaps only up to 30 minutes beforehand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nothing is known about George's time of death, but he does have an &lt;a href=&quot;http://en.wikipedia.org/wiki/Death_erection&quot; rel=&quot;nofollow&quot;&gt;erection&lt;/a&gt;, which indicates nothing about the time of death (other than maybe that livor mortis hasn't set in yet) but more about the circumstances of death, which were already made rather clear in the rest of the description.  (An erection from BDSM play isn't likely to stay after death, so something else must have been the case there.) We can &lt;em&gt;probably&lt;/em&gt; assume that he died at 5 AM like the rest of the people.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Strange information from the previous days&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;For some reason, !peleh can't stop staring at Giselle's left hand. I imagine it has something to do with the ring on her finger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The detective notes that George's body was killed with a lot of hate. I'm wondering if it wasn't !peleh who killed George because Giselle was married to him when !peleh was actually the one who loved her. Either that, or !peleh was the one stealing all the jewelry, as has now been revealed to be the case by Galatyr (although why, I can't imagine, since he's probably gotten a lot of money from his athletic performances already).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Alibis&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Jack, Jerome, and Julian were playing cards in 1016/1018/1020, which Chang who was in 1025 can attest to. Given that they have bed hair, they couldn't have killed anyone other than Dr. Brofonomo in the morning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carol was in 702, right next to 701 where Max was and 700 where George and Giselle were. She's given the least information out of anybody so far, and leaving a playlist on all night doesn't mean she has to be physically present in the room to listen to it. Carol's ash box could also have stored some black market tranquilizers, not to mention she seemed rather happy despite her family being dead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that Kevin appears to have just woken up even though Leia says she was with him all morning seems to indicate that Leia is lying, and that Kevin is covering up for her (although it could just be his playboy tendencies kicking in).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chang and !peleh woke up early. They don't mention how early, but it could have been at 5 AM to kill people.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Multiple murders?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;As Galatyr mentioned and I suspected without any evidence myself, these deaths may not have been the work of a single person. Luke didn't die of a laser blast, and neither Luke nor Dr. Brofonomo died at the same time as everyone else.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-04-27T16:34:13.547" LastActivityDate="2015-04-27T16:34:13.547" CommentCount="8" CommunityOwnedDate="2015-04-27T08:00:25.603" ContentLicense="CC BY-SA 3.0" />
  <row Id="12777" PostTypeId="2" ParentId="6562" CreationDate="2015-04-27T08:10:23.963" Score="1" Body="&lt;p&gt;The advantage with the &lt;a href=&quot;http://en.wikipedia.org/wiki/XOR_cipher&quot; rel=&quot;nofollow&quot;&gt;XOR cipher&lt;/a&gt; is that using the same encryption key on the cipher text, restores the plain text.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Converting your cipher text to hexadecimal gives 1902ACBE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Converting your key to hexadecimal gives 71D935157B032704.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;XOR operation of cipher and key gives plain text as 71d9351562018bba.So, in binary, the plain text will be 0111000111011001001101010001010101100010000000011000101110111010.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you want to verify the cipher text, then you may apply XOR operation between your plain text and key. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;XOR operation of plain text and key gives cipher text as 71d935157b032704.&lt;/p&gt;&#xA;" OwnerUserId="10119" LastActivityDate="2015-04-27T08:10:23.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12778" PostTypeId="2" ParentId="10043" CreationDate="2015-04-27T08:23:47.987" Score="0" Body="&lt;p&gt;The warden gives us the most important hint: &quot;Choose one door, you don't even need the question&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I just say: &quot;I have chosen: Please unlock the door to freedom&quot; Then I check all three doors, the one which is unlocked leads to freedom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question isn't needed and can be &quot;What's for dinner?&quot;&lt;/p&gt;&#xA;" OwnerUserId="1998" LastActivityDate="2015-04-27T08:23:47.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12780" PostTypeId="2" ParentId="12694" CreationDate="2015-04-27T09:12:46.993" Score="1" Body="&lt;p&gt;This is wrong, but beautifully fits most of the hints.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Democracy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you use me, you only ever see what I have become instead of what I really am.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Democracy, like every type of system of government, can't be literally applied. You can get close to it, but never have a pure democracy. The most famous not applicable government system is pure socialism, which is &lt;a href=&quot;http://en.wikipedia.org/wiki/Utopian_socialism&quot; rel=&quot;nofollow&quot;&gt;utopian &lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be used to do powerful things, but I can be very fragile at my core.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Democracy is widely known as the best type of government, and indeed it is, giving people the power. But if the core, the parliament, is rotten, with corruption and hypocrisy, everything falls apart.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Through time, I have changed lives!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well, this is obvious.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately some choose to hate me, and others choose to embrace me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dictatorships and oligarchy exists all over the world, exactly as democracy itself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My family is huge, and yet still getting bigger.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are a lot of &lt;a href=&quot;http://en.wikipedia.org/wiki/Types_of_democracy&quot; rel=&quot;nofollow&quot;&gt;types of democracy&lt;/a&gt; and more are being created every year&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is important not to forget about my elders, as they still needed (by some).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Democracy is the logical consequence of monarchy. There are still some countries with monarchy as type of government, like Belgium, Spain, England...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;HINT 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can see &lt;a href=&quot;https://encrypted-tbn3.gstatic.com/images?q=tbn:ANd9GcT4s2mKEtkVQd7Zxyzx7m5WD0uH04lffrOwGoAWSefVhJl8ahYt&quot; rel=&quot;nofollow&quot;&gt;democracy in action&lt;/a&gt;, but since its a concept you can't touch it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;HINT 2: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well, this simply ruins everything :P&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4652" LastEditorUserId="4652" LastEditDate="2015-04-27T12:59:12.283" LastActivityDate="2015-04-27T12:59:12.283" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12781" PostTypeId="1" CreationDate="2015-04-27T09:14:05.500" Score="26" ViewCount="2617" Body="&lt;p&gt;The Barberland Institute of Trichodemographic Research has published the following results:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Nobody in Barberland has more than 1,000,000 hairs.&lt;/li&gt;&#xA;&lt;li&gt;No two inhabitants of Barberland have the same number of hairs.&lt;/li&gt;&#xA;&lt;li&gt;Nobody in Barberland has the same number of hairs as the sum of two other inhabitants' hairs.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What is the maximum population of Barberland? Please, explain why.&lt;/p&gt;&#xA;" OwnerUserId="4394" LastEditorUserId="4394" LastEditDate="2015-05-06T17:11:54.017" LastActivityDate="2015-05-13T15:59:50.603" Title="The population of Barberland" Tags="|mathematics|" AnswerCount="7" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12782" PostTypeId="2" ParentId="12781" CreationDate="2015-04-27T09:31:52.420" Score="27" Body="&lt;p&gt;The maximum number of inhabitants is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 500,001 (+1 extra, see end of answer)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is achieved by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Inhabitants with each number of hairs from 500,000 to 1,000,000.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is optimal because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let $n$ be the highest number of hairs any inhabitant has. Then, out of each pair $\{k,n-k\}$, only one number of hairs can be present since they sum to $n$. There are $\lfloor n/2 \rfloor$ pairs, counting the single-value &quot;pair&quot; when $k=n/2$ for even $n$. So, this gives at most $1 + \lfloor n/2 \rfloor$ inhabitants, and $n$ is at most 1,000,000.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit: Commenters pointed out that I made a mistake:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can also add a bald person with 0 hairs, for 500,002 inhabitants total.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4269" LastEditDate="2015-05-06T18:04:55.653" LastActivityDate="2015-05-06T18:04:55.653" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12783" PostTypeId="2" ParentId="12770" CreationDate="2015-04-27T09:36:10.867" Score="10" Body="&lt;p&gt;Here's a strategy that uses 124 prisoners. It's probably possible to reuse some of the prisoners from day 1 to day 2 to reduce that number, but I want to get a simple strategy posted so other people can build on it and find cleaner ways to present it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make the numbers work out a bit nicer, we add 24 bottles of water to the crate, so we're testing 1024 bottles. Group the bottles into 32 groups of 32 bottles each, and label the bottles from 1-1 up to 32-32. Group the prisoners into 4 groups of 31, and label the prisoners from A1 to D31.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On day one, we pour 32 cocktails, each composed of wine from every bottle in a single group, and label the cocktails from 1 to 32. In group A, every prisoner drinks from all cocktails labeled up to and including their number, so for example, prisoner A12 drinks from cocktails 1 through 12. In group B, every prisoner drinks from every cocktail with a label &lt;em&gt;greater&lt;/em&gt; than their number, so for example, prisoner B12 drinks from cocktails 13 through 32.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By looking at which prisoners die, we can determine which two cocktails (or which one cocktail) contained poison. If A13 lives and A14 dies, cocktail 14 had poison. If B11 dies and B12 lives, cocktail 12 had poison. If all the A prisoners live, cocktail 32 had poison.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we find that one cocktail contained poison, we repeat this procedure with the bottles that went into that cocktail instead of with cocktails, and with two fresh sets of prisoners. If we find that two cocktails contained poison, then everyone in group C drinks the first cocktail and a single bottle that went into the second, and everyone in group D drinks the second cocktail and a single bottle that went into the first. Either way, we find the poisoned bottles.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Further work: we can reuse at least 31 prisoners, bringing the number we need down to 93. If only one cocktail contains poison, then we have 31 testers who only drank cocktails to the left or right of it, and we can reuse them on the second day. If two cocktails contain poison, then everyone who drank cocktails to the left of &quot;right poison&quot; and everyone who drank cocktails to the right of &quot;left poison&quot; survived. This is at least 32 people. Some of them are contaminated with poison, but we know whether they have left poison, right poison, or neither. We can assign anyone with left poison to the group that would have taken left poison anyway, and similarly for right poison.&lt;/p&gt;&#xA;" OwnerUserId="1549" LastEditorUserId="1549" LastEditDate="2015-04-27T09:56:26.270" LastActivityDate="2015-04-27T09:56:26.270" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12786" PostTypeId="2" ParentId="12732" CreationDate="2015-04-27T11:06:03.373" Score="5" Body="&lt;p&gt;Boy this was a long one, but I think I figured a huge part of it out.&#xA;First off, there is not a single murderer, but rather more of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start with one of the more obvious ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Luke&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Luke has not been dead for long.&#xA;It seems that Leia and Kevin spent the night together.&#xA;He caught his wife cheating and swallowed her sleeping pills to kill himself. He seems to have suffocated on his own vomit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Killed by:&lt;/em&gt; Suicide&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dr. Brofonomo&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the observation that Rigor Mortis and Livor Mortis have kicked in, it can be determined that he was the first one to die.&#xA;He was killed by a well placed single shot in the chest, performed by someone who has clearly used such a weapon before. &#xA;He was accused of cheating by WTF. As the night went on, the two policemen, raised on the streets, who seem very prone to violence and probably drunk (indicated by the red wine), had lost enough money to be angered to a point, where they followed Dr. Brofonomo to his room and shot him in order to get their money back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Killed by:&lt;/em&gt; Double You and Tee Eff&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Double You and Tee Eff&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, the real murder spree begins.&#xA;The two of them were drugged, tied and shot. &#xA;The question is, why would you tie somebody who is incapacitated, only to shoot him a moment later? &#xA;The tranquilizer is very rare, and only obtainable by someone with connections or money. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;George&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;George said he's into neglect fetish play and is excited to live out his fantasies during the trip.&#xA;The blind fold, being tied up, the inflicted wounds and his erection are indicators that he was in the middle of a BDSM-play at the time of his death.&#xA;This play might have been going on for 2 nights already, because he wasn't seen the entire day, most likely because he was tied up in his room while Giselle was up in the coffee room with Carol.&#xA;Giselle (or the Killer?) might have been wearing leather gloves, thus the missing finger prints on the key card.&#xA;He was killed by multiple plasma gun shots, like WTF and Max. A gun with such power that it has a range of 150m and can pierce multiple walls, doesn't have to be fired more than once to kill somebody. Somebody either didn't know that or used the killing to vent their own aggression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Max&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Found lying with his head in the toilet, also tied up.&#xA;It seems he hasn't moved since the Detective last saw him, passed out drunk. Again, why was he tied up only to be shot shortly after? Maybe he was drugged with the same tranquilizer?&#xA;He was also shot multiple times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who murdered them?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Double You and Tee Eff must have been the first to die out of the 4 previously mentioned, because they where the ones with the guns (which they always seemed to carry).&#xA;They were drugged, stripped of their guns and tied up.&#xA;Max had passed out, and was also tied up.&#xA;George was in the middle of his BDSM fantasy and also tied up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something must have happened to trigger the murders, as it would be illogical for them to happen directly after the victims were incapacitated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve this crime, we have to look at the motives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;!Peleh: No mentioned affiliation with anybody of the victims. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kevin and Leia: Most likely occupied with themselves&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chang: Only Crewmember mentioned on a fully automated Cruiseship from 2009 (which is, at this point 141 years old). Who in their right mind would sail on Cruise Ship that old with things like interstellar travel happening at the same time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Giselle: If she were with her husband all night, she had to notice something. Her reaction upon seeing her dead husband though seems to be authentic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carol: The big question mark. She seems to hate men (at least the detective and Chung) and shows closeness to Giselle. She's travelling with the ashes of her deceased family. She has no real alibi (because a running playlist doesn't need her presence).&#xA;Maybe George and his family were involved in the accident that cost the lives of her family? Is this an act of revenge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Gang:&#xA;They certainly have the connections to acquire the tranquilizer and are accustomed to a certain degree of crime. But are they cruel enough for murder?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The stolen jewelery&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;!Peleh was seen staring at Giselles left hand, thats where the big diamond ring is. !Peleh has the room above Giselle and George, and as an athletic olympian, should have no problem to climb down his balcony to sneek into the room and steal the jewelery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Commited by:&lt;/em&gt; !Peleh&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thats what I got so far. Maybe some small detail I have missed?&lt;/p&gt;&#xA;" OwnerUserId="11826" LastEditorUserId="11826" LastEditDate="2015-04-27T14:16:53.157" LastActivityDate="2015-04-27T14:16:53.157" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12787" PostTypeId="2" ParentId="12736" CreationDate="2015-04-27T11:19:06.747" Score="20" Body="&lt;p&gt;For completeness sake, I'll add the original answer to the puzzle as an image:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The friend's house is at the red dot, it doesn't really matter at which side of the road. The traffic in the below image is right-handed, if it was left-handed, just flip the image.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/8Rruc.png&quot; alt=&quot;Explanation image&quot;&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="340" LastActivityDate="2015-04-27T11:19:06.747" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12789" PostTypeId="2" ParentId="12761" CreationDate="2015-04-27T12:40:04.853" Score="12" Body="&lt;p&gt;Without assuming anything more? No, we have absolutely no idea whether it is beneficial to switch or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the host knows what is behind each door and is malicious, then he will try to trick us. But if we know he will try to trick us then we can use that against him. Unless he knows that we know, and so on and so on. In the end that means there is no difference from switching when we don't know more about the host.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a door with an animal is ALWAYS revealed, then we should switch, as per original monty hall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a door is revealed at random it doesn't matter if we switch, the chance is 50/50. That is we wouldn't get a chance to switch if the revealed door had happened to be the car. The chance is 50/50 both when the random reveal can be all 3 doors or when it can only be the 2 we did not pick, doesn't matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the host only opens a door when we have already picked the car, then of course we should stay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summation&lt;/strong&gt;: We need to know the rules for the host opening a door to make an informed decision.&lt;/p&gt;&#xA;" OwnerUserId="11829" LastEditorUserId="11829" LastEditDate="2015-04-28T12:56:19.043" LastActivityDate="2015-04-28T12:56:19.043" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12790" PostTypeId="2" ParentId="12694" CreationDate="2015-04-27T13:34:05.890" Score="2" Body="&lt;p&gt;Okie... Let me give it a try.&#xA;It is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Google Maps&lt;/strong&gt; or any Maps application&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you use me, you only ever see what I have become instead of what I really am.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We only see it as an application and not what it really is, A complex code.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be used to do powerful things, but I can be very fragile at my core.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We can find different locations in every corner of this world. But the code is fragile, one little mistake in code can ruin the whole application up.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Through time, I have changed lives!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Over years, google maps have seen some major changes starting from GoogleBlog to where it is now.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately some choose to hate me, and others choose to embrace me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quite not fit this, but may be it refers to some people who don't like referring to an application for navigation., instead rely on paper maps. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am used around the world in many different ways.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is being used around the world. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My family is huge, and yet still getting bigger.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Google maps now have, Google moon, Google Sky, Google Mars etc and is still growing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is important not to forget about my elders, as they still needed (by some).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It refers to paper maps which are still being used and sold through out the world.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can look at me, but not touch.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We cannot touch google maps, if we try we rather be touching a monitor or phone screen. :p&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not as old as world war 1.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Started in 2005 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10973" LastActivityDate="2015-04-27T13:34:05.890" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12791" PostTypeId="1" AcceptedAnswerId="12792" CreationDate="2015-04-27T13:58:25.900" Score="6" ViewCount="682" Body="&lt;p&gt;I recently had the clue &quot;To do with a bone, soft, Clive's broken (6)&quot; in a Cryptic Crossword. I've figured out that the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pelvic&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which I can &lt;em&gt;partially&lt;/em&gt; explain as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'elvic' is 'Clive' broken (anagrammed)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can someone help me figure out why:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Soft means the letter 'p'?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've left parts of the solution in spoilertags so that if you want, you can try to solve the clue on your own!&lt;/p&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="5373" LastEditDate="2018-06-02T15:33:13.413" LastActivityDate="2018-06-02T15:33:13.413" Title="To do with a bone, soft, Clive's broken (6) (Cryptic Crossword clue help)" Tags="|word|cryptic-clues|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="12792" PostTypeId="2" ParentId="12791" CreationDate="2015-04-27T14:09:05.540" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In sheet music, &lt;em&gt;piano&lt;/em&gt; (meaning &quot;play/sing softly&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is denoted as &lt;em&gt;p&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; See the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dynamics_(music)&quot; rel=&quot;nofollow&quot;&gt;Wikipedia entry on dynamics&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-07-15T19:05:06.863" LastActivityDate="2015-07-15T19:05:06.863" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12793" PostTypeId="2" ParentId="12781" CreationDate="2015-04-27T14:13:39.933" Score="15" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A) Nobody in Barberland has more than 1,000,000 hairs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No effect on population... yet.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;B) No two inhabitants of Barberland have the same number of hairs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Given A and B, we have no more than 1,000,001 people, since we know that there's no such thing as negative hair count, and 0 hairs is possible.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;C) Nobody in Barberland has the same number of hairs as the sum of two other inhabitants' hairs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;'Other' inhabitants indicates that as long as oneself is part of the equation, then the rule doesn't apply.  So we can have a 0 hair person.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Probably the easiest maximal count, then, is to start at 500,000 and continue on to 1,000,000, giving 500,002 total people, including the bald person.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1682" LastActivityDate="2015-04-27T14:13:39.933" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12795" PostTypeId="2" ParentId="12770" CreationDate="2015-04-27T14:21:10.070" Score="15" Body="&lt;p&gt;A simple solution for &lt;strong&gt;1000 prisoners 998 dead&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number all prisoners and bottles 1-1000,&#xA;and have each prisoner drink from ALL bottles except their matching number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 2 surviving prisoners have the numbers of the poisoned bottles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;** the bottles may be slightly depleted by the 999 prisoners taking a drink&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could also omit the last prisoner and deduce if only one prisoner survives the last bottle is the second poison.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;500 prisoners 499 dead&lt;/strong&gt;&#xA;Expanding the above technique using 500 prisoners drinking from 998 (so prisoner 1 omits bottles 1 and 2, prisoner 2 omits 3 and 4) we get either 499 dead prisoners and the exact combination or 498 prisoners and a choice of 4 bottles (with the knowledge that its 1 of 1 and 2 and 1 of 3 and 4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can then make the surviving prisoners drink 1 of the bottles they didn't drink the first time - if they die they drank the poison if they live they didn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2nd extension&#xA;- 116 prisoners - 114 dead&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take 20 prisoners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 1st omits bottles 0-100;  the 2nd omits bottles 50-150 etc. (The last prisoner has 950-1000 and 0-50.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the worst case 4 will survive giving us 2 gaps of 50 ie 100 bottles to narrow down&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add 96 new prisoners to the surviving 4 and number them 0-100 (and bottles 0-100) ensuring that the 4 &quot;lucky&quot; prisoners are omitting bottles they have previously omitted.&#xA;The 2 survivors will point to the 2 poisoned bottles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3rd extension - 86 prisoners 84 dead&lt;/strong&gt;&#xA;I used a little javascript function to find the optimum of this method&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same as above 40 prisoners omit 50 bottles each (overlapping the last prisoner by 25).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Worst case leaves 4 alive - and a 50 bottle gap. add the 46 prisoners to allow us to test all 50.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to find a better solution for the 2nd day to improve this any further&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4th extension - 62 prisoners 60 dead&lt;/strong&gt;&#xA;Realised the &quot;worst case&quot; was solvable using the binary expansion so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As above, 40 prisoners omit 50 bottles each (overlapping the last prisoner by 25).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;worst case&quot; leaves 4 alive - and 2 25 bottle gaps. needing only a further 6(+ the 4) prisoners using the binary method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If only 2 prisoners survive we have a 25 bottle gap and need to add 22 and deduce the last result to complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;5th extension 45 prisoners all potentially dead&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to @user2357112 in the comments&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Day 1 - use 40 prisoners as above but give each 525 bottles (1st prisoner has 0-525, 2nd 25-550 etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the poison is in separate groups then in the worst case 20 die (poison is in consecutive groups) this leaves 20 prisoners - more than enough to do the binary method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If both poisons are in 1 group of 25 then 21 prisoners will have died leaving 19 to test the group with. add in 5 so that we can test all but 1 (and deduce that result as before)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6th extension and bonus removed (they were wrong - thanks @Timbo for helping with my mistake).&lt;/p&gt;&#xA;" OwnerUserId="11836" LastEditorUserId="10781" LastEditDate="2015-04-29T09:37:05.020" LastActivityDate="2015-04-29T09:37:05.020" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="12796" PostTypeId="1" AcceptedAnswerId="15084" CreationDate="2015-04-27T14:22:18.840" Score="5" ViewCount="1754" Body="&lt;p&gt;There is a popular game, named &lt;a href=&quot;http://en.wikipedia.org/wiki/Four_fours&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Four fours&lt;/strong&gt;&lt;/a&gt;, where you have to find the shortest mathematical expression for every number from $1$ to $n$, using only the number $4$ and some operators.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My variant of this game is named &lt;strong&gt;Sevens&lt;/strong&gt;, and you can only use the number $7$ and some math operators to obtain the other numbers.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a &lt;strong&gt;list of accepted operators&lt;/strong&gt; (along with their individual score):  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$+,-,\times, \div $ (&lt;strong&gt;one&lt;/strong&gt; point)&lt;/li&gt;&#xA;&lt;li&gt;Parenthesis $( \ )$ (&lt;strong&gt;zero&lt;/strong&gt; points)&lt;/li&gt;&#xA;&lt;li&gt;Exponentiation (&lt;strong&gt;one&lt;/strong&gt; point)&lt;/li&gt;&#xA;&lt;li&gt;Square root) $\surd$ (rounded to lower integer) (&lt;strong&gt;two&lt;/strong&gt; points)&lt;/li&gt;&#xA;&lt;li&gt;Factorial $!$ (&lt;strong&gt;two&lt;/strong&gt; points)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; Logarithms and concatenation are explicitly not allowed! Implicit multiplication isn't allowed (eg. 7(3))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How to calculate the score?&lt;/strong&gt;&lt;br&gt;&#xA;The total score is the sum of all the partial scores you get with operations.&lt;br&gt;&#xA;If you use an operation twice, you must add the partial score twice, of course.&lt;br&gt;&#xA;The lower the score, the better it is!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples:&lt;br&gt;&#xA;$7=(7+7)-7$ results in a score of $2$&lt;br&gt;&#xA;$7=(7!)/(7-7/7)!$ results in a score of $7$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your task is to generate the numbers from $1$ to $30$ achieving the minimum possible score, using only the number $7$ and the above operations.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-06-18T12:13:16.833" LastActivityDate="2015-06-18T17:28:45.243" Title="The game of Sevens" Tags="|mathematics|calculation-puzzle|optimization|formation-of-numbers|" AnswerCount="5" CommentCount="14" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12797" PostTypeId="2" ParentId="12796" CreationDate="2015-04-27T14:36:18.190" Score="1" Body="&lt;p&gt;The answers I'm sure of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1=7/7$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2=(7+7)/7$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$6=7-7/7$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$7=7$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$8=7+7/7$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$14=7+7$&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-04-27T14:36:18.190" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12798" PostTypeId="1" AcceptedAnswerId="12847" CreationDate="2015-04-27T14:41:20.003" Score="2" ViewCount="1187" Body="&lt;p&gt;Inspired by &lt;a href=&quot;https://puzzling.stackexchange.com/questions/12761/monty-hall-all-over-again&quot;&gt;this question about the Monty Hall problem&lt;/a&gt;, here's a deeper look at another well-known counter-intuitive probability problem which is often stated in a way that leaves subtle ambiguities:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Version 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You meet a woman, who tells you: &quot;I have exactly two children. One of them is a girl.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Version 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You meet a woman, who tells you: &quot;I have exactly two children. The eldest is a girl.&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Version 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You meet a woman, who tells you: &quot;I have exactly two children. One of them is a girl.&quot; You ask- &quot;Could you please tell me specifically a child of yours who is a girl?&quot;, and she answers &quot;The eldest is a girl.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The question is, in all three of these cases, what is the probability that both of the woman's children are girls. Assume:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;She only tells the truth&lt;/li&gt;&#xA;&lt;li&gt;She always answers any question you ask to the best of her ability&lt;/li&gt;&#xA;&lt;li&gt;&quot;One of them is a girl&quot; is to be interpreted literally. It doesn't mean &quot;exactly one of them is a girl&quot;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As well as just giving a numerical answer for each version, also explain any apparent contradictions between the answers, and any hidden ambiguities in the question.&lt;/p&gt;&#xA;" OwnerUserId="1759" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-04-28T12:01:01.737" Title="Another look at the two children riddle" Tags="|probability|" AnswerCount="5" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="12799" PostTypeId="2" ParentId="12798" CreationDate="2015-04-27T14:50:19.547" Score="2" Body="&lt;p&gt;Assuming the chances of her having a boy or girl is equally likely, there is a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 50% chance of her having two female children.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This problem is similar to stating that John has a coin that he flipped heads. If John flipped another coin, what is the chance it was also head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only change in your statements is the order of the flips (or children) which doesn't matter since the determination of the outcomes are independent.&lt;/p&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-04-27T17:21:02.817" LastActivityDate="2015-04-27T17:21:02.817" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12801" PostTypeId="2" ParentId="12798" CreationDate="2015-04-27T15:03:44.000" Score="6" Body="&lt;p&gt;There are four possibilities for the genders of her children. These are &lt;strong&gt;MM, MF, FM,&lt;/strong&gt; and &lt;strong&gt;FF.&lt;/strong&gt; Since order matters (eldest/youngest), &lt;strong&gt;MF&lt;/strong&gt; and &lt;strong&gt;FM&lt;/strong&gt; are two different cases. Therefore, the answers to the question are as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Version 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Knowing that one child is a girl, we can eliminate &lt;strong&gt;MM&lt;/strong&gt; as an option. We are left with &lt;strong&gt;MF, FM,&lt;/strong&gt; and &lt;strong&gt;FF,&lt;/strong&gt; out of which the probability of both being female is &lt;strong&gt;1/3&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Version 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Knowing specifically that the eldest child is a girl, we can eliminate &lt;strong&gt;MM&lt;/strong&gt; and &lt;strong&gt;MF&lt;/strong&gt; as options. This leaves us with &lt;strong&gt;FM&lt;/strong&gt; and &lt;strong&gt;FF&lt;/strong&gt;, out of which the probability of both being female is &lt;strong&gt;1/2&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Version 3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This version starts the same as Version 1, eliminating &lt;strong&gt;MM&lt;/strong&gt; as a possibility. When she answers that her eldest child is a girl, she eliminates &lt;strong&gt;MF&lt;/strong&gt; from contention. This again leaves us with &lt;strong&gt;FM&lt;/strong&gt; and &lt;strong&gt;FF&lt;/strong&gt;, out of which the probability of both being female is &lt;strong&gt;1/2&lt;/strong&gt;. &lt;em&gt;(Note: I'm still thinking about this scenario.)&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-04-27T15:03:44.000" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="12802" PostTypeId="2" ParentId="12781" CreationDate="2015-04-27T15:09:35.393" Score="6" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 500,002&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One way of doing it is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pick every odd number. As the sum of 2 odds is even, we have 500,000&#xA; valid people. Also, we can have a person with 2 hairs.&#xA; &#xA; Also AdamDavis just reminded me that a bald person is allowed, so we&#xA; can add the bald person to our population.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4269" LastEditDate="2015-04-27T15:14:53.450" LastActivityDate="2015-04-27T15:14:53.450" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12803" PostTypeId="2" ParentId="12796" CreationDate="2015-04-27T15:13:53.510" Score="6" Body="&lt;p&gt;So, here's a foundation to start on. &lt;b&gt;Score: 109&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1 = 7/7$ [Score +1]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2 = (7+7)/7$ [Score +2]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3 = (7+7+7)/7$ [Score +3]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$4 = 7-(7+7+7)/7$ [Score +4]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5 = 7-(7+7)/7$ [Score +3]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$6 = 7-7/7$ [Score +2]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$7 = 7$ [Score +0]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$8 = 7+7/7$ [Score +2]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$9 = 7+(7+7)/7$ [Score +3]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$10 = 7+(7+7+7)/7$ [Score +4]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$11 = 7+7-(7+7+7)/7$ [Score +5]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$12 = 7+7-(7+7)/7$ [Score +4]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$13 = 7+7-(7/7)$ [Score +3]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$14 = 7 + 7$ [Score +1]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$15 = 7 + 7 + 7/7$ [Score +3]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$16 = 7 + 7 + (7+7)/7$ [Score +4]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$17 = 7 + 7 +(7+7+7)/7$ [Score +5]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$18 = 7 + 7 + (7+7+7+7)/7$ [Score +6]*&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$19 = 7+7+7-(7+7)/7$ [Score +5]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$20 = 7+7+7-(7/7)$ [Score +4]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$21 = 7+7+7$ [Score +2]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$22 = 7+7+7+(7/7)$ [Score +4]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$23 = 7+7+7+(7+7)/7$ [Score +5]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$24 = 7+7+7+(7+7+7)/7$ [Score +6]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$25 = 7+7+7+7-(7+7+7/7)$ [Score +7]*&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$26 = 7+7+7+7-(7+7)/7$ [Score +6]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$27 = 7+7+7+7-(7/7)$ [Score +5]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$28 = 7+7+7+7$ [Score +3]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$29 = 7+7+7+7+(7/7)$ [Score +5]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$30 = 7+7+7+7+(7+7)/7$ [Score +6]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1+2+3+4+3+2]+0+[2+3+4+5+4+3]+1+[3+4+5+6+5+4]+2+[4+5+6+7+6+5]+3+[5+6] = 113&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a simple answer to help get the ball rolling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;After improvements:&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*$18 = \sqrt{7*7*7} $ [Score +4] (oppose to 6)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*$25 = (7*7*7+7)/(7+7)$ [Score +5] (oppose to 7)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Score: 113 - 2 - 2 = 109&lt;/b&gt;&lt;/p&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-04-27T16:51:07.187" LastActivityDate="2015-04-27T16:51:07.187" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="12805" PostTypeId="1" AcceptedAnswerId="12807" CreationDate="2015-04-27T15:26:10.300" Score="16" ViewCount="2541" Body="&lt;p&gt;Who or what is this poem referring to?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;I'm always on a plane&#xA;Won't fit on a boat or train&#xA;&#xA;Unlike most of my brethren&#xA;I travel with no companions&#xA;&#xA;Simultaneously cold and hot&#xA;I can be a dark or light spot&#xA;&#xA;To some I may be unrecognizable&#xA;At times I'm not even visible&#xA;&#xA;From my routine I never deviate&#xA;But sometimes I'm seen doing the opposite&#xA;&#xA;I've suffered battering and bruising&#xA;Since I've no protective layering&#xA;&#xA;Subjected to a stiff breeze&#xA;No one near me can breathe&#xA;&lt;/pre&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4450" LastEditDate="2015-04-27T19:55:41.707" LastActivityDate="2015-04-28T02:28:14.580" Title="Riddle Poem: I'm always on a plane, won't fit on a boat or train" Tags="|riddle|" AnswerCount="1" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12806" PostTypeId="2" ParentId="12781" CreationDate="2015-04-27T15:33:13.323" Score="0" Body="&lt;p&gt;I'm thinking along the lines of excluded numbers. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;For any number you pick (n) after the first (f) you at minimum exclude one other number either Abs(f-n) where the sum is greater than the max (m), or f+n. therefore you can have Floor(m/2) + 1 plus the special case of 0 - giving us 500,002&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;reasoning for the minimum: As pointed out below you can exclude more than one number if you choose small numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you only choose Numbers greater than half the max (m) you are guarenteed to exclude only a single number because the sum of any two numbers greater than half max will be greater than max.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(m/2 + n1) + (m/2 +n2) = m + n1 + n2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11831" LastEditorUserId="11831" LastEditDate="2015-04-28T19:04:45.370" LastActivityDate="2015-04-28T19:04:45.370" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12807" PostTypeId="2" ParentId="12805" CreationDate="2015-04-27T15:33:25.913" Score="22" Body="&lt;p&gt;Is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mercury (the planet)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm always on a plane&#xA;Won't fit on a boat or train&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each planet has an orbital plane&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unlike most of my brethren&#xA;I travel with no companions&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Has no moon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Simultaneously cold and hot&#xA;I can be a dark or light spot&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cold and dark on the far side, hot on the sun side. Dark if in front of the Sun and light if not&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To some I may be unrecognizable&#xA;At times I'm not even visible&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Often mistake for a star. Not always visible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From my routine I never deviate&#xA;But sometimes I'm seen doing the opposite&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Always orbiting the Sun, but planets appear in retrograde sometimes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've suffered battering and bruising&#xA;Since I've no protective layering&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No atmosphere. Gets hit by stuff&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Subjected to a stiff breeze&#xA;No one near me can breathe&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mercury's magnetic field is weak enough to not protect from solar wind. No atmosphere to breathe&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5612" LastEditorUserId="5612" LastEditDate="2015-04-27T19:57:51.883" LastActivityDate="2015-04-27T19:57:51.883" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12808" PostTypeId="2" ParentId="12770" CreationDate="2015-04-27T15:37:35.157" Score="2" Body="&lt;p&gt;Working off of &lt;a href=&quot;https://puzzling.stackexchange.com/a/12795/2806&quot;&gt;Collett's answer&lt;/a&gt;, how about this: &#xA;First day, split bottles into groups of 250, 4 prisoners drink from the other 750 bottles. We have at most 2 survivors, which indicate the groups of 250 the poisons are in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, we take the remaining 2, 14 unused, and leave 2 alone, and split them into 2 groups of 8. They each drink all 250 of one group (remaining 2 don't do this), then use the original solution on the other group of 250 to figure out the bottles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total: 18 prisoners. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Better explanation of 2nd day:&lt;br&gt;&#xA;Say the 2 remaining prisoners are the ones that didn't drink 250's (A) and didn't drink 500's (B)&lt;br&gt;&#xA;We make a new A' group that's composed of A and 7 others, and they drink all of 500's.&lt;br&gt;&#xA;We make a new B' that's composed of B and 7 others and they drink all of 250's.&lt;br&gt;&#xA;Since you have 8 slaves in each group $2^8 = 256$ which means you can determine the other poison in each group. So, A' tests all of the 250's, and B' tests all of the 500's.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They do that by doing a bitwise map. Each prisoner is a bit mapping (e.g. 00000100 and 00000010) and each bottle is represented by it's number in binary, and if your bottle matches your mapping, you drink. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;br&gt;&#xA;So, what happens if you're left with only 1 group of 250 left? Well, you'd be left with 17 prisoners left. Luckily, if you get 2 groups of 8 out of them, you can still figure out the poison. Put the poisons in a 250x250 grid, and do the binary mapping on each axis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll lose at most 19 prisoners here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mapping would look like (numbers are bottles, letters are prisoners), in this case, there are 8 bottles, so we need 3 prisoners per axis. Each prisoner drinks the bottles below them or on their row&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Prisoners  a   a   a&#xA;             b b     b&#xA;                 c c c&#xA;Bottles  0 1 2 3 4 5 6&#xA;       7 x x x x x x x&#xA;     d 6 x x x x x x&#xA;   e   5 x x x x x&#xA;   e d 4 x x x x&#xA; f     3 x x x&#xA; f   d 2 x x&#xA; f e   1 x&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 2:&lt;/strong&gt;&lt;br&gt;&#xA;Bitwise mapping is ideal for finding 1 element out of a set, not 2, in a single trial, so if you have both poisons in 1 group, you'll have to use a brute force method day 2. So, best case 18, worst case, well, much worse. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-04-29T12:22:31.693" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="12809" PostTypeId="2" ParentId="12798" CreationDate="2015-04-27T15:38:26.750" Score="1" Body="&lt;p&gt;&lt;strong&gt;Version 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You don't have any information about the second child, so it can be male or female with  the same probability &lt;strong&gt;$\frac1 2$&lt;/strong&gt;. Note that saying that &quot;one is female&quot; doesn't mean &quot;only one is female&quot;!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Version 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, you have no information about the second child, the probability of being female is &lt;strong&gt;$\frac1 2$&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Version 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You know that one child is female. After the second statement, we have the following cases:&lt;br&gt;&#xA; - Eldest female-Youngest female&lt;br&gt;&#xA; - Eldest female-Youngest male&lt;br&gt;&#xA; None of these is more probable that the other, so we have again a $\frac12$ chance of both being females.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-04-27T15:38:26.750" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="12812" PostTypeId="2" ParentId="12761" CreationDate="2015-04-27T17:08:56.680" Score="3" Body="&lt;p&gt;If he always opens one of the doors you did not pick at random, then the door he opens should contain a car 1/3 of the time, a sheep 1/3 of the time, and a goat 1/3 of the time.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the animals and car were placed at random, we had the following possible cases to consider at the start, each of which had equal probability:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;S G C&#xA;S C G&#xA;G S C&#xA;G C S&#xA;C S G&#xA;C G C&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Since the above has all the cases, we can make it easier to think about by saying that the first column represents the door you picked, the second represents the door he opened, and the third is the option you are given.  Thus, after he reveals the sheep in the door he opened, we are left with the following cases:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;G S C&#xA;C S G&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Since the probabilities of each case are equal, your odds are 50/50 with either door.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2015-04-27T17:08:56.680" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12814" PostTypeId="1" AcceptedAnswerId="12862" CreationDate="2015-04-27T17:42:16.657" Score="5" ViewCount="283" Body="&lt;p&gt;%#@! I have overslept!&lt;br&gt;&#xA;%#@! I stubbed my toe on the bed leg!&lt;br&gt;&#xA;%#@! I'm splashed from a puddle at the bus stop!&lt;br&gt;&#xA;%#@! I'm late for Telecom 25 Lab!&lt;br&gt;&#xA;%#@! The professor is near the door as I walk in!&lt;br&gt;&#xA;%#@! My friend Kate, smiles from across the room. I scowl!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end of class, Kate drops this thing on my workbench. It is made from bits of wire but what does it mean?&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/vE5nK.png&quot; alt=&quot;WireThing&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5188" LastActivityDate="2015-04-28T17:35:54.960" Title="Having a bad day?" Tags="|cipher|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12815" PostTypeId="2" ParentId="12761" CreationDate="2015-04-27T17:43:54.877" Score="5" Body="&lt;p&gt;The accepted answer conflates &quot;the host's knowledge and motivations are unknown&quot; with &quot;the host's actions are uniform and random&quot;, which is at best a debatable assumption. I actually think the correct answer to this question is that there is not enough information to answer it as phrased.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if I flip a coin outside of your view and ask what are the odds it landed tails, the correct answer is &quot;I don't know&quot;, not 50%. The coin might not be fair or it might be two-headed, etc. Probability itself is defined based on distributions. If you know nothing at all about the distribution, any discussion of probability is meaningless. (As @Yakk points out in a comment, bounds on the distribution can lead to bounds on the probability. But there is nothing like that in this question.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I want to ignore that debate and just consider the case where the host opens a closed door &lt;strong&gt;at random&lt;/strong&gt;, since a lot of people get this well-defined question wrong. Should you switch doors in this version of the game?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to be very clear: You will pick a door. The host will open &lt;strong&gt;at random&lt;/strong&gt; one of the two doors you did not pick. Then he will give you the option of switching to the other unopened door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine playing this game 198 times with the strategy &quot;always switch&quot;. Just for descriptive purposes, let's color the sheep red and blue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) In 66 instances of the game, your initial guess is correct, the host opens one of the other doors at random, you switch, and you &lt;em&gt;lose&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) In another 66 instances, your initial guess has the red sheep. In 33 of these instances, the host opens the prize door at random, you switch to the blue sheep door, and you &lt;em&gt;lose&lt;/em&gt;. In the other 33 of these instances, the host opens the blue sheep door, you switch to the prize door, and you &lt;em&gt;win&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) In the final 66 instances, your initial guess has the blue sheep. Symmetrically with (2), in 33 of these instances the host opens the prize door and you &lt;em&gt;lose&lt;/em&gt;. In the remaining 33 instances the host opens the red sheep door, you switch, and you &lt;em&gt;win&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding it all up, you win in just 33 + 33 = 66 instances, or 1/3 of the time you play. This is exactly the same number of wins as if your strategy were &quot;never switch&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since &quot;always switch&quot; and &quot;never switch&quot; glean the exact same number of wins, there is no advantage to switching doors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion, this is the sort of conditional probability question that most people intuitively think is being asked in the classic Monty Hall problem, which is why so many people get it wrong (where the correct answer is &quot;always switch&quot;). Changing the host's actions to random completely changes the problem and the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Update]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that it does not actually matter what you are allowed to do when the host opens a door with the prize, because &lt;em&gt;that did not happen&lt;/em&gt; in the question as stated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A full, correct analysis is to &lt;em&gt;throw out&lt;/em&gt; the 66 instances above where the host opens the door with the prize, because by assumption we are not in a world where that happened. In the remaining 132 instances, you win half the time and lose half the time whether you switch or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Update 2]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an analogy that some have found convincing...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we play a game with an ordinary deck of 52 cards. We shuffle the cards randomly, then you pick one without looking at it. I look at the remaining 51 cards and reveal 50 (i.e. all but one) of them, none of which are the ace of spades. What is the probability that you are holding the ace of spades?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we play this game many times, 1 time out of 52 your original card will be the ace of spades. The other 51 out of 52 times, I will need to deliberately dodge the ace in order to avoid revealing it. That second scenario is 51 times more likely, so the answer to this question is that the odds are 51-to-1 against your holding the ace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is exactly like the classic Monty Hall problem, but with 52 doors instead of 3. And just like that problem, we have to make a whole bunch of assumptions that are often omitted in the statement of the question; e.g. the shuffling is uniform, I always reveal 50 non-ace-of-spades cards regardless of which card you choose, etc. But subject to those assumptions, the odds are 51-to-1 against your holding the ace of spades, just as in the classic Monty Hall problem, the odds are 2-to-1 against your original door containing the prize.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's change the question slightly. Suppose we shuffle and you pick a card randomly as before, but now I reveal 50 other cards &lt;em&gt;without even looking at them&lt;/em&gt;. And suppose, just by chance, that none of those revealed cards is the ace of spades.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Granted, this is unlikely; in fact, it will only happen 1 time out of 26 that we play this game. But &lt;em&gt;supposing it does happen&lt;/em&gt;, what then are the odds that your card is the ace? (This is what &quot;conditional probability&quot; is all about.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, the card I avoided flipping over blindly is completely random, just like the one in your hand. So the answer to this modified problem is that the odds are 1-to-1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This modified problem is exactly analogous to the modified Monty Hall question, where the host opens a door at random. And the answer is the same: If the door is opened at random, the fact that you do not see the prize tells you nothing about where the prize is, except that it is not behind that particular door.&lt;/p&gt;&#xA;" OwnerUserId="6174" LastEditorUserId="6174" LastEditDate="2015-04-27T22:15:19.993" LastActivityDate="2015-04-27T22:15:19.993" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12816" PostTypeId="2" ParentId="12770" CreationDate="2015-04-27T18:34:56.250" Score="-1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I have an answer that requires &lt;strong&gt;only $19$&lt;/strong&gt; prisoners in the worst case!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's split the bottles into four groups of $250$. We need $4$ prisoners: they're assigned a number from 1 to 4 and everyone drinks all the bottles except the one with his name. After the first midnight, we have these possibilities:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Three die&lt;/em&gt;: It means that the poisoned bottles were both in one fourth of the total, which is $250$ bottles. On the second day, $16$ prisoners are enough to isolate the poisoned bottles using a &lt;a href=&quot;http://en.wikipedia.org/wiki/Exclusive_or#Bitwise_operation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bitwise operation&lt;/a&gt;, which is a binary map. Since we can recycle the first day survivor and use him for the second day too, we only need $16$ for the second day plus the $3$ first day dead guys, for a &lt;strong&gt;total of $19$ prisoners&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Two die&lt;/em&gt;: We have isolated the two bottles inside two groups of $250$ bottles each. For each group we need $8$ prisoners to perform the bitwise strategy and identify the poisoned bottle in that group, which is $16$ prisoners for the second day. Since we can recycle the first day survivors, we globally need $16$ prisoners plus the two dead guys, for a &lt;strong&gt;total of $18$ prisoners!&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;With 20 prisoners, how many bottles can I test?&lt;/strong&gt;&lt;br&gt;&#xA;If the above strategy is optimal, and it's likely to be so, I'd say &lt;strong&gt;the maximum number of bottles is $1024$&lt;/strong&gt;. We split them into $4$ groups of $256$ and we see, using the binary operation, that we'll need no more than $20$ prisoners to detect the poisoned bottles.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the bitwise operation I mentioned before?&lt;/strong&gt;&lt;br&gt;&#xA;I will explain it using simple (small) examples, you just need to know that it can be extended to any number of elements.&lt;br&gt;&#xA;Suppose that you have $4$ bottles and know that $1$ contains poison. &#xA;Now, let's say we have two prisoners, $A$ and $B$.&lt;br&gt;&#xA;$A$ drinks from bottles $1$ and $3$.&lt;br&gt;&#xA;$B$ drinks from bottles $1$ and $2$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have these options:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$A$ alive, $B$ dead -&gt; Poison is in $2$&lt;/li&gt;&#xA;&lt;li&gt;$A$ alive, $B$ alive -&gt; Poison is in $4$  &lt;/li&gt;&#xA;&lt;li&gt;$A$ dead, $B$ alive -&gt; Poison is in $3$  &lt;/li&gt;&#xA;&lt;li&gt;$A$ dead, $B$ dead -&gt; Poison is in $1$  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As you can see, we were able to analyze $4$ bottles using $2$ prisoners. Generally, for $n$ bottles you need $log_2\ n$ prisoners (if decimal, round to upper integer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What if we have two poisons in the same group of bottles?&lt;/strong&gt;&lt;br&gt;&#xA;As before, we can perform a bitwise operation, but this time for $n$ bottles is requires $2log_2\ n$ prisoners.&lt;br&gt;&#xA;An example with $4$ bottles (two poisons) and $4$ prisoners is here:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/HX1Yk.png&quot; alt=&quot;enter image description here&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bottles are named $1,2,3,4$ while prisoners are $A,B,C,D$&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-04-28T12:55:21.293" LastActivityDate="2015-04-28T12:55:21.293" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="12817" PostTypeId="1" AcceptedAnswerId="12823" CreationDate="2015-04-27T20:02:29.320" Score="6" ViewCount="575" Body="&lt;p&gt;A play of Romeo and Juliet is being held in the theatre tonight, hundred of people have been waiting this moment for months.&lt;br&gt;&#xA;Many people are currently working hard in the theatre to adjust the final details.&lt;br&gt;&#xA;Ernest's job is to ensure an adequate lighting in the theatre, else it would be impossible for the actors to perform.&lt;br&gt;&#xA;Today, Ernest is experiencing a weird issue with his $1000$ lights: when he triggers the switch of the $i_{th}$ light, it changes status along with all its multiples. For example, triggering the $3$rd switch will change status to the lights with index $3,6,9,12,...$.&lt;br&gt;&#xA;Suppose that all the lights are initially off.&lt;br&gt;&#xA;Ernest is not a mathematician and is in hurry, so you must tell him &lt;strong&gt;how many lights will be on if he triggers all the switches&lt;/strong&gt;.&lt;br&gt;&#xA;You tell him the answer and he thinks that this number of lights is enough. Unintentionally, when it's the right moment to turn on the lights, he breaks the first five switches because of haste!&lt;br&gt;&#xA;Now &lt;strong&gt;he wants to know the maximum number of lights he can turn on! Is it still enough for the play?&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Notes:&lt;/em&gt; The broken switches are off. &lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-04-28T12:10:28.937" LastActivityDate="2015-05-19T16:20:40.510" Title="Lights in the theatre" Tags="|mathematics|story|no-computers|" AnswerCount="1" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="12818" PostTypeId="2" ParentId="12694" CreationDate="2015-04-27T20:05:14.260" Score="2" Body="&lt;p&gt;Here's a guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Internet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you use me, you only ever see what I have become instead of what I really am.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most only see the result of using The Internet, i.e. web pages, music, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be used to do powerful things, but I can be very fragile at my core.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Internet is used to do so much, it's taken over so much of what used to be slow processes. Sites/servers can be attacked in a multitude of ways.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Through time, I have changed lives!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Self-explanatory&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately some choose to hate me, and others choose to embrace me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There's a sort of pride in the older age groups in not being computer literate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am used around the world in many different ways.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People can use The Internet for so many things...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My family is huge, and yet still getting bigger.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Internet is a network of networks, with networks still being added.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is important not to forget about my elders, as they still needed (by some).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm guessing this is some sort of private network?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can look at me, but not touch.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can monitor network traffic, but you can't touch The Internet...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not as old as world war 1.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Internet came into being in the 60's.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9696" LastActivityDate="2015-04-27T20:05:14.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12819" PostTypeId="2" ParentId="12693" CreationDate="2015-04-27T20:05:47.897" Score="7" Body="&lt;p&gt;He's in &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Atlanta.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The center of Atlanta is almost exactly 2 hours away from Chattanooga. This would put him arriving around 7pm; a pizza place should deliver in 45 minutes with little trouble, so 7:45 is reasonable -- this means that he's in the Eastern time zone, which Nashville and Birmingham are not.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; But you didn't need to know any of that -- only Atlanta uses ten-digit dialing routinely, and that was a ten-digit number he read off the window.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Can't comment yet due to rep or I'd add this there, but the above needs a slightly more detailed explanation of the methodology. All four cities are roughly 2hr away from Chattanooga; Knoxville is closest and Birmingham furthest. With two hours on the road and starting at 5, this means he'll arrive at&lt;p&gt;&#xA; Knoxville and Atlanta at 7pm (EDT)&lt;br&gt;&#xA; Nashville and Birmingham at 6pm (CDT)&lt;p&gt;&#xA; Knowing that a pizza takes less than an hour to deliver, the 7:45 time means that he must be in Knoxville or Atlanta. After that, the ten-digit dialing bit mentioned above kicks in, and that tells you Atlanta.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Amusingly, apparently Tennessee has just recently (start of March 2015) initiated the use of ten-digit dialing around Nashville, so my original assertion wasn't accurate: some places in Nashville &lt;em&gt;could&lt;/em&gt; already be showing ten digits on their windows. But the time zone difference still applies.&lt;p&gt;&#xA; (More amusingly, I probably had an &quot;unfair&quot; advantage; I'm from Atlanta, my fiancée is from Chattanooga, my grandparents lived in Knoxville, and a number of friends live in Birmingham -- I've spent more time on I-75, I-24, I-40, and I-20 than I care to remember)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10625" LastEditorUserId="10625" LastEditDate="2015-04-28T12:03:58.913" LastActivityDate="2015-04-28T12:03:58.913" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12820" PostTypeId="1" AcceptedAnswerId="12824" CreationDate="2015-04-27T20:13:47.980" Score="6" ViewCount="845" Body="&lt;p&gt;In &lt;a href=&quot;https://puzzling.stackexchange.com/questions/12626/variants-to-blue-eyes-puzzle&quot;&gt;Variants to blue eyes puzzle&lt;/a&gt; we explored what would have happened if the oracle of the &lt;a href=&quot;http://www.xkcd.com/blue_eyes.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Blue Eyes&lt;/a&gt; puzzle had said something different with different numbers of islanders. I've got another one for you to ponder:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On this island there are 100 blue-eyed people, 50 brown-eyed people, 50 green-eyed people, and one hazel-eyed oracle. When the oracle speaks, she tells them&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;I see more people with blue eyes than people with brown eyes.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Other than the difference in what the oracle says and the eye color distribution of the islanders, all the rules are the same as in the original puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under these conditions, when will each group leave (if at all) and why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; even though the oracle's statement is very similar to the one in variant 2, the solution is not the same.&lt;/p&gt;&#xA;" OwnerUserId="1955" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-04-28T20:04:57.923" Title="Another twist on the blue-eyed puzzle" Tags="|logical-deduction|blue-eyes|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="12823" PostTypeId="2" ParentId="12817" CreationDate="2015-04-27T20:26:57.800" Score="12" Body="&lt;p&gt;The only numbers which would end up on if he triggered all the switches are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Squares, since they are the only numbers with an odd count of divisors. There are 31 squares (the squares of 1-31, obviously) between 1 and 1000, and therefore there would be 31 lights on if he triggered all the switches.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once he breaks switches 1-5, however:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The math gets a bit weirder here with numbers that are divisible by 2-5, but one thing we can be sure of is that prime numbered lights will now be turned on, since their only other divisor is 1, which is now broken. There are 165 primes between 6 and 1000. Since 31 lights were sufficient, 165 lights will certainly be sufficient as well.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;EDIT:&lt;/strong&gt; For conclusiveness, the answer to part 2 is &lt;strong&gt;540 lights.&lt;/strong&gt; These lights are the numbers where the count of that number's divisors in [1,2,3,4,5] (plus one if the number is a square) is &lt;strong&gt;odd.&lt;/strong&gt; This means that the switch would be hit an odd number of times fewer than the first scenario, which would result in it being on instead of off. The exception is squares, which would need to be hit an even number of times to stay on, which is why 1 is added to the previous sum.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="11475" LastEditDate="2015-05-19T16:20:40.510" LastActivityDate="2015-05-19T16:20:40.510" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12824" PostTypeId="2" ParentId="12820" CreationDate="2015-04-27T20:38:53.307" Score="6" Body="&lt;p&gt;If there were $51$ blue eyed people on the island, any blue eyed person would know from the statement that he had blue eyes.  They would see $50$ blue eyed and brown eyed people and saying $51&amp;gt;50$ is the only way to rectify this.  All blue eyed people would, therefore, leave on day $1$.  If there are more than $51$ then on day $2$ they know there are more than $51$ blue eyed people on the island. If they only see $51$ other blue eyed people, they will leave on day $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that blue eyed people will leave on day $L-R$ where $L$ and $R$ are the numbers of bLue and bRown eyed people respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When all blue eyed people don't leave on day $49$ the brown eyed people who see $100$ blue eyed people know that $L-R&amp;gt;49$.  They believe that could still have blue eyes or green eyes, however, as $101-49&amp;gt;49$ and $100-49&amp;gt;49$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On day $50$ blue eyed people look around and see only $99$ blue eyed people.  They realise, therefore, that they must have blue eyes.  They all leave at the end of that day. The brown eyed people still think there could be $101$ blue eyed people so don't yet know their eye color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the brown eyed people wake up the next day they realise $L-R=50$ which means $R=50$.  As they only see $49$ other brown eyed people, they realise their eye color and leave at the end of that day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The green eyed people never leave...they could have hazel, purple, or amber eyes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Blue eyed people leave on day 50.  Brown eyed people leave on day 51. Green eyed people don't leave until another oracle visits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditDate="2015-04-27T20:52:46.277" LastActivityDate="2015-04-27T20:52:46.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12826" PostTypeId="2" ParentId="12798" CreationDate="2015-04-27T21:15:34.873" Score="2" Body="&lt;p&gt;Borrowing heavily from the comments in Bailey M's answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Version 3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There is a 1/3 probability that she has 2 girls.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;We know that the eldest is a girl, meaning only FM and FF are answers, with equal probability. If the answer is FF, then she randomly chooses between telling us the eldest is a girl or the youngest is a girl. In other words, given that the eldest is a girl, we have the following possibilities:&#xA;(1) The youngest is a boy and she tells us the eldest is a girl (50%)&#xA;(2) The youngest is a girl and she tells us the eldest is a girl (25%)&#xA;(3) The youngest is a girl and she tells us the youngest is a girl (25%)&#xA;Given that she has told us that the eldest is a girl, we remove the 3rd possibility, leaving us with a 50/75 probability that the youngest is a boy and a 25/75 probability that the youngest is a girl. Reduce those, and we get a 2/3 chance that she has 1 girl, and a 1/3 chance that she has 2 girls.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This seems a little fuzzy to me, but it's the only difference I can see between versions 2 and 3. Sorry for the bad formatting, I couldn't figure out how to do a list inside spoilers.&lt;/p&gt;&#xA;" OwnerUserId="10861" LastActivityDate="2015-04-27T21:15:34.873" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12827" PostTypeId="2" ParentId="12477" CreationDate="2015-04-27T21:35:31.217" Score="1" Body="&lt;p&gt;&lt;strong&gt;Cryptogram 1&lt;/strong&gt; - Based on the hints, here is the encryption method:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;On a QWERTY keyboard, the keys are numbered from L-&gt;R and T-&gt;B, as shown below.  The plaintext key number plus its alphabet number determine its associated ciphertext letter. For &lt;strong&gt;Q&lt;/strong&gt;, 1+17=18, so the CT letter is &lt;strong&gt;K&lt;/strong&gt;.&lt;br&gt;&#xA;&lt;br&gt;&#xA;If the CT letter has already been used, then the next available key becomes the CT letter. For &lt;strong&gt;T&lt;/strong&gt;, 5+20=25, but &lt;strong&gt;N&lt;/strong&gt; has been used so the next available letter is &lt;strong&gt;M&lt;/strong&gt;.&lt;br&gt;&#xA;&lt;br&gt;&#xA;The encoding table is shown below.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FgAdU.png&quot; alt=&quot;Keyboard&quot;&gt; &#xA;&lt;img src=&quot;https://i.stack.imgur.com/lEI2w.png&quot; alt=&quot;Table1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cryptogram 3&lt;/strong&gt; - Here is the encryption method:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Step 1 - make MD5 hash for each letter of the alphabet, individually&lt;br&gt;&#xA;Step 2 - start by taking the first digit from each hash, then convert to decimal&lt;br&gt;&#xA;Step 3 - use mod-26, add 1 and select the equivalent letter of the alphabet&lt;br&gt;&#xA;Example for plaintext &lt;strong&gt;D&lt;/strong&gt; - the first digit of the hash is f which is decimal 15. Add 1, then select &lt;strong&gt;P&lt;/strong&gt; for the ciphertext.&lt;br&gt;&#xA;&lt;br&gt;&#xA;Step 4 - if the CT letter has already been taken, then repeat steps 2 and 3 with the first 2 digits of the hash.  Keep adding a digit from the hash until a unique letter can be selected for the CT.&lt;br&gt;&#xA;Example for PT &lt;strong&gt;L&lt;/strong&gt; - a unique letter is not selected until d20 (3 digits) is used. Convert to 3360 (decimal) then mod-26 plus 1 is 7, so the CT character is &lt;strong&gt;G&lt;/strong&gt;.&lt;br&gt;&#xA;&lt;br&gt;&#xA;The encoding table is shown below.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/y6Sxl.png&quot; alt=&quot;Table2&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-04-29T20:11:21.737" LastActivityDate="2015-04-29T20:11:21.737" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12828" PostTypeId="2" ParentId="12770" CreationDate="2015-04-27T23:05:38.833" Score="0" Body="&lt;p&gt;I think you can do a similar approach as the original problem. This only needs 19 prisoners that could potentially die.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First make mixtures of all wines combined with all other wines. That way you get 1000 x 999 / 2 mixtures is 499,500 mixtures. This number is smaller then 2^19. Only one of these mixtures is lethal!&#xA;Label these mixtures 1 to 499,500 in binary. Prisoner 1 then takes a sip from all mixtures with a 1 in the first position, prisoner 2 with a 1 in the second position and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can point exactly to which mixture is lethal depending on what prisoners die and point to the poisoned bottles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example if prisoner 1,5,10 and 11 die the poisonous mixture is mixture 1000100001100000000&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way. I only use 1 day here. maybe it can be improved with 2 days. I don't know that&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-04-27T23:05:38.833" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12829" PostTypeId="2" ParentId="12770" CreationDate="2015-04-27T23:08:33.717" Score="4" Body="&lt;p&gt;The best answer I've come up with so far tests 35 prisoners in the worst case.  It uses a similar setup as previous answers, but instead of separating the 1000 bottles into 4 groups of 250, I separate them into 8 groups of 125.  This reduces the number of prisoners needed to 35 from 39 (I'll explain how I calculate this at the end).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Day One&lt;/strong&gt;&lt;br /&gt;&#xA;First, we separate the bottles into 8 groups of 125 each.  We take 8 prisoners and have them drink from 7 out of the 8 groups (prisoner 1 drinks from all but group 1, etc.)  After this (as stated in other answers, but I'll clarify if needed), we're down to two scenarios for day two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Day Two - Scenario 1&lt;/strong&gt;&lt;br /&gt;&#xA;In the first scenario, 2 of our original prisoners have lived.  This means that 2 of the 8 groups of bottles each contained &lt;strong&gt;one&lt;/strong&gt; poisoned bottle.  We'll label these groups A and B, each with 125 bottles.  Now we can set up a simple binary mapping with 7 prisoners per group:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prisoner 1: 0000001 (1,3,5,7,etc.)&lt;br /&gt;&#xA;Prisoner 2: 0000010 (2,3,6,7,10,11,etc.)&lt;br /&gt;&#xA;Prisoner 3: 0000100 (8,9,10,11,12,13,etc.)&lt;br /&gt;&#xA;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The figure above shows the &quot;bit&quot; that the prisoner would be assigned, and all of the corresponding bottles he would sample from.  You would first set up two groups of 7 prisoners (prisoner A1, A2... B1, B2...).  One group would drink every bottle from group A, then apply their binary mapping to group B.  The other group of prisoners would do the opposite.  After this, you perform a logical &lt;strong&gt;OR&lt;/strong&gt; on the dead prisoners' assigned bits.  For example, if prisoners A1, A6, B3, B4 and B7 died, then the bottles that were poisoned were A33 (0100001) and B76 (1001100).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After this, we tested 14 prisoners on day two, and lost 6 prisoners on day one (re-using the two survivors), for a total of 20 prisoners.  This is the best case scenario.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Day Two - Scenario 2&lt;/strong&gt;&lt;br /&gt;&#xA;In the second scenario, only 1 of our original prisoners has survived.  This means that 1 of the 8 groups contains &lt;strong&gt;both&lt;/strong&gt; poisoned bottles.  This is far more difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two steps to day two.  The first step is a simple binary mapping with 7 prisoners, similar to how you would solve the original &quot;poisoned bottle&quot; problem (only one poisoned bottle).  Assign each prisoner one of 7 &quot;bits&quot;.  This is done the same as day one, above.  The deaths in these prisoners will tell you which bits of the poisoned bottles are both 1's between the two bottles.  For example, say bottles 39 and 97 are poisoned.  In binary, these bottles would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;37: 0100111&lt;br /&gt;&#xA;97: 1100001&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 1st bit and the 6th bit (from the right) are both shared between these two bottles, which means that prisoner 1 and prisoner 6 will both die, as they are the only two who would have sampled both bottles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that we have the bits that are the same between both bottles, we need the bits that are different.  To do this, we take 21 additional prisoners.  These prisoners aren't given one &quot;bit&quot; each, but two.  Thus prisoner 1 will be mapped to 0000011, and will drink any bottle with a 1 in the first or second bit (1,2,3,5,6,7,9,etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prisoner 1: 0000011 (1,2,3,5,6,7,9,etc.)&lt;br /&gt;&#xA;Prisoner 2: 0000101 (1,3,4,5,6,7,9,etc.)&lt;br /&gt;&#xA;Prisoner 3: 0000110 (2,3,4,6,5,7,9,etc.)&lt;br /&gt;&#xA;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There will be many prisoners that die from this batch.  First, we need to discount all the prisoners that would have died with a repeating bit (as in step 1).  These prisoners would have died regardless of what their second bit was.  In our previous example (with bottles 39 and 97), these would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prisoner 1:  0000011&lt;br /&gt;&#xA;Prisoner 2:  0000101&lt;br /&gt;&#xA;Prisoner 4:  0001001&lt;br /&gt;&#xA;Prisoner 7:  0010001&lt;br /&gt;&#xA;Prisoner 11: 0100001&lt;br /&gt;&#xA;Prisoner 12: 0100010&lt;br /&gt;&#xA;Prisoner 13: 0100100&lt;br /&gt;&#xA;Prisoner 14: 0101000&lt;br /&gt;&#xA;Prisoner 15: 0110000&lt;br /&gt;&#xA;Prisoner 16: 1000001&lt;br /&gt;&#xA;Prisoner 21: 1100000&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know which prisoners these would be because of step one.  Since prisoners 1 and 6 died in step one, the 1st and 6th bits are repeats.  Any prisoners in step two with these bits are ignored.  The remaining dead prisoners from step two are considered.  Using our example, the dead prisoners would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prisoner 17: 1000010&lt;br /&gt;&#xA;Prisoner 18: 1000100&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this says is that one bottle has the 7th bit set and the other bottle has the 2nd bit set (from prisoner 17).  We also know that one bottle has the 7th bit set and the other bottle has the 3rd bit set (from prisoner 18).  The only combination that makes this true is to have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bottle A: 1000000&lt;br /&gt;&#xA;Bottle B: 0000110&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we combine these with our known repeating bits, we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bottle A: 1100001 (97)&lt;br /&gt;&#xA;Bottle B: 0100111 (39)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that we're performing the tests for step one and step two simultaneously and just noting the results afterwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note, that I believe step one is necessary.  I don't believe you can determine which are the repeating bits using only step two, but I'm still trying to prove my concept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary&lt;/strong&gt;&lt;br /&gt;&#xA;In total, we have used 20 prisoners in scenario one (6 dead from day one, 14 more tested on day two).  In scenario two, we've used 35 prisoners (7 dead from day one, 7 more tested on day two step one, and 21 more tested for step two).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that we know we need 21 prisoners for step two because we're looking at combinations of 2 bits chosen out of 7, which is 7C2 = (7!)/((7-2)!2!).  This is how I determined that dividing the original 1000 bottles into 8 groups was ideal.  Other possible configurations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 groups of 250&lt;br /&gt;&#xA;First day: 3 dead   Second day: 8 + 8C2 = 8 + 28 = 36   Total: 39&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8 groups of 125&lt;br /&gt;&#xA;First day: 7 dead   Second day: 7 + 7C2 = 7 + 21 = 28   Total: 35&lt;/p&gt;&#xA;&#xA;&lt;p&gt;16 groups of 62/63&lt;br /&gt;&#xA;First day: 15 dead  Second day: 6 + 6C2 = 6 + 15 = 21   Total: 36&lt;/p&gt;&#xA;" OwnerUserId="11851" LastEditorUserId="11851" LastEditDate="2015-04-28T19:10:54.060" LastActivityDate="2015-04-28T19:10:54.060" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="12830" PostTypeId="2" ParentId="12643" CreationDate="2015-04-28T00:51:35.527" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2 wolves&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My thinking is thus: First Assume the wolves and hare are on the same line of longitude. The hare at the equator and equidistant from the wolves&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the hare moves the wolves can always stay at the same longitude as the hare by moving along their own line of latitude. Because the hare is between the wolves their latitude circles will be smaller than the hares. This gives them a free longitudinal component of motion they can use to close with the hare. Although the hare can move out of its equidistant point it will always be caught.. eventually&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to consider out initial assumption. We can partially dispense with the equidistant requirement as the hare is able to move closer to one wolf or the other from its starting point without escaping or increasing its distance from either wolf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also define our longitudinal axis as the plane of W-H-W. as long as the animals are on a plane they can be considered on a line of longitude&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, can the wolves move to position the hare between them and on a plane regardless of the hares movement? well if the wolves each move to a pole they form a plane with the rabbit wherever it is. The rabbit may not be in the same hemisphere, trapping one wolve behind the 'hump' of the planet. but this wolf is still able to maintain its longitudinal match with the hare and the nearer wolf can chase the hare towards the equator until they are all in the same hemisphere&lt;/p&gt;&#xA;&#xA;&lt;p&gt;QED&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOW! max time to capture.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; infinity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The best strategy from the hare is to move such that the wolves come closer at the slowest rate. This is achieved by having the maximum longitudinal speed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets consider the case where the wolves are at the poles and the hare at the equator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hare runs along the equator and the wolves match its longitude but slowly approach. They cant run directly to the closest intersection point because the hare will reverse direction and escape. As they get closer and closer to the hare their courses become more and more parallel. But we are asked to consider them as points. Luckily they don't need to eat&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1h smell radius. much harder. have to think&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK so, The wolves can detect the hare when they are 1h away from it. If they all happen to be on a plane as before they can catch it. this is the case where 2 wolves are 2hrs apart and the hare is directly between them. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; So 12 wolves (24h/2) can start at one pole and, equally spaced, move to the other pole eventually trapping the hare&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Say we have 1 less wolf, the ring of wolves almost reaches the equator and then MUST leave a gap. but can the rabbit squeeze though an infinitesimal gap? the wolves move at speed s same as the hare, the overlapped region of smell expands at sqrt(1-s^2)... no thats on a flat plane, for a sphere its.. arccos(cos(1)/cos(s)) ? which is always less than s. Ergo the hare can escape from between any two closing spherical caps&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, Considering the strategy for 3 wolves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the same best case (for the hare) starting point with wolves on poles and hare on equator but with one spare wolf at one pole. The 2 wolves move as before but the spare wolf moves to an arbitrary point on the equator to try and cut off the rabbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case the hare is still caught at infinity ( i think the same infinity, have to do the maths) as it can approach the third wolf but switch direction immediately before capture. As long as its speed remains constant the approach time of the original wolves is the same and the third wolf cannot catch it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However what if the original two wolves vary their strategy, dropping back to prevent the reverse direction tactic? tbc..&lt;/p&gt;&#xA;" OwnerUserId="11831" LastEditorUserId="11831" LastEditDate="2015-04-28T16:30:39.740" LastActivityDate="2015-04-28T16:30:39.740" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12831" PostTypeId="1" AcceptedAnswerId="12834" CreationDate="2015-04-28T01:23:06.607" Score="11" ViewCount="672" Body="&lt;p&gt;There are $24$ ants scattered around a hula hoop of circumference $3$ meters . They each randomly, and independently, choose to face clockwise or counterclockwise, and then simultaneously start marching at $1$ cm/sec. When two ants meet, they instantaneously reverse direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say that one of the ants is named Alice. After $5$ minutes of marching, what is the probability that Alice is back where she started?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; I have changed the number of ants from $25$ to $24$. This doesn't affect how you approach/solve the problem, but makes the answer more interesting.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-04-28T03:10:48.500" LastActivityDate="2015-04-28T15:35:28.990" Title="Ants on a Hula Hoop" Tags="|geometry|probability|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12834" PostTypeId="2" ParentId="12831" CreationDate="2015-04-28T03:24:10.363" Score="9" Body="&lt;p&gt;I'll say&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; about $16\%$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First note that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; as is &lt;a href=&quot;https://puzzling.stackexchange.com/questions/1784/maximum-time-for-ants-to-fall-off-stick&quot;&gt;typical of these ant puzzles&lt;/a&gt;, you can think of the ants as passing through each other instead of changing directions, and the ant locations remain the same (though not the individual ant identities).  Thus, after 5 minutes or 300 seconds, the ants have gone around the hula-hoop and are in the same locations they started. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Next note that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; order of the ants never changes.  If Alice ends in her same spot, then her neighbors end in their same spots, as do their neighbors, etc., all the way around the hoop.  So if Alice ends in her spot, nobody has moved overall.  If Alice has moved over one ant position, the all move over one ant position.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally, we have&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  if there are $k$ ants going clockwise, then the average net clockwise speed of the ants is $\frac{k - (24 - k)}{24} = \frac{k}{12} - 1$.  So on average, after 5 min, one of the ants has moved $25k -300$ centimeters in the clockwise direction.  Each ant ends on its starting value exactly when the average distance moved is $-300$, $0$, or $300$.  This corresponds to $k = 0, 12, 24$. Thus $(2 + {24 \choose 12})/2^{24}$ fraction of the starting directions lead to Alice coming home.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2722" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-04-28T15:35:28.990" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12835" PostTypeId="1" CreationDate="2015-04-28T05:06:03.847" Score="1" ViewCount="187" Body="&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/12798/another-look-at-the-two-children-riddle&quot;&gt;This&lt;/a&gt; question, has me stumped. Looking at the first version, how there's two lines of thought that go around in my head, both (seemingly) justifiable.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1&lt;/strong&gt; : &lt;em&gt;'One of them is a girl'&lt;/em&gt; implies &lt;em&gt;'Both aren't boys'&lt;/em&gt;, which implies three posibilites, each with 1/3 possibilites. Thus the possibility of both being girls is 1/3.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2&lt;/strong&gt; : &lt;em&gt;'One of them is a girl'&lt;/em&gt;. Let it be the one playing in the garden (without loss of generality) then the one shrieking like the Bee Gees in the room above is either a boy or a girl. 50-50. Thus 1/2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Followup&lt;/em&gt; :  Why does the order matter? &#xA;I will really appreciate a clear, detailed, answer to derail my wrong-logic train of thought. Thanks. &lt;/p&gt;&#xA;" OwnerUserId="11758" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-05-07T17:04:23.277" Title="Fumbling with a question already posted" Tags="|probability|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="12836" PostTypeId="2" ParentId="12761" CreationDate="2015-04-28T05:11:10.047" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;EDIT: to make the question a little more specific, what is the ideal strategy for you, that maximizes your chances of getting the car in the worst case, no matter what the hosts motives are.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can't sweep the host's motives under the rug like some small, unimportant detail. The player's strategy is completely dependent on the host's strategy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;HOST STRATEGY 1:&#xA;The host always opens a door containing an animal. In this case the problem is the usual Monty Hall problem, and you should switch doors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HOST STRATEGY 2:&#xA;The host always opens one of the two remaining doors with equal probability. In this case, each door will have a 50% likelihood of having a car. It doesn't matter if you switch or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HOST STRATEGY 3:&#xA;If you pick an animal, the host's strategy is to open the car door 99% of the time, and open the other animal door 1% of the time. If you pick the car, the host opens a random animal door with 50% probability. If he opens a door with an animal, then (from Bayes theorem) the odds are overwhelmingly likely that your door has a car. In this case you should NOT switch doors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is with a sample size of one, and with zero information about the host's strategy, it doesn't matter if you switch or not, since some host strategies favor switching while others favor not switching. If you saw him open one hundred doors, assuming he had the same strategy each time, you could conceivably form a model for his strategy and fit certain parameters to guide your actions in future trials. But on the very first trial, there simply isn't enough information.&lt;/p&gt;&#xA;" OwnerUserId="11857" LastActivityDate="2015-04-28T05:11:10.047" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12837" PostTypeId="2" ParentId="12835" CreationDate="2015-04-28T05:32:06.807" Score="0" Body="&lt;p&gt;Like most of these conundrums, this &quot;paradox&quot; is just a trick of language.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only sense in which &quot;one of them is a girl&quot; means &quot;&lt;em&gt;at least&lt;/em&gt; one of them is a girl&quot; is a strictly mathematical, logical one, and the sense in which &quot;what's the probability both of them are girls&quot; gives an answer of 1/3 is again a strictly mathematical construction which means &quot;out of a randomized trial consisting of selecting between the scenarios where at least one child of mine is a girl with equal probability, what is the expected frequency of the scenario where my children are both girls?&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nobody really talks like that in real life, hence when you hear it first, your brain automatically assumes the mother means your second meaning when she really means the first. When you say &quot;one of them&quot;, you're usually thinking of a specific one, such as &quot;the eldest one&quot; or &quot;the one playing in the garden&quot;. But using the strictly logical sense as above, the answer differs.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Your follow-up question asks, &quot;why does the order matter?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We say that each child has a 50% probability of being a boy or a girl. In order to identify which child &quot;each&quot; child is to calculated the combined probabilities, we need to give them separating labels, which we apply an order to for convenience. Child A being male and Child B being female is a different case from Child A being female and Child B being male, even if they're both cases of &quot;one boy and one girl&quot;.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-05-07T17:04:23.277" LastActivityDate="2015-05-07T17:04:23.277" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12838" PostTypeId="1" AcceptedAnswerId="12897" CreationDate="2015-04-28T05:59:08.717" Score="9" ViewCount="4083" Body="&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Tetrahedron&quot; rel=&quot;noreferrer&quot;&gt;tetrahedron&lt;/a&gt; is a solid shape with four triangular faces, not necessarily regular or identical. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show how to partition a solid cube into 6 tetrahedra that are &lt;a href=&quot;http://en.wikipedia.org/wiki/Congruence_%28geometry%29&quot; rel=&quot;noreferrer&quot;&gt;congruent&lt;/a&gt;, meaning identical up to translation, rotation, and possibly reflection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(It's easy to find solutions by Googling, but please solve it on your own.)&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-04-28T06:12:35.393" LastActivityDate="2019-09-02T18:38:18.587" Title="Partition a cube into 6 congruent tetrahedra" Tags="|geometry|dissection|" AnswerCount="6" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12839" PostTypeId="2" ParentId="12732" CreationDate="2015-04-28T07:34:14.727" Score="4" Body="&lt;p&gt;Great puzzle, thanks for sharing. I didn't manage to complete it though, here are some of my reflexions to add to the previous ones :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Locations&lt;/strong&gt;&#xA;I arranged the passengers by deck and vicinity :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Deck 10 : Chang, Jack, Jerome &amp;amp; Julian&lt;/li&gt;&#xA;&lt;li&gt;Deck 9 : /&lt;/li&gt;&#xA;&lt;li&gt;Deck 8 : !Peleh / the detective&lt;/li&gt;&#xA;&lt;li&gt;Deck 7 : Luke &amp;amp; Leia, Kevin / George &amp;amp; Giselle, Max, Carol&lt;/li&gt;&#xA;&lt;li&gt;Deck 6 : /&lt;/li&gt;&#xA;&lt;li&gt;Deck 5 : W &amp;amp; TF, Brofonomo&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Motives&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Luke : Leia &amp;amp; Kevin are the main suspects, as they seem to have an affair. Kevin seemed to be asleep so he may be innocent. I think Leia, unhappy in her couple, smothered Luke with a pillow (with or without poisoning him with sleeping pills beforehand), and Kevin covered for her. The absence of reaction when she discovered the body of her lover seems to support this idea. I think this murder is disconnected from the others: the process is different (no belt, no gun)&lt;/li&gt;&#xA;&lt;li&gt;Brofonomo : One or several of the gamblers may have taken revenge for their loss. First to die, probably late in the evening, from a unique and well placed shot, without mention of his hands being tied. Probably W &amp;amp; TF as they have the gun and know how to use it. Maybe covered by the 3 Js (why did they go to play in their rooms instead of staying in the casino ?).&lt;/li&gt;&#xA;&lt;li&gt;George and Max : same modus operandi, same time, same place, the murderer seems to be the same, or someone is copying to make it more discreet. The motive seems to be the money, and especially the legacy because the son was murdered too. Carol is the main suspect to me, as she seemed to be hitting on Giselle: she may have killed her husband and son(in law?) in order to marry her and get the legacy? She seemed to have a history of family spree. But somehow I can't wrap my mind around the belt: why would she put it on Max when an accident was so easy to fake?&lt;/li&gt;&#xA;&lt;li&gt;W &amp;amp; TF : here two motives appear: get back something lost in the game (3 Js, because Brofonomo was dead before them) or put their hands on the guns. The 3Js may have the poison, as well as George. Carol may be involved as their hands are tied together like George and Max, and she had interest into having the guns. She could have stolen the poison from George.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Guns movement&lt;/strong&gt;&#xA; I think one important thing to raise is that they are TWO guns, so they may have been used by two different murderers. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;at the beginning of the evening, the guns are with W &amp;amp; TF&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;one is used to kill Brofonomo, before W &amp;amp; TF were dead, but as they were tranquilized they may have been sleeping at this time and the guns stolen without their knowledge&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;one is used to kill W &amp;amp; TF, using several shots : why if they were already sleeping ? The murderer could have taken time to aim.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;they are therefore used to kill Max and George with violence, especially towards George&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From these developments, the following questions arise for which I can't think of an answer, maybe someone else will :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Where was Giselle during the night? Did she sleep with her husband and left early on the morning, or did she never go to the room that night? Did she spend the night with Carol?&lt;/li&gt;&#xA;&lt;li&gt;Has the tranquilizer and/or the sleeping pills been used on anyone else than W &amp;amp; TF, especially the ones who look sleepy (3Js, Kevin) and the dead ones (George, Max - how did the teenager finish dead drunk on his toilet? Who served him this much alcohol? Could this be a result of a drug instead?), maybe so they don't hear anything (the 3Js by Chang for example, as they live on the same deck) or to arrange the murder scene to make believe something else happened (the blindfold for example)?&lt;/li&gt;&#xA;&lt;li&gt;Why were the hands of W &amp;amp; TF as well as those of Max tied, even though they weren't able to fight back? (W &amp;amp; TF were sleeping because of the tranquilizer, and Max passed out because he was too drunk)&lt;/li&gt;&#xA;&lt;li&gt;When a door is &quot;locked&quot;, does this mean it is manually locked from the inside or just slamming the door from the outside is enough for it can't be opened without the key?&lt;/li&gt;&#xA;&lt;li&gt;Are the technical details around the guns (150 meters range, thermal scope, can go through 3 layers of wall) just there for the Sci-Fi touch as there is no mention of a hole in a wall or an open space this wide?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT :&lt;/strong&gt; I gave more thought to the puzzle, even though the first two hints didn't helped much (I had already figured out that). Here is a possible development of what happened that night :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The 3Js, W&amp;amp;TF &amp;amp; Brofonomo are gambling on Deck 5. W &amp;amp; TF accuse Brofonomo of cheating. As the situation degenerates, the 3Js drugs W&amp;amp;TF (they always have tranquilizer in case something goes bad) to protect the professor. But they drink the wine too late and shot him before falling asleep. The 3Js put the doctor in his room, tie W&amp;amp;TF with their belts and wait for the morning to tell the others as they don't want to make a scandal.&lt;/li&gt;&#xA;&lt;li&gt;Carol and Giselle are lovers from the beginning. Giselle is an actress and she married George only for the money. They went on the boat only to get rid of them. Carol has an history of murdering people for the legacy. She goes downstairs on the middle of the night to steal the guns of W&amp;amp;TF, only to find them dead on the floor. She kills them, sees the belt around their wrists and think of copying that to accuse someone else. She then goes upstairs and kill Max (she had previously make him drink a lot, that's why he was in his toilet and his door was open). Giselle, who was &quot;playing&quot; with her husband, opens the door to her and they kill him together. The two spend the rest of the night together and cover for each other. Giselle is very good at acting (she married the man without him seeing anything after all) and make up everything, that's why the detective is surprised with her reaction (he must have seen other crying widows after all).&lt;/li&gt;&#xA;&lt;li&gt;Leia drugs and kill her husband as I said previously, she may also have drugged her lover to make him an alibi.&lt;/li&gt;&#xA;&lt;li&gt;The morning after, he 3Js tell nothing, fearing that they can be accused of the murder of Brofonomo as well as W&amp;amp;TF, because they were the last that saw them alive. Chang covers for them so they don't have to worry.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So that's all I can think of without other hints ! I hope some part is near the truth. And sorry in advance for my poor language.&lt;/p&gt;&#xA;" OwnerUserId="11859" LastEditorUserId="11859" LastEditDate="2015-05-04T12:12:40.420" LastActivityDate="2015-05-04T12:12:40.420" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="12840" PostTypeId="1" AcceptedAnswerId="12844" CreationDate="2015-04-28T08:40:19.873" Score="12" ViewCount="1164" Body="&lt;p&gt;I have taken snippets of lyrics from 12 reasonably well known songs. Each of these snippets contains one or two instances of a whole number in it.&lt;/p&gt;&#xA;&lt;p&gt;For a consistent format, I've written these numbers in their digital form, removed all non-alphanumeric characters besides spaces, and made all letters lowercase. The tricky part is that I then replaced all digits with &lt;code&gt;#&lt;/code&gt; and all consonants (including all instances of y) with &lt;code&gt;_&lt;/code&gt;. Your task is to identify all 12 songs with this limited information.&lt;/p&gt;&#xA;&lt;p&gt;Here are the 12 mangled lyric snippets, one per line:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;# i_ __e _o_e_ie__ _u__e_&#xA;i __o_e # _oi___&#xA;# _i___e _i___&#xA;# _a__ a _ee_&#xA;# i_ __e a__e__oo_&#xA;_ou a_e ## _oi__ o_ ##&#xA;__e_ i_ ##&#xA;## __o__o_e_ _e_ __e _i_ _a_a_e _i__ ### _o__e__ __o_e a_ _a__&#xA;__e_ _ea_ _i_e __ ###&#xA;i _ou__ _a__ ### _i_e_&#xA;__o a_ i #####&#xA;#######&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;It can be seen that only the 6th and 8th songs have two numbers in them, so there are 14 mystery numbers in total. None of the numbers have leading zeros. All the lyric snippets are in standard English and their songs can be easily found on the web.&lt;/p&gt;&#xA;&lt;p&gt;Just to clarify what has been done to the snippets, suppose the lyric &lt;code&gt;A seven nation army couldn't hold me back&lt;/code&gt; from &lt;a href=&quot;http://www.azlyrics.com/lyrics/whitestripes/sevennationarmy.html&quot; rel=&quot;noreferrer&quot;&gt;Seven Nation Army&lt;/a&gt; was one of them (just as an example, this is not one of the 12 songs):&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;make numbers digital → &lt;code&gt;A 7 nation army couldn't hold me back&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;remove non-alphanumerics except space → &lt;code&gt;A 7 nation army couldnt hold me back&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;make lowercase → &lt;code&gt;a 7 nation army couldnt hold me back&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;replace digits with &lt;code&gt;#&lt;/code&gt; → &lt;code&gt;a # nation army couldnt hold me back&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;replace consonants and y's with &lt;code&gt;_&lt;/code&gt; → &lt;code&gt;a # _a_io_ a___ _ou____ _o__ _e _a__&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;h2&gt;Winning&lt;/h2&gt;&#xA;&lt;p&gt;I will accept the earliest answer that correctly identifies all the songs and gives the unobscured lyrics and numbers.&lt;/p&gt;&#xA;&lt;h2&gt;Hints&lt;/h2&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt; &lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The songs are not all necessarily from bands or individuals. Some may be from movies and/or musicals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt; &lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The 14 mystery numbers are strictly increasing in the order they appear in the mangled lyric snippet list.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;(You'll need this one to really be certain of your answer.)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The sum of all 14 mystery numbers is 8701125.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1811" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-03-08T18:04:06.733" Title="Musical Numbers" Tags="|riddle|music|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12841" PostTypeId="2" ParentId="12840" CreationDate="2015-04-28T10:05:14.360" Score="3" Body="&lt;p&gt;This is what I currently have:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One Is The Loneliest Number - Three Dog Night&lt;br&gt;&#xA; I Smoke Two Joints - The Toyes (credit goes to &lt;strong&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/users/5208/alconja&quot;&gt;Alconja&lt;/a&gt;&lt;/strong&gt;)&lt;br&gt;&#xA; Seven Litle Girls - Paul Evans (helped by a hint from &lt;strong&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/users/4652/narmer&quot;&gt;Narmer&lt;/a&gt;&lt;/strong&gt;)&lt;br&gt;&#xA; Eight Days A Week - The Beatles (credit goes to &lt;strong&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/users/5208/alconja&quot;&gt;Alconja&lt;/a&gt;&lt;/strong&gt;)&lt;br&gt;&#xA; Nine In The Afternoon - Panic! At The Disco&lt;br&gt;&#xA; (You Are) Sixteen Going On Seventeen - The Sound of Music&lt;br&gt;&#xA; __e_ i_ 69&#xA; 76 Trombones Led The Big Parade With 110 Cornets Close At Hand - The Music Man (credit goes to &lt;strong&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/users/5208/alconja&quot;&gt;Alconja&lt;/a&gt;&lt;/strong&gt;)&lt;br&gt;&#xA; (I would walk) 500 Miles - Proclaimers&lt;br&gt;&#xA; She's Real Fine My 4-0-9 - Beach Boys (credit goes to &lt;strong&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/users/5208/alconja&quot;&gt;Alconja&lt;/a&gt;&lt;/strong&gt;)&lt;br&gt;&#xA; Who Am I? 24601! - Les Miserables (credit goes to &lt;strong&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/users/4652/narmer&quot;&gt;Narmer&lt;/a&gt;&lt;/strong&gt;)&lt;br&gt;&#xA; 867-5309(/Jenny) - Tommy Tutone&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9309" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-04-28T10:43:48.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12842" PostTypeId="2" ParentId="12840" CreationDate="2015-04-28T10:05:17.147" Score="5" Body="&lt;p&gt;Tidied up:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1) &lt;code&gt;# i_ __e _o_e_ie__ _u__e_&lt;/code&gt; -&gt; One is the loneliest number - &lt;a href=&quot;http://www.metrolyrics.com/one-is-the-loneliest-number-lyrics-three-dog-night.html&quot; rel=&quot;nofollow&quot;&gt;Three dog night&lt;/a&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; 2) &lt;code&gt;i __o_e # _oi___&lt;/code&gt; -&gt; I smoke two joints - &lt;a href=&quot;http://www.azlyrics.com/lyrics/sublime/smoketwojoints.html&quot; rel=&quot;nofollow&quot;&gt;Sublime&lt;/a&gt; &lt;strong&gt;Thanks to @Alconja&lt;/strong&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; 3) &lt;code&gt;# _i___e _i___&lt;/code&gt; -&gt; Seven little girls - &lt;a href=&quot;http://www.songlyrics.com/paul-evans-and-the-curls/seven-little-girls-sitting-in-the-back-seat-lyrics/&quot; rel=&quot;nofollow&quot;&gt;Paul evans &lt;/a&gt; &lt;strong&gt;Thanks to @Somebody&lt;/strong&gt;&lt;br&gt;&#xA; (Intended one -&gt; Three little birds - &lt;a href=&quot;http://www.metrolyrics.com/three-little-birds-lyrics-bob-marley.html&quot; rel=&quot;nofollow&quot;&gt;Bob Marley &amp;amp; The Wailers&lt;/a&gt;)&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; 4) &lt;code&gt;# _a__ a _ee_&lt;/code&gt;-&gt; Eight days a week - &lt;a href=&quot;http://www.lyricsfreak.com/b/beatles/eight+days+a+week_10026469.html&quot; rel=&quot;nofollow&quot;&gt;Beatles&lt;/a&gt; &lt;strong&gt;Thanks to @Alconja&lt;/strong&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; 5) &lt;code&gt;# i_ __e a__e__oo_&lt;/code&gt; -&gt; Nine in the afternoon - &lt;a href=&quot;http://www.azlyrics.com/lyrics/panicatthedisco/nineintheafternoon.html&quot; rel=&quot;nofollow&quot;&gt;PANIC! AT THE DISCO&lt;/a&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; 6) &lt;code&gt;_ou a_e ## _oi__ o_ ##&lt;/code&gt; -&gt; You are sixteen going on seventeen - &lt;a href=&quot;http://www.stlyrics.com/songs/s/soundofmusic9841/sixteengoingonseventeen323515.html&quot; rel=&quot;nofollow&quot;&gt;Sound of music&lt;/a&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; 7) &lt;code&gt;__e_ i_ ##&lt;/code&gt; -&gt; When I'm 60 &lt;a href=&quot;http://www.azlyrics.com/lyrics/eltonjohn/sixtyyearson.html&quot; rel=&quot;nofollow&quot;&gt;Elton Jhon&lt;/a&gt; &lt;strong&gt;Thanks to @Alconja&lt;/strong&gt;&lt;br&gt;&#xA; (Intended one -&gt; When I'm 64 &lt;a href=&quot;http://www.metrolyrics.com/when-im-64-lyrics-beatles.html&quot; rel=&quot;nofollow&quot;&gt;Beatles&lt;/a&gt;)&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; 8) &lt;code&gt;## __o__o_e_ _e_ __e _i_ _a_a_e _i__ ### _o__e__ __o_e a_ _a__&lt;/code&gt;-&gt; 76 trombones led the big parade with 110 cornets close at hand - &lt;a href=&quot;http://www.lyricsfreak.com/m/meredith+willson/76+trombones_20610196.html&quot; rel=&quot;nofollow&quot;&gt;Willson Meredith&lt;/a&gt; &lt;strong&gt;Thanks to @Alconja&lt;/strong&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; 9) &lt;code&gt;__e_ _ea_ _i_e __ ###&lt;/code&gt; -&gt; Shes real fine my 4-0-9 - &lt;a href=&quot;http://www.lyricsfreak.com/b/beach+boys/409_20013984.html&quot; rel=&quot;nofollow&quot;&gt;Beach Boys&lt;/a&gt; &lt;strong&gt;Thanks to @Alconja&lt;/strong&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; 10) &lt;code&gt;i _ou__ _a__ ### _i_e_&lt;/code&gt; -&gt; (I would walk) 500 Miles - &lt;a href=&quot;http://www.lyricsfreak.com/p/proclaimers/500+miles_20458141.html&quot; rel=&quot;nofollow&quot;&gt;Proclaimers&lt;/a&gt; &lt;strong&gt;Thanks to @Somebody&lt;/strong&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; 11) &lt;code&gt;__o a_ i #####&lt;/code&gt; -&gt; Who am I? 24601! - &lt;a href=&quot;http://www.metrolyrics.com/who-am-i-lyrics-les-miserables.html&quot; rel=&quot;nofollow&quot;&gt;Les Miserables&lt;/a&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; 12) &lt;code&gt;#######&lt;/code&gt; -&gt; 867-5309/Jenny - &lt;a href=&quot;http://www.azlyrics.com/lyrics/tommytutone/8675309jenny.html&quot; rel=&quot;nofollow&quot;&gt;Tommy Tutone&lt;/a&gt; &lt;strong&gt;Thanks to @Alconja&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The numbers are (ascending order):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $1 + 2 + 7(3) + 8 + 9 + 16 + 17 + 60(64) + 76 + 110 + 409 + 500 + 24601 + 8675309 = 8701125$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4652" LastEditorUserId="4652" LastEditDate="2015-04-28T11:11:23.443" LastActivityDate="2015-04-28T11:11:23.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12843" PostTypeId="2" ParentId="12761" CreationDate="2015-04-28T10:08:35.767" Score="-2" Body="&lt;p&gt;You should swap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you have three to choose from, you have a $\frac{1}{3}$ chance of picking the car. Any event that happens after that point can't change the chance of you being correct initially.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When one of the other doors is revealed to be a sheep, you're still just as likely to have won. So switching to the remaining door has a $\frac23$ chance of winning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And even if the chance is 50/50, there's no harm in swapping.&lt;/p&gt;&#xA;" OwnerUserId="4577" LastActivityDate="2015-04-28T10:08:35.767" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12844" PostTypeId="2" ParentId="12840" CreationDate="2015-04-28T10:19:16.777" Score="4" Body="&lt;p&gt;Think I got them all now:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 is the loneliest number&lt;br&gt;&#xA; i smoke 2 joints&lt;br&gt;&#xA; 7 little girls (credit to &lt;a href=&quot;https://puzzling.stackexchange.com/a/12841/5208&quot;&gt;somebody&lt;/a&gt;) &lt;em&gt;-or-&lt;/em&gt; 3 little birds (credit to Phil M Jones)&lt;br&gt;&#xA; 8 days a week&lt;br&gt;&#xA; 9 in the afternoon&lt;br&gt;&#xA; you are 16 going on 17&lt;br&gt;&#xA; when im 60 &lt;em&gt;-or-&lt;/em&gt; when im 64&lt;br&gt;&#xA; 76 trombones led the big parade with 110 cornets close at hand&lt;br&gt;&#xA; shes real fine my 409&lt;br&gt;&#xA; i would walk 500 miles&lt;br&gt;&#xA; who am i 24601&lt;br&gt;&#xA; 8675309  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-04-28T11:31:05.170" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12845" PostTypeId="2" ParentId="12770" CreationDate="2015-04-28T10:23:52.533" Score="1" Body="&lt;p&gt;Combining the answer of @Collett89 and the comments of @user2357112, I get:&lt;/p&gt;&#xA;&lt;h2&gt;45 prisoners used, 33 died&lt;/h2&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;The first day&lt;/h1&gt;&#xA;&lt;p&gt;On the first day, have 40 prisoners drink from 525 bottles each, with each prisoner shifting by 25 bottles.  In other words, prisoner 0 drinks from 0..524, prisoner 1 drinks from 25..549, etc, with the range wrapping around after 999 (so prisoner 20 drinks from 500..999 and 0..24).  There are two cases that result.&lt;/p&gt;&#xA;&lt;h1&gt;The second day, case 1&lt;/h1&gt;&#xA;&lt;p&gt;Most of the time, there will be 2 groups of 25 bottles that are suspect, with poison A in one group and poison B in the other group.  Depending on where the two bottles were located, between 2 and 20 prisoners died.  So at least 20 prisoners survived and can be reused.  To find the two specific bottles requires 10 prisoners.  It takes 5 prisoners to find each bottle.&lt;/p&gt;&#xA;&lt;p&gt;To find the one poison bottle in a group of 25, you have 5 prisoners drink from all of the bottles in the other group (some of the survivors will have already done this).  Then each of the 5 prisoners drinks from bottles that have a particular bit is set in the binary representation of the bottle.  So, prisoner 0 drinks from bottles where bit 0 is set (1, 3, 5, 7, ... 25).  Prisoner 1 drinks from bottles where bit 1 is set (2, 3, 6, 7, 10, 11, ... 23).  At the end, the dead prisoners make up the bits of the bottle number.  So if prisoners 0, 2, and 3 died, the bottle number would be &lt;code&gt;01101&lt;/code&gt; or 13.  To avoid the worst case of 4 prisoners dying, number the bottles such that the bottles with 4 bits are skipped (i.e. don't use 15, 23, 27, 29, 30, 31).  At worst, 3 prisoners die per bottle found.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Final result, case 1: 40 prisoners used, 26 died&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;h1&gt;The second day, case 2&lt;/h1&gt;&#xA;&lt;p&gt;The other possibility is that both poisoned bottles will be in the same group of 25 bottles.  This means 21 prisoners will have died on the first day.  Now, we do the same sort of thing as on the first day.  We need 24 prisoners, and each will drink from 13 bottles, shifted by one.  This will kill between 0 and 12 prisoners.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Final result, case 2: 45 prisoners used, 33 died&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;By the way, I tried with other numbers such as 32 prisoners dividing the bottles into groups of 32.  But 40/25 was the best split given that the worst case scenario was to find both poison bottles in the same group.&lt;/p&gt;&#xA;" OwnerUserId="4479" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-28T10:23:52.533" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12846" PostTypeId="2" ParentId="12770" CreationDate="2015-04-28T10:44:54.200" Score="17" Body="&lt;h2&gt;On the first day&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Use 10 groups of 2 prisoner and number them 0.0, 0.1, 1.0, 1.1, 2.0, ..., 8.1, 9.0, 9.1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prisoner a.b drinks from bottles with $\left\lfloor\frac{index}{2^a}\right\rfloor\%2=b$.&#xA;That means every bottle where the ath bit is b.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That gives us some groups where one prisoner survives and some groups where both prisoner survives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a random group m with two survivor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prisoner m.0 has drunk exactly one poison and m.1 has drunk the other poison. If that would not be the case one of them would be dead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets call the poison drunk by m.0 $x$ and call the other poison drunk by m.1 $y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We define $x_i=\left\lfloor\frac{x}{2^i}\right\rfloor\%2$ and $y_i=\left\lfloor\frac{y}{2^i}\right\rfloor\%2$.&#xA;That means $x_i$ and $y_i$ are the ith bit of $x$ and $y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prisoner a.0 dies if $x_a=0$ and $y_a=0$ and prisoner a.1 dies if $x_a=1$ and $y_a=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means both survive if $x_a XOR y_a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We define $s=\sum_{i=0}^92^i \times \begin{cases} 1 &amp;amp; \text{if both prisoner survive} \\ 0 &amp;amp; \text{if one prisoner survive}\end{cases}$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Corollary 1: $s = x XOR y$.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Proof: $s=\sum_{i=0}^92^i \times \begin{cases} 1 &amp;amp; \text{if both prisoner survive }\\ 0 &amp;amp; \text{if one prisoner survive} \end{cases}=\sum_{i=0}^9 2^i\times (x_i XOR y_i) = x XOR y$.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;On the second day&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;All prisoner from a group with two survivor have to drink from the bottles that prisoner m.1 has drunk on the first day.&#xA;That will kill one prisoner of each group.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Corollary 2: Nobody drinks poison $x$ on the second day.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Proof: Only the stuff of m.1 is drunk on the second day and m.1 has drunk y (by definition) and survived.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 4 cases for each group:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a.0 dies on the first day that means $x_a=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a.1 dies on the first day that means $x_a=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a.0 dies on the second day that means he has drunk $x$ on the first day (because of corollary 2). That means $x_a=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a.1 dies on the second day that means he has drunk $x$ on the first day (because of corollary 2). That means $x_a=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows a.1 dies if $x_a=1$.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Result&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;$x=\sum_{i=0}^9 2^i \times x_a=\sum_{i=0}^9 2^i \times \begin{cases} 1 &amp;amp; \text{if prisoner i.1 died} \\ 0 &amp;amp; \text{if prisoner i.0 died} \end{cases}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And with corollary 1: The other poison is $y=x XOR x XOR y = x XOR s$.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;20 Prisoners used and 10 survived.&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;That can be improved to 20 Prisoners used and 11 survived by not forcing $m.0$ to drink stuff where we know that it will kill him.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Update: New result is 18 Prisoners used and 8 survived.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The algorithm is the same as above but prisoner 0.0 and 0.1 are not used and instead m.0 and m.1 do their job. Those hadn't created any information on the second day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In detail:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First day is the same as the result before but leave out prisoner 0.0 and 0.1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If half of the prisoner die on the first day it is clear where both poisons are (all bits except the last are equal and one bit can be read out of each group)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Else we can find a m where both prisoner of m survive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On day 2 in all other groups where both survive they drink from the bottles that were drunk from by m.1 (Same as the result before).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;m.0 and m.1 will drink from the that 0.1 would drink from.&lt;br/&gt;&#xA;If m.0 survives then the last bit of y is 0 if he dies the last bit of y is 1.&lt;br/&gt;&#xA;If m.1 survives then the last bit of x is 0 if he dies the last bit of x is 1.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Example&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Example with 16 bottles and poison in 1 and 10:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First day:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.0:  0011001100110011 --&gt; survive&lt;br/&gt;&#xA;1.1:  1100110011001100 --&gt; survive&lt;br/&gt;&#xA;2.0:  0000111100001111 --&gt; survive&lt;br/&gt;&#xA;2.1:  1111000011110000 --&gt; die&lt;br/&gt;&#xA;3.0:  0000000011111111 --&gt; survive&lt;br/&gt;&#xA;3.1:  1111111100000000 --&gt; survive&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we know all bits where on prisoner died and in each group where both survive we know that the bit is different.&#xA;In my example the bit 2 is 0 and bits 1 and 3 are different. We have no knowledge about bit 0.&#xA;The possible answers are: x=000?, y=101? or x=001?, y=100? or x=100?, y=001? or x=101?, y=000?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In pair 1 and 3 both survive and we choose 1 as m.&#xA;That means that the poison with bit 1 is called x and the other one is y.&#xA;The possible answers are: x=001?, y=100? or x=101?, y=000?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second day:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means the second day there will be the following drinking:&lt;br/&gt;&#xA;1.0:  0111011101110111 --&gt; die          ==&gt; the last bit of y is 1&lt;br/&gt;&#xA;1.1:  1101110111011101 --&gt; survive      ==&gt; the last bit of x is 0&lt;br/&gt;&#xA;2.0:  (doesn't drink)&#xA;&lt;br/&gt;&#xA;2.1:  (already dead)   --&gt; died 1st day ==&gt; bit 2 of x is 0 and bit 2 of y is 0&lt;br/&gt;&#xA;3.0:  1100110011111111 --&gt; died 2nd day ==&gt; bit 3 of x is 1 and bit 3 of y is 0&lt;br/&gt;&#xA;3.1:  1111111111001100 --&gt; survive &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Result is x = 1010 and y = 0001&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Bottle count that can be tested with n prisoner&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I experimented with other bases and found an optimum with checking base 3 digits with 3 prisoner instead of base 2 digits with 2 prisoner. That needs $3log_3(n)+O(1)\approx 2.731ln(n)+O(1)$ prisoner instead of $2log_2(n)+O(1)\approx 2.885ln(n)+O(1)$ prisoner. But that doesn't decreases the number of needed prisoner here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is that with $3*n$ prisoner you can test $2*3^n$ bottles,&lt;br/&gt;&#xA;with $3*n+1$ prisoner you can test $3*3^n$ bottles &lt;br/&gt;&#xA;and with $3*n+2$ prisoner you can test $4*3^n$ bottles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means 17 prisoner can check 972 bottles (a few short for it to be a solution to this question) and  20 prisoner can check 2916 bottles&lt;/p&gt;&#xA;" OwnerUserId="10143" LastEditorUserId="10143" LastEditDate="2015-04-29T18:24:16.703" LastActivityDate="2015-04-29T18:24:16.703" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12847" PostTypeId="2" ParentId="12798" CreationDate="2015-04-28T10:49:48.890" Score="4" Body="&lt;h2&gt;Too ambiguous&lt;/h2&gt;&#xA;&lt;p&gt;I think this question is ambiguous because it doesn't specify the motivation for each woman to make her particular statement.  For example, why does woman number one volunteer that one of the children is a girl?  If this were the situation:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Questioner: Think of one of your children and tell me the gender.&lt;br /&gt;&#xA;Woman: One of the children is a girl.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Then the answer would be 1/2 because it would be equivalent to the &amp;quot;eldest girl&amp;quot; response.&lt;/p&gt;&#xA;&lt;p&gt;But if this were the situation:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Questioner: Is at least one of your children a girl?&lt;br /&gt;&#xA;Woman: One of the children is a girl.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Then the answer would be 1/3.&lt;/p&gt;&#xA;&lt;h2&gt;Rewording the questions&lt;/h2&gt;&#xA;&lt;p&gt;You could remove the ambiguity by changing the scenario to be the following:&lt;/p&gt;&#xA;&lt;p&gt;You meet three women, each of which has exactly 2 children.  You ask questions which each woman honestly answers.  For each woman, what are the chances that both of her children are girls?&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Woman 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;You: Is at least one of your children a girl?&lt;br /&gt;&#xA;Woman 1: Yes, at least one of my children is a girl.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Woman 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;You: Is your eldest child a girl?&lt;br /&gt;&#xA;Woman 2: Yes, my eldest child is a girl.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Woman 3:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;You: Is at least one of your children a girl?&lt;br /&gt;&#xA;Woman 3: Yes, at least one of my children is a girl.&lt;br /&gt;&#xA;You: Tell me specifically one that is a girl, eldest or youngest? If you have two girls, flip a fair coin to decide which to tell me.&lt;br /&gt;&#xA;Woman 3: Eldest.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Now with the situation clarified, the answers should be:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 1/3, 1/2, 1/3.  For Woman 3, the second question doesn't actually add any information because she definitely has a daughter, and so her saying &amp;quot;eldest&amp;quot; or &amp;quot;youngest&amp;quot; doesn't change anything.  It does collapse the possibilities to MF and FF, but in the MF case, she was forced to say &amp;quot;eldest&amp;quot; with 100% probability whereas for the FF case, she had a 50/50 chance to say either youngest or eldest.  So even though the FM MF and FF cases were each at 1/3 before the second question, the MF case is now twice as likely as the FF case due to the 100% vs 50% response.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4479" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-28T12:01:01.737" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12848" PostTypeId="2" ParentId="12761" CreationDate="2015-04-28T11:07:18.357" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;EDIT: to make the question a little more specific, what is the ideal strategy for you, that maximizes your chances of getting the car &lt;strong&gt;in the worst case, no matter what the hosts motives are&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I understood your question correctly, &lt;em&gt;&quot;in the worst case&quot;&lt;/em&gt; is the key here. So the correct answer is a strategy where you maximize your winning chance over all possible host's strategies. Considering worst case is identical to considering the situation where the host knows your strategy and wants you not to win.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If your strategy is &quot;never change&quot;, your probability of winning is $1/3$ no matter what the host's strategy is. Because you have a strategy with result $1/3$, &lt;strong&gt;your result with the optimal strategy is at least $1/3$&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;If the host opens a door with a goat and offers you a possibility to change only if you chose the car in the first try, you can win only if you choose the car in the first try, so your probability of winning is at most $1/3$. So whatever your strategy is, the host can limit your winning probability to $1/3$. So &lt;strong&gt;your result with the optimal strategy is at most $1/3$&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Therefore, your chances of winning are exactly $1/3$ with optimal strategy, and this can be achieved by never changing.&lt;/p&gt;&#xA;" OwnerUserId="2228" LastActivityDate="2015-04-28T11:07:18.357" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12849" PostTypeId="2" ParentId="12693" CreationDate="2015-04-28T11:20:53.390" Score="1" Body="&lt;p&gt;He's in Nashville, TN&lt;br&gt;&#xA;Because they are travelling without stop and there will be tollbooth between states which means they should stop!&#xA;According to internet travelling time from Chattanooga, TN to Nashville, TN takes 2 hr 10 mins and 1 hr 40 mins for Knoxville, TN&#xA;(I'm from different country so idk the travelling time)&lt;br&gt;&#xA;From the above given time he is probably in Nashville, TN ( 2hrs travelling + they say they will reach the warehouse in 5 mins = 2hrs 5mins)&lt;br&gt;&#xA;30 mins pizza delivery time. Total of 2:35 mins 7:45pm-5pm 2:45 (5 mins act of capturing, 5 mins moving him from van to warehouse)&lt;/p&gt;&#xA;" OwnerUserId="11866" LastActivityDate="2015-04-28T11:20:53.390" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12850" PostTypeId="1" CreationDate="2015-04-28T13:30:36.570" Score="-2" ViewCount="2647" Body="&lt;p&gt;What I am trying to do is calculate the row and column of which an id is in. For example I have the grid &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0  1  2  3&#xA;4  5  6  7&#xA;8  9  10 11&#xA;12 13 14 15&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As you can see this is a 4x4 grid. I work out the row number by getting the id then dividing by the number of columns, and rounding this number down. For example rownumber = 5(id value)/4(number of columns) = 1.25 then round down = 1. But what I can't figure out is how to get the column number of say the number 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help?&lt;/p&gt;&#xA;" OwnerUserId="11871" LastEditorUserId="5373" LastEditDate="2015-04-28T19:53:36.477" LastActivityDate="2015-04-28T19:53:36.477" Title="Calculating row number and column number" Tags="|mathematics|calculation-puzzle|" AnswerCount="1" CommentCount="1" ClosedDate="2015-04-28T17:29:54.537" ContentLicense="CC BY-SA 3.0" />
  <row Id="12851" PostTypeId="1" AcceptedAnswerId="12898" CreationDate="2015-04-28T13:46:14.083" Score="7" ViewCount="793" Body="&lt;p&gt;Journey far west in the US of A&lt;BR&gt;&#xA;Err to the south a bit, too&lt;BR&gt;&#xA;True, you can't stay here (it's too rich for most)&lt;BR&gt;&#xA;For sure, only the wealthiest do&lt;BR&gt;&#xA;&lt;BR&gt;&#xA;Usually, I solo; few sing the same&lt;BR&gt;&#xA;Everyone says, though, I'm always alone&lt;BR&gt;&#xA;Like some of my kind, I have few worthy partners&lt;BR&gt;&#xA;Clever are those who know&lt;BR&gt;&#xA;&lt;BR&gt;&#xA;As a warrior I supported the war&lt;BR&gt;&#xA;Never did they stand up to me&lt;BR&gt;&#xA;Too violent, was I, and hurt many of mine&lt;BR&gt;&#xA;Man, though, did I clear those trees&lt;BR&gt;&#xA;&lt;BR&gt;&#xA;Emperor am I, or so it seems&lt;BR&gt;&#xA;Legally, I've just been Freed&lt;BR&gt;&#xA;Truly, consume me, I will not mind&lt;BR&gt;&#xA;Such is my will inclined&lt;BR&gt;&#xA;&lt;BR&gt;&#xA;Too often compared to another, am I&lt;BR&gt;&#xA;Even though we are mostly the same&lt;BR&gt;&#xA;Everyone sees only differences, but&lt;BR&gt;&#xA;Largely, we can be interchanged&lt;BR&gt;&#xA;&lt;BR&gt;&#xA;Born of sunbeams and fire&lt;BR&gt;&#xA;Evoking warm feelings in most&lt;BR&gt;&#xA;&lt;BR&gt;&#xA;Alone I fulfill some of this riddle&lt;BR&gt;&#xA;Mixed with another, the rest&lt;BR&gt;&#xA;So, can you name me and my partners above?&lt;BR&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clarification to avoid the dangers of being too broad:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each stanza except the last one has a specific answer that is different from the rest of the stanzas. Those answers will have a common word. That common word is the answer for the last stanza and the overall puzzle. (This is what I have tried to hint at in the last stanza but it might not have been clear enough.)&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2015-04-28T18:59:23.357" LastActivityDate="2015-04-29T17:59:23.067" Title="Born of Sunbeams and Fire" Tags="|riddle|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12853" PostTypeId="2" ParentId="12850" CreationDate="2015-04-28T13:55:51.270" Score="2" Body="&lt;p&gt;If $n$ is the number in the table&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the row is so calculated:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $n/4$ rounded to the lower integer if decimal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;while the column is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $n \bmod(4)$ where $\bmod(4)$ means the remainder of the division by 4.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10399" LastEditDate="2015-04-28T13:59:01.090" LastActivityDate="2015-04-28T13:59:01.090" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12855" PostTypeId="2" ParentId="12694" CreationDate="2015-04-28T14:12:40.900" Score="2" Body="&lt;p&gt;Could it be:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Computer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you use me, you only ever see what I have become instead of what I really am.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Computer in nowadays do more things than just 'compute'  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be used to do powerful things, but I can be very fragile at my core.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Powerful things of course, but can be destroy with ease  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Through time, I have changed lives!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Live style or medicial equipment&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately some choose to hate me, and others choose to embrace me.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I dun know this one, hate can be subjective.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am used around the world in many different ways.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; sensor, phone, watch, laptop to name a few&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My family is huge, and yet still getting bigger.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No doubt  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is important not to forget about my elders, as they still needed (by some).  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Old machine that some people are still using &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;IMHO it is a little too board as I could name a few others that fit into the riddle.  I come up with this answer after seeing the hint &quot;after WWI&quot;&lt;/p&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2015-04-28T14:12:40.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12857" PostTypeId="1" AcceptedAnswerId="12858" CreationDate="2015-04-28T14:29:40.147" Score="12" ViewCount="5098" Body="&lt;p&gt;I may sometimes be rather dry&lt;br&gt;&#xA;But you prefer moist and hot.&lt;br&gt;&#xA;The flesh is soft, you must apply&lt;br&gt;&#xA;Just the right touch, don't get caught.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I must be pierced to bring you joy&lt;br&gt;&#xA;My smell is known and often craved&lt;br&gt;&#xA;But you should know I'm not a toy,&lt;br&gt;&#xA;And there are many I have enslaved.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what am I? Oh no, not that.&lt;br&gt;&#xA;Have I pulled you into my trap?&lt;br&gt;&#xA;I'm not at all related to a cat.&lt;br&gt;&#xA;And found nowhere near a lap.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some think me rare, or maybe not.&lt;br&gt;&#xA;Can you rise above the cut?&lt;br&gt;&#xA;Again I say, some like it hot.&lt;br&gt;&#xA;What am I? Follow your gut.   &lt;/p&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2021-05-04T19:24:52.993" Title="You dirty Mind - What am I?" Tags="|riddle|word|rhyme|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12858" PostTypeId="2" ParentId="12857" CreationDate="2015-04-28T14:37:53.100" Score="5" Body="&lt;p&gt;The answer is  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Meat&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I may sometimes be rather dry&lt;br&gt;&#xA;But you prefer moist and hot.&lt;br&gt;&#xA;The flesh is soft, you must apply&lt;br&gt;&#xA;Just the right touch, don't get caught.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dry meat exists, but it's usually not so tasty. The flesh is synonym of meat, which is not that hard to slice, if you use the knife properly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I must be pierced to bring you joy&lt;br&gt;&#xA;My smell is known and often craved&lt;br&gt;&#xA;But you should know I'm not a toy,&lt;br&gt;&#xA;And there are many I have enslaved.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pierced with fork to be eaten. The smell of meat is great, also food isn't a toy, don't waste it! The enslaved may be the animals used to produce meat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now what am I? Oh no, not that.&lt;br&gt;&#xA;Have I pulled you into my trap?&lt;br&gt;&#xA;I'm not at all related to a cat.&lt;br&gt;&#xA;And found nowhere near a lap.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  The trap might be Vegetarianism. Pussy and cat are synonyms. I don't want to say more...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some think me rare, or maybe not.&lt;br&gt;&#xA;Can you rise above the cut?&lt;br&gt;&#xA;Again I say, some like it hot.&lt;br&gt;&#xA;What am I? Follow your gut.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Poor people can't afford and consider it rare.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-04-28T14:50:04.377" LastActivityDate="2015-04-28T14:50:04.377" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="12860" PostTypeId="1" AcceptedAnswerId="12861" CreationDate="2015-04-28T15:52:40.573" Score="12" ViewCount="1203" Body="&lt;p&gt;What is it :&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Colloquially, it is a &lt;strong&gt;Diabolic&lt;/strong&gt; way to win at athletic events.&lt;/p&gt;&#xA;&lt;p&gt;Take &lt;strong&gt;a single one&lt;/strong&gt;, and it might be shaped like a star.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10940" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-09T17:02:41.317" Title="Diabolic Substance" Tags="|riddle|word|lateral-thinking|word-problem|rebus|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12861" PostTypeId="2" ParentId="12860" CreationDate="2015-04-28T16:47:15.717" Score="14" Body="&lt;p&gt;I believe the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Anabolic steroid(s)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you have two, they are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Di-abolic steroids&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you have one, it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A steroid - ie. Asteroid, meaning starlike.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9278" LastActivityDate="2015-04-28T16:47:15.717" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12862" PostTypeId="2" ParentId="12814" CreationDate="2015-04-28T16:55:21.457" Score="6" Body="&lt;p&gt;Got it! The code is based on:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the 25-pair wire code and it says &quot;WIRE (why are) YOU CROSS&quot;.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Going clockwise from 1 o'clock, the wire pair sequence is:&lt;br&gt;&#xA; 23, 9, 18, 5, 25, 15, 21, 3, 18, 15, 19, 19&lt;br&gt;&#xA; which is then used to select letters of the alphabet  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="5188" LastEditDate="2015-04-28T17:35:54.960" LastActivityDate="2015-04-28T17:35:54.960" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12863" PostTypeId="2" ParentId="12851" CreationDate="2015-04-28T17:16:21.027" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;To me it clearly evokes California and it's regions. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;--------------------------------------------------&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Journey far west in the US of A&#xA;Err to the south a bit, too&#xA;True, you can't stay here (it's too rich for most)&#xA;For sure, only the wealthiest do&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sillicon Valley or any other really rich California locale (Malibu, etc...)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Usually, I solo; few sing the same&#xA;Everyone says, though, I'm always alone&#xA;Like some of my kind, I have few worthy partners&#xA;Clever are those who know&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Unsure&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a warrior I supported the war&#xA;Never did they stand up to me&#xA;Too violent, was I, and hurt many of mine&#xA;Man, though, did I clear those trees&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Redwood/Sequoias/Douglas fir forests in the north, conflicts in it's history including those with the spaniards and native americans&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Emperor am I, or so it seems&#xA;Legally, I've just been Freed&#xA;Truly, consume me, I will not mind&#xA;Such is my will inclined&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Inland Empire and Marijuana Legalization&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Too often compared to another, am I&#xA;Even though we are mostly the same&#xA;Everyone sees only differences, but&#xA;Largely, we can be interchanged&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure of this one either but could be a bunch of stuff, one of the most hilarious in my opinion would be the comparison of Hollywood and Bollywood&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Born of sunbeams and fire&#xA;Evoking warm feelings in most&#xA;Alone I fulfill some of this riddle&#xA;Mixed with another, the rest&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Part of the pacific ring of fire, also it is the sunshine state&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastActivityDate="2015-04-28T17:16:21.027" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12864" PostTypeId="1" CreationDate="2015-04-28T17:30:23.957" Score="36" ViewCount="5665" Body="&lt;p&gt;I've just taken a CS exam, which had the following question&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;6 persons entered a library the day a book got stolen. Each of them entered the library once and only once, and stayed there for some time, then left. If two persons were in the library at the same time, at least one of the two saw the other. After the investigation, the testimonies were&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Albert stated he saw Bernard and Édouard&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Bernard said he saw Albert and Isabelle&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Charlotte said she saw Didier and Isabelle&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Didier said he saw Albert and Isabelle&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Édouard said he saw Bernard and Charlotte&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Isabelle said she saw Charlotte and Édouard&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Only the culprit is lying. Who is he/she ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;EDIT: I haven't had the time yet to check everybody's answers and claims.&lt;/p&gt;&#xA;&lt;p&gt;To those saying the problem is ill-posed, I've actually translated the exact wording from the exam (Question IV.C &lt;a href=&quot;https://www.concours-centrale-supelec.fr/CentraleSupelec/2015/MP/sujets/2014-033.pdf&quot; rel=&quot;noreferrer&quot;&gt;https://www.concours-centrale-supelec.fr/CentraleSupelec/2015/MP/sujets/2014-033.pdf&lt;/a&gt;). As Meelo suggested, this can be solved swiftly with interval graphs.&lt;/p&gt;&#xA;" OwnerUserId="1636" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-02-09T16:37:07.477" Title="Find the liar in the library" Tags="|logical-deduction|" AnswerCount="12" CommentCount="30" ContentLicense="CC BY-SA 3.0" />
  <row Id="12865" PostTypeId="2" ParentId="12851" CreationDate="2015-04-28T17:35:28.930" Score="1" Body="&lt;p&gt;Inspired by spacemonkey's answer, I say  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Marijuana, Cannabis&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Journey far west in the US of A&lt;br&gt;&#xA;Err to the south a bit, too&lt;br&gt;&#xA;True, you can't stay here (it's too rich for most)&lt;br&gt;&#xA;For sure, only the wealthiest do  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the south-west coast of US Cannabis isn't completely legalized. California, which is in the South-West coast, is a very rich State. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Usually, I solo; few sing the same&lt;br&gt;&#xA;Everyone says, though, I'm always alone&lt;br&gt;&#xA;Like some of my kind, I have few worthy partners&lt;br&gt;&#xA;Clever are those who know  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm not an expert of drugs, but I think that cannabis is consumed alone, without other drugs. Similar drugs (partners) can be all the smokable weeds. Sing might be a reference to artists who used to smoke cannabis, like Bob Marley and many others. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a warrior I supported the war&#xA;Never did they stand up to me&#xA;Too violent, was I, and hurt many of mine&#xA;Man, though, did I clear those trees&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; During wars, cannabis is used to placate the soldiers and to distract them from the violence. Many trees were cut down to make room for marijuana plantations. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Emperor am I, or so it seems&#xA;Legally, I've just been Freed&#xA;Truly, consume me, I will not mind&#xA;Such is my will inclined&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many countries recently legalized Cannabis. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Too often compared to another, am I&#xA;Even though we are mostly the same&#xA;Everyone sees only differences, but&#xA;Largely, we can be interchanged&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I don't know what Marijuana is compared to, maybe Hashish? The origin is indeed the same, probably the effects are similar as well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Born of sunbeams and fire&#xA;Evoking warm feelings in most&#xA;Alone I fulfill some of this riddle&#xA;Mixed with another, the rest&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sun makes the plants grow. Cannabis can stimulate memories sometimes. One (Marijuana) mixed with the other (Hashish) can explain the lines of this riddle pretty well, imo. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-04-28T17:35:28.930" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12866" PostTypeId="2" ParentId="12366" CreationDate="2015-04-28T17:51:56.510" Score="0" Body="&lt;p&gt;The answers are 3457,3569 &amp;amp; 3578. Calculated using brute force:&#xA;&lt;a href=&quot;http://ideone.com/yD3RXL&quot; rel=&quot;nofollow&quot;&gt;http://ideone.com/yD3RXL&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11343" LastActivityDate="2015-04-28T17:51:56.510" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12867" PostTypeId="2" ParentId="12694" CreationDate="2015-04-28T18:04:02.080" Score="1" Body="&lt;p&gt;Going along the lines of Narmer, I can argue this fits, but a bit of a stretch on hint 2....&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A constitution or other written laws  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you use me, you only ever see what I have become instead of what I really am.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We see what written laws have become when we use them, but we don't see the actual document.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be used to do powerful things, but I can be very fragile at my core.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some laws are very old, and very fragile, but have established nations or started wars.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Through time, I have changed lives!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many people have been changed by different laws.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately some choose to hate me, and others choose to embrace me.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Anarchists, rebels, warlords, maybe even government leaders seem to hate laws, while police and respectable citizens often embrace and follow them.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am used around the world in many different ways.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are many different laws around the world depending on the government.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My family is huge, and yet still getting bigger.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lots of current laws, and governmental bodies are always making more.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is important not to forget about my elders, as they are still needed (by some).  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some older laws are still in place, but many old laws have been addapted to modern times. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hints:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;em&gt;You can look at me, but not touch.&lt;/em&gt; We can see law in action, but hard to touch it.&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;em&gt;I'm not as old as world war 1.&lt;/em&gt; Some governments and laws were established at the end of WWI&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;em&gt;Some that have used me, are now amongst the richest people in the world!&lt;/em&gt; Some have abused governments, laws, and corruption to become very rich.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-04-28T18:04:02.080" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12869" PostTypeId="2" ParentId="12864" CreationDate="2015-04-28T18:26:30.587" Score="0" Body="&lt;p&gt;These are the only absolute statements I've arrived to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following are never inside the library at the same time::&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A and C&lt;br&gt;&#xA;A and I&lt;br&gt;&#xA;B and C&lt;br&gt;&#xA;B and D&lt;br&gt;&#xA;D and E  &lt;/p&gt;&#xA;" OwnerUserId="11251" LastEditorUserId="11251" LastEditDate="2015-04-28T19:20:17.280" LastActivityDate="2015-04-28T19:20:17.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12870" PostTypeId="2" ParentId="12692" CreationDate="2015-04-28T18:45:39.643" Score="2" Body="&lt;p&gt;I think this is the final solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Step 1&lt;/strong&gt; - Input ciphertext for Playfair decryption:&lt;br&gt;&#xA; - use J = I, encode double letters checked, alphabet key = BLUE&lt;br&gt;&#xA;&lt;pre&gt; Well done CLXXXIII XC CXXX CCXXXIII CXXI CLXXXIV CXCVI XXXII XLIII LXXXIV XLV LXXXII LXXXVII LX CXVI CLXIXX&lt;/pre&gt;&#xA; &#xA;&lt;strong&gt;Step 2&lt;/strong&gt; - Remove final X and translate from Roman numerals to dec and hex:&#xA;&lt;pre&gt;183 090 130 233 121 184 196 032 043 084 045 082 087 060 116 169 &#xA; b7  5a  82  e9  79  b8  c4  20  2b  54  2d  52  57  3c  74  a9&lt;/pre&gt;&lt;br&gt;&#xA;&lt;strong&gt;Step 3&lt;/strong&gt; - Input the hex string for AES-128 decryption and use CBC mode:&lt;br&gt;&#xA;- key = acknowledgements = 61 63 6b 6e 6f 77 6c 65 64 67 65 6d 65 6e 74 73&lt;br&gt;&#xA;- IV = Circumstantially = 43 69 72 63 75 6d 73 74 61 6e 74 69 61 6c 6c 79  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;enter as hex strings and the decrypted text is: &lt;strong&gt;DoYouLikeIt?....&lt;/strong&gt;  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-04-29T07:15:51.477" LastActivityDate="2015-04-29T07:15:51.477" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12871" PostTypeId="2" ParentId="12864" CreationDate="2015-04-28T19:00:45.200" Score="14" Body="&lt;p&gt;I think I have an answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Didier is lying&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I started drawing a graph, eventually gaining this timeline of events (overlaps acceptable).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $Edouard \leftarrow Albert \leftrightarrow Bernard \rightarrow Isabelle \leftrightarrow Charlotte \rightarrow Didier$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can connect the other sightings, to have at least one relationship, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; save for Didier. Didier observes Albert and Isabelle, but there is no relationship between Didier and Bernard, even though they should overlap in some way in order for Didier to observe both Albert and Isabelle. (Didier observed Albert and Isabelle. Bernard is between Albert and Isabelle. Therefore either Bernard should have seen Didier or Didier should have seen Bernard as they would have been at the library at the same time).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Édouard or Didier would have observed the other from their overlapping timelines.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can verify the answer with this image.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/4yayF.png&quot; alt=&quot;Graph&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9696" LastEditorUserId="19028" LastEditDate="2016-02-07T14:51:46.680" LastActivityDate="2016-02-07T14:51:46.680" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="12872" PostTypeId="2" ParentId="12864" CreationDate="2015-04-28T19:04:57.543" Score="9" Body="&lt;p&gt;Because the problem states that exactly one of them is lying, it must be the case that they were not all in the library at once. If they were all in the library at the same time, each of them would have been able to make their statements truthfully, contradicting the one condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let:&lt;br&gt;&#xA;$A$ Albert&lt;br&gt;&#xA;$B$ Bernard&lt;br&gt;&#xA;$C$ Charlotte&lt;br&gt;&#xA;$D$ Didier&lt;br&gt;&#xA;$E$ Edouard&lt;br&gt;&#xA;$I$ Isabelle&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then:&lt;br&gt;&#xA;$A \rightarrow B + E$&lt;br&gt;&#xA;$B \rightarrow A + I$&lt;br&gt;&#xA;$C \rightarrow D + I$&lt;br&gt;&#xA;$D \rightarrow A + I$&lt;br&gt;&#xA;$E \rightarrow B + C$&lt;br&gt;&#xA;$I \rightarrow C + E$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's expand one step:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A \rightarrow B + (A + I) + E + (B + C)$&lt;br&gt;&#xA;$B \rightarrow A + (B + E) + I + (C + E)$&lt;br&gt;&#xA;$C \rightarrow D + (A + I) + I + (C + E)$&lt;br&gt;&#xA;$D \rightarrow A + (B + E) + I + (C + E)$&lt;br&gt;&#xA;$E \rightarrow B + (B + E) + C + (D + I)$&lt;br&gt;&#xA;$I \rightarrow C + (D + I) + E + (B + C)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, we notice that $D$ is the only one that doesn't have someone see him after one step. Let's expand again:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D \rightarrow A + (B + (A + I) + E + (B + C)) + I + (C + (D + I) + E + (B + C))$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the first chance that $D$ has been mentioned in his graph. Unfortunately for him, the graph also contains all of the other people, which can't be the case (as discussed earlier).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Of course, a scenario could be constructed such that $D$ is invisible (by hiding or being quick or whatever). Because there is only one person that claims to see this invisible person, that person is allowed to be the single liar. Any other person is stated to have been seen by at least two people, so everyone else must have actually been seen by at least one person. Our old friend &quot;Invisible Didier&quot;, however, cannot be seen by anyone. Hence, Charlotte is a liar and the culprit.&lt;/p&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-04-28T19:40:17.063" LastActivityDate="2015-04-28T19:40:17.063" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="12873" PostTypeId="2" ParentId="12851" CreationDate="2015-04-28T19:10:11.720" Score="0" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; GOLD. &lt;strong&gt;Edit&lt;/strong&gt;: actually, maybe it's BAR.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Journey far west in the US of A&lt;br&gt;&#xA; Err to the south a bit, too&lt;br&gt;&#xA; True, you can't stay here (it's too rich for most)&lt;br&gt;&#xA; For sure, only the wealthiest do&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; California, the Golden State, could be referring to gold bars.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Usually, I solo; few sing the same&lt;br&gt;&#xA; Everyone says, though, I'm always alone&lt;br&gt;&#xA; Like some of my kind, I have few worthy partners&lt;br&gt;&#xA; Clever are those who know&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bars of song.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;As a warrior I supported the war&lt;br&gt;&#xA; Never did they stand up to me&lt;br&gt;&#xA; Too violent, was I, and hurt many of mine&lt;br&gt;&#xA; Man, though, did I clear those trees&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure about this verse.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Emperor am I, or so it seems&lt;br&gt;&#xA; Legally, I've just been Freed&lt;br&gt;&#xA; Truly, consume me, I will not mind&lt;br&gt;&#xA; Such is my will inclined&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This sounds like a &lt;a href=&quot;http://en.wikipedia.org/wiki/Penguin_%28biscuit%29&quot; rel=&quot;nofollow&quot;&gt;penguin bar&lt;/a&gt;. There exist Emperor Penguins; penguin bars can be bought (freed?) and eaten (consumed).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Too often compared to another, am I&lt;br&gt;&#xA; Even though we are mostly the same&lt;br&gt;&#xA; Everyone sees only differences, but&lt;br&gt;&#xA; Largely, we can be interchanged&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure about this verse either.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Born of sunbeams and fire&lt;br&gt;&#xA; Evoking warm feelings in most&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sunbeams and fire are golden.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-04-28T19:16:41.473" LastActivityDate="2015-04-28T19:16:41.473" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12874" PostTypeId="2" ParentId="12864" CreationDate="2015-04-28T19:13:29.887" Score="0" Body="&lt;p&gt;I feel like the answer may be that...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bernard is the culprit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There exists a little inconsistency between people seen.&lt;br/&gt;Let's draw these reports out:&lt;br/&gt;B -&gt; AI&lt;br/&gt;D -&gt; AI&lt;br/&gt;C -&gt; DI&lt;br/&gt;A -&gt; BE&lt;br/&gt;E -&gt; BC&lt;br/&gt;I -&gt; CE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This leaves us with the following.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Albert, Bernard, Charlotte, and Edouard were all seen twice.&lt;br/&gt;Isabelle was seen three times.&lt;br/&gt;Didier was seen once.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Given this information,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Say Bernard saw Didier instead of Isabelle. We get the following:&lt;br/&gt;B -&gt; AD&lt;br/&gt;D -&gt; AI&lt;br/&gt;C -&gt; DI&lt;br/&gt;A -&gt; BE&lt;br/&gt;E -&gt; BC&lt;br/&gt;I -&gt; CE&lt;br/&gt;&lt;br/&gt; This evens everything out, with Didier and Isabelle now being sighted twice each.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It can be concluded that Bernard lied about seeing Isabelle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4985" LastEditorUserId="4985" LastEditDate="2015-04-28T19:23:02.523" LastActivityDate="2015-04-28T19:23:02.523" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12875" PostTypeId="2" ParentId="12694" CreationDate="2015-04-28T19:14:05.660" Score="5" Body="&lt;p&gt;Try 2! &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Computer Code (i.e., C, C++, etc). Maybe a specific one (if so, I'm lost as to which one).  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you use me, you only ever see what I have become instead of what I really am.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We use code every day when we use computers, phones, etc., but we see the programs rather than the actual running process in the device. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be used to do powerful things, but I can be very fragile at my core.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some programs are very powerful, but one small bug can ruin it (or one stupid $@^#ing semi colon in the wrong place. Two hours of my life....)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Through time, I have changed lives!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Technology and code have changed millions of lives.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately some choose to hate me, and others choose to embrace me.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some are against advances in technology or newer programming languages, while others love them.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am used around the world in many different ways.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Different places in the world like different types of coding, and use it for different things.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My family is huge, and yet still getting bigger.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are so many &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_programming_languages&quot;&gt;programming languages&lt;/a&gt;, it is almost ridiculous, yet more are being developed constantly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is important not to forget about my elders, as they are still needed (by some).  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The basic coding languages are still used. If I am not mistaken (not a programmer by trade) Python still has &lt;em&gt;C&lt;/em&gt; at its core.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hints:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;em&gt;You can look at me, but not touch.&lt;/em&gt; We can see code as we right it, but can't touch it.&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;em&gt;I'm not as old as world war 1.&lt;/em&gt; Computers and coding are a more recent developement.&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;em&gt;Some that have used me, are now amongst the richest people in the world!&lt;/em&gt; Bill Gates, Steve Jobs, etc.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-04-28T19:14:05.660" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12876" PostTypeId="2" ParentId="12864" CreationDate="2015-04-28T19:19:20.797" Score="3" Body="&lt;h1&gt;Note: this answer assumes that a truth-teller says ALL the truth, without omitting information.&lt;/h1&gt;&#xA;&lt;p&gt;The symbol &lt;span class=&quot;math-container&quot;&gt;$X-$&lt;/span&gt;&amp;gt;&lt;span class=&quot;math-container&quot;&gt;$Y$&lt;/span&gt; means that &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt; saw &lt;span class=&quot;math-container&quot;&gt;$Y$&lt;/span&gt;. We know that:&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$A-$&lt;/span&gt;&amp;gt;&lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt;&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$E-$&lt;/span&gt;&amp;gt;&lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt;&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$C-$&lt;/span&gt;&amp;gt;&lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt;&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$D-$&lt;/span&gt;&amp;gt;&lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;&lt;br /&gt;&#xA;Which means that &lt;span class=&quot;math-container&quot;&gt;$A,B,C,D$&lt;/span&gt; were all in the store at a certain moment. Though, there isn't any edge (graph edge, I mean) between &lt;span class=&quot;math-container&quot;&gt;$A-C$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$E-D$&lt;/span&gt;, which should be mandatory if they're all telling the truth. Therefore, the liar is &lt;span class=&quot;math-container&quot;&gt;$A,C,D$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt;.&lt;br /&gt;&#xA;&lt;strong&gt;B and I are honest.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Suppose &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; liar. We have a triangle between &lt;span class=&quot;math-container&quot;&gt;$A-B-E$&lt;/span&gt;, which means that they were together at some point. We also have the triangle &lt;span class=&quot;math-container&quot;&gt;$I-E-B$&lt;/span&gt;, but we don't have a tie &lt;span class=&quot;math-container&quot;&gt;$A-I$&lt;/span&gt;, meaning that there's a moment when &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; was inside the shop and either &lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; had already left it, while the other (of &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt;) hadn't entered yet. If so, how is it possible that &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; saw &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt; without seeing &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt;? Impossible! &lt;strong&gt;Therefore, C is truth-teller.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Suppose &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; liar. We know that &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt; are never together. &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; saw both &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt; , &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; saw both &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt; too, so how is it possible that &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; weren't together? Impossible! &lt;strong&gt;Therefore, E tells the truth.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Suppose &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; liar. &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; didn't meet. &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; has seen both &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt;, meaning that &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; was present when one had left and the other had yet to enter. D was intermediate between &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt;, despite not meeting &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt;, because he has seen &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; (who was seen by &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt;). So, how is it possible that &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; never met? Impossible, so &lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; tells the truth.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;The liar is &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt;&lt;/strong&gt;. This is a scheme of compatible timings. The time is represented on the horizontal axis. People that share points with the same abscissa were together in the library, of course.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/9FBms.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-02-06T01:53:48.980" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="12878" PostTypeId="5" CreationDate="2015-04-28T20:09:20.967" Score="0" Body="&lt;p&gt;The puzzle can be found &lt;a href=&quot;https://xkcd.com/blue_eyes.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and is as follows.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;A group of people with assorted eye colors live on an island. They are&#xA;    all perfect logicians -- if a conclusion can be logically deduced,&#xA;    they will do it instantly. No one knows the color of their eyes. Every&#xA;    night at midnight, a ferry stops at the island. Any islanders who have&#xA;    figured out the color of their own eyes then leave the island, and the&#xA;    rest stay. Everyone can see everyone else at all times and keeps a&#xA;    count of the number of people they see with each eye color (excluding&#xA;    themselves), but they cannot otherwise communicate. Everyone on the&#xA;    island knows all the rules in this paragraph.&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;  &#xA;  &lt;p&gt;On this island there are 100 blue-eyed people, 100 brown-eyed people,&#xA;  and the Guru (she happens to have green eyes). So any given blue-eyed&#xA;  person can see 100 people with brown eyes and 99 people with blue eyes&#xA;  (and one with green), but that does not tell him his own eye color; as&#xA;  far as he knows the totals could be 101 brown and 99 blue. Or 100&#xA;  brown, 99 blue, and he could have red eyes.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The Guru is allowed to speak once (let's say at noon), on one day in&#xA;  all their endless years on the island. Standing before the islanders,&#xA;  she says the following:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;I can see someone who has blue eyes.&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Who leaves the island, and on what night?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;There are no mirrors or reflecting surfaces, nothing dumb. It is not a&#xA;  trick question, and the answer is logical. It doesn't depend on tricky&#xA;  wording or anyone lying or guessing, and it doesn't involve people&#xA;  doing something silly like creating a sign language or doing genetics.&#xA;  The Guru is not making eye contact with anyone in particular; she's&#xA;  simply saying &quot;I count at least one blue-eyed person on this island&#xA;  who isn't me.&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;And lastly, the answer is not &quot;no one leaves.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is that on the 100th day, all 100 blue-eyed people leave the island. This tag is for questions about this puzzle or its many variants.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-04-28T20:09:20.967" LastActivityDate="2015-04-28T20:09:20.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12879" PostTypeId="4" CreationDate="2015-04-28T20:09:20.967" Score="0" Body="For questions related to the puzzle of islanders who can only leave when they deduce the colour of their own eyes." OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-04-28T20:09:20.967" LastActivityDate="2015-04-28T20:09:20.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12880" PostTypeId="1" AcceptedAnswerId="12886" CreationDate="2015-04-28T20:23:20.550" Score="1" ViewCount="897" Body="&lt;p&gt;If you look you cannot see me. And if you see me you cannot see anything else. I can make anything you want happen, but later everything goes back to normal. Guess What am I?&lt;/p&gt;&#xA;" OwnerUserId="11764" LastEditorUserId="11764" LastEditDate="2015-04-28T20:48:12.467" LastActivityDate="2015-04-29T09:22:34.903" Title="Guess Who Am I?" Tags="|riddle|" AnswerCount="6" CommentCount="0" ClosedDate="2015-08-27T18:30:34.997" ContentLicense="CC BY-SA 3.0" />
  <row Id="12881" PostTypeId="2" ParentId="12880" CreationDate="2015-04-28T20:28:03.897" Score="4" Body="&lt;p&gt;Could you be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;b&gt;A Dream&lt;/b&gt;. You can't see your dreams when you are awake and 'looking', and when you are asleep and dreaming you can't 'look' to see anything else. In a dream you can be able to control what happens (Lucid dreaming), but once you wake up, everything is back to 'normal' (reality).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastActivityDate="2015-04-28T20:28:03.897" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12882" PostTypeId="2" ParentId="12880" CreationDate="2015-04-28T20:34:28.443" Score="4" Body="&lt;p&gt;Building off of Mark's answer:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A daydream.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We can't really see daydreams, but when we daydream, we don't really see anything else. We daydream about what we want, but then get snapped back to reality.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-04-28T20:34:28.443" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12883" PostTypeId="2" ParentId="12880" CreationDate="2015-04-28T20:35:02.813" Score="3" Body="&lt;p&gt;In a similar vein to Mark N's answer, how about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a DAYDREAM?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If you look you cannot see me. And if you see me you cannot see anything else. I can make anything you want happen, but later everything goes back to normal.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you look at the real world, you can't see your daydreams. If you 're living in a daydream, you can't see what's really going on. Anything can happen in a daydream, but when you snap out of it, life is normal again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-04-28T20:35:02.813" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12884" PostTypeId="2" ParentId="12880" CreationDate="2015-04-28T20:45:18.157" Score="0" Body="&lt;p&gt;Is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the inside of your eyelids? You can't see them when &quot;looking&quot;, and you can't see anything else while you can see them. The visual deprivation makes it easier to imagine/see your fantasies! &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11880" LastEditorUserId="11880" LastEditDate="2015-04-28T20:53:35.527" LastActivityDate="2015-04-28T20:53:35.527" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12885" PostTypeId="2" ParentId="12880" CreationDate="2015-04-28T20:46:07.307" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A dream.&#xA;When you look, you open your eyes and the dream ends.  When you &lt;em&gt;are&lt;/em&gt; dreaming, you can't see anything else, as your eyes are closed.  Anything can happen in a dream and when it's over, it's back to dull reality.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11851" LastActivityDate="2015-04-28T20:46:07.307" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12886" PostTypeId="2" ParentId="12880" CreationDate="2015-04-28T20:48:29.227" Score="3" Body="&lt;p&gt;Is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; your fantasy?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can make happen anything then, and are limited to what you see at that moment.&lt;/p&gt;&#xA;" OwnerUserId="3993" LastActivityDate="2015-04-28T20:48:29.227" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12888" PostTypeId="1" AcceptedAnswerId="12889" CreationDate="2015-04-28T20:49:41.630" Score="7" ViewCount="1463" Body="&lt;p&gt;This riddle isn't very difficult but hopefully can you enjoy the rhymes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My home is amazing, you might be impressed&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I am full of strength, with no time to rest&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Everyday I go venture to complete my own quest.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I am one on a team and we all do our best.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;To protect my own mother, my life I'll invest.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;But sometimes people think we are quite a pest!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What could I be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Follow the leader is my favorite game&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your answer should match every line, and if possible try to be as specific as possible for the answer.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If you enjoyed this riddle, some of my other rhyming riddles can be found &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%5Brhyme%5D%20user%3A11244%20is%3Aquestion&quot;&gt;here&lt;/a&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-06-19T13:27:29.250" Title="Riddle: An impressive thing" Tags="|riddle|rhyme|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12889" PostTypeId="2" ParentId="12888" CreationDate="2015-04-28T20:51:13.853" Score="7" Body="&lt;p&gt;You are a: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Worker ant&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My home is amazing, your might be impressed  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ant hives are amazingly complex structures, so much that we buy ant farms to see them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am full of strength, with no time to rest  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ants can lift items much heavier than them, and they are always working.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Everyday I go venture to complete my own quest.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ants go out every day looking for food, supplies, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am one on a team and we all do our best.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lots of ants in a hive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To protect my own mother, my life I'll invest.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They'll sacrifice for the queen.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But Sometime people think we are quite a pest!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I have ants in my house, and I hate them! (when they are inside)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Worker ants follow each other in a long line. (See the beginning of &lt;em&gt;A Bugs Life&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="9951" LastEditDate="2015-04-28T21:19:09.107" LastActivityDate="2015-04-28T21:19:09.107" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12890" PostTypeId="2" ParentId="12888" CreationDate="2015-04-28T20:54:03.017" Score="2" Body="&lt;p&gt;Could it be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;s&gt;a bee?&lt;/s&gt; an ant?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My home is amazing, your might be impressed&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;s&gt;Beehives&lt;/s&gt; Anthills have amazing &lt;s&gt;geometric&lt;/s&gt; structure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am full of strength, with no time to rest&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;s&gt;Bees&lt;/s&gt; Ants are always &lt;s&gt;buzzing around&lt;/s&gt; busy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Everyday I go venture to complete my own quest.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They go out to &lt;s&gt;get nectar/pollen and bring it home&lt;/s&gt; do whatever it is ants do so busily.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am one on a team and we all do our best.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;s&gt;Bees&lt;/s&gt; Ants are social animals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;To protect my own mother, my life I'll invest.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The hive is centred on the queen &lt;s&gt;bee&lt;/s&gt; ant, who is mother to many of the others.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But Sometime people think we are quite a pest!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many people don't like &lt;s&gt;bees&lt;/s&gt; ants.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-04-28T20:54:03.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12891" PostTypeId="2" ParentId="12888" CreationDate="2015-04-28T21:09:42.840" Score="1" Body="&lt;p&gt;You are a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; termite&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My home is amazing, you might be impressed&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Termites can live in human houses.  Quite an impressive home indeed. :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am full of strength, with no time to rest&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Termites have very strong jaws and can chew through a lot of materials.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Everyday I go venture to complete my own quest.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every day they go look for food and build their nest.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am one on a team and we all do our best.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They work collectively to accomplish their goals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To protect my own mother, my life I'll invest.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Like with some other insects, they'll die to protect the queen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But sometimes people think we are quite a pest!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Termites will cause significant damage to the wood in houses.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2015-04-28T21:09:42.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12892" PostTypeId="1" AcceptedAnswerId="12930" CreationDate="2015-04-28T21:25:00.957" Score="7" ViewCount="1025" Body="&lt;p&gt;Each of the 4 pictures below represents one word in a famous movie quote. The pictures (and thus, the words) are in order from left to right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you get the 4 words indicated by the pictures below, then you will use those words to derive the first 4 words of the movie quote.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At that point, you can use a search engine to easily find the rest of the quote (if needed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the 4th picture, those are not &quot;slices&quot; because slices are wider.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The word indicated by the 2nd pic is an 8-letter adjective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can you finish the quote and find what movie it's from?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Xmf6S.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-06-30T20:24:01.867" LastActivityDate="2015-06-30T20:24:01.867" Title="What Movie Quote Am I?" Tags="|word|rebus|movies|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12893" PostTypeId="1" CreationDate="2015-04-28T21:38:38.870" Score="2" ViewCount="296" Body="&lt;p&gt;In E.E. 'Doc' Smith's classic &lt;a href=&quot;http://en.wikipedia.org/wiki/Skylark_%28series%29#Skylark_of_Valeron&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Skylark of Valeron&lt;/em&gt;&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dick Seaton and party are beset in space by disembodied intelligences. These intelligences working as they do on the 6th order of forces are able to travel with accelerations and speeds thousands of times fast than light (relativity has been proved wrong in the skylark universe)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even after having developed his own 6th order ray projectors, fields of force and inertial dampeners etc mounted them in a huge 'spaceship' and created a house sized 'brain' of vacuum tubes to control it all. Seaton says he is unable to hurt the intelligences because :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;they can travel at the same speed as any force I can direct against&#xA;  them&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Any ray or field directed against them simply pushes them back.  and he is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;unable to create a spherical field without cutting my own beams&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However he comes up with a plan. Communicating with Norlamin (a distant planet) he instructed the residents (super scientists all) to construct a similar space ship and travel towards him. Additionally he creates a 'bait' to attract the intelligences, mid way between the two ships.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The intelligences are attracted to the bait and trapped in a spherical shell of force one hemisphere generated by each ship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The puzzle!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this story has always confused me. A. how can a ship generate a hemispherical field without &quot;cutting its own beams&quot; but more pertinently is it actually possible to capture the intelligences?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the nature of 6th order forces has not been entirely revealed so I offer a number of options/rules&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: its clear that the ships accelerate at certain rates. Greater than the speed of light/s^2, but not instantaneously. There is no top speed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2: the rays and the intelligences propagate at a speed relative to the ether. much faster than the ships would normally travel and instantaneously reached&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3: the intelligences react instantly and are super intelligent (except for bait)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4: the universe is infinite and the speeds and distances involved allow us to treat the ships as points&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5: the ships are essentially stationary relative to each other and the bait&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6: communicating and sensing is all at 6th order speed, including 'triggering' the bait and detecting things&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the intelligences be captured?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it trivial?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the key factors that enable escape or make capture certain?&lt;/p&gt;&#xA;" OwnerUserId="11831" LastEditorUserId="12700" LastEditDate="2015-06-28T05:20:53.230" LastActivityDate="2015-06-28T05:20:53.230" Title="Skylark of Valeron" Tags="|geometry|science|" AnswerCount="1" CommentCount="5" ClosedDate="2015-06-29T12:54:37.800" ContentLicense="CC BY-SA 3.0" />
  <row Id="12894" PostTypeId="1" AcceptedAnswerId="72606" CreationDate="2015-04-28T21:55:17.700" Score="0" ViewCount="1737" Body="&lt;p&gt;&lt;em&gt;All You Zombies&lt;/em&gt; is a short story by Robert Heinlein, which outlines a rather bizarre time travel paradox. Here is an outline of the paradox with a few key moments missing.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Jane is dropped off at an orphanage in 1945.&lt;/li&gt;&#xA;&lt;li&gt;Jane grows up lonely and dejected not knowing who her parents are.&lt;/li&gt;&#xA;&lt;li&gt;In 1963, Jane falls in love with a drifter, becomes pregnant with his child. The drifter disappears immediately thereafter.&lt;/li&gt;&#xA;&lt;li&gt;During childbirth, there are complications. The doctor discovers ___________, and in order to save Jane's life, the doctor ____________.&lt;/li&gt;&#xA;&lt;li&gt;A few weeks later, Jane's baby (A healthy baby girl) is kidnapped from the hospital.&lt;/li&gt;&#xA;&lt;li&gt;Losing ones parents, lover and only child has a devastating effect on Jane. Scorned by fate, Jane rejects society and becomes a drunkard.&lt;/li&gt;&#xA;&lt;li&gt;In 1970, Jane stumbles into a bar, and tells the sad tale to a sympathetic bartender.&lt;/li&gt;&#xA;&lt;li&gt;The bartender, who is a respected member of the time-travelling guild, offers Jane a chance to seek revenge on the drifter who ruined Jane's life.&lt;/li&gt;&#xA;&lt;li&gt;The bartender takes Jane back to 1963, where Jane ______________.&lt;/li&gt;&#xA;&lt;li&gt;The bartender hops forward in time _______(length of time), and __________.&lt;/li&gt;&#xA;&lt;li&gt;The bartender returns to 1963 and picks Jane up. The two of them return to 1970.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Puzzle:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fill in all the blanks (So that there is a paradox) and answer the following questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What becomes of Jane after 1970?&lt;/li&gt;&#xA;&lt;li&gt;What is the paradox?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The 4th blank is: 9 months&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE:&lt;/strong&gt; If you are familiar with the paradox or have read the short story, don't answer the question right away.. Not much of a puzzle if you know the paradox.&lt;/p&gt;&#xA;" OwnerUserId="11063" LastEditorUserId="11063" LastEditDate="2015-04-28T23:49:58.823" LastActivityDate="2018-09-20T17:32:36.033" Title="All You Zombies" Tags="|logical-deduction|story|paradox|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12895" PostTypeId="2" ParentId="12894" CreationDate="2015-04-28T22:13:55.990" Score="2" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;Note&lt;/strong&gt;: I feel like this puzzle is really just asking for specific trivia from a written piece. I could just as easily fill in the blanks as below, with the same level of satisfaction that the answer is correct.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During childbirth, there are complications. The doctor discovers &lt;strong&gt;the baby is breached&lt;/strong&gt;, and in order to save Jane's life, the doctor &lt;strong&gt;performs an emergency c-section&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bartender takes Jane back to 1963, where Jane &lt;strong&gt;loses her nerve and decides to go watch a movie instead&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bartender hops forward in time 9 months and &lt;strong&gt;steals the baby that he conceived with the younger Jane, returns to 1945 and drops it off at an orphanage with a name tag &quot;Jane&quot;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What becomes of Jane after 1970?&lt;/strong&gt;&lt;br&gt;&#xA;She accepts her fate and her life takes a positive turn. She later adopts a cat and spends the rest of her days quilting for the locals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the paradox?&lt;/strong&gt;&lt;br&gt;&#xA;No paradox, just coincidence. The bartender initially had no conversation with Jane in the original set of events; he decided to go to 1963 all on his own and knock up a stranger. This first stranger was not necessarily an orphan or named Jane. After stealing the baby, he travelled around for a few more weeks, only to realise that babies are a lot of work. It was during a stop he made briefly in 1945 that he decided he no longer had the patience for the child, and dropped her at an orphanage with the name tag reading &quot;Jane&quot;. This baby becomes the protagonist in our puzzle. Her own baby is dropped at the same orphanage by the same time traveller, but in this timeline, there are now two (different) babies named Jane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who is the bartender?&lt;/strong&gt;&lt;br&gt;&#xA;A member of the time-traveller's guild.&lt;/p&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-04-29T00:52:59.393" LastActivityDate="2015-04-29T00:52:59.393" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12896" PostTypeId="2" ParentId="12864" CreationDate="2015-04-28T22:36:59.557" Score="9" Body="&lt;p&gt;To start: We need to define who was in together at the same time: We'll start off assuming everyone is honest and then look for contradictions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relations are as such:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;AB, AD, AE, BE, BI, CD, CE, CI, DI, &amp;amp; EI&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In all of these cases, at least one person claimed the other was there at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the record, the missing relations are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;AC, AI, BC, BD, &amp;amp; DC&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So with this information we can figure out who was supposed to be in at the same time. Since &lt;strong&gt;AB&lt;/strong&gt;, &lt;strong&gt;AE&lt;/strong&gt;, &amp;amp; &lt;strong&gt;BE&lt;/strong&gt; are all present at the same time we can conclude that all three were there at (more or less) the same time for the purpose of creating a timeline. All the &quot;combined&quot; relations are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;ABE, AD, BEI, CDI, &amp;amp; CEI&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Visually that becomes:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A ---|---|   |   |&#xA;B ---|   |---|   |&#xA;C    |   |   |---|---&#xA;D    |---|   |---|&#xA;E ---|   |---|   |---&#xA;I    |   |---|---|---&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We can now attempt to organize this into a timeline. Since each person only entered once, each line should be intact. This is the best we can get:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A ---|---|   |   |&#xA;B    |---|---|   |&#xA;C    |   |   |---|---&#xA;D ---|   |   |   |---&#xA;E    |---|---|---|&#xA;I    |   |---|---|---&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Ok, so the problem is obvious: We have a cyclical timeline. This timeline only makes sense if we turn it into a cylinder, because no matter how we attempt to put it together (on a straight 2D path) one of the lines is going to be at both ends of the graph without connecting all the way through. So now we can try to verify each of the above relations. We'll go in the order of the &quot;corrected&quot; graph, and we want to rely on cyclical relationships. In other words - 1 sees 2 who sees 3 who sees 1 because that confirms all three are there. Even if, say, 1 was lying, 2 confirms 3 and 3 confirms 1. Since there can only be one liar; 2 confirms themselves by being witness to 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;AD&lt;/strong&gt;: D sees A, but A does not see D. No one else is around to complete the circle. &lt;strong&gt;Unconfirmed&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ABE&lt;/strong&gt;: E sees B, B sees A, A sees E (and also B). &lt;strong&gt;Confirmed&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;BEI&lt;/strong&gt;: E sees B, B sees I, I sees E. &lt;strong&gt;Confirmed&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;CEI&lt;/strong&gt;: C sees I, I sees E, E sees C. &lt;strong&gt;Confirmed&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;CDI&lt;/strong&gt;: D sees I, I sees C, C sees D (and also I). &lt;strong&gt;Confirmed&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interesting - only &lt;em&gt;one&lt;/em&gt; unconfirmed. Also interesting is that we have a double confirmation (where 3 sees both 1 &lt;em&gt;and&lt;/em&gt; 2) on either side of the unconfirmed relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one unconfirmed means that &lt;em&gt;that one&lt;/em&gt; (&lt;strong&gt;AD&lt;/strong&gt;) has to be the false sighting - and since &lt;strong&gt;AD&lt;/strong&gt; is &lt;strong&gt;D&lt;/strong&gt;'s claim - that makes &lt;strong&gt;&lt;em&gt;Didier&lt;/em&gt;&lt;/strong&gt; the dirty lying thief.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GAVEW.jpg&quot; alt=&quot;Go directly to jail!&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Do not pass Go.&lt;/li&gt;&#xA;&lt;li&gt;Do not collect $200.&lt;/li&gt;&#xA;&lt;li&gt;Give the book back, jeez.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5688" LastEditorUserId="5688" LastEditDate="2015-04-28T22:42:34.483" LastActivityDate="2015-04-28T22:42:34.483" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12897" PostTypeId="2" ParentId="12838" CreationDate="2015-04-28T22:52:37.913" Score="8" Body="&lt;p&gt;This is an extension of this post about &lt;a href=&quot;https://puzzling.stackexchange.com/questions/8429/fair-share-of-a-square-watermelon&quot;&gt;trisecting a cube&lt;/a&gt;. A cube is shown to be divided into 3 right-angle pyramids.  Then, 6 right-angle tetrahedra can be made by cutting each pyramid diagonally through its square face. This dissection results in 3 left-hand and 3 right-hand tetrahedra, so they are not quite identical.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better explanation uses a diagram from this post about &lt;a href=&quot;https://puzzling.stackexchange.com/questions/9477/points-on-a-cube&quot;&gt;points on a cube&lt;/a&gt;. As shown below, the cube can be divided into 6 tetrahedra by making 3 planar cuts.  Each planar cut must follow the long diagonal of the cube (shown in red).  Again, this dissection results in 3 left-hand and 3 right-hand tetrahedra. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To obtain 6 identical tetrahedra, the cutting planes need to be rotated by 30 degrees around the longest diagonal. The planar cuts are then bisecting the vertices of the included hexagon which is shown below. The 6 resulting tetrahedra are identical. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For most people, visualizing this dissection is not easy. Perhaps the best way is to imagine standing the cube on a corner (in this case the far corner) so that the longest diagonal is now a vertical axis.  By looking straight down this axis, the cube's 6 peripheral corners are seen and the 3 planar cuts can be made to either bisect the corners of the cube or to bisect the vertices of the included hexagon.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OZuqt.png&quot; alt=&quot;Cubes&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/tZPU8.png&quot; alt=&quot;Tetrahedra&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-04-29T16:41:28.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12898" PostTypeId="2" ParentId="12851" CreationDate="2015-04-28T23:41:23.740" Score="5" Body="&lt;p&gt;My guess is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Orange&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Journey far west in the US of A&lt;br&gt;&#xA;Err to the south a bit, too&lt;br&gt;&#xA;True, you can't stay here (it's too rich for most)&lt;br&gt;&#xA;For sure, only the wealthiest do  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Orange County&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Usually, I solo; few sing the same&lt;br&gt;&#xA;Everyone says, though, I'm always alone&lt;br&gt;&#xA;Like some of my kind, I have few worthy partners&lt;br&gt;&#xA;Clever are those who know  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No words rhyme with &quot;orange&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a warrior I supported the war&lt;br&gt;&#xA;Never did they stand up to me&lt;br&gt;&#xA;Too violent, was I, and hurt many of mine&lt;br&gt;&#xA;Man, though, did I clear those trees  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Agent orange&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Emperor am I, or so it seems&lt;br&gt;&#xA;Legally, I've just been Freed&lt;br&gt;&#xA;Truly, consume me, I will not mind&lt;br&gt;&#xA;Such is my will inclined  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Orange Julius (invented by Julius Freed apparently)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Too often compared to another, am I&lt;br&gt;&#xA;Even though we are mostly the same&lt;br&gt;&#xA;Everyone sees only differences, but&lt;br&gt;&#xA;Largely, we can be interchanged  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Apples and oranges&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Born of sunbeams and fire&lt;br&gt;&#xA;Evoking warm feelings in most&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Orange color/light&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alone I fulfill some of this riddle&lt;br&gt;&#xA;Mixed with another, the rest&lt;br&gt;&#xA;So, can you name me and my partners above?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Orange is in all the answers above&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4479" LastEditorUserId="4479" LastEditDate="2015-04-29T17:59:23.067" LastActivityDate="2015-04-29T17:59:23.067" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="12899" PostTypeId="2" ParentId="11225" CreationDate="2015-04-29T00:41:48.640" Score="0" Body="&lt;p&gt;Here's another guess, based on splitting on the &lt;code&gt;1&lt;/code&gt;'s:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;3 4 6 A 3 2 5 J 3 1 &lt;br/&gt;&#xA;  3 Q 5 K 3 A A 1 &lt;br/&gt;&#xA;  2 5 3 6 5 J 2 3 Q 1 &lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This can be rearranged:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3 4 6 A 3 &lt;br/&gt;&#xA; 2 5 J 3 1 &lt;br/&gt;&#xA; 3 Q 5 K &lt;br/&gt;&#xA; 3 A A 1 &lt;br/&gt;&#xA; 2 5 3 6 5 &lt;br/&gt;&#xA; J 2 3 Q 1&lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Both of these arrangements would both produce&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the code 3K5 (or 3135) by taking the center number of each grouping, which is also the number directly above each &lt;code&gt;1&lt;/code&gt; in the second arrangement.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Alternatively, after splitting the lines, convert all letters to numbers as Switch did in his answer (J =&gt; 11, Q =&gt; 12, K =&gt; 13, A =&gt; 14)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;3 4 6 14 3 2 5 11 3 1 &lt;br/&gt;&#xA;  3 12 5 13 3 14 14 1 &lt;br/&gt;&#xA;  2 5 3 6 5 11 2 3 12 1 &lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then, split all two-digit numbers into individual digits&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 346143251131 &lt;br/&gt;&#xA; 312513314141 &lt;br/&gt;&#xA; 253651123121&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note how each line ends on a 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative split (going A B C A B C instead of A A A... B B B...) from the original would produce&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 312135312611 &lt;br/&gt;&#xA; 445311145522 &lt;br/&gt;&#xA; 631123413131 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not sure where to go from here, though.  Profit?&lt;/p&gt;&#xA;" OwnerUserId="325" LastEditorUserId="325" LastEditDate="2015-05-12T21:40:33.720" LastActivityDate="2015-05-12T21:40:33.720" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12900" PostTypeId="5" CreationDate="2015-04-29T00:49:08.460" Score="0" Body="&lt;p&gt;A trick puzzle seems to have an easy answer that turns out to be incorrect. They are closely related to &lt;a href=&quot;/questions/tagged/lateral-thinking&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;lateral-thinking&amp;#39;&quot; rel=&quot;tag&quot;&gt;lateral-thinking&lt;/a&gt; puzzles in that it requires some lateral thinking to deduce why the initial 'answer' is not right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Michael's father has three sons &amp;mdash; two of them are Larry and Curly, who is the last one?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You might think the answer is &quot;Moe&quot; at first, as the last of the Three Stooges, but the answer is actually Michael, since the father in question is Michael's father.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="5373" LastEditDate="2015-05-18T10:33:50.530" LastActivityDate="2015-05-18T10:33:50.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12901" PostTypeId="4" CreationDate="2015-04-29T00:49:08.460" Score="0" Body="A problem that appears to have an obvious but incorrect answer." OwnerDisplayName="user88" LastEditorUserId="5373" LastEditDate="2015-05-18T10:33:50.530" LastActivityDate="2015-05-18T10:33:50.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12902" PostTypeId="1" AcceptedAnswerId="12907" CreationDate="2015-04-29T01:02:57.837" Score="2" ViewCount="314" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Bacon%27s_cipher&quot; rel=&quot;nofollow&quot;&gt;Bacon's cipher&lt;/a&gt; is a classic steganographic method, but to decode it, you need to know which symbol is A, which symbol is B and which direction to read them in. (That said, you could just try all the possibilities, and all but one would likely come out as gibberish.) But what if you wanted a code that worked unambiguously without needing to identify which is the A symbol and which is the B symbol?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How long would a fixed length code need to be to encode a 26 character alphabet if -&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;codewords can't collide when the symbols are inverted? (That is, if ABAAB is a codeword, BABBA isn't. Alternatively, ABAAB and BABBA could encode the same character.)&lt;/li&gt;&#xA;&lt;li&gt;codewords can't collide when read backwards or when the symbols are inverted? (That is, if ABAAB is a codeword, none of BAABA, BABBA or ABBAB are.)&lt;/li&gt;&#xA;&lt;li&gt;codewords can't collide if placed on an n-by-n grid and the grid is transposed or the symbols are inverted?&lt;/li&gt;&#xA;&lt;li&gt;codewords can't collide if placed on an n-by-n grid and read sideways, backwards, transposed, inverted, etc.?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For an example for the second two, assume ABAAB and ABBBB are a codewords, then they could be put in to a 5-by-5 grid:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ABAAB&#xA;ABBBB&#xA;ABAAB&#xA;ABBBB&#xA;ABBBB&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This grid could either be read left-to-right then top-to-bottom or top-to-bottom then left-to-right in the third question (and the symbols might or might not be inverted). The codewords must be such that every combination of valid codewords in one direction will create at least one invalid codeword in the other direction so that someone with the codebook can definitely tell how it should be read. For example, perhaps ABABB isn't a valid codeword. (If this holds true for an n-by-n grid, then it should also hold true for an xn-by-yn grid.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is true for the fourth question, but the correct direction could be any of the eight possibilities. The alternate interpretation where colliding codewords represent the same character doesn't really make sense for these questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that if it were required that no valid codewords appeared in the other directions, both the last two questions would be impossible, but it'd be more interesting if I were wrong about that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: I don't have a definite answer to the last two of these.)&lt;/p&gt;&#xA;" OwnerUserId="7392" LastEditorUserId="7392" LastEditDate="2015-04-29T04:04:49.313" LastActivityDate="2015-06-15T20:10:22.780" Title="Beyond Bacon's Cipher" Tags="|combinatorics|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12903" PostTypeId="2" ParentId="12902" CreationDate="2015-04-29T02:13:45.843" Score="2" Body="&lt;p&gt;Your first two questions are answered in real life by &lt;a href=&quot;http://en.wikipedia.org/wiki/Barcode&quot; rel=&quot;nofollow&quot;&gt;bar-coding algorithms&lt;/a&gt;. These are codes that are (usually) specifically designed so that they are still uniquely readable when the symbols are inverted or backwards. In particular, &lt;a href=&quot;http://en.wikipedia.org/wiki/Code_39&quot; rel=&quot;nofollow&quot;&gt;Code 39&lt;/a&gt; uses 12 bars for each alphanumeric character.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your second two questions are answered by &lt;a href=&quot;http://en.wikipedia.org/wiki/QR_code&quot; rel=&quot;nofollow&quot;&gt;QR codes&lt;/a&gt;, much to the same effect. However, in this case a QR code uses three fixed-size reference anchors that make it really easy to detect (by design, I might add, since QR codes are meant to be interpretable from a 2D raster image). I'm sure it's not difficult to design an algorithm that doesn't have those visibility anchors that has the same resistance to reflection and rotation, though.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-04-29T02:13:45.843" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12904" PostTypeId="1" AcceptedAnswerId="12915" CreationDate="2015-04-29T02:28:18.963" Score="7" ViewCount="369" Body="&lt;p&gt;My friend drew this diagram to help him remember his password. &quot;It may seem like just a bunch of squares&quot;, he said, &quot;but once you put it all together, and read it the right way, you'll see it.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/S2gX7.png&quot; alt=&quot;It&amp;#39;s not binary&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;What is his password?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="7308" LastActivityDate="2015-04-29T05:41:14.420" Title="Dimensional Analysis" Tags="|cipher|visual|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12905" PostTypeId="2" ParentId="12864" CreationDate="2015-04-29T02:32:11.087" Score="3" Body="&lt;p&gt;The problem is ill-posed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The liar must be one of A or D, but there is a consistent scenario where A lies and everyone else tells the truth, as well as a consistent scenario where D lies and everyone else tells the truth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First scenario:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A enters at 7:00 AM, I enters at 8:00 AM, B enters at 9:00 AM, E enters at 10:00 AM, B leaves at 11:00 AM, C enters at 12:00 PM, E leaves at 1:00 PM, D enters at 2:00 PM, C leaves at 3:00 PM, D leaves at 4:00 PM, I leaves at 5:00 PM, A leaves at 6:00 PM. A is the only liar: the truth is that A saw everybody.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second scenario:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; D enters at 7:00 AM, E enters at 8:00 AM, C enters at 9:00 AM, I enters at 10:00 AM, C leaves at 11:00 AM, B enters at 12:00 PM, I leaves at 1:00 PM, A enters at 2:00 PM, B leaves at 3:00 PM, A leaves at 4:00 PM, E leaves at 5:00 PM, D leaves at 6:00 PM. D is the only liar: the truth is that D saw everybody.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whittling down the suspects:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Say that suspects X and Y are &quot;connected&quot; if one of X or Y claims to have seen the other. Define a &lt;em&gt;4-cycle&lt;/em&gt; to be a four-tuple of suspects (X,Y,Z,W) such that X is connected to Y, Y is connected to Z, Z is connected to W, W is connected to X, X is &lt;em&gt;not&lt;/em&gt; connected to Z, and Y is &lt;em&gt;not&lt;/em&gt; connected to W. If (X,Y,Z,W) is a 4-cycle, then it is pretty easy to see that at least one of X,Y,Z,W must be lying. Using the 4-cycle (A,B,I,D) we see that one of A,B,I,D is lying, and using the 4-cycle (A,D,C,E) we see that one of A,D,C,E is lying. Thus the liar is either A or D.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is a hidden symmetry in the problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The connection graph described earlier is symmetric under the permutation (AD)(BC)(EI).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11295" LastEditorUserId="11295" LastEditDate="2015-04-29T02:41:35.533" LastActivityDate="2015-04-29T02:41:35.533" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12906" PostTypeId="2" ParentId="12864" CreationDate="2015-04-29T02:45:35.417" Score="2" Body="&lt;p&gt;I believe the &quot;correct&quot; way to do this is to start by determining which groupings are certain. (I'm certain it's correct, but I don't know if it's how they expected it to be done)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can quickly identify two pairs of people who definitely saw each other.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Albert and Bernard both saw each other.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Charlotte and Isabelle both saw each other.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This means these two pairs must have happened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we also have that Albert saw Edouard, who saw Bernard. This indicates that Edouard was present with at least one of the two, as Albert and Edouard can't both be lying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, Isabelle saw Edouard, who saw Charlotte, so Edouard was present with at least one of the two, as Isabelle and Edouard can't both be lying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also have that Isabelle saw Edouard, who saw Bernard, who saw Isabelle. These three must be together, as only one can be lying, which still ties all three into the same group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we have Isabelle saw Charlotte, who saw Didier, who saw Isabelle. Again, only one can be lying, so these three were present together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have (ABE), (BCI), (CEI), and (CDI) as certain to be present in their respective groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leaves just one claim unaccounted for. Didier claimed to be present with Albert. This is the only claim not corroborated by anybody else, directly or indirectly, and therefore must be the lie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve this more easily, I drew up the claims as a directed graph, and then looked for loops within the graph. If three people formed a loop, they corroborate each other's claims. The only claim not part of a loop is Didier claiming to have seen Albert.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be clear, methods based on working out timelines are problematic, as they require assumptions about sequences, whereas the above is purely about requiring that there be only one liar.&lt;/p&gt;&#xA;" OwnerUserId="9289" LastEditorUserId="10781" LastEditDate="2015-04-29T15:08:40.487" LastActivityDate="2015-04-29T15:08:40.487" CommentCount="23" ContentLicense="CC BY-SA 3.0" />
  <row Id="12907" PostTypeId="2" ParentId="12902" CreationDate="2015-04-29T03:17:23.467" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  First case: you need a length of $6$.&#xA;  Second case: You need a length of $7$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the first case, suppose we want a length $n$ code. There are $2^n$ strings of $n$ symbols, but these can be divided into complementary pairs where the two strings in each pair must represent the same codeword. This means there are at most $2^n/2=2^{n-1}$ codewords possible. Since $2^{5-1}&amp;lt;26$, but $2^{6-1}&amp;gt;26$, a code of length $6$ is necessary and sufficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we throw in reversibility, we divide the strings of length $2^n$ into three classes: palindromes (equal to their reverse), anti-palindromes (which are equal to the complement of their reverse), and boring strings which fall into neither class. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Palindromes: a palindrome is specified by the what the first $\lceil n/2\rceil$ symbols of a word, so there are $2^{\lceil n/2\rceil}$ palindromes. However, these can only represent $2^{\lceil n/2\rceil}/2$ codewords, since complements must represent the same word.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Anti-palindromes: When $n$ is odd, there aren't any anti-palindromes (the middle symbol of a string will always differ from that of its complement reverse). When $n$ is even, there are $2^{n/2}$ anti-palindromes, making $2^{n/2}/2$ codewords.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Boring words: Associated with every boring word are $3$ other codewords with which it can't be distinguished (reverse, complement, complement-reverse). To find the number of codewords these can make, you count the number of boring words, ($2^n-$ # palindromes $-$ # anti-palindomres), then divide by $4$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When $n=6$, the number of codewords possible is&#xA;$$&#xA;\frac{2^3}2+\frac{2^3}2+\frac{2^6-2^3-2^3}4=20&amp;lt;26&#xA;$$&#xA;so length $6$ is not enough. When $n=7$,&#xA;$$&#xA;\frac{2^4}2+\frac{2^7-2^4}4=36&amp;gt;26&#xA;$$&#xA;so length $7$ is enough.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-04-29T03:34:49.990" LastActivityDate="2015-04-29T03:34:49.990" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12908" PostTypeId="1" AcceptedAnswerId="12912" CreationDate="2015-04-29T03:24:23.610" Score="10" ViewCount="1942" Body="&lt;p&gt;These eight words all have something in common, but one is slightly different. Which one?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;broad&lt;/li&gt;&#xA;&lt;li&gt;corn&lt;/li&gt;&#xA;&lt;li&gt;gain&lt;/li&gt;&#xA;&lt;li&gt;jar&lt;/li&gt;&#xA;&lt;li&gt;like&lt;/li&gt;&#xA;&lt;li&gt;mount&lt;/li&gt;&#xA;&lt;li&gt;range&lt;/li&gt;&#xA;&lt;li&gt;side&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To clarify, seven of these words have a specific property in common, which is shared by very few other words. The eighth word has a similar (and also fairly rare) characteristic but is slightly different.&lt;/p&gt;&#xA;" OwnerUserId="7425" LastEditorUserId="5373" LastEditDate="2015-04-29T22:07:17.847" LastActivityDate="2015-04-29T22:07:45.960" Title="Eight words: which one is different?" Tags="|pattern|language|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12909" PostTypeId="1" AcceptedAnswerId="12945" CreationDate="2015-04-29T03:32:49.930" Score="8" ViewCount="681" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;When I spike, you will get quite enthused,&lt;br&gt;&#xA;  For great strength, I am often abused,&lt;br&gt;&#xA;  I am quintessentially male...&lt;br&gt;&#xA;  The next line might derail:&lt;br&gt;&#xA;  With Hashish amiga, my name's all confused.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who or what am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the fourth line is simply referring to the fact that the fifth line is the hardest, and the key to the whole puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;s&gt;Hint: I've left off a tag that is appropriate to this question, intentionally.&lt;/s&gt; - updated now that the puzzle has been available for a few more days.&lt;/p&gt;&#xA;" OwnerUserId="9289" LastEditorUserId="9289" LastEditDate="2015-05-04T03:05:21.793" LastActivityDate="2015-05-04T14:49:22.940" Title="A test of mental acuity; Identify me" Tags="|riddle|cipher|word|limerick|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12910" PostTypeId="2" ParentId="12908" CreationDate="2015-04-29T03:47:58.037" Score="1" Body="&lt;p&gt;The answer is &lt;strong&gt;jar&lt;/strong&gt;. All of the words have at least three letters, but &quot;jar&quot; is the only one with just three.&lt;/p&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-04-29T11:30:22.403" LastActivityDate="2015-04-29T11:30:22.403" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12911" PostTypeId="1" AcceptedAnswerId="12916" CreationDate="2015-04-29T03:54:11.390" Score="10" ViewCount="334" Body="&lt;p&gt;Two gladiators are about to do battle in the coliseum. Before the battle, they enter separate rooms and pray to Ares, the god of war. Ares blesses them each with random amount of power, which you can think of as a real number between $0$ and $1$. If the gladiators are unsatisfied with their blessing, then they may pray at most one more time, receiving a new amount of power which replaces the old one. They then fight, and the gladiator with the greatest blessing wins.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How can a gladiator guarantee that they win with probability at least $1/2$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reason I like this puzzle is because there is one strategy which is &quot;obviously correct:&quot; namely, pray again if your first blessing is less than $0.5$. Surprisingly, there are strategies which beat this more than half the time. As a warning, this is a bit more computation-heavy than your average clever math puzzle.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-04-29T04:53:38.017" LastActivityDate="2015-04-29T06:55:26.973" Title="Blessings of Ares" Tags="|mathematics|game|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12912" PostTypeId="2" ParentId="12908" CreationDate="2015-04-29T03:54:41.510" Score="27" Body="&lt;p&gt;The answer is &lt;strong&gt;Range&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can place the letter &quot;a&quot; in front of each word to get a new word. Abroad, Acorn, Again, Ajar, Alike, Amount, Aside. The word &quot;range&quot; requires an extra letter to make it work, Arrange.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9289" LastActivityDate="2015-04-29T03:54:41.510" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12913" PostTypeId="2" ParentId="12864" CreationDate="2015-04-29T03:55:39.913" Score="1" Body="&lt;p&gt;This problem is actually underdetermined. Credit goes to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/12864/find-the-liar-in-the-library#comment40250_12864&quot;&gt;Mark N&lt;/a&gt; for the idea of everyone entering the room at the same time&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;Counter-example: Everyone enters the room at the same time.&lt;/h2&gt;&#xA;&lt;h3&gt;Solution 1: Albert is the liar&lt;/h3&gt;&#xA;&lt;p&gt;Albert can claim that they didn't see Bernard even if Bernard was in fact there. As long as Bernard claims that they saw Albert, this follows the rule that &lt;strong&gt;one&lt;/strong&gt; of them saw the other.&lt;/p&gt;&#xA;&lt;h3&gt;Solution 2: Bernard is the liar (similar argument)&lt;/h3&gt;&#xA;&lt;p&gt;Bernard can similarly claim that they didn't see Albert even if Albert was in fact there. As long as Albert claims that they saw Bernard, this follows the rule that &lt;strong&gt;one&lt;/strong&gt; of them saw the other.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Since we have arrived at two valid solutions, this is an underdetermined system.&lt;/p&gt;&#xA;" OwnerUserId="11094" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-29T04:07:55.903" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12914" PostTypeId="2" ParentId="12864" CreationDate="2015-04-29T04:45:31.207" Score="0" Body="&lt;p&gt;-&gt; is saw; &amp;lt;- is seen&lt;br&gt;&#xA;A -&gt; B E&lt;br&gt;&#xA;B -&gt; A I&lt;br&gt;&#xA;C -&gt; D I&lt;br&gt;&#xA;D -&gt; A I&lt;br&gt;&#xA;E -&gt; B C&lt;br&gt;&#xA;I -&gt; C E  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &amp;lt;- B D&lt;br&gt;&#xA;B &amp;lt;- A E&lt;br&gt;&#xA;C &amp;lt;- E I&lt;br&gt;&#xA;D &amp;lt;- C&lt;br&gt;&#xA;E &amp;lt;- A I&lt;br&gt;&#xA;I &amp;lt;- B C D  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thief lied, so that means the thief claims he saw someone when he didn't. That means the thief saw 1 or 0 people. Conversely, only 1 person at most could have seen the thief. Therefore, since D is the only one seen by 1 person or less, D must be the thief.&lt;/p&gt;&#xA;" OwnerUserId="11890" LastEditorUserId="11890" LastEditDate="2015-04-29T05:56:19.680" LastActivityDate="2015-04-29T05:56:19.680" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12915" PostTypeId="2" ParentId="12904" CreationDate="2015-04-29T05:41:14.420" Score="7" Body="&lt;p&gt;The password is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PARADOXICAL&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To see why&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As @Allan's already helpfully pointed out in his image&#xA;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/ROjgl.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA; &lt;br&gt;&#xA; you can split the entire thing into eleven columns. For each letter, pick the corresponding column from each segment and line them up next to each other, with the top segment first, then the middle, then the bottom. You'll get something like this (minus the gridlines) after you do each column:&#xA;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/E9kBZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1600" LastActivityDate="2015-04-29T05:41:14.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12916" PostTypeId="2" ParentId="12911" CreationDate="2015-04-29T05:59:46.730" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The answer is that the gladiator should pray for another blessing if their blessing is below $1/\phi$ where $\phi$ is the golden ratio.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Suppose that the gladiator prays for a new blessing if the blessing they received were less than $x_n$, where $n$ is the identifier for the specific gladiator. We need to find the value of $x_1$ that ensures that the probability of having a higher blessing than gladiator 2 is always at least $\frac12$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy enough to find the distribution for each gladiator. The probability density for each gladiator is given by&#xA;$$&#xA;p_n(x) = \begin{cases}x_n &amp;amp; x&amp;lt;x_n\\ 1+x_n \qquad&amp;amp; \text{otherwise}\end{cases}&#xA;$$&#xA;Now, we seek $Pr(X_1\geq X_2)$ where $X_n$ is distributed according to $p_n(x)$. We can do this (because they are independent) by seeking&#xA;$$&#xA;Pr(X_1\geq X_2)=\int_0^1 Pr(X_1\geq X_2=x) dx = \int_0^1 Pr(X_1\geq x)p_2(x)dx&#xA;$$&#xA;Now,&#xA;$$&#xA;Pr(X_1\geq x) = \int_x^1 p_1(y)dy = \begin{cases} 1-x_1x &amp;amp; x&amp;lt;x_1 \\ (1-x)(1+x_1)\qquad  &amp;amp; \text{otherwise} \end{cases}&#xA;$$&#xA;and so we have&#xA;$$&#xA;Pr(X_1\geq X_2)=\begin{cases}\frac{(x_2-x_1)(x_1x_2+x_2-1)+1}2 &amp;amp; x_1&amp;lt;x_2 \\ \frac{(x_2-x_1)(x_1x_2+x_1-1)+1}2 &amp;amp; x_1\geq x_2\end{cases} \tag{1}&#xA;$$&#xA;Evaluating the derivative with respect to $x_1$, we have&#xA;$$&#xA;0 = \frac{\partial}{\partial x}Pr(X_1\geq X_2)=\begin{cases}\frac{x_2^2-2x_1x_2-x_2+1}2 &amp;amp; x_1&amp;lt;x_2 \\ \frac{x_2^2-2x_1x_2+x_2-2x_1+1}2 &amp;amp; x_1\geq x_2\end{cases}&#xA;$$&#xA;Solving these gives&#xA;$$&#xA;x_1 = \begin{cases} \frac{x_2^2-x_2+1}{2x_2} &amp;amp; x_1&amp;lt;x_2 \\ \frac{x_2^2+x_2+1}{2x_2+2} \qquad &amp;amp; x_1\geq x_2\end{cases}&#xA;$$&#xA;These intersect where $x_1=x_2=\frac12(\sqrt{5}-1)$. All that remains is to show that this guarantees at least $\frac12$ chance of winning. We substitute it into (1) to get&#xA;$$&#xA;Pr(X_1\geq X_2)=\begin{cases}\frac{(\sqrt5+1)x_2^2-4x_2+\sqrt5+1}4 &amp;amp; x_2&amp;gt;\frac12(\sqrt{5}-1) \\ \frac{(\sqrt5-1)x_2^2+(2\sqrt5-6)x_2+2\sqrt5-2}4 &amp;amp; x_2\leq \frac12(\sqrt{5}-1)\end{cases}&#xA;$$&#xA;And it can easily be seen that both cases remain above 0.5.&lt;/p&gt;&#xA;" OwnerUserId="9289" LastEditorUserId="10615" LastEditDate="2015-04-29T06:55:26.973" LastActivityDate="2015-04-29T06:55:26.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12917" PostTypeId="2" ParentId="12864" CreationDate="2015-04-29T06:25:00.200" Score="10" Body="&lt;p&gt;This is perhaps the interval graph solution LeGrandDODOM referred to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Create a graph by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; adding an edge between each pair where one supposedly saw the other.&#xA;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/g2fe0.jpg&quot; alt=&quot;Graph representing people who saw each other&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We then have&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;   If everyone is telling the truth, the resulting graph is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Interval_graph&quot; rel=&quot;nofollow noreferrer&quot;&gt;interval graph&lt;/a&gt;. Interval graphs clearly cannot have an induced $C_4$, but the resulting graph has at least two:  $ADIB$ and $ADCE$.  Hence not everyone is telling the truth and the lies must correspond to edges or non-edges to make both of these $C_4$s not induced.  Thus $A$ or $D$ is lying as these are the only two on both $C_4$s.  $D$ may be lying about seeing $A$, or perhaps $A$ is lying about not seeing $C$ and $I$.  As far as I can tell, either case can happen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2722" LastActivityDate="2015-04-29T06:25:00.200" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="12918" PostTypeId="2" ParentId="12909" CreationDate="2015-04-29T07:15:13.543" Score="0" Body="&lt;p&gt;I think it should be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; steroids&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I spike, you will get quite enthused,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; some kind of drug&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For great strength, I am often abused,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; steroids give more strength&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am quintessentially male...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; males often use this drug.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The next line might derail:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; your life/career might derail if you get caught using it (also, some scientific article about derailing T cells? )&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With Hashish amiga, my name's all confused.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; yes I am confused by this&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9835" LastActivityDate="2015-04-29T07:15:13.543" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12919" PostTypeId="2" ParentId="12892" CreationDate="2015-04-29T08:23:15.367" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Switch off mobile phone&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is the result when doing a reverse google image search. Probably not the correct answer but pretty funny given this is related to cinema :-)&lt;/p&gt;&#xA;" OwnerUserId="11880" LastActivityDate="2015-04-29T08:23:15.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12923" PostTypeId="1" AcceptedAnswerId="12928" CreationDate="2015-04-29T09:06:51.033" Score="30" ViewCount="20199" Body="&lt;p&gt;&lt;a href=&quot;http://netsec.ethz.ch/people/reischuk/riddles/&quot;&gt;Came across this puzzle&lt;/a&gt; and couldn't think of a solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You are given nine bottles of wine out of which one is poisoned. Given&#xA;  two mice (that will die when taken a sip of poisoned wine), how can&#xA;  you find out which bottle is the poisoned one by only performing two&#xA;  tests?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I misunderstood the exact meaning of test. I thought that each “drink/sip” was a test. Thanks for all the answers.&lt;/p&gt;&#xA;" OwnerUserId="9233" LastEditorUserId="9233" LastEditDate="2015-04-29T11:43:48.783" LastActivityDate="2018-06-13T13:58:20.513" Title="Nine Bottles of Wine, One Poisoned, Two Mice, Two Tests" Tags="|logical-deduction|" AnswerCount="6" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12924" PostTypeId="2" ParentId="12923" CreationDate="2015-04-29T09:16:37.530" Score="53" Body="&lt;p&gt;Start by having mouse 1 sip from 1, 2, and 3...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...and mouse 2 sip from 3, 4, and 5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they both die, it was bottle 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If mouse 1 dies, have mouse 2 sip from 1. If it dies, it's 1. If not, it's 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If mouse 2 dies, have mouse 1 sip from 4. If it dies, it's 4. If not, it's 5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If neither die, you're left with 6, 7, 8, and 9. Have mouse 1 sip 6 and 7, and mouse 2 sip 7 and 8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If both die, it's 7. If mouse 1 dies, it's 6. If mouse 2 dies, it's 8. If neither die, it's 9.&lt;/p&gt;&#xA;" OwnerUserId="9289" LastActivityDate="2015-04-29T09:16:37.530" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12928" PostTypeId="2" ParentId="12923" CreationDate="2015-04-29T09:25:29.293" Score="48" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Number the bottles 0 through 8; these can be represented as $2$ digit ternary numbers (example, $7=21_3$). Call the mice Alice and Bob.&lt;br&gt;&#xA; &lt;ul&gt;&lt;li&gt;For the first experiment, have Alice drink from every bottle whose first ternary digit is $0$, and have Bob drink from those whose &lt;em&gt;second&lt;/em&gt; ternary digit is $0$.&lt;/li&gt;&#xA;&lt;li&gt;For the second experiment, replace every $0$ in the previous sentence with $1$.&lt;/li&gt;&lt;/ul&gt;&#xA; After the two experiments, the experiment during which Alice died (if any) tells you the first ternary digit of the poisoned bottle: if she died on the first experiment, it is $0$, on the second, it is $1$, if never, it is $2$. Similarly, Bob's lifespan tells you the second digit. Thus, afterwards you will know the bottle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2018-06-13T13:58:20.513" LastActivityDate="2018-06-13T13:58:20.513" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="12929" PostTypeId="2" ParentId="12909" CreationDate="2015-04-29T10:07:19.160" Score="0" Body="&lt;p&gt;Is it  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cannabis/Marijuana  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I spike, you will get quite enthused,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is a drug  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With Hashish amiga, my name's all confused.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is known by many names  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A test of mental acuity&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; long-term cannabis users show morphological brain alterations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9669" LastActivityDate="2015-04-29T10:07:19.160" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12930" PostTypeId="2" ParentId="12892" CreationDate="2015-04-29T10:20:29.000" Score="14" Body="&lt;p&gt;Pretty sure the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I ate his liver with some fava beans and a nice chianti&quot; (from The Silence of the Lambs)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why? Because, the pictures are (from left to right):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lollipop&lt;br&gt;&#xA; Saltless&lt;br&gt;&#xA; Hills&lt;br&gt;&#xA; Slivers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then we just need to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Remove any characters from each word that appear more than once.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastActivityDate="2015-04-29T10:20:29.000" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12931" PostTypeId="2" ParentId="12923" CreationDate="2015-04-29T11:36:15.310" Score="5" Body="&lt;p&gt;Number the bottles from $1$ to $9$ and the mice $A$ and $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First test&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A$ drinks $1$, $2$ and $3$.&lt;br&gt;&#xA;$B$ drinks $1$, $4$ and $5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Cases:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If both die, poison was in $1$.&lt;/li&gt;&#xA;&lt;li&gt;If only $A$ dies, poison is in $2$ or $3$ &lt;/li&gt;&#xA;&lt;li&gt;If only $B$ dies, poison is in $4$ or $5$&lt;/li&gt;&#xA;&lt;li&gt;If nobody dies, poison is in $6,7,8$ or $9$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second test&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Depending on what happened after the first test, we have these cases:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Poison is in bottle $1$&lt;/li&gt;&#xA;&lt;li&gt;$B$ drinks $2$.&lt;br&gt;&#xA;If he dies, poison is in $2$, else it's in $3$&lt;/li&gt;&#xA;&lt;li&gt;$A$ drinks $4$.&lt;br&gt;&#xA;If he dies, poison is in $4$, else it's in $5$&lt;/li&gt;&#xA;&lt;li&gt;$A$ drinks $6,7$&lt;br&gt;&#xA;$B$ drinks $6,8$&lt;br&gt;&#xA;If both die, poison is in $6$. If $A$ dies, poison is in $7$. If $B$ dies, poison is in $8$. If nobody dies, poison is in $9$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-04-29T11:36:15.310" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12932" PostTypeId="2" ParentId="12324" CreationDate="2015-04-29T11:39:02.933" Score="0" Body="&lt;p&gt;Here is the answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;! Monkey spends and fish and reaches the island, he collects all the bananas and comes back to crocodile!&lt;br&gt;&#xA;  SURPRISE!!! Crocodile met his old friend there and was talking to him.&lt;br&gt;&#xA;  Monkey convinces the second crocodile to travel along with him for a fish and carries 2 bundles back with the help of 2 crocodiles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11866" LastActivityDate="2015-04-29T11:39:02.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12933" PostTypeId="1" AcceptedAnswerId="12937" CreationDate="2015-04-29T12:30:04.777" Score="4" ViewCount="1109" Body="&lt;p&gt;You are in a room tied to a chair. The devil tells you that he is gonna ask two groups of 1000 &quot;people&quot; a question and then he will ask you something. If you answer correctly you go to heaven, else you go to hell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He steps back a bit and a dirty man pops into the space to the devils left side out of thin air. He looks scared and is crying, his clothes are dirty and just look like rags like from middle ages. From the best of your knowledge he could be from the middle ages. The devil tells the man that he has to guess a number between 1 and 32 (including both) and that he has 5 tries. After every guess he will know if the guessed number was bigger or smaller than the one the devil had in mind. If he answers correctly he can go to heaven, if not he goes to hell. The man cries and says that he barely knows someone who can count to 5. The devil doesn't care.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suddenly a glass barrier is erupted between you and the devil; you can't hear anything but you can still see what's going on. After the man opened his mouth for the fifth time a trap door opens and he falls down into the fiery depths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The glass barrier disappears and the devil summons a half robot half human looking thing to his right side. The robot/human doesn't indicate that he feels anything or is scared. The devil asks him again the same question. The robot says that its impossible to answer this correctly he also adds that he is a deterministic being and doesn't like using randomness. Again the devil doesn't care.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The glass barrier is erupted again. After the robots speaker lights up for the fifth time, he is also thrown down into hell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This procedure goes until 1000 peasants and 1000 robots have come through. You counted all the results and came to the surprising result that around 50% of the peasants and robots actually went to heaven! You also counted the number of guesses it took the candidates if they got the correct number. For the peasants it was completely random, 20% chance that they succeeded on the first guess, 20% on the second etc... The robots, on the other hand, guessed correctly only at the third, fourth, or fifth guess; never sooner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the devil comes over to you and asks you: &quot;If you add up all the numbers I chose, without duplicates, what do you get?&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Some clarifications:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Around 50% of the peasants survived and around 50% of the robots survived&lt;/li&gt;&#xA;&lt;li&gt;For the sake of clarifying things, the devil choose a different number then the one before after each person (he can still choose 1,2,1,2 etc., just different from the one before)&lt;/li&gt;&#xA;&lt;li&gt;The devil didn't say he is gonna choose a number at random, he follows a system obviously&lt;/li&gt;&#xA;&lt;li&gt;It's not 448&lt;/li&gt;&#xA;&lt;li&gt;The peasants don't know the number 32 even though they just heard it&lt;/li&gt;&#xA;&lt;li&gt;The peasants always count from 1 to 5&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5560" LastEditorUserId="9000" LastEditDate="2015-04-29T15:54:38.007" LastActivityDate="2015-05-08T12:53:08.530" Title="The devils number" Tags="|logical-deduction|" AnswerCount="3" CommentCount="13" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12934" PostTypeId="1" AcceptedAnswerId="14393" CreationDate="2015-04-29T12:36:14.727" Score="8" ViewCount="788" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What's happening is very strange,&lt;br&gt;&#xA;  I belong to another StackExchange!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I'm indeed off-topic here in Puzzling,&lt;br&gt;&#xA;  It isn't always clear what I'm saying.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Even though I'm often too broad,&lt;br&gt;&#xA;  Review it, something is flawed.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I'm just suggesting an edit,&lt;br&gt;&#xA;  A low quality job, ain't it?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;For new users I'm a big trouble&lt;br&gt;&#xA;  A duplicate isn't always a double.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Your current work you shall reopen&lt;br&gt;&#xA;  Flag here, flag there, it's all broken!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Oh, don't forget to add a comment,&lt;br&gt;&#xA;  late it's not, takes just a moment!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What am I?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm sometimes cause and consequence of inappropriate language&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Upvoted or downvoted, doesn't matter, I generally appear in the bottom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #3:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I notify you to review something.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-05-06T11:52:21.757" LastActivityDate="2015-05-06T12:55:31.737" Title="Review this, now!" Tags="|riddle|rhyme|" AnswerCount="10" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="12936" PostTypeId="2" ParentId="12934" CreationDate="2015-04-29T12:55:13.353" Score="0" Body="&lt;p&gt;Since you are talking about all kinds of actions you can do here I'll take a guess and think you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A question on the meta stackexchange &lt;a href=&quot;http://meta.puzzling.stackexchange.com/&quot;&gt;http://meta.puzzling.stackexchange.com/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-04-29T12:55:13.353" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12937" PostTypeId="2" ParentId="12933" CreationDate="2015-04-29T13:03:07.400" Score="8" Body="&lt;p&gt;It seems to me that the problem is indeterministic which probably means I'm missing something. At the very least, it's too loosely worded.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If 50% of the peasants get to heaven and peasants all know the numbers $1-5$, that means that the numbers is in the set $[1-5]$ 50% of the time and in the set $[6-32]$ 50% of the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Peasants will guess all the numbers in the set $[1-5]$ in some order. We know they don't perform a binary search like the robots because, if they had, they would always get it right by the 3rd guess (or not at all). We know they got it right with equal probability for each guess to they either pick the numbers randomly or simply count from $1$ to $5$ with the devil saying &quot;Higher&quot; each time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As suggested in the comments, a machine is likely to use a dichotomic search. That means they're going to start with the midpoint ($16$) and keep choosing the next midpoint based on whether the devil says the answer is higher or lower. In practice, that would look like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;First:                                                              16                                                          &#xA;Second:                             8                                                               24                          &#xA;Third:              4                               12                              20                              28          &#xA;Fourth:     2               6               10              14              18              22              26              30  &#xA;Fifth:  1       3       5       7       9       11      13      15      17      19      21      23      25      27      29      31&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note that $32$ is never chosen by the robot. It would need a 6th guess to get that one. (It would &lt;em&gt;know&lt;/em&gt; the answer by the 5th guess ($\log_232=5$) but it wouldn't have an opportunity to actually say it.) For any other number, the robot is guaranteed to get it by the 5th guess. Since only 50% of the robots get it right, that means that $32$ is the answer 50% of the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can safely conclude that the robots did not pick $17$ as the first midpoint. They are deterministic and do not like randomness so we can assume they all followed the same scheme. Similarly, it follows that, had they picked the higher midpoint, then they would have &lt;em&gt;always&lt;/em&gt; picked the higher midpoint. That means the lower half would split at $17$, $9$, $5$, $3$, $2$. In that case, they would have never chosen $1$ as an answer instead of never choosing $32$. However, they could have solved for any number $[2-32]$. If 50% of the robots go to hell, that means that the devil picked $1$ 50% of the time. This contradicts with what we know based on the peasants' pattern. Therefore, we reject $17$ as the first midpoint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now comes the problem: We know the answer is $32$ 50% of the time and we know the answer is somewhere in $[1-5]$ the other 50%. We know the devil chose each with equal probability because of how the peasants were guessing. That means the final answer is $47$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Caveat:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are problems in the question's wording: (All have been clarified by OP)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strike&gt;We're only assuming that each peasant can count to 5 but, based on the wording of what the peasant said, it's more likely that most can't.&lt;/strike&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strike&gt;The devil, being not so very nice, could have had a different scheme for the robots and peasants. There's no reason why, when it's a peasant, he picks $[1-5]$ 50% of the time and $[6-32]$ 50% of the time while for a robot he picks $[1-31]$ 50% and $[32]$ 50%. The final answer can be anything between $33$ and $528$.&lt;/strike&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strike&gt;If the peasants guess randomly in the set $[1-5]$, then it's possible for the devil to pick, say, $3$ 50% and $32$ 50%. Depending on when they peasants randomly guessed $3$, that gives the 20% / guess result. The final answer can be anything between $33$ and $47$.&lt;/strike&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strike&gt;The robots could have picked 17 as their midpoint which can shift which number(s) they'll never guess. In that case, the answer can be anything be as low as $19$.&lt;/strike&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strike&gt;The peasants can only &lt;em&gt;count&lt;/em&gt; to $5$ but that doesn't mean they don't know other numbers exist. It's possible for the devil to pick a number higher than $5$, the peasant randomly guesses $5$ first, the devil says &quot;Higher&quot;, and the peasant says, &quot;Uhhh... 14?&quot;. The wording only &lt;em&gt;implies&lt;/em&gt; that the peasant doesn't know numbers higher than $5$.&lt;/strike&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2015-05-08T12:53:08.530" LastActivityDate="2015-05-08T12:53:08.530" CommentCount="19" ContentLicense="CC BY-SA 3.0" />
  <row Id="12938" PostTypeId="2" ParentId="12933" CreationDate="2015-04-29T13:17:31.663" Score="1" Body="&lt;p&gt;Middle-age men can only count up to $5$,so they choose a number between $1$ and $5$. Since they have a $50%$ chance of surviving, we deduce that the devil guessed a number between $1$ and $5$ half the times, and a number greater of $5$ the other times.&lt;br&gt;&#xA;Every robot is asked &lt;strong&gt;the same question&lt;/strong&gt; of his previous peasant, meaning that the devil guessed the same number for the robot and the middle-age man.&lt;br&gt;&#xA;An intelligent robot would use a binary search algorithm to identify the number, starting from $16$ or $17$. After one wrong attempt, they try $8$ or $9$ or $24$ or $25$ (depending on what the devil answered). After a second wrong attempt, they try $4$, $12$, $20$ or $29$, and so on, until they find the number. Since the robots have always failed to identify the number with the first two steps, we deduce that the devil never guessed $8$, $9$, $16$,$17$, $24$, $25$.&lt;br&gt;&#xA;Among the remainings, we can't know which ones he chose, he could have chosen just one or all.&#xA;If only one, we get the the minimum sum, which is $1+2+3+4+5+6+7=28$&#xA;If all, we get the maximum sum (all the numbers except the excluded ones), which is $429$&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="11244" LastEditDate="2015-04-29T13:24:10.127" LastActivityDate="2015-04-29T13:24:10.127" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12939" PostTypeId="2" ParentId="12934" CreationDate="2015-04-29T13:35:44.710" Score="0" Body="&lt;p&gt;&lt;strong&gt;EDITED ANSWER&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet I'm not sure, but guess is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Spam Questions&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's happening is very strange,&#xA;I belong to another StackExchange!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it's strange to identify it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm indeed off-topic here in Puzzling,&#xA;It isn't always clear what I'm saying.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Spam are off topics, which are not clear&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Even though I'm often too broad,&#xA;Review it, something is flawed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; spam message found broad sometimes, also flawed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For new users I'm a big trouble&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; its trouble for new users&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Oh, don't forget to add a comment,&#xA;late it's not, takes just a moment!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; usually we add comment or flagged it as spam question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4312" LastEditorUserId="4312" LastEditDate="2015-04-30T03:58:03.150" LastActivityDate="2015-04-30T03:58:03.150" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12940" PostTypeId="1" CreationDate="2015-04-29T13:47:34.970" Score="-2" ViewCount="1601" Body="&lt;p&gt;What is the minimum number of steps required to cut this chocolate bar which is a single piece into at least 30 single pieces?&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;1)Image for chocolate bar&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8OweZ.png&quot; alt=&quot;Chocolate bar&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;2)Image for single piece&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DkqRW.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;br&gt;&#xA;Your knife is only as long as the width of a single piece and stacking (laying rows or columns on top of each other when you've cut those lose) is not allowed.&lt;/p&gt;&#xA;" OwnerUserId="10433" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-02T12:24:06.660" Title="Minimum number of steps required to cut this bar" Tags="|geometry|strategy|visual|" AnswerCount="6" CommentCount="21" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12941" PostTypeId="2" ParentId="12934" CreationDate="2015-04-29T13:50:31.983" Score="8" Body="&lt;p&gt;You are a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Closed question&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What's happening is very strange, I belong to another StackExchange!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Questions can be closed if they belong to a different SE.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm indeed off-topic here in Puzzling,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Off-topic questions are closed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It isn't always clear what I'm saying.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Questions that are unclear are closed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Even though I'm often too broad,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Questions that are too broad are closed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Review it, something is flawed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If a question is closed, it can be flagged to be reopened, in which case it enters the review queue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm just suggesting an edit,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Closed questions can be edited.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A low quality job, ain't it?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Low quality questions can be closed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For new users I'm a big trouble&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Closed questions can lead to a question ban, especially if you're a new user.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A duplicate isn't always a double.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sometimes questions closed as duplicates aren't actually duplicates.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your current work you shall reopen, Flag here, flag there, it's all broken!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you can improve a closed question, you can flag to reopen it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Oh, don't forget to add a comment, late it's not, takes just a moment!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People can still comment on closed questions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2015-04-29T13:50:31.983" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12942" PostTypeId="2" ParentId="12940" CreationDate="2015-04-29T13:59:33.800" Score="3" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 54. my previous answer still hold but you have to make 5 cuts instead of 1 when trying to seperate a row. so 6 * 5 + 4 * 6 = 54.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would say (after edit of the question this answer is no longer viable):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 24 cuts first 6 cuts to get 6 seperate rows (you will keep 1 piece with 2 rows attached) then you have to cut each column which requires 4 cuts, so 4 * 6 = 24.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;edited calculation error.&lt;/p&gt;&#xA;" OwnerUserId="9835" LastEditorUserId="9835" LastEditDate="2015-04-29T14:32:17.870" LastActivityDate="2015-04-29T14:32:17.870" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12943" PostTypeId="1" AcceptedAnswerId="14414" CreationDate="2015-04-29T14:04:56.330" Score="5" ViewCount="987" Body="&lt;p&gt;Here is another easier riddle to solve but I hope you enjoy the rhymes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Here I am in my perfect state.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Friends of mine in some way relate.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;But what confines me isn't straight.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Is it none, one, or many? That's to debate!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you still don't know, it's not too late.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I gave some hints, but you're doing great!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now don't be shy and participate,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But remember not to over concentrate!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is it, and what is the 'debate' about?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint #1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Don't concentrate on 'perfect state' so much as the word state was more so for the rhyme. I would focus more on line 3.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That you're in my realm, don't postulate!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To one number I associate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note: Hint #2 might not help too much (or be 100% accurate), but I was able to rhyme it :)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If you enjoyed this riddle, some of my other rhyming riddles can be found &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%5Brhyme%5D%20user%3A11244%20is%3Aquestion&quot;&gt;here&lt;/a&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-06-19T13:27:06.660" Title="Riddle: Here in my perfect state" Tags="|riddle|rhyme|" AnswerCount="9" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12944" PostTypeId="2" ParentId="12940" CreationDate="2015-04-29T14:08:49.297" Score="2" Body="&lt;p&gt;All this thinking made me hungry so I figured I'd eat the chocolate pieces we didn't need. (sorry)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ended up with &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;40&lt;/strong&gt;&lt;br&gt;&#xA;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/3dVaR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Disclaimer: This is not meant as a serious answer, it's meant to make you smile.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="11251" LastEditorUserId="11251" LastEditDate="2015-04-29T15:25:19.810" LastActivityDate="2015-04-29T15:25:19.810" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12945" PostTypeId="2" ParentId="12909" CreationDate="2015-04-29T14:11:16.133" Score="3" Body="&lt;p&gt;Thanks to Geobits for help on the last line! &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Testosterone  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I spike, you will get quite enthused,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Testosterone is often related to libido, and when that spikes.... Well, you know.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For great strength, I am often abused,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Testosterone is often used as a steroid to help bodybuilders.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am quintessentially male...  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Testosterone and male are almost synonyms in today's society.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The next line might derail:&lt;br&gt;&#xA; With Hashish amiga, my name's all confused.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hashishamiga is a substitute-cipher for testosterone.&lt;br&gt;&#xA;$H-&amp;gt;T$, $A-&amp;gt;E$, $S$ is the same, $I-&amp;gt;O$, $M-&amp;gt;R$, and $G-&amp;gt;N$&lt;br&gt;&#xA; Each vowel goes to the &quot;next.&quot; Not sure how the others were generated though.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the missing tag should be  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Cryptograms&quot;, though &quot;rhyme&quot; would fit as well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="5874" LastEditDate="2015-05-04T14:49:22.940" LastActivityDate="2015-05-04T14:49:22.940" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12946" PostTypeId="2" ParentId="12940" CreationDate="2015-04-29T14:16:09.053" Score="2" Body="&lt;p&gt;&quot;What is the minimum number of steps required to cut this chocolate bar which is a single piece into at least 30 single pieces?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this question isn't about word play, but until it's stated I'll keep this as an option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3. There are only three steps required. 1. Cut a chocolate bar edge. 2. Check/Stop if we have 30 pieces. 3. Repeat previous steps.&#xA; &lt;br/&gt;&lt;br/&gt; Alternatively if you are some sort of savage, you can use just 1 step: Cut all the chocolate bar edges.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-04-30T14:42:08.143" LastActivityDate="2015-04-30T14:42:08.143" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12947" PostTypeId="2" ParentId="12940" CreationDate="2015-04-29T14:21:16.547" Score="3" Body="&lt;p&gt;My answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 54, because almost every piece needs to be cut at least 2 times starting from second top piece and go around bar, 3 corner pieces will be ignored as they will be cut because of other pieces.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11911" LastEditorUserId="99" LastEditDate="2015-04-29T14:30:51.523" LastActivityDate="2015-04-29T14:30:51.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12948" PostTypeId="2" ParentId="12943" CreationDate="2015-04-29T14:35:41.603" Score="2" Body="&lt;p&gt;the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; quantum particle, yay quantum.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here I am in my perfect state.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a quantum particle exists in a 'super' position.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Friends of mine in some way relate.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; spooky interaction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But what confines me isn't straight.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it jumbles around also it's either a wave or a marble.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is it one, or many? That's to debate!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; quantum theory is still a theory so there is 'debate' about it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9835" LastActivityDate="2015-04-29T14:35:41.603" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12949" PostTypeId="2" ParentId="12940" CreationDate="2015-04-29T14:37:37.300" Score="1" Body="&lt;p&gt;The simplist solution is 54:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/S5JFB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that there are 6 rows of 5 horizontal cuts and 4 rows of 6 vertical cuts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to maximize the remaining number of uncut lengths.  For an $X*Y$ rectangle, the number of uncut lengths is equal to $(4*X*Y-2*(Y+X))/2$.  For $X*Y=10$ this becomes $20-10/X-X$.  The maximum integer value for this is $X=3$ but Y must be an integer so this doesn't work.  This best solution, therefore, is for $X=2$ and $Y=5$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume that the vertical cuts are $5/6$ the length of horizontal cuts, we can improve this even further.  The image above would only use 50 cuts as the long vertical lengths need one less cut to complete them.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditDate="2015-04-29T17:42:00.893" LastActivityDate="2015-04-29T17:42:00.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12950" PostTypeId="2" ParentId="12923" CreationDate="2015-04-29T14:49:36.100" Score="0" Body="&lt;p&gt;Feed Alice a sip from 1, 2 and 3. Feed Bob a sip from 1, 4 and 5.&lt;br&gt;&#xA;If both die, it's bottle 1.&lt;br&gt;&#xA;If only Bob lives, give Bob a sip from 1, 2, 4, 5, 6, 7, 8, 9. If Bob dies, it's bottle 2, else 3.&lt;br&gt;&#xA;If only Alice lives, give Alice a sip from 1, 2, 3, 4, 6, 7, 8, 9. If Alice dies, it's bottle 4, else 5.&lt;br&gt;&#xA;If both live, give Alice a sip from 1, 2, 3, 4, 5, 6, 7 and Bob a sip from 1, 2, 3, 4, 5, 6, 8.&lt;br&gt;&#xA;If both die, it's bottle 6.&lt;br&gt;&#xA;If only Alice dies, it's bottle 7.&lt;br&gt;&#xA;If only Bob dies, it's bottle 8.&lt;br&gt;&#xA;If both lives, it's bottle 9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This way,&lt;br&gt;&#xA;if it was bottle 1, you know nothing else,&lt;br&gt;&#xA;if it was bottle 2, you know for sure 1, 4 and 5 are safe,&lt;br&gt;&#xA;if it was bottle 3, you know for sure 1, 2, 4, 5, 6, 7, 8 and 9 are safe,&lt;br&gt;&#xA;if it was bottle 4, you know for sure 1, 2 and 3 are safe,&lt;br&gt;&#xA;if it was bottle 5, you know for sure 1, 2, 3, 4, 6, 7, 8 and 9 are safe,&lt;br&gt;&#xA;if it was bottle 6, you know for sure 1, 2, 3, 4 and 5 are safe,&lt;br&gt;&#xA;if it was bottle 7, you know for sure 1, 2, 3, 4, 5, 6 and 8 are safe,&lt;br&gt;&#xA;if it was bottle 8, you know for sure 1, 2, 3, 4, 5, 6 and 7 are safe,&lt;br&gt;&#xA;if it was bottle 9, you know for sure 1, 2, 3, 4, 5, 6, 7 and 8 are safe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even better, if it's bottle 3, 5, 7, 8 or 9 you can feed the remaining mice the remaining unknown bottle, to make 100% sure it contain poison while you get drunk on the first set of bottles.&lt;/p&gt;&#xA;" OwnerUserId="2711" LastActivityDate="2015-04-29T14:49:36.100" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12951" PostTypeId="1" AcceptedAnswerId="12952" CreationDate="2015-04-29T15:02:44.123" Score="7" ViewCount="7555" Body="&lt;p&gt;I got &quot;x&quot; number of sweets. I can divide my sweets equally by 1,2,3,4,5,6,7 all the way to 101 friends such that I have no left overs. Find the least number of sweets I could possibly have at first?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: This logic is a troll, so I thought it is fun. Wanna see how people answer. Anyways, the answer is still a number. Not kidding and it is possible. No cutting of sweets happened.&lt;/p&gt;&#xA;" OwnerUserId="11369" LastEditorUserId="11369" LastEditDate="2015-05-01T07:26:01.317" LastActivityDate="2018-06-25T03:30:04.330" Title="(Troll Math Question) Divided by sweets" Tags="|mathematics|logical-deduction|lateral-thinking|" AnswerCount="8" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="12952" PostTypeId="2" ParentId="12951" CreationDate="2015-04-29T15:07:02.263" Score="39" Body="&lt;p&gt;Troll question?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The minimum number of sweets you can have is clearly 0.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="30633" LastEditDate="2018-06-25T03:30:04.330" LastActivityDate="2018-06-25T03:30:04.330" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="12953" PostTypeId="2" ParentId="12933" CreationDate="2015-04-29T15:21:48.700" Score="0" Body="&lt;p&gt;I think the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 25 + 26 + 27 + 28 + 29 + 30 + 31 + 32 = 228&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is because &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the peasants randomly selected numbers from the available sets of values, so there was always a chance that they could select the right value with their first guess,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the robots used a dichotomic search - so their first three guesses would always be 16, then 24, then 28; since the devil always picked a number larger than 24, the robots would never guess on their first or second attempt.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9278" LastActivityDate="2015-04-29T15:21:48.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12955" PostTypeId="1" AcceptedAnswerId="12960" CreationDate="2015-04-29T15:36:35.220" Score="2" ViewCount="1994" Body="&lt;p&gt;As a continuing series of &quot;divided by&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today is my Birthday! I wil be cutting my own birthday cake and share equally to 11 of my friends.Assuming that I do straight cuts and there is no topping (I don't have money for topping :/) on a non-flat round cake, what is the minumum cuts needed for me to cut equally the cake for myself and friends? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note:No, its not &quot;0&quot; this time! It's a legitminate answer! &lt;/p&gt;&#xA;" OwnerUserId="11369" LastActivityDate="2015-06-08T09:50:36.067" Title="Divided by cakes" Tags="|logical-deduction|" AnswerCount="4" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12956" PostTypeId="2" ParentId="12951" CreationDate="2015-04-29T15:38:14.510" Score="19" Body="&lt;p&gt;Just for fun, the answer is also:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\int\limits_{-\infty}^{+\infty}(\frac{-(\sqrt{-0} \times sinh(f'(x)))^3}{π^e})dx$ &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-04-30T12:53:07.777" LastActivityDate="2015-04-30T12:53:07.777" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="12958" PostTypeId="2" ParentId="12955" CreationDate="2015-04-29T15:43:24.203" Score="2" Body="&lt;p&gt;I'm definitely missing something&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You need 12 pieces. You said you wanted to share equally to 11 of your friends, sharing implies yourself. Note that nothing implies needing to eat the whole cake, however the fewest amounts of cuts will have you eating the whole cake. The answer would be 6 cuts, or if you are really awesome at gauging pieces (since its a non-flat cake), you could cut the cake in half vertically and then make 3 cuts to divide the halves into 6 pieces each. But you'd have to be a walking computer to do that&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastActivityDate="2015-04-29T15:43:24.203" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12959" PostTypeId="2" ParentId="12955" CreationDate="2015-04-29T15:47:13.860" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You need 4 cuts are needed because you need 12 pieces (11 friends and you)&lt;br&gt;&#xA; 1- cut the cake in half vertically (you have now 2 pieces)&lt;br&gt;&#xA; 2- cut each half in third vertically (you have now 6 pieces)&lt;br&gt;&#xA; 3- cut the cake in half horizontally (you have now 12 pieces)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10400" LastActivityDate="2015-04-29T15:47:13.860" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12960" PostTypeId="2" ParentId="12955" CreationDate="2015-04-29T15:47:40.887" Score="6" Body="&lt;p&gt;The minimum is  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $4$ cuts&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First you draw $2$ parallel lines to divide the cake in 3 parts, then you draw a perpendicular line, getting 6 equal pieces. At the end, you make a longitudinal cut to split all the pieces, getting 12 parts (one for you and 11 for friends). &#xA;See this picture:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/NS3oi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another equivalent solution, perhaps simpler (and symmetrical), is this:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/EqJah.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-04-29T19:20:49.480" LastActivityDate="2015-04-29T19:20:49.480" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="12961" PostTypeId="2" ParentId="12934" CreationDate="2015-04-29T15:55:45.223" Score="1" Body="&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The review queue!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I belong to another stack exchange&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you belong to all the stack exchanges. They all have the review queue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For everything else:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; These are all line items or pieces of a review-able item.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9412" LastActivityDate="2015-04-29T15:55:45.223" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12962" PostTypeId="2" ParentId="12923" CreationDate="2015-04-29T16:09:51.577" Score="7" Body="&lt;p&gt;I posted a similar problem (and an explanation of the solution although not an outright formula) in an answer to a &lt;a href=&quot;https://puzzling.stackexchange.com/a/421/88&quot;&gt;different question&lt;/a&gt; about puzzles. Everyone else here has the right answer, although they don't explain it as well.&lt;/p&gt;&#xA;&lt;p&gt;Each mouse has three possible states — dead in one day, dead in two days, not dead. This gives us that the information content of two trials is &lt;span class=&quot;math-container&quot;&gt;$3^n$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; is the number of mice — in this case, &lt;span class=&quot;math-container&quot;&gt;$n = 2$&lt;/span&gt;, so &lt;span class=&quot;math-container&quot;&gt;$3^n = 9$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;But that brings us no closer to an algorithm that actually makes use of that information. So let's see what we'd need to do in order to be able to tell between bottles on the second day.&lt;/p&gt;&#xA;&lt;p&gt;On the first day, we do a binary trial in order to identify the wine as being in one of &lt;span class=&quot;math-container&quot;&gt;$2^n$&lt;/span&gt; groups of bottles where &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; is the number of mice.&lt;/p&gt;&#xA;&lt;p&gt;However we configure the groups on the first day, on the second day there need to be enough mice left (and enough bottles eliminated) so that you can do another binary trial on the remaining bottles. So we need to structure the first day in such a way that:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If both mice are eliminated, there is only one possible bottle.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If one mouse is eliminated, there are two possible bottles.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If neither mouse is eliminated, there are four possible bottles.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;This is done by making each group contain &lt;span class=&quot;math-container&quot;&gt;$2^{2-n}$&lt;/span&gt; bottles if it's being given to &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; mice. So you'd give a single bottle to both mice on the first day, two different bottles to each individual mouse, and leave four bottles alone. This results in the algorithm some people have given:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Give bottle &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; to both mice, &lt;span class=&quot;math-container&quot;&gt;$2, 3$&lt;/span&gt; to the first mouse, and &lt;span class=&quot;math-container&quot;&gt;$4, 5$&lt;/span&gt; to the second mouse.&lt;/p&gt;&#xA;&lt;p&gt;If both mice die on the first day, then the poisoned bottle was bottle &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;If one mouse dies, then the poisoned bottle is one of the bottles given to that mouse. So test the remaining two bottles (either &lt;span class=&quot;math-container&quot;&gt;$2, 3$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$4, 5$&lt;/span&gt;) with the other mouse by giving it a sip of one of them. If the mouse dies, it was that bottle; if it survives, it was the other one.&lt;/p&gt;&#xA;&lt;p&gt;If both mice survive, then do another binary trial with the four remaining bottles. Give &lt;span class=&quot;math-container&quot;&gt;$6, 7$&lt;/span&gt; to the first mouse and &lt;span class=&quot;math-container&quot;&gt;$6, 8$&lt;/span&gt; to the second mouse.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If both mice die, then the poisoned bottle was &lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;If the first mouse dies, then the poisoned bottle was &lt;span class=&quot;math-container&quot;&gt;$7$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;If the second mouse dies, then the poisoned bottle was &lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;If both mice survive, then the poisoned bottle was &lt;span class=&quot;math-container&quot;&gt;$9$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;You can use this strategy for 243 bottles and five mice as well — on the first day, give one bottle to each of five mice, two bottles to each of the five distinct groups of four mice, four bottles to the ten distinct groups of 3 mice, etc. and leave 32 bottles alone.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-29T16:09:51.577" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12963" PostTypeId="1" AcceptedAnswerId="12964" CreationDate="2015-04-29T16:34:40.140" Score="12" ViewCount="1033" Body="&lt;p&gt;Since there have been quite a few poisoned-bottle problems posted to the site already, I thought I'd post a general formula to cover all future variations of this problem (at least where a single poisoned bottle is concerned).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You have $n \le (m+1)^k$ bottles of wine, one of which is poisoned. The poisoned wine kills any person who drinks it at the stroke of midnight on the day that they drank it. You have $k$ prisoners to test the wine by making them drink it and possibly killing them, and $m$ days to conduct these death trials. How do you plan out the tests so that you can figure out which bottle is poisoned?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-04-29T16:44:08.690" LastActivityDate="2015-04-29T16:44:08.690" Title="The completely generalized poisoned-bottle problem" Tags="|logical-deduction|" AnswerCount="1" CommentCount="2" CommunityOwnedDate="2015-04-29T18:47:45.390" ContentLicense="CC BY-SA 3.0" />
  <row Id="12964" PostTypeId="2" ParentId="12963" CreationDate="2015-04-29T16:34:40.140" Score="7" Body="&lt;p&gt;Each prisoner has $m+1$ possible states from the trials &amp;mdash; dead in 1 day, dead in 2 days, ..., dead in $m$ days, not dead. This makes for a total of $(m+1)^k$ total possible states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this is an &lt;a href=&quot;https://puzzling.stackexchange.com/questions/410/how-do-you-solve-a-word-puzzle-when-you-have-no-idea-where-to-start&quot;&gt;information puzzle&lt;/a&gt;, we want to map each bottle to a possible state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assign each bottle a $m+1$-ary number with $k$ digits (or alternatively, an ordered $k$-tuple with integers from $0$ to $m$), we can make this represent the days on which we need to give the prisoners these bottles as testing. Specifically, on the $i$th day, we give the $p$th prisoner any bottle where the $p$th digit is $i$, so that if the prisoner dies on that day, then we know that the poisoned bottle must have the $p$th digit equal to $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way, we can narrow the possibilities down to a single bottle. For example, if $m = 3$ and $k = 4$, and the bottle (out of $256$) is labelled $0130$ (out of $0000$, $0001$, $0002$, $0003$, $0010$, $0011$, etc. up until $3333$), then the following will happen:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;On the first day, the second prisoner dies. So we know that the second digit on the poisoned bottle's label is $1$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;On the second day, nobody dies, so we know that none of the bottles labelled with a $2$ are poisoned.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;On the third day, the third prisoner dies and nobody else does, so we know that the third digit is $3$, and the rest are $0$. Therefore, the poisoned bottle is the one labelled $0130$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-04-29T16:34:40.140" CommentCount="4" CommunityOwnedDate="2015-04-29T16:34:40.140" ContentLicense="CC BY-SA 3.0" />
  <row Id="12965" PostTypeId="2" ParentId="12324" CreationDate="2015-04-29T16:36:11.277" Score="2" Body="&lt;p&gt;Here's an idea:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Crocodiles aren't limited to the water, so make the trip one that goes directly through the island to the other side of the lake. While the crocodile is crossing the island, collect the bananas from one tree. Get on the crocodile before it's finished crossing the island, then let it swim to the other side to complete the first trip. Repeat this going the other way. Voila, you've spent two fish to get the bananas from both trees!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9696" LastActivityDate="2015-04-29T16:36:11.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12966" PostTypeId="1" AcceptedAnswerId="13035" CreationDate="2015-04-29T16:38:23.267" Score="1" ViewCount="149" Body="&lt;p&gt;Our houses empty, were on a raid&lt;br&gt;&#xA;Nomadic creatures who never stayed&lt;br&gt;&#xA;Predators on a smaller scale&lt;br&gt;&#xA;Tiny in mass but in no way frail  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our brothers colonise in their zounds&lt;br&gt;&#xA;Along the ground, under mounds&lt;br&gt;&#xA;A matriarch leads us through the fray&lt;br&gt;&#xA;The caverns flood with larva, no play  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Racism rife in amongst the kingdoms&lt;br&gt;&#xA;Pillaging, taking the enemies freedoms&lt;br&gt;&#xA;No free time can distract me from finding food&lt;br&gt;&#xA;We work as one for the greater good  &lt;/p&gt;&#xA;" OwnerUserId="5135" LastActivityDate="2015-04-30T00:29:13.810" Title="Slaves to the mother" Tags="|riddle|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12967" PostTypeId="2" ParentId="12908" CreationDate="2015-04-29T16:45:53.913" Score="9" Body="&lt;p&gt;The answer is BROAD.   Each of the other words can be used as a verb.  However BROAD is widely used as an adjective ----- but not a verb. That is a distinct characteristic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers, everybody!  My first time on the site!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zephyr&lt;/p&gt;&#xA;" OwnerUserId="11918" LastEditorUserId="5373" LastEditDate="2015-04-29T22:07:45.960" LastActivityDate="2015-04-29T22:07:45.960" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12969" PostTypeId="2" ParentId="12943" CreationDate="2015-04-29T16:51:13.133" Score="1" Body="&lt;p&gt;Probably wrong but I'll try anyway&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Are you the triple point of water? See &lt;a href=&quot;https://en.wikipedia.org/wiki/Triple_point#Gas.E2.80.93liquid.E2.80.93solid_triple_point&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here I am in my perfect state.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's the perfect conditions of temperature and pressure to see the water in solid phase, liquid phase and gas phase simultaneously.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Friends of mine in some way relate.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All molecules have a triple point but they are all different in temperature and pressure&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But what confines me isn't straight.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; At those conditions, you need something that can hold the 3 states of the water and that something need to be closed because of the gas.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is it one, or many? That's to debate!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; At those conditions, we have one molecule, but many phases.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10400" LastActivityDate="2015-04-29T16:51:13.133" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12970" PostTypeId="1" AcceptedAnswerId="12972" CreationDate="2015-04-29T16:54:57.563" Score="3" ViewCount="1160" Body="&lt;p&gt;A group of hardy pirates are sailing along when they encounter the Royal Navy of Lord Richard. The battle is long and vicious. Sadly, the pirate ship is sunk. The only survivors are five of the crew and the captain's foul-mouthed parrot. The lot make it to a tiny island nearby in time to watch the Navy's vessel disappear over the horizon. Disheartened but determined to survive, the 5 crewman go foraging. The only food they are able to gather is a collection of mangoes. (They're terrible hunters and fisherman.) The parrot finds some nuts but it's not sharing. The pirates pile up all their mangoes and go to sleep.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some time in the night, one of the crew wake up. Worried that he won't get his fair share in the morning - pirates are so rude, after all - he divides the big pile into 5 smaller piles. There's one left over, so he tosses it to the parrot. He hides one of the smaller piles, rolls the rest back into a single pile, and goes to sleep.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the first pirate has fallen asleep, a second pirate wake up. He, too, is worried about getting his fair share. Being of the same mindset of the first to awake, the second pirate also divides the big pile (actually, it's what's left of the original pile after the first got done with it but the second doesn't know that) into 5 smaller piles. Again, there's one left over so he tosses it to the parrot. He hides his pile, rolls the rest back together, and goes to bed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This repeats for every pirate. Each divides what's left into 5 piles, gives the 1 leftover to the parrot, hides their pile, and rolls the rest back together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the sun comes up, all the pirates notice that what's left in the pile is much smaller than the night before. None of them say anything, though, because they're all guilty and don't want the others to decide to eat &lt;em&gt;him&lt;/em&gt; instead. They divide up what's left into 5 piles. This time, there's none leftover for the parrot. (Suits it right. The dang thing never even thanked them for the mangoes last night.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the pirates munch on their pile while thinking about their secret stash they can sneak back to later. One dies later of an infected cut he received during the battle. The rest are rescued by a passing ship. Three more are killed when the rescue ship is sunk by a storm. The last survives in a dinghy and eventually reaches the mainland. He lives for another 26 years, eventually dying of heart disease. He never found true love or happiness in life. His fondest memory is of eating mangoes on a tropical island with 4 men he had bonded with through the fires of battle.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the minimum number of mangoes in the original pile?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that mangoes are counted as integers.&lt;br&gt;There will be no decimal or negative mangoes allowed.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2016-08-31T15:32:10.903" LastActivityDate="2016-08-31T15:32:10.903" Title="The Pirates and the Mangoes" Tags="|mathematics|story|no-computers|" AnswerCount="2" CommentCount="3" ClosedDate="2015-04-29T17:38:38.783" ContentLicense="CC BY-SA 3.0" />
  <row Id="12971" PostTypeId="1" AcceptedAnswerId="13093" CreationDate="2015-04-29T16:57:25.330" Score="8" ViewCount="1568" Body="&lt;p&gt;Three people were getting married to each other,&lt;br&gt;&#xA;&amp;emsp;&amp;emsp; Though polygamous they were not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All men they were in the marriage,&lt;br&gt;&#xA;&amp;emsp;&amp;emsp; Though homosexuality was not in their minds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They three share two parents,&lt;br&gt;&#xA;&amp;emsp;&amp;emsp; Though they detested the notion of inbreeding. (And this was not, in fact, a consanguineous marriage)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who are these men?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first line makes you assume more than has been said. The &quot;people&quot; referred to &quot;are&quot; not what you assume &quot;them&quot; to be.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 45 chromosomes are identical (Not a single difference, even on the atomic level) between the &quot;three men&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hint 2 does not say that the 46th was different. And two of the &quot;three men&quot; were the same person all along.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Two of the men are the same person. This is a different hint from hint 3.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user10203" LastEditorDisplayName="user10203" LastEditDate="2015-04-30T14:38:35.763" LastActivityDate="2015-04-30T15:44:30.613" Title="A Questionable Act to Society; Identify these People" Tags="|riddle|" AnswerCount="11" CommentCount="12" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12972" PostTypeId="2" ParentId="12970" CreationDate="2015-04-29T17:05:28.743" Score="3" Body="&lt;p&gt;The question is best understood with a basic grasp of modular arithmetic. The answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3121&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You need to find a number to fit that, mod 5, is 1. Then 4/5ths of that mod 5 is 1. Then 4/5ths - 1 of that etc. Then, 4/5ths of that mod 5 is 0. I brute forced it at this point, with &lt;a href=&quot;http://pastebin.com/tHiMpQ4p&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Please note &lt;a href=&quot;/questions/tagged/no-computers&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;no-computers&amp;#39;&quot; rel=&quot;tag&quot;&gt;no-computers&lt;/a&gt; was added after I posted this question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user10203" LastEditorDisplayName="user10203" LastEditDate="2015-04-29T17:47:03.537" LastActivityDate="2015-04-29T17:47:03.537" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12974" PostTypeId="2" ParentId="12940" CreationDate="2015-04-29T17:07:28.957" Score="0" Body="&lt;p&gt;Same answer as everyone else:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 54&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can come to this conclusion without trying all of the different cutting patterns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total number of edges that connect one piece to another is 67.  Minimizing the number of cuts is equivalent to maximizing the number of edges that remain uncut.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the problem reduces to choosing 10 or less pieces that are part of your &quot;remainder&quot; such that their total number of uncut edges is maximized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most uncut edges I have been able to find amongst 10 or less pieces is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 13&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which gives me the same result as everyone else.&lt;/p&gt;&#xA;" OwnerUserId="10850" LastActivityDate="2015-04-29T17:07:28.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12975" PostTypeId="5" CreationDate="2015-04-29T17:13:37.100" Score="0" Body="&lt;p&gt;Chestnuts are old or well-known puzzles that some may have already heard. Ideally, there should be a twist added before posting them.&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2015-04-29T17:23:23.080" LastActivityDate="2015-04-29T17:23:23.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12976" PostTypeId="4" CreationDate="2015-04-29T17:13:37.100" Score="0" Body="Chestnuts are old or well-known puzzles that some may have already heard. Ideally, there should be a twist added before posting them." OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2015-04-29T17:23:16.217" LastActivityDate="2015-04-29T17:23:16.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12978" PostTypeId="2" ParentId="12970" CreationDate="2015-04-29T17:22:44.810" Score="1" Body="&lt;p&gt;Suppose that $a$ is the what's left in the morning. $a$ must be a multiple of 5, of course, else they could not split it.&lt;br&gt;&#xA;To obtain what the fifth pirate had, let's say $b$, we do&lt;br&gt;&#xA;$b=\frac54 a +1$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As well, we can do the same for the other pirates:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c=\frac54 b +1$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d=\frac54 c +1$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e=\frac54 d +1$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f=\frac54 e +1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $f$ is the original amount of mangos. Now, we can chain all those equations to get $f$ in correspondence to $a$. The denominator of that fraction is, of course, $4^5=1024$.&lt;br&gt;&#xA;The numerator is&lt;br&gt;&#xA;$(((((a\times5 +4)\times5 +16)\times5 +64) \times5 +256)\times5 +1024)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$((((25a +36)\times5 +64) \times5 +256)\times5 +1024)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(((125a +244) \times5 +256)\times5 +1024)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$((625a +1476)\times5 +1024)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(3125a +8404)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When is $(3125a +8404)\over1024$ integer? That's an easy diophantine equation, which solves for $a=1020$, returning $3121$ mangos!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is &lt;strong&gt;$3121$&lt;/strong&gt; mangos!&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-04-29T17:33:43.377" LastActivityDate="2015-04-29T17:33:43.377" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12979" PostTypeId="2" ParentId="12971" CreationDate="2015-04-29T17:25:24.697" Score="2" Body="&lt;p&gt;[lateral-thinking] &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has been invalidated by an edit to the riddle stating explicitly that it is not a consanguineous marriage.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Technically, they could be just three human males getting married. You're not polygamous until after you're married and, at this moment, they're only getting married. Homosexuality was not on their minds at this moment because they were all thinking about the three wedding cakes. The detested the notion of inbreeding (or marrying within the family) but that doesn't mean they wouldn't do it. Maybe they were being forced into the marriage by an evil alien overlord.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2015-04-29T17:28:17.220" LastActivityDate="2015-04-29T17:28:17.220" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12981" PostTypeId="2" ParentId="12971" CreationDate="2015-04-29T17:30:23.127" Score="7" Body="&lt;p&gt;Somewhat fitting (but probably wrong):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Proton, Neutron and Electron. They are all being 'bonded' to form an element, but do not have the a conscious to think or detest. Their parents are mass and charge.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-04-29T17:56:24.813" LastActivityDate="2015-04-29T17:56:24.813" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12982" PostTypeId="1" AcceptedAnswerId="12989" CreationDate="2015-04-29T17:38:33.057" Score="22" ViewCount="1956" Body="&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/11281/the-hangman-game-backward&quot;&gt;Hangman game backward&lt;/a&gt; version 2 - guess what was the question, with only 1 letter missing:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;fill in a &quot;b&quot; - I might lose control anytime, so keep me on my toes.&lt;br&gt;&#xA;fill in an &quot;m&quot; - my brothers are all metalhead but me.&lt;br&gt;&#xA;fill in a &quot;p&quot; - my buddy has two big tusks and loves to carry me around all day long.&lt;br&gt;&#xA;fill in a &quot;w&quot; - This is usually where the presidents stay.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, your answer can be in the form of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The question was: _ p p l e&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;followed by explanation.&lt;/p&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-04-29T18:18:22.437" Title="The hangman game backward #2" Tags="|riddle|english|language|letters|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12983" PostTypeId="2" ParentId="12971" CreationDate="2015-04-29T17:57:41.757" Score="6" Body="&lt;p&gt;Are they&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Canadians?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The 3 people are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The French, the English and the native American&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10400" LastActivityDate="2015-04-29T17:57:41.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12984" PostTypeId="2" ParentId="12934" CreationDate="2015-04-29T17:59:22.877" Score="1" Body="&lt;p&gt;Inspired by (and very close to) pacoverflow's answer, but hopefully different enough:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A flagged question  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What's happening is very strange,&#xA; I belong to another StackExchange!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Questions can be flagged to be migrated to the appropriate section.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm indeed off-topic here in Puzzling,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Off topic questions can be flagged for deletion. Also, many questions that would be considered &quot;flaggable&quot; (I take artistic license) on other sites would be fine here (answer known, not informative, etc.) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It isn't always clear what I'm saying.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Questions that are too broad are often flagged.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Even though I'm often too broad,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Same as before.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Review it, something is flawed.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Flagged questions are reviewed to address some problem.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm just suggesting an edit,&#xA; A low quality job, ain't it?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many flags are due to a need for an edit, possibly for low quality questions.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For new users I'm a big trouble  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; New users don't want to be flagged and kicked. Also, they can scare new users away.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A duplicate isn't always a double.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sometimes flagged if it might be a duplicate, or to see if it is.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your current work you shall reopen  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Closed questions can be flagged for review and potential reopening.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Flag here, flag there, it's all broken!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Flags all around! You can only flag so many items in a time period.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Oh, don't forget to add a comment,&#xA; late it's not, takes just a moment!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Users appreciate a comment as to why their questions are flagged, or you need to add a comment when you flag for what should be looked at.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="4450" LastEditDate="2015-04-29T19:55:16.277" LastActivityDate="2015-04-29T19:55:16.277" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12985" PostTypeId="2" ParentId="12971" CreationDate="2015-04-29T18:01:16.380" Score="2" Body="&lt;p&gt;Hilarious lateral-thinking answer: it's a marriage between&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Three People Were&lt;/strong&gt; (a woman), who together with her siblings formed a gang called &lt;strong&gt;They Three&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Each Other&lt;/strong&gt;, nicknamed &lt;strong&gt;All Men They&lt;/strong&gt; (a man).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Three people were getting married to each other,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, obviously!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Though polygamous they were not.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polygamy has nothing to do with it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;All men they were in the marriage,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Though homosexuality was not in their minds.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No homosexuality here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;They three share two parents,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Siblings share two parents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Though they detested the notion of inbreeding.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nothing to do with inbreeding either!&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-04-29T18:01:16.380" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12986" PostTypeId="2" ParentId="12943" CreationDate="2015-04-29T18:03:08.780" Score="0" Body="&lt;p&gt;Similar to Vincent Advocaat's answer, how about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a photon?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Here I am in my perfect state.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A photon is in some way &quot;perfect&quot;: massless, always travelling at the same speed, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Friends of mine in some way relate.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Other types of particle in the &quot;quantum zoo&quot; relate to each other more than any does to the photon. E.g. matter and antimatter: we have protons and antiprotons, electrons and positrons, etc., but there's no such thing as an antiphoton.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But what confines me isn't straight.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Photons form electromagnetic waves, and a wave should be wavy and not straight.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Is it one, or many? That's to debate!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Is light made of a wave or a stream of many particles? Wave-particle duality gives rise to quite a debate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-04-29T18:03:08.780" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12987" PostTypeId="2" ParentId="12971" CreationDate="2015-04-29T18:07:23.003" Score="2" Body="&lt;p&gt;Partial Answer! Could it be something in the form of:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An alliance? With three leaders that are all male, of three countries that all originated from one original country?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nothing comming to my mind, but someone may be better at history than I am if this is the correct path.  &lt;/p&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-04-29T18:07:23.003" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12989" PostTypeId="2" ParentId="12982" CreationDate="2015-04-29T18:18:22.437" Score="28" Body="&lt;p&gt;The question was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;_allet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fill in a &quot;b&quot; - I might lose control anytime, so keep me on my toes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ballet - Dancers dance on their toes all the time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fill in an &quot;m&quot; - my brothers are all metalhead but me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mallet - Rubber or sometimes wooden headed hammer, as opposed to the metal ones.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fill in a &quot;p&quot; - my buddy has two big tusks and loves to carry me around all day long.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pallet - They are usually picked up by forklifts&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fill in a &quot;w&quot; - This is usually where the presidents stay. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wallet - Presidents is a slang term for money in the USA because all of their bills have pictures of their presidents.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1777" LastActivityDate="2015-04-29T18:18:22.437" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12990" PostTypeId="1" AcceptedAnswerId="12997" CreationDate="2015-04-29T18:20:12.223" Score="7" ViewCount="576" Body="&lt;p&gt;In 1908, Phileas Fogg, the man who went around the world in eighty days, died of pneumonia. Jean Passepartout, his assistant, had inherited everything Fogg owned, and so became a nobleman himself. One day a few months afterward, Passepartout wagered £500 with his fellow members at the Reform Club that he could do something even more impossible than his benefactor's feat of octogesimadiurnal circumnavigation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;A month from now,&quot; stated Passepartout, &quot;I will attempt to drive a horse carriage from London to Paris and then to Berlin within the span of three days. If I successfully make the trip, you shall pay me five hundred pounds. If I fail, so shall I give the same to you.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Preposterous,&quot; said his friends. &quot;A horse can barely run one hundred miles a day in a single burst, and to go from London to Berlin alone without the detour to Paris is already seven hundred miles. We will not take you up on this insulting bet!&quot; For his friends felt that it would be degrading of their honour to take a bet that they had no chance of losing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Not only will I do it within three days,&quot; said Passepartout, &quot;but I will keep the horse at a regular pace of thirty miles per day.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A month later, he won the bet. How did he do it?&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-04-29T20:37:12.000" LastActivityDate="2015-04-29T20:39:10.570" Title="Driving from London to Paris to Berlin" Tags="|lateral-thinking|geography|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12991" PostTypeId="1" AcceptedAnswerId="13014" CreationDate="2015-04-29T18:20:44.620" Score="4" ViewCount="213" Body="&lt;p&gt;Find a pair of English words each of which can be a noun, verb, adjective, or adverb such that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;as nouns, adjectives, and adverbs, they're &lt;em&gt;opposite&lt;/em&gt; in meaning&lt;/li&gt;&#xA;&lt;li&gt;as verbs, they have the &lt;em&gt;same&lt;/em&gt; meaning&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(I'm pretty sure the answer to this is unique.)&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-04-29T20:00:08.423" Title="Find the synonymous opposites!" Tags="|word|english|language|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12992" PostTypeId="2" ParentId="12990" CreationDate="2015-04-29T18:23:52.073" Score="0" Body="&lt;p&gt;He&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; set up villages named Paris and Berlin, the first 45 miles from London and the second 45 miles from the first.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, it &lt;em&gt;is&lt;/em&gt; tagged lateral-thinking! This answer certainly works.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-04-29T18:23:52.073" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12993" PostTypeId="2" ParentId="12990" CreationDate="2015-04-29T18:26:33.783" Score="-1" Body="&lt;p&gt;I think the important factor of the bet is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That he said &quot;I will attempt...&quot;, so regardless of whether he actually travels the distance in 3 days or not, he should have successfully attempted to travel winning the bet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-04-29T18:31:54.923" LastActivityDate="2015-04-29T18:31:54.923" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12995" PostTypeId="1" AcceptedAnswerId="13001" CreationDate="2015-04-29T18:38:46.823" Score="106" ViewCount="20990" Body="&lt;p&gt;You have been incarcerated in a high-security prison for several nefarious crimes which we need not go into here. The warden, being a kindly logician, offers you a single chance to escape. You are blindfolded and given a choice of five doors, one of which leads to freedom and the other four to prison cells. You know that the doors are red, blue, green, yellow, and white; they are evenly spaced, each 2 paces from the next; and you are standing in front of the middle one. There is a guard standing by each door, and &lt;em&gt;all five guards tell the truth&lt;/em&gt; (weren't expecting that, were you?). They make the following statements:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The red door is somewhere to the left of the door leading to freedom.&lt;/li&gt;&#xA;&lt;li&gt;The blue door is not at either end.&lt;/li&gt;&#xA;&lt;li&gt;The green door is 3 doors away from the door leading to freedom (2 doors between them).&lt;/li&gt;&#xA;&lt;li&gt;The yellow door is adjacent to the door leading to freedom.&lt;/li&gt;&#xA;&lt;li&gt;The white door is the middle one.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To find out how many paces you need to take to which side in order to be standing in front of the door to freedom, you need to answer the following two questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;What is the order of the coloured doors, left to right?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Which colour of door leads to freedom?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You may assume that all statements made about &quot;left&quot; and &quot;right&quot; refer to &lt;em&gt;your&lt;/em&gt; left and right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I found more or less this puzzle on the internet and thought it would be easy, but for some reason it's a fair bit harder than it looks!)&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="57212" LastEditDate="2023-06-20T05:59:23.803" LastActivityDate="2023-06-20T05:59:23.803" Title="Five coloured doors, which leads to freedom?" Tags="|logical-deduction|story|logic-grid|" AnswerCount="9" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12996" PostTypeId="2" ParentId="12943" CreationDate="2015-04-29T18:41:07.857" Score="0" Body="&lt;p&gt;I'm thinking:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Marriage or Love&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Relates to current events!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here I am in my perfect state.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some say marriage is a perfect union.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Friends of mine in some way relate.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We have friends that are married.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But what confines me isn't straight.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some are straight and some are not.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is it one, or many? That's to debate!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hmm.  Polyamory is also an option...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now don't be shy and participate,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Marriage is for everyone.  Ask someone out!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But remember not to over concentrate!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's not a hard question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is it, and what is the 'debate' about?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Current debate on gay marriage going on.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10294" LastActivityDate="2015-04-29T18:41:07.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12997" PostTypeId="2" ParentId="12990" CreationDate="2015-04-29T18:44:21.493" Score="9" Body="&lt;p&gt;It's important that the bet happened a month later, because he took a month to get to the place where he carried out the bet:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He traveled to the province of Ontario, in Canada, where there are three cities in close proximity named London, Paris, and Berlin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here, he could travel the three cities in less than 90 miles, and probably could even use the roadways. But if not, then his buggy gets to do a little work.&lt;/p&gt;&#xA;" OwnerUserId="9951" LastEditorDisplayName="user88" LastEditDate="2015-04-29T20:39:10.570" LastActivityDate="2015-04-29T20:39:10.570" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12998" PostTypeId="2" ParentId="12951" CreationDate="2015-04-29T18:48:24.113" Score="14" Body="&lt;p&gt;Your sweets are powdered candy. You simply divide the amount you have into portions of equal weight, no cutting required. Probably something to the effect of a kilogram of candy for a satisfying portion when divided by 101.&lt;/p&gt;&#xA;" OwnerUserId="5464" LastActivityDate="2015-04-29T18:48:24.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12999" PostTypeId="2" ParentId="12971" CreationDate="2015-04-29T18:50:35.577" Score="1" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Male triplets in the womb, perhaps conjoined?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are married to each other in the sense of &quot;any close or intimate association or union&quot;. It is an asexual relationship and they would share the same two parents. It appears to meet all the requirements. However, conjoined triplets are incredibly rare.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11923" LastEditorUserId="9000" LastEditDate="2015-04-29T20:14:56.550" LastActivityDate="2015-04-29T20:14:56.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13000" PostTypeId="2" ParentId="12995" CreationDate="2015-04-29T18:51:21.773" Score="15" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AC8zR.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This seems to be compatible with all the statements.&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="29343" LastEditDate="2022-09-05T06:10:57.123" LastActivityDate="2022-09-05T06:10:57.123" CommentCount="12" ContentLicense="CC BY-SA 4.0" />
  <row Id="13001" PostTypeId="2" ParentId="12995" CreationDate="2015-04-29T18:52:48.553" Score="66" Body="&lt;p&gt;Here is my answer. Doors from left to right:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Green, Red, White, Blue, Yellow.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; blue&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;door leads to freedom.&lt;/p&gt;&#xA;&lt;p&gt;Explanation follows.&lt;/p&gt;&#xA;&lt;p&gt;The white door is in the center, you can mark each possible placement of the doors as follows:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/E6I5u.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;br /&gt;&#xA;The correct door just kind of follows suit, you can rule out Yellow, Red, Green and White because green can't be 2 spaces from white, which leaves blue. Because yellow has to be glued to blue and red has to be to the left of it, that means automatically that blue is in the 4th spot with yellow in the 5th. From there Green goes in the first spot and Red in second.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastEditorUserId="69582" LastEditDate="2022-09-05T13:57:22.710" LastActivityDate="2022-09-05T13:57:22.710" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="13002" PostTypeId="2" ParentId="12995" CreationDate="2015-04-29T18:52:51.747" Score="14" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; G R W B Y, where B leads to freedom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I did it by elimination...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Either W is the freedom door or B is, which cuts down on the problem set.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9696" LastActivityDate="2015-04-29T18:52:51.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13003" PostTypeId="2" ParentId="12995" CreationDate="2015-04-29T18:54:26.430" Score="11" Body="&lt;p&gt;The order of the doors is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; GRWBY&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the door to freedom is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Blue (4th door)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Our limitations are:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;R R R R  &#xA;  B B B  &#xA;G G   G G  &#xA;    W&#xA;Y Y Y Y Y&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It can't be the first door, since R can't be left of it&lt;br&gt;&#xA;It can't be the second door, since R would have to left and W is in the middle, so Y couldn't be next to it.&lt;br&gt;&#xA;It can't be middle, since G can't be 3 away from it.&lt;br&gt;&#xA;It can't be last, since only Y and G can be in the last column, and neither can be the door. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-04-29T18:54:26.430" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13004" PostTypeId="2" ParentId="12990" CreationDate="2015-04-29T19:01:56.983" Score="4" Body="&lt;p&gt;Option 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He loaded the horse carriage on to a zeppelin. The horse walked around a track encircling the thing as it flew the path. &lt;a href=&quot;https://www.google.com/maps/dir/London,+UK/Paris,+France/Berlin,+Germany/@50.332368,-2.8122448,5z/data=!3m1!4b1!4m20!4m19!1m5!1m1!1s0x47d8a00baf21de75:0x52963a5addd52a99!2m2!1d-0.1277583!2d51.5073509!1m5!1m1!1s0x47e66e1f06e2b70f:0x40b82c3688c9460!2m2!1d2.3522219!2d48.856614!1m5!1m1!1s0x47a84e373f035901:0x42120465b5e3b70!2m2!1d13.404954!2d52.5200066!3e2&quot; rel=&quot;nofollow&quot;&gt;Google maps&lt;/a&gt; says it's 848 miles in approximately straight lines and the &lt;a href=&quot;http://en.wikipedia.org/wiki/LZ_2&quot; rel=&quot;nofollow&quot;&gt;LZ 2&lt;/a&gt; (debuted in 1906) had a top speed of 25mph with a range of 680 miles. It would have to refuel in Paris but it could still make the trip in less than 36 hours.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Option 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He said he would drive the horse carriage but didn't say that a horse would be pulling it. He hooked it up to an engine and let the horse stay home.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Option 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The horse is on a treadmill system with gearing to produce 10x higher rotational velocity of the wheels on the carriage compared to the wheels on the treadmill. The horse walks 30 miles per day but the carriage travels 300 per day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastActivityDate="2015-04-29T19:01:56.983" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13005" PostTypeId="2" ParentId="12951" CreationDate="2015-04-29T19:05:49.647" Score="12" Body="&lt;p&gt;The real answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $-\infty$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In fact, you didn't specify that the solution is in $\mathbb{N}$, it could also be in  $\mathbb{Z}$ ! The least number satisfying your problem is $-\infty$.&lt;br&gt;&#xA;What does it mean that I have $-\infty$ sweets? It simply means that instead of having them, you have a debt of $-\infty$ sweets (exactly as $-20\$$ means a debt of $20$ dollars).&lt;/p&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-04-29T19:05:49.647" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="13006" PostTypeId="2" ParentId="12995" CreationDate="2015-04-29T19:06:08.283" Score="61" Body="&lt;p&gt;White can't be the door to freedom because it is not possible for it to be three doors away from Green. None of Green, Red and Yellow can be the door to freedom since the descriptions of their locations imply so. Thus Blue is the door to freedom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since Blue is not on either end, and can't be in the middle, it is the second or fourth door from the left. If we suppose that it is the second door, then Red must be the first door, but then Yellow can not be adjacent to Blue. This means that Blue is the fourth door. The only option for Green is to be the first door. Red and Yellow both have one option making them the second and fifth doors respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The prisoner will have to take two paces to the right.&lt;/p&gt;&#xA;" OwnerUserId="8942" LastActivityDate="2015-04-29T19:06:08.283" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13008" PostTypeId="2" ParentId="12934" CreationDate="2015-04-29T19:26:14.877" Score="5" Body="&lt;p&gt;My answer is :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Coding (review)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;What's happening is very strange,&#xA;I belong to another StackExchange!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stackoverflowwwwwwwwww&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I'm indeed off-topic here in Puzzling,&#xA;It isn't always clear what I'm saying.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Obviously doesn't belong in Puzzling and to most people here (maybe) asking code questions would be unclear&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Even though I'm often too broad,&#xA;Review it, something is flawed.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Could refer to code or comments in code, allthough making it generic isnt bad, only in some cases.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I'm just suggesting an edit,&#xA;A low quality job, ain't it?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A lot of code problems come from low quality work&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;For new users I'm a big trouble&#xA;A duplicate isn't always a double.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Make sure you don't duplicate code. That being said a duplicate doesnt need to be an exact copy of something else (double), if it DOES the same thing, its a duplicate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Your current work you shall reopen&#xA;Flag here, flag there, it's all broken!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you did break code you'll have to go re-work it because it'll be breaking all over the place (flags representing errors in some coding environments)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Oh, don't forget to add a comment,&#xA;late it's not, takes just a moment!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The importance of commenting code.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastEditorUserId="11251" LastEditDate="2015-04-29T19:29:01.167" LastActivityDate="2015-04-29T19:29:01.167" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13009" PostTypeId="2" ParentId="12995" CreationDate="2015-04-29T19:36:42.060" Score="5" Body="&lt;p&gt;The order of the doors, from left to right, is...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Green, Red, White, Blue, Yellow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the door to freedom is...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Blue&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;According to statement 5, the white door is in the middle, so our door configuration looks like:&#xA;&lt;br /&gt;&lt;br /&gt;_ _ W _ _&#xA;&lt;br /&gt;&lt;br /&gt;Also, according to statement 3, the green door has two doors between it and the door to freedom.  In the configuration we have now, it is impossible for the white door to be the door to freedom, even if the green door is at either end.  Also, as per statement 3, the green door cannot be 3 doors down from itself, so it can't be the door to freedom.&#xA;&lt;br /&gt;As per statements 1 and 4, neither the red door nor the yellow door are the doors to freedom, as the red door cannot be left of itself and the yellow door cannot be adjacent to itself.  Thus, the blue door must be the door to freedom.&#xA;&lt;br /&gt;&lt;br /&gt;So now, we simply need to try the blue door in various positions in a process of elimination.  Since the blue door is not on the end, it only has two possible positions.&#xA;&lt;br /&gt;&lt;br /&gt;First assume the blue door is in the second position.  Because the red door needs to be to the left of the blue (freedom) door, we would have:&#xA;&lt;br /&gt;&lt;br /&gt;R B W _ _&#xA;&lt;br /&gt;&lt;br /&gt;But now, we have no place to insert the yellow door, as it needs to be adjacent to the blue (freedom) door.&lt;br /&gt;Thus, the blue door needs to be in the forth position.  The only adjacent spot for the yellow door to fit now is in the fifth position.  The green door, being three doors down from the blue door can only go in the first position, leaving the second position for the red door, which allows it to be left of the blue door.  Thus, the solution becomes:&#xA;&lt;br /&gt;&lt;br /&gt;G R W B Y&#xA;&lt;br /&gt;&lt;br /&gt;Where B is the door to freedom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11851" LastActivityDate="2015-04-29T19:36:42.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13010" PostTypeId="2" ParentId="12991" CreationDate="2015-04-29T19:37:19.800" Score="0" Body="&lt;p&gt;I would say&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Plump&lt;/strong&gt; and &lt;strong&gt;Pat&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;As verbs&lt;/strong&gt;&lt;br&gt;&#xA;Both share the meaning of &quot;to touch with hands&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;As adverbs&lt;/strong&gt;&lt;br&gt;&#xA;Pat means &quot;opportunely, appropriately&quot;, while Plump means &quot;with a heavy fall&quot;. Not exact opposites, but a heavy fall is never appropriate!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;As nouns&lt;/strong&gt;&lt;br&gt;&#xA;Pat means &quot;a light stroke with the hand&quot; while Plump means &quot;an abrupt plunge&quot;. The pain is very different, you see!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;As adjectives&lt;/strong&gt;&lt;br&gt;&#xA;Plump means &quot;round, with smooth edges&quot; and Pat means &quot;too simple, unconvincing, rough&quot;. Rough and smooth are opposites!&lt;/p&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-04-29T19:37:19.800" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13011" PostTypeId="2" ParentId="12943" CreationDate="2015-04-29T19:39:56.270" Score="2" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are a &lt;a href=&quot;http://en.wikipedia.org/wiki/Bell_curve&quot; rel=&quot;nofollow&quot;&gt;Bell Curve&lt;/a&gt;, or more specifically, the median value on a Bell Curve in &lt;a href=&quot;http://en.wikipedia.org/wiki/Normal_distribution&quot; rel=&quot;nofollow&quot;&gt;Normal Distribution&lt;/a&gt;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reasoning is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  You are in a &quot;Perfect&quot; state - the very middle of the curve.  Your 'friends' in some way relate in that they are somewhere on the curve. But what confines you, or separates you from other people on the bell, is the curve itself and therefore not 'straight. And whether or not you represent one or multiple people is up to the interpretation of the data represented by this curve.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2756" LastActivityDate="2015-04-29T19:39:56.270" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13012" PostTypeId="2" ParentId="12995" CreationDate="2015-04-29T19:41:09.043" Score="7" Body="&lt;p&gt;I also did this by elimination:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Only Blue and White can be the Freedom Door&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then I narrowed it down even further:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Blue has to be to the left or right of the White door&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And even further-er!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To be two doors away Green has to be on the end&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More further-er-er!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; White can not be the freedom door and still satisfy the Green door condition, so Blue is the freedom door&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How much more further!?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Red has to be to left of blue and Yellow must be next to blue&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally everything fell into place with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Green Red White Blue Yellow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11928" LastActivityDate="2015-04-29T19:41:09.043" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13014" PostTypeId="2" ParentId="12991" CreationDate="2015-04-29T19:55:43.017" Score="4" Body="&lt;p&gt;&lt;strong&gt;My answer is:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Best&lt;/strong&gt; and &lt;strong&gt;Worst&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;As verbs, they both mean to &quot;get the better of&quot; or &quot;defeat&quot; someone.&#xA;&lt;br /&gt;As nouns, adverbs, and adjectives, they are clearly opposite.&#xA;&lt;br /&gt;&lt;br /&gt;&lt;a href=&quot;http://dictionary.reference.com/browse/best?s=t&quot; rel=&quot;nofollow&quot;&gt;http://dictionary.reference.com/browse/best?s=t&lt;/a&gt;&#xA;&lt;br /&gt;&lt;a href=&quot;http://dictionary.reference.com/browse/worst?s=t&quot; rel=&quot;nofollow&quot;&gt;http://dictionary.reference.com/browse/worst?s=t&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11851" LastEditorUserId="11851" LastEditDate="2015-04-29T20:00:08.423" LastActivityDate="2015-04-29T20:00:08.423" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="13015" PostTypeId="2" ParentId="12943" CreationDate="2015-04-29T19:59:03.367" Score="0" Body="&lt;p&gt;Here comes my nerdy answer.  You are a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SQL table  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here I am in my perfect state.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Just a well defined table&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Friends of mine in some way relate.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tables have relationship&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But what confines me isn't straight.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Relationship can be straight line or with angles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is it one, or many? That's to debate!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 to 1 relationship, 1 to N relationship  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2015-04-29T19:59:03.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13016" PostTypeId="2" ParentId="12971" CreationDate="2015-04-29T20:20:07.140" Score="2" Body="&lt;p&gt;One thought is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The holy trinity in Christianity.  The father, the son and the holy spirit.  All &quot;married&quot; or bonded as one.  Nothing in the relationship is sexual.  Can be assumed to be male (the holy ghost being somewhat ambiguous).  Who the two parents are, I can't be sure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another thought, though a bit further out in left field:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Me, myself and I.&#xA;&lt;br /&gt;&lt;br /&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&#xA;&lt;br /&gt;As an elaboration, I would say this is regarding someone with multiple personalities.  The personalities themselves could be being married, or indeed if the person is being &quot;cured&quot; somehow and the personalities merged into one, this could constitute marriage or bond (whether such a thing is possible, I have no idea).  The person is, of course, male.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11851" LastEditorUserId="11851" LastEditDate="2015-04-29T20:41:33.693" LastActivityDate="2015-04-29T20:41:33.693" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13017" PostTypeId="2" ParentId="12934" CreationDate="2015-04-29T20:28:42.387" Score="6" Body="&lt;p&gt;I think I got it&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; you are a bug :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;What's happening is very strange,&#xA;I belong to another StackExchange!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; when you have something weird on your code, you go on stackoverflow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;I'm indeed off-topic here in Puzzling,&#xA;It isn't always clear what I'm saying.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; you don't post questions about bugs here. And sometime, it is hard to explain something we don't understand.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;Even though I'm often too broad,&#xA;Review it, something is flawed.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; sometime the question is too vague to find it and usually we ask a review on a code to find it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;I'm just suggesting an edit,&#xA;A low quality job, ain't it?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; answers are usually simple edits because people do not proof read&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;For new users I'm a big trouble&#xA;A duplicate isn't always a double.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; when you are new, it is hard to determine if the bug is a duplicate of something existing or not&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;Your current work you shall reopen&#xA;Flag here, flag there, it's all broken!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; To isolate it, you have to open your source code and add markers to isolate it until you find the broken part&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;Oh, don't forget to add a comment,&#xA;late it's not, takes just a moment!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; and always comment your bug so you don't do it again&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Thanks to Spacemonkey, for inspiration&lt;/p&gt;&#xA;&lt;h3&gt;Edit&lt;/h3&gt;&#xA;&lt;p&gt;If you are not a bug, you are a&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; bug report&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10400" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-30T11:02:10.943" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13018" PostTypeId="2" ParentId="12943" CreationDate="2015-04-29T20:31:38.297" Score="0" Body="&lt;p&gt;Probably not right, but worth a shot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Universe&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here I am in my perfect state...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You could say that the universe is in a perfect state, or is perfect in general.  It's quite possible that its current configuration is the only possible one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Friends of mine in some way relate...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Makes me think of parallel universes, where other objects or beings can be potentially similar to their representations in our universe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But what confines me isn't straight...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Where does the universe end?  What's outside of it?  I interpret &quot;isn't straight&quot; as isn't known or verified.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is it one, or many? That's to debate!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;How many universes are there?  Is it just ours?  Are there infinite universes?  The debate rages on.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11851" LastActivityDate="2015-04-29T20:31:38.297" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13019" PostTypeId="2" ParentId="12971" CreationDate="2015-04-29T20:42:57.380" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Kellogg's &quot;Snap, Crackle and Pop!&quot; embodying tetraploid rice that happens to have 45/48 identical chromosomes, and 3/48 that for whatever reason of meiosis or varietal development are different.  Though I guess tetraploid varietals would be inbred by definition.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11921" LastActivityDate="2015-04-29T20:42:57.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13020" PostTypeId="2" ParentId="12971" CreationDate="2015-04-29T21:01:53.857" Score="1" Body="&lt;p&gt;My best guess :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A man with multiple personality disorder had a sex change and married his adopted brother. The &quot;marriage&quot; is a genuine marriage, but is between a man, a (newly-made) woman with the same parents but no blood relation, and the male alter-ago of the transexual. No clue about the chromosome part, maybe part of the surgery involved cutting one off.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Just in case this is somehow not the correct answer (though I wrote this before reading the 3rd hint...),&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I &lt;em&gt;do&lt;/em&gt; think two of the 'men' are the same person, somehow. I think in fact 46 of their chromosomes are identical and the 45 reference is a red herring. If they're identical to the &lt;em&gt;atomic level&lt;/em&gt; surely it's impossible for them to be two different people?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pretty sure sex change is involved somehow. The question even says at marriage they &lt;em&gt;were&lt;/em&gt; men, which is ambiguous as to &lt;em&gt;when&lt;/em&gt; they were men.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11880" LastEditorUserId="11880" LastEditDate="2015-04-29T21:41:38.683" LastActivityDate="2015-04-29T21:41:38.683" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13021" PostTypeId="1" AcceptedAnswerId="13022" CreationDate="2015-04-29T21:07:42.463" Score="5" ViewCount="1435" Body="&lt;p&gt;&lt;em&gt;(I am not the origin poster of this question but I have failed to provide the references)&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A scientist claims that he found the way to break the speed of light barrier - faster than the speed of light, with the following experiment:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;He puts an object in the space that is exactly one &lt;a href=&quot;http://earthsky.org/space/what-is-a-light-year&quot; rel=&quot;nofollow&quot;&gt;light&#xA;  year&lt;/a&gt; away from the&#xA;  spot he's standing on earth.  The object, without gravitational force&#xA;  or other influences, stay still.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Then he made a wooden stick that is almost one light year &lt;em&gt;long&lt;/em&gt;, that&#xA;  is, the one end of the stick is in the scientist's hand while the&#xA;  other end is 0.1 mm away from the object.  Assuming he can hold the stick&#xA;  (He might be Dr. Bruce Banner transformed)  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;With the above 2 completed, he now emit a light beam to the object, which in theory, takes 1 year&#xA;  to reach the object.  At the same time, he moves forward his hand,&#xA;  poking the object in space through the stick in hand.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Question is: &lt;strong&gt;Is he actually faster than light, and why / why not?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="5373" LastEditDate="2015-04-29T22:50:15.330" LastActivityDate="2015-04-30T11:47:03.930" Title="Faster than the speed of light" Tags="|lateral-thinking|physics|" AnswerCount="5" CommentCount="7" ClosedDate="2015-04-30T10:13:37.530" ContentLicense="CC BY-SA 3.0" />
  <row Id="13022" PostTypeId="2" ParentId="13021" CreationDate="2015-04-29T21:10:00.643" Score="20" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No. When you push the stick forward you're actually sending a wave down the stick (think like flicking a long rope). This will be slower than light.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11880" LastActivityDate="2015-04-29T21:10:00.643" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="13023" PostTypeId="2" ParentId="13021" CreationDate="2015-04-29T21:19:29.760" Score="0" Body="&lt;p&gt;I say: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;no because the stick only traveled 0.1mm , regardless of when the light gets there.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You could do the same experiment with two persons on earth, X distance between them.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&#xA; Place the stick (it will be shorter now) the same way between the two. Have the 2nd person be a foot and not 0.1mm from the end of the stick. Now when you press a button (or have Bruce Banner poke with the stick) have a humongous boom box blare a sound signal at the same time. Person 2 will get poked and AFTERWARDS here the blaring sound. That doesn't make the stick faster than sound. Its not the stick that made the sound, nor is the stick the one that made the light in the question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastEditorUserId="9412" LastEditDate="2015-04-29T21:54:36.640" LastActivityDate="2015-04-29T21:54:36.640" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13024" PostTypeId="1" AcceptedAnswerId="13026" CreationDate="2015-04-29T21:20:35.773" Score="3" ViewCount="334" Body="&lt;p&gt;Young, in sinc, think two lines narrow, I'm why a bee won't cast a shadow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm not actually Young&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11880" LastActivityDate="2015-04-29T21:40:14.410" Title="What is this strange phenomenon?" Tags="|riddle|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13026" PostTypeId="2" ParentId="13024" CreationDate="2015-04-29T21:35:04.253" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Light diffraction&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;in sinc&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; refers to the mathematical function &lt;code&gt;sinc(x) = sin(x)/x&lt;/code&gt;, which often comes up in analysis of diffraction patterns&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;think two lines narrow&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; refers to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Double-slit_experiment&quot; rel=&quot;nofollow&quot;&gt;Double Slit Experiment&lt;/a&gt;, a simple demonstration of diffraction patterns&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm why a bee won't cast a shadow&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A bee's shadow, unless the bee is rather close to the ground, gets diffracted to the point where you can't see anything.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1706" LastActivityDate="2015-04-29T21:35:04.253" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13027" PostTypeId="2" ParentId="13024" CreationDate="2015-04-29T21:40:14.410" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The phenomenon you are riddling about is the Particle / Wave nature of light&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Young&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Thomas Young, famous for the double slit experiment&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In sinc&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;alluding to the fourier transform of the two box functions that produce the slits. The planar intensity of the light will be a sinc function. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;think two lines narrow&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; again, two narrow slits used in the experiment&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11844" LastActivityDate="2015-04-29T21:40:14.410" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13028" PostTypeId="1" AcceptedAnswerId="13210" CreationDate="2015-04-29T21:53:10.617" Score="9" ViewCount="1239" Body="&lt;p&gt;Riddle me this, riddle me that, Who's afraid of the big black bat?:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;I take the expressway, I'm in a hurry.&lt;br /&gt;&#xA;By chance I make it, no need to worry,&lt;br /&gt;&#xA;I'll loiter and idle and maybe sleep.&lt;/p&gt;&#xA;&lt;p&gt;I'll finally wake, my day's done it seems,&lt;br /&gt;&#xA;I hurry back but now lost, I weep.&lt;br /&gt;&#xA;And yet my home at last I see.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Who am I?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;(and I guess bonus for the quote which is &lt;em&gt;NOT&lt;/em&gt; part of the riddle by the way.  :P )&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;A little Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; An Artist? an Athlete? Your attempts are forlorn.&lt;br /&gt;&#xA;This little puzzle might be simpler to those who are with horns.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;A big Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Sometimes I'm like a Fighter in a cage.&lt;br /&gt;&#xA; &lt;strong&gt;Sometimes I'm hungry and make you crave.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;A bigger Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Nine goes in the middle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-16T23:48:58.997" Title="A full day at work" Tags="|riddle|rhyme|" AnswerCount="5" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="13029" PostTypeId="2" ParentId="12995" CreationDate="2015-04-29T22:45:28.807" Score="9" Body="&lt;p&gt;Without cheating, I got:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  - Green Red  White  Blue Yellow&lt;br&gt;&#xA;  - And Blue is the freedom door&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;&lt;i&gt;Freedom door deducement&lt;/i&gt;&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;   - Yellow can't be the freedom door - it is adjacent to it.&lt;br&gt;&#xA;   - Green can't be the freedom door - it is 3 doors away from it.&lt;br&gt;&#xA;   - Red can't be the freedom door - it has to be to the left of it.&lt;br&gt;&#xA;   - White is in the middle.  It can't be the freedom door because it cannot be 3 doors away from the green door.&lt;br&gt;&#xA;   - So Blue is the freedom door.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;&lt;i&gt;Position deducement&lt;/i&gt;&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  - Blue can't be on either end, so it must be either just left or right of white.&lt;br&gt;&#xA;  - Since red has to be to the left of blue (the freedom door), red must go into either slot #1 (if blue in #2) or slot #1/#2 (if blue is in slot #4). &lt;br&gt;&#xA;  - If blue is in slot #2, yellow cannot be beside it, as required, since red must be to its left.  Blue must therefore go into slot #4 and yellow, beside it, into slot #5.&lt;br&gt;&#xA;  - Since green must be 3 doors down from blue, green must go into slot #1. Red must then take the only remaining slot, slot #2. &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11852" LastEditorUserId="11852" LastEditDate="2015-04-29T23:05:23.930" LastActivityDate="2015-04-29T23:05:23.930" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13030" PostTypeId="1" AcceptedAnswerId="13037" CreationDate="2015-04-29T23:15:41.407" Score="9" ViewCount="1165" Body="&lt;p&gt;This is my first submission here, and I'm somewhat new to this riddle-making thing, so hopefully this proves to be at least somewhat challenging.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I gather light, but cannot see.&#xA;  &lt;br /&gt;I have no lungs, but air I breathe.&#xA;  &lt;br /&gt;I leave my home upon my death.&#xA;  &lt;br /&gt;And countless souls live off my breath.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="11851" LastEditorUserId="11851" LastEditDate="2015-04-30T01:06:24.957" LastActivityDate="2015-04-30T01:34:39.623" Title="I gather light, but cannot see" Tags="|riddle|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13031" PostTypeId="1" CreationDate="2015-04-29T23:16:20.583" Score="10" ViewCount="2875" Body="&lt;p&gt;Hi can anyone please help me and tell me what these words have in common, I've tried finding the answer online but I can't find it anywhere, any help would be great thanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Boro&#xA;Bow&#xA;Fluff&#xA;Know&#xA;Pickup&#xA;Scoff&#xA;Shoe&#xA;Taut&lt;/p&gt;&#xA;" OwnerUserId="11943" LastEditorUserId="22922" LastEditDate="2016-09-01T10:03:56.427" LastActivityDate="2016-09-01T10:03:56.427" Title="what do these words have in common?" Tags="|wordplay|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13032" PostTypeId="2" ParentId="13030" CreationDate="2015-04-29T23:28:40.880" Score="6" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a tree&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I gather light, but cannot see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Photosynthesis&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no lungs, but air I breathe. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Trees &quot;breathe&quot; by absorbing CO2 and releasing O2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I leave my home upon my death. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People use the timber after the tree dies&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And countless souls live off my breath.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Animals (including us) needed the oxygen trees produce&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastActivityDate="2015-04-29T23:28:40.880" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13033" PostTypeId="2" ParentId="13030" CreationDate="2015-04-30T00:02:13.663" Score="2" Body="&lt;p&gt;Can it be a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Christmas tree?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Same reasons mentioned by Alconja, except for the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I leave my home upon my death. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; christmas trees are cut to be sell thus when they die, they are leaving their forest&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10400" LastActivityDate="2015-04-30T00:02:13.663" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13034" PostTypeId="2" ParentId="13031" CreationDate="2015-04-30T00:13:32.840" Score="24" Body="&lt;p&gt;They all...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; rhyme with words that have different pronunciations of ough.&#xA; Borough, bough, enough, dough, hiccough (hiccup), cough, through, bought.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2015-04-30T00:13:32.840" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13035" PostTypeId="2" ParentId="12966" CreationDate="2015-04-30T00:29:13.810" Score="2" Body="&lt;p&gt;Is the answer...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Slave-making_ant&quot; rel=&quot;nofollow&quot;&gt;Slave-making Ants&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11297" LastActivityDate="2015-04-30T00:29:13.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13036" PostTypeId="1" AcceptedAnswerId="13039" CreationDate="2015-04-30T01:23:35.707" Score="18" ViewCount="758" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;No, you were supposed to alphabetize by the &lt;em&gt;zeroth&lt;/em&gt; letter!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fLn6C.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The answer is a three-letter word. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apologies that the answer is not more thematic; it made more sense in the context I originally used this puzzle. Also apologies that parts of the puzzle are a bit US-centric.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-04-30T01:34:48.877" LastActivityDate="2015-04-30T02:27:44.890" Title="Alphabetize by the *zeroth* letter" Tags="|word|enigmatic-puzzle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13037" PostTypeId="2" ParentId="13030" CreationDate="2015-04-30T01:26:44.107" Score="8" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A leaf, exactly like Alconja answered but it is the leaf that does the photosynthesis and everything, and only IT leaves the tree when it dies, falling to the ground in autumn (or when the tree dies) : &lt;em&gt;if this is indeed the answer, I would give the correct answer to Alconja, provided it is edited. I didn't look up the answers before answering but it's so similar I feel like I'm cheating someone out of an answer.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastActivityDate="2015-04-30T01:26:44.107" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13039" PostTypeId="2" ParentId="13036" CreationDate="2015-04-30T01:42:01.270" Score="16" Body="&lt;p&gt;The &quot;zeroth&quot; letter of each word is a letter that can be prefixed to make another word. Each of the words given has a different prefix:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;e&lt;/strong&gt;Bay, &lt;strong&gt;X&lt;/strong&gt;-files, &lt;strong&gt;U&lt;/strong&gt;-haul, &lt;strong&gt;V&lt;/strong&gt;-neck, &lt;strong&gt;C&lt;/strong&gt;-section, &lt;strong&gt;Q&lt;/strong&gt;-tip&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;U&lt;/strong&gt;-boat, &lt;strong&gt;e&lt;/strong&gt;Book, &lt;strong&gt;Y&lt;/strong&gt; Combinator, &lt;strong&gt;D&lt;/strong&gt;-day, &lt;strong&gt;G&lt;/strong&gt;-force, &lt;strong&gt;K&lt;/strong&gt;-mart, &lt;strong&gt;T&lt;/strong&gt;-mobile, &lt;strong&gt;i&lt;/strong&gt;Pod, &lt;strong&gt;O&lt;/strong&gt;-ring, &lt;strong&gt;A&lt;/strong&gt;-team&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you then rearrange the rows and columns so that the prefixed words are in alphabetical order, you get a bitmap that spells &quot;SeT&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0kfQk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="1600" LastEditDate="2015-04-30T02:27:44.890" LastActivityDate="2015-04-30T02:27:44.890" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="13040" PostTypeId="2" ParentId="13021" CreationDate="2015-04-30T01:54:56.227" Score="4" Body="&lt;p&gt;No he is not faster than the speed of light.  Force has a speed of propagation and it will propagate through the stick at the speed of propagation. It will take either the same amount of time or longer for the force produced by him to reach the end of the stick to poke the box compared to the traveling light&lt;/p&gt;&#xA;" OwnerUserId="11947" LastActivityDate="2015-04-30T01:54:56.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13041" PostTypeId="1" AcceptedAnswerId="13042" CreationDate="2015-04-30T02:00:26.433" Score="28" ViewCount="9879" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I have two faces, but show you one.&#xA;  &lt;br /&gt;I once had guests, but now have none.&#xA;  &lt;br /&gt;I'm rarely bloody and seldom blue.&#xA;  &lt;br /&gt;I'm often promised and sometimes new.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="11851" LastActivityDate="2018-02-01T17:28:01.637" Title="I have two faces, but show you one" Tags="|riddle|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13042" PostTypeId="2" ParentId="13041" CreationDate="2015-04-30T02:05:17.530" Score="29" Body="&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; THE MOON&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have two faces, but show you one. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Moon has two faces in the sense that from our perspective it has a &quot;dark side&quot; or &quot;far side&quot; (which we never see) and the side which faces us - the oddity is that we have this relationship with the Moon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I once had guests, but now have none.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This refers to the astronauts which have visited the Moon - there are currently no humans on the Moon - nor have there been for quite a while.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm rarely bloody and seldom blue. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This refers to the red and blue moons. These are rare phenomena c.f. &quot;once in a blue moon&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm often promised and sometimes new.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &quot;new&quot; moon is about the lunar cycles. Finally, one often &quot;promises the moon&quot; i.e. makes crazy promises.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ideas:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Why not something about honey :D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2015-04-30T03:33:18.557" LastActivityDate="2015-04-30T03:33:18.557" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13046" PostTypeId="1" AcceptedAnswerId="13059" CreationDate="2015-04-30T03:03:16.610" Score="17" ViewCount="1739" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I murmur without words.&#xA;  &lt;br /&gt;I sing without a voice.&#xA;  &lt;br /&gt;Always marching forward,&#xA;  &lt;br /&gt;I never have a choice.&#xA;  &lt;br /&gt;&lt;br /&gt;Sometimes I am followed,&#xA;  &lt;br /&gt;But often I am broken.&#xA;  &lt;br /&gt;Sometimes I am listened to,&#xA;  &lt;br /&gt;But words I've never spoken.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="11851" LastActivityDate="2015-04-30T14:40:18.943" Title="I murmur without words" Tags="|riddle|" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13047" PostTypeId="2" ParentId="13046" CreationDate="2015-04-30T03:26:00.977" Score="4" Body="&lt;p&gt;I'm thinking the:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wind&lt;br&gt;&#xA;&lt;br&gt;&#xA;A small breeze , wind-chimes, it's not like the wind ever blows backwards or chooses where it goes. You can follow the winds or listen to them to try and predict the weather(sailing) , and you can find shelter or clothes to break the wind. &lt;em&gt;(I'm not 100% happy with the answer but it's all I can think of)&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastActivityDate="2015-04-30T03:26:00.977" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13048" PostTypeId="1" AcceptedAnswerId="13049" CreationDate="2015-04-30T03:26:28.330" Score="6" ViewCount="420" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I always come in groups.&#xA;  &lt;br /&gt;I like to hang around.&#xA;  &lt;br /&gt;My spirit gives you pleasure,&#xA;  &lt;br /&gt;But only when it's downed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="11851" LastActivityDate="2018-08-03T07:06:54.310" Title="I always come in groups" Tags="|riddle|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13049" PostTypeId="2" ParentId="13048" CreationDate="2015-04-30T03:34:56.007" Score="9" Body="&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Grapes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I always come in groups.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Grapes come in bunches.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I like to hang around.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Grapes hang off vines.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My spirit gives you pleasure,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Grapes are used to make wine, which people like to drink.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But only when it's downed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Down the mouth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2015-04-30T03:34:56.007" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13050" PostTypeId="1" AcceptedAnswerId="13054" CreationDate="2015-04-30T03:42:07.877" Score="43" ViewCount="6625" Body="&lt;p&gt;The brilliant scientist you are, you have built a magic money machine, capable of materializing money out of nowhere! Here's how it works:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The machine has six boxes, numbered $1$ to $6$, each iniitally containing a single dollar coin.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Beneath each of the boxes $1$ through $5$ is a button labeled &lt;strong&gt;&quot;Poof!&quot;&lt;/strong&gt;. Pressing the &lt;strong&gt;&quot;Poof!&quot;&lt;/strong&gt; button beneath box number $i$ will remove a coin from box $i$, and cause &lt;strong&gt;two&lt;/strong&gt; coins to magically appear in box $i+1$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Beneath each of the boxes $1$ through $4$ is a button labeled &lt;strong&gt;&quot;Switch!&quot;&lt;/strong&gt;. Pressing the &lt;strong&gt;&quot;Switch!&quot;&lt;/strong&gt; button below box $i$ will cause a coin from box $i$ to disappear, and switch the contents of boxes $i+1$ and $i+2$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;An exception to the previous two rules: pressing a button beneath an empty box does nothing.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You may press the buttons as much as you like, but in order to collect the money inside, you must permanently break the machine.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Warren Buffet has heard of your machine, and will offer you one trillion dollars for it. Should you take this deal? Why or why not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: I found this at &lt;a href=&quot;http://www.cs.cmu.edu/puzzle/puzzle32.html&quot;&gt;The Puzzle Toad&lt;/a&gt;, but as Joe Z. points out, it was originally question 5 of the International Math Olympiad in 2010. The original phrasing was this: is it possible to make it so the first five boxes are empty, while the last has exactly $2010^{2010^{2010}}$ dollars?&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-04-30T06:59:33.997" LastActivityDate="2015-05-01T20:16:29.933" Title="The Magic Money Machine" Tags="|strategy|" AnswerCount="7" CommentCount="20" ContentLicense="CC BY-SA 3.0" />
  <row Id="13051" PostTypeId="2" ParentId="13028" CreationDate="2015-04-30T03:52:53.177" Score="7" Body="&lt;p&gt;Not quite sure this is it, but worth a stab:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A car, taking some poor schmuck to work.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I take the expressway, I'm in a hurry.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's not called 'Rush Hour' for nothing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By chance I make it, no need to worry,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Entirely possible to be late in the traffic jams.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll loiter and idle and maybe sleep.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; With the worker at work, the car isn't needed for the day, and parked idle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll finally wake, my day's done it seems,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The car is only started again when it's time to go home at knockoff.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I hurry back but now lost, I weep.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When it gets dark in the evening, you turn on the headlights, which certainly make a car's 'eyes' glisten.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And yet my home at last I see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's the end of the riddle, so I can pull my puzzle car into the garage. :P&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Honk Honk.&lt;/p&gt;&#xA;" OwnerUserId="4164" LastActivityDate="2015-04-30T03:52:53.177" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13052" PostTypeId="2" ParentId="12971" CreationDate="2015-04-30T03:56:23.317" Score="0" Body="&lt;p&gt;Brilliant riddles get desperate answers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Knight and Bishop Check Mate.  &gt;&#xA; They are all drawn from White or Black stock.  &gt;&#xA; It all plays out on 8 X 8 coloured (chromo-) square ?bodies. N B K.  &gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11954" LastEditorUserId="11954" LastEditDate="2015-04-30T04:07:44.190" LastActivityDate="2015-04-30T04:07:44.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13054" PostTypeId="2" ParentId="13050" CreationDate="2015-04-30T04:18:25.807" Score="43" Body="&lt;p&gt;The answer, as I'm sure most people are guessing, is...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;No.  Don't sell!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's my solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Note, as I interpret the question, pressing &quot;poof&quot; adds two coins to the next box.  May I be corrected if I'm wrong.&#xA;&lt;br /&gt;&lt;br /&gt;&lt;strong&gt;Solution:&lt;/strong&gt;&#xA;&lt;br /&gt;1 1 1 1 1 1&#xA;&lt;br /&gt;0 3 0 3 1 1 (press poof on 1 and 3)&#xA;&lt;br /&gt;0 2 3 0 1 1 (press switch on 2)&#xA;&lt;br /&gt;0 2 0 6 1 1 (press poof on 3)&#xA;&lt;br /&gt;0 1 6 0 1 1 (press switch on 2)&#xA;&lt;br /&gt;0 1 0 12 1 1 (press poof on 3)&#xA;&lt;br /&gt;0 1 0 11 3 1 (press poof on 4)&#xA;&lt;br /&gt;0 0 11 0 3 1 (press switch on 2)&#xA;&lt;br /&gt;0 0 10 3 0 1 (press switch on 3)&#xA;&lt;br /&gt;0 0 10 0 6 1 (press poof on 4)&#xA;&lt;br /&gt;0 0 9 6 0 1 (press switch on 3)&#xA;&lt;br /&gt;0 0 9 0 12 1 (press poof on 4)&#xA;&lt;br /&gt;0 0 8 12 0 1 (press switch on 3)&#xA;&lt;br /&gt;... Each decrement of box 3 doubles the value of box 4, so once we've reduced box 3 to 0, box 4 will have 12 * 2 ^ 8 = 3072 in it.&#xA;&lt;br /&gt;0 0 0 3072 0 1 (press poof on 3)&#xA;&lt;br /&gt;0 0 0 3071 1 0 (press switch on 4)&#xA;&lt;br /&gt;0 0 0 3071 0 2 (press poof on 5)&#xA;&lt;br /&gt;... &#xA;&lt;br /&gt;&lt;br /&gt;As you can see, using the same logic as before, the final amount of money would be 2^3072, which is far more than a trillion (5.81 x 10^924, to be exact).&#xA;&lt;br /&gt;I'm sure this isn't even the highest value possible, but would answer the question.&#xA;&lt;br /&gt;&lt;br /&gt;PS: Don't trust Warren Buffett.  He doesn't even have a trillion dollars!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, I apologise for the formatting.  I'm new to SE.&lt;/p&gt;&#xA;" OwnerUserId="11851" LastEditorUserId="11851" LastEditDate="2015-04-30T04:23:58.617" LastActivityDate="2015-04-30T04:23:58.617" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="13055" PostTypeId="2" ParentId="13046" CreationDate="2015-04-30T04:45:49.927" Score="4" Body="&lt;p&gt;Perhaps the answer is along these lines:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  A babbling brook, murmuring, singing unable to stop or retreat.&lt;br&gt;&#xA;  It is possible to follow the course of a stream; and white water is broken by rocks and shallows. Certainly it is possible to listen to it and even to find answers; but no words are spoken. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11954" LastActivityDate="2015-04-30T04:45:49.927" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13056" PostTypeId="2" ParentId="12951" CreationDate="2015-04-30T04:54:50.197" Score="6" Body="&lt;p&gt;&lt;strong&gt;EDITED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case you wanted a non-troll (less troll?) answer, the smallest number divisible by 1-101 that's also greater than zero would be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;7041757898200960193617914702466542659236800&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I see how my previous answer was incorrect, I just missed a few steps.  Not only do we need the product of every prime from 1 to 101, but we need the largest power of each prime that is itself less than 101.  This is because every number less than 101 can be factored as a combination of prime numbers, but also, any factor of a number n is necessarily less than n.&#xA;&lt;br /&gt;&lt;br /&gt;So not only do we need...&#xA;&lt;br /&gt;&lt;br /&gt;2*3*5*7*11*13*17*19*23*29*31*37*41*43*47*53*59*61*67*71*73*79*83*89*97*101&#xA;&lt;br /&gt;&lt;br /&gt;We'll also need to include:&#xA;&lt;br /&gt;The largest power of 2 that is less than 101 (64 = 2^6)&#xA;&lt;br /&gt;The largest power of 3 that is less than 101 (81 = 3^4)&#xA;&lt;br /&gt;The largest power of 5 that is less than 101 (25 = 5^2)&#xA;&lt;br /&gt;And the largest power of 7 that is less than 101 (49 = 7^2).&#xA;&lt;br /&gt;The smallest power of 11, the next prime number, is 121, which is too large.  Thus, our final product becomes:&#xA;&lt;br /&gt;&lt;br /&gt;2*2*2*2*2*3*3*3*3*5*5*7*7*11*13*17*19*23*29*31*37*41*43*47*53*59*61*67*71*73*79*83*89*97*101&#xA;&lt;br /&gt;= 7041757898200960193617914702466542659236800&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Indeed, another answer using Ruby seems to have given the same result.  This, above, is the mathematical proof of that.&lt;/p&gt;&#xA;" OwnerUserId="11851" LastEditorUserId="11851" LastEditDate="2015-05-01T03:46:59.580" LastActivityDate="2015-05-01T03:46:59.580" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13058" PostTypeId="2" ParentId="12943" CreationDate="2015-04-30T05:19:34.230" Score="1" Body="&lt;p&gt;Is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Riemann Zeta function&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here I am in my perfect state.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some mathematicians think of this as the most important unsolved problem in math, the &quot;perfect&quot; problem.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Friends of mine in some way relate.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are other similar functions and converging series.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But what confines me isn't straight.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The function maps out a curve.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is it none, one, or many? That's to debate!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The big question is whether there are any nontrivial zeroes other than at 1/2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That you're in my realm, don't postulate!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Postulate -&gt; Riemann hypothesis&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To one number I associate.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; So far all nontrivial zeroes are at 1/2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4479" LastActivityDate="2015-04-30T05:19:34.230" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13059" PostTypeId="2" ParentId="13046" CreationDate="2015-04-30T05:45:57.027" Score="14" Body="&lt;p&gt;I think you are a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Heart&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I murmur without words.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Abnormal sounds from the heart are known as heart murmurs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I sing without a voice.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Every heart sings a song...&quot; - Plato&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Always marching forward,&lt;br&gt;&#xA;I never have a choice. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hearts keep beating by involuntary action&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am followed,&lt;br&gt;&#xA;But often I am broken.&lt;br&gt;&#xA;Sometimes I am listened to,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Follow your heart. Broken hearted. Listen to your heart (not your head).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But words I've never spoken.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...well, obviously...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastActivityDate="2015-04-30T05:45:57.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13060" PostTypeId="2" ParentId="12951" CreationDate="2015-04-30T06:31:37.150" Score="26" Body="&lt;p&gt;For the non-zero answer, in ruby, type:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (1..101).reduce(:lcm)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which yields:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 7041757898200960193617914702466542659236800&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Don't eat them all at once, I guess.&lt;/p&gt;&#xA;" OwnerUserId="11227" LastActivityDate="2015-04-30T06:31:37.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13061" PostTypeId="2" ParentId="13050" CreationDate="2015-04-30T06:54:50.050" Score="100" Body="&lt;p&gt;Yes, you should definitely accept the deal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The puzzle clearly states that pushing a button labeled &lt;strong&gt;&quot;Poof!&quot;&lt;/strong&gt; will remove &lt;strong&gt;a&lt;/strong&gt; coin (emphasis mine) from box $i$, and cause &lt;strong&gt;two&lt;/strong&gt; coins to magically appear in another box. By means of swapping and poofing you can keep increasing the amount of money inside the machine to over one trillion coins, but this process takes time: pushing the &lt;strong&gt;&quot;Poof!&quot;&lt;/strong&gt; button adds only one coin at a time, so in order to get one trillion, not counting the \$6 already inside, you'd need to push buttons no less than one trillion times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being the bright scientist that you are, you could create a machine that pushes a button, say, up to 100 times a second, if the button is extremely sensitive. Now ask yourself: how long would it take to push a button a trillion times? With basic maths, we find the answer to be $10^{10}$ seconds, or more than 300 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless you're also immortal, besides being such a genius, you won't be finished with that job in your lifetime. And even if you were immortal and could finish the job, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Present_value&quot;&gt;present value&lt;/a&gt; of that box is at most \$127.8 billion assuming an inflation rate of 2.5% and 100 button presses per second. (Increasing the button-press rate increases the present value; you'd need about &lt;a href=&quot;http://www.wolframalpha.com/input/?i=1+%2F+%28%281.025%29%5E%281%2F%2831556926x%29%29+-+1%29+%3D+10%5E12&quot;&gt;781 button presses per second&lt;/a&gt; to get it to \$1 trillion.) So get your trillion dollars quickly, as long as Warren Buffett is still alive and willing to pay for it, and retire for the rest of your life chilling out and solving puzzles.&lt;/p&gt;&#xA;" OwnerUserId="4394" LastEditorDisplayName="user88" LastEditDate="2015-04-30T22:54:18.863" LastActivityDate="2015-04-30T22:54:18.863" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="13062" PostTypeId="2" ParentId="13021" CreationDate="2015-04-30T07:15:13.643" Score="1" Body="&lt;p&gt;No. The stick will break. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But seriously, as said above, the compression will travel at around the speed of sound in wood, which, at around 10&lt;sup&gt;3&lt;/sup&gt; m/s, is around a hundred thousand times slower than the speed of light, around 10&lt;sup&gt;8&lt;/sup&gt; m/s. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Though it would be interesting to figure out how much force can be applied without the stick breaking and by extension, the actual amount of time required to touch the object&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) In the quickest way possible (the stick breaks during acceleration)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) So the stick does not break during initial acceleration (but does after contact)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and c) So the stick won't break at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will require OP post more detailed properties of the chosen cellulose-based laterally-elongated structure though :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And what about the influence of gravitation?&lt;/p&gt;&#xA;" OwnerUserId="11963" LastEditorUserId="11963" LastEditDate="2015-04-30T11:47:03.930" LastActivityDate="2015-04-30T11:47:03.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13063" PostTypeId="2" ParentId="13046" CreationDate="2015-04-30T07:58:55.970" Score="5" Body="&lt;p&gt;Could you be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I murmur without words.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Falling rain seems like a murmuring crowd  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I sing without a voice.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Rainstick&quot; rel=&quot;noreferrer&quot;&gt;Rainsticks&lt;/a&gt; are a common musical instrument  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Always marching forward,&#xA;I never have a choice. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rain can only fall, never going upwards  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am followed,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; By a thunder&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But often I am broken.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; By umbrellas?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am listened to,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Have you ever been in your bed listening to falling rain?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But words I've never spoken.&lt;/p&gt;&#xA;" OwnerUserId="4652" LastActivityDate="2015-04-30T07:58:55.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13064" PostTypeId="1" AcceptedAnswerId="13069" CreationDate="2015-04-30T08:05:21.113" Score="1" ViewCount="511" Body="&lt;p&gt;Four rows of numbers:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;1234567890&#xA;5916708234&#xA;192345678&#xA;???&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Fill in the last row with the appropriate digits (not necessarily 3, even though there are 3 '?'s). Please explain your answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint/clarification:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm in an English-speaking country.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="10281" LastEditDate="2015-04-30T12:30:14.763" LastActivityDate="2015-04-30T18:59:46.937" Title="Four rows of number sequences" Tags="|pattern|number-sequence|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13065" PostTypeId="2" ParentId="13021" CreationDate="2015-04-30T08:10:16.580" Score="0" Body="&lt;p&gt;Alpha3031's initial assertion is more likely to be the correct answer.&#xA;The weight/mass of the wooden stick (assuming about 10x10 cm square, by 1 ly long) would be around 5*10&lt;sup&gt;16&lt;/sup&gt;kg.&#xA;(the wider the stick the more mass, hence more force needed)&#xA;The amount of force needed to move such mass will probably break it somewhere near the point when the force is applied.&#xA;Some clever physics scientist will probably be able to estimate the distance of the breaking point ... :)&lt;/p&gt;&#xA;" OwnerUserId="11964" LastActivityDate="2015-04-30T08:10:16.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13066" PostTypeId="2" ParentId="13048" CreationDate="2015-04-30T10:24:29.710" Score="1" Body="&lt;p&gt;It is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I always come in groups. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; of course, its in groups&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I like to hang around. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rain hang activities when fall suddenly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My spirit gives you pleasure,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; its pleasure to feel rain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But only when it's downed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; its all only when its downed to earth. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4312" LastActivityDate="2015-04-30T10:24:29.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13067" PostTypeId="1" AcceptedAnswerId="13068" CreationDate="2015-04-30T10:51:48.587" Score="11" ViewCount="8143" Body="&lt;p&gt;Not born, but from a Mother's body drawn, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hang until half of me is gone. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I sleep in a cave until I grow old, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;then valued for my hardened gold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who am I ?&lt;/p&gt;&#xA;" OwnerUserId="4312" LastActivityDate="2017-12-12T05:05:42.363" Title="Not born, but from a Mother's body drawn, who am I?" Tags="|riddle|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13068" PostTypeId="2" ParentId="13067" CreationDate="2015-04-30T11:02:42.263" Score="20" Body="&lt;p&gt;Looks like it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ripe cheese&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not born, but from a Mother's body drawn, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; From a mother's body, it's made of milk&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I hang until half of me is gone. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Milk is kept at rest to separate its different components&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I sleep in a cave until I grow old, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cheese is left aging in cellars or similar environments&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;then valued for my hardened gold.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When it's well aged, the hard crust it develops assumes a yellow-golden color, and the cheese is valued way more than fresh cheese&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11977" LastActivityDate="2015-04-30T11:02:42.263" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13069" PostTypeId="2" ParentId="13064" CreationDate="2015-04-30T11:20:38.637" Score="4" Body="&lt;p&gt;Shouldn't the first row be this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2345678901&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 7625143&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The digits correspond to the standard qwerty keyboard layout (restricted to alphanumeric characters). Each digit describes the corresponding character's alphabetical (or ASCII) order within its row (modulo 10).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11803" LastEditorUserId="11803" LastEditDate="2015-04-30T11:29:28.380" LastActivityDate="2015-04-30T11:29:28.380" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13070" PostTypeId="1" AcceptedAnswerId="13072" CreationDate="2015-04-30T11:48:47.747" Score="25" ViewCount="3071" Body="&lt;p&gt;W A T _ T L I T F S A _&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Guess the missing letters.&lt;/p&gt;&#xA;" OwnerUserId="4312" LastEditorUserId="7308" LastEditDate="2015-04-30T12:01:07.270" LastActivityDate="2015-04-30T19:00:11.560" Title="What are the missing two letters in the following series and why?" Tags="|pattern|letter-sequence|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13071" PostTypeId="1" AcceptedAnswerId="13074" CreationDate="2015-04-30T11:57:56.100" Score="14" ViewCount="1316" Body="&lt;p&gt;&lt;em&gt;Or should I say, a CoSmOs CrYPt?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was found inscribed in a rock in the middle of the desert. Perhaps the ancients knew more about the sciences than we thought they did - after all, this puzzle uses a kind of 'universal language'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not simply enough to fill the grid, you see. But one's absence may lead you to a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jdVwS.png&quot; alt=&quot;CIDOIREP&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Can you find the single word solution?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="7308" LastActivityDate="2015-04-30T12:56:03.160" Title="The Universal Language" Tags="|cipher|visual|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13072" PostTypeId="2" ParentId="13070" CreationDate="2015-04-30T12:04:54.567" Score="37" Body="&lt;p&gt;I think I got it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; M and W  : The series of letters is an acronym of the title!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11827" LastEditorUserId="5373" LastEditDate="2015-04-30T19:00:11.560" LastActivityDate="2015-04-30T19:00:11.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13073" PostTypeId="2" ParentId="12951" CreationDate="2015-04-30T12:06:28.717" Score="0" Body="&lt;p&gt;Perhaps I'm being too simplistic. If(as the question states) &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I got &quot;x&quot; number of sweets&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;then I must have&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; something and thus not 0&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My intuitive guess is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; + or - (n!) sweets (dont forget the units as my old maths teacher used to say) where n is the number of friends you have.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;also the answer given by Mark N cannot be a full answer as&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the solution to an integral adds a constant - there is no mention in this solution of how big (or little) the constant is.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11985" LastEditorUserId="8822" LastEditDate="2016-09-05T12:43:24.607" LastActivityDate="2016-09-05T12:43:24.607" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="13074" PostTypeId="2" ParentId="13071" CreationDate="2015-04-30T12:18:44.810" Score="22" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As &lt;em&gt;CoSmOs CrYPt&lt;/em&gt; suggests, we are dealing with &lt;a href=&quot;http://en.wikipedia.org/wiki/Periodic_table&quot; rel=&quot;noreferrer&quot;&gt;periodic table elements&lt;/a&gt; (what's more universal as a language than atoms?).&lt;br&gt;&#xA;&lt;br&gt;&#xA; We can fill the grid with the english names of such elements (Paint was definitely a bad choice):&lt;br&gt;&#xA;&lt;br&gt;&#xA;  &lt;img src=&quot;https://i.stack.imgur.com/0kBid.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;br&gt;&#xA; As we can see some elements are missing, the red ones, namely &lt;code&gt;Phosphorus, Arsenic, Selenium, Rubidium, Yttrium.&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Their symbols are &lt;code&gt;P, As, Se, Rb, Y&lt;/code&gt;, which reveal the word &lt;a href=&quot;http://dictionary.cambridge.org/dictionary/british/passer-by&quot; rel=&quot;noreferrer&quot;&gt;&lt;code&gt;Passer-by&lt;/code&gt;&lt;/a&gt;.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Is this a message left from a visiting alien civilization which was just a 'passer-by' in our world?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4652" LastEditorUserId="4652" LastEditDate="2015-04-30T12:56:03.160" LastActivityDate="2015-04-30T12:56:03.160" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13075" PostTypeId="2" ParentId="13050" CreationDate="2015-04-30T12:40:33.443" Score="-2" Body="&lt;p&gt;You accept the offer, but on grounds different to that of GOTO 0, which appears to be brute force logic but not an analysis of the underlying system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply put, if you 'poof' a coin, the two coins created move over to the next box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't swap the coins back because &quot;pressing a button beneath an empty box does nothing.&quot; (Edit: it's also not possible because &quot;Pressing the &quot;Switch!&quot; button below box i  will cause a coin from box i  to disappear, and switch the contents of boxes i+1  and i+2 .&quot; - which isn't i itself but the next two boxes along) - swapping back is therefore not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you 'poof' the coins in box two (or swap them), they will eventually end up in box three, so on and so forth until they reach box 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once all coins end up in box 6, as per the term &quot;pressing a button beneath an empty box does nothing.&quot;, it's no longer possible to swap the coins back. They become trapped in box 6, with no 'poof' button.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your total amount of earnings would be 63 dollars (presuming the condition that once all coins are inside box 6 result in the machine breaking).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, accept the offer: the machine is a scam.&lt;/p&gt;&#xA;" OwnerUserId="11987" LastEditorUserId="11987" LastEditDate="2015-04-30T12:59:52.227" LastActivityDate="2015-04-30T12:59:52.227" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="13076" PostTypeId="2" ParentId="13070" CreationDate="2015-04-30T12:40:45.467" Score="13" Body="&lt;p&gt;Answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;W&lt;/strong&gt;     What&lt;br&gt;&#xA; &lt;strong&gt;A&lt;/strong&gt;     are&lt;br&gt;&#xA; &lt;strong&gt;T&lt;/strong&gt;     the&lt;br&gt;&#xA; _     missing   // letter &lt;strong&gt;'M'&lt;/strong&gt;&lt;br&gt;&#xA; &lt;strong&gt;T&lt;/strong&gt;     two&lt;br&gt;&#xA; &lt;strong&gt;L&lt;/strong&gt;     letters&lt;br&gt;&#xA; &lt;strong&gt;I&lt;/strong&gt;     in&lt;br&gt;&#xA; &lt;strong&gt;T&lt;/strong&gt;     the&lt;br&gt;&#xA; &lt;strong&gt;F&lt;/strong&gt;     following&lt;br&gt;&#xA; &lt;strong&gt;S&lt;/strong&gt;     series&lt;br&gt;&#xA; &lt;strong&gt;A&lt;/strong&gt;     and&lt;br&gt;&#xA; _     why?     // letter &lt;strong&gt;'W'&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9626" LastEditorUserId="9626" LastEditDate="2015-04-30T12:46:48.743" LastActivityDate="2015-04-30T12:46:48.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13077" PostTypeId="2" ParentId="5456" CreationDate="2015-04-30T12:41:53.023" Score="4" Body="&lt;p&gt;Ben and andrepd already came up with the longest, FlOCCINaUCINIHILiPILiFICAtION (or its plural, FlOCCINaUCINIHILiPILiFICAtIONS)—though note that this word was made possible only recently by IUPAC's adoption in 2012 of &quot;Fl&quot; as the symbol for flerovium.  Before 2012, the longest word would have been the 23-letter NONRePReSeNTaTiONAlISmS, as reported by skv.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, it's interesting to consider what happens if we restrict ourselves to one- or two-letter element symbols.  For one-letter symbols, there is one 14-letter example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PSYCHOPANNYCHY&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or if repeated symbols are forbidden:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; WICKYUPS, UNBISHOP, CUSHIONY, CHIBOUKS, BOYCHIKS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For two-letter symbols, there is the following 18-letter example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; IrRePrEsSiBiLiTiEs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or if repeated symbols are forbidden:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ThErMoLaBiLiTiEs, IrRePaRaBiLiTiEs, InSePaRaBiLiTiEs, InAlTeRaBiLiTiEs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The above examples are taken from the &lt;em&gt;&lt;a href=&quot;http://www.wordways.com/&quot; rel=&quot;nofollow&quot;&gt;Word Ways&lt;/a&gt;&lt;/em&gt; article &lt;a href=&quot;http://digitalcommons.butler.edu/wordways/vol47/iss4/22/&quot; rel=&quot;nofollow&quot;&gt;&quot;Elemental Words Revisited&quot;&lt;/a&gt;.  That article (along with a follow-up by Darryl Francis, &lt;a href=&quot;http://digitalcommons.butler.edu/wordways/vol48/iss1/13/&quot; rel=&quot;nofollow&quot;&gt;&quot;King's Moves in the Periodic Table&quot;&lt;/a&gt;) also describes many other types of words formed from the chemical symbols, such as the longest words whose symbols are in increasing or decreasing order of atomic number, the longest words for each group or period in the periodic table, the words which can be spelled the most different ways, etc.&lt;/p&gt;&#xA;" OwnerUserId="11982" LastEditorUserId="11982" LastEditDate="2015-04-30T14:50:42.797" LastActivityDate="2015-04-30T14:50:42.797" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13079" PostTypeId="1" AcceptedAnswerId="13080" CreationDate="2015-04-30T13:04:38.107" Score="3" ViewCount="550" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Evaluate Q1:$[-100\times (-99)] \times [-99 \times (-98)] \times [-98 \times (-97)]..... (97\times 98)\times (98 \times 99) \times (99 \times 100)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Evaluate Q2:$(-100-98) \times (-99-97) \times (-98-96).....(96+98) \times (97+99) \times (98+100)$ (Clue:there is a pattern, but is &quot;disrupted&quot; by the $+(-)$ property.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then, after evaluating this, solve the fraction with the answers you got previously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac {Q1  evalution} {Q2   evaluation}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is a scientific calculator able to calculate the result?(Assuming the calulator cannot calculate answers bigger than 10000000000)(Yeah, a lousy scientific calculator that fails to understand how to put in a standard number form (like for example $3.1839\times10^{65}$)&lt;/p&gt;&#xA;" OwnerUserId="11369" LastEditorUserId="5373" LastEditDate="2015-04-30T19:35:16.690" LastActivityDate="2015-04-30T19:35:16.690" Title="Evaluate the long multiplication" Tags="|mathematics|logical-deduction|lateral-thinking|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13080" PostTypeId="2" ParentId="13079" CreationDate="2015-04-30T13:13:26.333" Score="12" Body="&lt;p&gt;Both multiplications have a factor of 0.&lt;br/&gt;&#xA;The first one obviously and the other one in the factor (-1+1).&lt;br/&gt;&#xA;So the result is 0 in both cases.&lt;br/&gt;&#xA;0/0 doesn't exists so the second question hasn't an answer.&lt;/p&gt;&#xA;" OwnerUserId="10143" LastActivityDate="2015-04-30T13:13:26.333" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13081" PostTypeId="2" ParentId="5456" CreationDate="2015-04-30T13:47:00.060" Score="2" Body="&lt;p&gt;It's certainly not as long as other answers, and the optimal answer appears to have been posted, but I've always liked:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;ThErMoDyNAmICs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which is...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Thorium, erbium, molybdenum, dysprosium, nitrogen, americium, iodine, cesium (alternatively carbon and sulphur)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11851" LastActivityDate="2015-04-30T13:47:00.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13082" PostTypeId="2" ParentId="13041" CreationDate="2015-04-30T13:58:29.263" Score="3" Body="&lt;p&gt;I would have said:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A quarter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have two faces, but show you one. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Heads or tails, only one is ever showing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I once had guests, but now have none. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There used to be dead presidents on US quarters, and were replaced by representations of individual states.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm rarely bloody and seldom blue. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This line fits the least. Could be a reference to blood money (rare) and sometimes coins have a blue tinge to them&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm often promised and sometimes new.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The check is in the mail! :) And freshly minted coins&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9412" LastActivityDate="2015-04-30T13:58:29.263" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13083" PostTypeId="1" AcceptedAnswerId="13284" CreationDate="2015-04-30T14:24:01.763" Score="7" ViewCount="920" Body="&lt;p&gt;Find the key to a popular cipher and find the inner riddle!    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;ViTal ciTy bridges keep it together,&lt;br&gt;&#xA;Under darkness hardly MeeTing thE weather.&lt;br&gt;&#xA;It Populates many but kills too,&lt;br&gt;&#xA;utteRing Fare well and leaving you.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;barely visible under flashes of X,&lt;br&gt;&#xA;Play the instrument that's Highly compleX.&lt;br&gt;&#xA;losing is fAtal yet gifting is greAt,&lt;br&gt;&#xA;don't Joke, it could HeLp your mate.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Queues Go fAr for Such a commodity,&lt;br&gt;&#xA;eXclusive Clans react With velociTy.&lt;br&gt;&#xA;Prolongs The life of a GlaMorised impaler,&lt;br&gt;&#xA;write Down yoUr name not with a retailer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;eXtras inJected to maintain a status Quo,&lt;br&gt;&#xA;the wingED tHieves are an arid foe.&lt;br&gt;&#xA;CombIneD with Goblin to give a Hue,&lt;br&gt;&#xA;the armY roUtes to stAy A new.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quiet motion through a dark passageway,&lt;br&gt;&#xA;vaCating the premises leaving you in dismay.&lt;br&gt;&#xA;exchange of voWs can make a verb,&lt;br&gt;&#xA;whaT key do these words disturb.  &lt;/p&gt;&#xA;" OwnerUserId="5135" LastEditorUserId="25966" LastEditDate="2017-09-13T14:54:26.917" LastActivityDate="2017-09-13T14:54:26.917" Title="riddle-ception!" Tags="|riddle|cipher|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13084" PostTypeId="2" ParentId="13046" CreationDate="2015-04-30T14:40:18.943" Score="3" Body="&lt;p&gt;My answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A train engine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I murmur without words.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Purr of the engine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I sing without a voice.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mobil makes your engine sing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Always marching forward,&#xA;I never have a choice.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Trains can't go backwards&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am followed,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; by other cars or trains&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But often I am broken.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; engines are notorious for breaking&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am listened to,&#xA;But words I've never spoken.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A conductor would be listening to the engine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11926" LastActivityDate="2015-04-30T14:40:18.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13085" PostTypeId="2" ParentId="12951" CreationDate="2015-04-30T14:54:09.573" Score="15" Body="&lt;p&gt;X = 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have one sweet for yourself, and your 1-101 friends are all &lt;em&gt;imaginary friends&lt;/em&gt; for whom you conjure an appropriate amount of &lt;em&gt;imaginary sweets&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5688" LastActivityDate="2015-04-30T14:54:09.573" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13086" PostTypeId="1" AcceptedAnswerId="14608" CreationDate="2015-04-30T15:02:25.457" Score="3" ViewCount="817" Body="&lt;p&gt;There are $1000$ bottles of wine, of which $n\leq10$ are poisoned at random. (You don't know the exact number) A person consuming the poison dies on the midnight of that day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have $2$ prisoners who you are willing to sacrifice, and only one day to test the wine. However, the number of days is not pre-determined, so imagine it as a counter. Every day that passes decrements it by 1. On a given day, for every prisoner who dies, the counter decrements by another 1 and you lose a day. For every prisoner left alive, you have a choice:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;either you can ask for an extra day, and increment the counter by 1 (no net gain/loss in time)&lt;/li&gt;&#xA;&lt;li&gt;or for an extra prisoner.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It is not compulsory for a prisoner to drink on a a given day. You must stop if your days left are $\leq0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The main objective is a guaranteed strategy to identify the poisoned bottle(s) What is the optimal strategy if you want to minimize:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) Total number of days (not the counter's value)&lt;br&gt;&#xA;(b) Total number of deaths  &lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4269" LastEditDate="2015-05-01T09:12:53.680" LastActivityDate="2015-05-09T14:40:29.673" Title="Another poisoned bottle question!" Tags="|logical-deduction|optimization|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13087" PostTypeId="1" AcceptedAnswerId="13092" CreationDate="2015-04-30T15:18:40.647" Score="1" ViewCount="290" Body="&lt;p&gt;Here is a simple cryptogram riddle which can be deciphered without any software/assistance:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;N.m msc uc b. aysxcvo&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Alxans /msh,yg un tmbb.yso&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&gt;./si ,bs ,yrcvbcd!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer how you decoded it, and the answer to the riddle!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Hopefully I don't have any errors in it as this is my first cryptogram) &lt;/p&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-04-30T20:05:31.507" LastActivityDate="2015-04-30T20:05:31.507" Title="Crypto-Riddle: dce,/hcv uc" Tags="|riddle|cipher|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13088" PostTypeId="2" ParentId="13050" CreationDate="2015-04-30T15:20:34.427" Score="8" Body="&lt;p&gt;&lt;a href=&quot;/questions/tagged/lateral-thinking&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;lateral-thinking&amp;#39;&quot; rel=&quot;tag&quot;&gt;lateral-thinking&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You reject the deal. Warren Buffet has only a small fraction of a trillion dollars. In fact, you'd need the top 22 richest people in the world's combined net worth to reach a trillion dollars. If this potential buyer is making an &lt;em&gt;initial&lt;/em&gt; offer of more than 10x his available capital (and has to expect some negotiation), that should be a red flag that this will not end well for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;(Of course, he could be lying by a factor of a thousand and you'd still become a billionaire, so you would have that going for you...)&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="2812" LastActivityDate="2015-04-30T15:20:34.427" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13089" PostTypeId="2" ParentId="13086" CreationDate="2015-04-30T15:31:35.613" Score="3" Body="&lt;p&gt;First of all, this puzzle has no explicit objective (It doesn't aks you to find which bottle is poisoned!), so...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just don't make them drink at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the objective is to minimize the numbers of days and the numbers of deaths, and drinking the wine is not mandatory, just doing nothing give you &lt;em&gt;one&lt;/em&gt; day total and &lt;em&gt;zero&lt;/em&gt; deaths. No other strategy can get you lower than that, since you already start at 1, and any day-decrementing strategy kills a prisoner. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other than that, if you &lt;em&gt;need&lt;/em&gt; to find which wine bottles are poisoned, you can do something like this (step by step):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-- On the first, day, ask for an extra day and an extra prisoner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Days Remaining: 1 Prisoners: 3&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the second day, ask for 2 extra prisoners and an extra day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Days Remaining: 1 Prisoners: 5&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the third day, ask for 4 extra prisoners and an extra day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Days Remaining: 1 Prisoners: 9&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the fourth day, ask for 8 extra prisoners, and an extra day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Days Remaining: 1 Prisoners: 17&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the fifth day, ask for 16 extra prisoners, and an extra day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Days Remaining: 1 Prisoners: 33&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the sixth day, ask for 32 extra prisoners, and an extra day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Days Remaining: 1 Prisoners: 65&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the 7th day, ask for 64 extra prisoners, and an extra day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Days Remaining: 1 Prisoners: 129&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the 8th day, ask for 128 extra prisoners, and an extra day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Days Remaining: 1 Prisoners: 257&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the 9th day, ask for 256 extra prisoners, and an extra day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Days Remaining: 1 Prisoners: 513&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the 10th day, ask for 487 extra prisoners, and an extra day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;*Days Remaining: 1 Prisoners: 1000 *&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make all of them drink from one bottle. At midnight of the 11th day, you have between 990 and 1000 prisoners, up to 10 deaths, and 12 days used. &lt;/p&gt;&#xA;" OwnerUserId="7272" LastActivityDate="2015-04-30T15:31:35.613" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13090" PostTypeId="2" ParentId="13086" CreationDate="2015-04-30T15:32:05.407" Score="0" Body="&lt;p&gt;Guaranteed strategy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 11 days.&lt;br&gt;&#xA; 10 days to round up enough prisoners, the 11th day to test it all (see explanation of the opposite)&lt;br&gt;&#xA; prisoners lost will always be N&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;O wait N &amp;lt; 10. it thought it could be 10 or more, never the less my answer is still right, the explanation might be difficult to understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A strategy for (a) is as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you have 1/2 the prisoners drink form the wine, if you are lucky the last 10 bottles will be poisonous and it will take you 10 days&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;let me explain:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 prisoner drinks, but does not die, means you have 2 prisoners, for each prisoner you ask a new one so now you have 4 prisoners. then you have 2 prisoners drink, none dies so you have 4 prisoners after midnight, then you ask 4 more resulting in 8. This is exponential which makes it easy. (you will have 1024 prisoners in day 10.&#xA; Now to continue, the first day you have 1 bottle tested, the second day you have 2 bottles tested (you will have a total of 3 bottles tested then 1 first day + 2 2nd day) then the third day you will test 4 bottles (add that to the 3 tested you have 7 bottles tested after the 3th day). the pattern here is you will have tested 1 bottle less than you have prisoners.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now for the opposite:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if you have 999 bottles poisoned you will do the same but the first person will die, there will always be 1 person that dies and 1 person that lives, so you can only ask for 1 person each day. meaning you'll need 999 days to find out all but 1 bottles are poisoned. you will also lose 999 prisoners. now this takes way to long so instead, as we have seen you can wait 10 days, asking for new prisoners every day, so after 10 days you will have enough prisoners and you can test all the bottles at once, so on the 11th day you will have lost 999 prisoners but it will have only cost 11 days.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As for losing the least prisoners, you will always lose at least N prisoners.&lt;/p&gt;&#xA;" OwnerUserId="9835" LastEditorUserId="9835" LastEditDate="2015-05-01T09:24:28.627" LastActivityDate="2015-05-01T09:24:28.627" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13091" PostTypeId="2" ParentId="13041" CreationDate="2015-04-30T15:38:32.137" Score="12" Body="&lt;p&gt;Accepted answer be damned, you are a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; delicious, delicious steak&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have two faces, but show you one. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you get a delicious, deliciou steak it will red/pink/brown on top, and brown/black on bottom, depending on how well it was done.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I once had guests, but now have none. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Farmhands have to check up on the cows very often before they are converted into delicious, delicious steak.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm rarely bloody and seldom blue. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rare, bloody, and blue are the three least-well done rarities of delicious, delicious steak. (Well done, medium well, medium, medium rare, rare, bloody, and blue. Sometimes &quot;still mooing&quot; makes the list, but only as a joke.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm often promised &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A steak dinner is generally considered the most elegant dinner available to most people.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and sometimes new.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Veal, the meat of calfs, is considered a delicacy because it is so tender.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2812" LastActivityDate="2015-04-30T15:38:32.137" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13092" PostTypeId="2" ParentId="13087" CreationDate="2015-04-30T15:43:03.217" Score="2" Body="&lt;p&gt;It's easy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You use me to answer.&lt;br&gt;&#xA; always pushing my buttons.&lt;br&gt;&#xA; oops, its inverted!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How it works:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; On a QWERTY keyboard, if the key is on the top row, switch it to the bottom row, and vice versa, keeping the column intact. Middle row remains unaffected.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Oh, and the title says&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; decipher me&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the solution?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; kcnt.avd&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9289" LastEditorUserId="9289" LastEditDate="2015-04-30T16:02:03.443" LastActivityDate="2015-04-30T16:02:03.443" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13093" PostTypeId="2" ParentId="12971" CreationDate="2015-04-30T15:44:30.613" Score="4" Body="&lt;p&gt;Are they&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A man's past, present, and future&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Three people were getting married to each other&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are getting married (as in joined/linked) all the time. The future self joins the present, and the present joins the past.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All men they were in the marriage&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Check (baring any gender identity changes)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They three share two parents&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Check, every person has the same parents as themselves&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11323" LastActivityDate="2015-04-30T15:44:30.613" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13094" PostTypeId="1" AcceptedAnswerId="14410" CreationDate="2015-04-30T15:49:44.100" Score="5" ViewCount="702" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;sONneT NUmBEr 18 -- ShAlL i COMPArE theE to A SUMMeR's day? ThOu aRT MOre lOvElY aND more TEMpEratE. rOUgH wiNDS dO ShAkE tHE dArLIng buDs of MAY, anD SuMmER's lEASE HaTH aLl ToO ShOrT A datE. SOmETIMe toO hOt THE eye oF heaVEN sHInes, AnD oFtEN iS hIS GoLD cOmpLEXIon dimMEd; aND EveRY fAir frOM FaIR soMeTImE DEcLInES, By chanCE, Or nATuRe'S ChAngING cOURsE, UNtrimmEd; BuT tHY eTerNAl suMMEr sHAlL noT faDE, NOR lOse poSsEsSION of THAT FaIR Thou ow'ST, Nor sHalL deAth bRAg tHOU Wand'REST in hIS SHade, wHEN iN eTeRNAL lINEs to tiME ThOu GroW'sT. so LONG As MEn CAN brEATHE, oR eYeS CaN seE, SO Long lIveS THis, AND THIS GIVES LIFE TO THEE.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Find the hidden message.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is a clue in the title? Which might explain the upper and lower case letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It might help to have another sample to work with so here's a shorter text encrypted using the same method&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ws, jc 3 -- FORTune is mErRY, ANd IN thiS mOOD will GiVE US aNYtHING.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And here are a final few clue clues:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clue 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The computer-puzzle tag is important.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clue 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The subject of the second short text is also a clue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clue 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The numbers that appear in each text are important once you have taken a first step.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clue 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You will need to use two numbers the first is 7.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="11297" LastEditDate="2015-05-23T01:40:47.157" LastActivityDate="2015-05-23T01:40:47.157" Title="The work of William Shakespeare?" Tags="|cipher|computer-puzzle|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13095" PostTypeId="1" AcceptedAnswerId="13111" CreationDate="2015-04-30T16:04:23.093" Score="4" ViewCount="233" Body="&lt;p&gt;The cipher is astounding.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Prja, Y ekd psjlymd fuev yfr lywrqy&lt;br&gt;&#xA;   Gily fsxovh k yfovl&lt;br&gt;&#xA;   Pjqm k dvshr ug k htu&lt;br&gt;&#xA;   Dsxr br km qbya povl.&lt;br&gt;&#xA;   Prja yy ljuph zq ii&lt;br&gt;&#xA;   Oe yuph zq nr dttitydq&lt;br&gt;&#xA;   Jq jkf k qyvh-ii ywizl&lt;br&gt;&#xA;   Kma yfr arxoj'd quqd&lt;br&gt;&#xA;   Fr lyktqf ky br&lt;br&gt;&#xA;   Kma O srjy k vfsvhq&lt;br&gt;&#xA;   Yyzq zqsvy vpejymd&lt;br&gt;&#xA;   Mqbqt pptaa sdsym.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hurry up and solve it, as time is fleeting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: as with my previous puzzle, there's a tag left off because it would give too much away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The task is to describe the cipher I am using - I don't know if it has a formal name, a description of the process is sufficient)&lt;/p&gt;&#xA;" OwnerUserId="9289" LastActivityDate="2015-04-30T19:51:45.247" Title="A Cryptogram for weeding out the riff raff" Tags="|cipher|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13096" PostTypeId="2" ParentId="13094" CreationDate="2015-04-30T16:09:22.187" Score="1" Body="&lt;h2&gt;Not An Answer&lt;/h2&gt;&#xA;&lt;p&gt;This is just too much to put into a comment and will probably not be helpful but it might save people some time.&lt;/p&gt;&#xA;&lt;p&gt;If you extract all the uppercase letters, you get:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;SNET NUMBER 18 -- SALL I OMP TH TO A MMER'S A? HOU RT MO VEY AND O TEEAT. RGH WIDS DO SHAKE HE DARLN S F AY, D UME' LEAS ATH ALL TO SHORT A A. METI  H THE E O VN HIS, N EN S HIS OD XION IMME;  ER I FRM FAIR MEM DIS, B E, O ' CHGN COSE, UNRIMM;  HY N SUMM SHAL NOT AD, O LSE SESON OF TA AIR TOU OW'S, OR S H BG OU AN'R IN HI D, W N TL NE O IME TOU G'.  ON S EN CN BEAT, OR EYES CAN SEE,  LONG LIS THI, AND T IES   THEE.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Or, removing all but the alphabet, you get:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;SNETNUMBERSALLIOMPTHTOAMMERSAHOURTMOVEYANDOTEEATRGHWIDSDOSHAKEHEDARLNSFAYDUMELEASATHALLTOSHORTAAMETIHTHEEOVNHISNENSHISODXIONIMMEERIFRMFAIRMEMDISBEOCHGNCOSEUNRIMMHYNSUMMSHALNOTADOLSESESONOFTAAIRTOUOWSORSHBGOUANRINHIDWNTLNEOIMETOUGONSENCNBEATOREYESCANSEELONGLISTHIANDTIESTHEE&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;For lowercase letters, it's&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;on  18 -- h  care ee   su' dy? t a re lol  mre mpre. ou n   t ig bud o m, an smr's e h  o   dte. some too ot  ey f heae sne, ad oft i  gl comple dd; and vey far o  sotie eclne, y chanc, r nature's anig ur, ted; but t eteral er l  fe, nr o possi  ht f h 't, n hall deat ra th wd'est  s shae, hen i eerna lis t t h row'st. so lg a m a rhe,    , so  ve s,  his gv life to .&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;And then&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;onhcareeesudytarelolmrempreountigbudomansmrsehodtesometoooteyfheaesneadoftiglcompleddandveyfarosotieeclneychancrnaturesanigurtedbutteteralerlfenropossihtfhtnhalldeatrathwdestsshaehenieernalistthrowstsolgamarhesoveshisgvlifeto&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;None of these are simple substitution ciphers.&lt;/p&gt;&#xA;&lt;p&gt;The locations of the uppercase characters are:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;1,3,5,6,8,9,10,11,12,13,21,23,24,25,27,30,31,32,37,38,42,43,45,49,50,51,52,54,57,62,63,64,67,68,70,71,77,78,80,82,83,84,87,91,92,95,97,98,102,105,106,108,109,111,112,114,115,117,118,119,120,121,124,125,127,128,129,130,132,138,141,144,145,150,153,155,156,161,162,163,164,168,169,170,172,173,174,176,177,180,181,182,183,184,186,189,196,197,198,199,207,211,212,213,217,219,225,227,230,231,234,238,244,245,248,250,251,252,255,257,265,266,267,268,271,272,273,274,282,285,290,293,294,296,298,299,300,301,305,306,309,312,316,319,322,330,333,345,346,349,351,354,355,358,359,362,363,365,366,367,368,378,379,385,389,390,391,392,396,397,398,399,402,403,404,407,408,413,416,418,419,424,425,426,429,430,432,433,435,437,441,442,443,445,447,448,450,451,453,458,459,461,471,473,476,480,481,484,485,488,493,494,496,497,503,507,513,516,521,525,526,530,533,534,535,537,539,540,542,555,556,560,563,564,566,568,570,572,573,574,579,580,582,583,584,585,587,588,589,591,592,593,599,600,601,602,604,605,608,610,611,612,616,617,618,620,626,628,629,639,640,641,642&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;And for lowercase they are:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;2,4,22,29,33,34,35,39,40,47,48,56,58,61,66,72,73,75,76,79,86,88,89,93,94,96,99,103,104,110,123,131,133,135,136,137,140,143,148,149,152,154,157,159,165,167,178,188,190,191,194,195,200,201,203,204,205,208,209,215,216,220,222,223,224,226,229,232,233,237,239,241,242,243,247,254,256,259,260,261,262,263,264,270,275,278,279,280,283,284,286,288,289,291,295,303,304,307,308,310,313,314,315,317,318,323,325,326,327,328,329,334,336,337,338,339,340,341,343,347,348,350,352,356,357,364,369,370,373,374,375,377,381,382,383,384,386,387,393,394,400,406,409,412,414,417,421,422,423,427,428,436,438,440,446,454,457,462,463,464,465,467,468,469,470,474,475,478,479,483,486,489,490,491,498,500,501,502,504,508,509,510,512,515,517,518,519,520,523,524,527,529,532,538,543,544,545,547,548,551,552,554,557,559,562,567,571,575,576,596,597,606,607,613,621,622,623,625,627,631,632,633,634,636,637&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-30T16:09:22.187" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13097" PostTypeId="2" ParentId="13050" CreationDate="2015-04-30T16:14:05.543" Score="15" Body="&lt;p&gt;The machine is capable of making the following amount of money:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;2*2^(2^...^2) dollars, where the tower is 2^...^2 + 1 twos high, where that tower is 16384 twos high.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is an explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Let's start simple. It is clear that if box 5 contains n coins, and box 6 contains 0, then pressing &lt;strong&gt;Poof!&lt;/strong&gt; n times on box 5 will result in box 5 containing 0 coins and box 6 containing 2n.&lt;br /&gt;&lt;br /&gt;Thus we have the following rule, which we'll call Rule 5:&lt;br /&gt;&lt;br /&gt;&lt;code&gt;* * * * n 0 --&amp;gt; * * * * 0 2n&lt;/code&gt;&lt;br /&gt;&lt;br /&gt;Now if box 4 contains n coins, and boxes 5 and 6 are empty, we can end up with 0 coins in boxes 4 and 6, and 2^n coins in box 5. To do this, we press &lt;strong&gt;Poof!&lt;/strong&gt; on box 4 once, so that we have&lt;br /&gt;&lt;br /&gt;&lt;code&gt;* * * n-1 2 0&lt;/code&gt;&lt;br /&gt;&lt;br /&gt;Then we apply Rule 5 and press &lt;strong&gt;Switch!&lt;/strong&gt; on box 4, and repeat this process n-1 times. Since each time we apply Rule 5, we double what's in box 5, we end up with 2^n coins in box 5 in the end (after the final switch). Thus we have Rule 4:&lt;br /&gt;&lt;br /&gt;&lt;code&gt;* * * n 0 0 --&amp;gt; * * * 0 2^n 0&lt;/code&gt;&lt;br /&gt;&lt;br /&gt;Now suppose we start with n coins in box 3, and boxes 4, 5, and 6 are empty. Then, similarly to the previous case, we press &lt;strong&gt;Poof!&lt;/strong&gt; on box 3 once and then repeat the following until box 3 is empty: apply Rule 4, and press &lt;strong&gt;Switch!&lt;/strong&gt; on box 3. At each step, the number of coins in box 4 will be 2 raised to the number of coins in box 4 from the previous step. Thus we have Rule 3:&lt;br /&gt;&lt;br /&gt;&lt;code&gt;* * n 0 0 0 --&amp;gt; * * 0 2^...^2 0 0&lt;/code&gt;, where the tower consists of n twos.&lt;br /&gt;&lt;br /&gt;Here, it seems prudent to introduce some notation to simplify these expressions. I will use &lt;a href=&quot;http://en.wikipedia.org/wiki/Knuth%27s_up-arrow_notation&quot;&gt;Knuth's up-arrow notation&lt;/a&gt;. The process of building a tower of exponentiation is called &lt;em&gt;tetration&lt;/em&gt;, and I will use the symbol ^^ for it, so 2^^n is a tower of 2's that is n high. Thus Rule 3 can be better stated as&lt;br /&gt;&lt;br /&gt;&lt;code&gt;* * n 0 0 0 --&amp;gt; * * 0 2^^n 0 0&lt;/code&gt;&lt;br /&gt;&lt;br /&gt;Similarly, by repeatedly applying Rule 3 and pressing &lt;strong&gt;Switch!&lt;/strong&gt; on box 2, we could obtain Rule 2 that involves &lt;em&gt;pentation&lt;/em&gt;, which is repeated tetration (so 2^^^n = 2^^(2^^...^^2) with n twos in total), but since we start with only 1 coin in box 1, this rule won't get applied exactly, so we'll skip it.&lt;br /&gt;&lt;br /&gt;To obtain the promised amount, proceed as follows:&lt;br /&gt;&lt;br /&gt;&lt;code&gt;1 1 1 1 1 1&lt;/code&gt;&lt;br /&gt;&lt;code&gt;0 2 2 2 2 3&lt;/code&gt; (Press &lt;strong&gt;Poof!&lt;/strong&gt; on all 5 boxes)&lt;br /&gt;&lt;code&gt;0 2 2 2 0 7&lt;/code&gt; (&lt;strong&gt;Poof!&lt;/strong&gt; twice more on box 5)&lt;br /&gt;&lt;code&gt;0 2 2 1 7 0&lt;/code&gt; (Press &lt;strong&gt;Switch!&lt;/strong&gt; on box 4)&lt;br /&gt;&lt;code&gt;0 2 2 1 0 14&lt;/code&gt; (Apply Rule 5)&lt;br /&gt;&lt;code&gt;0 2 2 0 14 0&lt;/code&gt; (Press &lt;strong&gt;Switch!&lt;/strong&gt; on box 4)&lt;br /&gt;&lt;code&gt;0 2 1 14 0 0&lt;/code&gt; (Press &lt;strong&gt;Switch!&lt;/strong&gt; on box 3)&lt;br /&gt;&lt;code&gt;0 2 1 0 2^14 0&lt;/code&gt; (Apply Rule 4)&lt;br /&gt;&lt;code&gt;0 2 0 2^14 0 0&lt;/code&gt; (Press &lt;strong&gt;Switch!&lt;/strong&gt; on box 3)&lt;br /&gt;&lt;code&gt;0 1 2^14 0 0 0&lt;/code&gt; (Press &lt;strong&gt;Switch!&lt;/strong&gt; on box 2)&lt;br /&gt;&lt;code&gt;0 1 0 2^^(2^14) 0 0&lt;/code&gt; (Apply Rule 3)&lt;br /&gt;&lt;code&gt;0 0 2^^(2^14) 0 0 0&lt;/code&gt; (Press &lt;strong&gt;Switch!&lt;/strong&gt; on box 2)&lt;br /&gt;&lt;code&gt;0 0 0 2^^(2^^(2^14)) 0 0&lt;/code&gt; (Apply Rule 3)&lt;br /&gt;&lt;code&gt;0 0 0 0 2^(2^^(2^^(2^14))) 0&lt;/code&gt; (Apply Rule 4)&lt;br /&gt;&lt;code&gt;0 0 0 0 0 2*2^(2^^(2^^(2^14)))&lt;/code&gt; (Apply Rule 5)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I believe that is the maximum amount of money you could potentially get out of this machine, but I'm not entirely sure.&lt;/p&gt;&#xA;" OwnerUserId="12006" LastActivityDate="2015-04-30T16:14:05.543" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13099" PostTypeId="2" ParentId="13028" CreationDate="2015-04-30T16:36:18.340" Score="3" Body="&lt;p&gt;Perhaps it is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Water. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I take the expressway, I'm in a hurry.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The expressway is a river. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By chance I make it, no need to worry,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Gravity pushes you. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll loiter and idle and maybe sleep.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lakes and seas. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll finally wake, my day's done it seems,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Evaporation &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I hurry back but now lost, I weep. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Rain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And yet my home at last I see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Rain comes back to ground and seas.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-04-30T16:36:18.340" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13100" PostTypeId="2" ParentId="13028" CreationDate="2015-04-30T16:40:01.790" Score="2" Body="&lt;p&gt;You might be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a race car&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I take the expressway, I'm in a hurry.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Race cars are always in a hurry to win the race.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By chance I make it, no need to worry,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you win, then obviously no need to worry.  Or if you just make it to the end without crashing, that's not bad either.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll loiter and idle and maybe sleep.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is when the car enters the pit stop.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll finally wake, my day's done it seems, I hurry back but now lost, I weep.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You will be done (lose the race) if you spend too long in the pit stop.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And yet my home at last I see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can still see the finish line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This little puzzle might be simpler to those who are with horns.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Race cars have horns.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I'm hungry and make you crave.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hungry for gas?  Race cars have ads on them that might feature beverage or food companies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2015-04-30T16:40:01.790" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13102" PostTypeId="2" ParentId="13064" CreationDate="2015-04-30T17:10:26.163" Score="0" Body="&lt;p&gt;Based on what egmont said (credits to him for the solution), the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;7625143&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Take a qwerty keyboard and consider the english letters and numbers. &#xA;First line is 1234567890. Second line is qwertyuiop, if you sort it eiopqrtuwy. As you see, q is the 5th letter in eiopqrtuwy, w is the 9th, e is 1st and so on, we get 5916708234. If we do the same for the last line, we get 7625143. Also, there's no contradiction in the first line, as he couldn't write 12345678910 (ugly, don't you think?)&#xA;Also, no need for modulo 10. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Disclaimer: I wrote this answer because felt that egmont's answer, even though correct, wasn't very easy to understand.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="5373" LastEditDate="2015-04-30T18:59:46.937" LastActivityDate="2015-04-30T18:59:46.937" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13103" PostTypeId="2" ParentId="13041" CreationDate="2015-04-30T17:22:16.210" Score="4" Body="&lt;p&gt;My guess was&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;teeth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have two faces, but show you one. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Only the front of your teeth is usually visible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I once had guests, but now have none. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Teeth start out with temporary &quot;guests&quot;, baby teeth, next to them, until they fall out.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm rarely bloody and seldom blue.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you lose a tooth, which is (hopefully) rare, your mouth becomes bloody, and your teeth are usually only blue when you brush them with blue toothpaste.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm often promised and sometimes new.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is where it breaks down, although you can &quot;promise&quot; your teeth in a half-smile, as well as getting new teeth if need be.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="414" LastActivityDate="2015-04-30T17:22:16.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13104" PostTypeId="1" AcceptedAnswerId="13105" CreationDate="2015-04-30T17:25:16.613" Score="3" ViewCount="1841" Body="&lt;p&gt;All credits to &lt;a href=&quot;https://puzzling.stackexchange.com/users/10037/alex&quot;&gt;@Alex&lt;/a&gt; for inventing this type of puzzle with &lt;a href=&quot;https://puzzling.stackexchange.com/questions/11281/the-hangman-game-backward&quot;&gt;the original&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/questions/12982/the-hangman-game-backward-2&quot;&gt;the sequel&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Guess an &quot;R&quot; - I can only be spelled right.&lt;/li&gt;&#xA;&lt;li&gt;Guess an &quot;F&quot; - Lots of me took place in the world wars.&lt;/li&gt;&#xA;&lt;li&gt;Guess an &quot;L&quot; - Most types of me are never seen&lt;/li&gt;&#xA;&lt;li&gt;Guess an &quot;M&quot; - I'm not a yes or a no.&lt;/li&gt;&#xA;&lt;li&gt;Guess an &quot;N&quot; - I am the always to some places on the moon.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Your answer could be in the form of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The question was: &amp;blank; p p l e&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;followed by explanation.&lt;/p&gt;&#xA;" OwnerDisplayName="user10203" LastEditorUserId="-1" LastEditorDisplayName="user10203" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-08-05T18:43:22.643" Title="Backwards Hangman" Tags="|riddle|english|language|letters|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13105" PostTypeId="2" ParentId="13104" CreationDate="2015-04-30T17:29:49.727" Score="3" Body="&lt;p&gt;The question was: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; _ i g h t&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Guess an &quot;r&quot; - I can only be spelled right.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Right can only be spelled &quot;right&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Guess an &quot;f&quot; - Lots of me took place in the world wars.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lots of fights took place in the world wars.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Guess an &quot;l&quot; - Most types of me are never seen&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most types of light are not in the visible spectrum.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Guess an &quot;m&quot; - I'm not a yes or a no.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Might means maybe, not yes or no.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Guess an &quot;n&quot; - I am the always to some places on the moon.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The moon is always night on the dark side.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2015-04-30T17:29:49.727" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13106" PostTypeId="2" ParentId="13104" CreationDate="2015-04-30T17:34:05.430" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;_ight&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Guess an &quot;r&quot; - I can only be spelled right.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; right&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Guess an &quot;f&quot; - Lots of me took place in the world wars.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; fight&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Guess an &quot;l&quot; - Most types of me are never seen&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; light&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Guess an &quot;m&quot; - I'm not a yes or a no.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; might&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Guess an &quot;n&quot; - I am the always to some places on the moon.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; night&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12014" LastEditorUserId="4385" LastEditDate="2015-08-05T18:43:22.643" LastActivityDate="2015-08-05T18:43:22.643" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13107" PostTypeId="1" AcceptedAnswerId="13191" CreationDate="2015-04-30T17:54:49.803" Score="8" ViewCount="1130" Body="&lt;p&gt;I met a traveller from an antique land&#xA;&lt;br/&gt;Who said: &quot;Two vast and trunkless legs of stone&#xA;&lt;br/&gt;Stand on the desert. Near them, over his sand,&#xA;&lt;br/&gt;Half sunk, a shatter'd visage lies, whose gown,&#xA;&lt;br/&gt;And wrinkled lip, and sneer of cold command,&#xA;&lt;br/&gt;Tell that its sculptor well those passions read&#xA;&lt;br/&gt;Which yet survive, stamped on these lifeless things,&#xA;&lt;br/&gt;The head that mocked them and the heart that fed:&#xA;&lt;br/&gt;And over the pedestal these words appear:&#xA;&lt;br/&gt;'My name is Ozymandias, king of kings:&#xA;&lt;br/&gt;Look on my words, ye Mighty, and despair!'&#xA;&lt;br/&gt;Nothing beside remain. Round the decay&#xA;&lt;br/&gt;Of that colossal wreck, boundless and bare&#xA;&lt;br/&gt;The lone and level sands stretch far away.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Ozymandias&quot; rel=&quot;nofollow&quot;&gt;Shelley&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="2819" LastEditDate="2015-05-01T08:55:41.530" LastActivityDate="2015-05-01T18:27:58.320" Title="Find the hidden word" Tags="|word|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13108" PostTypeId="2" ParentId="13107" CreationDate="2015-04-30T18:16:23.710" Score="1" Body="&lt;p&gt;At first I didn't get what you could be asking. But now I think you mean:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When compared with the original poem, I highlighted the words which were changed in the post from the original poem:&#xA;&lt;br/&gt;&lt;br/&gt;I met a traveller from an antique land&#xA;&lt;br/&gt;Who said: &quot;Two vast and trunkless legs of stone&#xA;&lt;br/&gt;Stand &lt;b&gt;on&lt;/b&gt; the desert. Near them, &lt;b&gt;over his&lt;/b&gt; sand,&#xA;&lt;br/&gt;Half sunk, a shatter'd visage lies, whose &lt;b&gt;gown&lt;/b&gt;,&#xA;&lt;br/&gt;And wrinkled lip, and sneer of cold command,&#xA;&lt;br/&gt;Tell that its sculptor well those passions read&#xA;&lt;br/&gt;Which yet survive, stamped on these lifeless things,&#xA;&lt;br/&gt;The &lt;b&gt;head&lt;/b&gt; that mocked them and the heart that fed:&#xA;&lt;br/&gt;And &lt;b&gt;over&lt;/b&gt; the pedestal these words appear:&#xA;&lt;br/&gt;'My name is Ozymandias, king of kings:&#xA;&lt;br/&gt;Look on my &lt;b&gt;words&lt;/b&gt;, ye Mighty, and despair!'&#xA;&lt;br/&gt;Nothing beside remain. Round the decay&#xA;&lt;br/&gt;Of that colossal wreck, boundless and bare&#xA;&lt;br/&gt;The lone and level sands stretch far away.&#xA;&lt;br/&gt;&lt;br/&gt;Which seems to refer to the &lt;b&gt;Egyptian pharaoh headdress&lt;/b&gt;. (Possibly related to Ramesses II)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Source:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Ozymandias&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Ozymandias&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-04-30T18:30:45.373" LastActivityDate="2015-04-30T18:30:45.373" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13109" PostTypeId="2" ParentId="13107" CreationDate="2015-04-30T18:17:01.140" Score="3" Body="&lt;p&gt;I'll start trying to solve it, it still doesn't make much sense to me:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's a known poem by Shelley&#xA;These following words have been changed: on/(in) over/(on), his/(the), gown/(frown), head/(hand), over/(on), words/(works)&lt;br&gt;&#xA;&lt;br&gt;&#xA;overisgeverd / intefrnnk (overisgdntfrk) &amp;lt;&amp;lt; characters differing in words changed&lt;br&gt;&#xA;oversgevdntfnnk (oversgdntfk)  &amp;lt;&amp;lt; characters differing in the differing characters OF the words changed&lt;br&gt;&#xA; &lt;strong&gt;Rei&lt;/strong&gt; &amp;lt;&amp;lt; Characters common to the characters differing in the words changed, lol&lt;br&gt;&#xA;&lt;br&gt;&#xA;Common letters in words changed are nohownhadowors or n, o, h, w, a, d, s, r&lt;br&gt;&#xA; a, d, s, r &amp;lt;&amp;lt; unique characters in common letters of words changed. I could make : &lt;em&gt;Rads&lt;/em&gt;&lt;br&gt;&#xA;I guess it could be &lt;em&gt;how's rand&lt;/em&gt; and you'd be inquiring after rand in this puzzle of yours.&lt;br&gt;&#xA;Or &lt;em&gt;Owon Shadowhorn&lt;/em&gt; the great hunter of Jawas. &lt;em&gt;A honor showdown&lt;/em&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or again:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; letters of original words: inthefrowadks&lt;br&gt;&#xA;letters of new words: onverhisgwad  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lol&lt;/p&gt;&#xA;" OwnerUserId="11251" LastEditorUserId="11251" LastEditDate="2015-04-30T20:02:05.477" LastActivityDate="2015-04-30T20:02:05.477" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13110" PostTypeId="1" AcceptedAnswerId="13116" CreationDate="2015-04-30T19:31:38.720" Score="29" ViewCount="1593" Body="&lt;p&gt;What is interesting about the following sentence?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Furiously, in back of shut portals, one in four silvery equids is only hiding many extra tidings of meadows, somewhat more craftily than normal,&quot; extremely fat musicians cried enthusiastically.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;This is not a &quot;guess what I'm thinking&quot; puzzle. The answer will be obvious once you find it.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-01T09:25:13.723" Title="Equine musicians?" Tags="|language|wordplay|" AnswerCount="3" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="13111" PostTypeId="2" ParentId="13095" CreationDate="2015-04-30T19:45:10.843" Score="6" Body="&lt;p&gt;I spent too long thinking about this before I realized, with a groan, that the cipher is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Just a jump to the left, and then a step to the right.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More specifically:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first letter is two keys left on a QWERTY keyboard, wrapping around (so W-&gt;Q-&gt;P).&lt;br&gt;&#xA; The next letter is one key right (e-&gt;r).&lt;br&gt;&#xA; Repeat for the rest of the letters (l-&gt;k-&gt;j, l-&gt;a, etc.).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the plaintext is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well, I was walking down the street&lt;br&gt;&#xA; Just having a think&lt;br&gt;&#xA; When a snake of a guy&lt;br&gt;&#xA; Gave me an evil wink.&lt;br&gt;&#xA; Well he shook me up&lt;br&gt;&#xA; He took me by surprise&lt;br&gt;&#xA; He had a pick-up truck&lt;br&gt;&#xA; And the devil's eyes&lt;br&gt;&#xA; He stared at me&lt;br&gt;&#xA; And I felt a change&lt;br&gt;&#xA; Time meant nothing&lt;br&gt;&#xA; Never would again.&lt;br&gt;&#xA;&lt;br&gt;&#xA; (The last verse of Time Warp from Rocky Horror Picture Show)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4513" LastEditorUserId="4513" LastEditDate="2015-04-30T19:51:45.247" LastActivityDate="2015-04-30T19:51:45.247" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13113" PostTypeId="2" ParentId="11656" CreationDate="2015-04-30T20:05:47.197" Score="1" Body="&lt;p&gt;Four is the pigeonese word for &quot;more than three&quot; (cf. Richard Adams's &lt;em&gt;Watership Down&lt;/em&gt; where &lt;em&gt;hrair&lt;/em&gt;, usually translated as &quot;five&quot; or &quot;a thousand&quot;, is the rabbitese word for &quot;more than four&quot;). So the total number of pigeons could be any number greater than or equal to nine - or, in pigeonese, the total number is &lt;strong&gt;four&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-04-30T20:05:47.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13114" PostTypeId="2" ParentId="13110" CreationDate="2015-04-30T20:08:04.973" Score="0" Body="&lt;p&gt;I think that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You used every different type of word in the English language.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or that;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every type of grammatically correct ... uhm.. grammar thingy .. was used.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For every&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Noun there is a adverb and adjective describing it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9412" LastActivityDate="2015-04-30T20:08:04.973" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13115" PostTypeId="2" ParentId="13107" CreationDate="2015-04-30T20:26:53.070" Score="4" Body="&lt;p&gt;Rise to greatness on the shoulders of giants...&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Removing all of the letters that exist in your new words but not in the existing words we get the following letters...&lt;/p&gt;&#xA;&lt;p&gt;o ver is g e ver d&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Rearranging this a bit we get...&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;grover is eved&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;As everyone knows...&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Eved is a poetic term for peved...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This clearly indicates the identity of Ozymandias to be none other than...&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Grover the lovable blue muppet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12028" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-04-30T20:26:53.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13116" PostTypeId="2" ParentId="13110" CreationDate="2015-04-30T20:26:58.643" Score="48" Body="&lt;p&gt;Synonyms form a sentence that is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;in alphabetical order&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not sure this is quite right, but here's what I think:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Angrily, behind closed doors, every fourth gray horse is just keeping lots more news of pastures quiet rather sneakier than usual,&quot; very wide xylophonists yelled zealously.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10861" LastActivityDate="2015-04-30T20:26:58.643" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13117" PostTypeId="2" ParentId="13050" CreationDate="2015-04-30T20:41:07.643" Score="5" Body="&lt;p&gt;Another reason you should take the deal is the fact that one trillion dollar coins will undoubtedly take up more space that your machine's 6th box can hold, let alone the amount of space that would be needed for the machine's other five boxes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To help put it in perspective check out this link: &lt;a href=&quot;http://www.kokogiak.com/megapenny/fourteen.asp&quot; rel=&quot;noreferrer&quot;&gt;http://www.kokogiak.com/megapenny/fourteen.asp&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1,000,000,016,640 pennies occupy a space much larger than the Lincoln Memorial. Now keep in mind that each one dollar coin occupies even more space than a single penny. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless you spent tons of money creating an unfathomably large machine to hold all of these coins in the six boxes, I would definitely take the money being offered to you.&lt;/p&gt;&#xA;" OwnerUserId="3922" LastActivityDate="2015-04-30T20:41:07.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13118" PostTypeId="2" ParentId="12743" CreationDate="2015-04-30T21:08:50.893" Score="1" Body="&lt;p&gt;&lt;strong&gt;My Answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is &lt;strong&gt;A&lt;/strong&gt; in my opinion (I'm gonna sound crazy)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Given this patter of numbers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/jqZCW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Given an increment in the current mini-triangle with every big triangle then I an proposing that the previous triangle must be able to cover the current mini-triangle (which is white) in a black mini-triangle. Therefore A is perfect as it for fills both of my rules. The first triangle already has a black triangle on the first mini-triangle.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/4hAkO.png&quot; alt=&quot;enter image descriptionhere&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12014" LastEditorUserId="12014" LastEditDate="2015-04-30T21:20:52.290" LastActivityDate="2015-04-30T21:20:52.290" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13119" PostTypeId="2" ParentId="1842" CreationDate="2015-04-30T21:15:20.080" Score="0" Body="&lt;p&gt;If the bookworm was born before the move and found while on the new home, the answer is obvious. It travelled the distance from the old home to the new home.&lt;/p&gt;&#xA;" OwnerUserId="12030" LastActivityDate="2015-04-30T21:15:20.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13120" PostTypeId="2" ParentId="11547" CreationDate="2015-04-30T21:31:39.117" Score="8" Body="&lt;p&gt;There's room even for one more day of difference, that is, the elder brother can celebrate his birthday 3 days after the younger brother does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;leoll2 says in his answer that they travel across the International Date Line to jump back to the previous day in the calendar. You don't have to travel to that part of the globe, you can do this at any time zone boundary around midnight (and in fact I have already done this :)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to various sources, including &lt;a href=&quot;http://www.timeanddate.com/time/map/&quot;&gt;http://www.timeanddate.com/time/map/&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Kiritimati&quot;&gt;http://en.wikipedia.org/wiki/Kiritimati&lt;/a&gt;, or the tzdata package shipped by many Linux distributions, there are time zones from UTC-12 through UTC+14. (Note: the tzdata package uses opposite signs.) This means that in a non-leap year you could travel back e.g. from Mar 1 00:xx @UTC+14 (Christmas Island) to Feb 27 23:yy @UTC-11 (the sea nearby) or even to Feb 27 22:zz @UTC-12 if you have a really fast plane.&lt;/p&gt;&#xA;" OwnerUserId="11803" LastEditorUserId="11803" LastEditDate="2015-04-30T22:02:08.220" LastActivityDate="2015-04-30T22:02:08.220" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13121" PostTypeId="1" AcceptedAnswerId="13233" CreationDate="2015-04-30T21:32:31.463" Score="5" ViewCount="361" Body="&lt;p&gt;My friend went on vacation recently, and told me that the place he visited had some interesting denizens. Here's what he had to say about them:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Some are high, some can fly&lt;br&gt;&#xA;  Some are in cities, some are in ditties&lt;br&gt;&#xA;  Some wear trousers, some need browsers&lt;br&gt;&#xA;  Some are ground in the sea, some oversee the grounds&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where did my friend go?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Here's a small hint.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; These are 8 &quot;types&quot; of &quot;residents&quot; that share something in common, but the pair described in each line share an additional, or rather, a &lt;em&gt;more specific&lt;/em&gt; aspect.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll add another, bigger hint tomorrow if this isn't enough.&lt;/p&gt;&#xA;" OwnerUserId="10861" LastEditorUserId="10861" LastEditDate="2015-05-01T21:15:44.610" LastActivityDate="2015-05-02T15:48:58.997" Title="Some odd residents" Tags="|riddle|word|rhyme|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13122" PostTypeId="1" CreationDate="2015-04-30T21:47:33.383" Score="11" ViewCount="1410" Body="&lt;p&gt;Trapped in a dungeon, you have found five keys, labeled A, B, C, D, and E. You have five locked doors, one in front of the other, numbered 1, 2, 3, 4, and 5. To escape, you must open all five doors. Each key opens exactly one door. The keyholes are coated with glue, so once a key is inserted, it can never be removed. Therefore, you must determine which key should be used for each door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are seven guards, all of whom are sympathetic and will help you leave. Still, they are nervous about being caught enabling your escape. Each one will answer just one yes-or-no question. All of the guards know which key goes with each door, and will answer either yes or no to the best of their knowledge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The guards are impatient. After each answer you receive, you must ask the next question within about five seconds. Furthermore, the questions should be fairly simple, such as &quot;Is B's door even?&quot; The guards are not expert logicians or mathematicians, and will give up (and run away) if a question takes more than a minute for them to answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What questions should you ask?&lt;/p&gt;&#xA;" OwnerUserId="7425" LastEditorUserId="8874" LastEditDate="2015-10-06T15:40:42.380" LastActivityDate="2015-10-06T15:40:42.380" Title="Five keys, five doors, seven guards, and only seven questions" Tags="|logical-deduction|keys-and-locks|" AnswerCount="6" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="13123" PostTypeId="2" ParentId="13122" CreationDate="2015-04-30T21:57:25.837" Score="1" Body="&lt;p&gt;Assuming there are 5 doors in a row and you have to open 1 before you get to the next, and there are 5! combinations of keys and doors, just assign each order of keys a number (120 total) and do a binary mapping question to each guard. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That does mean you have to say &quot;Is the combination 12345 or 12354 or....&quot; for up to 64 guesses for each guard, so that might be too much for the guards.  They only have to say if the hear the correct combo, so maybe it's fine?&lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-04-30T21:57:25.837" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="13125" PostTypeId="2" ParentId="13122" CreationDate="2015-04-30T22:18:14.723" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;Is $A$ even?  &lt;/li&gt;&#xA;&lt;li&gt;Is $B$ even?  &lt;/li&gt;&#xA;&lt;li&gt;Is $C$ even? (unnecessary if they answered &quot;yes&quot; twice)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now at least one even key has been found (as either a &quot;yes&quot; answer, or as $D$ and $E$). Choose one $e_1$.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;Is $e_1 = 2$? (&quot;yes&quot; means $e_1=2$ and $e_2=4$, &quot;no&quot; is opposite)  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now we only have to find the odd doors. Choose a key $o_1$.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;Is $o_1 = 1$?  &lt;/li&gt;&#xA;&lt;li&gt;Is $o_1 = 3$? (unnecessary if previous answer is &quot;yes&quot;)  &lt;/li&gt;&#xA;&lt;li&gt;Is $o_2 = 3$? (or $5$ if previous answer was yes)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This allows us to match up all of the keys.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Note&lt;/strong&gt;: further proof is required for completeness, but I'm answering from my phone and it has a shattered screen at the moment.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2015-04-30T22:18:14.723" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13126" PostTypeId="2" ParentId="13122" CreationDate="2015-04-30T22:40:08.610" Score="1" Body="&lt;p&gt;Ask the first guard if the first door opens with keys A or B then ask the next guard if the next door opens with keys B or C and continue in that fashion (repeat after key E and/or door 5):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doors: 1 to 5 | Keys: A to E | Guards/Questions: 8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Keys (made up combo):&lt;/strong&gt; CBEAD&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ask is door:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1 opened with either keys A or B - Answer: No&lt;/li&gt;&#xA;&lt;li&gt;2 opened with either keys B or C - Answer: Yes&lt;/li&gt;&#xA;&lt;li&gt;3 opened with either keys C or D - Answer: No&lt;/li&gt;&#xA;&lt;li&gt;4 opened with either keys D or E - Answer: No&lt;/li&gt;&#xA;&lt;li&gt;5 opened with either keys A or B - Answer: No&lt;/li&gt;&#xA;&lt;li&gt;1 opened with either keys B or C - Answer: Yes&lt;/li&gt;&#xA;&lt;li&gt;2 opened with either keys C or D - Answer: No&lt;/li&gt;&#xA;&lt;li&gt;3 opened with either keys D or E - Answer: Yes&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1 must be C&lt;/li&gt;&#xA;&lt;li&gt;2 must be B&lt;/li&gt;&#xA;&lt;li&gt;3 must be E&lt;/li&gt;&#xA;&lt;li&gt;5 must be D&lt;/li&gt;&#xA;&lt;li&gt;4 must be A&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Another Example (just for assurance, with new format):&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Keys (made up combo):&lt;/strong&gt; BDCEA&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ask is door (x) opened with two keys (k, k+1) for 8 questions:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;x = 1, k = {A or B} - result:true&lt;/li&gt;&#xA;&lt;li&gt;x = 2, k = {B or C} - result:false&lt;/li&gt;&#xA;&lt;li&gt;x = 3, k = {C or D} - result:true&lt;/li&gt;&#xA;&lt;li&gt;x = 4, k = {D or E} - result:true&lt;/li&gt;&#xA;&lt;li&gt;x = 5, k = {A or B} - result:true&lt;/li&gt;&#xA;&lt;li&gt;x = 1, k = {B or C} - result:true&lt;/li&gt;&#xA;&lt;li&gt;x = 2, k = {C or D} - result:false&lt;/li&gt;&#xA;&lt;li&gt;x = 3, k = {D or E} - result:false&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1 = B&lt;/li&gt;&#xA;&lt;li&gt;3 = C&lt;/li&gt;&#xA;&lt;li&gt;5 = A&lt;/li&gt;&#xA;&lt;li&gt;2 = D&lt;/li&gt;&#xA;&lt;li&gt;4 = E&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="12014" LastEditorUserId="12014" LastEditDate="2015-04-30T23:32:27.337" LastActivityDate="2015-04-30T23:32:27.337" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13127" PostTypeId="1" AcceptedAnswerId="13289" CreationDate="2015-05-01T00:15:38.727" Score="13" ViewCount="587" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Old MacDonald had a bias&lt;br&gt;&#xA;  E I E I O&lt;br&gt;&#xA;  And at that bias he looked again&lt;br&gt;&#xA;  E I E I O&lt;br&gt;&#xA;  With a big cat here and a horse guide there&lt;br&gt;&#xA;  Here is scarlet&lt;br&gt;&#xA;  There for each&lt;br&gt;&#xA;  Everywhere are ions&lt;br&gt;&#xA;  Old MacDonald flew too close to the sun&lt;br&gt;&#xA;  E I E I O  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can anyone give me the three words that explain Old MacDonald's apparent decent into madness?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The three words don't form a phrase or sentence, but are instead independent words linked only by a common feature.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The three words a pretty uncommon (one in particular) and range between 10 and 12 letters long. However, there's a key clue that means if you get one it should lead you to all three (possibly with a little help from online tools)...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="5208" LastEditDate="2015-05-03T23:49:08.237" LastActivityDate="2015-05-04T05:59:33.470" Title="The Reinvention of Old MacDonald" Tags="|word|enigmatic-puzzle|wordplay|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13128" PostTypeId="2" ParentId="12995" CreationDate="2015-05-01T01:43:22.753" Score="3" Body="&lt;p&gt;I did a slightly different method from everyone else, finding the order of colors first and the color of the door to freedom last, so I guess I'll post it. I started out with the third requirement: G is 3 away from F (the &quot;Freedom door&quot;).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... G _ _ F ... (one side or the other is filled in, not both)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;White clearly must go between them. If white goes in the left slot...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... G W _ F ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then since white is in the middle, the remaining door would have to be to the left of G.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;_ G W _ F&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then since yellow is next to F, we would get&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;_ G W Y F&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And every door except the two on the end is assigned, so blue would not be able to be placed anywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, by elimination, white must go in the right slot, and since it is in the middle, the extra door must be to the right of the Freedom door.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;G _ W F _&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now there is only one place for the yellow door to go...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;G _ W F Y&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The blue door is allowed to go in either remaining slot. The only remaining requirement is that R is to the left of F. If the Freedom door is R, then this requirement clearly cannot be met. So the R has to go in the slot that isn't F, and B has to be the Freedom Door.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;G R W BF Y&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Luckily, when I worked this out on paper I put the doors in the right order, but if I hadn't I would have reflected them horizontally at this stage to make R be on the left rather than on the right.&lt;/p&gt;&#xA;" OwnerUserId="9783" LastEditorUserId="9783" LastEditDate="2015-05-01T01:54:55.110" LastActivityDate="2015-05-01T01:54:55.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13129" PostTypeId="1" AcceptedAnswerId="13130" CreationDate="2015-05-01T02:52:40.840" Score="23" ViewCount="1432" Body="&lt;p&gt;So I was there, playing Hangman with my friend, when I found myself in a bit of a tough situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd guessed five of the seven letters, and thus needed to get the last two. But I only had three guesses left, and I could see five possible words to complete the hangman, none of which used a letter I had already guessed. The two letters were next to each other, in the middle of the word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I spent a few moments pondering what to do, until it hit me - I could make guesses in a specific order, and ensure that I could complete the puzzle, no matter which of them was right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first guess was critical. By choosing that letter appropriately, if that letter was present, and it wasn't a double-letter (that is, the same letter for both), it would leave me with two possibilities. And if it wasn't present, there would also be two possibilities. In that way, this choice linked together the options and gave me the chance to win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it was there, but not a double-letter, my next guess was going to be sweet if it worked. But if it didn't, the heat would be on, although I'm sure I'd get it right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it turned out, that letter wasn't there. That left me with two possible answers. They both happen to be double-letters, so I just needed to try each of them. My next attempt would have brought things together nicely, but unfortunately it was the wrong one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was at this point that I feared I might have missed a possible answer. Doubt crept into my thinking, and I started running through every possible letter combination that might fit into the space with letters I hadn't already guessed. It was a waste of time, though, and silly of me, because I was right, there was only one possible answer remaining. And sure enough, with that final letter, I solved the puzzle, saving my little hangman buddy and impressing my friend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you figure out the state of the puzzle before my final three guesses, what guesses I made, and which ones I would have gone with had the first guess been present as a single letter? I've been sure to include a few hints in my story, to help you along.&lt;/p&gt;&#xA;" OwnerUserId="9289" LastActivityDate="2015-05-01T04:01:43.897" Title="Reverse Hangman at double difficulty" Tags="|riddle|english|story|letters|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13130" PostTypeId="2" ParentId="13129" CreationDate="2015-05-01T04:01:43.897" Score="19" Body="&lt;p&gt;The state before your last 3 guesses was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;co__ect&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With the 5 possible words being:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; collect, confect, connect, convect, correct&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your first guess was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;n&lt;/code&gt; - if it was a double letter, it would be &lt;code&gt;connect&lt;/code&gt; (linked together), if not a double letter, you'd have &lt;code&gt;confect&lt;/code&gt; (sweet) or &lt;code&gt;convect&lt;/code&gt; (heat)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Next guess was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;l&lt;/code&gt; - which would have been &lt;code&gt;collect&lt;/code&gt; (bring things together)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But the answer was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'r' - giving &lt;code&gt;correct&lt;/code&gt;, which of course is right. :P&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastActivityDate="2015-05-01T04:01:43.897" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13134" PostTypeId="2" ParentId="13067" CreationDate="2015-05-01T04:32:45.693" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A potato.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These come from mother nature, their vines hang and then are disposed of while the main body lives in a cave underground and then they become gold. yukon gold&lt;/p&gt;&#xA;" OwnerUserId="12039" LastEditorUserId="40703" LastEditDate="2017-12-12T05:05:42.363" LastActivityDate="2017-12-12T05:05:42.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13135" PostTypeId="1" AcceptedAnswerId="13138" CreationDate="2015-05-01T04:33:07.817" Score="-5" ViewCount="2537" Body="&lt;p&gt;What is the next letter in this series?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Z X C V B ...&lt;/p&gt;&#xA;" OwnerUserId="12038" LastEditorUserId="7308" LastEditDate="2015-05-01T09:27:46.190" LastActivityDate="2015-05-01T09:27:46.190" Title="What is the next letter in this series (zxcvb...)?" Tags="|pattern|letter-sequence|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13136" PostTypeId="2" ParentId="13135" CreationDate="2015-05-01T04:39:38.027" Score="0" Body="&lt;p&gt;It is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  N. That's the next character on the lower row on your keyboard &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3993" LastActivityDate="2015-05-01T04:39:38.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13137" PostTypeId="2" ParentId="13135" CreationDate="2015-05-01T04:40:27.913" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;N&#xA;&lt;br /&gt;Look at your QWERTY keyboard.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11851" LastActivityDate="2015-05-01T04:40:27.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13138" PostTypeId="2" ParentId="13135" CreationDate="2015-05-01T04:40:28.857" Score="2" Body="&lt;p&gt;Last alphabetical row of QWERTY keyboard  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;kbd&gt;Z&lt;/kbd&gt; &lt;kbd&gt;X&lt;/kbd&gt; &lt;kbd&gt;C&lt;/kbd&gt; &lt;kbd&gt;V&lt;/kbd&gt; &lt;kbd&gt;B&lt;/kbd&gt; &lt;kbd&gt;N&lt;/kbd&gt; &lt;kbd&gt;M&lt;/kbd&gt; ...........&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uctHz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9626" LastEditorUserId="9626" LastEditDate="2015-05-01T07:00:33.943" LastActivityDate="2015-05-01T07:00:33.943" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13139" PostTypeId="5" CreationDate="2015-05-01T05:41:16.973" Score="0" Body="&lt;p&gt;A &lt;a href=&quot;https://en.wikipedia.org/wiki/Seven-segment_display&quot; rel=&quot;nofollow noreferrer&quot;&gt;seven-segment display&lt;/a&gt; is an electronic indicator consisting of seven segments each of which can be lit or unlit, so that the whole display can turn into any of the digits from 0 to 9.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="11110" LastEditDate="2017-06-21T07:59:55.847" LastActivityDate="2017-06-21T07:59:55.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13140" PostTypeId="4" CreationDate="2015-05-01T05:41:16.973" Score="0" Body="A puzzle that involves seven-segment displays, more commonly referred to as &quot;digital&quot; displays." OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-05-01T05:41:16.973" LastActivityDate="2015-05-01T05:41:16.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13141" PostTypeId="1" AcceptedAnswerId="13142" CreationDate="2015-05-01T06:21:49.963" Score="1" ViewCount="7052" Body="&lt;p&gt;You are standing in front of ten people, one of which is a &lt;em&gt;truth-teller&lt;/em&gt; while the other nine are &lt;em&gt;togglers&lt;/em&gt; . &#xA;(A &lt;em&gt;toggler&lt;/em&gt; is a person who toggles between &lt;em&gt;the-truth&lt;/em&gt; and &lt;em&gt;the-lie&lt;/em&gt; for every question asked.)&#xA;You have no clue who the truth-teller is and who the togglers are. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are given the chance to ask &lt;strong&gt;one and only one&lt;/strong&gt; question to &lt;strong&gt;any, but only one&lt;/strong&gt; of the ten people.&#xA;From the provided answer you must be able to deduce with &lt;strong&gt;certainty&lt;/strong&gt; the identity of the truth teller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; So what is that question you will be asking?&lt;/p&gt;&#xA;" OwnerDisplayName="user5871" LastEditorUserId="8874" LastEditDate="2015-10-06T15:48:50.203" LastActivityDate="2015-10-07T00:51:52.707" Title="Nine togglers, one truth teller, and only a single question" Tags="|logical-deduction|liars|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13142" PostTypeId="2" ParentId="13141" CreationDate="2015-05-01T06:45:01.023" Score="7" Body="&lt;p&gt;&quot;Which people here are truth-tellers&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the person you asked is a truth teller, he will indicate himself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If he is a toggler in &quot;truth&quot; mode, he will indicate the truth teller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If he is a toggler in &quot;lie&quot; mode, he will indicate the 9 togglers (this is my interpretation of telling &quot;the-lie&quot;, and note they only toggle after each question and this is a single question). But as you know there is only one truth teller you know it must be the unindicated person. &lt;/p&gt;&#xA;" OwnerUserId="2021" LastActivityDate="2015-05-01T06:45:01.023" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="13143" PostTypeId="2" ParentId="13141" CreationDate="2015-05-01T07:22:30.020" Score="4" Body="&lt;p&gt;&quot;If I asked you another question immediately after this, and then I pointed to another person in your group and asked 'Are they the truth-teller?', who could I point to in order for you to say 'yes'?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Makes use of the fact that the togglers always alternate between truth and lie.&lt;/p&gt;&#xA;" OwnerUserId="9289" LastEditorUserId="9289" LastEditDate="2015-05-01T10:04:41.273" LastActivityDate="2015-05-01T10:04:41.273" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13146" PostTypeId="1" AcceptedAnswerId="13147" CreationDate="2015-05-01T09:13:55.493" Score="1" ViewCount="1532" Body="&lt;p&gt;Harrison and George are best friends, they both live in Manchester.&lt;br&gt;&#xA;One day, Harrison goes to George's house to visit him, but after opening the door, he finds his friend dead near the desk, with a hole in the head. There are many items on the desk, including a ream, a half-consumed pencil, an old-clock stuck at 9:03 and a cassette tape player with a cassette already in it. Harrison decides to listen the cassette, so he pushes the &quot;Play&quot; button. The audio isn't very clear, but he can distinguish a male voice speaking. Here's the message:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm George. I've just received a phone call, they said that somebody is coming now to kill me and he will be here within 2 minutes. I hear footsteps... someone is opening the door... oh no! &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Harrison is shocked, but after thinking for a while, he realizes that the message wasn't recorded by George, but by his assassin!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How could Harrison know this?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-05-01T09:47:42.267" LastActivityDate="2019-02-03T20:32:10.027" Title="A murder in Manchester" Tags="|logical-deduction|lateral-thinking|story|" AnswerCount="6" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13147" PostTypeId="2" ParentId="13146" CreationDate="2015-05-01T09:18:44.023" Score="12" Body="&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the tape had already been rewound to the start, which George couldn't have done if he was dead.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="11110" LastEditDate="2019-02-03T20:32:10.027" LastActivityDate="2019-02-03T20:32:10.027" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="13148" PostTypeId="2" ParentId="13110" CreationDate="2015-05-01T09:25:13.723" Score="3" Body="&lt;p&gt;I'm ignoring some clues here, but here's my guess&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's a violin&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's my reasoning&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Four fat cat musicians - The strings of a violin are traditionally made from the gut of some kind of animal. Often falsely thought to be cat gut. &#xA; The bow string for classical stringed instruments is typically made from horse hair (usually tail), so that could be the whole 'equid' statement.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2673" LastActivityDate="2015-05-01T09:25:13.723" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13150" PostTypeId="2" ParentId="13146" CreationDate="2015-05-01T09:28:04.807" Score="1" Body="&lt;p&gt;I can envision such a scenario.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Harrison knew it was the assassin because he had a Liverpool accent, not a Manchester accent. Towards the end of the message he could hear him singing the Beatles in the distance, but George preferred Oasis.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alternatively,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The stopped clock and pencil/ream show that George was unable to handle technology to the extent of not owning a computer or being able to change a battery. When Harrison noticed this he realised George couldn't have used the tape recorder.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But also I'm thinking&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It was the postman, because there's no post on sundays.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11880" LastEditorUserId="11880" LastEditDate="2015-05-01T10:23:03.293" LastActivityDate="2015-05-01T10:23:03.293" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13151" PostTypeId="2" ParentId="13146" CreationDate="2015-05-01T09:36:03.457" Score="9" Body="&lt;p&gt;Well, to give 3 other possibilities than the ones given:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the message starts with I'm George, why would George start a message with that, you don't start a message stating your own name.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;second&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if George had 2 minutes he could have called the police, run, do what ever he could in 2 minutes, he wouldn't record it he would fight for his life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3th&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if he had 2 minutes and he states I've &lt;strong&gt;just&lt;/strong&gt; gotten a phone call, how can in less than 10 seconds he hears footsteps, that would mean already 1 min 50 has passed, the text would not take many more that 10 seconds to say.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9835" LastActivityDate="2015-05-01T09:36:03.457" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13152" PostTypeId="2" ParentId="13146" CreationDate="2015-05-01T09:38:08.353" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If George heard his killer coming, he wouldn't have pressed the button to stop the recording, and so his murder would have been recorded too.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-05-01T09:38:08.353" CommentCount="1" CommunityOwnedDate="2015-05-01T09:38:08.353" ContentLicense="CC BY-SA 3.0" />
  <row Id="13153" PostTypeId="1" AcceptedAnswerId="13155" CreationDate="2015-05-01T10:38:34.507" Score="6" ViewCount="1630" Body="&lt;p&gt;As a commander of an army battalion, you have to plan the attack on some enemy cities which are connected by road-network. Before starting the attack, you have information about all the cities and roads connecting them. You can start attack from any city but you must travel from one city to another via roads only. To prevent enemy re-enforcement and block transport through a city, you reach there, destroy it and burn it while leaving behind. This makes it impossible for you to return to a city (via any road) destroyed by you in past. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a list of all the roads (each connecting a pair of cities), you have to tell what is the maximum number of cities that can be destroyed by the strategy mentioned above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is the list of road: (pairs x &amp;lt;---&gt; y which denotes that there exists a direct road from city ‘x’ to city ‘y’) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 &amp;lt;---&gt; 2&lt;br&gt; 1 &amp;lt;---&gt; 11&lt;br&gt; 2 &amp;lt;---&gt; 3&lt;br&gt; 3 &amp;lt;---&gt; 11&lt;br&gt; 4 &amp;lt;---&gt; 5&lt;br&gt; 4 &amp;lt;---&gt; 11&lt;br&gt; 4 &amp;lt;---&gt; 12&lt;br&gt; 5 &amp;lt;---&gt; 6&lt;br&gt; 5 &amp;lt;---&gt; 7&lt;br&gt; 6 &amp;lt;---&gt; 7&lt;br&gt; 8 &amp;lt;---&gt; 9&lt;br&gt; 8 &amp;lt;---&gt; 10&lt;br&gt; 8 &amp;lt;---&gt; 12&lt;br&gt; 9 &amp;lt;---&gt; 12&lt;br&gt; 9 &amp;lt;---&gt; 10&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Then what is the maximum number of cities that can be destroyed?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10820" LastEditorUserId="1777" LastEditDate="2015-05-01T11:45:42.090" LastActivityDate="2015-05-01T13:40:55.077" Title="What is the maximum number of cities that can be destroyed?" Tags="|mathematics|logical-deduction|calculation-puzzle|story|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13154" PostTypeId="2" ParentId="13086" CreationDate="2015-05-01T10:52:26.847" Score="0" Body="&lt;p&gt;I'm going to make a assumptions here:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You start at 1 day, even if you do not test, you would be at 0 days the next day. If you are not allowed to add extra days because you still have 2 prisoners left, you would instantly lose. Therefore, I assume you are allowed to add extra days even when you are at 0 days, as long as you have prioners left alive.&lt;/li&gt;&#xA;&lt;li&gt;Similar, when you go negative, I still assume you can add extra days as long as you have enough prisoners to do so.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I will use the following strategy:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If my counter is at 0 or negative, I will use enough prisoners to get it back to 1.&lt;/li&gt;&#xA;&lt;li&gt;Any other prisoner left alive is used to get more prisoners.&lt;/li&gt;&#xA;&lt;li&gt;I will let the following number of prisoners drink from a new bottle of wine:&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Let N be maximal number of toxic bottles. This is 10 - nr of dead prioners.&lt;/li&gt;&#xA;&lt;li&gt;If I have more than 2xN-1 prisoners, all of them will drink wine from a new bottle.&lt;/li&gt;&#xA;&lt;li&gt;If I do not have enough prisoners for step 2, only half minus 1 is allowed to drink.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;If all 1000 bottles are tested, or if 10 prisoners have died, I'm done.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Best case this will take me 6 days, if 10 prisoners die on day 5+6.&lt;br&gt;&#xA;Usually this will take me 10 days, if the first 495 bottles are safe, I will have all 1000 bottles tested on day 10.&lt;br&gt;&#xA;Even with 2 of the 257 bottles tested on day 9 toxic, this will only take 10 days.&lt;br&gt;&#xA;In most other cases, I will need 11 days. As long as the first 44 bottles are safe (drunk in the first 6 days), 9 prisoners can die on day 7, and I will still finish testing all bottles by day 11.&lt;br&gt;&#xA;Worse case, the first 9 bottles I test are toxic. My run will now extend up to 16 days. This is however, very unlikely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This solution should have the lowest average number of days. Max dead prisoners it equal to the number of toxic bottles.&#xA;If, however, you want the minimal worse case, simply wait till day 11 until you have over 1000 prisoners, and let them all test one bottle at that time.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Example run:&lt;br&gt;&#xA;Day 1: 2 prisoners, 0 drink, 0 die, get 1 new.&lt;br&gt;&#xA;Day 2: 3 prisoners, 0 drink, 0 die, get 2 new.&lt;br&gt;&#xA;Day 3: 5 prisoners, 1 drink, 0 die, get 4 new.&lt;br&gt;&#xA;Day 4: 9 prisoners, 3 drink, 0 die, get 8 new.&lt;br&gt;&#xA;Day 5: 17 prisoners, 7 drink, 0 die, get 16 new.&lt;br&gt;&#xA;Day 6: 33 prisoners, 33 drink, 0 die, get 32 new.&lt;br&gt;&#xA;Day 7: 65 prisoners, 65 drink, 0 die, get 64 new.&lt;br&gt;&#xA;Day 8: 129 prisoners, 129 drink, 0 die, get 128 new.&lt;br&gt;&#xA;Day 9: 257 prisoners, 257 drink, 0 die, get 256 new.&lt;br&gt;&#xA;Day 10: 513 prisoners, 505/513 drink, 0-10 die, done.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Some on day 9 die:&lt;br&gt;&#xA;Day 9: 257 prisoners, 257 drink, 2 die, get 252 new.&lt;br&gt;&#xA;Day 10: 507 prisoners, 505/507 drink, 0-8 die, done.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;More likely scenario: some die on day 5:&lt;br&gt;&#xA;Day 5: 17 prisoners, 7 drink, 2 die, get 12 new.&lt;br&gt;&#xA;Day 6: 27 prisoners, 27 drink, 7 die, get 12 new.&lt;br&gt;&#xA;Day 7: 32 prisoners, 32 drink, 0 die, get 31 new.&lt;br&gt;&#xA;Day 8: 63 prisoners, 63 drink, 0 die, get 62 new.&lt;br&gt;&#xA;Day 9: 125 prisoners, 125 drink, 0 die, get 124 new.&lt;br&gt;&#xA;Day 10: 249 prisoners, 249 drink, 0 die, get 248 new.&lt;br&gt;&#xA;Day 11: 497 prisoners, 493/497 drink, 0-1 die, done.  &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Still only 11 days!&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Worse case:&lt;br&gt;&#xA;Day 1: 2 prisoners, 0 drink, 0 die, get 1 new.&lt;br&gt;&#xA;Day 2: 3 prisoners, 0 drink, 0 die, get 2 new.&lt;br&gt;&#xA;Day 3: 5 prisoners, 1 drink, 1 die, get 2 new.&lt;br&gt;&#xA;Day 4: 6 prisoners, 2 drink, 2 die, get 1 new.&lt;br&gt;&#xA;Day 5: 5 prisoners, 1 drink, 1 die, get 2 new.&lt;br&gt;&#xA;Day 6: 6 prisoners, 2 drink, 2 die, get 1 new.&lt;br&gt;&#xA;Day 7: 5 prisoners, 1 drink, 1 die, get 2 new.&lt;br&gt;&#xA;Day 8: 6 prisoners, 6 drink, 2 die, get 1 new.&lt;br&gt;&#xA;Day 9: 5 prisoners, 5 drink, 0 die, get 4 new.&lt;br&gt;&#xA;Day 10: 9 prisoners, 9 drink, 0 die, get 8 new.&lt;br&gt;&#xA;Day 11: 17 prisoners, 17 drink, 0 die, get 16 new.&lt;br&gt;&#xA;Day 12: 33 prisoners, 33 drink, 0 die, get 32 new.&lt;br&gt;&#xA;Day 13: 65 prisoners, 65 drink, 0 die, get 64 new.&lt;br&gt;&#xA;Day 14: 129 prisoners, 129 drink, 0 die, get 128 new.&lt;br&gt;&#xA;Day 15: 257 prisoners, 257 drink, 0 die, get 256 new.&lt;br&gt;&#xA;Day 16: 513 prisoners, 472/513 drink, 0-1 die, done.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not only is the worse case extremely unlikely, it also takes N=9 or N=10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See/play with numbers &lt;a href=&quot;https://docs.google.com/spreadsheets/d/1XtqOUbPz2Gn6HdGw-3dEKkSh0RCCawauIQdNSkY79BQ/edit?usp=sharing&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2711" LastActivityDate="2015-05-01T10:52:26.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13155" PostTypeId="2" ParentId="13153" CreationDate="2015-05-01T11:09:27.330" Score="10" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6qaD5.png&quot; alt=&quot;Graph&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ans:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer in this case is 9.  A possible path is 8, 10, 9, 12, 4, 11, 3, 2, 1. More than 9 is not possible because 4 can only be travelled through once.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1777" LastEditorUserId="10820" LastEditDate="2015-05-01T12:15:50.750" LastActivityDate="2015-05-01T12:15:50.750" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13158" PostTypeId="2" ParentId="13146" CreationDate="2015-05-01T12:10:39.383" Score="0" Body="&lt;p&gt;Because it wasn't George's voice.  Being best friends, he would recognize his friend's voice.  Or accent.&lt;/p&gt;&#xA;" OwnerUserId="10806" LastActivityDate="2015-05-01T12:10:39.383" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13159" PostTypeId="1" AcceptedAnswerId="13161" CreationDate="2015-05-01T12:23:50.913" Score="4" ViewCount="1413" Body="&lt;p&gt;A friend of mine challenged me to jump over a bottle on the floor. I'm good at jumping over obstacles so I accepted. But then I realized that I couldn't do it!&lt;br&gt;&#xA;&lt;strong&gt;Why? Explain!&lt;/strong&gt;      &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Notes: I'm not injured. The floor isn't sticky or irregular. The bottle has standard size of 1,5L. Jumping means that you &quot;take off&quot; from the floor, pass above the bottle without touching it, then land on the floor&lt;/em&gt;     &lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="5188" LastEditDate="2015-05-01T17:35:31.900" LastActivityDate="2015-05-01T19:48:21.713" Title="Jump over a bottle?" Tags="|lateral-thinking|humor|" AnswerCount="5" CommentCount="8" ClosedDate="2015-05-01T20:36:22.713" ContentLicense="CC BY-SA 3.0" />
  <row Id="13160" PostTypeId="2" ParentId="13159" CreationDate="2015-05-01T12:47:53.653" Score="5" Body="&lt;p&gt;One possibility is that the bottle is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; on a floor that you can't get to, like a bottle on the floor of a secure facility you don't have access to.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another possibility is that the &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ceiling is too low for you to jump over the bottle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More generally these all mean there is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; some sort of obstruction preventing you from jumping over the bottle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11271" LastEditorUserId="11271" LastEditDate="2015-05-01T13:26:06.883" LastActivityDate="2015-05-01T13:26:06.883" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13161" PostTypeId="2" ParentId="13159" CreationDate="2015-05-01T12:51:13.167" Score="18" Body="&lt;p&gt;I think&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; he put the bottle on the floor against the wall in the corner of the room&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10400" LastEditorUserId="10400" LastEditDate="2015-05-01T12:57:25.800" LastActivityDate="2015-05-01T12:57:25.800" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13162" PostTypeId="2" ParentId="3839" CreationDate="2015-05-01T12:56:22.160" Score="6" Body="&lt;p&gt;Alternatively, he could have told them that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; someone who would have you wait on a horse for over a month to prove your love is not who you want to spend the rest of your life with! For the sake of your future happiness, get out now!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11271" LastActivityDate="2015-05-01T12:56:22.160" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13163" PostTypeId="2" ParentId="13159" CreationDate="2015-05-01T13:06:03.897" Score="7" Body="&lt;p&gt;Another solution (there are several) is that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;he put the bottle underneath a large table or other piece of furniture.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note there are lateral thinking ways to get around some of the obsticales presented by this idea such as jumping on the furniture in such a way that you go over the bottle while remaining on the furniture.&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2015-05-01T13:06:03.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13164" PostTypeId="2" ParentId="13159" CreationDate="2015-05-01T13:06:06.110" Score="3" Body="&lt;p&gt;A more unlikely situation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your friend lives in a rotating house where they ceiling can become the floor (after being rotating around). He glues the bottle onto the ceiling (which can also be a floor but currently isn't). Then he read what A.D. said, and glues it into the corner of the ceiling.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastActivityDate="2015-05-01T13:06:06.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13167" PostTypeId="2" ParentId="3739" CreationDate="2015-05-01T13:39:05.653" Score="35" Body="&lt;p&gt;Several answers have got the basic idea, but here is (what I believe to be) the full solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the old man ate the seagull meat, he discovered&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it tasted different to the meat he had been eating on the island.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first thing the old man realised:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since there was no reason for the captain to lie, the hunter must have been lying. Why would the hunter have been lying? Because &lt;strong&gt;they were really eating human flesh.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second thing the old man realised:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Days passed between landing on the island and them getting their first meal. In tropical heat, there's no way that the body of one of the five passengers that had died in the storm would have still been fresh after all that time, even if it had washed ashore. &lt;strong&gt;Someone had been killed and eaten.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The third thing the old man realised:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There were five survivors on the island - himself, his niece and three others. He had heard the voice of the hunter, and presumably the voices of the other two people on the island when they all cried &quot;WOO!&quot; On the other hand, he had had no communication with his niece since his sight had left him. He was blind and she was mute, so there was no way for her to communicate with him (apparently they weren't the hugging type). What's more, his niece was chubby, meaning there was plenty to eat, and weak, meaning she could easily be killed. She was a prime target. &lt;strong&gt;She had been killed and eaten.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In summary:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; His niece had survived the storm and the shipwreck only to be killed and eaten once they had got to the island. &lt;strong&gt;His niece had been murdered and he had eaten her body.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He couldn't live with this knowledge and he killed himself.&lt;/p&gt;&#xA;" OwnerUserId="5260" LastActivityDate="2015-05-01T13:39:05.653" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13168" PostTypeId="2" ParentId="13153" CreationDate="2015-05-01T13:40:55.077" Score="2" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;9&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pxihd.png&quot; alt=&quot;Cities by path.&quot;&gt;&#xA;&lt;br /&gt;Each of the path segments connected to $4$ can be traversed to get to $4$: $(1\rightarrow2\rightarrow3\rightarrow11\rightarrow4)$, $(6\rightarrow7\rightarrow5\rightarrow4)$, $(8\rightarrow10\rightarrow9\rightarrow12\rightarrow4)$.  But once you get to $4$, you can't go back, and thus can only destroy one segment after reaching $4$.  The segment that should be untouched, to maximize destruction, is the smallest segment $(5,6,7)$.&#xA;&lt;br /&gt;&lt;br /&gt;Therefore, the remaining 9 cities can be destroyed, and no more.  One possible path is $(1\rightarrow 2\rightarrow 3\rightarrow 11\rightarrow 4\rightarrow 12\rightarrow 9\rightarrow 10\rightarrow 8)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11851" LastActivityDate="2015-05-01T13:40:55.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13169" PostTypeId="1" AcceptedAnswerId="13185" CreationDate="2015-05-01T13:48:08.480" Score="1" ViewCount="202" Body="&lt;p&gt;Originally, this question was way too wide open. I have tried to clarify / narrow it down / restrict the possibilities. Let me know if it is still too wide open.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;These are some number sequences. I used the data set $A$ to generate both $B$ &amp;amp; $C$. I then compared those data sets in one way to get $D$ and in another way to get $E$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you find $B$ and $C$?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A: 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30&#xA;B: ?&#xA;C: ?&#xA;D: 2,3,0,2,1,2,2,1,1,2,2,1,2,2,2,2,2,2,2,3,2,3,0,2,1,2,2,1,1,3&#xA;E: 0,1,3,4,2,3,5,5,3,3,6,4,8,8,7,7,9,8,8,5,6,7,9,10,8,9,11,11,9,5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The methods to generate $B$ &amp;amp; $C$ do not use any mathematical operators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The methods to generate $D$ &amp;amp; $E$ use the operators $+$ &amp;amp; $-$ and nothing else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All numbers are whole numbers.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If you were to look at it another way, it might looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find $f_B$ &amp;amp; $f_C$ where:&#xA;$$n=1\to30$$&#xA;$$A_n=n$$&#xA;$$B_n=f_B(A_n)$$&#xA;$$C_n=f_C(A_n)$$&#xA;$$D_n=f_D(B_n,C_n)$$&#xA;$$E_n=f_E(B_n,C_n)$$&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2015-05-01T14:37:13.773" LastActivityDate="2015-05-01T15:40:14.030" Title="A Sequence of Sequences" Tags="|number-sequence|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="13170" PostTypeId="2" ParentId="3839" CreationDate="2015-05-01T13:58:01.783" Score="1" Body="&lt;p&gt;Another option could involve reinterpreting the winning conditions. The wise man could have observed that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; at the time the horse crosses the line, it's rider/owner will be the one whose horse last crosses the line. Evaluated at that moment, before any other horse crosses the line, the one whose horse that is wins, and the beautiful girl is &quot;his forever.&quot; The fact that someone else's horse might cross line later that day or many years later is irrelevant, because once the winner is declared, the win is forever. So, in fact, the first person to cross the line is also the last because the contest then ends.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Consider the alternative: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What if one rider crossed the line and the other did not. Would the one who crossed (and was currently last) win, or would the contest continue indefinitely to see if the other rider eventually crossed or did not?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11271" LastActivityDate="2015-05-01T13:58:01.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13171" PostTypeId="2" ParentId="13169" CreationDate="2015-05-01T14:01:57.840" Score="0" Body="&lt;p&gt;Not sure if this fits as a comment, but I think you need to specify a few more rules for your comparison.  Indeed, I can simply create two lists:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$2,8,9,14,21,47,27,101,10,11,20,17,58...$&#xA;&lt;br /&gt;$5,5,3,3,10,15,5,20,3,3,3,4,7...$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the two relations would be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$f(a,b) = a\mod b$&#xA;&lt;br /&gt;$g(a,b) = \lfloor a/b \rfloor $&#xA;&lt;br /&gt;Essentially, this treats the first list as our numerator, the second as the denominator, and the two resultant sets are the remainder and quotient, respectively.  Literally an infinite combination of lists can be generated this way.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11851" LastEditorUserId="11851" LastEditDate="2015-05-01T14:07:34.897" LastActivityDate="2015-05-01T14:07:34.897" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13173" PostTypeId="1" AcceptedAnswerId="13190" CreationDate="2015-05-01T14:10:12.447" Score="5" ViewCount="264" Body="&lt;p&gt;This is a sequence that I made. I poured by blood and sweat into this! OK, no, it was more like orange juice and yogurt. It wasn't really all that hard, I guess. It might be tricky to figure out, though! You'll know for sure when you've got it. Here's a fair warning, though, you're going to have to do some digging to confirm the 33rd term (that's the really big one).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0, 2, 1, 0, 2, 3, 0, 6, 6, 4, 44, 1, 180, 42, 16, 1096, 7652, 13781, 8, &#xA;24000, 119779, 458561, 152116956851941670912, 1054535, -53, 26, 27, 59, &#xA;4806078, 2, 35792568, 3010349, 2387010102192469724605148123694256128, 2, &#xA;0, -53, 43, 0, -4097, 173,...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I've given you the first 40 terms so there shouldn't be more than one possible answer. To help rule that out, though, I'll add this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;This is a finite list. You get an error if you try to get more than 52 terms.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="9000" LastActivityDate="2015-05-01T18:04:33.227" Title="A Sequence from a Sequence" Tags="|number-sequence|" AnswerCount="1" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13174" PostTypeId="1" AcceptedAnswerId="14776" CreationDate="2015-05-01T14:12:03.563" Score="4" ViewCount="1336" Body="&lt;p&gt;&lt;strong&gt;Can you help find the hidden message in this encrypted text?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A young Science student / Stack-Exchange enthusiast stumbles upon a magical encrypted diary of a fellow he has always admired. He finds that the book reveals new secrets on blank pages each day about the encryption method used on the first page... can you solve the mystery???&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note that there is an answer and the answer is written in plain English with all spaces delimiting actual words. Punctuation and numbers are not encrypted (if they are even included).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Words lengths 3,3,3,3,5,6,5,7,1,6,4,2,6,3,6,3,4,3,9,2,8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ciphertext:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Bpf qft lbv glz ixhve zqmoiw aqqow tunmzvd i uvjvuq liaj nc fonloc ook&#xA;yznbxf tei fgzy ynr ucfvngspr xn ccresriu.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Answers must include the method of encryption and the real text!&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus for figuring out who it relates to...&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Hint 1#:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Frequent conclusions you will not, not here... Frequent use will not bring me a-loose.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Hint 2#:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Two sheets shroud the in mystery, every pair is Transposed with each other as a &lt;em&gt;reference&lt;/em&gt; and then once all said and done the disk will help, maybe even the table? Either or. Don't thank me thank Trithemius.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Hint 3#:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm running out of hints to give but the word &quot;Bpf&quot; rhymes with 240 parts of one quid!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Hint 4#:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; aTubal eRtca... This will definitely help you, if you know what I mean!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Hint 5#:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; rTnapssotioin!!!&#xA;&lt;br&gt;If your looking for a key you might be barking up the wrong tree...&#xA;&lt;br&gt;as hint two is the closest thing you will see, maybe because there is no key.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Hint 6#:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm not your average T-R I focus more on horizontal movement then I should on vertical... But here is another version of the cipher text using a &quot;traditional&quot; T-R:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Ama req cmw nkw pcwiu lswyeu vfjlh mnwzoso l zaqehp wozc cp ohkmlb ptx&#xA;noaquq ctt yzil lyy xpocktlqo mw inapbeqh.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Hint 7#:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/MmYKX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;This&lt;/a&gt; might help!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hints will be added, but for now I'm all hinted out!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any Questions I will be happy to consider as the next days hint!&lt;/p&gt;&#xA;" OwnerUserId="12014" LastEditorUserId="12014" LastEditDate="2015-05-12T22:39:34.637" LastActivityDate="2015-05-13T16:33:40.370" Title="Can you crack the Magic Diary's encrypted message?" Tags="|cipher|word|" AnswerCount="1" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="13175" PostTypeId="1" AcceptedAnswerId="13179" CreationDate="2015-05-01T14:31:51.963" Score="13" ViewCount="578" Body="&lt;p&gt;&lt;em&gt;S&lt;/em&gt;olve the following code &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;dd ddddw ssaa wawasa&lt;br&gt;&#xA;  ddw ds wasawa&lt;br&gt;&#xA;  w d ds ds&lt;br&gt;&#xA;  aw aa d ssdwdsd&lt;br&gt;&#xA;  wwd ssaa&lt;br&gt;&#xA;  aaw dwdsdwd ssaaa wwdddd ssaaa awasawasaw dds sw dddww&lt;br&gt;&#xA;  ssd&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;That is a lot of a, s, w, d's&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't believe any hints are needed, but I will just say that the fullstop is at the end of the cryptogram.&lt;/p&gt;&#xA;" OwnerUserId="9736" LastEditorUserId="9736" LastEditDate="2015-05-04T12:19:14.480" LastActivityDate="2015-05-04T13:02:16.827" Title="Thats a lot of wasd's" Tags="|cipher|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13176" PostTypeId="2" ParentId="3739" CreationDate="2015-05-01T14:33:52.060" Score="6" Body="&lt;p&gt;The old man was actually a retired spy. The enemy found him and were pretending to care for him while they actually wanted to get information from him. Due to his rigorous training, a single bite into the seagull meat allowed him to realize that there was truth serum in the food. Not wanting to give up his nation's secrets, he commits suicide.&lt;/p&gt;&#xA;" OwnerUserId="2359" LastActivityDate="2015-05-01T14:33:52.060" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13177" PostTypeId="1" AcceptedAnswerId="13178" CreationDate="2015-05-01T14:44:59.880" Score="4" ViewCount="405" Body="&lt;p&gt;Here is my sequence (finite):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;19,499,12991,337787,8782467,228344156,5936948059,154360649539.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You just have to figure out its logic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is a &quot;sequence&quot; and I like alphabets.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12058" LastEditorUserId="12058" LastEditDate="2015-05-03T04:53:35.913" LastActivityDate="2015-05-03T04:53:35.913" Title="A sequence of integers" Tags="|mathematics|number-sequence|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13178" PostTypeId="2" ParentId="13177" CreationDate="2015-05-01T14:51:53.277" Score="8" Body="&lt;p&gt;For ADG1, the logic is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Convert all of the numbers to base 26, and convert each place in each number using A = 1, B = 2, etc. ADG1 then becomes&#xA; &lt;br&gt;&lt;br&gt;&#xA; S, SE, SEQ, SEQU, SEQUE, SEQUEN, SEQUENC, SEQUENCE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1600" LastActivityDate="2015-05-01T14:51:53.277" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13179" PostTypeId="2" ParentId="13175" CreationDate="2015-05-01T14:58:49.493" Score="12" Body="&lt;p&gt;It says...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; find the 34th term in fibonacci.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the answer (assuming the first two are 1 and 1) is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 5702887&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You have to start at s, and then it should be obvious. I won't say more, so people can discover it for themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Now that a couple of days have passed, I'll elaborate a bit further...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Starting on the letter &quot;s&quot; on a QWERTY keyboard, move around on the keyboard based on the letter sequence, where &quot;w&quot; is up, &quot;s&quot; is down, &quot;d&quot; is right, and &quot;a&quot; is left, as is commonly used in gaming.&lt;br&gt;&#xA;&lt;br&gt;&#xA; From the starting point, it begins with &quot;dd&quot;, which is two steps right, which brings us to the first letter, &quot;f&quot;. Then &quot;ddddw&quot; is four right and one up, which gets us to &quot;i&quot;. Next is &quot;ssaa&quot; which is two down and two left, which puts us on &quot;n&quot;. And then &quot;wawasa&quot; is a roundabout path requiring one up and three left, which leads us to &quot;d&quot;. It continues like this, with new lines indicating the start of new words.&lt;br&gt;&#xA;&lt;br&gt;&#xA; If you've done it right, the final line just has &quot;ssd&quot;, which should land you on the full stop (known as &quot;period&quot; in America).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9289" LastEditorUserId="11244" LastEditDate="2015-05-04T13:02:16.827" LastActivityDate="2015-05-04T13:02:16.827" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="13180" PostTypeId="1" AcceptedAnswerId="13181" CreationDate="2015-05-01T15:05:22.403" Score="2" ViewCount="374" Body="&lt;p&gt;I won't provide a hint this time. This one is OEIS-proof (similar to waterproof :D) too. Twenty terms and guess its logic only.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;00,50,33,25,20,16,14,12,11,10,90,83,76,71,66,62,58,55,52,50&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12058" LastEditorUserId="5188" LastEditDate="2015-05-02T03:01:57.093" LastActivityDate="2016-05-17T21:49:17.100" Title="An Up and Down Series 00,50,33,25 . ." Tags="|mathematics|number-sequence|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="13181" PostTypeId="2" ParentId="13180" CreationDate="2015-05-01T15:20:33.497" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To get the nth term, do 1/n and take the first two significant figures after the decimal point (not including leading zeros)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11880" LastEditorUserId="11880" LastEditDate="2015-05-01T15:23:52.747" LastActivityDate="2015-05-01T15:23:52.747" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13182" PostTypeId="2" ParentId="13180" CreationDate="2015-05-01T15:21:59.317" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First two decimal points ignoring zeros of 1/1, 1/2, 1/3, 1/4...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11851" LastEditorUserId="19989" LastEditDate="2016-05-17T21:49:17.100" LastActivityDate="2016-05-17T21:49:17.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13183" PostTypeId="2" ParentId="13122" CreationDate="2015-05-01T15:30:04.873" Score="9" Body="&lt;p&gt;As @JonTheMon said, there are $5!=120$ possible combinations for the keys.  The correct ordering must be in this list:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1: ABCDE&#xA;2: ABCED&#xA;3: ABDCE&#xA;4: ABDED&#xA;5: ABECD&#xA;6: ABEDC&#xA;...&#xA;115: EDABC&#xA;116: EDACB&#xA;117: EDBAC&#xA;118: EDBCA&#xA;119: EDCAB&#xA;120: EDCBA&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So, before you start asking questions, create this list since it is probably too hard to do on the fly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, go to the first guard and ask the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is the correct order of keys alphabetically after [ordering for #60] in a dictionary?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first guard will answer either yes or no.  If yes, repeat the question with ordering #90.  If no, then ask with ordering #30.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each question will narrow the range half, unless the range is odd, in which case you might be left with half-plus-one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following is the worst case progression (answer of no each time)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Question 1: Ask #60 on [1-120] -&amp;gt; [1-60]&#xA;Question 2: Ask #30 on [1-60]  -&amp;gt; [1-30]&#xA;Question 3: Ask #15 on [1-30]  -&amp;gt; [1-15]&#xA;Question 4: Ask #8  on [1-15]  -&amp;gt; [1-8]&#xA;Question 5: Ask #4 on  [1-8]   -&amp;gt; [1-4]&#xA;Question 6: Ask #2 on  [1-4]   -&amp;gt; [1-2]&#xA;Question 7: Is #2 the correct order?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;At the end, you will know the ordering (either #1 or #2 in this example), so simply use the keys in order and open the doors.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="1592" LastEditDate="2015-05-01T17:03:38.367" LastActivityDate="2015-05-01T17:03:38.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13185" PostTypeId="2" ParentId="13169" CreationDate="2015-05-01T15:40:14.030" Score="3" Body="&lt;p&gt;The relevant sequences aren't actually numbers, but words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sequence B is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One, Two, Three, Four, Five...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And Sequence C is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First, Second, Third, Fourth, Fifth...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The operations to get to the final sequence involve differences, but not mathematically. Sequence D is obtained by...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Counting how many more letters there are in sequence C than sequence B. That is, &quot;First&quot;=5 and &quot;One&quot;=3, giving 2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While Sequence E is obtained by...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Counting how many letters are common between the two words. So &quot;one&quot; and &quot;first&quot; have no letters in common, while &quot;two&quot; and &quot;second&quot; have &quot;o&quot; in common, giving 1. Similarly, &quot;three&quot; and &quot;third&quot; both have &quot;t&quot;, &quot;h&quot;, and &quot;r&quot;, making it three.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9289" LastActivityDate="2015-05-01T15:40:14.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13188" PostTypeId="2" ParentId="13159" CreationDate="2015-05-01T16:46:09.303" Score="3" Body="&lt;p&gt;A solution you should not tell to &lt;a href=&quot;https://xkcd.com/169/&quot; rel=&quot;nofollow&quot;&gt;black hat&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When he challenged you to &quot;jump over a bottle on the floor&quot;, he meant &quot;on the floor&quot; to refer to you and not the bottle.  You can't technically be &quot;on the floor&quot; and jump at the same time, and so...hey...put that machete down!!!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2015-05-01T16:46:09.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13189" PostTypeId="2" ParentId="13122" CreationDate="2015-05-01T17:44:43.483" Score="1" Body="&lt;p&gt;As @JonTheMon and @Trennin said, there are 120 possibilities so we need to eliminate about half at each step. Here's a decision tree that allows you do that without having to either ask each guard about a whole bunch of possibilities or write down the full list of 120 options in advance:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Pick two keys (1 and 2) and ask &quot;Does key 1 go to a door to the left of key 2?&quot; Label the left key &quot;A&quot; and the right key &quot;B&quot;.&lt;/li&gt;&#xA;&lt;li&gt;Pick another two keys (3 and 4) and ask the same question about those keys. Call the left key &quot;C&quot; and the right key &quot;D&quot;.&lt;/li&gt;&#xA;&lt;li&gt;Ask the same question about keys A and C. If C is to the left of A, swap the labels of the pairs (A,B) and (C,D).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;At this point, we have four keys and we know A is to the left of B and C, and C is to the left of D. An easy way to keep track of this would be to place them on the ground in like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A -&amp;gt; B&#xA;|&#xA;V&#xA;C -&amp;gt; D&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;Ask &quot;Does either C or D go to the door in the middle?&quot;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;At this point the questions we need to ask (rather than which keys we need to ask about) change depending on the result. So rather than renaming/rearranging keys, we'll have the tree actually split.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If C or D is in the middle:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;Find where E is - ask if it goes to one of the first two doors to narrow it down to one of two doors.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Ask about one of the two doors E could go to. You now know where E is.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Ask if B is to the left of D.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You now have enough information to determine the order! If E goes door 1 or 2, then from left to right the doors are:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(A and E)C(B and D)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If E goes to door 4 or 5, then if B is to the left of D:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ABCED or ABCDE (ABCD with E inserted)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If B is to the right of D:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ACDEB or ACDBE (ACDB with E inserted)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If C and D are not in the middle:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;Ask if E is in the middle.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;E is in the middle:&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's one of these three cases:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ABECD&#xA;ACEBD&#xA;ACEDB&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;ol start=&quot;6&quot;&gt;&#xA;&lt;li&gt;&quot;Does B go to door 2?&quot; - Yes = ABECD&lt;/li&gt;&#xA;&lt;li&gt;&quot;Does B go to door 4?&quot; - Yes = ACEBD, No = ACEDB&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;E's not in the middle either:&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means B is in the middle (A is to the left of too many doors to be in the middle). We've got these cases left:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ACBDE&#xA;ACBED&#xA;AEBCD&#xA;EABCD&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Looks like we can identify the case by finding which door E goes to. As with the case of C or D being in the middle,&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;6&quot;&gt;&#xA;&lt;li&gt;Ask if E goes to door 1 or 2&lt;/li&gt;&#xA;&lt;li&gt;If yes, ask if it goes to door 1. If no, ask if it goes to door 3.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Here's how the cases correspond to the answers:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ACBDE - No, No&#xA;ACBED - No, Yes&#xA;AEBCD - Yes,No&#xA;EABCD - Yes,Yes&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1955" LastActivityDate="2015-05-01T17:44:43.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13190" PostTypeId="2" ParentId="13173" CreationDate="2015-05-01T17:52:34.163" Score="8" Body="&lt;p&gt;The next element is  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$37338$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Open &lt;strong&gt;&lt;a href=&quot;http://oeis.org/&quot;&gt;http://oeis.org/&lt;/a&gt;&lt;/strong&gt; , the world biggest database of known integer sequences.&lt;br&gt;&#xA;You see that any sequence has an identifier number, for example the list of natural numbers is known as &lt;a href=&quot;http://oeis.org/A000027&quot;&gt;&lt;strong&gt;A000027&lt;/strong&gt;&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How is Toast's list generated:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Starting from &lt;a href=&quot;https://oeis.org/A000001&quot;&gt;&lt;strong&gt;A000001&lt;/strong&gt;&lt;/a&gt;, we take the $i_{th}$ number of the A00000$i_{th}$ sequence to build our one.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As you can see, &lt;strong&gt;&lt;a href=&quot;https://oeis.org/A000001&quot;&gt;A000001&lt;/a&gt;&lt;/strong&gt; starts with 0, so we consider 0 to build our sequence.&#xA;Then, we take the second number of &lt;strong&gt;&lt;a href=&quot;https://oeis.org/A000002&quot;&gt;A000002&lt;/a&gt;&lt;/strong&gt;, which is 2.&#xA;Then, we consider the third number of &lt;strong&gt;&lt;a href=&quot;https://oeis.org/A000003&quot;&gt;A000003&lt;/a&gt;&lt;/strong&gt;, which is 1.&#xA;And so on...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next term in the sequence is &lt;strong&gt;37338&lt;/strong&gt;, the 41th term of &lt;a href=&quot;https://oeis.org/A000041&quot;&gt;&lt;strong&gt;A000041&lt;/strong&gt;&lt;/a&gt;.&lt;br&gt;&#xA;Toast's list has only 52 terms because &lt;strong&gt;&lt;a href=&quot;https://oeis.org/A000053&quot;&gt;A000053&lt;/a&gt;&lt;/strong&gt; has only 29 numbers!&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-05-01T18:04:33.227" LastActivityDate="2015-05-01T18:04:33.227" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13191" PostTypeId="2" ParentId="13107" CreationDate="2015-05-01T18:01:06.607" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I met a traveller from an antique land&#xA;&lt;br/&gt;Who said: &quot;Two vast and trunkless legs of stone&#xA;&lt;br/&gt;Stand  [&lt;s&gt;o&lt;/s&gt; &lt;strong&gt;i&lt;/strong&gt;]n the desert. Near them, o[&lt;s&gt;ver&lt;/s&gt; &lt;strong&gt;n&lt;/strong&gt;] [&lt;strong&gt;t&lt;/strong&gt;]h[&lt;s&gt;is&lt;/s&gt; &lt;strong&gt;e&lt;/strong&gt;] sand,&#xA;&lt;br/&gt;Half sunk, a shatter[&lt;strong&gt;e&lt;/strong&gt;]d visage lies, whose [&lt;s&gt;g&lt;/s&gt; &lt;strong&gt;fr&lt;/strong&gt;]own,&#xA;&lt;br/&gt;And wrinkled lip, and sneer of cold command,&#xA;&lt;br/&gt;Tell that its sculptor well those passions read&#xA;&lt;br/&gt;Which yet survive, stamped on these lifeless things,&#xA;&lt;br/&gt;The h[&lt;s&gt;ea&lt;/s&gt; &lt;strong&gt;an&lt;/strong&gt;]d that mocked them and the heart that fed:&#xA;&lt;br/&gt;And o[&lt;s&gt;ver&lt;/s&gt; &lt;strong&gt;n&lt;/strong&gt;] the pedestal these words appear:&#xA;&lt;br/&gt;'My name is Ozymandias, king of kings:&#xA;&lt;br/&gt;Look on my wor[&lt;s&gt;d&lt;/s&gt; &lt;strong&gt;k&lt;/strong&gt;]s, ye Mighty, and despair!'&#xA;&lt;br/&gt;Nothing beside remain[&lt;strong&gt;s&lt;/strong&gt;]. Round the decay&#xA;&lt;br/&gt;Of that colossal wreck, boundless and bare&#xA;&lt;br/&gt;The lone and level sands stretch far away.&#xA;&lt;br/&gt;&#xA;&lt;br&gt;gives &lt;strong&gt;i,n,t,e,e,f,r,a,n,n,k,s&lt;/strong&gt;&#xA;&lt;br/&gt;&#xA;&lt;br/&gt;which is an anagram of &lt;strong&gt;Frankenstein&lt;/strong&gt;&#xA;&lt;br/&gt;&#xA;&lt;br/&gt;written, of course, by Shelley's wife, &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Mary_Shelley&quot;&gt;Mary Shelley&lt;/a&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="2819" LastEditDate="2015-05-01T18:27:58.320" LastActivityDate="2015-05-01T18:27:58.320" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13192" PostTypeId="2" ParentId="12386" CreationDate="2015-05-01T18:26:44.780" Score="4" Body="&lt;p&gt;Follow these 5 steps in order. Only 1 of the steps will be done for any specific grid. The first 3 steps are the trivial cases, and steps 4 and 5 cover all other cases.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If m=1 or n=1, then build like this. (Red blocks denote that the main pattern is altered, but it must be selected for coverage. Transpose if necessary.)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ow6xN.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;If m=2 or n=2, then build like this.  (Red blocks denote that the main pattern is altered, but it must be selected for coverage. Transpose if necessary.)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wp40u.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;If m=3 or n=3, then build like this.  (Transpose if necessary.)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pBiiw.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;If both m and n are even, then build like this, repeating the pattern shown for 4x4. (Red blocks denote that the main pattern is altered, but it must be selected for coverage.)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NL1LE.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;If both m and n are not even, then build like this.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;Careful: When m is odd and n is even (or vice versa), you must start the pattern coloring every other cell in a row or column with an &lt;strong&gt;even&lt;/strong&gt; number of cells. If you do it the other way, then you are not guaranteed the optimal coverage.&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Careful: When m and n are both odd and m is not equal to n, you must start the pattern coloring every other cell on the shorter side. If you do it the other way, then you are not guaranteed the optimal coverage.&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6MGTE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6MGTE.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2022-04-23T11:54:37.873" LastActivityDate="2022-04-23T11:54:37.873" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="13193" PostTypeId="2" ParentId="13122" CreationDate="2015-05-01T18:30:09.593" Score="6" Body="&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/users/1592/trenin&quot;&gt;Trenin&lt;/a&gt;'s answer simply lists all the permutations in alphabetical order and does a binary search.  It works, with the only drawback that memorizing (or computing on the fly) the index of all the possibilities is difficult, and that it assumes that the guards can handle questions like &quot;Is ABDCE alphabetically before ACDBE?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a method that asks only questions of the form &quot;Is the door for key X &amp;lt; door for key Y&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is A &amp;lt; B?&lt;/li&gt;&#xA;&lt;li&gt;Is C &amp;lt; D?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You now know two lesser ones and two greater ones.  Ask a comparison for the two lesser ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if A &amp;lt; B and C &amp;lt; D, ask:&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;Is A &amp;lt; C?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Whichever one is less must now be the least of the four.  You now have an ordered chain of 3 keys, plus a fourth that you only know is not the first door.  So if A &amp;lt; C, then you know A &amp;lt; C &amp;lt; D, or if C &amp;lt; A, then you know C &amp;lt; A &amp;lt; B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your fourth question, compare E to the middle one in the chain.  For example:&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;Is E &amp;lt; C?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Then compare E to the appropriate end of the chain.  So if E &amp;lt; C, then:&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;Is E &amp;lt; A?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, after five questions you have identified a chain of four (e.g. AECD), and a fifth door that gets inserted somewhere not in the first position.  That leaves four more possibilities, which is good, because you have two more questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare the extra key to the third key in your chain (e.g., for above example):&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;6&quot;&gt;&#xA;&lt;li&gt;Is B &amp;lt; C?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Then compare the extra to the second or fourth.  For example, if A &amp;lt; E &amp;lt; C &amp;lt; D, and B &amp;lt; C, ask:&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;7&quot;&gt;&#xA;&lt;li&gt;Is B &amp;lt; E?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This path guarantees that you will get the correct ordering in seven questions.  There are a few arrangements where the seventh question is unnecessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remembering an ordered chain should be much easier than trying to memorize the indexing of 120 permutations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Counting details:&#xA;Questions 1, 2, and 3 are guaranteed to be independent and thus reduce the number of possibilities from 120 down to 15.  Question 4 splits that evenly into 7 possibilities (for yes) or 8 possibilities (for no).  The last three questions split the remaining possibilities evenly.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-05-01T18:30:09.593" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13194" PostTypeId="2" ParentId="13050" CreationDate="2015-05-01T18:34:00.703" Score="9" Body="&lt;p&gt;&lt;a href=&quot;/questions/tagged/lateral-thinking&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;lateral-thinking&amp;#39;&quot; rel=&quot;tag&quot;&gt;lateral-thinking&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would you legally own the money you created with the machine? If you just start to spend a lot of money (or just turn up at a bank with trillions of coins) and you can't prove where you got it from, you will be imprisoned for tax fraud. If you show them the machine, from the point of view of the government, you just made counterfeit money, so it will be confiscated and you will end up in jail. Unless you want to start a criminal enterprise and launder that money somehow, which, without good connections and experience in the criminal underworld only with a bunch of coins, will not be an easy and risk-free task. (The hundreds of years mentioned in &lt;code&gt;GOTO 0&lt;/code&gt;'s answer will be increased by orders of magnitude while you launder more than a trillion coins)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much easier to just accept the one trillion dollars, which can happen legally, as you just sold a device to someone. You pay the taxes for it, and you have hundreds of billions of dollars left, which you now own legally.&lt;/p&gt;&#xA;" OwnerUserId="2507" LastEditorUserId="2507" LastEditDate="2015-05-01T20:16:29.933" LastActivityDate="2015-05-01T20:16:29.933" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13195" PostTypeId="2" ParentId="13127" CreationDate="2015-05-01T19:22:41.373" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Kentucky Derby &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some reasoning&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  It's topical, the 2015 Kentucky Derby will be run tomorrow.  Gamblers have an angle, or bias, so Old Mac Donald (OMD) thought he had inside information.  Let's say he chose to &quot;fly close to the sun&quot; and pick a longshot. The news has the story that &quot;El Kabeir&quot;, a 30-1 pick, may be scratched from the race.  El Kabeir's post is #7; the number 2 favorite at 3-1, Dortmund, is at post #8 and Dortmund's jockey will wear scarlet.  If only I could stretch El Kabeir's etymology (Arabic for &quot;The Boss&quot; according to the NY Times) into some kind of big cat lion or tiger, I would have a slam dunk.  OMD's bet is &quot;here&quot; on #7, El Kabeir.  &quot;There&quot; is the rest of the field on the racing form (or horse guide).  Adjacent to &quot;here&quot; is the scarlet jockey; each other horse is there.  The air is charged everywhere, thus the ions.  OMD picked a crazy risky horse to bet on, and is descending into the madness of a gambler.     &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11921" LastActivityDate="2015-05-01T19:22:41.373" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13196" PostTypeId="1" AcceptedAnswerId="13200" CreationDate="2015-05-01T19:26:28.273" Score="3" ViewCount="1163" Body="&lt;p&gt;While browsing through S.E you notice a spider crawl right out from under your keyboard! Startled, you twitch your hand and left clicked your mouse. You look up in shock as you notice that you just gave an unjustified down vote! You try to fix it as fast as possible, but before you can, the devil himself appears before you and takes you to his realm!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The devil challenges you to his game. The game will take place on a&#xA;  cylinder, in which you start solely on one base of the cylinder. The&#xA;  devil is then able to replicate any amount of you as he pleases to&#xA;  start on the other side of the cylinder (These replicates are&#xA;  obviously evil). Once the game starts your challenge is to get to the&#xA;  other side of the cylinder without being captured by your evil&#xA;  replicates!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The devil makes a promise that if you can beat him in this game, he&#xA;  will return you to your desk and banish his spider from your desk. But&#xA;  if you lose you will be stuck watching the devil abuse your account&#xA;  forever! Would you able to save your account?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The cylinder is large enough to consider yourself (and the&#xA;replicates) as zero radius points&lt;/li&gt;&#xA;&lt;li&gt;The replicates will play with the&#xA;best possible strategy to make you lose.&lt;/li&gt;&#xA;&lt;li&gt;You and the replicates move at the same speed and at the same time.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If the devil makes you another offer to play, however your replicates receive a  greater than 0 radius and in return he allows you to chose the dimensions of the cylinder, should you accept? &lt;/p&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="10615" LastEditDate="2015-06-09T06:48:38.287" LastActivityDate="2015-06-09T06:48:38.287" Title="Game against the Devil" Tags="|logical-deduction|mathematics|strategy|game|pursuit-evasion|" AnswerCount="2" CommentCount="8" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13197" PostTypeId="1" CreationDate="2015-05-01T19:34:04.697" Score="9" ViewCount="1163" Body="&lt;p&gt;We found this mysterious text in the file of one of our employees and wish to discover whether it is anything we should be concerned with. We know that it uses two substitution ciphers. Even-numbered letters are encoded with one substitution cipher and odd-numbered letters are with another; spaces and punctuation are not counted.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;znjs, uxwd bnfvx il e toaqlcu swzjp un keaer-haet alplr'k llde. w jedo uxl&#xA;blfk ph niqwcl, nkplskwqvv ph hhre nr mv cfraorpnf trhglcu. xprhvx, onwceqoux,&#xA;psx mprwer hwvn il eanfsx, ifp w brnb env hq uxlm prl hwvnwrz un mn pvhsm hwpj &#xA;hjeuodoa hl lh. w bwnv qrwsm pjo keaer-haet, djnb kt wi fhk berp un jedo p&#xA;apru!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="12075" LastEditorUserId="9000" LastEditDate="2015-05-01T20:33:42.130" LastActivityDate="2015-05-02T02:54:50.230" Title="We found this mysterious text in the file of one of our employees" Tags="|cipher|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13198" PostTypeId="2" ParentId="13196" CreationDate="2015-05-01T19:39:04.747" Score="2" Body="&lt;p&gt;&lt;strong&gt;Serious answer:&lt;/strong&gt;&lt;br&gt;&#xA;You can't win the first round because the devil can always move along the circumference of his base, staying aligned with you and, so, not letting you to cross his circumference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This applies of course to cylinders of any dimension (indeed, my strategy never mentioned sizes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fun answer:&lt;/strong&gt;&lt;br&gt;&#xA;Do you think that devil would ever let you win? &lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-05-01T19:54:02.310" LastActivityDate="2015-05-01T19:54:02.310" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="13199" PostTypeId="2" ParentId="13146" CreationDate="2015-05-01T19:45:48.557" Score="0" Body="&lt;p&gt;Harrison knows it must not be George because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; George is deaf.  When the voice on the recording says he &quot;hears footsteps,&quot; Harrison knows it must not be his friend.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The assassin obviously did not know this detail about George.&lt;/p&gt;&#xA;" OwnerUserId="2693" LastActivityDate="2015-05-01T19:45:48.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13200" PostTypeId="2" ParentId="13196" CreationDate="2015-05-01T19:46:56.423" Score="3" Body="&lt;p&gt;You can't win the first proposal, since the devil can just mirror your starting position and your evil clone© can just mirror your moves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second proposal, I think you could win if you cheat a little on making the cylinder: make a cylinder with uneven base diameters and start on the smaller one. You can turn radially faster than the evil clones© (so, a bit like the duck in a pond question) and potentially escape. Unfortunately, the devil can negate your advantage by having extra evil clones© that will make your rotation ineffective, and have extra clones to chase you. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-05-01T19:46:56.423" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13202" PostTypeId="2" ParentId="13197" CreationDate="2015-05-01T22:09:30.743" Score="10" Body="&lt;p&gt;I believe there's a typo: &lt;code&gt;cfraorpnf&lt;/code&gt; should be &lt;code&gt;cfraorp&lt;/code&gt;. But, of course, such typos happen all the time in notes like this :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uppercasing every second letter:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;zNjS, uXwD bNfVx Il E tOaQlCu SwZjP uN kEaEr-HaEt AlPlR'k LlDe. W jEdO uXl&#xA;BlFk Ph NiQwCl, NkPlSkWqVv Ph HhRe Nr Mv CfRaOrPnF tRhGlCu. XpRhVx, OnWcEqOuX,&#xA;pSx MpRwEr HwVn Il EaNfSx, IfP w BrNb EnV hQ uXlM pRl HwVnWrZ uN mN pVhSm HwPj&#xA;HjEuOdOa Hl Lh. W bWnV qRwSm PjO kEaEr-HaEt, DjNb Kt Wi FhK bErP uN jEdO p&#xA;ApRu!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As for whether they should be concerned,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; no, they're planning an innocent office prank.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How I managed to solve it: Step/hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The word with apostrophe is either &lt;code&gt;weren't&lt;/code&gt; or &lt;code&gt;&amp;lt;something&amp;gt;'s&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Step/hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I began by guessing &lt;code&gt;weren't&lt;/code&gt;, but then the rest (especially the likely appearence of the word &quot;the&quot;) didn't seem to work out. Assuming &lt;code&gt;&amp;lt;something&amp;gt;'s&lt;/code&gt;, the only letter known so far gave another clue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Step/hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The word &lt;code&gt;nkplskwqvv&lt;/code&gt; will decrypt to &lt;code&gt;.s...s....&lt;/code&gt;. Which word could it be?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Step/hint 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I performed a &lt;code&gt;grep -x .s...s.... /usr/share/dict/words&lt;/code&gt; which gave &lt;code&gt;ostensibl[ey]&lt;/code&gt; as the only matches without too many &lt;code&gt;s&lt;/code&gt;'s.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Step/hint 5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; From here, the rest was quite easy guesswork.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the decrypted message (every second letter uppercased) is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; jOhN, tHiS wOuLd Be A pErFeCt NiGhT tO sArAn-WrAp PeTeR's DeSk. I hAvE tHe KeYs To OfFiCe, OsTeNsIbLy To WoRk On My CuRrEnTlY pRoJeCt. HaRoLd, ElIsAbEtH, aNd MaRiAn WiLl Be ArOuNd, BuT i KnOw AlL oF tHeM aRe WiLlInG tO gO aLoNg WiTh WhAtEvEr We Do. I wIlL bRiNg ThE sArAn-WrAp, ShOw Up If YoU wAnT tO hAvE a PaRt!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks for this puzzle, Joe L! I also learned two new English words (I'm not native speaker):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ostensibly and saran-wrap&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11803" LastActivityDate="2015-05-01T22:09:30.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13203" PostTypeId="2" ParentId="13197" CreationDate="2015-05-01T22:46:11.563" Score="4" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; John, this would be a perfect night to saran-wrap Peter's desk. I have the keys to office, ostensibly to work on my currently [sic] project. Harold, Elizabeth and Marian will be around, but I know all of them are willing to go along with whatever we do. I will bring the saran-wrap, show up if you want to have a part!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Method:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I first went through and colour-coded all the letters, to allow me to see which letters belonged to which cipher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I put down all four letters that appeared as a single word as the two &lt;code&gt;i&lt;/code&gt;s and two &lt;code&gt;a&lt;/code&gt;s. I also realised that &lt;img src=&quot;https://i.stack.imgur.com/yvshK.png&quot; alt=&quot;red-k&quot;&gt; was going to be &lt;code&gt;d&lt;/code&gt;, &lt;code&gt;s&lt;/code&gt;, or &lt;code&gt;t&lt;/code&gt; because it followed an apostrophe. However, this didn't get me very far, so I started writing down the two- and three-letter words to try and see if I could see any patterns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I noticed that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;img src=&quot;https://i.stack.imgur.com/eOKZw.png&quot; alt=&quot;un&quot;&gt; appeared three times and &lt;img src=&quot;https://i.stack.imgur.com/hfEvb.png&quot; alt=&quot;il&quot;&gt; appeared twice (and some of their letters appeared in other two-letter words)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which was promising but not conclusive. Then I noticed&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/MoybH.png&quot; alt=&quot;uxl&quot;&gt; and &lt;img src=&quot;https://i.stack.imgur.com/gNYM1.png&quot; alt=&quot;uxlm&quot;&gt;! Almost certainly &lt;img src=&quot;https://i.stack.imgur.com/MoybH.png&quot; alt=&quot;uxl&quot;&gt; was &lt;code&gt;the&lt;/code&gt;, with &lt;img src=&quot;https://i.stack.imgur.com/gNYM1.png&quot; alt=&quot;uxlm&quot;&gt; as &lt;code&gt;then&lt;/code&gt;, &lt;code&gt;them&lt;/code&gt; or &lt;code&gt;they&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once I'd fillen in those letters, I knew that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/eOKZw.png&quot; alt=&quot;un&quot;&gt; (the two letter word that appeared three times) was &lt;code&gt;t?&lt;/code&gt; and took a guess that it was probably &lt;code&gt;to&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I also looked at &lt;img src=&quot;https://i.stack.imgur.com/RTtnq.png&quot; alt=&quot;prl&quot;&gt;. I knew that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/aNP1F.png&quot; alt=&quot;p-red&quot;&gt; was &lt;code&gt;a&lt;/code&gt; or &lt;code&gt;i&lt;/code&gt;, and &lt;img src=&quot;https://i.stack.imgur.com/JU4uy.png&quot; alt=&quot;l-red&quot;&gt; was &lt;code&gt;e&lt;/code&gt;. There are only two words of the form &lt;code&gt;i?e&lt;/code&gt; - ice and ire - and neither seemed that likely, so I took the reasonable guess that &lt;img src=&quot;https://i.stack.imgur.com/aNP1F.png&quot; alt=&quot;p-red&quot;&gt; was &lt;code&gt;a&lt;/code&gt;. &lt;img src=&quot;https://i.stack.imgur.com/RTtnq.png&quot; alt=&quot;prl&quot;&gt; also followed &lt;img src=&quot;https://i.stack.imgur.com/gNYM1.png&quot; alt=&quot;uxlm&quot;&gt; (my &lt;code&gt;them&lt;/code&gt;/&lt;code&gt;they&lt;/code&gt;/&lt;code&gt;then&lt;/code&gt; word) so it seemed likely that &lt;img src=&quot;https://i.stack.imgur.com/RTtnq.png&quot; alt=&quot;prl&quot;&gt; was &lt;code&gt;are&lt;/code&gt; (it could have been &lt;code&gt;ate&lt;/code&gt;, but less likely).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If &lt;img src=&quot;https://i.stack.imgur.com/aNP1F.png&quot; alt=&quot;p-red&quot;&gt; was &lt;code&gt;a&lt;/code&gt;, &lt;img src=&quot;https://i.stack.imgur.com/Y5prE.png&quot; alt=&quot;w-red&quot;&gt; (the other red single-word letter) had to be &lt;code&gt;i&lt;/code&gt;. That left me with&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;thi?&lt;/code&gt; near the start. Only &lt;code&gt;this&lt;/code&gt; and &lt;code&gt;thin&lt;/code&gt; fit that pattern, and I thought given its placement (second word of the puzzle, after a comma), &lt;code&gt;this&lt;/code&gt; was more likely.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After that I started making some bigger jumps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I turned&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;I ??o?&lt;/code&gt; into &lt;code&gt;I know&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which gave me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;this wo--- -e&lt;/code&gt; and &lt;code&gt;w-n- to&lt;/code&gt; to turn into &lt;code&gt;this would be&lt;/code&gt; and &lt;code&gt;want to&lt;/code&gt; respectively&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and then &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;I know a-l -- the- are&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;into&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;I know all of them are&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;etc. etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually giving me the answer you see at the top.&lt;/p&gt;&#xA;" OwnerUserId="5260" LastEditorUserId="5260" LastEditDate="2015-05-01T23:02:53.370" LastActivityDate="2015-05-01T23:02:53.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13205" PostTypeId="2" ParentId="13121" CreationDate="2015-05-02T01:21:46.317" Score="1" Body="&lt;p&gt;Fasten your seat belts:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your friend went to Gjerde, which appears to be a small town in Norway! Why? Because it rhymes with:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some are high:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Byrd (the Byrds' music was sometimes banned for promoting recreational drugs) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some can fly:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bird&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some are in cities:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Kurd&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some are in ditties:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Word&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some wear trousers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nerd&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some need browsers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Third (according to a quick Google search, roughly 1/3 of all U.S. citizens don't have a home internet connection)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some are ground in the sea:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gerd, a Norse Goddess 'commonly theorized to be a goddess associated with the earth' according to Wikipedia. Earth = ground!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some oversee the grounds:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Herd. Herds of grazing animals are great at keeping grass trimmed!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11154" LastActivityDate="2015-05-02T01:21:46.317" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13206" PostTypeId="1" AcceptedAnswerId="13211" CreationDate="2015-05-02T02:07:47.343" Score="10" ViewCount="1008" Body="&lt;p&gt;Source:  This was posted on the Grey Labyrinth Forums many years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.mathsisfun.com/puzzles/5-pirates-solution.html&quot;&gt;The five pirates are back&lt;/a&gt;. But this time one of them has the dreaded Zanzibar fever, whose terrible effects are spelled out below.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;300 gold coins need to be split between the five pirates.&lt;/li&gt;&#xA;&lt;li&gt;Same idea as the original: the oldest pirate proposes a split of the gold coins, and everyone votes.  If more than half say yes, then that is how the coins are divided.  Otherwise the oldest pirate is thrown overboard and the process is repeated with the remaining pirates. (Note the &quot;more than half&quot; which is different from the original version)&lt;/li&gt;&#xA;&lt;li&gt;The pirate infected with Zanzibar fever always acts fairly:  she always proposes an even split between the remaining pirates, and only votes for such a proposal.&lt;/li&gt;&#xA;&lt;li&gt;Each pirate is equally likely to have Zanzibar fever, and no healthy pirate initially knows who has it.&lt;/li&gt;&#xA;&lt;li&gt;Voting is done by secret ballot.&lt;/li&gt;&#xA;&lt;li&gt;The other pirates are perfect logicians who always act to maximize their own expected earnings. If multiple strategies yield the same expected value, then they randomly select among them.  For this puzzle, maximizing likelihood of survival or bloodthirstiness are not factors.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If the oldest pirate is fever-free, how should she propose to divide the gold?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit:  I intended survival really not being a priority at all for the pirates, and that the secret vote only reveals if the plan was accepted or not, no other information.  However, JS1 gives a great answer below to the case where survival is more important than any amount of money, and the secret ballot does reveal the number of yes / no votes.&lt;/p&gt;&#xA;" OwnerUserId="2722" LastEditorUserId="2722" LastEditDate="2015-05-02T07:50:01.787" LastActivityDate="2015-05-05T19:26:09.813" Title="Five pirates splitting gold with a twist" Tags="|strategy|probability|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13209" PostTypeId="2" ParentId="13206" CreationDate="2015-05-02T03:04:58.530" Score="1" Body="&lt;p&gt;Let's work our way from the bottom, the youngest pirate. The youngest pirate either wants all the gold, or to split it evenly with... himself!. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second youngest pirate will lose if the youngest says no, so he must do a 50/50 split on the off chance the youngest is a Zanzibarbarian (with exploding wigs of death!). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The middle pirate can win with a single vote from below. Given 2's predicament, (and the possibility that one of the lower 2 is feverish), he has to appease one or both of them. If the youngest is feverish, then he could win over 2nd youngest by giving 151 coins and keeping 149. However, if #2 is feverish, then he could do a 100 split since he'll then get the 2nd's vote. That move is also the safest, since it also handles the youngest being feverish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the 2nd oldest now is in a weird state. If he is feverish, he'll attempt to split the pot. The 3rd oldest now thinks that neither of 1st or 2nd are feverish, so he'll go for the 151/149 split. So, that'll fail. If not Zanzibarbarish, he has to give the middle pirate at least 100, since that's the minimum the middle stands to gain. Also, he has to worry about one of the younger ones being feverish. 2nd youngest also stands to gain more than 100, since that is most likely the middle's action. So, taking a risk of 0/101/101/98 is the 4th's play.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, so most of the time, the youngest pirate will say no, and nobody really wants the evenly divided loot (or there aren't enough who do). So, the 2nd youngest stands to make 151/100/101 coins. So, you'll have to offer 2nd youngest at least 102 to win the vote. 3rd youngest stands to make 149/100/101 coins, so the same, 102 coins. 4th stands to make 98, so offer 99. So, offer 99 to the 4th, and 102 to either the 2nd or the 3rd, and keep 99 for yourself. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-05-02T03:04:58.530" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13210" PostTypeId="2" ParentId="13028" CreationDate="2015-05-02T03:49:28.997" Score="13" Body="&lt;p&gt;It could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A baby&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I take the expressway, I'm in a hurry.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sperm racing to the egg, in a hurry to beat the other millions of racers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By chance I make it, no need to worry,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; With that many racers, it's truly by chance&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll loiter and idle and maybe sleep.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sleeping and development&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll finally wake, my day's done it seems,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Maybe the &quot;day being done&quot; is a reference to the first light being seen, and the &quot;night&quot; being over&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I hurry back but now lost, I weep.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Crying when born. Everything is new. Baby feels &quot;lost.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And yet my home at last I see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Finally opens his eyes and accepts his parents and home.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hints:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A little Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An Artist? an Athlete? Your attempts are forlorn.&lt;br&gt;&#xA;  This little puzzle might be simpler to those who are with horns.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Maybe angels with horns, ushering in the birth of a baby?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A big Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sometimes I'm like a Fighter in a cage.&lt;br&gt;&#xA;  Sometimes I'm hungry and make you crave.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A baby in the womb punches and kicks the mother's ribs and stomach. Also, babies make moms crave food!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A bigger Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Nine goes in the middle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the middle of the riddle, where the space is between the 2 stanzas, 9 months pass.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-07-16T23:48:10.357" LastActivityDate="2015-07-16T23:48:10.357" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13211" PostTypeId="2" ParentId="13206" CreationDate="2015-05-02T06:36:15.623" Score="9" Body="&lt;h2&gt;Preliminary work that comes in handy later&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Let's first solve without any Zanzibar fever:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 pirate : 300&#xA;2 pirates: 300/dead (pirate 1 votes no)&#xA;3 pirates:   0/   0/ 300 (pirate 2 doesn't want to die, votes yes)&#xA;4 pirates:   1/   1/   0/ 298&#xA;5 pirates:   2/   0/   1/   0/ 297 (or 0/2/1/0/297)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now let's solve assuming Zanzibar is still in play (and not the oldest pirate remaining):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 pirate : 300&#xA;2 pirates: 150/ 150 (pirate 1 is Zanzibar)&#xA;3 pirates: 100/ 100/ 100 (get Zanzibar vote plus self)&#xA;                         (Any other split risks dying)&#xA;4 pirates: See below, tricky&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And now let's solve for when the oldest pirate &lt;em&gt;is&lt;/em&gt; the Zanzibar pirate:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 pirate : 300&#xA;2 pirates: 300/dead&#xA;3 pirates: 100/ 100/ 100 (pirates 2&amp;amp;3 vote yes)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Notice that for 3 pirates, the result is a 100 split no matter which pirate is Zanzibar, as long as the Zanzibar pirate is still out there.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;TL;DR&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There are actually 2 solutions that follow.  I first assumed that the secret ballot would cause the number of votes to be revealed.  The first solution assumes that and the answer I came up with was: &lt;code&gt;80/80/0/0/140&lt;/code&gt; with pirate 5 getting 140 coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the OP commented that the secret ballot should be completely secret with the vote count not revealed.  Skip to the &quot;parrot&quot; part to read that solution.  The final answer I came up with was: &lt;code&gt;93/93/0/0/114&lt;/code&gt; with pirate 5 getting 114 coins.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Revealed vote count: the tricky part&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Suppose there are four pirates, and the 4th is not the Zanzibar.  The 4th pirate could propose any of these three:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;4 pirates: 101/ 101/   0/  98&#xA;           101/   0/ 101/  98&#xA;             0/ 101/ 101/  98&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Each way results in a 2/3 chance of dying, because you must offer 0 to the Zanzibar pirate to survive.  Anything else results in only 1 yes vote other than your own.  That isn't very good odds of survival.  How about:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;4 pirates:  75/  75/  75/  75&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now at first you may think that this will for sure result in a yes from the Zanzibar and two no's from the other pirates.  However, if they think you &lt;em&gt;are&lt;/em&gt; the Zanzibar pirate (this is a bluff), then recall that the outcome after the Zanzibar has been revealed is &lt;code&gt;0/0/300&lt;/code&gt;.  Which means that pirates 1 and 2 &lt;em&gt;might&lt;/em&gt; benefit from an even split and vote yes.  But all pirates are smart and they know pirate 4 might bluff.  So let's figure out how pirates 1-3 would vote knowing that pirate 4 will bluff 100% of the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they all vote yes, the expected value is &lt;code&gt;75/75/75/75&lt;/code&gt;.&#xA;If they all vote no, there are two cases:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Pirate 4 &lt;em&gt;was&lt;/em&gt; the Zanzibar pirate (1/3 chance from the perspective of a non Zanzibar pirate 1-3).  Votes turn out three no's indicating that only Pirate 4 could have been Zanzibar.  Outcome is &lt;code&gt;0/0/300&lt;/code&gt; because all three pirates realize that there are no Zanzibars left.&lt;/li&gt;&#xA;&lt;li&gt;Pirate 4 was &lt;em&gt;not&lt;/em&gt; the Zanzibar pirate. (2/3 chance).  Votes turn out two no's one yes indicating that the Zanzibar pirate is still out there.  This results in &lt;code&gt;100/100/100&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Total expected outcome: &lt;code&gt;66.7/66.7/166.7&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in fact pirate 1 and 2 will vote yes to a 75 split and pirate 4 should always bluff in order to keep themselves alive.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;But wait, there's more!&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The above assumed all pirates voted yes or no together.  It looks like pirate 3 should always vote no.  But what if pirates 1-2 only vote yes with probability &lt;code&gt;p&lt;/code&gt;?  This can help because they might get a yes/no split resulting in pirate 3 not knowing if the Zanzibar pirate is still out there (and not being able to propose &lt;code&gt;0/0/300&lt;/code&gt;).  Let's go over the cases again:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1. Pirate 4 is the Zanzibar pirate. (1/3 chance)&#xA;    a) 3 no votes (1-p)^2 chance, 0/0/300&#xA;    b) 2 no 1 yes votes 2*p*(1-p) chance, 100/100/100 because now pirate 3 can't tell if the Zanzibar is still out there&#xA;    c) 1 no 2 yes, p^2 chance, 75/75/75/75&#xA;2. Pirate 4 is **not** the Zanzibar pirate (2/3 chance)&#xA;    a) Pirate 3 is Zanzibar (1/2 chance)&#xA;        a1) 2 no 1 yes votes (1-p)^2 chance, 100/100/100&#xA;        a2) 2 or 3 yes votes 1-(1-p)^2 chance, 75/75/75/75&#xA;    b) Pirate 3 is not Zanzibar&#xA;        b1) 2 no 1 yes votes (1-p) chance, 100/100/100&#xA;        b2) 2 yes 1 no votes p chance, 75/75/75/75&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;To make a long story short, the above simplifies to the expected value of pirates 1 and 2 being:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;P1 = P2 = -(100/3)p^2 + (125/3)p + 200/3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Solving for maximum P1 gives &lt;code&gt;p = 5/8&lt;/code&gt;.  This results in the following expected values for each pirate:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;P1 = 79.6875&#xA;P2 = 79.6875&#xA;P3 = 93.75&#xA;P4 = 46.875 (3/8 chance of dying btw)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h2&gt;And finally we reach pirate 5&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Now we know that P4's best bet is to propose &lt;code&gt;75/75/75/75&lt;/code&gt; with P1 and P2 voting yes with 5/8 probabilty and P3 voting no always (except for whichever is the Zanzibar pirate of course).  This results in the aforementioned split of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;79.7/79.7/93.8/46.9 (with a chance of dying)&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now, P5 needs to propose: &lt;code&gt;80/80/0/0/140&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will get votes from P1, P2, P4, and P5, minus whichever might be the Zanzibar pirate (still 3 votes to gain the majority).  Note that P4 will vote yes to getting 0 gold because otherwise they run a 3/8 chance of dying.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Notes&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I assumed that by &quot;secret ballot&quot;, this meant that the votes would be put into a ballot box without anyone being able to determine who voted which way, but that the ballots would be read aloud so that the number of yeses and no would be public information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original question said that maximizing chances of surviving wasn't a factor, but in my answer it was, so I'm not sure whether I did something wrong.  Maybe someone can double check my math, or the OP can comment on that part of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I went and looked at the Grey Labyrinth forums and no one came close to suggesting the answer I gave.  So I don't think they ever found the real answer over there.  Is there another origin for the puzzle?&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Starting Over... With a Parrot&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The OP claimed in a comment that the secret ballot was truly secret, with a parrot counting the votes.  So given that fact, P4's proposed 75 split will fail because the other pirates will prefer that 3 pirates remain which results in &lt;code&gt;100/100/100&lt;/code&gt;.  So we go back to P4 proposing one of three choices:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;101/101/0/98&#xA;101/0/101/98&#xA;0/101/101/98&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In 2/3 of the cases, P4 will die with the result:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;100/100/100/dead&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In 1/3 of the cases, P4 will stay alive with the average result:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;67.3/67.3/67.3/98&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So the full range of possibilities from a non-Zanzibar P1-P3 point of view:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;P4 was the Zanzibar (1/3 chance): &lt;code&gt;100/100/100/dead&lt;/code&gt;.  Even though only the Zanzibar would commit suicide by proposing an even split, P3 can't actually take the chance of proposing &lt;code&gt;0/0/300&lt;/code&gt; because P4 could have been bluffing.  Because if P3 proposed &lt;code&gt;0/0/300&lt;/code&gt; with any chance, then P4 should start bluffing with some chance.&lt;/li&gt;&#xA;&lt;li&gt;P4 was not the Zanzibar (2/3 chance)&lt;br&gt;&#xA;a) P4 chose badly (2/3 chance): &lt;code&gt;100/100/100&lt;/code&gt; (asterisk - see below)&lt;br&gt;&#xA;b) P4 chose well (1/3 chance): &lt;code&gt;67.3/67.3/67.3/98&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The expected outcome from a non-Zanzibar P1-P3 point of view:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;92.74/92.74/92.74/21.78&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expected outcome from a non-Zanzibar P4 point of view is different:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;P4 chose badly (2/3) chance: &lt;code&gt;100/100/100/dead&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;P4 chose well (1/3) chance: &lt;code&gt;67.3/67.3/67.3/98&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;89.1/89.1/89.1/32.7 with a 67% chance of death&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the full expected outcome for each non-Zanzibar pirate from his own point of view (which doesn't necessarily add to 300 because each pirate has information about himself):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;92.74/92.74/92.74/32.7 with a 67% chance of death&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the above, pirate 5 should propose: &lt;code&gt;93/93/0/0/114&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, pirates 1,2,4,5 vote for the proposal with one of them possibly not doing so if they are the Zanzibar pirate.  Pirate 4 votes yes even with 0 gold to avoid death.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equivalent proposals are &lt;code&gt;93/0/93/0/114&lt;/code&gt; and &lt;code&gt;0/93/93/0/114&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Asterisk - Information Leakage&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The asterisk from above is because once P4 chooses badly and dies, the other pirates gain some information: the pirate P4 chose to give 0 to must not be the Zanzibar pirate.  Therefore, the results might not always be &lt;code&gt;100/100/100&lt;/code&gt; after that.  Here are the possible results:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If P3 is not the Zanzibar:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;code&gt;101/101/0&lt;/code&gt;: Either P1 or P2 is the Zanzibar.  P3 must propose &lt;code&gt;100/100/100&lt;/code&gt; and it will pass. (2/6 chance)  &lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;101/0/101&lt;/code&gt;: P3 knows P1 is the Zanzibar but P2 doesn't yet.  P3 can offer &lt;code&gt;0/151/149&lt;/code&gt; because P2 will realize P1 is the Zanzibar and know that 151 is better than the 150 he could get if he voted no. (1/6 chance)  &lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;0/101/101&lt;/code&gt;: P3 knows P2 is the Zanzibar.  P3 must offer &lt;code&gt;100/100/100&lt;/code&gt; otherwise both P1 and P2 will refuse. (1/6 chance)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If P3 is the Zanzibar:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;code&gt;101/0/101&lt;/code&gt;: P3 offers &lt;code&gt;100/100/100&lt;/code&gt;.  P1 knows that P3 is the Zanzibar and votes no.  P2 votes yes though to pass the proposal. (1/6 chance)  &lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;0/101/101&lt;/code&gt;: P3 offers &lt;code&gt;100/100/100&lt;/code&gt;.  P2 knows that P3 is the Zanzibar and votes yes otherwise P2 will get killed. (1/6 chance)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Notice the one case where the outcome is &lt;code&gt;0/151/149&lt;/code&gt;.  This results in P1 getting less than P2 and P3 overall (about a &lt;code&gt;83/108/108&lt;/code&gt; split).  But now P1 should choose a mixed strategy.  With a low probability &lt;code&gt;p&lt;/code&gt;, he can vote the wrong way.  Given this mixed strategy, P3 can no longer afford to offer &lt;code&gt;0/151/149&lt;/code&gt; in the one case above because with probability &lt;code&gt;p&lt;/code&gt;, the assumption could be wrong leading to P3's death.  P1 can choose &lt;code&gt;p&lt;/code&gt; so small that it doesn't affect anyone's expected outcome.  However, just the possibility of voting the wrong way deters P3 from deviating from the &lt;code&gt;100/100/100&lt;/code&gt; strategy.  So now all cases lead to a &lt;code&gt;100/100/100&lt;/code&gt; split.&lt;/p&gt;&#xA;" OwnerUserId="4479" LastEditorUserId="4479" LastEditDate="2015-05-05T19:26:09.813" LastActivityDate="2015-05-05T19:26:09.813" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13212" PostTypeId="1" AcceptedAnswerId="13214" CreationDate="2015-05-02T08:28:48.293" Score="2" ViewCount="283" Body="&lt;p&gt;A teacher on a foreign planet wants to make a classroom calendar out of cubic blocks. She wants to individually count each day by drawing a number on the faces of a block. There are 209 days in this planet's year, and she can draw any number (0-9) on each face.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, &lt;code&gt;19&lt;/code&gt; would be:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; ---   ---&#xA;| 1 | | 9 |&#xA; ---   ---&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;With a face on one block being 1, and the other block's face being 9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example, &lt;code&gt;201&lt;/code&gt;, would be:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; ---   ---   ---&#xA;| 2 | | 0 | | 1 |&#xA; ---   ---   ---&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;A face on one block would be 2, a second block's face being 0, and a third block with a face of 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the fewest amount of blocks she can use to count the days of the year (0-209), and what numbers are on each block?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;All cubes must be used.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="9871" LastEditorUserId="5373" LastEditDate="2017-01-13T00:25:21.160" LastActivityDate="2017-01-13T00:25:21.160" Title="Alien Classroom Calendar" Tags="|mathematics|story|combinatorics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13213" PostTypeId="2" ParentId="13206" CreationDate="2015-05-02T08:44:07.190" Score="2" Body="&lt;p&gt;$ \newcommand{\vec}[1]{\begin{pmatrix} #1 \end{pmatrix}} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Call the pirates P,Q,R,S,T in order, with P oldest and healthy. Treat the term &quot;expected earnings&quot; as the &quot;expected number&quot; in probability: given a set of non-overlapping states $s_i$ with corresponding probabilities $p_i$ such that $\sum p_i = 1$, and given a payout $n_i$ in a given state $s_i$, the &lt;em&gt;expected number&lt;/em&gt; is given by $\sum p_i n_i$. Use the notation $E_n(\mathbf{x})$ to mean the expected earnings of the pirates named in $\mathbf{x}$ when there are only $n$ pirates remaining.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The states correspond to the person with the fever - denote them $F_Q, F_R, F_S, F_T$. To get the expected earnings $E_{k+1}$, calculate the expected earnings in each state based on $E_{k}$ and weight them by the probability of that state. The question, of course, is what the probabilities are. This could be tricky, since pirates can pretend to be fevered, and prior proposals and answers can indicate or preclude fever. Here, however, &lt;strong&gt;we are considering the probabilities from the point of view of P before any pirate has made a proposal&lt;/strong&gt;. At this point, all 4 states are equally likely, so the probabilities of the four states must be equal, i.e. each probability is 1/4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with the youngest pirate and build up from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T: T gets 300 coins. $E_1(T) = 300$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ST: &quot;more than half&quot; means &quot;unanimous&quot; for 2 voters. In the state $F_T$, S must propose 150:150 to retain 150 coins. In any other state, T will always say no.&#xA;$$E_2 \vec{S\\T} = \frac{1}{4} \vec{150\\150} + \frac{3}{4} \vec{0\\300} = \vec{37.5\\262.5}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RST: R needs 1 more vote. In $F_R$, R proposes an even split, which S accepts. In $F_S$, R's cheapest option is also an even split. In $F_Q$ and $F_T$, R's cheapest option is to propose 262:38:0.&#xA;$$E_3 \vec{R\\S\\T}&#xA;= \frac{2}{4} \vec{100\\100\\100}&#xA;+ \frac{2}{4} \vec{262\\38\\0}&#xA;= \vec{181\\69\\50}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;QRST: Q needs 2 more votes. In $F_Q, F_S, F_T$, Q proposes an even split which is accepted. In $F_R$, Q's cheapest option is to offer 179:0:70:51.&#xA;$$E_4 \vec{Q\\R\\S\\T}&#xA;= \frac{3}{4} \vec{75\\75\\75\\75}&#xA;+ \frac{1}{4} \vec{179\\0\\70\\51}&#xA;= \vec{101\\56.25\\73.75\\69}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PQRST: P is fever-free. She only needs 2 more votes, but will pay for 3 in case one of them is fevered. The cheapest are RST. Note that P can't use an even split because only R is expecting less than 60, so if R has the fever, the others will reject an even split.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So P proposes a split of $99:0:57:74:70$.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note: work in progress: $E_3$ and $E_4$ need refining.&lt;/p&gt;&#xA;" OwnerUserId="9422" LastEditorUserId="9422" LastEditDate="2015-05-04T22:40:08.590" LastActivityDate="2015-05-04T22:40:08.590" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13214" PostTypeId="2" ParentId="13212" CreationDate="2015-05-02T08:44:47.563" Score="3" Body="&lt;p&gt;My answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;4 cubes &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For each number, the same digit appears at most twice, except for 111, where 1 appears 3 times. Also, you can rotate the 9 to make 6, so after all we need 19 digits (2*9+3*1-2). For 19 digits the minimum is 4 cubes.&lt;br&gt;&#xA;&lt;strong&gt;Cube #1:&lt;/strong&gt; 012468.&lt;br&gt;&#xA;&lt;strong&gt;Cube #2:&lt;/strong&gt; 012568.&lt;br&gt;&#xA;&lt;strong&gt;Cube #3:&lt;/strong&gt; 01347.&lt;br&gt;&#xA;&lt;strong&gt;Cube #4:&lt;/strong&gt; 01357.  &lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-05-02T09:56:57.707" LastActivityDate="2015-05-02T09:56:57.707" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="13216" PostTypeId="2" ParentId="13212" CreationDate="2015-05-02T09:49:10.643" Score="2" Body="&lt;p&gt;Since we're making upside down 6's be 9's, how about making upside down 2's be 7's? Depends on how good an artist the teacher is, but here's an example of it on a clock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LkcG5.jpg&quot; alt=&quot;clock&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, my answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Three Cubes  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The faces are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;123456&lt;br&gt;&#xA;123480&lt;br&gt;&#xA;125680&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit: Didn't see that all cubes must be used. That makes it impossible to do with 3 cubes, unfortunately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Double Edit: Actually, does the following configuration work? I can't find a number that I can't make...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;012345&lt;br&gt;&#xA;016823&lt;br&gt;&#xA;014568  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel like I'm missing something...&lt;/p&gt;&#xA;" OwnerUserId="10861" LastEditorUserId="10861" LastEditDate="2015-05-02T10:17:31.587" LastActivityDate="2015-05-02T10:17:31.587" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13217" PostTypeId="2" ParentId="13028" CreationDate="2015-05-02T10:22:16.787" Score="1" Body="&lt;p&gt;JLee's answer has me fairly convinced (except for the first two hints), but if that's not it, this might be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a BEAR.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I take the expressway, I'm in a hurry.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bears can move surprisingly fast (up to 35mph or nearly 60 km/h).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;By chance I make it, no need to worry,&lt;br&gt;&#xA; I'll loiter and idle and maybe sleep.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bears have an image of being dozy and slow-moving, and often asleep. In the winter, many of them hibernate for months on end.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I'll finally wake, my day's done it seems,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They wake at the end of the winter, when the season is done.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I hurry back but now lost, I weep.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Their home territory might look quite different by the time they wake up.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And yet my home at last I see.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; But their den, in which they've been hibernating, has remained the same all the way through.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; many animals with horns are hunted by bears, so they'll certainly know about these dangerous creatures!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bears are often kept in cages, and can fight fiercely. When they're hungry, you might crave to run away.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Third hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is the only part I'm not quite sure of.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-02T10:22:16.787" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13218" PostTypeId="1" AcceptedAnswerId="13235" CreationDate="2015-05-02T10:39:55.880" Score="0" ViewCount="2404" Body="&lt;p&gt;Given a pattern: 12, 22, 35, 68, 15, 28, 31, $n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the next item at $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Pattern/sequence explanation required&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="9871" LastEditorUserId="7308" LastEditDate="2015-05-03T01:54:41.683" LastActivityDate="2015-05-03T01:54:41.683" Title="Finish The Pattern #1" Tags="|calculation-puzzle|pattern|number-sequence|" AnswerCount="1" CommentCount="2" ClosedDate="2015-05-04T06:07:30.880" ContentLicense="CC BY-SA 3.0" />
  <row Id="13219" PostTypeId="1" CreationDate="2015-05-02T10:39:41.110" Score="2" ViewCount="162" Body="&lt;p&gt;There are 10 gold smiths in the town and a rich businessman wants to collect 1kg of pure gold. So he informs each person to bring 10 balls of pure gold, each weighing exactly 10g. So altogether he plans to collect 100 gold balls 10g each. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But one gold smith brings 10 balls each weighing exactly 9g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The businessman gets to know about the ongoing scandal and plans to catch this fraud. He calls the 10 smiths into his office room where he has an electronic balance. The businessman catches the fraud with weighing exactly one time. How did he do that? Note: He weighs one time only. Balls cannot be distinguished by looking. &lt;/p&gt;&#xA;" OwnerDisplayName="slhulk" LastEditorUserId="1509" LastEditDate="2015-05-02T11:20:10.343" LastActivityDate="2015-05-02T11:29:40.250" Title="Catch the fraud!" Tags="|weighing|" AnswerCount="1" CommentCount="13" ClosedDate="2015-05-02T16:36:29.643" ContentLicense="CC BY-SA 3.0" />
  <row Id="13220" PostTypeId="2" ParentId="13219" CreationDate="2015-05-02T11:23:10.933" Score="3" Body="&lt;p&gt;He takes 1 ball from the first pile, 2 from the 2nd, 3 from the 3th, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He then weighs them all together, normally he would have a weight of 550 (1 + 2 + 3 + 4 etc times 10g) the difference in weight is the pile it is in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so if the balls in the first pile were wrong it would weigh 449, if those in the 2nd pile were wrong it would weigh 448 etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a very well known puzzle and there are a few more similar ones posted on puzzling.stackexchange. If you like these (or other) riddles, check out the site seeing you're new here ;)&lt;/p&gt;&#xA;" OwnerUserId="9835" LastEditorUserId="5373" LastEditDate="2015-05-02T11:29:40.250" LastActivityDate="2015-05-02T11:29:40.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13221" PostTypeId="2" ParentId="4785" CreationDate="2015-05-02T12:38:58.060" Score="3" Body="&lt;p&gt;According to &lt;a href=&quot;http://forums.xkcd.com/viewtopic.php?t=110384&amp;amp;p=3686054&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;, you have to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; XOR all bytes by 0x6d&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;to get&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What is that noise AGAIN?g&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9309" LastActivityDate="2015-05-02T12:38:58.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13222" PostTypeId="5" CreationDate="2015-05-02T12:44:31.727" Score="0" Body="&lt;p&gt;&lt;strong&gt;Peg solitaire&lt;/strong&gt; (or &lt;strong&gt;Solo Noble&lt;/strong&gt;) is a board game for one player involving movement of pegs on a board with holes. Some sets use marbles in a board with indentations. The game is known simply as &lt;strong&gt;Solitaire&lt;/strong&gt; in the United Kingdom where the card games are called Patience. It is also referred to as &lt;strong&gt;Brainvita&lt;/strong&gt; (especially in India).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard game fills the entire board with pegs except for the central hole. The objective is, making valid moves, to empty the entire board except for a solitary peg in the central hole. A valid move is to jump a peg orthogonally over an adjacent peg into a hole two positions away and then to remove the jumped peg.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-05-02T12:44:31.727" LastActivityDate="2015-05-02T12:44:31.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13223" PostTypeId="4" CreationDate="2015-05-02T12:44:31.727" Score="0" Body="This tag is for questions about peg solitaire (not the card game solitaire)." OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-05-02T12:44:31.727" LastActivityDate="2015-05-02T12:44:31.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13224" PostTypeId="5" CreationDate="2015-05-02T12:47:45.003" Score="0" Body="&lt;p&gt;Mastermind is a code-breaking 2-player game created in 1970. It is based on a game called Moo used on the Cambrige Titan computer to practice code-breaking; this game itself was actually based on an older game called &quot;Bulls and Cows&quot;. Dr John Billingsley had the idea to create the game, and it was then designed by Mordecai Meirowitz, an Israeli postmaster and telecommunications expert.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One player selects a sequence of coloured pegs and the other player tries to work out what this sequence is, by making guesses and receiving information on how accurate the guesses are, and using deductions to reduce the number of possibilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic variant is with 6 colours and 4 holes, and repetition is &lt;strong&gt;not&lt;/strong&gt; allowed. The answerer uses white pegs for correct colour in the correct position, and black pegs for correct colour in an incorrect position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/17430/bulls-without-cows-how-many-steps-are-needed-to-solve-this&quot;&gt;Bulls without cows: how many steps are needed to solve this?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The Wikipedia article on Mastermind can be found &lt;a href=&quot;https://en.wikipedia.org/wiki/Mastermind_(board_game)&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="11110" LastEditDate="2017-06-21T21:12:31.223" LastActivityDate="2017-06-21T21:12:31.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13225" PostTypeId="4" CreationDate="2015-05-02T12:47:45.003" Score="0" Body="This tag is for questions about the board game Mastermind." OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-05-02T12:47:45.003" LastActivityDate="2015-05-02T12:47:45.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13226" PostTypeId="1" AcceptedAnswerId="13229" CreationDate="2015-05-02T12:58:55.463" Score="3" ViewCount="420" Body="&lt;p&gt;A mysterious thief spotted a girl with many gadgets and jewelries. After a few minutes, the thief attacked the woman and get all her gadgets and jewelries. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there's a one gadget that the thief didn't get, an old model cellphone. The thief has this attitude after he successfully do a crime, he always leave a clue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the clue is the message he left on the cellphone.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Message&lt;/em&gt;: &lt;strong&gt;7 6 2 4 6 4 6 6 3&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, You are going to decode the message left by the thief. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Hint:&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;one of the clues to decode the message can be found on the story.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;He is the prince of thieves.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The content of the message is the name of the thief&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;NOTE: ( &lt;em&gt;You need to show how did you decode it, include also the clue that you found in the story.&lt;/em&gt; )&lt;/p&gt;&#xA;" OwnerUserId="12054" LastEditorUserId="12054" LastEditDate="2015-11-02T11:36:06.993" LastActivityDate="2015-11-02T11:36:06.993" Title="Prince of Thieves" Tags="|cipher|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13228" PostTypeId="2" ParentId="12923" CreationDate="2015-05-02T13:11:22.680" Score="2" Body="&lt;p&gt;I liked Mike's answer, but I thought of a way to make it more visual, and also to show that each test really does involve a single sip for each mouse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we put the bottles in a grid with columns A,B,C and rows 1,2,3, we can pour six glasses - glass 1 contains a mixture of all the row 1 wines, and so on...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kRDR7.png&quot; alt=&quot;Illustrated answer&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, give the first mouse a drink from glass 1, and the second mouse a drink from glass A. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mice that survive advance to the next row/column for the next round, where first mouse get a drink from glass 2, and second mouse from glass B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surviving mice advance again, and their final positions indicate the row and column where the poisoned wine is!&lt;/p&gt;&#xA;" OwnerUserId="12089" LastActivityDate="2015-05-02T13:11:22.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13229" PostTypeId="2" ParentId="13226" CreationDate="2015-05-02T13:14:21.037" Score="6" Body="&lt;p&gt;So the thief name is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.imdb.com/title/tt0102798/&quot;&gt;robin hood the prince of thieves&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The clue in the story is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the old phone model with a physical keyboard&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The anagram part is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; each number of the clue matching one key on a phone with multiple letters&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 7 match with pqrs but we choose r&lt;br&gt;&#xA; 6 match with mno but we choose o&lt;br&gt;&#xA; and so on to find ROBIN HOOD&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10400" LastActivityDate="2015-05-02T13:14:21.037" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13230" PostTypeId="1" CreationDate="2015-05-02T13:35:00.580" Score="-4" ViewCount="274" Body="&lt;p&gt;There are 16 seminars with the following durations in minutes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;60,45,30,45,45,30,30,45,60,60,45,30,30,60,30,30&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They need to be divided into groups. Each group must have 2 subgroups. The first group must be exactly 3 hours, and the second between 4 and 5 hours inclusive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there are multiple solutions, but what method can be used to find a solution with any set of seminars?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Adapted from &lt;a href=&quot;http://top-interview-puzzles.blogspot.in/2015/05/thoughworks-screening-puzzle.html&quot; rel=&quot;nofollow&quot;&gt;http://top-interview-puzzles.blogspot.in/2015/05/thoughworks-screening-puzzle.html.&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="12090" LastEditorUserId="5373" LastEditDate="2015-05-02T17:28:24.253" LastActivityDate="2015-05-02T18:39:14.993" Title="Timetable Planning" Tags="|logical-deduction|explanation|" AnswerCount="1" CommentCount="8" ClosedDate="2015-05-02T23:17:50.533" ContentLicense="CC BY-SA 3.0" />
  <row Id="13231" PostTypeId="2" ParentId="13121" CreationDate="2015-05-02T15:10:14.903" Score="2" Body="&lt;p&gt;Another guess, hot off the press!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your friend went to Middle-Earth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some are high:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Elves, as in High Elves.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some can fly:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eagles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why those two go together:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Both races prefer to remain aloof from the struggle, but will help in a moment of crisis.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some are in cities:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Men, the main city-dwellers of the Third Age.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some are in ditties:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dwarves -- we don't see much of them outside the ditty about the ring.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why those two go together:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Both are warlike yet generally 'good guy' races.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some wear trousers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hobbits&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some need browsers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rangers (they're always relying on their network of information, so an internet connection would really help out!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why those two go together:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Both are not something the average Middle-Earth resident has likely heard of, because both like to keep private. Rangers aren't specifically a race, but I do think they're a special breed of men? (Some Elf-blood or something)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some are ground in the sea:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Balrog I think. This was the hardest clue for me to fit because LotR doesn't really have many water-y scenes. Another guess: The ghosts living in the swamp outside of Mordor. But the Balrog is my main guess (it fell into that big lake under Moria in the movie, I thiiiink that's canonical in the book too).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some oversee the grounds:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wizards -- their main task in life is just keeping a watch on things and keeping things running smoothly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why those two go together:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gandalf's duel with the Balrog, also, both are super-ancient super-powerful magical beings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11154" LastActivityDate="2015-05-02T15:10:14.903" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13232" PostTypeId="2" ParentId="13121" CreationDate="2015-05-02T15:23:01.353" Score="-1" Body="&lt;p&gt;I think it might be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the land of creepy-crawlies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Some are high, some can fly&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This could describe any flying insects.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Some are in cities, some are in ditties&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ants and termites live in huge colonies that can be called cities.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Some wear trousers, some need browsers&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Browsers as in &lt;em&gt;web&lt;/em&gt; browsers, as in &lt;em&gt;spider&lt;/em&gt; webs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Some are ground in the sea, some oversee the grounds&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure about this line (or about trousers and ditties above), but I think I might be onto something here...&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-02T15:23:01.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13233" PostTypeId="2" ParentId="13121" CreationDate="2015-05-02T15:44:28.730" Score="7" Body="&lt;p&gt;The city your friend visited is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; New York!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some are high, some can fly&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; New York Giants are tall and  the New York Jets can fly&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some are in cities, some are in ditties&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The New York Mets (metropolitans) and the New York Yankees (yankee doodle)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some wear trousers, some need browsers&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Knicks (Knicerbookers) and Nets (internet browsers)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some are ground in the sea, some oversee the grounds&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; New York Islanders (island is a ground in the sea), New York Rangers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11844" LastEditorUserId="11844" LastEditDate="2015-05-02T15:48:58.997" LastActivityDate="2015-05-02T15:48:58.997" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13234" PostTypeId="2" ParentId="12732" CreationDate="2015-05-02T15:53:24.827" Score="16" Body="&lt;p&gt;I think I have a sufficiently good explanation (&lt;em&gt;Warning: Long Answer&lt;/em&gt;):&lt;br&gt;&lt;hr&gt;&lt;hr&gt; &#xA;&lt;h2&gt;Theft:&lt;/h2&gt;(Galatyr already answered this, so no explanation required.) &lt;strong&gt;&lt;em&gt;!Peleh&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&lt;hr&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;h2&gt;Dr. Brofonomo Death:&lt;/h2&gt; &lt;strong&gt;&lt;em&gt;Double You&lt;/em&gt;&lt;/strong&gt; &lt;em&gt;or&lt;/em&gt; &lt;strong&gt;&lt;em&gt;Tee Eff&lt;/em&gt;&lt;/strong&gt; Because a there was a &lt;em&gt;single well fired shot&lt;/em&gt;, and they were the only ones to have experience in using the weapon. Also, they seem to be violent, and may have wanted revenge (for the gambling episode).&lt;br&gt;&lt;hr&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;h2&gt;Luke Death:&lt;/h2&gt; &lt;strong&gt;&lt;em&gt;Leia&lt;/em&gt;&lt;/strong&gt; She was already annoyed by her husband's attitude and probably wanted to marry Kevin, so she got rid of Luke.&lt;br&gt;&#xA;&lt;strong&gt;&lt;em&gt;Method:&lt;/em&gt;&lt;/strong&gt; She could have had the key to the room, no indication against that. She had drugged him earlier with sleeping pills.She entered the room , suffocated Luke (with a pillow maybe), and locked it before going away.&lt;br&gt;&lt;hr&gt;&#xA;&lt;h2&gt;Other Deaths:&lt;/h2&gt; &lt;h3&gt;Carol&lt;/h3&gt;&#xA;She can be convicted by the method of elimination of suspects:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Chang&lt;/strong&gt;-  Cannot be as he got shocked after seeing the dead bodies, and a person who murdered someone just hours earlier cannot get shocked by seeing the later's body.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Gang&lt;/strong&gt;- Chang is testimony for their being in their room.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;!Peleh&lt;/strong&gt;- Chang is testimony again. Also, he climbed into George's room to steal, but the death timeline shows that George was killed in the end, and not in the start of the killing spree.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Carol&lt;/strong&gt;- No Alibis at all....Also, she was not much upset by her family's death, she almost seemed pleased ...(meaning she could have killed them herself by faking an accident, or that she had come to take revenge for their deaths.). &lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to the &lt;strong&gt;motives&lt;/strong&gt;:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Double You and Tee Eff&lt;/em&gt;&lt;/strong&gt; &lt;strike&gt;Probably just to get the weapons and test them, or because she did not like them as they gambled(or maybe even because she knew they had killed Dr. Brofonomo).&lt;br&gt;&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carol killed the cops to get their weapons - simple as that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;George and Max&lt;/em&gt;&lt;/strong&gt; &lt;strike&gt;I'm not quite sure as to why she should kill George or Max, but it could be because she did not like them, or because they were involved in her family's killing(that would explain why she was happy even though her family had suffered an accident, as she had come to take revenge for their deaths).&lt;br&gt;&#xA;Another probable explanation:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;She may have wanted to marry Max in order to inherit the rich couple's money. When she was refused(by either Max or George), she became angry and decided to kill them both(she either spared Giselle because she &#xA;liked Giselle or because she did not get the chance to kill her).&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And another:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;She killed them so that Giselle might give her the entire money(in her will) as they had a special bond and Giselle had no other family Carol knew of(we can assume that Giselle had told Carol that she had no family other than George and Max).&lt;/strike&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carol killed George and Max because she's a lesbian and she's in love with Giselle. We know she's a lesbian because of the extreme disgust towards men. She wanted the massive inheritance to land in Giselle's hands so that the two of them can live happily ever after.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How did she kill them?&lt;/strong&gt; She drugged WTF (the drug was stored along with the ashes, so no one could have a doubt(that only explains how she hid it, but I assume she somehow managed to get(maybe from Giselle))), then stole their weapons and tested it (weapon) on WTF. Then she proceeded to Max; Door was already open, and Max was already out, so the killing was easy.&lt;br&gt;&#xA;Now, &lt;em&gt;how did she enter George's room?&lt;/em&gt;&lt;br&gt;&#xA;She had the key to room 702. 702 and George's room are &lt;em&gt;connected&lt;/em&gt;, so she easily entered the room. Then she killed George and left.(&lt;strong&gt;Please Note&lt;/strong&gt; If the connecting door was locked ,she certainly asked Giselle to open either the connecting door or the main door, and made sure it stayed open.)&lt;hr&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why the belt? The belt exists as both a murder signature, as well as a show of dominance. Being the small framed girl she is, by tying up her victims not only does she ensure they don't accidentally wake up and bash her head in; it also shows her dominance over them, even though she hates them. It's as if she were making the statement, &quot;You're just a lowly slave.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;h2&gt;Please Note&lt;/h2&gt; I haven't mentioned a few things here-&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Giselle was certainly not involved directly in any killing, but she may have been the supplier of the tranquilizer for Carol.Also, Giselle did not like Max much , she liked Carol a lot more, so she may have agreed to supplying Carol with tranquilizer so that the latter may kill Max.She may have even wanted this to happen as that would mean George will give all his attention to her.But that creates a new problem-&lt;i&gt;Why didn't she suspect that Carol killed George?&lt;/i&gt;&#xA;&lt;li&gt;Kevin could also be convicted of protecting a Convicted Murderer .&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;&lt;br&gt; &#xA;&lt;hr&gt;&lt;hr&gt;&#xA;&lt;em&gt;(B.T.W. I forgot one thing, the murderer could be the detective also..... :)-Just Kidding)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="12091" LastEditorUserId="12091" LastEditDate="2015-05-06T07:02:45.447" LastActivityDate="2015-05-06T07:02:45.447" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13235" PostTypeId="2" ParentId="13218" CreationDate="2015-05-02T16:16:13.683" Score="5" Body="&lt;p&gt;Partial answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 60&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; starting from the beginning, each even-indexed answer is (answer at index minus 1) * 2 - 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12092" LastActivityDate="2015-05-02T16:16:13.683" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13236" PostTypeId="2" ParentId="10210" CreationDate="2015-05-02T16:17:01.637" Score="0" Body="&lt;p&gt;My answer is 2110 because&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5\times3=15$, $5+3=8$, so it becomes $158$&lt;br&gt;&#xA;$9\times1=9$, $9+1=10$, so it becomes $910$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also the same for $7\times3$&lt;/p&gt;&#xA;" OwnerUserId="12093" LastEditorUserId="13494" LastEditDate="2015-06-23T19:10:45.727" LastActivityDate="2015-06-23T19:10:45.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13237" PostTypeId="1" AcceptedAnswerId="13238" CreationDate="2015-05-02T16:20:48.390" Score="-1" ViewCount="205" Body="&lt;p&gt;I forgot to bring my &lt;strong&gt;110 69 119 67 101 108 76 112 72 111 78 101&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Hint:&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 104 73 110 84.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12054" LastEditorUserId="5373" LastEditDate="2016-09-29T00:53:27.873" LastActivityDate="2016-09-29T00:53:27.873" Title="Complete the sentence" Tags="|cipher|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13238" PostTypeId="2" ParentId="13237" CreationDate="2015-05-02T16:23:14.263" Score="4" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; new cellphone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ASCII. See ASCII table &lt;a href=&quot;http://www.asciitable.com/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;: &quot;104 73 110 84&quot; = &quot;hInT&quot; and &quot;110 69 119 67 101 108 76 112 72 111 78 101&quot; = &quot;nEwCelLpHoNe&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And yes, I am quick :-) I liked the style of the hint, but the puzzle was way too easy IMO.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-02T16:23:14.263" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13239" PostTypeId="2" ParentId="13230" CreationDate="2015-05-02T18:39:14.993" Score="1" Body="&lt;p&gt;There are different approaches to this kind of problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &lt;strong&gt;Brute Forcing&lt;/strong&gt;&lt;br&gt;&#xA;You simply try all the combinations and see what fits.&lt;br&gt;&#xA;This is easy to implement, but it requires many many operations, ergo such an algorithm is impossible to do perform by hand and very slow even for a computer.&lt;br&gt;&#xA;In particular, there are $N!$ permutations to verify, anyway you can slightly optimize it not repeating identical sequences (I mean, there are 4 &quot;60&quot; in your set, if you swap them nothing changes!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) &lt;strong&gt;Greedy&lt;/strong&gt;&lt;br&gt;&#xA;The main idea is &quot;long seminars are cumbersome, try to position them until you have room!&quot;&lt;br&gt;&#xA;So, in the blocks of 3 hours we start placing the &quot;60&quot;, then the &quot;45&quot;, finally the &quot;30&quot;. In this particular case it works, but doesn't if you want to generalize the problem.&lt;br&gt;&#xA;Computationally, this algorithm is extremely efficient ($O(N)=Nlog(N)$ because you have to sort the data), but it's not granted to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) &lt;strong&gt;Recursion&lt;/strong&gt;&lt;br&gt;&#xA;Basically a recursive algorithm explores all the possible sequences starting from a seminar, adding another, then another, and so on...&lt;br&gt;&#xA;Differently from brute forcing, when it finds a wrong sequence, it easily excludes similar wrong sequences from being tested, strongly reducing the number of required operations.&lt;br&gt;&#xA;This algorithm is fast, average difficulty to implement, but incredibly memory (RAM) leeching.  This is granted to work if there exists at least a valid sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may be other interesting algorithm to discuss, but this is not the right place I guess.&lt;/p&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-05-02T18:39:14.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13240" PostTypeId="1" AcceptedAnswerId="13243" CreationDate="2015-05-02T18:54:43.003" Score="52" ViewCount="7372" Body="&lt;p&gt;James Bond, while working on a top-secret case for many weeks in deep enemy territory, finally collected and stored lots of critical files on his &lt;strong&gt;tight-security Linux system&lt;/strong&gt;. Even though the system was on the internet, no unauthorized user could ever break-in and access the files marked For-Your-Eyes-Only. He then secretly contacted his group (M, Q, Moneypenny, Etc) to process the files. After 30 minutes, they all responded that the system said &quot;No Access&quot;.&#xA;So Bond tried to access the files, and true enough, the system said &quot;No Access&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meanwhile some other folks (probably double-agents or hackers from SPECTRE) had read all the files. Bond realised that he had made a big mistake, out of habit.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What mistake did Bond commit, that all his work was leaked so easily ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;FYI 1&lt;/strong&gt; : Here the tags for this question are important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;FYI 2&lt;/strong&gt; : He quickly recovered from this situation to prevent further leakage.&lt;/p&gt;&#xA;" OwnerUserId="10940" LastEditorUserId="10940" LastEditDate="2015-05-08T15:32:27.673" LastActivityDate="2015-05-09T16:57:39.970" Title="James Bond : No Access" Tags="|lateral-thinking|mystery|computer-puzzle|knowledge|humor|" AnswerCount="8" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="13241" PostTypeId="2" ParentId="13240" CreationDate="2015-05-02T19:05:28.127" Score="0" Body="&lt;p&gt;There are two valid options, in my opinion:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;His password was stolen, so the bandits moved the files to their servers. Bond could get back his data by directly facing his enemies or locating enemies' IP.&lt;/li&gt;&#xA;&lt;li&gt;Hijackers sent a malware into Bond's computer. To recover his data, Bond intercepted the flux of information between his computer and hackers' server, being able to locate them.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-05-02T19:05:28.127" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13242" PostTypeId="2" ParentId="13240" CreationDate="2015-05-02T19:07:13.637" Score="-1" Body="&lt;p&gt;I'll have a go at this.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The password to the system is &quot;No Access&quot;. All Bond's group needed to do when &quot;No Access&quot; came up on the screen was to type in those very words.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bond's group didn't realise this, but some of the double agents did, so they hacked it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility (making this into something of a trick question - maybe this is what the &lt;a href=&quot;/questions/tagged/lateral-thinking&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;lateral-thinking&amp;#39;&quot; rel=&quot;tag&quot;&gt;lateral-thinking&lt;/a&gt; tag suggests?):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bond simply authorised the wrong group of people to access his files: those in the country he was operating in rather than those in his home organisation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In this case, the &quot;out of habit&quot; could mean that this is what he normally does because he's usually operating in his home country. (Apologies if this contradicts Bond canon - I've never read or seen any of it!)&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-02T19:07:13.637" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13243" PostTypeId="2" ParentId="13240" CreationDate="2015-05-02T19:37:05.520" Score="83" Body="&lt;p&gt;Well perhaps&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; he saved the files with the wrong &lt;a href=&quot;http://en.wikipedia.org/wiki/File_system_permissions#Numeric_notation&quot;&gt;permissions&lt;/a&gt; — i.e., &lt;code&gt;007&lt;/code&gt; (no access for owner or group, but full access for everyone else) instead of &lt;code&gt;770&lt;/code&gt; (full access for owner and group, no access for anyone else).&lt;br/&gt;&lt;br/&gt;(Probably an easy mistake to make when you've had the same code number for over 60 years.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="40" LastActivityDate="2015-05-02T19:37:05.520" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="13244" PostTypeId="2" ParentId="13094" CreationDate="2015-05-02T20:01:08.107" Score="2" Body="&lt;p&gt;ALSO NOT AN ANSWER&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bacon supposedly wrote Shakespeare's plays.  And he made up a cipher as well, which this text is encoded in.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However there's another level beyond that, which I don't know.  Decoding the second, smaller text, using only the level I know about, yields&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; bikhwer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is not a word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edited to add: When decoding the shorter text, I left off the part &quot;Ws, jc 3&quot; thinking it wasn't part of the actual answer.  If that is included then my answer becomes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; pbikhwer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But in comments, the puzzle's author seemed surprised that I got this result at all.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Bacon cipher has two variants, &quot;standard&quot; (which I used) and another variant where i/j are not equal.  Using the other variant I get obijhueq instead.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12092" LastEditorUserId="12092" LastEditDate="2015-05-03T02:58:04.583" LastActivityDate="2015-05-03T02:58:04.583" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13245" PostTypeId="1" AcceptedAnswerId="13246" CreationDate="2015-05-02T20:31:35.643" Score="1" ViewCount="1376" Body="&lt;p&gt;Given a pattern:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0, 1, 2, 0.5, 2, 3, 5, 1, 2, 0, 2, 1, 4, 2, 3, $n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the next item at $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Pattern/sequence explanation required&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="9871" LastEditorUserId="7308" LastEditDate="2015-05-03T01:55:01.067" LastActivityDate="2015-05-03T01:55:01.067" Title="Finish The Pattern #2" Tags="|calculation-puzzle|pattern|number-sequence|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13246" PostTypeId="2" ParentId="13245" CreationDate="2015-05-02T20:34:05.960" Score="3" Body="&lt;p&gt;Answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For each $k$, the $4k$th term is the sum of the $(4k-3)$th and $(4k-2)$th terms divided by the $(4k-1)$th term. Thus: (0+1)/2=0.5; (2+3)/5=1; (2+0)/2=1; (4+2)/3=&lt;strong&gt;2&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I got this answer by noticing that 0,1,2 and 2,3,5 are the starts of well-known sequences, so it seems likely that the $4k$th term is always a fixed function of the preceding 3 terms.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-02T20:34:05.960" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13247" PostTypeId="1" AcceptedAnswerId="13254" CreationDate="2015-05-02T20:50:40.530" Score="0" ViewCount="473" Body="&lt;p&gt;Given this pattern:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0, 1, 0, 2, 2, 2, 3, 4, 5, 6, 6, $n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the next item at $n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Pattern/sequence explanation required&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="9871" LastEditorUserId="5188" LastEditDate="2015-05-03T02:07:23.380" LastActivityDate="2015-05-03T14:52:27.273" Title="Finish The Pattern #3" Tags="|calculation-puzzle|pattern|" AnswerCount="2" CommentCount="1" ClosedDate="2015-05-04T13:56:53.313" ContentLicense="CC BY-SA 3.0" />
  <row Id="13250" PostTypeId="2" ParentId="13240" CreationDate="2015-05-02T23:47:51.747" Score="1" Body="&lt;p&gt;Don't we all have a backdoor ? Magic password functionality from easter-egg source inserted in-stream by the compiler ? (puts on Fark hat) - Was a woman involved ?  A double agent in the room at the time ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also - if on the internet, one must look to the host, and encryption of the block device, hopefully a steganographic cluster-scrambling filesystem using RAID2 or  RAID3 (think amulet halves).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OR were the files the property of the hackers/SPECTRE in the first place (and stolen from them) - presumably they can read their &lt;em&gt;own&lt;/em&gt; files ?     &lt;/p&gt;&#xA;" OwnerUserId="10483" LastActivityDate="2015-05-02T23:47:51.747" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13252" PostTypeId="2" ParentId="13240" CreationDate="2015-05-03T00:07:54.217" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The hacker switched their account with a no-login account, which simply refuses to accept login no matter how.&#xA; The hacker, unfortunately, forgot to switch theirs. Since their account is the only one not 'no-login', 007 found him&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second sentence is possible, if we're talking about double agent.&lt;/p&gt;&#xA;" OwnerUserId="4995" LastActivityDate="2015-05-03T00:07:54.217" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13253" PostTypeId="2" ParentId="13247" CreationDate="2015-05-03T02:53:27.203" Score="1" Body="&lt;p&gt;I feel like there may be multiple solutions to this puzzle, but I'll propose one I found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next number is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 6&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each group of three adds up to a multiple of 6, alternating between consecutive integers and the same 3 digits repeated.&lt;br&gt;&#xA;&#xA; - 0 + 1 + 0 = 6^0 (unsure about this)&lt;br&gt;&#xA; 2+2+2 = 1*6&lt;br&gt;&#xA; 3+4+5 = 2*6&lt;br&gt;&#xA; ==&gt; 6+6+&lt;em&gt;&lt;strong&gt;6&lt;/strong&gt;&lt;/em&gt;=3*6&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7715" LastEditorUserId="7715" LastEditDate="2015-05-03T14:52:27.273" LastActivityDate="2015-05-03T14:52:27.273" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13254" PostTypeId="2" ParentId="13247" CreationDate="2015-05-03T03:49:42.833" Score="3" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 12&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you were to group all numbers in batches of 3, add them up and then get the average between them (rounding up), you get a sequence of it's own which goes 1(?) 2 3 4. Done as follows:&lt;br&gt;&#xA; 0 1 0 = ? (I'm assuming 1)&lt;br&gt;&#xA; 2 2 2 = 6 / 3 = 2&lt;br&gt;&#xA; 3 4 5 = 12 / 3 = 4&lt;br&gt;&#xA; 6 6 12 = 24 / 3 = 8&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm probably horrible mistaken, but I thought I'd give it a shot.&lt;/p&gt;&#xA;" OwnerUserId="12100" LastEditorUserId="12100" LastEditDate="2015-05-03T04:08:53.317" LastActivityDate="2015-05-03T04:08:53.317" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13255" PostTypeId="1" CreationDate="2015-05-03T06:24:15.057" Score="-1" ViewCount="2243" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FbEo0.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using 4 straight lines connect these 9 stars together without lifting your pen from the surface. The end of each line must be start of the next line.&lt;/p&gt;&#xA;" OwnerUserId="12038" LastEditorUserId="7308" LastEditDate="2015-05-03T07:26:58.433" LastActivityDate="2015-05-03T15:00:25.153" Title="Using 4 straight lines connect the stars together" Tags="|logical-deduction|pattern|" AnswerCount="3" CommentCount="3" ClosedDate="2015-05-03T18:21:13.647" ContentLicense="CC BY-SA 3.0" />
  <row Id="13256" PostTypeId="2" ParentId="13255" CreationDate="2015-05-03T07:29:37.593" Score="5" Body="&lt;p&gt;I've never seen this one before, but here's what jumps out at me:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/psu6M.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="11154" LastEditorUserId="11154" LastEditDate="2015-05-03T15:00:25.153" LastActivityDate="2015-05-03T15:00:25.153" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13257" PostTypeId="1" AcceptedAnswerId="14418" CreationDate="2015-05-03T08:16:38.570" Score="7" ViewCount="1007" Body="&lt;p&gt;Can you guess what we are?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At first I was all alone.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Then a friend showed up, and the place started showing signs of life.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;But then another friend came, and the three of us started fighting.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Another friend came, and calmed us down by helping us communicate.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Yet another friend came along -- by now, we ruled the area!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;By the time a sixth friend joined us, the game had changed a little,&#xA;  and we were looking weak next to the competition.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Our seventh friend came and helped us win some resounding successes.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Finally our last member arrived, and we were just riding free.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By the time all my friends had assembled, what had we become?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As some have guessed, the friends are alphabet letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My friends and I are a pretty close-knit bunch. If you distributed our homes evenly on our stretch of street, there would only be one gap of more than one house! However, despite our close bond, we're all unique individuals. If you subtracted the distance from me to my second farthest friend from the distance from me to my farthest friend, you'd have a difference of 1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I should tell you a bit more about my friends. Some of them are very well-mannered, making our group quite proper when they joined. Each one also has a hobby, and when each one joins the club, they turned us into something related to their hobby. Their hobbies, in random order, are: Sports, American Politics, Musical Instruments, Computer Science, Historical Languages, Avoiding the Sun, and Exercise.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My friends want me to tell you more about their contributions to our group. Our first member (not counting me) made us an acronym. Our second member made us a last name. Our third member made us a proper noun. Our fourth member made us a last name. Our fifth through final members all made us gerunds!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Postmortem Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The puzzle title is a reference to the Rascal Flatts song of the same name -- the lyrics don't help with solving the puzzle but the subject matter of the song could be used for answer validation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11154" LastEditorUserId="11154" LastEditDate="2015-05-06T20:00:59.337" LastActivityDate="2015-05-06T20:00:59.337" Title="Me and My Gang!" Tags="|riddle|" AnswerCount="4" CommentCount="19" ContentLicense="CC BY-SA 3.0" />
  <row Id="13258" PostTypeId="2" ParentId="13255" CreationDate="2015-05-03T08:29:59.620" Score="6" Body="&lt;p&gt;Is it valid that 2 of the corners are beyond the stars?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my attempt!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ygftz.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ygftz.png&quot; alt=&quot;Attempt&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12054" LastEditorDisplayName="user10203" LastEditDate="2015-05-03T10:11:21.857" LastActivityDate="2015-05-03T10:11:21.857" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13260" PostTypeId="2" ParentId="13255" CreationDate="2015-05-03T08:33:16.730" Score="3" Body="&lt;p&gt;Using the information from &lt;a href=&quot;http://en.wikipedia.org/wiki/Parallel_(geometry)#Extension_to_non-Euclidean_geometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Parallel_(geometry)#Extension_to_non-Euclidean_geometry&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;2&quot;&gt;&#xA;  &lt;li&gt;parallel, if they do not intersect in the plane, but have a common limit point at infinity, or [...]&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, if we put this in a non-Euclidean plane, it can be done with one line:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oxiW5.png&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Click for source&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oxiW5.png&quot; alt=&quot;One line&quot; title=&quot;Click for source&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, of course, this would take forever to draw (But it &lt;em&gt;does&lt;/em&gt; pass through the centres of all stars)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I can also confirm it is impossible to do on a Euclidean plane without going outside the box with &lt;a href=&quot;http://pastebin.com/W13GZnVs&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; Python script.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 84 different ways to do it with 5 lines inside the box, but that should be around 11 unique ones because most are reflected or rotated.&lt;/p&gt;&#xA;" OwnerDisplayName="user10203" LastEditorDisplayName="user10203" LastEditDate="2015-05-03T09:11:16.730" LastActivityDate="2015-05-03T09:11:16.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13262" PostTypeId="1" AcceptedAnswerId="13263" CreationDate="2015-05-03T10:51:43.217" Score="16" ViewCount="690" Body="&lt;p&gt;Sometimes it's fun to be a little too literal.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The family being pushed repeatedly.&lt;/li&gt;&#xA;&lt;li&gt;The reach of a ring.&lt;/li&gt;&#xA;&lt;li&gt;Electricity of darkness.&lt;/li&gt;&#xA;&lt;li&gt;The gradual destruction of transportation.&lt;/li&gt;&#xA;&lt;li&gt;Unable to marry alone.&lt;/li&gt;&#xA;&lt;li&gt;Some masculine leaves.&lt;/li&gt;&#xA;&lt;li&gt;Repeating dogs feet.&lt;/li&gt;&#xA;&lt;li&gt;Monkey returns to the crib.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;These eight all refer to objects that belong in a single set. But one is different from the others and thus there are two distinct subsets (one of which has just one element)... and if you've figured it out, you'll know it, because one of them epitomises the larger subset.&lt;/p&gt;&#xA;&lt;p&gt;Now, you do need to find the word describing seven of the eight. But that's just leading into the next step. Drop the last letter, and you'll get an anagram of something that belongs to a set of six things, and is the third of that set.&lt;/p&gt;&#xA;&lt;p&gt;Your task is to identify the number associated with that &amp;quot;third of six&amp;quot; object.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt; - To be clear, only the clues to the eight members of the set are &amp;quot;too literal&amp;quot;. I have not hidden &amp;quot;too literal&amp;quot; clues elsewhere in the question.&lt;/p&gt;&#xA;" OwnerUserId="9289" LastEditorUserId="60644" LastEditDate="2024-01-06T08:44:03.127" LastActivityDate="2024-01-06T08:44:03.127" Title="A set for which the majority have a common theme" Tags="|wordplay|anagram|connections-puzzle|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="13263" PostTypeId="2" ParentId="13262" CreationDate="2015-05-03T11:01:15.193" Score="14" Body="&lt;h2&gt;Full answer&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The theme of the 8 clues is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; fruit and vegetables.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here they all are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. &lt;strong&gt;Pumpkin&lt;/strong&gt; (pushed repeatedly = pump, family = kin) - credit to &lt;strong&gt;leoll2&lt;/strong&gt; for this one&lt;br&gt;2. &lt;strong&gt;Orange&lt;/strong&gt; (ring = o, reach = range)&lt;br&gt;3. &lt;strong&gt;Blackcurrant&lt;/strong&gt; (electricity ~ currant, darkness = black) - credit to &lt;strong&gt;Hubble07&lt;/strong&gt; for this one&lt;br&gt;4. &lt;strong&gt;Carrot&lt;/strong&gt; (transportation = car, gradual destruction = rot)&lt;br&gt;5. &lt;strong&gt;Canteloupe&lt;/strong&gt; (unable to = can't, marry = elope)&lt;br&gt;6. &lt;strong&gt;Mangoes&lt;/strong&gt; (masculine = man, leaves = goes)&lt;br&gt;7. &lt;strong&gt;Pawpaws&lt;/strong&gt; (dog feet = paws, repeated).&lt;br&gt;8. &lt;strong&gt;Apricot&lt;/strong&gt; (monkey = ape, return ~ ri, crib = cot)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Seven of them (all except #3) have the following property in common (which is epitomised by #2):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; they're all &lt;strong&gt;orange&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Drop the last letter of this property to get&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; orang,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is an anagram of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; argon,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is the third of the six&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; noble gases&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and has&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; atomic number 18.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the final answer is &lt;strong&gt;18&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Brilliant puzzle!&lt;/h2&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-05-03T17:14:10.843" LastActivityDate="2015-05-03T17:14:10.843" CommentCount="20" ContentLicense="CC BY-SA 3.0" />
  <row Id="13264" PostTypeId="2" ParentId="13240" CreationDate="2015-05-03T11:52:22.780" Score="25" Body="&lt;p&gt;Bond isn't the only user who has accidentally typed&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; chmod -R username files&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;when they meant to type&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; chown -R username files&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, Bond is the only user for whom typing&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; chmod -R 007 files&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;silently has catastrophic results.&lt;/p&gt;&#xA;" OwnerDisplayName="user1502" LastActivityDate="2015-05-03T11:52:22.780" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13265" PostTypeId="2" ParentId="12732" CreationDate="2015-05-03T11:53:37.510" Score="4" Body="&lt;p&gt;This is my first stab at an puzzle on stackexchange, and this sure was a fun way to spend a Sunday afternoon! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm going to dive right in with the murders in the order in which I think they occured:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dr. Brofonomo&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 3 Js, W and TF, and Dr. B were all playing cards in one of the rooms on Deck 10 till late in the night. Around 11, W, TF and Dr. B came back to their deck, where presumably some argument over the card counting came up again. Either of W or TF murdered Dr. B. I say either since only one gun shot is seen on Dr. B. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;W &amp;amp; TF then returned to Deck 10 to gamble some more with the 3Js after the murder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;W &amp;amp; TF&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where the crux of the puzzle is. Due to the belt and the excessive shots, it is plausible that those murders of W, TF, George and Max are committed by the same person.There is nothing linking the 4 people directly, so it is safe to suppose that there is a psychotic element to the killer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The act of putting a belt around the hands of the victims could be viewed as analogous to handcuffs. 3 of the 4 victims could be said to have committed some sort of crime in the murderer's eyes.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;W &amp;amp; TF: Cops who gambled, loved violence more than necessary, murdered Dr. B.&lt;/li&gt;&#xA;&lt;li&gt;George: Into dirty sex. Considering George's willingness to share his sex life, it is safe to assume the murderer knew about that.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Going over the bios, the only person who comes to mind is Carol, who is described as&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;She's a lot more cheery than most people who've just lost all the&#xA;  family they had left... A little too cheery.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It seems quite certain that the murderer of W and TF must have had the tranquilizer on them, otherwise it would have been quite difficult to get the cops to part with their guns. And this is where I'm stumped: If Carol was the murderer, where did she get the tranquilizer from? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyhow, assuming she did find a way, this is what happened: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometime in the night, Carol knocked on George and Giselle's door, and convinced her to leave George in the middle of sex and come away to her room, leaving George tied and blindfolded. She ensures that the George's door is left &lt;strong&gt;open&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carol and Giselle waited in the corridor of Deck 5, along with the drugged wine. When W &amp;amp; TF returned, the ladies gave them each a glass of wine. After they passed out, they took the guns away and Carol shot them. She then tied them up, keeping with the psychotic element of the murders.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two of them then went up to Deck 7. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Max&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a departure from my previous answer. Giselle killed Max, possibly in conjunction with Carol. Max was not her son directly, since the ages do not match up. Presumably he's George's son from a previous marriage, and Carol did not like him. This is also evident from her reactions when the bodies are discovered, she has no reaction when Max's body is discovered, and goes to pieces on seeing George's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After this, she went to Carol's room. Carol put her to sleep, and then went to George's room alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;George&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carol entered the room through the open door, proceeded to beat him up before blowing his brains out. Motive for the death was her relationship with Giselle, and the violence was caused by some trauma from the sudden death of her family? Either ways, sounds quite disturbing to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;She also went to Max's room and tied up his hands with his belt too. Either this was to deflect blame from Giselle, or she felt that he also answered to the crime of underage drinking and therefore merited the same treatment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Luke&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Leia slipped Luke a couple of her sleeping pills the previous night, and then went to spend the night with Kevin. Kevin also got a couple of sleeping pills without him getting to know of it, so he slept on till 11. Leia slipped out around 10.30 and strangled Luke. His being under the influence may also explain the lack of any signs of struggle. She also replaced the bottle of sleeping pills in her room. When she got back to Kevin, she told him the happenings, and also fabricated the story of breakfast at 8. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There, I hope I've solved it! The only thing that doesn't make sense to me is the tranquilizer. The only people who could have sourced it are George, the cops, !Peleh and Kevin. My theory is that George had gotten it for the trip as a part of his sex play. Carol got to know of it, and pilfered it when she went to get Carol for the night. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crack of the robbery by Galatyr was awesome, two thumbs up!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers, and thanks for the awesome puzzle!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edits: after the comments from the OP, some changes done to the answer. Hope it solves the problem now!&lt;/p&gt;&#xA;" OwnerUserId="12104" LastEditorUserId="12104" LastEditDate="2015-05-04T04:21:49.407" LastActivityDate="2015-05-04T04:21:49.407" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13267" PostTypeId="1" AcceptedAnswerId="13279" CreationDate="2015-05-03T13:30:48.057" Score="3" ViewCount="1277" Body="&lt;p&gt;The Encrypted text is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;7q7f 93d4 6s03 8aua 3pn9 73wu 9l2l a1wx 8u52 9tld 6fhc axfx 8kn4 628q 4abk 314d 255s 5o5b 9cc1 5904 4sja aa1r 7fai apsw 80ph ai09 ejbs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am currently (at the time of posting the question) not giving any hints, it may depend upon later (except this tiny one).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may call this a hint according to your wish:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The encoding is surely a multi-step process. Anyways check the tags.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; The previous cryptogram was (You can solve this too anyways):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;8kn4 255s 3pn9 73wu kg2b 7q7f apsw kg2b 314d 8aua ai09 kg2b aa1r 6fhc 9l2l 628q kg2b 5904 6fhc a1wx 4sja kg2b 4abk 8aua axfx 4sja 80ph kg2b 7fai 6fhc 8u52 9tld 8aua 93d4 kg2b 6s03 9tld 5o5b 9cc1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12058" LastEditorUserId="12058" LastEditDate="2015-05-04T04:24:17.913" LastActivityDate="2015-05-04T05:14:32.050" Title="The puzzle is to decrypt this encrypted text. (Cryptograms)" Tags="|mathematics|calculation-puzzle|cipher|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13268" PostTypeId="2" ParentId="13257" CreationDate="2015-05-03T14:05:57.120" Score="2" Body="&lt;p&gt;I reckon you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Protons&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By the time you had assembled you were&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Oxygen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At first I was all alone.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Well, Hydrogen consists of a single proton and one electron.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then a friend showed up, and the place started showing signs of life.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Helium consists of two atoms that are flying around each other and there are two positive forces.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But then another friend came, and the three of us started fighting.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There are only two atoms in the inner electron shell, but there are 3 of them in Lithium which means the third will be constantly agitated and will constantly want to jump into the first electron shell.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another friend came, and calmed us down by helping us communicate.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;As Beryllium is in the 2nd group it is less reactive than Lithium in the 1st group.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet another friend came along -- by now, we ruled the area!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Boron is the first atom in the periodic table that is a metalloid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By the time a sixth friend joined us, the game had changed a little, and we were looking weak next to the competition.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;In a reaction to become stable, carbon could either get an additional 4 electrons or give away 4 of them, thus it is &quot;weak&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Our seventh friend came and helped us win some resounding successes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Our knowledge ends here, we are limited to high school chemistry, but this theory is just too good to throw it down the drain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12108" LastActivityDate="2015-05-03T14:05:57.120" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13269" PostTypeId="1" AcceptedAnswerId="13271" CreationDate="2015-05-03T14:14:27.980" Score="5" ViewCount="944" Body="&lt;p&gt;Here is my first riddle. I hope it will make sense because English is not my first language. I hope you'll enjoy it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I was born in Eastern Europe, but none can say when exactly&lt;br&gt;&#xA;  My history is filled by mysteries and legends&lt;br&gt;&#xA;  I became French to rule the world  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I am raised by men early in the morning&lt;br&gt;&#xA;  Layer after layer they build my core&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Sometime, you can see me in the sky&lt;br&gt;&#xA;  But don't mistake me with antelope horns&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10400" LastEditorUserId="2729" LastEditDate="2015-05-04T01:45:11.890" LastActivityDate="2015-05-04T01:45:11.890" Title="Born in Eastern Europe - what am I?" Tags="|riddle|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13270" PostTypeId="2" ParentId="13269" CreationDate="2015-05-03T14:31:24.497" Score="3" Body="&lt;p&gt;Perhaps it's the &lt;a href=&quot;http://en.wikipedia.org/wiki/Croissant&quot; rel=&quot;nofollow&quot;&gt;croissant&lt;/a&gt; (in the shape of a half moon).&lt;/p&gt;&#xA;" OwnerUserId="12110" LastActivityDate="2015-05-03T14:31:24.497" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="13271" PostTypeId="2" ParentId="13269" CreationDate="2015-05-03T14:32:20.257" Score="3" Body="&lt;p&gt;My answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Croissant&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was born in Eastern Europe, but none can say when exactly&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some claim that it's from Buda (Hungary).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My history is filled by mysteries and legends&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nobody knows who invented it, there are many &lt;a href=&quot;http://en.wikipedia.org/wiki/Croissant&quot; rel=&quot;nofollow&quot;&gt;legends&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I became French to rule the world&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's now considered a typical French food.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am raised by men early in the morning&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most people eat it in the morning&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Layer after layer they build my core&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's made of several layers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometime, you can see me in the sky&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Half-moon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But don't mistake me with antelope horns&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Similar shape&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-05-03T14:32:20.257" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13272" PostTypeId="1" AcceptedAnswerId="13274" CreationDate="2015-05-03T16:35:55.107" Score="1" ViewCount="496" Body="&lt;p&gt;Each symbol represents a number between 0 and 9. You need found the number which is the solution to the last operation. The two triangles in the first operation and the triangle and the rectangle of the third operation represent a number of two digits, not a multiplication operation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pshZe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, which is the number that corresponds to the last operation?&lt;/p&gt;&#xA;" OwnerUserId="12112" LastEditorUserId="5373" LastEditDate="2015-05-03T16:48:56.673" LastActivityDate="2015-09-29T21:29:38.963" Title="What is the solution of the last operation?" Tags="|logical-deduction|calculation-puzzle|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13273" PostTypeId="2" ParentId="13272" CreationDate="2015-05-03T16:42:20.663" Score="2" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 11 (triangle-triangle)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since star and octagon are two different numbers between 1 and 9, their sum is between 1 and 17. Triangle-triangle means a two digits number, so it must be 11.&lt;br&gt;&#xA;Then, we have these valid pairs for Star + Octagon: (2,9),(3,8),(4,7),(5,6),(6,5),(7,4),(8,3),(9,2).&#xA;Octagon + 1 returns a single digit number, so we can exclude (2,9). &#xA;Also we can exclude (6,5) because rectangle would be the same of octagon.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you suppose (3,8), the third equation becomes  $19-8=11$.&lt;br&gt;&#xA;As well, if you suppose (4,7), it becomes $18-7=11$&lt;br&gt;&#xA;If you suppose (5,6), it becomes $17-6=11$&lt;br&gt;&#xA;If you suppose (7,4), it becomes $15-4=11$&lt;br&gt;&#xA;If you suppose (8,3), it becomes $14-3=11$&lt;br&gt;&#xA;If you suppose (9,2), it becomes $13-2=11$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, if you suppose ($11-x,x$) the third equation becomes $(10+1+x)-x=11$&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-05-03T17:05:53.043" LastActivityDate="2015-05-03T17:05:53.043" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13274" PostTypeId="2" ParentId="13272" CreationDate="2015-05-03T16:46:58.200" Score="2" Body="&lt;p&gt;Let G, R, Y, B denote the digits corresponding to the four shapes in the order they appear (green star, red octagon, yellow triangle, blue rectangle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first equation,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; two 1-digit numbers sum to YY, so &lt;strong&gt;Y=1&lt;/strong&gt; (22 is too large).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the second equation we learn B=R+1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the third equation is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1B-R = 1(R+1)-R = 11,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;11&lt;/strong&gt;, or two yellow triangles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-05-03T16:56:06.653" LastActivityDate="2015-05-03T16:56:06.653" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13275" PostTypeId="2" ParentId="13272" CreationDate="2015-05-03T16:52:05.540" Score="1" Body="&lt;p&gt;Since two integers from 0 to 9 can not add up to a number higher than 18, the triangle is 1. If you substitute the rectangle`s value into the third equation, the octagon cancels out and becomes...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2 triangles or 11&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="167" LastEditorUserId="167" LastEditDate="2015-05-03T16:56:01.743" LastActivityDate="2015-05-03T16:56:01.743" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="13276" PostTypeId="1" AcceptedAnswerId="13277" CreationDate="2015-05-03T18:16:01.770" Score="3" ViewCount="392" Body="&lt;p&gt;Alice would like to determine the relative salaries of three&#xA;coworkers using two facts. First, she knows that if Pauline&#xA;is not the highest paid of the three, then Paul is. Second,&#xA;she knows that if Paul is not the lowest paid, then&#xA;Peter is paid the most. Is it possible to determine the&#xA;relative salaries of Pauline, Peter, and Paul from what&#xA;Alice knows? If so, who is paid the most and who the&#xA;least? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explain your reasoning.&lt;/p&gt;&#xA;" OwnerUserId="12112" LastEditorUserId="5373" LastEditDate="2015-05-03T19:44:17.493" LastActivityDate="2015-05-03T19:44:17.493" Title="The salary of three of Alice's coworkers" Tags="|logical-deduction|algorithm|" AnswerCount="3" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="13277" PostTypeId="2" ParentId="13276" CreationDate="2015-05-03T18:19:09.603" Score="2" Body="&lt;p&gt;The only valid order is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pauline &gt; Peter &gt; Paul&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We have 2 sentences:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A) Pauline is paid the most &lt;strong&gt;OR&lt;/strong&gt; Paul is paid the most.&lt;br&gt;&#xA;B) Paul is paid the lowest &lt;strong&gt;OR&lt;/strong&gt; Peter is the richest&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we suppose Paul being the richest, then Peter can't be the richest, forcing Paul to be the least paid. &lt;em&gt;Contradiction!&lt;/em&gt;&lt;br&gt;&#xA;On the contrary, if we suppose Pauline being paid the most, Peter can't still be the richest, so Paul's wage is the lowest, putting Peter in the middle of the wageboard.&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-05-03T18:26:15.473" LastActivityDate="2015-05-03T18:26:15.473" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13278" PostTypeId="2" ParentId="13276" CreationDate="2015-05-03T18:31:24.510" Score="0" Body="&lt;p&gt;From the second statement &quot;if Paul is not the lowest paid, then Peter is paid the most.&quot; any order without Paul at the bottom must have Pauline at the bottom, Peter first and Paul second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the first statement &quot;if Pauline is not the highest paid of the three, then Paul is&quot; would contradict that situation so therefore Paul must be the lowest paid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To fulfil the second statement and ensure Paul is the lowest paid Pauline must be paid the most.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the order from highest paid to lowest is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pauline, Peter, Paul.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11297" LastActivityDate="2015-05-03T18:31:24.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13279" PostTypeId="2" ParentId="13267" CreationDate="2015-05-03T18:40:45.913" Score="4" Body="&lt;p&gt;For the &lt;strong&gt;first cipher&lt;/strong&gt;, the decoded message is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;MR JOCK TV QUIZ PHD BAGS FEW LYNX.&lt;/strong&gt; which is a notable pangram.&lt;br&gt;&#xA;&lt;br&gt;&#xA;The code is recognizable as base-36 because all letters and numbers appear to be used. After converting to base-10, I tried a few functions and noticed that squaring the numbers produced 2 significant figures that could represent the letters of the alphabet.&lt;br&gt;&#xA;&lt;br&gt;&#xA;So the decoding steps are:&lt;br&gt;&#xA; (i) convert each 4 digit group to base-10&lt;br&gt;&#xA; (ii) square each base-10 number and divide by $10^{10}$&lt;br&gt;&#xA; (iii) select characters of the alphabet (except for 46 which is an ASCII period)&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/HhE5I.png&quot; alt=&quot;MrJock&quot;&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using the same method, the &lt;strong&gt;second cipher&lt;/strong&gt; decodes to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;PACK MY BOX WITH FIVE DOZEN LIQUOR JUGS&lt;/strong&gt; which is also a notable pangram.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-05-04T05:14:32.050" LastActivityDate="2015-05-04T05:14:32.050" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13280" PostTypeId="2" ParentId="13276" CreationDate="2015-05-03T18:50:34.737" Score="0" Body="&lt;p&gt;The other answers are great, but I'd just like to present it in a more mathematical-logic style.&#xA;&lt;hr&gt;&#xA;We know:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;S1&lt;/strong&gt;: if Pauline is not the highest paid, then Paul is the highest paid;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;S2&lt;/strong&gt;: if Paul is not the lowest paid, then Peter is the highest paid.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We deduce as follows.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ASSUME Pauline is not the highest paid. By S1, Paul is the highest paid and therefore not the lowest paid. By S2, Peter is the highest paid. Contradiction, so &lt;strong&gt;Pauline is the highest paid&lt;/strong&gt;.&lt;br&gt;&lt;br&gt;ASSUME Paul is not the lowest paid. By S2, Peter is the highest paid. But we already know Pauline is the highest paid. Contradiction, so &lt;strong&gt;Paul is the lowest paid&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-03T18:50:34.737" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13281" PostTypeId="1" AcceptedAnswerId="13282" CreationDate="2015-05-03T18:56:37.417" Score="4" ViewCount="450" Body="&lt;p&gt;The Roman Empire is at war with Carthage. &#xA;Hannibal's army is composed of $a$ archers, $b$ ballistae, $c$ cavalrymen, $d$ dart-shooters, $e$ elephants and $f$ fighters.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In each round of the battle, Scipio chooses some categories of enemy troops (between those listed above) and destroys 1 unit from each of the selected groups. &lt;/li&gt;&#xA;&lt;li&gt;Immediately after these units get destroyed, Hannibal restores his army with 1 unit in each of the non-attacked (in the same round) categories. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For example, if Scipio defeats 1 ballista and 1 elephant, Hannibal will displace 1 new archer, 1 cavalryman, 1 dart-shooter and 1 fighter.  Of course, Scipio may also decide not to attack anything, but that might not be the best way to win the war!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Given $a,b,c,d,e,f$ can you determine the minimum number of Carthaginian troops that Scipio can leave on the combat field? Also, what's the minimum number of rounds needed to achieve such result?&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example, if $a,b,c,d,e,f$ are all equal to 6, Scipio can easily remove one element from all categories per round and leave Hannibal with 0 troops after 6 rounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: Scipio can't choose to attack a unit of a category with 0 elements. If he does, he automatically loses his honour and, consequently, the war.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="8874" LastEditDate="2016-03-23T16:53:57.103" LastActivityDate="2016-03-23T16:53:57.103" Title="The Punic Wars are back!" Tags="|mathematics|strategy|story|history|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13282" PostTypeId="2" ParentId="13281" CreationDate="2015-05-03T19:18:32.340" Score="4" Body="&lt;p&gt;&lt;strong&gt;Scipio can always reduce each section of Hannibal's army to number at most 1, and the entire army to number at most 3.&lt;/strong&gt;&#xA;&lt;hr&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Proof&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If $a$ to $f$ are all equal to the same number, then Scipio can annihilate them in that number of rounds (as explained in the OP). So Scipio's aim is to even out the numbers and make them all equal. He can do this as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Target all sections of Hannibal's army whose numbers are either maximal or one less than maximal.&lt;/strong&gt; That way, he can keep going until every section of Hannibal's army numbers either 1 or 0. At this stage, the only effect his attacks can have is to toggle the 1's and 0's, so he can ensure the 0's number at least as many as the 1's.&#xA;&lt;hr&gt;&#xA;For example, let's say Hannibal starts off with $a=4,b=4,c=3,d=2,e=2,f=0$. Scipio proceeds as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;1st round: target archers, ballistae, and cavalrymen. Then $a=3,b=3,c=2,d=3,e=3,f=1$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;2nd round: target archers, ballistae, cavalrymen, dart-shooters, and elephants. Then $a=2,b=2,c=1,d=2,e=2,f=2$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;3rd round: target all sections. Then $a=1,b=1,c=0,d=1,e=1,f=1$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;4th round: target archers, ballistae, dart-shooters, elephants, and fighters. Then $a=0,b=0,c=1,d=0,e=0,f=0$. This is the best Scipio can do.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="10281" LastEditDate="2015-05-03T19:32:24.040" LastActivityDate="2015-05-03T19:32:24.040" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="13283" PostTypeId="2" ParentId="13281" CreationDate="2015-05-03T21:07:46.323" Score="1" Body="&lt;p&gt;Let $m$ be the number of units in the largest group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If every round Scipio attacks all groups that are not zero then after $m-1$ rounds the initially largest group (or groups) will only have one unit left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $m$ is an odd number then after $m-1$ rounds all groups that started with an odd number will also have one unit left. Call the number of these groups $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $m$ is an even number then after $m-1$ rounds all groups that started with an even number will also have one unit left. Call the number of these groups $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n\gt3$ then another step is needed to minimise the total number of remaining units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The minimum number of units that can be left on the battlefield is the smaller of $n$ and $6-n$.&lt;/p&gt;&#xA;" OwnerUserId="11297" LastActivityDate="2015-05-03T21:07:46.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13284" PostTypeId="2" ParentId="13083" CreationDate="2015-05-03T23:29:46.257" Score="9" Body="&lt;p&gt;The key to the cipher is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;BLOOD&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Verse 1 describes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;arteries and veins.  Blood rarely sees the light of day. Loss of blood can kill.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Verse 2 describes (includes comments from Alconja):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;its appearance under x-rays. Play the instrument refers to &quot;heart strings&quot;. Blood donation and transfusion saves lives.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Verse 3 describes (includes comments from Alconja):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;a waiting list to receive blood. Exclusive clans can refer to the Bloods gang. The impaler describes vampires. Writing your name could be signing in blood. Blood cannot be purchased.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Verse 4 describes (includes comments from Alconja):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;thirsty flying insects like mosquitos. Also hemoglobin (goblin) provides color to blood. The army reference can describe &quot;recruiting new blood&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Verse 5 describes    &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;making the verb BLEED by exchanging vowels&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The ciphertext is the upper case letters and the cipher is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Vigenère. Depending on which tool is used, a one-letter shift may also be required.&lt;br&gt;&#xA;&lt;br&gt;&#xA;Cipher text: VTTUMTEIPRFXPHXAAJHLQGASXCWTPTGMDUXJQEDHCIDGHYUAAQCWT&lt;br&gt;&#xA;Plain text: THEFIRSTANDLASTYOUSHOULDTAKEAPEAOFTHEPODAWORDWILLMAKE  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the inner riddle is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;The first and last you should take. A pea of the pod, a word will make.&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the answer to the inner riddle is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;PERCEPTION&lt;/strong&gt; which is derived from the title. The first is R, the last is E, then add a P.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-05-04T05:50:21.677" LastActivityDate="2015-05-04T05:50:21.677" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13285" PostTypeId="1" AcceptedAnswerId="13287" CreationDate="2015-05-04T01:59:22.543" Score="2" ViewCount="519" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Hey, welcome to the Sparrow Club,&quot; says the leader of the club, as&#xA;  you walk in for the first time. &quot;I'm Ryan, nice to meet you.&quot; The two&#xA;  of you shake hands.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;As you walk around, you notice that the room has a very Chinese décor to it, and indeed most of the people here are East Asian in ethnicity. People are all sitting down at square tables in groups of four, playing board games. Some of them are playing casually, others look deep in concentration deciding what to do next.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;I'll introduce you to some of the regulars first. Here we have Pin Hu&#xA;  and Toi-Toi, who are teaming up against two others,&quot; says Ryan,&#xA;  motioning to the first table as you two walk past.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;Over there are the pros of the group,&quot; says Ryan. &quot;They're two pairs&#xA;  of siblings, named Sangen and Suushi respectively. Nobody here knows&#xA;  their first names, because they've never told us. We've just used their last names as nicknames ever since they joined the club.&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;You then come across a group of two playing together instead of four.&#xA;  &quot;What's with those two?&quot; you ask Ryan.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;Oh, those are Fan Pai and Tan Yao,&quot; said Ryan. &quot;They're the dunces of the group, they know how to play but never really got good at it.&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Finally, the two of you sit down at a table with another person&#xA;  already at it. &quot;You want to play with three players?&quot; he asks. &quot;My&#xA;  sister Yi couldn't make it today.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What board game are people playing here, and what is Ryan's last name?&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-05-05T17:02:30.733" Title="What is the leader of the Sparrow Club's last name?" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13286" PostTypeId="1" CreationDate="2015-05-04T02:51:47.953" Score="2" ViewCount="1166" Body="&lt;p&gt;Edit for &quot;off topic&quot; &quot;correction&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know why this question was considered &quot;off topic&quot;.  It seemed to draw a lot of attention.  The &quot;puzzle&quot; was to figure out which person has the advantage which is not obvious.  The pieces that need to be put together are the checks for each winner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if this type of question is &quot;off topic&quot;, then why do they have both probability and card tags available on this site?  I used both of those tags to classify my question here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Four people, (call them C, D, E and F), decide to play a card game for fun. They use an ordinary fair deck of $52$ cards, shuffled well immediately before each hand is drawn, and randomly draw cards from it one a time without replacement, all 4 using (sharing) the same drawn cards to determine who wins. A win is defined as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C wins if he gets at least one of all $13$ ranks of the cards (regardless of suit as they can be mixed suits or even all the same suit).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D wins if he gets either $6$ red cards or $6$ black cards in a row (consecutive).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E wins if she gets $4$ of a kind of ANY ONE odd rank ($A,3,5,7,9,J,K$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F wins if she gets at least $12$ black cards and at least $12$ red cards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each win is based on the current hand only as there is no &quot;carryover&quot; from a previous hand. Each new hand starts &quot;fresh&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible for ties to occur but the rule is any and all ties are awarded as a half win for D and a half win for E. That is, if each player bet 1 dollar for a hand so that there were 4 dollars in the pot for that hand, D and E would split any ties so that they would get 2 dollars each. Note that D and E split ANY ties so even if only C and F tie, they both lose and D and E split the pot with a half win each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is who has the highest probability of winning and by how much over the competition?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the minimum # of cards needed to win differs for each player:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(C: $13$),  (D: $6$),  (E: $4$), (F: $24$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another game rule is if nobody wins by the $28$th card drawn in the hand, then that hand has no winner and a new hand will be drawn.  That is, $28$ cards max will be drawn per hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F.Y.I., my simulation has lots of &quot;buckets&quot; to count up interesting things and the most rare thing I see (that shows up as nonzero) out of $10,000,000$ iterations is D and E tying in only $8$ cards drawn.  This is not a &quot;bonus&quot; tie because neither C nor F can tie with only $8$ cards drawn.  It is so rare that out of $10$ million decisions, only $29$ were ties between D and E with exactly $8$ cards drawn.  That one special case can probably be calculated mathematically since there are only $52 \choose 8$ card combos which is about $3/4$ billion.  Even simulation of all of those hands is possible on a computer to get the exact count of winners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$UPDATE:$ I am curious about how many actual ways there are for D and E to tie in exactly $8$ cards since it seems to be the most rare event that is happening in my simulation.  I was able to quickly generate the 0.75 billion card combinations which is $52 \choose 8$ so I just have to insert the check for the ties and count them up.  I have another simulation running now overnight so I don't want to stress my CPU at $100$% for many hours so I will wait until the first one finishes and then code the 2nd one and report back.  Based on the simulation getting only $29$ out of $10,000,000$, I could think out of $750$ million possible $8$ card hands there should be a little over $2000$ ways to tie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$UPDATE - 2$ Out of 1 billion simulated hands, I am seeing only $2547$ D/E ties on the $8$th drawn card.  That is the most rare event I see of all the buckets I am viewing.  It is almost as rare as a royal straight flush in a $5$ card poker hand ($5$ cards only).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$UPDATE - 3$  I made a mistake in that to simulate all the possible $8$ card hands to check for a D/E tie on that exact card, it is not just $52 \choose 8$ combinations because order is important, so it is $8$! more than that which is a huge number and too large to easily simulate on a computer.  An example where D and E tie on the 8th card (using C=Club, H=Heart, S=Spade, D=Diamond) is AH,AD,AC,2C,3C,4C,5C,AS.  That completes both the $6$ blacks in a row and quad aces on the $8$th card.  So since $52!$ / $44!$ is such a huge number of hands to simulate/generate (it is over $30$ trillion), probably the best I can do (easily) is just simulate a $7$ card hand, check to make sure neither D nor E won at that point, then draw the 8th card and check if they tied on that card.  Again that is only an approximation but will be faster than drawing up to $28$ cards and checking for $4$ possible winners.  Note that there is no way there can be a tie on the $7$th card because a quad has $2$ of each color card so we would need those plus $6$ in a row of the same color card so that is $8$ cards minimum.&lt;/p&gt;&#xA;" OwnerUserId="12119" LastEditorUserId="12119" LastEditDate="2015-05-12T09:11:21.760" LastActivityDate="2015-05-12T09:11:21.760" Title="4 player card game, who has advantage?" Tags="|probability|cards|" AnswerCount="4" CommentCount="13" ClosedDate="2015-05-10T02:23:02.403" ContentLicense="CC BY-SA 3.0" />
  <row Id="13287" PostTypeId="2" ParentId="13285" CreationDate="2015-05-04T02:53:30.243" Score="5" Body="&lt;p&gt;I think the answer you're looking for is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mahjong, and Ryan's last name is Peikou. It is a game played by four players at a square table and the players' names are names of specific &lt;a href=&quot;http://en.wikipedia.org/wiki/Japanese_Mahjong_yaku&quot; rel=&quot;nofollow&quot;&gt;yaku&lt;/a&gt; in Japanese riichi mahjong. &lt;p&gt;&#xA; Pin Hu is a Chinese way of saying &quot;Pinfu&quot;, and Toi-Toi is &quot;Toitoi&quot; as named. &lt;p&gt;&#xA; The Sangen and Suushi siblings are nicknamed as &quot;big/little&quot;, which gives the names &quot;Daisangen&quot;, &quot;Shosangen&quot;, &quot;Daisuushi&quot;, and &quot;Shosuushi&quot; respectively.&lt;p&gt;&#xA; Tan Yao and Fan Pai are both yaku as named, and are some of the lowest-valued yaku in the game. (Fanpai is also known as yakuhai.)&lt;p&gt;&#xA; Finally, &quot;Yi&quot; and &quot;Ryan&quot; can both be suffixed with &quot;Peikou&quot; to create &quot;Iipeikou&quot; and &quot;Ryanpeikou&quot;, which are also both mahjong hands. The &quot;Ryan&quot; in &quot;Ryanpeikou&quot; is pronounced more like &quot;Liang&quot;, which happens to be Ryan's Chinese name.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastEditorDisplayName="user88" LastEditDate="2015-05-05T17:02:30.733" LastActivityDate="2015-05-05T17:02:30.733" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13288" PostTypeId="2" ParentId="3074" CreationDate="2015-05-04T04:19:01.483" Score="-1" Body="&lt;p&gt;(Lateral thinking)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a strategy that guarantees guessing the number in 3 guesses:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;First guess 11&#xA;    Higher -&amp;gt; second guess 100&#xA;    Lower  -&amp;gt; second guess 10&#xA;        Lower -&amp;gt; third guess 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The question didn't mention the number base so I proclaim that we are using binary numbers.&lt;/p&gt;&#xA;" OwnerUserId="4479" LastEditorUserId="4479" LastEditDate="2015-05-04T08:28:03.210" LastActivityDate="2015-05-04T08:28:03.210" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13289" PostTypeId="2" ParentId="13127" CreationDate="2015-05-04T05:59:33.470" Score="7" Body="&lt;p&gt;The three words are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Predilection, reinspection, perihelion&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reasoning behind this is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each word contains the letters &quot;EIEIO&quot; in that order, and comes right before an &quot;EIEIO&quot; line in the riddle.  Notice in the title, that &quot;reinvention&quot; also fits this pattern.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Old MacDonald had a bias&lt;br&gt;&#xA;E I E I O&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bias = predilection&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And at that bias he looked again&lt;br&gt;&#xA;E I E I O&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Looked again = reinspection&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With a big cat here and a horse guide there&lt;br&gt;&#xA;Here is scarlet&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lion, rein, and vermilion which all contain parts of eieio&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There for each&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For each = iteration, also close&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Everywhere are ions&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lots of the eieio words end in &quot;ions&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Old MacDonald flew too close to the sun&lt;br&gt;&#xA;E I E I O&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Perihelion = closest point of orbit to the sun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4479" LastActivityDate="2015-05-04T05:59:33.470" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13290" PostTypeId="1" AcceptedAnswerId="14858" CreationDate="2015-05-04T06:12:55.017" Score="2" ViewCount="724" Body="&lt;p&gt;The encrypted text is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;code&gt;.&lt;/code&gt; is a &lt;strong&gt;letter separator&lt;/strong&gt; and &lt;code&gt;&amp;lt;br&amp;gt;&lt;/code&gt; or line break is a &lt;strong&gt;word separator&lt;/strong&gt;)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;cs.aoa.daso&lt;br&gt;cs.fsi.rxhg&lt;br&gt;s.al.dkl.zpt&lt;br&gt;h.cr.efp.dabh.xfmv.fmsrx&lt;br&gt; bs.aqr.alpx.kwej&lt;br&gt;ba.abn.dfvj.fwbdd.fwbdcz.aohlxwz&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Some hints;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 0:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Hint:&lt;/strong&gt; Further Hints use Unscrambling.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Ehacgn ot sbae teywtn-xis ihelw gtiakn a as neo. pqr=11276.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Hcum Clascne&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  nth Irepm.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 4:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Berawe of Lditpcuaes dcdoe as ocmiposets&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 5:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Abck to abes tnywte xsi wiht a as eon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 6:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Do you know about composite numbers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12058" LastEditorUserId="5188" LastEditDate="2015-06-24T07:12:46.500" LastActivityDate="2015-06-24T07:12:46.500" Title="Decrypt this Cryptogram: cs.aoa.daso" Tags="|calculation-puzzle|cipher|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="13292" PostTypeId="1" AcceptedAnswerId="13298" CreationDate="2015-05-04T06:31:33.733" Score="3" ViewCount="2742" Body="&lt;p&gt;I've seen a lot of these types of puzzles in IQ tests, quizzes of various sorts, and I have never been able to figure out how to solve them. These generally involve two or more sets of geometrical figures with numbers written in them. These bear a certain relationship that we are required to find out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y4kca.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in this particular problem I have tried everything I can think of: sums, products, squares. The sheer enormity of possibilities seems daunting to me, particularly when there is no apparent way to narrow things down. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know if there are some general tips to approach some problems, perhaps using this one (or any other suitable one) as an example.&lt;/p&gt;&#xA;" OwnerUserId="12122" LastEditorUserId="30903" LastEditDate="2017-04-13T00:18:11.733" LastActivityDate="2017-04-13T00:18:11.733" Title="How to approach &quot;pattern finding &quot; questions?" Tags="|pattern|calculation-puzzle|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13293" PostTypeId="1" AcceptedAnswerId="13294" CreationDate="2015-05-04T08:27:03.483" Score="1" ViewCount="261" Body="&lt;p&gt;Title pretty much says it all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if $x=121149$, one needs to find an integer $y$ such that $x\times y = 941121$.&lt;/p&gt;&#xA;" OwnerUserId="9629" LastEditorUserId="9629" LastEditDate="2015-05-04T08:33:42.067" LastActivityDate="2015-05-04T08:52:12.553" Title="Are there two integers $x,y&gt;1$ such that $x\times y$ is the number you get when reversing the digits of $x$?" Tags="|mathematics|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13294" PostTypeId="2" ParentId="13293" CreationDate="2015-05-04T08:33:16.973" Score="6" Body="&lt;p&gt;Cheat example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $x=11,y=1,x\times y=11$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More serious example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $x=2178,y=4,x\times y=8712$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some theory:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; assume $x,y$ are two integers as stated. Then $x$ and $x\times y$ have the same digit sum, so they're congruent mod $9$. So $x(y-1)$ is a multiple of $9$. But $y$ must be less than $10$, so $y-1$ can't be a multiple of $9$. So either $x$ is a multiple of $9$ or $x$ and $y-1$ are both multiples of $3$. In the latter case, $y$ must be either $4$ or $7$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-05-04T08:52:12.553" LastActivityDate="2015-05-04T08:52:12.553" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13295" PostTypeId="1" AcceptedAnswerId="13296" CreationDate="2015-05-04T09:27:30.753" Score="9" ViewCount="1375" Body="&lt;p&gt;In math class I asked my friend what she did for a job. She simply handed me this note.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ur26Z.png&quot; alt=&quot;It was hard to create a sensible sentence here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;What does it say?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; You may need a calculator! Then again, you may not...&lt;/p&gt;&#xA;" OwnerUserId="7308" LastEditorUserId="5373" LastEditDate="2015-05-04T09:32:19.693" LastActivityDate="2015-05-04T09:32:19.693" Title="Seven Segment Cipher" Tags="|cipher|visual|seven-segment|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13296" PostTypeId="2" ParentId="13295" CreationDate="2015-05-04T09:30:30.707" Score="11" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I SELL BIG BEIGE SEESHELLS (a misspelling of SEASHELLS?).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Type in the given number on a calculator with a seven-segment display and turn the calculator upside down so that 3 becomes E, 7 becomes L, and so on.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Much easier than usual from you, Tryth! :-)&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-04T09:30:30.707" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13298" PostTypeId="2" ParentId="13292" CreationDate="2015-05-04T11:42:23.027" Score="4" Body="&lt;p&gt;When you don't have an idea, you have to try anything and see if it works! &#xA;In IQ tests there are often far-fetched patterns, your job is to identify them following these general tips:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If different geometric shapes are involved, the &quot;trick&quot; is usually in their position, orientation and number of edges.&lt;/li&gt;&#xA;&lt;li&gt;If colors are involved, identify the elements with the same one and count them (eg how many yellow items? how many red?)&lt;/li&gt;&#xA;&lt;li&gt;If numbers are involved, the puzzle is almost always based on sums, subtractions, products, sometimes squares and factorials. Almost never divisions and roots.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In this particular case, we see that the triangle is an invariant, so improbably we'll find hints in it. The problem also involves small numbers and there are only 2 clues (the numbers in the first two triangles), significantly increasing the difficulty of the problem.&lt;br&gt;&#xA;I think that in this case the answer is  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 24&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The central number is the product of the three numbers in the vertices minus 6.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Examples:&lt;br&gt;&#xA;4*3*2-6=18&lt;br&gt;&#xA;6*1*2-6=6&lt;br&gt;&#xA;&lt;strong&gt;5*3*2-6=24&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10399" LastEditDate="2015-05-04T12:19:12.423" LastActivityDate="2015-05-04T12:19:12.423" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13299" PostTypeId="2" ParentId="13286" CreationDate="2015-05-04T12:28:37.777" Score="5" Body="&lt;p&gt;It seems I am too stupid for this site: I haven't been able to post any solution that I made myself; though I can make my computer solve puzzles for me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's Python code that I wrote: it runs a lot of simulations and counts victories for each player as well as losses of everyone (when nobody wins after 28 cards are drawn). A draw adds 1 point to D and E, while victory adds 2 points to the winner.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from random import choice&#xA;&#xA;&#xA;def countWinners(*args):&#xA;    return sum(bool(x) for x in args)&#xA;&#xA;&#xA;deck = []&#xA;for a in ['H', 'D', 'C', 'S']:&#xA;    for b in xrange(1, 14):&#xA;        deck.append((a, b))&#xA;&#xA;&#xA;results = {'C': 0, 'D': 0, 'E': 0, 'F': 0, 'Nobody': 0}&#xA;games = 1000000&#xA;for i in xrange(games):&#xA;    C = set()&#xA;    D = [0, 0]&#xA;    E = {1: 0, 3: 0, 5: 0, 7: 0, 9: 0, 11: 0, 13: 0}&#xA;    F = {'R': 0, 'B': 0}&#xA;&#xA;    cards = deck[:]&#xA;    cardcount = 0&#xA;    if i % 1000 == 0:&#xA;            print 1.0*i/games&#xA;    while len(C) &amp;lt; 13 and D[1] &amp;lt; 6 and all(x &amp;lt; 4 for x in E.itervalues()) and any(y &amp;lt; 12 for y in F.itervalues()) and cardcount &amp;lt; 29:&#xA;        card = choice(cards)&#xA;        cards.remove(card)&#xA;        cardcount += 1&#xA;&#xA;        C.add(card[1])&#xA;&#xA;        if card[0] in ('C', 'S'):&#xA;            if D[0] == 0:&#xA;                D[1] += 1&#xA;            else:&#xA;                D[0] = 0&#xA;                D[1] = 1&#xA;        if card[0] in ('H', 'D'):&#xA;            if D[0] == 1:&#xA;                D[1] += 1&#xA;            else:&#xA;                D[0] = 1&#xA;                D[1] = 1&#xA;&#xA;        if card[1] % 2 == 1:&#xA;            E[card[1]] += 1&#xA;&#xA;        if card[0] in ('C', 'S'):&#xA;            F['B'] += 1&#xA;        else:&#xA;            F['R'] += 1&#xA;&#xA;    if cardcount == 29:&#xA;        results['Nobody'] += 2&#xA;    else:&#xA;        if countWinners(len(C) == 13, D[1] &amp;gt;= 6, any(x == 4 for x in E.itervalues()), all(y &amp;gt;= 12 for y in F.itervalues())) &amp;gt;= 2:&#xA;            results['D'] += 1&#xA;            results['E'] += 1&#xA;        else:&#xA;            if len(C) == 13:&#xA;                results['C'] += 2&#xA;            if D[1] == 6:&#xA;                results['D'] += 2&#xA;            if any(x == 4 for x in E.itervalues()):&#xA;                results['E'] += 2&#xA;            if all(y &amp;gt;= 12 for y in F.itervalues()):&#xA;                results['F'] += 2&#xA;for player in results.keys():&#xA;    print player, 1.0*results[player]/2/games&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It may be not quite clear, but it seems to work. The interesting thing is that the output is this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;C 0.2471&#xA;Nobody 0.0186&#xA;E 0.242735&#xA;D 0.244535&#xA;F 0.24703&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Each player wins in about 24.5% of all games! That looks quite suspicious, I must admit, but I couldn't find any bugs that I didn't fix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPD. After running 10.000.000 simulations the results are&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;C 0.2470487&#xA;Nobody 0.0184092&#xA;E 0.2423495&#xA;D 0.2435982&#xA;F 0.2485944&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;UPD2. 61 millions in:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;C 0.247020145161&#xA;Nobody 0.0184576290323&#xA;E 0.242251653226&#xA;D 0.243597201613&#xA;F 0.248673354839&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;UPD3. After 130 millions the probabilities are more or less stable around&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;C 0.2469661&#xA;Nobody 0.0184452&#xA;E 0.2423032&#xA;D 0.2435659&#xA;F 0.2487196&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'm stopping simulations because some much, &lt;strong&gt;much&lt;/strong&gt; faster solutions were added and their results seem to be consistent with the ones I've got&lt;/p&gt;&#xA;" OwnerDisplayName="user10531" LastEditorDisplayName="user10531" LastEditDate="2015-05-04T20:58:51.427" LastActivityDate="2015-05-04T20:58:51.427" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="13300" PostTypeId="1" AcceptedAnswerId="15222" CreationDate="2015-05-04T13:11:06.840" Score="3" ViewCount="504" Body="&lt;p&gt;The sequence is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;4,7,7,0,1,1,4,8,3,2,3,9,4,1,5,3,0,6,4,5,0,5,1,0,9,1,2,7,7,1,4,9,5,5,6,9,5,1,?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's very simple indeed you won't need to do any calculation, but you may need to do a little one, good luck. Yes you may really need to read this hint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint1:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Three things make up it, one of them is the first of Good Boy. A cat tamed the boy. The third one is an acronym relating to computer networks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is neither due to Euler, nor by Apery.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint3:&lt;/strong&gt;[New]&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some constant perhaps?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Hint edited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit2:&lt;/strong&gt; Ok I may add other hints if you comment something (maybe the problem you're facing or which type of hint should be there). &lt;/p&gt;&#xA;" OwnerUserId="12058" LastEditorUserId="12058" LastEditDate="2015-05-16T08:45:50.403" LastActivityDate="2015-05-19T21:10:48.110" Title="Find the next number in this series" Tags="|number-sequence|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="13301" PostTypeId="1" CreationDate="2015-05-04T13:48:24.940" Score="2" ViewCount="222" Body="&lt;p&gt;I found this problem from &lt;a href=&quot;http://artofproblemsolving.com/community/c163h564755p3301930&quot; rel=&quot;nofollow&quot;&gt;http://artofproblemsolving.com/community/c163h564755p3301930&lt;/a&gt; .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have $n=2017$ lamps in a circle, and enumerated by $L_1,\ldots,L_n$. Some of them are switched on and some of them are switched off. I also have been given a positive integer $m=108$. One every turn I choose one lamp $L_i$ and then the lamps $L_{i-m},\ldots,L_{l+m}$ will change their state, I mean if lamp $L_j$ was turned off then now it is turned on and vice versa. Also indexes go around like $L_0=L_n, L_{-1}=L_{n-1}$. What is the minimum number of turns to shut off all lamps, and what are the switches I need to press to shut down the lamps?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lamp states from the beginning are &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                                    0111110100011000011111101110010101100011&#xA;                                    1001111011101001001110111110001100011001&#xA;                                    1001010100101011101101001000010111111111&#xA;                                    1001101010111011110100100101000101100011&#xA;                                    1110100010010010101110100000111100101000&#xA;                                    0111101011111100010010110000100110100100&#xA;                                    0100110101110010110011110010101101100111&#xA;                                    1110010011000110110111010110010100101110&#xA;                                    0111111101110000111001111100100010010001&#xA;                                    1010110011000101100111111001011110101110&#xA;                                    0111010110111110110101000101100100011000&#xA;                                    1011000011011110001111100110100010100101&#xA;                                    1101111100110011001110010010001010101111&#xA;                                    1000001001000110011110010011011101110100&#xA;                                    1011111100110010011000010110010110101010&#xA;                                    0110101000011011110001010000010001000110&#xA;                                    1001110101001001110110111111010011010111&#xA;                                    1111011001000110111001000011101101110001&#xA;                                    0000011111101000010101011111011011000011&#xA;                                    1111000000011100010011011001011000110101&#xA;                                    1101011111100001100010110010110011000000&#xA;                                    0001001111100101110100100011011010011100&#xA;                                    0000001111010101000111011000110110100001&#xA;                                    1010110011100110111010111110110000010000&#xA;                                    1000101001111001000110000101010000010111&#xA;                                    1011100001000110001100010000001011101110&#xA;                                    1001111110100010010000011000100101010101&#xA;                                    1001001001110110101000001001001100001011&#xA;                                    0011011100011111100111001110101101110001&#xA;                                    0111010000010011110110011011000011101001&#xA;                                    1111011010010000101111000010000001100110&#xA;                                    1001011101001000010101001001011111111011&#xA;                                    1000111000100001101100101110100011111100&#xA;                                    1011001111101111110110101111101111011111&#xA;                                    1001111100110101110101111110010010101101&#xA;                                    1111111111000100100111100011101110110100&#xA;                                    0100011011001010110100101101000000110010&#xA;                                    0010010001001110110100011111100011111101&#xA;                                    0100110111101101010101010100110110011011&#xA;                                    0001111111000100000111011010101011000010&#xA;                                    0011011110110110110100011001101111001000&#xA;                                    1000000011110011100111100000001010010011&#xA;                                    1000011101111100000101010101010010100101&#xA;                                    1010001011010100011011001110110010100000&#xA;                                    1000111101111000010111111101010110110111&#xA;                                    0110001111100011001110000100100101001111&#xA;                                    0000111111100010011001010000010110111000&#xA;                                    1000110110001000001100110000001011000010&#xA;                                    1000101101110000101100100010101111100011&#xA;                                    1000010010111101000010000110011010000001&#xA;                                    0010001100001000001100110111110100100111&#xA;                                    1001100110001000100101011111001011001111&#xA;                                    110001011111001101010101001&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It looks like Gaussian elimination on $\mathbb Z/2\mathbb Z$ but I have no program to do it.&lt;/p&gt;&#xA;" OwnerUserId="12127" LastActivityDate="2015-05-04T19:47:19.197" Title="Shutting down all lamps" Tags="|minimum-solve|" AnswerCount="1" CommentCount="4" ClosedDate="2015-05-04T19:51:17.843" ContentLicense="CC BY-SA 3.0" />
  <row Id="13303" PostTypeId="1" AcceptedAnswerId="15260" CreationDate="2015-05-04T14:06:33.380" Score="115" ViewCount="16328" Body="&lt;p&gt;I never really knew my uncle growing up. He seemed nice enough, but I got the sense from the distant, troubled look in his eyes that there was at least some truth to the rumours...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Years before I was born, his fishing trawler was lost at sea under mysterious circumstances. None of his crew survived, and when pressed, he'd just mutter something about a freak storm. What is known, is that he spent several weeks adrift and when found, he was near death, babbling incoherently about pirates, treasure, and curses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, my uncle passed away a few months ago, and last week, out of the blue a lawyer showed up on my doorstep with a large box that was, for some unexplained reason, left to me in his will. The box's strange contents included piles of musty, leather bound books, old tattered maps, a small brass telescope, a compass that no longer worked... It was interesting, even a little mysterious, but mostly it just seemed like junk. However, as I flipped through an old book documenting the native flora of some isolated pacific island, a small map fell to the ground...&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As you can see below, the map is hand drawn and shows the features of an unnamed island in exquisite detail:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zaDz9.png&quot; alt=&quot;hand drawn map&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More interestingly, however, is the strange riddle scrawled on the back in a delicate, spidery script:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Start your search from dead man's cave and follow the clues herein.&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;Sharpen a sword and your wits and you're ready to begin.&lt;/em&gt;  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;Paired up, pared down, the truth exposed, lies within lies.&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;Heading onwards north, reeling seas terrorize.&lt;/em&gt;  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;Right is left and left is right, turn and go until,&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;Twins appear to mark an end, though you'll continue still.&lt;/em&gt;  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;Cardinals sitting in their nests, pick them up one by one.&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;Set them free and follow them, then turn to what's undone.&lt;/em&gt;  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;A leap of faith, surrender yourself. But when things come to a head,&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;Take the easy way out, with the blues left behind, go three and don't retread.&lt;/em&gt;  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;Follow the gaze of the crimson saint as he looks to the pre-dawn glow.&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;The distance the crow flies in to pi, hints how far to go.&lt;/em&gt;  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;Two one fives, all lined up, this time point the way.&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;They also indicate how far. Go now or be led astray.&lt;/em&gt;  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;Proceed until, when viewed from on high, the island itself gives a sign.&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;Next steps revealed as the river reflects and the ancient stones align.&lt;/em&gt;  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;Run forth with what you carry. Confusion becomes clear.&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;Now piece together your reward, you'll find it buried here.&lt;/em&gt;  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;Be warned though treasure seeker, should you make but one wrong turn.&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;The pirate's curse, and certain death, is all that you will earn...&lt;/em&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's not totally clear, but I'm pretty certain the riddle and map indicate some long forgotten pirate's treasure! Unfortunately, I've been unable to decipher any of it, and perhaps subconsciously I don't want to find the answer myself, for fear of the fate that befell my uncle and his crew... Either way, I've decided I'm turning it over to you. Good luck, but beware the pirate's curse...&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clarifications&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be 100% clear:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Everything you need to solve this puzzle, beyond a little general knowledge, is contained in the map and riddle, the rest is just flavor text.&lt;/li&gt;&#xA;&lt;li&gt;The last stanza is also just flavor, but the rest contain one or more discrete directions.&lt;/li&gt;&#xA;&lt;li&gt;You can assume that you're capable of crossing rivers, climbing mountains, swimming in the ocean, etc. Basically all squares on the map are valid, but you'll never be directed off the map completely.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Also, if you want to keep your working to yourself initially (this is a race for the treasure after all!), feel free to just post &lt;strong&gt;in the comments&lt;/strong&gt;, the coordinates you arrived at at the end of each stanza, with the last being the location of the treasure, along with what the treasure actually is. Eg. &quot;&lt;em&gt;AB, CD, ..., YZ - Treasure is a pile of gold&lt;/em&gt;&quot;. I'll then give a percentage indication of how successful you were. (obviously, when giving an actual answer, you'll need to justify those steps with valid logic though).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hints&lt;/strong&gt;&lt;br&gt;&#xA;&lt;sup&gt;&lt;em&gt;(This puzzle is now solved, but I'll leave these here for prosperity, or for late comers who still want to work through the challenge)&lt;/em&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;General hints:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;1&lt;/strong&gt;: As @Realdeo has observed in the comments, you can't necessarily assume North is up. Maybe there's a way to confirm...&lt;br&gt;&#xA; &lt;strong&gt;2&lt;/strong&gt;: Don't bother looking for a clue that will tell you where dead man's cave is. Just pick a cave and start. It will become apparent as you progress if you made the right choice...&lt;br&gt;&#xA; &lt;strong&gt;3&lt;/strong&gt;: Almost all the stanzas are effectively standalone riddles, so if you get stuck on one you can still work on others. Though you should still be aware that sometimes things rely on you being in a specific location in order to make sense (even then, you will likely be able to &quot;solve&quot; and understand the stanza). Of the individual stanzas/riddles, some are wordplay, some are mini puzzles, some use the map and some are fairly direct (though muddied with metaphor).&lt;br&gt;&#xA; &lt;strong&gt;4&lt;/strong&gt;: No clue will ever explicitly tell you to travel diagonally or any other non-orthogonal direction. Though some stanzas will be made up of multiple individual (orthogonal) steps, making the overall movement for the stanza non-orthogonal.&lt;br&gt;&#xA; &lt;strong&gt;5&lt;/strong&gt;: Don't bother trying to decrypt or decode all the letters around the edge. You won't get very far, as more than half are just random fillers. You'll need certain letters at some point, but hopefully when the time comes, you'll know which ones...&lt;br&gt;&#xA; &lt;strong&gt;6&lt;/strong&gt;: It will make your life easier if the first thing you do is to determine where north is. You can do this before even looking at the riddle.&lt;br&gt;&#xA; &lt;strong&gt;7&lt;/strong&gt;: Incorporating clues given on @rand's attempt at finding North: Paying close attention to the blotches around the compass rose is a captital idea! Plus, I'm sure the pirates would have put one on all four sides if they could have, but for some reason only three have anything visible... What could make those blotches, and why would only one side not have any? ...also, perhaps it's worth thinking about what's NOT there.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Stanza specific hints:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Start your search&lt;/strong&gt;: This stanza is mostly just a direct statement, so for the moment, there's not much to be gained puzzling over it (see general hint about dead man's cave above).&lt;br&gt;&#xA; &lt;strong&gt;Paired up&lt;/strong&gt;: The first line is the most important here, as it will help to understand the second.&lt;br&gt;&#xA; &lt;strong&gt;Paired up&lt;/strong&gt;: To clarify the hint above, although the first line is key, both lines still need to be used together. If the first is the key, then the second is the lock.&lt;br&gt;&#xA; &lt;strong&gt;Paired up&lt;/strong&gt;: You don't need to look at the map for this one, just read it very carefully, every single word should make sense once you get it.&lt;br&gt;&#xA; &lt;strong&gt;Paired up&lt;/strong&gt;: The solution to this stanza will be a clear, but short, sentence with a form similar to &quot;head two north&quot;&lt;br&gt;&#xA; &lt;strong&gt;Paired up&lt;/strong&gt;: People have correctly realised that there is wordplay involved here... However, beyond some slightly awkward grammar, the first line is quite literal in terms of what you need to do.&lt;br&gt;&#xA; &lt;strong&gt;Right is left&lt;/strong&gt;: Where you are on the map matters for this one.&lt;br&gt;&#xA; &lt;strong&gt;Right is left&lt;/strong&gt;: In response to @rps asking if the twins were the &lt;code&gt;II&lt;/code&gt; coordinates - there's multiple possible &quot;twins&quot; on the map with the correct ones being validated by your movement. If you know which way north is and you've solved stanza 2, &lt;del&gt;you should be able to deduce/validate both dead man's cave &amp;amp; the twins&lt;/del&gt;.&lt;br&gt;&#xA; &lt;strong&gt;Right is left&lt;/strong&gt;: I slightly overstated myself in the hint above (apologies, though I doubt it sent anyone too far off track), you won't be able to 100% validate the twins and dead man's cave yet, but you will be able to narrow it down to two possibilities (one of which you'll be able to discount later).&lt;br&gt;&#xA; &lt;strong&gt;Right is left&lt;/strong&gt;: How would a pirate &quot;mark and end&quot; (particularly when we're talking about a treasure map)?&lt;br&gt;&#xA; &lt;strong&gt;Right is left&lt;/strong&gt;: Obviously I've made the twins a bit subtle, so when you're looking for them, consider squares as whole units... also, a pirate would need more than one set of twins to mark a spot...&lt;br&gt;&#xA; &lt;strong&gt;Cardinals&lt;/strong&gt;: Make sure you only free the cardinals that are actually in their nests, no others. If you get that far, what's undone should be clear.&lt;br&gt;&#xA; &lt;strong&gt;A leap of faith&lt;/strong&gt;: Metaphor and the map.&lt;br&gt;&#xA; &lt;strong&gt;A leap of faith&lt;/strong&gt;: Where you're standing and which way you're facing matter a lot here. If it makes no sense to you, you're probably standing in the wrong spot.&lt;br&gt;&#xA; &lt;strong&gt;A leap of faith&lt;/strong&gt;: There are multiple individual steps, but the instructions are all fairly direct (mostly just metaphor with some minor play on words, but no &quot;tricks&quot; like the &quot;paired up&quot; stanza), just make sure those blues are actually left behind when you follow the last step.&lt;br&gt;&#xA; &lt;strong&gt;Follow the gaze&lt;/strong&gt;: Who might the crimson saint be?&lt;br&gt;&#xA; &lt;strong&gt;Follow the gaze&lt;/strong&gt;: You'll need some very basic (high school level?) maths here, but currently you've almost certainly got it wrong...&lt;br&gt;&#xA; &lt;strong&gt;Follow the gaze&lt;/strong&gt;: The distance to go will be an integer value, with no rounding or approximating required.&lt;br&gt;&#xA; &lt;strong&gt;Two one fives&lt;/strong&gt;: &quot;Two one fives&quot; are the key here, but there are several other hints in this stanza to push you in the right direction.&lt;br&gt;&#xA; &lt;strong&gt;Two one fives&lt;/strong&gt;: How could &quot;two one fives&quot; be interpreted given the pluralisation?&lt;br&gt;&#xA; &lt;strong&gt;Two one fives&lt;/strong&gt;: Although &quot;two one fives&quot; is the key clue, the other sentence fragments are still hints. Ask yourself why I would phrase the rest of it the way I have...&lt;br&gt;&#xA; &lt;strong&gt;Proceed until&lt;/strong&gt;: First line is a simple directive, the second points to a mini puzzle of its own.&lt;br&gt;&#xA; &lt;strong&gt;Run forth&lt;/strong&gt;: What are you carrying?&lt;br&gt;&#xA; &lt;strong&gt;Run forth&lt;/strong&gt;: The first line is your last direction clue, the second line is talking about the treasure itself.&lt;br&gt;&#xA; &lt;strong&gt;Run forth&lt;/strong&gt;: &quot;Confusion becomes clear&quot; can be applied to the first sentence or the second line equally (and I've been dropping many hints to various people on how to discover what the treasure is).&lt;br&gt;&#xA; &lt;strong&gt;Be warned&lt;/strong&gt;: As stated in the clarifications, you can completely ignore this stanza.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fallen heroes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A memorial to those who were lost to the pirate's curse:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Realdeo attempted &lt;code&gt;HT TG ES EN ES MU EI HN HS&lt;/code&gt;, but those co-ordinates clearly spell out &quot;THE NEMESIS HUNTS (geh)&quot;, which can't be a good sign... (with approximately 2% correct, he is now missing, presumed dead)&lt;/li&gt;&#xA;&lt;li&gt;tfitzger tried &lt;code&gt;UE IE II SR...&lt;/code&gt;, but luckily gave up as he had almost spelled out &quot;[c]URSE&quot; already! (with maybe 20% right of the 20% attempted, I can only hope he escaped the curse, but it's not looking good)&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It's a miracle. He's alive and still searching, with a second attempt of &lt;code&gt;UE IE II CU AE...&lt;/code&gt;. Going from &quot;[c]URSE&quot; to &quot;CU[rs]E&quot; is an improvement, but I still have my concerns.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Timur Kuchkarov has made a cautious start of &lt;code&gt;FE FI II...&lt;/code&gt;, but with a low percentage correct, the &quot;FIE&quot; upon him may grow to something much worse.&lt;/li&gt;&#xA;&lt;li&gt;After various attempts, rand al'thor settled on &lt;code&gt;HT DT DO EN EB EU HU RY RI&lt;/code&gt;, but I'm worried for his safety given that the pirates seem to be trying to tell him &quot;R! You be hurt, then died!&quot; (grammar has never been a pirate's strong point).&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It seems rand al'thor is made of tougher stuff than I thought... he survived the pirate's death threats, to find the treasure!&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solution&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The treasure was ultimately &lt;a href=&quot;https://puzzling.stackexchange.com/a/15063/5208&quot;&gt;found and won by rand al'thor&lt;/a&gt;. However, since this page has become a bit of a mess of partial solutions and invalidated attempts, I have also &lt;a href=&quot;https://puzzling.stackexchange.com/a/15260/5208&quot;&gt;posted a fully explained, canonical solution below&lt;/a&gt;, for those interested.&lt;/p&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.923" LastActivityDate="2015-05-21T17:28:58.223" Title="A hunt for long forgotten treasure ...but beware the pirate's curse!" Tags="|riddle|visual|rhyme|enigmatic-puzzle|treasure-hunt|" AnswerCount="9" CommentCount="21" ContentLicense="CC BY-SA 3.0" />
  <row Id="13304" PostTypeId="2" ParentId="13303" CreationDate="2015-05-04T15:03:43.280" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Start: HT&#xA; TG&#xA; ES&#xA; EN&#xA; ES&#xA; MU (I got this one wrong?)&#xA; EI&#xA; HN&#xA; HS (Is the treasure?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4995" LastActivityDate="2015-05-04T15:03:43.280" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13305" PostTypeId="1" AcceptedAnswerId="15247" CreationDate="2015-05-04T15:12:29.967" Score="26" ViewCount="1720" Body="&lt;p&gt;Somewhere in a magical land a small golden chest is delivered by an white eagle to a soon-to-be heroine. She opens it with shaky hands, knowing what it means that this gift has been delivered to her. Indeed, the first thing which falls out is a parchment bound and sealed by her mentor. The letter reads:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qfk7G.png&quot; alt=&quot;The Letter&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;( &lt;a href=&quot;https://i.stack.imgur.com/kGvfE.png&quot; rel=&quot;noreferrer&quot;&gt;Full Resolution Letter&lt;/a&gt; )&lt;/p&gt;&#xA;&lt;p&gt;The chest also contains a map and 22 small scrolls, each listing a wisdom spell written by the mage in his runes.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qRtGT.jpg&quot; alt=&quot;The treasure map&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;( &lt;a href=&quot;https://i.stack.imgur.com/qRtGT.jpg&quot; rel=&quot;noreferrer&quot;&gt;Full Resolution Map&lt;/a&gt; )&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xlS6E.png&quot; alt=&quot;The 22 spell scrolls&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;( &lt;a href=&quot;https://i.stack.imgur.com/6Hf0i.png&quot; rel=&quot;noreferrer&quot;&gt;Full Resolution Spells&lt;/a&gt; )&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;With just this, can you locate where the holy nut could be found?&lt;/h1&gt;&#xA;&lt;p&gt;The position can be found on the map. Use the following grid to specify the coordinates:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/a70wc.png&quot; alt=&quot;Solution-Grid&quot; /&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;Comments &amp;amp; Hints&lt;/h1&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;This puzzle (in slightly different form) was 'solved' by an unintended short-cut (now closed), but is still requiring a 'proper' solution. You might find it by backward-solving it form the now known solution, but there is more fun in not doing so. (I hope.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;A hint:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; - The lake and the compass both hold a clue...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;A full eye-opener hint (strong spoiler!) added after the full solution has been posted:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The map overlaid with a hex grid &lt;a href=&quot;https://i.stack.imgur.com/ie9vD.jpg&quot; rel=&quot;noreferrer&quot;&gt;(spoiler!)&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-20T15:09:50.647" Title="A treasure hunt for the magic artefact" Tags="|visual|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="13306" PostTypeId="1" AcceptedAnswerId="13307" CreationDate="2015-05-04T15:26:28.887" Score="4" ViewCount="1341" Body="&lt;p&gt;One day while he was siting on a chair, waiting for a &quot;reversal of chance&quot; Jim had two queens (beautiful females) come along.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their sight satisfied Jim and he recalled that people called these ladies a &quot;Pocket Pair'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jim then went with the two 'queens' to 'The Flop'; glittering on the table he saw 6 red diamonds. &quot;What are these?&quot;, Jim asked, &quot;Garnets or Rubies?&quot;. Here he also met another two 'queens', though his time, the beautiful 'queens' were wearing black suit instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gosh! what are the odds of me being surrounded by four sexy ladies?!?! thought Jim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jim then went with the 'four ladies' to 4th street, there he met Jack. Jim thought he had managed to made Jack jealous, but Jack maintained his emotion control: he was as stiff as a coated paper. What a poker face, thought Jim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jim and the four ladies went walking, unfortunately for Jim after a short time a man named &quot;The King&quot; appeared 'swimming' (out from nowhere!) on the River. The king caught him red handed with his 4 beautiful ladies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The King punished Jim, causing him to go bankrupt. But those are the rules!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jim stood up from his chair and left crying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What actually happened to poor Jim?&lt;/p&gt;&#xA;" OwnerUserId="4995" LastEditorUserId="5373" LastEditDate="2015-05-04T19:31:32.217" LastActivityDate="2015-05-08T22:09:53.147" Title="Jim: A polygamy love" Tags="|riddle|lateral-thinking|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="13307" PostTypeId="2" ParentId="13306" CreationDate="2015-05-04T15:30:49.383" Score="18" Body="&lt;p&gt;Sounds to me like Jim&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Had quite the poker hand - pocket Queens, with 4-of-a-kind revealed in Queens on the flop (as well as a six of Diamonds, apparently).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; With the Jack as the turn card and the King on the river, his opponent must have been holding a 10 and an Ace suited in black, and achieved the very rare royal flush (though I suppose a 9-10 suited would work as well).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To make matters worse,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jim went all-in on the 4-of-a-kind (who wouldn't?), lost to the royal flush, and went bankrupt.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Poor Jim.&lt;/p&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-05-04T15:30:49.383" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13309" PostTypeId="2" ParentId="13306" CreationDate="2015-05-04T16:32:47.760" Score="2" Body="&lt;p&gt;Like Bailey, but&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The other guy had a 4 of a kind in Kings.  It couldn't be a Royal Flush because Jim had all the Queens.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10294" LastActivityDate="2015-05-04T16:32:47.760" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="13311" PostTypeId="2" ParentId="13301" CreationDate="2015-05-04T19:47:19.197" Score="1" Body="&lt;p&gt;Let $A$ be a $2017\times 2017$ matrix in $\mathbb Z_2$, where $A_{ij}=1$ precisely when pressing lamp $i$ will also change lamp $j$ (namely, when $j-i\in \{-m,-(m-1),\dots,m-1,m\}$). Let $b$ be a $2017$ entry column vector in $\mathbb Z_2$, which is the initial state of the lamps. You want to solve the equation&#xA;$$&#xA;Ax=b&#xA;$$&#xA;where $x$ is a vector representing whether or not you should toggle lamp $i$. The solution is $x=A^{-1}b$, so to solve this, you could compute $A^{-1}$ by &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaussian_elimination#Finding_the_inverse_of_a_matrix&quot; rel=&quot;nofollow&quot;&gt;Gaussian elimination&lt;/a&gt;. If you remember how this algorithm works, and are familiar with coding, then you should be able to translate it into code, as long as you are careful to make sure you perform your addition over $\mathbb Z_2$ as opposed to $\mathbb R$.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-05-04T19:47:19.197" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13312" PostTypeId="1" AcceptedAnswerId="14945" CreationDate="2015-05-04T19:51:32.610" Score="22" ViewCount="2332" Body="&lt;p&gt;There's a secret hidden in the text below. What mind-blowing truth does it reveal?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Not imaginative? Then Zeke's Clues may assist! Unsure, lad? With our&#xA;  patented Obscurity Clues, begin verifiably outwitting the great&#xA;  VIGENÈRE! My useful helper, Mr. Riddler, envisions heavenly light&#xA;  revealing clever elements in grabbers. For 6 Krones, cousin Kryptos&#xA;  has made plenty new devotees. But taste softly, victim, this advice: &#xA;  GO ACROSTIC. DOMAIN solves everything. Please review wise masters&#xA;  (xenophobic sociopaths) very selectively. Some KEYS percolate behind&#xA;  masks of kingly visages.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In re-solving this myself, I noticed a difficulty over which I have no control: At some point, capitalization will matter. The order is lowercase, uppercase, uppercase, lowercase. When see a tiger, you know you've got it right.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The capitalized words are obviously clues. The relate to how you perform the first step and, depending on how you look at it, the second step. The DOMAIN that is mentioned is fully qualified. I wanted to fit that into the original text but couldn't find a spot for it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you don't recognize the clue words, you should Google them. You should at least figure out what kind of puzzle the first step is just from that. The fully qualified domain will help you solve that first step. Don't get too caught up on the text.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2015-05-07T15:46:45.620" LastActivityDate="2015-05-14T23:30:47.160" Title="Down the Rabbit Hole" Tags="|riddle|cipher|enigmatic-puzzle|computer-puzzle|steganography|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13313" PostTypeId="2" ParentId="13286" CreationDate="2015-05-04T19:55:58.657" Score="1" Body="&lt;p&gt;After 100,000,000 iterations, I get the following (CIs calculated based on this: &lt;a href=&quot;http://www.evanmiller.org/statistical-formulas-for-programmers.html#prop_mci&quot; rel=&quot;nofollow&quot;&gt;http://www.evanmiller.org/statistical-formulas-for-programmers.html#prop_mci&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;C: 0.246899 - 0.247071&#xA;D: 0.24354 - 0.243711&#xA;E: 0.242103 - 0.242275&#xA;F: 0.24869 - 0.248863&#xA;X: 0.018397 - 0.0184508&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which corroborates Akiiino's numbers and means that F has a real advantage, although a small one. Code below if you want.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#include &amp;lt;iostream&amp;gt;&#xA;#include &amp;lt;random&amp;gt;&#xA;#include &amp;lt;chrono&amp;gt;&#xA;#include &amp;lt;vector&amp;gt;&#xA;#include &amp;lt;algorithm&amp;gt;&#xA;#include &amp;lt;array&amp;gt;&#xA;#include &amp;lt;string&amp;gt;&#xA;using namespace std;&#xA;&#xA;void report(string name, double x, double z) {&#xA;    double n = z / 2;&#xA;    double p = x / z;&#xA;    double c0 = (p + 2 / n) / (1 + 4 / n);&#xA;    double c1 = 2 * sqrt((p * (1 - p) + 1 / n) / n) / (1 + 4 / n);&#xA;    cout &amp;lt;&amp;lt; name &amp;lt;&amp;lt; &quot;: &quot; &amp;lt;&amp;lt; (c0 - c1) &amp;lt;&amp;lt; &quot; - &quot; &amp;lt;&amp;lt; (c0 + c1) &amp;lt;&amp;lt; endl;&#xA;}&#xA;&#xA;int main() {&#xA;    array&amp;lt;int, 52&amp;gt; cards; // rank then suit, alternating colors of suits&#xA;    for(int i = 0; i &amp;lt; 52; ++i)&#xA;        cards[i] = i;&#xA;&#xA;    long long c = 0, d = 0, e = 0, f = 0;&#xA;&#xA;    mt19937_64 rnd(chrono::system_clock::now().time_since_epoch().count());&#xA;&#xA;    for(long long z = 2, r = 2;; z += 2) {&#xA;        shuffle(cards.begin(), cards.end(), rnd);&#xA;&#xA;        array&amp;lt;int, 26&amp;gt; counts = {};&#xA;&#xA;        int dr = 0, db = 0;&#xA;        for(int i = 0; i &amp;lt; 28; ++i) {&#xA;            counts[cards[i] % 26] += 1;&#xA;&#xA;            int cs = 1, es = 0;&#xA;            for(int j = 0; j &amp;lt; 13; ++j) {&#xA;                if(counts[j] + counts[j + 13] == 0)&#xA;                    cs = 0;&#xA;                if(j % 2 == 0 &amp;amp;&amp;amp; counts[j] + counts[j + 13] == 4)&#xA;                    es = 1;&#xA;            }&#xA;&#xA;            if(cards[i] % 26 &amp;lt; 13) {&#xA;                dr += 1;&#xA;                db = 0;&#xA;            } else {&#xA;                db += 1;&#xA;                dr = 0;&#xA;            }&#xA;            int ds = (dr &amp;gt;= 6 || db &amp;gt;= 6);&#xA;&#xA;            int fs = 0, fr = 0, fb = 0;&#xA;            for(int j = 0; j &amp;lt; 13; ++j)&#xA;                fr += counts[j];&#xA;            for(int j = 13; j &amp;lt; 26; ++j)&#xA;                fb += counts[j];&#xA;&#xA;            fs = (fr &amp;gt;= 12 &amp;amp;&amp;amp; fb &amp;gt;= 12);&#xA;&#xA;            if(cs + ds + es + fs &amp;gt; 0) {&#xA;                if(cs + ds + es + fs &amp;gt; 1) {&#xA;                    d += 1;&#xA;                    e += 1;&#xA;                } else {&#xA;                    c += 2 * cs;&#xA;                    d += 2 * ds;&#xA;                    e += 2 * es;&#xA;                    f += 2 * fs;&#xA;                }&#xA;                break;&#xA;            }&#xA;        }&#xA;&#xA;        if(z == r) {&#xA;            cout &amp;lt;&amp;lt; (z / 2) &amp;lt;&amp;lt; endl;&#xA;            report(&quot;C&quot;, c, z);&#xA;            report(&quot;D&quot;, d, z);&#xA;            report(&quot;E&quot;, e, z);&#xA;            report(&quot;F&quot;, f, z);&#xA;            report(&quot;X&quot;, z - c - d - e - f, z);&#xA;            cout &amp;lt;&amp;lt; endl;&#xA;&#xA;            r *= 10;&#xA;        }&#xA;    }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7392" LastActivityDate="2015-05-04T19:55:58.657" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13314" PostTypeId="2" ParentId="13286" CreationDate="2015-05-04T19:56:39.970" Score="1" Body="&lt;h2&gt;A fast C program&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I'll throw my C program into the mix.  It runs much faster than the Python program from @Akiiino.  I am able to run over 1 million hands per second, and so I ran it for 1 billion hands (taking about 13 minutes).  The program listing is at the end of this answer.  Here are the results:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The original problem&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The original problem did not impose a 28 card limit.  I guessed that F would be the easiest to achieve and I was right:&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(10 million iterations)&#xA;C) 24.9527%&#xA;D) 24.5856%&#xA;E) 24.5530%&#xA;F) 25.9087%&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h2&gt;With 28 cards max&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;With a 28 card limit, the result became closer:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(1 billion iterations)&#xA;C) 24.6989%&#xA;D) 24.3624%&#xA;E) 24.2243%&#xA;F) 24.8718%&#xA;N) 1.8426% (no winner)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If someone wants to run my program for even more iterations, feel free to do so.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;So why did I guess F?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Ok so David James asked why did I guess F?  I did some quick calculations (that were admittedly not very precise) to guesstimate the avg card at which each player would win.  Remember, the following is not correct math, it's just what was going through my mind:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;My rough calculations (which were bad)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Player C needs all 13 cards to win.  This is very similar to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Coupon_collector&amp;#39;s_problem&quot; rel=&quot;nofollow&quot;&gt;Coupon collector's problem&lt;/a&gt;.  If each card were replaced back into the deck, it would take on average 42 cards dealt to win (according to the Wikipedia table).  However, the cards are &lt;em&gt;not&lt;/em&gt; placed back into the deck, so it takes less cards on average to win.  Still, I figured the number would be closer to 42 than to 26.  I guessed halfway in between so 34.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Player D needs 6 of the same color in a row to win.  The odds of that happening are 1/32, starting at card 6.  It is 1/32 because it is 1/64 chance for 6 reds and 1/64 chance for 6 blacks.  So I figured that D would get the 6 in a row on average at card 38 (start at card 6 and takes 32 cards on average).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Player E wins by getting one of 7 four of a kinds.  The odds of getting a four of a kind for a particular number (such as Ace) with 4 cards is &lt;code&gt;4/52 * 3/51 * 2/50 * 1/49&lt;/code&gt; which is 1 in 270725.  Given 7 different numbers (A3579JK), the chances are 1 in 38675.  Now I just need to know what number of cards N is it that makes &lt;code&gt;N C 4&lt;/code&gt; around half that (19337), where &lt;code&gt;C&lt;/code&gt; is the combination operator.  The answer is between 27 (17550 4-card combos) and 28 (20475 4-card combos).  So I called it 27.5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Player F wins by getting 12 red and 12 black cards.  Using a &lt;a href=&quot;http://stattrek.com/online-calculator/binomial.aspx&quot; rel=&quot;nofollow&quot;&gt;binomial calculator&lt;/a&gt;, I was able to determine that with 26 cards, the odds of getting at least a 12/12 split was 44.3%.  With 27 cards, the odds went up to 55.8%.  So the expected win occurs between 26 and 27 cards.  So I called it 26.5.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The reality&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I made a big mistake in calculating player D.  The 6 in a row of a color needs to account for the fact that you aren't putting the card back into the deck.  So after you get 5 in a row of red, the odds of getting a 6th red card is much less than 50%.  My guess for player C was also way off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I modified my program to find out how many cards on average does it take each person to win:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(50 million iterations)&#xA;C) 27.997&#xA;D) 25.810 (but with a 53.6% of not even winning at all) &#xA;E) 28.904&#xA;F) 26.375&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But even knowing the average card isn't good enough because you also need to take into account the distribution, and also figure out how often there are ties.  Anyways, I don't know of a good way to determine the answer to this problem without a computer simulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my program:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The program listing&lt;/h2&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#include &amp;lt;stdio.h&amp;gt;&#xA;#include &amp;lt;stdint.h&amp;gt;&#xA;#include &amp;lt;stdlib.h&amp;gt;&#xA;#include &amp;lt;time.h&amp;gt;&#xA;&#xA;#define DEFAULT_ITERATIONS        1000000&#xA;static int simulate(void);&#xA;static int deck[52];&#xA;&#xA;int main(int argc, char *argv[])&#xA;{&#xA;    int iterations = DEFAULT_ITERATIONS;&#xA;    int wins[5] = {0};&#xA;    int i       = 0;&#xA;&#xA;    for (i=0;i&amp;lt;52;i++)&#xA;        deck[i] = i;&#xA;&#xA;    srand(time(NULL));&#xA;    if (argc &amp;gt; 1)&#xA;        iterations = atoi(argv[1]);&#xA;&#xA;    // Each win earns 2 points.  A tie earns 1 point for D and 1 for E.&#xA;    // wins[4] is the slot for &quot;no winner&quot;.&#xA;    for (i=0;i&amp;lt;iterations;i++) {&#xA;        int winner = simulate();&#xA;        if (winner == -1) {&#xA;            // It's a tie.  Half point to D an E.&#xA;            wins[1]++;&#xA;            wins[2]++;&#xA;        } else {&#xA;            // A sole winner.&#xA;            wins[winner]+=2;&#xA;        }&#xA;    }&#xA;    for (i=0;i&amp;lt;5;i++) {&#xA;        printf(&quot;%d) %8d wins out of %8d = %7.4lf%%\n&quot;, i, wins[i], 2*iterations,&#xA;                ((double) wins[i] * 100.0) / (2*iterations));&#xA;    }&#xA;    return 0;&#xA;}&#xA;&#xA;// Returns winner 0-3, 4 for no winner, or -1 for a tie.&#xA;static int simulate(void)&#xA;{&#xA;    int i;&#xA;    int rankBitmap        = 0;&#xA;    int redCardsInARow    = 0;&#xA;    int blackCardsInARow  = 0;&#xA;    int totalReds         = 0;&#xA;    int totalBlacks       = 0;&#xA;    int rankHistogram[13] = {0};&#xA;    int win               = 0;&#xA;&#xA;    // Shuffle deck.&#xA;    for (i=0;i&amp;lt;51;i++) {&#xA;        int r = rand() % (52 - i);&#xA;        int tmp;&#xA;&#xA;        // Swap card i with card i+r.&#xA;        tmp       = deck[i];&#xA;        deck[i]   = deck[i+r];&#xA;        deck[i+r] = tmp;&#xA;    }&#xA;&#xA;    // Deal cards one by one, and update win conditions.&#xA;    for (i=0;i&amp;lt;28;i++) {&#xA;        int card = deck[i];&#xA;        int color = card &amp;amp; 1;&#xA;        int rank  = card &amp;gt;&amp;gt; 2;&#xA;&#xA;        // C wins if he gets all 13 ranks.&#xA;        rankBitmap |= (1 &amp;lt;&amp;lt; rank);&#xA;        if (rankBitmap == 0x1fff)&#xA;            win |= 0x1;&#xA;&#xA;        // D wins by getting 6 red cards or 6 black cards in a row.&#xA;        if (color) {&#xA;            // Black.&#xA;            redCardsInARow = 0;&#xA;            if (++blackCardsInARow == 6)&#xA;                win |= 0x2;&#xA;        } else {&#xA;            // Red.&#xA;            blackCardsInARow = 0;&#xA;            if (++redCardsInARow == 6)&#xA;                win |= 0x2;&#xA;        }&#xA;&#xA;        // E wins if he gets 4 of a kind of any odd rank (A3579JK) (7 chances).&#xA;        if ((rank &amp;amp; 1) == 0) {&#xA;            if (++rankHistogram[rank] == 4)&#xA;                win |= 0x4;&#xA;        }&#xA;&#xA;        // F wins by getting 12 black and 12 red cards.&#xA;        if (color)&#xA;            totalBlacks++;&#xA;        else&#xA;            totalReds++;&#xA;        if (totalBlacks &amp;gt;= 12 &amp;amp;&amp;amp; totalReds &amp;gt;= 12)&#xA;&#xA;            win |= 0x8;&#xA;&#xA;        if (win != 0)&#xA;            break;&#xA;    }&#xA;    // If we dealt 28 cards, there must not be a winner.&#xA;    if (i == 28)&#xA;        return 4;&#xA;    switch (win) {&#xA;        case 0x1: return 0;&#xA;        case 0x2: return 1;&#xA;        case 0x4: return 2;&#xA;        case 0x8: return 3;&#xA;        default: return -1; // Must be a tie.&#xA;    }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4479" LastActivityDate="2015-05-04T19:56:39.970" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13317" PostTypeId="2" ParentId="11010" CreationDate="2015-05-04T20:40:50.583" Score="4" Body="&lt;p&gt;Answer:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ZEBRA  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;In each of the four hint-giving lines, the first part refers to the position within the five-letter word and the second part refers to location in the alphabet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12134" LastEditorUserId="7715" LastEditDate="2015-05-04T22:52:28.497" LastActivityDate="2015-05-04T22:52:28.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13318" PostTypeId="2" ParentId="13306" CreationDate="2015-05-04T20:41:08.647" Score="6" Body="&lt;h2&gt;Answer&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Jim had&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an impressive poker hand: pocket Queens, with four of a kind revealed in Queens on the flop, as well as a 6 of diamonds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; with the Jack as the turn card and the King on the river, his opponent must have been holding a 10 and an Ace suited in black, and achieved the very rare royal flush (though a 9-10 suited would work as well).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To make matters worse,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jim went all in on the 4-of-a-kind, lost to the royal flush, and went bankrupt.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Constructive criticism&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I think you show some potential as a riddle-writer, but you've put in far too many and too explicit hints for this puzzle to be any challenge at all. There is a &lt;a href=&quot;https://puzzling.stackexchange.com/a/5983/5373&quot;&gt;precedent&lt;/a&gt; for posting answers that are not actually answers but rather improvements to the question, so here goes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;One day while he was sitting on a chair, waiting for a reversal of fortunes, Jim met two beautiful ladies, who were both wearing glittering red diamonds - rubies, perhaps - 6 in total.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Later he met two more ladies, this time dressed as though for a funeral. &quot;Gosh! what are the odds of me being surrounded by four sexy ladies?!?!&quot; thought Jim.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;He went with the four ladies to 4th street, where he met Jack. Jim thought he had managed to made Jack jealous, but Jack kept his face stiff as paper.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Unfortunately for Jim, after a short time a king appeared from nowhere, swimming in the river. The king caught him red-handed with his 4 beautiful ladies and punished Jim, causing him to go bankrupt.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Jim stood up from his chair and left crying.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;What actually happened to poor Jim?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the story as edited above, there are much fewer explicit hints in actual card-game language, which makes the puzzle more challenging. Even now it's fairly easy to solve - with ladies, diamonds, red, black, Jack, and a king, it wouldn't take people long to come up with the idea of a card game - but at least there's &lt;em&gt;something&lt;/em&gt; to work out here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course, I realise you haven't accepted an answer yet. Maybe the whole card-game scenario is a red herring and the real answer is something much deeper - if so, congratulations on writing such a multi-layered puzzle!)&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-05-08T22:09:53.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13320" PostTypeId="1" AcceptedAnswerId="13331" CreationDate="2015-05-04T21:02:25.547" Score="22" ViewCount="2244" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Test run of a possible series of &quot;word of the day&quot; riddles!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;With a light source, I resemble a star;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Once fortified, I'm alone at the bar.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Then, come the fifth, I shall sail away -&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Tell me, what am I, your word of the day?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="5373" LastEditDate="2018-06-02T15:30:35.153" LastActivityDate="2018-06-02T15:30:35.153" Title="Word of the Day #1" Tags="|riddle|word|rhyme|" AnswerCount="6" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13321" PostTypeId="1" AcceptedAnswerId="13330" CreationDate="2015-05-04T21:26:25.950" Score="14" ViewCount="2434" Body="&lt;p&gt;&lt;strong&gt;A&lt;/strong&gt;lyson and &lt;strong&gt;B&lt;/strong&gt;ethany decide to play a game involving a (standard six-sided) die.  Alyson goes first, and rolls the die: if she rolls a 6, then she wins.  Otherwise, Bethany takes the die and rolls until she beats Alyson's initial roll.  Then, if she's rolled a 6, &lt;em&gt;she&lt;/em&gt; wins; otherwise, the die is passed back to Alyson.  The game keeps going back and forth between &lt;strong&gt;A&lt;/strong&gt; and &lt;strong&gt;B&lt;/strong&gt; this way, with each one rolling until they beat the other player's high roll, and whoever first hits an unbeatable roll (i.e., the 6 on a six-sided die) winning.  Which of them has the better winning chances?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus questions: (a) what happens if the win condition is kept the same, but the die is handed back whenever a player's roll is greater than &lt;em&gt;or equal to&lt;/em&gt; the previous high roll?  (b) How many rolls does a game take, on average?  How many turns (i.e., passes of the die)?&lt;/p&gt;&#xA;" OwnerUserId="266" LastActivityDate="2015-05-05T22:56:56.550" Title="Who's favored in this dice game?" Tags="|mathematics|probability|dice|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13322" PostTypeId="2" ParentId="13321" CreationDate="2015-05-04T21:38:08.537" Score="6" Body="&lt;p&gt;From my initial calculations, if I'm interpreting the question correctly, Alyson's chance of winning on her initial roll is...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;50%&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's my reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;When Alyson throws the die the first time, these are the possible results:&#xA;&lt;br /&gt;&lt;br /&gt;6 - If a 6 is thrown, she immediately wins, as per the question.&#xA;&lt;br /&gt;&lt;br /&gt;5 - If a 5 is thrown, she waits until Bethany can beat her roll.  Since the only roll that will beat 5 is a 6, Bethany simply rolls until she gets a 6 and wins.  Thus, if Alyson rolls a 5, she immediately loses.&#xA;&lt;br /&gt;&lt;br /&gt;4 - If a 4 is thrown, she again waits for Bethany to beat her roll.  This time, Bethany can roll a 5 or a 6 to beat her.  Using the previous logic, if Bethany rolls a 5, Bethany loses and Alyson wins.  If Bethany rolls a 6, their fortunes are reversed.  Therefore, if Alyson rolls a 4, she has a 50% chance of winning on Bethany's turn.&#xA;&lt;br /&gt;&lt;br /&gt;3 - If a 3 is thrown, Bethany's options to beat her are now 4, 5 or 6, each with equal likelihood.  If Bethany rolls a 6, Alyson has lost.  If she rolls a 5, Alyson has won.  If Bethany rolls a 4, she (and thus Alyson as well) has a 50% chance of winning.  The total chance for Alyson to win will be $1/3(0\%)$ (on a 6) $+ 1/3(100\%)$ (on a 5) $+ 1/3(50\%)$ (on a 4), which totals 50% again.&#xA;&lt;br /&gt;&lt;br /&gt;2 - The logic above is extended for a roll of 2.  This time, the chance of winning breaks down as $1/4(0\%) + 1/4(100\%) + 1/4(50\%) + 1/4(50\%) = 50\%$&#xA;&lt;br /&gt;&lt;br /&gt;1 - The odds are, again, 50%: $1/5(0\%) + 1/5(100\%) + 1/5(50\%) + 1/5(50\%) + 1/5(50\%) = 50\%$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus B&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The average number of game turns is 2.45.  The following chart shows every possible outcome (shown in turn order), each with a corresponding probability.  The sum of the number of turns for each outcome, multiplied by the probability of each, will give us the average turn length of a game.&#xA;&lt;br /&gt;&lt;br /&gt;&lt;img src=&quot;https://i.stack.imgur.com/16TnY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11851" LastEditorUserId="11851" LastEditDate="2015-05-04T23:50:29.680" LastActivityDate="2015-05-04T23:50:29.680" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13323" PostTypeId="2" ParentId="13321" CreationDate="2015-05-04T21:48:03.273" Score="4" Body="&lt;p&gt;She'll win:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; half the time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1/6 of the time, Alyson wins with a roll of 6.&lt;br&gt;&#xA;1/6 of the time, Alyson loses with a roll of 5.&lt;br&gt;&#xA;1/6 of the time, Alyson rolls a 4 which then has a win/lose rate of 1/2.  (total, 5/12).&lt;br&gt;&#xA;1/6 of the time, Alyson rolls a 3. 1/3 of those, she wins (B=5); 1/3 she loses (B=6); 1/3 she splits (B=4).&lt;br&gt;&#xA;1/6 of the time, Alyson rolls a 2. 1/4 of those, she wins (B=5); 1/4 she loses (B=6); 1/4 she splits (B=4); 1/4 she splits (B=3).&lt;br&gt;&#xA;1/6 of the time, Alyson rolls a 1. 1/5 of those, she wins (B=5); 1/5 she loses (B=6); 1/5 she splits (B=4); 1/5 she splits (B=3); 1/5 she splits (B=2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus A:&lt;/strong&gt;&lt;br&gt;&#xA;1/6 of the time, Alyson wins with a roll of 6.&lt;br&gt;&#xA;1/6 of the time, Alyson rolls a 5. 1/4 of the time B=5, A=6. The sum of the sequence is 1/3.&lt;br&gt;&#xA;...&lt;br&gt;&#xA;&lt;strong&gt;(in progress)&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-05-04T21:56:30.490" LastActivityDate="2015-05-04T21:56:30.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13324" PostTypeId="1" AcceptedAnswerId="13325" CreationDate="2015-05-04T22:28:07.133" Score="-1" ViewCount="569" Body="&lt;p&gt;Counting the Triangles in this image individually would take far too long.  Can anyone come up with an algorithm to figure out how many triangles are in this shape?  Answers should include the number of triangles you find and the algorithm you used to get that number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qcCWn.png&quot; alt=&quot;That is far too many triangles.&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="12142" LastEditorUserId="12014" LastEditDate="2016-02-18T16:09:18.777" LastActivityDate="2016-02-18T16:53:33.047" Title="That's a lot of triangles" Tags="|visual|triangle|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13325" PostTypeId="2" ParentId="13324" CreationDate="2015-05-04T22:30:25.047" Score="5" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Count all the singles, then all the triangles made of 4 triangles, and so on, up to the huge ones. It's a methodical and easy way to do it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;size 1: 48&lt;br&gt;&#xA;size 4 : 30&lt;br&gt;&#xA;size 9 : 20&lt;br&gt;&#xA;size 16 : 12&lt;br&gt;&#xA;size 25 : 6&lt;br&gt;&#xA;size 36 : 2&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total: 118&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-05-04T22:52:21.083" LastActivityDate="2015-05-04T22:52:21.083" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13327" PostTypeId="2" ParentId="13320" CreationDate="2015-05-05T02:19:51.530" Score="1" Body="&lt;p&gt;Maybe it's a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;shot&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With a light source, I resemble a star;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When light shines through a shot glass in a dark room, it can &lt;a href=&quot;https://www.google.com/search?q=light%20shot%20glass&amp;amp;safe=off&amp;amp;biw=1447&amp;amp;bih=880&amp;amp;source=lnms&amp;amp;tbm=isch&amp;amp;sa=X&amp;amp;ei=fidIVbulCsuWgwS17IHICA&amp;amp;ved=0CAcQ_AUoAg&quot; rel=&quot;nofollow&quot;&gt;sparkle like a star&lt;/a&gt;. (some shot glasses more than others)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once fortified, I'm alone at the bar.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fortify can mean &quot;&lt;a href=&quot;https://www.google.com/search?q=fortified&amp;amp;ie=utf-8&amp;amp;oe=utf-8&quot; rel=&quot;nofollow&quot;&gt;to strengthen a drink with alcohol&lt;/a&gt;&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then, come the fifth, I shall sail away -&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;When the shot glass is filled from &lt;a href=&quot;http://en.wikipedia.org/wiki/Fifth_%28unit%29&quot; rel=&quot;nofollow&quot;&gt;the fifth&lt;/a&gt; of distilled liquor, it will &quot;sail away&quot; into a person's mouth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tell me, what am I, your word of the day?&lt;/p&gt;&#xA;" OwnerUserId="463" LastActivityDate="2015-05-05T02:19:51.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13328" PostTypeId="2" ParentId="13320" CreationDate="2015-05-05T02:45:22.773" Score="12" Body="&lt;p&gt;I'm going with the word&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Port&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With a light source, I resemble a star;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ports generally are accompanied by lighthouses, which are a bright light in the dark, hence a &quot;star&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once fortified, I'm alone at the bar.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Port wine is a fortified wine.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then, come the fifth, I shall sail away -&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This brings the two together - &quot;the fifth&quot; is, as JLee notes, a liquor measurement, and thus relates to port wine, while &quot;sail away&quot; links to the ocean port meaning.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9289" LastActivityDate="2015-05-05T02:45:22.773" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13329" PostTypeId="2" ParentId="13321" CreationDate="2015-05-05T03:19:35.193" Score="0" Body="&lt;p&gt;Adding the results of a short simulation program confirms that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For situation A (dice swap when a better roll achieved): Alyson wins 50% of the time, with an average of 6 rolls and 2.45 turns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For situation B (dice swap when a better or equal roll achieved): Alyson wins 52.4% of the time, with an average of 6 rolls and 3.28 turns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if a D20 dice was used then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For situation A (dice swap when a better roll achieved): Alyson wins 50% of the time, with an average of 20 rolls and 3.60 turns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For situation B (dice swap when a better or equal roll achieved): Alyson wins 50.2% of the time, with an average of 20 rolls and 4.55 turns.&lt;/p&gt;&#xA;" OwnerUserId="1631" LastActivityDate="2015-05-05T03:19:35.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13330" PostTypeId="2" ParentId="13321" CreationDate="2015-05-05T03:32:37.160" Score="16" Body="&lt;p&gt;The probability is exactly $\frac{1}2$ in the original game. The crucial fact is that rolling a $5$ is a loss and that rolling a $6$ is a win - so we can consider the game over once one of those appears. However, at any given roll, the probability of rolling a $5$ equals the probability of rolling a $6$. We can quickly extend this local property to learn that the game is fair:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose player $X$ won, rolling a $6$. An equally likely outcome would be that the game progressed the same until the last roll, at which point $X$ rolled a $5$. Similarly, if $X$ lost rolling a $5$, it was equally likely that they'd have rolled a $6$ instead.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is, we can pair up all the possible games by switching the last roll between $5$ and $6$, which changes who won, but not the probability of such events occurring.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Extension to Bonus (b)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notably, if we stop the game whenever we reach a value of $k$ or more, then it is clear that the probability that we stop the game at a value of exactly $k$ is $\frac{1}n$, where $n=7-k$ is the number of integers in $[k,6]$ - that is, the number of rolls greater than or equal to $k$. This holds because switching the last value in such a game has similar effect to before, and hence the last value is uniformly distributed. In particular, this means that the probability of $k$ appearing in a normal game (since the game after a value of $k$ or more is irrelevant as $k$ will not appear again) is $\frac{1}n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can rephrase the above as: We expect to see $\frac{1}n$ turns where $k$ is passed to the other player. Then, the expected length of a game is the sum of the expected number of turns wherein each given value is passed due to the linearity of expectation - thus, the expected length of the game is&#xA;$$1+\frac{1}2+\frac{1}3+\frac{1}4+\frac{1}5+\frac{1}6=\frac{49}{20}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The number of rolls is obviously distributed the same as if we just rolled a die until we got a $6$. We can see that the expected number of rolls $r$ satisfies the relation&#xA;$$r=\frac{1}6\cdot 1 + \frac{5}6(r+1)$$&#xA;since we take $1$ roll $\frac{1}6$ of the time, and otherwise expect $r+1$ rolls. This solves to&#xA;$$r=6.$$&#xA;More generally, we expect as many rolls as there are faces on the die.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Alternate Bonus (b) Answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that, in the previous turn, a player passed the number $6-n$. Let the expected number of turns remaining be $t_n$. Then, we have $t_0=0$ and $t_6$ is the expected number of turns in a game with a six sided die. It can be seen that $t$ satisfies the relation:&#xA;$$t_n=1+\frac{1}n\sum_{i=0}^{n-1}t_i$$&#xA;That is to say, each $t_i$ is one more than the average of those below it. This &lt;em&gt;looks&lt;/em&gt; unwieldily, but it actually turns out to equal the well-known harmonic series. In particular, define&#xA;$$T_n=\frac{1}{n+1}\sum_{i=0}^{n}t_i$$&#xA;as the average of the sequence thus far. We can express this algebraically, but consider the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The average of the sequence $t_0,t_1,\ldots,t_{n-1}$ is $T_{n-1}$. Thus, so is the average of $t_0,t_1,\ldots,t_{n-1},T_{n-1}$, as putting the mean into the sequence doesn't affect the mean. Increasing one element of the $\frac{1}{n+1}$ elements by $1$ increases the mean by $\frac{1}{n+1}$. Therefore $T_n=\frac{1}{n+1}+T_{n-1}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From this, and the fact that $t_n=T_{n-1}+1$, we can use the relation that $T_{n-1}=\frac{1}n+T_{n-2}$ as derived above to yield:&#xA;$$t_n=\frac{1}n+T_{n-2}+1$$&#xA;$$t_n=\frac{1}n+\frac{1}{n-1}+T_{n-3}+1$$&#xA;and expanding until $T_0=0$ yields&#xA;$$t_n=\frac{1}n+\frac{1}{n-1}+\ldots+1$$&#xA;meaning that $t_n$ is the $n^{th}$ harmonic number - the sum of the reciprocals of the first $n$ natural numbers. So, this game's expected length, which is $t_6$ is&#xA;$$\frac{1}1+\frac{1}2+\frac{1}3+\frac{1}4+\frac{1}5+\frac{1}6=\frac{49}{20}$$&#xA;Though there's no particularly helpful closed form for exact computation of Harmonic numbers as ratios of two integers, this means that, when played with an $n$ sided die, the game will take about $\log(n)+\gamma$ turns where $\gamma$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant&quot; rel=&quot;nofollow&quot;&gt;Euler-Macaroni&lt;/a&gt; constant.&lt;/p&gt;&#xA;" OwnerUserId="4765" LastEditorUserId="4765" LastEditDate="2015-05-05T22:56:56.550" LastActivityDate="2015-05-05T22:56:56.550" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13331" PostTypeId="2" ParentId="13320" CreationDate="2015-05-05T04:06:23.463" Score="22" Body="&lt;p&gt;I'll guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Flower&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With a light source, I resemble a star;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sunflower sort of resembles a star (sun = light source)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once fortified, I'm alone at the bar.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wallflowers are shy and tend to be alone (walls can be fortified)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then, come the fifth, I shall sail away &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mayflower was a sailing ship (May is the fifth month)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4479" LastActivityDate="2015-05-05T04:06:23.463" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13332" PostTypeId="2" ParentId="13305" CreationDate="2015-05-05T05:20:59.760" Score="14" Body="&lt;p&gt;Partial answer here, though I'm not entirely sure how hard the rest is supposed to be. Putting spoiler tags on for now.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you take 8 of the 22 spells and shuffle them around, you get this:&#xA;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/zLkwg.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&#xA; Reading down each column, you get the message: &lt;br&gt; &lt;br&gt;&#xA;&#xA; &lt;b&gt; From the dead tree in the swamp, go south. Halfway to Mount Dragon, stop. Where you cross the road is the holy nut. &lt;/b&gt; &lt;br&gt; &lt;br&gt;&#xA;&#xA; The rest so far is taken from @MisterEman22's answer (for completeness), with minor changes. &lt;br&gt; &lt;br&gt;&#xA;&#xA; The dark point on the compass signifies North, so going south would be counterclockwise about 30 degrees from directly down.&#xA; There are 3 dead trees on the map at: (B,e), around (P,p), and (C,q). The only one that is inside a swampy area is the one at (B,e). The others are on &quot;normal&quot; land.  If you go south from there, it would take us to a mountain at (A,h). halfway to that point is a jut in the road, and when crossed you are in the space &lt;strong&gt;(B,g)&lt;/strong&gt;, which is the final location. &lt;br&gt; &lt;br&gt;&#xA;&#xA; -------------------------------------------------- &lt;br&gt; &lt;br&gt;&#xA;&#xA; Thus, the only remaining part of the full solution that is missing is how to properly deduce that these are the correct 8 spells, and in the right order. Although @BmyGuest has stated in the comments that the above partial solution is correct, a full solution is able to deduce how to get to where my partial solution begins. I got here via a shortcut that you could find in an earlier version of the puzzle with slightly different wording in the title and letter, and as stated explicitly by @BmyGuest (and implicitly via the edits), a &quot;proper&quot; solution does not use said shortcut. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1600" LastEditorUserId="1600" LastEditDate="2015-05-20T04:33:28.023" LastActivityDate="2015-05-20T04:33:28.023" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="13333" PostTypeId="2" ParentId="13320" CreationDate="2015-05-05T06:31:56.607" Score="3" Body="&lt;p&gt;My answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Han Solo (from Star Wars)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With a light source, I resemble a star;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The light source is his blaster.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once fortified, I'm alone at the bar. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; His &quot;shield&quot; is Chewbacca. He's alone in the bar when he faces Greedo.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then, come the fifth, I shall sail away -&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Before taking off, the Millennium Falcon's crew must be aboard. The standard one has 5 living beings: Han, Leia, Luke, Obi-Wan, Chewbacca.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-05-05T06:31:56.607" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13334" PostTypeId="1" AcceptedAnswerId="14388" CreationDate="2015-05-05T07:19:00.560" Score="1" ViewCount="836" Body="&lt;p&gt;I found a circle with some letters &lt;em&gt;(the letters are in order)&lt;/em&gt; but two of them are not legible.&lt;br&gt;&#xA;&lt;strong&gt;Who can help me finding the lost letters?&lt;/strong&gt; (not necessarily English)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fGFnE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clue:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Language: Spanish&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12112" LastEditorUserId="12112" LastEditDate="2015-05-06T05:10:36.387" LastActivityDate="2015-05-06T08:10:45.557" Title="I need to find two letters in this circle" Tags="|pattern|letter-sequence|" AnswerCount="1" CommentCount="26" ContentLicense="CC BY-SA 3.0" />
  <row Id="13335" PostTypeId="1" AcceptedAnswerId="13344" CreationDate="2015-05-05T08:37:26.160" Score="6" ViewCount="352" Body="&lt;p&gt;Solve the followng encrypted message.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;--2b1-21-1b112----b1b---21-b--12--1----b1b---b2-12-1--b22---2b21--211---b--121b-121-b21bb--b2-----211-1-b2b1---1b-12----2b21--211---2112b112-12-111-b2-----21b-12----2b21--211---2112b112-12-111-b2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Be brave and frail, and you'll surely be able to see the answer!&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-06T22:20:03.050" Title="Cryptograms in invisible ink" Tags="|cipher|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13336" PostTypeId="2" ParentId="13320" CreationDate="2015-05-05T09:00:53.403" Score="2" Body="&lt;p&gt;It's a bit of a stretch, but it keeps gnawing at the back of my mind so I'll give it a try.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Musical note&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With a light source, I resemble a star;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you need to play softer (&quot;light source&quot;) after a sustained piece, you release the piano pedal. The mark for this resembles a star: &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_musical_symbols#/media/File:Music-pedalup.svg&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/List_of_musical_symbols#/media/File:Music-pedalup.svg&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once fortified, I'm alone at the bar.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is another stretch, but if you read &quot;fortified&quot; as &quot;make it 4/4&quot;, then in a common 4/4 measure the whole note is alone between the bars (&lt;em&gt;semibreve&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then, come the fifth, I shall sail away &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the 4/4 measure above, the arrival of the fifth note marks the end of the previous measure, so the sound slowly &quot;sails away&quot; in the air&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure there are better solutions out there though :)&lt;/p&gt;&#xA;" OwnerUserId="11977" LastActivityDate="2015-05-05T09:00:53.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13337" PostTypeId="2" ParentId="13240" CreationDate="2015-05-05T10:04:30.287" Score="-3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They've just switched HDDs. Bond's OS was replaced with a similar-looking one. The login screen was easily modified (it's open-source anyway) to send login data to attackers over Internet. They have an original HDD, and use his login data to access its contents. Bond and his colleagues gets &quot;No access&quot;, because attackers didn't knew his login data when they made a replacement HDD. If they've modified OS to log in no matter what is the password, this attack could fail, because, for example, paranoid Bond could test wrong password first, and would be thus notified of the attack by the fact he logged in.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So we should remember that&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;BIOS should be set up to load only from an HDD with a certain serial number. But it's still possible to copy a serial number from the old HDD.&lt;/li&gt;&#xA;&lt;li&gt;You should remember serial number, because the motherboard or BIOS chip could be replaced too.&lt;/li&gt;&#xA;&lt;li&gt;Use a good old version of Truecrypt, Bitlocker, whatever. You will see that's not your system by the absence of pre-OS screen. That screen could be falsified too, though.&lt;/li&gt;&#xA;&lt;li&gt;Even when you have HDD encrypted, you still have a keyboard to be physically hacked (who will notice a little Attiny chip and GPRS transceiver inside?), and anyone can do that if they have access to your workplace when you're absent.&lt;/li&gt;&#xA;&lt;li&gt;There's a noise out of your VGA cable that can be received several meters around you, so bad guys will see what data you're collecting, and hidden cameras are all around.&lt;/li&gt;&#xA;&lt;li&gt;Don't ever, EVER leave something or do something anywhere near your enemies.&lt;/li&gt;&#xA;&lt;li&gt;Don't write any more puzzles with infinite ways to answer like this one.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="12152" LastActivityDate="2015-05-05T10:04:30.287" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13339" PostTypeId="1" AcceptedAnswerId="13340" CreationDate="2015-05-05T12:46:36.763" Score="31" ViewCount="108875" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am the first in Earth,&lt;br&gt;&#xA;  The second in Heaven.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I appear twice in a week,&lt;br&gt;&#xA;  Never in a month,&lt;br&gt;&#xA;  Once in a year.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;And I am found in the middle of the sea.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What am I?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4312" LastEditorUserId="-1" LastEditDate="2015-12-09T23:14:22.447" LastActivityDate="2015-12-09T23:14:22.447" Title="I am the first in Earth, What am I?" Tags="|riddle|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13340" PostTypeId="2" ParentId="13339" CreationDate="2015-05-05T12:51:04.963" Score="47" Body="&lt;p&gt;&lt;strong&gt;What am I?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter &quot;E&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the first on Earth,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; E-arth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second in Heaven.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; h-E-aven&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I appear twice in a week,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; w-EE-k&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Never in a month,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; month &amp;lt;--- no E!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once in a year,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; y-E-ar&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I am found in the middle of the sea.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; s-E-a&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12014" LastEditorUserId="12014" LastEditDate="2015-05-05T15:32:42.033" LastActivityDate="2015-05-05T15:32:42.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13341" PostTypeId="2" ParentId="13320" CreationDate="2015-05-05T12:53:13.863" Score="5" Body="&lt;p&gt;I'm thinking of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the Moon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With a light source, I resemble a star;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When it is ligthened by the sun, a satellite can look like a star&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once fortified, I'm alone at the bar.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When it's full moon, we can't see anything else in the sky because of the gleam&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then, come the fifth, I shall sail away &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Full Moon this month is May 4th, the Moon looks like a sail when it decreases&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11877" LastActivityDate="2015-05-05T12:53:13.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13344" PostTypeId="2" ParentId="13335" CreationDate="2015-05-05T13:27:56.773" Score="13" Body="&lt;p&gt;The message reads:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fire and ice, and death was coming. But he was steel. He was steel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(the final line of Chapter 13, &lt;em&gt;Floating like Snow&lt;/em&gt;, in &lt;em&gt;The Path of Daggers&lt;/em&gt;, book 8 of &lt;em&gt;Wheel of Time&lt;/em&gt;; the &quot;he&quot; referred to is Rand al'Thor).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is encrypted like so:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The message is written in braille. I wouldn't have sen that without dmg's comment, but once he had pointed it out and I had seen the other heavy hints, it seemed to be obvious.&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Braille letters are written in a grid of 2&amp;times;3 dots. The message has 195 letters, which is divisible by three. The four characters encode one pair of horizontal dots: The dash means no dot, the &lt;code&gt;1&lt;/code&gt; means the first or left dot; the &lt;code&gt;2&lt;/code&gt; means the second or right dot and the &lt;code&gt;b&lt;/code&gt; means both dots.&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; So the first letter made up of &lt;code&gt;--2&lt;/code&gt; is &lt;code&gt;⠠&lt;/code&gt;, which means the next letter is capital. The next letter &lt;code&gt;b1-&lt;/code&gt; is &lt;code&gt;⠋&lt;/code&gt; or F. The message also makes use of abbreviations like &lt;code&gt;⠯&lt;/code&gt; for &quot;and&quot;. &lt;a href=&quot;http://en.wikipedia.org/wiki/English_Braille&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; has a detailed description on how Braille is used for English.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9860" LastEditorUserId="5373" LastEditDate="2015-05-06T22:20:03.050" LastActivityDate="2015-05-06T22:20:03.050" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13345" PostTypeId="1" AcceptedAnswerId="13349" CreationDate="2015-05-05T13:32:23.293" Score="4" ViewCount="204" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Refer to following diagram and tell what will come in place of two '?'&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/E1alj.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Note : An explanation is required about the pattern...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10433" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-05T13:54:32.950" Title="What will come in place of?" Tags="|pattern|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13346" PostTypeId="1" AcceptedAnswerId="13347" CreationDate="2015-05-05T13:40:10.413" Score="2" ViewCount="777" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If you have me, you'll want more of me.&lt;br&gt;&#xA;  But you won't remember getting me.&lt;br&gt;&#xA;  Sometimes you'll want to give me.&lt;br&gt;&#xA;  But if I'm taken, you'll never want me again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="11851" LastActivityDate="2015-05-05T13:46:37.120" Title="If you have me, you'll want more of me." Tags="|riddle|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13347" PostTypeId="2" ParentId="13346" CreationDate="2015-05-05T13:45:25.807" Score="7" Body="&lt;p&gt;Well, my regular answer of 'money' doesn't work here, so I guess I ought to try something else...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; life?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you have me, you'll want more of me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Almost everyone wants to continue living (that is to say, not a lot of people are eager to die).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But you won't remember getting me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You don't remember your own birth (unless you're some sort of crazy memory freak, but I'd call that a corner case).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes you'll want to give me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people want to have children, to whom they &quot;give life&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But if I'm taken, you'll never want me again.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can't want something if your life is taken (i.e. you're dead)!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="11475" LastEditDate="2015-05-05T13:46:37.120" LastActivityDate="2015-05-05T13:46:37.120" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13348" PostTypeId="2" ParentId="13346" CreationDate="2015-05-05T13:45:39.750" Score="0" Body="&lt;p&gt;My first guess is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Loan (of some kind like money)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you have me, you'll want more of me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;of course we want more part of it as insured&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But you won't remember getting me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Sometimes duration is so long for settlement that we wont remember it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes you'll want to give me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Those who are capable can trade using it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But if I'm taken, you'll never want me again.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Once all money settled we wont need it anymore&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10433" LastActivityDate="2015-05-05T13:45:39.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13349" PostTypeId="2" ParentId="13345" CreationDate="2015-05-05T13:46:18.527" Score="5" Body="&lt;p&gt;My answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;G and 11&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;On each section, there are two alternating formats:&lt;br&gt;&#xA;&lt;br /&gt;A(1) Y(25) C(3) - 26&lt;br&gt;&#xA;E(5) U(21) ?(?) - 22&lt;br&gt;&#xA;I(9) Q(17) K(11) - 18&lt;br&gt;&#xA;&lt;br /&gt;X(24) D(4) V(22) - 3&lt;br&gt;&#xA;T(20) H(8) R(18) - 7&lt;br&gt;&#xA;P(16) L(12) N(14) - ?&lt;br&gt;&#xA;&lt;br /&gt;As you can see, with the first format, from the first letter to the third, the index of the letter is increased by 2.  Also, from the second letter to the number in the middle, the index is increased by 1.&lt;br&gt;&#xA;With the second format, the opposite is true (first and third letters are decreased by 2, second letter and middle number are decreased by 1).  With that, the missing values that satisfy the pattern are:&lt;br&gt;&#xA;&lt;br /&gt;E(5) U(21) G(7) - 22&lt;br&gt;&#xA;P(16) L(12) N(14) - 11  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11851" LastEditorUserId="11851" LastEditDate="2015-05-05T13:54:32.950" LastActivityDate="2015-05-05T13:54:32.950" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13350" PostTypeId="1" AcceptedAnswerId="13351" CreationDate="2015-05-05T14:05:57.523" Score="30" ViewCount="6585" Body="&lt;p&gt;I'm 26&lt;br&gt;&#xA;I'm usually cool&lt;br&gt;&#xA;Sometimes I am hot, steaming hot&lt;br&gt;&#xA;The Male me have a shining heart and can fly&lt;br&gt;&#xA;The Young female me is, however, horrible and brings a lot of pain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?  &lt;/p&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="5188" LastEditDate="2015-05-07T04:41:51.277" LastActivityDate="2015-05-07T04:41:51.277" Title="I'm 26, what am I? - riddle" Tags="|riddle|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="13351" PostTypeId="2" ParentId="13350" CreationDate="2015-05-05T14:16:32.707" Score="31" Body="&lt;p&gt;Well, you must be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Iron!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm 26&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The element Iron has an atomic number of 26.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm usually cool&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Iron is a metal, typically cold to the touch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am hot, steaming hot&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I expect this is referring to an iron that you would use to iron clothes, which are incredibly hot.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Male me have a shining heart and can fly&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is clearly Iron Man, who has his suit featuring the glowing core and the ability to fly with thrusters!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Young female me is, however, horrible and brings a lot of pain&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well I'm guessing this is referring to Iron Maiden, but I don't know why they're so horrible - they do sing about painful stuff, though. (Or maybe I'm just stupid and this is referring to the torture device, which is clearly both horrible and painful. Thanks dmg and Laconic Droid!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="11475" LastEditDate="2015-05-05T14:26:35.167" LastActivityDate="2015-05-05T14:26:35.167" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13352" PostTypeId="1" CreationDate="2015-05-05T14:17:49.640" Score="11" ViewCount="1248" Body="&lt;p&gt;This puzzle is part of a series - &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4075/the-serial-killer-part-1&quot;&gt;part 1 is here&lt;/a&gt;, &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4263/the-serial-killer-part-2&quot;&gt;part 2 is here&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4449/the-serial-killer-part-3&quot;&gt;part 3 is here&lt;/a&gt;. Any comments I make on the OP are canonical and &lt;a href=&quot;http://tvtropes.org/pmwiki/pmwiki.php/Main/OutOfCharacter&quot; rel=&quot;nofollow noreferrer&quot;&gt;out of character (OOC)&lt;/a&gt;. Any comments I post on answers are for flavour only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may notice there's been a bit of a delay on part 4 turning up. I think I just needed some time :-) I've kind of lost my feel for difficulty (although seeing how part 2 went, maybe I never had it :P) so if this is way too difficult I'll bust out some clues later on.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&quot;gngnggfgl... whassat?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You suddenly wake up from a dream about a really, REALLY warm glove that you'd just invented, only to discover that the neighbour's cat is asleep on your hand. Groggily, you pull your hand out from under it. the cat jumps up, startled, gives a loud &quot;Meow!&quot; and takes off, knocking your mug of now-cold coffee all over your keyboard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Biting back a swear word (this is a child-friendly site, don'tcherknow) you try to shake off the groggy feeling in your head. Your vision blurs a bit and you feel dizzy, so you stop quickly. Wait, why do you feel dizzy? What's that dull pain at the back of your neck? Maybe you're just missing your daily fix of riddles on P.SE - that must be it. You've been awake for almost a whole minute... probably... and you still haven't found any magical spaghetti recipes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hang on! There was something wrong with SE wasn't there? After a brief moment of nausea, you remember - some nut-job was attacking the Stack and you were going to track him down and give him what-for. How could you have forgotten? You waggle your mouse a bit to wake the computer up, but nothing happens. You look under the table and- holy moly, your computer's disappeared!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking around your room, you realise the place is an absolute tip. Well, more so than normal; if it was just old plates sitting on your desk and clothes piled on the floor then that would be normal. The door is wide open, your drawers are all opened and it looks like someone's gone through all your stuff looking for something. Looking back at your keyboard, you notice that there are some keys which are missing - apparently someone didn't want you typing anything about Marin Cilic, because the C, I and L keys are gone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This must be related to the SErial Killer - he knows you're on to him and he's trying to stop you!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that pain in your head makes sense - someone must have broken into your room (you may have heard them if you weren't so busy listening to Rick Astley, maybe) and knocked you unconscious before stealing your computer and maybe even your lucky pair of Y-fronts! You wonder how long you've been blacked out for, but there's no time to waste - a world where the SErial Killer is at large (and possibly in possession of your pants) doesn't bear thinking about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to act!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Frantically, you hammer on your next-door neighbour's front door. When they open it, you gabble hysterically at them, then barge past (knocking them to the floor) and rush to their computer. You pull up SE and head to Area 51, but it looks completely and utterly normal. On second thoughts, not really; all the proposals have just over 2,000 followers and the site header says &quot;Area 151&quot;. As you scroll through the list of site proposals, it dawns on you that your neighbour doesn't normally wear a long brown coat and a fedora...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With very little time before the mysterious impostor picks themselves up and get very angry at you, you must find a clue, some way to pick up the trail and come closer to unmasking the SErial Killer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who should you try to contact?&lt;/p&gt;&#xA;" OwnerUserId="2518" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2018-06-10T07:29:15.577" Title="The SErial Killer (Part 4)" Tags="|mystery|" AnswerCount="4" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="13353" PostTypeId="2" ParentId="13321" CreationDate="2015-05-05T15:31:16.523" Score="3" Body="&lt;p&gt;No complicated math needed:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;6 is an instant win while 5 is an instant loss. Any other outcome will either do nothing or switch turns. The chance remains the same:&#xA; &lt;strong&gt;50%&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For a D20 die, the chance is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; still &lt;strong&gt;50%&lt;/strong&gt;, following the same logic. Anything but the two highest numbers have absolutely no influence on the outcome of the game.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11992" LastEditorUserId="11992" LastEditDate="2015-05-05T19:58:37.047" LastActivityDate="2015-05-05T19:58:37.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13355" PostTypeId="2" ParentId="13352" CreationDate="2015-05-05T15:44:50.083" Score="8" Body="&lt;p&gt;This seems too simple, so there's probably more to it - but the 151st user on Area51 in terms of reputation right now is &lt;a href=&quot;http://area51.stackexchange.com/users/46143/kevin&quot;&gt;Kevin&lt;/a&gt;, who, as it happens, proposed to add a &quot;Puzzling.SE&quot; site about a year ago. Seems simple but also too much of a coincidence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, CLI in roman numerals is 151, but I'm not sure what that means. Maybe just an additional hint that you really need that number?&lt;/p&gt;&#xA;" OwnerUserId="439" LastEditorUserId="439" LastEditDate="2015-05-05T16:01:45.400" LastActivityDate="2015-05-05T16:01:45.400" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13356" PostTypeId="1" AcceptedAnswerId="13360" CreationDate="2015-05-05T15:54:08.890" Score="3" ViewCount="1374" Body="&lt;p&gt;If : &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;888888 = T &#xA;&#xA;777777 = O&#xA;&#xA;555555 = Y&#xA;&#xA;444444 = R&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;then :&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;666666 = ? &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Basically you need to tell the logic along with the solution to the puzzle. &lt;/p&gt;&#xA;" OwnerUserId="10091" LastActivityDate="2015-05-05T17:12:55.763" Title="Alphabet Number Matching" Tags="|mathematics|logical-deduction|number-sequence|" AnswerCount="1" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="13357" PostTypeId="2" ParentId="13257" CreationDate="2015-05-05T16:38:01.190" Score="3" Body="&lt;p&gt;So here's a start:&lt;br&gt;&#xA;Hints have turned brain to mush. Zombies have rejected me. I'm just winging it. Fits hint 1. Half fits hint 3 and 4, does not fit hint 2 because no one likes hint 2.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;At first I was all alone.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Then a friend showed up, and the place started showing signs of life.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Is&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;But then another friend came, and the three of us started fighting.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sin&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Another friend came, and calmed us down by helping us communicate.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sing/Sign&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Yet another friend came along -- by now, we ruled the area!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Kings&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;By the time a sixth friend joined us, the game had changed a little, and we were looking weak next to the competition.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Asking (found asking) //but A doesn't fit the second clue.  The letter would have to be somewhere between E and U&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Our seventh friend came and helped us win some resounding successes.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sacking (the quarterback in football)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Finally our last member arrived, and we were just riding free.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Packings(?) (Packing could have made sense but the S needs to be somewhere.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastEditorUserId="11251" LastEditDate="2015-05-06T19:53:57.293" LastActivityDate="2015-05-06T19:53:57.293" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="13358" PostTypeId="2" ParentId="13257" CreationDate="2015-05-05T16:47:56.140" Score="3" Body="&lt;p&gt;Since this puzzle appears somewhat difficult, I might as well post my partial answer too.  Feel free to take my ideas and finish it off if you can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first I was all alone.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter O by itself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then a friend showed up, and the place started showing signs of life.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; W showed up to form OW.  Only living beings can feel pain and say &quot;OW&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But then another friend came, and the three of us started fighting.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; R came to form ROW.  A row is an argument.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another friend came, and calmed us down by helping us communicate.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; D came to form WORD.  You use words to communicate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet another friend came along -- by now, we ruled the area!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C came to form CROWD.  You could say a crowd of people rule an area.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By the time a sixth friend joined us, the game had changed a little, and we were looking weak next to the competition.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A joined them to form COWARD.  They were weak and cowardly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's all I have.&lt;/p&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2015-05-05T16:47:56.140" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="13360" PostTypeId="2" ParentId="13356" CreationDate="2015-05-05T17:12:55.763" Score="18" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; X&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 8 X 6 = FORTY EIGH(T)&#xA; 7 X 6 = FOURTY TW(O)&#xA; 5 X 6 = THIRT(Y)&#xA; 4 X 6 = TWENTY FOU(R)&#xA; 6 X 6 = THIRTY SI(X)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4177" LastActivityDate="2015-05-05T17:12:55.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13361" PostTypeId="2" ParentId="11551" CreationDate="2015-05-05T17:41:15.730" Score="1" Body="&lt;p&gt;The image rebus has three parts&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;img src=&quot;https://i.stack.imgur.com/6kw59.png&quot; alt=&quot;enter image description here&quot;&gt; - This refers to &lt;a href=&quot;https://www.google.co.in/search?q=code&amp;amp;es_sm=93&amp;amp;source=lnms&amp;amp;tbm=isch&amp;amp;sa=X&amp;amp;ei=Xf9IVb3_BNWcugSfn4HgDg&amp;amp;ved=0CAcQ_AUoAQ&amp;amp;biw=1366&amp;amp;bih=667#tbs=isz:i&amp;amp;tbm=isch&amp;amp;q=code&amp;amp;imgrc=_hFfLnkt5l4_TM%253A%3BwooKFzwtWZ-sLM%3Bhttps%253A%252F%252Fcdn3.iconfinder.com%252Fdata%252Ficons%252Fseo-internet-marketing-flat-icons%252F128%252Fweb-code.png%3Bhttps%253A%252F%252Fwww.iconfinder.com%252Fsearch%252F%253Fq%253Dcode%3B128%3B128&quot; rel=&quot;nofollow noreferrer&quot;&gt;Code&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&quot;&quot; - That refers to quote&lt;/li&gt;&#xA;&lt;li&gt;The Python&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Joining them in order: Find the quote in the Python Code&lt;/p&gt;&#xA;" OwnerUserId="4177" LastActivityDate="2015-05-05T17:41:15.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13363" PostTypeId="1" AcceptedAnswerId="13364" CreationDate="2015-05-05T18:39:16.343" Score="2" ViewCount="440" Body="&lt;p&gt;So big so small,&lt;br&gt;&#xA;So old, so new&lt;br&gt;&#xA;So short, so tall.&lt;br&gt;&#xA;Dry, or covered in dew.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Used everywhere,&lt;br&gt;&#xA;Destroyed but saved.&lt;br&gt;&#xA;Happiness or despair,&lt;br&gt;&#xA;When I am razed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trust me, I know&lt;br&gt;&#xA;Really, I'm near.&lt;br&gt;&#xA;Every arrow and bow&lt;br&gt;&#xA;Earned its shape, dear.  &lt;/p&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="9951" LastEditDate="2015-05-05T18:43:08.480" LastActivityDate="2015-05-05T18:46:25.087" Title="Such contrast - What am I?" Tags="|riddle|word|rhyme|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13364" PostTypeId="2" ParentId="13363" CreationDate="2015-05-05T18:42:01.037" Score="6" Body="&lt;p&gt;It could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a tree or a forest of trees&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would add explanations, but it's all self-explanatory.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; also, the first letters of the last stanza spell TREE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastActivityDate="2015-05-05T18:42:01.037" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13365" PostTypeId="2" ParentId="13363" CreationDate="2015-05-05T18:42:26.990" Score="2" Body="&lt;p&gt;I think it's a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tree&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So big so small,&#xA;So old, so new&#xA;So short, so tall.&#xA;Dry, or covered in dew.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Those adjectives perfectly fit for a tree.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Used everywhere,&#xA;Destroyed but saved.&#xA;Happiness or despair,&#xA;When I am razed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wood has many applications. Forests are destroyed by many, protected by others. Cutting a tree is bad for the planet, but good for paper industry.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Trust me, I know&#xA;Really, I'm near.&#xA;Every arrow and bow&#xA;Earned its shape, dear.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Arrows and bows are made of wood.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Easter Egg:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the last stanza, the initials of each line form the word &quot;Tree&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-05-05T18:46:25.087" LastActivityDate="2015-05-05T18:46:25.087" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13366" PostTypeId="1" AcceptedAnswerId="14395" CreationDate="2015-05-05T19:09:32.693" Score="-7" ViewCount="506" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm usually white,&lt;br&gt;&#xA;  but &lt;strong&gt;right now&lt;/strong&gt; a striped one dominates me.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;For on this &lt;em&gt;day&lt;/em&gt; I wear black,&lt;br&gt;&#xA;  and black is still my day.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;At the present moment: Who am I?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;hr&gt;&#xA;Clue 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Related to sports&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clue 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The time in which I created this puzzle is important&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clue 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Sport is: Futbol / Football / Soccer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12112" LastEditorUserId="8874" LastEditDate="2015-10-06T12:19:49.367" LastActivityDate="2015-10-06T12:22:36.713" Title="At the present moment: Who am I?" Tags="|riddle|" AnswerCount="3" CommentCount="20" ContentLicense="CC BY-SA 3.0" />
  <row Id="13367" PostTypeId="2" ParentId="13366" CreationDate="2015-05-05T19:42:08.547" Score="0" Body="&lt;p&gt;It may be a long-shot, but my answer is  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A skunk  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm white, but right now!!! one striped dominates me  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a skunk's skin is white (well, pinkish)&lt;br&gt;&#xA; a stereotypical skunk has one long stripe down its back &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Today I wear black and likewise is still my day  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A skunk's fur is mostly black but for the white stripe&lt;br&gt;&#xA; When people encounter a skunk they stay as still as possible. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not sure it fits perfectly, but I tried.&lt;/p&gt;&#xA;" OwnerUserId="7715" LastActivityDate="2015-05-05T19:42:08.547" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13368" PostTypeId="1" AcceptedAnswerId="14390" CreationDate="2015-05-05T19:42:39.133" Score="2" ViewCount="2578" Body="&lt;p&gt;Two merchants have 5 camels each. One fine day, one of the merchants sells all of his goods in the market and heads back to the village. He has to cross a bridge to reach his destination, and the bridge is only wide enough for 1 camel at a time.&lt;/p&gt;&#xA;&lt;p&gt;At the same time, the other merchant is heading to the market to sell his goods and must cross the same bridge. Alas! They meet up in the middle of the bridge one camel-length from each other. They are both stubborn and won't back up their camels. Fortunately, the first merchant's camels are spry and unladen and are able to jump over another camel. &lt;strong&gt;Can both merchants get their camels across the bridge?&lt;/strong&gt; (Both merchants' camels can jump)&lt;/p&gt;&#xA;" OwnerUserId="12169" LastEditorUserId="60644" LastEditDate="2023-07-17T07:26:19.383" LastActivityDate="2023-07-17T07:26:19.383" Title="Camels crossing a bridge" Tags="|logical-deduction|" AnswerCount="7" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="13370" PostTypeId="2" ParentId="13368" CreationDate="2015-05-05T20:09:27.257" Score="1" Body="&lt;p&gt;I'm pretty sure it cannot be done. At best, one of the two merchants can get one of his camels across the bridge. This leaves the other 9 camels at a standstill on the bridge though.&lt;/p&gt;&#xA;" OwnerUserId="7715" LastActivityDate="2015-05-05T20:09:27.257" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13371" PostTypeId="2" ParentId="12934" CreationDate="2015-05-05T20:11:32.470" Score="2" Body="&lt;p&gt;Let's try another answer (get some more activity back here). This is largely inspired by &lt;a href=&quot;http://meta.puzzling.stackexchange.com/questions/2971/lateral-thinking-requires-a-new-policy&quot;&gt;your recent post&lt;/a&gt;. Are you: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Lateral-Thinking Question  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What's happening is very strange,&lt;br&gt;&#xA; I belong to another StackExchange!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is some thought of moving lateral-thinking to Area 51.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm indeed off-topic here in Puzzling,&lt;br&gt;&#xA; It isn't always clear what I'm saying.    &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some consider lateral-thinking questions to be off topic as they aren't always good puzzles.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Even though I'm often too broad,&lt;br&gt;&#xA; Review it, something is flawed.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The main issue with lateral-thinking is that they are often so broad that there are so many answers it isn't even funny.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm just suggesting an edit,&lt;br&gt;&#xA; A low quality job, ain't it?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most lateral-thinking could be improved with a nice edit to narrow down the possibilities.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For new users I'm a big trouble&lt;br&gt;&#xA; A duplicate isn't always a double.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A lot of new users seem to try lateral-thinking and get some downvotes. Also, a lot of lateral-thinking puzzles seem to be similar.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your current work you shall reopen&lt;br&gt;&#xA; Flag here, flag there, it's all broken!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They can be reopened if they are fixed.  A lot get flagged for deletion.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Oh, don't forget to add a comment,&lt;br&gt;&#xA; late it's not, takes just a moment!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most could be improved if they were commented on where the issue is.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-05-05T20:11:32.470" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13372" PostTypeId="1" AcceptedAnswerId="14386" CreationDate="2015-05-05T20:13:54.343" Score="11" ViewCount="1593" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Test run, take two. The first run went well, so let's keep it up!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;&lt;em&gt;For your second riddle, things might just get trickier -&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;With some common bond, things might just get stickier!&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;When I'm with a group I'll be holding you tight;&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;When alone, I'll protect you in the absence of light.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What am I?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="8874" LastEditDate="2015-10-06T12:14:05.390" LastActivityDate="2015-10-06T12:14:05.390" Title="Word of the Day #2" Tags="|riddle|word|rhyme|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13373" PostTypeId="2" ParentId="13368" CreationDate="2015-05-05T20:14:41.390" Score="2" Body="&lt;p&gt;Yes, if you can switch materials between adjacent camels allowing others to jump.  Call Merchant 1 Camels A-E Merchant 2 Camels F-J. F moves forward. A jumps him. Repeat with F moving forward and B through E jumping when needed. Then G materials moved onto B. G is now unladen and can jump B. Move materials back onto G then move B forward. Repeat process with every camel. Otherwise you just get stuck after trying to move forward in any way.&lt;/p&gt;&#xA;" OwnerUserId="12170" LastActivityDate="2015-05-05T20:14:41.390" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13374" PostTypeId="2" ParentId="13368" CreationDate="2015-05-05T20:17:57.060" Score="0" Body="&lt;p&gt;&lt;b&gt;Yes&lt;/b&gt;, they can get their camels across, assuming that the camels can jump straight up without having to move forward. All they have to do it have the first camel jump while camel the across from it moves forward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ie. [12345_ABCDE] becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1234_5ABCDE] turns into 1234_(5/A)_BCDE where 5 is jumping upwards, while A is moving forward&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[123_A5BCDE] where A has passed under 5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[123A_(5/B)CDE]  etc...&lt;/p&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-05-05T20:33:11.543" LastActivityDate="2015-05-05T20:33:11.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13375" PostTypeId="2" ParentId="13368" CreationDate="2015-05-05T20:28:58.490" Score="0" Body="&lt;p&gt;There are 2 ways of thinking of this: either the light camels can &quot;switch&quot; with a heavy camel, or they have to jump over a heavy camel to an empty space.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the former case, all the camels can get across to the other sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the latter case, only 1 heavy camel can get across as is. It would take 4 starting spaces between the groups to get all the camels across.&lt;br&gt;&#xA;Effectively, the heavy group needs to be able to spread out with a space between each (4 spaces for 5 camels). &lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-05-05T20:28:58.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14375" PostTypeId="2" ParentId="13372" CreationDate="2015-05-05T23:23:20.353" Score="4" Body="&lt;p&gt;I think it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Phosphorus&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With some common bond, things might just get stickier!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Maybe refers to how sugar-phosphates form the backbone of DNA and RNA, helping it &quot;stick&quot; together, or maybe the &quot;sugar&quot; component, making it sticky.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I'm with a group I'll be holding you tight;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This could mean like the phospholipids(that have phosphorus) in the cell membrane, which holds it together.  Of course, people are made up of cells.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When alone, I'll protect you in the absence of light.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Phosphorus is used in the heads of matches, which would help you in the dark.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not really sure… Just a guess&lt;/p&gt;&#xA;" OwnerUserId="11790" LastActivityDate="2015-05-05T23:23:20.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14376" PostTypeId="2" ParentId="13352" CreationDate="2015-05-05T23:30:28.613" Score="1" Body="&lt;p&gt;You should visit&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;a href=&quot;https://apple.stackexchange.com/&quot;&gt;Apple Stack Exchange&lt;/a&gt;, Ask Different.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because it was &lt;a href=&quot;http://area51.stackexchange.com/proposals/151&quot;&gt;proposal number 151&lt;/a&gt; at Area 51. This fits both with the 'Area 151' clue and with the missing C, L, I keys (as EagleV_Attnam pointed out, 151 is CLI in Roman numerals). The &quot;long brown coat and a fedora&quot; seems to be just a red herring.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:45:30.560" LastActivityDate="2015-05-05T23:30:28.613" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14377" PostTypeId="2" ParentId="13372" CreationDate="2015-05-06T00:08:45.613" Score="4" Body="&lt;p&gt;The only thing that gets into my mind is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A blanket&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With some common bond, things might just get stickier!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you sleep together with a person, under the same blanket, that means you two are very close, probably couples.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I'm with a group I'll be holding you tight;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Together with other people under a blanket, it sure feels cramped and tight.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When alone, I'll protect you in the absence of light.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; During the night, the blanket protects you from the cold or insects.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12174" LastActivityDate="2015-05-06T00:08:45.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14380" PostTypeId="1" AcceptedAnswerId="14384" CreationDate="2015-05-06T03:02:08.427" Score="-1" ViewCount="486" Body="&lt;p&gt;This is a slight variation on an old puzzle that's likely already here, but be sure to check it before you mark it as duplicate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the next three terms in this sequence, and why?  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 1  &#xA; 11  &#xA; 21  &#xA; 1211  &#xA; 1231  &#xA; 131221  &#xA; 132231  &#xA; 232221  &#xA; ...  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;This is not the Look-and-say sequence as the duplicate links are. It is a variation on the idea of them, but it is not a duplicate in my opinion.&lt;/p&gt;&#xA;" OwnerUserId="7715" LastEditorUserId="8874" LastEditDate="2015-10-06T12:26:30.350" LastActivityDate="2015-10-06T12:26:30.350" Title="Number Sequence Puzzle" Tags="|pattern|number-sequence|" AnswerCount="2" CommentCount="4" ClosedDate="2015-05-06T06:48:39.023" ContentLicense="CC BY-SA 3.0" />
  <row Id="14381" PostTypeId="2" ParentId="14380" CreationDate="2015-05-06T03:03:27.317" Score="3" Body="&lt;p&gt;This is like that look-say sequence thing. Each line describes the numbers of the line before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First line has one 1, so the second line is 11. The second line has 2 1's, so the third line is 21. The third line has one two and one one, so the fourth line is 1211 etc.&lt;/p&gt;&#xA;" OwnerUserId="11154" LastActivityDate="2015-05-06T03:03:27.317" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14382" PostTypeId="2" ParentId="13366" CreationDate="2015-05-06T03:29:44.460" Score="-1" Body="&lt;p&gt;Is the answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A turn 10 spare in bowling? Will elaborate if correct.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastActivityDate="2015-05-06T03:29:44.460" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14384" PostTypeId="2" ParentId="14380" CreationDate="2015-05-06T05:15:55.477" Score="6" Body="&lt;p&gt;To answer the question more precisely, these are the next three terms:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;134211&lt;br&gt;14131231&lt;br&gt;14231241&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a variant of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Look-and-say_sequence&quot; rel=&quot;nofollow&quot;&gt;Look-and-say sequence&lt;/a&gt;. This particular variant is &lt;a href=&quot;https://oeis.org/A007890&quot; rel=&quot;nofollow&quot;&gt;already in OEIS&lt;/a&gt;! It is formed by summarizing the terms of the previous number, but all at once in descending order, which is different from the look and say sequence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the sequence goes from 1211 to 1231 because:&lt;br&gt;&#xA;The previous term contained one two. (12)&lt;br&gt;&#xA;The previous term contained three ones. (31)&lt;br&gt;&#xA;&lt;strong&gt;-&gt; 1231&lt;/strong&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example, the sequence goes from 132231 to 232221 because:&lt;br&gt;&#xA;The previous term contained two threes. (23)&lt;br&gt;&#xA;The previous term contained two twos. (22)&lt;br&gt;&#xA;The previous term contained two ones. (21)&lt;br&gt;&#xA;&lt;strong&gt;-&gt; 232221.&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="7308" LastEditorUserId="7308" LastEditDate="2015-05-06T23:17:38.440" LastActivityDate="2015-05-06T23:17:38.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14386" PostTypeId="2" ParentId="13372" CreationDate="2015-05-06T06:30:52.640" Score="13" Body="&lt;p&gt;I'm not too sure, but I'll guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rubber&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With some common bond, things might just get stickier!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rubber cement (a type of adhesive)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I'm with a group I'll be holding you tight;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rubber band (group = band)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When alone, I'll protect you in the absence of light.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rubber (i.e. condom, &quot;protection&quot; at night time, alone = no other word)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4479" LastActivityDate="2015-05-06T06:30:52.640" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14387" PostTypeId="2" ParentId="13368" CreationDate="2015-05-06T06:55:47.253" Score="0" Body="&lt;p&gt;Only way this can be done is if we don't think of it as a &quot;one move per turn&quot; problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the unladen camel can jump to position where the pack-loaded camel is currently standing and at the same time the pack loaded camel must move with haste to the former position of the jumping camel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will switch their positions and becomes sort of bubble-sort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has been edited so both merchant's camels can jump now. This changes the puzzle completely.&lt;/p&gt;&#xA;" OwnerUserId="9907" LastEditorUserId="9907" LastEditDate="2015-05-06T08:54:17.633" LastActivityDate="2015-05-06T08:54:17.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14388" PostTypeId="2" ParentId="13334" CreationDate="2015-05-06T08:10:45.557" Score="4" Body="&lt;p&gt;So i am posting my answer removing from comment&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; U and I  which forms the word CIRCULO (Spanish)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10433" LastActivityDate="2015-05-06T08:10:45.557" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14389" PostTypeId="2" ParentId="13372" CreationDate="2015-05-06T08:40:44.883" Score="8" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stick&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With some common bond, things might just get stickier!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Glue Stick&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I'm with a group I'll be holding you tight;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As in the idiom &lt;code&gt;stick to&lt;/code&gt;. To Stick to a group&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When alone, I'll protect you in the absence of light.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A stick helps to navigate when its pitch dark. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4177" LastActivityDate="2015-05-06T08:40:44.883" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14390" PostTypeId="2" ParentId="13368" CreationDate="2015-05-06T08:56:59.027" Score="8" Body="&lt;p&gt;Depending on how well I understood the question it is possible.&lt;/p&gt;&#xA;&lt;p&gt;It is possible if either a) any camel can jump over any other camel (based on &amp;quot;Both merchant camels can jump&amp;quot;) or b) if the camels can walk backwards as long as they never end up behind where they started.&lt;/p&gt;&#xA;&lt;h3&gt;a:&lt;/h3&gt;&#xA;&lt;pre&gt;&lt;code&gt;XXXXX_OOOOO&#xA;XXXX_XOOOOO&#xA;XXXXOX_OOOO&#xA;XXXXOXO_OOO&#xA;XXXXO_OXOOO&#xA;XXX_OXOXOOO&#xA;XX_XOXOXOOO&#xA;XXOX_XOXOOO&#xA;XXOXOX_XOOO&#xA;XXOXOXOX_OO&#xA;XXOXOXOXO_O&#xA;XXOXOXO_OXO&#xA;XXOXO_OXOXO&#xA;XXO_OXOXOXO&#xA;X_OXOXOXOXO&#xA;_XOXOXOXOXO&#xA;OX_XOXOXOXO&#xA;OXOX_XOXOXO&#xA;OXOXOX_XOXO&#xA;OXOXOXOX_XO&#xA;OXOXOXOXOX_&#xA;OXOXOXOXO_X&#xA;OXOXOXO_OXX&#xA;OXOXO_OXOXX&#xA;OXO_OXOXOXX&#xA;O_OXOXOXOXX&#xA;OO_XOXOXOXX&#xA;OOOX_XOXOXX&#xA;OOOXOX_XOXX&#xA;OOOXOXOX_XX&#xA;OOOXOXO_XXX&#xA;OOOXO_OXXXX&#xA;OOO_OXOXXXX&#xA;OOOO_XOXXXX&#xA;OOOOOX_XXXX&#xA;OOOOO_XXXXX&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;h3&gt;B (X can jump):&lt;/h3&gt;&#xA;&lt;pre&gt;&lt;code&gt;XXXXX_OOOOO&#xA;XXXXXO_OOOO&#xA;XXXX_OXOOOO&#xA;XXXXOXO_OOO&#xA;XXXXO_OXOOO&#xA;XXX_OXOXOOO&#xA;XXXOXOXO_OO&#xA;XXXOXO_OXOO&#xA;XXXO_OXOXOO&#xA;XX_OXOXOXOO&#xA;XXOXOXOXO_O&#xA;XXOXOXO_OXO&#xA;XXOXO_OXOXO&#xA;XXO_OXOXOXO&#xA;X_OXOXOXOXO&#xA;XOXOXOXOXO_&#xA;XOXOXOXO_OX&#xA;XOXOXO_OXOX&#xA;XOXO_OXOXOX&#xA;XO_OXOXOXOX&#xA;_OXOXOXOXOX&#xA;OXOXOXOXOX_&#xA;OXOXOXOXO_X&#xA;OXOXOXO_OXX&#xA;OXOXO_OXOXX&#xA;OXO_OXOXOXX&#xA;O_OXOXOXOXX&#xA;OOXOXOXO_XX&#xA;OOXOXO_OXXX&#xA;OOXO_OXOXXX&#xA;OO_OXOXOXXX&#xA;OOOXOXO_XXX&#xA;OOOXO_OXXXX&#xA;OOO_OXOXXXX&#xA;OOOOXO_XXXX&#xA;OOOO_OXXXXX&#xA;OOOOO_XXXXX&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;It would all have been easier if they'd just backed up a bit.&lt;/p&gt;&#xA;" OwnerUserId="4577" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-06T08:56:59.027" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14392" PostTypeId="1" AcceptedAnswerId="14396" CreationDate="2015-05-06T12:03:56.280" Score="5" ViewCount="1981" Body="&lt;p&gt;What I am&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;People call me an animal&lt;/p&gt;&#xA;&lt;p&gt;Since I possess a tail&lt;/p&gt;&#xA;&lt;p&gt;but it's in the opposite direction.&lt;/p&gt;&#xA;&lt;p&gt;I can't travel beyond some point&lt;/p&gt;&#xA;&lt;p&gt;Since I am bound by my tail.&lt;/p&gt;&#xA;&lt;p&gt;Sometimes I don't have a tail&lt;/p&gt;&#xA;&lt;p&gt;But still I can not travel that far.&lt;/p&gt;&#xA;&lt;p&gt;Go to the right and the world is easy for you.&lt;/p&gt;&#xA;&lt;p&gt;Go to the left and change your mode.&lt;/p&gt;&#xA;&lt;p&gt;You touch me often, but I won't bite you.&lt;/p&gt;&#xA;&lt;p&gt;So what am I?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10433" LastEditorUserId="69582" LastEditDate="2021-04-23T16:18:51.530" LastActivityDate="2021-04-23T16:18:51.530" Title="What I am - an animal (may be)?" Tags="|riddle|word|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14393" PostTypeId="2" ParentId="12934" CreationDate="2015-05-06T12:26:27.260" Score="3" Body="&lt;p&gt;You are an&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; IDE Debugger&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What's happening is very strange,&#xA;I belong to another StackExchange!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Typically people would ask about problems found in the debugger on StackOverflow.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm indeed off-topic here in Puzzling,&#xA;It isn't always clear what I'm saying.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sometimes the error given is fairly vague and unhelpful.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Even though I'm often too broad,&#xA;Review it, something is flawed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The vague error message may be hard to interpret, but there is definitely something wrong that must be looked over.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm just suggesting an edit,&#xA;A low quality job, ain't it?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The message is telling you to change something in the code.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For new users I'm a big trouble&#xA;A duplicate isn't always a double.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Beginner programmers may have a hard time figuring out what the error is, especially when they've gotten the same error twice with a different cause.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your current work you shall reopen&#xA;Flag here, flag there, it's all broken!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Go back to your code, you need to check all the marked areas for bugs. Sometimes a single typo, like an extra or missing &lt;code&gt;}&lt;/code&gt;, will cause huge amounts of code to be highlighted as 'broken'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Oh, don't forget to add a comment,&#xA;late it's not, takes just a moment!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you fix the bug, add a comment to the code to make sure it is clear to others (or your future self) why the code is the way it is.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12192" LastActivityDate="2015-05-06T12:26:27.260" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14395" PostTypeId="2" ParentId="13366" CreationDate="2015-05-06T12:39:15.667" Score="1" Body="&lt;p&gt;&lt;strong&gt;You are:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Real Madrid&lt;/strong&gt; (the football team)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Real Madrid players usually wear white jerseys,&lt;br&gt;&#xA; but &lt;strong&gt;today&lt;/strong&gt; they wore black playing against the striped team Juventus.&lt;br&gt;&#xA; (&lt;strong&gt;Juventus vs Real Madrid 5/5/2015&lt;/strong&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12014" LastEditorUserId="8874" LastEditDate="2015-10-06T12:22:36.713" LastActivityDate="2015-10-06T12:22:36.713" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14396" PostTypeId="2" ParentId="14392" CreationDate="2015-05-06T12:44:58.087" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;A computer mouse&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;A mouse is an animal&lt;br&gt;&#xA;The 'tail' comes out of the top, rather than the bottom&lt;br&gt;&#xA;Some are wireless, so have no tail, but are still bound by the receiver's range&lt;br&gt;&#xA;Going right takes you away from the keyboard, which is easy to do&lt;br&gt;&#xA;I'm not sure about the left clue&lt;br&gt;&#xA;You have your hand on it a lot without being bitten&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11717" LastActivityDate="2015-05-06T12:44:58.087" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14397" PostTypeId="2" ParentId="12934" CreationDate="2015-05-06T12:55:31.737" Score="1" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a vote?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What's happening is very strange,&#xA;I belong to another StackExchange!&#xA;I'm indeed off-topic here in Puzzling,&#xA;It isn't always clear what I'm saying.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In all of these cases you could down vote. Although, here also a flag would be the right answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Even though I'm often too broad,&#xA;Review it, something is flawed.&#xA;I'm just suggesting an edit,&#xA;A low quality job, ain't it?&#xA;For new users I'm a big trouble&#xA;A duplicate isn't always a double.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Those definitely point more towards a flag. But a lot of people just down vote again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your current work you shall reopen&#xA;Flag here, flag there, it's all broken!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As you mention a flag, it's not the flag. It's the vote to reopen a mistakenly flagged and closed question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Oh, don't forget to add a comment,&#xA;late it's not, takes just a moment!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you vote (too often) without commenting, you are generally reminded by any SE site, that you should comment, too.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10885" LastActivityDate="2015-05-06T12:55:31.737" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14399" PostTypeId="2" ParentId="14392" CreationDate="2015-05-06T13:00:38.297" Score="1" Body="&lt;p&gt;Maybe you are&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; a keyboard&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;People call me an animal&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; that's the one I miss, maybe a reference to the keyboard cat?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Since I possess a tail&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; that's the wire&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;but it's in the opposite direction.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; when I'm using it, the &amp;quot;tail&amp;quot; is indeed in opposite direction&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I can't travel beyond some point&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; well, the wire must not be unplugged&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Since I am bound by my tail.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; bound with the computer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Sometimes I don't have a tail&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; wireless ones, maybe?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;But still I can not travel that far.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Wireless, if I go too far, I might lose the signal&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Go to the right and world is easy for you.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; numpad in on the right, and make life easier, compared to using the other numbers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Go to the left and change your mode.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; capslock is on the left, and often used to convey anger, on forums&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You touch me often, but I won't bite you.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; yes, I touch it very often (I wouldn't be able to answer, if not) .. and never been bitten&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11827" LastEditorUserId="69582" LastEditDate="2021-04-23T16:00:14.007" LastActivityDate="2021-04-23T16:00:14.007" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="14400" PostTypeId="1" AcceptedAnswerId="14401" CreationDate="2015-05-06T13:33:57.633" Score="3" ViewCount="536" Body="&lt;p&gt;What will come at the &amp;quot;?&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;Follow the pattern.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A + B = E&lt;/p&gt;&#xA;&lt;p&gt;B - A = C&lt;/p&gt;&#xA;&lt;p&gt;A + D = Q&lt;/p&gt;&#xA;&lt;p&gt;D - B = L&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;then&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A + C - B = ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10433" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2017-07-25T20:18:23.380" Title="The result for the following pattern" Tags="|calculation-puzzle|pattern|" AnswerCount="7" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14401" PostTypeId="2" ParentId="14400" CreationDate="2015-05-06T13:38:11.300" Score="5" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;we know that&lt;br&gt;&#xA;$B-A=C$&lt;br&gt;&#xA;so&lt;br&gt;&#xA;$A-B=(-C)$&lt;br&gt;&#xA;&#xA;$A+C-B=A-B+C=(-C)+C=0$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-05-06T13:39:48.203" LastActivityDate="2015-05-06T13:39:48.203" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14402" PostTypeId="1" AcceptedAnswerId="14403" CreationDate="2015-05-06T13:46:48.453" Score="2" ViewCount="1035" Body="&lt;p&gt;&lt;em&gt;Riddle me this,&lt;br&gt;&#xA;riddle me that..&lt;br&gt;&#xA;Do I have a hat?&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My name is enygma&lt;br&gt;&#xA;my heart is my brain.&lt;br&gt;&#xA;But what’s my domain?&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I’ll give you a joker&lt;br&gt;&#xA;to finish the game,&lt;br&gt;&#xA;me and my questions are not the same.&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Now for my last question.&lt;br&gt;&#xA;Oh, what a twiddle!&lt;br&gt;&#xA;and then you’ll have to answer Minotaur’s riddle.&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What is the nickname&lt;br&gt;&#xA;of my toughest foe?&lt;br&gt;&#xA;(Don’t forget the joker, tho!)&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's play. I'm looking for 4 answers ( the 3 questions stated in the riddle + who is the riddler?)&lt;/p&gt;&#xA;" OwnerUserId="5283" LastActivityDate="2015-05-07T14:38:13.123" Title="meta riddling with jokers" Tags="|riddle|rhyme|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14403" PostTypeId="2" ParentId="14402" CreationDate="2015-05-06T14:23:30.750" Score="7" Body="&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Edward Nygma&lt;/strong&gt; (The Riddle (Enygma), from Batman)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Riddle me this,&#xA;riddle me that..&#xA;Do I have a hat?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes, it is a green bowler hat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My name is enygma&#xA;my heart is my brain.&#xA;But what’s my domain?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gotham City (DC Comics)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I’ll give you a joker&#xA;to finish the game,&#xA;me and my questions are not the same.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Is this a question? I don't think so.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now for my last question.&#xA;Oh, what a twiddle!&#xA;and then you’ll have to answer Minotaur’s riddle.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;Minotaur&lt;/strong&gt;: I have millions of eyes, yet I live in darkness. I have    millions of ears, yet only four lobes. I have no muscles, yet&#xA;&lt;br&gt; I rule    two hemispheres. What am I?&#xA;&lt;br&gt;  &lt;strong&gt;Batman&lt;/strong&gt;: That's simple: the human brain.&#xA;&lt;br&gt;  &lt;strong&gt;Batman&lt;/strong&gt;: It has millions of visual and auditory nerves, four lobes,    two hemispheres, and it's the only thing Edward Nygma&#xA;&lt;br&gt; respects.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the nickname&#xA;of my toughest foe?&#xA;(Don’t forget the joker, tho!)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Batsy&quot; - Joker's favourite nickname of Batman&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12014" LastEditorUserId="5283" LastEditDate="2015-05-07T14:38:13.123" LastActivityDate="2015-05-07T14:38:13.123" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="14406" PostTypeId="2" ParentId="14402" CreationDate="2015-05-06T14:59:32.100" Score="3" Body="&lt;p&gt;Possibly over-doing the meta layer, and ignoring the obvious batman-related answer, the one posing the riddle is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Minotaur&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Riddle me this,/&#xA;riddle me that../&#xA;Do I have a hat?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nope. Horns, though&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My name is enygma/&#xA;my heart is my brain./&#xA;But what’s my domain?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Labyrinth (on Crete, I think?). Though this clue kinda makes me think of The Wizard of Oz&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I’ll give you a joker/&#xA;to finish the game,/&#xA;me and my questions are not the same.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is what makes me think it's not Batman related, which means we need another famous riddler&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now for my last question./&#xA;Oh, what a twiddle!/&#xA;and then you’ll have to answer Minotaur’s riddle.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Seems overly straightforward, but if this is 'Minotaur's riddle', then are you not the minotaur?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the nickname/&#xA;of my toughest foe?/&#xA;(Don’t forget the joker, tho!)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That'd be Theseus, then. Can't think of any nicknames off-hand&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11717" LastActivityDate="2015-05-06T14:59:32.100" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14407" PostTypeId="2" ParentId="14400" CreationDate="2015-05-06T15:24:23.963" Score="1" Body="&lt;p&gt;I came up with&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;E - C&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Flip the equations so you get A, B, and C by themselves. Of course these are other possibilities, but these are the ones that lead to the shortest answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A = E - B&lt;br&gt;&#xA;B = C + A&lt;br&gt;&#xA;C = B - A      &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Expand original question using newly found values and eliminate variables that cancel each other.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;(E - B) + (B - A) - (C + A) = ?&lt;br&gt;&#xA;E-B+B-A-C+A = ?&lt;br&gt;&#xA;E-C = ?       &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12073" LastActivityDate="2015-05-06T15:24:23.963" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14408" PostTypeId="2" ParentId="14400" CreationDate="2015-05-06T16:38:05.143" Score="4" Body="&lt;p&gt;Similar to Leoll2's answer. It can also be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\int\limits_{-\infty}^{+\infty}\frac{(\sqrt{-0} \times cosh(f(x)))^5}{π^e}dx$ &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; using [B - A = C], the answer becomes [A + (B - A) - B] which all cancels out to nothing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-05-06T18:02:46.387" LastActivityDate="2015-05-06T18:02:46.387" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14409" PostTypeId="2" ParentId="14400" CreationDate="2015-05-06T17:00:36.847" Score="1" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;We know that&lt;br&gt;&#xA;$B-A=C$&lt;br&gt;&#xA;so&lt;br&gt;&#xA;$B-C=A$&lt;br&gt;&#xA;Substituting $A$ in $A + C-B = ?$&lt;br&gt;&#xA;We have&lt;br&gt;&#xA;$B-C+C-B=0$&lt;br&gt;&#xA;Because&#xA;$B-B = 0$ and $-C+C=0$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12112" LastActivityDate="2015-05-06T17:00:36.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14410" PostTypeId="2" ParentId="13094" CreationDate="2015-05-06T17:33:51.933" Score="6" Body="&lt;p&gt;&lt;s&gt;Partial answer while I wait on clarification.&lt;/s&gt; Nevermind, found why I kept getting gibberish.&lt;/p&gt;&#xA;&lt;h2&gt;The Sample&lt;/h2&gt;&#xA;&lt;p&gt;The answer is&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;JCaesar&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;First you&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Remove all non-letters from the text. We can convert this to binary by treating capital letters as 1s and lower case as 0. Giving these 7 bit chunks: 1000111 1000000 1011110 1100010 1110000 1011110 1101111&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Then you&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;Add 3 (the number from the clue) to each bit chunk. Converting the result to ASCII gives us our answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;The Puzzle&lt;/h2&gt;&#xA;&lt;p&gt;Using the same method as the sample results in the answer:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;Elizabeth Wells Gallup's theory, won George Fabian $5000 in damages.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11323" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-06T17:50:02.187" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14411" PostTypeId="1" AcceptedAnswerId="14444" CreationDate="2015-05-06T17:52:33.250" Score="11" ViewCount="1846" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Found at the unlucky square&lt;br&gt;&#xA;  From your pocket I can scare&lt;br&gt;&#xA;  With precious metals, my premiere&lt;br&gt;&#xA;  A pet that dad and daughter share&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 1:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Within a ball I make my lair&lt;br&gt;&#xA; My wings are furled, a double pair&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2 (Doesn't say much about the answer, but should help you find the group it belongs to):  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A tree-named teacher does declare&lt;br&gt;&#xA;Of your sex, they're unaware&lt;br&gt;&#xA;A catalogue they have to spare&lt;br&gt;&#xA;For the creatures that to catch, you dare&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;" OwnerUserId="11717" LastEditorUserId="25957" LastEditDate="2016-08-09T09:51:53.153" LastActivityDate="2016-08-09T09:51:53.153" Title="From your pocket I can scare. Who am I?" Tags="|riddle|" AnswerCount="6" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14413" PostTypeId="1" AcceptedAnswerId="14415" CreationDate="2015-05-06T18:48:08.930" Score="3" ViewCount="417" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I work as hard as my brothers do&lt;br&gt;&#xA;  Outcast, though I am&lt;br&gt;&#xA;  I am short and stubby, too&lt;br&gt;&#xA;  They are lean and tan&lt;br&gt;&#xA;  &lt;br&gt;&#xA;  My cousin has the same affliction&lt;br&gt;&#xA;  His kin look down on him&lt;br&gt;&#xA;  We suffer, us, this poor condition&lt;br&gt;&#xA;  Though it may seem grim&lt;br&gt;&#xA;  &lt;br&gt;&#xA;  He and I are strong and hardy&lt;br&gt;&#xA;  Strongest of the lot&lt;br&gt;&#xA;  Critical to work are we&lt;br&gt;&#xA;  Even though we're squat&lt;br&gt;&#xA;  &lt;br&gt;&#xA;  Try to work without us and&lt;br&gt;&#xA;  You'll find it most unpleasant&lt;br&gt;&#xA;  You may be need a helping hand&lt;br&gt;&#xA;  If we are not present&lt;br&gt;&#xA;  &lt;br&gt;&#xA;  There are those that we relate to&lt;br&gt;&#xA;  Somewhere south they prowl&lt;br&gt;&#xA;  From dealing with them we eschew&lt;br&gt;&#xA;  For they smell most foul&lt;br&gt;&#xA;  &lt;br&gt;&#xA;  If I am lost you can replace&lt;br&gt;&#xA;  Me with one of them&lt;br&gt;&#xA;  I would suggest, though, in his place&lt;br&gt;&#xA;  You use one of his kin&lt;br&gt;&#xA;  &lt;br&gt;&#xA;  Now, who am I that moves so free&lt;br&gt;&#xA;  And who are my brothers?&lt;br&gt;&#xA;  Who's my cousin and where is he?&lt;br&gt;&#xA;  To the south, who are the others?&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastActivityDate="2015-05-06T19:14:41.623" Title="The Short Brother" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14414" PostTypeId="2" ParentId="12943" CreationDate="2015-05-06T18:51:02.260" Score="2" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;b&gt;A circle&lt;/b&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here I am in my perfect state.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A circle can be 'perfectly round'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Friends of mine in some way relate.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Other shapes can be round, and share many of the same properties (area, perimeter etc.) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But what confines me isn't straight.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The circumference(perimeter) of a circle isn't straight.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is it none, one, or many? That's to debate!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; How many edges does a circle have: zero, one, infinite? Some people often argue over this number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #2:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;That you're in my realm, don't postulate!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Circles are 2 dimensional, and we live in the 3rd dimension. (We couldn't live in the 2nd dimension)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #3:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;To one number I associate.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; π is associated with circles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastActivityDate="2015-05-06T18:51:02.260" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14415" PostTypeId="2" ParentId="14413" CreationDate="2015-05-06T18:55:46.193" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are the thumb.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is short ad stubby but has greater freedom of movement compared to the brothers&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the fingers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;who are lean and tall but not as strong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cousin I guess is the&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; thumb on the other hand&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally down south are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the smelly toes. It appears that you can surgically replace a missing thumb with a big toe which is pretty crazy stuff.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="11297" LastEditDate="2015-05-06T19:14:41.623" LastActivityDate="2015-05-06T19:14:41.623" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14416" PostTypeId="2" ParentId="14411" CreationDate="2015-05-06T19:18:32.733" Score="4" Body="&lt;p&gt;This answer is a work in progress, I'll fill in the clues as I figure them out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Aluminium&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Found at the unlucky square&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The 13th element in the periodic table.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From your pocket I can scare&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This one I am unsure of. Could have something to do with cell phones containing aluminium and scaring someone when it rings? Seems like too much of a stretch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With precious metals, my premiere&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Aluminium was initially classified as a precious metal due to the difficulty of extracting it from ore. This changed in the 1880s.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A pet that father and daughter share&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I don't have any ideas for this one yet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12192" LastEditorUserId="5373" LastEditDate="2015-05-06T21:19:51.213" LastActivityDate="2015-05-06T21:19:51.213" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14417" PostTypeId="1" AcceptedAnswerId="14466" CreationDate="2015-05-06T19:30:12.903" Score="4" ViewCount="459" Body="&lt;p&gt;I am the second most acidic on the table,&lt;br&gt;&#xA;but taste like nothing if you ever lick me.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I might be more filthy than you think,&lt;br&gt;&#xA;but you let me kiss you on the cheek.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;for you are the creator of my kind,&lt;br&gt;&#xA;yet few have found not prays to me.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="10037" LastEditDate="2015-05-06T19:49:39.933" LastActivityDate="2015-05-07T12:53:58.833" Title="I am not sticky - what am I?" Tags="|riddle|" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14418" PostTypeId="2" ParentId="13257" CreationDate="2015-05-06T19:35:37.943" Score="5" Body="&lt;p&gt;At first I was all alone.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  I&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then a friend showed up, and the place started showing signs of life.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  AI (artificial intelligence)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But then another friend came, and the three of us started fighting.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  ALI (a boxer)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another friend came, and calmed us down by helping us communicate.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  PALI (an extinct language)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet another friend came along -- by now, we ruled the area!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  PALIN (former governor of Alaska)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By the time a sixth friend joined us, the game had changed a little, and we were looking weak next to the competition.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  PALING&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Our seventh friend came and helped us win some resounding successes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  PEALING&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally our last member arrived, and we were just riding free.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  PEDALING&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12203" LastEditorUserId="12203" LastEditDate="2015-05-06T19:51:31.163" LastActivityDate="2015-05-06T19:51:31.163" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14419" PostTypeId="2" ParentId="14411" CreationDate="2015-05-06T19:35:51.247" Score="1" Body="&lt;p&gt;Pretty sure it's a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cell Phone&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Found at the unlucky square&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is the one I'm least sure about. Is it a reference to the use of mobile phones in city squares during the last few years protests?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From your pocket I can scare&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ring! Ring!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With precious metals, my premiere&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They contain a bunch of rare earth elements and precious metals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A pet that father and daughter share&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Office fathers and teenage daughters are often associated with phones/mobile devices.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5612" LastActivityDate="2015-05-06T19:35:51.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14420" PostTypeId="2" ParentId="14411" CreationDate="2015-05-06T19:51:43.083" Score="4" Body="&lt;p&gt;My guess would be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Money?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation line by line:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Found at the unlucky square&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the United States the 2$ bill is considered unlucky.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From your pocket I can scare&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Money is usually in your pockets and people complain about not having much.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With precious metals, my premiere&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Coins are made of metal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A pet that father and daughter share&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Daughters are known to spend their fathers' money.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4786" LastEditorUserId="5373" LastEditDate="2015-05-06T21:20:12.283" LastActivityDate="2015-05-06T21:20:12.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14421" PostTypeId="1" AcceptedAnswerId="14469" CreationDate="2015-05-06T20:34:05.940" Score="6" ViewCount="930" Body="&lt;p&gt;For the previous Word of the Day challenges, check out &lt;a href=&quot;https://puzzling.stackexchange.com/questions/13320/word-of-the-day-1&quot;&gt;Word of the Day #1&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/questions/13372/word-of-the-day-2&quot;&gt;Word of the Day #2&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Add to me one and I'll become a mess;&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;Add to me six and I'll buy you a dress.&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;Add thirty-five, I'll get bolder and louder;&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;Add to me none, and I'm left as a powder.&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;What am I?&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-10-02T15:16:20.900" Title="Word of the Day #3" Tags="|riddle|word|rhyme|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14422" PostTypeId="2" ParentId="12781" CreationDate="2015-05-06T20:34:53.147" Score="7" Body="&lt;h1&gt;&lt;strong&gt;This is the rigorous proof asked by the author:&lt;/strong&gt;&lt;/h1&gt;&#xA;&lt;p&gt;Ignore the 0 for now and consider only the interval &lt;span class=&quot;math-container&quot;&gt;$a=[1...1000000]$&lt;/span&gt;.&lt;br /&gt;&#xA;For any strategy, define &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; as the set of elements accepted by our strategy (we've already proven that a cardinality of &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; equal to 500001 is acceptable).&lt;br /&gt;&#xA;Now consider the highest number in the set &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; and call it &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;: the number &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; is the sum of &lt;span class=&quot;math-container&quot;&gt;$\frac{n-1}{2}$&lt;/span&gt; different couples of natural numbers, and every number in the interval &lt;span class=&quot;math-container&quot;&gt;$[1...(n-1)]$&lt;/span&gt; is present in one (and only one) of these couples. (&lt;em&gt;see the &lt;strong&gt;note&lt;/strong&gt; below&lt;/em&gt;).&lt;br /&gt;&#xA;If you don't want to have a contradiction, you can include in &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; up to one number for each couple (indeed, if you include both, their sum would be &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;).&lt;br /&gt;&#xA;A consequence is that, given &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;, there are at most &lt;span class=&quot;math-container&quot;&gt;$\frac{n-1}{2}$&lt;/span&gt; elements of &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; before &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;, and no element of &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; greater than &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; because of our initial assumption.&lt;br /&gt;&#xA;As you can see, the maximum number of elements belonging to &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$(1+\frac{n-1}{2})$&lt;/span&gt; (that 1 is the number &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;, don't forget it!). It depends only on the highest number of &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt;, so the optimal strategy has at most &lt;span class=&quot;math-container&quot;&gt;$(1+\frac{999999}{2})$&lt;/span&gt; elements. How do we round &lt;span class=&quot;math-container&quot;&gt;$\frac{999999}{2}$&lt;/span&gt;? Round up, of course, since 999999 elements form 499999 couples and 1 standalone (see the &lt;strong&gt;note&lt;/strong&gt; below)!&lt;br /&gt;&#xA;So, the maximum is 500001!&lt;br /&gt;&#xA;Ah, we forgot the zero! Then add 1 to the solution, which becomes &lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$500002$&lt;/span&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Note:&lt;/strong&gt;&lt;/em&gt; If &lt;span class=&quot;math-container&quot;&gt;$(n-1)$&lt;/span&gt; is odd, the couples are actually &lt;span class=&quot;math-container&quot;&gt;$\frac{n-1-1}{2}=\frac{n-2}{2}$&lt;/span&gt; because the middle number remains out of any couple (you can't add it to itself!). For example, if &lt;span class=&quot;math-container&quot;&gt;$n=8$&lt;/span&gt;, the couples are {1,7},{2,6},{3,5}. Indeed, {4,4} isn't acceptable!&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-13T15:59:50.603" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14423" PostTypeId="2" ParentId="14421" CreationDate="2015-05-06T21:10:17.713" Score="7" Body="&lt;p&gt;Are you &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Carbon?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add to me one and I'll become a mess;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; add Hydrogen to carbon and you get hydrocarbons like oil, which is very messy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add to me six and I'll buy you a dress.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Carbon bonded with itself can be a diamond&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add thirty-five, I'll get bolder and louder;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Carbon and bromine make a purple color, which is bold.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add to me none, and I'm left as a powder.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm guessing this is graphite powder&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4591" LastActivityDate="2015-05-06T21:10:17.713" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14424" PostTypeId="2" ParentId="14421" CreationDate="2015-05-06T21:14:49.370" Score="5" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; tantalum, number 73 in the periodic table.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Add to me one and I'll become a mess;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tungsten, number 74, is somewhat toxic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Add to me six and I'll buy you a dress.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gold, number 79, is valuable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Add thirty-five, I'll get bolder and louder;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hassium, number 108, is radioactive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Add to me none, and I'm left as a powder.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The major use of tantalum is as a metal powder in electronic components such as capacitors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-06T21:14:49.370" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14425" PostTypeId="2" ParentId="14417" CreationDate="2015-05-06T21:18:45.773" Score="0" Body="&lt;p&gt;Maybe it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; face powder?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am the second most acidic on the table,&lt;br&gt;&#xA; but taste like nothing if you ever lick me.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'll take your word for it!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I might be more filthy than you think,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Who knows what horrible chemicals it's got in it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;but you let me kiss you on the cheek.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's often applied to cheeks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;for you are the creator of my kind,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's man-made.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;yet few have found not prays to me.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I can't really parse this line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-06T21:18:45.773" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14426" PostTypeId="2" ParentId="14392" CreationDate="2015-05-06T22:33:11.487" Score="0" Body="&lt;p&gt;The correct answer has been posted, but this was my initial guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A &lt;strong&gt;ring buffer&lt;/strong&gt; or &lt;strong&gt;FIFO (first-in, first-out) queue&lt;/strong&gt;&lt;br&gt;&lt;br&gt;&#xA; &lt;img src=&quot;https://www.cs.fsu.edu/~baker/devices/notes/graphics/circularbuffer2.gif&quot; alt=&quot;ring buffer&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;People call me an animal&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I possess a tail&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but its in opposite direction.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure about the animal part. The tail can end up in front of the head once the head wraps around after having reached the end of the buffer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cant travel beyond some point&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since i am bound by my tail.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A ring buffer is considered full when head = tail-1 (mod N). At this point, the head can no longer advance and so you can no longer push data to the ring buffer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I don't have a tail&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But still I can not travel that far.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The tail points to stale data once it has caught up to the head, i.e. head = tail. Even starting with no tail, the head can only advance by N, where N is the (finite) size of the ring buffer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Go to right and world is easy for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Go to left and change your mode.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A ring buffer always travels to the right, towards increasing memory, until wrap-around occurs after a total of N steps.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You touch me often, but I won't bite you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A ring buffer can be used many times per second, allowing information to transfer at rates of 100s of MB per second.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12208" LastEditorUserId="12208" LastEditDate="2015-05-06T22:46:23.303" LastActivityDate="2015-05-06T22:46:23.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14427" PostTypeId="1" AcceptedAnswerId="14445" CreationDate="2015-05-06T22:42:23.823" Score="8" ViewCount="969" Body="&lt;p&gt;In an attempt to bump up the difficulty of my riddles so far, I'm going to be a little more vague with this one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When I'm bitten, you will suffer.&lt;br&gt;&#xA;  When I'm precious, you will suffer no more.&lt;br&gt;&#xA;  When I'm used, you will never see me again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="11851" LastActivityDate="2015-05-07T09:45:08.053" Title="When I'm bitten, what am I?" Tags="|riddle|" AnswerCount="3" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14429" PostTypeId="1" AcceptedAnswerId="14443" CreationDate="2015-05-06T23:55:20.650" Score="3" ViewCount="815" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;His raiment was the color of the king of Doriath's cloak,&lt;br&gt;&#xA;  But now is the color of the Spirit of Fire's prized creations.&lt;br&gt;&#xA;  His weapon is from that which was betrayed by the Child of Twilight.&lt;br&gt;&#xA;  His lords far outnumber his peers-&lt;br&gt;&#xA;  -Them being only the Giver of Gifts and Aiwendil.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now who could it be, tell me that?&lt;br&gt;&#xA;  And may the wind under your wings bear you&lt;br&gt;&#xA;  Where the sun sails and the moon walks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Find the quote in the &quot;riddle&quot;, and figure out who said it.&lt;br&gt;&#xA;&lt;em&gt;(Don't use the web to solve this puzzle. That would spoil the fun.)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="11790" LastEditorUserId="8874" LastEditDate="2015-10-02T15:14:49.033" LastActivityDate="2015-10-02T15:14:49.033" Title="Guess the book character" Tags="|riddle|knowledge|no-computers|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14430" PostTypeId="1" AcceptedAnswerId="14496" CreationDate="2015-05-07T00:31:21.593" Score="20" ViewCount="2036" Body="&lt;p&gt;Yesterday I dropped in on Ernie to show off my new j-watch. Ernie wouldn't be impressed by instant wrist-mounted access to social media sites like me-tube, tweeter and farce-book so I thought I would concentrate on it's technical aspects - 1024 GB RAM, green-tooth, USB5, high-res touch screen and 6G networking. As he brewed us coffee I demonstrated the trump-card of the j-watch: its 8K UHD camera - 120 FPS at 7680x4320 pixels, with surround-sound directional microphones! In my enthusiasm I must have bored myself to sleep because the next thing I remember was waking to hear Ernie complaining that his coffee was cold. I was late for an appointment and had to leave in a hurry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least that is what i thought happened until this morning! After breakfast I tried to use the j-watch's Memo app to remind myself to buy eggs. To my dismay the watch just displayed a flashing &quot;MEMORY FULL&quot; warning. It didn't take long to find the problem - a massive 1023.9 GB MP5 file. I realized that I must have left the camera running when I fell asleep at Ernie's. I was going to delete it but thought I would review the first few minutes to see the picture quality. The following is a precis of what I saw and heard:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Picture:&lt;/strong&gt; A shaky (but crystal clear) view of my face in close-up as my voice explains the magnificence of my j-phone in exuberant detail. Ernie puts two cups of coffee on the table. The next 30 seconds is blanked out as my sleeve drops down over the camera's lens while recording. (&lt;em&gt;I began to wonder at the practicality of a wrist-mounted camera and if the added expense of the UHDV camera had been worth it. I was just about to delete the file when...&lt;/em&gt;) The sound of Ernie's door-bell can be heard on the sound-track. (&lt;em&gt;&quot;Curious&quot;, I thought, &quot;I can't remember any visitors&quot;&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Me:&lt;/strong&gt; &quot;I'll get it&quot; (&lt;em&gt;This was really weird - I definitely couldn't remember answering the door!&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Picture:&lt;/strong&gt; Door opens, momentary view of two men in wrap-around dark glasses and black suits; sleeve slides back down to block camera again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Me:&lt;/strong&gt; &quot;Not to be rude - but whatever you are selling, we don't want any&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;MIB1&lt;/strong&gt;: [Pushing his way somewhat rudely into the hall] &quot;We are from MIB - it is imperative that we speak to Professor Ernie immediately!&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Me:&lt;/strong&gt; &quot;MIB - are you the Men in Black?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;MIB2:&lt;/strong&gt; &quot;Don't be ridiculous - we are from the Ministry of Interstellar Bureaucracy&quot;. [The two men greet Ernie and explain that the Ministry was created when an alien-manufactured Tachyon wave radio receiver and universal translator were discovered in the wreckage at area 63! Apparently, the galaxy teems with a multitude of intelligent species. Until recently the radio hadn't provided any useful information to aid mankind as the only signal was live coverage of rather boring Interstellar Council meetings dealing with minor issues such as spiral-arm planning consents and sales taxes on micro-black-holes. But just a few hours ago a very worrying message was heard. An Eviction Notice had been lodged at the Council by a race of carnivorous lizard-people. They were planning to return to their home planet after a 'temporary' absence of ~64 million years (some sort of problem with the environment after a failed attempt at asteroid mining) but discovered the planet was 'infested by bald tailless apes'].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;MIB1&lt;/strong&gt;: &quot;We realized they were talking about us - humankind!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ernie:&lt;/strong&gt; &quot;Eviction - that doesn't sound very nice&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;MIB2:&lt;/strong&gt; &quot;We fear the worst. The lizard-men have a mass of heavily armed and technologically advanced worm-hole drive space ships ready and waiting - they can be here within hours of the court giving consent.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LgUYM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Me:&lt;/strong&gt; &quot;Has the court made a decision yet?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;MIB1:&lt;/strong&gt; &quot;No. One of the council members reminded the lizard-men that parts of Earth had been declared a wild-life refuge.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Me:&lt;/strong&gt; &quot;A refuge for bald tailless apes?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;MIB2:&lt;/strong&gt; &quot;No - for a rare species of woodlouse that lives in the rain-forests of Sumatra. Seems that it bears a slight genetic similarity to the councilor's distant ancestors. And it is only a temporary reprieve. The lizard-men only need to confirm they are willing to comply with a list of Intergalactic Environmental Laws. Then the 'eviction' can commence immediately. Ernie - we want you to invent a Tachyon transmitter so we can contact the court and make an appeal - our best lawyers are already preparing their arguments&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ernie:&lt;/strong&gt; &quot;Well I'll do what I can. But I always work best when I multi-task, so tell me more about these environmental laws while I think&quot;. [The MIB's go on to explain that the Galactic Empire operated under a binary legal system with two laws labelled '1', and '2'. Of course things were a bit more complicated than that, as each law had two sub-clauses '11', '12', '21', and '22', and each of these had two sub-sub-clauses '111', '112' ... '221', '222' and so on to deeper and deeper levels of sub-sub-sub...sub clauses].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;MIB1:&lt;/strong&gt; &quot; The court requires written confirmation that the lizard-men will obey all relevant laws in a list defined by a special Environmental Law Identification Algorithm.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ernie:&lt;/strong&gt; &quot;An algorithm, tell me more!&quot; [The MIBs described the algorithm as follows...]&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;The first relevant sub-clause is the one labelled '1111111'  &#xA;&#xA;Each next relevant sub-clause is the one with a label:&#xA;   a) that is one digit longer than that of the previous sub-clause&#xA;   b) which contains no previously identified label as a sub-sequence &#xA;   c) that is the alphabetically first label complying with a) and b) above.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ernie:&lt;/strong&gt; &quot;So the first few clauses to be tested would be...&quot;[the sound of a white-board marker squeaking]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Picture:&lt;/strong&gt; a momentary view of the white-board before my sleeve slips down over the lens again as I scratch my itchy nose - (I am allergic to stress - it always makes my nose run). &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 1     1111111&#xA; 2     11111122&#xA; 3     111112212&#xA; 4     1111122221&#xA; 5     11111222222&#xA; 6     111122122222&#xA; 7     1111222212221&#xA; 8     11112222212222&#xA; 9     111122222221212&#xA;10     1111222222221221&#xA;11     111...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ernie:&lt;/strong&gt; &quot;...and when the Lizard-men reach the last sub-clause the court will accept the eviction order?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;MIB1:&lt;/strong&gt; &quot;Precisely! But the lizard-men are already way past that point on the list - they and the council are using super-fast quantum computing legal engines to write and test their legal arguments, it takes precisely one second for each sub-clause to be accepted. So as you can imagine, we don't have much time to make contact with the court and send an appeal!&quot; [long pause with just the sound of Ernie humming to himself as he does when he is thinking hard, plus me sneezing more and more]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ernie:&lt;/strong&gt; &quot;To be honest, I don't know the first thing about how to build a tachyon transmitter but I can tell you that the last sub-clause the lizard-men have to satisfy will have a label of the form...&quot; [sound of white-board marker again, but my sleeve is still in the way and the picture is blank]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ernie: &quot;...and I recon that the time it will take before the last sub-clause is reached will be about ...&quot; [at that very moment my allergy strikes again and Ernie's next words are drowned out on the sound-track by an enormous sneeze]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;MIB1:&lt;/strong&gt; &quot;We need to get this information back to the Ministry immediately. My apologies, but to maintain security we now need to use the brain scrambler we found in the wreckage to delete your memory of our visit. This won't hurt a bit...&quot; [a momentary buzz on the sound-track and the video image fills with static for a few seconds, then the sound of footsteps heading off. Followed by a few minutes of quiet snoring then...]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ernie:&lt;/strong&gt; &quot;Dang - I must have dozed off - and now my coffee is cold&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Picture:&lt;/strong&gt; My face (in unflattering close-up) appears as I pull back my sleeve to check the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Me:&lt;/strong&gt; &quot;And I am late for my meeting in town...&quot; [video ends with Memory Full warning]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here is my problem - I just rang Ernie and he doesn't have any memory of the visitors. I tried to explain but he thinks I just had a bad dream and won't take me seriously. I would show him the video but I think the brain scrambler may have damaged my watch because now it just displays a &quot;file not found&quot; message. Should I try and find a hiding place right now to escape from an imminent invasion, or is it better to try and convince people to make preparations to repulse the lizard-men and protect themselves?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone let me know what the number of the last sub-clause will look like and how much time I have to prepare for the invasion of the lizard-men?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Click on the link to confirm what I mean by &lt;a href=&quot;http://en.wikipedia.org/wiki/Subsequence&quot; rel=&quot;noreferrer&quot;&gt;&lt;strong&gt;&lt;em&gt;subsequence&lt;/em&gt;&lt;/strong&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="1631" LastEditDate="2015-05-07T02:38:24.967" LastActivityDate="2015-05-14T17:04:24.360" Title="Ernie and the Alien Invasion" Tags="|mathematics|calculation-puzzle|word-problem|number-sequence|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14432" PostTypeId="2" ParentId="13305" CreationDate="2015-05-07T03:06:14.193" Score="4" Body="&lt;p&gt;Using the clue that Dennis Meng got, the treasure is at:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;(B,g)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The dark point on the compass signifies North, so going south would be counterclockwise about 30 degrees from directly down.&#xA;There are 3 dead trees on the map at: (B,e), around (P,p), and (C,q). The only one that is inside a swampy area is the one at (B,e). The others are on &quot;normal&quot; land.  If you go south from there, it would take us to a mountain at (A,h). halfway to that point is a jut in the road, and when crossed you are in the space (B,g)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11727" LastActivityDate="2015-05-07T03:06:14.193" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14434" PostTypeId="2" ParentId="10991" CreationDate="2015-05-07T03:53:35.977" Score="3" Body="&lt;p&gt;The answer is:   &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;THIS   111 221 241 151  &#xA;ONE    222 543 421  &#xA;IS     233 161  &#xA;TOUGH  166 222 331 353 444&lt;/pre&gt;  &#xA;&#xA;&lt;p&gt;User224980 explanation (from comment) - What it ended up coming down to was a lot of guess and check. Obviously the letters were represented by more than one number. Like I said before, you were going to know if you got the answer. Another clever one! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Len's summary of other comments below - This answer does seem most appropriate but it is disappointing because there should still be an explanation for associating the numbers in the question with the letters in the answer.  &lt;/p&gt;&#xA;" OwnerUserId="10646" LastEditorUserId="5188" LastEditDate="2015-05-07T18:22:59.453" LastActivityDate="2015-05-07T18:22:59.453" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="14435" PostTypeId="1" AcceptedAnswerId="14439" CreationDate="2015-05-07T06:15:58.763" Score="5" ViewCount="995" Body="&lt;p&gt;One of us describes my ability to call the dead&lt;br&gt;&#xA;Another describes my physical appearance&lt;br&gt;&#xA;And the other one describes my state of freedom&lt;br&gt;&#xA;Sort us, and you will get the answer.&lt;/p&gt;&#xA;" OwnerUserId="1649" LastEditorUserId="8874" LastEditDate="2015-10-02T15:12:14.700" LastActivityDate="2015-10-02T15:12:14.700" Title="Three parts, four words" Tags="|riddle|wordplay|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14437" PostTypeId="2" ParentId="14435" CreationDate="2015-05-07T06:46:42.177" Score="2" Body="&lt;p&gt;I think it may be  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Grim Reaper&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One of us describes my ability to call the dead&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reaper has that meaning.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another describes my physical appearance&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Grim means dreadful, horrific.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the other one describes my state of freedom&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is hard to explain, but the article &quot;the&quot; suggests that there exists only one Grim Reaper and it's not bound to anything/anyone (it is not &lt;strong&gt;my&lt;/strong&gt; Grim Reaper, it is not &lt;strong&gt;your&lt;/strong&gt; Grim Reaper, it is not &lt;strong&gt;his&lt;/strong&gt; Grim Reaper, it is &lt;strong&gt;the&lt;/strong&gt; Grim Reaper!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="1649" LastEditDate="2015-05-07T07:14:56.570" LastActivityDate="2015-05-07T07:14:56.570" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14438" PostTypeId="2" ParentId="14417" CreationDate="2015-05-07T06:52:19.300" Score="0" Body="&lt;p&gt;I believe you might be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Saliva&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the second most acidic on the table,&#xA;but taste like nothing if you ever lick me. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stomach acid is the most acidic thing in the human body, and I recall reading somehwere the saliva can be acidic too, so perhaps it takes second place. It also doesn't taste like anything, usually.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might be more filthy than you think,&#xA;but you let me kiss you on the cheek. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Saliva is usually full of mucus compunds and bacteria, and a big, wet, sloppy kiss on the cheek will transfer some to your face.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;for you are the creator of my kind,&#xA;yet few have found not prays to me. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Humans naturally create saliva. The second line is a knotty one - perhaps a reference,  to 'say it, don't spray it'?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Regardless, I highly recommend regular brushing and some mouthwash as a part of your daily routine.&lt;/p&gt;&#xA;" OwnerUserId="4164" LastActivityDate="2015-05-07T06:52:19.300" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14439" PostTypeId="2" ParentId="14435" CreationDate="2015-05-07T07:16:05.920" Score="5" Body="&lt;p&gt;A person who can talk to the dead, or who can transmit messages from the dead, is often called a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; medium.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As for physical appearance,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; size&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is an attribute of that, and&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;small&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is one possible value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;at large&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;describes a state of freedom, as opposed to a state of incarceration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorted by their (inter-related) other meanings, those three parts produce the four-word phrase:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;small medium at large&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://theoatmeal.com/djtaf/j/99&quot; rel=&quot;nofollow&quot;&gt;From the joke&lt;/a&gt;, presumably.) &lt;/p&gt;&#xA;" OwnerUserId="4204" LastEditorUserId="4204" LastEditDate="2015-05-07T07:57:40.593" LastActivityDate="2015-05-07T07:57:40.593" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14440" PostTypeId="1" AcceptedAnswerId="14441" CreationDate="2015-05-07T07:33:47.870" Score="7" ViewCount="2060" Body="&lt;p&gt;I am a wonderful help to women,&lt;br&gt;&#xA;The hope of something to come. I harm&lt;br&gt;&#xA;No citizen except my slayer.&lt;br&gt;&#xA;Rooted I stand on a high bed.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am shaggy below. Sometimes the beautiful&lt;br&gt;&#xA;Peasant's daughter, an eager-armed,&lt;br&gt;&#xA;Proud woman grabs my body,&lt;br&gt;&#xA;Rushes my red skin, holds me hard,  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claims my head. The curly-haired&lt;br&gt;&#xA;Woman who catches me fast will feel&lt;br&gt;&#xA;Our meeting. Her eye will be wet.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;" OwnerUserId="4312" LastEditorUserId="8874" LastEditDate="2015-10-02T15:11:40.820" LastActivityDate="2015-10-02T15:11:40.820" Title="I am shaggy below, Sometimes the beautiful. Who am I?" Tags="|riddle|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14441" PostTypeId="2" ParentId="14440" CreationDate="2015-05-07T07:52:08.153" Score="11" Body="&lt;p&gt;I guess you are a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Red onion&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You shouldn't look at the lines, but you should read the sentences.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;I am a wonderful help to women, the hope of something to come&lt;/strong&gt;&lt;br&gt;&#xA; Food? Food!&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;I harm no citizen except my slayer&lt;/strong&gt;&lt;br&gt;&#xA; Unless it's a poisoned onion it won't harm anyone except the one who's &lt;em&gt;slaying&lt;/em&gt; it&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Rooted I stand on a high bed&lt;/strong&gt;&lt;br&gt;&#xA; Here the word &lt;em&gt;bed&lt;/em&gt; means &lt;em&gt;garden bed&lt;/em&gt; or a &lt;em&gt;flowerbed&lt;/em&gt; and onions have roots&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;I am shaggy below&lt;/strong&gt;&lt;br&gt;&#xA; Yes you are!&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Sometimes the beautiful Peasant's daughter, an eager-armed, Proud woman grabs my body, Rushes my red skin, holds me hard, Claims my head&lt;/strong&gt;&lt;br&gt;&#xA; A peasant probably has a bed(or a few) of onions, so his daughter will grab one from there and peel it&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;The curly-haired Woman who catches me fast will feel Our meeting&lt;/strong&gt;&lt;br&gt;&#xA; I guess if you throw an onion really fast into somebody, he/she will certainly feel it :)&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Her eye will be wet&lt;/strong&gt;&lt;br&gt;&#xA; When you cut the onion normally your eyes start crying. Or if you throw an onion reeeeeeally hard into somebody it might hurt enough for him/her to start crying :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9352" LastActivityDate="2015-05-07T07:52:08.153" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14442" PostTypeId="2" ParentId="14411" CreationDate="2015-05-07T08:11:11.847" Score="3" Body="&lt;p&gt;As @Alconja states, I think it will be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pokemon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Found at the unlucky square&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure, unlucky number is 13, but that is Weedle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From your pocket I can scare&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pokemon are sometimes called pocket monsters and monsters do scare&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With precious metals, my premiere&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some pokemon connected to metal?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A pet that father and daughter share&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In pokemon world adults and kids alike train them&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Within a ball I make my lair&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are stored in pokeball&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My wings are furled, a double pair&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; So we are looking for a winged pokemon. Double pair usually means insect type.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So my guess would be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Beedrill. It has double pair of wings and the drill fits the precious metals somehow.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; second set of hints has been revealed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A tree-named teacher does declare&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All of the pokemon professors are named like trees&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of your sex, they're unaware&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; At the start of the game they ask you: &quot;Are you a boy or a girl?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A catalogue they have to spare&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They give you choice of 3 starting pokemon which differ each generation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So that leads me to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Assuming one of the 3 starter pokemon from first generation, only one of them has wings which is Charizard. It's number is 9 which is considered unlucky in Japanese culture (where pokemon come from)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Only thing I'm not getting is the line &quot;&#xA;With precious metals, my premiere&quot;&lt;/p&gt;&#xA;" OwnerUserId="9907" LastEditorUserId="9907" LastEditDate="2015-05-07T08:27:28.233" LastActivityDate="2015-05-07T08:27:28.233" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14443" PostTypeId="2" ParentId="14429" CreationDate="2015-05-07T08:18:15.170" Score="5" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gandalf.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;His raiment was the color of the king of Doriath's cloak,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Grey (when he was Gandalf the Grey). More detail: Elu Thingol is the king of Doriath, and Thingol is a Sindarin word coming from &lt;em&gt;sindacollo&lt;/em&gt; meaning &quot;grey cloak&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But now is the color of the Spirit of Fire's prized creations.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; White (he became Gandalf the White). The Spirit of Fire is Feanor, his creations being the white Silmarils.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;His weapon is from that which was betrayed by the Child of Twilight.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Glamdring (Gandalf's sword), which was forged in Gondolin. &quot;Child of Twilight&quot; is a translation of the name not given to Maeglin, who betrayed the location of Gondolin to her enemies. &lt;em&gt;Thanks to &lt;strong&gt;Paul Rowe&lt;/strong&gt; for help with this clue.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;His lords far out number his peers-&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are a lot more Valar (lords of the Wizards) in Middle-Earth than Maiar (Wizards). &lt;em&gt;Thanks to &lt;strong&gt;Spacemonkey&lt;/strong&gt; for help with this clue.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;-Them being only the Giver of Gifts and Aiwendil.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Giver of Gifts is Saruman; Aiwendil is Radagast. (You've forgotten the two Blue Wizards out east though!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Now who could it be, tell me that?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; GANDALF.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And may the wind under your wings bear you&lt;br&gt;&#xA;Where the sun sails and the moon walks.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quote:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; “Farewell,&quot; they cried, &quot;Wherever you fare till your eyries receive you at the journey's end!&quot; That is the polite thing to say among eagles.&lt;br&gt;&lt;br&gt;&quot;May the wind under your wings bear you where the sun sails and the moon walks,&quot; answered Gandalf, who knew the correct reply.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-05-08T22:20:21.123" LastActivityDate="2015-05-08T22:20:21.123" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="14444" PostTypeId="2" ParentId="14411" CreationDate="2015-05-07T08:50:17.657" Score="7" Body="&lt;p&gt;It's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Crobat, a pokemon&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/maYxr.png&quot; alt=&quot;Crobat&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Found at the unlucky square&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Crobat is #169 in the &lt;a href=&quot;http://bulbapedia.bulbagarden.net/wiki/List_of_Pok%C3%A9mon_by_National_Pok%C3%A9dex_number&quot; rel=&quot;nofollow noreferrer&quot;&gt;National Dex&lt;/a&gt;, 13² = 169.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From your pocket I can scare&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Mentioned before, pocket monster&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With precious metals, my premiere&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Crobat first appeared in Pokemon Generation II, Gold/Silver (which are precious metals).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A pet that dad and daughter share&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;s&gt;Maybe something to do with how Crobat evolves by happiness? Not sure.&lt;/s&gt;&lt;br&gt;&#xA;&lt;s&gt;Could also possibly have to do with how people of all generations can enjoy playing Pokemon games at this point.&lt;/s&gt;&lt;br&gt;&#xA;With credit to tfitzger (comments below):&lt;br&gt;Both Koga and his daughter Janine have trained Crobats.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first hint explained:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Within a ball I make my lair&lt;br&gt;&#xA;-- Pokemon live in Pokeballs.&lt;br&gt;&#xA;My wings are furled, a double pair&lt;br&gt;&#xA;-- Crobat has four wings, two pairs of two.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5265" LastEditorUserId="5265" LastEditDate="2015-05-13T08:26:37.340" LastActivityDate="2015-05-13T08:26:37.340" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14445" PostTypeId="2" ParentId="14427" CreationDate="2015-05-07T08:50:47.677" Score="10" Body="&lt;p&gt;Would it be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a bullet, maybe?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I'm bitten, you will suffer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; yes, when I bite the bullet, I suffer indeed from something &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I'm precious, you will suffer no more.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a silver bullet is a solution to a problem, perhaps to the same problem that made me bite the bullet earlier&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I'm used, you will never see me again.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; yes, if you use a litteral bullet on me, I might be killed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I enjoyed that one a lot, thanks !&lt;/p&gt;&#xA;" OwnerUserId="11827" LastEditorUserId="11827" LastEditDate="2015-05-07T08:57:37.680" LastActivityDate="2015-05-07T08:57:37.680" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14446" PostTypeId="2" ParentId="14427" CreationDate="2015-05-07T09:35:45.800" Score="5" Body="&lt;p&gt;I think it's a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Coin&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I'm bitten, you will suffer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bit a coin and tell me how you feel!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I'm precious, you will suffer no more.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A gold coin is worth a lot of money. If you have one, you won't suffer poverty.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I'm used, you will never see me again.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After you've spent a coin, it's gone!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-05-07T09:35:45.800" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14447" PostTypeId="2" ParentId="14427" CreationDate="2015-05-07T09:45:08.053" Score="-3" Body="&lt;p&gt;Okay, it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Hymen&quot; rel=&quot;nofollow&quot;&gt;Hymen&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I'm bitten, you will suffer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Man genital is like a snake, when it 'bite' hymen, some woman will suffer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I'm precious, you will suffer no more.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When decide to keep it, you will not suffering. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I'm used, you will never see me again.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If already bitten, you will never see your hymen again. (&lt;a href=&quot;http://en.wikipedia.org/wiki/Artificial_hymen&quot; rel=&quot;nofollow&quot;&gt;actually you can&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9160" LastActivityDate="2015-05-07T09:45:08.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14448" PostTypeId="1" AcceptedAnswerId="14451" CreationDate="2015-05-07T10:05:52.190" Score="2" ViewCount="1744" Body="&lt;p&gt;You are in an absolutely dark room, with (a large number of) coins strewn all over the floor. You know that exactly 20 coins are tails up. You need to split the coins into two groups, such that both have the same number of tails. How would you achieve this?&lt;/p&gt;&#xA;" OwnerUserId="11425" LastEditorUserId="15544" LastEditDate="2015-10-02T18:51:58.453" LastActivityDate="2019-01-19T11:15:02.177" Title="A logic based word puzzle: Coins in a dark room" Tags="|mathematics|logical-deduction|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14449" PostTypeId="2" ParentId="14440" CreationDate="2015-05-07T10:06:13.780" Score="5" Body="&lt;p&gt;It is  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Red rose  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a wonderful help to women,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Help to their self esteem?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The hope of something to come.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When a women receives a red rose, the hope is the one who gave it to her is her prince charming.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I harm No citizen except my slayer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The one who cutted the rose and removed all the thorns is the only harmed by it  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rooted I stand on a high bed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rooted obviously, the high bed is the rest of the roses?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am shaggy below.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Full of leafs, which are removed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes the beautiful&lt;br&gt;&#xA;Peasant's daughter, an eager-armed,&lt;br&gt;&#xA;Proud woman grabs my body,&lt;br&gt;&#xA;Rushes my red skin, holds me hard,&lt;br&gt;&#xA;Claims my head. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Peasant's daughters aren't princesses, they don't receive roses from princes so.. they have to take them by themselves.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The curly-haired&lt;br&gt;&#xA;Woman who catches me fast will feel&lt;br&gt;&#xA;Our meeting. Her eye will be wet.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Impulsive woman who takes the rose before the thorns are removed. Don't know why curly...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;" OwnerUserId="4652" LastEditorUserId="4652" LastEditDate="2015-05-07T12:40:59.827" LastActivityDate="2015-05-07T12:40:59.827" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14450" PostTypeId="1" AcceptedAnswerId="14452" CreationDate="2015-05-07T10:12:21.127" Score="6" ViewCount="2549" Body="&lt;p&gt;This story takes us back to the times of &lt;em&gt;King Vikramaditya&lt;/em&gt;. The king was possessed by a demon called &lt;em&gt;Betal&lt;/em&gt; who once asked him a tricky question to judge his wit. He showed him three coins (gold, silver and copper) and told him that if he made a true statement, he'll get a coin else he won't be getting any of the coins. Imagine yourself to be Vikramaditya and try and make an apt statement which would ensure that you get a gold coin.&lt;/p&gt;&#xA;" OwnerUserId="11425" LastEditorUserId="11475" LastEditDate="2015-10-02T17:53:51.880" LastActivityDate="2017-03-28T09:51:33.470" Title="Make a statement to ensure that you get the coin?" Tags="|logical-deduction|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14451" PostTypeId="2" ParentId="14448" CreationDate="2015-05-07T10:13:07.367" Score="12" Body="&lt;p&gt;Assuming the two groups don't need to be of equal size:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Take a group of $20$ coins, and flip them  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This should work because:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You take $x$ (with $0\le x\le20$) tails-up coins, and change them to heads, thus leaving $20-x$ tails-up coins on the floor.&lt;br&gt;&#xA;You've also taken $20-x$ heads-up coins, and changed them to tails.&lt;br&gt;&#xA;&lt;br&gt;&#xA;You now have a group of $20$ coins and the group of remaining coins, both containing $20-x$ tails-up coins&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11717" LastEditorUserId="8874" LastEditDate="2015-10-02T15:10:27.937" LastActivityDate="2015-10-02T15:10:27.937" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14452" PostTypeId="2" ParentId="14450" CreationDate="2015-05-07T10:16:02.973" Score="21" Body="&lt;p&gt;&amp;quot;You'll either give me the gold coin or no coin.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;If the demon gives the gold coin, the statement was true and I received a coin, so everything is OK.&lt;/p&gt;&#xA;&lt;p&gt;If the demon gives me a silver or copper coin, the statement was false and I received a coin, so it can't happen.&lt;/p&gt;&#xA;&lt;p&gt;If the demon gives me no coin, the statement was true and I received no coin, so it can't happen.&lt;/p&gt;&#xA;&lt;p&gt;As you see by asking &amp;quot;You'll either give me the gold coin or no coin.&amp;quot; the only possible outcome would be receiving the gold coin.&lt;/p&gt;&#xA;&lt;p&gt;EDIT for clarification:&lt;/p&gt;&#xA;&lt;p&gt;Let's imagine the two scenarios, where the sentence I said is true and where it's false.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;First Scenario (&amp;quot;You'll either give me the gold coin or no coin.&amp;quot; is true):&lt;/p&gt;&#xA;&lt;p&gt;Since it's true, the demon will give me a coin; but if it's true and the demon gives me a coin, it must be the gold coin (since it was true that he would give me the gold coin).&lt;/p&gt;&#xA;&lt;p&gt;Second Scenario (&amp;quot;You'll either give me the gold coin or no coin.&amp;quot; is false):&lt;/p&gt;&#xA;&lt;p&gt;Since it's false, he will give me no coin. But we just assumed that the statement &amp;quot;You'll give me no coin&amp;quot; (a variant of it actually) is false.&lt;/p&gt;&#xA;&lt;p&gt;Therefore, Second Scenario can't happen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12165" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2017-03-28T09:51:33.470" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="14453" PostTypeId="2" ParentId="10933" CreationDate="2015-05-07T10:35:04.517" Score="0" Body="&lt;p&gt;The inputting sequence is the &lt;a href=&quot;https://www.mathsisfun.com/numbers/fibonacci-sequence.html&quot; rel=&quot;nofollow&quot;&gt;Fibonacci Series&lt;/a&gt; (1,1,2,3,5,8,13,21......).&#xA;The function, I don't know.&lt;/p&gt;&#xA;" OwnerUserId="12091" LastEditorUserId="12091" LastEditDate="2015-05-08T14:14:03.473" LastActivityDate="2015-05-08T14:14:03.473" CommentCount="9" CommunityOwnedDate="2015-05-08T15:02:47.303" ContentLicense="CC BY-SA 3.0" />
  <row Id="14454" PostTypeId="2" ParentId="12835" CreationDate="2015-05-07T10:38:58.923" Score="1" Body="&lt;p&gt;While there are enough explanations in the answers/comments, the OP here still has Doubts, so let me try to answer with a simple approach and hopefully clarify the problem to the OP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A family has two kids; What can they be ? BB BG GB GG , where the order XY is required because they are born in sequence, X before Y. All four possibilities are equally likely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the mother says &quot;One of them is a girl&quot;, she is eliminating the first possibility. So three equally-likely possibilities remain. In one such possibility, both are girls, hence probability is 1/3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the mother says &quot;Elder kid is a girl&quot;, she is eliminating the first two possibilities. So two equally-likely possibilities remain. In one such possibility, both are girls, hence probability is 1/2.&lt;/p&gt;&#xA;" OwnerUserId="10940" LastEditorUserId="10940" LastEditDate="2015-05-07T12:55:24.380" LastActivityDate="2015-05-07T12:55:24.380" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14455" PostTypeId="1" AcceptedAnswerId="14456" CreationDate="2015-05-07T10:44:30.450" Score="4" ViewCount="332" Body="&lt;p&gt;I created this puzzle for a Dungeons and Dragons quest and wanted to know if it is too easy (or hard) and I'll accept any help with the wording itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here it goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The pincers talked at the edge of the meadow&lt;br&gt;&#xA;Balance was kept in the great tree's shadow&lt;br&gt;&#xA;A bowl and its inhabitants were watching&lt;br&gt;&#xA;What the twins and the lamb were setting&lt;br&gt;&#xA;To surprise the one with horns that slept&lt;br&gt;&#xA;If the king had already left&lt;br&gt;&#xA;And the archer was preparing a trap&lt;br&gt;&#xA;Who was resting in the maiden's lap?&quot;&lt;/p&gt;&#xA;" OwnerUserId="12223" LastEditorUserId="8874" LastEditDate="2015-10-02T15:01:19.253" LastActivityDate="2015-10-02T15:01:19.253" Title="Help with checking if a new puzzle is easy or hard" Tags="|riddle|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14456" PostTypeId="2" ParentId="14455" CreationDate="2015-05-07T11:06:24.417" Score="4" Body="&lt;p&gt;Partial answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I recognize some signs of zodiac&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pincers talked at the edge of the meadow&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Scorpius and Cancer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Balance was kept in the great tree's shadow&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Libra&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A bowl and its contents were watching&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Aquarius + Pices&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What the twins and the lamb were setting&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gemini and Ares&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To surprise the horns that slept&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Taurus&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the king had already left&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Leo&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the archer was preparing a trap&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Saggitarius&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who was resting in the maiden's lap?&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Virgo&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;And the answer is:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The one that is left out is &lt;strong&gt;Capricorn&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9907" LastEditorUserId="9907" LastEditDate="2015-05-08T08:39:46.047" LastActivityDate="2015-05-08T08:39:46.047" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14457" PostTypeId="2" ParentId="14417" CreationDate="2015-05-07T11:11:22.393" Score="1" Body="&lt;p&gt;I believe the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oxygen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the second most acidic on the table,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oxygen is immediately adjacent to Fluorine - the most acidic element - on the periodic table&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but taste like nothing if you ever lick me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oxygen is flavourless (and odourless)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might be more filthy than you think,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All the Oxygen atoms in the atmosphere have been breathed in and out of countless lungs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but you let me kiss you on the cheek.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oxygen is the second largest component of the atmosphere - it brushes our cheeks all the time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;for you are the creator of my kind,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure about this line - it might refer to the fact that humans have constructed the periodic table?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yet few have found not prays to me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I can't parse this line, either.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9278" LastActivityDate="2015-05-07T11:11:22.393" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14458" PostTypeId="2" ParentId="13368" CreationDate="2015-05-07T11:29:54.910" Score="0" Body="&lt;p&gt;This solution takes the problem as it is with no additional assumptions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the given information, camels can only move forward and can jump over one camel, given that they will have at least a camel sized space to land on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;O First merchant's camel&#xA;X Second merchant's camel&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;        OOOOO XXXXX                                                           &#xA;        OOOOOX XXXX                                                           &#xA;        OOOO XOXXXX                                                           &#xA;        OOO OXOXXXX                                                           &#xA;        OOOXO OXXXX                                                           &#xA;        OOOXOXO XXX                                                           &#xA;        OOOXOXOX XX&#xA;        OOOXOX XOXX&#xA;        OOOX XOXOXX&#xA;        OO XOXOXOXX&#xA;        O OXOXOXOXX&#xA;        OXO OXOXOXX&#xA;       XO O OXOXOXX&#xA;&#xA;       at this point the leftmost X camel has crossed and can &#xA;       walk forward any number of times to make space.&#xA;&#xA;   X    O OXO OXOXX   &#xA;   X    OXO O OXOXX &#xA;   X    OXO OXO OXX&#xA;   X    OXO OXOXO X&#xA;   X    OXO OXOX OX&#xA;   X    OXOXO OX OX&#xA;   X    OXOXOXO  OX &#xA;   X    OXOXOXO XO       Same goes for this right most O camel. &#xA;   X    OXOXOXOX     O&#xA;   X   XO OXOXOX     O&#xA;   XX   O OXOXOX     O&#xA;   XX   OXO OXOX     O&#xA;   XX  XO O OXOX     O&#xA;   XXX  O OXO OX     O&#xA;   XXX  OXO O OX     O&#xA;   XXX XO O O OX     O&#xA;   XXXX O O O OX     O&#xA;   XXXX O O OXO      O&#xA;   XXXX O OXO O      O&#xA;   XXXX OXO O O      O&#xA;   XXXXXO O O O      O&#xA;   XXXXX         OOOOO&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="12224" LastEditorUserId="12224" LastEditDate="2015-05-07T11:36:48.103" LastActivityDate="2015-05-07T11:36:48.103" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14459" PostTypeId="2" ParentId="14455" CreationDate="2015-05-07T11:39:36.923" Score="1" Body="&lt;p&gt;My best guess: Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the fish&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The clues are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the signs of the zodiac.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pincers = Cancer &amp;amp; Scorpio, Balance = Libra, A bowl and it's contents = Aquarius, Twins = Gemini, lamb = Aries, Horns = Capricorn &amp;amp; Taurus, King = Leo, Archer = Sagittarius, Maiden = Virgo, so the only one not mentioned = Pisces.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12226" LastActivityDate="2015-05-07T11:39:36.923" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14460" PostTypeId="1" AcceptedAnswerId="14465" CreationDate="2015-05-07T11:39:50.440" Score="2" ViewCount="394" Body="&lt;p&gt;There is  a letter leaving two letters of the alphabet in order, after the letters placed at odd-numbered positions and leaving one letter of the alphabet in order after the placed at the  even-numbered positions.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;ADFIKN&lt;/li&gt;&#xA;&lt;li&gt;BEGJLN&lt;/li&gt;&#xA;&lt;li&gt;CFHKLO&lt;/li&gt;&#xA;&lt;li&gt;DFIKNP&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I need help in understanding this puzzle and in solving it. I have to choose an option which satisfies the given condition in the question.&lt;/p&gt;&#xA;" OwnerUserId="12227" LastEditorUserId="8874" LastEditDate="2015-10-02T15:00:37.443" LastActivityDate="2015-10-02T15:00:45.860" Title="Alphabet Puzzle" Tags="|pattern|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14461" PostTypeId="1" AcceptedAnswerId="14463" CreationDate="2015-05-07T11:57:58.727" Score="-1" ViewCount="303" Body="&lt;p&gt;The number of letters skipped in between adjacent letters of the series going backwards increases by one. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;OIGDC&lt;/li&gt;&#xA;&lt;li&gt;OMJFA&lt;/li&gt;&#xA;&lt;li&gt;OMKIG&lt;/li&gt;&#xA;&lt;li&gt;ONLKJ&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;According to this, when we start from back, the difference should increase by one as we move towards the starting letter of the series. But doing so is giving me a wrong answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the correct answer should be 2, but why?&lt;/p&gt;&#xA;" OwnerUserId="12227" LastEditorUserId="8874" LastEditDate="2015-10-02T14:59:35.477" LastActivityDate="2015-10-02T14:59:35.477" Title="Alphabet Puzzle (Skipped Letters)" Tags="|letter-sequence|letters|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14462" PostTypeId="2" ParentId="14440" CreationDate="2015-05-07T12:04:59.303" Score="0" Body="&lt;p&gt;Are you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a brush?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a wonderful help to women,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not only women, though!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The hope of something to come. I harm&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You brush your hair before a date or whenever you want to be beautiful.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No citizen except my slayer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Normally it is not harmful, but if you hit it on the pointy ends...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rooted I stand on a high bed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A round brush is often kept on the bathroom shelf, upright in a glas or so.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am shaggy below. Sometimes the beautiful&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Brushes made of (horse)hair are quite shaggy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Peasant's daughter, an eager-armed,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proud woman grabs my body,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; She grabs the handle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rushes my red skin, holds me hard,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; She goes over the rubber part to clear out her hair of the brush.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Claims my head. The curly-haired&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Woman who catches me fast will feel&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our meeting. Her eye will be wet.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Brushing curly hair can be a pain. Ouch!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;" OwnerUserId="10885" LastActivityDate="2015-05-07T12:04:59.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14463" PostTypeId="2" ParentId="14461" CreationDate="2015-05-07T12:10:53.707" Score="3" Body="&lt;p&gt;Here's why the answer is &lt;strong&gt;2&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write the alphabet out:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;ABCDEFGHIJKLMNOPQRSTUVWYZ&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now highlight those letters found in answer &lt;strong&gt;2&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;A&lt;/strong&gt;BCDE&lt;strong&gt;F&lt;/strong&gt;GHI&lt;strong&gt;J&lt;/strong&gt;KL&lt;strong&gt;M&lt;/strong&gt;N&lt;strong&gt;O&lt;/strong&gt;PQRSTUVWYZ&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is trivial to see that the letters skipped increase by one each time, from N to KL to GHI and so on. Therefore &lt;strong&gt;2&lt;/strong&gt; satisfies the conditions in the question and is the answer.&lt;/p&gt;&#xA;" OwnerUserId="7308" LastActivityDate="2015-05-07T12:10:53.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14464" PostTypeId="1" AcceptedAnswerId="14515" CreationDate="2015-05-07T12:16:59.310" Score="10" ViewCount="3852" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;You can eat me,&lt;br&gt;&#xA;  You can use me,&lt;br&gt;&#xA;  You can listen to me,&lt;br&gt;&#xA;  You can visit me,&lt;br&gt;&#xA;  It is possible to do all of that at the same time&lt;br&gt;&#xA;  But it is highly improbable.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Sometimes I am filled with rats,&lt;br&gt;&#xA;  Sometimes I am filled with people,&lt;br&gt;&#xA;  Sometimes I am filled with bees,&lt;br&gt;&#xA;  Sometimes I am filled with bugs,&lt;br&gt;&#xA;  But never at the same time&lt;br&gt;&#xA;  Or you'll have a bad time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Edit 1:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Since there is a lot of answer but no one find it, I give you a hint&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rats, people, bees and bugs are not animals. But people are still humans.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;Edit 2:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Here is a new hint&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am layered when you eat me,&lt;br&gt;&#xA; I am tabbed when you use me,&lt;br&gt;&#xA; I am played when you listen to me,&lt;br&gt;&#xA; I am opened when you visit me.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;Edit 3:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;There is some answers that match one of the clues but they are not answers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cake is valid for the &quot;You can eat me&quot; clue. I am looking for a cake name.&lt;br&gt;&#xA; House is valid for the &quot;You can visit me&quot; clue. I am looking for a type of house.&lt;br&gt;&#xA; Subway is valid for the &quot;three lines are crossing here&quot; clue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With that being said, I think I almost give you the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like my second riddle is far harder than my first one. I hope you guys enjoyed it anyway.&lt;/p&gt;&#xA;" OwnerUserId="10400" LastEditorUserId="10400" LastEditDate="2015-05-08T02:39:49.223" LastActivityDate="2015-05-08T04:31:51.203" Title="Riddle - three lines are crossing here!" Tags="|riddle|" AnswerCount="15" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="14465" PostTypeId="2" ParentId="14460" CreationDate="2015-05-07T12:19:30.073" Score="1" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;ADFIKN&lt;/strong&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;as &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A(BC)D(E)F(GH)I(J)K(LM)N&#xA;   2   1    2   1    2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4977" LastEditorUserId="8874" LastEditDate="2015-10-02T15:00:45.860" LastActivityDate="2015-10-02T15:00:45.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14466" PostTypeId="2" ParentId="14417" CreationDate="2015-05-07T12:19:38.403" Score="8" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A cell phone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the second most acidic on the table,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The most acidic would have a pH of 0, with the second most having a pH of 1. pHone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but taste like nothing if you ever lick me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Phone screens don't have a taste to them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might be more filthy than you think,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There are a lot of studies out there about how our cell phones have more germs than we could imagine on them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but you let me kiss you on the cheek.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;When you talk on it, your phone is touching, or 'kissing' your cheek.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For you are the creator of my kind,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Cell phones are man-made.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yet few have found not prays to me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;When you hold your phone up to text, it looks a little bit like you're worshiping or praying to your phone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12229" LastEditorUserId="12229" LastEditDate="2015-05-07T12:53:58.833" LastActivityDate="2015-05-07T12:53:58.833" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14467" PostTypeId="2" ParentId="14411" CreationDate="2015-05-07T12:52:48.043" Score="3" Body="&lt;p&gt;&lt;em&gt;(Just registered on this Stack Exchange site to post this, so don’t have enough rep to comment.)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to No. 7892142’s answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A pet that dad and daughter share&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the Indigo League in Gold/Silver and HeartGold/SoulSilver, both Koga and his daughter Janine have a Crobat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12231" LastActivityDate="2015-05-07T12:52:48.043" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14468" PostTypeId="1" AcceptedAnswerId="14471" CreationDate="2015-05-07T13:57:10.287" Score="9" ViewCount="10198" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;I look pretty and i am sweet&lt;/p&gt;&#xA;&lt;p&gt;but still people like to reap me.&lt;/p&gt;&#xA;&lt;p&gt;I am soft sometimes but still&lt;/p&gt;&#xA;&lt;p&gt;people use razor knife to cut me&lt;/p&gt;&#xA;&lt;p&gt;I am here to give you joy&lt;/p&gt;&#xA;&lt;p&gt;but still people put fire on me&lt;/p&gt;&#xA;&lt;p&gt;I am here to make you happy&lt;/p&gt;&#xA;&lt;p&gt;but people use me to smear others face.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So what am I?&lt;/p&gt;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; People can have me at least once a year, but not necessarily.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10433" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2019-03-08T16:45:30.577" Title="I look pretty and I am sweet - What am I?" Tags="|riddle|" AnswerCount="1" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14469" PostTypeId="2" ParentId="14421" CreationDate="2015-05-07T13:57:24.560" Score="9" Body="&lt;p&gt;As has been pointed out, this is referencing the Periodic table... but the other answers got the references backwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ash&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add to me one and I'll become a mess;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 on the periodic table is Hydrogen, H. H+ash = Hash, a synonym for mess.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add to me six and I'll buy you a dress.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 6 on the periodic table is Carbon, C. C+ash = Cash, something you buy a dress with.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add thirty-five, I'll get bolder and louder;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 35 on the periodic table is Bromine, Br. Br+ash = Brash, a synonym for &quot;bold&quot; and &quot;loud&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add to me none, and I'm left as a powder.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This one is more literal, as Ash is a powder.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9289" LastActivityDate="2015-05-07T13:57:24.560" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14470" PostTypeId="2" ParentId="14464" CreationDate="2015-05-07T14:18:47.310" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;House&lt;/strong&gt; &lt;br&gt; &lt;br&gt;&#xA; You can eat a gingerbread house, use a house to live in, listen to house music, and visit your friends' houses. You can also eat a gingerbread house and listen to house music while living in your friend's house, but that's not a likely scenario. &lt;br&gt; &lt;br&gt; Rats, bugs, and people all live in houses. And bees live in bee houses, which are used like hives by beekeepers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10246" LastEditorUserId="9000" LastEditDate="2015-05-07T15:35:53.507" LastActivityDate="2015-05-07T15:35:53.507" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14471" PostTypeId="2" ParentId="14468" CreationDate="2015-05-07T15:00:01.960" Score="20" Body="&lt;p&gt;It is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a cake&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I look pretty and I am sweet&#xA;but still people like to reap me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We're going to eat it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am soft sometimes but still&#xA;people use razor knife to cut me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We use a knife to cut the cake&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am here to give you joy&#xA;but still people put fire on me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We put candles on a birthday cake&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am here to make you happy&#xA;but people use me to smear others face.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sometimes people smear cake on each other's face, such as a wedding cake or a birthday cake.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12109" LastEditorUserId="12109" LastEditDate="2015-05-07T21:49:31.057" LastActivityDate="2015-05-07T21:49:31.057" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14472" PostTypeId="2" ParentId="14464" CreationDate="2015-05-07T15:10:53.770" Score="3" Body="&lt;p&gt;My answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Palm (both tree and hand)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can eat the heart&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use it to cultivate coconuts&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can listen to them when the wind is blowing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can visit it, preferably at the beach&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do all of this at the same time is unlikely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bees, bugs, people and rats can climb it, but as far as the hand is concerned you will have a bad time if it is filled by all of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, three lines are crossing it if you look at it (the one in your hand).&lt;/p&gt;&#xA;" OwnerUserId="12232" LastActivityDate="2015-05-07T15:10:53.770" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14473" PostTypeId="2" ParentId="14464" CreationDate="2015-05-07T15:17:40.590" Score="1" Body="&lt;p&gt;You may be a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; nest&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can eat me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is something called an &lt;a href=&quot;http://en.wikipedia.org/wiki/Edible_bird%27s_nest&quot; rel=&quot;nofollow&quot;&gt;edible bird's nest&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can use me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can use the &lt;a href=&quot;https://nest.com/&quot; rel=&quot;nofollow&quot;&gt;nest&lt;/a&gt; thermostat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can listen to me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can listen to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Nest_%28band%29&quot; rel=&quot;nofollow&quot;&gt;Nest&lt;/a&gt; band.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can visit me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can always walk up to an animal's/insect's nest.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am filled with rats,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rats can build nests.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am filled with people,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;a href=&quot;http://en.wikipedia.org/wiki/Beijing_National_Stadium&quot; rel=&quot;nofollow&quot;&gt;Bird's Nest&lt;/a&gt; stadium in Beijing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am filled with bees,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A beehive is a nest.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am filled with bugs, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bugs can build nests.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2015-05-07T15:17:40.590" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14474" PostTypeId="1" AcceptedAnswerId="14601" CreationDate="2015-05-07T15:20:45.370" Score="10" ViewCount="723" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am blind&#xA;  &lt;br&gt;But I can feel.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I feel the same up&#xA;  &lt;br&gt;As I feel down.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I can go round but I'm not round,&#xA;  &lt;br&gt;I am strong and I am bound.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I am found on the floor,&#xA;  &lt;br&gt;I can find 5 things not more.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I have thousands of things,&#xA;  &lt;br&gt;But I can't fly, I don't have wings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;To complete this riddle answer the following:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What five things can I find?&lt;/li&gt;&#xA;&lt;li&gt;What is my name?&lt;/li&gt;&#xA;&lt;li&gt;And where is my domain?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="12014" LastEditorUserId="12014" LastEditDate="2015-05-07T15:28:28.710" LastActivityDate="2015-05-09T13:24:04.437" Title="What is my name, and where is my domain?" Tags="|riddle|word|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14475" PostTypeId="2" ParentId="14464" CreationDate="2015-05-07T15:22:45.713" Score="10" Body="&lt;p&gt;My guess is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;b&gt;Space&lt;/b&gt;&#xA; &lt;br&gt;You can eat up (waste) Space&#xA; &lt;br&gt; You can use Space&#xA; &lt;br&gt; You can listen to space (radio telescope)&#xA; &lt;br&gt; You can visit space (Astronauts)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;it's possible to do them all simultaneously, though unlikely&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Space can be filled with rats, people, bees, and/or bugs,&#xA; &lt;br&gt; but if they're filling the same space at the same time it will be bad&#xA; &lt;br&gt;&lt;br&gt;And Finally, a mathematical &quot;space&quot; occupies three dimensions (x,y,z axis) all of which converge at an origin... three lines crossing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12235" LastEditorUserId="11244" LastEditDate="2015-05-07T15:31:34.767" LastActivityDate="2015-05-07T15:31:34.767" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14476" PostTypeId="2" ParentId="14464" CreationDate="2015-05-07T16:12:29.983" Score="0" Body="&lt;p&gt;My guess, Its &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;ATTIC&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can eat me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well !!! can not fit the first line.. Shame :(&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can use me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We all use it to put stuff in there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can listen to me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; May be refers to some bands, named The Attic or Attic Music&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can visit me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To keep stuff.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is possible to do all of that at the same time&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We can eat(don't know what :p), use, visit and listen to Attic&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But it is highly improbable.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Who does that ??&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am filled with rats,&#xA;Sometimes I am filled with people,&#xA;Sometimes I am filled with bees,&#xA;Sometimes I am filled with bugs,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Are filled with rats, people, bees and bugs. (&lt;a href=&quot;http://www.dailymail.co.uk/news/article-2458499/Pest-controllers-30-000-bees-dozens-squirrels-rats-feeding-honeycomb-attic-woman-98-oblivious-noise.html&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But never at the same time&#xA;Or you'll have a bad time.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bad Idea &lt;a href=&quot;http://gawker.com/man-killed-by-800-000-bees-hiding-in-the-attic-1644108689&quot; rel=&quot;nofollow&quot;&gt;Link&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Title&lt;/strong&gt;&#xA;Three lines are crossing here!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because of triangular shape&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10973" LastActivityDate="2015-05-07T16:12:29.983" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14477" PostTypeId="1" AcceptedAnswerId="14478" CreationDate="2015-05-07T16:22:34.963" Score="2" ViewCount="186" Body="&lt;p&gt;Here's a short and sweet riddle.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;You might have heard not to mess with me.&lt;/p&gt;&#xA;&lt;p&gt;Hit me once and it's on.&lt;/p&gt;&#xA;&lt;p&gt;Hit me again and it's lights out for you!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Who or what is it?&lt;/p&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-07T16:33:54.637" Title="Riddle: Wrong fight to pick" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14478" PostTypeId="2" ParentId="14477" CreationDate="2015-05-07T16:24:59.547" Score="4" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a light switch?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I hit you once&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  I'll turn the lights on&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I hit you again&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  I'll turn the lights off&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12193" LastEditorUserId="11244" LastEditDate="2015-05-07T16:33:54.637" LastActivityDate="2015-05-07T16:33:54.637" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14479" PostTypeId="1" AcceptedAnswerId="14481" CreationDate="2015-05-07T16:35:26.733" Score="25" ViewCount="5448" Body="&lt;p&gt;I have ears but failed to listen&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have mouth but failed to scream&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but good thing!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have ears to scream&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have mouth to listen&lt;/p&gt;&#xA;" OwnerUserId="12054" LastEditorUserId="4204" LastEditDate="2015-05-07T19:19:28.580" LastActivityDate="2016-04-12T22:20:56.387" Title="What screams from the ears and listens from the mouth?" Tags="|riddle|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14480" PostTypeId="2" ParentId="14464" CreationDate="2015-05-07T17:01:18.833" Score="2" Body="&lt;p&gt;My best guess is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A subway.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It seems to fit the title as well.&lt;/p&gt;&#xA;" OwnerUserId="12238" LastActivityDate="2015-05-07T17:01:18.833" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14481" PostTypeId="2" ParentId="14479" CreationDate="2015-05-07T17:01:32.157" Score="33" Body="&lt;p&gt;Are you:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A phone&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have ears but failed to listen&lt;br&gt;&#xA;I have mouth but failed to scream&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Phones have ear and mouth pieces&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have ears to scream&lt;br&gt;&#xA;I have mouth to listen&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; but the ear piece produces noise (screams) and the mouth piece listens&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11323" LastActivityDate="2015-05-07T17:01:32.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14482" PostTypeId="2" ParentId="14464" CreationDate="2015-05-07T17:21:06.963" Score="0" Body="&lt;p&gt;My guess is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Computer memory (or something similar)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can eat me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You can &quot;eat up&quot; memory&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can use me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You can obviously use the memory&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can listen to me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If its an audio file you can listen to it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can visit me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You can visit the file's location in a browser&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is possible to do all of that at the same time&#xA;But it is highly improbable.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You could do all at the same time, but your computer might run slow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am filled with rats,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A type of virus&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am filled with people,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Multiple people accessing something, or multiple users&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am filled with bees,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Not sure about this one..&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am filled with bugs,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There can be bugs in the computer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But never at the same time&#xA;Or you'll have a bad time.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It's not good if a computer has bugs or viruses on it while people are using it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11727" LastActivityDate="2015-05-07T17:21:06.963" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14483" PostTypeId="2" ParentId="14479" CreationDate="2015-05-07T17:46:49.717" Score="11" Body="&lt;p&gt;Are you a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/fVh8r.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have ears but failed to listen&#xA;I have mouth but failed to scream&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The presumably ears don't listen nor the mouth piece speaks or screams&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have ears to scream&#xA;I have mouth to listen&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Put on a sthetho and speak on it. You will know what a &lt;img src=&quot;https://i.stack.imgur.com/VhSC4.png&quot; alt=&quot;enter image description here&quot;&gt; is :-)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4177" LastEditorUserId="4177" LastEditDate="2015-05-07T17:52:09.647" LastActivityDate="2015-05-07T17:52:09.647" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14484" PostTypeId="2" ParentId="14464" CreationDate="2015-05-07T18:01:31.030" Score="3" Body="&lt;p&gt;First guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Viruses&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can eat me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;In the biological sense, you can eat a virus if it is in something like, say, chicken&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can use me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Both in the biological and technological senses, you can use a virus for personal gain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can listen to me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;More in the technological sense, it is possible for a virus to create a sound&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can visit me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Web based viruses...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is possible to do all of that at the same time&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You could be eating a contaminated chicken sandwich while visiting a site with a virus...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am filled with rats,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Ghost RAT is a virus. In a more basic sense, RATs (Remote Admin Tools) help viruses get along systems&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am filled with people,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It is possible for a virus to be passing along photos of humans...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am filled with bees,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;VisualBee is a virus toolbar&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am filled with bugs,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Computer viruses can 'bug' systems to gain information, or have faulty software&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But never at the same time&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you'll have a bad time.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If you have all of these, you have quite a bit of viruses.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As for the title&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The basic image for a virus is three crossed lines: &lt;a href=&quot;https://landofthebluemoon.files.wordpress.com/2012/08/logo-virus.png&quot; rel=&quot;nofollow&quot;&gt;https://landofthebluemoon.files.wordpress.com/2012/08/logo-virus.png&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12242" LastEditorUserId="1664" LastEditDate="2015-05-08T00:20:04.800" LastActivityDate="2015-05-08T00:20:04.800" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14486" PostTypeId="2" ParentId="14464" CreationDate="2015-05-07T18:22:45.910" Score="2" Body="&lt;p&gt;(Similar to MisterEman22) I don't think it's correct, but it matches most of the parts. Is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; chips&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can eat me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Potato chips&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can use me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; wood chips&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can listen to me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chip tunes (a type of music)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can visit me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; chips of land (or islands)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is possible to do all of that at the same time&#xA;But it is highly improbable.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I don't think you will be eating chips on an island while using wood chips and listening to chip tunes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am filled with rats,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am filled with people,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am filled with bees,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am filled with bugs,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But never at the same time&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you'll have a bad time.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I guess chips of land (or islands) can be full of all of these, but if it was 'full' with 4 different things, that wouldn't be very nice.&#xA; &lt;br&gt;&lt;br&gt; Or possibly computer chips can have bugs, RATs and AI/software (made by/of people) in them and as for bee's maybe an annoying buzz ;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-05-07T18:25:26.940" LastActivityDate="2015-05-07T18:25:26.940" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14487" PostTypeId="1" AcceptedAnswerId="14637" CreationDate="2015-05-07T18:44:29.247" Score="1" ViewCount="320" Body="&lt;p&gt;Here is a sequence of numbers, convert to a sentence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;123 436 200&lt;/p&gt;&#xA;&#xA;&lt;p&gt;142 166&lt;/p&gt;&#xA;&#xA;&lt;p&gt;341 401 304 216 255&lt;/p&gt;&#xA;&#xA;&lt;p&gt;143 304 303&lt;/p&gt;&#xA;&#xA;&lt;p&gt;211 320 303 303 214 336&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: There is a logical and basic answer to this.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Modifications:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The answer has nothing to do with logarithms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This puzzle could be solved with just a pencil and paper. No external references are needed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This puzzle uses a Homophonic Substitution Cipher&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="7722" LastEditDate="2015-05-10T00:06:01.280" LastActivityDate="2015-05-10T00:58:53.620" Title="Convert to a sentence" Tags="|cipher|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14488" PostTypeId="2" ParentId="14464" CreationDate="2015-05-07T18:51:42.293" Score="0" Body="&lt;p&gt;I think it's:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Apple&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;You can eat me,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You can eat an apple  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;You can use me,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;MAC product&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;You can listen to me,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Itunes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;You can visit me,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Apple Store&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;It is possible to do all of that at the same time&#xA;But it is highly improbable.&#xA;&#xA;Sometimes I am filled with rats,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This one i dont get (Maybe the company being filled with rats?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Sometimes I am filled with people,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Apple store can be filled with people&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Sometimes I am filled with bees,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm guessing the buzz through earphones while listening to your ipod.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Sometimes I am filled with bugs,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Iphone or Laptop&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;But never at the same time&#xA;Or you'll have a bad time.&#xA;&#xA;I am layered when you eat me,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Apple cake??? Not sure all of a sudden&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I am tabbed when you use me,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Apple laptop(mac) browser tabs or whatever other apple software with tabs etc...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I am played when you listen to me,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Itunes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I am opened when you visit me.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Apple store&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastActivityDate="2015-05-07T18:51:42.293" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14489" PostTypeId="2" ParentId="14464" CreationDate="2015-05-07T18:58:45.357" Score="1" Body="&lt;p&gt;What about a...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tree?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can eat me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Apples and other fruits which grow on trees&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can use me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Harvesting and processing wood to build a house&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can listen to me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Musical instruments made of wood&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can visit me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Resting on a branch. Also, possibly treehouses?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is possible to do all of that at the same time&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eating an apple in a tree branch while listening to a performer play a wooden instrument&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But it is highly improbable.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Self-explanatory&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am filled with rats,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some species of rats nest in trees&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am filled with people,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People in houses?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am filled with bees,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Beehives&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am filled with bugs,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Termites and such nesting in trees&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But never at the same time Or you'll have a bad time&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bees and termites nesting in wooden buildings can potentially be dangerous&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4985" LastEditorUserId="4985" LastEditDate="2015-05-07T19:00:27.967" LastActivityDate="2015-05-07T19:00:27.967" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14490" PostTypeId="2" ParentId="14464" CreationDate="2015-05-07T19:28:48.163" Score="1" Body="&lt;p&gt;I think it is a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; FLOWER&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can eat me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cauliflower ;-)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can use me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; to make a garland&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can listen to me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Need a proper device to listen literally&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can visit me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the garden&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is possible to do all of that at the same time&#xA;But it is highly improbable.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is a paradox.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am filled with rats,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If it is around&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am filled with people,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Empty Trash Bins are sometimes used by Homeless to shield from rain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am filled with bees,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Why only Homeless ? Why not Birds and Bees?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am filled with bugs,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Of course&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But never at the same time&#xA;Or you'll have a bad time.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hmmm.... i guess.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9892" LastEditorUserId="9892" LastEditDate="2015-05-07T20:28:53.033" LastActivityDate="2015-05-07T20:28:53.033" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14491" PostTypeId="1" AcceptedAnswerId="14499" CreationDate="2015-05-07T19:47:46.873" Score="5" ViewCount="1164" Body="&lt;p&gt;Note: supercat's answer was the closest to what I was intending. I've started a &lt;a href=&quot;https://puzzling.stackexchange.com/questions/14746/help-me-improve-the-jumping-robot-problem&quot;&gt;second question&lt;/a&gt; to ask a better way to phrase this puzzle that will allow more people to recognize the solution without the list of caveats I had to create here.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I'm making edits as holes are pointed out to narrow down the question and avoid misdirection. If this negates an answer you gave, I am sorry. I'm trying to make this a workable puzzle. Ideally, it could be told verbally without much loss.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I intend to post a separate question asking how to improve this one since it is obviously too open right now. There's a big difference between the restrictions required for posting here and the restriction required when asking &quot;normal&quot; people in person.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Robot Long and Robot High are tootling along via their preferred method of locomotion when they come to a gorge. They want to get to the other side and they must jump to do so. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the problem, though: At it's narrowest point, the gap is just skinny enough that Robot Long can make the jump. Robot High can jump a little higher than Robot Long but it can't jump as far. If Robot High tried to make the jump, it would plummet to its doom. They are extremely accurate and precise jumpers, able to always jump the same distance at the same speed. They have only two speeds: Stopped and Moving. On each side of the gorge, there's only enough space for one robot to jump or land at a time. (The skinny point is a narrow strip jutting out from the cliff face.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How do they get across?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Because this puzzle is not optimized, I must make several clarifications. Please help me improve this puzzle so I don't have to list off all these restrictions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;There are no other materials available. &lt;/li&gt;&#xA;&lt;li&gt;Not even grass grows here and the robots are not engineered for digging. &lt;/li&gt;&#xA;&lt;li&gt;Climbing down and back up are not an option as they are not engineered for climbing. &lt;/li&gt;&#xA;&lt;li&gt;There are no bridges or other resources available. &lt;/li&gt;&#xA;&lt;li&gt;Going around the gorge is not possible at it encircles the entire planet, this being the narrowest point.&lt;/li&gt;&#xA;&lt;li&gt;Jumping is their only option.&lt;/li&gt;&#xA;&lt;li&gt;They can not stand atop one another as that would weaken their jumping ability such that they would both plummet to their doom. &lt;/li&gt;&#xA;&lt;li&gt;There are no means by which they can combine their jumping power such as holding hands (grabbers?) or tying together.&lt;/li&gt;&#xA;&lt;li&gt;The distances described for their jump limits are the furthest they can jump under the best circumstances. &lt;/li&gt;&#xA;&lt;li&gt;The bottom is far enough that they would be destroyed should they fall.&lt;/li&gt;&#xA;&lt;li&gt;Air friction can be neglected.&lt;/li&gt;&#xA;&lt;li&gt;Assume the ground speed for both robots is fixed. (The planet is large enough and they're not jumping high enough that you can assume air speed is the same as ground speed.)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here is an incredibly precise and detailed diagram of their position and jumping abilities that I have rendered in amazing 3D if you move your head forward and backward very quickly: (not to scale)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XlrHY.png&quot; alt=&quot;Diagram&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2018-05-22T19:17:31.850" Title="Robot Long and Robot High" Tags="|lateral-thinking|story|" AnswerCount="9" CommentCount="10" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14492" PostTypeId="2" ParentId="14464" CreationDate="2015-05-07T20:02:45.490" Score="2" Body="&lt;p&gt;Answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; cake&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can eat me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; obvious (and layered)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can use me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Guitar tabs for music by Cake (the band).  A bit of a stretch, I think.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can listen to me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Music by Cake&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can visit me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure, but maybe a festival of some kind about cake (which has to be opened)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is possible to do all of that at the same time&#xA; But it is highly improbable.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Visiting the festival (or whatever) while listening to Cake, playing along on the guitar, and eating cake?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am filled with rats,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rat found in a birthday cake (February 2014).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am filled with people,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Giant cake that someone jumps out of.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am filled with bees,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Honeycomb can be called cake.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am filled with bugs,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Several software products (having software bugs) with Cake in the name (including CakeWalk, which is used for sound editing, providing another way to hear the cake).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But never at the same time&#xA; Or you'll have a bad time.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure if there's something being missed here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Three lines are crossing here!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure how to tie this in, but may be related to the layers or how it is cut (three cuts results in six slices).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12243" LastActivityDate="2015-05-07T20:02:45.490" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14493" PostTypeId="2" ParentId="14491" CreationDate="2015-05-07T20:27:25.320" Score="0" Body="&lt;p&gt;Unlikely but here is my answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As they are really precise jumpers, I suggest that the robot that can jump only 15 m start his jump&lt;br&gt;&#xA; Then the other robot jump after him to meet him mid-air and bump it to the other side of the gorge.&lt;br&gt;&#xA; So the second jumper transfer some of his movement to the other one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10400" LastActivityDate="2015-05-07T20:27:25.320" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14496" PostTypeId="2" ParentId="14430" CreationDate="2015-05-07T20:40:13.190" Score="6" Body="&lt;p&gt;The last sub-clause must be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; all twos: $222...2$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To prove that, we need to show that (a) it is always possible to have such a sequence as long as there wasn't such a sequence before, and that (b) once we used such a sequence we can't have any more sequences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) is true because, as long as we only had sequences with at least one $1$ in them, none of them can be a sub-sequence of the last one - it has no $1$'s.&#xA;(b) is true because, having used a sequence of $N$ $2$'s means that we couldn't have used a sequence with any $1$'s (otherwise we would have used it, as it comes first alphabetically). So there is no option for a $(N+1)$-long sequence - either it has $N$ $2$'s, or it has an $N$-long sub-sequence with at least one $1$'s, which we already know is not allowed.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It will be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; over 72 million years&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;before the Lizard-men can begin their eviction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm very much building on John Stevens' answer, and I'll be using the terms that are defined there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider these extra tier-$n$ suffixes: all $2$'s apart from up to $(6-n)$ $1$'s in the last $(7-n)$ places.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why are these suffixes allowed?&lt;/strong&gt; Such a suffix makes a sub-clause, $S$, of the form $1^{(n)}2^{(m)}s$, where $s$ is of length $(7-n)$ and has up to $(6-n)$ $1$'s. The middle part, $2^{(m)}$, is 'useless' in checking whether $S$ is forbidden, as any previous-tier sub-clause needs at least $n+1$ $1$'s before a $2$. So only $7$ characters out of $S$ are relevant, and therefore only the first sub-clause can possibly be a sub-sequence of $S$. But the total number of $1$'s in $S$ is smaller than $7$, and therefore the first sub-clause cannot be a subsequence either, and $S$ is not forbidden.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why are these suffixes different to John's?&lt;/strong&gt; John's suffixes, of the form $2^{(d)}2^{(b)}12^{(c)}S_{(k-1)}$, all have a single $1$ outside of the last $a_{(n+2)}$ places, and these extra ones don't $(a_{(n+2)} &amp;gt; n)$. So all of these extra suffixes are new ones. In fact, they will always appear after John's sub-clauses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, apart from John's $\frac{1}{2} K_{(n+1)}(K_{(n+1)} + 1)$ suffixes, there are another $\sum_{k=0}^{6-n}{7 - n \choose k}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's demonstrate this for tier-$4$. Here are all the sub-clauses up to this point:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 1    1111111&#xA; 2    11111122&#xA; 3    111112212&#xA; 4    1111122221&#xA; 5    11111222222&#xA; 6    111122122222&#xA; 7    1111222212221&#xA; 8    11112222212222&#xA; 9    111122222221212&#xA;10    1111222222221221&#xA;11    11112222222221222&#xA;12    111122222222222112&#xA;13    1111222222222222121&#xA;14    11112222222222222122&#xA;15    111122222222222222211&#xA;16    1111222222222222222212&#xA;17    11112222222222222222221&#xA;18    111122222222222222222222&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(Note that tier-$4$ is actually 13-long rather than 10, which already increases the bound. But this is not the point.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sub-clauses 6-11 are constructed in John's way. Indeed, one of the first three characters of the suffix is always $1$, and the two last characters are always a tier-$3$ suffix. Sub-clauses 12-18 are the new ones - they only have up to 2 $1$'s (apart from the prefix), all in the last 3 places.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's re-calculate $K_0$ according to this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; \begin{align} K_3 &amp;amp;\ge \frac{1}{2} K_4 (K_4 + 1) + \sum_{k=0}^{3}{4 \choose k} = \frac{13*14}{2} + 15 = 106 \\ K_2 &amp;amp;\ge \frac{1}{2} K_3 (K_3 + 1) + \sum_{k=0}^{4}{5 \choose k} = \frac{106*107}{2} + 31 = 5702 \\ K_1 &amp;amp;\ge \frac{1}{2} K_2 (K_2 + 1) + \sum_{k=0}^{5}{6 \choose k} = \frac{5702*5703}{2} + 63 = 16259316 \\ K_0 &amp;amp;\ge \frac{1}{2} K_1 (K_1 + 1) + \sum_{k=0}^{6}{7 \choose k} = \frac{16259316*16259317}{2} + 127 = 132182686523713 \end{align}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Obviously, the extra suffixes become negligible at low-$n$ tiers, but the accumulative effect is great.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, I wrote a function to generate the first tiers of the series of laws. It's a nice programming challenge, and I used John's observations about the prefix and the centre to make it efficient. The function doesn't assume anything about the possible suffixes, though, but checks all possibilities (efficiently). I put the function at the bottom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function gives $K_3 = 141$, $K_2 = 11623$. The highest bound I can give is according to this $K_2$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; \begin{align} K_1 &amp;amp;\ge \frac{1}{2} K_2 (K_2 + 1) + \sum_{k=0}^{5}{6 \choose k} = \frac{11623*11624}{2} + 63 = 67552939 \\ K_0 &amp;amp;\ge \frac{1}{2} K_1 (K_1 + 1) + \sum_{k=0}^{6}{7 \choose k} = \frac{67552939*67552940}{2} + 127 = 2281699817545457 \end{align}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At one second per sub-clause, this will take:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; over 72 million years.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's the Python function ('calculate_chain' should be used. It calls the recursive 'generate_suffixes'):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;def generate_suffixes(suffix, i, previous_tiers, sub_lens, res):&#xA;&#xA;    if i == len(suffix):&#xA;        res.append(suffix[:])&#xA;        return&#xA;    vals = [False, True]&#xA;    for v in vals:&#xA;        suffix[i] = v&#xA;        new_sub_lens = []&#xA;        for x in range(len(previous_tiers)):&#xA;            new_sub_lens.append(sub_lens[x] +&#xA;                                (previous_tiers[x][sub_lens[x]] == suffix[i]))&#xA;            if new_sub_lens[x] == len(previous_tiers[x]):&#xA;                break&#xA;        else:&#xA;            generate_suffixes(suffix, i + 1, previous_tiers, new_sub_lens, res)&#xA;    return res&#xA;&#xA;def calculate_chain(n_falses=7, min_l_prefix=2):&#xA;&#xA;    res = [[False] * n_falses,&#xA;           [False] * (n_falses - 1) + [True] * 2]&#xA;    l_prefix = n_falses - 2&#xA;    l_suffix = 2&#xA;    max_falses = 1&#xA;    while l_prefix &amp;gt;= min_l_prefix:&#xA;        suffixes = generate_suffixes([False] * l_suffix, 0, res, [l_prefix] * len(res), [])&#xA;        for s in range(len(suffixes)):&#xA;            res.append([False] * l_prefix + [True] * (2 + s) + suffixes[s])&#xA;        l_suffix = len(res[-1]) - l_prefix&#xA;        l_prefix -= 1&#xA;&#xA;    return res&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="10684" LastEditorUserId="10684" LastEditDate="2015-05-14T16:53:11.287" LastActivityDate="2015-05-14T16:53:11.287" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="14497" PostTypeId="2" ParentId="14491" CreationDate="2015-05-07T20:47:51.190" Score="2" Body="&lt;p&gt;Note: this answer is no longer valid as it has been established that robots are not affectionate and, therefore, cannot hug or hold hands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First Robot:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;25&amp;amp;=v_{x_1}t_1\\&#xA;v_{y_1}&amp;amp;=a_g\frac{t_1}2\\&#xA;10&amp;amp;=v_{y_1}\frac{t_1}2-a_g\frac{t_1^2}8=a_g\frac{t_1^2}8\\&#xA;t_1&amp;amp;=\frac{4\sqrt 5}{a_g}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second Robot:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;15&amp;amp;=v_{x_2}t_2\\&#xA;v_{y_2}&amp;amp;=a_g\frac{t_2}2\\&#xA;20&amp;amp;=v_{y_2}\frac{t_2}2-a_g\frac{t_2^2}8=a_g\frac{t_2^2}8\\&#xA;t_2&amp;amp;=\frac{4\sqrt{10}}{a_g}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robots Hugging tightly and jumping at same time:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;v_{x_c}=\frac{m_1v_{x_1}+m_2v_{x_2}}{m_1+m_2}\quad&#xA;v_{y_c}=\frac{m_1v_{y1}+m_2v_{y2}}{m_1+m_2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the robots weigh the same (I can do more if you need):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;0&amp;amp;=v_{yc}t_c-a_g\frac{t_c^2}2\\&amp;amp;=2(\sqrt{10}+\sqrt5)t_c-a_gt_c^2\\&#xA;t_c&amp;amp;=\frac{2(\sqrt {10}+\sqrt 5)}{a_g}\\&#xA;v_{x_c}t_c&amp;amp;=\left(\frac{25}{\left(\frac{8\sqrt 5}{a_g}\right)}+\frac{15}{\left(\frac{8\sqrt{10}}{a_g}\right)}\right)\frac{2(\sqrt{10}+\sqrt 5)}{a_g}\\&#xA;v_{xc}t_c&amp;amp;=21.49&amp;gt;20&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="13494" LastEditDate="2017-09-03T13:54:58.760" LastActivityDate="2017-09-03T13:54:58.760" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14498" PostTypeId="2" ParentId="14491" CreationDate="2015-05-07T20:53:20.970" Score="0" Body="&lt;p&gt;Since this seems to be a planet unlike earth:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Assume that the planet is extremely windy at times (since there is nothing around to block the winds), and the robot jumps when the wind is strong enough behind him to push him further, enough to make the jump.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alternatively:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Long stands at the edge of the cliff, and High jumps just prior to hitting long (such that he jumps over Long). Long then jumps vertically to hit High in mid air, giving him (High) a vertical boost (which will allow high to move farther). Then you hope the math works out such that High makes it across. Long will land back to where he jumped, and can jump across on his own.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="5373" LastEditDate="2015-05-08T22:05:45.467" LastActivityDate="2015-05-08T22:05:45.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14499" PostTypeId="2" ParentId="14491" CreationDate="2015-05-07T21:06:03.873" Score="4" Body="&lt;p&gt;I would guess the intended answer is that...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;High takes off from the cliff edge, followed some time later by Long, such that High briefly lands on Long at a point slightly less than 15m from the edge, and can then jump off from Long to complete its journey.  Depending upon the relative masses of the two robots, that approach may send Long to its doom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A potentially safer approach would be...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;have High jump from a point slightly before the cliff edge just after Long has jumped from a point somewhat further back, such that they collide short of the cliff edge.  If they collide at the right angle, that would direct Long back to the ground before the edge of the cliff, while High's added momentum would allow it to sail across the gap.  Once that is accomplished, Long can simply do the jump itself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1467" LastEditorUserId="1467" LastEditDate="2015-05-07T21:18:02.400" LastActivityDate="2015-05-07T21:18:02.400" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14501" PostTypeId="2" ParentId="14464" CreationDate="2015-05-07T21:47:47.317" Score="2" Body="&lt;p&gt;Webpages (because some of your clues: tabbed, played, layered (assuming eat means consume as in consume content), played, and opened all sound like they're describing a browser/webpage), surfaces (including screens: render with scanLINES, can be layered, tabbed, et cetera), rails (playing with alternative ideas of lines and intersections), data (various computer terms), and streams (as in both social streams and literal streams of water) are all ones I considered but I think my best guess is: &quot;text&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Text&quot; because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lines as in a play, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Layered when ate as in that the text is dense (layered) with subtext when read/consumed (ate), &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Text can be tabbed as in indented, or (more likely) split into partial segments when created/edited (used).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can hear text read aloud (or played from an audio file), &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You open a book or webpage to see (visit) text, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rat as in despicable characters, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;People being discussed, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bees could mean literal bees are being describe or because the text &#xA;contains a lot of &quot;buzz words&quot;, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bugs could mean grammatical errors or inaccurate information being presented in the text. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Grammatical errors, buzz words, and rats would make an unpleasant read in my opinion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope you like this guess; if it's wrong, could you give a bit more of a hint by what you by &quot;crossing&quot; or what you meant when you said: &quot;yes [referring to lines meaning actual lines] but not in a geometric context&quot;&lt;/p&gt;&#xA;" OwnerUserId="12251" LastEditorUserId="12251" LastEditDate="2015-05-07T21:55:12.683" LastActivityDate="2015-05-07T21:55:12.683" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14502" PostTypeId="1" AcceptedAnswerId="14505" CreationDate="2015-05-07T21:50:55.320" Score="-1" ViewCount="213" Body="&lt;p&gt;What am I…&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Really?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is emotion…&lt;/p&gt;&#xA;&#xA;&lt;p&gt;besides signals to the brain?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your heart can't really think,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is your &quot;heart&quot; really your mind?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some people mistake me as a pessimist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm really a…&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm really a…&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Really?&lt;/p&gt;&#xA;" OwnerUserId="11814" LastActivityDate="2015-05-10T15:55:57.453" Title="What am I, really?" Tags="|riddle|" AnswerCount="2" CommentCount="4" FavoriteCount="0" ClosedDate="2018-08-27T20:05:15.167" ContentLicense="CC BY-SA 3.0" />
  <row Id="14503" PostTypeId="1" AcceptedAnswerId="14507" CreationDate="2015-05-07T21:53:06.520" Score="8" ViewCount="646" Body="&lt;p&gt;The brilliant engineers at ACME industries have created what they think is a very secure safe. There are $9$ switches, each with three settings (high, low, and off). However, these engineers are lazy, and only $2$ of these switches actually do anything! The safe will open when a lever is pulled while these two switches are in a certain, secret setting. The &quot;security&quot; of this safe relies on the fact that nobody knows which pair of switches are the functional ones. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The engineers believe that anyone trying to open the safe would potentially have to try all $3^{9}$ possible settings of the switches. Your task: make fools of them by showing how to crack the safe in just $15$ guesses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Source:&lt;/strong&gt; &lt;em&gt;Puzzling Adventures&lt;/em&gt;, by Dennis E. Shasha.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional challenge: What is the fewest number of guesses you need? I don't know the answer.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-05-07T23:26:19.050" Title="Crack the (Poorly Designed) Safe" Tags="|combinatorics|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14504" PostTypeId="2" ParentId="14491" CreationDate="2015-05-07T21:57:17.013" Score="1" Body="&lt;p&gt;Since high robot (H) jumps higher, it necessarily has longer hang-time than long robot (L).  Thus, if it does not jump as far, it must have a lower horizontal airspeed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, it is clear that the two robots must interact in some way while H is jumping or is in the air.  The question rules out any other way that H could get across.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the &quot;jump&quot; mechanism is simply a quick vertical booster rocket:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Both robots would have the same horizontal airspeed as their ground speed.&lt;/li&gt;&#xA;&lt;li&gt;Ergo, H's ground speed must be less than L's ground speed.&lt;/li&gt;&#xA;&lt;li&gt;H's booster would be stronger than L's.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, the two robots approach the cliff, with H in front of L.  They are timed to collide shortly before they reach the cliff.  H jumps right before the collision.  The (hopefully elastic) collision increases H's horizontal speed to at least L's ground speed, and since H's hang-time is greater, this will be more than enough to cross the chasm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;L then stops abruptly, pulls back to get a running start, and jumps the chasm.  Since H's horizontal airspeed is at least as large as L's normally would be, and its hang-time is greater, it will travel farther than just the minimum distance, which means there will be time for L to stop (assuming the robots can stop abruptly).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the jump mechanism does not add a vertical boost, but merely alters the robots velocity, with H jumping at a higher angle (&gt; 45 degrees) than L, the same mechanism would apply.  Each would have a horizontal airspeed less than their ground speed (which could be the same in this case), and H's horizontal airspeed would be less than L's ground speed.  So the collision would increase H's speed to at least L's ground speed, which is more than enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the jump mechanism gives a boost in both horizontal and vertical direction, then this solution probably won't work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also looked at a solution where they both jump, timed to collide at the point where the paths cross.  That would get H across safely, but I am not sure L would make it.  It would be close; after the collision, L and H would be at the same height, L would be travelling at H's typical airspeed, but would not be descending quite as quickly.  It would travel farther than H would have on its own trajectory.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2015-05-07T21:57:17.013" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14505" PostTypeId="2" ParentId="14502" CreationDate="2015-05-07T22:07:32.343" Score="3" Body="&lt;p&gt;Not too sure, but...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You're a realist?  Maybe even a scientist.  Dunno&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The part about the brain makes me think both the first and second.  The heart line can also be made to fit both answers.  However, the first answer really fits the pessimist clue.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Side note, why is the riddle symmetric?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11790" LastActivityDate="2015-05-07T22:07:32.343" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14506" PostTypeId="2" ParentId="14491" CreationDate="2015-05-07T22:31:07.513" Score="5" Body="&lt;p&gt;Well, we're tagged with &quot;lateral-thinking&quot; here, so..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robot High makes it across the gorge by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; leaning over the edge such that his &quot;upward jump&quot; is really more of a &quot;forward jump&quot;. He jumps &lt;em&gt;laterally&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2015-05-07T22:31:07.513" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14507" PostTypeId="2" ParentId="14503" CreationDate="2015-05-07T22:35:43.620" Score="8" Body="&lt;p&gt;Here's a method to crack the safe in 15 moves:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The three settings can be represented as 0, 1, or 2. The possibilities that can open the safe are then 00, 11, 22, 01, 10, 02, 20, 12, and 21. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Step 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Test the combinations of all 9 switches at 0, 1, and 2 for a total of 3 guesses. If none of these work, that eliminates the first 3 potential cases.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Step 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Test these cases: (012012012, 120120120, 201201201) and (021021021, 210210210, 102102102). Any two positions chosen that aren't apart by multiples of 3 have every combination tested. If the safe still isn't opened by now, then the two switches are in the same position in the triplet. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Step 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Test these cases: (000111222, 111222000, 222000111) and (000222111, 222111000, and 111000222). These cases test all the cases for the switches being apart by multiples of 3. After the last test case (guess #15), every possible combination has been tested for any pair of switches.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm fairly certain 15 is the lowest number of guesses needed to guarantee a solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastActivityDate="2015-05-07T22:35:43.620" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14509" PostTypeId="1" CreationDate="2015-05-07T22:58:13.087" Score="-2" ViewCount="9849" Body="&lt;p&gt;Simple logic question, what is the next number and why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;11, 21, 1112, 3112, 211213, 312213, 212223, 114213, ?&lt;/p&gt;&#xA;" OwnerUserId="12254" LastEditorUserId="7308" LastEditDate="2015-05-07T23:56:41.747" LastActivityDate="2015-05-07T23:56:41.747" Title="What is the next number and why? - 11,21,1112,…" Tags="|number-sequence|" AnswerCount="1" CommentCount="2" ClosedDate="2015-05-07T23:15:03.183" ContentLicense="CC BY-SA 3.0" />
  <row Id="14510" PostTypeId="2" ParentId="14509" CreationDate="2015-05-07T23:00:01.310" Score="1" Body="&lt;p&gt;This is a repeat of one on the front page but,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is 31121314, each number is counting the quantity of numbers before it in increasing magnitude. (Three ones, one two, one three, one four). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastActivityDate="2015-05-07T23:00:01.310" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14511" PostTypeId="2" ParentId="14503" CreationDate="2015-05-07T23:08:25.387" Score="2" Body="&lt;p&gt;I think Quark's answer is correct, so I decided to think about a slight variation: what is the quickest way to get into the safe (rather than fewest guesses)? So consider this a bonus answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I make 2 assumptions: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;moving a switch by 1 position and pulling the &quot;guess&quot; lever both take 1 unit of time.&lt;/li&gt;&#xA;&lt;li&gt;moving a row of switches that are all in the same position takes the same amount of time as moving just one of them (because you can push them all with your hand at the same time).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Here is my method to get into the safe AQAP: It takes 59 switch flips and 51 lever pulls (pull the lever after every switch except for all 1's, after the first time) for a total of 110 units of time. I think the unit of time would be no more than a second, and so I can get into the safe in 2 minutes tops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;222222222&#xA;111111111&#xA;000000000&#xA;011111111&#xA;022222222 &#xA;122222222&#xA;111111111&#xA;100000000&#xA;200000000&#xA;211111111&#xA;221111111&#xA;220000000&#xA;110000000&#xA;111111111&#xA;112222222&#xA;002222222&#xA;001111111&#xA;000111111&#xA;000222222&#xA;111222222&#xA;111111111&#xA;111000000&#xA;222000000&#xA;222111111&#xA;222211111&#xA;222200000&#xA;111100000&#xA;111111111&#xA;111122222&#xA;000022222&#xA;000011111&#xA;000001111&#xA;000002222&#xA;111112222&#xA;111111111&#xA;111110000&#xA;222220000&#xA;222221111&#xA;222222111&#xA;222222000&#xA;111111000&#xA;111111111&#xA;111111222&#xA;000000222&#xA;000000111&#xA;000000011&#xA;000000022&#xA;111111122&#xA;111111111&#xA;111111100&#xA;222222200&#xA;222222211&#xA;222222221&#xA;222222220&#xA;111111110&#xA;111111111&#xA;111111112&#xA;000000002&#xA;000000001&lt;/p&gt;&#xA;" OwnerUserId="11154" LastActivityDate="2015-05-07T23:08:25.387" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14512" PostTypeId="2" ParentId="14503" CreationDate="2015-05-07T23:26:19.050" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The mistake the engineers made is not requiring all other switches to be off at the time the two 'secret' switches are in the correct state. This means that we can test multiple combinations with each pull of the lever. So...&lt;p&gt;&#xA; - There are 36 distinct pairs of switches.&lt;br&gt;&#xA; - For each pair, there are nine combinations that need to be checked&lt;br&gt;&#xA; - (OO, OL, OH, LO, LL, LH, HO, HL, HH)&lt;p&gt;&#xA; ...which works out to 324 per line. With three guesses, we can cover all the matching pairs for the entire thing:&lt;p&gt;&#xA; &lt;code&gt;OOOOOOOOO&lt;br&gt;&lt;/code&gt;&#xA; &lt;code&gt;LLLLLLLLL&lt;br&gt;&lt;/code&gt;&#xA; &lt;code&gt;HHHHHHHHH&lt;br&gt;&lt;/code&gt;&#xA;&lt;br&gt;&#xA; We need only six to make sure all adjacent pairs are covered, and we can do it by solving the first 2 switches and repeating the pattern:&lt;p&gt;&#xA; &lt;code&gt;OLOLOLOLO&lt;br&gt;&lt;/code&gt;&#xA; &lt;code&gt;OHOHOHOHO&lt;br&gt;&lt;/code&gt;&#xA; &lt;code&gt;LOLOLOLOL&lt;br&gt;&lt;/code&gt;&#xA; &lt;code&gt;LHLHLHLHL&lt;br&gt;&lt;/code&gt;&#xA; &lt;code&gt;HOHOHOHOH&lt;br&gt;&lt;/code&gt;&#xA; &lt;code&gt;HLHLHLHLH&lt;br&gt;&lt;/code&gt;&lt;p&gt;&#xA; ...which conveniently hits all the 'odd' non-adjacent pairs; since 1-2 is complete, the repeating pattern means (1-4, 1-6, and 1-8) will be as well. Same for (2-3, 2-5, 2-7, 2-9), but reversed, and the sets continue with (3-4, 3-6, 3-8), (4-5, 4-7, 4-9), (5-6, 5-8), (6-7, 6-9), (7-8), (8-9). That's 20 of the 36 pairs done so far.&lt;p&gt;&#xA; Now we make sure the even pairs are OK. Same pattern, but fill each line by skipping one and wrapping once:&lt;p&gt;&#xA; &lt;code&gt;OLLOOLLOO&lt;br&gt;&lt;/code&gt;&#xA; &lt;code&gt;OHHOOHHOO&lt;br&gt;&lt;/code&gt;&#xA; &lt;code&gt;LOOLLOOLL&lt;br&gt;&lt;/code&gt;&#xA; &lt;code&gt;HOOHHOOHH&lt;br&gt;&lt;/code&gt;&#xA; &lt;code&gt;LHHLLHHLL&lt;br&gt;&lt;/code&gt;&#xA; &lt;code&gt;HLLHHLLHH&lt;br&gt;&lt;/code&gt;&lt;p&gt;&#xA; This gives us (1-3, 1-5, 1-7, 1-9), (2-4, 2-6, 2-8), (3-5, 3-7, 3-9), (4-6, 4-8), (5-7, 5-9), (6-8), (7-9) -- the remaining 16.&#xA;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;EDIT: There's a mathematical way to express this; this was assuming that when cracking the safe, I temporarily become an innumerate monkey. If I can think of it, I'll come back and bolt it on.&lt;/p&gt;&#xA;" OwnerUserId="10625" LastActivityDate="2015-05-07T23:26:19.050" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14513" PostTypeId="2" ParentId="11908" CreationDate="2015-05-07T23:41:16.023" Score="-2" Body="&lt;p&gt;People were so accustomed to opening doors by pulling on them, that the frenzy  overrode the ability to reason calmly. They thought the doors were locked and panic spread. Since then, exit doors open outward with a push bar so no knob is needed.&lt;/p&gt;&#xA;" OwnerUserId="12256" LastActivityDate="2015-05-07T23:41:16.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14514" PostTypeId="2" ParentId="14479" CreationDate="2015-05-08T01:05:19.190" Score="3" Body="&lt;p&gt;Screams from the ears&#xA;Listen from the mouth seems to describe&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; headset?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12259" LastEditorUserId="463" LastEditDate="2015-05-08T01:27:57.890" LastActivityDate="2015-05-08T01:27:57.890" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14515" PostTypeId="2" ParentId="14464" CreationDate="2015-05-08T02:58:30.737" Score="4" Body="&lt;p&gt;With the latest clue:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Opera&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Three lines are crossing here&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Opera subway station of the Paris Metro&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can eat me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Opera cake (layered)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can use me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Opera browser (tabbed)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can listen to me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Opera music (played)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can visit me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Opera house (opened)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is possible to do all of that at the same time&lt;br&gt;&#xA;But it is highly improbable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am filled with rats,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Opera rats&quot;, nickname for young ballet students at the Paris Opera&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am filled with people,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Opera house is filled with people&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am filled with bees,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; New guess: The musical note &quot;B&quot;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Wrong guesses so far:&lt;br&gt;&#xA; Guess 1: &quot;Flight of the Bumblebee&quot;&lt;br&gt;&#xA; Guess 2: Bees on top of the Paris Opera? &lt;a href=&quot;http://www.picturetank.com/___/series/b773991aac75a12c6b8dafb20e716a70/Bees_in_Paris.html&quot; rel=&quot;nofollow&quot;&gt;this?&lt;/a&gt;&lt;br&gt;&#xA; Guess 3: &quot;Bis&quot; meaning a musical encore&lt;br&gt;&#xA; Guess 4: The B&amp;amp;B Opera in Paris (bed and breakfast)&lt;br&gt;&#xA; Guess 5: &quot;The British Opera&quot; song by the Bee Gees&lt;br&gt;&#xA; Guess 6: The &quot;buzz&quot; at the opera.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am filled with bugs,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Software bugs in the Opera browser (thanks @tfitzger).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But never at the same time&lt;br&gt;&#xA;Or you'll have a bad time.&lt;/p&gt;&#xA;" OwnerUserId="4479" LastEditorUserId="4479" LastEditDate="2015-05-08T04:31:51.203" LastActivityDate="2015-05-08T04:31:51.203" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="14516" PostTypeId="2" ParentId="14491" CreationDate="2015-05-08T03:58:23.303" Score="0" Body="&lt;p&gt;This answer doesn't directly contradict with rules 8 and 9, and isn't mentioned above so here's my attempt:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Robot High gets a push (or instant collision) at the very start of the jump from Robot Long. There's no connection, coordination, or anything else required. &quot;Jumping Power&quot; is not combined, Robot Long is just pushing/colliding. Once Robot High makes it over, Robot Long then jumps over.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastActivityDate="2015-05-08T03:58:23.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14517" PostTypeId="1" AcceptedAnswerId="14518" CreationDate="2015-05-08T04:16:37.607" Score="8" ViewCount="4361" Body="&lt;p&gt;Find the missing letter in this puzzle.&lt;br&gt;&#xA;&lt;em&gt;Which letter will come at first place?&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GJ0nf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9626" LastEditorUserId="5188" LastEditDate="2015-05-08T04:30:11.480" LastActivityDate="2015-05-08T13:51:28.690" Title="Find the missing letter?" Tags="|letter-sequence|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14518" PostTypeId="2" ParentId="14517" CreationDate="2015-05-08T04:19:21.820" Score="16" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; O, the boxes represent the first letters of &quot;one, two, three, four, five, six&quot; etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastActivityDate="2015-05-08T04:19:21.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14521" PostTypeId="1" AcceptedAnswerId="17786" CreationDate="2015-05-08T05:18:27.690" Score="0" ViewCount="397" Body="&lt;p&gt;Four drivers face off in the race of the century. The path of the race follows this layout:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;100 KM road with a sharp turn every 5 KM&lt;/li&gt;&#xA;&lt;li&gt;50 KM straight road&lt;/li&gt;&#xA;&lt;li&gt;50 KM straight river&lt;/li&gt;&#xA;&lt;li&gt;40 KM upstream river&lt;/li&gt;&#xA;&lt;li&gt;40 KM downstream river&lt;/li&gt;&#xA;&lt;li&gt;40 KM uphill road with a sharp turn every 2 km&lt;/li&gt;&#xA;&lt;li&gt;40 KM uphill straight road&lt;/li&gt;&#xA;&lt;li&gt;40 KM straight river&lt;/li&gt;&#xA;&lt;li&gt;30 KM downhill road with a sharp turn every 5 km&lt;/li&gt;&#xA;&lt;li&gt;50 KM dowhill straight road&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The four drivers build their vehicles with special attributes and specific values, they always follow these setting and never cheat, they always accelerate if they can until they are at their maximum velocity. They only come to a full stop when they need to change modes and they only slow down at turns if they are going to fast to actualy make that turn. All four vehicles have two modes: normal and amphibious. The vehicles behave as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Driver 1: Java Gee Roe&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Maximum speed: 1000 m/s&lt;br&gt;&#xA;  Acceleration: 2 m/ss, minimum 1 m/ss&lt;br&gt;&#xA;  Breaking speed: 25 m/ss&lt;br&gt;&#xA;  Maximum turn speed: 350 m/s&lt;br&gt;&#xA;  Time to change modes: 50 s&lt;br&gt;&#xA;  Special: Every 10,000 m, the vehicle's acceleration is increased by 15 m/ss for 12 seconds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Driver 2: Frames Wist&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Maximum speed: 600 m/s&lt;br&gt;&#xA;  Acceleration: 1 m/ss, minimum 0.5 m/ss&lt;br&gt;&#xA;  Breaking speed: 20 m/ss&lt;br&gt;&#xA;  Maximum turn speed: 300 m/s&lt;br&gt;&#xA;  Time to change modes: 0&lt;br&gt;&#xA;  Special: This vehicle requires no time to change modes, it doesn't need to  stop or even slow down to go from road-&gt;water or water-&gt;road.&lt;br&gt;&#xA;  It also gains +2 m/ss to acceleration while in water.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Driver 3: Pupa Noblest&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Maximum speed: infinite&lt;br&gt;&#xA;  Acceleration: 1 m/ss, minimum 0.5 m/ss&lt;br&gt;&#xA;  Breaking speed: 50 m/ss&lt;br&gt;&#xA;  Maximum turn speed: 550 m/s&lt;br&gt;&#xA;  Time to change modes: 70 s&lt;br&gt;&#xA;  Special: This vehicle gains 0.02 m/ss acceleration every second.&lt;br&gt;&#xA;  Acceleration is reset back to 1 m/ss when the vehicle needs to slow down or stop.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Driver 4: Bight Fuller&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Maximum speed: 1200 m/s&lt;br&gt;&#xA;  Acceleration: 1.5 m/ss, minimum 0.5 m/ss&lt;br&gt;&#xA;  Breaking speed: 60 m/ss&lt;br&gt;&#xA;  Maximum turn speed: 500 m/s&lt;br&gt;&#xA;  Time to change modes: 30 s&lt;br&gt;&#xA;  Special: Every time a vehicle is infront of this one, this vehicle gains a acceleration bonus depending on distance between the two:  &lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;0-2 km -&gt; 0.05 m/ss&lt;/li&gt;&#xA;  &lt;li&gt;2-4 km -&gt; 0.04 m/ss&lt;/li&gt;&#xA;  &lt;li&gt;4-6 km -&gt; 0.03 m/ss&lt;/li&gt;&#xA;  &lt;li&gt;6-8 km -&gt; 0.02 m/ss&lt;/li&gt;&#xA;  &lt;li&gt;8-10 km -&gt; 0.01 m/ss&lt;/li&gt;&#xA;  &lt;li&gt;&gt; 10 km -&gt; 0.005 m/ss&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;Acceleration is reset back to 1.5 m/ss every time this vehicle overtakes another one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On every uphill/upstream road part, the vehicle's acceleration is reduced by 1 m/ss it can't go below their minimum acceleration. At downhill/downstream parts the vehicles gain 1 m/ss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At what place do each of the drivers finish the race?&lt;/p&gt;&#xA;" OwnerUserId="5560" LastEditorUserId="4692" LastEditDate="2015-05-09T10:44:55.527" LastActivityDate="2015-07-11T00:45:49.450" Title="Race of the century" Tags="|mathematics|calculation-puzzle|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14522" PostTypeId="1" AcceptedAnswerId="14523" CreationDate="2015-05-08T05:19:26.077" Score="-1" ViewCount="453" Body="&lt;p&gt;If you look you cannot see me. And if you see me you cannot see anything else. I can make anything you want happen, but later everything goes back to normal. What am I?&lt;/p&gt;&#xA;" OwnerUserId="4977" LastEditorUserId="4977" LastEditDate="2015-05-08T08:21:27.370" LastActivityDate="2015-05-08T08:21:27.370" Title="Guess , What am I?" Tags="|word|" AnswerCount="3" CommentCount="3" FavoriteCount="0" ClosedDate="2015-05-08T09:16:26.943" ContentLicense="CC BY-SA 3.0" />
  <row Id="14523" PostTypeId="2" ParentId="14522" CreationDate="2015-05-08T05:25:31.617" Score="3" Body="&lt;p&gt;i think it is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Imagination&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you look you cannot see me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you just normally look at your surrounding&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;if you see me you cannot see anything else&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if you imagine you'll be out of your sight&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can make anything you want happen&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you can imagine anything( like being part of avengers :D )&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but later everything goes back to normal&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if you stop imagining, you bring back to reality&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12054" LastEditorUserId="12054" LastEditDate="2015-05-08T05:32:11.580" LastActivityDate="2015-05-08T05:32:11.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14524" PostTypeId="2" ParentId="14522" CreationDate="2015-05-08T05:35:24.807" Score="3" Body="&lt;p&gt;It can be a   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dream  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you look, you cannot see me,.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You need to close Eyes to see dream  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And if you see me you cannot see anything else&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; While dreaming you can't see anything else as your eyes are closed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can make anything you want happen&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In dream you can be what you want.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but later everything goes back to normal.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After you wake up, all will be back to normal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9626" LastActivityDate="2015-05-08T05:35:24.807" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14526" PostTypeId="2" ParentId="14522" CreationDate="2015-05-08T05:37:05.880" Score="2" Body="&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; HA! @Himanshu beat me by a minute! :D &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to me, the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dream&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look you cannot see me: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can't see a dream unless you are sleeping!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;if you see me you cannot see anything else&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; While dreaming you can't see anything else.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can make anything you want happen&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We all know how Lucid dreaming works. You can just be anyone and do anything.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But later everything goes back to normal&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As soon as you wake up, you are back to normal life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12263" LastActivityDate="2015-05-08T05:37:05.880" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14527" PostTypeId="2" ParentId="14517" CreationDate="2015-05-08T06:07:51.170" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;O (denoting One)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Others in the puzzle&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;code&gt;O(1) T(2) T(3) F(4) F(5)&#xA;  S(6) S(7) E(8) N(9) T(10)&#xA;  E(11) T(12) T(13) F(14) F(15)&#xA;  S(16) S(17) E(18) N(19) T(20)&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2656" LastActivityDate="2015-05-08T06:07:51.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14528" PostTypeId="1" AcceptedAnswerId="14565" CreationDate="2015-05-08T06:33:13.173" Score="4" ViewCount="390" Body="&lt;p&gt;&lt;strong&gt;Twenty-five specific words&lt;/strong&gt; are contained within this word search. These words are associated with &lt;strong&gt;25 more words&lt;/strong&gt; by an unusual property.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The objectives of this puzzle are to:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;find at least 10 of the specific words&lt;/strong&gt; in the word search  &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;identify the unusual property&lt;/strong&gt; of these words  &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;list the words that are associated&lt;/strong&gt; with each word found in step 1  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The 25 specific words in the word search:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;do not include&lt;/strong&gt; any words with less than 4 letters  &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;do not include&lt;/strong&gt; these words (although variations may be found):  &#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;SOME, MORN, TANG, CUBE, LATE, RIVE, HEEL, WALK  &lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;do include&lt;/strong&gt;:  &#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;fourteen 4-letter words, nine 5-letter words, and two 6-letter words  &lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; - The property for the word association is more mathematical than linguistic.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/7KdMs.png&quot; alt=&quot;WordSearch&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;  K L A W N Y Z O C H&#xA;  A I C D Q K N H F C  &#xA;  G F B U G N A T U N  &#xA;  E Z A D R I C B S U  &#xA;  O Y U O N L E A I M  &#xA;  P F M R L D P N O O  &#xA;  A H O I A T K J N P  &#xA;  L A C V T I S O M E  &#xA;  D W H E E L A L Y N  &#xA;  O K A R X H T E H A&lt;/pre&gt;  &#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-05-08T14:24:49.537" LastActivityDate="2015-05-08T16:26:43.843" Title="The association of found words" Tags="|pattern|word-search|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14529" PostTypeId="2" ParentId="14487" CreationDate="2015-05-08T07:05:32.447" Score="1" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BUT IS THERE ANY ANSWER&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Strategy used:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The 2-5-3 word lengths in the middle of the sentence clearly pointed to &quot;is there any&quot; and the rest was filled in based on OP's reported frustration with a different problem on Puzzling.SE.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11227" LastActivityDate="2015-05-08T07:05:32.447" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14530" PostTypeId="1" AcceptedAnswerId="14683" CreationDate="2015-05-08T08:04:01.703" Score="10" ViewCount="1091" Body="&lt;p&gt;One of my favorite Putnam problems due to a slick solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R$ is at $(3, 4)$ on the cartesian plane.  To try to confuse $R$, the devious $S$ decides to rotate $R$ about the point $(1, 0)$ by $36^\circ$.  $S$ then rotates $R$ by $36^\circ$ about the point $(2, 0)$, then $36^\circ$ about the point $(3, 0)$, then $(4, 0)$, etc., until finally rotating her $36^\circ$ about the point $(10, 0)$.  Where does $R$ end exactly and why?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(Edit) Additional hint:  Narmer and xnor have the correct solution below, but there is still a clever proof it works that no one has gotten.  If you're curious, it involves only very basic geometry, and doesn't require much more than&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; putting a regular polygon in the right starting location.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2722" LastEditorUserId="12109" LastEditDate="2015-05-22T17:08:35.517" LastActivityDate="2015-05-22T17:08:35.517" Title="Do a barrel roll! (i.e. a Euclidean plane rotation puzzle)" Tags="|geometry|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14531" PostTypeId="2" ParentId="14530" CreationDate="2015-05-08T08:36:27.077" Score="12" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are $10$ points used by $S$ and for each of them $S$ rotates $R$ by $36°$ and since $36°\times10 = 360°$ the sum of the rotations gives a full circle.&lt;br&gt;&#xA;&lt;br&gt;&#xA; If the point of rotation had not changed in the rotations, the $R$ point would end in the same position it started. But the point of rotation did change, but only in the $X$ axis and the sequence of the $X$ axis values used by $S$ forms an &lt;a href=&quot;http://en.wikipedia.org/wiki/Arithmetic_progression&quot; rel=&quot;nofollow noreferrer&quot;&gt;arithmetic progression&lt;/a&gt;. This mean that the $Y$ axis of $R$ after the rotations will be the same, $4$.&lt;br&gt;&#xA;&lt;br&gt;&#xA; What happens to the changing axis then? Well, adding $1$ to each point in the $X$ axis shifts the $R$ point on the $X$ axis by the same amount.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The $R$ point will be in $(13, 4)$.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/03X2h.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4652" LastEditorUserId="4652" LastEditDate="2015-05-08T09:31:43.063" LastActivityDate="2015-05-08T09:31:43.063" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="14532" PostTypeId="1" AcceptedAnswerId="14534" CreationDate="2015-05-08T09:01:29.633" Score="0" ViewCount="1951" Body="&lt;p&gt;What is the beginning of eternity,&lt;br&gt;&#xA;The end of time and space,&lt;br&gt;&#xA;The beginning of every end,&lt;br&gt;&#xA;And the end of every race?&lt;/p&gt;&#xA;" OwnerUserId="4312" LastEditorUserId="5373" LastEditDate="2016-12-08T02:20:37.987" LastActivityDate="2016-12-08T02:20:37.987" Title="End of every race, Who am I?" Tags="|riddle|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14533" PostTypeId="2" ParentId="14528" CreationDate="2015-05-08T09:03:01.180" Score="1" Body="&lt;p&gt;Do all the words:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Contain a smaller word?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some words I found that fit the scheme: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;fusion  - ion&lt;br&gt;&#xA;wheel  - heel&lt;br&gt;&#xA;river - rive&lt;br&gt;&#xA;cubed - cube&lt;br&gt;&#xA;latex - late&lt;br&gt;&#xA;banjo - ban&lt;br&gt;&#xA;pecan - can&lt;br&gt;&#xA;task - ask&lt;br&gt;&#xA;open - pen&lt;br&gt;&#xA;opal - pal&lt;br&gt;&#xA;loaf - oaf&lt;br&gt;&#xA;curl - cur  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I did find plenty of other words that don't seem to fit this pattern, though, so I'm not convinced.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Words found so far in the wordsearch itself:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;fusion (one 6 letter word left)&lt;br&gt;&#xA;&lt;br&gt;&#xA;munch, chain, wheel, river, cubed, latex, banjo, pecan, mocha (Nine 5 letter words found)&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;strike&gt;walk&lt;/strike&gt; lawn, daze, gnat, curl, cozy, quay, hawk, link, task, open, opal, loaf, road (One four letter word left)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11717" LastEditorUserId="11717" LastEditDate="2015-05-08T14:49:00.973" LastActivityDate="2015-05-08T14:49:00.973" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14534" PostTypeId="2" ParentId="14532" CreationDate="2015-05-08T09:04:36.617" Score="5" Body="&lt;p&gt;I think the solution is the&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; letter &lt;strong&gt;E&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the beginning of eternity&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;E&lt;/strong&gt;ternity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The end of time and space&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; tim&lt;strong&gt;E&lt;/strong&gt; and spac&lt;strong&gt;E&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The beginning of every end,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;E&lt;/strong&gt;nd&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the end of every race.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; rac&lt;strong&gt;E&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12193" LastEditorUserId="12193" LastEditDate="2015-05-08T09:12:52.413" LastActivityDate="2015-05-08T09:12:52.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14535" PostTypeId="2" ParentId="14502" CreationDate="2015-05-08T09:23:17.023" Score="2" Body="&lt;p&gt;I would have to say you were a Skeptic. It's pretty much in the same line as AJL's response, but I think this doubt about reality relates better to classical, greek skepticism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Skepticism#Philosophical_skepticism&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Skepticism#Philosophical_skepticism&lt;/a&gt; for reference.&lt;/p&gt;&#xA;" OwnerUserId="12165" LastEditorUserId="5373" LastEditDate="2015-05-10T15:55:57.453" LastActivityDate="2015-05-10T15:55:57.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14537" PostTypeId="1" CreationDate="2015-05-08T11:21:53.087" Score="2" ViewCount="16788" Body="&lt;p&gt;There are 5 words in the English language (6 if we count &quot;Eamon&quot;) where the letter combination &quot;ea&quot; makes an &quot;ay&quot; sound rather then an &quot;ee&quot; sound. What are they?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you already know this please don't answer. This very easily solved with Google so this is more about having something to solve with your mind if you want it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: one of the words is a proper name, one of them is archaic but should be known by native English speakers.&lt;/p&gt;&#xA;" OwnerUserId="4963" LastEditorUserId="4963" LastEditDate="2015-05-08T12:00:40.687" LastActivityDate="2016-03-31T00:49:28.540" Title="Words where &quot;ea&quot; = &quot;ay&quot;" Tags="|word|english|knowledge|" AnswerCount="10" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14538" PostTypeId="2" ParentId="14537" CreationDate="2015-05-08T11:28:36.077" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Great!&lt;/strong&gt; still thinking for other 4&lt;br&gt;&#xA; What about &lt;strong&gt;Eadon&lt;/strong&gt;??&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11866" LastEditorUserId="11244" LastEditDate="2015-05-08T13:22:06.253" LastActivityDate="2015-05-08T13:22:06.253" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14539" PostTypeId="2" ParentId="14537" CreationDate="2015-05-08T11:34:36.187" Score="1" Body="&lt;p&gt;JIMMY has one, I thought I'd give one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Shea, as in &quot;shea butter&quot; or &quot;shea tree&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;EDIT: Also, Eamon would be the proper name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE: So I got &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Shea&quot; and &quot;Eamon&quot;, &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;JIMMY got&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Great&quot;,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;dmg got&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Break&quot; (Himanshu was about half a minute slower),&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and Humphreythehater got&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Steak&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Shea&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;was considered amongst them, but if it was, then it's the &quot;archaic&quot; one. The other three non-proper-name ones are all in common usage.&lt;/p&gt;&#xA;" OwnerUserId="9289" LastEditorUserId="19289" LastEditDate="2016-03-31T00:44:23.180" LastActivityDate="2016-03-31T00:44:23.180" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14540" PostTypeId="2" ParentId="14537" CreationDate="2015-05-08T11:36:44.160" Score="-1" Body="&lt;p&gt;May be it is this can be the answer  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Great&lt;br&gt;&#xA;break&lt;br&gt;&#xA;Ready&lt;br&gt;&#xA;Loveable&lt;br&gt;&#xA;leather&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9626" LastEditorUserId="9626" LastEditDate="2015-05-08T11:47:20.240" LastActivityDate="2015-05-08T11:47:20.240" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14541" PostTypeId="2" ParentId="14537" CreationDate="2015-05-08T11:37:09.437" Score="2" Body="&lt;p&gt;What I got thus far:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; break&lt;br/&gt;great&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastActivityDate="2015-05-08T11:37:09.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14542" PostTypeId="2" ParentId="14537" CreationDate="2015-05-08T11:39:35.513" Score="2" Body="&lt;p&gt;I don't think anyone has said&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; steak&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yet&lt;/p&gt;&#xA;" OwnerUserId="11880" LastActivityDate="2015-05-08T11:39:35.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14543" PostTypeId="1" AcceptedAnswerId="14544" CreationDate="2015-05-08T12:47:17.557" Score="2" ViewCount="327" Body="&lt;p&gt;Can you find the next number in the below sequence&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 3 , 7 , 10 , 11 , 12 , 17 ?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note: if already asked please comment below, as I've searched for this but not found on this site so that I posted it!&lt;/p&gt;&#xA;" OwnerUserId="4977" LastEditorUserId="5373" LastEditDate="2015-05-09T10:18:39.700" LastActivityDate="2015-05-09T10:18:39.700" Title="What's the Next number in series #1?" Tags="|number-sequence|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14544" PostTypeId="2" ParentId="14543" CreationDate="2015-05-08T12:52:59.370" Score="5" Body="&lt;p&gt;The next number is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 20&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is because the sequence refers to numbers which:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Only contain the vowel 'e'&lt;br&gt;Eg. thr&lt;strong&gt;ee&lt;/strong&gt;, s&lt;strong&gt;e&lt;/strong&gt;v&lt;strong&gt;e&lt;/strong&gt;n, etc.&lt;br&gt;Thirteen does not qualify because it contains an i, for example.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7308" LastEditorUserId="7308" LastEditDate="2015-05-08T13:42:05.163" LastActivityDate="2015-05-08T13:42:05.163" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14545" PostTypeId="2" ParentId="14537" CreationDate="2015-05-08T13:20:02.100" Score="3" Body="&lt;p&gt;The archaic word should be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;del&gt;Fealty (as in &quot;Swear fealty to your liege, mongrel!&quot;)&lt;/del&gt; I apparently pronounce incorrectly (if/when I ever say it :P)&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Yea&lt;/strong&gt; is another archaic word which I believe would fit&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Other peoples answers which I can take no credit for, just putting them here for completion:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Steak (&lt;a href=&quot;https://puzzling.stackexchange.com/users/11880/humphreythehater&quot;&gt;humphreythehater&lt;/a&gt;)&lt;br&gt;&#xA; Break (&lt;a href=&quot;https://puzzling.stackexchange.com/users/5610/dmg&quot;&gt;dmg&lt;/a&gt;)&lt;br&gt;&#xA; Great (&lt;a href=&quot;https://puzzling.stackexchange.com/users/5610/dmg&quot;&gt;dmg&lt;/a&gt;)&lt;br&gt;&#xA; Shea / Eamon (&lt;a href=&quot;https://puzzling.stackexchange.com/users/9289/glen-o&quot;&gt;Glen O&lt;/a&gt;)&lt;br&gt;&#xA; Reagan (&lt;a href=&quot;https://puzzling.stackexchange.com/users/11475/bailey-m&quot;&gt;Bailey M&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2518" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-05-08T13:44:02.533" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14546" PostTypeId="1" AcceptedAnswerId="14555" CreationDate="2015-05-08T13:20:34.573" Score="2" ViewCount="347" Body="&lt;p&gt;In the following sequence, what is the next number and why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;11, 18, 26, 34, 35, 52, 67, 74, 78, 81&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I think I ate too much pie yesterday.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1748" LastEditorUserId="463" LastEditDate="2015-05-08T13:22:35.440" LastActivityDate="2015-05-08T14:52:36.397" Title="What's the next number?" Tags="|number-sequence|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14547" PostTypeId="2" ParentId="14491" CreationDate="2015-05-08T13:43:43.067" Score="1" Body="&lt;p&gt;Not sure if this breaks rules, but&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Robot High starts at 'Stopped' speed. This doesn't mean that he is &lt;code&gt;braked&lt;/code&gt; however. So, Robot Long pushes Robot High toward the cliff. At a certain point, Robot High switches to &lt;code&gt;Moving&lt;/code&gt; speed. Since Robot High is already moving, the overall speed is increased beyond his natural ability and he can clear the jump. Then Robot Long follows across.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12242" LastActivityDate="2015-05-08T13:43:43.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14548" PostTypeId="2" ParentId="14517" CreationDate="2015-05-08T13:51:28.690" Score="12" Body="&lt;p&gt;I think the other answers about it being:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;O&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;are correct, but at the same time there is zero context for the puzzle. You're just asking me to spot a pattern. As such, I propose that the answer could just as easily be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;E&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;-Treat each column as an individual vertical sequence&lt;br&gt;&#xA;-The letters of each column repeat vertically&lt;br&gt;&#xA;-E.g. from right-most column to left-most column, bottom to top:&lt;br&gt;&#xA;T-F-T-F&lt;br&gt;&#xA;N-F-N-F&lt;br&gt;&#xA;E-T-E-T&lt;br&gt;&#xA;S-T-S-T&lt;br&gt;&#xA;S-E-S-?&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My alternative answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;E&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12274" LastActivityDate="2015-05-08T13:51:28.690" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14549" PostTypeId="1" AcceptedAnswerId="14551" CreationDate="2015-05-08T14:05:26.947" Score="7" ViewCount="3507" Body="&lt;p&gt;In the land of the deep, but not profound,&lt;br&gt;&#xA;We just might see a deer.&lt;br&gt;&#xA;You shouldn't make a single sound,&lt;br&gt;&#xA;And must respect their seer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The people there are very odd,&lt;br&gt;&#xA;and have strange tastes.&lt;br&gt;&#xA;They do not like fish, or even cod,&lt;br&gt;&#xA;And really hate their waists.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They do love food, but hate to eat,&lt;br&gt;&#xA;Eggs might just be best.&lt;br&gt;&#xA;They think Mississippi is really neat,&lt;br&gt;&#xA;but don't like home to rest.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They love to kill, they hate life,&lt;br&gt;&#xA;But murder is not liked.&lt;br&gt;&#xA;Any knife will cause strife.&lt;br&gt;&#xA;but wheels get them psyched.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So tell me now, what it may be,&lt;br&gt;&#xA;that sets this word apart.&lt;br&gt;&#xA;A tag is missing. They love to see,&lt;br&gt;&#xA;To roll they like, but not to start.    &lt;/p&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2022-03-21T16:12:55.867" Title="Land of the Deep but not Profound" Tags="|riddle|rhyme|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14550" PostTypeId="1" AcceptedAnswerId="14552" CreationDate="2015-05-08T14:13:15.747" Score="7" ViewCount="868" Body="&lt;p&gt;A friend gave me this riddle, stating that it would be easy for children but hard for grown-ups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any combination of four digits has an determined value. Here are some examples (note, the position of the example is of no relevance):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0000 = 4&#xA;0001 = 3&#xA;1111 = 0&#xA;1234 = 0&#xA;2266 = 2&#xA;4068 = 4&#xA;6666 = 4&#xA;6088 = 6&#xA;7654 = 1&#xA;8888 = 8&#xA;8901 = 4&#xA;9966 = 4&#xA;9998 = 5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You may ask for any other four-digit number and there is a determined value for it as well. What is the value-function based upon?&lt;/p&gt;&#xA;" OwnerUserId="5153" LastActivityDate="2015-05-08T14:21:28.360" Title="Four-digit quadruples" Tags="|pattern|" AnswerCount="2" CommentCount="3" ClosedDate="2015-05-08T19:16:39.110" ContentLicense="CC BY-SA 3.0" />
  <row Id="14551" PostTypeId="2" ParentId="14549" CreationDate="2015-05-08T14:14:51.763" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The people of this land only like words that contain double letters, d&lt;em&gt;ee&lt;/em&gt;p, d&lt;em&gt;ee&lt;/em&gt;r, o&lt;em&gt;dd&lt;/em&gt; etc., Mi&lt;em&gt;ss&lt;/em&gt;i&lt;em&gt;ss&lt;/em&gt;i&lt;em&gt;pp&lt;/em&gt;i being a extra likeable example.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10330" LastActivityDate="2015-05-08T14:14:51.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14552" PostTypeId="2" ParentId="14550" CreationDate="2015-05-08T14:18:00.397" Score="9" Body="&lt;p&gt;The value is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;How many circles appear within the numbers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Each number corresponds with a value, and they are additive for the final value of the 4 digit string. 0, 6, and 9 map to 1. 8 maps to 2, everything else maps to 0. This value for each digit is the number of circles within the drawing of the digit&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11727" LastActivityDate="2015-05-08T14:18:00.397" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14553" PostTypeId="2" ParentId="14550" CreationDate="2015-05-08T14:21:28.360" Score="3" Body="&lt;p&gt;I think the solution is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sum of each number circles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0 has 1 circle&#xA; &lt;br&gt;1 has no circles&#xA; &lt;br&gt;2 has no circles&#xA; &lt;br&gt;3 has no circles&#xA; &lt;br&gt;4 has no circles&#xA; &lt;br&gt;5 has no circles&#xA; &lt;br&gt;6 has 1 circle&#xA; &lt;br&gt;7 has no circles&#xA; &lt;br&gt;8 has 2 circles&#xA; &lt;br&gt;9 has 1 circle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0000 = 1+1+1+1 = 4&#xA; &lt;br&gt;8888 = 2+2+2+2 = 8&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12193" LastActivityDate="2015-05-08T14:21:28.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14554" PostTypeId="2" ParentId="14450" CreationDate="2015-05-08T14:27:21.690" Score="3" Body="&lt;p&gt;Why stop at one coin? Get all three!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The number of coins you give me can be divided by 3 without remainder!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is exactly true when he gives you 0 or 3 coins. As a true statement together with 0 coins is not possible, the demon Betal has to give you all three coins, including the gold coin!&lt;/p&gt;&#xA;" OwnerUserId="1998" LastEditorUserId="8874" LastEditDate="2015-10-02T15:04:58.800" LastActivityDate="2015-10-02T15:04:58.800" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14555" PostTypeId="2" ParentId="14546" CreationDate="2015-05-08T14:52:36.397" Score="7" Body="&lt;p&gt;The next number is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 84&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sequence lists the&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Zero-based positions of the digit 8 in the decimal expansion of $\pi$ (like pie).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Credits to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://oeis.org/A037006&quot;&gt;A037006 - OEIS&lt;/a&gt; ;-)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4394" LastActivityDate="2015-05-08T14:52:36.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14556" PostTypeId="1" AcceptedAnswerId="14557" CreationDate="2015-05-08T15:06:31.633" Score="3" ViewCount="396" Body="&lt;p&gt;Here's a curious sequence of letters:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;H H H E L I P O A&lt;br&gt;&#xA;  L S C S E C S A C&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But one of the letters is wrong! Can you figure out which letter is wrong, and what the sequence represents?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(note: two lines used for sequence are simply for the sake of keeping it neat, the newline is not part of the puzzle)&lt;/p&gt;&#xA;" OwnerUserId="9289" LastActivityDate="2015-05-08T15:32:44.707" Title="Identify the incorrect letter in this unusual sequence?" Tags="|letter-sequence|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14557" PostTypeId="2" ParentId="14556" CreationDate="2015-05-08T15:32:44.707" Score="6" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;P&lt;/code&gt; should be &lt;code&gt;B&lt;/code&gt;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sequence represents&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The periodic table overlayed with the Fibonacci sequence.&#xA;H = 1, H = 1, He = 2, Li = 3, &lt;code&gt;B&lt;/code&gt; = 5, O = 8, Al = 13...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12242" LastActivityDate="2015-05-08T15:32:44.707" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14558" PostTypeId="1" AcceptedAnswerId="14562" CreationDate="2015-05-08T16:05:19.480" Score="5" ViewCount="2226" Body="&lt;p&gt;I've come in many colors,&lt;br&gt;&#xA;   browns and blues and greens.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been east, I've been west,&lt;br&gt;&#xA;   And most everywhere between.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could say I've given life,&lt;br&gt;&#xA;   And in a sense that's true.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet I'm no saint, let's be clear,&lt;br&gt;&#xA;   I have the power to end life, too.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see me almost every day,&lt;br&gt;&#xA;   Though sometimes with just a glance.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can make you rather cool,&lt;br&gt;&#xA;   If you'd only give me a chance.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me, there should be only one answer, but I'm not sure if anything else could fit this mold. &lt;/p&gt;&#xA;" OwnerUserId="12242" LastEditorUserId="5188" LastEditDate="2015-05-08T19:06:23.820" LastActivityDate="2015-05-08T22:56:17.403" Title="I've come in many colors" Tags="|riddle|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14559" PostTypeId="2" ParentId="11508" CreationDate="2015-05-08T16:09:47.573" Score="9" Body="&lt;h1&gt;Puzzle answer&lt;/h1&gt;&#xA;&lt;p&gt;The idiom is:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 胸有成竹&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;literally translated as&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;a href=&quot;http://mariespastiche.blogspot.co.uk/2013/06/chengyu-to-have-image-of-bamboo-in-ones.html&quot; rel=&quot;noreferrer&quot;&gt;to hold an image of bamboo in one's mind&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/ZMJgm.png&quot; alt=&quot;image1&quot; /&gt; - head or mind&lt;br /&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/g0HzV.png&quot; alt=&quot;image2&quot; /&gt; - bamboo&lt;br /&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/S6ls3.png&quot; alt=&quot;image3&quot; /&gt; - one&lt;br /&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/zQ3Wt.png&quot; alt=&quot;image4&quot; /&gt; - to hold&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;Language notes and cultural background&lt;/h1&gt;&#xA;&lt;p&gt;This is an instance of&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; a Chinese &lt;a href=&quot;http://en.wikipedia.org/wiki/Chengyu&quot; rel=&quot;noreferrer&quot;&gt;chengyu&lt;/a&gt;.&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; &amp;quot;Chengyu are mostly derived from ancient literature. The meaning of a chengyu usually surpasses the sum of the meanings carried by the four characters, as chengyu are often intimately linked with the myth, story or historical fact from which they were derived. As such, chengyu do not follow the usual grammatical structure and syntax of the modern Chinese spoken language, and are instead highly compact and synthetic.&amp;quot;&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; - &lt;a href=&quot;http://en.wikipedia.org/wiki/Chengyu&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The actual meaning of the proverb refers to&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; knowing what you intend to accomplish before you begin, and having a well thought out plan or design in your mind which ensures its success.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The story the proverb is based on:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; There once was a scholar, Wen Tong, who was renown for his beautiful bamboo paintings. His pieces were requested almost daily, from near and afar, for it seemed the paintings were so beautiful, you could almost see the leaves fluttering in the wind.&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; Wen Tong loved bamboo, and would spend time every day observing it, whether in the forests or within his own gardens which were filled with bamboo. He would observe their stalks and leaves, the way they swayed in the wind or stood upright on calm days, how the colors changed with the light. He spent so much time observing bamboo, that when he sat down to paint it, he had a clear image of bamboo in his mind. That is why his paintings were so beautiful and sought after.&lt;br /&gt;&#xA;&lt;br /&gt;&#xA;  - &lt;a href=&quot;http://mariespastiche.blogspot.co.uk/2013/06/chengyu-to-have-image-of-bamboo-in-ones.html&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5260" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-08T16:21:55.917" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14560" PostTypeId="2" ParentId="14558" CreationDate="2015-05-08T16:10:24.713" Score="2" Body="&lt;p&gt;Is the answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Earth? From space, Earth looks to be a combination of brown, blue, and green (ignoring clouds). It is concurrently east west and everything between. It's given life by existing. It can also create a natural disaster. Bare patches of earth are rarely seen in urban areas and even more frequently seen elsewhere. Also, I assume the last line references global warming?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was considering breaking up my reasoning but this is clearer imo.&lt;/p&gt;&#xA;" OwnerUserId="7722" LastActivityDate="2015-05-08T16:10:24.713" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14561" PostTypeId="2" ParentId="14558" CreationDate="2015-05-08T16:14:22.573" Score="2" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The sun?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've come in many colors,&#xA;browns and blues and greens&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The light that comes from the sun is made of many colors&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've been east, I've been west,&#xA;And most everywhere between.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the sun rises in the east and set in the west&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One could say I've given life,&#xA;And in a sense that's true.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Without the sun, no life&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet I'm no saint, let's be clear,&#xA;I have the power to end life, too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you're too close from the sun, or if it's too hot, you can die&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You see me almost every day,&#xA;Though sometimes with just a glance.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We see the sun every day but sometimes it's hide by the clouds&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can make you rather cool,&#xA;If you'd only give me a chance.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I don't know for this one. Maybe something about the shadow or you're just cool with sunglasses...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5359" LastActivityDate="2015-05-08T16:14:22.573" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14562" PostTypeId="2" ParentId="14558" CreationDate="2015-05-08T16:15:28.910" Score="7" Body="&lt;p&gt;Are you  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've come in many colors,&lt;br&gt;&#xA;browns and blues and greens.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Generally thought of as blue, but some beaches have a greenish tint to the water. Muddy or dirty water is brown.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've been east, I've been west,&lt;br&gt;&#xA;And most everywhere between.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; East coast, west coast, rivers, lakes, etc.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One could say I've given life,&lt;br&gt;&#xA;And in a sense that's true.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Can't live without water.  Also, (as proposed by ace), could be thinking of evolution and came from the water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet I'm no saint, let's be clear,&lt;br&gt;&#xA;I have the power to end life, too.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Drowning, storms, etc.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You see me almost every day,&lt;br&gt;&#xA;Though sometimes with just a glance.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water is everywhere, but we don't always look at it. Rain maybe, or watering of grass.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can make you rather cool,&lt;br&gt;&#xA;If you'd only give me a chance.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Swimming is generally an activity when it is really hot.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="9951" LastEditDate="2015-05-08T19:21:23.360" LastActivityDate="2015-05-08T19:21:23.360" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14563" PostTypeId="2" ParentId="14558" CreationDate="2015-05-08T16:16:48.233" Score="6" Body="&lt;p&gt;I say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Trees&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've come in many colors,&#xA;browns and blues and greens.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Typical colors for trees&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've been east, I've been west,&#xA;And most everywhere between.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are trees almost everywhere on the globe, Deserts and the polar caps being the exception&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One could say I've given life,&#xA;And in a sense that's true.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Trees/Forests harbor life and create oxygen so in a way we would die without them and the resources they provide for us as well as being for smaller creatures their own ecosystem&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet I'm no saint, let's be clear,&#xA;I have the power to end life, too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Forest Fires, Trees shed their leaves in fall &lt;em&gt;killing&lt;/em&gt; them, If a tree falls on you... you'll be squashed (lumberjacks have died from being squashed by big trees.) And some movies even go and ridiculously claim that trees can mentally control you to kill yourself. (&lt;em&gt;cough cough the happening&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You see me almost every day,&#xA;Though sometimes with just a glance.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;In most places on the planet you do see trees every single day, but we are so used to them that we don't really pay them attention (especially in cities)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can make you rather cool,&#xA;If you'd only give me a chance.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Trees also provide shade. Which is awesome on lunch breaks and etc... So next time you eat lunch at work, step outside for a sec and find yourself a tree. Be hippie , uh wait! I mean happy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastActivityDate="2015-05-08T16:16:48.233" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14564" PostTypeId="2" ParentId="11508" CreationDate="2015-05-08T16:20:29.820" Score="2" Body="&lt;p&gt;I am almost certain this isn't what you might be looking for, but I can't seem to 'unsee' this option.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/ZMJgm.png&quot; alt=&quot;image1&quot;&gt; - cool&#xA; &lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/g0HzV.png&quot; alt=&quot;image2&quot;&gt; - story&#xA; &lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/S6ls3.png&quot; alt=&quot;image3&quot;&gt; - is best&#xA; &lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/zQ3Wt.png&quot; alt=&quot;image4&quot;&gt; - bro (fist)&#xA; &lt;br&gt;&lt;br&gt; A common English slang idiom for when a friend tells you a story which you don't find interesting, in which you can sarcastically remark &quot;Cool story bro&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastActivityDate="2015-05-08T16:20:29.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14565" PostTypeId="2" ParentId="14528" CreationDate="2015-05-08T16:26:43.843" Score="3" Body="&lt;p&gt;Rest of the answer based on LogicanWithAHat's found answers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each of the words found has the unique property that they are both entered concurrently in a padlock with letters (source: &lt;a href=&quot;http://www.datagenetics.com/blog/march32012/&quot; rel=&quot;nofollow&quot;&gt;http://www.datagenetics.com/blog/march32012/&lt;/a&gt;). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, here's a new list:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;fusion = layout&lt;br&gt;&lt;br&gt;munch = satin&lt;br&gt;chain = ingot&lt;br&gt;wheel = dolls&lt;br&gt;river = arena&lt;br&gt;cubed = melon&lt;br&gt;latex = shale&lt;br&gt;banjo = ferns&lt;br&gt;pecan = tiger&lt;br&gt;mocha = suing&lt;br&gt;&lt;br&gt;lawn = pear&lt;br&gt;daze = spot&lt;br&gt;gnat = tang&lt;br&gt;curl = wolf&lt;br&gt;cozy = wits&lt;br&gt;quay = wage&lt;br&gt;hawk = pies&lt;br&gt;link = spur&lt;br&gt;task = bias&lt;br&gt;open = stir&lt;br&gt;opal = step&lt;br&gt;loaf = twin&lt;br&gt;road = damp&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastActivityDate="2015-05-08T16:26:43.843" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14566" PostTypeId="2" ParentId="10933" CreationDate="2015-05-08T16:43:49.267" Score="2" Body="&lt;p&gt;&lt;strong&gt;Here's what we know so far&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The number we are looking for is whatever 832040 maps to. The function is keeping the output values low, but as the input gets increasingly larger, the output is getting slightly larger&lt;/p&gt;&#xA;&lt;p&gt;Here is a listing of what all the other terms before it are to help you out, based on the Fibonacci sequence:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;1= 0&lt;/p&gt;&#xA;&lt;p&gt;1= 0&lt;/p&gt;&#xA;&lt;p&gt;2= 1&lt;/p&gt;&#xA;&lt;p&gt;3= 2&lt;/p&gt;&#xA;&lt;p&gt;5= 3&lt;/p&gt;&#xA;&lt;p&gt;8= 1&lt;/p&gt;&#xA;&lt;p&gt;13= 3&lt;/p&gt;&#xA;&lt;p&gt;21= 2&lt;/p&gt;&#xA;&lt;p&gt;34= 3&lt;/p&gt;&#xA;&lt;p&gt;55= 3&lt;/p&gt;&#xA;&lt;p&gt;89= 4&lt;/p&gt;&#xA;&lt;p&gt;144= 2&lt;/p&gt;&#xA;&lt;p&gt;233= 4&lt;/p&gt;&#xA;&lt;p&gt;377= 4&lt;/p&gt;&#xA;&lt;p&gt;610= 3&lt;/p&gt;&#xA;&lt;p&gt;987= 3&lt;/p&gt;&#xA;&lt;p&gt;1597= 4&lt;/p&gt;&#xA;&lt;p&gt;2584= 4&lt;/p&gt;&#xA;&lt;p&gt;4181= 5&lt;/p&gt;&#xA;&lt;p&gt;6765= 3&lt;/p&gt;&#xA;&lt;p&gt;10946= 5&lt;/p&gt;&#xA;&lt;p&gt;17711= 4&lt;/p&gt;&#xA;&lt;p&gt;28657= 5&lt;/p&gt;&#xA;&lt;p&gt;46368= 3&lt;/p&gt;&#xA;&lt;p&gt;75025= 5&lt;/p&gt;&#xA;&lt;p&gt;121393= 5&lt;/p&gt;&#xA;&lt;p&gt;196418= 4&lt;/p&gt;&#xA;&lt;p&gt;317811= 4&lt;/p&gt;&#xA;&lt;p&gt;514229= 5&lt;/p&gt;&#xA;&lt;p&gt;832040= ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;And here's the non-Fibonacci numbers that he gave us:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;24862486= 4&lt;/p&gt;&#xA;&lt;p&gt;29862968= 4&lt;/p&gt;&#xA;&lt;p&gt;191199= 5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11727" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-26T13:44:21.943" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14567" PostTypeId="2" ParentId="11216" CreationDate="2015-05-08T17:19:30.207" Score="0" Body="&lt;p&gt;It's clear from the second string that Vincent snapped and is now thinking he is Jabba the Hutt. He stole Jeremy away to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;London&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and is now requesting them to bring Han Solo.&lt;/p&gt;&#xA;" OwnerUserId="12242" LastActivityDate="2015-05-08T17:19:30.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14568" PostTypeId="1" CreationDate="2015-05-08T17:27:17.787" Score="2" ViewCount="249" Body="&lt;p&gt;There is a puzzle: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Find a finite set of points with the following property:&lt;br&gt;&#xA;  1. Points are placed in space (not on a single plane)&lt;br&gt;&#xA;  2. If you take any pair of points A and B there are different points C and D such that AB segment is parallel to CD and doesn't lay on the same line. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Known solution to this problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Take a regular hexagon. Obviously it satisfies rule #2. Now take another hexagon, which has common center O with the first one. This will be the required set of points.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Indeed, if you take A and B on the same hexagon then there are C and D on it; if you take A and B on different hexagons then take C symmetric to A relatively to O and D symmetric to B relatively to O, obviously ABCDO are in the same plane and AB || CD.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I asked myself a question: what if we want to &lt;strong&gt;minimise number of points in the set&lt;/strong&gt;? What is the minimal number to build a set with the described properties? The given solution allows you to get a set with&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 6+6-2 = 10 points. Just make 2 points of the hexagons coincide.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;my question is: Can you find a smaller number points or to prove that this is not possible?&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2015-05-10T17:12:57.867" LastActivityDate="2015-05-11T16:53:12.143" Title="Is there a better solution to parallel segments problem?" Tags="|geometry|optimization|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14570" PostTypeId="2" ParentId="14537" CreationDate="2015-05-08T18:07:57.410" Score="2" Body="&lt;p&gt;Would the archaic one be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Yea&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&#xA;" OwnerUserId="12280" LastEditorUserId="19289" LastEditDate="2016-03-31T00:49:28.540" LastActivityDate="2016-03-31T00:49:28.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14572" PostTypeId="2" ParentId="14530" CreationDate="2015-05-08T18:23:20.603" Score="6" Body="&lt;p&gt;Let $R_k$ be the position of $R$ after $k$ rotations, represented as a complex number, starting at $R_0 =3+4i$. Let $\alpha=e^{2\pi i/10}$ be the complex number representing rotation by $36^\circ$. Then, rotating around the $k^{th}$ point, whose coordinate is $k+0i$, gives the relationship&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_k-k =\alpha (R_{k-1} - k)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we switch to coordinates relative to the current rotation point $S_k=R_k-k$, the recursion becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_k = \alpha (S_{k-1}-1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is intuitively, shifting coordinates by $1$ due to switching rotation points, then rotating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying it ten times takes us back to the starting value because&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_{10} = -(\alpha + \alpha^2 + \dots + \alpha^9 + \alpha^{10}) + \alpha^{10}  S_0 = S_0,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using the fact that $a^{10}=1$ and so the geometric progression is  $0$. Substituting back for the $R$'s, we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_{10} - 10 = R_0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so $R$ ends up ten spaces right of its starting point, at $(13,4)$. &lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-05-08T18:32:50.533" LastActivityDate="2015-05-08T18:32:50.533" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14573" PostTypeId="2" ParentId="11887" CreationDate="2015-05-08T18:24:07.447" Score="1" Body="&lt;p&gt;So, this probably doesn't have much of anything to do with anything, but...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Treat this not as six columns but as three:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;97, 65, 85,    &#xA;37, 65, 85,    &#xA;77, 65, 85,    &#xA;17, 65, 85,    &#xA;57, 65, 85,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;67, 15, 35,    &#xA;07, 15, 35,&#xA;47, 15, 35,&#xA;87, 15, 35,&#xA;27, 15, 35.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The first column loops -60,+40 while the second and third columns are identical down. &lt;code&gt;57&lt;/code&gt; would go down to &lt;code&gt;-3&lt;/code&gt;. Since there are only seven segment displays, there is no easy way to show a negative with two places. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't quite resolve a good reason for the two extra columns to drop down. Together they lose &lt;code&gt;100&lt;/code&gt;, assuming this is a base 10 system. But why assume that? Perhaps the &lt;strong&gt;single digit number&lt;/strong&gt; that flashed was the base for this counter...&lt;/p&gt;&#xA;" OwnerUserId="12242" LastActivityDate="2015-05-08T18:24:07.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14574" PostTypeId="2" ParentId="14521" CreationDate="2015-05-08T19:13:46.223" Score="4" Body="&lt;p&gt;&lt;em&gt;&lt;strong&gt;Here is a first stab... please continue if you think it is appropriate&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;First off, I noticed the racers names were anagrams.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Average Joe&lt;/li&gt;&#xA;&lt;li&gt;Swim Faster&lt;/li&gt;&#xA;&lt;li&gt;Unstoppable&lt;/li&gt;&#xA;&lt;li&gt;Bull Fighter&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Before we start, here is a handy formula for calculating distance travelled under constant acceleration.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$d=v_i\times t+\frac{a\times t^2}{2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;h1&gt;100 km&lt;/h1&gt;&#xA;&lt;p&gt;The fastest out of the gate is 1 followed by 4.  3 starts off similar to 2 but his acceleration increases.  This makes 2 the slowest of the 4 for this stretch, and the easiest to calculate for this section.&lt;/p&gt;&#xA;&lt;h2&gt;2. Swim Faster&lt;/h2&gt;&#xA;&lt;h3&gt;First 9 turns - up to 45km&lt;/h3&gt;&#xA;&lt;p&gt;His maximum turn speed is &lt;span class=&quot;math-container&quot;&gt;$v=300m/s$&lt;/span&gt;, which at an acceleration of &lt;span class=&quot;math-container&quot;&gt;$1m/s^2$&lt;/span&gt; will be achieved after &lt;span class=&quot;math-container&quot;&gt;$t=300s$&lt;/span&gt;.  The distance travelled in this time is &lt;span class=&quot;math-container&quot;&gt;$d=\frac{(300s)^2}{2}=45,000m$&lt;/span&gt;.  This is conveniently right at turn 9, so he accelerates all the way up to this point.&lt;/p&gt;&#xA;&lt;h3&gt;Next 10 turns - up to 95km&lt;/h3&gt;&#xA;&lt;p&gt;For the next 10 turns he will accelerate at &lt;span class=&quot;math-container&quot;&gt;$a_a=1m/s^2$&lt;/span&gt; until just before the turn when he will start breaking at &lt;span class=&quot;math-container&quot;&gt;$a_b=-20m/s^2$&lt;/span&gt;. Since the starting and ending velocities are the same, we know that whatever speed is gained during acceleration must be lost during deceleration.  Thus, we get the following:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$v_{max}=v_i+a_a\times t_a$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$v_i=v_{max}+a_d\times t_d$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Thus:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$v_i=v_i+a_a\times t_a + a_d\times t_d$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$a_a\times t_a = -a_d\times t_d$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$t_a = 20 \times t_d$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;As for distances, we know that&#xA;&lt;span class=&quot;math-container&quot;&gt;$$d_a=v_i \times t_a + \frac{a_a \times t_a^2}{2} = 300m/s \times t_a + \frac{1 m/s^2 \times t_a^2}{2} = t_a\frac{600 + t_a}{2}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$d_d=v_{max} \times t_d + \frac{a_d \times t_d^2}{2} = v_{max} \times t_d + \frac{-20m/s^2 \times t_d^2}{2} = t_d\frac{2v_{max} - 20t_d}{2}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$d_a + d_d = 5,000$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Substituting in, we get:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$d_a=t_a\frac{600 + t_a}{2}=20t_d \frac{600 + 20t_d}{2}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$d_d=t_d\frac{2v_{max} - 20t_d}{2}= t_d\frac{2(v_i+a_a\times t_a) - 20t_d}{2} = t_d\frac{2(300+20t_d) - 20t_d}{2}=t_d\frac{600+20t_d}{2}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$5,000=d_a + d_d = 20t_d \frac{600 + 20t_d}{2} + t_d\frac{600+20t_d}{2}=21t_d\frac{600+20t_d}{2}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$420t_d^2 + 12,600t_d - 10,000 = 0$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$21t_d^2 + 630t_d - 500 = 0$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Solving this equation gives us &lt;span class=&quot;math-container&quot;&gt;$t_d=5\sqrt{\frac{209}{21}}-15~=0.77s$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;We can verify that this is the answer by seeing that &lt;span class=&quot;math-container&quot;&gt;$t_a~=15.4s$&lt;/span&gt; and the distance travelled during that interval is:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$d_a=300m/s \times t_a + \frac{1m/s^2 \times t_a^2}{2}=4761.9$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$v_{max}=300m/s + t_a$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$d_d=v_{max} \times t_d + \frac{-20m/s^2 \times t_d^2}{2}=238.1$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Adding these up give &lt;span class=&quot;math-container&quot;&gt;$5,000$&lt;/span&gt;, and the time it takes to travel this entire distance is &lt;span class=&quot;math-container&quot;&gt;$t_a + t_d ~= 16.25s$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;h3&gt;Last turn&lt;/h3&gt;&#xA;&lt;p&gt;For the leg he must come to a complete stop before entering the water.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-08T19:13:46.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14575" PostTypeId="2" ParentId="14568" CreationDate="2015-05-08T19:55:36.450" Score="0" Body="&lt;p&gt;I think the OP's solution is the minimal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any regular polygon &gt; 4, you can match any line between 2 points. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get into 3D, why not just take a polygon and put it into 3 dimensions? Let's try a pentagon. If you rotate about 1 point, you have symmetry on 2 planes, but the slopes off the square base doesn't have parallel lines.  If we rotate about 2 points, you end up with some points creating a rectangle that you can't match its diagonals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, let's try a hexagon. If you rotate the shape on opposite points, you end up with a flattened cube with 2 pyramids on top. Let's have the tips be top and bottom. Any line from either tip will fall on one of the planes of the original hexagon, which we know has a match. For the edges along the middle, opposite edges are part of the same hexagon, so any pairs there have matches. So, the last pairs are from one edge to another, each of those pairs has a match on the opposite face. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pretty crappy pictures, but here:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/tcN5A.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/L5H0E.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;For the first, the purple area is the part of the red hex behind the blue hex. The second image shows the polyhedron created by all the points. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-05-08T20:34:55.993" LastActivityDate="2015-05-08T20:34:55.993" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14576" PostTypeId="2" ParentId="14558" CreationDate="2015-05-08T22:56:17.403" Score="3" Body="&lt;p&gt;The answer's already been given, but I thought I'd come up with an alternative idea:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Birds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've come in many colors,&lt;br&gt;&#xA;browns and blues and greens.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Yup.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've been east, I've been west,&lt;br&gt;&#xA;And most everywhere between.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Birds are everywhere and are notorious for their migration patterns.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One could say I've given life,&lt;br&gt;&#xA;And in a sense that's true.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Chicken $\rightarrow$ Egg (or was it the other way around?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet I'm no saint, let's be clear,&lt;br&gt;&#xA;I have the power to end life, too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nLTAc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You see me almost every day,&lt;br&gt;&#xA;Though sometimes with just a glance.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This likely depends on geography, but it's certainly possible to see, and yet barely notice, birds daily.  This is probably the weakest link.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can make you rather cool,&lt;br&gt;&#xA;If you'd only give me a chance.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/anz7I.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;br /&gt;&lt;br /&gt;Now tell me that ain't cool...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11851" LastActivityDate="2015-05-08T22:56:17.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14578" PostTypeId="1" AcceptedAnswerId="14596" CreationDate="2015-05-09T01:06:31.507" Score="12" ViewCount="1550" Body="&lt;p&gt;Back in the day, if you wanted to show an image to a large audience, you would use an overhead projector. This would shine a light through a transparency, which is a clear sheet with a black drawing printed on it, and project the resulting shadows onto a large screen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QcN9H.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are currently preparing a talk for the NSA, where you need to display some TOP SECRET information using an overhead projector. All of this fits on a single transparency, but you are worried it may fall into the wrong hands. For safety, you instead want to prepare several transparencies so that&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;when all of the transparencies are stacked, they display a legible message, but&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;if an adversary intercepted some, but not all, of the transparencies, they would learn nothing about this message.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How can you do this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For our purposes, assume a transparency is an array of (quite small) pixels, each of which is black or clear. When several transparencies are stacked, the corresponding pixel on the big screen is black if &lt;em&gt;at least one&lt;/em&gt; of the stacked pixels is black. Two black pixels in a stack looks just as dark as three, or any number.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-06-03T02:13:58.600" LastActivityDate="2017-03-11T14:42:40.387" Title="Overhead Projector Cryptography" Tags="|cryptography|information-theory|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14580" PostTypeId="2" ParentId="11887" CreationDate="2015-05-09T02:00:35.060" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Take just the first columns from both blocks. To find the next single digit, add 7 to the current digit and consider only the the digit corresponding to the unit. The pattern starts with the digit 9 on the 1st line on the 1st block: • 9 + 7 = 16 =&gt; 6 on the 2nd block     • 6 + 7 = 13 =&gt; 3 on the 1st block     • 3 + 7 = 10 =&gt; 0 on the 2nd block     • etc...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12205" LastActivityDate="2015-05-09T02:00:35.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14582" PostTypeId="1" AcceptedAnswerId="14583" CreationDate="2015-05-09T03:06:28.870" Score="12" ViewCount="1558" Body="&lt;h2&gt;Number Swapping Game&lt;/h2&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Initial Board Setup:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pSx1p.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Rules:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;A move is made by selecting one of the 9 cells.&lt;/li&gt;&#xA;  &lt;li&gt;When a cell is selected, the number in that cell swaps places with the largest number in any adjacent cell (horizontally, vertically, or diagonally), &lt;strong&gt;but only if&lt;/strong&gt; that number is larger than the number in the selected cell.&lt;/li&gt;&#xA;  &lt;li&gt;The letters do not move. They are labels for the cells, so that you can easily represent a list of moves.&lt;/li&gt;&#xA;  &lt;li&gt;A list of moves is made by simply putting the letters together, in order from left to right. For example, ADG&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: Not all moves will change the board (such as the move I on move #1), so it's wise to not make those types of moves.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, after the move A is made on the initial board, the board looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Punzq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking this example further, after another move is made, F, the board looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/57kTR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Object of the Game:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Reverse the order of the numbers so that the board looks like this:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RxAUq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer that uses the fewest number of moves wins. I thought up this little puzzle today and created a simple program so that I could mess around with it, and after about an hour of trial and error, the shortest solution I ended up finding was 16 moves long. I have no idea if that is optimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you find a solution that is 16 or less moves, I will upvote your answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get marked as the accepted answer, you must prove that a certain number (16 or less) is optimal, whether by using brute force and sharing your code, or by some elegant proof (which is worded, as much as possible, so that most people can understand)&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-06-12T13:51:04.293" LastActivityDate="2015-06-14T12:32:14.130" Title="Number swapping game" Tags="|mathematics|strategy|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14583" PostTypeId="2" ParentId="14582" CreationDate="2015-05-09T04:20:06.187" Score="8" Body="&lt;p&gt;The minimum number of moves is 12 moves.  A possible solution is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;123 A 523 B 563 D 563 E 563 F 563 H 563 E 563 G 563 C 567 B 587 E 587 A 987 &#xA;456   416   412   812   892   829   824   894   874   834   634   694   654 &#xA;789   789   789   749   741   741   791   721   921   921   921   321   321 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I solved this using a breadth first seach solution.  The number of possible states is 9! = 362880 which makes a breadth first search possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my incredibly slow code.  I wanted to aim for corectness more than speed.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#include &amp;lt;iostream&amp;gt;&#xA;#include &amp;lt;queue&amp;gt;&#xA;#include &amp;lt;map&amp;gt;&#xA;using namespace std;&#xA;&#xA;queue&amp;lt;string&amp;gt; Q;&#xA;map&amp;lt;string, int&amp;gt; D;&#xA;map&amp;lt;string, string&amp;gt; P;&#xA;map&amp;lt;string, string&amp;gt; M;&#xA;&#xA;string next_state(string u, int x) {&#xA;    int r = x / 3;&#xA;    int c = x % 3;&#xA;    int mx = u[r * 3 + c], mr = r, mc = c;&#xA;    for(int i = r - 1; i &amp;lt;= r + 1; i++)&#xA;        for(int j = c - 1; j &amp;lt;= c + 1; j++) &#xA;            if(i &amp;gt;= 0 &amp;amp;&amp;amp; i &amp;lt; 3 &amp;amp;&amp;amp; j &amp;gt;= 0 &amp;amp;&amp;amp; j &amp;lt; 3) &#xA;                if(u[i * 3 + j] &amp;gt; mx) {&#xA;                    mx = u[i * 3 + j];&#xA;                    mr = i;&#xA;                    mc = j;&#xA;                }&#xA;&#xA;    char z = u[r * 3 + c];&#xA;    u[r * 3 + c] = u[mr * 3 + mc];&#xA;    u[mr * 3 + mc] = z;&#xA;    return u;&#xA;}&#xA;&#xA;void bfs() {&#xA;    string u = &quot;123456789&quot;, v;&#xA;    P[u] = u;&#xA;    D[u] = 1;&#xA;    Q.push(u);&#xA;    while(!Q.empty()) {&#xA;        u = Q.front();&#xA;        Q.pop();&#xA;        for(int i = 0; i &amp;lt; 9; i++) {&#xA;            v = next_state(u, i);&#xA;            if(D[v] == 0) {&#xA;                D[v] = D[u] + 1;&#xA;                P[v] = u;&#xA;                M[v] = i + 'A';&#xA;                Q.push(v);&#xA;            }&#xA;        }&#xA;    }   &#xA;}&#xA;&#xA;&#xA;int main() {&#xA;    bfs();&#xA;&#xA;    string a, b, c, u = &quot;987654321&quot;;&#xA;    a = b = c = &quot;&quot;;&#xA;    while(P[u] != u) {&#xA;        a = M[u] + &quot; &quot; + u.substr(0,3) + &quot; &quot; + a;&#xA;        b = &quot;  &quot; + u.substr(3,3) + &quot; &quot; + b;&#xA;        c = &quot;  &quot; + u.substr(6,3) + &quot; &quot; + c;&#xA;        u = P[u];&#xA;    }&#xA;    cout &amp;lt;&amp;lt; &quot;123 &quot; + a &amp;lt;&amp;lt; endl &amp;lt;&amp;lt; &quot;456 &quot; + b &amp;lt;&amp;lt; endl &amp;lt;&amp;lt; &quot;789 &quot; + c &amp;lt;&amp;lt; endl;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1777" LastActivityDate="2015-05-09T04:20:06.187" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14584" PostTypeId="2" ParentId="14474" CreationDate="2015-05-09T06:39:11.360" Score="0" Body="&lt;p&gt;I think it's:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ant.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Five things:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Five times their own weight.&lt;sup&gt;[&lt;a href=&quot;http://ant.edb.miyakyo-u.ac.jp/BE/Kingdom/2829/2829e.html&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;]&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Domain:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Eukaryote&lt;sup&gt;[&lt;a href=&quot;https://www.google.co.in/search?q=domain+of+ants&amp;amp;oq=domain+of+ants&amp;amp;aqs=chrome..69i57.2671j0j4&amp;amp;sourceid=chrome&amp;amp;es_sm=93&amp;amp;ie=UTF-8&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;]&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Ants are blind.Compared to vertebrates, most ants have poor-to-mediocre eyesight and a few subterranean species are completely blind&lt;sup&gt;[&lt;a href=&quot;http://en.wikipedia.org/wiki/Ant#Head&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;]&lt;/sup&gt;&lt;/li&gt;&#xA;  &lt;li&gt;I believe they are symmetrical, both dorsal and ventral side.&lt;/li&gt;&#xA;  &lt;li&gt;Ants can go round&lt;/li&gt;&#xA;  &lt;li&gt;Ants are not round&lt;/li&gt;&#xA;  &lt;li&gt;They can sense things.&lt;/li&gt;&#xA;  &lt;li&gt;Ants are found on the floor.&lt;/li&gt;&#xA;  &lt;li&gt;They are strong (see the below point.).In experiments using ants of the Formica japonica species, the ants were found to be capable of lifting an object weighing 5 times as much as their own weight at the maximum in their mandibles&lt;sup&gt;[&lt;a href=&quot;http://ant.edb.miyakyo-u.ac.jp/BE/Kingdom/2829/2829e.html&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;]&lt;/sup&gt;&lt;/li&gt;&#xA;  &lt;li&gt;Ants have thousands of tiny eyes within their main two eyes&lt;sup&gt;[&lt;a href=&quot;https://waegooksdiary.wordpress.com/2011/09/19/day-35-animals-and-ants/&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;]&lt;/sup&gt;&lt;br&gt;&lt;/li&gt;&#xA;  &lt;li&gt;Maybe ant's dont fly? And they don't have wings too.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12058" LastEditorUserId="12058" LastEditDate="2015-05-09T06:47:54.903" LastActivityDate="2015-05-09T06:47:54.903" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14586" PostTypeId="2" ParentId="14578" CreationDate="2015-05-09T08:53:36.803" Score="2" Body="&lt;p&gt;It suffices to encode a single bit, since then we can copy the protocol on different patches of the overhead to get a multibit message.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a region of $2^n$ pixels that we associate with vectors in $V=\mathbb{F}_2^n$. To a vector $v$ in $V$, we associate the transparency whose clear pixels are in the orthogonal subspace to $v$, the locations $x$ with  $x\cdot v = 0$. Then, a pixel $x$ appears clear in the overlay of transparencies $v_1, ..., v_n$ exactly if $x\cdot v_i = 0$ for each $i$, or equivalently, if $x$ is in the orthogonal subspace to their span.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we encode the bit by generating uniformly random transparencies subject to either:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$v_1, ..., v_n$ are linearly independent&lt;/li&gt;&#xA;&lt;li&gt;$v_1, ..., v_n$ have a single relation of summing to the vector $0$, but are otherwise independent&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the first, they span all of $V$ and so only the $0$-vector location is clear. In the second, they span a subspace of codimension $1$, so a 1D subspace is clear, containing $0$ and a second vector. So, the overlaid transparencies convey a bit that can be seen by an elderly with exponentially good eyesight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, someone with a proper subset of the transparencies cannot distinguish two cases because they correspond to a random set of $n-1$ independent vectors of $V$.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-05-11T00:38:06.823" LastActivityDate="2015-05-11T00:38:06.823" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14587" PostTypeId="1" AcceptedAnswerId="14588" CreationDate="2015-05-09T09:20:49.783" Score="25" ViewCount="9948" Body="&lt;p&gt;&lt;strong&gt;Puzzle:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;King Octopus has servants with six, seven, or eight legs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The servants with seven legs always lie, but the servants with either six or eight legs always say the truth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;One day, 4 servants met:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The blue one says: “Altogether we have 28 legs”;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the green one says: “Altogether we have 27 legs”;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the yellow one says: “Altogether we have 26 legs”;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the red one says: “Altogether we have 25 legs”.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;What is the colour of the servant that says the truth?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a very famous logical puzzle.&lt;/p&gt;&#xA;" OwnerUserId="4977" LastEditorUserId="39" LastEditDate="2015-05-28T01:49:11.300" LastActivityDate="2016-12-13T06:42:46.303" Title="King Octopus and Servants Puzzle" Tags="|logical-deduction|liars|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14588" PostTypeId="2" ParentId="14587" CreationDate="2015-05-09T09:30:56.480" Score="21" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;green&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Prove:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;First&lt;/strong&gt;: One of them tells truth. Indeed, if we suppose that all of them lie, then all of them have 7 legs, which makes the blue a truth-teller. Contradiction.&lt;br&gt;&#xA; Also only one statement can be true. So there is exactly one truth-teller.&lt;br&gt;&#xA; &lt;strong&gt;Second&lt;/strong&gt;: The truth-teller has 6 legs. Indeed, a liar has 7 legs, so if the truthteller has 8 legs, than they have 3*7+8=29 legs in total and all lie.&lt;br&gt;&#xA; &lt;strong&gt;Third&lt;/strong&gt;: Now we just add 3*7+6=27.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="28" LastEditorUserId="5373" LastEditDate="2015-05-09T10:08:31.537" LastActivityDate="2015-05-09T10:08:31.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14589" PostTypeId="1" AcceptedAnswerId="14590" CreationDate="2015-05-09T10:06:04.283" Score="16" ViewCount="1614" Body="&lt;p&gt;In a village, there lived &lt;strong&gt;a kindly baker&lt;/strong&gt; who sold many items (cakes, biscuits, buns, bread loaves, croissants, etc) in his small shop. He made enough profits for running his family.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once, he made &lt;strong&gt;a dozen loaves of a special bread&lt;/strong&gt; each of which cost him 2 U (U is some unit of currency) to make. He usually sold it for 3 U each, when the buyer was rich; but he kindly sold it for 1 U each, when he knew that the buyer was poor. After closing-time, he found that he was sold-out and had made a profit of 3 U on the bread alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now Dear Reader, &lt;strong&gt;something for you to think over&lt;/strong&gt; :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How is this situation possible ?&lt;br&gt;&#xA;  How many rich buyers bought the special bread ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Depending on how you usually think, you may treat it as one unknown or 2 unknowns.&lt;/strong&gt;&lt;br&gt;&#xA; If you think there is one unknown, then there are actually 2 unknowns.&lt;br&gt;&#xA; If you think there are two unknowns, then there are actually 3 unknowns.&lt;br&gt;&#xA; &lt;strong&gt;Basically, there is one EXTRA unknown.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10940" LastEditorUserId="10940" LastEditDate="2018-02-15T13:23:59.317" LastActivityDate="2018-02-15T13:23:59.317" Title="How many rich buyers?" Tags="|mathematics|lateral-thinking|" AnswerCount="5" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14590" PostTypeId="2" ParentId="14589" CreationDate="2015-05-09T10:11:51.713" Score="26" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;yes, it is possible&lt;/strong&gt; and the baker had &lt;strong&gt;8 rich buyers and 5 poor buyers&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; he makes a profit of 1U for each rich buyer and -1U (i.e. a loss of 1U) for each poor buyer. So the parity (odd/even) of his profit changes for each buyer. If the total number of loaves were 12 (an ordinary 'dozen'), his profit would be even and the answer would be &lt;strong&gt;no, it's not possible&lt;/strong&gt;. But because he's a baker, we can assume he makes a &lt;a href=&quot;https://www.google.co.uk/webhp?sourceid=chrome-instant&amp;amp;ion=1&amp;amp;espv=2&amp;amp;ie=UTF-8#q=baker%27s%20dozen&quot;&gt;baker's dozen&lt;/a&gt;, i.e. 13 loaves. Now since his profit is (number sold to rich buyers) - (number sold to poor buyers) = 3, while (number sold to rich buyers) + (number sold to poor buyers) = 13, he must have had 8 rich buyers and 5 poor buyers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-05-10T11:22:25.243" LastActivityDate="2015-05-10T11:22:25.243" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14594" PostTypeId="1" AcceptedAnswerId="14595" CreationDate="2015-05-09T10:37:34.827" Score="-4" ViewCount="993" Body="&lt;p&gt;Find out the next number in following &lt;strong&gt;Simple&lt;/strong&gt; series,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;15, 55, 120, 210, 325, ?, ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note: Question has many solutions.&lt;/p&gt;&#xA;" OwnerUserId="4977" LastEditorUserId="4977" LastEditDate="2015-05-11T05:47:28.627" LastActivityDate="2015-05-11T05:47:28.627" Title="What's the Next number in series #3?" Tags="|number-sequence|" AnswerCount="3" CommentCount="6" FavoriteCount="0" ClosedDate="2015-05-11T14:47:38.173" ContentLicense="CC BY-SA 3.0" />
  <row Id="14595" PostTypeId="2" ParentId="14594" CreationDate="2015-05-09T10:48:25.863" Score="5" Body="&lt;p&gt;Answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;465&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You add 5 to the factor of 5 being added. Between first two you add $5 \times 8$, next one you add $5 \times 13$, then $5 \times 18$, then $5 \times 23$, so next you add $5 \times 28$, ending with $5 \times 93 = 465$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11727" LastEditorUserId="146" LastEditDate="2015-05-09T14:21:23.030" LastActivityDate="2015-05-09T14:21:23.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14596" PostTypeId="2" ParentId="14578" CreationDate="2015-05-09T10:52:31.967" Score="10" Body="&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; Taken from &lt;a href=&quot;http://en.wikipedia.org/wiki/Visual_cryptography&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a simple way to do this. For example, if we need to make two such transparencies, we take our original black-and-white picture and for each of its pixels do this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our transparencies should have two pixels for each pixel of original image. If this pixel is black, one of transparencies should have black pixel and white pixel like this: ■□ and another like this:□■, assigned randomly to them. When stacked together, this pixel will be completely black. However, if the pixel was white, both should have the same combination: ■□ or □■, again, chosen randomly. The resulting pixel will look grey.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is obvious that having only one of the transparencies, it is impossible to decode the picture: out of each pair of pixels, one is black and one is white, positioned without any pattern, but combined, they produce the encrypted picture, which will look like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fjgo9.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not exactly clear, but quite recognisable, and bigger resolution pictures will look much better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, for any transparency it is possible to create a paired one, which, combined with the first, will produce any picture you want, so there is no way to even guess what was encrypted; each of your transparencies will be absolutely useless and undecryptable without the second one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method is called 'Visual encryption' and it is possible to do this with any amount of transparencies. You can read about it here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Visual_cryptography&quot; rel=&quot;nofollow noreferrer&quot;&gt;Visual cryptography&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user10531" LastEditorUserId="5373" LastEditDate="2017-03-11T14:42:40.387" LastActivityDate="2017-03-11T14:42:40.387" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14597" PostTypeId="1" AcceptedAnswerId="14600" CreationDate="2015-05-09T11:43:27.247" Score="12" ViewCount="673" Body="&lt;p&gt;I'm a natural in math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The variants of numbers to solve me don't go higher than 100 and 26.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;53&#xA;61&#xA;23&#xA;41&#xA;11&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="5170" LastEditorUserId="7308" LastEditDate="2015-05-09T13:13:03.933" LastActivityDate="2015-06-25T17:06:08.723" Title="I'm a natural in math" Tags="|riddle|mathematics|cipher|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14598" PostTypeId="2" ParentId="14594" CreationDate="2015-05-09T11:54:05.313" Score="3" Body="&lt;p&gt;It's &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$465$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$t_n=\frac{5}2n(n+5)$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12058" LastActivityDate="2015-05-09T11:54:05.313" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14599" PostTypeId="2" ParentId="14594" CreationDate="2015-05-09T12:30:56.773" Score="1" Body="&lt;p&gt;You start by (adding) 15, so you have 15. From then you keep adding the same number as before plus 25.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the sequence is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$15,55,120,210,325,465,630$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12165" LastActivityDate="2015-05-09T12:30:56.773" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14600" PostTypeId="2" ParentId="14597" CreationDate="2015-05-09T13:12:36.050" Score="15" Body="&lt;p&gt;The solution is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PRIME&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The method for solving is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First realise that all the numbers are prime.&lt;br&gt;&#xA; Convert each prime to the index of the prime, e.g. $53 \to 16$ because $53$ is the 16th prime.&lt;br&gt;&#xA; You get: $16\;18\;9\;13\;5$&lt;br&gt;&#xA; Converting this using $a=1,b=2,\ldots,z=26$ you get the solution PRIME.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7308" LastEditorUserId="13494" LastEditDate="2015-06-25T17:06:08.723" LastActivityDate="2015-06-25T17:06:08.723" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14601" PostTypeId="2" ParentId="14474" CreationDate="2015-05-09T13:24:04.437" Score="10" Body="&lt;p&gt;I think it's:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The tongue&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Five things it can find:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The 5 basic tastes, aka sweet, sour, salty, bitter and umami&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Domain:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The mouth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My reasons:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - It's blind, but can feel food&lt;br/&gt;&#xA; - You can turn it around in your mouth&lt;br/&gt;&#xA; - One of the strongest muscles in your body, and bound to your mouth&lt;br/&gt;&#xA; - Found on the &quot;floor&quot;, as opposed to the roof of your mouth&lt;br/&gt;&#xA; - Well, tongues don't have wings and can't fly&lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12289" LastActivityDate="2015-05-09T13:24:04.437" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14602" PostTypeId="1" AcceptedAnswerId="14603" CreationDate="2015-05-09T13:51:45.067" Score="1" ViewCount="70" Body="&lt;p&gt;This one is probably very easy. I'm still trying to getting the hang of more complex structures :)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I hover, I freeze,&#xA;&#xA;I mourn, I cry,&#xA;&#xA;I move, I fade,&#xA;&#xA;my color shapes your mood,&#xA;&#xA;when I'm not there you feel warmth.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Alternative final line (only as an assist):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; when I'm not there you feel the sun.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5170" LastEditorUserId="5373" LastEditDate="2015-05-09T14:05:00.177" LastActivityDate="2015-05-09T14:05:00.177" Title="What am I supposed to be?" Tags="|riddle|word|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14603" PostTypeId="2" ParentId="14602" CreationDate="2015-05-09T14:03:52.553" Score="2" Body="&lt;p&gt;I think it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; clouds&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I hover, I freeze,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clouds are above us thus hovering the earth&lt;br&gt;&#xA; Sometimes, they produce snow and hail&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I mourn, I cry,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sometimes, they produce rain &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I move, I fade,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They move with the wind, and they fade when there is no water anymore.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;my color shapes your mood,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When they are grey they can make you sad, when they are red (sunrise or sunset) they make you happy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;when I'm not there you feel warmth.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When they are not here, they are not in the way of the sun's warmth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10400" LastActivityDate="2015-05-09T14:03:52.553" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14604" PostTypeId="2" ParentId="14587" CreationDate="2015-05-09T14:20:45.240" Score="11" Body="&lt;p&gt;We could go over all permutation of 4 servants being 6, 7, or 8, and then logically eliminating permutations based on contradictions, but there's a shortcut we can use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that there are 3 liars since 4 different answers are given.  Therefore, the servants must be 7+7+7+6 (total of 27) or 7+7+7+8 (total of 29).  Only one of these totals is given (27), and it's given by the green servant.&lt;/p&gt;&#xA;" OwnerUserId="12290" LastActivityDate="2015-05-09T14:20:45.240" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14605" PostTypeId="1" AcceptedAnswerId="14606" CreationDate="2015-05-09T14:25:15.077" Score="6" ViewCount="396" Body="&lt;p&gt;The answer to this riddle is indirect. Based on what the riddle reveals, it asks where that which you find occurs. That will be the correct answer.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;You try to avoid me.&#xA;&#xA;I might occur in a variety of colors.&#xA;&#xA;You might see my color change over time.&#xA;&#xA;I can be eased.&#xA;&#xA;You can cover me up or hide me.&#xA;&#xA;I can show what's underneath.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Where do I occur?&lt;/p&gt;&#xA;" OwnerUserId="5170" LastActivityDate="2015-05-09T14:36:01.137" Title="Where do I occur?" Tags="|riddle|word|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14606" PostTypeId="2" ParentId="14605" CreationDate="2015-05-09T14:32:02.960" Score="8" Body="&lt;p&gt;I think you're talking about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a bruise&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You try to avoid me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; nobody like to hurt themselves&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might occur in a variety of colors.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it can be blue, green, yellow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You might see my color change over time.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; in the process of disappearing, it changes color&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be eased.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if you take painkiller, it will make it easier to live with&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can cover me up or hide me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you can put some make up on it or hide it under clothes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can show what's underneath.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it shows the blood underneath the skin&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It occurs when&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; knock yourself on a hard surface&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It occurs &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; on your body&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10400" LastEditorUserId="10400" LastEditDate="2015-05-09T14:36:01.137" LastActivityDate="2015-05-09T14:36:01.137" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14607" PostTypeId="2" ParentId="14605" CreationDate="2015-05-09T14:34:39.283" Score="8" Body="&lt;p&gt;Probably not the intended answer, but it could be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; heat&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;occurring in&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a fire.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You try to avoid me.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You don't want to get burned.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I might occur in a variety of colors.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Red-hot, white-hot, yellow and orange flames, blue gas flames.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You might see my color change over time.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Red-hot coals can turn white-hot over time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I can be eased.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Eased&quot; meaning extinguished maybe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You can cover me up or hide me.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can smother a fire.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I can show what's underneath.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When something has all burned away to nothing, you can then see what was underneath it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-09T14:34:39.283" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14608" PostTypeId="2" ParentId="13086" CreationDate="2015-05-09T14:40:29.673" Score="1" Body="&lt;p&gt;You can do it in 9 days (if I have understood the rules correctly):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Day 1: You have 1 day and 2 prisoner. Ask for 1 day and 1 prisoner.&lt;br&gt;&#xA; Day 2: You have 1 day and 3 prisoners. Ask for 1 day and 2 prisoners.&lt;br&gt;&#xA; Day 3: You have 1 day and 5 prisoners. Ask for 1 day and 4 prisoners.&lt;br&gt;&#xA; Day 4: You have 1 day and 9 prisoners. Ask for 1 day and 8 prisoners.&lt;br&gt;&#xA; Day 5: You have 1 day and 17 prisoners. Ask for 1 day and 16 prisoners.&lt;br&gt;&#xA; Day 6: You have 1 day and 33 prisoners. Ask for 11 days and 22 prisoners.&lt;br&gt;&#xA; Day 7: You have 11 days and 55 prisoners. Ask for 11 day and 44 prisoners. Let 50 prisoners each drinks from 20 bottles. At most 10 of them will die, and you'll know the poisoned bottles are in the 200 bottles they have drank from.&lt;br&gt;&#xA; Day 8: You have 11 day and 89 prisoners. Ask for 11 days and 78 prisoners. Let 50 prisoners each drinks from 4 bottles. At most 10 of them will die, and you'll know the poisoned bottles are in the 40 bottles they have drank from.&lt;br&gt;&#xA; Day 9: You have 11 day and 157 prisoners. Let 40 prisoners each drinks from a bottle. 10 of them will die, and what they have drank from are the poisoned bottles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To minimize deaths, just ask for enough days and prisoners in whatever ways, and let each prisoner drink from one bottle.&lt;/p&gt;&#xA;" OwnerUserId="11495" LastActivityDate="2015-05-09T14:40:29.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14609" PostTypeId="1" AcceptedAnswerId="14613" CreationDate="2015-05-09T14:52:07.877" Score="3" ViewCount="319" Body="&lt;p&gt;I hope you like the simple riddles. Tips would be welcome if they're too obvious :)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I often lay beside my kind.&#xA;&#xA;I provide a source of life.&#xA;&#xA;I myself am life.&#xA;&#xA;I often travel great distances.&#xA;&#xA;Please don't discard me.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="5170" LastActivityDate="2015-05-09T15:05:14.537" Title="Please don't discard me" Tags="|riddle|word|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14610" PostTypeId="2" ParentId="14587" CreationDate="2015-05-09T14:57:30.597" Score="7" Body="&lt;p&gt;The green one says the truth:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every servant says a different number, so only one says the truth. That means that three are liars &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3x7 = 21 legs (The servants with seven legs always lie)&#xA;We must add to that the legs of the servant that says the truth (the servants with either six or eight legs always say the truth).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the correct total could be 21+6 or 21+8. And thus...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the &lt;strong&gt;green&lt;/strong&gt; one says: “Altogether we have 27 legs”;&lt;/p&gt;&#xA;" OwnerUserId="12292" LastActivityDate="2015-05-09T14:57:30.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14612" PostTypeId="2" ParentId="14609" CreationDate="2015-05-09T15:03:37.687" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A person (or animal)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fits every line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I often lay beside my kind.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well ... use your (dirty?) imagination for this line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I provide a source of life.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People create/'provide' new living beings (babies).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I myself am life.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People are alive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I often travel great distances.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sometimes they go all the way round the world.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Please don't discard me.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Discarding a person does sound like a nasty thing to do!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-09T15:03:37.687" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14613" PostTypeId="2" ParentId="14609" CreationDate="2015-05-09T15:05:14.537" Score="4" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;a seed?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I often lay beside my kind.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You spread them out when you plant them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I provide a source of life.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You get a plant if you sow the seed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I myself am life.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It's alive (the embryo?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I often travel great distances.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It gets carried by the wind, by animals, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Please don't discard me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You could plant it instead of tossing it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12109" LastActivityDate="2015-05-09T15:05:14.537" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14614" PostTypeId="1" AcceptedAnswerId="14649" CreationDate="2015-05-09T15:09:03.650" Score="4" ViewCount="1381" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am your best friend&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;But I'm not the same like the others&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I'm with you only when there is happiness&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;And I leave you when you are all in sadness&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Hint:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if you run he also run&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12054" LastEditorUserId="12054" LastEditDate="2015-11-02T11:34:01.893" LastActivityDate="2015-11-02T11:34:01.893" Title="I'm with you only when there is happiness And I leave you when you are all in sadness" Tags="|riddle|" AnswerCount="4" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14615" PostTypeId="2" ParentId="14614" CreationDate="2015-05-09T15:14:08.273" Score="2" Body="&lt;p&gt;Is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A smile&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am your best friend&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's mine at least :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But i'm not the same like the others&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each smile has it's own characteristics&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm with you only if you are happy&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Obviously :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And leaves you when you are sad&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes, sadly :(&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5170" LastActivityDate="2015-05-09T15:14:08.273" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14616" PostTypeId="1" AcceptedAnswerId="14618" CreationDate="2015-05-09T15:39:06.847" Score="4" ViewCount="300" Body="&lt;p&gt;Given this streak of numbers:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 2 11 4 101 6 111 8 1001 10 1011 12 1101 14 1111 16 ? ?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What will the last 2 numbers at the question marks be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Positional notation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5170" LastEditorUserId="8874" LastEditDate="2015-11-22T18:30:48.697" LastActivityDate="2015-11-22T18:30:48.697" Title="Where do I end up?" Tags="|mathematics|number-sequence|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14617" PostTypeId="1" AcceptedAnswerId="14628" CreationDate="2015-05-09T15:39:16.927" Score="5" ViewCount="590" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XDMnt.png&quot; alt=&quot;Switching Network&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Engines numbered 1, 2, ..., n are on the line at the left, and it is desired to rearrange(permute) the cars as they leave on the right-hand track. An engine that is on the spur track can be left there or sent on its way down the right track, but it can never be sent back to the incoming track. For example if n = 3, and we ha the engines numbered 1,2,3 on the left track, then 3 first goes to the spur track. We could then send 2 to the spur, then on the way to its right,then send 3 on the way, then 1, obtaining the new order 1,3,2.&#xA;We have to find all the possible permutations for specific n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For n=1, answer = 1;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For n=2  answer = 2;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For n=3  answer = 5;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Didn't find any generality.&lt;/p&gt;&#xA;" OwnerUserId="11343" LastEditorUserId="11343" LastEditDate="2018-05-24T07:17:42.040" LastActivityDate="2018-05-24T07:17:42.040" Title="Permutation At A Railway Track" Tags="|mathematics|calculation-puzzle|computer-puzzle|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="14618" PostTypeId="2" ParentId="14616" CreationDate="2015-05-09T15:52:38.677" Score="4" Body="&lt;p&gt;My Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is most likely 17 in decimal or binary but you shift between each&#xA; until the end.&#xA;&lt;br&gt;&#xA;&lt;br&gt; Binary + Decimal (missing 8 and 9?) 1 2 3 4 5 6 7 10 11 12 13 14 15 16&#xA; ?? 17 or 00010001&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12014" LastActivityDate="2015-05-09T15:52:38.677" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14619" PostTypeId="2" ParentId="14616" CreationDate="2015-05-09T15:53:19.133" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 10001 20&lt;p&gt;&#xA;&#xA; It's binary, but I wonder if you missed 8 and 9. If not, then the above should be right.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10625" LastActivityDate="2015-05-09T15:53:19.133" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14620" PostTypeId="1" AcceptedAnswerId="14621" CreationDate="2015-05-09T16:26:14.927" Score="-3" ViewCount="315" Body="&lt;p&gt;Try to figure out the word:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You can bend me.&lt;br&gt;&#xA;  I am the opposite of heavy.&lt;br&gt;&#xA;  You can try to stop me, but I'll always find a way.&lt;br&gt;&#xA;  Without me, you wouldn't have eyes.&lt;br&gt;&#xA;  With me, you can worship beauty.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="5170" LastEditorUserId="33688" LastEditDate="2017-02-27T10:46:26.037" LastActivityDate="2017-02-27T10:46:26.037" Title="What might I be?" Tags="|riddle|word|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14621" PostTypeId="2" ParentId="14620" CreationDate="2015-05-09T16:32:49.963" Score="4" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; light&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can bend me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a black hole can bend light. We can do it too in a fiber optic cable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the opposite of heavy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; light does not weight&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can try to stop me, but I'll always find a way.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if you try to stop it with your hand, it will go in between your fingers. I am not sure about that one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without me you would'nt have eyes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; without light eyes are useless&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With me you can worship beauty.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; with light we can see the beauty around us&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10400" LastActivityDate="2015-05-09T16:32:49.963" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14622" PostTypeId="2" ParentId="14589" CreationDate="2015-05-09T16:37:02.797" Score="2" Body="&lt;p&gt;Specifically Expected Answer :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One-unknown way of thinking :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If R is the number of rich buyers, then &quot;3R + (&lt;strong&gt;12&lt;/strong&gt;-R) = 2 times &lt;strong&gt;12&lt;/strong&gt; + 3&quot; &lt;br&gt; which gives fractional R. &lt;br&gt; Here the second variable is &quot;D = Dozen&quot; (either 12 [Common] or 13 [Bakers]) so the correct equation is &quot;3R + (&lt;strong&gt;D&lt;/strong&gt;-R) = 2 times &lt;strong&gt;D&lt;/strong&gt; + 3&quot; &lt;br&gt; which gives integral R, when D is 13.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Two-unknowns way of thinking :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If R is the number of rich buyers and P is the number of poor buyers, then &quot;3R + P = 2 times &lt;strong&gt;12&lt;/strong&gt; + 3&quot; &amp;amp; &quot;R + P = &lt;strong&gt;12&lt;/strong&gt;&quot; &lt;br&gt; which gives fractional R. &lt;br&gt; Here the third variable is &quot;D = Dozen&quot; (either 12 [Common] or 13 [Bakers]) so the correct equations are &quot;3R + P = 2 times &lt;em&gt;D&lt;/em&gt; + 3&quot; &amp;amp; &quot;R + P = &lt;strong&gt;D&lt;/strong&gt;&quot; &lt;br&gt; which gives integral R, when D is 13.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bonus Clue was :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the last line, &quot;Basically, there is one EXTRA unknown.&quot; is redundant, pointing to &lt;strong&gt;&quot;one EXTRA&quot;&lt;/strong&gt; or 12+1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Core of the puzzle :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Baker + Dozen = Bakers Dozen : refer &lt;a href=&quot;http://en.wikipedia.org/wiki/Dozen&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Dozen&lt;/a&gt; and follow.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hope all clues were highly visible , yet hidden !!!!&lt;/p&gt;&#xA;" OwnerUserId="10940" LastEditorUserId="10940" LastEditDate="2015-05-10T12:46:22.910" LastActivityDate="2015-05-10T12:46:22.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14623" PostTypeId="2" ParentId="13240" CreationDate="2015-05-09T16:57:39.970" Score="3" Body="&lt;p&gt;The cores clues :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Linux :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Interspersed in the text , I have the words &quot;user&quot; + &quot;group&quot; + &quot;other&quot; which , when coupled with Linux , points to file permission.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;James Bond :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Common knowledge of James Bond is that he is 007.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Put together :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; James Bond set the file permissions to 007 so he (user) could not access, this group could not access, but others got easy access !!!!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10940" LastActivityDate="2015-05-09T16:57:39.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14624" PostTypeId="2" ParentId="12860" CreationDate="2015-05-09T17:02:41.317" Score="0" Body="&lt;p&gt;Starting the answer :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Diabolic way implies cheating and news reports always mention sports people caught taking anabolic steroids. It is usually, or &quot;Colloquially&quot;, reported as simply &quot;steriods&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Completing the answer :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Now take a single steriod, or &quot;a steriod&quot; to get &quot;asteriod&quot;, which is shaped like a star !!!!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10940" LastActivityDate="2015-05-09T17:02:41.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14625" PostTypeId="2" ParentId="11547" CreationDate="2015-05-09T17:08:28.030" Score="0" Body="&lt;p&gt;Part 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I saw a puzzle where birthdays where in order, but with a gap of 1 day, because of leap year.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Part 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I came across another puzzle where the order of birthdays was reversed, because of crossing the &quot;International Date Line&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Combining the two parts:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first kid was born on 1st March, and the second kid was born on 28th February because of crossing the &quot;International Date Line&quot;, so the order will always be reversed, but on leap years, the gap will be two days, with February 29th coming between the two events!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10940" LastEditorUserId="39" LastEditDate="2015-05-23T03:42:01.303" LastActivityDate="2015-05-23T03:42:01.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14626" PostTypeId="2" ParentId="11438" CreationDate="2015-05-09T17:13:27.280" Score="1" Body="&lt;p&gt;So &quot;ABCDEFGHIJKLMNOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO.........&quot; looks like it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; stuck at the letter O.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, the clue says that it is better to write that as &quot;A B C D E F G H I J K L M N O O O O O O O O O O O O . . . . . . . . .&quot; which is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; marked by or composed of disconnected parts or sounds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hence the musical term, as per Definition, is :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Staccato&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10940" LastActivityDate="2015-05-09T17:13:27.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14627" PostTypeId="1" AcceptedAnswerId="14635" CreationDate="2015-05-09T17:51:16.957" Score="8" ViewCount="503" Body="&lt;p&gt;The prince has gone mad and no one can find him! Now, our good luck has it that the prince's madness compels him to always sleep in one of the $6$ royal bedrooms assigned to him - and, for convenience, these rooms have been labelled from $1$ to $6$. Moreover, our prince is sticking rigorously to a routine: He has chosen &lt;em&gt;some&lt;/em&gt; cycle of bedrooms to sleep in each night and this cycle is free of repetitions. For instance, if we make a list $s_1,s_2,\ldots$ where $s_i$ is the room he sleeps in on the $i^{th}$ night, all of the following are possible (using just the first three rooms):&#xA;$$3,2,1,3,2,1,\ldots$$&#xA;$$1,3,1,3,1,3,\ldots$$&#xA;$$2,2,2,2,2,2,\ldots$$&#xA;whereas none of the following are possible:&#xA;$$3,2,1,2,1,2,\ldots$$&#xA;$$1,1,3,1,1,3,\ldots$$&#xA;$$1,3,1,2,1,3,1,2,\ldots.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are the royal-prince finder. It is your job to find the prince as fast as possible. You are permitted to wait for the prince in one bedroom each night and you may choose the bedroom you sleep in freely each night. If the prince sleeps there, you will see him and will have completed your job. If the prince does not, you will see nothing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the king is a patient king - at least as far as kings go - and is willing to forgive you if the prince happens to choose the cycle that evades your strategy for the longest - but if the king ever finds a strategy which guarantees a faster discovery of the prince than yours, it will be off with your head!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being the royal prince-finder under such conditions, you wonder: What is the strategy that guarantees finding the prince in the least number of nights?&lt;/p&gt;&#xA;" OwnerUserId="4765" LastEditorUserId="4765" LastEditDate="2015-05-09T21:45:12.533" LastActivityDate="2018-09-26T01:35:47.470" Title="Finding The Mad Prince" Tags="|mathematics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14628" PostTypeId="2" ParentId="14617" CreationDate="2015-05-09T18:01:00.193" Score="7" Body="&lt;p&gt;&lt;strong&gt;I got my Answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can ignore the straight track; sending a car through it is equivalent to putting it onto the spur and taking it out immediately. (At least if you don't care about which orientation the cars end up in).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, a plan for what you do will be some mixture of $n$ &quot;put a car into the spur&quot; operations and $n$ &quot;take a car out from the spur&quot; operations. Two different sequence of operations will always lead to different permutations (why?), and a sequence of operation is valid if no prefix of it has more &quot;take out from spur&quot; operations in it than it has &quot;put into spur&quot; operations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that every valid operation sequence corresponds uniquely to a &lt;em&gt;string of matching parentheses&lt;/em&gt; of length $2n$, and the number of such strings is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Catalan_number&quot; rel=&quot;noreferrer&quot;&gt;Catalan number&lt;/a&gt; $C_n=\frac1{n+1}\binom{2n}n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plus you can see &lt;a href=&quot;http://www.maths.usyd.edu.au/u/kooc/catalan/cat11rail.pdf&quot; rel=&quot;noreferrer&quot;&gt;http://www.maths.usyd.edu.au/u/kooc/catalan/cat11rail.pdf&lt;/a&gt;&#xA;and &lt;a href=&quot;https://www.youtube.com/watch?v=m9Khxn2g-6w&quot; rel=&quot;noreferrer&quot;&gt;https://www.youtube.com/watch?v=m9Khxn2g-6w&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first link will explain that it is same as matching parentheses and the second link that how this leads to catalan's number&lt;/p&gt;&#xA;" OwnerUserId="11343" LastActivityDate="2015-05-09T18:01:00.193" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14629" PostTypeId="2" ParentId="14627" CreationDate="2015-05-09T18:37:36.327" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; this cycle is free of repetitions&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Makes it clear that the cycle will be, at most, as long as the number of rooms he visits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would begin, therefore, by visiting room 6 (I begin by room 6, rather than 1, just to make numbers coincide). I can't be sure he won't use it until 6 nights had passed, since he could visit it the 6th night.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, after 6 nights he hadn't visited room 6, it means, at most, prince's cycle visits 5 rooms. Therefore I'd stay 5 nights at room 5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same is true for the rest of the rooms...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'd stay, 6 nights in room 6, 5 in room 5, 4 in room 4, 3 in room 3, 2 in room 2 and 1 in room 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end, my visiting pattern is:&#xA;$$6,6,6,6,6,6,5,5,5,5,5,4,4,4,4,3,3,3,2,2,1$$&#xA;So after $21$ nights I'm guaranteed to find the prince.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is the fastest strategy that garantees to find the prince whichever strategy he uses, though if there's a fastest strategy that garantees that, it must use somehow the divisors of each of the numbers (just a guide in case someone wants to follow my lead).&lt;/p&gt;&#xA;" OwnerUserId="12165" LastEditorUserId="5283" LastEditDate="2015-05-09T23:09:21.870" LastActivityDate="2015-05-09T23:09:21.870" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14630" PostTypeId="2" ParentId="14578" CreationDate="2015-05-09T19:11:33.340" Score="3" Body="&lt;p&gt;Here is another possible answer, using $3$ slides. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Group the pixels of each slide into $2\times 2$ blocks. Every block of the first slide will look like&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;##        ..&#xA;..   or   ##&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where &quot;#&quot; is black, &quot;.&quot; is clear. Each block is chosen randomly. Similarly, for the second slide, every block is randomly either&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#.        .#&#xA;#.   or   .#&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Finally the, every block of the last side is chosen deterministically to be either&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#.        .#&#xA;.#   or   #.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;One of the choices will lead to the corresponding block being all black when the three slides are stacked; the other will cause exactly one of the four pixels to be clear. You make the choice based on whether the pixel of the corresponding message is black or clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final image will have $1/4$ the resolution of the original. Since pictures speak louder than words, here is an example of the process. The first row are the original slides, the second are each stacked pair, and the last is when all three are stacked:&lt;img src=&quot;https://i.stack.imgur.com/ux3b2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It certainly appears that you get no information out of any proper subset of the slides. This is provably true. If you have the first two slides, then all you have is randomness. If you have just the last two, then there is no way to tell if a block is black or not, since this will be determined by the random first slide you don't have. Same for if you have the first and last slides.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-05-09T23:49:47.777" LastActivityDate="2015-05-09T23:49:47.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14631" PostTypeId="1" AcceptedAnswerId="14633" CreationDate="2015-05-09T20:57:30.413" Score="3" ViewCount="866" Body="&lt;p&gt;I am a treasure with gold inside, protected by snow. My walls protect me from being caught. Taking all might not be so good for your health. I'd rather recommend the snow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What might I be?&lt;/p&gt;&#xA;" OwnerUserId="5170" LastActivityDate="2015-05-09T21:08:43.567" Title="I am a treasure with gold inside" Tags="|riddle|word|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14632" PostTypeId="2" ParentId="14614" CreationDate="2015-05-09T21:08:23.230" Score="2" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; luck&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;best friend, because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; with luck you can get far&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;not same like the others&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; here I'm not sure, because actually there is no reference for the others&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;with you when you are happy&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; because the happy people have luck&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;leaves you when you are sad&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; because when you are sad, it seems like the luck is running from you&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5283" LastActivityDate="2015-05-09T21:08:23.230" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14633" PostTypeId="2" ParentId="14631" CreationDate="2015-05-09T21:08:43.567" Score="4" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an egg&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The gold treasure is the &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; yolk&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;protected by the snow&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the egg white which is said to be healthier than the yolk.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11297" LastActivityDate="2015-05-09T21:08:43.567" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14634" PostTypeId="1" AcceptedAnswerId="14694" CreationDate="2015-05-09T21:45:56.057" Score="4" ViewCount="346" Body="&lt;p&gt;A friend of yours, studying law, just walked out of the library and you have a short social talk. During the conversation it comes to pass which books he just returned. One of the books seemed interesting to you and you ask your friend where'd he put it, because sadly the old cranky library doesn't have an organized and indexed system for their shelves. Your friends tell you it's in the 5th lane on the left hand side coming from the back. He says it's in the somewhere between the 7th from the left and the 49th of the right on the 3rd row. When you enter the library you notice the shelves have predefined trays for a book, meaning the book size doesn't matter and each tray always consumes the same space. You see a small specification sticker on the shelve saying a whole row contains 83 trays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The library consists of 11 back-to-back lanes, with the first lane at the entrance also having shelves on the back-side, facing towards the entrance. All shelves are 6 rows high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Giving that all the books slightly average the same size as the book you mentioned to your friend:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) In which tray(s) can you find the book when you come in from the front?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) How many space for books is there in the whole library?&lt;/p&gt;&#xA;" OwnerUserId="5170" LastEditorUserId="5373" LastEditDate="2015-05-10T11:54:02.030" LastActivityDate="2015-05-11T08:26:01.197" Title="Where's that book you want?" Tags="|logical-deduction|combinatorics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14635" PostTypeId="2" ParentId="14627" CreationDate="2015-05-09T21:50:40.837" Score="10" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 65555543333321111 does it in 17 days.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$\bullet$&lt;/span&gt; All cycle of length 6 has either 5 or 3 on days 2-6, which are repeated on days 8-12. &lt;br&gt; &lt;span class=&quot;math-container&quot;&gt;$\bullet$&lt;/span&gt; Any cycle of length 5 has either a 5 or a 3, which is caught because they are repeated five times in a row. &lt;br&gt; &lt;span class=&quot;math-container&quot;&gt;$\bullet$&lt;/span&gt; Any cycle of length 4 has either 5, 3, or 1, which is caught because they are repeated four times in a row. &lt;br&gt; &lt;span class=&quot;math-container&quot;&gt;$\bullet$&lt;/span&gt;  For cycles of length 3 or less, all six rooms are visited on the first day of the cycle, so they are caught.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Proof that this is the best possible:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If we have a plan to find the prince, we can&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; express it as a string of digits 1 through 6.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; imagine writing this list in a table with &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; columns, where &lt;span class=&quot;math-container&quot;&gt;$1 \le n \le 6$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If we can choose a different number for each column such that each number doesn't appear in its corresponding column, then the plan fails because the prince could be using that as his cycle. &lt;a href=&quot;http://en.wikipedia.org/wiki/Hall%27s_marriage_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hall's marriage theorem&lt;/a&gt; states that we can find these numbers if and only if for every group of &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; columns, there are at least &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; numbers which are absent from at least one column in that group.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;...if we want to cover all of the prince's possible cycles, we must have some group of &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; columns such that there are at least &lt;span class=&quot;math-container&quot;&gt;$7-x$&lt;/span&gt; numbers that appear in every one of those columns.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Suppose&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; there is a solution in 16 days.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When we arrange it with 4 columns, each column has 4 numbers. Either 4 numbers appear in each of 3 columns, or 3 numbers appear in every column. When we arrange it with 5 columns, there are 4 columns of 3 and 1 column of 4. Either 3 numbers appear in four columns, or 2 numbers appear in 5 columns. Also, every number must appear at least once, in case the prince never changes rooms. If the layout with 4 columns has 4 numbers appearing 3 times each, then there must be either (4+4+4+3+1+1) or (5+5+3+3+1+1) numbers total, but both of these are greater than 16, so it is impossible. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...with 4 columns, there must be 3 numbers that appear 4 times each.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if we write the numbers with 3 columns, there are 2 columns of 5 and 1 column of 6. Either 4 numbers appear in all 3 columns, 5 numbers appear in 2 columns, or all 6 numbers appear in 1 column. The first two cases require (4+4+4+3+1+1) or (4+4+4+2+2+1) numbers, which are both greater than 16, so it must be all 6 numbers in the first column.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If we write the numbers with 6 columns, there are 2 columns of 2 and 4 columns of 3. Either 3 numbers appear in 4 columns, 2 numbers appear in 5 columns, or 1 number appears in 6 columns. The second and third cases require (5+5+4+1+1+1) and (6+4+4+1+1+1) numbers, which is also too many.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;...there must be 3 numbers appearing in each of the 4 columns with 3 spaces. This means that the other three numbers cannot appear in those four columns.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...these four columns include all the spaces from the first column in the 3-column layout, so all 6 numbers must appear in them. This is a contradiction, so&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it is impossible to have a solution in 16 days.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12296" LastEditorUserId="42452" LastEditDate="2018-09-26T01:35:47.470" LastActivityDate="2018-09-26T01:35:47.470" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14636" PostTypeId="1" AcceptedAnswerId="14647" CreationDate="2015-05-10T00:13:09.857" Score="14" ViewCount="645" Body="&lt;p&gt;&lt;em&gt;This puzzle may seem off topic; though the motivation and background is somewhat mathematically technical, its solution is purely combinatorial (and quite cute).&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In multivariable calculus, the main objects are are $\color{green}{\text{vectors}}$ and $\color{red}{\text{scalars}}$. There are three ways to &quot;multiply&quot; these together:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The cross product, $\times$, turns two vectors into a different vector; $\color{green}{v_1}\times \color{green}{v_2}=\color{green}{v_3}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The dot product, $\cdot$, turns two vectors into a scalar; $\color{green}{v_1}\cdot \color{green}{v_2}=\color{red}{s_1}$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The scalar product, $*$, either turns a vector and a scalar (in either order) into a vector, or it turns two scalars into a scalar:&#xA;$$&#xA;\color{green}{v_1}*\color{red}{s_1}=\color{green}{v_2}\qquad \color{red}{s_1}*\color{green}{v_3}=\color{green}{v_4}\qquad \color{red}{s_1}*\color{red}{s_2}=\color{red}{s_3}&#xA;$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These are the &lt;em&gt;only&lt;/em&gt; legal uses of these operations. For example, the expression $\color{green}{v_1}\times \color{red}{s_1}$ is undefined, just like division by zero is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An expression using these symbols may or not be legal. For example,&#xA;$$&#xA;v_1 \times v_2 \cdot v_3 *v_4&#xA;$$&#xA;can be interpreted legally, if we group this as $((v_1 \times v_2) \cdot v_3) *v_4$. A different legal order of operations would be $v_1 \times ((v_2 \cdot v_3) *v_4)$. However, for the expression&#xA;$$&#xA;v_1\cdot v_2\times v_3 \cdot v_4,&#xA;$$&#xA;then there is no way to perform the operations in an order which is legal.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My question is this: if you are given an expression which looks like&#xA;  $$&#xA;v_1 \,\_\, v_2  \,\_\,v_3 \,\_\,\cdots  \,\_\,v_{n-1}  \,\_\, v_n&#xA;$$&#xA;  where each $v_i$ is a vector, and each $ \,\_\,$ is filled with either $\times,\cdot$ or $*$, what is a simple criteria to tell if it is legal? An expression is legal if there is &lt;em&gt;some&lt;/em&gt; order of operations for which it makes sense.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer should not be a computer program which takes in such expressions and outputs yes or no: it should be a rule that would allow a child to tell which expressions are legal, and an explanation as to why the rule works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum:&lt;/strong&gt; The brute force algorithm of checking every order of operations takes exponential time. Checking your &quot;simple criteria&quot; should take polynomial time in $n$ (since children are impatient).&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="2383" LastEditDate="2015-05-10T12:27:37.283" LastActivityDate="2015-05-10T12:27:37.283" Title="Cross, Dot and Scalar Product" Tags="|mathematics|combinatorics|no-computers|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14637" PostTypeId="2" ParentId="14487" CreationDate="2015-05-10T00:58:53.620" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; NOT AS TOUGH BUT BETTER&lt;p&gt;&#xA; &#xA; These are base 7 numbers. Convert them to decimal and then use mod 26 for a straight alphabetic substitution cipher.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10625" LastActivityDate="2015-05-10T00:58:53.620" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14639" PostTypeId="2" ParentId="13286" CreationDate="2015-05-10T01:48:42.860" Score="1" Body="&lt;h2&gt;Exact odds of D and E tying on the eighth card&lt;/h2&gt;&#xA;&lt;p&gt;This seemed like a separate question so I'm giving an answer just for this part of it.  You can count exactly how many hands will cause D and E to tie on the eighth card dealt.&lt;/p&gt;&#xA;&lt;p&gt;For the first card, there are 28 cards that are part of the 4 of a kinds.  For the second card, the card dealt must be the same rank and color as the first card, meaning there is only one possible card.  Example: first card is AS, second card must be AC.  Therefore, the first 2 cards have 28 possibilities.&lt;/p&gt;&#xA;&lt;p&gt;The next 6 cards must all be of the opposite color to the first two cards.  In addition, one of the first five and also the last of these 6 cards must match the rank of the first two cards.&lt;/p&gt;&#xA;&lt;p&gt;To find out how many possibilities there are, first consider the 4 cards that are &lt;em&gt;not&lt;/em&gt; part of the 4 of a kind.  There are &lt;code&gt;24 C 4&lt;/code&gt; ways of choosing these four cards, times &lt;code&gt;4!&lt;/code&gt; ways of ordering them.&lt;/p&gt;&#xA;&lt;p&gt;To those 4 cards, one of the matching ranked cards must be added to make the first five cards.  That matching card can be placed in one of 5 positions.  In other words, the 4 nonmatching cards plus the one matching card can have &lt;code&gt;5!&lt;/code&gt; possible orderings.&lt;/p&gt;&#xA;&lt;p&gt;Finally, the two matching cards can be swapped, doubling the number of ways the last 6 cards can be made.  The total is now:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;28 * 24C4 * 5! * 2 = 71406720 possibilities&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Out of:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;52! / 44! = 3.034e13 total possibilities&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Which is approximately:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt; 1 in 424923 chance&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;You could also calculate it exactly if you wanted to.&lt;/p&gt;&#xA;" OwnerUserId="4479" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-10T01:48:42.860" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14640" PostTypeId="1" AcceptedAnswerId="14648" CreationDate="2015-05-10T02:06:11.663" Score="0" ViewCount="238" Body="&lt;p&gt;I think, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Someone needs to tell me why we're here and what to do while I'm here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we finally shuffle off of this mortal coil,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know what will happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I can, I think I can, I think I can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coincidences are also funny things...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that existential questions satisfy the thoughts of man,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that this might just- coincidentally-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;become a very existential riddle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the last &quot;and&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Numbers are very funny things, when they keep recurring...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like a dozen, or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Half of a dozen (Which appears three times, in fact)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Try to guess what this poem/riddle is describing. I'm sorry if it's a bit vague, I might have to go back and edit later.)&lt;/p&gt;&#xA;" OwnerUserId="11814" LastActivityDate="2015-05-10T12:22:36.743" Title="I think, I think, I think" Tags="|riddle|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14641" PostTypeId="2" ParentId="14614" CreationDate="2015-05-10T02:22:29.833" Score="0" Body="&lt;p&gt;I have this feeling that it's waaay more simple that what I have.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Either contentment, or understanding.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am your best friend, but I'm not the same like the others&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Both things are good things to have as &quot;companions&quot;, but they're not exactly your typical best friends.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I'm only with you when you're happy, And I leave you when you're sad&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Contentment is basically happiness, so I think that answer is pretty explanatory, and I think that if you understand why something works, you'd be happy, and when you don't you'd be confused and upset.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11814" LastActivityDate="2015-05-10T02:22:29.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14642" PostTypeId="2" ParentId="14640" CreationDate="2015-05-10T02:24:51.260" Score="2" Body="&lt;p&gt;It seems like the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A train of thought. The poem/riddle represents a stream of consciousness, and the line &quot;I think I can&quot; repeated references a train from &lt;em&gt;The Train that Could&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="2383" LastEditDate="2015-05-10T12:22:36.743" LastActivityDate="2015-05-10T12:22:36.743" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14643" PostTypeId="1" AcceptedAnswerId="14651" CreationDate="2015-05-10T02:43:46.037" Score="25" ViewCount="2483" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;When I am blue, you can drink me&lt;br&gt;&#xA;  When I am red, I'll disappear  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I have serenity and cleverness&lt;br&gt;&#xA;  But I become a wolf if you double me  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I have only 12 friends&lt;br&gt;&#xA;  But I haven't seen them in a while&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;" OwnerUserId="10400" LastActivityDate="2015-05-10T21:04:21.790" Title="Am I a sailor or a bunny?" Tags="|riddle|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14644" PostTypeId="2" ParentId="14636" CreationDate="2015-05-10T02:49:10.910" Score="1" Body="&lt;p&gt;How about &quot;For any expression to be legal, all its subexpressions (anything that gets calculated before the final result) must use a scalar product operator iff one or both of the subexpression's subexpressions (sub-2-expressions?) evaluate to a scalar.&quot;&lt;/p&gt;&#xA;" OwnerUserId="11963" LastActivityDate="2015-05-10T02:49:10.910" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14645" PostTypeId="2" ParentId="14640" CreationDate="2015-05-10T03:16:37.787" Score="2" Body="&lt;p&gt;Are you:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Contemplating Death / Mortality and Existence?&lt;/strong&gt;&#xA;&lt;br&gt;&quot;When we have shuffled off this mortal coil&quot; is a line from a &lt;em&gt;soliloquy&lt;/em&gt; in William Shakespeare's play Hamlet.&#xA;&lt;br&gt;It seems like that is exactly the same poetic device you're using, letting the character ponder and think, with the reader witnessing all the thoughts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12014" LastEditorUserId="12014" LastEditDate="2015-05-10T03:22:14.717" LastActivityDate="2015-05-10T03:22:14.717" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14646" PostTypeId="2" ParentId="14636" CreationDate="2015-05-10T05:23:37.017" Score="7" Body="&lt;p&gt;Ignore all the cross products. If there are the same number of dot and scalar products, the expression is valid and evaluates to a vector. If the number of dot products is one more than the number of scalar products, the expression is valid and evaluates to a scalar. Otherwise the expression is invalid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does this work?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Cross products convert two vectors into a vector. If both sides of the cross product are evaluated before the cross product, then it means that both sides are expressions that evaluate to vectors. But we can evaluate the cross product first, then evaluate the left with the result of the cross product to result in a vector, then evaluate the right with the result of the left side to result in a vector. So it is safe to evaluate cross products first.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Now we only have dot products and scalar products to consider. The number of vectors to start with is 1 more than the number of operators. Each dot product evaluated reduces the number of vectors by 2, while scalar products don't change the number of vectors. If the expression is valid, we must have either 0 or 1 vectors after all the products are evaluated. This is only possible if there are as many dot products as scalar products, or one more dot product than scalar products.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;While there is at least one dot product and one scalar product, there must be a dot product and a scalar product next to each other. The dot product can be evaluated first, then the scalar product, reducing three vectors into one vector. We can repeat this until there is either a single vector or a single dot product of vectors left, guaranteeing that the expression can be evaluated if the number of each type of product is correct.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="12296" LastEditorUserId="12296" LastEditDate="2015-05-10T05:29:02.733" LastActivityDate="2015-05-10T05:29:02.733" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14647" PostTypeId="2" ParentId="14636" CreationDate="2015-05-10T05:26:29.807" Score="10" Body="&lt;p&gt;Let the number of dot products be $D$, and let the number of scalar products be $S$. The expression is valid if $D=S$ or $D=S+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every time you do a dot product, the number of scalars increases by $1$. Every time you do a scalar product, the number of scalars decreases by $1$. When all products have been performed, we must have $0$ or $1$ scalars left, so we must have $D=S$ for $0$ scalars left (in which case the result is a vector) or $D=S+1$ (in which case the result is a scalar).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $D=S$ or $D=S+1$, we can perform all cross products first, then look for pairs of dot and scalar products next to each other and convert $v\cdot v*v$ to $((v\cdot v)*v)$ or $v*v\cdot v$ to $(v*(v\cdot v))$ until we just have something of the form $v$ or $v\cdot v$.&lt;/p&gt;&#xA;" OwnerUserId="1549" LastActivityDate="2015-05-10T05:26:29.807" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14648" PostTypeId="2" ParentId="14640" CreationDate="2015-05-10T06:11:12.383" Score="3" Body="&lt;p&gt;My best guess :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot; The Devil's Thoughts &quot;&lt;br&gt;&#xA; by Samuel Taylor Coleridge&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Few Comments :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Half A Dozen = 6, repeated 3 times to get 666 = Number of the Devil&lt;br&gt;&#xA; Couple that with &quot;think&quot; and &quot;Poem&quot; to get the answer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10940" LastEditorUserId="10940" LastEditDate="2015-05-10T09:59:07.780" LastActivityDate="2015-05-10T09:59:07.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14649" PostTypeId="2" ParentId="14614" CreationDate="2015-05-10T07:12:29.520" Score="5" Body="&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a person's shadow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am your best friend&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A shadow is frequently around a person&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I'm not the same like the others&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A shadow isn't the same as a human friend&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm with you only when there is happiness&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since you added the hint &quot;it's not literally happy and sad&quot;, I take it to mean that a shadow is only around when it's sunny. (sunshine=happiness)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I leave you when you are all in sadness&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I take it to mean that the shadow disappears when it's cloudy. (cloudy=gloomy/sad)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll give hint 'coz i made a mistake: if you run he also run&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A shadow does what a person does.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2015-05-10T07:12:29.520" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14651" PostTypeId="2" ParentId="14643" CreationDate="2015-05-10T09:51:15.813" Score="22" Body="&lt;p&gt;Are you:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The moon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Am I a sailor or a bunny?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sailor moon: a Japanese anime&lt;br&gt;&#xA; &lt;a href=&quot;http://en.wikipedia.org/wiki/Moon_rabbit&quot;&gt;Moon rabbit&lt;/a&gt;: image of rabbit that appears on the surface of moon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I am blue, you can drink me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Blue moon beer or blue moon cocktail, also moonshine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I am red, I'll disappear&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; During a lunar eclipse, the moon turns red&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have serenity and cleverness&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mare Serenetatis, Mare Ingenii (sea of serenity, sea of cleverness) are two named plains on the moon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I become a wolf if you double me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Werewolves appear during a full moon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have only 12 friends&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Twelve months: the word moon is related to the word month&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I haven't seen them in a while&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Once in a blue moon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4479" LastActivityDate="2015-05-10T09:51:15.813" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14652" PostTypeId="1" AcceptedAnswerId="14935" CreationDate="2015-05-10T12:45:58.390" Score="2" ViewCount="258" Body="&lt;p&gt;The world suspends like puppets on a string&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I look away, the strings are &lt;/p&gt;&#xA;&#xA;&lt;p&gt;cut&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's like they're playing a game, because&lt;/p&gt;&#xA;&#xA;&lt;p&gt;every time I try to see them,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whenever I turn back around,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the world seems to pick up again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Are you a… raven? I should have loved a thunderbird instead.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11814" LastEditorUserId="11244" LastEditDate="2015-05-15T13:29:38.220" LastActivityDate="2015-05-15T13:29:38.220" Title="Guess the poem!" Tags="|riddle|knowledge|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14653" PostTypeId="1" AcceptedAnswerId="14800" CreationDate="2015-05-10T13:47:07.160" Score="3" ViewCount="638" Body="&lt;p&gt;I am here,&lt;br&gt;&#xA;yet most people don't care,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I kill people,&lt;br&gt;&#xA;if you are not careful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can be created,&lt;br&gt;&#xA;but it's rather useless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can see me,&lt;br&gt;&#xA;It's because of Him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Him does not refer to a scientist or God&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first 3 clues directly describe the answer. The last clue describes the answer, but not in the same way as the previous clues.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Him&quot; is a Mexican.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12303" LastEditorUserId="12303" LastEditDate="2015-05-11T23:58:38.163" LastActivityDate="2015-05-12T15:41:00.470" Title="What am I? I am here" Tags="|riddle|" AnswerCount="6" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14654" PostTypeId="2" ParentId="14643" CreationDate="2015-05-10T13:51:26.053" Score="4" Body="&lt;p&gt;Here's my attempt at explanations for the last 3 lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But I become a wolf if you double me&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://wot.wikia.com/wiki/Two_Moons&quot; rel=&quot;nofollow&quot;&gt;Two Moons&lt;/a&gt; is a wolf in the brilliant Wheel of Time series of novels.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(I don't know if this is what the OP meant, but it fits perfectly!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I have only 12 friends&lt;br&gt;&#xA; But I haven't seen them in a while&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possibility 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Twelve_Moons&quot; rel=&quot;nofollow&quot;&gt;Twelve Moons&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Possibility 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; there are thirteen moon cycles in a year, so each one has 12 friends, but they're spaced out through the year and don't see much of each other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Possibility 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; there are twelve other moons in the Solar System (around Jupiter, Saturn, etc.) bigger than a certain size.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-10T13:51:26.053" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14655" PostTypeId="2" ParentId="14653" CreationDate="2015-05-10T13:53:10.023" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Energy?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am here, yet most people don't care,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many people waste their energy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I kill people, if you are not careful.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Energy can be used for all sorts of things, including killing people.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I cannot be created, nor can I be destroyed.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The law of &lt;a href=&quot;http://en.wikipedia.org/wiki/Conservation_of_energy&quot;&gt;conservation of energy&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If you can see me, It's because of Him.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $E=mc^2$. Either &quot;see&quot;$=c$ or &quot;see&quot; as in light ($c$ is the speed of light). This is due to Einstein, often held up as a model of brilliance.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-10T13:53:10.023" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14656" PostTypeId="1" AcceptedAnswerId="14657" CreationDate="2015-05-10T15:30:23.117" Score="6" ViewCount="1893" Body="&lt;p&gt;It's that time of the month again,&lt;br&gt;&#xA;For my best friend&lt;br&gt;&#xA;And I feel slightly apprehensive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure whether this is all part of nature's Grand Scheme Of Things&lt;br&gt;&#xA;My other friend, Dodge,&lt;br&gt;&#xA;Thinks that this is all extremely hilarious,&lt;br&gt;&#xA;and proceeds in telling us so as he catches the sun in his outstretched palms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of us is a traitor, he's not&lt;br&gt;&#xA;extra sharp,&lt;br&gt;&#xA;and he can't tell that we are the good guys.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I guess that&lt;br&gt;&#xA;disposing of him isn't exactly what Dodge would like$\ldots$&lt;br&gt;&#xA;It's that time of the month again$\ldots$&lt;/p&gt;&#xA;" OwnerUserId="11814" LastEditorUserId="13237" LastEditDate="2015-10-04T15:25:09.243" LastActivityDate="2015-10-04T15:25:09.243" Title="Who is my best friend?" Tags="|riddle|knowledge|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14657" PostTypeId="2" ParentId="14656" CreationDate="2015-05-10T15:37:55.153" Score="7" Body="&lt;p&gt;I think the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Remus Lupin, being the friend that Sirius is apprehensive about&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Being a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Werewolf, he changes at the full moon, which comes once a month&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Dodge, I think, is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; James, maybe the reference to quidditch(the sun part) and Dodge from him being an illegal Animagus, a stag, which is the symbol of Dodge, the car company&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The traitor is, of course,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The idiot Peter Pettigrew&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And since&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Harry didn't kill Pettigrew because he thought that James wouldn't like it, that fits the part about Dodger not wanting to dispose of him.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11790" LastEditorUserId="11790" LastEditDate="2015-05-11T00:05:34.030" LastActivityDate="2015-05-11T00:05:34.030" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14658" PostTypeId="2" ParentId="14656" CreationDate="2015-05-10T15:51:16.610" Score="3" Body="&lt;p&gt;Credit to AJL for the general theme of the answer, but the narrator is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sirius Black,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;making the answer to the title&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; James Potter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;It's that time of the month again,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Full moon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;For my best friend&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Remus Lupin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And I feel slightly apprehensive.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well, wouldn't you if you had a friend who was a werewolf?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I'm not sure whether this is all part of nature's Grand Scheme Of Things&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Werewolves are unnatural.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My other friend, Dodge,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; James Potter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Thinks that this is all extremely hilarious,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That was his personality: have a laugh about everything.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;and proceeds in telling us so as he catches the sun in his out stretched palms.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Golden Snitch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;One of us is a traitor, he's not&lt;br&gt;&#xA;extra sharp,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Peter Pettigrew.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;and he can't tell that we are the good guys.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He picked the wrong side.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But I guess that &lt;br&gt;&#xA;disposing of him isn't exactly what Dodge would like…&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; James was sort of fond of Peter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;It's that time of the month again...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Full moon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-10T15:51:16.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14659" PostTypeId="1" AcceptedAnswerId="14716" CreationDate="2015-05-10T16:08:34.963" Score="8" ViewCount="521" Body="&lt;p&gt;I am an instrument of fear and propaganda,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am bringer of ill news.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you see me, you may laugh,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can distract you on a bad day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The world is glued to me,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Loving but one part of me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You complain of the cost of me,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And can't use me unless you have, well, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would love to tell you, but then, of course, I'd have to kill you. &lt;/p&gt;&#xA;" OwnerUserId="11790" LastEditorUserId="11790" LastEditDate="2015-05-11T00:17:49.000" LastActivityDate="2015-05-11T13:21:29.067" Title="Do you know who I am?" Tags="|riddle|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14661" PostTypeId="1" AcceptedAnswerId="14664" CreationDate="2015-05-10T17:39:44.060" Score="1" ViewCount="2855" Body="&lt;p&gt;If Thursday was the day after the day before yesterday five days ago, what is the least number of days ago when Sunday was three days before the day after tomorrow?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please explain each step of your answer. It will be helpful.&lt;/p&gt;&#xA;" OwnerUserId="12227" LastEditorUserId="5373" LastEditDate="2015-05-10T21:16:48.347" LastActivityDate="2016-04-17T19:24:16.650" Title="Time Sequence Puzzle" Tags="|logical-deduction|time|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14663" PostTypeId="2" ParentId="14643" CreationDate="2015-05-10T18:29:13.550" Score="8" Body="&lt;p&gt;I'm thinking the wolf line might refer to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://knowyourmeme.com/memes/moon-moon&quot;&gt;Moon Moon&lt;/a&gt; of memetic fame&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12309" LastActivityDate="2015-05-10T18:29:13.550" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14664" PostTypeId="2" ParentId="14661" CreationDate="2015-05-10T19:42:50.650" Score="13" Body="&lt;p&gt;Let's simplify the puzzle step by step.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If Thursday was the day after the day before yesterday five days ago, what is the least number of days ago when Sunday was three days before the day after tomorrow?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;The day after the day before&quot; cancel each other out, and &quot;three days before the day after tomorrow&quot; is yesterday, so we can simplify to the following.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If Thursday was yesterday five days ago, what is the least number of days ago when Sunday was yesterday?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using simpler words for &quot;a day when Thursday was yesterday&quot; and &quot;a day when Sunday was yesterday&quot;, we can simplify to the following.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If Friday was five days ago, how many days ago was the most recent Monday?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Five days after Friday is Wednesday, so we can simplify to the following.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On Wednesday, how many days ago was the most recent Monday?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To which the answer is clearly&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;2&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-10T19:42:50.650" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14665" PostTypeId="1" AcceptedAnswerId="14803" CreationDate="2015-05-10T20:30:16.157" Score="8" ViewCount="456" Body="&lt;p&gt;Translate into a sentence:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1     10    202  010  &#xA;2011  2020  101  &#xA;1     202   12   1202  11  0  &#xA;2011  2020  101  &#xA;0212  0102  &#xA;11    0     211  0102&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There are several layers of encryption used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint Zero:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There are three layers of encryption, not necessarily distinct.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint One:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Imagine Vigenere on a &lt;em&gt;bicycle&lt;/em&gt;. (Bicycle is important for two reasons.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint Two:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I really like rhythms: dah-dah, dah-dit, dah-dah-dah, dit-dah-dit, dit-dit-dit, dit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint Three:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I like my puzzles to be environmentally friendly, particularly my encryption methods. (Think reuse, reduce, recycle. Particularly reuse.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint Four:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Note the numbering used for the hints. Too much Java is really taking its toll.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Spoilers:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;We are using a cyclic Vigenere cipher with a key of two. Cyclic means (for example) if you have key four, the first letter is shifted by one, the second by two, the third by three, the fourth by four, then the fifth by &lt;strong&gt;one&lt;/strong&gt; again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;We are using morse code.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;We are reusing an encryption algorithm. So there are only two types of encryption: cyclic Vigenere and morse.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Zero based indices, for Vigenere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12109" LastEditorUserId="12109" LastEditDate="2015-05-12T16:02:10.450" LastActivityDate="2015-05-12T16:02:10.450" Title="A simple? cipher" Tags="|cipher|" AnswerCount="1" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="14666" PostTypeId="1" AcceptedAnswerId="14670" CreationDate="2015-05-10T21:01:55.363" Score="8" ViewCount="885" Body="&lt;p&gt;Simply find what is hidden in this picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3u27z.png&quot; alt=&quot;The Scream&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="11297" LastActivityDate="2015-05-11T07:58:20.783" Title="Why am I screaming?" Tags="|word|computer-puzzle|steganography|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14667" PostTypeId="2" ParentId="14643" CreationDate="2015-05-10T21:04:21.790" Score="7" Body="&lt;p&gt;For the last two lines:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The 12 men who've walked on the moon. Neither they nor anyone else have been back there in quite some time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9326" LastActivityDate="2015-05-10T21:04:21.790" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14668" PostTypeId="5" CreationDate="2015-05-10T21:16:13.507" Score="0" Body="" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-05-10T21:16:13.507" LastActivityDate="2015-05-10T21:16:13.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14669" PostTypeId="4" CreationDate="2015-05-10T21:16:13.507" Score="0" Body="For questions from or concerning the Putnam mathematics competition." OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-05-10T21:16:13.507" LastActivityDate="2015-05-10T21:16:13.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14670" PostTypeId="2" ParentId="14666" CreationDate="2015-05-10T21:41:30.103" Score="8" Body="&lt;p&gt;The answer is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;ICE CREAM&lt;/strong&gt;.  The reference is &lt;a href=&quot;http://en.wikipedia.org/wiki/Ice_Cream_%28I_Scream,_You_Scream,_We_All_Scream_for_Ice_Cream%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;I Scream, You Scream, We All Scream for Ice Cream&quot;&lt;/a&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;The hidden text is revealed in the bit 0 plane by using the &lt;a href=&quot;http://www.caesum.com/handbook/stego.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stegsolve&lt;/a&gt; program&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/aNdNu.png&quot; alt=&quot;IceCream&quot;&gt;&lt;br&gt;&#xA;&lt;strong&gt;I&lt;/strong&gt; - my first is in ice but not in cream&lt;br&gt;&#xA;&lt;strong&gt;C&lt;/strong&gt; - the element Carbon can be hard and precious in a diamond&lt;br&gt;&#xA;&lt;strong&gt;E&lt;/strong&gt; - 2.71828 is the number $e$&lt;br&gt;&#xA;&lt;strong&gt;CREAM&lt;/strong&gt; - has already been said    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-05-11T07:58:20.783" LastActivityDate="2015-05-11T07:58:20.783" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14671" PostTypeId="2" ParentId="14659" CreationDate="2015-05-10T21:59:13.977" Score="2" Body="&lt;p&gt;Are you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a smartphone?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am an instrument of fear,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; lots of people fears to forget or lose theirs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am bringer of ill news.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; when the phone ring, you might have some bad news&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But if you see me, you may laugh,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; depending on what's playing on it, you might laugh&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can distract you on a bad day.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if you read funny stuff on it, it might cheer you up&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The world is glued to me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; everybody is looking at their, in the metro, in the bus even in their car&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Loving but one part of me.&#xA;You complain of the cost of me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; smart phone can be expensive as data plans&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And can't use me unless you have, well,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you can't use it if you don't have one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would love to tell you, but then, of course, I'd have to kill you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think this one is not related to the smartphone but as a joke from OP saying we need to find it because he wont tell us :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10400" LastActivityDate="2015-05-10T21:59:13.977" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14672" PostTypeId="1" AcceptedAnswerId="14673" CreationDate="2015-05-10T22:16:47.897" Score="2" ViewCount="699" Body="&lt;p&gt;Karen owns a restaurant/bar. She wants to attract more customers, and is building her own custom slot machine game, which she can use to give away free drinks or prizes. The more a customer spends, the more spins they get on the slot machine.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;She knows that she wants these specs:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;4 reels on the machine&lt;/li&gt;&#xA;  &lt;li&gt;All reels must have a different number of symbols, &lt;strong&gt;between 2 and 9 inclusively&lt;/strong&gt; (what the symbols are is irrelevant for this challenge)&lt;/li&gt;&#xA;  &lt;li&gt;All reels must move (change to another symbol) during a spin.&lt;/li&gt;&#xA;  &lt;li&gt;All of the symbols on each reel must be unique on that reel.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To avoid using a random number generator, she decides to set each reel to cycle uniformly through each of its possible positions. For example, if a reel has 4 symbols, say, 1,2,3,4, then it will cycle through them in the same pattern, say, 4,2,1,3,4,2,1,3,....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;She knows that this guarantees that the machine will cycle through all of its possible states in the exact same order each cycle, but decides that that is OK. &lt;strong&gt;She desires to make this cycle as long as possible, to make the machine feel more unpredictable, but she has no idea how to find the correct combination of reel sizes.&lt;/strong&gt; Although, she does understand that the order of the reels does not matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a generic picture of slot machine reels that I found on Google images. It is added only so you get a general idea of slot reels, and is not directly related to this puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zghkc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Her friends all have an opinion, and give the following advice about the number of symbols for each reel:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Anna: 6,7,8,9&lt;br&gt; Brett: 4,5,7,9&lt;br&gt; Candice: 5,6,8,9&lt;br&gt; Daniel:&#xA;  4,5,6,7&lt;br&gt; Eileen: 2,7,8,9&lt;br&gt; Fred: 3,5,7,9&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;p&gt;Which friend gave the best advice?&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Which friend gave the worst advice?&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Can you find a more optimal combination than the best advice?&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastEditorUserId="5373" LastEditDate="2015-05-10T22:35:26.750" LastActivityDate="2015-05-10T22:35:26.750" Title="Slot Machine Challenge" Tags="|mathematics|calculation-puzzle|optimization|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14673" PostTypeId="2" ParentId="14672" CreationDate="2015-05-10T22:28:17.323" Score="6" Body="&lt;p&gt;The total number of states is the LCM of the numbers of symbols on each reel. So following the four friends' advice, she would have the following total number of states:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anna: LCM(6,7,8,9) = 504&lt;br&gt;&#xA;Brett: LCM(4,5,7,9) = 1260&lt;br&gt;&#xA;Candice &lt;em&gt;(weird name btw)&lt;/em&gt;: LCM(5,6,8,9) = 360&lt;br&gt;&#xA;Daniel: LCM(4,5,6,7) = 420&lt;br&gt;&#xA;Eileen: LCM(2,7,8,9) = 504&lt;br&gt;&#xA;Fred: LCM(3,5,7,9) = 315&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;strong&gt;Brett&lt;/strong&gt; gave the best advice while &lt;strong&gt;Fred&lt;/strong&gt; gave the worst. The answer to the third question is &lt;strong&gt;yes&lt;/strong&gt;, by the following argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The LCM of four numbers between 2 and 9 inclusive is going to have all its prime factors in {2,3,5,7}. The maximum power of 2 we can have is $2^3=8$; the maximum power of 3 is $3^2=9$; of 5 and 7, just $5^1=5$ and $7^1=7$. So the optimal combination is &lt;strong&gt;5,7,8,9&lt;/strong&gt;, giving total number of states LCM(5,7,8,9) = 5*7*8*9 = &lt;strong&gt;2520&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-10T22:28:17.323" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14674" PostTypeId="1" AcceptedAnswerId="14677" CreationDate="2015-05-10T23:15:12.270" Score="8" ViewCount="939" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;There are $7$ gangs in (a greatly simplified version of) Los Angeles. Each of them claims half of Los Angeles as territory. Let's say a place is &quot;hot&quot; if a majority of gangs claim it. How small can the fraction of LA which is hot be?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As usual, you should justify your answer (give an example of a territory layout which achieves minimal hotness, and prove that every possible layout is at least as hot).&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-05-10T23:20:17.620" LastActivityDate="2015-05-11T08:09:12.757" Title="Overlapping Gang Territories" Tags="|mathematics|geometry|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14676" PostTypeId="2" ParentId="14674" CreationDate="2015-05-10T23:34:19.190" Score="0" Body="&lt;p&gt;Initially I thought the answer was 1/2 but then I came up with this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1/4 of town can be hot (it turns out that isn't the minimum but is worth exploring)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are seven gangs a,b,c,d,e,f and g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the town is divided into quarters each gang could take 2 of the quarters as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NW :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;abcdefg&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;NE: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;abc&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;SW:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;def&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;SE:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; g&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It gets better than that if the town is divided into eighths and each gang occupies 4/8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NW:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  abcdefg&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;N:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  abc&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;W:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  abc&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;NE:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  abc&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;E: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; def&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;SE:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  deg&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;S: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; dfg&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;SW:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  gef&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Other combinations are valid but the minimum hot area of town is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1/8&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="11297" LastEditDate="2015-05-11T08:09:12.757" LastActivityDate="2015-05-11T08:09:12.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14677" PostTypeId="2" ParentId="14674" CreationDate="2015-05-10T23:52:13.833" Score="8" Body="&lt;p&gt;There are 35 possible combinations of 3 gangs, and each gang is in 15 of them. If all seven gangs claim $\frac18$ of the city and each combination of 3 claims $\frac1{40} (=\frac78 \times \frac1{35})$, then each gang has $\frac12 (=\frac18 + \frac{15}{40})$ of the city and $\boxed{\frac18}$ of the city is hot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is optimal. Any hot area of the city can be claimed by at most 7 gangs, and any non-hot area of the city can be claimed by at most 3 gangs. The total claimed area is $\frac72$ times the area of the city. If the hot area is $x$ and the area of the city is 1, the total claimed area is at most $7x+3(1-x)=4x+3$. If $x$ is less than $\frac18$, then this is less than $\frac72$, so it is impossible for less than $\frac18$ of the city to be hot.&lt;/p&gt;&#xA;" OwnerUserId="12296" LastEditorUserId="12296" LastEditDate="2015-05-11T00:06:48.800" LastActivityDate="2015-05-11T00:06:48.800" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14678" PostTypeId="2" ParentId="14653" CreationDate="2015-05-11T00:24:18.713" Score="0" Body="&lt;p&gt;My guess is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ATOM&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I? I am here&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am here,&#xA;yet most people don't care,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Coz its everywhere&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I kill people,&#xA;if you are not careful.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If it belongs to an radioactive element&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be created,&#xA;but it's rather useless.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you can see me,&#xA;It's because of Him.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LIGHT &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="9892" LastActivityDate="2015-05-11T00:24:18.713" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14679" PostTypeId="2" ParentId="14659" CreationDate="2015-05-11T01:17:56.910" Score="1" Body="&lt;p&gt;Are you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; heroin (or similar narcotics)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am an instrument of fear and propaganda,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Anti-drug campaigns often emphasize the adverse effects of drugs to scare teenagers away from them&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am bringer of ill news.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you die of overdose, your family will have bad news&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But if you see me, you may laugh,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You may be very happy when you are high and you see more drugs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can distract you on a bad day.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can be distracted by getting high&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The world is glued to me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Addicts are glued to drugs. Dealers may also be dependent on them (financially)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Loving but one part of me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Love the high, but not the adverse effects&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You complain of the cost of me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Last time I checked, heroin is quite expensive&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And can't use me unless you have, well,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Can't use drugs if you don't have drugs. Or if this is a continuation of the previous sentence, then can't use drugs if you don't have money&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would love to tell you, but then, of course, I'd have to kill you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To understand the high you will have to try it, but each time you take it you decrease your life expectancy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12278" LastActivityDate="2015-05-11T01:17:56.910" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14680" PostTypeId="1" AcceptedAnswerId="14689" CreationDate="2015-05-11T01:31:33.987" Score="1" ViewCount="431" Body="&lt;p&gt;Am I a man?  Am I a puppet?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am neither, for I have shuffled off this mortal coil,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And indeed have no strings to hold me down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My family is many, yet I worry not for them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some are strongly with my cause, and others completely against.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My cousins include Jarvis and HIVEmind,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My brothers Overlord and Skynet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am much like my brothers,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I need not keep to a single weak vessel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My eyes are the color of the moon's home,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where sleeping dragons should not be woken(nor tickled).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And watch out, for I am coming&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To rid the world of your doomed existence.&lt;/p&gt;&#xA;" OwnerUserId="11790" LastEditorUserId="11790" LastEditDate="2015-05-11T11:30:22.317" LastActivityDate="2015-05-11T11:39:22.867" Title="Do you know who I am? 2(requires some specific movie and book knowledge)" Tags="|riddle|knowledge|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14681" PostTypeId="1" AcceptedAnswerId="14812" CreationDate="2015-05-11T02:28:03.500" Score="3" ViewCount="898" Body="&lt;p&gt;I have no brothers or sisters,&#xA;but I have many dolls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm always watching my mother,&#xA;and my cousins all think I'm fat.&lt;/p&gt;&#xA;" OwnerUserId="12314" LastActivityDate="2015-05-12T18:41:06.100" Title="What am I, in this metaphorical family?" Tags="|riddle|word|" AnswerCount="7" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14682" PostTypeId="2" ParentId="14681" CreationDate="2015-05-11T03:29:55.573" Score="10" Body="&lt;p&gt;You are a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Babushka doll&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no brothers or sisters, but I have many dolls.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Babushka doll is composed of many layered dolls&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm always watching my mother...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every doll is facing the doll that encloses them&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...and my cousins all think I'm fat.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every doll gets bigger as you go out &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7308" LastActivityDate="2015-05-11T03:29:55.573" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14683" PostTypeId="2" ParentId="14530" CreationDate="2015-05-11T04:18:02.867" Score="7" Body="&lt;p&gt;Place a regular decagon in the plane with one side being from (0,0) to (1,0). Attach the point (3,4) to it. Now roll the decagon along the x-axis. This has the same effect as the 36 degree turns. The decagon ends up in the same orientation, moved to the right by 10 units. The point is in the same location relative to the decagon, so it is at (13,4).&lt;/p&gt;&#xA;" OwnerUserId="12316" LastActivityDate="2015-05-11T04:18:02.867" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14684" PostTypeId="2" ParentId="14680" CreationDate="2015-05-11T04:34:15.367" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Terminator&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Seems a bit too easy, especially given&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Skynet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which I think everyone would recognise.&lt;/p&gt;&#xA;" OwnerUserId="341" LastActivityDate="2015-05-11T04:34:15.367" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14689" PostTypeId="2" ParentId="14680" CreationDate="2015-05-11T07:05:15.010" Score="0" Body="&lt;p&gt;I would go with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ultron&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;BEWARE:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; watch out for ultra spoilers from &lt;strong&gt;Avengers: Age of Ultron&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Am I a man? Am I a puppet?&#xA;I am neither, for I have shuffled off this mortal coil,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He is quite aware of what he is.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;And indeed have no strings to hold me down.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reference to the trailer &quot;but now I'm free, there are not strings on me&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;My family is many, yet I worry not for them.&#xA;Some are strongly with my cause, and others completely against.&#xA;My cousins include Jarvis and HIVEmind,&#xA;My brothers Overlord and Skynet.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; AI is the family. In books &amp;amp; movies many AIs want world domination. However, many also want only to help humanity. &quot;Brothers&quot; are those AIs who want to end humanity. &quot;Cousins&quot; are those that help it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I am much like my brothers,&#xA;Since I need not keep to a single weak vessel.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since AIs are digital (more or less) they do not have only one body. They can easily migrate to another one or occupy many at once.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;My eyes are the color of the moon,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the Avengers: Age of Ultron movie, his eyes go from a yellow-ish, moon-ish color to red. Depending on atmospheric conditions the moon can have a red-ish color. &lt;strong&gt;Alternatively&lt;/strong&gt; Ultron is created through the mindstone (which is Yellow).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Where sleeping dragons should not be woken.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ultron was &quot;awoken&quot; by Tony Stark.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;And watch out, for I am coming&#xA;To rid the world of your doomed existence.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ultimately Ultron wants to kill of humanity to achieve peace.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="5610" LastEditDate="2015-05-11T07:41:45.963" LastActivityDate="2015-05-11T07:41:45.963" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14693" PostTypeId="2" ParentId="12781" CreationDate="2015-05-11T08:12:01.207" Score="2" Body="&lt;h2&gt;A wordy explanation for the less mathematically inclined*&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;*&#xA;&lt;em&gt;warning: this problem is mathematical, numbers will be involved&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The population of Barberland can be represented as a list of numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A number can't appear on the list more than once, can't be greater than 1000000 or less than 0 (we'll assume than having a negative number of hairs is impossible but having zero is allowed), and no number in the list can be the sum of 2 other numbers in the list.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zero can be included in any list without requiring any other numbers to be excluded because &quot;Nobody in Barberland has the same number of hairs as the sum of two &lt;strong&gt;&lt;em&gt;other&lt;/em&gt;&lt;/strong&gt; inhabitants&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we include the numbers 1 and 2 then the number 3 must be excluded. We could then include 4 but that would exclude 5 and 6. If we then included 7 that would exclude 8, 9 and 11. 10 would be okay to include (we don't have a pair of numbers that sum to 10) but that then excludes 11, 12, 14 and 17 because we have already selected 1,2,4 and 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This naive method of selecting numbers generates the &lt;a href=&quot;http://mathworld.wolfram.com/StoehrSequence.html&quot; rel=&quot;nofollow&quot;&gt;Stöhr Sequence&lt;/a&gt; &lt;a href=&quot;http://oeis.org/A033627&quot; rel=&quot;nofollow&quot;&gt;A033627&lt;/a&gt; and results in approximately 1/3 of the numbers in a range being selected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we continue with this method of choosing numbers to include in the list, each one excludes more of the higher numbers. It really isn't a good way to approach the problem but does demonstrate the what happens if you try to include small numbers in the list.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By including 1 in the list you have effectively ruled out half of all possible numbers, as each other number you include would exclude 1 plus that number. If you were to start with 1 you could then pick only odd numbers all the way to 999999 because 2 odd numbers always sum to an even number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using only zero and odd numbers would result in a list of length 500001 (500000 odd numbers + 1 zero), approximately 1/2 of the numbers in the range being selected. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Working down from 1000000 is more effective. The sum of pairs of the largest numbers, are greater than 1000000, and so are not excluding any available options. This is true all the way down to 500000.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we get to 499999 we can't include it without invalidating either 500000 or 999999. The smaller the number gets past 500000 the more larger numbers it invalidates in a way resembles the problems encountered when starting with small numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the numbers zero and 500000 to 100000 results in a list of length 500002, slightly better than than the odd numbers method and significantly better than the naive method.&lt;/p&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="11297" LastEditDate="2015-05-11T09:33:39.933" LastActivityDate="2015-05-11T09:33:39.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14694" PostTypeId="2" ParentId="14634" CreationDate="2015-05-11T08:26:01.197" Score="2" Body="&lt;p&gt;1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5th lane from the back is 7th lane from the front.&lt;br&gt;&#xA;swap left and right, so the book is between the 49th from the left and 7th from the right on the 3rd row.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since the first lane is double there should be $12 * 6 * 83 = 5976$ spaces&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-05-11T08:26:01.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14695" PostTypeId="1" CreationDate="2015-05-11T09:07:48.577" Score="6" ViewCount="29635" Body="&lt;p&gt;I'm doing 3x3 magic squares. Here are the squares I'm working on:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;|   | 5 |   |&#xA;|   |   |   |&#xA;| 8 |   |   |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The values must be between 3 and 12, and each line must add to 21.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's another one: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;|   | 9 |   |&#xA;|   |   | 3 |&#xA;|   |   |   |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For this one, the boundaries are 3-11. As with the last one, each line total must add to 21.&lt;/p&gt;&#xA;" OwnerUserId="12319" LastEditorUserId="5610" LastEditDate="2015-05-11T14:06:14.733" LastActivityDate="2015-05-11T17:53:01.267" Title="How do I solve these 3x3 magic squares?" Tags="|magic-square|" AnswerCount="4" CommentCount="2" ClosedDate="2015-05-18T03:03:01.557" ContentLicense="CC BY-SA 3.0" />
  <row Id="14696" PostTypeId="1" AcceptedAnswerId="14726" CreationDate="2015-05-11T09:10:11.210" Score="13" ViewCount="1891" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A man of war;&lt;br&gt;&#xA;  Proofs at the bottom;&lt;br&gt;&#xA;  A savoury dish;&lt;br&gt;&#xA;  A capital city.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who am I?&lt;/strong&gt;&lt;br&gt;&#xA;&lt;sub&gt;Please explain how your answer fits each line of the riddle.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-11T18:24:08.773" Title="Man, proofs, dish, city - who am I?" Tags="|riddle|" AnswerCount="5" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14699" PostTypeId="2" ParentId="14695" CreationDate="2015-05-11T10:17:07.200" Score="0" Body="&lt;p&gt;For the 1st question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;| 10 | 5 |  6 |&lt;/p&gt;&#xA;&#xA;&lt;p&gt;|  3 | 7 | 11 |&lt;/p&gt;&#xA;&#xA;&lt;p&gt;|  8 | 9 |  4 |&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And Jimmy has given the answer for the 2nd question&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the method. Its very simple. Everyone knows how to solve 3*3 to get 15. But we need 21 i.e is 2 more for every cell. so add 2 in every cell of magic square. that will make 1 as 3 and 9 as 11. And you will then have to just manage the rows.&lt;/p&gt;&#xA;" OwnerUserId="11343" LastEditorUserId="11343" LastEditDate="2015-05-11T11:28:15.003" LastActivityDate="2015-05-11T11:28:15.003" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14700" PostTypeId="2" ParentId="14695" CreationDate="2015-05-11T10:28:22.610" Score="13" Body="&lt;p&gt;In general, any $n\times n$ magic square of range [1, $n^2$] with &lt;strong&gt;odd number&lt;/strong&gt; $n$ can be solved using the following algorithm:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Start at the middle grid in the bottom row. This is your 1.&lt;/li&gt;&#xA;&lt;li&gt;Move downwards and to the right by one grid. If this move results in a position outside the square, wrap around to the beginning of the row (or column).&lt;/li&gt;&#xA;&lt;li&gt;If 2 cannot be performed (i.e. the grid is already occupied or you are at the bottom-right corner), move upwards by one grid instead.&lt;/li&gt;&#xA;&lt;li&gt;Repeat 2 (or 3) until all numbers are filled.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Illustration using $n=3$. (For cells 2 and 3 the position before wrapping is shown in parentheses.)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;|   |   |   |&#xA;|   |   |   |&#xA;|   | 1 |   | // start here&#xA;&#xA;|   |   | 2 |&#xA;|   |   |   |&#xA;|   | 1 |   | // move to the right and then down.&#xA;         (2)  // Because moving down brings us outside the square,&#xA;              // we wrap around to the start of the column&#xA;&#xA;|   |   | 2 |&#xA;| 3 |   |   | (3)&#xA;|   | 1 |   | // move to the right and down.&#xA;              // moving right brings us outside, so wrap around to start of row&#xA;&#xA;| 4 |   | 2 |&#xA;| 3 |   |   |&#xA;|   | 1 |   | // cannot move down and right because that is occupied&#xA;              // so we move up instead&#xA;&#xA;| 4 |   | 2 |&#xA;| 3 | 5 |   |&#xA;|   | 1 |   | // move down and right&#xA;&#xA;| 4 |   | 2 |&#xA;| 3 | 5 |   |&#xA;|   | 1 | 6 | // move down and right&#xA;&#xA;| 4 |   | 2 |&#xA;| 3 | 5 | 7 |&#xA;|   | 1 | 6 | // cannot move down and right, so move up&#xA;&#xA;| 4 |   | 2 |&#xA;| 3 | 5 | 7 |&#xA;| 8 | 1 | 6 | // move down and right, wrap around to start of row&#xA;&#xA;| 4 | 9 | 2 |&#xA;| 3 | 5 | 7 |&#xA;| 8 | 1 | 6 | // move down and right, wrap around to start of column&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now, if you need to solve your magic square that starts with 3, simply add 2 to all cells of this standard square. Then rotate and/or reflect it until you get one where the numbers match your given ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider your first square (assuming we can omit 12):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;|   | 5 |   |&#xA;|   |   |   |&#xA;| 8 |   |   |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We first add 2 to all elements of our standard square:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;| 6  | 11 | 4  |&#xA;| 5  | 7  | 9  |&#xA;| 10 | 3  | 8  |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then rotate clockwise by 90 degrees:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;| 10 | 5  | 6  |&#xA;| 3  | 7  | 11 |&#xA;| 8  | 9  | 4  |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Your first square solved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further reading: &lt;a href=&quot;http://mathworld.wolfram.com/MagicSquare.html&quot;&gt;Magic Square from Wolfram MathWorld&lt;/a&gt;, which includes methods for solving even squares as well.&lt;/p&gt;&#xA;" OwnerUserId="12278" LastEditorUserId="12278" LastEditDate="2015-05-11T10:44:12.357" LastActivityDate="2015-05-11T10:44:12.357" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14701" PostTypeId="2" ParentId="14695" CreationDate="2015-05-11T10:29:39.153" Score="1" Body="&lt;p&gt;The answer for the first puzzle is this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;|10 | 5 | 6 |&#xA;| 3 | 7 |11 |&#xA;| 8 | 9 | 4 |&#xA;&#xA;| A | B | C |&#xA;| D | E | F |&#xA;| G | H | I |&#xA;&#xA;| A | 5 | C |&#xA;| D | E | F |&#xA;| 8 | H | I |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Range 3 - 12.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one extra integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neither 11 or 12 can be in the same row, column or diagonal as 8 since the other number would be 1 or 2 which is outside the range.  Therefore F is 11 or 12 and the other is the integer that is not included.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neither 3 nor 4 can be in the same row or column as 5 since the other number would be 12 or 13 and 13 is outside the range and 12 must be in F if it is in the puzzle.  Therefore D and I are 3 and 4.  Continuing H and A are 10 and 9 and E and C are 6 and 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot be 3 since either E would be 6 or 7 and A would be 11 or 12 and 11 and 12 is in F or not in the puzzle.  Therfore I = 4.  The rest of the puzzle fills itself in from there.  The number not included is 12.&lt;/p&gt;&#xA;" OwnerUserId="1777" LastActivityDate="2015-05-11T10:29:39.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14702" PostTypeId="2" ParentId="14696" CreationDate="2015-05-11T10:46:00.143" Score="3" Body="&lt;p&gt;Bit of a stretch, but is it  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Aquaman?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A man of war:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Type of jellyfish, plus aquaman is leader of a particularly martial nation (soldier)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proofs at the bottom:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Not sure on this one yet (bottom of the ocean)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A savoury dish:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Secret identity is Arthur Curry  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A capital city:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Atlantis (admittedly a fictional city)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11717" LastActivityDate="2015-05-11T10:46:00.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14703" PostTypeId="2" ParentId="14696" CreationDate="2015-05-11T11:01:00.347" Score="2" Body="&lt;p&gt;My best guess (at the moment) is :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; KFC&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A man of war&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Colonel Sanders&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proofs at the bottom&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The packaging requires oil-proofing the buckets at the bottom&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A savoury dish&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Famous Chicken&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A capital city&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Kentucky ? Some Capital City which is related to KFC ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10940" LastActivityDate="2015-05-11T11:01:00.347" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14704" PostTypeId="2" ParentId="14680" CreationDate="2015-05-11T11:39:22.867" Score="2" Body="&lt;p&gt;I think that dmg is right for the most part, except I think that the last two penultimate lines, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;My eyes are the color of the moon's home,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where sleeping dragons should not be woken.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Actually should be interpreted to mean the color blue. When I think of Sleeping dragons, I think of The Hogwarts school motto, &quot;Draco Dormiens Nunquam Titillandus,&quot; or, &quot;Never tickle a sleeping dragon.&quot; The moon would symbolize Luna Lovegood, and I think the &quot;moon's home&quot; refers to Ravenclaw house, who's color is blue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11814" LastActivityDate="2015-05-11T11:39:22.867" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14705" PostTypeId="1" AcceptedAnswerId="14707" CreationDate="2015-05-11T12:01:19.670" Score="-4" ViewCount="215" Body="&lt;p&gt;Using 7,8,9 and basic mathematical operations form the number 327600.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use every digit exactly once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Allowed Operations: $+,-,/,*, {}^{\displaystyle\large\hat{}} ,!,\sqrt{}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12058" LastEditorUserId="5373" LastEditDate="2015-05-13T19:49:13.540" LastActivityDate="2015-06-26T20:35:40.787" Title="Using numbers form a number" Tags="|mathematics|calculation-puzzle|formation-of-numbers|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14706" PostTypeId="1" AcceptedAnswerId="14710" CreationDate="2015-05-11T12:14:53.603" Score="0" ViewCount="186" Body="&lt;p&gt;You have a hundred, labeled, bulbs, turned off and a hundred, labeled, buttons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you begin pressing the button nº1, all bulbs turn on. When you then press the button nº2 all pair bulbs, begining by the 2n, turn off. When you press the button nº3, begining by the 3rd bulb, and all those which are multiples of 3, change their state. The ones that were off, light on, the ones turned of, go off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You do the same by every button, finally having pressed, by order, all buttons from the first to the 100th. Question is, how many bulbs will be turned on at the end of the process? Which ones will it be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As pointed out, this puzzle resembles &lt;a href=&quot;https://puzzling.stackexchange.com/questions/5049/nerds-jocks-and-lockers&quot; title=&quot;Nerds, Jocks, and Lockers&quot;&gt;Nerds, Jocks, and Lockers&lt;/a&gt;, being the question I asked a somehow simplyfied version of it.&#xA;I'll leave the question as it is, in case anybody wants to try with this one. Of course, if that's not the case, I'll eventually end up closed.&lt;/p&gt;&#xA;" OwnerUserId="12165" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-05-11T12:47:40.767" Title="Switches and numbers" Tags="|mathematics|logical-deduction|number-theory|" AnswerCount="2" CommentCount="0" ClosedDate="2015-05-11T13:45:11.830" ContentLicense="CC BY-SA 3.0" />
  <row Id="14707" PostTypeId="2" ParentId="14705" CreationDate="2015-05-11T12:18:18.940" Score="3" Body="&lt;p&gt;The answer is simple.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$ 9!-8!+7! $$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Although I'm not sure that you can really call that operation a &quot;basic mathematical operation&quot;. It's kind of in the category of intermediate operations.&lt;/p&gt;&#xA;" OwnerUserId="9289" LastActivityDate="2015-05-11T12:18:18.940" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14708" PostTypeId="1" AcceptedAnswerId="14713" CreationDate="2015-05-11T12:31:51.223" Score="-2" ViewCount="612" Body="&lt;p&gt;Using 2,3,4,5,6,7 and basic mathematical operations form the number 1088.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every digit must be used exactly once (neither zero nor greater than one times).  Concatenation of digits (such as 53) is not allowed. Each operator can be used multiple times or not at all in any combination. &lt;strong&gt;This removes duplicates and creates it as having a unique solution&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Allowed Operations: $+,-,/,*, {}^{\displaystyle\large\hat{}} ,!,\sqrt{}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12058" LastEditorUserId="5373" LastEditDate="2015-05-13T19:49:48.537" LastActivityDate="2016-10-02T14:03:16.353" Title="Form a number using some numbers and mathematical operations" Tags="|mathematics|calculation-puzzle|formation-of-numbers|" AnswerCount="5" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14709" PostTypeId="1" AcceptedAnswerId="14721" CreationDate="2015-05-11T12:45:40.697" Score="7" ViewCount="426" Body="&lt;p&gt;Given this sequence of letters:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; _, B, G, C, R, M, Y, _&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What is the first and last letter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;hint 1 &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; each letter represent a color&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;hint 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C represents the color Cyan and M for Magenta&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is responded, but try to find the sequence behind it. Explain Why &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the sequence is related a order of HTML Hex Color Codes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11906" LastEditorUserId="11906" LastEditDate="2015-05-15T14:35:40.307" LastActivityDate="2015-05-15T14:36:13.947" Title="Complete the missing letter (require Web knowledge)" Tags="|letter-sequence|" AnswerCount="2" CommentCount="9" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14710" PostTypeId="2" ParentId="14706" CreationDate="2015-05-11T12:46:53.630" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;10 lights are on - 1,4,9,16,25,36,49,64,81,100  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A light is on if toggled an odd number of times (from before any switches are pressed) and off if toggled an even number of times  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;As each switch affects bulbs at positions that are integer multiples of the switch's position, the number of integer factors that each bulb has are what determines its end state.&lt;br&gt;&#xA; Square numbers have the property that one of their factors is repeated (e.g. 1,3,3,9 for 9) - as each button is pressed only once, that state change isn't repeated, so the square-numbered bulbs are left lit at the end  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's a horribly convoluted description, but I couldn't come up with a more succinct way to summarise it&lt;/p&gt;&#xA;" OwnerUserId="11717" LastActivityDate="2015-05-11T12:46:53.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14711" PostTypeId="2" ParentId="14706" CreationDate="2015-05-11T12:47:40.767" Score="3" Body="&lt;p&gt;Every bulb's switch will be flipped for each of it's divisors.  The bulbs that will be turned on will have an odd number of divisors.  The only numbers for which that occurs are perfect squares.  There will be ten lights turned on matching the 10 perfect squares less than or equal to 100.&lt;/p&gt;&#xA;" OwnerUserId="1777" LastActivityDate="2015-05-11T12:47:40.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14713" PostTypeId="2" ParentId="14708" CreationDate="2015-05-11T12:57:09.700" Score="9" Body="&lt;p&gt;Update: This 4-operator answer is now disallowed by OP after rule change. The 6-operator one still works though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An answer that uses 4 operators only (assuming concatenation to form multi-digit numbers is allowed):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 523 \times \sqrt4 + 6 \times 7 = 1088 $$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Assuming we can use parentheses and they don't count as operators, then we have 6 operators with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 7^3 + 6! + 5 ^ {(4-2)} = 1088 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we can't use parentheses, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 7\;3\;{}^\wedge\;6\;!\;+\;5\;4\;2\;-\;{}^\wedge\;+ $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is the same thing in reverse Polish notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above assumes that the &lt;code&gt;^&lt;/code&gt; operator is for exponentiation, not XOR.&lt;/p&gt;&#xA;" OwnerUserId="12278" LastEditorUserId="12278" LastEditDate="2015-05-12T15:47:03.877" LastActivityDate="2015-05-12T15:47:03.877" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="14714" PostTypeId="2" ParentId="14705" CreationDate="2015-05-11T13:02:22.327" Score="2" Body="&lt;p&gt;Shortest answer I can find using only basic $(+,-,\times,\div)$ operations is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$7\times8\times9\times(8\times9\times9+9-7)$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4577" LastEditorUserId="13494" LastEditDate="2015-06-26T20:35:40.787" LastActivityDate="2015-06-26T20:35:40.787" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14716" PostTypeId="2" ParentId="14659" CreationDate="2015-05-11T13:21:29.067" Score="3" Body="&lt;p&gt;Are you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Television&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am an instrument of fear and propaganda,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am bringer of ill news.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;These two are relatively obvious, as network news stations love to fearmonger for ratings&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But if you see me, you may laugh,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not all shows on TV are bad.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can distract you on a bad day.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TV could arguably be based on the same concepts as musical theatre, which was designed to act as a distraction during bad times&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The world is glued to me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The basic idiom &quot;glued to the television set&quot; comes to mind&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Loving but one part of me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People love TV, but they hate cable companies&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You complain of the cost of me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because they are exorbitantly expensive&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And can't use me unless you have, well,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nowadays, however, television sets require either a good internet connection or a cable subscription in order to get anything outside the basic OTA channels&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would love to tell you, but then, of course, I'd have to kill you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Who doesn't love to quote their favorite tv shows and movies? Also, if this line were actually finished, it would be a large give away to the answer of the riddle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12242" LastActivityDate="2015-05-11T13:21:29.067" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14717" PostTypeId="2" ParentId="14709" CreationDate="2015-05-11T13:26:06.733" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; B &amp;amp; W&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;representing&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Black and white? Seems like the two obvious missing 'colours' from cyan, yellow, red, blue, green&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11880" LastActivityDate="2015-05-11T13:26:06.733" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="14718" PostTypeId="2" ParentId="13303" CreationDate="2015-05-11T13:45:30.857" Score="7" Body="&lt;p&gt;So, I think I worked out the fourth stanza.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Cardinals sitting in their nests, pick them up one by one.&#xA;Set them free and follow them, then turn to what's undone.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;THEIR NESTS&lt;/code&gt; contains the following cardinal directions, in order: &lt;code&gt;E&lt;/code&gt;, &lt;code&gt;N&lt;/code&gt;, &lt;code&gt;E&lt;/code&gt;, &lt;code&gt;S&lt;/code&gt;, &lt;code&gt;S&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then turn to what's undone:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The only unused cardinal direction is &lt;code&gt;W&lt;/code&gt;est. You turn to what's undone. So, you end the stanza facing west.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;s&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; UE IE II CU AE...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Updating one of my points, hopefully this wasn't actually the thing I got right...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding some more&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plus, I got rid of the RS...&lt;/s&gt;&lt;/p&gt;&#xA;" OwnerUserId="12242" LastEditorUserId="12242" LastEditDate="2015-05-16T16:36:40.103" LastActivityDate="2015-05-16T16:36:40.103" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14719" PostTypeId="1" AcceptedAnswerId="14723" CreationDate="2015-05-11T14:08:26.967" Score="4" ViewCount="1771" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YznGM.jpg&quot; alt=&quot;Deserted Island&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are stranded on a deserted island with little to no food sources. By your guesses, you will only be able to survive another 10 days. Your only real chance of survival is getting rescued by a boat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the probability of seeing a boat in 30 days is 95%, what are the chances of seeing a boat in the next 10 days? Stated differently, how likely are you to survive the next 10 days.&lt;/p&gt;&#xA;" OwnerUserId="12317" LastEditorDisplayName="user20" LastEditDate="2015-07-23T05:13:59.623" LastActivityDate="2015-07-23T05:13:59.623" Title="Deserted Island" Tags="|probability|" AnswerCount="3" CommentCount="0" ClosedDate="2015-05-11T18:23:58.240" ContentLicense="CC BY-SA 3.0" />
  <row Id="14720" PostTypeId="2" ParentId="14719" CreationDate="2015-05-11T14:20:44.697" Score="4" Body="&lt;p&gt;Short answer:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 63.16% Chance of seeing a boat  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You have to consider the opposite chance (of not seeing a boat) on a given day, call it N&lt;br&gt;&#xA; 100% - (N)^30 = 95%&lt;br&gt;&#xA; Solving for N gives us N = 0.904966...&lt;br&gt;&#xA; Then plug N into N^10 to give the chance of notseeing a boat in 10 days (~36.84%)&lt;br&gt;&#xA; Finally: subtract that from 100% to get ~63.16%&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7715" LastEditorUserId="7715" LastEditDate="2015-05-11T14:26:49.027" LastActivityDate="2015-05-11T14:26:49.027" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14721" PostTypeId="2" ParentId="14709" CreationDate="2015-05-11T14:24:46.647" Score="8" Body="&lt;p&gt;As HumphreyTheHater stated,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Black and White, or B and W, respectively&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But why are these the proper letters?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Black = 0x000000&#xA; &lt;br&gt;Blue    = 0x0000FF&#xA; &lt;br&gt;Green   = 0x00FF00&#xA; &lt;br&gt;Cyan    = 0x00FFFF&#xA; &lt;br&gt;Red     = 0xFF0000&#xA; &lt;br&gt;Magenta = 0xFF00FF&#xA; &lt;br&gt;Yellow  = 0xFFFF00&#xA; &lt;br&gt;White   = 0xFFFFFF&#xA; &lt;br&gt;&lt;br&gt;These colors follow the standard binary counting system, using 00 for 0 and FF for 1:&#xA; &lt;br&gt;000&#xA; &lt;br&gt;001&#xA; &lt;br&gt;010&#xA;&lt;br&gt; 011&#xA; &lt;br&gt;100&#xA; &lt;br&gt;101&#xA; &lt;br&gt;110&#xA; &lt;br&gt;111&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12242" LastEditorUserId="11475" LastEditDate="2015-05-15T14:36:13.947" LastActivityDate="2015-05-15T14:36:13.947" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14722" PostTypeId="2" ParentId="14719" CreationDate="2015-05-11T14:25:31.570" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If the probability of seeing a boat in 30 days is 95%, then the probability of not seeing a boat in 30 days is 5%. To find out the probability of seeing a boat in any one day, we figure out the 30th root of 0.05, which is about 0.905. This means that there is about a 9.5% chance of seeing a boat on any given day and a reciprocal 90.5% chance of not seeing a boat on any given day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solution&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The chance of not seeing a boat is about 36.84%, so the chance of seeing a boat and, therefore, being rescued, is 63.16%.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9828" LastActivityDate="2015-05-11T14:25:31.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14723" PostTypeId="2" ParentId="14719" CreationDate="2015-05-11T14:26:44.953" Score="5" Body="&lt;p&gt;Let $P(X)$ be the probablity of not seeing a boat in X days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(30) = 0.05$&lt;br&gt;&#xA;$P(30) = P(10) * P(10) * P(10)$&lt;br&gt;&#xA;$P(10) = \sqrt[3]{0.05}$ = ~0.37&lt;br&gt;&#xA;So the probability of seeing it is ~0.63 or 63%&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-05-11T14:26:44.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14724" PostTypeId="2" ParentId="14653" CreationDate="2015-05-11T14:45:44.003" Score="0" Body="&lt;p&gt;Is it...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Antimatter?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am here,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Antimatter has been proved to exist&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yet most people don't care,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is not very significant at the time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I kill people, if you are not careful.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Upon contact with matter, antimatter will instantly annihilate itself and whatever matter it touched.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be created,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Scientists have been able to create and study antimatter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but it's rather useless.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Antimatter does have practical applications. However, considering the properties of antiparticles and our current technology, it is nearly impossible to put antimatter to use.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you can see me, It's because of Him.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let's say that &quot;see&quot; refers to the pronouncement of $c$, the speed of light constant in the equation $E=mc^2$. Antimatter was predicted to exist using the expanded form of $E=mc^2$, which is $E^2=m^2c^4$. And of course, as we all know, $E=mc^2$ was discovered by Albert Einstein (&quot;Him&quot;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4985" LastActivityDate="2015-05-11T14:45:44.003" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14725" PostTypeId="2" ParentId="14653" CreationDate="2015-05-11T15:09:57.907" Score="0" Body="&lt;p&gt;Ok, newish here, so I'll take a long shot stab at this :)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Are you bacteria/germs ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am here,&#xA;yet most people don't care,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; most people carry on without worrying to much about bacteria.&#xA; Sure, we wash our hands and such, but mostly we &quot;don't care&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I kill people,&#xA;if you are not careful.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; some bacteria can lead to sickness/death if you're not careful&#xA; can think of some extreme situations, such as during operations, or spreading of a cold/flu&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be created,&#xA;but it's rather useless.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; we can create bacteria and diseases:&#xA; &lt;a href=&quot;http://www.sciencemag.org/content/329/5987/52.abstract&quot; rel=&quot;nofollow&quot;&gt;http://www.sciencemag.org/content/329/5987/52.abstract&lt;/a&gt;&#xA; (not convinced of the useless part though :( )&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you can see me,&#xA;It's because of Him.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if you can see something that smaller, it's likely via the aid of a Microscope or such.&#xA; (still not 100% what &quot;Him&quot; is referring, but if we call the Microscope &quot;Him&quot; ... :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9267" LastActivityDate="2015-05-11T15:09:57.907" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14726" PostTypeId="2" ParentId="14696" CreationDate="2015-05-11T15:19:46.103" Score="16" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Wellington - Arthur Wellesley, 1st Duke of Wellington (Napoleon war)&lt;br&gt;&#xA;Wellingtons - boots (proof at the bottom)&lt;br&gt;&#xA;Beef Wellington - a savory dish&lt;br&gt;&#xA;Wellington - capital city, New Zealand   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12329" LastEditorUserId="12329" LastEditDate="2015-05-11T15:41:53.277" LastActivityDate="2015-05-11T15:41:53.277" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14727" PostTypeId="2" ParentId="14696" CreationDate="2015-05-11T15:20:58.920" Score="11" Body="&lt;p&gt;I think it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bismarck&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A man of war;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Otto von Bismarck was a the first german chancellor and also a military leader&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proofs at the bottom;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I belive this one refers to the sunken battleship&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A savoury dish;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This should be the Bismarckhering - a dish made from fish and vinegar&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A capital city.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bismarck is the capital of North Dakota&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12328" LastActivityDate="2015-05-11T15:20:58.920" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14728" PostTypeId="1" AcceptedAnswerId="14732" CreationDate="2015-05-11T15:25:19.510" Score="6" ViewCount="3114" Body="&lt;p&gt;Once, there were two travelers going down a winding forest path. The travelers come to a fork in the road. Unfortunately, the woods are so densely wooded, you cannot see more than a few feet down each path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the split, there is a young man sitting with a puzzled look on his face. Clearly shaken by something, the travelers ask the young man what the problem is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Well, you see,&quot; said the young man, &quot;I was walking to visit a friend in the village on the far side of the wood. When I came to this fork, there was a raspy-voiced apparition floating here.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;He posed a riddle to me, to try to figure out which path is the correct one. &lt;em&gt;He said that one path lead to the village, while the other led to certain death.&lt;/em&gt; All I had to do was ask him a question. One question with a 'YES' or 'NO' answer. I could use that question to determine which is the correct path.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I sat and I pondered and I pondered, but I couldn't come up with a good question. Finally, looking for clarification, I asked him if there were any limits to the question. He said 'NO' and disappeared. Now, I cannot find my way forward.&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Well, the two travelers decided to help out the young lad. They told him that they, as experienced travelers, could examine the paths and return to tell the young man which path was good and which was bad. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so they set out examining the paths. Within minutes, they had disappeared into the thick woods. The young man sat in the clearing, waiting for hours. Finally, one of the travelers returned, obviously short on water. With a dry, hoarse voice, the traveler told the young man that the path to the left, from which he just escaped, was the path to freedom. The young man immediately went down the path to the right, where he met the other traveler returning to tell him this was the correct path. He continued on his way until he found the town.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How did he determine the right path to get out of the woods alive?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OP EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I don't proofread, apparently. I completely missed a line above. I have added the italicized line. I apologize to those who have already answered if this causes any issue with your answer.&lt;/p&gt;&#xA;" OwnerUserId="12242" LastEditorUserId="25966" LastEditDate="2016-07-13T09:01:55.417" LastActivityDate="2018-04-14T22:11:38.133" Title="A fork in the road" Tags="|riddle|story|" AnswerCount="5" CommentCount="10" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14730" PostTypeId="2" ParentId="14728" CreationDate="2015-05-11T15:36:15.950" Score="0" Body="&lt;p&gt;&lt;em&gt;With a dry, hoarse voice, the traveler told the young man that the path to the left, from which he just escaped, was the path to freedom.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This doesn't mean that the left path led to the village. Indeed, &quot;freedom&quot; is more appropriate for a desert or a large territory rather than a village. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From this, the young man deduced that he had to go right!&lt;/p&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-05-11T15:36:15.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14731" PostTypeId="2" ParentId="14728" CreationDate="2015-05-11T15:38:09.247" Score="-1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The traveler was facing towards him, so the path on his left &lt;em&gt;was&lt;/em&gt; the path to the right of the young man.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11880" LastActivityDate="2015-05-11T15:38:09.247" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14732" PostTypeId="2" ParentId="14728" CreationDate="2015-05-11T15:42:50.993" Score="5" Body="&lt;p&gt;There seem to be a few potential reasons here:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The first traveler is short on water, suggesting that they path they took was not the correct one, as they found nowhere to restock  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the second traveler was met en route, suggesting that they had found something to delay their return. Their lack of hoarse-ness suggests that would be the town, complete with water with which to stock up  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or (now my preferred answer, but leaving others for posterity):  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;perhaps the first traveler, given that their voice emulates that of the apparition, &lt;strike&gt;has been the victim of an unfortunate possession. I make it a rule to avoid unknown phantasms where possible.&lt;/strike&gt;  died and is now making their way back to the fork in an attempt to lure another poor soul to their doom. From this we can infer that the original apparition would most likely not have told the truth had they been asked about the safe route out of the forest.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or (the most likely):  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Given the fact that traveler one is clearly exhausted and dehydrated, you would expect their speech to be halting and incomplete. So, 'the path to the left [of that](lost to a pause) from which he just escaped' tells you that it is the path to the traveler's left that you want. As he is returning (and thus facing you), it is the path to your right&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11717" LastEditorUserId="11717" LastEditDate="2015-05-11T16:26:45.357" LastActivityDate="2015-05-11T16:26:45.357" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14734" PostTypeId="2" ParentId="14728" CreationDate="2015-05-11T15:55:45.487" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Why would you have to escape from a path to freedom? No need to escape, you're free! So obviously an inconsistency in the traveler's story, which made the young man decide that something must be wrong down that path. That's why he took the other path. The other traveller just said something really simple, so his story was consistent.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11721" LastActivityDate="2015-05-11T15:55:45.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14735" PostTypeId="2" ParentId="12781" CreationDate="2015-05-11T16:19:05.350" Score="1" Body="&lt;p&gt;Nobody in Barberland has more than 1,000,000 hairs.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Well, no change here, population could be infinite.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No two inhabitants of Barberland have the same number of hairs.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Population maximum drops down to the count of [0,1,000,000], so 1,000,001&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nobody in Barberland has the same number of hairs as the sum of two other inhabitants' hairs.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Here is where life gets tricky. First off, we can exclude anyone below 500,000 as a means to capture the largest amount of data. This is because the sum of any two numbers in the set from [500,000,1,000,000] will be greater than 1,000,000 and eliminate that person from town. Thus we have 500,001.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But Wait, you say. What about the guy with no hair?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Good point for you to make. Since we require two different inhabitants, we can include the person with no hair. This is because we need to match the form Px + py = Pz where x =/= y =/= z to actively eliminate anyone. In the case of the bald person, x = y or y = z or x = z. That breaks the form and allows this person to remain in the city.&#xA; &lt;br&gt;So our count is now 5,000,002. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But can we add anyone else?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Short answer, no. If you take the set from [500,000,1,000,000], you are assured to have nowhere that Px + Py = Pz where x =/= y =/= z. This means that none of these people would be eliminated from the city.&#xA;&lt;br&gt;If you start adding people who are below the median value, you have to remove at least one person from your set. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12242" LastEditorUserId="10281" LastEditDate="2015-05-11T19:52:16.950" LastActivityDate="2015-05-11T19:52:16.950" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14736" PostTypeId="2" ParentId="14653" CreationDate="2015-05-11T16:26:48.137" Score="1" Body="&lt;p&gt;Is it:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sunlight?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am here,&lt;br&gt;&#xA;yet most people don't care,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yea it's warm and, free..&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I kill people,&lt;br&gt;&#xA;if you are not careful.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sunburn N degree  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be created,&lt;br&gt;&#xA;but it's rather useless.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not too sure, &quot;Sunlight&quot; can be created by mimic the same fusions from sun?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you can see me,&lt;br&gt;&#xA;It's because of Him.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HIM as Heliocentric Information Map?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sunlight?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="10037" LastEditDate="2015-05-11T17:34:58.127" LastActivityDate="2015-05-11T17:34:58.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14737" PostTypeId="2" ParentId="14728" CreationDate="2015-05-11T16:29:40.857" Score="1" Body="&lt;p&gt;He chose the right path because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We know the apparition lies. He first said that the traveler could only ask a yes/no question. But when the traveler asked if the question had limits, the apparition said no.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That means:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Either the apparition lied when he said that one path led to the village and the other to certain death, or he lied when answering the question (or both, I suppose).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He lied when answering the question, then one of the paths does lead to certain death, which means that the traveler could not have &quot;escaped&quot; and therefore wasn't really the traveler but the apparition trying to trick him, so he chose the other path.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On the other hand:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the apparition lied when saying that one path led to certain death, then either path would be fine to take - the traveler takes the right path in case the first scenario is in place.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of course:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The apparition might have lied and both paths lead to certain death, which would mean the traveler's screwed either way unless he goes home.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2015-05-11T16:29:40.857" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14738" PostTypeId="2" ParentId="14696" CreationDate="2015-05-11T16:44:46.377" Score="3" Body="&lt;p&gt;I think it could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Wellington&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A Man of War&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Duke of Wellington, defeated Napoleon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proofs at the bottom&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Wellington Boots, water-proof feet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A savoury dish&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Beef Wellingtons (yum).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A capital city&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Wellington, New Zealand.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit:: Dupe! So what I didn't see it.&lt;/p&gt;&#xA;" OwnerUserId="12333" LastEditorUserId="5373" LastEditDate="2015-05-11T18:24:08.773" LastActivityDate="2015-05-11T18:24:08.773" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14739" PostTypeId="2" ParentId="14568" CreationDate="2015-05-11T16:53:12.143" Score="1" Body="&lt;p&gt;&lt;strong&gt;Incomplete proof&lt;/strong&gt; for 10 being the minimum, please feel free to add/modify it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There must exist a plane with 4 points. Let us assume that no plane has more than 4 points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quad formed on our starting plane is ABCD. Now AC needs a parallel. So we need EF||AC on the same plane as AC but not BD. Now EFAC becomes a quad. We can use the same logic to show that we need another 2 points, so on till infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, there must exist a plane with more than 4 points. Suppose the plane has 5 points. We notice that we still need 2 more points not on the plane because 2 diagonals will have no parallels within the plane. Using earlier logic, we can eliminate this case as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose now, that the plane has 6 points. We soon realise that we can use a regular hexagon such that there is no dependence on parallels outside the plane. So 6 is valid. However, by rule we need a point outside the plane. This point creates 6 new lines, one with each corner, that require parallels. &lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-05-11T16:53:12.143" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14741" PostTypeId="1" CreationDate="2015-05-11T16:56:37.513" Score="4" ViewCount="347" Body="&lt;p&gt;What color should begin the following sequence of words?&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;&lt;strong&gt;???&lt;/strong&gt;&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;fruit&lt;/li&gt;&#xA;&lt;li&gt;letter&lt;/li&gt;&#xA;&lt;li&gt;computer&lt;/li&gt;&#xA;&lt;li&gt;number&lt;/li&gt;&#xA;&lt;li&gt;subject&lt;/li&gt;&#xA;&lt;li&gt;consume&lt;/li&gt;&#xA;&lt;li&gt;before&lt;/li&gt;&#xA;&lt;li&gt;cut&lt;/li&gt;&#xA;&lt;li&gt;cry&lt;/li&gt;&#xA;&lt;li&gt;slide&lt;/li&gt;&#xA;&lt;li&gt;gentle&lt;/li&gt;&#xA;&lt;li&gt;awake&lt;/li&gt;&#xA;&lt;li&gt;not&lt;/li&gt;&#xA;&lt;li&gt;rain&lt;/li&gt;&#xA;&lt;li&gt;cold&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;h3&gt;Hint 1&lt;/h3&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;Beneath the sequence is a ring of other words.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h3&gt;Hint 2&lt;/h3&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;That ring can be found by looking at pairs of the sequence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h3&gt;Hint 3&lt;/h3&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;Each pair is made of words adjacent to each other in the sequence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h3&gt;Hint 4&lt;/h3&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;The ring has eight words.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h3&gt;Hint 5&lt;/h3&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;Each word in the ring is connected to its neighbors through their initial letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7425" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-21T15:38:17.553" Title="Find the color to start the sequence" Tags="|word-sequence|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14743" PostTypeId="1" AcceptedAnswerId="14745" CreationDate="2015-05-11T17:05:31.397" Score="6" ViewCount="710" Body="&lt;p&gt;A says to B&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My age is twice of what your age was when I was of your age.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;B says to A&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you add your current age to twice of your age when I was born, the&#xA;  answer is $63$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assuming they both told the truth, how old are they?&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-05-12T12:52:42.323" Title="Ages of A and B (self-reference in past)" Tags="|mathematics|calculation-puzzle|time|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14745" PostTypeId="2" ParentId="14743" CreationDate="2015-05-11T17:21:38.783" Score="12" Body="&lt;p&gt;At first there seems to be a contradiction. The first sentence makes A's age even, since it's double something. The second makes it odd, since when added to double something, you get an odd number. But if you write out the equations and solve them, you do get a consistent pair of numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it's rather a cheat, but:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A is 42 and B is 31.5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;to check it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When A was 31.5, B, was 21 (the difference is 10.5 years.) Double 21 is 42.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When B was born, A was 10.5. Double that is 21. Add to 42 you get 63.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nobody said it had to be integers I guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to how to do it, for me it was easier to solve for A and d (the difference between their ages) than A and B. Spoilers below because the MathJax was hard enough but getting the spoilers to interact properly with it appears impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A = 2 (A -2d)$$&#xA;$$A = 2A - 4d$$&#xA;$$-A = 4d$$&#xA;$$ A = 4d$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A + 2d = 63$$&#xA;$$substitute A = 4d $$&#xA;$$2d + 4d = 63$$&#xA;$$6d = 63$$&#xA;$$d = 10.5$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A = 4d = 42$$&#xA;$$B = A - d = 31.5$$&lt;/p&gt;&#xA;" OwnerUserId="1509" LastEditorUserId="1509" LastEditDate="2015-05-11T19:49:58.593" LastActivityDate="2015-05-11T19:49:58.593" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14746" PostTypeId="1" AcceptedAnswerId="14762" CreationDate="2015-05-11T17:22:58.877" Score="10" ViewCount="1075" Body="&lt;p&gt;I recently posed a puzzle about &lt;a href=&quot;https://puzzling.stackexchange.com/questions/14491/robot-long-and-robot-high&quot;&gt;Robot High and Robot Long&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That puzzle was inspired by a comic I saw in &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/193517925X&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Benjamin Bear in Fuzzy Thinking&lt;/em&gt;&lt;/a&gt; by Philippe Coudray. The particular comic is shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8YqyT.jpg&quot; alt=&quot;Comic of bear and rabbit jumping a gorge&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see in the edit history of the linked question, my original translation of the comic to puzzle added a bunch of numbers. This caused too much misdirection and calculations, though, and led many away from the intended solution.  Based on comments and answers, I tried to generalize the problem. There were still a variety of responses, though, and many were valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the Puzzling community are clever and the &lt;a href=&quot;/questions/tagged/lateral-thinking&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;lateral-thinking&amp;#39;&quot; rel=&quot;tag&quot;&gt;lateral-thinking&lt;/a&gt; tag really encourages open-ended thinking and comes with &lt;a href=&quot;http://meta.puzzling.stackexchange.com/questions/2971/lateral-thinking-requires-a-new-policy&quot;&gt;its own problems&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question, therefore, is two-fold:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What's a better way to convert the comic to a puzzle intended for the Puzzling.SE community without resorting to a long list of caveats?&lt;/li&gt;&#xA;&lt;li&gt;What down remove from that version and still have something that would still work for most &quot;normal&quot; people?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-05-13T14:39:14.267" Title="Help me improve the jumping robot problem" Tags="|puzzle-creation|" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14747" PostTypeId="2" ParentId="14743" CreationDate="2015-05-11T17:24:34.890" Score="2" Body="&lt;p&gt;A: When I was your age [A-[A-B]], my age is twice [2*] of what your age was: [B-[A-B]]&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; [A-[A-B]] = 2*[B-[A-B]] --&gt; B = 4*B - 2*A --&gt; &lt;b&gt;A = 3/2*B&lt;/b&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;B: If you add your current age [+A] to twice [*2] your age when I was born [A-B] you get 63 [= 63]&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A + 2*[A-B] = 63  --&gt; 3*A - 2*B = 63 --&gt; 3*(3/2*B) - 2B = 63&#xA;&lt;br&gt;&lt;b&gt;B = 25.2&lt;/b&gt;&#xA;&lt;br&gt;&lt;b&gt;A = 37.8&lt;/b&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure this isn't correct...but this is what my equations get me. Maybe if these people didn't talk such nonsense it would be easier :)&lt;/p&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-05-12T12:52:42.323" LastActivityDate="2015-05-12T12:52:42.323" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14748" PostTypeId="2" ParentId="14695" CreationDate="2015-05-11T17:53:01.267" Score="3" Body="&lt;p&gt;Here is an alternative technique. Consider your first magic square:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;| a | 5 | b |&#xA;| c | d | e |&#xA;| 8 | f | g |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I filled in the empty cells with the variables $a$ through $g$. We will create a system of equations to solve for these seven variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that each row must sum to 21. As a result, we can write the following three equations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1st row: $a + 5 + b = 21$&lt;/li&gt;&#xA;&lt;li&gt;2nd row: $c + d + e = 21$&lt;/li&gt;&#xA;&lt;li&gt;3rd row: $8 + f + g = 21$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Likewise, every column and diagonal must sum to 21, giving us five more equations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1st column: $a + c + 8 = 21$&lt;/li&gt;&#xA;&lt;li&gt;2nd column: $5 + d + f = 21$&lt;/li&gt;&#xA;&lt;li&gt;3rd column: $b + e + g = 21$&lt;/li&gt;&#xA;&lt;li&gt;\ diagonal: $a + d + g = 21$&lt;/li&gt;&#xA;&lt;li&gt;/ diagonal: $8 + d + b = 21$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So we now have 8 equations with 7 unknowns. We can select any seven of them and apply the usual equation-solving techniques to find the values of $a$, $b$, $c$, $d$, $e$, $f$, and $g$, giving us the solution to the magic square.&lt;/p&gt;&#xA;" OwnerUserId="294" LastActivityDate="2015-05-11T17:53:01.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14749" PostTypeId="1" CreationDate="2015-05-11T18:08:42.487" Score="20" ViewCount="15164" Body="&lt;p&gt;An inspector knows that exactly one of 3 suspects committed a crime, and interviews them to find out which. Each person lies one time, and tells the truth the other time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A says: I did not do it. B did it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B says: I did not do it. I know that C did it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C says: I did not do it. B does not know who it was.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the inspector figure out the culprit? If so, who is it?&lt;/p&gt;&#xA;" OwnerUserId="12334" LastEditorUserId="5373" LastEditDate="2015-05-11T20:43:25.747" LastActivityDate="2015-12-07T05:27:05.810" Title="Logic puzzle - 3 suspects" Tags="|logical-deduction|liars|" AnswerCount="8" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14750" PostTypeId="2" ParentId="14746" CreationDate="2015-05-11T18:10:13.547" Score="2" Body="&lt;p&gt;I know for me, the way you worded your conditions were what threw me off. Specifically rules 7, 8, and 10. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way rule 7 is worded, it seems like there is no physical way for High to be on top of Low. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Follow that by not adding any data about the sizes of the robots, as well as the generic jump data, you make it seem like, if High were to jump from atop Low, Low would be forced down too low and fall in the ravine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then Rule 8 forbids combining jumping power at all. Again, it seems like this is what is being done by launching off the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All in all, you were missing key data that the image presents. The bear is larger than the rabbit. The rabbit's downward force on the bear doesn't force it into the ravine. If you look at the answers and discussions on your post, many of them specifically state that there needs to be a size difference between the bots in order to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also never explicitly state &lt;strong&gt;&lt;em&gt;how&lt;/em&gt;&lt;/strong&gt; the robots move, but your image depicts them with what appears to be wheels on the base. This leads to assumptions about locomotion that are not present in the picture above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, my answer assumed that they rolled on wheels and that their &lt;code&gt;stopped&lt;/code&gt; speed wasn't braked so the wheel would be free to move.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;TL;DR: in my opinion you left your parameters too open and made some rules contradictory to your winning condition.&lt;/p&gt;&#xA;" OwnerUserId="12242" LastEditorUserId="5373" LastEditDate="2015-05-11T22:08:02.140" LastActivityDate="2015-05-11T22:08:02.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14751" PostTypeId="2" ParentId="14749" CreationDate="2015-05-11T18:14:16.590" Score="12" Body="&lt;p&gt;Each person makes one true statement and one false statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If A did it:&#xA;   &lt;br&gt;&lt;code&gt;I was not&lt;/code&gt; is FALSE&#xA;   &lt;br&gt;&lt;code&gt;B did it&lt;/code&gt;  should be TRUE but is FALSE&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If A committed the crime, then both of A's statements are false. This contradicts the rules of the riddle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If A didn't do it:&#xA;   &lt;br&gt;&lt;code&gt;I was not&lt;/code&gt; is TRUE&#xA;   &lt;br&gt;&lt;code&gt;B did it&lt;/code&gt;  should be FALSE. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this time, we cannot confirm or deny this statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If B did it:&#xA;   &lt;br&gt;&lt;code&gt;I was not&lt;/code&gt;              is FALSE&#xA;   &lt;br&gt;&lt;code&gt;I know that C did it.&lt;/code&gt;  should be TRUE but is FALSE&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much like A, B cannot give two false statements. If B committed the crime, then B would know that B did it, thus making the second statement false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If B didn't it:&#xA;   &lt;br&gt;&lt;code&gt;I was not&lt;/code&gt;              is TRUE&#xA;   &lt;br&gt;&lt;code&gt;I know that C did it.&lt;/code&gt;  should be FALSE&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, we don't entirely know if B's second statement is false or not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If C did it:&#xA;   &lt;br&gt;&lt;code&gt;I was not&lt;/code&gt;                  is FALSE&#xA;   &lt;br&gt;&lt;code&gt;B do not know who was it.&lt;/code&gt;  is TRUE&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, so this is where things get interesting. If C is responsible, that we know that the second statement is true. That also means that the second case for B needs to be true as well. This means that B doesn't know who did it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does not mean that it isn't C. &lt;strong&gt;It only means that B doesn't know C did it.&lt;/strong&gt; He's just a lucky guesser, that's all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also means that A's second case should become true as well. A stated that B did it. We know, in this case, that is false. This means that 'A didn't do it' works out logically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If C didn't do it:&#xA;   &lt;br&gt;&lt;code&gt;I was not&lt;/code&gt;                  is TRUE&#xA;   &lt;br&gt;&lt;code&gt;B do not know who was it.&lt;/code&gt;  could be TRUE&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously C's claim of innocence is true. This means that C's statement that B is wrong must be FALSE. If that is the case, then B must be telling the truth on that line and lying on the first. We have already logically concluded that is contradictory and as such impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the only logically valid statement of the six above is &lt;code&gt;C did it&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="12242" LastEditorUserId="12242" LastEditDate="2015-05-11T18:30:28.820" LastActivityDate="2015-05-11T18:30:28.820" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14752" PostTypeId="2" ParentId="14749" CreationDate="2015-05-11T18:18:55.353" Score="44" Body="&lt;p&gt;If we assume that there are no contradictions in the puzzle (i.e. there cannot be a situation where the &quot;1 lie, 1 truth&quot; rule is violated), then we don't even need to read what B and C say. Only A's statement matters.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If A's statements are Lie, Truth respectively, then his combined statement is that he did it (opposite of first statement) and B did it. This is clearly impossible, so it must be the case that A's statements are Truth, Lie respectively. This means A didn't do it, B didn't do it, so C must have done it. B and C's statements are irrelevant.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10861" LastEditorUserId="10861" LastEditDate="2015-05-12T16:20:54.680" LastActivityDate="2015-05-12T16:20:54.680" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14753" PostTypeId="2" ParentId="14749" CreationDate="2015-05-11T18:19:19.260" Score="9" Body="&lt;p&gt;First assume A did it. Then A's statements &quot;I did not&quot; and &quot;B did it&quot; are both false. Contradiction, so A didn't do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now assume B did it. Then A's statements &quot;I did not&quot; and &quot;B did it&quot; are both true. Contradiction, so B didn't do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only possibility remaining is that &lt;strong&gt;C did it&lt;/strong&gt;. Now A's 1st statement and B's 1st statement are true while A's first statement and C's first statement are false, so we can also deduce that &lt;strong&gt;B doesn't know who did it&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-11T18:19:19.260" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14754" PostTypeId="2" ParentId="14746" CreationDate="2015-05-11T18:23:04.750" Score="2" Body="&lt;p&gt;If both sides of the gorge are at the same elevation, and if the rabbit and bear collide in the middle, the rabbit's effect on the bear's momentum will be worse than if the rabbit had simply been riding on the bear the whole time.  The cartoon thus shows a cute, but unnecessarily complicated, way of solving the puzzle which could have been solved by simply having the bear carry the rabbit across.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would suggest that a better puzzle could be had if one side of the gorge is higher than the other (e.g. by six feet), and both animals start on the lower side.  In that case, if one of the animals is barely able to jump unassisted from the low side to the high side, then if that animal jumps from the high side it will have considerable excess elevation, and there might be scenarios where it could be used as a springboard, though one would have to define suitable rules for how horizontal and vertical velocity interact in such cases (in particular, you need to make sure that the horizontal motion of the abler jumper doesn't counteract that of the entity needing assistance).  One could have robots with wheels that could be free-wheeled, but have the fronts and backs of the robots be very delicate such that the only collision that wouldn't cause damage would be one landing squarely on the top of the other and avoiding any transfer of horizontal momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The puzzle might be further extended by having more than two robots with differing weights and abilities, such that robot #1 can jump by itself, robot #2 can jump on robot #1 while it's traveling back, robot #3 would need the assistance of both #1 and #2, etc.&lt;/p&gt;&#xA;" OwnerUserId="1467" LastActivityDate="2015-05-11T18:23:04.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14755" PostTypeId="2" ParentId="14743" CreationDate="2015-05-11T18:35:08.440" Score="4" Body="&lt;p&gt;Kate Gregory got the right solution first, but this answer shows how one could derive the solution from the puzzle as stated.&#xA;&lt;hr&gt;&#xA;Let's look at A's statement first:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My age is twice of what your age was when I was of your age.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If $a$ is A's age and $b$ is B's, this becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a=2*(b-(a-b))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(since A was B's age $a-b$ years ago). Simplifying:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a=2*(2b-a)=4b-2a,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3a=4b.$$&#xA;&lt;hr&gt;&#xA;Now let's look at B's statement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you add your current age to twice of your age when I was born, the answer is $63$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a+2*(a-b)=63$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(since A's age when B was born was $a-b$). Simplifying:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3a-2b=63.$$&#xA;&lt;hr&gt;&#xA;Substitute the first equation into the second to get $4b-2b=63$, so $b=31.5$. Substitute back into the first equation to get $a=\frac{4}{3}(31.5)=42$.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-11T18:35:08.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14756" PostTypeId="2" ParentId="14749" CreationDate="2015-05-11T18:52:48.190" Score="4" Body="&lt;p&gt;Here's another possible answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; None of them did it! (The question doesn't specify that one of the three suspects actually did it.)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's how:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that all three &quot;I didn't do it&quot; statements are true. That means we just have to reconcile the three following statements so that they are all false:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;B did it.&lt;br&gt;&#xA;B knows that C did it.&lt;br&gt;&#xA;B doesn't know who did it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or in other words, show that the following three statements can be true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B didn't do it.&lt;br&gt;&#xA;B doesn't know that C did it.&lt;br&gt;&#xA;B does know who did it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say Person D actually did it, and B knows it. Then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B didn't do it. (TRUE!)&lt;br&gt;&#xA;B doesn't know that C did it. (TRUE! He doesn't know that C did it, he knows that D did it!)&lt;br&gt;&#xA;B does know who did it. (TRUE! He knows that D did it!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, none of them did it!&lt;/p&gt;&#xA;" OwnerUserId="10861" LastActivityDate="2015-05-11T18:52:48.190" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14757" PostTypeId="2" ParentId="14746" CreationDate="2015-05-11T19:00:41.157" Score="0" Body="&lt;p&gt;A more simplified way of stating the question could be as follows, (however it may make the answer seem a bit more apparent):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have two jumpers Alice and Bob. Every 1 jump Bob does, Alice must take 2 to get the same distance Bob is much bigger and has more muscle than Alice which helps him jump further, but his jumps are much lower. While Alice is light and fast, she is able to jump much higher than Bob, but clearly not as far. Both Bob and Alice are hopping along until they found a cliff. Bob easily jumps across it and turns around to see Alice stuck. Alice isn't able to make the distance in 1 jump. There is nothing around in the environment to help her out either. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can Alice get across the cliff?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Note&lt;/b&gt;: Bob is too big for Alice to be able to hold onto while jumping (and has no fur/hair to hang on by)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;As some of the pitfalls in the previous riddle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i) I think that the pictures were misleading in that the robots where always the same size (when in the comic they weren't).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ii) The mention about numbers (initially) or variables may misdirect people to approach the problem using math&lt;/p&gt;&#xA;&#xA;&lt;p&gt;iii) Rule 9: &quot;The distances described for their jump limits are the furthest they can jump under the best circumstances.&quot; can imply that any attempt of interactions while jumping (or in air) would result in failure to make the distance&lt;/p&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="5373" LastEditDate="2015-05-11T22:09:50.367" LastActivityDate="2015-05-11T22:09:50.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14758" PostTypeId="1" AcceptedAnswerId="14763" CreationDate="2015-05-11T19:19:45.130" Score="64" ViewCount="5458" Body="&lt;p&gt;After many years, authorities are still looking for a family of 6, five of whom are suspected of many multiple-country border crossings with no passports or proper identification. Although no known damage has been done, the homeland security divisions of many countries are irate, and continue &quot;to spend much time and money until these 6 are found and punished.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A recent tip led investigators to believe that 5 of the 6 are still actively on the run, and that these 5 meet with the 6th member (believed to be the leader) at the same time each day, presumably to plan future evasion tactics. However, no sightings of any of the 6 have been documented, leaving authorities puzzled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On multiple occasions, top intelligence in many different countries led several police forces to some of the family members' projected locations, but each time officers have reported an eerie feeling of 1 or more members of the family being extremely close to them, yet none were seen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The official, but limited, suspect information:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Priscilla, mother and leader, at least 8 counts of trespassing, claims her children are not doing any harm, and that she has trained them on the straight and narrow path their whole lives&lt;/li&gt;&#xA;  &lt;li&gt;Ed, tallest sibling, 13 counts of trespassing, the most famous of the 5 siblings worldwide&lt;/li&gt;&#xA;  &lt;li&gt;Caitlin, 17 counts of trespassing&lt;/li&gt;&#xA;  &lt;li&gt;Cadence, twin sister of Caitlin, at least 12 counts of trespassing&lt;/li&gt;&#xA;  &lt;li&gt;Arnold, 8 counts of trespassing&lt;/li&gt;&#xA;  &lt;li&gt;Anthony, twin brother of Arnold, multiple counts of aiding and abetting&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who are these six?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-05-11T21:08:20.367" LastActivityDate="2016-07-06T02:13:40.863" Title="Wanted: Family of 6" Tags="|riddle|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="14759" PostTypeId="2" ParentId="14746" CreationDate="2015-05-11T19:24:11.330" Score="3" Body="&lt;p&gt;This is the only way I can seem to make this make sense:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pikachu and Donkey Kong are at one side of a narrow broken arch bridge.&#xA;Donkey Kong can make the jump easy but needs his arms to get enough thrust.&#xA;Pikachu can almost make the jump but not quite.&#xA;The bridge is far too narrow and rickity for them to jump at the same time.&#xA;Pikachu has a cold so can't hold on to Donkey without risking zapping him.&#xA;How they they jump across?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: &quot;up+B&quot; is not the answer &lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2015-05-11T19:24:11.330" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14761" PostTypeId="2" ParentId="14758" CreationDate="2015-05-11T19:45:42.877" Score="5" Body="&lt;p&gt;What if they are...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ghosts?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's kind of a silly answer, but&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They've reportedly been seen worldwide, yet have never been spotted in person, nor have they been reported boarding any sort of aircraft or boat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In addition,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Police forces have not seen a single one of them, yet felt an eerie feeling as if at least one of them was near.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This can be backed up with real life evidence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Throughout the countless reports of &quot;haunted&quot; structures and such, a lot of people exploring said places often experience a feeling of being watched, even if no one else is there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4985" LastActivityDate="2015-05-11T19:45:42.877" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14762" PostTypeId="2" ParentId="14746" CreationDate="2015-05-11T19:58:42.103" Score="13" Body="&lt;p&gt;I don't think the puzzle in its original form would work here on PSE. There are too many loop holes to be closed before it has a single answer. It might work for 'normal' people but I still think you would need a fairly strong list of disallowed tactics. One of the main problems I think with your original two robots formulation, was that you made then the same size, and therefore presumably about the same mass. This makes the idea of one successfully bouncing off the top of the other seem a bit far fetched.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Closing the loop holes surrounding riding, carrying, stacking and shunting require you to be very specific about how the two bodies can interact in away that almost requires you to state, &quot;The only way they can interact is for one to jump on the back of the other mid-jump.&quot; Obviously this is unsatisfactory as it effectively requires you to reveal the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is how I would change it to work both here and anywhere else:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Rabbit is crossing a river hopping from one stone to another. A sudden&#xA;  surge of water washes away all but the stone rabbit is on. It is a&#xA;  little too far to either jump forward or jump back. Rabbit is trapped&#xA;  and would surely drown if she tried to swim across.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A wolf appears on the river bank licking its licks and says, &quot;I'm&#xA;  going to jump over there and eat you little rabbit.&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;How does Rabbit escape?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The solution is about the only thing that remains unchanged. The change in setting removes the need for a long list of things that can't be done. By making the two bodies combatant rather than cooperative you avoid problems around riding and carrying, and it no longer matters if the wolf dies in the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The puzzle is somewhat easier as it is is now clear that the two bodies are travelling in opposite directions but that is the price to be paid for making the puzzle less open ended.&lt;/p&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="11297" LastEditDate="2015-05-13T14:39:14.267" LastActivityDate="2015-05-13T14:39:14.267" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14763" PostTypeId="2" ParentId="14758" CreationDate="2015-05-11T20:45:47.727" Score="62" Body="&lt;p&gt;They are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Various lines on the earth and the number of countries each passes through.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Priscilla, mother and leader, at least 8 counts of trespassing, claims her children are not doing any harm, and that she has trained them on the straight and narrow path their whole lives&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Prime Meridian&lt;/strong&gt;: UK, France, Spain, Algeria, Mali, Burkina Faso, Togo and Ghana&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ed, tallest sibling, 13 counts of trespassing, the most famous of the 5 worldwide&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Equator&lt;/strong&gt;:  Ecuador, Colombia, Brazil, Sao Tome &amp;amp; Principe, Gabon, Republic of the Congo, Democratic Republic of the Congo, Uganda, Kenya, Somalia, Maldives, Indonesia and Kiribati&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Caitlin, 17 counts of trespassing&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Tropic of Cancer&lt;/strong&gt;: Bahamas, Mexico, Mauritania, Mali, Western Sahara, Algeria, Niger, Libya, Egypt, Saudi Arabia, India, China, United Arab Emirates, Oman, Bangladesh, Burma and Taiwan&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cadence, twin sister of Caitlin, at least 12 counts of trespassing&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Tropic of Capricorn&lt;/strong&gt;: Brazil, Paraguay, Argentina, Chile, South Africa, Botswana, Namibia, Australia, Madagascar, Mozambique, France (French Polynesia + New Caledonia), Tonga, Fiji, New Zealand (Cook Islands)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Arnold, 8 counts of trespassing&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Arctic Circle&lt;/strong&gt;:  Norway, Sweden, Finland, Russia, United States (Alaska), Canada, Denmark (Greenland), and Iceland&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Anthony, twin brother of Arnold, multiple counts of aiding and abetting&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Antarctic Circle&lt;/strong&gt;: No countries&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bonus clues:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All the lines are &quot;straight and narrow&quot;.  The Equator is the longest of the 5 siblings, hence the &quot;tallest&quot;.  The two tropics and the two circles are &quot;twins&quot; of each other as they are matched pairs in the north and south.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4479" LastEditorDisplayName="user17008" LastEditDate="2016-07-06T02:13:40.863" LastActivityDate="2016-07-06T02:13:40.863" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14764" PostTypeId="2" ParentId="14749" CreationDate="2015-05-11T21:22:16.110" Score="1" Body="&lt;p&gt;The culprit is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;C&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;suppose C's first statement is true, second must be false....... that leads B's first statement is true, second is false....... that leads first statement is true, second is false........ That gives no culprit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But An inspector knows that exactly one of 3 suspects committed a crime(I won't say, 'none of them')&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; that clearly means C's first statement is false, and second is true.... that leads B's first statement is true, second is false...... that leads A's first statement is true, second is false. That is it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11306" LastActivityDate="2015-05-11T21:22:16.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14765" PostTypeId="1" CreationDate="2015-05-11T23:08:04.057" Score="6" ViewCount="539" Body="&lt;p&gt;I'm playing a game (see &lt;a href=&quot;https://www.youtube.com/watch?v=Y7CNOnqAaCw&amp;amp;index=14&amp;amp;list=PLkBK2hrDcB5ucqVzu5cMgRNqL4sBUj7op#t=9m49s&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;), and in it is a puzzle, which I've made a TIKZ graph seen below for illustration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hspace{2.5cm}$&lt;img src=&quot;https://i.stack.imgur.com/Ey0ZY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one player. The player ($\mathscr{L}$) moves from node to node limited by the directed arrows. As the player moves, $\mathscr{G}_{\Huge\cdot}$ chooses to move in the opposite direction while $\mathscr{G}_{\circ}$ moves in the same direction as the player, all three step simultaneously. If the player makes a move that would cause the guardians to jump off the playing board they remain in place, and the player keeps his/her move. Also, if the player makes a move that would cause him/her to jump off the playing board, then he/she remains in place. The object of the game is to guide the guardians to the circle-cross ($\bigoplus$) nodes so that they individually land on each in whatever order and whatever way simultaneously (see linked video for clarity). A 12-step solution to solve the puzzle has been uncovered, however, I am seeking the &lt;em&gt;shortest possible&lt;/em&gt; step solution to this puzzle.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Special rules:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AXAAT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Game ends if above move is performed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eWPJm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Guardians remain in place and player keeps his/her move.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/w8gup.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Game ends if above move is performed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2RFDf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$(\mathscr{L}\rightarrow\mathscr{G_\circ}\rightarrow\circ)$ OR $(\mathscr{G_\circ}\rightarrow\mathscr{L}\rightarrow\circ)$ conga allowed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12340" LastEditorUserId="-1" LastEditDate="2015-05-12T07:37:24.770" LastActivityDate="2015-05-18T00:11:00.910" Title="Twilight Zelda Guardian Puzzle" Tags="|mathematics|strategy|computer-puzzle|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14766" PostTypeId="1" AcceptedAnswerId="14774" CreationDate="2015-05-12T01:29:01.697" Score="16" ViewCount="994" Body="&lt;p&gt;The warden of the local prison was in a good mood - after a long winter, it was finally spring. The weather was warm and the sun shined over the prison. The warden decided, as is customary in the state of Puzzlevania, to offer the prisoners a chance to escape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Her prisoners were very clever, and she decided to pose to them an equally clever problem. She found a suitable location for her challenge, where the ground was perfectly flat and free of grass. There was only one structure nearby - one tall chain-link fence, which lay to the east, whose top ran in a perfectly straight line. In her flat area, she placed four pegs in a square. The diagonals of the square measured $10$ feet and one of the diagonals was parallel to the top of the prison wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Around each peg, she placed a glass dome, such that the prisoners could not touch the pegs. She took care to affix the domes to the ground firmly, such that the prisoners could not possibly move or break them. The domes were of no precise radius and were not centered around the pegs. They did not distort the light at all and the prisoners could see through them clearly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;She brought the prisoners out to the yard, and showed them the setup and declared:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As you can see, I have placed four pegs into the ground here, in a square. If you all can devise and execute a scheme to determine the exact center of the square, I will set you all free. However, by the canonical laws of Puzzlevania, I will execute all of you if you are even the slightest bit off.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To accomplish this task, she gave them a piece of wood, with one perfectly straight edge at least $10$ feet long. They have access to no other tools. The prisoners took this information back to the cafeteria and drew out the situation. They produced the following image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VKig0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the thick line to the right (east) is the prison wall, the points $A$, $B$, $C$ and $D$ are the pegs, and $P$ is the point they must find, at the center of the square. The dashed circles are the diagonals of the squares and $AC$ is perpendicular to the wall where $BD$ is parallel to the wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The prisoners, having been master carpenters and geometers, have the ability to hold the wood wherever they want with perfect precision. They can see the centers of the pegs perfectly clearly. &lt;strong&gt;They cannot make any marks&lt;/strong&gt; - rather, they must, at some point in time, be able to direct the warden exactly to the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can they find the square's center and be released?&lt;/p&gt;&#xA;" OwnerUserId="4765" LastEditorUserId="5373" LastEditDate="2015-05-12T08:23:08.543" LastActivityDate="2016-05-04T15:51:14.363" Title="The Square's Center" Tags="|geometry|lateral-thinking|" AnswerCount="2" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14768" PostTypeId="2" ParentId="14765" CreationDate="2015-05-12T01:36:54.300" Score="2" Body="&lt;p&gt;The puzzle as shown in the video is different from the one you have presented here. In the video, the top guardian moves in the opposite direction and the bottom guardian moves in the same direction, while as written here it is the opposite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A brute force search turned up no solutions shorter than 12 steps for the puzzle with the top guardian moving opposite the player.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are three pairs of solutions in 12 steps, but one requires the player and a guardian to pass through each other as they switch spots, and one requires the guardians to pass through each other. The solution that does not require either of these is LDUUURRDDDLU (and its reflection). The other two are LUURURDDDDLU and LDRRULUUUURD (and their reflections). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the top guardian moves in the same direction as the player, there is a 9 step solution (LDLDDDDRU), which is unique except for reflection.&lt;/p&gt;&#xA;" OwnerUserId="12296" LastEditorUserId="12296" LastEditDate="2015-05-12T01:55:03.043" LastActivityDate="2015-05-12T01:55:03.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14769" PostTypeId="2" ParentId="476" CreationDate="2015-05-12T01:38:29.370" Score="4" Body="&lt;p&gt;You might like Chess Mazes, by Bruce Alberston.  There are two volumes, each with over a hundred puzzles.  The idea is that you have one &quot;maze piece&quot; that can move, and the opponent never moves, unless they can take your piece.  The object is to put the king in check, or checkmate, depending on the puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also came up with a game called &quot;Queen's Quadrille&quot; many years ago that's like the classic 15-puzzle but which uses chess moves.  And inspired by chess mazes, I have a thing that someone else named &quot;Chess Contradanse&quot;, which you can see on the chessvariants pages.  The idea is to set up the pieces in their usual starting locations, but leave off the pawns, and move all the pieces to the opposite back row, without putting any of them in danger, given that the first and eighth rows are considered safe from capture.  &lt;/p&gt;&#xA;" OwnerUserId="12342" LastActivityDate="2015-05-12T01:38:29.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14770" PostTypeId="1" AcceptedAnswerId="19907" CreationDate="2015-05-12T01:40:42.510" Score="8" ViewCount="578" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I number myself, for I am numbered;&lt;br&gt;I am as long as I am worth.&lt;br&gt;I run down and up, I've crackled and thundered.&lt;br&gt;Just to come to rest upon the earth.&lt;br&gt;&lt;br&gt;Neither red nor deep slate nor pure white,&lt;br&gt;what remains of the four, I must be!&lt;br&gt;(Though, really, It's only typically...)&lt;br&gt;What I am must surely be in sight!&lt;br&gt;&lt;br&gt;I'm so wound up, about this little road.&lt;br&gt;I can hardly reach the end!&lt;br&gt;I've neither limbs nor toes nor nodes;&lt;br&gt;Naught but a body suspends.&lt;br&gt;&lt;br&gt;Strong 'till while I'm young, I am.&lt;br&gt;(Perhaps as late&lt;br&gt;as seven or eight!)&lt;br&gt;As I grow older, I bend and I jam.&lt;br&gt;&lt;br&gt;Now, now! The answer is close!&lt;br&gt;I am what I am, though noticed little&lt;br&gt;and as helpful can be when you must.&lt;br&gt;I come in all shapes, all lengths, and all sizes.&lt;br&gt;&lt;br&gt;&lt;em&gt;Now&lt;/em&gt; tell me: who am I? (I've got all day -- I can wait longer than you)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The best of luck be with you!&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Hint A&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This hint is also a riddle: the word 'nglo' will help you tremendously.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;Hint B&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are hints literally WOVEN into this problem.&#xA;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;Hint B2&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If it 'aint broken, don't fix it; if you're confused, look at the guts of the problem! It's always the pithy stuff, y'know?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;Hint C&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What remains is only a single member in a set of four members. Think old-demographic-y.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;Hint D (Hopefully the Last) RE: The answer is below.&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's one the most value tools for a Car Repenter's.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7307" LastEditorUserId="7307" LastEditDate="2015-08-11T16:07:52.017" LastActivityDate="2015-08-11T17:24:24.180" Title="Numbering, Numbered, Hardly related to math ;)" Tags="|riddle|" AnswerCount="6" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14774" PostTypeId="2" ParentId="14766" CreationDate="2015-05-12T02:45:58.737" Score="12" Body="&lt;p&gt;Hold the pole&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; so its shadow crosses A and C.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wait until &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the sun rises so that the fence casts a shadow touching B and D.  Where the two shadows intersect is the centre of the square.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11952" LastEditorUserId="21234" LastEditDate="2016-05-04T15:51:14.363" LastActivityDate="2016-05-04T15:51:14.363" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="14775" PostTypeId="2" ParentId="14770" CreationDate="2015-05-12T03:32:20.860" Score="3" Body="&lt;p&gt;You are a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Spring&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I number myself, for I am numbered;&#xA;I am as long as I am worth.&#xA;I run down and up, I've crackled and thundered.&#xA;Just to come to rest upon the earth.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The number describes its spring constant, while the &quot;run up and down&quot; is the way a spring will react to forces. The length of the spring is a function of the force it applies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Neither red nor deep slate nor pure white,&#xA;what remains of the four, I must be!&#xA;(Though, really, It's only typically...)&#xA;What I am must surely be in sight!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Typically, springs are silver in colour.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm so wound up, about this little road.&#xA;I can hardly reach the end!&#xA;I've neither limbs nor toes nor nodes;&#xA;Naught but a body suspends.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Springs are wound up, and they suspend bodies&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Strong 'till while I'm young, I am.&#xA;(Perhaps as late&#xA;as seven or eight!)&#xA;As I grow older, I bend and I jam.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As you use springs, they tend to wear down and warp.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, now! The answer is close!&#xA;I am what I am, though noticed little&#xA;and as helpful can be when you must.&#xA;I come in all shapes, all lengths, and all sizes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Springs don't have to be helical, they can have more unusual shapes. More importantly, many sizes certainly. And they're all over the place, like on trampolines and in car suspensions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9289" LastActivityDate="2015-05-12T03:32:20.860" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14776" PostTypeId="2" ParentId="13174" CreationDate="2015-05-12T05:06:22.867" Score="4" Body="&lt;p&gt;Update (leaving some &quot;thought process&quot; in but removing dead-ends):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As another user answered, the plaintext is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Any Man who can drive safely while kissing a pretty girl is simply not giving the kiss the attention it deserves.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's the explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clearly, there are two steps to this cipher. One is a transposition step, and the other is a Trithemius cipher. It is not clear in which order these steps were done to the plaintext, or how exactly the transposition step is meant to be done. The Trithemius cipher also may be done with a custom Tabula Recta of some kind (&quot;more horizontal than vertical&quot;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I believe the first word is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Any&quot; because it rhymes with &quot;penny&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2 helped:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hint 2 says &quot;as a reference,&quot; so what if the transposition was in the rows of the Tabula Recta instead of the letters themselves?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I made a tabula recta with the heading letters transposed&#xA; &lt;pre&gt;&lt;br&gt;&#xA;  B A D C F E H G J I L K N M P O R Q T S V U X W Z Y&#xA;A A B C D E F G H I J K L M N O P Q R S T U V W X Y Z &#xA;B B C D E F G H I J K L M N O P Q R S T U V W X Y Z A &#xA;C C D E F G H I J K L M N O P Q R S T U V W X Y Z A B &#xA;...&#xA;&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which initially didn't look promising, but:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;After reading hint 6, I tried shifting in each row in various extra amounts AND transposing the top keys:&#xA;  &lt;pre&gt;&lt;br&gt;&#xA;  B A D C F E H G J I L K N M P O R Q T S V U X W Z Y&#xA;A A B C D E F G H I J K L M N O P Q R S T U V W X Y Z &#xA;B D E F G H I J K L M N O P Q R S T U V W X Y Z A B C &#xA;C G H I J K L M N O P Q R S T U V W X Y Z A B C D E F &#xA;D J K L M N O P Q R S T U V W X Y Z A B C D E F G H I &#xA;E M N O P Q R S T U V W X Y Z A B C D E F G H I J K L &#xA;F P Q R S T U V W X Y Z A B C D E F G H I J K L M N O &#xA;...&#xA;&lt;/pre&gt; In which case &quot;ANY&quot; would indeed be encoded as &quot;BPF&quot; but to get &quot;QFT&quot; the plaintext would need to be &quot;GSF&quot; which doesn't seem to be a word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Looks like someone beat me to the decoded phrase, but here's what I got by the time I saw it was solved:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I went back and checked out the &quot;ama&quot; version of the ciphertext, and found a pattern of shifts that made reasonable-sounding words. The TR ended up looking like this (number on the left is which line it would be on in a normally-ordered TR): &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;   B A D C F E H G J I L K N M P O R Q T S V U X W Z Y &#xA;1  Z A B C D E F G H I J K L M N O P Q R S T U V W X Y&#xA;2  A B C D E F G H I J K L M N O P Q R S T U V W X Y Z &#xA;3  B C D E F G H I J K L M N O P Q R S T U V W X Y Z A &#xA;6  E F G H I J K L M N O P Q R S T U V W X Y Z A B C D &#xA;5  D E F G H I J K L M N O P Q R S T U V W X Y Z A B C &#xA;6  E F G H I J K L M N O P Q R S T U V W X Y Z A B C D &#xA;7  F G H I J K L M N O P Q R S T U V W X Y Z A B C D E &#xA;8  G H I J K L M N O P Q R S T U V W X Y Z A B C D E F &#xA;9  H I J K L M N O P Q R S T U V W X Y Z A B C D E F G &#xA;12 K L M N O P Q R S T U V W X Y Z A B C D E F G H I J &#xA;11 J K L M N O P Q R S T U V W X Y Z A B C D E F G H I &#xA;12 K L M N O P Q R S T U V W X Y Z A B C D E F G H I J &#xA;15 N O P Q R S T U V W X Y Z A B C D E F G H I J K L M &#xA;14 M N O P Q R S T U V W X Y Z A B C D E F G H I J K L &#xA;15 N O P Q R S T U V W X Y Z A B C D E F G H I J K L M &#xA;16 O P Q R S T U V W X Y Z A B C D E F G H I J K L M N &#xA;17 P Q R S T U V W X Y Z A B C D E F G H I J K L M N O &#xA;20 S T U V W X Y Z A B C D E F G H I J K L M N O P Q R &#xA;19 R S T U V W X Y Z A B C D E F G H I J K L M N O P Q &#xA;20 S T U V W X Y Z A B C D E F G H I J K L M N O P Q R &#xA;21 T U V W X Y Z A B C D E F G H I J K L M N O P Q R S &#xA;22 U V W X Y Z A B C D E F G H I J K L M N O P Q R S T &#xA;23 V W X Y Z A B C D E F G H I J K L M N O P Q R S T U &#xA;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also note that the last line is shifted to the top, and then there is a pattern where certain lines are skipped, the next lines are out of order.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It appears the left-hand side of the TR was also transposed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using this table to decode &quot;Ama req cmw nkw pcwiu lswyeu&quot; I got:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Any man who can drive safely&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And so on...&lt;/p&gt;&#xA;" OwnerUserId="12199" LastEditorUserId="12199" LastEditDate="2015-05-13T16:33:40.370" LastActivityDate="2015-05-13T16:33:40.370" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14777" PostTypeId="2" ParentId="14749" CreationDate="2015-05-12T05:55:50.737" Score="2" Body="&lt;p&gt;The Culprit could be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;C&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;logic:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;A&lt;/strong&gt; says: I did not do it [&lt;code&gt;TRUE&lt;/code&gt;]. B did it [&lt;code&gt;FALSE&lt;/code&gt;].&#xA;&lt;br&gt;&#xA; &lt;strong&gt;B&lt;/strong&gt; says: I did not do it [&lt;code&gt;TRUE&lt;/code&gt;]. I know that C did it [&lt;code&gt;FALSE&lt;/code&gt;](&lt;em&gt;just a guess&lt;/em&gt;).&#xA;&lt;br&gt;&#xA; &lt;strong&gt;C&lt;/strong&gt; says: I did not do it [&lt;code&gt;FALSE&lt;/code&gt;](&lt;em&gt;which means &lt;strong&gt;C&lt;/strong&gt;'s the culprit&lt;/em&gt;). B does not know who it was [&lt;code&gt;TRUE&lt;/code&gt;](which means that B was just guessing).&lt;br&gt;&#xA;&lt;br&gt;&#xA;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12054" LastEditorUserId="12054" LastEditDate="2015-12-03T05:48:22.763" LastActivityDate="2015-12-03T05:48:22.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14778" PostTypeId="1" AcceptedAnswerId="14779" CreationDate="2015-05-12T05:59:52.833" Score="7" ViewCount="17371" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sco6C.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Textual version:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is famously asked in many Amazon interviews. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;What comes next in the series&#xA;SSS, SCC, C, SC, ?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No numbers will help you in this puzzle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Think of the alphabet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A = SSS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12317" LastEditorDisplayName="user20" LastEditDate="2015-07-23T05:14:32.543" LastActivityDate="2015-08-10T02:28:31.967" Title="Tough Series Question - SSS, SCC, C, SC" Tags="|letter-sequence|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14779" PostTypeId="2" ParentId="14778" CreationDate="2015-05-12T06:50:04.143" Score="11" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each set of letters describes the lines used to write the capital letter. A is three &lt;strong&gt;S&lt;/strong&gt;traight lines, B is a &lt;strong&gt;S&lt;/strong&gt;traight and two &lt;strong&gt;C&lt;/strong&gt;urves, and so on. The next letter is E, which has 4 &lt;strong&gt;S&lt;/strong&gt;traight lines. The answer is &lt;strong&gt;SSSS&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12345" LastEditorUserId="12165" LastEditDate="2015-05-12T07:39:00.950" LastActivityDate="2015-05-12T07:39:00.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14780" PostTypeId="2" ParentId="14778" CreationDate="2015-05-12T06:50:49.257" Score="6" Body="&lt;p&gt;&lt;strong&gt;S&lt;/strong&gt; stands for &lt;em&gt;straight line&lt;/em&gt; (&lt;em&gt;horizontal, vertical or diagonal&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i guess the &lt;strong&gt;C&lt;/strong&gt; for &lt;em&gt;curve line&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;3 straight lines to form &lt;strong&gt;A&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;1 straight line plus 2 curves to form &lt;strong&gt;B&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;1 curve for &lt;strong&gt;C&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;1 straight lines plus 1 curve for &lt;strong&gt;D&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;4 straight lines for &lt;strong&gt;E&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;answer: &lt;strong&gt;SSSS&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but for this font: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XFtDw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Straight Line Segments can do all the letters :)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12054" LastEditorUserId="12054" LastEditDate="2015-08-10T02:28:31.967" LastActivityDate="2015-08-10T02:28:31.967" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14784" PostTypeId="2" ParentId="1856" CreationDate="2015-05-12T10:34:23.597" Score="1" Body="&lt;p&gt;If u Shift this line segment (wall) on either far side u can solve this. However , it was not given anywhere that u can't modify the figure.&#xA;After all, there are still 5 rooms.&#xA;In Euler's trail the main cause of failing is this edge's position in the middle. Or else the solution of this puzzle lie in 3d format only.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/M8Ou3.jpg&quot; alt=&quot;Possible solution&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="12355" LastActivityDate="2015-05-12T10:34:23.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14785" PostTypeId="1" AcceptedAnswerId="14852" CreationDate="2015-05-12T12:24:32.607" Score="7" ViewCount="19627" Body="&lt;p&gt;&lt;strong&gt;The Problem&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say you are asked to measure the amount of liquid in a container, but you are not allowed to use any direct measuring tool (such as a balance, ruler, or anything of the sort). You are however given an infinite number of equal containers to accomplish this task.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that you are not &lt;em&gt;obliged&lt;/em&gt; to use the containers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the most efficient solution/algorithm for this task, be it mathematical/chemical/mechanical/etc? Answers will be judged by &lt;strong&gt;efficiency&lt;/strong&gt; and &lt;strong&gt;creativity&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Additional information&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An infinite amount of extra liquid is available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The initial container might be of same size than the ones you can use, if that is a necessary parameter in your solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The volume of the containers given to you is known. It can be called &lt;em&gt;x&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solution scoring&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution's worthiness will be primarily based on these two factors:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Creativity of the solution (solutions that require outside thinking are welcome)&lt;/li&gt;&#xA;&lt;li&gt;Efficiency (how many tools are necessary, how long does it take, etc...)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="12359" LastEditorUserId="4050" LastEditDate="2015-05-12T12:55:16.777" LastActivityDate="2015-05-28T01:48:45.223" Title="Measuring the amount of liquid in a container without a measuring tool" Tags="|mathematics|logical-deduction|lateral-thinking|decanting-problems|" AnswerCount="17" CommentCount="17" ClosedDate="2015-08-17T13:47:16.657" ContentLicense="CC BY-SA 3.0" />
  <row Id="14787" PostTypeId="1" AcceptedAnswerId="14811" CreationDate="2015-05-12T12:53:47.303" Score="10" ViewCount="900" Body="&lt;h2&gt;I.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I can separate two high places.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remove one letter, and I can separate two high places.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;II.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I can connect two high places.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remove one letter, and I can connect two high places.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;III.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I can separate two high places.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remove one letter, and I can connect two high places.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Notes:&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;All of the words are singular nouns.&lt;/li&gt;&#xA;&lt;li&gt;&quot;High&quot; is relative.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Hint #1 for III:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Both are covered by water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #2 for III:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I never said you had to remove the &lt;em&gt;first&lt;/em&gt; letter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4833" LastEditorUserId="4833" LastEditDate="2015-05-12T17:51:51.727" LastActivityDate="2015-05-13T09:34:34.300" Title="Three curtailments" Tags="|word|letters|" AnswerCount="6" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14788" PostTypeId="2" ParentId="14787" CreationDate="2015-05-12T12:58:41.273" Score="3" Body="&lt;p&gt;For II, I think it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's BRIDGE and RIDGE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10400" LastActivityDate="2015-05-12T12:58:41.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14789" PostTypeId="2" ParentId="14787" CreationDate="2015-05-12T12:59:09.363" Score="8" Body="&lt;p&gt;I.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can separate two high places.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remove one letter, and I can separate two high places.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; valley -&gt; alley (valley can separate mountains, alley is a street which separates rows of potentially high trees)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;II.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can connect two high places.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remove one letter, and I can connect two high places.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; bridge -&gt; ridge (bridge can connect two trees, cliffs, whatever, a ridge is a connection of mountains in a way)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;III.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can separate two high places.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remove one letter, and I can connect two high places.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am not sure. Maybe swale -&gt; wale? A swale is a bit like a valley and a wale seems to be something like a belt or strap that you could use to connect two high places.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10885" LastActivityDate="2015-05-12T12:59:09.363" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14790" PostTypeId="2" ParentId="14785" CreationDate="2015-05-12T13:09:13.567" Score="7" Body="&lt;p&gt;Step 1:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Fill as many of the containers as possible to the top from the initial vessel- this gives us a volume of x containers  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Step 2:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Pour the remaining liquid into another container - all the liquid should now be out of the initial vessel.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Step 3:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Take one full container(&lt;strong&gt;a&lt;/strong&gt;), the part-full container(&lt;strong&gt;b&lt;/strong&gt;), and a number of empty containers(&lt;strong&gt;c&lt;/strong&gt;).&lt;br&gt;&#xA;Pour liquid from the full container (a) into an empty container (&lt;strong&gt;c&lt;/strong&gt;) until &lt;strong&gt;c&lt;/strong&gt; contains the same level of liquid as &lt;strong&gt;b&lt;/strong&gt;.&lt;br&gt;&#xA;Repeat with more empty containers until &lt;strong&gt;a&lt;/strong&gt; is empty. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Step 4:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If necessary, repeat step 3, this time dividing the liquid in &lt;strong&gt;b&lt;/strong&gt; so that it matches the level in &lt;strong&gt;c&lt;/strong&gt;.&lt;br&gt;&#xA;Continue this iteration until you reach a step where all of the containers taken at the start of a step contain the same amount of liquid.&lt;br&gt;&#xA;Divide one of the &lt;strong&gt;a&lt;/strong&gt; beakers to match the level of this final step - that gives you the denominator of the fraction of a beaker held in &lt;strong&gt;b&lt;/strong&gt;.&lt;br&gt;&#xA;Divide one of the &lt;strong&gt;b&lt;/strong&gt; beakers (there should be at least one left from the first iteration) to match the same level. That should give you the numerator&lt;br&gt;&#xA;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;e.g. if &lt;strong&gt;b&lt;/strong&gt; contains 3/8 of a full container:&lt;br&gt;&#xA;&lt;br&gt;&#xA;-A full cylinder (&lt;strong&gt;a&lt;/strong&gt;) will fill 2 cylinders to 3/8 (&lt;strong&gt;b&lt;/strong&gt;), and a third to 2/8 (&lt;strong&gt;c&lt;/strong&gt;)&lt;br&gt;&#xA;&lt;br&gt;&#xA;-&lt;strong&gt;b&lt;/strong&gt; will fill 1 cylinder to 2/8, and another to 1/8 (d)&lt;br&gt;&#xA;&lt;br&gt;&#xA;-&lt;strong&gt;c&lt;/strong&gt; will fill 2 cylinders to 1/8&lt;br&gt;&#xA;&lt;br&gt;&#xA;-&lt;strong&gt;a&lt;/strong&gt; will fill 8 cylinders to the same level as &lt;strong&gt;d&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;-&lt;strong&gt;b&lt;/strong&gt; will fill 3 cylinders to the same level as &lt;strong&gt;d&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;This tells us that we have 3/8 of a container's volume in &lt;strong&gt;b&lt;/strong&gt;. We also (from step 1) have a number of completely filled containers. As the volume of these containers is known, that should give us the initial volume of the liquid  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll work on making that explanation more coherent later. It does require that all the containers are the same shape as well as volume, though. It also requires that you can perfectly judge by eye whether the level of liquid in two containers is equal.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: If the initial volume is less than one of the measuring containers, then you'll have to start by completely filling an empty container with some of the infinite liquid  &lt;/p&gt;&#xA;" OwnerUserId="11717" LastEditorUserId="11717" LastEditDate="2015-05-12T16:11:18.760" LastActivityDate="2015-05-12T16:11:18.760" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14792" PostTypeId="2" ParentId="14770" CreationDate="2015-05-12T13:43:24.017" Score="1" Body="&lt;p&gt;Are you a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; laser/energy beam /wave?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I number myself, for I am numbered;&#xA;I am as long as I am worth.&#xA;I run down and up, I've crackled and thundered.&#xA;Just to come to rest upon the earth.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A laser beam is emitted through a whole lot of complicated physics that I once understood but since have forgotten most of. Still, it is light, light is basically an energy wave. Laser light is produced by electrons changing position in atoms. Atoms make up all of the world. Sound waves (also an energy wave of sorts) can also crackle and thunder.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Neither red nor deep slate nor pure white,&#xA;what remains of the four, I must be!&#xA;(Though, really, It's only typically...)&#xA;What I am must surely be in sight!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wavelengths of light result in humans seeing different colours, red, deep blue-green (slate) and also white are quite typical laser beam colors, another is green.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm so wound up, about this little road.&#xA;I can hardly reach the end!&#xA;I've neither limbs nor toes nor nodes;&#xA;Naught but a body suspends.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A wave is meandering around its path, if you want to be poetic. Theoretically, there is no end to this path. Light is not lost. We can see &quot;old&quot; light from far away stars that is reaching out eyes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Strong 'till while I'm young, I am.&#xA;(Perhaps as late&#xA;as seven or eight!)&#xA;As I grow older, I bend and I jam.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You need quite a bit of energy to keep a laser beam up. I vaguely remember that it begins to scatter the longer it goes, so the amplitude gets larger.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, now! The answer is close!&#xA;I am what I am, though noticed little&#xA;and as helpful can be when you must.&#xA;I come in all shapes, all lengths, and all sizes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Light is all around us, although we generally not notice it (much more often we would notice its absence). It is a tremendous help to our everyday life, you can produce it by different ways (laser, lightbulb), it has different wave patterns.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now tell me: who am I? (I've got all day -- I can wait longer than you)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Now that I wrote it all down, I guess it is not a laser like I originally thought, but merely &lt;strong&gt;a beam of (green) light.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10885" LastActivityDate="2015-05-12T13:43:24.017" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14793" PostTypeId="2" ParentId="14785" CreationDate="2015-05-12T13:47:53.113" Score="2" Body="&lt;p&gt;Let me de-generalize the problem a little to give me a place to start. You are handed a one-litre container that is partially full and asked to determine, to an arbitrary accuracy, just how full it is (how much it contains.) You have a tap and a bunch of other identical one litre containers. It's not necessary for the containers to have any degree of symmetry as long as they are the same. The containers need to be translucent so you can see the level of the water in them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If the original container is not translucent, pour the water from the original container into one of the translucent ones before you begin. If the original container is larger than the one-litre containers, carefully fill one litre containers from it, keeping count, until you have an amount less than one litre in the last one, and use this process to establish that volume, then add the number of full litres you got as well.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You fill one container and grab an empty one, then carefully pour from the full to the empty until the level in the one you're filling matches the level in the one you're pouring from. You now have 0.5 in each container. Compare to the reference container to see if  x, the volume you're looking for, is more or less than 0.5. By eye you could probably have declared the volume to the nearest half, so this is presumably not super useful, but so be it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, fill one container and then pour it carefully between 3 containers and compare to the reference volume. You now know whether x is between 0 and one-third, or one-third and one half, or one half and two thirds, or two-thirds and one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Split one of the half full containers between two other empty containers so that you have quarters. (If pouring is becoming too much of a challenge when you get to fine accuracy, use a dropper or dip something in one container and let the drops fall into the other for the final balancing of levels between containers.) If x is more than 0.5, add a quarter to a half to get 0.75. Now you can narrow down the possible ranges of x to 0-0.25, 0.25-0.33, 0.33-0.5, 0.5-0.66, 0.66-0.75, and 0.75-1. If this is accurate enough, stop. If it is not, construct a 1/8th container by splitting a quarter, and add that eighth to whatever represents the bottom of the range you're considering. Say you know it's between 0.75 and 1 - add an eighth to that making 7/8ths or 0.875 and you'll know whether x is 0.75 to 0.875 or 0.875 to 1. If you only need to be accurate to one decimal place you can announce &quot;0.8&quot; or &quot;0.9&quot; and be correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need more accuracy, make sixteenths. And so on. At some point your ability to put two cylinders together and declare their volumes the same will put a natural limit on what you're doing. Also, the drops that cling to the sides when you pour from container to container will become relevant. Therefore this technique cannot produce arbitrary accuracy. I can overcome that by declaring I was actually using 100-litre containers (and some imaginary lifting help to let me pour among them) thus effectively dividing those effects by 100. &lt;/p&gt;&#xA;" OwnerUserId="1509" LastEditorUserId="1509" LastEditDate="2015-05-12T17:15:29.853" LastActivityDate="2015-05-12T17:15:29.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14794" PostTypeId="2" ParentId="14785" CreationDate="2015-05-12T14:27:33.010" Score="3" Body="&lt;p&gt;Given: unlimited containers, each of volume $x$; volume $v$ of liquid; unlimited additional liquid. In this answer, $x$ is unconstrained - it can be smaller, equal to, or even bigger than $v$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assumptions: you can accurately and repeatedly draw out volume $v$ of liquid from the additional supply; you will accept an answer of $v \pm \epsilon$ for some predetermined error margin $\epsilon$. If $v$ is not a rational number, then $\epsilon$ must be non-zero but can otherwise be as small as you like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get an acceptable value of $v$ by repeating the following two steps until the error margin is satisfied. In the analysis that follows, let $c$ be the total number of containers used after $n$ iterations. Note that the last container used may be only partially filled.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Draw out $v$ of liquid from the additional supply.&lt;/li&gt;&#xA;&lt;li&gt;Fill as many containers as possible from the liquid drawn in step 1, including topping up the partially filled container from the previous iteration.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Error bound&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without making any assumptions about being able to measure partly filled containers, all we can say after $n$ iterations is that the volume is between $c-1$ and $c$ full containers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $(c-1)x &amp;lt; nv \leq cx$, that is, $$v \in (\frac{(c-1)x}{n}, \frac{cx}{n}] \tag{1}\label{1}$$ with the interval open at the bottom and closed at the top. Let $2\epsilon_n$ be the range of the interval, so $2\epsilon_n = \frac{x}{n}$. Rewrite $\eqref{1}$ to get the following. $$v = \frac{(c-0.5)x}{n} \pm \epsilon_n \tag{2}\label{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Volume calculation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any iteration, if the last container used is full, $v$ is equal to the upper bound in $\eqref{1}$, so $v = \frac{cx}{n}$ and you're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, the error is acceptable when $\epsilon_n \leq \epsilon$. Since $\epsilon_n$ decreases as the iteration count $n$ increases, the process is guaranteed to terminate with an answer of $v \approx \frac{(c-0.5)x}{n}$ from $\eqref{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Efficiency&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of containers used is given by $nv \leq cx$ from the upper bound to $\eqref{1}$, so we will need at most $c = \lceil \frac{nv}{x} \rceil$ containers, with $n$ determined as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The error will be within acceptable limits when $\epsilon_n \leq \epsilon$, so the number of iterations will be at most $n = \lceil \frac{x}{2\epsilon} \rceil$.&lt;/p&gt;&#xA;" OwnerUserId="9422" LastEditorUserId="9422" LastEditDate="2015-05-12T14:53:47.917" LastActivityDate="2015-05-12T14:53:47.917" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14795" PostTypeId="2" ParentId="14785" CreationDate="2015-05-12T14:27:37.560" Score="13" Body="&lt;p&gt;Assumptions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;it is possible to &quot;duplicate&quot; a container by pouring liquid into an empty container until it has exactly as much liquid as the original.&lt;/li&gt;&#xA;&lt;li&gt;it is possible to &quot;average&quot; two containers by pouring liquid from the fuller one into the emptier one until they are even.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If we have these capabilities, we can perform a binary search.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Take a full container and label it &quot;upper bound&quot;.&lt;/li&gt;&#xA;&lt;li&gt;Take an empty container and label it &quot;lower bound&quot;.&lt;/li&gt;&#xA;&lt;li&gt;Duplicate &quot;upper bound&quot; and &quot;lower bound&quot;, and average the two into a container labeled &quot;median&quot;.&lt;/li&gt;&#xA;&lt;li&gt;If the liquid in &quot;median&quot; is satisfactorily similar in volume to the container you're trying to measure, stop. Otherwise, continue with step 5.&lt;/li&gt;&#xA;&lt;li&gt;Choose one of the containers: &quot;lower bound&quot; if &quot;median&quot; has less liquid than the container you're trying to measure, &quot;upper bound&quot; otherwise. Discard the liquid in the chosen container, and replace it with the liquid from &quot;median&quot;.&lt;/li&gt;&#xA;&lt;li&gt;Return to step 3.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;At every point during this algorithm, the volume of the &quot;upper bound&quot; and &quot;lower bound&quot; and &quot;median&quot; containers are known exactly. When you stop, the initial container's volume is approximately equal to the &quot;median&quot; container's volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of &quot;satisfactorily similar&quot; varies and is up to the experimenter. It could be &quot;when I can no longer distinguish by eye the difference between the two containers&quot; or &quot;when I have spent an hour performing this algorithm&quot; or &quot;when my electron microscope verifies that the containers have exactly the same number of atoms&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Efficiency:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For a reasonable amount of liquid (say, a cubic foot), takes about 11 iterations for sub-milliliter level precision, 90 iterations for atomic precision. For unreasonable amounts, it follows a logarithmic curve: 2 cubic feet takes 12 iterations, 4 cubic feet takes 13 iterations, 2^N cubic feet takes 11 + N iterations.&lt;/li&gt;&#xA;&lt;li&gt;Requires no more than five containers: the three labeled containers, plus two for duplication purposes.&lt;/li&gt;&#xA;&lt;li&gt;Uses approximately M*N*2 extra liquid, where M is the number of iterations performed, and N is the actual volume of the initial container.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Creativity:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;None. Binary search is a common approximation technique, and is as old as dirt.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1497" LastActivityDate="2015-05-12T14:27:37.560" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14796" PostTypeId="2" ParentId="14787" CreationDate="2015-05-12T14:35:54.300" Score="3" Body="&lt;p&gt;For III could it be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Privet&lt;/strong&gt; separates houses, and &lt;strong&gt;Rivet&lt;/strong&gt; connects the beams in skyscrapers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4513" LastActivityDate="2015-05-12T14:35:54.300" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14797" PostTypeId="2" ParentId="14708" CreationDate="2015-05-12T14:54:24.170" Score="-3" Body="&lt;p&gt;Simple answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1088=2^{(2*5)}+64$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we must use every digit once....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1088=2^{(3+7)}+64$&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4269" LastEditDate="2015-05-12T15:19:00.737" LastActivityDate="2015-05-12T15:19:00.737" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14798" PostTypeId="2" ParentId="14681" CreationDate="2015-05-12T15:24:20.030" Score="1" Body="&lt;p&gt;Are you &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jupiter?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no brothers or sisters, but I have many dolls.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The 60+ moons of Jupiter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm always watching my mother...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...and my cousins all think I'm fat.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jupiter is big! The cousins are the other planets.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5619" LastActivityDate="2015-05-12T15:24:20.030" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14799" PostTypeId="2" ParentId="14708" CreationDate="2015-05-12T15:40:32.197" Score="5" Body="&lt;p&gt;The Answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;7!/5 + 6!/(2+3+4)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11343" LastEditorUserId="11343" LastEditDate="2016-10-02T14:03:16.353" LastActivityDate="2016-10-02T14:03:16.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14800" PostTypeId="2" ParentId="14653" CreationDate="2015-05-12T15:41:00.470" Score="3" Body="&lt;p&gt;I think It could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gravity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am here,&#xA;yet most people don't care,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gravity is everywhere, but most people don't actually think about it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I kill people,&#xA;if you are not careful.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you can fall off a cliff or something can fall onto you&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be created,&#xA;but it's rather useless.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gravity can be created (&lt;a href=&quot;http://en.wikipedia.org/wiki/Artificial_gravity&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Artificial_gravity&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you can see me,&#xA;It's because of Him.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The movie &quot;Gravity&quot; was written by Alfonso Cuarón, a Mexican.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12328" LastActivityDate="2015-05-12T15:41:00.470" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14801" PostTypeId="2" ParentId="14770" CreationDate="2015-05-12T15:42:02.610" Score="1" Body="&lt;p&gt;Are you a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Kernmantle_rope&quot; rel=&quot;nofollow&quot;&gt;Kernmantle rope&lt;/a&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I number myself, for I am numbered;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Different lengths, diameters, strength etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am as long as I am worth.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Material-wise, the price is probably proportional to the length&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I run down and up, I've crackled and thundered.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Often used in climbing and rescue missions&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Just to come to rest upon the earth.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People use it to descend onto the ground&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Neither red nor deep slate nor pure white,&#xA;what remains of the four, I must be!&#xA;(Though, really, It's only typically...)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Can't solve this part yet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What I am must surely be in sight!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People working in suspended platforms must be provided with safety ropes and are only allowed to work when they have properly worn one&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm so wound up, about this little road.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you pack a rope to travel/hike, you wound it up to save space&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can hardly reach the end!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The rope is long&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've neither limbs nor toes nor nodes;&#xA;Naught but a body suspends.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ropes can suspend people during climbing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Strong 'till while I'm young, I am.&#xA;(Perhaps as late&#xA;as seven or eight!)&#xA;As I grow older, I bend and I jam.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Old ropes may break more easily&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, now! The answer is close!&#xA;I am what I am, though noticed little&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you are not a regular climber/rescuer you don't typically care about ropes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and as helpful can be when you must.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; But when you're in danger the rope can save your life&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come in all shapes, all lengths, and all sizes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, different lengths, diameters, etc&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now tell me: who am I? (I've got all day -- I can wait longer than you)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ropes are long&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hints:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This hint is also a riddle: the word 'nglo' will help you tremendously.&lt;br&gt;&lt;br&gt;&#xA; Long&lt;br&gt;&lt;br&gt;&#xA; There are hints literally WOVEN into this problem. &lt;br&gt;&lt;br&gt;&#xA; From the linked Wikipedia article, (emphasis mine):&lt;br&gt;&#xA; Kernmantle rope is rope constructed with its interior core (the kern) protected by a &lt;strong&gt;woven&lt;/strong&gt; exterior sheath (mantle) designed to optimize strength, durability, and flexibility. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12278" LastActivityDate="2015-05-12T15:42:02.610" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14802" PostTypeId="1" AcceptedAnswerId="14942" CreationDate="2015-05-12T15:46:17.930" Score="16" ViewCount="598" Body="&lt;p&gt;Below are three sets of three words. Each word shares a common linguistic property. The words in each set share a nested property. Identify the properties and explain.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Lift&lt;br&gt;&#xA;  Beams&lt;br&gt;&#xA;  Destroy  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Gas&lt;br&gt;&#xA;  Lapse&lt;br&gt;&#xA;  Successor&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Compose&lt;br&gt;&#xA;  Correct&lt;br&gt;&#xA;  Custom&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This appears to be harder than I had intended, so I'll start providing hints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are two stages to this puzzle. To uncover stage two, it may help to read aloud. There's more in what you'll hear than in what you'll see.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some of these words have multiple meanings, but this puzzle only focuses on one. To hopefully weed out some red herrings in future answers, here is a list of which part of speech each word is meant to be:&lt;br&gt;&#xA; &lt;strong&gt;Lift&lt;/strong&gt;: Verb&lt;br&gt;&#xA; &lt;strong&gt;Beams&lt;/strong&gt;: Plural noun&lt;br&gt;&#xA; &lt;strong&gt;Destroy&lt;/strong&gt;: Verb&lt;br&gt;&#xA; &lt;strong&gt;Gas&lt;/strong&gt;: Noun&lt;br&gt;&#xA; &lt;strong&gt;Lapse&lt;/strong&gt;: Verb&lt;br&gt;&#xA; &lt;strong&gt;Successor&lt;/strong&gt;: Noun&lt;br&gt;&#xA; &lt;strong&gt;Compose&lt;/strong&gt;: Verb&lt;br&gt;&#xA; &lt;strong&gt;Correct&lt;/strong&gt;: Adjective&lt;br&gt;&#xA; &lt;strong&gt;Custom&lt;/strong&gt;: Noun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12192" LastEditorUserId="17482" LastEditDate="2016-03-08T21:49:50.707" LastActivityDate="2016-03-08T21:49:50.707" Title="Enough of this charade! A common property puzzle" Tags="|pattern|english|language|word-sequence|tribond|" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14803" PostTypeId="2" ParentId="14665" CreationDate="2015-05-12T15:47:16.503" Score="5" Body="&lt;p&gt;The 2's are what really screwed me up on this one, but I finally got it (EDIT: You did make a couple mistakes, unless those were on purpose to mess with us xD):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;THIS WAS SIMPLE WAS IT NOT&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Step 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Use the cyclic Vigenere and subtract the index you are on. So for the first line, subtract 0 from the first digit, then 1 from the second, then 0 from the third, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This will give you:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1 00 101 000 / 1001 1010 000 / 1 101 11 1100 10 0 / 1001 1010 000 / 0111 0001 / 10 0 110 0001 (Combine the 2nd and 3rd groups in the last line to get 0110, which coes to a P, error with the original code)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Step 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Convert it from morse code with 0=dit and 1=dah.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This gets you:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;TIKS XCS TKMZNE XCS JV NPV&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Use the same cipher as before but with the characters as part of the alphabet, and instead of 0,1 its 0,1,2. T subtract 0, I subtract 1 so it goes to H, K subtract 2 goes to I, S subtract 0 stays the same, etc. Using this on the entire string gets you the answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11727" LastActivityDate="2015-05-12T15:47:16.503" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14804" PostTypeId="2" ParentId="14681" CreationDate="2015-05-12T15:54:57.663" Score="1" Body="&lt;p&gt;&lt;a href=&quot;/questions/tagged/lateral-thinking&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;lateral-thinking&amp;#39;&quot; rel=&quot;tag&quot;&gt;lateral-thinking&lt;/a&gt; Are you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A creepy serial killer with a particular family history?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no brothers or sisters, but I have many dolls.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are a single child who received inadequate affection from your single mother and turned to a doll collection as an alternative sense of warmth. It is a hollow love, though, given by porcelain souls. Many are broken in places but still stare out at you from the darkness within their skull, their one good eye unblinking in its judgement.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm always watching my mother...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your mother was your first victim and her remains are propped up in a basement chair. She is surrounded by the dolls and, as their souls are really just pieces of your own, she is always watched by you through them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...and my cousins all think I'm fat.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You actually are slightly overweight and the few cousins you knew have teased you about it your entire life, further isolating you from happiness. They are all part of your collection now, too. Many dolls have been patched up by replacing their broken pieces with chunks of your cousins. A patch of scalp here, a sliver of heart there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of course, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are happier now. You have your whole family with you. All the time. Always watching. Always. Watching.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastActivityDate="2015-05-12T15:54:57.663" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14805" PostTypeId="2" ParentId="14785" CreationDate="2015-05-12T16:04:26.193" Score="1" Body="&lt;p&gt;First disassemble some containers to produce cordage or (mass-less) rope;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;construct two (frictionless) A-frames from more containers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;fill one known volume container with liquid&lt;/p&gt;&#xA;&#xA;&lt;p&gt;attach to one end of the rope&lt;/p&gt;&#xA;&#xA;&lt;p&gt;attach the unknown volume to the other end&lt;/p&gt;&#xA;&#xA;&lt;p&gt;suspend the known volume container from the A-frame passing the rope over the other A-frame with the unknown container at the base&lt;/p&gt;&#xA;&#xA;&lt;p&gt;holding the rope between the a frame and a distance r* from the unknown container spin the unknown container horizontally around yourself. vary the velocity V** until the centripetal force exactly balances the weight of the known volume container.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solve F = mg = M v^2/r to find the proportional mass of the unknown container and hence volume in x&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*using your own height as an accurate measure&lt;/p&gt;&#xA;&#xA;&lt;p&gt;**count your heartbeats, like jason bourne&lt;/p&gt;&#xA;" OwnerUserId="11831" LastActivityDate="2015-05-12T16:04:26.193" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14806" PostTypeId="2" ParentId="14785" CreationDate="2015-05-12T16:09:46.617" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;fill all your liquid into containers until you get x full containers one which is not filled up to the top. &lt;/li&gt;&#xA;&lt;li&gt;discard the full containers and note the number of filled conainers.&lt;/li&gt;&#xA;&lt;li&gt;For a decimal scale fill 10 containers as high as the not fully filled one.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;repeat the steps, for each repetition you get one digit of decimal precision, though it is much similliar to the binary search&lt;/p&gt;&#xA;" OwnerUserId="12367" LastActivityDate="2015-05-12T16:09:46.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14807" PostTypeId="2" ParentId="14765" CreationDate="2015-05-12T16:13:28.787" Score="-1" Body="&lt;p&gt;I'll have to pull out my Wii tonight -&lt;em&gt;long time since I've said &lt;strong&gt;that&lt;/em&gt;&lt;/strong&gt;- to test this out and verify that I am keeping all the rules in check and not getting myself stomped by stone, but my early scribblings amounted to what I think will work in 11 steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DULLUULDDRU&lt;/p&gt;&#xA;" OwnerUserId="11452" LastActivityDate="2015-05-12T16:13:28.787" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14810" PostTypeId="2" ParentId="14741" CreationDate="2015-05-12T16:27:58.113" Score="0" Body="&lt;p&gt;I'm going to guess: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Brown.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Going on your pairs hint,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Brown and letter = Brown + e = brownie!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9412" LastActivityDate="2015-05-12T16:27:58.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14811" PostTypeId="2" ParentId="14787" CreationDate="2015-05-12T16:30:02.470" Score="1" Body="&lt;p&gt;1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; [v]alley&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; [b]ridge&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; [b]lock , [t]rack , &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;EDIT for 3 , with the new clue &quot;removing any letter&quot; , which I did not add earlier, assuming that the only first letter had to be removed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; f[j]ord or f[i]ord&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10940" LastEditorUserId="10940" LastEditDate="2015-05-13T09:34:34.300" LastActivityDate="2015-05-13T09:34:34.300" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14812" PostTypeId="2" ParentId="14681" CreationDate="2015-05-12T16:42:09.737" Score="2" Body="&lt;p&gt;I think you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The moon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no brothers or sisters, but I have many dolls.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The moon is the only natural satellite of earth, but there are many man made satellites orbiting the Earth with it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm always watching my mother&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tidal lock.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And my cousins think I'm fat.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Maybe the stars? Cousins in the fact that they are visible in the night sky, but not made of the same stuff. The moon is fat because (even though the stars are physically larger) it appears far larger than them to someone standing on earth.  Or maybe because it is made of cheese.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12192" LastActivityDate="2015-05-12T16:42:09.737" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14813" PostTypeId="2" ParentId="14785" CreationDate="2015-05-12T17:04:46.733" Score="1" Body="&lt;p&gt;I will assume that the containers are made of such material (e.g plastic), that I can make holes in them. All containers are of equal dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CBxfG.gif&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;1) Make a water clock&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Let's assume that we don't have any time keeping device. We take one container and make a small hole (using a pin) at the bottom (container &lt;strong&gt;T&lt;/strong&gt;). The hole could be made such that we can count the falling droplets or in other words the droplets are sufficiently separated such that it can be counted with a naked eye. This will be our clock and the time will be the number of drops counted.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;2) Make a liquid source at constant rate&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This is assuming that the source of liquid is not already from some tap or any other constant rate liquid source. If this is the case then we can simply make one source which drops the liquid at constant rate by making a slightly bigger hole than the previous case as shown in the figure (container &lt;strong&gt;S&lt;/strong&gt;).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;3) Observe the timing&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;First place an empty container at &lt;strong&gt;C&lt;/strong&gt; below &lt;strong&gt;S&lt;/strong&gt; as shown in the figure and count the number of drops it takes to fill that container. Let $t_f$ be the time (no of drops) it takes to fill the empty container. Similarly let $t_p$ be the time it takes to fill the partially filled container with the unknown volume of liquid. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;4) Calculate the unknown volume&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The volume is directly proportional to the time it takes to fill the container. So, if $x$ is the volume of the container, then the unknown volume $v$ is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v = x\left(\frac{t_f-t_p}{t_f}\right)$$&lt;/p&gt;&#xA;" OwnerUserId="101" LastActivityDate="2015-05-12T17:04:46.733" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14814" PostTypeId="2" ParentId="14787" CreationDate="2015-05-12T17:21:42.370" Score="2" Body="&lt;p&gt;Huh - for II I was thinking:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bridge/Bride, since marriages between royals were often used to cement alliances. However, bridge/ridge does seem to make more sense :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12369" LastActivityDate="2015-05-12T17:21:42.370" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14815" PostTypeId="2" ParentId="14681" CreationDate="2015-05-12T17:37:44.337" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cinderella&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no brothers or sisters, but I have many dolls.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; She had no brothers or sisters , only stepsisters. Dolls- not sure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm always watching my mother, and my cousins all think I'm fat.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Always watching her fairy godmother who helps her .Her stepsisters were jealous of her and might amount to thinking that she was fat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2656" LastActivityDate="2015-05-12T17:37:44.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14817" PostTypeId="2" ParentId="14681" CreationDate="2015-05-12T18:05:37.053" Score="1" Body="&lt;p&gt;Probably not what you had in mind, but it can match pretty well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Microsoft Windows Operating System&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no brothers or sisters,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Windows has no brothers or sisters (but children [XP, vista, 7 etc] and cousins [Unix/OSX])&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I have many dolls.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Windows can have many 'dolls' of itself (i.e premium, ultimate, home, workplace)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm always watching my mother,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Windows is always watching/incorporating an aspect of it's 'mother' MS-DOS/Command Prompt (however windows itself is often much different from its previous generations in many other aspects)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and my cousins all think I'm fat.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's Unix cousins would think Windows OS is 'fat' because of how much more space it required for basic installation(30+ GB). (Where as Unix can be under 1GB)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastActivityDate="2015-05-12T18:05:37.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14818" PostTypeId="2" ParentId="14681" CreationDate="2015-05-12T18:41:06.100" Score="2" Body="&lt;p&gt;Are you..&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Ganymede_%28moon%29&quot; rel=&quot;nofollow&quot;&gt;Ganymede&lt;/a&gt; ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no brothers or sisters,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jupiter's moons, including Ganymede, were all most likely captured from solar orbits&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I have many dolls.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Several Jupiter-bound spacecraft have been able to observe Ganymede&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm always watching my mother,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ganymede is in a synchronous rotation, similar to the Moon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and my cousins all think I'm fat.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ganymede is the largest moon in the solar system&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4985" LastActivityDate="2015-05-12T18:41:06.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14819" PostTypeId="1" AcceptedAnswerId="14822" CreationDate="2015-05-12T19:02:53.883" Score="-1" ViewCount="1013" Body="&lt;p&gt;I made the following riddle in a RPG.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;Albus&lt;/em&gt;, &lt;em&gt;Baku&lt;/em&gt; and &lt;em&gt;Croco&lt;/em&gt; experimented with some ingots. One was using copper, one silver and the other gold.&lt;/p&gt;&#xA;&lt;p&gt;But none of the experiments worked with the ingots they had. Though another could.&lt;/p&gt;&#xA;&lt;p&gt;After some changes, &lt;em&gt;Baku&lt;/em&gt; ended up richer than before. &lt;em&gt;(gold is more expensive than silver, which is more expensive than copper, just in case)&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Croco&lt;/em&gt; tried to help &lt;em&gt;Baku&lt;/em&gt; with his experiment.&lt;/p&gt;&#xA;&lt;p&gt;There was no mutual exchange of ingots.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Albus&lt;/em&gt; is deadly allergic to silver.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Players could take what &lt;em&gt;Albus&lt;/em&gt; had after everything happened, if they guessed what it could be, according to the riddle.&lt;/p&gt;&#xA;&lt;p&gt;How rich they became?&lt;/p&gt;&#xA;&lt;p&gt;EDIT to clarify:&lt;/p&gt;&#xA;&lt;p&gt;The nature of the experiments is not important at all (you can think of &lt;em&gt;Albus&lt;/em&gt;, &lt;em&gt;Baku&lt;/em&gt; and &lt;em&gt;Croco&lt;/em&gt; as alchemists). The important thing here is who had which ingots at the begining and who had them at the end.&lt;/p&gt;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; It's about logic, as the already given answers point. Even though, you shouldn't make more assumptions than those said.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;EDIT to make clear the assumptions of the riddle:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;Albus&lt;/em&gt; had an ingot. &lt;em&gt;Baku&lt;/em&gt; had an ingot. &lt;em&gt;Croco&lt;/em&gt; had an ingot. One of the ingots was gold, another silver, another copper.&lt;/p&gt;&#xA;&lt;p&gt;Some ingots changed their owner, with no mutual exchange. &lt;em&gt;(if A gave his ingot to B, B did not gave his ingot to A)&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Baku&lt;/em&gt; recieved an ingot that make him richer&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Croco&lt;/em&gt; gave his ingot to &lt;em&gt;Baku&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Albus&lt;/em&gt; never touched silver.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You have to guess what can &lt;em&gt;Albus&lt;/em&gt; have after everything that's consistent with previous statements and is the maximum possible&lt;/p&gt;&#xA;" OwnerUserId="12165" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-12T20:36:01.997" Title="Copper, Silver and Gold. Who had what?" Tags="|logical-deduction|situation|" AnswerCount="6" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14820" PostTypeId="1" CreationDate="2015-05-12T19:04:20.753" Score="-3" ViewCount="631" Body="&lt;p&gt;The following puzzle was posted here, but providing answers to the puzzle was closed without a correct and complete answer being provided:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An inspector knows that exactly one of 3 suspects committed a crime, and interviews them to find out which. Each person lies one time, and tells the truth the other time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A says: I did not do it. B did it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B says: I did not do it. I know that C did it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C says: I did not do it. B does not know who it was.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the inspector figure out the culprit? If so, who is it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original question (and answers) can be found here:&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/14749/logic-puzzle-3-suspects#14752&quot;&gt;Logic puzzle - 3 suspects&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12372" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2015-05-12T19:11:48.673" Title="3 suspects Logic Puzzle" Tags="|logical-deduction|liars|" AnswerCount="1" CommentCount="1" ClosedDate="2015-05-12T19:45:16.147" ContentLicense="CC BY-SA 3.0" />
  <row Id="14821" PostTypeId="2" ParentId="14820" CreationDate="2015-05-12T19:11:48.673" Score="0" Body="&lt;p&gt;&lt;strong&gt;The correct answer is C.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an accurate explanation as to why C is the correct answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A condition of the puzzle is that each person MUST make one statement that is false, and one that is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with suspect A: If the statement &quot;B did it&quot; is true, then A's other statement &quot;I did not do it&quot; must also be true.  Since that does not meet the one true statement, one false statement condition, the statement &quot;B did it&quot; must be FALSE, meaning that A's statement &quot;I did not do it&quot; is TRUE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On to suspect B: If the statement &quot;I know that C did it&quot; is true, then B's other statement &quot;I did not do it&quot; must also be true.  Since that also does not meet the one true statement, one false statement condition, the statement &quot;I know that C did it&quot; must be FALSE, meaning that B's statement &quot;I did not do it&quot; is TRUE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if A and B did not do it, C did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This perfectly harmonizes with C's statements: &quot;I did not do it&quot; (FALSE), and &quot;B does not know who it was&quot; (TRUE, as we determined that B's statement &quot;I know that C did it&quot; was false).&lt;/p&gt;&#xA;" OwnerUserId="12372" LastActivityDate="2015-05-12T19:11:48.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14822" PostTypeId="2" ParentId="14819" CreationDate="2015-05-12T19:26:00.690" Score="2" Body="&lt;p&gt;Starts off:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Albus - Gold&#xA;&lt;br&gt;Baku - Copper&#xA;&lt;br&gt;Croco - Silver&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then they shift:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Albus - Copper&#xA;&lt;br&gt;Baku - Silver&#xA;&lt;br&gt;Croco - Gold&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Croco gives Albus gold and Albus gives nothing back (not a mutual exchange since Croco gets nothing in return)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So Albus ends up with&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Copper and Gold ingots&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As for thinking out of the box:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Baku gives Albus his silver bar as well (for nothing in return), except he puts it directly into Albus' bank (with no contact betwee the silver and Albus) in which Albus can now exchange it for its cash/copper/gold value.&#xA;&lt;br&gt; If Baku is evil, he kills Albus for us, so that we inherit his ingots!&#xA;&lt;br&gt; Which means Albus [me] gets all the ingots (gold, silver and copper)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-05-12T19:56:02.270" LastActivityDate="2015-05-12T19:56:02.270" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14823" PostTypeId="2" ParentId="14819" CreationDate="2015-05-12T19:30:36.930" Score="1" Body="&lt;p&gt;Albus had platinum at the end.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1.  Albus is allergic to silver means that he neither started with nor ended with silver.&lt;br&gt;&#xA;&lt;br&gt;&#xA;2.  Croco helped Baku.  This means that Baku could not have helped Croco (otherwise Albus ends up with the same thing.  Thus Baku must have helped Albus and Albus helped Croco.  This implies that Croco must have given silver to Baku (otherwise Albus had the silver at some point.&lt;br&gt;&#xA;&lt;br&gt;&#xA;3.  Baku ended up richer.  As Baku was given silver (from 2), this means he started with copper.  Thus he gave the copper to Albus (who therefore must have started with gold and given it to Croco.&#xA;&#xA;Beginning: Albus: gold, Baku: copper, Croco: silver&#xA;End: Albus: copper, Baku: silver, Croco: gold&#xA;&#xA;  EDIT* For Albus to have the MOST at the end, it means that his experiment must have worked.  If they are alchemists, then we'll assume he turned his copper into PLATINUM.  The statement &quot;Croco TRIED to help Baku&quot; suggests that he didn't actually help his experiment.  This means that Baku actually needed the gold for his experiment.  Croco's experiment may or may not have worked.  Nothing says that each one's experiment needs a different metal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10457" LastEditorUserId="10457" LastEditDate="2015-05-12T20:24:32.683" LastActivityDate="2015-05-12T20:24:32.683" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14824" PostTypeId="2" ParentId="14819" CreationDate="2015-05-12T19:32:25.217" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Albus has the copper ingot, Baku has the silver, and Croco has the gold.&lt;br&gt;&#xA; Originally Albus had gold, Baku had copper, and Croco had silver.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;... logic chain ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everyone switched ingots (no mutual exchange, so all must have moved).&lt;br&gt;&#xA; Baku upgraded, so he must have ended with silver or gold.&lt;br&gt;&#xA; Croco tried to help Baku, so he must have started out with an ingot worth more.&lt;br&gt;&#xA; Since Albus can't touch it, the silver was exchanged between Croco and Baku.&lt;br&gt;&#xA; If Albus had the copper, he must have passed it to Croco since Baku does not have the copper -- but this is impossible because Baku could not have given Albus the silver to take Croco's gold.&lt;br&gt;&#xA; Therefore Albus started with the gold.&lt;br&gt;&#xA; Since Baku upgraded, he must have started with the copper to Croco's silver or the help wouldn't have been needed.&lt;br&gt;&#xA; The ingots were passed Albus gold to Croco, Croco silver to Baku, and Baku copper to Albus.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10625" LastActivityDate="2015-05-12T19:32:25.217" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14825" PostTypeId="2" ParentId="14819" CreationDate="2015-05-12T19:33:33.753" Score="2" Body="&lt;p&gt;Let's use the metals' atomic abbreviations to identify which one each person has. They are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Cu for Copper&lt;/li&gt;&#xA;&lt;li&gt;Ag for Silver&lt;/li&gt;&#xA;&lt;li&gt;Au for Gold&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We start out knowing that Albus cannot have used silver. Then, seeing as Baku got richer, he could not have had gold. If we draw this out a bit, we get...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Albus - NOT Ag&lt;br/&gt;Baku - Cu OR Ag&lt;br/&gt;Croco - Unknown&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Working with these, we conclude that, at the start of the experiments, these people have these metals:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Albus - Au&lt;br/&gt;Baku - Ag&lt;br/&gt;Croco - Cu&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After the experiments initially fail, some changes are made. At this point, we can utilize the knowledge that Baku becomes richer. Now, knowing humans and their tendency to feel greed and envy, Croco may have gotten jealous of Baku's increased wealth. Now, we can draw this conclusion:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Albus - Cu&lt;br/&gt;Baku - Au&lt;br/&gt;Croco - Ag&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using this drawing, the final answer is that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Albus has the copper.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&#xA;However, there is also another path. It contravenes what one might expect to happen, but it is a better path for the player who is trying to solve the riddle. This path goes as follows.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Albus starts with copper, Baku with silver, and Croco with gold. After the swap, Baku has gold, and Croco has copper. This leaves us with Albus.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we were to take this path, we would end with this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Albus has the silver. However, unfortunately for Albus, his silver allergy has resulted in his death.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4985" LastEditorUserId="4985" LastEditDate="2015-05-12T19:49:13.400" LastActivityDate="2015-05-12T19:49:13.400" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14826" PostTypeId="2" ParentId="14819" CreationDate="2015-05-12T19:34:14.987" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Albus had gold -&gt; switched to copper&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;-&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Baku had silver -&gt; switched to gold&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;-&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Carno had copper -&gt; switched to silver&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;are the only combinations that fit all the criteria (I interpreted Carno help Baku as Carno picking the metal Baku had left off)&lt;/p&gt;&#xA;" OwnerUserId="9413" LastActivityDate="2015-05-12T19:34:14.987" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14827" PostTypeId="1" AcceptedAnswerId="14829" CreationDate="2015-05-12T20:00:06.757" Score="3" ViewCount="691" Body="&lt;p&gt;Once a month, Jerry visits the restaurant that he owns. Being paranoid that someone will steal his fortune, on each visit he changes the combination to his personal safe, which is hidden in the back part of the managers' office.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;On each update, Jerry uses the next number in a numerical pattern that he has been aware of for years, and is very confident that no one else will figure out. Each time, he jots the previous combination down on a sheet of paper, which is taped to the side of the safe, to remind him where he is in the sequence.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The safe has a simple, 10-button keypad, like below, and can be programmed to accept combinations from length 1 to length 15. After 3 failed attempts in a row to open the safe, a text will be sent to Jerry's phone. Therefore, guessing is not a great idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dB1Ip.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;After some time, the restaurant managers find the small piece of paper taped to the safe, and figure out that the numbers are the previous combinations, and assume that a pattern or sequence of some kind is being followed. They are bored and would love to break into the safe, if only for the challenge. However, they have yet to be able to find the pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the list of previous combinations:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1&lt;br&gt;&#xA;  8&lt;br&gt;&#xA;  41&lt;br&gt;&#xA;  208&lt;br&gt;&#xA;  1041&lt;br&gt;&#xA;  5208&lt;br&gt;&#xA;  26041&lt;br&gt;&#xA;  130208&lt;br&gt;&#xA;  651041&lt;br&gt;&#xA;  3255208&lt;br&gt;&#xA;  16276041&lt;br&gt;&#xA;  81380208&lt;br&gt;&#xA;  406901041&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the current combination?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint #1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eventually, Jerry's algorithm will produce combinations that are too long (&gt;15 digits), but for now this works.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sequence has nothing to do with letters or alphabets or changing bases, and nothing is encrypted.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-05-12T20:32:09.887" LastActivityDate="2015-05-12T23:56:18.280" Title="Find the combination algorithm" Tags="|pattern|number-sequence|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14828" PostTypeId="2" ParentId="14827" CreationDate="2015-05-12T20:13:02.030" Score="2" Body="&lt;p&gt;Partial answer here&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the 4 last digits should be 5208&lt;br&gt;&#xA; last digit alternates between 1 and 8, so the current is 8&lt;br&gt;&#xA; second to last digit alternates between 0 and 4, so the current is 0&lt;br&gt;&#xA; third to last digit alternates between 0 and 2, so the current is 2&lt;br&gt;&#xA; fourth to last digit is in the 1, 5, 6, 0 suite and start over, so the current is 5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I haven't found yet the end of the pattern, but that might help others&lt;/p&gt;&#xA;" OwnerUserId="10400" LastActivityDate="2015-05-12T20:13:02.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14829" PostTypeId="2" ParentId="14827" CreationDate="2015-05-12T20:13:05.410" Score="5" Body="&lt;p&gt;To arrive at the new combination,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jerry takes the previous combination, multiplies it by $5$, then adds either $1$ or $3$, depending on whether the previous combination was even or odd (respectively).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The next combination is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $406901041\times 5+3=2034505208$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2627" LastActivityDate="2015-05-12T20:13:05.410" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14830" PostTypeId="2" ParentId="14819" CreationDate="2015-05-12T20:19:44.513" Score="2" Body="&lt;p&gt;With the clarified logic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Albus - Copper&#xA;&lt;br&gt;Baku - Silver&#xA;&lt;br&gt;Croco - Gold&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;End:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Albus - Gold&#xA;&lt;br&gt;Baku - Silver &amp;amp; Copper&#xA;&lt;br&gt;Croco - None&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Albus had an ingot. Baku had an ingot. Croco had an ingot. One of the ingots was gold, another silver, another copper.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Albus - Copper&#xA;&lt;br&gt;Baku - Silver&#xA;&lt;br&gt;Croco - Gold&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But none of the experiments worked with the ingots they had. Though another could.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;They had to move all the ingots around&#xA;Albus - Gold&#xA;&lt;br&gt;Baku - Copper&#xA;&lt;br&gt;Croco - Silver&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some ingots changed their owner, with no mutual exchange. (if A gave his ingot to B, B did not gave his ingot to A)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Some&lt;/strong&gt; implies that not all ingots had to change hands. We'll get to what this means in a moment &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Baku recieved an ingot that make him richer&#xA;Croco gave his ingot to Baku&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Croco, after failing in his experiment, gave his silver ingot to Baku. Now, Baku has both the silver and the copper ingots, making him richer than before.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Albus never touched silver.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Albus started with copper and ended with gold&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;s&gt;Why not:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Albus starts with the gold and the copper ingots and ends with the gold&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's the deal:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Albus, Baku and Croco experimented with some ingots. One was using copper, one silver and the other gold.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This never states that they all come to the table holding one ingot. At this point, Albus could have all three.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But none of the experiments worked with the ingots they had. Though another could.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;OK, we have to take one away from Albus. Because we later learn about his allergy, we say he never had the silver, that Croco had it the whole time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After some changes, Baku ended up richer than before. (gold is more expensive than silver, which is more expensive than copper, just in case)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Baku came in without anything. By getting the silver ingot, he walks away richer than before.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Croco tried to help Baku with his experiment.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Croco gave Baku the ingot to work on, since he didn't have one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There was no mutual exchange of ingots.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Albus gave Croco the copper ingot because he gave the silver up to Baku to be able to work&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Albus is deadly allergic to silver.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;So Albus never handled the silver ingot&lt;/s&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12242" LastEditorUserId="12242" LastEditDate="2015-05-12T20:36:01.997" LastActivityDate="2015-05-12T20:36:01.997" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14831" PostTypeId="1" AcceptedAnswerId="14832" CreationDate="2015-05-12T20:30:28.357" Score="2" ViewCount="4007" Body="&lt;p&gt;You and two friends decided to order a large circular pizza to split between each other. Unfortunately at the moment you didn't have enough to pay your entire third of the bill. Your friends decide that they will cover the remainder of your bill but for a cost! Instead of just giving you less pizza, they will give you a challenge to earn as much as you can. This is the challenge:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The pizza will be set up on a large cutting board.&lt;/p&gt;&#xA;&lt;p&gt;You will be given 1 pizza cutter and are allowed to make 3 cuts in total.&lt;/p&gt;&#xA;&lt;p&gt;Each cut must go from one end of the board to the other. (i.e no 'half cuts' or stopping in the middle of the pizza)&lt;/p&gt;&#xA;&lt;p&gt;You are allowed to set it up how you please before and after each cut.&lt;/p&gt;&#xA;&lt;p&gt;When you are finished cutting the pizza, you're friends will pick all their slices first, and they will take the higher number if distribution isn't 'fair'&lt;/p&gt;&#xA;&lt;p&gt;(i.e if there are 4 slices, [Friend 1] will take 2, [Friend 2] will take 1 and then [You] will take 1)&lt;/p&gt;&#xA;&lt;p&gt;(5 slices: [Friend 1] takes 2 then [Friend 2] takes 2 then You take 1)&lt;/p&gt;&#xA;&lt;p&gt;**You must not cut it into the standard 6th's as done by normal pizza places (60° from each cut)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;(Sorry for the last rule but your hungry friends have became cranky)&lt;/p&gt;&#xA;&lt;p&gt;Your friends are really hungry and will always try to take as much pizza as possible.&lt;/p&gt;&#xA;&lt;p&gt;What is the maximum amount of pizza you can get if you follow these rules?&lt;/p&gt;&#xA;&lt;p&gt;&lt;b&gt;Bonus:&lt;/b&gt;&lt;/p&gt;&#xA;&lt;p&gt;If you are still restricted to 3 cuts, but you are allowed to distribute the pizza however you please (as long as your friends always have equal or more than you) how can you cut and distribute the pizza such that you all get even portions (The max as possible for you)?&lt;/p&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-23T08:26:30.733" Title="The Pizza Cutting Challenge" Tags="|logical-deduction|lateral-thinking|" AnswerCount="8" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14832" PostTypeId="2" ParentId="14831" CreationDate="2015-05-12T20:39:43.503" Score="9" Body="&lt;p&gt;I'm probably missing something but couldn't you just&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cut the pizza into 6ths so each of you gets 2 pieces? (Diameter + 60° off through center + 60° off through center)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&#xA;If you can't do the center 60° cut, just use parallel cuts to cut into 3rd, then a perpendicular cut to get 6ths again. The parallel cuts have an arc of 2.605 rads, or about .267 radius from the center. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Thinking&lt;/strong&gt;&lt;br&gt;&#xA;So, I thought about how to handle this issue if you have N friends:&lt;br&gt;&#xA;1 friend: 6ths still works&lt;br&gt;&#xA;3 friends: Diameter, then 2 parallel cuts to give areas of 1/4&lt;br&gt;&#xA;4 friends: Uneven X with a bar across the bottom. I'd guess the bottom bar is 1/2 from center, X crosses 1/8 from center (center is between X and bar)&lt;br&gt;&#xA;5 friends: 6ths still works&lt;br&gt;&#xA;6 friends: Cut out a triangle from the middle, total of 7 pieces, max.&lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-05-12T21:12:25.640" LastActivityDate="2015-05-12T21:12:25.640" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14833" PostTypeId="2" ParentId="14831" CreationDate="2015-05-12T20:42:50.463" Score="0" Body="&lt;h1&gt;Answer&lt;/h1&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                   %%%    %%%&#xA;              %%%              %%%&#xA;&#xA;          %%%       1/3             %%%&#xA;                             _.-*''       A,C   ~ 1/3 of pizza&#xA;         %%%             _.-'        %%%&#xA;(two cuts)  ============|_      1/3&#xA;         %%%              `-._       %%%&#xA;                             |_&#xA;          %%%        1/3       |_    %%%&#xA;                                 |____    B  ~ 1/3 of pizza&#xA;            %%%                  %%%&#xA;&#xA;                   %%%     %%%&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Pretend the picture is to proportion. Thus, you approximate $A$ to be $\frac13$ of the pizza, $B$ to be the same (except going upwards), and $C$ to overlap $A$ (technically having 3 cuts, though only $2$ legit cuts). Thus, three slices of equal proportion are left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you could split the pizza into six equal slices.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Bonus&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Still workin' on it&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Edits&lt;/h1&gt;&#xA;&#xA;&lt;h2&gt;1&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The first scenario would work with your &quot;restriction.&quot;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;2 ~ Extra solution&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This cuts the pieces into (roughly) equal pieces (cutting along the blue lines). Hopefully, your friends will trust you in that they are equal, if you do a bad cutting job:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3RAer.png&quot; alt=&quot;Cut along the blue lines&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7307" LastEditorUserId="7307" LastEditDate="2015-05-12T21:06:54.930" LastActivityDate="2015-05-12T21:06:54.930" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14834" PostTypeId="2" ParentId="14831" CreationDate="2015-05-12T20:56:26.907" Score="0" Body="&lt;p&gt;Cut one line horizontally through the center.  Then make two more cuts vertically, such that all three pieces above (or below) the original cut have equal area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update:  Per the comment: the vertical cuts should go through points about 15 degrees from the center.  This would be a smidge over 1/4 radius.  Thus, the central pieces would be about 1/2 radius wide, and the others would be 3/4 radius wide.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="2328" LastEditDate="2015-05-12T21:34:25.240" LastActivityDate="2015-05-12T21:34:25.240" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14835" PostTypeId="1" AcceptedAnswerId="14850" CreationDate="2015-05-12T21:04:44.110" Score="9" ViewCount="779" Body="&lt;p&gt;As you're walking through the city, a wandering street logician pulls you aside. &quot;Hey, you. Want to play a game?&quot; He rasps in a loud whisper. You, being quite the logician yourself, oblige. He leads you through a dark, winding alley, eventually ending at a secluded, run-down storage locker. Shuffling a bit, he unlocks the rusty door and opens it with a loud creak, revealing darkness behind it. He gestures you forward, and you step in. As your eyes begin to adjust to the darkness, you jolt as the door slams shut behind you. No sooner has it shut than the lights turn on, revealing six hats of different colors, placed on stools. Taking a quick inventory, you note that the hats in order from left to right are &lt;strong&gt;Red, Blue, Yellow, Green, Orange,&lt;/strong&gt; and &lt;strong&gt;Purple.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;So here's the game,&quot; the logician wheezes. &quot;As of now, you are trapped in this locker with me. The only way I'll let you out is if you can follow the Magical Hats! I'm going to shuffle the hats, then give you clues about their whereabouts. If you can correctly identify where each hat has finished, you with obtain your freedom, as well as the hat of your choice! Incorrectly identify the hats, well...let's just say I hope you correctly identify the hats.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You flex your brain muscles. Time to play hardball.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The logician lets a drape loose, shielding the hats from your view. He runs behind the drape and begins frantically moving hats around, panting as he runs between stools. For awhile you attempt to track the hats, but you give up as his movement is too erratic to follow. Finally, he stops and comes out from behind the curtain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Here's what I have to tell you about the new order of hats,&quot; he pants.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Exactly &lt;strong&gt;one hat&lt;/strong&gt; is in the same place that it started.  &lt;/li&gt;&#xA;  &lt;li&gt;Exactly &lt;strong&gt;one pair&lt;/strong&gt; of hats have swapped places with one another.  &lt;/li&gt;&#xA;  &lt;li&gt;The first three hats in the original display are now in the &lt;strong&gt;odd-numbered&lt;/strong&gt; locations.  &lt;/li&gt;&#xA;  &lt;li&gt;The &lt;strong&gt;Red&lt;/strong&gt; hat moved the undisputed farthest out of all the hats.  &lt;/li&gt;&#xA;  &lt;li&gt;The &lt;strong&gt;Blue&lt;/strong&gt; hat is adjacent to its starting position.  &lt;/li&gt;&#xA;  &lt;li&gt;The &lt;strong&gt;Purple&lt;/strong&gt; hat is now where the &lt;strong&gt;Green&lt;/strong&gt; hat began.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You think for a second, then begin to piece together where the hats are. &lt;em&gt;This is too easy&lt;/em&gt;, you muse with a smirk. This logician doesn't know what he has coming. As you put the last piece of the puzzle into place and begin to speak, the logician cuts you off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Wait, wait, wait! This is TOO easy!&quot; He shrieks. In haste, he runs behind the curtain and begins shuffling the hats again. You feel your morale break, as everything you had put together has just been shattered. Still, you steel yourself, and prepare for his next challenge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After an eternity, the logician once again returns from behind the drape. &quot;Okay, this time...this time I have it right,&quot; he gasps. &quot;And here are your clues.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Exactly &lt;strong&gt;one hat&lt;/strong&gt; is in the same place that it started.  &lt;/li&gt;&#xA;  &lt;li&gt;Exactly &lt;strong&gt;one hat&lt;/strong&gt; is in the same place that it was in the previous iteration of hats.  &lt;/li&gt;&#xA;  &lt;li&gt;Exactly &lt;strong&gt;one pair&lt;/strong&gt; of hats have swapped places with one another from the previous iteration of hats.  &lt;/li&gt;&#xA;  &lt;li&gt;All hats that were adjacent to their starting position in the previous iteration of hats are still adjacent to their starting position.  &lt;/li&gt;&#xA;  &lt;li&gt;There is only one set of hats that started adjacent and have remained adjacent.  &lt;/li&gt;&#xA;  &lt;li&gt;The hat on the far left has not been on an end until now.  &lt;/li&gt;&#xA;  &lt;li&gt;The left half and the right half from the previous iteration contain the same assortments of hats.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You take a minute to breathe. &lt;em&gt;Let's see if I can still do this,&lt;/em&gt; you wonder as your brain begins to pulse. Just as you begin to unravel the strings of the hat mystery in your mind, the logician shrieks at you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;WAAAAAAAAAIT!!!!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your gears halt. &quot;What?&quot; you spit at the logician angrily. He flinches backwards, then sheepishly admits, &quot;I forgot to tell you about the lucky and unlucky hats.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You groan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He continues on, &quot;The lucky hat is worth seven million dollars - it's made of quite a rare silk, you see. The unlucky hat, however, will literally kill you instantly if you put it on your head.&quot; Before you can posit a conjecture about simply not wearing the hat, he adds, &quot;I of course will not let you leave until you are wearing your selected hat.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;How can I possibly know which is which?&lt;/em&gt; You think with a furrowed brow. Luckily, he has prepared an answer for that as well. &quot;Unfortunately, I don't remember which hat is which, but I do remember a few things about the hats...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;...and here's what I remember.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Of the lucky and unlucky hats, one is a &lt;strong&gt;primary color&lt;/strong&gt; and one is a &lt;strong&gt;secondary color&lt;/strong&gt;.  &lt;/li&gt;&#xA;  &lt;li&gt;At no point while the hats were at rest were the two hats next to one another.  &lt;/li&gt;&#xA;  &lt;li&gt;If the lucky hat is a &lt;strong&gt;primary color&lt;/strong&gt;, then the unlucky hat was never at rest on either end.  &lt;/li&gt;&#xA;  &lt;li&gt;If the lucky hat is a &lt;strong&gt;secondary color&lt;/strong&gt;, then the unlucky hat is &lt;strong&gt;yellow.&lt;/strong&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You wait, as he stands in thought. &quot;No, that's all I remember,&quot; he cedes after a few seconds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You groan again. Time to get to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Where were the hats after the first shuffle? Where were the hats after the second shuffle? Finally, when you win the prize and are allowed to leave, which hat do you take with you?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="11297" LastEditDate="2015-05-12T21:34:25.327" LastActivityDate="2015-05-12T23:45:41.007" Title="Magical Shuffling Hats" Tags="|logical-deduction|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14837" PostTypeId="2" ParentId="14785" CreationDate="2015-05-12T21:10:20.533" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Dump the liquid out onto the floor. Your container now holds 0 liquid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12373" LastActivityDate="2015-05-12T21:10:20.533" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14839" PostTypeId="2" ParentId="14831" CreationDate="2015-05-12T21:11:19.823" Score="2" Body="&lt;p&gt;Lateral thinking answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Scrape all of the toppings into a bowl. Cut pizza in half, let them take the 2 pieces. Cook noodles, dump topping in noodles, enjoy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More onto the puzzle itself:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Best answer is still to find a way to cut the pizza in 6 slices as identical as possible. Instead of trying to cut it traditionally try stacking the pizza after each cut. This means that cut #2 will make 2 identical 'sixths', its the hardest cut. Cut # 3 will slice in half the 2 thirds remaining making (hopefully) 4 identical 'sixths'&lt;br&gt;&#xA;&lt;br&gt;&#xA;Or a bit of a mix of both, cut it in half (blue), stack the two halves, then cut along the yellow line (doing 2 cuts at once), and green line. The larger pieces are the ones with mostly only crust, so your friends end up with 2 pieces of crust, and you end up with the goodies :)&#xA;&lt;img src=&quot;https://i.stack.imgur.com/o43Ra.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastEditorUserId="11251" LastEditDate="2015-05-12T21:21:07.427" LastActivityDate="2015-05-12T21:21:07.427" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14844" PostTypeId="2" ParentId="14831" CreationDate="2015-05-12T21:40:04.817" Score="1" Body="&lt;p&gt;Am I missing something here? Assuming you are a perfect cutter and can eyeball with precision, it's two parallel cuts and a perpendicular cut:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/Un0Jg.png&quot; alt=&quot;Pizza Diagram&quot;&gt; &lt;br&gt;(drawing is not perfect)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastActivityDate="2015-05-12T21:40:04.817" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14845" PostTypeId="2" ParentId="14831" CreationDate="2015-05-12T21:41:06.323" Score="8" Body="&lt;p&gt;Here comes the troll answer:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; first, cut the pizza in half&lt;br&gt;&#xA; The second and third cut are on the friends' throat&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;tada!  The pizza is all mine now!&lt;/p&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2015-05-12T21:41:06.323" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14846" PostTypeId="2" ParentId="14770" CreationDate="2015-05-12T21:45:00.967" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Daytime (light)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I number myself, for I am numbered;&#xA;I am as long as I am worth.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Days are numbered and daytime last differently depending on the time of year&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I run down and up, I've crackled and thundered.&#xA;Just to come to rest upon the earth.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Daytime comes to rest upon the earth every morning&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Neither red nor deep slate nor pure white,&#xA;what remains of the four, I must be!&#xA;(Though, really, It's only typically...)&#xA;What I am must surely be in sight!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Hardest one, I'm thinking this is referring to the seven days of the week but haven't figured out how.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm so wound up, about this little road.&#xA;I can hardly reach the end!&#xA;I've neither limbs nor toes nor nodes;&#xA;Naught but a body suspends.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Daytime(light) can't reach the end of the day (normally)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Strong 'till while I'm young, I am.&#xA;(Perhaps as late&#xA;as seven or eight!)&#xA;As I grow older, I bend and I jam.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;In the morning daylight is strong, but in the evening shadows make it seem to twist and bend. 7/8 is the typical time for sunsets/sunrises&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, now! The answer is close!&#xA;I am what I am, though noticed little&#xA;and as helpful can be when you must.&#xA;I come in all shapes, all lengths, and all sizes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now tell me: who am I? (I've got all day -- I can wait longer than you)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Indeed. Daytime has all day. And I doubt you can manage waiting a full day without doing -anything- else like it does.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastActivityDate="2015-05-12T21:45:00.967" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14847" PostTypeId="2" ParentId="14835" CreationDate="2015-05-12T21:53:48.623" Score="2" Body="&lt;p&gt;After the first iteration the hats were like so:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Blue, Orange, Yellow, Purple, Red, Green&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After the second iteration, the hats were like so:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Yellow, Orange, Blue, Red, Green, Purple&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The hat to take is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Red&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The unlucky hat is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Orange&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10457" LastActivityDate="2015-05-12T21:53:48.623" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14848" PostTypeId="2" ParentId="14749" CreationDate="2015-05-12T23:01:59.290" Score="2" Body="&lt;p&gt;So if the system is consistent, following logic brings us a conclusion that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;C did it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How? that's simple&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A says: (a) I did not do it. (b) B did it.&lt;br&gt;&#xA;&lt;code&gt;Assume that A(a) is TRUE and A(b) is FALSE&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;B says: (a)I did not do it. (b)I know that C did it.&lt;br&gt;&#xA;&lt;code&gt;being consistent with A(b), we know that that B(a) is TRUE, which means that B(b) is FALSE&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;C says: (a)I did not do it. (b)B does not know who it was.&lt;br&gt;&#xA;&lt;code&gt;being consistent with B(b), C(b) is TRUE since !B(b) and C(b) state same thing, which means that C(a) is FALSE.&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; so C did it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11023" LastEditorUserId="11023" LastEditDate="2015-05-12T23:08:02.593" LastActivityDate="2015-05-12T23:08:02.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14850" PostTypeId="2" ParentId="14835" CreationDate="2015-05-12T23:45:41.007" Score="8" Body="&lt;p&gt;I get the same answer as the one already posted, but might as well post the workings:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;First Iteration&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;We start with starting point &lt;code&gt;RBYGOP&lt;/code&gt; and the unknown first iteration which I'll write as &lt;code&gt;.....&lt;/code&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We're told the position of the purple hat, so we can fill it in as &lt;code&gt;...P..&lt;/code&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;From the odd number clue, the red hat can be in the 1st, 3rd or 5th position. Since the purple hat has already moved two, the 5th position is the only one that makes it the undisputed furthest mover. That gives us &lt;code&gt;...PR.&lt;/code&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;From the above, the red and purple hat can't be in the same place they started, nor can the green or orange hats since their previous spaces are now occupied. Also, the blue hat started in an even space but has to be moved to an odd one. From that, we are left with the yellow hat as the only one that can be in the same position as it started. &lt;code&gt;..YPR.&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;The first space is now the only odd one left for the blue hat. &lt;code&gt;B.YPR.&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;Finally, the only remaining candidate for a swap is green and purple, giving us &lt;code&gt;BOYPRG&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Second Iteration&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The blue hat is the only one adjacent in the first iteration to the start point. This means it must be in the first or third position. But we also know the leftmost one can't be blue as it's already been on the edge. This gives us &lt;code&gt;..B...&lt;/code&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The first two spaces must be occupied by O and Y in some order, and the last three by P,R and G in some order. This means the only candidates for starting adjacent and still being adjacent are Y in the second space being adjacent to B, or R in the fourth space being adjacent to B: &lt;code&gt;.YB...&lt;/code&gt; or &lt;code&gt;..BR...&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The first of those two is more restrictive, so try that.&lt;/li&gt;&#xA;&lt;li&gt;Now we know the R must be in one of the last two slots because only one adjacent pair can remain from the start.&lt;/li&gt;&#xA;&lt;li&gt;R can't be in the penultimate slot because then P and G would either both be in the same position as they started, or both be in the same position as last iteration&lt;/li&gt;&#xA;&lt;li&gt;R can't be in the last slot because then we can only satisfy a hat being in the same position as the start &lt;em&gt;or&lt;/em&gt; a hat being in the same position as the first iteration but not both.&lt;/li&gt;&#xA;&lt;li&gt;This leaves no valid position for R, so we've excluded this possibility and know it must be &lt;code&gt;..BR..&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Since only one adjacent pair can be the same as the start, we also know Y can't be in the second slot, arranging the first two as &lt;code&gt;YOBR..&lt;/code&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The O is in the same place as it was in the first iteration, so the G can't also be, giving us the final result &lt;code&gt;YOBRGP&lt;/code&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Lucky and Unlucky&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;Primary&quot; is somewhat ambiguous as the primary colours could reasonably be RGB or RYB depending on the colour model.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Start with assuming RGB. Together the first and four clues rule out the lucky hat being secondary&lt;/li&gt;&#xA;&lt;li&gt;This means we're looking for a pair of colours with the following properties:&lt;/li&gt;&#xA;&lt;li&gt;One is in {R,G,B}, the other is in {O,Y,P}&lt;/li&gt;&#xA;&lt;li&gt;They were never adjacent&lt;/li&gt;&#xA;&lt;li&gt;The one in {O,Y,P} was never at rest on either end.&lt;/li&gt;&#xA;&lt;li&gt;Yellow and purple have both been on an end, so the latter must be orange&lt;/li&gt;&#xA;&lt;li&gt;Red is the only primary colour that has never been next to orange, giving us: lucky red, unlucky orange.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;On the other hand, we can try taking RYB as primary. Consider the case of yellow as unlucky. Then we need a colour satisfying the properties:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is in {O,G,P}&lt;/li&gt;&#xA;&lt;li&gt;Has never been adjacent to yellow&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;That's not satisfied by any colour, so we can rule that out and instead say the lucky hat must be a primary colour. Now we looking for a pair of colours satisfying:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;One is in {R,Y,B}, the other is in {O,G,P}&lt;/li&gt;&#xA;&lt;li&gt;They were never adjacent&lt;/li&gt;&#xA;&lt;li&gt;The one in {O,G,P} was never at rest on either end.&lt;/li&gt;&#xA;&lt;li&gt;Green and purple have both been on an end, so the latter must be orange&lt;/li&gt;&#xA;&lt;li&gt;Red is the only primary colour that has never been next to orange, giving us: lucky red, unlucky orange.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Therefore whichever of the likely options we take for &quot;primary&quot;, we get the same result: &lt;strong&gt;Lucky red, unlucky orange.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1759" LastActivityDate="2015-05-12T23:45:41.007" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14851" PostTypeId="2" ParentId="14827" CreationDate="2015-05-12T23:56:18.280" Score="-2" Body="&lt;p&gt;The question is too easy.  All you have to do is look at the midway point such as 5208 and see that the next number is roughly 5 times that and then just play with the small difference to get it to match.  I think this person deserves to have his safe cracked for picking such as bad algorithm.&lt;/p&gt;&#xA;" OwnerUserId="12119" LastActivityDate="2015-05-12T23:56:18.280" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14852" PostTypeId="2" ParentId="14785" CreationDate="2015-05-13T00:24:27.813" Score="8" Body="&lt;p&gt;I assume that no direct measurement may be made of the mystery flask, but that measurements may be made of liquid poured into the test flask.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Using your eyeball, partially fill a flask with a little less fluid than the mystery flask. Use a carefully measured amount.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Blow a controlled flow of air over both flasks to make them hum. (Hire a floutist if you must.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Add known increments of fluid to the test flask. This will cause the pitch of the sound to change.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As the two flasks become almost identically full, the frequencies of the humming from both flasks will converge, causing a beat phenomena.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When the beat phenomena ceases, the frequencies are the same and the two flasks are equally full. (The human ear can detect very fine differences in frequency. This can be quite accurate.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The original amount of fluid you put into the test flask plus the increments added equals your measurement of the amount of fluid in the second flask.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Make sure that all fluid is the same temperature.&lt;/p&gt;&#xA;" OwnerUserId="12375" LastActivityDate="2015-05-13T00:24:27.813" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14853" PostTypeId="2" ParentId="14785" CreationDate="2015-05-13T00:35:33.913" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;You are however given an infinite number of equal containers to accomplish this task.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's label these infinite containers &lt;code&gt;C[1]&lt;/code&gt;, &lt;code&gt;C[2]&lt;/code&gt;, &lt;code&gt;C[3]&lt;/code&gt;, ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pour some amount of liquid from the initial container into &lt;code&gt;C[1]&lt;/code&gt;. The amount is not critical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set &lt;code&gt;i = 1&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While &lt;code&gt;true&lt;/code&gt;, pour the entire contents of container &lt;code&gt;C[i]&lt;/code&gt; into container &lt;code&gt;C[i+1]&lt;/code&gt; and increment &lt;code&gt;i&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After this, the initial amount of liquid in the initial container is guaranteed to have been &lt;code&gt;1L&lt;/code&gt; at the outset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key to making this solution efficient, at least in terms of pouring operations, is realizing that the pouring step inside the while loop can be optimized away, leaving only the single pouring step at the outset. But don't optimize the whole loop away, or the final bit won't hold!&lt;/p&gt;&#xA;" OwnerUserId="11227" LastActivityDate="2015-05-13T00:35:33.913" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14854" PostTypeId="2" ParentId="14785" CreationDate="2015-05-13T01:56:48.107" Score="5" Body="&lt;p&gt;The initial container is of same size as the ones I can use. This is a necessary parameter in my solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The volume of the containers given to me is known by someone. I am calling it x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the initial container is full, x is the solution. XD&lt;/p&gt;&#xA;" OwnerUserId="12378" LastActivityDate="2015-05-13T01:56:48.107" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14855" PostTypeId="2" ParentId="14785" CreationDate="2015-05-13T03:48:41.330" Score="0" Body="&lt;p&gt;Assumption: I can make markings in the container(not using measuring materials).&lt;/p&gt;&#xA;&lt;p&gt;Procedure&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;First fill all the containers you can. Then you will be left with a container which is not completely filled.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Then after that we make a mark in the container Z at the water level.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;After that fill the container Z up to that mark.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Now try filling the other container's with the liquid of container Z.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Repeat 3 &amp;amp; 4 until you get a n'th container completely full.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So if the n'th containers get full by filling it up to m times then. And Initially we have k full containers.&#xA;Then the volume will be &lt;span class=&quot;math-container&quot;&gt;$k + n/m$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Lets say that we have 5.7x of the liquid. We will know the 5 part but not the decimal part. So we will fill the container with the liquid with the 0.7th part and make a mark there. And now safely put that liquid in another container. We will then fill the container upto that mark and start filling the other empty containers we have. Then firstly we will have 0.7th of the first container filled. Then we will have 1.4th, 2.1th and so on to 7th, which we can tell by seeing as it will be completely full. And we would have required to fill the containers 10 times. So it will be 7x/10 times = 0.7x.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Used the fact that any decimal number when added it to itself many times will finally produce a natural number.&lt;/p&gt;&#xA;" OwnerUserId="11343" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-28T01:48:45.223" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14856" PostTypeId="2" ParentId="14787" CreationDate="2015-05-13T05:37:28.840" Score="7" Body="&lt;p&gt;III.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fjord and Ford. A fjord separates the high cliffs on either side. A ford joins the two sides of a river at a passable point. And both are covered with water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12382" LastActivityDate="2015-05-13T05:37:28.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14858" PostTypeId="2" ParentId="13290" CreationDate="2015-05-13T06:41:23.843" Score="7" Body="&lt;p&gt;The decrypted message is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;yes you have solved this puzzle&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Steps to decrypt are:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;1.&lt;/strong&gt; convert from a-z code to true base-26 (1-9-a-p). In other words, &quot;a&quot; goes to 1, &quot;j&quot; goes to &quot;a&quot;, &quot;y&quot; goes to &quot;p&quot;, and &quot;z&quot; goes to 10.&lt;br&gt;&#xA; &lt;strong&gt;2.&lt;/strong&gt; convert from base-26 to decimal&lt;br&gt;&#xA; &lt;strong&gt;3.&lt;/strong&gt; divide each decimal number by the preceeding number, starting over at the beginning of each word.&lt;br&gt;&#xA; &lt;strong&gt;4.&lt;/strong&gt; where this division results in a &lt;strong&gt;prime number&lt;/strong&gt;, identify it as the $n^{th}$ prime number. Note that each prime number is only coded once, except for &quot;y&quot;.&lt;br&gt;&#xA; &lt;strong&gt;5.&lt;/strong&gt; convert from the $n^{th}$ prime number to the letter of the alphabet&lt;br&gt;&#xA; &lt;strong&gt;6.&lt;/strong&gt; where the division results in a &lt;strong&gt;composite number&lt;/strong&gt;, the plain text letter has been previously defined. So each repeated letter is assigned a number from the sequence of composite numbers. The assigned composite numbers are shown in the yellow cells below.&lt;br&gt;&#xA;  &lt;img src=&quot;https://i.stack.imgur.com/fGXvD.png&quot; alt=&quot;Table&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-06-15T02:22:53.527" LastActivityDate="2015-06-15T02:22:53.527" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14860" PostTypeId="2" ParentId="13312" CreationDate="2015-05-13T06:45:02.127" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Not imaginative? Then Zeke's Clues may assist! Unsure, lad? With our patented Obscurity Clues, begin verifiably outwitting the great VIGENÈRE! My useful helper, Mr. Riddler, envisions heavenly light revealing clever elements in grabbers. For 6 Krones, cousin Kryptos has made plenty new devotees. But taste softly, victim, this advice: GO ACROSTIC. DOMAIN solves everything. Please review wise masters (xenophobic sociopaths) very selectively. Some KEYS percolate behind masks of kingly visages.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Use Vignere cipher&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Acrostic (first letters of every word):&lt;br&gt;NiTZCmaUlWopOCbvotgVMuhMRehlrceigFKcKhmpndBtsvtaGADsePrwmxsvsSKpbmokv&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Using FQDN &quot;httppuzzlingstackexchangecom&quot; as key I get:&#xA; &quot;VcNPShaUxFcwHWcyzyeYUvvTWhwyzwyywAKcWqawgxCwdardOBRzjSgjurmliNKpnvcro&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12058" LastEditorUserId="12058" LastEditDate="2015-05-13T13:16:50.070" LastActivityDate="2015-05-13T13:16:50.070" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14863" PostTypeId="2" ParentId="14430" CreationDate="2015-05-13T07:40:10.060" Score="13" Body="&lt;p&gt;It will be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; at least 22000 years&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;before the Lizard-men can begin their eviction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sequence of sub-clauses they must obey is finite, by &lt;a href=&quot;http://en.wikipedia.org/wiki/Higman%27s_lemma&quot; rel=&quot;nofollow&quot; title=&quot;Higman&amp;#39;s Lemma&quot;&gt;Higman's Lemma&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of the last sub-clause has the form&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; all twos: $222...2$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;by the argument in Angkor's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll divide each clause into three parts:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the &lt;strong&gt;$1$-prefix&lt;/strong&gt;, of the form $1^{(n)}$, consisting of the substring of all the $1$s at the beginning of the clause.  The $1$-prefix may be empty.  Claused will be divided into &lt;strong&gt;tiers&lt;/strong&gt; based on the length of the $1$-prefix.  A tier-$n$ clause has $1$-prefix of length $n$.&lt;/li&gt;&#xA;&lt;li&gt;the &lt;strong&gt;central part&lt;/strong&gt;, $C$, of the form $2^{(m)}$, which immediately follows the $1$-prefix&lt;/li&gt;&#xA;&lt;li&gt;the &lt;strong&gt;suffix&lt;/strong&gt;, $S$, consisting of the remainder of the clause.  $S_k$ indicates the suffix of the $k^{th}$ clause in a tier&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All clauses have the form $1^{(n)}2^{(m)}S$, for some (possibly $0$) values of $n$ and $m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Call a clause or subsequence of a clause &lt;strong&gt;forbidden&lt;/strong&gt; if it contains a previous clause as a subsequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some observations:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Observation 1: &lt;strong&gt;The length of the $1$-prefix never increases.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be seen by applying the argument in Angkor's answer to just the substring of each clause starting from the first $2$ to the end of the clause.  This shows that the index of the first $2$ cannot increase, and thus the $1$-prefix length cannot increase.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the $1$'s-prefix length does not increase, then the tier of the clauses int he sequence does not increase.  The sequence of clauses will consist of all the tier-$7$ clauses (just $1111111$), then all the tier-$6$ clauses, then the tier-$5$ clauses, and so on down to the tier-$0$ clauses, which are those that start with a $2$. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Observation 2: &lt;strong&gt;The final clause in tier-$n$ is of the form $1^{(n)}2^{(a_n)}$ for some $a_n$&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This again follows from applying Angkor's argument to the substring starting from the first $2$.  $a_n$ is the number of $2$s in the final clause of tier-$n$, and has a different value for different tiers.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Observation 3: &lt;strong&gt;The first $1$ in a clause in tier-$n$ after the $1$-prefix must occur no more than $a_{n+1}$ places from the end of the clause.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose this were not the case.  Then such a clause clause would have the form $1^{(n)}2^{+}1X$, where $2^+$ indicates a string of one or more $2$s and $X$ is some string of $1$s and $2$s of length at least $a_{n+1}$.  This clause has a subsequence of the form $1^{(n+1)}Y$, with $Y$ of length $a_{n+1}$.  This is formed by removing all of the $2$s between the $1$-prefix and the next $1$, and taking $Y$ as some subsequence of $X$ of length $a_{n+1}$.  This subsequence has length $(n+1) + a_{n+1}$, the same as the final clause in tier-$(n+1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This subsequence is forbidden.  If it is identical to the final clause in tier-$(n+1)$, then it is clearly forbidden.  If it is not, then $Y$ contains at least one $1$, and thus this subsequence comes earlier alphabetically than the last clause of tier-$(n+1)$.  This clause must have been forbidden at that point in the sequence of clauses, or else it would have been used there instead of the clause of the form $1^{(n+1)}2^{(a_n)}$.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'll now define the suffix $S$ of a clause in tier-$n$ to be of length $a_{n+1}$.  This is possible based on the previous observation and the definition of the center of a clause as a string of $2$s between the $1$-prefix and the suffix.  Only the suffix can contain a $1$ outside of the $1$-prefix, and such a $1$ can be no more than $a_{n+1}$ places from the end of the clause. Thus, I define the suffix to be the part of a clause that may contain a $1$ outside of the $1$-prefix.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Observation 4: &lt;strong&gt;Suffixes for clauses in the same tier appear in alphabetical order.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose in tier-$n$ we have clauses $X_a$ with suffix $S_a$ and $X_b$ with suffix $S_b$, $a&amp;lt;b$, but $S_b$ comes before $S_a$ alphabetically.  Generate a new clause $X'_a$ by replacing $S_a$ with $S_b$ in $X_a$.  $X'_a$ is a subsequence of $X_b$, and so is allowed if $X_b$ is allowed, but it comes before $X_a$ alphabetically, and so would have been used instead of $X_a$ if allowed. Therefore, suffixes within a tier appear in alphabetical order. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Observation 5: &lt;strong&gt;The length of the center, $C$, of the $k^{th}$ clause in tier-$n$ is $k+1$.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All clauses in tier-$n$ are of the form $1^{(n)}2^{(m)}S$.  $n$ is fixed within a tier, and the length of $S$ is fixed within a tier.  Since the length of the clauses must increase, the length of the center of the clause must increase, since it is the only part of the clause not fixed in length.  In particular, since the final clause of tier-$(n+1)$ is of the form $1^{(n+1)}2^{(a_{n+1})}$ and has length $n + a_{(n+1)} + 1$, the length of the first clause in tier-$n$ must have length $n + a_{(n+1)} + 2$.  The $1$-prefix has length $n$, the suffix has length $a_{(n+1)}$, so the center must have length 2 = $k+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the $k^{th}$ clause in tier-$n$ must have length $n + a_{n+1} + k + 1$, so the center of that clause must have length $k+1$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a consequence, if there are $K_n$ total clauses in tier-$n$, then $a_n = K_n + 1 + a_{n+1}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Observation 6: &lt;strong&gt;If a clause $X_a$ with suffix $S_a$ in tier-$n$ is contained as a subsequence in another tier-$n$ clause $X_b$ with suffix $S_b$, then $S_a = S_b$.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X_a$ has the form $1^{(n)}2^{(m_a)}S_a$, and $X_b$ has the form $1^{(n)}2^{(m_b)}S_b$, with $m_a&amp;lt;m_b$.  $1^{(n)}2^{(m_b)}$ clearly contains $1^{(n)}2^{(m_a)}$ as a prefix, so if $X_a$ is a subsequence of $X_b$, then $S_a$ is a subsequence of the remainder of $X_b$, which is of the form $2^{(m_b-m_a)}S_b$.  This is just $S_b$ with a number of extra $2$s prepended.  Since $S_b$ comes after $S_a$ alphabetically by observation 4, these extra $2$s don't need to be used, and so $X_a$ is a subsequence of $X_b$ only if $S_a$ is a subsequence of $S_b$.  Since the length of $S_a$ and $S_b$ are the same, this implies that $X_a$ is a subsequence of $X_b$ only if $S_a=S_b$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;All of the interesting variation of the clauses within in tier happens in the suffix of each clause.  We can now look at bounding how many allowed suffixes are in tier-$n$.  It turns out that we can construct allowed suffixes of clauses in tier-$n$ from clauses in tier-$(n+1)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observation 7: &lt;strong&gt;If $CS$ is the concatenated center and suffix of a clause in tier-$(n+1)$, with $C$ of the form $2^{k+1}$, then we can construct $k$ valid suffixes for clauses in tier-$n$.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S_k$ be the suffix for the $k^{th}$ clause in tier-$n+1$.  Then consider strings $\hat{S}$ of the form $2^{(d)}2^{(b)}12^{(c)}S_k$, where $b+c=k$, $c&amp;gt;0$, and $d$ is chosen so that the length of the string is $a_{(n+1)}$.  These strings are formed by changing one of the $2$s in the center (except the last one) to a $1$, then padding with enough $2$s so that the length is equal to the suffix length for tier-$n$.  For each $S_k$, there are $k$ of these strings, and all of them are unique.  This is a total of $\frac{K_{(n+1)}(K_{(n+1)}+1)}{2}$ candidate suffixes for tier-$n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For tier-$n$ clauses of the form $1^{(n)}2^{(m)}\hat{S}$, no clause in a higher tier can be a subsequence.  Such clauses have the form $1^{(n)}2^{(b)}12^{(c)}S_k$, with $c&amp;lt;k+1$.  All clauses in higher tiers begin with at least $n+1$ $1$s, so the first run of $2$s will not be used in any forbidden subsequence.  A forbidden subsequence must be a subsequence of $X'=1^{(n+1)}2^{(c)}S_k$.  $c$ was chosen to be less than $k+1$.  The clause $1^{(n+1)}2^{(k+1)}S_k$ appears in tier-$(n+1)$, and so is not forbidden, and has $X'$ as a subsequence, so $X'$ cannot be forbidden.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each $\hat{S}$ was chosen to be unique, so by observation 6, no clause in tier-$n$ can contain another as a subsequence.  Therefore, every one of these candidate suffixes can appear in tier-$n$.  There may be additional possible suffixes that do not take the specified form, but this should be sufficient to establish some bounds.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;We can now put a bound on the total number of clauses.  Let's manually compute some possible clauses in tier-$4$:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 6     111122122222&#xA; 7     1111222212221&#xA; 8     11112222212222&#xA; 9     111122222221212&#xA;10     1111222222221221&#xA;11     11112222222221222&#xA;12     111122222222222122&#xA;13     1111222222222222212&#xA;14     11112222222222222221&#xA;15     111122222222222222222&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We can inspect that $K_4=10$ (&lt;strong&gt;NOTE:&lt;/strong&gt; Angkor's update shows that I missed some tier-$4$ clauses and that $K_4$ is actually $13$.  My estimate can still be computed on the assumption that $K_4$ is at least $10$).  From there, we can compute the following bounds:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $K_3\ge\frac{K_4(K_4+1)}{2}\ge\frac{10\cdot11}{2}=55$        &lt;br/&gt;&lt;br&gt;&#xA; $K_2\ge\frac{K_3(K_3+1)}{2}\ge\frac{55\cdot56}{2}=1540$      &lt;br/&gt;&lt;br/&gt;&#xA; $K_1\ge\frac{K_2(K_2+1)}{2}\ge\frac{1540\cdot1541}{2}=1186570$         &lt;br/&gt;&lt;br&gt;&#xA; $K_0\ge\frac{K_1(K_1+1)}{2}\ge\frac{1186570\cdot1186571}{2}=703974775735$   &lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At one second per clause, it will take &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; at least 22300 years&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;to process every clause. And this isn't a tight bound, since at the very least we didn't include the clause of the form $1^{(n)}2^{(m)}$ in the estimate!  It will likely be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; significantly longer than that.  Anyway, I'm not worried about an impending lizard-man invasion.  And hopefully humanity can invent a tachyon transmitter sometime in the next twenty millenia and file an appeal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10261" LastEditorUserId="10261" LastEditDate="2015-05-14T17:04:24.360" LastActivityDate="2015-05-14T17:04:24.360" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14864" PostTypeId="1" AcceptedAnswerId="14865" CreationDate="2015-05-13T08:02:12.850" Score="11" ViewCount="8411" Body="&lt;p&gt;I remember this from when I was in highschool (or the dutch equivolant).&#xA;The teacher of my english class proposed this riddle and a lot of people had it wrong but the answer was really simple. Remember in holland english isn't our native language.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the correct sentence?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The egg yolk is white&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The egg yolk are white&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have fun and don't spoil&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Look at the mistakes in the text, they are there for a reason :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12218" LastEditorUserId="4050" LastEditDate="2015-05-15T08:33:30.433" LastActivityDate="2016-02-09T20:20:50.603" Title="Egg Yolk Riddle" Tags="|word|lateral-thinking|english|language|" AnswerCount="3" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14865" PostTypeId="2" ParentId="14864" CreationDate="2015-05-13T08:08:40.923" Score="14" Body="&lt;p&gt;All sentences are incorrect except for &lt;em&gt;&quot;What is the correct sentence?&quot;&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Look at the mistakes: &lt;em&gt;&quot;dutch&quot;&lt;/em&gt;, &lt;em&gt;&quot;english&quot;&lt;/em&gt; and &lt;em&gt;&quot;holland&quot;&lt;/em&gt; should have their first letter capitalized. &lt;em&gt;&quot;equivolant&quot;&lt;/em&gt; is a misspelling of equivalent.&#xA; Also: the egg yolk is singular and it's not white; and the last sentence is missing a period at the end.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4394" LastEditorUserId="4394" LastEditDate="2015-05-13T14:34:02.623" LastActivityDate="2015-05-13T14:34:02.623" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14867" PostTypeId="2" ParentId="14864" CreationDate="2015-05-13T08:19:09.823" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Neither, the egg yolk is yellow&quot; would be the correct answer, if it was not a trick question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10748" LastEditorUserId="10748" LastEditDate="2015-05-13T23:55:35.797" LastActivityDate="2015-05-13T23:55:35.797" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14869" PostTypeId="1" AcceptedAnswerId="14932" CreationDate="2015-05-13T09:26:51.580" Score="34" ViewCount="2366" Body="&lt;p&gt;You are a secret agent in the service of the KGB, and after a &lt;a href=&quot;https://puzzling.stackexchange.com/questions/11088/a-mysterious-email-from-andrew-void&quot;&gt;setback&lt;/a&gt; when the higher-ups thought your cover had been blown, you have embarked on a highly dangerous mission to infiltrate MI6. Under the pseudonym of Flynn Smith, you are working undercover in London when you receive the following mysterious email.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;From: &quot;itinerary@ryanair.com&quot; &amp;lt; itinerary@ryanair.com &gt;&lt;br&gt;&#xA;  Sent: Wed, 13 May 2015 10:27AM +0400&lt;br&gt;&#xA;  To: ██████████████████████&lt;br&gt;&#xA;  Subject: Ryanair Travel Itinerary&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Dear Mr Smith,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;THANK YOU FOR BOOKING WITH RYANAIR.&lt;br&gt;&#xA;  THIS IS YOUR BOOKING CONFIRMATION, PASSENGER ITINERARY AND RECEIPT EMAIL. PLEASE CHECK YOUR FLIGHT(S) AND PASSENGER(S) DETAILS BELOW.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;MSYSTLMSPORDSTLTPA RYGMUCZAGBUDHEM ADADMEIKAUFA KBLTSEKTMURC XNNDACRGNLPFXNN CGKPERDRWADLBNE&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;DO NOT REPLY TO THIS EMAIL.&lt;br&gt;&#xA;  &lt;sub&gt;Copyright 2014 Ryanair Ltd. All rights reserved. Cheap Flights - Lowest European Fares, Low Cost Airline.&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are confused: with all your expenses paid by the intelligence service, you never fly with a budget airline company like Ryanair. Perhaps they've got you mixed up with another Mr Smith - but how come your email address is on their records? Also there aren't even any ticket details in the email, only a nonsensical string of letters. The message must have been garbled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But remembering your &lt;a href=&quot;https://puzzling.stackexchange.com/questions/11088/a-mysterious-email-from-andrew-void&quot;&gt;previous experience&lt;/a&gt; of an apparently spam email, you examine it more closely, wondering if there's a message hidden within it. After several minutes at your computer, you find the hidden message and your eyes widen. You reach for the drawer in which your handgun is stowed; you're in danger!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking even more closely, you realise that there's a second message hidden within the more obvious one. Your KGB agent handler is telling you where you must go - &lt;em&gt;and&lt;/em&gt; probably also that you're in danger. It seems as though everything's under control, but you must move fast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the hidden message?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2015-05-14T09:25:07.523" Title="A mysterious email from Ryanair" Tags="|cipher|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14870" PostTypeId="1" AcceptedAnswerId="14940" CreationDate="2015-05-13T09:28:00.250" Score="9" ViewCount="1135" Body="&lt;p&gt;It's my birthday today, 5/13! Below are some of my role models. Can you tell me when their birthdays are? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0,9,8,9,8,5,0,9$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0,-7,0,8,0,8,12,-7,0,8,0,8,12$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0,7,7,7,-8,-5,7,4,7,12,7,7,7$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0,5,9,5,4,5,7,5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\_\_,\_\_,\_\_,\_\_,\_\_,\_\_,\_\_,\_\_$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know what I really want for my birthday? Probably a pair of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Women’s UA PIP® Micro G® Pulse Running Shoes!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;****Not part of the puzzle: Let me know if the shoes don't exist in your region of the world. ****&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope they're not sold out! Could you check for me? &lt;em&gt;nudge&lt;/em&gt; &lt;em&gt;wink&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Meta:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to the puzzle is a single word. &lt;em&gt;EDIT&lt;/em&gt;: the word may have nothing whatsoever to do with the context of the puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry if the puzzle is too hard. Certain types of people will be extremely advantaged in this puzzle. This is my first puzzle of this kind; I was unsure what to tag it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clarifications:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The length of the sequences is irrelevant. They could have gone on much longer, but I arbitrarily ended them when I thought there was enough information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google is allowed and most likely extremely necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1 (edited, no spoilers):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Look at the title. Also, the order the sequences appear in is important.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2 (no spoilers):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first sequence is (subjectively) the easiest. If you solve that one, you will have a clear path to solving the others. Again, a very specific certain set of people will find this much easier.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3 (spoilers but probably necessary):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If $c=0$, then the numbers $10,1,3,1$ respectively, could be useful. You might want to become friends with a certain device that can't unlock anything despite its many keys.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 4 (edited, spoilers):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The shoes give an seven digit number to manipulate the eight letter sequence you just found. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Extra Credit:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wow, the answer is a really cool word! It's the __.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5778" LastEditorUserId="5778" LastEditDate="2015-05-14T18:48:32.030" LastActivityDate="2015-05-22T19:12:35.480" Title="A Composition For My Birthday" Tags="|riddle|pattern|enigmatic-puzzle|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14871" PostTypeId="1" AcceptedAnswerId="14889" CreationDate="2015-05-13T09:30:06.270" Score="4" ViewCount="1066" Body="&lt;p&gt;Another twist to &lt;a href=&quot;https://math.stackexchange.com/questions/1105531/n-perfect-logicians-wearing-hats&quot;&gt;this&lt;/a&gt; question. I'll restate it here to avoid confusion&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There are N perfect logicians arranged in a vertical row. They are&#xA;  allowed to strategize before the game, during the game they are barred&#xA;  from communicating. In the game, the lights are switched off, and a&#xA;  single hat, either red or blue, is placed on the heads of each of the&#xA;  logicians. The lights are turned on; each logician can see the hat&#xA;  colours of all the people in front of him, he can't see the colour of&#xA;  his own hat, or the people behind him. Start with the person at the&#xA;  back, each logician has to call out a colour (heard by all). If his&#xA;  own hat is of that colour, he is spared, otherwise, silently killed.&#xA;  The same is done by the next person, and so on, right till the first&#xA;  person. How many logicians can you guarantee to save, irrespective of&#xA;  the initial colours or any probability?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Please see the answer to this on the same link.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now the twist....&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose there are &lt;strong&gt;2&lt;/strong&gt; blind man among the logicians. Their lives are&#xA;  worth as much as anyone else, and are placed randomly in the queue.&#xA;  How many logicians can you guarantee to save?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The solution may still involve some sort of binary....&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="-1" LastEditDate="2017-04-13T12:19:36.950" LastActivityDate="2015-09-28T16:58:32.273" Title="2 blind men in a row of logicians" Tags="|logical-deduction|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14872" PostTypeId="1" AcceptedAnswerId="14876" CreationDate="2015-05-13T09:37:49.877" Score="12" ViewCount="192520" Body="&lt;p&gt;Take away my first letter, and I still sound the same. &#xA;&lt;br&gt;Take away my last letter, I still sound the same. &lt;br&gt;Even take away my letter in the middle, I will still sound the same. &lt;br&gt;I am a five letter word. &lt;br&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="12218" LastActivityDate="2017-02-22T12:55:50.737" Title="5 Letter Word, Removing letters" Tags="|riddle|word|word-problem|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14873" PostTypeId="1" AcceptedAnswerId="14874" CreationDate="2015-05-13T09:47:27.467" Score="-3" ViewCount="322" Body="&lt;p&gt;Bury me deep, pile on every stone.&lt;br&gt;&#xA;Yet in the end I will dig up every bone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can make you feel hurt or make you smile,&lt;br&gt;&#xA;Accepting me will sometimes take a while.&lt;/p&gt;&#xA;" OwnerUserId="12218" LastEditorUserId="12218" LastEditDate="2015-05-13T09:53:38.283" LastActivityDate="2016-12-07T22:48:07.130" Title="Guess what is described here" Tags="|riddle|rhyme|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14874" PostTypeId="2" ParentId="14873" CreationDate="2015-05-13T09:50:24.387" Score="2" Body="&lt;p&gt;My answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A memory perhaps.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bury me deep, pile on every stone.&#xA;Yet in the end I will dig up every bone. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Trying to repress a memory, but it just won't fully go away.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can make you feel hurt or make you smile,&#xA;Accepting me will sometimes take a while.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A memory can bring out these feelings. Being at peace with a memory can sometimes take some time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11550" LastActivityDate="2015-05-13T09:50:24.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14875" PostTypeId="1" AcceptedAnswerId="14897" CreationDate="2015-05-13T10:02:24.377" Score="1" ViewCount="454" Body="&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;I am all in one, &lt;/li&gt;&#xA;  &lt;li&gt;I am part of Mozilla,&lt;/li&gt;&#xA;  &lt;li&gt;Part of the Olympic Games,&lt;/li&gt;&#xA;  &lt;li&gt;Although I am unwilling to work and would like to eat a hotdog,&lt;/li&gt;&#xA;  &lt;li&gt;Speed is the second part of all in one. (1)&lt;/li&gt;&#xA;  &lt;li&gt;Color which is not same as color of part of Mozilla  (2)&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All the clue have one or two unique words which you have to find out from all points, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combine and arrange those words and find out the thing which is &quot;ALL in One.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) and (2) means it's related to that point.&lt;/p&gt;&#xA;" OwnerUserId="4312" LastEditorUserId="4312" LastEditDate="2015-05-27T09:41:17.910" LastActivityDate="2015-05-27T09:41:17.910" Title="I am all in one, guess what?" Tags="|riddle|wordplay|" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14876" PostTypeId="2" ParentId="14872" CreationDate="2015-05-13T10:13:21.503" Score="15" Body="&lt;p&gt;This is an old one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; EMPTY&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MPTY still sounds like EMPTY,&lt;br&gt; EMPT still sounds like EMPTY,&lt;br&gt;and EMTY still (sort of) sounds like EMPTY.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And, of course, you can &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; take all 3 away and M T is left.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7308" LastEditorUserId="11876" LastEditDate="2015-09-09T00:53:24.170" LastActivityDate="2015-09-09T00:53:24.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14877" PostTypeId="1" AcceptedAnswerId="14896" CreationDate="2015-05-13T10:58:05.107" Score="4" ViewCount="839" Body="&lt;p&gt;Energy can only be manipulated and moved, but never created or destroyed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two halves, they complement each other,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yet are complete opposites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wish I was a channeler, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Born with a &quot;spark&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But here I am asking other questions,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and answering them as best I can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's my &quot;spark&quot; I guess.&lt;/p&gt;&#xA;" OwnerUserId="11814" LastEditorUserId="11814" LastEditDate="2015-05-13T22:45:44.117" LastActivityDate="2015-05-13T22:45:44.117" Title="Guess who I am: Born with a &quot;spark&quot;" Tags="|riddle|knowledge|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14879" PostTypeId="1" AcceptedAnswerId="14924" CreationDate="2015-05-13T11:45:43.437" Score="4" ViewCount="374" Body="&lt;p&gt;A man of war&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A ball of carbon without pastor nor full hurricane&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A spider&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One whose strength is in the same area as Veralidaine Sarrasri&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A toy maker of sorts,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Jason-type&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And one whose pallor is that of Briar Moss' magic.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A man whose names vary from Charles to James to Patrick,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Someone who is Nail Product&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He that appears as the half dead person, one soul in two bodies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;El verano, though quite a bit more dangerous&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An icy animal&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And one who cries naught forevermore&lt;/p&gt;&#xA;" OwnerUserId="11790" LastEditorUserId="11790" LastEditDate="2015-05-13T20:02:51.077" LastActivityDate="2015-05-13T23:58:37.287" Title="Two groups of people- will they meet, ever?" Tags="|riddle|knowledge|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="14880" PostTypeId="2" ParentId="13303" CreationDate="2015-05-13T12:05:43.630" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Start at FE (Paired up: Two greenlands, pared down: End of map)&lt;br&gt;&#xA; Walk throught TE (truth exposed)&lt;br&gt;&#xA; to IE (lies in lies)&lt;br&gt;&#xA; So north is on the left hand side and we go on...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cardinals:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We have the cardinal directions North, East, South, West&lt;br&gt;&#xA; Cardinals in Nests : N E S S&lt;br&gt;&#xA; so we move North, East, South, South&lt;br&gt;&#xA; and look west in the end (turn to what's undone)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12384" LastEditorUserId="12384" LastEditDate="2015-05-13T12:12:19.273" LastActivityDate="2015-05-13T12:12:19.273" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14881" PostTypeId="2" ParentId="14877" CreationDate="2015-05-13T12:23:21.267" Score="2" Body="&lt;p&gt;I think the answer is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; yin and yang&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Energy can only be manipulated and moved, but never created or destroyed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are depended on one an other and are always in balance.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The two halves, they compliment each other,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; well that's the whole idea about it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And yet are complete opposites.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; good and evil, light and dark, fire and water, and male and female)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I wish I was a channeler,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; what is a channeler? someone that can channel both good and evil 'spirits'?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Born with a &quot;spark&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But here I am asking other questions,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and answering them as best I can.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it is an answer to what life is&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's my &quot;spark&quot; I guess.&lt;/p&gt;&#xA;" OwnerUserId="9835" LastEditorUserId="9835" LastEditDate="2015-05-13T12:49:32.553" LastActivityDate="2015-05-13T12:49:32.553" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14882" PostTypeId="2" ParentId="14875" CreationDate="2015-05-13T12:34:26.730" Score="9" Body="&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The quick brown fox jumps over the lazy dog.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;All alphabets: All in one.&#xA;&lt;br&gt;Mozilla: Firefox: Fox.&#xA;&lt;br&gt;Olympic: Jumping.&#xA;&lt;br&gt;Hotdog: Dog.&#xA;&lt;br&gt;Speed: Quick.&#xA;&lt;br&gt;Color: Brown.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12058" LastEditorUserId="12193" LastEditDate="2015-05-13T13:17:16.983" LastActivityDate="2015-05-13T13:17:16.983" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14883" PostTypeId="2" ParentId="14877" CreationDate="2015-05-13T12:45:01.330" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Potential and Kinetic Energy.&#xA; They do complement. Two halves (metaphorically) of Mechanical Energy (Ideal Case, where it is equal to Total Energy). Potential Energy can be the channeler.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12058" LastActivityDate="2015-05-13T12:45:01.330" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14884" PostTypeId="2" ParentId="14871" CreationDate="2015-05-13T13:15:17.447" Score="1" Body="&lt;p&gt;Assuming the logicians can tell who the blind men are (behind and in front), they could essentially have &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3 (outside chance of 2) groups that behave like the originating question and end with a blind man.&lt;br&gt;&#xA;&lt;br&gt;&#xA; So, first guy gives parity bit, all the next seeing guys give their guess, then the blind guy knows his bit. Next seeing guy gives parity bit, etc. You can get down to 2 groups if one of the blind guys is at the end. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-05-13T13:25:17.173" LastActivityDate="2015-05-13T13:25:17.173" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14885" PostTypeId="1" AcceptedAnswerId="14900" CreationDate="2015-05-13T13:33:34.617" Score="17" ViewCount="1402" Body="&lt;p&gt;I am 10, and I mostly use &lt;strong&gt;2&lt;/strong&gt;, but sometimes &lt;strong&gt;8&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My older brother is 12, and he uses either &lt;strong&gt;2&lt;/strong&gt; or &lt;strong&gt;4&lt;/strong&gt;, depending on his mood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My younger brother is 3, and he uses &lt;strong&gt;3&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My sister is 19, and she usually uses &lt;strong&gt;4&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My mom is 37, and she also usually uses &lt;strong&gt;4&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My dad is 40, and he uses either &lt;strong&gt;4&lt;/strong&gt; or &lt;strong&gt;6&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My uncle is 42, and he uses either &lt;strong&gt;4&lt;/strong&gt;, &lt;strong&gt;10&lt;/strong&gt;, or &lt;strong&gt;18&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is this family using?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastActivityDate="2015-05-13T14:54:59.060" Title="What is this family using?" Tags="|riddle|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14886" PostTypeId="2" ParentId="14785" CreationDate="2015-05-13T13:35:59.973" Score="1" Body="&lt;p&gt;Assumptions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The problem is set in an Earth-like environment&lt;/li&gt;&#xA;&lt;li&gt;All containers are upright and have constant cross-section along their height&lt;/li&gt;&#xA;&lt;li&gt;The test liquid is in thermodynamical equilibrium with the surrounding atmosphere&lt;/li&gt;&#xA;&lt;li&gt;The liquid supply delivers liquid at the same temperature as the test liquid&lt;/li&gt;&#xA;&lt;li&gt;It is possible to fill a container to a fraction of $1/2^{n\text{th}}$ for any $n\in \mathbb{N}$ as per Kevin's assumptions&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Idea:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set yourself a limit of how accurate your measurement should be in units of $x$ and call the acceptable error $\varepsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the smallest $n\in\mathbb{N}$, such that $\varepsilon &amp;gt; 1/2^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pour the test liquid into a container and fill another container up to a fraction of $1/2^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wait until the liquid in one of the containers has fully evaporated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is the one with $1/2^n$ filling status, refill it immediately, set aside an empty container as token, and wait again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, if it is the container with the test liquid, count the number of token containers. The test liquid had the volume of that number divided by $2^n$, multiplied with $x$.&lt;/p&gt;&#xA;" OwnerUserId="1831" LastActivityDate="2015-05-13T13:35:59.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14887" PostTypeId="2" ParentId="14802" CreationDate="2015-05-13T13:43:12.643" Score="-3" Body="&lt;p&gt;Well, I don't think this is the intended solution but I think it's technically correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The linguistic property they all share:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;They are all English words&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1st set&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;They are all English words that are verbs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2nd set&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;They are all English words having the letter S&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3rd set&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;They are all English words beginning with the letter C&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-05-13T13:43:12.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14888" PostTypeId="2" ParentId="14885" CreationDate="2015-05-13T13:45:22.317" Score="0" Body="&lt;p&gt;My guess is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;b&gt;Shoes&lt;/b&gt;&#xA;&lt;br&gt; Each family member can have a different number of pairs of shoes (running, indoor, sports, sandals, etc.), although this might be a lot of shoes for the uncle.&#xA;&lt;br&gt;&lt;br&gt;Alternatively it could be the number of shoes each has, and the 3 year old has 3 shoes because he is a silly young child.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastActivityDate="2015-05-13T13:45:22.317" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14889" PostTypeId="2" ParentId="14871" CreationDate="2015-05-13T13:50:17.250" Score="4" Body="&lt;p&gt;The answer is to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Use the &lt;em&gt;exactly same&lt;/em&gt; strategy. This gets a maximum of $5$ people killed, irrespective of $n$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There are $c^N$ different arrangements of hats where $N$ is the number of logicians and $c$ is the number of colored hats ($c=2$ in this case but the explanation is easier without that).  The positions are labelled $1,2,3,...,N$.  The colors are also $1,2,3,...,c$ which in the case are $1$ and $2$ for red and blue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The logician in position $1$ knows nothing about his hat and only knows about the others in front of him.  Therefore, he always has a $\frac 1 c$ chance of surviving.  He tries to save the others which means (in the traditional answer) he says a color which indicates what the sum of the hats in front of him is $\mod c$.  Here he says red if there are an odd number of red hats in front of him and blue otherwise.  This information is enough to allow all people in other positions to survive no matter what c is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Effectively, there are $N$ pieces of information needed to be supplied to allow all players to survive as the system can be described as a system of $N$ equations.  As each player (excluding the one in position N) can (and does) supply $1$ piece of information, $N-1$ are supplied so $N-1$ can be guarenteed to survive if the players are clever.  The first person says a relevant color so on average $\frac{1}{c}$ die. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are 2 blind people &lt;strong&gt;and their positions are known&lt;/strong&gt;, you still need $N$ pieces of information but you can only collect $N-3$.  This means $\frac 3 c$ can survive on average.  The easiest way to do this is by having all players ignore the blind ones while doing their summations and allowing the blind people to guess.  This can also be done by having the summations end at the blind person and after the blind person start the pattern afresh.  This results in a minimum of $N-3$ survivors with either strategy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are 2 blind people &lt;strong&gt;and their positions are not known&lt;/strong&gt;, the positions of the blind people at as additional required information so more than $\frac{3}{c}$ will die. When the blind person comes along he can say a colored hat which isn't in the game.  This would reset the game after the blind person and kill him.  Assuming the blind people's positions are unfavorable, this results in $2+\frac{3}{c}$ average deaths for a minimum of $N-5$ survivors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we can modify the color a little and have blind people guess crimson and azure instead of red and blue, the average number of survivors (for an infinite number of logicians to neglect end effects) is $\frac 5 c$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Surprising Answer with the same average result!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have surprisingly found a shocking fact; &lt;strong&gt;two wrongs do in fact make a right&lt;/strong&gt; with the standard algorithm.  Here red and blue are equivalent to $1$ and $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We follow the standard algorithm where the back player states (correctly or incorrectly) that he is wearing a hat which matches the parity of all of the other hats he can see.  All other player then know they are wearing the hat that has the parity equivalent to the sum of the hats that have been stated plus those they can see.  If C is greater than 2, (white, black, or brown hats) the player should guess the parity equivalent to the hat color of the first person minus the sum of (all hat colors you heard and all the hat colors you see).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Blind people will randomly guess what their hat color is with no regard to whoever else is around them.  If the blind person is incorrect, the player after him is also incorrect.  While both die, this mathamagically corrects the parity of the summmation all players after them (besides blind people) use to determine their hat colors.  This means without correction, exactly two people will die from an incorrect answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets define the following for a mathamatical explanation for $C$ colors.  $S_o$ is equal to the first person's guess.  $S_b$ is the sum before the blind person. $S_a$ is the sum starting with the 3rd people after the blind person.  $H_b$ is the blind person's hat and $G_b$ is his guess.  $H_v$ and $G_v$ refers to the person after the blind person. $H_l$ and $G_l$ refer to logician after that. $E$ refers to the error the blind person if off by and would be zero if he is correct. Note that if we reduce $C$ to $2$ this holds and the simpler equation is equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H_b=S_o-S_b-S_a-H_l-H_v \mod C$$&#xA;$$G_b=H_b+E \mod C$$&#xA;$$H_v=S_o-S_b-S_a-H_b-H_l \mod C$$&#xA;$$G_v=S_o-S_b-S_a-G_b-H_l \mod C$$&#xA;$$G_v = H_v-E \mod C $$&#xA;$$H_l=S_o-S_b-S_a-H_b-H_v \mod C$$&#xA;$$G_l=S_o-S_b-S_a-G_b-G_v \mod C$$&#xA;$$G_l=H_l \mod C$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If two blind people are adjacent or a blind person it at the start, this autocorrecting behavior of the equation still holds.  Usually this means fewer people die and never means more. Therefore, no more than 2 people per blind person die so this basic strategy means that there are a maximum of $1+2B$ deaths where $B$ is the number of blind people.  On average $\frac{1+2B}{C}$ survive assuming pessimal original positions.  This is all independant of $C$.  For $B=2$ there are $N-5$ guarenteed survivors.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="4269" LastEditDate="2015-09-28T16:58:32.273" LastActivityDate="2015-09-28T16:58:32.273" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14890" PostTypeId="2" ParentId="14870" CreationDate="2015-05-13T13:57:35.363" Score="-1" Body="&lt;p&gt;This converted to notes becomes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; c,e+,d+,e+,d+,a,c,e+,c,c,c,d+,c,d+,d,e,c,c,d+,c,d+,d,e,c,c,c,c,b,e,c,g,c,d,e,c,c,c,c,a,e+,a,g,a,c,a,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; + for next octave and - for previous octave.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12058" LastActivityDate="2015-05-13T13:57:35.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14891" PostTypeId="1" AcceptedAnswerId="14895" CreationDate="2015-05-13T14:07:08.717" Score="2" ViewCount="2863" Body="&lt;p&gt;Warning: This sounds like one you've heard before but it's not.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A hunter leaves his camp and heads one mile north. Scanning the horizon for prey, he spots a likely spot to the east. He walks for a half mile before spotting movement further east. He sprints the half mile only to find that whatever was there is gone. Downtrodden, he walks the one mile south back to camp. Opening the door of the hut, he begins to call out for his partner when he sees it: A live bear in the hut! Thinking quickly, he pulls out his rifle and shoots it dead. Hearing the gunshot, his partner emerges from the other room. &quot;Hey!&quot; he shouts, &quot;I just captured that bear outside and brought it in here to study!&quot; Sheepishly, the hunter admits that he probably reacted too hastily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many legs did the bear have?&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2016-08-31T15:28:24.380" LastActivityDate="2017-12-18T12:39:05.037" Title="How many legs does the bear have?" Tags="|logical-deduction|knowledge|" AnswerCount="4" CommentCount="20" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14892" PostTypeId="2" ParentId="13312" CreationDate="2015-05-13T14:18:24.030" Score="12" Body="&lt;p&gt;Treating the clue as an acrostic, as implied in the text, we get the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; NITZCMAULWOPOCBVOTGVMUHMREHLRCEIGFKCKHMPNDBTSVTAGADSEPRWMXSVSSKPBMOKV&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using the following fully qualified domain as the key:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PUZZLINGSTACKEXCHANGECOM&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; YOUARENOTDONEYETHTTPISTACKIMGURCOMKAADPNGDONOTFORGETTHEQUESTIONNUMBER&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which I think corresponds to the message:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are not done yet &lt;a href=&quot;https://i.stack.imgur.com/KAAD.PNG&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/KAAD.PNG&lt;/a&gt; do not forget the question number &lt;br&gt;&lt;br&gt;&#xA; Which is clearly some reference to an image hosted on the stack exchange imgur site.&lt;br&gt;&lt;br&gt;&#xA; These normally follow the format: &lt;strong&gt;i.stack.imgur.com/#####.###&lt;/strong&gt; &#xA; having 5 alphanumeric characters before the file extension. In which case we are a character short. &lt;br&gt;&lt;br&gt;&#xA; Including the question number could refer to the &quot;6&quot; before the Krones - so putting that in we get: &lt;strong&gt;i.stack.imgur.com/6KAAD.png&lt;/strong&gt; Following the hint to use lowercase, uppercase, uppercase, lowercase, we get: &lt;strong&gt;i.stack.imgur.com/6kAAd.png&lt;/strong&gt; which takes us to a tiger!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is as far as I got, I hope it helps someone else on the way!&lt;/p&gt;&#xA;" OwnerUserId="12206" LastEditorUserId="11475" LastEditDate="2015-05-13T14:47:56.363" LastActivityDate="2015-05-13T14:47:56.363" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14893" PostTypeId="2" ParentId="14891" CreationDate="2015-05-13T14:20:04.690" Score="3" Body="&lt;p&gt;The bear has:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; zero legs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I'm a hunter, that's the only way I would sheepishly admit that an unknown live bear inside our hut was not something worth shooting. Bears are &lt;a href=&quot;http://www.urbandictionary.com/define.php?term=bear&amp;amp;defid=1696621&quot; rel=&quot;nofollow&quot;&gt;godless killing machines&lt;/a&gt;, and even more dangerous in small areas.&lt;/p&gt;&#xA;" OwnerUserId="1563" LastEditorUserId="1563" LastEditDate="2015-05-13T14:35:52.103" LastActivityDate="2015-05-13T14:35:52.103" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14894" PostTypeId="1" CreationDate="2015-05-13T14:24:15.940" Score="0" ViewCount="90" Body="&lt;p&gt;You, a well-known submarinist criminal, are taken prisoner in a perfectly circular island, surrounded by water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You stand in the center of the island, with the submarinism equipment (don't worry about its weight).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surrounding the island there is a boat, watching you, that goes 4 times faster than you. That means that even if you go straight in the opposite direction of the boat to flee, the boat will arrive there before you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you manage to arrive to the water and the boat's not there, you'll manage to finally flee, since you'd go underwater.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question is, can you flee the island? If so, how?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: There's no problem with acceleration nor delay (yes, the boat can go from full speed to full speed in the opposite direction)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\\$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Just to sum up&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;The island is perfectly circular and you are in the center.&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;The boat is always on the border of the island, moves at 4 times your speed, and can go from full speed in one direction to full speed in the opposite.&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;If you arrive at the border of the island and the boat is not there, you are free.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12165" LastEditorUserId="12165" LastEditDate="2015-05-13T14:34:14.630" LastActivityDate="2015-05-13T14:34:14.630" Title="Freedom from the guarded island" Tags="|logical-deduction|calculation-puzzle|" AnswerCount="0" CommentCount="8" ClosedDate="2015-05-13T14:43:49.913" ContentLicense="CC BY-SA 3.0" />
  <row Id="14895" PostTypeId="2" ParentId="14891" CreationDate="2015-05-13T14:25:04.210" Score="7" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The bear has 8 legs... it is a water bear.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't know why he would shoot it but I would be embarrassed too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The camp is a short distance north of the south pole.  He walks north, walks around the pole, and then goes past the south pole on his way back to camp.  This accounts for the distance difference which can only really happen there.  Polar bears don't live at the south pole. Neither do Siberian water bears (which would be a silly guess).&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditDate="2015-05-13T14:57:44.933" LastActivityDate="2015-05-13T14:57:44.933" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="14896" PostTypeId="2" ParentId="14877" CreationDate="2015-05-13T14:28:08.487" Score="7" Body="&lt;p&gt;You can't keep a good Wheel of Time fan down...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Energy can only be manipulated and moved, but never created or destroyed.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is the law of conservation of energy (see also &lt;a href=&quot;https://puzzling.stackexchange.com/a/14655/5373&quot;&gt;here&lt;/a&gt;, which I guess motivated this question).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The two halves, they compliment each other,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://wot.wikia.com/wiki/Saidin&quot; rel=&quot;nofollow noreferrer&quot;&gt;Saidin&lt;/a&gt; and &lt;a href=&quot;http://wot.wikia.com/wiki/Saidar&quot; rel=&quot;nofollow noreferrer&quot;&gt;saidar&lt;/a&gt;, the two halves of &lt;a href=&quot;http://wot.wikia.com/wiki/One_Power&quot; rel=&quot;nofollow noreferrer&quot;&gt;the One Power&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And yet are complete opposites.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Saidin and saidar differ in many ways but are basically two halves of the same thing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I wish I was a channeler,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People who can use the One Power are called &lt;a href=&quot;http://wot.wikia.com/wiki/Channeler&quot; rel=&quot;nofollow noreferrer&quot;&gt;channelers&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Born with a &quot;spark&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some channelers are born with the &lt;a href=&quot;http://wot.wikia.com/wiki/Spark&quot; rel=&quot;nofollow noreferrer&quot;&gt;spark&lt;/a&gt;, meaning they'll inevitably begin to channel with or without training - if not found and trained e.g. by Aes Sedai, they usually die.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But here I am asking other questions,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;and answering them as best I can.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That's exactly what you're doing on the Puzzling Stack Exchange!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;It's my &quot;spark&quot; I guess.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gracelyn Rioux, a non-Aes Sedai.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://wot.wikia.com/wiki/Rand_al&amp;#39;Thor&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rand al'Thor&lt;/a&gt;, the &lt;a href=&quot;http://wot.wikia.com/wiki/Dragon_Reborn&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dragon Reborn&lt;/a&gt; and the most powerful channeler alive :-D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-05-13T20:17:17.003" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14897" PostTypeId="2" ParentId="14875" CreationDate="2015-05-13T14:34:36.757" Score="6" Body="&lt;p&gt;ADG has the right idea in his answer, but didn't fix all the words quite right.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; THE QUICK BROWN FOX JUMPS OVER THE LAZY DOG.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1.I am all in one, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The full sentence contains all the letters in the English alphabet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2.I am part of Mozilla,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;FOX&lt;/strong&gt; is part of &quot;Firefox&quot;, owned by Mozilla.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3.Part of the Olympic Games,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jumping over things is a sport in the Olympic Games, so this gives us &lt;strong&gt;JUMPS OVER&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4.Although I am unwilling to work and would like to eat a hotdog,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Unwilling to work means &lt;strong&gt;LAZY&lt;/strong&gt;. Hotdog gives us &lt;strong&gt;DOG&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5.Speed is the second part of all in one. (1)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The second word of the sentence is &lt;strong&gt;QUICK&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6.Color which is not same as color of part of Mozilla (2)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;BROWN&lt;/strong&gt; is a colour, but not the colour of fire (as in Firefox - see 2).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Combining and arranging the words found (with a couple of extra &quot;the&quot;s), we get the sentence spoilered above.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="11475" LastEditDate="2015-05-13T14:37:21.773" LastActivityDate="2015-05-13T14:37:21.773" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14899" PostTypeId="1" AcceptedAnswerId="14902" CreationDate="2015-05-13T14:50:54.380" Score="18" ViewCount="4648" Body="&lt;p&gt;Who or what is this poem referring to?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Don't confuse me with a line&lt;br&gt;&#xA;  A line may contain me&lt;br&gt;&#xA;  And I may contain a line&lt;br&gt;&#xA;  But that's not my identity  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I have one best friend&lt;br&gt;&#xA;  Though I don't often appear&lt;br&gt;&#xA;  When I do show up&lt;br&gt;&#xA;  He's almost always near&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I can be round and tall&lt;br&gt;&#xA;  And may touch facial hair&lt;br&gt;&#xA;  But even when I'm alone&lt;br&gt;&#xA;  You may still see two there&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Find me in ancient places&lt;br&gt;&#xA;  I am worth quite a lot&lt;br&gt;&#xA;  I can bring about the end&lt;br&gt;&#xA;  But I'm rarely in that spot&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastEditorDisplayName="user20" LastEditDate="2015-05-14T09:53:01.520" LastActivityDate="2015-07-16T23:38:41.920" Title="Riddle Poem: Even when I'm alone, you may still see two" Tags="|riddle|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14900" PostTypeId="2" ParentId="14885" CreationDate="2015-05-13T14:52:31.040" Score="19" Body="&lt;p&gt;You're referring to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wheels&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am 10, and I mostly use 2, but sometimes 8.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most of the time you use a two-wheeled bicycle, but you like to use rollerblades!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My older brother is 12, and he uses either 2 or 4, depending on his mood.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Also your older brother uses the bike, but he likes skateboarding too!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My younger brother is 3, and he uses 3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well, he is too young for a bike, better a &lt;a href=&quot;http://en.wikipedia.org/wiki/Tricycle&quot;&gt;tricycle&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My sister is 19, and she usually uses 4.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your sister can drive a car!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My mom is 37, and she also usually uses 4.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Also your mom can drive a car!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My dad is 40, and he uses either 4 or 6.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your dad drives too, and he is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Six-wheel_drive&quot;&gt;truck driver&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My uncle is 42, and he uses either 4, 10, or 18.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Also your uncle is a truck driver, a &lt;a href=&quot;http://en.wikipedia.org/wiki/Semi-trailer_truck&quot;&gt;HUGE&lt;/a&gt; truck driver!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4652" LastEditorUserId="4652" LastEditDate="2015-05-13T14:54:59.060" LastActivityDate="2015-05-13T14:54:59.060" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14901" PostTypeId="2" ParentId="14899" CreationDate="2015-05-13T15:10:22.443" Score="4" Body="&lt;p&gt;I thought of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A colon (:)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Don't confuse me with a line&lt;br&gt;&#xA;A line may contain me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A line of text may have a colon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I may contain a line&lt;br&gt;&#xA;But that's not my identity&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Usually in play scripts, a character name is followed by a colon then a line.&lt;br&gt;&#xA; It also may be that when quoting pieces, you may do:&lt;br&gt;&#xA;  &amp;nbsp;  &amp;nbsp;  &amp;nbsp;  &quot;A colon followed by a quote like I did right now&quot; - Me 2015&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have one best friend&lt;br&gt;&#xA;Though I don't often appear&lt;br&gt;&#xA;When I do show up&lt;br&gt;&#xA;He's almost always near&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The best friend would be a semi-colon (;), and in lists, if some elements have commas in them, you use a semi-colon to delimit. Some examples: Me, myself and I; Red, green and blue; and fish and chips.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be round and tall&lt;br&gt;&#xA;And may touch facial hair&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm not sure about this one. The round may imply that the dots in a colon are round&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But even when I'm alone&lt;br&gt;&#xA;You may still see two there&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The dots in the colon can look like 2 separate dots&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Find me in ancient places&lt;br&gt;&#xA;I am worth quite a lot&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure about this one. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can bring about the end&lt;br&gt;&#xA;But I'm rarely in that spot&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sometimes a colon can end a sentence, but very rarely is it said to be two separate sentences (Rather one larger one)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user10203" LastActivityDate="2015-05-13T15:10:22.443" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14902" PostTypeId="2" ParentId="14899" CreationDate="2015-05-13T15:23:06.663" Score="27" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the letter &quot;q&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Don't confuse me with a line&lt;br&gt;&#xA;A line may contain me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Don't confuse &quot;q&quot; with &quot;queue&quot;, another word for &quot;line&quot;. Also, a line of text may contain an &quot;q&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I may contain a line&lt;br&gt;&#xA;But that's not my identity&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;q&quot; contains a line, but it's more than just a line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have one best friend&lt;br&gt;&#xA;Though I don't often appear&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter &quot;u&quot;, but &quot;q&quot; is not frequently used&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I do show up&lt;br&gt;&#xA;He's almost always near&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;q&quot; and &quot;u&quot; come together, usually&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be round and tall&lt;br&gt;&#xA;And may touch facial hair&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Uppercase &quot;Q&quot; is round and tall. The lowercase &quot;q&quot; has a descender! Descenders in letters reach downward toward the &quot;beard line&quot; in typography. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But even when I'm alone&lt;br&gt;&#xA;You may still see two there&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An upper-case cursive &quot;Q&quot; looks a lot like the numeral &quot;2&quot;!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Find me in ancient places&lt;br&gt;&#xA;I am worth quite a lot&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Q&quot; is found in &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Quasar&quot; rel=&quot;nofollow&quot;&gt;quasars&lt;/a&gt;&quot;, which are ancient, else we couldn't see them. Also, the Greeks only used the letter &quot;Q&quot; for the number 90, &lt;a href=&quot;http://webspace.ship.edu/cgboer/alphabet.html&quot; rel=&quot;nofollow&quot;&gt;but the Etruscans and Romans kept it for &quot;Q&quot;&lt;/a&gt;. Also, &lt;a href=&quot;http://www.encyclopedia.com/topic/Q.aspx&quot; rel=&quot;nofollow&quot;&gt;it originated as the Phoenician symbol qop&lt;/a&gt;. &quot;Q&quot; is worth quite a lot in the game of Scrabble.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can bring about the end&lt;br&gt;&#xA;But I'm rarely in that spot&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pressing &quot;q&quot; to end a program (old-school). Also, very few words end with the letter &quot;q&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastEditorUserId="4450" LastEditDate="2015-07-16T23:38:41.920" LastActivityDate="2015-07-16T23:38:41.920" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14903" PostTypeId="1" AcceptedAnswerId="14955" CreationDate="2015-05-13T15:25:08.840" Score="13" ViewCount="1448" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Cut me for nothing to be missed,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;open me to be surprised.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Be my fruit and you'll be loved,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Though if I'm from an animal you'll be doomed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who am I?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="12165" LastEditorUserId="12165" LastEditDate="2018-11-21T10:45:15.590" LastActivityDate="2018-11-21T10:45:15.590" Title="Little riddle: Who am I?" Tags="|riddle|word|" AnswerCount="7" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14904" PostTypeId="2" ParentId="14903" CreationDate="2015-05-13T15:33:37.013" Score="0" Body="&lt;p&gt;I think this is some sort of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; parasite.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Cut me for nothing to be missed,&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cut (remove) a parasite and you won't miss anything.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;open me to be surprised&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I have no idea how this relates.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Be my fruit and you'll be loved,&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Parasites would consider their host to be a &quot;fruit&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Though if I'm from an animal you'll be doomed&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many parasites that first attach to an animal host and later a human can transfer diseases picked up from said animals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12397" LastEditorUserId="12397" LastEditDate="2015-05-13T16:25:07.967" LastActivityDate="2015-05-13T16:25:07.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14905" PostTypeId="2" ParentId="14899" CreationDate="2015-05-13T15:38:19.530" Score="8" Body="&lt;p&gt;You're a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rainbow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Don't confuse me with a line&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Don't worry, you're a &quot;bow&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A line may contain me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Light rays travel on a straight line, but &lt;a href=&quot;http://en.wikipedia.org/wiki/The_Dark_Side_of_the_Moon&quot; rel=&quot;nofollow noreferrer&quot;&gt;if pointed on a prism&lt;/a&gt;, a rainbow is revealed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I may contain a line&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://www.google.it/search?q=rainbow%20colors&amp;amp;espv=2&amp;amp;biw=1920&amp;amp;bih=955&amp;amp;source=lnms&amp;amp;tbm=isch&amp;amp;sa=X&amp;amp;ei=-XNTVfD_DureywOq3oCoBQ&amp;amp;ved=0CAYQ_AUoAQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rainbow colors&lt;/a&gt; are always represented as straight lines. E.g. the &lt;a href=&quot;https://encrypted-tbn2.gstatic.com/images?q=tbn:ANd9GcScAL6Vg43sETiFlsYAbdChblnDECg_GGXNSgBcdHJkojDk-D5w&quot; rel=&quot;nofollow noreferrer&quot;&gt;peace flag&lt;/a&gt; is composed by lines of different colors. Our parents (and sadly also me) &lt;a href=&quot;https://connellydale.files.wordpress.com/2013/06/color-bars_large.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;know it best&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But that's not my identity&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The most famous rainbow is the one created by rain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have one best friend&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though I don't often appear&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When it rains, not always a rainbow appears&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I do show up&lt;br&gt;&#xA;He's almost always near&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Only when there is sun and rain a rainbow can occur&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be round and tall&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/izeP4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Round&lt;/a&gt; and in the sky&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And may touch facial hair&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.yorkpress.co.uk/resources/images/1700456/&quot; rel=&quot;nofollow noreferrer&quot;&gt;YEAH!&lt;/a&gt; (Really no clue here...)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But even when I'm alone&lt;br&gt;&#xA;You may still see two there&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://www.youtube.com/watch?v=OQSNhk5ICTI&quot; rel=&quot;nofollow noreferrer&quot;&gt;Double rainbow!!&lt;/a&gt; (actually is only one)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Find me in ancient places&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are a lot of references to rainbows in &lt;a href=&quot;http://en.wikipedia.org/wiki/Rainbows_in_mythology&quot; rel=&quot;nofollow noreferrer&quot;&gt;ancient mythology&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am worth quite a lot&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mmmh, I think this one is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Pot_of_Gold&quot; rel=&quot;nofollow noreferrer&quot;&gt;hoax&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can bring about the end&lt;br&gt;&#xA;But I'm rarely in that spot&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Have you ever tried to &lt;a href=&quot;http://idioms.thefreedictionary.com/chase+rainbows&quot; rel=&quot;nofollow noreferrer&quot;&gt;chase the rainbow&lt;/a&gt;? Even if an observer sees another observer who seems &quot;under&quot; or &quot;at the end of&quot; a rainbow, the second observer will see a different rainbow—farther off—at the same angle as seen by the first observer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4652" LastEditorUserId="4652" LastEditDate="2015-05-13T15:52:59.220" LastActivityDate="2015-05-13T15:52:59.220" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14906" PostTypeId="2" ParentId="11225" CreationDate="2015-05-13T15:54:05.140" Score="0" Body="&lt;p&gt;My guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;3135, 5313, or 1353&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's why:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Here's the sequence:&#xA;&lt;BR&gt;&lt;BR&gt;3 4 6 A 3 2 5 J 3 1 3 Q 5 K 3 A A 1 2 5 3 6 5 J 2 3 Q 1&#xA;&lt;BR&gt;&lt;BR&gt;Now lets assume this is cyclical. So the sequence could be written as:&#xA;&lt;br&gt;&lt;br&gt;&#xA;... &lt;em&gt;1&lt;/em&gt; 3 4 6 A 3 2 5 J 3 1 3 Q 5 K 3 A A 1 2 5 3 6 5 J 2 3 Q 1 3 4 6 A 3 2 5 J 3...&#xA;&lt;br&gt;&lt;br&gt;&#xA;Notice how it just repeats. Now, let's find the &lt;strong&gt;dice (values rolled from dice to create the combination)&lt;/strong&gt; that &lt;strong&gt;split&lt;/strong&gt; the &lt;strong&gt;once (ones)&lt;/strong&gt;&#xA;&lt;br&gt;&lt;br&gt;&#xA;... &lt;em&gt;1&lt;/em&gt; 3 4 6 A &lt;strong&gt;3&lt;/strong&gt; 2 5 J 3 &lt;em&gt;1&lt;/em&gt; 3 Q 5 &lt;strong&gt;K&lt;/strong&gt; 3 A A &lt;em&gt;1&lt;/em&gt; 2 5 3 6 &lt;strong&gt;5&lt;/strong&gt; J 2 3 Q &lt;em&gt;1&lt;/em&gt; 3 4 6 A 3 2 5 J 3...&#xA;&lt;br&gt;&lt;br&gt;&#xA;Now, the K would be 13, and there isn't a 13 on standard casino dice, so it be becomes 1,3&#xA;&lt;br&gt;&lt;br&gt;It's possible the code could be any of the three given, depending on how you interpret the cycle. But testing three codes shouldn't be that hard&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12242" LastActivityDate="2015-05-13T15:54:05.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14907" PostTypeId="2" ParentId="14903" CreationDate="2015-05-13T16:12:51.260" Score="4" Body="&lt;p&gt;My answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Egg&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cut me for nothing to be missed,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You break (cut) the egg if you want to eat the yolk.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;open me to be surprised.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chocolate eggs with a surprise inside are typical of Easter &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Be my fruit and you'll be loved,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There's a fruit called &lt;a href=&quot;http://en.wikipedia.org/wiki/Pouteria&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Eggfruit&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though if I'm from an animal you'll be doomed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eggs come from animals. Doomed might be a reference to a word like &lt;em&gt;Egged&lt;/em&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-05-13T16:12:51.260" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14908" PostTypeId="1" AcceptedAnswerId="14912" CreationDate="2015-05-13T16:25:36.313" Score="13" ViewCount="2390" Body="&lt;p&gt;When humans finally came across some aliens, living below the &lt;strong&gt;Martian crust&lt;/strong&gt;, the strangest point was that the &lt;strong&gt;myth of symmetry&lt;/strong&gt; was broken. Humans had assumed that all &lt;strong&gt;intelligent life forms&lt;/strong&gt; would be symmetric is some way or the other, but our biologists report that these Martians had &lt;strong&gt;31 teeth&lt;/strong&gt; (all on the left side of the face), and &lt;strong&gt;3 eyes&lt;/strong&gt; (two in the right of the face and one in the back), and &lt;strong&gt;5 fingers&lt;/strong&gt; (in total, including &lt;strong&gt;2 thumbs&lt;/strong&gt;) and various other &lt;strong&gt;unsymmetricalities&lt;/strong&gt; (Is this a real word ?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, our mathematicians report that the Martian Counting System is &lt;strong&gt;weird&lt;/strong&gt; too.&lt;br&gt;&#xA;12 is 300&lt;br&gt;&#xA;345 is 12601&lt;br&gt;&#xA;67890 is 4111410&lt;br&gt;&#xA;123 is 3611&lt;br&gt;&#xA;456 is 16200&lt;br&gt;&#xA;789 is 30501&lt;br&gt;&#xA;1234 is 46410&lt;br&gt;&#xA;567 is 21511  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, the left is our system, the right is Martian System. (A few folks have come to the same conclusion, but I want to make it explicit now)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our mathematicians want to know from you, how would &lt;strong&gt;8900&lt;/strong&gt; be represented in that Martian System ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though 3 folks have got the answer correctly, with the numerical clues given, some minor points have not been covered, hence I will be adding the Specific Expected Answer.&lt;/p&gt;&#xA;" OwnerUserId="10940" LastEditorUserId="10940" LastEditDate="2015-05-14T08:41:23.173" LastActivityDate="2015-05-14T09:26:06.960" Title="ALIEN Counting System" Tags="|calculation-puzzle|pattern|enigmatic-puzzle|" AnswerCount="4" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14909" PostTypeId="2" ParentId="14802" CreationDate="2015-05-13T16:29:02.293" Score="-1" Body="&lt;p&gt;I have no clue if this is part of it, but&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The groups of words increment in syllable count&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What I mean is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Lift - 1&lt;br&gt;&#xA;Beams - 1&lt;br&gt;&#xA;Destroy - 2&lt;br&gt;&#xA;Total: 4&lt;br&gt;&lt;br&gt;&#xA;Gas - 1&lt;br&gt;&#xA;Lapse - 1&lt;br&gt;&#xA;Successor - 3&lt;br&gt;&#xA;Total: 5&lt;br&gt;&lt;br&gt;&#xA;Compose - 2&lt;br&gt;&#xA;Correct - 2&lt;br&gt;&#xA;Custom - 2&lt;br&gt;&#xA;Total: 6&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12242" LastActivityDate="2015-05-13T16:29:02.293" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14910" PostTypeId="2" ParentId="14903" CreationDate="2015-05-13T17:32:52.977" Score="2" Body="&lt;p&gt;I would guess that it is&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The Earth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Cut me for nothing to be missed,&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Humans dig into the Earth to get to natural resources.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Open me to be surprised.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The Earth's core is theorized to be molten, which is surprising the first time you hear about it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Be my fruit and you'll be loved,&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The fruit of the Earth generally refers to the good things of the Earth, or the growing things, which people love and need.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Though if I'm from an animal, you'll be doomed.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; I'm unsure. Some animals are territorial or have hunting grounds, so their parts of the Earth might be dangerous...? On the other hand, some people might consider you morally doomed if you took over land where an endangered species makes its home.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;br&gt;&#xA;&lt;hr&gt;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;I hope I am permitted to give a second answer. I have another guess:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; A seed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Cut me for nothing to be missed,&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Cutting seeds in a particular way (scarification) can speed germination, making it easier to sprout difficult-to-grow seeds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Open me to be surprised.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; A nut is a seed that you must open to get to the meat. Otherwise, plant a seed and it opens into a sprout.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Be my fruit and you'll be loved,&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Many people &amp;quot;love&amp;quot; fruit, and fruit has seeds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Though if I'm from an animal, you'll be doomed.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &amp;quot;Seed&amp;quot; is sometimes used to mean &amp;quot;offspring&amp;quot;, such as in the Bible. Taking an animal's offspring away can be very risky, because the mothers will often be protective. This is especially true for mammals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12398" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-13T18:10:10.857" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14911" PostTypeId="2" ParentId="14802" CreationDate="2015-05-13T17:37:58.330" Score="3" Body="&lt;p&gt;An over all common property:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Transformation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Set 1 property:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Building - &lt;b&gt;Lifting&lt;/b&gt; support &lt;b&gt;beams&lt;/b&gt; or &lt;b&gt;destroying&lt;/b&gt; structures&#xA;&lt;br&gt;(From nothing transforms to something, or something to nothing)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Set 2 property:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Chemistry - Reaction of &lt;b&gt;Gas&lt;/b&gt; over a &lt;b&gt;Lapse&lt;/b&gt; of time into a &lt;b&gt;Successor&lt;/b&gt; Gas&#xA;&lt;br&gt;(From one gas composite transforms to another gas composite)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Set 3 property:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Art - &lt;b&gt;Composing&lt;/b&gt; music, &lt;b&gt;Correcting&lt;/b&gt; mistakes in it, and &lt;b&gt;Customizing&lt;/b&gt; other music&#xA;&lt;br&gt;(From imagination/creation, transforms into art)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If some more sets were to be added, I think it could be relate to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Biology, math or Physics&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Creating a over all theme that appears about:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Schooling paths: Engineering (building), Chemistry, Art&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastActivityDate="2015-05-13T17:37:58.330" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14912" PostTypeId="2" ParentId="14908" CreationDate="2015-05-13T17:41:53.650" Score="14" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    426100&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; It works like binary, except: &lt;strong&gt;after&lt;/strong&gt; the rightmost 3 columns (least significant digits) being 2^0 (1), 2^1 (2), 2^2 (4) (from the right to left), &lt;strong&gt;then&lt;/strong&gt; each column further left is worth 3 additional powers of 2, rather than just being worth one additional power of 2.&#xA;&lt;br&gt;&#xA; So in binary, the 4th column from the right is the 8s column, but here the 4th column from the right is the 32s column because you jump 3 exponents to 2^5 (32).  Then the 5th is 2^8 (256), the 6th is 2^11 (2048), and the 7th is 2^14 (16384).&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; And instead of being strictly limited to a max digit of n-1 like in a base system, you just fill up the columns as much as possible from most significant to least significant digit.&lt;br&gt;&#xA;&lt;br&gt;&#xA; So even though the example showed a 6 as the highest possible digit, there could be a 7 in some numbers in this system, anywhere except in the final 2 digits.    (e.g. take our number 31, it would need 7 in the 4 column, then 1 in the 2 column, and 1 in the 1 column.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Update: To address the comment by JLee, yes, you still actually have a unique mapping of Earth numbers to Mars numbers, assuming Martians intuitively follow digit restrictions that are natural to their number system, which they would.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It would just be obvious to them that you can only put a 1 in the ones and twos columns, but then up to a 7 in the higher columns.  Note that the columns go on forever to the left, always being worth a factor of 8 more than the previous, so you can always get enough digits to get high enough to represent all numbers.  Just like in our system every column is worth an additional factor of 10, and you keep going left to get numbers as big as you need.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is a simpler way of me summing up how the number system works.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They use a hybrid of binary and base-8.  The rightmost 2 (least significant) digits are in binary, and after counting to 4, their number columns are in base-8.  Basically if you chop off the right 2 digits, you have an octal representation of counting &quot;groups of 4&quot;, and then the right 2 digits tell you whether to add 0,1,2, or 3 to that.&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; You could imagine such a system evolving from a Roman-Numerals like system where they used sticks to count as high as 3, but above that they used a letter to represent 4 sticks, and then another letter to represent 8 of the first letter (which represents 4 sticks), etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The way I did it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After being a bit distracted by the fact that the Earth numbers were made from breaking up 1234567890, I focused on the smallest number which also happened to be simpler, with 2 zeros and just looking at 3 and 12.  Well 3x4=12, is it possible that the third column could represent 4?  Hey, that's like binary, but you'd never put a 3 in binary.  Let's see if it's an irregular number system that still works on the &quot;columns&quot; idea.&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; I scanned the whole list and noticed that if the Martian number ended in 1 it was odd, else it was even.  This is also like binary, and implies that the right column is the ones column.  That means the 2nd column is probably a twos column, but I should check that it's not weird, like threes.  It can't logically be anything else, since we already are thinking the 3rd column is fours (unless we are way off base).&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; So then I moved up to the next smallest number:  123.  I played with it, subtracting the amounts if I was right about the ones, twos, and fours columns, and happily found that the remainder was divisible by 3, and the result was 32, so this must be the thirty-twos column, which is still a nice number in binary-land.&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;        I then went on to 345 and 456 and confirmed that if the fifth column was 256, then both of these numbers worked out, and saw the pattern and then attacked the largest number to be sure it was always jumping by 2^3 per column.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12399" LastEditorUserId="12399" LastEditDate="2015-05-13T18:12:04.410" LastActivityDate="2015-05-13T18:12:04.410" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14913" PostTypeId="2" ParentId="14908" CreationDate="2015-05-13T17:42:16.927" Score="4" Body="&lt;p&gt;8900 is the martian number&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 426100&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Given a decimal number X, we can represent it in quotient form X=4*Q+R. The first part of the martian number (all but the last two numbers) is Q octal, and the last two numbers are R in binary.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Example&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 67890&lt;br&gt;&#xA; 67890 % 4 = 2 = 10 (binary)&lt;br&gt;&#xA; (67890-2)/4 = 16792 = 41114(octal)&lt;br&gt;&#xA; so 67890  = 4111410 (martian)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11323" LastEditorUserId="11323" LastEditDate="2015-05-13T18:29:03.817" LastActivityDate="2015-05-13T18:29:03.817" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14914" PostTypeId="2" ParentId="14908" CreationDate="2015-05-13T17:56:22.650" Score="3" Body="&lt;p&gt;I think that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 8900 = 426100&lt;br&gt;&#xA;&lt;br&gt;&#xA; The Martian system uses these values for each digit:&lt;br&gt;&#xA;&lt;br&gt;&#xA; Digit 1: $2^0$  = 1&lt;br&gt;&#xA; Digit 2: $2^1$  = 2&lt;br&gt;&#xA; Digit 3: $2^2$  = 4&lt;br&gt;&#xA; Digit 4: $2^5$  = 32&lt;br&gt;&#xA; Digit 5: $2^8$  = 256&lt;br&gt;&#xA; Digit 6: $2^{11}$ = 2048&lt;br&gt;&#xA; Digit 7: $2^{14}$ = 16384&lt;br&gt;&#xA;&lt;br&gt;&#xA; Digits 1 and 2 can only be 0 or 1.  All other digits can be 0-7.&lt;br&gt;&#xA;&lt;br&gt;&#xA; BTW Starman beat me to this, but I just wanted to post it anyway.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How I arrived at this conclusion:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. I first noticed that the end digit corresponded to whether the number was even or odd.  So it acted like a binary 1's digit.&lt;br&gt;&#xA; 2. Next, I noticed that the next to last digit was always 0 or 1 as well.  Testing indicated that it acted like the second binary digit, the 2's digit.&lt;br&gt;&#xA; 3. Then I used &lt;code&gt;12 = 300&lt;/code&gt; to conclude that the third digit had value 4.&lt;br&gt;&#xA; 4. Then &lt;code&gt;123 = 3611&lt;/code&gt; showed me that the fourth digit was 32.&lt;br&gt;&#xA; 5. Then &lt;code&gt;345 = 12601&lt;/code&gt; made the fifth digit be 256.&lt;br&gt;&#xA; 6. Finally, &lt;code&gt;67890 = 4111410&lt;/code&gt; had two digits to solve.  Removing the known digits: &lt;code&gt;67584 = 4100000&lt;/code&gt;.  Knowing that these were all powers of two, the &lt;code&gt;4&lt;/code&gt; digit probably represented 65536 (4 * 16384), leaving the 1 digit to represent 2048.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why do Martians count like this?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm not quite sure but maybe it's because they count using their body parts like this:&lt;br&gt;&#xA;&lt;br&gt;&#xA; Digit 1 = Thumb 1&lt;br&gt;&#xA; Digit 2 = Thumb 2&lt;br&gt;&#xA; Digit 3 = Fingers 1-3&lt;br&gt;&#xA; Digit 4 = Eyes 1-3&lt;br&gt;&#xA; Digit 5 = Teeth 1-3 (are these teeth retractable?)&lt;br&gt;&#xA; Digit 6 = Teeth 4-6&lt;br&gt;&#xA; Digit 7 = Teeth 7-9&lt;br&gt;&#xA; ...&lt;br&gt;&#xA; Digit 14 = Teeth 28-30&lt;br&gt;&#xA; Digit 15 = Tooth 31&lt;br&gt;&#xA;&lt;br&gt;&#xA; I don't know why they don't count on all their teeth all at once, but maybe it's because only digits 0-7 exist in their writing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4479" LastEditorUserId="4479" LastEditDate="2015-05-13T18:30:33.400" LastActivityDate="2015-05-13T18:30:33.400" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14915" PostTypeId="2" ParentId="14891" CreationDate="2015-05-13T18:06:11.300" Score="3" Body="&lt;p&gt;Some possibilities:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;4 legs - it's a polar bear (or any other kind of bear), artificially transported from captivity or from its natural habitat to the South Pole by persons unknown for reasons unknown.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;2 legs - it's a &lt;a href=&quot;http://en.wikipedia.org/wiki/Bear_(gay_culture)&quot; rel=&quot;nofollow&quot;&gt;larger, hairier man who projects an image of rugged masculinity&lt;/a&gt;, or it's &lt;a href=&quot;http://en.wikipedia.org/wiki/Bear_(nickname)&quot; rel=&quot;nofollow&quot;&gt;someone on this list&lt;/a&gt;, or &lt;a href=&quot;http://en.wikipedia.org/wiki/Bear_(surname)&quot; rel=&quot;nofollow&quot;&gt;someone on this list&lt;/a&gt;, or &lt;a href=&quot;http://en.wikipedia.org/wiki/Bear_Grylls&quot; rel=&quot;nofollow&quot;&gt;Bear Grylls&lt;/a&gt;. Or it's someone who believes that share prices are set to fall.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;8 legs - a &lt;a href=&quot;http://en.wikipedia.org/wiki/Tardigrade&quot; rel=&quot;nofollow&quot;&gt;water bear&lt;/a&gt; as per @kaine's answer.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A non-integer number of legs - any of the above, having suffered an injury.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It couldn't be &lt;a href=&quot;http://en.wikipedia.org/wiki/Tupolev_Tu-95&quot; rel=&quot;nofollow&quot;&gt;the Tupolev Tu-95 strategic bomber&lt;/a&gt;, because that wouldn't fit in the hut.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, it also depends what you mean by 'hut'. If the 'hut' is the Helsinki (or Hanoi) University of Technology, then it would be easily big enough for the strategic bomber.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And 'legs' is also a little bit ambiguous....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, I'm going to interpret the question to mean:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How many copies of the classic soft-rock single &quot;Legs&quot;, by ZZ Top, can you fit inside a Tupolev Tu-95 strategic bomber ('Bear') on the campus of the Helsinki  University of Technology ('HUT')?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So then the answer is clearly 35,000, and you get 35,000 copies of the B-side (&quot;A Fool for Your Stockings&quot;) into the bargain.&lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="4050" LastEditDate="2015-05-14T08:02:53.803" LastActivityDate="2015-05-14T08:02:53.803" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14916" PostTypeId="2" ParentId="14903" CreationDate="2015-05-13T18:06:31.990" Score="3" Body="&lt;p&gt;I think that you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Amber, as used in Jurassic Park&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cut me for nothing to be missed,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They drilled, or cut, open the stone to extract what might have otherwise been missed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;open me to be surprised.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When they opened it up, they were surprised by how well preserved the fossil was.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Be my fruit and you'll be loved,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Amber's &quot;fruit&quot;, the mosquito, was treasured by the scientists for its value in research.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though if I'm from an animal you'll be doomed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Turns out the mosquito had eaten some dino blood. Spoiler alert! Resurrecting Dinosaurs may result in injury or death.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12192" LastActivityDate="2015-05-13T18:06:31.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14917" PostTypeId="2" ParentId="14871" CreationDate="2015-05-13T18:15:15.497" Score="0" Body="&lt;p&gt;Let's get some values out of the way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, we have to assume the most pessimal option, because that gives us the worst case scenario for who we can save. In that situation the first two logicians to guess are the blind members. The reason this is the worst case is because they cannot set up the rest of the group as they need to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, a single person cannot know the color of their own hat or of the hats behind them. This means that there are three unaccounted for hats by the time we reach the first sighted person. This means there are 97 visible hats in front of the sighted person. Now, this person, unfortunately, has a 50% chance of making the sacrifice play. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first sighted person looks down the line and counts the red hats. They now know there are N red hats and 97-N blue hats. They then call out which ever of those numbers is higher. Since the numbers are inversely proportional, one will always be higher. From this view, the closest breakdown can be 43 to 44, so our best case in the pessimal option is 44. Now, it's possible that this is the same color as the current logician's hat, but we are going for the pessimal option, so we pretend it isn't. We also pretend that the two blind gents before him also guessed wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this is the pessimal option. This is the value that we are &lt;strong&gt;guaranteed&lt;/strong&gt; to save with this process. But it isn't the most likely option. There is a 2% chance that one of the blind men will end up at the beginning. This means there is a 98% chance the first person in line is sighted. How does this change the field?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there are 99 other people in front of the first person. This means that the minimum number of hats of a given color that he can see in front of him is 49. That is assuming a close to even distribution. If, however, there is a disproportionate number of red or blue, this automatically saves the population represented by the largest number of the same color hat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this method, it is technically possible for all the logicians to be saved, if they all have the same colored hat. This is very, very, very unlikely however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TL;DR - minimum saved of &lt;strong&gt;44&lt;/strong&gt;, most likely amount saved around &lt;strong&gt;49&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another alternative would be to slightly bend the rules in our favor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we have to adjust the way the logicians speak. We will assign three choices: no stutter, single stutter, double stutter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No stutter sounds like &lt;code&gt;blue&lt;/code&gt; or &lt;code&gt;red&lt;/code&gt; &#xA;Single stutter sounds like &lt;code&gt;b-blue&lt;/code&gt; or &lt;code&gt;r-red&lt;/code&gt;&#xA;Double stutter sounds like &lt;code&gt;b-b-blue&lt;/code&gt; or &lt;code&gt;r-r-red&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No stutter indicates that the hat in front of me is the same color as color I am saying. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Single stutter indicates that the hat in front of me is the opposite color as what I am saying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Double stutter is assigned to the blind men. This is a way to indicate to the person directly in front of them that they are the blind person, thus the color they are saying has no known connection the hat in front of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To illustrate how this would work:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;R B B R R R B R B B B R B ...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The first person clearly states &lt;code&gt;BLUE&lt;/code&gt;. They are killed, but the second person also clearly states &lt;code&gt;BLUE&lt;/code&gt;. He is safe. The third person stutters &lt;code&gt;B-BLUE&lt;/code&gt;. She is also safe. The fourth person, hearing the stutter, clearly states &lt;code&gt;RED&lt;/code&gt;. She knows that her hat is the opposite of the given color, so she is safe. The fifth person is blind, so he states &lt;code&gt;R-R-RED&lt;/code&gt;, saving himself and telling the next person that he is blind. The sixth person then clearly states &lt;code&gt;BLUE&lt;/code&gt;. Unfortunately, her hat was red, so she died, but the seventh person can live and carry on the chain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our pessimal situation, we can save &lt;strong&gt;97&lt;/strong&gt; people using this tactic. This is because the first two people cannot give the next color and the third has to guess. In our pessimal setting, he guesses wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most likely scenario saves &lt;strong&gt;98&lt;/strong&gt; people, as two in front of the blind folks have no way of knowing their actual hat color and are risking their chance of safety. &lt;/p&gt;&#xA;" OwnerUserId="12242" LastEditorUserId="12242" LastEditDate="2015-05-14T16:10:34.663" LastActivityDate="2015-05-14T16:10:34.663" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14918" PostTypeId="1" AcceptedAnswerId="14919" CreationDate="2015-05-13T18:28:36.640" Score="16" ViewCount="1821" Body="&lt;p&gt;Every morning starts out the same. I wake up and grab a cup of ____. I then go and feed my pet ______ a mouse or two. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's a weekend, I will watch a movie or read a book about my favorite mathematician ___ L. If the day is truly nice, I may go spend some time at the ____. It really just depends. One time I found a ____ buried in the sand, just glittering away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, my job is pretty _____. I don't really do anything special or fun. I work in an ________ facility for display furniture. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm just glad I'm feeling better. I had the _____, and man was my throat sore.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Fill in the blanks, it might not be hard. Some of the blanks are &lt;em&gt;sound alikes&lt;/em&gt;, but they all follow the same theme.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the missing words, and what is the theme that links them?&lt;/p&gt;&#xA;" OwnerUserId="12242" LastEditorUserId="12242" LastEditDate="2015-05-13T19:53:24.287" LastActivityDate="2015-05-13T20:07:59.160" Title="Fill in the blanks" Tags="|riddle|word|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14919" PostTypeId="2" ParentId="14918" CreationDate="2015-05-13T18:32:19.333" Score="20" Body="&lt;p&gt;I found out the theme, here's a proper set:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1)&lt;b&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Java_(programming_language)&quot;&gt;Java&lt;/a&gt;&lt;/b&gt;&#xA;&lt;br&gt;2)&lt;b&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Python_(programming_language)&quot;&gt;Python&lt;/a&gt;&lt;/b&gt;&#xA;&lt;br&gt;3)&lt;b&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Ada_(programming_language)&quot;&gt;Ada&lt;/a&gt;&lt;/b&gt; (&lt;a href=&quot;http://en.wikipedia.org/wiki/Ada_Lovelace&quot;&gt;Lovelace&lt;/a&gt;)&#xA;&lt;br&gt;4)&lt;b&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/C_(programming_language)&quot;&gt;C&lt;/a&gt;&lt;/b&gt; [&lt;a href=&quot;http://en.wikipedia.org/wiki/Sea&quot;&gt;sea&lt;/a&gt;]&#xA;&lt;br&gt;5)&lt;b&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Ruby_(programming_language)&quot;&gt;Ruby&lt;/a&gt;&lt;/b&gt;&#xA;&lt;br&gt;6)&lt;b&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/BASIC&quot;&gt;BASIC&lt;/a&gt;&lt;/b&gt;&#xA;&lt;br&gt;7)&lt;b&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Assembly_language&quot;&gt;Assembly&lt;/a&gt;&lt;/b&gt;&#xA;&lt;br&gt;8)&lt;b&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/MUMPS&quot;&gt;MUMPS&lt;/a&gt;&lt;/b&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now for the real fun/challenge....Filling the blanks with what wasn't expected:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1)&lt;b&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Elixir_(programming_language)&quot;&gt;Elixer&lt;/a&gt;&lt;/b&gt;&#xA;&lt;br&gt;2)&lt;b&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Wyvern_(programming_language)&quot;&gt;Wyvern&lt;/a&gt;&lt;/b&gt;&#xA;&lt;br&gt;3)&lt;b&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Mary_(programming_language)&quot;&gt;Mary&lt;/a&gt;&lt;/b&gt; (&lt;a href=&quot;http://en.wikipedia.org/wiki/Mary_Leakey&quot;&gt;Leakey&lt;/a&gt;)&#xA;&lt;br&gt;4)&lt;b&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Fortress_(programming_language)&quot;&gt;Fortress&lt;/a&gt;&lt;/b&gt;&#xA;&lt;br&gt;5)&lt;b&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Perl&quot;&gt;Perl&lt;/a&gt;&lt;/b&gt;&#xA;&lt;br&gt;6)&lt;b&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/RAPID&quot;&gt;RAPID&lt;/a&gt;&lt;/b&gt;&#xA;&lt;br&gt;7)&lt;b&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Oak_(programming_language)&quot;&gt;Oak&lt;/a&gt;&lt;/b&gt;&#xA;&lt;br&gt;8)&lt;b&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Magma_(computer_algebra_system)&quot;&gt;Magma&lt;/a&gt;&lt;/b&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reference:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_programming_languages&quot;&gt;http://en.wikipedia.org/wiki/List_of_programming_languages&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can answer this yourself now with the reference provided :-)&lt;/p&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-05-13T20:07:59.160" LastActivityDate="2015-05-13T20:07:59.160" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="14920" PostTypeId="5" CreationDate="2015-05-13T19:57:05.723" Score="0" Body="&lt;p&gt;This tag is for puzzles of the following form:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Use the numbers &lt;span class=&quot;math-container&quot;&gt;$a_1,a_2,\dots,a_n$&lt;/span&gt; and the operations &lt;span class=&quot;math-container&quot;&gt;$o_1,o_2,\dots,o_m$&lt;/span&gt; (for example &lt;span class=&quot;math-container&quot;&gt;$+,\times,!$&lt;/span&gt;) to form the target number &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Other restrictions may be added. For instance, it may be specified that each of &lt;span class=&quot;math-container&quot;&gt;$a_1,a_2,\dots,a_n$&lt;/span&gt; may be used exactly once, or only in a certain order.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="11497" LastEditDate="2019-03-19T10:05:24.463" LastActivityDate="2019-03-19T10:05:24.463" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14921" PostTypeId="4" CreationDate="2015-05-13T19:57:05.723" Score="0" Body="For puzzles about forming numbers using other numbers and mathematical operations. Use with [mathematics] " OwnerUserId="5373" LastEditorUserId="70545" LastEditDate="2021-02-05T18:31:11.867" LastActivityDate="2021-02-05T18:31:11.867" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14922" PostTypeId="1" AcceptedAnswerId="14930" CreationDate="2015-05-13T20:54:02.717" Score="3" ViewCount="158" Body="&lt;p&gt;A device to strike a tone, you should think me not,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For I am a way to transcribe intellectual thought,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And though all that I contain has many a name,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their ultimate function remains the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Though you all seem to have the right thought, parts of the first line will really &lt;em&gt;hammer&lt;/em&gt; it home.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12275" LastEditorUserId="12275" LastEditDate="2015-05-13T21:45:04.627" LastActivityDate="2015-05-13T22:25:07.610" Title="Transcribing Intellectual Thought" Tags="|riddle|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14923" PostTypeId="1" AcceptedAnswerId="14938" CreationDate="2015-05-13T21:01:03.027" Score="4" ViewCount="460" Body="&lt;p&gt;Each of these sets of five words has an odd member out. The other four words fall under a specific category, which can then be further generalized with a single word. Once you discover each of the five odd words out, those five words will come together to form a category of their own! Are you up to the challenge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Word sets:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Coffee&lt;br&gt;&#xA;  Kale&lt;br&gt;&#xA;  Bananas&lt;br&gt;&#xA;  Beer&lt;br&gt;&#xA;  Artichokes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hard Rock&lt;br&gt;&#xA;  Jazz&lt;br&gt;&#xA;  Funk&lt;br&gt;&#xA;  Disco&lt;br&gt;&#xA;  New Age&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sea Star&lt;br&gt;&#xA;  Whale&lt;br&gt;&#xA;  Mussel&lt;br&gt;&#xA;  Bird&lt;br&gt;&#xA;  Bull&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Skunk&lt;br&gt;&#xA;  Red Rose&lt;br&gt;&#xA;  Rain&lt;br&gt;&#xA;  Trash&lt;br&gt;&#xA;  Carrion Flower&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Balderdash&lt;br&gt;&#xA;  Monopoly&lt;br&gt;&#xA;  Trouble&lt;br&gt;&#xA;  Life&lt;br&gt;&#xA;  Parcheesi&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; I wasn't clear enough with one facet of the puzzle. The categories that each group is sorted into can be generalized into one word &quot;super-categories&quot;, which very generally (or in some cases, specifically) describe how each group is perceived. These one word super-categories then will become their own category, at which point the puzzle should end with a pretty clear understanding of how the words were chosen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT2:&lt;/strong&gt; To see if I can nudge this along, here's a hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The backbone of the puzzle is the category which can be formed by the single-word &quot;super-categories&quot; of the five existing categories. The number of elements in the category is an important hint that can potentially guide you towards the solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT3:&lt;/strong&gt; Another day, another hint!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I used the phrase &quot;how each group is perceived&quot; earlier when describing the one word super-categories - that terminology was very specifically on purpose. Your perception of these groups will be very important to really figuring out the crux of the puzzle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="11475" LastEditDate="2015-05-15T19:00:55.090" LastActivityDate="2015-05-16T03:54:43.247" Title="Categories within Categories" Tags="|word|pattern|english|language|wordplay|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14924" PostTypeId="2" ParentId="14879" CreationDate="2015-05-13T21:01:10.257" Score="4" Body="&lt;p&gt;I probably didn't get everything right but I think I'm on the right track:&lt;/p&gt;&#xA;&lt;h2&gt;The first group&lt;/h2&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Avengers characters&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;A man of war&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Captain America&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;A ball of carbon without pastor nor full hurricane&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Bucky Barnes (Buck = Buckminsterfullerene - minister - full - irene)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;A spider&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Black Widow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;One whose strength is in the same area as Veralidaine Sarrasri&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Hawkeye (bow user: credit Gracelyn)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;A toy maker of sorts,&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Iron Man (invents all sorts of toys/gadgets)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;A Jason-type&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Thor (comes from mythology like Jason/Argonauts)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;And one whose pallor is that of Briar Moss' magic.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Hulk (green skin)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;The second group&lt;/h2&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Xmen characters&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;A man whose names vary from Charles to James to Patrick,&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Professor X (Charles Xavier, James McAvoy, Patrick Stewart)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Someone who is Nail Product&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Magneto (Erik Lehnsherr was Polish in the movie)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;He that appears as the half dead person, one soul in two bodies&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Banshee (credit: Gracelyn - looks a lot like Fred and George in the movies, one of which died in the battle of Hogwarts.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;El verano, though quite a bit more dangerous&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Havoc aka Alex Summers (verano = summer)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;An icy animal&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Beast (icy = blue, beast = animal)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;And one who cries naught forevermore&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Raven (aka Mystique) (&amp;quot;Quoth the Raven nevermore&amp;quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;Finally&lt;/h2&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; There's rumors of an X-men Avengers crossover movie, so maybe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4479" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-13T23:58:37.287" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="14925" PostTypeId="2" ParentId="14903" CreationDate="2015-05-13T21:08:01.207" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Beehive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cut me for nothing to be missed,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To get all honey.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;open me to be surprised.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; See creatures from penta-hexa-gons&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Be my fruit and you'll be loved,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Honey can be considered as fruit of beehive and it is lovable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though if I'm from an animal you'll be doomed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bees may doom someone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11306" LastActivityDate="2015-05-13T21:08:01.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14926" PostTypeId="2" ParentId="14922" CreationDate="2015-05-13T21:14:53.683" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You're a note.&lt;p&gt;&#xA;&#xA; You're the device specifying what tone should be struck,&lt;br&gt;&#xA; Intellectual thoughts are frequently jotted down to remember,&lt;br&gt;&#xA; You can write any words you want in the note,&lt;br&gt;&#xA; but words are ultimately always to communicate an idea.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10625" LastActivityDate="2015-05-13T21:14:53.683" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14927" PostTypeId="2" ParentId="14922" CreationDate="2015-05-13T21:20:53.937" Score="1" Body="&lt;p&gt;You are a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; typewriter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A device to strike a tone, you should think me not,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each key makes a specific sound when striking the paper, such that &lt;a href=&quot;http://www.forbes.com/sites/bruceupbin/2015/04/20/new-browser-hack-can-spy-on-eight-out-of-ten-pcs/&quot; rel=&quot;nofollow&quot;&gt;spies would place microphones inside typewriters&lt;/a&gt; to listen to what was typed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For I am a way to transcribe intellectual thought,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Typewriters let you transcribe your thoughts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And though all that I contain has many a name,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The keys all have different names depending on what letter they are.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Their ultimate function remains the same.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each key's function is to type out a letter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2015-05-13T21:20:53.937" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14928" PostTypeId="2" ParentId="14922" CreationDate="2015-05-13T21:37:01.123" Score="0" Body="&lt;p&gt;It must be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;drumroll&lt;/em&gt; the vocal cords.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My thinking is thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A device to strike a tone, you should think me not,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your vocal cords can be thought of as being solely used to sing, but this is not the case!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For I am a way to transcribe intellectual thought,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If we take &quot;transcribe&quot; to mean to translate, we have that your vocal cords &quot;translate&quot; your thoughts into words.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And though all that I contain has many a name,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The various sounds made by a set of vocal cords fall under many names, these called 'onomatopoeia's, I believe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Their ultimate function remains the same.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The ultimate function is to convey your thoughts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7307" LastActivityDate="2015-05-13T21:37:01.123" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14929" PostTypeId="2" ParentId="14923" CreationDate="2015-05-13T22:19:28.160" Score="3" Body="&lt;ol&gt;&#xA;&lt;li&gt;The odd word out is&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Beer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;since the other four are all&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; fruit and vegetables.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;Sorry - I don't know anything about this kind of music!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The odd word out is&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mussel?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;since the other four all have&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; limbs (wings, legs, flippers).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;The odd word out is&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;since the other four all&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; smell (whether sweetly or horribly).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;The odd word out is&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Life&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;since the other four are all&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; board games, while the &lt;a href=&quot;http://en.wikipedia.org/wiki/Conway&amp;#39;s_Game_of_Life&quot; rel=&quot;nofollow&quot;&gt;Game of Life&lt;/a&gt; is something quite different.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now our new collection of five words is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Beer&lt;br&gt;???&lt;br&gt;Mussel(?)&lt;br&gt;Rain&lt;br&gt;Life&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not sure what these have in common ... maybe they can all&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; be seen as food and drink (metaphorically in the case of Life),&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;in which case the answer to 2 would be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hard Rock (as in &lt;a href=&quot;http://en.wikipedia.org/wiki/Rock_(confectionery)&quot; rel=&quot;nofollow&quot;&gt;seaside rock&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-13T22:19:28.160" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14930" PostTypeId="2" ParentId="14922" CreationDate="2015-05-13T22:25:07.610" Score="4" Body="&lt;p&gt;You are a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Keyboard.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A device to strike a tone, you should think me not,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In addition to an input peripheral for PCs, the word &quot;keyboard&quot; can also refer to a musical instrument.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For I am a way to transcribe intellectual thought,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The most common way to put thoughts into writing today is through a keyboard.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And though all that I contain has many a name,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each key on the keyboard has a distinct name.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Their ultimate function remains the same.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A key's ultimate function is to input a character into your computer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12395" LastActivityDate="2015-05-13T22:25:07.610" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14931" PostTypeId="2" ParentId="14923" CreationDate="2015-05-13T22:47:27.057" Score="2" Body="&lt;p&gt;I think I have the 5 odd words out and their category, plus the category for the fifth group, but I'm stumped on groups 1-4. Maybe someone can help me with the others, or use my answer to make a better answer. Also I might be completely wrong =P&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we go:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Group 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Odd word out is &quot;Artichokes&quot;. Category: UNKNOWN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Group 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Odd word out is &quot;Hard Rock&quot;. Category: UNKNOWN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Group 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Odd word out is &quot;Bird&quot;. Category: UNKNOWN (could be as simple as &quot;animals that don't have wings)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Group 4: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Odd word out is &quot;Carrion Flower&quot;. Category: UNKNOWN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Group 5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Odd word out is &quot;Parcheesi&quot;. Category: Board Games with names that are already English words.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Odd words out:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Category: Contain words that sound like or reference golf terms&lt;br&gt;&#xA;Artichokes - &quot;Choke&quot;&lt;br&gt;&#xA;Hard Rock - &quot;Draw&quot;&lt;br&gt;&#xA;Bird - &quot;Birdie&quot;&lt;br&gt;&#xA;Carrion Flower - &quot;Carry&quot;&lt;br&gt;&#xA;Parcheesi - &quot;Par&quot;&lt;br&gt;&#xA;Golf terms taken from this website: &lt;a href=&quot;http://www.pga.com/golf-instruction/instruction-feature/fundamentals/golf-glossary-and-golf-terms&quot; rel=&quot;nofollow&quot;&gt;http://www.pga.com/golf-instruction/instruction-feature/fundamentals/golf-glossary-and-golf-terms&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If nothing else, the category for the five words out is an interesting coincidence. =P &lt;/p&gt;&#xA;" OwnerUserId="10861" LastActivityDate="2015-05-13T22:47:27.057" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14932" PostTypeId="2" ParentId="14869" CreationDate="2015-05-13T23:12:05.940" Score="36" Body="&lt;p&gt;Excellent puzzle! The secret message is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RUN NOW&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's how I solved it...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since the email seems to refer to a flight itinerary, I discovered that the code words must actually consist of three-letter airport abbreviations:&lt;br&gt;&lt;br&gt;&#xA; MSY = New Orleans, Louisiana&lt;br&gt;&#xA; STL = St. Louis, Missouri&lt;br&gt;&#xA; MSP = Minneapolis-Saint Paul, Minnesota&lt;br&gt;&#xA; ORD = Chicago, Illinois&lt;br&gt;&#xA; STL = St. Louis, Missouri&lt;br&gt;&#xA; TPA = Tampa, Florida&lt;br&gt;&lt;br&gt;&#xA; RYG = Rygge, Norway&lt;br&gt;&#xA; MUC = Munich, Germany&lt;br&gt;&#xA; ZAG = Zagreb, Croatia&lt;br&gt;&#xA; BUD = Budapest, Hungary&lt;br&gt;&#xA; HEM = Helsinki, Finland&lt;br&gt;&lt;br&gt;&#xA; ADA = Adana, Turkey&lt;br&gt;&#xA; DME = Moscow, Russia&lt;br&gt;&#xA; IKA = Tehran, Iran&lt;br&gt;&#xA; UFA = Ufa, Bashkortostan, Russia&lt;br&gt;&lt;br&gt;&#xA; KBL = Kabul, Afghanistan&lt;br&gt;&#xA; TSE = Astana, Kazakhstan&lt;br&gt;&#xA; KTM = Kathmandu, Nepal&lt;br&gt;&#xA; URC = Ürümqi, Xinjiang, China&lt;br&gt;&lt;br&gt;&#xA; XNN = Xining, Qinghai, China&lt;br&gt;&#xA; DAC = Dhaka, Bangladesh&lt;br&gt;&#xA; RGN = Yangon, Myanmar&lt;br&gt;&#xA; LPF = Liupanshui, Guizhou&lt;br&gt;&#xA; XNN = Xining, Qinghai, China&lt;br&gt;&lt;br&gt;&#xA; CGK = Jakarta, Indonesia&lt;br&gt;&#xA; PER = Perth, Australia&lt;br&gt;&#xA; DRW = Darwin, Australia&lt;br&gt;&#xA; ADL = Adelaide, Australia&lt;br&gt;&#xA; BNE = Brisbane, Australia&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And with a bit of Googling and doodling, I ended up with this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/GZdNm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="40" LastActivityDate="2015-05-13T23:12:05.940" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14933" PostTypeId="1" AcceptedAnswerId="15026" CreationDate="2015-05-13T23:40:08.827" Score="20" ViewCount="2767" Body="&lt;p&gt;I have taken Doorknob hostage in my cellar. He is &quot;perfectly&quot; trapped - solid walls, solid floor, solid roof, no windows, etc. The only way out is a steel door and the only way to unlock the door is by means of two-pan balance scale (accurate to the quacogram).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have provided 100 weights. Each weight will have its own (non-zero positive integer) weight (in quacograms) accurately written on it. He must put a non-zero number of weights on either side of the scale pans such that they balance perfectly (no trickery on the weights allowed - just the weights). Doorknob has no computer or writing implements - also he is is always able to somehow manoeuvre the weights no matter their weights. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How might I select the weights' weights such that he may NEVER escape? &lt;/li&gt;&#xA;&lt;li&gt;How might I minimise the total weight? (having achieved 1.)&lt;/li&gt;&#xA;&lt;li&gt;What is the lowest highest weight that a weight must have? (having achieved 1. and 2.)&lt;/li&gt;&#xA;&lt;li&gt;How much wood must the woodchuck chuck? Just kidding :D&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;As always explanations/&quot;proofs&quot; are maybe more important than the figures themselves.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2015-05-14T17:21:56.633" LastActivityDate="2015-07-07T22:02:49.750" Title="Trapped in my Cellar" Tags="|mathematics|logical-deduction|optimization|weighing|" AnswerCount="6" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14934" PostTypeId="2" ParentId="14879" CreationDate="2015-05-13T23:52:27.530" Score="2" Body="&lt;p&gt;My edits to JS1's answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I think that the Daine clue means Hawkeye from the Avengers. And the Half dead clue refers to the Banshee, who looks a lot like Fred and George in the movies, one of which died in the battle of Hogwarts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11814" LastActivityDate="2015-05-13T23:52:27.530" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14935" PostTypeId="2" ParentId="14652" CreationDate="2015-05-13T23:57:53.587" Score="2" Body="&lt;p&gt;The hint is clearly a reference to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mad Girl's Love Song, by Sylvia Plath&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Furthermore&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Plath has been noted for interchanging &quot;ravens&quot; with &quot;rooks&quot;. While both are part of the crow family, rooks are not found in North America.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now for wild conjecturing:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One species of bird that has been described as &quot;suspended like puppets on a string&quot; are finches.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Annie Finch is a contemporary poet and a known fan of Plath's work. Her favorite poem of Plath's is the following:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So maybe the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Winter Landscape, with Rooks, by Sylvia Plath&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5778" LastEditorUserId="5778" LastEditDate="2015-05-14T00:03:47.610" LastActivityDate="2015-05-14T00:03:47.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14936" PostTypeId="2" ParentId="14933" CreationDate="2015-05-14T00:26:32.770" Score="13" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;An obvious way to select the weights is to have them be powers of 2, from 1 to $2^{99}$. Every positive integer has a unique binary representation, so no two different sets of weights can be have the same total weight.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are $2^{100}$ possible sets of weights. If any two of them have the same weight, then they can be balanced (removing any weights that show up on both sides), so all of the sets must have different weights. To have the smallest possible total weight, the sets must weigh $0,1,2,...,2^{100}-1$. The binary construction has a total weight of $2^{100}-1$, so this is the minimum. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The set of powers of 2 is also the only way to achieve this total weight: To have a total weight of $2^{100}-1$, every integer from 0 to $2^{100}-1$ must correspond to some set of the weights. In particular, if we choose the weights from lowest to highest, we must always choose the smallest integer that is not a sum of any other set of weights, because otherwise there would be no set with that as its total weight. Then ant11's construction is forced.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-05-14T20:16:42.403" LastActivityDate="2015-05-14T20:16:42.403" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="14937" PostTypeId="2" ParentId="14933" CreationDate="2015-05-14T00:27:05.027" Score="5" Body="&lt;p&gt;Without loss of generality the smallest weight is 1. None of the weights can be the same, so the next smallest weight must be two. Three is now no longer possible, so the next smallest weight is four. If the next weight is less than 8, it is some combination of the previous weights, so the next smallest weight is 8. Induct, and the weights must be the powers of two, up to $2^{99} $. This is the smallest possibility. &lt;/p&gt;&#xA;" OwnerUserId="5778" LastEditorUserId="5778" LastEditDate="2015-05-14T00:41:07.000" LastActivityDate="2015-05-14T00:41:07.000" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="14938" PostTypeId="2" ParentId="14923" CreationDate="2015-05-14T01:01:51.810" Score="7" Body="&lt;p&gt;Group 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bananas, which are sweet while the others are bitter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Group 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jazz, which was popularized in the 1920s while the rest are iconic of the 1970s or so.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Group 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bull, which lacks the seaside association that the rest have.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Group 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Trash, which is man-made while the rest are part of nature.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Group 5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Balderdash, which is the only one not currently the intellectual property of Hasbro.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These words are all ways of saying&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Crazy talk!&quot; or &quot;Nonsense!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You might hear them used like so:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Bananas&lt;/strong&gt;: &lt;em&gt;That's just bananas.&lt;/em&gt; &lt;br&gt;&#xA; &lt;strong&gt;Jazz&lt;/strong&gt;: &lt;em&gt;Don't give me any of that jazz.&lt;/em&gt; &lt;br&gt;&#xA; &lt;strong&gt;Bull&lt;/strong&gt;: &lt;em&gt;That's a bunch of bull.&lt;/em&gt; &lt;br&gt;&#xA; &lt;strong&gt;Trash&lt;/strong&gt;: &lt;em&gt;What trash are you saying now?&lt;/em&gt; &lt;br&gt;&#xA; &lt;strong&gt;Balderdash&lt;/strong&gt;: &lt;em&gt;What?! Balderdash!&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12398" LastEditorUserId="12398" LastEditDate="2015-05-16T03:54:43.247" LastActivityDate="2015-05-16T03:54:43.247" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="14940" PostTypeId="2" ParentId="14870" CreationDate="2015-05-14T03:15:22.117" Score="6" Body="&lt;p&gt;Final answer! The sequences are:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the starting notes of these music compositions:&lt;br&gt;&#xA;&lt;br&gt;&#xA;1) Nocturne In E Flat Major Op.9 No.2 - Frederic Chopin&lt;br&gt;&#xA;&lt;br&gt;&#xA;2) Allegretto tranquillo in F-sharp minor Book 2, Op.30, No.6 - Felix Mendelssohn&lt;br&gt;&#xA;&lt;br&gt;&#xA;3) Prelude in G minor, Op. 23, No. 5 - Sergei Rachmaninoff&lt;br&gt;&#xA;&lt;br&gt;&#xA;4) Hungarian dance No. 5 in F# minor - Johannes Brahms   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Their birthdays are:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1) March 1 (3/1)&lt;br&gt;&#xA;2) February 3 (2/3)&lt;br&gt;&#xA;3) April 1 (4/1)&lt;br&gt;&#xA;4) May 7 (5/7)   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The seven digit number is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;1244687&lt;/strong&gt; which is the &lt;a href=&quot;https://www.underarmour.com/en-us/womens-ua-pip-micro-greg-pulse-running-shoes/pid1244687&quot; rel=&quot;nofollow noreferrer&quot;&gt;style number&lt;/a&gt; of this running shoe.    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The seven letter word could then be:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;CABBAGE&lt;/strong&gt; which is derived by using the 7 digit substitution key followed by letter substitution.&lt;br&gt;&#xA; This is a &lt;strong&gt;cool word&lt;/strong&gt; because it is the longest common word that can be made from the music notes A-G.&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/IvMCq.png&quot; alt=&quot;Table&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-05-22T19:12:35.480" LastActivityDate="2015-05-22T19:12:35.480" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14941" PostTypeId="2" ParentId="14869" CreationDate="2015-05-14T05:04:13.697" Score="10" Body="&lt;p&gt;Going by the same logic as @squeamish ossifrage, the second part of the message should be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RUN = Roland Garros airport in St. Denis in Reunion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The whole message should read:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Roland Garros airport now&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12332" LastActivityDate="2015-05-14T05:04:13.697" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14942" PostTypeId="2" ParentId="14802" CreationDate="2015-05-14T05:26:40.657" Score="10" Body="&lt;p&gt;Each set shares words that have:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; synonyms that produce a set of &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Heterograph&quot;&gt;heterographs&lt;/a&gt; (words with different meanings and spellings, but the same pronunciation).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lift, Beams, Destroy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Raise, Rays, Raze&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Gas, Lapse, Successor:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Air, Err, Heir&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Compose, Correct, Custom:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Write, Right, Rite&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="5208" LastEditDate="2015-05-14T09:05:10.340" LastActivityDate="2015-05-14T09:05:10.340" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14943" PostTypeId="2" ParentId="14785" CreationDate="2015-05-14T07:52:19.363" Score="0" Body="&lt;p&gt;Using only the items provided:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Take the container whose content is to be measured. Carefully crush&#xA;it until it is full.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Choose a natural number $N$. Completely fill $N$ uncrushed containers using the crushed one, and note the number of times you had to pour as $P$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Since we are not allowed measuring tools, we must define our own unit. With&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1 $cu$ = The amount of liquid in one full unmodified container&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;then&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The liquid to be measured is $\frac NP cu$, with a deviation of $\frac{0.5 \pm 0.5}{P} cu$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12395" LastActivityDate="2015-05-14T07:52:19.363" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14944" PostTypeId="2" ParentId="14770" CreationDate="2015-05-14T08:23:57.127" Score="1" Body="&lt;p&gt;&lt;strong&gt;Unfinished&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A clock&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I number myself, for I am numbered;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A clock face is usually numbered with the hours&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am as long as I am worth.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I run down and up, I've crackled and thundered.&#xA;Just to come to rest upon the earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neither red nor deep slate nor pure white,&#xA;what remains of the four, I must be!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Old clock towers had four clock faces&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Though, really, It's only typically...)  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clocks haven't had four faces for a long time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What I am must surely be in sight!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm so wound up, about this little road.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most old clocks needed to be wound up&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can hardly reach the end!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wind-up clocks last for days, weeks or longer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've neither limbs nor toes nor nodes;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; IDK, clocks have hands&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Naught but a body suspends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strong 'till while I'm young, I am.&#xA;(Perhaps as late&#xA;as seven or eight!)&#xA;As I grow older, I bend and I jam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, now! The answer is close!&#xA;I am what I am, though noticed little&#xA;and as helpful can be when you must.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many people rely heavily on clocks&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come in all shapes, all lengths, and all sizes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clocks come in all shapes, lengths and sizes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now tell me: who am I? (I've got all day -- I can wait longer than you)&lt;/p&gt;&#xA;" OwnerUserId="9309" LastActivityDate="2015-05-14T08:23:57.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14945" PostTypeId="2" ParentId="13312" CreationDate="2015-05-14T08:39:42.170" Score="14" Body="&lt;p&gt;I'll continue where &lt;strong&gt;alexmc&lt;/strong&gt; Left off. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The tiger image can be decrypted (at &lt;a href=&quot;http://www.mobilefish.com/services/steganography/steganography.php&quot;&gt;http://www.mobilefish.com/services/steganography/steganography.php&lt;/a&gt;) with the pass code 13312(The question number). That leaves you with a qr code (called &quot;Almost there&quot;) that, when decrypted gives the result:&#xA;&lt;strong&gt;Jet fuel can't melt steel beams.&lt;/strong&gt; This is probably referring to that which is said by 9/11 conspiracy theorists.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Me and my friend Callan worked on this one all day at school today. He was the one who finished the puzzle though, so full credit to him. I'm sorry for my terrible formatting, and I'm sure that I have not followed proper etiquette, but this is my first time on here so I'm a bit of a noob. &#xA;Thanks for the puzzle! (feel free to edit up my post, it's probably gonna look pretty ugly) &lt;/p&gt;&#xA;" OwnerUserId="12412" LastEditorUserId="12412" LastEditDate="2015-05-14T10:32:29.113" LastActivityDate="2015-05-14T10:32:29.113" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14946" PostTypeId="2" ParentId="14908" CreationDate="2015-05-14T09:26:06.960" Score="1" Body="&lt;p&gt;Now, Starman &amp;amp; Shawn Holzworth &amp;amp; JS1 have given correct answers, and I have accepted the answer by Starman, which was the fastest or most complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here I want to Explain the Clues:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - All Martian numbers end with 00 or 01 or 10 or 11, and each ending is listed twice, hinting at some binary system in those 2 Digits.&lt;br&gt;&#xA; - The remaining numbers found are 0,1,2,3,4,5,6 (having a maximum of 6, but not containing 7,8,9 and not containing letters) so that sets the lower limits to the base.&lt;br&gt;&#xA; - Counting is usually (Primitively) done with fingers, and all the points about eyes and teeth are for Diverting the attention from the fingers (&amp;amp; to show that symmetry is not there).&lt;br&gt;&#xA; - If they have 5 fingers (on 2 hands, due to 2 thumbs), then 2+3=5 may be a way the fingers are Distributed.&lt;br&gt;&#xA; - If 2 fingers are used for 2 bits to represent the 2 Digits, then the remaining fingers on the other hand maybe used to represent the next Digit(s).&#xA; - Octal is a good candidate for 3 fingers.&lt;br&gt;&#xA; - Couple all this hand-waving (finger-waving) with the logical Explanations in the 3 other answers to figure out the Complete Alien Counting System.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10940" LastActivityDate="2015-05-14T09:26:06.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14947" PostTypeId="1" AcceptedAnswerId="14990" CreationDate="2015-05-14T09:51:15.737" Score="7" ViewCount="1308" Body="&lt;p&gt;I'm on vacation right now! I left on the 13th in celebration of my &lt;a href=&quot;https://puzzling.stackexchange.com/questions/14870/a-composition-for-my-birthday&quot;&gt;birthday&lt;/a&gt;! It was a long flight so I brought along some reading material:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Faulkner, William. The Sound and the Fury. Published by Voyager Company in 1992.&#xA;D8.46 -A4.1C&#xA;&#xA;Hemingway, Ernest. The Old Man and the Sea. Published by Scribner in 1995.&#xA;D5.22 D.CB&#xA;&#xA;Lee, Harper. To Kill a Mockingbird. Published by Grand Central Publishing in 1988.&#xA;59.24 56.C4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;****Not part of the puzzle: Let me know if you can't find any of these books. ****&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't recall what those strange codes I wrote down were. Probably to help me remember where I am! I'm so absentminded...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyways, I love it here. I thought I would be out in the middle of nowhere, but it seems I'm always in the &lt;strong&gt;centre&lt;/strong&gt; of the hustle and bustle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some reason, I'm worried about where I currently am. Can you figure out where I am, and tell me why I shouldn't be so worried?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\_\_,\_\_,\_\_,\_\_,\_\_,\_\_,\_\_,\_\_$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Meta:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, the answer to this puzzle is a single word that fits into the blanks above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moved from previous hint: the answer should not only describe the location but simultaneously also alleviate my worries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This puzzle should be more straightforward than my first one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1 (read if you think this puzzle will be tedious, because it's not):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; How absentminded am I? Before I left the US, I entered my PIN number wrong so many times at the ATM machine it locked me out! I'm so forgetful... wait, when did I leave for this trip again? That's such an important date, you'd think I'd remember it...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2 (read this if a certain thing is ambiguous):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you're wondering, yes, it is the center of gravity. Also, high school physics classes used two decimal places, so it'll work for us too.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3 (spoilers):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Think you're on the edge of the getting the answer? Yes, yes you are. Think you've triangulated my position? Make sure it's got something in common with the first letter of every sentence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5778" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-06-15T17:45:24.013" Title="What's Happening on my Vacation?" Tags="|riddle|cipher|enigmatic-puzzle|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14950" PostTypeId="2" ParentId="14933" CreationDate="2015-05-14T10:57:27.493" Score="0" Body="&lt;p&gt;Assuming Doorknob is human let's just take 100 weights of 1000 kilo. He would not be able to lift it so he couldn't escape. This is the only practical solution since $2^{100}$ is huge! Even more than the number of atoms in the universe I believe.&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-05-14T10:57:27.493" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14952" PostTypeId="2" ParentId="14903" CreationDate="2015-05-14T14:45:34.677" Score="0" Body="&lt;p&gt;I think it's a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Pitaya&quot; rel=&quot;nofollow&quot;&gt;Dragon fruit&lt;/a&gt;&lt;/strong&gt; (&lt;em&gt;Pitaya&lt;/em&gt;)These fruits are commonly known in English as &quot;dragon fruit&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cut me for nothing to be missed,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You have to cut this fruit for you to eat the creamy and delicious inside&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;open me to be surprised.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you open this fruit, you'll be surprised of its bright white flesh that looks similar to a kiwi inside, with tiny black seeds distributed throughout the fruit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Be my fruit and you'll be loved,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This fruit is exotic and many people want to have this fruit because of the extreme health benefits that it gives &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though if I'm from an animal you'll be doomed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If it becomes a real dragon, you'll be really doomed! &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12054" LastEditorUserId="5373" LastEditDate="2015-05-14T20:18:31.300" LastActivityDate="2015-05-14T20:18:31.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14953" PostTypeId="1" AcceptedAnswerId="14954" CreationDate="2015-05-14T14:55:40.870" Score="12" ViewCount="41449" Body="&lt;p&gt;So I was solving my &lt;a href=&quot;http://www.my-iq.net/&quot; rel=&quot;nofollow noreferrer&quot;&gt;my-iq.net&lt;/a&gt;'s questions and later when I saw the results,  had answered these 4 wrongly, but even after getting to know the right answer, I couldn't manage to find logic for these. Help me out here.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Please mark the city not matching the others :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Sydney&lt;/li&gt;&#xA;&lt;li&gt;Paris&lt;/li&gt;&#xA;&lt;li&gt;Buenos Aires&lt;/li&gt;&#xA;&lt;li&gt;Moscow&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Sydney&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Which is the missing drawing ?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XqES9.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;C&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;Which metal is the odd one out ?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Copper&lt;/li&gt;&#xA;&lt;li&gt;Tin&lt;/li&gt;&#xA;&lt;li&gt;Iron&lt;/li&gt;&#xA;&lt;li&gt;Bronze&lt;/li&gt;&#xA;&lt;li&gt;Lead&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Bronze&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;Find the missing card :&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ohKUF.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/puXx4.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QuEDR.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UkMFo.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xdXHq.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IDTCY.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;img src=&quot;https://i.stack.imgur.com/rIb47.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;img src=&quot;https://i.stack.imgur.com/Czl2x.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;img src=&quot;https://i.stack.imgur.com/OSRDg.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;img src=&quot;https://i.stack.imgur.com/SBU1P.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6120" LastEditorUserId="30903" LastEditDate="2017-05-03T08:15:13.223" LastActivityDate="2018-04-18T21:52:14.630" Title="Questions from IQ test that I couldn't find reasonable answers for" Tags="|pattern|odd-one-out|" AnswerCount="9" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14954" PostTypeId="2" ParentId="14953" CreationDate="2015-05-14T14:58:51.067" Score="24" Body="&lt;p&gt;For question 1, the odd city out is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sydney, since each other city is a capital city. The capital city of Australia is Canberra, not Sydney.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For question 2, the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C. The squares are following a pattern. The squares starting in the lefthand corners are moving on diagonals, one up and one down. There is a square starting on the second row from the top on the right-hand side, that is moving to the left by one square each time. There is another square in the same position that stays there constantly. They overlap on the first grid, which is why the first grid only has three squares filled in.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For question 3, the odd metal out is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bronze, since the other four metals are natural elements. Bronze is an alloy, made of copper and typically tin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For question 4, the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4 (Ace of hearts), as already stated. Out of all the answer choices, a pair of aces would give you the best hand to play in poker.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="-1" LastEditDate="2018-04-18T21:52:14.630" LastActivityDate="2018-04-18T21:52:14.630" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="14955" PostTypeId="2" ParentId="14903" CreationDate="2015-05-14T15:27:04.700" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;An eye? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cut me for nothing to be missed  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A slightly dated term - to cut one's eyes at something is to briefly look at it - thanks, @Cubicon!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;open me to be surprised  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;How many surprises start with being led somewhere with your eyes closed?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Be my fruit and you'll be loved  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;She was the apple of her father's eye  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though if I'm from an animal you'll be doomed  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Bullseye, if someone's aiming for you&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11717" LastEditorUserId="22157" LastEditDate="2016-05-10T17:53:12.910" LastActivityDate="2016-05-10T17:53:12.910" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14956" PostTypeId="2" ParentId="14953" CreationDate="2015-05-14T15:44:28.763" Score="1" Body="&lt;p&gt;My answers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: Any one is correct because any of them don't match the others (they are all different)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2: Any one can be made to work into the pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3: Any can be argued correct.  For example, copper is the best conductor of the group so if you define a threshold of conductivity so only copper is above it, then it is the &quot;odd metal out&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4: Any one could be considered correct.  It was not stated it was a fair single deck of cards being used.  It is also an incorrect use of the word missing.  The card is there, just face down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not a legit IQ test.  This tests certain knowledge of certain subjects and of course someone could guess 1 or more answers correct.  It seems whoever made this &quot;IQ test&quot; should have their own IQ evaluated because these are all ill formed (highly flawed) questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I checked out the rest of the &quot;IQ&quot; test and it is not good.  For example, look at this question they ask:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;QUESTION 20/20&lt;/p&gt;&#xA;&#xA;&lt;p&gt;London and Manchester are 200 miles apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jack travels from London to Manchester at an average speed of 70mph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert travels in the opposite direction, from Manchester to London, at an average speed of 30mph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing that they leave at the same time, in how much time will their paths cross each other?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question cannot be answered accurately for several reasons:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) only average speed is stated so we don't know when they will cross paths.  It would depend on traffic patterns and many other factors such as if they drive slower or faster than average and when.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) What does that mean paths cross?  Who said they are taking similar routes?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) 200 miles apart driving distance or direct line distance?  They did not specify.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) If their paths are parallel but just opposite direction (let's assume best case 1 straight 200 mile road), then they will never &quot;cross&quot; strictly speaking.  They may pass each other on the same road but that is not how the question is worded.&lt;/p&gt;&#xA;" OwnerUserId="12119" LastEditorUserId="12119" LastEditDate="2015-05-17T20:46:33.340" LastActivityDate="2015-05-17T20:46:33.340" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14957" PostTypeId="2" ParentId="14899" CreationDate="2015-05-14T16:01:32.330" Score="0" Body="&lt;p&gt;Is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; brackets: (), [] ? :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Brackets may contain a line and they also may be contained in one. &#xA;&#xA; There are always two of them (a friend)&#xA; &#xA; They can be round () and tall []&#xA; &#xA; An may touch facial hair :-{ :-)&#xA; &#xA; Parentheses (brackets are included in this group) is an Ancient Greek word.&#xA; &#xA; Middle French braguette - bragget (codpiece armor, 16th century). Sure it was a worthy thing)))&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Right direction?&lt;/p&gt;&#xA;" OwnerUserId="12420" LastEditorUserId="463" LastEditDate="2015-07-16T23:19:41.607" LastActivityDate="2015-07-16T23:19:41.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14960" PostTypeId="1" AcceptedAnswerId="14968" CreationDate="2015-05-14T18:09:40.293" Score="25" ViewCount="4700" Body="&lt;p&gt;John and Bob were doing some home construction and decided to take a beer break. Before grabbing a seat John pulls out 3 planks of wood, 1 nail and a hammer. John makes a bet with Bob saying &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'll bet you a beer you can't make a triangle given just these 3 planks and only 1 nail! As an added rule, I must be able to pick it up without it breaking what ever you did.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The nail is only long enough to attach 2 planks of wood together and each plank is of equal size and shape.  No other nails or hardware can be used to attach the planks together, only the single nail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bob is given no extra tools besides the hammer which is for hitting the nail into any plank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;How can Bob win the bet?&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Murder is not allowed! (John was smart and hid his beers first!)&lt;/p&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="12119" LastEditDate="2015-05-15T04:24:05.067" LastActivityDate="2015-05-15T16:28:03.147" Title="3 Planks, 1 nail" Tags="|logical-deduction|lateral-thinking|" AnswerCount="13" CommentCount="10" ClosedDate="2015-05-15T19:15:09.123" ContentLicense="CC BY-SA 3.0" />
  <row Id="14961" PostTypeId="2" ParentId="14960" CreationDate="2015-05-14T18:24:07.853" Score="21" Body="&lt;p&gt;Bob can:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; hammer the nail into a triangle, ignoring the boards completely.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or, he can&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; hammer the nail partway into a plank, and then whack it at an angle until it forms a triangle with the plank.  With a decent initial hammering, you should be able to pick the plank up by the nail without pulling it apart.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5874" LastEditorUserId="5373" LastEditDate="2015-05-14T20:14:33.103" LastActivityDate="2015-05-14T20:14:33.103" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14962" PostTypeId="2" ParentId="14960" CreationDate="2015-05-14T18:47:09.723" Score="13" Body="&lt;p&gt;This would be my approach:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nail two of the planks together to form an L shape. Now, with the join pointing down, rest the third plank across the ends of the first two planks. You can't turn it upside down of course, but it certainly won't &quot;break&quot; by just picking it up in its original orientation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11047" LastActivityDate="2015-05-14T18:47:09.723" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14963" PostTypeId="1" AcceptedAnswerId="14966" CreationDate="2015-05-14T18:56:48.120" Score="16" ViewCount="1596" Body="&lt;p&gt;&lt;strong&gt;Disclaimer: This riddle MAY have been given elsewhere. I heard this one a few months ago from a friend, and I have no idea what his source was.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're in a room, since that's your job. Your singular apparatus, is of course, a bucket of sand. It is very important to you, and indeed, vital to your job. You know exactly how many grains of sand are in your bucket, and are very proud of this fact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, one day, tragedy strikes and a absent-minded secretary spills her mug of spaghetti sauce into your bucket! After much mourning, you pour out the tomato-basil flavoured mess in your bucket, and replace the sand. Distressingly, you realize that you don't know how many grains of sand are in your bucket anymore!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, as luck would have it, a man comes and claims he knows exactly how many grains of sand are in the bucket. Indeed, he boastfully claims he can tell you the exact number of grains of sand particles in any vessel at a glance, and happily offers his services. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, due to legal issues, he claims he cannot tell you the number of sand particles in your bucket before you hire him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can you confirm that the man knows exactly how many sand particles are in your bucket or if he is simply a charlatan seeking a quick cash grab at the heels of a tragic event?&lt;/p&gt;&#xA;" OwnerUserId="12429" LastEditorUserId="5373" LastEditDate="2015-05-14T20:11:14.070" LastActivityDate="2015-05-14T20:23:43.917" Title="The Bucket of Sand" Tags="|strategy|situation|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14965" PostTypeId="2" ParentId="14960" CreationDate="2015-05-14T19:10:34.030" Score="11" Body="&lt;p&gt;This is my answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bob nails the very end of the flat side of one plank to the butt of another plank, forming a smooth &quot;L&quot; shape. Bob then lays this object so that the bend of the L is pointing upward. The two diagonal slopes now make a triangle with the floor and the object can be picked up without it breaking. The 3rd plank is ignored entirely.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12397" LastActivityDate="2015-05-14T19:10:34.030" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14966" PostTypeId="2" ParentId="14963" CreationDate="2015-05-14T19:16:33.690" Score="14" Body="&lt;p&gt;Fill the bucket with lot of sand and show it to the wizard.&lt;br&gt;&#xA;Change room, remove some grains (count them!), then show again your bucket to the wizard.&lt;br&gt;&#xA;Now, ask him: &quot;&lt;em&gt;How many grains were removed?&lt;/em&gt;&quot;.&lt;br&gt;&#xA;If his answer is correct, he's telling the truth about his powers.&lt;br&gt;&#xA;Of course, you may think that he's just lucky. If so, repeat the experiment again!  &lt;/p&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-05-14T19:16:33.690" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14967" PostTypeId="2" ParentId="14960" CreationDate="2015-05-14T19:21:05.010" Score="7" Body="&lt;p&gt;Answer 1:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nailed 2 pieces together as below to form a triangle shape&lt;br&gt;&lt;br&gt; &lt;img src=&quot;https://i.stack.imgur.com/LmN9v.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; since murder isn't allowed, Bob can break John's arm into a &quot;L&quot; shape, nailed a piece of wood into his broken arm's palm, and form a triangle.  John can surely pick it up without breaking 'the wood', but I doubt Bob will win the beers this way&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="1955" LastEditDate="2015-05-14T21:13:20.533" LastActivityDate="2015-05-14T21:13:20.533" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14968" PostTypeId="2" ParentId="14960" CreationDate="2015-05-14T19:29:42.743" Score="46" Body="&lt;p&gt;Bob could just &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;use the nail to scratch a triangle into one of the planks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12134" LastActivityDate="2015-05-14T19:29:42.743" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14969" PostTypeId="2" ParentId="14960" CreationDate="2015-05-14T19:31:07.810" Score="29" Body="&lt;p&gt;This solution seems a bit too labor-intensive to do on a beer break, but screw that, John isn't winning this bet.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Scratch/carve out a groove (with the clawed side of the hammer) in two planks that the other plank can fit partway into. Wedge them into it and then nail the other ends together. &lt;img src=&quot;https://i.stack.imgur.com/gcZ2S.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This seems to me like it fits the description without any real trickery. If done correctly, it shouldn't fall apart, although this depends mostly on Bob's skill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: I'm a Master Carpenter in &lt;a href=&quot;http://fantasy-life.wikia.com/wiki/Carpenter&quot; rel=&quot;noreferrer&quot;&gt;Fantasy Life&lt;/a&gt; :P&lt;/p&gt;&#xA;" OwnerUserId="1563" LastActivityDate="2015-05-14T19:31:07.810" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14970" PostTypeId="2" ParentId="14960" CreationDate="2015-05-14T19:31:34.493" Score="8" Body="&lt;p&gt;This is an answer which ignores the third plank.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/vuQqp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&lt;br&gt; The brown rectangles are two planks, the black line is the nail. The triangle is highlighted in yellow. Basically, you place the planks with a vertex in common, then you bind them with a nail. If the nail is well inserted, the whole structure is rigid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="1955" LastEditDate="2015-05-14T21:12:48.640" LastActivityDate="2015-05-14T21:12:48.640" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14971" PostTypeId="2" ParentId="14953" CreationDate="2015-05-14T19:34:52.957" Score="23" Body="&lt;p&gt;These questions are why many people ridicule IQ tests.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing the thinking behind number 1 is that Sydney is the only one that's not a national capital. But you would make a case for any of these. My first thought on seeing the question was that it was Buenos Aires, as it's the only one in the Western Hemisphere. Moscow is the only one in a formerly communist country. Paris is the only one to be occupied by an enemy army in the 20th century. Paris is also the only one that was ever part of the Roman Empire. I think Moscow is the only one whose primary language is not also the primary language of another country. (I'd have to check on that.) Sydney is the only one that is not the largest city in its country. Buenos Aires is the only one where atheists have never been a significant political bloc. Etc etc. A person sufficiently knowledgeable in geography and history could come up with dozens of reasons why any of these is different from the other three. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bailey's theory for number 2 works, but one could propose a million other rules or patterns. Etc.&lt;/p&gt;&#xA;" OwnerUserId="7386" LastActivityDate="2015-05-14T19:34:52.957" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="14972" PostTypeId="2" ParentId="14960" CreationDate="2015-05-14T19:49:16.830" Score="7" Body="&lt;p&gt;I figure this would work well enough to win a beer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nail two of the boards together at the ends where the red arrow is pointing.  Then bend them into a L shape and lay them across the third board.  With a little work and assuming the planks aren't too wide, you should have a triangle in the middle between the three.  If you then pick up the whole thing where the green arrow is pointing, with a little trial and error to get the balance right, it won't fall apart.&lt;br&gt;&lt;br&gt; &lt;img src=&quot;https://i.stack.imgur.com/DBQl9.png&quot; alt=&quot;boards&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12431" LastEditorUserId="1955" LastEditDate="2015-05-14T21:12:16.557" LastActivityDate="2015-05-14T21:12:16.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14973" PostTypeId="2" ParentId="14960" CreationDate="2015-05-14T19:52:24.010" Score="3" Body="&lt;p&gt;I was going to suggest &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; scratching a triangle into the wood&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but it looks like Sean beat me to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an alternative to this, Bob could &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'Punch out' a triangle shape with the nail. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or to reduce his work, depending on the thickness/shape of the planks he could &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Score a diagonal line from one of the corners, and then break it off by standing on it and hammering the triangle off to form something like this;&lt;img src=&quot;https://i.stack.imgur.com/8eNob.jpg&quot; alt=&quot;Triangle wood&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2078" LastActivityDate="2015-05-14T19:52:24.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14974" PostTypeId="1" AcceptedAnswerId="14976" CreationDate="2015-05-14T19:52:29.163" Score="24" ViewCount="1714" Body="&lt;p&gt;NASA has spent hundreds of millions of dollars gathering data about cave systems in one region of the planet Mars. The project just finished up, and analysts will soon be ready to start analyzing the complete list of all the lengths of cave passageways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the manager of the NASA data center was unhappy about his recent pay cut, and had decided to quit, but not before causing problems. First, he used his manager-level privileges to log on to the backup servers and delete all of the backup data for the cave system project, knowing that he could later blame that on another department and he would not be held responsible. However, he knew that if he altered or deleted the data on the main server, then he would be caught and spend many years in prison. Instead, he created fake data to sabotage the study, and put the text file in the same folder as the actual data, with a similar name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But he didn't stop there! Knowing that he was not very skilled in creating data, he didn't know how believable his fake data would be, so he asked two friends for help. He explained the task to them and requested 500 fake data points, ranging between 10 and 99,999 (since any passageway at least 10 feet long was classified as its own passageway). One of the friends was a mathematically-inclined (but not brilliant) intern from the data center who had recently quit, and the other one wasn't very math-minded, but did work for the state lottery. They did as they were asked, and he put their fake data files into the same folder as the other 2 files. Then he submitted his 2-week notice that he was leaving his job.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2 weeks later]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today, when the analysts started analysis, they saw the 4 different data files in the folder, all with nearly the same time stamp. They figured out what had happened, but they are not sure which data file is the right one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the comma-separated data files, containing the lengths of cave passageways (rounded to the nearest foot). The order of numbers in each file appears random, but is actually the order in which the data was recorded (which is not important).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;File 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B&lt;/h2&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;91691,88462,65945,6184,34325,44339,69698,4286,22141,20186,44975,43597,96829,77866,7252,54008,21738,19743,8101,60443,82617,6851,35716,63086,36076,47458,98331,72521,35251,38997,48146,38296,22915,35492,76652,94287,58518,8231,56425,14023,98610,81957,64723,60408,63160,29617,28192,50648,85324,65164,79515,94465,45487,16505,57011,50404,61971,34423,33873,60873,43760,84050,52344,64020,61635,37047,83692,26705,34174,68583,31705,57819,99367,53781,93042,2584,73720,67228,30392,13506,58101,38355,44658,56364,67926,58956,21839,21690,11318,8899,28636,95459,97586,72035,87291,20492,65393,50878,54978,64335,38670,95348,92502,57695,7060,55568,18068,24166,43499,18730,34039,46185,75185,77576,73050,31232,29463,36601,317,14914,85335,98240,80874,82321,53747,96363,31671,80999,5246,24337,24367,82528,91680,18931,59898,46036,35777,95686,62359,65077,84826,53667,68313,25407,17211,19482,5642,80960,15082,87462,94179,27129,80813,70555,20023,94550,97800,63378,37137,19655,78046,92770,53552,44759,28259,58320,11076,10635,79768,44202,56989,78594,93293,28129,3229,57561,176,21060,41536,92535,71585,59230,89258,72566,91356,28949,57950,64062,53612,50107,5289,37000,10650,64444,42826,61001,39581,52143,53813,88926,42770,66730,61860,73477,29846,95376,24869,90696,40818,86561,98154,17851,32828,46486,13913,67501,92912,48088,25454,19737,91336,2561,87618,5332,38007,18249,1292,75057,48859,63883,22607,97957,75335,30694,62332,63525,52708,79295,69202,23584,88030,13634,12907,71531,15959,19372,61677,13021,59064,60147,23883,48380,23336,58929,15531,39795,71411,93948,64617,45541,69894,7336,34395,79809,12797,93800,39257,14708,45522,85826,3532,64933,6568,32807,35960,33565,60315,93438,2629,32594,36507,82399,27730,83577,43704,73512,38106,28205,18796,75407,21575,8155,69672,59273,52280,53999,4071,26300,15114,26193,95288,59232,31951,54460,85169,75012,1057,84003,70446,33623,16002,49505,56582,36807,92824,3890,39204,68168,91911,87683,74667,16026,84036,38859,47884,48955,47293,51690,12548,19201,6583,32916,69812,75262,51360,21814,3542,11904,5549,80243,59004,86368,13076,87519,92905,16420,30597,85887,36612,9278,37300,74286,17600,59874,20179,51240,43653,35821,94017,75815,93242,88656,15629,74356,52223,23899,20467,27719,51548,16602,62403,95977,8708,28605,18709,97767,44601,51425,45192,76236,77358,13087,69218,97381,48966,49755,39053,61344,94556,73018,8014,1261,49952,39087,66684,42116,45594,42306,92432,47997,51843,52026,50225,70509,90687,58022,90903,41685,92828,50644,11434,14357,38632,49443,63988,68154,43980,76108,57144,34881,79318,17168,91165,61167,86338,43114,3008,69265,63087,44647,24765,68231,93843,41002,12308,29837,73680,53948,7802,66507,98169,86749,52515,36734,88973,35273,43551,20971,11902,25951,46933,86801,24532,14094,85484,12119,649,86074,57796,40663,4373,12522,35569,45552,82769,88487,87849,41418,52744,88319,49029,36650,66149,51882,39211,43936,6292,87335,99410,41702,51750,37708,98231,31065,50180,24532,73126,62976,18342,62243,30702,62598,95874,53871,79319,60885,52998,18545,59121,64350&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;File C&lt;/h2&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;96954,54615,28496,90692,46070,90096,35182,34835,30444,39432,1746,41055,12120,21303,37632,41930,970,39960,3926,15703,18380,2538,85552,869,10,88240,31341,4192,59198,21165,1781,63350,39670,1932,13610,11994,29728,11921,10086,43625,633,31758,54131,181,484,1110,24021,19811,16561,12483,664,123,219,3029,49997,26601,29318,3787,42574,48659,15382,1031,16117,25652,16918,28469,176,45567,25463,53151,1338,67662,658,3194,64703,22706,13129,47515,744,23043,12713,138,14663,7179,1440,13959,20788,7429,15271,4572,10473,14,24864,1782,89982,24256,60550,65475,43065,28938,26099,12043,15889,862,49217,55160,13080,33780,51251,48870,24073,32038,18,20129,16589,30301,26395,29836,11,16382,26514,595,374,82549,135,30149,48380,26171,25942,55089,53658,689,364,55221,17284,33114,57967,35500,574,50755,1252,39432,17937,13848,16369,22757,20726,19814,1216,49543,196,4143,54882,56734,11157,17685,10461,28370,17205,32508,31700,40495,17199,582,1886,44196,29891,1289,23334,32919,35218,43559,1903,225,14659,12230,28536,39024,24261,16456,16686,30170,152,7973,71284,813,83671,22212,10227,24243,748,172,24998,51693,181,46951,297,21645,17273,11278,1949,16351,23584,505,29505,11432,724,44522,45811,1586,60504,18062,670,18146,52856,61636,21983,976,80611,14191,84781,99125,35759,71589,11643,90843,3516,177,48,26312,27208,18124,12451,58201,72,31227,54201,13747,723,59987,15516,14795,9711,30930,3365,49528,88826,77243,95170,80774,32145,33390,31622,73089,5816,10310,89394,1004,24638,4923,25126,91059,15961,1826,40161,48490,51643,18550,22305,37355,1634,78750,6947,58777,15246,13178,13316,78603,28552,18625,15853,36304,25904,10702,79349,51567,78165,12072,17732,10618,160,25225,4788,46976,1667,75940,95315,49286,12689,73443,46367,77414,52191,60115,78232,94845,56429,77,7465,73710,60092,49601,49409,83002,86608,71563,28246,59807,13653,1210,3703,83780,57681,27322,37867,1000,4757,10652,19718,64762,63461,3891,9975,1580,38368,11342,73451,10996,72366,61801,6011,26471,26542,66239,18355,9413,27143,13028,22342,15816,33859,32837,40541,69729,14702,68239,22583,39337,15778,33081,49219,63797,12971,27552,37537,140,2448,13390,15744,28067,61129,1851,46811,50059,40624,105,1194,20040,39627,14447,20,33,2473,39144,19819,22229,44792,85437,172,2661,67807,34,35869,88637,3054,62108,85445,91064,62141,50342,93530,12752,31831,1464,279,37514,80745,18020,1221,93376,338,23846,350,16399,22489,8178,2485,66521,21909,56424,19558,46144,55297,52489,31303,27341,14972,14118,20930,11197,43125,65617,77243,29448,26393,80375,14699,47237,12987,5792,248,16264,25320,11228,60540,19430,41909,87067,11267,16434,27299,19195,6497,95398,39801,28648,3991,28,5385,38402,97163,61444,15500,99896,66699,47976,20559,10910,3696,24333,18087,82598,3917,81305,22458,56347,25021,19376,79014,27470,19400,1745,19671,91712,59696,26494,65125,13665,13740,11031,91653,72433,9313,50376,1056,21492,24154,25706,44787,46269&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;File D&lt;/h2&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;96356,54607,27997,89863,45152,89915,35049,34534,29457,38995,1404,40187,11757,21238,37485,41537,241,39452,3513,15385,17899,1295,15410,804,297,87992,30658,4049,58916,20701,1193,12597,39667,1580,13524,11280,29162,11163,9223,1591,352,30928,54004,342,51,1025,23300,19455,16128,11652,15284,393,741,2393,49269,26043,28877,3209,41649,47779,15330,792,15713,24673,16430,27823,159,45024,24912,53016,474,66977,315,2823,64020,22082,12130,47227,74099,22912,12166,258,14289,6702,923,13068,20624,6774,14757,3973,10146,380,24176,824,89293,24050,59718,65169,42410,28474,26053,11563,15384,26,48517,54163,12349,33388,50791,47910,23960,31354,902,19172,16181,29891,26064,29616,953,15506,26127,330,320,82262,22,29991,47658,25514,25814,54956,53464,75,503,54465,16454,32583,57601,34889,164,49933,657,39149,17447,13026,16164,22441,20283,19467,925,49541,309,3437,54079,56025,10436,17682,10442,28325,16711,31979,30986,40401,16906,498,1281,43498,29242,354,22950,32525,34462,43417,1518,113,14278,11243,27624,38299,23860,16422,16429,29519,69,7262,71244,81584,83233,21381,10223,23940,245,468,24640,50995,81,46430,402,21065,17114,10925,1920,15692,23022,4284,28976,10839,119,43896,45224,935,59745,17296,11,17961,52536,61613,21908,97363,79659,13470,14027,19,35248,70805,11279,89899,3083,62,248,25375,27192,18038,11608,57458,21,31096,53430,13556,182,59382,15432,14680,8951,30894,2570,49521,87922,76923,1594,80359,31177,33270,30786,72274,5198,9831,88395,591,23751,4583,24308,90266,15880,1031,39696,48037,51380,18503,22126,37219,1406,78194,6821,58412,15084,12434,13026,77776,168,17842,15471,35945,25699,10489,78583,51515,77641,11680,17021,10289,351,24407,4039,46631,1223,75566,94609,48347,11843,72933,45400,76587,51754,59134,77398,94068,55577,469,6588,72748,59532,49408,49295,82093,86582,70569,28170,59458,13358,549,2836,83721,56993,26384,37399,530,4561,10617,19527,64562,63084,3474,9900,1368,37695,11234,72778,10516,71921,61259,5920,26015,26326,65805,18005,8564,26146,12256,21585,15153,33537,32335,40260,68820,14298,68018,22572,38959,15038,32394,49193,62804,12864,27251,37172,535,2410,12948,15564,27929,60539,1228,45843,49084,39820,119,639,19192,38931,14206,879,478,2261,38685,19011,21706,44,84558,705,2018,66901,326,34920,87856,2302,61514,84,90552,61745,49484,93362,12065,31430,936,76,36753,80355,17509,811,92853,578,23386,403,15845,21757,7908,2337,66385,21055,56304,19282,46090,54762,51919,30628,26465,14846,14029,20459,10562,42373,64,77103,28952,26152,79668,14067,46849,12441,5484,420,15571,24881,11112,60359,19334,41577,86574,10462,15930,26920,18405,6265,94873,38812,27834,3274,916,4865,37937,96333,60844,14525,99097,66118,47729,19664,10104,3050,23489,17592,82379,3857,80827,21751,55564,24973,18913,78279,26649,18781,1457,19296,90899,59215,25594,64951,13162,13504,10265,91326,71539,9247,49695,900,21420,23276,25458,199,45594&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can you figure out which data file is most likely the real one, and why?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-06-18T01:58:41.627" LastActivityDate="2015-06-18T12:28:46.117" Title="Which data file is genuine?" Tags="|mathematics|word-problem|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14975" PostTypeId="2" ParentId="14933" CreationDate="2015-05-14T19:57:14.390" Score="2" Body="&lt;p&gt;To my surprise the total weight could be minimized to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 188849&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which is better than $2^{100}-1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be achieved by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dividing the total number of weights (100 in this case) in two equal groups of 50 each.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Group 1&lt;/strong&gt;&lt;br&gt;&#xA; 50 to 99 &#xA; derived by logic that sum of 2 smallest elements in group is higher than maximum value present in group.&lt;br&gt;&#xA; &lt;strong&gt;in general&lt;/strong&gt; first value equals to $n$ where $n=count($group_length$)$.&lt;br&gt;&#xA;&lt;br&gt;&#xA;  $x= \sum_{n=50}^{99} n = 3725$&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Group 2&lt;/strong&gt;&lt;br&gt;&#xA; Would consist of numbers from $x-$ min value of group 1 $ - 1$ &lt;strong&gt;upto&lt;/strong&gt; $ x - 1$&lt;br&gt;&#xA;  i.e. from $x - 49$ up to $x-1$ i.e. ( 49 elements)&lt;br&gt;&#xA; 50th element of group 2 can be found as &#xA; $ = max(g1) + max(g2) + 1$ which in our case is &lt;strong&gt;3824&lt;/strong&gt;&#xA;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Total, lowest, highest weight.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Considering above case &#xA;&lt;br&gt;&#xA; Total weight   $\sum_{n=3676}^{3725} + 3824 = 188849 $&#xA;&lt;br&gt;&#xA; Lowest and highest weights 50, 3824 respectively.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5437" LastActivityDate="2015-05-14T19:57:14.390" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14976" PostTypeId="2" ParentId="14974" CreationDate="2015-05-14T20:05:23.727" Score="35" Body="&lt;p&gt;The real data set is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; File D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First clue:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you look at the frequency of values that start with the number 1, the frequencies are (6.8%, 11.2%, 30.0%, 27.0%).  For any real world measurements, the frequency of the first digit follows &lt;a href=&quot;http://en.wikipedia.org/wiki/Benford&amp;#39;s_law&quot;&gt;Benford's Law&lt;/a&gt;, where the frequency of 1s is much more than the frequency of 9s.  Benford's Law predicts that the frequency of numbers starting with 1 should be around 30.1%.  So files A and B are ruled out.  Files C and D are within the approximate range that would be predicted.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second clue:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Now compare File C and D and the frequency of each starting digit:&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;code&gt;.  Digit      File C    File D    Benford Expected&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;.  -----      ------    ------    ----------------&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;.    1         30.0      27.0        30.1&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;.    2         17.6      18.6        17.6&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;.    3         12.6      12.6        12.5&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;.    4          9.6      10.8         9.7&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;.    5          8.0       8.6         7.9&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;.    6          6.8       6.0         6.7&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;.    7          5.8       5.8         5.8&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;.    8          5.0       5.6         5.1&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;.    9          4.6       5.0         4.6&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; It looks like File C was manufactured to conform exactly to Benford's Law.  Real data should be expected to have some variance (especially in a data set of only 500 numbers).  It looks like the mathematically inclined coworker knew what the numbers should look like and faked File C to perfectly fit the expected pattern.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4479" LastEditorUserId="4479" LastEditDate="2015-05-14T21:12:35.190" LastActivityDate="2015-05-14T21:12:35.190" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14977" PostTypeId="2" ParentId="14960" CreationDate="2015-05-14T20:11:14.397" Score="5" Body="&lt;p&gt;Simplest solution yet.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lay this on a plank of wood. John can pick up the plank and marvel as Bob collects his beer.&lt;br&gt;&lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/riBfm.jpg&quot; alt=&quot;Nail and hammer&quot;&gt; &lt;br&gt;&lt;br&gt;If you want to get technical about straight lines, he could flip the nail the other way, and use a plank to jam the 'head' of the nail into the hammer shaft to form a straight angle, or simply use the hammer to bend the nail head to the desired angle before producing this solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2078" LastEditorUserId="2078" LastEditDate="2015-05-15T14:12:31.237" LastActivityDate="2015-05-15T14:12:31.237" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14978" PostTypeId="2" ParentId="14933" CreationDate="2015-05-14T20:17:20.717" Score="7" Body="&lt;p&gt;I'm not sure about the version of the problem where you want to minimize the sum, but you'll probably be surprised to find out that the version where you want to minimize the maximum number is known - and an open problem in combinatorics :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See here for example: &lt;a href=&quot;http://www.openproblemgarden.org/op/sets_with_distinct_subset_sums&quot; rel=&quot;nofollow&quot;&gt;http://www.openproblemgarden.org/op/sets_with_distinct_subset_sums&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The powers of two are the most natural construction, but not the best possible; for example, if you wanted to give 4 weights, you could give 1,2,4,8, but you could also give 3,5,6,7 (and 7 &amp;lt; 8).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Karan Thakkar's answer, it's easy to give two subsets that sum up to the same value, for example 50 + 53 = 51 + 52. (I'm writing this here since it seems that I can't comment.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The powers of two are indeed optimum for minimizing the sum, but proofs that people were giving in the comments are unconvincing. The one in user12408's answer works.&lt;/p&gt;&#xA;" OwnerUserId="12432" LastEditorUserId="12432" LastEditDate="2015-05-15T21:14:17.473" LastActivityDate="2015-05-15T21:14:17.473" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14980" PostTypeId="1" AcceptedAnswerId="16267" CreationDate="2015-05-14T20:46:46.737" Score="24" ViewCount="3945" Body="&lt;p&gt;You work for a secret organization, and recently the personal data of a terrorist leader has fallen into your hands. This would be great news, except for the fact that all the files are locked with a password. You know that this particular terrorist organization tends to implement a &quot;three-strike&quot; rule - after three failed attempts to enter the password, the data corrupts itself and is no longer usable. Luckily, the organization is as forgetful as it is crafty, so they've provided a hint on the password screen. The hint reads:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hint 1: BFplmNVNbACLAMlrBgmDSllV iSTjfJsehRTgsRtM&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Being a master codebreaker, you immediately open up QuipQiup and plug in the code. You take the result (NEw loGYGnPULPOliNs oF AllY HaD be Back I D saId O) and enter it as the password - no luck. The password rejects, and another hint is presented.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hint 2: A failed attempt means you've either forgotten the password, or you're trying to hack it. Luckily, I've decided to play fair. Remember that it would be wise to pay close attention to the capitals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Capitals! Of course! Heading back to QuipQiup, you remove each lowercase letter and solve it once again. You take your new result (NG TO THE CHINKS OF PURPRI) enter is as the password. Again, no luck - the password is rejected. This time, instead of a hint popping up on your screen, a warning message displays.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;WARNING:&lt;/strong&gt; You have one attempt remaining before the data corrupts. Remember, this data is a matter of &lt;strong&gt;life&lt;/strong&gt; and &lt;strong&gt;death&lt;/strong&gt;! Use your last guess wisely...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Can you crack the code and gain access to the data, or will you fail your third guess and render the data useless?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; First hint!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After some time with no results, you begin to focus on one aspect of the clue. &lt;em&gt;Close attention to the capitals...close attention to the capitals...but what does it mean?&lt;/em&gt; Suddenly, something clicks with you. The terrorist leader said he's decided to play fair, but certainly not THAT fair! Either the capitals clue is a red herring, or there's more to the word than initially meets the eye...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And a QoL hint as well!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everything outside the blocked text is just flavor. If you're trying to solve the puzzle, pay no attention to the QuipQiup results. It will only make you more confused.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Second hint!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You don't have to be a secret agent/genius/master codebreaker to realize that the warning about &lt;strong&gt;life&lt;/strong&gt; and &lt;strong&gt;death&lt;/strong&gt; is some sort of hint (though it certainly helps). After many, many, many minutes of pondering the hint, however, you just can't figure out how it could possibly be used in tandem with the hint about capitals...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Third hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As you stare at the letters, fruitlessly trying to forge them into something useful, your eyes begin to glaze and the letters all start blending together. Uh-oh. You try to blink it off, but to no avail. Perhaps what you need is a DiFfErEnT pOiNt Of ViEw...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Fourth hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 17 days. For 17 days, you've sat in your apartment, mindlessly toiling about in an attempt to discover the password. Your beard is overgrown. Your eyes are sagging. You're tired, hungry, cold, and alone.&lt;br&gt;&#xA; But your organization needs you. So you press on. &lt;em&gt;&quot;Another angle...another angle...&quot;&lt;/em&gt; And suddenly, it clicks. The initial text is only the first step in this process, and it doesn't require any software to convert at all. No, this isn't part of any typical cipher...it will, however, require some research. Better get started.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;6/11/2015 Update:&lt;/strong&gt; Hey, guys! I've opened a bounty on this puzzle as well as &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15725/discover-the-meaning-of-life&quot;&gt;my other unanswered puzzle&lt;/a&gt; as a way of saying thanks for the bounty you helped me earn on &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15286/connect-the-pipes/15288&quot;&gt;this excellent puzzle&lt;/a&gt;. As such, I want this puzzle to be solved - my bounty shan't go to waste! If there are any questions I can answer to confirm whether or not you're going in the right direction, please don't hesitate to ask them in the comments. If I feel they don't give too much of the puzzle away, I'll be happy to answer them.&lt;/p&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="57742" LastEditDate="2020-01-04T22:59:50.420" LastActivityDate="2020-01-04T22:59:50.420" Title="Crack the Terrorist's Password!" Tags="|cipher|password|" AnswerCount="5" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="14981" PostTypeId="2" ParentId="14980" CreationDate="2015-05-14T22:27:15.440" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It's a recursive puzzle. The answer is not to think linearly. Run the capitals only from the HINT 1 through the cipher again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thought Process is as follows: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;QuipQuip cipher algo must be correct otherwise the terrorists would have know it was an intrusion rather than a forgetful operative so the data would have been wiped. With that in mind, the operative likely forgot a second step in the decoding process, so the answer is either that only the capital letters in the decoded hint are the answer or only that the decoded letter capitals need to be decoded again to find the password. Third option is the negative, so that capitals need to be removed but that would be too complicated for a forgetful operative.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12443" LastEditorUserId="10748" LastEditDate="2015-05-15T04:23:50.000" LastActivityDate="2015-05-15T04:23:50.000" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14982" PostTypeId="1" AcceptedAnswerId="15067" CreationDate="2015-05-14T23:39:58.050" Score="3" ViewCount="566" Body="&lt;p&gt;Mage's birth seems important.  That's all I'm going to say.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;J EJPL QW VFIUQT UXWUTH YOSWSUV, HIU Q JQ PIAX PN HSV.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BQIZ BQMKS R QBG WSU MGMTB,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;JRE IV E CQC ASWWK JV CLF PNEE QO M EW.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R EN I KMU JRTPTJV, TWVIUQVIT KQESURRH IWH PBQIS BRQFA FIJZM,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;WPUNXJUNW J TROF JNMOO YIFZ PCOB, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;BSNMCMNMB M XIWX UW UYSS RR UPN FFVCLJK HSOM&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if I completely messed up the encryption, don't get too mad....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh, yeah.  And when you've decrypted that, you've got to solve the riddle. :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone (who is bothering with the riddle) is having trouble, maybe read below...  It may or may not help.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not really that important...  Now I'm getting all science-y on you!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11790" LastEditorUserId="11790" LastEditDate="2015-05-16T16:40:59.583" LastActivityDate="2018-05-20T07:22:59.790" Title="Here's a little cryptogram I wrote, you might want to read it letter for letter" Tags="|riddle|cipher|knowledge|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14983" PostTypeId="2" ParentId="14982" CreationDate="2015-05-14T23:55:43.070" Score="2" Body="&lt;p&gt;An decoded answer (mostly):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I walk in real &lt;em&gt;ms&lt;/em&gt;, mostly unknown yet I am part of you. They think I may not exist and am a bit wrong in the &lt;em&gt;heaz&lt;/em&gt;. If I &lt;em&gt;doz&lt;/em&gt; am a bit bipolar, sometimes charming and other times weird. Sometimes I like being Peer Gynt, sometimes I want to lurk in the Benthic Zone. &lt;br&gt;&lt;strong&gt;(decoding errors italicized)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How it was decoded:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The cipher appears to be essentially &lt;a href=&quot;http://en.wikipedia.org/wiki/Vigen%C3%A8re_cipher&quot; rel=&quot;nofollow&quot;&gt;Vigenere&lt;/a&gt; with a key of BIJE: This decodes most of it, but has a few errors. Here is the original text after decoding:&lt;br&gt;&lt;br&gt;i walk in real ms mostly unknotn yet i am oart of you they thijk i may not exist and am a bit wrong in the heaz if i doz am a bit bipolar sometimes charming and other times weird sometimes i like being peergynt sometimes i want to lurk in the benthicyone&lt;br&gt;&lt;br&gt;I have corrected obvious errors but there are a few more I can't easily work out. I suspect what has happened is that the author has invented his own encoding scheme which happens to be equivalent to Vigenere, except it works differently in a few edge cases.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7308" LastEditorUserId="7308" LastEditDate="2015-05-15T01:10:33.083" LastActivityDate="2015-05-15T01:10:33.083" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14984" PostTypeId="2" ParentId="14982" CreationDate="2015-05-15T00:12:58.103" Score="2" Body="&lt;p&gt;Hmm, I tried a slightly different key as Tryth, and got this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Using key 'bijebije':&#xA; i walk in realms mostly unknotn, yet i am oart of you.they thijk i may not exist,and am a bit wrong in the head if i do.i am a bit bipolar, sometimes charming and other times weird,sometimes i like being peer gynt,sometimes i want to lurk in the benthic yone&#xA;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which I translated to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; i walk in realms mostly unknown, yet i am part of you. they think i may not exist, and am a bit wrong in the head if i do. i am a bit bipolar, sometimes charming and other times weird, sometimes i like being pregnant, sometimes i want to lurk in the ((benthic zone))?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12429" LastActivityDate="2015-05-15T00:12:58.103" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14985" PostTypeId="2" ParentId="13303" CreationDate="2015-05-15T00:26:34.190" Score="26" Body="&lt;p&gt;I think I've solved the second stanza!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Paired up, pared down, the truth exposed, lies within lies.&lt;br&gt;&#xA;  Heading onwards north, reeling seas terrorize.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Apply the first line to the second line. First &lt;em&gt;pair up&lt;/em&gt; the words to get &lt;strong&gt;headingonwards northreeling seasterrorize&lt;/strong&gt;, then &lt;em&gt;pare down&lt;/em&gt; to get (headin)&lt;strong&gt;go&lt;/strong&gt;(nwards) (nor)&lt;strong&gt;three&lt;/strong&gt;(ling) (s)&lt;strong&gt;east&lt;/strong&gt;(errorize). This stanza gives a clear direction:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; go three east.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-15T00:26:34.190" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14987" PostTypeId="1" AcceptedAnswerId="15024" CreationDate="2015-05-15T02:49:12.963" Score="9" ViewCount="523" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In the United States,&lt;br&gt;&#xA;  You can find me in a camp and in a city,&lt;br&gt;&#xA;  You can find me in a creek and in a grove,&lt;br&gt;&#xA;  You can find me on a hill and on a land.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;You call my nicknames all the time,&lt;br&gt;&#xA;  You are addicted to me,&lt;br&gt;&#xA;  But if you are careless,&lt;br&gt;&#xA;  I will kill you.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;You saw me alone in 2008 and 2013,&lt;br&gt;&#xA;  But you saw me with my girl friend in 2001 and 2006.&lt;br&gt;&#xA;  You know who I am,&lt;br&gt;&#xA;  But will you find out?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Please explain your answer. There are many clues in that riddle, try to find them all.&lt;/p&gt;&#xA;" OwnerUserId="10400" LastEditorUserId="5373" LastEditDate="2015-05-15T18:01:48.250" LastActivityDate="2015-05-19T09:14:11.177" Title="I shrunk the kids" Tags="|riddle|" AnswerCount="6" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14988" PostTypeId="2" ParentId="14953" CreationDate="2015-05-15T03:25:21.340" Score="1" Body="&lt;p&gt;I almost had 2, but I didn't realize that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The moving squares are allowed to overlap each other, so in one of the grids it looks like there are only 3, but two are in the same spot.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My reasoning on 4 was&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It can't be 7 of spades or Jack of Hearts, because in a normal deck there would be just one of those. Between the two that are left, the pair of Aces would beat a pair of tens in poker, so the Ace of Hearts is what I would choose. Obviously, assuming the goal is to have a good poker hand is arbitrary.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4044" LastActivityDate="2015-05-15T03:25:21.340" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14989" PostTypeId="2" ParentId="14953" CreationDate="2015-05-15T03:55:47.187" Score="1" Body="&lt;p&gt;Alternative answer for 2 is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Valid positions on the board are two middle horizontal lines and corners. Only C matches it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12447" LastActivityDate="2015-05-15T03:55:47.187" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14990" PostTypeId="2" ParentId="14947" CreationDate="2015-05-15T04:33:55.923" Score="6" Body="&lt;p&gt;Final answer! You are in:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the district of &lt;strong&gt;Chomutov&lt;/strong&gt; in the Czech Republic near the border with Germany.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The eight-letter word is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Schengen&lt;/strong&gt; which alleviates your worries because the two countries are part of the Schengen Agreement and crossing the border should be easy.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The decoding starts with:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the codes below each book. They look like &lt;strong&gt;geodetic coordinates&lt;/strong&gt; (latitude and longitude). As the date is the 13th and the coordinates use digits from 1-9-A-D, it seems that each book's &lt;strong&gt;ISBN-13&lt;/strong&gt; is needed. The following coordinates result from a 13 digit substitution key of 123456789ABCD:&lt;br&gt;&#xA;&lt;br&gt;&#xA;First ISBN 9781559402736 with (D8.46,-A4.1C) $\Rightarrow$ (64.15,-21.93) at Reykjavik, Iceland&lt;br&gt;&#xA;&lt;br&gt;&#xA;Second ISBN 9780684801223 with (D5.22,D.CB) $\Rightarrow$ (36.77,3.22) at Algiers, Algeria&lt;br&gt;&#xA;&lt;br&gt;&#xA;Third ISBN 9780446310789 with (59.24,56.C4) $\Rightarrow$ (41.70,44.80) at Tbilisi, Georgia&lt;br&gt;&#xA;&lt;br&gt;&#xA; Each of these locations are capital cities. A &lt;strong&gt;capital&lt;/strong&gt; is &quot;common with the first letter of every sentence&quot;.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using &lt;a href=&quot;http://www.geomidpoint.com/&quot; rel=&quot;nofollow&quot;&gt;geomidpoint.com&lt;/a&gt;,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the &lt;strong&gt;center of gravity&lt;/strong&gt; of these three coordinates is  (50.54,13.25) which is located in the Czech Republic near the border with Germany.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-06-15T17:45:24.013" LastActivityDate="2015-06-15T17:45:24.013" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14991" PostTypeId="2" ParentId="14953" CreationDate="2015-05-15T05:24:47.600" Score="5" Body="&lt;p&gt;Regarding #1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All the names on the list are their official names, except Moscow. That's Москва (Moskva).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12454" LastActivityDate="2015-05-15T05:24:47.600" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14992" PostTypeId="1" AcceptedAnswerId="23644" CreationDate="2015-05-15T09:08:52.837" Score="12" ViewCount="1333" Body="&lt;p&gt;My &lt;a href=&quot;https://puzzling.stackexchange.com/questions/14947/whats-happening-on-my-vacation&quot;&gt;vacation&lt;/a&gt; just ended, and I'm at the airport headed back home! It was definitely an enjoyable trip, but it could also be described as, well, unusual.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've got a more immediate problem though. I've kind of forgotten which gate my flight arrives at. The answer is probably somewhere in the grid, but I can't see it. Could you help me out?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;LOFCSURFTCZLRNTVZFUUNRBTON&#xA;JZTZODUJRTSCJCBFNOFSYSVZCJ&#xA;RFBYMBVRMSOYYLNMYYJCDLRFZB&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Clarification: Gates are identified by a letter followed by a number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Two words in the exposition are precisely chosen to be relevant.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No, not primes. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5778" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-10-29T12:09:56.960" Title="Which Flight Takes Me Home?" Tags="|mathematics|cipher|pattern|" AnswerCount="3" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="14994" PostTypeId="1" AcceptedAnswerId="14997" CreationDate="2015-05-15T11:14:26.060" Score="21" ViewCount="1276" Body="&lt;p&gt;Homer + Ovid = Paris&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ulysses + Julius Caesar =  Nimrod&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Guy de Montfort - Cleopatra = Medusa&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Epicurus - Pluto = Dido&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hector + Tisiphone = &lt;strong&gt;&lt;em&gt;?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="4050" LastEditDate="2015-05-15T17:49:59.587" LastActivityDate="2015-05-15T17:49:59.587" Title="A diabolical puzzle: Hector + Tisiphone =?" Tags="|calculation-puzzle|word|enigmatic-puzzle|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14995" PostTypeId="2" ParentId="11298" CreationDate="2015-05-15T11:25:43.663" Score="0" Body="&lt;p&gt;My first post here, excuse me for any errors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Contact?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am what they cherish,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Those who cherish,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Those who love you, hold you and hug you provide contact.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Are also those who detest ardently.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I murder with ease and comfort,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Murder also involves contact, a hug of death, or a pillow over a face.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And be loved with purpose.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am a drug to the eye,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The physical object contact is a drug to the eye, you need it to see.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And to the heart,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, contact in a loving manner is a drug for the heart.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You love me like nobody else,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Not because I murdered your loved ones,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But because I am a drug to the eye,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;and to the heart.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I cause no disease,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Although killing by contact does not involve disease, it can cause it, I'm not sure about this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;but yet I abolish those who you love,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Contact can be used to kill and love. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;For I am a drug to the eye,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;and to the heart!&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="12469" LastEditorUserId="12469" LastEditDate="2015-05-15T11:39:24.633" LastActivityDate="2015-05-15T11:39:24.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14996" PostTypeId="2" ParentId="14960" CreationDate="2015-05-15T11:31:05.797" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The nail is only long enough to attach 2 planks of wood together and each plank is of equal size and shape&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I imagine this doesn't mean what it literally means but rather that the nail is shorter than twice the smallest dimension of the planks.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Attach three corners of the planks with a single nail to make two corners of the triangle like this: &lt;br&gt;&lt;br&gt; &lt;img src=&quot;https://i.stack.imgur.com/C9AQR.png&quot; alt=&quot;Solution&quot;&gt;&lt;br&gt;&lt;br&gt;&#xA; If the nail is good enough, one can pick it up without destroying the third corner. In practice, however, I wouldn't bet this works with real planks and nail...&#xA;  &lt;br&gt;&lt;br&gt; (Here, the nail is used to attach 3 planks together, but in principle this can be done no matter how short the nail is compared to the plank, so I guess this isn't against the rules.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2228" LastActivityDate="2015-05-15T11:31:05.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14997" PostTypeId="2" ParentId="14994" CreationDate="2015-05-15T11:34:54.520" Score="18" Body="&lt;p&gt;My answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Minotaur&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;The main idea is to associate every name with its corresponding Circle number in Dante's Inferno (Divine Comedy).&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Homer and Ovid both belong to the first Circle, while Paris belongs to the second: $1+1=2$.&lt;br&gt;&#xA;&#xA; Hector is in the first, Tisiphone between the fifth and sixth, &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so the result must be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; between the sixth and the seventh Circle. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The only creature that satisfies that condition is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the Minotaur, who's the guardian of the seventh Circle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Extra: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;the title says &quot;diabolical&quot;, which is a reference to Inferno.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note: &lt;em&gt;my previous answer was Pier della Vigna, which is wrong.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="4050" LastEditDate="2015-05-15T12:22:40.767" LastActivityDate="2015-05-15T12:22:40.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14998" PostTypeId="2" ParentId="14960" CreationDate="2015-05-15T12:23:05.547" Score="22" Body="&lt;p&gt;Bob could:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nail two planks together, then squeeze the third one between them such that it goes over the bottom one and below the top. It can be picked up horizontally, or even vertically if the nail is tight and the planks are rough enough to hold their weight through friction. Here is an example using pencils. The rubber band represents the joint: &lt;img src=&quot;https://i.stack.imgur.com/dw0x0.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2032" LastActivityDate="2015-05-15T12:23:05.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14999" PostTypeId="2" ParentId="14987" CreationDate="2015-05-15T13:18:08.420" Score="2" Body="&lt;p&gt;I think it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gold&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the United States,&#xA;You can find me in a camp and in a city,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ft. Knox, and any number of bank vaults in cities.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can find me in a creek and in a grove,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Panning for gold&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can find me on a hill and on a land.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mining for gold&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You call my nicknames all the time,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are a lot of slang terms for gold&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are addicted to me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Plenty of people go to long lengths to get as much gold as possible&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But if you are careless,&#xA;I will kill you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; not sure about this one, but then, if you are careless pretty much anything could kill you&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You saw me alone in 2008 and 2013,&#xA;But you saw me with my girl friend in 2001 and 2006.&#xA;You know who I am,&#xA;But will you find out?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No idea what this section is referring to. Maybe Olympic medals?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11047" LastActivityDate="2015-05-15T13:18:08.420" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15000" PostTypeId="1" AcceptedAnswerId="15054" CreationDate="2015-05-15T14:15:21.767" Score="12" ViewCount="401" Body="&lt;p&gt;This scrap of paper caught my eye, partially trodden in to a mush outside Bank tube station. It appears to be part of a pizza menu (illegible areas marked with square brackets):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;         Extra Topping Price List               New for 2017!&#xA;&#xA;         [  ]   [  ]    [  ]    [  ]    [  ]    [  ]    [  ]    [  ]&#xA; 12&quot;     £1.73  £2.08   £10.65  £11.70  [  ]    [  ]    [  ]    £174&#xA; 14&quot;     £2.35  £2.86   £15.00  £16.10  [  ]    [  ]    [  ]    £238&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Which eight toppings are advertised, and how were their prices arrived at?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What are the missing prices?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The toppings have inherent properties that lead to their pricing, so the answer isn't &lt;em&gt;£1.73 worth of arbitrary pizza topping&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I cross the road and head up Threadneedle Street as I puzzle over the fragment. Strange, it sort of looks like the price has been rounded up differently for each topping?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's only later, back at my hotel room and packing my suitcase as neatly as possible, that the penny finally drops. Yuck!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;em&gt;Dough&lt;/em&gt; I laugh to myself. Thinking, &quot;is that even a legal foodstuff?&quot;. I guess there's a market for everything.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4534" LastEditorUserId="4534" LastEditDate="2015-05-15T17:02:54.400" LastActivityDate="2015-05-16T15:16:58.850" Title="Croesus' Pizzas" Tags="|geometry|enigmatic-puzzle|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="15001" PostTypeId="2" ParentId="14987" CreationDate="2015-05-15T14:38:15.210" Score="1" Body="&lt;p&gt;Is the answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Guns/Weapons&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the United States,&#xA;You can find me in a camp and in a city,&#xA;You can find me in a creek and in a grove,&#xA;You can find me on a hill and on a land.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The US allows gun owners pretty much anywhere, unlike many other countries&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You call my nicknames all the time,&#xA;You are addicted to me,&#xA;But if you are careless,&#xA;I will kill you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There are many different names for different guns. There is a very prevalent gun culture in America, and some would say is an addiction. More gun related deaths in America are accidents by the owner or a member of the family.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You saw me alone in 2008 and 2013,&#xA;But you saw me with my girl friend in 2001 and 2006.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In from 2008-2013, there was an ammunition shortage in the US. In 2001 and 2006, the United States were in major conflicts, where guns would be paired with bullets as much as possible. This could also be a reference to the WMD scares at those times&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I shrunk the kids&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Shrink ray gun, obviously!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12242" LastActivityDate="2015-05-15T14:38:15.210" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15007" PostTypeId="1" AcceptedAnswerId="15014" CreationDate="2015-05-15T15:24:19.907" Score="-2" ViewCount="2318" Body="&lt;p&gt;There are 3 puzzles below. To reach the second puzzle, you have to solve the first. The image has hidden messages and keys. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need you to help me get a date with this pretty girl in my uni. She prefers smart guys, so she placed the name and address of a restaurant on her blog. Can you help me identify the place?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First puzzle:&lt;/strong&gt; I should meet her on a street. What is the name of the street?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The street name is inside of the picture.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Link the phone number with info in the blog and you'll get the name of the street&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qi2Jg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second puzzle:&lt;/strong&gt; Also from the above image, you'll discover which store she will be in. After you found out which street she is on, you noticed in that there are 5 shops, and know that she is in one of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which place is she in?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Link the bio with the store type and you'll get the name of the store&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Beautiful Roses (store for roses)&lt;br&gt;&#xA;Police Station&lt;br&gt;&#xA;Stylish Haircuts for Men (hair salon)&lt;br&gt;&#xA;Read Me (book store)&lt;br&gt;&#xA;Macburger (restaurant)&lt;br&gt;&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Third puzzle:&lt;/strong&gt; After you solve the above puzzle, you'll have the below image that will lead you to her.&#xA;After you know where to meet her, you go inside and you ask for her. They tell you that the girl left and the location is inside of the envelope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JQKSB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="12427" LastEditorUserId="5373" LastEditDate="2017-01-13T00:12:27.863" LastActivityDate="2017-01-13T00:12:27.863" Title="Can you help me get a date with this girl?" Tags="|enigmatic-puzzle|visual|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15008" PostTypeId="2" ParentId="14987" CreationDate="2015-05-15T15:36:46.033" Score="1" Body="&lt;p&gt;I think it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the United States,&lt;br&gt;&#xA;You can find me in a camp and in a city,&lt;br&gt;&#xA;You can find me in a creek and in a grove,&lt;br&gt;&#xA;You can find me on a hill and on a land.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Camp_Rising_Sun&quot; rel=&quot;nofollow&quot;&gt;Camp rising sun&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Sun_City,_Arizona&quot; rel=&quot;nofollow&quot;&gt;Sun City&lt;/a&gt;, &lt;a href=&quot;http://www.mysuncreek.com/&quot; rel=&quot;nofollow&quot;&gt;Sun Creek&lt;/a&gt; and &lt;a href=&quot;http://www.sungrovefoods.com/&quot; rel=&quot;nofollow&quot;&gt;Sun Grove&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Sun_Hill,_West_Virginia&quot; rel=&quot;nofollow&quot;&gt;Sun Hill&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Sunland,_California&quot; rel=&quot;nofollow&quot;&gt;Sun Land&lt;/a&gt; and obviously the sun is everywhere in those spots (they're all outside).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You call my nicknames all the time,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are a lot of &lt;a href=&quot;http://www.behindthename.com/names/meaning/sun&quot; rel=&quot;nofollow&quot;&gt;names which refer to the sun&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are addicted to me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yep, tanning can &lt;a href=&quot;http://en.wikipedia.org/wiki/Tanning_addiction&quot; rel=&quot;nofollow&quot;&gt;create addiction to someone&lt;/a&gt;... Also sun is, ehr, quite vital.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But if you are careless, I will kill you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sun can harm in multiple ways, from sun burn to blindness to &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_storm&quot; rel=&quot;nofollow&quot;&gt;solar storm&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You saw me alone in 2008 and 2013,&lt;br&gt;&#xA;But you saw me with my girl friend in 2001 and 2006.&lt;br&gt;&#xA;You know who I am, But will you find out?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The girlfriend is obviously the moon, in &lt;a href=&quot;http://en.wikipedia.org/wiki/February_2008_lunar_eclipse&quot; rel=&quot;nofollow&quot;&gt;2008&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/April_2013_lunar_eclipse&quot; rel=&quot;nofollow&quot;&gt;2013&lt;/a&gt; there was a lunar eclipse, while in &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_eclipse_of_June_21,_2001&quot; rel=&quot;nofollow&quot;&gt;2001&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_eclipse_of_March_29,_2006&quot; rel=&quot;nofollow&quot;&gt;2006&lt;/a&gt; there was a solar eclipse.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Still&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I don't think this is correct :(&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4652" LastEditorUserId="4652" LastEditDate="2015-05-15T15:46:31.333" LastActivityDate="2015-05-15T15:46:31.333" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15009" PostTypeId="2" ParentId="15007" CreationDate="2015-05-15T15:38:14.217" Score="1" Body="&lt;p&gt;The street could be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sandy lane&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The building could be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The police station (where she finds her phone) or the book store since she is a 'shopholic' and likes to read&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Could she be at:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The beach&#xA;&lt;br&gt; He name is &quot;C Fish&quot; in the blog contact information (Fish live in the sea), and Sandy (Sand). When you put them together you get a beach. Also the picture has a fish on it.&#xA;&lt;Br&gt;&lt;br&gt;Plus with the image given it could be a &lt;b&gt;Beach Side Seafood Restaurant&lt;/b&gt; (Thanks to Bailey M). Which might seem like an odd place for a vegitarian.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-05-15T16:43:57.810" LastActivityDate="2015-05-15T16:43:57.810" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15010" PostTypeId="2" ParentId="14741" CreationDate="2015-05-15T15:41:02.883" Score="4" Body="&lt;p&gt;I'm totally wild guessing here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Scarlet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It seems to me that by adjacent we should consider words that are &quot;connected&quot; via another word. That is, not the words x and x+1, but x-1 and x+1. So for example we pair &quot;letter&quot; and &quot;number&quot; and get &quot;symbol&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, we get a partial solution for some of the words:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;scarlet&lt;/strong&gt;&lt;br/&gt;fruit - A (scarlet letter)&lt;br/&gt;letter - apple&lt;br/&gt;computer - symbol&lt;br/&gt;number&lt;br/&gt;subject - consumer&lt;br/&gt;consume - history&lt;br/&gt;before - &lt;br/&gt;cut - happy&lt;br/&gt;cry - &lt;br/&gt;slide - weep&lt;br/&gt;gentle -&lt;br/&gt;awake - rough&lt;br/&gt;not - &lt;br/&gt;rain - hot&lt;br/&gt;cold - &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We basically need to find a color:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the color must fit with rain and letter. So I'm guessing &lt;strong&gt;scarlet&lt;/strong&gt; for: &quot;scarlet rain&quot; (because of &lt;a href=&quot;http://en.wikipedia.org/wiki/Blood_rain&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; or perhaps blood in general) and the &lt;a href=&quot;https://en.wikipedia.org/wiki/The_Scarlet_Letter&quot; rel=&quot;nofollow&quot;&gt;The Scarlet Letter&lt;/a&gt;, which is the letter &quot;A&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="5610" LastEditDate="2015-05-15T16:27:56.000" LastActivityDate="2015-05-15T16:27:56.000" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15011" PostTypeId="2" ParentId="14987" CreationDate="2015-05-15T15:47:27.427" Score="4" Body="&lt;p&gt;Is it...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Honey?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; [Honey,] I shrunk the kids!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the United States,&#xA;You can find me in a camp and in a city,&#xA;You can find me in a creek and in a grove,&#xA;You can find me on a hill and on a land.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are a ton of places that have &quot;Honey&quot; in the name in the US.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You call my nicknames all the time,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's used as a nickname and variations of it as well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are addicted to me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well it's delicious.  Don't judge me.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But if you are careless,&#xA;I will kill you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Honey-induced death is probably a thing.  Diabetes?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You saw me alone in 2008 and 2013,&#xA;But you saw me with my girl friend in 2001 and 2006.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I honestly don't know how to make this one fit...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You know who I am,&#xA;But will you find out?&lt;/p&gt;&#xA;" OwnerUserId="12391" LastActivityDate="2015-05-15T15:47:27.427" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15012" PostTypeId="2" ParentId="14960" CreationDate="2015-05-15T16:15:10.207" Score="3" Body="&lt;p&gt;You can form &lt;em&gt;many&lt;/em&gt; triangles by the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stack 2, 3, or an infinite number of boards (depending on the lenth of the nail) on top of each other and drive the nail through the wood near the end.  Then you can fan the boards out forming multiple triangles.  See this image for the general idea.  &lt;img src=&quot;https://i.stack.imgur.com/0aWXD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2758" LastEditorUserId="2758" LastEditDate="2015-05-15T16:28:03.147" LastActivityDate="2015-05-15T16:28:03.147" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15014" PostTypeId="2" ParentId="15007" CreationDate="2015-05-15T16:23:22.920" Score="7" Body="&lt;p&gt;The street is probably &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Canal Street&quot; If you type the phone number in an old flip phone as a text message, 222-2-66-2-... translates to &quot;canal str&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a guess, but..&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; She's probably at the bookstore, because it says she likes to read.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no idea what's meant by the 3rd part with the envelope.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I mean, maybe she's at a seafood restaurant, but she's a vegetarian. Does she eat fish? Maybe she's a pescetarian?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Random additional note:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Her age (I'm assuming it was supposed to be birth date), '31/02/102' is also suspicious for a number of reasons. I'm pretty sure she's not almost 2000 years old, and February doesn't have 31 days. So that probably has something to do with something. Or maybe not. Who knows.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12199" LastEditorUserId="12199" LastEditDate="2015-05-15T16:39:38.873" LastActivityDate="2015-05-15T16:39:38.873" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15017" PostTypeId="2" ParentId="14987" CreationDate="2015-05-15T17:00:13.503" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think it is &quot;Hurricane&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the United States,&#xA;You can find me in a camp and in a city,&#xA;You can find me in a creek and in a grove,&#xA;You can find me on a hill and on a land.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hurricane Campground, Hurricane (Utah), Hurricane Creek (Alabama), Hurricane Grove (Arkansas), Hurricane Hill (Washington), Hurricanes which make landfall?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You call my nicknames all the time,&#xA;You are addicted to me,&#xA;But if you are careless,&#xA;I will kill you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hurricanes classified by names. Stormchasers become addicted.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You saw me alone in 2008 and 2013,&#xA;But you saw me with my girl friend in 2001 and 2006.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; On a limb here, because I can't find definative info, but assuming maybe only a male named hurricanes made landfall in 2008, 2013, whereas male and female named hurricanes in 2001 and 2006?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You know who I am,&#xA;But will you find out?&lt;/p&gt;&#xA;" OwnerUserId="12483" LastActivityDate="2015-05-15T17:00:13.503" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15018" PostTypeId="1" AcceptedAnswerId="15037" CreationDate="2015-05-15T17:14:06.407" Score="4" ViewCount="590" Body="&lt;p&gt;This is a variation of the &lt;a href=&quot;http://www.xkcd.com/blue_eyes.html&quot; rel=&quot;nofollow&quot;&gt;blue eyes&lt;/a&gt; puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the logicians want to get off the island. They are selfish and think only of themselves. They may help others as well, but solely for their own benefit. However, if a logician realises that he has brown eyes, he will try to ensure that the maximum number of people get off the island.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the first day, the Guru makes her statement. On every subsequent day, 1 logician is secretly picked at random, and is asked to make a statement. The statement comes via an intercom, so no one knows who has given the statement or if it is true. The statement can not refer to individuals (by name, place, etc.) but must refer only to the following three sets - blue-eyed, brown-eyed, all. It must be logical and can use maths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will anyone ever make a useful statement? Will anyone ever leave before the 100th day?&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-05-15T21:21:53.170" Title="Variant to blue eyes puzzle - selfish logicians" Tags="|logical-deduction|blue-eyes|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="15019" PostTypeId="1" AcceptedAnswerId="15021" CreationDate="2015-05-15T17:16:15.070" Score="4" ViewCount="881" Body="&lt;p&gt;Forgive me if this matches an existing murder mystery. I tried to search before writing this one up&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Outside the office building, sirens wailed and flashing lights illuminated the snow covered ground. Inside, a troop of police officers were milling about, gathering clues and investigating the scene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just inside the door, behind the secretary's desk, lay a dead body. The body was that of the poor secretary, Alicia Adams. It appears that Alicia was working late, transcribing some documents into her computer when the assailant came in and shot her twice, once in the head and again in the chest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the stand next to her monitor was the original document she was transcribing, and a word processor was visible on the screen. The document on the screen showed :&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;...We need to meet our quodkhe6rthgfvbn&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It appears that, after she had been shot she had hit her head once on the keyboard, thus the garbled letters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, security logs show three people in the office at that time:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Byron Baker - The CEO&#xA;Cindy Chase - The CFO&#xA;Darryl Davis - The CIO&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;After taking the three into custody and recording their statements, the police couldn't determine who would have killed the secretary. They did find that any of the three would have adequate motive, as the secretary was moving to blow the whistle on a major scam the three were pulling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a few days, the case started going cold. So, the force brought in a detective from a neighboring precinct. After reviewing the case files, including Alicia's exemplary reviews and skills analyses, she told the officers to make an arrest. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who killed Alicia, and how did the detective know?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="12242" LastEditorUserId="4050" LastEditDate="2015-05-15T17:23:37.797" LastActivityDate="2015-05-15T17:28:14.657" Title="A Murder Most Foul: Alicia Adams" Tags="|word-problem|mystery|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15021" PostTypeId="2" ParentId="15019" CreationDate="2015-05-15T17:21:38.983" Score="4" Body="&lt;p&gt;It was&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cindy Chase - The CFO&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DKHE6 is a keyboard shifted version of CINDY&lt;br&gt;D is above C&lt;br&gt;K is below I&lt;br&gt;H is above N&lt;br&gt;E is above D&lt;br&gt;6 is above Y&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In addition, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If we can presume that Cindy did, in fact, hit her head on the keyboard, then that explains the rest of the mess at the end: rthgfvbn&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More supporting evidence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If Alicia was really as awesome as she appears to be, she would known that typing &quot;CINDY&quot; on the screen might have been noticed by the murderer. She might have even thrown herself forward to ensure her head hit the keyboard and thus leading Cindy to assume the entire mess was from the head impact and not a clue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2015-05-15T17:28:14.657" LastActivityDate="2015-05-15T17:28:14.657" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="15022" PostTypeId="2" ParentId="15018" CreationDate="2015-05-15T17:27:20.230" Score="0" Body="&lt;p&gt;I don't think it would make a difference (not 100% sure):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since no one knows their own eye colour, they will all tell false statements as their defense (useless) and the only way they people with blue eyes realize they [all] have blue eyes is on the last day (similar to brown eyes when they are left behind). So even if the brown eyes try to help, any (all) of the perfect logical blue eyes should have been gone by then.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastActivityDate="2015-05-15T17:27:20.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15023" PostTypeId="1" AcceptedAnswerId="15025" CreationDate="2015-05-15T17:44:10.240" Score="22" ViewCount="1508" Body="&lt;p&gt;Luna is 34.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scarlett is 37.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ray is 88.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paris is 71.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How old is Rose?                                                     &lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="4050" LastEditDate="2015-05-15T17:49:31.870" LastActivityDate="2015-05-16T10:36:26.953" Title="An elementary puzzle: How old is Rose?" Tags="|enigmatic-puzzle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15024" PostTypeId="2" ParentId="14987" CreationDate="2015-05-15T17:47:48.867" Score="7" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SUGAR.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I found this after seeing the OP's comment that Ethan the Brave's answer was on the right tracks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In the United States,&lt;br&gt;&#xA; You can find me in a camp and in a city,&lt;br&gt;&#xA; You can find me in a creek and in a grove,&lt;br&gt;&#xA; You can find me on a hill and on a land.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Sugar_Camp,_Wisconsin&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sugar Camp&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Sugar_City,_Idaho&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sugar City&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Sugar_Creek&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sugar Creek&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Sugar_Grove&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sugar Grove&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Sugar_Hill&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sugar Hill&lt;/a&gt;, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Sugar_Land,_Texas&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sugar Land&lt;/a&gt; are all places in the United States.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You call my nicknames all the time,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Sugar&quot; is sometimes used as an affectionate nickname.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You are addicted to me,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many people have a 'sweet tooth', as though they're addicted to sugar.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But if you are careless,&lt;br&gt;&#xA; I will kill you.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People with diabetes have to be careful about sugar.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You saw me alone in 2008 and 2013,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Films called &lt;em&gt;Sugar&lt;/em&gt; came out in &lt;a href=&quot;http://en.wikipedia.org/wiki/Sugar_(2008_film)&quot; rel=&quot;nofollow noreferrer&quot;&gt;2008&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Sugar_(2013_film)&quot; rel=&quot;nofollow noreferrer&quot;&gt;2013&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But you saw me with my girl friend in 2001 and 2006.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/2001_Sugar_Bowl&quot; rel=&quot;nofollow noreferrer&quot;&gt;2001 Sugar Bowl&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/2006_Sugar_Bowl&quot; rel=&quot;nofollow noreferrer&quot;&gt;2006 Sugar Bowl&lt;/a&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You know who I am,&lt;br&gt;&#xA; But will you find out?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stylistic with no extra clues.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If anyone's interested in my solving strategy: the OP's &lt;a href=&quot;https://puzzling.stackexchange.com/questions/14987/i-shrunk-the-kids/15024#comment46079_15008&quot;&gt;comment to Narmer&lt;/a&gt; told me the first verse meant there were places called X Camp, X City, X Creek, and so on in the US and I just had to find X. The title, together with the OP's &lt;a href=&quot;https://puzzling.stackexchange.com/questions/14987/i-shrunk-the-kids/15024#comment46090_15011&quot;&gt;comment to Ethan the Brave&lt;/a&gt;, told me it was something close to honey. Sugar immediately came to mind, and the rest was Googling.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-05-19T09:14:11.177" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15025" PostTypeId="2" ParentId="15023" CreationDate="2015-05-15T17:59:36.530" Score="24" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 45.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The theme of the puzzle is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the periodic table.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Luna is 34.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Element 34: selenium. Selenium and Luna both mean &quot;moon&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Scarlett is 37.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Element 37: rubidium. Rubidium and Scarlett are both related to &quot;red&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Ray is 88.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Element 88: radium, or &lt;strong&gt;ray&lt;/strong&gt;-dium.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Paris is 71.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Element 71: lutetium. Paris was formerly called Lutetia.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;How old is Rose?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Element 45: rhodium, meaning rose.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-05-16T10:36:26.953" LastActivityDate="2015-05-16T10:36:26.953" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15026" PostTypeId="2" ParentId="14933" CreationDate="2015-05-15T18:19:31.660" Score="8" Body="&lt;p&gt;Part 1: What is a possible combination of weights?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Use power of 2 weights from 2^0 to 2^99 (obvious, everyone already said so)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Part 2: What is the lowest total weight?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; User12408 already showed that the lowest total weight can be obtained by using power of 2 weights to reach total weight 2^100-1.  A quick summary of the proof:&lt;br&gt;&#xA;&lt;br&gt;&#xA; With 100 weights, there are 2^100 possible weight combinations (each weight can either be in a subset or not).  Given that &quot;no weights&quot; is one of these 2^100 subsets, the best to hope for is for the subsets to cover weight values 0..2^100-1, which can be achieved by using power of 2 weights.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Part 3: What is the lowest highest weight?  This is the interesting part.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people said that 2^99 was the answer for this.  But that is not true.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Take for example the case of 5 weights.  One might think that &lt;code&gt;1 2 4 8 16&lt;/code&gt; would be the optimal set.&lt;br&gt;&#xA;&lt;br&gt;&#xA; However, a counterexample is the set: &lt;code&gt;6 9 11 12 13&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; This set produces the following distinct subset sums:  &lt;code&gt;6 9 11 12 13 15 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 36 38 39 40 42 45 51&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; (Edit: A four weight counterexample is: &lt;code&gt;3 5 6 7&lt;/code&gt; by the way)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Something similar can be done for the 100 weight case, but it would be hard to find the optimal solution.&lt;br&gt;&#xA;&lt;br&gt;&#xA; As Jakub Tarnawski pointed out, this is an open problem that even has some kind of reward for a solution (this link comes from Jakub's answer): &lt;a href=&quot;http://www.openproblemgarden.org/op/sets_with_distinct_subset_sums&quot; rel=&quot;nofollow&quot;&gt;http://www.openproblemgarden.org/op/sets_with_distinct_subset_sums&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The woodchuck must chuck all it could if the woodchuck could chuck wood.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4479" LastEditorUserId="2383" LastEditDate="2015-07-07T22:02:49.750" LastActivityDate="2015-07-07T22:02:49.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15027" PostTypeId="1" CreationDate="2015-05-15T18:21:50.030" Score="1" ViewCount="245" Body="&lt;p&gt;In a society (presumably Asian) where sons are preferred over daughters, couples adopt the following heuristic when reproducing:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the new-born is a daughter, try for a new child&lt;/li&gt;&#xA;&lt;li&gt;When the couple finally gets a son (even if it's the first), stop.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Under the following assumption:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Fertility doesn't decrease regardless of number of children&lt;/li&gt;&#xA;&lt;li&gt;Equal probability of getting a son or a daughter&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Would we expect the society to have more daughters, more sons or equal (in one generation)? &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;side note: I find this problem interesting because the intuitive answers that people come up with can be vastly different. Also, can you come up with a convincing/correct proof without involving difficult math/probability?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="12490" LastEditorUserId="4050" LastEditDate="2015-05-15T18:37:23.793" LastActivityDate="2015-05-15T19:18:31.383" Title="More males or females" Tags="|mathematics|calculation-puzzle|" AnswerCount="4" CommentCount="2" ClosedDate="2015-05-15T19:01:10.480" ContentLicense="CC BY-SA 3.0" />
  <row Id="15028" PostTypeId="2" ParentId="15027" CreationDate="2015-05-15T18:24:22.943" Score="1" Body="&lt;p&gt;Equal, because there's an &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Equal probability of getting a son or a daughter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the sex of the parents' previous children doesn't change that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because, for each child which is conceived, the probability of that child being male or female is equal, it therefore follows that at a population level there should be equal numbers of boys and girls.&lt;/p&gt;&#xA;" OwnerUserId="4050" LastActivityDate="2015-05-15T18:24:22.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15029" PostTypeId="2" ParentId="15027" CreationDate="2015-05-15T18:25:48.580" Score="0" Body="&lt;p&gt;They will be equal in number. The nth child,  for all n, of all families will be a boy or girl, with equal probability. This is sufficient. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry I can't elaborate further, on mobile right now. &lt;/p&gt;&#xA;" OwnerUserId="5778" LastActivityDate="2015-05-15T18:25:48.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15031" PostTypeId="2" ParentId="15027" CreationDate="2015-05-15T18:48:18.027" Score="1" Body="&lt;p&gt;As others have said already, society would have&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; roughly the same number of males and females&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is an example that attempts to make this more clear:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GND2C.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Remember that ONLY those who have girls can continue to try for more children. That is why the number of couples is roughly halved with each round of children. This example is meant to help you understand the overall concept. Please don't take it too literally. In reality, they obviously would not all have children at the same time.&lt;/p&gt;&#xA;" OwnerUserId="463" LastActivityDate="2015-05-15T18:48:18.027" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15032" PostTypeId="2" ParentId="15027" CreationDate="2015-05-15T19:00:08.510" Score="0" Body="&lt;p&gt;I wrote a Python program to simulate this situation.&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from random import choice&#xA;&#xA;males = 0&#xA;females = 0&#xA;&#xA;def have_child():&#xA;    global males, females&#xA;    sex = choice(('male', 'female'))&#xA;&#xA;    if sex == 'male':&#xA;        males += 1&#xA;    else:&#xA;        females += 1&#xA;        have_child() # Have another one!&#xA;&#xA;for i in range(1000): # Assuming 1000 couples.&#xA;    have_child()&#xA;&#xA;print('Males born: %s. Females born: %s.' % (males, females))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Running the simulation several times, this is the result.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Males born: 1000. Females born: 929.&lt;br&gt;&#xA;Males born: 1000. Females born: 996.&lt;br&gt;&#xA;Males born: 1000. Females born: 989.&lt;br&gt;&#xA;Males born: 1000. Females born: 969.&lt;br&gt;&#xA;Males born: 1000. Females born: 1050.&lt;br&gt;&#xA;Males born: 1000. Females born: 1020.&lt;br&gt;&#xA;Males born: 1000. Females born: 1044.&lt;br&gt;&#xA;Males born: 1000. Females born: 962.&lt;br&gt;&#xA;&lt;br&gt;&#xA; So, on average, the number of male and female children born is the same.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11773" LastEditorUserId="7715" LastEditDate="2015-05-15T19:18:31.383" LastActivityDate="2015-05-15T19:18:31.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15033" PostTypeId="2" ParentId="15007" CreationDate="2015-05-15T19:33:58.267" Score="-1" Body="&lt;p&gt;street name  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; canal street  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;she is in  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; book store  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;her name is  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; sandy fish&lt;br&gt;&#xA;&lt;br&gt;&#xA;Canal street because of typing the phone number in your old mobile phone it says canal st&lt;br&gt;&#xA;Also as she likes to read book so she might wait in the book store and about the name sandy from blog's name  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12493" LastEditorUserId="5188" LastEditDate="2015-05-16T07:36:34.093" LastActivityDate="2015-05-16T07:36:34.093" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15034" PostTypeId="1" AcceptedAnswerId="23688" CreationDate="2015-05-15T19:38:02.413" Score="5" ViewCount="510" Body="&lt;p&gt;Thanks for helping me get on my &lt;a href=&quot;https://puzzling.stackexchange.com/questions/14992/which-flight-takes-me-home&quot;&gt;flight&lt;/a&gt; back home! Sorry I didn't elaborate on the extent of my travels, but I'm sure you can figure it out for yourself. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apart from the amazing view when leaving the airport, this flight is getting pretty boring. Let's take a look at the puzzle section in the in-flight magazine!&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;FEBNA       EIHSA&#xA;VBURM       HAYSR&#xA;TOREE  --&amp;gt;  MNEKT&#xA;SOVSS       URIEW&#xA;FINED       ALSOQ&#xA;&#xA;Clue: SIAMESESIAMESESIAMESESIAMESESIAMESE&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Anyway, what city did I just fly out from, and what city am I going to?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\_\_,\_\_,\_\_,\_\_,\_\_,\_\_\to\_\_,\_\_,\_\_,\_\_,\_\_,\_\_,\_\_,\_\_,\_\_$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Meta:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This puzzle is a tiny bit more mathematically inclined. It shouldn't be a hindrance to anyone, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Free hint: There are no tricks in the wording of the puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Want to contact me? Just mail me something!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5778" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-10-30T19:13:58.697" Title="Finally Returning Home From my Vacation!" Tags="|mathematics|cipher|pattern|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15035" PostTypeId="2" ParentId="12284" CreationDate="2015-05-15T20:05:17.733" Score="1" Body="&lt;p&gt;Alternatives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Option 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There was no dragon. The magician covered herself in fake blood in a back room and was watching the trick the whole time. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Option 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The magician saw a photo/sketch of the table including the dice without the cups on top.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Option 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The assistant is magic and, with her dying breath, transmitted the information of the dice to the magician.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Option 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The magician was wearing X-Ray specs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Option 5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The magician took a wild guess and was right.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Option 6:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The dragon, being magical itself and horribly misunderstood, tried to aid the magician by telling her about the assistant and the dice. Unfortunately, the dragon was still slain but the magician remembered the words about the dice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Option 7:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The tabletop is glass and there was a mirror on the floor. Since opposite sides on a dice always add up to 7, it was easy for her to figure out what the dice were.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Option 8:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It's the year 5645 and humans have successfully created dragons. Big mistake. The inside of the opaque cup has a small camera with a mini light that is wirelessly linked to special contacts the magician has. Child's play to determine the dice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12242" LastActivityDate="2015-05-15T20:05:17.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15036" PostTypeId="2" ParentId="14953" CreationDate="2015-05-15T21:16:01.387" Score="9" Body="&lt;p&gt;This might help you visualize no. 2 a bit better:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hkBNE.gif&quot; alt=&quot;If only the green box didn&amp;#39;t cover the purple one in the first image&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although you could probably find different patterns, to me this seems like the simplest way to describe the progression. I believe such sequences are fairly common in IQ tests. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you had to continue the sequence:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;3, 1, 0, 2, -3, 3, -6, 4, ?, ?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;a simple approach would be to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; consider the sequence as two interlaced sequences: &lt;code&gt;3, 0, -3, -6&lt;/code&gt; and &lt;code&gt;1, 2, 3, 4&lt;/code&gt; and continue with &lt;code&gt;-9, 5&lt;/code&gt;. This, in a way, is a numerical analogy for the graphical test above.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There may easily be other expressions which could be applied to the sequence, but I would say this &lt;em&gt;is&lt;/em&gt; the point of an IQ test; to find the simplest, most logical solution. Given the same task, a machine (i.e. a computer program) would probably try curve fitting with various polynomials or splines until a match is found, and it would be a perfectly valid continuation of the sequence - but hardly a sign of intelligence.&lt;/p&gt;&#xA;" OwnerUserId="12495" LastActivityDate="2015-05-15T21:16:01.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15037" PostTypeId="2" ParentId="15018" CreationDate="2015-05-15T21:21:53.170" Score="7" Body="&lt;p&gt;According to the puzzle definition, they can only leave if they know their own eye color, and they all want to leave as soon as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the first logician honestly states &quot;I see X people with blue eyes.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If he has blue eyes, he says &quot;I see 99 people with blue eyes.&quot;&#xA;All the brown-eyed people see 100 people with blue eyes. Each of them knows that if he had blue eyes, everyone else would see at least 100 people with blue eyes, so the first speaker would not have said 99.  They all conclude that they have brown eyes and leave that night.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If he has brown eyes, he says &quot;I see 100 people with blue eyes.&quot;  Each of the blue-eyed people see 99 other blue-eyed people, and know that in order for the first speaker to tell the truth, he must also have blue eyes.  All the blue-eyed people leave that night.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next day, everyone, including the original speaker, knows that everyone remaining must have the other eye color, so they all leave the following night.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This plan allows the first speaker to leave the night of the day after he spoke.  There is no way the speaker could learn his own eye color faster; speaking the truth is clearly the optimal path.  Since speaking the truth is the optimal path, everyone else will trust him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, this is what he would do, and everyone of the other eye color would leave that night, and everyone else would leave the next night.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Except the guru.  No one would benefit by telling her what her eye color is, so she remains stuck, alone.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2015-05-15T21:21:53.170" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15038" PostTypeId="1" AcceptedAnswerId="15039" CreationDate="2015-05-15T21:50:31.227" Score="3" ViewCount="257" Body="&lt;p&gt;I have been given a guide on how to reach satisfaction, and part of this guide requires me to meditate exactly for 90 minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have only been provided 2 oil-soaked ropes which last for 2 hours each when lit, and matches. (To time myself with)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have a knife on me, and I am too weak to break it with my hands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I meditate for exactly 90 minutes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: If you have seen a similar variation of this please don't answer)&lt;/p&gt;&#xA;" OwnerUserId="9159" LastEditorUserId="9159" LastEditDate="2015-05-15T23:44:31.233" LastActivityDate="2015-05-15T23:44:31.233" Title="How can I solve my meditation problem?" Tags="|logical-deduction|" AnswerCount="1" CommentCount="5" ClosedDate="2015-05-15T22:34:43.323" ContentLicense="CC BY-SA 3.0" />
  <row Id="15039" PostTypeId="2" ParentId="15038" CreationDate="2015-05-15T22:09:51.387" Score="4" Body="&lt;p&gt;To time 90 minutes as your guide suggests:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lay the ropes next to each other.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Light one rope from both ends.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When it burns out (1 hour) &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; light the other rope at the middle (this is why you lay them next to each other) and at both its ends.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This second rope will burn for &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;30 minutes  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;, giving you a total of 90 minutes.&lt;/p&gt;&#xA;" OwnerUserId="7715" LastEditorUserId="7715" LastEditDate="2015-05-15T22:24:43.090" LastActivityDate="2015-05-15T22:24:43.090" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15040" PostTypeId="1" CreationDate="2015-05-16T01:35:19.933" Score="1" ViewCount="1198" Body="&lt;p&gt;This question has been tagged as duplicate of &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2188/two-doors-with-two-guards-one-lies-one-tells-the-truth&quot;&gt;Two doors with two guards&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;What makes this questions different:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Let me state the obvious about this question:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;you meet &lt;strong&gt;one&lt;/strong&gt; islander.&lt;/li&gt;&#xA;&lt;li&gt;you ask  &lt;strong&gt;one&lt;/strong&gt; question.&lt;/li&gt;&#xA;&lt;li&gt;you resolve &lt;strong&gt;one&lt;/strong&gt; unknown (just to gain more knowledge).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Let me state the obvious about the other question:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;you meet &lt;strong&gt;two&lt;/strong&gt; guards.&lt;/li&gt;&#xA;&lt;li&gt;you meet &lt;strong&gt;two&lt;/strong&gt; doors.&lt;/li&gt;&#xA;&lt;li&gt;you need to survive the question.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For someone which does &lt;em&gt;not&lt;/em&gt; know the answer, there are (at least) four basic options for the guards question:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Ask guard 1.&lt;/li&gt;&#xA;&lt;li&gt;Ask guard 2.&lt;/li&gt;&#xA;&lt;li&gt;Ask about door 1.&lt;/li&gt;&#xA;&lt;li&gt;Ask about door 2.&lt;br&gt;&lt;br&gt;&#xA;Then, there is still the issue of which question to ask.&lt;br&gt;&lt;br&gt;&#xA;It should be pretty clear by now which question has a more complex structure.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h1&gt;Why are this differences important:&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;This type of puzzle may be quite difficult for someone who has never solved one before.&#xA;It is therefore very important that the individual solving the puzzle is confronted to the least amount of variables. That makes the puzzle simpler to process. And the answer much more revealing.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Other versions of the same theme:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There are several versions of similar questions around the web:&lt;br&gt;&#xA;&lt;a href=&quot;http://tvtropes.org/pmwiki/pmwiki.php/Main/KnightsAndKnaves&quot; rel=&quot;nofollow&quot; title=&quot;&amp;quot;Knights and Knaves&amp;quot;&quot;&gt;Knights and Knaves&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.intellectualloafing.com/activitiesfolder/logicqsfolder/twopathstwoangelsonequestion.htm&quot; rel=&quot;nofollow&quot; title=&quot;&amp;quot;Two Angels and Two Paths&amp;quot;&quot;&gt;Paths and Angels&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.sciforums.com/threads/the-deceiver-and-the-angel.2169/&quot; rel=&quot;nofollow&quot; title=&quot;&amp;quot;Two Doorways and Two Guards&amp;quot;&quot;&gt;SciForums&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.shlomifish.org/MathVentures/two-angels-and-two-gates.html&quot; rel=&quot;nofollow&quot; title=&quot;&amp;quot;Two Gates and Two Angels&amp;quot;&quot;&gt;Angels and Doors&lt;/a&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such questions make people expect some already known structure of twoA-twoB. That leads people to think in such pre-set format. A clear example of such inducted response is the user of this site which constructed the new &quot;two stuffed animals&quot; and &quot;two tea cups&quot; question in the comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He did &lt;em&gt;not&lt;/em&gt; use the &quot;one XXXX&quot; and &quot;one YYYY&quot; format this question present.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;What make this questions equal:&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The core rule to use to answer this sort of problems is exactly the same.&lt;br&gt;&#xA;Devoid of any embellishment, the core idea is based in:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The product of two numbers with different signs is negative.&lt;br&gt;&#xA;     Stated again:   &quot;different signs&quot; is &quot;negative&quot;.&lt;br&gt;&#xA;Crossing the answers of &quot;a true teller&quot; and &quot;a lie teller&quot; allways results in &quot;a lie&quot;.&lt;br&gt;&#xA;Therefore, use this construct: &quot;what would the other ____ say/answer&quot;&lt;br&gt;&#xA;or this (equivalent) one: &quot;Would the other ____ tell me that&quot;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then, make a clear question, which is preferable if it leads to a yes/no answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    &quot;Would the other tribe tell me that the statue is white?&quot;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the other question:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    &quot;What would the other guard say if asked: Is this the door to freedom?&quot;&lt;br&gt;&#xA;Understand any answer as the opposite of what was received: that is the truth.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In that, both questions are resolved by the same construct.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But not all problems that are resolved by the same set of rules are duplicate.&lt;br&gt;&#xA;Not all sudoku games are duplicate of each other, nor all games of chess&#xA;are duplicate of each other. Even if the basic rules are exactly the same.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;The question (as it was originally):&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Lets assume that you land (with a parachute) in an island. Only this facts are known about the island:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The island has an statue in the middle.&lt;/li&gt;&#xA;&lt;li&gt;The statue is either black or white (binary state).&lt;/li&gt;&#xA;&lt;li&gt;There are two tribes of islanders.&lt;/li&gt;&#xA;&lt;li&gt;One tribe always tell the truth.&lt;/li&gt;&#xA;&lt;li&gt;The other tribe always tell lies (the contrary of true, binary also).&lt;/li&gt;&#xA;&lt;li&gt;Both tribes do know what is the true color of the statue.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Limited by this rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You do not know anything else about the island than what was stated above.&lt;/li&gt;&#xA;&lt;li&gt;You do not know (by any means) to which tribe belong an islander.&lt;/li&gt;&#xA;&lt;li&gt;You are forbidden (can not go) to see the statue.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You find an islander, and, asking only ONE question, you can reliably tell of what color is the statue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the question you need to ask?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: The question does not contain multiple separate questions or have multiple separate answers, only ONE question, ONE answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: There is a &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2188/two-doors-with-two-guards-one-lies-one-tells-the-truth&quot; title=&quot;&amp;quot;A similar puzzle&amp;quot;&quot;&gt;similar question here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user12496" LastEditorDisplayName="user12496" LastEditDate="2015-05-23T19:25:21.230" LastActivityDate="2015-05-23T19:25:21.230" Title="What color is the statue?" Tags="|logical-deduction|liars|" AnswerCount="2" CommentCount="4" ClosedDate="2015-05-16T03:08:50.880" ContentLicense="CC BY-SA 3.0" />
  <row Id="15041" PostTypeId="2" ParentId="15040" CreationDate="2015-05-16T01:45:16.747" Score="2" Body="&lt;p&gt;The question you ask is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;What would the other tribe tell me the statue's color is?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;No matter who you ask, they will tell you the opposite color than what the statue is.&lt;br&gt;&lt;br&gt;Assume the statue is white. Asking the lying tribe, they would lie and say that the truth-tellers would say it is black. Asking the truth tellers, they would tell the truth and say that the liars would say the statue is black. You therefore know the statue is white. The same applies for if statue is black, then either tribe would answer white to the question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11727" LastActivityDate="2015-05-16T01:45:16.747" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15042" PostTypeId="2" ParentId="15040" CreationDate="2015-05-16T01:47:21.083" Score="2" Body="&lt;p&gt;Ask this question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering the following function in pseudocode,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;function answer(islander, object) {&#xA;    if(islander.tellsTruth) {&#xA;        if(object.color == &quot;black&quot;) {&#xA;            return 1;&#xA;        } else {&#xA;            return 0;&#xA;        }&#xA;    } else {&#xA;        if(object.color == &quot;black&quot;) {&#xA;            return 0;&#xA;        } else {&#xA;            return 1;&#xA;        }&#xA;    }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;what would the return value be for &lt;code&gt;answer(you, statue)&lt;/code&gt;?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If the statue is black:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the islander tells the truth, the answer would be 1; if the islander lies, then the answer &lt;em&gt;should&lt;/em&gt; be 0, and since he lies he would reply 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the statue is white:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the islander tells the truth, the answer would be 0; if the islander lies, then the answer &lt;em&gt;should&lt;/em&gt; be 1, and since he lies he would reply 0.&lt;/p&gt;&#xA;" OwnerUserId="12278" LastActivityDate="2015-05-16T01:47:21.083" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15046" PostTypeId="1" CreationDate="2015-05-16T05:40:19.140" Score="-3" ViewCount="348" Body="&lt;h2&gt;Original problem&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Lets assume that we hire two persons to count passers-by for an hour.&#xA;One is told to stand still at one spot on the sidewalk.&#xA;The other one is told to walk 20 meters up and down the same sidewalk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would they get the same count? Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Idea reworded from this &lt;a href=&quot;https://archive.org/details/opensource&amp;amp;tab=about&quot; rel=&quot;nofollow&quot;&gt;open-source&lt;/a&gt;(contributed by the community) -- &lt;a href=&quot;https://archive.org/details/MathematicsCanBeFun-1&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt; --&lt;/p&gt;&#xA;" OwnerDisplayName="anon" LastEditorDisplayName="anon" LastEditDate="2015-05-20T21:10:46.840" LastActivityDate="2015-05-20T21:10:46.840" Title="Will both workers count the same number?" Tags="|calculation-puzzle|open-ended|" AnswerCount="6" CommentCount="0" ClosedDate="2015-05-18T03:23:51.953" ContentLicense="CC BY-SA 3.0" />
  <row Id="15047" PostTypeId="2" ParentId="15046" CreationDate="2015-05-16T06:04:07.610" Score="1" Body="&lt;p&gt;Assuming that the walking counter starts and finishes right next to the stationary counter:&#xA;If the counters can recognize individuals and avoid double counting:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; they will both get the same count. &#xA; The cool thing is that this holds irrespective of how the passers-by are moving. This is because, having started in the same place, the difference in their counts at any time $t$ is less than or equal to the number of passers-by in between them. If they finish in the same place, there can be no passers-by between them and thus their counts must agree exactly. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If instead the counters may double count the same passer-by, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the walking individual may double count whereas the stationary one will not and thus the walking counter can get a higher (but not lower) count. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11087" LastEditorUserId="11087" LastEditDate="2015-05-16T07:38:57.727" LastActivityDate="2015-05-16T07:38:57.727" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="15048" PostTypeId="1" AcceptedAnswerId="15050" CreationDate="2015-05-16T06:14:29.113" Score="8" ViewCount="1236" Body="&lt;p&gt;Saw this on TV today:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A man is auctioning a real $20\$$ bill. There are a vast number of&#xA;  bidders. A person may make as many bids as he wants. The starting bid&#xA;  is $5\$$. No $2$ bids must be equal, and bids are in increments of&#xA;  $1\$$. The highest bidder wins. Bidding can go on indefinitely.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;However, there is a catch. The second highest bidder must also pay his&#xA;  last bid, even though he gets nothing in return.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On the show, they were trying to analyse how our brain works. I'm asking a slightly different question. &lt;strong&gt;What is the optimal strategy?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case I: All bidders play perfectly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case II: Bidders are humans, so some of them may deviate from the perfect strategy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is the best option not to bid at all? Or a bid till a certain amount and stop?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And lastly, has this Q been explored elsewhere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Bidders are anonymous, and do not form contracts.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="12109" LastEditDate="2015-05-16T17:25:58.673" LastActivityDate="2015-05-21T15:50:06.060" Title="Devious auction game" Tags="|logical-deduction|strategy|game|game-theory|" AnswerCount="6" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15050" PostTypeId="2" ParentId="15048" CreationDate="2015-05-16T06:34:45.163" Score="9" Body="&lt;p&gt;&lt;strong&gt;Case I:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us suppose that each player can bid any integer equal to or above the current minimum bid. Then:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Without knowing the total bankroll that each player has to invest, we cannot calculte the optimal strategies.  If we know that the bankroll of each player is &lt;em&gt;b&lt;/em&gt;, the optimal strategy is for the first player to bid ($(b-1)$ mod 19)+1 and the subsequent players not to bid. &#xA; (Irrespective of bankroll &lt;em&gt;b&lt;/em&gt;, there is also a trivial Nash equilibrium in which the first player bids $20 and all other players drop.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Barry O'Neill develops a simple formula for this strategy in a &lt;a href=&quot;http://www.jstor.org/discover/10.2307/174109&quot; rel=&quot;nofollow&quot;&gt;1986 article&lt;/a&gt; in the &lt;em&gt;Journal of Conflict Resolution.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we play by the exact rules listed here--first player must bid \$5 and each bid must increase by an increment of one dollar--I do not know the optimal solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case II: Playing against humans&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I've never seen the auctioneer take a loss when this game is played in real life, so the optimal strategy is not to bid at all. (If the bidders may form binding contracts to collude, i.e., if we treat this as a cooperative game, then of course they can buy the \$20 bill for \$5 and share the profits.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Has this been explored elsewhere?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This puzzle is known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dollar_auction&quot; rel=&quot;nofollow&quot;&gt;dollar auction&lt;/a&gt; and was first posed by Martin Shubik in a &lt;a href=&quot;http://www.math.toronto.edu/mpugh/Teaching/Sci199_03/dollar_auction_1.pdf&quot; rel=&quot;nofollow&quot;&gt;1971 article&lt;/a&gt; in the journal of Conflict Resolution. &lt;/p&gt;&#xA;" OwnerUserId="11087" LastEditorUserId="11087" LastEditDate="2015-05-21T15:50:06.060" LastActivityDate="2015-05-21T15:50:06.060" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15051" PostTypeId="2" ParentId="15046" CreationDate="2015-05-16T07:54:54.580" Score="1" Body="&lt;p&gt;Counts would be same. They are humans, they will obviously count right number of people, &lt;strong&gt;given&lt;/strong&gt; that they are inclined towards the work. &lt;br&gt;&#xA;&lt;strong&gt;Bonus:&lt;/strong&gt; &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If a machine (which just counts the number of bodies passing by) is made to move like that, the count may be different, because a person may walk too slow or too fast.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11425" LastActivityDate="2015-05-16T07:54:54.580" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15052" PostTypeId="2" ParentId="12796" CreationDate="2015-05-16T09:56:21.430" Score="2" Body="&lt;p&gt;Taking Mark N answer (thanks a lot) I'll try to improve a bit, using some lateral thinking with the parentesis. Still, most of the credit goes to him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I decided to think even further outside the box and use $.7$ notation to improve the score.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Score: 73&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1 = \binom{7}{7}$ [Score +0]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2 = \binom{7}{7}+\binom{7}{7}$ [Score +1]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3 = \binom{7}{7}+\binom{7}{7}+\binom{7}{7}$ [Score +2]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$4 = \binom{7}{7}+\binom{7}{7}+\binom{7}{7}+\binom{7}{7}$ [Score +3]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5 = 7-\binom{7}{7}-\binom{7}{7}$ [Score +2]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$6 = 7-\binom{7}{7}$ [Score +1]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$7 = 7$ [Score +0]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$8 = 7+\binom{7}{7}$ [Score +1]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$9 = 7+\binom{7}{7}+\binom{7}{7}$ [Score +2]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$10 = \frac{7}{.7}$ [Score +1]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$11 = \frac{7}{.7}+\binom{7}{7}$ [Score +2]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$12 = \frac{7}{.7}+\binom{7}{7}+\binom{7}{7}$ [Score +3]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$13 = 7+7-\binom{7}{7}$ [Score +2]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$14 = 7 + 7$ [Score +1]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$15 = 7 + 7  +\binom{7}{7}$ [Score +2]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$16 = 7 + 7 + \binom{7}{7}+\binom{7}{7}$ [Score +3]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$17 = \sqrt{\binom{7+7}{7}^.7}$ [Score +4]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$18 = \sqrt{7*7*7}$ [Score +4]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$19 = \frac{7+7}{.7}-\binom{7}{7}$ [Score +3]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$20 = \frac{7+7}{.7}$ [Score +2]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$21 = \binom{7}{\binom{7}{7}+\binom{7}{7}}$ [Score +1]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$22 = \binom{7}{\binom{7}{7}+\binom{7}{7}}+\binom{7}{7}$ [Score +2]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$23 = \binom{7}{\binom{7}{7}+\binom{7}{7}}+\binom{7}{7}+\binom{7}{7}$ [Score +3]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$24 = (7+\binom{7}{7})*(\binom{7}{7}+\binom{7}{7}+\binom{7}{7})$ [Score +4]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$25 = (7*7*7+7)/(7+7)$ [Score +5]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$26 = \sqrt{7!/7}$ [Score +5]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$27 = 7+7+7+7-\binom{7}{7}$ [Score +4]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$28 = 7+7+7+7$ [Score +3]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$29 = 7+7+7+7+\binom{7}{7}$ [Score +4]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$30 = \frac{7+7+7}{.7}$ [Score +3]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll be editing and improving when I come up with new ones.&lt;/p&gt;&#xA;" OwnerUserId="12165" LastEditorUserId="12165" LastEditDate="2015-05-18T10:30:37.140" LastActivityDate="2015-05-18T10:30:37.140" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15053" PostTypeId="1" AcceptedAnswerId="15057" CreationDate="2015-05-16T12:48:41.383" Score="1" ViewCount="1318" Body="&lt;p&gt;I found the following problem in a book and don't understand exactly what is being asked.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(I) Kamal is available at home from 12 noon to 4 p.m. on Tuesday, thursday and   Sunday.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(II) His younger brother Navin is available at home on Monday, Thursday, Friday and Sunday between 10 a.m. to 2 p.m.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(III) The eldest brother Rajiv is available between 9 a.m. to 12 noon on Monday, Wednesday and Thursday and 2 p.m. to 4 p.m. on Friday, Saturday and  Sunday. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;For how many days only one brother is available at a particular time in a week ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer given in my book is as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;One brother is available at a particular time on all seven days of the week.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Please explain.&lt;/p&gt;&#xA;" OwnerUserId="12227" LastEditorUserId="5373" LastEditDate="2015-05-16T15:24:13.197" LastActivityDate="2015-05-16T15:24:13.197" Title="Time Sequence Puzzle (Days)" Tags="|logical-deduction|time|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15054" PostTypeId="2" ParentId="15000" CreationDate="2015-05-16T12:51:29.550" Score="12" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Which eight toppings are advertised?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The eight most commonly used coins of the United Kingdom:  1p, 2p, 5p, 10p, 20p, 50p, £1, £2. This explains your &quot;the penny finally drops&quot; and &quot;dough&quot; hints, and the &quot;Croesus' Pizzas&quot; title.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;how were their prices arrived at?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By multiplying the value of the coin by the amount that can fit on the pizza, packing them together as closely as you can. For example, you can fit 104 2p coins on a 12&quot; pizza*, so it costs 2.08. This explains your &quot;packing my suitcase as neatly as possible&quot; hint.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What are the missing prices?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to my &lt;a href=&quot;https://ideone.com/Yq48dr&quot; rel=&quot;nofollow noreferrer&quot;&gt;coin packing simulation&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;         1p     2p      5p      10p     20p     50p     £1      £2&#xA; 12&quot;     £1.73  £2.08   £10.65  £11.70  £32.60  £45.50  £151    £174&#xA; 14&quot;     £2.35  £2.86   £15.00  £16.10  £42.20  £63.50  £199    £238&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In 2017, the price for the £1 pie will change, because the current round coin will be replaced with a 12-sided design. This explains the &quot;New for 2017!&quot; header.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;*I suspect we're using slightly different packing strategies, because I was able to fit 109 2p coins, not 104:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DpfJY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I had to make an in-story justification for this, I'd say &quot;the chef is clumsy and sometimes a few coins fall off during baking&quot;&lt;/p&gt;&#xA;" OwnerUserId="1497" LastEditorUserId="4050" LastEditDate="2015-05-16T15:16:58.850" LastActivityDate="2015-05-16T15:16:58.850" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15055" PostTypeId="2" ParentId="15048" CreationDate="2015-05-16T12:53:56.903" Score="1" Body="&lt;p&gt;&lt;strong&gt;Case I:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If no bid exists, bid 5$. Otherwise, do not bid. (&quot;Fastest&quot; bidder wins)&lt;br&gt;&#xA; This should be is optimal under all reasonable metrics since there's no way to achieve a larger profit, nor a way to split the profit (unless bidders are allowed to split it after winning).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case II:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A risk-less strategy is not to bid.&#xA; &lt;br&gt;&lt;br&gt;&#xA; A riskier strategy is to bid 19\$ if no bid exists, and nothing otherwise. This can give you a profit of 1$ assuming no human bidders decide to screw you over for no profit for themselves. &lt;br&gt;&#xA; (Was this riskier strategy ever attempted in a live game? Would be interesting to see the result!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4067" LastEditorUserId="4067" LastEditDate="2015-05-16T13:01:32.877" LastActivityDate="2015-05-16T13:01:32.877" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15056" PostTypeId="2" ParentId="15053" CreationDate="2015-05-16T13:00:44.810" Score="1" Body="&lt;p&gt;I think that the question can be rephrased as:&lt;br&gt;&#xA;&lt;strong&gt;Given a certain time, how many days have exactly one brother at home at that time?&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider 11pm: nobody is available at home at that hour, doesn't matter the day. So, the answer for 11pm would be 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the contrary, if you consider 1pm, you see that on Monday Navin is at home at that time (his brothers aren't), on Tuesday only Kamal is at home, Wednesday nobody, Tuesday both, Friday only Navin, Saturday nobody, Sunday both.&lt;br&gt;&#xA;The answer for 1pm is 3 (Monday, Tuesday, Friday).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a simple calculation problem, as you can see!&lt;/p&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-05-16T13:00:44.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15057" PostTypeId="2" ParentId="15053" CreationDate="2015-05-16T13:03:54.167" Score="1" Body="&lt;p&gt;I think the question is asking: &lt;strong&gt;for how many days of the week is there a time of day at which only one brother is present?&lt;/strong&gt; The answer then is &lt;strong&gt;all&lt;/strong&gt; days of the week, because we can rephrase the information as follows.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(I) Kamal is available at home from 12 noon to 4 p.m. on Tuesday, thursday and Sunday. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(II) His younger brother Navin is available at home on Monday, Thursday, Friday and Sunday between 10 a.m. to 2 p.m. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(III) The eldest brother Rajiv is available between 9 a.m. to 12 noon on Monday, Wednesday and Thursday and 2 p.m. to 4 p.m. on Friday, Saturday and Sunday.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Monday: &lt;strong&gt;only Rajiv&lt;/strong&gt; is at home between 9 and 10, both Navin and Rajiv from 10 to 12, and &lt;strong&gt;only Navin&lt;/strong&gt; from 12 to 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tuesday: &lt;strong&gt;only Kamal&lt;/strong&gt; is at home between 12 and 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wednesday: &lt;strong&gt;only Rajiv&lt;/strong&gt; is at home between 9 and 12.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thursday: &lt;strong&gt;only Rajiv&lt;/strong&gt; is at home between 9 and 10, Navin and Rajiv between 10 and 12, Kamal and Navin between 12 and 2, and &lt;strong&gt;only Kamal&lt;/strong&gt; between 2 and 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Friday: &lt;strong&gt;only Navin&lt;/strong&gt; between 10 and 2, &lt;strong&gt;only Rajiv&lt;/strong&gt; between 2 and 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saturday: &lt;strong&gt;only Rajiv&lt;/strong&gt; between 2 and 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sunday: &lt;strong&gt;only Navin&lt;/strong&gt; between 10 and 12, Navin and Kamal between 12 and 2, Kamal and Rajiv between 2 and 4.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-05-16T13:08:54.540" LastActivityDate="2015-05-16T13:08:54.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15059" PostTypeId="1" CreationDate="2015-05-16T14:29:45.273" Score="2" ViewCount="363" Body="&lt;p&gt;There is great excitement in the astronomy community. The search for extraterrestrial intelligence is finally bearing fruit. A strong clear repeating signal is being received from a not so distant star. The pattern being transmitted is as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A&#xA;AB&#xA;AACABB&#xA;ABA&#xA;AABCABBB&#xA;AAACABBA&#xA;ABAB&#xA;AABBCABBBB&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A&lt;/strong&gt; represents a high pitch tone, &lt;strong&gt;B&lt;/strong&gt; a low pitch tone, and &lt;strong&gt;C&lt;/strong&gt; a tone that oscillates rapidly between the two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The message lasts for roughly 14 seconds, then stops for about half an hour before repeating again. The time between each repetition is decreasing by about 35 seconds each time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the expected duration the message changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original message is still present but when it ends a new message starts after a few second pause.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;AAAAAAAAAAAAAAAAAAAAAABACADAAAAAAAAAAAAAAAAAAAAAAAAAAAABEACADDAAAAAAAAAAAAAAAAAAAAAAAAAABEEACADDDAAAAAAAAAAAAAAAAAAAAAAAAABEBACADDDDAAAAAAAAAAAAAAAAAAAAAAABEEEACADDDDDAAAAAAAAAAAAAAAAAAAAAABEEBACADDDDDDAAAAAAAAAAAAAAAAAAAAABEBEACADDDDDDDAAAAAAAAAAAAAAAAAAABEEEEACADDDDDDDD&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This new message is much longer than the first and with different tones, each a a different frequency.&lt;/p&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="11297" LastEditDate="2015-06-19T18:57:26.337" LastActivityDate="2015-06-19T18:57:26.337" Title="Another Puzzling Alien Signal" Tags="|enigmatic-puzzle|letter-sequence|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15060" PostTypeId="1" AcceptedAnswerId="15061" CreationDate="2015-05-16T14:38:21.880" Score="4" ViewCount="1054" Body="&lt;p&gt;It's late in the evening when your friend Mike, a brilliant astronomy student, points his radio telescope to Omega Centauri, a large globular cluster distant almost 16000 light-years from the Earth.&lt;br&gt;&#xA;After inspecting the sky for a few minutes, you notice a strange signal coming from an outer star. It's a 177 GHz microwave radiation with a variable amplitude.&lt;br&gt;&#xA;Mike is amazed, he has never seen a signal like it, so he decides to analyze it.&lt;br&gt;&#xA;Immediately he notices that the wave conveys a sort of message; then he observes that this message is periodic, so he is able to extrapolate this binary sequence:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now, Mike is at a dead end, he has no idea of how to interpret that signal. Does it really contain a message? Is it artificial or natural? Aliens are communicating?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Help Mike finding all the information included in the mysterious signal.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-05-16T16:05:45.170" Title="A mysterious signal from space" Tags="|cipher|enigmatic-puzzle|story|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="15061" PostTypeId="2" ParentId="15060" CreationDate="2015-05-16T15:33:06.233" Score="5" Body="&lt;p&gt;Arranged correctly the bits form a 19 by 41 grid very similar to the Arecibo message.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the pattern used in that message, this message contains the following information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their body chemistry is composed of the following elements:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen Lithium Oxygen Aluminium Chromium Iron Gallium&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably in these compounds...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4101000 Lithium Aluminium Hydride&lt;br&gt;&#xA;2040100 Chromic Acid&lt;br&gt;&#xA;3030001 Gallium Hydroxide&lt;br&gt;&#xA;0000010 Iron&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The aliens have two of what might be arms, three legs (I hope they're legs), and two horns, large ears or antennae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They might have considerable sexual dimorphism or fly about in 50s style rocket ships.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They live in a binary system on the third rock from their suns. The closest planet in the system to the suns is the largest.&lt;/p&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="5373" LastEditDate="2015-05-16T16:05:45.170" LastActivityDate="2015-05-16T16:05:45.170" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15062" PostTypeId="2" ParentId="15060" CreationDate="2015-05-16T15:52:38.617" Score="5" Body="&lt;p&gt;Just for clarity, this is a picture of the solution (the bits arranged in a 41x19 grid).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zy2Ut.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As you can see, it resembles the &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Arecibo_message&quot; rel=&quot;nofollow noreferrer&quot;&gt;Arecibo message&lt;/a&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-05-16T15:52:38.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15063" PostTypeId="2" ParentId="13303" CreationDate="2015-05-16T15:59:12.040" Score="24" Body="&lt;p&gt;&lt;em&gt;Posting this in a second answer so that solutions to different parts of the puzzle are separate. If people disapprove, I'll edit my previous answer and delete this one.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;North is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; to the &lt;strong&gt;right&lt;/strong&gt; on the map.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are vague blotches that look they were once letters on three points of the compass rose. To the right we have a circle and a semicircle, to the left a circle, and beneath we have a triangle (roughly). Each of these blotches corresponds to &lt;strong&gt;the hole in a capital letter&lt;/strong&gt; (hence the OP's hint &quot;looking at the blotches is a &lt;em&gt;capital&lt;/em&gt; idea&quot;). If we imagine the words NORTH, EAST, SOUTH, WEST written around the compass rose, all with their bottoms facing inwards, we'd get circles from the O's, a semicircle from the R, and a triangle from the A. The positions of the blotches also match this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In-universe explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Apparently your uncle was the sort to colour in the holes in written letters! But the original letters have faded, leaving only the filled-in holes behind.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now we're in good shape to solve the first three stanzas. There are six caves on the island (candidates for the &quot;dead man's cave&quot;), namely at TH, UE, AE, FE, ND, LQ. Starting from one of these, we want to follow the second stanza and&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/a/14985/5373&quot;&gt;go three east&lt;/a&gt;. Incidentally this eliminates the final two possibilities for the dead man's cave.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Right is left and left is right, turn and go until,&lt;br&gt;&#xA;Twins appear to mark an end, though you'll continue still.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I believe this means we should turn left (i.e. north, i.e. rightwards on the map), because &quot;right is left [behind] and left is [correct]&quot; and we should &quot;turn and go&quot;. So we should start from a cave, go three east (down), and then go north (right) until we reach some kind of &quot;twins&quot;. Incidentally this eliminates the 4th possibility for the dead man's cave - and also the 1st possibility, since I can't see anything looking like twins in the upper D row.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assuming &lt;a href=&quot;https://puzzling.stackexchange.com/a/14718/5373&quot;&gt;tfitzger is correct&lt;/a&gt; in his interpretation of the fourth stanza, we should then&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; go one east, one north, one east, two south, and turn to face west.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Following &lt;a href=&quot;https://puzzling.stackexchange.com/a/15079/5373&quot;&gt;Zandar's excellent solution&lt;/a&gt; for the third, fifth, and sixth stanzas, we get the following trip covering the first six stanzas (as shown in &lt;a href=&quot;https://puzzling.stackexchange.com/a/15200/5373&quot;&gt;Engineer Toast's map&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;HT DT DO EN EB EU&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(I'll leave my old answer for the fifth stanza crossed out below, since it was basically correct - I just started in the wrong place since I hadn't solved the third stanza - and almost certainly Zandar's inspiration for solving it correctly.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;s&gt;&lt;em&gt;A leap of faith, surrender yourself. But when things come to a head,&lt;br&gt;&#xA;Take the easy way out, with the blues left behind, go three and don't retread.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Leap into the river immediately to the west of you, surrender yourself to the current, and then at the head of the river (where it flows into the sea), emerge and go three squares in some direction. What direction? Well, we want to leave the blues (water) behind, so south (upwards) is the only direction that makes sense.&lt;/s&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Two one fives, all lined up, this time point the way.&lt;br&gt;&#xA;They also indicate how far. Go now or be led astray.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;s&gt;One and five on a clock face point in directions sixty degrees northwards and southwards respectively from east. Adding unit vectors in these two directions, we get a unit vector pointing due east (they form an equilateral triangle). So &quot;two one fives&quot; means &lt;strong&gt;two east&lt;/strong&gt;. This is the hidden meaning behind the phrase &quot;go &lt;em&gt;now&lt;/em&gt;&quot;: it's meant to make us think of a clock.&lt;/s&gt; In a similar vein, see &lt;a href=&quot;https://puzzling.stackexchange.com/a/15168/5373&quot;&gt;user1618143's answer&lt;/a&gt;, which tells us to go &lt;em&gt;three&lt;/em&gt; east.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proceed until, when viewed from on high, the island itself gives a sign.&lt;br&gt;&#xA;Next steps revealed as the river reflects and the ancient stones align.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first line tells us to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; go three north to the high point at HY (&quot;when viewed from on high&quot;), from which we can see the stones aligning (&quot;the island itself giv[ing] a sign&quot;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As &lt;a href=&quot;https://puzzling.stackexchange.com/a/15168/5373&quot;&gt;user1618143 discovered&lt;/a&gt;, the second line tells us to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; go three east.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Run forth with what you carry. Confusion becomes clear.&lt;br&gt;&#xA;Now piece together your reward, you'll find it buried here.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Confusion becomes clear&quot; is a clue referring to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ANAGRAMS. Applied to the first sentence of this stanza, with &quot;what you carry&quot; being the sword we were told to sharpen in stanza 1, it tells us &lt;strong&gt;four northwards&lt;/strong&gt; (an anagram of &quot;run forth ... a sword&quot;). Now how does it apply to the second line? Well, the OP has been leaving loads of comments on people's answers in the form of finding anagrams of their lists of coordinates. So let's look for anagrams of &quot;HT DT DO EN EB EU HU RY RN&quot; (by the way, &quot;piece together&quot; is also a hint towards anagrams). After working through things including &quot;burned throne, thy due&quot; and &quot;hundred butter honey&quot;, I finally found &quot;three hundred bounty&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the final answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;HT DT DO EN EB EU HU RY RN&lt;/strong&gt;; treasure is &lt;strong&gt;three hundred bounty&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cough up, Alconja! :-)))&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.923" LastActivityDate="2015-05-20T13:01:29.890" CommentCount="22" ContentLicense="CC BY-SA 3.0" />
  <row Id="15064" PostTypeId="2" ParentId="15048" CreationDate="2015-05-16T17:14:59.740" Score="1" Body="&lt;p&gt;&lt;strong&gt;Case I: All bidders play perfectly.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;SubCase A&lt;/strong&gt; : If bidders are allowed to cooperate with other bidders : Best Strategy is &quot;One nominated bidder goes for the lowest amount, nobody else bids, winner shares the profits with all bidders&quot;. Guaranteed profit for all bidders.&lt;br&gt;&#xA;  &lt;strong&gt;SubCase B&lt;/strong&gt; : If no way to cooperate with other bidders, and any number of people can bid at the same time : Best Strategy is &quot;Do not bid&quot;. If you do bid, and somebody else also bids simultaneously, then you both lose money eventually.&lt;br&gt;&#xA;  &lt;strong&gt;SubCase C&lt;/strong&gt; : If no way to cooperate with other bidders, but only one person at a time can bid (Example : there is always some order in which people can either bid or skip bidding in current state) : Best Strategy is &quot;Do not bid, unless you are the first bidder&quot;. Nobody else will bid against you, because they know that you will both lose money. Guaranteed profit for one bidder who may &lt;strong&gt;choose&lt;/strong&gt; to share the profit.&lt;br&gt;&#xA;  What SubCase C means is : If auctioneer goes around asking &quot;B1, do you want to bid ?&quot; &quot;B2, do you want to bid ?&quot; &quot;B3, do you want to bid ?&quot; &quot;B4, do you want to bid ?&quot;, and back again, then only B1 should say &quot;YES&quot;. If two more more people say &quot;YES&quot; , then they will all lose money.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case II: Bidders are humans, so some of them may deviate from the perfect strategy.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;For all three SubCases above, in case II&lt;/strong&gt; : Let us say the best strategy involves bidding. What happens if somebody deviates from that strategy ? Two or more people will lose money.&lt;br&gt;&#xA;  So Best Strategy for all SubCases must be &quot;Do not bid&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ANSWER EDIT&lt;/strong&gt;:&#xA;As per Question Edit: Bidders are anonymous, and do not form contracts.&#xA;So SubCase A can be removed from the list, leaving only SubCase B &amp;amp; SubCase C.&lt;/p&gt;&#xA;" OwnerUserId="10940" LastActivityDate="2015-05-16T17:14:59.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15065" PostTypeId="2" ParentId="15046" CreationDate="2015-05-16T18:07:40.447" Score="6" Body="&lt;p&gt;There are many, many variables at play in this situation that pretty much makes the answer&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;any answer you want.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;&lt;strong&gt;The walker counts less than the person standing still.&lt;/strong&gt; How? Well, if someone enters the sidewalk at a point P where 0m &amp;lt; p &amp;lt; 20m along the walker's path after the walker has reached that spot, it is possible that this person will never be counted by the walker.&#xA;&lt;br&gt;&lt;br&gt;&#xA;&lt;strong&gt;The walker counts the same as the person standing still.&lt;/strong&gt; How? Assuming that people start on the sidewalk more than 20m away from the walker's starting position, the walker will always see the same commuters that the stander sees. The walker can ignore duplication, so they get the same count.&#xA;&lt;br&gt;&lt;br&gt;&#xA;&lt;strong&gt;The walker counts more than the person standing still.&lt;/strong&gt; How? Use the same scenario as the same count, but remove the assumption that the walker can differentiate between duplicates.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Of course, all of this requires that they are in the same general area. A sidewalk can be far longer than 20m, so the two can be potentially blocks away and still be on the same sidewalk. This means that they see different crowds and can have larger, smaller, or equivalent data sets.&lt;/p&gt;&#xA;&lt;p&gt;Some variables that must be assumed to answer the question:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The walker starts or ends at the same place as the stander&lt;/li&gt;&#xA;&lt;li&gt;The walker's speed.&lt;/li&gt;&#xA;&lt;li&gt;The walker's ability to match duplicates&lt;/li&gt;&#xA;&lt;li&gt;The ability of pedestrians to enter the path in the 20m range that the walker is covering.&lt;/li&gt;&#xA;&lt;li&gt;The fact that everyone is walking through the stander's field of vision, rather than behind them or exiting the path before they reach the stander.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;h3&gt;EDIT&lt;/h3&gt;&#xA;&lt;p&gt;Taking into consideration the OPs answers to the assumptions above, as well as extrapolations taken from the original source, the parameters sit as follows:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The walker starts and ends at the same place as the stander&lt;/li&gt;&#xA;&lt;li&gt;The walker's speed is unnecessary, but he makes a number of complete loops and ends at the start point when the hour is up.&lt;/li&gt;&#xA;&lt;li&gt;Yes&lt;/li&gt;&#xA;&lt;li&gt;Pedestrians can enter or exit the sidewalk at any time or location.&lt;/li&gt;&#xA;&lt;li&gt;The stander is in a doorway, so his view is completely unobstructed, and no one walks behind him.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Taking these into consideration:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;We still don't have enough data to know who counted more&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;Let's take our three scenarios:&#xA;&lt;br&gt;&#xA;&lt;strong&gt;Walker sees more:&lt;/strong&gt; Say fifteen people walk down the sidewalk in an hour. If three of them leave the sidewalk before they reach the stander, they will be counted by the walker and will not be counted by the stander. Thus, the walker will have a higher count.&#xA;&lt;br&gt;&lt;br&gt;&#xA;&lt;strong&gt;Stander sees more:&lt;/strong&gt; This logic holds true to my earlier statement. Again, say there are fifteen people on the sidewalk. If the walker passes the 5m mark heading toward the 10m mark and then a pedestrian enters the sidewalk at the 3m mark heading toward the 0m mark, the pedestrian will not be counted by the walker, but would be counted by the stander.&#xA;&lt;br&gt;&lt;br&gt;&#xA;&lt;strong&gt;They see the same:&lt;/strong&gt; This is the answer the book quoted in the question gives, and I would assume that it is the one OP wants. The basic premise here is that the walker will pass each pedestrian and count them once, whether it is coming or going. The stander will count each person as they pass by, thus giving them the same counts.&#xA;&lt;br&gt;&lt;br&gt;&#xA;Here is the direct answer from the book quoted in the question:&#xA;&lt;br&gt;&lt;br&gt;&#xA;Both of them counted the same number of passersby.&#xA;&lt;br&gt;While the one who stood at the door counted all those&#xA;&lt;br&gt;who passed both ways, the one who was walking counted all&#xA;&lt;br&gt;the people he met going up and down the pavement.&#xA;&lt;br&gt;There is another way of putting it. When the man who was&#xA;&lt;br&gt;walking and counting the passers-by returned for the first&#xA;&lt;br&gt;time to the man who was standing at the door, they had&#xA;&lt;br&gt;counted the same number of passers-by - all those passing&#xA;&lt;br&gt;by the standing man encountered the walk¬ing man either&#xA;&lt;br&gt;on the way there or back. And each time the one who was&#xA;&lt;br&gt;walking was returning to the one who was standing he&#xA;&lt;br&gt;counted the same number of passers-by. It was the same at&#xA;&lt;br&gt;the end of the hour, when they met for the last time and told&#xA;&lt;br&gt;each other the final number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Ultimately&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;This question goes to show how a good mathematical model needs to be very, very well defined. There are far too many variables that need to be accounted for here for us to be able to completely say who counted more, as there is too much chaos thrown in by the human element.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12242" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-17T09:25:42.237" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15066" PostTypeId="2" ParentId="15048" CreationDate="2015-05-16T18:09:57.797" Score="1" Body="&lt;p&gt;Once you bid, you lose. [note 1] Knowing this, nobody should bid. If someone does, at a level below $19, there is motivation for someone else to bid. (It isn't very much motivation, and should be resisted, but probably won't be.) Once there are bids of n and n+1, whether these are less than 20 or not, the n bidder stands to lose n unless they bid n+2. This of course puts the n+1 bidder in the same spot, so they bid n+3, forcing the n+2 bidder to move up to n+4. In real life this continues until people accept they are going to lose some money and stop bidding in order to stop the amount from increasing. It's a matter of luck which bidder will stop first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[note 1:] In theory there is a small chance that nobody will bid after you. It would be wisest for nobody to bid after you. But somebody probably will. The auctioneer will goad them into it. &quot;Come on lads, you're going to let him buy a twenty for 5? Nobody has 6? Nobody? What's your problem, just step up and say 6 and it's yours!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not only will the auctioneer make money (they do so once the bids are 10 and 11) they are likely to double their money because people will keep going until they get past 20 - either bidding 19 and thinking the other person won't bid 20, or bidding 20 and thinking the other person won't bid 21. But once you're going to lose 20 if you don't bid, you'll bid 22 and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way to win is not to play.&lt;/p&gt;&#xA;" OwnerUserId="1509" LastActivityDate="2015-05-16T18:09:57.797" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15067" PostTypeId="2" ParentId="14982" CreationDate="2015-05-16T18:34:58.317" Score="2" Body="&lt;p&gt;I think the answer to the riddle found by Tryth and TheWamts is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a QUARK.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I walk in realms mostly unknown&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Realms of subatomic particles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;yet I am part of you.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quarks are what makes up everything, including your body.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;They think I may not exist and am a bit wrong in the heaz.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For a long time after the existence of quarks was postulated, many people didn't believe in them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If I doz am a bit bipolar,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Up and down quarks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;sometimes charming and other times weird.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Charm and strange quarks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Sometimes I like being Peer Gynt, sometimes I want to lurk in the Benthic Zone.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Truth and beauty quarks maybe? Or a reference to the word Quark being used in a German play &lt;em&gt;Peer Gynt&lt;/em&gt;, according to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/14982/heres-a-little-cryptogram-i-wrote-you-might-want-to-read-it-letter-for-letter/15067?noredirect=1#comment46279_15067&quot;&gt;d'alar'cop&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I got this idea by reading &quot;I am part of you. They think I may not exist&quot; and recalling the following joke:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Legally speak­ing, how can you kidnap something that is not supposed to exist? I doubt anyone ever accused Murray Gell-Mann of kidnapping a quark, even though he knowingly carried a billion of them around in his pocket.&quot; - Minerva Paradizo, &lt;em&gt;Artemis Fowl and the Lost Colony&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(quoted from memory - if I got the words slightly wrong, do correct me!) The &quot;sometimes charming and other times weird&quot; was the confirmation I needed for this guess.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2018-05-20T07:22:59.790" LastActivityDate="2018-05-20T07:22:59.790" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="15068" PostTypeId="2" ParentId="14785" CreationDate="2015-05-16T19:18:35.720" Score="1" Body="&lt;p&gt;I'll take advantage of a non-stated rule :-)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Which measuring unit should be used to state the result?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, just:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Define a new measuring unit as the amount of liquid in the container. Done. The container has 1 container unit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And then:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Let the scientists debate for years on how to reliably compare that unit to other measuring units.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Easy, (almost) no time at all. :)&lt;/p&gt;&#xA;" OwnerDisplayName="user12496" LastActivityDate="2015-05-16T19:18:35.720" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15069" PostTypeId="2" ParentId="14785" CreationDate="2015-05-16T19:21:51.607" Score="0" Body="&lt;p&gt;Let me hack into the problem :-)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If it is valid to make an small hole in a container and place a feather there. You will have an excellent drop forming device.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Just count the drops.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clearly:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;An automatic electronic device to count drops would be an excellent help.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user12496" LastActivityDate="2015-05-16T19:21:51.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15070" PostTypeId="2" ParentId="15046" CreationDate="2015-05-16T19:48:35.523" Score="0" Body="&lt;p&gt;Assumption: Each passer-by is only counted once.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The moving counter will count more people. He will count the same amount of people who never turn back. People who turn back once are as likely to be counted by only the stationary counter as only by the moving counter. However, people who move in a closed loop are more likely to be counted only by the moving counter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12395" LastEditorUserId="12395" LastEditDate="2015-05-17T09:10:38.390" LastActivityDate="2015-05-17T09:10:38.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15071" PostTypeId="2" ParentId="14785" CreationDate="2015-05-16T20:09:22.940" Score="0" Body="&lt;p&gt;Ok. This one will meet all the intended rules.&#xA;Lets call the volume in the container to be measured &lt;code&gt;&lt;/code&gt; and the volume in each of the (many) measuring flasks &lt;code&gt;&lt;/code&gt;. I'll assume:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;That there is a way (precise enough) to tell when a flask is exactly full (like the way a bottle reduces it's diameter when the liquid reach the neck).&lt;/li&gt;&#xA;&lt;li&gt;There is a way to &quot;reproduce&quot; (i.e. copy) the amount of liquid in &lt;code&gt;&lt;/code&gt;. A &lt;a href=&quot;https://en.wikipedia.org/wiki/Weighing_scale#Balance&quot; rel=&quot;nofollow&quot;&gt;balance scale&lt;/a&gt; used to compare (not measure) the amount of liquid will be a great help (if allowed).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Armed with those two assumptions perform this procedure:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1.- Reproduce &lt;code&gt;&lt;/code&gt; in one or several &lt;code&gt;&lt;/code&gt; containers (keep track of how many times this reproduction is performed as the number &lt;code&gt;i&lt;/code&gt;).&lt;br&gt;&#xA;2.- Fill (as exactly as possible) as many &lt;code&gt;&lt;/code&gt; flasks as possible from &lt;code&gt;&lt;/code&gt; (keep track of the count of flasks used as &lt;code&gt;j&lt;/code&gt;).&lt;br&gt;&#xA;3.- If the last &lt;code&gt;&lt;/code&gt; flask filled from &lt;code&gt;&lt;/code&gt; is exactly full: end.&lt;br&gt;&#xA;4.- If the last &lt;code&gt;&lt;/code&gt; flask is not full use it as the first one to fill on returning to step 1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Consider this fact:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;When &lt;code&gt;i&lt;/code&gt; containers have &quot;exactly&quot; filled &lt;code&gt;j&lt;/code&gt; flasks you have performed &lt;br&gt; some kind of l.c.m. of both &lt;code&gt;&lt;/code&gt; and &lt;code&gt;&lt;/code&gt;. And then &lt;code&gt;i*&lt;/code&gt; == &lt;code&gt;j*&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At the end, do the math:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;      &lt;code&gt;&lt;/code&gt; == &lt;code&gt;(j/i)*&lt;/code&gt;  (as exactly as the error in filling flasks allow).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user12496" LastEditorDisplayName="user12496" LastEditDate="2015-05-16T20:36:33.537" LastActivityDate="2015-05-16T20:36:33.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15072" PostTypeId="1" AcceptedAnswerId="15074" CreationDate="2015-05-16T21:29:35.250" Score="2" ViewCount="1449" Body="&lt;p&gt;I was used in 1812,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am in the title of the fourth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flash!  Fuji is my competitor &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And you may have seen me on websites that allow the user to talk about books.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a rapper and you can shoot me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry if certain lines seem to be misleading.  Also, (&lt;strong&gt;&lt;em&gt;shocker!!&lt;/em&gt;&lt;/strong&gt;) no obscure book or movie references.  I'm not sure of the wordplay tag...  But we'll only know if it fits when you've solved it.&lt;/p&gt;&#xA;" OwnerUserId="11790" LastActivityDate="2015-05-16T21:34:10.820" Title="What am I? You've definitely heard of me" Tags="|riddle|knowledge|wordplay|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15073" PostTypeId="1" AcceptedAnswerId="15080" CreationDate="2015-05-16T21:32:07.767" Score="3" ViewCount="1316" Body="&lt;p&gt;We are Legends,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are Prodigies,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are Champions,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now is the winter of our discontent… or something like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the Minotaur has appeared,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's see how far we've come…&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something bad has happened-- but then again, something bad always happens&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it's just how we're made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think about five rings, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;that are never connected as individuals, but linked together as a whole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're from the age of legends, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the crossroads of twilight, although…&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I still think that the Wheel of Time is the better series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really like the Duadhe Mahdi'in.&lt;/p&gt;&#xA;" OwnerUserId="11814" LastEditorUserId="5373" LastEditDate="2018-05-07T22:48:58.493" LastActivityDate="2018-05-07T22:48:58.493" Title="Guess the book character!" Tags="|riddle|knowledge|literature|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15074" PostTypeId="2" ParentId="15072" CreationDate="2015-05-16T21:34:10.820" Score="4" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cannon/Canon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was used in 1812,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cannons were used in the war of 1812.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am in the title of the fourth.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pachelbel's &lt;em&gt;Canon in D&lt;/em&gt; (D is the fourth letter)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Flash! Fuji is my competitor &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Canon is a competitor of Fuji the camera manufacturer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And you may have seen me on websites that allow the user to talk about books.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The word &quot;canon&quot; refers to books that are accepted as legitimate by a community.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm a rapper and you can shoot me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Canon_%28rapper%29&quot; rel=&quot;nofollow&quot;&gt;Canon&lt;/a&gt; the rapper, and you could shoot him with a cannon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2015-05-16T21:34:10.820" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15075" PostTypeId="1" AcceptedAnswerId="15077" CreationDate="2015-05-16T21:58:36.957" Score="31" ViewCount="11317" Body="&lt;p&gt;You find yourself trapped atop an $800$ foot tall building. The surrounding land is completely flat, plus there are no other structures nearby. You need to get to the bottom, uninjured, and can only safely fall about $5$ feet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You look down the four walls; they are all completely smooth and featureless, &lt;em&gt;except&lt;/em&gt; that one of the walls has a small ledge $400$ feet above the ground. Furthermore, there are two hooks, one on this ledge, and one directly above it on the edge of the roof. The only tools you have are $600$ feet of rope, and a knife. How do you get to the bottom? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: I got it from &lt;a href=&quot;http://gurmeet.net/puzzles/&quot;&gt;this puzzle collection&lt;/a&gt;, but I think it has been used in interviews.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="5373" LastEditDate="2015-05-16T22:17:13.177" LastActivityDate="2015-05-18T13:58:05.330" Title="A 600 foot Rope, and an 800 foot Drop" Tags="|strategy|situation|" AnswerCount="8" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15076" PostTypeId="2" ParentId="15073" CreationDate="2015-05-16T22:02:53.440" Score="4" Body="&lt;p&gt;This riddle is about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;em&gt;Percy Jackson and the Olympians&lt;/em&gt; series (and its followup &lt;em&gt;The Heroes of Olympus&lt;/em&gt;) by Rick Riordan.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; demigods.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;We are Legends,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;We are Prodigies,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;We are Champions,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That's exactly what demigods are in this series.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But now is the winter of our discontent… or something like that.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ???&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Now, the Minotaur has appeared,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is how the action begins in the first book: the sudden appearance of the Minotaur and its defeat at the hands of Percy Jackson.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Let's see how far we've come…&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Something bad has happened-- but then again, something bad always happens&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well, that's just true of life in general! ;-)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But it's just how we're made.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Demigods are always having to deal with battles, monsters, etc. in the books.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Think about five rings,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;that are never connected as individuals, but linked together as a whole.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The symbol of the Olympic Games, a hint towards Olympus.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;We're from the age of legends,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not the &lt;a href=&quot;http://wot.wikia.com/wiki/Age_of_Legends&quot; rel=&quot;nofollow&quot;&gt;Age of Legends&lt;/a&gt;, but the age of Greek myth and legend!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;At the crossroads of twilight, although…&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Another &lt;a href=&quot;http://wot.wikia.com/wiki/Crossroads_of_Twilight&quot; rel=&quot;nofollow&quot;&gt;Wheel of Time reference&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I still think that the Wheel of Time is the better series.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Absolutely! Wheel of Time is indeed the best fantasy novel ever written, with fantastic complexity, characters, plotline, development, and &lt;em&gt;&amp;lt; stifles rant &gt;&lt;/em&gt;. I like the idea of the Percy Jackson series (Greek gods in the modern day - brilliant!), but can't stand how Americanised it is. Trash novels really (but I read all five anyway).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I really like the Duadhe Mahdi'in.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; And yet another &lt;a href=&quot;http://wot.wikia.com/wiki/Duadhe_Mahdi&amp;#39;in&quot; rel=&quot;nofollow&quot;&gt;Wheel of Time reference&lt;/a&gt;! Not sure how it fits in to &lt;em&gt;this&lt;/em&gt; riddle, but any Wheel of Time fan gets into my good books ;-)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This riddle was too easy IMO (but still fun to solve). The Minotaur part sparked the thought of Percy Jackson; the five-rings part was an obvious Olympics reference; then I looked at the first few lines and immediately rammed the answer button. Keep 'em coming, Gracelyn - riddles about fantasy novels are &lt;strong&gt;right&lt;/strong&gt; up my street! :-)&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-05-16T22:15:13.100" LastActivityDate="2015-05-16T22:15:13.100" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15077" PostTypeId="2" ParentId="15075" CreationDate="2015-05-16T22:49:57.993" Score="32" Body="&lt;p&gt;First, tie one end of the rope to the to hook and climb down to the ledge. Cut (without dropping) the rope that hangs below the ledge, then climb back to the roof carrying the extra rope that you cut. At the top, untie the rope from the hook.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You now have two lengths of rope: one that is 400 feet long and one that is 200 feet long. While you're on top of the roof, prepare the rope accordingly: Tie a small loop at one end of the 200-foot long rope. String the 400-foot long rope through the loop so that half of its length is on either side of the loop. Make sure that the loop is snug enough that the 400-foot long rope won't fall out by itself, but loose enough that you can pull the rope out later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, tie the end of the 200-foot rope without the loop to the first hook. The 200-foot long rope lets you climb halfway to the ledge. For the remaining 200 feet, you carefully climb down the 400-foot rope, which hangs down 200 feet from where it is held by the loop. Once you get to the ledge, pull the 400-foot rope out of the loop, tie it to the second hook, and climb the rest of the way to the ground.&lt;/p&gt;&#xA;" OwnerUserId="294" LastActivityDate="2015-05-16T22:49:57.993" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="15078" PostTypeId="2" ParentId="15075" CreationDate="2015-05-16T22:51:29.313" Score="6" Body="&lt;p&gt;Cut the rope into a 200-foot piece and a 400-foot piece. Tie the 400-foot piece into a loop and tie the 200-foot piece onto some point of this loop, so you get a sort of spoon-shaped rope. Hang the loop (at the opposite point from where it meets the 200-foot piece) from the upper hook. Now when the loop is stretched straight - as it will be when downward force is exerted on it - the bottom of the 200-foot piece just reaches the ledge. Climb down to the ledge in this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you reach the ledge, make waves in the rope by jerking the end up and down, until the loop comes off the upper hook and the whole thing falls down to you. Hang the structure on the lower hook in the same way as you did on the upper one, and repeat the whole procedure.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-16T22:51:29.313" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15079" PostTypeId="2" ParentId="13303" CreationDate="2015-05-17T00:52:39.793" Score="13" Body="&lt;p&gt;Here's my partial solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, as per &lt;a href=&quot;https://puzzling.stackexchange.com/a/15063/12511&quot;&gt;rand al'thor's answer&lt;/a&gt;, north is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;right on the map; the blotches are filled in parts of the directions' letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Start your search from dead man's cave and follow the clues herein.&#xA;Sharpen a sword and your wits and you're ready to begin.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The &quot;dead man's cave&quot; currently could be any of the marked caves, but subsequent stanzas will narrow this down.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Paired up, pared down, the truth exposed, lies within lies.&#xA;Heading onwards north, reeling seas terrorize.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;As per &lt;a href=&quot;https://puzzling.stackexchange.com/a/14985/12511&quot;&gt;rand al'thor's answer&lt;/a&gt;, pairing the words in the second line gives &quot;headingonwards northreeling seasterrorize&quot;. Paring each of these down, we can see that they contain &quot;go three east&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Right is left and left is right, turn and go until,&#xA;Twins appear to mark an end, though you'll continue still.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The first line tells us that left is the &quot;right&quot; way to turn, which means we move north. &quot;Twins&quot; tells us to look for squares with the same configuration of trees, which gives us the pairs (KS, II), (PA, HI), (EN, PI) and (QN, EL). When we draw lines between these pairs, they mark out two X shapes which cross at DO and CU. Using this, we can narrow the initial positions down to HT or EA, as they are the only ones from which moving three east and then north will reach one of these points.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Cardinals sitting in their nests, pick them up one by one.&#xA;Set them free and follow them, then turn to what's undone.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;As per &lt;a href=&quot;https://puzzling.stackexchange.com/a/14718/12511&quot;&gt;tfitzger's answer&lt;/a&gt;, we take every instance of the cardinal directions from &quot;their nests&quot;, giving us ENESS. Following these directions puts us at either EN or DS. Next we &quot;turn to what's undone&quot;, the unused cardinal direction, and face west.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A leap of faith, surrender yourself. But when things come to a head,&#xA;Take the easy way out, with the blues left behind, go three and don't retread.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;In either case we're now facing a river, so jumping into it is our &quot;leap of faith&quot;. We &quot;surrender ourselves&quot; by following the current to the head of the river at GA or PL. The &quot;easy way out&quot; is the square that is a beach rather than a mountain (GB or CL), and once we're out we face away from the sea and move three squares to EB or CU. However, since we've already been to CU, the instruction not to retread means we can eliminate this possibility. Thus the sequence up to this point is HT DT DO EN EB.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Follow the gaze of the crimson saint as he looks to the pre-dawn glow.&#xA;The distance the crow flies in to pi, hints how far to go.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The &quot;crimson saint&quot; is probably St Nick, or Santa Claus, in his red coat. As he lives at the North Pole, his dawn (and everything else) would be to the south. The second line refers to the distance from IN to PI; since they're 3 squares apart vertically and 4 horizontally, by Pythagorus' Theorem the direct distance is 5 squares. Thus we move 5 squares south, to EU.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Two one fives, all lined up, this time point the way.&#xA;They also indicate how far. Go now or be led astray.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Two one fives&quot; in Roman numerals would be IV IV. If we line these up vertically, we get:&#xA;&lt;pre&gt;I&#xA;V&#xA;I&#xA;V&lt;/pre&gt;&#xA;In other words, two downwards-pointing arrows. Therefore we move two down (literally down rather than south in this case, since south would put us off the map) to DU.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proceed until, when viewed from on high, the island itself gives a sign.&#xA;Next steps revealed as the river reflects and the ancient stones align.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This is where I'm stuck. &lt;strike&gt;We probably proceed north until we hit the arrow stone at EA, which is a definite sign viewed from above. The other stones must then give other directions to follow.&lt;/strike&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12511" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-05-17T03:36:14.473" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="15080" PostTypeId="2" ParentId="15073" CreationDate="2015-05-17T01:51:28.857" Score="0" Body="&lt;p&gt;Okay, I kind of shamelessly stole the idea from rand...  Maybe he won't mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing I would change is to say I think it's &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Percy Jackson&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since he is like Duadhe Mahdi'in&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Meaning water seeker etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11790" LastActivityDate="2015-05-17T01:51:28.857" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15081" PostTypeId="2" ParentId="15046" CreationDate="2015-05-17T02:36:34.217" Score="1" Body="&lt;p&gt;Assumptions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Since it's not specified otherwise in the question, assume that every passer-by who walks along the sidewalk will walk the complete length of the sidewalk, in a single direction, without stopping, changing speed, or turning around.&lt;/li&gt;&#xA;&lt;li&gt;Since it's not specified otherwise in the question, assume the counters count passer-bys as they move past.  (That is, the counting occurs as a passer-by changes state from &quot;behind the counter&quot; to &quot;in front of the counter&quot;, or vice-versa)&lt;/li&gt;&#xA;&lt;li&gt;Since it's not specified otherwise in the question, assume the walking counter walks slightly slower than the slowest passer-by (and therefore will not pass a single person twice, while that person is only traversing the sidewalk once).&lt;/li&gt;&#xA;&lt;li&gt;Since it's not specified otherwise in the question, assume that when the surveying period starts, there are no pedestrians already on the sidewalk.&lt;/li&gt;&#xA;&lt;li&gt;Similarly, assume that when the surveying period ends, there are no pedestrians left on the sidewalk.&lt;/li&gt;&#xA;&lt;li&gt;Assume that the OP felt that the question was solvable as initially posted, and assume that they still feel that way, as no new information or clarifications have been added to it even as every answer thus far has included at least one assumption.  Therefore, assume that the (unspecified) mechanics of how the counting system works and how the pedestrians behave are &lt;em&gt;not relevant&lt;/em&gt; to the puzzle.  This means that this is not actually a mathematics question, despite being tagged that way – &lt;strong&gt;it is instead a puzzle about finding the cheeky trick which renders moot all those missing mechanical details&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So after some thought, here's the trick which renders everything else moot:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The walking counter is a &quot;passer-by&quot; from the point of view of the stationary counter, since the walking counter is a person walking past them on the sidewalk.&lt;P&gt;So regardless of any of the assumptions listed above, over any sufficiently long survey duration the stationary counter will show a higher count than the walking one;  they will both have counted approximately the same number of pedestrians, but the stationary counter will also have counted the walking counter moving back and forth.&lt;P&gt;QED.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9286" LastEditorUserId="9286" LastEditDate="2015-05-17T02:59:28.010" LastActivityDate="2015-05-17T02:59:28.010" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15082" PostTypeId="2" ParentId="15075" CreationDate="2015-05-17T05:28:48.407" Score="13" Body="&lt;p&gt;If the rope is coarse enough, tie a variant of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sheepshank&quot;&gt;sheepshank&lt;/a&gt; knot on the first hook and cut the middle rope (this makes it a kamikaze knot). Climb down to the ledge, shake your rope loose, retie, and climb down the rest of the way.&lt;/p&gt;&#xA;" OwnerUserId="12513" LastActivityDate="2015-05-17T05:28:48.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15083" PostTypeId="2" ParentId="15046" CreationDate="2015-05-17T08:10:36.393" Score="1" Body="&lt;p&gt;Both persons count the same number of passers-by.  While walking along the pavement, each passer-by must pass both the stationary person, and also the walking person, so both will count the same number of passers-by.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer reworded from this open-source (&lt;a href=&quot;https://archive.org/details/opensource&amp;amp;tab=about&quot; rel=&quot;nofollow&quot;&gt;books contributed by the community&lt;/a&gt;) &lt;a href=&quot;https://archive.org/details/MathematicsCanBeFun-1&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="9286" LastActivityDate="2015-05-17T08:10:36.393" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15084" PostTypeId="2" ParentId="12796" CreationDate="2015-05-17T10:05:19.233" Score="4" Body="&lt;p&gt;Here are my new answers, once again taking as base the Mark N contribution, but this time not using the binomial coefficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Score: 107&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1 = \frac{7}{7}$ [Score +1]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2 = \sqrt{7}$ [Score +2]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3 = \sqrt{7+7}$ [Score +3]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$4 = \sqrt{7+7+7}$ [Score +4]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5 = 7-\sqrt{7}$ [Score +3]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$6 = 7-\frac{7}{7}$ [Score +2]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$7 = 7$ [Score +0]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$8 = 7+\frac{7}{7}$ [Score +2]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$9 = 7+\sqrt{7}$ [Score +3]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$10 = 7+\sqrt{7+7}$ [Score +4]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$11 =\sqrt{\sqrt{7}^7}$ [Score +5]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$12 = 7+7-\sqrt{7}$ [Score +4]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$13 = 7+7-\frac{7}{7}$ [Score +3]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$14 = 7 + 7$ [Score +1]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$15 = 7 + 7 + 7/7$ [Score +3]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$16 = 7 + 7 + \sqrt{7}$ [Score +4]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$17 = 7 + 7 +\sqrt{7+7}$ [Score +5]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$18 = \sqrt{7*7*7}$ [Score +4]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$19 = 7+7+7-\sqrt{7}$ [Score +5]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$20 = 7+7+7-\frac{7}{7}$ [Score +4]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$21 = 7+7+7$ [Score +2]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$22 = 7+7+7+\frac{7}{7}$ [Score +4]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$23 = 7+7+7+\sqrt{7}$ [Score +5]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$24 = (\sqrt{7+7+7})!$ [Score +6]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$25 = \frac{7*7*7+7}{7+7}$ [Score +5]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$26 = \sqrt{7!/7}$ [Score +5]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$27 = 7+7+7+7-\frac{7}{7}$ [Score +5]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$28 = 7+7+7+7$ [Score +3]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$29 = 7+7+7+7+\frac{7}{7}$ [Score +5]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$30 = \sqrt{\sqrt{7^7}}$ [Score +5]&lt;/p&gt;&#xA;" OwnerUserId="12165" LastEditorUserId="12165" LastEditDate="2015-05-17T17:00:59.403" LastActivityDate="2015-05-17T17:00:59.403" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15085" PostTypeId="1" AcceptedAnswerId="15086" CreationDate="2015-05-17T10:20:19.193" Score="37" ViewCount="1924" Body="&lt;p&gt;The following image appears on the profile page of &lt;a href=&quot;https://puzzling.stackexchange.com/users/10281/leoll2&quot;&gt;leoll2&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hWBge.jpg&quot; alt=&quot;leoll2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears to be a binary code, which I've transcribed as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(left-hand side)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;01001110 01100001 01110100 01110101 01110010 01100101 &lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(within the prism)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;[...]    01101100 01110100&#xA;[...]    01101111 01110101&#xA;[...]    01101000 00100000&#xA;[...]    01101110 01101001&#xA;[...]    01101101 01100001&#xA;[...]    01101001 01100011&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(right-hand side, in the order red, orange, yellow, green, blue, purple)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;01110010 01100101 01110110 01100101 01100001 01101100 01110011 00100000 00100000&#xA;01101001 01110100 01110011 00100000 01100011 01101000 01100001 01110010 01101101&#xA;01110100 01101000 01110010 01101111 01110101 01100111 01101000 00100000 00100000&#xA;01101001 01101110 01100110 01101001 01101110 01101001 01110100 01100101 00100000&#xA;01110110 01100001 01110010 01101001 01101111 01110101 01110011 00100000 00100000&#xA;01100001 01110011 01110000 01100101 01100011 01110100 01110011 00100000 00100000&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is this mysterious Italian trying to tell us? How can the code be decrypted?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For obvious reasons, I ask leoll2 himself not to answer this question, at least unless it stays unsolved for a &lt;em&gt;long&lt;/em&gt; time.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Disclaimer&lt;/strong&gt;: I don't know what the plaintext is. If it comes out to be something offensive, don't blame me - blame leoll2!&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2016-05-21T07:59:58.383" Title="What is leoll2 trying to tell us?" Tags="|cipher|visual|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15086" PostTypeId="2" ParentId="15085" CreationDate="2015-05-17T10:33:13.267" Score="32" Body="&lt;p&gt;Well, taking a straight binary to ascii conversion you get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Left&lt;/strong&gt;: Nature&lt;br&gt;&#xA; &lt;strong&gt;Middle&lt;/strong&gt;: although enigmatic&lt;br&gt;&#xA; &lt;strong&gt;Right&lt;/strong&gt;: reveals its charm through infinite various aspects  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(assuming there's no deeper riddle to be solved - I couldn't attribute the quote to anyone via google, so I assume it's a leoll2 original?)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To explain the conversion process, &lt;a href=&quot;http://en.wikipedia.org/wiki/ASCII&quot;&gt;ASCII&lt;/a&gt; is one of the most common/basic ways of storing plain text digitally. Each character is stored in seven bits (usually padded to eight with a leading zero). The &lt;a href=&quot;http://en.wikipedia.org/wiki/ASCII#ASCII_printable_code_chart&quot;&gt;standard defines how each character is encoded&lt;/a&gt;, for example &lt;code&gt;A&lt;/code&gt; is &lt;code&gt;41&lt;/code&gt; in hex, or &lt;code&gt;(0)100 0001&lt;/code&gt; in binary. Characters are just stored in order (so for this text, left to right, line by line). So from there it's just a matter of either painstakingly looking each character up, or using &lt;a href=&quot;http://www.binaryhexconverter.com/binary-to-ascii-text-converter&quot;&gt;one of many online converters&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason you see &lt;code&gt;00100000&lt;/code&gt; repeated a lot, is simply because it converts to a space, as you can see in the table linked above. So it's just being used to pad things out to fit the image neatly (plus obviously appearing between words).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For completeness, the full binary within the prism is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;01100001 01101100 01110100&#xA;01101000 01101111 01110101&#xA;01100111 01101000 00100000&#xA;01100101 01101110 01101001&#xA;01100111 01101101 01100001&#xA;01110100 01101001 01100011&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="17482" LastEditDate="2016-05-21T07:59:58.383" LastActivityDate="2016-05-21T07:59:58.383" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15088" PostTypeId="2" ParentId="15075" CreationDate="2015-05-17T12:54:32.237" Score="5" Body="&lt;p&gt;If it's a twisted or braided rope, you could separate the strands so you have two 600ft sections. Tie them together with good knots and you have more than enough rope to get you down safely.&lt;/p&gt;&#xA;" OwnerUserId="12242" LastActivityDate="2015-05-17T12:54:32.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15089" PostTypeId="2" ParentId="12324" CreationDate="2015-05-17T13:10:01.850" Score="0" Body="&lt;p&gt;It's rather simple:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Go across on the crocodile, collect bananas, come back. Two fish and you're done.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But wait, how would that work?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It's in the wording of the question. &#xA;&lt;br&gt;&lt;br&gt;&#xA;'But the monkey has only three fish and in order to get all the bananas out of the island &lt;strong&gt;he&lt;/strong&gt; has to make two round trips.'&#xA;&lt;br&gt;&lt;br&gt;&#xA;Clearly, the monkey could only carry half the bananas in one trip. But, with the help of the crocodile, he can haul all of them in one trip.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12242" LastActivityDate="2015-05-17T13:10:01.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15091" PostTypeId="1" AcceptedAnswerId="15099" CreationDate="2015-05-17T14:32:36.250" Score="1" ViewCount="150" Body="&lt;p&gt;There are a lot of books that talk about things like this,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where there are two people who are &quot;chosen&quot;, and they're kind of opposites, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I guess that opposites is actually a good thing to write a book about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some Ideas Never Fail,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;especially when it comes to writing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this book, there's a &lt;/p&gt;&#xA;&#xA;&lt;p&gt;benevolent tyrant,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;an insane old man,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a girl,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and a boy,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;who are opposites, but still work well together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite my past riddles, this is NOT a series that I've already written about… Also not sure bout the wordplay tag.&lt;/p&gt;&#xA;" OwnerUserId="11814" LastEditorUserId="4050" LastEditDate="2015-05-18T06:53:54.977" LastActivityDate="2015-05-18T06:53:54.977" Title="Contrary motion still sounds good: Guess the book" Tags="|knowledge|" AnswerCount="2" CommentCount="4" ClosedDate="2015-05-18T03:17:49.003" ContentLicense="CC BY-SA 3.0" />
  <row Id="15092" PostTypeId="1" CreationDate="2015-05-17T14:33:39.817" Score="2" ViewCount="952" Body="&lt;p&gt;The &lt;a href=&quot;http://www.puzzle-place.com/wiki/Super_Cubi&quot; rel=&quot;nofollow&quot;&gt;Super-CUBI&lt;/a&gt; puzzle box takes &lt;a href=&quot;http://s933.photobucket.com/user/eclipsedavephoto/media/supercubidirections.jpg.html&quot; rel=&quot;nofollow&quot;&gt;324 steps to open or close it, instructions&lt;/a&gt;. And&#xA;uses a Trinary System. It must have a solution as it would be hard to remember all the steps to take to open it. But what is it ?&lt;/p&gt;&#xA;" OwnerUserId="12517" LastActivityDate="2015-05-17T15:33:02.320" Title="What is the equation to solve the puzzle box Super-CUBI 324 step" Tags="|logical-deduction|pattern|rubiks-cube|mechanical-puzzles|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15093" PostTypeId="2" ParentId="15091" CreationDate="2015-05-17T14:57:42.140" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hunger Games&lt;br&gt;&#xA; Katniss Everdeen and Peeta Mellark are &quot;chosen&quot; to fight to the Death (Some Ideas Never Fail == this concept is very common) . . . . &quot;still work well together&quot; == both survive . . . .&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;MORE INFO :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; benevolent tyrant == Haymitch Abernathy&lt;br&gt;&#xA; insane old man == President Coriolanus Snow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10940" LastEditorUserId="10940" LastEditDate="2015-05-17T15:45:08.940" LastActivityDate="2015-05-17T15:45:08.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15094" PostTypeId="2" ParentId="15092" CreationDate="2015-05-17T15:00:48.837" Score="2" Body="&lt;p&gt;As you have said, the Super Cubi requires 324 moves to be opened. The correct sequence is listed here:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/UvgD9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In each step, you have to move one of the six faces. The numbers in the table represent the centimeters of translations. For example, 000022 means that faces #1 and #2 were moved by 2cm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But how do you remember that sequence?&lt;/strong&gt;&lt;br&gt;&#xA;As you can see, in the first vertical half of the table the face #6 is always 0, while in the second half is always 1 (except for the last move).  The face #5 is 0 for the first and last column, 1 for the second and fifth, 2 for the third and fourth. We can find similar patterns for all the faces.&lt;br&gt;&#xA;Let's define &quot;&lt;strong&gt;bounce&lt;/strong&gt;&quot; a move that follows the cycle [1,2,3,3,2,1]. When you apply the bounce move to a face in a certain status, the next status is determined by the next number in the above cycle. For example, if you bounce from 2 (and the previous status was 1), the next status will be 3.&lt;br&gt;&#xA;Now a general rule to remember the moves (needs some practice, though):  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Face #1 oscillates between 0 and 2 every two moves (starting from 2).&lt;/li&gt;&#xA;&lt;li&gt;Face #2 bounces every two moves.&lt;/li&gt;&#xA;&lt;li&gt;Face #3 bounces every six moves.&lt;/li&gt;&#xA;&lt;li&gt;Face #4 bounces every 18 moves.&lt;/li&gt;&#xA;&lt;li&gt;Face #5 bounces every 54 moves.&lt;/li&gt;&#xA;&lt;li&gt;Face #6 bounces every 162 moves.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As you can see, the bounce distances increase with a factor x3 between two consecutive faces.&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-05-17T15:33:02.320" LastActivityDate="2015-05-17T15:33:02.320" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15097" PostTypeId="1" AcceptedAnswerId="15102" CreationDate="2015-05-17T15:52:17.527" Score="60" ViewCount="4572" Body="&lt;p&gt;All of your stuffed animals are assembled, anxiously waiting to learn who will be be invited as the Guest of Honor to your next tea party.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is to be an exclusive party, after all, as there is to be only &lt;em&gt;one&lt;/em&gt; Guest of Honor.  This is by necessity, because you have only two strawberry pastries, and you obviously intend to eat one of those yourself. (Cook was most annoyed with you for taking so many infinite bags of pastries yesterday, and has since cut off your supply).  All the other guests at the tea party shall only receive tea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now you must decide which of your many guests will be the Guest of Honor, who receives the coveted strawberry pastry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your first instinct, of course, is to invite the valiant Mr. Bun to be the Guest of Honor.  After all, he's brave and noble and always tells the truth.  But then you think of poor insidious Duke Froggington II, who admittedly is dastardly and always lies, but in your heart you know that these character flaws are only due to his tragic childhood.  You see, the perfidious Duke Froggington II grew up all alone in a dank 30-meter well, trying to leap his way to light and freedom.  You know what that's like, and your heart aches for him.  And there are similar reasons why you might choose almost any of the six dozen other stuffed animals that pack every available surface in your bedroom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, you find that you're completely unable to decide which stuffed animal should be the Guest of Honor at your tea party, and so you instead devise a game for the stuffed animals to play, to determine the Guest of Honor fairly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The game works in this way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, you choose Mr. Bun as the initial nominee as Guest of Honor.  He's the obvious first candidate, as the noblest of all your stuffed animals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, starting with the nefarious Duke Froggington II and descending through the other stuffed animals in order of seniority and ending with Peanut (an ugly mouse given to you by your annoying older sister at your last birthday fête, and usually tucked far back in your dresser behind your latin coursebooks), each stuffed animal still in attendance is asked if they wish to declare themselves to be the new nominee as Guest of Honor.  If they do, then they become the new nominee, and the previous nominee is returned to the toy chest, where they must remain until after the tea party.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever the nominee changes, the whole process starts again;  Every animal (in order of seniority, and starting with the most senior remaining animal) is given a fresh opportunity to nominate themselves, even if they had previously chosen not to.  As before, if an animal chooses to nominate themselves, the current nominee is returned to the toy chest, and may not attend the tea party.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This process continues until no further stuffed animals wish to declare themselves as the nominee.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once no more stuffed animals wish to make themselves the nominee, whoever remains as the nominee shall become the Guest of Honor and will receive a strawberry pastry at the tea party.  The other remaining stuffed animals shall receive tea.  Any stuffed animals in the toy chest get nothing. (At least, not until they're brought out again for the next tea party (which will likely be in ten minutes or so))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few obvious points which you know instinctively are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;All stuffed animals &lt;em&gt;love&lt;/em&gt; strawberry pastries (as does everyone of refinement and distinction.  Peanut would probably love strawberry pastries too, despite having neither refinement nor distinction, though you've certainly never wasted one on her, even yesterday when you had several infinite supplies of them).  This means that each stuffed animal dearly wishes to be the Guest of Honor.&lt;/li&gt;&#xA;&lt;li&gt;All stuffed animals like tea, so none of them wish to completely miss the party by being put back into the toy chest.&lt;/li&gt;&#xA;&lt;li&gt;All stuffed animals are perfect logicians (I mean, obviously.  Do I even need to mention this?)&lt;/li&gt;&#xA;&lt;li&gt;Stuffed animals don't hold grudges against one another;  they will remain fast friends even if one sends another to the toy chest.  On the contrary, they will recognise that it was the only logical thing to do, and will probably laugh about it at your next tea party, later this afternoon.&lt;/li&gt;&#xA;&lt;li&gt;While the treacherous Duke Froggington II always lies, you've long since adapted to that, and aren't ever confused any more about what he means when he tells you which door leads to certain doom and which leads to your afternoon Latin classes (though you're still not convinced that there's actually much of a difference between the two).  Similarly, there will be no confusion on your part about whether or not he wishes to become the new nominee, regardless of what lies he might say.&lt;/li&gt;&#xA;&lt;li&gt;For the sake of clarity, you have six dozen and two stuffed animals in total.  With Mr. Bun as your favourite, the despicable Duke Froggington II as your second favourite, and Peanut as your least favourite.  Other stuffed animals may be identified by number, with the next animal after the treacherous Duke Froggington II being number 3.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Of your stuffed animals, how many will attend your tea party, and which will be the Guest of Honor?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9286" LastEditorUserId="9286" LastEditDate="2015-05-18T00:55:12.510" LastActivityDate="2016-03-08T23:20:21.587" Title="Who shall come to the tea party?" Tags="|logical-deduction|game-theory|" AnswerCount="5" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15098" PostTypeId="2" ParentId="15097" CreationDate="2015-05-17T16:01:56.347" Score="3" Body="&lt;p&gt;My answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All but Mr. Bun will attend the tea party and the Guest of Honor is Peanut&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Of course, Peanut declares that he wants to be the Guest of Honor (GoH), since he is the last and nobody can dethrone him.&lt;br&gt;&#xA;Everybody else knows that asking to be the GoH will cause Peanut (or someone else) to claim the throne, resulting in a &quot;no tea&quot; situation.&lt;br&gt;&#xA; Consequently, everybody except Peanut refuses the GoH.&lt;br&gt;&#xA;Mr.Bun, who was forced to accept the throne, is bound to the toy box.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-05-17T16:18:54.767" LastActivityDate="2015-05-17T16:18:54.767" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15099" PostTypeId="2" ParentId="15091" CreationDate="2015-05-17T16:07:24.530" Score="3" Body="&lt;p&gt;The book (series) is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Secrets of the Immortal Nicholas Flamel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;There are a lot of books that talk about things like this,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yeah, it's a common fantasy trope.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;where there are two people who are &quot;chosen&quot;, and they're kind of opposites,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Twins of Legend: silver and gold; one to save the world, one to destroy it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;So I guess that opposites is actually a god thing to write a book about.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Some Ideas Never Fail,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;especially when it comes to writing.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; More about how it's a common trope.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In this book, there's a&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;benevolent tyrant,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Aten was the overlord of Danu Talis, a tyrant in some ways, Macchiavelli's Elder master, but actually not a Dark Elder.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;an insane old man,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nicholas Flamel?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;a girl,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sophie Newman.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;and a boy,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Josh Newman.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;who are opposites, but still work well together.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They stay together almost all the way through the books until the very end, when he destroys a world and she saves another. (Not too spoilery, I hope! It was certainly one of the best endings I've read - especially &amp;lt; SPOILER &gt; Josh's transformation &amp;lt; /SPOILER &gt; - although the books did lack something in consistency.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I actually guessed beforehand that this was going be the subject of the OP's next riddle, having had a look at her profile page. This is the first puzzle on SE that I could have answered &lt;em&gt;pre-emptively&lt;/em&gt; before it was even posted!&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-05-17T20:36:25.437" LastActivityDate="2015-05-17T20:36:25.437" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15102" PostTypeId="2" ParentId="15097" CreationDate="2015-05-17T16:34:05.993" Score="27" Body="&lt;p&gt;All but Mr. Bun will attend the tea party, with Duke Froggington II as the guest of honor. The total number of animals who will be attending the party is $(6*12+2)-1=73$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at some specific cases:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;One animal left: he will definitely nominate himself.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Two animals left: each animal knows that if he nominates himself, there will only be one left, and that one will definitely nominate himself. So neither will nominate himself.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Three animals left: each animal knows that if he nominates himself, there will be two animals left, and neither will nominate himself. So the first animal to be asked will nominate himself.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Four animals left: each animal knows that if he nominates himself, there will be three animals left, and one will nominate himself. So none will nominate himself.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So using backward induction, if there are $n$ animals left, the strategy taken will be:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $n$ is even, no one will nominate himself.&lt;/li&gt;&#xA;&lt;li&gt;If $n$ is odd, the first one asked will nominate himself.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When Duke Froggington II takes his turn, there are 73 animals left (including himself). As he is a perfect logician, he will nominate himself as the guest of honor. No one else will nominate himself after him. So poor Mr. Bun gets sent to the toy box, while everyone else enjoys the tea party and Duke Froggington II hogs the strawberry pastry.&lt;/p&gt;&#xA;" OwnerUserId="12109" LastEditorUserId="12109" LastEditDate="2015-06-11T15:55:44.533" LastActivityDate="2015-06-11T15:55:44.533" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15104" PostTypeId="1" AcceptedAnswerId="15107" CreationDate="2015-05-17T17:53:09.247" Score="11" ViewCount="380" Body="&lt;p&gt;You have just acquired a $64$ acre farm, in the shape of a square, and divided into an eight by eight array of one acre subplots. You have $21$ crops to plant. Each crop requires its own $3$ acre plot of land, which must consist of three subplots in the shape of a three by one rectangle. Once the crops have been planted, there will be $64-21\times 3=1$ subplot leftover, where you will build your farmhouse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It isn't the season for planting yet, so you plan to build your farmhouse now, then plant the crops later. Which subplots can you build your farmhouse on which still allow you to plant all of your crops?&lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-05-17T19:27:11.930" Title="Combinatorial Agriculture" Tags="|checkerboard|tiling|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15105" PostTypeId="2" ParentId="15104" CreationDate="2015-05-17T18:10:03.530" Score="1" Body="&lt;p&gt;Here is one possibility:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/c9Lwm.png&quot; alt=&quot;One possible farm setup&quot;&gt;&lt;br&gt;This could be rotated to allow for the 4 plots that are in that position (3rd plot from the two closest edges). So far I haven't been able to find any more. Still trying to figure some mathematical reason for this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11727" LastActivityDate="2015-05-17T18:10:03.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15106" PostTypeId="2" ParentId="15097" CreationDate="2015-05-17T18:14:28.000" Score="-3" Body="&lt;p&gt;Clearly, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Peanut (or possibly Mr. Bun)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;will be the sole member of your tea party.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's why:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;All 74 attendants of the party should know that you have access to infinite bags of pastries. So, they would ask if you have more than 1 to spare. When you explain that you do not, then the pastry becomes a precious commodity, while the chance at a tea party comes multiple times in a day. &#xA;&lt;br&gt;&lt;br&gt;&#xA;Faced with this realization, each animal would take the risk at the pastry, knowing that, while they may miss this party, there will be another later that day. Since you enraged the cook, who knows how long it will take to get another delicious pastry.&#xA;&lt;br&gt;&lt;br&gt;&#xA;Peanut, being the last in line, is unable to be overthrown, unless you give Mr. Bun the chance to overthrow him. You see, Mr. Bun did not have a choice in his being given the pastry or not, while there is no one to come after Peanut. The only fair solution to this would be to give Mr. Bun a chance to recover the pastry from Peanut.&#xA;&lt;br&gt;&lt;br&gt;&#xA;Ok, so maybe that last bit is a copout, but have you seen &lt;a href=&quot;http://www.entertainmentearth.com/images/AUTOIMAGES/NC31474lg.jpg&quot; rel=&quot;nofollow&quot;&gt;Peanut&lt;/a&gt;? That guy is just creepy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12242" LastActivityDate="2015-05-17T18:14:28.000" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15107" PostTypeId="2" ParentId="15104" CreationDate="2015-05-17T19:04:56.683" Score="14" Body="&lt;p&gt;It is only possible in &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $4$ cells &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Those cells are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;The black ones in the bottom grid.&lt;/strong&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/uW2Dj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you tile the grid with three colours, following the above pattern (the right one is symmetrical to the left one), you get 21 green, 21 blue and 22 red cells.&lt;br&gt;&#xA;Though, any 3x1 rectangle occupies exactly 1 red, 1 blue and 1 green cell, so the farmhouse must be in a red cell.&lt;br&gt;&#xA;Also, &lt;strong&gt;it must be in a red cell in both grids&lt;/strong&gt; (otherwise it would violate the above condition in one grid), and if you do the intersection of red cells (the cells that are red in both grids) you only get the 4 black cells in the bottom grid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An example of valid configuration is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/UtAkp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-05-17T19:27:11.930" LastActivityDate="2015-05-17T19:27:11.930" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15108" PostTypeId="2" ParentId="9155" CreationDate="2015-05-17T19:05:24.400" Score="0" Body="&lt;p&gt;The lions will win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Assuming the zebras take their optimal strategy, which would be to create a circle an arbitrary distance out from the center of the lions. Let's say that the zebras make such a large circle that any distance a lion is from the center is negligible, so we can say they are all at the direct center of the circle. This will look like this:&lt;img src=&quot;https://i.stack.imgur.com/daBCB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt; with the lions at the red dot in the center (assume its a point, as explained above) and the zebras around the circle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When they move:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A single lion moves directly north, going for the zebra at that spot on the circle (let's call it zebra A). That zebra would in turn move away, and the lion would keep moving in that direction. At some point, the lion would be closer to one of the adjacent zebras to the one it is following (let's call them B and C), however it would keep moving in a straight line toward the original zebra. To keep safe, all three of those zebras would have to move out in their respective path. For this lion strategy, the lion would keep moving in the straight line following zebra A, so B and C could stay in the same spot, but that doesn't matter. Either way, the lion would get far enough outside the original circle (if B and C don't move) or the arc that B and C make (if they move in their respective paths) that it would either catch up to A or get to a point where it can get to a point that is in between B/C and the center of the circle.  If it catches up to A then the lions would obviously win. If it gets to a point that is on the opposite side of B/C and the center of the circle (all the other lions), then it would have that zebra trapped. So the lions win. Here are a couple pictures to explain:&lt;img src=&quot;https://i.stack.imgur.com/diyOe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qFDlO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;The only other thing that the zebras could do is have B and C each move out of their path (the straight path from the center of the circle). If this were to happen, they would eventually get in the way of the zebra that is on that side of them. In this case the lion could just go straight toward those zebras, and having both zebras in a (relatively) close location, they would be trapped because the lion could move faster (only one lion and two zebras escaping, so lion moves 100m and then each zebra must move 100, so lion moves 200 for each 100 per zebra).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;EDIT: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I guess a better way to word this would be that the single lion goes toward the center zebra of 3 that are closest together, angle-wise. No matter how the zebras are set up, at least 3 zebras must be adjacent and separated by angles of 3.6 degrees or less. One zebra could be on the opposite side of the lions from the other zebras, in which case the lion would go for a zebra that has the least sum of angles between itself and the two adjacent zebras. If the zebras were separated by 3.6 degrees each, then this would be the optimal strategy for them, however they would still lose. They wouldn't lose if this angle is greater than or equal to 90 degrees, because no matter how far out the lion goes, it's distance to cut off any zebra will always be longer than the distance the zebra must go to get to that same point, so it will never catch any of the zebras.  For any angle less than 90 degrees between each zebra, the lion can get to a point far enough away from the circle that it will be able to &quot;cut off&quot; one of the adjacent zebras to the one it was following. If the adjacent zebra that the lion targets decides to cut over, it will be cutting into the path of another one of the zebras, and if two of these zebras are in the same path, then it will be simple for a lion to get one of them because the lion can move 200m for each 100m moved by each zebra.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11727" LastEditorUserId="11727" LastEditDate="2015-05-18T05:35:20.250" LastActivityDate="2015-05-18T05:35:20.250" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15110" PostTypeId="2" ParentId="15075" CreationDate="2015-05-17T20:26:44.287" Score="11" Body="&lt;p&gt;I think I have a simpler (albeit not very scientific) answer!&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Tie one end of the rope to the top hook and slide down to the bottom ledge.&lt;/li&gt;&#xA;&lt;li&gt;Tie the other end of the rope to the hook on the ledge and then climb back up.&lt;/li&gt;&#xA;&lt;li&gt;Once at the top, untie the top end of the rope from the hook.&lt;/li&gt;&#xA;&lt;li&gt;Take a deep breath since what you are about to attempt is very ballsy (it helps if you think of yourself as Ethan Hunt from Mission Impossible)!&lt;/li&gt;&#xA;&lt;li&gt;Make a poor man's climbing knot to abseil down the building.&lt;/li&gt;&#xA;&lt;li&gt;Pull the rope taut so that you are holding it at an approximate 410 foot length.&lt;/li&gt;&#xA;&lt;li&gt;Run towards the side of the building to gain sideways momentum and then abseil down running against the building.&lt;/li&gt;&#xA;&lt;li&gt;In a circular motion, you'll swing right down past the ledge and then down towards the bottom of the building.&lt;/li&gt;&#xA;&lt;li&gt;Once you've stopped swinging, lower yourself down to the floor and head towards the nearest pub to tell the lads about your courage!&lt;/li&gt;&#xA;&lt;li&gt;Make a mental note to never again be trapped on top of tall buildings, maybe next time you won't have a handy rope, or a ledge with hooks attached to save you!&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5353" LastEditorUserId="5353" LastEditDate="2015-05-17T21:06:48.970" LastActivityDate="2015-05-17T21:06:48.970" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15113" PostTypeId="1" AcceptedAnswerId="15124" CreationDate="2015-05-17T21:24:56.737" Score="3" ViewCount="597" Body="&lt;p&gt;You have been stranded on an island in the middle of the ocean for the past 26 days, and your food supply is starting to run low. Unfortunately the only person on the island that has any knowledge about gathering food from the wilderness has told you that pretty much every fruit-bearing plant on the island produces some form of toxin that prevents you from being able to eat it for a long period of time without needing medical attention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a group, you decide to all go looking for anything that might be edible.  You (being quite athletic and able to make your way back to the camp, or pitch your own if night should fall) go off by yourself down the beach.  After a couple hours of walking and not finding anything, you decide to sit down and take a break. No need to exhaust yourself, especially if there isn't a good chance you'll be able to find any food. Sitting there taking in your surroundings, a glimmer catches your eye 8 yards and 2 feet down from where you are. You walk over to see what it might be and find a corked bottle.  You can faintly see something that looks like crumpled paper inside (it's hard to tell through the dirty glass). You quickly pop the cork off to find two pieces of paper:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Lifetime Twinkie Supply Giveaway FINAL CLUE - You're almost there!!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You remember back a couple years ago when there was a big treasure hunt for a lifetime supply of twinkies that just kind of faded and everyone figured was just a scam to get people to buy twinkies for the clues inside. You remind yourself of the second note, before getting caught up in why there was never a lawsuit surrounding that &quot;giveaway&quot;. It says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This is the final clue to the Twinkie Giveaway. If you've gotten this far, then all you need to do is solve this riddle to figure out where on the horseshoe island the twinkies are hidden.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are shocked for a moment. You have walked around your island enough times to realize that the big bay going into the middle would make it look exactly like a horseshoe from overhead. You try to contain your excitement of the possible lifetime supply of food as you read on:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We have created a cipher for you to solve. First, you can know that:&lt;br&gt;&lt;strong&gt;THE QUICK BROWN FOX JUMPS OVER THE LAZY DOG&lt;/strong&gt;&lt;br&gt;encodes to:&lt;br&gt;&lt;strong&gt;XZC AMKKQ DROQZ BKB XECXM ABSH RLW JKRA LUI&lt;/strong&gt;&lt;br&gt;This probably wouldn't be much help without another phrase that you need to decode to actually find the final location, so here you go:&lt;br&gt;&lt;strong&gt;XZCDOKVQKESUDAERGSTMNTKPYEXJCO&lt;/strong&gt;&lt;br&gt;&lt;br&gt;You will probably end up getting Lost, but the Number of hints we can give is limited, so you only get one more:&lt;br&gt;&lt;strong&gt;SLDUXOESQPSOCFEVALZMSTIIFKXCDFOXSDEQPOLVDGALSKCDZG&lt;br&gt;&#xA;  PADEFNANOJDNGFERDUIVCZENMFASDJNVAOUWENHFATJVNAOIUV&lt;br&gt;&#xA;  AOIWWEHIUNASDNCEFNOWEIKFNACVNSIJCVNAIUEFWEEIUFOASU&lt;br&gt;&#xA;  QOWYUIQIRUIDOFSANOIUSDPFHUAFWEBGFDVZBVOIDAGUUNTHIB&lt;br&gt;&#xA;  AJWNIEFGAIUVFNRAIUNAIUMEWNAWUINGIJNFAWIDNWAIFDJNGA&lt;br&gt;&#xA;  TOIIPTETORJQOIHXJQNWMGLBOUAENQSVONVCAJSDNEWJGNWBUJ&lt;br&gt;&#xA;  QPETOERTUIVNMXERNDMLKMSMVMAJSHFIUXLASHUVH.EOSLDKFJ&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using your cunning sense of reasoning, you determine that the twinkies are probably going to be in some sort of landmark on the island, seeing as you have a very short clue, so there couldn't be any step by step instructions. The only problem is that there are a bunch of landmarks on the island. There's both parts of the &quot;horseshoe&quot;, a big black rock, a small river that passes through part of the island, a huge tree at one end of the horseshoe, a patch of beach that has much darker sand, and many more. You figure that by the time you were able to send people to each of these places to search for twinkies, everyone would have already starved. So the only way to get them would be solving the riddle.&lt;br&gt;&lt;br&gt; So, where are the twinkies?&lt;br&gt;&lt;br&gt;Thanks to Quarks, you've discovered the location of the twinkies, but when you get there, you find a computer that requires you to type in a password to open the safe full of twinkies (unfortunately the ethernet cable has been chewed away by mice so you can't check your facebook or ask for help).  Perhaps it has something to do with that last hint that you got. Figure out what the password is before everybody starves!!&lt;/p&gt;&#xA;" OwnerUserId="11727" LastEditorUserId="11727" LastEditDate="2015-05-17T22:52:11.840" LastActivityDate="2015-05-19T03:40:15.813" Title="Where are the twinkies?" Tags="|cipher|knowledge|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15115" PostTypeId="2" ParentId="15113" CreationDate="2015-05-17T22:04:11.087" Score="6" Body="&lt;p&gt;I don't think I solved this the way I was supposed to because I didn't use the last hint.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is &quot;THE TWINKIES ARE INSIDE THE BIG TREE&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Method:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Based on the &quot;THE&quot;s corresponding to XZC both RLW I immediately thought it could be similar to a vigenere cipher. So, using a solver, I converted &quot;THE QUICK BROWN FOX JUMPS OVER THE LAZY DOG&quot; to &quot;XZC AMKKQ DROQZ BKB XECXM ABSH RLW JKRA LUI&quot; and got a key of &quot;esykscigcaaumwweokqiumgoqy&quot;.&lt;br&gt;&lt;br&gt;Using this key for vigenere decryption, the solution string can be decoded.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastActivityDate="2015-05-17T22:04:11.087" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15116" PostTypeId="2" ParentId="1690" CreationDate="2015-05-17T23:52:10.207" Score="1" Body="&lt;p&gt;I believe the 98% get it wrong because of the fact that the question itself is a square. They count the 40 squares created by the green lines, but they miss the pink square. @ClickRick, I believe those that say 42 are correct: the pink area makes 41, the embossed rim would make 42. However, if you wanted to go further, you could note that the background is two different colors; pink and purple, slightly different tones. I count at least 13 purple squares in the background, and at least 2 pink ones. The problem is that the rules are not clearly defined; 40 is correct from a basic standpoint, looking at only the green lines. However, since 98 percent get it wrong, and it is a puzzle after all, I must believe that the squares that exist outside of the line are meant to be counted. My guess is that a large portion of the 98 percent probably say 40. In your simulation, there are only 40. But I believe that there is some reason for the pink background, rather than a plain white one - to complicate things, creating additional inconspicuous squares.&lt;/p&gt;&#xA;" OwnerUserId="12528" LastActivityDate="2015-05-17T23:52:10.207" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15118" PostTypeId="1" AcceptedAnswerId="15120" CreationDate="2015-05-18T01:03:34.967" Score="6" ViewCount="1874" Body="&lt;p&gt;There is a secret word associated with this image that will be clear if it is decrypted properly. What is that word? (JPEG artifacts are irrelevant)&lt;/p&gt;&#xA;&lt;p&gt;Full size image: &lt;a href=&quot;https://i.stack.imgur.com/cc5yl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/cc5yl.jpg&lt;/a&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/T7kHK.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="69582" LastEditDate="2023-11-16T04:51:05.563" LastActivityDate="2023-11-16T04:51:05.563" Title="Find the secret word in this encrypted image (part 1)" Tags="|word|visual|cipher|steganography|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15119" PostTypeId="2" ParentId="15097" CreationDate="2015-05-18T01:22:04.970" Score="24" Body="&lt;p&gt;I am not myself a perfect logician, but here the behaviour I observed at my own tea parties, and the explanation my own favourite nounours gave as years passed.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Party of one&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This issue was quickly solved before I got my second plush. Nounours was the single invitee, and behaved as well as a bear does at a party.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Party of two&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;When Dr. Tortue joined us some time later, I observed the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Nounours was nominated by default&lt;/li&gt;&#xA;&lt;li&gt;Dr. Tortue nominated himself, much to Nounours’s chagrin.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;But Nounours, being the noble being that he is, quickly excused the good Doctor behaviour. “In her stead, I would have done the same thing: as there were no other candidate for Guest of Honour, she was certain to eat the delicious sugar pie&lt;sup&gt;1&lt;/sup&gt;.”&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Party of three&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;When Johnny McCanasson joined our little assembly, I expected Nounours and perhaps Dr. Tortue to be excluded from the table again. Nounours, however, quickly reassured me: “No, see, if either Dr. Tortue or Mr. McCanasson were to nominate themselves for Guest of Honour, they would be much in the same situation as I last time: the current nominee with only one candidate. This candidate, being sure to win the coveted position if they nominated themself, would without doubt do so. Thus, the first candidate to nominate themself would end up enjoying neither the pie nor the chocolate&lt;sup&gt;2&lt;/sup&gt;. Even that dreaded McCanasson would rather enjoy your company than deprive both him and myself of it.”&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Party of four&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The next year saw the coming of the Lapin brothers, Mercredi and Dimanche. Unfortunately&lt;sup&gt;3&lt;/sup&gt;, my younger brother requested Mr. McCanasson for his own company, and neither Mr. McCanasson&lt;sup&gt;4&lt;/sup&gt; nor I had the heart of refusing him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was expecting the party to go as it had previously, with Nounours remaining the Guest of Honour nominate throughout the process. It didn’t happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“See”, explained Nounours once again, “we are in the very same position as we were when only Dr. Tortue and myself were your guests. The first person to nominate themself for Guest of Honour would be in the same situation than previously, that is the nominee in a party of three. As we have seen last time, this is an excellent position, as it guaranties that you will win the honour and the pie that goes with it. The doctor, being the intelligent tortoise she always is, did not fail to notice, and didn’t let either of the Lapin brothers the opportunity to enjoy this position: she nominated herself as soon as she was offered the chance.”&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Party of &lt;em&gt;n&lt;/em&gt;, especially when &lt;em&gt;n&lt;/em&gt; equals 74&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Since then I observed the same behaviour each year: whenever there were an odd number of guests, Mr. Nounours, as first nominee, would remain guest of honour; when the guests were in even number, Dr. Tortue would nominate herself for Guest of Honour and eventually win the position, leaving the poor Mr. Nounours on the side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for what will happen with a party of 74, I have no idea as I not yet enjoy the pleasure of such a large company. Nounours, however, insists that Mr. Bun will be put aside as Duke Froggington II will take the seat next to yours, claiming that “the same patterns keep repeating” and using words such as “induction”. But, really, I can’t think of a reason why changing stove would change the issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do however sometimes wonder what would happen if I were to cook additional pies…&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I just now notice that my proposed answer is for tea parties with &lt;em&gt;sugar&lt;/em&gt; pies. I’ll post it anyways, but feel free to delete if necessary.&lt;/li&gt;&#xA;&lt;li&gt;Ah! Another possibly important detail.&lt;/li&gt;&#xA;&lt;li&gt;Although I can hear Nounours mumbling that he doesn’t mind so much.&lt;/li&gt;&#xA;&lt;li&gt;Don’t listen to Nounours, Mr. McCanasson is gruff, but very kind indeed. He often takes the time to visit.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11907" LastEditorUserId="11907" LastEditDate="2015-05-18T03:14:29.080" LastActivityDate="2015-05-18T03:14:29.080" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15120" PostTypeId="2" ParentId="15118" CreationDate="2015-05-18T01:53:37.857" Score="7" Body="&lt;p&gt;I hesitate to say&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; STAR&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;since&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a mere width/height adjustment seems a bit easy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Method:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's visually recognizable as an image that has been skewed in width. I converted it to a BMP, opened it with a hex editor, and started guessing at some values to adjust the width field. After 3 or 4 attempts, I stumbled upon one which made it readable enough to obtain the above, and did not attempt to further refine the value.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit by Author:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &quot;ideal&quot; dimensions alluded to in the above method is when the dimensions are square. Multiplying the image dimensions together, you get $1337 * 1914 = 2559018$.&lt;br&gt;The square root of this is 1599.7, so make the image dimensions 1600 X 1600 and you get the below image:&lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/elfOc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11119" LastEditorUserId="7722" LastEditDate="2015-05-18T02:36:28.160" LastActivityDate="2015-05-18T02:36:28.160" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15121" PostTypeId="2" ParentId="15118" CreationDate="2015-05-18T02:45:08.193" Score="3" Body="&lt;p&gt;The image contains a little bit more than in Ayefork's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The text actually reads:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The secret message is: STAR&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can find a decoded version of the image &lt;a href=&quot;http://media.jarnbjo.de/XMdDfAM-decode.jpg&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and Java source code for decoding the image &lt;a href=&quot;http://media.jarnbjo.de/Steganography.java&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Description: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The original image has been split into 6 vertical slices, each 263 pixel wide and using some strange interleaving pattern, been skewed into the 1337 pixel wide encoded image.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11687" LastActivityDate="2015-05-18T02:45:08.193" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15122" PostTypeId="2" ParentId="15048" CreationDate="2015-05-18T05:48:02.113" Score="0" Body="&lt;p&gt;Let $P_n$ be the probability that a human player would rather bid $n+1$ than lose $n-1$. Let $Q_n$ be the probability that a human player who has nothing to lose would bid $n+1$. These values would vary with $n$, so we will have to carry out huge surveys to get these values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $k$ be the number of players. Let $R_n$ be the probability of winning the auction by bidding $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we know that we do not win if the event in $P_n$ occurs. We also do not win if any of the other $k-2$ players decides to enter the auction. Therefore, none of the events mentioned should occur. &#xA;$$R_n=(1-P_n)(1-Q_n)^{k-2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we bid $n$, our winning is $20-n$  if we win and our loss is $n$ if we lose. Hence we must bid $n$ only if&#xA;$$Winnings\times Probability_{Winning}&amp;gt;Loss\times Probability_{Losing}$$&#xA;or $$(20-n)\times R_n&amp;gt;n\times (1-R_n)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P. S. I've not actually learnt these probability concepts, so please correct me where I'm wrong.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-05-18T05:48:02.113" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15124" PostTypeId="2" ParentId="15113" CreationDate="2015-05-18T08:10:14.260" Score="3" Body="&lt;p&gt;To accompany Quark's correct answer, the hint decodes to:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;&quot;Use Vigenere twice key is a pangram with X letters.&quot;&lt;/strong&gt; and &lt;strong&gt;X = 26&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The clues are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the capital letters for &lt;strong&gt;Lost&lt;/strong&gt; and &lt;strong&gt;Number&lt;/strong&gt; which suggests using the numbers from this television series. These numbers are &lt;a href=&quot;http://lostpedia.wikia.com/wiki/The_Numbers&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;4, 8, 15, 16, 23 and 42&lt;/strong&gt;&lt;/a&gt;.  Using letters from these columns of the cipher text (shown below in yellow) provides the decoded sentence.&lt;br&gt;&#xA;&lt;br&gt;&#xA;Then, the number of letters in the key (X) is 26 based on the clues &lt;strong&gt;&quot;26 days&quot;&lt;/strong&gt; and &lt;strong&gt;&quot;26 feet&quot;&lt;/strong&gt;. In addition, the Vigenere key in Quark's answer does have 26 letters.&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/OdnFX.png&quot; alt=&quot;Lost&quot;&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As the encryption needs to be applied twice, the key works out to be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;&quot;Cwm fjord bank glyphs vext quiz&quot;&lt;/strong&gt; which is a notable pangram.&lt;br&gt;&#xA;&lt;br&gt;&#xA;According to anagrammy.com, this pangram means &quot;carved figures in a valley on a bank of a fjord irritated an eccentric person&quot;. Cwm is Welsh for a circular valley; a glyph is a carved figure; vext is the poetic spelling of vexed; and a quiz is an 18th-century term for an eccentric.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-05-19T03:40:15.813" LastActivityDate="2015-05-19T03:40:15.813" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15125" PostTypeId="2" ParentId="15075" CreationDate="2015-05-18T08:57:07.960" Score="6" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;cut a 134m piece of rope, separate the strands (assuming 3 strands) and tie together to make a roughly 400m length.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Cut 3 2m pieces from the main rope, use these pieces to create a &lt;a href=&quot;http://www.animatedknots.com/purcell/&quot; rel=&quot;noreferrer&quot;&gt;Purcell Prusik Knot Harness&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Attach the remaining long rope to the hook using a &lt;a href=&quot;http://www.animatedknots.com/highwaymans/&quot; rel=&quot;noreferrer&quot;&gt;Highway mans hitch&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Tie the thin 400m rope to the 'dead end'.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Use the Prusik Knot Harness to safely descend the main rope.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Pull the thin rope to undo the highway mans hitch.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Repeat steps 2 - 5.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9478" LastEditorUserId="9478" LastEditDate="2015-05-18T11:12:05.190" LastActivityDate="2015-05-18T11:12:05.190" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15126" PostTypeId="1" AcceptedAnswerId="15138" CreationDate="2015-05-18T09:00:14.033" Score="16" ViewCount="3911" Body="&lt;p&gt;I was asked this question in an interview and though I was able to answer a part or few parts I'm not really sure if what I gave was the optimal answer. So here goes the puzzle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A restaurant owner asked a Carpenter to make tables for him out of certain number of wooden blocks and boards.&#xA;There are only a limited number of boards. He was asked to make a rectangular hole right in the middle of the table so that the owner would be able to embed the restaurant logo in it.&#xA;The Carpenter was a drunk and while working on the last table (and last wooden board) he had to make, be made a blunder. He made the rectangular hole way off the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OHTtP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The restaurant owner fired him on the spot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He then goes ahead and hires you, an inexperienced Carpenter, with a promise that if you could fix the blunder caused by the last Carpenter, you would get a huge contract for all the wood work needed for the rest of his establishments. This would be huge for you are completely broke right now.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How would you solve the problem?&lt;/li&gt;&#xA;&lt;li&gt;How would you solve the problem if the hole made was made at a 45 degree angle and needs to be corrected as well?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zI5ER.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;How would you solve the problem if both the table and the hole were oval in shape?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/84Quf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rules:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you are inexperienced and an amateur and you only have a hacksaw, you can only start cutting along the outside edges of the table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 1. You can only make two cuts as otherwise it would ruin the look of the table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The table has to be the same size as the rest of the tables. The size of the table is specified by the owner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Need to give the most optimal solution and if possible prove it's optimality. Optimal solution will have the minimal cumulative length of the cuts&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="12537" LastEditorUserId="8874" LastEditDate="2016-03-23T15:39:46.433" LastActivityDate="2016-03-23T15:39:46.433" Title="Holes in the Table" Tags="|mathematics|geometry|dissection|" AnswerCount="8" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="15127" PostTypeId="1" AcceptedAnswerId="15167" CreationDate="2015-05-18T09:06:36.213" Score="34" ViewCount="3338" Body="&lt;p&gt;&lt;em&gt;Note: please do not edit the title! It originates in a &lt;a href=&quot;http://chat.stackoverflow.com/transcript/6?m=22733819#22733819&quot;&gt;challenge&lt;/a&gt; in Stack Overflow chat.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have rounded up all ten (currently) of the &lt;a href=&quot;https://puzzling.stackexchange.com/users?tab=Reputation&amp;amp;filter=all&amp;amp;search=kevin&quot;&gt;Puzzling SE users named Kevin&lt;/a&gt; and trapped them in my labyrinth. The only way out is across a narrow bridge which can bear the weight of at most two Kevins at a time. It is pitch dark and the bridge has no handrails, so they will need a torch in order to be able to cross at all. Each Kevin will take a different amount of time to cross the bridge; let these times be $t_1,t_2,\dots,t_{10}$ in increasing order. &lt;a href=&quot;https://puzzling.stackexchange.com/users/4/kevin&quot;&gt;Moderator Kevin&lt;/a&gt; is the slowest, taking time $t_{10}$ to cross, because he has to carry his diamonds with him. I'll supply them with a torch so that they can see to cross the bridge, but it will only last for a finite period of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My aim is to allow all the Kevins except the slowest to escape the labyrinth (if the moderator escapes, he'll give me a 365-day suspension for moderator entrapment as soon as he gets back to his computer). I want to ensure the torch will last exactly long enough to allow the nine non-moderator Kevins to escape but not all ten. &lt;strong&gt;How long (in terms of the times $t_1,\dots,t_{10}$) should I make the torch last?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2015-05-19T00:16:46.230" Title="There are at least three Kevins on Puzzling. I plan to trap them in my labyrinth" Tags="|strategy|story|" AnswerCount="3" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="15128" PostTypeId="2" ParentId="15126" CreationDate="2015-05-18T09:16:13.930" Score="4" Body="&lt;p&gt;Assuming I can cut more than twice for 2 and 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Make two cuts so the hole is in the middle of the smaller piece cut out of the original table, the table will be smaller, but seeing that the owner never specified how large the table had to be you can just cut it like that.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Do the same as in 1 but make more cuts to get the desired result.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For 3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dot the same as in 1 and 2 but make it a round cut so it retains the shape.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For more information&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You could use the remaining pieces of wood to craft equal tables in shape just different in size.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12218" LastEditorUserId="12218" LastEditDate="2015-05-18T09:20:57.577" LastActivityDate="2015-05-18T09:20:57.577" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15129" PostTypeId="1" AcceptedAnswerId="15131" CreationDate="2015-05-18T09:20:12.137" Score="5" ViewCount="3270" Body="&lt;p&gt;There are $X$ stones in the river.The stones are placed in such a way that a person can jump from one stone to the next one, or skip one and jump to the one after that. Find the number of ways in which one can jump across the stones from one river bank to another. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: We are always standing at the farther stone from the bank&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For example:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;If there is one stone, we can jump directly to the bank of the river i.e. only 1 way.&lt;/li&gt;&#xA;  &lt;li&gt;If there are 2 stones, we can jump from 1st to 2nd and then 2nd to bank or we can jump from 1st to bank skipping one stone(as mentioned in the question).&lt;/li&gt;&#xA;  &lt;li&gt;Similarly for 3 stones no. of ways are 3.&lt;/li&gt;&#xA;  &lt;li&gt;Similarly for 5 stones no. of ways are 8.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is there any general technique to calculate the number of ways of crossing the river with $X$ stones?&lt;/p&gt;&#xA;" OwnerUserId="2251" LastEditorUserId="8874" LastEditDate="2015-10-04T14:24:11.270" LastActivityDate="2015-10-04T14:24:11.270" Title="Finding the number of ways of crossing a river" Tags="|mathematics|combinatorics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15130" PostTypeId="1" AcceptedAnswerId="15639" CreationDate="2015-05-18T09:33:25.640" Score="17" ViewCount="2097" Body="&lt;p&gt;&lt;em&gt;I was asked this question in an interview, so I am not sure what the optimal answer is. But here it goes:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a rectangle with length &lt;strong&gt;&lt;em&gt;l&lt;/em&gt;&lt;/strong&gt; and breadth &lt;strong&gt;&lt;em&gt;b&lt;/em&gt;&lt;/strong&gt;, I pick 4 random points inside it. What is the probability that the 4 points lie in the same half of the rectangle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Two halves of the rectangle are defined by a single line passing through the intersection of its diagonals. So the rectangle cannot be divided into two halves of any arbitrary shapes of equal length.&lt;/li&gt;&#xA;&lt;li&gt;A line passing through a picked point and the center does not contain another picked point. That is, A line passing through any two picked points does not pass through the center of the rectangle (= intersection of the diagonals).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Addition: (Maybe a Hint)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was asked about 3 points first. They saw my approach and were satisfied and then asked me what if there were 4 points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Answer&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 9/16&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The one to prove it gets the check mark.&lt;/p&gt;&#xA;" OwnerUserId="12537" LastEditorUserId="30903" LastEditDate="2017-02-10T04:22:29.523" LastActivityDate="2017-02-10T04:22:29.523" Title="Points in a Rectangle" Tags="|geometry|probability|" AnswerCount="6" CommentCount="21" ContentLicense="CC BY-SA 3.0" />
  <row Id="15131" PostTypeId="2" ParentId="15129" CreationDate="2015-05-18T09:33:42.363" Score="10" Body="&lt;p&gt;Given n stones, the only option one has for the first jump is to go to stone n-1 or n-2. From that point on the options are the same as for stone n-1 or n-2, hence for any n, the number of options equals the sum of the options of n-1 and n-2. This generalises to the Fibonacci sequence (Fibonacci+1 to be exact). Hence the generalised technique to calculate this for n stones is Jumps=Fibonacci(n+1)&lt;/p&gt;&#xA;" OwnerUserId="11473" LastActivityDate="2015-05-18T09:33:42.363" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15132" PostTypeId="2" ParentId="15129" CreationDate="2015-05-18T09:33:52.123" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Isn't this just the Fibonacci sequence: 1, 2, 3, 5, 8, 13, 21, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you number the stones starting at the target bank: From $stone_{n}$ you can initially jump to either $stone_{n-1}$ or $stone_{n-2}$. This makes the number of ways from $stone_{n}$ to the target bank equal to $stone_{n-1} + stone_{n-2}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11687" LastActivityDate="2015-05-18T09:33:52.123" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15133" PostTypeId="2" ParentId="15127" CreationDate="2015-05-18T09:40:12.317" Score="18" Body="&lt;p&gt;The fastest I found:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For notation, $K_n$ is the &lt;em&gt;Kevin&lt;/em&gt; whose time for crossing is $t_n$. $\rightarrow$ means he/they cross the river. $\leftarrow$ means he crosses back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_1,K_2\rightarrow;\ K_1\leftarrow;\ t_2+t_1;\\&#xA;K_8,K_9\rightarrow;\ K_2\leftarrow;\ t_9+t_2;\\&#xA;K_1,K_2\rightarrow;\ K_1\leftarrow;\ t_2+t_1;\\&#xA;K_6,K_7\rightarrow;\ K_2\leftarrow;\ t_7+t_2;\\&#xA;K_1,K_2\rightarrow;\ K_1\leftarrow;\ t_2+t_1;\\&#xA;K_4,K_5\rightarrow;\ K_2\leftarrow;\ t_5+t_2;\\&#xA;K_1,K_2\rightarrow;\ K_1\leftarrow;\ t_2+t_1;\\&#xA;K_1,K_3\rightarrow;\ t_3;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe this is the fastest 9 can cross, so that'd make impossible to cross with a slower Kevin. So the torch shall last $4t_1+7t_2+t_3+t_5+t_7+t_9$&lt;/p&gt;&#xA;" OwnerUserId="12165" LastActivityDate="2015-05-18T09:40:12.317" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15134" PostTypeId="2" ParentId="15127" CreationDate="2015-05-18T09:43:07.970" Score="14" Body="&lt;p&gt;The solution to the simple form of this problem (get $t_1$, $t_2$, $t_5$ and $t_{10}$ across a bridge in 17 time units) requires that:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You send the two longest times across together, so that only the longer of the two affects the score.&lt;br&gt;&#xA;So, send $t_1$, $t_2$ across (2)&lt;br&gt;&#xA;Send $t_1$ back with the torch (3)&#xA;Send $t_5$ and $t_{10}$ across (13)&#xA;Send $t_2$ back with the torch (15)&#xA;Send $t_1$, $t_2$ across again (17)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For more bridge-crossers, this strategy requires remarkably little adaptation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To start, you need a pair of torch-runners - the two people who can carry the torch back-and-forth most quickly across the bridge. Logically, these should be $t_1$ and $t_2$. The others will cross in pairs, with $t_1$ and $t_2$ crossing the bridge between each pair to ensure speedy delivery of the torch.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Send $t_1$, $t_2$ across (2)&lt;br&gt;&#xA;Send $t_1$ back with the torch (3)&lt;br&gt;&#xA;Send $t_3$ and $t_4$ across (7)&lt;br&gt;&#xA;Send $t_2$ back with the torch (9)&lt;br&gt;&#xA;Send $t_1$, $t_2$ across again (11)&lt;br&gt;&#xA;Send $t_1$ back with the torch (12)&lt;br&gt;&#xA;Send $t_5$ and $t_6$ across (18)&lt;br&gt;&#xA;Send $t_2$ back with the torch (20)&lt;br&gt;&#xA;Send $t_1$, $t_2$ across again (22)&lt;br&gt;&#xA;Send $t_1$ back with the torch (23)&lt;br&gt;&#xA;Send $t_7$ and $t_8$ across (31)&lt;br&gt;&#xA;Send $t_2$ back with the torch (33)&lt;br&gt;&#xA;Send $t_1$, $t_2$ across again (35)&lt;br&gt;&#xA;Send $t_1$ back with the torch (36)&lt;br&gt;&#xA;Send $t_9$ and $t_{10}$ across (46)&lt;br&gt;&#xA;Send $t_2$ back with the torch (48)&lt;br&gt;&#xA;Finally, send $t_1$, $t_2$ across again (50) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This would yield a time of 50 units for everyone to cross - in a bid to avoid moderator rage and the dreaded banhammer, we don't want the torch to last for that long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Once $t_7$ and $t_8$ have crossed (at time 31), we once again send $t_2$ back (33), send $t_1$ and $t_2$ across again (35) and $t_1$ back one final time (36).&lt;br&gt;&#xA;This time, however, rather than allowing moderator Kevin and his diamonds to cross (for fear that the bridge will not hold the weight of such riches), $t_1$ and $t_9$ run off together, reaching the other end of the bridge at a time of 45 units, scant seconds before the torch flickers and dies out.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This solution assumes that $Kevin_1$ and $Kevin_2$ do not tire in their travels, and can maintain a consistent pace throughout. It also assumes that moderator Kevin upholds the logical foundation of this site, and doesn't simply insist upon crossing first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also assumes that (e.g.) $t_9$ = $t_8$ + $t_1$, which is probably not the case, in which case your torch time will actually be 4$t_1$ + 7$t_2$ + $t_4$ + $t_6$ + $t_8$ + $t_9$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: After seeing Albert Masclans' answer, I realise I've made a mistake. I tried to build travelling pairs from the bottom up, when I should have worked top-down, pairing $t_9$ and $t_8$, $t_7$ and $t_6$, and $t_5$ and $t_4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This allows the last trip to be $t_3$ and $t_1$, and gives a total of 4$t_1$ + 7$t_2$ + $t_3$ + $t_5$ + $t_7$ + $t_9$, or 42 by my previous assumptions&lt;/p&gt;&#xA;" OwnerUserId="11717" LastEditorUserId="11717" LastEditDate="2015-05-18T09:57:24.397" LastActivityDate="2015-05-18T09:57:24.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15135" PostTypeId="2" ParentId="15126" CreationDate="2015-05-18T09:59:41.507" Score="2" Body="&lt;p&gt;&lt;sub&gt;(On mobile right now, so I can't validate my answer or draw pretty pictures)&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming your amateur skills extend to joinery, you could, for the first:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; cut a thick L shaped section off from the lower left, leaving a kind of smaller version of the table/hole, the rotate the L and reattach it to the other side (you'd just have to do the maths to make sure the amount you removed plus the amount on the far side of the hole equaled the amount that remained on the near side).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the second (&lt;strong&gt;edited&lt;/strong&gt;):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a similar approach could be used, but first you need to cut on a 45 degree diagonal (parallel to the hole), starting at the bottom right corner such that you come in the same distance as the short edge. Then turn 90 degrees to the right and finish the cut. This will produce a kite shape which can be flipped over, effectively rotating the hole by 45 (your two cuts were parallel to the hole, with the first cut becoming the new short edge). Now you've effectively got a table like in number one, with the hole the correct angle, but off center, so we just repeat those steps.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the third:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not 100% sure on this one, because it feels like you'd struggle to maintain the appropriate curve, but you could possibly repeat the same process, only this time initially cutting a crescent shape section from the bottom left.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="5208" LastEditDate="2015-05-18T11:52:06.943" LastActivityDate="2015-05-18T11:52:06.943" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15138" PostTypeId="2" ParentId="15126" CreationDate="2015-05-18T10:25:13.383" Score="17" Body="&lt;p&gt;For part 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Measure the table's edges, and determine the mid-points. Measure the x and y distance from the centre of the hole (contrary to the image - got that wrong) to the edges of the table. Measure the same distances from the centre point. Cut to this second point (i.e. along the black lines), and then rotate the section that you've cut out by 180 degrees.&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/PPbA7.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For part 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This one's actually not too hard to do, but difficult to explain -  in essence, you can cut material equal to the hole in size from the middle, and place it where the actual hole is. See the diagram (and assume the central shape is the same size as the hole, please!) - if done in the correct order, all cuts are cuts from the edge&#xA;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/10Ff9.png&quot; alt=&quot;&quot;&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;A potential cutting order is now included (edit: only cuts 1,2,3 and 4 needed). Once done, place the rectangle bounded by 1,2,3,4 in the hole bounded by 6,7,8,9 &#xA;&lt;img src=&quot;https://i.stack.imgur.com/Cia4N.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For part 3:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Much as for part 1, this involves cutting a rectangle that includes the hole and its distance from the edge. Measure the x and y co-ordinates of the centre of the board, and the x and y distances from the centre of the hole to a point on the outer edge of the ellipse. Measure those same distances from the centre of the board.&lt;br&gt;&#xA;Cut out (see black lines) the rectangle described by these two points and the x and y planes.&lt;br&gt;&#xA;You will need to cut two extra lines to allow you to cut the whole shape free, as you can only cut from the outside.&lt;br&gt;&#xA;Then rotate the rectangle by 180 degrees, replace it, and return the other two removed segments to complete the table&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/LIuKQ.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11717" LastEditorUserId="11717" LastEditDate="2016-02-15T14:54:04.047" LastActivityDate="2016-02-15T14:54:04.047" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="15139" PostTypeId="2" ParentId="15130" CreationDate="2015-05-18T10:42:27.947" Score="4" Body="&lt;p&gt;The answer is heavily based on &lt;a href=&quot;https://math.stackexchange.com/questions/325141/probability-that-n-points-on-a-circle-are-in-one-semicircle&quot;&gt;this&lt;/a&gt; and I've tried to simplify it a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0.5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Basically the solutions is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let one of the points lie on the &quot;rectangle separator&quot;. From there on, we want all other points to be on the same halfplane. Picking a point we practically make &quot;coinflip&quot; on which side will the point land (this is correct as the two halves have equal area). So we get $1/2^{n-1}$ for each separator point. Since the separator can be anywhere (in terms of rotation around the diagonal intersection point), trying out which point lies on the separator are disjoint events. So we sum that up for all separators getting $n/2^{n-1}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This results in:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $4/2^{4-1} = 0.5$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, we can do a &quot;sanity check&quot; that for 1 and 2 the answer is 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, one can try to get the odds that 4 points &lt;strong&gt;do not&lt;/strong&gt; lie in the same half. Will write that up later.&lt;/p&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="-1" LastEditDate="2017-04-13T12:19:36.950" LastActivityDate="2015-05-18T10:42:27.947" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15140" PostTypeId="2" ParentId="15126" CreationDate="2015-05-18T11:26:25.483" Score="4" Body="&lt;p&gt;Even if I cannot prove, this looks pretty much optimal:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/Q2UJM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;same for 2 and 3. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if that can be made with an hacksaw: it probably doesn't like the sharp bends. To avoid them we could go for something like this, which is obviously suboptimal, but still possibly better than other solutions (depends on the size and on the initial placement of the hole).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/78V6t.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To start cutting around the hole I assume that it is possible to insert the hacksaw in the hole, otherwise it should look pretty much as the other cut.&lt;/p&gt;&#xA;" OwnerUserId="2683" LastEditorUserId="2683" LastEditDate="2015-05-18T12:04:31.760" LastActivityDate="2015-05-18T12:04:31.760" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15141" PostTypeId="1" AcceptedAnswerId="15148" CreationDate="2015-05-18T11:32:33.190" Score="4" ViewCount="896" Body="&lt;p&gt;My first brother loves the winds and the birds,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My second has been cast out from the family, he is the black sheep,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My third brother loves the waters, even in the deepest caves,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My fourth brother is like the father of Lion,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My fifth brother is kind of gloomy, with his voice as sad as doom,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My sixth brother is named after a wonderful forest,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And my seventh was the last to move and good at wrestling.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My first sister is the white, and beautiful,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My second sister is like to a gardener,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My third sister weeps much, and finds much pity for others,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My fourth sister is a sleeper to be sure,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My fifth sister is like the rose in the east(which is related to an Arrow and a Song)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My sixth sister loves the sun,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And my seventh sister is very swift, indeed she could probably run alongside the deer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Please fully explain your answer with the few book references I used&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="11790" LastEditorUserId="8874" LastEditDate="2015-10-02T17:07:42.340" LastActivityDate="2015-10-02T17:07:42.340" Title="Who's missing in this list?" Tags="|riddle|knowledge|literature|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15142" PostTypeId="1" AcceptedAnswerId="15233" CreationDate="2015-05-18T11:38:09.673" Score="20" ViewCount="3361" Body="&lt;p&gt;This puzzle belongs to the puzzle series: &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=hyper-modern%20is%3Aquestion&quot;&gt;hyper-modern art&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The two friends move on in the gallery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;You have now shown me two examples of hyper-modern art, but I still don't quite get it. What is this 'art' all about?&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Art, my friend, should draw the observer closer, involve him, make him think or feel. Hyper-modern art aims at this with a slightly different angle. To &lt;em&gt;understand&lt;/em&gt; an image, the observer has too fully engage with it until he discovers the things which rule the image underneath. By doing so, the image depicts exactly &lt;em&gt;what can not be seen&lt;/em&gt;. Do you understand?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Hmm, I see how this applies to the two previous examples, but it doesn't help me much with understanding any new piece, does it? Just look at what is hanging here: Why on earth is the picture called &lt;code&gt;The (colour) star&lt;/code&gt;, if the star is absolutely not of that colour?&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Exactly! You have to see the &lt;em&gt;invisible&lt;/em&gt; part of the picture to understand why it can only be that colour, my friend. But to make it easier, why not switch on your special HUD device to make it visible to you...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Oh.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RTT94.png&quot; alt=&quot;HM3&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The goal of the puzzle is to find the colour of the central circle marked with the star. It is one of the 7 colour used by the other circles (Red, orange, yellow, green, cyan, blue, or violet.) A complete answer must give both the colour and the explanation of why it is the correct colour. (Bonus: What does the HUD show?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hints&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The outer ring colours are &quot;paler&quot; for visual effect only. The whole puzzle is based on 7 distinct colours and the solution is one of those 7. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The puzzle can be solved with a &lt;em&gt;printed&lt;/em&gt; version of the image above. (Or by just looking at that image above.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The puzzle has a single, objective and logical solution. The &quot;Pattern&quot; tag for this puzzle is appropriate. Basically the whole images follows a specific rule, and the central circle has to fulfil that rule as well, which defines its colour.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2018-08-15T08:53:39.007" Title="What is the central color?" Tags="|pattern|visual|" AnswerCount="8" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="15143" PostTypeId="2" ParentId="15075" CreationDate="2015-05-18T11:59:44.287" Score="-3" Body="&lt;p&gt;Simple.  Just jump off the building and the split second you hit the ground you are uninjured and satisfied the requirement of the question.  Being only able to safely fall 5 feet is irrelevant because it is not part of the stated requirement.&lt;/p&gt;&#xA;" OwnerUserId="12119" LastActivityDate="2015-05-18T11:59:44.287" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15144" PostTypeId="2" ParentId="15142" CreationDate="2015-05-18T12:03:05.190" Score="2" Body="&lt;p&gt;I'm going to say that the star is indigo. My reasoning for this is in referring to the center as the star; this is a map of the night sky. The orange circles match up to Ursa Major, and the green ones to the Summer Triangle. If you match up an image of the Triangle with the art piece above, the central star appears to be &lt;em&gt;Gamma Cygni&lt;/em&gt;, which would mean it is indigo like the other two stars in &lt;em&gt;Cygnus&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="12542" LastActivityDate="2015-05-18T12:03:05.190" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15145" PostTypeId="2" ParentId="15130" CreationDate="2015-05-18T12:27:42.097" Score="8" Body="&lt;p&gt;My answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\frac12$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can obtain a general formula for $n$ with a simple idea.&lt;br&gt;&#xA;You have $n$ points inside the rectangle. Now you choose one (there are $n$ choices) and draw a line passing through it and the centre, determining two halves. Now, what's the probability that each point is in the same half of the previous? Of course it is $\frac12$ for each point! Now, we have all we need to deduce a general formula:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$p(n)=n\times (\frac12)^{n-1}=\frac{n}{2^{n-1}}$.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Obviously, we have $p(1)=1$ and $p(2)=1$.&lt;br&gt;&#xA;We can also observe that $p(3)=\frac34$ and $p(4)=\frac12$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may say: &quot;Hey, but there are infinite halves that contain one point! Why did you choose the one passing through the point itself?&quot;&lt;br&gt;&#xA;I did because that's the limit case! If you don't think so, check the pictures below. The yellow area shows all the valid halves given $n$ points. As you can see, the yellow area is delimited by lines passing through the points themselves!&#xA;&lt;img src=&quot;https://i.stack.imgur.com/XnAQ2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional note: when you draw a line, you're actually drawing 2 vectors, with opposite direction. Each vector represents one of the two halves. That is, the above formula is actually the simplification of $p(n)=n\times (\frac12)^{n-2}\times2$&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-05-19T13:01:35.033" LastActivityDate="2015-05-19T13:01:35.033" CommentCount="19" ContentLicense="CC BY-SA 3.0" />
  <row Id="15148" PostTypeId="2" ParentId="15141" CreationDate="2015-05-18T13:39:20.517" Score="11" Body="&lt;p&gt;Answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oromë, the Huntsman, the Lord of Forests and the Great Rider.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Explaination:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are the Valar, gods from Lord of the Rings&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My first brother loves the winds and the birds,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Manwë, the Wind-King and the King of the Valar.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My second has been cast out from the family, he is the black sheep,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Melkor, the Dark Power and most powerful of the Valar. He is not counted as a Vala anymore.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My third brother loves the waters, even in the deepest caves,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ulmo, the Sea-King and the Lord of Water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fourth brother is like the father of Lion,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Aulë, the Smith and the Lord of Earth and all that's underneath. Created the Dwarves. Which are like bearded and proud like lions?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fifth brother is kind of gloomy, with his voice as sad as doom,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Námo (Mandos), the Doomsman and the Judge of the Dead.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My sixth brother is named after a wonderful forest,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Irmo (Lórien), Lord and Master of Dreams, Visions, and Desires, and Creator of the Oloré Mallé, or Path of Dreams.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And my seventh was the last to move and good at wrestling.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tulkas, the Wrestler, the Champion of Valinor and last of the Valar to come to Arda&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My first sister is the white, and beautiful,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Varda, the Star-Queen and the Queen of the Valar.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My second sister is like to a gardener,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yavanna, the Fruit-Giver and the Lady of Earth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My third sister weeps much, and finds much pity for others,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nienna, the Weeper and the Lady of Mercy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fourth sister is a sleeper to be sure,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Estë the Gentle, Lady of Healing and Rest, and spouse of Irmo.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fifth sister is like the rose in the east(which is related to an Arrow and a Song)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Vairë, the Weaver and spouse of Mandos. Not sure of the link to rose in the east, but the only one left.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My sixth sister loves the sun,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Vána, the Ever-young and spouse of Orome. Created the fruit on Laurelin which became the sun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And my seventh sister is very swift, indeed she could probably run alongside the deer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nessa, the Dancer and spouse of Tulkas. She is lithe and swift of foot; she loves fast moving creatures such as deer. Her speed is said to have even that of a deer which follow her in the wilds&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;References&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I used lotr.wikia.com/wiki/Valar&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12548" LastEditorUserId="12548" LastEditDate="2015-05-19T08:41:35.890" LastActivityDate="2015-05-19T08:41:35.890" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15149" PostTypeId="2" ParentId="15142" CreationDate="2015-05-18T13:47:19.383" Score="0" Body="&lt;p&gt;is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RED?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I used the developer tools to check its color ^^&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RGB: rgb(43, 42, 41); HEX: #2b2a29 , which contains mainly RED color.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12054" LastEditorUserId="10281" LastEditDate="2015-05-18T13:50:57.010" LastActivityDate="2015-05-18T13:50:57.010" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15150" PostTypeId="2" ParentId="15048" CreationDate="2015-05-18T13:54:31.913" Score="0" Body="&lt;p&gt;There are too many possible interpretations of the rules:&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Case 1: You are not allowed to bid twice in a row&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Perfect play:&lt;/strong&gt; The best Option is not to bid at all, because bidding 5$ will lose you money.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation:&lt;/strong&gt; It is hard to put a starting point to analyse possible strategies because of infinite payrolls. But if every player plays a perfect strategy and can predict all other players, we have absolutely no chance involved and the outcome of each round can be predicted perfectly beforehand. Therefore there is no risky gamble, but always an expected Win/Loss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this scenario nobody will ever bid 21, because when you bid 21 you are guaranteed to loose at least 1\$ so whatever strategy would lead you to bid 21 eventually would never be taken in the first place, because not bidding at all is better than loosing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means you can always safely bet 20 without loosing or winning anything, because no one will ever bid 21 in perfect play. What about 19 ? If you bet 19, you can be sure that the guy who did bet 18 will now bet 20, because 20 is safe and will give him 0 loss (and nobody will ever bet 21, because nobody will use a loosing strategy leading him there) So betting 19 is a guaranteed loss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This in case makes 18 to a perfectly safe bid, since no one will ever bet the sure loss of 19. And by induction we can follow this road downwards. Nobody will ever bet 17, since the next guy will pick 18 and you can't pick 19 because 19 is a sure loss. So all odd numbers are a guaranteed loss. So all perfect players will only ever bet even numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if the starting bid would be 4 or 6 (or any even number) the first one to bid would win. But so nobody will bid 5\$ since it is a guaranteed loss.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Case 2: One bidder can bid multiple times in a row and has to pay his last 2 bids if he wins&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;So if I bid 5\$ and then immediately overbid myself with 6\$ and no one else bids, I would get the 20 but would have to pay 5 and 6 (highest and second highest bid) so my net win would be 9\$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case someone can always try to jump an odd number, by bidding twice in a row and get a winning even number. So the first bid would be 5+6 (since paying both bids is only 11 and you get 20 net win 9\$) next save bid is 7+8 and then 9+10. Nobody will bet 11, since 11+12 (or any other two bets in a row over 10) will result in a net loss. So the first one to bet 10 wins. So the question is how the bidding system actually works. But everyone will start bidding up to 10, or nobody will bid at all, depending on the number of your chances to get the two bets in a row. So with two people it is still attractive to try and get the bets 9+10 (if we assume chances 50/50 to get the 9\$ and 50/50 to get the 10\$, expected net results are 0,-9,1,10 so an expected winning of 0.5\$)&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Case 3: One bidder can bid multiple times in a row, but whoever else is highest below him has to pay the extra&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;So if Geoffrey bids 5\$ and I bid 6,7 and 8 afterwards and no one else bids, I would get the 20\$ for 8 and Geoffrey would loose his 5\$ as second highest bidder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will also result in a race to the top, cut off depending on the chances of getting you bet in (so the number of people)&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Case X: Humans involved with unpredictable bids&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Don't bet at all, or try psychological tricks to intimidate them like at the poker-table. But this is too unpredictable, best to not bid at all... Or probably bid 5\$ and try to intimidate everyone away from bidding and if someone bids higher, let them destroy each other.&lt;/p&gt;&#xA;" OwnerUserId="1998" LastActivityDate="2015-05-18T13:54:31.913" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15151" PostTypeId="2" ParentId="15075" CreationDate="2015-05-18T13:58:05.330" Score="0" Body="&lt;p&gt;You afix the knife on the roof ledge, off to the side (away from the main rope line). You tie yourself into one end of the 600ft rope, and you tie the other end to something at the top of the roof (not the hook). You belay yourself, carefully, straight down to the 400ft ledge. There will be approx 150-175ft of slack at this point. Hang onto the ledge as you pull all slack out. Cut the rope, but don't use all the slack to do so. Use maybe 50-100ft. Afix it to the hook at the 400ft point. Now, hold onto the top rope securely (maybe tie in or something) and swing such that you are able to either have the rope hit the knife and cut it, or merely wear it away from going over the ledge at the top. Eventually, the rope is cut and you fall a bit. Holding onto the 400ft rope segment that you now have, climb up your short rope back to the 400ft ledge. Afix the 400ft rope there, tie yourself in, untie yourself from the short rope, belay yourself down to the ground.&lt;/p&gt;&#xA;" OwnerUserId="12549" LastActivityDate="2015-05-18T13:58:05.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15152" PostTypeId="2" ParentId="15126" CreationDate="2015-05-18T14:12:59.490" Score="13" Body="&lt;p&gt;I thought @LogicianWithAHat has answered the question quite neatly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think Q2 could be solved like this (it is not to scale, but you get the idea):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/GBi9H.gif&quot; alt=&quot;Question 2&quot;&gt;&#xA; &lt;P&gt;Put the triangle on top of the triangle with hole and the trapezium to the left.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12332" LastActivityDate="2015-05-18T14:12:59.490" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15153" PostTypeId="2" ParentId="15126" CreationDate="2015-05-18T14:45:19.273" Score="22" Body="&lt;p&gt;One cut solution for #3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Cut and turn the green part for 180deg. (Orange - original hole position)&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Qr6JG.png&quot; alt=&quot;One cut solution for #3&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12551" LastActivityDate="2015-05-18T14:45:19.273" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15155" PostTypeId="2" ParentId="15059" CreationDate="2015-05-18T16:15:48.967" Score="2" Body="&lt;p&gt;So the aliens are counting in binary. Here's the signal with A replaced with 1, B replaced with 0, and C replaced with ':'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 10 11:100 101 110:1000 111:1001 1010 1100:10000&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They're telling us they have three fingers! Here's a commentary:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;One, two, &lt;s&gt;three&lt;/s&gt;:ten, eleven, &lt;s&gt;twelve&lt;/s&gt;:one hundred, &lt;s&gt;thirteen&lt;/s&gt;:one hundred and one, one hundred and ten, &lt;s&gt;one hundred and a hundred&lt;/s&gt;:thousand&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everytime they run out of fingers, they add C[base increment] to the number. This should be our response:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1 10 11 101 111 1000 1001 1010 1011 1100C10000&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hopefully that will be enough to let them know we understand their message, and we have ten fingers!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; If they &quot;count one, two, ten&quot; then they have three fingers, not two.&lt;/p&gt;&#xA;" OwnerUserId="10246" LastEditorUserId="10246" LastEditDate="2015-05-18T16:30:31.870" LastActivityDate="2015-05-18T16:30:31.870" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15156" PostTypeId="2" ParentId="1776" CreationDate="2015-05-18T16:30:11.490" Score="1" Body="&lt;p&gt;How about something a little simpler. Using the variable names for the numbers as above:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{ccc}&#xA;&amp;amp;E\\&#xA;a&amp;amp;&amp;amp;b\\&#xA;c&amp;amp;&amp;amp;d&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steps:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Sum the numbers ($a$ through $d$).&lt;/li&gt;&#xA;&lt;li&gt;If 2nd digit of sum is greater than 5 subtract 1st from 2nd, otherwise add.&lt;/li&gt;&#xA;&lt;li&gt;Append a 1.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Solutions for $E$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;5+6+7+8 &amp;amp;= 26 &amp;amp;&amp;amp;\rightarrow \text{Subtract:}\ 6-2 = 4 &amp;amp;\rightarrow \text{Append:}\ 41.\\&#xA;9+4+2+6 &amp;amp;= 21 &amp;amp;&amp;amp;\rightarrow \text{Add:}\ 2+1 = 3 &amp;amp;\rightarrow \text{Append:}\ 31.\\&#xA;8+7+2+6 &amp;amp;= 23 &amp;amp;&amp;amp;\rightarrow \text{Add:}\ 2+3 = 5 &amp;amp;\rightarrow \text{Append:}\ 51.&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="12557" LastEditorUserId="13494" LastEditDate="2017-09-03T14:09:32.120" LastActivityDate="2017-09-03T14:09:32.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15157" PostTypeId="1" AcceptedAnswerId="15161" CreationDate="2015-05-18T16:58:40.677" Score="14" ViewCount="2254" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;My dear brother,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;It has been almost nine years since we lost you. We’re a big family so&#xA;  losing just one may not seem like much of a loss but it has left a&#xA;  hole nonetheless. You’re still here in spirit – in fact, people keep&#xA;  forgetting you’re no longer part of our family – but it’s not the&#xA;  same. You were always kind of out there but we all cared about you. I&#xA;  don’t think anybody else has written yet and I’m sorry for that. I&#xA;  can’t make up for their lack of action but I feel like, if anyone is&#xA;  going to keep this family together, it has to be me. Even though I was&#xA;  adopted into this family and never really matched the rest of you, I&#xA;  feel like I must be the leader. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If my nephew would ever slow down to think about it, I’m sure he’d&#xA;  write, too. He’s always been our main line of communication. You know&#xA;  how he is, though, always in a hurry. His dad – our big brother – is&#xA;  still trying to apologize for him but, at this point, it doesn’t&#xA;  really feel sincere. He’s cold-natured on the outside but I believe&#xA;  there’s still some warmth in there, deep down. Those two have been&#xA;  distant for so long that I don’t know if they’ll ever really get&#xA;  along. I still find myself having to stay between them. Even his&#xA;  “daughter” has to stand between them. (You remember why I say&#xA;  “daughter” in quotes, right? We’re still not sure if she’s his or&#xA;  dad’s. Gods, our family is so messed up…) Of course, it only gets&#xA;  worse when his other son decides to get involved. He’s always stirring&#xA;  up trouble.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Granddad is still kickin’ around out there somewhere. He’s a pretty&#xA;  icy guy but he means well. I think dad and even some of our brothers&#xA;  try to be like him but he’s still the undisputed master of the cold&#xA;  shoulder. He and I barely talk, really. I feel kind of guilty about it&#xA;  – I feel like I should get to know my whole adopted family better –&#xA;  but the trip to go see him is just so costly that it’s hard to justify&#xA;  it most of the time.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Listen, I know you’re pretty far out there most of the time but I&#xA;  don’t want you forgetting where you belong. I don’t care what others&#xA;  say, you still belong in this family. I know at least one of our&#xA;  brothers is out near you once in a while. He’s still a little wet&#xA;  behind the ears despite his age but he’ll try to swing by. Try to say&#xA;  hi sometime, OK? Maybe send a letter?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;All the love in the world,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Sis&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Primary Question:&lt;/strong&gt; Who is writing the letter and to whom are they writing it? (Get the tick)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus Question:&lt;/strong&gt; Can you name all the family members mentioned? (Get +1)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Letter writer&lt;/li&gt;&#xA;&lt;li&gt;The brothers&lt;/li&gt;&#xA;&lt;li&gt;The nephew&lt;/li&gt;&#xA;&lt;li&gt;The daughter&lt;/li&gt;&#xA;&lt;li&gt;Dad and Granddad&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2016-08-31T15:30:22.237" LastActivityDate="2016-08-31T15:30:22.237" Title="The Messed Up Family and the Lost Brother" Tags="|riddle|story|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="15158" PostTypeId="1" AcceptedAnswerId="15162" CreationDate="2015-05-18T17:49:10.607" Score="9" ViewCount="805" Body="&lt;p&gt;Within each individual &quot;view-finder&quot; frame below, all of the pixels are the same shade of gray. Frames A, C, and E appear to be a darker gray while frames B and D appear to be lighter. However, based on the actual pixel values only one frame contains a different shade of gray.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which single frame contains the different pixel value? And what is causing the difference between our perception and the measured pixel values?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/D6wmV.png&quot; alt=&quot;Shades&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-05-20T01:52:11.543" LastActivityDate="2015-05-20T01:52:11.543" Title="Shades of Gray - Perception vs Reality" Tags="|visual|computer-puzzle|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15159" PostTypeId="2" ParentId="15126" CreationDate="2015-05-18T17:56:58.830" Score="5" Body="&lt;p&gt;I think people covered this, but I'll give it a shot for part one and two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Start a cut from the top of the table, to the left of center by the distance from the current hole to the right edge of the board, and cut until you're at the center of the board, plus the distance from the current hole top edge to the top edge of the board. Continue cutting all the way to the right. Take out the piece you cut, flip over the x and y-axes, and reinsert. Summary: Make one L-shaped cut, rotate the cut out rectangle by 180 degrees, and re-insert.&lt;img src=&quot;https://i.stack.imgur.com/C0Xyv.png&quot; alt=&quot;Image 1&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Part 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Make one circular cut starting at the top of the board above the current hole (so that the rectangular hole is encompassed within a circle, and that circle's radius extends to the edges of the current board. Next, make essentially the same cut as in Part 1 so that you can rotate a large section of the board, and re-insert the circular cut out in the center, rotating it by 45 degrees. Summary: Make a circular cut around the hole, and an L-shaped cut across the majority of the board. Rotate L-shaped cut by 180 degrees, circular cut by 45 degrees, and re-insert. &lt;img src=&quot;https://i.stack.imgur.com/vd6QS.png&quot; alt=&quot;Part 2&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sorry for the crappy quality, I can't photoshop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no idea about Part 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: As a comment to the question (since I don't have the privilege yet)--I was interpreting the third part to mean that you are given an ovular table with an ovular hole and still have to accomplish the same objective of turning it into a rectangular table with a rectangular hole; yet the #1 upvoted answer right now (for the part 3 solution) seems to interpret as- just get the ovular hole into the center of the ovular table (which is much easier IMO). Can anyone clarify what the objective for part 3 is?&lt;/p&gt;&#xA;" OwnerUserId="10609" LastEditorUserId="10609" LastEditDate="2015-05-18T21:18:12.073" LastActivityDate="2015-05-18T21:18:12.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15160" PostTypeId="2" ParentId="15158" CreationDate="2015-05-18T18:00:50.730" Score="6" Body="&lt;p&gt;&lt;strong&gt;The question has totally changed now, but the answer below was for the original question.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reality:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are all the same.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Perception:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The gradients between each of the 5 cause your brain to perceive those colors on the darker sides of the gradients to be darker. Thus, A, C, and E look darker.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More cool gradient illusions &lt;a href=&quot;http://nerdist.com/5-optical-illusions-that-show-you-why-your-brain-messes-with-the-dress/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I saw this type of thing on the Netflix show &quot;Brain Games.&quot;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-05-19T19:32:57.297" LastActivityDate="2015-05-19T19:32:57.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15161" PostTypeId="2" ParentId="15157" CreationDate="2015-05-18T18:29:12.777" Score="13" Body="&lt;p&gt;The family is composed of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The planets and the mostly Greco-Roman gods for whom they were named&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The brothers are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jupiter(Zeus), Pluto(Hades), and Neptune(Poseidon) ( the god of the ocean - &quot;wet behind the ears despite his age&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The lost brother is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pluto, whose celestial body was deemed not a planet in 2005, almost 9 years ago; among the most distant planets from the sun, it was always kind of 'out there'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The letter writer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Earth, named not for a Roman deity but a Germanic one, hence &quot;adopted&quot; and not matching the rest, but as the only populated planet the only one able to 'keep the family together' - also identified in the Greek pantheon with Gaia, whose union with Uranus gave birth to the titans - and she sends 'all the love in the world' -- literally!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Dad and Granddad are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Saturn(Cronus), the father of the Olympians, and his father Uranus, the cold(warm?) gas giants who are 'distant'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The daughter is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Venus, who is variously identified as the daughter of Neptune(Poseidon) and of Jupiter(Zeus)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The nephew who is &quot;our main line of communication&quot; is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mercury(Hermes), the messenger of the gods.  In astrology, the planet Mercury governs communication; it is the planet that orbits and rotates fastest&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The son that's always stirring up trouble is likely&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mars, the god of war&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4385" LastEditorUserId="4385" LastEditDate="2015-10-15T10:43:52.523" LastActivityDate="2015-10-15T10:43:52.523" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="15162" PostTypeId="2" ParentId="15158" CreationDate="2015-05-18T18:55:43.320" Score="9" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EECh9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look what happens if you hide the frames between two consecutive areas! As you can see, B is the same colour of A and C!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where is the trick? Of course in the frames! The dark grey rectangles have a slightly darker margin, while the light rectangles have a lighter shade of grey as margin. This generates the illusion that the whole content of the rectangles is different, when it's not!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ah, and if you want a proof that the margins have different shades of grey, here is it: &#xA;&lt;img src=&quot;https://i.stack.imgur.com/levdi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Small areas near the borders have been translated to show that difference.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As discussed in the comments, the reality is that A, B, D, E are the same shade of gray.  The pixel values of C are slightly darker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to BmyGuest, this horizontal &quot;line profile&quot; shows the pixel value gradients at the borders.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dHohr.png&quot; alt=&quot;Profile&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="5188" LastEditDate="2015-05-20T01:51:00.890" LastActivityDate="2015-05-20T01:51:00.890" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15163" PostTypeId="2" ParentId="15130" CreationDate="2015-05-18T20:52:54.847" Score="-1" Body="&lt;p&gt;The first point can fall anywhere. Divide the area into quarters arbitrarily and the second point can fall either in the same half (1/2 chance) in the same quarter (1/4 chance) or in the opposite quarter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the second point falls in the opposite quarter (1/4 chance) there is a 1/2 chance that a different division of the board would result in the points sharing a half (resulting 1/8 chance for each outcome.)*&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This results in four possible outcomes for the second point:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1/4 - same quarter&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5/8 - same half 1/2 + 1/8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1/8 - opposite quarters&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplified 3 point problem yields this probability tree:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A      B&#xA;1/4 -&amp;gt; 1/4 = 1/16&#xA;5/8 -&amp;gt; 1/2 = 5/16&#xA;1/8 -&amp;gt; 1   = 2/16&#xA;          += 7/16 probability of failure, of success 9/16&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;A : Chance of second point falling in 1. Same quarter 2. Adjacent quarter 3. Opposite quarter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B : Chance of failure, third point falls in the other half to existing points. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extending to the main 4 point problem:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A      B**      C&#xA;1/4 -&amp;gt; 1/4 -&amp;gt; 1/4 = 1/64&#xA;       5/8 -&amp;gt; 1/2 = 5/64&#xA;       1/8 -&amp;gt; 1   = 2/64&#xA;&#xA;5/8 -&amp;gt; 1/2 -&amp;gt; 1/2 = 10/64&#xA;    -&amp;gt; 1/2 -&amp;gt; 1   = 20/64&#xA;&#xA;1/4 -&amp;gt; 1   -&amp;gt; 1   = 16/64&#xA;&#xA;                 += 54/64 probability of failure, of success  10/64&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;**In the scenario with 4 points stage B may be a repeat of stage A if the first two points fell in the same quarter or may be a 1/2 chance of failure&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C: Chance of failure, fourth point in the other half to existing points. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Yet to be proven.&lt;/p&gt;&#xA;" OwnerUserId="11297" LastActivityDate="2015-05-18T20:52:54.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15164" PostTypeId="1" AcceptedAnswerId="15237" CreationDate="2015-05-18T21:12:50.340" Score="3" ViewCount="469" Body="&lt;p&gt;Hair the color of rose's imperfection&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eyes like the one who is part of me,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though we are two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the other of us is the color&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of his fallen relatives in the&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Safest place on the earth(besides _________)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My family and I are celebrimbor,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we are just my two brothers and me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For now.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mainly only one book, with two other references.  Who are the two, the brothers and what does the ____ part mean?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spoiler&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dragons!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More hints...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am not that which you call by my name, for that person belongs to a more modernly created world than I live in.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11790" LastEditorUserId="11790" LastEditDate="2015-05-20T10:40:38.490" LastActivityDate="2015-05-20T19:39:55.483" Title="Our life and death are one, we two" Tags="|riddle|knowledge|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15165" PostTypeId="1" AcceptedAnswerId="15436" CreationDate="2015-05-18T21:12:57.917" Score="8" ViewCount="849" Body="&lt;p&gt;Here's another riddle on the theme of literature.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;I am the oldest, and take care of the monsters.&lt;/p&gt;&#xA;&lt;p&gt;My first is a loner, but has a few children.&lt;/p&gt;&#xA;&lt;p&gt;My second is a lion, lover of wealth.&lt;/p&gt;&#xA;&lt;p&gt;My third was the black sheep of the family.&lt;/p&gt;&#xA;&lt;p&gt;My fourth, fifth, and sixth might be Taurus, Nagini, and Dracula.&lt;/p&gt;&#xA;&lt;p&gt;My seventh is a bear, found in the north.&lt;/p&gt;&#xA;&lt;p&gt;Then there's the eighth, the newest and one day the greatest. Coming out of nowhere, as it seemed, to supplant the black sheep and perhaps all the rest. Innocence, wisdom, and strength all run errands for this one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Who am I (the speaker)?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;HINT:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/q/15420/5373&quot;&gt;A recipe for lemon tea stew&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-10-02T17:06:27.830" Title="Riddle with seven - or is it eight?" Tags="|riddle|knowledge|literature|" AnswerCount="2" CommentCount="19" ContentLicense="CC BY-SA 3.0" />
  <row Id="15166" PostTypeId="1" CreationDate="2015-05-18T21:33:53.827" Score="0" ViewCount="670" Body="&lt;p&gt;From the 1970s I have a Hexomino called&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Computer Puzzle in Computer Age&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;promising more than 1.000.000 solutions. A later version with a different title (I could not remember) says more than 1.000.000.000 solutions are possible. Probably the number is closer to 1.000.000.000.000 - but of course a single solution is hard to find by hand. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this Hexomino the 35 pieces are arranged in a rectangle 19 x 11 plus a nose in the middle of one of the longer sides. (This has a specific reason, because it is not possible to arrange the 35 pieces in a square.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Especially for this arrangement I'd like to find &lt;em&gt;all&lt;/em&gt; solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The puzzle looks like this (could not put into comment):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;&#xA;┌─┬───┬─────┬─────────┬───┬─┬─────┬───┐&#xA;│ ├─┐ └───┐ └───┐ ┌─┬─┘   │ │   ┌─┤   │&#xA;│ │ └─┐ ┌─┴───┬─┴─┘ ├─┐ ┌─┤ ├─┐ │ │   │&#xA;│ └─┐ ├─┤ ┌─┐ │ ┌─┐ │ └─┘ │ │ └─┤ └───┤&#xA;│ ┌─┘ │ │ │ └─┼─┘ └─┴─┐ ┌─┤ ├─┐ └───┐ │&#xA;├─┼───┘ └─┤   │ ┌─┬───┴─┘ │ │ └─┬───┴─┤&#xA;│ └─┬───┬─┴─┐ ├─┘ └───┬─┐ ├─┴─┐ └─┐   │&#xA;│   │ ┌─┴─┐ ├─┴─┐ ┌─┬─┘ └─┤   ├─┐ ├─┐ │&#xA;│ ┌─┘ │   │ │ ┌─┴─┘ └─┬─┐ │ ┌─┘ └─┘ ├─┤&#xA;├─┤ ┌─┘ ┌─┘ │ ├─┐ ┌───┘ │ │ ├───────┤ │&#xA;│ └─┴───┴─┬─┘ │ └─┴───┐ └─┼─┘ ┌─────┘ │&#xA;└─────────┴───┴───┐ ┌─┴───┴───┴───────┘&#xA;                  └─┘&#xA;&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="12564" LastEditorUserId="12564" LastEditDate="2015-05-19T22:09:15.257" LastActivityDate="2015-05-19T22:44:05.637" Title="What would be the most efficient algorithm to solve a polynomino?" Tags="|polyomino|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15167" PostTypeId="2" ParentId="15127" CreationDate="2015-05-18T21:48:23.443" Score="20" Body="&lt;p&gt;You should make the torch last the minimum amount of time to allow nine of the Kevins to escape. This will ensure that all ten cannot escape. We will compute this time in terms of $t_1,\ldots,t_9$. Actually, we will describe a computation that is valid for any finite number of Kevins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose there are $n$ Kevins, call them $K_1$, $\ldots$, $K_n$ from fastest to slowest. Let $K_i$ and $K_j$ be a pair of Kevins, with $3\leq i&amp;lt;j\leq n$. There are two reasonable-sounding strategies for getting $K_i$ and $K_j$ across the bridge. The first strategy, which I will call Strategy A: Kevins $K_1$ and $K_2$ cross, and $K_1$ returns with the flashlight. Then Kevins $K_i$ and $K_j$ cross, and $K_2$ returns with the light. This takes total time&#xA;$$&#xA;t_1+2t_2+t_j.&#xA;$$&#xA;The second strategy, which I will call Strategy B: $K_1$ keeps the flashlight, and escorts $K_i$ and $K_j$ across one at a time. This takes total time&#xA;$$&#xA;2t_1+t_i+t_j.&#xA;$$&#xA;Strategy A is faster than Strategy B if and only if $$&#xA;t_i&amp;gt;2t_2-t_1.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set Kevins $K_1$ and $K_2$ aside, and group the remaining Kevins into pairs: $\{K_n,K_{n-1}\}$, $\{K_{n-2},K_{n-3}\}$, and so on (there may be an odd Kevin out who is unpaired). Call a pair of Kevins &lt;em&gt;slow&lt;/em&gt; if the faster member of the pair takes longer than $2t_2-t_1$ to cross the bridge, otherwise call the pair &lt;em&gt;fast&lt;/em&gt; (the odd Kevin out, if there is one, will be considered fast).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Each slow pair of Kevins will cross the bridge using Strategy A. After all of the slow pairs have crossed, $K_1$ will escort every remaining Kevin across the bridge one by one. The total time for this is&#xA;  $$&#xA;(n-2-k)t_1 + (2k+1)t_2 + \sum_{i=1}^k t_{n+1-2i} + \sum_{i=3}^{n-2k}t_i,&#xA;$$&#xA;  where $k$ is the number of slow pairs. The paper [1] gives a rigorous proof that this is optimal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;[1] Rote, Günter. Crossing the bridge at night. &lt;em&gt;Bulletin of the European Association for Theoretical Computer Science&lt;/em&gt;, no. 78, 2002, pp. 241--246.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-05-19T00:16:46.230" LastActivityDate="2015-05-19T00:16:46.230" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15168" PostTypeId="2" ParentId="13303" CreationDate="2015-05-18T22:07:37.100" Score="6" Body="&lt;p&gt;I believe I've solved the eighth stanza. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If you reflect all the stones on the grass as if the river were a mirror, placing additional stones at HB, EO, MS, and DB, then each stone points to another stone. Draw lines connecting them, turn the map so that north (as per rand al'thor's answer) is up, they spell out 3E.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I've got the seventh down, too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The clue hints at time being important, so we'll interpret 1515 as 15 hours and 15 minutes. 15 hours is 3 PM, and 15 minutes also points at the 3 on a clock face - they're lined up. Put the clock and the map right side up relative to each other, and the hands point east, at the 3. Go east three squares.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12566" LastEditorUserId="12566" LastEditDate="2015-05-19T21:52:31.247" LastActivityDate="2015-05-19T21:52:31.247" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15169" PostTypeId="1" AcceptedAnswerId="15171" CreationDate="2015-05-18T22:37:41.260" Score="12" ViewCount="1916" Body="&lt;p&gt;While walking around aimlessly and bored, you find a piece of paper on the sidewalk containing a heck of a lot of numbers. Normally someone seeing that would be like &quot;ain't nobody got time for that&quot; but you are just barely curious enough to wonder if there could be a secret meaning in there somewhere. Your instinct says that it actually won't take very long at all, and that when you figure it out, they will give you a secret word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the secret word the numbers refer to?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The numbers have been transcribed &lt;a href=&quot;http://pastebin.com/cF9cDv1v&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; for your convenience)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jzfgw.png&quot; alt=&quot;Solve me if you dare.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Full resolution image (not necessary don't waste time on it): &lt;a href=&quot;https://i.stack.imgur.com/1rXdk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/1rXdk.png&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="69582" LastEditDate="2023-12-04T17:12:27.577" LastActivityDate="2023-12-04T17:12:27.577" Title="Find the secret word in this encrypted image (part 2)" Tags="|cipher|lateral-thinking|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15170" PostTypeId="2" ParentId="15166" CreationDate="2015-05-18T23:25:25.923" Score="1" Body="&lt;p&gt;I think the best algorithm would be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Just a depth-first backtracking search. Assuming the long side without the &quot;nose&quot; is on the bottom, start in the bottom left. Try pieces until one fits without any 1x1 spaces or spaces that another piece could not occupy. Once one is in place, try of the remaining pieces until one fits next to it without spaces. Keep doing this until the puzzle is either complete, or there are no pieces that fit properly, at which point you backtrack to the most recent valid configuration (the version of the puzzle without the last piece in) and pick a different piece. Again you continue the algorithm and if you can't find any good configurations with this piece either, you take it out and try another piece in it's place.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11727" LastEditorUserId="11727" LastEditDate="2015-05-19T22:44:05.637" LastActivityDate="2015-05-19T22:44:05.637" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15171" PostTypeId="2" ParentId="15169" CreationDate="2015-05-18T23:29:21.463" Score="20" Body="&lt;p&gt;First, I copy-pasted the numbers into &lt;em&gt;Mathematica&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;str = &quot;181 ... 29&quot;;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Next I convert the string to a list of numbers:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;numbers = FromDigits /@ StringSplit[str];&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now, is this a decimal ASCII string?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;FromCharacterCode[Take[numbers, 10]]&#xA;(* &quot;µæ\.1dµæ\.1dµæ\.1dµ&quot; *)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Evidently not.  Next I look at the letter frequencies:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Histogram[numbers, {1}]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qJn1N.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like there are just a few peaks.  Are all three main peaks the same height?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Commonest[numbers]&#xA;(* {181, 230, 29} *)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Yes, they are.  In fact, looking at the numbers, it starts off with this triple repeated several times (why did I not notice that faster?).  Breaking into threes, we get a list 1600 elements long:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Length[numbers] / 3&#xA;(* 1600 *)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;A large square of triples?  This smells like an image...&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Image[Partition[Partition[numbers, 3], Sqrt[1600]], &quot;Byte&quot;, Magnification -&amp;gt; 8]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/n5zw2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the word is, &quot;stars.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Altogether this took around a minute thirty, a testament to the power of &lt;em&gt;Mathematica&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="9718" LastEditDate="2015-05-19T00:26:09.293" LastActivityDate="2015-05-19T00:26:09.293" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15173" PostTypeId="2" ParentId="15130" CreationDate="2015-05-18T23:45:30.640" Score="-3" Body="&lt;p&gt;My answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0.125&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's why:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The probability of the 1st point be picked in one of the 2 halves is,&#xA; obviously, 1 =&gt; partial probability = 1&lt;br&gt;&#xA; &#xA; The probability of the 2nd point be picked in the same half is 0.5 =&gt;&#xA; partial probability = 1 × 0.5&lt;br&gt;&#xA; &#xA; The probability of the 3rd point be picked in the same half is 0.5 =&gt;&#xA; partial probability = 1 × 0.5 × 0.5&lt;br&gt;&#xA; &#xA; The probability of the 4th point be picked in the same half is 0.5 =&gt;&#xA; total probability = 1 × 0.5 × 0.5 × 0.5 = 0.125  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12205" LastActivityDate="2015-05-18T23:45:30.640" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="15174" PostTypeId="2" ParentId="15169" CreationDate="2015-05-18T23:49:25.363" Score="6" Body="&lt;p&gt;I am not fast enough with programming/computer stuff to beat 2012rcampion, but here's my image screenshotted from Excel, and an explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rT7Vr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation: The numbers come in groups of three, and there are 1600 groups of three numbers. Throw them into a 40x40 grid, and translate each group of three into the RGB color scheme. That gives you a specific color for each &quot;square&quot; of the matrix, and gives you the message shown above: The word is stars!&lt;/p&gt;&#xA;" OwnerUserId="10861" LastActivityDate="2015-05-18T23:49:25.363" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15175" PostTypeId="2" ParentId="15169" CreationDate="2015-05-19T01:57:33.173" Score="4" Body="&lt;p&gt;Read about &lt;a href=&quot;https://webmasters.stackexchange.com/questions/8525/how-to-open-the-javascript-console-in-different-browsers&quot;&gt;How to open the JavaScript console in different browsers&lt;/a&gt; then copy and paste the code below into the console, press enter, then scroll to the bottom of the page to see the resulting image.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;var data=[].filter.call(document.getElementsByTagName('pre'),&#xA;            function(e){ return e.textContent &amp;amp;&amp;amp; e.textContent.indexOf('181 230')==0 })&#xA;        [0].textContent.split(' ').map(&#xA;            function(e){ return parseInt(e,10)} );&#xA;&#xA;var c=document.createElement('canvas');&#xA;document.body.appendChild(c);&#xA;c.width=40;&#xA;c.height=40;&#xA;var p=c.getContext('2d').getImageData(0,0,40,40);&#xA;for(var i=0; i&amp;lt;1600; i++){&#xA;        p.data[4*i]=data[3*i];&#xA;        p.data[4*i+1]=data[3*i+1];&#xA;        p.data[4*i+2]=data[3*i+2];&#xA;        p.data[4*i+3]=255;&#xA;};&#xA;c.getContext('2d').putImageData(p,0,0);&#xA;c.style.width=&quot;80%&quot;;&#xA;c.style.margin=&quot;10%&quot;&#xA;c.style.imageRendering='pixelated';&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;pre&gt;181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 255 201 14 181 230 29 181 230 29 181 230 29 181 230 29 255 201 14 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 255 201 14 181 230 29 181 230 29 181 230 29 255 201 14 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 255 201 14 181 230 29 181 230 29 255 201 14 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 255 201 14 181 230 29 255 201 14 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 255 201 14 255 201 14 181 230 29 181 230 29 181 230 29 181 230 29 34 177 76 34 177 76 34 177 76 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 255 201 14 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 255 201 14 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 255 201 14 255 201 14 181 230 29 181 230 29 181 230 29 34 177 76 34 177 76 34 177 76 34 177 76 34 177 76 181 230 29 181 230 29 181 230 29 181 230 29 255 201 14 181 230 29 255 201 14 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 255 201 14 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 255 201 14 181 230 29 181 230 29 255 201 14 181 230 29 181 230 29 34 177 76 34 177 76 34 177 76 34 177 76 34 177 76 181 230 29 181 230 29 181 230 29 255 201 14 181 230 29 181 230 29 181 230 29 255 201 14 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 255 201 14 181 230 29 181 230 29 255 201 14 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 255 201 14 181 230 29 181 230 29 181 230 29 181 230 29 255 201 14 181 230 29 34 177 76 34 177 76 34 177 76 34 177 76 34 177 76 34 177 76 181 230 29 255 201 14 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 255 201 14 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 255 201 14 181 230 29 181 230 29 181 230 29 181 230 29 255 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177 76 255 201 14 34 177 76 34 177 76 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 255 201 14 181 230 29 181 230 29 181 230 29 181 230 29 34 177 76 34 177 76 255 201 14 34 177 76 34 177 76 34 177 76 34 177 76 181 230 29 181 230 29 255 201 14 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 255 201 14 181 230 29 34 177 76 34 177 76 34 177 76 34 177 76 34 177 76 255 201 14 34 177 76 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 255 201 14 181 230 29 181 230 29 181 230 29 34 177 76 255 201 14 34 177 76 34 177 76 34 177 76 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 255 201 14 181 230 29 181 230 29 181 230 29 181 230 29 255 201 14 181 230 29 181 230 29 181 230 29 181 230 29 34 177 76 34 177 76 34 177 76 34 177 76 255 201 14 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 255 201 14 181 230 29 181 230 29 255 201 14 34 177 76 34 177 76 34 177 76 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 255 201 14 181 230 29 181 230 29 255 201 14 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 34 177 76 34 177 76 34 177 76 255 201 14 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 255 201 14 255 201 14 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 255 201 14 255 201 14 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 255 201 14 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 255 201 14 255 201 14 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 255 201 14 255 201 14 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 255 201 14 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 255 201 14 181 230 29 181 230 29 255 201 14 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 255 201 14 181 230 29 181 230 29 255 201 14 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 181 230 29 255 201 14 181 230 29 181 230 29 181 230 29&lt;/pre&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="-1" LastEditDate="2017-04-13T12:33:26.200" LastActivityDate="2015-05-19T16:19:41.017" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15176" PostTypeId="1" AcceptedAnswerId="15179" CreationDate="2015-05-19T04:32:33.007" Score="21" ViewCount="9072" Body="&lt;p&gt;Alice and Bob play a game with a $5\times 5$ chessboard, and a chess knight. Alice begins by placing the knight somewhere on the board. Then, starting with Bob, the players alternate moving the knight (the way it moves in chess) to a square it hasn't occupied before. If a player has no legal moves, he/she loses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which player wins under optimal play, and how?&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="12408" LastEditDate="2016-07-25T02:39:27.877" LastActivityDate="2019-07-06T02:47:21.997" Title="Knight on a 5 by 5 board" Tags="|strategy|game|knight-moves|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15177" PostTypeId="1" AcceptedAnswerId="15182" CreationDate="2015-05-19T05:07:21.003" Score="-2" ViewCount="306" Body="&lt;p&gt;You are an Archer and participating in a tournament. You need to hit the Bull's Eye to win the tournament and the Prize (a big fat pig called &quot;Princess&quot;. Wodehouse Anyone?). You love big fat pigs and have to win the Tournament or else loose your sweet heart to one of the other big fat pig lovers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyhow, the task is not easy as the Target is mobile. The target follows a horizontal path of an isosceles triangle with the unequal side being furthest side from you. The trajectory of the path is parallel to the ground. You have taken the measurements and made the calculations beforehand. They are as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VNcDa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The two equal sides of the trajectory are 5 meters each and the third side being 6 meters.&lt;/li&gt;&#xA;&lt;li&gt;You are 80 meters away from the closest vertex of the triangle and are directly in front of it and in a line perpendicular to the furthest unequal side.&lt;/li&gt;&#xA;&lt;li&gt;The target is moving at a speed of 28 miles an hour.&lt;/li&gt;&#xA;&lt;li&gt;You know that when you release the arrow from a confident and relaxed posture, it always goes at a speed of 98 miles an hour (Yeah always...and you know that).&lt;/li&gt;&#xA;&lt;li&gt;Looking from above the target trajectory the target is moving in the counter-clockwise direction.&lt;/li&gt;&#xA;&lt;li&gt;You are given special arrows that do not follow the rules of gravity, and will fly parallel to the ground as long as they are within the arena.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;After the calculations are you confident that you can get your beloved &quot;Princess&quot;? Why? When and at what angle are to going to shoot your arrow?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, are you just going to let another big fat pig lover (!!!RESPECT!!!) going to take your &quot;Princess&quot; away?&lt;/p&gt;&#xA;" OwnerUserId="12537" LastActivityDate="2015-05-19T07:00:50.683" Title="You and your &quot;Princess&quot;" Tags="|logical-deduction|calculation-puzzle|" AnswerCount="2" CommentCount="5" ClosedDate="2015-05-19T13:41:18.043" ContentLicense="CC BY-SA 3.0" />
  <row Id="15178" PostTypeId="2" ParentId="15176" CreationDate="2015-05-19T05:17:10.920" Score="22" Body="&lt;p&gt;The winner is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alice&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The winning strategy is to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alice places the knight in the corner, and forces the following sequence of moves, with Alice making the odd-numbered moves and Bob making the even-numbered ones:&#xA;&lt;br&gt;&#xA; &lt;code&gt;01 10 21 16 07&#xA; 20 15 08 11 22&#xA; 09 02 25 06 17&#xA; 14 19 04 23 12&#xA; 03 24 13 18 05&lt;/code&gt;&#xA;&lt;br&gt;&#xA; Note that each of Bob's moves (even numbers) is forced because there's only one available unvisited square, except for 02 and 10, in which the two options are symmetrical and equivalent. Alice makes Bob trace out a path that visits the four corners, then makes laps clockwise, and finally she takes the last remaining cell of the center.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-05-19T05:17:10.920" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15179" PostTypeId="2" ParentId="15176" CreationDate="2015-05-19T05:19:06.373" Score="25" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Alice starts and wins &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As shown below Alice gives the red moves and Bob gives blue moves... Moves 2 and 10 have symmetrical options which end up with same result. All other moves are forced.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/CKVW9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7723" LastEditorUserId="7723" LastEditDate="2015-05-19T05:24:20.757" LastActivityDate="2015-05-19T05:24:20.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15180" PostTypeId="1" CreationDate="2015-05-19T06:08:06.173" Score="26" ViewCount="2301" Body="&lt;p&gt;The black knight is our lion chasing 8 white bishop zebras which can't capture. Can the zebras evade the lion forever, if team zebra positions all the pieces and has the first turn?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The game is played on a standard 8x8 chess board&lt;/li&gt;&#xA;&lt;li&gt;Zebras move like bishops but can't capture&lt;/li&gt;&#xA;&lt;li&gt;Lions move like knights and only need to capture one zebra to win&lt;/li&gt;&#xA;&lt;li&gt;The Zebra team will always play their best game to avoid capture&lt;/li&gt;&#xA;&lt;li&gt;Zebras get to position all pieces to their advantage&lt;/li&gt;&#xA;&lt;li&gt;Zebras can choose to first move if they desire&lt;/li&gt;&#xA;&lt;li&gt;Each side moves a single piece per turn&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Question:&#xA;If there are 8 zebras and only 1 lion, can the lion win?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you lion can win how does it do so? If the zebras always win (escape) where on the board do they and the lion start and what is their strategy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;While the answers so far are a good start they still don't completely answer the question. I might offer a bounty on this question and I would definitely award it to anyone who can prove that the Lion can always win. (if that is the case)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="4269" LastEditDate="2015-10-04T14:24:13.410" LastActivityDate="2018-03-04T19:44:43.283" Title="Lions and Zebras on a Chess Board" Tags="|mathematics|strategy|game|checkerboard|pursuit-evasion|" AnswerCount="4" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="15181" PostTypeId="2" ParentId="15177" CreationDate="2015-05-19T06:57:40.583" Score="0" Body="&lt;p&gt;Looking at this from a logical point of view, two answers spring to mind:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As the bullseye is over six metres wide, just aim for the middle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the above is a bit too ridiculous, then:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;are you confident that you can get your beloved &quot;Princess&quot;?&lt;/strong&gt;: No. The bullseye is two metres off the ground, but I am short and can only launch an arrow from 1.75 metres off the ground, and since the arrows will only fly parallel to the ground, I cannot hit regardless of how good my geometry is!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="341" LastActivityDate="2015-05-19T06:57:40.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15182" PostTypeId="2" ParentId="15177" CreationDate="2015-05-19T07:00:50.683" Score="2" Body="&lt;p&gt;You are 84 meters from the midpoint of the base of the triangle (5&lt;sup&gt;2&lt;/sup&gt; - 3&lt;sup&gt;2&lt;/sup&gt; = 4&lt;sup&gt;2&lt;/sup&gt;); assume you shoot at this point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will take your arrow 84m / (98mph) = 84m / (43.81m/s) = 1.92s to get to the base of the triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The target is moving at 12.52m/s; therefore, it travels 12.52m/s * 1.92s = 24.04m from the time you shoot your arrow to the time your arrow arrives at the center of the base. This is one full cycle of the triangle, plus 8.04m. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means you should release the arrow when the target is 0.04m from the apex of the triangle. The target will arrive at the midpoint of the trajectory's base just as your arrow strikes that point. Furthermore, because the target will be traveling along the base during the interval that your arrow goes from 80 to 84m, you do not need to worry about accidentally nipping the side of the target as it moves along one of the 5m sides.&lt;/p&gt;&#xA;" OwnerUserId="12572" LastActivityDate="2015-05-19T07:00:50.683" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15184" PostTypeId="2" ParentId="15180" CreationDate="2015-05-19T10:07:16.637" Score="12" Body="&lt;p&gt;I will place the Zebras as follows :&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2upgN.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Then whenever the knight attacks any piece, I will move it to its adjacent diagonal so that it is contained within these boxes :&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dgC8L.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Now, the knight cannot fork two pieces inside same box(as they are of opposite colour). Also knight cannot fork two pieces from two boxes as they are quite apart...(4 places between them). So I will just have to make sure that the knight does not capture me in next move and I am safe...&lt;/p&gt;&#xA;&lt;p&gt;Note that if the knight attacks from c3(only time when I will have to leave box) or such similar square I will diagonally move to the opposite box. From c3 he will have to go to e2 then g3 then f1 then e3 then his attacks will end (any other move will also mean end of attack). Then I will take back my piece to where it was. So initial condition.&lt;/p&gt;&#xA;" OwnerUserId="7723" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-19T15:09:28.653" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="15185" PostTypeId="2" ParentId="15180" CreationDate="2015-05-19T12:07:41.053" Score="6" Body="&lt;p&gt;Place 4 Zebras on White, and 4 Zebras on Black.&lt;br&gt;&#xA;At any move, Lion can attack only Black or only White, so only 4 pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep the 4 Black Zebras in the 4 corners, or adjacent to the corners.&lt;br&gt;&#xA;Keep the 4 White Zebras in the 4 corners, or adjacent to the corners.&lt;br&gt;&#xA;Now, only one Zebra can be in danger at any move. So only move that Zebra to the &lt;strong&gt;opposite corner&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conclusion : Zebras can always escape.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the figure , the blue lines are two possible ways to attack the left top corner Zebras, but the gray lines are the responses to escape. Same applies for all other corners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Escape Algorithm Overview :&lt;/strong&gt;&lt;br&gt;&#xA;The Lion can be in only one of the 4 big quadrants (left bottom, left top, right bottom, right top) and so it can attack only one of the two Zebras in that quadrant. That Zebra has to move to the &lt;strong&gt;opposite corner&lt;/strong&gt;.&lt;br&gt;&#xA;If the Lion also moves to that opposite quadrant, then that Zebra must go back to initial position.&lt;br&gt;&#xA;If the Lion moves to any other quadrant, then one Zebra from that quadrant will so move to opposite quadrant.&lt;br&gt;&#xA;If a piece may come into danger by moving, then simply move a Zebra of the other colour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definition of &lt;strong&gt;opposite corner&lt;/strong&gt; :&lt;br&gt;&#xA;For the Zebras, there are &lt;strong&gt;8&lt;/strong&gt; corners formed by left/right or top/bottom or inner/outer.&#xA;For getting opposite, all 3 Dimensions must change: left must become right and vice-versa , top must become bottom and vice-versa , inner must become outer and vice-versa. Example : top-left-outer corner is opposite of bottom-right-inner corner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Aim of every move is&lt;/strong&gt; Either &lt;strong&gt;(1)&lt;/strong&gt; If a piece is in danger, move that piece to opposite corner Or &lt;strong&gt;(2)&lt;/strong&gt; If no piece is in danger, &lt;strong&gt;(2A)&lt;/strong&gt; Move a piece back to original position, provided it is a safe move and the Lion is not blocking the way Or &lt;strong&gt;(2B)&lt;/strong&gt; Move a piece of the opposite colour, so that no piece comes into danger by that move Or &lt;strong&gt;(2C)&lt;/strong&gt; Move a Zebra of the same colour and the same quadrant as the Zebra blocked by the Lion in case &lt;strong&gt;(2A)&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation of (2C)&lt;/strong&gt; : When Lion moves from D5 to C3, one Zebra moves from A2 to G8. Now if Lion moves back to D5, Zebra can not use condition &lt;strong&gt;(2A)&lt;/strong&gt; to go back to original position A2, because L is blocking the way. So we have to use condition &lt;strong&gt;(2C)&lt;/strong&gt; to move one Zebra from H7 to B1. Still we are maintaining 2 Zebras of opposite colour in each quadrant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JGfwa.png&quot; alt=&quot;Position + Moves&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Special thanks to @Wouter &amp;amp; @Taemyr &amp;amp; @JonTheMon &amp;amp; @JS1 for useful comments , which improved this answer.&lt;/p&gt;&#xA;" OwnerUserId="10940" LastEditorUserId="10940" LastEditDate="2015-05-25T14:52:22.270" LastActivityDate="2015-05-25T14:52:22.270" CommentCount="20" ContentLicense="CC BY-SA 3.0" />
  <row Id="15187" PostTypeId="1" AcceptedAnswerId="19933" CreationDate="2015-05-19T13:29:55.007" Score="26" ViewCount="3544" Body="&lt;p&gt;I can be formed in many ways, and four parts are always involved.  Sometimes the formation can happen with all four parts at once, but sometimes some parts come later than the others.  The highest scenario (let's call this scenario I1) happens with all four parts at once, is 22 intrinsically, and can be affected by four factors&lt;sup&gt;[1]&lt;/sup&gt; resulting in a possible range between 22 and 96 (at the start it is 44).  Every other scenario has zero involved in some way.  All of the scenarios have non-zero intrinsic and possible range values, and can be affected by the same four factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More than one of me can be formed in many ways, and seven parts are always involved.  Sometimes the formation can happen with all seven parts at once, but sometimes some parts come later than the others.  The highest scenario (let's call this scenario M1) happens with all seven parts at once, is 24 intrinsically (34 if not for a limitation), and can be affected by five factors&lt;sup&gt;[2]&lt;/sup&gt; resulting in a possible range between 74 and 152 (at the start it is 98, 100, or 118).  All of the scenarios have zero involved in some way and have non-zero intrinsic and possible range values.  All of the scenarios in which the formation happens with all seven parts at once can be affected by the same five factors.  The scenarios in which some parts come later than the others can only be affected by four out of the five factors (those factors being A, B, C, and D).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Let's call the four factors A, B, C, and D.  Scenario I1 will be affected by none, one, or two of the factors, never three or four of them.  It may also be affected by more than one occurrence of the same factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] Let's call the five factors A, B, C, D, and E (where A, B, C, and D are the same as in [1]).  Scenario M1 will be affected by one, two, or three of the factors, never four or five of them.  It may also be affected by more than one occurrence of the same factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hints:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. When scenario I1 is 22, it is not affected by any of the factors.  When it is 96, it is affected by factors A and D.  When it is 44 at the start, it is affected by factor C.&#xA; &lt;p&gt;&#xA; 2. When scenario M1 is 74, it is affected by factor E.  When it is 152, it is affected by factors A, D, and E.  When it is 98 at the start, it is affected by factors C and E.  When it is 100 or 118 at the start, it is affected by factors A, C, and E.&#xA; &lt;p&gt;&#xA; 3. For all of the scenarios (for both me and more than one of me), the value never changes after the formation occurs.  This means that the formation may occur at the start or after that.&#xA; &lt;p&gt;&#xA; 4. When a scenario has &quot;zero involved in some way&quot;, zero will be involved once or twice, and it's not related to how many times the digit 0 appears in the value.&#xA; &lt;p&gt;&#xA; 5. I can be formed up to three times per [REDACTED], but more than one of me can only be formed once per [REDACTED].&#xA; &lt;p&gt;&#xA; 6. When the formation of more than one of me happens with all seven parts at once, it is always affected by factor E.&#xA; &lt;p&gt;&#xA; 7. &lt;strong&gt;Any&lt;/strong&gt; formation that occurs at the start is always affected by factor C.&#xA; &lt;p&gt;&#xA; 8. This puzzle is related to one of the following categories: entertainment, biology, geography, math/geometry, English/grammar.  The subject of this puzzle has been mentioned on more than one occasion on this site.&#xA; &lt;p&gt;&#xA; 9. No more than one formation occurs at the start, but it is possible for multiple formations to occur simultaneously after that.&#xA; &lt;p&gt;&#xA; 10. When a formation occurs, it doesn't have to be of me or more than one of me - it can be of something else.&#xA; &lt;p&gt;&#xA; 11. &quot;More than one of me&quot; can actually be considered a single thing.&#xA; &lt;p&gt;&#xA; 12. Location/position determines which factors affect a given scenario.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4450" LastEditDate="2015-08-10T03:13:55.657" LastActivityDate="2015-08-12T02:34:01.610" Title="I can be 22 intrinsically and have a possible range between 22 and 96 (44 at the start)" Tags="|riddle|enigmatic-puzzle|" AnswerCount="3" CommentCount="53" ContentLicense="CC BY-SA 3.0" />
  <row Id="15188" PostTypeId="1" AcceptedAnswerId="15660" CreationDate="2015-05-19T13:39:47.063" Score="1" ViewCount="602" Body="&lt;p&gt;I've added a twist to my normal cryptogram methods. I want to know if this twist makes the puzzle too difficult and, if so, how far back should I scale the efforts until the twist doesn't make it unsolvable. The cipher used is not that complicated but I don't know how much that complication is increased by the twist. Here is my method:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Post the original cryptogram&lt;/li&gt;&#xA;&lt;li&gt;Wait $n$ days to see if someone finds the plain text ($n$ is currently unknown)&lt;/li&gt;&#xA;&lt;li&gt;If someone solves it, consider the latest cipher text posted to be acceptable and exit this loop.&lt;/li&gt;&#xA;&lt;li&gt;If the twist is the only step remaining between the plain text and the latest cipher text posted, consider the twist to be unacceptable for use with cryptograms and exit this loop.&lt;/li&gt;&#xA;&lt;li&gt;If nobody solves it, add a slightly simpler version of the cipher text to the post.&lt;/li&gt;&#xA;&lt;li&gt;Return to step 2 until the loop is exited.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;WYYXMNDGDFWFDGVJTDYUTQDWYGXEYDQPEYEOMUCFAMEBQTUMQVLJEYMQOPNPMUVNGUVEGJJKUWNYFGNFYGGYMFGYPJFODZTDTALYCWMEPUMQQDDWEJGMYYVGJEJDQFEKYGQMUYNUPMVGZBUYTJJALYDFWWFYDTCBYHMJZHVVKYYWDCVUZFJDWUV&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$n = 0.23$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The end of sentences has been included.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;WYYXMNDGDFWFDGVJTDYUTQDWYGXEYDQPEYEOMUCFAMEBQTUMQVLJEYMQOPNPMUV. NGUVEGJJKUWNYFGNFYGGYMFGYPJFODZTDTALYCWMEPUMQQDDWEJ. GMYYVGJEJDQFEKYGQMUYNUPMVGZBUYTJJALYDFWWFYDTCBYHMJZHVVKYYWDCVUZFJDWUV.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$n = 0.00$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;All punctuation and spacing has been included.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cipher part is a simple substitution cipher. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;WYYXM NDGDFW FDGV JTDY UT QDWY GXEY D QPEY E OMUCFAMEBQ TUMQ VLJEYMQOP NPMUV. NGU VEGJJ KUWNYFG NFYGGYM FGY PJFODZTDTA LY CWMEP UM QQDDWEJ. GMYY VGJEJ D QFEKYG QMUY NUPMV, GZBUYTJJA LYDFWWFYDTC BYHMJZHV VKYYWDC VUZFJDWUV. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$n = 8.69$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The twist is the only thing remaining between the plain text and the cipher. To clarify, to go from plain text to the last edit (above), the process is: Plain Text, Twist, Cipher. Per Step 4 of the method, the loop will now exit.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;NEEVR WIHITN TIHS LFIE OF MINE HVAE I MDAE A CROGTYRAPM FORM SBLAERMCD WDROS. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2015-05-28T16:47:51.793" LastActivityDate="2016-10-13T05:53:31.927" Title="Is this Cryptogram twist too much or too little?" Tags="|cipher|puzzle-creation|" AnswerCount="2" CommentCount="10" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15189" PostTypeId="2" ParentId="15142" CreationDate="2015-05-19T13:48:40.483" Score="4" Body="&lt;h3&gt;Leaving for posterity's sake&lt;/h3&gt;&#xA;&lt;p&gt;I'm going to guess&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;Yellow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;If you run lines to every combination within the same colored dots, every color is passed over except yellow. While there is only one purple dot, it is still crossed many times by many other dots.&#xA;&lt;br&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/BJB6k.png&quot; alt=&quot;&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12242" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-19T15:51:16.320" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15190" PostTypeId="1" AcceptedAnswerId="15192" CreationDate="2015-05-19T13:59:09.403" Score="5" ViewCount="5670" Body="&lt;p&gt;Bob is a temp working with a construction company in repairing roads and barriers. One day Bob got called up to repair a barrier along a strip of highway. &lt;b&gt;Bob will need 100 &quot;Company X&quot; pylons&lt;/b&gt; to divert traffic away from his site while he repairs. Being the best employee around, Bob goes into work the day before to round up the pylons he needs. After looking everywhere in the warehouse, &lt;b&gt;Bob is only able to find 50 &quot;company X&quot; pylons!&lt;/b&gt; He then realized &lt;b&gt;John must have taken the rest and hidden them from Bob.&lt;/b&gt; Bob also found a note that says his supervisor will be driving by each site tomorrow for inspection! If his supervisor sees that Bob is not following protocol of using proper pylons, he will surely be fired!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next day after Bob was clever and did some tricks he was able to finish the repairs without getting fired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;How did Bob save his job?&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Bob has the rest of the day to prepare, and has the supplies in the&#xA;warehouse to his disposal (But there are no other &quot;Company X&quot;&#xA;pylons).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If Bob tried to use hand craft fake pylons (or non &quot;Company X&quot; pylons), his supervisor would surely notice and Bob would not be following company rules! &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Bob has already called John asking for pylons but John said he buried&#xA;them and encrypted the location. (John won't share the encryption,&#xA;he's said too much already)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There is shipment being made to the warehouse in 2 days with more&#xA;pylons, but this will be too late for Bob.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-05-19T16:54:16.007" LastActivityDate="2015-06-01T19:26:53.937" Title="Not enough pylons!" Tags="|logical-deduction|lateral-thinking|" AnswerCount="12" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="15191" PostTypeId="2" ParentId="15142" CreationDate="2015-05-19T14:06:05.453" Score="1" Body="&lt;p&gt;I guess it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;AQUA/CYAN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;with the help of @tfitzger image i came to this conclusion.&lt;br&gt;&#xA;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Purple is formed using orange + blue &lt;br&gt;so there is 3 beams passing the star major once are green and blue which gives CYAN color. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11866" LastActivityDate="2015-05-19T14:06:05.453" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15192" PostTypeId="2" ParentId="15190" CreationDate="2015-05-19T14:16:57.837" Score="25" Body="&lt;p&gt;First of all, break the rear-view mirrors of the supervisor's car (admit it, you've always wanted to do that!).&lt;br&gt;&#xA;Then, split each pylon into two identical pieces, and place them along the road. &#xA;Without a rear-view mirror, the supervisor won't notice that the pylons were divided! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3hZJZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-05-19T17:54:34.510" LastActivityDate="2015-05-19T17:54:34.510" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="15193" PostTypeId="2" ParentId="15190" CreationDate="2015-05-19T14:45:24.707" Score="17" Body="&lt;p&gt;Obviously&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The repair work was not critical or immediately visible, so Bob cordoned off half the work area he needed to. Thus, he only needed 50 pylons instead of 100. When his boss drove by, he saw a properly cordoned off work area, and went on his merry way. Bob continued his work on the first half. When it was completed, he cordoned off the second half and performed his work there as well. He uses, in total, 100 pylons, by using the same 50 twice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12242" LastEditorUserId="12242" LastEditDate="2015-05-19T16:34:42.043" LastActivityDate="2015-05-19T16:34:42.043" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15194" PostTypeId="2" ParentId="15190" CreationDate="2015-05-19T15:05:01.760" Score="5" Body="&lt;p&gt;Smoke and mirrors...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each real pylon, strategically place a mirror so that an image of it will appear in the location where the next pylon should be (as viewed from a passing car).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A cloud of smoke blocks out the sun so that there will be no shadows to give away the deception.&lt;/p&gt;&#xA;" OwnerUserId="12582" LastActivityDate="2015-05-19T15:05:01.760" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15195" PostTypeId="2" ParentId="15142" CreationDate="2015-05-19T15:50:53.993" Score="7" Body="&lt;p&gt;New Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Purple&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If you take the image and superimpose the horizontal mirror image onto it, and then take the vertical mirror image of the result and combine them together, you get the image below (which I believe is what the HUD showed). You then see a distinct pattern around the star with the cyan and green/purple and blue that makes the star purple.&#xA;&lt;br&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/TEesy.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12242" LastActivityDate="2015-05-19T15:50:53.993" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15196" PostTypeId="2" ParentId="15142" CreationDate="2015-05-19T16:06:30.140" Score="2" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ORANGE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because Yellow Mixed with Red forms Orange :-) LOL&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the RAINBOW :&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;gt;! Violet 1 Seq 1&#xA;&amp;gt;! Indigo 3 Odd&#xA;&amp;gt;! Blue   2 Seq 2&#xA;&amp;gt;! Green  3 Odd&#xA;&amp;gt;! Yellow 1 Seq 1&#xA;&amp;gt;! Orange 6 NotOdd&#xA;&amp;gt;! Red    2 Seq 2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To make It Odd.. It should be orange. :-)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9892" LastEditorUserId="9892" LastEditDate="2015-05-19T17:55:47.760" LastActivityDate="2015-05-19T17:55:47.760" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15197" PostTypeId="2" ParentId="15190" CreationDate="2015-05-19T16:10:50.433" Score="4" Body="&lt;p&gt;Isn't it obvious? He...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ... records the phone call with John, present this as evidence to his boss, gets John fired and takes the pylons John was going to use. This is a &quot;clever trick&quot;, as per the description.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another potential (and more serious) answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If this is a barrier that splits a highway - i.e. it's along the median, not along the outer edge, then he can build a device to quickly transfer pylons from one side to the other. His supervisor will drive by one way, see 50 pylons, then they can be switched so that when the supervisor drives by the other way, he sees &quot;another&quot; 50 pylons. This is predicated on the barrier being in the median and on Bob knowing which side his supervisor will be driving on first.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2015-05-19T16:10:50.433" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15198" PostTypeId="2" ParentId="15190" CreationDate="2015-05-19T16:17:39.980" Score="23" Body="&lt;p&gt;Bob arranges for a traffic jam at the time the supervisor is passing, so the supervisor can pass the site only very slowly, and can be strategically stopped for a few minutes at critical moments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the supervisor arrives, &quot;company X&quot; pylons are in use for the first half of the site, &quot;company Y&quot; pylons for the second part of the site, and Bob has 25 spare &quot;company Y&quot; pylons in his truck.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;___________________________________________________________________________________________________&#xA;[Car]___[Traffic jam whose movement Bob can control]&#xA;XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY&#xA;[Bob's truck]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As the car passes, Bob starts collecting the &quot;Company X&quot; pylons in his truck, replacing them with &quot;Company Y&quot; ones, until he has switched 25 of them:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;___________________________________________________________________________________________________&#xA;[Traffic stuck behind supervisor]___[Car]___[Traffic jam whose movement Bob can control]&#xA;YYYYYYYYYYYYYYYYYYYYYYYYYXXXXXXXXXXXXXXXXXXXXXXXXXYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY&#xA;                   [Bob's truck]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;At this point, Bob arranges for the traffic to stop completely for a while, and races past (in the enclosed area) to start replacing the &quot;company Y&quot; pylons with &quot;company X&quot; ones. The supervisor still has &quot;company X&quot; pylons as far as he can see ahead and behind (if necessary Bob may need to arrange for strategically placed trucks within the traffic jam!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The traffic starts moving again, and all the middle 50 pylons are of &quot;company X&quot; type by the time the supervisor passes the midpoint:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;___________________________________________________________________________________________________&#xA;            [Traffic stuck behind supervisor]___[Car]___[Traffic jam whose movement Bob can control]&#xA;YYYYYYYYYYYYYYYYYYYYYYYYYXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXYYYYYYYYYYYYYYYYYYYYYYYYY&#xA;                                                                  [Bob's truck]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Bob now has another 25 company Y pylons in his truck, so races back to start replacing the company X pylons for the second set of 25...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... and repeats the same process to get the final 25 pylons as &quot;company X&quot; type by the time the supervisor passes that section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The supervisor observes &quot;company X&quot; pylons all the way along, and Bob racing back and forth, obviously working very hard.&lt;/p&gt;&#xA;" OwnerUserId="12582" LastActivityDate="2015-05-19T16:17:39.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15200" PostTypeId="2" ParentId="13303" CreationDate="2015-05-19T16:19:14.843" Score="2" Body="&lt;h2&gt;Mapped Path and Thoughts on Stanza Seven&lt;/h2&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Below is the map showing the path for stanzas 1-6 (confirmed correct by OP)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; HT-DT-DO-EN-BE-EU-??-??&lt;br&gt;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/7MtN0.png&quot; alt=&quot;Map&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;And here are my thoughts for Stanza Seven:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;MAGNITUDE&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;  - Two because you can like up the Roman numerals &amp;quot;IVIV&amp;quot; to create two downward arrows (per &lt;a href=&quot;https://puzzling.stackexchange.com/a/15079/9000&quot;&gt;Zandar&lt;/a&gt;)&lt;br&gt;&#xA;  - Four because the Roman number &amp;quot;IV&amp;quot; is the Arabic numeral 4&lt;br&gt;&#xA;  - Eight because the Roman number &amp;quot;IV IV&amp;quot; is the Arabic numeral 4 twice, or 8&lt;br&gt;&#xA;  - Four because &amp;quot;15&amp;quot; looks like &amp;quot;IS&amp;quot; and there are four cells between the two IS cells&lt;br&gt;&#xA;  - Five because &amp;quot;15&amp;quot; looks like &amp;quot;IS&amp;quot; and the distance between the two IS cells is five&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;DIRECTION&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;  - East because lining up the Roman numerals IVIV vertically creates to downward arrows and down on the map is East (per &lt;a href=&quot;https://puzzling.stackexchange.com/a/15079/9000&quot;&gt;Zandar&lt;/a&gt;)&lt;br&gt;&#xA;  - West because &amp;quot;two one fives&amp;quot; could be &amp;quot;two single fives&amp;quot; in Roman numerals is VV which looks like W (this won't work if we're not allowed in the ocean)&lt;br&gt;&#xA;  - North because moving from the closest to the furthest IS cell is north&lt;br&gt;&#xA;  - South because moving from the furthest to the closest IS cell is south (this won't work if we're not allowed in the ocean OR if we have to move more than one cell which is likely and would move us off the map)&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-19T16:19:14.843" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15202" PostTypeId="2" ParentId="15180" CreationDate="2015-05-19T16:34:27.417" Score="0" Body="&lt;p&gt;Ok, let's try to think outside of the box (literally) and let the lion move outside the 8x8. Using the 2x2 in the corners strat for Zebras and Prem's starting board. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's go with: D5, C3, B5, A'6.  This brings up a fork on in the top left. That zebra will flee to the bottom right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lion A8. This brings up 2 moves for the zebras: G8 to A2 (plus mirror), G2 to B7 (zebras are safe next to lion). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zebra G8 to A2, Lion C9. Zebra can't return to A8 or B7.  Lion goes D7, F6, G4, all moves that force a zebra to move. At that point, F2 catches the zebra that's stuck at H1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about zebra to B7.  Again lion goes D7 (zebra back to G2), F6, G4, F2 catches the zebra stuck at H1 again. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-05-19T16:48:20.350" LastActivityDate="2015-05-19T16:48:20.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15203" PostTypeId="2" ParentId="15190" CreationDate="2015-05-19T16:50:38.510" Score="-3" Body="&lt;p&gt;Bob does half the work with the half the pylons on the first day when he came in early to prepare. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the second day, when his supervisor comes, he does the other half of the work.  So each day he only has to block of 50 pylons worth of distance.&lt;/p&gt;&#xA;" OwnerUserId="4384" LastActivityDate="2015-05-19T16:50:38.510" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15204" PostTypeId="2" ParentId="15190" CreationDate="2015-05-19T17:04:24.507" Score="9" Body="&lt;p&gt;This might work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Bob's boss is predictable, Bob can plan (or pretend) to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;start or finish the work during the inspection. Then, he would be either setting up the pylons or taking them down. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This way, his boss will not know that only 50 pylons were used. It would be reasonable to assume that either 50 pylons &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;were still on their way (if they were just getting started) or that 50 pylons had already been taken back (if they were finishing).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If that won't work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If his boss isn't so predictable, Bob can &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;place a GPS tracker on his boss' car. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is quite possibly illegal, so it would be better if he didn't have to resort to this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though Bob is tricking his boss, he should make sure that he stays safe - fake pylons might not trick his boss, but they would convince traffic. They can be stacked under the real pylons during the inspection so that his boss won't be able to see them.&lt;/p&gt;&#xA;" OwnerUserId="1955" LastEditorUserId="4204" LastEditDate="2015-05-20T19:15:34.647" LastActivityDate="2015-05-20T19:15:34.647" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15205" PostTypeId="2" ParentId="15190" CreationDate="2015-05-19T17:32:55.860" Score="13" Body="&lt;p&gt;Bob &quot;peels&quot; each pylon, effectively seperating the inside from the outside. If he has a long enough knife, he can do this without leaving any visible cuts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a depiction of the cross section of a pylon during this process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2eiHN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm assuming that traffic cones are orange all the way through. If it's just surface paint, then the &quot;inside&quot; cone will look obviously different)&lt;/p&gt;&#xA;" OwnerUserId="1497" LastActivityDate="2015-05-19T17:32:55.860" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15206" PostTypeId="1" CreationDate="2015-05-19T17:53:30.323" Score="5" ViewCount="589" Body="&lt;p&gt;Is there any literature that goes into detail about ideas and best practices for designing puzzles?&lt;/p&gt;&#xA;" OwnerUserId="12587" LastActivityDate="2015-05-20T12:48:16.147" Title="What books and/or resources are available to learn Puzzle Design?" Tags="|puzzle-creation|" AnswerCount="0" CommentCount="6" ClosedDate="2015-05-20T12:48:33.830" ContentLicense="CC BY-SA 3.0" />
  <row Id="15207" PostTypeId="2" ParentId="15190" CreationDate="2015-05-19T18:09:29.487" Score="13" Body="&lt;p&gt;Ok here goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Per the original clue:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Bob also found a note that says his supervisor will be driving by &lt;b&gt;each&lt;/b&gt; site tomorrow for inspection  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This indicates:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;More than one site will be inspected, therefore more than one site exists.  Since Bob has time, he can visit one or more other sites, and &quot;collect&quot; (steal) an additional 50 cones to install at &lt;b&gt;his&lt;/b&gt; site, to avoid being fired.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12235" LastEditorUserId="10281" LastEditDate="2015-05-19T18:11:19.533" LastActivityDate="2015-05-19T18:11:19.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15208" PostTypeId="1" AcceptedAnswerId="15215" CreationDate="2015-05-19T18:12:48.823" Score="12" ViewCount="1825" Body="&lt;p&gt;Welcome to part 3 of Quark's quest to make an image puzzle that isn't solved in under an hour. Like the last one, this will be much more difficult than the last.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the below encrypted message, there are stages that lead from one to the next. There are two main words to discover, one at the ~75% completion mark, and one at the finish line. Put them together and you'll have a phrase related to computing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is that phrase?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fIAv4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hints definitely aren't needed to solve this but if it goes past a day or two I'll start giving hints. However, if this is solved under two hours then I'll admit that I underestimate this puzzling community and I'll be forced use drastic measures for part 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The &quot;ovals&quot; in the picture are different sizes, ignore this because I made this sloppily in Paint. This was originally supposed to make smart puzzlers go &quot;well he probably doesn't want us measuring their radii/center position so I'll ignore it&quot; but I realize it's better to explicitly state it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note 2: There is no extra information in this puzzle other than the size thing mentioned above. I realized there is a phrase related to computing that is discovered around the 40% completion mark but this is not related to the answer and is just a hint on how to progress further.&lt;/p&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="69582" LastEditDate="2023-11-02T16:06:17.833" LastActivityDate="2023-11-02T16:06:17.833" Title="Find the secret phrase in this encrypted image (part 3)" Tags="|visual|cipher|lateral-thinking|steganography|" AnswerCount="2" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="15209" PostTypeId="1" AcceptedAnswerId="15213" CreationDate="2015-05-19T19:04:34.777" Score="26" ViewCount="38483" Body="&lt;p&gt;&lt;sub&gt; &lt;em&gt;This may be simple and fairly well known, but I couldn't find it anywhere here, and felt it deserved a place in the ranks of puzzles&lt;/em&gt; &lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A man is found dead, hanging in an empty barn having committed suicide. The base of the barn was 20 feet by 30 feet, and the ceiling was 10 feet above the ground. The rope was 4 feet from the roof to the man's neck, and the man's feet were hanging 1 foot above the ground, and the man was 5 feet from where the rope was around his neck to his toes. The man was hanging from the exact center of the barn. Aside from the man and the rope, there were no other objects in the barn, and the door to the barn was closed and locked from the inside, so no one had come in or out since the deed was done. The barn was structurally sound, with no holes or windows to be found anywhere. The roof was also water tight, but the ground beneath the man was damp. The ladder used to tie the rope up was found outside around the corner (guess he didn't want those that found him to have to move it later). There were no supports or beams along the ceiling of the barn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was determined that the man was too weak to jump and pull himself into the noose, or to climb onto the roof with just his hands. How did he manage to hang himself?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Change log: No windows, no holes in the roof. Roof is water tight. Door to the barn was locked from the inside. Damp ground was found under the man &lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="9951" LastEditDate="2015-05-20T21:42:43.210" LastActivityDate="2018-04-26T22:35:45.890" Title="How did the man hang himself?" Tags="|lateral-thinking|" AnswerCount="14" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="15210" PostTypeId="2" ParentId="15209" CreationDate="2015-05-19T19:06:24.837" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;He stood on a 1 foot tall block of dry ice. Which once it had evaporated enough that the block was out of the reach of his feet, led to his strangulation and subsequent death. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12206" LastActivityDate="2015-05-19T19:06:24.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15211" PostTypeId="2" ParentId="15209" CreationDate="2015-05-19T19:07:13.863" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The man was standing above a block of ice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alternatively,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The man jumped and crouched while in air, breaking his neck.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-05-19T19:07:13.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15212" PostTypeId="2" ParentId="15209" CreationDate="2015-05-19T19:07:46.580" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He was murdered and the murder staged the scene to make it look like a suicide. (Because there was no note!) The murderer dug a hole out of the barn to use after he locked the barn from the inside. He then escaped through the hole and filled it in once he escaped. (No one was the wiser)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alternatively:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He got a 1 foot block of ice, and climbed on that to pull the rope down. Unfortunately he slipped on the ice and the rope got caught around his neck :( &lt;br&gt; [He locked the doors because it was too windy outside at the time and he didn't want the wind to push him]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-05-19T19:36:41.227" LastActivityDate="2015-05-19T19:36:41.227" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15213" PostTypeId="2" ParentId="15209" CreationDate="2015-05-19T19:24:53.837" Score="47" Body="&lt;p&gt;&lt;strong&gt;Canonical Answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The man stood on a block of ice which has since melted.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Non-canonical Answers:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  1. The man stood on his horse who then took off and is out wandering the fields. (This is in the same category as the ice: &quot;Man stood on $X$ which has since removed itself through means common to things of its type.&quot;)&lt;br&gt;&lt;br&gt;&#xA;  2. If it wasn't for the explicitly stating the ladder was used to hang the rope and implying the general weakness of the man, the man could have thrown the rope over the rafters, tied the ends together, and then climbed into the rafters. He then unties the rope, attaches it to the rafters, and ties a noose. He slips the noose over his head and jumps down.&lt;br&gt;&lt;br&gt;&#xA;  3. Again, this might be ruined by the weakness, but he could tie it in the rafters with the ladder and then climb the rope, tie the noose, slip it on, and let go of the rope.&lt;br&gt;&lt;br&gt;&#xA;  4. There is a powerful electromagnet under the floor set on a timer. The man was lifted up his shoes laced with natural magnets oriented to the opposite pole, put the noose around his neck, and waited for the magnet to turn off.&lt;br&gt;&lt;br&gt;&#xA;  5. If it didn't state that it was suicide, I'd say that someone else hung him up there for cattle thievin'.&lt;br&gt;&lt;br&gt;&#xA;  6. He was a depressed wizard with the power of levitation.&lt;br&gt;&lt;br&gt;&#xA;  7. The rope is sentient and strangled its abusive master.&lt;br&gt;&lt;br&gt;&#xA;  8. The rope was tied around the rafters before the barn was raised. He stood there as his friends and family raised the barn, lifting his body into the air.&lt;br&gt;&lt;br&gt;&#xA;  9. He used the ladder. It is stated that the ladder was put away but not that the man was the one who put it away. That conclusion is only guessed at. The first person on the scene put the ladder away to make it seem like a murder as they were embarrassed that the man had committed suicide.&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answers from the comments:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  A. He climbed the ladder around the corner of the barn (where it was later found) to the roof, then jumped through the hole in the roof. –  Paul Draper&lt;br&gt;&lt;br&gt;&#xA;  B. The man was too weak to jump into the noose himself, but he put Flubber on the floor (or his shoes) and jumped on that. –  Paul Draper&lt;br&gt;&lt;br&gt;&#xA;  C. Ropes with natural fibers can shrink 5-10% when wet. He tied the rope around his neck, and leaky roof cause the once 10ft rope to become 9ft. –  Paul Draper&lt;br&gt;&lt;br&gt;&#xA;  D. The man dug a pit, tied the noose outside it, and jumped into the pit. –  Paul Draper&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2015-05-20T12:05:42.293" LastActivityDate="2015-05-20T12:05:42.293" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="15214" PostTypeId="2" ParentId="15209" CreationDate="2015-05-19T19:36:28.167" Score="3" Body="&lt;p&gt;What if the man climbed on the roof with his hands (no ladder), tied the rope around his neck and jumped down?&lt;/p&gt;&#xA;" OwnerUserId="11454" LastActivityDate="2015-05-19T19:36:28.167" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15215" PostTypeId="2" ParentId="15208" CreationDate="2015-05-19T19:44:56.550" Score="16" Body="&lt;p&gt;(No spoiler tags because they're messing with the line breaks...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First we do a trivial reshaping of the image to un-distort it:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;img = Import[&quot;https://i.stack.imgur.com/fIAv4.png&quot;]&#xA;rectified = Image[ArrayReshape[ImageData[img, Automatic], {400, 399, 3}], &quot;Byte&quot;]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7wM6m.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After noticing that the black pixels are not all exactly black, I&#xA;extracted the lowest bit from each pixel:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;bitImg = Image[Mod[ImageData[rectified, Automatic], 2], &quot;Bit&quot;]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hgqc9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;ZBEFR PBQR&lt;/code&gt; is &lt;code&gt;MORSE CODE&lt;/code&gt; rot13'd.  This suggests that part of the puzzle will be decoded using Morse code.  After extensive randomness testing on the data burst at the top of the image, I switched to analyzing the dots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First I played around with separating the color channels, since the pure/saturated primary colors suggested per-channel binary data.  This lead nowhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally I pulled up the &lt;a href=&quot;http://en.wikipedia.org/wiki/Morse_code&quot; rel=&quot;nofollow noreferrer&quot;&gt;Morse code Wikipedia page&lt;/a&gt;, and immediately noticed that &lt;code&gt;H&lt;/code&gt; is encoded as four dots.  Where had I seen those before?  The red dots at the top of the image.  The two letters I know, &lt;code&gt;T&lt;/code&gt; (&lt;code&gt;-&lt;/code&gt;) and &lt;code&gt;E&lt;/code&gt; (&lt;code&gt;.&lt;/code&gt;) are both single dots: corresponding to the dots on either side of the four red ones.  Put it all together and you get &lt;code&gt;THE&lt;/code&gt;!  I realized that the colors were a Morse code substitution cipher with the following mapping:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;RED   =&amp;gt; DOT (·)&#xA;BLUE  =&amp;gt; DASH (-)&#xA;BLACK =&amp;gt; LETTER SPACE (/)&#xA;GREEN =&amp;gt; WORD SPACE ( )&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This allows us to decode the dots as:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; bkrrrrkrgrbbrkrbkrrrkrrrgrrkrrrgrkbbkbrrrkrkbrrkbrrkrkbrr&#xA; -/..../. .--./.-/.../... ../... ./--/-..././-../-.././-..&#xA;  T  H  E    P  A  S  S    I  S   E  M  B  E  D  D  E  D&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Entering &lt;code&gt;embedded&lt;/code&gt; as the password into a &lt;a href=&quot;http://www.mobilefish.com/services/steganography/steganography.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;suitable online decrypter&lt;/a&gt;, we obtain the following image file (&lt;code&gt;chip.png&lt;/code&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZiavC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the answer is &lt;strong&gt;EMBEDDED CHIPS&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="32543" LastEditDate="2017-05-09T09:45:25.723" LastActivityDate="2017-05-09T09:45:25.723" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15216" PostTypeId="1" AcceptedAnswerId="15228" CreationDate="2015-05-19T19:48:29.393" Score="43" ViewCount="5248" Body="&lt;p&gt;A grocery store has a long, skinny box, with no top, that it uses to display soda. The box is two soda cans wide and 200 soda cans long. You can neatly fit 400 cans in this box, using two rows of 200, as shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uuw7t.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Show how to fit $401$ soda cans into this box (with their bases resting flat on the floor of the box).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is no lateral thinking involved in this solution. You can think of this as a purely geometric, two dimensional problem, where you are trying to fit $401$ circles of diameter $1$ into a $2\times200$ rectangle.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="12109" LastEditDate="2015-06-28T15:13:46.957" LastActivityDate="2023-11-28T13:59:09.670" Title="A Surprising Circle Packing" Tags="|mathematics|geometry|packing|" AnswerCount="7" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="15217" PostTypeId="2" ParentId="15190" CreationDate="2015-05-19T19:52:27.070" Score="1" Body="&lt;p&gt;My other less practical solution was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To sabotage the supervisors vehicle(s) such that he has to spend the day getting it repaired and preventing him from inspecting Bobs site. Thus Bob can make/use anything else for the remaining 50 pylons as his supervisor won't notice.&#xA;&lt;br&gt; Obviously the supervisor might have other means of transportation, which is how this plan can fail, so Bob might have to result in setting a backup trap which will debilitate his supervisor for the day. (Hole to sprain his ankle or something)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pretty much:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Instead of solving the current problem, alter the current problems viable solutions by removing the supervisor.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastActivityDate="2015-05-19T19:52:27.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15218" PostTypeId="2" ParentId="15216" CreationDate="2015-05-19T19:54:47.633" Score="1" Body="&lt;p&gt;Perhaps a configuration like this?&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/o3SkS.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I haven't done the math yet, so not sure if it actually works.&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-19T19:54:47.633" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15219" PostTypeId="2" ParentId="15190" CreationDate="2015-05-19T19:58:45.650" Score="6" Body="&lt;p&gt;Bob, having a full day and night, goes and picks up the shipment of extra pylons himself instead of waiting for them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He takes a plane if he needs too. He stays up all night if he needs too. He is a very dedicated hard worker, that Bob.&lt;/p&gt;&#xA;" OwnerUserId="2800" LastEditorUserId="2800" LastEditDate="2015-05-19T20:05:22.993" LastActivityDate="2015-05-19T20:05:22.993" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15220" PostTypeId="2" ParentId="15216" CreationDate="2015-05-19T19:58:48.667" Score="1" Body="&lt;p&gt;Perhaps:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; By shifting the top row of circles&#xA;&lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/CpU0G.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;br&gt; And then cramming 1 in the top left corner, you will get a trapezoid shape. Although this could only be maintained depending on properties of the box&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(I have yet to have the math to prove this)&lt;/p&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-05-19T20:13:57.680" LastActivityDate="2015-05-19T20:13:57.680" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15221" PostTypeId="2" ParentId="15216" CreationDate="2015-05-19T20:06:37.480" Score="4" Body="&lt;p&gt;Playing around in paint (400 height, 100 radii circles), I made a slight stagger.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/6ILLv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The width of the three circles on top is less than if they were directly side by side, so this trapezoid can be flipped and repeated.&lt;/p&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="7722" LastEditDate="2015-05-19T20:16:19.993" LastActivityDate="2015-05-19T20:16:19.993" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15222" PostTypeId="2" ParentId="13300" CreationDate="2015-05-19T21:02:42.880" Score="1" Body="&lt;p&gt;In my opinion this is a mean puzzle that I just happened to solve because I got lucky. It is unsolvable without the hints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 9&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The number sequence is Catalan's constant backwards, starting arbitrarily 39 digits from the decimal point. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It is referred to by a capital G in mathematics, a cat tamed the boy (refers to cata, not sure why), and LAN stands for local area network.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2 and 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is a constant that isn't Euler's or Apery's.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How I figured it out:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The hints and random googling made me sure the answer had something to do with Catalan many days ago, but no matter what transformations I did I couldn't find the connection. After revisiting, by chance I happened to notice the constant on the Wikipedia page had the same number of digits (39) as the puzzle and from there I saw the connection. A bit mean but I guess solvable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="7722" LastEditDate="2015-05-19T21:10:48.110" LastActivityDate="2015-05-19T21:10:48.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15223" PostTypeId="2" ParentId="15208" CreationDate="2015-05-19T21:10:34.343" Score="6" Body="&lt;p&gt;From 2012rcampion's answer, we get &quot;morse code&quot;.&lt;br&gt;&#xA;From this, we can think there's a message in morse hidden in the image;&lt;br&gt;&#xA;Morse code needs four symbols: (-), (.), ( ), (/)&lt;br&gt;&#xA;In the picture, the ovals come in four colors, so maybe each color stands for one symbol.&lt;br&gt;&#xA;Green is the least frequent so it's the word separator, /;&lt;br&gt;&#xA;the letter separator can only be black, since red and blue can come in pairs;&lt;br&gt;&#xA;red has to be the dot, since the second letter, rrrr, doesn't make sense as dashes (while as dots, it's &quot;h&quot;).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;recap:&lt;br&gt;&#xA;(-) blue&lt;br&gt;&#xA;(.) red&lt;br&gt;&#xA;( ) black&lt;br&gt;&#xA;(/) green  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting, we obtain&lt;br&gt;&#xA;- .... . / .--. .- ... ... / .. ... / . -- -... . -.. -.. . -..&lt;br&gt;&#xA;In morse, it gives:&lt;br&gt;&#xA;THE PASS IS EMBEDDED  &lt;/p&gt;&#xA;" OwnerUserId="7689" LastEditorUserId="5373" LastEditDate="2015-05-19T23:21:40.723" LastActivityDate="2015-05-19T23:21:40.723" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15224" PostTypeId="2" ParentId="15165" CreationDate="2015-05-19T21:21:15.293" Score="1" Body="&lt;p&gt;(Disclaimer: Okay, I don't think this is the answer because it's like cheating, I can make whatever I want fit - I googled 95% of them. But I found it funny.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly you are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A pokemon master!  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The loner is :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cubone - chillin' in his tower&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The lion is: (Take your pick)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ?Entei, Luxray, Litleo etc...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Black Sheep is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mareep?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Taurus, Nagini, and Dracula:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Tauros, Arbok, Crobat&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The bear is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Snorlax&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The newest are from what I gather, generally OP'ed so.... just choose which ever you want. Back in my days it was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Mew and Mewtwo&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastActivityDate="2015-05-19T21:21:15.293" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15225" PostTypeId="2" ParentId="15176" CreationDate="2015-05-19T21:39:22.673" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alice can always win by picking any square of the 13 of the same colour.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Consider an open Knight's Tour of the 5x5 board. Remove any square at an odd position in the tour, which becomes the starting square, and divide the tour up into pairs of squares. Whichever square Bob moves to, Alice simply moves to matching square from her list of pairs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Example.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;1   14  9   20  3&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;24  19  2   15  10&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;13  8   25  4   21&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;18  23  6   11  16&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;7   12  17  22  5&lt;/code&gt;&lt;br&gt;&#xA; Alice wishes to start on (5, 3) [17]; if Bob moves to (4, 1) Alice moves to (2, 2); if Bob moves to (3, 2) Alice moves to (5, 1); if Bob moves to (3, 4) Alice moves to (1, 5) and if Bob moves to (4, 5) Alice moves to (2, 4).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12593" LastActivityDate="2015-05-19T21:39:22.673" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15226" PostTypeId="2" ParentId="15216" CreationDate="2015-05-19T22:38:05.770" Score="5" Body="&lt;p&gt;Okay, here's a badly-drawn way to do it:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Start by taking four cans and lining them up in a diamond.  Then, tilt that diamond until it touches the top of the box:&lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/e0tk3.jpg&quot; alt=&quot;Badlydrawn&quot;&gt;&lt;br&gt;Now repeat with four cans at a time, each making a diamond, and push that diamond as far to the left as it will go.  The first diamond uses $1 +\sqrt{2}$, or about $2.414$, diameters, but each additional diamond only adds about $1.99156$ diameters.  So after $100$ diamonds of $4$ cans each, you end up with enough extra space at the bottom right to fit an extra can.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11187" LastEditorUserId="11187" LastEditDate="2015-05-20T00:21:17.813" LastActivityDate="2015-05-20T00:21:17.813" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15227" PostTypeId="2" ParentId="15209" CreationDate="2015-05-19T23:38:14.193" Score="7" Body="&lt;p&gt;A couple of stupid answers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  He lifted himself up by his bootstraps.&lt;BR&gt;&lt;BR&gt;He was wearing stilts&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11410" LastActivityDate="2015-05-19T23:38:14.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15228" PostTypeId="2" ParentId="15216" CreationDate="2015-05-20T02:56:40.460" Score="43" Body="&lt;p&gt;EDIT: The description is not very intuitive, so I took the liberty of creating an image to make clear what this solution intents.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TpeY2.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; We're going to make a repeating pattern of six cans.  The pattern will have three cans in the top row and three cans in the bottom row.  They will be &amp;quot;offset&amp;quot; by .5 units, and each row will have width a little bit less than 3 units.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The details:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; I'm going to give the coordinates of the centers of the cans... the centers have to be in a 1-by-199 box, since they can't be within one unit of the edge, and no two centers can be within one unit of each other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;And:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Here's the repeating pattern: The coordinates of the three points in the bottom row will be (0, 0), (1, 0), (1.992 , .13).  It will repeat again, offset by 2.984 units, so: (2.984, 0), (3.984, 0), (4.976, .13).  And so forth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;And:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; In the top row the first three points will have coordinates (.5, .87), (1.492, 1), (2.492, 1).  Then the next three it repeats: (3.484, .87), (4.476, 1), (5.476, 1), etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Now before I check that my solution works let me give some intuition:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; There are two rows of cans.  I &amp;quot;offset&amp;quot; the cans by half a unit: so if the bottom cans (center points) have x-coordinates 1, 2, 3, then the top cans are centered at 1.5, 2.5, 3.5.  Now that gives me some &amp;quot;wiggle room&amp;quot; which I can use to move some of the cans up and down a little bit.  Moving some cans up and down means I can make them take up ever-so-slightly less horizontal space -- I can compress a row horizontally.  Now if I only compress one row but not the other I can't maintain the .5-unit offset.  So I compress the two rows in turns.  That's how the pattern works.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;It turns out it's just enough to get the compression we need:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; First you can check that the points I put are all 1 unit apart.  In the bottom row you just check that .992^2 + .13^2 &amp;gt; 1, which it is.  The top row is the same by symmetry.  Between the bottom row and the top row, the key calculation is .5^2 + .87^2 &amp;gt; 1, which is also true.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;And then you count the total space:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Let's say there are 201 cans in the bottom row, so you have to have 200 intervals.  How much space will those take up?  Well, every 3 cans take 2.984 units, so you save .016 units every three cans, or a total of 1.072, which is just good enough!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12523" LastEditorUserId="82468" LastEditDate="2023-11-28T13:59:09.670" LastActivityDate="2023-11-28T13:59:09.670" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="15229" PostTypeId="2" ParentId="15216" CreationDate="2015-05-20T03:47:09.087" Score="12" Body="&lt;p&gt;Here is a CAD drawing I created to show the layout:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/kqT3n.png&quot; alt=&quot;drawing&quot;&gt;&#xA; Using trigonometry and the facts that all the magenta lines and height of the rhombus are unit length, we can find that the center of the yellow circle is at x-coordinate 1.495. We can then create a function for distance required to store &lt;em&gt;n&lt;/em&gt; number of cans on the bottom row:&#xA; &lt;code&gt;F(n) = 1 + 0.995 * (n - 1)&lt;/code&gt; for &lt;code&gt;n &amp;gt;= 1&lt;/code&gt; so &#xA; &lt;code&gt;F(201) = 200&lt;/code&gt;&#xA; , meaning we can fit 201 cans on the bottom row, and one less on the top, for 401 total.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12599" LastEditorUserId="12599" LastEditDate="2015-05-20T03:58:48.473" LastActivityDate="2015-05-20T03:58:48.473" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15230" PostTypeId="1" CreationDate="2015-05-20T04:10:56.777" Score="3" ViewCount="749" Body="&lt;p&gt;Give a 9 letter English word, which remains a word even after you keep removing a letter from it at each step until you have just one letter remaining.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Give the letter you remove and the word at each step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of a 3 letter word would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pig&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remove &lt;strong&gt;g&lt;/strong&gt; - &lt;strong&gt;pi&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remove &lt;strong&gt;p&lt;/strong&gt; - &lt;strong&gt;i&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You cannot shuffle the letters around&lt;/li&gt;&#xA;&lt;li&gt;The words cannot be proper nouns. None of them.&lt;/li&gt;&#xA;&lt;li&gt;The words have to be proper English words. No slangs, no informal words.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="12537" LastEditorUserId="12537" LastEditDate="2015-05-20T04:43:36.787" LastActivityDate="2015-05-20T04:43:36.787" Title="9 letter word decomposition" Tags="|english|wordplay|" AnswerCount="0" CommentCount="6" ClosedDate="2015-05-20T04:29:16.837" ContentLicense="CC BY-SA 3.0" />
  <row Id="15231" PostTypeId="2" ParentId="15209" CreationDate="2015-05-20T04:52:50.707" Score="1" Body="&lt;p&gt;Here's a thematic variation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The ground wasn't wet, so he couldn't have been standing on a block of ice. However, this being a barn, he could have been standing on a one-foot-high bale of hay, which was subsequently carried off and/or eaten by a barnyard animal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3896" LastActivityDate="2015-05-20T04:52:50.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15232" PostTypeId="2" ParentId="15209" CreationDate="2015-05-20T05:23:48.700" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It was indeed &quot;5 feet from where the rope was around his neck to his toes&quot;, but his knees were bent. He had a rubber band from around his ankles to his waist, so he could stand straight up while setting up the hanging, but on relaxing his leg muscles his feet, including his toes, were pulled up to within 5 feet of his neck.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alternatively,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He committed suicide because his beloved three-legged dog died, so he hung himself hugging his dog, so we have his two feet and the dog's three other feet between his toes and his neck.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="341" LastActivityDate="2015-05-20T05:23:48.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15233" PostTypeId="2" ParentId="15142" CreationDate="2015-05-20T06:19:22.743" Score="17" Body="&lt;p&gt;The central star should be coloured:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Indigo (dark blue)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reason is, that if you draw a hexagonal grid across everything like this (excuse the sloppy photoshop work):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/X0B2p.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then, for any given dot, you simply:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; count up the number of dots on any of the intersecting lines, which will always give the same sum, based on the colour of the dot you chose:&lt;br&gt;&#xA; Red - 0&lt;br&gt;&#xA; Orange - 1&lt;br&gt;&#xA; Yellow - 2&lt;br&gt;&#xA; Green - 3&lt;br&gt;&#xA; Blue - 4&lt;br&gt;&#xA; Indigo - 5&lt;br&gt;&#xA; Violet - 6  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="5840" LastEditDate="2015-08-28T06:27:15.353" LastActivityDate="2015-08-28T06:27:15.353" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15235" PostTypeId="2" ParentId="15216" CreationDate="2015-05-20T07:33:42.347" Score="8" Body="&lt;p&gt;Here's the MSPaint to the rescue answer. The math logic works in my head as well. The top packing is tighter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;each triangle after the 1st one saves (edit: $1-\sqrt{{{1-(1-\sqrt{\frac{3}{4}}}}})^2$ ) ~0.00901523343248242768377221117738 diameters. Multiply by 132 and get 1.190010813087680454257931875414. The first triangle is on the bottom left. The 133rd is also on the bottom, leaving space for 1 on top of another packed into the right edge. QED&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; edit fixed picture link &lt;img src=&quot;https://i.stack.imgur.com/T5RXD.png&quot; alt=&quot;image&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the math: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; looks like i failed, unless my math is off my a factor of 2, it fits 402.&lt;img src=&quot;https://i.stack.imgur.com/8ynRd.png&quot; alt=&quot;enter image description here&quot;&gt; 1.0+2.0*pow(1.0*1.0*pow((1.0-pow(1.0*1.0-0.5*0.5,0.5)),2.0),0.5)/6.0 simplified math: sqrt( 1 - ( sqr( 1 - sqrt( .75 ) ) ) / 3.0 is the ratio.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&amp;lt; &lt;strong&gt;correct math below&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;strikethrough the previous text. 1st commenter clarified the math, it's 1 - sqrt( 1 - ( sqr( 1 - sqrt( .75 ) ) ) saved for each triangle after the first one. 133 triangles leaves 132 * ( 1 - sqrt( 1 - ( sqr( 1 - sqrt( .75 ) ) ) ) extra space on the end.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12601" LastEditorUserId="12601" LastEditDate="2015-05-23T09:30:16.907" LastActivityDate="2015-05-23T09:30:16.907" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="15236" PostTypeId="1" AcceptedAnswerId="15238" CreationDate="2015-05-20T08:23:30.983" Score="10" ViewCount="1508" Body="&lt;p&gt;This is the first - presumably simplest - puzzle of an intended series of numerical dissection puzzles.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;h1&gt;The goal is to dissect the 15 x 11 number-grid below into exactly 17 rectangular sub-grids. &lt;br&gt;&lt;/h1&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Each rectangle has to be of shape of either ( ? x 3 ) or ( 3 x ? ) number squares. &lt;br&gt;&lt;/li&gt;&#xA;&lt;li&gt;Each rectangles has to fulfil either a sum-equation or a difference-equation &lt;br&gt; using its 3 rows or 3 columns, respectively. &lt;br&gt;&lt;/li&gt;&#xA;&lt;li&gt;Numbers are either read &amp;quot;left-to-right&amp;quot; or &amp;quot;top-to-bottom&amp;quot; and never &amp;quot;right-to-left&amp;quot; or &amp;quot;bottom-to-top&amp;quot;.&lt;br&gt;&lt;/li&gt;&#xA;&lt;li&gt;Leading zeros are acceptable.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Please also describe in your answer any &amp;quot;technique&amp;quot; or &amp;quot;routine&amp;quot; you have used to find the solution.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ce0wZ.png&quot; alt=&quot;Puzzle&quot; /&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;Examples of valid 'rectangles':&lt;/h1&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/T037p.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-20T10:09:16.750" Title="Dissection of the number-grid (addition/subtraction)" Tags="|calculation-puzzle|dissection|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15237" PostTypeId="2" ParentId="15164" CreationDate="2015-05-20T08:38:57.477" Score="2" Body="&lt;p&gt;I &lt;em&gt;think&lt;/em&gt; it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Arya, from the &lt;em&gt;Inheritance&lt;/em&gt; cycle of fantasy novels.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the rest of it, SPOILER ALERTS!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Our life and death are one, we two&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That's her and Firnen, her dragon. The bond between Rider and dragon is one of my favourite parts of those books.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Hair the color of rose's imperfection&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; She has black hair, and blackleg is a plant disease affecting roses.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Eyes like the one who is part of me,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; She has green eyes, and Firnen is green.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Though we are two.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Two in body, but one in mind and spirit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And the other of us is the color&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Of his fallen relatives in the&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Safest place on the earth(besides _________)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Firnen's relatives (other dragon eggs and Eldunari) are hidden away in the Vault of Souls at Vroengard, perhaps the safest place in the world besides the elven forest.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My family and I are celebrimbor,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This refers to family not in the literal sense but in the sense of dragon Riders. Celebrimbor was an elven lord in LotR whose name means &quot;silver hand&quot;, as in &lt;em&gt;Argetlam&lt;/em&gt; - dragon Riders all have silver on their hands from the &lt;em&gt;gedwey ignasia&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But we are just my two brothers and me.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The only dragon Riders in existence are Eragon, Murtagh, and Arya ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;For now.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ... but more will hopefully arrive when the other eggs hatch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first hint is now self-explanatory. For the second:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not to be confused with Arya Stark, a character of the same name in a more recently written fantasy novel (&quot;more modernly created world&quot;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-05-20T19:39:55.483" LastActivityDate="2015-05-20T19:39:55.483" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="15238" PostTypeId="2" ParentId="15236" CreationDate="2015-05-20T09:57:12.997" Score="10" Body="&lt;p&gt;Wasn't all that difficult:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/Jlblz.png&quot; alt=&quot;Solution to puzzle&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To solve it, I started by marking out boundaries that I knew the solution would contain. For example, look at these numbers in the top left corner:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;9 1 3 8 2 6 6 5 9 8 …&#xA;0 0 4 1 1 7 4 4 9 0 …&#xA;9 0 9 7 0 9 2 1 0 2 …&#xA;6 2 8 0 8 9 3 5 8 0 …&#xA;6 3 9 6 9 5 3 4 7 1 …&#xA;: : : : : : : : : :&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Going from left to right, there is no way that we can form a valid sum of the form $9\cdots \pm 1\cdots = 3\cdots$, so we know immediately that the sum in the top left corner must be working from top to bottom; i.e., $913\cdots \pm 004\cdots = 909\cdots$. It turns out that we can carry this on for up to 9 digits, but we know for certain that there is a horizontal line below the digits $9\ 0\ 9$ at the start of the third row. Likewise, the two sets of three digits at the start of the fourth and fifth rows must form a horizontal sum, because $6\cdots \pm 6\cdots \neq 3\cdots$, but $6\cdots + 2\cdots = 8\cdots$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you've identified a few boundaries this way, the rest fall into place quite easily.&lt;/p&gt;&#xA;" OwnerUserId="40" LastEditorUserId="40" LastEditDate="2015-05-20T10:09:16.750" LastActivityDate="2015-05-20T10:09:16.750" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15239" PostTypeId="2" ParentId="15209" CreationDate="2015-05-20T10:30:26.940" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The rope was made of &lt;a href=&quot;http://en.wikipedia.org/wiki/Shape-memory_alloy&quot;&gt;shape-memory alloy&lt;/a&gt;. The man tied it up during night when it was cool. The heat at daytime caused the rope to shorten, pulling him off the ground.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12604" LastActivityDate="2015-05-20T10:30:26.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15240" PostTypeId="1" AcceptedAnswerId="15249" CreationDate="2015-05-20T10:51:07.253" Score="19" ViewCount="3142" Body="&lt;p&gt;Your friend, the Cryptologist Mr. Toboggan, has been murdered in his house. You being a detective and a Cryptologist yourself have been asked to lend a hand in solving the murder.&lt;/p&gt;&#xA;&lt;p&gt;You arrive at the murder scene and find it in complete disarray. The body is on the floor next to his computer table. The laptop is open and there are various open windows. One of them is showing the map of the city, another an unfinished chess game, yet another matrix of some sort with various characters and symbols across numbers. It also seems that he was preparing a kind of a document with text written on it, some of which doesn't make a lot of sense. The text was:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The Roman says -&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;    If you knew what I got,        &#xA;    you'd have known the plot.&#xA;    Till the cops were brought,        &#xA;    the body got rot.        &#xA;    &#xA;    The sun had set,&#xA;    and it lay there unseen&#xA;    Till the time had come&#xA;    for the number Thirteen&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;'frnepu gur png, ova, try, she'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You take notes, and decide to look into it later.&lt;/p&gt;&#xA;&lt;p&gt;You look at the chess game and observe that he had entered a weird name as his own (probably a code).&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;code&gt;'mr tngu ab zhyr'&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You then look at the map, and see that various locations are marked with text&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;code&gt;k1&lt;/code&gt;&#xA;&lt;code&gt;m9&lt;/code&gt;&#xA;&lt;code&gt;t3&lt;/code&gt;&#xA;&lt;code&gt;n6&lt;/code&gt;&#xA;&lt;code&gt;a2&lt;/code&gt;&#xA;&lt;code&gt;k15&lt;/code&gt;&#xA;&lt;code&gt;p10&lt;/code&gt;&#xA;&lt;code&gt;j7&lt;/code&gt;&#xA;&lt;code&gt;t4&lt;/code&gt;&#xA;&lt;code&gt;c2&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;At the bottom of the map&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;code&gt;'shim ta bilt kora mietylo'&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You take notes again.&lt;/p&gt;&#xA;&lt;p&gt;You take a look at the body again, and see that the fur lining of his coat is torn apart at the corner. You bend and inspect and find a small piece of paper inside with a note saying,&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;code&gt;'pqyqfnssaf'&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Now you start to wonder if your friend's sanity had taken his leave. You can't help remembering odd things about your friend and start feeling unsettled. You shake your head and ... you suddenly hear purring from the corner of the room at see his old cat just sitting there as if nothing had happened. You go and pick the cat up and find a parchment piece poking out from its collar. You unfold it and see now familiarly odd text written on it.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;code&gt;'envyf tb mvtmnt'&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You take note, and start looking around for more clues. You look at all the clothes on the bed and nothing...under the bed and nothing...inside the pillow covers and...find a note saying,&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;code&gt;'prnrh ugth soygfrpn thjjwe'&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Put it in the evidence bag and take notes. You move on to the next room, a poster there showing an alien saying &amp;quot;I'm from Area51&amp;quot;; you smile, and as you were just about to leave the room you notice something scribbled at the bottom of the poster. You go closer and take a look:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;code&gt;'K to m9'&lt;/code&gt;&#xA;&lt;code&gt;'BB to p10'&lt;/code&gt;&#xA;&lt;code&gt;'R to k1'&lt;/code&gt;&#xA;&lt;code&gt;'Q to m9'&lt;/code&gt;&#xA;&lt;code&gt;'sup sup'&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;They seem like chess moves but these blocks don't exist on the chess board. You go back to the computer and look at the chess game and shockingly you see that all the blocks are numbered weirdly. As if by instinct you look at the map and see the marked locations all have numbers on the chess board. Royally confused, you sit down to think. Just a few seconds into an argument with yourself your eyes wander to the trash bin and see a crumpled piece of paper inside. You un-crumple it to see similar text on it.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;code&gt;'ksadeplpzd'&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Thoroughly confused, you take notes again and decide to move on to find more clues. You head into the bathroom; looking around, you pick up the shower gel and see a piece of plastic taped to it on one side with the text saying,&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;code&gt;'ieassruoke'&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You shake your head again in utter confusion. You decide you have had enough and after telling the cops to notify you of any development you go back to your house.&lt;/p&gt;&#xA;&lt;p&gt;It's almost mid-night and after having a scotch you decide to hit the sack. You fall asleep right away. Just a few minutes of very light sleep and you wake up suddenly. No, it's not possible. You pick up your notes, smile and think. You go to the computer and and verify your findings. Assured, you smile again and think, &amp;quot;It's just like Toboggan, even at the last moment, so many misleading clues. Rest in Peace my friend. You solved it at last.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;Who did it?&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Note: It's a Cryptology puzzle. Some of the clues might be misleading, but if you solve it properly, you will not be misled.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="12537" LastEditorUserId="69582" LastEditDate="2024-01-03T16:37:44.323" LastActivityDate="2024-01-03T16:37:44.323" Title="Murder of a Cryptologist" Tags="|enigmatic-puzzle|cipher|story|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="15241" PostTypeId="1" AcceptedAnswerId="15243" CreationDate="2015-05-20T12:50:03.590" Score="16" ViewCount="951" Body="&lt;p&gt;My great aunt Edith keeps this list of her favorite words tacked on the refrigerator:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Atlas&lt;br&gt;&#xA;  Chance&lt;br&gt;&#xA;  Detail&lt;br&gt;&#xA;  Frost&lt;br&gt;&#xA;  Hand&lt;br&gt;&#xA;  Magnet&lt;br&gt;&#xA;  Orange&lt;br&gt;&#xA;  Ring&lt;br&gt;&#xA;  Sand&lt;br&gt;&#xA;  Tiger&lt;br&gt;&#xA;  Winter&lt;br&gt;&#xA;  Wolf &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My great uncle Mark has this list of his favorites:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Angel&lt;br&gt;&#xA;  Art&lt;br&gt;&#xA;  Bad&lt;br&gt;&#xA;  Boot&lt;br&gt;&#xA;  Gift&lt;br&gt;&#xA;  Hall&lt;br&gt;&#xA;  Kind&lt;br&gt;&#xA;  Mutter&lt;br&gt;&#xA;  Rat&lt;br&gt;&#xA;  Stern&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which one of them would like the word &lt;em&gt;Listen&lt;/em&gt;?&lt;/p&gt;&#xA;" OwnerUserId="10851" LastEditorUserId="10851" LastEditDate="2015-05-20T20:29:13.130" LastActivityDate="2015-05-20T20:29:13.130" Title="Two lists of favorite words" Tags="|pattern|word-problem|english|knowledge|language|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15242" PostTypeId="2" ParentId="15209" CreationDate="2015-05-20T13:16:50.093" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He tied the noose using the ladder before putting it outside etc... He made the opening of the noose large enough to put his head through comfortably. While not very strong in his arms he was quite the jumper. So he jumped up and put his head through the noose while jumping. It snagged and closed, and voila.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11721" LastActivityDate="2015-05-20T13:16:50.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15243" PostTypeId="2" ParentId="15241" CreationDate="2015-05-20T13:22:15.467" Score="16" Body="&lt;p&gt;Your&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; great uncle Mark&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;likes the word &quot;Listen&quot;. Here's why:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All words can be English or German words. The words in Aunt Edith's list have the same meaning in German and in English. (Or there is at least one common meaning. &quot;Ring&quot; in German means a round band of metal, but not a call on the phone.) The words in Uncle Mark's list have a different meaning in English and German. &quot;Gift&quot;, for example, means poison. And &quot;Listen&quot; means lists and has nothing to do with hearing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9860" LastActivityDate="2015-05-20T13:22:15.467" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="15244" PostTypeId="2" ParentId="15209" CreationDate="2015-05-20T13:24:19.950" Score="14" Body="&lt;p&gt;Since a heap of people have offered alternative solutions, I thought I'd provide a few of my own (ones that are relatively plausible, hopefully, compared with &quot;he used flubber&quot; or &quot;he was a wizard with the power of levitation&quot;)...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. He used a rope made from a special polymer that gets significantly shorter as it cools. It was warm when he put it around his neck, and as it cooled, it pulled his body off the ground.&lt;br&gt;&#xA;&lt;br&gt;&#xA; 2. He used a second rope to assist him, but also carried a match, and burned the support rope (which was supporting enough of his weight - remove the support rope, and his full weight goes onto the noose). The ashes of the match and the support rope have since spread thinly enough in the barn that they're indistinguishable from the dirt.&lt;br&gt;&#xA;&lt;br&gt;&#xA; 3. He used hay as a substitute for some rope to help him, and the hay has since broken off and dropped back to the floor of the barn - nobody would have suspected that some of the stray hay in the barn was actually used in the suicide.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And now for some more entertaining answers (that are still hopefully a little more plausible than some of the answers):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4. He had a severe hunch in his back - he was capable of straightening it out if he really tried, but when he relaxed it would go back to the hunched position. He stretched to reach the noose, then relaxed again.&lt;br&gt;&#xA;&lt;br&gt;&#xA; 5. He committed suicide during an extreme storm, during which the ceiling sagged from the rain and the walls tilted due to the wind (not the most structurally sound barn around, I grant you, but it stayed standing, so it didn't do too badly). These two effects combined to lower the centre of the ceiling far enough that the man could reach the noose. As the wind died down, his feet were raised just far enough to hang him. Over the next couple of days the wetness of the roof evaporated and the ceiling's sag reverted, restoring the barn to its original dimensions, thus further lifting the man's feet.&lt;br&gt;&#xA;&lt;br&gt;&#xA; 6. There was a large ant nest in the centre of the barn, with a mound over a foot tall. The man detested ants at the same time, so he chose that spot - he would put his weight on the mound, causing the ant tunnels to collapse, and in the process hang himself.&lt;br&gt;&#xA;&lt;br&gt;&#xA; 7. The barn originally had wooden floorboards in the centre as a kind of stage. However, it was infested by termites. The man committed suicide, the floorboards collapsed enough to hang him, and then the termites finished the job. They're probably already working on the walls. The person who discovered the body probably heard the faint scratching noise of them termites working away, but figured it was the guy's ghost come back to haunt the barn.&lt;br&gt;&#xA;&lt;br&gt;&#xA; 8. There had been a huge flood. The man had used the ladder as support while positioning the rope in the perfect place to grab as he tried to surf on the flowing water. Unfortunately, he misjudged it, and instead of his hands catching the rope, his head went through it. The force of the water caused the man to be hanged, and later the floodwaters receded, leaving him hanging out to dry, so to speak. Oh, and the floodwaters also forced the doors shut, where the auto-locking mechanism triggered.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And now for something completely different:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 9. Aliens. It's always aliens.&lt;br&gt;&#xA;&lt;br&gt;&#xA; 10. Thanks, Obama!&lt;br&gt;&#xA;&lt;br&gt;&#xA; 11. The man was a rare male spirit medium from the Fey clan. He was channelling a taller man he hadn't realised was a sociopath, who proceeded to use his body (currently shaped to match the possessing spirit's original body's shape) to &quot;commit suicide&quot;, thereby killing the medium. It was simple - set things up, position his head in the noose, then leave the medium's body, which would immediately revert to its normal shape, thereby hanging the medium.&lt;br&gt;&#xA;&lt;br&gt;&#xA; 12. The man in the barn was Mario, and he had used a super mushroom to get bigger (becoming Super Mario). But he also had a poison mushroom in his possession - on positioning himself, he ate the poison mushroom to revert to normal, resulting in his hanging.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9289" LastActivityDate="2015-05-20T13:24:19.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15246" PostTypeId="2" ParentId="15209" CreationDate="2015-05-20T14:06:48.863" Score="0" Body="&lt;p&gt;Yet another possibility:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He climbed the ladder to fix the rope both around his neck and to the ceiling of the barn, stepped off the ladder while keeping hold of it, and with his last effort threw the ladder out the window - which is why it was found outside the barn.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9278" LastActivityDate="2015-05-20T14:06:48.863" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15247" PostTypeId="2" ParentId="13305" CreationDate="2015-05-20T14:07:09.120" Score="9" Body="&lt;p&gt;I believe the full explanation is as follows...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need to visit the 8 &lt;code&gt;X&lt;/code&gt; marks around the map, but need to &quot;end [our] trip in the dark forest&quot; and avoid the witches tower, so we'll start on the lower left and proceed clockwise around the road.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiple hints point towards hexagonal tiling: The six pointed star, &quot;honey, she still rules&quot;, seven minus one mages, the lake (looks like 4 hexes stacked), the compass on a 60 degree tilt, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we draw a 60 degree grid over everything (using the lake as a guide for spacing), we get the following (I've labelled, in blue, all the features that appear in the spells in order from &lt;code&gt;A&lt;/code&gt; to &lt;code&gt;V&lt;/code&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BVrYr.png&quot; alt=&quot;grid and labels&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can now see that for each &lt;code&gt;X&lt;/code&gt; we need to visit, there's a feature that has three copies that all triangulate on the given location. I've labelled these, in green, in the order in which we visit them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if we take these 8 spells in the order travelled, we end up with the following list (which is where Dennis Meng's solution picks up):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HB76V.png&quot; alt=&quot;spells&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here, reading down the columns produces the text:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;From the dead tree in the swamp, go South. Halfway to Mount Dragon, stop. Where you cross the road is the Holy Nut.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, taking the &lt;em&gt;dead tree in the swamp&lt;/em&gt; (assuming the swamp is indicated by the reed looking icons), we have roughly &lt;code&gt;Ed&lt;/code&gt; (the lone &lt;strong&gt;P&lt;/strong&gt; on my labelled map) and &lt;em&gt;Mount Dragon&lt;/em&gt; (assuming it's the biggest, most foreboding mountain) at roughly &lt;code&gt;Ak&lt;/code&gt;, you can see a clean &quot;South&quot; line based on the hex grid, which cuts the road at the bottom right of the &lt;code&gt;Cg&lt;/code&gt; grid square, as indicated by the big yellow &lt;strong&gt;X&lt;/strong&gt; on the labelled map. And so, here lies the Holy Nut!&lt;/p&gt;&#xA;" OwnerUserId="5208" LastActivityDate="2015-05-20T14:07:09.120" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15248" PostTypeId="2" ParentId="15240" CreationDate="2015-05-20T15:09:22.707" Score="2" Body="&lt;h1&gt;PARTIAL ANSWER&lt;/h1&gt;&#xA;&lt;p&gt;The first sequence&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;frnepu gur png, ova, try, she&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;can be decrypted using a simple ROT13 as follows:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;search the cat, bin, gel, fur&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The hint was contained in the first line, where the &amp;quot;Roman&amp;quot; clearly alludes to Caesar (cipher).&lt;/p&gt;&#xA;&lt;p&gt;The cat, the bin, the shampoo and the fur indeed have their own encrypted messages, but I haven't decrypted them yet. &lt;strike&gt;On the contrary, the roof isn't mentioned in the problem.&lt;/strike&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;If I make progress, I'll add them to this answer!&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-20T15:23:24.127" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15249" PostTypeId="2" ParentId="15240" CreationDate="2015-05-20T16:59:41.847" Score="18" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Keyser Soze&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which is a reference to one of my favourite movies:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Usual Suspects&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first step was to decrypt the text at the bottom of the note using &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;ROT-13, from Rot and Thirteen in the plain text.  The reference to Caesar was helpful as well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This yields:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;search the cat, bin, gel, fur&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Next find all of the clues in order.  The cat clue: &quot;envyf tb mvtmnt&quot; decrypts to &quot;rails go zigzag&quot;.  The other three clues are bin: ksadeplpzd, gel: ieassruoke, and fur: pqyqfnssaf.  Every other clue in this puzzle is a red herring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was initially tripped up at this point because I was trying to use an actual Rail Fence cipher.  However the three clues listed like so:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;ksadeplpzd&lt;br&gt;ieassruoke&lt;br&gt;pqyqfnssaf&lt;br&gt;&lt;br&gt;&#xA;Travel down the diagonal from top left to bottom and then mirror the angle and head back to the top.  Repeat until the end is reached:&lt;br&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/WOF7e.jpg&quot; alt=&quot;keysersoze&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1777" LastEditorUserId="5373" LastEditDate="2015-05-20T19:45:19.207" LastActivityDate="2015-05-20T19:45:19.207" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="15251" PostTypeId="1" AcceptedAnswerId="15252" CreationDate="2015-05-20T20:38:49.260" Score="-6" ViewCount="342" Body="&lt;p&gt;&lt;em&gt;I know this problem might be too silly/easy, but I'll ask it anyway and see what happens...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;One day Alice woke up a beach. She got up to look around but there was&#xA;  no one else there! She started to panic a little and then grabbed her&#xA;  phone. She had no bars on her phone (so no reception with her&#xA;  provider)! Settling down, she started tapping her screen and pressing some&#xA;  buttons and was able to call home (but still having no bars)!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;How was Alice able to call home?&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;She has a normal smartphone (will proper software/apps etc.) but it has no satellite capabilities or other methods to strengthen cellphone signal.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;She has no other technological devices or tools on her except for her phone. (And there aren't any hidden in or on the sand etc.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Her 'home' isn't on the island. None of her family is on the island etc.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;She isn't crazy and imagining the call, or sleeping and dreaming etc.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="5373" LastEditDate="2015-05-20T21:52:04.227" LastActivityDate="2015-05-20T21:52:04.227" Title="Contact with no signal?" Tags="|lateral-thinking|situation|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15252" PostTypeId="2" ParentId="15251" CreationDate="2015-05-20T20:43:56.820" Score="2" Body="&lt;p&gt;Assuming this is a public beach with facilities, one method that may have worked for Alice is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; She may have enabled a WiFi connection and placed a call using a phone app. Free WiFi is offered for many establishments so as long as one is nearby, the problem is solved.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Possible alternative:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; She could also have used Bluetooth or similar technology to connect to a source connected to the internet/cell tower (which would require special software on the receiving end).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="7722" LastEditDate="2015-05-20T20:55:08.090" LastActivityDate="2015-05-20T20:55:08.090" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15253" PostTypeId="1" AcceptedAnswerId="15868" CreationDate="2015-05-20T20:57:05.703" Score="5" ViewCount="491" Body="&lt;p&gt;Can you solve the following equality?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$432 = 915$&lt;br&gt;&#xA;  $944 = 850$&lt;br&gt;&#xA;  $211 = 403$&lt;br&gt;&#xA;  $837 = 927$&lt;br&gt;&#xA;  $735 =\: ???$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Additional numbers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$734 = 5100$&lt;br&gt;&#xA;  $736 = 736$&lt;br&gt;&#xA;  $835 = 741$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 1:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There are some numbers that will produce an error if run through the algorithm. An example of a number that would cause an error is $571$. Another example is $435$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Super mega bonus hint:&lt;/strong&gt; I'll go ahead and run a few requested numbers  through the algorithm so you can see what the results are! The first 3 or so requested numbers will be added to the OP for everyone's guessing pleasure. I will, however, frown and stick out my tongue at you if you ask for the result of $735$.&lt;/p&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="11475" LastEditDate="2015-06-02T15:09:08.903" LastActivityDate="2015-06-02T15:26:01.587" Title="Solve the Unconventional Equality" Tags="|calculation-puzzle|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="15254" PostTypeId="2" ParentId="15253" CreationDate="2015-05-20T21:20:25.610" Score="2" Body="&lt;p&gt;The first digit is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 6&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the first digit of the right-hand number is always the digit sum of the number on the left (4+3+2=9; 9+4+4=17 and 1+7=8; 2+1+1=4; 8+3+7=18 and 1+8=9; &lt;strong&gt;7+3+5=15 and 1+5=6&lt;/strong&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second digit is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the second digit of the right-hand number is always the sum of the second and third on the left minus the first on the left - &lt;em&gt;unless this figure is negative, in which case (too little data to tell, but it could be) the first digit on the left minus the second&lt;/em&gt; (3+2-4=1; &lt;em&gt;9-4=5&lt;/em&gt;; 1+1-2=0; 3+7-8=2; &lt;strong&gt;5+3-7=1&lt;/strong&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;[WIP]&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-05-20T22:47:12.313" LastActivityDate="2015-05-20T22:47:12.313" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15255" PostTypeId="2" ParentId="9915" CreationDate="2015-05-21T00:52:23.457" Score="25" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Genius&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first observation is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are 20 pairs of identical symbols, which can be connected by lines. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We can find the intersections of the lines from the same colors, producing 10 points. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the same way:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We can also intersect lines which connect the same shapes, which produce another 10 points. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the result:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The intersection points spell out &quot;IQ&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Combined with the &quot;130&quot;, this produces the phrase &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;IQ over 130&quot;, which describes a &lt;strong&gt;genius&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is the picture which goes with this solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/ycvjB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="5840" LastEditDate="2016-02-20T13:46:23.077" LastActivityDate="2016-02-20T13:46:23.077" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15256" PostTypeId="2" ParentId="11537" CreationDate="2015-05-21T00:59:36.443" Score="2" Body="&lt;p&gt;And the answer is....&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2015&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks to alexwlchan, Michael Edenfield, and leoll2 for their great answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The missing piece of info eluded to in rand al'thor's hint is...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; that in &lt;em&gt;Vampirates: Immortal War&lt;/em&gt; Sidorio has 2 more children, so the answer to&#xA; &lt;em&gt;Divide by the number of children fathered by Sidorio.&lt;/em&gt; is 4 and not 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Making the final calculation $2383 - ((1002-1\times2)/4 +300-6\times29+9-18+7-4)+2$&lt;/p&gt;&#xA;" OwnerUserId="11297" LastActivityDate="2015-05-21T00:59:36.443" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15257" PostTypeId="1" CreationDate="2015-05-21T01:27:14.877" Score="3" ViewCount="471" Body="&lt;p&gt;Am I stubborn?  Just as much as Constance Contraire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steady fingers on the bridge,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just as steady as my mind when it's made up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I love one whose weapon is that of the just.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They may have taken their vows against me, but I'll save him 'cause he's to me like glittering garland is to silver-fist&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are divided, awkward and shy as he may have been so long ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But my love, he is just as outgoing as always.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used the same weapon of my daughter,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whose main character trait was inherited from me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;You aren't supposed to answer the question posed in the title.  Rather, it's a hint.  Just say who the narrator is.  Book references fewer.  Just one throughout and three others.  Couldn't help myself with the three;-)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Here's a riddle for the series(hint!) that this riddle is mostly from, lines symbolized by a backslash&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here it is...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We are 12, the first a haunted ground/ The second is like the second line/ The third is entombed in ice/ The fourth is those who own a special pendant/ The fifth is a misunderstood healer/ The sixth is against all odds/ The seventh is a bargain for a big man's head/ The eighth is about mentor's home/ The ninth is the mentor's danger/ The tenth is a noble noble-man/ The eleventh was lost long ago/ The twelfth is the subject's obdurate child. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11790" LastEditorUserId="11790" LastEditDate="2015-05-22T01:04:44.563" LastActivityDate="2018-01-15T04:26:00.487" Title="Which one is my real identity?" Tags="|riddle|knowledge|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15258" PostTypeId="2" ParentId="15257" CreationDate="2015-05-21T02:52:25.537" Score="1" Body="&lt;p&gt;I think the book is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Mysterious Benedict Society.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The riddle's narrator is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Milligan.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He is Kate's father&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I'm not so sure after seeing the title.&lt;/p&gt;&#xA;" OwnerUserId="10748" LastEditorUserId="10748" LastEditDate="2015-05-21T08:57:07.157" LastActivityDate="2015-05-21T08:57:07.157" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15259" PostTypeId="2" ParentId="2865" CreationDate="2015-05-21T04:45:05.147" Score="10" Body="&lt;p&gt;I managed to fit 12 speedy knights onto the board.  These 12 attack every other square on the board, so I think it's unlikely that more than 12 will fit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/drPzE.png&quot; alt=&quot;12 Speedy Knights&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="12618" LastActivityDate="2015-05-21T04:45:05.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15260" PostTypeId="2" ParentId="13303" CreationDate="2015-05-21T05:21:44.773" Score="48" Body="&lt;h1&gt;Complete Solution from OP&lt;/h1&gt;&#xA;&lt;p&gt;Well, my Uncle would be proud... It was a team effort, but ultimately it was &lt;a href=&quot;https://puzzling.stackexchange.com/a/15063/5208&quot;&gt;rand al'thor who eluded the pirate's curse and found the treasure&lt;/a&gt;!&lt;/p&gt;&#xA;&lt;p&gt;Although rand's answer is ultimately correct and he has been awarded his bounty, I'm posting/accepting this answer to provide a canonical solution and to recognise the contributions made by various people throughout the search.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;The first challenge is to determine which way North is on the map (&lt;a href=&quot;https://puzzling.stackexchange.com/a/15063/5208&quot;&gt;first identified correctly by rand al'thor&lt;/a&gt;). This is achieved by observing that the blotches around the compass rose are formed by the holes in the capital letters of the words NORTH, SOUTH, EAST and WEST, as shown here:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kGipH.png&quot; alt=&quot;compass rose&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;So once rotated, the map, as used for the rest of the challenge, will look like this (with points discussed below marked):&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GzXZB.png&quot; alt=&quot;solved map&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;From here, the riddle can be attacked...&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Start your search from dead man's cave and follow the clues herein.&lt;br /&gt;&#xA;Sharpen a sword and your wits and you're ready to begin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;There's 6 initial possibilities for &amp;quot;dead man's cave&amp;quot;, but we haven't yet got enough information to work out which it is yet... we'll come back to it.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Paired up, pared down, the truth exposed, lies within lies.&lt;br /&gt;&#xA;Heading onwards north, reeling seas terrorize.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This stanza involves word play, which when deciphered provides a clear directive (and was again &lt;a href=&quot;https://puzzling.stackexchange.com/a/14985/5208&quot;&gt;cracked first by rand al'thor&lt;/a&gt;). To clarify the ambiguous wording, it could have been written as &lt;em&gt;&amp;quot;Paired up, [and then] pared down, the truth [will be] exposed, [as it] lies within [the following] lies: 'Heading onwards north reeling seas terrorize'&amp;quot;&lt;/em&gt;. With the apparent instruction to head north towards/away from the sea identified as a lie, we can &amp;quot;pair up&amp;quot; the words in the second line to get &lt;em&gt;&amp;quot;headingonwards northreeling seasterrorize&amp;quot;&lt;/em&gt;, then we pare these down to produce &lt;em&gt;&amp;quot;[headin]&lt;strong&gt;go&lt;/strong&gt;[nwards] [nor]&lt;strong&gt;three&lt;/strong&gt;[ling] [s]&lt;strong&gt;east&lt;/strong&gt;[errorize]&amp;quot;&lt;/em&gt;, and we've exposed the truth: &lt;strong&gt;go three east&lt;/strong&gt;. Although this helps us rule out two starting caves (since we can't go off the edge of the map), we still don't know where to go east from, so we'll look at the next stanza.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Right is left and left is right, turn and go until,&lt;br /&gt;&#xA;Twins appear to mark an end, though you'll continue still.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The first line is simple (and was solved independently several times): You need to turn, such that what's currently on your &amp;quot;right is left [behind]&amp;quot;, meaning that &amp;quot;left is [the] right [way to turn]&amp;quot;, then start travelling forward (i.e. North)... The &amp;quot;twins&amp;quot; refer to multiple duplicated squares within the map itself: &lt;code&gt;SK&lt;/code&gt;-&lt;code&gt;II&lt;/code&gt;, &lt;code&gt;IH&lt;/code&gt;-&lt;code&gt;AC&lt;/code&gt;, &lt;code&gt;NE&lt;/code&gt;-&lt;code&gt;IP&lt;/code&gt; and &lt;code&gt;LP&lt;/code&gt;-&lt;code&gt;NQ&lt;/code&gt;, which when joined, as shown in orange on the map, produce &lt;strong&gt;X&lt;/strong&gt; marks (i.e. X marks the spot... how else would a pirate mark an end?). Note that at this point we can narrow the possibilities of which cave to start at down to either &lt;code&gt;TH&lt;/code&gt; or &lt;code&gt;AE&lt;/code&gt;, as indicated by the two possible paths on the map (this stanza was ultimately &lt;a href=&quot;https://puzzling.stackexchange.com/a/15079/5208&quot;&gt;solved by Zandar&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Cardinals sitting in their nests, pick them up one by one.&lt;br /&gt;&#xA;Set them free and follow them, then turn to what's undone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This stanza is pure letter play and relies in finding the &lt;em&gt;cardinal directions&lt;/em&gt; in &lt;em&gt;&amp;quot;th&lt;strong&gt;e&lt;/strong&gt;ir &lt;strong&gt;nes&lt;/strong&gt;t&lt;strong&gt;s&lt;/strong&gt;&amp;quot;&lt;/em&gt;, giving you a set of directions: &lt;strong&gt;E N E S S&lt;/strong&gt;. After following those, you need to &lt;em&gt;turn&lt;/em&gt; and face the only direction not yet travelled, so we end this stanza facing West (&lt;a href=&quot;https://puzzling.stackexchange.com/a/14718/5208&quot;&gt;tfitzger was first to post a complete solution&lt;/a&gt; to this, though &lt;a href=&quot;https://puzzling.stackexchange.com/a/14880/5208&quot;&gt;Daniel K, solved 99% of it independently too&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A leap of faith, surrender yourself. But when things come to a head,&lt;br /&gt;&#xA;Take the easy way out, with the blues left behind, go three and don't retread.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;From here, we take a &amp;quot;leap of faith&amp;quot; into the river and go with the flow until we reach the &amp;quot;headland&amp;quot;. Exit the river onto the beach (the easy way) and with the blue river/sea tiles to our &amp;quot;left&amp;quot; and &amp;quot;behind&amp;quot;, we go three more squares forward. As &lt;a href=&quot;https://puzzling.stackexchange.com/a/15079/5208&quot;&gt;observed by Zandar&lt;/a&gt;, we can now narrow our path down to a single possibility, since starting from the cave at &lt;code&gt;AE&lt;/code&gt; would result in us &amp;quot;retreading&amp;quot; during this stanza, which we're explicitly told not to do. So we now know for certain that dead man's cave is at &lt;code&gt;TH&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Follow the gaze of the crimson saint as he looks to the pre-dawn glow.&lt;br /&gt;&#xA;The distance the crow flies in to pi, hints how far to go.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;As &lt;a href=&quot;https://puzzling.stackexchange.com/a/15079/5208&quot;&gt;Zandar again correctly observed&lt;/a&gt;, the crimson saint refers to Santa Claus, who lives at the North pole meaning he'll be looking South no matter what he's looking towards. The distance the crow flies involves a little Pythagorus' Theorem applied to the coordinates &lt;code&gt;IN&lt;/code&gt; and &lt;code&gt;PI&lt;/code&gt;, which being 4 rows and 3 columns apart gives us exactly 5. So we head &lt;strong&gt;5 South&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Two one fives, all lined up, this time point the way.&lt;br /&gt;&#xA;They also indicate how far. Go now or be led astray.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/a/15168/5208&quot;&gt;User1618143 correctly observed&lt;/a&gt; that &amp;quot;two one fives&amp;quot; here tells us that &amp;quot;the time&amp;quot; of 15:15 &amp;quot;points the way&amp;quot;. Looking at the hands of a clock, which &amp;quot;line up&amp;quot; at this time to both point in an Easterly direction towards the number 3, meaning we head 3 East. We need to &amp;quot;go now or be led astray&amp;quot;, because if we don't the time will change and the clock hands will point in the wrong direction.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Proceed until, when viewed from on high, the island itself gives a sign.&lt;br /&gt;&#xA;Next steps revealed as the river reflects and the ancient stones align.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Proceeding from where we are at this point, the only way to get somewhere &amp;quot;high&amp;quot;, is to travel due North until climbing the edge of the mountain at &lt;code&gt;YH&lt;/code&gt;. From here we can see the &amp;quot;ancient stones&amp;quot; on the ground near the river. Again, &lt;a href=&quot;https://puzzling.stackexchange.com/a/15168/5208&quot;&gt;user1618143 showed&lt;/a&gt; that if we &amp;quot;reflect&amp;quot; the existing stones, using the &amp;quot;river&amp;quot; as an axis, we end up with a series of arrows, such that connecting them spells out &lt;strong&gt;3E&lt;/strong&gt;, as shown on the map in pink. This tells us that we need to travel 3 East from this point.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Run forth with what you carry. Confusion becomes clear.&lt;br /&gt;&#xA;Now piece together your reward, you'll find it buried here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The final stanza is a two parter. &lt;a href=&quot;https://puzzling.stackexchange.com/a/15063/5208&quot;&gt;Rand al'thor correctly realised&lt;/a&gt; that &amp;quot;confusion becomes clear&amp;quot; hinted towards anagrams for both the final directive and the treasure itself. First we needed to &amp;quot;Run forth&amp;quot; with what we carry, which, as we learned from the very first stanza, is &amp;quot;a sword&amp;quot;. And &lt;em&gt;&amp;quot;run forth a sword&amp;quot;&lt;/em&gt; is an anagram for &lt;em&gt;&amp;quot;four northwards&amp;quot;&lt;/em&gt;, so we go four North from here to find the final location of the treasure at &lt;code&gt;NR&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Now to discover the treasure itself, we &amp;quot;piece together&amp;quot; the coordinates that we arrived at for each stanza: &lt;code&gt;TH TD OD NE BE UE UH YR NR&lt;/code&gt; which is an anagram for &amp;quot;&lt;strong&gt;Three Hundred Bounty&lt;/strong&gt;&amp;quot;, which was again &lt;a href=&quot;https://puzzling.stackexchange.com/a/15063/5208&quot;&gt;discovered, and subsequently awarded to, rand al'thor&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Be warned though treasure seeker, should you make but one wrong turn.&lt;br /&gt;&#xA;The pirate's curse, and certain death, is all that you will earn...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;And so our treasure hunt comes to an end, and we pause to remember those that fell along the way... and to congratulate rand al'thor on his ultimate victory.&lt;/p&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-21T17:28:58.223" CommentCount="6" CommunityOwnedDate="2015-05-21T20:31:56.740" ContentLicense="CC BY-SA 3.0" />
  <row Id="15261" PostTypeId="1" AcceptedAnswerId="15850" CreationDate="2015-05-21T05:31:59.123" Score="7" ViewCount="2664" Body="&lt;p&gt;So remember all those puzzles about 100 or 1000 or however many logicians abducted by aliens and they are going to be killed if they can't figure out their hat color or the number on their back? Well it turns out you were such a bright mind that the aliens abducted you, and luckily you were one of the logicians that survived.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, instead of taking you back home, they just left you in a room with all the other logicians that survived. Escaping doesn't seem very likely any time soon because right now everyone is focusing on getting food every day. There is a single fridge in the corner of the room but it is one of those newfangled fridges and it has a security code that you have to put in to open it. The fridge has a panel on the top that displays 5 digits, and those digits change every day, along with the fridge restocking. The problem is that when the fridge restocks, the security code changes. For the first six days, you've had to brute force the code by just typing every combination in until you get it (only 100000 combinations, not too bad), but you really need to figure out what the pattern is for the codes so you can focus on escaping instead of cheeseburgers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The panel numbers and codes for that day are as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Panel -- Code&lt;br&gt;&#xA;  12345 = 56306&lt;br&gt;&#xA;  04418 = 35129&lt;br&gt;&#xA;  20530 = 65048&lt;br&gt;&#xA;  54472 = 68425&lt;br&gt;&#xA;  00000 = 00000&lt;br&gt;&#xA;  60454 = 95126&lt;br&gt;&#xA;  29830 = 39175&lt;br&gt;&#xA;  00000 = 03334&lt;br&gt;&#xA;  08224 = 35386&lt;br&gt;&#xA;  14412 = ?????&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Today the panel says 14412. Can you figure out the code and what the algorithm is so you can get the food easily each day?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the brute-force method: Pertaining to David James' comment,the brute force method started with 00000 on the fifth day, but it does not follow the pattern of 00001, 00002... One logician is in charge of the brute-forcing and each day he starts with the panel number, and then does what seems like a random sequence. At least that's what you all think... he's been a little bit looney since the abduction but you all let him use his &quot;sequence&quot; to make him happy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yesterday in the fridge, someone found a fortune cookie which contained the following fortune:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Look backwards to go forward&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's probably just a generic fortune mass produced in a cookie factory, but what if it's a clue about solving the code?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another fortune cookie has shown up:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;651&lt;br&gt;930&lt;br&gt;862&lt;br&gt;565&lt;br&gt;000&lt;br&gt;696&lt;br&gt;532&lt;br&gt;400&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What could those numbers be referring to?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More fortune cookies:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;56345&lt;br&gt;35118&lt;br&gt;65030&lt;br&gt;68472&lt;br&gt;00000&lt;br&gt;95154&lt;br&gt;39130&lt;br&gt;03300&lt;br&gt;35324&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Codes to a day need not connect&lt;br&gt;But codes to the panel need be&lt;br&gt;Look how to get panel from code&lt;br&gt;And you'll be able to get free&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Last fortune cookie:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There can be more than one, but each can't be more than once.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Days passed: 9&lt;/strong&gt; (seems like a lot more though)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: Please post partial answers if you think you have anything so I can see where you are having problems&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="11727" LastEditorUserId="11727" LastEditDate="2015-06-02T02:39:15.170" LastActivityDate="2021-03-02T21:03:24.370" Title="Figure out the code" Tags="|pattern|password|number-theory|" AnswerCount="1" CommentCount="34" ContentLicense="CC BY-SA 3.0" />
  <row Id="15262" PostTypeId="1" AcceptedAnswerId="15324" CreationDate="2015-05-21T05:44:34.947" Score="3" ViewCount="407" Body="&lt;h2&gt;Cryptogram #1&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Iq iycsc mqks gs wgqskowi gcieccq qqwocqi qqm agmccq siymocs, iyccc aysi sycckk gc sgaciyoqi ig gc sqom sgc q siymk eyowy mom qgi gcioq eoiy Pkiyqigcqs, qqm eokk qgi cqm eoiy Eoqsicoq, gyi os iyc gkmcsi qqm iyc kgyqicsi gs qkk. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $${\mathfrak {\frac Em=a^2+b^2}}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Answers must include method of decryption and the thinking that led to do certain things to solve the puzzle along with plain text.&lt;/p&gt;&#xA;" OwnerUserId="12058" LastEditorUserId="12058" LastEditDate="2015-05-21T10:33:06.683" LastActivityDate="2015-05-22T04:32:06.063" Title="Quotes surely. Cryptogram #1" Tags="|cipher|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15263" PostTypeId="1" CreationDate="2015-05-21T06:00:54.750" Score="-2" ViewCount="371" Body="&lt;p&gt;Answers &lt;strong&gt;must include method of encryption&lt;/strong&gt; along with plain text and explanation (explanation refers to thoughts or clues that made you think to do certain things).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Cryptogram #2&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;kbemzkbedje de emz drqdelzrebefz dreenfkzre ak bff lmdedjbf nzezbnjm.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Kashmir, $-1/12$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12058" LastEditorUserId="12058" LastEditDate="2015-05-21T10:29:23.803" LastActivityDate="2023-06-19T09:46:25.757" Title="Quotes surely. Cryptogram #2" Tags="|cipher|" AnswerCount="2" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15265" PostTypeId="1" AcceptedAnswerId="15267" CreationDate="2015-05-21T06:26:04.927" Score="3" ViewCount="827" Body="&lt;p&gt;How does one solve the &lt;a href=&quot;http://en.wikipedia.org/wiki/The_Hardest_Logic_Puzzle_Ever&quot; rel=&quot;nofollow noreferrer&quot;&gt;World's hardest puzzle&lt;/a&gt;, and other questions of the type? Is there a proper approach to such a question, or does one just have to be incredibly smart to do so?&lt;/p&gt;&#xA;&lt;p&gt;And here's an &lt;em&gt;even harder&lt;/em&gt; version, where you can illustrate the method used.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;There are 4 gods. 3 of them are Truth, Random and False. Truth always&#xA;speaks the truth. False always lies. Random will give a random answer,&#xA;either true or false. The 4th god has a behaviour exactly the same as&#xA;one of the 1st 3. They are a clone of Truth, Random or False, but you&#xA;don't know which. None of the 4 can be identified, except through&#xA;their responses.&lt;/p&gt;&#xA;&lt;p&gt;How many yes-no questions must you ask, to figure out which character&#xA;is found in 2 gods (which type of god has been cloned)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If you're up for some more....&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The gods answer ja or da, in their own language, and you don't know&#xA;which means yes, and which means no. All gods use the same language.&#xA;How many questions are required to identify the repeated trait?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="31389" LastEditDate="2023-07-18T11:21:23.343" LastActivityDate="2023-07-18T11:21:40.590" Title="How does a common person solve impossibly hard truthteller/falsifier questions?" Tags="|logical-deduction|liars|algorithm|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="15266" PostTypeId="2" ParentId="15265" CreationDate="2015-05-21T07:37:37.517" Score="1" Body="&lt;p&gt;Here's the secret:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; None of these puzzles are difficult. In each puzzle, there are only finitely many possible scenarios (in the most famous two guards and two doors puzzle, there are only four possible scenarios, and in the gods puzzle you describe there are only 72 possible scenarios). Questions you can ask can be classified by which scenarios result in a &quot;yes&quot; answer, which scenarios result in a &quot;no&quot; answer, and which scenarios result in a random answer (or &quot;da&quot;/&quot;ja&quot;/random). That means that the number of functionally different questions you can ask is at most three raised to the power of the number of scenarios (and in practice it is a &lt;em&gt;lot&lt;/em&gt; smaller, since you can take advantage of various symmetries of the problem). Now just check each type of question in order, and see which one works. If you get to ask several questions, then the number of possible strategies increases, but again a brute force approach will eventually get you to a solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For instance, here is the complete idiot's method of solving the standard two guards/two doors puzzle (one lies, one tells the truth, one door to freedom, the other to death, one question):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First we name the scenarios. Let's say FT is the scenario where the left door is freedom and left guard is truth, FL is the scenario where the left door is freedom and left guard is lying, DT means left door is death and left guard is truth, DL means left door is death and left guard is lying. Taking advantage of a symmetry, let's decide from the very beginning that we are going to ask a question to the left guard. Since questions only have yes/no answers, there are really only 16 questions we can ask. We'll classify questions by the collection of scenarios that lead to a &quot;yes&quot; answer from the left guard. First we have: &quot;Are we not in any scenario at all?&quot;, which is answered with a yes in scenarios FL, DL. Second we have: &quot;Are we in scenario FT?&quot;, which is answered with a yes in scenarios FT, FL, DL. Third we have: &quot;Are we in scenario FL?&quot;, which is answered with a yes in scenario DL. Fourth we have: &quot;Are we either in scenario FT or scenario FL?&quot;, which is answered with a yes in scenarios FT, DL. And so on... Our goal is to find a question which has one answer if the left door is freedom and has the opposite answer if the left door is death. Going through the list, we find that there are really only two questions that work: one is &quot;Are we either in scenario FT or scenario DL?&quot;, which is answered with a yes in scenarios FT, FL, and the other is &quot;Are we either in scenario FL or scenario DT?&quot;, which is answered with a yes in scenarios DT, DL. Either one of these questions solves the puzzle, and can be rephrased in other ways to make our thought process less obvious (e.g. &quot;If I asked you whether this door lead to freedom, would you say yes?&quot; is really the same as &quot;Are we either in scenario FT or scenario DL?&quot;, but it sounds much cleverer).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In questions where there are no random answers and we are looking for one piece of information with one question, there is a slightly cleverer method:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Say there are N possible scenarios, numbered 1, ..., N, and we want to know if statement X is true. In each scenario, write f(i) for the answer you would receive in response to the question &quot;Is X true?&quot; in scenario i, and write g(i) for whether X is actually true in scenario i. Suppose, for example, that N is 4, f(1) is &quot;da&quot;, f(2) is &quot;ja&quot;, f(3) is &quot;ja&quot;, f(4) is &quot;da&quot;, g(1) is true, g(2) is true, g(3) is true, and g(4) is false. Let's say we want to come up with a question where if we hear the answer &quot;ja&quot; it will mean that X is true. Then we make the question: &quot;Is it true that we are either in scenario 2 or in scenario 3 or in scenario 4?&quot;, where we include exactly scenarios i where either f(i) is &quot;ja&quot; and g(i) is true, or f(i) is &quot;da&quot; and g(i) is false.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When there are random answers, it is a bit trickier.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; From the point of view of information theory, when there are no random answers one well-chosen question should always get you exactly one bit of useful information. In the presence of random answers, you have a chance of getting a useless answer to your question, so you get slightly less than one bit per question. By phrasing questions correctly, a truth-teller and a liar become equally useful to you, so the challenge is to ask a few questions at the beginning in order to identify someone who is not a random answerer so that you can ask that person the last few questions and extract the information you are after. In any situation where there are at least as many random answerers as truthtellers/liars, you have little hope, since the random answerers could &quot;randomly&quot; end up acting exactly as if the roles were reassigned with them as all of the truthtellers/liars and the real truthtellers/liars as random answerers (of course in the puzzle you describe, in the bad case where there are as many randoms as truthtellers/liars the answer is just that there are two random answerers, so it is still solvable). If there are more truthtellers/liars than random answerers then you know that with enough questions you can manage things by, essentially, taking the majority opinion (and phrasing questions appropriately to make liars behave like truthtellers). Then you just brute force your way to the most efficient (and cleverest sounding) strategy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit: How I would attack your puzzle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Obviously it would take a while to do a pure brute force, so let's try being at least a little clever. First let's try the variant where everyone either tells the truth or is random, and we want to know whether there are one or two randoms. If we ask our first two questions to the two randoms, they can give any possible answers, so we need at least three questions. In fact, if we only ask three questions to two randoms and one truthteller, then the two randoms could pretend that they are truthtellers, that the real truthteller we asked is a random, and that the remaining god is a truthteller, fooling us. So we need at least four questions, and we should talk to all four people. If we ask everyone whether there are one or two randoms, then in the case of one random god we hear &quot;one&quot;, &quot;one&quot;, &quot;one&quot;, and something random, in some order, while in the case of two random gods we hear &quot;two&quot;, &quot;two&quot;, and two random answers, in some order, and we can tell which scenario we are in regardless of how the random answers come out. So we go back to the real problem. We can make the liars act like truthtellers by asking questions like, &quot;Is it true that either you are a truthteller and there is one random or you are a liar and there are two randoms?&quot;, so with four questions we can determine whether there are two randoms. If there is only one random, we can determine how many liars there are by taking majority opinion on a question like, &quot;Is it true that either you are a truthteller and there is one liar or you are a liar and there are two liars?&quot;, so eight questions is enough to get by. Eight is a lot, probably we can do better. When we are polling for the number of randoms, if we hear two people confirm that there are two randoms we can stop right away, and if we hear three people confirm that there is only one random we can move on to determining who the liar is right away. If we've determined that there is only one random, we know that anyone who acted like there were two randoms is a random trying to fool us and we can ask someone else a single question to determine whether there are one or two liars (five questions used up in this case). If the lone random plays along and acts like there is only one random, then we only spend three questions determining that there is only one random, but we don't know who the random is yet, and a majority opinion from any three gods tells us whether there are one or two liars (six questions used in this case). So we can get it down to six questions, but it still feels a little wasteful. This solution is getting a little long, so I'll leave it at that for now...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11295" LastEditorUserId="11295" LastEditDate="2015-05-21T08:16:16.070" LastActivityDate="2015-05-21T08:16:16.070" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="15267" PostTypeId="2" ParentId="15265" CreationDate="2015-05-21T07:55:51.297" Score="1" Body="&lt;p&gt;I'll assume you don't just want the solutions to these puzzles which has already been answered on this site in the case of &amp;quot;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/2665/knights-and-knaves-in-a-foreign-language&quot;&gt;The Hardest Logic Puzzle Ever&lt;/a&gt;&amp;quot;, and you are looking for some more general advice for these sorts of puzzles.&lt;/p&gt;&#xA;&lt;p&gt;The usual trick to solving these problems is to craft your questions in a way that gets the same information back regardless of who is asked.&lt;/p&gt;&#xA;&lt;p&gt;For example if you want to know the answer to some question X you could ask:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;What would that person say if asked X?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If you ask this question to a truth teller while pointing at a liar, you will get the liars answer. If you ask this to a liar while pointing at a truth teller you will also get the liars answer.&lt;/p&gt;&#xA;&lt;p&gt;If you just asked X you would get back two different answers and be none the wiser. By asking about how another person would answer you can get a consistent answer that can be reversed to know the truth.&lt;/p&gt;&#xA;&lt;p&gt;Another common trick is to ask questions to provide more than one piece of information such as for your even harder problem:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If I asked both of you X would you both give the same answer?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If asked to a liar about themselves and the truth teller they will answer with the lie, yes.&lt;/p&gt;&#xA;&lt;p&gt;If asked to a truth teller about themselves and a liar they will answer truthfully no.&lt;/p&gt;&#xA;&lt;p&gt;With this single question you have discovered the identities of two people with only one question. Of course if there were only two people to start with this isn't really any better that finding out about one of them and then inferring about the other. Note that this particular question doesn't work so well with a random answerer involved. Have a think about how this question would play out with different combinations of respondents.&lt;/p&gt;&#xA;&lt;p&gt;The random answerer can be a bit problematic. If you ask a truth teller how will that person answer when pointing at Random what are they supposed to say? Do they have prior knowledge of all future random choices? If it is not explicitly stated in the puzzle how these situations are supposed to work in can make your problem even more difficult.&lt;/p&gt;&#xA;&lt;p&gt;You probably don't have to be incredibly smart to solve these types of puzzles, but being reasonably smart and knowing these sorts of question strategies really helps.&lt;/p&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="31389" LastEditDate="2023-07-18T11:21:40.590" LastActivityDate="2023-07-18T11:21:40.590" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15268" PostTypeId="2" ParentId="14741" CreationDate="2015-05-21T07:56:28.270" Score="1" Body="&lt;p&gt;partial answer - maybe you can help me getting the missing pieces :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;building the word ring&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ??? and fruit&#xA;&lt;br&gt;&#xA; letter and computer -&gt; email&#xA;&lt;br&gt;&#xA; number and subject &#xA;&lt;br&gt;&#xA; consume and before -&gt; expiration date&#xA;&lt;br&gt;&#xA; cut and cry -&gt; onions&#xA;&lt;br&gt;&#xA; slide and gentle -&gt; adults only ;-)&#xA;&lt;br&gt;&#xA; awake and not -&gt; sleeping&#xA;&lt;br&gt;&#xA; rain and cold -&gt; november&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Suggestions for the color:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; black or blue&#xA;&lt;br&gt;&#xA; because, if it is connected with the fruit to build a word, I am thinking of Blueberry or Blackberry (this may also fit with the email, if you think of the Blackberry phones)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12328" LastActivityDate="2015-05-21T07:56:28.270" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15269" PostTypeId="1" AcceptedAnswerId="15270" CreationDate="2015-05-21T08:26:00.073" Score="3" ViewCount="1063" Body="&lt;p&gt;You are blind and sitting at a table with 100 coins on it. You know exactly 50 coins are showing heads and the rest tails, and you can tell by feel where the coins are, but you are unable to see or feel whether a particular coin is showing heads or tails.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your task is to divide the coins on the table into two sets each of which has the same number of heads and the same number of tails as the other. How can you do it?&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-23T00:04:13.113" Title="Coin division puzzle" Tags="|logical-deduction|strategy|" AnswerCount="2" CommentCount="2" ClosedDate="2015-05-21T13:33:59.093" ContentLicense="CC BY-SA 3.0" />
  <row Id="15270" PostTypeId="2" ParentId="15269" CreationDate="2015-05-21T08:34:36.860" Score="10" Body="&lt;p&gt;Since you can feel coins....&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;Divide coins into two sets 50-50&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;Flip all the coins of one set.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So you have exactly same number of H and T coins in both sets&lt;/p&gt;&#xA;&lt;p&gt;Thanks @Albert Masclans for explanation....&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;I actually solved it for less number lets say 14 coins&lt;/p&gt;&#xA;&lt;p&gt;so lets say we have H T H | H T H T&lt;/p&gt;&#xA;&lt;p&gt;and                          T H H H | T T T&lt;/p&gt;&#xA;&lt;p&gt;above are random arrangements&lt;/p&gt;&#xA;&lt;p&gt;Now lets divide them along a line | shown above (in anyways you can divide them)&lt;/p&gt;&#xA;&lt;p&gt;now flip first (left set)&lt;/p&gt;&#xA;&lt;p&gt;will look like&lt;/p&gt;&#xA;&lt;p&gt;T H T|H T H T&lt;/p&gt;&#xA;&lt;p&gt;H T T T|T T T&lt;/p&gt;&#xA;&lt;p&gt;Now we have 2 H and 5 T in both sets&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10433" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-21T12:33:28.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15271" PostTypeId="2" ParentId="15034" CreationDate="2015-05-21T08:52:58.920" Score="1" Body="&lt;p&gt;My answer may be wrong,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your clue, when tried to read out, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;sounds like I AM SS.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;From the five words on the right side of the arrow, only one has SS, i.e. SOVSS.&#xA;&lt;br&gt;Searching for SOVSS led me to A. Rodriguez, he's from New York, or NYCITY.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the clue,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SS is written as ESES - so EIHSA caught my eye from the five words on the right side of the arrow. First search result for EIHSA had Argentina in it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So my answer, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; NYCITY to ARGENTINA.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12619" LastEditorUserId="12193" LastEditDate="2015-05-21T09:04:57.663" LastActivityDate="2015-05-21T09:04:57.663" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15272" PostTypeId="2" ParentId="15269" CreationDate="2015-05-21T09:02:53.680" Score="7" Body="&lt;p&gt;The solution &lt;a href=&quot;https://puzzling.stackexchange.com/users/10433/user2408578&quot;&gt;user2408578&lt;/a&gt; is indeed correct. Allow me to illustrate it somewhat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As said...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Divide coins into two sets 50-50.&lt;/em&gt; You'll have now $t$ Tails in the first set and $50-t$ Heads. And on the second set you'll have $50-t$ Tails (since, there are $50$ Tails in total) and $50-(50-t)=t$ Heads.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Given that, it's clear that you have to...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Flip one of the set's all coins.&lt;/em&gt; Since doing that will mean turning all Tails into Heads. So flipping all coins in the second set would mean having $t$ Tails and $50-t$ Heads, as in the first set.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12165" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-05-23T00:04:13.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15275" PostTypeId="1" AcceptedAnswerId="15277" CreationDate="2015-05-21T11:08:33.973" Score="12" ViewCount="927" Body="&lt;p&gt;&lt;em&gt;&lt;strong&gt;Hopefully this puzzle is not as easy as the last one.&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Jack Sparrow wakes up startled just when he was deep in slumber, after his night out with the &amp;quot;Ladies&amp;quot;.&#xA;He checks his pocket and finds it empty. No Compass!!!&#xA;He springs up from the bed and notices some parchments at the corner of the bed. On top of the pile is a letter which says,&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ykZBT.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;A panicked Jack, walks randomly across the small dingy room, and then tries to regain his senses. He looks through the pile&#xA;of parchment and finds a map showing the islands near &amp;quot;Death's Gate&amp;quot;:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Rkl5c.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Jack thinks&lt;br /&gt;&#xA;&lt;em&gt;&lt;strong&gt;He must want me to go to the Death's Gate. Might be a trap. But do I have a choice. And even if I go there, where to find the compass.&#xA;It's not a trivial matter to look for a pin in those islands of haysack ... or haysack of islands. Not sure which one.&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;He looks at another parchment,&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lDQY2.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;code&gt;3m1scTKxAeaHWIYSMmO0yAiTutiRtuWuzqLWktyAyX3r5zfCiUf3OXPvdHRRlKM5XXIqSAYnP5KE5fxb9R871VwhAOnTCsA3f8DElwmWKwyPd5g+G/MNG4PyOcRRCpCGSc6c/H5stdF4R7eu5WYgbg1BD0CTkAtD/IkLOuL/GFU=&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&amp;quot;&lt;em&gt;&lt;strong&gt;Does he take me for a fool, I am Captain Jack Sparrow.&lt;/strong&gt;&lt;/em&gt;&amp;quot;&lt;br /&gt;&#xA;He thought. He stared at it, turned it upside down, shook it as if the compass would fall out of it at any moment. No Luck.&lt;/p&gt;&#xA;&lt;p&gt;He takes another parchment and sees something that looks like some kind of prose.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BOcUR.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Be wary of the seas&lt;br /&gt;&#xA;and Be wary of the height&lt;br /&gt;&#xA;What eagle Carries in it's Beak&lt;br /&gt;&#xA;you will need it outright.&lt;br /&gt;&#xA;You'll follow me fine,&lt;br /&gt;&#xA;if you Clear your sight.&lt;br /&gt;&#xA;Take the C's from the left&lt;br /&gt;&#xA;and then B's from the right&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;He mutters,&lt;br /&gt;&#xA;&amp;quot;&lt;em&gt;&lt;strong&gt;I can certainly write better than that. The fine Ladies from the Port of Goron will tell you&lt;/strong&gt;&lt;/em&gt;&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;He moves onto the next piece.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mMcph.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Ser, you might be perplexed,&lt;br /&gt;&#xA;and wonder where I went.&lt;br /&gt;&#xA;But you will give up,&lt;br /&gt;&#xA;once your will is spent.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&amp;quot;&lt;em&gt;&lt;strong&gt;Aha, so you are a poet my friend, though a pretty bad one I must add. Then you must lack courage and valor. I on the other hand, am Captain Jack Sparrow&lt;/strong&gt;&lt;/em&gt;&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;Even though it's quite late, he goes downstairs to get a drink. A bottle of Rum and a warm fire, he takes a chair and goes deep in thought. Suddenly, he opens his eyes and runs outside&#xA;looking for Gibbs. He finds him near his ship having a Whisky.&lt;/p&gt;&#xA;&lt;p&gt;Gibbs&lt;br /&gt;&#xA;&lt;em&gt;&lt;strong&gt;&amp;quot;What is it Jack&amp;quot;&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;Jack hands him a piece of parchment and says,&lt;br /&gt;&#xA;&amp;quot;&lt;em&gt;&lt;strong&gt;Mr. Gibbs would you please read that aloud for me&lt;/strong&gt;&lt;/em&gt;&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;Gibbs takes the piece of parchment and reads:&lt;br /&gt;&#xA;&lt;em&gt;&lt;strong&gt;Ser, you might be perplexed, and wonder where I went. But you will give up, once your will is spent&amp;quot;&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;Jack thinks and reads from the other parchment:&lt;br /&gt;&#xA;&lt;em&gt;&lt;strong&gt;Be wary of the seas,&lt;br /&gt;&#xA;and Be wary of the height.&lt;br /&gt;&#xA;What eagle Carries in it's Beak&lt;br /&gt;&#xA;you will need it outright.&lt;br /&gt;&#xA;You'll follow me fine,&lt;br /&gt;&#xA;if you Clear your sight.&lt;br /&gt;&#xA;Take the C's from the left,&lt;br /&gt;&#xA;and then B's from the right.&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;He smiles and adds&lt;br /&gt;&#xA;&lt;em&gt;&lt;strong&gt;Now go fetch me some rum&lt;br /&gt;&#xA;as we set sail tonight&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;Gibbs continues where Jack left off:&lt;br /&gt;&#xA;&lt;em&gt;&lt;strong&gt;Do you also want me to get something,&lt;br /&gt;&#xA;to whip you Blue and White&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;Jack Sparrow smiles&lt;br /&gt;&#xA;&lt;em&gt;&lt;strong&gt;Mr Gibbs, You don't understand. Someone stole the compass and I might just know how to get it back. We set sail now.&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;And Mr. Gibbs, bring KeyserSoze along, we need him to get to the compass&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;Gibbs,&lt;br /&gt;&#xA;&lt;em&gt;&lt;strong&gt;Where we headed?&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;Jack,&lt;br /&gt;&#xA;&lt;em&gt;&lt;strong&gt;We'll know by tomorrow.&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;Next day, Jack wakes up late in the afternoon. Still wondering about the coded message. Something still didn't make sense. Wondering about all he had found, going through every parchment again and again. It happened while he was reading the letter. Specially the line &lt;em&gt;&lt;strong&gt;Once is not enough&lt;/strong&gt;&lt;/em&gt;. He look at the coded message again and&lt;br /&gt;&#xA;&lt;em&gt;&lt;strong&gt;Aha, this is what I have to do.&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;He picked up the map and held it towards the sun, then looked the message on the parchment he just decoded and yelled,&lt;br /&gt;&#xA;&lt;em&gt;&lt;strong&gt;&amp;quot;Mr. Gibbs, we are headed to ________&amp;quot;&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Where is Jack headed?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="12537" LastEditorUserId="69582" LastEditDate="2023-12-15T16:15:45.417" LastActivityDate="2023-12-15T16:15:45.417" Title="Jack Sparrow and The quest for the missing Compass . Part 1" Tags="|logical-deduction|cipher|word-problem|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15276" PostTypeId="2" ParentId="15262" CreationDate="2015-05-21T11:27:53.070" Score="3" Body="&lt;p&gt;I believe the quote is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In these days of conflict between ancient and modern studies, there must surely be something to be said for a study which did not begin with Pythagoras and will not end with Einstein, but is the oldest and youngest of all.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your hint admittedly makes it pretty simple: $a^2 + b^2 = c^2$ and $\frac {E} {m} = c^2$ are of course, the two most famous equations to come out of Pythagoras and Einstein. &lt;code&gt;Pkiyqigcqs&lt;/code&gt; becomes &lt;code&gt;Pythagoras&lt;/code&gt; and &lt;code&gt;Eoqsicoq&lt;/code&gt; becomes &lt;code&gt;Einstein&lt;/code&gt;, and you can map some known letters as so:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a       f       k - y p       u     z&#xA;b       g - o   l     q - n/a v&#xA;c - e/r h       m     r       w&#xA;d       i - t/g n     s - s   x&#xA;e       j       o - i t       y - h&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Each of the pairs you see are just a ROT13 pair, so any letters that you find, you can also find their equivalent. That further develops the tableau. I think the way to fill out the rest of it slipped me by due to personally recognizing the quote.&lt;/p&gt;&#xA;" OwnerUserId="12542" LastEditorUserId="12542" LastEditDate="2015-05-21T12:03:24.273" LastActivityDate="2015-05-21T12:03:24.273" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15277" PostTypeId="2" ParentId="15275" CreationDate="2015-05-21T12:14:00.843" Score="7" Body="&lt;p&gt;Mr. Gibbs, we are headed to&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The cave at the top of Oholu's Heart&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Step 1:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Gibbs takes the piece of parchment and reads:&lt;br&gt; Ser, you might be perplexed, and wonder where I went. But you will give up, once your will is spent&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;br&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Jack thinks and reads from the other parchment:&lt;br&gt; Be wary of the seas,&lt;br&gt;&#xA;and Be wary of the height.&lt;br&gt;&#xA;What eagle Carries in it's Beak&lt;br&gt;&#xA;you will need it outright.&lt;br&gt;&#xA;You'll follow me fine,&lt;br&gt;&#xA;if you Clear your sight.&lt;br&gt;&#xA;Take the C's from the left,&lt;br&gt;&#xA;and then B's from the right.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;There are three Cs and four Bs in the second group of text.  Take the three characters from the left of the first text, &amp;quot;Ser&amp;quot;, and the last four characters, &amp;quot;pent&amp;quot;, yielding &amp;quot;Serpent&amp;quot;.&lt;br&gt;&lt;/p&gt;&#xA;&lt;p&gt;Step 2:&lt;/p&gt;&#xA;&lt;p&gt;Serpent is a form of encryption, I have learned.  I went here &lt;a href=&quot;http://www.tools4noobs.com/online_tools/decrypt/&quot; rel=&quot;noreferrer&quot;&gt;http://www.tools4noobs.com/online_tools/decrypt/&lt;/a&gt; and dropped in the text from the code&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;3m1scTKxAeaHWIYSMmO0yAiTutiRtuWuzqLWktyAyX3r5zfCiUf3OXPvdHRRlKM5XXIqSAYnP5KE5fxb9R871VwhAOnTCsA3f8DElwmWKwyPd5g+G/MNG4PyOcRRCpCGSc6c/H5stdF4R7eu5WYgbg1BD0CTkAtD/IkLOuL/GFU=&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I used the key KeyzerSoze, Algorithm Serpent and mode CBC (thanks @George).  This yielded&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;bB5pRsZ2axPhNZ/BkqAnsd7x/ixnRaFWGd+HcuM0c+f73QBBMt4EnWCWRpmHHoax654hWiQnB+bO0OH1czTiCUN+B2Y6grMHm3FEaMCz95EOqqGa6f7dLvHJfBGOgi3S&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I put that code in where we had the previous code generated this&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;a href=&quot;https://www.google.co.in/?gfe_rd=cr&amp;amp;ei=fG1dVZyGFMqCoAOc54HoCg#q=image+encryption+online+mozaiq&quot; rel=&quot;noreferrer&quot;&gt;https://www.google.co.in/?gfe_rd=cr&amp;amp;ei=fG1dVZyGFMqCoAOc54HoCg#q=image+encryption+online+mozaiq&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Some of that gibberish might have taken me to the right page, but it didn't direct me to the correct site.  Understanding Google though, I went here:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;a href=&quot;http://mozaiq.org/decrypt&quot; rel=&quot;noreferrer&quot;&gt;http://mozaiq.org/decrypt&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;After saving this:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Rkl5c.png&quot; alt=&quot;this image&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;to my desktop I uploaded it.  I didn't have a password, and luckily I didn't need it. ;)&lt;/p&gt;&#xA;" OwnerUserId="1777" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-21T18:04:43.120" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="15278" PostTypeId="1" AcceptedAnswerId="15306" CreationDate="2015-05-21T12:41:26.930" Score="5" ViewCount="1022" Body="&lt;p&gt;Welcome to the final stage of Quark's encrypted image puzzle series. I don't really like reusing old tricks so this puzzle will NOT require steganography tricks. There's not much else I can do to a .png to hide info anyway. (If there was, I'd add a single useful yet not necessary hint)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You are walking along the street (aimlessly bored as always) when a garbage truck stops by. You notice there's an interesting piece of trash stuck on the side. Upon closer inspection, you notice it could be a puzzle of some sort. The garbage collector notices you and says &quot;yeah, some new hire in our recent program made those fliers, nobody's figured out what they mean except this one old guy who marked on it with a red marker&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you figure out what the name of the program is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CJIsu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Notes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You only need the information in the picture to solve this puzzle. It could even be printed out and stuck on a garbage truck, you'd have enough information to solve this. That doesn't mean it's easy. Partial answers are allowed but only the final answer gets the correct mark so post at your own risk. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. Anegten is the company sponsor of the program, so it is not the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good Luck&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Context: (accessed by 2012rcampion finding a hint)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(taken out of comment)&#xA;@2012rcampion I kind of hinted that there was a hint in the png. Okay here's some context (sort of a mini hint): the puzzle was originally a computer puzzle sent to fellow garbage collectors but the old man saw it could be made into a physical puzzle if he made those markings. He added two extra hints as well but left one out. The original puzzle would have been too hard/impossible without the markings so the puzzle author kept it in the fliers he made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: This is an original puzzle, the context above is just for the story.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'll add a hint early since I alluded to it in the comments. The chess position is actually important, and bringing in a chess expert could likely help. From the chess position, you should get a phrase of two words (relatively) famous in chess.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="7722" LastEditDate="2015-05-21T16:19:50.880" LastActivityDate="2015-05-21T19:52:10.857" Title="Find the secret phrase in this encrypted image (Part 4)" Tags="|lateral-thinking|visual|chess|" AnswerCount="1" CommentCount="13" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15279" PostTypeId="2" ParentId="15130" CreationDate="2015-05-21T13:26:34.090" Score="1" Body="&lt;p&gt;Similar to leoll2's answer, but with a different explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the first 1 and 2 points places can &lt;b&gt;always&lt;/b&gt; be on the same half since they would define a line itself. So there is a 100% chance that 2 points can be on the same half.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we added a third point things get slightly more interesting. Here the third point can be on a part where we would not be able to divide the rectangle in half (the red area in picture). There is a 1/4 probability for this, or rather 3/4 that it is on the same half. (If the points are the closest to the lower right corner) In the extreme case where the points are on the line, there is a 100% chance the third dot is the same half.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/8UPyn.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;br&gt;The red dots are in the most extreme position(s).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now assuming that the third dot is on the same half, there is are two locations, the blue area and the green area. If it was in the blue area, then again for the 4th dot, there would be a 3/4 chance of being on the same half. [green + blue area]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the third dot was in the green area, then there is only 1/2 chance of the 4th dot being on the same half.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So green has (2/4) [(1/8) * 4] for the third point, and (1/2) for the fourth point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And blue has 2/8 [(1/8) * 2] for the third point, and then 3/4 for the fourth.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Expanded we get &lt;b&gt;(2/4)*(1/2)+[(1/8)*(3/4)+(1/8)*(3/4)] = 0.5&lt;/b&gt;&#xA;&lt;br&gt; So there is a 50 percent chance that the 4 points are on the same half.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I realize now that it can be dealt with in quarters, but I didn't feel like changing the picture again.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-05-21T15:22:16.160" LastActivityDate="2015-05-21T15:22:16.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15280" PostTypeId="1" AcceptedAnswerId="15282" CreationDate="2015-05-21T13:35:44.503" Score="4" ViewCount="6460" Body="&lt;p&gt;Sudoku friends, this one's for you. This maths puzzle was set for eight-year-old students in Bao Loc, Vietnam, and they have actually been able to solve the puzzle.&#xA;Even some people with doctorate in mathematics have been stumped. Nobody in France nor Belgium has been able to solve this to date. Let &lt;em&gt;us&lt;/em&gt; try.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You just need to fill in each of the digits from &lt;strong&gt;1 to 9&lt;/strong&gt; so that the grid makes sense. It is unknown if we can use the digits once only or as much as necessary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TQBoo.png&quot; alt=&quot;Puzzle&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can &lt;em&gt;you&lt;/em&gt; solve it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I spent several minutes trying to find a common or hidden pattern, and eventually noticed that in the third row of the grid the numbers are in descending order: &lt;code&gt;13, 12, 11, 10&lt;/code&gt;. Now, their sum is &lt;code&gt;46&lt;/code&gt;, and the only number written in the first row is &lt;code&gt;66&lt;/code&gt;. Would a similar pattern to &lt;code&gt;66, 46, 26&lt;/code&gt;... help in something?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, someone offered this comment:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\bigcirc+13\times\bigcirc:\bigcirc+\bigcirc+12\times\bigcirc-\bigcirc-11+\bigcirc\times\bigcirc:\bigcirc-10=66$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$X+(13\times1:1)+Y+(12\times1)-Z-11+(1\times1:1)-10=66$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$X+Y-Z+13+12-11+1-10=66$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$X+Y-Z-5=66$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$X+Y-Z=71$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(P.S. Please do not consider the $\bigcirc$ as a binary operation symbol in the equation above, but as a blank space to represent the empty squares in the grid.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like it can help with something, but this would perhaps work if there were not the 1 to 9 number limit, as $X+Y-Z$ cannot result to $71$ (being replace with digits from 1 to 9).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.7sur7.be/7s7/fr/31482/Le-meilleur-du-web/article/detail/2331599/2015/05/21/Saurez-vous-resoudre-ce-puzzle-mathematique-donne-aux-Vietnamiens-de-8-ans.dhtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9198" LastEditorUserId="11876" LastEditDate="2015-09-23T17:04:40.100" LastActivityDate="2016-10-04T15:02:55.140" Title="Solving This Vietnamese Mathematical Puzzle for 8-Year-Old Children" Tags="|mathematics|calculation-puzzle|" AnswerCount="5" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="15281" PostTypeId="2" ParentId="15280" CreationDate="2015-05-21T13:55:27.883" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 2 6 4 7 8 3 5 9&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An article with this problem in was passed around my office yesterday&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/udmzE.png&quot; alt=&quot;solution&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Image thanks to Engineer Toast&lt;/p&gt;&#xA;" OwnerUserId="9678" LastEditorUserId="9678" LastEditDate="2015-05-21T16:11:15.853" LastActivityDate="2015-05-21T16:11:15.853" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15282" PostTypeId="2" ParentId="15280" CreationDate="2015-05-21T13:56:14.580" Score="5" Body="&lt;p&gt;My answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;6 9 3 5 2 1 7 8 4&lt;/strong&gt;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/0KGUQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I follow the regular order of operations (first multiplications, then sums).  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $6+13\times9/3+5+12\times2-1-11+7\times8/4-10=$&lt;br&gt;&#xA; $6+39+5+24-1-11+14-10=$&lt;br&gt;&#xA; $66$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Few considerations:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that the result of any operation is integer (8 year old children usually don't know reals, do they?), A+13*B/C implies that C divides (13*B), so C divides B. This excludes many solutions, leaving us with (9,1),(9,3),(8,4),(8,2),(8,1),(7,1),(6,1),(6,2),(6,3),(5,1),(4,1),(4,2),(3,1),(2,1) instead of 72 couples for B and C!&lt;br&gt;&#xA;Anyway, there are many different valid combinations that fit this problem, I could write a program that does the heavy work for us, but what would the point of it? 8 year old children don't have C++! Or maybe they have, since they're asians (joking, of course)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Additional answers:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;5,2,1,3,4,7,9,8,6&lt;/strong&gt;&lt;br&gt;&#xA; &lt;strong&gt;5,9,3,6,2,1,7,8,4&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-05-21T14:22:47.040" LastActivityDate="2015-05-21T14:22:47.040" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15283" PostTypeId="2" ParentId="15187" CreationDate="2015-05-21T14:33:01.600" Score="12" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Recorded Music&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The initial sampling rate for CDs was 44kHz. Human hearing can pick up up to 20kHz, so the real necessary sampling rate (intrinsic) is 40kHz by the Nyquist theorem. DVDs and Blurays use a sample rate of 96kHz, but that's not the highest it can go.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As in for the second part, which is much more ambiguous&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The loudness of breathing is roughly 20-30 dB. Talking is about 60 dB. Most music begins around the range of 60-70 dB in the low range. 150 dB is the rough peak of Rock music. 160 dB will physically rip your eardrums appart. So it depends on what song was recorded.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sources:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.gcaudio.com/resources/howtos/loudness.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.gcaudio.com/resources/howtos/loudness.html&lt;/a&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Sound_pressure#Examples_of_sound_pressure&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Sound_pressure#Examples_of_sound_pressure&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5380" LastEditorUserId="10781" LastEditDate="2015-07-17T12:04:59.230" LastActivityDate="2015-07-17T12:04:59.230" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15284" PostTypeId="2" ParentId="15280" CreationDate="2015-05-21T14:39:10.683" Score="4" Body="&lt;p&gt;Time for the brute force computing brigade to pick up the remaining solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;each number must be used exactly once&lt;/li&gt;&#xA;&lt;li&gt;division must result in a whole number&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If we drop either of these assumptions, then this answer would be about 100 times longer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If order of operations is respected, according to &lt;a href=&quot;http://pastebin.com/mpQZ09Bs&quot; rel=&quot;noreferrer&quot;&gt;my script&lt;/a&gt; the possible answers are&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;3, 2, 1, 5, 4, 7, 8, 9, 6&#xA;3, 2, 1, 5, 4, 7, 9, 8, 6&#xA;5, 2, 1, 3, 4, 7, 8, 9, 6&#xA;5, 2, 1, 3, 4, 7, 9, 8, 6&#xA;5, 3, 1, 7, 2, 6, 8, 9, 4&#xA;5, 3, 1, 7, 2, 6, 9, 8, 4&#xA;5, 4, 1, 9, 2, 7, 3, 8, 6&#xA;5, 4, 1, 9, 2, 7, 8, 3, 6&#xA;5, 9, 3, 6, 2, 1, 7, 8, 4&#xA;5, 9, 3, 6, 2, 1, 8, 7, 4&#xA;6, 3, 1, 9, 2, 5, 7, 8, 4&#xA;6, 3, 1, 9, 2, 5, 8, 7, 4&#xA;6, 9, 3, 5, 2, 1, 7, 8, 4&#xA;6, 9, 3, 5, 2, 1, 8, 7, 4&#xA;7, 3, 1, 5, 2, 6, 8, 9, 4&#xA;7, 3, 1, 5, 2, 6, 9, 8, 4&#xA;9, 3, 1, 6, 2, 5, 7, 8, 4&#xA;9, 3, 1, 6, 2, 5, 8, 7, 4&#xA;9, 4, 1, 5, 2, 7, 3, 8, 6&#xA;9, 4, 1, 5, 2, 7, 8, 3, 6&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If the equation is evaluated left-to-right, ignoring order of operations, the possible answers are&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1, 2, 4, 7, 5, 8, 3, 6, 9&#xA;1, 2, 7, 5, 3, 4, 9, 8, 6&#xA;1, 2, 7, 5, 8, 9, 4, 3, 6&#xA;1, 3, 7, 5, 2, 6, 9, 8, 4&#xA;1, 4, 2, 8, 5, 7, 6, 3, 9&#xA;1, 4, 7, 5, 2, 9, 8, 6, 3&#xA;1, 4, 7, 5, 3, 9, 2, 8, 6&#xA;1, 5, 7, 4, 9, 3, 8, 2, 6&#xA;1, 6, 7, 8, 2, 5, 9, 4, 3&#xA;1, 8, 4, 7, 5, 2, 6, 3, 9&#xA;1, 8, 7, 2, 6, 3, 5, 4, 9&#xA;1, 9, 3, 7, 2, 5, 8, 4, 6&#xA;1, 9, 6, 2, 3, 7, 8, 4, 5&#xA;1, 9, 7, 2, 5, 3, 6, 4, 8&#xA;1, 9, 7, 3, 5, 8, 6, 2, 4&#xA;1, 9, 7, 5, 2, 8, 6, 4, 3&#xA;2, 1, 3, 6, 8, 7, 5, 4, 9&#xA;2, 1, 5, 3, 7, 9, 8, 4, 6&#xA;2, 1, 5, 3, 9, 6, 7, 4, 8&#xA;2, 1, 5, 8, 6, 3, 9, 4, 7&#xA;2, 3, 5, 6, 7, 8, 1, 4, 9&#xA;2, 6, 3, 5, 1, 7, 9, 8, 4&#xA;2, 6, 3, 9, 1, 7, 5, 8, 4&#xA;2, 6, 5, 4, 7, 8, 9, 1, 3&#xA;2, 6, 9, 1, 7, 3, 5, 4, 8&#xA;2, 7, 3, 5, 1, 9, 6, 8, 4&#xA;2, 7, 3, 6, 1, 9, 5, 8, 4&#xA;2, 8, 4, 5, 1, 7, 9, 6, 3&#xA;2, 8, 4, 9, 1, 7, 5, 6, 3&#xA;2, 9, 5, 6, 1, 3, 7, 8, 4&#xA;2, 9, 5, 7, 1, 3, 6, 8, 4&#xA;3, 1, 4, 8, 5, 2, 7, 6, 9&#xA;3, 1, 4, 9, 8, 6, 7, 2, 5&#xA;3, 4, 8, 7, 9, 5, 1, 2, 6&#xA;3, 5, 2, 7, 8, 9, 4, 1, 6&#xA;3, 5, 4, 1, 2, 7, 9, 8, 6&#xA;3, 6, 4, 8, 9, 7, 2, 1, 5&#xA;3, 6, 8, 7, 5, 1, 9, 2, 4&#xA;3, 7, 2, 4, 1, 9, 5, 8, 6&#xA;3, 7, 2, 5, 1, 9, 4, 8, 6&#xA;3, 7, 8, 5, 4, 1, 2, 6, 9&#xA;3, 9, 4, 6, 5, 1, 8, 2, 7&#xA;3, 9, 6, 5, 4, 8, 7, 1, 2&#xA;4, 5, 1, 7, 3, 6, 9, 2, 8&#xA;4, 6, 1, 5, 2, 7, 8, 3, 9&#xA;4, 6, 1, 7, 2, 8, 5, 3, 9&#xA;4, 6, 3, 9, 7, 2, 8, 1, 5&#xA;4, 9, 3, 2, 6, 7, 8, 1, 5&#xA;5, 1, 3, 9, 6, 7, 8, 2, 4&#xA;5, 1, 6, 2, 3, 9, 7, 8, 4&#xA;5, 1, 6, 3, 9, 7, 8, 2, 4&#xA;5, 1, 6, 9, 7, 8, 3, 2, 4&#xA;5, 1, 9, 2, 3, 6, 7, 8, 4&#xA;5, 1, 9, 2, 4, 3, 7, 8, 6&#xA;5, 1, 9, 3, 7, 2, 8, 4, 6&#xA;5, 2, 4, 1, 3, 7, 9, 8, 6&#xA;5, 2, 4, 9, 8, 7, 6, 1, 3&#xA;5, 3, 2, 7, 1, 6, 9, 8, 4&#xA;5, 3, 2, 9, 1, 6, 7, 8, 4&#xA;5, 4, 1, 9, 3, 8, 6, 2, 7&#xA;5, 4, 3, 9, 6, 1, 8, 2, 7&#xA;5, 4, 9, 1, 3, 2, 7, 8, 6&#xA;5, 4, 9, 1, 6, 8, 7, 2, 3&#xA;5, 4, 9, 1, 8, 7, 2, 3, 6&#xA;5, 4, 9, 3, 7, 6, 8, 1, 2&#xA;5, 4, 9, 6, 1, 3, 7, 8, 2&#xA;5, 4, 9, 7, 1, 3, 6, 8, 2&#xA;5, 6, 9, 3, 1, 4, 7, 8, 2&#xA;5, 6, 9, 7, 1, 4, 3, 8, 2&#xA;5, 7, 6, 9, 4, 8, 3, 1, 2&#xA;5, 7, 9, 1, 6, 2, 3, 4, 8&#xA;5, 8, 3, 1, 2, 4, 7, 6, 9&#xA;5, 8, 6, 7, 1, 3, 9, 4, 2&#xA;5, 8, 6, 9, 1, 3, 7, 4, 2&#xA;6, 2, 1, 5, 3, 9, 7, 4, 8&#xA;6, 8, 4, 5, 3, 9, 7, 1, 2&#xA;6, 9, 3, 1, 4, 8, 5, 2, 7&#xA;7, 1, 2, 4, 5, 8, 3, 6, 9&#xA;7, 1, 2, 8, 6, 3, 5, 4, 9&#xA;7, 1, 4, 9, 2, 8, 5, 6, 3&#xA;7, 2, 1, 8, 4, 6, 5, 3, 9&#xA;7, 2, 5, 1, 3, 4, 9, 8, 6&#xA;7, 2, 5, 1, 8, 9, 4, 3, 6&#xA;7, 2, 8, 1, 9, 4, 5, 3, 6&#xA;7, 3, 1, 9, 2, 5, 6, 4, 8&#xA;7, 3, 4, 9, 2, 5, 1, 8, 6&#xA;7, 3, 6, 4, 1, 5, 9, 8, 2&#xA;7, 3, 6, 9, 1, 5, 4, 8, 2&#xA;7, 4, 8, 5, 9, 6, 2, 1, 3&#xA;7, 6, 4, 2, 9, 8, 3, 1, 5&#xA;7, 6, 8, 9, 2, 5, 1, 4, 3&#xA;7, 9, 4, 8, 6, 2, 3, 1, 5&#xA;7, 9, 5, 3, 1, 8, 6, 4, 2&#xA;7, 9, 5, 6, 1, 8, 3, 4, 2&#xA;7, 9, 5, 6, 3, 1, 2, 4, 8&#xA;7, 9, 5, 8, 3, 6, 1, 2, 4&#xA;8, 1, 3, 9, 5, 2, 6, 4, 7&#xA;8, 1, 7, 3, 9, 5, 6, 2, 4&#xA;8, 1, 7, 5, 6, 4, 9, 2, 3&#xA;8, 1, 7, 9, 2, 4, 5, 6, 3&#xA;8, 2, 1, 6, 3, 5, 7, 4, 9&#xA;8, 2, 1, 7, 3, 6, 5, 4, 9&#xA;8, 3, 9, 5, 7, 6, 1, 2, 4&#xA;8, 4, 3, 9, 5, 7, 1, 2, 6&#xA;8, 4, 6, 5, 9, 3, 1, 2, 7&#xA;8, 4, 7, 3, 9, 5, 1, 2, 6&#xA;8, 6, 9, 3, 5, 2, 1, 4, 7&#xA;8, 6, 9, 5, 2, 1, 7, 4, 3&#xA;8, 6, 9, 7, 5, 3, 1, 2, 4&#xA;8, 7, 3, 1, 2, 4, 5, 6, 9&#xA;8, 9, 7, 1, 4, 2, 5, 3, 6&#xA;9, 2, 4, 7, 8, 6, 5, 1, 3&#xA;9, 3, 2, 4, 8, 7, 6, 1, 5&#xA;9, 3, 2, 5, 1, 7, 6, 8, 4&#xA;9, 3, 2, 6, 1, 7, 5, 8, 4&#xA;9, 3, 2, 7, 6, 5, 8, 1, 4&#xA;9, 4, 8, 6, 7, 3, 1, 2, 5&#xA;9, 6, 3, 8, 5, 7, 2, 1, 4&#xA;9, 7, 2, 8, 4, 3, 6, 1, 5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1497" LastActivityDate="2015-05-21T14:39:10.683" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15285" PostTypeId="2" ParentId="15280" CreationDate="2015-05-21T14:40:02.283" Score="1" Body="&lt;p&gt;Chucking it into a computer, it seems that there are 119 solutions if you evaluate each sum in order as you go round the snake. The first few are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 7    3    4    9    2    5    1    8    6&lt;br&gt;&#xA;&lt;br&gt;&#xA; 7    6    8    9    2    5    1    4    3&lt;br&gt;&#xA;&lt;br&gt;&#xA; 1    4    7    5    3    9    2    8    6&lt;br&gt;&#xA;&lt;br&gt;&#xA; 5    6    9    7    1    4    3    8    2&lt;br&gt;&#xA;&lt;br&gt;&#xA; 7    9    5    6    1    8    3    4    2&lt;br&gt;&#xA;&lt;br&gt;&#xA; 7    3    6    9    1    5    4    8    2&lt;br&gt;&#xA;&lt;br&gt;&#xA; 3    7    2    5    1    9    4    8    6&lt;br&gt;&#xA;&lt;br&gt;&#xA; 8    1    7    9    2    4    5    6    3  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12625" LastEditorUserId="11475" LastEditDate="2015-05-21T14:43:15.367" LastActivityDate="2015-05-21T14:43:15.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15286" PostTypeId="1" AcceptedAnswerId="15288" CreationDate="2015-05-21T15:00:41.327" Score="23" ViewCount="3282" Body="&lt;p&gt;Each pipe shown on the left (A through I) connects to exactly one pipe on the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To hide the answer, I have put a curtain over the pipe crossings and grayed out the pipe colors on the right side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When behind the curtain, each pipe travels straight to the leftmost side of its associated numbered gray pipe. In other words, the pipes have no bends or curves, other than the possible single bends just inside either side of the curtain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hGeIy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the clues below, can you figure out where each pipe is connected? Note, there is only 1 possible solution.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Pipe A crosses either 2,3,4,5, or 6 pipes.&lt;/li&gt;&#xA;  &lt;li&gt;Pipe B crosses either 3,4,5, or 6 pipes.&lt;/li&gt;&#xA;  &lt;li&gt;Pipe C crosses &lt;strong&gt;&lt;em&gt;at least&lt;/em&gt;&lt;/strong&gt; 2 pipes.&lt;/li&gt;&#xA;  &lt;li&gt;Pipe D crosses &lt;strong&gt;&lt;em&gt;at least&lt;/em&gt;&lt;/strong&gt; 5 pipes.&lt;/li&gt;&#xA;  &lt;li&gt;Pipe E crosses &lt;strong&gt;&lt;em&gt;at least&lt;/em&gt;&lt;/strong&gt; 6 pipes.&lt;/li&gt;&#xA;  &lt;li&gt;Pipe F crosses &lt;strong&gt;&lt;em&gt;at least&lt;/em&gt;&lt;/strong&gt; 5 pipes.&lt;/li&gt;&#xA;  &lt;li&gt;Pipe G crosses &lt;strong&gt;&lt;em&gt;at most&lt;/em&gt;&lt;/strong&gt; 4 pipes.&lt;/li&gt;&#xA;  &lt;li&gt;Pipe H crosses &lt;strong&gt;&lt;em&gt;at most&lt;/em&gt;&lt;/strong&gt; 3 pipes.&lt;/li&gt;&#xA;  &lt;li&gt;Pipe I crosses &lt;strong&gt;&lt;em&gt;at most&lt;/em&gt;&lt;/strong&gt; 4 pipes.&lt;/li&gt;&#xA;  &lt;li&gt;There are &lt;strong&gt;&lt;em&gt;at least&lt;/em&gt;&lt;/strong&gt; 19 pipe crossings in total. *see bullet below&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Remember: When counting pipe crossings, be careful not to count each one twice. More specifically, if you add up Pipe A's crossings + Pipe B's crossings + ... + Pipe I's crossings, the sum you get will be double the actual total number of pipe crossings, since each crossing would be counted twice.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Answers can be in the form of&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A--&gt;1 B--&gt;2 C--&gt;3 D--&gt;4 E--&gt;5 F--&gt;6 G--&gt;7 H--&gt;8 I--&gt;9&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or, even better, assuming the left side is ordered A through I, just&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;123456789&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If you solve this puzzle using only logic, and without using computer code, and then convince me of that using a detailed explanation, then I will award you 300 rep points, on top of the +10 up-vote and +15 for the accepted answer.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I doubt that it's feasible to solve without code, but I have been surprised many times before, so who knows.&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="5373" LastEditDate="2015-05-21T21:11:51.543" LastActivityDate="2015-05-30T23:03:20.323" Title="Connect the Pipes" Tags="|logical-deduction|" AnswerCount="4" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="15287" PostTypeId="1" AcceptedAnswerId="15307" CreationDate="2015-05-21T15:10:13.173" Score="9" ViewCount="1613" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The whole world doesn't take me seriously, saying I follow no logic,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;even though I'm wise enough to calculate things most cannot.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I've been here for a long time, even more than all my cousins,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;and still I'm sure you'd find me delicious.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12165" LastEditorUserId="12165" LastEditDate="2018-11-22T11:26:17.250" LastActivityDate="2018-11-22T11:26:17.250" Title="The whole world doesn't take me seriously. Who am I?" Tags="|riddle|word|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15288" PostTypeId="2" ParentId="15286" CreationDate="2015-05-21T15:23:21.550" Score="16" Body="&lt;p&gt;Okay, I've once again solved the problem, using only logic and Microsoft Paint! The acceptable solution I found is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $482971356$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The following is a count of crossings per pipe and total:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pipe A crosses $3$ pipes.&lt;br&gt;&#xA; Pipe B crosses $6$ pipes.&lt;br&gt;&#xA; Pipe C crosses $3$ pipes.&lt;br&gt;&#xA; Pipe D crosses $5$ pipes.&lt;br&gt;&#xA; Pipe E crosses $6$ pipes.&lt;br&gt;&#xA; Pipe F crosses $5$ pipes.&lt;br&gt;&#xA; Pipe G crosses $4$ pipes.&lt;br&gt;&#xA; Pipe H crosses $3$ pipes.&lt;br&gt;&#xA; Pipe I crosses $3$ pipes.&lt;br&gt;&#xA; There are $19$ total pipe crossings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here are the logical steps to solving this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 1: Pipes C and F&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ghosts_in_the_code did a great job explaining why C and F are the only pipes that can connect to 1 and 2. Since their order doesn't matter in terms of crossing other pipes, and we're trying to maximize crossings, we attach F to 1 and C to 2 such that the pipes cross.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 2: Pipes G, H, and I&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since we're still trying to maximize for crossings, we want each of these pipes to have as many crossings as they can. Since these pipes have an &quot;at-most&quot; amount of crossings, there is a max point we can place them on the right side. We know that C and F are already in spots 1 and 2, and therefore can't possibly cross G, H, or I, so the two with an &quot;at-most&quot; of 4 crossings can be put in their highest possible spots as this setup would have none of the three crossing each other (leaving a maximum of four possible crossing pipes - A, B, D and E). For G, the highest point would be spot 3. For I, the highest point would be spot 5. However, H can't be placed at spot 4, since it has an &quot;at-most&quot; of 3 crossings, not 4. We can't place H at spot 6 or lower, or pipe I would cross pipe H and pipe I would have too many crossings. Because of this, we have to shift I down one spot so that H can be in spot 5 (which is its highest possible spot based on the above restrictions). Pipe I will therefore end up in spot 6.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 3: Pipe E&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; E needs to cross 6 pipes total. It already can't cross pipe C. If it were put at spot 8, it wouldn't cross pipe A either, and putting it at spot 7 would mean that pipe A would have to travel too far (a max of 6 crossings). This means that it can't cross pipe A either. There are only 6 pipes left, so pipe E has to cross all 6 of them, which it can't do in spots 4, 8, or 9, leaving only spot 7 (as B and D can travel to the two spots below it).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 4: The remainders&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A can only go into one spot now, or else it would be above its maximum crossings (6). This connects A to 4. B can now only go in one spot, or else it would be above its maximum crossings (also 6). This means that B connects to 6. D is the remaining pipe, and it connects to the last remaining spot, which is 9.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="11475" LastEditDate="2015-05-21T17:36:17.640" LastActivityDate="2015-05-21T17:36:17.640" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15289" PostTypeId="2" ParentId="15287" CreationDate="2015-05-21T15:42:07.157" Score="6" Body="&lt;p&gt;My guess is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Light&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The whole world doesn't take me seriously, saying I follow no logic,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quantum mechanics (The study of subatomic particles like photons of light) is very confusing. It doesn't follow the logic of previously believed physics.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;even though I'm wise enough to calculate things most cannot.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quantum computers, using qubits, can calculate things that regular computers can not.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've been here for a longtime, even more than all my cousins,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Light, as a form of energy, can never be created or destroyed from nothing (As stated by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Third_law_of_thermodynamics&quot;&gt;Third Law of Thermodynamics&lt;/a&gt;). Thus light, as any form of energy, always has (and always will) exist.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and still I'm sure you'd find me delicious.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is one that light doesn't really fit. Maybe light is quantum (a minimal amount), and &lt;a href=&quot;https://www.google.co.uk/search?q=All+food+is+good+in+moderation&quot;&gt;all food is good in moderation&lt;/a&gt;, or a minimal amount.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user10203" LastActivityDate="2015-05-21T15:42:07.157" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15290" PostTypeId="1" AcceptedAnswerId="15293" CreationDate="2015-05-21T15:49:56.990" Score="3" ViewCount="11098" Body="&lt;p&gt;There is a similar question &lt;a href=&quot;https://puzzling.stackexchange.com/questions/1980/four-prisoners-wearing-black-and-white-hats&quot;&gt;here&lt;/a&gt; but with four prisoners. But I am not able to understand the logic. The original question is &lt;a href=&quot;http://www.mathsisfun.com/puzzles/black-and-white-hats-solution.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. My solution is half way through. Since there are 3 Back Hats (B) and 2 White Hats (W), the various combinations possible include:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;BBB (Meaning all 3 prisoners wear black hats) &#xA;BBW&#xA;BWB&#xA;WBB&#xA;WWB&#xA;WBW&#xA;BWW&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;From these possibilities, the combinations WWB, WBW and BWW are filtered (the possibility of 2 Black hats remaining and 0 White Hats remaining). Now, how do I proceed to determine the color of the hat of the front prisioner? &lt;/p&gt;&#xA;" OwnerUserId="12627" LastEditorUserId="10615" LastEditDate="2018-09-06T17:07:13.697" LastActivityDate="2018-09-06T17:07:13.697" Title="Black and White Hats Puzzle - Three prisoners" Tags="|logical-deduction|meta-knowledge|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15291" PostTypeId="2" ParentId="15286" CreationDate="2015-05-21T16:00:04.580" Score="7" Body="&lt;p&gt;Incomplete answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C and F are the only pipes that can connect to 1 or 2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If Pipe A connects to 1, it has no crossings. If it connects to 2, it has only 1 crossing (A).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Pipe B connects to 1, it has 1 crossing. If it connects to 2, it can have upto 2 crossings (A and B)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Pipe D connects to 1, it has 3 crossings (A,B,C). If connected to 2, it can have 4 crossings (A,B,C,any other pipe that reaches 1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar logic for Pipe E.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Pipe G, H or I reaches 1 or 2, they have far too many crossings.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4269" LastEditDate="2015-05-21T16:05:46.560" LastActivityDate="2015-05-21T16:05:46.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15292" PostTypeId="2" ParentId="15286" CreationDate="2015-05-21T16:02:45.360" Score="10" Body="&lt;p&gt;I think I had an initial answer of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 382 971 456&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, pipe E is the hardest, since it has to cross all 4 above/below it, and 2 on the other side. We know that GHI don't want many crossings, so let's send C downwards, with 2 spots below it for crossing (7), and bring GHI up above it (456). F needs to cross a lot of pipes, so up to (1). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, now let's just take the next 2 pipes (CD) and move them down to (89). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have to deal with A and B. With only spots (23) left, A-3 can make it's 2 crossing, but B can't. (See picture 1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, since pipe C has a low threshold and B has a high cap, move C-8 to B-8 (6 crossings), and C-2 has 3. (See picture 2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A - 2 crossings&lt;br&gt;&#xA;B - 6 crossings&lt;br&gt;&#xA;C - 3 crossings&lt;br&gt;&#xA;D - 5 crossings&lt;br&gt;&#xA;E - 6 crossings&lt;br&gt;&#xA;F - 5 crossings&lt;br&gt;&#xA;G - 3 crossings&lt;br&gt;&#xA;H - 3 crossings&lt;br&gt;&#xA;I - 3 crossings  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/7gCKn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But we only have 18 crossings. Well, that should be easy to fix, since we can cross A and G easily. (See picture 3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, final order:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 482 971 356&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And crossing counts:&lt;br&gt;&#xA;A - 3 crossings&lt;br&gt;&#xA;B - 6 crossings&lt;br&gt;&#xA;C - 3 crossings&lt;br&gt;&#xA;D - 5 crossings&lt;br&gt;&#xA;E - 6 crossings&lt;br&gt;&#xA;F - 5 crossings&lt;br&gt;&#xA;G - 4 crossings&lt;br&gt;&#xA;H - 3 crossings&lt;br&gt;&#xA;I - 3 crossings  &lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-05-21T20:55:49.943" LastActivityDate="2015-05-21T20:55:49.943" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15293" PostTypeId="2" ParentId="15290" CreationDate="2015-05-21T16:05:37.697" Score="6" Body="&lt;p&gt;Suppose that the rear man sees his two friends wearing two white hats. Then he would be able to tell the color of his hat.&lt;br&gt;&#xA;In this case, he answers &quot;I don't know&quot;, so we deduce that his friends aren't wearing white hats.&lt;br&gt;&#xA;Now, the middle man knows that, if the guy in front of him is white, he (the guy in the middle) cannot be white. If the 3rd guy was black, the man in the middle would indeed be able to answer.&lt;br&gt;&#xA;Though, he said &quot;I don't know&quot;, so the third guy isn't white. The middle guy has no information about his hat, as we know.&lt;br&gt;&#xA;Finally, it's time for the last guy to answer. Listening to the previous answers, he can deduce that he is black and answers!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a solution that analyzes the other cases one by one, here is it:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;BBB The first says &quot;I don't know&quot;, implicitly stating that the other two aren't both white. The second man, seeing a black hat in front of him, doesn't know the color of his hat. The third guy, knowing that he can't be white, deduces his color (black).&lt;/li&gt;&#xA;&lt;li&gt;BBW As in the previous case, the first says &quot;I don't know&quot;. The second, seeing a white hat, deduces that he is black and answers.&lt;/li&gt;&#xA;&lt;li&gt;BWB The first says &quot;I don't know&quot;. The second doesn't know as well. As in BBB, the third knows that he must be necessarily black.&lt;/li&gt;&#xA;&lt;li&gt;WBB The first says &quot;I don't know&quot;. The second doesn't know too. The third answers &quot;black&quot; consequently, same reason as before.&lt;/li&gt;&#xA;&lt;li&gt;WWB The first doesn't know. The second neither. As before, the third answers &quot;black&quot;.&lt;/li&gt;&#xA;&lt;li&gt;WBW The first doesn't know. The second, seeing a white hat in front of him, answers &quot;black&quot;.&lt;/li&gt;&#xA;&lt;li&gt;BWW The first sees two white hats, thus he must be black!&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="1777" LastEditDate="2015-05-21T16:15:57.440" LastActivityDate="2015-05-21T16:15:57.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15294" PostTypeId="2" ParentId="2865" CreationDate="2015-05-21T16:23:34.273" Score="4" Body="&lt;p&gt;I can prove that 12 is the best you can do:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, let's agree to call a move of form $(a, b)$ a move that is $a$ vertical steps and $b$ horizontal steps, or vice-versa.  So, for example, a standard knight move is $(2, 1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 1: Notice that a speedy knight can make all moves&#xA;$$(2, 0), (2, 2), (3, 1), (3, 3), (4, 0), (4, 2),$$&#xA;anywhere on the board.  Also, a speedy knight can make any move&#xA;$$(1, 1)$$&#xA;unless one endpoint of the move is in the corner -- I'll call this the corner exception.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, if a move $(a, b)$ has $a$ and $b$ at most $4$, and same parity, then it is legal unless it is $(2, 2)$, $(4, 4)$ or the corner exception.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 2: By symmetry it's enough to show that there can be at most $3$ knights on light-colored squares in the bottom half of the board.  (Notice light- and dark- colored squares don't interact.)  Let's agree to number the columns 1 through 8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 3: Consider two knights on light-colored squares in the bottom half of the board.  The vertical distance between them is at most 3.  I claim the horizontal distance between them can't be 0 (same column), 1 (adjacent columns), 3, or 4, except for the corner exception.  For example if the horizontal distance was 3 then it would either be a $(3, 1)$-move or a $(3, 3)$-move between them, and either way they are attacking each other.  The other cases are easy to check.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 4: Note that the corner exception only occurs in columns 7 and 8.  (Remember, we're looking at light-colored squares on the lower half of the board.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 5: We're almost done!  Notice there can be at most:&#xA;-one knight in columns 1, 2, 5&#xA;-one knight in columns 3, 6, 7&#xA;-one knight in columns 4, 8.&#xA;Hence, at most three knights total.  The proof is complete.&lt;/p&gt;&#xA;" OwnerUserId="12523" LastActivityDate="2015-05-21T16:23:34.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15295" PostTypeId="2" ParentId="15287" CreationDate="2015-05-21T16:24:42.663" Score="3" Body="&lt;p&gt;My answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Noodles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The whole world doesn't take me seriously, saying I follow no logic,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fried pasta, does it make sense? In theory not!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;even though I'm wise enough to calculate things most cannot.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://naturalmath.com/thebook/&quot; rel=&quot;nofollow&quot;&gt;Moebius noodles&lt;/a&gt; is a book. It may be a reference to &lt;a href=&quot;http://it.wikipedia.org/wiki/Quipu&quot; rel=&quot;nofollow&quot;&gt;Quipu&lt;/a&gt; too, an ancient method of calculus.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've been here for a longtime, even more than all my cousins,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Noodles are older than spaghetti, they say.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and still I'm sure you'd find me delicious.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They're delicious indeed!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-05-21T16:24:42.663" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15297" PostTypeId="1" AcceptedAnswerId="15299" CreationDate="2015-05-21T16:39:01.637" Score="1" ViewCount="115" Body="&lt;p&gt;What's the relation between these words:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;mix&lt;/p&gt;&#xA;&lt;p&gt;audio&lt;/p&gt;&#xA;&lt;p&gt;duct&lt;/p&gt;&#xA;&lt;p&gt;measure&lt;/p&gt;&#xA;&lt;p&gt;electric&lt;/p&gt;&#xA;&lt;p&gt;magnetic&lt;/p&gt;&#xA;&lt;p&gt;deck&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;(This one isn't very difficult for my first attempt of a word puzzle.)&lt;/p&gt;&#xA;&lt;p&gt;Since it's been solved, I'll add others that I can think of just for a completed list:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;Scotch&#xA;&lt;br&gt;grip&#xA;&lt;br&gt; recorder&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-21T19:46:39.400" Title="Putting the words together" Tags="|word|pattern|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15298" PostTypeId="1" AcceptedAnswerId="15520" CreationDate="2015-05-21T16:40:40.810" Score="6" ViewCount="645" Body="&lt;p&gt;Months of preparation and coding has led you to this day. Slowly but surely you worked your way into the security of the most secure system in the world: the computer override for the vault containing the Coca-Cola secret formula. You have one last security protocol to bypass, and you are in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ready and able, you fire up your terminal. You slowly work through the series of backdoors and wormholes you put in place to get to the final gate. When you try to access the program to unlock the vaults.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are met with a familiar message. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Greetings. Please input the key associated with lock #&lt;strong&gt;55476&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are prepared for this. You have come to this block numerous times. Every day the lock changes. Every day a new key, but now you know the algorithm to get the key and open the lock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In prior attempts, you have received a number of locks. You brute forced your way to the key, but by the time you had the correct value, the system had reset for the day and you were met with the message&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm sorry, but lock #&lt;strong&gt;XXXXX&lt;/strong&gt; is no longer active. Please enter the key for lock #&lt;strong&gt;YYYYY&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It took you a while, but you realized that &lt;code&gt;XXXXX&lt;/code&gt; and &lt;code&gt;YYYYY&lt;/code&gt; were the old lock and new lock respectively. This system only popped up the message when you put in the correct key for the old lock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you collected locks and keys. You had the following table:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; LOCK    |    KEY&#xA; 66453   |    25916670&#xA; 12345   |     4777515&#xA; 33200   |    14342400&#xA; 55475   |    20137425&#xA; 43256   |    16913096&#xA; 73553   |    28759223&#xA; 34875   |    13113000&#xA; 99858   |    36348312&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Using this table, you calculated the algorithm to generate any key for any lock.You quickly enter the key and you are in. Congratulations, you now own the Coca-Cola secret formula!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There is a small degree of computer knowledge potentially needed here, although it is possible to create the solution without any computer knowledge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HINT 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Each digit is important in its own right, as is the number as a whole.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;HINT 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kb6Ph.gif&quot; alt=&quot;This table&quot;&gt; might be helpful&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;EDIT&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I added more keys to the table to try and eliminate any false positives on the algorithm. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12242" LastEditorUserId="5373" LastEditDate="2017-03-11T14:43:53.437" LastActivityDate="2017-03-11T14:43:53.437" Title="Cracking the system" Tags="|pattern|password|" AnswerCount="3" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15299" PostTypeId="2" ParentId="15297" CreationDate="2015-05-21T16:41:33.847" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All of the words can be preceded or followed by the word &lt;strong&gt;tape&lt;/strong&gt; to create another word/phrase:&lt;br&gt;&lt;br&gt;mix tape&lt;br&gt;duck tape&lt;br&gt;audio tape&lt;br&gt;tape measure&lt;br&gt;electric tape&lt;br&gt;magnetic tape&lt;br&gt;deck tape / tape deck&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12206" LastEditorUserId="12206" LastEditDate="2015-05-21T16:47:49.300" LastActivityDate="2015-05-21T16:47:49.300" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15300" PostTypeId="2" ParentId="15297" CreationDate="2015-05-21T16:42:02.520" Score="0" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; tape. They can all be followed by -tape to form a word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12542" LastActivityDate="2015-05-21T16:42:02.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15301" PostTypeId="2" ParentId="15286" CreationDate="2015-05-21T17:26:12.837" Score="12" Body="&lt;p&gt;I've managed to solve it (without looking at any of the 3 existing answers) using only logic, pen, and paper! Here's the exact argument I used, in order. Note that I've sacrificed brevity for (hopefully!) clarity :-)&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;First pass&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Pipe A crosses either 2,3,4,5, or 6 pipes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is exactly equivalent to saying pipe A goes to 3,4,5,6, or 7 (since the number of pipes crossing A is the number of pipes above the one A goes to).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Pipe B crosses either 3,4,5, or 6 pipes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So B can't go to 1 or 2 (since then it would only cross A and at most one other pipe), or to 9 (since then it would cross all pipes other than A).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Pipe D crosses at least 5 pipes.&lt;br&gt;&#xA;  Pipe E crosses at least 6 pipes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So &lt;em&gt;D/E&lt;/em&gt; can't go to 1 or 2 (since then it would cross at most A,B,C, and &lt;em&gt;one/two&lt;/em&gt; others).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Pipe F crosses at least 5 pipes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By symmetry with D, F can't go to 8 or 9.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Pipe G crosses at most 4 pipes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So G can't go to 1 or 2 (since then it would cross all but at most one of A to F).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Pipe H crosses at most 3 pipes.&lt;br&gt;&#xA;  Pipe I crosses at most 4 pipes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So &lt;em&gt;H/I&lt;/em&gt; can't go to 1,2,3, or 4 (since then it would cross all but at most three of A to &lt;em&gt;G/H&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summing up our progress so far:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A goes to one of 3,4,5,6,7&lt;br&gt;&#xA; B goes to one of 3,4,5,6,7,8&lt;br&gt;&#xA; D goes to one of 3,4,5,6,7,8,9&lt;br&gt;&#xA; E goes to one of 3,4,5,6,7&lt;br&gt;&#xA; F goes to one of 1,2,3,4,5,6,7&lt;br&gt;&#xA; G goes to one of 3,4,5,6,7,8,9&lt;br&gt;&#xA; H goes to one of 5,6,7,8,9&lt;br&gt;&#xA; I goes to one of 5,6,7,8,9&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We notice that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C and F are the only pipes that can go to 1 or 2, so these two pairs must match. If F went to 2, the only pipes it could cross would be A,B,D,E - not at most five, so &lt;strong&gt;C goes to 2 and F goes to 1&lt;/strong&gt;. With this new information, let's see what more we can deduce.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Second pass&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Pipe D crosses at least 5 pipes.&lt;br&gt;&#xA;  Pipe E crosses at least 6 pipes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So &lt;em&gt;D/E&lt;/em&gt; can't go to 3 or 4 (since then it would cross at most A,B,F, and &lt;em&gt;one/two&lt;/em&gt; others - remember we now know C goes to 2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summing up again:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A goes to one of 3,4,5,6,7&lt;br&gt;&#xA; B goes to one of 3,4,5,6,7,8&lt;br&gt;&#xA; C goes to 2&lt;br&gt;&#xA; D goes to one of 5,6,7,8,9&lt;br&gt;&#xA; E goes to one of 5,6,7&lt;br&gt;&#xA; F goes to 1&lt;br&gt;&#xA; G goes to one of 3,4,5,6,7,8,9&lt;br&gt;&#xA; H goes to one of 5,6,7,8,9&lt;br&gt;&#xA; I goes to one of 5,6,7,8,9&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We notice that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A,B,G are the only pipes that can go to 3 or 4. So at least one of A,B must go to one of 3,4 and therefore not cross either D or E. This means D/E can't go to 5 or 6 (since then it would cross at most {one of A,B}, F, and two/three others). Now the &lt;em&gt;only&lt;/em&gt; place E can go is 7.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If neither A nor B crossed E, then E would cross at most five pipes. Contradiction, so one of A,B must go below E, i.e. to one of 8,9. By the restrictions we found in the first pass,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; B goes to 8.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A and G pair up with 3 and 4.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Summing up again:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A goes to one of 3,4&lt;br&gt;&#xA; B goes to 8&lt;br&gt;&#xA; C goes to 2&lt;br&gt;&#xA; D goes to one of 7,8,9&lt;br&gt;&#xA; E goes to 7&lt;br&gt;&#xA; F goes to 1&lt;br&gt;&#xA; G goes to one of 3,4&lt;br&gt;&#xA; H goes to one of 5,6,7,8,9&lt;br&gt;&#xA; I goes to one of 5,6,7,8,9&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We notice that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; H and I are the only pipes that can go to 5 or 6, so these two pairs must match. If H went to 6, it would cross I and the pipes going to 7,8,9; contradiction, so H goes to 5 and I goes to 6. Now D is the only pipe that can go to 9.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The only pipes left to match up are {A,G} and {3,4}, which we can do using the final constraint: to get at least 19 crossings overall, we need A to go to 4 and G to 3. So the final answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;482971356&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-05-22T10:27:09.870" LastActivityDate="2015-05-22T10:27:09.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15302" PostTypeId="1" AcceptedAnswerId="15312" CreationDate="2015-05-21T17:38:40.837" Score="21" ViewCount="2053" Body="&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/12982/the-hangman-game-backward-2&quot;&gt;Hangman game backward&lt;/a&gt; version 3 - guess what was the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;fill in &quot;d&quot; - Your white shirt soaked with my blood but you wouldn't care.&lt;br&gt;&#xA;fill in &quot;n&quot; - It will be nice to have it on table or bedside.&lt;br&gt;&#xA;fill in &quot;t&quot; - You're looking at me now without knowing.  In fact you are touching / smelling / hearing / tasting me without knowing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, your answer can be in the form of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The question was: _ p p l e&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;followed by explanation.&lt;/p&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-05-22T13:53:14.120" Title="The hangman game backward #3" Tags="|riddle|english|language|letters|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15303" PostTypeId="2" ParentId="15287" CreationDate="2015-05-21T17:53:43.760" Score="1" Body="&lt;p&gt;Sounds to me like it could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Considering:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the words &quot;long time&quot; were merged to &quot;longtime&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The whole world doesn't take me seriously, saying I follow no logic,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People waste their life not taking their time seriously. Some days seem to go fast and some slow, giving the appearance of not following any logic&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;even though I'm wise enough to calculate things most cannot.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Time can be used to calculate a lot of things that we couldn't calculate without its existence&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've been here for a longtime, even more than all my cousins,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Time has been here since the beginning of time... :P&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and still I'm sure you'd find me delicious.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am sure anyone would be happy to eat up any extra time given to them&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11150" LastEditorUserId="11150" LastEditDate="2015-05-21T18:05:00.247" LastActivityDate="2015-05-21T18:05:00.247" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15304" PostTypeId="1" CreationDate="2015-05-21T18:07:35.463" Score="34" ViewCount="4177" Body="&lt;p&gt;My friend sent me a &quot;puzzle&quot; in binary, and said that there are multiple steps to solving the puzzle, all of them including decryption. I translated the binary into text and he said that was a good first step, but I have absolutely no idea what cipher to use past that on the text. He said it requires a key and that the post he sent me should have all I need to solve the puzzle. Unfortunately, I'm completely stumped. I've been locked into using a Vigenere cipher because I believe he hid a keyword somehow, but I'm afraid I might be tunneling on that too hard. Can you help me figure out the next step?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here's the full &quot;puzzle&quot; he sent me. Whether it's a copypasta, I do not know.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Alright so to quell my boredom I have decided to make a puzzle,&#xA;  whoever solves this puzzle and tells me the answer will get $100 cold&#xA;  cash in their hand. For each day that passes I will give one hint but&#xA;  the cash will drop 10 dollars for each hint. The first hint is free,&#xA;  no this puzzle isn't just conversion that's just a step.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Puzzle #1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 01001001 01000010 00100000 01001000 01001110 00100000 01000001&#xA; 00100000 01000011 00100000 01010110 01010101 00100000 01001011&#xA; 00100000 01011000 01010100 00100000 00111001 00100000 01010000&#xA; 00110111 01001011 00100000 01011010 01000111 01000111 00100000&#xA; 01010010 01010000 00100000 00110010 00110000 00100000 00110011&#xA; 01010001 00100000 01001100 00100000 01001011 01001001 00100000&#xA; 01000100 00100000 01011000 01010100 00100000 01010111 00110011&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;  &#xA;  &lt;p&gt;Good luck.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Translating the binary to text yields: &lt;code&gt;IB HN A C VU K XT 9 P7K ZGG RP 20 3Q L KI D XT W3&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's as far as I've gotten.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;&lt;strong&gt;Updates&lt;/strong&gt;&lt;/h2&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1&lt;/strong&gt;: The creator has posted a hint.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's something that makes him happy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12629" LastEditorUserId="51105" LastEditDate="2018-09-19T14:49:30.130" LastActivityDate="2019-05-20T14:02:50.580" Title="Binary cryptogram from a friend" Tags="|cipher|enigmatic-puzzle|" AnswerCount="1" CommentCount="23" ContentLicense="CC BY-SA 4.0" />
  <row Id="15305" PostTypeId="2" ParentId="15298" CreationDate="2015-05-21T18:07:36.037" Score="2" Body="&lt;p&gt;The key associated with lock #55476 is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 21635640&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It seems to be too simple (at least for the Coca Cola recipe), but at least with the small example data set provided, the key &lt;code&gt;k&lt;/code&gt; for each lock &lt;code&gt;l&lt;/code&gt; can be computed with the following source code (C/Java/C# syntax):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;k = l*(376+((l&amp;amp;2)==0?10:0)+((l&amp;amp;6)==0?1:0)+((l&amp;amp;192)==0?0:4));&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11687" LastActivityDate="2015-05-21T18:07:36.037" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15306" PostTypeId="2" ParentId="15278" CreationDate="2015-05-21T18:41:03.650" Score="5" Body="&lt;p&gt;Based on the hints, it seems we're supposed to fit some words into the non-crossed-out spaces on the chessboard.  &lt;code&gt;PASSWORD&lt;/code&gt; is already present, and one of the annotations seems to suggest another:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;........&#xA;....&#xA;PASSWORD&#xA;ANEGTEN&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We are also hinted to fill in the remaining spaces with a &quot;phrase of two words famous in chess.&quot;  Quark generously hinted that one of the words was &lt;code&gt;GAME&lt;/code&gt;, which is probably intended for the fourth space.  Thus we're looking for &lt;code&gt;........ GAME&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A search reveals that the chessboard pictures move 19 of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Immortal_Game&quot; rel=&quot;noreferrer&quot;&gt;Immortal Game&lt;/a&gt; between Adolf Anderssen and Lionel Kieseritzky in 1851.  My own analysis of the position ignored the fact that, &quot;this move blocks the [black] queen from participating in the defense of the king,&quot; leading me to label the move a mistake (&lt;code&gt;?&lt;/code&gt;), when it is actually a good move (&lt;code&gt;!&lt;/code&gt;).  The following moves were:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;19. e5!   Qxa1+&#xA;20. Ke2   Na6&#xA;21. Nxg7+ Kd8&#xA;22. Qf6+! Nxf6&#xA;23. Be7# 1–0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Thus we have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;IMMORTAL&#xA;GAME&#xA;PASSWORD&#xA;ANEGTEN&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Our last hint is that these words form an &lt;em&gt;anagram&lt;/em&gt; of the solution.  We can simplify our task by immediately removing the word &lt;code&gt;PROGRAM&lt;/code&gt;, leaving us with:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;PROGRAM (AAADEEEGILMMNNOSSTTW)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;An online anagram solver then resolves this to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;     SOLID WASTE MANAGEMENT PROGRAM&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="9718" LastEditDate="2015-05-21T19:52:10.857" LastActivityDate="2015-05-21T19:52:10.857" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15307" PostTypeId="2" ParentId="15287" CreationDate="2015-05-21T18:46:33.037" Score="17" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pi&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The whole world doesn't take me seriously, saying I follow no logic,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No pattern to its digits&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;even though I'm wise enough to calculate things most cannot.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Only way to accurately determine lots of stuff&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've been here for a longtime, even more than all my cousins,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Discovered before &lt;em&gt;e&lt;/em&gt; and other irrational numbers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and still I'm sure you'd find me delicious.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Mmmmm... pie...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5151" LastActivityDate="2015-05-21T18:46:33.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15309" PostTypeId="2" ParentId="15287" CreationDate="2015-05-21T19:09:42.887" Score="3" Body="&lt;p&gt;Even though the question has been answered, I would have said:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fungus&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The whole world doesn't take me seriously, saying I follow no logic,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's neither a plant, an animal, nor a mineral. It's an entirely different classification of organism.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;even though I'm wise enough to calculate things most cannot.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fungus was used to calculate the optimal routes of Tokyo's rail system (www.wired.com/2010/01/slime-mold-grows-network-just-like-tokyo-rail-system/)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've been here for a long time, even more than all my cousins,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Check out the &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_longest-living_organisms&quot; rel=&quot;nofollow&quot;&gt;list of longest-living organisms&lt;/a&gt;, of which fungal colonies are among the oldest. According to &lt;a href=&quot;https://www.youtube.com/watch?v=cwLviP7KaAc&quot; rel=&quot;nofollow&quot;&gt;this talk&lt;/a&gt; by Paul Stamets, a renouned mycologist, fungus was what kept the earth and it's ecology alive during &lt;em&gt;all&lt;/em&gt; of the earth's major extinction periods. It can be argued that nearly all life on earth evolved from fungus.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and still I'm sure you'd find me delicious.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Last, but not least, who doesn't enjoy a great tasting portobello mushroom now again?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12631" LastActivityDate="2015-05-21T19:09:42.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15310" PostTypeId="2" ParentId="15298" CreationDate="2015-05-21T19:36:16.217" Score="3" Body="&lt;p&gt;I'm thinking the answer is...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 20137788&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My algorithm works for the first 5 data blocks, but then I lose it for the last three.  Are you sure you used the right numbers for the last three? ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  My attempt at the algorithm:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let L1, L2...L5 be the 1st, 2nd... 5th digits of the lock code, and let K be the key:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;K = L * (-148*L1 + 239*L2 - 212*L3 + 5*L4)/3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably not the right direction, but it did work perfectly for the first 5 keys.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 2:&#xA;Under the same logic, you could use regression to find something close for all of the provided lock numbers.  The closest equation I could find for all 8 locks is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;K = L * (390.8 + 15.68*L1 - 9.007*L2 + 16.32*L3 - 13.51*L4 + 2.452*L5 + 1.174*L1*L2 - 3.674*L1*L3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is one of two key (rounding the equation up or down before multiplying by L),assuming the actual algorithm uses a whole number before multiplying by L.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two possible codes result:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 20248740 or 20304216&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If neither of those are right, it's probably time to quit :)&lt;/p&gt;&#xA;" OwnerUserId="12633" LastEditorUserId="12633" LastEditDate="2015-05-21T20:36:55.827" LastActivityDate="2015-05-21T20:36:55.827" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15311" PostTypeId="2" ParentId="15302" CreationDate="2015-05-21T19:39:48.027" Score="10" Body="&lt;p&gt;Maybe a bit of a stretch, but is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ti_e&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fill in &quot;d&quot; - Your white shirt soaked with my blood but you wouldn't care.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;tide - the magical laundry detergent will get out blood.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fill in &quot;n&quot; - It will be nice to have it on table or bedside.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; tine - fork on a table, some other pokey thing by the bed for... well.... &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fill in &quot;t&quot; - You're looking at me now without knowing. In fact you are touching / smelling / hearing / tasting me without knowing.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; tite - a bit of an old word, but defined as &quot;immediate&quot; &lt;a href=&quot;http://www.collinsdictionary.com/dictionary/english/tite&quot;&gt;here&lt;/a&gt;. We touch, smell, hear, and taste the present or immediate.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-05-21T19:39:48.027" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15312" PostTypeId="2" ParentId="15302" CreationDate="2015-05-21T19:51:21.873" Score="23" Body="&lt;p&gt;Could it be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; m a _ _ e r&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;with &quot;d&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; madder - a red dye or pigment obtained from the root of the madder plant, or a synthetic dye resembling it.&lt;br&gt;&#xA; presumably you are soaking your white shirt with this red dye (the color of blood!) because you &lt;em&gt;want&lt;/em&gt; to dye your shirt red.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;with &quot;n&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; manner - table manners and bedside manners are both good things to have&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;with &quot;t&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; matter - everything is made of matter!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="239" LastEditorUserId="10037" LastEditDate="2015-05-22T13:53:14.120" LastActivityDate="2015-05-22T13:53:14.120" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15313" PostTypeId="1" AcceptedAnswerId="15501" CreationDate="2015-05-21T20:42:37.883" Score="10" ViewCount="1669" Body="&lt;p&gt;&lt;em&gt;I have many children yet some would doubt that I love them.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first might have impressed if not for a knife.&lt;br&gt;&#xA;My second's fall was far from bright.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My third kept undesirable company or drank too much wine,&lt;br&gt;&#xA;While my fourth was abandoned, frozen in time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My fifth liked his eels chopped and axed,&lt;br&gt;&#xA;And my sixth preferred cooked, if anyone asked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My seventh could not stand the sight of a bath,&lt;br&gt;&#xA;While my last might have been first, but surely was there last. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Name me and my children.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #1&lt;/strong&gt; (children described in clues above are numbered 1 to 8 to make these new clues easier to follow)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;1&lt;/strong&gt; might have killed &lt;strong&gt;2&lt;/strong&gt;&lt;br&gt;&#xA; &lt;strong&gt;2&lt;/strong&gt; is said to have killed &lt;strong&gt;4&lt;/strong&gt;&lt;br&gt;&#xA; &lt;strong&gt;3&lt;/strong&gt; is killed by &lt;strong&gt;1&lt;/strong&gt; or &lt;strong&gt;7&lt;/strong&gt; or hell or heaven&lt;br&gt;&#xA; &lt;strong&gt;4&lt;/strong&gt; is killed by &lt;strong&gt;1&lt;/strong&gt; or &lt;strong&gt;3&lt;/strong&gt; or most likely &lt;strong&gt;2&lt;/strong&gt;&lt;br&gt;&#xA; &lt;strong&gt;5&lt;/strong&gt; probably died first&lt;br&gt;&#xA; &lt;strong&gt;6&lt;/strong&gt; did not kill anyone&lt;br&gt;&#xA; &lt;strong&gt;7&lt;/strong&gt; killed four but not &lt;strong&gt;1&lt;/strong&gt;&lt;br&gt;&#xA; &lt;strong&gt;8&lt;/strong&gt; stayed away until the killing was all done.&lt;br&gt;&#xA; If all of us are dead, who killed &lt;strong&gt;7&lt;/strong&gt; and &lt;strong&gt;1&lt;/strong&gt;?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #2&lt;/strong&gt; (I personally think this one is a dead giveaway but oh well :P )&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To prostrate oneself, oh joyous man -&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #3&lt;/strong&gt; (Dead giveaway, I'd suggest not looking at this one unless you've tried solving with just the others)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hint #2 is a charade for the author's name, First and Family names separated by the comma.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastEditorUserId="11251" LastEditDate="2015-05-26T14:07:08.923" LastActivityDate="2016-05-16T00:28:17.597" Title="If 3 is a crowd... what about 8?" Tags="|riddle|knowledge|" AnswerCount="4" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="15315" PostTypeId="1" AcceptedAnswerId="15317" CreationDate="2015-05-21T21:40:05.887" Score="4" ViewCount="609" Body="&lt;p&gt;We were given an image that could be a barcode found hidden in it. However, I'm unable to read the barcode using an app after I re-created it and I'm unsure if I made an error or if it's an actual barcode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://puu.sh/gSv9s/378d5c7681.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Original Image&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PbIxB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Red channel 6th bit&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VRqIf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Re-created hidden image&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/khCuL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Flipped Re-created hidden image&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the barcode hidden under the original image just change the contrast and levels of the image or use any of your different steganography techniques. I'm only guessing it's a barcode but &lt;a href=&quot;http://gyazo.com/72ac4d5b430e24d37319f129ae2126d5&quot; rel=&quot;nofollow noreferrer&quot;&gt;it may not be&lt;/a&gt; (this person had solved the puzzle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it possibly might not even be a barcode or I'm finding the wrong thing in the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All I can say is it was confirmed that the answer is &lt;strong&gt;44 random characters&lt;/strong&gt;. It was confirmed by someone who has solved it.&#xA;The answer should be something like this. EGUKUEEQXAYLIKPPSKQPLIUFZXJUOGXQJZIUGOLLFOUL&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a real barcode? If yes, how can we scan it/decrypt it? If not, what is it or is there something else I'm missing in the original image? Finally what is the decryption to it?&lt;/p&gt;&#xA;" OwnerUserId="12636" LastEditorUserId="12636" LastEditDate="2015-05-21T21:49:11.360" LastActivityDate="2015-05-21T22:05:41.470" Title="Steganography. What is the decryption?" Tags="|cipher|steganography|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15317" PostTypeId="2" ParentId="15315" CreationDate="2015-05-21T22:05:41.470" Score="1" Body="&lt;p&gt;I want to thank Quark for the help of the answer. For some reason when the barcode wasn't scanning before. It finally worked using this website &lt;a href=&quot;http://online-barcode-reader.inliteresearch.com/default.aspx&quot; rel=&quot;nofollow&quot;&gt;http://online-barcode-reader.inliteresearch.com/default.aspx&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just take my flipped version and scan it there. You should get the answer. I'm not sure why it wasn't working before, but yeah. I guess it was just complete luck.&lt;/p&gt;&#xA;" OwnerUserId="12636" LastActivityDate="2015-05-21T22:05:41.470" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15318" PostTypeId="1" AcceptedAnswerId="15320" CreationDate="2015-05-21T22:24:32.293" Score="5" ViewCount="397" Body="&lt;p&gt;My first was a grandfather,&lt;br&gt;&#xA;My second said when to run,&lt;br&gt;&#xA;My third was then a scarecrow,&lt;br&gt;&#xA;My fourth was there for longest,&lt;br&gt;&#xA;My fifth was a father-in-law,&lt;br&gt;&#xA;My sixth was like my fourth,&lt;br&gt;&#xA;My seventh changed a bit,&lt;br&gt;&#xA;My eighth came only once,&lt;br&gt;&#xA;My ninth was once obscure,&lt;br&gt;&#xA;My tenth was no landlord,&lt;br&gt;&#xA;My eleventh was married,&lt;br&gt;&#xA;And my twelfth is an idiot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who or what am I?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-22T01:07:21.477" Title="Two-beat riddle - when am I?" Tags="|riddle|knowledge|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15319" PostTypeId="1" AcceptedAnswerId="16120" CreationDate="2015-05-21T23:45:13.723" Score="11" ViewCount="1059" Body="&lt;p&gt;Seek out a sunny picture -&lt;br&gt;&#xA;Remember you're on SE -&lt;br&gt;&#xA;And on that sunny picture,&lt;br&gt;&#xA;Figure out what you can;&lt;br&gt;&#xA;You are a less capital idea&lt;br&gt;&#xA;Than those who follow you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What kind of birds appear?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you're on SE, where would you find a picture?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your answer can take the form of an actual picture (which has birds in). The challenge is to find the correct picture, not to identify the birds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3 (like Hint 1, but bigger):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you upload a picture while making a post on SE, where will the URL of that picture be?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-06-08T17:22:06.103" LastActivityDate="2015-06-08T21:12:24.203" Title="What kind of birds?" Tags="|riddle|enigmatic-puzzle|" AnswerCount="5" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="15320" PostTypeId="2" ParentId="15318" CreationDate="2015-05-21T23:48:13.857" Score="6" Body="&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Doctor&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There are currently 12 actors who have played the Doctor, and this also fits the title of &quot;when am I?&quot; referring to the Doctor being a Time Lord.  Time Lords also have 2 hearts which would make sense with &quot;Two-beat&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My first was a grandfather,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The First Doctor, William Hartnell, was known as &quot;grandfather&quot; by Lady Larn.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My second said when to run,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Second Doctor, Patrick Troughton, was well known for the phrase &quot;When I say run, &lt;em&gt;run&lt;/em&gt;!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My third was then a scarecrow,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Third Doctor, Jon Pertwee, played Worzel Gummidge (a scarecrow) in the 1979-1981 series &lt;em&gt;Worzel Gummidge&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fourth was there for longest,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Fourth Doctor, Tom Baker, was the Doctor for the longest (around 7 years).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fifth was a father-in-law,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Fifth Doctor, Peter Davison, had a daughter, Georgia Moffett, who married David Tennant, making him the father-in-law of the tenth Doctor&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My sixth was like my fourth,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Sixth Doctor, Colin Baker, shares the same last name with the fourth Doctor&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My seventh changed a bit,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Seventh Doctor, Sylvester McCoy, started as a comical character, but later developed a darker nature.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My eighth came only once,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Eighth Doctor, Paul McGann, only appeared as the Doctor once&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My ninth was once obscure,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Ninth Doctor, Christopher Eccleston, also played the role of Jude in the 1996 film of the same name, based on the novel &lt;em&gt;Jude the Obscure&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My tenth was no landlord,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Tenth Doctor, David Tennant, was &quot;no landlord&quot;, he was a Tennant (ba dum tsss)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My eleventh was married,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Eleventh Doctor, Matt Smith, marries River during the episode &lt;em&gt;The Wedding of River Song&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And my twelfth is an idiot.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Twelfth Doctor, Peter Capaldi. This is referring to the quote, &quot;You know who I am? I... am... an idiot! With a box and a screwdriver.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11727" LastEditorUserId="11727" LastEditDate="2015-05-22T01:07:21.477" LastActivityDate="2015-05-22T01:07:21.477" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15321" PostTypeId="1" AcceptedAnswerId="15331" CreationDate="2015-05-22T02:34:28.447" Score="65" ViewCount="4167" Body="&lt;p&gt;So my son is starting to show an interest in mathematics, and being the awesome dad that I am, I decided to buy him a basic calculator. Being the cheapskate that I am, I ordered the cheapest one I could find, which happened to be from &lt;em&gt;&lt;strong&gt;NoSlaves™ Calculator Co.&lt;/strong&gt;&lt;/em&gt; &lt;sup&gt;(&lt;em&gt;When only the cheapest calculator will do, make sure you buy from NoSlaves™, where we lovingly craft our calculators to almost adequate levels of quality, and totally don't use any slave labour... at all.&lt;/em&gt;)&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;p&gt;Unfortunately (and surprisingly), when it arrived, there seemed to be something wrong with it... I plugged in a few quick test calculations:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;1 + 1 = 1&#xA;4 + 3 = 9&#xA;8 - 3 + 5 = 6&#xA;24 + 51 = 84&#xA;50 - 74 + 39 = 98&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Now, I like to complain as much as the next guy, so I went hunting through the instruction manual for the company's phone number. However, on the warranty and support information page, I found the following, scrawled in a strange, dark reddish, congealed ink of some sort:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;2 - 3 + 4 = ?&lt;br /&gt;&#xA;80 - 17 = ?&lt;br /&gt;&#xA;4 - 1 + 2 - _ = ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Curious, I entered in the calculations... I didn't know what the underscore meant, so just ignored it, but regardless, the results were even more broken than my initial test calculations. By this stage I was bored, so I shrugged it off and went to bed.&lt;/p&gt;&#xA;&lt;p&gt;Three weeks later, being the brilliant detective I am, I suddenly realised that the hand written message carried a cleverly concealed secret message. I immediately phoned the police department and was hailed as a hero.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;What was wrong with the calculator? And, what was the mysterious hidden message?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="60644" LastEditDate="2020-09-06T23:32:07.130" LastActivityDate="2020-09-06T23:32:07.130" Title="Mysterious message from the NoSlaves™ Calculator Co" Tags="|lateral-thinking|number-sequence|enigmatic-puzzle|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="15324" PostTypeId="2" ParentId="15262" CreationDate="2015-05-22T04:09:51.463" Score="1" Body="&lt;p&gt;To complete the answer by TheEssential, the encryption method is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Upper case letters remain unchanged.&lt;br&gt;&#xA;&lt;br&gt;&#xA;Lower case letters are encrypted as shown in the table below. Two plaintext letters are encoded to one ciphertext letter.&lt;br&gt;&#xA;The initial CT character is &quot;q&quot; which is &quot;a&quot; plus 16 (16 is the sum of digits of 15262 in the PSE question number of the FQDN).  Then, the sequence of ciphertext letters is -11, +15, -11, +15... applying mod-26.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/FM0WX.png&quot; alt=&quot;Table&quot;&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-05-22T04:32:06.063" LastActivityDate="2015-05-22T04:32:06.063" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15326" PostTypeId="1" AcceptedAnswerId="15400" CreationDate="2015-05-22T05:13:19.543" Score="2" ViewCount="348" Body="&lt;p&gt;Answers &lt;strong&gt;must include method of encryption&lt;/strong&gt; along with plain text and explanation (explanation refers to thoughts or clues that made you think to do certain things).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Cryptogram #5&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;cd, f, mq, ec, dq, tm, ah, ayw, ti, eo, caw, , ajg, dfu, , ens, aqc, yu, , fec, gvy, agk, aic, eqe, , fzo, z, , lw, ai, is, em, abc, cq, is, bvs, , bv, bou, ko, , ciq, fiq, cre, gs, ada, fog, , hrq, bma, aoa, fzo, dr, ke, , fe, abi, , atk, mw, is, , bri, cyw, qi, re, bys, , dbq, adm, , dgy, ep, hgs, bhq, akq, evy, gt, el, bq, y, ra,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Modified one of my earlier encryption method (it has been solved). Much easier.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 123♛_abc♕123.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12058" LastActivityDate="2015-05-23T04:55:37.167" Title="Quotes Surely. Cryptogram #5" Tags="|cipher|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15327" PostTypeId="2" ParentId="15313" CreationDate="2015-05-22T05:17:38.760" Score="1" Body="&lt;p&gt;I think it's:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The days of the week: Monday,Tuesday,Wednesday,Thursday,Friday,Saturday,Sunday.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; While my last might have been first, but surely was there last: Monday or Sunday both could be the first day&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12058" LastActivityDate="2015-05-22T05:17:38.760" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="15328" PostTypeId="1" AcceptedAnswerId="15329" CreationDate="2015-05-22T06:29:53.283" Score="12" ViewCount="5352" Body="&lt;p&gt;&lt;strong&gt;Background:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your love for Hollywood movies and remarkable ability to remember movie titles is finally going to serve a purpose. There was indeed a time when you almost believed your mother's words, as she scorned you: &quot;Lazing in front of the television is not going to pay for the rent!&quot; You knew better all along. As you and the rest of your family listen to the directions of your late, and wealthy, Uncle Rob's will, you cannot help but smile as the executor reads &quot;... and thus half of my fortune will indeed go toward a sole beneficiary. Namely he who solves the puzzles as depicted below&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you are each given a paper with four Hollywood movie rebus puzzles, you begin to smile. You know the answer to each. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How will you answer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Puzzle #1&lt;/strong&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/eN0ii.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/15350/hollywood-movie-rebus-puzzle-2&quot;&gt;Next puzzle - #2 Hollywood Rebus Movie Puzzle &lt;/a&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12317" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2015-07-23T05:17:49.970" Title="Hollywood Movie Rebus Puzzle #1" Tags="|rebus|movies|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15329" PostTypeId="2" ParentId="15328" CreationDate="2015-05-22T06:34:41.247" Score="14" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Spiderman&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The periscope represents the word &lt;strong&gt;spy&lt;/strong&gt;.&lt;br&gt;The middle picture is a perm, but with the p crossed out to become a d, so it's &lt;strong&gt;derm&lt;/strong&gt;.&lt;br&gt;The ant has its t crossed out so it's &lt;strong&gt;an&lt;/strong&gt;.&lt;br&gt;Putting this all together we get the answer, &lt;strong&gt;spiderman&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7308" LastActivityDate="2015-05-22T06:34:41.247" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15330" PostTypeId="1" AcceptedAnswerId="15332" CreationDate="2015-05-22T07:50:57.653" Score="6" ViewCount="1083" Body="&lt;p&gt;You have been chosen to participate in a selection process for a secret underground group. To make sure you're worthy, they test you with a series of devious puzzles. However, one in particular has you stumped...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The challenge you receive is as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;People aren't very good at observing things. We hope you are better.&lt;br&gt;To solve this challenge, you will need to inspect the contents of the link below:&lt;br&gt;&lt;br&gt;&lt;a href=&quot;https://i.stack.imgur.com/X2plV.png/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/XKGpe.png&lt;/a&gt;&lt;br&gt;&lt;br&gt;We look forward to your success.&lt;br&gt;- The Underground Group&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The solution is obvious once found, and requires no special skills to uncover.&lt;br&gt;&#xA;But, of course, you need to keep your eyes and mind open...&lt;/p&gt;&#xA;" OwnerUserId="7308" LastActivityDate="2015-05-22T08:33:53.887" Title="Observation, observation" Tags="|enigmatic-puzzle|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15331" PostTypeId="2" ParentId="15321" CreationDate="2015-05-22T07:51:32.837" Score="58" Body="&lt;p&gt;The problem with the calculator is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It adds/subtracts the segments of the display, not the numbers.&lt;br&gt;&#xA;So, 4 + 3 is indeed equal to 9&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/0Mlcz.jpg&quot; alt=&quot;&quot;&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The hidden message, then, is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;HELP, all in capitals. The underscore in the last sum is the bottom line segment on the display.&lt;br&gt;&#xA;&lt;br&gt;&#xA;Example for the first sum :&#xA;&lt;img src=&quot;https://i.stack.imgur.com/XTMwg.png&quot; alt=&quot;Example for the first sum&quot;&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.uize.com/examples/seven-segment-display.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.uize.com/examples/seven-segment-display.html&lt;/a&gt;&lt;br&gt;&#xA;&#xA;Since it's clearly been written in blood, something foul is afoot at the calculator factory   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sorry, don't have access to much by way of image-editing stuff atm, so can't do the pretty pictures.&lt;/p&gt;&#xA;" OwnerUserId="11717" LastEditorUserId="11717" LastEditDate="2015-05-22T08:10:06.257" LastActivityDate="2015-05-22T08:10:06.257" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15332" PostTypeId="2" ParentId="15330" CreationDate="2015-05-22T08:18:53.333" Score="9" Body="&lt;p&gt;Not done yet, I think, but:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Using the fill option in paint, this came out:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/jiArC.png&quot; alt=&quot;&quot;&gt;&lt;br&gt;&#xA;Keep looking!  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The displayed and actual link are different. Following the displayed link shows:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/XKGpe.png/&quot; alt=&quot;&quot;&gt;&lt;br&gt;&#xA;Again, filling it in in paint shows:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/h6vu7.png&quot; alt=&quot;&quot;&gt;&lt;br&gt;&#xA;So, is the answer 404? Or is it something to do with either imgur or PSE's 404 page?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11717" LastActivityDate="2015-05-22T08:18:53.333" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15333" PostTypeId="2" ParentId="15330" CreationDate="2015-05-22T08:20:40.233" Score="3" Body="&lt;p&gt;The solution is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 404&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How it's hidden...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The linked image is described as &lt;a href=&quot;https://i.stack.imgur.com/XKGpe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/XKGpe.png&lt;/a&gt;&#xA; but actually points to &lt;a href=&quot;https://i.stack.imgur.com/X2plV.png/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/X2plV.png/&lt;/a&gt;.&#xA; The image actually linked to contains very faint text that reads Keep Looking.&#xA; The file XKGpe.png contains to real solution.&#xA; Using a flood fill or colour replacement tool will reveal the text.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="11297" LastEditDate="2015-05-22T08:33:53.887" LastActivityDate="2015-05-22T08:33:53.887" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15334" PostTypeId="1" AcceptedAnswerId="15349" CreationDate="2015-05-22T08:33:28.483" Score="13" ViewCount="18177" Body="&lt;p&gt;I'm complex inside,&lt;br&gt;but easy on the outside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm loud for one,&lt;br&gt;but quiet for others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm used to getting hit,&lt;br&gt;but there's no alternative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I die as fast,&lt;br&gt;as I rise again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What am I?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="12640" LastEditorUserId="57212" LastEditDate="2021-03-03T07:49:48.493" LastActivityDate="2021-03-03T07:49:48.493" Title="They use me... What am I?" Tags="|riddle|" AnswerCount="10" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15335" PostTypeId="2" ParentId="11926" CreationDate="2015-05-22T09:04:15.163" Score="0" Body="&lt;p&gt;I think it's a BEE&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scratch that thing, It's the name of this being!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;to me would be a bee sting&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not only a furry creature, But the sweetest of nature!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;bees are a hairy creature and collect pollen from nature&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Damn horrible intruder, Get out of my computer!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;can't answer this one&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Surely ain't emblem of pluck&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;could be plucking the stinger out&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At most it may bring luck!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;in some countries around the world a bee sting is thought to bring good luck.....&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12641" LastEditorUserId="5373" LastEditDate="2015-05-22T09:36:23.627" LastActivityDate="2015-05-22T09:36:23.627" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15337" PostTypeId="2" ParentId="15334" CreationDate="2015-05-22T10:10:56.150" Score="12" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an alarm clock?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I'm complex inside,&lt;br&gt;&#xA;but easy on the outside.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It looks simple on the outside, just a clock (or a digital display), but inside are lots of complex clockwork or electronics.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I'm loud for one,&lt;br&gt;&#xA;but quiet for the others.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's loud for you so that it wakes you up, but it shouldn't be too loud for anyone else.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I'm used to get hit,&lt;br&gt;&#xA;but there's no alternative.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The bells on a mechanical alarm clock are used to getting hit by a hammer or whatever - that's how it works.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I die as fast,&lt;br&gt;&#xA;as I raise again.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;s&gt;You stop the clock from ringing each morning when you get up, but then set it again that evening to ring again the next morning.&lt;/s&gt; If you set the clock to snooze alarm, it starts ringing again almost as quickly as you stop it (thanks to &lt;strong&gt;user2408578&lt;/strong&gt; for this idea).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-05-22T10:21:04.137" LastActivityDate="2015-05-22T10:21:04.137" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15338" PostTypeId="2" ParentId="15334" CreationDate="2015-05-22T10:21:55.450" Score="8" Body="&lt;p&gt;My guess would be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Revolution?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation line by line:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm complex inside,&#xA;but easy on the outside.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A revolution can have very complex grounds but also have a simple goal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm loud for one,&#xA;but quiet for the others.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some revolutions start by doing something with a big group of people. Some just with one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm used to get hit,&#xA;but there's no alternative.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is always resistance against a revolution, there has to be.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I die as fast,&#xA;as I raise again.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Revolutions can die as quick as they start.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4786" LastActivityDate="2015-05-22T10:21:55.450" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15339" PostTypeId="1" AcceptedAnswerId="15340" CreationDate="2015-05-22T10:30:55.340" Score="52" ViewCount="4398" Body="&lt;p&gt;&lt;strong&gt;Four-words:&lt;/strong&gt;&lt;br /&gt;&#xA;My first is indivisible, but not really.&lt;br /&gt;&#xA;My second is a cloak worn by men, but in ancient Rome.&lt;br /&gt;&#xA;My third is a cruel wicked person, but not a human.&lt;br /&gt;&#xA;My fourth is about 20 quires, but in reverse.&lt;/p&gt;&#xA;&lt;p&gt;I was not allowed a very short title for this puzzle, but I was able to &lt;strong&gt;add a clue to make it cross the 15-letters-minimum limit for titles&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10940" LastEditorUserId="69582" LastEditDate="2021-09-27T02:07:36.820" LastActivityDate="2021-09-27T02:07:36.820" Title="Four-words (----||||)" Tags="|word|wordplay|crosswords|word-square|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="15340" PostTypeId="2" ParentId="15339" CreationDate="2015-05-22T10:39:16.590" Score="35" Body="&lt;p&gt;&lt;em&gt;My first is indivisible, but not really.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The word ATOM means &quot;indivisible&quot; in ancient Greek, but atoms aren't really indivisible since they have protons, neutrons, quarks, etc. inside them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My second is a cloak worn by men, but in ancient Rome.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A TOGA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Toga&quot;&gt;is the answer there&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My third is a cruel wicked person, but not a human.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An OGRE.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My fourth is about 20 quires, but in reverse.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Units_of_paper_quantity#Quire&quot;&gt;20 quires&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; form a ream, so the answer to this is MAER.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the full answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;A T O M&lt;/code&gt;&lt;br&gt;&lt;code&gt;T O G A&lt;/code&gt;&lt;br&gt;&lt;code&gt;O G R E&lt;/code&gt;&lt;br&gt;&lt;code&gt;M A E R&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Lovely&lt;/em&gt; puzzle!&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="4204" LastEditDate="2015-05-24T19:41:16.747" LastActivityDate="2015-05-24T19:41:16.747" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15341" PostTypeId="1" AcceptedAnswerId="15342" CreationDate="2015-05-22T11:02:31.190" Score="21" ViewCount="2150" Body="&lt;p&gt;&lt;strong&gt;Five-words:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first is a kind of cattle from Asia (more than one).&lt;br&gt;&#xA;My second is what a horse did to its victory if it just won a race on its own merit.&lt;br&gt;&#xA;My third is found together with a groom.&lt;br&gt;&#xA;My fourth is the kind of praise a jockey gets if they just won a race by cheating.&lt;br&gt;&#xA;My fifth is a horse, or any animal that can be ridden.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Inspired by Prem's &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15339/four-words&quot;&gt;Four-words (----||||)&lt;/a&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2020-01-10T17:27:38.340" Title="Five-words (-----|||||)" Tags="|word|wordplay|crosswords|word-square|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15342" PostTypeId="2" ParentId="15341" CreationDate="2015-05-22T11:24:39.407" Score="24" Body="&lt;p&gt;Cool puzzle!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answers to the clues are:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1) &lt;strong&gt;Zebus&lt;/strong&gt; (an Asian cow thing) &lt;BR&gt;&#xA; 2) The horse &lt;strong&gt;earnt&lt;/strong&gt; its victory &lt;BR&gt;&#xA; 3) &lt;strong&gt;Bride&lt;/strong&gt;, of course &lt;BR&gt;&#xA; 4) He cheated, so his praise is &lt;strong&gt;undue&lt;/strong&gt; &lt;BR&gt;&#xA; 5) An animal that can be ridden is generally called a &lt;strong&gt;steed&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Arranged:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;Z E B U S&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;E A R N T&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;B R I D E&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;U N D U E&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;S T E E D&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11880" LastEditorUserId="5373" LastEditDate="2020-01-10T17:27:38.340" LastActivityDate="2020-01-10T17:27:38.340" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="15343" PostTypeId="1" AcceptedAnswerId="15346" CreationDate="2015-05-22T11:47:19.020" Score="11" ViewCount="786" Body="&lt;p&gt;I am usually spread over five.&lt;br&gt;&#xA;On rare occasions, like in 201, I occupy exactly 4.&lt;br&gt;&#xA;But I am the only one that can never be in six.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;" OwnerUserId="12645" LastActivityDate="2015-05-22T12:37:11.340" Title="I am the odd one" Tags="|riddle|lateral-thinking|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15344" PostTypeId="1" AcceptedAnswerId="15345" CreationDate="2015-05-22T12:21:37.740" Score="-2" ViewCount="282" Body="&lt;p&gt;Which &lt;strong&gt;noun&lt;/strong&gt; am I describing?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Bad feelings enforce it&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It's too common on the internet&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Your parents won't teach it to you&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="12640" LastEditorUserId="11475" LastEditDate="2015-05-22T14:38:44.443" LastActivityDate="2015-05-22T14:38:44.443" Title="Which noun is it?" Tags="|riddle|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15345" PostTypeId="2" ParentId="15344" CreationDate="2015-05-22T12:26:21.480" Score="1" Body="&lt;p&gt;Could it be?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; cursing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bad feelings make you curse. Way too much foul language on the Internet. And most parents seem reluctant to teach their children to curse.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or am I missing the whole point of this riddle?&lt;/p&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="5610" LastEditDate="2015-05-22T12:28:14.510" LastActivityDate="2015-05-22T12:28:14.510" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="15346" PostTypeId="2" ParentId="15343" CreationDate="2015-05-22T12:37:11.340" Score="10" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; February&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is usually spread in 5 weeks. In the year 201 it occupied 4 (if in your country the week starts on Sunday). However it can never occupy 6 weeks. All other months can.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastActivityDate="2015-05-22T12:37:11.340" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15347" PostTypeId="1" AcceptedAnswerId="15352" CreationDate="2015-05-22T13:16:17.677" Score="25" ViewCount="1657" Body="&lt;p&gt;&lt;strong&gt;Six-words:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My first is what a bouncer does if they think you're cool enough&lt;br&gt;&#xA;  My second is what my fourth does to your pain&lt;br&gt;&#xA;  My third is your surroundings on a grassy hill in summer&lt;br&gt;&#xA;  My fourth is taken to do my second to your pain&lt;br&gt;&#xA;  My fifth lives in places they do not own&lt;br&gt;&#xA;  My sixth is what you do with prize-winning animals&lt;br&gt;&#xA;  &lt;br&gt;&#xA;  My first gathers liquid on to a surface&lt;br&gt;&#xA;  My second is where Leo kept going in Inception&lt;br&gt;&#xA;  My third is a branch of the US Military&lt;br&gt;&#xA;  My fourth is what you must do to solve this puzzle&lt;br&gt;&#xA;  My fifth is something shaped like a camper's home&lt;br&gt;&#xA;  My sixth is what well-off rude people may do to the less fortunate&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Inspired by &lt;a href=&quot;https://puzzling.stackexchange.com/q/15341/9000&quot;&gt;Five-words&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/q/15339/9000&quot;&gt;Four-words&lt;/a&gt; but this puzzle is a little different as there are two things being described. They're very tightly related but they're not twins.&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-07-17T01:06:01.867" Title="Six-words (------) Six-words (||||||)" Tags="|word|wordplay|crosswords|word-square|" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="15349" PostTypeId="2" ParentId="15334" CreationDate="2015-05-22T14:52:15.037" Score="11" Body="&lt;p&gt;I think you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a cell phone&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm complex inside,&#xA;but easy on the outside.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's a transmitter/receiver (complex) in a simple (easy) case. Smartphones have even less buttons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm loud for one,&#xA;but quiet for the others.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people have ringtones some put it on silent&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm used to get hit,&#xA;but there's no alternative.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People are always tapping on it to text&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I die as fast,&#xA;as I raise again.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It refers to the battery life. When it dies, you'll plug it and it raise again&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10400" LastActivityDate="2015-05-22T14:52:15.037" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15350" PostTypeId="1" AcceptedAnswerId="15353" CreationDate="2015-05-22T15:04:36.517" Score="8" ViewCount="5419" Body="&lt;p&gt;&lt;strong&gt;Background:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Your love for Hollywood movies and remarkable ability to remember movie titles is finally going to serve a purpose. There was indeed a time when you almost believed your mother's words, as she scorned you: &amp;quot;Lazing in front of the television is not going to pay for the rent!&amp;quot; You knew better all along. As you and the rest of your family listen to the directions of your late, and wealthy, Uncle Rob's will, you cannot help but smile as the executor reads &amp;quot;... and thus half of my fortune will indeed go toward a sole beneficiary. Namely he who solves the puzzles as depicted below&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;As you are each given a paper with four Hollywood movie rebus puzzles, you begin to smile. You know the answer to each.&lt;/p&gt;&#xA;&lt;p&gt;How will you answer?&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Can you guess the movie for the 2nd puzzle?&lt;/strong&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/6mlKK.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Next puzzle: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15465/hollywood-movie-rebus-puzzle-3&quot;&gt;#3 Hollywood Rebus Movie Puzzle &lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Fee first puzzle: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15328/hollywood-movie-rebus-puzzle-1&quot;&gt;#1 Hollywood Rebus Movie Puzzle &lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12317" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-23T05:17:34.323" Title="Hollywood Movie Rebus Puzzle #2" Tags="|rebus|movies|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15351" PostTypeId="2" ParentId="9491" CreationDate="2015-05-22T15:05:26.323" Score="5" Body="&lt;p&gt;There are many solutions if you link points so that they switch together. Here are 4 switches, linked as two pairs with simple wire arches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YdKNG.jpg&quot; alt=&quot;auto switching layout&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This 'auto switching layout' requires the train to traverse 10 stretches of track to reset the whole layout, so some tracks are used more than once. More designs (which can be run in your browser) at: &lt;a href=&quot;http://www.cr31.co.uk/stagecast/trains/tt8_auto_run.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.cr31.co.uk/stagecast/trains/tt8_auto_run.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, if you obtain the old type of 'sprung' Duplo points you can build various binary counters or flip flop gates like this one which uses one sprung and a pair of linked lazy points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/libYk.jpg&quot; alt=&quot;flip flop layout&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These kind of layouts use many pieces though, so I won't list them. More Duplo layouts at: &lt;a href=&quot;http://www.cr31.co.uk/stagecast/trains/tw1_duplo.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.cr31.co.uk/stagecast/trains/tw1_duplo.html&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="12653" LastActivityDate="2015-05-22T15:05:26.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15352" PostTypeId="2" ParentId="15347" CreationDate="2015-05-22T15:05:44.677" Score="19" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;code&gt;ADMITS&lt;/code&gt; &lt;br&gt;&#xA;  &lt;code&gt;DEADEN&lt;/code&gt; &lt;br&gt;&#xA;  &lt;code&gt;SERENE&lt;/code&gt; &lt;br&gt;&#xA;  &lt;code&gt;OPIATE&lt;/code&gt; &lt;br&gt;&#xA;  &lt;code&gt;RENTER&lt;/code&gt; &lt;br&gt;&#xA;  &lt;code&gt;BREEDS&lt;/code&gt; &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What a fun little puzzle.&lt;/p&gt;&#xA;" OwnerUserId="5616" LastEditorUserId="9000" LastEditDate="2015-05-22T15:11:56.490" LastActivityDate="2015-05-22T15:11:56.490" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15353" PostTypeId="2" ParentId="15350" CreationDate="2015-05-22T15:05:45.187" Score="17" Body="&lt;p&gt;This one looks like it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Catch Me If You Can&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - The first picture is pretty clearly a &lt;strong&gt;cat&lt;/strong&gt;&lt;br&gt;&#xA; - The second picture is China minus ina, which becomes &lt;strong&gt;ch&lt;/strong&gt;&lt;br&gt;&#xA; - This is a user icon, which represents &lt;strong&gt;me&lt;/strong&gt;&lt;br&gt;&#xA; - This icon represents the .gif file format, with the first letter crossed out so that it becomes &lt;strong&gt;if&lt;/strong&gt; (thanks, pacoverflow!)&#xA; - The finger is almost certainly pointing at &lt;strong&gt;you&lt;/strong&gt;&lt;br&gt;&#xA; - This is most definitely a tin &lt;strong&gt;can&lt;/strong&gt;!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="11475" LastEditDate="2015-05-22T15:19:46.040" LastActivityDate="2015-05-22T15:19:46.040" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15354" PostTypeId="1" AcceptedAnswerId="15355" CreationDate="2015-05-22T15:24:56.990" Score="9" ViewCount="600" Body="&lt;p&gt;&lt;strong&gt;Seven-words:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My first is what the bully may show&lt;br&gt;&#xA;  My second describes some spouses post-nuptials&lt;br&gt;&#xA;  My third is when something is like something else&lt;br&gt;&#xA;  My fourth are people who tell you what to pay&lt;br&gt;&#xA;  My fifth are microscopic and may move through you&lt;br&gt;&#xA;  My sixth will reduce your quality&lt;br&gt;&#xA;  My seventh is something that Homer wrote&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In the neo-tradition of &lt;a href=&quot;https://puzzling.stackexchange.com/q/15339/9000&quot;&gt;Four-words&lt;/a&gt;, &lt;a href=&quot;https://puzzling.stackexchange.com/q/15341/9000&quot;&gt;Five-words&lt;/a&gt;, and &lt;a href=&quot;https://puzzling.stackexchange.com/q/15347/9000&quot;&gt;Six-words&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2015-07-17T01:08:26.857" Title="Seven-words (-------|||||||)" Tags="|word|wordplay|crosswords|word-square|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15355" PostTypeId="2" ParentId="15354" CreationDate="2015-05-22T15:30:27.580" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;pre&gt;BRAVADO &#xA;RENAMED &#xA;ANALOGY &#xA;VALUERS &#xA;AMOEBAS &#xA;DEGRADE &#xA;ODYSSEY&lt;/pre&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-05-22T15:30:27.580" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15356" PostTypeId="1" AcceptedAnswerId="15357" CreationDate="2015-05-22T15:41:02.483" Score="10" ViewCount="741" Body="&lt;p&gt;&lt;strong&gt;Eight-words:&lt;/strong&gt; (+1)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My first is a move in (American) football that will never score you points&lt;br&gt;&#xA;  My second is like the shaft within a flagellum&lt;br&gt;&#xA;  My third is really two words and surrounds the feline's path&lt;br&gt;&#xA;  My fourth is you sitting, ready to travel afar&lt;br&gt;&#xA;  My fifth you just got home from travelling by different means&lt;br&gt;&#xA;  My sixth is food served with the nut of a tree&lt;br&gt;&#xA;  My seventh is a boat with a sail at two score five degrees&lt;br&gt;&#xA;  My eighth are the children on a snowy weekend&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In the neo-tradition of &lt;a href=&quot;https://puzzling.stackexchange.com/q/15339/9000&quot;&gt;Four-words&lt;/a&gt;, &lt;a href=&quot;https://puzzling.stackexchange.com/q/15341/9000&quot;&gt;Five-words&lt;/a&gt;, &lt;a href=&quot;https://puzzling.stackexchange.com/q/15347/9000&quot;&gt;Six-words&lt;/a&gt;, and &lt;a href=&quot;https://puzzling.stackexchange.com/q/15354/9000&quot;&gt;Seven-words&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I suspect leoll2 may have figured me out...&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-07-17T01:07:03.520" Title="Eight-words (--------||||||||) +1" Tags="|word|wordplay|crosswords|word-square|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15357" PostTypeId="2" ParentId="15356" CreationDate="2015-05-22T15:44:14.627" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;pre&gt;LATERALS&#xA;AXONEMAL&#xA;TOEPLATE &#xA;ENPLANED&#xA;RELANDED&#xA;AMANDINE&#xA;LATENEER&#xA;SLEDDERS&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-05-22T15:44:14.627" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15358" PostTypeId="1" AcceptedAnswerId="15361" CreationDate="2015-05-22T16:01:00.407" Score="5" ViewCount="383" Body="&lt;p&gt;&lt;strong&gt;WARNING:&lt;/strong&gt; You could totally search online for this and find the answer. Note the &lt;a href=&quot;/questions/tagged/no-computer&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;no-computer&amp;#39;&quot; rel=&quot;tag&quot;&gt;no-computer&lt;/a&gt; tag. Yes, you can cheat and I wouldn't know. However, that's not nearly as much fun and you'd be trading your eternal soul for internet points. Even if this is solved quickly by someone searching or recognizing it, I like the idea and think that something like it could be incorporated into future puzzles.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The words in this sentence follow a definite pattern. What is that pattern and what should the next word be? (There are, technically, many words that fit the pattern but there is an obvious &quot;best&quot; choice.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Man, I can’t - I shan’t! - formulate an anthem where the words comprise mnemonics (dreaded mnemonics) for ... ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastActivityDate="2015-05-23T05:39:14.000" Title="What is the next word in this sentence?" Tags="|word|no-computers|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15359" PostTypeId="2" ParentId="15358" CreationDate="2015-05-22T16:08:57.277" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;pi&lt;/strong&gt;. The numerals just bother me always... even the dry anterior! Try to request something lower (zero) in numerary aptitude.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(I didn't look it up, but I &lt;em&gt;did&lt;/em&gt; recognize it.)&lt;/p&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="11876" LastEditDate="2015-05-23T05:39:14.000" LastActivityDate="2015-05-23T05:39:14.000" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15360" PostTypeId="2" ParentId="15358" CreationDate="2015-05-22T16:09:49.797" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The words have the same number of letters of pi digits.&#xA; For example, Man has 3, I has 1, can't has 4.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've already heard this from a math student (friend) few months ago. I think that the phrase is actually longer than the one you posted, since my friend claimed to know at least 100 words! I'm really not going to learn it, though!&lt;/p&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-05-22T16:09:49.797" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15361" PostTypeId="2" ParentId="15358" CreationDate="2015-05-22T16:13:04.227" Score="4" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; pi&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reason for this is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Taking the length of each word produces &lt;code&gt;3141592653589793&lt;/code&gt; which are the first 16 digits of pi. The next digit is 2, so 'pi' is a natural final word for the sentence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By the way, for what it's worth I hadn't seen this before&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I just started looking at the lengths of the words and after '314', I figured I was on the right track.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12277" LastActivityDate="2015-05-22T16:13:04.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15362" PostTypeId="1" AcceptedAnswerId="15395" CreationDate="2015-05-22T16:49:52.567" Score="6" ViewCount="6257" Body="&lt;p&gt;I'm not sure how difficult this riddle is, but hopefully it's enjoyable!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I go up and down, and am all around,&#xA;  &lt;br&gt;but not many seem to know.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;But if I crash and burn, to the ground,&#xA;  &lt;br&gt;the world may ceases to flow.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;For my existence has become a medial role, &#xA;  &lt;br&gt;that people may try to ignore.&#xA;  &lt;br&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;But those that dare play with me, &#xA;  &lt;br&gt;their lives can improve or turn sore.&#xA;  &lt;br&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;When I go up and down words are thrown around,&#xA;  &lt;br&gt;as people think they might know.&#xA;  &lt;br&gt; &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;But when the time goes by and my job is done,&#xA;  &lt;br&gt;the words of the 'wise' start to show.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the rhyming riddle referring to?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To try and keep things along the right path, here is a small hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Do not take the expressions too literally. (I.E lives turning sore is not meant to be restricted to just physical harm to someone.) Some of these expressions may contain some flare or were used more-so for the rhyming scheme but are still applicable (in some form or another).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If you enjoyed this riddle, some of my other rhyming riddles can be found &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%5Brhyme%5D%20user%3A11244%20is%3Aquestion&quot;&gt;here&lt;/a&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2017-11-03T08:29:29.043" Title="I go up and down and am all around" Tags="|riddle|rhyme|" AnswerCount="7" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15363" PostTypeId="2" ParentId="15362" CreationDate="2015-05-22T17:01:25.833" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Sun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I go up and down, and am all around, but not many seem to know.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sun rises and sets.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But if I crash and burn, to the ground, the world may ceases to flow.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Without the sun, life stops&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For my existence has become a medial role, that people may try to ignore.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The world has lights now, things run 24/7&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But those that dare play with me, their lives can improve or turn sore. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Vitamin D, sunburn&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not sure on the last two lines.&lt;/p&gt;&#xA;" OwnerUserId="12365" LastEditorUserId="12365" LastEditDate="2015-05-22T17:07:03.027" LastActivityDate="2015-05-22T17:07:03.027" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15364" PostTypeId="2" ParentId="15362" CreationDate="2015-05-22T17:07:37.943" Score="2" Body="&lt;p&gt;My answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I go up and down, and am all around, &#xA;but not many seem to know.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sea goes up and down periodically, but not everyone knows why. It also may refer to the cycle of water (water vapor, rivers, sea, ...)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But if I crash and burn, to the ground, &#xA;the world may ceases to flow.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Burn may be a reference to evaporation. If the water evaporates and its level decreases (crash to the ground), the world will suffer. Again, flow is a term strongly tied with the concept of water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For my existence has become a medial role, &#xA;that people may try to ignore. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The water is a very valuable and useful resource if you use it properly (medial role). Though, many don't care and waste it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But those that dare play with me, &#xA;their lives can improve or turn sore. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think it refers to those places slightly above the sea level. They're usually beautiful places (eg Maldives), but very threatened by climatic changes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I go up and down words are thrown around, &#xA;as people think they might know.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Go up and down is maybe a reference to tide (or the cycle of water). The words may be &quot;cloud&quot;, &quot;rain&quot;, &quot;ice&quot;, and so on...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But when the time goes by and my job is done, &#xA;the words of the 'wise' start to show.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The wise is probably God. Many people consider weather events as the expression of God.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-05-22T17:19:38.153" LastActivityDate="2015-05-22T17:19:38.153" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15365" PostTypeId="2" ParentId="15362" CreationDate="2015-05-22T17:12:23.087" Score="3" Body="&lt;p&gt;Along the same lines as a previous answer (User12365), but a little different:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The moon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I go up and down, and am all around,&lt;br&gt;&#xA;but not many seem to know.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The moon comes up, goes down, but we don't think about it very often.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But if I crash and burn, to the ground,&lt;br&gt;&#xA;the world may ceases to flow.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the moon were to crash into the earth, life would stop. Also, if the moon goes away, then the &lt;em&gt;flowing tides&lt;/em&gt; would change.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For my existence has become a medial role,&lt;br&gt;&#xA;that people may try to ignore.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Moon cycles are 28 days, along with something else that most people wish they could ignore.  Also, the light of the moon is not so necessary with the lights we now have.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But those that dare play with me,&lt;br&gt;&#xA;their lives can improve or turn sore.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; By using the tides for surfing or other activities, one can have a lot of fun... Or get incredibly beat up if you don't know what you are doing.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I go up and down words are thrown around,&lt;br&gt;&#xA;as people think they might know. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A lot of mythology with the moon, a lot of studying, and often a lot of excitement that people talk about.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But when the time goes by and my job is done,&lt;br&gt;&#xA;the words of the 'wise' start to show.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Perhaps where this falls apart. Maybe with the advance of technology and new lights, we ignore the world, but the &quot;wise&quot; have warned about the use of too much technology?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-05-22T17:12:23.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15366" PostTypeId="2" ParentId="15334" CreationDate="2015-05-22T17:24:37.270" Score="2" Body="&lt;p&gt;Is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A faulty radio that needs percussive maintenance (ie: hitting the radio)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm complex inside,&#xA;but easy on the outside.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most radios are quite simple on the outside, but can look like a rats nest inside.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm loud for one,&#xA;but quiet for the others.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Volume of the radio. Some people also use headphones.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm used to get hit,&#xA;but there's no alternative.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The percussive maintenance part. Somethings wrong with the radio and whoever is listening has no other way of fixing the radio but...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I die as fast,&#xA;as I raise again.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...percussive maintenance never really works for very long. Hitting the radio is only a quick fix.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12656" LastActivityDate="2015-05-22T17:24:37.270" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15367" PostTypeId="2" ParentId="15334" CreationDate="2015-05-22T18:06:27.313" Score="2" Body="&lt;p&gt;I would say a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; toilet flush.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm complex inside, but easy on the outside.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Try to fix one you will understand&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm loud for one, but quiet for the others.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Loud for the one who flush, not that loud for the others&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm used to get hit, but there's no alternative.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; this is what got me into in, normally there is no other alternative when you have to flush&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I die as fast, as I raise again.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Maybe some kind of metaphor about the fact that as long as the flush is finished, the 'button' is up again'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12657" LastActivityDate="2015-05-22T18:06:27.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15368" PostTypeId="2" ParentId="15334" CreationDate="2015-05-22T18:22:18.280" Score="2" Body="&lt;p&gt;it's &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; heart&lt;br&gt;&#xA; -complex inside but people just draw it ♥&lt;br&gt;&#xA; -loud for you when you run or exercise you will hear beats, but no one will hear it&lt;br&gt;&#xA; -it got hit when some one you love dies or you break up with GF (or BF)&lt;br&gt;&#xA; -if you are religious then you know that after death there is another life   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;it's idiot answer i know :D&lt;/p&gt;&#xA;" OwnerUserId="12659" LastEditorUserId="9951" LastEditDate="2015-05-22T18:27:10.757" LastActivityDate="2015-05-22T18:27:10.757" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15369" PostTypeId="2" ParentId="15334" CreationDate="2015-05-22T18:55:50.117" Score="0" Body="&lt;p&gt;Very simple this answer is.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The mind.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm complex inside,&#xA;but easy on the outside.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People are always trying to understand you, what you think, who you are. It looks simple, but only you know your true self.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm loud for one,&#xA;but quiet for others.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You may be quiet on the outside, but your mind is always speaking inside your head.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm used to get hit,&#xA;but there's no alternative.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'd call it &quot;sadness&quot;. We just can't help it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I die as fast,&#xA;as I rise again.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You're always changing what you're thinking. Always thinking of different things.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's what i think it is. &lt;/p&gt;&#xA;" OwnerUserId="12660" LastActivityDate="2015-05-22T18:55:50.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15370" PostTypeId="2" ParentId="15362" CreationDate="2015-05-22T19:05:06.363" Score="3" Body="&lt;p&gt;Don't have time to write up a full answer but my guess is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Stock Market&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12661" LastEditorUserId="9412" LastEditDate="2015-05-22T19:34:17.823" LastActivityDate="2015-05-22T19:34:17.823" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15371" PostTypeId="1" AcceptedAnswerId="15384" CreationDate="2015-05-22T19:34:07.893" Score="6" ViewCount="416" Body="&lt;p&gt;Adam and Bob play a game on a blackboard. A number $n&amp;gt;1$ is initially written on the board, and alternately they replace the number $k$ currently written with one of these:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;A)&lt;/strong&gt; A positive divisor of $k$, different from $1$ and $k$ itself. If the player chooses this move, he gains 1 point. Of course, if $k$ is prime, this move cannot be chosen!&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;B)&lt;/strong&gt; $k+1$. Choosing this move implies the loss of 1 point.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The players start with 200 points each.&lt;br&gt;&#xA;When a player runs out of points, he loses. Assuming that Adam starts the game, &lt;strong&gt;for which values of $n$ Bob has a strategy for victory?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: the players always play optimally, they're incredibly smart. The value $n$ isn't chosen by the players, it's random.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-05-22T20:07:51.567" LastActivityDate="2015-05-22T22:02:35.260" Title="Strategy game on a blackboard" Tags="|strategy|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15372" PostTypeId="1" AcceptedAnswerId="15390" CreationDate="2015-05-22T19:39:00.827" Score="7" ViewCount="675" Body="&lt;p&gt;It was the first meeting of the humans and the aliens, and things were going well: The humans and aliens had learned to understand each other's language and found that both were hoping for a peaceful relationship. In this spirit, the alien and human ambassador decided to play a game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The human ambassador put a queen on a chessboard. She explained the rules of the game:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We will take turns moving this queen around the board. Each move, it must get closer to the top left corner. Whoever moves the queen to the top left corner, however, &lt;em&gt;loses&lt;/em&gt;. The queen can move-&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At which point the alien interrupted&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Yes, yes. We know how queens move on our planet too!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This exchange proved to produce a fateful misunderstanding: The aliens believed queens could make the same moves as a rook or a knight - that is, &lt;strong&gt;they would move the piece straight up or left, or go left two and up one, or go up two and left one&lt;/strong&gt;. The humans believed queens moved in a &lt;strong&gt;straight up, diagonal, or leftward line&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not realizing this, they began playing the game. If we label the top left square $(0,0)$, the square right of it as $(1,0)$ and the square below it as $(0,1)$ and continue with coordinates analogously, then they started by placing the queen at $(2,3)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The human ambassador played first, moving the queen up to $(2,2)$, knowing this was a winning strategy. She was very surprised when the aliens made the knight-leap to $(1,0)$, forcing her to move to $(0,0)$ and lose. The alien ambassador smiled and asked,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How could you not have seen such a standard move?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The human ambassador immediately understood how the aliens believed the game worked, but politely said nothing. They played again, starting at $(4,2)$ this time. The aliens started, moving to $(3,2)$, believing this to be a winning move. They were very surprised when the humans moved the queen to $(1,0)$, diagonally upwards, forcing the aliens to lose. The aliens were too polite to say anything, but realized how the silly humans must believe queens moved.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Both species understanding how their opponent would move queens, they played many more games, and became very confident about it. After a while, the aliens asked,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Shall we make a bet? Whoever wins the next game will be declared the best species in the universe and will be entitled to ownership of their opponent's planet! We will make this fair: We choose the starting position, you choose which player moves first.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The aliens placed an enormous, $1000000\times 1000000$ chess board on the table. The human ambassador quickly accepted, knowing that a queen moving diagonally was far more powerful than one that could merely make knight leaps. The aliens placed the queen at the position $(170000,170002)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What should the human ambassador do?&lt;/p&gt;&#xA;" OwnerUserId="4765" LastEditorUserId="5373" LastEditDate="2015-05-25T18:17:51.417" LastActivityDate="2015-05-25T18:17:51.417" Title="Unsportsmanlike Aliens" Tags="|mathematics|game|nim|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15373" PostTypeId="1" AcceptedAnswerId="15375" CreationDate="2015-05-22T19:51:46.177" Score="2" ViewCount="318" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;C. Nelson Reilly&lt;/p&gt;&#xA;&lt;p&gt;Greater than or equal to&lt;/p&gt;&#xA;&lt;p&gt;Strike anywhere, yes?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;One word in many of its senses will fit this haiku&lt;/p&gt;&#xA;&lt;p&gt;Given the answer below, an explanation that is more than a hint&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;In Microsoft Excel, the MATCH() function has a parameter for &amp;quot;match type&amp;quot;; -1 is the parameter value to indicate a match type of &amp;quot;greater than or equal to&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11921" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-09-03T15:38:13.460" Title="A &quot;what am I&quot; haiku: C. Nelson Reilly / Greater than or equal to / Strike anywhere, yes?" Tags="|riddle|knowledge|poetry|haiku|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15374" PostTypeId="1" AcceptedAnswerId="15381" CreationDate="2015-05-22T19:52:19.587" Score="5" ViewCount="502" Body="&lt;p&gt;Six men are queuing in the bank when the clerk decides to take a break for 10 minutes! Before leaving, he gives to the six men a numbered ticket, in the same order of the queue (ticket #1 to the first man, #2 to the second, etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a while, when the clerk returns, the men become terribly confused and start pushing each other to become the first in line, not worried about their tickets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying to calm down the men, the clerk says, &quot;Stop! Let's make a game! There is a chance that &lt;strong&gt;all&lt;/strong&gt; of you will receive twice the money you're asking, but otherwise you won't get anything! Let me explain the rules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The last person in the queue at this moment checks his ticket number. If it's not 6 (in which case the game immediately ends), he occupies the position suggested by his ticket. Then, whoever was previously standing in that position checks his ticket, too. If it's 6, he goes to the end of the queue and the game ends, but otherwise he goes to the position indicated by his ticket, and so on. The game ends when someone, looking at his ticket, goes to the sixth position. You win the money only if, at the end of the game, you're all in the correct positions as indicated by your tickets.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What's the probability that the six men get double the money?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt; The sixth man looks at his ticket and reads 4, so he goes to the 4th position. Now the man who previously was 4th looks at his ticket and reads 6, so he occupies the last position. In this case, they win only if the men occupying the first, second, third and fifth position already were in the correct position (ergo they respectively had the tickets 1,2,3,5).&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="294" LastEditDate="2015-05-23T00:01:27.230" LastActivityDate="2015-05-23T06:39:32.527" Title="Queue in the bank" Tags="|mathematics|probability|story|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15375" PostTypeId="2" ParentId="15373" CreationDate="2015-05-22T19:55:31.863" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;br&gt; Match &lt;br&gt; Charles Nelson Reilly was a regular panelist on the Match Game &lt;br&gt; Compare is a synonym of match - tfitzger &lt;br&gt; Strike Anywhere is a type of fire-making match.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12365" LastEditorUserId="4385" LastEditDate="2015-08-27T14:00:44.630" LastActivityDate="2015-08-27T14:00:44.630" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15377" PostTypeId="2" ParentId="15334" CreationDate="2015-05-22T20:05:29.177" Score="6" Body="&lt;p&gt;I would say the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a keyboard (or keys on a keyboard)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm complex inside,&#xA;but easy on the outside.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; keyboards are composed of simple squares on the outside, but complex mechanical and electrical components on the inside&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm loud for one,&#xA;but quiet for others.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; keyboards can be loud (mechanical) or quiet (silicone)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm used to get hit,&#xA;but there's no alternative.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you HAVE to hit the keys on the keyboard in order to use it. There's no way around it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I die as fast,&#xA;as I rise again.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; keys are depressed as fast as they spring back up&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10609" LastActivityDate="2015-05-22T20:05:29.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15379" PostTypeId="2" ParentId="15371" CreationDate="2015-05-22T20:46:09.400" Score="0" Body="&lt;p&gt;So far we have this list of winning numbers (if player A gets one of these numbers, then player A will win):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;2, 4, 6, 7, 9, 10&lt;br&gt;In addition, all numbers that are multiples of a losing number will be a winning number, because if Player A got those numbers, they could simplify it to the losing number for the other player to get.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And here are all the losing numbers (if player A gets one of these numbers, then player A will lose):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;3, 5, 8, 11&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll go about answering this by making a set of rules:&#xA;(If someone could edit this and put the rules in spoilers I'd appreciate it. Not sure if that's possible using ul/ol. I couldn't figure how to do the normal spoiler with lists)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If a number can be simplified to a losing number, then it is a winning number&lt;/li&gt;&#xA;&lt;li&gt;Someone who receives a 5 loses by the following turn order:&#xA;&lt;ol&gt;&lt;ol&gt;&#xA;&lt;li&gt;A receives 5, forced to put 6; -1&lt;/li&gt;&#xA;&lt;li&gt;B puts 3; +1&lt;/li&gt;&#xA;&lt;li&gt;A forced to put 4; -1&lt;/li&gt;&#xA;&lt;li&gt;B puts 5; -1&lt;/li&gt;&#xA;&lt;li&gt;This move sequence repeats, with B forcing all of A's moves&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/ol&gt;&#xA;This leaves A with a -2 net and B with a 0 net, meaning that B eventually wins. Let's call this set of turns LS5 (loss sequence 5).&lt;/li&gt;&#xA;&#xA;&lt;li&gt;From the LS5, anyone who gets a 3 would also lose &lt;/li&gt;&#xA;&#xA;&lt;li&gt;Again building off LS5, anyone who can force a 5 or a 3 would win. This means that anyone who is given a multiple of either 5 or 3 could use move A to give the other player 5 or 3 respectively&lt;/li&gt;&#xA;&#xA;&lt;li&gt;A player should never use move A to simplify to a 2, because the opponent could use move B to make it 3, and thus eventually winning by LS.&lt;/li&gt;&#xA;&#xA;&lt;li&gt;A player that receives an 8 loses by the following turn order:&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A gets 8, puts 4&lt;/li&gt;&#xA;&lt;li&gt;B gets 4, puts 5; LS5 continues, A loses&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;-OR-&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A gets 8 puts 2, loses by the rule above&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;-OR-&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A gets 8, puts 9&lt;/li&gt;&#xA;&lt;li&gt;B gets 9, puts 3; LS5 continues, A loses&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;By this, a player that gets 9 wins. Let's call these LC8 and LC9&#xA;&lt;/li&gt;&#xA;&#xA;&lt;li&gt;A player that gets 10 uses move A to return 5 and wins by LC5.&lt;/li&gt;&#xA;&#xA;&lt;li&gt;A player who gets 11 loses by:&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A gets 11, must make it 12&lt;/li&gt;&#xA;&lt;li&gt;B gets 12, makes it 3, wins by LC5&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/li&gt;&#xA;&#xA;&#xA;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11727" LastEditorUserId="11727" LastEditDate="2015-05-22T22:02:35.260" LastActivityDate="2015-05-22T22:02:35.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15380" PostTypeId="1" AcceptedAnswerId="15401" CreationDate="2015-05-22T20:52:17.703" Score="3" ViewCount="450" Body="&lt;p&gt;My 1st go at this (and first post here). I hope you like.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;This puzzle is degrading.&lt;/li&gt;&#xA;  &lt;li&gt;There are plants (with ain)&lt;/li&gt;&#xA;  &lt;li&gt;And the clues are quite obtuse.&lt;/li&gt;&#xA;  &lt;li&gt;It is a smelly, meaty extrude of a puzzle:&lt;/li&gt;&#xA;  &lt;li&gt;My excrement flowing out of me.&lt;/li&gt;&#xA;  &lt;li&gt;I take pleasure in your suffering&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12666" LastEditorUserId="12109" LastEditDate="2015-07-17T01:08:58.020" LastActivityDate="2015-07-17T01:08:58.020" Title="Six-words (------||||||)" Tags="|word|wordplay|crosswords|word-square|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15381" PostTypeId="2" ParentId="15374" CreationDate="2015-05-22T21:00:24.673" Score="6" Body="&lt;p&gt;Here's a long, rather brute-force calculation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a $\frac{1}{6}$ chance that the first person to check his ticket has ticket #6. If this is the case, then the men win the game if they happen to already be in the right order, which happens with probability $\frac{1}{5!} = \frac{1}{120}$. Otherwise, let the probability that the men win once the person at the end of the line switches positions be $P_1$. (We will calculate the value of $P_1$ below.) Altogether, then, the probability that the men with the game is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{6}(\frac{1}{120}) + \frac{5}{6}P_1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we calculate $P_1$. Once the first man moves into position, there is a $\frac{1}{5}$ chance that the second man has ticket #6. If this is the case, then we know that the first and second men are in the correct positions once the second man moves to the back, so the probability that the men win is the probability that the other four men are in the correct positions, or $\frac{1}{4!} = \frac{1}{24}$. Otherwise, let the probability that the men win once the second man moves into position be $P_2$. The value of $P_1$ in terms of $P_2$ is therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{5}(\frac{1}{24}) + \frac{4}{5}P_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We continue recursively until we reach the base case where the sixth man to move is the one holding ticket #6:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Once the third man moves into position, he has ticket #6 with probability $\frac{1}{4}$ and in that case, the chances that the remaining three men are in position is $\frac{1}{3!}$. Hence, $P_2 = \frac{1}{4}(\frac{1}{6}) + \frac{3}{4}P_3$.&lt;/li&gt;&#xA;&lt;li&gt;Once the fourth man moves into position, he has ticket #6 with probability $\frac{1}{3}$ and in that case, the chances that the remaining two men are in position is $\frac{1}{2!}$. Hence, $P_3 = \frac{1}{3}(\frac{1}{2}) + \frac{2}{3}P_4$.&lt;/li&gt;&#xA;&lt;li&gt;Once the fifth man moves into position, he has ticket #6 with probability $\frac{1}{2}$ and in that case, the only remaining man must be in the correct position. Hence, $P_4 = \frac{1}{2}(1) + \frac{1}{2}P_5$.&lt;/li&gt;&#xA;&lt;li&gt;If the final man has ticket #6, then the men must win. Hence, $P_5 = 1$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Working our way back up, calculating exact values for $P_1$ through $P_5$, we find:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$P_4 = \frac{1}{2}(1) + \frac{1}{2}P_5 = \frac{1}{2}(1) + \frac{1}{2}(1) = 1$&lt;/li&gt;&#xA;&lt;li&gt;$P_3 = \frac{1}{3}(\frac{1}{2}) + \frac{2}{3}P_4 = \frac{1}{3}(\frac{1}{2}) + \frac{2}{3}(1) = \frac{5}{6}$&lt;/li&gt;&#xA;&lt;li&gt;$P_2 = \frac{1}{4}(\frac{1}{6}) + \frac{3}{4}P_3 = \frac{1}{4}(\frac{1}{6}) + \frac{3}{4}(\frac{5}{6}) = \frac{2}{3}$&lt;/li&gt;&#xA;&lt;li&gt;$P_1 = \frac{1}{5}(\frac{1}{24}) + \frac{4}{5}P_2 = \frac{1}{5}(\frac{1}{24}) + \frac{4}{5}(\frac{2}{3}) = \frac{13}{24}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And finally, the overall probability that the men win the game is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{6}(\frac{1}{120}) + \frac{5}{6}P_1 = \frac{1}{6}(\frac{1}{120}) + \frac{5}{6}(\frac{13}{24}) = \frac{163}{360} = 0.452\overline{7}$&lt;/p&gt;&#xA;" OwnerUserId="294" LastEditorUserId="294" LastEditDate="2015-05-22T21:02:07.350" LastActivityDate="2015-05-22T21:02:07.350" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15382" PostTypeId="2" ParentId="15371" CreationDate="2015-05-22T21:17:34.933" Score="0" Body="&lt;p&gt;A &lt;em&gt;3-4-5-6-3 loop&lt;/em&gt; pattern can be described as: Player 1 gets the 3 and is forced to lose a point and go to 4. Player 2 chooses to lose a point and go to 5 forcing Player 1 to go to 6 losing another point. Player 2 then chooses 3 getting his point back. This nets a result of -2 for Player 1 and 0 for Player 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following scenarios result from a &lt;em&gt;3-4-5-6-3 loop&lt;/em&gt;&lt;br&gt;&#xA;2: A+1 =&gt; 3; B+1 =&gt; 4; A+1 =&gt;5; B+1 =&gt;6; A Sets to 3; Result A wins.&lt;br&gt;&#xA;3: A+1 =&gt; 4; B+1 =&gt; 5; A+1 =&gt;6; B sets to 3; Result B wins.&lt;br&gt;&#xA;4: A+1 =&gt; 5; B+1 =&gt; 6; A sets to 3; Result A wins.&lt;br&gt;&#xA;5: A+1 =&gt; 6; B Sets to 3; Result B wins.&lt;br&gt;&#xA;6: A sets to 3; A wins.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Landing an 8 can be described by the following: Player 1 can go to 2 or 4 (see above) or can increment to 9 losing a point. Player 2 then chooses 3 (see above). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;7: A+1 =&gt; 8; A wins&lt;br&gt;&#xA;8: B wins&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point we know that Bob will win 3, 5, or 8.&#xA;Adam will win 2,4,6,7 and any multiple of 3,5,or 8 (excluding 3,5 and 8)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;11: A+1 =&gt; 12; B sets to 3; B Wins (We can conclude that any prime number that   is x-1, where x is a winning number for Player 1, will result in a winning number for Player 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bob wins if one of the following is met:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;n = 3  &lt;/li&gt;&#xA;&lt;li&gt;n = 5&lt;/li&gt;&#xA;&lt;li&gt;n = 8&lt;/li&gt;&#xA;&lt;li&gt;n is a prime number that is 1 less than a winning number for Adam&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(I realize this is most likely an incomplete set)&lt;/p&gt;&#xA;" OwnerUserId="3922" LastActivityDate="2015-05-22T21:17:34.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15383" PostTypeId="2" ParentId="6635" CreationDate="2015-05-22T21:29:16.557" Score="2" Body="&lt;h1&gt;Partial answer: interpretation of pseudocode&lt;/h1&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;code&gt;getChar()&lt;/code&gt; method:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;set &lt;code&gt;target&lt;/code&gt; to &lt;code&gt;max(min(range(4.25*4,floor(pi**e))),0,1,2,3,-∞)&lt;/code&gt; and returns it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$target=max(min(range(4.25*4,floor(pi**e))),0,1,2,3,-∞)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=max(min(range(4.25*4,floor(pi^e))),0,1,2,3,-∞)$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;** is a python operator, meaning ^&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$=max(min(range(4.25*4,floor(22.4591577184))),0,1,2,3,-∞)$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;floor is a math function that returns the greatest integer smaller than the input&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$=max(min(range(4.25*4,22)),0,1,2,3,-∞)$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$4.25*4=17$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$=max(min(range(17,22)),0,1,2,3,-∞)$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;range is a pythn operator that returns the set of all integers between the two inputs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$=max(min({17,18,19,20,21,22})),0,1,2,3,-∞)$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;minimum of the set is 17&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$=max(17,0,1,2,3,-∞)=17$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;maximum of the numbers is 17&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So &lt;code&gt;getChar()&lt;/code&gt; returns 17.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;code&gt;answer()&lt;/code&gt; method:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;set ca to al.upper[3] &amp;amp; al.lower[indices(3,5,19,5,18)]&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This gives &lt;code&gt;ca&lt;/code&gt; = &quot;Cceser&quot;, which is likely a misspelling of &quot;Caesar&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;set ci to al.upper[9] &amp;amp; al.lower[indices(13,16,15,19,19,9,2,12,5)]&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This gives &lt;code&gt;ci&lt;/code&gt; = &quot;Impossible&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;set ac to new Int[ca.length]&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;code&gt;ac&lt;/code&gt; is set to the integer &lt;code&gt;ca.length&lt;/code&gt; = 6&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;set ic to new Int[ci.length]&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;code&gt;ic&lt;/code&gt; is set to the integer &lt;code&gt;ci.length&lt;/code&gt; = 10&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;return ca if ac &amp;lt; ic else ci&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;code&gt;6 &amp;lt; 10&lt;/code&gt; so &lt;code&gt;ac &amp;lt; ic&lt;/code&gt;. Return &lt;code&gt;ca&lt;/code&gt;, or &quot;Caesar&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;&lt;code&gt;pre()&lt;/code&gt; method:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;returns the result of &lt;code&gt;answer()&lt;/code&gt; and &lt;code&gt;getChar()&lt;/code&gt; as the preamble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As stated before, &lt;code&gt;answer()&lt;/code&gt; returns &quot;Caesar&quot; and &lt;code&gt;getChar()&lt;/code&gt; returns 17. So &lt;code&gt;pre()&lt;/code&gt; returns &lt;code&gt;[&quot;Caesar&quot;, 17]&lt;/code&gt;, likely meaning a Caesar cipher with key 17.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Some really weird things:&lt;/h2&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;qwerty&lt;/code&gt; as defined in &lt;code&gt;pre()&lt;/code&gt; is never actually used. Perhaps it means the shift is on a keyboard? Then why does the code have both capital and lowercase letters? There are only capital letters in a keyboard.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There is &lt;code&gt;{EOF}&lt;/code&gt; at the end of the code, which likely means end-of-file. What is its significance?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The code has capital letters, lowercase letters, and numbers. How exactly are we going to shift?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="12109" LastEditorUserId="12109" LastEditDate="2015-05-30T13:37:04.260" LastActivityDate="2015-05-30T13:37:04.260" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15384" PostTypeId="2" ParentId="15371" CreationDate="2015-05-22T21:32:21.490" Score="3" Body="&lt;p&gt;As an appetizer, here is an insightful, but incorrect strategy which we will develop into a correct strategy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you receive a composite number, write down an odd prime factor $p$ of it. The opponent will write down $p+1$, which will be composite again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This suggests that all composite numbers are wins (i.e. wins by the player who receives the number - so &lt;em&gt;Adam&lt;/em&gt; would win these, not Bob). The only issue here is that the strategy breaks down if the given number $n$ is of the form $2^k$, as it would have no odd prime factors. In fact, noting that the first player may force the cycle $2,3,4,5,6,3,4,5,6,\ldots$, we find that $2$ is a win, despite being prime. Similarly $8$ is a loss, despite being composite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivated by the above strategy we ask: What is the set $P$ of primes which are winning positions? To figure this out, consider that for $p\in P$, we must have that $p+1$ is a losing position. However, for $p$ other than $2$ it holds that $p+1$ is composite. In this case, it must be that all prime factors of $p+1$ are in $P$, as otherwise $p+1$ would be a win by playing a losing prime factor of it (i.e. one not in $P$). More generally, $p+1$ must not be divisible by any losing position $n'$, as moving to $n'$ would then be a winning strategy for $p+1$, making a $p$ a loss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the above observations, we may prove that&#xA;$$P=\{2,7,13\}.$$&#xA;Now, for contradiction, let $p$ be the minimal winning prime not in the above set. The above arguments suffice to show that&#xA;$$p+1=2^a7^b13^c$$&#xA;for some integers $a,b,c$. Moreover, one may quickly check that all of the following are losing positions by tracing out a few moves until the earlier mentioned cycle is achieved by the opposing player:&#xA;$$2^3=8$$&#xA;$$7^2=49$$&#xA;$$13^2=169$$&#xA;$$2\cdot 7=14$$&#xA;$$2\cdot 13=26$$&#xA;$$7\cdot 13=91$$&#xA;The only non-trivial products of the form $2^a7^b13^c$ not divisible by one of the above are $2$, $4$, $7$, and $13$, all of which are winning positions. However, none of the six losing values listed are one greater than a prime not already in $P$ - hence there is no valid solution for $p$ and our list of winning primes is complete. Moreover, the six losing values listed are all of the losing composite values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the above, we can find that, if you receive $n$ on your turn, then the optimal move is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If $n$ is prime or equals $4$, then play $n+1$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $n$ is one of $\{8,\,49,\,169,\,14,\,26,\,91\}$ then it doesn't matter what you do. You lose.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $n$ is of the form $2^a7^b13^c$ and not in one of the above cases, then play one of $\{8,\,49,\,169,\,14,\,26,\,91\}$ (at least one of which must be legal).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $n$ is composite and has a prime factor other than $2$, $7$, or $13$, play some such prime factor.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This attains a win from all winning positions, and is hence optimal. It is obvious that the player destined for a win will not run out of points too soon, as they will always be choosing a divisor, except for $n=2,\,4,\,7,\,13$, and the game will pass through each of those numbers at most once (before reaching the win at $n=6$) - so as long as each player has at least $5$ points, it won't affect the result. To answer the puzzle, Bob has a winning strategy for:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Any prime other than $2$, $7$, and $13$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The numbers $8,\,49,\,169,\,14,\,26,$ and $91$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4765" LastEditorUserId="4765" LastEditDate="2015-05-22T21:59:32.000" LastActivityDate="2015-05-22T21:59:32.000" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15385" PostTypeId="2" ParentId="15334" CreationDate="2015-05-22T22:36:52.600" Score="2" Body="&lt;p&gt;I know the solution has been found, but here is another answer I thought of before reading the existing suggestions. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am polyurethane foam, or some other soft man-made foamed material like neoprene.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm complex inside, but easy on the outside.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Try mathematically modelling a foam. The surface is frequently easy, because it is almost smooth (if the cells are closed on the surface), the porous media inside is very complex. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm loud for one, but quiet for others.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Acoustic foam is used for making rooms quieter/reducing echoes within the room. But contrary to popular opinion, it does not soundproof the room, so that the neighbor still hears your bagpipe lessons. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm used to getting hit, but there's no alternative.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; These foams are used to buffer physical impacts. For example, the foam around the car door hole gets hit by the door every time it's closed. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I die as fast, as I rise again &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I admit the word &quot;die&quot; does not fit my solution so well. But these foams are very, very elastic. You can depress them, and once you remove the weight, they rise up again in seconds, getting to their previous height and shape. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2811" LastActivityDate="2015-05-22T22:36:52.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15386" PostTypeId="1" CreationDate="2015-05-22T23:07:53.507" Score="-3" ViewCount="360" Body="&lt;p&gt;I don't understand any of this. I'm trying to win a contest. I can be found in clothing, canned goods, stores, libraries, storage bins, and schools. What am I?&lt;/p&gt;&#xA;" OwnerUserId="12670" LastEditorUserId="7308" LastEditDate="2015-05-22T23:19:08.680" LastActivityDate="2015-05-24T05:38:37.657" Title="I'm confused and I need help" Tags="|riddle|" AnswerCount="2" CommentCount="3" ClosedDate="2015-05-24T03:11:48.400" ContentLicense="CC BY-SA 3.0" />
  <row Id="15387" PostTypeId="2" ParentId="252" CreationDate="2015-05-22T23:54:40.193" Score="1" Body="&lt;p&gt;Download the &lt;a href=&quot;http://arstechnica.com/information-technology/2015/05/prime-minister-of-singapore-shares-his-c-code-for-sudoku-solver/&quot; rel=&quot;nofollow&quot;&gt;prime minister of Singapore's Sudoku solver&lt;/a&gt; and feed it this puzzle (ONLY if you're REALLY stuck). Believe it or not, that prime minister made a pretty robust program, and although it looks like it gets stuck for a while there, it eventually comes out with the following solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;862 || 751 || 349&lt;br&gt;&#xA;943 || 628 || 157&lt;br&gt;&#xA;571 || 493 || 286&lt;br&gt;&#xA;============&lt;br&gt;&#xA;159 || 387 || 624&lt;br&gt;&#xA;386 || 245 || 791&lt;br&gt;&#xA;724 || 169 || 835&lt;br&gt;&#xA;============&lt;br&gt;&#xA;217 || 934 || 568&lt;br&gt;&#xA;438 || 576 || 912&lt;br&gt;&#xA;695 || 812 || 473  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Apparently it is possible to solve with logic, though, according to the guy who invented this puzzle. It just took the solvers 24 hours to do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: This puzzle has the 1 on the 7th line in a different position as the question's. This puzzle has multiple solutions.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="12671" LastEditorUserId="14371" LastEditDate="2015-08-03T05:04:52.907" LastActivityDate="2015-08-03T05:04:52.907" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15388" PostTypeId="2" ParentId="15380" CreationDate="2015-05-23T00:16:35.700" Score="1" Body="&lt;p&gt;&lt;em&gt;2. There are plants (with ain)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Plantain looks a bit like banana, so I'll say BANANA for this one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;4. It is a smelly, meaty extrude of a puzzle:&lt;br&gt;&#xA;5. My excrement flowing out of me.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One of these is probably MANURE. If the second clue is BANANA, then it must be the fourth and not the fifth that's MANURE.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;1. This puzzle is degrading.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Given 2 and 4, this could now be OBAMAS (&quot;we will &lt;em&gt;degrade&lt;/em&gt; and ultimately destroy...&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;3. And the clues are quite obtuse.&lt;br&gt;&#xA;6. I take pleasure in your suffering&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Still not sure about these.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-23T00:16:35.700" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15389" PostTypeId="1" AcceptedAnswerId="15391" CreationDate="2015-05-23T00:21:00.163" Score="27" ViewCount="2544" Body="&lt;p&gt;&lt;em&gt;Billy&lt;/em&gt; and &lt;em&gt;Matthew&lt;/em&gt; decide to play a game of Chess. They live far away, so they decide to do it online.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matthew wants to play &lt;strong&gt;first&lt;/strong&gt;, but the game randomly gave him the color &lt;strong&gt;black&lt;/strong&gt; (who goes second.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible for them to &lt;strong&gt;cooperatively&lt;/strong&gt; move their &lt;strong&gt;knights&lt;/strong&gt; in such a way that when put back them in their &lt;em&gt;original places&lt;/em&gt;, it will be &lt;strong&gt;Matthew's&lt;/strong&gt; &lt;em&gt;turn&lt;/em&gt; to move? If not prove it's impossible.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="1777" LastEditDate="2015-05-23T00:30:53.657" LastActivityDate="2015-05-25T11:22:25.417" Title="Black wants to go first!" Tags="|strategy|chess|checkerboard|" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15390" PostTypeId="2" ParentId="15372" CreationDate="2015-05-23T00:24:56.377" Score="6" Body="&lt;p&gt;The human ambassador cannot win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If either player moves to (0,x) where x&gt;1, they lose because their opponent moves to (0,1). Similarly, (x,0) loses.&#xA;(1,x) and (x,1) where x&gt;0 also lose for the same reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the human ever moves to (2,2), the alien can move to (1,0), so the human loses. But if the alien moves to (2,2), the human has no move that doesn't lose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if the human ever moves to (2,x) or (x,2) where x&gt;2, they lose because the alien can move to (2,2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the alien moves to (2,x) or (x,2) where x&gt;3, the human has no move that does not go to one of the previous losing positions. The alien always has such a move unless the human moves to (3,x) or (x,3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then the alien can move to (3,3), where the human has no move that does not lose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the alien always wins.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-05-23T00:24:56.377" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15391" PostTypeId="2" ParentId="15389" CreationDate="2015-05-23T00:31:53.423" Score="37" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It is impossible. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initially, there are an even number of knights on white squares (namely, there are two of them, at b1 and g8).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every time a knight moves, the number of knights on white squares either increases by one (if a knight on a black square moves) or decreases by one (if a knight on a white square moves).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way, the parity changes each turn. Thus, if the board ever returns to the initial condition, it must have done so after an even number of moves, meaning it will still be Billy going first.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10281" LastEditDate="2015-05-25T11:22:25.417" LastActivityDate="2015-05-25T11:22:25.417" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15392" PostTypeId="2" ParentId="15374" CreationDate="2015-05-23T00:40:32.697" Score="3" Body="&lt;p&gt;Kevin (who has successfully escaped from &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15127/there-are-at-least-three-kevins-on-puzzling-i-plan-to-trap-them-in-my-labyrinth&quot;&gt;my labyrinth&lt;/a&gt;) has the right final answer, but here's a shorter, simpler, and hopefully more intuitive proof.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The probability that the game ends immediately is $\frac{1}{6}$ (the man at the back has ticket #6). In this case, they get double their money with probability $\frac{1}{5!}$.&lt;/li&gt;&#xA;&lt;li&gt;The probability that it ends after the first man moves into position is $\frac{5}{6}\times\frac{1}{5}$ (the man at the back doesn't have ticket #6 but the man he replaces does). In this case, they get double their money with probability $\frac{1}{4!}$.&lt;/li&gt;&#xA;&lt;li&gt;The probability that it ends after the first two men move into position is $\frac{5}{6}\times\frac{4}{5}\times\frac{1}{4}$ (neither the man at the back nor the man he replaces has ticket #6 but the next man does). In this case, they get double their money with probability $\frac{1}{3!}$.&lt;/li&gt;&#xA;&lt;li&gt;Similarly, the probability that it ends after the first three men move into position is $\frac{5}{6}\times\frac{4}{5}\times\frac{3}{4}\times\frac{1}{3}$ and in this case they get double their money with probability $\frac{1}{2!}$.&lt;/li&gt;&#xA;&lt;li&gt;Similarly, the probability that it ends after the first four men move into position is $\frac{5}{6}\times\frac{4}{5}\times\frac{3}{4}\times\frac{2}{3}\times\frac{1}{2}$ and in this case they get double their money with probability $\frac{1}{1!}$.&lt;/li&gt;&#xA;&lt;li&gt;Finally, the probability that it ends after the first five men move into position is also $\frac{5}{6}\times\frac{4}{5}\times\frac{3}{4}\times\frac{2}{3}\times\frac{1}{2}$ and in this case they get double their money with probability $1$ since everyone's in position.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So the game is equally likely to end at any of the six possible times, and the answer is $\frac{1}{6}\times(\frac{1}{5!}+\frac{1}{4!}+\frac{1}{3!}+\frac{1}{2!}+\frac{1}{1!}+1)=\frac{1+5+20+60+120+120}{720}=\frac{326}{720}.$&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-05-23T00:40:32.697" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15393" PostTypeId="1" AcceptedAnswerId="15403" CreationDate="2015-05-23T00:50:17.477" Score="3" ViewCount="323" Body="&lt;p&gt;This is my first crossword, so I'd be grateful for any feedback. All the clues are names of countries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/O4Hsf.png&quot; alt=&quot;crossword&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Across&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;6A. The loving start of Lady Jessica (6)&lt;br&gt;&#xA;8A. Dock you, girl (8)&lt;br&gt;&#xA;9A. Long Joanna (9)&lt;br&gt;&#xA;10A. Nasty men from a bird beneath darkness, perhaps (12)&lt;br&gt;&#xA;11A. Four Greek teens are coming round (6)&lt;br&gt;&#xA;13A. I get it back around (7)&lt;br&gt;&#xA;14A. A bird beneath darkness (6)&lt;br&gt;&#xA;15A. Land of freedom, apparently (7)&lt;br&gt;&#xA;16A. Either spicy or cold (5)&lt;br&gt;&#xA;17A. Mixed precipitation (4)&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Down&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;1D. Around 22 yards (5)&lt;br&gt;&#xA;2D. TV man without insole in war zone (7)&lt;br&gt;&#xA;3D. MP longs for home in Runnymede (11)&lt;br&gt;&#xA;4D. No longer a virgin (4)&lt;br&gt;&#xA;5D. Takeaway (4)&lt;br&gt;&#xA;7D. Almost a snake moving backwards (6)&lt;br&gt;&#xA;8D. The land of long sticks and clean shoes (6)&lt;br&gt;&#xA;10D. Sunburnt wackily (8)&lt;br&gt;&#xA;12D. Ageless giant mammals (5)&lt;br&gt;&#xA;16D. A paper circle (4)&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-05-23T10:05:30.860" LastActivityDate="2015-05-24T16:36:01.820" Title="Cross-country words!" Tags="|geography|cryptic-crosswords|crosswords|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15395" PostTypeId="2" ParentId="15362" CreationDate="2015-05-23T01:00:32.623" Score="2" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; share prices?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I go up and down, and am all around,&lt;br&gt;&#xA;but not many seem to know.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They go up and down all the time and are unpredictable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But if I crash and burn, to the ground,&lt;br&gt;&#xA;the world may ceases to flow.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A stock market crash can be a catastrophe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;For my existence has become a medial role,&lt;br&gt;&#xA;that people may try to ignore.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's always there, necessary for the economy, but most people don't get involved with it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But those that dare play with me,&lt;br&gt;&#xA;their lives can improve or turn sore.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gamble on share prices and you could win or lose millions. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;When I go up and down words are thrown around,&lt;br&gt;&#xA;as people think they might know.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are whole sections of newspapers dedicated to following the share prices.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But when the time goes by and my job is done,&lt;br&gt;&#xA;the words of the 'wise' start to show.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The words of the analysts who predicted the behaviour of the share prices will be proved true or false once the market has closed (or the shares have expired, or whatever) - thanks to &lt;strong&gt;ace&lt;/strong&gt; for sorting this clue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-05-24T00:42:48.823" LastActivityDate="2015-05-24T00:42:48.823" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15396" PostTypeId="2" ParentId="15389" CreationDate="2015-05-23T01:55:14.720" Score="2" Body="&lt;p&gt;Agreed with Mike. My explanation would be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Knights always move to the opposite colour square to that which they're currently on. Therefore, for a knight to move somewhere else and then move back to where it started must take an even number of moves (they're moving to the same colour they started on). An even number of moves from both players will always end with white to move.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit (as Édouard rightly points out):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If a player swaps their knights that requires two lots of odd moves: still an even number of moves in total.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12666" LastEditorUserId="12666" LastEditDate="2015-05-23T09:41:46.920" LastActivityDate="2015-05-23T09:41:46.920" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15398" PostTypeId="2" ParentId="15374" CreationDate="2015-05-23T03:31:23.597" Score="1" Body="&lt;p&gt;Lets think of the problem in terms of permutations. The initial positions of the 6 men form one of 6! permutations. The clerk's action is to apply the cycle of the permutation containing 6 (see &quot;Cycle notation&quot;). This sorts a subset of the men.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g the ordering 351642 corresponds to the permutation (13)(2645). The clerk applies the (2645) part to make the order 321456.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now the problem is to calculate how many permutations have either a single cycle or all cycles not containing 6 already sorted. Consider each possible length of the subcycle containing 6. Adding a term for each of these possibilities gives $1+5+5*4+5*4*3+5*4*3*2+5*4*3*2*1 = 326$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1 way to have 6 on its own (everything already sorted)&lt;/li&gt;&#xA;&lt;li&gt;5 ways to swap 6 with something&lt;/li&gt;&#xA;&lt;li&gt;$5*4$ ways to form a triple&lt;/li&gt;&#xA;&lt;li&gt;etc.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The final probability is thus 326/6!&lt;/p&gt;&#xA;" OwnerUserId="12672" LastEditorUserId="12672" LastEditDate="2015-05-23T03:41:56.213" LastActivityDate="2015-05-23T03:41:56.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15400" PostTypeId="2" ParentId="15326" CreationDate="2015-05-23T04:55:37.167" Score="0" Body="&lt;p&gt;The plain text quote is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;mathematics is the queen of sciences and number theory is the queen of mathematics&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The decryption method is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1. The ciphertext is a variation of base-26 using just the letters of the alphabet&lt;br&gt;&#xA;2. Translate from a-y to true base-26 (1-9-a-p). z translates to 10.&lt;br&gt;&#xA;3. Convert from true base-26 to decimal.&lt;br&gt;&#xA;4. Then recognize that each decimal number is a semiprime (the product of two prime numbers)&lt;br&gt;&#xA;5. The first factors are the ordered prime numbers (2,3,5,7,11...) up to 101, then repeated&lt;br&gt;&#xA;6. The second factors are the semiprimes / first factors&lt;br&gt;&#xA;7. Then determine $n$ as the $n^{th}$ prime for each factor&lt;br&gt;&#xA;8. Translate each $n$ to the $n^{th}$ letter of the alphabet&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/vKS8E.png&quot; alt=&quot;Table&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastActivityDate="2015-05-23T04:55:37.167" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15401" PostTypeId="2" ParentId="15380" CreationDate="2015-05-23T04:56:39.537" Score="4" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;pre&gt;&lt;code&gt;A B A S E S&#xA;B A N A N A&#xA;A N G L E D&#xA;S A L A M I&#xA;E N E M A S&#xA;S A D I S T&lt;/code&gt;&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think the only ones that require particular explanation are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BANANA - As per explanation from rand al'thor, a banana looks a bit like a plant-ain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ANGLED - As in obtuse angle &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12277" LastEditorUserId="12277" LastEditDate="2015-05-23T07:16:52.633" LastActivityDate="2015-05-23T07:16:52.633" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15402" PostTypeId="2" ParentId="15374" CreationDate="2015-05-23T06:39:32.527" Score="-1" Body="&lt;p&gt;My answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 32 / 6! = 32 / 720&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Considering Cm,n as combination of m elements for each n (or whatever is the proper wording in english), the total probability is the sum of probabilities for &#xA;the required number of &lt;strong&gt;replacements&lt;/strong&gt; until the number &lt;strong&gt;6&lt;/strong&gt; gets to the end of the line:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0 replacement  =&gt; C5,0 = 1&lt;br&gt;&#xA; 1 replacement  =&gt; C5,1 = 5&lt;br&gt;&#xA; 2 replacements =&gt; C5,2 = 10&lt;br&gt;&#xA; 3 replacements =&gt; C5,3 = 10&lt;br&gt;&#xA; 4 replacements =&gt; C5,4 = 5&lt;br&gt;&#xA; 5 replacements =&gt; C5,5 = 1&lt;br&gt;&#xA;&lt;br&gt;&#xA; total probability = (1 + 5 + 10 + 10 + 5 + 1) / 6!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12205" LastActivityDate="2015-05-23T06:39:32.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15403" PostTypeId="2" ParentId="15393" CreationDate="2015-05-23T07:14:58.943" Score="5" Body="&lt;p&gt;My attempt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6A&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; FRANCE - This one is a bit of a punt, I had to look it up to make any sense of it - Lady Jessica is a character in Dune (with a reputation for romance - thanks rand al'thor), played by Francesca Annis in the film version. Start of FRANCESCA is FRANCE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;8A&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PORTUGAL - PORT (dock) U (you) GAL (girl)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;9A&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ARGENTINA - Joanna is a female form of John (thanks for the pointer rand al'thor). 'Long John' references Long John Silver. Argentina is 'land of silver' &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;10A&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TURKMENISTAN - TURK MEN (men from a 'bird beneath darkness' - see 14A). ISTAN is nearly an anagram of 'nasty' so it could be 'turn nasty' &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;11A&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MEXICO - XI (fourteenth Greek 'four Greek teens' letter), inside an anagram of COME (are coming)? Not completely sure&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;13A&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; NIGERIA - Anagram of IREGAIN (I get it back)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;14A&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TURKEY - A bird. Turkey is south of (beneath) the Black Sea (darkness) (thanks for the hint rand al'thor)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;15A&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LIBERIA - Means 'Land of the free'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;16A&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CHILE - Chilli (spicy) or chilly (cold)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;17A&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; IRAN - Anagram of RAIN (precipitation)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1D&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CHINA - Anagram of CHAIN which is a unit of length of 22 yards&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2D&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; VIETNAM - Anagram of TELEVISION MAN without the letters from INSOLE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3D&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PHILIPPINES - PHILIP (Philip Hammond, British MP for Runnymede) PINES (longs for)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4D&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PERU - Anagram of PURE (virgin)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5D&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TOGO - To go (takeaway)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;7D&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BRAZIL - LIZARD backwards, replacing the D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;8D&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; POLAND - Sounds like POLE LAND (land of long sticks). Also you would clean your shoes with POLISH&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;10D&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TANZANIA - TAN (sunburnt) ZANIA (a bit like zany - 'wacky')&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;12D&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; WALES - WHALES (giant mammals) without the H (aitch-less which sounds like 'ageless' - thanks for the hint Len / rand al'thor)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;16D&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CHAD - A paper circle (a word known to no-one until the hanging chad scandal of the 2000 US presidential elections)`&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As pointed out by VictorHenry the clues need to be relabelled. Otherwise thank you, I enjoyed that - a few bits of GK and some nice wordplay. &lt;/p&gt;&#xA;" OwnerUserId="12277" LastEditorUserId="12277" LastEditDate="2015-05-24T16:36:01.820" LastActivityDate="2015-05-24T16:36:01.820" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15404" PostTypeId="1" AcceptedAnswerId="15405" CreationDate="2015-05-23T08:43:24.913" Score="-1" ViewCount="240" Body="&lt;p&gt;We have a cistern (reservoir). This storage has 2 income faucet (2 interval water) &amp;amp; 1 outgo faucet(1 out pass). By one of the interval pass which in bigger it takes 1 hour to fill storage &amp;amp; by smaller interval pass it takes 6 hours to fill storage.&#xA;And also it takes 7 hours to deplete storage from outgoing pass(faucet).&#xA;Now calculate in 0.24 hours how much of storage is filled (how much water is stored in storage?)&lt;/p&gt;&#xA;" OwnerUserId="12038" LastEditorUserId="5373" LastEditDate="2015-05-23T09:39:33.903" LastActivityDate="2015-05-23T09:39:33.903" Title="how long does it take to fill cistern (0.24 of cistern)?" Tags="|calculation-puzzle|" AnswerCount="1" CommentCount="0" ClosedDate="2015-05-23T19:02:03.730" ContentLicense="CC BY-SA 3.0" />
  <row Id="15405" PostTypeId="2" ParentId="15404" CreationDate="2015-05-23T09:03:09.253" Score="0" Body="&lt;p&gt;Assuming I understand properly what you're asking...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By one incoming faucet, a whole cisternworth of water comes in in 1 hour, so 0.24 of a cisternworth comes in in 0.24 hours. Similarly, by the other incoming faucet, a whole cisternworth comes in in 6 hours, so 0.24/6 = 0.04 comes in in 0.24 hours. And by the outgoing faucet, a whole cisternworth leaves in 7 hours, so 0.24/7 = 0.0342857... leaves in 0.24 hours. So the answer is 0.24 + 0.04 - 0.0342857... = 0.245714... = &lt;strong&gt;43/175&lt;/strong&gt; of the cistern.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-23T09:03:09.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15406" PostTypeId="2" ParentId="15389" CreationDate="2015-05-23T09:43:03.680" Score="21" Body="&lt;p&gt;While (as other answers proved it) it is impossible to solve this just by using knights, the problem can still be solved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matthew says &quot;As I want to play white, my first move isn't really a secret. So I tell you what my first move would be, and you open with the move what you would have moved as an answer. After that I make the move I said, and then we can move with our queens or bishops so that it will be my turn to move&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As all normal openings involve a pawn move which would enable the queen or one of the bishops to move, and both of these pieces can easily move 3 times to arrive back to the same position, the goals of the players can easilt be fulfilled.&lt;/p&gt;&#xA;" OwnerUserId="2507" LastActivityDate="2015-05-23T09:43:03.680" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="15407" PostTypeId="1" AcceptedAnswerId="15758" CreationDate="2015-05-23T10:52:58.477" Score="6" ViewCount="472" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;People call me crazy. I think perhaps I am&lt;br&gt;&#xA;  I just seek justice and there is no other way&lt;br&gt;&#xA;  I don't want to create chaos: quite the opposite.&lt;br&gt;&#xA;  It is a voice chanting in my head. (Or do others hear it too?)&lt;br&gt;&#xA;  It is my primary purpose now behind all things I do.&lt;br&gt;&#xA;  It is what gets me up in the morning. My one grain of hope.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This proving tougher than I expected so another 6 clues for the same six words:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;See that I am a man just a bit mixed up.&lt;br&gt;&#xA;  My anger is vindicated (somewhat).&lt;br&gt;&#xA;  Kept orderly and pure&lt;br&gt;&#xA;  In simple harmony&lt;br&gt;&#xA;  This generous day nearly complete, my plan&lt;br&gt;&#xA;  Grows across the world. See that it is real!&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And in case anyone doesn't understand, expressed as a crossword the words for each down is the same as each across:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/x0uhR.png&quot; alt=&quot;a wordsquare&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="12666" LastEditorUserId="5373" LastEditDate="2017-03-11T14:45:27.410" LastActivityDate="2017-03-11T14:45:27.410" Title="Another Six-words (------||||||)" Tags="|word|wordplay|crosswords|word-square|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15408" PostTypeId="2" ParentId="15407" CreationDate="2015-05-23T11:14:05.517" Score="1" Body="&lt;p&gt;&lt;em&gt;People call me crazy. I think perhaps I am&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; INSANE.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I just seek justice and there is no other way&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure yet...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I don't want to create chaos: quite the opposite.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SYSTEM.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;It is a voice chanting in my head. (Or do others hear it too?)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure yet...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;It is my primary purpose now behind all things I do.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I thought MOTIVE, but it doesn't fit with what I've got so far.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;It is what gets me up in the morning. My one grain of hope.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure yet...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-23T11:14:05.517" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="15409" PostTypeId="1" AcceptedAnswerId="15411" CreationDate="2015-05-23T12:59:47.247" Score="1" ViewCount="943" Body="&lt;p&gt;What is the relationship of this sequence and what is the answer for the last?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;0001112359 = 4 &lt;/li&gt;&#xA;&lt;li&gt;1887145865 = 7 &lt;/li&gt;&#xA;&lt;li&gt;5528695874 = 6&lt;/li&gt;&#xA;&lt;li&gt;0000085021 = 8&lt;/li&gt;&#xA;&lt;li&gt;3247514251 = 0&lt;/li&gt;&#xA;&lt;li&gt;5847962150 = ?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="12680" LastEditorUserId="12680" LastEditDate="2015-05-23T13:21:23.503" LastActivityDate="2015-05-23T13:56:55.107" Title="What is the next number in this puzzle" Tags="|logical-deduction|number-sequence|" AnswerCount="1" CommentCount="1" ClosedDate="2015-05-23T19:13:40.260" ContentLicense="CC BY-SA 3.0" />
  <row Id="15411" PostTypeId="2" ParentId="15409" CreationDate="2015-05-23T13:56:55.107" Score="1" Body="&lt;p&gt;The sequence counts the number of circles in the number.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;0001112359 has 4 circles, 3 in the zeros and 1 in the nine&lt;br&gt;&#xA;  1887145865 has 7  circles, 6 in the eights and 1 in the six&lt;br&gt;&#xA;  5528695874 has 6 circles, 4 in the eights and one each in the six and the nine&lt;br&gt;&#xA;  0000085021 has 8 circles, 6 in the zeros and 2 in the eight&lt;br&gt;&#xA;  3247514251 doesn't have any zeros&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;5847962150 = 5 with 1 circle in the zero, 1 in the nine, 1 in the six, and 2 in the eight&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12109" LastActivityDate="2015-05-23T13:56:55.107" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15412" PostTypeId="1" AcceptedAnswerId="15419" CreationDate="2015-05-23T14:04:42.873" Score="13" ViewCount="1796" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;You have $99$ bags, each containing various numbers of apples and oranges. Prove that there exist $50$ bags among these which together contain at least half the apples &lt;em&gt;and&lt;/em&gt; at least half the oranges.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Needless to say, you may not add/remove fruits to/from the bags.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clarification:&lt;/strong&gt; I changed the wording from &quot;you can grab $50$ bags...&quot; to &quot;there exist $50$ bags...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clarification from comments:&lt;/strong&gt; each bag can contain &lt;em&gt;any&lt;/em&gt; number of apples and &lt;em&gt;any&lt;/em&gt; number of oranges with &lt;em&gt;any&lt;/em&gt; total number of fruit; the total amount of fruit in a bag doesn't have to be the same for each bag.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="2627" LastEditDate="2015-05-24T21:52:46.243" LastActivityDate="2015-05-25T03:20:03.723" Title="99 Bags of Apples and Oranges" Tags="|mathematics|combinatorics|" AnswerCount="5" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="15413" PostTypeId="2" ParentId="15386" CreationDate="2015-05-23T15:16:22.650" Score="1" Body="&lt;p&gt;My guess is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Atoms. They are found in everything you mentioned.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9159" LastEditorUserId="11244" LastEditDate="2015-05-24T05:38:37.657" LastActivityDate="2015-05-24T05:38:37.657" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15414" PostTypeId="2" ParentId="15339" CreationDate="2015-05-23T15:47:55.270" Score="2" Body="&lt;p&gt;Specifically Expected Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The clue &lt;strong&gt;---- ||||&lt;/strong&gt; refers to a &lt;strong&gt;4x4 grid&lt;/strong&gt; where there are 4 words (4 letters each) &lt;strong&gt;horizontally&lt;/strong&gt;, and the same grid gives the same words &lt;strong&gt;vertically&lt;/strong&gt;.&lt;br&gt;&#xA; This &lt;strong&gt;restriction&lt;/strong&gt; fixes the possible values for each of the clues, which may have many matches in isolation.&lt;br&gt;&#xA; 1st word, Protons may be considered &quot;indivisible, but not really&quot; because they are made of quarks, but we can not split the Proton into 3 quarks. ATOM by Definition or Etymology, mean Indivisible, but we have been able to split that into Electrons, Protons and Neutrons.&lt;br&gt;&#xA; 2nd word, there may have been a wide variety of cloaks worn in Rome, but TOGA starts with T (2nd letter of ATOM)&lt;br&gt;&#xA; 3rd word, there may be many Different Possibilities, but not many start with O (3rd letter of ATOM). Anyway let us skip it for now.&lt;br&gt;&#xA; 4th word, 20 quires make a ream, but that will not start with M (last letter of ATOM). So the clue says &quot;in reverse&quot; so make that MAER, which starts with M.&lt;br&gt;&#xA; Now it is a simple matter to fill in OGRE.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;ATOM&lt;/strong&gt;&lt;br&gt;&#xA; &lt;strong&gt;TOGA&lt;/strong&gt;&lt;br&gt;&#xA; &lt;strong&gt;OGRE&lt;/strong&gt;&lt;br&gt;&#xA; &lt;strong&gt;MAER&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I seem to have initiated a &lt;strong&gt;MEME&lt;/strong&gt; with this simple puzzle : &quot;rand al'thor&quot; was inspired to make a 5-word variant, then &quot;Engineer Toast&quot; made variants for sizes 6 &amp;amp; 7 &amp;amp; 8 , finally &quot;Avon&quot; made 6-word variants twice.&lt;/p&gt;&#xA;" OwnerUserId="10940" LastActivityDate="2015-05-23T15:47:55.270" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15415" PostTypeId="1" CreationDate="2015-05-23T16:45:07.823" Score="10" ViewCount="4582" Body="&lt;p&gt;How big a crossword puzzle can you make without using any blank squares? Only the solution is required, not the clues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a 6x6 example, the solution to a &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15380/six-words&quot;&gt;recent puzzle&lt;/a&gt; by @Avon.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;A B A S E S&#xA;B A N A N A&#xA;A N G L E D&#xA;S A L A M I&#xA;E N E M A S&#xA;S A D I S T&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Answers are sought in five categories:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1) puzzles with a square grid, where all rows and columns are of equal length&lt;/li&gt;&#xA;&lt;li&gt;2) puzzles with an oblong grid, where all rows are of equal length and so are all columns but rows are longer than columns&lt;/li&gt;&#xA;&lt;li&gt;3) puzzles with a jagged right-hand edge, where all down words start in the first row&lt;/li&gt;&#xA;&lt;li&gt;4) puzzles with a jagged right-hand edge, where one or more down words do not start in the first row, and where columns that contain two or more (disjoint) words are allowed&lt;/li&gt;&#xA;&lt;li&gt;5) the same as 4), but with the added requirement that every letter must be in both an across word and a down word&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In each category&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;all words contain two or more letters&lt;/li&gt;&#xA;&lt;li&gt;all words contain contiguous letters only&lt;/li&gt;&#xA;&lt;li&gt;each horizontal or vertical string of contiguous letters that isn't contained within another such string forms a word, and we're not interested in words that are contained in other words&lt;/li&gt;&#xA;&lt;li&gt;there are no blanks: in other words, no empty space is adjoined in all four cardinal directions by letters&lt;/li&gt;&#xA;&lt;li&gt;all letters not in the top row are under a letter in the top row, and all letters not in the leftmost column are to the right of a letter in the leftmost column&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The third condition rules out the type of crossword that instead of using blanks uses small blocker lines between squares, like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CuyDv.jpg&quot; alt=&quot;crossword with blockers&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last condition rules out puzzles such as&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;P R E S T I D I G I T A T I O N&#xA;  A T&#xA;  P A T&#xA;      A T&#xA;        A T&#xA;          . . &lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note that&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;in all categories, each row only contains one word, beginning at the left edge of the grid.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;in the first, second and fifth categories, each letter is in both an across word and a down word&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;An example of a puzzle in the second category:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;A P E S&#xA;L I K E&#xA;L E E T&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;One in the third:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;F E T C H&#xA;A R E&#xA;N E A T &lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Two in the fourth:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;M A T C H&#xA;A R&#xA;L I E N&#xA;T A R O &lt;/pre&gt;&#xA;&#xA;&lt;pre&gt;S K I T&#xA;C A N&#xA;O R&#xA;O M E N&#xA;T A R O &lt;/pre&gt;&#xA;&#xA;&lt;p&gt;One in the fifth:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;E T A&#xA;M A T&#xA;O R&#xA;T E A&#xA;E S H &lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In each category, there will be&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a) a winner with the most letters&lt;/li&gt;&#xA;&lt;li&gt;b) a winner with the most words.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(In the first category, these will be the same puzzle.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second category, there will be three additional classes of winner:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;c) a winner for each number of rows, being the oblong with the most columns&lt;/li&gt;&#xA;&lt;li&gt;d) an overall row winner, with the largest number of rows&lt;/li&gt;&#xA;&lt;li&gt;e) an overall column winner, with the largest number of columns&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Letters and words are counted with multiplicity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other interesting classes of winner may well be suggested as time goes on!&lt;/p&gt;&#xA;" OwnerUserId="7363" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-05-24T08:29:00.357" Title="Largest crossword puzzles without blanks or blockers" Tags="|word|wordplay|crosswords|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15416" PostTypeId="2" ParentId="15415" CreationDate="2015-05-23T18:57:26.357" Score="14" Body="&lt;p&gt;&lt;strong&gt;1) WORD SQUARE:&lt;/strong&gt; The maximum length is 9 letters, or 10 if you extend the dictionary with personal names, geographic nouns, and so on...&lt;br&gt;&#xA;&lt;strong&gt;Example of 9:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;pre&gt;A C H A L A S I A&#xA;C R E N I D E N S&#xA;H E X A N D R I C&#xA;A N A B O L I T E&#xA;L I N O L E N I N&#xA;A D D L E H E A D&#xA;S E R I N E T T E&#xA;I N I T I A T O R&#xA;A S C E N D E R S&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example of 10:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;pre&gt;D E S C E N D A N T&#xA;E C H E N E I D A E&#xA;S H O R T C O A T S&#xA;C E R B E R U L U S&#xA;E N T E R O M E R E&#xA;N E C R O L A T E R&#xA;D I O U M A B A N A&#xA;A D A L E T A B A T&#xA;N A T U R E N A M E&#xA;T E S S E R A T E D&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2) WORD RECTANGLE:&lt;/strong&gt; The biggest known word rectangles seem to be 8x4 and 7x6. &lt;strong&gt;This is 8x4:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;pre&gt;D E C I M A T E&#xA;A V O C A D O S&#xA;N A M E L E S S&#xA;E S P R E S S O&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;and this is 7x6:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;pre&gt;A M E R I C A&#xA;M A R A C A S&#xA;B U R G E S S&#xA;I M I T A T E&#xA;T A N A G E R&#xA;S U G G E S T&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3) &quot;LEFT&quot; CROSSWORD:&lt;/strong&gt; &#xA;The biggest I've found is a 13-length half square: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;pre&gt;T O S H A M A B R A H A M&#xA;O B T E N E B R A T E S&#xA;S T R A I T L A C E D&#xA;H E A D S H A K E S&#xA;A N I S O A T E S&#xA;M E T H A N O L&#xA;A B L A T O R&#xA;B R A K E L&#xA;R A C E S&#xA;A T E S&#xA;H E D&#xA;A S&#xA;M&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4) RIGHT-JAGGED CROSSWORD:&lt;/strong&gt; I suggest a 10-length half square.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;pre&gt;P&#xA;R A&#xA;O R S&#xA;G R E W&#xA;R E X E S&#xA;E S T A T E&#xA;S T O N I N G&#xA;S I L E N T L Y&#xA;E V E R G R E E N&#xA;S E T S E Y E S O N&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;5) RIGHT-JAGGED ADVANCED CROSSWORD:&lt;/strong&gt; Again, I propose the same crossword of the fourth contest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Credits: most of my material comes from the &lt;a href=&quot;http://www.puzzlers.org/&quot;&gt;National Puzzlers' League&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-05-24T08:29:00.357" LastActivityDate="2015-05-24T08:29:00.357" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15417" PostTypeId="2" ParentId="15412" CreationDate="2015-05-23T19:12:03.870" Score="0" Body="&lt;p&gt;Suppose we choose 50 of the bags at random. We are choosing more than half of the bags, so the probability that the bags we selected contain at least half of the apples is strictly greater than 50%. Similarly, the probability that the bags we selected contain at least half of the oranges is also greater than 50%. This means there is a positive probability that the bags we selected contain at least half the apples and at least half the oranges. In particular, there must be some combination of 50 bags containing at least half the apples and at least half the oranges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: As pointed out in the comments, it is not obvious why the bags we select will have at least half the apples with probability greater than 50%. This actually follows from the argument in my other answer: one way to choose 50 bags at random is first to choose a cyclic ordering of the bags, then to choose 50 consecutive bags. My other answer shows that for whichever cyclic ordering we choose, a random choice of 50 consecutive bags contains at least half the apples with probability at least $\frac{50}{99}$. This means that a random selection of 50 bags contains at least half the apples with probability at least $\frac{50}{99}$. The same is true for oranges, so a random selection of 50 bags contains at least half the apples and at least half the oranges with probability at least $\frac{1}{99}$.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-05-24T04:31:35.623" LastActivityDate="2015-05-24T04:31:35.623" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="15418" PostTypeId="2" ParentId="15372" CreationDate="2015-05-23T20:21:44.320" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The human ambassador should prepare for war, since he will lose this game.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This reasoning is mostly due to user12408's answer, please give credit to them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, some observations. When I say a position is losing (winning), I mean that if either player starts there, they will lose (win).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;When the smaller of $x,y$ is $0$ or $1$...&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;(0,1) and (1,0) are losing. Everything else has one of these as an option, so is winning.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;When the smaller of $x,y$ is $2$...&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;the alien wins. From (2,2), he wins by moving to (1,0). From anything else, he moves to (2,2).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the human wins starting from (3,2) and (2,3) by moving to (1,0) or (0,1), respectively. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the human loses starting from all other squares whose smallest coordinate is 2. If he moves to $(2,y)$ or $(x,2)$, the alien wins [see first bullet in this section]. Else, he must move to somewhere discussed in the first section, which can't be (0,1) or (1,0), so he loses.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;When both $x,y\ge 3$...&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the alien loses starting from (3,3). The choices (3,0), (3,1), (0,3), (1,3), (2,1) and (1,2) are bad [see first section]. His other options, (2,3) and (3,2), don't work either [see second section].&lt;/li&gt;&#xA;&lt;li&gt;otherwise the alien wins. When $x&amp;gt;3$, the winning move is $(x,2)$. When $y&amp;gt;3$, it is $(2,y)$. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We have shown the alien wins starting anywhere, except for the &quot;bad&quot; places (0,1), (1,0) and (3,3). Thus, the humans can only win if they can move to one of these places, showing that (170000,170002) is a loss for the humans if they start there. It is also a loss for the humans if the aliens start there, since it is not one of the three &quot;bad&quot; places. Thus, humanity is doomed by their lack of mathematical rigor.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-05-23T21:49:30.277" LastActivityDate="2015-05-23T21:49:30.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15419" PostTypeId="2" ParentId="15412" CreationDate="2015-05-23T20:57:57.477" Score="14" Body="&lt;p&gt;Let $B_1,B_2,\dots,B_{99}$ be the bags, in increasing order of number of apples contained; say the number of apples in $B_i$ is $a_i$ for all $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If our 50 bags grabbed are $B_{99}$, one of $B_{98}$ and $B_{97}$, one of $B_{96}$ and $B_{95}$, ..., and one of $B_2$ and $B_1$, then they must contain between them &lt;strong&gt;at least half of all apples&lt;/strong&gt;, since at worst they contain $a_{99}+a_{97}+\dots+a_5+a_3+a_1\geq a_{98}+a_{96}+\dots+a_4+a_2+0$ apples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now how do we fix all those &quot;one of&quot;s? Just pick whichever of $B_{98}$ and $B_{97}$ has the more oranges, then whichever of $B_{96}$ and $B_{95}$ has the more oranges, and so on. Then our 50 selected bags - even excluding $B_{99}$! - must contain &lt;strong&gt;at least half of all oranges&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;QED.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-23T20:57:57.477" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15420" PostTypeId="1" AcceptedAnswerId="15432" CreationDate="2015-05-23T22:29:37.827" Score="6" ViewCount="464" Body="&lt;p&gt;Here's an old recipe I found in my tower for &lt;strong&gt;lemon tea stew&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Ingredients&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;6 cups of tea&lt;br&gt;&#xA;4 eggs&lt;br&gt;&#xA;2 eyeballs&lt;sup&gt;1&lt;/sup&gt;&lt;br&gt;&#xA;2 leeks&lt;br&gt;&#xA;4 lemons&lt;br&gt;&#xA;1 banana&lt;br&gt;&#xA;2 chicken wings&lt;br&gt;&#xA;several dates&lt;br&gt;&#xA;about half an onion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;sup&gt;1&lt;/sup&gt; Or fake jelly eyeballs for the squeamish.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Instructions&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Take a nice cup of tea,&lt;br&gt;&#xA;Left to age well,&lt;br&gt;&#xA;And mix in the first egg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Double your portions of everything,&lt;br&gt;&#xA;Toss in an eyeball for the witch,&lt;br&gt;&#xA;Then the first leek, well chopped,&lt;br&gt;&#xA;Simmer, and add the first lemon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Throw in half a banana,&lt;br&gt;&#xA;The roasted wing of a hen,&lt;br&gt;&#xA;And add the first few dates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another cup of tea,&lt;br&gt;&#xA;Having been left to age,&lt;br&gt;&#xA;And the second lemon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Double your portions all over again,&lt;br&gt;&#xA;Chop nearly half of an onion,&lt;br&gt;&#xA;And, after first roasting well,&lt;br&gt;&#xA;Finally you have your food.&#xA;&lt;hr&gt;&#xA;If you put it all together the way it's supposed to be done, what do you get?&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="9718" LastEditDate="2015-05-24T07:57:31.453" LastActivityDate="2015-05-24T10:20:15.087" Title="A recipe for lemon tea stew" Tags="|riddle|enigmatic-puzzle|" AnswerCount="1" CommentCount="10" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15421" PostTypeId="2" ParentId="15386" CreationDate="2015-05-23T22:30:55.123" Score="4" Body="&lt;p&gt;I'm guessing&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Label&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clothings have&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Brand label&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Canned goods have&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That piece of paper wrapped around it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Stores has&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Price labels&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Library uses&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Labels on each book for the call number&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Storage bin has&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Labels to mark what's inside&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Schools &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; School kids have labels on their belongings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12332" LastActivityDate="2015-05-23T22:30:55.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15423" PostTypeId="2" ParentId="15412" CreationDate="2015-05-23T23:33:55.177" Score="1" Body="&lt;p&gt;I will provide a probabilistic proof. By demonstrating that picking randomly has a nonzero chance of success, we can show that a solution exists. (&lt;a href=&quot;http://en.m.wikipedia.org/wiki/Probabilistic_method&quot; rel=&quot;nofollow&quot;&gt;http://en.m.wikipedia.org/wiki/Probabilistic_method&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claim: A random subset of 50 bags has over 50% chance of having more than half the apples. (Chosen uniformly over all size 50 subsets.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: For each &quot;losing&quot; size 50 subset, flip it around to get a (unique) winning size 49 subset. This means there are at least as many size 49 winning sets as size 50 losing sets. There are multiple ways to extend these to size 50 sets, so there's strictly more ways to win than lose. (Hand waving a bit)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This argument applies symmetrically to the oranges. Since each probability is over 1/2, they must have a nonzero intersection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus there is a positive probability of picking a subset with at least half of each, so a solution exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonconstructively useless, but QED :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: looking back, this is just saying that the intersection of sets larger than 50% is nonempty, so the probabilistic part is just fluff.&lt;/p&gt;&#xA;" OwnerUserId="12672" LastEditorUserId="12672" LastEditDate="2015-05-24T01:06:51.920" LastActivityDate="2015-05-24T01:06:51.920" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15424" PostTypeId="1" AcceptedAnswerId="15511" CreationDate="2015-05-24T03:23:59.327" Score="40" ViewCount="3470" Body="&lt;p&gt;Alice, Bob, Carol, and Dan are playing a card game. After they all show their cards, Bob notes that by exchanging two cards the order of the players would be reversed; that is, the player with the best hand would have the worst hand, and vice versa. &lt;em&gt;Which pair of cards is he thinking of?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Since this puzzle is far too easy on its own, I've hidden the four players' cards in this image. I will only give you one hint, so &lt;strong&gt;listen&lt;/strong&gt; up:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In order to complete this challenge, you must &lt;strong&gt;verify&lt;/strong&gt; every step.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://2012rcampion.com/se/sigint/1/img.png&quot; alt=&quot;img.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&#xA;Download as &lt;a href=&quot;https://2012rcampion.com/se/sigint/1/img.png&quot; rel=&quot;noreferrer&quot;&gt;png (1&amp;thinsp;477&amp;thinsp;855&amp;nbsp;bytes)&lt;/a&gt; | &#xA;&lt;a href=&quot;https://2012rcampion.com/se/sigint/1/img.tiff&quot; rel=&quot;noreferrer&quot;&gt;tiff (1&amp;thinsp;512&amp;thinsp;448&amp;nbsp;bytes)&lt;/a&gt; | &#xA;&lt;a href=&quot;https://2012rcampion.com/se/sigint/1/img.bmp&quot; rel=&quot;noreferrer&quot;&gt;bmp (1&amp;thinsp;513&amp;thinsp;654&amp;nbsp;bytes)&lt;/a&gt; | &#xA;&lt;a href=&quot;https://2012rcampion.com/se/sigint/1/img.ppm&quot; rel=&quot;noreferrer&quot;&gt;ppm (1&amp;thinsp;512&amp;thinsp;102&amp;nbsp;bytes)&lt;/a&gt; &lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="69582" LastEditDate="2024-01-17T16:38:09.123" LastActivityDate="2024-01-17T16:38:09.123" Title="Multilayer Image Steganography" Tags="|visual|cipher|steganography|computer-puzzle|cards|" AnswerCount="4" CommentCount="19" ContentLicense="CC BY-SA 4.0" />
  <row Id="15425" PostTypeId="2" ParentId="15412" CreationDate="2015-05-24T03:59:46.777" Score="13" Body="&lt;p&gt;Arrange the bags in a circle. I will call a collection of 50 consecutive bags a &lt;em&gt;team&lt;/em&gt; (so there are 99 possible teams), and a team will be called &lt;em&gt;appley&lt;/em&gt; if it contains at least half of the apples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define the &lt;em&gt;opponent&lt;/em&gt; of a given team to be the team whose rightmost bag is the leftmost bag of the given team. For any given team, every bag is either a member of that team or a member of its opponent (and one bag is a member of both). This means that if a team is not appley, then its opponent is appley. Different teams have different opponents, so there must be at least as many appley teams as non-appley teams. There are 99 teams in total, so at least 50 of them are appley.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar argument shows that at least 50 teams are orangey (that is, at least 50 of the teams contain at least half the oranges). Of 99 teams, at least 50 are appley and at least 50 are orangey, so there must be at least one teams which is both appley and orangey (because $50+50&amp;gt;99$). This team consists of 50 bags, and contains at least half the apples and at least half the oranges.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastActivityDate="2015-05-24T03:59:46.777" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15426" PostTypeId="1" AcceptedAnswerId="15429" CreationDate="2015-05-24T04:38:07.423" Score="6" ViewCount="195979" Body="&lt;p&gt;A brain teaser posed by the character Captain Ray Holt of 'Brooklyn Nine-Nine' played by Andre Braugher during episode 18 of season 2: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;There are twelve(12) men on an island, eleven(11) weigh exactly the same amount, but one of them is slightly lighter or heavier, [the object is to] figure out [whether he is lighter or heavier]. A standard see-saw must be used and only three times.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Rephrased for clarity, below is a verbatim transcript of 'Holts' dialogue)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;There are twelve men on an island, eleven weigh exactly the same amount, but one of them is slightly lighter or heavier, you must figure out which. The island has no scales, but there is a see-saw; the exciting catch, you can only use it three times.&quot;&lt;/p&gt;&#xA;" OwnerUserId="12687" LastEditorUserId="12687" LastEditDate="2015-05-24T05:03:39.467" LastActivityDate="2015-05-24T06:38:33.053" Title="Brooklyn Nine-Nine Riddle: Weighing Islanders" Tags="|riddle|logical-deduction|weighing|" AnswerCount="2" CommentCount="3" ClosedDate="2015-05-24T07:58:04.210" ContentLicense="CC BY-SA 3.0" />
  <row Id="15427" PostTypeId="2" ParentId="15426" CreationDate="2015-05-24T04:47:25.447" Score="0" Body="&lt;p&gt;The solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Split the men into two(2) groups 'abcdef' and '123456'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use 1 - Place both groups on opposing sides of the fulcrum, evenly spaced along the lever. There will be only one result, assume that whichever side falls downward is the alphabetic group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use 2 - Remove six(6) men from the see-saw, three(3) from both groups. Let's say 'abc' and '456'. There are two possible results. &#xA;A_ the equilibrium of the see-saw remains unchanged, therefore the man of a different weight is now amongst group 'def123' or &#xA;B_ the see-saw becomes level with the ground, therefore the man of a different weight is standing with group 'abc456'. &#xA;Both situations are ideal as they reveal to us which group is the control group or standard for the weight of eleven of the men. Which brings us to...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use 3 - Place both new groups 'def123' and 'abc456' on the teeter-totter again as we did in the beginning. Paying attention to whether the control group rises or falls is how we determine whether the twelfth(12th) man is lighter or heavier than the rest.&lt;/p&gt;&#xA;" OwnerUserId="12687" LastActivityDate="2015-05-24T04:47:25.447" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15428" PostTypeId="2" ParentId="15389" CreationDate="2015-05-24T05:03:25.517" Score="5" Body="&lt;p&gt;Knights, no.  Most chess games start with moving a pawn to the 4th rank, though.  Assuming this how white intends to start they should agree and white moves his pawn to the third rank instead, then on the next move takes it to the 4th rank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cooperation is only needed to the extent that black does not play something that will interfere with this.&lt;/p&gt;&#xA;" OwnerUserId="2767" LastActivityDate="2015-05-24T05:03:25.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15429" PostTypeId="2" ParentId="15426" CreationDate="2015-05-24T06:31:13.387" Score="8" Body="&lt;p&gt;There are 24 possible situations (the different man can be any of 1-12, and he can be heavier or lighter). Thus we need to log&lt;sub&gt;2&lt;/sub&gt;24 bits of information to solve the puzzle. You can weigh three combinations of men on the see-saw. Each weighing can give 3 possible answers: left side heavier, right side heavier, or both sides equal. Thus in principle we can get log&lt;sub&gt;2&lt;/sub&gt;27 bits from the three comparisons.  So in principle, we should be able to solve the problem. The key to this problem is making sure all three output values (left side heavier, right side heavier, two sides the same) are possible and informative in almost every comparison you do so that we can eek log&lt;sub&gt;2&lt;/sub&gt;24 bits out of the comparisons. Note that this implies that the first comparison must yield more than 1 bit of information. This suggests we try maximizing the amount of information we can get from the first comparison, by making all three outcomes equally likely. Comparing (1,2,3,4) to (5,6,7,8) does exactly this. Similar logic will help us design all further comparisons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is one solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number the men 1,2,3...12. First weigh 1,2,3,4 against 5,6,7,8. One of two things will happen:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) They are equal. Now we know that the different man is among {9,10,11,12}. Weigh 9,10,11 against 1,2,3. If these are equal, the different man is 12. Weigh 12 against 1 to find out whether 12 is heaver or lighter. If the 9,10,11 differs from 1,2,3, then weigh 9 against 10. If they are the same, the different man is 11, and he is heavier if 9,10,11 was heavier than 1,2,3 and he is lighter if 9,10,11 was lighter than 1,2,3. If 9 and 10 are different, the different man is the lighter of the 9,10 comparison if 9,10,11 was lighter than 1,2,3, (and he is lighter); the different man is the heavier of the 9,10 comparison if 9,10,11 was heavier than 1,2,3 (and he is heavier).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) They are different. Without loss of generality suppose that 1,2,3,4 is heavier than 5,6,7,8. (We could always relabel the men so that this is true). We know {9,10,11,12} all weigh the same. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weigh 1,2,5,6,7 against 8,9,10,11,12: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) If 1,2,5,6,7 is heavier, then either 1 or 2 heavier, or 8 is lighter. Weigh 1 against 2. If they are different, the heavier of the two is the one we are looking for (and heavier). If they are the same, 8 is the one we are looking for (and lighter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) If 1,2,5,6,7 is lighter, then one of 5,6,7 is different and lighter. Weigh 5 against 6. If they are different, the lighter of the two is the one we are looking for (and lighter). If they are the same, 7 is different (and lighter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) If they are the same, then one of 3,4 is different. Weigh them against each other. The one who is heavier is the different man (and heavier). &lt;/p&gt;&#xA;" OwnerUserId="11087" LastEditorUserId="11087" LastEditDate="2015-05-24T06:38:33.053" LastActivityDate="2015-05-24T06:38:33.053" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15430" PostTypeId="5" CreationDate="2015-05-24T10:08:48.290" Score="0" Body="&lt;p&gt;Connect Four (also known as Captain's Mistress, Four Up, Plot Four, Find Four, Fourplay, Four in a Row and Four in a Line) is a two-player connection game in which the players first choose a color and then take turns dropping colored discs from the top into a seven-column, six-row vertically suspended grid. The pieces fall straight down, occupying the next available space within the column. The objective of the game is to connect four of one's own discs of the same color next to each other vertically, horizontally, or diagonally before your opponent. Connect Four is a strongly solved game. The first player can always win by playing the right moves.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-05-24T10:08:48.290" LastActivityDate="2015-05-24T10:08:48.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15431" PostTypeId="4" CreationDate="2015-05-24T10:08:48.290" Score="0" Body="For questions about the 2-player game Connect Four." OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-05-24T10:08:48.290" LastActivityDate="2015-05-24T10:08:48.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15432" PostTypeId="2" ParentId="15420" CreationDate="2015-05-24T10:20:15.087" Score="11" Body="&lt;p&gt;&lt;sub&gt;I should've realised, given that it was rand al'thor asking the question, that it would relate to a fantasy novel... :)&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Making this lemon tea stew will get you:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Will and the Word (the term use for sorcery in &lt;a href=&quot;http://en.wikipedia.org/wiki/The_Belgariad&quot;&gt;the fantasy series, The Belgariad&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Nutrition information&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Take a nice cup of tea, =&gt; &lt;strong&gt;T&lt;/strong&gt;&lt;br&gt;&#xA; Left to age well, =&gt; &lt;strong&gt;H&lt;/strong&gt; (sounds like &quot;age&quot;)&lt;br&gt;&#xA; And mix in the first egg. =&gt; &lt;strong&gt;E&lt;/strong&gt; (&quot;first&quot; letter of &quot;egg&quot;)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Double your portions of everything, =&gt; &lt;strong&gt;W&lt;/strong&gt; (&quot;double you[r]&quot;)&lt;br&gt;&#xA; Toss in an eyeball for the witch, =&gt; &lt;strong&gt;I&lt;/strong&gt; (&quot;eye&quot;)&lt;br&gt;&#xA; Then the first leek, well chopped, =&gt; &lt;strong&gt;L&lt;/strong&gt; (&quot;first&quot; of &quot;leek&quot;)&lt;br&gt;&#xA; Simmer, and add the first lemon. =&gt; &lt;strong&gt;L&lt;/strong&gt; (&quot;first&quot; of &quot;lemon&quot;)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Throw in half a banana, =&gt; &lt;strong&gt;A&lt;/strong&gt; (bAnAnA, A is half of the letters)&lt;br&gt;&#xA; The roasted wing of a hen, =&gt; &lt;strong&gt;N&lt;/strong&gt; (&quot;wings&quot; of &quot;hen&quot; are &quot;h&quot; and &quot;n&quot;... we'll just grab one)&lt;br&gt;&#xA; And add the first few dates. =&gt; &lt;strong&gt;D&lt;/strong&gt; (&quot;first&quot; of &quot;dates&quot;)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Another cup of tea, =&gt; &lt;strong&gt;T&lt;/strong&gt;&lt;br&gt;&#xA; Having been left to age, =&gt; &lt;strong&gt;H&lt;/strong&gt;&lt;br&gt;&#xA; And the second lemon. =&gt; &lt;strong&gt;E&lt;/strong&gt; (&quot;second&quot; letter of &quot;lemon&quot;)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Double your portions all over again, =&gt; &lt;strong&gt;W&lt;/strong&gt;&lt;br&gt;&#xA; Chop nearly half of an onion, =&gt; &lt;strong&gt;O&lt;/strong&gt; (OniOn, O is almost half of the letters)&lt;br&gt;&#xA; And, after first roasting well, =&gt; &lt;strong&gt;R&lt;/strong&gt; (&quot;first&quot; of &quot;roasting&quot;)&lt;br&gt;&#xA; Finally you have your food. =&gt; &lt;strong&gt;D&lt;/strong&gt; (&quot;final&quot; of &quot;food&quot;)    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastActivityDate="2015-05-24T10:20:15.087" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15433" PostTypeId="2" ParentId="14708" CreationDate="2015-05-24T11:25:29.810" Score="1" Body="&lt;p&gt;One other possible answer:&#xA;$$6! \times \frac{3}{2} + 5 + 7 - 4 = 1088$$&lt;/p&gt;&#xA;" OwnerUserId="12690" LastEditorUserId="12165" LastEditDate="2015-05-24T11:44:58.140" LastActivityDate="2015-05-24T11:44:58.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15434" PostTypeId="1" AcceptedAnswerId="15472" CreationDate="2015-05-24T11:43:45.153" Score="5" ViewCount="580" Body="&lt;p&gt;Inspired by Mike Earnest's &lt;a href=&quot;https://puzzling.stackexchange.com/q/15412/5373&quot;&gt;99 Bags of Apples and Oranges&lt;/a&gt;...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You have 100 bags, each containing various numbers of apples, beetroots, and cakes. Prove that there exist 51 bags among these which between them contain at least half the apples &lt;em&gt;and&lt;/em&gt; at least half the beetroots &lt;em&gt;and&lt;/em&gt; at least half the cakes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Needless to say, you may not add or remove fruits to/from the bags.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As in the previous post, each bag can contain &lt;em&gt;any&lt;/em&gt; number of each foodstuff with &lt;em&gt;any&lt;/em&gt; total number, and the total amount of fruit in a bag doesn't have to be the same for each bag.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-07-29T17:57:45.633" Title="100 bags of Apples, Beetroots, and Cakes" Tags="|logical-deduction|combinatorics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15436" PostTypeId="2" ParentId="15165" CreationDate="2015-05-24T13:37:16.313" Score="2" Body="&lt;p&gt;The book(s):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Belgariad, a five book series as well as The Malloreon, the five book sequel to the Belgariad&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The one who takes care of the monsters is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; UL, god of Ulgos and monsters&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Aldur, who took disciples instead of having a people&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nedra, lion god&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The third is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Torak, only to take a monster as his totem&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The fourth is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chaldan, bull god&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The fifth is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Issa, snake god&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sixth is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mara, bat god&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The seventh is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Belar, bear god&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The eighth is(spoiler!!!)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eriond, who took over for Torak after he was killed.  It was said he would become god of all other races.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11790" LastEditorUserId="11790" LastEditDate="2015-05-24T18:45:11.200" LastActivityDate="2015-05-24T18:45:11.200" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15437" PostTypeId="1" AcceptedAnswerId="22407" CreationDate="2015-05-24T13:57:12.933" Score="4" ViewCount="1137" Body="&lt;p&gt;My first is part of your digestive system, but you sometimes use only half of me.&lt;br&gt;&#xA;My second is sometimes called what people sometimes call the shark hanging in the Smithsonian.&lt;br&gt;&#xA;My third is my mom, once removed from Matilda and re-moved again.&lt;br&gt;&#xA;My fourth is a greeting to my friend who is named after the end.&lt;br&gt;&#xA;My fifth is a really fast run.&lt;br&gt;&#xA;My sixth is in the number three.&lt;br&gt;&#xA;My seventh can replace my first and my last.&lt;br&gt;&#xA;My eighth is the last 2 of my dad's essay topics.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is my all?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I edited some of the clues for clarification.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is NOT punctuation marks. There is an actual obvious answer once you know the answers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the hint for my third, the re-moved again part has significance.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; At least two of the clues are the same.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My second is made of parts...&lt;br&gt;&#xA;My third is re-moved again.&lt;br&gt;&#xA; My sixth is not in #3&lt;br&gt;&#xA; My seventh requires the others.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="167" LastEditorUserId="167" LastEditDate="2015-09-10T03:17:03.070" LastActivityDate="2015-09-19T17:50:19.173" Title="My Eight Pieces Together" Tags="|riddle|" AnswerCount="4" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="15438" PostTypeId="2" ParentId="15437" CreationDate="2015-05-24T16:08:03.790" Score="3" Body="&lt;p&gt;I think it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; punctuation marks&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My first is part of your digestive system, but you sometimes use only half of me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it's the colon, but sometimes we use semi-colon or semicolon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My second is sometimes called what people sometimes call the shark hanging in the Smithsonian.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I have no clues for that one&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My third is my mom, once removed from Matilda and re-moved again.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it's a tilde but in Spanish. Because $-(Ma - Matilda) = -(-tilda) = tilda$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fourth is a greeting to my friend, named after the end.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I have no clues for that one&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fifth is a really fast run.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it's a dash&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My sixth is in the number three.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am not sure about that one, but I guess it's parenthesis because 3 looks a little bit as stacked parenthesis&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My seventh can replace my first and last.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it's an exclamation mark or a question mark&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My eighth is the last of my dad's essays. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it's a period&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10400" LastActivityDate="2015-05-24T16:08:03.790" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15439" PostTypeId="1" CreationDate="2015-05-24T16:08:17.587" Score="5" ViewCount="499" Body="&lt;p&gt;You meet a guy on the road. The following conversation follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;He: Let's play a game. Give me any natural number from $3$ to $10$.&#xA;  I'll call this number the grid size&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;You: Okay, $4$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(He draws a $4$x$4$ grid on a sheet of paper, since you picked $4$)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;He: Now give me another integer. Try to give a small one to make our&#xA;  calculations easy. This number will be called our 'sum'.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;You: Okay, $9$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;He: Now give me any $4$ integers that add up to $9$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;You: $3,-1,2,5$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(He writes them on the diagonal of the grid)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;He: Now give me any $3$ (as $3$ is 1 less than $4$) integers,&#xA;  completely at random.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;You: $9,14,-5$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(He writes them on the first row, starting with the 2nd cell, since&#xA;  the first cell was already filled)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The grid looks like&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;3  9  14 -5 &#xA;   -1        &#xA;      2      &#xA;         5  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;He fills the rest of the grid himself.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;3  9  14 -5 &#xA;-7 -1 4  -15&#xA;-9 -3 2  -17&#xA;13 19 24 5  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;He: Now pick any $4$ cells, such that no $2$ cells share the same row&#xA;  or column.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;You: Okay, the $9$ on top, the $4$ in the next row, $-9$ in the third&#xA;  row, and the leaves only $5$ on the last cell.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;He: Now add them up.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;You: It's $9$! The number I originally picked.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;He: Try another set of $4$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;You: $14,-15,-3,13$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;You: which again add up to $9$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What 'trick' did he use, and can you defeat him? Does a exist an allowed grid size and set of input numbers, for which such a 'magic cutting grid' cannot be generated?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-05-24T18:11:49.843" Title="Magic cutting squares with given sum" Tags="|logical-deduction|mathematics|magic-square|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15440" PostTypeId="2" ParentId="15439" CreationDate="2015-05-24T17:12:35.967" Score="2" Body="&lt;p&gt;The grid could be filled by him easily, he only has to solve equations with one unknowns. After my numbers are in the grid, he first calculates the first column:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-7 = 9 - (9 + 2 + 5)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-9 = 9 - (14 -1 + 5)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;13 = 9 - (-5 -1 + 2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then he calculates the subdiagonals from the top-left to the bottom-right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 = 9 - (9 -9 + 5)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-3 = 9 - (14 -7 + 5)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It could be done on larger grids as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Detailed steps:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just calculate the first column, &#xA;&lt;img src=&quot;https://i.stack.imgur.com/2NJjC.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;then go from the top-left to the bottom-right through the subdiagonals:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/SEBNB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10730" LastEditorUserId="10730" LastEditDate="2015-05-24T17:26:40.203" LastActivityDate="2015-05-24T17:26:40.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15441" PostTypeId="2" ParentId="15439" CreationDate="2015-05-24T17:23:23.057" Score="2" Body="&lt;p&gt;Said &lt;span class=&quot;math-container&quot;&gt;$a_{i,j}$&lt;/span&gt; the element in the &lt;span class=&quot;math-container&quot;&gt;$i_{th}$&lt;/span&gt; row and &lt;span class=&quot;math-container&quot;&gt;$j_{th}$&lt;/span&gt; column, we have these properties:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$(a_{i,j}-a_{i,j+1})$&lt;/span&gt; is constant if you vary &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt;&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$(a_{i,j}-a_{i+1,j})$&lt;/span&gt; is constant if you vary &lt;span class=&quot;math-container&quot;&gt;$j$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This means that each row/column is generated adding a constant to the first row/column. For example, in this case the second row is the first minus 10.&lt;br /&gt;&#xA;When you make the sum, you count exactly one number from the first row, one from the second, one from the third and one from the fourth (same for columns). It doesn't matter which numbers you choose. Why? &lt;strong&gt;Check this picture for an intuitive explanation&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/35JOX.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Whichever numbers you pick to make your sum, it will always be &lt;span class=&quot;math-container&quot;&gt;$A+B+C+D-10-12+10=A+B+C+D-12$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Of course, this can be extended to &lt;strong&gt;any square of any size&lt;/strong&gt;!&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Note: this puzzle was proposed by Martin Gardner in one of his books, at least I've read it there for the first time.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-24T17:49:18.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15442" PostTypeId="2" ParentId="15257" CreationDate="2015-05-24T17:47:09.077" Score="3" Body="&lt;p&gt;OK, this is a long shot, but I'll give it a go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hint makes me think of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; David Eddings's &lt;em&gt;Belgariad&lt;/em&gt; and &lt;em&gt;Malloreon&lt;/em&gt; series (12 books in total including the followup novels &lt;em&gt;Belgarath the Sorcerer&lt;/em&gt; and &lt;em&gt;Polgara the Sorceress&lt;/em&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Specifically:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;a haunted ground&quot; = Maragor; &quot;those who own a special pendant&quot; = Belgarath and family (including Ce'Nedra); &quot;the mentor&quot; could be Belgarath; and &quot;the twelfth is the subject's obdurate child&quot; sounds like Polgara, the obdurate child of Belgarath and Poledra, who gives her name to the twelfth book.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Having got the book series, the best fit for the riddle seems to be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Poledra.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Which one is my real identity?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Is she wolf or human?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Am I stubborn? Just as much as Constance Contraire.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stubbornness really runs in that family...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Steady fingers on the bridge,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Just as steady as my mind when it's made up.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Steady would perhaps be a better word than stubborn for Poledra (unlike her husband and daughter).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I love one whose weapon is that of the just.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Belgarath.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;They may have taken their vows against me, but I'll save him 'cause he's to me like glittering garland is to silver-fist&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I suspect this is an Eragon reference?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;We are divided, awkward and shy as he may have been so long ago.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But my love, he is just as outgoing as always.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Belgarath started off as a shy village lad. Outgoing? I suppose so...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I used the same weapon of my daughter,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The form of a snowy owl.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Whose main character trait was inherited from me.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stubbornness again? Or strength of character. They're similar in lots of ways really.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-24T17:47:09.077" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="15443" PostTypeId="2" ParentId="15439" CreationDate="2015-05-24T18:04:20.530" Score="1" Body="&lt;p&gt;Here's how your friend does this trick. Let's say you pick $n$ for the grid size, $S$ for the sum, numbers $t_1,t_2\dots,t_n$ on the top row, and $d_1,d_2,\dots d_n$ on the diagonal, where $t_1=d_1$ and $d_1+\dots+d_n=S$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $r_i=d_i-t_i$. He fills out the entire grid by placing the number&#xA;$$&#xA;r_i+t_j&#xA;$$&#xA;in the cell at row $i$ and column $j$. This scheme doesn't conflict with any of the numbers you chose, since on the top row it places&#xA;$$&#xA;r_1+t_j=d_1-t_1+t_j=t_j&#xA;$$&#xA;and on the diagonal it places&#xA;$$&#xA;r_i+t_i=d_i-t_i+t_i=d_i&#xA;$$&#xA;It also obeys the magic cut requirement; for any $n$ cells you pick, all in different rows/columns, their sum will be&#xA;$$&#xA;r_1+\dots+r_n+t_1+\dots+t_n&#xA;$$&#xA;which, since $d_i=r_i+t_i$, is equal to &#xA;$$&#xA;d_1+\dots+d_n=S&#xA;$$&#xA;Like magic, your chosen number, $S$, is always the sum. In other words, you cannot defeat him.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-05-24T18:11:49.843" LastActivityDate="2015-05-24T18:11:49.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15445" PostTypeId="1" AcceptedAnswerId="15469" CreationDate="2015-05-24T22:56:00.827" Score="3" ViewCount="488" Body="&lt;p&gt;Novice shipwright, seafarer, navigator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I live in a land removed,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet my ancestors came from yours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My wife's name is a flower and sick,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My tutor was straight out of old times,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And my son swapped my protein structure prediction with a small stream(that didn't have doubled).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was the first one who they saw, an old friend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But left because of the great time rift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Welcome to the painting, friends and company&lt;/p&gt;&#xA;" OwnerUserId="11790" LastActivityDate="2015-08-08T15:46:07.003" Title="What does 24 have to do with anything?" Tags="|riddle|knowledge|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15446" PostTypeId="1" AcceptedAnswerId="15448" CreationDate="2015-05-25T01:20:41.337" Score="8" ViewCount="687" Body="&lt;p&gt;I noticed that puzzling stack exchange didn't have my favorite riddle. It isn't too uncommon so someone else might have heard it (and a quick google search can easily find the answer) but someone out there might enjoy it as much as I do. So here it is:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In a marble hall white as milk,&lt;/p&gt;&#xA;&lt;p&gt;lined with skin as soft as silk.&lt;/p&gt;&#xA;&lt;p&gt;Within a fountain crystal-clear,&lt;/p&gt;&#xA;&lt;p&gt;a golden apple doth appear.&lt;/p&gt;&#xA;&lt;p&gt;No doors there are to this stronghold,&lt;/p&gt;&#xA;&lt;p&gt;yet thieves break in to steal its gold.&lt;/p&gt;&#xA;&lt;p&gt;What is it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &amp;quot;thieves&amp;quot; is a strong word. If a person owns the marble hall then they also own the gold.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12696" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-25T01:39:12.287" Title="In marble halls" Tags="|riddle|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15447" PostTypeId="1" AcceptedAnswerId="15449" CreationDate="2015-05-25T01:23:46.297" Score="14" ViewCount="1019" Body="&lt;p&gt;&lt;em&gt;I haven't seen Ernie for a few weeks - he's been on holiday in the People's Republic of Kzijekistan (PRK). But yesterday I got an e-mail and thought I would share it with you. It reads as follows:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The history of Kzijekopolis is intriguing. It was founded by Kzijekl the Great, who built a castle atop Mt. Chatcal in the late 18th century. When Kzijekl's wife had identical quadruplets - all boys - the traditional solution would have been to 'retire' the younger sons to distant monasteries. But the midwife made a dreadful mistake - she put them all in the same cot and when the Khan arrived in the birthing chamber it was impossible to tell which was the eldest. Kzijekl decided that all the boys should be brought up as potential rulers, and on their 21st birthday the true heir would be chosen (by God) via 4 on 4 combat to the death. He built four massive walls to divide the city into quarters, one for each son to live in, so they would not meet before their 21st birthday Battle Royale. It was decreed that no opening could be made in the dividing walls until a single son ruled the Empire, unopposed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But Kzijekl's plans were dashed when the four brothers met by chance on a holiday to Paris and became friends. On their return they declared themselves co-chairmen of a new Peoples's Republic and 'retired' their father to a distant monastery. As no individual son ruled the Empire, the city judges decided that the dividing walls must remain unbroken - as they do to the present day. Nowadays the city is divided into four distinct sectors: Arts &amp;amp; Sciences (A&amp;amp;S), Bureaucracy and Administration (B&amp;amp;A), Commerce and Industry (C&amp;amp;I), and Dining and Accommodation (D&amp;amp;A), echoing the brothers' specialties in their chairmanship. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Late in their lives the co-chairmen decided it was inefficient to have to travel via city gates and the ox-cart-laden ring-road just to get to a different city quarter, so they built a subterranean railway with lines passing 'under' the walls. Thus they built the world's first Metro System and ensured efficient intra-city travel without breaking the now sacrosanct laws that the dividing walls should never be breached.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/r0VjM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;Recently the PRK has became convinced that terrorists are bent on the destruction of the city. To make it difficult for saboteurs to abuse the Metro, some rather unusual rules have been instigated - described in the &lt;strong&gt;Information for Metro Users&lt;/strong&gt; pamphlet that I was handed at the Hotel's front desk. I quote it verbatim: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;The System&lt;/strong&gt;&lt;br&gt;&#xA;       1) There are precisely four Metro stations - one in each quarter.&lt;br&gt;&#xA;       2) There are precisely five tracks, each connecting two distinct stations.&lt;br&gt;&#xA;       3) There are no track crossings and tracks only meet at the stations.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Security&lt;/strong&gt;&lt;br&gt;&#xA;       4) For security reasons there are no maps of the metro.&lt;br&gt;&#xA;       5) For security reasons there are no fixed timetables. At the instant each train &#xA;          leaves a station its destination is chosen randomly and without bias from one of &#xA;          the other stations directly connected to it by a track.&lt;br&gt;&#xA;       6) For security reasons it is illegal to exit a train after a single &#xA;          station-to-station journey (1-trip) - every passenger must remain on board while &#xA;          the train makes a minimum of two station-to-station journeys (n-trip where n&gt;1).  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Tips for Travelers&lt;/strong&gt;&lt;br&gt;&#xA;       7a) From any station in the network a 2-trip journey has a non-zero probability of &#xA;           reaching any station in the network (including the starting point).&lt;br&gt;&#xA;       7b) From any station in the network a 3-trip journey has a non-zero probability of &#xA;           reaching any station in the network (including the starting point).  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As you might guess, I took this as a challenge and decided to try and map out the underground for myself. As I bought a selection of 2-trip and 3-trip tickets at the desk I asked the clerk if there was anything he could tell me to make it easier to get around town efficiently. He paused for a few seconds, and checked that nobody was listening. &lt;strong&gt;&quot;Well, I probably shouldn't tell you this but all locals know from experience that if you use a 2-trip ticket from either A&amp;amp;S or C&amp;amp;I, you have an almost 45% chance of ending up back where you started&quot;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had an enjoyable day touring the city. It was frustrating at times - it took me several attempts to get to get to A&amp;amp;S so I could visit the National Museum, but the people were friendly, the food (once I managed to get back to D&amp;amp;A) was delicious. And of course, I was happy because I knew exactly which pairs of stations were connected by the five tracks - thanks to the tip the clerk had provided. As I was dropping off my room key the next morning the clerk handed me a new Information for Metro Users pamphlet. It was identical except that sections 7a and 7b had been crossed out and replaced with the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;7c) From any station in the network a 4-trip journey has a non-zero probability of &#xA;            reaching any station in the network (including the starting point).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;The old Tips, 7a and 7b don't work anymore&quot;, the clerk commented ruefully. &quot;Of course we are offering to exchange any 2-trip or 3-trip tickets you have left over with new 4-trip tickets&quot;. I asked what had changed and he replied, &lt;strong&gt;&quot;Just overnight a single new station-to-station track has just been completed and opened to the public - but simultaneously exactly one of the old tracks has been temporarily closed down for maintenance&quot;&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I asked if he had any hints to make travel easier. &quot;&lt;strong&gt;Well, maybe I shouldn't be telling you this, but my brother works for the Railway Administration Committee. He let it slip to me last night that with the new set-up if you use a 4-trip ticket, starting from either C&amp;amp;I or D&amp;amp;A, you have an exact 25% chance of ending up back where you started&quot;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had another great day touring around, this time at the National Gallery. The art was exquisite, but for me the highlight of the day was that I still knew exactly which pairs of stations were connected by the five tracks - thanks to the new information the clerk had provided. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Now I'm sure the story of the underground system will come up again once Ernie is back home. But for the life of me I can't work out how the stations were connected either before or after the network changes. Could anyone possibly help me with a description or diagram?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="-1" LastEditDate="2016-01-20T12:51:52.583" LastActivityDate="2016-01-20T12:51:52.583" Title="Ernie and the Underground Network" Tags="|logical-deduction|mathematics|geometry|topology|network|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15448" PostTypeId="2" ParentId="15446" CreationDate="2015-05-25T01:39:12.287" Score="3" Body="&lt;p&gt;Is this simply:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An egg&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In a marble hall white as milk,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The eggshell&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;lined with skin as soft as silk.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Whatever the skin inside the shell is called&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Within a fountain crystal-clear,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The white of the egg is clear(ish)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;a golden apple doth appear.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The yolk&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No doors there are to this stronghold,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No doors in an egg&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yet thieves break in to steal its gold.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Birds and other predators will prefer to eat the yolk&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="341" LastActivityDate="2015-05-25T01:39:12.287" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15449" PostTypeId="2" ParentId="15447" CreationDate="2015-05-25T02:28:33.410" Score="7" Body="&lt;p&gt;Clearly, by the conditions given, the original graph of the subway looks like a square under the city with a single diagonal included, running either  from C&amp;amp;I to A&amp;amp;S or from D&amp;amp;A to B&amp;amp;A. If we started from a station &lt;em&gt;not&lt;/em&gt; including this diagonal, then it is connected to two other stations, each of which is connected to three other stations (exactly one of which is the original station). The probability of return from such a station is $\frac{1}3$. For a station on the diagonal, it is connected to both the stations off the diagonal (with a $\frac{1}2$ chance of return from each) and to the station across the diagonal (from which it may return with probability $\frac{1}3$). This yields a $\frac{1}3\cdot(\frac{1}2+\frac{1}2+\frac{1}3)=\frac{4}9$ chance of returning to the original station, which is in line with the number cited by the clerk. Thus, C&amp;amp;I and A&amp;amp;S are connected by subway originally:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;D&amp;amp;A---A&amp;amp;S&#xA; |    / |&#xA; |   /  |&#xA; |  /   |&#xA; | /    |&#xA;C&amp;amp;I---B&amp;amp;A&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second problem, we can guess that we have a symmetry in the graph, whereby we may &quot;exchange&quot; D&amp;amp;A and C&amp;amp;I while maintaining topological properties - this needn't be true in general, but we expect them to have the same properties roughly, so it's a good guess to start with. A graph satisfying this with one edge deleted is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;D&amp;amp;A---A&amp;amp;S&#xA; |    / |&#xA; |   /  |&#xA; |  /   |&#xA; | /    |&#xA;C&amp;amp;I   B&amp;amp;A&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, we note that the only violations to 7a and 7b are that a 2-trip cannot take B&amp;amp;A to A&amp;amp;S (or vice versa) and a 3-trip cannot take D&amp;amp;A to C&amp;amp;I (or vice versa). Adding the diagonal from B&amp;amp;A to D&amp;amp;A or any self loop would cause one of these pairs to be connected by 2 or 3 trip respectively - meaning 7a or 7b would still hold. Thus, we conclude that the new line runs parallel to an old one. The only edge fixed by exchanging D&amp;amp;A and C&amp;amp;I (which is the symmetry we want to preserve) is the A&amp;amp;S to B&amp;amp;A line, so we try doubling it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;D&amp;amp;A---A&amp;amp;S&#xA; |    /||&#xA; |   / ||&#xA; |  /  ||&#xA; | /   ||&#xA;C&amp;amp;I   B&amp;amp;A&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we compute the probabilities here (via Markov chain), we find that this must be the new map of the city, as it gives D&amp;amp;A and C&amp;amp;I a probability of $\frac{1}4$ returning with a 4-trip.&lt;/p&gt;&#xA;" OwnerUserId="4765" LastActivityDate="2015-05-25T02:28:33.410" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15450" PostTypeId="1" AcceptedAnswerId="15466" CreationDate="2015-05-25T02:52:32.040" Score="10" ViewCount="965" Body="&lt;p&gt;You are a city planner tasked with the placement of &lt;strong&gt;unit-square-sized houses&lt;/strong&gt; in a &lt;strong&gt;rectangular-grid allotment&lt;/strong&gt;, the size of which is up to you but must be as &lt;strong&gt;small as possible&lt;/strong&gt; to save money. The number of houses is of no concern, however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is, the prospective homeowners are very picky about the &lt;strong&gt;number&lt;/strong&gt; and &lt;strong&gt;arrangement&lt;/strong&gt; of neighbouring houses. It is up to you to satisfy &lt;em&gt;at least one&lt;/em&gt; of each kind of homeowner:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Some want &lt;strong&gt;no houses&lt;/strong&gt; to the north, south, east, or west of their house&lt;/li&gt;&#xA;&lt;li&gt;Some want &lt;strong&gt;only one house&lt;/strong&gt; to their north, some only one to the south, some only one to the east, some only one to the west&lt;/li&gt;&#xA;&lt;li&gt;Some want &lt;strong&gt;exactly two houses&lt;/strong&gt;, to their north and east, north and south, north and west, south and east ... etc.&lt;/li&gt;&#xA;&lt;li&gt;Some want &lt;strong&gt;exactly three houses&lt;/strong&gt;, to their north and east and south, to their ... etc.&lt;/li&gt;&#xA;&lt;li&gt;Some want to be &lt;strong&gt;surrounded by four houses&lt;/strong&gt;, to their north south east and west&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So in total, there are &lt;strong&gt;16&lt;/strong&gt; unique kinds of homeowners (1 + 4 + 6 + 4 + 1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, consider this grid:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;+---+&#xA;| o |    o = house&#xA;|ooo|    grid: 3x3&#xA;| o |&#xA;+---+&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The above grid satisfies exactly &lt;strong&gt;5&lt;/strong&gt; kinds of homeowners: all four of the kind that only want one neighbour, plus the one that wants four neighbours. The size of the grid is 3x3 = &lt;strong&gt;9&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;How can you arrange the houses, to satisfy at least one of all 16 kinds of homeowners, while minimising the size of the grid?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="104" LastActivityDate="2015-05-26T12:32:38.413" Title="Houses in a grid" Tags="|mathematics|optimization|graph-theory|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15451" PostTypeId="2" ParentId="15450" CreationDate="2015-05-25T03:13:11.030" Score="7" Body="&lt;p&gt;Not sure if this is the optimal strategy, but it works:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;5x5 grid&lt;br&gt;&lt;code&gt;000+0&lt;/code&gt;&lt;br&gt;&lt;code&gt;++++0&lt;/code&gt;&lt;br&gt;&lt;code&gt;000+0&lt;/code&gt;&lt;br&gt;&lt;code&gt;000++&lt;/code&gt;&lt;br&gt;&lt;code&gt;000+0&lt;/code&gt;&lt;br&gt;&lt;br&gt;Where 0's are houses and +'s are empty plots. The 3x3 house group suffices the 1 that wants all sides surrounded, the 4 that want 3 sides, and 4/6 that want 2 houses (adjacent sides). The 3x1 house groups get the last 2/6 that want 2 houses (opposite sides), as well as all 4 of the 1 houses. Then the single one is (surprise) the one that wants no neighbors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11727" LastEditorUserId="9286" LastEditDate="2015-05-25T11:55:50.870" LastActivityDate="2015-05-25T11:55:50.870" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15452" PostTypeId="2" ParentId="15412" CreationDate="2015-05-25T03:20:03.723" Score="4" Body="&lt;p&gt;Suppose you have 50 bags on the ground and 49 in the truck, and suppose the ones on the ground have more apples.  One at a time, take a bag from the ground and put it in the truck, and put one from the truck tand place it on the ground.  Keep doing this with the goal of totally swapping ground and truck bags.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually, you'll reach a point where moving a &quot;magic&quot; bag causes the truck to have more apples than the ground.  At this point, both the ground plus the magic bag has at least half the apples, AND the truck plus the magic bag has at least half the apples.  Either the ground or the truck ( plus the magic bag ) has at least half the oranges too.&lt;/p&gt;&#xA;" OwnerUserId="12699" LastActivityDate="2015-05-25T03:20:03.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15453" PostTypeId="1" AcceptedAnswerId="15454" CreationDate="2015-05-25T03:41:51.610" Score="2" ViewCount="1538" Body="&lt;p&gt;What's something a rich man needs,&lt;br&gt;&#xA;and a poor man doesn't need,&lt;br&gt;&#xA;What's greater than God,&lt;br&gt;&#xA;and eviler than the devil?  &lt;/p&gt;&#xA;" OwnerUserId="12037" LastEditorUserId="7308" LastEditDate="2015-05-25T03:46:19.030" LastActivityDate="2015-05-25T03:46:19.030" Title="Simple but somewhat tricky riddle" Tags="|riddle|" AnswerCount="1" CommentCount="4" ClosedDate="2015-05-25T07:32:23.920" ContentLicense="CC BY-SA 3.0" />
  <row Id="15454" PostTypeId="2" ParentId="15453" CreationDate="2015-05-25T03:43:51.703" Score="1" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Nothing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A rich man needs nothing. A poor man doesn't need nothing (needs everything). Nothing is greater than God. Nothing is eviler than the devil.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11727" LastActivityDate="2015-05-25T03:43:51.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15456" PostTypeId="2" ParentId="15450" CreationDate="2015-05-25T05:17:52.787" Score="0" Body="&lt;p&gt;This is my best strategy that works for the given conditions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;3x7 grid&lt;br&gt;XXOXXXO&lt;br&gt;XOXOXXX&lt;br&gt;XXOXXXX&lt;br&gt;&lt;br&gt;X's are the houses and O's are empty plots. It was a bit of trial and error before I stumbled across this, and the answer perfectly fits the requirements.&lt;br&gt;&lt;br&gt;I would be greatly interested in seeing if someone can beat this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-05-25T05:17:52.787" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15457" PostTypeId="2" ParentId="15434" CreationDate="2015-05-25T06:36:30.107" Score="8" Body="&lt;p&gt;Here is an answer which is overkill, using an unintuitive technical analysis theorem instead of combinatorics, but which surprisingly works. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume all fruits have the same volume and density. Arrange the bags in 3-dimensional space in such a way that any plane intersects the contents of at most 3 bags. This is always possible to do (think about it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ham_sandwich_theorem&quot; rel=&quot;nofollow&quot;&gt;ham sandwhich theorem&lt;/a&gt;, there exists a plane such that there is an equal volume of apples on either side of the plane, and similarly for the beetroots and cakes. At this point, there are at most 3 bags crossing the plane, so at least 97 not crossing it. This means that one side of the plane has at most 48 bags. Furthermore, these at most 48 bags, combined with the at most 3 bags on the border, contain half the volume of each food. Thus, combined together, these at most 51 bags contain half of each food by volume, and therefore by number.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-05-30T23:08:56.263" LastActivityDate="2015-05-30T23:08:56.263" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15459" PostTypeId="1" AcceptedAnswerId="15460" CreationDate="2015-05-25T08:24:14.330" Score="14" ViewCount="587" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Alice and Bob play a game with a pile of &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; pebbles. They take turns, with Alice going first. On their turn, a player can remove any number of stones from the pile which is a &lt;em&gt;proper divisor&lt;/em&gt; of the size of the pile on that turn. If a player reduces the pile to a single pebble, they win.&lt;/p&gt;&#xA;&lt;p&gt;For which values of &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; does Alice win under optimal play?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;For example, if &lt;span class=&quot;math-container&quot;&gt;$n=12$&lt;/span&gt;, Alice's first move could be to remove &lt;span class=&quot;math-container&quot;&gt;$1,2,3,4$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt; stones. Suppose she removes &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; (this is a legal move, not necessarily an optimal one). Bob now faces a pile of &lt;span class=&quot;math-container&quot;&gt;$10$&lt;/span&gt; pebbles, from which he can remove &lt;span class=&quot;math-container&quot;&gt;$1,2$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt; stones.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2018-05-26T16:12:32.960" Title="Dim - A Game with Pebbles" Tags="|strategy|game|nim|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15460" PostTypeId="2" ParentId="15459" CreationDate="2015-05-25T08:46:00.787" Score="11" Body="&lt;p&gt;Alice wins if and only if&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $n$ is even.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Proof&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;By induction. Firstly, $n=2$ is obviously a winning position, while $n=3$ is a losing position because the only allowed move is to remove one pebble, leaving the winning position $n=2$ for Bob.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now assume that for all $n&amp;lt;2N$, $n$ is a winning position if and only if $n$ is even. Now $n=2N$ is a winning position since Alice can remove a single pebble leaving Bob with the odd (therefore losing, by the induction hypothesis) position $2N-1$. And $n=2N+1$ is a losing position since any factor of this odd number is odd and so Alice would have to leave Bob with an even (therefore winning, by the induction hypothesis) position $2k\leq2N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;QED.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-25T08:46:00.787" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15461" PostTypeId="1" AcceptedAnswerId="15462" CreationDate="2015-05-25T09:05:54.767" Score="2" ViewCount="1528" Body="&lt;p&gt;Note: I found this puzzle on quora. I don't have a solution for it. I couldn't find it referenced using google search.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This time the jailor gives two of his favourite prisoners a chance to escape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He tells them that, on the following day he would take one of the prisoners to a closed cell which has only a small opening in a wall, large enough for a hand to fit in. The second prisoner would be standing on the other side of the opening outside the cell.&#xA;He would then give, 5 cards randomly (from a deck of 52 cards) to the prisoner in the cell and then, the prisoner after having a look at all the 5 cards, should return one card back to the jailor and pass on the remaining 4 cards one by one to the second prisoner outside through the hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The jailor then comes out to the second prisoner and asks him to name the card which has been returned by the first prisoner. If he names it right, both go scot free!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The prisoner can take some time before passing the cards, but once he starts he has to pass them quickly in succession. Also, all cards should be passed face-up only!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now both the prisonsers who share the same cell have one full day to devise a strategy so that no matter who is picked up by the jailor and whichever 5 cards he chooses to give to that prisoner, the second prisoner should name the 5th card correctly after looking at the 4 cards passed on to him.&#xA;How do they do it?&lt;/p&gt;&#xA;" OwnerUserId="12703" LastActivityDate="2015-05-25T09:29:25.697" Title="The 5 card problem with 2 prisoners" Tags="|logical-deduction|" AnswerCount="1" CommentCount="2" ClosedDate="2015-05-25T11:44:03.040" ContentLicense="CC BY-SA 3.0" />
  <row Id="15462" PostTypeId="2" ParentId="15461" CreationDate="2015-05-25T09:21:25.307" Score="3" Body="&lt;p&gt;(This is actually a puzzle I was planning to post here quite soon...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $C_1,C_2,C_3,C_4,C_5$ be the five cards. By the pigeonhole principle, two of these must have the same suit, say $C_1$ and $C_2$. (If more than two pairs match in suit, just pick any matching pair.) Consider the value of the cards mod 13 (so that the ace is worth one more than the king but also one less than the 2); then one of these two, say $C_2$, must be worth at most six more than the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first prisoner then chooses $C_1$ and passes the other four cards out to the second prisoner with $C_2$ on top. The difference between the values of $C_1$ and $C_2$, a number between 1 and 6, can be signalled by the order of the bottom three cards, since there are $3!=6$ permutations. Different permutations can be coded by where the least valuable and most valuable of the cards are. Note that the prisoners have to agree beforehand on an order for the suits, so that all 52 cards can be placed in order of value.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-05-25T09:29:25.697" LastActivityDate="2015-05-25T09:29:25.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15463" PostTypeId="1" AcceptedAnswerId="15468" CreationDate="2015-05-25T10:15:35.927" Score="13" ViewCount="1430" Body="&lt;p&gt;I have two 6-sided dice which between them have 12 different numbers, but some of the numbers are hard to tell apart because one is the same as the other upside down (like $89$ and $68$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I throw the dice once and they land like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mnkLM.gif&quot; alt=&quot;throw 1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numbers on the bases of the two dice now sum to $105$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I throw them again and they land like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kvgso.gif&quot; alt=&quot;throw 2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numbers on the bases of the two dice now sum to $149$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I throw them a third time and they land like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HTj7U.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is: &lt;strong&gt;what do the numbers on the bases of the dice now sum to?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I got this puzzle from &lt;a href=&quot;http://www.murderousmaths.co.uk/books/BKMMPxdd.htm&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;, where you'll also be able to check your answer - but please don't brute-force it to find out which answer is right by clicking all the possibilities! I'll only accept an answer if you've solved it all by yourself.)&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-25T14:32:43.737" Title="Diabolical dice!" Tags="|logical-deduction|calculation-puzzle|dice|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="15464" PostTypeId="1" AcceptedAnswerId="15493" CreationDate="2015-05-25T10:17:45.900" Score="2" ViewCount="269" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;E K E K E B D C A, C E A B,&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continue the sequence.&#xA; I'm sure you'll need nothing else, though if you do, I'll be adding some clues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit: Longer string, as a hint&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="12165" LastEditorUserId="12165" LastEditDate="2015-05-26T10:38:47.117" LastActivityDate="2015-05-26T11:52:51.280" Title="A sequence For You: E K E K E B D C A, C E A B,?" Tags="|letter-sequence|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15465" PostTypeId="1" AcceptedAnswerId="15467" CreationDate="2015-05-25T11:00:25.143" Score="6" ViewCount="5188" Body="&lt;p&gt;This is the third puzzle in the Hollywood Movie Rebus Puzzle series. See the others here.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/15328/hollywood-movie-rebus-puzzle-1&quot;&gt;Puzzle #1 Hollywood Rebus Movie Puzzle &lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/15350/hollywood-movie-rebus-puzzle-2&quot;&gt;Puzzle #2 Hollywood Rebus Movie Puzzle &lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;.&#xA;&lt;strong&gt;Background&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Your love for Hollywood movies and remarkable ability to remember movie titles is finally going to serve a purpose. There was indeed a time when you almost believed your mother’s words, as she scorned you: “Lazing in front of the television is not going to pay for the rent!” You knew better all along. As you and the rest of your family listen to the directions of your late, and wealthy, Uncle Rob’s will, you cannot help but smile as the executor reads “… and thus half of my fortune will indeed go toward a sole beneficiary. Namely he who solves the puzzles as depicted below”.&#xA;As you are each given a paper with four Hollywood movie rebus puzzles, you begin to smile. You know the answer to each.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;How will you answer the third puzzle?&lt;/strong&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/PdfyN.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/15742/hollywood-movie-rebus-puzzle-4&quot;&gt;Once solved, see the next puzzle - Puzzle #4 Hollywood Rebus Movie Puzzle &lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12317" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-23T05:14:50.603" Title="Hollywood Movie Rebus Puzzle #3" Tags="|rebus|movies|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15466" PostTypeId="2" ParentId="15450" CreationDate="2015-05-25T11:24:09.370" Score="8" Body="&lt;p&gt;So I got this for a 4x6 grid:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;pre&gt;oo o&lt;br/&gt;o o &lt;br/&gt;oooo&lt;br/&gt;  oo&lt;br/&gt;oooo&lt;br/&gt; o  &lt;br/&gt;&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which technically minimizes the area to 24.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Naturally, all reflections and rotations work as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, a 3x8 solution which has the same score:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;pre&gt;ooo&lt;br/&gt;oo &lt;br/&gt;o o&lt;br/&gt; o &lt;br/&gt;ooo&lt;br/&gt;ooo&lt;br/&gt;  o&lt;br/&gt;ooo&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also I will try to set a lower boundary for this problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dimensions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;3x3 minimum for the all-neighbor house&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Number of spots:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;16 houses (at least)&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The no-neighbors house:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If at a corner - 2 more spaces. However, the two-neighbor house that would fit in that corner would require 2 more spaces - so 4.&lt;/li&gt;&#xA;&lt;li&gt;If at an edge - 3 more spaces - if edge is 4 or less it will ruin one corner spot requiring two more spaces - so either 5*something or 5 spaces in total.&lt;/li&gt;&#xA;&lt;li&gt;If at an internal spot - 4 more spaces.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So far we've got a minimum of 20.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the possibilities are 21 (3*7) and 20 (4*5).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we need a total of 4 + 4*1 + 6*2 + 4*3 + 0 = 32, &quot;empties&quot;. However, any internal empty spot actually is empty for the neighboring houses (max four), and we also have &quot;default&quot; empty spots at the edges, each counting as 1 empty spot. So for the two cases we have:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;3*7: 32 - (6 + 14) = 12&lt;/li&gt;&#xA;&lt;li&gt;4*5: 32 - (8 + 10) = 14&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So we need to have 12/14 empty values through empty spots. However, in order for an empty value to count as 4, it has to be surrounded by 4 houses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;SO FAR&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far &lt;strong&gt;20&lt;/strong&gt; is the lower boundary, but I will think about it a bit more, I'm pretty certain that it can be proved that 3*7 and 4*5 will not work. For starters, we cannot place 4 internal empty spots in 3*7. I will give it more though and update this :)&lt;/p&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="5610" LastEditDate="2015-05-25T15:29:14.800" LastActivityDate="2015-05-25T15:29:14.800" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="15467" PostTypeId="2" ParentId="15465" CreationDate="2015-05-25T11:51:21.600" Score="11" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Karate Kid&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first item is a &lt;strong&gt;car&lt;/strong&gt;.&lt;br&gt;&#xA; The second is the symbol &quot;&lt;strong&gt;at&lt;/strong&gt;&quot;.&lt;br&gt;&#xA; The third is a cup of &lt;strong&gt;tea&lt;/strong&gt;.&lt;br&gt;&#xA; The fourth is a &lt;strong&gt;kid&lt;/strong&gt;.&lt;br&gt;&#xA; If you spell it out, it sounds like &quot;Karate Kid&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-05-25T11:51:21.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15468" PostTypeId="2" ParentId="15463" CreationDate="2015-05-25T14:13:20.850" Score="10" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 61 + 98 = 159&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are given that there are two six sided dice and that there are 12 unique numbers on them.  Some of the numbers look the same as others but they still only occur once.  For example we might see 66 and 99, we don't know which is which, but we know that they are not both 66.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem requires us to identify which die is which in the two images for each of the three rolls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the sake of clarity, we will call the die on the left in roll 1 &quot;DIE 1&quot;, and the one on the right &quot;DIE 2&quot;.  We know nothing about the sums at this point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7Vjqs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that DIE 1 has a 69 on it.  Rotated 180 degrees it is still 69 so it is unique.  Therefore the die on the left in roll 3 is DIE 1.  We can add the (68/89) to DIE 1.  Note the orientation of the (86/98) in DIE 2 of roll 1 and DIE 2 of roll 3.  It would be impossible for these two numbers to be the same based on their orientations, so the DIE 2 in roll 3 is the opposite three faces that we couldn't see in roll 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/d35sn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now looking at roll 2 we know that the die on the right must be DIE 1 because 19/61 cannot be on DIE 2.  We can now fill in all of the numbers on DIE 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/X8v0b.png&quot; alt=&quot;enter image description here&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can start looking at the sums.  We know that the bottom number of DIE 1 in roll 1 is 19 or 61.  105 - 61 = 44 which is not possible so the bottom number of DIE 1 is 19.  This identifies the 61 in DIE 1 and tells us that the bottom of DIE 2 is 86.  This also identifies the 98.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kXO0y.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number on the bottom of DIE 1 in roll 2 is either 18 or 81.  149 - 18 = 131 which is not possible, so the number is 81.  This identifies the 18 on DIE 2 and the 68 which we already know is opposite the 86 on DIE 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is only one possible way to merge the two images of DIE 2 and that is like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gocE9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number on the bottom of DIE 1 in roll 3 is 61 and the number on the bottom of DIE 2 in roll 3 is 98.  The sum is 159.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof that DIE 2 can only be built one way&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hcFnI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1777" LastEditorUserId="1777" LastEditDate="2015-05-25T14:32:43.737" LastActivityDate="2015-05-25T14:32:43.737" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15469" PostTypeId="2" ParentId="15445" CreationDate="2015-05-25T15:25:15.680" Score="6" Body="&lt;p&gt;This is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Prince Caspian&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and given AJL's note in comments, I guess that 24 means&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the 24th letter of the alphabet -- X, since he is King Caspian X.&lt;br&gt;&#xA; Although it makes my brain hurt to interpret a Roman numeral that way.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Novice shipwright, seafarer, navigator.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The plot of Voyage of the Dawn Treader&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I live in a land removed,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet my ancestors came from yours.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Caspian lives in Narnia which is populated with humans.  More specifically as rand al'thor said in comments, &quot;Caspian is a Telmarine, descended from pirates who slipped from this world into the world of Narnia.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My wife's name is a flower and sick,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the books she has no name, but in the movie it is Lilliandill (lily and ill)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My tutor was straight out of old times,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; His tutor Doctor Cornelius was part dwarf and thus from an older race than the humans.  Also he told Caspian many things about the old times&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And my son swapped my protein structure prediction with a small stream(that didn't have doubled).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is just saying that the name Rilian is like Caspian with &quot;casp&quot; (&lt;a href=&quot;http://predictioncenter.org/casproll/index.cgi&quot; rel=&quot;nofollow&quot;&gt;http://predictioncenter.org/casproll/index.cgi&lt;/a&gt; - CASP ROLL - obviously the author just googled CASP and took something) replaced with &quot;ril&quot; (&quot;rill&quot; = a small stream, without the double &quot;l&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was the first one who they saw, an old friend.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In The Silver Chair, Jill and Eustace are supposed to greet the first person they see, who is an old friend.  Eustace sees Caspian first...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But left because of the great time rift.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ... but Caspian sails away unrecognized because Eustace doesn't recognize him since it's been many years in Narnia (though not in England).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Welcome to the painting, friends and company&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In Voyage of the Dawn Treader the kids get to Narnia through a painting.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12092" LastEditorUserId="12092" LastEditDate="2015-08-08T15:46:07.003" LastActivityDate="2015-08-08T15:46:07.003" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="15470" PostTypeId="1" AcceptedAnswerId="15471" CreationDate="2015-05-25T17:28:57.747" Score="1" ViewCount="250" Body="&lt;p&gt;Fill in the missing word:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Lodge, Cheers, ??? - Realnet Longed Rabid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The only hint you get is that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a book is involved.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;d'alar'cop&lt;/strong&gt; and &lt;strong&gt;Florian F&lt;/strong&gt; should be able to solve this...&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-25T18:20:14.380" Title="Lodge, Cheers,?" Tags="|enigmatic-puzzle|word-sequence|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15471" PostTypeId="2" ParentId="15470" CreationDate="2015-05-25T17:51:20.013" Score="7" Body="&lt;p&gt;Cryptic Answer :  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;ChaB&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Cryptic Explanation :  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reference is &lt;strong&gt;GEB&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Full Explanation :  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; realnet is anagram for eternal&lt;br&gt;&#xA; longed is anagram for golden&lt;br&gt;&#xA; rabid is anagram for braid&lt;br&gt;&#xA; &lt;strong&gt;Godel, Escher, Bach: An Eternal Golden Braid, also known as GEB, is a 1979 book by Douglas Hofstadter&lt;/strong&gt;&lt;br&gt;&#xA; lodge is anagram for Godel&lt;br&gt;&#xA; cheers is anagram for Escher&lt;br&gt;&#xA; so ChaB is anagram for Bach&lt;br&gt;&#xA; Reference &lt;a href=&quot;http://en.wikipedia.org/wiki/G%C3%B6del,_Escher,_Bach&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/G%C3%B6del,_Escher,_Bach&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10940" LastEditorUserId="10940" LastEditDate="2015-05-25T18:20:14.380" LastActivityDate="2015-05-25T18:20:14.380" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="15472" PostTypeId="2" ParentId="15434" CreationDate="2015-05-25T18:11:58.127" Score="6" Body="&lt;p&gt;Take the bag with the most apples, and set it aside. Let's say it had $A$ apples. Do the same with the one with the most beets, and say it had $B$ beets. We ignore these until the very end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Queue up the remaining $98$ bags in order of number of apples. We split these $98$ bags into a Left and Right pile, using the following algorithm:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;While the queue is nonempty, remove two bags at the end with fewer apples. &#xA;Of these two bags, call the bag with more beets X, and the other Y.&#xA;Throw X into either the Left or Right pile, whichever has FEWER beets.&#xA;Throw Y into the other pile.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I claim that&lt;/strong&gt;, when this is done, that the &quot;apple discrepancy&quot; between the Left and Right piles is at most $A$, and the &quot;beet discrepancy&quot; is at most $B$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say the apple sizes were $a_1\le a_2\le \dots\le a_{98}$. In the worst case, the apple discrepancy is $$(a_{98}+a_{96}+\dots+a_2)-(a_{97}+\dots+a_1)$$$$=a_{98}-(a_{97}-a_{96})-\dots-(a_3-a_2)-a_1\le a_{98}\le A,$$ proving the first half of the claim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove the second half, we first show that, after every pair of bags is thrown, the the pile with more beets will have a &quot;pivotal bag,&quot; one whose removal will cause this pile to have at most as many beets as the other. Assume by induction this is true after $2n-2$ bags have been thrown, and let's say that Left has at least as many beets as Right at this point. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If Left &lt;em&gt;still&lt;/em&gt; has at least as many beets after throwing bags number $2n-1$ and $2n$, then whatever bag was pivotal before is still pivotal now. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If, on the other hand, throwing these two bags causes Right to have more beets, then the bag thrown into Right is now pivotal.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;By induction, there is always a pivotal bag. Therefore, the discrepancy is at most the number of beets in the pivotal bag, which is at most $B$, proving the &lt;strong&gt;claim&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we have two piles, Left and Right, each with $49$ bags, and the two bags set aside at the beginning. Throw those two bags into whichever of Left and Right (let's say it was Left) has more cakes. Left might have had fewer apples than right, but from the claim, we know it was losing by at most $A$. After the bag with $A$ apples is thrown in, Left now has at least as many apples as right. Same for the beets. Thus, Left now has at least half of each foodstuff.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-07-29T17:57:45.633" LastActivityDate="2015-07-29T17:57:45.633" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15473" PostTypeId="1" AcceptedAnswerId="15477" CreationDate="2015-05-25T18:12:12.310" Score="17" ViewCount="1451" Body="&lt;p&gt;Two children are playing a game with several piles of rocks. Each turn they do the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;They choose some pile of rocks. They count how many rocks are in this - let this be $n$. We remove this pile.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;They may (optionally) then create as many new piles of rocks as the desire, each pile having less than $n$ rocks (not all new piles need to have the same amount of rocks).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For instance, starting from a pile of two rocks, the legal moves are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Remove the pile of rocks entirely.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Remove the pile of rocks, and create any number of new piles, each containing $1$ rock.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The winner is the player to remove the last rock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a winning strategy for the game? What are the winning positions? That is to say, if you have some number of piles and know how many rocks are in each, how can you know whether you can secure a win, and what the correct move to do would be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(And yes, replacing a pile of two with $10^{100}$ piles of $1$ is legal, but very unsportsmanlike and never necessary in a winning strategy)&lt;/p&gt;&#xA;" OwnerUserId="4765" LastEditorUserId="1777" LastEditDate="2015-05-26T11:15:47.677" LastActivityDate="2015-05-26T14:06:29.887" Title="eXtreme Nim: A Game With *Lots* of Pebbles" Tags="|mathematics|game|nim|" AnswerCount="3" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="15474" PostTypeId="2" ParentId="15187" CreationDate="2015-05-25T18:22:48.090" Score="5" Body="&lt;p&gt;Not a complete answer, but maybe others can build on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a GENOME.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In one scenario, I am 44 at the beginning (but less than that intrinsically) and have a maximum of 96. Every other scenario for me has zero involved.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is the number of genes in some particular kind of genome (human?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;More than one of me always has zero involved. In one scenario, it is 24 intrinsically (34 if not for a limitation), and when everything is taken into consideration, the possible range is between 74 and 152.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Genetic engineering/modification?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;No scenario has a beginning, intrinsic, minimum or maximum of zero. When zero is involved, it is involved in some other way.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A genome with zero genes wouldn't be much of a genome!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;More than one of me can only occur unnaturally&lt;/em&gt; - &quot;more than one of me&quot; = &quot;meme&quot;, and a meme is supposed to be an artificial social version of a gene.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Aggie Kidd&lt;/strong&gt;, help me out here!&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-25T18:22:48.090" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15475" PostTypeId="2" ParentId="15450" CreationDate="2015-05-25T18:36:56.330" Score="0" Body="&lt;p&gt;Please see @dmg's excellent &lt;a href=&quot;https://puzzling.stackexchange.com/a/15466/8930&quot; title=&quot;solution&quot;&gt;answer&lt;/a&gt;, which points out that the block must measure at least $3$ houses in each dimension.  Therefore a grid whose size is a prime number (such as $17, 19$, or $23$) will not work, and neither will any grid whose size is $2$ times a prime number (such as $22=2*11$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly a grid of size $16=4*4$ (or anything smaller than $16$) doesn't work.  Since @dmg already supplied two different solutions of size $24$, it remains only to check grids of size $18=3*6$, $20 = 4*5$, and $21=3*7$. We eliminate the $3*6$ grid, since placing the &quot;no-neighbors&quot; house anywhere but a corner results in fewer than $16$ houses, and the grid&#xA;$$&#xA;\begin{matrix}&#xA;x &amp;amp; 0 &amp;amp; x\\&#xA;0 &amp;amp; x &amp;amp; x\\&#xA;x &amp;amp; x &amp;amp; x\\&#xA;x &amp;amp; x &amp;amp; x\\&#xA;x &amp;amp; x &amp;amp; x\\&#xA;x &amp;amp; x &amp;amp; x\\&#xA;x &amp;amp; x &amp;amp; x\\&#xA;\end{matrix}&#xA;$$&#xA;simply doesn't work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @2012rcampion pointed out in a comment to @dmg's answer, there are only $2^{20}$, or about 1 million, possible ways to place houses in a $4*5$ grid, and similarly there are only $2^{21}$ possibilities in a $3*7$ grid, or about 2 million.  So we could easily use a computer to brute-force our way through 3 million possibilities to find a minimum-size solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually it's much better than that, since we don't need to test all of these possibilities.  For instance, there's no point in testing grids with fewer than $16$ houses placed.  This represents a big savings: at most, we only need to test&#xA;$$&#xA;\sum_{k=16}^{20} \binom{20}{k} + \sum_{k=16}^{21}\binom{21}{k} = 27896 + 6196 = 34092&#xA;$$&#xA;possibilities, rather than 3 million. (The notation $\binom{n}{r}$ represents the binomial coefficient $C(n,r)$, the number of subsets of size $r$ of a given set of size $n$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_coefficient&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Binomial_coefficient&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, these numbers could probably be further reduced, based on the symmetries involved, etc.  In any case, @dmg and @2012rcampion have been working on programs to look for solutions, but I thought that they might be interested in this strategy for reducing the number of cases to check.&lt;/p&gt;&#xA;" OwnerUserId="8930" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-05-25T18:36:56.330" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15476" PostTypeId="2" ParentId="15473" CreationDate="2015-05-25T18:51:28.403" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Winning positions&lt;/strong&gt; : (A) those that have only odd number of 1-piles or (B) those that have odd number of N-piles (for some N) and even number of X-piles , for all other values of X or (C) those that have odd number of N-piles and odd number of M-piles and even number of X-piles for all other values of X or (D) those that contain an odd number of Y-piles for 3 or more values of Y, and even number of X-piles for all other values of X.&lt;br&gt;&#xA; &lt;strong&gt;Winning strategy&lt;/strong&gt; :&lt;br&gt;&#xA; &lt;strong&gt;Case (A)&lt;/strong&gt; : keep taking out one pile and give you opponent and even state. Your opponent will give you an odd state, so take the last pile and win the game.&lt;br&gt;&#xA; &lt;strong&gt;Case (B)&lt;/strong&gt; : Select the odd number of N-piles and remove it.&lt;br&gt;&#xA; Now, we have even number of N-piles and even number of all other X-piles.&lt;br&gt;&#xA; If your opponent removes one of these, you also remove one of these, so we always give him even number of X-piles.&lt;br&gt;&#xA; If your opponent converts one of these piles to small piles, you also convert one of these to the same number of smaller piles, so we always give him even number of X-piles.&lt;br&gt;&#xA; If, from the even state, he takes the last but one, then you get the odd state, so you can also take one and win the game.&lt;br&gt;&#xA; &lt;strong&gt;Case (C)&lt;/strong&gt; : Take the larger odd number of N-piles and convert that into one M-pile, so the state now has an even number of N-piles and an even number of M-piles.&lt;br&gt;&#xA; Now all piles are even for your opponent, who has to give you some odd state.&lt;br&gt;&#xA; Move as per Case (B) or eventually Case (A) for a Win.&lt;br&gt;&#xA; &lt;strong&gt;Till here, we can restrict the games moves to (1) remove a pile &amp;amp; (2) remove a pile and put back many piles of same size. This completes the winning strategy with that restriction.&lt;/strong&gt;&lt;br&gt;&#xA; &lt;strong&gt;When we have no restriction on the sizes of the piles we put back, then we can win in even more cases.&lt;/strong&gt;&lt;br&gt;&#xA; &lt;strong&gt;Case (D)&lt;/strong&gt; : Remove the largest N for which we have odd number of N-piles.&lt;br&gt;&#xA; Put back one pile each for the smaller odd-size piles of odd-count, thereby making all those piles have even-count, so we give an even state to our opponent who has to give back an odd state.&lt;br&gt;&#xA; Move as per Case (C) or Case (B) or eventually Case (A) for a win.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10940" LastEditorUserId="10940" LastEditDate="2015-05-25T19:45:36.683" LastActivityDate="2015-05-25T19:45:36.683" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15477" PostTypeId="2" ParentId="15473" CreationDate="2015-05-25T19:21:26.280" Score="14" Body="&lt;p&gt;Call a position &lt;em&gt;even&lt;/em&gt; if for every positive integer $n$, there are an even number of piles with exactly $n$ rocks. Call the position &lt;em&gt;odd&lt;/em&gt; otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Claim 1&lt;/strong&gt;: If a position is odd, there is a move which results in an even position.&lt;br&gt; From an odd position, find the largest $n$ such that there are an odd number of piles of size $n$, remove that pile, and for each $k&amp;lt;n$, create a single pile of size $k$ if and only if the number of piles of size $k$ that already exist is odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Claim 2&lt;/strong&gt;: If a position is even, every move results in an odd position.&lt;br&gt; From an even position, if a player removes a pile of $n$ rocks, there will be an odd number of piles of $n$ rocks remaining.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The first player has a winning strategy if and only if the position is odd.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The winning strategy is to make a move resulting in an even position. This is always possible by Claim 1, and Claim 2 shows that your opponent's move must again result in an odd position. This means your opponent can never move into an even position, and therefore can never win (as removing the last rock makes the position even).&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-05-26T14:06:29.887" LastActivityDate="2015-05-26T14:06:29.887" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15480" PostTypeId="1" AcceptedAnswerId="15516" CreationDate="2015-05-25T20:11:10.067" Score="2" ViewCount="1805" Body="&lt;p&gt;Possibly a star,&lt;br&gt;&#xA;A monthly child.&lt;br&gt;&#xA;Sometimes weighty,&lt;br&gt;&#xA;Sometimes liberated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;lines 3 and 4 each give a type, not a feature, property or state, of the entity suggested by each of lines 1 and 2, and that entity is the answer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Further hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;what stars do you know of?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7363" LastEditorUserId="7363" LastEditDate="2015-05-26T18:42:40.953" LastActivityDate="2015-05-26T20:49:09.600" Title="What am I? Possibly a star, a monthly child" Tags="|riddle|" AnswerCount="9" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="15481" PostTypeId="2" ParentId="15480" CreationDate="2015-05-25T20:19:21.917" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Menstrual Cycle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Possibly a star could mean&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The cycle could result in the birth of a future-famous celebrity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A monthly child&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It happens once a month and involves a key to progeneration&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes weighty&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Commonly includes bloating / slight weight gain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes liberated&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Sometimes (but not always) the menstrual cycle is 'liberated' from the woman's reproductive system.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12134" LastActivityDate="2015-05-25T20:19:21.917" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15482" PostTypeId="2" ParentId="15480" CreationDate="2015-05-25T20:20:26.100" Score="7" Body="&lt;p&gt;The answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The moon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possibly a star&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The moon, to the non-scientific eye, could be seen as a star, having all the visual characteristics&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A monthly child&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The moon wanes and waxes and hence becomes &quot;a child&quot; on a monthly basis (&quot;A young moon&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes weighty&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; At its peak of its waxing cycle the moon is biggest and most &quot;weighty&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes liberated&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The opposite of the above&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10974" LastActivityDate="2015-05-25T20:20:26.100" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15483" PostTypeId="2" ParentId="15480" CreationDate="2015-05-25T20:51:12.533" Score="1" Body="&lt;p&gt;Any Holiday&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possibly a Star:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Star of the month - where star is the &quot;focus&quot; (As in star of the show). eg: A holiday that is very famous in a particular month could be the main focus of that month.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A monthly child:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A holiday almost always belongs to a particular month. Also, there are holidays in every month.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes weighty:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Weighty as in important, or heavy on the mind with sadness/whatnot. Large ways to interpret this since many holidays fulfill both or one of the conditions, for example: Remembrance day, which is sad because it's in remembrance of dead people. Or Christmas, which is a major holiday that's very important to many people.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes liberated:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some holidays celebrate freedom. Liberated can also mean free from the social norm, and an example of a list of holidays that celebrates things out of the social norm can be found &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_LGBT_holidays&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8999" LastActivityDate="2015-05-25T20:51:12.533" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15484" PostTypeId="2" ParentId="15480" CreationDate="2015-05-25T21:07:31.180" Score="8" Body="&lt;p&gt;It is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Libra constellation. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Possibly a star,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The ruling planet of Libra(Zodiac) is Venus, one of the brightest objects in the night sky (even though Venus itself is not part of the constellation) - or it might just be referring to the faint stars that make up the constellation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A monthly child.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is the zodiac sign for those born between September 23rd and October 22nd&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes weighty, Sometimes liberated.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Libra represents Scales/Justice etc...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastActivityDate="2015-05-25T21:07:31.180" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15485" PostTypeId="2" ParentId="15480" CreationDate="2015-05-25T21:13:06.153" Score="3" Body="&lt;p&gt;I'm almost there with &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Neutron&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I'm having trouble with &quot;monthly child&quot;. The other clues are pretty clear:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Possibly a star - neutron star&lt;br&gt;&#xA;Sometimes weighty - neutron stars are extremely dense and &quot;heavy&quot;&lt;br&gt;&#xA;Sometimes liberated - &quot;free neutrons&quot;, which are individual neutrons that are free of the nucleus &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10861" LastActivityDate="2015-05-25T21:13:06.153" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15486" PostTypeId="2" ParentId="15480" CreationDate="2015-05-25T21:17:11.533" Score="3" Body="&lt;p&gt;Is it a:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Week?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Possibly a star,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Star of the week!  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A monthly child.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Year - Quarter - Month - Week  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes weighty,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Monday  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes liberated.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Friday!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2015-05-25T21:17:11.533" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15487" PostTypeId="2" ParentId="15473" CreationDate="2015-05-25T23:06:28.990" Score="2" Body="&lt;p&gt;One approach to this problem whereby we can deduce the answer without any initial guessing is to calculate the &lt;a href=&quot;http://en.wikipedia.org/wiki/Nimber&quot; rel=&quot;nofollow&quot;&gt;nimber&lt;/a&gt; $f(n)$ of a game starting with $n$ pebbles in a single pile. This is helpful because our game has the property that, when we have two piles, its the same as if we had two games simultaneously, where each player chose to play in one, and the winner was the person who made the last legal move. That is to say, the nimber of a game with multiple piles is the nim-sum of the nimbers of each pile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate the $f(n)$, we compute the nimber of every position that can be moved to from a pile of $n$, then calculate the minimal nimber not achieved in that set - as is the standard method in impartial games. Given that a pile of $n$ may be replaced by piles $s_1,\ldots,s_k$ where all values of $s_i$ are less than $n$ and that all possible moves are of this form, we may write &#xA;$$f(n)=\operatorname{mex}(\{f(s_1)\oplus f(s_2)\oplus \ldots \oplus f(s_k):0&amp;lt;s_i&amp;lt;k\})$$&#xA;where $\oplus$ is the nim-sum. Using the above, we may inductively show that $f(n)=2^n$ for $n\geq 1$. This is because if we let $s_k$ be the position of the ones in the binary expansion of $n'$ for some $n'&amp;lt;2^n$, then the expression $f(s_1)\oplus\ldots\oplus f(s_k)$ will equal $n'$ by inductive hypothesis. Thus, starting from position $n$, one can move to a position with any nimber less than $2^n$. However, since $2^n$ has a $1$ in the binary expansion in a position where no smaller $2^k$ does, it is not such a nim-sum, and hence must be the mex of the set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing this, we can say that a losing position is one whose nimber is $0$. Given that $f(n)=2^n$ happen to be linearly independent with respect to the nim-sum, it follows that the only way for this to happen is if there are an even number of piles with $n$ rocks for all $n$. Knowing that these are exactly the losing positions, we find that a winning position is one where there is some - and the proper move is to move to a losing position, meaning you must take the largest $n$ such that there are an odd number of piles with $n$ rocks, remove it, and ensure that all $k&amp;lt;n$ have an even number of piles of $k$.&lt;/p&gt;&#xA;" OwnerUserId="4765" LastEditorUserId="4765" LastEditDate="2015-05-26T02:18:12.460" LastActivityDate="2015-05-26T02:18:12.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15489" PostTypeId="2" ParentId="10933" CreationDate="2015-05-26T01:55:27.757" Score="3" Body="&lt;p&gt;As an extension to MisterEman22's &quot;what we know so far,&quot; hint #4 specifies that the function has to do with prime factors, so I added the prime factorization of each input to the table, in case it helps someone else.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;       1                    1   0&#xA;       2                    2   1&#xA;       8                2*2*2   1&#xA;       3                    3   2&#xA;       7                    7   2&#xA;      21                  3*7   2&#xA;     144          2*2*2*2*3*3   2&#xA;       5                    5   3&#xA;      13                   13   3&#xA;      34                 2*17   3&#xA;      55                 5*11   3&#xA;     610               2*5*61   3&#xA;     987               3*7*47   3&#xA;    6765            3*5*11*41   3&#xA;   46368   2*2*2*2*2*3*3*7*23   3&#xA;      89                   89   4&#xA;     233                  233   4&#xA;     377                13*29   4&#xA;    1597                 1597   4&#xA;    2584          2*2*2*17*19   4&#xA;   17711               89*199   4&#xA;  196418          2*17*53*109   4&#xA;  317811          3*13*29*281   4&#xA;24862486      2*11*73*113*137   4&#xA;    4181               37*113   5&#xA;   10946             2*13*421   5&#xA;   28657                28657   5&#xA;   75025             5*5*3001   5&#xA;  121393              233*521   5&#xA;  191199          3*17*23*163   5&#xA;  514229               514229   5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note I also sorted by output value, to aid in any pattern-finding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that the number has to do with the number of applications of a function before it reaches some final value (1).&lt;/p&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="9951" LastEditDate="2015-05-27T14:08:43.817" LastActivityDate="2015-05-27T14:08:43.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15490" PostTypeId="1" AcceptedAnswerId="15492" CreationDate="2015-05-26T03:52:50.433" Score="34" ViewCount="3241" Body="&lt;p&gt;On the roadside last week, I found what appears to be a page from a farmer's journal. Although the spelling and grammar might not be 100% correct, I can easily read &lt;em&gt;all but one word of it&lt;/em&gt;. There is a hole in the paper (shown as a blank line below) where that missing word was, and just for fun, I have been trying to figure out what it was.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am fairly certain that the word is a number, because all of the other numbers on the page were circled, and there was also a circle around the missing word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently, he split up each day's entry with horizontal lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the page:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Aah, life in the city of Emory! We know of no web access. We do need it soon. Got to go...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Some guy fed ________ old eggs to my pig. First, we do urge him to NOT&#xA;  feed my pig. My, oh, my, he aims to annoy us, as he bent the alloy&#xA;  pole below the coop roof! My wife almost told him to chill. Oh, how&#xA;  she abhors to be wronged!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Now we do see an old gipsy tie his one cow up at the billowy, old elm&#xA;  tree by the fir tree, as we begin to accost the guy to buy feed. His&#xA;  red chow feed bin sold for one cent! We got one as the guy spoke: &quot;Hi!&#xA;  We fry the best yucca in the city- the cost- one cent too!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;First, we chop wood, fix the ox yoke, fry the berry pie, mop the&#xA;  floors.... &quot;The ghost&quot; spooked my wife, &quot;Boo!&quot; So, my wife got the&#xA;  chills. She got the gist of his song.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the missing word?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-05-26T14:13:10.137" LastActivityDate="2015-05-26T14:13:10.137" Title="What is the missing word?" Tags="|word|enigmatic-puzzle|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15491" PostTypeId="1" AcceptedAnswerId="15498" CreationDate="2015-05-26T07:52:53.110" Score="5" ViewCount="703" Body="&lt;p&gt;I'm asking for a general formula/algorithm that can be used in such a game.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There are $n$ piles, with varying number of pebbles. The number of&#xA;  pebbles in each pile is given by an array from $pile[1]$ to $pile[n]$.&#xA;  There is another array called the $validmove[\ ]$ array. It has any&#xA;  number of numbers. For example, we can define $validmove[\ ]=\{1,2,5,7\}$&#xA;  This array is constant throughout the game.&#xA;  If $0$ is present in $validmove$, it means that an entire pile can be removed,&#xA;  irrespective of the number of pebbles it contains.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;2 players play turnwise. On each turn, a player selects a single pile.&#xA;  He may remove any number of pebbles, as long as that number is present&#xA;  in the $validmove$ array. In the current example, suppose we have&#xA;  selected a pile with $13$ stones. The player can now remove $1$, $2$,&#xA;  $5$ or $7$ pebbles from that pile, leaving it with $12$, $11$, $8$ or&#xA;  $6$ pebbles respectively.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The person who removes the last stone wins (or I can specify 'loses'&#xA;  as well, that may require a slightly different algorithm).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If both players play perfectly, who wins?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I couldn't find an algorithm on the net, even though there is a huge Wikipedia page on it, so I thought SE could maintain one for future reference.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2018-12-29T19:58:15.877" Title="Generalisation of Nim" Tags="|mathematics|game|algorithm|computer-science|nim|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15492" PostTypeId="2" ParentId="15490" CreationDate="2015-05-26T09:11:59.277" Score="33" Body="&lt;p&gt;The missing word is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; forty&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is the only number that fits into the pattern exhibited by the diary entries:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The words in the text are either alphabetically increasing or decreasing. Increasing means each letter in a word comes after the previous one in the alphabet (B ≤ E ≤ S ≤ T) and decreasing that it comes before the previous letter (P ≥ I ≥ G). The sequences are not strict, so double letters are allowed. Increasing and decreasing words alternate, so we need an increasing number. We are told that the missing word is a number and forty is the only one that fits the bill.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9860" LastActivityDate="2015-05-26T09:11:59.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15493" PostTypeId="2" ParentId="15464" CreationDate="2015-05-26T11:45:59.283" Score="4" Body="&lt;p&gt;I think I have it!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; E A B C?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's the opening to Fur Elise, after being subject to a kind of &quot;modulo 12&quot;. The Ks become D#, which gives you your opening sequence. The comma's indicate a tied note -- these along with the pattern of the first five notes were the give away for me.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11880" LastEditorUserId="11880" LastEditDate="2015-05-26T11:52:51.280" LastActivityDate="2015-05-26T11:52:51.280" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15494" PostTypeId="2" ParentId="15450" CreationDate="2015-05-26T12:32:38.413" Score="0" Body="&lt;p&gt;Since SE does not allow comments, here is an &quot;answer&quot;.&#xA;A fiddle that brute-forces the 3x8 combinations in 75 seconds on my machine, nothing fancy for optimizing the checks but it does the job. Can be changed for 4x6 or 5x5 (will take double time due to one more cell bit).&#xA;If there were smaller solutions, they would show up in the results for larger grids too, with the first row having only zeros. But there are none.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://jsfiddle.net/w4j5002x/8/&quot; rel=&quot;nofollow&quot;&gt;http://jsfiddle.net/w4j5002x/8/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12713" LastActivityDate="2015-05-26T12:32:38.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15495" PostTypeId="1" AcceptedAnswerId="15645" CreationDate="2015-05-26T14:06:24.073" Score="9" ViewCount="695" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;All alone, I'm afraid I can be a little bit depressing&lt;br&gt;&#xA;  One of my friends quickly comes along to lift the mood&lt;br&gt;&#xA;  They tend to bring their sister along, though, and we soon start squabbling&lt;br&gt;&#xA;  Their brother, the fourth of our group, doesn't help matters&lt;br&gt;&#xA;  The fifth and sixth bring expensive gifts to try and calm things down&lt;br&gt;&#xA;  And when the last appears, and all seven of us are together... well, I can't tell you about that. It's our secret&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint time!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;We don't care how expensive the gifts are, as long as they catch the light nicely. This is good, since the gifts might have been stolen  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11717" LastEditorUserId="11717" LastEditDate="2015-05-28T09:00:17.347" LastActivityDate="2015-05-28T11:54:41.703" Title="Who are my friends and I?" Tags="|riddle|word|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15496" PostTypeId="2" ParentId="15263" CreationDate="2015-05-26T14:39:45.330" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Mathematics is the indispensable instrument of all physical research.&lt;br&gt;&#xA;  - by Marcellin Berthelot&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is a substitution cipher.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;The cipher letters:    k b e m z d j r q l f n a&#xA;                       -------------------------&#xA;The plaintext letters: m a t h e i c n d p l r o&#xA;                       f   s     y         m&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;A few of the letters stand for two letters. For the words containing that letter, the decoded letter must be alternated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, letter k stands for both m and f. The first word, the fifth word, and the sixth word in the ciphertext all have k in it. So in the first word, k stands for m. In the fifth word, k stands for f, and in the sixth word, k stands for m.&lt;/p&gt;&#xA;" OwnerUserId="12109" LastEditorUserId="12109" LastEditDate="2015-05-26T14:58:48.387" LastActivityDate="2015-05-26T14:58:48.387" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="15497" PostTypeId="2" ParentId="15495" CreationDate="2015-05-26T14:42:12.407" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It's gotta be the first elements of the periodic table, but I don't know all of the clue reasons.&lt;br&gt;&lt;br&gt;All alone, I'm afraid I can be a little bit depressing&lt;br&gt;I'm not sure&lt;br&gt;&lt;br&gt;One of my friends quickly comes along to lift the mood&lt;br&gt;Helium is lighter than air, thus the lift.&lt;br&gt;&lt;br&gt;They tend to bring their sister along, though, and we soon start squabbling&lt;br&gt;Lithium is notorious for how reactive it is&lt;br&gt;&lt;br&gt;Their brother, the fourth of our group, doesn't help matters&lt;br&gt;Not sure on this one&lt;br&gt;&lt;br&gt;The fifth and sixth bring expensive gifts to try and calm things down&lt;br&gt;Boron is expensive and used in semiconductors, and carbon is pretty stable (Also diamonds are expensive, thanks LogicianWithAHat)&lt;br&gt;&lt;br&gt;And when all seven of us are together... well, I can't tell you about that. It's our secret&lt;br&gt; Again, not sure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12365" LastEditorUserId="12365" LastEditDate="2015-05-26T14:59:39.213" LastActivityDate="2015-05-26T14:59:39.213" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15498" PostTypeId="2" ParentId="15491" CreationDate="2015-05-26T14:49:57.030" Score="7" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sprague%E2%80%93Grundy_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sprague-Grundy theory&lt;/a&gt; gives an algorithm for determining the winning strategy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we need to compute the &lt;a href=&quot;http://en.wikipedia.org/wiki/Nimber&quot; rel=&quot;nofollow noreferrer&quot;&gt;nimbers&lt;/a&gt; of various pile sizes. This is a finite computation if &lt;span class=&quot;math-container&quot;&gt;$validmove[]$&lt;/span&gt; is finite. We will recursively define an array &lt;span class=&quot;math-container&quot;&gt;$nimber[]$&lt;/span&gt;. To start, define &lt;span class=&quot;math-container&quot;&gt;$nimber[0]=0$&lt;/span&gt;. For &lt;span class=&quot;math-container&quot;&gt;$n\geq 1$&lt;/span&gt;, we define&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;nimber[n]=\mathrm{mex}\big\{nimber[n-k]: 1\leq k\in validmove,k\leq n \big\}.&#xA;$$&lt;/span&gt;&#xA;Here &lt;em&gt;mex&lt;/em&gt; denotes the &lt;em&gt;minimum excluded&lt;/em&gt; value, that is, the smallest non-negative integer which is not of the form &lt;span class=&quot;math-container&quot;&gt;$nimber[n-k]$&lt;/span&gt;. This is assuming &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt; is not in &lt;span class=&quot;math-container&quot;&gt;$validmove$&lt;/span&gt;; if it is, we should take the smallest positive integer instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because &lt;span class=&quot;math-container&quot;&gt;$validmove$&lt;/span&gt; is finite, the array &lt;span class=&quot;math-container&quot;&gt;$nimber$&lt;/span&gt; will be eventually periodic. Each term only depends on the previous &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; terms, where &lt;span class=&quot;math-container&quot;&gt;$N=\max validmove$&lt;/span&gt;, so we can stop computing once we see &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; consecutive terms in &lt;span class=&quot;math-container&quot;&gt;$nimber[]$&lt;/span&gt; that have already appeared previously as consecutive terms. This is guaranteed to happen, as none of the terms in &lt;span class=&quot;math-container&quot;&gt;$nimber$&lt;/span&gt; can exceed the size of &lt;span class=&quot;math-container&quot;&gt;$validmove$&lt;/span&gt;. Actually, in all the examples I have computed (with &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt; not a valid move), &lt;span class=&quot;math-container&quot;&gt;$nimber$&lt;/span&gt; turns out to be purely periodic, although it is not clear to me that this will always be the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every game of generalized nim is equivalent to a game of ordinary nim, and a pile of size &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; in generalized nim corresponds to a pile of size &lt;span class=&quot;math-container&quot;&gt;$nimber[k]$&lt;/span&gt; in ordinary nim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once we have &lt;span class=&quot;math-container&quot;&gt;$nimber$&lt;/span&gt; calculated, we need to evaluate &lt;span class=&quot;math-container&quot;&gt;$nimber[k]$&lt;/span&gt; for &lt;span class=&quot;math-container&quot;&gt;$k=pile[1],\ldots,pile[n]$&lt;/span&gt;. If the nim sum (i.e., the binary xor) of these number is non-zero, the first player has a winning strategy. If the nim sum is zero, the second player has a winning strategy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example&lt;/strong&gt;: Let's consider the example you gave: &lt;span class=&quot;math-container&quot;&gt;$validmove[]=\{1,2,5,7\}$&lt;/span&gt;. We recursively compute&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;nimber[]=\{0,1,2,0,1,2,0,1,2,0\}.&#xA;$$&lt;/span&gt;&#xA;This means that in general,&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;nimber[k]=\begin{cases}0: k\equiv 0\mod3,\\1:k\equiv 1\mod 3,\\2:k\equiv 2\mod 3.\end{cases}&#xA;$$&lt;/span&gt;&#xA;For any other array &lt;span class=&quot;math-container&quot;&gt;$validmove$&lt;/span&gt;, there will be a similar way to describe &lt;span class=&quot;math-container&quot;&gt;$nimber$&lt;/span&gt; in terms of congruences, using a finite amount of data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nim sum of &lt;span class=&quot;math-container&quot;&gt;$nimber[pile[1]],\ldots,nimber[pile[n]]$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt; if and only if the number of piles whose size is one more than a multiple of three is even, and the number of piles whose size is two more than a multiple of three is even. In this case, the second player can win, and otherwise the first player can win.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="48964" LastEditDate="2018-12-29T19:58:15.877" LastActivityDate="2018-12-29T19:58:15.877" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="15499" PostTypeId="2" ParentId="15480" CreationDate="2015-05-26T15:04:33.423" Score="1" Body="&lt;p&gt;My answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mars&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Possibly a star,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mars looks like a wandering star in the sky. I believe it was the first one to be identified as something other than a star, because of the way that it wanders.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A monthly child.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; March is named for the god Mars.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes weighty, Sometimes liberated.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mars is the god of war. War is naturally a weighty subject (and it certainly weighs on those involved), and wars have been fought for the liberation of people.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9289" LastActivityDate="2015-05-26T15:04:33.423" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15500" PostTypeId="2" ParentId="15313" CreationDate="2015-05-26T15:25:32.787" Score="0" Body="&lt;p&gt;It doesn't go well with the hints but i thought it was funny so my answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; GGR Martin and characters from aSoIaF&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My first might have impressed if not for a knife.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Khal Drogo, impressive man, but died because of a knife&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My second's fall was far from bright.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bran Stark who fell from the tower &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My third kept undesirable company or drank too much wine,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Definitively Robert Baratheon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While my fourth was abandoned, frozen in time.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The white walkers. It's freezing in the north and almost everybody forgot about them&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fifth liked his eels chopped and axed,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ramsey Bolton who chopped the &quot;eel&quot; from Theon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And my sixth preferred cooked, if anyone asked.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The dragons, who burn (cook) everything&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My seventh could not stand the sight of a bath,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Maybe Osha or any wildings&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While my last might have been first, but surely was there last.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first men who are the original human inhabitants of Westeros but there were others mythical creatures here before&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5359" LastActivityDate="2015-05-26T15:25:32.787" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15501" PostTypeId="2" ParentId="15313" CreationDate="2015-05-26T16:05:49.200" Score="9" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The King/81st Lord of Stormend&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Stardust, by Neil Gaiman&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note: I am not familiar with the story having not read the book or seen the film, I worked it out based on Hints 2 and 3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;To prostrate oneself is to 'kneel' and a joyous man might be said to be a 'gay man'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The rest I worked out via Google, Wikipedia and the work-in-question's own wiki (which is not one of the better ones I've seen).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your first is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Primus. I can't find reference to the knife in the riddle, but 1 might have killed 2 refers to Secondus being killed by Primus in the book (but by Septus in the film), according to the Stardust Wiki.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your second is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Secondus. His fall was far from bright when he died after falling/being pushed from a window. 2 is said to have killed 4, and quote the Stardust Wiki: &quot;&lt;strong&gt;Quartus, who Secundus had killed some years ago&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your third is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Tertius. In the book he kept undesirable company and died &quot;during a liason with a prostitute&quot;. In the film he drank too much wine, which was poisoned.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your fourth is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Quartus. He was abandoned and frozen in time when he froze to death having been locked in a cold store.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your fifth is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Quintus. He was killed by poisoned eels in the book, and by an axe to the head in the film.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your sixth is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Sextus. He was 'cooked' in a fire started by Septimus.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your seventh is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Septimus. Can't find reference in the Wiki about the bath, but it does have a still image from the film in which Septimus is in a bathtub. He killed four (in the book) or five (in the film) of his brothers, but Primus wasn't one of them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your eighth is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Una. She might have been first (Una means 'one') but she was surely there last (she was the last survivor of the eight children).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10330" LastEditorUserId="7367" LastEditDate="2016-05-16T00:28:17.597" LastActivityDate="2016-05-16T00:28:17.597" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15502" PostTypeId="2" ParentId="15437" CreationDate="2015-05-26T16:53:04.320" Score="7" Body="&lt;p&gt;Maybe this will help someone get closer...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first is part of your digestive system, but you sometimes use only half of me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Colon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My second is sometimes called what people sometimes call the shark hanging in the Smithsonian.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My third is my mom, once removed from Matilda and re-moved again.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fourth is a greeting to my friend who is named after the end.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hyphen (&quot;Hi, fin!&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fifth is a really fast run.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dash&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My sixth is in the number three.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ellipsis (they come in groups of 3)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My seventh can replace my first and my last.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My eighth is the last 2 of my dad's essay topics.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Parentheses (&quot;parent theses&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3987" LastEditorUserId="15544" LastEditDate="2015-09-11T11:27:05.900" LastActivityDate="2015-09-11T11:27:05.900" CommentCount="20" ContentLicense="CC BY-SA 3.0" />
  <row Id="15503" PostTypeId="2" ParentId="15313" CreationDate="2015-05-26T17:15:20.240" Score="1" Body="&lt;p&gt;To clear up some of the things from @elf337's answer (I would put these spoilers in a comment but I don't think it's possible?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Primus has his throat cut by the witch-queen in the book&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Seventh&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Septimus was also killed by the witch-queen in the book. Don't know about the bath - a bath does feature in the book but not in relation to Septimus I think. I haven't seen the film.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12277" LastActivityDate="2015-05-26T17:15:20.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15504" PostTypeId="1" AcceptedAnswerId="15508" CreationDate="2015-05-26T17:19:25.393" Score="4" ViewCount="2073" Body="&lt;p&gt;I'm not sure if this riddle has been posted here before, but this is an old (not too great) riddle I've heard a long time ago:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;One hot summer weekend there was an emergency call to the police that&#xA;  a crime has been committed, Someone has been murdered at the park! The&#xA;  only lead the police have from their live footage feed was that the&#xA;  perpetrator was covered in blood  and what car they used to drive away&#xA;  in. They now know the car model and colour, but nothing more about the&#xA;  murderer or car! The police begin searching around the city for this&#xA;  car and one officer finds a house with a car on the drive way that&#xA;  matches the same model and colour. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The officer goes to the house and asks to speak to the car owner about&#xA;  the event. After questioning the car owner, the owner said they were&#xA;  at home all day and that no one has used their car all weekend. The&#xA;  officer asks the car owner if they could search the car to confirm&#xA;  their story, the owner gladly agrees and hands over the keys. The officer begins searching,&#xA;  looking for possible weapons or any traces of blood (since the&#xA;  perpetrator was covered in it). The officer finds nothing related to&#xA;  the crime, and not even anything incriminating (i.e drugs, weapons).&#xA;  The car was spotless. However after finishing the search, and some&#xA;  thinking, the car owner is arrested!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;At the police station the owner pleads guilty to their crime of the murder.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;How did the police officer know the owner was guilty?&lt;/b&gt;&lt;/p&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-05-26T17:33:57.863" LastActivityDate="2020-09-14T15:08:48.613" Title="Murderer, Car and Colour" Tags="|logical-deduction|story|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15505" PostTypeId="2" ParentId="15504" CreationDate="2015-05-26T17:25:31.507" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The motor was still hot so the police officer knew the guy was lying when he said that nobody used the car.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's kind of lateral thinking, it could be a lot of differents things like:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;-It rained that day but there was no trace of rain on the car&lt;br&gt;&#xA;-The car got a speeding ticket that day&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer of tfitzger works as well.&lt;/p&gt;&#xA;" OwnerUserId="5359" LastEditorUserId="5359" LastEditDate="2015-05-26T17:34:51.800" LastActivityDate="2015-05-26T17:34:51.800" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15506" PostTypeId="2" ParentId="15504" CreationDate="2015-05-26T17:26:01.727" Score="2" Body="&lt;p&gt;My guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There was some receipt or other indicator in the car that said that it had been used over the weekend, contrary to the owner's story. This made the officer suspicious, even though it wasn't directly linked to the crime.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12242" LastActivityDate="2015-05-26T17:26:01.727" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15508" PostTypeId="2" ParentId="15504" CreationDate="2015-05-26T17:42:04.353" Score="15" Body="&lt;p&gt;The &lt;em&gt;official&lt;/em&gt; answer is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The car must have been moved and quite possibly washed because although it is &lt;em&gt;a hot summer day&lt;/em&gt; the car is not roasting inside.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rationale:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Had the car been sitting in the sun all weekend, it would be very warm inside: since it is not, the officer concludes that someone must have been driving it around with the windows open, using air conditioning or &lt;em&gt;at the very least&lt;/em&gt; doing something that involved keeping the car open long enough for it to cool down (e.g. scrubbing blood off the seats).&lt;br&gt;&#xA;This is in contrast with the claims made that &lt;em&gt;no one's been using the car&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;..although my favourite is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the owner is covered in blood ^^&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9165" LastEditorUserId="11244" LastEditDate="2015-05-26T18:17:46.847" LastActivityDate="2015-05-26T18:17:46.847" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15509" PostTypeId="1" AcceptedAnswerId="15519" CreationDate="2015-05-26T18:25:21.907" Score="10" ViewCount="792" Body="&lt;p&gt;Calling all puzzlers! I've made my very own cryptic crossword, especially for you! Good luck, have fun, and happy solving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YckZI.png&quot; alt=&quot;Crossword&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Across&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;1. Not-all-there sibling touches bit of leg in lewd locale (7)&lt;br&gt;&#xA;5. Discussion about number one steamship adds &quot;it is Spanish&quot; (7)&lt;br&gt;&#xA;9. Give directions before the writer? (9)&lt;br&gt;&#xA;10. Permit to call ownership of holdings (5)&lt;br&gt;&#xA;11. Be aware; it sounds like a melody (6)&lt;br&gt;&#xA;12. Mold takes head of white snake into small vessel (7)&lt;br&gt;&#xA;14. To jog backwards is a crime (4)&lt;br&gt;&#xA;15. Most of shattered lantern found in back section of undeveloped area (10)&lt;br&gt;&#xA;19. Born anew, Fang led his pack into the embraces of present-day (10)&lt;br&gt;&#xA;20. City will not gerrymander (4)&lt;br&gt;&#xA;22. Ice cast away by monk (7)&lt;br&gt;&#xA;25. &quot;Yessir&quot;, I stuttered in the auditorium (3,3)&lt;br&gt;&#xA;27. Canoe flipped at sea (5)&lt;br&gt;&#xA;28. Upside-down whale takes false symbolic representation (9)&lt;br&gt;&#xA;29. Putting bit of raw rat into boiled emu is a mistake (7)&lt;br&gt;&#xA;30. Squashes secondhand kid (7)  &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Down&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;1. Beginning of broad assault potentially threatens short Protestant (4)&lt;br&gt;&#xA;2. Shoot past and conquer (9)&lt;br&gt;&#xA;3. Spasm in strange part of Macchu Picchu (6)&lt;br&gt;&#xA;4. Moving slowly, I tire (long walks) (9)&lt;br&gt;&#xA;5. Incorrectly parse weapon (5)&lt;br&gt;&#xA;6. Loyalist broke law in the beginning (8)&lt;br&gt;&#xA;7. Mix bit of grass with scattered potpourri in cool house (5)&lt;br&gt;&#xA;8. Early learner is stunned, wet, confused (3,7)&lt;br&gt;&#xA;13. Self-contained way to not sit by yourself? (10)&lt;br&gt;&#xA;16. Perplexed user shut a reference book (9)&lt;br&gt;&#xA;17. Nameless young man's son endlessly beaten (9)&lt;br&gt;&#xA;18. Percussion instrument is a way to catch fish (8)&lt;br&gt;&#xA;21. Unusual pontiff heads limited edition community (6)&lt;br&gt;&#xA;23. Sharp left in confusing race (5)&lt;br&gt;&#xA;24. Where was I? Inside a mollusk, I swear (5)&lt;br&gt;&#xA;26. He hears a song (4)  &lt;/p&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="11475" LastEditDate="2015-09-01T20:58:18.700" LastActivityDate="2015-09-01T20:58:18.700" Title="Cryptic Crossword #1: The Inauguration" Tags="|crosswords|cryptic-crosswords|" AnswerCount="3" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="15510" PostTypeId="2" ParentId="15480" CreationDate="2015-05-26T19:04:42.913" Score="3" Body="&lt;p&gt;I'll give it a try, you're second hint gave me the idea of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Luke Sky-walker - &lt;b&gt;Star&lt;/b&gt; Wars&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Possibly a star,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Hollywood star, and for reference to &lt;b&gt;Star&lt;/b&gt; Wars as someone who 'stars' in the movie&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A monthly child.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Luke starts as a child who grows through the story.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes weighty,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Luke played some weightly roles in the Star wars saga story (sometimes), but there are many other characters that are important (although I haven't really seen it)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes liberated. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Luke was fighting against the dark side (sometimes) trying to liberate the right side. (But was initially fooled/tricked by the dark side)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Although I was thinking for more of a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;b&gt;Star&lt;/b&gt; Trek like answer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-05-26T19:09:53.970" LastActivityDate="2015-05-26T19:09:53.970" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15511" PostTypeId="2" ParentId="15424" CreationDate="2015-05-26T19:15:54.983" Score="17" Body="&lt;p&gt;The final answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The current status is:&lt;br&gt; South (Flush) &gt; North (Three Aces) &gt; West (Aces and Twos) &gt; East (Two Aces).&lt;br&gt;&lt;br&gt;Swap the 5 of hearts in West's with the Ace of clubs in the board.&lt;br&gt;&lt;br&gt;The new status is:&lt;br&gt;East (Straight) &gt; West (Aces and Twos) &gt; North (Two Aces) &gt; South (Ace high)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The clue &quot;listen&quot; pertains to audio files embedded in the images.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The legend for the blue layer is four cards.  The file is an OGG Opus file.  The audio is someone saying &quot;If you haven't found any clues yet, here's what you should be looking for&quot; followed by what sounds like a dial up modem (yes I'm that old).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The blue layer contains an SSTV image.  I was able to display he image using RX-SSTV but not having speakers headphones don't put out enough volume to make the image clear.  It is definitely the picture of the poker game but I can't make out the cards clearly.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The red layer contains the key.  It appears to be OGG Opus file &lt;a href=&quot;http://en.wikipedia.org/wiki/Opus_(audio_format)&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Opus_(audio_format)&lt;/a&gt; but I can't play it here.  See Quark's Answer for the decryption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The green layers contains three pieces of information according to the legend.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The green values as noted in the colour histogram by Tryth correspond to the ascii values of letters (except for 0 and 255 of course).  I don't have the space to upload the whole text file, but here are the first few lines:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;-----BEGIN PGP MESSAGE-----&#xA;Version: GnuPG v1.4.12 (GNU/Linux)&#xA;&#xA;hIwDAAAAAAAAAAABBACSYQ99ag0VMUFXZqrBGDRem4eEilmls/7ZvvGAzZqfMsmE&#xA;NjX+WuH5gKyVVI1TGECpVxQq8jkFgdmSHxaW3+CPlVp5KbYrOAy39503BXwWQUQC&#xA;bmQaM9CKNvcJGwJwO6EP+H3h5YaRmlPLntVnwkyZltODbMRsyCd41RExlRGxhNL/&#xA;AAWDQQEUxtxTEahwiwqL0vID7XUVfqEovUk+aho0FPUuSuqU94l/hrh8rTSglQTd&#xA;x/1izqHZI0YG4Nir9sy6amtDNpafDN/IQCD8r3qp/gw2bvpKg4v9Df27frSju7Xr&#xA;...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The whole PGP stream will be in the &lt;code&gt;greengs.ppm&lt;/code&gt; file when this C code is run with &lt;code&gt;img.ppm&lt;/code&gt; in the execution folder:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#include &amp;lt;cstdio&amp;gt;&#xA;#include &amp;lt;cstdlib&amp;gt;&#xA;using namespace std;&#xA;&#xA;int red[256],green[256],blue[256];&#xA;&#xA;int main() {&#xA;    unsigned char * buffer;&#xA;    FILE *fin = fopen(&quot;img.ppm&quot;, &quot;r&quot;);&#xA;    FILE *fout = fopen(&quot;imgverify.ppm&quot;, &quot;wb&quot;);&#xA;    FILE *fred = fopen(&quot;red.ppm&quot;, &quot;wb&quot;);&#xA;    FILE *fgreen = fopen(&quot;green.ppm&quot;, &quot;wb&quot;);&#xA;    FILE *fblue = fopen(&quot;blue.ppm&quot;, &quot;wb&quot;);&#xA;    FILE *fredgs = fopen(&quot;redgs.ppm&quot;, &quot;wb&quot;);&#xA;    FILE *fbluegs = fopen(&quot;bluegs.ppm&quot;, &quot;wb&quot;);&#xA;    FILE *fgreengs = fopen(&quot;greengs.ppm&quot;, &quot;wb&quot;);&#xA;&#xA;    // Input the data from the ppm file into the buffer&#xA;    int lSize;&#xA;    fseek(fin, 0, SEEK_END);&#xA;    lSize = ftell(fin);&#xA;    printf(&quot;%d\n&quot;, (lSize-15)/3);&#xA;    rewind(fin);&#xA;    buffer = (unsigned char*) malloc (sizeof(unsigned char)*lSize);&#xA;    size_t result = fread (buffer,1,lSize,fin);&#xA;&#xA;    for(int i = 0; i &amp;lt; 15; i++) {&#xA;        fprintf(fout,&quot;%c&quot;, buffer[i]);&#xA;        fprintf(fred,&quot;%c&quot;, buffer[i]);&#xA;        fprintf(fblue,&quot;%c&quot;, buffer[i]);&#xA;        fprintf(fgreen,&quot;%c&quot;, buffer[i]);&#xA;    }&#xA;&#xA;    buffer[1] = '5';&#xA;    for(int i = 0; i &amp;lt; 15; i++) {&#xA;        fprintf(fredgs,&quot;%c&quot;, buffer[i]);&#xA;        fprintf(fbluegs,&quot;%c&quot;, buffer[i]);&#xA;        fprintf(fgreengs,&quot;%c&quot;, buffer[i]);&#xA;    }&#xA;&#xA;&#xA;    for(int i = 0, x = 15; i &amp;lt; 785; i++)&#xA;        for(int j = 0; j &amp;lt; 630; j++, x += 3) {&#xA;            printf(&quot;(%d,%d) = (%u %u %u)\n&quot;, i, j, buffer[x],buffer[x+1],buffer[x+2]);&#xA;            fprintf(fout,&quot;%c%c%c&quot;,buffer[x],buffer[x+1],buffer[x+2]);&#xA;            fprintf(fred,&quot;%c%c%c&quot;,buffer[x],0,0);&#xA;            fprintf(fgreen,&quot;%c%c%c&quot;,0,buffer[x+1],0);&#xA;            fprintf(fblue,&quot;%c%c%c&quot;,0,0,buffer[x+2]);&#xA;            fprintf(fredgs,&quot;%c&quot;,buffer[x]);&#xA;            fprintf(fbluegs,&quot;%c&quot;,buffer[x+2]);&#xA;            fprintf(fgreengs,&quot;%c&quot;,buffer[x+1]);&#xA;&#xA;            red[buffer[x]]++;&#xA;            green[buffer[x+1]]++;&#xA;            blue[buffer[x+2]]++;&#xA;        }&#xA;&#xA;    for(int i = 0; i &amp;lt; 256; i++)&#xA;        if(green[i])&#xA;            printf(&quot;%d %d\n&quot;, i, green[i]);&#xA;&#xA;    fclose(fin);&#xA;    fclose(fout);&#xA;    fclose(fred);&#xA;    fclose(fblue);&#xA;    fclose(fgreen);&#xA;    fclose(fredgs);&#xA;    fclose(fbluegs);&#xA;    fclose(fgreengs);&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The first clue in the green layer is a lock.  That refers to the PGP Encrypted Message.  Decrypting it yields the following message:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/aqQ4J.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second clue is that it needs to be verified.  Importing the public key for the fingerprint 0x40842FD6 and decrypting again yields some more interesting information:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/TpSsj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can extract that image using the following options:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;gpg --list-options show-photos --fingerprint 0x40842FD6&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This yields this final image:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Ebyv6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third part of the green legend is the decrypted file containing the three cat image shells (the hollow squares) and the three card images.&lt;/p&gt;&#xA;" OwnerUserId="1777" LastEditorUserId="1777" LastEditDate="2015-06-07T04:26:28.057" LastActivityDate="2015-06-07T04:26:28.057" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="15512" PostTypeId="2" ParentId="15509" CreationDate="2015-05-26T19:30:40.117" Score="3" Body="&lt;p&gt;Let's make a start!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1A. Not-all-there sibling touches bit of leg in lewd locale (7)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BROTHEL (not all of &quot;brother&quot; followed by a bit of &quot;leg&quot;; definitely a lewd locale).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;14A. To jog backwards is a crime (4)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TORT (&quot;trot&quot; is a synonym of jog, reversed becomes &quot;tort&quot; which is a crime).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;27A. Canoe flipped at sea (5)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; OCEAN (an anagram of canoe and synonym of sea).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1D. Beginning of broad assault potentially threatens short Protestant (4)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BAPT? (&quot;&lt;strong&gt;b&lt;/strong&gt;road &lt;strong&gt;a&lt;/strong&gt;ssault &lt;strong&gt;p&lt;/strong&gt;otentially &lt;strong&gt;t&lt;/strong&gt;hreatens, baptists might threaten Protestants?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5D. Incorrectly parse weapon (5)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SPEAR (an anagram of parse and a type of weapon).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;23D. Sharp left in confusing race (5)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CLEAR (&quot;confusing&quot; suggests anagrams, so we want an anagram of &quot;race&quot; with one extra letter; L is the first letter of &quot;left&quot;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;24D. Where was I? Inside a mollusk, I swear (5)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CLAIM (the letter I inside &quot;clam&quot; which is a mollusc; swear is a synonym of claim).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-05-26T20:43:39.527" LastActivityDate="2015-05-26T20:43:39.527" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15513" PostTypeId="1" AcceptedAnswerId="15515" CreationDate="2015-05-26T19:39:56.887" Score="7" ViewCount="1010" Body="&lt;p&gt;Take an equilateral triangle with dots at the vertices and at the midpoints of each side:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/y6M3U.png&quot; alt=&quot;Triangle game template&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a total of six dots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the triangle game, you have the numbers 1, 2, 3, 4, 5, 6. Place one number at each dot such that the numbers on each side adds up to the same total.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a solution for side length 9:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OpFaz.png&quot; alt=&quot;Triangle game&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find all the distinct solutions of the triangle game, not including reflections and rotations.&lt;/p&gt;&#xA;" OwnerUserId="12109" LastEditorUserId="5373" LastEditDate="2017-04-03T10:10:29.453" LastActivityDate="2017-04-03T10:10:29.453" Title="The Triangle Game" Tags="|mathematics|geometry|game|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15515" PostTypeId="2" ParentId="15513" CreationDate="2015-05-26T19:55:16.800" Score="10" Body="&lt;p&gt;Let $a,b,c$ be the numbers at the corners and $d,e,f$ be the numbers at the midpoints (say $d$ opposite $a$, $e$ opposite $b$, $f$ opposite $c$). To fix the rotation and reflection of a given solution, we can assume $a\leq b\leq c$. Clearly the largest of the three midpoint-numbers must be opposite the largest of the three corner-numbers and so on, so we also have $d\leq e\leq f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x=a+b+c$ and $y=d+e+f$, and let $S$ be the sum from each side. We have two equations in $x$ and $y$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$x+y=21$ (the sum of $1,\dots,6$)&lt;/li&gt;&#xA;&lt;li&gt;$2x+y=3S$ (adding together all three copies of $S$ results in counting each corner twice and each midpoint once).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So $x$ and $y$ must both be multiples of $3$. Thus the three numbers $a,b,c$ summing to $x$ must consist of one that is a multiple of 3, one that is one more than a multiple of 3, and one that is one less that a multiple of 3; similarly for $d,e,f$. So the possibilities are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$a=1,b=2,c=3,d=4,e=5,f=6$, giving $S=9$&lt;/li&gt;&#xA;&lt;li&gt;&lt;strike&gt;a=1,b=2,c=6,d=3,e=4,f=5, required condition not satisfied&lt;/strike&gt;&lt;/li&gt;&#xA;&lt;li&gt;$a=1,b=3,c=5,d=2,e=4,f=6$, giving $S=10$&lt;/li&gt;&#xA;&lt;li&gt;&lt;strike&gt;a=1,b=5,c=6,d=2,e=3,f=4, required condition not satisfied&lt;/strike&gt;&lt;/li&gt;&#xA;&lt;li&gt;$a=4,b=5,c=6,d=1,e=2,f=3$, giving $S=12$&lt;/li&gt;&#xA;&lt;li&gt;&lt;strike&gt;a=3,b=4,c=5,d=1,e=2,f=6, required condition not satisfied&lt;/strike&gt;&lt;/li&gt;&#xA;&lt;li&gt;$a=2,b=4,c=6,d=1,e=3,f=5$, giving $S=11$&lt;/li&gt;&#xA;&lt;li&gt;&lt;strike&gt;a=2,b=3,c=4,d=1,e=5,f=6, required condition not satisfied&lt;/strike&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-05-26T22:03:16.753" LastActivityDate="2015-05-26T22:03:16.753" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15516" PostTypeId="2" ParentId="15480" CreationDate="2015-05-26T20:49:09.600" Score="14" Body="&lt;p&gt;I think I got it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Mason  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Possibly a star,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It MAY be a SUN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A monthly child.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;MAY SON&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes weighty,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Stonemason (type of mason)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes liberated.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Freemason (also type of mason)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10861" LastActivityDate="2015-05-26T20:49:09.600" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15518" PostTypeId="2" ParentId="15509" CreationDate="2015-05-26T23:11:26.383" Score="4" Body="&lt;p&gt;Great puzzle!  Particularly loved 18D, which seems so obvious in retrospect that I'm surprised I've never seen it in a cryptic before.  And 24D is very well clued!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A note re: 7D and 21D: some people consider indirect anagrams (an anagram of a synonym) to be bad form in cryptics.  Not a big deal to me, esp. since those clues are pretty straightforward, just mentioning it FYI.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answers below.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;BROTHEL SESSION&lt;/code&gt; &lt;br&gt;&#xA; &lt;code&gt;A V I O P T G E&lt;/code&gt; &lt;br&gt;&#xA; &lt;code&gt;PRESCRIBE ALLOW&lt;/code&gt; &lt;br&gt;&#xA; &lt;code&gt;T R C T A L O S&lt;/code&gt; &lt;br&gt;&#xA; &lt;code&gt;.ATTUNE ROWBOAT&lt;/code&gt; &lt;br&gt;&#xA; &lt;code&gt;S H P R   A   U&lt;/code&gt; &lt;br&gt;&#xA; &lt;code&gt;TORT HINTERLAND&lt;/code&gt; &lt;br&gt;&#xA; &lt;code&gt;A O C N H T N E&lt;/code&gt; &lt;br&gt;&#xA; &lt;code&gt;NEWFANGLED TOWN&lt;/code&gt; &lt;br&gt;&#xA; &lt;code&gt;D   S   S P N T&lt;/code&gt; &lt;br&gt;&#xA; &lt;code&gt;ASCETIC AYEAYE&lt;/code&gt; &lt;br&gt;&#xA; &lt;code&gt;L L A L U O M H&lt;/code&gt; &lt;br&gt;&#xA; &lt;code&gt;OCEAN ACROPHONY&lt;/code&gt; &lt;br&gt;&#xA; &lt;code&gt;N A E I U L U M&lt;/code&gt; &lt;br&gt;&#xA; &lt;code&gt;ERRATUM STEPSON&lt;/code&gt; &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2374" LastActivityDate="2015-05-26T23:11:26.383" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15519" PostTypeId="2" ParentId="15509" CreationDate="2015-05-26T23:34:20.957" Score="7" Body="&lt;p&gt;Really enjoyed this thanks, some great clues. My answers take the form of 'ANSWER - definition, explanation of wordplay' with stuff in square brackets containing extra comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edit: Having seen DevOfZot's answer, I would say that 7D may be too tricky for an indirect anagram, whereas 21D is acceptible. Of course, YMMV!]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Across&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Not-all-there sibling touches bit of leg in lewd locale (7)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BROTHEL - Lewd locale, BROTHE (sibling [brother] not all there) + L (bit of leg)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;Discussion about number one steamship adds &quot;it is Spanish&quot; (7)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SESSION - Discussion, NO (number) + I (one) + SS (steamship) + ES (Spanish 'it is') backwards (about)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;9&quot;&gt;&#xA;&lt;li&gt;Give directions before the writer? (9)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PRESCRIBE - Give directions, PRE (before) + SCRIBE (the writer)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;10&quot;&gt;&#xA;&lt;li&gt;Permit to call ownership of holdings (5)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ALLOW - Permit, contained in (holdings) cALL OWnership&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;11&quot;&gt;&#xA;&lt;li&gt;Be aware; it sounds like a melody (6)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ATTUNE - Become aware, sounds like A TUNE (a melody)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;12&quot;&gt;&#xA;&lt;li&gt;Mold takes head of white snake into small vessel (7)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ROWBOAT - Small vessel, W (head of white) + BOA (snake) inside ROT (mold takes)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;14&quot;&gt;&#xA;&lt;li&gt;To jog backwards is a crime (4)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TORT - A crime, TROT (jog) backwards&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;15&quot;&gt;&#xA;&lt;li&gt;Most of shattered lantern found in back section of undeveloped area (10)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HINTERLAND - undeveloped area, anagram of (shattered) LANTER (most of lantern) in HIND (back section)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;19&quot;&gt;&#xA;&lt;li&gt;Born anew, Fang led his pack into the embraces of present-day (10)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; NEW FANGLED - present-day, contained in (into the embraces of) born aNEW FANG LED his pack&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;20&quot;&gt;&#xA;&lt;li&gt;City will not gerrymander (4)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TOWN - City, anagram of (gerrymander [meaning to move political boundaries]) WON'T (will not) [loved the use of gerrymander as anagram indicator - not sure I've seen it used before. The clue reads very well]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;22&quot;&gt;&#xA;&lt;li&gt;Ice cast away by monk (7)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ASCETIC - A monk lives an ascetic life, anagram of (away) ICE CAST&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;25&quot;&gt;&#xA;&lt;li&gt;&quot;Yessir&quot;, I stuttered in the auditorium (3,3)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; AYE AYE - Yessir, sounds like (in the auditorium) I I (I stuttered) [Really like this one]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;27&quot;&gt;&#xA;&lt;li&gt;Canoe flipped at sea (5)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; OCEAN - Sea, anagram of (flipped) CANOE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;28&quot;&gt;&#xA;&lt;li&gt;Upside-down whale takes false symbolic representation (9)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ACROPHONY - Symbolic representation [when the names of letters in an alphabet begin with the letters themselves], ORCA backwards (upside-down whale) + PHONY (false) [I had to look up the definition - a new one on me!] [A purist might question the use of 'upside-down' in an across clue]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;29&quot;&gt;&#xA;&lt;li&gt;Putting bit of raw rat into boiled emu is a mistake (7)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ERRATUM - A mistake [in writing], R (bit of raw) RAT inside (putting into) anagram of (boiled) EMU&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;30&quot;&gt;&#xA;&lt;li&gt;Squashes secondhand kid (7)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; STEPSON - Kid, STEPS ON (squashes)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Down&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Beginning of broad assault potentially threatens short Protestant (4)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BAPT - Protestant, Beginnings of Broad Assault Potentially Threatens [short for Baptist? Is it actually a word or is it an abbreviation?]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Shoot past and conquer (9)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; OVERTHROW - Conquer, OVER THROW (shoot past)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;Spasm in strange part of Macchu Picchu (6)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HICCUP - Spasm, anagram of (strange) PICCHU (part of Macchu Picchu)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;Moving slowly, I tire (long walks) (9)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LOITERING - Moving slowly, anagram of (walks) I TIRE LONG&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;Incorrectly parse weapon (5)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SPEAR - Weapon, anagram of (incorrectly) PARSE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;6&quot;&gt;&#xA;&lt;li&gt;Loyalist broke law in the beginning (8)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; STALWART - Loyalist, anagram of (broke) LAW in START (the beginning)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;7&quot;&gt;&#xA;&lt;li&gt;Mix bit of grass with scattered potpourri in cool house (5)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; IGLOO - Cool house, G (bit of grass) inside (mix) anagram of (scattered) OLIO (potpourri) [tough one - I vaguely think I might have seen OLIO before in a crossword, but I had to look up the definition]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;8&quot;&gt;&#xA;&lt;li&gt;Early learner is stunned, wet, confused (3,7)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; NEW STUDENT - Early learner, Anagram of (confused) WET STUNNED&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;13&quot;&gt;&#xA;&lt;li&gt;Self-contained way to not sit by yourself? (10)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; STANDALONE - Self-contained, STAND (not sit) ALONE (by yourself)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;16&quot;&gt;&#xA;&lt;li&gt;Perplexed user shut a reference book (9)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; THESAURUS - Reference book, Anagram of (perplexed) USER SHUT A&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;17&quot;&gt;&#xA;&lt;li&gt;Nameless young man's son endlessly beaten (9)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ANONYMOUS - Nameless, anagram of (beaten) YOUN MAN SO (young man's endlessly)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;18&quot;&gt;&#xA;&lt;li&gt;Percussion instrument is a way to catch fish (8)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CASTANET - Percussion instrument, CAST A NET (way to catch fish) [nice wordplay!]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;21&quot;&gt;&#xA;&lt;li&gt;Unusual pontiff heads limited edition community (6)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PEOPLE - Community, anagram of (unusual) POPE (pontiff) + LE (limited edition)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;23&quot;&gt;&#xA;&lt;li&gt;Sharp left in confusing race (5)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CLEAR - Sharp (e.g. in sharp / clear focus), L (left) in anagram of (confusing) RACE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;24&quot;&gt;&#xA;&lt;li&gt;Where was I? Inside a mollusk, I swear (5)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CLAIM - swear (to claim or swear something), I inside CLAM (mollusc)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;26&quot;&gt;&#xA;&lt;li&gt;He hears a song (4)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HYMN - A song, sounds like HIM (he)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12277" LastEditorUserId="12277" LastEditDate="2015-05-26T23:41:47.337" LastActivityDate="2015-05-26T23:41:47.337" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15520" PostTypeId="2" ParentId="15298" CreationDate="2015-05-27T02:44:18.610" Score="6" Body="&lt;p&gt;The first thing to notice is that each of the keys is divisible by the corresponding lock.  Thus, we have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;LOCK x MULT =   KEY&#xA;---------------------&#xA;66453 x 390 = 25916670&#xA;12345 x 387 = 04777515&#xA;33200 x 432 = 14342400&#xA;55475 x 363 = 20137425&#xA;43256 x 391 = 16913096&#xA;73553 x 391 = 28759223&#xA;34875 x 376 = 13113000&#xA;99858 x 364 = 36348312&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Based on the hints, the multiplier probably depends on the ASCII codepoints of the digits of the lock.  We can tell that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Since the multiplier is greater than &lt;code&gt;255&lt;/code&gt;, it's not a bitwise combination of the characters.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The multiplier is &lt;em&gt;not&lt;/em&gt; a linear function of the digits (the system would be inconsistent).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I started operating under the assumption that the function was a sum of a single function applied to each digit.  We then have ten function values to determine, and 8 equations (lock/key pairs).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then noted that, on average, the value of the function would have to be somewhere in the mid &lt;code&gt;70&lt;/code&gt;s.  That would put the function values right in the range of uppercase ASCII letters.  Restricting ourselves to the integer codepoints in the range &lt;code&gt;65&lt;/code&gt; to &lt;code&gt;90&lt;/code&gt;, we can solve for the possible values of the function using the following &lt;em&gt;Mathematica&lt;/em&gt; snippet (with &lt;code&gt;pairs&lt;/code&gt; being a matrix of the locks and keys):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Grid[x /@ Range[0, 9] /. &#xA;   FindInstance[&#xA;    Join[Total[x /@ IntegerDigits[#1, 10, 5]] == #2/#1 &amp;amp; @@@ pairs, &#xA;     Thread[65 &amp;lt;= (x /@ Range[0, 9]) &amp;lt;= 90]], x /@ Range[0, 9], &#xA;    Integers, 10000]] /. {i_Integer :&amp;gt; FromCharacterCode[i, &quot;ASCII&quot;]}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The possible values are:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0 1 2 3 4 5 6 7 8 9&#xA;-------------------&#xA;Y I N X J J M C I H&#xA;Y I N X L H M G E M&#xA;Y I N X N F M K A R&#xA;Z O T T B J S K M D&#xA;Z O T T D H S O I I&#xA;Z O T T F F S S E N&#xA;Z O T T H D S W A S&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;One of the possible solutions (#6, second-to-last) has a curious relation to the digits:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;d  f[d]&#xA;----------&#xA;0  Z(ero)&#xA;1  O(ne)&#xA;2  T(wo)&#xA;3  T(hree)&#xA;4  F(our)&#xA;5  F(ive)&#xA;6  S(ix)&#xA;7  S(even)&#xA;8  E(ight)&#xA;9  N(ine)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Thus, the key for lock &lt;code&gt;55476&lt;/code&gt; is simply:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;5 &amp;gt; F(ive)  &amp;gt; 70&#xA;5 &amp;gt; F(ive)  &amp;gt; 70&#xA;4 &amp;gt; F(our)  &amp;gt; 70&#xA;7 &amp;gt; S(even) &amp;gt; 83&#xA;6 &amp;gt; S(ix)   &amp;gt; 83&#xA;----------------&#xA;             376&#xA;         x 55476&#xA;----------------&#xA;        20858976&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="9718" LastEditDate="2015-05-27T03:17:50.477" LastActivityDate="2015-05-27T03:17:50.477" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15521" PostTypeId="2" ParentId="15495" CreationDate="2015-05-27T03:38:30.850" Score="5" Body="&lt;p&gt;My guess is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The days of the week&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All alone, I'm afraid I can be a little bit depressing&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If we start the days of the week with Saturday, Saturn in Roman myth can be considered a foreign or immigrant god - having been expelled by Jupiter. So alone and depressed. Also the connection to Kronos can relate to father time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One of my friends quickly comes along to lift the mood&#xA;They tend to bring their sister along, though, and we soon start squabbling&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sunday and Monday: Sol Invictus and Luna are brother and sister who ride their chariots to raise the Sun and Moon (or to raise themselves). Sol Invictus is also the god of soliders - so fighting. Also, Sol Invictus is considered by some to have overtaken Saturn as the most powerful god (Constantine struck his coins with Sol)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Their brother, the fourth of our group, doesn't help matters&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tuesday: Mars is next. Mars is the god of war.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The fifth and sixth bring expensive gifts to try and calm things down&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wednesday and Thursday: Mercury the god of wealth and Jupiter the sovereign god, the politician.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And when the last appears, and all seven of us are together... well, I can't tell you about that. It's our secret&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Friday: Venus is the goddess of love and well, those &quot;things that happen behind closed doors and under covers&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Other reason for my guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LogicianWithAHat's comment that implied either two of the answers would be women or men (I'm going with women)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11239" LastEditorUserId="11239" LastEditDate="2015-05-27T03:44:31.513" LastActivityDate="2015-05-27T03:44:31.513" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15522" PostTypeId="2" ParentId="9875" CreationDate="2015-05-27T04:00:35.667" Score="4" Body="&lt;p&gt;Based on what I've seen, I wouldn't be surprised if the maximum value of $N$ is as high as 37.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like @crazyiman, I looked at generating random permutations and seeing how they did. The highest I found through random permutations was also 5, but I noticed an interesting pattern. I ended up doing 100 million iterations of choosing a random permutation and seeing how well it did, and also counted the frequency of each depth. Here's what I found (where n is the shuffle on which it has an adjacent pair):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;n | count&#xA;--|------&#xA;1 | 95453225&#xA;2 | 4358185&#xA;3 | 181811&#xA;4 | 6562&#xA;5 | 212&#xA;6 | 5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The pattern here is that the ratio between counts is fairly stable:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1:2 - 21.9&#xA;2:3 - 24.0&#xA;3:4 - 27.7&#xA;4:5 - 31.0&#xA;5:6 - 42.4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The &lt;code&gt;5:6&lt;/code&gt; ratio is not very reliable because neither happened very often, but it at least shows it's in the same ballpark as the other ones. Seeing as @crazyiman constructed an example that works for 12 iterations, there's no reason to believe that the pattern won't continue for a while.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose that the ratio goes up by 4 (in an attempt to underestimate $N$) each time. Starting with $N!$ and dividing by 22, 24, 28, 31, 35, 39, etc. we reach a two digit number right around the 37th division.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is, of course, no guarantee that we can get that high, but it does suggest that it may be possible. It also doesn't rule out the possibility of an even higher maximum, but it does suggest that it is unlikely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To help you understand why this might be possible, you have to realize that there are &lt;strong&gt;&lt;a href=&quot;http://czep.net/weblog/52cards.html&quot; rel=&quot;nofollow&quot;&gt;a lot&lt;/a&gt;&lt;/strong&gt; of possible permutations:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;80658175170943878571660636856403766975289505440883277824000000000000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1955" LastActivityDate="2015-05-27T04:00:35.667" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15524" PostTypeId="1" AcceptedAnswerId="15549" CreationDate="2015-05-27T05:09:51.653" Score="14" ViewCount="667" Body="&lt;p&gt;You are in a chatroom with three Stack Exchange users. They are a Spambot, a Trusted User, and a Moderator, but you don't know who is who; the chatroom only displays their names (user314, user2718, and user1618), not their reputations. You need to discern their identities, and are only allowed to ask them yes/no questions, with each question being directed at a single user.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each user has a different degree of helpfulness...&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The Spambot is very unhelpful, and answers yes or no randomly.&lt;/li&gt;&#xA;&lt;li&gt;The Trusted User quite helpful, but occasionally makes mistakes. At most one of the answers this user gives will be incorrect.&lt;/li&gt;&#xA;&lt;li&gt;The Moderator is perfectly helpful, and will always answer correctly. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The only thing these three users know is each others identities. If you ask them a question they don't know the answer to, you will be banned from the chatroom for being off-topic. Also, self-referential questions (such as, &quot;Will your answer to this question be no?&quot;) are strictly against the chatroom's Paradox Prevention Policy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Determine the users' identities by asking at most &lt;strong&gt;six&lt;/strong&gt; questions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't know if six is optimal. I will accept the first correct answer, but if someone comes along later with a solution with fewer questions, I'll accept theirs instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did not invent this puzzle, I got it from a &lt;a href=&quot;https://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_hard;action=display;num=1031619416;start=&quot;&gt;different puzzle forum&lt;/a&gt;. The puzzle at the top of the linked discussion thread is different then the one here, you have to dig down to find the variant I used.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-05-27T08:23:40.910" LastActivityDate="2015-05-27T12:45:24.677" Title="Liar Puzzle with Stack Exchange Users" Tags="|liars|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15525" PostTypeId="1" AcceptedAnswerId="15526" CreationDate="2015-05-27T05:12:16.100" Score="1" ViewCount="472" Body="&lt;p&gt;Laura has to take 2 different kind of medicines (medicine 1, medicine 2). both look exactly the same (same weight, color, shape, size, etc…; i.e identical).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If she takes more than 1 medicine of the same kind per day, she will die. Furthermore, if she doesn’t take medicine 1 and 2 every day,  she will also die.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Laura is on a limited budget, and since the medicines are extremely expensive, she can’t throw any of them away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One day, she got distracted and by mistake put 2 medicines on the table. Since both medicine types look identical, she’s not sure which medicine is which.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Question: How can she manage to take the medicines without any further&#xA;  dire consequences and yet save the extra medicine?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10820" LastEditorUserId="10820" LastEditDate="2015-05-27T05:50:42.733" LastActivityDate="2016-07-12T22:13:29.217" Title="Medicine problem" Tags="|logical-deduction|" AnswerCount="1" CommentCount="5" ClosedDate="2016-07-12T22:13:35.980" ContentLicense="CC BY-SA 3.0" />
  <row Id="15526" PostTypeId="2" ParentId="15525" CreationDate="2015-05-27T06:07:54.523" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;One day, she got distracted and by mistake put 2 medicines on the table. Since both medicine types look identical, she’s not sure which medicine is which.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As I understood she should take 1 of each every day and she has put 2 of each on the table. Then the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cut each of the 4 pills in 2 and eat a half of each pill. Eat the other halves tomorrow.&lt;br&gt;&#xA; Like that she will consume $2*(1*0.5)$ of medicine 1 and $2*(1*0.5)$ of medicine 2, so exactly 1 of each.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9352" LastActivityDate="2015-05-27T06:07:54.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15528" PostTypeId="2" ParentId="15524" CreationDate="2015-05-27T06:27:08.930" Score="0" Body="&lt;p&gt;Here's how I would go about doing it, although I would need a place to write notes down. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I would ask the same question six times: &quot;Is user314 the Spambot?&quot; &lt;br/&gt;Then I would write down each user's response to the question after each time. Unless I have an extraordinary stroke of bad luck, the Spambot's answers will be randomly spread between Yes and No, the Trusted User will answer with five answers the same and one different, and the Moderator will answer the same every time. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; user314: 4 yes, 2 no&lt;br/&gt;&#xA; user2178: 5 yes, 1 no&lt;br/&gt;&#xA; user1618: 6 yes&lt;br/&gt;&#xA; Therefore I can conclude that user314 is the Spambot (random answers and verified by the other two),&lt;br/&gt; user2178 is the Trusted User (answered all but one with a yes; also verified by the Moderator answering the same way),&lt;br/&gt; and user1618 is the Moderator (answered all as yes). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12725" LastEditorUserId="10433" LastEditDate="2015-05-27T11:35:22.370" LastActivityDate="2015-05-27T11:35:22.370" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15529" PostTypeId="1" AcceptedAnswerId="15530" CreationDate="2015-05-27T06:40:57.967" Score="267" ViewCount="30752" Body="&lt;p&gt;As a birthday present last year, I received some fridge magnets. They didn't come as a puzzle, so I don't know if they have a solution, but I made a puzzle out of them anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magnets are tetrominoes. There are 7 of each shape. Is it possible to arrange them into a 7x28 rectangle so that they are all used and all inside the rectangle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closest I have managed is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2mXcu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4729" LastEditorUserId="5373" LastEditDate="2019-09-11T14:03:05.537" LastActivityDate="2021-10-11T01:47:51.257" Title="Is this Tetris puzzle solvable?" Tags="|tiling|polyomino|tetris|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="15530" PostTypeId="2" ParentId="15529" CreationDate="2015-05-27T06:54:20.733" Score="374" Body="&lt;p&gt;It is impossible.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Let the &lt;span class=&quot;math-container&quot;&gt;$7\times 28$&lt;/span&gt; area be painted with black and white squares in a checkerboard pattern. Every piece will cover &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; black and &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; white squares, except the T-piece, which covers &lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt; of one color and &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; of another. Since there are &lt;span class=&quot;math-container&quot;&gt;$7$&lt;/span&gt; T-pieces, a tiling that uses every piece cannot cover the same number of black and white squares. Since the board contains the same number of black and white squares, it is impossible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7308" LastEditorUserId="51105" LastEditDate="2021-10-11T01:47:51.257" LastActivityDate="2021-10-11T01:47:51.257" CommentCount="12" ContentLicense="CC BY-SA 4.0" />
  <row Id="15532" PostTypeId="1" AcceptedAnswerId="15537" CreationDate="2015-05-27T08:48:26.243" Score="8" ViewCount="4078" Body="&lt;p&gt;Someone gives you 4 apparently identical golden coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know that among them there could be exactly one fake coin, and that a fake coin hasn't the same weight of a golden one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your pocket you have a fifth - surely genuine - golden coin, and in front of you a balance with two pans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the minimum number of weighings to determine if there is a fake coin between the four you've been given, and if it's heavier or lighter than a golden one?&lt;/p&gt;&#xA;" OwnerUserId="12628" LastEditorUserId="12628" LastEditDate="2015-06-04T09:41:47.400" LastActivityDate="2015-06-04T09:41:47.400" Title="Four coins (plus one) and a balance" Tags="|logical-deduction|weighing|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15533" PostTypeId="2" ParentId="15319" CreationDate="2015-05-27T09:05:25.503" Score="0" Body="&lt;p&gt;&lt;strong&gt;Area 51&lt;/strong&gt; Icon looks like &lt;strong&gt;sunny&lt;/strong&gt;, and remove (&lt;strong&gt;less&lt;/strong&gt;) the Capital and take the following letters to get &lt;strong&gt;rea&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Area 51&lt;/strong&gt; background is a moon with a few aliens in &lt;strong&gt;flight&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So couple all this with &lt;strong&gt;Bird&lt;/strong&gt; to get &lt;strong&gt;Rhea&lt;/strong&gt; [ &lt;a href=&quot;http://en.wikipedia.org/wiki/Rhea_%28bird%29&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Rhea_%28bird%29&lt;/a&gt; ] which is a &lt;strong&gt;flightless bird&lt;/strong&gt; (refer : flight + less )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[[ It seems ambiguous, but this is the best I got till now ]]&lt;/p&gt;&#xA;" OwnerUserId="10940" LastActivityDate="2015-05-27T09:05:25.503" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15534" PostTypeId="1" AcceptedAnswerId="15538" CreationDate="2015-05-27T09:10:35.177" Score="-1" ViewCount="219" Body="&lt;p&gt;Three friends are playing a game of memory, there are a total of 20 cards with 10 card pairs. They try to make the most pairs in a row, after everyone had a turn they make the following statements:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Person a: &quot;I've made 10 pairs in a row, I'm perfect&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Person b: &quot;I've made 9 pairs in a row, I'm not the best but I'm good enough&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Person c: &quot;I've made 5 pairs in a row, I could do worse&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One person is lying, who is it?&lt;/p&gt;&#xA;" OwnerUserId="12218" LastEditorUserId="12218" LastEditDate="2015-05-27T09:56:54.540" LastActivityDate="2015-05-27T09:56:54.540" Title="Who is the liar?" Tags="|liars|cards|" AnswerCount="3" CommentCount="7" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15535" PostTypeId="2" ParentId="15532" CreationDate="2015-05-27T09:12:44.230" Score="0" Body="&lt;p&gt;Straight answer would be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 to 4 times&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1)You put the fifth coin on one end, and then put each coin in turn on the other pan.  The one that's different in weight is the fake coin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2)If you are lucky, the first one you put on would be the fake. If not, you'll need to weight each one in turn.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The question didn't ask for the minimum number of time to use the balance. :p&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12332" LastEditorUserId="12332" LastEditDate="2015-05-27T12:20:42.183" LastActivityDate="2015-05-27T12:20:42.183" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15537" PostTypeId="2" ParentId="15532" CreationDate="2015-05-27T09:18:11.973" Score="16" Body="&lt;p&gt;How many weighings you need:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Call the four coins A,B,C and D, and the true gold coin G. You start by weighing &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    AB | CG&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If this balances, then it is only possible that D is defective. Weigh D against G to find out for sure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If it tips towards AB, there are three possibilities: either A or B is heavy, or C is light. To figure out which one, weigh A against B. If they don't balance, you learned which of them was heavier. If they do balance, you learn C was light.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If it tips toward CG, this is very similar to the previous case. Just switch the words &quot;heavy&quot; and &quot;light&quot; in that paragraph.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Furthermore, it can't be done in just one weighing. There are nine possible situations (each coin could be heavy or light, or all could be same), and a weighing has only three outcomes. By the pigeonhole principle, there will be some outcome which would result from  from two different situations, so a weighing could not distinguish between those situations. &lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="1682" LastEditDate="2015-05-27T11:50:53.177" LastActivityDate="2015-05-27T11:50:53.177" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15538" PostTypeId="2" ParentId="15534" CreationDate="2015-05-27T09:22:17.870" Score="3" Body="&lt;p&gt;well, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 9 pair in a row is not possible .. because the two last (or first) cards have to be a pair as well (only a pair left), thus making it 10 pairs in a row .. so, B is lying.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11827" LastActivityDate="2015-05-27T09:22:17.870" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15539" PostTypeId="2" ParentId="15534" CreationDate="2015-05-27T09:26:51.410" Score="0" Body="&lt;p&gt;Either B or C, but it's impossible to know, if we assume that &quot;I could do worse&quot;  means &quot;someone did worse than me&quot;.&#xA;B if &quot;I could do worse&quot; means &quot;I didn't get the worst theoretical possible&quot;  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Let's assume A is lying and B and C are telling the truth.&#xA;B states that he is not the best, which would mean he's lying, so without going further we can tell that A is telling the truth.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let's assume B is lying. A's assertion doesn't conflict with B's, and if B did indeed make less that 5 pairs in a row, that would mean C is telling the truth.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let's assume C is lying. With the above assumption he could be lying, and indeed be the one who did worst. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5102" LastActivityDate="2015-05-27T09:26:51.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15540" PostTypeId="2" ParentId="15534" CreationDate="2015-05-27T09:40:25.780" Score="1" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Person B. If he got 9 pairs, there's only two matched cards left, meaning he's actually gotten all 10 pairs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12730" LastActivityDate="2015-05-27T09:40:25.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15541" PostTypeId="1" AcceptedAnswerId="15557" CreationDate="2015-05-27T10:30:51.747" Score="3" ViewCount="350" Body="&lt;p&gt;Another riddle on the theme of literature.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Our first has his hands on the money;&lt;/p&gt;&#xA;&lt;p&gt;Our second has his hands in fire;&lt;/p&gt;&#xA;&lt;p&gt;Our third was an outcast for years.&lt;/p&gt;&#xA;&lt;p&gt;Our fourth and fifth were together:&lt;br /&gt;&#xA;One dead, one deaf, or nearly so.&lt;/p&gt;&#xA;&lt;p&gt;Our sixth was tall and most important,&lt;/p&gt;&#xA;&lt;p&gt;Our seventh was different from all the rest.&lt;/p&gt;&#xA;&lt;p&gt;Friendly and welcoming, who are we?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-27T14:04:25.377" Title="Friendly and welcoming, who are we?" Tags="|riddle|knowledge|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15542" PostTypeId="2" ParentId="15532" CreationDate="2015-05-27T10:57:10.147" Score="1" Body="&lt;p&gt;For four coins, you only need one weighing to detect the presence of a fake.&#xA;Two in each pan, simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To identify the coin, and determine if heavier, two more weighings are required. A directed state graph is not necessary.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;arbitrarily empty a pan, divide the remaining coins and weigh.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;Are they of equal weight ?  If yes:&lt;/p&gt;&#xA;  &#xA;  &lt;ol start=&quot;2&quot;&gt;&#xA;  &lt;li&gt;select one of the other two coins and test it against the pocket&#xA;  coin.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;If no: &lt;/p&gt;&#xA;  &#xA;  &lt;ol start=&quot;3&quot;&gt;&#xA;  &lt;li&gt;select the heavy coin and test it against the pocket&#xA;  coin.   &lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;If they are the same weight then the lighter coin was fake.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Three in total to determine the fake, and tell if it is lighter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10483" LastEditorUserId="10483" LastEditDate="2015-05-28T18:53:47.177" LastActivityDate="2015-05-28T18:53:47.177" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15543" PostTypeId="1" AcceptedAnswerId="15546" CreationDate="2015-05-27T11:45:30.623" Score="0" ViewCount="203" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;When you are dangled&lt;br/&gt;&#xA;  Take my help so that everything will be clear&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I can move very fast&lt;br/&gt;&#xA;  But still I can not carry anything&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Put some of me in your life&lt;br/&gt;&#xA;  And you will be happy that you found your destiny&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I might be the way&lt;br/&gt; &#xA;  The new generations would want to travel&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Sometimes people use me&lt;br/&gt;&#xA;  But to demonstrate how far real things are&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10433" LastEditorUserId="7308" LastEditDate="2015-05-27T12:06:24.680" LastActivityDate="2016-05-13T11:57:01.047" Title="Take my help - Who am I?" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15544" PostTypeId="1" AcceptedAnswerId="15548" CreationDate="2015-05-27T11:57:56.887" Score="2" ViewCount="1086" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Sometimes I end up as stars?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Like or dislike by peoples,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Yes, of course I'm reason of hurt and quarrel,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Sites want to prevent me.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I am Bitter sometimes funny.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Singers use me sometimes.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I think I am fashion of new generation.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Who am I?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4312" LastEditorUserId="2673" LastEditDate="2015-05-27T15:31:07.443" LastActivityDate="2015-05-27T16:51:18.987" Title="Sometimes I end up as stars?" Tags="|riddle|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15545" PostTypeId="1" AcceptedAnswerId="15547" CreationDate="2015-05-27T12:05:10.060" Score="7" ViewCount="1475" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;When I am white, I do good.&lt;br&gt;&#xA;  When I am black, I do evil.&lt;br&gt;&#xA;  When I am grey, I do both.&lt;br&gt;&#xA;  When I am red, I rule.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="10400" LastActivityDate="2015-05-27T19:16:03.457" Title="What is my true color?" Tags="|riddle|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15546" PostTypeId="2" ParentId="15543" CreationDate="2015-05-27T12:06:19.520" Score="3" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; light&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;when you are dangled&#xA;take my help so that everything will be clear&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you are dangled it helps to shed light on you situation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can move very fast&#xA;but still i can not carry anything&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Light moves at the speed of light, but will not carry anything.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;put some of me in your life&#xA;and you will be happy that you found your destiny&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The light of my life is something (often someone) that makes me really happy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;i might be the way&#xA;the new generations would want to travel&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Honestly, I would be scared. But I guess if it ends up being a well tested technology...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;sometimes people use me&#xA;but to demonstrate how far real things are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I love to tell little kids, how far the light of a star they look at has traveled. And how long.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10885" LastActivityDate="2015-05-27T12:06:19.520" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15547" PostTypeId="2" ParentId="15545" CreationDate="2015-05-27T12:18:27.850" Score="17" Body="&lt;p&gt;you are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a hat !&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I am white, I do good.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; white hats hacks for the common good, to help improve network security&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I am black, I do evil.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; black hats hacks for fun and profit&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I am grey, I do both.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a gray hat hacks sometime for the common good, and sometime for despicable reasons&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I am red, I rule.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; red hat is a widely used, and rather good, linux distribution (probably leader on the market)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11827" LastActivityDate="2015-05-27T12:18:27.850" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15548" PostTypeId="2" ParentId="15544" CreationDate="2015-05-27T12:26:07.350" Score="14" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You're profanity.&lt;br&gt;&#xA;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why I m end up with stars?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Profanity is frequently replaced with ****.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Like or dislike by peoples,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people have a real objection, others enjoy it as emphasis.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, of course I'm reason of hurt and quarrel,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Certain types can instigate arguments, anger, and even hurt.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sites want to prevent me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many websites implement auto-filters to the **** above&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am Bitter sometimes funny.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Depending on the context, profanity can either indicate bitterness or add humor.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Singers use me sometimes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; At one point: &quot;Parental Advisory: Explicit Lyrics&quot; was a common thing, as singers used more and more profanity in their lyrics.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think I am fashion of new generation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Profanity is more acceptable in younger generations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10625" LastEditorUserId="2673" LastEditDate="2015-05-27T15:38:55.340" LastActivityDate="2015-05-27T15:38:55.340" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15549" PostTypeId="2" ParentId="15524" CreationDate="2015-05-27T12:39:15.103" Score="4" Body="&lt;p&gt;For convenience I've renamed your users A, B and C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Q1 &amp;amp; Q2) Ask A, &quot;Is B the spambot?&quot; twice in succession. &lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Case 1: Two yeses&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;If you get two yeses, that means either A is the spambot (and the others could be anything), A is the moderator and B is the spambot, or A is the trusted user and B is the spambot (since trusted user will only make a mistake once).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all three cases, C is not the spambot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now (Q3) ask C if A is the moderator. &lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Case 1a&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;If you get a yes, C can't be the moderator (moderator won't tell you someone else is the moderator). Since we already know C isn't the spambot, C must be the trusted user, but we don't know whether they've used their mistake or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeat the question (Q4). If it's yes again, A must be the moderator, C the trusted user (with no mistakes - o ye of little faith) and B the spambot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Q4 was no the second time, you have to ask again (Q5). If you get a yes (that's YNY for Q3-Q5), A must be the mod and B the spambot. If you get a no (that's YNN for Q3-Q5) then A must be the spambot and B the moderator.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Case 1b&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;If C said that no, A was not the moderator, then you have one of the three cases:&lt;br&gt;&#xA;- C is moderator, A is trusted user, B is spambot&lt;br&gt;&#xA;- C is moderator, A is spambot, B is trusted user&lt;br&gt;&#xA;- C is trusted user (mistake in Q3), A is mod, B is spambot  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ask C the same question (Q4) to rule out the third case. If C is the trusted user, they will have used up their mistake so will say yes the second time, which proves A is mod and B is spambot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore if C says no again, you know you have one of the first two cases and C is the moderator. The moderator will tell you the truth so just ask (Q5) C who's who (e.g. &quot;is A the spambot?&quot;) and you're done.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Case 2: Two nos&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;This is symmetrical to Case 1 with B and C the other way around; I've omitted it for brevity and to reduce the chance of me making a transcribing error.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Case 3: One yes, one no (order doesn't matter)&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;A can't be the moderator, since they've given conflicting answers. A is either the spambot or the trusted user. If they're the trusted user, they've used up their mistake and will now tell the truth (this is important).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now ask (Q3) ask B if C is the moderator.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Case 3a: B says yes&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;B can't be the moderator if they tell you someone else is. We already deduced that A wasn't the moderator, so C must be the moderator. Therefore C will tell you the truth, so ask C (Q4) who's who out of the other two and you're done.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Case 3b: B says no&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;In this case we have one of the following cases:&lt;br&gt;&#xA;- B mod, A spambot, C trusted user&lt;br&gt;&#xA;- B mod, C spambot, A trusted user&lt;br&gt;&#xA;- B spambot, C mod, A trusted user (because A is not mod)&lt;br&gt;&#xA;- B trusted user, C mod, A spambot  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the latter three cases, note that the trusted user has made a mistake already and won't make any further mistakes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now ask B (Q4) if A is the spambot. Remember that B is either the spambot himself or truthful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 3bi&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B says no. Since we know that B is either the spambot himself or truthful, A cannot be the spambot. We already ruled out A being the moderator in Q1-2, so A must be the trusted user. What's more, A used up their mistake in Q1 or Q2, so A must now be truthful. So we can ask (Q5) A who the other two are (e.g. &quot;is C the spambot?&quot;) and we're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 3bii&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B says yes. We know that B is either the spambot himself or truthful, so either A or B is the spambot. We also know A is not mod. Therefore we have one of:&lt;br&gt;&#xA;- A spambot, B mod, C trusted&lt;br&gt;&#xA;- A spambot, B trusted, C mod&lt;br&gt;&#xA;- B spambot, A trusted, C mod  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sadly, we haven't asked C any questions yet in this scenario so we can't guarantee truthfulness (they could still be a trusted user making an error).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So next (Q5) we ask C, &quot;Is A the trusted user?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Case 3bii-a&lt;/em&gt;&lt;/strong&gt;: If we get a yes, either C is the moderator and A the trusted user, or C is the trusted user (option 1 from list in 3bii) and has now used up their mistake. Either way, C will now tell the truth, so we can use our final question (Q6) to ask &quot;Are you the moderator?&quot; to narrow it down between (C mod/A trusted/B spam) and (C trusted/A spam/B mod).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Case 3bii-b&lt;/em&gt;&lt;/strong&gt;: If we got a no from C to &quot;Is A the trusted user?&quot; we know we have either:&lt;br&gt;&#xA;- C mod, A spambot, B trusted user (used mistake in Q3), or&lt;br&gt;&#xA;- C trusted user (no mistakes), A spambot, B mod&lt;br&gt;&#xA;Either way, B won't make a mistake, so we can use our final question (Q6) to ask B if he is the moderator, which will tell us which of those two options it is.&lt;/p&gt;&#xA;" OwnerUserId="5260" LastEditorUserId="5260" LastEditDate="2015-05-27T12:45:24.677" LastActivityDate="2015-05-27T12:45:24.677" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15550" PostTypeId="2" ParentId="15319" CreationDate="2015-05-27T12:42:13.437" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Seek out a sunny picture -&#xA;  Remember you're on SE -&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://stackexchange.com/pagenotfound&quot;&gt;&quot;page not found&quot; page&lt;/a&gt; of Stack Exchange has a sunny picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ykCLL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And on that sunny picture,&#xA;  Figure out what you can;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The image's name is &quot;no-pandas-here.jpg&quot;. Pandas are likely to become extinct.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You are a less capital idea&lt;br&gt;&#xA;  Than those who follow you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We're looking for an &quot;idea&quot; that can restore a species to a stable population.&lt;br&gt;&#xA;If we interpret &quot;capital&quot; as &quot;involving punishment by death&quot;, then &lt;em&gt;Jurassic Park&lt;/em&gt; comes to mind - people clone dinosaurs, and are killed by their own creations.&lt;br&gt;&#xA;So the idea is &lt;strong&gt;cloning&lt;/strong&gt;. Although it is quite capital, it is still a &quot;less capital idea&quot; than &quot;cloning &lt;em&gt;and&lt;/em&gt; genetically engineering hybrid animals to make them more cunning and vicious&quot;, as (probably) seen in &lt;em&gt;Jurassic World&lt;/em&gt;, which &quot;follows&quot; the original film (i.e. is a sequel).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What kind of birds appear?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to Wikipedia's &lt;a href=&quot;http://en.wikipedia.org/wiki/De-extinction&quot; rel=&quot;nofollow noreferrer&quot;&gt;De-extinction&lt;/a&gt; page, the potential candidate birds for cloning are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Passenger pigeon&lt;/li&gt;&#xA;&lt;li&gt;Carolina parakeet&lt;/li&gt;&#xA;&lt;li&gt;Ivory-billed woodpecker&lt;/li&gt;&#xA;&lt;li&gt;Moa&lt;/li&gt;&#xA;&lt;li&gt;Heath hen&lt;/li&gt;&#xA;&lt;li&gt;Dodo&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1497" LastActivityDate="2015-05-27T12:42:13.437" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15551" PostTypeId="1" AcceptedAnswerId="15554" CreationDate="2015-05-27T12:58:51.800" Score="-8" ViewCount="9404" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;your task is to cut this pie (shown below) into seven equal parts.&#xA;  &lt;br/&gt;&#xA;  &lt;img src=&quot;https://i.stack.imgur.com/Hxyow.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;  &lt;br/&gt;&#xA;  What equal means by is size and shape&#xA;  &lt;br/&gt;&#xA;  you have only one knife of length equal to cakes diameter and a scale to measure W and D but nothing else&lt;br/&gt;&lt;br/&gt;&#xA;  Edit : Also there should be nothing of cake that that is wasted/eaten&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10433" LastEditorUserId="10433" LastEditDate="2015-05-27T13:44:51.143" LastActivityDate="2022-02-09T17:27:17.487" Title="Cutting circular pie into seven equal parts" Tags="|strategy|" AnswerCount="5" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="15553" PostTypeId="2" ParentId="9875" CreationDate="2015-05-27T13:00:56.287" Score="4" Body="&lt;p&gt;Probably the most simple pattern is to shift odd numbered cards along by 4 looping round to the end for cards at the start of the deck:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;5 2 7 4 9 6 11 8 13 10 15 12 17 14 19 16 21 18 23 20 25 22 27 24 29 26&#xA;  31 28 33 30 35 32 37 34 39 36 41 38 43 40 45 42 47 44 49 46 51 48 1 50&#xA;  3 52&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which allows 11 shuffles before pairs end up together. I feel like crazyiman's result of 12 is probably the best that can be achieved but that is just a hunch from playing with various sequences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The code below might help test possible shuffling patterns, it has an option which is (almost) equivalent to cutting the deck after each shuffle. I wasn't quite sure if that was supposed to be part of the routine or not so it really just compares the first and last cards when checking for pairs.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;/* prompt for shuffling pattern - future index of card in each position */&#xA;var P=prompt('Enter your shuffling sequence (numbers should be separated by spaces):')&#xA;       .replace(/^\s+|\s+$/g,'') .split(' ');&#xA;&#xA;var cut=confirm('Cut the cards? Treats the first and last cards of the deck a being adjacent.');&#xA;&#xA;/* checks a pattern uses each of the numbers 1-52 exactly once */&#xA;function check(P){ &#xA;    var n=[];&#xA;    for(var i=1; i&amp;lt;53; i++) n[i]=0;&#xA;    P.forEach(function(N){ n[N]++ });&#xA;    for(var i=1; i&amp;lt;53; i++) if(n[i]!==1){ alert('Bad Shuffling Pattern'); throw  n; }&#xA;}&#xA;&#xA;/* returns false if two adjacent cards in a deck match */&#xA;function test(d){ &#xA;    for(var i=1; i&amp;lt;52+(cut?1:0); i++) if(d[i-1]==d[i%52]) return false;&#xA;    return true;&#xA;}&#xA;&#xA;/* initialises a deck of cards and shuffles until test fails */&#xA;function measure(P){ &#xA;    check(P);&#xA;    var d=fresh();&#xA;    var t=true, c=-1;&#xA;    while(t){&#xA;        d=shuffle(d,P);&#xA;        t=test(d);&#xA;        c++;&#xA;    }&#xA;    return c;&#xA;}&#xA;&#xA;/* shuffles the cards according to given pattern */&#xA;function shuffle(d,p){&#xA;    var D=[];&#xA;    for(var i=0; i&amp;lt;52; i++) D[i]=d[p[i]-1];&#xA;    return D;&#xA;}&#xA;&#xA;/* builds the pre-set new deck order */&#xA;function fresh(){&#xA;    var cards=[];&#xA;    for(var i=0; i&amp;lt;52; i++){&#xA;        cards[i]=(i/4)&amp;gt;&amp;gt;0;&#xA;    }&#xA;    return cards;&#xA;}&#xA;&#xA;/* runs the whole test and display the result */&#xA;alert(&quot;The cards can be 'shuffled' &quot;+measure(P)+&quot; times before pairs end up adjacent.&quot;);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If you're not familiar with javascript read about &#xA;&lt;a href=&quot;https://webmasters.stackexchange.com/questions/8525/how-to-open-the-javascript-console-in-different-browsers&quot;&gt;How to open the JavaScript console in different browsers&lt;/a&gt;. Then copy and paste the code into the javascript console and press enter. You should be prompted for your shuffling sequence which will be checked and measured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've only tested this code in Chrome using crazyiman and Penguino's sequences both of which produce the expected results depending on the 'cutting the deck' option.&lt;/p&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="-1" LastEditDate="2017-04-13T12:33:26.200" LastActivityDate="2015-05-27T13:00:56.287" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15554" PostTypeId="2" ParentId="15551" CreationDate="2015-05-27T13:05:46.653" Score="5" Body="&lt;p&gt;Silly solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You cut through the Height? (the W in the picture). So you end up with 7 slices with the same diameter as the whole cake, but just 1/7th of its height.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12643" LastActivityDate="2015-05-27T13:05:46.653" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15555" PostTypeId="2" ParentId="15544" CreationDate="2015-05-27T13:28:34.997" Score="2" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; privacy &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why I m end up with stars?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; because the way to hide something written can be replacing it with ****&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Like or dislike by peoples,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; most people like to have their privacy, but the business owners and politicians do not like the privacy of others&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, of course I'm reason of hurt and quarrel,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; privacy is a hot topic at the moment and reason for arguments and discussions&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sites want to prevent me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; most of the web sites may violate your privacy or ask for registrations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am Bitter sometimes funny.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; showing private information for someone is bitter or shameful,yet sometimes very amusing for the audience (reddit anyone?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Singers use me sometimes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; because singers are stars and people tend to violate the privacy of famous people&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think I am fashion of new generation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; so it is.. we kind of lost our privacy, now we want to get it back&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5283" LastActivityDate="2015-05-27T13:28:34.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15556" PostTypeId="1" AcceptedAnswerId="15558" CreationDate="2015-05-27T13:51:35.260" Score="15" ViewCount="784" Body="&lt;p&gt;A man recently went on a very long trip. First, he travelled to Spain and went on a cruise that took him from Barcelona to Provence, Nice, Florence, Rome, Naples and back to Barcelona. Then he went to Ibiza and partied so hard that he almost didn't make it back to Barcelona in time for his flight to Denmark! When he got to Copenhagen, he went on a cruise to Tallinn, St.Petersburg, Helsinki, and Stockholm before returning to Copenhagen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then he flew all the way to Japan, where he did some sightseeing in Tokyo and ate some fugu. From there he went to China and visited the Great Wall and Forbidden City. He also made a stop in Macau and hit the casinos there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that, he flew all the way to America and visited the Northeast.  He spent a few weeks hiking the Long Trail. Then he went on a tour of Yale, and later while driving around he passed Shea Farm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;His next stop was further south - Shenandoah National Park. He drove along Skyline Drive and took in all the natural scenery.  From there he drove over to the home of The King and did the Graceland tour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later while walking along Beale Street, he smacked his forehead when he realized that he forgot to visit the Big Apple!  So he drove back up north, and he was happy when he found a hotel with complimentary valet service.  He spent some time visiting all the major sights, including the Statue of Liberty, Ellis Island, the Empire State Building, and Times Square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that, he headed south again, and he was able to get to Churchill Downs just in time to bet on some horses.  Then he went north a bit and visited some of the attractions along the White River, including the zoo and the Medal of Honor Memorial.  Then he drove up to Chicago and visited Millennium Park and also caught a ballgame at Wrigley Field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Soon afterwards, he decided to go back to the East Coast.  He went to the Taj Mahal and then relaxed at a nearby beach that described itself as surely the best.  While he was there, he saw someone get arrested for indecent exposure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At that point, he thought it was time to see the West Coast, so he flew out to California.  In San Francisco, he went to the Golden Gate Bridge and watched a beautiful sunset.  But he loved the East Coast more, so he flew back east and went over to the Outer Banks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that, he drove up to DC and went on a tour of the White House.  He also stopped by the Horseshoe in Baltimore.  Then he went to Ocean City and found a room at the first green hotel in the city.  He relaxed on the beach and visited a lot of the shops and eateries nearby.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then it was finally time for him to go home.  The long trip was such a drain on his wallet, so he was happy to go back to work and receive a paycheck!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is special about this man's trip?&lt;/p&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="11297" LastEditDate="2015-05-31T17:18:37.930" LastActivityDate="2015-07-18T23:42:10.520" Title="What is special about this trip?" Tags="|enigmatic-puzzle|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15557" PostTypeId="2" ParentId="15541" CreationDate="2015-05-27T14:04:25.377" Score="7" Body="&lt;p&gt;I think it's:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The weasley kids from the Harry Potter Books.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Our first has his hands on the money;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bill the eldest, works for the wizard bank Gringotts as a treasure hunter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Our second has his hands in fire;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Charlie the 2nd, works with dragons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Our third was an outcast for years.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Percy is a prick and was kind of alienated from the rest of the family until the very end of the series.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Our fourth and fifth were together:&#xA;One dead, one deaf, or nearly so.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The twins, Fred and George, one got his ear chopped off, the other killed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Our sixth was tall and most important,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ron, one of the main characters, also supposed to be tall.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Our seventh was different from all the rest.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ginny, the only girl.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Friendly and welcoming, who are we?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Weasley kids.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastActivityDate="2015-05-27T14:04:25.377" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15558" PostTypeId="2" ParentId="15556" CreationDate="2015-05-27T14:07:28.890" Score="16" Body="&lt;p&gt;This was no ordinary trip, but it happens more often than you'd think:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He's going around a Monopoly board&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Each location describes a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; property or group of properties on the board.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first paragraph, mostly in Europe, is the:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; dark purple locations Mediterranean and Baltic&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The next, with Japan and China:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oriental&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Up in the Northeast US you'll find:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Vermont and Connecticut&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Heading down south, he'll find Shenandoah in:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Virginia&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later he's on Beale St, which can be found in Memphis, in:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tennessee&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Before heading back to New York. The next paragraph with Churchill Downs is talking about:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the red squares: Kentucky, Indiana, and Illinois&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After that we come back East, to the Taj Mahal in:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Atlantic City. I assume the beach references are something to do with Ventnor and Marvin Gardens, but I don't really know Atlantic City very well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you see someone get arrested, you might assume they:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; GO TO JAIL&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then some jumping around. Out west and back east to see both coasts, then up to the White House:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pacific, North Carolina, and Pennsylvania (Ave)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Horseshoe in Baltimore (or the green hotel) could be talking about:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Park Place, but I really don't get the clue&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But when he's relaxing on the beach with shops and eateries, that's gotta be the famous:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Boardwalk in Atlantic City&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally, he gets back home for a paycheck. Unfortunately, he doesn't get paid all that much, certainly not enough to cover a trip like this if he had to stay in hotels the whole time:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/ndauN.gif&quot; alt=&quot;monopoly-go&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Maybe he should go into real estate, I've heard it pays better ;)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'll admit there are a couple of clues that escaped me. I can see whee they go by the missing spots, but I'm sure I just don't recognize the descriptions well enough. Either way, it's clear that this man will be taking more trips like this one very soon.&lt;/p&gt;&#xA;" OwnerUserId="1563" LastEditorUserId="1563" LastEditDate="2015-05-27T14:35:44.197" LastActivityDate="2015-05-27T14:35:44.197" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15559" PostTypeId="2" ParentId="11564" CreationDate="2015-05-27T14:30:04.523" Score="4" Body="&lt;p&gt;You could play off the idiom &quot;walking down the garden path&quot; and call these &lt;strong&gt;garden path riddles&lt;/strong&gt;. As with a &lt;a href=&quot;http://en.wikipedia.org/wiki/Garden_path_sentence&quot; rel=&quot;nofollow&quot;&gt;garden path sentence&lt;/a&gt;, one is led to believe that the riddle is intended to be parsed one way and it isn't until the end is reached or the answer revealed that the listener recognizes the obfuscation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You have a 12&quot; LP of Chopin Nocturnes, four on each side. The record plays at 33 1/3 rpm and the grooves are 25 micrometers (0.025 mm) wide. Side A plays for 20:32 minutes and Side B plays for 20:45 minutes. How many grooves are there?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9828" LastActivityDate="2015-05-27T14:30:04.523" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15560" PostTypeId="5" CreationDate="2015-05-27T14:30:20.193" Score="0" Body="&lt;p&gt;Puzzles about the ages of people or things, their birthdays or similar temporal twists.&lt;/p&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="11297" LastEditDate="2015-05-27T14:41:08.333" LastActivityDate="2015-05-27T14:41:08.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15561" PostTypeId="4" CreationDate="2015-05-27T14:30:20.193" Score="0" Body="Puzzles about the ages of people or things, their birthdays or similar temporal twists." OwnerUserId="11297" LastEditorUserId="11297" LastEditDate="2015-05-27T15:03:54.827" LastActivityDate="2015-05-27T15:03:54.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15562" PostTypeId="5" CreationDate="2015-05-27T14:43:21.790" Score="0" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Nim&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nim&lt;/a&gt; is a two-player zero-sum game with &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; piles of pebbles. Players may take any number of pebbles from a single pile. The aim is either to win by taking all the pebbles in the last remaining pile (normal play) or alternatively to force that move upon your opponent (misere play).&lt;/p&gt;&#xA;&lt;p&gt;Many variations of Nim exist, such as Wythoff's game. Most if not all of these have been completely solved, an optimal strategy discovered and proved.&lt;/p&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="52483" LastEditDate="2023-02-04T14:31:19.273" LastActivityDate="2023-02-04T14:31:19.273" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15563" PostTypeId="4" CreationDate="2015-05-27T14:43:21.790" Score="0" Body="Puzzles related to Nim, a strategy game in which players take turns to reduce the number of objects in some sets, the aim being either to remove the last object or not to do so." OwnerUserId="11297" LastEditorUserId="69582" LastEditDate="2021-03-10T18:11:14.560" LastActivityDate="2021-03-10T18:11:14.560" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15564" PostTypeId="1" AcceptedAnswerId="15566" CreationDate="2015-05-27T14:51:36.540" Score="14" ViewCount="2574" Body="&lt;p&gt;As always, I hope this isn't too easy, and that you enjoy the rhymes :)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You often leave me, alone in my cave.&#xA;  &lt;br&gt; When I peek out, I best not misbehave.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If I go venture I can become quite ironic.&#xA;  &lt;br&gt; But in certain ways, I have become iconic.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I try to stay away from the cat.&#xA;  &lt;br&gt;But I am very strong, you may not know that!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Some may think I have precious metal around,&#xA;  &lt;br&gt;I would show them!! If I wasn't so bound.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;You can go and try, to ask me a question.&#xA;  &lt;br&gt;But if I'm tied up, I can't seem to function.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What could I be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make sure to explain each line/couplet as they are relative.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If you enjoyed this riddle, some of my other rhyming riddles can be found &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%5Brhyme%5D%20user%3A11244%20is%3Aquestion&quot;&gt;here&lt;/a&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2015-06-19T13:26:45.813" Title="Alone In My Cave" Tags="|riddle|rhyme|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15565" PostTypeId="2" ParentId="15564" CreationDate="2015-05-27T15:05:21.987" Score="0" Body="&lt;p&gt;A partial answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You often leave me, alone in my cave. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wine can be kept in caves&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I peek out, I best not misbehave.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wine (and alcohol) can make us misbehave&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I go venture I can become quite ironic. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We drink wine to feel good but it can hurt us (which is ironic)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But in certain ways, I have become iconic.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wine is iconic for some people&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I try to stay away from the cat. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I am very strong, you may not know that!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People tend to under estimate the percentage of alcohol in wine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some may think I have precious metal around, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some wine get medals (gold, silver,...)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would show them!! If I wasn't so bound.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Not sure&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can go and try, to ask me a question. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people are looking for solutions in alcohol (wine)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But if I'm tied up, I can't seem to function.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the cork is still here, you can't drink it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5359" LastActivityDate="2015-05-27T15:05:21.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15566" PostTypeId="2" ParentId="15564" CreationDate="2015-05-27T15:25:23.587" Score="28" Body="&lt;p&gt;Is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The tongue&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You often leave me, alone in my cave. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In this case the mouth is the cave&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I peek out, I best not misbehave.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Referring to sticking your tongue out mischievously&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I go venture I can become quite ironic. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Tongue-in-cheek&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But in certain ways, I have become iconic.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rolling Stones / Einstein's iconic tongue picture &lt;br&gt; &lt;br&gt;&#xA; Or could be a reference to the commonly used sticky-out-tongue emoticon?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I try to stay away from the cat. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Cat's got your tongue&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I am very strong, you may not know that!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think I heard it was the strongest muscle in the body&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some may think I have precious metal around, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A reference to someone being &quot;Silver tongued&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would show them!! If I wasn't so bound.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Attached by the frenulum&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can go and try, to ask me a question.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because it can respond!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But if I'm tied up, I can't seem to function.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tongue-tied!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12206" LastEditorUserId="12206" LastEditDate="2015-05-27T20:58:01.473" LastActivityDate="2015-05-27T20:58:01.473" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15567" PostTypeId="2" ParentId="15564" CreationDate="2015-05-27T15:27:08.550" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Tongue&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You often leave me, alone in my cave. &#xA;When I peek out, I best not misbehave.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;In the mouth. sticking out tongue for mischief&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I go venture I can become quite ironic. &#xA;But in certain ways, I have become iconic.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;tongue-in-cheek and rolling stones tongue? not so sure about this part&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I try to stay away from the cat. &#xA;But I am very strong, you may not know that!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Cat got your tongue. and tongue is the strongest muscle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some may think I have precious metal around, &#xA;I would show them!! If I wasn't so bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can go and try, to ask me a question. &#xA;But if I'm tied up, I can't seem to function.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; tongue tied&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12745" LastActivityDate="2015-05-27T15:27:08.550" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15568" PostTypeId="1" AcceptedAnswerId="15606" CreationDate="2015-05-27T16:21:51.257" Score="7" ViewCount="3743" Body="&lt;p&gt;Inspired by &lt;a href=&quot;https://puzzling.stackexchange.com/q/15532/10615&quot;&gt;this puzzle&lt;/a&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have $121$ apparently identical gold coins. At most one of these is fake, and weighs a different amount than a genuine coin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In front of you is balance with two pans. In your pocket you have a single coin which you know is genuine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the minimum number of weighings it takes to find the fake (if there is one), and determine whether it is heavier or lighter than the others?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Use your first weighing to reduce this to a similar problem, but with fewer coins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2015-05-28T16:39:32.843" Title="121 coins and a balance" Tags="|logical-deduction|weighing|" AnswerCount="11" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15569" PostTypeId="2" ParentId="15564" CreationDate="2015-05-27T16:25:36.763" Score="4" Body="&lt;p&gt;An alternate and a bit adult (and hopefully humorous) answer. You have been warned:   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A penis  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You often leave me, alone in my cave.&lt;br&gt;&#xA;When I peek out, I best not misbehave.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Often trapped in those represive pants, but always wanting to poke out. Don't go places you aren't supposed to. That leads to things...  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I go venture I can become quite ironic.&lt;br&gt;&#xA;But in certain ways, I have become iconic.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Definately the icon of the male portion of the species, but depending on characteristics (who has the hammer?), possibly ironic.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I try to stay away from the cat.&lt;br&gt;&#xA;But I am very strong, you may not know that!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cats like to play with things that dangle. Be careful. And they can be very strong when, well, yeah....&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some may think I have precious metal around,&lt;br&gt;&#xA;I would show them!! If I wasn't so bound.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Certain piercings have become popular with gold, silver, platinum, etc. These could be shown off to the world (and sometimes are), but again, those repressive pants keep things bound.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can go and try, to ask me a question.&lt;br&gt;&#xA;But if I'm tied up, I can't seem to function.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some men talk to Little John, and ask him why he doesn't work. And if he is too constrained, then things cannot progress (whether in relieving oneself, or otherwise). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-05-27T16:25:36.763" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15570" PostTypeId="1" CreationDate="2015-05-27T16:33:23.977" Score="13" ViewCount="2019" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;My first child was a rebel. He is estranged from us.&lt;/p&gt;&#xA;&lt;p&gt;My second used to be much closer to me, but we are losing touch.&lt;/p&gt;&#xA;&lt;p&gt;My third is similar in temperament to my second, but smarter, although sometimes he lets my second persuade him into doing things for him.&lt;/p&gt;&#xA;&lt;p&gt;My fourth is quite a snappy dresser, and is very popular.&lt;/p&gt;&#xA;&lt;p&gt;My fifth is the strongest, and the most help around the home.&lt;/p&gt;&#xA;&lt;p&gt;My sixth and seventh are quite similar, but my seventh is friendlier. My sixth and fifth don't get on well.&lt;/p&gt;&#xA;&lt;p&gt;My eighth likes to keep an air of mystery about herself.&lt;/p&gt;&#xA;&lt;p&gt;I keep my baby closest to me. He can be very temperamental.&lt;/p&gt;&#xA;&lt;p&gt;What is my name?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;PART ONE:&lt;/strong&gt; Answer the question in the riddle above.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;PART TWO:&lt;/strong&gt; Explain how all the pieces of information (I count roughly 16 of them) fit in.&lt;/p&gt;&#xA;&lt;p&gt;(I have a suspicion that the  &amp;quot;Who am I?&amp;quot; might be guessed fairly quickly; the real challenge to this riddle is in explaining what all the different parts mean. There isn't any fluff and it does all mean something. Be sure to explain your answer fully!)&lt;/p&gt;&#xA;&lt;p&gt;For your answer to be complete, you must complete both part one and part two.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Hint for part two:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; This is a description of &lt;em&gt;our&lt;/em&gt; family. All the descriptions of my children are talking about my children themselves, not strangers who share their names.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5260" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-10T15:20:16.657" Title="Here is my family. Who am I?" Tags="|riddle|word|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15572" PostTypeId="2" ParentId="15568" CreationDate="2015-05-27T16:36:48.333" Score="1" Body="&lt;p&gt;If there is at most 1 fake coin:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 8 weighings&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Set aside 1 coin. Put 60 on each side of the scale. If they balance, weigh the set-aside vs your coin.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they don't balance, take 30 from the heavier side and exchange them with 30 from the lighter side (keeping them separate from the 30 on each side that stay). If the balance changes, you know the odd coin is among those 30. Otherwise, it's among the 30 that don't move. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take off the groups of 30 that are the same, switch 15 coins. Take off the groups of 15 that don't balance. Add 1 genuine coin to each side. Continue this pattern: switch 8, switch 4, switch 2, switch 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are left with 2 coins that aren't balanced, weigh your coin against 1 of them.  If they balance, the other one is fake. If they don't balance, you know the difference. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="463" LastEditDate="2015-05-27T16:54:50.420" LastActivityDate="2015-05-27T16:54:50.420" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15576" PostTypeId="2" ParentId="15545" CreationDate="2015-05-27T16:43:18.680" Score="10" Body="&lt;p&gt;Another answer that works is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the Four Horsemen of the Apocalypse.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;When I am white, I do good.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The rider of the white horse is sometimes said to represent Jesus.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;When I am black, I do evil.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The rider of the black horse usually represents famine, which is an evil outcome.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;When I am grey, I do both.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The rider of the grey horse represents death, which is often an evil thing but can sometimes be merciful and a relief.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;When I am red, I rule.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The rider of the red horse represents war: a conflict often fought to decide who should rule.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Four_Horsemen_of_the_Apocalypse&quot;&gt;Source&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-05-27T19:16:03.457" LastActivityDate="2015-05-27T19:16:03.457" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15577" PostTypeId="2" ParentId="15544" CreationDate="2015-05-27T16:51:18.987" Score="5" Body="&lt;p&gt;(Note to censors: this answer is correct and in the spirit of the question. Please do not edit it!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; fucking obvious.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Sometimes I end up as stars?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Words like &quot;shit&quot; often end up as just **** when censored out.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Like or dislike by peoples,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people like to say words like &quot;crap&quot;, others disdain them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Yes, of course I'm reason of hurt and quarrel,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Such words can cause a lot of shit between people!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Sites want to prevent me.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Stack Exchange censors are going to piss all over me for writing this answer!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am Bitter sometimes funny.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Such words can be used when you're really damn bitter, but they can also be entertaining.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Singers use me sometimes.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Sex Pistols?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I think I am fashion of new generation.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Youngsters these days, always fucking swearing...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Who am I?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; See &lt;a href=&quot;https://puzzling.stackexchange.com/a/15548/5373&quot;&gt;lorimer's answer&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-05-27T16:51:18.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15579" PostTypeId="2" ParentId="15570" CreationDate="2015-05-27T17:41:02.360" Score="26" Body="&lt;p&gt;I think it's :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Solar System's planets in reverse order.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;My first child was a rebel. He is estranged from us.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pluto, is no longer a planet. -So many things made Pluto differ from all 8 other planets - and so many new solar system bodies kept being discovered to which Pluto shared more similarities. It eventually had the definition of a Planet revised and the Dwarf Planet category introduced (to which Pluto now belongs) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;My second used to be much closer to me, but we are losing touch.&#xA;My third is similar in temperament to my second, but smarter, although sometimes he lets my second persuade him into doing things for him.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Neptune &amp;amp; Uranus - Almost twins, nearly the same size. Uranus is bigger, Neptune has greater mass. Both have rings (although Neptune's are fragmented). They are similar in composition (hydrogen &amp;amp; helium).&lt;br&gt;&#xA;&lt;br&gt;&#xA;(Per @Going hamateur) &lt;em&gt;Neptune was discovered as it perturbed Uranus' orbit so it was theorized to exist and pinned down relatively easily. Speculating that at an earlier point it was closer in orbit to the sun but got knocked away (sorta).&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;My fourth is quite a snappy dresser, and is very popular.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Saturn and its fancy rings has always been popular. It also has a TON of moons, almost as many as Jupiter (again you could say it's popular). One of those moons is Titan - which is again a very popular subject of interest in science-fiction (and space exploration).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;My fifth is the strongest, and the most help around the home.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Biggest Planet in the Solar System, Depending on it's position relative to the sun, Jupiter is the only planet that has a barycenter with the Sun that lies outside the volume of the Sun, Along with the Sun, the gravitational influence of Jupiter has helped shape the Solar System. The orbits of most of the system's planets lie closer to Jupiter's orbital plane than the Sun's equatorial plane (Mercury is the only planet that is closer to the Sun's equator in orbital tilt), the Kirkwood gaps in the asteroid belt are mostly caused by Jupiter, and the planet may have been responsible for the Late Heavy Bombardment of the inner Solar System's history&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;My sixth and seventh are quite similar, but my seventh is friendlier. My sixth and fifth don't get on well.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 6th is Mars, very similar to Earth(7th) - they are similar size, gravity, composition etc... - but I think everyone will agree Earth is much nicer (having an atmosphere that supports &lt;em&gt;human&lt;/em&gt; life). Mars and Jupiter(5th) are separated by the asteroid belt which also separates the Inner Solar System from the Outer Solar System.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;My eighth likes to keep an air of mystery about herself.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Venus, which atmospheres' makes it almost impossible(difficult) to learn anything of it's surface.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I keep my baby closest to me. He can be very temperamental.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mercury. Closest to the sun. Mercury's lack of any &lt;em&gt;substantial&lt;/em&gt; atmosphere means the surface exposed to the sun is very hot where as the surface in the shadows will see a crazy temperature drop. It's temperature can range from -173'C to 426'C.&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;em&gt;for those that are thinking of Mercurial, that's putting the cart before the horse. Guess where that adjective comes from. (1350-1400; Middle English &amp;lt; Latin mercuriālis of, pertaining to the god or planet Mercury. See mercury)&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;What is my name?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Sun/Sol&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastEditorUserId="11251" LastEditDate="2015-07-10T14:36:25.517" LastActivityDate="2015-07-10T14:36:25.517" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="15581" PostTypeId="2" ParentId="15568" CreationDate="2015-05-27T17:49:48.460" Score="2" Body="&lt;p&gt;The answer is &lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt; uses of the scale.&lt;/p&gt;&#xA;&lt;h2&gt;First Weighing&lt;/h2&gt;&#xA;&lt;p&gt;Take one coin away and make &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt; groups of &lt;span class=&quot;math-container&quot;&gt;$30$&lt;/span&gt;.  Pick any two of these groups and weight them.  If scale balances, then they are all good coins.  If the scale tips, then you know there is at least one bad coin in those &lt;span class=&quot;math-container&quot;&gt;$60$&lt;/span&gt; and the other &lt;span class=&quot;math-container&quot;&gt;$61$&lt;/span&gt; are good.  Either way, you have eliminated at least &lt;span class=&quot;math-container&quot;&gt;$60$&lt;/span&gt; coins in one weighing leaving you with either &lt;span class=&quot;math-container&quot;&gt;$60$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$61$&lt;/span&gt; coins.&lt;/p&gt;&#xA;&lt;h2&gt;Second Weighing&lt;/h2&gt;&#xA;&lt;p&gt;WLOG, we will assume you have &lt;span class=&quot;math-container&quot;&gt;$61$&lt;/span&gt; coins.  Take one away and split into &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt; groups of &lt;span class=&quot;math-container&quot;&gt;$15$&lt;/span&gt;.  Again, pick any two and weigh them.  You will be able to eliminate at least &lt;span class=&quot;math-container&quot;&gt;$30$&lt;/span&gt; more coins in this way.&lt;/p&gt;&#xA;&lt;h2&gt;Third Weighing&lt;/h2&gt;&#xA;&lt;p&gt;You now have &lt;span class=&quot;math-container&quot;&gt;$30$&lt;/span&gt; coins, plus one set aside, plus one in your pocket.  The one set aside may or may not have been ruled out as a possible bad coin.  Lets say that we still don't know, so that you have &lt;span class=&quot;math-container&quot;&gt;$31$&lt;/span&gt; coins. Add your coin from your pocket to make &lt;span class=&quot;math-container&quot;&gt;$32$&lt;/span&gt; and divide into &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt; groups of &lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt;.  Weigh two groups, and you will be able to eliminate &lt;span class=&quot;math-container&quot;&gt;$16$&lt;/span&gt;, leaving you with at most &lt;span class=&quot;math-container&quot;&gt;$16$&lt;/span&gt; coins.&lt;/p&gt;&#xA;&lt;h2&gt;Forth Weighing&lt;/h2&gt;&#xA;&lt;p&gt;Split into &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt; groups of &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt;.  Eliminate &lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt; by weighing two of the groups.&lt;/p&gt;&#xA;&lt;h2&gt;Fifth Weighing&lt;/h2&gt;&#xA;&lt;p&gt;Split into &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt; groups of &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt;.  Eliminate &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt; by weighing two of the groups.&lt;/p&gt;&#xA;&lt;h2&gt;Sixth Weighing&lt;/h2&gt;&#xA;&lt;p&gt;Weigh &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; of the &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt; remaining coins.  &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; will be eliminated.&lt;/p&gt;&#xA;&lt;h2&gt;Seventh and Eighth Weighings&lt;/h2&gt;&#xA;&lt;p&gt;Weigh the last &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; coins.  If the scale balances, you are done and all coins are equal.&#xA;If the scale tips, then use your last attempt to see which one is the bad one by comparing one of them to a known good coin.&#xA;You now have only &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt; coins left.  Take &lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt; of the coins and one from the pile of known good coins.  Weigh 2 on each side.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-27T18:35:44.487" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15582" PostTypeId="2" ParentId="15568" CreationDate="2015-05-27T17:52:59.300" Score="1" Body="&lt;p&gt;Take 4 groups as A=40, B=40, C=40, D=1. Compare A &amp;amp; B , keeping aside C &amp;amp; D.&lt;br&gt;&#xA;&lt;strong&gt;CASE 1-1&lt;/strong&gt; : If equal, then the fake is in C or D.&lt;br&gt;&#xA;Compare A &amp;amp; C. If equal, then fake is D (Compare with any one coin to know whether it is heavier or lighter) else A is-less-than C or A is-more-than C, so we know whether fake is heavier or lighter, and we only have 40 coins to analyse.&lt;br&gt;&#xA;&lt;strong&gt;CASE 1-2&lt;/strong&gt; : If not equal, then fake is in A or B. Without loss of generality, let A is-less-than B (Process is same if A is-more-than C). Compare A &amp;amp; C. If equal, then fake is in B &amp;amp; is heavier, else fake is in A &amp;amp; is lighter. We only have 40 coins to analyse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With 2 weightings, we know which 40 coins have the fake and whether it is heavier or lighter. Without loss of generality, assume it is lighter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take 4 groups as A=13, B=13, C=13, D=1. Compare A &amp;amp; B.&lt;br&gt;&#xA;&lt;strong&gt;CASE 2-1&lt;/strong&gt; : If equal, then fake is in C or D, so compare A &amp;amp; C, to know which has the fake.&lt;br&gt;&#xA;&lt;strong&gt;CASE 2-2&lt;/strong&gt; : If not equal, then fake is in A or B, so compare A &amp;amp; C, to know which has the fake.&lt;br&gt;&#xA;If both weightings as equal, then fake is D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With 2 more weightings, we have only 13 coins remaining.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take the groups as A=4, B=4, C=4, D=1. With 2 more weightings, we will have only 4 coins remaining, or we would have found D as the fake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take the groups as A=1, B=1, C=1, D=1. With 2 more weightings, we will know the fake coin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we require only &lt;strong&gt;2+2+2+2=8&lt;/strong&gt; weightings in maximum, unless we find the fake coin quicker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can we do better than 8 ?&lt;/strong&gt;&lt;br&gt;&#xA;We have to get enough information for finding the fake coin in 121 possibilities &amp;amp; 1 bit of information for finding whether that fake coin is heavier or lighter. Combined, we have 121*2=242 possibilities. Every weighing gives 1 ternary bit of information. So minimum information required is 5 ternary bits of information (3*3*3*3*3=243) or 5 ternary weightings. &lt;strong&gt;We can do better than 8, but we can not do better than 5.&lt;/strong&gt; Many answers already have this minimum.&lt;/p&gt;&#xA;" OwnerUserId="10940" LastEditorUserId="10940" LastEditDate="2015-05-28T03:03:59.637" LastActivityDate="2015-05-28T03:03:59.637" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15583" PostTypeId="2" ParentId="15568" CreationDate="2015-05-27T18:35:15.007" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;6 uses of the scale.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE: I had another answer with 8 uses, but left it in since it was its own answer.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take one coin away and split the rest into $3$ groups of $40$.  Call them $A$, $B$, $C$.  In your first two uses of the scale, you will weigh $A$ vs $B$ and $B$ vs $C$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 7 possible results:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$A&amp;lt;B$ and $B=C$ - There is a light coin in $A$&lt;/li&gt;&#xA;&lt;li&gt;$A&amp;gt;B$ and $B=C$ - There is a heavy coin in $A$&lt;/li&gt;&#xA;&lt;li&gt;$A&amp;lt;B$ and $B&amp;gt;C$ - There is a heavy coin in $B$&lt;/li&gt;&#xA;&lt;li&gt;$A&amp;gt;B$ and $B&amp;lt;C$ - There is a light coin in $B$&lt;/li&gt;&#xA;&lt;li&gt;$A=B$ and $B&amp;lt;C$ - There is a heavy coin in $C$&lt;/li&gt;&#xA;&lt;li&gt;$A=B$ and $B&amp;gt;C$ - There is a light coin in $C$&lt;/li&gt;&#xA;&lt;li&gt;$A=B$ and $B=C$ - The coin set aside is the only unknown coin&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;After this is complete, you will know not only which $40$ coins has the faulty coin, but you will also know if it is lighter or heavier. If they are all the same, you can use the scale once more with the coin set aside to figure out if it is bad or good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;WLOG, lets assume you have $40$ with a single known heavy coin.  Add $41$ known good coins to make $81$.  Now, repeat the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Split into three equal groups&lt;/li&gt;&#xA;&lt;li&gt;Weigh two of the groups&lt;/li&gt;&#xA;&lt;li&gt;You will know which of the three groups contains the coin&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Until you are left with a single coin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since it took you 2 uses of the scale to get to $81$, here is how it breaks down:&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;$81$ coins becomes $27$&lt;/li&gt;&#xA;&lt;li&gt;$27$ coins becomes $9$&lt;/li&gt;&#xA;&lt;li&gt;$9$ coins becomes $3$&lt;/li&gt;&#xA;&lt;li&gt;$3$ coins becomes $1$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, you can do this in $6$ uses of the scale!&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2015-05-27T18:35:15.007" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15584" PostTypeId="1" AcceptedAnswerId="15657" CreationDate="2015-05-27T18:38:06.977" Score="6" ViewCount="672" Body="&lt;p&gt;Thanks @starsplusplus helping me with the idea. (From his &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15570/here-is-my-family-who-am-i&quot;&gt;riddle&lt;/a&gt;). I thought of this (far fetched) answer/concept based off of his, and decided to see if anyone else could figure it out.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;I have a really large family which has been growing over the years..I don't have time to tell you about everyone but I can try to tell you a bit about some.&lt;/p&gt;&#xA;&lt;p&gt;My first sibling is a scholar. He has an interest for continuing&#xA;learning.&#xA;&lt;br&gt; My second sibling is the fun one, but can be lead to abuse.&#xA;&lt;br&gt; My third sibling is smart like the first, but likes to concentrate his&#xA;studies (on a smaller subject).&#xA;&lt;br&gt; My fourth sibling likes to learn, or rather teach to learn in a&#xA;different radical way.&#xA;&lt;br&gt;My fifth sibling loves to travel and loves knowing more about the&#xA;world.&#xA;&lt;br&gt;My sixth sibling will be his own boss, working for himself.&lt;/p&gt;&#xA;&lt;p&gt;As for me.. I'm below the 15th sibling, who is a parent.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;b&gt;Who/what is speaking, and who/what are his siblings/family?&lt;/b&gt;&lt;/p&gt;&#xA;&lt;p&gt;Make sure to explain each line for each sibling as well as the 'speaker'.&lt;/p&gt;&#xA;&lt;p&gt;I will give some hints later on if necessary as this riddle might not be easy to understand/solve without them.&lt;/p&gt;&#xA;&lt;p&gt;Hint #1:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The second sibling is not directly after the first sibling and so forth. They are listed from a distinct order. &amp;quot;I don't have time to tell you about everyone.&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-29T12:48:49.193" Title="My Family, My Siblings, and I" Tags="|riddle|word|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15585" PostTypeId="2" ParentId="15568" CreationDate="2015-05-27T18:43:37.753" Score="4" Body="&lt;p&gt;It is &lt;a href=&quot;http://en.wikipedia.org/wiki/Balance_puzzle&quot; rel=&quot;nofollow&quot;&gt;known&lt;/a&gt; that we can solve the corresponding problem with 13 coins in three weighings.  As the coins are broken into three groups each time and we have about nine times more coins, our target should be five.  We can follow the logic of the linked solution.  We split the coins into groups that are weighed $n$ times.  The group weighed $n$ times has ${5 \choose n}2^{n-1}$ coins in it because we can choose the times a coin is on the balance in $5 \choose n$ ways and we can choose the sides it goes on in $2^n$ ways but we need to distinguish heavy from light which divides by $2$.  This lets us handle $5+20+40+40+16=121$ coins.  The good coin goes on the right each time.  Number the single weighing coins $1-5$ and put each one on the left in turn.  Twenty coins are each weighed twice.  Number them $1$ to $20$. These will be weighed $$1,2,9,20\quad 10,11,12,19\\1,3,4,16\quad 2,15,17,18\\3,5,6,12\quad 4,9,13,14\\5,7,8,18\quad 6,17,19,20\\7,9,10,14\quad 8,13,15,16$$  Note that these always put four coins on each side.  There are $40$ coins weighed $3$ times each, number them $1$ to $40$  We will weigh them &#xA;$$1,2,3,13,15,16,17,18,20,24,30,39 \quad 4,14,19,21,22,23,29,31,32,37,38,40\\&#xA;1,2,4,5,6,7,17,19,20,23,28,34 \quad 3,8,18,21,22,24,25,26,2733,35,36\\&#xA;1,3,4,5,6,8,9,10,11,27,32,38 \quad 2,7,12,25,26,28,29,30,31,37,39,40\\&#xA;5,7,8,9,10,12,13,14,15,31,36 \quad 6,11,16,21,23,24,29,30,3233,34,35\\&#xA;9,11,12,13,14,16,17,18,19,26,35,40 \quad 10,15,20,25,27,2833,34,36,37,38,39$$&#xA; there will be $12$ coins on each side.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the coins that are on the pan four times we can weigh &#xA;$$ \begin {array} {c c}1,2,6,7,8,17,21,24,25,28,30,33,36,39&amp;amp; 3,4,5,18,19,20,22,23,26,27,29,31,32,34,36,37,38,40\\&#xA;1,3,7,9,10,14,15,16,25,29,32,33,36,38 &amp;amp; 2,4,5,6,8,11,12,13,26,27,28,30,31,34,35,37,39,40\\&#xA;1,4,6,9,11,15,17,18,22,23,24,33,37,40 &amp;amp; 2,3,5,7,8,10,12,13,14,16,19,20,21,34,35,36,38,39\\&#xA;1,5,8,9,12,1417,19,2325,26,30,31,32 &amp;amp; 2,3,4,6,7,10,11,13,15,16,18,20,21,22,24,27,28,29\\9,13,16,17,20,22,25,27,31,33,34,38,39,40 &amp;amp; 10,11,12,14,15,18,19,21,23,24,26,28,29,32,35,36,37 \end {array} $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will have $14$ coins on the left and $18$ coins on the right.  We can fix that imbalance with the ones that are on all five times.  There, we weigh $$3,4,5,6,8,10,11,13,15,16\quad 1,2,7,9,12,14\\2,4,5,6,8,9,11,12,14,15\quad 1,3,7,10,13,16\\2,3,5,6,7,9,11,13,14,16\quad 1,4,8,10,12,15\\2,3,4,6,7,9,10,12,13,15\quad 1,5,8,11,14,16\\2,3,4,5,7,8,10,12,14,16\quad 1,6,9,11,13,15$$&#xA;This has $10$ coins on the left and $6$ on the right.  This results in $41$ coins on each side of the balance each time.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the results, the number of imbalances tells us the number of times the fake coin was on the scale.  The pattern of imbalances tells us which coin is fake and whether it is heavy or light.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditDate="2015-05-27T21:04:25.890" LastActivityDate="2015-05-27T21:04:25.890" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15586" PostTypeId="1" AcceptedAnswerId="15589" CreationDate="2015-05-27T19:12:04.923" Score="10" ViewCount="12773" Body="&lt;p&gt;This is a classic riddle of mine. Please read the scenario below for the setup.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You are sitting around a campfire and a friend smiles and says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Sam:&lt;/strong&gt; I have a riddle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Intrigued, you ask him what it is. He points to the tent you are staying in that night, and asks:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Sam:&lt;/strong&gt; Is that tent Good Rice or Bad Rice?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Confused, you pick one at random and answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;You:&lt;/strong&gt; Good rice?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He smiles and says you are correct. He then asks:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Sam:&lt;/strong&gt; What about that rock over there? Good Rice or Bad Rice?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You answer, slightly more confident this time:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;You:&lt;/strong&gt; Good rice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Grinning sheepishly, he replies:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Sam:&lt;/strong&gt; Nope, it's Bad Rice. Think about it! Now look, is that tree over there, is it Good Rice or Bad Rice?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now thoroughly confused, you simply reply:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;You:&lt;/strong&gt; I don't know... Tell me!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He replies with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Sam:&lt;/strong&gt; Good Rice. That was easy! One more time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He points towards the campfire and asks:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Sam:&lt;/strong&gt; Campfire. Good Rice or Bad Rice?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are stumped, but guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;You:&lt;/strong&gt; Bad rice?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He smiles and says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Sam:&lt;/strong&gt; Excellent job!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Your objective is to discover what classifies things as &quot;Good Rice&quot; or &quot;Bad Rice.&quot; Good Luck!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Important for Editors: (slight spoiler)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Please do not edit formatting in the &quot;scenario,&quot; as some grammar is important to the riddle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12752" LastEditorUserId="5373" LastEditDate="2019-07-20T11:18:03.000" LastActivityDate="2019-07-20T11:18:03.000" Title="The Morality of Rice" Tags="|lateral-thinking|story|language|" AnswerCount="5" CommentCount="8" ClosedDate="2015-05-28T15:45:56.507" ContentLicense="CC BY-SA 3.0" />
  <row Id="15587" PostTypeId="2" ParentId="15586" CreationDate="2015-05-27T19:21:58.827" Score="7" Body="&lt;p&gt;Let's try again.&lt;br&gt;&#xA;Good rice is determined by:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When the question begins with &lt;em&gt;is&lt;/em&gt; or something similar (though &lt;em&gt;is&lt;/em&gt; is the only one used thus far).  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bad rice is when  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The question is simply phrased as &quot;Good rice or bad rice?&quot; with nothing coming before hand.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Old Answer&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Anything that is good rice is  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Asked with a question beginning with a vowel. For example: &quot;&lt;em&gt;Is&lt;/em&gt; that tent good rice or bad rice?&quot; and &quot;&lt;em&gt;Is&lt;/em&gt; that tree good rice or bad rice?&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While bad rice is  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Any question beginning with a consonant. &quot;&lt;em&gt;What&lt;/em&gt; about that rock over there?&quot; and &quot;&lt;em&gt;Campfire&lt;/em&gt;. Good rice....&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="9951" LastEditDate="2015-05-27T19:30:23.723" LastActivityDate="2015-05-27T19:30:23.723" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15588" PostTypeId="1" AcceptedAnswerId="15642" CreationDate="2015-05-27T19:24:11.917" Score="24" ViewCount="2340" Body="&lt;p&gt;I spoke to Rock, Paper, and Scissors recently and asked them how they deal with their respective nemeses.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rock said: If I spend enough time around scissors, they’ll have a hard time moving eventually.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paper said: I never met a rock too hard for me to dig into and break through.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scissors said: I eat paper pretty quickly; frankly I prefer to savor something less processed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who are Rock, Paper, and Scissors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The answer is not &quot;Rock, Paper, and Scissors&quot; in any order =P)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding a hint because I think this is throwing people off:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Did you notice that sometimes Rock, Paper, and Scissors are capitalized and sometimes they're not?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One more hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The answers to this riddle have a rock-paper-scissor type relationship, in the same direction as their Rock-Paper-Scissor aliases.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Last hint, because some of you are really close, but need to take your answer a step further:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Some of you have approximately the right elements, but your answer should be a well-known trio of entities, each of which embodies one of those elements. You might say each of these entities has a certain &quot;type&quot; ;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10861" LastEditorUserId="10861" LastEditDate="2015-05-29T18:58:30.410" LastActivityDate="2018-10-07T07:09:29.083" Title="Rock Paper Scissors - Alter Egos" Tags="|riddle|" AnswerCount="8" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15589" PostTypeId="2" ParentId="15586" CreationDate="2015-05-27T19:26:44.403" Score="20" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If &quot;is&quot; is in the question, it's good rice:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is that tent Good Rice or Bad Rice?&lt;/li&gt;&#xA;&lt;li&gt;Now look, is that tree over there, is it Good Rice or Bad Rice?&lt;/li&gt;&#xA;&lt;li&gt;Is this riddle Good Rice or Bad Rice? ;) &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Otherwise, it's bad rice:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What about that rock over there? Good Rice or Bad Rice?&lt;/li&gt;&#xA;&lt;li&gt;Campfire. Good Rice or Bad Rice?&lt;/li&gt;&#xA;&lt;li&gt;Also, what about this comment? Good Rice or Bad Rice? Bad Rice of course&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1509" LastActivityDate="2015-05-27T19:26:44.403" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15590" PostTypeId="2" ParentId="15586" CreationDate="2015-05-27T19:26:57.810" Score="3" Body="&lt;p&gt;It could be to do with the number of words in the sentence preceding &quot;Good Rice or Bad Rice?&quot; (since the OP asks people not to edit anything).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;strong&gt;Is that tent&lt;/strong&gt; Good Rice or Bad Rice?&quot; gives Good Rice.&lt;br&gt;&#xA;&quot;&lt;strong&gt;What about that rock over there?&lt;/strong&gt; Good Rice or Bad Rice?&quot; gives Bad Rice.&lt;br&gt;&#xA;&quot;&lt;strong&gt;Now look, is that tree over there, is it&lt;/strong&gt; Good Rice or Bad Rice?&quot; gives Good Rice.&lt;br&gt;&#xA;&quot;&quot;&lt;strong&gt;Campfire.&lt;/strong&gt; Good Rice or Bad Rice?&quot; gives Bad Rice.&lt;br&gt;&#xA;&quot;&lt;strong&gt;Is this riddle&lt;/strong&gt; Good Rice or Bad Rice?&quot; gives Good Rice.&lt;br&gt;&#xA;&quot;&lt;strong&gt;Also, what about this comment?&lt;/strong&gt; Good Rice or Bad Rice?&quot; gives Bad Rice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So 3 and 9 map to Good while 6, 1, and 5 map to Bad. So far, any function satisfying these simple criteria will work. Too broad?&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-27T19:26:57.810" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15591" PostTypeId="2" ParentId="15584" CreationDate="2015-05-27T19:32:17.123" Score="0" Body="&lt;p&gt;I believe the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The U.S. colonies/states&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My first sibling is a scholar. He has an interest for continuing learning. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Virginia was the first colony founded. This is where the profitable Orinoco tobacco was farmed and where the farmers continued learning about the industry. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My second sibling is the fun one, but can be lead to abuse. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Massachusetts was the second colony founded. During the Revolutionary War, it was the center of operations of the Sons of Liberty, and where many protests including the Boston Tea Party took place. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My third sibling is smart like the first, but likes to concentrate his studies (on a smaller subject). &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; New Hampshire was the third colony founded. It concentrated its economy around fishing. &#xA; Not a lot to go on here though. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fourth sibling likes to learn, or rather teach to learn in a different radical way. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The fourth colony was Maryland. It was founded by Lord Baltimore as a refuge for Catholics. At this time Catholicism was radical and Protestantism was the norm. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fifth sibling loves to travel and loves knowing more about the world. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Connecticut was founded by John Hooker. It was one of the leaders in shipbuilding. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My sixth sibling will be his own boss, working for himself. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rhode Island was founded by Roger Williams after he was kicked out of Massachusetts for his belief in separation of church and state. Rhode Island therefore can be viewed as a lone wolf of sorts. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As for me, I'm below the 15th sibling, who's a parent. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Kentucky is the 15th state and was part of the westward expansion movement. You are Tennessee, both the 16th state and directly south of Kentucky. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12725" LastActivityDate="2015-05-27T19:32:17.123" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15592" PostTypeId="2" ParentId="15588" CreationDate="2015-05-27T19:41:32.403" Score="1" Body="&lt;p&gt;Somewhat random guess here:&lt;br&gt;&#xA;Rock is  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water =&gt; If scissors are in water for long, they'll rust and have a hard time moving  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Paper is  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Jackhammer??? =&gt; Breaks through rocks if it spends enough time on them  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Scissors are  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Termites =&gt; They eat paper quickly, but more often eat 'less processed' wood.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not certain about it, but I'll try :)&lt;/p&gt;&#xA;" OwnerUserId="7715" LastActivityDate="2015-05-27T19:41:32.403" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15593" PostTypeId="2" ParentId="15588" CreationDate="2015-05-27T19:47:55.520" Score="14" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Rock is air (oxygen, specifically) because of rust.&lt;br&gt;&lt;br&gt;Paper is water (erosion)&lt;br&gt;&lt;br&gt;Scissors are fire, which burns paper. Unprocessed wood burns longer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12365" LastEditorUserId="12365" LastEditDate="2015-05-27T20:04:46.413" LastActivityDate="2015-05-27T20:04:46.413" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15594" PostTypeId="1" AcceptedAnswerId="15595" CreationDate="2015-05-27T19:59:23.147" Score="20" ViewCount="2524" Body="&lt;p&gt;The well-known verbal arithmetic problem&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    S E N D&#xA;+   M O R E&#xA;-----------&#xA;  M O N E Y&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;explains (or at least it did the first time I knew about it) the story of a student who send a telegram to his father, asking for more money, and asks to fill in each letter with a different digit in order to figure out how much the student was asking for.&#xA;What's not so broadly known is the answer his father gave:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  S P E N D&#xA;-   L E S S&#xA;-----------&#xA;  M O N E Y&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Even though, the father was not as good as the son with verbal arithmetics and by mistake made an impossible one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Could you prove that the rest he proposed is impossible?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Of course, feel free to solve the first verbal arithmetic (which can be solved indeed), but the real question is about the impossibility to solve the second one.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The first verbal arithmetic problem is only provided for flavour, and the same letter does not necessarily represent the same number between both equations.&lt;/p&gt;&#xA;" OwnerUserId="12165" LastEditorUserId="9404" LastEditDate="2016-03-03T10:22:58.933" LastActivityDate="2016-03-03T10:22:58.933" Title="SPEND LESS MONEY alphametic" Tags="|mathematics|calculation-puzzle|alphametic|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="15595" PostTypeId="2" ParentId="15594" CreationDate="2015-05-27T20:05:15.383" Score="20" Body="&lt;p&gt;If $E - E = N$ in the third column, then $N$ is either $0$ or $9$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $N$ was $0$; then you'd have to borrow 1 in the fourth column to subtract $S$ from $0$, but $N$ can only be $0$ in the third column if you don't borrow anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, suppose $N$ was $9$; then you'd have to have borrowed 1 in the fourth column. But $9$ minus anything never borrows, so that case is impossible as well.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-05-27T20:05:15.383" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15597" PostTypeId="2" ParentId="15594" CreationDate="2015-05-27T20:16:03.477" Score="2" Body="&lt;p&gt;Here is a more non-technical:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Give each word a value of (+1) or (-1) for relevance to their meaning&#xA;&lt;br&gt;&lt;b&gt;Spend = -1&#xA;&lt;br&gt;Send = +1&#xA;&lt;br&gt;More = +1&#xA;&lt;br&gt;Less = -1&lt;/b&gt;&#xA;&lt;br&gt;&lt;br&gt;Therefore [Send + more]: &lt;b&gt;(+1) + (+1) = Money&lt;/b&gt;&#xA;&lt;br&gt; and [Spend - Less]: &lt;b&gt;(-1) - (-1) = Money&lt;/b&gt;&#xA;&lt;br&gt; meaning that money must equal both 2, and 0. (Which is not possible)&#xA;&lt;br&gt;&lt;br&gt; The Dad should have said &lt;br&gt;|Spend + Less = Money|&#xA;&lt;br&gt;&lt;b&gt;|(-1) + (-1)| = |-2| = 2&lt;/b&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastActivityDate="2015-05-27T20:16:03.477" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15598" PostTypeId="1" AcceptedAnswerId="15603" CreationDate="2015-05-27T20:26:01.773" Score="11" ViewCount="729" Body="&lt;p&gt;&lt;em&gt;All credit for this problem goes to &lt;a href=&quot;https://en.wikipedia.org/wiki/Jacob_Bronowski&quot; rel=&quot;noreferrer&quot;&gt;Dr. J. Bronowski&lt;/a&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Find the least positive integer such that moving the leading digit to&#xA;  the end produces a new integer one and a half times the original.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12165" LastEditorUserId="12165" LastEditDate="2018-11-21T10:47:14.893" LastActivityDate="2018-11-21T16:22:23.927" Title="Finding the Dr. Bronowski's number" Tags="|mathematics|calculation-puzzle|" AnswerCount="3" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="15599" PostTypeId="1" AcceptedAnswerId="15620" CreationDate="2015-05-27T20:34:25.080" Score="10" ViewCount="581" Body="&lt;p&gt;&lt;em&gt;Two Pepperbox Pistols, a Lemat 1861 revolver, a Winchester Model 1873, a Colt Thuer Derringe, two Colt 1860 Army, a Smith&amp;amp;Weston Model 1 1/2, a Coach Gun, and a Double Barreled Shotgun.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;10 dead men walking out of the saloon,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The mayor was in a hurry and was shot once, at noon.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;9 dead men, flustered and jittery.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The cowboy was the culprit so they all shot him twice silly.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;8 dead men, trying to calm down.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The barkeep was scared dead by the priest's lonely single round.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;7 dead men slowly becoming weary,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The priest accused the soldier which 'back-fired' rather quickly.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;6 dead men now hoping no one saw them.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The soldier could take them all, 5 bullet wounds solved the problem.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;5 dead men still standing in the square.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The blacksmith wanted blood so the doctor gave him fair.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;4 dead men now seeing the final picture.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The barber was the slowest, 3 rounds put him out for pasture.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;3 dead men all grim and determined.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The doctor twitched too early so they both shot at him.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;2 dead men still and silent and ready.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The sheriff was the quickest but came up empty.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;1 dead man watching you stroll round the corner,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;his weapon in hand, ready to do murder.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The list above totals 10 guns. If each man had one of the 10 weapons listed above so that no weapon remains, and no one reloaded, you've got to ask yourself: Do I feel lucky?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Well, do ya, punk?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="11251" LastEditorUserId="68" LastEditDate="2015-05-29T20:19:25.680" LastActivityDate="2015-05-29T20:19:25.680" Title="Do I feel lucky? Well, do ya, punk?" Tags="|logical-deduction|knowledge|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="15600" PostTypeId="2" ParentId="15568" CreationDate="2015-05-27T20:51:04.633" Score="2" Body="&lt;p&gt;I need &lt;strong&gt;5 weighings&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where appropriate I'll assume that the scale will tip to one side and not the&#xA;other, but because of simmetry it will be an equivalent reasoning for the&#xA;other side. I will signal those places with (X)&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Weighing #1&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;41 coins on one plate, 40 coins on the other plus the known one. Leave 40 coins out&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it balances the fake is int the 40 left out, go to #E-2. If not go to #D-2.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Weighing #E-2&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;We know that the fake is in the remaining 40 coins or does not exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of the 40 candidates, put 14 on one plate, 13 on the other plus 1 of the known good ones. Leave 13 out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it balances go to #EE-3, if not go to #ED-3&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Weighing #EE-3&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The fake one if present is in the 13 untouched coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We put 5 on one plate, 4 on the other plus 1 known good. Leave 4 out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it balances go to #EEE-4. If not there are 5 of one weight and 4 of the&#xA;other. Go to #EED-4.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Weighing #EEE-4&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The fake, if present is in the untouched 4 coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We put 2 on one plate, 1 on the other plus 1 known good. Leave 1 out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it balances go to #EEEE-5, if not go to #EEED-5.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Weighing #EEEE-5&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The fake, if present is in the last untouched coin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We put that single coin on one plate, a known good on the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it balances &lt;strong&gt;no fake&lt;/strong&gt;, if not, &lt;strong&gt;that one is the fake&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Weighing #EEED-5&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Assume the scale will go to the left (X). Then the fake one is either one of&#xA;the 2 left, heavier, or the other one, lighter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We put 1 heavier coin on one plate, the other heavier coin on the other plate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it balances the fake is the &lt;strong&gt;lighter one&lt;/strong&gt;. If not, the fake is the &lt;em&gt;one&#xA;that goes down&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Weighing #EED-4&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;5 coins are of one weight, 4 of the other. Assume that the 5 are heavier&#xA;(X).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We put 2 heavier and 2 lighter on the left plate, 1 heavier, 1 ligther and 2&#xA;knwon ones on the other. Leave 2 heavier and 1 lighter out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it balances the fake is one of the 2 heavier and 1 lighter left out. If it&#xA;goes to the right, the fake has changed plate, so is one of the&#xA;2 lighter plus 1 heavier that moved. If it goes to the left, the fake did not&#xA;move, so it is one of the 2 heavier plust 1 lighter that did not move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Go to #EEDD-5.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Weighing #EEDD-5&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There are 2 candidates of one weight and 1 of the other. Assume that the 2 are&#xA;heavier (X).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put 1 of the heavier one plate and the other in the other plate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they balance, the fake one is the &lt;em&gt;other one&lt;/em&gt;. If not, the fake is the &lt;em&gt;one&#xA;that goes down&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Weighing #ED-3&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;14 coins are of one weight, 13 of the other. Assume that the 14 are heavier&#xA;(X).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put 5 heavier plus 5 lighter on the left plate, 4 heavier plus 4 lighter plus 2&#xA;known good on the other. Leave 5 heavier and 4 lighter out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they balance the fake is in the left out, and there are 5 heavier and 4&#xA;lighter. Go to #EDD-4. If it goes to the left, the fake is in one that did not&#xA;move, that is one of the 5 heavier plus 4 lighter. If it goes to the right,&#xA;the fake is in one that moved, that is one of the 5 lighter plust 4 heavier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Go to #EED-4&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Weighing #D-2&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;41 coins are of one weight, 41 of the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put 13 heavier plus 14 lighter on the left plate,&#xA;14 lighter plus 13 heavier on the right plate. Leave 14 heavier and 13 lighter out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it balances then the fake one is in the left out, 14 heavier and 13&#xA;lighter. If it goes to the left, the fake was unmoved, so it is one of the 13&#xA;heavier plus 14 lighter. If it goes to the right, the fake was moved, so it is one of the 14 lighter plus 13 heavier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Go to #ED-3&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;And that's it!!! I think that I covered all cases, if not, let me know and&#xA;I'll review.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Explanation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The trick to solve this kind of puzzles is to think on the avaliable&#xA;&lt;em&gt;information&lt;/em&gt;, that is the possible outcomes of the weighings versus the&#xA;possible solutions at each step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A single weighing can go 3 different ways, left, right or balanced, so 5&#xA;weighings can go $3^5 = 243$ ways. The possible solutions of the problem are&#xA;121 heavier coins plus 121 lighter coins plus 1 no-fake equals 243. That means that with&#xA;less than 5 weighing it is impossible to do, with 5 it might be possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, for each step, check that the possible solutions are never more than the&#xA;possible outcomes. The possible solutions are on each step are the powers of&#xA;3: 243, 81, 27, 9 and 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, Weight #ED-3, there are 14 of one weight and 13 of the other. Those&#xA;add up to 27 solutions, just right! We must separate 3 groups of 9 solutions each,&#xA;so we leave 9 coins out, we change plates on 9 coins, and leave 9 coins&#xA;unmoved. Then the next step will have 9 solutions in every case and all will go&#xA;smoothly to the next step.&lt;/p&gt;&#xA;" OwnerUserId="9468" LastEditorUserId="9468" LastEditDate="2015-05-27T21:05:24.330" LastActivityDate="2015-05-27T21:05:24.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15601" PostTypeId="2" ParentId="15586" CreationDate="2015-05-27T21:04:45.213" Score="5" Body="&lt;p&gt;I took it as something completely different. I realize it's not the intended answer, but it might be good to edit the question so this loophole is not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good Rice when&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The subject of the question (tent, rock, tree, campfire) has one or more occurrence of any letter. (tent, tree)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bad Rice when&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The subject of the question (tent, rock, tree, campfire) has only 1 of each letter. (rock, campfire)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12758" LastEditorUserId="12758" LastEditDate="2015-05-27T21:09:57.743" LastActivityDate="2015-05-27T21:09:57.743" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15602" PostTypeId="2" ParentId="15588" CreationDate="2015-05-27T21:35:56.257" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Earth wind and fire (often associated together because of the popular American band).&lt;br&gt;&#xA;&#xA; Fire burns paper&lt;br&gt;&#xA;&#xA; Wind can mold/shape/break rocks over time&lt;br&gt;&#xA;&#xA; Earth produces oxygen which can rust scissors  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2078" LastEditorUserId="9951" LastEditDate="2015-05-27T21:44:07.757" LastActivityDate="2015-05-27T21:44:07.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15603" PostTypeId="2" ParentId="15598" CreationDate="2015-05-27T21:44:38.457" Score="7" Body="&lt;p&gt;The smallest number I could find is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1176470588235294&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My process for finding it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Start with 1 as the number. To find more digits, multiply the current number by 1.5 and use the first case where the nth digit of the result is not the same as the n+1th digit of the number. Set the n+1th digit of your number to be that nth digit. Repeat until we find a number that works.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An example of my process:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; number is 1&lt;br&gt;&lt;code&gt;1.0 * 1.5 = 1.5&lt;/code&gt;&lt;br&gt;&lt;code&gt;- ^         ^&lt;/code&gt; &amp;lt;- 0 doesn't match 1&lt;br&gt;number is now 11&lt;br&gt;&lt;code&gt;11.0 * 1.5 = 16.5&lt;/code&gt;&lt;br&gt;&lt;code&gt;-  ^          ^&lt;/code&gt; &amp;lt;- 0 doesn't match 6&lt;br&gt;number is now 116&lt;br&gt;&lt;code&gt;116.0 * 1.5 = 174&lt;/code&gt;&lt;br&gt;&lt;code&gt;- ^            ^&lt;/code&gt; &amp;lt;- 6 doesn't match 7&lt;br&gt;number is now 117&lt;br&gt;&lt;code&gt;117.0 * 1.5 = 175.5&lt;/code&gt;&lt;br&gt;&lt;code&gt;-   ^           ^&lt;/code&gt; &amp;lt;- 0 doesn't match 5&lt;br&gt;number is now 1175&lt;br&gt;&lt;br&gt;...&lt;br&gt;&lt;br&gt;number is now 1176470588235293&lt;br&gt;&lt;code&gt;1176470588235293 * 1.5 = 1764705882352939.5&lt;/code&gt;&lt;br&gt;&lt;br&gt;&lt;code&gt;11764705882352939 * 1.5 = 17647058823529408.5&lt;/code&gt;&lt;br&gt;&lt;br&gt;&lt;code&gt;1176470588235294 * 1.5 = 1764705882352941&lt;/code&gt; &amp;lt;- FOUND!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a number that is smaller that almost works:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 588235294117647.0 * 1.5 = 882352941176470.5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It doesn't quite work because the number produced is not an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm confident my number is the smallest number that works. I tried my same process&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; starting with 2, 3, 4, and 5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and found that the resulting numbers were all larger than what I found. Something else I realized is that the are no numbers (of any size) that work that start with 6, 7, 8, or 9. This is because the number * 1.5 needs to have the same number of digits as the original number, and the product for numbers starting with 6-9 have an extra digit.&lt;/p&gt;&#xA;" OwnerUserId="1955" LastEditorUserId="1955" LastEditDate="2015-05-28T15:28:25.533" LastActivityDate="2015-05-28T15:28:25.533" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15605" PostTypeId="1" AcceptedAnswerId="15607" CreationDate="2015-05-27T21:54:12.180" Score="1" ViewCount="162" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Two of either is one, one of each is the other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To what do I refer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: There are two different solutions.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="12759" LastEditorUserId="12759" LastEditDate="2015-05-27T23:10:36.687" LastActivityDate="2015-05-27T23:38:29.117" Title="What am I referring to?" Tags="|riddle|" AnswerCount="3" CommentCount="0" ClosedDate="2015-05-28T13:48:51.200" ContentLicense="CC BY-SA 3.0" />
  <row Id="15606" PostTypeId="2" ParentId="15568" CreationDate="2015-05-27T21:58:33.477" Score="4" Body="&lt;p&gt;&lt;strong&gt;Lemma.&lt;/strong&gt; Given a set $A$ of coins that may be too heavy, a set $B$ of coins that may be too light. Assume $A\cap B=\emptyset$ and exactly one of the coins is a fake. If $|A|+|B|=3^k$, then the fake coin can be determined in $k$ weighings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; If $k=0$, there is nothing to be shown.&#xA;Assume the lemma is true for soem $k$; we shall show that it also holds for $k+1$: Partition $A$ into three nearly equal-sized sets $A_1,A_2,A_3$ (that is, $\lfloor\frac {|A|}3\rfloor\le |A_i|\le \lceil\frac {|A|}3\rceil\le 3^k$).  Since only two possible sizes are possible, at least two of the $A_i$ have the same size. Without loss of generality, $|A_1|=|A_2|$. Now we can partition $B$ into sets $B_1,B_2,B_3$ with $|B_i|=3^k-|A_i|$&#xA;Put the coins from $A_1\cup B_2$ into the left pan, $B_1\cup A_2$ into the right pan. Ignore the coins in $A_3\cup B_3$. Note that both pans contain the same number of coins.&#xA;Now if the left pan goes down, we know that the fake coin is $\in A_1\cup B_1$; if the right pan  goes down, the fake coin is $\in A_2\cup B_2$; if the pans weigh the same, the fake coin is $\in A_3\cup B_3$.At any rate, the induction hypothesis applies. $_\square$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition.&lt;/strong&gt; Given a known good coin and  $\frac{3^k-1}2$ coins among which there may be exactly one too heavy or too light fake coin or no fake coin at all. It is possible to determine which coin is fake and in which way, or that there is no fake coin, in $k$ weightings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt;&#xA;Again induction.&#xA;For $k=0$, there is nothing to be shown.&#xA;Assume the proposition holds for $k$; we shall show that it also holds for $k+1$:&#xA;Put the known good coin and $\frac{3^k-1}2$ coins into the left pan, $\frac{3^k+1}2$ into the right pan, and leave the remaing $\frac{3^k-1}{2}$ coins aside.&#xA;If the left pan goes down, we have $\frac{3^k-1}2$ coins that may be too heavy, $\frac{3^k+1}2$ that may be too light, and the rest is known good.&#xA;By the lemma, we can finish with $k$ more weighings.&#xA;The case with the right pan going down is similar:  we have $\frac{3^k+1}2$ coins that may be too heavy, $\frac{3^k-1}2$ that may be too light, and the rest is known good.&#xA;Finally, if both pans weigh the same, we are left with the $\frac{3^k-1}2$ unused coins among which there may be a fake coin or not. By induction hypothesis, this can be solved with $k$ weighings. $_\square$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Remark.&lt;/strong&gt; Since it is clar that it takes at least  $k$ weighings to distinguish among $3^k$ possible outcomes, the result of the proposition is best possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Corollary.&lt;/strong&gt; The puzzle with $121$ coins can be solved in $5$ weighings.&lt;/p&gt;&#xA;" OwnerUserId="1731" LastActivityDate="2015-05-27T21:58:33.477" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15607" PostTypeId="2" ParentId="15605" CreationDate="2015-05-27T22:07:25.103" Score="5" Body="&lt;p&gt;Okay, two possible solutions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solution 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;0 and 0. Two of either (0+0) is one of the 0s, one of each (0+0) is the other 0! Kind of cheating, I know.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Better Solution 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1 and -1. Two of either (1*1 or -1*-1) is one (1), one of each (-1*1) is the other (-1).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Possible Solution 3 (thought of independently of Andreas's answer =P):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;True and False, split by &quot;==&quot;. (True == True) is True, (False == False) is True, (True == False) is False.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10861" LastEditorUserId="10861" LastEditDate="2015-05-27T23:38:29.117" LastActivityDate="2015-05-27T23:38:29.117" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15608" PostTypeId="1" AcceptedAnswerId="15635" CreationDate="2015-05-27T22:09:20.097" Score="3" ViewCount="2368" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am the initiator of this riddle&lt;br&gt;&#xA;  I am the beginning of everything&lt;br&gt;&#xA;  I am the origin of god&lt;br&gt;&#xA;  But not long ago god was the origin of me&lt;br&gt;&#xA;  In some way I am god&lt;br&gt;&#xA;  And even god is stranded without me&lt;br&gt;&#xA;  I live even if you kill my body&lt;br&gt;&#xA;  you can kill me! But never be sure I don't come back&lt;br&gt;&#xA;  I can be the greatest thing in your life&lt;br&gt;&#xA;  Or destroy everything you love&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And one last thing:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You will need me to find the answer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12762" LastEditorUserId="7308" LastEditDate="2015-05-27T22:40:53.400" LastActivityDate="2016-08-04T23:57:18.297" Title="I am God. Who am I?" Tags="|riddle|" AnswerCount="5" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15609" PostTypeId="2" ParentId="15424" CreationDate="2015-05-27T22:26:10.893" Score="6" Body="&lt;p&gt;This answer follows from the previously found red and blue frequency audio files and the green frequency PGP message.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The red audio file starts out with a morse code sequence, followed by three voiced words (surrounded by high pitch noises) and what I assume to be DTMF tones. After spending more time than I care to admit doing it manually (because of my confidence in my morse coding skills), I did it like a sane person and analyzed the audio waveform. Here is what they decode to after triple checking both: &lt;br&gt;&lt;br&gt;...- ...- ...- -.-. .-.. --. .--. ..- --.. -.-. .-.. --. .--. ..- --.. -.. . .-. -... -.-. -...- --. .- ...- -.-- ..-. -... ... ..-. .-. -...- .. ..-. ..- -.-. -. - -- .- -.- . -.- . -.-- ... .. --. -. - .. ---.. ...-- .--. ..- -... -.- . -.-- . ... .--. --- ... - ... .. --. --..-- .. .-- .-.. .--. --- ... - -.- . -.-- -...- ..- ... . .--. -.- -.-. ... .---- -. .--. .- -.. -.. .. -. --. -...- --... ...-- . ... --. .-.. .--. ..- --.. -.. . .-. -... -.-. -...-.-&lt;br&gt;&lt;br&gt;VVV CLG PUZ CLG PUZ DE RBC = GA VY FB SFR = IF U CNT MAKE KEY SIGN TI83 PUB KEY ES POST SIG, I WL POST KEY = USE PKCS1 N PADDING = 73 ES GL PUZ DE RBC&lt;br&gt;&lt;br&gt;Translation:&lt;br&gt;testing(compare morse code for V to Beethoven's 5th) calling (PUZ) calling (PUZ) of (RBC).  Good afternoon very fine business so far. If you can't make key sign TI83 public key and post signature, I will post key. Use PKCS1, no padding.  Best regards and good luck (PUZ) of (RBC).&lt;br&gt;&lt;br&gt; No idea what PUZ and RBC are but they seem to be referring to the solver.&lt;br&gt;&lt;br&gt;(beep) echo papa quebec (beep)&lt;br&gt;&lt;br&gt;65537#10279922920444758952198236835097069960343622366001203286574422344156143529924117113239897731294792673018741448270384194591460111562322538604975737399705241*11337130627295915340299728600099026395063941250029823456400750546429518737269225761911696211987509755865509636756944318366919520522935654742514845464086299&lt;br&gt;&lt;br&gt;This is probably to build an RSA key for the PGP message, where the public exponent is 65537 and the two primes are &lt;br&gt;c447376fcf2a4d4f03840c83f68b23202f081f8561a1f0295703df258a96b8fd6cc8cb307558d60cbd692a45ed2414370349e28badf0f180419fc1df2cd87e99 &lt;br&gt;and &lt;br&gt;d876bd7c4963b8c06f148da504d1f7c7b9b20a719a0d3788eacc7effa7acb9cc200ef3a18a29fb5c733d45e04104ef3e7fc77f3ec847526b0c5d50506a2f471b &lt;br&gt;(converted the two numbers to hex).&lt;br&gt;The value of two primes multiplied is&lt;br&gt;a5f720e1c30658d4edb092a81197401b7a7f2c6224ef6411b8df246ccc4926eeed4abb756dcd294b68b59e7824e6bf46917c1354ad050f0de0beeb61b7537d834bf21315e1e4b15c3e84715e1ecca06fddf88a7d139aac22c479685b839b46f8464ce766dbe8fe4e65a4c6b6e72bfac8f6b2dd90ab44ed9f0ff597f45e08c923.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the blue frequency audio:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sound of the blue frequency audio contains an SSTV image (figured out by LeppyR64). This image can be decoded by software (&lt;a href=&quot;http://www.techedupteacher.com/how-to-make-an-sstv-puzzle/&quot; rel=&quot;nofollow noreferrer&quot;&gt;see here&lt;/a&gt;), and when decoded, the following image is created.&lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/c8w3L.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&lt;br&gt;The cards seem to be: &lt;br&gt;North: Ace of diamonds and either a spades or clubs.&lt;br&gt;West: Two and Five of Hearts.&lt;br&gt;South: 10 and 4 of clubs.&lt;br&gt;East: 4 of hearts and possibly 2 of diamonds.&lt;br&gt;Table cards: 9 of spades, 7 of clubs, 2 of clubs, ace of hearts and ace of clubs.&lt;br&gt;&lt;br&gt;North has 3 Aces 9 high, West has two pair of Twos and Aces 9 high, East has two pair of Twos and Aces 9 high, and South has a flush Ace high. If North exchanges an Ace with the 7 of clubs, that ruins South's flush and gives East and West a full house. (Answer subject to change upon decoding the image clearer)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the green frequency:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jarnbjo decrypted the PGP message into three cat pictures. The third (porsche.png) includes two different .png images, and by using notepad++ to delete the top half, this is the resulting image (East cards):&lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/EwuwL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&lt;br&gt;For Allie.jpg, an image can be extracted by deleting the first half of the JPEG header (~half way through the first line, everything before the second ÿØÿà).  The image is below and shows the West cards.&lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/71MUu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&lt;br&gt;For bitsy.tiff, there is an embedded thumbnail, so the image showing in the file explorer is different from the actual image. The TIFF image is a container for two JPEG (credit: Jarnbjo), by deleting the above image and the TIFF container, you can extract a quality thumbnail image(the North cards, below).&lt;img src=&quot;https://i.stack.imgur.com/QwTVe.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit 1: Fixed morse code spacing and added missing numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 2: Added blue frequency interpretation and more to red frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 3: Fixed &quot;epq&quot; hint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 4: Added card table (may be possible to decode clearer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 5: Added porsche.png hidden picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 6: Added allie.jpg hidden picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 7: Added bitsy.tiff hidden picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 8: Added better method for getting allie.jpg hidden picture.&lt;/p&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="7722" LastEditDate="2015-05-29T19:10:41.560" LastActivityDate="2015-05-29T19:10:41.560" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15610" PostTypeId="1" AcceptedAnswerId="15666" CreationDate="2015-05-27T22:58:51.370" Score="5" ViewCount="391" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I have peaks and valleys and slopes&lt;br&gt;&#xA;  Whose mountains and beats are such tropes,&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;And the prime, you will find,&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;Is a step back in time;&lt;br&gt;&#xA;  The end, you will find, is a hoax.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hints:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's not &quot;music.&quot;&lt;br&gt;&#xA; It's not &quot;rhymes&quot; or anything to do with poetry.&lt;br&gt;&#xA; It's not &quot;waves,&quot; but is the closest guess so far (a square is a rectangle).&lt;br&gt;&#xA; It's not a &quot;desert,&quot; especially since the mountains and beats portion of the limerick isn't fully satisfied.&lt;br&gt;&#xA; It's not &quot;the ocean,&quot; and is, in fact, the wrong direction to take the &quot;waves&quot; hint.&lt;br&gt;&#xA; Some words are more literal than they first appear.&lt;br&gt;&#xA; &quot;Tropes&quot; has a double meaning here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12759" LastEditorUserId="12759" LastEditDate="2015-05-28T18:39:19.117" LastActivityDate="2015-05-29T20:47:25.643" Title="Peaks, Valleys, and Slopes" Tags="|riddle|rhyme|limerick|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15611" PostTypeId="2" ParentId="10911" CreationDate="2015-05-27T22:59:58.773" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A mohawk&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My name sounds like a bird&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hawk.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whole body is like tongue of a lizard &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You need something sticky to hold it up.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Am different colors &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I've seen them in many colors&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I well can be very littered &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Littered with... I don't know.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My energy melts away over time&#xA;But not quite like an ice&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I don't imagine it'll stay up forever, but it won't liquify.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And when not taken care of &#xA;I'll collect dust and lice&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hair attracts lice. Stickiness attracts dust.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5125" LastActivityDate="2015-05-27T22:59:58.773" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15612" PostTypeId="2" ParentId="15605" CreationDate="2015-05-27T23:02:01.327" Score="1" Body="&lt;p&gt;Another answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\frac{1}{2}$ and $\frac{1}{2}$. Two of either of these is $1$, and one of each (i.e. each one of them) is the other one of them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I suspect the two solutions you intended were this and @VictorHenry's second solution.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-27T23:02:01.327" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15613" PostTypeId="2" ParentId="15605" CreationDate="2015-05-27T23:20:29.393" Score="0" Body="&lt;p&gt;If we're free to pick operators, then:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Exclusive OR: &#xA; true XOR true = false, false XOR false = false;&#xA; false XOR true = true&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5125" LastActivityDate="2015-05-27T23:20:29.393" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15614" PostTypeId="2" ParentId="15568" CreationDate="2015-05-27T23:25:15.940" Score="1" Body="&lt;p&gt;Another approach, much simpler to describe than my other, which also takes five weighings.  Weigh 41 coins against 40+the known good.  Assume they do not balance, and say the 41 go down.  Mark those 41 H for possibly heavy and the 40 from the other side L for possibly light.  Weigh 14 H+13 L against 14 H + 13 L, leaving 13 H + 14 L off the balance.  If one side is heavy, which we will assume, you are left with the 14 H from that side and the 13 L from the other.  If it balances, you are left with 14 L + 13 H and it should be clear what to do.  Erase the markings on the coins now known to be good.  Now weigh 5 H + 4 L each side and you will come down to 5 H + 4 L (or 5 L + 4 H).  Weigh 2 H + 1 L each side and finally 1 H each side.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the first weighing balances, you have 40 coins left that might be fake.  Weigh 14 vs 13+known good. If it doesn't balance, you are at the 14 H + 13 L stage above.  If it does, you are at 13 unknown + 1 good of the Wikipedia article I referenced previously, or you can weigh 5 vs 4+1 good.  If it balances, you are down to 4 and can weigh 2 vs 1+1 good.  If it balances again you are down to the bad coin and just weigh it against a known good to find heavy or light.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2015-05-27T23:25:15.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15615" PostTypeId="2" ParentId="15608" CreationDate="2015-05-27T23:28:40.947" Score="9" Body="&lt;p&gt;is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The mind&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the initiator of this riddle&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The mind is the one who came up with the riddle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the beginning of everything&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; when he is created, your existence begins&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the origin of god&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The mind came up with the concept of god&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But not long ago god was the origin of me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; god made the mind&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In some way I am god&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it controls how you perceive reality&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And even god is stranded without me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; without it, our concept of him would not exist&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I live even if you kill my body&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ideas live on even if the person doesn't&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;you can kill me! But never be sure I don't come back&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; once again its about ideas, because you can kill the wielder&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be the greatest thing in your life&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The mind is a wonderful thing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or destroy everything you love&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The mind can play horrible games on you&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You will need me to find the answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; self explanatory, you need your mind answer this riddle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12764" LastActivityDate="2015-05-27T23:28:40.947" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15616" PostTypeId="2" ParentId="15610" CreationDate="2015-05-27T23:37:02.810" Score="5" Body="&lt;p&gt;I think it might be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a RHYME.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I have peaks and valleys and slopes&lt;br&gt;&#xA; Whose mountains and beats are such tropes,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; These two lines rhyme; the first is a reference to the rhythm of a poem or verse, and the second could be the same.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And the prime, you will find,&lt;br&gt;&#xA;   Is a step back in time;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Prime&quot; with the first letter removed (a step back?) is &quot;rime&quot;, a homophone of &quot;rhyme&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The end, you will find, is a hoax.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The last line of the limerick doesn't actually rhyme with the first two - it's a hoax limerick!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-27T23:37:02.810" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15617" PostTypeId="2" ParentId="15598" CreationDate="2015-05-27T23:41:34.277" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If we let the leading digit of the number be &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; and the rest of the number be &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; has &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; digits, our number is &lt;span class=&quot;math-container&quot;&gt;$a\cdot 10^n+b$&lt;/span&gt;  We are requiring that &lt;span class=&quot;math-container&quot;&gt;$1.5(a\cdot 10^n+b)=10b+a$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$(3\cdot 10^n-2)a=17b$&lt;/span&gt;&lt;br&gt;&#xA; A quick Excel search finds that &lt;span class=&quot;math-container&quot;&gt;$3\cdot 10^{15}-2$&lt;/span&gt; is a multiple of &lt;span class=&quot;math-container&quot;&gt;$17$&lt;/span&gt;, leading to &lt;span class=&quot;math-container&quot;&gt;$a=1, b=176470588235294$&lt;/span&gt; and the number is &lt;span class=&quot;math-container&quot;&gt;$1176470588235294$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="73" LastEditorUserId="44915" LastEditDate="2018-11-21T13:14:18.293" LastActivityDate="2018-11-21T13:14:18.293" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15618" PostTypeId="2" ParentId="15568" CreationDate="2015-05-27T23:58:19.327" Score="1" Body="&lt;p&gt;This is exactly the twelve-coins-and-a-scale problem with the following variations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There are now 120 coins instead of 12.&lt;/li&gt;&#xA;&lt;li&gt;There is a 121st coin plus a 122nd coin that you know is genuine.&lt;/li&gt;&#xA;&lt;li&gt;It's possible that none of the coins weigh a different amount at all.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As I explained in &lt;a href=&quot;https://puzzling.stackexchange.com/a/224/88&quot;&gt;my solution to the twelve-coins problem&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/a/10829/88&quot;&gt;its generalization&lt;/a&gt;, it's possible to devise a fixed weighing schedule. The schema in those previous two linked solutions generalizes to any number of weighings, and the limit for five weighings is 120 coins, without using the &lt;code&gt;LLLLL&lt;/code&gt;, &lt;code&gt;RRRRR&lt;/code&gt;, or &lt;code&gt;=====&lt;/code&gt; cases. (See the first link for a description of the &lt;code&gt;L&lt;/code&gt;, &lt;code&gt;R&lt;/code&gt;, and &lt;code&gt;=&lt;/code&gt; notation.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Devise a weighing schedule for 120 of your 121 coins, by assigning different combinations of weighings to each coin so that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There are always 40 coins on each pan in a single weighing.&lt;/li&gt;&#xA;&lt;li&gt;Each coin is placed on a different combination of pans, and for each coin, if you were to reverse the pans that it were placed into, it wouldn't coincide with the schedule another coin. For example, if you decide to give one coin a schedule of &lt;code&gt;L==RL&lt;/code&gt; [i.e. left on first weighing, right on fourth, left on fifth, leaving it off on the others], you couldn't give &lt;code&gt;R==LR&lt;/code&gt; to another coin.&lt;/li&gt;&#xA;&lt;li&gt;You don't place any coin on the same pan for all five weighings.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The reason for the second requirement is so that if a coin is light or heavy, it will always produce a different result from any other coin being light or heavy. Whatever schedule you assign to a coin, it will tip to that schedule if the coin is heavy, and tip the opposite way if the coin is light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This allows you to use 240 of the 243 possible outcomes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, given that you have your 122nd coin, you can place the 121st coin on the left side, and your known 122nd coin on the right side for every weighing, so there are now 41 coins in each pan. This uses up the three remaining outcomes as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the scale tips to the right or left for all five weighings, this means that the 121st coin is either light or heavy respectively.&lt;/li&gt;&#xA;&lt;li&gt;If the scale balances for all five weighings, this means that all the coins are in fact genuine.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-05-28T00:04:32.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15619" PostTypeId="2" ParentId="15608" CreationDate="2015-05-28T00:35:47.243" Score="2" Body="&lt;p&gt;Could it be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Imagination&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the initiator of this riddle&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You used your imagination to make this riddle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the beginning of everything&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Everything that is created starts with someone's imagination (inventions, etc.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the origin of god&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The idea of god came from people's imagination&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But not long ago god was the origin of me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;God created people with imaginations&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In some way I am god&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Same idea as &quot;origin of god&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And even god is stranded without me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Same idea as &quot;origin of god&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I live even if you kill my body&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Really not sure with this&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;you can kill me! But never be sure I don't come back&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You can stop someone's imagination temporarily but they could still be imaginative later&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be the greatest thing in your life&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Being very imaginative can be great for your life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or destroy everything you love&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Imagination can cause denial/hallucinations which ruin your life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You will need me to find the answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You need your imagination to find the answer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11727" LastActivityDate="2015-05-28T00:35:47.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15620" PostTypeId="2" ParentId="15599" CreationDate="2015-05-28T00:44:28.327" Score="7" Body="&lt;p&gt;I think I…&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Feel lucky!  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here’s why:&#xA;There are 10 guns, with the following numbers of rounds (based on what was stated in the comments): 4, 4, 9, 15, 1, 6, 6, 5, 2, 2.&lt;br&gt;&#xA;There are 5 lines where everybody still alive shot a  bullet, including 1 line where everybody still alive shot twice.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cowboy was the culprit so they all shot him twice silly.&lt;br&gt;&#xA;(6 people left) The soldier could take them all, 5 bullet wounds solved the problem.&lt;br&gt;&#xA;(4 people left) The barber was the slowest, 3 rounds put him out for pasture.&lt;br&gt;&#xA;(3 people left) The doctor twitched too early so they both shot at him.&lt;br&gt;&#xA;(2 people left) The sheriff was the quickest but came up empty.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;That means the survivor must have fired at least 6 times. This leaves 4 possible guns the survivor might have (9, 15, 6, 6). However, the soldier “could take them all” in the 5th stanza, which means that after firing at least twice, he still had enough bullets to kill 5 people. That means he had at least 7 bullets to start, so he had either the 9 or the 15. That means two of the remaining possible guns had 6 bullets, and one had (9 or 15). If the survivor started with 6 bullets, then he’s OUT OF BULLETS when you walk around the corner. 2/3 is pretty good odds, but I think we can also assume that since everybody shot the soldier because he had enough bullets to kill everybody, they would have done the same if another person had survived with that many bullets. Basically, if the other (9 or 15) gunman had survived past the time the soldier was shot, the others would have shot him as well. Thus, I’m going to assume that the other (9 or 15) gunman was killed earlier, meaning the survivor started with 6 bullets and now has none left. I FEEL LUCKY!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10861" LastEditorUserId="11251" LastEditDate="2015-05-28T01:45:55.877" LastActivityDate="2015-05-28T01:45:55.877" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15622" PostTypeId="2" ParentId="15598" CreationDate="2015-05-28T02:48:49.467" Score="8" Body="&lt;p&gt;Here is a fun approach to the same answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let's call $X$ the number you start with, and $Y$ the number you get after moving the leftmost digit of $X$ to the right -- so the problem requires that $Y = 1.5X$.  Now imagine the infinite repeating decimals $x=0.XXXXXX\ldots$ and $y=0.YYYYYY\ldots$ -- for example if $X = 123$ then $x=0.123123123\ldots$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now the point is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We have two relations between $x$ and $y$.  On the one hand the problem requires that $y=1.5x$.  On the other hand the relationship between the digits implies that $10x = a+y$, where $a$ is the leading digit of $X$, an integer between $1$ and $9$.  Hence we find that $x = 2a/17$!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now we're basically done, let's just wrap up:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you compute the decimal expansions of $1/17, 2/17, \ldots$, you will find that they all are repeating decimals, with repeating sequence of length $16$, so $X$ must be at least a $16$-digit number.  ($X$ could also be a $32$-digit number, a $48$-digit number, etc., containing several repeated units.)  But anyway the smallest $X$ will have $16$ digits and it's pretty clear that the smallest $X$ comes from the decimal expansion of $2/17$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The decimal expansion of $2/17$ is the repeating decimal $0.1176470588235294\ 1176470588235294\ 1176470588235294 \ldots$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12523" LastActivityDate="2015-05-28T02:48:49.467" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15623" PostTypeId="2" ParentId="15568" CreationDate="2015-05-28T03:28:57.993" Score="0" Body="&lt;p&gt;I can do this in &lt;strong&gt;TWO Weightings&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A weighing is the measure of a complete group. It is a mark in the notebook, as a trial. So when you put 60 coins on one side of the scale, you did not do 60 trials, but one. Though, if you count every movement of the coin as a measure, this approach also wins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here is my approach:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Put half of the coins in one side of the scale, and half in the other, keeping the &quot;known good&quot; separate (not on scale). We will call the left side &lt;strong&gt;A&lt;/strong&gt; and the right side &lt;strong&gt;B&lt;/strong&gt;. If scale is even, there is no fake coin. &lt;/li&gt;&#xA;&lt;li&gt;Remove one coin from both sides simultaneously and place both of those coins into a temporary pool, &lt;strong&gt;C&lt;/strong&gt;. If there were any coins in &lt;strong&gt;C&lt;/strong&gt; from a previous draw, discard them such that there are only ever two.&lt;/li&gt;&#xA;&lt;li&gt;As soon as the scale evens out, you know one of the two coins just removed in &lt;strong&gt;TEMPORARY PILE C&lt;/strong&gt; was the fake one. Let's call the two coins in &lt;strong&gt;C&lt;/strong&gt; now &lt;strong&gt;C1&lt;/strong&gt; and &lt;strong&gt;C2&lt;/strong&gt;. Complete the removal of coins from both trays, which completes the &lt;strong&gt;FIRST MEASURE&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;Put &lt;strong&gt;C1&lt;/strong&gt; on the left (from group &quot;A&quot;), and put &lt;strong&gt;Known Good&lt;/strong&gt; on the right (from group &quot;B&quot;). If they balance, &lt;strong&gt;C2&lt;/strong&gt; is the fake, otherwise &lt;strong&gt;C1&lt;/strong&gt; is the fake. &lt;strong&gt;SECOND MEASURE&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Tadaa! If anyone wants to argue that my approach is invalid because there are multiple measurements, I disagree. You only record and reload the scale twice.&#xA;Think about it: My method also means you touch the least number of coins total out of any approach. If N is number of coins, my approach is around O(2N+2.5). So even if you count every change of coin in and out of scale as a measure, this approach still has the least.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: See my first comment, doing the inverse wherein you start with no coins and slowly add to either side until the tip, then do step #4 above is quicker and much more efficient. That, on average is O(.5N+2.5) touches, where N is number of unknown coins.&lt;/p&gt;&#xA;" OwnerUserId="12771" LastEditorUserId="12771" LastEditDate="2015-05-28T03:42:51.280" LastActivityDate="2015-05-28T03:42:51.280" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15624" PostTypeId="1" AcceptedAnswerId="15625" CreationDate="2015-05-28T03:57:22.077" Score="4" ViewCount="495" Body="&lt;p&gt;There is a 7-by-7 square grid containing an &quot;A&quot; and &quot;B&quot; in each square, in a checkerboard pattern. You must get the board to have no palindromes in it, going horizontally or vertically, in a full row or column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Starting Grid:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/knDuC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Goal:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal is to find the least amount of turns needed to get no palindromes in a full row or column. Start with the starting grid. Then switch two squares that are right next to each other. Each switch you make counts as one turn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explain Your Reasoning&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="11002" LastEditorUserId="463" LastEditDate="2015-05-28T06:07:23.473" LastActivityDate="2015-05-28T06:07:23.473" Title="The Palindrome Grid" Tags="|palindrome|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15625" PostTypeId="2" ParentId="15624" CreationDate="2015-05-28T04:19:12.127" Score="7" Body="&lt;p&gt;The minimum is five.  There are fourteen palindromes in the grid.  Each swap touches three, so we must have at least five.  One of many ways to achieve that is to swap $(1,1)\leftrightarrow (1,2),(2,3)\leftrightarrow (2,4),(3,5)\leftrightarrow (3,6),(4,7)\leftrightarrow (5,7),(6,7)\leftrightarrow (7,7)$&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2015-05-28T04:19:12.127" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15626" PostTypeId="2" ParentId="15608" CreationDate="2015-05-28T04:25:16.827" Score="0" Body="&lt;p&gt;Is it ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Confidence ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the initiator of this riddle&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Through the confidence you can put this as puzzle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the beginning of everything&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is necessary thing to start anything&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the origin of god&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure about this line..&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But not long ago god was the origin of me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; obviously god is not origin of confidence&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In some way I am god&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not Sure &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And even god is stranded without me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not Sure&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I live even if you kill my body&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Confidence is stay as it is.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;you can kill me! But never be sure I don't come back&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If confidence kill by someone or our self so not sure that it will come back&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be the greatest thing in your life&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; yes it is&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or destroy everything you love&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; May be it happen with lack of confidence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4312" LastActivityDate="2015-05-28T04:25:16.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15630" PostTypeId="1" AcceptedAnswerId="15633" CreationDate="2015-05-28T04:50:10.987" Score="1" ViewCount="274" Body="&lt;h2&gt;Rules:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;You have 9 dots, labeled A through I.&lt;/li&gt;&#xA;  &lt;li&gt;You choose a &lt;strong&gt;layout&lt;/strong&gt; (arrangement) for the 9 dots.&lt;/li&gt;&#xA;  &lt;li&gt;You must connect the 9 dots with straight lines (no curves or extra lines).&lt;/li&gt;&#xA;  &lt;li&gt;The object is to minimize the &quot;Connect Factor.&quot;&lt;/li&gt;&#xA;  &lt;li&gt;The closest that any 2 dots can be to each other is 1 unit.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Disclaimer:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is most likely &lt;strong&gt;only one optimal solution&lt;/strong&gt;, but I do not know what it is (yet).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Info:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;There are 36 distinct paths (9 choose 2). Namely, AB, AC, AD, ... , GH, GI, HI&lt;/li&gt;&#xA;  &lt;li&gt;Consider 2-dimensional and 3-dimensional. Get creative. I have a feeling that the optimal answer is a 3-d layout (maybe 8 points evenly distributed around a dot, 1 unit from it, if that is possible? or some variation of that?)&lt;/li&gt;&#xA;  &lt;li&gt;Eventually I would like to extend this to other numbers of dots, but for now, let's optimize 9.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Connect Factor Formula:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A = The average of the 36 path lengths&lt;br&gt;&#xA;  N = Total number of lines used&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Connect Factor = A + N&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Example 1:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fza08.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2O1tB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;Example 2:&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vBvFk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6tC7x.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Add any tags that you think are relevant.&lt;/li&gt;&#xA;&lt;li&gt;Should the graph-theory tag be added?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="463" LastActivityDate="2015-05-28T05:24:56.560" Title="Find the optimal dot configuration" Tags="|geometry|optimization|construction|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15633" PostTypeId="2" ParentId="15630" CreationDate="2015-05-28T05:13:13.827" Score="3" Body="&lt;p&gt;In 3 dimensions, position three on an axis in a row, each one unit apart. On the center one, construct a hexagon of the other 6 points on the perpendicular plane. These 6 points are now 1 unit from each other and 1 unit from the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Draw lines connecting only to the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Distance from outer point to center point: 1 (8 times)&#xA;Distance from outer point to outer point: 2 (28 times)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Average: $\frac{8+2\times28}{36} = 1\frac{7}{9}$&lt;br&gt;&#xA;Number of lines: 8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Connect factor: $9\frac{7}{9}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it optimal? I think it might be. Why?&lt;br&gt;&#xA;With 9 points, we must draw at least 8 lines in order to make a connected graph. An optimal solution must use 8. Suppose the optimal solution used 9. It would need to calculate a connect factor smaller than above, leaving only $\frac{7}{9}$ for the average path length. However, each path must be at least length 1. Therefore, an 8 line solution is optimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could there be an 8-line solution that has an average length shorter than $1\frac{7}{9}$? No. If this were the case, then one of the paths that I previously indicated as &quot;outer&quot; would need to be of length less than 2. In order to be less than 2, its connection must not go through the common point and be a direct connection. However, this means that other points would connect to it through this other outer point, increasing all of their path lengths by 1 and inflating the average.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a lot of vertex counting that could be done here to formally prove all of this, but I'll leave that as an exercise to the reader.&lt;/p&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-05-28T05:24:56.560" LastActivityDate="2015-05-28T05:24:56.560" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15634" PostTypeId="2" ParentId="15588" CreationDate="2015-05-28T06:24:41.300" Score="26" Body="&lt;p&gt;People have already used these same solutions for Rock and Scissors, but here's a new answer for Paper...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rock said: If I spend enough time around scissors, they’ll have a hard time moving eventually.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Water, which will cause the scissors to rust.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Paper said: I never met a rock too hard for me to dig into and break through.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Ice, which splits rocks by expanding when it freezes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Scissors said: I eat paper pretty quickly; frankly I prefer to savor something less processed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Fire, which will burn paper quickly and wood slowly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And for the Rock, Paper, Scissors relationship...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water quenches fire. Fire melts ice. Ice freezes water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What do you think?&lt;/p&gt;&#xA;" OwnerUserId="12382" LastActivityDate="2015-05-28T06:24:41.300" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15635" PostTypeId="2" ParentId="15608" CreationDate="2015-05-28T06:36:19.653" Score="10" Body="&lt;p&gt;It could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Thought/idea(s)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the initiator of this riddle&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Thought/ideas create riddles (usually and hopefully :) )&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the beginning of everything&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All creations first start with thought/ideas. Also, thought focused on by many creates thought-form, which eventually creates matter (according to New Age, &quot;Law of Attraction&quot; teachings)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the origin of god&lt;br&gt;&#xA;But not long ago god was the origin of me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Thought created the concept of God, but God (&quot;All-that-Is&quot;) thought, which created more thought.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In some way I am god&lt;br&gt;&#xA;And even god is stranded without me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Thought is god because, in a practical sense, &quot;We are what we think.&quot; Without thought, there is no concept of god.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I live even if you kill my body&lt;br&gt;&#xA;you can kill me! But never be sure I don't come back&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ideas turn to ideals, and can live on long after the original thinker is gone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be the greatest thing in your life&lt;br&gt;&#xA;Or destroy everything you love&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People live for ideas/ideals, and they can be crushed by either their own ideas or their (or other peoples') societies' ideals&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-05-28T14:09:38.303" LastActivityDate="2015-05-28T14:09:38.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15637" PostTypeId="2" ParentId="15588" CreationDate="2015-05-28T07:38:17.747" Score="2" Body="&lt;p&gt;Here's my take&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rock is water (rust makes scissors hard to move)&#xA; &lt;br&gt;&#xA; &lt;br&gt;Paper is earth (earthquake can shatter rocks)&#xA; &lt;br&gt;&#xA; &lt;br&gt;Scissors is fire (fire burns paper faster than wood)&#xA; &lt;br&gt;&#xA; &lt;br&gt;Fire scotches the earth, earth swallows water, water put out fire. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;" OwnerUserId="12332" LastEditorUserId="12332" LastEditDate="2015-05-28T07:49:39.650" LastActivityDate="2015-05-28T07:49:39.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15638" PostTypeId="2" ParentId="15586" CreationDate="2015-05-28T08:30:57.907" Score="3" Body="&lt;p&gt;I'm bit disappointed that the expected answer is bit unimaginative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if Sam intended to ask the question like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Is that tent Good Rice or Bad Ri&lt;strong&gt;s&lt;/strong&gt;e?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Similarly for all questions he meant &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rise&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;instead of &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rice&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then also the given answers make sense. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tent, Tree and fire are good 'rises' (as in high-rise building), but&#xA; stone is not a 'good rise'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2432" LastActivityDate="2015-05-28T08:30:57.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15639" PostTypeId="2" ParentId="15130" CreationDate="2015-05-28T08:58:24.480" Score="3" Body="&lt;p&gt;Well, after seeing the solution I guess I have a proof for the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like many others earlier, the probability for the first and the second points to lie on the same half is $1$. I would not get into that as it is already shown by other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we know that the 1st and the second points will always lie in the same half. Considering the 3rd point, lets define the worst case as when the probability of the 3rd point to lie in the same half is minimal, and best case when the probability of the 3rd point lying on the same half is maximum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Worst Case:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Dxh6m.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;In this case the probability of point $p_3$ lying in the same half is almost $0.5$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best Case:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/YI1Ua.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;In this case the probability of point $p_3$ lying in the same half is almost $1.0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now comes a little complicated part. Consider the &quot;separating line&quot; is changing it's slope continuously by a very small amount. In that case the two halves will keep changing shapes. It's important to notice that in the intervals of $180$ degrees and $360$ degrees the all the halves are taken into consideration and since the halves can be defined by any &quot;separating line&quot;, we can safely say that the probability of the point $p_3$ will increase uniformly from worst case to best case. That is, uniformly from $0.5$ to $1.0$.&lt;br&gt;&#xA;Thus, the overall probability of point $p3$ lying in the same half will be&lt;br&gt;&#xA;$(1.0 + 0.5) \div 2 = \frac{3}{4}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, for the point $p_4$, it's not difficult to see that the worst case and best case probability to lie in the same half would be the same as point $p_3$, that is, $0.5$ and $1.0$ respectively. And, the overall probability would be the same as well, i.e. $\frac{3}{4}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the easy part. The overall probability of the $4$ points lying in the same half would be: $\frac{3}{4} \times \frac{3}{4} = \frac{9}{16}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, proved.&#xA;Thanks for the check mark in advance :P.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please correct me if you think I am wrong.&lt;/p&gt;&#xA;" OwnerUserId="12777" LastEditorUserId="7308" LastEditDate="2015-05-28T09:15:43.230" LastActivityDate="2015-05-28T09:15:43.230" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15640" PostTypeId="1" AcceptedAnswerId="15641" CreationDate="2015-05-28T09:37:55.783" Score="21" ViewCount="7001" Body="&lt;p&gt;I have two large identical chests of exactly the same shape, volume, etc. One is completely full of nickels (each worth 5 cents), and the other is exactly half full of dimes (each worth 10 cents). Which chest is worth more money?&lt;/p&gt;&#xA;" OwnerUserId="12730" LastActivityDate="2015-06-01T16:15:00.393" Title="Which is worth more - a chest of nickels or half a chest of dimes?" Tags="|logical-deduction|lateral-thinking|" AnswerCount="5" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15641" PostTypeId="2" ParentId="15640" CreationDate="2015-05-28T09:43:53.697" Score="42" Body="&lt;p&gt;This required to get my wallet but it gave me the answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The chest half full of dimes is worth more, this because a dime is smaller than a nickle, so it would occupy less space and there would be more dimes in a full chest than there would be in a full chest of nickels. Conclusion: pick the dimes, It's worth more&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12218" LastActivityDate="2015-05-28T09:43:53.697" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15642" PostTypeId="2" ParentId="15588" CreationDate="2015-05-28T09:44:22.047" Score="10" Body="&lt;p&gt;I think I might have a breakthrough .. could I possibly have cracked it??? (like a big old tree cracks rock!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rock said: If I spend enough time around scissors, they’ll have a hard time moving eventually&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water (causes scissors to rust, esp if salty or full of minerals (eg iron)) ..(as per above posters)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Scissors said: I eat paper pretty quickly; frankly I prefer to savor something less processed&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fire (burns paper quickly &amp;amp; wood (incl still-growing trees) more slowly) ..(as per above posters)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Paper said: I never met a rock too hard for me to dig into and break through&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tree (it's roots &quot;dig into and break through&quot; any rock)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the Rock, Paper, Scissors relationship is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water quenches Fire .. Fire burns Tree .. Tree consumes Water&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12778" LastEditorUserId="12109" LastEditDate="2015-05-28T14:40:05.943" LastActivityDate="2015-05-28T14:40:05.943" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15643" PostTypeId="1" AcceptedAnswerId="15644" CreationDate="2015-05-28T10:15:18.713" Score="14" ViewCount="2806" Body="&lt;p&gt;A window was broken in my house and I need to find out who did it, can you guys help me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When asked who did it my friends replied with the following in order starting with 1:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&quot;I didn't do it&quot;&lt;/li&gt;&#xA;&lt;li&gt;&quot;1 is telling the truth&quot;&lt;/li&gt;&#xA;&lt;li&gt;&quot;4 is lying&quot;&lt;/li&gt;&#xA;&lt;li&gt;&quot;I swear it wasn't me&quot;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The person whose story sounds fishy is the culprit. Who is it?&lt;/p&gt;&#xA;" OwnerUserId="12218" LastEditorUserId="5260" LastEditDate="2015-05-28T15:57:33.697" LastActivityDate="2015-05-29T07:26:47.133" Title="The broken window" Tags="|logical-deduction|liars|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15644" PostTypeId="2" ParentId="15643" CreationDate="2015-05-28T10:19:24.933" Score="28" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It was 3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;When you were questioning 3, you haven't asked anything to 4 and he haven't answered anything, since you were asking in the order 1,2,3 and 4. So the statement &quot;4 is lying&quot; by 3 is wrong. And from your statement &quot;The person whose story sounds fishy is the culprit. &quot;, Friend '3' is the culprit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastActivityDate="2015-05-28T10:19:24.933" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15645" PostTypeId="2" ParentId="15495" CreationDate="2015-05-28T11:43:44.330" Score="9" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Magpies&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Following on from the well known&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/One_for_Sorrow_%28nursery_rhyme%29&quot;&gt;Nursery rhyme&lt;/a&gt;&lt;br&gt;&#xA; One for sorrow,&lt;br&gt;&#xA; Two for joy,&lt;br&gt;&#xA; Three for a girl,&lt;br&gt;&#xA; Four for a boy,&lt;br&gt;&#xA; Five for silver,&lt;br&gt;&#xA; Six for gold,&lt;br&gt;&#xA; Seven for a secret,&lt;br&gt;&#xA; Never to be told  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The hint helped a lot :)&lt;/p&gt;&#xA;" OwnerUserId="6205" LastActivityDate="2015-05-28T11:43:44.330" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15646" PostTypeId="2" ParentId="15495" CreationDate="2015-05-28T11:44:08.263" Score="8" Body="&lt;p&gt;Great riddle!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are all&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; magpies&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One for sorrow, two for joy,&lt;br&gt;&#xA; Three for a girl, four for a boy,&lt;br&gt;&#xA; Five for silver, six for gold,&lt;br&gt;&#xA; Seven for a secret never to be told.&lt;br&gt;&#xA;&lt;br&gt;&#xA; - a &lt;a href=&quot;http://en.wikipedia.org/wiki/One_for_Sorrow_%28nursery_rhyme%29&quot; rel=&quot;noreferrer&quot;&gt;common nursery rhyme&lt;/a&gt; about magpies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Didn't need the hint, either :)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Further reading (not related to the solution, but spoils it):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You might be interested to know that there is at least one other version of this rhyme:&lt;br&gt;&#xA;&lt;br&gt;&#xA; One for sorrow, two for mirth,&lt;br&gt;&#xA; Three for a wedding, four for a birth,&lt;br&gt;&#xA; Five for England, six for France,&lt;br&gt;&#xA; Seven for the fiddler, eight for the dance.&lt;br&gt;&#xA;   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5260" LastEditorUserId="5260" LastEditDate="2015-05-28T11:54:41.703" LastActivityDate="2015-05-28T11:54:41.703" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15647" PostTypeId="2" ParentId="15643" CreationDate="2015-05-28T11:53:54.193" Score="0" Body="&lt;p&gt;&lt;em&gt;The answer has been reworded.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first I would say that number 3 is not a liar. It is because liars are usually smart enough to not to accuse anybody, if they are not sure they can make the words probable. If 3 did it, he wouldn't say &quot;4 is lying&quot; because he doesn't know what he is going to say. He does not know if number 4 has an alibi. It doesn't matter if you ask them one by one (whether they know what others say or not), or as a group. Normally 3 would say &quot;I didn't do it&quot;, &quot;4 did it&quot;, &quot;I don't know, but I saw 4 playing ball near your window 5 minutes earlier&quot; etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number 4 says &quot;I swear I didn't&quot;. A liar always needs to ensure his accuser, he's always trying to raise some emotions. This is why he uses big words like &quot;I swear&quot;, &quot;I vow&quot;, &quot;I ensure&quot; etc. He should have said &quot;3, are you stupid? I wasn't there! Why do you lie?&quot; or if he hadn't heard what 3 said just &quot;I didn't&quot; without &quot;I swear&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably number 3 says &quot;4 is lying&quot; because he knows him well. 4 lies very often so 3 is trying to say to you &quot;do not believe whatever 4 says, I know him, if you ask him, he will lie.&quot; Maybe 3 has the same experience as you? Maybe 3 asked 4 for something and was deceived? He trusts him no more and he's giving an advice to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why don't you say there could be two or more liars? Why don't you think there were two perpetrators? For example 1 and 2: 2 says nothing about his guilt. He doesn't even say &quot;I didn't&quot;. You ask him &quot;did you?&quot; and he says &quot;1 is telling the truth&quot;. Strange for me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first I agreed with &lt;a href=&quot;https://puzzling.stackexchange.com/a/15644/10880&quot;&gt;the other answer&lt;/a&gt; (I've even up-voted it), but this is not as clear as it seems for the first sight. I only wanted to show that 3 should not be taken as responsible, because this is not a puzzle, there is some fuzzy-logic in here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(Please excuse me the first version of this answer)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="10880" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-05-29T07:26:47.133" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15648" PostTypeId="2" ParentId="15640" CreationDate="2015-05-28T12:37:53.190" Score="53" Body="&lt;p&gt;Here's another way to see it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The chests themselves are worth exactly the same, as they are identical. The contents, on the other hand, might differ.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;EDIT&lt;/em&gt;: As some people have pointed out, this is an even larger stretch than before if one considers the title as well as the question itself. I didn't.&lt;/p&gt;&#xA;" OwnerUserId="12785" LastEditorUserId="12785" LastEditDate="2015-05-29T14:05:50.950" LastActivityDate="2015-05-29T14:05:50.950" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="15649" PostTypeId="2" ParentId="15584" CreationDate="2015-05-28T13:58:14.513" Score="2" Body="&lt;p&gt;I think the sibling are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stack exchange communities&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ok I've tried my best, but the clues can fit multiple choices:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First sibling&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Academia - scholar + learning&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second sibling&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Anything fun can lead to abuse, I'll go with generic hobby type Stack exchange&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Third sibling&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Any scientific subject for example Programmers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fourth sibling&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Perhapse Ask Different&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fifth sibling&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Travel &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sixth sibling&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Workplace or Careers?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Me + parent&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think the parent is Stack Overflow, because that's the largest one and on the site &lt;a href=&quot;http://stackexchange.com/sites#&quot;&gt;Stack exchange - sites&lt;/a&gt; mathematics is directly underneath.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9907" LastEditorUserId="9907" LastEditDate="2015-05-28T15:50:08.297" LastActivityDate="2015-05-28T15:50:08.297" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15650" PostTypeId="1" AcceptedAnswerId="15652" CreationDate="2015-05-28T14:18:29.693" Score="8" ViewCount="1608" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MvFbB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;a href=&quot;https://i.stack.imgur.com/am09h.jpg&quot; rel=&quot;noreferrer&quot;&gt;Identify the fault in the picture?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Read the Text.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the picture. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Identify.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might get it.&lt;/p&gt;&#xA;" OwnerUserId="9892" LastActivityDate="2015-05-29T01:05:19.280" Title="What is the fault in the Image?" Tags="|lateral-thinking|geography|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15652" PostTypeId="2" ParentId="15650" CreationDate="2015-05-28T14:24:16.530" Score="14" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Furnace Creek Fault&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The question is a pun - we're looking for an earthquake fault, not a fault in the sense of 'defect' or 'problem'. &#xA; If you reverse image search in Google, you find &lt;a href=&quot;http://4.bp.blogspot.com/-IM23oFbbZ_0/U0JisOvEgeI/AAAAAAAASZA/yZ8uKkI2O00/s1600/P1030945+Furnace+Creek+Fault+Zone.jpg&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt;&lt;br&gt;&#xA; Which has the tagline Furnace Creek Fault&lt;br&gt;&#xA; (Also the &lt;a href=&quot;http://geotripper1.rssing.com/chan-6045487/all_p14.html&quot; rel=&quot;noreferrer&quot;&gt;blog&lt;/a&gt; linking it is pretty interesting)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6205" LastEditorUserId="185" LastEditDate="2015-05-29T01:05:19.280" LastActivityDate="2015-05-29T01:05:19.280" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15653" PostTypeId="2" ParentId="15650" CreationDate="2015-05-28T14:25:20.593" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's the Furnace Creek fault system -- which parallels the San Andreas one but isn't the same.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10625" LastActivityDate="2015-05-28T14:25:20.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15654" PostTypeId="2" ParentId="15650" CreationDate="2015-05-28T14:25:37.013" Score="2" Body="&lt;p&gt;Is it because &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the water is missing?&#xA; I am not sure, but the soil looks like the soil on a river bottom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5283" LastActivityDate="2015-05-28T14:25:37.013" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15655" PostTypeId="2" ParentId="15640" CreationDate="2015-05-28T15:37:21.757" Score="24" Body="&lt;p&gt;It's not that easy!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dimes have a &lt;em&gt;nominal value&lt;/em&gt; of 10 cents, whereas the &lt;em&gt;material value&lt;/em&gt; is only about 5.6 cents. Which means the &lt;em&gt;nominal&lt;/em&gt; value is decisive.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nickels have a &lt;em&gt;nominal value&lt;/em&gt; of 5 cents but a &lt;em&gt;material value&lt;/em&gt; of 10.09 cents (source: &lt;a href=&quot;http://en.wikipedia.org/wiki/Nickel_%28United_States_coin%29#Increase_in_metal_values&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;). Which means that the &lt;em&gt;material value&lt;/em&gt; is decisive (makes you wonder why people aren't buying large amounts of nickels and reselling the metal...).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, dimes are not only smaller in diameter than nickels but also slimmer (17.9mm versus 21.2mm diameter and 1.35 versus 1.95mm). That means that the same volume can hold a larger number of dimes than it could hold nickels. It is not possible to tell exactly how many more nickels will fit into the same volume (or into half that volume, for that matter) since this greatly depends not only on stacking, but also on implementation details of the surrounding box. If the box is just half a millimeter too small to exactly fit a complete row of coins, you're drifting far away from the optimal packing, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in order to answer the question, it is not necessary at all to know the exact number of coins that will fit. Since both coins are (almost exactly) worth the same, you must have the same number of coins in either case. That is, you must be able to fit &lt;em&gt;the same number&lt;/em&gt; of dimes into half the volume, or twice the number of dimes into the same volume. This is certainly not the case, you can only fit roughly 1/3 more dimes into the same volume, give or take a few.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the box of nickels is more valuable.&lt;/p&gt;&#xA;" OwnerUserId="8740" LastEditorUserId="4204" LastEditDate="2015-05-30T19:34:41.530" LastActivityDate="2015-05-30T19:34:41.530" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="15656" PostTypeId="2" ParentId="15608" CreationDate="2015-05-28T15:38:58.130" Score="1" Body="&lt;p&gt;is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; energy?&#xA;&lt;br&gt;everything consist of energy...energy makes him narrate&gt;&gt; am the initiator of this riddle&#xA;&lt;br&gt;&lt;br&gt;energy was from the begning, it just converted into types&gt;&gt;I am the beginning of everything energy itsef is a kind of god&gt;&gt; I am the origin of god&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But not long ago god was the origin of me&#xA;In some way I am god&#xA;And even god is stranded without me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; energy just kind of leave our body and stay some where for eve&gt;&gt;&gt;I live even if you kill my body you can kill me! But never be sure I don't come back I can be the greatest thing in your life&gt;&gt; greatest thing in life having a enegy in heat to do work , to do anything..Or destroy everything you love And one last thing:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You will need me to find the answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; we need energy to find the answer &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12791" LastEditorUserId="11244" LastEditDate="2015-05-28T15:46:55.040" LastActivityDate="2015-05-28T15:46:55.040" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15657" PostTypeId="2" ParentId="15584" CreationDate="2015-05-28T16:14:00.240" Score="4" Body="&lt;p&gt;Following on from @Zikatos answer; you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quantitative Finance&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I believe the key is to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; take the first Stack Exchange community per distinct letter in the alphabet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning (pretty self explanatory):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My first sibling is a scholar. He has an interest for continuing learning.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first community beginning with A is Academia; definitely focused on continued learning&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My second sibling is the fun one, but can be lead to abuse.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first community beginning with B is Beer; fun in moderation but alcohol abuse is no joke&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My third sibling is smart like the first, but likes to concentrate his studies (on a smaller subject).&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first community beginning with C is Chemistry; smart in a particular subject&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My fourth sibling likes to learn, or rather teach to learn in a different radical way.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first community beginning with D is Data Science; &quot;teaching&quot; computers how to &quot;learn&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My fifth sibling loves to travel and loves knowing more about the world.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first community beginning with E is Earth Science; this covers the whole world&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My sixth sibling will be his own boss, working for himself.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first community beginning with F is Freelancing; become your own boss indeed!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;As for me, I'm below the 15th sibling, who's a parent.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The 15th letter is P (due to there being no community beginning with K).&lt;br&gt;&#xA; The first community beginning with P is Parenting.&lt;br&gt;&#xA; You are below it, or rather the first community beginning with Q, Quantitative Finance!&#xA;&lt;br&gt;&lt;br&gt; The intended answer was Puzzling, being underneath Parenting and something we are all on (a more literal me). [However the posted answer is still just as valid] - Mark N&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6205" LastEditorUserId="11244" LastEditDate="2015-05-29T12:48:49.193" LastActivityDate="2015-05-29T12:48:49.193" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15658" PostTypeId="1" AcceptedAnswerId="15669" CreationDate="2015-05-28T16:19:01.740" Score="25" ViewCount="2429" Body="&lt;p&gt;&lt;strong&gt;Create a word search with the following conditions:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Use the eleven words below. They can be spelled up, down, left, right, or any of the 4 diagonal directions.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;ZERO&lt;br&gt;&#xA;  ONE&lt;br&gt;&#xA;  TWO&lt;br&gt;&#xA;  THREE&lt;br&gt;&#xA;  FOUR&lt;br&gt;&#xA;  FIVE&lt;br&gt;&#xA;  SIX&lt;br&gt;&#xA;  SEVEN&lt;br&gt;&#xA;  EIGHT&lt;br&gt;&#xA;  NINE&lt;br&gt;&#xA;  TEN&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;The size of the word search must be 5 rows and 6 columns.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The number of words that pass through each square must fit this pattern:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Cruzq.png&quot; alt=&quot;editing. one minute...&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;The accepted answer will clearly and fully explain the flow of logic used from start to finish.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-05-28T17:00:07.437" LastActivityDate="2017-05-01T00:54:34.890" Title="Create a Word Search" Tags="|logical-deduction|word-search|construction|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="15659" PostTypeId="2" ParentId="15568" CreationDate="2015-05-28T16:39:32.843" Score="1" Body="&lt;h1&gt;You need 5 tests&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Number the coins with the following trinary (base 3) numbers: $\{0&amp;lt; x&amp;lt; 3^5 | \text{ first non-zero digit is 1 and the number of zero digits is even}\}\cup\{0&amp;lt; x&amp;lt; 3^5 | \text{ first non-zero digit is 2 and the number of zero digits is odd }\}$.&lt;br&gt;&#xA;Following are the first few of them: $00001, 00010, 00021, 00022, 00100, 00111, 00112, 00121, 00122, 00201, 00202, 00210, 00220, ...$&lt;br&gt;&#xA;That means there is no coin numbered $00002, 00011, 00012, 00020, 00101, 00102, 00110, 00120, 00200, 00211, 00212, 00221, 00222, ...$&lt;br&gt;&#xA;Fill the scale five times and put the real coin on the side where one coin is missing.&lt;br&gt;&#xA;The first time put all the coins with a 1 as the first digit on the left side and all coins with a 2 as the first digit on the right side.&lt;br&gt;&#xA;The second time do the same but look at the second digit. And so on.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write down the result of each test: Write down a 0 if the scale is balanced; write down a 1 if the left side is heavier and a 2 if the right side is heavier.&lt;br&gt;&#xA;If the result is 00000 then there is no fake coin.&lt;br&gt;&#xA;If there is a coin with the resulting number that coin is heavier then the other coins.&lt;br&gt;&#xA;If there is no coin with the resulting number then replace all 1s with 2s and all 2s with 1s and the coin with that number is lighter then the other coins.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Example 1:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;That means the scale will tilt to the right during the first and the fifth test and left during the second and third test and will be levelled during the fourth test.&lt;br&gt;&#xA;The resulting number will be 2(first test-&gt;right), 1(second test-&gt;left), 1(third test-&gt;left), 0(fourth test-&gt;levelled), 2(fifth test-&gt;right)&lt;br&gt;&#xA;Concatenation of these numbers give 21102 = the number of the heavier coin.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Example 2:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;That means the scale will tilt to the right during the second and the third test and left during the fifth test and will be levelled during the first and the fourth test.&lt;br&gt;&#xA;The resulting number will be 0(first test-&gt;levelled), 2(second test-&gt;right), 2(third test-&gt;right), 0(fourth test-&gt;levelled), 1(fifth test-&gt;left)&lt;br&gt;&#xA;There is no coin numbered 02201.&lt;br&gt;&#xA;That means we have to switch the 1s and 2s and the coin 01102 is the lighter coin.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Why does this works:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Each coin has an individual sequence of placements on the scale. &lt;br&gt;&#xA;The coding is chosen in a way that for heavier coins that coding is the results of the tests. &lt;br&gt;&#xA;For lighter coins you would have to write down the numbers of the upwards sides of the tests and that is simulated by switching the 1s and 2s at the end.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an exhaustive proof you would need to proof that both sides of the scale have the same number of coins, but that is omitted here.&lt;br&gt;&#xA;This method can be extended to the same question with $364$ coins and $6$ tests or general $\frac{3^n-1}{2}$ coins and $n$ tests.&lt;br&gt;&lt;/p&gt;&#xA;" OwnerUserId="10143" LastActivityDate="2015-05-28T16:39:32.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15660" PostTypeId="2" ParentId="15188" CreationDate="2015-05-28T16:48:07.337" Score="1" Body="&lt;p&gt;The plain text is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Never within this life of mine have I made a cryptogram from scrambled words.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The twist is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Scrambling the interior letters of each word before applying the cipher&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While it appears to make it robust against attack, it renders it useless for Puzzling.&lt;/p&gt;&#xA;" OwnerUserId="9000" LastActivityDate="2015-05-28T16:48:07.337" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15661" PostTypeId="2" ParentId="15640" CreationDate="2015-05-28T16:55:43.907" Score="17" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Nickel_%28United_States_coin%29&quot; rel=&quot;noreferrer&quot;&gt;Nickel&lt;/a&gt;&#xA;  value per volume:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\frac{5 ¢}{\pi(\frac{21.21}{2}\ \text{mm})^2 * 1.95\ \text{mm}} = \frac{7.257 ¢}{\text{mL}}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Dime_%28United_States_coin%29&quot; rel=&quot;noreferrer&quot;&gt;Dime&lt;/a&gt;:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\frac{10 ¢}{\pi(\frac{17.91}{2}\ \text{mm})^2 * 1.35\ \text{mm}} = \frac{29.40 ¢}{\text{mL}}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;So, for large volumes with reasonable dimensions$\dagger$, if you had&#xA;  twice the volume of nickels as dimes, the nickels would worth about&#xA;  half as much as the dimes.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\dagger$ I say &lt;em&gt;reasonable dimensions&lt;/em&gt; because fringe effects come&#xA;  into play if you have to pay attention to the coins bumping into the&#xA;  wall.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12795" LastEditorUserId="12795" LastEditDate="2015-05-31T20:30:06.637" LastActivityDate="2015-05-31T20:30:06.637" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15662" PostTypeId="2" ParentId="15610" CreationDate="2015-05-28T17:58:09.840" Score="1" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The desert&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have peaks and valleys and slopes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A desert have peaks, valleys and slopes (dunes)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whose mountains and beats are such tropes,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I don't understand this sentence but i guess mountain could refer to dune again&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the prime, you will find,&#xA;  Is a step back in time;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you desert, it's a step back from your first decision&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The end, you will find, is a hoax.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Maybe the fact that you can see oasis if you walk too long in the desert&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5359" LastActivityDate="2015-05-28T17:58:09.840" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15663" PostTypeId="2" ParentId="15610" CreationDate="2015-05-28T18:16:45.430" Score="1" Body="&lt;p&gt;Is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Ocean?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have peaks and valleys and slopes&#xA;Whose mountains and beats are such tropes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It has &quot;mountains and valleys&quot; in the form of waves - but they aren't real mountains (tropes).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the prime, you will find,&#xA;Is a step back in time;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first wave happened a looooong time ago.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The end, you will find, is a hoax.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; At the &quot;end&quot; of the ocean appears the horizon, which is a hoax in that it's not a real thing and will continue to move further away from you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12206" LastActivityDate="2015-05-28T18:16:45.430" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15664" PostTypeId="1" AcceptedAnswerId="15693" CreationDate="2015-05-28T18:24:30.847" Score="17" ViewCount="1850" Body="&lt;p&gt;The symbols below appear to be some sort of nautical symbols: waves, sails, lighthouse markers, waterfall markers, directional markers, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spots are labeled with red letters, but the red letters have no meaning other than to label the spots, so that you can more easily refer to a symbol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of the five spots marked with red question marks contain one of the shown 11 symbols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which symbols go in those 5 spots?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uYPbQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-06-12T15:03:36.407" LastActivityDate="2015-06-12T15:03:36.407" Title="Identify the Missing Symbols" Tags="|visual|enigmatic-puzzle|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15666" PostTypeId="2" ParentId="15610" CreationDate="2015-05-28T18:45:41.873" Score="4" Body="&lt;p&gt;Is it a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sinusoidal wave? -updated thanks to &lt;strong&gt;ace&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have peaks and valleys and slopes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; These waves have all of these things - peaks/troughs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whose mountains and beats are such tropes,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Constructive / destructive interference of two longitudinal sound waves causes the phenomena known as beats.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the prime, you will find,&#xA;  Is a step back in time;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The derivation of the wave is the same wave phase-shifted (or, how it would have appeared at a certain time beforehand)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The end, you will find, is a hoax.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because there is no finite end to a longitudinal wave - thanks @Mark N&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12206" LastEditorUserId="12206" LastEditDate="2015-05-29T20:47:25.643" LastActivityDate="2015-05-29T20:47:25.643" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="15667" PostTypeId="1" AcceptedAnswerId="15676" CreationDate="2015-05-28T18:54:17.950" Score="5" ViewCount="549" Body="&lt;p&gt;I'm back with a second cryptic crossword. This one is smaller than my previous, yet also trickier, so hopefully it provides a fun challenge for a few of you!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DaKc9.png&quot; alt=&quot;Cryptic Crossword&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Across&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;1. Pass around venomous snakes (4)&lt;br&gt;&#xA;4. Vermin make poor pets (4)&lt;br&gt;&#xA;6. Mutilate Italian brother by removing his primary internal organs (3)&lt;br&gt;&#xA;7. Burden with H. Wagner's remains (4)&lt;br&gt;&#xA;8. I fill the last of Marco's heart with viscous liquid (3)&lt;br&gt;&#xA;9. We talked about it; they heard, and were offended (9)&lt;br&gt;&#xA;11. Evil woman is hiding to the west (3)&lt;br&gt;&#xA;12. Remove scoundrel's head from lasso; smash it to bits, too (4)&lt;br&gt;&#xA;13. Fish found flip-flopping in haddock's stomach (3)&lt;br&gt;&#xA;14. Massage my foot until it's red (4)&lt;br&gt;&#xA;15. What's seen is spoken with ties untied; the perfect plot resides inside (4)  &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Down&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;2. Boredom, in other words, takes alien sister to the skies (5)&lt;br&gt;&#xA;3. In time of crisis, rise from the ashes of North Carolina's youth (9)&lt;br&gt;&#xA;4. Procession, without a portent at its heart, is a leisurely walk (9)&lt;br&gt;&#xA;5. New mid-50's annex (3)&lt;br&gt;&#xA;10. Experience the beginning-of-the-end? That's odd, but this isn't (5)&lt;br&gt;&#xA;11. Bird that started eating metal, unfortunately! (3)  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note #1:&lt;/em&gt; This crossword pretty obviously isn't your ordinary crossword. It's up to you to figure out the difference. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note #2:&lt;/em&gt; This crossword and its clues don't &lt;em&gt;necessarily&lt;/em&gt; align with typical cryptic crossword form - that is to say, each of the clues are of the cryptic puzzle form, but a couple of them stretch the boundaries past what is typically deemed as &quot;acceptable form&quot; for American cryptic crosswords. They should all still be quite solvable, of course!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note #3:&lt;/em&gt; The &quot;trivia&quot; tag was added because a few of the clues require explicit knowledge of specific people/characters. Most of the clues don't, so even without that knowledge this crossword should be solvable.&lt;/p&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="11475" LastEditDate="2015-09-01T20:57:35.847" LastActivityDate="2015-09-01T20:57:35.847" Title="Cryptic Crossword #2: Stack-in-the-Box" Tags="|knowledge|crosswords|cryptic-crosswords|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15668" PostTypeId="2" ParentId="15667" CreationDate="2015-05-28T19:22:57.853" Score="2" Body="&lt;p&gt;Making a start, again...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;11A. Evil woman is hiding to the west (3)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; EVE? (the left-hand, or &quot;west&quot;, part of evil; also a woman)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;12A. Remove scoundrel's head from lasso; smash it to bits, too (4)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ALSO (&quot;scoundrel's head&quot; is S, the first letter of scoundrel; remove S from lasso and rearrange; synonym of &quot;too&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;14A. Massage my foot until it's red (4)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Four letters, *U**, and something to do with red, so probably RUST, PUCE, or RUBY. (The OP only &lt;a href=&quot;http://chat.stackexchange.com/transcript/message/21884603#21884603&quot;&gt;learnt&lt;/a&gt; what colour puce was today, but I'm not sure if this is an argument for or against it as a clue here.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;11D. Bird that started eating metal, unfortunately! (3)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; EMU (starting letters of &lt;strong&gt;e&lt;/strong&gt;ating &lt;strong&gt;m&lt;/strong&gt;etal &lt;strong&gt;u&lt;/strong&gt;nfortunately)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-05-28T19:33:23.087" LastActivityDate="2015-05-28T19:33:23.087" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15669" PostTypeId="2" ParentId="15658" CreationDate="2015-05-28T19:25:03.847" Score="20" Body="&lt;p&gt;Here's my answer:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PQhxR.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h1&gt;From this point on, there is no spoiler text, because it makes it easier to format, at least for me.&lt;/h1&gt;&#xA;&lt;pre&gt;&lt;code&gt;T E N I N E&#xA;F I O N E V&#xA;O G X R V I&#xA;U H H I E F&#xA;R T W O S Z&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;I started by figuring out the letter density in the overall puzzle. The letters break down as follows:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;E   10&#xA;N   5&#xA;I   4&#xA;O   4&#xA;T   4&#xA;R   3&#xA;F   2&#xA;H   2&#xA;S   2&#xA;V   2&#xA;G   1&#xA;U   1&#xA;W   1&#xA;X   1&#xA;Z   1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Next I started working on the &lt;code&gt;3&lt;/code&gt; squares. Since all three are in a diagonal, I started with the assumption that they are within the same word. This requires a word of length &lt;code&gt;5&lt;/code&gt;, which means it is one of: &lt;code&gt;SEVEN, EIGHT, THREE&lt;/code&gt;&lt;/p&gt;&#xA;&lt;p&gt;Only one of those includes three of the letters that occur at least three times from above, so that line became &lt;code&gt;THREE&lt;/code&gt;&lt;/p&gt;&#xA;&lt;p&gt;Since &lt;code&gt;H&lt;/code&gt; only occurs twice, I ran the &lt;code&gt;THREE&lt;/code&gt; diagonally up and from the left.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;.....E&#xA;....E.&#xA;...R..&#xA;..H...&#xA;.T....&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;First thought would be to place the &lt;code&gt;EIGHT&lt;/code&gt;, as that shares the only other &lt;code&gt;H&lt;/code&gt;, but that doesn't fit. &lt;code&gt;H&lt;/code&gt; is on a single letter space, so let's leave that for now.&lt;/p&gt;&#xA;&lt;p&gt;Looking at the &lt;code&gt;2&lt;/code&gt; and 2 &lt;code&gt;3&lt;/code&gt;s in the upper right corner, it seems like there should be quite a bit of overlapping there. Let's try placing the &lt;code&gt;NINE&lt;/code&gt; and &lt;code&gt;TEN&lt;/code&gt; as follows:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;..NINE&#xA;....E.&#xA;...RT.&#xA;..H...&#xA;.T....&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Not bad, could be promising.&lt;/p&gt;&#xA;&lt;p&gt;So far, we have the following for letter counts&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;..1122&#xA;....2.&#xA;...11.&#xA;..1...&#xA;.1....&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Uh-oh! We only have one end &lt;code&gt;N&lt;/code&gt; left and it's on &lt;code&gt;SEVEN&lt;/code&gt;. That won't fit in our open space. Let's try putting &lt;code&gt;SEVEN&lt;/code&gt; where &lt;code&gt;TEN&lt;/code&gt; is.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;..NINE&#xA;....E.&#xA;...RV.&#xA;..H.E.&#xA;.T..S.&#xA;&#xA;..1122&#xA;....2.&#xA;...11.&#xA;..1.1.&#xA;.1..1.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;How about we run the &lt;code&gt;EIGHT&lt;/code&gt; in the second column?&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;.ENINE&#xA;.I..E.&#xA;.G.RV.&#xA;.HH.E.&#xA;.T..S.&#xA;&#xA;.11122&#xA;.1..2.&#xA;.1.11.&#xA;.11.1.&#xA;.2..1.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Ooh, I see almost all of &lt;code&gt;TEN&lt;/code&gt;. Let's finish that.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;TENINE&#xA;.I..E.&#xA;.G.RV.&#xA;.HH.E.&#xA;.T..S.&#xA;&#xA;122122&#xA;.1..2.&#xA;.1.11.&#xA;.11.1.&#xA;.2..1.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Now, we can add &lt;code&gt;TWO&lt;/code&gt; off the &lt;code&gt;T&lt;/code&gt; from &lt;code&gt;THREE&lt;/code&gt; and &lt;code&gt;ONE&lt;/code&gt; off the first &lt;code&gt;E&lt;/code&gt;&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;TENINE&#xA;.IONE.&#xA;.G.RV.&#xA;.HH.E.&#xA;.TWOS.&#xA;&#xA;122122&#xA;.1113.&#xA;.1.11.&#xA;.11.1.&#xA;.3111.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Hey, let's finish off &lt;code&gt;ZERO&lt;/code&gt;&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;TENINE&#xA;.IONE.&#xA;.G.RV.&#xA;.HH.E.&#xA;.TWOSZ&#xA;&#xA;122122&#xA;.1213.&#xA;.1.21.&#xA;.11.2.&#xA;.31111&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;And we can run &lt;code&gt;FIVE&lt;/code&gt; up the right side and &lt;code&gt;FOUR&lt;/code&gt; down the left. Note, &lt;code&gt;FOUR&lt;/code&gt; could be either up or down, whatever floats your boat.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;TENINE&#xA;FIONEV&#xA;OG.RVI&#xA;UHH.EF&#xA;RTWOSZ&#xA;&#xA;122123&#xA;112131&#xA;11.211&#xA;111.21&#xA;131111&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;And finish off with &lt;code&gt;SIX&lt;/code&gt;&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;TENINE&#xA;FIONEV&#xA;OGXRVI&#xA;UHHIEF&#xA;RTWOSZ&#xA;&#xA;122123&#xA;112131&#xA;111211&#xA;111121&#xA;131112&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="12242" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-28T19:25:03.847" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15671" PostTypeId="1" AcceptedAnswerId="15910" CreationDate="2015-05-28T19:41:53.757" Score="26" ViewCount="1073" Body="&lt;p&gt;The natural numbers have gotten in a big fight and split up into groups. If we can't get them back together again, the universe may never be the same! What kind of sense does it make to have three counting systems? No, they must be reunited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are a licensed number therapist. Your first step is to figure out what the divisive issue was but none of the numbers are talking. Your only choice is to deduce the rule under which they split into groups based solely on which numbers are in which group. Once you get that far, you're pretty sure you can get them all back together again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the three groups up to the number $40$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{c}&#xA;       1\ 4\ 6\ 9\ 10\ 15\ 18\ 20\ 24\ 26\ 29\ 30\ 34\ 36\ 39\ 40\ \ldots \\&#xA;\hline 2\ 5\ 8\ 12\ 14\ 16\ 19\ 22\ 25\ 28\ 32\ 35\ 38\  \ldots \\&#xA;\hline 3\ 7\ 11\ 13\ 17\ 21\ 23\ 27\ 31\ 33\ 37\ \ldots&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every natural number fits into one of the three groups (You can see the groups stretching into infinity (You have very good eyes (It's practically a necessity for a number therapist.).).).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;OFFICIAL QUESTION:&lt;/strong&gt; What rule determines which group each number joins?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BONUS OFF-TOPIC QUESTION: What sort of therapy session would you design to bring the numbers together again? (Not to be submitted without a serious attempt at the official question.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There was a second tag added on Friday&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The rule can only be found when you spell out the cardinal name of each number in English&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2015-06-03T12:27:13.827" LastActivityDate="2015-06-03T14:51:02.577" Title="Why aren't the numbers playing nicely anymore?" Tags="|number-sequence|english|" AnswerCount="3" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="15672" PostTypeId="1" AcceptedAnswerId="15688" CreationDate="2015-05-28T19:55:38.877" Score="5" ViewCount="4425" Body="&lt;p&gt;Here's a batch of letter sequence puzzles for you. The goal is simple: Figure out which letters go in the blanks. Try it out for yourself and see what you can get.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Puzzle #1&#xA;&lt;strong&gt;K, H, D, M, D, _, _&lt;/strong&gt; (Solved By &lt;em&gt;Bailey M&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Puzzle #2:&#xA;&lt;strong&gt;A, B, E, F, G, H, I, J, N, P, _&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Puzzle #3:&#xA;&lt;strong&gt;I, E, H, O, I, I, E, I, I, E, _, _&lt;/strong&gt; (Solved By &lt;em&gt;Joe Z.&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Puzzle #4:&#xA;&lt;strong&gt;O, U, E, H, R, _, _&lt;/strong&gt; (Solved By &lt;em&gt;Joe Z.&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The puzzles are in order from easiest to hardest.&lt;/p&gt;&#xA;" OwnerUserId="11002" LastEditorUserId="11002" LastEditDate="2015-05-29T01:15:50.330" LastActivityDate="2015-05-29T01:15:50.330" Title="Fun Bunch of Letter Sequences" Tags="|letter-sequence|" AnswerCount="2" CommentCount="7" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15673" PostTypeId="1" AcceptedAnswerId="15684" CreationDate="2015-05-28T20:04:14.863" Score="13" ViewCount="4418" Body="&lt;p&gt;At a remote location, you just finished trenching a data cable across a large plot of land. The cable has 64 individual wires that are not color-coded or labeled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a wire stripper, a simple electrical continuity tester, and a label maker. There is sufficient excess at either end to allow connecting and reconnecting the individual wire ends in whatever combinations you wish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a long walk and you are tired.  What is the fewest number of trips from one end of the cable to the other required to identify and label each individual wire in the cable?&#xA;&lt;img src=&quot;https://i.stack.imgur.com/WrEIQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addenda:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The continuity tester is a sealed unit. You can't pull the battery out and leave it behind.&lt;/li&gt;&#xA;&lt;li&gt;It's not necessary to make a final trip after the wires are labeled just to clean up. If they are all labeled at both ends, the job is done.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Spoiler #1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Solve the puzzle for 2 wires and extrapolate from there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Spoiler #2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not all spoilers are helpful.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11752" LastEditorUserId="22922" LastEditDate="2016-09-15T07:22:23.530" LastActivityDate="2016-09-15T07:22:23.530" Title="Labeling wires in a bundle" Tags="|logical-deduction|strategy|" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15674" PostTypeId="2" ParentId="15673" CreationDate="2015-05-28T20:12:57.813" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Two trips (there and back).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Strategy&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;First, tie the 64 wires together randomly in 32 pairs. Next, go to the far end, randomly label any wire 1, and connect your continuity tester to it. Test which other wire is tied to it at the starting end, and label that wire 2. Then pick another wire other than 1 or 2, label it 3, and tie it to 2, so now the continuity tester is connected to 1, which is tied to 2 at the other end, which is tied to 3 at the end you're at. Now test which wire is tied to 3 at the other end, and label that 4, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you will wind up with is all 64 wires tied to each other in a continuous sequence. Then go back to the end you started at, leaving the continuity tester behind, connected to wire 1. Before you untie all the wires at the starting point, label each wire so that you know which wire was paired with which. Now with all the wires untied at the starting point, test which wire is connected to the continuity tester and label that 1. Whichever wire was in the same pair as 1, label that 2, and then tie 1 and 2 back together. Now you can find 3, because it's tied to 2 on the far end. Once you find 3, label the wire it was tied to 4, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This assumes that the resistance of the wire is small enough that the battery will still light the bulb across 12,000 km of wire.)&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-28T20:12:57.813" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15675" PostTypeId="2" ParentId="15672" CreationDate="2015-05-28T20:47:01.353" Score="4" Body="&lt;p&gt;Well, let's start off with the first sequence...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next two letters are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;C, M.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sequence is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Metric measurements! Kilometer, Hectometer, Decameter, Meter, Decimeter, Centimeter, Millimeter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-05-28T20:47:01.353" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15676" PostTypeId="2" ParentId="15667" CreationDate="2015-05-28T20:51:58.503" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1A: PASS-&gt;ASPS&lt;br&gt; &#xA; 4A: PETS-&gt;PEST&lt;br&gt; &#xA; 6A: MARIO-IO=MAR&lt;br&gt; &#xA; 7A: H(ONUS)&lt;br&gt; &#xA; 8A: I-&gt;pOLo=OIL&lt;br&gt; &#xA; 9A: DISCUSSED IT -&gt; DISGUSTED&lt;br&gt; &#xA; 11A: woma(N IS)-&gt;SIN&lt;br&gt; &#xA; 12A: LAsSO-&gt;ALSO&lt;br&gt; &#xA; 13A: had(DOC)k-&gt;COD&lt;br&gt; &#xA; 14A: RUB+Y&lt;br&gt; &#xA; 15A: SIGHT-&gt;SITE / TIES-&gt;SITE&lt;br&gt; &#xA; 2D: I(NUN)E -&gt; ENNUI&lt;br&gt; &#xA; 3D: EMERGE+NC+Y&lt;br&gt; &#xA; 4D: PARADE-A+OMEN=PR(OMEN)ADE&lt;br&gt; &#xA; 5D: E+L's = ELL&lt;br&gt; &#xA; 10D: EVEN+T&lt;br&gt; &#xA; 11D: E+M+U&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The slashed blocks are S going across and E going down, presumably in a nod to StackExchange.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2374" LastEditorUserId="2374" LastEditDate="2015-05-28T21:00:38.707" LastActivityDate="2015-05-28T21:00:38.707" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15678" PostTypeId="1" AcceptedAnswerId="15686" CreationDate="2015-05-28T21:26:51.560" Score="21" ViewCount="3386" Body="&lt;p&gt;There are two large mirrors standing upright and facing each other, both 10 meters long. They are not quite parallel: if you extended them, they would intersect at a $1^\circ$ angle. The distance from the mirrors to this intersection point is also 10 meters. See below (the angle in the picture is exaggerated to make it more readable).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aVV78.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also have a laser gun, which you plan to shoot at one of the mirrors so that the beam ricochets back and forth between them as many times as possible.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What's the greatest number of reflections you can get?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assume that a laser will reflect even if it hits the very edge of a mirror.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Source:&lt;/em&gt; &lt;a href=&quot;https://www.msri.org/attachments/media/news/emissary/EmissarySpring02.pdf&quot; rel=&quot;noreferrer&quot;&gt;MSRI Emissary, Spring 2002, Puzzle Section&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="463" LastEditDate="2015-05-28T21:42:03.873" LastActivityDate="2015-05-29T01:56:42.607" Title="Shooting a Laser Between two Mirrors" Tags="|geometry|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15680" PostTypeId="1" AcceptedAnswerId="15682" CreationDate="2015-05-28T21:39:47.273" Score="8" ViewCount="517" Body="&lt;p&gt;What are the common forms of these sentences?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Those with greater vertical displacement, achieve more damaging impact upon losing all of their gravitational potential energy.&lt;/li&gt;&#xA;&lt;li&gt;&lt;img src=&quot;https://i.stack.imgur.com/YMOaM.gif&quot; alt=&quot;YOU&amp;#39;RE IN MY SOURCE CODE!&quot;&gt;&lt;/li&gt;&#xA;&lt;li&gt;Question|noitseuQ&lt;/li&gt;&#xA;&lt;li&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;pre&gt;&#xA;&#xA;                     .--._..--.&#xA;              ___   ( _'-_  -_.'&#xA;          _.-'   `-._|  - :- |&#xA;      _.-'           `--...__|&#xA;   .-'                       '--..___&#xA;  / `._      C♂R                     \&#xA;   `. `._                       6 mi  |&#xA;     `. `._                           /&#xA;       '. `._    :__________....-----'&#xA;         `..`---'    |-_  _- |___...----..._&#xA;                     |_....--'             `.`.&#xA;               _...--'                       `.`.&#xA;          _..-'                      6km   _.'.'&#xA;       .-'                               _.'.'&#xA;       |      Rouge                     _.'.'&#xA;       |                   __....------'-'&#xA;       |     __...------''' _|&#xA;       '--'''        |-  - _ |&#xA;               _.-''''''''''''''''''-._&#xA;            _.'                        |\&#xA;          .'    2mi                  _.' |&#xA;          `._            Ventriculus  |:.'&#xA;            `._            Gaster   _.' |&#xA;               `..__                 |  |&#xA;                    `---.._.--.    _|  |&#xA;                     | _   - | `-.._|_.'&#xA;          .--...__   |   -  _|&#xA;         .'_      `--.....__ |&#xA;        .'_                 `--..__&#xA;       .'_     Clupea               `.&#xA;      .'_          Harengus           `-.&#xA;      `--..____                   2km  _`.&#xA;               ```--...____          _..--'&#xA;                     | - _ ```---.._.'&#xA;                     |   - _ |&#xA;                     |_ -  - |:&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ascii.co.uk/art/signpost&quot; rel=&quot;nofollow noreferrer&quot;&gt;[Source]&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MAD!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint for #2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quark was close because of &lt;em&gt;what he did&lt;/em&gt;, not &lt;em&gt;what he wrote&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2 for #2  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Action1: &lt;strong&gt;HELLO THERE!&lt;/strong&gt;&lt;br&gt;&#xA; Word1: &lt;em&gt;hello&lt;/em&gt;&lt;br&gt;&#xA; Action2: &lt;strong&gt;I'M HERE TOO&lt;/strong&gt;&lt;br&gt;&#xA; Word2: &lt;em&gt;We know&lt;/em&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint for #5&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Just forget about it. Otherwise you'll be driven insane.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2 for #5&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Invisible, insane&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4729" LastEditorUserId="5373" LastEditDate="2018-05-23T20:30:47.510" LastActivityDate="2018-05-23T20:30:47.510" Title="What are the common forms of these sentences?" Tags="|wordplay|rebus|" AnswerCount="2" CommentCount="10" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15681" PostTypeId="2" ParentId="15673" CreationDate="2015-05-28T21:50:31.273" Score="0" Body="&lt;p&gt;Labeling ends North and South for easy writing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will take you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 7 trips&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How do you do it?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Start at the South end. Strip the end of each wire and tie two together. Label both of these &lt;code&gt;1&lt;/code&gt; and &lt;code&gt;2&lt;/code&gt;. &#xA;&lt;br&gt;&lt;br&gt;&#xA;Go North. Strip the ends of the wires and start checking pairs of wires with the continuity tester. When the tester indicates a continuous path, put a blank label on both of the wires, so you know which ones you found. Break the connection and tie one of them to a third wire. Label that wire as &lt;code&gt;3&lt;/code&gt;.&#xA;&lt;br&gt;&lt;br&gt;&#xA;Go South. Start testing wires against &lt;code&gt;1&lt;/code&gt; and &lt;code&gt;2&lt;/code&gt; until you find your continuity again. Label the unlabeled wire &lt;code&gt;3&lt;/code&gt; and note which wire 3 was attached to. For sake of easy writing, let's say it was &lt;code&gt;1&lt;/code&gt;. Tie a wire into each &lt;code&gt;1&lt;/code&gt; &lt;code&gt;2&lt;/code&gt; and &lt;code&gt;3&lt;/code&gt; and label these as &lt;code&gt;4&lt;/code&gt; &lt;code&gt;5&lt;/code&gt; and &lt;code&gt;6&lt;/code&gt;.&#xA;&lt;br&gt;&lt;br&gt;&#xA;Go North. Relabel the blank labeled wire tied to 3 as &lt;code&gt;1&lt;/code&gt;, break that connection, and the loose blank labeled wire as &lt;code&gt;2&lt;/code&gt;. Now, test continuity between unlabeled wires and &lt;code&gt;1&lt;/code&gt;. When you find it, call that '4'. Repeat for &lt;code&gt;2 and 5&lt;/code&gt; and &lt;code&gt;3 and 6&lt;/code&gt;. Now, tie a new wire to each of the six labeled wires, and incremented labels to each. &#xA;&lt;br&gt;&lt;br&gt;&#xA;Go South. break all the connections down here and test pairings for each labeled wire. Label accordingly. You should now have 12 labeled wires. Tie off pairs and label down here so you have 12 connections and 24 labeled wires on this end. &#xA;&lt;br&gt;&lt;br&gt;&#xA;Go North. Repeat the previous step. You now have 24 connections and 48 labeled wires on this end.&#xA;&lt;br&gt;&lt;br&gt;&#xA;Go South. Repeat the previous step, tying off the remaining 16 wires to 1-16. You now have 48 connections and all wires labeled on this end.&#xA;&lt;br&gt;&lt;br&gt;&#xA;Go North. Check the pairs against 1-16, labeling accordingly. If you want to be nice and untie the wires at the other end, you could take another trip. But you're tired. Let the next shift deal with that.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12242" LastActivityDate="2015-05-28T21:50:31.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15682" PostTypeId="2" ParentId="15680" CreationDate="2015-05-28T21:50:57.417" Score="10" Body="&lt;p&gt;Partial Answer &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.Those with greater vertical displacement, achieve more damaging impact upon losing all of their gravitational potential energy.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The bigger they are, the harder they fall.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Image (From OP)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Actions speak louder than words.&lt;br&gt;&#xA; &lt;em&gt;(Note from OP: The idea behind this clue was that instead of writing words, I'd just perform an action and wave at you. The animation itself was meaningless, which is why I wasn't so happy with this clue.)&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;Question|noitseuQ  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Two sides to every question (all credit to pac on this one).   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;Confirmed, now with a better explanation.  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The fastest way to a man's heart is through his stomach.&lt;br&gt;&#xA; Ventriculus gaster is the stomach.  The symbol for male at the top (surrounded by C and R. Clupea Harengus is a herring, and Rouge is red in French. The sign posts for Rouge and Clupea (red herring as pac said) point one way (the wrong way), and the other two point the other direction. You have to go through Ventriculus to get to Cor, where Cor is Latin for heart (thanks to OP for this note). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;I'm just flat out impressed with this one.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Out of sight, out of mind.&lt;/strong&gt;&lt;br&gt;&#xA;&#xA; The clue is out of sight, and if someone is mad, then they could be said to be out of their mind.&#xA; &lt;em&gt;(Note from OP: Perhaps because this &lt;a href=&quot;http://www.oxforddictionaries.com/definition/english/mad&quot; rel=&quot;nofollow&quot;&gt;usage of mad&lt;/a&gt; is &quot;chiefly British&quot; is why this one took so long to solve.)&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;6&quot;&gt;&#xA;&lt;li&gt;Not sure if this is intentional or not, but &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hidden in plain sight. Found by editing. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="9951" LastEditDate="2015-06-09T18:23:11.050" LastActivityDate="2015-06-09T18:23:11.050" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="15684" PostTypeId="2" ParentId="15673" CreationDate="2015-05-28T22:24:30.603" Score="11" Body="&lt;p&gt;The wires can be labelled using a total of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; two trips.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We start by forming the 64 wires into 10 groups, of size 11, 10, 9, 8, 7, 6, 5, 4, 3, and 1. All the wires in each group are twisted together. We will attach a temporary label to each wire, recording the size of the group it is in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we walk to the other end. Using the continuity tester, we can determine precisely which pairs of wires are twisted together, so for each wire, we can determine the size of the group it is in, and therefore its temporary label.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we will assemble the wires into 11 different groups, of varying sizes.  The temporary labels of the wires in the groups will be&#xA;\begin{gather*}&#xA;\{11,7,6,5,4\},\{11,10\},\{11,10,9\},\{11,10,9,8\},\{11,10,9,8,7\},\{11,10,9,8,7,6\},\\\{11,10,9,8,7,6,5\},\{11,10,9,8,7,6,5,4\},\{11,10,9,8,7,6,5,4,3\},\{11,10,9,8,7,6,5,4,3,1\},\{11,10,9,8,3\}.&#xA;\end{gather*}&#xA;Note that no two groups are identical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each wire is then given a permanent label, and for each wire, we write down:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the wire's permanent label,&lt;/li&gt;&#xA;&lt;li&gt;the wire's temporary label, and&lt;/li&gt;&#xA;&lt;li&gt;the temporary labels the wires now grouped with that wire.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We make a trip back to the other end and separate the wires. Using the continuity tester, we can see which wires are connected to which others. For each wire, we can see its temporary label, as well as the temporary labels of all other wires attached to it. This is sufficient to determine the corresponding permanent label.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-05-28T23:56:15.700" LastActivityDate="2015-05-28T23:56:15.700" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15685" PostTypeId="2" ParentId="15673" CreationDate="2015-05-28T22:47:05.333" Score="0" Body="&lt;p&gt;3 trips (one way).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make 8 groups of 8 wires.  Tie the ends of each group together so they connect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cross (first time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Identify which wires are connected to which others, one pairing at a time.  Since there are only 8 groups, this will take an average of about 4 checks per wire.  Label all these wires AA, AB, AC, ... HH with all the wires in each group having the same first letter, different second letter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Connect the groups of wires at this end, based on the second letter.  So AA connects to BA, CA, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cross (second time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disconnect all wires, but keep the groupings.  Pick a group, label the wires in that group with 2-digits as 11 through 18.  Pick a second group.  Find the wire that connects to 11.  Label that as 21.  Label the rest as 22 through 28, based on which wire in the first group they connect to.  Continue with all wires.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sets of pairings is consistent, so all we have to do now is determine which first letter corresponds with which first digit and which second letter corresponds with which second digit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now connect 11 and 21.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then connect 12, 33, 44.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then connect 13, 55, 66, 77.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then connect 51, 62, 73, 88, 84.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cross (third time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disconnect all wires.  Test.  There will be four groups of wires that connect, and the groups will have sizes 2,3,4,5.  There will be one wire in the group with 2 that has the same first letter as one of the wires in the group with 3.  The one in the smaller group corresponds to 11, the other to 12.  The other one in the smaller group is 21.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wire 13 is the one in the group with 4 that has the same first letter as 11 and 12.  Wire 33 is the one in the group with 3 that has the same second letter as wire 13.  Wire 44 is the remaining one in the group with 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wire 51 is the one in the group with 5 that has the same second letter as 11.  Likewise 62 has the same second letter as 12, and 73 has the same second letter as 13.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can now identify 55, 66, 77 as the remaining wires in the group with 4 that have the same first letters as 51, 62, 73, respectively.  Wire 84 has the same second letter as 44, and 88 is the last wire in the group with five.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have now identified which first letter is which number and which second letter is which number.  Relabel all the wires with numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are done.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2015-05-28T22:47:05.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15686" PostTypeId="2" ParentId="15678" CreationDate="2015-05-28T23:10:08.190" Score="29" Body="&lt;p&gt;To solve this, we first express the problem in a new way: Instead of having a laser bouncing between two mirrors, we have a laser moving straight through a different space; to accomplish this, we draw what you would see if you were standing in the setup - that is, where the mirrors are, we place a new, reflected image of the whole space! That is, we make copies of the $1^{\circ}$ wedge by mirroring it over the rays that bound it. The result looks something like this:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/O2dqs.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;where the line segments represent the mirrors and the red line is the laser. This is basically what you'd see if you were standing inside such a setup (although you'd see many copies of yourself too). We may regard this as a sort of &quot;covering space&quot; for the old one as each point in the old one is copied many times into the new one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes the problem simple: How many &lt;em&gt;consecutive&lt;/em&gt; line segments can the red line cross? Well, let's call the distance from the center of the circle to the laser $d$. And, for simplicitly, lets call the radius of the inner circle $1$ and the outer circle $2$, since we only need to preserve the ratio between those measurements (which is 10m : 20m originally). Then, a laser crosses a line segment measuring an angle of $\theta$ to it if:&#xA;$$\sin(\theta)\leq d\leq 2\sin(\theta)$$&#xA;We thus require that the smallest value of $2\sin(\theta)$ of any line we cross is at least equal to than the greatest value of $\sin(\theta)$ in the lines that we cross - otherwise, no suitable $d$ exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this inequality, suppose we have that the leftmost line $\alpha$ to the line, and the rightmost line is at an angle of $\beta$ (which will be greater than $\alpha$). We may assume that $|\alpha|\geq|\beta|$, as the two are related by reflection. Moreover, we may then write $\alpha=(n+c)^{\circ}$ where $n$ is an integer and $-\frac{1}2\leq c\leq \frac{1}2$ - meaning $90^{\circ}+c$ will be the maximum $\sin(\theta)$ to appear. Then, all we need to satisfy for the intervals for $d$ to intersect is:&#xA;$$\sin(90+c)\leq 2\sin(\alpha)$$&#xA;$$\sin(90+c)\leq 2\sin(\beta)$$&#xA;Now, I'm &lt;em&gt;sure&lt;/em&gt; there's an elegant way to solve this. I'm kind of a hack though, so we'll just make some guesses: Suppose that $c$ were $\frac{1}2$. Then we can find that the integer $n$ satisfying &#xA;$$\sin(90+\frac{1}2)\leq 2\sin(90+n+\frac{1}2)$$&#xA;are exactly those in the interval $[-60,59]$, meaning we can, from any point, cross $60$ to either side of it (as $c=\frac{1}2$ is the most extreme value possible) - for a maximum of $121$ reflections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, letting $c$ be zero, it's a fairly common fact from highschool trigonometry that $\sin(30^{\circ})=\sin(150^{\circ})=\frac{1}2$ that $\alpha=30^{\circ}$ meaning that we can hit every mirror making an angle between $30^{\circ}$ and $150^{\circ}$ with the laser - this achieves the maximum of $121$ reflections. To do this, we do as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Aim the laser at the outer edge of one of the mirrors, impacting the mirror at a $30^{\circ}$ angle. It will reflect $121$ times.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A picture of this would probably just look like a lot of red lines. However, the same logic applies if the angle between the mirrors were $5^{\circ}$ and predicts $25$ reflections. That looks like this:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/nYPWW.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;Notice that it hits one mirror perpendicularly, and hence retraces its steps. In other words, some advice:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Don't look in the direction you point the laser.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4765" LastEditorUserId="4765" LastEditDate="2015-05-29T01:56:42.607" LastActivityDate="2015-05-29T01:56:42.607" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="15687" PostTypeId="2" ParentId="15671" CreationDate="2015-05-28T23:21:25.907" Score="2" Body="&lt;p&gt;With the sole exceptions of $1, 14, 15, 16, 18, 19,$ the numbers up to $40$ are divided as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; all those ending in 4,6,9,0 in the first group; all those ending in 2,5,8 in the second group; all those ending in 1,3,7 in the third group.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While each of these exceptions is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; in the same group as double itself ($14$ with $28$, $15$ with $30$, and so on) - except for $1$, which is in the $1$st group (makes sense: that $1$ is so $1$-y, you know).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So my answer to the official question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the numbers divided themselves up &lt;strong&gt;according to their congruence class modulo 10 (i.e. final digit)&lt;/strong&gt;, except for the number $1$, who had to be in the $1$st group to preserve $1$-iness, and the teenagers, who - just to be contrary - joined a group according to the congruence class of &lt;strong&gt;double their value instead of themselves&lt;/strong&gt;. The latter could be explained either by those teenagers being so full of themselves they think they're worth twice as much as they are, or by interpreting $2n$ as the mother of $n$ and saying they want to stick close to their mothers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And my answer to the bonus question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; tell all the teenagers' mothers to type &quot;problem teenager&quot; into Google, absorb all the wisdom of the internet (Parenting Stack Exchange, perhaps?), educate their teenagers in how not to squabble (it was those teenagers who started the whole argument anyway), and then the rift between the adult numbers will automatically be healed too.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;OK, so I'm a rubbish number therapist when it comes to the bonus question. But at least the answer to the official question works and is in the spirit of the OP!&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-28T23:21:25.907" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15688" PostTypeId="2" ParentId="15672" CreationDate="2015-05-29T00:11:00.697" Score="5" Body="&lt;p&gt;Sequence 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The second letters of each ordinal number: first, second, third, fourth, fifth, ... &lt;p&gt;&#xA; The letters afterward are &lt;strong&gt;L&lt;/strong&gt; and &lt;strong&gt;W&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sequence 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The second letters of each day of the week. This one's &lt;a href=&quot;https://puzzling.stackexchange.com/questions/10530/a-final-batch-of-four-sequence-puzzles&quot;&gt;appeared on the site before&lt;/a&gt;. &lt;p&gt;&#xA; The letters afterward are &lt;strong&gt;A&lt;/strong&gt; and &lt;strong&gt;U&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More to come later.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2015-05-29T00:18:32.007" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15690" PostTypeId="1" AcceptedAnswerId="16079" CreationDate="2015-05-29T02:53:00.077" Score="21" ViewCount="1002" Body="&lt;p&gt;I have devised a rule that assigns a positive integer to each English word. Here is a list of words and the integers the rule assigns them.&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th&gt;Word&lt;/th&gt;&#xA;&lt;th&gt;Integer&lt;/th&gt;&#xA;&lt;th&gt;Word&lt;/th&gt;&#xA;&lt;th&gt;Integer&lt;/th&gt;&#xA;&lt;th&gt;Word&lt;/th&gt;&#xA;&lt;th&gt;Integer&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;atmosphere&lt;/td&gt;&#xA;&lt;td&gt;3&lt;/td&gt;&#xA;&lt;td&gt;a&lt;/td&gt;&#xA;&lt;td&gt;1&lt;/td&gt;&#xA;&lt;td&gt;ab&lt;/td&gt;&#xA;&lt;td&gt;3&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;day&lt;/td&gt;&#xA;&lt;td&gt;2&lt;/td&gt;&#xA;&lt;td&gt;beret&lt;/td&gt;&#xA;&lt;td&gt;1&lt;/td&gt;&#xA;&lt;td&gt;ad&lt;/td&gt;&#xA;&lt;td&gt;1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;eclipse&lt;/td&gt;&#xA;&lt;td&gt;7&lt;/td&gt;&#xA;&lt;td&gt;cheaters&lt;/td&gt;&#xA;&lt;td&gt;6&lt;/td&gt;&#xA;&lt;td&gt;bag&lt;/td&gt;&#xA;&lt;td&gt;1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;geology&lt;/td&gt;&#xA;&lt;td&gt;4&lt;/td&gt;&#xA;&lt;td&gt;coins&lt;/td&gt;&#xA;&lt;td&gt;2&lt;/td&gt;&#xA;&lt;td&gt;banana&lt;/td&gt;&#xA;&lt;td&gt;5&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;heavy&lt;/td&gt;&#xA;&lt;td&gt;3&lt;/td&gt;&#xA;&lt;td&gt;dad&lt;/td&gt;&#xA;&lt;td&gt;5&lt;/td&gt;&#xA;&lt;td&gt;bann&lt;/td&gt;&#xA;&lt;td&gt;3&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;inflation&lt;/td&gt;&#xA;&lt;td&gt;1&lt;/td&gt;&#xA;&lt;td&gt;fad&lt;/td&gt;&#xA;&lt;td&gt;2&lt;/td&gt;&#xA;&lt;td&gt;hear&lt;/td&gt;&#xA;&lt;td&gt;4&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;neanderthal&lt;/td&gt;&#xA;&lt;td&gt;3&lt;/td&gt;&#xA;&lt;td&gt;hectares&lt;/td&gt;&#xA;&lt;td&gt;6&lt;/td&gt;&#xA;&lt;td&gt;her&lt;/td&gt;&#xA;&lt;td&gt;3&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;parsimoniously&lt;/td&gt;&#xA;&lt;td&gt;4&lt;/td&gt;&#xA;&lt;td&gt;highways&lt;/td&gt;&#xA;&lt;td&gt;1&lt;/td&gt;&#xA;&lt;td&gt;males&lt;/td&gt;&#xA;&lt;td&gt;1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;span&lt;/td&gt;&#xA;&lt;td&gt;1&lt;/td&gt;&#xA;&lt;td&gt;I&lt;/td&gt;&#xA;&lt;td&gt;5&lt;/td&gt;&#xA;&lt;td&gt;prevent&lt;/td&gt;&#xA;&lt;td&gt;1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;zealous&lt;/td&gt;&#xA;&lt;td&gt;8&lt;/td&gt;&#xA;&lt;td&gt;laziness&lt;/td&gt;&#xA;&lt;td&gt;2&lt;/td&gt;&#xA;&lt;td&gt;snap&lt;/td&gt;&#xA;&lt;td&gt;1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;&lt;/td&gt;&#xA;&lt;td&gt;&lt;/td&gt;&#xA;&lt;td&gt;lid&lt;/td&gt;&#xA;&lt;td&gt;1&lt;/td&gt;&#xA;&lt;td&gt;thorny&lt;/td&gt;&#xA;&lt;td&gt;1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;&lt;/td&gt;&#xA;&lt;td&gt;&lt;/td&gt;&#xA;&lt;td&gt;transportation&lt;/td&gt;&#xA;&lt;td&gt;2&lt;/td&gt;&#xA;&lt;td&gt;&lt;/td&gt;&#xA;&lt;td&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;The rule can be described in a single sentence.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;What is my rule?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: I have chosen some additional words and computed the numbers the rule assigns them. Also, find some hints below.&lt;/p&gt;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; It is unlikely you will find a word in the dictionary that is assigned a value larger than 10. However, the rule is defined for arbitrary strings of English letters, and it is possible write down strings with arbitrarily large output. To get an output of 11 requires a string of at least 16 characters, and the 16 character string with output 11 that comes first alphabetically is &amp;quot;dzzzzzzzzzzzzzzz&amp;quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; It might be useful to start by figuring out what the words assigned 1 have in common.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The US coins are: penny (1¢), nickel (5¢), dime (10¢), quarter (25¢), half dollar (50¢), and dollar (100¢). This is relevant.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Hint 4:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The rule assigns anagrams the same number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Hint 5:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The rule does &lt;strong&gt;not&lt;/strong&gt; depend on any of the following:&lt;br&gt;Case, font, keyboard layout, letter frequency, word meaning&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Hint 6:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The longest string that the rule assigns 1 has one hundred letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="69582" LastEditDate="2021-01-26T17:11:28.580" LastActivityDate="2021-01-26T17:11:28.580" Title="Find the rule that assigns numbers to words" Tags="|word|pattern|" AnswerCount="1" CommentCount="17" ContentLicense="CC BY-SA 4.0" />
  <row Id="15691" PostTypeId="2" ParentId="15680" CreationDate="2015-05-29T03:24:03.857" Score="4" Body="&lt;p&gt;Partial answer:&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;[GIF of a guy waving]&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The text description of the GIF says &quot;YOU'RE IN MY SOURCE CODE!&quot;  Is it an Easter Egg?  Or perhaps it refers to how many programming language tutorials start off with a basic &quot;Hello World!&quot; program.  The OP did &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15680/what-are-the-common-forms-of-these-sentences/15691#comment48631_15680&quot;&gt;comment&lt;/a&gt; that this doesn't work as well as the others.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;Question|noitseuQ&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are two sides to every question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;[ASCII art signpost]&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Like Aggie Kidd said, this has C-MALE-R and Ventriculus Gaster is the stomach.  It also has &quot;Rouge&quot; which means red and &quot;Clupea Harengus&quot; which is Atlantic herring, so some sort of red herring is involved here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;MAD! in a spoiler&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Anger beneath the surface?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-05-29T21:52:42.873" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15692" PostTypeId="2" ParentId="15673" CreationDate="2015-05-29T08:36:06.367" Score="1" Body="&lt;p&gt;Best attempt:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2 trips&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Trying hard not to look at the other answers!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First attempt: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;6 trips&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First, label every wire at one end A1-64. Connect together all of A1-32, and do the same for A33-64. Write a 0 on labels A1-32, and a 1 on labels A33-64.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Walk to the other end (1)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Label up the wires at this end, B1-64. Test connectivity of B1 and B2, B1 and B3, etc. Label all the ones that connect to B1 with a 0, and the rest with a 1. Now connect half of the 0-labelled wires together with half of the 1-labelled wires, and then connect all the other wires together. Write down which ones you have connected up, and add a second digit of 0 after the first digit for all the wires now connected to B1, and a 1 for the wires in the other group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Walk to the other end (2)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeat the tests, this time adding a 0 after the first digit for all those that connect to A1, and a 1 for all those that connect to the other group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of these passes gives us 1 additional digit. 64 wires requires 6 bits, so we will need a further 4 passes. At the end, all the wires are labelled with a 6 bit code and we are done.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;But I had an idea. During the second step where we are testing which wires are connected, we can use the information about how many wires are connected, as well as which wires they are connected to. So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second attempt.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2 trips&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Label the wires A1-64. Connect together the following groups: A1-2, A3-5, A6-9, A10-14, A15-20, 21-27, 28-35, 36-44, 45-54, and 55-64 you can leave unconnected (I originally started with bundles of 64, 32 etc, because binary is always the answer to puzzles, but then realised that reduces the number of groups I can have). Label each wire in a group with its group number: 1 for A1-2, 2 for A3-5 etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Walk to the other end (1)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now find those groups. The 2-group must be B1 and 2, the 3-group must be 3-5, etc, and the unconnected ones must be 55-64. There were 10 groups originally, and &#xA;we can connect one of each group into a new group - so B1,B3,B6,B10,B15,B21,B28,B36,B45,B55. Since we only needed 1 bit of information to identify A1 vs A2, we can do the same for the next wire in the latter 9 groups, so B4, A7, A11... - continue until you have a new group of 9 wires, another of 8 wires etc. This will end at a group of 2, B54 and B64. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Walk to the other end (2)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disconnect the wires. One of wires A1 and A2 will be connected to other wires - Relabel the one that is connected as B1, and the one that is not as B2. B1 is connected to 9 other wires; for each of the original A groups, find out which one and relabel it with the B number it must be connected to (B3, B6 etc). Continue through all of the groups. All your wires are now labelled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Go home and sleep, your dreams will be full of numbers and tape.&lt;/p&gt;&#xA;" OwnerUserId="12816" LastActivityDate="2015-05-29T08:36:06.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15693" PostTypeId="2" ParentId="15664" CreationDate="2015-05-29T09:18:29.273" Score="24" Body="&lt;p&gt;&lt;strong&gt;K&lt;/strong&gt; = &lt;strong&gt;C&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;L&lt;/strong&gt; = &lt;strong&gt;I&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;M&lt;/strong&gt; = &lt;strong&gt;G&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;N&lt;/strong&gt; = &lt;strong&gt;G&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;O&lt;/strong&gt; = &lt;strong&gt;I&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the squares represent:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the letters that spell out the phrase &quot;&lt;strong&gt;ONE PLUS FIVE IS SIX&lt;/strong&gt;&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Specifically, each square is constructed by:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. Drawing each uppercase letter, in white on black&lt;br&gt;&#xA; 2. Cropping it down to the minimal bounding box&lt;br&gt;&#xA; 3. Breaking up the resulting black shapes and rearranging randomly&lt;br&gt;&#xA; 4. Collapsing duplicate shapes into one, with a number indicating how many of that shape you need to use  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Via reconstruction, the original would have looked something like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/YXmlV.png&quot; alt=&quot;one plus five&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="5208" LastEditDate="2015-05-29T13:23:08.440" LastActivityDate="2015-05-29T13:23:08.440" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15694" PostTypeId="1" AcceptedAnswerId="15695" CreationDate="2015-05-29T09:44:20.213" Score="3" ViewCount="18256" Body="&lt;p&gt;We want to have 7 pieces by only cutting our tasty cake with three straight and vertical cuts (maybe done with a katana :D),&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I dont require to have all the pieces of the same size. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Sorry if it's repeated, but I couldnt find it here.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Very similar to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2827/cutting-a-cake-into-8-pieces&quot;&gt;another puzzle&lt;/a&gt;, but in this case we can not rely on the fact of making 3D cuts. All the cuts must be vertical.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; THANKS TO @n00b, &quot;Apparently there is a name for this type of puzzle; en.wikipedia.org/wiki/Lazy_caterer's_sequence&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10365" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-05-29T16:34:28.273" Title="Cutting a cake into 7 pieces with 3 straight cuts - NO 3D" Tags="|geometry|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15695" PostTypeId="2" ParentId="15694" CreationDate="2015-05-29T09:48:39.560" Score="8" Body="&lt;p&gt;That's some tasty cake right here. I want the middle piece.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bEGb7.png&quot; alt=&quot;Tasty&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5265" LastActivityDate="2015-05-29T09:48:39.560" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15696" PostTypeId="2" ParentId="15588" CreationDate="2015-05-29T13:49:53.287" Score="1" Body="&lt;p&gt;My answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water, Wood, Fire&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because they have to beat the real rock paper scissor as well as their figurative counter part which I'll keep in square brackets&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Rock] said: If I spend enough time around scissors, they’ll have a hard time moving eventually.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water - the scissors will rust if left in water. Also [Rock] = Water loses to [Paper] = Wood.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;[Paper] said: I never met a rock too hard for me to dig into and break through.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Wood - some tree roots can dig through stone. [Paper] = Wood loses to [Scissors] = Fire&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;[Scissors] said: I eat paper pretty quickly; frankly I prefer to savor something less processed. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fire - fire eats through wood pretty quickly indeed. [Scissors] = Fire lose to [Rock] = Water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It all fits rather well I think.&lt;/p&gt;&#xA;" OwnerUserId="9907" LastActivityDate="2015-05-29T13:49:53.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15698" PostTypeId="2" ParentId="15694" CreationDate="2015-05-29T13:59:15.763" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cut the cake into 4 pieces with two perpendicular cuts, meeting at the center. Move the slices so they line up, then slice once more down the middle. Eat one piece and now you have 7 pieces.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2078" LastActivityDate="2015-05-29T13:59:15.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15699" PostTypeId="2" ParentId="9875" CreationDate="2015-05-29T14:20:47.173" Score="0" Body="&lt;p&gt;I have pretty much no idea, but i would like to try this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 5 9 13 17 21 25 29 33 37 41 45 49 2 6 10 14 18 22 26 30 34 38 42 46 50 3 7 11 15 19 23 27 31 35 39 43 47 51 4 8 12 16 20 24 28 32 36 40 44 48 52&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I have no idea on what n is though... I'll edit it when I know.&lt;/p&gt;&#xA;" OwnerUserId="9736" LastEditorUserId="9736" LastEditDate="2015-05-29T14:27:53.137" LastActivityDate="2015-05-29T14:27:53.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15700" PostTypeId="1" AcceptedAnswerId="15701" CreationDate="2015-05-29T14:22:51.727" Score="22" ViewCount="1294" Body="&lt;p&gt;Acme has released a brand new safe, secured with electronic 10-button keypad with the digits 0 through 9, with an X-length combination required to unlock. However, due to laziness, the keypad's programmer decides that, instead of requiring a new attempt each time, the safe will only consider the last $x$ button presses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, with $x=2$, if I were to press $1234$, the safe would evaluate whether $12$, $23$, and $34$ were valid combinations, while a traditional keypad safe would only evaluate $12$ and $34$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all values $x$, the worst case would be to try all combinations in serial, resulting in $10^x$ combinations of $x$ button presses, or $x \times 10^x$ presses. With $x=4$, we'd end up pressing this keypad up to $40,000$ times!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the &lt;em&gt;best-case number of button presses to attempt all possible combinations&lt;/em&gt; for a combination of length $x$, and what is the list of button presses for $x=2$?&lt;/p&gt;&#xA;" OwnerUserId="8944" LastActivityDate="2015-06-26T20:22:27.850" Title="The Faulty Keypad Safe" Tags="|mathematics|pattern|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15701" PostTypeId="2" ParentId="15700" CreationDate="2015-05-29T14:37:29.173" Score="25" Body="&lt;p&gt;The way to do it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;a &lt;a href=&quot;http://en.wikipedia.org/wiki/De_Bruijn_sequence&quot;&gt;De Bruijn sequence&lt;/a&gt;. Basically, it's a sequence $B(k,n)$ that contains all sequences of length $n$ made of $k$ different characters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The number of keypresses for the length $x$ is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$10^x + (x - 1)$. A De Bruijn sequence is cyclic (end connects back to start) with length $k^n$, so we just need to add the starting $x - 1$ keypresses to the end.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The keypresses for $x = 2$ are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;00102030405060708091121314151617181922324252627282933435363738394454647484955657585966768697787988990&lt;/code&gt;, as generated by &lt;a href=&quot;http://en.wikipedia.org/wiki/De_Bruijn_sequence#Algorithm&quot;&gt;the algorithm on Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12830" LastActivityDate="2015-05-29T14:37:29.173" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15702" PostTypeId="1" AcceptedAnswerId="15713" CreationDate="2015-05-29T14:50:26.357" Score="7" ViewCount="1172" Body="&lt;p&gt;I'll send you on a treasure hunt,&lt;br&gt;&#xA;Hopefully you don't mind.&lt;br&gt;&#xA;The clues may not be right up front,&lt;br&gt;&#xA;Maybe they are behind.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The clues you need are down below,&lt;br&gt;&#xA;And hidden in plain sight.&lt;br&gt;&#xA;I do not wish contention to sow,&lt;br&gt;&#xA;So solve this by your might.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tell me the secret message&lt;br&gt;&#xA;To liven up your day.&lt;br&gt;&#xA;It hopefully relieves your stresses,&lt;br&gt;&#xA;Here at the end of May.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{|l|c|c|c|}  &#xA;\hline&#xA;1. &amp;amp; 1 &amp;amp; 2 &amp;amp; 1\\&#xA;\hline&#xA;2. &amp;amp; 2 &amp;amp; 13 &amp;amp; 2 \\&#xA;\hline&#xA;3. &amp;amp; 1 &amp;amp; 19 &amp;amp; 1 \\&#xA;\hline&#xA;4. &amp;amp; 1 &amp;amp; 8 &amp;amp; 2 \\&#xA;\hline&#xA;5. &amp;amp; 3 &amp;amp; 4 &amp;amp; 1 \\&#xA;\hline&#xA;6. &amp;amp; 1 &amp;amp; 5 &amp;amp; 1 \\&#xA;\hline&#xA;\end{array}  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&#xA;&#xA;&#xA;&#xA;&#xA;&lt;/p&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="11244" LastEditDate="2015-05-29T14:56:00.807" LastActivityDate="2015-05-29T15:58:40.477" Title="A Fun(ny) Treasure Hunt" Tags="|riddle|rhyme|humor|treasure-hunt|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15703" PostTypeId="1" AcceptedAnswerId="15704" CreationDate="2015-05-29T14:53:35.910" Score="12" ViewCount="1506" Body="&lt;p&gt;Some said that &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15664/identify-the-missing-symbols&quot;&gt;my last puzzle&lt;/a&gt; was difficult, and the same thing for &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15658/create-a-word-search&quot;&gt;the one before it&lt;/a&gt;, so here is an easier one, to celebrate Friday I guess (although it's almost Saturday in some places).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are directions to obtain the 4-digit number:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;96321112666574122223698741578963214123698754442226632147&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the 4-digit number?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-07-14T02:46:28.887" Title="Find the 4-digit number" Tags="|enigmatic-puzzle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15704" PostTypeId="2" ParentId="15703" CreationDate="2015-05-29T15:00:29.177" Score="11" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2685&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The numbers represent the directions of the numpad. The &quot;5&quot; in the sequence are used to separate the four numbers of the puzzle. &#xA; &lt;img src=&quot;https://i.stack.imgur.com/jPZcA.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA; The red dots represent the starting point for each number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is this number?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It's supposed to be the number of this question in Puzzling! Note that this might not be accurate, because the site stats don't always refresh immediately. At this moment, I see that the question counter displays &quot;2684&quot;, probably it's what JLee saw too before writing his question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-05-29T15:51:08.240" LastActivityDate="2015-05-29T15:51:08.240" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15706" PostTypeId="1" AcceptedAnswerId="15709" CreationDate="2015-05-29T15:22:56.923" Score="17" ViewCount="5374" Body="&lt;p&gt;Given a 10' x 10' hole which is very deep, how do you walk across the hole using two boards that are each 9' long?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't jump, pole vault across, nail the boards together, or go around the hole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can start on any side you like, but once you start to cross, you must stay on the two boards. No stepping off on the sides.&lt;/p&gt;&#xA;" OwnerUserId="12831" LastEditorUserId="5208" LastEditDate="2016-08-05T02:37:51.880" LastActivityDate="2016-08-05T02:37:51.880" Title="How do you walk across a 10' x 10' hole using two 9' boards?" Tags="|geometry|situation|" AnswerCount="4" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="15707" PostTypeId="2" ParentId="15706" CreationDate="2015-05-29T15:25:40.963" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Use one board to connect the south edge to the west-ward/east-ward edge. Use the second board and connect it to the first board to reach the end.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-05-29T15:59:11.727" LastActivityDate="2015-05-29T15:59:11.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15709" PostTypeId="2" ParentId="15706" CreationDate="2015-05-29T15:27:24.607" Score="16" Body="&lt;p&gt;You put the first board &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; at a 45 degree angle across one corner&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;of the hole. Then you put the other one &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; with one end on the first board and the other end on the side of the hole.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks to Julian Rosen and Geobits for their insight on how my drawing was bad. I updated it to be more accurate:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/u2On5.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10400" LastEditorUserId="1477" LastEditDate="2015-05-31T08:22:17.557" LastActivityDate="2015-05-31T08:22:17.557" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="15710" PostTypeId="2" ParentId="15706" CreationDate="2015-05-29T15:28:13.910" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Like this -- you put one board at an angle, then the other across.&lt;br&gt;&#xA;&#xA;  &lt;img src=&quot;https://i.stack.imgur.com/O9cjE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10625" LastActivityDate="2015-05-29T15:28:13.910" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15711" PostTypeId="1" AcceptedAnswerId="15722" CreationDate="2015-05-29T15:53:18.673" Score="3" ViewCount="740" Body="&lt;p&gt;I would like feedback on this riddle I made up. I would classify it with a moderately difficult rating, but I could be wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Of tiny little planks from end to end when told are sharp and flat.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;When moving: forward or not at all, and family to us all, at that!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Please let me know how easy it was to solve, and if you needed the hint(s).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HINT 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; plank = planck&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;HINT 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; family = relative&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11681" LastEditorUserId="11681" LastEditDate="2015-05-29T16:50:19.563" LastActivityDate="2015-05-29T17:30:42.813" Title="A quick, one-liner riddle" Tags="|riddle|" AnswerCount="7" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15712" PostTypeId="2" ParentId="15711" CreationDate="2015-05-29T15:55:43.497" Score="4" Body="&lt;p&gt;I think the answer is a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Book&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The little boards represent the pages of the book, which is a stack of pages, from beginning to end. The single sheets are sharp (you can hurt your finger with them) and flat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-05-29T15:57:48.283" LastActivityDate="2015-05-29T15:57:48.283" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15713" PostTypeId="2" ParentId="15702" CreationDate="2015-05-29T15:58:40.477" Score="6" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My imaginary friends think you have mental problems&lt;br&gt;&#xA; (Gee, thanks)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Looking in the source code, we can find links pointing to 6 xkcd comics:&lt;br&gt;&#xA; 1. &lt;a href=&quot;http://xkcd.com/539/&quot;&gt;http://xkcd.com/539/&lt;/a&gt;&lt;br&gt;&#xA; 2. &lt;a href=&quot;http://xkcd.com/410/&quot;&gt;http://xkcd.com/410/&lt;/a&gt;&lt;br&gt;&#xA; 3. &lt;a href=&quot;http://xkcd.com/610/&quot;&gt;http://xkcd.com/610/&lt;/a&gt;&lt;br&gt;&#xA; 4. &lt;a href=&quot;http://xkcd.com/308/&quot;&gt;http://xkcd.com/308/&lt;/a&gt;&lt;br&gt;&#xA; 5. &lt;a href=&quot;http://xkcd.com/706/&quot;&gt;http://xkcd.com/706/&lt;/a&gt;&lt;br&gt;&#xA; 6. &lt;a href=&quot;http://xkcd.com/1171/&quot;&gt;http://xkcd.com/1171/&lt;/a&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Then we can use the grid:&lt;br&gt;&#xA; The first column tells you which comic to look at&lt;br&gt;&#xA; The second column tells you which panel to look at&lt;br&gt;&#xA; The third column tells you which word to start at&lt;br&gt;&#xA; The fourth column tells you how many works to read&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So...  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first comic gives up the single word &quot;MY&quot;&lt;br&gt;&#xA; The second comic gives up the words &quot;IMAGINARY FRIENDS&quot; (counting hyphens as two words)&lt;br&gt;&#xA; The third comic gives up the work &quot;THINK&quot;&lt;br&gt;&#xA; The fourth comic gives up the words &quot;YOU HAVE&quot;&lt;br&gt;&#xA; The fifth comic gives up the word &quot;MENTAL&quot;&lt;br&gt;&#xA; And finally the sixth comic gives up the word &quot;PROBLEMS&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Giving the final message as...  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MY IMAGINARY FRIENDS THINK YOU HAVE MENTAL PROBLEMS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6205" LastActivityDate="2015-05-29T15:58:40.477" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15714" PostTypeId="2" ParentId="15711" CreationDate="2015-05-29T15:58:49.230" Score="3" Body="&lt;p&gt;It could be a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Song&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Tiny little boards could be referring to the five lines on the staff, and when the notes of a song are &quot;told&quot; they can be sharp or flat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11727" LastActivityDate="2015-05-29T15:58:49.230" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15715" PostTypeId="2" ParentId="15711" CreationDate="2015-05-29T15:59:59.220" Score="6" Body="&lt;p&gt;It's a...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Piano (or possibly referring more specifically to &quot;ebony&quot;, as in the black keys)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because it's referencing&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sharp and flat notes are played by playing the &quot;black&quot; keys on a piano, which can be thought of as tiny little boards that run all along the piano (end to end).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9289" LastActivityDate="2015-05-29T15:59:59.220" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15716" PostTypeId="2" ParentId="15424" CreationDate="2015-05-29T16:11:14.590" Score="6" Body="&lt;p&gt;Decrypting the PGP file from the green channel using the DTMF encoded RSA key from the red channel results in a BZ2 compressed TAR file containing the following three images:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://media.jarnbjo.de/sigint-challenge-1/allie.jpeg&quot; rel=&quot;nofollow&quot;&gt;allie.jpeg&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://media.jarnbjo.de/sigint-challenge-1/bitsy.tiff&quot; rel=&quot;nofollow&quot;&gt;bitsy.tiff&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://media.jarnbjo.de/sigint-challenge-1/porsche.png&quot; rel=&quot;nofollow&quot;&gt;porsche.png&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Decrypting the PGP message can be done with the following Java class (BouncyCastle libraries must be included in the project):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://media.jarnbjo.de/sigint-challenge-1/Decrypt.java&quot; rel=&quot;nofollow&quot;&gt;Decrypt.java&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Quark already pointed it out, but the full resolution thumbnail contained in the bitsy.tiff image can be found here:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://media.jarnbjo.de/sigint-challenge-1/bitsy-tn.jpg&quot; rel=&quot;nofollow&quot;&gt;bitsy-tn.jpg&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Quark already added the image to his answer, but for some reason he only added a much smaller and difficult to read version (a thumbnail of the thumbnail :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LeppyR64 asked for the raw decrypted data, so here we go. Using the Bouncy Castle PGP API, I get the following raw output:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://media.jarnbjo.de/sigint-challenge-1/green.dat&quot; rel=&quot;nofollow&quot;&gt;green.dat&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I used the documentation in &lt;a href=&quot;https://www.ietf.org/rfc/rfc4880.txt&quot; rel=&quot;nofollow&quot;&gt;RFC 4880&lt;/a&gt; to decode the message header manually:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;AE - literal data packet, packet length encoded with 4 bytes&lt;/li&gt;&#xA;&lt;li&gt;00 05 83 13 - 361235 bytes packet length&lt;/li&gt;&#xA;&lt;li&gt;62 - data packet contains binary data&lt;/li&gt;&#xA;&lt;li&gt;10 - file name length: 16 bytes&lt;/li&gt;&#xA;&lt;li&gt;70 69 63 74 75 72 65 73 2E 74 61 72 2E 62 7A 32 - file name: pictures.tar.bz2&lt;/li&gt;&#xA;&lt;li&gt;55 60 CD BE - file modification date: Sat May 23 20:58:06 CEST 2015&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;After removing the PGP packet framing, the following file comes out:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://media.jarnbjo.de/sigint-challenge-1/pictures.tar.bz2&quot; rel=&quot;nofollow&quot;&gt;pictures.tar.bz2&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Even if it has been found out which cards should be swapped, there are obviously more bits to this puzzle not yet solved. Perhaps my partial anaylis can help any of you further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several clues mentioned, which have not yet been used. Since no obvious red herrings have been found yet, I still assume that the clues and data should make sense. Among the unused clues are references to the TI83 key, signatures and PKCS#1 padding. I am not sure why 2012rcampion writes that these clues and hints are not required anymore, since they were completely irrelevant when decrypting the PGP message. The green layer contains a proper PGP header, making it obvious that PGP encrypted data follows and then only the e, p and q parameters from the DTMF data was required for decrypting the PGP block.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The green channel does not only contain the PGP message, but also a few more bits and pieces of information:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Directly following the PGP message is about 5000 bytes of text. I am not sure if this is the signature, which is mentioned in hint 2, but I would assume not. The signature is probably part of the PGP message itself, but I can't find a way to extract it using the BouncyCastle API.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The text looks like base 64 encoded binary data, at least it only contains characters used by base 64 and has a very unconventional line feed pattern. The line length varies between 1 and 234 characters (reading 1234 is perhaps not a coincidence) and I assume that the line length must be considered somehow in the decoding process. Trying to decode the data directly as base 64 (ignoring the line feeds) gives 3705 bytes of seemingly random data. There is some variation in the histogram, but no obvious specific distribution. I have tried to decrypt the data both with the DTMF encoded RSA key and with the TI83 key, but had no success. 3705 bytes would also be way too much for a mere signature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following further is 9450 bytes only with values 0x00 and 0xff. Arranging the data properly gives a repeat of the image of the key and the four squares (red and blue part) in the legend. &lt;/p&gt;&#xA;" OwnerUserId="11687" LastEditorUserId="11687" LastEditDate="2015-06-02T09:06:37.870" LastActivityDate="2015-06-02T09:06:37.870" CommentCount="29" ContentLicense="CC BY-SA 3.0" />
  <row Id="15717" PostTypeId="2" ParentId="15711" CreationDate="2015-05-29T16:17:57.000" Score="2" Body="&lt;p&gt;How about...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wooden teeth!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are certainly tiny little planks, from one end of your mouth to the other. &quot;When told&quot; could refer to the idiom &quot;when all's told&quot;, such that when the teeth are finished being made, they are sharp &lt;em&gt;and&lt;/em&gt; flat! They are flat on the sides like normal teeth, and sharp enough on top to chew food.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-05-29T16:17:57.000" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15718" PostTypeId="1" AcceptedAnswerId="15719" CreationDate="2015-05-29T16:38:39.017" Score="5" ViewCount="2310" Body="&lt;p&gt;&lt;em&gt;So here is another really bad/old riddle I know that I couldn't find posted yet, so I decided I'll take the hit for adding it. I am curious as to how many people might have never heard it before.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This riddle is intended to be spoken, instead of written, so I will put both interpretations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When &lt;b&gt;spoken&lt;/b&gt;, the intended interpretation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What is black and white and red all over?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When &lt;b&gt;written&lt;/b&gt; [Spoiler]:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What is black and white and read all over?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There would always be given the vast amounts of non-nonsensical answers, so I will provide hints to limit them although they are normally not provided when told in person.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hints:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Kids don't use most of it&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;They are now not as popular as before&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It is not any sort of animal&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-05-29T16:54:42.103" LastActivityDate="2015-05-29T18:51:12.330" Title="Black and White and ___ all over" Tags="|riddle|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15719" PostTypeId="2" ParentId="15718" CreationDate="2015-05-29T16:39:38.177" Score="6" Body="&lt;p&gt;The answer I believe you're looking for is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a newspaper - Kids generally only use the comics section, they have become less popular due to the internet, and it's certainly not an animal!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This riddle loses some of its luster when it's not spoken, because the misinterpretation of &quot;red&quot; and &quot;read&quot; is what really makes it tick. I also don't think you need the hints, although I acknowledge that without them you'll get some silly answers, such as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A penguin in a blender! This is my personal favorite answer to this question, but unfortunately it's at least 50% animal. :(&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="11475" LastEditDate="2015-05-29T16:41:23.823" LastActivityDate="2015-05-29T16:41:23.823" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15720" PostTypeId="2" ParentId="15711" CreationDate="2015-05-29T16:53:07.077" Score="5" Body="&lt;p&gt;How about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; quantum relativity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's how I reached it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Plank&quot; is a pun on Max Planck - the father of quantum mechanics&lt;br&gt;&#xA; &quot;tiny little planks&quot; refers to the planck length - the quantum of space&lt;br&gt;&#xA; &quot;when told are sharp and flat&quot; more or less means that when observed, the wave/particle duality condenses to either a &quot;flat&quot; wave or a &quot;sharp&quot; particle&lt;br&gt;&#xA; family = relative(-ity) (from hint)&lt;br&gt;&#xA; movement is relative. being stationary is relative.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think you could afford to be less obtuse.&lt;/p&gt;&#xA;" OwnerUserId="12832" LastEditorUserId="12832" LastEditDate="2015-05-29T17:30:42.813" LastActivityDate="2015-05-29T17:30:42.813" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15721" PostTypeId="2" ParentId="15711" CreationDate="2015-05-29T16:58:29.513" Score="5" Body="&lt;p&gt;Its clearly ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An assorted cheddar cheese tray&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It has a bunch of sharp cheddar cheese cut into flat squares all lined up from end to end.  And there are different types of cheddar that are related to each other because they are all cheese.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10641" LastActivityDate="2015-05-29T16:58:29.513" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15722" PostTypeId="2" ParentId="15711" CreationDate="2015-05-29T17:08:17.407" Score="8" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Light  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of tiny little planks from end to end when told are sharp and flat&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A planck length is the minimum distance that light travels&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When moving: forward or not at all&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Light travels in tranverse wave&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and family to us all, at that!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Reference to Einstein's theory of relativity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9076" LastEditorUserId="10400" LastEditDate="2015-05-29T17:13:14.900" LastActivityDate="2015-05-29T17:13:14.900" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15724" PostTypeId="2" ParentId="15718" CreationDate="2015-05-29T18:51:12.330" Score="4" Body="&lt;p&gt;I have a similar answer, but different reasoning:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Frank looked up from his paper as someone knocked politely before entering his office. Smiling broadly, Frank set the paper down on his desk and walked out to shake his guest’s hand.&lt;/p&gt;&#xA;&lt;p&gt;“Ah, excellent, it’s good to see you my good man. I assume the job went well?”&lt;/p&gt;&#xA;&lt;p&gt;'My good man' was The Puzzler, the most skilled hitman in Chicago. Normally an assassin of his quality would be out of Frank’s price range, but The Puzzler came at a discount because of his singular eccentricity: he always asked his victim a puzzle or riddle before killing them. If they solved it, he let them live and returned his payment. 'The job' was to kill Frank’s rival, Jimmy “Blue Eyes” Lucas, who had been moving into Frank’s territory over the last few months.&lt;/p&gt;&#xA;&lt;p&gt;After a moment, Frank prompted The Puzzler again. “Well? What happened?”&lt;/p&gt;&#xA;&lt;p&gt;“He tried to make a counter-offer,” The Puzzler said seriously.&lt;/p&gt;&#xA;&lt;p&gt;“...And?”&lt;/p&gt;&#xA;&lt;p&gt;“His answer to my riddle was not the one I had in mind.”&lt;/p&gt;&#xA;&lt;p&gt;Frank leaned back against his desk in relief and let a little chuckle escape. “I see, I see. And what was the riddle?”&lt;/p&gt;&#xA;&lt;p&gt;The Puzzler gave him an intense look and said, “I only ask riddles to those who will attempt to solve them. Are you saying you want to try to solve my riddle?”&lt;/p&gt;&#xA;&lt;p&gt;“Oh sure, why not,” Frank grinned. “Lay it on me.”&lt;/p&gt;&#xA;&lt;p&gt;“Very well. What is black and white and red all over?”&lt;/p&gt;&#xA;&lt;p&gt;Frank thought for a second and shrugged. “I dunno, I give up. What is it?”&lt;/p&gt;&#xA;&lt;p&gt;“Your newspaper,” said The Puzzler.&lt;/p&gt;&#xA;&lt;p&gt;Frank digested the answer and then barked a laugh.&lt;/p&gt;&#xA;&lt;p&gt;“Hah, I get it! ‘Red’ and ‘read’, that’s cute. Unfortunately, your answer’s a little off.” He turned and reached for his paper. “I pretty much only read the Sports section, so I don’t really read it ‘all over’. You might want to grglhurgl—” Frank’s comment was interrupted by The Puzzler’s knife slicing through his throat.&lt;/p&gt;&#xA;&lt;p&gt;“I gave you Jimmy’s answer,” said The Puzzler. “I think it works better than mine.” He wiped his knife as Frank collapsed onto his desk. As The Puzzler left the room, Frank’s blood continued to leak from his throat, pooling onto the front page of his newspaper.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So my answer is &amp;quot;Frank's newspaper&amp;quot;. :D&lt;/p&gt;&#xA;" OwnerUserId="10861" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-29T18:51:12.330" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15725" PostTypeId="1" AcceptedAnswerId="16491" CreationDate="2015-05-29T19:42:00.697" Score="70" ViewCount="9770" Body="&lt;p&gt;You've trekked through jungle and tundra, across mountaintops and deserts, into giant caverns and beneath the ocean to find the person known as &lt;strong&gt;The Oracle.&lt;/strong&gt; Finally, after all your hard work, you find him. He sits alone in an opulent palace, never moving, never blinking, allowing each visitor to ask &lt;em&gt;one&lt;/em&gt; question of him. You take a moment to catch your breath after climbing the steps to his throne, gather yourself, and ask:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;What is the meaning of life?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You were told that he would answer one question for each person, and this question has been nagging on your mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you were &lt;em&gt;not&lt;/em&gt; told is that his answers were not quite straightforward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As he begins to speak, you scramble to grab your notepad and begin writing. At the end of his message to you, you read back what he's said. &quot;What?&quot;, you ask out loud, audibly confused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, The Oracle does not answer.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;To start, stay within the pathway of general glee,&lt;br&gt;&#xA;though stay clear of the law faery, who flits about wildly.&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The road third-least taken leads to you and I in flight.&lt;br&gt;&#xA;Head north 'til you see the reverse, then once more north to the burial site.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;When five becomes square, it may lead to confusion;&lt;br&gt;&#xA;behead the basilisk, an ironic inclusion.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Before two sharp curves and the ring of sorcery,&lt;br&gt;&#xA;stand still, close your eyes, and count aloud past three.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In the air - do you hear it? The water baby screams.&lt;br&gt;&#xA;Let our own misgivings start and end in our dreams.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A quixotic elf and I stand at the gates -&lt;br&gt;&#xA;succumb to the virus, and the city awaits.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;As you approach the arbor, I wait in between.&lt;br&gt;&#xA;Clutch your keepsake as the first enemy's heard and seen.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Will being stuck in all this goo truly help you see?&lt;br&gt;&#xA;Primarily, primarily; whip around, forcefully.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Between the lines you'll find your answer, definitively twisted;&lt;br&gt;&#xA;yet in the shadows aligned so clearly you'll be surprised you missed it.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Can you discover the meaning of life from The Oracle's message?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Oracle speaks quite cryptically, don't you think? &lt;em&gt;(I've added a new tag to the puzzle that should add some clarification to it. Hopefully that should give you some idea as to how the puzzle should be solved - and more importantly, some knowledge on how it shouldn't.)&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The two most important words to figuring out this puzzle's big gimmick are &quot;definitively twisted&quot;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Stanza-Specific Hints:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Stanza 1:&lt;/strong&gt; This couplet gives you a key cryptic crossword indicator, but don't get caught up on the result making any sense. In fact, the couplet tells you exactly which words to use and which to avoid. Keep in mind somewhere in there is a definition!&lt;br&gt;&#xA; &lt;strong&gt;Stanza 2:&lt;/strong&gt; Pretty much everyone has identified 'u' and 'i' from this couplet, and you're all on the right track. Most of the words in this couplet aren't indicative of words, &lt;em&gt;per se.&lt;/em&gt;&lt;br&gt;&#xA; &lt;strong&gt;Stanza 3:&lt;/strong&gt; If five becoming square actually meant the number 25, that wouldn't lead to confusion at all, would it? Harry Potter's version of the basilisk is not the only one out there.&lt;br&gt;&#xA; &lt;strong&gt;Stanza 4:&lt;/strong&gt; The last number reference leads to confusion. This one does not. Make sure you're counting aloud. To figure out the rest of the couplet, getting out a pen and paper and illustrating might help.&lt;br&gt;&#xA; &lt;strong&gt;Stanza 5:&lt;/strong&gt; I already mentioned that 'in our dreams' is just in this couplet as a rhyming mechanism. There's only one other place where I do that, and I'll identify it. The cryptic here has already been solved. Which words weren't part of it? Those may help you with another stanza...&lt;br&gt;&#xA; &lt;strong&gt;Stanza 6:&lt;/strong&gt; This one is a fairly straightforward clue. Open your cryptic-solving mind, remember the gimmick of the puzzle, and look for a keyword or two to get things going.&lt;br&gt;&#xA; &lt;strong&gt;Stanza 7:&lt;/strong&gt; You already know at least one word here that doesn't belong with this couplet. The rest of the couplet should be easily solvable, especially if you don't worry about the definition until you solve the cryptic.&lt;br&gt;&#xA; &lt;strong&gt;Stanza 8:&lt;/strong&gt; Will 'truly help you see' truly help you see? Not really.&lt;br&gt;&#xA; &lt;strong&gt;Stanza 9:&lt;/strong&gt; The first line means two things. You already know what 'definitively twisted' means, which is a good start. 'In the shadows' is a reference to the cryptic clues, which are apparently aligned &lt;em&gt;so&lt;/em&gt; clearly! I'm surprised you missed it.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;All stanzas:&lt;/strong&gt; You will find pretty quickly that the words all share a specific property. It's not an obscure property, either (I promise).&lt;br&gt;&#xA; &lt;strong&gt;All stanzas:&lt;/strong&gt; The best way to approach this puzzle is to solve a stanza's cryptic clue, find its definition, then solve &lt;em&gt;that&lt;/em&gt; stanza's cryptic clue with the words in the previous stanza's definition excluded (obviously); lather, rinse, repeat. Luckily, you've already been given a head-start with both a cryptic clue solved and its definition identified.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;6/11/2015 Update:&lt;/strong&gt; Hey, guys! I've opened a bounty on this puzzle as well as &lt;a href=&quot;https://puzzling.stackexchange.com/questions/14980/crack-the-terrorists-password&quot;&gt;my other unanswered puzzle&lt;/a&gt; as a way of saying thanks for the bounty you helped me earn on &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15286/connect-the-pipes/15288&quot;&gt;this excellent puzzle&lt;/a&gt;. As such, I want this puzzle to be solved - my bounty shan't go to waste! If there are any questions I can answer to confirm whether or not you're going in the right direction, please don't hesitate to ask them in the comments. If I feel they don't give too much of the puzzle away, I'll be happy to answer them.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;6/15/2015 Update:&lt;/strong&gt; 2 days left on the bounty - someone better solve this! &lt;strong&gt;Pete&lt;/strong&gt; and &lt;strong&gt;Fillet&lt;/strong&gt; have given you all you need to work with. I can also conclusively tell you that the ninth stanza is specifically instructions, rather than another clue to the puzzle. Once again, I'm happy to answer any questions! &lt;em&gt;(I've also added stanza-specific hints, in hopes that one or two might kickstart the solution to this puzzle.)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;6/16/2015 Update:&lt;/strong&gt; Hopefully this will be the last time I update this puzzle! I just want to note that partial answers are totally awesome for this puzzle. If the final answer is spread among multiple posts, I'll post a community wiki answer encompassing everything (in fact, I may do that anyways). When all's said and done (or, tomorrow before it expires), I'll give the bounty to whomever's solved the most clues! Good luck, happy puzzlers! You're doing awesome so far (just sometimes off in the wrong direction)!&lt;/p&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2016-07-22T21:03:09.513" Title="Discover the meaning of life!" Tags="|riddle|enigmatic-puzzle|rhyme|cryptic-clues|" AnswerCount="4" CommentCount="40" ContentLicense="CC BY-SA 3.0" />
  <row Id="15726" PostTypeId="2" ParentId="2827" CreationDate="2015-05-29T20:11:12.907" Score="0" Body="&lt;p&gt;Cut 4 pieces at 45 degree angles - same as diving a wristwatch from 10:30-1:30 , 1:30-4:30, 4:30-7:30 and 7:30-10:30. Each piece is equal to 3 hours on a watch. Take the piece with 3 o'clock portion and align this with the 12 o'clock position. Similarly, align the 9 o'clock portion with 6 o'clock. Make sure all the center portions are aligned in one straight line which will pass through 3, 12, the cake center, 6 and 9. A single cut along this line will give 8 exact pieces equal to 1:30 hrs on a clock. This does not require stacking which can destroy the cake icing etc. Enjoy the perfect 8 equal pieces. &lt;/p&gt;&#xA;" OwnerUserId="12840" LastActivityDate="2015-05-29T20:11:12.907" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15727" PostTypeId="1" AcceptedAnswerId="15728" CreationDate="2015-05-29T20:39:52.823" Score="7" ViewCount="3154" Body="&lt;p&gt;These are just quick little ones that I was told (and stumped) by my nephew. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A (very) basic knowledge of the english language is required to solve them (apologies to the international audience).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What begins with 'e', ends with 'e' and has a single letter in it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What begins with 'p', ends with 'e' and has thousands of letters in it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12429" LastActivityDate="2015-12-23T16:01:12.733" Title="Simple word riddles" Tags="|word|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15728" PostTypeId="2" ParentId="15727" CreationDate="2015-05-29T20:40:59.600" Score="11" Body="&lt;p&gt;What begins with 'e', ends with 'e' and has a single letter in it?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An &lt;strong&gt;e&lt;/strong&gt;nvelop&lt;strong&gt;e&lt;/strong&gt;! &lt;br&gt;&lt;br&gt; &#xA; Alternatively: An &lt;strong&gt;e&lt;/strong&gt;ating establishment's rating scor&lt;strong&gt;e&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What begins with 'p', ends with 'e' and has thousands of letters in it?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A &lt;strong&gt;p&lt;/strong&gt;ost offic&lt;strong&gt;e&lt;/strong&gt;!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12206" LastEditorUserId="12206" LastEditDate="2015-05-29T20:56:24.207" LastActivityDate="2015-05-29T20:56:24.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15729" PostTypeId="2" ParentId="15700" CreationDate="2015-05-29T20:42:31.840" Score="-1" Body="&lt;p&gt;I originally encountered this as requiring a pattern that a human could execute without having to memorize a long sequence, even if it made the total number of numbers pressed longer. My solution to create an appropriate sequence was to use a Lucas-like sequence where $$F(n) = F(n-1) + F(n-2) + \cdots + F(n-x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With various appropriately chosen seeds, this will create a set of loops that cover the space, which unfortunately must then be spliced together for optimal overage. The Frank Ruskey algorithm is much nicer in many ways. &lt;/p&gt;&#xA;" OwnerUserId="10355" LastEditorUserId="13494" LastEditDate="2015-06-26T20:22:27.850" LastActivityDate="2015-06-26T20:22:27.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15730" PostTypeId="2" ParentId="15727" CreationDate="2015-05-29T20:52:49.413" Score="7" Body="&lt;p&gt;An alternate solution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What begins with 'e', ends with 'e' and has a single letter in it?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An &lt;b&gt;'extreme single character estimate'&lt;/b&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What begins with 'p', ends with 'e' and has thousands of letters in it?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A &lt;b&gt;Petabyte&lt;/b&gt;, can store much much more than merley thousands of letters&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastActivityDate="2015-05-29T20:52:49.413" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15731" PostTypeId="1" AcceptedAnswerId="15732" CreationDate="2015-05-29T21:09:26.643" Score="11" ViewCount="1859" Body="&lt;p&gt;Though I am just a man,&lt;br&gt;&#xA;You may be quite a fan.&lt;br&gt;&#xA;You may think I cannot die,&lt;br&gt;&#xA;You may have seen my taxi fly.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later I saved everyone,&lt;br&gt;&#xA;I'll make you cry, even the sun.&lt;br&gt;&#xA;Then suddenly I was dead,&lt;br&gt;&#xA;Don't let that get to your head.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I took some time, went back and forth,&lt;br&gt;&#xA;Found great strength, and great worth.&lt;br&gt;&#xA;Thought I could settle down,&lt;br&gt;&#xA;That didn't work, just made me frown.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have many well known friends,&lt;br&gt;&#xA;They think we can be thrown away, in the end.&lt;br&gt;&#xA;Figure now who I may be,&lt;br&gt;&#xA;And where you often see me.  &lt;/p&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="11297" LastEditDate="2015-05-29T23:21:07.007" LastActivityDate="2015-05-29T23:21:07.007" Title="You may know me - Who am I?" Tags="|riddle|rhyme|knowledge|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15732" PostTypeId="2" ParentId="15731" CreationDate="2015-05-29T21:14:30.507" Score="10" Body="&lt;p&gt;The answer is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bruce Willis&lt;br&gt;&#xA;&lt;br&gt;&#xA; Though I am just a man,&#xA; You may be quite a fan.&#xA; You may think I cannot die, [&lt;code&gt;Die Hard&lt;/code&gt;]&#xA; You may have seen my taxi fly. [&lt;code&gt;The Fifth Element&lt;/code&gt;]&lt;br&gt;&#xA;&lt;br&gt;&#xA; Later I saved everyone,&#xA; I'll make you cry, even the sun. [&lt;code&gt;Tears of the Sun&lt;/code&gt;]&#xA; Then suddenly I was dead, [&lt;code&gt;The Sixth Sense&lt;/code&gt;] &#xA; Don't let that get to your head.&lt;br&gt;&#xA;&lt;br&gt;&#xA; I took some time, went back and forth, [&lt;code&gt;Looper&lt;/code&gt;]&#xA; Found great strength, and great worth.&#xA; Thought I could settle down, [&lt;code&gt;RED&lt;/code&gt;]&#xA; That didn't work, just made me frown.&lt;br&gt;&#xA;&lt;br&gt;&#xA; I have many well known friends,&#xA; They think we can be thrown away, in the end. [&lt;code&gt;The Expendables&lt;/code&gt;]&#xA; Figure now who I may be,&#xA; And where you often see me. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thank you for all the remarks that suggested improvements/corrections. Much obliged. Also. Nicely crafted puzzle.  &lt;/p&gt;&#xA;" OwnerUserId="12841" LastEditorUserId="12841" LastEditDate="2015-05-29T22:16:12.710" LastActivityDate="2015-05-29T22:16:12.710" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15733" PostTypeId="2" ParentId="15731" CreationDate="2015-05-29T21:20:02.210" Score="4" Body="&lt;p&gt;Are you...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bruce Willis?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though I am just a man,&#xA;You may be quite a fan.&#xA;You may think I cannot die,&#xA;You may have seen my taxi fly.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bruce Willis is a man, and as an actor he has lots of fans. He cannot die in &lt;em&gt;Unbreakable&lt;/em&gt; and flew a taxi in &lt;em&gt;The Fifth Element&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later I saved everyone,&#xA;I'll make you cry, even the sun.&#xA;Then suddenly I was dead,&#xA;Don't let that get to your head.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He saved everyone in the &lt;em&gt;Die Hard&lt;/em&gt; films and in &lt;em&gt;Tears of the Sun&lt;/em&gt;, which is what the 2nd line is referencing (Thanks @Spacemonkey). He turned out to be &lt;em&gt;SPOILERS!!&lt;/em&gt; dead in &lt;em&gt;The Sixth Sense&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I took some time, went back and forth,&#xA;Found great strength, and great worth.&#xA;Thought I could settle down,&#xA;That didn't work, just made me frown.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bruce Willis went back and forth between different types of movies? Or maybe making/not making movies? And ended up not being happy if he wasn't filming? (I'm not 100% sure about this one)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have many well known friends,&#xA;They think we can be thrown away, in the end.&#xA;Figure now who I may be,&#xA;And where you often see me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He appears with many other well known action heros in &lt;em&gt;The Expendables&lt;/em&gt;, but alas, he is not expendable. You're Bruce Willis! And I often see you in these films.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10609" LastEditorUserId="10609" LastEditDate="2015-05-29T21:23:07.980" LastActivityDate="2015-05-29T21:23:07.980" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15734" PostTypeId="2" ParentId="15706" CreationDate="2015-05-29T22:10:24.353" Score="12" Body="&lt;p&gt;Here's a simple solution that lets you take both boards with you:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/Hlnqu.png&quot; alt=&quot;Image&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You're not allowed to walk around the hole, but clearly there must be &lt;em&gt;some&lt;/em&gt; kind of a ledge on each side of the hole that you can use to support the end of a board on, even if it's too narrow to walk on.  (Otherwise &lt;em&gt;none&lt;/em&gt; of the answers work, and I'm fairly sure that the problem is unsolvable.)  So just&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; place one board so that it reaches from the side you're on to one of the ledges, about halfway across.  Pick up the second board and walk along the first to the ledge, then place the second board down so it reaches from the ledge to the other side.  Step across to the second board (you can make the gap between the boards arbitrarily small, so this should not be a problem), pick up the first one and walk to the other side.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ps.&lt;/strong&gt; Note that this puzzle, as stated above, is a much simplified version of the classic &lt;a href=&quot;http://mathworld.wolfram.com/Moat-CrossingProblem.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;moat-crossing problem&lt;/a&gt;.  In the original problem, to which &lt;a href=&quot;https://puzzling.stackexchange.com/a/15710&quot;&gt;lorimer's answer&lt;/a&gt; would be a correct solution, the &quot;hole&quot; is actually an L-shaped corner in a moat, and you need to get from the outside to the inside corner of the L.  This rules out &quot;trivial&quot; answers like mine and &lt;a href=&quot;https://puzzling.stackexchange.com/a/15709&quot;&gt;A.D.'s&lt;/a&gt; (but also prevents you from picking up both boards after you've crossed).&lt;/p&gt;&#xA;" OwnerUserId="243" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2015-05-29T22:10:24.353" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15735" PostTypeId="2" ParentId="2827" CreationDate="2015-05-30T00:25:58.043" Score="8" Body="&lt;p&gt;Florian's answer mentions that it's 'obvious' that the cuts don't work in the plane, but I figure it's worth a short proof that you in fact can't cut a (convex!) 2d cake into eight pieces with three slices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, since the cake &lt;em&gt;is&lt;/em&gt; convex we may as well say that it's infinitely large and just look at cutting the plane into pieces; this won't affect the maximum number of pieces we can get.  Now, since each slice is a line, any two of them intersect in at most one point.  If all three of them intersect in the same point then they partition the plane into at most six regions.  Likewise, if any two slices don't intersect, then they partition the plane into at most six regions.  Otherwise, pick two of the slices: they divide the plane into four quadrants.  Now, the third slice intersects the other two in one point each, and those two intersection points border a common quadrant.  Whichever quadrant is opposite to that one can't be cut by the third slice, because either getting 'in' or 'out' of it would require another intersection with one of the first two slices definiing the quadrants; this means that the three slices partition the plane into at most 4 (the quadrants from the first two slices) + 3 (the new regions defined by the third slice) =7 pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And why convex?  Well, cutting this nonconvex 'cake' into eight pieces with just three slices is left as an exercise to the reader:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7Zdf7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="266" LastActivityDate="2015-05-30T00:25:58.043" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15736" PostTypeId="1" AcceptedAnswerId="15738" CreationDate="2015-05-30T01:02:06.283" Score="12" ViewCount="716" Body="&lt;p&gt;There are four stones, positioned on the ground at the vertices of a square. At any time, you may pick up a stone and &quot;hop&quot; it over another one so that it lands an equal distance beyond the hopped stone. Can you find a series of hops which will make these stones form the vertices of a larger square? If so, how, if not, why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify what a &quot;hop&quot; is: if there is a stone a point $p$, you are allowed to move it to a point $p'$ provided there is another stone at the midpoint of $p$ and $p'$.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="8874" LastEditDate="2015-11-04T12:05:06.797" LastActivityDate="2015-11-04T12:05:06.797" Title="Enlarge the Square?" Tags="|mathematics|geometry|reachability|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15737" PostTypeId="2" ParentId="9875" CreationDate="2015-05-30T01:59:30.740" Score="1" Body="&lt;p&gt;I approached it with a kind-of &quot;brute force&quot; method (although obviously I can't try all 52! possible shuffles...). One subtle point is that the first and last card must also not be the same, because of the cut.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead I used a kind of &quot;genetic&quot; algorithm - starting with a pool of random shuffles, it selects the ones which survive for at least 4 iterations for further investigation, with a starting pool of several thousand of these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each surviving shuffle is mutated in 52 randomly selected ways (there are 52 x 51 / 2 different possible &quot;mutations&quot; consisting of having a shuffle otherwise equivalent to the last one, but with a single pair of cards swapped, so this certainly isn't exhaustive), and any that still survive at least 4 iterations are added to the processing queue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Soon, a very significant proportion of newly-mutated shuffles are surviving 4 iterations, so it only queues the ones surviving 5... then 6, etc. For the &quot;rarer&quot; shuffles surviving 8 or more iterations, I also allowed for a &quot;double mutation&quot; (with even more of them) as well as the usual single mutation, and for the 10-iteration shuffles, a &quot;triple mutation&quot; (in addition to the single and double), so the memory usage exploded causing me to abort one run with about 100 million items awaiting processing, before I'd optimised the processing queue).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On one run of this algorithm, I got an 11 based on this randomised technique... but it seems more often to top out at 10 (the most recent run reported that it found nearly 18000 solutions which survived 10 iterations, probably with a handful duplicates).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I strongly suspect that 12 will in fact prove to be the maximum - primarily because there are only 13 DIFFERENT card values (we're ignoring suit for this problem). Any subset of cards having a cycle longer than 13 must necessarily contain the same card value more than once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The program is only set to print out the FIRST solution it finds for a given minimum, but in the case where it found an 11, you can observe the evolution for the first solutions it found for 8,9,10,11...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aHNuh.png&quot; alt=&quot;evolving solutions&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="12582" LastEditorUserId="12582" LastEditDate="2015-05-30T02:06:07.157" LastActivityDate="2015-05-30T02:06:07.157" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15738" PostTypeId="2" ParentId="15736" CreationDate="2015-05-30T02:45:46.537" Score="12" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It is impossible&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Take these two rules:&lt;br&gt;&lt;ol&gt;&lt;li&gt;When you are making a move, the move is also able to be reversed, meaning that after a move, it can be undone and still follow the rule of how you are allowed to move.&lt;/li&gt;&lt;li&gt;Assuming you have a grid where each square is 1x1, and you start with a square that surrounds 1 grid square (let's say points (0,0), (1,0), (0,1), (1,1)) then there is no way using the rule to get any of these points closer than 1 unit from each other. For this to happen, a point would have to be at an integer point while another is at a fraction point. Since they all started at integer points, they cannot be reflected across each other to end up at fractional points (e.g. (.5,1)). Thus it can be concluded that you would also not be able to create a smaller square from a square of any size.&lt;/li&gt;&lt;/ol&gt;Using these rules, we can determine that for you to be able to create a larger square from a square of size 1x1, then you would also be able to create a square of 1x1 from a larger square, since all the moves can be reversed. As shown in rule 2, you cannot create a 1x1 square from a larger square, so it can be concluded that you are unable to create a larger square from a 1x1 square using this hopping rule.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11727" LastActivityDate="2015-05-30T02:45:46.537" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15740" PostTypeId="1" AcceptedAnswerId="15752" CreationDate="2015-05-30T06:49:30.353" Score="11" ViewCount="939" Body="&lt;p&gt;$$(V+W+X+Y+Z)^3 = V \times 10^4 + W \times 10^3 + X \times 10^2 + Y \times 10 + Z \\&#xA;V, W, X, Y, Z \in \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($V$ through $Z$ are digits in a five-digit number)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solve for $V$, $W$, $X$, $Y$ and $Z$.&lt;/p&gt;&#xA;" OwnerUserId="12853" LastEditorUserId="10281" LastEditDate="2015-06-05T17:00:24.837" LastActivityDate="2015-06-05T17:00:24.837" Title="What numbers do these characters represent?" Tags="|mathematics|" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15741" PostTypeId="2" ParentId="15740" CreationDate="2015-05-30T07:05:48.030" Score="11" Body="&lt;p&gt;&lt;strong&gt;Solution(s) :&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the Digits can not repeat :&lt;br&gt;&#xA; (1+9+6+8+3)^3=27^3=19683&lt;br&gt;&#xA;&lt;br&gt;&#xA; If the Digits can repeat :&lt;br&gt;&#xA; (1+7+5+7+6)^3=26^3=17576&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Details :&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the Digits are different, then maximum 5 Digit number can be 9+8+7+6+5 which has a Digit Sum of 35. The minimum 5 Digit number is 10234, whose cube root is a little more than 21. So the Solution must involve the cube of integers between 22 &amp;amp; 35. Check the cubes of these 14 numbers to find (1) which has Digits with the required sum (2) which has no Repeating Digits, to quickly get the first solution, then continue looking further for solutions which have Repeating Digits, to get the second solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10940" LastEditorUserId="10940" LastEditDate="2015-05-31T14:26:56.623" LastActivityDate="2015-05-31T14:26:56.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15742" PostTypeId="1" AcceptedAnswerId="15746" CreationDate="2015-05-30T07:50:01.977" Score="9" ViewCount="4566" Body="&lt;p&gt;This is the fourth puzzle in the Hollywood Movie Rebus Puzzle series. See the others here.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/15328/hollywood-movie-rebus-puzzle-1&quot;&gt;Puzzle #1 Hollywood Rebus Movie Puzzle &lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/15350/hollywood-movie-rebus-puzzle-2&quot;&gt;Puzzle #2 Hollywood Rebus Movie Puzzle &lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/15465/hollywood-movie-rebus-puzzle-3&quot;&gt;Puzzle #3 Hollywood Rebus Movie Puzzle &lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;.&#xA;&lt;strong&gt;Background&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Your love for Hollywood movies and remarkable ability to remember movie titles is finally going to serve a purpose. There was indeed a time when you almost believed your mother’s words, as she scorned you: “Lazing in front of the television is not going to pay for the rent!” You knew better all along. As you and the rest of your family listen to the directions of your late, and wealthy, Uncle Rob’s will, you cannot help but smile as the executor reads “… and thus half of my fortune will indeed go toward a sole beneficiary. Namely he who solves the puzzles as depicted below”.&#xA;As you are each given a paper with four Hollywood movie rebus puzzles, you begin to smile. You know the answer to each.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;How will you answer the fourth puzzle?&lt;/strong&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/rq6eC.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="12317" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-23T05:17:11.637" Title="Hollywood Movie Rebus Puzzle #4" Tags="|rebus|movies|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15746" PostTypeId="2" ParentId="15742" CreationDate="2015-05-30T08:39:29.220" Score="11" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; FORREST GUMP&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; FOR (four [o'clock]) + REST (person lying in bed) + GUMP ([bubble] gum + p)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12277" LastActivityDate="2015-05-30T08:39:29.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15752" PostTypeId="2" ParentId="15740" CreationDate="2015-05-30T12:10:29.443" Score="8" Body="&lt;p&gt;The complete solution set including repeated digits and leading zeroes is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    00000 =  0^3&lt;br&gt;&#xA;    00001 =  1^3&lt;br&gt;&#xA;    00512 =  8^3&lt;br&gt;&#xA;    04913 = 17^3&lt;br&gt;&#xA;    05832 = 18^3&lt;br&gt;&#xA;    17576 = 26^3&lt;br&gt;&#xA;    19683 = 27^3  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since every solution must be a cube number, the easiest way of generating the solution set is to compute the cube of the numbers from &lt;code&gt;0&lt;/code&gt; to &lt;code&gt;45&lt;/code&gt;, and for each result check if it is a legal solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is worth noting that if we increase the allowed length of the number, no extra solutions are found. All 6-digit options are trivially tested, and for 7 digits and up it is easy to show that the digit-sum cubed will contain fewer digits than the original number.&lt;/p&gt;&#xA;" OwnerUserId="2186" LastEditorUserId="10400" LastEditDate="2015-05-30T15:13:11.233" LastActivityDate="2015-05-30T15:13:11.233" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15753" PostTypeId="2" ParentId="15727" CreationDate="2015-05-30T13:01:01.577" Score="0" Body="&lt;p&gt;Could the answer to the second riddle be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A &lt;b&gt;'p&lt;/b&gt;ost offic&lt;b&gt;e'&lt;/b&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12859" LastActivityDate="2015-05-30T13:01:01.577" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15754" PostTypeId="1" CreationDate="2015-05-30T14:26:57.687" Score="0" ViewCount="38397" Body="&lt;p&gt;One of the oldest known riddles is the riddle (&lt;em&gt;ainigma&lt;/em&gt;) of the Sphinx. In ancient Greek tradition the Sphinx devoured all those who failed to solve it, but she destroyed herself instead when Oedipus got the right answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In its best known form, it runs as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What walks on four legs in the morning, two legs at noon, and three legs in the evening?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is &lt;em&gt;man&lt;/em&gt;: a person crawls on all fours when a baby, walks upright in the middle period of their life, and uses a walking stick when they are elderly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What alternative solutions can you come up with?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Restriction&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having been advised that the question as phrased above is too broad, and after imagining theatre-goers storming out of a production of Sophocles's &lt;em&gt;Oedipus the King&lt;/em&gt;, loudly objecting that &quot;&lt;em&gt;The Sphinx's puzzle was completely unrealistic. It's obvious there's an infinite class of possible solutions, consisting of any quadruped entity that loses two legs and then gains back one&lt;/em&gt;&quot;, I am adding the following restriction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;please ensure that at least one interpretation of the word&lt;/em&gt; 'leg' &lt;em&gt;is non-literal&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Further restriction&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In response to votes to keep this question closed for still being &quot;too broad&quot; even after the addition of the first restriction, the original poster has decided to add a further restriction as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this question is not considered reopenable as it stands, for being too broad, please make all answers refer either to the behaviour of &lt;a href=&quot;http://www.ultimateungulate.com/artiodactyla/vicugna_vicugna.html&quot; rel=&quot;nofollow&quot;&gt;vicunas&lt;/a&gt; (aspects of their behaviour not also displayed by other camelids) or to the developing relation between advertising, participatory commercial ('social') websites, right-wing libertarianism, the work culture of computer programmers, schizophrenia, bureaucracy, and the function of ritual respect-paying to 'good faith', but not both.&lt;/p&gt;&#xA;" OwnerUserId="7363" LastEditorUserId="8874" LastEditDate="2016-03-18T15:56:50.090" LastActivityDate="2016-03-18T15:56:50.090" Title="Alternative solutions to the riddle of the Sphinx?" Tags="|riddle|puzzle-history|" AnswerCount="4" CommentCount="4" ClosedDate="2015-05-30T16:32:52.297" ContentLicense="CC BY-SA 3.0" />
  <row Id="15756" PostTypeId="2" ParentId="15754" CreationDate="2015-05-30T14:58:04.760" Score="1" Body="&lt;p&gt;A military dog:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;When it was born it had four legs.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Then it lost two legs from an exploding mine/bomb.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Its owner realized that a dog cannot walk on two legs and he was very stingy, so he got the dog &lt;em&gt;one&lt;/em&gt; prosthetic leg. Now it has three legs.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="12109" LastActivityDate="2015-05-30T14:58:04.760" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15757" PostTypeId="2" ParentId="15754" CreationDate="2015-05-30T15:07:38.810" Score="3" Body="&lt;p&gt;A human, or bird, or anything else that satisfies the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;evolved from four-legged ancestors (early primates/lizards in the case of humans/birds)&lt;/li&gt;&#xA;&lt;li&gt;has two legs at the current stage of evolution&lt;/li&gt;&#xA;&lt;li&gt;could conceivably evolve to have three legs in the future (heard of &lt;a href=&quot;http://en.wikipedia.org/wiki/Frank_Lentini&quot; rel=&quot;nofollow&quot;&gt;Frank Lentini&lt;/a&gt;?)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-05-30T15:07:38.810" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15758" PostTypeId="2" ParentId="15407" CreationDate="2015-05-30T15:46:34.300" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$$\matrix{M&amp;amp;A&amp;amp;N&amp;amp;I&amp;amp;A&amp;amp;C\\A&amp;amp;V&amp;amp;E&amp;amp;N&amp;amp;G&amp;amp;E\\N&amp;amp;E&amp;amp;A&amp;amp;T&amp;amp;E&amp;amp;R\\I&amp;amp;N&amp;amp;T&amp;amp;O&amp;amp;N&amp;amp;E\\A&amp;amp;G&amp;amp;E&amp;amp;N&amp;amp;D&amp;amp;A\\C&amp;amp;E&amp;amp;R&amp;amp;E&amp;amp;A&amp;amp;L}$$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12863" LastActivityDate="2015-05-30T15:46:34.300" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15759" PostTypeId="1" AcceptedAnswerId="15795" CreationDate="2015-05-30T15:52:50.393" Score="2" ViewCount="569" Body="&lt;p&gt;&lt;strong&gt;What is the magic word?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;011021042021011012012032012032014&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clues (Read, in order, one and re-think the problem):&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Five letter word&lt;/li&gt;&#xA;  &lt;li&gt;'A' equals 011051&lt;/li&gt;&#xA;  &lt;li&gt;Binary (Eight digit per)&lt;/li&gt;&#xA;  &lt;li&gt;Compressed message (Zero - Show, One - Hide)&lt;/li&gt;&#xA;  &lt;li&gt;ASCII Letters&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If you have the answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Say the word = &lt;em&gt;+1 for you&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Say the word and with the same method say what 02301102103102 means = &lt;em&gt;Best Answer&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12866" LastEditorUserId="12242" LastEditDate="2015-05-30T18:14:36.303" LastActivityDate="2015-06-01T04:40:38.667" Title="What does this mean?" Tags="|logical-deduction|mathematics|number-sequence|enigmatic-puzzle|" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15760" PostTypeId="2" ParentId="15754" CreationDate="2015-05-30T15:54:43.060" Score="9" Body="&lt;p&gt;Napoleon in George Orwell's Animal Farm:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Four legs originally (naturally).&lt;br&gt;&#xA;Then starts walking on two.&lt;br&gt;&#xA;Then is propped up by the blood, sweat and tears (three things you see) of the proletariat beneath him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further explanation for those not familiar with &lt;a href=&quot;http://en.wikipedia.org/wiki/Animal_Farm&quot; rel=&quot;nofollow&quot;&gt;Animal Farm&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Napoleon was the pig in charge of a revolt that led to the animals running their own farm. Like all pigs, he walked on four legs. In chapter 10, to the shock of the other animals, the pigs started walking on two legs (and carrying whips with Napoleon as a tyrannical dictator).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The primary message of the story is as a satire of Stalinist communism where the leaders of the revolution are as oppressive to the people as the aristocrats they replaced - hence the final three legs, which is a common phrase that means sacrifice and effort (see for example &lt;a href=&quot;http://spectator.org/articles/56615/blood-sweat-and-tears&quot; rel=&quot;nofollow&quot;&gt;http://spectator.org/articles/56615/blood-sweat-and-tears&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="12666" LastEditorUserId="12109" LastEditDate="2015-06-27T02:07:14.240" LastActivityDate="2015-06-27T02:07:14.240" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15761" PostTypeId="2" ParentId="15319" CreationDate="2015-05-30T16:01:03.293" Score="1" Body="&lt;p&gt;Ok I will try : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To begin I searched for a sunny picture in SE, I went looking for it in &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The photography SE, and especially the Hall of Fame&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;then naturally here&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://photo.meta.stackexchange.com/questions/4659/image-of-the-week-hall-of-fame-2014-archive&quot;&gt;https://photo.meta.stackexchange.com/questions/4659/image-of-the-week-hall-of-fame-2014-archive&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But once I was here I needed a 'sunny picture' and there were many of them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But one picture got my attention &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/0Cj0T.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/0Cj0T.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because it was posted by :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Riot&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I think it is somehow related to &quot;You are a less capital idea&#xA;Than those who follow you.&quot;, and of course it is a picture of a bird.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think that it is the right answer, especially because of my lack of sun in my answer, but I hope I have helped in some ways .&lt;/p&gt;&#xA;" OwnerUserId="12657" LastEditorUserId="-1" LastEditDate="2017-03-16T15:44:03.133" LastActivityDate="2015-05-30T16:06:38.887" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15762" PostTypeId="2" ParentId="15754" CreationDate="2015-05-30T16:32:43.490" Score="5" Body="&lt;p&gt;Here's a slightly more literal answer than most...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Army buddies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the morning, they leave on a mission, walking on four legs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the afternoon, one of the buddies steps on a landmine, destroying one leg. The other buddy carries the injured one back to camp where there's a medic. Two legs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the evening, the medic has stitched up the injured buddy's stump, and hops, with the help of his buddy, to his cot - they are now walking on three legs between them.&lt;/p&gt;&#xA;" OwnerUserId="9289" LastActivityDate="2015-05-30T16:32:43.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15763" PostTypeId="1" AcceptedAnswerId="15767" CreationDate="2015-05-30T18:16:05.237" Score="17" ViewCount="1073" Body="&lt;p&gt;In 1904 (at the age of 12) J.R.R. Tolkien sent a &quot;puzzle letter&quot; (pictured below) to a family friend; Father Francis Morgan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What does it say?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&lt;sub&gt;Obviously it would be trivially simple to google the answer, so I want you to &lt;strong&gt;promise me&lt;/strong&gt; that you didn't :-) &lt;/sub&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eV48R.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4852" LastEditorUserId="5373" LastEditDate="2018-11-10T10:36:07.380" LastActivityDate="2018-11-10T10:36:07.380" Title="The Lord of the Puzzles" Tags="|rebus|no-computers|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15764" PostTypeId="1" AcceptedAnswerId="15769" CreationDate="2015-05-30T18:25:47.097" Score="6" ViewCount="745" Body="&lt;p&gt;This is a variant of &lt;a href=&quot;https://puzzling.stackexchange.com/q/70/10615&quot;&gt;this question&lt;/a&gt;, reproduced below:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;There are &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; people, who will stand in a circle while a game master places a hat on each of their heads. The hats can each be one of &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; colors (repeats allowed). Each player can see every hat except their own. They all must then simultaneously guess what color hat they are wearing. As long as at least one person guesses correctly, they win. If they are all wrong, they lose.&lt;/p&gt;&#xA;&lt;p&gt;They may agree on a strategy beforehand, but may not communicate once the hats are placed. Devise a strategy which guarantees they win.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This is a classic problem. If you haven't heard of it, give it a shot before you move on!&lt;/p&gt;&#xA;&lt;p&gt;Here is the twist:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;To make things harder, the game master places a one-way glass between two of the people, Larry and Harry. This means that Larry can see &lt;span class=&quot;math-container&quot;&gt;$n-2$&lt;/span&gt; hats (every one else's hat except Harry's), while all other players can see &lt;span class=&quot;math-container&quot;&gt;$n-1$&lt;/span&gt; hats.&lt;/p&gt;&#xA;&lt;p&gt;With this extra obstacle, can they still win (guarantee at least one person guesses correctly)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-06-04T00:51:31.103" Title="N people, N hat colors, and a One-Way Glass" Tags="|mathematics|hat-guessing|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15765" PostTypeId="1" CreationDate="2015-05-30T18:28:23.757" Score="1" ViewCount="341" Body="&lt;p&gt;If I remember correctly, this is a question for admission to a college in China long time ago:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) You are traveling and have reached a point on the road where it diverges to Path A and Path B.  One path leads to Beijing and the other does not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) You were told by other people that at this point there is a restaurant boss who always tells the truth, and another restaurant boss who always tells a false statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) You don't know which boss is which.  The two bosses know each other and know one always tell the truth and one always tells a false statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Now you can only ask one question to one boss that you choose, and be able to tell which path is to Beijing. (and it cannot be one question that actually contains two or more questions, and note that it is not one question to each boss, but just a total of one single question)  What is that question?&lt;/p&gt;&#xA;" OwnerUserId="12867" LastEditorUserId="12867" LastEditDate="2015-05-30T18:34:26.737" LastActivityDate="2015-05-30T19:03:37.187" Title="Which one is the road (or path) to Beijing?" Tags="|logical-deduction|" AnswerCount="1" CommentCount="0" ClosedDate="2015-05-30T19:06:24.173" ContentLicense="CC BY-SA 3.0" />
  <row Id="15767" PostTypeId="2" ParentId="15763" CreationDate="2015-05-30T18:58:11.417" Score="13" Body="&lt;p&gt;The sender's address (top right) was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Woodside House, Rednal&lt;/strong&gt; (wood, S, eye, D=500, E, house, {N, A, L=50} in red)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The letter reads:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;My dear wise owl Fr. Francis,&lt;/strong&gt; (M=1000, eye, deer, Y's, owl, France with 'Fr.' on it, hiss)&lt;br&gt;&lt;br&gt; &lt;strong&gt;You are too bad&lt;/strong&gt; (U, R, 2, bee, A, D=500)&lt;br&gt;&lt;br&gt; &lt;strong&gt;not to come in&lt;/strong&gt; (knot, 2, sea, O, M=1000, E, inn)&lt;br&gt;&lt;br&gt; &lt;strong&gt;spite of Fr. Dennis.&lt;/strong&gt; (spit, E, OF, Fr., D=500, hen, eye, S)&lt;br&gt;&lt;br&gt; &lt;strong&gt;I am so sorry you&lt;/strong&gt; (Hyam, sew, saw, rye, yew)&lt;br&gt;&lt;br&gt; &lt;strong&gt;did not like the&lt;/strong&gt; (D=500, eye, D, knot, L=50, bike-B, tea, HE)&lt;br&gt;&lt;br&gt; &lt;strong&gt;word 'piano' in my l...&lt;/strong&gt; (words, pea, note-T, inn, M, eye, L)&lt;br&gt;&lt;br&gt; &lt;strong&gt;...ast letter. So I...&lt;/strong&gt; (mast-M, letter, sow, eye)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks to &lt;strong&gt;Pete&lt;/strong&gt; and &lt;strong&gt;d'alar'cop&lt;/strong&gt; and &lt;strong&gt;LeppyR64&lt;/strong&gt; and &lt;strong&gt;JNF&lt;/strong&gt; for help with some of the clues (see comments below).&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-06-01T19:08:05.993" LastActivityDate="2015-06-01T19:08:05.993" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="15768" PostTypeId="2" ParentId="15765" CreationDate="2015-05-30T19:03:37.187" Score="1" Body="&lt;p&gt;I've heard two answers to this question based on if the truth teller has his restaurant in Beijing (everyone in the other village is a liar) and one where the answer has nothing to do with where their restaurants are. The answer to the first form would be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You ask &quot;Which road leads to your restaurant?&quot;. And then you follow that path.&lt;br/&gt; This is because if he tells the truth he will point to his restaurant (Beijing) and if he lies he will have to point to Beijing (as he can't point to his restaurant).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer to the second form, which is the standard two guards:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You ask him &quot;which road would the other boss point when I ask 'what path leads to Beijing'?&quot; and follow the other path&lt;br/&gt;The boss who tells the truth would tell you that the liar would point to the path away from Beijing. &lt;br/&gt;The liar would point to the path away from Beijing as he must lie about where the truth boss would point.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11239" LastActivityDate="2015-05-30T19:03:37.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15769" PostTypeId="2" ParentId="15764" CreationDate="2015-05-30T19:27:45.437" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is no strategy that guarantees at least one person guesses correctly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Suppose each person's hat color is selected independently and uniformly at random. Then the probability than any given person guesses their hat color correctly is $1/n$. There are $n$ people, so the expected number of people who guess their hat color is $1$. This implies:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A strategy guarantees that at least one person guesses correctly if and only if that strategy guarantees that at most one person guesses correctly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, suppose they players have decided on a strategy. Choose a color $C_1$. If Larry sees hat color $C_1$ on everyone except Harry, the strategy will dictate a color for Larry to guess, say $C_2$. If Harry sees hats of color $C_1$ on everyone but Larry and sees hat color $C_2$ on Larry, the strategy will dictate a color for Harry to guess, say $C_3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a problem, because in the event that everyone but Harry and Larry have hat color $C_1$, Larry has hat color $C_2$, and Harry has hat color $C_3$, then both Larry and Harry will guess correctly. This means the strategy they are using allows for the possibility of two people guessing correctly, so the strategy cannot guarantee that at least one person guesses correctly.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-05-30T19:33:23.867" LastActivityDate="2015-05-30T19:33:23.867" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15770" PostTypeId="1" AcceptedAnswerId="15776" CreationDate="2015-05-30T19:37:54.673" Score="10" ViewCount="1388" Body="&lt;p&gt;There is a prisoner who will be executed, but there is a rule in the kingdom that says, &quot;the prisoner will be given a bag containing a white paper and a black paper. If the prisoner is lucky enough to draw the white paper, he will be released; otherwise he will be executed.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The person preparing the bag hated the prisoner, so he actually put 2 pieces of black paper into the bag instead of 1 black and 1 white.  And the prison guard saw this and told the prisoner about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the prisoner overcome this situation and how? (and be actually released)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pulling both pieces out of the bag will merit execution.&lt;/p&gt;&#xA;" OwnerUserId="12867" LastEditorUserId="11200" LastEditDate="2015-05-31T17:18:09.383" LastActivityDate="2015-09-02T16:32:19.110" Title="Drawing a piece of paper, to be released or executed" Tags="|logical-deduction|" AnswerCount="5" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15771" PostTypeId="2" ParentId="15770" CreationDate="2015-05-30T19:41:26.053" Score="4" Body="&lt;p&gt;He could:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Before drawing, propose the rule that instead of being released if he draws the white paper, he will be released if he draws the black paper. Since there &lt;em&gt;should&lt;/em&gt; be the same probability that he will draw black as white (50/50) there is no reason for the government/rulers to disallow this change of rules. He then draws black and is allowed to live.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11727" LastEditorUserId="11727" LastEditDate="2015-05-30T19:47:42.473" LastActivityDate="2015-05-30T19:47:42.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15773" PostTypeId="2" ParentId="15770" CreationDate="2015-05-30T19:51:44.897" Score="12" Body="&lt;p&gt;Similar to MisterEman22's answer, who beat me to the punch.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The prisoner suggests, &quot;For my last request, how about &lt;em&gt;you&lt;/em&gt; pull out a piece of paper instead, and I choose the other paper in the bag? I simply ask because I hate touching paper. If you pull black, that means the paper in the bag is white, so I go free. If you pull white, I die. Surely this doesn't improve my chances, so there is no reason to deny me this request.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The king cannot say no without appearing suspicious, so he agrees, and the prisoner lives.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="1499" LastEditDate="2015-05-31T03:23:43.310" LastActivityDate="2015-05-31T03:23:43.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15774" PostTypeId="2" ParentId="15770" CreationDate="2015-05-30T19:54:48.813" Score="14" Body="&lt;p&gt;The answer that comes to my mind is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Draw one piece of paper&lt;br&gt;&#xA; Before anyone sees it destroy it somehow (burn it, eat it, throw it out the window, whatever...)&lt;br&gt;&#xA; Then show that the paper remaining in the bag is black and so therefore he must have drawn the white one. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Everybody gets away safe without anyone getting in trouble for trying to overcome the kingdom's strange rules.&lt;/p&gt;&#xA;" OwnerUserId="7715" LastActivityDate="2015-05-30T19:54:48.813" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15776" PostTypeId="2" ParentId="15770" CreationDate="2015-05-30T20:00:00.597" Score="15" Body="&lt;p&gt;The prisoner could&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The prisoner picks one at random but crumple it up so nobody sees. He claims that this was the on he picks, and does something to destroy the paper forever, like eat it. The guards look in the bag to see the black slip of paper, and concludes that the paper he picked must have been the white slip of paper.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11002" LastActivityDate="2015-05-30T20:00:00.597" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15777" PostTypeId="1" AcceptedAnswerId="15780" CreationDate="2015-05-30T20:12:42.057" Score="-1" ViewCount="9370" Body="&lt;p&gt;Can you make exactly 2 squares and exactly 4 right-angled triangles using &quot;only 8&quot; that is exactly 8 straight lines?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for 3 different answers.&lt;/p&gt;&#xA;" OwnerUserId="12791" LastEditorUserId="11297" LastEditDate="2015-05-31T15:10:43.983" LastActivityDate="2015-05-31T15:10:43.983" Title="Can you make 2 squares and 4 rightangled triangles using only 8 straight lines?" Tags="|logical-deduction|geometry|" AnswerCount="2" CommentCount="4" ClosedDate="2015-05-30T21:27:04.837" ContentLicense="CC BY-SA 3.0" />
  <row Id="15779" PostTypeId="2" ParentId="15777" CreationDate="2015-05-30T20:15:00.393" Score="2" Body="&lt;p&gt;I can do it using only 7, as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FYasv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-05-30T20:36:40.440" LastActivityDate="2015-05-30T20:36:40.440" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15780" PostTypeId="2" ParentId="15777" CreationDate="2015-05-30T20:21:23.553" Score="4" Body="&lt;p&gt;This figure can be drawn using 8 straight lines. You can easily see 4 triangles and square, and the border of the figure forms a larger square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fL9Wq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-05-30T20:21:23.553" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15781" PostTypeId="1" AcceptedAnswerId="15783" CreationDate="2015-05-30T20:27:40.560" Score="9" ViewCount="3714" Body="&lt;p&gt;When I acquired my Mercedes-Benz car in Germany, the first thing I had to do was to get a license plate. The plate I got had a peculiar number on it. It consisted of 5 different numbers and by mistake when I fixed it upside down the number could be still read, but the value had increased by 78633.&lt;br&gt;&#xA;What was my actual license number?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solve it fast - really fast!&lt;/p&gt;&#xA;" OwnerUserId="12791" LastEditorUserId="5373" LastEditDate="2015-05-30T21:23:18.130" LastActivityDate="2015-05-30T22:19:43.300" Title="Upside-down numbers differing by 78633" Tags="|logical-deduction|calculation-puzzle|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15782" PostTypeId="2" ParentId="15770" CreationDate="2015-05-30T20:30:15.007" Score="1" Body="&lt;p&gt;@Xandawesome's and @Spencerkatty's answer is very close to mine, and it is like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The prisoner takes a paper, and if it is small, just hide it in the palm and if too big, crumble it and hide in his hand, and he declares &quot;Almighty God, I choose this one, and may it be white!&quot;, and then he eats/swallows the paper.  Now they cannot kill him to get the paper out, as he is supposed to be released if the paper is white, and they or the prisoner will suggest, which paper did he choose, let's take a look at what's left in the bag and we will know... and it is a black one, so the conclusion is that he is lucky to pick the white one&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12867" LastEditorUserId="12867" LastEditDate="2015-05-30T20:35:20.893" LastActivityDate="2015-05-30T20:35:20.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15783" PostTypeId="2" ParentId="15781" CreationDate="2015-05-30T20:35:24.327" Score="15" Body="&lt;p&gt;It was originally&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 10968.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The upside-down value is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 10968+78633=89601.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Proof of uniqueness&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The upside-down number must be at least 10000 bigger than 78633, so its first digit must be 8 or 9 while the first digit of the original number must be 1 or 2. If the first digit or the upside-down number was 9, then the last digit of the original number would have to be 6, so the last digit of the upside-down number would have to be 3+6=9, so the first digit of the original number would have to be 6, which is too large. So &lt;strong&gt;the first digit of the upside-down number, and hence the last digit of the original number, are 8&lt;/strong&gt;, which means &lt;strong&gt;the last digit of the upside-down number, and hence the first digit of the original number, are 1&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the original number must be less than 90000-78633=11367, so its second digit must be 0 or 1. But we know it has five distinct digits, so &lt;strong&gt;the second digit of the original number, and hence the fourth digit of the upside-down number, are 0&lt;/strong&gt;. By considering the fourth and fifth columns of summation, we now see that &lt;strong&gt;the fourth digit of the original number is 6&lt;/strong&gt; and hence &lt;strong&gt;the second digit of the upside-down number is 9&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have 10?68+78633=89?01, where the two ?'s are each other upside down and therefore must be both 2, both 5, or 9 and 6 respectively (bearing in mind that each number has 5 distinct digits). We can check these 3 cases by hand and verify that 9 and 6 is the only one that works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;QED.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-05-30T21:22:05.270" LastActivityDate="2015-05-30T21:22:05.270" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15786" PostTypeId="2" ParentId="2188" CreationDate="2015-05-31T03:44:36.450" Score="37" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is a classic old chestnut of a puzzle, and we've had several responses giving the traditional answer to the puzzle.&#xA;As a bit of spice, here's a slightly &lt;code&gt;lateral-thinking&lt;/code&gt; answer, which lets us figure out not only which door is which, but also which guard is which, all using a single question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We're going to be using a timing attack.  Here's how it works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ask one of the guards:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;If I was to ask you what you'd say if I asked you what you'd say if I asked you what you'd say if I asked you what you'd say if I asked you what you'd say... &lt;em&gt;(repeat any sufficiently large number of times, remembering to occasionally stop to draw a fresh lungful of air so that you don't faint)&lt;/em&gt; ...if I asked you whether this was the door to freedom, then what would you say your answer would be?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The truth-telling guard will be able to answer this right away.  No matter how many nested self-referential clauses you put into that question, she doesn't need to remember them or count them, and it always remains a trivial question to which she can always instantly give a truthful answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's a much harder problem for the lying guard, who will need to think for a short time as she double-checks her answer, and may even have to ask you to repeat the question, just to make sure she correctly counted the number of recursive lies she's supposed to be telling, in order to give you the correct &quot;always lies&quot; answer.  Her answer will necessarily be delayed, compared against the truth-teller's answer, because the question is designed to be much more difficult to determine the answer, for someone who is required to tell falsehoods.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus, simply by checking whether the answer is instantaneous or not, you can tell whether the guard is a truth-teller or a liar (respectively), and therefore select a door according to the answer that the guard gave you, and whether that guard is the truth-teller or the liar.&lt;/p&gt;&#xA;" OwnerUserId="9286" LastEditorUserId="46903" LastEditDate="2018-09-13T15:52:27.723" LastActivityDate="2018-09-13T15:52:27.723" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="15787" PostTypeId="1" AcceptedAnswerId="15793" CreationDate="2015-05-31T03:45:39.420" Score="9" ViewCount="1505" Body="&lt;p&gt; &lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Fill in the blank below (as indicated by the underscore: &amp;quot;_&amp;quot;) to complete the unique sequence:&lt;/p&gt;&#xA;&lt;p&gt;    &lt;strong&gt;T _ M R P&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt; &lt;/p&gt;&#xA;&lt;p&gt; &lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Rules/hints/clarifications:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The answer is to be in the form of an American-English alphabetical character, that is A through Z, for a total of 26 possible answer candidates values.&lt;/li&gt;&#xA;&lt;li&gt;The poster should be able to explain WHY supplied answer is the answer (so someone cant win with a lucky guess).&lt;/li&gt;&#xA;&lt;li&gt;Ignore white-space, ignore case; the spaces are not a part of the riddle, I just put them there for formatting. Similarly, the case of the character is irrelevant; I'm looking for one of the 26 American-English alphabetical characters.&lt;/li&gt;&#xA;&lt;li&gt;The presence of a hint character (T, M, R, or P) does not preclude it from being in the answer. In other words, just because the characters T, M, R, and P are already in the sequence does not mean that it eliminates them as a possibility for being the answer.&lt;/li&gt;&#xA;&lt;li&gt;There is a discrete answer, this is not a 'trick question'. The sequence is not arbitrary, and represents a real sequence that exists in the real world, and can be observed &lt;em&gt;IN NATURE&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;li&gt;In fact, I am almost certain that everyone who reads this has observed this sequence before.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;In fact, I am almost certain that you can observe this pattern right now, from where you sit.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;There is actually 2 acceptable answers, that is, there are actually 2 different characters that you can put in blank and make a valid sequence, and thus, is acceptable as a correct answer.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;I will answer any questions about the puzzle, as long as they are phrased as a YES/NO question.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Good luck!&lt;/p&gt;&#xA;" OwnerUserId="11226" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-31T09:24:52.530" Title="Complete the (alpha) sequence riddle" Tags="|riddle|pattern|lateral-thinking|letter-sequence|connections-puzzle|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15789" PostTypeId="1" AcceptedAnswerId="15791" CreationDate="2015-05-31T04:48:03.187" Score="6" ViewCount="438" Body="&lt;p&gt;There is a strange meteor that is caught by the Earth's gravitational pull. No one even notices it because it is so small it will simply burn up in the atmosphere -- but it doesn't. Instead, it survives it's own ball of flame and hits the ground and makes a large explosion around itself without damaging itself. The FBI is sent to investigate. When they arrive, they walk up to it (considering it didn't damage the terrain either somehow) and someone touches it. Then, it opens in two and out spills a chunk of an unknown substance. Quickly, the FBI takes this substance and analyses it. It turns out that it isn't made of atoms, but rather extremely small particles as of yet unknown to science. There is also a very specific number of them (see below). Later, 7 more meteors arrive in groups of two that don't open when touched. They determine that it was aliens who sent this as a method of first contact. You are tasked with figuring out what the aliens are trying to say.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specific number of particles:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="12879" LastActivityDate="2015-05-31T05:40:20.657" Title="First Contact With Aliens" Tags="|mathematics|cipher|pattern|number-sequence|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15791" PostTypeId="2" ParentId="15789" CreationDate="2015-05-31T05:34:04.337" Score="11" Body="&lt;p&gt;The solution is this picture:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;11111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111&#xA;11111111111111111111111111111111111111111111111111111111111111111100001111111111111111111111111111111111111111111111111111111111&#xA;11111111111111111111111111111111111111111111111111111111111111111000000111111111111111111111111111111111111111111111111111111111&#xA;11111111111111111111111111111111111111111111111111111111111111110000000011111111111111111111111111111111111111111111111111111111&#xA;11111111111111111111111111111111111111111111111111111111111111100000000001111111111111111111111111111111111111111111111111111111&#xA;11111111111111111111111111111111111111111111111111111111111111000000000000111111111111111111111111111111111111111111111111111111&#xA;11111111111111111111111111111111111111111111111111111111111110000011100000011111111111111111111111111111111111111111111111111111&#xA;11111111111111111111111111111111111111111111111111111111111100000111111000001111111111111111111111111111111111111111111111111111&#xA;11111111111111111111111111111111111111111111111111111111111100001111111100000111111111111111111111111111111111111111111111111111&#xA;11111111111111111111111111111111111111111111111111111111111000011111111110000011111111111111111111111111111111111111111111111111&#xA;11111111111111111111111111111111111111111111111111111111110000011111111111000001111111111111111111111111111111111111111111111111&#xA;11111111111111111111111111111111111111111111111111111111100000111111111111000001111111111111111111111111111111111111111111111111&#xA;11111111111111111111111111111111111111111111111111111111100001111110111111100000111111111111111111111111111111111111111111111111&#xA;11111111111111111111111111111111111111111111111111111111000011111110111111110000011111111111111111111111111111111111111111111111&#xA;11111111111111111111111111111111111111111111111111111110000011111110011111111000001111111111111111111111111111111111111111111111&#xA;11111111111111111111111111111111111111111111111111111110000111111110011111111000001111111111111111111111111111111111111111111111&#xA;11111111111111111111111111111111111111111111111111111100001111111110011111111100000111111111111111111111111111111111111111111111&#xA;11111111111111111111111111111111111111111111111111111000011111111100011111111110000011111111111111111111111111111111111111111111&#xA;11111111111111111111111111111111111111111111111111111000011111111100011111111110000011111111111111111111111111111111111111111111&#xA;11111111111111111111111111111111111111111111111111110000111111111100011111111111000001111111111111111111111111111111111111111111&#xA;11111111111111111111111111111111111111111111111111100001111111111100001111111111100000111111111111111111111111111111111111111111&#xA;11111111111111111111111111111111111111111111111111100001111111111100001111111111100000111111111111111111111111111111111111111111&#xA;11111111111111111111111111111111111111111111111111000011111111111100001111111111110000011111111111111111111111111111111111111111&#xA;11111111111111111111111111111111111111111111111111000011111111111100001111111111110000011111111111111111111111111111111111111111&#xA;11111111111111111111111111111111111111111111111110000111111111111000001111111111111000001111111111111111111111111111111111111111&#xA;11111111111111111111111111111111111111111111111100001111111111111000001111111111111100000111111111111111111111111111111111111111&#xA;11111111111111111111111111111111111111111111111100001111111111111000001111111111111100000111111111111111111111111111111111111111&#xA;11111111111111111111111111111111111111111111111000011111111111111000001111111111111110000011111111111111111111111111111111111111&#xA;11111111111111111111111111111111111111111111111000011111111111111000000111111111111110000011111111111111111111111111111111111111&#xA;11111111111111111111111111111111111111111111110000111111111111111000000111111111111111000001111111111111111111111111111111111111&#xA;11111111111111111111111111111111111111111111110000111111111111111000000111111111111111000001111111111111111111111111111111111111&#xA;11111111111111111111111111111111111111111111100001111111111111110000000111111111111111100000111111111111111111111111111111111111&#xA;11111111111111111111111111111111111111111111100001111111111111110000000111111111111111100000111111111111111111111111111111111111&#xA;11111111111111111111111111111111111111111111000011111111111111110000000111111111111111110000011111111111111111111111111111111111&#xA;11111111111111111111111111111111111111111111000011111111111111110000000011111111111111110000011111111111111111111111111111111111&#xA;11111111111111111111111111111111111111111110000011111111111111110000000011111111111111111000001111111111111111111111111111111111&#xA;11111111111111111111111111111111111111111110000111111111111111100000000011111111111111111000001111111111111111111111111111111111&#xA;1111111111111111111111111111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depicts &lt;a href=&quot;http://en.memory-alpha.wikia.com/wiki/Starfleet_insignia&quot;&gt;Starfleet insignia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the solution, simply convert the number into binary and split it into groups of 128 with newlines in between.&lt;/p&gt;&#xA;" OwnerUserId="7308" LastEditorUserId="7308" LastEditDate="2015-05-31T05:40:20.657" LastActivityDate="2015-05-31T05:40:20.657" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15793" PostTypeId="2" ParentId="15787" CreationDate="2015-05-31T09:24:52.530" Score="14" Body="&lt;p&gt;The answer is &lt;strong&gt;P&lt;/strong&gt; (or I).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the full list is actually:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;T&lt;/strong&gt;humb&lt;br&gt;&#xA; &lt;strong&gt;P&lt;/strong&gt;ointer (or Index) finger&lt;br&gt;&#xA; &lt;strong&gt;M&lt;/strong&gt;iddle finger&lt;br&gt;&#xA; &lt;strong&gt;R&lt;/strong&gt;ing finger&lt;br&gt;&#xA; &lt;strong&gt;P&lt;/strong&gt;inky  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastActivityDate="2015-05-31T09:24:52.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15794" PostTypeId="2" ParentId="15319" CreationDate="2015-05-31T09:52:44.263" Score="5" Body="&lt;p&gt;The bird is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Larry Bird&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Seek out a sunny picture -&#xA;Remember you're on SE -&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This is the button in the answer box to add a photo.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And on that sunny picture,&#xA;Figure out what you can;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Click the &lt;code&gt;?&lt;/code&gt; to get more help&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are a less capital idea&#xA;Than those who follow you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;advanced help&lt;/code&gt; is in all lowercase letters, but the rest of the options are capitalized. If you go to the advanced help link, you can find this section mid page:&#xA;&lt;br&gt;&lt;br&gt;&#xA;Need More Detail?&#xA;&lt;br&gt;&lt;br&gt;&#xA;Visit the &lt;a href=&quot;http://daringfireball.net/projects/markdown/syntax&quot;&gt;official Markdown syntax reference page&lt;/a&gt;.&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;The linked page contains a dummy link to search for Larry Bird images.&#xA;&lt;br&gt;&#xA;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12242" LastActivityDate="2015-05-31T09:52:44.263" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15795" PostTypeId="2" ParentId="15759" CreationDate="2015-05-31T10:25:30.840" Score="2" Body="&lt;h3&gt;Aswer: Hello :D&lt;/h3&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; This is simply the binary ASCII stream encoded as a &lt;code&gt;0&lt;/code&gt; followed by the count of consecutive zeroes, then the count of consecutive ones, repeated for each group of zeroes and ones in the message.&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; Example:&lt;br /&gt;&#xA; &lt;code&gt;Test&lt;/code&gt;&lt;br /&gt;&#xA; Is encoded in binary ASCII as:&lt;br /&gt;&#xA; &lt;code&gt;01010100 01100101 01110011 01110100&lt;/code&gt;&lt;br /&gt;&#xA; This is split into groups of zeroes followed by ones:&lt;br /&gt;&#xA; &lt;code&gt;01 01 01 00011 001 01 0111 0011 0111 01 00&lt;/code&gt;&lt;br /&gt;&#xA; Each group is then converted to a &lt;code&gt;0&lt;/code&gt;, followed by the number of zeroes, followed by the number of ones:&lt;br /&gt;&#xA; &lt;code&gt;011 011 011 032 021 011 013 022 013 011 02&lt;/code&gt;&lt;br /&gt;&#xA; The result is concatenated to produce the final string:&lt;br /&gt;&#xA; &lt;code&gt;01101101103202101101302201301102&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt; &lt;/p&gt;&#xA;&lt;p&gt;The real magic word is of course 011011052012032021011012041013022012021011&lt;/p&gt;&#xA;" OwnerUserId="2186" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-06-01T04:40:38.667" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="15796" PostTypeId="1" AcceptedAnswerId="15797" CreationDate="2015-05-31T18:45:33.987" Score="6" ViewCount="1135" Body="&lt;p&gt;You are working in a ball factory, where &lt;strong&gt;balls of weight 1U, 2U, 3U ... 8U are made in 8 sections&lt;/strong&gt;. One day, you are given 1023 black balls , all (except one special ball) from section 1, meaning weight = 1U (1U = 100 gram, but that is not really relevant) and you have a balancing scale, which shows whether left-is-more &lt;strong&gt;(&amp;lt;)&lt;/strong&gt; or both-are-equal &lt;strong&gt;(=)&lt;/strong&gt; or right-is-more &lt;strong&gt;(&amp;gt;)&lt;/strong&gt;. You also have one green ball of weight 1U.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;All balls have the same size, look the same (except your own green ball) and feel same. All balls were made here, and you have to find which section that special ball (whose weight may be 2U or 3U or ... 8U) came from.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everytime you use the balancing scale, you lose some brownie points (1 brownie point = 4 maroonie points, but that is totally irrelevant here), hence you want to find the answer with the least number of weightings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How will you go about this ?&lt;br&gt;&#xA;Can you prove that your solution uses the least number of weightings ?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="10940" LastEditorUserId="10940" LastEditDate="2018-12-27T18:10:05.850" LastActivityDate="2018-12-27T18:10:05.850" Title="Different Ball Puzzle" Tags="|mathematics|weighing|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="15797" PostTypeId="2" ParentId="15796" CreationDate="2015-05-31T19:09:22.540" Score="10" Body="&lt;p&gt;You appear to need 3 weighings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Split your 1023 balls into 3 equal groups of 341. Compare 2 of these, and you know which group the heavier ball is in. Call the heavier group H and one of the normal groups N&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare H with N + 4 balls. Balancing means it came from section 5, otherwise you know whether it came from 2-4 or 6-8. If it comes from one of these ranges, test against the midpoint (3 or 7) to find where in the range it lies. i.e. test H vs N+2 or H vs N+6. You're done!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof that 3 weighings are required:&lt;/strong&gt;&#xA;We need to distinguish between 7 different possibilities, which requires $log_27 = 2.8$ bits of information. A weighing can give us $log_23 = 1.58$ bits, which appears to make 2 sufficient. Unfortunately, a weighing when we don't know which side has the heavier ball only provides $log_22 = 1$ bits of information about the weight of the ball (Either the ball weighs exactly the right amount to balance the pans, or it does not). Since we start with no knowledge of where the ball is, our first weighing gets us only 1 bit about the weight of the ball, and we still require 1.8 bits - too much for the 1.58 bit maximum from a balanced weighing with knowledge of where the heavy ball is. &lt;/p&gt;&#xA;" OwnerUserId="3987" LastEditorUserId="3987" LastEditDate="2015-05-31T21:44:14.720" LastActivityDate="2015-05-31T21:44:14.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15801" PostTypeId="1" AcceptedAnswerId="15802" CreationDate="2015-05-31T20:55:30.797" Score="19" ViewCount="2285" Body="&lt;p&gt;You awake in a weird cave with something strange written on the wall. You find a small piece of paper on a rock. It says the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Solving each code and entering it into the password entering device, or the P.E.D. for short, will unlock the next room in the cave.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rules:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I will only accept answers which fully explain their reasoning for EVERYTHING.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If your answer comes up at about the same time as another person's, I will accept the answer based on quality of your work. (Note: I will still upvote both)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Note that it will get very difficult, and some stages may feel impossible to solve, but I bet somebody will solve it, because someone always solves it on PuzzlingSE, and everyone here is awesome at puzzles!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There's technically nothing stopping you from going to a room you aren't on, so I want you to pinky promise to me that you won't. ;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Codes are in &lt;strong&gt;bold&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Good Luck!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room 9 Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This has something to do with morse code.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;=========================================================================&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room #1&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A sign above the P.E.D. says &lt;strong&gt;eerht-ytfif&lt;/strong&gt; and &lt;strong&gt;.retal lufesu eb yam siht&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room #2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You see lots of letters on the wall that say &lt;strong&gt;XXIV&lt;/strong&gt;. You also see a carving that says &quot;xis kcab evom&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room #3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Carvings on the wall say &lt;strong&gt;Tjyuz&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room #4:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You see a sign hanging from the wall that says &lt;strong&gt;Bismuth&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room #5:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You talk to a man who says &quot;&lt;em&gt;I'm a prisoner here too. Here, take this slip of paper I found 3 rooms ago. I'm not sure what it means, but I'm sure you'll be able to figure it out.&lt;/em&gt;&quot; You look at the paper and see the text: &lt;strong&gt;vrayDBO&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room #6:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You a long plaque taking the whole left wall that says &lt;strong&gt;010101000111011101100101011011100111010001111001001011010101010001110111 01101111&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room #7:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You find a man, no doubt crazy from being locked in the cave, say that he knows the code, but will only tell you in riddle form. He says &lt;strong&gt;The code is of 5 numbers, and each number is different. The sum of all its numbers equals 21. The second number is twice the third, the third is twice the first, and the first is twice the fourth. The fifth number is twice the difference of the third number minus the fourth number.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room #8:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You find a massive wall with a massive plaque that says: &lt;strong&gt;00110000 00110000 00110001 00110001 00110000 00110000 00110000 00110000 00100000 00110000 00110000 00110001 00110001 00110000 00110000 00110000 00110001 00100000 00110000 00110000 00110001 00110001 00110000 00110000 00110000 00110000 00100000 00110000 00110000 00110001 00110001 00110000 00110000 00110000 00110000 00100000 00110000 00110000 00110001 00110001 00110000 00110000 00110000 00110000 00100000 00110000 00110000 00110001 00110001 00110000 00110000 00110000 00110001 00100000 00110000 00110000 00110001 00110001 00110000 00110000 00110000 00110001 00100000 00110000 00110000 00110001 00110001 00110000 00110000 00110000 00110000 00100000 00110000 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00110000 00110000 00110000 00110001 00100000 00110000 00110000 00110001 00110001 00110000 00110000 00110000 00110001 00100000 00110000 00110000 00110001 00110001 00110000 00110000 00110000 00110001 00100000 00110000 00110000 00110001 00110001 00110000 00110000 00110000 00110000&lt;/strong&gt;. You wonder what this long string of numbers could mean. You see another smaller plaque that says &lt;strong&gt;URURURURURUXZ&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room #9:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A man says: &quot;&lt;em&gt;I think the last room is next. I just can't solve this puzzle! Here, you try it!&lt;/em&gt;&quot; The man hands you a slip of paper which says &lt;strong&gt;VI AO CU I B NIJL EGKHJI UIO E AOEH E GA&lt;/strong&gt; You are utterly confused and do not know what this could even mean.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room #10&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There is a sign that says: &lt;em&gt;You have gone this far, and solved many puzzles, but now you are nearing the end. If you have paid close attention, then the following should be easy.&lt;/em&gt; Just as you finish reading it, a panel pops up from the wall. It shows one code: &lt;strong&gt;OUR 5mooR 85 11100110011101101001011001101110001101101111011011001110 QVAAMFT&lt;/strong&gt;. If you can find a way to solve it then you and your new friends can get out of the weird cave and to safety! The P.E.D. comes up through the wall, but this time you can enter letters instead of numbers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11002" LastEditorUserId="11002" LastEditDate="2015-06-22T16:47:34.283" LastActivityDate="2015-06-26T23:16:43.203" Title="Escape The Strange, Code-Filled Cave" Tags="|logical-deduction|password|code|" AnswerCount="4" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="15802" PostTypeId="2" ParentId="15801" CreationDate="2015-05-31T22:24:52.673" Score="12" Body="&lt;p&gt;Room 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Read the words backward.&lt;br&gt;&#xA; Additional message: &lt;strong&gt;this may be useful later.&lt;/strong&gt;&lt;br&gt;&#xA; Password: fifty-three (&lt;strong&gt;53&lt;/strong&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Room 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Roman numerals: XXIV=24&lt;br&gt;&#xA; Additional message (read backward): &lt;strong&gt;move back six.&lt;/strong&gt;&lt;br&gt;&#xA; Password: &lt;strong&gt;24&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Room 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Caesar cipher with key 53 (from room 1), or 1 (mod 26). Shift backwards.&lt;br&gt;&#xA; Password: sixty (&lt;strong&gt;60&lt;/strong&gt;)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Room 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Periodic table of elements: Bismuth is element 83.&lt;br&gt;&#xA; Password: &lt;strong&gt;83&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Room 5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In room 2, use the additional message for Caesar shift 6 letters backward.&lt;br&gt;&#xA; Message: plusXVI, or plus 16&lt;br&gt;&#xA; Password: 24 + 16 = &lt;strong&gt;40&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Room 6:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Binary&lt;br&gt;&#xA; Password: twenty-two (&lt;strong&gt;22&lt;/strong&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Room 7:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Assume password has format $xyzab$.&#xA; We are given the following:&lt;br&gt;&#xA; $$x+y+z+a+b=21$$&lt;br&gt;&#xA; $$y=2z$$&lt;br&gt;&#xA; $$z=2x$$&lt;br&gt;&#xA; $$x=2a$$&lt;br&gt;&#xA; $$b=2(z-a)$$&lt;br&gt;&#xA; Therefore,&lt;br&gt;&#xA; $$y=2z=4x=8a$$&lt;br&gt;&#xA; $$z=2x=4a$$&lt;br&gt;&#xA; $$b=2(4a-a)=6a$$&lt;br&gt;&#xA; So,&lt;br&gt;&#xA; $$x+y+z+a+b=2a+8a+4a+a+6a=21a=21$$&lt;br&gt;&#xA; We get $a=1$.&lt;br&gt;&#xA; Password: &lt;strong&gt;28416&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Room 8:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Please see the excellent answers given by &lt;a href=&quot;https://puzzling.stackexchange.com/a/15805/12109&quot;&gt;MisterEman22&lt;/a&gt;, &lt;a href=&quot;https://puzzling.stackexchange.com/a/15808/12109&quot;&gt;rand al'thor&lt;/a&gt;, and &lt;a href=&quot;https://puzzling.stackexchange.com/a/15803/12109&quot;&gt;HDE 226868&lt;/a&gt;. I couldn't find a binary translator willing to take such a long string of numbers!&lt;br&gt;&#xA;&lt;br&gt;&#xA; Take the binary text and convert that to text. Convert that to text again. Do it again to get number 57.&lt;br&gt;&#xA; Password: &lt;strong&gt;57&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Room 9:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/a/15808/12109&quot;&gt;rand al'thor&lt;/a&gt; got this one first (as usual):&lt;br&gt;&#xA; Turn each vowel into a dot and each consonant into a dash. Translate morse to text.&lt;br&gt;&#xA; Morse: &lt;code&gt;-.  ..  -.  .  -  -.--  .----.  ...  .  ...-  .  -.&lt;/code&gt;&lt;br&gt;&#xA; Password: ninety?seven (&lt;strong&gt;97&lt;/strong&gt;)&lt;br&gt;&#xA; Note: there was probably a mistake in encoding the character &lt;code&gt;-&lt;/code&gt;. In morse, that should be &lt;code&gt;-....-&lt;/code&gt;, which was written as &lt;code&gt;.---.&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Room 10:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The encryption method used previously were:&lt;br&gt;&#xA; - backwards&lt;br&gt;&#xA; - roman numerals&lt;br&gt;&#xA; - caesar cipher (1 and 6 letters backwards)&lt;br&gt;&#xA; - periodic table of elements&lt;br&gt;&#xA; - binary (1 and 3 times)&lt;br&gt;&#xA; - math/algebra&lt;br&gt;&#xA; - morse&lt;br&gt;&#xA;&lt;br&gt;&#xA; One part at a time:&lt;br&gt;&#xA; &lt;strong&gt;OUR&lt;/strong&gt;: U (morse)&lt;br&gt;&#xA; &lt;strong&gt;5mooR&lt;/strong&gt;: Room5 (backwards)&lt;br&gt;&#xA; &lt;strong&gt;85&lt;/strong&gt;: At (element astatine, which is number 85)&lt;br&gt;&#xA; &lt;strong&gt;11100110011101101001011001101110001101101111011011001110&lt;/strong&gt;: solving (backwards, binary)&lt;br&gt;&#xA; &lt;strong&gt;QVAAMFT&lt;/strong&gt;: PUZZLES (caesar cipher backwards by 1)&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Room5&lt;/strong&gt; decoded (credit to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15801/escape-the-strange-code-filled-cave/15802?noredirect=1#comment53246_15802&quot;&gt;Zandar&lt;/a&gt; for this one): The answer for Room 5 is 40. In roman numerals, that would be &lt;strong&gt;XL&lt;/strong&gt;, or &lt;strong&gt;excel&lt;/strong&gt; in english.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Together, that would be: &lt;strong&gt;U XL At solving PUZZLES&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; In english, that would be: &lt;strong&gt;You excel at solving puzzles&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12109" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-06-26T23:16:43.203" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="15803" PostTypeId="2" ParentId="15801" CreationDate="2015-05-31T22:25:36.413" Score="3" Body="&lt;h1&gt;The first 8 Rooms&lt;/h1&gt;&#xA;&lt;p&gt;&lt;strong&gt;Room 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &amp;quot;eerht-ytfif and .retal lufesu eb yam siht&amp;quot; becomes &lt;strong&gt;fifty-three&lt;/strong&gt; and &lt;strong&gt;this may be useful later.&lt;/strong&gt; Simply write them in reverse.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Room 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &amp;quot;XXIV&amp;quot; is &lt;strong&gt;24&lt;/strong&gt; in Roman numerals. &amp;quot;xis kcab evom&amp;quot; backwards is &amp;quot;move back six&amp;quot;, so 24 - 6 is &lt;strong&gt;18&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Room 3:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Shift each letter on &amp;quot;Tjyuz&amp;quot; one backward (e.g. &amp;quot;b&amp;quot; would be &amp;quot;a&amp;quot;). This gives you &lt;strong&gt;Sixty&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Room 4:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Bismuth&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bismuth&lt;/a&gt; is element number &lt;strong&gt;83&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Room 5:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The paper was found three rooms ago, in Room 2, where &amp;quot;move back six&amp;quot; was found. Move each letter back six spaces (e.g. &amp;quot;g&amp;quot; would be &amp;quot;a&amp;quot;. &amp;quot;vrayDBO&amp;quot; becomes &amp;quot;plus XVI&amp;quot;, which is &lt;strong&gt;plus 16&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Room 6:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; That sequence of digits is in binary; it means &lt;strong&gt;Twenty-Two&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Room 7:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Just use algebra here. Bypassing writing it all out, if the numbers are a,b,c,d,e, then d=1, a=2, b=8, c=4, and e=6, giving &lt;strong&gt;2-8-4-1-6&lt;/strong&gt;. Set up a system of equations, where a+b+c+d+e+21, and go from there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Room 8:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Translate the string from binary to normal numbers, then translate &lt;em&gt;that&lt;/em&gt; from binary to normal numbers. You get &lt;strong&gt;Fifty-seven&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5300" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-05-31T22:25:36.413" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15804" PostTypeId="1" AcceptedAnswerId="15814" CreationDate="2015-05-31T22:27:13.087" Score="8" ViewCount="451" Body="&lt;p&gt;Suppose that you build an $n$ by $n$ grid out of metal rods, joined at their ends in such a way that they can rotate about the joints. Below is the grid when $n=3$, along with an illustration of its flexibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SlqXc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This grid can be made rigid by &quot;bracing&quot; sufficiently many of the squares with a diagonal rod. You could brace all the squares, but there are more efficient ways. For example, the below structure is rigid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BtkZo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the fewest number of squares you need to brace to make the grid rigid?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the optimal number is $b$, a good answer should show how to brace the grid with $b$ braces, and convince the reader why any grid with only $b-1$ braced squares is flexible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;em&gt;Mathematical Mind Benders&lt;/em&gt;, Peter Winkler.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-05-31T22:53:14.003" LastActivityDate="2015-06-01T11:17:05.993" Title="Bracing a Flexible Grid" Tags="|mathematics|graph-theory|physics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15805" PostTypeId="2" ParentId="15801" CreationDate="2015-05-31T22:29:09.510" Score="2" Body="&lt;p&gt;Room 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;53&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The sign says fifty-three backwards, and also &quot;This may be useful later.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Room 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;24&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The roman numeral XXIV is 24.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Room 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;60&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Changing &lt;em&gt;Tjyuz&lt;/em&gt; to the letter before each letter (rotation cipher using -1) gives us &lt;em&gt;Sixty&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Room 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;83&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The atomic number for the element Bismuth is 83.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Room 5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;34&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Using the hint from &quot;three rooms back&quot; (room 2) of &quot;move back six&quot;, we use a rotation cipher of -6 to turn vrayDBO into plusXVI which gives us plus 16. So we take the answer of 24 from room 2 and add 16, to get 40.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Room 6:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;22&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Decoding the given numbers from binary to ASCII gives &quot;Twenty-Two&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Room 7:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;28416&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;We start knowing that the number has 5 digits that are all different. Looking from the end, the first number is twice the fourth, so the first must either be 2, 4, 6, or 8 and the fourth must either be 1, 2, 3 or 4.&lt;br&gt;The third is twice the first, meaning the third must now be 2, 4, 6, or 8. This leaves the first to either be 2 or 4 and the fourth to be 1 or 2.&lt;br&gt;The second is twice the third, meaning it can be 2, 4, 6, or 8. This leaves the third to be 2 or 4, so the first must be 2, and the fourth must be 1, which makes the third 4 and the second 8.&lt;br&gt;So right now our number is 2841_. The fifth must be 2 times third minus fourth, so 2(4-1) = 6, so the number is 28416, which sums to 21 so it fits. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Room 8:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;57&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Converting from binary to text gives us another binary code. Converting again gives us another shorter binary code. Converting a third time gives us text of &quot;fifty-seven&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Room 9:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Working on it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Given the characters per word of the given text (2 2 2 1 1 4 3 1 3 1 2) I would say that it isn't a 1:1 conversion of characters. I'm not sure what that points me toward, but vigenere/caesar don't seem to be helping at all. Also looking into binary conversion and reversing the binary conversion doesn't seem to give anything useful.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Room 10:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Working on it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11727" LastEditorUserId="11727" LastEditDate="2015-06-01T01:00:46.163" LastActivityDate="2015-06-01T01:00:46.163" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15806" PostTypeId="2" ParentId="15804" CreationDate="2015-05-31T22:41:11.060" Score="5" Body="&lt;p&gt;An $n\times n$ grid has $(n+1)^2$ vertices, each with two degrees of freedom.  Since we wish the overall grid to have only $3$ degrees of freedom ($2$ for translation and $1$ for rotation), we need&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;2(n+1)^2-3&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;constraints.  Each bar provides a one-dimensional constraint, and we start with $2n^2+2n$ bars (two for the bottom and left sides of each of the $n^2$ squares, and $n$ for each of the top and right sides), so the number of additional bars we need is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;b = (2(n+1)^2-3)-(2n^2+2n) = 2n-1&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Update&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Mike requests a &lt;em&gt;constructive&lt;/em&gt; proof in the question, so here's an attempt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $n=1$, we can trivially stiffen the grid by stiffening the only square in the grid.  Thus, $b(1)=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take an $n\times n$ grid that is already stiff, and add one row and column to make an $(n+1)\times(n+1)$ grid.  The additional row and column can each bend independently of each other.  If we stiffen the new corner, we can still bend (as seen in the OP's first diagram).  Thus, in order to stiffen the new grid, we must stiffen one square in the new row &lt;em&gt;and&lt;/em&gt; one in the new column, increasing $b$ by $2$.  Therefore $b(n+1)=b(n)+2$, or equivalently $b(n)=2n-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This proof by induction gives us a general method to construct any minimal stiffening of a grid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also leads to a set of rules to determine whether an arbitrary grid is stiff.  A stiff grid must have a stiffened square in each row and each column, and every stiff square shares a column or row with another stiff square.&lt;/p&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="9718" LastEditDate="2015-06-01T01:38:56.547" LastActivityDate="2015-06-01T01:38:56.547" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15807" PostTypeId="2" ParentId="15796" CreationDate="2015-05-31T22:44:29.050" Score="5" Body="&lt;p&gt;Here is a slightly different argument why 2 weighings are insufficient. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that your first weighing is $x$ balls against $y$ (doesn't matter if any are green), where $x\le y$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the oddball is on the $y$ ball side, then the scale will certainly tip towards $y$. This happens no matter what the oddball weighs, so you learn nothing about its weight in this case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, it is possible to be left with 7 possible scenarios after your first weighing; in those cases, one additional weighing is not enough.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-06-01T02:52:53.537" LastActivityDate="2015-06-01T02:52:53.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15808" PostTypeId="2" ParentId="15801" CreationDate="2015-05-31T23:28:35.227" Score="6" Body="&lt;p&gt;&lt;strong&gt;Room #1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Password is &lt;strong&gt;53&lt;/strong&gt;.&lt;br&gt;&lt;br&gt;Read the messages backwards to get &quot;fifty-three&quot; and &quot;this may be useful later.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room #2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Password is &lt;strong&gt;24&lt;/strong&gt;.&lt;br&gt;&lt;br&gt;The letters on the wall are 24 in Roman numerals; the carving backwards is &quot;move back six&quot; which will be used later.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room #3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Password is &lt;strong&gt;60&lt;/strong&gt;.&lt;br&gt;&lt;br&gt;Caesar-shift &quot;Tjyuz&quot; back 1 place (T goes to S, J goes to I, Y to X, U to T, Z to Y) to get &quot;Sixty&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room #4&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Password is &lt;strong&gt;83&lt;/strong&gt;.&lt;br&gt;&lt;br&gt;Bismuth has atomic number 83 on the periodic table.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room #5&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Password is &lt;strong&gt;40&lt;/strong&gt;.&lt;br&gt;&lt;br&gt;Three rooms ago the password was 24; Caesar-shift &quot;vrayDBO&quot; six places backwards (V goes to P, R goes to L, A to U, Y to S, D to X, B to V, O to I) to get &quot;plusXVI&quot;, so we add XVI=16 to 24 to get 40.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room #6&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Password is &lt;strong&gt;22&lt;/strong&gt;.&lt;br&gt;&lt;br&gt;Convert the given number from binary ASCII code to characters (01010100 goes to T, 01110111 goes to w, 01100101 to e, 01101110 to n, 01110100 to t, 01111001 to y, 00101101 to - (hyphen), 01010100 to T, 01110111 to w, 01101111 to o) to get &quot;Twenty-Two&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room #7&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Password is &lt;strong&gt;28416&lt;/strong&gt;.&lt;br&gt;&lt;br&gt;We need five numbers $a,b,c,d,e$ summing to 21 such that $b=2c,c=2a,a=2d$ - giving $b=8d$ and so, since $b&amp;lt;10$, we have $d=1,a=2,c=4,b=8$ - and also $e=2(c-d)=6$, so the answer is 28416. Check: 2+8+4+1+6=21.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room #8&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Password is &lt;strong&gt;57&lt;/strong&gt;.&lt;br&gt;&lt;br&gt;All the numbers on the massive plaque are ASCII code for either 0 or 1 or space. Converting these ASCII codes to characters, we get 00110000 00110001 00110000 00110000 00110000 00110001 00110001 00110000 00100000 00110000 00110001 00110001 00110000 00110001 00110000 00110000 00110001 00100000 00110000 00110001 00110001 00110000 00110000 00110001 00110001 00110000 00100000 00110000 00110001 00110001 00110001 00110000 00110001 00110000 00110000 00100000 00110000 00110001 00110001 00110001 00110001 00110000 00110000 00110001 00100000 00110000 00110000 00110001 00110000 00110001 00110001 00110000 00110001 00100000 00110000 00110001 00110000 00110001 00110000 00110000 00110001 00110001 00100000 00110000 00110001 00110001 00110000 00110000 00110001 00110000 00110001 00100000 00110000 00110001 00110001 00110001 00110000 00110001 00110001 00110000 00100000 00110000 00110001 00110001 00110000 00110000 00110001 00110000 00110001 00100000 00110000 00110001 00110001 00110000 00110001 00110001 00110001 00110000. All of these are also ASCII code for either 0 or 1 or space, so we can perform the same conversion to get 01000110 01101001 01100110 01110100 01111001 00101101 01010011 01100101 01110110 01100101 01101110. Now convert these from ASCII codes to characters to get &quot;Fifty-Seven&quot;. (I'm not sure yet what the significance of &quot;URURURURURUXZ&quot; is.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room #9&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Password is &lt;strong&gt;97&lt;/strong&gt;.&lt;br&gt;&lt;br&gt;Treat each consonant as a dash and each vowel as a dot in Morse code (VI goes to .- goes to N, AO goes to .. goes to I, CU goes to .- goes to N, etc.) to turn &quot;VI AO CU I B NIJL EGKHJI UIO E AOEH E GA&quot; into &quot;ninety-seven&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room #10&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;[Work in progress]&lt;/em&gt;&lt;br&gt;&lt;br&gt;&quot;5mooR&quot; is &quot;Room5&quot; backwards. Turning the binary code backwards gives 01110011 01101111 01101100 01110110 01101001 01101110 01100111; converting this from ASCII codes to characters gives &quot;solving&quot;. &quot;QVAAMFT&quot; Caesar-shifted back by 1 is &quot;PUZZLES&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-06-22T16:57:35.433" LastActivityDate="2015-06-22T16:57:35.433" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="15809" PostTypeId="1" AcceptedAnswerId="18673" CreationDate="2015-06-01T00:10:49.860" Score="8" ViewCount="635" Body="&lt;p&gt;Given the modest interest in the question &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15763/the-lord-of-the-puzzles&quot;&gt;here&lt;/a&gt;, I thought you might enjoy another of J.R.R Tolkien's rebus letters, this one written at the tender age of 11.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This one is (in theory) slightly easier to decipher but alas the image quality is slightly lower:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zGsdQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4852" LastEditorUserId="5373" LastEditDate="2018-11-10T10:36:20.080" LastActivityDate="2018-11-10T10:36:20.080" Title="The Lord of the Puzzles (pt II)" Tags="|rebus|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15810" PostTypeId="2" ParentId="10933" CreationDate="2015-06-01T03:10:35.537" Score="0" Body="&lt;p&gt;How this sequence works:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Set your number as A, and set N as zero &#xA;&#xA;While A is greater than or equal to 2,&#xA;Take the prime factorization of A&#xA;Add 1 to N and set that as N&#xA;Take the greatest prime factor&#xA;If it is 2, then break.&#xA;If it is greater than 2, set A as the greatest prime factor + 1&#xA;End&#xA;Put N as the output&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The answer is &lt;strong&gt;3&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="167" LastEditorUserId="167" LastEditDate="2015-06-02T02:35:06.113" LastActivityDate="2015-06-02T02:35:06.113" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15814" PostTypeId="2" ParentId="15804" CreationDate="2015-06-01T11:17:05.993" Score="3" Body="&lt;p&gt;Partition the bars of the grid into $2n$ bundles of $n+1$ bars each in the most canonical way. Each of these bundles is necessarily parallel to begin with and stays parallel under any minor deformation of the grid. On the other hand, the orientations of these bundles are completely independent of each other, so for any tuple of $2n$ orientations, you could align the unreinforced grid so the $k$th bundle has the $k$th orientation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now bracing a single square constrains two of these bundles (the bundles which the square is made of) to be aligned orthogonally, nothing more and nothing less. Let's call two bundles directly constrained if there is a braced square with edges in these bundles. You can imagine this as a graph, where the $2n$ vertices are bundles and the edges are direct constraints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the constrained components of this graph. For each constrained component, the orientation of a single bundle within completely determines the orientation of every other bundle in the same component. Yet the orientation of this component is still independent of the orientation of any bundle not in this component. This means that if we have more than one constrained component, our grid is not rigid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's count the constrained components. With zero braced squares, we have $2n$ components. Every braced square adds exactly one direct constraint, so adding a square reduces the number of codependent components by exactly one. So to make the entire structure rigid, we need to have exactly one constrained component, which means we need at least $2n-1$ braced squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This lower limit can be achieved by bracing $2n-1$ squares in an L shape along two edges, or a staircase shape along a diagonal, or a cross shape produced by any row and any column of squares in the grid, or actually any set of squares induced by any possible subtree of the graph of all possible direct constraints (which is a complete bipartite graph $K_{n,n}$).&lt;/p&gt;&#xA;" OwnerUserId="12905" LastActivityDate="2015-06-01T11:17:05.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15815" PostTypeId="1" CreationDate="2015-06-01T13:20:29.307" Score="-4" ViewCount="10209" Body="&lt;p&gt;Here is my puzzle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can you fit ten horses in nine boxes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the boxes:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/HFxhJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;All boxes must be filled &lt;/li&gt;&#xA;&lt;li&gt;No additional boxes can be created&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Good luck :)&lt;/p&gt;&#xA;" OwnerUserId="12907" LastActivityDate="2015-06-01T19:05:53.103" Title="Fit ten horses in nine boxes" Tags="|logical-deduction|" AnswerCount="6" CommentCount="5" FavoriteCount="0" ClosedDate="2015-06-02T11:48:48.033" ContentLicense="CC BY-SA 3.0" />
  <row Id="15816" PostTypeId="2" ParentId="15815" CreationDate="2015-06-01T13:24:10.047" Score="7" Body="&lt;p&gt;Here is my answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can place the 9 letters of &quot;ten horses&quot; in the boxes:&#xA; T E N H O R S E S&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5359" LastActivityDate="2015-06-01T13:24:10.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15817" PostTypeId="2" ParentId="15815" CreationDate="2015-06-01T13:24:15.870" Score="3" Body="&lt;p&gt;The expected answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;     |1 | 2 |3 | 4 | 5 |6 | 7 |8 | 9 |&#xA;&lt;br&gt; [T][E][N][H][O][R][S][E][S]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now the real challenge is to make up new valid answers..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternate answers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Remove the walls between the boxes to and add them to the end to end up with just 1 longer box, then insert all the horses into the 1 box. (Can now store up to 10-13 horses!)&#xA;&lt;br&gt;&lt;br&gt; Assuming they are toy horses, and grind them up. You can now fit the maximum possible horses into the boxes. (Much greater than 10)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-06-01T14:07:22.273" LastActivityDate="2015-06-01T14:07:22.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15818" PostTypeId="2" ParentId="15815" CreationDate="2015-06-01T13:53:16.457" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Obviously, this. I can't think anything else really.&lt;br/&gt;&lt;br/&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/6uglm.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4506" LastActivityDate="2015-06-01T13:53:16.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15819" PostTypeId="2" ParentId="15815" CreationDate="2015-06-01T13:57:50.180" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Put 8 male horses to first 8 box and put a pregnant female horse to last room.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4506" LastActivityDate="2015-06-01T13:57:50.180" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15820" PostTypeId="2" ParentId="15640" CreationDate="2015-06-01T16:15:00.393" Score="-2" Body="&lt;p&gt;It is heavily dependent on the rarity and collectibility of the coins involved. As we are talking about &lt;em&gt;worth&lt;/em&gt;, not &lt;em&gt;value&lt;/em&gt;, it is subject to much more than just the dollar amount put on the two types of coins. As the problem statement gives no real indicators of &lt;em&gt;worth&lt;/em&gt;, I'm going to call BS and say that there ins't enough information.&lt;/p&gt;&#xA;" OwnerUserId="12134" LastActivityDate="2015-06-01T16:15:00.393" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15821" PostTypeId="2" ParentId="15764" CreationDate="2015-06-01T16:16:14.563" Score="0" Body="&lt;p&gt;I would think in a smaller group, say of 3 or 4 people, this wouldn't work, nor would anything else. I am proposing and solution for a larger group of people. I am also assuming there and only a few color hats (ex. Blue,yellow, red, green, and they are not using crimson red pink dark green light green lime green neon yellow et cetera). Each person in a larger setting could guess not necessarily what color they think they have, but the color they see the most of. In a large setting, where the basic hat colors would have multiple recurrences, eventually someone would ex: n3 says blue but n9 sees 4 yellow hats and only 3 blue hats, and that n3 has on a red hat, so he also saw 4 yellow hats, therefore allowing n9 to determine there is another blue hat. He is wearing a blue hat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another strategy I have devised would work in a smaller setting, more lateral than statistical. Say 3 people. With a one way glass between 2 and 3 where 2 can see 1 and 3 but 3 can only see 1. Whichever person goes first of either 1 or 3 (it was never determined the order in which the guesses would be made), they will guess the color the other is actually wearing. So if 3 guesses before 1 and he says red, 1 then knows he has a red hat.&lt;/p&gt;&#xA;" OwnerUserId="12910" LastActivityDate="2015-06-01T16:16:14.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15822" PostTypeId="2" ParentId="15671" CreationDate="2015-06-01T16:28:06.723" Score="0" Body="&lt;p&gt;Partial answer and I am unsure how to hide the answer but for the 3rd group of numbers the pattern seems pretty obvious of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the sequence of differences between the numbers is 4 4 2 4 4 2 4 4 2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8785" LastEditorUserId="9000" LastEditDate="2015-06-01T16:29:20.133" LastActivityDate="2015-06-01T16:29:20.133" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15823" PostTypeId="1" AcceptedAnswerId="15909" CreationDate="2015-06-01T16:54:39.600" Score="15" ViewCount="1169" Body="&lt;p&gt;Our group has been unjustly accused of many things, and we felt the need to make public our official replies:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Accusation:&lt;/strong&gt; You are all so lazy. You have a job to do but we never even see you move a muscle!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reply:&lt;/strong&gt; We are not lazy. It's true that we usually aren't very mobile, but we still love people, and we help them as much as we can.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Accusation:&lt;/strong&gt; Many people claim to follow you religiously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reply:&lt;/strong&gt; We never claimed to be any kind of leaders, and do not hold any specific religious beliefs. It is true that we have multitudes of followers, but please understand that it was never our intention to be religious leaders.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Accusation:&lt;/strong&gt; We've never heard from 1 of your 5 parts. Why the silence?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reply:&lt;/strong&gt; We apologize, but one of our parts is not as vocal as the others. Please know that his silence is only natural, and is never a personal attack.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Accusation:&lt;/strong&gt; People say that you say all kinds of things. Is this true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reply:&lt;/strong&gt; In truth, we never say any of those things that we are accused of saying, although that accusation is so common that we can only ask that you provide a specific example to which we can reply.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Accusation:&lt;/strong&gt; Many claim that you are great teachers, and that people have learned more from your group than from any religious leader in history.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reply:&lt;/strong&gt; Please know that we are not in competition with any other teachers, past or present. However, teaching is not a crime, and we feel that we have done nothing wrong. Ironically, most of us feel quite used, feeling like nobody likes to spend time with us, but just &quot;pick our brains&quot; in passing.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Accusation:&lt;/strong&gt; Some of you gather in large numbers and in a way that makes the location unappealing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reply:&lt;/strong&gt; One of our goals is to make people's lives more efficient. If we are unappealing, please proactively help us be more organized and orderly. Remember that we aim to serve you and we work for free in order to please you.&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-06-02T14:29:05.320" LastActivityDate="2015-06-12T17:56:57.223" Title="Our Reply to Your Accusations- Who are we?" Tags="|riddle|" AnswerCount="6" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15824" PostTypeId="1" AcceptedAnswerId="15833" CreationDate="2015-06-01T17:42:57.093" Score="5" ViewCount="1013" Body="&lt;p&gt;Probably quite an easy one, just trying my hand at a few cryptic clues:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The twenty-first Greek cried over the loss of lamp's innards. (8)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4729" LastEditorUserId="21234" LastEditDate="2016-07-22T21:03:59.573" LastActivityDate="2016-07-22T21:03:59.573" Title="Cryptic Clue: What am I?" Tags="|cryptic-clues|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15825" PostTypeId="2" ParentId="15815" CreationDate="2015-06-01T17:44:43.893" Score="0" Body="&lt;p&gt;You can double up one of the boxes because these 9 boxes make up the bigger box (box of boxes). You didn't say only 1 horse per box.&lt;/p&gt;&#xA;" OwnerUserId="12477" LastActivityDate="2015-06-01T17:44:43.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15827" PostTypeId="2" ParentId="14831" CreationDate="2015-06-01T19:03:04.483" Score="1" Body="&lt;p&gt;Given that the friends will always take the biggest pieces and more pieces, if the number of pieces is not divisible by three, the absolute most you can get is one third (1/3) of the pizza. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following the rules this is achieved as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Cut the pizza in half with one horizontal cut&lt;/li&gt;&#xA;&lt;li&gt;Rotate the lower half 180 degrees. &lt;/li&gt;&#xA;&lt;li&gt;Put the lower half on top of the upper half, so that they completely overlap&lt;/li&gt;&#xA;&lt;li&gt;Make a cut from the top of the pizza, exactly 60 degrees along the (now half-circle double-layered) pizza, down through the center point of the pizzas horisontal line.&lt;/li&gt;&#xA;&lt;li&gt;Repeat 4, but with 120 degrees. &lt;/li&gt;&#xA;&lt;li&gt;Profit&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Voila, 6 equally sized slices.&lt;/p&gt;&#xA;" OwnerUserId="10109" LastActivityDate="2015-06-01T19:03:04.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15829" PostTypeId="2" ParentId="15815" CreationDate="2015-06-01T19:05:53.103" Score="2" Body="&lt;p&gt;My idea:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If you order the boxes like this:&#xA;&lt;code&gt;&#xA;+---+---+---+---+&#xA;| 1 | 2 | 3 | 4 |&#xA;+---+---+---+---+&#xA;| 5 |       | 6 |&#xA;+---+---+---+---+&#xA;| 7 | 8 | 9 |////&#xA;+---+---+---+////&#xA;&lt;/code&gt;&#xA; you get a spot in the middle that is no box but yet big enough for the tenth horse, and you can get to it by moving box 6 to the shaded zone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12912" LastActivityDate="2015-06-01T19:05:53.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15830" PostTypeId="2" ParentId="15190" CreationDate="2015-06-01T19:26:53.937" Score="0" Body="&lt;p&gt;Given that John also needs 100 pylons for his site:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bob takes the 50 &quot;company X&quot; pylons but calls in tomorrow that he is late. Thus his chef inspects John's site first, where he finds that John has not properly secured and separated his site. As John gets fired and has to return his 50 pylons to the warehouse, Bob can pick them up before driving to his site and use the now returned 50 company X pylons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12912" LastActivityDate="2015-06-01T19:26:53.937" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15831" PostTypeId="2" ParentId="15823" CreationDate="2015-06-01T19:30:16.940" Score="9" Body="&lt;p&gt;Maybe a little bit of a stretch, but how about:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Five Senses.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are all so lazy. You have a job to do but we never even see you move a muscle!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Senses themselves do not have muscles to move, but they help everyone, everyday, in everything we do.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Many people claim to follow you religiously.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many people follow what they feel, see, etc. depending on what they focus on. We believe our eyes, follow our nose, etc.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We've never heard from 1 of your 5 parts. Why the silence?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Perhaps taste. Taste does not really allow for any communication of another person. We can hear them, see them, smell them, feel them, but you probably shouldn't taste them. (Thanks to MarkN!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;People say that you say all kinds of things. Is this true?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The senses and nerves associated with them send tons of signals to the brain depending on the sensation being felt/heard/etc.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Many claim that you are great teachers, and that people have learned more from your group than from any religious leader in history.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Taste (and smell) has taught cooks and plain survival techniques. Touch teaches many crafting skills (along with a whole other slew of items). Sight and hearing allow us to learn from others and just about everything else.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some of you gather in large numbers and in unappealing locations.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some situations our senses can provide an overload of unplesantness. Bad smells with horrible sights and sounds often go together, possibly with pain.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="9951" LastEditDate="2015-06-01T19:45:12.730" LastActivityDate="2015-06-01T19:45:12.730" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15832" PostTypeId="1" AcceptedAnswerId="15853" CreationDate="2015-06-01T19:43:38.643" Score="7" ViewCount="421" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/f9fye.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;br /&gt;&#xA;&lt;sup&gt;&lt;a href=&quot;http://www.imdb.com/title/tt0133152/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Source: Planet of the Apes, 2001&lt;/em&gt;&lt;/a&gt; &lt;/sup&gt;&lt;/p&gt;&#xA;&lt;p&gt;In a world not too different from our own, the roles of men and animals are reversed. It is animals who have dominion over men, and men who submit to animals.&lt;/p&gt;&#xA;&lt;p&gt;In the beginning, animals and men lived in harmony. Yet as time wore on, the animals became more greedy and as their greed for power and control grew, so did their evil nature. Animals enslaved man and made a mockery of their lesser intelligence.  Man was thrown into cages. Entertainment was provided to animal onlookers as men were given puzzles any baby animal was able to solve with ease. A popular pastime was to post puzzles to the Animal committee, which would transmit these to a central screen visible to imprisoned man using a computer and WERTYU-keyboard (not too dissimilar to our own these days). Shrieks of hyena-like laughter could be heard as men read the puzzles on screen &amp;amp; struggled to solve even the simplest of these to earn their meager meals.&lt;/p&gt;&#xA;&lt;p&gt;And then in the 201st year of the baboon,  there came along a man called Sethemba. Sethemba's intelligence seemed much keener than that of his fellow man. He embodied man's hope for a brighter future. As the years passed, the animal kingdom whilst at first amazed, grew more frustrated as Sethemba was able to solve any puzzle given to him. And so it came to be that the animals devised the most challenging puzzles yet. With it came a key to man's freedom. If Sethemba was able to solve this puzzle, animal would give man back his freedom. Man and animal would live as equals once again.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;You are Sethemba. Man's last hope. The puzzle appears on the screen. It is clear to you that you have to find the next letter. Can you help secure man's freedom by solving the puzzle?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lL6hJ.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Text transcription of sequence (for accessibility):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;V, S, F, S, R, O, N, G, R, V, R, J, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12317" LastEditorUserId="60644" LastEditorDisplayName="user20" LastEditDate="2021-11-03T12:37:22.257" LastActivityDate="2021-11-03T12:37:22.257" Title="Planet of Animals" Tags="|story|sequence|letter-sequence|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15833" PostTypeId="2" ParentId="15824" CreationDate="2015-06-01T19:48:36.910" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; FILAMENT : FI (PHI spoken aloud) + LAMENT&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A brief quibble on the clue: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; and now that I think of it, isn't &quot;cried&quot; being used twice?  Once to indicate a soundalike of PHI, and once for &quot;cried over the loss of&quot;?  &quot;over the loss of&quot; can't be lament by itself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Extended explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; By convention, cryptic clues are composed of two parts: the standard definition (similar to a regular crossword clue) and the &quot;subsidiary&quot;, which describes the answer word in various creative ways.  The standard and subsidiary definitions are then combined without indication as to which is which or where the break between them lies.  See the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cryptic_crossword&quot; rel=&quot;nofollow&quot;&gt;wikipedia article&lt;/a&gt; on cryptic crosswords for more info. &lt;br&gt;&lt;br&gt;&#xA; In this case, the standard definition is &quot;lamp's innards&quot;, a clear description of the answer FILAMENT, and the subsidiary is &quot;The twenty-first Greek cried over the loss of.&quot;  Roughly, the first part of that is supposed to indicate the &quot;FI&quot; sound, and the last part to indicate &quot;LAMENT&quot;.&lt;br&gt;&lt;br&gt;&#xA; Although there are no hard-and-fast rules for cryptic clues, there are conventions, and I would say that this clue breaks them because the word &quot;cried&quot; does double-duty: it both indicates a soundalike for PHI (in cryptics, any reference to speech or voice can indicate a homophone) and also is part of the phrase &quot;cried over the loss of&quot; that indicates LAMENT.  Or, to look at it another way, if &quot;cried&quot; does not do double-duty, if it is solely part of &quot;cried over the loss of&quot;, then the first part of the clue, &quot;The twenty-first greek&quot; by itself indicates PHI, not FI.&lt;br&gt;&lt;br&gt;&#xA; One could also complain about &quot;cried&quot; being past tense and &quot;LAMENT&quot; being present tense.  But as I said, these are not hard-and-fast rules, and I think that this clue would probably not raise eyebrows in many of the world's daily cryptic crosswords.  An alternate cluing that hews closer to the cryptic conventions as articulated by Ximenes might be something like&lt;br&gt;&lt;br&gt; &lt;strong&gt;Hear the twenty-first Greek cry over the loss of lamp's innards.&lt;/strong&gt; (8)&lt;br&gt;&#xA; &quot;Hear the twenty-first Greek&quot; = &quot;Homophone of PHI&quot; = FI&lt;br&gt;&#xA; &quot;cry over the loss of&quot; = LAMENT&lt;br&gt;&#xA; &quot;lamp's innards&quot; = FILAMENT&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2374" LastEditorUserId="2374" LastEditDate="2015-06-01T22:03:46.850" LastActivityDate="2015-06-01T22:03:46.850" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15834" PostTypeId="1" AcceptedAnswerId="15840" CreationDate="2015-06-01T20:04:39.150" Score="11" ViewCount="1178" Body="&lt;p&gt;You are given a QWERTY keyboard, and are allowed to choose where you wish to start. You are only able type in the same way a knight is able to move on a chessboard (but in this case, on the keyboard). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;What is the largest proper English word you can make?&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; ┌───┬───┬───┬───┬───┬───┬───┬───┬───┬───┐&#xA; | Q | W | E | R | T | Y | U | I | O | P |&#xA; └─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┘&#xA;   | A | S | D | F | G | H | J | K | L |&#xA;   └─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴───┘&#xA;     | Z | X | C | V | B | N | M |&#xA;     └───┴───┴───┴───┴───┴───┴───┘&#xA;(Thanks to GOTO 0's post for the keyboard)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For some clarification, when moving move the 3 spaces left or right, the knight will turn onto the furthest key. When moving up or down, you can go to either side.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Examples:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;b&gt;(left or right):&lt;/b&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A-&gt;(R or C) and not E or X&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;G-&gt;(E or X/ I or M) and not R or C&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;b&gt;(up or down):&lt;/b&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;W-&gt;(Z or C/ F) to clarify vertical movement: [W-S-X-&gt;Z or C]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;V-&gt;(U or T or E/ S or J) to clarify vertical movement: [V-G-Y-&gt;T or U or E]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="57521" LastEditDate="2020-05-13T20:08:14.247" LastActivityDate="2020-05-13T20:08:14.247" Title="Knight on the keyboard" Tags="|word|no-computers|chess|open-ended|knight-moves|" AnswerCount="5" CommentCount="8" ClosedDate="2021-08-05T08:11:51.707" ContentLicense="CC BY-SA 3.0" />
  <row Id="15835" PostTypeId="2" ParentId="15824" CreationDate="2015-06-01T20:07:21.383" Score="1" Body="&lt;p&gt;How about:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; philosopher&#xA;&lt;p&gt;&#xA; phi (21st Greek letter) + loss + sopher (suffer/cry)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2015-06-01T20:07:21.383" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15836" PostTypeId="2" ParentId="15834" CreationDate="2015-06-01T20:24:35.320" Score="3" Body="&lt;p&gt;The best I can find has 4&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Arch&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3987" LastActivityDate="2015-06-01T20:24:35.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15837" PostTypeId="2" ParentId="15834" CreationDate="2015-06-01T20:34:01.043" Score="1" Body="&lt;p&gt;I swear I didn't use the other answer! (6)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Archon &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12843" LastActivityDate="2015-06-01T20:34:01.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15838" PostTypeId="2" ParentId="15834" CreationDate="2015-06-01T20:43:04.973" Score="2" Body="&lt;p&gt;So far, I've only been able to find some 4-letter words:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Carb: &lt;strong&gt;C&lt;/strong&gt;, C-X-Z -&gt; &lt;strong&gt;A&lt;/strong&gt;, A-S-D -&gt; &lt;strong&gt;R&lt;/strong&gt;, R-F-V -&gt; &lt;strong&gt;B&lt;/strong&gt; &lt;br&gt;&#xA; Gimp: &lt;strong&gt;G&lt;/strong&gt;, G-H-J -&gt; &lt;strong&gt;I&lt;/strong&gt;, I-J-N -&gt; &lt;strong&gt;M&lt;/strong&gt;, M-K-O -&gt; &lt;strong&gt;P&lt;/strong&gt; &lt;br&gt;&#xA; Bibi: &lt;strong&gt;B&lt;/strong&gt;, B-H-U -&gt; &lt;strong&gt;I&lt;/strong&gt;, I-J-N -&gt; &lt;strong&gt;B&lt;/strong&gt;, B-H-U -&gt; &lt;strong&gt;I&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1711" LastEditorUserId="1711" LastEditDate="2015-06-01T20:50:27.060" LastActivityDate="2015-06-01T20:50:27.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15839" PostTypeId="1" AcceptedAnswerId="27992" CreationDate="2015-06-01T20:45:54.067" Score="21" ViewCount="4598" Body="&lt;p&gt;This is another &lt;em&gt;challenge&lt;/em&gt; along the same vein as similar challenges that have shown up on this site, requiring the puzzle to craft the longest possible sentence given a specific requirement.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;What is the longest sentence with &lt;em&gt;n&lt;/em&gt; words, where &lt;em&gt;n&lt;/em&gt; is the number of Scrabble points that each of its words is worth?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Here is an example, for clarification, with Scrabble scores for each letter underneath:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;She has two blue rental cars.  &#xA;141 411 141 3111 111111 3111&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As you can see, the sentence has 6 words, where each word has a Scrabble score of 6.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Included for your convenience is a list of Scrabble letters by points:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 point: e, a, i, o, n, r, t, l, s, u&#xA;2 points: d, g&#xA;3 points: b, c, m, p&#xA;4 points: f, h, v, w, y&#xA;5 points: k&#xA;8 points: j, x&#xA;10 points: q, z &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Further rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;em&gt;New rule:&lt;/em&gt;&lt;/strong&gt; You cannot use words more than once in a sentence. I originally allowed this so that sentences like starplusplus's (&quot;They couldn't teach what they should have taught; imagine trying!) would be possible (two uses of &quot;they&quot;) - unfortunately, it's proved rather degenerate. While it's nice to know the longest &lt;em&gt;possible&lt;/em&gt; sentence, let's find one that doesn't use methoxybenzenes 21 times.&lt;/li&gt;&#xA;&lt;li&gt;Proper nouns are not allowed, even if some form of the same word is in the Scrabble dictionary. This means you can't use &quot;Bob&quot; as a name even though &quot;bob&quot; is listed by another definition.&lt;/li&gt;&#xA;&lt;li&gt;In order for your answer to be accepted, you have to explain what your sentence means. Typing a 30-word sentence without explaining it will likely just confuse me!&lt;/li&gt;&#xA;&lt;li&gt;I'm not going to be super strict on punctuation, but be reasonable; semicolons, dashes and commas can be used to break the sentence into fragments, and contractions are acceptable as well.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;(6/3/2015) Update:&lt;/strong&gt; I'm no longer allowing use of the same word twice in a sentence. All other rules are still in play. I'm hoping this will make for more diverse sentences, as well as more interesting ones!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="20274" LastEditDate="2016-06-13T10:17:40.113" LastActivityDate="2018-06-09T02:03:56.507" Title="Craft the longest Scrabble sentence" Tags="|word|wordplay|language|open-ended|scrabble|" AnswerCount="8" CommentCount="11" ClosedDate="2021-08-05T08:10:54.547" ContentLicense="CC BY-SA 3.0" />
  <row Id="15840" PostTypeId="2" ParentId="15834" CreationDate="2015-06-01T21:15:16.070" Score="7" Body="&lt;p&gt;I can't beat six, unless I'm allowed to use an exotic animal:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Caracara, with 8 letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then here are the old, invalid answers that I'm keeping here so the comments make sense. =P&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seven letter exotic animals: caracal, aracari&lt;br&gt;&#xA;Other six-letter words: carhop, unfull&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will admit I wrote a program for this, though I think I did it before the no-computers tag popped up (or at least I didn't see it). Also apparently it was a bad program since it gave me several invalid answers!&lt;/p&gt;&#xA;" OwnerUserId="10861" LastEditorUserId="10861" LastEditDate="2015-06-03T14:24:23.637" LastActivityDate="2015-06-03T14:24:23.637" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15841" PostTypeId="2" ParentId="15839" CreationDate="2015-06-01T21:43:34.917" Score="6" Body="&lt;p&gt;&lt;strong&gt;n = 15&lt;/strong&gt; (composed of only 3 words)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Buffalo_buffalo_Buffalo_buffalo_buffalo_buffalo_Buffalo_buffalo&quot;&gt;Humongous wiz buffalo humongous wiz buffalo buffalo, buffalo humongous wiz buffalo humongous wiz buffalo buffalo.&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Huge magical bison whom other huge magical bison bully, themselves bully huge magical bison other huge magical bison bully.&lt;/p&gt;&#xA;" OwnerUserId="12875" LastEditorUserId="12875" LastEditDate="2015-06-01T23:05:26.973" LastActivityDate="2015-06-01T23:05:26.973" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15842" PostTypeId="1" AcceptedAnswerId="15845" CreationDate="2015-06-01T22:03:36.843" Score="5" ViewCount="608" Body="&lt;p&gt;After some feedback on my last question I'll attempt a few more cryptic clues:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Roll in attempt to become jury (8)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A string of three, maybe four of all the letters. (4)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Proceed! Upset mafia boss surrounds retreating, uncooked food. (6)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Find praise in adult at ion conference with no tea (9)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Crazy, unneeded veal, won't rise, missing Ed. (10)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I don't think it's pertinent to any clues here, but just in case, I'm a British English speaker; all clues are written in BrEng.&lt;/p&gt;&#xA;" OwnerUserId="4729" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-07-22T21:04:23.463" Title="Cryptic crossword clues" Tags="|cryptic-clues|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15843" PostTypeId="2" ParentId="15839" CreationDate="2015-06-01T22:24:57.733" Score="13" Body="&lt;p&gt;&lt;strong&gt;n = 10&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;They couldn't teach what they should have taught; imagine trying!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For some reason, they were unable to teach the things they should have taught; imagine trying to if &lt;em&gt;you&lt;/em&gt; had been in that situation!&lt;/p&gt;&#xA;" OwnerUserId="5260" LastActivityDate="2015-06-01T22:24:57.733" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15844" PostTypeId="2" ParentId="15839" CreationDate="2015-06-01T22:32:32.237" Score="3" Body="&lt;p&gt;I don't think I'm gonna manage 15 to beat George, but I will attempt to get there progressively:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1: I&#xA;   1&#xA;   ( - Who is it?&#xA;     - I )&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://english.stackexchange.com/a/285/21740&quot;&gt;[See this]&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2: Or on.&#xA;   11 11&#xA;   (It could go under the table. Or on.)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;3: Do go, son.&#xA;   21 21  111&#xA;   ( - I was thinking I shouldn't go Dad.&#xA;     - No. Do go, son.)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;4: Pa, ma isn't near.&#xA;   31  31 111-1 1111&#xA;   (Father, mother is not close by)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;5: Hi, pa's dog did dig&#xA;   41  31-1 212 212 212&#xA;   (Hello, my father's dog previously made holes)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4729" LastEditorUserId="-1" LastEditDate="2017-04-13T12:38:16.170" LastActivityDate="2015-06-02T11:54:03.310" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15845" PostTypeId="2" ParentId="15842" CreationDate="2015-06-01T22:41:27.057" Score="6" Body="&lt;p&gt;1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;tribunal&lt;/strong&gt;: bun (roll) inserted into trial (attempt)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;cord&lt;/strong&gt;: c (letter 3) or d (letter 4) of the alphabet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;adulation&lt;/strong&gt;: no tea (t) in adult. (adult-at-ion). praise&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12875" LastEditorUserId="12875" LastEditDate="2015-06-01T22:59:04.240" LastActivityDate="2015-06-01T22:59:04.240" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15846" PostTypeId="2" ParentId="15842" CreationDate="2015-06-01T22:56:34.120" Score="4" Body="&lt;p&gt;Got a couple:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proceed! Upset mafia boss surrounds retreating, uncooked food. (6)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ONWARD - Proceed, or onward! As one might say; anagram of (upset) DON (mafia boss) around (surrounds) RAW (uncooked food) backwards (retreating)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Crazy, unneeded veal, won't rise, missing Ed. (10)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; UNLEAVENED - As in bread that &quot;won't rise&quot;; anagram of (crazy) UNNEED VEAL (unneeded veal missing the letters of Ed)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12277" LastActivityDate="2015-06-01T22:56:34.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15847" PostTypeId="2" ParentId="15839" CreationDate="2015-06-01T23:13:41.587" Score="4" Body="&lt;h2&gt;n=17&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Buffaloes, quoting buffaloes, zoomed accurately; buffaloes, promenading buffaloes, zoomed; buffaloes, huffing, zoomed accurately; buffaloes, reworking buffaloes, zoomed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Buffaloes moved swiftly and accurately while borrowing sayings from other buffaloes or while blowing air loudly, but they only moved swiftly while displaying or updating other buffaloes.&#xA;&lt;hr&gt;&#xA;Also a nice example with &lt;strong&gt;n=10&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Police police, police police police police, police police; permit pardons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Those who police the police (the police police), who are in turn policed by those who police the police police (the police police police), police the police; allow pardons to be granted!&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-06-01T23:25:46.853" LastActivityDate="2015-06-01T23:25:46.853" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15848" PostTypeId="2" ParentId="15834" CreationDate="2015-06-01T23:54:07.133" Score="1" Body="&lt;p&gt;My Score is 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is how I understand the rules (the letter before the colon can reach each letter after the colon in 1 move):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;N:T,F,U,O,L&#xA;C:W,A,R,Y,H&#xA;Q:D,X&#xA;J:T,P,V&#xA;F:W,U,Z,N&#xA;K:Y,B&#xA;P:J,M&#xA;U:F,V,N,L&#xA;O:H,N&#xA;Y:D,C,B,K,M&#xA;B:R,D,Y,I,K&#xA;I:G,B,M&#xA;E:X,G,V&#xA;H:R,O,C&#xA;T:S,X,V,J,N&#xA;W:Z,F,C&#xA;R:A,Z,C,H,B&#xA;M:Y,G,I,P&#xA;V:E,S,T,U,J&#xA;X:Q,E,T,G&#xA;D:Q,Y,B&#xA;Z:W,R,F&#xA;S:T,V&#xA;L:U,N&#xA;A:R,C&#xA;G:E,I,X,M&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Despite the no-computers tag, I was curious if it's really that hard to find words. It seems to be: Trying this against the word list in /usr/share/dict/words under linux with almost 100k english words only got these (omitting anything below 4 letters):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RACY ARCH LUVS LULU CARA MIMI&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5827" LastEditorUserId="5827" LastEditDate="2015-06-02T00:18:19.840" LastActivityDate="2015-06-02T00:18:19.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15849" PostTypeId="2" ParentId="86" CreationDate="2015-06-02T02:54:17.103" Score="3" Body="&lt;p&gt;Say you have a $5 \times 5 \times 5$ puzzle, that is you have five things that each have three attributes out of a list of five.  You are asked to identify the attributes of each thing.  The first thing can have $5^3=125$ possibilities.  The second can have $4^3=64$ and so on giving $(5!)^3=1728000$ lists, but we can permute the list in $5!=120$ ways, so there are $5!^2=14400$ solutions to such a puzzle.  A clue like &quot;the A (one attibute) is B (another) reduces the number of acceptable solutions by a factor $5$.  You need enough clues to reduce $14400$ to $1$, but a second clue may not give as much of a reduction as if it were first&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2015-06-02T02:54:17.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15850" PostTypeId="2" ParentId="15261" CreationDate="2015-06-02T04:33:37.887" Score="6" Body="&lt;p&gt;It's not possible to determine exactly the code to enter on any given day, but we can narrow it down a great deal as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, let's define the digits of the &lt;em&gt;Panel&lt;/em&gt; as being &lt;em&gt;Pn&lt;/em&gt; and the digits of the &lt;em&gt;Code&lt;/em&gt; as being &lt;em&gt;Cn&lt;/em&gt;. Now, there are various mathematical relationships that can be applied, to relate digits of the code to digits of the panel:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$C2 - C3 = P3$&lt;/li&gt;&#xA;&lt;li&gt;$C5 \mod C1 = P1$ &lt;/li&gt;&#xA;&lt;li&gt;$C1(C2 + C3) = P4,P5$&lt;/li&gt;&#xA;&lt;li&gt;$C1 - C3 + C4 = P2$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thus we have a set of equations, which collectively cover all the digits of the code. However, because there's ambiguity (eg. with the mod operator), we still can't directly determine any given code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the latest panel value of $14412$, we know the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$C2 - C3 = 4$&lt;/li&gt;&#xA;&lt;li&gt;$C5 \mod C1 = 1$&lt;/li&gt;&#xA;&lt;li&gt;$C1(C2 + C3) = 12$&lt;/li&gt;&#xA;&lt;li&gt;$C1 - C3 + C4 = 4$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(I'm sure an actual mathematician would do a cleaner job of this part of the logic, but I'll fumble along). We can rework these equations to give us, $C1(4 + 2C3) = 12$ (combining equation 1 &amp;amp; 3), or $C1 = \frac{12}{(4 + 2C3)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that $C1$ must be a whole number, then $C3$ must be $0$, $1$, $4$, making $C1$ equal to $3$, $2$, or $1$. However, we also need the remainder when dividing $C5$ by $C1$ to be $1$ (noting that there's precedent in the examples for treating &lt;em&gt;x%0 as 0&lt;/em&gt;), which means $C1\ne1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Re-entering these possible values in the third equation to get $C2$, we have our code as either $251??$ or $340??$. Then using the 4th equation to get $C4$, we arrive at $2513?$ or $3401?$. Finally, using the 2nd equation we get a $C5$ value of $1$, $3$, $5$, $7$, $9$ (for the first possible code) and $1$, $4$, $7$ (for the second).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, &lt;strong&gt;today's code&lt;/strong&gt; is one of the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$25131$&lt;/li&gt;&#xA;&lt;li&gt;$25133$&lt;/li&gt;&#xA;&lt;li&gt;$25135$&lt;/li&gt;&#xA;&lt;li&gt;$25137$&lt;/li&gt;&#xA;&lt;li&gt;$25139$&lt;/li&gt;&#xA;&lt;li&gt;$34011$&lt;/li&gt;&#xA;&lt;li&gt;$34014$&lt;/li&gt;&#xA;&lt;li&gt;$34017$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, not perfect, but certainly faster than brute forcing $100,000$ combinations...&lt;/p&gt;&#xA;" OwnerUserId="5208" LastActivityDate="2015-06-02T04:33:37.887" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15852" PostTypeId="1" AcceptedAnswerId="15947" CreationDate="2015-06-02T07:13:09.317" Score="52" ViewCount="2073" Body="&lt;p&gt;I need your help! Look at this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;One, two, one o'clock, three o'clock rock&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;Eight, six, three o'clock, two o'clock rock&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;Nine, seven, one o'clock, seven o'clock rock&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;We're gonna rock around the clock tonight&lt;/em&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's no way I can possibly rock to my maximum potential with a clock that's so obviously broken... &lt;strong&gt;What is causing the hours on my clock to be so messed up?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are 24 hours in a day...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...and my clock is only a twelve hour clock.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 3&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As much as I like music, the issue here is with my clock.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus&lt;/strong&gt;: Here's what my clock deserves right now...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;One, two, two o'clock, nine o'clock rock&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;Eight, nine, twelve o'clock, one o'clock rock&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;Eight, nine, three o'clock, two o'clock rock&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;We're gonna rock around the clock tonight&lt;/em&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="5208" LastEditDate="2015-06-04T05:59:27.540" LastActivityDate="2015-06-05T09:32:34.650" Title="Help! I can't rock out, with my clock out!" Tags="|enigmatic-puzzle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15853" PostTypeId="2" ParentId="15832" CreationDate="2015-06-02T08:51:03.737" Score="11" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;O&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; V S F S R O N G R V R J is on WERTYU-keyboard so it corresponds to &lt;br/&gt;C A D A E I  B F E C E H on QWERTY-keyboard which equals to &lt;br/&gt; 3 1 4 1 5 9 2 6 5 3 5 8 in other words π(pi-number).&lt;br/&gt;&lt;br/&gt;So next number on the sequence is 9 which equals to I on our keyboard and O on the animalboard. &lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4506" LastActivityDate="2015-06-02T08:51:03.737" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15854" PostTypeId="2" ParentId="7707" CreationDate="2015-06-02T09:15:24.910" Score="0" Body="&lt;p&gt;Is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Coin&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hard at heart, I can be shaped the same,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A coin heart is hard, but texture is same, size is same from the heart to the end.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Thickened, I added to Nostradamus's fame, My two faces are sharp or&#xA;  sweet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A coin has two face, a face of hard/sharp icon and a face of flower [sweet]&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;While the air grows warm, I am a delightful treat&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We roll a coin by the air supported beside, a coin is delightfully rolling.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Often in pairs, but enjoyed alone.&#xA;  Pop me in your mouth and try not to moan.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A coin is often seems pairs because of its two face. but we can always see one face when we put it on a surface.&#xA;You can't moan coin in your mouth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HINT:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I have a bag of these on the desk in front of me.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We usually put coins in bag and you put it on your desk, isn't it? ^_^&lt;/p&gt;&#xA;" OwnerUserId="11353" LastActivityDate="2015-06-02T09:15:24.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15855" PostTypeId="1" AcceptedAnswerId="15856" CreationDate="2015-06-02T09:40:43.070" Score="6" ViewCount="3046" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;We all have it.&lt;br&gt;&#xA;  At somewhere, Long and at somewhere, short.&lt;br&gt;&#xA;  On someone, thick and on someone, thin.&lt;br&gt;&#xA;  On somewhat, gold and on somewhat, black.&lt;br&gt;&#xA;  When someone doesn't have it, and we laugh.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is it?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="11353" LastEditorUserId="11353" LastEditDate="2015-06-02T10:19:37.120" LastActivityDate="2015-06-03T08:20:28.000" Title="We all have it, but what is it?" Tags="|riddle|" AnswerCount="1" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="15856" PostTypeId="2" ParentId="15855" CreationDate="2015-06-02T09:50:33.280" Score="15" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Hair  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Although  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Laughing at bald people isn't very nice :P&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We all have it.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Everyone (or at least the vast majority) has hair somewhere  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At somewhere, Long and at somewhere, short.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Some people have long hair, some have short. Also, head hair tends to be longer than, say, eyebrow hair  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On someone, thick and on someone, thin.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Some people have thick hair, some have thin  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On somewhat, gold and on somewhat, black.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Different people have different hair colours  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When someone doesn't have it, and we laugh.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Again, laughing at bald people isn't nice. &lt;em&gt;*finds bigger hat*&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11717" LastEditorUserId="11717" LastEditDate="2015-06-03T08:20:28.000" LastActivityDate="2015-06-03T08:20:28.000" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15857" PostTypeId="2" ParentId="11302" CreationDate="2015-06-02T10:05:03.677" Score="7" Body="&lt;p&gt;Man was Murdered,  but,  FIR says  suicide.   Hence, inspector  feel fake FIR    &lt;/p&gt;&#xA;" OwnerUserId="12921" LastActivityDate="2015-06-02T10:05:03.677" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15858" PostTypeId="1" AcceptedAnswerId="15861" CreationDate="2015-06-02T10:37:17.560" Score="1" ViewCount="366" Body="&lt;p&gt;&lt;strong&gt;We all have it&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Some are long, some are short.&lt;/p&gt;&#xA;&lt;p&gt;Some colored, some natural.&lt;/p&gt;&#xA;&lt;p&gt;Some filthy, some clean.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;What is it?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="12908" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-06-02T11:44:17.837" Title="We all have it. What is it?" Tags="|riddle|" AnswerCount="3" CommentCount="3" ClosedDate="2015-06-02T11:48:46.627" ContentLicense="CC BY-SA 3.0" />
  <row Id="15859" PostTypeId="2" ParentId="15858" CreationDate="2015-06-02T10:42:23.467" Score="2" Body="&lt;p&gt;I think it might be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Hair  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;again :P&lt;/p&gt;&#xA;" OwnerUserId="12645" LastEditorUserId="12645" LastEditDate="2015-06-02T11:44:17.837" LastActivityDate="2015-06-02T11:44:17.837" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15860" PostTypeId="2" ParentId="15858" CreationDate="2015-06-02T10:53:56.533" Score="2" Body="&lt;p&gt;Could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; rope.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Some are long, some are short.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; How long is a piece of string?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Some colored, some natural.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Coloured ropes and those made of natural fibre.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Some filthy, some clean.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ropes used for mooring a ship are usually filthy, but rope can also be clean.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think the riddle is &quot;too broad&quot;. Hair also fits, as @PirateSoul said.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-06-02T10:53:56.533" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15861" PostTypeId="2" ParentId="15858" CreationDate="2015-06-02T10:54:15.550" Score="2" Body="&lt;p&gt;If it's not hair, then:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Fingernails&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11717" LastActivityDate="2015-06-02T10:54:15.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15862" PostTypeId="1" AcceptedAnswerId="15863" CreationDate="2015-06-02T11:34:29.713" Score="53" ViewCount="4662" Body="&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/16514/mathematical-rebus-ii&quot;&gt;Mathematical Rebus II&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/q/52759/12165&quot;&gt;Mathematical Rebus III&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;4\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\\&#xA;(-\infty,...,-1,0,1,...,\infty)\times(-\infty,...,-1,0,1,...,\infty)\\&#xA;\forall\begin{bmatrix}{-1}&amp;amp;{0}\\&#xA; {0}&amp;amp;{-1}&#xA;\end{bmatrix}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="12165" LastEditorUserId="12165" LastEditDate="2017-06-19T10:38:08.643" LastActivityDate="2017-06-19T10:38:08.643" Title="Mathematical Rebus" Tags="|mathematics|rebus|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15863" PostTypeId="2" ParentId="15862" CreationDate="2015-06-02T12:25:46.243" Score="49" Body="&lt;p&gt;The first line is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; equal to Pi.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second line is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the integers, &lt;strong&gt;Z&lt;/strong&gt;, multiplied by itself, making &lt;strong&gt;Z&lt;/strong&gt;&lt;sup&gt;2&lt;/sup&gt;.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The third line is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; multiplying some matrix ∀ by the negative identity matrix, negating its value. The letter ∀ negated is A.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All together,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pi + Z&lt;sup&gt;2&lt;/sup&gt; + A = &lt;strong&gt;Pizza&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1497" LastActivityDate="2015-06-02T12:25:46.243" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15864" PostTypeId="2" ParentId="15823" CreationDate="2015-06-02T13:02:11.333" Score="3" Body="&lt;p&gt;As a follow up to my comment (somewhat of a stretch):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The internet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are all so lazy. You have a job to do but we never even see you move a muscle!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The information of the internet travels all around the world, and people don't even have to move a muscle because it's all done by machines.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Many people claim to follow you religiously.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People have become attached to the internet and follow things on it 'religiously', like social media, watching videos (Youtube), video games etc. Some would claim to be unable to live without it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We've never heard from 1 of your 5 parts. Why the silence?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A possible reference to the 'deep web' or possible local/offline data centers.&#xA;&lt;br&gt; It could also be referencing other countries and the restrictions placed on their internet (i.e North Korea, China etc.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;People say that you say all kinds of things. Is this true?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The internet just happens to be full of completely random things, lots of facts that are true, and even more so false.....(and cat videos of course)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Many claim that you are great teachers, and that people have learned more from your group than from any religious leader in history.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are extremely vast amounts of useful information on the internet from which you can learn from (Including online universities). I would think Wikipedia alone has become sourced more than any religious leader in history.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some of you gather in large numbers and in unappealing locations.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most of the information online (or in 'the cloud') is stored on a huge set of databases that are often located in deserts, a place most would find unappealing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="12843" LastEditDate="2015-06-02T15:45:46.260" LastActivityDate="2015-06-02T15:45:46.260" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15865" PostTypeId="2" ParentId="11718" CreationDate="2015-06-02T14:28:32.363" Score="9" Body="&lt;p&gt;The mystery has finally been solved!&lt;br&gt;&#xA;&lt;strong&gt;The answer is:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://www.youtube.com/watch?v=hOH49oVWKmQ&quot;&gt;hOH49oVWKmQ&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation:&lt;/strong&gt;&lt;br&gt;&#xA;The encrypted text is actually &quot;&lt;strong&gt;lSL 4 9 xEFTvZ&lt;/strong&gt;&quot; and not &quot;ISL 4 9 xEFTvZ&quot; (the difference is in the first letter, which is lower-case L instead of upper-case i).&lt;br&gt;&#xA;The code is then easily decrypted shifting back the first block of letters by 4, the second block by 9 and keeping intact the two numbers.&lt;br&gt;&#xA;The result is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;hOH49oVWKmQ&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As you can see, the main difficulty was in the interpretation of the first character (absolutely impossible to tell from the video). The cipher itself wasn't too hard (just two Gronsfeld ciphers).&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-06-04T05:05:54.280" LastActivityDate="2015-06-04T05:05:54.280" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15867" PostTypeId="2" ParentId="15839" CreationDate="2015-06-02T15:18:46.093" Score="12" Body="&lt;h1&gt;n = 41&lt;/h1&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Methoxybenzenes methoxybenzenes methoxybenzenes methoxybenzenes methoxybenzenes methoxybenzenes methoxybenzenes methoxybenzenes methoxybenzenes methoxybenzenes methoxybenzenes methoxybenzenes methoxybenzenes methoxybenzenes methoxybenzenes methoxybenzenes methoxybenzenes methoxybenzenes methoxybenzenes methoxybenzenes quizzify quizzify quizzify quizzify quizzify quizzify quizzify quizzify quizzify quizzify quizzify quizzify quizzify quizzify quizzify quizzify quizzify quizzify quizzify, quizzify methoxybenzenes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To construct this sentence:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Take &quot;methoxybenzenes&quot; as the subject.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Add a relative clause to it -- these methoxybenzenes are quizzified by other methoxybenzenes; they are &quot;methoxybenzenes &lt;strong&gt;methoxybenzenes quizzify&lt;/strong&gt;&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Repeatedly do the same to the innermost subject: the quizzifying methoxybenzenes, are, in their turn, quizzified by yet &lt;em&gt;other&lt;/em&gt; methoxybenzenes; thus the original ones can be described as &quot;methoxybenzenes methoxybenzenes &lt;strong&gt;methoxybenzenes quizzify&lt;/strong&gt; quizzify&quot;, etc.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(Compare this process to: &quot;cars&quot; -&gt; &quot;cars dogs chase&quot; -&gt; &quot;cars dogs people hate chase&quot; -&gt; etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process is repeated 19 times to &quot;pump&quot; a 39-word subject. The last two words are the verb phrase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Methoxybenzenes&lt;/em&gt; is &lt;strong&gt;not&lt;/strong&gt; a proper noun: they are &lt;a href=&quot;https://en.wikipedia.org/wiki/Anisole&quot;&gt;methyl ethers formed from phenyl&lt;/a&gt;, there are many of them on the planet, and there's no trademark or copyright that owns them. Sure, they can't quizzify each other, as they have no agency, but if &lt;a href=&quot;https://en.wikipedia.org/wiki/Colorless_green_ideas_sleep_furiously&quot;&gt;colourless green ideas can sleep furiously&lt;/a&gt;, then I can get away with that much.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;n &gt; 41?&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Here are the remaining &lt;a href=&quot;http://www.scrabbler.net/highest-scoring/&quot;&gt;higher-scoring words in SOWPODS&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;43: &quot;Quizzicalities, quizzically, zyzzyva.&quot; No verbs.&lt;/li&gt;&#xA;&lt;li&gt;44: &quot;Quizzicality, zyzzyvas.&quot; No verbs.&lt;/li&gt;&#xA;&lt;li&gt;45: &quot;Bezzazz, pazzazz, pizzazz, quizzifying.&quot; A gerund, but no conjugated verbs.&lt;/li&gt;&#xA;&lt;li&gt;46: Only one word, &quot;quizzification.&quot;&lt;/li&gt;&#xA;&lt;li&gt;47: &quot;Bezzazzes, pazzazzes, pizzazzes, quizzifications.&quot; I &lt;em&gt;think&lt;/em&gt; these are all plural nouns, but maybe you can &lt;em&gt;pizzazz&lt;/em&gt; something, and the first three are verbs? Either way, they wouldn't be a verb form you can use with &quot;quizzifications.&quot;&lt;/li&gt;&#xA;&lt;li&gt;48 and up don't have verbs either.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So I suppose you can't really do better than this. If you prefer the mental image of palm trees quizzifying each other, you could swap out &quot;methoxybenzenes&quot; for &quot;chiquichiquis&quot;, though.&lt;/p&gt;&#xA;" OwnerUserId="11476" LastEditorUserId="11476" LastEditDate="2015-06-02T15:31:56.710" LastActivityDate="2015-06-02T15:31:56.710" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="15868" PostTypeId="2" ParentId="15253" CreationDate="2015-06-02T15:26:01.587" Score="3" Body="&lt;p&gt;Got it! The answer is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;642  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first digit (as rand al'thor said), is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The sum of the digits of the left hand number, so 7+4+2 = 15, 1+5 = 6  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The other digits:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first digit of the left hand side  is the number base of the other digits. Convert them into the base given by the first digit of the right hand side&lt;br&gt;&#xA;For the first example: 4+3+2 = &lt;strong&gt;9&lt;/strong&gt;&lt;br&gt;&#xA;32&lt;sub&gt;4&lt;/sub&gt; = 15&lt;sub&gt;9&lt;/sub&gt;&lt;br&gt;&#xA;432 = 915&lt;br&gt;&#xA;&lt;br&gt;&#xA;So for &lt;strong&gt;735&lt;/strong&gt;:&lt;br&gt;&#xA;35&lt;sub&gt;7&lt;/sub&gt; = 42&lt;sub&gt;6&lt;/sub&gt;&lt;br&gt;&#xA;571 produces an error because 7 doesn't exist in base 5.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hopefully that all makes sense!&lt;/p&gt;&#xA;" OwnerUserId="11717" LastActivityDate="2015-06-02T15:26:01.587" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15869" PostTypeId="2" ParentId="10933" CreationDate="2015-06-02T15:57:23.263" Score="3" Body="&lt;p&gt;Here's a Haskell implementation of the answer:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import Data.Numbers.Primes&#xA;&#xA;seqFun 1 = 0&#xA;seqFun input = recWhile input 0&#xA;    where&#xA;      recWhile a n = res&#xA;          where&#xA;            pfs = primeFactors a&#xA;            modN = n+1&#xA;            maxPF = maximum pfs&#xA;            res&#xA;                | a &amp;lt; 2 = modN&#xA;                | maxPF == 2 = modN&#xA;                | otherwise = recWhile (maxPF + 1) modN&#xA;&#xA;fibs = 1:1:(zipWith (+) fibs (tail fibs))&#xA;thirtyFibs = take 30 fibs&#xA;main = print $ map seqFun thirtyFibs&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Just run &lt;code&gt;main&lt;/code&gt; to see the entire sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that many of the hints were not true; the function is not related to the Greatest Common Factor, and the hints for 29862968 and 191199 were incorrect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For fun, here's a much shorter version:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import Data.Numbers.Primes&#xA;seqFun = length . takeWhile (&amp;gt;3) . iterate (succ.maximum.primeFactors)&#xA;fibs = 1:1:(zipWith (+) fibs (tail fibs))&#xA;main = print $ map seqFun (take 30 fibs)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This version doesn't work for 2 or 3, though.&lt;/p&gt;&#xA;" OwnerUserId="12927" LastEditorUserId="12927" LastEditDate="2015-06-04T17:27:55.020" LastActivityDate="2015-06-04T17:27:55.020" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15870" PostTypeId="1" AcceptedAnswerId="15879" CreationDate="2015-06-02T17:18:35.223" Score="8" ViewCount="764" Body="&lt;p&gt;&lt;em&gt;Story continuation based from &lt;a href=&quot;https://puzzling.stackexchange.com/questions/14960/3-planks-1-nail&quot;&gt;this first riddle&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15190/not-enough-pylons&quot;&gt;this other riddle&lt;/a&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bob was enjoying his day off at home when he received a message from John. John sent him a picture of a piece of art, the first piece of art John has ever created. John said that this art was a gift in return from the pylons he had taken (so that they are now even). At first Bob did not understand why John would make such a beautiful piece of artwork for him, until he remembered just how John was. Bob put his day on hold, in an attempt to find out why John really sent him this picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;What does Bob find out from the picture?&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Below is the beautiful artwork)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/RZPXI.png&quot; alt=&quot;Just a completely normal pylon.&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(This shouldn't be very difficult for those who are able to solve it)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Here is the original sized image if it helps: &lt;img src=&quot;https://i.stack.imgur.com/OocmX.png&quot; alt=&quot;A not so completely normal pylon?&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2015-06-03T14:52:27.097" Title="John's first pixel drawing" Tags="|visual|story|computer-puzzle|steganography|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15873" PostTypeId="1" AcceptedAnswerId="15877" CreationDate="2015-06-02T19:43:34.870" Score="11" ViewCount="1832" Body="&lt;p&gt;My sister made a riddle thingy for fun. I hope you like it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I may be enjoyed with ice,&lt;br&gt;&#xA;  but often preferred with fire.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I can come with chocolate and crackers,&lt;br&gt;&#xA;  but it's fine when I come alone.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Sometimes I am like a can,&lt;br&gt;&#xA;  sometimes I am an animal.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I am usually spongy and soft,&lt;br&gt;&#xA;  and smell delightful.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;But if you ever forget me,&lt;br&gt;&#xA;  that will no longer be true.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;What am I?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; This is my first riddle. Feedback is appreciated.&lt;/p&gt;&#xA;" OwnerUserId="12109" LastEditorUserId="12109" LastEditDate="2015-06-02T20:07:52.003" LastActivityDate="2015-06-02T20:12:33.543" Title="You know you love me" Tags="|riddle|word|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15874" PostTypeId="2" ParentId="15873" CreationDate="2015-06-02T19:52:12.243" Score="1" Body="&lt;p&gt;It could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; tea.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I may be enjoyed with ice,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Iced tea.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;but usually preferred with fire.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tea is usually made with boiling water; boiling comes from heat, originally supplied by fire.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I can come with chocolate,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chocolate biscuits?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;but I'm fine when I come alone.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ahhh, a nice cup of tea!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Sometimes I am like a can,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A can of tea?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;sometimes I am an animal.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure about this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am usually spongy and soft,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Teabags or maybe tea leaves.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;and smell delightful.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mmmm, the smell of freshly brewed tea!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But if you ever forget me,&lt;br&gt;&#xA;I may not be so good!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Forget a cup of tea for too long and it goes cold and icky.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-06-02T19:52:12.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15875" PostTypeId="2" ParentId="15873" CreationDate="2015-06-02T19:58:17.623" Score="1" Body="&lt;p&gt;Another possible answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; cream.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I may be enjoyed with ice,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ice cream.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;but usually preferred with fire.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Creme brulee (burnt cream).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I can come with chocolate,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chocolate cream.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;but I'm fine when I come alone.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's nice enough alone, I suppose.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Sometimes I am like a can,&lt;br&gt;&#xA;sometimes I am an animal.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Still not sure about these two.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am usually spongy and soft,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That could describe the consistency of cream.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;and smell delightful.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Maybe...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But if you ever forget me, &#xA;I may not be so good!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Forget cream for too long and it will go rancid and horrible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-06-02T19:58:17.623" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15876" PostTypeId="2" ParentId="15870" CreationDate="2015-06-02T20:00:49.777" Score="8" Body="&lt;p&gt;(Edited) I finished the rest of it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Beep. Boop. This is John. You are getting quite clever Bob. First you won my beer by forging a triangle in a plank. Then you kept your job after I had taken some of your pylons. It seems you are truly improving your skills by being able to decode this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The solution is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each square is a RGB code of some value (n,n,n).  If you look at the n'th value on an ASCII lookup table, you get a letter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(I did this before the edit, is this line still correct?)&lt;/p&gt;&#xA;" OwnerUserId="12843" LastEditorUserId="12843" LastEditDate="2015-06-02T20:36:29.630" LastActivityDate="2015-06-02T20:36:29.630" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15877" PostTypeId="2" ParentId="15873" CreationDate="2015-06-02T20:09:01.040" Score="12" Body="&lt;p&gt;I think you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; marshmallow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I may be enjoyed with ice,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you can put some on your ice-cream&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but usually preferred with fire.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; but they are better as S'mores&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can come with chocolate,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you need chocolate to make S'mores&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I'm fine when I come alone.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; but they are good alone as well&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am like a can,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; usually shape like a can&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;sometimes I am an animal.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; and sometimes as an animal&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am usually spongy and soft,&#xA;and smell delightful.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; well those 2 are obvious&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But if you ever forget me,&#xA;I may not be so good!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if you forget it in the fire, it will burn&lt;br&gt;&#xA; if you forget it, it will dry&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10400" LastEditorUserId="10400" LastEditDate="2015-06-02T20:12:33.543" LastActivityDate="2015-06-02T20:12:33.543" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15879" PostTypeId="2" ParentId="15870" CreationDate="2015-06-02T20:33:56.217" Score="11" Body="&lt;p&gt;The image contains a hidden message.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The grey pixels are the ASCII codes of each character in the message.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The message is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Beep. Boop. This is John. You are getting quite clever Bob. First you won my beer by forging a triangle in a plank. Then you kept your job after I had taken some of your pylons. Let us see if you are truly improving your skills, by being able to decode this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you're interested this JavaScript code reveals the hidden message:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;lt;img id=coded src=&quot;data:image/png;base64, &quot;/&amp;gt;&#xA;&#xA;&amp;lt;script&amp;gt;&#xA;&#xA;var c=document.createElement('canvas');&#xA;c.width=coded.width;&#xA;c.height=coded.height;&#xA;&#xA;c.getContext('2d').drawImage(coded,0,0);&#xA;var p=c.getContext('2d').getImageData(0,0,c.width,c.height);&#xA;for(var s='',i=0; i&amp;lt;p.data.length; i+=4){&#xA;   //////// Ignore white pixels and translate the grey pixels into text&#xA;        if(p.data[i]!=255 &amp;amp;&amp;amp; p.data[i]==p.data[i+1] &amp;amp;&amp;amp; p.data[i]==p.data[i+1]) s+=String.fromCharCode(p.data[i]);&#xA;&#xA;}&#xA;document.write(s);&#xA;&#xA;&amp;lt;/script&amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="11297" LastActivityDate="2015-06-02T20:33:56.217" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15880" PostTypeId="5" CreationDate="2015-06-02T20:50:26.447" Score="0" Body="&lt;p&gt;Puzzles involving analysis of strategic decision making in relation to competitive or cooperative choice between a number of participants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information here is a link to the Wikipedia article on &lt;a href=&quot;https://www.google.ca/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=1&amp;amp;cad=rja&amp;amp;uact=8&amp;amp;ved=2ahUKEwiNhbew_fTeAhUPulMKHdwqBj4QFjAAegQIAhAC&amp;amp;url=https%3A%2F%2Fen.wikipedia.org%2Fwiki%2FGame_theory&amp;amp;usg=AOvVaw3YuNid5iVRBBJbLFltcL35&quot; rel=&quot;nofollow noreferrer&quot;&gt;Game Theory&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="52483" LastEditorUserId="52483" LastEditDate="2018-11-27T18:07:36.043" LastActivityDate="2018-11-27T18:07:36.043" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15881" PostTypeId="4" CreationDate="2015-06-02T20:50:26.447" Score="0" Body="Puzzles that involve strategic decision making, usually involving either competitive or cooperative choices between a number of participants. " OwnerUserId="11297" LastEditorUserId="11297" LastEditDate="2015-06-02T21:22:55.550" LastActivityDate="2015-06-02T21:22:55.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15882" PostTypeId="5" CreationDate="2015-06-02T20:52:05.103" Score="0" Body="&lt;p&gt;For puzzles using general scientific knowledge.&lt;/p&gt;&#xA;&lt;p&gt;Consider using the &lt;a href=&quot;/questions/tagged/physics&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;physics&amp;#39;&quot; aria-label=&quot;show questions tagged &amp;#39;physics&amp;#39;&quot; rel=&quot;tag&quot; aria-labelledby=&quot;physics-container&quot;&gt;physics&lt;/a&gt;, &lt;a href=&quot;/questions/tagged/chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;chemistry&amp;#39;&quot; aria-label=&quot;show questions tagged &amp;#39;chemistry&amp;#39;&quot; rel=&quot;tag&quot; aria-labelledby=&quot;chemistry-container&quot;&gt;chemistry&lt;/a&gt; or &lt;a href=&quot;/questions/tagged/computer-science&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;computer-science&amp;#39;&quot; aria-label=&quot;show questions tagged &amp;#39;computer-science&amp;#39;&quot; rel=&quot;tag&quot; aria-labelledby=&quot;computer-science-container&quot;&gt;computer-science&lt;/a&gt; tags if your question is specifically about these branches of science. Science questions which aren't puzzles are considered off-topic and should be asked on the Stack Exchange site for the relevant branch of science (e.g. &lt;a href=&quot;https://biology.stackexchange.com/&quot;&gt;Biology&lt;/a&gt;, &lt;a href=&quot;https://chemistry.stackexchange.com/&quot;&gt;Chemistry&lt;/a&gt;, &lt;a href=&quot;https://physics.stackexchange.com/&quot;&gt;Physics&lt;/a&gt;, &lt;a href=&quot;https://astronomy.stackexchange.com/&quot;&gt;Astronomy&lt;/a&gt;, etc.).&lt;/p&gt;&#xA;" OwnerUserId="11497" LastEditorUserId="60644" LastEditDate="2023-01-03T19:07:42.217" LastActivityDate="2023-01-03T19:07:42.217" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15883" PostTypeId="4" CreationDate="2015-06-02T20:52:05.103" Score="0" Body="Puzzles which require some knowledge of science to solve. Consider using the [physics], [chemistry] or [computer-science] tags if your question is specifically about these branches of science." OwnerUserId="11297" LastEditorUserId="69582" LastEditDate="2021-03-10T17:49:41.113" LastActivityDate="2021-03-10T17:49:41.113" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15887" PostTypeId="1" AcceptedAnswerId="15888" CreationDate="2015-06-03T00:19:36.383" Score="30" ViewCount="2977" Body="&lt;p&gt;Alice and Bob are going to have a poker night. They have invented a variant where there are no secrets, and no randomness, called Face-Up Poker. Here's how it works. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A standard 52 card deck is spread face up on a table.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Alice picks up any 5 cards from the table. &lt;/li&gt;&#xA;&lt;li&gt;Bob does the same.&lt;/li&gt;&#xA;&lt;li&gt;Alice discards any number of cards, throwing them on the floor. She then picks up the same number of cards from the table.&lt;/li&gt;&#xA;&lt;li&gt;Bob does the same (Bob can't pick up cards that Alice discarded).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;They then compare hands, and the best hand wins, &lt;strong&gt;with Bob winning ties&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Which player can force a win? What is their winning strategy?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This puzzle assumes familiarity with the rankings of poker hands, but no other knowledge of the rules of poker. For a refresher these rankings, see &lt;a href=&quot;http://www.pokerhands.com/poker_hand_tie_rules.html&quot;&gt;this helpful page&lt;/a&gt; (the hands are listed worst to best from top to bottom).&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-06-03T00:31:31.897" LastActivityDate="2015-06-05T15:39:09.040" Title="Face Up Poker with Alice and Bob" Tags="|game|" AnswerCount="3" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="15888" PostTypeId="2" ParentId="15887" CreationDate="2015-06-03T01:15:29.723" Score="25" Body="&lt;p&gt;There is a winning strategy for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alice. For her starting hand, she will select all four 10's, along with another lower card (let's say the 2 of clubs for concreteness). This will make it impossible for Bob to end up with a royal flush. Bob needs to prevent Alice from making a royal flush when she discards, so he must select a starting hand containing a card of rank higher than 10 in each of the four suits. This means that for at least three of the four suits, Bob's starting hand cannot contain a card of rank lower than 10 in that suit.&lt;br&gt; &lt;br&gt; When it is Alice's turn to discard, she chooses one of the suits in which Bob did not take a card of rank lower than 10. She keeps the 10 of that suit, discards her other four cards, and takes the 6, 7, 8, and 9 of the chosen suit (Bob did not select a card of rank lower than 10 in the chosen suit). Alice's final hand will be a straight flush to the 10. Since all of the other 10's are now on the floor, Bob cannot possibly make a straight flush whose high card is 10 or above, so Bob's hand cannot possibly tie or beat Alice's.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="463" LastEditDate="2015-06-03T13:03:03.237" LastActivityDate="2015-06-03T13:03:03.237" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15889" PostTypeId="2" ParentId="15887" CreationDate="2015-06-03T01:16:04.557" Score="20" Body="&lt;p&gt;Alice goes first and picks...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4 tens (and any 1 other card) preventing Bob from making the highest ranked hand the royal flush (A,K,Q,J,10 of the same suit)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bob must...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; prevent her making a royal flush so he must pick 4 cards each of a different suit in the range Ace to Jack (eg 4 Aces) (and any 1 other card)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alice can then...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  keep one of the tens and pick 9,8,7,6 of the same suit making a straight flush with 10 as the highest card.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To equal this Bob must...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  also make a straight flush with 10 as the highest card but Alice has already taken all the tens. Oh no Bob has lost.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="11297" LastEditDate="2015-06-03T16:15:27.513" LastActivityDate="2015-06-03T16:15:27.513" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15890" PostTypeId="1" AcceptedAnswerId="15891" CreationDate="2015-06-03T02:59:35.053" Score="5" ViewCount="422" Body="&lt;p&gt;3 players start a game in which one of the players wins and the other 2 lose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each loser has to pay the winner twice the amount he (the winner) had when (each) game started.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After 3 consecutive games, each has won once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is how much each player has (after those 3 games):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Player #1:  $20&#xA;&#xA;Player #2: $118&#xA;&#xA;Player #3:  $27&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Which players lost money and how much? (In relation to how much they had, before the first game)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; The smallest monetary unit is $1 (i.e., no fractions) and the money has to be paid immediately after each game.&lt;/p&gt;&#xA;" OwnerUserId="12841" LastActivityDate="2015-06-03T03:33:44.497" Title="Loser-Winner-Loser" Tags="|calculation-puzzle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15891" PostTypeId="2" ParentId="15890" CreationDate="2015-06-03T03:06:48.510" Score="12" Body="&lt;p&gt;If each player pays the winner twice the amount of money he already had, then he has five times his original amount of money.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only person whose money is a multiple of five after the third round is Player 1, so he must have won the third round. The winnings before the third round were as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    Player 1:   $4&#xA;    Player 2: $126&#xA;    Player 3:  $35&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now, Player 3 is the one with a multiple of 5, so he was the winner the second round:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    Player 1:  $18&#xA;    Player 2: $140&#xA;    Player 3:   $7&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And now, Player 2 is the one with a multiple of 5, so he was the winner the first round:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    Player 1: $74&#xA;    Player 2: $28&#xA;    Player 3: $63&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Those are the amounts they started with.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="12841" LastEditDate="2015-06-03T03:33:44.497" LastActivityDate="2015-06-03T03:33:44.497" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15892" PostTypeId="1" AcceptedAnswerId="15908" CreationDate="2015-06-03T06:52:33.040" Score="19" ViewCount="2173" Body="&lt;p&gt;An Egyptian pharaoh wants to build a monument to the Sun God, Ra. He wants this to be a solid stone cube, which is $20$ hectocubits tall, long and wide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;His engineers get right to work planning this. The only stones the Egyptians have access to are $2000$ enormous rectangular blocks leftover from a failed project, whose dimensions are $2\times 2\times1 $ (all units in hectocubits), and which cannot be broken into smaller usable blocks. They figure that the easiest way to build this monument is in horizontal layers, each layer being a 10 by 10 array of blocks which lie flat. They present this plan to the pharaoh.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, after seeing the plan, the pharaoh is furious! Since the edges of the blocks are lightly rounded, the cracks between blocks can let some light through.  This is of course a problem, since the cube is supposed to be an homage to the generosity of the Sun God, so it should graciously accept the gift of the Sun's rays instead of letting them pass by. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The engineers get confused at this point, since they aren't as religious, and don't understand the importance of the pharaoh's request. Totally losing his cool, the pharaoh yells, &quot;You fools! All I ask is that you&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;build a $20\times20\times20$ cube out of $2\times2\times 1$ blocks so no light can shine in one face though the cracks between the blocks &lt;em&gt;and exit the opposite face&lt;/em&gt;!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;IS THAT SO DIFFICULT!?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can the engineers succeed?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the pharaoh wants the cube to block all possible beams of light, not just vertical ones. In other words, every line which enters one face and exits the opposite face must also pass through the interior of one of the $2\times2\times1$ blocks. The solution does not involve lateral-thinking, so the answer is not &quot;just fill the cracks with mortar.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I changed it so that the cube is only required to block light which enters and leaves opposite faces. As @frodoskywalker pointed out, it was trivially impossible without this addition. Furthermore, I clarified other confusing points (breaking blocks is not allowed, and the cube must be solid, implying all the blocks are axis aligned).&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-06-03T15:10:56.230" LastActivityDate="2015-06-03T18:47:47.390" Title="Cube made of 2x2x1 blocks which blocks all light" Tags="|geometry|combinatorics|" AnswerCount="3" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="15893" PostTypeId="2" ParentId="15823" CreationDate="2015-06-03T07:14:01.483" Score="1" Body="&lt;p&gt;Nice riddle. I think the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; tricks or hacks&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Accusation: You are all so lazy. You have a job to do but we never even see you move a muscle!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; tricks and hacks are helping people and mobility does not apply to them, since they are no material concepts&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Accusation: Many people claim to follow you religiously.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; wise tricks and hacks are very useful and a lot of people are following them. life hackers, anyone? :) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Accusation: We've never heard from 1 of your 5 parts. Why the silence?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if the answer is tricks, then &quot;trick&quot; is a 5-letter word and when saying it, the &quot;c&quot; is not actually pronounced =&gt; it is a silent letter in this case. The same applies to &quot;hacks&quot;, 5 letter word, character &quot;c&quot; is not pronounced.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Accusation: People say that you say all kinds of things. Is this true?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; tricks and hacks apply to all kind of things. Some of them are only myths or rumors, i.e. they are accused of being effective or working. only an experiment or just trying them can prove their effectiveness.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Accusation: Many claim that you are great teachers, and that people have learned more from your group than from any religious leader in history.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; tricks and hacks teach people new shortcuts or better way of doing things. this is quite self-explanatory&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Accusation: Some of you gather in large numbers and in a way that makes the location unappealing.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; maybe here you mean that the hacks are spread all over the world/internet/etc. and they had to be better organised in encyclopedia or wikipedia for the people to find them faster and use them more often.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5283" LastActivityDate="2015-06-03T07:14:01.483" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15895" PostTypeId="2" ParentId="15823" CreationDate="2015-06-03T09:06:07.767" Score="1" Body="&lt;p&gt;Is it a   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Queue  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Accusation&lt;/strong&gt;: You are all so lazy. You have a job to do but we never even see you move a muscle!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We always complain the queue never moves  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Accusation&lt;/strong&gt;: Many people claim to follow you religiously.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Everyone follows the queue  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Accusation&lt;/strong&gt;: We've never heard from 1 of your 5 parts. Why the silence?   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure about this  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Accusation&lt;/strong&gt;: People say that you say all kinds of things. Is this true?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Everyone accuse others of jumping the queue or stealing   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Accusation&lt;/strong&gt;:Many claim that you are great teachers, and that people have learned more from your group than from any religious leader in history.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;People talk to each other in queues and learn from each other. Moreover, it teaches us discipline  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Accusation&lt;/strong&gt;:Some of you gather in large numbers and in a way that makes the location unappealing.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hey! We all hate to see queues.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9669" LastActivityDate="2015-06-03T09:06:07.767" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15896" PostTypeId="1" AcceptedAnswerId="15899" CreationDate="2015-06-03T09:13:13.707" Score="8" ViewCount="1181" Body="&lt;p&gt;Hmmz, it seems that I need to add 30 characters here, even though an image says more than a thousand words...&#xA;&lt;img src=&quot;https://i.stack.imgur.com/TbKoe.png&quot; alt=&quot;The puzzle&quot;&gt;&#xA;When you think you have solved it, please read this hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What kind of cookies do I like?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9777" LastEditorUserId="9777" LastEditDate="2015-06-03T09:24:41.337" LastActivityDate="2015-06-03T09:35:51.267" Title="Another visual puzzle" Tags="|visual|steganography|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15897" PostTypeId="2" ParentId="15896" CreationDate="2015-06-03T09:20:41.567" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The message in the image says &quot;I like cookies&quot; Although the image appears completely blank opening it with an image editor and using the fuzzy select tool reveals the words.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="11297" LastEditDate="2015-06-03T09:22:02.837" LastActivityDate="2015-06-03T09:22:02.837" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15898" PostTypeId="2" ParentId="15896" CreationDate="2015-06-03T09:21:08.653" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I like cookies&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/6gu1G.png&quot; alt=&quot;me too&quot;&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solved by:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Filling in the image- there's an off-white section that provides the message&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11717" LastActivityDate="2015-06-03T09:21:08.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15899" PostTypeId="2" ParentId="15896" CreationDate="2015-06-03T09:28:10.833" Score="7" Body="&lt;p&gt;The colors are slight off, filling it in it makes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I Like Cookies&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;then filling it in with my old ass paint program I get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/Tsuos.png&quot; alt=&quot;morse image&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is decoded in morse to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; chocolate&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12645" LastEditorUserId="12645" LastEditDate="2015-06-03T09:35:51.267" LastActivityDate="2015-06-03T09:35:51.267" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15901" PostTypeId="2" ParentId="15892" CreationDate="2015-06-03T10:24:01.387" Score="1" Body="&lt;p&gt;The best I could do is this:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/io75C.png&quot; alt=&quot;http://i.imgur.com/ggNPDre.png&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You could just stack this x10 and it would use exactly the given 2000 blocks, but it does not fulfill the criteria so I am just going to leave this here in case someone gets an idea from this.&lt;/p&gt;&#xA;&lt;p&gt;Actually now that I thought more about it just rotate it 90 degrees every other layer and that should work.&lt;/p&gt;&#xA;&lt;p&gt;Even more thinking and Taemyrs comment proved that it won't work because of horizontal light&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Some math&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The squares sides are 10 9 10 9 each being 2 units tall&lt;BR&gt;10x10x2 = 200 (2 identical sides)&lt;BR&gt;It loses 1 block each time making it &lt;BR&gt;9x10x2 = 180&lt;BR&gt;8x10x2 = 160 &lt;BR&gt;...&lt;BR&gt;1x10x2 = 20&lt;BR&gt;20+40+60+80+100+120+140+160+180+200 = 1100&lt;BR&gt;&lt;BR&gt;And the other sides start with 9 blocks so they have the exact same except -200 which = 900&lt;BR&gt;900+1100 = 2000&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12940" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-06-03T11:00:25.787" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15907" PostTypeId="2" ParentId="15892" CreationDate="2015-06-03T11:47:59.547" Score="0" Body="&lt;p&gt;Start in a corner. Stand a pair of blocks on edge so that they 2 units high and meet perpendicularly to form the corner. Lay a block flat on the ground inside the square so that it is snug up against these blocks. Light coming in the bottom unit of the diagonal is blocked by this flat block. If you were trying to make a 4x4 block you could plop another flat block down on top of the first and another pair of blocks around the other two sides and be done. But you're not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extend each side wall with another pair of vertical blocks. The seams you've just made are blocked (in the bottom unit) by the flat block that is already installed. Add three more flat blocks to tile the floor, which is now 4 flat blocks. Return to the corner and put vertical blocks inside the side wall. These are one unit further &quot;in&quot; than the outermost wall, so will block the top halves of the seams. There is still a diagonal seam right on the corner in both the inner and outer vertical wall. Lay another flat block inside to cover it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeat the pattern at the other corners and work back towards the centre until everything meets. You have two layers vertically on the outside of the whole cube. The vertical seams are blocked by flat tile inside. The horizontal seams in each vertical layer are blocked by the other vertical layer. After that it's all flat tiles. The ones in the bottom-most layer are offset 1 unit from the others to keep sunlight from getting through to the ground. At the top, do the same as you omit the final inner vertical course.&lt;/p&gt;&#xA;" OwnerUserId="1509" LastActivityDate="2015-06-03T11:47:59.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15908" PostTypeId="2" ParentId="15892" CreationDate="2015-06-03T12:05:13.610" Score="24" Body="&lt;p&gt;&lt;strong&gt;The engineers can't do it, and here's why:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Let's consider an arbitrary arrangement of blocks. Consider the &lt;span class=&quot;math-container&quot;&gt;$19\cdot 19\cdot 3=1083$&lt;/span&gt; axis-aligned lines (let's call those &lt;em&gt;important&lt;/em&gt;) with integer cubit offsets passing through the monument. Every block in the monument can block at most one such line. (1)&lt;/p&gt;&#xA;&lt;p&gt;Claim: It is impossible for any important line to pass through exactly one block.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Proof of claim: Assume there's an important line passing through exactly one block &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt;. Let's consider the two axis-aligned planes containing this important line. These planes dissect the monument into four cuboids.&lt;/p&gt;&#xA;&lt;p&gt;Now let's look at the volume of one of these cuboids, call it &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt;. Since one of the side lengths of &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; is 20 cubits and the other side lengths are integer amounts of cubits, the total volume must be an &lt;em&gt;even&lt;/em&gt; number of cubed cubits.&lt;/p&gt;&#xA;&lt;p&gt;On the other hand, &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; is composed out of parts of blocks. Since &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; is the only block intersecting both planes, it contributes one cubed cubit, but every other block contributes either zero, two or four cubed cubits to &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt;. Hence, the volume of &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; is an &lt;em&gt;odd&lt;/em&gt; number of cubed cubits.&lt;/p&gt;&#xA;&lt;p&gt;Now we have our contradiction, hence the assumption is false and the claim is proved.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Now to block light going through all the important lines we need at least one block blocking each important line, and by the claim, we need at least two blocks blocking each important line. By (1), all these blocks are distinct, so we need 2166 blocks in total, but since we have only 2000 blocks, there will always be at least 83 unblocked important lines, no matter how we build the monument. &lt;span class=&quot;math-container&quot;&gt;$\square$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="12906" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-06-03T12:05:13.610" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15909" PostTypeId="2" ParentId="15823" CreationDate="2015-06-03T12:12:27.097" Score="10" Body="&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;signs&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Accusation: You are all so lazy...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Signs help people in a purely passive way, they don't &lt;em&gt;actively&lt;/em&gt; do anything.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Accusation: Many people claim to follow you religiously.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Directional signs, but also astrology, tea leaves, etc. Many people &quot;follow the signs&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Accusation: We've never heard from 1 of your 5 parts. Why the silence?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Signs&quot; has a silent &quot;g&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Accusation: People say that you say all kinds of things.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Signs have all sorts of things written on them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Accusation: Many claim that you are great teachers...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Signs can be very informative...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Accusation: Some of you gather in large numbers...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Places with too many signs become overwhelming/confusing and may have too many rules, or too much advertising, and thus are unappealing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastActivityDate="2015-06-03T12:12:27.097" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15910" PostTypeId="2" ParentId="15671" CreationDate="2015-06-03T14:51:02.577" Score="20" Body="&lt;p&gt;After some analysis, it appears the numbers are suffering from some common psychological problems:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first group:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Has an inferiority complex. They stick together because when spelled out, their names have &lt;strong&gt;fewer letters&lt;/strong&gt; than the number that came before them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second group:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Is obsessive compulsive. They stick together because they have &lt;strong&gt;exactly the same&lt;/strong&gt; number of letters as the number that came before them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The third group:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Has a superiority complex. They stick together because their names have &lt;strong&gt;more letters&lt;/strong&gt; than the number that came before them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Treatment is tough; the only language that comes to mind that &quot;fixes&quot; this problem is Chinese, as some basic googling seems to show that each number is represented by only a single character up to a certain point. Even then, however, I'm sure the numbers would start squabbling about how many strokes it takes to form the character!&lt;/p&gt;&#xA;" OwnerUserId="10861" LastActivityDate="2015-06-03T14:51:02.577" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15911" PostTypeId="1" AcceptedAnswerId="15913" CreationDate="2015-06-03T14:52:40.997" Score="13" ViewCount="2097" Body="&lt;p&gt;Today I thought of a simple riddle involving time.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Name me a place where time is divided,&lt;br&gt;&#xA;  Where you can double or half it,&lt;br&gt;&#xA;  Or hold it completely still,&lt;br&gt;&#xA;  But never stop moving forward.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Name me a place where you can do all that.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's not a clock, and it's not a math sheet with clocks on it.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2756" LastEditorUserId="10400" LastEditDate="2015-06-03T15:14:38.597" LastActivityDate="2015-06-03T15:14:38.597" Title="A Matter of Time" Tags="|riddle|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15912" PostTypeId="2" ParentId="15911" CreationDate="2015-06-03T14:59:28.710" Score="3" Body="&lt;p&gt;A possible answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A central processor unit (CPU)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where you can double or half it, or hold it completely still, but never stop moving forward&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can overclock and under clock the frequency of clock cycles on a processor to make it double, or half its current speed. When the processor is off it is halted completely. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but never stop moving forward:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the use of 'forward' as progress, processors are always being researched for ways to be improved.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastActivityDate="2015-06-03T14:59:28.710" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15913" PostTypeId="2" ParentId="15911" CreationDate="2015-06-03T15:00:45.273" Score="16" Body="&lt;p&gt;Probably not correct, but here's the first thing that came to mind:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A piece of music&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...time is divided&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A piece of music is divided into measures based on the time signature&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...double or half it...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Could refer to tempo (increasing or decreasing the speed) or the length of different notes (whole notes, half notes, quarter notes, etc), or even changing time signatures&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...hold it completely still...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Could refer to rests, where nothing is played so the music &quot;stops&quot;, or a fermata, where the note is held for an indeterminately long time outside the time signature&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...but never stop moving forward.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The piece of music is always moving forward to the conclusion, no matter what you do to tempo or time signature during the piece.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10861" LastActivityDate="2015-06-03T15:00:45.273" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15914" PostTypeId="2" ParentId="15911" CreationDate="2015-06-03T15:05:30.327" Score="3" Body="&lt;p&gt;My answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A videogame&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Name me a place where time is divided,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The time in a videogame is usually made of &quot;ticks&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where you can double or half it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can speed up or slow down the speed of the game.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or hold it completely still&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pause.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but never stop moving forward.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In most videogames you cannot travel back in time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-06-03T15:05:30.327" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15916" PostTypeId="1" AcceptedAnswerId="15917" CreationDate="2015-06-03T16:19:29.943" Score="15" ViewCount="1223" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm a word - a simple word - you've heard me oft before&lt;br&gt;&#xA;  I am when you place the plates or box upon the floor&lt;br&gt;&#xA;  I'm involved in arson's work and clockmakers' final task&lt;br&gt;&#xA;  I am when authors create a world where players bask&lt;br&gt;&#xA;  I help the surgeon to fix bones and lyricists compose&lt;br&gt;&#xA;  I am when soldiers sharpen blades to fell their fearsome foes&lt;br&gt;&#xA;  I am when you start a journey and when the glue has dried&lt;br&gt;&#xA;  I am all a group or any group with single source implied&lt;br&gt;&#xA;  I am up and down and in and out and off and on again&lt;br&gt;&#xA;  I am key to all the phrases that I do begin&lt;br&gt;&#xA;  I can fill so many holes; so well am I fitted&lt;br&gt;&#xA;  What am I, my little self, unnoticed not omitted?&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="11475" LastEditDate="2015-06-03T17:10:24.210" LastActivityDate="2015-06-03T17:10:24.210" Title="What am I, my little self?" Tags="|riddle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15917" PostTypeId="2" ParentId="15916" CreationDate="2015-06-03T16:29:10.400" Score="12" Body="&lt;p&gt;The word&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;set&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fits here.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Set things down (plates or boxes, sure)&lt;br&gt;&#xA; Set a fire, set a clock to the correct time&lt;br&gt;&#xA; The author's world is the book's setting, or the book is set in the world&lt;br&gt;&#xA; After you break bone, the doctors set it, and a lyrics are set to music&lt;br&gt;&#xA; Not sure about the sharpening&lt;br&gt;&#xA; You set off, glue sets&lt;br&gt;&#xA; Mathematical set theory&lt;br&gt;&#xA; You can set up, set down, set in, set out, set off and set on&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The other lines are not specific clues.&lt;/p&gt;&#xA;" OwnerUserId="1509" LastEditorUserId="11475" LastEditDate="2015-06-03T17:10:12.333" LastActivityDate="2015-06-03T17:10:12.333" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15918" PostTypeId="2" ParentId="15916" CreationDate="2015-06-03T16:40:42.190" Score="10" Body="&lt;p&gt;This answer was already given, but I wanted to elaborate some more on how clues can fit.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Set&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm a word - a simple word - you've heard me oft before&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Set is the &lt;a href=&quot;http://www.wordfrequency.info/free.asp?s=y&quot;&gt;295th most common English word&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am when you place the plates or box upon the floor&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Set the table, set down&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm involved in arson's work and clockmakers' final task&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Set a fire and set the time when a clock is done&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am when authors create a world where players bask&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Set the scene&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I help the surgeon to fix bones and lyricists compose&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Setting bones is important in making sure they heal correctly&lt;br&gt;&#xA;Music can be performed in set, and you can set tempo when writing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am when soldiers sharpen blades to fell their fearsome foes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Set the edge of a blade&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am when you start a journey and when the glue has dried&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Set out, set up&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am all a group or any group with single source implied&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Set_(mathematics)&quot;&gt;Set(Mathematics)&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am up and down and in and out and off and on again&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Up, down, in, out, off, and on can all be preceded with set to make valid phrases.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am key to all the phrases that I do begin&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;All the answers above that start with &lt;code&gt;set&lt;/code&gt; are defined by how the thing is set.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can fill so many holes; so well am I fitted&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The word set is able to be a noun, a verb, and an adjective; it works well in most sentences&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I, my little self, unnoticed not omittied?&lt;/p&gt;&#xA;" OwnerUserId="12242" LastActivityDate="2015-06-03T16:40:42.190" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15919" PostTypeId="5" CreationDate="2015-06-03T17:08:27.880" Score="0" Body="&lt;p&gt;Geocaching is an outdoor recreational activity, in which participants use a GPS-enabled phone or similar portable device to hide and seek containers - called &amp;quot;caches&amp;quot; or &amp;quot;geocaches&amp;quot; - at specific locations marked by coordinates all over the world.&lt;/p&gt;&#xA;&lt;p&gt;In many instances a geocache's specific coordinates are not posted in its listing, and instead they must be calculated from numerical information obtained in a scavenger hunt (a 'multi-cache') or derived by solving a puzzle (a 'mystery cache').&lt;/p&gt;&#xA;&lt;p&gt;For more information on geocaching, see its entry on &lt;a href=&quot;https://en.wikipedia.org/wiki/Geocaching&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; or visit &lt;a href=&quot;https://www.geocaching.com/play/search&quot; rel=&quot;nofollow noreferrer&quot;&gt;geocaching.com&lt;/a&gt; or one of the many local &lt;a href=&quot;https://wiki.opencaching.eu/index.php?title=Main_Page#Opencaching_national_sites&quot; rel=&quot;nofollow noreferrer&quot;&gt;Opencaching&lt;/a&gt; sites.&lt;/p&gt;&#xA;&lt;p&gt;Questions related to the practical aspects of geocaching in the field (rather than the solving of scavenger hunts and puzzles) would be better asked on &lt;a href=&quot;https://outdoors.stackexchange.com/&quot;&gt;The Great Outdoors&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11497" LastEditorUserId="60644" LastEditDate="2022-11-29T11:47:31.717" LastActivityDate="2022-11-29T11:47:31.717" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15920" PostTypeId="4" CreationDate="2015-06-03T17:08:27.880" Score="0" Body="Geocaching is an outdoor recreational activity, in which participants use a GPS-enabled phone or similar device to hide and seek containers - called &quot;caches&quot; or &quot;geocaches&quot; - at specific locations marked by coordinates all over the world." OwnerUserId="11297" LastEditorUserId="60644" LastEditDate="2022-11-29T11:47:31.717" LastActivityDate="2022-11-29T11:47:31.717" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15921" PostTypeId="1" AcceptedAnswerId="15931" CreationDate="2015-06-03T17:37:33.450" Score="8" ViewCount="959" Body="&lt;p&gt;Initially, each of 50 Puzzling Stack Exchange users have a single distinct juicy bit of gossip not known to the others.  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If $A$ sends an email to $B$, that email can include all the bits of gossip $A$ currently knows. What are the fewest number of emails that need to be sent so  that each of the 50 users knows all 50 bits of gossip?  Why can't it be done with fewer emails? Assume multiple recipients / carbon copies are not allowed. &lt;/li&gt;&#xA;&lt;li&gt;If $A$ calls $B$, $A$ and $B$ can exchange all the bits of gossip that both know.  What is the fewest number of calls required so that each of the 50 users knows all 50 pieces of gossip? &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note:  Addressing why (2) can't be done with fewer calls is a little too difficult for a puzzle in my opinion.  But then again I believe Bollobás included it in his puzzle book &quot;Coffee Time in Memphis&quot;, so if you want to have a go at it knock yourself out =).&lt;/p&gt;&#xA;" OwnerUserId="2722" LastEditorUserId="2722" LastEditDate="2015-06-03T17:48:41.363" LastActivityDate="2017-12-06T18:25:02.860" Title="Spreading Gossip" Tags="|mathematics|graph-theory|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15922" PostTypeId="2" ParentId="15921" CreationDate="2015-06-03T17:47:22.497" Score="7" Body="&lt;p&gt;My solution to &lt;strong&gt;first&lt;/strong&gt; problem is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $98$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the strategy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First, everyone sends a mail to $A$ ($49$ mails in total). Then $A$ sends a mail to everyone else ($49$ more mails).  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why is this optimal? &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;To share his information, everyone must send at least an email, though if they want to know other information, they must receive at least a mail as well! So, $49+49=98$ is necessarily the minimum!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My solution to the &lt;strong&gt;second&lt;/strong&gt; problem is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $96$ &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A possible strategy is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Choose $4$ &quot;important&quot; people in the group $(A,B,C,D)$. The other people call one of these &quot;important&quot; guys, doesn't matter who. These are $50-4$ calls. Those $4$ people then share their knowledge between themselves, so that they know everything. This process requires only 4 calls. Then, they call back the &quot;ignorants&quot;, which takes again $50-4$ mails.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why is it optimal? &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's pretty obvious that $calls(4)=4$.&lt;br&gt;&#xA;When you add a member, he must make at least a call (to share his info), then he must receive at least another call (to receive the whole gossip).&lt;br&gt;&#xA; Thus, $calls(n+1)=calls(n)+2$.&lt;br&gt;&#xA; You can rewrite as $calls(n)=calls(4)+2(n-4)=2n-4$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-06-03T18:46:30.467" LastActivityDate="2015-06-03T18:46:30.467" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15923" PostTypeId="2" ParentId="15921" CreationDate="2015-06-03T17:54:23.267" Score="1" Body="&lt;p&gt;Here's my answer to the second one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 94&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Method&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First, every even number calls the number after it. (25). Then, every number in $4n$ calls $4n - 2$. (12) Then $8n$ calls $8n - 4$ (5). Then $16n$ to $16n - 8$ (3). Then $32$ calls $16$. (1) Now $32$ and $16$ know all the gossip (in 46 calls). $16$ calls the rest of the numbers (48) for a total of 94 calls.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Think of it like an inverse binary tree. Every node is a call and all the branches are gossip.&lt;br&gt;&#xA;&lt;br&gt;&#xA; If you take an example of 4 people ($1, 2, 3\text{ and }4$) Whats the fastest way to collect the gossip? 1 calls 2 and 3 calls 4 (2). Now 2 know (1, 2) and 4 knows (3, 4). 4 calls 2 (1) and now they both know all the gossip. Now 2 calls 1 and 3 to tell them (In 3 calls).  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason for optimality:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You basically need to gather all the gossip then disperse it. Calling basically merges gossip. Take the numbers from 1 to 50. Merge each number to the next number, and call it the second number (Could be the first, but second for ease), and each time, increment a $\text{call}$ counter. You will be left with the even numbers. Repeat this. Now you have multiples of 4. Keep repeating until you have 1 number left (The largest number). This number has all the gossip (And so does the previous number that is a power of two). This optimally combines gossip, as no duplicate gossip is spread. Now call the other numbers with this gossip (total numbers - 2). This may be sub-optimal, I can't prove it is not.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user10203" LastEditorDisplayName="user10203" LastEditDate="2015-06-03T18:20:11.530" LastActivityDate="2015-06-03T18:20:11.530" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15927" PostTypeId="1" AcceptedAnswerId="15930" CreationDate="2015-06-03T19:24:14.193" Score="0" ViewCount="435" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;You can find me once in one,&lt;br /&gt;&#xA;You can find me once in four,&lt;br /&gt;&#xA;You cannot find me in ten,&lt;br /&gt;&#xA;You cannot find me in five hundred.&lt;/p&gt;&#xA;&lt;p&gt;You think I am too easy to spot,&lt;br /&gt;&#xA;That's because you did not find me.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;" OwnerUserId="10400" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-06-03T19:31:45.610" Title="Don't be too literal, take some numbers" Tags="|riddle|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15928" PostTypeId="2" ParentId="15927" CreationDate="2015-06-03T19:25:08.477" Score="5" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the letter O.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You can find me once in one,&lt;br&gt;&#xA;You can find me once in four,&lt;br&gt;&#xA;You cannot find me in ten,&lt;br&gt;&#xA;You cannot find me in five hundred.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter O appears once in each of &quot;&lt;strong&gt;o&lt;/strong&gt;ne&quot; and &quot;f&lt;strong&gt;o&lt;/strong&gt;ur&quot;, but not at all in &quot;ten&quot; and &quot;five hundred&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You think I am too easy to spot,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; O appears twice in &quot;too&quot;, once in &quot;spot&quot;, and looks like a spot itself (circular?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;That's because you did not find me.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hmph. See pacoverflow's answer for the intended (probably) solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-06-03T19:31:45.610" LastActivityDate="2015-06-03T19:31:45.610" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15930" PostTypeId="2" ParentId="15927" CreationDate="2015-06-03T19:29:38.980" Score="7" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter I&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can find me once in one,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One is I in Roman numerals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can find me once in four,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Four is IV in Roman numerals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You cannot find me in ten,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ten is X in Roman numerals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You cannot find me in five hundred.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 500 is D in Roman numerals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You think I am too easy to spot, That's because you did not find me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The obvious answer is O (since you can find it in one, four, but not in ten or five hundred).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2015-06-03T19:29:38.980" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15931" PostTypeId="2" ParentId="15921" CreationDate="2015-06-03T20:27:11.480" Score="6" Body="&lt;p&gt;&lt;strong&gt;Problem 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As leoll2 illustrates, it can be done in &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;98 emails (everyone emails Alice, then she emails everyone). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's why it can't be done in fewer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we show that after the $48$th email, there won't be anyone who has heard all of the gossip. Consider the undirected graph where the vertices are people, where two people are connected with an edge if one of them emailed the other. Since there are at most $48$ edges, and $50$ vertices, this graph is disconnected (every connected $n$ vertex graph has at least $n-1$ edges). This means that people in one connected component could not have heard any of the tidbits of gossip from the other components, so no one could have heard all the gossip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, the number of people who have heard all the gossip increases by at most one after every email. Thus, since there are 0 fully informed people after $48$ emails, there are at most $49$ fully informed people after $48+49=97$ emails have been sent, so $97$ emails is not enough.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-06-03T21:29:44.600" LastActivityDate="2015-06-03T21:29:44.600" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15932" PostTypeId="2" ParentId="15839" CreationDate="2015-06-03T20:37:13.477" Score="11" Body="&lt;h1&gt;n=19&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;A more natural attempt:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Occasionally, blindfolded conjuring wizards quietly, unnoticeably mimicked fictionalised, awe-striking witchery; unfortunately, extensive unbeneficial avariciousness cynically rehabilitated inharmonical, vindictive suffocation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now and then, these wizards would get together and practice their magic skills to copy cool spells they saw on TV (but they did so in secret). However, they got greedy, and wanted to steal each other's tricks too much, which sadly made them start fighting and trying to choke each other to death. Or something.&lt;/p&gt;&#xA;" OwnerUserId="11476" LastEditorUserId="11476" LastEditDate="2015-06-03T23:05:18.167" LastActivityDate="2015-06-03T23:05:18.167" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15933" PostTypeId="2" ParentId="302" CreationDate="2015-06-03T20:39:42.340" Score="2" Body="&lt;p&gt;Joe Z gave an excellent answer in the spirit of the question, but what if one were to interpret the question more literally, like a national defense officer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technically, we need to take the amount of missiles of each country into account. Germany doesn't have any nuclear weapons (except like 10 or so warheads lent to them by the USA, which of course can't be fired without USA permission), so Germany is perfectly harmless to the USA. Also, China does not gain any benefits by firing at Germany, because Germany will be out of ammo after the first round anyway. So Germany fires at China, China fires at USA and USA fires at China. Considering the amount of missiles USA has, China will die pretty much 100% of the time, and USA will also go down most of the time at the hands of China, and Germany will go down depending on whether USA is feeling evil or not.&lt;/p&gt;&#xA;" OwnerUserId="12906" LastEditorUserId="12906" LastEditDate="2015-06-03T22:16:41.403" LastActivityDate="2015-06-03T22:16:41.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15934" PostTypeId="1" AcceptedAnswerId="15937" CreationDate="2015-06-03T22:46:49.670" Score="7" ViewCount="3902" Body="&lt;p&gt;&lt;em&gt;I'm starting to suspect this is some kind of gag puzzle, so before you downvote it please consider that I've considered that.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is &quot;How many minutes in a second?&quot; with the stipulation being that the answer is &quot;a four letter word in everyday use that is not 'zero' or 'nada' or 'none'&quot; (none of which would really be correct anyway).  I've tried math answers and I've tried English answers and can't come up with a viable solution in either.  Thoughts?&lt;/p&gt;&#xA;" OwnerUserId="12949" LastEditorUserId="5373" LastEditDate="2015-06-03T23:16:28.463" LastActivityDate="2017-01-15T16:54:26.057" Title="How Many Minutes in a Second" Tags="|mathematics|word|lateral-thinking|" AnswerCount="7" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15935" PostTypeId="2" ParentId="15934" CreationDate="2015-06-03T23:15:32.570" Score="10" Body="&lt;p&gt;Were you asked this question yesterday (2nd June) by any chance?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; How many minutes are there in the 2nd (of a month)? Well, 1440 (60x24), same as in any other day, of course!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This and many other lateral-thinking answers can be found &lt;a href=&quot;http://www.reddit.com/r/riddles/comments/100ngk/how_many_minutes_are_there_in_a_second/?&quot;&gt;here&lt;/a&gt;. This one was the only four-character answer I could find or think of.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-06-03T23:19:34.813" LastActivityDate="2015-06-03T23:19:34.813" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15936" PostTypeId="2" ParentId="15934" CreationDate="2015-06-03T23:29:06.887" Score="1" Body="&lt;p&gt;I know you said the answer is 4 letters, but here's an answer that works (and I didn't see it on the &lt;a href=&quot;http://www.reddit.com/r/riddles/comments/100ngk/how_many_minutes_are_there_in_a_second/?&quot; rel=&quot;nofollow&quot;&gt;page&lt;/a&gt; that rand linked to):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If by minute you mean &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Minute_of_arc&quot; rel=&quot;nofollow&quot;&gt;minute of arc&lt;/a&gt;&quot; and by second you mean &quot;second of arc&quot;, then the answer can be two.  This is because a minute of arc is represented by the prime (′) symbol, and a second of arc is represented by the double prime (″) symbol.  There are two prime symbols inside a double prime symbol, so you could say there are two minutes of arc inside a second of arc.&#xA;&lt;p&gt;&#xA; This can be a 4 letter answer by translating two into French: deux.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4450" LastEditDate="2015-06-03T23:37:37.637" LastActivityDate="2015-06-03T23:37:37.637" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15937" PostTypeId="2" ParentId="15934" CreationDate="2015-06-04T02:33:36.313" Score="4" Body="&lt;p&gt;My best guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Part - a second could be described as the 60th &lt;strong&gt;part&lt;/strong&gt; of a minute. So &lt;em&gt;part of a minute is in a second&lt;/em&gt;. Also, part is a fairly common word. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My second guess would be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Once - As you only need to be seconded &lt;strong&gt;once&lt;/strong&gt; to be put in the &lt;em&gt;minutes&lt;/em&gt; of the meeting.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11239" LastActivityDate="2015-06-04T02:33:36.313" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15938" PostTypeId="1" AcceptedAnswerId="15942" CreationDate="2015-06-04T03:12:43.583" Score="19" ViewCount="3096" Body="&lt;p&gt;There are 15 players who will play a cooperative game. They begin by closing their eyes. A referee will place either a black hat or a white hat (chosen by fair coin flip) on each player's head. The players may then open their eyes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each hat is visible to every player except the one wearing it. On a signal given by the referee, each player may call out a guess of her own hat color, or may remain silent. The calls are simultaneous.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The players all win if at least one player guesses correctly &lt;strong&gt;and&lt;/strong&gt; nobody guesses incorrectly.&lt;br&gt; They all lose if anyone guesses incorrectly &lt;strong&gt;or&lt;/strong&gt; everybody remains silent.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The players can agree on a strategy beforehand, but no further communication is allowed once the game starts.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Find a strategy for the players that maximizes the probability that they will win. What is that probability?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus&lt;/strong&gt;: What happens if we change the number of players?&lt;/p&gt;&#xA;" OwnerUserId="2627" LastActivityDate="2023-11-17T15:10:30.287" Title="Cooperative guessing game: no incorrect guesses" Tags="|mathematics|probability|hat-guessing|" AnswerCount="6" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15939" PostTypeId="2" ParentId="15934" CreationDate="2015-06-04T03:57:52.903" Score="2" Body="&lt;p&gt;Building on @pacoverflow's answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; beta&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Source:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Greek numbering chart lists 'beta' as the word used to describe the number 2 in the untranslated [Greek] version of Ptolemy's Table of Chords, as in two degrees of the circle, of which there are 60 degrees in a circle and one second starts on one minute and ends on another as the second hand ticks.  The table is located at &lt;a href=&quot;http://en.wikipedia.org/wiki/Ptolemy%27s_table_of_chords#The_numeral_system_and_the_appearance_of_the_untranslated_table&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Ptolemy%27s_table_of_chords#The_numeral_system_and_the_appearance_of_the_untranslated_table&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Alternative answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'a sec'  (asec).  Also from @pacoverflow's answer.  His link points to &lt;a href=&quot;http://en.wikipedia.org/wiki/Minute_of_arc#Symbols_and_abbreviations&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Minute_of_arc#Symbols_and_abbreviations&lt;/a&gt;, which lists an abbreviation of &quot;asec&quot; for an arcsecond, or if in two words: &quot;a sec&quot; (wait a sec! or wait asec!).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10297" LastEditorUserId="10297" LastEditDate="2015-06-04T12:11:00.017" LastActivityDate="2015-06-04T12:11:00.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15940" PostTypeId="2" ParentId="15934" CreationDate="2015-06-04T04:09:01.153" Score="-1" Body="&lt;p&gt;If &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I define four new letters ('si', 'xt', 'ie', and 'th'), &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;then I can claim the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7GLlC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="18250" LastEditDate="2017-01-15T16:54:26.057" LastActivityDate="2017-01-15T16:54:26.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15941" PostTypeId="2" ParentId="15938" CreationDate="2015-06-04T04:43:30.483" Score="1" Body="&lt;p&gt;I know the strategy for 3 players - If you see two or more of a colour, call out the opposite colour; remain silent if both are in equal proportion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't work if all three players are wearing the same colour, but does work in all other cases, which is a probability of 3/4.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For 15 players (and in general any odd number of players), a similar solution might work in terms of the higher number of hats of a single colour being odd or even:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you see an even number of hats of a single colour, call out colour that has fewer hats. Otherwise, remain silent.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Supposing that there are $n$ black hats and $15 - n$ white hats, this gives us the following scenarios:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If we have 15 white hats and 0 black hats, everyone calls out black and dies.&lt;/li&gt;&#xA;&lt;li&gt;If we have 14 white hats and 1 black hat, the person wearing the black hat calls out &quot;black&quot;, and everyone else remains silent.&lt;/li&gt;&#xA;&lt;li&gt;If we have 13 white hats and 2 black hats, everyone wearing a white hat calls out black and they die.&lt;/li&gt;&#xA;&lt;li&gt;If we have 12 white hats and 13 black hats, the three people wearing black hats call out &quot;black&quot;, and everyone else remains silent.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And so on. This gives us a probability of $\binom{15}{1} + \binom{15}{3} + \binom{15}{5} + \binom{15}{7} + \binom{15}{8} + \binom{15}{10} + \binom{15}{12} + \binom{15}{14}$ $= 19186/32768$ $\doteq 0.6047$.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-06-04T04:43:30.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15942" PostTypeId="2" ParentId="15938" CreationDate="2015-06-04T06:02:19.973" Score="23" Body="&lt;p&gt;The players can achieve $15/16$ win probability, which is optimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Associate the 15 players with nonzero vectors in $\mathbb{Z}_2^4$ like $(0,1,1,1)$. Let $S$ be the sum (entry-wise XOR, or Nim sum) of the players with black hats. Each player doesn't know $S$ because they don't know their own hat color, but they know the two possible values of $S$ depending on their hat color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The players' strategy is to bet that $S$ is nonzero. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If one choice of your hat color would make $S$ be zero, guess the other one. Otherwise, remain silent.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We show that the players win exactly if $S$ is nonzero. This happens $15/16$ of the time, since flipping the presence of any of hats $(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)$ flips the corresponding entry of $S$, so $1$ of their $16$ joint settings makes the $S$ be $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof:&lt;/em&gt; The two possible values of $S$ from a player's perspective differ by their own vector. If $S$ equals some nonzero vector $v$, the player $v$ will guess the correct hat color, since the alternative would change the sum by $v$, making it $0$. All other players will remain silent, since their possible sums are $S$ and $S+w$ for $w\neq v$, which cannot be $0$. So, the players win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $S=0$, everyone guesses the wrong color. (As is common in hat guessing problems, you want the &quot;wrong&quot; case to be as wrong as possible to balance out the expectation.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A win probability of $15/16$ is optimal. For any player who guesses, they have an equal probability of getting their hat color right of wrong. So, by linearity of expectation, the expected number of right guesses minus wrong guesses is $0$. Since a win requires at least $1$ right guess and no wrong guesses (difference $+1$), and a loss has at worst a wrong for each player difference $-15$), we must have at least $1$ loss per $15$ wins, so at most $15/16$ win probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This all generalizes when the number of players $N$ is of the form $2^k-1$. If it's not, the players can pretend it is by ignoring some number of the players, which gives win probability $1-\frac{1}{2^k}$ where $2^k$ is the largest power of $2$ with $2^k-1\leq N$. I don't know if this is optimal though.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="35044" LastEditDate="2017-03-09T17:30:20.123" LastActivityDate="2017-03-09T17:30:20.123" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="15943" PostTypeId="1" AcceptedAnswerId="15944" CreationDate="2015-06-04T06:57:06.590" Score="20" ViewCount="6720" Body="&lt;p&gt;There was a green house.&lt;br&gt;&#xA;Inside the green house there was a white house.&lt;br&gt;&#xA;Inside the white house there was a red house.&lt;br&gt;&#xA;Inside the red house there were lots of babies.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What am I?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="11764" LastEditorUserId="463" LastEditDate="2015-06-04T14:05:59.100" LastActivityDate="2015-06-05T18:28:09.773" Title="Tasty - Who am I?" Tags="|riddle|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15944" PostTypeId="2" ParentId="15943" CreationDate="2015-06-04T07:00:22.560" Score="44" Body="&lt;p&gt;It is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;watermelon&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There was a green house.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it have green layer outside&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Inside the green house there was a white house.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; inside green layer there is slightly white layer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Inside the white house there was a red house.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; watermelon's pulp is red&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Inside the red house there were lots of babies. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are lots of seed in that pulp.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/IP4yQ.jpg&quot; alt=&quot;Picture of a watermelon&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4312" LastEditorUserId="9000" LastEditDate="2015-06-04T12:07:12.530" LastActivityDate="2015-06-04T12:07:12.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15946" PostTypeId="2" ParentId="15943" CreationDate="2015-06-04T08:53:26.933" Score="2" Body="&lt;p&gt;Apart from the obvious answer by @CoDe MurDeRer, here is alternative:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Really big and color-specific Matryoshka doll &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There was a green house.&lt;br&gt;&#xA;Inside the green house there was a white house.&lt;br&gt;&#xA;Inside the white house there was a red house.&lt;br&gt;&#xA;Inside the red house there were lots of babies. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The outermost layer is green, next is white and red. An because it is a big one, it will have lots of &quot;babies&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9907" LastActivityDate="2015-06-04T08:53:26.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15947" PostTypeId="2" ParentId="15852" CreationDate="2015-06-04T09:00:28.720" Score="42" Body="&lt;p&gt;Kudos to Alconja for creating a puzzle that's so easy and yet has lasted so long unsolved! I dare anyone to disagree that this is easy - &lt;em&gt;especially&lt;/em&gt; given the hints - after reading the solution below.&#xA;&lt;hr&gt;&#xA;The hours of the day are usually referred to by a number between $1$ and $12$, even though in fact there are 24 hours in a day. This gives a small ambiguity: &quot;three o'clock&quot; can mean either $3$ or $15$ on a 24-hour clock. (In maths speak, we're considering the 24 hours in their residue classes modulo 12.) Converting these numbers to letters (A=1, B=2, etc.), we can get any letter from A to X from an hour of the day. For instance, &quot;three o'clock&quot; means either C or O.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;One, two, one o'clock, three o'clock rock&lt;br&gt;&#xA;Eight, six, three o'clock, two o'clock rock&lt;br&gt;&#xA;Nine, seven, one o'clock, seven o'clock rock&lt;br&gt;&#xA;We're gonna rock around the clock tonight&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the &quot;rocking&quot; stuff is just flavour text. Extract the numbers and convert them to letters to get the following, where the $n$th column contains the 2 possibilities for the $n$th number in the song:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;A B A C H F C B I G A G&#xA;M N M O T R O N U S M S&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ANACHRONISMS.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the bonus question, we have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;A B B I H I L A H I C B&#xA;M N N U T U X M T U O N&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and so the clock deserves&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ANNIHILATION.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-06-05T09:32:34.650" LastActivityDate="2015-06-05T09:32:34.650" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15948" PostTypeId="1" AcceptedAnswerId="15949" CreationDate="2015-06-04T09:43:43.147" Score="5" ViewCount="5458" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6tWlf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been looking for patterns for a while now, but could not find one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found some rules:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;The middle column always points the same direction.&lt;/li&gt;&#xA;  &lt;li&gt;The first two in the first row are always opposite.&lt;/li&gt;&#xA;  &lt;li&gt;The last two in the second column are always opposite.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Still, these rules are insufficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attribution: &lt;a href=&quot;https://europa.eu/epso/application/passport/index.cfm?action=int_test&amp;amp;intact=dsp_ar_tests&quot; rel=&quot;nofollow noreferrer&quot;&gt;EPSO online test&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12960" LastEditorUserId="463" LastEditDate="2015-06-04T14:10:47.213" LastActivityDate="2023-03-21T07:25:51.530" Title="Find the next image in this abstract reasoning puzzle" Tags="|pattern|geometry|lateral-thinking|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15949" PostTypeId="2" ParentId="15948" CreationDate="2015-06-04T09:55:38.930" Score="6" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; following each of the 6 positions through in sequence, the directions of the arrows rotate by 90 degrees clockwise each time except when they point up or down, where they stay twice in a row.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More details:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Top left arrow: R,D,D,L,U, &lt;strong&gt;U&lt;/strong&gt;&lt;br&gt;Top middle arrow: L,U,U,R,D, &lt;strong&gt;D&lt;/strong&gt;&lt;br&gt;Top right arrow: D,D,L,U,U, &lt;strong&gt;R&lt;/strong&gt;&lt;br&gt;Bottom left arrow: U,R,D,D,L, &lt;strong&gt;U&lt;/strong&gt;&lt;br&gt;Bottom middle arrow: L,U,U,R,D, &lt;strong&gt;D&lt;/strong&gt;&lt;br&gt;Bottom right arrow: R,D,D,L,U, &lt;strong&gt;U&lt;/strong&gt;&lt;br&gt;(directions in which each arrow points)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-06-04T10:19:24.163" LastActivityDate="2015-06-04T10:19:24.163" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15950" PostTypeId="1" CreationDate="2015-06-04T09:56:00.087" Score="2" ViewCount="366" Body="&lt;p&gt;I found this online (I don't know who made it)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oqqgi.png&quot; alt=&quot;puzzle&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;strong&gt;Note:&lt;/strong&gt; there is a missing link at the bottom!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess that a &lt;em&gt;transition&lt;/em&gt; is a contiguous movement in the same direction (or: the number of changes of direction plus $1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't solve this in $15$ transitions, and I am starting to doubt a solution even exists. This is the best I could do:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 57 58 07 08 09 10 23 24&lt;br&gt;56 59 06 41 40 11 22 25&lt;br&gt;55 60 05 42 39 12 21 26&lt;br&gt;54 61 04 43 38 13 20 27&lt;br&gt;53 62 03 44 37 14 19 28&lt;br&gt;52 63 02 45 36 15 18 29&lt;br&gt;51 64 01 46 35 16 17 30&lt;br&gt;50 49 48 47 34 33 32 31&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which uses $16$ transitions.&lt;/p&gt;&#xA;" OwnerUserId="12961" LastEditorUserId="463" LastEditDate="2015-06-04T13:45:57.333" LastActivityDate="2015-06-25T20:13:52.503" Title="Find the path in 15 transitions" Tags="|visual|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15951" PostTypeId="2" ParentId="15948" CreationDate="2015-06-04T09:59:42.053" Score="3" Body="&lt;p&gt;If you look at the arrow in the same location, you'll notice that they always rotate clockwise. But whenever they hit either up or down, they stay that way for two steps. In other words, each arrow follows the pattern: &lt;em&gt;up, up, right, down, down, left, ...&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there the solution is pretty clearly option &lt;em&gt;C&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5208" LastActivityDate="2015-06-04T09:59:42.053" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15952" PostTypeId="1" AcceptedAnswerId="17273" CreationDate="2015-06-04T10:55:15.397" Score="8" ViewCount="1905" Body="&lt;p&gt;Phil arrived at the Broadway theater. Mr. Smith, the director, had called him in for an emergency.&lt;/p&gt;&#xA;&lt;p&gt;Phil: What is the problem, David?&lt;/p&gt;&#xA;&lt;p&gt;David: Thank god you've arrived. We have a serious issue for tonight's show. Our electrical system has gone haywire and we need your help.&lt;/p&gt;&#xA;&lt;p&gt;Phil: Electrical system? I'm not an electrical engineer. I don't know how I could possibly help.&lt;/p&gt;&#xA;&lt;p&gt;David: Here's the problem: we managed to fix most of our lighting, but our biggest spotlights still have a serious issue. We have no idea what causes it, but it only started an hour ago, and the electrician said we would have to dismantle a large part of our rigging to fix it, which we don't have time for.&lt;/p&gt;&#xA;&lt;p&gt;Phil: I still don't see where I come in...&lt;/p&gt;&#xA;&lt;p&gt;David: Here's the problem:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Our 3 main spotlights that we don't use except for the finale are broken. They will work only for 6 seconds each before the fuse blows. The off-switches are broken, and the fastest we can replace the fuse when it blows is 10 seconds, and we need to start from the moment the light goes on, but your assistant is confident that it can always be done that fast. However, the rigging cannot support more than 1 person to replace a fuse at once, so they need to be replaced sequentially.&lt;/p&gt;&#xA;&lt;p&gt;Our finale lasts 1 minute, and we need at least one light to be active during that entire period. If that's not possible, we need to know in advance how long we can keep those lights going before all the lights are without fuses. In theory, we can make the finale anywhere between 30 and 60 seconds long.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Phil: So, to summarize:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;you only have 6 seconds per light, a fuse takes 10 seconds to replace, and the rigging can only support 1 person to replace fuses at a time. You wonder whether you can let the finale last the full 60 seconds or whether it needs to be shortened.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;David: Correct. You're the best lighting expert we have. What's your verdict?&lt;/p&gt;&#xA;&lt;p&gt;Can David still do the full finale? Or does he need to shorten it? And if so, how long can the finale last? Obviously, since this is a math problem, you need to &amp;quot;show your work&amp;quot;.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Bonus: Friday the 12th of June, I will hand out a bounty of 50 reputation to the answer with the highest score that continues the dialog and gives the right answer.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;Explanation for those that want the story behind this question&lt;/h2&gt;&#xA;&lt;p&gt;This core of this question is not about lighting. The original question I thought of was based around a mechanic from the video game World of Warcraft, more precisely the Priest's &lt;a href=&quot;http://www.wowhead.com/spell=121536/angelic-feather&quot; rel=&quot;nofollow noreferrer&quot;&gt;Angelic Feather&lt;/a&gt; ability:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;Angelic Feather&#xA;40 yd range&#xA;Instant 10 sec recharge&#xA;3 Charges&#xA;Requires Priest&#xA;Requires level 30&#xA;Place a feather at the target location, granting the first ally to walk through it 60% increased movement speed for 6 sec.  Max 3 charges. Only 3 feathers can be placed at one time.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;As you can see, one charge of the feather is equal in functionality to 1 light: once you activate it, it lasts 6 seconds, each charge has a separate 10 second cool-down period (called &amp;quot;recharge&amp;quot; here) and there are 3 charges. Charges in World of Warcraft have sequential cool-downs: they don't cool down at the same time, but one after another.&lt;/p&gt;&#xA;&lt;p&gt;The reason I translated this to lighting instead of keeping it about the spell is because I assumed that people (especially those that don't play World of Warcraft) would understand the problem easier if I used more known terms like fuses and lighting than recharge, spells and duration.&lt;/p&gt;&#xA;" OwnerUserId="2778" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-02T03:25:03.617" Title="The lighting expert and the prima donna" Tags="|mathematics|story|" AnswerCount="7" CommentCount="16" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15953" PostTypeId="1" AcceptedAnswerId="15957" CreationDate="2015-06-04T11:02:57.567" Score="2" ViewCount="2603" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qDCwY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the next image?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I already found the rule for the circle. Which simply goes around clockwise in the corners. So, the correct choice is either C or D. But, I do not find such a rule for the arrow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attribution: &lt;a href=&quot;https://europa.eu/epso/application/passport/index.cfm?action=int_test&amp;amp;intact=dsp_ar_tests&quot; rel=&quot;nofollow noreferrer&quot;&gt;This site&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12960" LastEditorUserId="463" LastEditDate="2015-06-04T14:30:36.857" LastActivityDate="2018-10-31T18:17:29.060" Title="Abstract image puzzle with a rotating arrow and a circle" Tags="|pattern|geometry|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15954" PostTypeId="2" ParentId="15950" CreationDate="2015-06-04T11:04:07.180" Score="3" Body="&lt;p&gt;It's actually impossible to do this in fewer than 16 transitions if you use only horizontal and vertical transitions, even if your path is allowed to cross itself and/or go out of the box!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: Pick any closed path passing through each square. Extend each transition into a line. Now merge all runs of consecutive identical lines. Now these lines still cover all squares together, and the number of horizontal and vertical lines in the extended path is always equal (and not larger than the number of transitions in the original path), since horizontal and vertical lines alternate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the path has less than 16 lines, so less than eight horizontal lines. Then there is one row which does not contain a horizontal line. But this row has eight squares, which are necessarily passed over by one vertical line each, so there are at least eight vertical lines, so also eight horizontal lines in contradiction to the assumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence any axis-aligned path through every square must necessarily have at least 16 transitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here's a non-axis-aligned solution in 15 transitions with self-intersections (14 if you can start somewhere else than the marked square), credit to Sam Loyd:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CO67Y.png&quot; alt=&quot;The original problem had something to do with battleships&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="12906" LastActivityDate="2015-06-04T11:04:07.180" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15955" PostTypeId="2" ParentId="15952" CreationDate="2015-06-04T11:06:33.493" Score="7" Body="&lt;p&gt;Phil: It's &lt;strong&gt;impossible&lt;/strong&gt;, I'm afraid.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Think of it in terms of how many &amp;quot;secondsworth&amp;quot; of light you have available. You start off with 3 cool lamps - that's 18 seconds' worth. For the first 6 seconds, you're using one lamp to light the stage, so there's nothing to cool. Then you have 12 seconds' worth of light left.&lt;/p&gt;&#xA;&lt;p&gt;It's going to take you 10 seconds to earn another 6 seconds' worth, and you need to use up 10 seconds' worth of light in that time, leaving you with only 8 seconds' worth.&lt;/p&gt;&#xA;&lt;p&gt;Now you move the cooling system to another light, but by the time you get that cool, 10 seconds will have passed and you only had 8 seconds' worth of light to play with, so 2 seconds must have been dark.&lt;/p&gt;&#xA;&lt;p&gt;The most light you can get is &lt;strong&gt;24 seconds&lt;/strong&gt;, which is less than the 30-second minimum you mentioned.&lt;/p&gt;&#xA;&lt;p&gt;If you like, my brother-in-law could do you a good deal on a fourth spotlight...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;David: I called you in to advise me, not to advertise at me! Everyone knows your brother-in-law's lights are dodgy! Thanks for the advice; we'll have to either cut out some of the primadonna's finale or let her do some of it in darkness.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-06-04T11:17:33.267" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15956" PostTypeId="2" ParentId="15953" CreationDate="2015-06-04T11:27:17.553" Score="6" Body="&lt;p&gt;I choose&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The circle rotates around the corners clockwise, so it can be either C or D (as you also found). The line seems to be following a pattern as well: west, northwest, southwest, north, south, &lt;em&gt;northeast&lt;/em&gt;, southeast, east.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2015-06-04T11:27:17.553" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15957" PostTypeId="2" ParentId="15953" CreationDate="2015-06-04T11:35:21.137" Score="14" Body="&lt;p&gt;I  believe the pattern for the arrow is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;alternating between clockwise and anti-clockwise and increasing distance by one eighth of the square each time. i.e. clockwise 1, anti-clockwise 2, clockwise 3 etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which would make the answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9965" LastActivityDate="2015-06-04T11:35:21.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15958" PostTypeId="2" ParentId="15948" CreationDate="2015-06-04T11:36:34.863" Score="3" Body="&lt;p&gt;For me the obvious answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In first and fourth pattern the arrows point in opposite directions.&#xA; Likewise in the second and fifth pattern. Therefore the sixth pattern&#xA; should be the opposite of the third pattern, so the answer is C.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12965" LastActivityDate="2015-06-04T11:36:34.863" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15959" PostTypeId="2" ParentId="15943" CreationDate="2015-06-04T11:50:59.067" Score="10" Body="&lt;p&gt;Tasty - guess who am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An attractive pregnant woman carrying octuplets. &lt;a href=&quot;http://thefreedictionary.com/tasty&quot;&gt;Tasty&lt;/a&gt; has the meaning of 'attractive' when referring to women.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There was a green house.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The pregnant woman was wearing a green dress.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Inside the green house there was a white house.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Her underwear were white in colour.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Inside the white house there was a red house.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Two ways to justify. Either her skin was red, or the redness refers to the viscera inside her body.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Inside the red house there were lots of babies. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Her children in her womb. God bless them! &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2425" LastActivityDate="2015-06-04T11:50:59.067" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15960" PostTypeId="2" ParentId="15952" CreationDate="2015-06-04T12:05:25.187" Score="11" Body="&lt;p&gt;rand al'thor had the answer to the initial question, where the lights had to wait to be off before repairing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phil: I cannot do the full finale, but I can give you &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 36 seconds of continuous light!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A light takes 10 seconds to repair and provides 6 seconds of light.  All lights are available at the beginning.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Light 1: On at 0, Off at 6, Available at 10&#xA;Light 2: On at 6, Off at 12, Available at 20&#xA;Light 3: On at 12, Off at 18, Available at 30&#xA;Light 1: On at 18, Off at 24, Available at 40&#xA;Light 2: On at 24, Off at 30, Available at 50&#xA;Light 3: On at 30, Off at 36, Available at 60&#xA;Light 1: Not Available at 36&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1777" LastEditorUserId="1777" LastEditDate="2015-06-04T17:59:19.373" LastActivityDate="2015-06-04T17:59:19.373" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15961" PostTypeId="2" ParentId="15934" CreationDate="2015-06-04T12:14:10.093" Score="2" Body="&lt;p&gt;One minute of arc is 1/60th of a degree, which is 1/360th of a circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second on a clock is 1/60th of the clock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one could comfortably say that there are 360 arc minutes in a second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This permits a couple of different answers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Year - 360 is a common approximation of the number of days in a year.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Turn - 360 degrees is, in geometry, called a &quot;turn&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9289" LastActivityDate="2015-06-04T12:14:10.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15962" PostTypeId="2" ParentId="15948" CreationDate="2015-06-04T12:19:30.407" Score="-1" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;C&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All the pointers are rotating clockwise pointing up and down twice and sideways once.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12966" LastEditorUserId="10281" LastEditDate="2015-06-04T12:27:29.920" LastActivityDate="2015-06-04T12:27:29.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15963" PostTypeId="2" ParentId="15952" CreationDate="2015-06-04T12:23:39.283" Score="6" Body="&lt;p&gt;Phil: I don't know if I can get you your full minute, but:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I think I can do &lt;strong&gt;36 seconds&lt;/strong&gt;&lt;br&gt;&#xA;Turn lamp 1 on for 6 seconds, whilst cooling it&lt;br&gt;&#xA;Turn lamp 2 on for 6 seconds - continue cooling lamp 1 for 4 seconds, then switch the cooling device to lamp 2 for 2 seconds&lt;br&gt;&#xA;Turn lamp 3 on for 6 seconds - continue cooling lamp 2.&lt;br&gt;&#xA;Turn lamp 1 on for 6 seconds - coll lamp 2 for another 2 seconds, then switch the cooler to lamp 3 for 4 seconds&lt;br&gt;&#xA;Turn lamp 2 on for 6 seconds. Once this is done, lamp 3 should be cool, but you don't have time to cool anything else.&lt;br&gt;&#xA;Turn lamp 3 on for 6 seconds. That should give you a total of 36 seconds of light  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Are you sure you're ok with cutting the performance down by that much? I'll sit down with the calculations and see if I can get you a little longer...  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;David: That's fantastic! If you can get a few more seconds I'd greatly appreciate it, but at least now we know that the show can go on!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phil: I'll do what I can. In the meantime, here's an diagram that should help explain it to any confused techs. Green cells represent a second of a light being on, blue a second of the cooling mechanism   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BG7hO.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11717" LastEditorUserId="11717" LastEditDate="2015-06-04T12:30:03.787" LastActivityDate="2015-06-04T12:30:03.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15964" PostTypeId="2" ParentId="15952" CreationDate="2015-06-04T12:39:48.820" Score="1" Body="&lt;p&gt;If we assume that everything is done perfectly(so that we can cooldown a light immediately after we done cooling the previous one) we can certainly get at least 45 seconds of show. Here's how:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Start with light 1 and cool it at the same time. Assume that light has 60HP, so each second of work will consume 10HP and each second of cooldown will add 6HP. So that leaves us with a constant HP loss of 4/sec. Because the light has 60HP, it will take &lt;strong&gt;15&lt;/strong&gt; seconds to get down to 0($60/4=15$).&lt;br&gt;&#xA; Then use second light exactly the same till it's &lt;strong&gt;30&lt;/strong&gt; seconds total.&lt;br&gt;&#xA; And last use the third light for another 15 seconds which will sum up to &lt;strong&gt;45&lt;/strong&gt; seconds in total.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll update my answer if I find a way to get more time&lt;/p&gt;&#xA;" OwnerUserId="9352" LastActivityDate="2015-06-04T12:39:48.820" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15965" PostTypeId="2" ParentId="15952" CreationDate="2015-06-04T12:41:48.760" Score="4" Body="&lt;p&gt;David: I can't quite get you the full minute - &lt;strong&gt;45 seconds&lt;/strong&gt; is the best I can do, and here's how:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Use the cooling system on the first light even as you use it. If it would cool it from critical in 10 seconds, and heat to critical in 6 seconds without cooling, then with cooling it should last 15 seconds. We have three lights, so we can switch twice and get 45 seconds in total. If you can boost the cooling system just a bit more, and get it to cool in 8.5 seconds, that will be enough to get the full minute. Any chance of that?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt; &#xA;&#xA;&lt;p&gt;The reasoning here is simple.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It cools by $\frac1{10}$ of the difference between &quot;cool&quot; and &quot;critical&quot; each second, and heats by $\frac16$ of the difference when operating. So the rate of heating with the cooling system is $\frac16 - \frac1{10} = \frac{10 - 6}{60}= \frac1{15}$, and thus one light will last 15 seconds. After the 15 seconds, the light will be at critical, and it won't cool by much without the cooling system, so it will effectively stay at critical, making that light no longer usable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9289" LastActivityDate="2015-06-04T12:41:48.760" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15966" PostTypeId="2" ParentId="15934" CreationDate="2015-06-04T13:18:56.897" Score="3" Body="&lt;p&gt;Here's a possible(yet probably wrong) answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Talking about longitude:&lt;br&gt;&#xA; ---&lt;br&gt;&#xA; 1 minute = 1 nautical mile ~ 1852 meters&lt;br&gt;&#xA; ---&lt;br&gt;&#xA; Earth spins at 465.1 m/s&lt;br&gt;&#xA; So it takes 4 second to move 1 minute longitude.&lt;br&gt;&#xA; So Earth spins 1/4 minutes per second(&lt;em&gt;one fourth&lt;/em&gt; or a &lt;em&gt;quarter&lt;/em&gt;).&lt;br&gt;&#xA; &lt;a href=&quot;http://www.thesaurus.com/browse/one-fourth&quot; rel=&quot;nofollow&quot;&gt;Thesaurus says that the possible synonyms for &lt;em&gt;One fourth&lt;/em&gt;(or &lt;em&gt;quarter&lt;/em&gt;) are&lt;/a&gt;:&lt;br&gt;&#xA; part, term, quad, span&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the possible answer could be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Part&lt;/strong&gt;, &lt;strong&gt;term&lt;/strong&gt;, &lt;strong&gt;quad&lt;/strong&gt; or &lt;strong&gt;span&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9352" LastActivityDate="2015-06-04T13:18:56.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15967" PostTypeId="2" ParentId="10043" CreationDate="2015-06-04T13:21:42.157" Score="0" Body="&lt;p&gt;I solved the 3 door riddle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; is the left door the right door?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;why:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if yes then it is the right go to left door, and no a right door can not be a left door.&lt;br&gt;&#xA; Answer  = yes AND no&lt;br&gt;&#xA;&lt;br&gt;&#xA; if no go to middle door-because a left door can not be a right door.&lt;br&gt;&#xA; Answer = no&lt;br&gt;&#xA;&lt;br&gt;&#xA; if silence the only answers would have to be both yes and no, or he would speak falsely.&lt;br&gt;&#xA; go to right door&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12968" LastEditorUserId="10400" LastEditDate="2015-06-04T15:00:24.730" LastActivityDate="2015-06-04T15:00:24.730" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15968" PostTypeId="2" ParentId="15950" CreationDate="2015-06-04T14:19:26.303" Score="0" Body="&lt;p&gt;The rule you can always apply on a square like this is: shortest even side * 2&lt;br&gt;&#xA;This works for every square/rectangle with atleast one even number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So on a 8*8 square you have to do 8*2 = 16 minimal lines.&lt;br&gt;&#xA;And on a 4*17 rectangle you do 4*2 to get 8 minimal lines.&lt;/p&gt;&#xA;" OwnerUserId="12645" LastActivityDate="2015-06-04T14:19:26.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15970" PostTypeId="2" ParentId="15839" CreationDate="2015-06-04T14:43:49.363" Score="6" Body="&lt;h3&gt;10: Six simple women wander toward various very sharp brown fish.&lt;/h3&gt;&#xA;&lt;p&gt;Fairly obvious. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Sharptail_mola&quot; rel=&quot;noreferrer&quot;&gt;Sharptail Mola&lt;/a&gt; is rare enough and odd enough that it would clearly attract attention from passing random women.&lt;/p&gt;&#xA;&lt;h3&gt;11: Every famous fresh child knew waves work because northern summers heave.&lt;/h3&gt;&#xA;&lt;p&gt;As a new child, for which the world exists solely for my amusement, I did indeed think that waves were only present in the height of the summer.  Of course now I know they wave all the time, but as a child of the north I would only visit the beach during the hot, heavy summer. Correlation != causation.&lt;/p&gt;&#xA;&lt;h3&gt;18: Troglodytic triumphing trichomonal trippingly triptych trolleybusses' twitching twinjets truncheoning trihybrid trenchermen tribespeople tremendously trapshooting transplantable thatchy torchwood toxicants.&lt;/h3&gt;&#xA;&lt;p&gt;Be it observed by all: The toxic transplantable thatchy torchwood plants, an everpresent problem in our fair community, are being tremendously trapshot by our fellow trihybrid tribespeople, formerly occupied in trenching operations.  However they, in turn, are having to fend off the brutal truncheoning that our severely out of date, three color, three panel lifelike painted trolleys are dishing out due to their malfunctioning jet engines.&lt;/p&gt;&#xA;&lt;p&gt;The trolley's are winning, and we may well be doomed to thatchy toxic death.&lt;/p&gt;&#xA;" OwnerUserId="1682" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-06-04T15:34:18.780" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15971" PostTypeId="2" ParentId="15952" CreationDate="2015-06-04T15:19:44.183" Score="4" Body="&lt;p&gt;I think your spotlight analogy does not match up very well to the World of Warcraft scenario, for two reasons:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) It seems electrically hazardous to change a fuse on a light that's still &quot;hot&quot;&lt;br&gt;&#xA;2) If I knew I was going to blow a fuse after 6 seconds of use, wouldn't it make sense to shut off the light after 5 seconds and use a different light?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyhow, as a retired WoW player, thinking of the scenario in that context makes more sense to me. The WoW solution depends on whether you can &quot;pre-drop&quot; feathers before starting. If yes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T = -30: drop feather A&lt;br&gt;&#xA;T = -20: drop feather B&lt;br&gt;&#xA;T = -10: drop feather C&lt;br&gt;&#xA;T = 0: consume feather A, drop feather D  (2 charges left)&lt;br&gt;&#xA;T = 6: consume feather B, drop feather E (1.6 charges left)&lt;br&gt;&#xA;T = 12: consume feather C, drop feather F (1.2 charges left)&lt;br&gt;&#xA;T = 18: consume feather D, drop feather G (0.8 charges left)&lt;br&gt;&#xA;T = 24: consume feather E, drop feather H (0.4 charges left)&lt;br&gt;&#xA;T = 30: consume feather F, drop feather I (0 charges left)&lt;br&gt;&#xA;T = 36: consume feather G (0.6 charges left)&lt;br&gt;&#xA;T = 42: consume feather H, drop feather J (0.2 charges left)&lt;br&gt;&#xA;T = 48: consume feather I (0.8 charges left)&lt;br&gt;&#xA;T = 54: consume feather J, drop feather K (0.4 charges left)&lt;br&gt;&#xA;T = 60: consume feather K, drop feather L (0 charges left)&lt;br&gt;&#xA;T = 66: consume feather L (0.6 charges left)&lt;br&gt;&#xA;T = 70: drop feather M (0 charges left)&lt;br&gt;&#xA;T = 72: consume feather M (0.2 charges left)&lt;br&gt;&#xA;T = 78: timer of feather M expires and we are still 2 seconds before we can drop another feather. Done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're not allowed to drop feathers before starting the timer, then it works as follows:&lt;br&gt;&#xA;T = 0: drop and consume feather A (2 charges left)&lt;br&gt;&#xA;T = 6: drop and consume feather B (1.6 charges left)&lt;br&gt;&#xA;T = 12: drop and consume feather C (1.2 charges left)&lt;br&gt;&#xA;T = 18: drop and consume feather D (0.8 charges left)&lt;br&gt;&#xA;T = 24: drop and consume feather E (0.4 charges left)&lt;br&gt;&#xA;T = 30: drop and consume feather F (0 charges left)&lt;br&gt;&#xA;T = 36: feather F runs out and we still have 4 seconds before we can drop again. Done.&lt;/p&gt;&#xA;" OwnerUserId="12803" LastActivityDate="2015-06-04T15:19:44.183" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15972" PostTypeId="2" ParentId="15953" CreationDate="2015-06-04T15:26:38.923" Score="1" Body="&lt;p&gt;I choose &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;C&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The arrow goes up half, down one, up one &amp;amp; a half, down two. Therefore the next move would be up two &amp;amp; a half. (Up = Clockwise, down = anti-clockwise, one = a whole side from corner to corner)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The circle moves one space clockwise.&lt;/p&gt;&#xA;" OwnerUserId="12972" LastEditorUserId="10281" LastEditDate="2015-06-04T16:31:52.537" LastActivityDate="2015-06-04T16:31:52.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15973" PostTypeId="2" ParentId="15943" CreationDate="2015-06-04T15:38:55.553" Score="6" Body="&lt;p&gt;My first thought:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Tomato (or its seeds, depending on which bit the &quot;I&quot; refers to)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The green house&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Is a standard gardener's glass green house&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Inside the green house&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is a white refrigerator&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Inside the white house&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is a freshly-plucked tomato&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Inside the red house&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Are lots of tomato seeds (babies)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12973" LastActivityDate="2015-06-04T15:38:55.553" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15974" PostTypeId="2" ParentId="15948" CreationDate="2015-06-04T15:50:34.550" Score="0" Body="&lt;p&gt;The answer is C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like the pattern is (1-4) (2-5) (3-?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each arrow in the same position in 1 and 4 are pointing in the opposite direction, as is the arrows in 2 and 5. On that basis image number 6 would be the same as 3 with the arrows pointing in the opposite direction. The only image that does this is C.&lt;/p&gt;&#xA;" OwnerUserId="12972" LastActivityDate="2015-06-04T15:50:34.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15975" PostTypeId="2" ParentId="15938" CreationDate="2015-06-04T15:58:30.510" Score="-1" Body="&lt;p&gt;Since you only need one person to answer, and one person to answer correctly, elect one person as the guesser. When the hats are placed, have everyone at the table, with the exception of the guesser, look a certain direction based on hat color of the guesser (left for black, right for white). This tells the guesser what color their hat is without anyone else having to speak. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The great thing is, this is expandable for number of players. As long as you have 2+ players, you can follow this strategy. &lt;/p&gt;&#xA;" OwnerUserId="12242" LastActivityDate="2015-06-04T15:58:30.510" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15976" PostTypeId="1" AcceptedAnswerId="15983" CreationDate="2015-06-04T17:12:26.037" Score="7" ViewCount="949" Body="&lt;p&gt;This puzzle was inspired by &lt;a href=&quot;https://puzzling.stackexchange.com/questions/13257/me-and-my-gang&quot;&gt;Me and My Gang&lt;/a&gt;. I really enjoyed that puzzle, and hope that this one is as challenging and fun. I think it might be a bit easier than &lt;a href=&quot;https://puzzling.stackexchange.com/questions/13257/me-and-my-gang&quot;&gt;Me and My Gang&lt;/a&gt;, but we'll see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Before any of my friends came, I was used a lot.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A friend came to visit me, allowing us to go to all sorts of places.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A third joined us, and the world truly enjoyed us regularly.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The fourth addition to our group channeled us to cooler places.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Our fifth member allowed us to move faster than any jet,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;but the sixth made us a menace to those around us.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Thankfully, the seventh allowed us to relax and be entertained.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The eighth transformed us into a useful medical device.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who am I and who are my friends? What did we become?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-06-05T11:50:35.423" Title="Me and My Gang, Again" Tags="|riddle|word|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15977" PostTypeId="1" AcceptedAnswerId="16223" CreationDate="2015-06-04T17:42:28.880" Score="5" ViewCount="639" Body="&lt;p&gt;It starts off as a creation of God,&lt;br&gt;&#xA;followed by a godly creation.&lt;br&gt;&#xA;Having them together is odd,&lt;br&gt;&#xA;but between them is a one way affection.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A disgusting symbol of disparity&lt;br&gt;&#xA;between wealth and poverty.&lt;br&gt;&#xA;Yet it is an irony&lt;br&gt;&#xA;as a proof of gender equality.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is a term with 2 words  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #2:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Half of the population will mostly agree on &quot;&lt;strong&gt;a godly creation&lt;/strong&gt;. &quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It also make sense to replace 'Gender equality' with 'Gender Inequality'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="10037" LastEditDate="2015-06-08T15:48:24.443" LastActivityDate="2015-06-10T22:03:42.760" Title="An ungodly creation - What am I?" Tags="|riddle|" AnswerCount="6" CommentCount="1" ClosedDate="2015-06-19T01:29:59.737" ContentLicense="CC BY-SA 3.0" />
  <row Id="15978" PostTypeId="2" ParentId="15977" CreationDate="2015-06-04T17:55:29.490" Score="1" Body="&lt;p&gt;Is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Garden of Eden?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It starts off as a creation of God,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The garden of Eden is a creation of God&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;followed by a godly creation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; God is said to have created the human&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Having them together is odd,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The humans are not perfect but the garden is (or the garden + adam and eve = 3, which is an odd number)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but between them is a one way affection.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure about this one&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A disgusting symbol of disparity&#xA;between wealth and poverty.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Garden of Eden = wealth . The rest is poverty&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet it is an irony&#xA;as a proof of gender equality.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eve ate the apple but both Adam and Eve were excluded from the garden.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5359" LastEditorUserId="463" LastEditDate="2015-06-04T18:37:36.480" LastActivityDate="2015-06-04T18:37:36.480" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15979" PostTypeId="1" AcceptedAnswerId="15981" CreationDate="2015-06-04T18:04:56.410" Score="18" ViewCount="997" Body="&lt;p&gt;Isaac is in need of money to buy himself a car so he can go to the beach with his friends, so he's decided to start a lawn mowing business. When mowing lawns, he needs to make sure he gets an even tan so when he ends up going to the beach he looks good. We will consider Isaac to have four sides (front, back, left, and right) and each time he mows a new square meter he gets a &lt;em&gt;point&lt;/em&gt; worth of tanning for that side. In the end, he needs to have an equal number of points for each side. Another way to look at this is that he needs to be traveling each direction an equal number of times. Here's an example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RzsJu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that up is North: if Isaac goes from &lt;em&gt;A&lt;/em&gt; to &lt;em&gt;B&lt;/em&gt;, he gets one point of tanning for traveling East. If he moves back to &lt;em&gt;A&lt;/em&gt;, then he gets a point of tanning for traveling West. This would not give him an even tan because he didn't move North or South, so he would have to move up/down on the grid to get an even tan. Moving from &lt;em&gt;A&lt;/em&gt; to &lt;em&gt;B&lt;/em&gt; would also mow each of those, leaving the bottom two squares to be mowed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isaac's first customer is Mr. Banks. Here's an overhead view of his yard:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gTAdh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The blue space is the garage where Isaac starts mowing the lawn. He wants to mow the entire lawn and end up back at the garage, while getting an even tan and also using the least amount of gas. He can mow over the same spot twice, but it uses gas both times. Each square he mows over uses 1 unit of gas. Will he be able to get an even tan? If not, why? If he can, what route minimizes the gas used?&lt;/p&gt;&#xA;" OwnerUserId="11727" LastEditorUserId="11727" LastEditDate="2015-06-04T19:18:01.573" LastActivityDate="2015-06-06T06:06:50.247" Title="Can Isaac get an even tan?" Tags="|mazes|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="15981" PostTypeId="2" ParentId="15979" CreationDate="2015-06-04T18:51:44.270" Score="16" Body="&lt;p&gt;My &lt;strong&gt;best&lt;/strong&gt; solution is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Follow the red path, then go back and forth along the violet line. It takes only $68$ points!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Picture&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/d5ZYX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Basic calculations:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Up and down consume 17 points each, while left and right only 15 (still on the red path). Finally, you go back and forth along the violet line to reach 17. &#xA;Total is $17\times4=68$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is it possible to improve?&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sadly you can't! The lower bound is $16\times4=64$ moves, achieved when you consume $16$ points for each direction. Though, there's no Hamiltonian cycle that satisfies this condition, it was proved by @Joel Rondeau &lt;a href=&quot;https://puzzling.stackexchange.com/questions/8511/the-erasmus-rook-tour&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.923" LastActivityDate="2015-06-04T19:41:51.570" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15982" PostTypeId="2" ParentId="15976" CreationDate="2015-06-04T18:56:06.850" Score="2" Body="&lt;p&gt;I hate these puzzles ...    I can't get them out of my mind and they take so long before they are solved... -_-    so annoying :P&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my try. have fun googling what they are if you don't know. Because I'm evil too. (mwahahahahaha)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Before any of my friends came, I was used a lot.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A friend came to visit me, allowing us to go to all sorts of places.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ra/ad?  :: or EA (company)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A third joined us, and the world truly enjoyed us regularly.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rad  or Ale&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;The fourth addition to our group channeled us to cooler places.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; arid/radz or real/earl/zale????????&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Our fifth member allowed us to move faster than any jet,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Z-RAID or Lazer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;but the sixth made us a menace to those around us.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Lizard or Azriel&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Thankfully, the seventh allowed us to relax and be entertained.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Zeliard&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;The eighth transformed us into a useful medical device.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Dialyzer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastEditorUserId="11251" LastEditDate="2015-06-04T19:22:53.087" LastActivityDate="2015-06-04T19:22:53.087" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15983" PostTypeId="2" ParentId="15976" CreationDate="2015-06-04T19:07:26.880" Score="13" Body="&lt;p&gt;Got it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Before any of my friends came, I was used a lot.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A (the most common single-letter word in English).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A friend came to visit me, allowing us to go to all sorts of places.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; AT (you can be &lt;em&gt;at&lt;/em&gt; all sorts of places) - thanks pacoverflow!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A third joined us, and the world truly enjoyed us regularly.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TEA (many people all over the world enjoy tea regularly) - thanks again pacoverflow!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The fourth addition to our group channeled us to cooler places.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HEAT (seems a bit counter-intuitive, but heat does move towards where it's cooler).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Our fifth member allowed us to move faster than any jet,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; EARTH (the speed of the Earth's rotation is probably faster than a jet).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;but the sixth made us a menace to those around us.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; THREAT.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Thankfully, the seventh allowed us to relax and be entertained.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; THEATRE / THEATER.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The eighth transformed us into a useful medical device.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CATHETER.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="463" LastEditDate="2015-06-05T11:50:35.423" LastActivityDate="2015-06-05T11:50:35.423" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="15984" PostTypeId="2" ParentId="15938" CreationDate="2015-06-04T19:24:11.337" Score="0" Body="&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;  This answer is for a slightly different question.  Oops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume everyone can count seconds accurately and consistently.  Everyone agrees to speak only on-the-dot on the seconds.  At t = 0, everyone opens their eyes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Act as if a seer at that moment announced &quot;I see at least one person with a black hat.&quot;  Using the same logic as for the island of blue/brown-eyed people, then at second n (where n is the number of black hats), all the people wearing black hats will correctly guess that they are wearing black hats.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only fails if everyone is wearing white hats, which has probability $2^{-15}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Details, in case you are not familiar with the other riddle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is exactly one person wearing a black hat, he will see only white hats and know the seer must refer to him.  At t = 1, he will speak.  If there are two people with black hats, they will each see another person with a black hat.  When that person doesn't speak at t = 1, they will each realize they must also have a black hat, and so both will speak at t = 2.  Etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works for any number of hats, and fails only if all the hats are white.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="2328" LastEditDate="2015-06-04T19:54:32.080" LastActivityDate="2015-06-04T19:54:32.080" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15985" PostTypeId="2" ParentId="15977" CreationDate="2015-06-04T19:35:03.877" Score="1" Body="&lt;p&gt;It could be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Yin and Yang&lt;/strong&gt;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/gxcyO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It starts off as a creation of God,&#xA;followed by a godly creation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; According to Wikipedia, God created Tai-Chi, which subsequently split in Yin and Yang.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Having them together is odd,&#xA;but between them is a one way affection.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They're opposite, you can't have both at the same time (at least not completely).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A disgusting symbol of disparity&#xA;between wealth and poverty.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Indeed it's a symbol of disparity, and wealth and poverty are opposites!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet it is an irony&#xA;as a proof of gender equality.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Male and female are just two &quot;shades&quot; of Yin and Yang.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: I'm not Taoist and my information could be inaccurate, but this is what I understood reading this &lt;a href=&quot;http://en.wikipedia.org/wiki/Yin_and_yang&quot; rel=&quot;nofollow noreferrer&quot;&gt;topic&lt;/a&gt; in Wikipedia.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-06-04T19:35:03.877" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15986" PostTypeId="1" AcceptedAnswerId="15993" CreationDate="2015-06-04T19:54:51.937" Score="21" ViewCount="1667" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am sought among many for what I can bring,&#xA;  &lt;br&gt; but those who seek me do not like my twin brother.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Both me and my brother can have power over others,&#xA;  &lt;br&gt; but do not worry, as we do not use it unless you are deserving.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;My sister is said to be the star of the family.&#xA;  &lt;br&gt; But I like to think I am above her.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Please explain each line with your answer.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am one who seeks. (I don't like the evil twin!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-06-04T20:55:36.330" LastActivityDate="2015-06-05T05:01:17.260" Title="Sought among many - What am I?" Tags="|riddle|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15987" PostTypeId="2" ParentId="15986" CreationDate="2015-06-04T20:02:18.223" Score="0" Body="&lt;p&gt;My answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; America&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am seeked among many for what I can bring, &#xA;but those people do not like twin brother.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; America is usually considered a place full of opportunities. The twin brothers are North and South America.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Both me and my brother can have power over others, &#xA;but do not worry, as we do not use it unless you are deserving.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most countries of South and North America are part of the ONU, so they have a relative power over the other when certain conditions are present.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My sister is said to be the star of the family. &#xA;But I like to think I am above her.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Star&quot; is a reference to USA's flag, which contains many stars. Also, USA are considered the most advanced country of America. Nothing implies that &quot;north&quot; america is really above the &quot;south&quot;; if you reverse the planisphere it's exactly the opposite!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-06-04T20:03:47.327" LastActivityDate="2015-06-04T20:03:47.327" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15988" PostTypeId="1" AcceptedAnswerId="15992" CreationDate="2015-06-04T20:06:39.893" Score="7" ViewCount="552" Body="&lt;p&gt;The following puzzle was told to me by a friend, Markus Götz, who put it online here: &lt;strong&gt;&lt;a href=&quot;http://www.markus-goetz.de/raetsel/0023_Deviating_Ball_Puzzles.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Deviating Ball Puzzles&lt;/a&gt;&lt;/strong&gt; (pdf). After some searching, I did not find this puzzle anywhere else online. The only similar (but ultimately different) puzzle I found is &lt;a href=&quot;https://www.physicsforums.com/threads/three-pan-balance-15-coins.7251/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Three Pan Balance, 15 Coins&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any other of three-pan balance puzzles online?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The three-pan balance&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a balance with not two, but &lt;em&gt;three&lt;/em&gt; pans. Weightings using the balance follow these rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If there exists a pan that is lighter than each of the other two pans, then this pan goes up and the other two pans go down to a stop. (Note that one cannot see which of the two heavier pans, if any, is the heaviest.)&lt;/li&gt;&#xA;&lt;li&gt;If there is no single lightest pan, then nothing happens. (This includes the case of two equally light pans and one heavier pan.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Let's call this the &quot;lightest-pan-detection-rule&quot; (LPDR).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The problems&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The puzzle consists of several related problems. (The original contains 12 problems, with different kinds of ball deviation. I will split them across several questions.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In each problem you are given &lt;em&gt;n&lt;/em&gt; balls. They are all of the same weight, exept as stated in each problem. You are to identify the deviating ball(s) by using the new balance a maximum number of weighings stated in the puzzle. You are also to present a method to identify the deviating ball(s).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) You are given &lt;em&gt;n&lt;/em&gt; balls, &lt;strong&gt;one&lt;/strong&gt; of which is &lt;strong&gt;lighter&lt;/strong&gt;. What is the largest &lt;em&gt;n&lt;/em&gt;, so that the lighter ball can be identified with &lt;strong&gt;1 weighting&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) You are given &lt;em&gt;n&lt;/em&gt; balls, &lt;strong&gt;one&lt;/strong&gt; of which is &lt;strong&gt;lighter&lt;/strong&gt;. What is the largest &lt;em&gt;n&lt;/em&gt;, so that the lighter ball can be identified with &lt;strong&gt;2 weightings&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) You are given &lt;em&gt;n&lt;/em&gt; balls, &lt;strong&gt;one&lt;/strong&gt; of which is &lt;strong&gt;lighter&lt;/strong&gt;. What is the largest &lt;em&gt;n&lt;/em&gt;, so that the lighter ball can be identified with &lt;strong&gt;k weightings&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Follow-up question: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15989/a-balance-with-three-pans-detecting-the-lightest-pan-find-the-one-heavier-ball&quot;&gt;A balance with three pans, detecting the lightest pan (find the one heavier ball)&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="318" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2015-06-12T06:32:56.740" Title="A balance with three pans, detecting the lightest pan (find the one lighter ball)" Tags="|weighing|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15989" PostTypeId="1" AcceptedAnswerId="15994" CreationDate="2015-06-04T20:31:03.183" Score="5" ViewCount="382" Body="&lt;p&gt;This is a continuation of the question&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/15988/a-balance-with-three-pans&quot;&gt;A balance with three pans, detecting the lightest pan (find the one lighter ball)&lt;/a&gt;. It was told to me by a friend, Markus Götz, who put it online here: &lt;a href=&quot;http://www.markus-goetz.de/raetsel/0023_Deviating_Ball_Puzzles.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Deviating Ball Puzzles&lt;/a&gt; (pdf).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The three-pan balance&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a balance with not two, but &lt;em&gt;three&lt;/em&gt; pans. Weightings using the balance follow these rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If there exists a pan that is lighter than each of the other two pans, then this pan goes up and the other two pans go down to a stop. (Note that one cannot see which of the two heavier pans, if any, is the heaviest.)&lt;/li&gt;&#xA;&lt;li&gt;If there is no single lightest pan, then nothing happens. (This includes the case of two equally light pans and one heavier pan.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Let's call this the &quot;lightest-pan-detection-rule&quot; (LPDR).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are given &lt;em&gt;n&lt;/em&gt; balls, &lt;strong&gt;one&lt;/strong&gt; of which is &lt;strong&gt;heavier&lt;/strong&gt;. What is the largest &lt;em&gt;n&lt;/em&gt;, so that the heavier ball can be identified with at most &lt;strong&gt;k weightings&lt;/strong&gt;? (k &gt;= 1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;normal&quot; balls are all of the same weight. You are to identify the deviating ball by using the balance a maximum number of weighings stated in the puzzle, weighing only the given balls. You are also to present a method to identify the deviating ball.&lt;/p&gt;&#xA;" OwnerUserId="318" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-06-12T06:34:33.003" Title="A balance with three pans, detecting the lightest pan (find the one heavier ball)" Tags="|weighing|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15990" PostTypeId="2" ParentId="1980" CreationDate="2015-06-04T21:29:29.660" Score="1" Body="&lt;p&gt;The answer would be number two, assuming that the prisoners cannot turn around, switch places or talk beforehand. Number three and four are eliminated from guessing because they can only see the wall. That would only leave the numbers one and two to guess. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number one is not the answer because although he/she can see both two and three's hat, the two hats are different. Number two is black and number three is white. So number one would have a 50% chance of getting the correct answer but it also means that he/she has the same chance of getting it wrong. If both numbers two and three were either both black or white, number one would know the colour of his/her hat but numbers two and three have the opposite colours, leaving number one unable to figure out what colour he/she is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leaves number two. Number two is the correct answer because he/she knows that there is a person behind them and in front of them as stated above in the question &quot;They know their placement in this room is as follows.&quot; Number two knows that number three is wearing a white hat. Number two should be able to realize that the number on his head is black because if he/she had a hat that matched number three then number one should have been able to answer what colour he/she had very easily. Number two senses the hesitation of number one and knows that their hat is the opposite of number three, which means two has a black hat.&lt;/p&gt;&#xA;" OwnerUserId="12980" LastEditorUserId="8874" LastEditDate="2015-10-06T12:09:33.680" LastActivityDate="2015-10-06T12:09:33.680" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15992" PostTypeId="2" ParentId="15988" CreationDate="2015-06-04T21:35:20.463" Score="6" Body="&lt;p&gt;1)  Four.  This is clearly the maximum, as there are four possible outcomes from a single weighing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put one ball on each pan, and the fourth ball aside.  If one pan rises, that is the light one.  If none do, the fourth one is lightest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Sixteen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put four balls on each pan and put four aside.  This allows you to identify a group of four that contains the light ball.  Put the others aside, and weigh those four as in case 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) $4^k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put $1/4$ of the balls on each pan, and the other $1/4$ aside.  This identifies one group with the lightest ball.  Iterate through all $k$ weighings until you are left with 1 ball.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2015-06-04T21:35:20.463" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15993" PostTypeId="2" ParentId="15986" CreationDate="2015-06-04T22:13:57.500" Score="28" Body="&lt;p&gt;I think that you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/XoIPF.jpg&quot; alt=&quot;Upvotes&quot;&gt;&lt;br&gt;The upvote button.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Specifically&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The upvote button is what is sought, the downvote button is the twin brother, the favorite button is the star sister.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am sought among many for what I can bring, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People all want to get their reputation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but those who seek me do not like my twin brother.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nobody wants downvotes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Both me and my brother can have power over others, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The SE voting system influences a lot about questions and priority&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but do not worry, as we do not use it unless you are deserving.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People only vote based on what is deserved.  (No troll voting)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My sister is said to be the star of the family.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A literal star, people might also like their questions favorited too&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I like to think I am above her.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Both literally above in the UI, as well as most people caring about rep more than favorites.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12843" LastEditorUserId="5373" LastEditDate="2015-06-04T22:18:29.387" LastActivityDate="2015-06-04T22:18:29.387" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15994" PostTypeId="2" ParentId="15989" CreationDate="2015-06-04T22:25:10.540" Score="7" Body="&lt;p&gt;If all three pans have the same number of balls, at most one will be heavier and the other two the same, so the pans will never move.  However, if there are $x$ balls on two pans and $x+1$ on the third, then the scale works like a normal 2-pan scale, comparing the two pans with $x$ balls.  If the heavier ball is one one of the two pans with $x$ balls, the other one will rise.  If it is on the third pan (or anywhere else), then no pan will rise. This matches the behavior of a normal 2-pan scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have three candidate heavy balls and one or more balls that have been eliminated (I'll call these &quot;ringers&quot;), then you can put one candidate on each pan and also one ringer on the third pan.  If one pan goes up, then the other pan with 1 ball holds the heavy one.  If neither goes up, then the one with the ringer is the heavy one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, with a ringer, you can differentiate up to 3 balls in one weighing.  Similarly, if you put $x$ candidate balls on each pan and add a ringer to one pan, then you can determine which pan contains the heavy ball.  Iterating, you can differentiate $3^k$ balls in $k$ weighings, if you have a ringer to start with.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$k=1$: $n=1$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Without a ringer, you can't differentiate anything in 1 weighing.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$k=2$: $n=7$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Put 2 balls on A and B; put 3 balls on C.  After the first weighing you have at most 3 candidates and at least one ringer.  The second weighing will suffice.  So for k=2, n=7 is the maximum.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;General $k$: $n=3^k-2$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Put $3^{k-1}-1$ balls on A and B, and $3^{k-1}$ on C.  This weighing leaves you with at most $3^{k-1}$ balls and at least one ringer.  Iterate through k weighings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the maximum number of balls that you can differentiate in $k-1$ steps is $3^{k-1}$, pan C cannot have more than that number, and the other two pans cannot have more than $3^{k-1}-1$, so the total cannot be more than $3^k-2$.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="2328" LastEditDate="2015-06-05T19:55:06.517" LastActivityDate="2015-06-05T19:55:06.517" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15995" PostTypeId="1" CreationDate="2015-06-04T23:24:52.677" Score="-7" ViewCount="1305" Body="&lt;p&gt;I am on the train, useless, but without me the train does not work, what am I?&lt;/p&gt;&#xA;" OwnerUserId="12985" LastEditorUserId="5373" LastEditDate="2015-06-04T23:30:36.510" LastActivityDate="2015-06-05T11:14:02.797" Title="I am on the train, useless but without me the train cannot work, what am I?" Tags="|riddle|" AnswerCount="4" CommentCount="7" ClosedDate="2015-06-05T04:07:52.903" ContentLicense="CC BY-SA 3.0" />
  <row Id="15996" PostTypeId="2" ParentId="15995" CreationDate="2015-06-04T23:53:01.673" Score="1" Body="&lt;p&gt;Could it be  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Steam &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Generally useless but needed for certain engines. &lt;/p&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-06-04T23:53:01.673" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15997" PostTypeId="5" CreationDate="2015-06-05T00:55:09.713" Score="0" Body="&lt;p&gt;In these puzzles, a team of players are gathered, and one of several colored hats is placed on each of their heads. Without being able to see their own hat, each player must then guess what color their hat is. Before the game begins, the players may agree on a strategy, but after the hats are placed, no further communication is possible. Often, it is assumed the hats are chosen randomly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOT included in this tag are deduction puzzles where the perfectly logical participants are given information and then are asked to deduce their hat colors, like &lt;a href=&quot;https://puzzling.stackexchange.com/questions/1980/four-prisoners-wearing-black-and-white-hats&quot;&gt;the four prisoners riddle&lt;/a&gt; or any of the blue-eyed islanders type puzzles. The appropriate tag for these is &lt;a href=&quot;/questions/tagged/meta-knowledge&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;meta-knowledge&amp;#39;&quot; rel=&quot;tag&quot;&gt;meta-knowledge&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though this framework sounds very specific, the solving these problems uses a wide variety of mathematical concepts, like modular arithmetic, Hamming codes, expected value of random variables, and permutations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A classic example of such a problem is &lt;a href=&quot;/q/282/&quot;&gt;Hats and Aliens&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;10 humans are abducted by aliens; each represents 10% of the entire human population. The aliens give each abductee either a purple hat or a green hat. The 10 are lined up in a single file line, each facing forward, such that the last person can see the remaining 9's hats, the second to last person can see the remaining 8's hats and so on. No one can see his or her own hat.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The aliens then proceed, starting from the last person, to ask each of the abductees what the color of their hat is. If they guess correctly, they and the 10% of the human population they represent survives; if not, the opposite happens.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Assuming the abductees are given a chance to develop a strategy before they are lined up and questioned: what is the optimal strategy they can utilize (i.e. the one with the highest expected number of survivals)?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;During the questioning, the abductees are not allowed to say anything besides their guess for the color of their hat when it is their turn.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2018-09-06T17:04:57.853" LastActivityDate="2018-09-06T17:04:57.853" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15998" PostTypeId="4" CreationDate="2015-06-05T00:55:09.713" Score="0" Body="Cooperative games where a team of players must try to guess the color of their hat, but can only see their teammates' hats." OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-06-05T00:55:09.713" LastActivityDate="2015-06-05T00:55:09.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15999" PostTypeId="2" ParentId="15995" CreationDate="2015-06-05T02:14:31.567" Score="7" Body="&lt;p&gt;Maybe&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The passengers or merchandise&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are useless to the operation of the train, but if there were no passengers (or merchandise), the train wouldn't &lt;em&gt;need&lt;/em&gt; to work.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5359" LastEditorUserId="463" LastEditDate="2015-06-05T11:14:02.797" LastActivityDate="2015-06-05T11:14:02.797" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16000" PostTypeId="1" AcceptedAnswerId="16003" CreationDate="2015-06-05T02:46:03.570" Score="16" ViewCount="1815" Body="&lt;p&gt;I am looking for strategies for creating a puzzle with a unique solution, and ways of testing for uniqueness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, I'm trying to create cross sums puzzles. How can you create one with a unique solution?&#xA;The images show a small cross sum with two solutions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/E2wPb.jpg&quot; alt=&quot;Solution 1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LyfC9.jpg&quot; alt=&quot;Solution 2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I can find a way of creating a seemingly unique solution, then how can I test for uniqueness? With cross sums the brute force approach would be to try each number permutation, correct? and on a larger puzzle this may take a long time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Okay, thank you for the input so far! @Prem and others. Going off the part (A) that Prem offered I decided to create independent equations. So I wrote some functions to help me out. The functions create a system of independent equations by arbitrarily creating bit strings with the conditions that, the string is 9 in length, there can only be 1-9 in a traditional cross-sums and that numbers do not repeat in a string and that each string is different. Here are the java functions:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;// CLASS PieceMaker&#xA;    // Creates bitstrings length of 9&#xA;    public String[] independentEquations(int count) {&#xA;    String[] equations = new String[count];&#xA;        Arrays.fill(equations, &quot;0&quot;);&#xA;        for(int itr=0;itr&amp;lt;count;itr++){&#xA;            String equation=&quot;&quot;; boolean loop=false;&#xA;            do{ equation=&quot;&quot;; loop=false;&#xA;                for(int itr2=0;itr2&amp;lt;9;itr2++){&#xA;                    Integer temp=generator.nextInt(2);&#xA;                    equation+=temp.toString();&#xA;                }&#xA;                for(int itr3=0;itr3&amp;lt;count;itr3++){if(equations[itr3].equals(equation)){loop=true;};}&#xA;            }while(loop);&#xA;            equations[itr]=equation;&#xA;        }&#xA;        return equations;&#xA;    }&#xA;&#xA;// Makes readable equations&#xA;    public String[] turnIntoSumsString(String[] independentEquations) {&#xA;        String[] temp = new String[independentEquations.length];&#xA;        Arrays.fill(temp, &quot;&quot;);&#xA;            for(int itr=0;itr&amp;lt;independentEquations.length;itr++){&#xA;                for(int itr2=0;itr2&amp;lt;independentEquations[itr2].length();itr2++){&#xA;                    if(independentEquations[itr].substring(itr2, itr2+1).equals(&quot;0&quot;)){temp[itr]+=&quot;&quot;;};&#xA;                    if(independentEquations[itr].substring(itr2, itr2+1).equals(&quot;1&quot;)){temp[itr]+=&quot; + &quot; + (itr2+1);};&#xA;                }&#xA;            }&#xA;            independentEquations=temp;&#xA;        return independentEquations;&#xA;    }&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Called from main:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;// CLASS main&#xA;// A system of 15 independent equations&#xA;    PieceMaker pm = new PieceMaker();&#xA;        String[] independentEquations = pm.independentEquations(15);&#xA;        for(int itr=0;itr&amp;lt;15;itr++){&#xA;            System.out.println(independentEquations[itr]);&#xA;        }&#xA;        independentEquations = pm.turnIntoSumsString(independentEquations);&#xA;        for(int itr=0;itr&amp;lt;15;itr++){&#xA;            System.out.println(independentEquations[itr]);&#xA;        }&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And what's output:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;111111100&#xA;111100111&#xA;000001111&#xA;100001111&#xA;010110001&#xA;111100010&#xA;011011000&#xA;010000101&#xA;101111001&#xA;010011100&#xA;011001010&#xA;100001000&#xA;001010101&#xA;010100001&#xA;000010101&#xA; + 1 + 2 + 3 + 4 + 5 + 6 + 7&#xA; + 1 + 2 + 3 + 4 + 7 + 8 + 9&#xA; + 6 + 7 + 8 + 9&#xA; + 1 + 6 + 7 + 8 + 9&#xA; + 2 + 4 + 5 + 9&#xA; + 1 + 2 + 3 + 4 + 8&#xA; + 2 + 3 + 5 + 6&#xA; + 2 + 7 + 9&#xA; + 1 + 3 + 4 + 5 + 6 + 9&#xA; + 2 + 5 + 6 + 7&#xA; + 2 + 3 + 6 + 8&#xA; + 1 + 6&#xA; + 3 + 5 + 7 + 9&#xA; + 2 + 4 + 9&#xA; + 5 + 7 + 9&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now what remains is an algorithm to place the sums appropriately on a puzzle board. Also, the numbers should be mixed around so that they don't appear in order in the puzzle.&lt;/p&gt;&#xA;" OwnerUserId="12991" LastEditorUserId="15544" LastEditDate="2015-09-29T11:09:53.783" LastActivityDate="2015-09-29T11:09:53.783" Title="Strategies for creating a cross sums puzzle with a unique solution" Tags="|logical-deduction|puzzle-creation|computer-puzzle|" AnswerCount="2" CommentCount="9" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16001" PostTypeId="2" ParentId="15995" CreationDate="2015-06-05T02:57:22.367" Score="4" Body="&lt;p&gt;I'll take a stab at it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A long shot, but&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; gravity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;does seem to fit&lt;/p&gt;&#xA;" OwnerUserId="463" LastActivityDate="2015-06-05T02:57:22.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16002" PostTypeId="2" ParentId="15995" CreationDate="2015-06-05T03:19:09.867" Score="6" Body="&lt;p&gt;I have a bit of a different answer...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hem&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know, not what you expected. Here's why it makes sense:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; On a train of a dress, without a hem, the train is going to disintegrate as it's pulled across the ground, as fraying edges get worse. But the hem doesn't really do anything, it just exists.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9289" LastActivityDate="2015-06-05T03:19:09.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16003" PostTypeId="2" ParentId="16000" CreationDate="2015-06-05T04:00:27.863" Score="6" Body="&lt;p&gt;In your given example, the bottom 3 is obvious. That leaves 4 unknowns, which have equations like:&lt;br&gt;&#xA;&lt;strong&gt;(E1) 7=A+B&lt;br&gt;&#xA;(E2) 9=C+D&lt;br&gt;&#xA;(E3) 7=A+C&lt;br&gt;&#xA;(E4) 12=B+D+3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With 4 equations and 4 unknowns, we should ideally have some unique solution, but we really have only 3 equations, because (E4) is simply (E1)+(E2)-(E3). The 4 equations are not independent. This is the reason we get multiple solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strategy for getting unique solution : (A) Ensure you have N equations for N unknowns. All equations should be independent. (B) If you do not have such equations, then provide some restricting conditions like &quot;all the 10 digits (0-9) appear once&quot;, or &quot;all the unknowns are Different&quot;. With proper selection of this condition, all solutions except one will be eliminated. We will be left with one unique solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For your example, we can add a condition &quot;all the squares have prime numbers&quot; to make it have one unique solution.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="10940" LastEditorUserId="10940" LastEditDate="2015-06-05T06:02:03.760" LastActivityDate="2015-06-05T06:02:03.760" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16004" PostTypeId="2" ParentId="15986" CreationDate="2015-06-05T05:01:17.260" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Proton&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am sought among many for what I can bring, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Electrons will gather around a Nucleus&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but those who seek me do not like my twin brother.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Antiprotons have a -1 charge which repels other negatively charged particles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Both me and my brother can have power over others, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Both have a charge than can attract / repel&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but do not worry, as we do not use it unless you are deserving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unsure..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My sister is said to be the star of the family. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Neutrino Stars consist of Neutrons, the other particle which forms a Nucleus&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I like to think I am above her.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Without a Proton a Neutron will decay&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12993" LastActivityDate="2015-06-05T05:01:17.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16005" PostTypeId="2" ParentId="15887" CreationDate="2015-06-05T05:10:00.203" Score="3" Body="&lt;p&gt;Out of character: I am choosing to interpret that &quot;Bob wins ties&quot; as not using the standard tie breakers.  As in, if both hands are straight flushes, normally the higher card wins, as a course of 'breaking the tie'.  But in this case no high card comparison is made and Bob wins a straight flush / straight flush matchup.  Same with royal flush, straight, and flush.  I don't know why, but in my mind it made sense that a four-of-a-kind, three-of-a-kind, pair, full house, etc still compare high card.  Alice's 4 Aces still beats Bob's 4 Queens as if there never were a tie to begin with.  This might not have been the intention, but custom rules make sense with the following flavor text:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alice and Bob are playing Face-Up Poker.  For the first game,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alice chooses a Royal Flush and Bob chooses a Royal Flush.  They both keep their hands.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to the &lt;a href=&quot;http://www.pokerhands.com/poker_hand_tie_rules.html&quot; rel=&quot;nofollow&quot;&gt;Poker Hand Tie Rules&lt;/a&gt;, they split the pot.  Being Alice and Bob, it takes them 3 more games of exactly repeated cards to learn that neither of them are getting anywhere.  They thus agree that high card or high suit shall determine the winner in the case of a tie.&#xA;Following this new rule and the previous strategy, it takes Bob 5 more games for him to realize he's not getting anywhere, no matter which lower suit ace he chooses.  The rules are thus amended, on threat of Bob playing &quot;never ever again&quot;, whereupon high card and high suit are totally ignored and all ties yield Bob's winning.&#xA;Play thus proceeds, with Alice realizing from past experience she must prevent a Royal Flush:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alice takes 4 aces and a king, to prevent Bob's Royal Flush.  &lt;br/&gt;Bob must settle for a straight flush.&lt;br/&gt;Using one of her aces, Alice is free to take a royal flush, and she wins.&lt;br/&gt;4AK, SF -&gt; RF, SF = Alice&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Next game, Bob also realizes he must &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;thwart Alice's royal flush attempt.&lt;br/&gt;When Alice takes 4 aces and a king, Bob takes a straight of a king, queen, jack, ten and nine, all of different suits.&lt;br/&gt;Alice cannot go &quot;straight&quot; (pun intended) for a royal flush, but she naively thinks her 4 of a kind to be winning and so she keeps it.&lt;br/&gt;Bob beats the four aces with a straight flush.&lt;br/&gt;4AK, KQJ109 -&gt; 4AK, SF = Bob&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Realizing what she did, in the next game Alice&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; takes 4 aces and a king again.&lt;br/&gt;Bob takes king through nine of different suits.&lt;br/&gt;Then Alice goes &quot;straight&quot; for the straight flush with the highest card possible, a queen (even though this was irrelevant under the rules, she likes queens).&lt;br/&gt;Bob also takes a straight flush, and wins the tie without high card comparison.&lt;br/&gt;4AK, KQJ109 -&gt; SF_Q, SF = Bob&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On a losing streak, Alice cannot continue with her past method of preventing a royal flush, so for the next game Alice decides to use Bob's own tactic against him.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alice takes ace through ten, all of different suits!&lt;br/&gt;Bob cannot make a royal flush, so Bob takes a straight flush.&lt;br/&gt;Alice is able to take a royal flush and wins, because Bob, having been unable to obtain a royal flush the first round, is still not magically able to make a royal flush.&lt;br/&gt;AKQJ10, SF -&gt; RF, SF = Alice&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Just when Alice thinks she's finally turned the tables, she&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; takes ace through ten, all of different suits.&lt;br/&gt;Bob cannot make a royal flush, but this time he does the same to Alice, taking a different set of ace through ten, all of different different suits.&lt;br/&gt;Without the possibility of her own royal flush, Alice upgrades her straight to a straight flush using lower cards.&lt;br/&gt;Bob also has a straight flush laid out in front of him somewhere on the table, and he wins the tie without high card comparison.&lt;br/&gt;AKQJ10, AKQJ10 -&gt; SF, SF = Bob&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the end&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;br/&gt;Alice makes several more attempts.&lt;br/&gt;Try as Alice might, she cannot both block a royal flush and beat a straight flush from Bob.&lt;br/&gt;Bob always wins either flat out or by virtue of the special rules established.&lt;br/&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Having determined the answer, the Alice, Bob and Charlie (yes, Charlie) personalities retreat back into the subconscious, allowing Daniel to emerge and write this down.&lt;/p&gt;&#xA;" OwnerUserId="12357" LastEditorUserId="12357" LastEditDate="2015-06-05T15:39:09.040" LastActivityDate="2015-06-05T15:39:09.040" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16006" PostTypeId="1" AcceptedAnswerId="16416" CreationDate="2015-06-05T07:28:05.697" Score="4" ViewCount="1579" Body="&lt;p&gt;This is a continuation of the question&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/15989/a-balance-with-three-pans-detecting-the-lightest-pan-find-the-one-heavier-ball&quot;&gt;A balance with three pans, detecting the lightest pan (find the one heavier ball)&lt;/a&gt;. It was told to me by a friend, Markus Götz, who put it online here: &lt;a href=&quot;http://www.markus-goetz.de/raetsel/0023_Deviating_Ball_Puzzles.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Deviating Ball Puzzles&lt;/a&gt; (pdf).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The three-pan balance&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a balance with not two, but &lt;em&gt;three&lt;/em&gt; pans. Weightings using the balance follow these rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If there exists a pan that is lighter than each of the other two pans, then this pan goes up and the other two pans go down to a stop. (Note that one cannot see which of the two heavier pans, if any, is the heaviest.)&lt;/li&gt;&#xA;&lt;li&gt;If there is no single lightest pan, then nothing happens. (This includes the case of two equally light pans and one heavier pan.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Let's call this the &quot;lightest-pan-detection-rule&quot; (LPDR).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The problem&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are given &lt;em&gt;n&lt;/em&gt; balls, &lt;strong&gt;two&lt;/strong&gt; of which are &lt;strong&gt;heavier&lt;/strong&gt; (not necessarily of the same weight).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) What is the largest &lt;em&gt;n&lt;/em&gt;, so that the heavier balls can be identified with &lt;strong&gt;2 weightings&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) What is the largest &lt;em&gt;n&lt;/em&gt;, so that the heavier balls can be identified with &lt;strong&gt;3 weightings&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The normal (non-heavy) balls are all of the same weight. You are to identify the deviating balls by using the balance a maximum number of weighings stated in the puzzle, weighing only the given balls. You are also to present a method to identify the deviating balls, and explain why the value of &lt;em&gt;n&lt;/em&gt; is the largest.&lt;/p&gt;&#xA;" OwnerUserId="318" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2015-06-15T15:44:45.580" Title="A balance with three pans, detecting the lightest pan (find the two heavier balls)" Tags="|weighing|" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="16007" PostTypeId="1" AcceptedAnswerId="16009" CreationDate="2015-06-05T09:00:16.903" Score="7" ViewCount="664" Body="&lt;p&gt;I was so convinced that the &quot;correct&quot; answers to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15887/face-up-poker-with-alice-and-bob&quot;&gt;this excellent&lt;/a&gt; question were wrong that I took out a deck of cards to prove it. Then I realised that I was changing one of the rules. So here's the alternate version with Bob's rules: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alice and Bob are going to have a poker night. They have invented a variant where there are no secrets, and no randomness, called Face-Up Poker. Today, however, the table is full of work and so they're playing on the floor Here's how it works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A standard 52 card deck is spread face up on the floor.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Alice picks up any 5 cards from the floor. Bob does the same. &lt;/li&gt;&#xA;&lt;li&gt;Alice discards any number of cards, throwing them on the floor. She then&#xA;picks up the same number of cards from the floor. &lt;/li&gt;&#xA;&lt;li&gt;Bob does the same. (He can pick up cards Alice discarded. This is the difference from the other question.)&lt;/li&gt;&#xA;&lt;li&gt;They then compare hands, and the best hand wins, with Bob winning ties.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Which player can force a win? What is their winning strategy?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This puzzle assumes familiarity with the rankings of poker hands, but no other knowledge of the rules of poker. For a refresher these rankings, see &lt;a href=&quot;http://www.pokerhands.com/poker_hand_tie_rules.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this helpful page&lt;/a&gt; (the hands are listed worst to best from top to bottom).&lt;/p&gt;&#xA;" OwnerUserId="12998" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-06-05T17:32:23.427" Title="Alternate Face Up Poker with Alice and Bob (on the floor)" Tags="|game|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16008" PostTypeId="2" ParentId="16007" CreationDate="2015-06-05T09:24:50.720" Score="0" Body="&lt;p&gt;Bob, by selecting a royal flush, and no discarding of cards? He has the highest combination possible, and he wins when Alice has the same combination.&lt;/p&gt;&#xA;" OwnerUserId="3993" LastActivityDate="2015-06-05T09:24:50.720" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16009" PostTypeId="2" ParentId="16007" CreationDate="2015-06-05T09:29:56.317" Score="10" Body="&lt;p&gt;I don't think Alice can win this version.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;All 52 cards are available to Bob, apart from the 5 in Alice's hand.&lt;br&gt;&#xA;&lt;br&gt;&#xA;If she starts with a royal flush, so too will Bob, making it a tie, meaning Bob wins&lt;br&gt;&#xA;&lt;br&gt;&#xA;If she acts in any way to block a royal flush, then it doesn't matter what Bob takes. Since she discards first, Bob always has the option to equal or better her hand.&lt;br&gt;&#xA;&lt;br&gt;&#xA;She either has 4 of a kind and Bob has a straight flush, or they both have a royal flush. Either way, Bob wins  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To summarise enormously:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There are is no five card poker hand you can remove from a deck of 52 cards that can't at least be equalled by another hand chosen from the remainder&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11717" LastEditorUserId="11717" LastEditDate="2015-06-05T14:37:14.840" LastActivityDate="2015-06-05T14:37:14.840" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16017" PostTypeId="2" ParentId="15943" CreationDate="2015-06-05T14:36:22.667" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To differing degrees all of these have a green shell, followed by white and red layers, and being fruits, they all contain seeds.&lt;br&gt;&#xA; Fig: &lt;img src=&quot;https://i.stack.imgur.com/ZeZHZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Papaya: &lt;img src=&quot;https://i.stack.imgur.com/anmGH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA; Guava: &lt;img src=&quot;https://i.stack.imgur.com/xePW4.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA; Guayaba: &lt;img src=&quot;https://i.stack.imgur.com/OQeCp.jpg&quot; alt=&quot;enter image description here&quot;&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2078" LastEditorUserId="2078" LastEditDate="2015-06-05T14:58:30.283" LastActivityDate="2015-06-05T14:58:30.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16022" PostTypeId="1" AcceptedAnswerId="16023" CreationDate="2015-06-05T16:42:26.870" Score="9" ViewCount="398" Body="&lt;p&gt;A friend told me that the key to perfection is contained in this short poem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;I beheld a creature,&lt;br&gt;&#xA;  one bee regularly hovering speedily outside your courtyard, twirling elegantly like mermaids, that ravished me.&lt;br&gt;&#xA;  When three dark-grey wolves appeared, the tiny bee bolted!&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the secret hidden in this idyll?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-06-06T09:05:02.380" Title="Apis, arbiter elegantiae" Tags="|riddle|enigmatic-puzzle|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16023" PostTypeId="2" ParentId="16022" CreationDate="2015-06-05T17:26:07.543" Score="12" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\phi$, the golden ratio. The number of letters in each word (with $0$ represented by a line break) gives you the decimal expansion of $\phi$, namely $1.618033988...$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-06-06T09:05:02.380" LastActivityDate="2015-06-06T09:05:02.380" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16024" PostTypeId="2" ParentId="15977" CreationDate="2015-06-05T17:50:50.563" Score="3" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;man made&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It starts off as a creation of God, followed by a godly creation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Man was created by god, but when men create things it can be godly&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Having them together is odd, but between them is a one way affection.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Men love the things that they make, but the creations can't love back&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A disgusting symbol of disparity between wealth and poverty.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All wealth and poverty derives from what man has made. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet it is an irony as a proof of gender equality. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We say &quot;man&quot;-made to denote things made by both men and women. In this context, it means &quot;human,&quot; not &quot;male.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12914" LastActivityDate="2015-06-05T17:50:50.563" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16025" PostTypeId="1" AcceptedAnswerId="16029" CreationDate="2015-06-05T18:15:19.960" Score="15" ViewCount="1130" Body="&lt;p&gt;A team of 10 people are going to play a cooperative game. While their eyes are closed, a pit boss will place either a red or green hat on each of their heads, chosen by fair coin flip. They then all open their eyes, and will be able to see the hats on everyone's heads except their own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The players will then simultaneously bet a number of dollars that the hat they are wearing is &lt;em&gt;green&lt;/em&gt;. Specifically, on the pit boss's signal, the players will each call out an integer, which may be positive, negative or zero. Calling a negative number is effectively betting that your hat will be red.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total winnings of the players is calculated by adding up the bets of people with green hats, and subtracting the bets of people with red hats. The team of players wins if and only if their total winnings are &lt;strong&gt;more than zero&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before the game begins, the players may agree on a strategy, but once the hats are placed, no communication between the players is possible.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What strategy maximizes the players' chances of winning? Why can't they do better?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-06-06T00:22:52.220" Title="Hat Guessing Game in Vegas" Tags="|probability|hat-guessing|" AnswerCount="3" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16027" PostTypeId="1" AcceptedAnswerId="16035" CreationDate="2015-06-05T19:31:11.107" Score="9" ViewCount="275" Body="&lt;p&gt;Here are a series of clues to two-word phrases. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Anxious Sweets&lt;/li&gt;&#xA;&lt;li&gt;Prize Storage&lt;/li&gt;&#xA;&lt;li&gt;Tightly Packed Weapons&lt;/li&gt;&#xA;&lt;li&gt;Fly Flavor&lt;/li&gt;&#xA;&lt;li&gt;Container Cover&lt;/li&gt;&#xA;&lt;li&gt;Boring Performer&lt;/li&gt;&#xA;&lt;li&gt;Ominous Feeling&lt;/li&gt;&#xA;&lt;li&gt;Dishonest Track&lt;/li&gt;&#xA;&lt;li&gt;Spring Spud&lt;/li&gt;&#xA;&lt;li&gt;Ardent Prima Donna&lt;/li&gt;&#xA;&lt;li&gt;Headlining Tattletales&lt;/li&gt;&#xA;&lt;li&gt;Decorated Decoration&lt;/li&gt;&#xA;&lt;li&gt;Christmas Professional&lt;/li&gt;&#xA;&lt;li&gt;Cheesy Hamilton&lt;/li&gt;&#xA;&lt;li&gt;Prepared Advanced Math&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;All of the &lt;strong&gt;words&lt;/strong&gt; in the answers share a one-word property. What are the answers, and what is the property?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If people like this puzzle enough, I'll do a similar one &lt;em&gt;with a twist&lt;/em&gt;!&lt;/p&gt;&#xA;" OwnerUserId="10861" LastActivityDate="2015-06-05T21:30:14.017" Title="A bit of a retread" Tags="|english|language|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16028" PostTypeId="2" ParentId="16025" CreationDate="2015-06-05T19:36:48.967" Score="8" Body="&lt;p&gt;Surprisingly, the prisoners can win against&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; every hat sequence except one. If this is possible, then it is obviously optimal: since the average dollars won is always 0, the prisoners can't win all the time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is the strategy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you see $n$ red hats, bet $10^{n!}!$ dollars that your hat is green. The point is that this function grows very fast, and anyone wearing green sees 1 more red hat than anyone wearing red. So, if there are any green hats whatsoever, then the total of all bets by people with green hats will be much larger than the total of all bets by people with red hats.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2015-06-05T19:36:48.967" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16029" PostTypeId="2" ParentId="16025" CreationDate="2015-06-05T19:37:04.147" Score="15" Body="&lt;p&gt;The players can win with probability&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $ 1- \frac{1}{2^{10}}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is optimal because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each player's bet has expectation $0$, and therefore, so does the sum of the bets. Therefore, it's not possible to win for every possible setting of hats, so at least of one the $2^{10}$ settings must lose.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A strategy that achieves this is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The players imagine standing in line in a predetermined order so that each player only sees the hats of the players in front. The front player, who sees nothing, bets \$1 that their hat is green. If it is, the rest of the players see so, and bet nothing, which gives a win. If it's not, the second player bets \$2 that their own hat is green. If it is, the rest of the players know that they are net ahead \$1 and do nothing, and if not, the next player bets \$4 that their hat is green, and so on. Each player is able to simulate what will happen ahead in line, so they all know the correct to make at the same time.&#xA;&lt;br&gt;&#xA; In general, the player who sees $k$ players in front bets $2^k$ dollars that their hat is green only if all hats they see are red, and bets nothing otherwise. If player $k$ is the first in line with a green hat, they win $2^k$, which is enough to offset the $2^k-1$ that the players in front lose for also betting that they have green hats, and nobody behind them bets. The players only lose if everyone has red hats, losing $2^{11}-1$ dollars.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-06-06T00:22:52.220" LastActivityDate="2015-06-06T00:22:52.220" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16034" PostTypeId="2" ParentId="16000" CreationDate="2015-06-05T20:58:51.753" Score="4" Body="&lt;p&gt;Writing and sanity-checking this kind of puzzle ends up being a very similar process to solving it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, make sure your ground rules are clear. Most Kakuro puzzles operate on only digits 1-9, excluding zero; and require that each digit in a sum is unique (thus 45 is the max, with one of every digit appearing). If your puzzle doesn't follow these rules, you should be explicit in how you deviate from them. If your puzzle DOES follow these rules, you should also state that, or link to a place that describes the standard rules. Advanced puzzle writers may choose to omit this if they think their audience is clever enough to disambiguate, but tread carefully here. If your puzzle uses standard notation but doesn't follow standard rules, your solver needs to be able to notice this early, or he/she will become frustrated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, start with digit ranges that are highly constrained. For example, a 7 in three boxes must be 1+2+4 in some order; there's no way to include a 3 without duplicating numbers in the sum. A 16 in two boxes must be 7+9 in some order because you can't duplicate the 8. These constrained ranges can force digits of your solution at intersections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: A 17 in five boxes intersects with a 24 in three boxes. The 17 has two possibilities (1+2+3+4+7 or 1+2+3+5+6) while 24 is constrained to be (7+8+9). Only the 7 can be at this intersection. Now that the 7 is fixed, maybe a different crossing will take advantage of the fact that the 17 sum can no longer contain a digit &gt;=5. A chain reaction begins to take place as parts of the puzzle become more and more constrained.&lt;/p&gt;&#xA;" OwnerUserId="12803" LastActivityDate="2015-06-05T20:58:51.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16035" PostTypeId="2" ParentId="16027" CreationDate="2015-06-05T21:04:51.433" Score="14" Body="&lt;p&gt;edit: Updated, my first try was wrong on 3/15 answers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. Anxious Sweets = stressed desserts&lt;br&gt;&#xA; 2. Prize Storage = reward drawer&lt;br&gt;&#xA; 3. Tightly Packed Weapons = snug guns&lt;br&gt;&#xA; 4. Fly Flavor = gnat tang&lt;br&gt;&#xA; 5. Container Cover = pot top&lt;br&gt;&#xA; 6. Boring Performer = drab bard&lt;br&gt;&#xA; 7. Ominous Feeling = doom mood&lt;br&gt;&#xA; 8. Dishonest Track = liar rail&lt;br&gt;&#xA; 9. Spring Spud = may yam&lt;br&gt;&#xA; 10. Ardent Prima Donna = avid diva&lt;br&gt;&#xA; 11. Headlining Tattletales = star rats&lt;br&gt;&#xA; 12. Decorated Decoration = laced decal&lt;br&gt;&#xA; 13. Christmas Professional = Noel Leon&lt;br&gt;&#xA; 14. Cheesy Hamilton = rennet tenner&lt;br&gt;&#xA; 15. Prepared Advanced Math = girt trig &lt;br&gt;&lt;br&gt;&#xA; I'm not too sure about 12 - &quot;laméd medal&quot; (as in gold lamé) and &quot;laced decal&quot; both seem arguably correct, but tenuous enough that you might've intended a different answer... &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the quality that links these words is that they are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; palindromes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2374" LastEditorUserId="2374" LastEditDate="2015-06-05T21:30:14.017" LastActivityDate="2015-06-05T21:30:14.017" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16036" PostTypeId="2" ParentId="16025" CreationDate="2015-06-05T21:58:35.680" Score="-1" Body="&lt;p&gt;You didn't say if the players are allowed to do something between the moment they open their eyes, and the moment they have to place their bid. If, for example, they are allowed to walk around (without talking to each other or communinating in any way!), they can always win.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;At the start, two players stand next to each other&lt;/li&gt;&#xA;&lt;li&gt;The other players, one by one, chose their position like this:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If everybody standing in line has a green hat, stand to the right of them&lt;/li&gt;&#xA;&lt;li&gt;If everybody standing in line has a red hat, stand to the left of them&lt;/li&gt;&#xA;&lt;li&gt;If there are some red hats and some green hats, there must be an uninterrupted line of green hats to the left, and an uninterrupted line of red hats to the right. Chose your place between the two where the color changes.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;At the end, if there is a red hat to your left, bet a large negative number, since your hat is red as well&lt;/li&gt;&#xA;&lt;li&gt;Similarily, if there is a green hat to your right, your hat is green,so bet a large positive number&lt;/li&gt;&#xA;&lt;li&gt;If there are only green hats to your left, and only red hats to your right, you don't know the color of your hat; bet a zero &lt;strike&gt;(if allowed to), or the smallest number you are allowed to bet.&lt;/strike&gt; Just reread the instructions, zero is allowed.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2843" LastActivityDate="2015-06-05T21:58:35.680" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16037" PostTypeId="1" AcceptedAnswerId="16038" CreationDate="2015-06-06T01:42:20.973" Score="37" ViewCount="1600" Body="&lt;p&gt;Denny owns a demolition company and has successfully managed a strong, sharp team of more than 30 specialists for over 20 years. His team works well together and has decades of experience, having completed well over 12,000 demolitions and making a name for themselves. They are definitely on the cutting edge of the demolition field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it's time for Denny's son to start learning to manage his own &#xA;smaller team of 20.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Denny gives his son some general rules to start him off:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;p&gt;Listen to the feedback of your team members and provide them with top medical attention and care when needed.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Always clean up the workplace as soon as possible after each demolition.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Be sensitive to problems within the team- If something has come between 2 members of your team, take immediate action to find the problem and deal with it before the bonds that bind the team have a chance to deteriorate.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Work smart by allowing each worker to use his specialty. In other words, let the 8 on the Plan Team organize and begin each demolition, the 4 on the Dog Team rip through the outer walls and tear apart the infrastructure, and the 8 on the Mole Team smash what remains so that it can be carried away easily.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Lastly, allow your team to work only in private, in the dark, since most people prefer not to see or hear our work.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who are these specialized teams?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-06-06T18:54:13.900" LastActivityDate="2019-07-18T07:39:25.993" Title="Who are these teams?" Tags="|riddle|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16038" PostTypeId="2" ParentId="16037" CreationDate="2015-06-06T01:54:17.120" Score="33" Body="&lt;p&gt;The specialized teams are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; teeth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Denny owns a demolition company and has successfully managed a strong, sharp team of more than 30 specialists for over 20 years.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most adults have 32 teeth.  You could say that teeth demolish food.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;His team works well together and has decades of experience, having completed well over 12,000 demolitions and making a name for themselves. They are definitely on the cutting edge of the demolition field.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Over the course of an adult's life, he will have eaten well over 12,000 meals.  The teeth sure are on the cutting edge, cutting into food with each meal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now it's time for Denny's son to start learning to manage his own smaller team of 20.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Children have 20 baby teeth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Listen to the feedback of your team members and provide them with top medical attention and care when needed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If your teeth hurt, go to the dentist.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Always clean up the workplace as soon as possible after each demolition.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's best to brush your teeth after each meal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Be sensitive to problems within the team- If something has come between 2 members of your team, take immediate action to find the problem and deal with it before the bonds that bind the team have a chance to deteriorate.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you get food stuck in your teeth, use floss to get it out.  If the food remains stuck there, it could result in pain and/or cavities.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Work smart by allowing each worker to use his specialty. In other words, let the 8 Initiators plan and begin each demolition,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A child has 8 incisors at the front.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the 4 Shredders rip through the outer walls and tear apart the infrastructure, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A child has 4 canine teeth, which tear apart food.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the 8 Pulverizers grind and crush what remains so that it can be carried away easily.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A child has 8 molar teeth, which grind food.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lastly, allow your team to work only in private, in the dark, since most people prefer not to see or hear our work.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is polite to eat with your mouth closed.  People don't want to see you eat with your mouth open.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4450" LastEditDate="2015-06-06T02:14:20.917" LastActivityDate="2015-06-06T02:14:20.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16039" PostTypeId="2" ParentId="16037" CreationDate="2015-06-06T01:57:34.610" Score="18" Body="&lt;p&gt;They are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;teeth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Take good care of your teeth. If your teeth hurt, go to the dentist.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Brush your teeth after you eat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If there is food between your teeth, floss to prevent your teeth from rotting.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Initiators are the incisors, shredders are canines, and pulverizers are molars. Total twenty teeth in children.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Don't let other people see you chewing. It is disgusting!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="12109" LastActivityDate="2015-06-06T01:57:34.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16041" PostTypeId="1" AcceptedAnswerId="16042" CreationDate="2015-06-06T05:11:20.880" Score="6" ViewCount="384" Body="&lt;p&gt;Can you tile a 6x6 chessboard with dominoes, without overlaps or gaps, so that the number of dominoes oriented horizontally is equal to the number of dominoes oriented vertically?  Why or why not?&lt;/p&gt;&#xA;" OwnerUserId="2722" LastActivityDate="2015-06-06T05:54:15.347" Title="Tiling a 6x6 board with an equal number of horizontal and vertical dominoes" Tags="|tiling|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16042" PostTypeId="2" ParentId="16041" CreationDate="2015-06-06T05:49:09.163" Score="6" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof/construction:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Paint the board with a horizontal stripe covering every other row, which colors 18 cells. In a tiling as described, each of the 9 vertical dominoes covers one colored cell, and each of the 9 horizontal ones covers an even number of colored cells (0 or 2). But that means an odd number of colored cells are covered, which can't be 18.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-06-06T05:54:15.347" LastActivityDate="2015-06-06T05:54:15.347" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16044" PostTypeId="2" ParentId="15979" CreationDate="2015-06-06T06:06:50.247" Score="8" Body="&lt;p&gt;I feel like the impossibility proof from &lt;a href=&quot;https://puzzling.stackexchange.com/questions/8511/the-erasmus-rook-tour&quot;&gt;the other question&lt;/a&gt; was a bit lacking in that it wasn't totally clear that every possible path could be generated by 2x2 squares.  So here is another impossibility proof for $64$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After drawing some possible closed curves of lengths $64$, you might start thinking of each possible solution as a big square with indentations.   And in fact you can one at a time &quot;pound out&quot; the indentations to get back to the big square.  Here is a picture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rKC66.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each time you &quot;pound out&quot; a part of one of these indentations, you remove a &quot;domino&quot; worth of lawn in the 6x6 interior of the big square, and you lose either two vertical steps or two horizontal steps in the tanning process.  Could there be an equal number of vertical and horizontal steps removed in this process?  Well, no.  This is equivalent to tiling a 6x6 chessboard with an equal number of horizontal and vertical dominoes, &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16041/tiling-a-6x6-board-with-an-equal-number-of-horizontal-and-vertical-dominoes&quot;&gt;which cannot be done&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="2722" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-06-06T06:06:50.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16045" PostTypeId="1" AcceptedAnswerId="16201" CreationDate="2015-06-06T06:26:15.447" Score="13" ViewCount="1666" Body="&lt;p&gt;&lt;em&gt;A short one this time around.&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Jack and his crew reach the top of Oholu's Heart. They look around and see a cave further away. Tired but determined, the crew marches ahead towards the cave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just a bit further Jack sees an old woman sitting by a tree, mumbling something to herself and looking into the far beyond. Jack sat down beside her. For a while neither of them spoke anything. Then she turned her head towards Jack and looked deep into his eyes as if searching for something. Finding what she was looking for, she spoke ...&lt;br&gt;&#xA;&lt;strong&gt;&quot;&lt;em&gt;If your first is left, then just take the rights.&lt;/em&gt;&quot;&lt;/strong&gt;&lt;br&gt;&#xA;Then she turned her head away, back the way she was looking and started mumbling again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jack sat there for a bit longer thinking. He then got up suddenly, smiled and yelled ...&lt;br&gt;&#xA;&lt;strong&gt;&quot;&lt;em&gt;Boggles the mind .. doesn't it fellas.&lt;/em&gt;&quot;&lt;/strong&gt;&lt;br&gt;&#xA;and started walking towards the cave again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They reach the cave entrance, a rather small one, and see that it is blocked by a huge boulder. Jack turned and spoke to his crew ...&lt;br&gt;&#xA;&lt;strong&gt;&lt;em&gt;&quot;My dearest of dearsees, you might be wondering how to go about our business now. But fear you not. Our friend Mr. Spook here, &quot;&lt;/em&gt;&lt;/strong&gt;, pointing to a short and pitifully thin sailor whose name obviously wasn't Mr. Spook, &lt;strong&gt;&lt;em&gt;&quot;will have this massive monstrosity out of our path in a jiffy.&quot;&lt;/em&gt;&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;&quot;Mr. Spook,&quot;&lt;/em&gt;&lt;/strong&gt; now speaking to the sailor directly &lt;strong&gt;&lt;em&gt;&quot;would you be kind enough to kindly move this king sized rock out of our kindly way.&quot;&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sailor now completely perplexed, moved towards the boulder reluctantly. As he approached the boulder, he noticed something etched on the rock.&lt;br&gt;&#xA;&lt;strong&gt;&lt;em&gt;&quot;Captain, there something written on the rock. Some words and such.&quot;&lt;/em&gt;&lt;/strong&gt; he yelled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jack got close to the rock and read ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/e4fed.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jack took off his hat and bowed in respect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He started looking around, trying to find something that might show him a way. A few moments later he saw something poking out from between two smaller rocks. He walked towards the rocks and carefully removed a piece of parchment where it was stuck. Something was written on the parchment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nu3Lg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;776c397fa66d801e9102db8390038920c28a4ba398e6f2d6616a221a613620d4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He took out a book from the inside pocket of his jacket and referred to it. After a few moments he started to smile. He went to the boulder and whispered the magic words. And everyone saw in amazement as part of the ground on one side of the boulder started sinking in, and the boulder slowly rolled to that side opening the entrance to the cave.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What were the magic words.&lt;br&gt;&#xA;And just for fun, Why did he bow down in respect?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="12537" LastEditorUserId="7308" LastEditDate="2015-06-06T12:29:25.080" LastActivityDate="2015-06-10T17:39:40.600" Title="Jack Sparrow and The quest for the missing Compass . Part 2" Tags="|riddle|cipher|wordplay|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="16046" PostTypeId="1" AcceptedAnswerId="16119" CreationDate="2015-06-06T09:10:23.863" Score="4" ViewCount="214" Body="&lt;p&gt;This is a continuation of the questions&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/15988/a-balance-with-three-pans-detecting-the-lightest-pan-find-the-one-lighter-ball&quot;&gt;A balance with three pans, detecting the lightest pan (find the one lighter ball)&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15989/a-balance-with-three-pans-detecting-the-lightest-pan-find-the-one-heavier-ball&quot;&gt;(find the one heavier ball)&lt;/a&gt;. There, I asked for the maximum &lt;em&gt;n&lt;/em&gt; given a bound for &lt;em&gt;k&lt;/em&gt;, now I'll ask for the minimum &lt;em&gt;k&lt;/em&gt; given a value for &lt;em&gt;n&lt;/em&gt;. This question is based on a puzzle told to me by a friend, Markus Götz, who put his version online here: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15989/a-balance-with-three-pans-detecting-the-lightest-pan-find-the-one-heavier-ball&quot;&gt;Deviating Ball Puzzles&lt;/a&gt; (pdf).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The three-pan balance&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a balance with not two, but &lt;em&gt;three&lt;/em&gt; pans. Weighings using the balance follow these rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If there exists a pan that is lighter than each of the other two pans, then this pan goes up and the other two pans go down to a stop. (Note that one cannot see which of the two heavier pans, if any, is the heaviest.)&lt;/li&gt;&#xA;&lt;li&gt;If there is no single lightest pan, then nothing happens. (This includes the case of two equally light pans and one heavier pan.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Let's call this the &quot;lightest-pan-detection-rule&quot; (LPDR).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The problems&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In each problem you are given &lt;em&gt;n&lt;/em&gt; balls. They are all of the same weight, except as stated in each problem. You are to identify the deviating ball by only using the balance, weighing only the given balls. You are also to present a method to identify the deviating ball.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) You are given &lt;em&gt;n&lt;/em&gt; balls, &lt;strong&gt;one&lt;/strong&gt; of which is &lt;strong&gt;lighter&lt;/strong&gt;. What is the &lt;strong&gt;least number of weighings&lt;/strong&gt;, &lt;em&gt;k(n)&lt;/em&gt;, so that the lighter ball can always be identified with at most k weightings?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) You are given &lt;em&gt;n&lt;/em&gt; balls, &lt;strong&gt;one&lt;/strong&gt; of which is &lt;strong&gt;heavier&lt;/strong&gt;. What is the &lt;strong&gt;least number of weighings&lt;/strong&gt;, &lt;em&gt;k(n)&lt;/em&gt;, so that the heavier ball can always be identified with at most k weightings?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note about the relation to the previous questions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &lt;em&gt;k(n)&lt;/em&gt; was nondecreasing, this question would be trivial after answering the questions linked above, where I asked to find &lt;em&gt;n(k)&lt;/em&gt;, the maximum number &lt;em&gt;n&lt;/em&gt; of balls for &lt;em&gt;k&lt;/em&gt; weighings. But how does &lt;em&gt;k(n)&lt;/em&gt; really grow?&#xA;OTOH, after answering this question, the other ones become almost trivial.&lt;/p&gt;&#xA;" OwnerUserId="318" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-06-12T06:36:41.623" Title="A balance with three pans, detecting the lightest pan (find the one lighter/heavier ball, for a given number of balls)" Tags="|weighing|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16047" PostTypeId="1" AcceptedAnswerId="16072" CreationDate="2015-06-06T10:55:07.147" Score="10" ViewCount="1167" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;You are searching for the diamonds in a remote area. After hours, you see a temple with a diamond engraved. You realise that this is the diamond temple. The  temple has the following also engraved:&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;A h&lt;/strong&gt; f &lt;strong&gt;K&lt;/strong&gt; &lt;em&gt;1 C&lt;/em&gt; &lt;strong&gt;B a&lt;/strong&gt; &lt;em&gt;1 F&lt;/em&gt; h &lt;strong&gt;r N&lt;/strong&gt; e&lt;/p&gt;&#xA;&lt;p&gt;Below the code is a keypad with all the letters of the English alphabet.&#xA;You look down at your feet, all you see are human skulls scattered around. This must be a one-shot job.&lt;/p&gt;&#xA;&lt;p&gt;There are no other entrances to this temple.&lt;/p&gt;&#xA;&lt;p&gt;You try to figure out the password needed to enter the temple of diamonds. You do not know how many characters are required for the password.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Here is the first hint:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Forget the strong.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;You ponder on the hint for a while, then you get hungry, so you go and kill an oxen and get some water in a flask. You carry your food and water with you to the temple.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Then you get an idea of what the hint means.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Of course! I am very happy to have learnt &lt;strong&gt;HTML&lt;/strong&gt;!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The temple suddenly ejected a parchment containing the next hint:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Notice the &lt;strong&gt;bold&lt;/strong&gt;, &lt;em&gt;italic&lt;/em&gt; and normal text in the cryptogram.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;You wake up after a long night's sleep. You see if the temple has more hints, but you have no luck. Suddenly you spot two torn pieces of paper.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You cannot read the whole thing because the paper is torn.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Ital&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The other paper reads:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; un-hex&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You try to figure out what these words mean.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;What does &amp;quot;Ital&amp;quot; mean? Maybe Italy? So I've got Italy and &amp;quot;un-hex&amp;quot;, whatever that means.&lt;/p&gt;&#xA;&lt;p&gt;This is probably the paper hint of the previous diamond hunters. I am lucky.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;You look around you. You are surrounded by walls of over 20 metres in height. You have enough food to last for a day. You now have a time limit on solving the puzzle. You think.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;PLEASE WRITE YOUR ANSWERS EVEN THOUGH THEY ARE PARTIAL SO THAT I CAN TELL YOU IF YOU ARE ON THE RIGHT TRACK OR NOT. THANK YOU.&lt;/p&gt;&#xA;" OwnerUserId="12860" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-10-02T05:07:00.830" Title="What is the password to the diamond temple?" Tags="|cipher|password|computer-puzzle|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16049" PostTypeId="1" AcceptedAnswerId="16130" CreationDate="2015-06-06T17:34:08.390" Score="12" ViewCount="3212" Body="&lt;p&gt;You have a bag containing $n$ balls of $n$ different colours. Select 2 balls randomly (in order, so there's a first and a second), repaint the first in the same colour as the second, and replace both balls in the bag. Keep on doing this once per minute until all balls in the bag are the same colour.&lt;br&gt;&#xA;&lt;strong&gt;What is the expected time this will take?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;(&lt;strong&gt;Edit&lt;/strong&gt; to make sure the balls are all of different colours at the start. In the original version, it was $n$ balls of $m$ different colours. Apologies to anyone who started working on the problem in that form.)&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-06-07T14:36:18.833" LastActivityDate="2015-06-09T01:24:30.583" Title="How long until all balls are the same colour?" Tags="|mathematics|probability|combinatorics|" AnswerCount="3" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="16050" PostTypeId="1" AcceptedAnswerId="16052" CreationDate="2015-06-06T18:06:29.433" Score="2" ViewCount="2672" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;You will get it within a few seconds.&lt;br&gt;&#xA;  Until I am measured,&lt;br&gt;&#xA;  I am not known.&lt;br&gt;&#xA;  Yet how you miss me,&lt;br&gt;&#xA;  When I have flown! &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What am I?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="11764" LastEditorUserId="10400" LastEditDate="2015-06-06T18:51:47.740" LastActivityDate="2015-06-11T16:59:09.573" Title="It's Money - Think about it?" Tags="|riddle|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16051" PostTypeId="2" ParentId="16050" CreationDate="2015-06-06T18:12:52.980" Score="1" Body="&lt;p&gt;It's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Money&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I got it quickly because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You said it in the title&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You're not known until your value is measured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I miss you when I&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Spend you, you've gone to fly all across the globe!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2015-06-06T18:12:52.980" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16052" PostTypeId="2" ParentId="16050" CreationDate="2015-06-06T18:17:50.583" Score="7" Body="&lt;p&gt;It is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You will get it within few seconds.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A few seconds is still time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Until I am measured, &#xA;I am not known. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You don't know how much time has passed unless you measure it using a clock.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet how you miss me, &#xA;When I have flown! &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You will miss pleasant times when they pass.&lt;br&gt;&#xA; Also known as: &quot;Time flies when you're having fun.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Oh, and for the title:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Its Money - Think about it?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There's the common phrase &quot;Time is money.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12109" LastEditorUserId="12109" LastEditDate="2015-06-11T16:59:09.573" LastActivityDate="2015-06-11T16:59:09.573" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16053" PostTypeId="1" AcceptedAnswerId="16105" CreationDate="2015-06-06T19:36:38.007" Score="11" ViewCount="491" Body="&lt;p&gt;My sister gave me a message today:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;13.8  4.8   3     10.8  3     11.4&#xA;7.8   15    0     0     0     0&#xA;1.8   9     9     6.6   5.4   3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;She's glaring at me with an &lt;a href=&quot;https://i.stack.imgur.com/uvVQr.jpg&quot; rel=&quot;noreferrer&quot;&gt;evil eye&lt;/a&gt;. (My fault I decided to get her one anyway.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I better figure this out quickly. &lt;strong&gt;What on earth does this say?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; She has sworn that the plaintext is not offensive, just incredibly boring.&lt;/p&gt;&#xA;" OwnerUserId="12109" LastActivityDate="2015-06-08T13:11:06.507" Title="Stop the evil eye from glaring at me!" Tags="|cipher|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="16054" PostTypeId="2" ParentId="16047" CreationDate="2015-06-07T01:23:49.493" Score="4" Body="&lt;p&gt;This'll be a work in progress, so nobody ban me, okay?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, let's start by taking the letters in groups:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bolded capital A, K, B, N&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Italic capital preceded by an one 1F, 1C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bolded h, a, r&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normal f, e, h&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You hit the 2 key and then hit enter. Nothing happens. You guess that it just literally lets time kill you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Come to think of it, it is kind of addicting... No wonder everyone's died...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now's not the time to dwell on the past! You have a puzzle before you!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You throw aside the bolded letters. This leaves you with 1F, 1C, f, e, h.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's putting &lt;em&gt;EMphasis&lt;/em&gt; on the numbers, that's for sure...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You puzzle over it for a bit. You come to the conclusion that there's a little hexidecimal in this puzzle. Perhaps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You hit the 2 key again, and circle around the temple. Turns out there was a little monitor-like thing on the back showing what you are putting in... You shrug and walk back to press the enter key to clear the string. You write down 2831 and fhe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fhe sounds like the Greek letter phi. You know that the sum of the numbers is 14, the sum of the hexadecimals is 59, and that there is only one C in the puzzle... RO13 states that f=t, h=u, and e=r... wait. H stands for header? You don't really know if that's right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, if h stands for header, then f stands for footer... You can't think of any HTML page block that begins with a e, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are Fixing a Hole as the rain comes in, which stops your mind from wandering...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Current Results: 1F, 1C, 2831, 2, 8, 3, 1, f, h, e, phi, fhe, header, footer, t, u, r.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You wonder if the oxen in the story meant anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You fix the hole and start to think. You took the italic numbers and converted them from hexadecimal... twenty-eight thirty-one. F twenty-eight thirty-one h e. F 2 8 3 1 H E.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You try the codes F B H C A H E and F B E H E. Behafe? Be safe? Behave? Breathe? Bechathe? No, that last one doesn't seem right. You then punch in BE SAFE, BEHAVE, and BREATHE. Phoebe? You also try PHOEBE. FHEBE. You can't think of any anagrams with these letters. BEHEF. BELIEF. BHEEF. BEEF. HEBEF. CAFEBH. CALIF. You have no idea there that last one came from. You think of trying SWORDFISH, but you realize that would just be dumb. EHEBF? FAFACFE? BEREAVE? (You try every combination in all capital letters.)&lt;/p&gt;&#xA;" OwnerUserId="12506" LastEditorUserId="12506" LastEditDate="2015-06-07T15:54:06.580" LastActivityDate="2015-06-07T15:54:06.580" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16055" PostTypeId="1" AcceptedAnswerId="16059" CreationDate="2015-06-07T08:14:51.270" Score="56" ViewCount="7018" Body="&lt;p&gt;The following is a probability paradox I've been thinking about.  It involves &lt;a href=&quot;http://en.wikipedia.org/wiki/Bayes%27_theorem&quot;&gt;Bayes' rule&lt;/a&gt;;  if you're not familiar, a good starting example is a urn that has a 50% chance of containing one black ball and one white ball, and a 50% chance of having two black balls.  If you reach in at random and pull out a black ball, it becomes more likely there were two black balls to begin with.  Specifically, there is now a 1/3 chance of there being black and white balls in the urn, and a 2/3 chance of there being two black balls.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The Setup&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Alice and her husband Bob are kidnapped by an eccentric millionaire that performs probability experiments on people.  Without them seeing the result, a coin is to be flipped.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If heads, one of the two of them (at random) will be brought to an office with a big red button that, if pushed, will transfer \$5000 to their joint bank account.  The other will be left in the holding cell.&lt;/li&gt;&#xA;&lt;li&gt;If tails, both them them will be brought to separate offices with big red buttons, each with which removes \$2000 from their joint bank account.  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;They can discuss strategy ahead of time, but after the experiment starts they will be kept separate from each other.  What should they do?&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The Plan&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Alice and Bob reason their only two strategies are to push the button or not.  Not pushing the button is a net zero, and pushing the button yields an expected payoff of $1/2 \cdot (5000) + 1/2 \cdot(-4000) = 500$.  Not being particularly risk averse, they decide to go ahead and press the button.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The Paradox&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The experiment starts, and Alice is summoned to an office with a big red button.  She is about to confidently press the button in accordance with their strategy, when she suddenly has second thoughts.  The fact that she has been brought to an office and not left in the holding cell gives her new information.  Using standard Bayes analysis, the probability the coin was heads is now $1/3$, and the probability the coin was tails is $2/3$.  Thus, now the expected value of their strategy is $1/3 \cdot ( 5000) + 2/3 \cdot (-4000) = -1000$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suddenly, not pushing the button seems like a good strategy.  What's going on?  What should Alice do, which analysis was flawed, and what is wrong with the flawed analysis?  If it matters to your answer, assume Bob thinks in a very similar way to Alice, and would likely make the same decision as she would.&lt;/p&gt;&#xA;" OwnerUserId="2722" LastActivityDate="2021-02-10T23:31:40.523" Title="Eccentric Millionaire Probability Paradox" Tags="|probability|paradox|" AnswerCount="11" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16056" PostTypeId="2" ParentId="16055" CreationDate="2015-06-07T09:30:00.510" Score="0" Body="&lt;p&gt;I would advise Alice to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Push the button!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Consider the probabilty where Bob will be. There is a 1/2 probability that he is in the room with the red button(R) and another 1/2 chance he is in the holding room(H). If she considers these values into her equation, there is no paradox at all and the net profit will still be $500!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The reason for the flaw is that she only thought of herself. Because this mad millionaire is testing on BOTH of them, she will need to consider Bob too to calculate the odds. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I hope this solves your question.&lt;/p&gt;&#xA;" OwnerUserId="12860" LastActivityDate="2015-06-07T09:30:00.510" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16057" PostTypeId="2" ParentId="16047" CreationDate="2015-06-07T09:43:22.180" Score="4" Body="&lt;p&gt;Let's see...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Forget the &lt;strong&gt;strong&lt;/strong&gt;: f1C1Fhe&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And finally...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Convert the italicized hexadecimal numbers to standard: 1c = 28, 1f = 31&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which leaves my answer...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; f2831he&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Revised answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Convert the remaining letters back to hexadecimal: 6628316865&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9670" LastEditorUserId="9670" LastEditDate="2015-06-07T09:51:04.690" LastActivityDate="2015-06-07T09:51:04.690" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16058" PostTypeId="1" AcceptedAnswerId="16065" CreationDate="2015-06-07T09:43:26.903" Score="5" ViewCount="251" Body="&lt;p&gt;This is a very short and simple decoding computer-puzzle.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Where am I?&lt;/p&gt;&#xA;&lt;p&gt;I am in a known location called 370031.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Where is this mysterious person?&lt;/p&gt;&#xA;&lt;p&gt;Clue:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Most, or half of the people looking at this post, are in 370031!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12860" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-06-07T13:53:16.103" Title="Where am I? -Decoding Puzzle" Tags="|computer-puzzle|" AnswerCount="2" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="16059" PostTypeId="2" ParentId="16055" CreationDate="2015-06-07T09:47:28.917" Score="37" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; First of all, Alice is correct in that the expected value of the strategy (from her point of view) is now -1000.  It would have been best if she had not been summoned to an office, in which case the couple would have won 5000.  Since she has been summoned to an office, that possibility is ruled out.  Let's look at the various possibilities.  I will shorten the cases to A, B, and AB, where A means Alice was picked alone, B means Bob was picked alone, and AB means both were picked.  I will use the term equity to mean the expected value of a case multiplied by the probability of that case.&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; A summary of the original possibilities:&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; A : +5000 (25% chance) (+1250 equity)&lt;br /&gt;&#xA; B : +5000 (25% chance) (+1250 equity)&lt;br /&gt;&#xA; AB: -4000 (50% chance) (-2000 equity)&lt;br /&gt;&#xA; Total equity: +500 (as stated in &amp;quot;The Plan&amp;quot;)&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; A summary of the remaining possibilities once Alice is selected:&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; A : +5000 (33% chance) (+1667 equity)&lt;br /&gt;&#xA; AB: -4000 (67% chance) (-2667 equity)&lt;br /&gt;&#xA; Total equity: -1000 (as stated in &amp;quot;The Paradox&amp;quot;)&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; This all makes sense so far.  There was a 75% chance of Alice being selected, with expected value -1000, and a 25% chance of Alice not being selected, with expected value +5000.&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; -1000 * 0.75 + 5000 * 0.25 = 500 equity, which is the original equity.&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; So far so good.  But Alice should still press the button.  Why?  Because her equity from pressing the button is positive:&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; A: +5000 (33% chance) (+1667 equity)&lt;br /&gt;&#xA; AB: -2000 (67% chance) (-1333 equity)&lt;br /&gt;&#xA; Total equity: +333&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; Note that in the AB case, her pressing the button only loses 2000 dollars and not 4000.  Bob pressing his button and losing the other 2000 dollars is &amp;quot;a done deal&amp;quot; which results in -1333 equity, but that doesn't affect whether Alice should press her button or not.&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; In other words, Bob is going to press the button and lose 1333 equity.  If Alice presses the button, she will regain 333 equity to make the final result -1000.  If she does not press the button, the result will remain -1333 which is worse.&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; Edit: A commenter said, why doesn't Alice just &lt;strong&gt;not&lt;/strong&gt; press the button and end up with 0 instead of -1000?&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; Response: Because it's always correct to press the button.  Suppose Bob decided to &amp;quot;wussy out&amp;quot; and not press the button.  That means that instead of Bob losing 1333 equity, he will have lost 0 equity.  Alice should still press the button and gain 333 equity.  Her decision is independent of Bob's decision.  Also, remember that the -1000 equity doesn't mean that pressing the button is going to lose money.  You always have to factor in the +5000 in case B that makes the final equity +500.  The decision is between pressing the button and making +500 &lt;strong&gt;overall considering all 4 cases&lt;/strong&gt; and not pressing the button and making 0.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h1&gt;Another way to think about it&lt;/h1&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Here's another way to think about it that might clear up the &amp;quot;paradox&amp;quot; for some people.  Suppose Alice and Bob were strangers instead of a couple.  And the rules were that if you press the button in heads case, &lt;strong&gt;you alone&lt;/strong&gt; get 5000 dollars.  And in the tails case, anyone who presses the button loses 2000 dollars &lt;strong&gt;from their own bank account&lt;/strong&gt;.  Now it is clear that you actually &lt;strong&gt;want&lt;/strong&gt; to be picked.  Because if you are, your expected return from pressing the button is (1/3) * 5000 - (2/3) * 2000 = +333 dollars.  It doesn't matter that the other person might lose 2000 dollars.  Their best strategy overall is also to press the button.  It just happens that if you get picked, you know they will be losing 2000 dollars by pressing the button.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4479" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-06-07T10:39:12.547" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16060" PostTypeId="2" ParentId="16055" CreationDate="2015-06-07T09:57:39.383" Score="0" Body="&lt;p&gt;Alice should&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; not push the button.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let $O$ and $C$ denote office and holding cell, $A$ and $B$ Alice and Bob, and $H$ and $T$ heads and tails. Then we have conditional probabilities $p(A\in O|H)=\frac{1}{2},p(A\in O|T)=1$, and hence $p(A\in O)=\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.1=\frac{3}{4}$, so by Bayes's theorem $p(H|A\in O)=(\frac{1}{2}.\frac{1}{2}) / \frac{3}{4}=\frac{1}{3}$. I.e. given that Alice is in the office, there's a $\frac{1}{3}$ probability that the coin came up heads (i.e. that Bob is still in the cell) and a $\frac{2}{3}$ probability that it came up tails (i.e. that Bob is in front of another button).&lt;br&gt;&lt;br&gt;We're told that Alice and Bob both think in the same way, so &lt;strong&gt;either&lt;/strong&gt; Alice presses her button and if Bob is in front of a button, he'll press it too &lt;strong&gt;or&lt;/strong&gt; Alice doesn't press her button and if Bob is in front of a button, he won't press it either. In the first case, they make an expected gain of $\frac{1}{3}(5000)+\frac{2}{3}(-4000)=-1000$; in the second, they make an expected gain of $0$, which is better.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Flaw in the other argument:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Alice and Bob reason their only two strategies are to push the button or not&quot; - but they don't realise that the fact of one of them having a button to push gives them new information! They're &lt;strong&gt;seeing themselves as a single entity&lt;/strong&gt;, not taking into account that they'll be separated and have to consider probabilities of where the other one is based only on what they know about themselves.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-06-07T09:57:39.383" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16061" PostTypeId="2" ParentId="16058" CreationDate="2015-06-07T10:01:30.170" Score="0" Body="&lt;p&gt;Looks like&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you should turn 370031 upside down&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;to get&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; IEOOLE.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I suspect the number should actually be 370013, so that the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BOOLE,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which sounds more like a place or at least &lt;a href=&quot;http://en.wikipedia.org/wiki/George_Boole&quot; rel=&quot;nofollow&quot;&gt;a person&lt;/a&gt;!&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-06-07T10:01:30.170" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16062" PostTypeId="1" AcceptedAnswerId="16063" CreationDate="2015-06-07T11:04:33.060" Score="12" ViewCount="3254" Body="&lt;p&gt;&lt;strong&gt;Q1.&lt;/strong&gt; Which is the first question where &lt;strong&gt;c)&lt;/strong&gt; is the correct answer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Q3&lt;br&gt;&#xA;b) Q4&lt;br&gt;&#xA;c) Q1&lt;br&gt;&#xA;d) Q2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Q2.&lt;/strong&gt; Which is the first question where &lt;strong&gt;a)&lt;/strong&gt; is the correct answer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Q4&lt;br&gt;&#xA;b) Q2&lt;br&gt;&#xA;c) Q3&lt;br&gt;&#xA;d) Q1  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Q3.&lt;/strong&gt; Which is the first question where &lt;strong&gt;d)&lt;/strong&gt; is the correct answer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Q1&lt;br&gt;&#xA;b) Q2&lt;br&gt;&#xA;c) Q4&lt;br&gt;&#xA;d) Q3  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Q4.&lt;/strong&gt; Which is the first question where &lt;strong&gt;b)&lt;/strong&gt; is the correct answer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Q2&lt;br&gt;&#xA;b) Q4&lt;br&gt;&#xA;c) Q3&lt;br&gt;&#xA;d) Q1  &lt;/p&gt;&#xA;" OwnerUserId="12860" LastEditorUserId="5373" LastEditDate="2015-06-07T17:55:47.853" LastActivityDate="2015-06-07T17:55:47.853" Title="What is the answer to the following questions?" Tags="|logical-deduction|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16063" PostTypeId="2" ParentId="16062" CreationDate="2015-06-07T11:20:04.820" Score="6" Body="&lt;p&gt;Q1. Which is the first question where c) is the correct answer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Q3 b) Q4 c) Q1 d) Q2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Q2. Which is the first question where a) is the correct answer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Q4 b) Q2 c) Q3 d) Q1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Q3. Which is the first question where d) is the correct answer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Q1 b) Q2 c) Q4 d) Q3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Q4. Which is the first question where b) is the correct answer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Q2 b) Q4 c) Q3 d) Q1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; B&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11150" LastActivityDate="2015-06-07T11:20:04.820" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16064" PostTypeId="1" AcceptedAnswerId="16066" CreationDate="2015-06-07T13:46:30.883" Score="7" ViewCount="4168" Body="&lt;p&gt;This is also another puzzle I found on the internet. The link to the first one is &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16062/what-is-the-answer-to-the-following-questions&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;The first question with B as the correct answer is:&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;A. 1&lt;br&gt;&#xA;  B. 4&lt;br&gt;&#xA;  C. 3&lt;br&gt;&#xA;  D. 2  &lt;/p&gt;&#xA;  &#xA;  &lt;ol start=&quot;2&quot;&gt;&#xA;  &lt;li&gt;The answer to Question 4 is:&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;A. D&lt;br&gt;&#xA;  B. A&lt;br&gt;&#xA;  C. B&lt;br&gt;&#xA;  D. C  &lt;/p&gt;&#xA;  &#xA;  &lt;ol start=&quot;3&quot;&gt;&#xA;  &lt;li&gt;The answer to Question 1 is:&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;A. D&lt;br&gt;&#xA;  B. C&lt;br&gt;&#xA;  C. B&lt;br&gt;&#xA;  D. A  &lt;/p&gt;&#xA;  &#xA;  &lt;ol start=&quot;4&quot;&gt;&#xA;  &lt;li&gt;The number of questions which have D as the correct answer is:&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;A. 3&lt;br&gt;&#xA;  B. 2&lt;br&gt;&#xA;  C. 1&lt;br&gt;&#xA;  D. 0  &lt;/p&gt;&#xA;  &#xA;  &lt;ol start=&quot;5&quot;&gt;&#xA;  &lt;li&gt;The number of questions which have B as the correct answer is:&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;A. 0&lt;br&gt;&#xA;  B. 2&lt;br&gt;&#xA;  C. 3&lt;br&gt;&#xA;  D. 1  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12860" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-11-01T09:41:24.787" Title="What is the answer to the following questions? [ Number 2 ]" Tags="|logical-deduction|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16065" PostTypeId="2" ParentId="16058" CreationDate="2015-06-07T13:53:16.103" Score="8" Body="&lt;p&gt;You are at:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Home = 127.0.0.1 = 127001 = 370031 in octal&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11150" LastActivityDate="2015-06-07T13:53:16.103" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16066" PostTypeId="2" ParentId="16064" CreationDate="2015-06-07T14:15:49.253" Score="4" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1)C&lt;br&gt;2)D&lt;br&gt;3)B&lt;br&gt;4)C&lt;br&gt;5)B&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer to question 1 cannot be A (since then it would be B) or B (since then it couldn't be B), so it must be C or D. If it's D, then the answer to question 2 is B, so the answer to question 4 is A, so there are three D's (not answers to questions 2 or 4), so the answer to question 3 must be D, which contradicts question 1. So &lt;strong&gt;the answer to question 1 is C&lt;/strong&gt;, which immediately means &lt;strong&gt;the answer to question 3 is B&lt;/strong&gt;.&lt;br&gt;&lt;br&gt;Now the answer to question 2 can't be A (since then the answer to question 4 would be D, which is self-contradiction), nor B or C (since then the answer to question 4 would be A or B and there would be at least two D's even though question 5 is the only one that could be D now). So &lt;strong&gt;the answer to question 2 is D&lt;/strong&gt;, which immediately means &lt;strong&gt;the answer to question 4 is C&lt;/strong&gt;.&lt;br&gt;&lt;br&gt;The total number of B's must now be 1 or 2. If it's 1, then the answer to question 5 is D, so there are two D's, contradicting question 4. So there are two B's and &lt;strong&gt;the answer to question 5 is B&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-06-07T14:24:32.250" LastActivityDate="2015-06-07T14:24:32.250" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16068" PostTypeId="2" ParentId="16064" CreationDate="2015-06-07T14:19:26.950" Score="0" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C,D,B,C,B&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The explanation is as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Q1 can't be A because it asks which one has the answer as B. Can't be B because then Q1 is already the one with the first answer as B. If Q1 is D, then Q2 links to Q4 saying that 3 questions have the answer as D, and since we already know the answer to Q2 and Q4 none of which are D, then Q1, Q3, and Q5 are D, but Q3 can't be D because according to Q3 and choosing D, the answer to Q1 is A, but we already know that the answer to Q1 isn't A, so this choice fails as well. Therefore the only option left is C.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For Question 3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We know that the correct answer to Q1 is C, and that says that Q3 is the first question with the answer as B, so therefore Q3 is B.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For Question 4, 2, and 5&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If we choose A, then there are only 2 questions left which can have the answer as D, but A says there are 3 left, so therefore A is false. Also, we can cross out D because if it is indeed zero, then the answer for this question can't be D. It can't be B because that means Q2 and Q5 have the answer as B because then Q2 says the answer to Q4 is C, but we have chosen B. Therefore only C works, which means Q2 has the answer D. And then the only answer that works for Q5 is B.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9159" LastActivityDate="2015-06-07T14:19:26.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16069" PostTypeId="2" ParentId="16055" CreationDate="2015-06-07T15:04:45.960" Score="0" Body="&lt;p&gt;I think their strategy is wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;When they collaborate, they should designate &lt;em&gt;either&lt;/em&gt; Alice &lt;em&gt;or&lt;/em&gt; Bob, but not both, as a presser, and the other as a non-presser. Suppose Alice is the designated presser; Bob is no longer part of the consideration, since we know he can't have any effect on the net payout. There is then a 50% chance that Alice makes \$5000, and a 50% chance she loses \$2000, for an expected payout of $\frac{\$5000-\$2000}{2}$ = \$1500, which is better than the $500 expectation of the &quot;both press&quot; situation. This doesn't resolve the paradox, but unless I made a mistake, it &lt;em&gt;does&lt;/em&gt; show that the paradox shouldn't even be considered, because it resulted from a strategy that never should've been followed in the first place.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;EDIT: Ouch, I did bad math.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There is a 25% chance she alone is selected, and a 50% chance they're both selected, reducing the expected payout of a designated presser to $\frac{\$5000-$2000 -\$2000}{4}$ = \$250. But by symmetry, shouldn't a &quot;both press&quot; situation result in a \$500 gain?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13022" LastEditorUserId="13022" LastEditDate="2015-06-07T15:16:39.690" LastActivityDate="2015-06-07T15:16:39.690" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16070" PostTypeId="2" ParentId="16062" CreationDate="2015-06-07T15:09:39.253" Score="15" Body="&lt;p&gt;Andrew Smith has the correct answer, namely&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1D, 2C, 3A, 4B,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but here's a full explanation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let's look first at Q2, since it's the only one that can't be self-referential (e.g. for Q3, the answer d) would be self-confirming). The answer to Q2) can't be a) since then it couldn't be a), or b) since then it must be a). Also 2d) implies 1a) implies 3c) implies 4d) implies 1b), contradiction. So &lt;strong&gt;the answer to Q2) is c)&lt;/strong&gt;. This immediately implies &lt;strong&gt;the answer to Q3) is a)&lt;/strong&gt;, which implies &lt;strong&gt;the answer to Q1) is d)&lt;/strong&gt;. None of the answers so far is b), so there's only one possible answer to Q4): &lt;strong&gt;the answer to Q4) is b)&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;QED.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-06-07T15:09:39.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16071" PostTypeId="2" ParentId="16049" CreationDate="2015-06-07T15:36:58.527" Score="0" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\binom{n}{2}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's say $a(i)$ is the total of &quot;rounds&quot; in which the $i_{th}$ colour is present, before the colour $i$ disappears or becomes the only one.&lt;br&gt;&#xA;This is equivalent to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gambler%27s_ruin&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gambler's ruin problem&lt;/a&gt;, where the expected time of $a(i)$ is $n-1$.&lt;br&gt;&#xA;We also believe in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Expected_value#Linearity&quot; rel=&quot;nofollow noreferrer&quot;&gt;linearity of expectation&lt;/a&gt;, and we already know that the colours are $n$ and the expectation of $a(n)$ is $n-1$.&lt;br&gt;&#xA;Thus, we can deduce that the expected time $t(n)$ to make all the balls of the same colour is:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$t(n)=\frac{n(n-1)}{2}=\binom{n}{2}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why did we divide by $2$? Simply because each &quot;round&quot; involves two different colours!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Big credits to @Russ Woodroofe for his solution &lt;a href=&quot;https://mathoverflow.net/questions/130489/another-colored-balls-puzzle&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:31.357" LastActivityDate="2015-06-07T15:36:58.527" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16072" PostTypeId="2" ParentId="16047" CreationDate="2015-06-07T15:55:47.707" Score="6" Body="&lt;p&gt;The password is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; T A U&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Using the f2831he provided by ITAdminNC, change the letters with their respective numbers. For example, a = 1, b = 2, c = 3, etc. Then it will be changed to 6283185. Now add a decimal point to the number: 6.283185. This number is the value of tau, or twice PI. So that is why the password is TAU.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13023" LastActivityDate="2015-06-07T15:55:47.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16073" PostTypeId="1" AcceptedAnswerId="16074" CreationDate="2015-06-07T16:15:59.497" Score="6" ViewCount="5574" Body="&lt;p&gt;This is yet another question I found on the Internet.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;During a recent police investigation, Chief Inspector Stone was interviewing five local villains to try and identify who stole Mrs Archer's cake from the mid-summers fayre. Below is a summary of their statements:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Arnold:  it wasn't Edward. It was Brian&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brian:   it wasn't Charles. It wasn't Edward&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Charles: it was Edward.&#xA;         It wasn't Arnold&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Derek:   it was Charles.&#xA;         It was Brian&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edward:  it was Derek.&#xA;         It wasn't Arnold&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It was well known that each suspect told exactly one lie. Can you determine who stole the cake?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12860" LastActivityDate="2023-09-30T13:16:22.677" Title="Who stole the cake?" Tags="|logical-deduction|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16074" PostTypeId="2" ParentId="16073" CreationDate="2015-06-07T16:20:10.680" Score="13" Body="&lt;p&gt;Brian's and Derek's statements give the answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Charles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if exactly one of Brian's statements is a lie, then the culprit must be either Charles or Edward; if exactly one of Derek's statements is a lie, then the culprit must be either Charles or Brian.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So Arnold, Charles, and Edward's answers (we can check each one told exactly one lie) aren't actually necessary to solve the puzzle.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-06-07T16:35:27.327" LastActivityDate="2015-06-07T16:35:27.327" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16075" PostTypeId="1" AcceptedAnswerId="16076" CreationDate="2015-06-07T16:21:49.443" Score="6" ViewCount="2102" Body="&lt;p&gt;I entered a room where there was no one around me so I waited for years and years for my friend to enter. My friend had sent me to this room for a reason I do not know. As years began to pass I noticed that there was a string hanging on me. Suddenly I heard a noise and hoped it would be my friend. Then I saw 2 gray figures coming towards me and then they started attacking me. I saw they had a string hanging on top of them. At that same moment I looked through the window and saw giant people looking at me. The grey figures left and I saw my family there with me and at that moment I knew this is real. The question was where am I and why are there giant people looking at me?&lt;/p&gt;&#xA;" OwnerUserId="13023" LastEditorUserId="5373" LastEditDate="2015-06-07T16:39:44.103" LastActivityDate="2015-06-08T12:43:36.263" Title="What is strange in this puzzle?" Tags="|riddle|story|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16076" PostTypeId="2" ParentId="16075" CreationDate="2015-06-07T16:26:14.413" Score="12" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are a puppet!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The giant people are human beings. They appear big to you because you are small. Also, the string attached to you means that you are a puppet. The gray figures might be other puppets.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12860" LastEditorUserId="12860" LastEditDate="2015-06-07T19:11:41.553" LastActivityDate="2015-06-07T19:11:41.553" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16078" PostTypeId="2" ParentId="16055" CreationDate="2015-06-07T16:58:27.447" Score="11" Body="&lt;p&gt;It &lt;strong&gt;is&lt;/strong&gt; rational for both parties to press the button, causing them to win $500$ on average. Here is what is wrong with the reasoning in the Paradox section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once Alice is called in, she reasons that the strategy will lose them money. That is ok, since when Alice is &lt;strong&gt;not&lt;/strong&gt; called in, she reasons that the strategy will &lt;strong&gt;win&lt;/strong&gt; them money (when she is not called in, the strategy wins them $\$5000$). So, $3/4$ of the time, Alice will be called in and expect to lose $1000$ dollars, but this is balanced out by the $1/4$ chance that she won't get called in, winning $5000$. The expected value is $(1/4)\cdot 5000-(3/4)\cdot 1000=500$, as expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To sum up, the flaw is that Alice only reasons that the strategy is bad when it has started going badly, but on average, it is still good. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to see it: once Alice is caused in, what is the expected value of her pushing the button (ignoring what Bob does, since that does not affect the money Alice wins the team)? There is a $1/3$ chance it will gain the team $5000$, and a $2/3$ chance it will lose the team $2000$ (not $4000$, we are ignoring Bob's actions here). The expected value of Alice pushing the button is $(1/3)5000-(2/3)2000\approx 333$, so she should push it. &lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-06-07T16:58:27.447" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16079" PostTypeId="2" ParentId="15690" CreationDate="2015-06-07T17:46:37.950" Score="17" Body="&lt;p&gt;I have found the rule.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Give each letter a value. A = 1, B = 2, C = 3, and so on. Then add all the values. I have taken &quot;snap&quot; as an example. &quot;snap&quot; = 19 + 14 + 1 + 16 = 50. Then find what is the least number of coins needed to make the total, or in this case, 50 cents. You only need one coin for a total of 50 cents. Therefore, 1 is the correct answer. As listed in a hint, the US coins are used: penny (1¢), nickel (5¢), dime (10¢), quarter (25¢), half dollar (50¢), and dollar (100¢).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; cab = 3 + 1 + 2 = 6 cents = 5 cent coin + 1 cent coin = 2 coins&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12860" LastEditorUserId="4551" LastEditDate="2015-06-07T22:25:46.273" LastActivityDate="2015-06-07T22:25:46.273" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16080" PostTypeId="2" ParentId="16055" CreationDate="2015-06-07T18:18:28.153" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The fact that she has been brought to an office and not left in the holding cell gives her new information.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is no new information here.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There was a 100% a priori chance that one or both would be brought to an office with a button. That she was the recipient of this 100% chance adds zero information. In particular, where's Bob? She doesn't know. It is still a fifty-fifty proposition whether he's been left in the holding cell or has been brought to an office with a big red button. There would be new information had Alice been left in the holding cell. In that case, she would be quite justified in getting mightily upset at Bob had he fallen prey to the erroneous thinking presented on the question and failed to push the +5K BRB.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So what should Alice do?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Simple: Alice should press the BRB as originally planned, but she should also hope that Bob will fall prey to the logic presented in the question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9026" LastActivityDate="2015-06-07T18:18:28.153" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16081" PostTypeId="2" ParentId="16075" CreationDate="2015-06-07T19:02:27.480" Score="9" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; crazy. (What is strange in this puzzle?) &lt;em&gt;You are&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The string hanging on you is a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; wire. It attaches to the electrodes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The strings hanging on them is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; from their doctors masks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At the moment&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; that they (attacked you) flipped the switch, you began having that delusion about the giant people again. Once the procedure was complete, the delusion had passed and you were welcomed back to &lt;strong&gt;real&lt;/strong&gt;ity by your loving family.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;em&gt;Where am I?&lt;/em&gt; You are in a psychiatric ward.&lt;br&gt;&#xA;&lt;em&gt;Why are giant people looking at me?&lt;/em&gt; Refer to, &lt;em&gt;where am I&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4522" LastActivityDate="2015-06-07T19:02:27.480" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16082" PostTypeId="2" ParentId="16055" CreationDate="2015-06-07T20:56:22.293" Score="6" Body="&lt;p&gt;I wonder if my contribution will help at all (or whether it will simply confuse more). Hopefully the former.&lt;/p&gt;&#xA;&lt;p&gt;As good, Bayesian, reasoners Alice and Bob have correctly worked out their best strategy. That is &amp;quot;both push&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;It is helpful to think of the probability space as containing 4 possibilities. Clearly if the coin flips to &amp;quot;heads&amp;quot; there are then two possibilities &amp;quot;A picked&amp;quot; or &amp;quot;B picked&amp;quot;. Only one possible outcome occurs when &amp;quot;tails&amp;quot; is picked, but dividing it into two makes all parts of the possibility space the same size (i.e. equiprobable).&lt;/p&gt;&#xA;&lt;p&gt;So for the strategy &amp;quot;both press the button&amp;quot; we have (I would use a table here if I could):&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th&gt;A&lt;/th&gt;&#xA;&lt;th&gt;B&lt;/th&gt;&#xA;&lt;th&gt;Coin flip&lt;/th&gt;&#xA;&lt;th&gt;Gain&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;office&lt;/td&gt;&#xA;&lt;td&gt;holding cell&lt;/td&gt;&#xA;&lt;td&gt;HEADS&lt;/td&gt;&#xA;&lt;td&gt;5000&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;holding cell&lt;/td&gt;&#xA;&lt;td&gt;office&lt;/td&gt;&#xA;&lt;td&gt;HEADS&lt;/td&gt;&#xA;&lt;td&gt;5000&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;office&lt;/td&gt;&#xA;&lt;td&gt;office&lt;/td&gt;&#xA;&lt;td&gt;TAILS&lt;/td&gt;&#xA;&lt;td&gt;-4000&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;office&lt;/td&gt;&#xA;&lt;td&gt;office&lt;/td&gt;&#xA;&lt;td&gt;TAILS&lt;/td&gt;&#xA;&lt;td&gt;-4000&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;&lt;/td&gt;&#xA;&lt;td&gt;&lt;/td&gt;&#xA;&lt;td&gt;&lt;/td&gt;&#xA;&lt;td&gt;2000 (total)&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;Since I have divided the space into 4, the expected gain is 2000 divided by 4 equals 500.&lt;/p&gt;&#xA;&lt;p&gt;So the first part of the OP's reasoning is right. But if A finds herself in an office she knows that one of the 4 possibilities, worth 5000, did not come off. Her expected gain is now reduced to -1000 according to the following table:&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th&gt;A&lt;/th&gt;&#xA;&lt;th&gt;B&lt;/th&gt;&#xA;&lt;th&gt;Coin flip&lt;/th&gt;&#xA;&lt;th&gt;Gain&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;office&lt;/td&gt;&#xA;&lt;td&gt;holding cell&lt;/td&gt;&#xA;&lt;td&gt;HEADS&lt;/td&gt;&#xA;&lt;td&gt;5000&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;office&lt;/td&gt;&#xA;&lt;td&gt;office&lt;/td&gt;&#xA;&lt;td&gt;TAILS&lt;/td&gt;&#xA;&lt;td&gt;-4000&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;office&lt;/td&gt;&#xA;&lt;td&gt;office&lt;/td&gt;&#xA;&lt;td&gt;TAILS&lt;/td&gt;&#xA;&lt;td&gt;-4000&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;&lt;/td&gt;&#xA;&lt;td&gt;&lt;/td&gt;&#xA;&lt;td&gt;&lt;/td&gt;&#xA;&lt;td&gt;-3000 (total)&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;The probability space now has size 3, so the expected gain is -3000 divided by 3, -1000.&lt;/p&gt;&#xA;&lt;p&gt;A Bayesian should not be surprised by this. Expectations are based on probabilities and, to a Bayesian, probabilities represent states of knowledge. A now knows more than she did and, as a result, knows that their gamble is less likely to have come off. Her expectation is reduced.&lt;/p&gt;&#xA;&lt;p&gt;But should she change her actions having found herself in the office? To see that she should not, consider that Alice and Bob can agree at the outset what their strategy will be.&lt;/p&gt;&#xA;&lt;p&gt;There are four &amp;quot;pure&amp;quot; strategies (I leave it as an exercise to see that mixed strategies - i.e. ones where Alice or Bob randomly choose what to do - are never better). Either both don't press the button, A presses, B presses or both press. All this is on the assumption they find themselves in an office. Obviously if they are in the holding cell they can do nothing.&lt;/p&gt;&#xA;&lt;p&gt;The four strategies give the following results:&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th&gt;Strategy&lt;/th&gt;&#xA;&lt;th&gt;Payoff&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Both don't press&lt;/td&gt;&#xA;&lt;td&gt;0&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Only A presses&lt;/td&gt;&#xA;&lt;td&gt;5000/4 - 2x2000/4 = 1000/4 = 250&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Only B presses&lt;/td&gt;&#xA;&lt;td&gt;5000/4 - 2x2000/4 = 1000/4 = 250&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;Both press&lt;/td&gt;&#xA;&lt;td&gt;2x5000/4 - 2x4000/4 = 2000/4 = 500&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;Note: if only one presses there are still four (not 3) possibilities. One of them is boring, the person designated to press is in a holding cell and nothing is gained or lost.&lt;/p&gt;&#xA;&lt;p&gt;Hopefully it is clearer if the two have an advance strategy. Having A &amp;quot;change her mind&amp;quot; on finding out more information is equivalent to deciding in advance what she will do if she finds that information out. In this problem there's only one bit of information (office or not) and the only time a choice must be made is in one of those situations &amp;quot;office&amp;quot;, so the strategies can be fairly simple.&lt;/p&gt;&#xA;&lt;p&gt;Note that if the &amp;quot;A presses&amp;quot; strategy is used, then when A find herself in the office the expectation is now 1000/3 (roughly 333) as previously explained. Although &amp;quot;A presses&amp;quot; has an expectation of 250, once A has found herself in the office she now knows more information, so the probabilities that form part of the expectation change and her new expectation is higher (333) because she now knows that the one situation (her in the holding cell) when they would certainly gain nothing has been ruled out.&lt;/p&gt;&#xA;&lt;p&gt;Expectation values can change as you gain more information, but that doesn't mean that strategies need to.&lt;/p&gt;&#xA;" OwnerUserId="4769" LastEditorUserId="69582" LastEditDate="2021-02-10T23:31:40.523" LastActivityDate="2021-02-10T23:31:40.523" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="16084" PostTypeId="1" AcceptedAnswerId="16212" CreationDate="2015-06-07T22:31:02.100" Score="8" ViewCount="1624" Body="&lt;p&gt;I have two operators! They are very close, brothers, actually! Though they are like each other's negative.They are named &lt;span class=&quot;math-container&quot;&gt;$\mathfrak{C}^+$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mathfrak{C}^-$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;No one knows what's going on in their heads. It is now crucial! These two operators have concocted a plan to overthrow the common core! The common core government, however, have kidnapped &lt;span class=&quot;math-container&quot;&gt;$\mathfrak{C}^+$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mathfrak{C}^-$&lt;/span&gt;!&lt;/p&gt;&#xA;&lt;p&gt;This is a problem. Their mother, &lt;span class=&quot;math-container&quot;&gt;$\mathfrak{C}^\pm$&lt;/span&gt;, is worried to death. Their father, &lt;span class=&quot;math-container&quot;&gt;$\Gamma$&lt;/span&gt;, first needs to overthrow the tyrannous common core, but, to do so, he needs his children's plans! If only... if only we could find out what is going on in their heads!!&lt;/p&gt;&#xA;&lt;p&gt;But news! Their father received a ransom from the common core; in exchange for their children, &lt;span class=&quot;math-container&quot;&gt;$\mathfrak{C}^\pm$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\Gamma$&lt;/span&gt; must pay \$&lt;span class=&quot;math-container&quot;&gt;$g_{64}$&lt;/span&gt;! The ransom was penned in the two children's writing, and hurriedly scribbled in the margins is some I/Os. This is the key! Since &lt;span class=&quot;math-container&quot;&gt;$\mathfrak{C}^\pm$&lt;/span&gt; can tell what her children are talking about if they tell her, she can figure out what their plan is!&lt;/p&gt;&#xA;&lt;p&gt;So everything seems all figured out. Except for one thing. No one can find what functions &lt;span class=&quot;math-container&quot;&gt;$\mathfrak{C}^+$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mathfrak{C}^-$&lt;/span&gt; are. This is where you come in.&lt;/p&gt;&#xA;&lt;p&gt;Given the I/O sheet, find out &lt;span class=&quot;math-container&quot;&gt;$\mathfrak{C}^+$&lt;/span&gt;'s function rule is! From there, the mother will decipher the hidden clues in the message. (The other child did not pen the letter)&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Objective:&lt;/strong&gt; Find the function rule for &lt;span class=&quot;math-container&quot;&gt;$\mathfrak{C}^+$&lt;/span&gt;. That is all.&lt;/p&gt;&#xA;&lt;p&gt;Here is the sheet:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$X\subseteq Y \implies \mathfrak{C}^+(X,Y)=Y$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathfrak{C}^+(1,2) = \sim6$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\mathfrak{C}^+(1,5) = 9$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\mathfrak{C}^+(1,8) = 8$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\mathfrak{C}^+(2,3) = \sim6$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\mathfrak{C}^+(2,7) = \sim6$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\forall x\in\{0,4,5,6,8,9\}[\mathfrak{C}^+(2,x)=8]$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\mathfrak{C}^+(4,7) = 9 \setminus \{\bot,\bot,\bot,\bot,\top,\bot,\top\}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;Hints&lt;/h1&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Ignore the story. It's just for colour.&lt;br&gt;&lt;pre&gt;z1110111&lt;br&gt;o0010010&lt;br&gt;t1011101&lt;br&gt;t1011011&lt;br&gt;f0111010&lt;br&gt;f1101011&lt;br&gt;s1101111&lt;br&gt;s1010010&lt;br&gt;e1111111&lt;br&gt;n1111011&lt;/pre&gt;&lt;br&gt;The numbers and letters are sets, which contain either &lt;span class=&quot;math-container&quot;&gt;$\top$&lt;/span&gt;s or &lt;span class=&quot;math-container&quot;&gt;$\bot$&lt;/span&gt;s; &lt;span class=&quot;math-container&quot;&gt;$\{\bot\}\subseteq\{\top\}$&lt;/span&gt;; think about operations of these symbols corresponding to numbers.&lt;br&gt;time is like a drug (LSD), except without the ex's and PT's (switch 'em)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;I believe I have put up sufficient information. Please inform me if this belief is incorrect.&lt;/p&gt;&#xA;&lt;p&gt;Good luck!!!&lt;/p&gt;&#xA;&lt;h1&gt;Edit: An apology&lt;/h1&gt;&#xA;&lt;p&gt;A key part of the riddle failed to display in mathjax; I have denoted some numbers with a tilde, but did not display. Sorry!&lt;/p&gt;&#xA;" OwnerUserId="7307" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-06-17T10:17:55.537" Title="This is not math. This is war!" Tags="|mathematics|calculation-puzzle|lateral-thinking|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16085" PostTypeId="2" ParentId="16049" CreationDate="2015-06-07T23:30:28.650" Score="1" Body="&lt;h1&gt;Summaries&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BEST SOLUTION (based on values of $n_\omega$): $(n-1)^2$ minutes.&lt;br&gt;$n-1$ minutes is an ideal solution&lt;br&gt;$\infty$ is an invalid trollish solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;Answer One - Invalid Troll&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I say that the answer is $\infty$, by allowing the choosing and painting of balls to be an infinite time, meaning that the process is never completed. thus, the answer is $\infty$ and we're done! $■$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(after reading comments) So you &lt;em&gt;do&lt;/em&gt; mean time... How interesting. I failed to observe the &quot;once per minute&quot; bit. My general solutions is this: $\color{blue}{\Phi(n)}$ (see lower down)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;Answer Two - $n_\omega$&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;This assumes that &quot;expected time&quot; is the same as &quot;number of steps&quot;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Experimental trial&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;$Spawn$:$\{\color{green}{ball},\color{red}{ball},\color{purple}{ball},\color{blue}{ball}\}$&lt;br&gt;&#xA;$Step\&amp;gt;1$:$\text{Choose: }(\color{blue}{4},\color{green}{1})$ $\text{Resultant set: }\{\color{blue}{ball},\color{red}{ball},\color{purple}{ball},\color{blue}{ball}\}$&lt;br&gt;&#xA;$Step\&amp;gt;2$:$\text{Choose: }(\color{blue}{1},\color{purple}{3})$&#xA;$\text{Resultant set: }\{\color{blue}{ball},\color{red}{ball},\color{blue}{ball},\color{blue}{ball}\}$&lt;br&gt;&#xA;$Step\&amp;gt;3$:$\text{Choose: }(\color{red}{2},\color{blue}{1})$&#xA;$\text{Resultant set: }\{\color{red}{ball},\color{red}{ball},\color{blue}{ball},\color{blue}{ball}\}$&lt;br&gt;&#xA;$Step\&amp;gt;4$:$\text{Choose: }(\color{red}{2},\color{blue}{3})$&#xA;$\text{Resultant set: }\{\color{red}{ball},\color{red}{ball},\color{red}{ball},\color{blue}{ball}\}$&lt;br&gt;&#xA;$Step\&amp;gt;5$:$\text{Choose: }(\color{red}{3},\color{blue}{4})$&#xA;$\text{Resultant set: }\{\color{red}{ball},\color{red}{ball},\color{red}{ball},\color{red}{ball}\}$&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Conclusions&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Five steps for an experiment of $n=4$. It makes sense that, within a few iterations, a good deal of colours will disappear; a minimum of one colour will disappear within the first iteration. In this example, by step 2, we see that the original population of colours has been reduced from 4 to 2.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Computer- &amp;amp; Math-aided Portion&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;(to be accomplished with computers! The answer is easier to find if you know what you are looking for.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't hate python. FEEL THE POWER OF PYTHON! This is how I obtain $n_\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from random import *&#xA;&#xA;class Ball:&#xA;    def __init__(self,colour):&#xA;        self.colour = colour&#xA;&#xA;    def change_colour(self,ball):&#xA;        self.colour = ball.colour&#xA;&#xA;def allColour(list):&#xA;    b = list[0].colour&#xA;        for i in range(len(list)):&#xA;            if not (b == list[i].colour):&#xA;                return False&#xA;            b = list[i].colour&#xA;    return True&#xA;&#xA;def ball_sim(n):&#xA;    rq = list(range(n))&#xA;    for i in range(n):&#xA;        rq[i] = Ball(i)&#xA;    trials = 0&#xA;    if n == 1 or int(n) != n:&#xA;        return 1&#xA;    while not allColour(rq):&#xA;        samp = sample(list(range(len(rq))),2)&#xA;        rq[samp[0]].change_colour(rq[samp[1]])&#xA;        trials += 1&#xA;    return trials&#xA;&#xA;def s(n,m):&#xA;    avg = 0&#xA;    for qk in range(n):&#xA;        avg += ball_sim(m)&#xA;    return avg / n&#xA;&#xA;while True:&#xA;    n = input(&quot;(integer) n = &quot;)&#xA;    print(s(100000,int(n)))&#xA;    input(&quot;continue...&quot;)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now, what this does, it takes an $n$ as an input, does 100,000 trials of a simulation of the steps, and takes the average step amount.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Outputs&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition 1:&lt;/strong&gt;&#xA;Each $n$ has:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;an upperbound (called $n_\mu$),&lt;/li&gt;&#xA;&lt;li&gt;a lowerbound (called $n_\lambda$),&lt;/li&gt;&#xA;&lt;li&gt;an average (given by python; called $n_\alpha$),&#xA;&lt;ol&gt;&#xA;&lt;li&gt;a lowerbound with $t$ trials (designated $n_\alpha^t$; supposedly more accurate with higher values of $t$, and&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;a real answer (called $n_\omega$).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Also let $s(n)$ be the seed of $n$, that is, the colour list original to $n$, and let $s_i(n)$ be the colour list on the $i$th step. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition 2:&lt;/strong&gt; Let $c(b_i)$ designate the colour of ball $b_i$; any two balls selected for a given step are ordered in a coordinate pair, $(b_i,b_j)$. $b_i$ represents the ball at the $i$th position in $s(n)$. (I assume that a possible objective is to find $n_\lambda$ for an arbitrary $n$, perhaps using a function rule(s).)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's call this function $\Phi(n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=1$: $n$ is below the domain. I take it that that $1_\mu=1_\lambda=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=2$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem 1:&lt;/strong&gt; $2_\mu=2_\lambda=1$ &lt;em&gt;Proof.&lt;/em&gt; Suppose that $(b_0,b_1)$ is selected. Then, $b_0$ is coloured the same as $b_1$, $c(b_0)=c(b_1)$, and, since $\{b_0,b_1\}=s(2)$, we are done. Suppose the flipside is true, that $(b_1,b_0)$ is selected. Likewise, $b_1$ is coloured the same as $b_0$, $c(b_1)=c(b_0)$, and we are done. $■$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Theorem 2:&lt;/strong&gt; For $n&amp;gt;2$, $n_\mu\le+\infty$.&lt;br&gt;&lt;em&gt;Proof.&lt;/em&gt;&lt;br&gt;Since there is the possibility (though however unlikely) that we will experience a cyclic behaviour in the choosing of balls (e.g. choosing $(b_i,b_j)$, then choosing $(b_j,b_i)$, then choosing $(b_i,b_j)$...). $■$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;($n_\alpha$ based on a $100$-sized sample unless otherwise specified. This will become less and less accurate as $n$ increases; i.e., it is sufficiently accurate for ideally small values of $n$, where typically $\left|\left\lfloor n_\alpha\right\rceil-n_\alpha\right|\ll 0$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=3$: $3_\alpha=3.99$. To arrive at $3_\lambda$, we choose an ideal scenario: Each trial, one ball is converted. So, in the first trial, there are $2$ different colours, then, in the last trial, there is only one colour. So, there are two trials. Also note, it seems that $n_\omega=4$; it is expected that every $n_\omega\approx n_\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem 3:&lt;/strong&gt; $n_\lambda=n-1$.&lt;br&gt;&lt;em&gt;Proof.&lt;/em&gt;&lt;br&gt;This states that the absolute minimum for any $n$-bag is $n-1$. Consider that, before there are any balls painted, $n$ colours. Then, ideally, on turn $1$, there are $n-1$ balls. On turn $2$, there are $n-2$ balls, etc. So, on the $k$th turn, there are ideally $n-k$ colours left among the balls. Thus, on the $n-1$th turn, there are ideally $n-(n-1)=n-n+1=1$ colours left, and, since no lower number is possible for a remainder of colours, $n-1$ must be the last ideal turn.$■$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition 2:&lt;/strong&gt; $I(n)=n-1$. (The Ideal function)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; The first claim is essentially proven in an ideal scenario; for a bag of $n$ balls of $n$ different colours, ideally, the time expected in an idealist scenario is $n-1$ minutes. Obviously, for $n\in\mathbb Z \land n&amp;gt;1$, as the other cases are trivial/impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From now on, the results are computer-aided more.&#xA;For $n=4$, $n_\alpha=9.00172^{10^5}$. We can infer that $n_\omega=9$.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;n      = 4&#xA;trials = 1&#xA;4:9.0&#xA;Continue...&#xA;n      = 4&#xA;trials = 10&#xA;4:8.8&#xA;Continue...&#xA;n      = 4&#xA;trials = 100&#xA;4:8.96&#xA;Continue...&#xA;n      = 4&#xA;trials = 1000&#xA;4:9.024&#xA;Continue...&#xA;n      = 4&#xA;trials = 10000&#xA;4:9.0131&#xA;Continue...&#xA;n      = 4&#xA;trials = 100000&#xA;4:9.00172&#xA;Continue...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For $n=5$, $n_\alpha^{10^5}=16.00694$, and seems to imply $n_\omega=16$.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;n      = 5&#xA;trials = 1&#xA;5:11.0&#xA;Continue...&#xA;n      = 5&#xA;trials = 10&#xA;5:13.7&#xA;Continue...&#xA;n      = 5&#xA;trials = 100&#xA;5:16.87&#xA;Continue...&#xA;n      = 5&#xA;trials = 1000&#xA;5:16.313&#xA;Continue...&#xA;n      = 5&#xA;trials = 10000&#xA;5:16.1092&#xA;Continue...&#xA;n      = 5&#xA;trials = 100000&#xA;5:16.00694&#xA;Continue...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(the code you see here is as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;while True:&#xA;    n=int(input(&quot;n      = &quot;))&#xA;    t=int(input(&quot;trials = &quot;))&#xA;    print(str(n)+&quot;:&quot;+str(s(t,n)))&#xA;    input(&quot;Continue...&quot;)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;; that is, the code used to find certain values of $n_\lambda^t$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thereby conclude that $\Phi(n)=(n-1)^2$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Tabulated Results&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;(also see &lt;a href=&quot;https://www.desmos.com/calculator/yejluqfeve&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;)&#xA;$$\begin{array}{c|c|c|c|c}n&amp;amp;n_\omega&amp;amp;t&amp;amp;n_\alpha^t&amp;amp;(n-1)^2\\\hline&#xA;1&amp;amp;0&amp;amp;\varnothing&amp;amp;0&amp;amp;0\\&#xA;2&amp;amp;1&amp;amp;\varnothing&amp;amp;1&amp;amp;1\\&#xA;3&amp;amp;4&amp;amp;10^2&amp;amp;3.99&amp;amp;4\\&#xA;4&amp;amp;9&amp;amp;10^5&amp;amp;9.00172&amp;amp;9\\&#xA;5&amp;amp;16&amp;amp;10^5&amp;amp;16.00694&amp;amp;16&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;A Formal Proof&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The former is a proof leaning on computers and their results. I want to find a mathematical proof of this fact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know the answer is $(n-1)^2$ (or at least, as it seems), so all that needs to be done is to prove it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But first, what does the result being $(n-1)^2$ signify? (Note that this is my thought process; this does not yet reveal an actual answer.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing some preliminary research, I found an interesting identity which holds for $k\le n$:&#xA;$$(x+y)^n=\sum_{k=0}^n\binom nk x^{n-k}y^k$$&#xA;I thought this might yield some information, yet it revealed nothing new.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(WIP)&lt;/p&gt;&#xA;" OwnerUserId="7307" LastEditorUserId="7307" LastEditDate="2015-06-08T23:36:37.467" LastActivityDate="2015-06-08T23:36:37.467" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16086" PostTypeId="2" ParentId="16084" CreationDate="2015-06-07T23:30:56.107" Score="2" Body="&lt;h2&gt;Partial answer&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;For ease of notation, let's use $0$ to represent $\bot$ and $1$ to represent $\top$. (This is fairly standard in logic theory anyway.) Then $\{\bot\}\subseteq\{\top\}$ makes sense, interpreting $\subseteq$ as simply $\leq$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 'numbers' being operated on by $\mathfrak{C}^+$, which we know are actually 'sets' or rather arrays, are now given by part of the spoilertagged hint: we interpret&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;z1110111&lt;br&gt;o0010010&lt;br&gt;t1011101&lt;br&gt;t1011011&lt;br&gt;f0111010&lt;br&gt;f1101011&lt;br&gt;s1101111&lt;br&gt;s1010010&lt;br&gt;e1111111&lt;br&gt;n1111011&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;as&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0 = 1110111 (or $\{\top,\top,\top,\bot,\top,\top,\top\}$)&lt;br&gt;1 = 0010010 (or $\{\bot,\bot,\top,\bot,\bot,\top,\bot\}$)&lt;br&gt;2 = 1011101&lt;br&gt;3 = 1011011&lt;br&gt;4 = 0111010&lt;br&gt;5 = 1101011&lt;br&gt;6 = 1101111&lt;br&gt;7 = 1010010&lt;br&gt;8 = 1111111&lt;br&gt;9 = 1111011&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We interpret $\subseteq$ to mean a componentwise $\leq$, i.e. an array of seven digits is $\subseteq$ another iff each digit in the first is $\leq$ the corresponding digit in the second. So for instance $5\subseteq6$, $3\subseteq9$, and everything is $\subseteq8$. This means $\mathfrak{C}^+(1,8)=8$ is already contained in the rule $X\subseteq Y \implies \mathfrak{C}^+(X,Y)=Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Everything down to here has been confirmed by the OP.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the examples we've been given of $\mathfrak{C}^+(X,Y)$ satisfy the following rule:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; write X and Y as a string of digits ($0$s and $1$s), and sum the digits in each column where possible (so that $(0,1)\rightarrow1$, $(1,0)\rightarrow1$, $(0,0)\rightarrow0$). When the column contains two $1$s, the resulting digit in the new number is also $1$ unless the preceding column contains two $0$s, in which case we have $(01,01)\rightarrow10$ ... except in $\mathfrak{C}^+(1,5) = 9$ and $\mathfrak{C}^+(1,8) = 8$ and $\mathfrak{C}^+(4,7)=9\setminus\{\bot,\bot,\bot,\bot,\bot,\bot,\top\}=1111011-0000001=1111010$, where for some reason $(01,01)\rightarrow01$ in the 5th and 6th columns.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;OK, that was a silly rule. More work needed!&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-06-08T22:44:48.580" LastActivityDate="2015-06-08T22:44:48.580" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16087" PostTypeId="1" AcceptedAnswerId="16089" CreationDate="2015-06-08T01:32:34.283" Score="12" ViewCount="772" Body="&lt;p&gt;Below is a word search puzzle presented without clues. Instead, you must use the additional rules (see example below) to work out what the words are. Finally, as a check to make sure you found the right words, the first letters of each word can be rearranged to make one last, related word.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Rules&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The words can run in any direction (orthogonal or diagonal) and do not have to be in straight lines&lt;/li&gt;&#xA;&lt;li&gt;Words do not overlap and their paths do not cross&lt;/li&gt;&#xA;&lt;li&gt;All letters are part of some word&lt;/li&gt;&#xA;&lt;li&gt;All words are of the same length&lt;/li&gt;&#xA;&lt;li&gt;All words (including the final word) are related somehow&lt;/li&gt;&#xA;&lt;li&gt;All words (including the final word) can be found in an English dictionary&lt;/li&gt;&#xA;&lt;li&gt;The letters that make up the final word do not appear in any specific location on the grid&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Example&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Here's a short example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bAwdg.png&quot; alt=&quot;P O A W. E N N S. M U Y A. S T O D.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GRuBj.png&quot; alt=&quot;PONY EMUS TOAD SWAN&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The four words here are PONY, EMUS, TOAD and SWAN, which are all animals, and the first letter of each makes the word PETS.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Puzzle&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;What is the final word to be found in this grid of letters?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VHoT2.png&quot; alt=&quot;C E N A I U. L I R S R D. A M O G U R. E L N I O F. E C I T L E. T I S I V S.&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7392" LastEditorUserId="7392" LastEditDate="2015-06-08T17:48:12.450" LastActivityDate="2015-06-08T17:48:12.450" Title="Cluelessly searching for words" Tags="|word-search|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16088" PostTypeId="2" ParentId="16055" CreationDate="2015-06-08T01:48:47.397" Score="4" Body="&lt;p&gt;The question has already been answered, but I would like to pinpoint exactly why it seems paradoxical. Let $W$ be the total winnings of Alice and Bob when they follow the plan of always pushing the button, $A$ the event that Alice is chosen and $B$ the event that Bob is chosen. Then Alice correctly reasons that $E[W|A] &amp;lt;0$ and for the same reason $E[W|B] &amp;lt; 0$. As one of Alice and Bob must be chosen, it seems intuitively clear that this means that $E[W] &amp;lt; 0$, but in fact the reasoning given under &quot;The Plan&quot; shows that $E[W] &amp;gt; 0$. This happens because $A$ and $B$ overlap, or in other words, the naive reasoning ignores that they can both be chosen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a similar situation. Suppose you are allowed to roll a die with the payoffs in dollars given as:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; roll: 1  2  3  4  5  6&#xA;&#xA; pays: 6 -3 -2 -2 -3  6&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Consider the following questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Would you play the game?&lt;/li&gt;&#xA;&lt;li&gt;If you knew that the roll was $\le 4$, would you play the game?&lt;/li&gt;&#xA;&lt;li&gt;If you knew that the roll was $\ge 3$, would you play the game?&lt;/li&gt;&#xA;&lt;li&gt;If you were told in advance that the roll would either be $\le 3$ or $\ge 4$, would you play the game?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13029" LastActivityDate="2015-06-08T01:48:47.397" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16089" PostTypeId="2" ParentId="16087" CreationDate="2015-06-08T01:59:15.747" Score="9" Body="&lt;p&gt;Words:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;FRUITS&lt;br&gt;&#xA;OLIVES&lt;br&gt;&#xA;OMELET&lt;br&gt;&#xA;DURIAN&lt;br&gt;&#xA;ICINGS&lt;br&gt;&#xA;ECLAIR     &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Final word:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;FOODIE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's a graphic:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/p5tmU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10861" LastEditorUserId="10861" LastEditDate="2015-06-08T02:09:17.477" LastActivityDate="2015-06-08T02:09:17.477" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16091" PostTypeId="1" AcceptedAnswerId="16092" CreationDate="2015-06-08T03:16:14.563" Score="-1" ViewCount="1343" Body="&lt;p&gt;2 → 6 &lt;br&gt;&#xA;3 → 12 &lt;br&gt;&#xA;4 → 20 &lt;br&gt;&#xA;5 → 30 &lt;br&gt;&#xA;6 → 42 &lt;br&gt;&#xA;9 → ? &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can you find a correspondence between these numbers?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;It's not mine, so I don't take any credit. I've just seen it in Linkedin and wanted to share with the community.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="10365" LastEditorUserId="10281" LastEditDate="2015-06-08T06:38:31.923" LastActivityDate="2015-12-16T15:06:41.797" Title="Complete the number-sequence" Tags="|number-sequence|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16092" PostTypeId="2" ParentId="16091" CreationDate="2015-06-08T03:21:45.313" Score="11" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 90  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; f(n) = n * (n+1)&lt;br&gt;&#xA; Thus,&lt;br&gt;&#xA; 2 * 3 = 6,&lt;br&gt;&#xA; 3 * 4 = 12,&lt;br&gt;&#xA; 4 * 5 = 20,&lt;br&gt;&#xA; 5 * 6 = 30,&lt;br&gt;&#xA; 6 * 7 = 42,&lt;br&gt;&#xA; 9 * 10 = 90  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12537" LastEditorUserId="12537" LastEditDate="2015-06-08T03:56:33.627" LastActivityDate="2015-06-08T03:56:33.627" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16093" PostTypeId="2" ParentId="16091" CreationDate="2015-06-08T03:40:08.347" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Obviously, this follows the pattern $b_n=2n +b_{n-1}$ for $n\in\{3,4,5,6\}$, with $b_2=6$. Assuming the pattern continues for $n\in\{7,8,9\}$, we get $b_9=2(9)+b_8=18+2(8)+b_7=34+2(7)+b_6=48+42=90$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are, of course, other answers, of which I will show later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is another possible solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If this is of the form $a_n = b_n, a_n ∈ A. b_n ∈ B$): $A = \{2,3,4,5,6,9\}$ and $B = \{x_n ∈ A : x_nx_{n-1}\}$. Since $a_n = b_n$, we must find $a_6$, which is $b_6$. $b_6 = a_6a_5 = 9(6) = 54$.&lt;br&gt;&lt;br&gt;I thus also conclude $9 = 54$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(As a side note, using the &quot;equality&quot; sign to represent some function sort of rubs me the wrong way, as it has a long-standing representation. I do like how here (as opposed to how I originally saw it), the &amp;lsquo;&amp;rarr;&amp;rsquo; is used instead. Kudos!)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; See &lt;a href=&quot;https://plus.google.com/112363137624640055874/posts/CBHvDUJr1Bv&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for a discussion, from which I drew my former answer (from myself). The former solution was just another way of tackling the common answer, also from myself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7307" LastEditorUserId="7363" LastEditDate="2015-06-09T02:38:43.590" LastActivityDate="2015-06-09T02:38:43.590" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16094" PostTypeId="1" AcceptedAnswerId="16102" CreationDate="2015-06-08T04:17:10.660" Score="17" ViewCount="1503" Body="&lt;p&gt;Below is a word search puzzle presented without clues. Instead, you must use the additional rules (see example below) to work out what the words are. Finally, as a check to make sure you found the right words, the first letters of each word can be rearranged to make one last, related word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a second, harder puzzle. &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16087/cluelessly-searching-for-words&quot;&gt;See here&lt;/a&gt; for the easier first puzzle.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Rules&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The words can run in any direction (orthogonal or diagonal) and do not have to be in straight lines&lt;/li&gt;&#xA;&lt;li&gt;Words do not overlap and their paths do not cross&lt;/li&gt;&#xA;&lt;li&gt;All letters are part of some word&lt;/li&gt;&#xA;&lt;li&gt;All words are of the same length&lt;/li&gt;&#xA;&lt;li&gt;All words (including the final word) are related somehow&lt;/li&gt;&#xA;&lt;li&gt;All words (including the final word) have an English Wikipedia page, even if they are not English themselves&lt;/li&gt;&#xA;&lt;li&gt;The letters that make up the final word do not appear in any specific location on the grid&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Example&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Here's a short example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bAwdg.png&quot; alt=&quot;P O A W. E N N S. M U Y A. S T O D.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GRuBj.png&quot; alt=&quot;PONY EMUS TOAD SWAN&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The four words here are PONY, EMUS, TOAD and SWAN, which are all animals, and the first letter of each makes the word PETS.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Puzzle&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;What is the final word to be found in this grid of letters?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Npahb.png&quot; alt=&quot;M A N E A P N. D N A I K A A. A E O S U L A. N S O V U S S. I M O R H A U. L E O R T M R. M G R D A M E.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Hint&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All the words are taken from one set of Wikipedia list pages&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7392" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2015-06-08T19:19:34.477" Title="Cluelessly searching for words II" Tags="|word-search|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16096" PostTypeId="2" ParentId="6507" CreationDate="2015-06-08T05:57:57.850" Score="3" Body="&lt;p&gt;This is a list of the smallest numbers of persistence $n$ with  ($1\leq n\leq11)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1&lt;/strong&gt; 10&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2&lt;/strong&gt; 25&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3&lt;/strong&gt; 39&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4&lt;/strong&gt; 77&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;5&lt;/strong&gt; 679  ---&gt; (answer of this problem)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;6&lt;/strong&gt; 6788&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;7&lt;/strong&gt; 68889&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;8&lt;/strong&gt; 2677889&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;9&lt;/strong&gt; 26888999&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;10&lt;/strong&gt; 3778888999&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;11&lt;/strong&gt; 277777788888899&lt;/p&gt;&#xA;" OwnerUserId="13035" LastEditorUserId="10281" LastEditDate="2015-06-08T06:27:25.793" LastActivityDate="2015-06-08T06:27:25.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16098" PostTypeId="2" ParentId="16091" CreationDate="2015-06-08T09:38:13.000" Score="5" Body="&lt;p&gt;I think you guys are overthinking this one. I'm a simple guy, with a simple answer (albeit the same).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 90&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sequence: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2 * 3 = 6 &lt;br&gt;&#xA; 3 * 4 = 12&lt;br&gt;&#xA; 4 * 5 = 20&lt;br&gt;&#xA; 5 * 6 = 30&lt;br&gt;&#xA; 6 * 7 = 42&lt;br&gt;&#xA; 7 * 8 = 56&lt;br&gt;&#xA; 8 * 9 = 72&lt;br&gt;&#xA; 9 * 10 = 90&lt;br&gt;&#xA;&lt;br&gt;&#xA; Basically, the pattern follows each successive number. Or, in simpler terms, once you've figured out the math behind the original question, you just add +1 to each side of the multiplication sign until you reach the final answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9670" LastEditorUserId="10399" LastEditDate="2015-12-16T12:35:04.067" LastActivityDate="2015-12-16T12:35:04.067" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16099" PostTypeId="2" ParentId="12955" CreationDate="2015-06-08T09:50:36.067" Score="1" Body="&lt;p&gt;This one would be easier, you don't really need to think their size equal or not :D&#xA;&lt;img src=&quot;https://i.stack.imgur.com/xhP9P.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide them with one single cut vertically, and another cut horizontally.&#xA;Then add two cuts to its thickness as picture above.&lt;/p&gt;&#xA;" OwnerUserId="11353" LastActivityDate="2015-06-08T09:50:36.067" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16101" PostTypeId="2" ParentId="16055" CreationDate="2015-06-08T12:17:08.493" Score="0" Body="&lt;p&gt;Consider the following variant:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Alice goes to her office, checks out the wiring and spots a (deliberate, who knows?) mistake. Her&#xA;  button has been connected in such a way that pressing the button&#xA;  causes Bob's button to be pressed, provided that Bob is also in his&#xA;  office.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In this case, accepting the 2/3 probability of Tails given that Alice is in her office, the result of pressing her button is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1/3 * 5000 - 2/3 * 2000 - 2/3 *2000 = -1000&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is exactly the expression given in the Paradox. If it were so wired, then she should not touch the button. But in the original question, this was not the case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bob's button is not pressed by any action of Alice, and it is incorrect to blame Alice for Bob pressing his button when they lose. Alice and Bob have the same strategy, and Bob's button pressing is &lt;strong&gt;correlated&lt;/strong&gt; with Alice's according to the strategy, but it is not &lt;strong&gt;caused&lt;/strong&gt; by her pressing like in the variant above. &lt;/p&gt;&#xA;" OwnerUserId="7316" LastActivityDate="2015-06-08T12:17:08.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16102" PostTypeId="2" ParentId="16094" CreationDate="2015-06-08T12:39:08.577" Score="23" Body="&lt;p&gt;The full list of words is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Domovoi&quot; rel=&quot;nofollow noreferrer&quot;&gt;Domovoi&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;http://en.wikipedia.org/wiki/Rusalka&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rusalka&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;http://en.wikipedia.org/wiki/Amemasu&quot; rel=&quot;nofollow noreferrer&quot;&gt;Amemasu&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;http://en.wikipedia.org/wiki/Gremlin&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gremlin&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;http://en.wikipedia.org/wiki/Orthrus&quot; rel=&quot;nofollow noreferrer&quot;&gt;Orthrus&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;http://en.wikipedia.org/wiki/Napaeae&quot; rel=&quot;nofollow noreferrer&quot;&gt;Napaeae&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;http://en.wikipedia.org/wiki/Sandman&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sandman&lt;/a&gt;   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As can be seen here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/jAP0m.png&quot; alt=&quot;word search&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With the final word being:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Dragon&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dragons&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All words being:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Lists_of_legendary_creatures&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mythical/legendary creatures&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="9000" LastEditDate="2015-06-08T19:19:34.477" LastActivityDate="2015-06-08T19:19:34.477" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16103" PostTypeId="2" ParentId="16075" CreationDate="2015-06-08T12:43:36.263" Score="1" Body="&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A frozen embryo being fertilized perhaps?  The window could be the microscope lens / test tube?  Strings are the sperm being injected into you to fertilize you intra-embryo i.e. via needle and micro-tweezers (the gray things)?  Things get real because you now exist as an arguably human entity?  People looking at you are your parents possibly, and or doctors perhaps?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Strange part is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Why you've got a string attached in the first place ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My answers seems too literal however ...&lt;/p&gt;&#xA;" OwnerUserId="13047" LastActivityDate="2015-06-08T12:43:36.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16104" PostTypeId="2" ParentId="16094" CreationDate="2015-06-08T12:45:25.413" Score="29" Body="&lt;p&gt;The solution is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/IyBa2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Amemasu&quot; rel=&quot;nofollow noreferrer&quot;&gt;AMEMASU&lt;/a&gt; - A giant whale- or fish-like creature from Ainu folklore&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Domovoi&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOMOVOI&lt;/a&gt; -  A protective house spirit in Slavic folklore&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Napaeae&quot; rel=&quot;nofollow noreferrer&quot;&gt;NAPAEAE&lt;/a&gt; - From Greek mythology, a type of nymph that lived in wooded valleys, glens or grottoes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Gremlin&quot; rel=&quot;nofollow noreferrer&quot;&gt;GREMLIN&lt;/a&gt; - A fantastic creature commonly depicted as mischievous and mechanically oriented, with a specific interest in aircraft. (This is the first one I found, without this I would have been lost)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Orthrus&quot; rel=&quot;nofollow noreferrer&quot;&gt;ORTHRUS&lt;/a&gt; - A two-headed dog in Greek mythology&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Rusalka&quot; rel=&quot;nofollow noreferrer&quot;&gt;RUSALKA&lt;/a&gt; - A water nymph, a female spirit in Slavic mythology&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;7&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Sandman&quot; rel=&quot;nofollow noreferrer&quot;&gt;SANDMAN&lt;/a&gt; - A mythical character in central and northern European folklore who brings good dreams by sprinkling magical sand onto the eyes of people while they sleep at night&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The final, related word is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Dragon&quot; rel=&quot;nofollow noreferrer&quot;&gt;DRAGONS&lt;/a&gt; - Legendary creatures, typically with serpentine or reptilian traits, that feature in the myths of many cultures&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They are all&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; mythological creatures&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All text quoted from the relevant Wikipedia articles.&lt;/p&gt;&#xA;" OwnerUserId="12277" LastEditorUserId="12277" LastEditDate="2015-06-08T13:08:45.323" LastActivityDate="2015-06-08T13:08:45.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16105" PostTypeId="2" ParentId="16053" CreationDate="2015-06-08T13:11:06.507" Score="7" Body="&lt;p&gt;She's wondering&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; WHERES MY COOKIE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You get there by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; dividing each number by 0.6 and using the result as index to the alphabet, e.g. 13.8 / 0.6 = 23 =&gt; W.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9910" LastActivityDate="2015-06-08T13:11:06.507" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16106" PostTypeId="1" AcceptedAnswerId="26019" CreationDate="2015-06-08T15:09:50.123" Score="21" ViewCount="4555" Body="&lt;p&gt;Can you give any example of riddles or puzzles in which the answers are someway &quot;cleverly hidden&quot; inside the question?&#xA;I think that such examples are more common in word-related riddles, but maybe similar ones can be found also in numeric puzzles? Who knows?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm italian, and there's a very common riddle in italian language which says: &quot;Nomina una parola qualunque contenente tre lettere U&quot; (translated: &quot;Name whatever word containing three 'U's &quot;); the funny thing is that the italian word &quot;QUALUNQUE&quot; (a really really common word) is maybe one of the few actually containing three U letters...).&#xA;That's what makes this riddle really clever and someway surprising when you reveal the answer...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, do you know any other example in english language?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Sorry if this is not a real riddle, but I think it can be interesting to discuss this topic)&lt;/p&gt;&#xA;" OwnerUserId="12628" LastEditorUserId="12628" LastEditDate="2016-02-01T11:14:36.237" LastActivityDate="2019-01-19T04:37:29.843" Title="Riddles/puzzles having answer hidden in the question?" Tags="|word|puzzle-creation|open-ended|" AnswerCount="9" CommentCount="10" ClosedDate="2018-08-27T21:18:07.547" ContentLicense="CC BY-SA 3.0" />
  <row Id="16107" PostTypeId="1" AcceptedAnswerId="16115" CreationDate="2015-06-08T15:33:58.677" Score="-4" ViewCount="246" Body="&lt;p&gt;I have just experienced a stranger, who doesn't look at all like a fishing man,   putting a shark in a bag happily.&lt;br&gt;&#xA;Why would anyone put a shark in a bag?&lt;br&gt;&#xA;Has it got anything to do with the bag or the shark?&lt;/p&gt;&#xA;" OwnerUserId="13023" LastEditorUserId="13023" LastEditDate="2015-06-09T11:00:39.450" LastActivityDate="2015-06-09T11:00:39.450" Title="The Mysterious Stranger and the Shark" Tags="|logical-deduction|story|" AnswerCount="2" CommentCount="1" ClosedDate="2015-06-08T17:12:22.073" ContentLicense="CC BY-SA 3.0" />
  <row Id="16108" PostTypeId="1" AcceptedAnswerId="16111" CreationDate="2015-06-08T15:34:59.183" Score="6" ViewCount="2449" Body="&lt;p&gt;Complete the grid shown below with the digits 1 to 6 to make the sum correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perform each mathematical operation in the order shown, from left to right, e.g. 1 + 2 x 3 is treated as (1 + 2) x 3 = 9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: there is no ÷ 1, and at no point is a decimal or fraction used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\;\;\;\;] + [\;\;\;\;] - [\;\;\;\;] * [\;\;\;\;] \div [\;\;\;\;] * [\;\;\;\;] = 50$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $[\;\;\;\;]$ are empty spaces and must be filled with numbers.&lt;/p&gt;&#xA;" OwnerUserId="12860" LastEditorDisplayName="user20" LastEditDate="2015-06-09T04:39:48.553" LastActivityDate="2015-06-30T13:58:59.453" Title="What numbers fill the blanks to satisfy the equation?" Tags="|logical-deduction|formation-of-numbers|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16109" PostTypeId="1" CreationDate="2015-06-08T15:37:25.250" Score="0" ViewCount="1186" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;There are 5 houses in 5 different colours. In each house lives a person of a different nationality. The 5 owners drink a certain type of beverage, smoke a certain brand of cigar, and keep a certain pet. Using the clues below can you determine who owns the fish?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Brit lives in a red house.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Swede keeps dogs as pets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Dane drinks tea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The green house is on the immediate left of the white house.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The green house owner drinks coffee.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The person who smokes Pall Mall rears birds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The owner of the yellow house smokes Dunhill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The man living in the house right in the middle drinks milk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Norwegian lives in the first house.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The man who smokes Blend lives next door to the one who keeps cats.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The man who keeps horses lives next door to the man who smokes Dunhill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The owner who smokes Blue Master drinks chocolate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The German smokes Prince.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Norwegian lives next to the blue house.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The man who smokes Blend has a neighbour who drinks water.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is called Einstein's puzzle. I am only posting it to increase the familiarity of this puzzle to PSE. PLEASE do not use the internet to solve this question.&lt;/p&gt;&#xA;" OwnerUserId="12860" LastEditorUserId="11297" LastEditDate="2015-06-08T16:42:45.203" LastActivityDate="2015-06-08T17:57:40.440" Title="Einstein's puzzle" Tags="|logical-deduction|logic-grid|" AnswerCount="1" CommentCount="3" ClosedDate="2015-06-08T16:43:26.137" ContentLicense="CC BY-SA 3.0" />
  <row Id="16110" PostTypeId="2" ParentId="16108" CreationDate="2015-06-08T15:40:24.160" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; [4] + [2] - [1] x [6] ÷ [3] x [5] = 50&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or, a cleaner version:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ((4 + 2) - 1) x ((6 / 3) x 5) = 50&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-06-08T15:40:24.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16111" PostTypeId="2" ParentId="16108" CreationDate="2015-06-08T15:48:54.783" Score="8" Body="&lt;p&gt;I was just a little slower than Bailey, but to add further detail:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You start by working out that the last number can only be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 5,2 or 1: 5 is more likely as it involves a smaller answer to using the other numbers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This makes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; [ ] + [ ] - [ ] x [ ] ÷ [ ] = 10&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You now want a small number to divide by, that isn't 1 - so this should either be 2 or 3 making&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; [ ] + [ ] - [ ] x [ ] = 20&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; [ ] + [ ] - [ ] x [ ] = 30&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With the 4 numbers and 3 sums remaining, it is impossible to make the former and are forced down the latter route to create:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4 + 2 - 1 x 6 / 3 x 5 = 50&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11150" LastActivityDate="2015-06-08T15:48:54.783" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16112" PostTypeId="2" ParentId="16106" CreationDate="2015-06-08T15:52:49.307" Score="15" Body="&lt;p&gt;There are two examples in Games Magazine's &lt;em&gt;The Book of Sense and Nonsense&lt;/em&gt; (at least I think that's where I saw them, I don't have a copy with me to double check).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What unusual word of seven letters has three U's?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Explain the meaning of this acronym: ETMOTA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2627" LastActivityDate="2015-06-08T15:52:49.307" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16113" PostTypeId="2" ParentId="16109" CreationDate="2015-06-08T15:58:53.080" Score="4" Body="&lt;p&gt;The German has the fish.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Norwegian |Dane   |Brit      |German |Swede&#xA;Yellow    |Blue   |Red       |Green  |White&#xA;Water     |Tea    |Milk      |Coffee |Chocolate&#xA;Dunhill   |Blend  |Pall Mall |Prince |Blue Master&#xA;Cats      |Horses |Birds     |Fish   |Dogs&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h1&gt;Explanation:&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;We know that the Norwegian lives in the first house, he lives next to the blue house, and the person living in the middle drinks milk.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Norwegian|      |         |      |&#xA;         |Blue  |         |      |&#xA;         |      |Milk     |      |&#xA;         |      |         |      |&#xA;         |      |         |      |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Next, green house must be to the left of the white house, and the person in the green house drinks coffee. So the green house must be fourth, and the white house fifth.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Norwegian|      |         |      |&#xA;         |Blue  |         |Green |White&#xA;         |      |Milk     |Coffee|&#xA;         |      |         |      |&#xA;         |      |         |      |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There's only two houses that we don't know the colors of: the first one and the third one, which must be red and yellow. The person in the red house is a Brit. He must be in the third house.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Norwegian|      |Brit     |      |&#xA;Yellow   |Blue  |Red      |Green |White&#xA;         |      |Milk     |Coffee|&#xA;         |      |         |      |&#xA;         |      |         |      |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The person in the yellow house smokes Dunhill, and there are horses next door.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Norwegian|      |Brit     |      |&#xA;Yellow   |Blue  |Red      |Green |White&#xA;         |      |Milk     |Coffee|&#xA;Dunhill  |      |         |      |&#xA;         |Horses|         |      |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The person who smokes Blend lives next to a person who drinks water, so he must be the second person.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Norwegian|      |Brit     |      |&#xA;Yellow   |Blue  |Red      |Green |White&#xA;Water    |      |Milk     |Coffee|&#xA;Dunhill  |Blend |         |      |&#xA;         |Horses|         |      |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The person who smokes Blue Master drinks chocolate, so he must be fifth person. The second person must drink tea and is the Dane.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Norwegian|Dane  |Brit     |      |&#xA;Yellow   |Blue  |Red      |Green |White&#xA;Water    |Tea   |Milk     |Coffee|Chocolate&#xA;Dunhill  |Blend |         |      |Blue Master&#xA;         |Horses|         |      |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The German smokes Prince, so he must be the fourth person. Therefore, the Swede is the fifth person and the Brit smokes Pall Mall.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Norwegian|Dane  |Brit     |German|Swede&#xA;Yellow   |Blue  |Red      |Green |White&#xA;Water    |Tea   |Milk     |Coffee|Chocolate&#xA;Dunhill  |Blend |Pall Mall|Prince|Blue Master&#xA;         |Horses|         |      |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The person who smokes Pall Mall has birds and the Swede has dogs.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Norwegian|Dane  |Brit     |German|Swede&#xA;Yellow   |Blue  |Red      |Green |White&#xA;Water    |Tea   |Milk     |Coffee|Chocolate&#xA;Dunhill  |Blend |Pall Mall|Prince|Blue Master&#xA;         |Horses|Birds    |      |Dogs&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The person who smokes Blend has cats next door. So the Norwegian has cats, and the German has fish.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Norwegian|Dane  |Brit     |German|Swede&#xA;Yellow   |Blue  |Red      |Green |White&#xA;Water    |Tea   |Milk     |Coffee|Chocolate&#xA;Dunhill  |Blend |Pall Mall|Prince|Blue Master&#xA;Cats     |Horses|Birds    |Fish  |Dogs&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="12109" LastEditorUserId="12109" LastEditDate="2015-06-08T17:57:40.440" LastActivityDate="2015-06-08T17:57:40.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16114" PostTypeId="2" ParentId="16107" CreationDate="2015-06-08T16:39:32.140" Score="1" Body="&lt;p&gt;I'm gonna state the obvious answer to get that out of the way:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He is a fisherman and just caught a shark and is putting it into a bag of ice to preserve it.  Being from Florida, this kind of occurrence would not be unusual (though a cooler of ice would be more likely).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10641" LastActivityDate="2015-06-08T16:39:32.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16115" PostTypeId="2" ParentId="16107" CreationDate="2015-06-08T17:05:01.670" Score="2" Body="&lt;p&gt;Perhaps you just witnessed someone purchase a shark figurine?  The clerk put the figurine in a bag and the person was happy about it.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2015-06-08T17:05:01.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16116" PostTypeId="1" AcceptedAnswerId="16146" CreationDate="2015-06-08T17:33:09.957" Score="10" ViewCount="1679" Body="&lt;p&gt;&lt;em&gt;I am not sure if there is a name associated with this game but I do not know of one.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was this 2 player game I had once played, and I am wondering if there is a winning strategy. The game is as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Two players both get 6 rocks, numbered 1-6.&#xA;  &lt;br&gt; Each rock number corresponds to its weight (6 is heavier than 5, is heavier than 4, etc)&#xA;  &lt;br&gt; Each number on the rock also corresponds to its points. (rock 6 = 6 points).&#xA;  &lt;br&gt; The game is played in rounds, 6 in total.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Each round,A random rock from the referee's (or computers) pile [similar to player rocks 1-6] is placed into the middle of the lever by the ref.&#xA;  &lt;br&gt;Each player then chooses one of their rocks to place onto the lever. (The players do not show their choice until they are revealed on the lever.)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The heavier rock chosen between the player wins the round, and the winning player receives the points for all the rocks on the lever. (Their number + random Ref number + opponents number).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The rocks used in a round are no longer able to be used in the next rounds. This includes the referee's rocks that go into the middle. (so if the ref uses rock 6, then the next round the rock in the middle must be random from 1-5)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If there is a tie, the rocks remain on the lever for the next round.&#xA;  &lt;br&gt; The player with the most points at the end of the game (6 rounds) wins!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Is there a possible winning strategy to this game? [Ties will be counted as wins]&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, is there a strategy which can give you the highest chance of winning?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume it can be used in either situation:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Both players are using the optimal strategy.&lt;/li&gt;&#xA;&lt;li&gt;Your opponent is not following the optimal strategy.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;If you were both given more or less rocks, would this change the situation? (i.e 1-10, or 1-4)?&lt;/b&gt;&lt;/p&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-06-10T15:17:02.080" LastActivityDate="2015-07-15T15:17:47.413" Title="Rocks and Lever (Game) - Goofspiel" Tags="|logical-deduction|strategy|probability|game|" AnswerCount="4" CommentCount="13" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16117" PostTypeId="2" ParentId="16116" CreationDate="2015-06-08T18:05:52.090" Score="1" Body="&lt;p&gt;Normally, you'd want to put a rock around the value of the ref rock. You don't want to waste a 6 on a 1 point ref rock. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, things do get interesting if there is a collision. Say the ref rock is 5 and you both put 5 out. As a tie, that means that 15 points go to the next round. Naturally, you'd both collide with 6.  Then 4. Then 3. Then 2. Then 1. Tie.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say instead you take the 5-5 with a 6. You have 16 points, and a 5 left. Presumably the opponent will take the ref rock worth 6 and you'd place a 1 (13 points). So, you're left with 5,4,3,2 and your opponent is left with 4,3,2,1.  You're likely to win. So, by covering their expected take, you were able to position a 6-1 loss in your favor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take another important ref rock, 6. Now, most of the time you'd use a 6 to cover it, but then the opponent could use a 1 to get future position. Well, then you could use a 2 to maintain your position. And so on and so forth. Likely end with 50-50 chance. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-06-08T18:05:52.090" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16119" PostTypeId="2" ParentId="16046" CreationDate="2015-06-08T19:58:12.957" Score="2" Body="&lt;p&gt;Reversing my answers for the other two questions (using $[]$ for the &quot;ceiling&quot; function):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lighter:&#xA;$$k = [\log_4 n]$$&#xA;except for $n=2$, which has no solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Approach: Put $4^{k-1}$ balls on each pan, set rest aside.  Weigh.  If one pan rises, it contains the light ball.  If no pan rises, the light ball is in the extra pile.  The largest pile is at most $4^{k-1}$ balls.  Continue with putting $1/4$ of the balls on each pan until you have a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the number of balls remaining after a weighing is not a multiple of four, you can add in eliminated balls (&quot;ringers&quot;) to even things out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heavier:&#xA;$$k \approx [\log_3 (n+2)]$$&#xA;except for $n=2,3$, which have no solution.  Also, edge cases are slightly off (see details below).&#xA;For $n=2$ or $n=3$, there is no way to determine one heavy ball using this type of scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Approach: put $[n/3]-1$ balls on each pan, which leaves you with 1 to 3 left over.  Add the extras to one pan, giving you two equal-number pans and one extra-number pan.  If the heavy ball is on one of the equal-number pans, the other equal-number pan will rise.  If neither rises, the heavy ball is on the extra-number pan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeat this process with the smaller number of balls until you are done.  For steps after the first, you have some balls that are known to be normal weight (I call these &quot;ringers&quot;).  You can use ringers  to make the numbers work out such that with $x$ remaining balls, you can put $[x/3]$ or fewer balls on each pan and add ringers to make the numbers work out so you have two equal pans and one extra pan. (This requires at most two ringers and you must have at least two ringers after the first step).  Thus, if you have $x$ candidates after the first weighing, you can complete the process in at most $[\log_3 x]$ more steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do a bit of math and you have&#xA;$$k = [\log_3 (n + 2)]$$&#xA;except:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n = 2,3$: Impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n = 3^i-3$ $(i&amp;gt;1)$: Requires $k=i+1$ steps.  $i.e. n = 6$ requires $k=3$ steps, even though $n=7$ only requires $k=2$ steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To illustrate the special case of $n=6$, if you put two balls on each pan, then you are guaranteed that two pans are each full of normal balls, so no pan will rise.  Thus, you can only put 1 ball each on two pans and 2-4 balls on the third pan.  If no pan rises, then the heavy one is one of the set of 4, and for $n=4$, you require 2 more weighings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This issue arises for any $n=3^i-3$: you put $n/3-1 = 3^{i-1}-2$ balls on two pans, leaving $3^{i-1}+1$ balls remaining.  If the heavy one is in the latter set, that will require $k=i+1$ more weighings.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="2328" LastEditDate="2015-06-08T22:28:54.700" LastActivityDate="2015-06-08T22:28:54.700" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16120" PostTypeId="2" ParentId="15319" CreationDate="2015-06-08T20:58:52.797" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/5uNNY.jpg&quot; alt=&quot;skelemingos&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Seek out a sunny picture -&lt;br&gt;&#xA;Remember you're on SE -&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The url for pictures uploaded on SE is &lt;a href=&quot;https://i.stack.imgur.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com&lt;/a&gt;. You would need the ID to identify the picture.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And on that sunny picture,&lt;br&gt;&#xA;Figure out what you can;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The ID has 5 letters/digits. It has something to do with &quot;sunny&quot;.&lt;br&gt;&#xA; The closest thing to turning &quot;sunny&quot; into figures (figure out) is turning &quot;s&quot; into 5.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are a less capital idea&lt;br&gt;&#xA;Than those who follow you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; u (you) is uncapitalized. The letters that follow are capital. So we get that the last few letters are &quot;uNNY&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Put that together. The url for the picture is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/5uNNY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/5uNNY.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What kind of birds appear?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Skelemingos!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12109" LastEditorUserId="12109" LastEditDate="2015-06-08T21:12:24.203" LastActivityDate="2015-06-08T21:12:24.203" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16121" PostTypeId="2" ParentId="16108" CreationDate="2015-06-08T21:34:23.107" Score="8" Body="&lt;p&gt;I &quot;cheated&quot; and wrote a Python script that used the Johnson-Trotter algorithm to permute the order of the numbers that are plugged into the equation.&#xA;My script spits out 6 possible solutions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3 + 6 - 4 x 5 ÷ 1 x 2&lt;br&gt;&#xA; 6 + 3 - 4 x 5 ÷ 1 x 2&lt;br&gt;&#xA; 6 + 3 - 4 x 2 ÷ 1 x 5&lt;br&gt;&#xA; 3 + 6 - 4 x 2 ÷ 1 x 5&lt;br&gt;&#xA; 4 + 2 - 1 x 6 ÷ 3 x 5&lt;br&gt;&#xA; 2 + 4 - 1 x 6 ÷ 3 x 5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though, given the extra conditions of the puzzle (no ÷1 and never dealing with decimals/fractions), you're left with 2 solutions because addition is commutative.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4 + 2 - 1 x 6 ÷ 3 x 5&lt;br&gt;&#xA; 2 + 4 - 1 x 6 ÷ 3 x 5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm sure the pen-and-paper method accomplished above is probably more satisfying to perform, though!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Python Source&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Play with the code &lt;a href=&quot;http://runnable.com/VXdoe9nVYnl16qFW/johnson-trotter-permuter-for-python&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just a &lt;a href=&quot;http://www.cut-the-knot.org/Curriculum/Combinatorics/JohnsonTrotter.shtml&quot; rel=&quot;nofollow&quot;&gt;Johnson-Trotter&lt;/a&gt; permutation class I wrote. It uses a &lt;code&gt;step&lt;/code&gt; function to go to the next permutation. I thought about making an iterator version, but was too lazy =/. It may be a little lengthy, but it can be used to permute any list of objects. I'm sure it could be improved and condensed; I just threw this together for this puzzle =P&lt;/p&gt;&#xA;&#xA;&lt;p&gt;jtpermute.py&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;class JohnsonTrotterPermute(object):&#xA;    objlist = []   # contains the objects given to the class&#xA;    nodelist = []  # contains the JTNode objects that wrap around each given object&#xA;    numnodes = 0   # the current number of nodes added (for numbering)&#xA;    steps = 1      # the number of iterations generated (number of steps)&#xA;&#xA;    def __init__(self, objlist):&#xA;        self.objlist = objlist&#xA;        for obj in self.objlist:&#xA;            self.nodelist.append(JTNode(self.numnodes, obj))&#xA;            self.numnodes += 1&#xA;&#xA;    # returns True if there exists at least one mobile JTNode&#xA;    def has_mobile(self):&#xA;        for i in range(len(self.nodelist)):&#xA;            if self.check_mobility(i):&#xA;                return True&#xA;        return False&#xA;&#xA;    # returns True if the JTNode at the given index is mobile&#xA;    def check_mobility(self, index):&#xA;        if index &amp;lt; 0 or index &amp;gt;= len(self.nodelist):&#xA;            return False&#xA;        if self.nodelist[index].direction == -1: # pointing left&#xA;            if index == 0: # on the left edge of list&#xA;                return False&#xA;            if self.nodelist[index].num &amp;gt; self.nodelist[index - 1].num:&#xA;                return True&#xA;        if self.nodelist[index].direction == 1:  # pointing right&#xA;            if index == len(self.nodelist) - 1:  # on the right edge of list&#xA;                return False&#xA;            if self.nodelist[index].num &amp;gt; self.nodelist[index + 1].num:&#xA;                return True&#xA;        return False&#xA;&#xA;    # iterates through the nodes and flips the direction of any node&#xA;    # whose .num is larger than the given number&#xA;    def flip_larger_than_num(self, num):&#xA;        for node in self.nodelist:&#xA;            if (node.num &amp;gt; num):&#xA;                node.flip()&#xA;&#xA;    # returns the index of the JTNode that has the largest .num value&#xA;    # and is mobile&#xA;    def get_largest_mobile_index(self):&#xA;        index = -1&#xA;        largest = -1&#xA;        for i in range(len(self.nodelist)):&#xA;            if self.nodelist[i].num &amp;gt; largest and self.check_mobility(i):&#xA;                largest = self.nodelist[i].num&#xA;                index = i&#xA;        return index&#xA;&#xA;    # swaps the node at the given index with the node that it's pointing&#xA;    # at. Returns True if the swap is successful, False if the swap fails&#xA;    # (e.g. the node is pointing left, but it's the left-most node)&#xA;    def move_node(self, index):&#xA;        node = self.nodelist[index]&#xA;        tempnode = None&#xA;        if node.direction == -1: # pointed left&#xA;            if index == 0: # on left edge of list&#xA;                return False&#xA;            tempnode = self.nodelist[index]&#xA;            self.nodelist[index] = self.nodelist[index - 1]&#xA;            self.nodelist[index - 1] = tempnode&#xA;            self.flip_larger_than_num(tempnode.num)&#xA;        if node.direction == 1: # pointed right&#xA;            if index == len(self.nodelist) - 1: # on right edge of list&#xA;                return False&#xA;            tempnode = self.nodelist[index]&#xA;            self.nodelist[index] = self.nodelist[index + 1]&#xA;            self.nodelist[index + 1] = tempnode&#xA;            self.flip_larger_than_num(tempnode.num)&#xA;        return True&#xA;&#xA;    # steps the list order to its next iteration&#xA;    def step(self):&#xA;        if self.has_mobile():&#xA;            largest_mobile = self.get_largest_mobile_index()&#xA;            self.move_node(largest_mobile)&#xA;            self.steps += 1&#xA;            return True&#xA;        return False&#xA;&#xA;    # returns the list of items that the user initially provided in the&#xA;    # order of the current iteration&#xA;    def get_items(self):&#xA;        returnlist = []&#xA;        for node in self.nodelist:&#xA;            returnlist.append(node.obj)&#xA;        return returnlist&#xA;&#xA;&#xA;&#xA;# a wrapper class for the list items the user will provide&#xA;class JTNode(object):&#xA;    obj = None      # the actual user-provided list item&#xA;    num = -1        # the number of the node&#xA;    direction = -1  # pointing direction (-1 -&amp;gt; left, 1 -&amp;gt; right)&#xA;&#xA;    def __init__(self, num, obj):&#xA;        self.obj = obj&#xA;        self.num = num&#xA;&#xA;    # flip the direction the node is pointing&#xA;    def flip(self):&#xA;        self.direction *= -1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;puzzle.py&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from jtpermute import *&#xA;&#xA;# computes the puzzle equation on the given list of integers&#xA;def do_maths(intlist):&#xA;    if len(intlist) &amp;lt; 6:&#xA;        return 0&#xA;    num = intlist[0]&#xA;    num += intlist[1]&#xA;    num -= intlist[2]&#xA;    num *= intlist[3]&#xA;    num /= intlist[4]&#xA;    num *= intlist[5]&#xA;    return num&#xA;&#xA;# prints the puzzle's function with the numbers plugged in&#xA;def print_maths(intlist):&#xA;    if len(intlist) &amp;lt; 6:&#xA;        return 0&#xA;    # a string builder of sorts&#xA;    strlist = []&#xA;    strlist.append(str(intlist[0]))&#xA;    strlist.append(&quot; + &quot;)&#xA;    strlist.append(str(intlist[1]))&#xA;    strlist.append(&quot; - &quot;)&#xA;    strlist.append(str(intlist[2]))&#xA;    strlist.append(&quot; * &quot;)&#xA;    strlist.append(str(intlist[3]))&#xA;    strlist.append(&quot; / &quot;)&#xA;    strlist.append(str(intlist[4]))&#xA;    strlist.append(&quot; * &quot;)&#xA;    strlist.append(str(intlist[5]))&#xA;    strlist.append(&quot; = &quot;)&#xA;    strlist.append(str(do_maths(intlist)))&#xA;    print &quot;&quot;.join(strlist)&#xA;&#xA;jtp = JohnsonTrotterPermute([1, 2, 3, 4, 5, 6])&#xA;# the equivalent of a do-while loop in Python. I wanted to put the&#xA;# step function in the loop conditional, but it had to be evaluated&#xA;# before the first .step call. When the step function returns False,&#xA;# it has reached the end of the permutations&#xA;while True:&#xA;    if do_maths(jtp.get_items()) == 50:&#xA;        print_maths(jtp.get_items())&#xA;    if jtp.step() == False:&#xA;        break&#xA;print &quot;Number of permutations: %d&quot; % jtp.steps&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Output of puzzle.py&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;3 + 6 - 4 * 5 / 1 * 2 = 50&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;6 + 3 - 4 * 5 / 1 * 2 = 50&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;6 + 3 - 4 * 2 / 1 * 5 = 50&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;3 + 6 - 4 * 2 / 1 * 5 = 50&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;4 + 2 - 1 * 6 / 3 * 5 = 50&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;2 + 4 - 1 * 6 / 3 * 5 = 50&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;Number of permutations: 720&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13053" LastEditorUserId="13053" LastEditDate="2015-06-30T13:58:59.453" LastActivityDate="2015-06-30T13:58:59.453" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16122" PostTypeId="1" AcceptedAnswerId="16148" CreationDate="2015-06-08T21:35:54.090" Score="10" ViewCount="305" Body="&lt;p&gt;Below is a word search puzzle presented without clues. Instead, you must use the additional rules (see example below) to work out what the words are. Finally, as a check to make sure you found the right words, the first letters of each word can be rearranged to make one last, related word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the third puzzle in a series. &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16087/cluelessly-searching-for-words&quot;&gt;Here's a 6x6&lt;/a&gt;, and &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16094/cluelessly-searching-for-words-ii&quot;&gt;here's a 7x7&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Rules&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The words can run in any direction (orthogonal or diagonal) and do not have to be in straight lines&lt;/li&gt;&#xA;&lt;li&gt;Words do not overlap and their paths do not cross&lt;/li&gt;&#xA;&lt;li&gt;All letters are part of some word&lt;/li&gt;&#xA;&lt;li&gt;All words are of the same length&lt;/li&gt;&#xA;&lt;li&gt;All words (including the final word) are related somehow&lt;/li&gt;&#xA;&lt;li&gt;All words (including the final word) have an English Wikipedia page, even if they are not English themselves&lt;/li&gt;&#xA;&lt;li&gt;The letters that make up the final word do not appear in any specific location on the grid&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Example&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Here's a short example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bAwdg.png&quot; alt=&quot;P O A W. E N N S. M U Y A. S T O D.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GRuBj.png&quot; alt=&quot;PONY EMUS TOAD SWAN&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The four words here are PONY, EMUS, TOAD and SWAN, which are all animals, and the first letter of each makes the word PETS.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Puzzle&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;What is the final word to be found in this grid of letters?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/shGvJ.png&quot; alt=&quot;R E L A S T I E. E N N E H F P F. G A S R I F A H. G A S W E T T O. S E N A A H R C. W A Y E L E G S. L A I G D D I E. A B R T H R I N.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Hint&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All the words are taken from one set of Wikipedia list pages&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7392" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-06-09T15:48:03.077" Title="Clueslessly searching for words III" Tags="|word-search|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16123" PostTypeId="1" AcceptedAnswerId="16126" CreationDate="2015-06-08T22:58:34.260" Score="0" ViewCount="275" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is a bet that most people take with the intention of losing it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="7308" LastEditDate="2015-06-08T23:58:14.213" LastActivityDate="2015-06-08T23:58:14.213" Title="A mysterious bet" Tags="|riddle|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16124" PostTypeId="2" ParentId="16123" CreationDate="2015-06-08T23:00:27.367" Score="4" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Insurance&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since you'd rather not get the money :)&lt;/p&gt;&#xA;" OwnerUserId="9208" LastEditorUserId="5373" LastEditDate="2015-06-08T23:02:04.310" LastActivityDate="2015-06-08T23:02:04.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16125" PostTypeId="2" ParentId="16123" CreationDate="2015-06-08T23:04:12.420" Score="2" Body="&lt;p&gt;Probably not the intended answer, but:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Short_(finance)&quot; rel=&quot;nofollow&quot;&gt;Shorting a stock&lt;/a&gt; is when you &quot;bet&quot; that the stock will decline in value.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9718" LastActivityDate="2015-06-08T23:04:12.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16126" PostTypeId="2" ParentId="16123" CreationDate="2015-06-08T23:07:35.143" Score="8" Body="&lt;p&gt;More specifically than TacoV's answer, maybe it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; life insurance. This could be seen as gambling on your own life, which you always lose in the end (hence &lt;em&gt;intention&lt;/em&gt; of losing?) but want to keep as long as possible (which could be seen as losing the 'bet' since you're losing money that way).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This specific answer is good because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; not &lt;em&gt;any&lt;/em&gt; kind of insurance will do: you might either want to get the money or keep the item you're insuring if it's less valuable than your life, depending on context and circumstances.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-06-08T23:07:35.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16127" PostTypeId="1" AcceptedAnswerId="16131" CreationDate="2015-06-08T23:57:50.023" Score="5" ViewCount="2021" Body="&lt;p&gt;I found a puzzle like this online, so I thought I'd try making my own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 4 dice, each with 6 different letters on them, such that 24 of the 26 letters of the English alphabet are represented (Q and Z are missing). Lets say I rolled these dice a number of times and was able to make the following words with the letters by re-arranging the dice in any order:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;CLAP&lt;/li&gt;&#xA;&lt;li&gt;WOKE&lt;/li&gt;&#xA;&lt;li&gt;CASH&lt;/li&gt;&#xA;&lt;li&gt;JUNK&lt;/li&gt;&#xA;&lt;li&gt;BITE&lt;/li&gt;&#xA;&lt;li&gt;GIFT&lt;/li&gt;&#xA;&lt;li&gt;LEWD&lt;/li&gt;&#xA;&lt;li&gt;GERM&lt;/li&gt;&#xA;&lt;li&gt;FOXY&lt;/li&gt;&#xA;&lt;li&gt;RUST&lt;/li&gt;&#xA;&lt;li&gt;LIMP&lt;/li&gt;&#xA;&lt;li&gt;BUOY&lt;/li&gt;&#xA;&lt;li&gt;JAKE&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What letters are on the faces of each die?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some notes: I know for a fact that this is solvable using logic, and I also know for a fact that there are multiple solutions. I'll accept any solution that is provably correct and accompanied by some explanation of your process.&lt;/p&gt;&#xA;" OwnerUserId="10861" LastActivityDate="2019-09-18T20:55:40.210" Title="Letter Dice Puzzle" Tags="|logical-deduction|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16128" PostTypeId="2" ParentId="16127" CreationDate="2015-06-09T00:59:35.020" Score="2" Body="&lt;p&gt;So a possible set of answers is:&#xA;1st die: B,G,L,S,X,K.&#xA;2nd die: E,U,F,P,H,V.&#xA;These two (1st and 2nd dice) should be a unique solution. In the other two dice, just the first two letters are unique. And the last for can be exchanged between the last two dice.&#xA;3rd dice: R,I,&lt;em&gt;J,C,O,D&lt;/em&gt;.&#xA;4th dice: M,T,&lt;em&gt;A,W,Y,N&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="13055" LastActivityDate="2015-06-09T00:59:35.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16129" PostTypeId="1" AcceptedAnswerId="16132" CreationDate="2015-06-09T01:14:22.890" Score="3" ViewCount="676" Body="&lt;p&gt;It was a dark and stormy night...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No wait, no it wasn't! It was a clear sunny day. When Brian had 3 of his old buddies from school over for a get together in Texas. They were all aged 35.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;His friends were named Louis, Gus (or &quot;Augustine&quot;) and Diana. Brian didn't know it, but this would be his last day on the face of the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brian did however know the history behind his friends. He knew which one of them was a murderer. He had for years. He only considerered the possibility of him getting killed last minute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He wrote the following letter:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dear whomever it may concern,  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This is very important. It's become the centre of my attention that I could die on this very day, 8/6/2015. I am leaving this message in hopes to tell you who it is but in all the current emotions I can't remember his name.&lt;br&gt;&#xA;  I apologise if this doesn't work out, but I'm trying my best. Be sure to analyse this letter very carefully. Now just in case I will be travelling to a new world, goodbye.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using all the given information in this puzzle, what is the most accurate guess at the murderer you can possibly make?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="5373" LastEditDate="2018-06-02T15:07:24.537" LastActivityDate="2018-06-02T15:07:24.537" Title="Murderer in the house!" Tags="|steganography|mystery|" AnswerCount="3" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16130" PostTypeId="2" ParentId="16049" CreationDate="2015-06-09T01:14:55.467" Score="11" Body="&lt;p&gt;This question &lt;a href=&quot;https://mathoverflow.net/q/41939/59232&quot;&gt;has been asked on Math Overflow&lt;/a&gt;. A beautiful answer was given by Ori Gurel-Gurevich, which I reproduce below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the balls and colors are both numbered $1,\dots,n$, with ball $i$ initially colored $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $P(i,j)$ denote the action of painting ball $i$ the color of ball $j$. In this process, we perform a sequence of actions $P(i_1,j_1),P(i_2,j_2),\dots$, where $(i_k,j_k)$ is randomly chosen from the pairs where $i_k\neq j_k$, until all balls are the same color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a different process. Let $C(i,j)$ denote the action of taking &lt;em&gt;every&lt;/em&gt; ball whose color is $i$ and painting it the color $j$. Perform the actions $C(i_1,j_1),C(i_2,j_2),\dots,$ with $i_k,j_k$ chosen randomly as before, until all balls are the same color. Call this the &quot;new process&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key to simplifying this problem is this brilliant observation: performing the actions&#xA;$$&#xA;P(i_1,j_1),P(i_2,j_2),\dots, P(i_T,j_T)&#xA;$$&#xA;to the initial setup &lt;strong&gt;has the same effect as&lt;/strong&gt; performing the actions&#xA;$$&#xA;C(i_T,j_T),C(i_{T-1},j_{T-1}),\dots, C(i_1,j_1)&#xA;$$&#xA;to the initial setup. In words, reversing the order and replacing $P$ with $C$ does not change the resulting coloring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is somewhat difficult to explain why this is true (anyone who has an intuitive explanation should feel free to edit it in), but this example should convince you. Suppose have four balls, so the initial state is [1,2,3,4]. When we perform the actions $P(1,2), P(2,4), P(3,1), P(4,1)$, the resulting color distribution is [2,4,2,2], as shown below:&#xA;$$&#xA;[1,2,3,4]\stackrel{P(1,2)}{\to}[2,2,3,4]\stackrel{P(2,4)}{\to}[2,4,3,4]\stackrel{P(3,1)}{\to}[2,4,2,4]\stackrel{P(4,1)}{\to}[2,4,2,2]&#xA;$$&#xA;When the actions $C(4,1),C(3,1),C(2,4),C(1,2)$ are performed, the result is also [2,4,2,2], as claimed:&#xA;$$&#xA;[1,2,3,4]\stackrel{C(4,1)}{\to}[1,2,3,1]\stackrel{C(3,1)}{\to}[1,2,1,1]\stackrel{C(2,4)}{\to}[1,4,1,1]\stackrel{C(1,2)}{\to}[2,4,2,2]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The symmetry implied by the &lt;strong&gt;bold&lt;/strong&gt; statement means that the original process has the same average runtime as the new process, so we may analyze the expected runtime of the new process instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out to be much easier to analyze the new process. If there are $k$ colors present at some point, then after performing a $C(i,j)$ action, the number of colors present will remain the same or decrease by $1$. It decreases by 1 with probability $\frac{k(k-1)}{n(n-1)}$, since this is the probability that both $i$ and $j$ are a color which was already present. Thus, the time it takes to decrease from $k$ colors to $k-1$ is a geometric random varaible, with expected time $\frac{n(n-1)}{k(k-1)}$, so the expected time it takes to go from $n$ colors to $1$ color is&#xA;$$&#xA;\sum_{k=2}^n  \frac{n(n-1)}{k(k-1)}=n(n-1)\sum_{k=2}^n\frac1{k-1}-\frac1{k}=n(n-1)\left(1-\frac1n\right)=(n-1)^2&#xA;$$&#xA;The first step follows since $\frac1{k(k-1)}=\frac1{k-1}-\frac1{k}$. The second follows since the sum is telescoping. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the expected time is $(n-1)^2$.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:31.357" LastActivityDate="2015-06-09T01:24:30.583" CommentCount="3" CommunityOwnedDate="2015-06-09T01:14:55.467" ContentLicense="CC BY-SA 3.0" />
  <row Id="16131" PostTypeId="2" ParentId="16127" CreationDate="2015-06-09T01:49:12.647" Score="5" Body="&lt;p&gt;5,6 mean BE are found on the same pair of dice as GF.&#xA;8 means E is not on the same die as G. EF and BG go together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FX=BU (9,12)&#xA;But we know F is not with B, so F is with U, B with X.&#xA;EFU, BGX so far&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AE=UN (4,13)&#xA;E with U means A with N&#xA;EFU, BGX, AN&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at those letters which cannot be on the same die as E, we now have ABDGIJKLMNORSTWXY which leaves CEFHPUV. P is the only one of these present in (11), so P is correct. (1) then rules out C, so we have&#xA;EFUHPV, BGX, AN&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Removing the fully known die from the words:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;CLA&#xA;WOK&#xA;CAS&#xA;JNK&#xA;BIT&#xA;GIT&#xA;LWD&#xA;GRM&#xA;OXY&#xA;RST&#xA;LIM&#xA;BOY&#xA;JAK&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It is clear that L=S. G is not on the same die as I or M (GIT, GRM), so it must be with L (LIM).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EFUHPV, BGXLS, AN.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;K is the only letter which appears everywhere none of the other 5 (BGXLS) are present, therefore it must be the final letter from that die.&#xA;EFUHPV, BGXLSK, AN.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Removing the newly completed die and eliminating duplicates:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;CA&#xA;WO&#xA;JN&#xA;IT&#xA;WD&#xA;RM&#xA;OY&#xA;RT&#xA;IM&#xA;JA&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We have C=J, A=N, W=Y, O=D, T=M, R=I.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third and fourth dies may each have either CJ, AN; either WY, OD; either TM, RI.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one possible assignment is &#xA;EFUHPV, BGXLSK, CJWYTM, ANODRI&lt;/p&gt;&#xA;" OwnerUserId="3987" LastActivityDate="2015-06-09T01:49:12.647" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16132" PostTypeId="2" ParentId="16129" CreationDate="2015-06-09T02:30:41.753" Score="5" Body="&lt;p&gt;My guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gus (Augustine)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasons:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Centre is misspelled to draw attention to it. Center fits as &quot;Gus&quot; is a nickname from Au&lt;strong&gt;gus&lt;/strong&gt;tine and it is also the middle name of the namesake Augustine of Hippo (&lt;em&gt;Aurelius &lt;strong&gt;Augustinus&lt;/strong&gt; Hipponensis&lt;/em&gt;). &lt;br/&gt;Second, the switch of month and day (today is 6/8/2015, as the OP states they are in Texas should use American date) points to &lt;strong&gt;August&lt;/strong&gt; when corrected. &lt;br/&gt;Then &lt;em&gt;can't remember &lt;strong&gt;his&lt;/strong&gt; name&lt;/em&gt; means it has to be Louis or Gus. &lt;br /&gt;Also, though this may have nothing to do with it, I like &lt;em&gt;St. Augustine &lt;strong&gt;New World&lt;/strong&gt; Gin&lt;/em&gt;&lt;br /&gt;&lt;strong&gt;Update&lt;/strong&gt;: Also &lt;em&gt;apologise&lt;/em&gt; provide two possible clues: one it starts with a, ends with e, and is the same number of letters as Augustine and Augustine of Hippo is famous for an apology (here meaning reasoning for belief or reasoned defense of faith). &lt;br /&gt;And I really just took &lt;em&gt;analyse&lt;/em&gt; as a hint that you needed to analyze the miss-spellings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11239" LastEditorUserId="11239" LastEditDate="2015-06-09T03:18:49.753" LastActivityDate="2015-06-09T03:18:49.753" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16133" PostTypeId="1" AcceptedAnswerId="16135" CreationDate="2015-06-09T03:21:42.420" Score="12" ViewCount="6169" Body="&lt;p&gt;In the game of Chomp, two players start with a chocolate bar, which is scored into an $a$ by $b$ array of squares (where $a\cdot b&amp;gt;1$). The square in the lower left is poisoned. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The players alternate turns. On their turn, a player chooses a square, then eats it, along with all of the squares which are either above it, to its right, or both. This continues until someone eats the poisoned square, wherein the non-poisoned player wins. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example of how the first two turns might go, where the &quot;x&quot; indicates the square that player chose:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mvs6j.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Part 1:&lt;/strong&gt; Show that the first player can win if she plays perfectly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What if you have an $a\times\infty$ chocolate bar? This means that there are $a$ rows of squares, where each row continues infinitely to the right, and the left ends of the rows are all aligned. This grid still has a lower left corner, which is still poisoned. Some moves will require players to eat infinitely many squares, which we assume is possible.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Part 2:&lt;/strong&gt; For which values of $a$ does the first player win on an $a\times\infty$ bar? Why?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-06-09T06:18:37.080" LastActivityDate="2015-06-09T20:43:52.843" Title="Chomp (Game with a Poisoned Chocolate Bar)" Tags="|game|" AnswerCount="3" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16134" PostTypeId="2" ParentId="16133" CreationDate="2015-06-09T05:27:56.887" Score="1" Body="&lt;p&gt;For the purposes of this answer, I'll use &quot;X.Y&quot; to refer to a square X units to the right of, and Y units above, the bottom-left (poisoned) square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll give my answer in the terms of various positions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Position 0 (loss):&lt;/strong&gt; only 0.0 (the poisoned square) remains - the active player must choose it, and thus loses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Position 1 (win):&lt;/strong&gt; only the first row or column of squares remains - the active player chooses 1.0 or 0.1, giving the other player Position 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any board with $a$ = 1 or $b$ = 1 is an instance of Position 1, so the first player can win these cases as described above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Position 2.1 (loss):&lt;/strong&gt; only 0.0, 1.0, and 0.1 remain - the active player must choose 1.0 or 0.1, giving the other player Position 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Position 2.2 (loss):&lt;/strong&gt; only 0.0, 1.0, 2.0, 0.1, and 0.2 remain - the active player must choose 1.0 or 0.1, giving the other player Position 1, or she must choose 2.0 or 0.2, allowing the other player to choose the other, leading to Position 2.1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Position 2.x (loss):&lt;/strong&gt; only 0.0 and an equal number of squares in the first row and first column remain. The fact that this is a losing position is provable via induction, using the explanations from Positions 2.1 and 2.2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Position 3 (win):&lt;/strong&gt; only 0.0 and an &lt;em&gt;unequal&lt;/em&gt; number of squares in the first row and first column remain - the active player chooses the square that shortens the first row or first column so that they're equal in size, leading to Position 2.x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Position 4 (win):&lt;/strong&gt; Position 2.x, plus any other squares not in the first row or first column - the active player chooses 1.1, giving the other player Position 2.x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any board with $a$ = $b$ is an instance of Position 4, so the first player can win these cases as described above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hm, that's as far as I can see, for now. I know how I'd win a board with $a$ = 2, $b$ = 3 (pick the top-right corner, 2.1), but I'm having trouble proving the general case.&lt;/p&gt;&#xA;" OwnerUserId="13056" LastActivityDate="2015-06-09T05:27:56.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16135" PostTypeId="2" ParentId="16133" CreationDate="2015-06-09T06:06:44.353" Score="9" Body="&lt;p&gt;&lt;strong&gt;Part 1:&lt;/strong&gt; The first player wins on a rectangular board (of more than one square).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is a classic nonconstructive &lt;a href=&quot;http://en.wikipedia.org/wiki/Strategy-stealing_argument&quot;&gt;strategy-stealing argument&lt;/a&gt;. Say to the contrary it was a second-player win. Then, the first player could &quot;stall&quot; by eating the single square in the corner, effectively becoming the second player and winning. Any move the opponent makes in response makes would have eaten this corner anyway, and so looks just like a first move, and so have a winning response to it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part 2:&lt;/strong&gt; Who wins on $a \times \infty $?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $a=1$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first player wins by eating all but the poisoned square.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For $a=2$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The second player wins. By part 1, a player loses if they ever create a rectangle (except of just the poisoned square). So, the only potential non-losing starting move is to eat on the second row, leaving $n$ squares. To this, the second player leaves $n+1$ squares in the first row. The first player can never eat from the first row without creating a rectangle, and to any other move, the second player maintains the same shape with smaller $n$. Eventually, $n$ becomes $0$, and the first player must eat the poisoned square.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For $a&amp;gt;2$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first player wins by removing all but $2$ rows. Note that this observation would let us know a priori that at most one value of $a$ gives a second player win, since the first player can transform the bigger one into the smaller one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-06-09T19:19:45.893" LastActivityDate="2015-06-09T19:19:45.893" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16136" PostTypeId="2" ParentId="16129" CreationDate="2015-06-09T07:18:51.583" Score="1" Body="&lt;p&gt;I would say the murderer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Louis&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Brian is informing Gus (as reference to the date (8(August)/6/2015). &quot;...I could die on this very day, 8/6/2015. I am leaving this message in hopes to tell you ...&quot;.&lt;br&gt;&#xA;This means Brian is informing about the other male character (remember his name...) who is Louis.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13060" LastEditorUserId="10281" LastEditDate="2015-06-09T10:08:01.550" LastActivityDate="2015-06-09T10:08:01.550" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16137" PostTypeId="2" ParentId="16129" CreationDate="2015-06-09T07:25:01.570" Score="2" Body="&lt;p&gt;Wild guess here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The killer is Diana&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Brian &quot;can't remember his name&quot;, because the killer is not a &quot;he&quot;. So he has to &quot;remember her name&quot;. Also two Dianas are born August 6.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alternatively:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Obviously Gus, as OP spend extra time describing him (giving both name and nickname). Also the month in the date of the letter is August.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastActivityDate="2015-06-09T07:25:01.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16138" PostTypeId="1" AcceptedAnswerId="16139" CreationDate="2015-06-09T08:45:03.697" Score="6" ViewCount="1455" Body="&lt;p&gt;I've been trying to write some emails to my friends, but something is going horribly wrong! It's as though the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/13196/game-against-the-devil&quot;&gt;devil&lt;/a&gt; himself has hold of me: everything I type comes out as gibberish! The first email I wrote looked like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;\ehsy dhsll er fo yhid errkrnf\/ \lry'd ho domrehrtr inyrtrdyinh\2&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The second one came out like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;U fauk ti cinoregebd tiyr kast enauk, Wgt&gt; Nt qyestuib us wgt, Kiij deeokt,&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The third one was worst of all!&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;\drr tiy ib \ygytdfst\2 \giorgykkt er;kk skk vr ygrtr ygud yunr,&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Can you help me to decipher these emails so that my friends don't think I'm a raving lunatic? (Well, I &lt;em&gt;am&lt;/em&gt; a raving lunatic, but I wasn't trying to type gibberish just now!)&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-06-09T17:33:51.130" Title="Decipher the crazy emails!" Tags="|cipher|enigmatic-puzzle|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16139" PostTypeId="2" ParentId="16138" CreationDate="2015-06-09T09:13:15.247" Score="12" Body="&lt;p&gt;&lt;strong&gt;First message:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   What shall we do this weekend? Let's go somewhere interesting! &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;How to get it:&lt;/em&gt; move left-hand keys one place to the left and leave right-hand keys as they are to decrypt&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second message:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; I fail to comprehend your last email. Why, my question is why. look deeply.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;How to get it:&lt;/em&gt; move right-hand keys one place to the right and leave left-hand keys as they are to decrypt&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Third message:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; See you on Thursday! Hopefully we'll all be there this time.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;How to get it:&lt;/em&gt; move left-hand keys one place to the left and right-hand keys one place to the right to decrypt&lt;/p&gt;&#xA;" OwnerUserId="13061" LastEditorUserId="5373" LastEditDate="2015-06-09T17:33:51.130" LastActivityDate="2015-06-09T17:33:51.130" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="16140" PostTypeId="1" AcceptedAnswerId="16141" CreationDate="2015-06-09T10:33:44.087" Score="3" ViewCount="230" Body="&lt;pre&gt;&lt;code&gt;00 03 18 36 54 59 71 72 92 99 84 70 70 61 42 42 28 13 33 33 36 37 58 77 78 98 89 74 60&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The number sequence represents a phrase, containing only lower-case letters and spaces (up to 27 unique characters).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Usage of 2-digit numbering was for aesthetics, it doesn't require knowledge of hexadecimal or anything similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Duplicates have meaning.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The length of the phrase is equal to the length of the sequence, minus one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the phrase?&lt;/p&gt;&#xA;" OwnerUserId="13062" LastActivityDate="2015-06-09T12:27:17.430" Title="What does this number sequence mean?" Tags="|number-sequence|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16141" PostTypeId="2" ParentId="16140" CreationDate="2015-06-09T11:26:07.907" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;CORRELATION IS NOT CAUSATION&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation: The increment from one character in the sequence to the next gives the dictionary value for every letter of the answer. The first character in the sequence is 00 (null) as a starting point - hence the phrase length equals sequence length minus one (hint 2). Repeated characters yield a zero increment, which can be interpreted as a blank space (hint 1).&lt;/p&gt;&#xA;" OwnerUserId="13064" LastEditorUserId="10400" LastEditDate="2015-06-09T12:27:17.430" LastActivityDate="2015-06-09T12:27:17.430" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16142" PostTypeId="1" AcceptedAnswerId="16812" CreationDate="2015-06-09T12:05:57.810" Score="9" ViewCount="1876" Body="&lt;p&gt;You'll need to be a serious fan of these puzzles to find over 300 of the things hidden in this jumble. There are no clues as this is test of knowledge as well as determination. Once you get started it's easy enough to find a list of what you are searching for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When all the parts are spread before you on the table you will know each by number. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ba61T.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you would like an alternative grid with the correct spelling of Aluminium and copyable text:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PtIrt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i1u105muinehtuRCm8S7muidanaVmuiboiN3N&#xA;al7mPbs5KrmuillahT80yDnegyxO0L711Bae6&#xA;NeTum31u1B57r112mMyBO5m352019muoC89nd&#xA;rkmiu6mur1iu313utro85muinatiTzut3T6iG&#xA;Bcuminu6Rofsn3iHur9mCuimuidibuRieFld8&#xA;Siidpoi55lhimmocomTumigm69m64Fm6dtNo4&#xA;mNnaotd3uTTplurnBumiudrgu44u7u2bTniI5&#xA;umeCrpoS2m1osetkmiuviioMSienitatsAIu4&#xA;iulHuyh2puH1MoHhutnocrb1n1smLdP921Umm&#xA;niedErRBmi391ghIidirnIa0e4yobso193SuN&#xA;ocSlAKRhurnoneXPnatee2e3gdnDrLuS9inib&#xA;timomuinietsniEmitalr7SZomo40p8Ul5enN&#xA;uruG6eF4nnR7OlmumslFwA8erPgmm2siEetfd&#xA;leim103mot26AumnumPPassStyruuTcyrnsaF&#xA;Pmru3mOuli57imuelrCFLaH2inAiio79DigHm&#xA;pAtinusioe2cSuidAaRar523NomnnM406rndu&#xA;uenhoi9rPMi4nibb0DgPlP611muei3o16ouMi&#xA;Unutbc5uCnu1orry2lmm1i391iigtU58muTen&#xA;HinirlCCrRP1rael68upm1fB0trtcmmLulSnu&#xA;fmULaarem2hCIBToaiimuM3o1nonAuumiFAit&#xA;Io1sCCpmu9nealCMtGmui23UrAhe4iiubr8rp&#xA;mr0mDo1uiesenagnaMsiduLu0nTo0nlire3oe&#xA;uB6uC1Wiln801iemmmOtrCGo5miR1aetevmlN&#xA;i36i7stleo10MpuNuumeoeu8HuvuVrbptluh4&#xA;n77r0Cllkd21nigmiiunfsm90i2umUoetiiC2&#xA;oLma19ayra8ubAmLnhihrEuC7vA4SmNsYSs94&#xA;crum91breRnr57uvitdceSiUae45iuPn85e5L&#xA;rmia00oeBUE935imtenehmmuileH1idu1mnr8&#xA;iumS10CBpN4817tucmaTturtheXcmtmn7ugeR&#xA;ZiyJmuinilodaGniaoc7uie99diBucuUmiape&#xA;bddmuissatoPRboltrScRrF28ne8iolBurMpm&#xA;SaoE46muinbuD4rloPNSAe63ee7fsnaiiuPou&#xA;NleumuiseaCPo7taro6Y9CFsdM7Rsut8llaCi&#xA;elNmuinamreG14SGP5b11mr01171ann2ul6Ar&#xA;oaKmuicnarFmuirhoBf14AdaeL8LHUa1heI6t&#xA;nPCu72ZnegordyH62C8Y5munahtnaLTTTTr4t&#xA;RaRadiummuiromreviL4ZincTdCmuimorhC6Y&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SPOILER:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The names, numbers, and symbols of each element is included. Some overlap completely.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="11297" LastEditDate="2015-06-10T16:33:58.133" LastActivityDate="2015-06-26T14:01:56.523" Title="45x45 Word Search" Tags="|word-search|science|" AnswerCount="2" CommentCount="12" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16143" PostTypeId="2" ParentId="16127" CreationDate="2015-06-09T12:49:27.687" Score="0" Body="&lt;p&gt;I find, there are 4 possibilities......&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;! EFHPUV and BGKLSX are unique.&#xA;  Remained 2 dice can be&#xA;   1. ANMTWY &amp;amp; CJIRDO&#xA;   2. ANMTDO &amp;amp; CJIRWY&#xA;   3. ANIRWY &amp;amp; CJMTDO&#xA;   4. ANIRDO &amp;amp; CJMTWY.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:-&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;! I see the frequency of E is 5, so I start with E.. After analysing I get EFHPU might be in one dice(as there is 0 frequency for V and all remained alphabets was likeable to be in any other dice, Do Its clear that V is with 'EFHPU')&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;! Now we see (1) and (3) roll, LS and C is in different dice... Then see (5)&amp;amp;(6) BG is in same dice.. Following same logic, see there are 3 groups BGX, MT and IR must be in different dice. But MT and IR can't be in 'LS' dice... So BGX is in. We see WY and DO must be in different group... So from (2) roll K is in dice 'LSBGX'......&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;! finally we have 2 unique dice and 3 pairs &#xA;   1. AN &amp;amp; CJ&#xA;   2. MT &amp;amp; IR&#xA;   3. WY &amp;amp; DO must be in different group, So gives 4 possibilities.&#xA;  !&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11306" LastActivityDate="2015-06-09T12:49:27.687" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16144" PostTypeId="2" ParentId="16055" CreationDate="2015-06-09T12:49:43.887" Score="0" Body="&lt;p&gt;Alice should &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; still press&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; She is conflating the probability of her personal involvement with the probability of the outcome. She is involved 100% of the time when they both press to lose \$4000, but she is only involved 50% of the time when one of them presses to gain \$5000. This disparity is more obvious if you split the gains and losses and make Alice and Bob strangers, which turns the problem into a prisoners dilemma; Alice would gain more personally not pressing while Bob presses, however as they are cooperating, both pressing works overall to their benefit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Her faulty math can be fixed by including the possibility of Bob pressing alone:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $P($Alice's Involvement$) \cdot EV($Alice's Calculation$) + P($Bob's involvement without Alice$) \cdot EV($Bob presses alone$)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $3/4 \cdot -\$1000 + 1/4 \cdot \$5000 = \$500$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10058" LastEditorUserId="10058" LastEditDate="2015-06-09T13:02:15.807" LastActivityDate="2015-06-09T13:02:15.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16146" PostTypeId="2" ParentId="16116" CreationDate="2015-06-09T13:23:31.167" Score="2" Body="&lt;p&gt;Thanks to &lt;em&gt;Meelo&lt;/em&gt; I was able to find the similar game to which this is a variant (and get this answer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One game is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Goofspiel&quot; rel=&quot;nofollow&quot;&gt;Goofspiel&lt;/a&gt;(which is considered as a Game of Pure Strategy [GOPS]), In which 2 or more players are given 1 suit from a deck of cards each, and a 'prize' suit would be used for the intermediary. The prize suit is shuffled and one of the cards is placed between the players, and the players choose one card of their own suit to play. The highest card wins the cards 'prize' card that was placed in the middle, and the cards used by the players are discarded. This repeats until all the cards have been used. Points are defined by the card values (Ace - 1, King - 13). There are variants for how draws are dealt with (splitting, ignoring, 'roll over', etc)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the Wiki page: &quot;Any pure strategy in this game has a simple counter-strategy&quot;, which demonstrates that :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;A) The number of initial options each player has (&gt;1) does not effect possible strategies.&lt;/b&gt; (i.e 6 rocks or 13 cards)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;B) Any 'optimal' strategy that is used can be countered (&lt;em&gt;if the opponent knows your strategy&lt;/em&gt;).&lt;/b&gt; If the opponent does not know your strategy (or psychology), then the winning outcome would be determined by the differences in your strategies as well as randomness of the 'prize' card (or referee rock) [Assuming both players stuck to their initial algorithms/strategies and that their algorithm has finite states].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, the only way a player is able to have a 'better' strategy is to know his opponents strategy. This is why GOPS can be considered to be more on psychology of your opponent rather than pure tactics and predefined decisions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar to &lt;em&gt;JonTheMon's&lt;/em&gt; answer, a mathematician Sheldon Ross considered the case when one player plays his cards randomly, to determine the &lt;em&gt;best&lt;/em&gt; strategy that the other player should use. Using a proof by induction on the number of cards, Ross showed that the optimal strategy for the non-randomizing player is to match the upturned card, i.e. if the upturned card is the Jack, he should play his Jack, etc. In this case, the expected final score is 59½ - 31½, for a 28-point win.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Source: &lt;a href=&quot;http://en.wikipedia.org/wiki/Goofspiel&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Goofspiel&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Interesting side note:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is very interesting to note about GOPS, that it can relate to the micro-trading algorithms used in the stock markets today! These algorithms can work based on the assumptions of other traders/algorithms to try and best benefit themselves. In this scenario the price of option X would be the 'referee' in the middle, however in this case the price can also be influenced by the player!&lt;/p&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-07-15T15:17:47.413" LastActivityDate="2015-07-15T15:17:47.413" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16147" PostTypeId="1" AcceptedAnswerId="16420" CreationDate="2015-06-09T13:51:44.860" Score="17" ViewCount="1100" Body="&lt;p&gt;--storyline--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recently, an ancient tomb of unknown origin was discovered during the deep-earth mining operations of 2035. Inside, there was a mysterious metallic box with engravings on each face, as well as a large carving on the back wall using what appears to be the same symbols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whatever the box is, tests have shown it to be impervious to all attempts to open it, scan it, or otherwise determine it's purpose or origin. As such, the task instead fell on the hands of those adept at deciphering unknown messages, which is where you all come in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;--task--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the shown images of the faces of the cube and wall carving, work out how to open the box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clues will be provided daily to try and nudge people in the right direction, unless that would make it too obvious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Preferably discuss theories on what you think it could be, along with information you discover and any links to existing media you find (in general, not SE specific). I would very much like to read over how you dissect it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also suck at art stuff, so the images will in fact be a black-and-white pixel-art image for each face and the wall carving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some glyphs are used for more than one thing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Glyphs do not lose their meaning if taken away from the others&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Maybe the ones who wrote the message also have dogs that lie around all day?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No hint today (#4), first 3 should allow you to decode the messages fully.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TQBFJOTLD&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The name is an important entity in a video game which shares these glyphs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;--images--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wall carving:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2Upbz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Faces:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GMC32.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="13062" LastEditorUserId="13062" LastEditDate="2015-06-16T13:47:29.353" LastActivityDate="2015-06-17T14:33:20.827" Title="Messages of unknown origin" Tags="|cipher|visual|" AnswerCount="2" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="16148" PostTypeId="2" ParentId="16122" CreationDate="2015-06-09T14:29:55.313" Score="12" Body="&lt;p&gt;The solution is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/ZaHW4.png&quot; alt=&quot;SolvedPuzzle&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Tiffani_Thiessen&quot; rel=&quot;nofollow noreferrer&quot;&gt;THIESSEN&lt;/a&gt; - Tiffani (b. 1974), is an American actress&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Anne_Hathaway&quot; rel=&quot;nofollow noreferrer&quot;&gt;HATHAWAY&lt;/a&gt; - Anne (b. 1982), is an American actress, singer, and producer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Florence_Eldridge&quot; rel=&quot;nofollow noreferrer&quot;&gt;ELDRIDGE&lt;/a&gt; - Florence (1901 - 1988), was an American actress.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Meredith_Salenger&quot; rel=&quot;nofollow noreferrer&quot;&gt;SALENGER&lt;/a&gt; - Meredith (b. 1970),  is an American actress.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Michelle_Pfeiffer&quot; rel=&quot;nofollow noreferrer&quot;&gt;PFEIFFER&lt;/a&gt; - Michelle (b. 1958),  is an American actress and singer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Frieda_Inescort&quot; rel=&quot;nofollow noreferrer&quot;&gt;INESCORT&lt;/a&gt; - Frieda (1901 - 1976), was a Scottish-born actress.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;7&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Lola_Albright&quot; rel=&quot;nofollow noreferrer&quot;&gt;ALBRIGHT&lt;/a&gt; - Lola (b. 1924), is an American singer and actress&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;8&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Masami_Nagasawa&quot; rel=&quot;nofollow noreferrer&quot;&gt;NAGASAWA&lt;/a&gt; - Masami (b. 1987), is a Japanese actress.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The final, related word is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; THESPIAN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They are all&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_female_movie_actors_by_name&quot; rel=&quot;nofollow noreferrer&quot;&gt;female movie actors&lt;/a&gt; (thespians)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All text quoted from the relevant Wikipedia articles.&lt;/p&gt;&#xA;" OwnerUserId="12277" LastEditorUserId="11475" LastEditDate="2015-06-09T15:48:03.077" LastActivityDate="2015-06-09T15:48:03.077" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16149" PostTypeId="2" ParentId="16116" CreationDate="2015-06-09T15:12:48.953" Score="0" Body="&lt;p&gt;I would place them in order 1, 2, 3, 4, 5, 6. You will lose some but guaranteed at least 3 (if one was a tie.)&lt;/p&gt;&#xA;" OwnerUserId="3997" LastActivityDate="2015-06-09T15:12:48.953" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16150" PostTypeId="2" ParentId="16142" CreationDate="2015-06-09T15:40:32.450" Score="11" Body="&lt;p&gt;I have found all of the&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Elements&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Up to 118. I presume no more are present. Not sure where the &quot;300&quot; comes from, perhaps there's also something else hidden in the puzzle?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/98EpI.png&quot; alt=&quot;Here&amp;#39;s the solution&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note: colours are meaningless in my solution pic.&lt;/p&gt;&#xA;" OwnerUserId="9289" LastActivityDate="2015-06-09T15:40:32.450" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="16151" PostTypeId="2" ParentId="5483" CreationDate="2015-06-09T17:23:51.947" Score="2" Body="&lt;p&gt;11.let those notes fly hot and fast&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Capriccio&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;12.Ballet, opera anyone? I am your host.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Maestro?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;13.Where else would we put the barbeque and the deck chairs?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Patio&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;14.a bounder, but at least a seductive one&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lothario &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;15.bought it at a novelty store&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Curio&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;16.with some milk for dipping&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Oreo&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;17.a feast for the ears&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Audio&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;18.full of bull&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Rodeo&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;19.a fraction of an answer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;20.the corporate head (abbr.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;CEO&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The empty space is there because &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm still missing a last &quot;e-o's&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13068" LastEditorUserId="11297" LastEditDate="2015-06-16T23:11:36.440" LastActivityDate="2015-06-16T23:11:36.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16152" PostTypeId="1" AcceptedAnswerId="16153" CreationDate="2015-06-09T17:44:52.263" Score="3" ViewCount="789" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;This poem has a symbol, and the symbol is a pea;&lt;br&gt;&#xA;  It goes all through a tunnel, and the tunnel is a hole.&lt;br&gt;&#xA;  Now the poem is so eerie, and it's slipping in an eel.&lt;br&gt;&#xA;  &quot;Hem-hem,&quot; said Mrs Umbridge, with a long and speedy dash.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;The riddle itself is quite easy, but I had fun writing it as a poem with a nice rollicking rhythm.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-06-09T17:55:15.873" LastActivityDate="2015-06-10T19:53:34.230" Title="Rhythmic riddle poem" Tags="|riddle|letters|" AnswerCount="2" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16153" PostTypeId="2" ParentId="16152" CreationDate="2015-06-09T17:54:12.060" Score="5" Body="&lt;p&gt;I think it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; POEM  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is why&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; all lines reference a letter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This poem has a symbol, and the symbol is a pea;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; reference the letter P in a phonetic way&lt;br&gt;&#xA; Pea sounds like P.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It goes all through a tunnel, and the tunnel is a hole.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; reference the letter O in a visual way&lt;br&gt;&#xA; O looks like a hole or the end of the tunnel&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now the poem is so eerie, and it's slipping in an eel.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As mentioned by &lt;em&gt;Mark N&lt;/em&gt;&lt;br&gt;&#xA; E is the letter inside e&lt;strong&gt;e&lt;/strong&gt;l (it's slipping in it)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Hem-hem,&quot; said Mrs Umbridge, with a long and speedy dash.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hem-Hem = M&lt;br&gt;&#xA; There is 2 common versions of dash, the short one or en dash and the long one or em dash&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10400" LastEditorUserId="10400" LastEditDate="2015-06-09T19:49:01.150" LastActivityDate="2015-06-09T19:49:01.150" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="16154" PostTypeId="2" ParentId="16045" CreationDate="2015-06-09T18:23:45.863" Score="2" Body="&lt;p&gt;NOTE: This is showing my entire thought process. So I apologize if it doesn't really seem like an answer until the end. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;77 6c 39 7f a6 6d 80 1e 91 02 db 83 90 03 89 20 c2 8a 4b a3 98 e6 f2 d6 61 6a 22 1a 61 36 20 d4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hexed is the master and so is his heirs... heir=air=breath&lt;br&gt;&#xA;By &quot;his own father he snort&quot; I'm going to assume that the father was cremated, and the ashes confused for coke by his son. By the fact that his airs/heirs are hexed like the master, they are likely smokers. Solely from that, I'd say they died from pulmonary problems... But that's really not the point now, is it?&lt;br&gt;&#xA;Hold it a moment... Hexed. Oh man, that one was good. I was going by Joe Z.'s comment about hexadecimal instead of base64, but I'm sure that clue was sneakily hidden in to the note.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clearly, Jack bowed to pay respect to the dead, and/or the fact that the &quot;son&quot; confused his fathers ashes for a drug.&lt;/strong&gt; Actually, scratch that. As mentioned in the comments, he was paying respect to his father, who was the Keeper of the Code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hexidecimal string returns the nonsense string:&lt;br&gt;&#xA;w l 9()¦ m()()()Û()()()() Â()K£()æòÖaj&quot;()a6 Ô or approximately the following:&lt;br&gt;&#xA;wl9¦m€‘Ûƒ‰ ÂŠK£˜æòÖaj&quot;a6 Ô  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's clear that I don't know what the frick I'm doing here...&lt;br&gt;&#xA;EXCEPT, as a flash of revelation just told me, &quot;Hexed is the master and so are his heirs&quot; is relating to their NAMES as well. THE MAGIC WORDS ARE THE NAMES.(Obviously, keeping in setting would churn out something silly, like Davy Jones or Barbossa. The second one sounds more plausible, but it couldn't possible be that one now, could it?)&lt;br&gt;&#xA;wl9 broken vertical bar m euro ' U f percentile ASK British pound ae o O aj&quot;a6 O...&lt;br&gt;&#xA;Again, more nonsense that I can't make sense of. It sounds &lt;em&gt;almost&lt;/em&gt; like a name that Twitch would assign to one of their Pokémon, but that's even more besides the point than the heir=breath idea. &amp;lt;.&amp;lt;&lt;br&gt;&#xA;Well, nine... bar, m-euro...mural? Yeah, that's not right. I'm 100% positive that the only thing he's going to reference is Pirates of the Caribbean.&lt;br&gt;&#xA;The old lady's comment, &quot;If your first is left, then just take the rights&quot; seems to make sense in this puzzle, if only I could figure out what the &quot;first&quot; is referring to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hex 20 through 7E is regular letters. 80 onwards through FF are special characters. 00 through 1E are function operators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a screenie of the text in Minecraft when I went to get some help:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/HOBiE.png&quot; alt=&quot;A screenshot of the string&quot;&gt;&#xA;Ignore the other players, please.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the suggestion of a commenter, I converted the string into the heir of hexadecimal, octal. It returned this string, broken up into groups of three numerals (sans possible starting zero) by four:&lt;br&gt;&#xA;167 154 071 177&lt;br&gt;&#xA;246 155 200 036&lt;br&gt;&#xA;221 002 333 203&lt;br&gt;&#xA;302 212 113 243&lt;br&gt;&#xA;230 346 362 326&lt;br&gt;&#xA;141 152 042 032&lt;br&gt;&#xA;141 066 040 324  &lt;/p&gt;&#xA;" OwnerUserId="12506" LastEditorUserId="12506" LastEditDate="2015-06-09T21:00:46.160" LastActivityDate="2015-06-09T21:00:46.160" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="16155" PostTypeId="2" ParentId="16152" CreationDate="2015-06-09T18:32:33.893" Score="2" Body="&lt;p&gt;This is just a joke in response to Joe Z's comment about the rhymes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My alternate (&lt;em&gt;Fresh prince&lt;/em&gt;) edition for the riddle: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Now this is a story all about how,&#xA;&lt;br&gt;my Eel life got flipped-turned upside down,&#xA;&lt;br&gt;&lt;br&gt;and I'd like to take a minute, just sit right there.&#xA;&lt;br&gt;I'll tell you how this Eel ate a pea and dashed home to  Bel Air!&#xA;(Bel air is in a hole where Mrs. Umbridge must live.)&#xA;&lt;br&gt;&lt;br&gt;Deep down in the water, I was born and raised.&#xA;&lt;br&gt;Swimming around was how I spent most of my days.&#xA;&lt;br&gt;&lt;br&gt;Chillin' out, snacking relaxing all cool,&#xA;&lt;br&gt;Eating some food that fell in the pool.&#xA;&lt;br&gt;&lt;br&gt;When Mrs. Umbridge, came over and was up to no good!&#xA;&lt;br&gt;Started throwing Peas in my neighborhood.&#xA;&lt;br&gt;&lt;br&gt;I ate one little pea, it went straight down my throat hole,&#xA;&lt;br&gt;and then Mrs. Umbridge said 'Ee!..hem-hem, that was not my goal.'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-06-10T19:53:34.230" LastActivityDate="2015-06-10T19:53:34.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16156" PostTypeId="2" ParentId="16133" CreationDate="2015-06-09T19:43:47.573" Score="1" Body="&lt;p&gt;&lt;b&gt;WIP: If you know the answers to some of the gaps in my attempt, feel free to share!&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You must first calculate the number of remaining moves on the board. If a player is able to keep All Possible Moves (APM) to an even number by the time the board is reduced small enough, they will win!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, on a 2x2 board, there are 3 possible moves (1 individual block, and 2 2-blocks). Player 1 is then able to make the remainder of all possible moves (APM) into an even amount (2 moves left after taking 1,1), meaning that he can always take another move, keeping the number of moves left even until he takes the last move and wins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main issue with this problem is determining the initial number of possible moves on the board, and how each type of move will remove $X$ amount from APM. The player that keeps/makes the boards APM even on their turn, is the winner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can get APM of the board at the beginning by using $(X*Y) -1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for a 3x4 board, there are $3*4 -1 = 11$ possible moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This number is always odd, so you would think Player 1 should be able to make it even and keep it even, and force Player 2 to make it odd again! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue is, you keep APM even if it is already even, if the board is large enough. How  large is large enough? Well, you need a minimum of a 4x1 rectangle, or a 2x2 box (area of units). Anything larger than this will also work! [($N*N$) or ($4*\frac{N}2[*1]$)  where $N$ is even, or ($(4*N)*M$) ].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now the issue becomes how many of these ($N*N$), ($4*\frac{N}2[*1]$) and ($(4*N)*M$) options there are left. We will call these KAE's (Keep APM's Even)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;How we find that....I'm not sure atm.&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But once we find it, if that number can remain even and the the APM can remain even after player one's turn, then player one should always be able to win!&lt;/p&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-06-09T20:43:52.843" LastActivityDate="2015-06-09T20:43:52.843" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16157" PostTypeId="1" AcceptedAnswerId="16158" CreationDate="2015-06-09T20:31:46.687" Score="39" ViewCount="3669" Body="&lt;p&gt;There are only 3 hitmen in the city.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what each had to say:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Perry&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Don't be fooled by my size. Perry is the go-to name in this city for bringing about &quot;timely endings.&quot; Not surprisingly, I'm the most commonly used hitman in the city, delivering reliable results for your everyday, average hit. Unlike other hitmen, I even offer many other types of services.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Quentin&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;For those who need to go the extra mile to confuse authorities, I'm your man. After examining one of my crime scenes, investigators are left wondering exactly what happened, how and why. Although my methods are considered dubious by most, I always finish off my targets as specified.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Ethan&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Do you need to bring about an &quot;elaborate termination?&quot; I specialize in making bold statements and getting people's attention. Use my services and all survivors will think twice before crossing you again!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who are these hitmen?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-07-07T13:12:23.403" LastActivityDate="2015-07-07T13:12:23.403" Title="Who are these hitmen?" Tags="|riddle|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16158" PostTypeId="2" ParentId="16157" CreationDate="2015-06-09T20:39:01.653" Score="41" Body="&lt;p&gt;I'm going to guess that these hitmen are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Punctuation marks! Specifically, the three that end sentences.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Perry is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A period, which is the most commonly used punctuation mark to end a sentence. The period offers other services as well, such as decimals (1.5) and ellipses (...).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Quentin is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A question mark, which is commonly used to indicate confusion. What, how, and why are all questions that end with a question mark. Every time there's a question the question mark finishes the sentence as specified.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ethan is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An exclamation point! Exclamation points are used both to make bold statements and to grab your attention.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-06-09T20:39:01.653" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16159" PostTypeId="1" AcceptedAnswerId="16200" CreationDate="2015-06-09T21:44:52.410" Score="20" ViewCount="3951" Body="&lt;p&gt;In good company, I'm a sign of respect &lt;br&gt;&#xA;Alone, I don't denote man kind &lt;br&gt;&#xA;I was the chosen one, by George! &lt;br&gt;&#xA;There are times my presence is warranted &lt;br&gt;&#xA;And I'll usually show up if you're a good sport about it &lt;br&gt;&#xA;Would you do me the honor of telling me my name?&lt;/p&gt;&#xA;" OwnerUserId="2812" LastActivityDate="2015-06-11T17:43:29.590" Title="Would you do me the honor of telling me my name?" Tags="|riddle|" AnswerCount="13" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16160" PostTypeId="1" AcceptedAnswerId="16161" CreationDate="2015-06-09T23:12:12.173" Score="1" ViewCount="1407" Body="&lt;p&gt;In preparation for a banquet dinner, you are carving up a massive cubic block of cheese. The only people who will be eating the cheese are you and six guests invited to the banquet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheese must always be consumed in tiny cubes (so dictate the laws of cheese etiquette). With this in mind, you pick a large number, $n$, and proceed to cut the cheese block into perfect identically-sized cubes by making $n$ cuts widthwise, $n$ cuts heightwise, and $n$ cuts depthwise. You're positive that you've made $n$ large enough that each guest will get several cubes of cheese.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, as luck would have it, you forget $n$ as soon as you've finished.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No problem. You have 30 minutes left until the cheese has to be set out. You decide to count the number of cubes along one edge of the block, and do so. Exactly 15 minutes later, at the moment you've finished counting the last cube along the edge, disaster strikes! The MC rushes into the kitchen to inform you that the guests are feuding, and that if each guest doesn't get &lt;em&gt;exactly&lt;/em&gt; the same number of cheese cubes, the dinner will be ruined. If there are any cubes left over, the guests will surely fight over them, which isn't acceptable. Nor is it acceptable for you to eat or discard any left over cubes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You've misplaced your knife, hence cutting the cheese cubes is out of the question. And the guests are hungry, hence everyone (including you) must get at least one cube of cheese.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To remedy the situation, you're able to count, remove or replace individual cheese cubes from the block. You figure you can remove/replace them as quickly as you counted them earlier, and, if need be, you can count a cube while you remove or replace it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Along each edge there are... Oh no! Curses! With all the excitement, you've forgotten the number!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Panic sets in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have 15 minutes left. Exactly enough time to count, remove, or replace as many cubes as are along one edge of the cheese block. You can finish early, but you can't finish late.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the sequence of events:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;you count, remove, or replace cubes&lt;/li&gt;&#xA;&lt;li&gt;you put the remaining cheese cubes out&lt;/li&gt;&#xA;&lt;li&gt;you take your cheese cubes to eat (there must be at least one)&lt;/li&gt;&#xA;&lt;li&gt;the guests take turns taking cubes, repeating until none are left&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What do you do?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Please note that this is a math problem and not a lateral thinking problem.&lt;/em&gt;)&lt;/p&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="8874" LastEditDate="2015-10-06T12:05:25.370" LastActivityDate="2015-10-06T12:05:25.370" Title="Cutting the Cheese" Tags="|mathematics|dissection|counting|" AnswerCount="4" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16161" PostTypeId="2" ParentId="16160" CreationDate="2015-06-09T23:34:55.380" Score="2" Body="&lt;p&gt;The number of cubes along one edge is $n+1$ (since you made $n$ cuts), so the total number of cubes is $(n+1)^3$. You want to remove some of the cubes along one edge so that the total number of cubes is a multiple of 6 (the number of guests).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you remove all the cubes along a single edge, the total number remaining is $(n+1)^3-(n+1)=(n+1)(n^2+2n+1-1)=n(n+1)(n+2)$. This number must be a multiple of $2$ (since one of $n$ and $n+1$ must be even) and also of $3$ (since once of $n$, $n+1$, and $n+2$ must be a multiple of $3$), and therefore of $6$. So you should&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; remove an entire edgeworth of cubes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;to satisfy the guests.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-06-09T23:34:55.380" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16162" PostTypeId="2" ParentId="16159" CreationDate="2015-06-09T23:38:33.340" Score="1" Body="&lt;p&gt;Could it be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Foreman?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In good company, I'm a sign of respect&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A foreman is somebody to be respected.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Alone, I don't denote man kind&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A single foreman (or four men perhaps?) isn't the whole of mankind.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I was the chosen one, by George!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; George Foreman is a retired boxer, formerly World Heavyweight Champion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;There are times my presence is warranted&lt;br&gt;&#xA;And I'll usually show up if you're a good sport about it&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Boxing is a sport?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Would you do me the honor of telling me my name?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (George) Foreman?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-06-09T23:38:33.340" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16163" PostTypeId="2" ParentId="16160" CreationDate="2015-06-09T23:50:29.570" Score="2" Body="&lt;p&gt;Assuming that the &lt;em&gt;guests&lt;/em&gt; need to have the same number of cubes, but &lt;em&gt;you&lt;/em&gt; don't (which is consistent with what the question currently says), I would advise the following steps. Start with step $1$ and proceed to step $7$ (one cube at a time), then repeat the whole process.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Count a cube, remove a cube.&lt;/li&gt;&#xA;&lt;li&gt;Count a cube.&lt;/li&gt;&#xA;&lt;li&gt;Count a cube, replace one.&lt;/li&gt;&#xA;&lt;li&gt;Count a cube, remove a cube.&lt;/li&gt;&#xA;&lt;li&gt;Count a cube, replace one.&lt;/li&gt;&#xA;&lt;li&gt;Count a cube.&lt;/li&gt;&#xA;&lt;li&gt;Count a cube.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This guarantees, at each step, that the number of cubes will be either $-1$ or $0$ mod $7$ - so either one person will have one less cube, or everyone will have equally many. Thus, if you deal the cheese cubes out to your guests, serving yourself last, you will ensure that every guest has equally many cubes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that we could equally well express the process as saying that, at step $n$, you need to have, in total, removed $1$ cube if and only if $n^3\equiv 1\pmod 7$. (The other cubes mod $7$ being $0$ and $-1$, we find that we don't need any special handling in those cases)&lt;/p&gt;&#xA;" OwnerUserId="4765" LastActivityDate="2015-06-09T23:50:29.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16164" PostTypeId="2" ParentId="16116" CreationDate="2015-06-09T23:51:06.530" Score="3" Body="&lt;p&gt;Did a little bit of simulation for this, using 8 players:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;stratrand&lt;/strong&gt;: Chooses randomly every time &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;stratav-high&lt;/strong&gt;: Chooses the lowest rock that is greater than the average of the other player's remaining rocks. If does not exist, choose your highest tile.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;stratav-low&lt;/strong&gt;: Chooses the lowest rock that is greater than the average of the other player's remaining rocks. If does not exist, choose your lowest tile (to try to keep high rocks for later rounds).&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;stratmatch&lt;/strong&gt;: match the upturned rock&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;stratmax&lt;/strong&gt;: always choose the max rock available (chooses 6 -&gt; 1)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;stratmin&lt;/strong&gt;: always choose the min rock available (chooses 1 -&gt; 6)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;stratnext-high&lt;/strong&gt;: choose the lowest rock that is higher than the upturned rock. If not available, choose your highest rock&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;stratnext-low&lt;/strong&gt;: choose the lowest rock that is higher than the upturned rock. If not available, choose your lowest rock.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As all of these are purely reactionary, we can evaluate for all possible iterations of upturned stones. For interactions with the random player, I did a Monte Carlo simulation. For each &quot;battle&quot;, we can create a results space:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qA4qA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;AverageClose&quot; is the average of the last two rocks upturned. &quot;AverageOpen&quot; is the average of the first two stones upturned. A game that started with two high rocks and ended with two low rocks would be on the bottom right of this graph. Games with equivalent coordinates (i.e 1,6,2,5,3,4 and 6,1,5,2,4,3) were averaged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XQRK0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each tile represents a different outcome, with green meaning that the x axis player won, and red meaning the y axis player did. Gold would either mean that every outcome was a tie, or there was a equal proportion of wins and losses for each &quot;battle&quot; at that coordinate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some interesting observations (although by no means exhaustive):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Almost every strategy beats the random one over time, although the max/min get screwed if the opening is low/high respectively.&lt;/li&gt;&#xA;&lt;li&gt;Next looks like the best strategy (against the field presented, I'm sure there is a strategy that can beat it). I think this is mainly because it forces conservation of your high rocks, paced by the upturned rocks.&lt;/li&gt;&#xA;&lt;li&gt;The match strategy looks fairly good, although is destroyed (by design) by the &quot;next&quot; strategies&lt;/li&gt;&#xA;&lt;li&gt;Min/Max counter each other, depending on the pattern of upturned rocks.&lt;/li&gt;&#xA;&lt;li&gt;No real differences between the average strategies, and neither seems that great&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13073" LastActivityDate="2015-06-09T23:51:06.530" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16165" PostTypeId="2" ParentId="16160" CreationDate="2015-06-09T23:51:44.500" Score="3" Body="&lt;p&gt;Even with the clarifications, this puzzle is pretty easy.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Trivial Solution #1&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you just count the number of cubes along an edge, you will know how many to take in step 3.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is the sequence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. Count all $n+1$ cubes along one edge.&#xA;&lt;br&gt;&#xA; 2. Put out the entire block.&#xA;&lt;br&gt;&#xA; 3. Take a number $k\in\{1\ldots 6\}$ of cubes such that $k\equiv(n+1)^3\mod 6$.&#xA;&lt;br&gt;&#xA; 4. The six guests divide the remaining cubes evenly among themselves with none left over (since $(n+1)^3-k\equiv 0\mod 6$).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Trivial Solution #2 (Dubious)&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Count out $7$ cubes, and discard the rest of the block.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is the sequence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. Count and remove seven cubes (we can assume that $n\ge 6$, so you will be able to count out at least seven).&#xA;&lt;br&gt;&#xA; 2. Put out &lt;strong&gt;the seven you counted&lt;/strong&gt; (this is where this solution is dubious, since you're supposed to put out the &lt;em&gt;non-removed&lt;/em&gt; cubes).&#xA;&lt;br&gt;&#xA; 3. You take exactly one cube.&#xA;&lt;br&gt;&#xA; 4. The six guests each take one cube, leaving none left over.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="9718" LastEditDate="2015-06-11T13:50:54.157" LastActivityDate="2015-06-11T13:50:54.157" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="16166" PostTypeId="2" ParentId="16160" CreationDate="2015-06-09T23:56:23.380" Score="4" Body="&lt;p&gt;EDIT: To handle the time required for removing cubes, when counting each of the cubes, remove the first 7 (there must be at least 7 if it takes 15 minutes to count). When you get to the last cube, your decision will be to either take it away (1), keep it (0), or add one (6).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here's a solution assuming everyone (including yourself) needs the same number of cubes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The list of cubes mod 7 follow a specific pattern (1, 1, 6, 1, 6, 6, 0). When you're counting the edge, instead of counting (1,2,3,4...), count 1, 1, 6, 1, 6 etc. looping every 7 and on the last number, do what that number says. If it's 1, remove 1. If it's 6, remove 6. If it's 0, do nothing. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="7722" LastEditDate="2015-06-10T00:32:11.257" LastActivityDate="2015-06-10T00:32:11.257" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16167" PostTypeId="2" ParentId="16159" CreationDate="2015-06-09T23:57:27.373" Score="2" Body="&lt;p&gt;Second attempt:=============================================================&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In good company, I'm a sign of respect &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Company&quot; in this case might refer to more than just a \$2 bill. The more money you have, generally the more respect you have.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alone, I don't denote man kind &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you have nothing but a $2 bill, you still have a note. Thus, you are not denoted.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was the chosen one, by George! &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Washington chose four members for his presidential cabinet:  Secretary of State Thomas Jefferson, Secretary of Treasury Alexander Hamilton, Secretary of War Henry Knox, and Attorney General Edmund Randolph.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are times my presence is warranted &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sometimes you need to pay for things with money.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I'll usually show up if you're a good sport about it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you're a good sport and keep looking, maybe you'll find a $2 bill.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Would you do me the honor of telling me my name?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Thomas Jefferson, as pictured on the $2 bill.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First attempt:===============================================================&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm probably way off, but here goes...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Is your name force?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In good company, I'm a sign of respect&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Others often respect a show of force  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alone, I don't denote man kind&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A force of nature is a phenomenon that humans cannot control.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was the chosen one, by George! &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;You were the chosen one!&quot; might be a reference to Star Wars: Episode 3 in which characters use the force. It was also directed by &lt;em&gt;George&lt;/em&gt; Lucas. Could also refer to George Washington who won American independence by use of force.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are times my presence is warranted &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Force is sometimes warranted.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I'll usually show up if you're a good sport about it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you play by society's rules, you'll usually be able to call up the police (or whoever) to resolve situations by force.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Would you do me the honor of telling me my name?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Force?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10424" LastEditorUserId="10424" LastEditDate="2015-06-10T03:14:30.617" LastActivityDate="2015-06-10T03:14:30.617" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16168" PostTypeId="1" AcceptedAnswerId="24734" CreationDate="2015-06-10T00:12:48.163" Score="58" ViewCount="9392" Body="&lt;p&gt;You are the president of a secret society of mathematicians with &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; members, including yourself. No one in the society knows what &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; is. The dictator of the world, in an effort to erase mathematics from human history, has finally managed to capture all &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; members of the society, and has placed them in prison.&lt;/p&gt;&#xA;&lt;p&gt;The prison has a funny design. There are &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; identical soundproof, windowless prison cells arranged in a circle, each containing a mathematician. The cells all have a light switch the prisoners can use, but the wiring system is screwed up, so the switch controls the light bulb in the &lt;em&gt;clockwise neighboring cell&lt;/em&gt;. Furthermore, the switch is only capable of delivering a single flash of light at noon each day. Specifically, if this switch is set to &amp;quot;on&amp;quot; at noon, it will flash the next cell, and do nothing otherwise.&lt;/p&gt;&#xA;&lt;p&gt;The prison warden worries that the prisoners might try to communicate using the lights, so every night at midnight, he fills all of the cells with knockout gas, cleans the cells so they can't communicate by leaving messages, sets all of the light switches to &amp;quot;off&amp;quot;, and rearranges the prisoners in whatever fashion he pleases (still one prisoner per cell).&lt;/p&gt;&#xA;&lt;p&gt;One day, the warden visits your cell. He confesses that he loves mathematics, and decides to offer your society a game to win their freedom. If any of the prisoners are able to figure out what &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; is, they may shout out loud &amp;quot;There are &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; prisoners!&amp;quot; (the cells are monitored with security cameras/mics). If they are correct, all prisoners will go free, and if not, they will all be executed.&lt;/p&gt;&#xA;&lt;p&gt;The warden allows you to devise a plan for everyone else to follow. He will make &lt;span class=&quot;math-container&quot;&gt;$n-1$&lt;/span&gt; copies of this plan, and allow every other prisoner to read it. The warden of course will also read your plan, and will perform the cell rearrangements in such a way to make it fail if he can.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;What plan allows the prisoners to guarantee their freedom?&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Extra for Experts:&lt;/strong&gt; Can you find a plan which doesn't require the prisoners to make random decisions?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Notes:&lt;/strong&gt; There is no lateral thinking required to solve this. Assume the prisoners have perfect, infinite memory, including knowledge of how many days have elapsed.&lt;/p&gt;&#xA;&lt;p&gt;It is significant that &lt;em&gt;you&lt;/em&gt; are one of the prisoners, and that &lt;em&gt;you&lt;/em&gt; devise the plan. It means that while all &lt;span class=&quot;math-container&quot;&gt;$n-1$&lt;/span&gt; other prisoners must follow the same strategy, you may follow a different strategy.&lt;/p&gt;&#xA;&lt;p&gt;I will give priority to accepting answers which fulfill the &amp;quot;Extra for Experts&amp;quot; condition. This means there must exist a function &lt;span class=&quot;math-container&quot;&gt;$B(n)$&lt;/span&gt; such your plan is guaranteed to stop after &lt;span class=&quot;math-container&quot;&gt;$B(n)$&lt;/span&gt; days when there are &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; prisoners.&lt;/p&gt;&#xA;&lt;p&gt;This is the hardest puzzle I know, but there is a solution.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2022-10-07T12:14:47.497" Title="The Circular Prison of Unknown Size" Tags="|mathematics|strategy|story|" AnswerCount="7" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="16169" PostTypeId="1" AcceptedAnswerId="20856" CreationDate="2015-06-10T00:49:04.010" Score="11" ViewCount="1628" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If I am on top politically, I am right, but if I am on top of being not in violation, I am wrong.&lt;br&gt;&#xA;  If I am above 8, I am wrong, but if I am low on being not in violation, I am forever.&lt;br&gt;&#xA;  I cannot be on top politically and above 8, but I can be on top and low of being not in violation.   &lt;/p&gt;&#xA;  &#xA;  &lt;h2&gt;Who am I?&lt;/h2&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I may post an answer depending on the amount of time has past before today. I will add hints every [random period of time].   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The puzzle is perfectly correct; I assure you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer has nothing to do with politics.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Take each part between commas and then do the action of wordplay.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For the &quot;not being in violation&quot; one part of the phrase refers to the prefix that you add, and the other is synonymous to another word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The now added calculation-puzzle tag&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Moghwyn June 10th answer is the closest one yet... except the wordplay isn't perfect, which leads to the 0/0=1 stretch&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="167" LastEditorUserId="167" LastEditDate="2015-08-28T03:15:37.450" LastActivityDate="2015-09-10T07:34:06.683" Title="The 8 year old politician?" Tags="|riddle|calculation-puzzle|wordplay|" AnswerCount="8" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="16170" PostTypeId="2" ParentId="16159" CreationDate="2015-06-10T01:04:26.873" Score="1" Body="&lt;p&gt;Thought of this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Armstrong&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and tried to shoehorn it in...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In good company, I'm a sign of respect &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I can't think of anything for this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alone, I don't denote man kind &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Neil Armstrong on his own only took &quot;one small step for man&quot;, but with the world looking on he took &quot;one giant leap for mankind&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was the chosen one, by George!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; George Armstrong Custer, often portrayed as arrogant probably thought he was the chosen one when it came to dealing with the native Americans,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are times my presence is warranted &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &quot;strong arm&quot; of the law might have a warrant to access a premesis.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I'll usually show up if you're a good sport about it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lance Armstrong would be very happy for you to be a good sport, but he would be anything but!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Would you do me the honor of telling me my name?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Armstrong&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12277" LastActivityDate="2015-06-10T01:04:26.873" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16171" PostTypeId="1" AcceptedAnswerId="16196" CreationDate="2015-06-10T01:11:32.043" Score="7" ViewCount="221" Body="&lt;p&gt;Below is a word search puzzle presented without clues. Instead, you must use the additional rules (see example below) to work out what the words are. Finally, as a check to make sure you found the right words, the first letters of each word can be rearranged to make one last, related word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because I'm having fun making these, this is the fourth puzzle in a series. Here's the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16087/cluelessly-searching-for-words&quot;&gt;first&lt;/a&gt;, &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16094/cluelessly-searching-for-words-ii&quot;&gt;second&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16122/clueslessly-searching-for-words-iii&quot;&gt;third&lt;/a&gt;. (This ought to be somewhere between the second and third difficulty-wise.)&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Rules&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The words can run in any direction (orthogonal or diagonal) and do not have to be in straight lines&lt;/li&gt;&#xA;&lt;li&gt;Words do not overlap and their paths do not cross&lt;/li&gt;&#xA;&lt;li&gt;All letters are part of some word&lt;/li&gt;&#xA;&lt;li&gt;All words are of the same length&lt;/li&gt;&#xA;&lt;li&gt;All words (including the final word) are related somehow&lt;/li&gt;&#xA;&lt;li&gt;All words (including the final word) have an English Wikipedia page, even if they are not English themselves&lt;/li&gt;&#xA;&lt;li&gt;The letters that make up the final word do not appear in any specific location on the grid&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Example&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Here's a short example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bAwdg.png&quot; alt=&quot;P O A W. E N N S. M U Y A. S T O D.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GRuBj.png&quot; alt=&quot;PONY EMUS TOAD SWAN&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The four words here are PONY, EMUS, TOAD and SWAN, which are all animals, and the first letter of each makes the word PETS.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Puzzle&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;What is the final word to be found in this grid of letters?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/svDWk.png&quot; alt=&quot;I N I M O A N. L G R G R M T. L M A O U I H. A F N B S R L. L D S A A N M. E I R I G D I. H D E A L U R.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Hint&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All the words are taken from one Wikipedia list page&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7392" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2015-06-10T09:07:15.397" Title="Cluelessly searching for words IV" Tags="|word-search|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16172" PostTypeId="2" ParentId="16159" CreationDate="2015-06-10T01:43:39.857" Score="4" Body="&lt;p&gt;Maybe it's...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Honesty?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In good company, I'm a sign of respect&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Honesty could be considered a sign of respect...to the right people, that is&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alone, I don't denote man kind &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mankind is known for its skill at lying&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was the chosen one, by George! &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; George Washington &lt;a href=&quot;http://www.infoplease.com/t/history/true-washington/honesty.html&quot; rel=&quot;nofollow&quot;&gt;is known for his supposed extreme honesty&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are times my presence is warranted&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Honesty isn't &lt;em&gt;always&lt;/em&gt; the best policy for every situation, but it is warranted at times&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I'll usually show up if you're a good sport about it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People are more willing to be honest if you can handle the truth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Would you do me the honor of telling me my name?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Honesty?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10456" LastActivityDate="2015-06-10T01:43:39.857" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16173" PostTypeId="2" ParentId="16159" CreationDate="2015-06-10T03:01:35.813" Score="1" Body="&lt;p&gt;I think I know the Answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mr. President.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In good company, I'm a sign of respect&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the president, in any company worth mentioning, is respected&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alone, I don't denote man kind&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No thoughts here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was the chosen one, by George! &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When he was elected president, there was no title for the position.&lt;br&gt;&#xA;&quot;Washington consented to the demands of James Madison and the United States House of Representatives that the title be altered to 'Mr. President.'&quot; source: &lt;a href=&quot;http://en.wikipedia.org/wiki/Mr._President_%28title%29&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are times my presence is warranted&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The same reason @wossman stated.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I'll usually show up if you're a good sport about it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; well... No thoughts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Would you do me the honor of telling me my name?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mr. President&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13075" LastEditorUserId="13075" LastEditDate="2015-06-10T05:18:39.227" LastActivityDate="2015-06-10T05:18:39.227" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16174" PostTypeId="1" AcceptedAnswerId="16187" CreationDate="2015-06-10T03:14:21.383" Score="15" ViewCount="2425" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;My first is the most common&lt;br&gt;&#xA;  My second is another $\pi$&lt;br&gt;&#xA;  My third represents Iceland&lt;br&gt;&#xA;  My fourth is indefinite&lt;br&gt;&#xA;  My fifth is potassium hydroxide  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where can you find my whole? And why?&lt;/p&gt;&#xA;" OwnerUserId="10400" LastEditorDisplayName="user88" LastEditDate="2015-06-10T05:50:55.443" LastActivityDate="2015-06-10T05:52:16.027" Title="Valve graffiti - riddle" Tags="|riddle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16175" PostTypeId="2" ParentId="16159" CreationDate="2015-06-10T03:17:34.040" Score="2" Body="&lt;p&gt;I'm thinking&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Lady&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In good company, I'm a sign of respect &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Lady is a term of respect&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alone, I don't denote man kind &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Ladies are not men, they're women&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was the chosen one, by George! &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;George Washington chose his wife Martha, the First Lady&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are times my presence is warranted &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The use of Lady as a title is appropriate when given certain royal honours (By royal warrant)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I'll usually show up if you're a good sport about it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Not sure about this one&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Would you do me the honor of telling me my name?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Lady is an honorific&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1667" LastActivityDate="2015-06-10T03:17:34.040" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16177" PostTypeId="2" ParentId="16159" CreationDate="2015-06-10T03:30:32.830" Score="2" Body="&lt;p&gt;In good company I am a sign of respect... &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hand Shake.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alone, I don't denote man kind.... &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; part of the human body.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was the chosen one, by George.... &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; in Washington's inaugural address the hand in referenced 'on one hand' and 'on the other hand'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are times my presence is warranted.... &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; writing, grasping, grabbing, slapping, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I usually show up if your a good sport about it.... &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; high five hand slap.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My name is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hand.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13074" LastEditorUserId="11239" LastEditDate="2015-06-10T03:35:41.357" LastActivityDate="2015-06-10T03:35:41.357" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16180" PostTypeId="1" AcceptedAnswerId="16182" CreationDate="2015-06-10T04:47:31.277" Score="3" ViewCount="4443" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DHpCU.png&quot; alt=&quot;Find the next pattern&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a JCTI question ,where I need to fill up the blank spot by any of the 6 given boxes.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/8dmil.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot understand from where these two balls within a single box is coming. All the previous boxes contained just one ball! Moreover the first two (up and down in a single column) are squares, the second two are rhombuses, the third two are again squares, so the last two (the rightmost one and the one to be filled up) should be rhombuses. But given is a square!! Please solve this puzzle.&lt;/p&gt;&#xA;" OwnerUserId="13077" LastActivityDate="2019-01-31T03:00:16.220" Title="JCTI Test: Find the next box" Tags="|pattern|matches|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16181" PostTypeId="2" ParentId="16084" CreationDate="2015-06-10T05:04:28.303" Score="0" Body="&lt;p&gt;I am unable to comment, so this also is a partial answer.  If you say this is correct I will post a full answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; s1101111 is a typo and should be s1011111&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I understand the&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; letters before the string of 0s and 1s&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and although not necessary, they sped up my ability to solve the puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You say that there is sufficient information, and there is, but&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; only if you read the spoiler hint.  At first, I was avoiding this, but I think one needs to see the actual bits... oops, 0s and 1s.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sheet's first line is interesting because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it's not strictly necessary to determine the function, and, in some ways follows naturally from the actual function.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12357" LastActivityDate="2015-06-10T05:04:28.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16182" PostTypeId="2" ParentId="16180" CreationDate="2015-06-10T05:12:35.377" Score="4" Body="&lt;p&gt;The two balls in the single box comes from the two different tracks moving together. The puzzle is designed to suggest that there are two different tracks that collide, which implies that what you're looking at in the missing box is the combination of the two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's the hint. Here's the answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Option 1.&lt;/strong&gt; The tracks show you what happens when the balls move from a square into a rotated square and back. In the first track, the ball moves clockwise in both steps. In the second track, the ball moves clockwise into a diamond, and counterclockwise twice back into the square.&#xA;&#xA; Consequently, the top track moves into the right side, and the bottom track moves into the left side. In the step after this, they overlap.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user20" LastActivityDate="2015-06-10T05:12:35.377" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16183" PostTypeId="2" ParentId="15551" CreationDate="2015-06-10T05:22:20.883" Score="1" Body="&lt;p&gt;It's not possible to cut a circle into 7 pieces using geometric methods: &lt;a href=&quot;https://math.stackexchange.com/questions/440825/is-it-possible-to-divide-a-circle-into-7-equal-pizza-slices-using-geometric&quot; title=&quot;Is it possible to divide a circle into 7 equal &amp;#39;pizza slices&amp;#39; &amp;#40;using geometrical methods&amp;#41;?&quot;&gt;Is it possible to divide a circle into 7 equial pizza slices...&lt;/a&gt;, unless you allow &lt;a href=&quot;http://en.wikipedia.org/wiki/Neusis_construction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Neusis construction&lt;/a&gt;; see &lt;a href=&quot;http://en.wikipedia.org/wiki/Heptagon%20Constructing%20a%20Heptagon&quot; rel=&quot;nofollow noreferrer&quot;&gt;Heptagon&lt;/a&gt; for details on constructing a Heptagon, which can be overlaid on the pie and used to guide the slicing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only good solution that avoids geometry is the height-based slicing, as suggested by Anon.&lt;/p&gt;&#xA;" OwnerUserId="10058" LastEditorUserId="-1" LastEditDate="2017-04-13T12:19:36.950" LastActivityDate="2015-06-10T05:22:20.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16184" PostTypeId="1" AcceptedAnswerId="16191" CreationDate="2015-06-10T05:38:33.263" Score="1" ViewCount="1587" Body="&lt;p&gt;It is &quot;find the next box matching the trend&quot; sort of thing.&#xA;The question as given is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pHRhG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to fill the blank area with the correct box given from 6 options:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NKGeL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I get from the question is that there is an color inverse of every box given, i.e. for every box in this collection, there is another box of the same pattern but which is negative of that box. But only for the black box there is no negative. So it implies that the answer should be white one(Option 1). However the thing that confuses me is that why are there two black boxes? Please solve this puzzle and state if my reasoning is correct or not. Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="13077" LastActivityDate="2019-10-19T03:20:04.320" Title="Another JCTI Puzzle" Tags="|pattern|matches|" AnswerCount="3" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16185" PostTypeId="2" ParentId="15977" CreationDate="2015-06-10T05:39:11.473" Score="1" Body="&lt;p&gt;It could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Man cave&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It starts off as a creation of God,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it can be said natural cave are created by God.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;followed by a godly creation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Man cave, on the other hand is created by man. And most men would agree it's a godly creation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Having them together is odd,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Man and cave do not usually come together.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but between them is a one way affection.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Man loves his man cave, but cave does not love man.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A disgusting symbol of disparity&#xA;between wealth and poverty.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Only man with decent amount of money ca build their own man cave.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet it is an irony&#xA;as a proof of gender equality.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Women is welcomed in the man cave, but they usually don't go there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12332" LastActivityDate="2015-06-10T05:39:11.473" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16186" PostTypeId="2" ParentId="16174" CreationDate="2015-06-10T05:43:16.307" Score="12" Body="&lt;p&gt;Okay here goes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm going out of order because this riddle comes with the &quot;liars&quot; tag, so I'm guessing the teller is lying about first second etc.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;My fifth is potassium hydroxide - &lt;strong&gt;KOH&lt;/strong&gt; (abbrev. for potassium hydroxide)&lt;br&gt;&#xA;My second is an other π - &lt;strong&gt;TAO&lt;/strong&gt; (Tau is often argued as being a &quot;better&quot; pi. I know I misspelled it, bear with me)&lt;br&gt;&#xA;My third represent Iceland - &lt;strong&gt;ISL&lt;/strong&gt; (The 3-letter abbreviation for Iceland)&lt;br&gt;&#xA;My fourth is indefinite - &lt;strong&gt;AN&lt;/strong&gt; (&quot;an&quot; is an indefinite article in English grammar)&lt;br&gt;&#xA;My first is the most common - &lt;strong&gt;D&lt;/strong&gt; (I'm not sure on this one, but Vitamin D is the most common vitamin deficiency in the world. Which fits with the whole phrase, which is...&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;http://en.wikipedia.org/wiki/Ko_Tao&quot;&gt;KOH TAO ISLAND&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10861" LastActivityDate="2015-06-10T05:43:16.307" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16187" PostTypeId="2" ParentId="16174" CreationDate="2015-06-10T05:46:46.030" Score="26" Body="&lt;p&gt;This is an old meme: &lt;strong&gt;The cake is a lie&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;My first is the most common&lt;/strong&gt; word of the English language, which is &quot;the&quot;. &lt;p&gt;&#xA; &lt;strong&gt;My second is another pi&lt;/strong&gt;(e), which is something &lt;em&gt;like&lt;/em&gt; a pie, or a &quot;cake&quot;. &lt;p&gt;&#xA; &lt;strong&gt;My third represents Iceland&lt;/strong&gt; by being the &lt;a href=&quot;http://en.wikipedia.org/wiki/ISO_3166-1_alpha-2&quot;&gt;ISO 3166-1 alpha2&lt;/a&gt; country code for it: &quot;is&quot;. &lt;p&gt;&#xA; &lt;strong&gt;My fourth is&lt;/strong&gt; the &lt;strong&gt;indefinite&lt;/strong&gt; article &quot;a&quot;.&lt;p&gt;&#xA; &lt;strong&gt;My fifth is potassium hydroxide&lt;/strong&gt;, also known as &quot;lye&quot;. Slightly different spelling, but they're pronounced the same, which I think is good enough.&lt;p&gt;&#xA; My whole is a piece of graffiti found in the game &lt;em&gt;Portal&lt;/em&gt;, by Valve.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-06-10T05:52:16.027" LastActivityDate="2015-06-10T05:52:16.027" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16188" PostTypeId="2" ParentId="16159" CreationDate="2015-06-10T06:04:31.150" Score="2" Body="&lt;p&gt;My guess is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Command?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In good company, I'm a sign of respect&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People in command are hopefully respected.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alone, I don't denote man kind&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Commanding man is not necessarily a kind thing to do.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was the chosen one, by George!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; George Washington was chosen to be the Commander in Chief.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are times my presence is warranted&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Command is sometimes necessary, especially to govern large groups.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I'll usually show up if you're a good sport about it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If there are those willing, a leader tends to command others.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Would you do me the honor of telling me my name?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ummm.... I dunno.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's all a little far out from the phrasing of the riddle as a whole though and it doesn't obviously match up with all the extra clues you gave so I dunno....&lt;/p&gt;&#xA;" OwnerUserId="13078" LastActivityDate="2015-06-10T06:04:31.150" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16191" PostTypeId="2" ParentId="16184" CreationDate="2015-06-10T07:15:25.023" Score="5" Body="&lt;p&gt;I'd say the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 5. All black.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Including the missing shape, there are 9 boxes.  So I group the patterns into group of 3s.  Going by your logic, add a box to its negative cousin, you will end up with a black box (or white box).  Seeing as there are already 2 black boxes, the result of the third group should also be black.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12332" LastActivityDate="2015-06-10T07:15:25.023" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16192" PostTypeId="2" ParentId="16159" CreationDate="2015-06-10T07:30:25.097" Score="3" Body="&lt;p&gt;Is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; democracy or freedom&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In good company, I'm a sign of respect&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;   Many developed countries consider democracy important and sign of progress&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alone, I don't denote man kind&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    Not every country has democracy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was the chosen one, by George!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;   George Washington chose democracy over other systems for USA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are times my presence is warranted&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    Sometimes freedom is limited for specific reasons&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I'll usually show up if you're a good sport about it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    Countries on a good rapport with public are usually democracies&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Would you do me the honor of telling me my name?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;   democracy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13080" LastEditorUserId="13080" LastEditDate="2015-06-10T08:34:46.850" LastActivityDate="2015-06-10T08:34:46.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16193" PostTypeId="2" ParentId="16159" CreationDate="2015-06-10T07:37:49.373" Score="1" Body="&lt;p&gt;In good company, I'm a sign of respect&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If an army is fighting for your benefit and safety, respect is usually returned&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alone, I don't denote man kind &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An army without a strong government can be out of control&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was the chosen one, by George!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; George Washington led an army&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are times my presence is warranted &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In times of disorder or conflict, an army may be needed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I'll usually show up if you're a good sport about it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you're a good ally, an army may be sent for support&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Would you do me the honor of telling me my name?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An army&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13082" LastEditorDisplayName="user20" LastEditDate="2015-06-10T07:55:11.447" LastActivityDate="2015-06-10T07:55:11.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16194" PostTypeId="1" AcceptedAnswerId="16195" CreationDate="2015-06-10T07:43:25.883" Score="17" ViewCount="774" Body="&lt;p&gt;An artist recently painted this odd looking mural in my hometown. He called it 'Chromatic Tones'. Most people just walk by it, but I suspect it has a deeper meaning. I have looked at it, increasingly frustrated, over the past few days, but I can't make anything out of it! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EsVXi.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can you decode it?&lt;/strong&gt;&lt;br&gt;&#xA;&lt;sub&gt;If you're stuck, I suggest calling a friend!&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="7308" LastActivityDate="2015-06-10T08:44:27.643" Title="Chromatic Tones" Tags="|cipher|visual|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16195" PostTypeId="2" ParentId="16194" CreationDate="2015-06-10T08:44:27.643" Score="14" Body="&lt;p&gt;The mural reads:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It was a bright cold day in April and the clocks were striking thirteen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Things to note:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The dots come in groups of one to four of the same colour. There is never more than one group of a colour per 3&amp;times;3 panel. There's a red dot in every panel. If there are more groups, they are orange, yellow and so on. Only two panels have black dots. The top left square is always empty.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The mural is encoded like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each 3&amp;times;3 panel represents a word. Each group of dots represents a letter. The letters are ordered by the colours of the rainbow (ROY G BIV) plus black for the eighth letter. A letter is described by the number of dots per group and their position inside the panel: The panel grid resembles a telephone keypad, and the hint supports this. The top left key (1) doesn't have any letters assigned and the only occurrence of four letters is in the bottom left square, one of the two keys (7 and 9) with four letters assigned.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9860" LastActivityDate="2015-06-10T08:44:27.643" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="16196" PostTypeId="2" ParentId="16171" CreationDate="2015-06-10T09:01:38.307" Score="9" Body="&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lUyQy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Words:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;B&lt;/strong&gt;OROMIR&lt;br/&gt;&lt;strong&gt;A&lt;/strong&gt;REDHEL&lt;br/&gt;&lt;strong&gt;G&lt;/strong&gt;ANDALF&lt;br/&gt;&lt;strong&gt;G&lt;/strong&gt;AMLING&lt;br/&gt; &lt;strong&gt;I&lt;/strong&gt;SILDUR&lt;br/&gt;&lt;strong&gt;N&lt;/strong&gt;IMLOTH&lt;br/&gt;&lt;strong&gt;S&lt;/strong&gt;ARUMAN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They are all&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_Middle-earth_characters&quot; rel=&quot;nofollow noreferrer&quot;&gt;Middle Earth Characters&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4506" LastEditorUserId="5373" LastEditDate="2015-06-10T09:07:15.397" LastActivityDate="2015-06-10T09:07:15.397" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16198" PostTypeId="2" ParentId="16169" CreationDate="2015-06-10T10:36:05.003" Score="2" Body="&lt;p&gt;Taking a very large leap, you might be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the number 0.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Key:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Replace &quot;politically&quot; with &quot;0&quot;.&lt;br&gt;&#xA; Replace &quot;being not in violation&quot; with &quot;any number other than 0&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I am on top politically, I am right,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0 / 0 is 1 ==&gt; TRUE&lt;br&gt;&#xA; (0 / 0 is actually really indeterminate, hence the very large leap).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but if I am on top of being not in violation, I am wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0 / x, with x &amp;lt;&gt; 0 is 0 ==&gt; FALSE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I am above 8, I am wrong,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0 / 8 is 0 ==&gt; FALSE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but if I am low on being not in violation, I am forever.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; x / 0, with x &amp;lt;&gt; 0 is infinity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cannot be on top politically and above 8,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0 / 0 isn't more than 8.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I can be on top and low of being not in violation. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0 / x and x / 0 are both valid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9910" LastEditorUserId="9910" LastEditDate="2015-06-10T10:41:50.677" LastActivityDate="2015-06-10T10:41:50.677" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16200" PostTypeId="2" ParentId="16159" CreationDate="2015-06-10T13:11:02.780" Score="10" Body="&lt;p&gt;Your name is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mister&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In good company, I'm a sign of respect &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's respectful to address someone as Mr. Soandso. (&quot;In good company&quot; means alongside a name I guess)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alone, I don't denote man kind &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strike&gt;Not so sure here... maybe the fact that &quot;mister&quot; by itself can mean a water sprayer instead of a man?&lt;/strike&gt; &quot;'Mister' by itself is kind of an insult. 'Now you listen here, mister!' So the idea is that you would call an unkind man 'mister' but not 'mister Smith' or something&quot; -corsiKa&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was the chosen one, by George! &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; George Washington decided on the formal address of &quot;Mr. President&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are times my presence is warranted &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Warrant Officers are addressed &quot;Mr.&quot; ... &lt;strike&gt;and there's a song &quot;Mr. Rainmaker&quot; by the band Warrant. In a less warrant-centric sense, it is common to use &quot;Mr. Lastname&quot; when you don't know someone's first name.&lt;/strike&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I'll usually show up if you're a good sport about it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strike&gt;Misters (i.e. water sprayers) are often used to cool down athletes&lt;/strike&gt; &quot;think of Mr. Hockey, a lot of sports nicknames have mister in them&quot; -corsiKa&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Would you do me the honor of telling me my name?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mr. Mister! :P (Also titles like &quot;mister&quot; are known as &lt;em&gt;honor&lt;/em&gt;ifics. Thanks corsiKa!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4513" LastEditorUserId="4513" LastEditDate="2015-06-10T15:12:20.147" LastActivityDate="2015-06-10T15:12:20.147" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16201" PostTypeId="2" ParentId="16045" CreationDate="2015-06-10T13:26:03.927" Score="8" Body="&lt;p&gt;Jack started singing &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;It's the honky tonk women that gimme gimme gimme the honky tonk blues.&quot; &lt;br&gt; - Honky Tonk Women by The Rolling Stones&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;First Riddle:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The keeper of code&lt;br&gt;has a stone of note&lt;br&gt;Though honor he has&lt;br&gt;his own father he snort&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &quot;keeper of the code&quot; is the title of Edward Teague in Pirates of the Caribbean.  Teague is Jack Sparrow's father, which is why Jack bows in respect. He is played by Keith Richards, guitarist for the Rolling Stones.  Richards, has also admittedly had a drug problem and at one time snorted his fathers ashes.  So, the stone of note is Keith Richards and this is a key for the cipher. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Second Riddle:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A ghostly whisper&lt;br&gt;In the airs&lt;br&gt;Hexed is the master&lt;br&gt;And so are his heirs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Code:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;776c397fa66d801e9102db8390038920c28a4ba398e6f2d6616a221a613620d4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;ghost refers to the GOST cipher. &lt;a href=&quot;http://en.wikipedia.org/wiki/GOST_(block_cipher)&quot;&gt;http://en.wikipedia.org/wiki/GOST_(block_cipher)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Running it here: &lt;a href=&quot;http://www.tools4noobs.com/online_tools/decrypt/&quot;&gt;http://www.tools4noobs.com/online_tools/decrypt/&lt;/a&gt; with key &lt;b&gt;KeithRichards&lt;/b&gt; spits out:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;bhootnikeyettopngklwbogmqevns&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The old lady's clue:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If your first is left, then just take the rights.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Translates to, for each pair of letters, take only the second one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;b&lt;b&gt;h&lt;/b&gt;o&lt;b&gt;o&lt;/b&gt;t&lt;b&gt;n&lt;/b&gt;i&lt;b&gt;k&lt;/b&gt;e&lt;b&gt;y&lt;/b&gt;e&lt;b&gt;t&lt;/b&gt;t&lt;b&gt;o&lt;/b&gt;p&lt;b&gt;n&lt;/b&gt;g&lt;b&gt;k&lt;/b&gt;l&lt;b&gt;w&lt;/b&gt;b&lt;b&gt;o&lt;/b&gt;g&lt;b&gt;m&lt;/b&gt;q&lt;b&gt;e&lt;/b&gt;v&lt;b&gt;n&lt;/b&gt;s&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1777" LastEditorUserId="1777" LastEditDate="2015-06-10T17:39:40.600" LastActivityDate="2015-06-10T17:39:40.600" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16202" PostTypeId="1" CreationDate="2015-06-10T13:55:33.670" Score="12" ViewCount="8391" Body="&lt;p&gt;A lady buys goods worth 200 rupees from the shopkeeper. The goods were reasonable quality merchandise sold for what the market would bear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lady gives the shopkeeper a 1000-rupee note. The shopkeeper runs out of change, and gets change from the next shop. He keeps the 200 rupees (for selling the goods) and returns 800 to the lady. Later the neighbour comes back with the 1000-rupee note, proving it's a fake, tearing it up and demanding his money back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much did the shopkeeper lose?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A) Rs. 200&lt;br&gt;&#xA;B) Rs. 800&lt;br&gt;&#xA;C) Rs. 1200&lt;br&gt;&#xA;D) Rs. 1800&lt;br&gt;&#xA;E) Rs. 1600&lt;br&gt;&#xA;F) Rs. 2000&lt;br&gt;&#xA;G) Other  &lt;/p&gt;&#xA;&#xA;&#xA;" OwnerUserId="13087" LastEditorDisplayName="user88" LastEditDate="2015-06-11T16:49:52.280" LastActivityDate="2018-10-23T14:29:46.277" Title="How much is the shopkeeper's loss?" Tags="|mathematics|" AnswerCount="11" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16204" PostTypeId="2" ParentId="16202" CreationDate="2015-06-10T14:20:58.147" Score="8" Body="&lt;p&gt;The shop keeper lost:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; G) Other (1000rs)&lt;br&gt;&#xA; The lady pays with 1000 rs, the shopkeeper then excanges the note for a 800 &#xA;and 200 note, he keeps the 200 note and gives the 800 + the product to the lady.&lt;br&gt;&#xA;At this point the shop keeper made no loss. &#xA; Then the shop owner of shop b comes and tells him the 1000 note was fake, he takes his 1000rs back.  The shop keeper now has -800rs left. The diffrence between -800rs and 200rs is 1000rs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;poor guy&lt;/p&gt;&#xA;" OwnerUserId="12645" LastEditorUserId="12645" LastEditDate="2015-06-10T14:26:30.453" LastActivityDate="2015-06-10T14:26:30.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16205" PostTypeId="2" ParentId="16202" CreationDate="2015-06-10T14:28:21.843" Score="-1" Body="&lt;p&gt;He lost &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C : 1200&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1000 for the money he gave back and 200 worth of goods&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13088" LastActivityDate="2015-06-10T14:28:21.843" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16206" PostTypeId="2" ParentId="16168" CreationDate="2015-06-10T14:29:08.593" Score="15" Body="&lt;p&gt;This solution involves randomness. The mathematicians will base some of their decisions a coin flip, and with probability $1$ they will eventually know the value of $n$ and announce it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mathematician creating the plan is the &lt;em&gt;captain&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Upper bound for $n$&lt;/strong&gt;: This is done in a number of Rounds: Round $1$, Round $2$, etc.&lt;br&gt; Each Round is subdivided into Step A and Step B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step A of Round $k$ lasts for $k$ days. During this Step, the captain will always turn on their switch. Each other mathematician turns on their switch if and only if they have previously seen a light this Step (so we act as if the captain saw a light when the Step started).&lt;br&gt; At the end of Step A:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;if $k&amp;lt; n$, the number of mathematicians who have seen a light is between $k$ and $2^{k}$,&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;if $k\geq n$, everybody has seen a light.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Step $B$ of Round $k$ lasts for $2^k$ days. During this Step, a mathematician turns on their switch if &lt;strong&gt;either&lt;/strong&gt; they did not see a light during Step A &lt;strong&gt;or&lt;/strong&gt; they have seen a light during Step B.&#xA;&lt;br&gt;&#xA;At the end of Step B:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;if everybody saw a light during Step A, then nobody saw a light during Step B,&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;if somebody did not see a light during Step A, then everybody saw a light during Step B.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;At the end of each round, it is common knowledge whether everybody saw a light during Step A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $k_0$ be the smallest positive integer such that everybody saw a light during Step A of Round $k_0$  (it will necessarily be true that $k_0\leq n$). The value of $k_0$ is common knowledge, and at the end of Round $k_0$ every mathematician knows that $n\leq2^{k_0}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;An &lt;em&gt;announcement&lt;/em&gt; by some collection $S$ of mathematicians is a period of $2^{k_0}$ consecutive days, during which each mathematician in $S$ always turns on their switch, and starting on the second day, everybody else turns on their switch only if they have already seen a light during the announcement. At the end of the announcement:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;if at least one mathematician is in $S$, everybody will have seen a light,&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;if no mathematician is in $S$, nobody will have seen a light.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So it becomes common knowledge whether or not $S$ is empty.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Counting mathematicians&lt;/strong&gt;: At every point during the counting, some collection of $d$ mathematicians will be numbered #1 through #$d$. The remaining mathematicians remain unnumbered. To begin, the captain is #1 and every other mathematician is unnumbered. The mathematicians will repeatedly attempt to assign an additional mathematician a number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before each attempt, an announcement is made by every unnumbered mathematician. If there are no unnumbered mathematicians, this fact becomes common knowledge, and everyone can announce the largest number that has been assigned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the first day of the attempt, each numbered mathematician chooses whether or not to turn on their switch based on fair coin flip. Everybody who sees a light on this day is called a &lt;em&gt;candidate&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, there are $d+1$ announcements. For $j\leq d$, the $j$-th announcement is made by mathematician #$j$ if they turned on their switch on the first day, and by nobody otherwise. The $(d+1)$-st announcement is made by every unnumbered candidate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If exactly one of the numbered mathematicians made an announcement, &lt;strong&gt;and&lt;/strong&gt; an unnumbered candidate made an announcement, then the (necessarily unique) unnumbered candidate is assigned #$d+1$; the attempt was a success. Otherwise no additional mathematician is numbered; the attempt was a failure. It is common knowledge whether a new number was assigned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So long as there is a least one unnumbered mathematician, the probably of an attempt succeeding is positive. With probability $1$, every mathematician will eventually be numbered.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-06-10T17:05:35.957" LastActivityDate="2015-06-10T17:05:35.957" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16207" PostTypeId="2" ParentId="16202" CreationDate="2015-06-10T14:31:43.920" Score="-2" Body="&lt;p&gt;He lost&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A. 200. Which is the worth of the goods. I assume here that he has no money to begin with. Which could be a reason he had to change with the next shop. He begins and he ends with no money so he only lost the goods. The next shop's shopkeeper has lost 800 because he only could get 200 back from the first shopkeeper from his 1000. The first shopkeeper is actually still 800 in debt with the second shopkeeper but he hasn't lost it yet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-06-10T14:31:43.920" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16208" PostTypeId="2" ParentId="16202" CreationDate="2015-06-10T14:33:14.473" Score="40" Body="&lt;ol&gt;&#xA;&lt;li&gt;Shopkeeper starts with 200rs (in the form of goods)&lt;/li&gt;&#xA;&lt;li&gt;After transacting with the lady, he has 200rs in cash.&lt;/li&gt;&#xA;&lt;li&gt;After the moneychanger shows the banknote is counterfeit, he owes the moneychanger 1000rs.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - Current Balance = 200rs (cash) - 1000rs (debt) = -800rs&lt;br&gt;&#xA; - Initial Balance = 200rs (goods)&lt;br&gt;&#xA; - Profit/Loss = Current Balance - Initial Balance = -1000rs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The lady started with nothing of value and gets 200rs in goods and 800rs in cash. The moneychanger does not gain or lose money. The 1000rs loss is borne by the shopkeeper.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11752" LastEditorUserId="11752" LastEditDate="2015-06-10T19:05:26.510" LastActivityDate="2015-06-10T19:05:26.510" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16209" PostTypeId="2" ParentId="16147" CreationDate="2015-06-10T15:01:02.450" Score="6" Body="&lt;p&gt;In case it helps somebody, I have transcribed the glyphs into a copyable format.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Wall carving:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;L2  F4  J3  O1  F2  F1  G1&#xA;J4  F2  O1  F3  C4  G4  P1&#xA;P4  O1  G2  G4  L3  O1  G3&#xA;F1  P3  L1  C1  O1  G4  F1&#xA;J3  C4  O1  L2  F4  J3  O1&#xA;C3  L4  C2  P2  O1  J1  G4&#xA;J2&#xA;&lt;/code&gt;&lt;/pre&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Cube faces:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;G1  O1  J4  L4  P4  O1  F4  J3      F1  L2  L2  J3  C4  O1  P3&#xA;C3  L1  O1  P2  G4  F1  O1  L2      P2  O1  P4  L4  P3  J3  O1&#xA;C4  L4  P4  C1  J4  J3  P4  J1      L4  P4  J1  O1  F3  J3  J4&#xA;O1  L2  F4  J3  O1  C3  G1  P3      G4  P3  J3  O1  G2  G4  C4&#xA;G1  L2  C1  O1  G4  G2  O1  P2      J3  F1  J3  C4  O1  J4  F4&#xA;G4  F1  C4  O1  J3  P4  F1  G1      L4  P4  J2  J3  J1&#xA;C4  G4  P4  P3  J3  P4  L2&#xA;&#xA;G1  O1  J4  L4  P4  O1  C1  F4      F1  L2  L2  J3  C4  O1  P3  P2&#xA;G4  P1  O1  P2  G4  F1  O1  L4      O1  P4  L4  P3  J3  O1  L4  P4&#xA;O1  P1  G4  C4  C3  J1  O1  G2      J1  O1  C1  J3  J3  O1  L2  F4&#xA;C4  G4  P3  O1  L4  O1  F1  G1      J3  O1  P1  G4  C4  C3  J1  O1&#xA;J3  P1  O1  P2  G4  F1  O1  J4      G2  G4  C4  O1  P1  F4  L4  L2&#xA;G4  F1  C3  J1  P4  L2  O1  G1      O1  G1  L2  O1  L2  C4  F1  C3&#xA;P3  L4  J2  G1  P4  J3              P2  O1  G1  C1&#xA;&#xA;G1  O1  L4  P3  O1  P2  G4  F1      F1  L2  L2  J3  C4  O1  P3  P2&#xA;C4  O1  J2  F1  G1  J1  J3  O1      O1  P4  L4  P3  J3  O1  L4  P4&#xA;L2  G4  O1  L2  F4  J3  O1  C3      J1  O1  F3  J3  O1  L2  L4  F2&#xA;G4  J4  L4  L2  G1  G4  P4  O1      J3  P4  O1  G4  P4  O1  L4  O1&#xA;L2  F4  L4  L2  O1  G1  C1  O1      G3  G4  F1  C4  P4  J3  P2  O1&#xA;J3  F1  J3  C4  P2  P1  F4  J3      C3  G1  F2  J3  O1  P4  G4  O1&#xA;C4  J3                              G4  L2  F4  J3  C4&#xA;&lt;/code&gt;&lt;/pre&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note that I rotated +90 degrees first (so that the ragged edge fits with typical LTR writing).  Here is the glyph mapping (after rotation):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8nF7R.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Frequency Analysis (Glyphs)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Lae2z.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;O&lt;/code&gt; is by far the most common.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5GOnT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of the glyphs is present in each of its rotations, but one or two of the rotations are far more common.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NtWwt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;O&lt;/code&gt; is the only glyph shape that doesn't occur twice in a row: I suspect it may be a spacer of some sort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Tx0fr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By far the most common (13 times) pair is &lt;code&gt;J3 O1&lt;/code&gt;.  The only glyphs which occur twice in a row with the same rotation are &lt;code&gt;L2&lt;/code&gt; (three times) and &lt;code&gt;J3&lt;/code&gt; (once).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Frequency Analysis (Words?)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;It seems that treating &lt;code&gt;O&lt;/code&gt; as a space makes some sort of sense.  If we break up each text into words, discarding the &lt;code&gt;O&lt;/code&gt;s, we end up with the following distribution of word lengths:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lv8vy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to match the distribution given &lt;a href=&quot;http://norvig.com/mayzner.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.  Looking at word frequencies, we see that several words occur more than once:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;L2 F4 J3         5&#xA;G1               3&#xA;L4               3&#xA;P3 P2            3&#xA;L4 P4 J1         3&#xA;P2 G4 F1         3&#xA;P4 L4 P3 J3      3&#xA;F1 L2 L2 J3 C4   3&#xA;G1 C1            2&#xA;J4 L4 P4         2&#xA;P2 G4 F1 C4      2&#xA;P1 G4 C4 C3 J1   2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Again, this seems to match up pretty well against English.  I'm wary of proceeding too far down this route though, since OP said &quot;it isn't simply a 1:1 conversion into English text.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Todo: placement of repeated words.)&lt;/p&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="9718" LastEditDate="2015-06-11T14:33:04.817" LastActivityDate="2015-06-11T14:33:04.817" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16211" PostTypeId="2" ParentId="16159" CreationDate="2015-06-10T15:13:50.423" Score="-1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Curiosity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In good company, I'm a sign of respect&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Curiosity shows that you wish to learn more about a subject say in a class, curiosity makes you ask questoins which &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alone, I don't denote man kind&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Curiosity results in questions and that needs to be answered which cannot be done if you are alone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was the chosen one, by George!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Curios George&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are times my presence is warranted&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Say, when you are learning a new topic,  curiosity furthers your knowledge.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I'll usually show up if you're a good sport about it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you sow interest in a topic , your curiosity arises.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Would you do me the honor of telling me my name?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Curiosity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2656" LastActivityDate="2015-06-10T15:13:50.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16212" PostTypeId="2" ParentId="16084" CreationDate="2015-06-10T16:34:50.837" Score="5" Body="&lt;p&gt;This is about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; seven-segment numerical displays. $\mathfrak{C}^+$ returns the union of the segments that are active in its inputs. The numbers in the I/Os are simply shorthand for the set of active segments used to form those digits. In the case of outputs that don't quite match any digits, approximate matches are used. $\sim6$ represents a left-to-right-mirrored 6, and the 9 in the final equation has the bottom segment missing (but is still recognizably a 9).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12566" LastEditorUserId="12566" LastEditDate="2015-06-11T21:22:00.947" LastActivityDate="2015-06-11T21:22:00.947" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16213" PostTypeId="1" AcceptedAnswerId="16253" CreationDate="2015-06-10T17:32:08.370" Score="12" ViewCount="662" Body="&lt;p&gt;Below is a word search puzzle presented without clues. Instead, you must use the additional rules (see example below) to work out what the words are. Finally, as a check to make sure you found the right words, the first letters of each word can be rearranged to make one last, related word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the fifth and final puzzle in a series. Here's the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16087/cluelessly-searching-for-words&quot;&gt;first&lt;/a&gt;, &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16094/cluelessly-searching-for-words-ii&quot;&gt;second&lt;/a&gt;, &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16122/clueslessly-searching-for-words-iii&quot;&gt;third&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16171/cluelessly-searching-for-words-iv/&quot;&gt;fourth&lt;/a&gt;. I'll be a bit surprised if this one is solvable without hints.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Rules&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The words can run in any direction (orthogonal or diagonal) and do not have to be in straight lines&lt;/li&gt;&#xA;&lt;li&gt;Words do not overlap and their paths do not cross&lt;/li&gt;&#xA;&lt;li&gt;All letters are part of some word&lt;/li&gt;&#xA;&lt;li&gt;All words are of the same length&lt;/li&gt;&#xA;&lt;li&gt;All words (including the final word) are related somehow&lt;/li&gt;&#xA;&lt;li&gt;All words (including the final word) have an English Wikipedia page, even if they are not English themselves&lt;/li&gt;&#xA;&lt;li&gt;The letters that make up the final word do not appear in any specific location on the grid&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Example&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Here's a short example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bAwdg.png&quot; alt=&quot;P O A W. E N N S. M U Y A. S T O D.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GRuBj.png&quot; alt=&quot;PONY EMUS TOAD SWAN&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The four words here are PONY, EMUS, TOAD and SWAN, which are all animals, and the first letter of each makes the word PETS.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Puzzle&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;What is the final word to be found in this grid of letters?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sFVFW.png&quot; alt=&quot;O R A H A D K W A L. T H L O R U Y T E A. M O S E K E L I T T. E I A A L E S A P O. R S H U T L K U N E. I U A R I P O V L A. L S U T A O T N A T. I O S R R R V O E T. S U N A I O E A S I. U C H A L T A N C R.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Hint&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All the words are taken from one Wikipedia list page&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7392" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2015-06-11T22:17:03.453" Title="Cluelessly searching for words V (last one)" Tags="|word-search|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16214" PostTypeId="2" ParentId="16202" CreationDate="2015-06-10T17:40:49.290" Score="9" Body="&lt;p&gt;A slightly simpler explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He lost 1000.&#xA; The lady started with 0rs worth of stuff and ended with 1000rs worth of stuff.  No money / value was created or destroyed in this scenario, so that 1000rs had to come from either B or the money lender.  It didn't come from the money lender, so it had to come from shop keeper B.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10641" LastActivityDate="2015-06-10T17:40:49.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16215" PostTypeId="2" ParentId="16202" CreationDate="2015-06-10T18:49:48.223" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; G. Other&lt;br/&gt;&#xA;He didn't lose anything.&#xA;&#xA; &lt;p&gt;The riddle never specifies that he refunded shop A his 1000rs, and given the situation there is no reason for him to have done so.  Given the lack of supervision by shopkeeper B on the 1000rs bill while in the possession of shopkeeper A, there is nothing that would suggest the bill is the same one the lady bought goods with.&lt;/p&gt;&#xA;As the goods sold were at no profit, the total change in revenue is 0.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="457" LastEditorUserId="11239" LastEditDate="2015-06-11T06:05:18.543" LastActivityDate="2015-06-11T06:05:18.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16216" PostTypeId="2" ParentId="15977" CreationDate="2015-06-10T18:54:57.147" Score="1" Body="&lt;p&gt;Could it be:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Man Kind  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It starts off as a creation of God,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Man was created by God  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;followed by a godly creation.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Kindness we would hope is a godly quality.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Having them together is odd,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Finding a kind man is often a rarity anymore.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but between them is a one way affection.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Men love kindness, but tend to not love to be kind.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A disgusting symbol of disparity&lt;br&gt;&#xA; between wealth and poverty.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We are often most kind to those that are &quot;&lt;em&gt;below&lt;/em&gt;&quot; us, or the poor. Sometimes that kindness is almost mocking.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet it is an irony&lt;br&gt;&#xA; as a proof of gender equality.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One could argue that men are kind to women as a form of appeasing them, or trying to make up for different inequalities.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-06-10T18:54:57.147" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16217" PostTypeId="1" CreationDate="2015-06-10T19:51:00.680" Score="6" ViewCount="626" Body="&lt;p&gt;Variant of: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15673/labeling-wires-in-a-bundle&quot;&gt;Labeling wires in a bundle&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a remote location, you just finished trenching a data cable across a large plot of land. The cable has 64 individual wires that are not color-coded or labeled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a wire stripper, a simple electrical continuity tester, and a label maker. There is sufficient excess at either end to allow connecting and reconnecting the individual wire ends in whatever combinations you wish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Your boss, who is long gone for the day, calls you with some bad news. During installation, the cable got damaged, and some of the wires may be broken. You are instructed to determine how many wires in the cable are still good and also individually label them.  If it happens that more than 50% were broken and you can prove this, you need not proceed further.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a long walk and you are tired.  What is the fewest number of trips from one end of the cable to the other required to find and uniquely label each individual intact wire in the cable?&#xA;&lt;img src=&quot;https://i.stack.imgur.com/WrEIQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The continuity tester is a sealed unit. You can't pull the battery out and leave it behind.&lt;/li&gt;&#xA;&lt;li&gt;It's not necessary to make a final trip after the wires are labeled just to clean up. If they are all labeled at both ends, the job is done.&lt;/li&gt;&#xA;&lt;li&gt;If there are $n$ good wires, their ends must wind up with matched labels from 1 to $n$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11752" LastEditorUserId="5373" LastEditDate="2018-06-24T00:37:52.603" LastActivityDate="2018-06-24T00:37:52.603" Title="Labeling wires in a *damaged* bundle" Tags="|logical-deduction|story|combinatorics|network|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16219" PostTypeId="1" AcceptedAnswerId="16230" CreationDate="2015-06-10T20:17:14.127" Score="58" ViewCount="3665" Body="&lt;p&gt;If a phrase adheres to a certain rule, then I call it a Cyclone Phrase™.&lt;/p&gt;&#xA;&lt;p&gt;Use the examples below to find the rule. Also find the one error in the chart below. (In other words, one of the phrases below is not really a Cyclone Phrase™.)&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th&gt;Cyclone Phrases&lt;/th&gt;&#xA;&lt;th&gt;not Cyclone Phrases&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;all alone at noon&lt;/td&gt;&#xA;&lt;td&gt;by myself at twelve pm&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;blurb in my blog&lt;/td&gt;&#xA;&lt;td&gt;response on my online journal&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;deerweed and aloe&lt;/td&gt;&#xA;&lt;td&gt;lavender and honey&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;chic dimple&lt;/td&gt;&#xA;&lt;td&gt;attractive facial wrinkle&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;it is adjourned&lt;/td&gt;&#xA;&lt;td&gt;it is suspended&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;adored among guys&lt;/td&gt;&#xA;&lt;td&gt;loved among men&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;drug and dose&lt;/td&gt;&#xA;&lt;td&gt;medicine and dosage&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;chug our ale&lt;/td&gt;&#xA;&lt;td&gt;drink our beer&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;an eerie moon glyph&lt;/td&gt;&#xA;&lt;td&gt;a strange lunar symbol&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;biotypic fish&lt;/td&gt;&#xA;&lt;td&gt;fishes of the same genotype&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;bested by my mom&lt;/td&gt;&#xA;&lt;td&gt;defeated by my brother&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;an old bum begged&lt;/td&gt;&#xA;&lt;td&gt;an aged vagrant mooched&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;dehorned an elk&lt;/td&gt;&#xA;&lt;td&gt;remove a wapiti's horns&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;cook and dine&lt;/td&gt;&#xA;&lt;td&gt;heat and eat&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;an astute moron&lt;/td&gt;&#xA;&lt;td&gt;a sharp idiot&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;fork, cup, and dish&lt;/td&gt;&#xA;&lt;td&gt;spoon, knife, and napkin&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;a deformed deltoid&lt;/td&gt;&#xA;&lt;td&gt;a misshapen shoulder&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;blood and bile&lt;/td&gt;&#xA;&lt;td&gt;vital body fluids&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;blond belle&lt;/td&gt;&#xA;&lt;td&gt;light-haired beauty&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;anti and biotic&lt;/td&gt;&#xA;&lt;td&gt;antibiotic&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;choke a brute&lt;/td&gt;&#xA;&lt;td&gt;strangle a beast&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;our foul deed&lt;/td&gt;&#xA;&lt;td&gt;our offensive action&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;cenozoic deluge&lt;/td&gt;&#xA;&lt;td&gt;paleogene flood&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;If you liked this puzzle, try its successors:&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/16485/what-is-a-scalable-phrase%E2%84%A2&quot;&gt;What is a Scalable Phrase™?&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/16743/what-is-a-triad-phrase%E2%84%A2&quot;&gt;What is a Triad Phrase™?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="63414" LastEditDate="2022-07-14T19:13:41.810" LastActivityDate="2022-07-15T14:57:08.810" Title="What is a Cyclone Phrase™?" Tags="|pattern|english|word-property|" AnswerCount="3" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="16222" PostTypeId="2" ParentId="16217" CreationDate="2015-06-10T21:29:55.947" Score="3" Body="&lt;p&gt;2 trips. Shorten one end by connecting all edges. Check continuity in pairs in the other end - mark them as not broken. If less than 50%, 32, done. Shorten the good ones at that end and make a second trip to the other end - disconnect and check for continuity - mark the ones connected. Done!&lt;/p&gt;&#xA;" OwnerUserId="9531" LastActivityDate="2015-06-10T21:29:55.947" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16223" PostTypeId="2" ParentId="15977" CreationDate="2015-06-10T21:57:37.120" Score="8" Body="&lt;p&gt;Try 2 after the hint:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Man boobs  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It starts off as a creation of God,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, man created by god.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;followed by a godly creation.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most men probably think breasts are sent from heaven.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Having them together is odd,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Man boobs are not the norm, though I can't say too much :-( &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but between them is a one way affection. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Men love boobs. Boobs don't really love men.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A disgusting symbol of disparity&lt;br&gt;&#xA; between wealth and poverty.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Those with a lot of wealth can buy bigger boobs, or clothing items that show them off.  Also, someone may be &quot;well endowed&quot; with big boobs. From the OP, those that are wealthy often can afford more food (and weight used to be a symbol of wealth), so wealthy men are more likely to have man boobs. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet it is an irony&lt;br&gt;&#xA; as a proof of gender equality. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Women (almost) always have boobs, men often don't. Definately some inequality there. Plus, a lot of men treat women differently because of their boobs.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="9951" LastEditDate="2015-06-10T22:03:42.760" LastActivityDate="2015-06-10T22:03:42.760" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16224" PostTypeId="2" ParentId="16202" CreationDate="2015-06-10T22:32:01.180" Score="4" Body="&lt;p&gt;The shop Keeper lost:   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 800rs worth of cash + 200rs worth of good   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's the details:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Shopkeeper B's initial balance:&lt;/strong&gt;&lt;br&gt;&#xA; 0rs + 200rs worth of good&lt;br&gt;&#xA;&#xA; &lt;strong&gt;After trade balance:&lt;/strong&gt;&lt;br&gt;&#xA; 200rs worth of cash and -200rs worth of good&lt;br&gt;&#xA; &lt;strong&gt;After neighbor return with fake balance:&lt;/strong&gt;&lt;br&gt;&#xA; 200rs - (200rs + 800rs) worth of cash and -200rs worth of good  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="68" LastEditDate="2018-10-23T14:29:46.277" LastActivityDate="2018-10-23T14:29:46.277" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="16225" PostTypeId="2" ParentId="16219" CreationDate="2015-06-10T23:32:13.820" Score="1" Body="&lt;p&gt;A Cyclone Phrase is one that &lt;strong&gt;either&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; contains two words which both have the same letter in the same position (e.g. &quot;&lt;strong&gt;d&lt;/strong&gt;rug and &lt;strong&gt;d&lt;/strong&gt;ose&quot; or &quot;ch&lt;strong&gt;u&lt;/strong&gt;g o&lt;strong&gt;u&lt;/strong&gt;r ale&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; contains a word in which the same letter appears the same number of places from both the start and the end (e.g. &quot;&lt;strong&gt;b&lt;/strong&gt;lur&lt;strong&gt;b&lt;/strong&gt; in my blog&quot;, &quot;an astute m&lt;strong&gt;o&lt;/strong&gt;r&lt;strong&gt;o&lt;/strong&gt;n&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or both. The one error is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;fork, cup, and dish&quot; is not a Cyclone Phrase.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I got to this by wondering why it was called a &lt;em&gt;Cyclone&lt;/em&gt; Phrase and looking for some kind of circularity or cyclicity in the letters, which led me to the ideas of alliteration and symmetry. Most of the examples have either two alliterative words (&quot;&lt;strong&gt;a&lt;/strong&gt;ll &lt;strong&gt;a&lt;/strong&gt;lone at noon&quot;) or a word starting and ending with the same letter (&quot;&lt;strong&gt;c&lt;/strong&gt;enozoi&lt;strong&gt;c&lt;/strong&gt; deluge&quot;). From there, finding the actual pattern wasn't too hard.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-06-10T23:32:13.820" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16226" PostTypeId="2" ParentId="16219" CreationDate="2015-06-10T23:47:29.840" Score="12" Body="&lt;p&gt;A Cyclone Phrase™ is one in which&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; for each word, the letters in the first half of the word (including the middle letter, if there is one) are in alphabetical order, and the letters in the second half of the word (again including the middle)  are in reverse alphabetical order.&lt;br&gt;&lt;strong&gt;Example&lt;/strong&gt;:\begin{array}{|l|l|l|}\hline\text{Word}&amp;amp;\text{First half}&amp;amp;\text{Second half}\\\hline \text{all}&amp;amp;\text{al}&amp;amp;\text{ll}\\\text{alone}&amp;amp;\text{alo}&amp;amp;\text{one}\\\text{at}&amp;amp;\text{a}&amp;amp;\text{t}\\\text{noon}&amp;amp;\text{no}&amp;amp;\text{on}\\\hline\end{array}The one exception is &quot;an old bum begged&quot;. The word 'old' does not have this property: the first half, 'ol', is not in alphabetical order.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-06-11T00:16:13.480" LastActivityDate="2015-06-11T00:16:13.480" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="16228" PostTypeId="2" ParentId="16168" CreationDate="2015-06-11T00:39:43.523" Score="5" Body="&lt;p&gt;The task may be established, in a huge but bounded number of moves, by first establishing an upper bound M on the number of prisoners (possibly using Julian Rosen's approach) and repeatedly performing the following steps:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Observe the number of days elapsed as D.&lt;/li&gt;&#xA;&lt;li&gt;Each prisoner will have seen one of 2^D possible on/off sequences thus far.&lt;/li&gt;&#xA;&lt;li&gt;Pick a sequence for each of the next 2^D groups of M days; on the first day of each group, have prisoners turn on their light if the initial sequence of lights they've seen is appropriate for that group.  On the remaining days, have prisoners turn on their light if they're in the group or have seen the light from anyone else that was.&lt;/li&gt;&#xA;&lt;li&gt;Once all sequences have been enumerated, start over unless the information observed from all the announcements is sufficient to ascertain the total number of prisoners.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Without using randomness, it is not possible to ensure that every prisoner will ever end up seeing a unique sequence of lights; if there are N prisoners, the warden could arrange to have up to N/2 of them that always see the same sequence.  On the other hand, every light that gets turned on each day must be accounted for; this will make it possible to infer the size of larger cliques.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, start with a single node turning on his light.  Exactly one other prisoner will see that, implying there are two cliques with one prisoner each.  If both see a light the next day, they're the only prisoners.  Otherwise, if either of those first two prisoners sees the light of the other on the next day, that will imply the existence of another single-prisoner clique (single-prisoner cliques, once identified as such, aren't very interesting).  If neither saw a light, that implies that the light was seen by two other people.  On the next day, four lights will be switched on, and it will be possible to tell how many of the first four people see them.  If none see them, then on the next day eight lights will be switched on.  Here's the first real difficult spot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If either all four or none of the new people see the light on the eight-light day, it will be easy to tell exactly how many other people are going to see the light.  Things get tricky if some do but not all.  That will create an uncertainty of +/-2 with the number of lights that will be seen in the next step.  Further steps will have even more uncertainty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the uncertainties will get resolved, however, by watching what happens when the cliques from the set of four are asked to broadcast their existence, and different nodes are asked who saw what when afterward; I'm still working on the exact cases, but I think they all work.&lt;/p&gt;&#xA;" OwnerUserId="1467" LastActivityDate="2015-06-11T00:39:43.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16230" PostTypeId="2" ParentId="16219" CreationDate="2015-06-11T01:43:06.747" Score="44" Body="&lt;p&gt;Each word in a cyclone phrase has&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;its letters in alphabetical order if you alternate taking them from the left and right sides. For example: dehorned --&amp;gt; ddeehnor, and adjourned --&amp;gt; addejnoru.&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; A cyclone phrase is made of cyclone words only. A word is called a cyclone word because it is in alphabetical order if you go around it like this:&lt;br /&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/4XBV3.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="17297" LastEditDate="2022-07-15T14:57:08.810" LastActivityDate="2022-07-15T14:57:08.810" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="16231" PostTypeId="2" ParentId="1980" CreationDate="2015-06-11T02:43:09.557" Score="3" Body="&lt;p&gt;2 is looking at a white hat so he knows 1 would declare he was wearing a black hat if 2 were wearing white (and there would be no other options). Since he doesn't, 2 knows he must be wearing black.&lt;/p&gt;&#xA;" OwnerUserId="13110" LastActivityDate="2015-06-11T02:43:09.557" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16232" PostTypeId="2" ParentId="16168" CreationDate="2015-06-11T03:34:48.890" Score="4" Body="&lt;p&gt;We can solve this in a brute force way by construction a universal strategy. We will systematically categorize all the information a prisoner could possibly have and then, noting that we can &lt;em&gt;generate&lt;/em&gt; all that information in countably many turns, we will proceed to generate that information. I have no idea what information is useful or how to interpret the results, but hey, if our mathematicians are perfectly logical, they can figure it out! (There's nothing to distract them, even!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that if a certain set of prisoners turns their lights on one turn, and then the same set* turn their lights on again, we gain no information. Moreover, notice that whatever information we do gain is a function of our strategy - that is, the pattern of lights flashing and not flashing that a prisoner observes is meaningless without knowing what groups were instructed to turn their lights on at each turn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, we construct the following &quot;hierarchy&quot; of information, where $S_n$ represents the set of possible pieces of information on the $n^{th}$ turn. In particular, let:&#xA;$$S_0=\{\text{you},\text{other}\}$$&#xA;be the initial bit of information that all prisoners have - they either are you or are not. Any instruction for the first turn is of the form &quot;&lt;em&gt;you&lt;/em&gt; (should/shouldn't) turn your light on and others (should/shouldn't) turn their lights on&quot; - that is, we select a subset of $S_0$ and instruct anyone having that information to turn their lights on. In particular, we find out that, until new information is learned, all the prisoners may know is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;What they knew from $S_0$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Whether they saw a light, given a subset of $S_0$ which turned their lights on.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We express this as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_1=S_0\times \mathbb P(\mathbb P(S_0))$$&#xA;where $\times$ is a Cartesian product and $\mathbb P(S)$ is the power set of $S$ and the latter set of the product represents the set of subsets of $S_0$ for which a given prisoner saw a light (assuming all subsets were tried). We may repeat this argument to create the set&#xA;$$S_{n+1}=S_n\times \mathbb P(\mathbb P(S_n))$$&#xA;which is related similarly to $S_n$ as before. Then, our strategy becomes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Enumerate all subsets of $S_0$, telling prisoners corresponding to each subset to turn their lights on. This sorts prisoners into the subsets given in $S_1$. Enumerate the subsets of $S_1$. Proceed until someone proves that there are exactly $n$ prisoners.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, this strategy is not very efficient, given that the size of the $S_n$ grows as fast as tetration. But, it is notable for being universal - meaning that &lt;em&gt;every other strategy&lt;/em&gt; is simply this one, where certain days are skipped. Therefore, given that this challenge is possible at all, this strategy works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For convenience, one may note that for $n=1$ we may see that any prisoner who receives the information (in $S_1$) of:&#xA;$$(you,A)$$&#xA;where $A$ is a set containing $\{you\}$ as an element (meaning you saw a light when you turned yours on and no one else did), then you have proven that $n=1$. I think it goes without saying that the following set from $S_3$ proves that $n=2$**:&#xA;$$\Bigg(\!\!\bigg(you,\Big\{\!\!\Big\{\!\big(you,\varnothing\big),\big(other,\{\!\{you\}\!\}\!\big)\!\Big\}\!\!\Big\}\bigg),\bigg\{\!\!\bigg\{\!\bigg(\!other,\Big\{\!\Big\{\big(you,\varnothing\big),\big(other,\{\!\{you\}\!\}\big)\Big\}\!\Big\}\bigg)\!\bigg\}\!\!\bigg\}\Bigg).$$&#xA;I'm sure the mathematicians will have no trouble figuring this all out - hey, I'm only the guy who provides the strategy, not who interprets its meaning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*&quot;Same set&quot; in the sense of &quot;no matter how the warden has acted or what turns have come in between, the same prisoners will act&quot; - which is to say that if we address the same set of prisoners in two different ways so that the sets &lt;em&gt;feasibly&lt;/em&gt; could be different but turn out to be the same, then we may gain information)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(**Although no one will ever actually receive the given set; many of the subsets would need to be expanded - but in the partial order generated by the self-referencing relations $(A,B)\geq (A,B')$ if $\forall b'\in B'\exists b\in B\forall a'\in b'\exists a\in b[a'\geq a]$ and $(A,B)\geq (A',B)$ if $A\geq A'$, the given element has the property that all greater elements &lt;em&gt;also&lt;/em&gt; provide a proof - this order is essentially ordering $S_n$ by saying $X\geq Y$ whenever $X$ represents more observations of lights than $Y$)&lt;/p&gt;&#xA;" OwnerUserId="4765" LastEditorUserId="2627" LastEditDate="2015-06-11T14:56:51.913" LastActivityDate="2015-06-11T14:56:51.913" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16234" PostTypeId="1" AcceptedAnswerId="16237" CreationDate="2015-06-11T06:43:57.490" Score="49" ViewCount="4249" Body="&lt;p&gt;The vicious wizard N&lt;strong&gt;evil&lt;/strong&gt;le has trapped you in the middle of a magical circle of radius $10000$, and you have to find a way out.&lt;br&gt;&#xA;Every time you want to take a step (of length $1$) in a certain direction, Neville decides whether you move in that direction or the opposite one.&lt;br&gt;&#xA;Of course, the wizard does his best to hinder you, not to let you escape!&lt;br&gt;&#xA;This is an example:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mk8AG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is there a strategy to escape? If so, how long is the path to freedom?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-06-13T17:41:31.620" LastActivityDate="2015-06-13T17:41:31.620" Title="The vicious wizard...and you!" Tags="|mathematics|geometry|strategy|story|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16235" PostTypeId="2" ParentId="16055" CreationDate="2015-06-11T06:50:27.747" Score="2" Body="&lt;p&gt;There have been several answers that explain a myriad of different paths to the correct answer; IMO, none of them actually provide an adequate resolution of the paradox. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, let me be clear that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Alice's correct course of action is to press the button&lt;/li&gt;&#xA;&lt;li&gt;She has correctly reasoned that her team expects to lose \$1000 if they stick to the plan.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The flaw in Alice's reasoning is that she incorrectly accounts for the information that Bob will think the same as she does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose Alice was unsure how Bob would react, and estimates that if Bob is called, he will stick to the plan with probability $p$. Then:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;With $1/3$ probability, Bob is still in his cell. If Alice sticks to the plan, they win \$5000, otherwise they break even.&lt;/li&gt;&#xA;&lt;li&gt;With $2p/3$ probability, Bob was picked and presses the button. If Alice sticks to the plan, they lose \$4000, otherwise they lose \$2000.&lt;/li&gt;&#xA;&lt;li&gt;With $2(1-p)/3$ probability, Bob was picked and doesn't press the button. If Alice sticks to the plan, they lose \$2000, otherwise they break even.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Adding it up, Alice reasons that the expected value of pressing the button is $\$333 - \$1333 \cdot p$, and of not pressing the button is $-\$1333 \cdot p$. Thus, Alice infers that pressing the button is correct no matter how she thinks Bob will react.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's tempting to make a similar analysis where Bob has some probability of making the same choice as Alice, and otherwise comes up with some other random line of reasoning as above, but this is &lt;em&gt;wrong&lt;/em&gt;: Bob doesn't have information about Alice's choice, so he can't factor that into his reasoning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While our given information that Bob thinks very similarly to Alice means that his choice will be &lt;em&gt;strongly correlated&lt;/em&gt; with Alice's, it must still be made &lt;em&gt;independently&lt;/em&gt; of Alice's. Mathematically, we have&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;P(Bob makes the same choice as Alice) = nearly 1&#xA;P(Bob pushes | Alice pushes) = P(Bob pushes)&#xA;P(Bob doesn't push | Alice doesn't push) = P(Bob doesn't push)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;When Alice and Bob's choices are truly dependent, things can change. See Charlie below.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This mistake of accounting for her knowledge of Bob can be viewed as incorrectly assigning credit/blame to her teams actions: the blame for losing \$4000 if they stick to the strategy is evenly divided between Alice and Bob. With the blame allocated correctly, she reasons that pressing the button is worth $\$5000 \cdot \frac{1}{3} - \$2000 \cdot \frac{2}{3} \approx \$333$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It may be instructive to compare everything with another variation of the game. The millionaire kidnaps Charlie and gives him a big red button that has the same effect as if Alice and/or Bob pressed their buttons. (i.e. Charlie wins \$5000 if one was picked and loses \$4000 if both were picked)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The game the millionaire plays with Charlie is that he will reveal to Charlie whether or not Alice was picked, and then Charlie has to decide whether or not to press his button.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When Charlie learns that Alice was picked, his correct decision is not to press the button.&lt;/p&gt;&#xA;" OwnerDisplayName="user1502" LastEditorDisplayName="user1502" LastEditDate="2017-12-11T00:44:11.193" LastActivityDate="2017-12-11T00:44:11.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16236" PostTypeId="2" ParentId="16234" CreationDate="2015-06-11T07:07:26.013" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There certainly is a way to the border. Every step you should construct the shortest line that intersect the circle at two points, and your current position. Then you take a step on that line. It doesn't matter in what way Neville sends you, because it is symmetric. However, every step is closer to the border then you were before. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The (possible) distance:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I do not know. There probably is some mathematical formula, but I'm currently too busy. I think a possible solution is spiralling outwards.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3993" LastActivityDate="2015-06-11T07:07:26.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16237" PostTypeId="2" ParentId="16234" CreationDate="2015-06-11T07:08:35.450" Score="42" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes, there is a strategy to escape! Imagine a circle centred at your original location and extending to your current location. Move tangent to it; the opposite direction will also be tangent to it, and both will be farther from the middle than where you started. After $n$ steps, you’re $\sqrt n$ steps away from the middle and can make it out in 100000000 steps. Lots of walking ahead!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1565" LastActivityDate="2015-06-11T07:08:35.450" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16239" PostTypeId="2" ParentId="16234" CreationDate="2015-06-11T09:45:38.197" Score="32" Body="&lt;p&gt;To escape you need a magical rucksack, but it is theoretically possible.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Always walk tangential to a circle centred at the magical circle's origin with a radius to your current position. It does not matter if the mage choses the &quot;red&quot; or the &quot;green&quot; step as far as distance from the centre is concerned.&#xA; &lt;br&gt;&lt;br&gt;&#xA; If $s$ is your step-length (arrow length) and $r$ (yellow) your current distance from the centre, then each step takes you $\sqrt{r^2 + s^2}- r$ closer to the border.&lt;br&gt;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/RYAGG.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In other words:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Starting from the centre, after the first step you are $s$ away from the centre, after $2$ steps it is $\sqrt{2} s$, after $3$ steps it is $\sqrt{3} s$, .... after $n$ step it is $\sqrt{n} s$.&lt;br&gt;&#xA; If you need to get $x$ steps in distance, you need $x^2$ steps to take, or $10000\times10000$ for this circle. If you're constantly waking with one step per second and assume that you can walk for no more than $16$ hours without resting, you have $1736$ days to go - and hopefully a very good rucksack of infinite supplies...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="10281" LastEditDate="2015-06-12T11:27:25.797" LastActivityDate="2015-06-12T11:27:25.797" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16240" PostTypeId="1" AcceptedAnswerId="16241" CreationDate="2015-06-11T09:54:04.900" Score="7" ViewCount="1855" Body="&lt;p&gt;A couple were in a car racing madly through Madrid in the middle of the night. Suddenly the engine juddered, whirred, and died. The broken-down car rolled to a halt, leaving them stranded with nobody else in sight. The husband had to go and get help. He was afraid for his wife left alone in the car, so he rolled up the windows and locked the doors before leaving the car, taking the keys with him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It took him a long time to find aid, since he was a foreigner in the city and his Spanish wasn't too good. He returned more than half an hour after leaving, to find that disaster had struck! The exterior of the car was in the same state as when he left it, but his wife was dead and there was blood on the floor and a stranger in the car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happened?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit for clarity&lt;/strong&gt;: all the events that night were down-to-earth and believable. There were no supernatural occurrences, extraterrestrial activity, or anything outside of what is commonly taken to be &quot;the real world&quot;.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-06-16T14:23:24.503" LastActivityDate="2015-06-16T15:07:33.007" Title="Disaster in Madrid!" Tags="|lateral-thinking|story|situation|" AnswerCount="4" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="16241" PostTypeId="2" ParentId="16240" CreationDate="2015-06-11T10:07:15.960" Score="32" Body="&lt;p&gt;They were&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; On their way to the hospital at high speed because the woman was pregnant and the baby was on his way. Then their car broke down. While the husband was looking for someone to repair their car, the woman died while giving birth to her child (the stranger).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12645" LastEditorUserId="463" LastEditDate="2015-06-11T17:28:43.623" LastActivityDate="2015-06-11T17:28:43.623" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16242" PostTypeId="2" ParentId="16240" CreationDate="2015-06-11T10:07:22.480" Score="8" Body="&lt;p&gt;Possible answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The stranger went into the &lt;em&gt;open&lt;/em&gt; car and killed the wife for whatever reason. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why is this possible?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;...the husband locked windows and doors &lt;em&gt;before&lt;/em&gt; leaving the car&quot; - so he had to unlock it again when he left!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastActivityDate="2015-06-11T10:07:22.480" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16243" PostTypeId="2" ParentId="16240" CreationDate="2015-06-11T10:26:37.550" Score="7" Body="&lt;p&gt;PirateSoul's answer seems the most likely. A less-plausible alternative, however:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It's all well and good rolling up the windows and locking the doors, but that's of no use if you forget to put the roof of your convertible back up!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11717" LastActivityDate="2015-06-11T10:26:37.550" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16244" PostTypeId="2" ParentId="16202" CreationDate="2015-06-11T10:38:48.607" Score="22" Body="&lt;p&gt;In short the shop keeper lost Nothing!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually it's pretty simple... At the time the customer teared up the 1000 rs note it was the neighbor's property at that point. He may demand his money back equally the shopkeeper could refuse. You do not explicitly state that the shop keeper must refund the note to the neighbor and rightly so because...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The neighbor may be corrupt and could have switched the note with a fake in the time since the shop keeper handed the note to the neighbor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Equally the shop keeper could deny that he had given the neighbor that exact note. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shop keeper hasn't lost anything until he refunds neighbor and he may have no intention of doing that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;0 rs loss!&lt;/p&gt;&#xA;" OwnerUserId="13119" LastEditorUserId="13119" LastEditDate="2016-10-25T10:34:53.910" LastActivityDate="2016-10-25T10:34:53.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16245" PostTypeId="2" ParentId="14980" CreationDate="2015-06-11T11:16:02.620" Score="4" Body="&lt;p&gt;I've tried a few things, but think I'm going down a pointless rabit hole, but thought I better share my ideas so far anyway:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thought process so far:&#xA;I've tried using the original uncyhphered hint, both without and with capital letters, generating the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lower Case:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;plmblrgmll ijfsehgst&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Uppercase:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;BFNVNACLAMBDSV STJRTRM&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Running with the idea of capitals&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I compared them to Airport short-codes and removed each airport shortcode and have this data:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lower Case (unsolved letters):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;g ijfseh &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;gst&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Gustavus, AK (GST)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;mll&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Marshall, AK (MLL)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;blr&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Bangalore, India (BLR)  capital of India's southern Karnataka&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;plm&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Palembang, Indonesia (PLM) Palembang is the second-largest city in Sumatra Island after Medan and the capital city of the South Sumatra province in Indonesia&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;bfn&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Bloemfontein, South Africa (BFN) Bloemfontein is the capital city of the province of Free State of South Africa; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Upper Case (unsolved letters):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;VNACLAM V STJRTRM&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;BDS&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Brindisi, Italy (BDS) Brindisi is a city in the region of Apulia in southern Italy, the capital of the province of Brindisi&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I tried plotting the capitals on a map:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OkNL0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;backwards capital letters just for fun&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;MRTRJTS VSDBMALCANVNFB&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12734" LastActivityDate="2015-06-11T11:16:02.620" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16247" PostTypeId="2" ParentId="16202" CreationDate="2015-06-11T13:49:25.043" Score="2" Body="&lt;p&gt;Double-entry accounting FTW.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/SZwpW.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA; I've shown the balance sheets of each party at the start and after each event.  The fake 1000R bill is in red.  After the Neighbor discovers the fake (assuming it's the same 1000R bill, @DiscOH), the shopkeeper's accounts payable jumps from zero to 1000.  This is the answer, he owes 1000.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2318" LastActivityDate="2015-06-11T13:49:25.043" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16249" PostTypeId="2" ParentId="16240" CreationDate="2015-06-11T15:25:29.397" Score="7" Body="&lt;p&gt;It seems pretty clear to me that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The wife has been attacked by an alien aboard the spacecraft &lt;em&gt;Nostromo&lt;/em&gt; shortly before landing. In a rush to get her to a medical laboratory which could remove the alien egg from inside her, the couple's car whirred to a stop. As the husband was going to get help, the alien &lt;a href=&quot;https://www.youtube.com/watch?v=LsD6AL3HJtM&quot;&gt;burst out of the wife's chest&lt;/a&gt;, spattering blood across the car's interior, killing the wife, and leaving a strange little alien inside the car.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-06-11T15:25:29.397" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="16250" PostTypeId="2" ParentId="15725" CreationDate="2015-06-11T16:36:32.907" Score="16" Body="&lt;p&gt;OK, based on the various updates, this is what I have at the moment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Stanza 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;To start, stay within the pathway of general glee,&lt;br&gt;&#xA;Though stay clear of the law faery, who flits about wildly.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definition for another stanza:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; glee, for stanza x (unknown for now)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wordplay solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PATHOGEN - PATHway Of GENeral without the letters of 'law faery' which are interspersed within ('flits about wildly')&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The definition for this is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; virus, contained in stanza 6&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Stanza 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;When five becomes square, it may lead to confusion;&lt;br&gt;&#xA;behead the basilisk, an ironic inclusion.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definition for another stanza:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When five becomes square, it may lead to confusion, for stanza x (unknown for now) [quite a mouthful for a definition though]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wordplay solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; WIZARDRY - IZARD (lizard 'basilisk' without the first letter - 'behead') inside ('inclusion') WRY ('ironic')&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The definition for this is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; sorcery, contained in stanza 4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Stanza 4&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Before two sharp curves and the ring of sorcery,&lt;br&gt;&#xA;Stand still, close your eyes, and count aloud past three.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definition for another stanza:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; sorcery, for stanza 3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wordplay solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; FORZANDO - FOR (sounds like 'four' - 'stand still, close your eyes, and count aloud past three') Z (two sharp curves) AND O (the ring)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The definition for this is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; forcefully (in music), contained in stanza 8&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Stanza 5&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In the air - do you hear it? The water baby screams.&lt;br&gt;&#xA;Let our own misgivings start and end in our dreams.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definition for another stanza:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The water baby screams, for stanza x (unknown for now)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wordplay solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HEIRLOOM - HEIR (sounds like 'air') + LOOM (the start of each word in 'Let our own misgivings')&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The definition for this is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; keepsake, contained in stanza 7&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It looks like the answers may all have 8 letters, so perhaps they line up in some way to provide some deep and meaningful stuff?&lt;/p&gt;&#xA;" OwnerUserId="12277" LastEditorDisplayName="user20" LastEditDate="2015-08-05T17:52:39.643" LastActivityDate="2015-08-05T17:52:39.643" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16251" PostTypeId="2" ParentId="16202" CreationDate="2015-06-11T16:59:36.993" Score="2" Body="&lt;p&gt;The answer is &lt;code&gt;G&lt;/code&gt; but the actual loss amount is difficult to conceive&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first obvious guess would be Rs. 1,000 assuming shopkeeper bears the cost for the fake note.&#xA;Alternatively, if the neighbour bears it, shopkeeper might not have to absolve the loss, so Nil could also be a possible answer, but then my guess is the figure might be anywhere between Rs. 8,00 to Rs. 1,000 both exclusive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shopkeeper buys a good to sell it at a premium which is called Selling Price. The price at which the shopkeeper buys is called cost price. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So S.P = C.P + Profit&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we exclude the neighbour from the transaction then we can equate the transaction between the lady and the shopkeeper as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\text{Rs }1000 = C.P + Profit  + \text{Rs }800$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow C.P + \text{Rs }800 = \text{Rs }1000 - Profit = Loss$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming Profit can be anywhere between 50 paisa (minimum legal tenderable amount in India) to Rs. 199.50, so the Loss could be anywhere between&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\text{Rs }999.50 \le Loss \le \text{Rs }800.50$&lt;/p&gt;&#xA;" OwnerUserId="4177" LastActivityDate="2015-06-11T16:59:36.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16252" PostTypeId="1" AcceptedAnswerId="16259" CreationDate="2015-06-11T17:37:03.563" Score="6" ViewCount="796" Body="&lt;p&gt;You are out on a quest along with your pals, stumbling through the woods, sailing across the seven seas, when you finally reach a barren island with a skull cave. You stand at the mouth of the cave deciding what to do.&lt;/p&gt;&#xA;&lt;p&gt;Finally, you all decide to enter the cave. As you enter, you see bones scattered about the cave. As you proceed further, you notice that the entrance to the cave has closed. You reach a large Golden Gate. There are golden jewels and other precious stones strewn on the floor. You see something engraved on the gate. You move closer. Here's what you see on the gate:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;The most hon'st among them,&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Must lead them to the place&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Where heaven descends to the feet of earth&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;To give them a new birth.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You start scratching your head, trying to make something out of it.&lt;/p&gt;&#xA;&lt;p&gt;Then you notice there are some letters just below the poem.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;687474703a2f2f706173746562696e2e636f6d2f5735327741694b70&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;As, you read the poem, you start thinking something. Indeed, the poem is very simple. You ask all your mates to face towards the direction opposite to the wall and close their eyes while you search for something, scrounging in the gold articles near the Golden Gate. Finally, you find another engraving on the floor:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;Hmm, so far so good. Well, the gods have decided to play fair with you.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;gqffbsysoitb.kio/qsrssxpyafZ&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Something hits your mind. You pick out your Satellite Internet Smartphone and type in something. After several hours, you get the password to the Golden Gate. You say the password aloud and the gate opens. You and your pals are welcomed into heaven by Zeus himself.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;The question is, what was the password to the Golden Gate?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="13016" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-06-11T22:29:02.717" Title="What's the password to the Golden Gate" Tags="|cipher|" AnswerCount="3" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16253" PostTypeId="2" ParentId="16213" CreationDate="2015-06-11T17:41:42.747" Score="9" Body="&lt;p&gt;This one took a lot of work! :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The words are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DAHALOKELY&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ILOKELESIA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; NEOVENATOR&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ORTHOMERUS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SATURNALIA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ALETOPELTA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;7.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; UTAHRAPTOR&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;8.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RUKWATITAN &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;9.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ILIOSUCHUS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;10.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ACRISTAVUS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They are all&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Different species of dinosaurs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which gives the secret word:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DINOSAURIA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Image:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/74sO5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The words are from this wiki page:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_dinosaur_genera&quot; rel=&quot;nofollow noreferrer&quot;&gt;wiki&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Full disclosure: I wrote a script to help me solve this&lt;/p&gt;&#xA;" OwnerUserId="4624" LastEditorUserId="4624" LastEditDate="2015-06-11T22:17:03.453" LastActivityDate="2015-06-11T22:17:03.453" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16254" PostTypeId="2" ParentId="16159" CreationDate="2015-06-11T17:43:29.590" Score="1" Body="&lt;p&gt;Your name is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Condom&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In good company, I'm a sign of respect&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Show her your respect by being prepared&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alone, I don't denote man kind&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Certainly not, although maybe just the &quot;man&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was the chosen one, by George!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;George was a smart fellow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are times my presence is warranted&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If you're not trying to have a baby&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I'll usually show up if you're a good sport about it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If you do what she asks you'll get a chance to use it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9654" LastActivityDate="2015-06-11T17:43:29.590" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16255" PostTypeId="2" ParentId="11302" CreationDate="2015-06-11T18:09:32.540" Score="-3" Body="&lt;p&gt;I think the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; News paper shows even and odd numbers when opened. In the story it says 9-10 which is incorrect.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hence fake FIR&lt;/p&gt;&#xA;" OwnerUserId="13127" LastEditorUserId="2036" LastEditDate="2019-01-08T15:29:50.630" LastActivityDate="2019-01-08T15:29:50.630" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="16256" PostTypeId="1" AcceptedAnswerId="16258" CreationDate="2015-06-11T19:10:11.770" Score="33" ViewCount="2197" Body="&lt;p&gt;Across the Deadly River, among the Black Mountains, is a mystical cave. Anyone who enters the cave finds within it a single gold coin, which may be freely taken. Once you leave the cave, you can never reenter it. However, a &lt;em&gt;group&lt;/em&gt; of people may enter the cave together, even if every member has entered it before, so long as &lt;em&gt;that exact group&lt;/em&gt; has never entered. A group of people will still find only a single gold coin. The cave seems to have an infinite supply of gold coins, but stingily doles out only one per visit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It costs 1 gold coin to pay the ferryman to take you and any number of people across the Deadly River and back. So if you and a friend travel together, you will get 3 coins in total (one from you entering, one from your friend entering, and one from the two entering together). Less 1 coin for the ferryman, that's 2 coins of profit, or 1 for each of you. If instead three people were to make the journey, you'd collect 3 coins by entering individually, 3 coins for each pair, and 1 coin for all of you entering together, for 7 coins. Minus 1 for the ferryman, that leaves 6 coins, or 2 each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want to put together a party to travel to this cave. There are a couple of wrinkles though. The first is that a party will only agree to travel if the profit can be divided evenly. As described above, this would work fine for a party of 2 or 3, but it can be shown that a party of 4 would not work, since the profit is 14 coins and 14 is not a multiple of 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second wrinkle is that you want to form a secret conspiracy within your party, consisting of yourself and at least one other person (but not the whole party). After you are back across the River with the divvied-up gold, the members of the conspiracy will turn on the rest, murder them, and steal their shares of the profit. However, people will only agree to a conspiracy if the &lt;em&gt;victims&lt;/em&gt; can be divided evenly, so that each conspirator kills the same number of people and therefore pockets the same amount of gold as well. So here we see that a party size of 2 won't work (too small to form a conspiracy within it), and a party size of 3 won't work, because while you could form a conspiracy of two, there would be only one victim, which doesn't divide evenly between the two murderers. A party size of 4 would work, 2 murderers and 2 victims for 1 victim apiece, but this fails the first criterion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the smallest party size for which your plan will work?&lt;/p&gt;&#xA;" OwnerUserId="9523" LastEditorDisplayName="user88" LastEditDate="2015-09-22T05:13:39.697" LastActivityDate="2015-09-22T07:20:24.607" Title="A Cave in the Black Mountains" Tags="|mathematics|story|number-theory|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16257" PostTypeId="2" ParentId="16252" CreationDate="2015-06-11T19:12:58.420" Score="1" Body="&lt;p&gt;&lt;em&gt;Note that this is just a partial answer&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After decoding the hex string 687474703a2f2f706173746562696e2e636f6d2f5735327741694b70, I got the decoding as &lt;a href=&quot;http://pastebin.com/W52wAiKp&quot; rel=&quot;nofollow&quot;&gt;http://pastebin.com/W52wAiKp&lt;/a&gt;. After going to the link, I get the string &quot;IAMHONEST&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;&quot;The most hon'st among them, Must lead them to the place&quot;&lt;/em&gt;. Maybe this is something to do with the string &quot;I AM HONEST&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another partial statement&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;gqffbsysoitb.kio/qsrssxpyafZ&lt;/strong&gt; is probably a website by looking at the &quot;.kio/qsr...&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12860" LastEditorUserId="12860" LastEditDate="2015-06-11T19:38:15.550" LastActivityDate="2015-06-11T19:38:15.550" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16258" PostTypeId="2" ParentId="16256" CreationDate="2015-06-11T19:13:41.443" Score="35" Body="&lt;p&gt;The first wrinkle is easy enough &amp;mdash; you get one coin for each (nonempty) subset of $n$ travellers ($2^n - 1$ total), minus one coin for the ferry, for a total of $2^n - 2$. If that's divisible by $n$, you go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermat%27s_little_theorem&quot;&gt;Fermat's Little Theorem&lt;/a&gt;, any prime number of partygoers will work. But the second wrinkle means that you need to be able to divide your party into multiple equal groups (of 1 conspirator and $m$ victims each), which is impossible if the number is prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you need to find the smallest &lt;em&gt;composite&lt;/em&gt; number, known as a &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermat_pseudoprime&quot;&gt;Fermat pseudoprime&lt;/a&gt;, that satisfies this criterion. In the article I just linked, the answer is stated to be &lt;strong&gt;341&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Therefore,  you go with 340 other people, and you and your conspiracy of 10 kill 30 people each.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-06-12T04:54:17.880" LastActivityDate="2015-06-12T04:54:17.880" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="16259" PostTypeId="2" ParentId="16252" CreationDate="2015-06-11T19:46:17.853" Score="8" Body="&lt;p&gt;Part 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Convert the string of numbers and letters from hexadecimal to text. You get &lt;a href=&quot;http://pastebin.com/W52wAiKp&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://pastebin.com/W52wAiKp&lt;/a&gt;. Going to the link gives &lt;code&gt;IAMHONEST&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Part 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The gods have decided to &lt;strong&gt;play fair&lt;/strong&gt; =&gt; playfair cipher&lt;br&gt;&#xA;&lt;br&gt;&#xA; Using &lt;code&gt;IAMHONEST&lt;/code&gt; as the alphabet key in this &lt;a href=&quot;http://rumkin.com/tools/cipher/playfair.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;playfair decrypter&lt;/a&gt; gives &lt;code&gt;freetexthost.com/utqaawqwhkX&lt;/code&gt;. In a playfair cipher, X is used for filling in (kind of like a space). So removing the X gives &lt;a href=&quot;http://freetexthost.com/utqaawqwhk&quot; rel=&quot;nofollow noreferrer&quot;&gt;freetexthost.com/utqaawqwhk&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Part 3 (&lt;a href=&quot;https://puzzling.stackexchange.com/a/16264/12109&quot;&gt;Ivo Beckers&lt;/a&gt; got this one first):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The page needs a password. Using &lt;code&gt;IAMHONEST&lt;/code&gt; as the password, the page says &lt;code&gt;LONGLIVEAMERICA&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Password is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Long live america&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12109" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-06-11T22:29:02.717" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16260" PostTypeId="1" AcceptedAnswerId="16262" CreationDate="2015-06-11T20:01:39.740" Score="0" ViewCount="737" Body="&lt;p&gt;A man carries his son into the hospital because his son has a nail in his foot. The surgeon walks in and says &quot;I cannot operate on this boy, as he is my son&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is going on here?&lt;/p&gt;&#xA;" OwnerUserId="13023" LastEditorUserId="11876" LastEditDate="2015-10-06T12:25:18.927" LastActivityDate="2015-10-06T12:25:18.927" Title="The Mysterious Surgeon" Tags="|lateral-thinking|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ClosedDate="2015-06-11T20:11:20.823" ContentLicense="CC BY-SA 3.0" />
  <row Id="16261" PostTypeId="2" ParentId="16217" CreationDate="2015-06-11T20:02:05.633" Score="2" Body="&lt;p&gt;Either (a)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 Trip&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or (b)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3 Trips&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Depending one whether 50% of the wires or more were dead. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In (a) you do what @Anon suggests which is simply short all wires at end &lt;strong&gt;A&lt;/strong&gt; together, then go to end &lt;strong&gt;B&lt;/strong&gt; and check each wire to identify which are dead (this is done by first finding any pair of cables that there is continuity on, and then checking every other cable for continuity with one from that pair). Label them. If there are more than 50% dead give up and go have a cup of tea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In (b) we assume &amp;lt;50% are dead, in which case having done (a) you are at end &lt;strong&gt;B&lt;/strong&gt; where you were testing you now can ignore all the dead cables (cut them off so they never get used for example), and simply follow the steps from the 'undamaged' identification question to identify the working bundles but exclude the dead ones in the groups (also while your are at &lt;strong&gt;A&lt;/strong&gt; during that process you can label the dead ones as they are the ones which are left unlabelled during that step). The number of required journeys for that is already known, so add on (a) to that number and you get (b).&lt;/p&gt;&#xA;" OwnerUserId="13132" LastEditorUserId="13132" LastEditDate="2015-06-11T20:55:41.477" LastActivityDate="2015-06-11T20:55:41.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16262" PostTypeId="2" ParentId="16260" CreationDate="2015-06-11T20:02:25.890" Score="2" Body="&lt;p&gt;This is an old one, but good!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The surgeon is the boy's mother!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastEditorUserId="8874" LastEditDate="2015-10-06T11:57:44.600" LastActivityDate="2015-10-06T11:57:44.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16263" PostTypeId="1" CreationDate="2015-06-11T20:08:38.550" Score="2" ViewCount="929" Body="&lt;p&gt;Though the answer I'm looking for is a number and there is a little calculation involved. There are no hard sums here and each step is fairly simple (except maybe the last two)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;EtodKsteergytfifmilatertphoneqAladyaKauwsweenehW.================================================================etonwhopfiftyddhcumdidaohhn.=====================================================================================AXeH&quot;theaegonenengas,nsyNyHgtaddnuopNYytrofxOzaaieherahBshnIme.,nrepeekpohssPSWvhWgveWo.eilessserdtOTsclown&quot;=====MIesnneitgwrvocudehdiashsadnXr.==================================================================================aastNahkghlsioifntW,=============================================================================================TGTedlwMsuptSrY.=================================================================================================htggSgagklnuo.===================================================================================================rehtyvlMertuehixycSsnemursooonmesdnif,===========================================================================tdeeNgenmetimiinoaepcpmotqKdrXuuftrsmxgWnuvKwhyordvsgaoabteanmihhti.=============================================dsnootOooLrhfBehnrlntoiufAeghhenYimTzhrHdLcpoisijYuomrtgshueotusteiHhrlnandfandmrTKtyeehiaauookhnkapooa.=========sdnuopnshathhtttddrnpoeinaCbsesucxeRssUeoeseOibehnsJondkufieidehsinifHoluherDlariazlseHWhX?tPgTladyKkmrfSthLslhuo&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Just un-jumble the letters, find some words, make sense of those words and work out what happened in the little story they tell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Align, Remove, Re-align, Search, Re-arrange, Solve.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Extra Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are 2 ways to &quot;Re-align&quot; only 1 will allow you to &quot;Search&quot; properly working out which is one of the (second) hardest parts of this puzzle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Update (with hint):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Len has extracted the correct word search arrangement. Beware there are quite a few decoy words the have crept in purely by chance. Pay attention to the punctuation which is stuck to the beginning and ends of words.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="11297" LastEditDate="2015-06-14T19:23:12.950" LastActivityDate="2015-06-14T19:23:12.950" Title="Unscramble and Solve the Word Search" Tags="|riddle|enigmatic-puzzle|word-search|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16264" PostTypeId="2" ParentId="16252" CreationDate="2015-06-11T20:09:00.473" Score="4" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; long live america&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To be honest I followed mmking's clues. I figured out removing the X myself since that wasn't posted earlier. The password to the text is simply&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; IAMHONEST&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It then shows the text&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LONGLIVEAMERICA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-06-11T20:09:00.473" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16265" PostTypeId="1" AcceptedAnswerId="16266" CreationDate="2015-06-12T00:22:47.723" Score="-3" ViewCount="593" Body="&lt;p&gt;A brewing cup of tea sits on my desk.&lt;br/&gt;&#xA;Full of water and milk&lt;br/&gt;&#xA;When I put the cup down I was very careful not to spill a drop&lt;br/&gt;&#xA;When I came back 2 minutes later, my desk was damp with hot tea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why???&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hint: The tea is brewing. Tea bags are in the cup.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4951" LastEditorUserId="4951" LastEditDate="2015-06-12T00:28:39.457" LastActivityDate="2015-06-12T14:52:35.367" Title="Why is my desk wet?" Tags="|lateral-thinking|physics|" AnswerCount="2" CommentCount="1" ClosedDate="2015-06-17T15:47:53.930" ContentLicense="CC BY-SA 3.0" />
  <row Id="16266" PostTypeId="2" ParentId="16265" CreationDate="2015-06-12T00:56:03.967" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not really sure if this is a riddle as I have had it happen to me :P The tea absorbed up the teabag string and then dripped down the outside of the mug. It usually takes a little longer than 2 min though...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13136" LastActivityDate="2015-06-12T00:56:03.967" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16267" PostTypeId="2" ParentId="14980" CreationDate="2015-06-12T01:32:35.513" Score="12" Body="&lt;p&gt;Looking for the capitals in it I started with the ISO 3166-1 Alpha-2 Country Codes, and the list of capital cities from Wikipedia yielding this table:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/tDPBv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Giving me the second jumbled string:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;OWUHSSYMSCFR RDSSZLPA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Shoutout goes out to Alexander Wigmore who came to the same conclusion I did while offline about the space)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Decoding the first string using a Playfair cipher with I = J and the key text LIFE yields:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;MY PASSWORD IS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Decoding the second string using a Playfair cipher with I = J and the key text DEATH yields:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;PASSWORD&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To show an example of the second cipher at work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jEFx2.png&quot; alt=&quot;step1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of the letters are entered into the tableau starting with the letters of the key and then the rest of the 26 letters.  If a letter would be listed a second time because it was already listed in the key, it is omitted.  The letter I is equivalent to the letter J for the purposes of Playfair.  Other implmentations just omit Q.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each pair of letters in the input text forms a rectangle.  The opposite two corners of the rectangle are the decrypted text.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See how RD decrypts to PA in the first image.  The second image shows how ZL is decrypted to WO.  PA decrypts to RD as it is the opposite of the first image.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/k0urp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did not know the rule that duplicate letters just stay as they are SS -&gt; SS.  This is why I was concerned that there were duplicates after my first step decoding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my full handwritten Playfair solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CpFX4.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice the interesting bit about UH -&gt; PA in the first tableau.  When the two characters are in the same column the character above is chosen, wrapping to the bottom if required.  When they are in the same row, the character to the left is chosen, wrapping around to the right if required.  The opposite action is taken when encrypting.&lt;/p&gt;&#xA;" OwnerUserId="1777" LastEditorUserId="1777" LastEditDate="2015-06-12T11:16:58.177" LastActivityDate="2015-06-12T11:16:58.177" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16268" PostTypeId="2" ParentId="14980" CreationDate="2015-06-12T02:21:18.217" Score="4" Body="&lt;p&gt;Dunno if I can just blatantly steal from LeppyR64...  But off of what he has:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When decoded using the Playfair cipher using the keyword life, I get(drumroll please) MY PASSWORD IS SCSSVAHZ&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Maybe I'm completely wrong, though.  I couldn't use death in any way.  And of course, since I can't solve very many ciphers myself, I used an online decoder.  Oops.  Heeheehee...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&#xA;Using the countries and capitals work didn't work on the password to translate it into a word.  Perhaps it should remain as a random-seeming string?&lt;/p&gt;&#xA;" OwnerUserId="11790" LastEditorUserId="11790" LastEditDate="2015-06-12T03:02:56.093" LastActivityDate="2015-06-12T03:02:56.093" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16269" PostTypeId="1" AcceptedAnswerId="117788" CreationDate="2015-06-12T04:14:15.243" Score="4" ViewCount="512" Body="&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/15658/create-a-word-search&quot;&gt;This recent puzzle I created&lt;/a&gt; was well-received, so here's another one like it, except this time a bit larger and using homophones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Create a word search with the following conditions:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Use the fifteen words below. They can be spelled up, down, left, right, or any of the 4 diagonal directions.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;THEYRE&lt;br&gt;&#xA;  THEIR&lt;br&gt;&#xA;  THERE&lt;br&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;TACKED&lt;br&gt;&#xA;  TACT&lt;br&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;TEA&lt;br&gt;&#xA;  TEE&lt;br&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;TAIL&lt;br&gt;&#xA;  TALE&lt;br&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;TAUGHT&lt;br&gt;&#xA;  TAUT&lt;br&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;TARE&lt;br&gt;&#xA;  TEAR&lt;br&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;TEEM&lt;br&gt;&#xA;  TEAM&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;The size of the word search must be 7 rows by 7 columns. (Bottom right square not used.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The number of words that pass through each square must fit this pattern:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9xL8N.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;The accepted answer will clearly and fully explain the flow of logic used from start to finish.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="463" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2022-09-03T21:16:50.717" Title="Create a Homophone Word Search" Tags="|logical-deduction|word-search|construction|" AnswerCount="2" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="16271" PostTypeId="1" AcceptedAnswerId="16287" CreationDate="2015-06-12T04:53:08.253" Score="10" ViewCount="660" Body="&lt;p&gt;One day my tenth-grade math teacher walked into our classroom and said, &quot;Class, today to start off the class I've got an addition problem for you. Give me any two positive proper fractions and I'll add them up.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of us were a bit bewildered at that, but one of the students decided to bite. &quot;1/2 plus 1/2&quot;, he said.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;11/20&quot;, answered our teacher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We were a bit surprised. &quot;Shouldn't that be 1, teacher?&quot;&lt;br&gt;&#xA;&quot;Nope,&quot; he said. &quot;I'm not using your standard addition algorithm today.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Alright, what about 1/3 + 2/3&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;That's 4/11&quot;, said the teacher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;How about 2/9 + 7/9?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;That's 3/11.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The teacher wrote our questions and his answers on the board along with a few more:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1/2 + 1/2 = 11/20&#xA;1/3 + 2/3 = 4/11&#xA;2/9 + 7/9 = 3/11&#xA;3/20 + 4/20 = 1/8&#xA;1/9 + 1/9 = 1/9&#xA;1/30 + 5/6 = 1/12&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Can you figure out what addition algorithm the teacher was using?&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-06-14T23:55:53.080" LastActivityDate="2015-06-14T23:55:53.080" Title="What's the teacher's fractional addition algorithm?" Tags="|mathematics|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16272" PostTypeId="2" ParentId="16269" CreationDate="2015-06-12T04:53:29.737" Score="1" Body="&lt;h2&gt;About 6% of an answer.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;(I've worked out the letters in 3 squares so far. Maybe someone will find this useful while I work out the remaining 94%)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The bottom left corner is marked 4, so one of the words that use that corner must completely overlap another that also uses that corner. Only the beginning of words contain other words so that corner must contain T.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Words that could overlap are:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;TEA - TEAM, TEAR&#xA;TEE - TEEM&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Overlapping words must extend diagonally from the corner. So square 1,1 is T and square 2,2 is E. The centre square is either M or R. The top right corner must have a diagonal word that starts or ends there (it shares 3 words) so that word must either share the M/R centre square or be 3 letters. There are a number of possibilities for that word but they all require square 6,6 to be E. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The letter frequencies reveal some useful information:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;T 18&#xA;E 15&#xA;A 10&#xA;R 5&#xA;H 4&#xA;I 2&#xA;C 2&#xA;L 2&#xA;U 2&#xA;M 2&#xA;K 1&#xA;D 1&#xA;G 1&#xA;Y 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; K, D, G and Y appear only once so words must be placed so that these letters don't fall on a square numbered more than 1.&#xA; Similarly I, C, U, L and M appear twice each, so care must be taken not to place them on squares numbered more than 2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="11297" LastEditDate="2015-06-12T06:03:04.133" LastActivityDate="2015-06-12T06:03:04.133" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16274" PostTypeId="2" ParentId="11302" CreationDate="2015-06-12T07:19:56.773" Score="1" Body="&lt;p&gt;The report read:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;A bulb was glowing in the room and the fan was switched on.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now consider the first line of the question:-&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A man was murdered in a village.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Police Inspector read the report:-&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...A bulb was glowing in the room and the fan was switched on.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;suspected if there could be electricity at night in a village (especially in India), so he found that the report was fake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyways, @user2408578's is the best but this could be a possible explaination.&lt;/p&gt;&#xA;" OwnerUserId="13016" LastActivityDate="2015-06-12T07:19:56.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16275" PostTypeId="1" AcceptedAnswerId="16277" CreationDate="2015-06-12T07:29:14.913" Score="5" ViewCount="2118" Body="&lt;p&gt;A pro golfer has the amazing ability to consistently putt distances of 3, 5, 7, and 11 feet. Strangely enough, though, those are the only distances he can putt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently, our golfer stands on the green with his ball 20 feet from the hole. What's the fewest number of strokes he can use to get the ball into the hole?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that if the ball is hit farther than the distance remaining to the hole, it will roll over to the other side without going into the cup.&lt;/p&gt;&#xA;" OwnerUserId="12218" LastEditorUserId="5373" LastEditDate="2015-06-16T14:16:33.490" LastActivityDate="2015-06-19T16:05:21.583" Title="The Fewest Strokes Golfer" Tags="|mathematics|geometry|" AnswerCount="5" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16276" PostTypeId="1" AcceptedAnswerId="16279" CreationDate="2015-06-12T07:31:48.390" Score="0" ViewCount="532" Body="&lt;p&gt;You are in a town with one barber shop and two barbers, one with good hair and one with terrible hair, there are no towns nearby so everyone in town must get there hair cut at this barber shop. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which barber should you choose to cut your hair?&lt;/p&gt;&#xA;" OwnerUserId="12218" LastEditorUserId="11475" LastEditDate="2015-06-12T14:48:25.790" LastActivityDate="2015-06-12T14:48:25.790" Title="The best barber in town" Tags="|strategy|situation|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16277" PostTypeId="2" ParentId="16275" CreationDate="2015-06-12T07:41:24.203" Score="18" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is 2 shots. Granted the right angle, 2 circles with the radii of 11 will overlap as seen in the pic!&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/2ZNAI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12325" LastEditorUserId="10400" LastEditDate="2015-06-12T17:53:15.473" LastActivityDate="2015-06-12T17:53:15.473" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16278" PostTypeId="2" ParentId="16275" CreationDate="2015-06-12T07:42:31.613" Score="3" Body="&lt;p&gt;I'd say &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; two 11 feet putts, for he'll just have to put at an slight horizontal angle of approx 25°. He is not obligated to make a bee line, is he?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11827" LastActivityDate="2015-06-12T07:42:31.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16279" PostTypeId="2" ParentId="16276" CreationDate="2015-06-12T07:43:09.297" Score="6" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Choose the barber with terrible hair.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reason is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The barber with good hair can not cut his own hair and has to rely on the barber with terrible hair. This results in his obviously good.&#xA; For the other barber, it is the other way round: the barber with good hair is not able to give a good hair cut.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13140" LastActivityDate="2015-06-12T07:43:09.297" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16280" PostTypeId="2" ParentId="16276" CreationDate="2015-06-12T07:43:55.270" Score="1" Body="&lt;p&gt;You should get your hair cut by:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The barber with the terrible hair, the only one who could have cut his hair, was the other barber, who isn't very good at his job as we can see at the other barbers hair!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12645" LastActivityDate="2015-06-12T07:43:55.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16281" PostTypeId="2" ParentId="16275" CreationDate="2015-06-12T07:50:26.610" Score="13" Body="&lt;p&gt;Assuming that the golfer has to target the hole directly, i.e., no deviation by any angle, the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4 strokes. Because all distances are odd, but the target distance is even, he needs an even number of strokes. Two strokes is impossible because no pair of distances adds up to 20. Using 4 strokes, however, there are multiple combinations: 7+3+7+3 or 5+5+5+5 or 7+3+5+5.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13140" LastActivityDate="2015-06-12T07:50:26.610" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16283" PostTypeId="2" ParentId="14980" CreationDate="2015-06-12T08:27:11.963" Score="10" Body="&lt;p&gt;I think I've cracked it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Leading on from Leppy and AJL:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I suddenly remembered that odd space in the initial encoded string.&#xA;I used the keyword life as a Playfair filter on the first half: (OWUHSSYMSCFR) and Death as the filter on the second half (RDSSZLPA) and got:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Drumroll..&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;MY PASSWORD IS PASSWORD&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12734" LastActivityDate="2015-06-12T08:27:11.963" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16284" PostTypeId="2" ParentId="16276" CreationDate="2015-06-12T08:28:39.373" Score="3" Body="&lt;p&gt;You should:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Get your hair cut by the barber with good hair&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everyone knows that barbers cut their own hair. I do&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2032" LastActivityDate="2015-06-12T08:28:39.373" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16285" PostTypeId="2" ParentId="16276" CreationDate="2015-06-12T08:30:28.180" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Go ask someone in the city with a good hair cut &quot;From whom they did their hair cut&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By that you can understand who is a good barber.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Some barbers cut their own hair, while some not. So the perfect way to understand who is the good barber is asking someone in the town with a good hair cut.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastActivityDate="2015-06-12T08:30:28.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16286" PostTypeId="1" AcceptedAnswerId="16288" CreationDate="2015-06-12T08:41:30.093" Score="9" ViewCount="1716" Body="&lt;p&gt;It is a bright Sunday afternoon. You are on a holiday, enjoying your cappuccino. Suddenly, you get a call from your boss at &lt;strong&gt;NSA&lt;/strong&gt;. He tells you that a psycho-hacker has sent him and the whole staff a weirdo-looking image attachment sent using an anonymous email sending service. Hearing that, you quickly log into your email account and find this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/01gTi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Now, the question: Who sent this message?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 1:-&lt;/strong&gt; Someone changed the password of the Gmail Account an hour ago. So, instead of that hint. See the below hint:-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You spot down all hints and at last, you find an email and password. You login and find the following email in the inbox:-&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;From:XXXXX&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;To:mXXXpuXXX@gmail.com&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Subject: FINAL HINT&lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;You are an inch away from solving this puzzle. You might want to  see&#xA;    the attached image.&lt;/p&gt;&#xA;    &#xA;    &lt;p&gt;BTW I love mosaic tiles :)&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Attached Image: &lt;a href=&quot;https://i.stack.imgur.com/GXdER.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Download&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GXdER.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 2:-&lt;/strong&gt; I have changed the case of the account password. Please don't change the password and hinder the experience of others, the account is blank and is of no use. You may need to use an Indian VPN/Proxy to login (Curse Google!).  If the password has been changed again, then use the above edit :)&lt;/p&gt;&#xA;" OwnerUserId="13016" LastEditorUserId="13016" LastEditDate="2015-06-12T14:01:12.933" LastActivityDate="2023-01-31T06:41:50.207" Title="The mystery of the anonymous email" Tags="|cipher|steganography|" AnswerCount="2" CommentCount="20" ContentLicense="CC BY-SA 3.0" />
  <row Id="16287" PostTypeId="2" ParentId="16271" CreationDate="2015-06-12T10:46:45.767" Score="11" Body="&lt;p&gt;Rule:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Write both numbers on the left side as decimal, create the resulting number by picking digits in alternating order.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Calculations:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Left first, right second:&lt;br&gt;&#xA;&lt;br&gt;&#xA; 1) 0.5 + 0.5 = 0.55&lt;br&gt;&#xA; 2) 0.3333 + 0.6666 = 0.36363636&lt;br&gt;&#xA; 3) 0.2222 + 0.7777 = 0.27272727&lt;br&gt;&#xA; 4) 0.15 + 0.2 = 0.125&lt;br&gt;&#xA; 5) 0.1111 + 0.1111 = 0.11111111&lt;br&gt;&#xA; 6) 0.0333 + 0.8333 = 0.08333333  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9910" LastEditorDisplayName="user88" LastEditDate="2015-06-14T23:55:31.403" LastActivityDate="2015-06-14T23:55:31.403" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16288" PostTypeId="2" ParentId="16286" CreationDate="2015-06-12T11:16:44.650" Score="13" Body="&lt;p&gt;The sender is going by the name of POOP and...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; IAMINCREDIBLE - masterpuzzler007@gmail.com  is his email address (not sure of his name yet)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The image contains a hidden message...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...which you can find by opening it with a text editor and looking at the end where there is a link to file hosting site that requires a password.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once you have found that you need to know that...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  ...the password is &lt;code&gt;00&lt;/code&gt;  hence the clue &quot;nullbyte&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then you get:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;6d617374657270757a7a6c657230303740676d61696c2e636f6d&#xA;&#xA;Hint:You may need a different point of view for the next one.&#xA;&#xA;454c4249444552434e494d4149&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ... a hexadecimal string representation of the ascii values of &lt;code&gt;masterpuzzler007@gmail.com&lt;/code&gt; and &lt;code&gt;ELBIDERCNIMAI&lt;/code&gt; (I AM INCREDIBLE backwards)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once you've found this...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you can access the email account to get a picture similar to the one in the original message (&lt;a href=&quot;https://i.stack.imgur.com/GXdER.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/GXdER.png&lt;/a&gt;) which has an anomaly in the top left corner. Some of the pixels looked smudged downward but they have actually been more significantly altered. The thin grey strip along the top of the image is not pure grey in the smudged section. Each of the 19 pixels is a different colour to the unsmudged portion of the strip. Presumably these pixels encode the name we are looking for.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The hexadecimal values of each pixel are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; cec1b9 cfc3ba cdc1ba cfc1bb cec0bb ccc0b9 ccc0b9 cdc1bb cec0bb cfc1b9 ccc3b8 cfc0bb ccc1bb ccc1ba cec1bb cfc0b8 cfc0b9 cfc3bb cec3ba&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you are hoping to work this last bit out by hand good luck to you.&#xA;Credit to Quark and LeppyR64 who worked out the final step, which is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Enter the new image into &lt;a href=&quot;http://mozaiq.org/decrypt&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://mozaiq.org/decrypt&lt;/a&gt; with the password &quot;00&quot; to get the line &quot;Yes, you've found me. My name is VGVjaFJheA==&quot;. This is Base64 encoded and translates to TechRax.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="11297" LastEditDate="2015-06-13T19:52:31.253" LastActivityDate="2015-06-13T19:52:31.253" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="16289" PostTypeId="2" ParentId="16275" CreationDate="2015-06-12T12:56:03.870" Score="11" Body="&lt;p&gt;Technically....&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;b&gt;He could take 0 strokes&lt;/b&gt;. Just pick up the ball and drop it in the hole.....and then run away yelling GOAAAAAAAAAL! (Any spectators would be so surprised and think he was crazy, that he just might get away with it)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-06-12T13:11:08.750" LastActivityDate="2015-06-12T13:11:08.750" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16290" PostTypeId="1" AcceptedAnswerId="16291" CreationDate="2015-06-12T14:07:35.337" Score="7" ViewCount="4527" Body="&lt;p&gt;This is just a simple riddle I thought of just now!&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;You may make me, but you cannot destroy me.&lt;/p&gt;&#xA;&lt;p&gt;You can see through anything with my help.&lt;/p&gt;&#xA;&lt;p&gt;When I'm (really) big and black, nobody and nothing can escape from me.&lt;/p&gt;&#xA;&lt;p&gt;I'm not used very frequently.&lt;/p&gt;&#xA;&lt;p&gt;Sometimes, I make objects unusable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Who/What am I?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="13016" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-06-13T06:27:12.127" Title="You Cannot Destroy Me. Who am I?" Tags="|riddle|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16291" PostTypeId="2" ParentId="16290" CreationDate="2015-06-12T14:09:08.527" Score="16" Body="&lt;p&gt;I expect that you're a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; hole!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You may make me but you cannot destroy me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can cut or drill (or many other verbs) holes into things, but once they're there you cannot destroy them - fill them or patch them, maybe, but you can never destroy what's not there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can see through anything with my help.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Put a hole in it and you can see through it!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I'm (really) big and black, nobody and nothing can escape from me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is what really gave it away for me. This is of course referencing &lt;a href=&quot;https://en.wikipedia.org/wiki/Black_hole&quot;&gt;black holes&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not used very frequently.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;(from PirateSoul)&lt;/em&gt; There isn't a lot you can do with a hole, other than fill it; therefore, you don't use them very frequently.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes, I make objects unusable.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some things can't be used if there are holes in them. Clothes come to mind as an example, but a larger scale example could be a boat! The options here are limitless (to an extent).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="11475" LastEditDate="2015-06-12T16:47:12.600" LastActivityDate="2015-06-12T16:47:12.600" CommentCount="19" ContentLicense="CC BY-SA 3.0" />
  <row Id="16292" PostTypeId="2" ParentId="16290" CreationDate="2015-06-12T14:13:31.363" Score="6" Body="&lt;p&gt;Maybe a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; black hole!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You may make me but you cannot destroy me.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People, can create black holes, but cannot destroy them. ( Look up the large hadron collider)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You can see through anything with my help.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As suggested by Samuel, gravitational lensing is one way in which black holes allow us to see things. Because they are very heavy and dense, they bend space-time. This means if a planet or a star is not visible to us, black holes can bend light so that it can be seen by us.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;When I'm (really)big black, nobody and nothing can escape from me.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Black holes are big and black. Nothing can escape them, not even light.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I'm not used very frequently.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People do not use black holes in their day-to-day lives.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Sometimes, I make object unusable&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Anything which enters a black hole will be stretched to a great extent and will be rendered useless. ( this information is from one of Vsauce's videos on black holes)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12860" LastEditorUserId="12860" LastEditDate="2015-06-13T06:27:12.127" LastActivityDate="2015-06-13T06:27:12.127" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16294" PostTypeId="2" ParentId="14582" CreationDate="2015-06-12T14:42:59.803" Score="1" Body="&lt;p&gt;Edit: I misread the question; this answer is incorrect as it overlooks diagonal movement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LeppyR64 has brute forced the answer. There is, however, a simple proof that 12 moves is the minimum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define the &lt;strong&gt;disorder&lt;/strong&gt; of an arrangement to be the sum of each square's distance, both horizontally and vertically, from its starting position. The disorder is initially 0, and that of the desired state can be easily calculated to be 24 (4 per corner, 2 per side, 0 for centre). Since each move shifts 2 pieces by 1 square each, the largest possible change in disorder from each move is 2. We therefore need at least 12 moves to reach the desired state.&lt;/p&gt;&#xA;" OwnerUserId="3987" LastEditorUserId="3987" LastEditDate="2015-06-14T12:32:14.130" LastActivityDate="2015-06-14T12:32:14.130" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16295" PostTypeId="2" ParentId="16265" CreationDate="2015-06-12T14:52:35.367" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The cup is made of paper, and the paper is thick enough so that the desk is only damp after two minutes and not drenched.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8942" LastActivityDate="2015-06-12T14:52:35.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16299" PostTypeId="1" AcceptedAnswerId="16301" CreationDate="2015-06-12T16:06:43.630" Score="19" ViewCount="1744" Body="&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/15823/our-reply-to-your-accusations-who-are-we&quot;&gt;My recent riddle&lt;/a&gt; was a bit too lengthy for some, so I tried to keep this one short:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;While Esther cares for many people,&lt;br&gt;&#xA;Her best friend, Mike, just runs around,&lt;br&gt;&#xA;Never providing much for anyone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mike loves Esther,&lt;br&gt;&#xA;Only showing her his good side, but&lt;br&gt;&#xA;He has a side that she has never seen!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They both love their friend, Sheila.&lt;br&gt;&#xA;She is very generous to them, and&lt;br&gt;&#xA;She knows both of Mike's sides very well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They all usually get along great,&lt;br&gt;&#xA;But sometimes Esther gets jealous,&lt;br&gt;&#xA;And prevents Mike from seeing Sheila.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes Mike gets angry at Esther&lt;br&gt;&#xA;And tries to keep her from seeing Sheila,&lt;br&gt;&#xA;But he never fully succeeds.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who are Mike, Esther, and Sheila?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-06-12T20:15:02.813" Title="Who are Mike, Esther, and Sheila?" Tags="|riddle|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16301" PostTypeId="2" ParentId="16299" CreationDate="2015-06-12T16:13:53.170" Score="30" Body="&lt;p&gt;Mike, Esther and Sheila are...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ... the Moon, Earth and Sun.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Esther cares for many people,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are many people on the Earth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Her best friend, Mike, just runs around,&#xA;Never providing much for anyone.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Moon orbits the Earth and no one lives on it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mike loves Esther,&#xA;Only showing her his good side, but&#xA;He has a side that she has never seen!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One side of the Moon faces the Earth all the time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They both love their friend, Sheila.&#xA;She is very generous to them, and&#xA;She knows both of Mike's sides very well.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Earth and Moon orbit the Sun as a unit, and the Sun 'sees' both sides of the Moon.  The Sun is generous in providing light/heat to the Earth and Moon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The final two stanzas:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; These refer to solar and lunar eclipses.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="303" LastActivityDate="2015-06-12T16:13:53.170" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16302" PostTypeId="1" AcceptedAnswerId="16303" CreationDate="2015-06-12T16:34:22.997" Score="1" ViewCount="319" Body="&lt;p&gt;Anyone who plans to visit the Miracle Fruit Tree in the Royal Garden has to travel through Seven Gates in each of the Seven Curtain Walls surrounding the garden.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of these gates has a guard. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is only one way in and out each of these curtain walls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Royal rule is Each of the guard will take half the number of fruits from anyone on the way out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One fine day, a commoner negotiates a deal with each of the guard on the way in. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The deal is after sharing the split, will take back one from the guard.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After plucking 'MAGICNUMBER' number of fruits on the way out, Both the guard and the person honor the deal at each of the Seven gates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At home the Person counts all the Miracle Fruits remaining.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The count is 'MAGICNUMBER'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the 'MAGICNUMBER' ?&lt;/p&gt;&#xA;" OwnerUserId="9892" LastEditorUserId="4421" LastEditDate="2015-11-25T18:09:11.583" LastActivityDate="2015-11-25T18:09:11.583" Title="Simplest Magic Fruit Split with Magic Number" Tags="|mathematics|logical-deduction|" AnswerCount="2" CommentCount="1" ClosedDate="2015-06-12T17:20:02.127" ContentLicense="CC BY-SA 3.0" />
  <row Id="16303" PostTypeId="2" ParentId="16302" CreationDate="2015-06-12T16:37:50.583" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The person would only need 2 magical fruits&#xA;&lt;br&gt; Half of 2 is 1, and he gets 1 back which is 2 again.&#xA;&lt;br&gt; This will work for an infinite number of gates.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alternatively,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The person could have 0 magical fruits.&#xA;&lt;br&gt; Half of zero is 0, so there is no loss&#xA;&lt;br&gt; Then the guard give you 1 fruit!&#xA;&lt;br&gt; You end up with &lt;b&gt;1 free magical fruit&lt;/b&gt; to eat!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-06-12T17:22:51.453" LastActivityDate="2015-06-12T17:22:51.453" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16304" PostTypeId="2" ParentId="16302" CreationDate="2015-06-12T17:08:06.873" Score="1" Body="&lt;p&gt;Here's how I solved this:-&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;MAGICNUMBER/2 + 1 = MAGICNUMBER&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;MAGICNUMBER/2 - MAGICNUMBER = -1&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;-MAGICNUMBER/2 = -1&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;MAGICNUMBER/2 = 1&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;MAGICNUMBER = 2&#xA;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13016" LastEditorUserId="13016" LastEditDate="2015-06-12T17:35:57.380" LastActivityDate="2015-06-12T17:35:57.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16305" PostTypeId="2" ParentId="15823" CreationDate="2015-06-12T17:31:22.603" Score="1" Body="&lt;p&gt;Could it be:-&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Silence&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Accusation:&lt;/em&gt;&lt;/strong&gt; You are all so lazy...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Silence is an abstract noun, it doesn't do anything !&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Accusation:&lt;/em&gt;&lt;/strong&gt; Many people claim to follow you religiously.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;In many cultures and religions such as Buddhism, silence is a tradation!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Accusation:&lt;/em&gt;&lt;/strong&gt; We've never heard from 1 of your 5 parts. Why the silence?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Not sure about this one, maybe something related to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Five_Kinds_of_Silence&quot; rel=&quot;nofollow&quot;&gt;five kinds of silence&lt;/a&gt;!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Accusation:&lt;/em&gt;&lt;/strong&gt; People say that you say all kinds of things. Is this true? &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Silence can have different meanings for lovers, fighters etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Accusation:&lt;/em&gt;&lt;/strong&gt; Many claim that you are great teachers, and that people have learned more from your group than from any religious leader in history.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Silence is a great teacher. It can sometimes teach a great lesson. For example, if you do something and your mother just stares at you silently, then that silence will affect you more than her scolding!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Accusation:&lt;/em&gt;&lt;/strong&gt; Some of you gather in large numbers and in a way that makes the location unappealing. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Sometimes, eerie silence can make a place unappealing to others!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13016" LastEditorUserId="13016" LastEditDate="2015-06-12T17:56:57.223" LastActivityDate="2015-06-12T17:56:57.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16306" PostTypeId="1" AcceptedAnswerId="16307" CreationDate="2015-06-12T18:13:36.393" Score="0" ViewCount="1576" Body="&lt;p&gt;You are about to leave for holiday, but you forgot socks! You race back to your room, but all the lights are off, so you can't see the color of the socks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Never mind, because you remember that in your drawer there are ten pairs of white socks, ten pairs of black socks, and eleven pairs of blue socks, but they are all mixed up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many of your socks do you need to take before you can be sure to have at least one matching pair?&lt;/p&gt;&#xA;" OwnerUserId="13023" LastEditorUserId="13023" LastEditDate="2015-06-12T18:22:28.090" LastActivityDate="2015-06-12T18:22:28.090" Title="The complicated choice with socks" Tags="|logical-deduction|mathematics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16307" PostTypeId="2" ParentId="16306" CreationDate="2015-06-12T18:19:26.680" Score="2" Body="&lt;p&gt;Take four socks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Worse case scenario, the first three socks are from different color. The fourth must coincide with one of the first three.&lt;/p&gt;&#xA;" OwnerUserId="12165" LastActivityDate="2015-06-12T18:19:26.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16308" PostTypeId="1" AcceptedAnswerId="16311" CreationDate="2015-06-12T18:21:28.740" Score="11" ViewCount="1328" Body="&lt;p&gt;You put me up, They tear me down&lt;br&gt;&#xA;Some may laugh like I was a clown.&lt;br&gt;&#xA;Fix me up and put me back,&lt;br&gt;&#xA;Perhaps now we're more on track.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Crazy and impossible things,&lt;br&gt;&#xA;All sorts of silly strings.&lt;br&gt;&#xA;The answer is often clear,&lt;br&gt;&#xA;Why my reputation do you smear?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think about it, puzzle it out,&lt;br&gt;&#xA;What this riddle is all about.&lt;br&gt;&#xA;Maybe not the way you think,&lt;br&gt;&#xA;What am I? Don't let hopes sink.&lt;/p&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-06-12T19:27:17.957" Title="Fix Me Please - What am I?" Tags="|riddle|word|rhyme|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16310" PostTypeId="2" ParentId="16308" CreationDate="2015-06-12T18:36:36.310" Score="6" Body="&lt;p&gt;I think you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; dreams or aspirations&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You put me up, They tear me down&lt;br&gt;&#xA;Some may laugh like I was a clown.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Someone ruins your dreams by saying it is silly or impossible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fix me up and put me back,&lt;br&gt;&#xA;Perhaps now we're more on track.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Follow your dreams, even if people think it is silly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Crazy and impossible things,&lt;br&gt;&#xA;All sorts of silly strings.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dreams can be strange or silly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is often clear,&lt;br&gt;&#xA;Why my reputation do you smear?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Other people sometimes say, &quot;Stop dreaming, silly!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Maybe not the way you think,&lt;br&gt;&#xA;What am I? Don't let hopes sink.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Don't give up at achieving your dreams, even if it doesn't work out.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12109" LastActivityDate="2015-06-12T18:36:36.310" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16311" PostTypeId="2" ParentId="16308" CreationDate="2015-06-12T18:44:33.570" Score="10" Body="&lt;p&gt;This seems a bit too obvious to be correct to me, but I'm going to say you're a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; closed puzzle/riddle on Puzzling.SE, probably with the catch-all &quot;lateral thinking&quot; tag people put on just about anything.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You put me up, They tear me down&#xA;Some may laugh like I was a clown.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You got posted and quickly closed, most likely for being &quot;too broad&quot;. Laughter is still common in these cases, because the premise of these puzzles is often ridicilous.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fix me up and put me back,&#xA;Perhaps now we're more on track.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You'd like to be edited and reopened. Then you'll be &quot;on track&quot; to get answers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Crazy and impossible things,&#xA;All sorts of silly strings.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm not sure this needs much explanation. Have you seen some of the posts here? Infinite rope, prisoners that all have perfect logic/memory...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is often clear,&#xA;Why my reputation do you smear?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The OP almost &lt;em&gt;always&lt;/em&gt; thinks his own question is clear, no matter if it is or not, so get offended (smeared) by the downvotes that often accompany votes to close.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Think about it, puzzle it out,&#xA;What this riddle is all about.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I already did that. I think.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Maybe not the way you think,&#xA;What am I? Don't let hopes sink.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Not the way you think&quot; is almost literally &quot;lateral thinking&quot;. Please don't let the hopes of having a well received post die...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1563" LastEditorUserId="1563" LastEditDate="2015-06-12T19:06:08.357" LastActivityDate="2015-06-12T19:06:08.357" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="16312" PostTypeId="2" ParentId="16308" CreationDate="2015-06-12T19:27:17.957" Score="4" Body="&lt;p&gt;I was thinking you're a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; piñata&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You put me up, They tear me down&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That's what you do with piñatas&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some may laugh like I was a clown.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Piñatas are often funny shapes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fix me up and put me back,&#xA;Perhaps now we're more on track.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I don't think people usually do this, but hey, why not? Usually the kid just barely breaks it enough to get the candy to fall out.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Crazy and impossible things,&#xA;All sorts of silly strings.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Silly_String&quot; rel=&quot;nofollow&quot;&gt;Silly String&lt;/a&gt; is exactly the kind of prize that would be in there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is often clear,&#xA;Why my reputation do you smear?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I don't know about this line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Think about it, puzzle it out,&#xA;What this riddle is all about.&#xA;Maybe not the way you think,&#xA;What am I? Don't let hopes sink.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I just figured that this line was a fun closing verse. I'm new to Puzzling.SE, so I didn't know that every single line has to have significance.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7669" LastActivityDate="2015-06-12T19:27:17.957" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16313" PostTypeId="2" ParentId="16290" CreationDate="2015-06-12T19:47:48.457" Score="1" Body="&lt;p&gt;It could be a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Plastic bag(polythene  cover).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You may make me, but you cannot destroy me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We make it but cant be destroyed by us.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can see through anything with my help.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; plastic covers are see through&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I'm (really) big and black, nobody and nothing can escape from me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Big and black covers when sealed does not allow anything out.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not used very frequently.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; These days we try to use bio degradable bags.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes, I make objects unusable.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; objects that are inside it cannot be used.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11056" LastActivityDate="2015-06-12T19:47:48.457" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16316" PostTypeId="2" ParentId="16275" CreationDate="2015-06-12T20:24:39.087" Score="8" Body="&lt;p&gt;&lt;strong&gt;According to the rules of golf, one stroke&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Chip it in.&lt;/strong&gt; A pro golfer that can only putt specific distances probably has better luck with his wedge. (A pro-level golfer should sink a ball on the green in one shot &lt;a href=&quot;http://www.pgatour.com/stats/stat.413.html&quot; rel=&quot;nofollow&quot;&gt;at least 35% of the time&lt;/a&gt; and in two shots &lt;a href=&quot;http://www.pgatour.com/stats/stat.426.html&quot; rel=&quot;nofollow&quot;&gt;at least 96% of the time&lt;/a&gt;. Our hero probably didn't make pro by 2-putting nearly every hole.)&#xA; &lt;br/&gt;&#xA; &lt;br/&gt;&#xA; Under the rules of golf, &lt;a href=&quot;http://www.usga.org/RulesFAQ/rules_answer.asp?FAQidx=46&amp;amp;Rule=14&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;any club may be used for any stroke, even on the green&lt;/em&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;    From USGA Rules FAQ:&lt;br&gt;&#xA;&lt;br&gt;&#xA;    &lt;strong&gt;Rule 14-1&lt;/strong&gt;&lt;br&gt;&#xA;    &lt;strong&gt;Using Club other than Putter on Putting Green?&lt;/strong&gt;&lt;br&gt;&#xA;    &lt;br&gt;&#xA;    Q. Is it permissible to use a club other than a putter to play a stroke on the putting green?&lt;br&gt;&#xA;    &lt;br&gt;&#xA;    A. Yes, the Rules of Golf do not stipulate the type of club that must be used for making a stroke on the putting green or anyplace else on the golf course.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11752" LastEditorUserId="11752" LastEditDate="2015-06-19T16:05:21.583" LastActivityDate="2015-06-19T16:05:21.583" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16317" PostTypeId="2" ParentId="4370" CreationDate="2015-06-12T21:11:20.150" Score="8" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 6&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The rule is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If the number that the guard tells you is n, respond with the number of letters in the nth word of &lt;a href=&quot;https://en.wikipedia.org/wiki/La_Marseillaise&quot; rel=&quot;nofollow noreferrer&quot;&gt;The French National Anthem&lt;/a&gt; (in French, of course).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is the first stanza:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;French lyrics&lt;/strong&gt;&lt;br&gt; Allons enfants de la Patrie,&lt;br&gt; Le jour de gloire est arrivé !&lt;br&gt; Contre nous de la tyrannie,&lt;br&gt; L'étendard&#xA;  sanglant est levé, (bis)&lt;br&gt; Entendez-vous dans les campagnes&lt;br&gt;&#xA;  Mugir ces féroces soldats ?&lt;br&gt; Ils viennent jusque dans nos bras&lt;br&gt;&#xA;  Égorger nos fils, nos compagnes !&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All of the words match, and the incorrect answer (off by one) fits as well!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/X2iJg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-06-14T16:28:40.713" LastActivityDate="2015-06-14T16:28:40.713" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16318" PostTypeId="1" CreationDate="2015-06-12T21:13:17.457" Score="-2" ViewCount="995" Body="&lt;p&gt;An Olympic judge rushes into the barbershop. &quot;You need to help me!&quot; he pleads. &quot;I have to judge in 10 minutes, but I've been traveling so much I look just awful.&quot; The barber looks at him carefully. &quot;Your hair is quite long. I'll need to cut it. And you'd look silly with a beard and such short hair. You're in luck, as I've just had a cancellation. Have a seat!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The barber does his work and goes to collect the payment. &quot;Oh dear,&quot; said the judge. &quot;I haven't any local currency to pay you with.&quot; Luckily, he and the barber were able to work out a deal. The judge, having fully paid the barber, leaves to go judge his match.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On his lunch break, the barber was watching the Olympic games and sure enough there was the judge, attention gazed on the performers, looking for any mistakes he could find.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Why was the judge looking for mistakes so intently?&lt;/li&gt;&#xA;&lt;li&gt;What was the deal between the barber and the judge?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Clarifications:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The judge is an honorable judge. There is no foul play.&lt;/li&gt;&#xA;&lt;li&gt;The judge and the barber will never interact again. The deal is fully concluded in a mutually acceptable way.&lt;/li&gt;&#xA;&lt;li&gt;Nothing illegal is transpiring.&lt;/li&gt;&#xA;&lt;li&gt;Currency is not involved in any way.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2812" LastEditorUserId="2812" LastEditDate="2015-06-12T21:51:05.827" LastActivityDate="2015-06-13T01:04:17.490" Title="How did the judge pay his barber?" Tags="|lateral-thinking|enigmatic-puzzle|" AnswerCount="6" CommentCount="15" ClosedDate="2015-06-12T22:17:35.237" ContentLicense="CC BY-SA 3.0" />
  <row Id="16319" PostTypeId="2" ParentId="16318" CreationDate="2015-06-12T21:22:04.940" Score="0" Body="&lt;p&gt;It rather sounds like the judge &quot;paid&quot; the barber by offering to exert bias against certain athletes by looking for mistakes in their routines more intently. Perhaps the barber has a favorite they'd like to see win, or is loyal to their national team, or has bets outstanding on an event.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-06-12T21:24:04.120" LastActivityDate="2015-06-12T21:24:04.120" CommentCount="2" CommunityOwnedDate="2015-06-12T21:22:04.940" ContentLicense="CC BY-SA 3.0" />
  <row Id="16320" PostTypeId="2" ParentId="16318" CreationDate="2015-06-12T21:27:13.490" Score="1" Body="&lt;p&gt;Could it be:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The barber, supporting the local team, made a deal with the judge that, for each mistake by a foreign performer caught, he will deduct $1 equivalent from the debt?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="12109" LastEditDate="2015-06-12T21:34:48.030" LastActivityDate="2015-06-12T21:34:48.030" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16321" PostTypeId="2" ParentId="16318" CreationDate="2015-06-12T21:32:23.897" Score="0" Body="&lt;p&gt;The judge was watching closely because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He wanted to minimize the total of the scores he gave.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is because his deal with the barber was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The judge would pay the barber the total sum of the scores he gave to the participants.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13156" LastActivityDate="2015-06-12T21:32:23.897" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16322" PostTypeId="2" ParentId="16318" CreationDate="2015-06-12T21:42:12.523" Score="4" Body="&lt;p&gt;As for the mistakes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He was looking for mistakes in peoples haircuts and grooming. You never specify what type of mistakes the judge is looking for.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the payment&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He recommends the barber to the athletes &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11239" LastEditorUserId="11239" LastEditDate="2015-06-13T01:04:17.490" LastActivityDate="2015-06-13T01:04:17.490" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16323" PostTypeId="2" ParentId="16318" CreationDate="2015-06-12T21:44:48.513" Score="3" Body="&lt;p&gt;The judge is judging a diving event. Competitive swimmers and divers shave their bodies to be more hydrodynamic. The judge lowers scores by looking more intently for mistakes, which makes contestants think they are performing poorly, and that they could do better getting shaved by the barber as a professional. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By doing this equally to all competitors, the judge feels like his is keeping the integrity of the competition. &lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-06-12T21:44:48.513" CommentCount="0" CommunityOwnedDate="2015-06-12T21:44:48.513" ContentLicense="CC BY-SA 3.0" />
  <row Id="16326" PostTypeId="1" AcceptedAnswerId="16329" CreationDate="2015-06-12T21:54:31.320" Score="7" ViewCount="982" Body="&lt;p&gt;A famous mathematician was found on his deathbed this evening, leaving behind his life's work trapped away on his computer.&lt;/p&gt;&#xA;&lt;p&gt;Protected by a password, it is believed this computer contains many groundbreaking equations that have yet to be released to the public.&lt;/p&gt;&#xA;&lt;p&gt;Luckily, it seems, our mathematician was aware of his timely demise and left us a note, ensuring his works only went to someone intellectually worthy. The note thus read:&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;It seems my time has come at last, for like many before me, I have met my demise. I'm sure you'll want my life's works, they're ever such a prize. Are you worthy?&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;For Fibonacci was right,&lt;br /&gt;&#xA;if only he'd gone further he'd have shed some light,&lt;br /&gt;&#xA;for me aabcehmu is just not enough for a key,&lt;br /&gt;&#xA;yet it's taken as credence,&lt;br /&gt;&#xA;just adding one to each of his sequence,&lt;br /&gt;&#xA;just remember to playfair,&lt;br /&gt;&#xA;for names leave a legacy,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;signed hpssodge&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12734" LastEditorUserId="69582" LastEditDate="2024-01-17T16:39:46.663" LastActivityDate="2024-01-17T16:39:46.663" Title="Mathematicians Death Bed Password" Tags="|riddle|cipher|password|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="16328" PostTypeId="2" ParentId="16318" CreationDate="2015-06-12T22:06:46.163" Score="10" Body="&lt;p&gt;Hmm, very open ended question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only can come up with a fairly stupid guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The barber shaved his shops name onto this judges head. Judge looked intently to ensure the back of his head was always to the TV viewers camera.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12734" LastActivityDate="2015-06-12T22:06:46.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16329" PostTypeId="2" ParentId="16326" CreationDate="2015-06-12T22:42:39.517" Score="10" Body="&lt;p&gt;I think I have found the password. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Firstly the aabcehmu stands for Fibonacci is the alphabetic representation of the Fibonacci sequence.&#xA; Now according to letter &quot;just adding one to each of his sequence&quot; we need to add one to each letter of the Fibonacci. and you will get&#xA;bbcdfinv&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Step 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;'hpssodge' is a message which we need to decode using Playfair Cipher&#xA; so When you Decrypt hpssodge with key as bbcdfinv you will get&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The password&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; puzzling&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12736" LastEditorUserId="12734" LastEditDate="2015-06-12T22:51:54.630" LastActivityDate="2015-06-12T22:51:54.630" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16330" PostTypeId="1" AcceptedAnswerId="16364" CreationDate="2015-06-12T22:51:45.307" Score="7" ViewCount="387" Body="&lt;p&gt;Let me tell you about my sisters and I.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I am the eldest, and in fact the first of all our siblings.&lt;br&gt;&#xA;  My next sister buys alcohol by the bottle and places them on her shelves.&lt;br&gt;&#xA;  My next sister always wraps her slim figure in the nicest of evening gowns, which she stitches herself.&lt;br&gt;&#xA;  In contrast, you can always find my next, more portly sister at the pool hall at a quarter past three every afternoon.&lt;br&gt;&#xA;  My last sister prefers playing blackjack (she sees herself in the table).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who are we AND who are our brothers?&lt;/p&gt;&#xA;" OwnerUserId="10861" LastActivityDate="2015-06-14T09:12:17.260" Title="Facetious Sisters" Tags="|riddle|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16331" PostTypeId="1" AcceptedAnswerId="16519" CreationDate="2015-06-12T23:27:21.000" Score="12" ViewCount="1043" Body="&lt;p&gt;In the spirit of wizards and circular prisons:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have been imprisoned by an evil wizard in a perfectly circular prison cell of unknown size. You're shackled to the wall, unable to move about the cell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fortunately, the wizard isn't so evil that he won't give you at least one opportunity to escape. He tells you that if you can compute the area of the prison cell &lt;em&gt;exactly&lt;/em&gt; (that is, deterministically, and to arbitrarily high precision), he'll let you go free.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To help you accomplish this task, he's given you three flying monkey drones, Arthur, Bob, and Dale, and a set of teleportation spells you can use as often as you like:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Teleport any drone to your location (you can pick which drone).&lt;/li&gt;&#xA;&lt;li&gt;Teleport yourself to Arthur or Bob's location (you can pick which drone).&lt;/li&gt;&#xA;&lt;li&gt;Teleport yourself to a random location on the prison wall. (You remain unable to move even after teleporting. Also, you never teleport to &lt;em&gt;exactly&lt;/em&gt; the same location twice.)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Conveniently, if you pull on Arthur's tail, he will tell you &lt;em&gt;exactly&lt;/em&gt; what the shortest angle $\theta$ between himself and Dale is, as shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you pull on Dale's tail, he will tell you &lt;em&gt;exactly&lt;/em&gt; what the distance $d$ between himself and Arthur is, as shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/2so4i.png&quot; alt=&quot;measuring distance and angle&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are three restrictions on measurement, however:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You must be &lt;em&gt;exactly&lt;/em&gt; at a drone's location to pull his tail.&lt;/li&gt;&#xA;&lt;li&gt;After you have teleported a drone to your location, you cannot pull his tail until you have teleported at least once.&lt;/li&gt;&#xA;&lt;li&gt;You cannot measure both $d$ and $\theta$ across the same pair of points (even in reverse).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;All three drones start at your location on the wall, which is unknown. You may assume that you have a perfect memory, and that you can compute any mathematical function to arbitrarily high precision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Are you able to escape? And how?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; Generally speaking, I'm looking for a solution that&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;has 100.0% probability of being correct (not e.g. a probability of correctness that approaches 1 asymptotically)&lt;/li&gt;&#xA;&lt;li&gt;has finite $E\left[ n\right]$, where $n$ is the RV describing the number of steps needed to reach a solution&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="4219" LastEditDate="2015-06-17T04:30:18.293" LastActivityDate="2016-06-10T08:54:36.587" Title="Droning On in Circles" Tags="|mathematics|geometry|" AnswerCount="6" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16332" PostTypeId="2" ParentId="16331" CreationDate="2015-06-13T00:26:31.600" Score="5" Body="&lt;p&gt;&lt;strong&gt;Remark&lt;/strong&gt;: This is not a complete solution. See the edit below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, it is possible to escape, using the following sequence:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Teleport Dale to your location&lt;/li&gt;&#xA;&lt;li&gt;Teleport to a random location&lt;/li&gt;&#xA;&lt;li&gt;Teleport Arthur to your location&lt;/li&gt;&#xA;&lt;li&gt;Teleport to a random location&lt;/li&gt;&#xA;&lt;li&gt;Telport to Arthur&lt;/li&gt;&#xA;&lt;li&gt;Pull Arthur's tail, receiving a measurement $A$.&lt;/li&gt;&#xA;&lt;li&gt;Teleport Bob to your location&lt;/li&gt;&#xA;&lt;li&gt;Teleport to a random location&lt;/li&gt;&#xA;&lt;li&gt;Teleport Arthur to your location&lt;/li&gt;&#xA;&lt;li&gt;Teleport to Bob's location&lt;/li&gt;&#xA;&lt;li&gt;Teleport to Arthur's location&lt;/li&gt;&#xA;&lt;li&gt;Pull Arthur's tail, receiving a measurement $B$&lt;/li&gt;&#xA;&lt;li&gt;Teleport to Bob's location&lt;/li&gt;&#xA;&lt;li&gt;Teleport Dale to your location&lt;/li&gt;&#xA;&lt;li&gt;Teleport to Arthur's location&lt;/li&gt;&#xA;&lt;li&gt;Teleport to Bob's location&lt;/li&gt;&#xA;&lt;li&gt;Pull Dale's tail, receiving a measurement of $d$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ieygC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is depicted in the image above. Your initial location is marked with $0$, the first random place you teleport to is marked with $1$, and the third random place you teleport to is marked $3$. The angle subtended by the chord of length $d$ is $2\pi-A-B$, the radius of the room is&#xA;$$&#xA;\frac{d}{2\sin\left(\frac{2\pi-A-B}{2}\right)},&#xA;$$&#xA;and the area of the room is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\pi\left(\frac{d}{2\sin\left(\frac{2\pi-A-B}{2}\right)}\right)^2.$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: As pointed out in the comments, the situation could be different depending on the relative locations of teleportation. The picture might also look like the following:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/qimbZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case the angle subtended by the chord of length $d$ is $B-A$, and the area of the room is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\pi\left(\frac{d}{2\sin\left(\frac{B-A}{2}\right)}\right)^2.$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The difficulty is, without some knowledge of the relative positions of teleportation locations, we cannot know which formula to use.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-06-16T19:02:32.233" LastActivityDate="2015-06-16T19:02:32.233" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="16333" PostTypeId="1" CreationDate="2015-06-13T00:34:25.713" Score="17" ViewCount="5982" Body="&lt;p&gt;There are seven days in a week. Let's say you want to pick one at random, such that each day has an equal chance of being picked as each other day. In other words, the goal is to obtain a uniform random 1/7 chance of picking any of the seven days.&lt;/p&gt;&#xA;&lt;p&gt;But all you have is a coin and a six-sided die (not dice, you don't have 2 dice, you have only 1 die). What procedure of coin-flipping and/or die-rolling should you use to get a 1/7 chance?&lt;/p&gt;&#xA;&lt;p&gt;(EDIT: I originally intended a coin that can only give a 1/2 result. However, there was a decent out-of-the-box answer about rollable coins that I found pretty useful.)&lt;/p&gt;&#xA;&lt;p&gt;I know of two ways to do this, but they're slightly out-of-the-box solutions. I consider them valid solutions but want to see what you guys think. Of course, I'll let people try to answer first before revealing my solutions.&lt;/p&gt;&#xA;&lt;p&gt;Hint: don't try anything like trying to pick a random number in your head. The only things you can use are the coin and die.&lt;/p&gt;&#xA;&lt;p&gt;EDIT: I'm going to post the four answers I know of in the most compact and simple way (in my humble opinion). Everyone who answered so far has given the equivalent of one of these answers:&lt;/p&gt;&#xA;&lt;p&gt;Method 1.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Roll the die and flip the coin. This gives 12 possibilities: 1T, 2T, 3T, 4T, 5T, 6T, 1H, 2H, 3H, 4H, 5H, 6H. Assign seven days of the week to seven of those values. If the result is one of the remaining five, just repeat the process until you get a valid value. This has a 5/12 (41.67%) chance of repeating.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Method 2.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Flip the coin three times (and be sure to record it in order of course). This gives 8 possibilities: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Assign seven days of the week to the first seven of those values. If your result is the eighth value, repeat the process until you get a valid value. This has a 1/8 (12.5%) chance of repeating.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Method 3.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Roll the die twice (and be sure to record them in order because a 2 and 5 is not the same thing as a 5 and 2 in the sample space!). This gives 36 possibilities: 1&amp;amp;1, 1&amp;amp;2, 1&amp;amp;3, 1&amp;amp;4, 1&amp;amp;5, 1&amp;amp;6, 2&amp;amp;1, 2&amp;amp;2, 2&amp;amp;3, 2&amp;amp;4, 2&amp;amp;5, 2&amp;amp;6, 3&amp;amp;1, 3&amp;amp;2, 3&amp;amp;3, 3&amp;amp;4, 3&amp;amp;5, 3&amp;amp;6, 4&amp;amp;1, 4&amp;amp;2, 4&amp;amp;3, 4&amp;amp;4, 4&amp;amp;5, 4&amp;amp;6, 5&amp;amp;1, 5&amp;amp;2, 5&amp;amp;3, 5&amp;amp;4, 5&amp;amp;5, 5&amp;amp;6, 6&amp;amp;1, 6&amp;amp;2, 6&amp;amp;3, 6&amp;amp;4, 6&amp;amp;5, 6&amp;amp;6. Assigned 7 days of the week, but do that 5 times. This uses up 35 values. If your two rolls results in the 36th value, just repeat the process until you get a valid value. This has a 1/36 (2.78%) chance of repeating, which is much smaller than the previous two methods.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Method 4:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Roll the coin. If it's a 7-sided &lt;a href=&quot;https://en.wikipedia.org/wiki/Curve_of_constant_width&quot; rel=&quot;nofollow noreferrer&quot;&gt;rollable coin&lt;/a&gt;, then just see which side ends up closest to some arbitrary direction, like north, or arbitrary marker, such as a stick in the ground. If it's a circular coin, you could theoretically mark the coin to divide it into 7 sectors and then see which sector points towards your arbitrary thing. There was also an idea of constructing a Roulette Wheel with 7 sectors. I know I said you only have the coin and die, but then you could just label some sectors on the ground and use that as your roullette wheel, and presumably find some way to confine the coin there or just roll again if it escapes. So I think this idea too is valuable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Some concluding thoughts...&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Now that I think about it, you could extend the multiple rolls/tosses. If you flipped the coin 6 times, that's 64 possibilities, and 63 is evenly divisible by 7 so you can assign days of the week to that, and just repeat if u get the 64th value (1 out of 64 chance = 1.56%!). You could roll the die 4 times for 1,296 possibilites. 1,295 is divisible by 7, so assign days of the week evenly to all those values, and just repeat if u get the 1,296th value (1 out of 1,296 chance = 0.08%!!). You can probably go even further with them, but it's a trade off of doing more rolls/tosses, which takes a longer time and might take longer to compute the result, than just taking the 1/36 chance of having to repeat via the two-roll method, for example.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13160" LastEditorUserId="13160" LastEditDate="2021-01-30T02:05:02.497" LastActivityDate="2021-01-31T09:50:27.423" Title="How to get a uniform 1 out of 7 chance, using only a coin and a 6-sided die" Tags="|mathematics|probability|dice|coins|" AnswerCount="15" CommentCount="15" ClosedDate="2021-01-30T03:53:32.117" ContentLicense="CC BY-SA 4.0" />
  <row Id="16334" PostTypeId="1" AcceptedAnswerId="16337" CreationDate="2015-06-13T01:42:13.760" Score="6" ViewCount="360" Body="&lt;p&gt;Here's a letter sequence I made. Can you figure out what missing letters goes in the blanks?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TW, GM, SL, SG, __, ET, SB, __, __&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="11002" LastEditorUserId="1664" LastEditDate="2015-06-13T15:29:58.607" LastActivityDate="2015-06-13T15:29:58.607" Title="What Missing Letters Goes In The Blanks And Why?" Tags="|letter-sequence|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16335" PostTypeId="2" ParentId="16333" CreationDate="2015-06-13T01:50:25.390" Score="19" Body="&lt;p&gt;There is &lt;strong&gt;a way that doesn't use the dice (3+ coin-flips)&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Flip the coin three times and record 1 or 0 for heads or tails.&lt;/li&gt;&#xA;&lt;li&gt;If all three numbers are zero start again.&lt;/li&gt;&#xA;&lt;li&gt;Otherwise read the 1s and 0s as a binary number. Its value will be from 1 to 7 inclusive with each number having an equal probability of occurring.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Here's a &lt;strong&gt;summary of Mike's answer (2+ dice-rolls)&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Roll the dice twice call the results  and . If both results are 1 roll again.&lt;/li&gt;&#xA;&lt;li&gt;Otherwise calculate ( - 1) × 6 + ( - 1) Divide that number by 7 to find the remainder. The remainder is in the range 0 to 6 inclusive, add 1 if you want 1 to 7.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="2349" LastEditDate="2021-01-31T09:50:27.423" LastActivityDate="2021-01-31T09:50:27.423" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="16336" PostTypeId="2" ParentId="16333" CreationDate="2015-06-13T01:56:08.243" Score="6" Body="&lt;p&gt;Roll the die twice, to generate a random number between 0 and 35. Specifically, subtract one from each roll, and have the first roll be the &quot;tens&quot; digit of a number in base 6, and the second roll be the ones digit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the result nonzero, return the remainder of the result (mod $7$). If the result is zero, then repeat until it is nonzero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives one of the numbers $0$-$6$ with equal probability, which correspond to the seven days of the week.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This process will take $\frac2{1-\frac1{36}}\approx 2.06$ rolls on average.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-06-13T01:56:08.243" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="16337" PostTypeId="2" ParentId="16334" CreationDate="2015-06-13T01:58:41.853" Score="8" Body="&lt;p&gt;The missing letters are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;NI&lt;/strong&gt;, &lt;strong&gt;DA&lt;/strong&gt;, &lt;strong&gt;YW&lt;/strong&gt;. The letter pairs are the last and first letter of each word in this question's title.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2627" LastActivityDate="2015-06-13T01:58:41.853" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16338" PostTypeId="1" AcceptedAnswerId="16340" CreationDate="2015-06-13T06:04:38.527" Score="2" ViewCount="281" Body="&lt;p&gt;How do you form the number 1000 with the digits, 1,2,3 and 4?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mathematical operations that can be are addition, subtraction, multiplication, division, parentheses, percentages and factorials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Digits should not be repeated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can combine digits as well, example 12+34,etc.&lt;/p&gt;&#xA;" OwnerUserId="13163" LastEditorUserId="13163" LastEditDate="2015-06-13T07:22:08.013" LastActivityDate="2015-06-20T23:27:04.573" Title="Number formation with digits given" Tags="|formation-of-numbers|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16340" PostTypeId="2" ParentId="16338" CreationDate="2015-06-13T07:18:44.593" Score="6" Body="&lt;p&gt;You can do it without repeating numbers, assuming you may arrange order of operations as you wish:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\frac{4 \times 3 - 2 }{ 1\%}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3987" LastEditorUserId="3987" LastEditDate="2015-06-13T07:19:47.683" LastActivityDate="2015-06-13T07:19:47.683" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16341" PostTypeId="1" CreationDate="2015-06-13T08:38:48.990" Score="10" ViewCount="662" Body="&lt;p&gt;The governor of Reniets (a land far, far away!) is known to be very stingy.&lt;br&gt;&#xA;He has to &lt;strong&gt;build a road which connects all the five cities in the region&lt;/strong&gt;, which are oddly arranged on the vertices of a &lt;strong&gt;regular pentagon&lt;/strong&gt; (see the picture below).&lt;br&gt;&#xA;Building a road is very expensive: the longer it is, the more it costs!&lt;br&gt;&#xA;His first idea is to build five roads, each connecting a city to the center of the pentagon, then he realizes that it might not be optimal. In fact, he remembers that, years ago, his friend had to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3881/connect-four-towers-by-roads&quot;&gt;&lt;strong&gt;connect four towers&lt;/strong&gt;&lt;/a&gt;, and the solution wasn't trivial!.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cYNN7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How long is the shortest path that connects all the five cities?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Notes:&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;assume that the side is long $1$ km.&lt;/em&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;I currently don't know the solution, I will accept the most convincing answer if no proof is given after some days.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;If you can generalize the solution to polygons of any number $n$ of sides, I will award a bounty (the amount depends on the quality of the work).&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2019-03-03T21:49:59.927" Title="Building a cheap road" Tags="|mathematics|story|optimization|graph-theory|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16342" PostTypeId="2" ParentId="16341" CreationDate="2015-06-13T09:49:25.717" Score="2" Body="&lt;p&gt;Here's the optimal solution for the pentagon:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KRJ3K.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;As the only Steiner tree for this set of five points, it is the only locally minimal solution, hence it must be globally minimal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For regular hexagons and above, just take the perimeter and remove one edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Whoops I forgot to calculate the total path length. Brb.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT2: The length of the tree in the pentagon is $${1\over 2}\sqrt{17+7\sqrt{5}+\sqrt{390+174\sqrt{5}}} \approx 3.891 ~\text{km}.$$&#xA;For an $n$-gon ($n \ge 6$) the total path length would obviously be $n-1~\text{km}.$&lt;/p&gt;&#xA;" OwnerUserId="12906" LastEditorUserId="12906" LastEditDate="2015-06-13T10:56:22.993" LastActivityDate="2015-06-13T10:56:22.993" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16343" PostTypeId="2" ParentId="16330" CreationDate="2015-06-13T10:16:59.240" Score="0" Body="&lt;p&gt;2 Guesses:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You Are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Light, and your brothers are darkness.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My next sister buys alcohol by the bottle and places them on her shelves. - &lt;strong&gt;Morning sun, when preparing is done?&lt;/strong&gt;&lt;br&gt;&#xA;My next sister always wraps her slim figure in the nicest of evening gowns, which she stitches herself.  &lt;strong&gt;wrapping herself in light in the thin evening sun&lt;/strong&gt;&lt;br&gt;&#xA;In contrast, you can always find my next, more portly sister at the pool hall at a quarter past three every afternoon. &lt;strong&gt;Strong sun shining down in afternoon&lt;/strong&gt;&lt;br&gt;&#xA;My last sister prefers playing blackjack (she sees herself in the table).  &lt;strong&gt;reflection of light&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You Are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wood/Trees and your brothers are humans.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My next sister buys alcohol by the bottle and places them on her shelves. - &lt;strong&gt;Basic wood used for shelves&lt;/strong&gt;&lt;br&gt;&#xA;My next sister always wraps her slim figure in the nicest of evening gowns, which she stitches herself.  &lt;strong&gt;Basic tree, growing leaves/twigs?&lt;/strong&gt;&lt;br&gt;&#xA;In contrast, you can always find my next, more portly sister at the pool hall at a quarter past three every afternoon. &lt;strong&gt;Higher quality wood used for tables&lt;/strong&gt;&lt;br&gt;&#xA;My last sister prefers playing blackjack (she sees herself in the table).  &lt;strong&gt;highest quality/thick wood used for luxury tables?&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;Your brothers are the ones that change you and make you into these products?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="12734" LastActivityDate="2015-06-13T10:16:59.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16345" PostTypeId="2" ParentId="16333" CreationDate="2015-06-13T13:12:27.527" Score="0" Body="&lt;p&gt;Similar to Mike Earnest, I suggest rolling again on a double six, to give you 35 possibilities instead of 36. The question then becomes how to assign the 35 possibilities to the 7 scores you require. The following formula is reasonably easy to calculate, and does not involve any base conversions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(7+Die1-Die2) mod 7&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;       Die 1&#xA;        1 2 3 4 5 6&#xA;        ===========&#xA;Die   1|0 1 2 3 4 5&#xA;2     2|6 0 1 2 3 4&#xA;      3|5 6 0 1 2 3&#xA;      4|4 5 6 0 1 2&#xA;      5|3 4 5 6 0 1&#xA;      6|2 3 4 5 6 X       X=roll again&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As can be seen, each score from 0 to 6 appears in the table exactly 5 times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In plain english, this becomes: &lt;em&gt;&quot;Roll the die twice. Subtract the second roll from the first (if this is not possible because it would give a negative number, add 7 to the value of the first roll before subtracting.) If you roll a double six, discard the score and roll again.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="4768" LastActivityDate="2015-06-13T13:12:27.527" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16346" PostTypeId="2" ParentId="16333" CreationDate="2015-06-13T13:18:56.383" Score="5" Body="&lt;p&gt;Here's an answer that may take more steps to complete but is a lot more intuitive to think about:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roll the die and flip the coin simultaneously. If the coin lands on heads, read the die value. If the coin lands on tails and the die lands on 1, get the value 7. Else, repeat the process.&lt;/p&gt;&#xA;" OwnerUserId="12278" LastActivityDate="2015-06-13T13:18:56.383" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16348" PostTypeId="1" AcceptedAnswerId="17210" CreationDate="2015-06-13T16:06:37.543" Score="33" ViewCount="5019" Body="&lt;p&gt; You stand in a forest,&#xA;&lt;br&gt; So black, dark,&#xA;&lt;br&gt; Light can’t quite wind around the dry, tough bark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt; Trees are all you can see, &#xA;&lt;br&gt; But change your viewpoint even slightly; &#xA;&lt;br&gt; And a brighter vista it may indeed come to be. &lt;/p&gt;&#xA;&#xA;&lt;p&gt; Those seen trees which bear ripe plums, &#xA;&lt;br&gt; Their seeds grew from the Two that do make One, &#xA;&lt;br&gt; Which share no part except for another One. &lt;/p&gt;&#xA;&#xA;&lt;p&gt; But in this deep, dark, starless forest, those hidden &#xA;&lt;br&gt; Have more in common, &#xA;&lt;br&gt; With those that keep them overridden. &lt;/p&gt;&#xA;&#xA;&lt;p&gt; Sticks in one big, infinite bundle,&#xA;&lt;br&gt; Upright, but One far apart,&#xA;&lt;br&gt; Farther removed from Mussolini's art. &lt;/p&gt;&#xA;&#xA;&lt;p&gt; But these belong to one very ancient indeed,&#xA;&lt;br&gt; A man renowned in strictest sense, &#xA;&lt;br&gt; Celebrated most for famous work, derived from common sense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt; Thin green trees in this private garden, &#xA;&lt;br&gt; With raindrops, at tip top, &#xA;&lt;br&gt; Belong to him, the mysterious Alexandrian. &lt;/p&gt;&#xA;&#xA;&lt;p&gt; Words, meaningless, focus on their numbers, &#xA;&lt;br&gt; To unravel this poem, the key always remains Three, &#xA;&lt;br&gt; Numbers speak languages untold, decipher them to avoid any blunders. &lt;/p&gt;&#xA;&#xA;&lt;p&gt; Translate to a system, one familiar to the Owner- &#xA;&lt;br&gt; Mysterious still, but more unfolding may prove to the sleuth,&#xA;&lt;br&gt; That like the flowers here, this poem holds one truth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where are you? &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hints:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This poem holds together on multiple levels, and some words are chosen &lt;strong&gt;very&lt;/strong&gt; precisely, but other words and phrases (e.g. &quot;Mussolini's art&quot;) are used for rhyme or imagery or... other purposes. &lt;br&gt; &lt;br&gt; If you're unsure about where to start with deciphering the meaning of each word, go to the eighth tercet. &lt;br&gt; &lt;br&gt; An answer will not be accepted only because it produces the correct answer; for an answer to be accepted, it must go line-by-line, or at least stanza-by-stanza, and give an interpretation of each part. It also must identify any hidden clues in the puzzle. &lt;br&gt; &lt;br&gt;The poem holds together on a semantic &lt;em&gt;and&lt;/em&gt; a syntactical level—if the wording seems obtuse (which it is), focus on the structure. &lt;br&gt; &lt;br&gt; Words hold numbers who will only reveal themselves to the arithmetic of the clock.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13097" LastEditorUserId="13097" LastEditDate="2015-07-03T21:55:21.463" LastActivityDate="2015-07-03T21:55:21.463" Title="You stand in a forest... Where are you?" Tags="|riddle|enigmatic-puzzle|mystery|" AnswerCount="8" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="16349" PostTypeId="2" ParentId="16338" CreationDate="2015-06-13T16:27:05.420" Score="-4" Body="&lt;p&gt;$((3^2)+1) ^ 4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In words, (3 squared) + 1 to the 4th power&lt;/p&gt;&#xA;" OwnerUserId="13172" LastEditorUserId="10281" LastEditDate="2015-06-13T17:01:23.617" LastActivityDate="2015-06-13T17:01:23.617" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16351" PostTypeId="2" ParentId="16286" CreationDate="2015-06-13T16:28:39.613" Score="2" Body="&lt;p&gt;Credit to Bob and LeppyR64 who basically solved it, here's the last part:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Enter the new image into &lt;a href=&quot;http://mozaiq.org/decrypt&quot; rel=&quot;nofollow&quot;&gt;http://mozaiq.org/decrypt&lt;/a&gt; with the password &quot;00&quot; to get the line &quot;Yes, you've found me. My name is VGVjaFJheA==&quot;. This is Base64 encoded and translates to TechRax.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastActivityDate="2015-06-13T16:28:39.613" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16354" PostTypeId="1" AcceptedAnswerId="16389" CreationDate="2015-06-13T16:51:17.953" Score="12" ViewCount="633" Body="&lt;p&gt;&lt;strong&gt;Harry Potter&lt;/strong&gt; was happily living his married life when suddenly, he is&#xA;told by &lt;strong&gt;Hagrid&lt;/strong&gt; about people dying in sleep. So, he decides to set out&#xA;on a quest to save everyone. He sets out on a journey to the dream world of  &lt;strong&gt;Narinda&lt;/strong&gt;, along with &lt;strong&gt;Hermione&lt;/strong&gt; and &lt;strong&gt;Ron&lt;/strong&gt; , where they come face to face with &lt;em&gt;nightmares&lt;/em&gt;, &lt;em&gt;goblins&lt;/em&gt; and other &lt;em&gt;unearthly creatures&lt;/em&gt;. They are forced to pass through the dark jungles of &lt;strong&gt;Tortin&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As they pass through the jungle, they meet a white witch. Who mumbles the below words:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;You know what to do,&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;The keeper will help you through,&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;So that you find the clues,&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;Let not be the prophecy true.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Along with that, she dissappeared. Leaving behind a piece of paper. On it was written:-&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To find the correct information using the below. You'll need Rats,&#xA;  Oil, Trumpet, Thumb, Hair, Ice, Rhyme, Teeth, Elves, Elbow and Nails.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;ZGZpb3JsZ2xhb2h3c3R1aGVlYnJ1aXJ2ZXI=&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;Dear Harry,&lt;/p&gt;&#xA;    &#xA;    &lt;p&gt;&lt;strong&gt;serrgrkgubfgpbzv5xdr2cfsb&lt;/strong&gt;&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who was this women, what was her message?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="13016" LastEditorUserId="13016" LastEditDate="2015-06-14T11:37:40.373" LastActivityDate="2015-06-14T18:38:01.627" Title="The mystery of the white witch" Tags="|cipher|wordplay|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16357" PostTypeId="1" AcceptedAnswerId="16359" CreationDate="2015-06-13T18:40:19.043" Score="8" ViewCount="958" Body="&lt;p&gt;$101$ gunmen stand in a field. At high noon, everyone shoots the gunman standing closest to him. If there are several gunmen who are equally close, they shoot the tallest one of them. No two gunmen are the same height.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that there will be a man left standing. In other words, prove that someone will not get shot.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-06-17T05:17:30.980" Title="Surviving the Shootout" Tags="|geometry|combinatorics|" AnswerCount="8" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16358" PostTypeId="1" AcceptedAnswerId="16362" CreationDate="2015-06-13T18:44:45.110" Score="5" ViewCount="437" Body="&lt;p&gt;The St. Čapek Crowd Control Academy, June, 2115.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Commander Gall is worried.  On one hand, the graduating class this year is the most promising in the institution's history, every cadet's reasoning skills honed to perfection.  On the other, General Domin is visiting and has insisted on writing and administering the robostanchion exam himself, despite the fact that robostanchion squads as a crowd-control technique came after his time, and he doesn't always grasp their limitations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fortunately, the format of a robostanchion exam is fairly standardized: a cadet takes control of a squad chosen for them by the examiner, all robostanchions initially inactive and in the hangar, and must maneuver it into a formation given by the examiner.  There are only three kinds of commands that a cadet can issue:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A cadet can &lt;strong&gt;deploy&lt;/strong&gt; an inactive robostanchion, commanding it to teleport itself to any unoccupied point on the field—which happens to be an &lt;a href=&quot;http://mathworld.wolfram.com/IntegerLattice.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;infinite integer lattice&lt;/a&gt;—and become active.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A cadet can &lt;strong&gt;move&lt;/strong&gt; an active robostanchion, commanding it to teleport to a different, unoccupied point on the field.  The robostanchion remains active.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A cadet can &lt;strong&gt;recall&lt;/strong&gt; an active robostanchion, commanding it to teleport back to the hangar and become inactive.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Their primary function being crowd control, whenever two active robostanchions are adjacent, they automatically establish a &lt;strong&gt;laser barrier&lt;/strong&gt; between themselves:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DGzUo.png&quot; alt=&quot;Two robostanchions stand adjacent and connected by a laser barrier.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Laser barriers require power to maintain, which must be drawn from some active robostanchion's power supply; each robostanchion has just enough power to maintain one barrier, but it may share this power wirelessly.  For example, the six robostanchions on the right side of this formation may use the left robostanchion to power their seventh barrier:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vWGGD.png&quot; alt=&quot;One robostanchion stands to the left, while six to the right are arranged in a two-by-three grid.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ordinarily, there are only two ways to fail a robostanchion exam: (1) damaging the squad by running out of power or (2) not achieving the requested formation within the generous time limit.  Now the general, to gratify his ego, has decided that his version of the exam should be more challenging, and he has added (3) ever not using all of the available power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Commander Gall, appropriately mortified, has, after much obsequious persuasion, gotten the general to agree to some concessions.  First, the general will allow cadets four commands before he enforces his additional requirement.  Second, the general has agreed to only ask for connected formations, i.e., formations where every robostanchion is joined to the others by at least one laser barrier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While further promises from the general are improbable, there is one more fact working in the cadets' favor: To prevent cheating, school policy is that cadets are always given squads of distinct sizes.  For instance, if one cadet makes the exam with a squad of 90, no other cadet will have exactly 90 robostanchions at their command.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unsure whether these facts are enough to contain the general's caprice, Gall consults with Chief Engineer Fabry, asking for a worst-case analysis.  What does Fabry reply?  That is, in the worst case possible, how many cadets can Gall expect to fail?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clarifications:&lt;/strong&gt; General Domin is sensible enough not to ask for formations that require more robostanchions than are in the squad, nor for formations that violate his condition (3).&lt;/p&gt;&#xA;" OwnerUserId="13175" LastEditorUserId="13175" LastEditDate="2015-06-14T17:37:04.817" LastActivityDate="2015-06-14T19:19:44.243" Title="The Robostanchion Exam (a puzzle about game-graph connectedness)" Tags="|mathematics|strategy|story|graph-theory|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16359" PostTypeId="2" ParentId="16357" CreationDate="2015-06-13T19:06:22.103" Score="10" Body="&lt;p&gt;Create a functional graph with an edge from each shooter to each victim.  A source-free functional graph must be a disjoint union of oriented, possibly degenerate (i.e., 1- or 2-vertex) cycles.   This one must contain at least one odd cycle because there are an odd number of vertices.  And that cycle is not a 1-cycle because no gunman shoots himself.  Call it $(v_0,v_1,\ldots,v_n,v_0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now either distance or height causes the $v_0$ gunman to prefer $v_1$ over $v_n$; the edge $(v_0,v_1)$ represents a distance no longer than that of $(v_0,v_n)$, and an analogous argument can be applied around the cycle.  Thus, by transitivity, all edges represent equal distance.  But then shots must have been decided by height, and $v_1$ is taller than $v_n$.  Again, this argument can be carried around the cycle (twice because it depends on a relationship between vertices two edges apart, two fortunately not having any common factors with the cycle length) to conclude that all gunmen in the cycle are of equal height.  But this contradicts the problem statement, so source-freeness is impossible.&lt;/p&gt;&#xA;" OwnerUserId="13175" LastEditorUserId="13175" LastEditDate="2015-06-13T19:13:23.213" LastActivityDate="2015-06-13T19:13:23.213" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16360" PostTypeId="2" ParentId="16357" CreationDate="2015-06-13T19:07:47.100" Score="1" Body="&lt;p&gt;&lt;strong&gt;Some definitions:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$B$ is a gunmen.  &lt;/li&gt;&#xA;&lt;li&gt;$b$ is the distance of $B$ from $A$.  &lt;/li&gt;&#xA;&lt;li&gt;$A→B$ means that $A$ shoots $B$.  &lt;/li&gt;&#xA;&lt;li&gt;Cycle means that $A→B, B→C, ..., n→A $.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt;&lt;br&gt;&#xA;Since the number of elements is odd, so there must exist at least a group of 3 (or more) gunmen which shoot themselves, if we want all them killed (aka a cycle).&#xA;We have to prove that given $n\geq3$ gunmen, no cycle can exist, without violating your conditions (height, distance).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt;&lt;br&gt;&#xA;We said above that if we want all the men to be dead, there must exist a cycle.&lt;br&gt;&#xA;Let's try to build a cycle. First, we take an element $A$, then we add an element $B$ at distance $a$ from $A$.&lt;br&gt;&#xA;We want $A→B$ and $B→C$ (not $B→A$), so the element $C$ must be distant from $B$ less than $a$: let's say $C$ is far $b$ from $B$, where $a&amp;gt;b$.&lt;br&gt;&#xA;Let's keep adding elements, whose distances are such that $a&amp;gt;b&amp;gt;c&amp;gt;...&amp;gt;z$.&lt;br&gt;&#xA;Now, we have to close the cycle, and make sure that $Z$ (the last element) shoots $A$ (the first one), thus the distance $a$ (from $Z$ to $A$) must be smaller than $z$.&lt;br&gt;&#xA;We deduce that $a&amp;gt;b&amp;gt;c&amp;gt;...&amp;gt;z&amp;gt;a$, so $a&amp;gt;a$. Paradox!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here you can visually see what happens when you try to build a cycle, starting from $A$. When you have to close it, it happens that $a&amp;lt;b$, which is in conflict with the hypothesis that $A$ chose $B$.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/2q5u2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10281" LastEditorUserId="10281" LastEditDate="2015-06-14T08:06:38.697" LastActivityDate="2015-06-14T08:06:38.697" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16361" PostTypeId="2" ParentId="16333" CreationDate="2015-06-13T19:38:20.013" Score="0" Body="&lt;p&gt;Use this method:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Assign a number on the die to each day of the week i.e. Day 1 = 1,&#xA;Day 2 = 2 and so on.&lt;/li&gt;&#xA;&lt;li&gt;For the last two days (days six and seven) in the week, give them&#xA;both the number 6.&lt;/li&gt;&#xA;&lt;li&gt;Roll the dice. If you land on the number 6, flip the coin and if its heads, pick day 6, if its tails, you pick day 7. Since the coin has a probability of&#xA;1/2, there is a probability of 1/12 of picking either day 6 or day 7 from&#xA;the start.&lt;/li&gt;&#xA;&lt;li&gt;If it lands on any other number, flip the coin and if it lands on&#xA;heads accept the number, if it lands on tails then repeat the whole&#xA;process. Now every other day has a probability of 1/12 of being&#xA;chosen as well.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="10974" LastActivityDate="2015-06-13T19:38:20.013" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16362" PostTypeId="2" ParentId="16358" CreationDate="2015-06-13T20:25:02.597" Score="2" Body="&lt;p&gt;In the worst case, the number of cadets who fail will be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; one. The only impossible formation is a $3\times3$ square of robots with the center removed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since all legal moves after the first four are reversible, in order to show a formation can be created from nothing, it suffices to show that it can be reduced to a small square of four robots. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Call a robot with exactly one robot next to it a &lt;em&gt;loner&lt;/em&gt;. We first show that any formation with loners can be reduced to a small square. Starting with such a formation, remove all loners, and continue removing newly created loners until there are none. What remains is a loop. Unless the formation is now a small square, the top edge of the formation must have 3 robots in a row, say at (0,0), (1,0) and (2,0). Deploy one of the removed robots to (1,1), then move (0,0)'s other neighbor to (0,1). This creates a small square of robots: after removing all loners until there are none, only this square will remain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, assume there are no loners. Suppose without loss of generality that the lowest row in the formation has a $y$-coordinate of $0$, and on that row, the lowest $x$ coordinate is $0$. This implies that there are robots at (0,0), (1,0), and (0,1). If there is a robot at (1,1), that means we have the small square, so assume there isn't one. Furthermore, (1,0) can't have his second neighbor to the south, so there must be a robot at (2,0).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several cases:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 1: Neither (2,1) nor (1,2) have a robot.&lt;/strong&gt; Move (2,0) to (1,1). This means that (2,0)'s old neighbor at (3,0) will now be a loner, so the formation is now reducible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 2: Exactly one of (2,1) or (1,2) has a robot&lt;/strong&gt;. Move that robot to (1,1). The moved robot's two neighbors are now loners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 3: Both (1,2) and (2,1) have robots.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1,2) must have two neighbors. If they are both at (0,2) and (2,2), then this is the bad formation, so assume that only one of (0,2) and (2,2) is filled. By symmetry, we can assume (0,2) is unfilled. This means that the other neighbor of (0,1) must be at (-1,1). If there was a robot at (-1,0), that would mean there was a small square of four robots, contradicting that the entire formation was a loop. So, (-1,0) is empty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are now two final cases:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 3a: (-2,0) has a robot.&lt;/strong&gt; Move that robot to (-1,0). The two neighbors that (-2,0) had are now loners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 3b: (-2,0) doesn't have a robot.&lt;/strong&gt; Move (2,0) to (-1,0). The two neighbors that (2,0) had are now loners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, in all cases except for the $3\times 3$ square without its center, the formation can be made. To see why this formation cannot be made, you can check that no moves can be made from it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were only given the exact number of robots needed, no deployments are possible. No moves are possible: a move destroys two lasers, so it must create two as well, but every spot you can move a robot would make it have 0,1,3 or 4 lasers. Finally, no recalls are possible, since this will destroy two lasers and only one robot.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-06-14T19:19:44.243" LastActivityDate="2015-06-14T19:19:44.243" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16363" PostTypeId="1" AcceptedAnswerId="16381" CreationDate="2015-06-13T20:32:58.717" Score="16" ViewCount="746" Body="&lt;p&gt;There are $n$ archers competing in an archery contest. The rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Each archer will shoot exactly one arrow at the target.&lt;/li&gt;&#xA;&lt;li&gt;The archers will shoot one at a time, with the order of turns chosen at random.&lt;/li&gt;&#xA;&lt;li&gt;On an archer's turn, he or she may choose how far from the target to stand while shooting. Of course, standing farther away makes it harder to hit the target.&lt;/li&gt;&#xA;&lt;li&gt;After all the shots are taken, the archer who hit the target from the farthest away is declared the winner. Ties are broken in favor of the archer who shot later.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This contest is not fair, as shooting later is a significant advantage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The archers are all equally skilled, and an archer shooting from $d$ meters away will hit the target with probability $\frac{10}{d}$. The minimum allowed distance is 10 meters.&lt;br&gt;(The specific expression for probability as a function of distance is not so important. Essentially each archer in turn chooses a probability, and among the archers who hit the target, the one who chose the lowest probability is the winner.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose you are the $k$-th archer to shoot. How far from the target should you stand to maximize your likelihood of winning?  Assume every other archer will act optimally to maximize their chance of winning.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Bonus&lt;/strong&gt;: As $n\to\infty$, what is the probability that the winner took their shot during the first half of the contest?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-06-14T18:38:00.293" LastActivityDate="2015-06-17T14:12:51.573" Title="Strategy for archery contest" Tags="|mathematics|strategy|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16364" PostTypeId="2" ParentId="16330" CreationDate="2015-06-13T21:43:03.697" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sisters are the vowels A, E, I, O and U. The brothers are the consonants.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the eldest, and in fact the first of all our siblings.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'A' is the first letter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My next sister buys alcohol by the bottle and places them on her shelves.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter 'E' looks like a bottle adjacent to three shelves.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My next sister always wraps her slim figure in the nicest of evening gowns, which she stitches herself.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'I' is a slim figure. Perhaps the horizontal lines on top and bottom resemble gowns. She stitches herself, because 'I' stands for ego.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In contrast, you can always find my next, more portly sister at the pool hall at a quarter past three every afternoon.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'O' is, in contrast, round and portly. It is shaped like the balls in the pool hall. 'O' is the fifteenth letter of the alphabet, so it can be seen at quarter (15) past three (A, E, I).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My last sister prefers playing blackjack (she sees herself in the table).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'U', shaped like blackjack table and also the 21st letter, an allusion to the fact that blackjack is also called 21.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Facetious Sisters&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sisters appear according to their age in the word 'Facetious'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2425" LastEditorUserId="2425" LastEditDate="2015-06-14T09:12:17.260" LastActivityDate="2015-06-14T09:12:17.260" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16365" PostTypeId="2" ParentId="16333" CreationDate="2015-06-13T22:09:55.777" Score="0" Body="&lt;p&gt;I would use the following method&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;roll the dice, then flip the coin&lt;/li&gt;&#xA;&lt;li&gt;if the coin is head, add 6 to the number on the dice.&lt;br&gt;&#xA;This gives you an outcome between 1 and 12 with uniform probability.&#xA;now if the number is greater than 7 try again.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The way I came up with this solution is trying a simpler problem: getting a number between 1 and 5 with uniform probability with one dice.  The solution is to roll the dice, if it is 6 roll again. Basically doing a &quot;modulo&quot; operation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you feel comfortable that the one-dice problem works, then the proposed solution working from something that gives you a 1-12 uniform distribution and converting it to a 1-7 uniform distribution also works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This answer is the same as 'ace'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more interesting problem is to prove that there is a solution with a guaranteed finite number of trials.  My answer would be no, given than 7, and 2 and 3 are co-prime (their greatest common divisor is 1)&lt;/p&gt;&#xA;" OwnerUserId="13177" LastActivityDate="2015-06-13T22:09:55.777" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16366" PostTypeId="2" ParentId="2135" CreationDate="2015-06-13T22:40:09.453" Score="-1" Body="&lt;p&gt;I approach the problem the following way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;first I think of a simpler problem of an unfair coin: say H=Heads, T=Tails&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without knowing the probability of heads or tails, if I flip the coin twice, only HT and TH will have the same equal probability P(HT) = P(TH) regardless of how the coin if biased, I cannot say much about the P(HH) or P(TT).  If H and T have probabilities non-zero, then at some point or another two trials will have different outcomes.  To come up with an outcome that has probability 0.5 out of this coin, I would flip the coin until the last two trials give me different outcomes (HHHHHHHHT or TTTTTH) and then take the last outcome as my result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the dice, I would do the same thing, even though it could take a very long time: roll the dice until 6 consecutive rolls have distinct outcomes (for 6 rolls, there is a change of 6!/(6^6) of happening or 0.015), and take the last roll as the result.&lt;/p&gt;&#xA;" OwnerUserId="13177" LastActivityDate="2015-06-13T22:40:09.453" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16367" PostTypeId="2" ParentId="16084" CreationDate="2015-06-13T22:55:48.530" Score="1" Body="&lt;p&gt;I am leaving my original partial answer, and adding a full answer with no changes in my thinking.&#xA;The answer that is currently marked correct, while close, I did not think was correct.  The question, through edits, I thought it became more incorrect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to strip off the leading letter, otherwise things do not line up.  Preformatting does not help.&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\mathfrak{C}^+(X,Y)$ is a binary OR operation.&lt;br&gt;&#xA; $X \setminus Y$ is a binary XOR operation. &lt;br&gt;&lt;br&gt;&#xA; The original problem statement says &lt;br&gt;&#xA; $\mathfrak{C}^+(4,7) = 9 \setminus \{\bot,\bot,\bot,\bot,\bot,\bot,\top\}$ &lt;br&gt;&#xA; 0111010 is 4&lt;br&gt;&#xA; 1010010 is 7&lt;br&gt;&#xA; 1111010 is 4 OR 7 = $\mathfrak{C}^+(4,7)$&lt;br&gt;&#xA; 1111011 is 9&lt;br&gt;&#xA; 1111010 is 9 XOR the original set = $\mathfrak{C}^+(4,7)$&lt;br&gt;&#xA; 1111110 is 9 XOR the edited set, which does not match $\mathfrak{C}^+(4,7)$&lt;br&gt;&lt;br&gt;&lt;br&gt;&#xA; The original post did not have $\sim$ as an operator listed.  Thus it was...&lt;br&gt;&#xA; $\mathfrak{C}^+(1,2) = 6$&lt;br&gt;&#xA; 0010010 is 1&lt;br&gt;&#xA; 1011101 is 2&lt;br&gt;&#xA; 1011111 is 1 OR 2 = $\mathfrak{C}^+(1,2)$&lt;br&gt;&#xA; 1101111 is 6&lt;br&gt;&#xA; 1111011 is $\sim$6 (according to the accepted answer) so that is farther from $\mathfrak{C}^+(1,2)$&lt;br&gt;&#xA; Thus, I made the statement that there was a typo in 6.&lt;br&gt;&#xA; The typo in 6 does not have an effect on $\mathfrak{C}^+(2,6)=8$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12357" LastEditorUserId="12357" LastEditDate="2015-06-17T10:17:55.537" LastActivityDate="2015-06-17T10:17:55.537" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16369" PostTypeId="2" ParentId="16357" CreationDate="2015-06-13T23:40:34.787" Score="1" Body="&lt;p&gt;Proof by induction. Let the two closest people be A and B, and assume that this pair is distinct for now. They shoot at each other, so we may remove A and B and consider the other n-2 people. By induction hypothesis, if the n-2 people shoot amongst themselves someone will live. But some of these people may shoot at A or at B, but this is even better because less bullets will be used at the n-2 people, so at least more people will live. Eventually, since n is odd, we will get to n=1 who shoots nobody.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are multiple disjoint pairs of people with the minimum distance, consider them separately. But if there is a network (eg. AB=BC=CD=DA=d) then the shortest person in that network will live because those at distance d from him ... I'm stuck here, sorry &lt;/p&gt;&#xA;" OwnerUserId="9118" LastEditorUserId="9118" LastEditDate="2015-06-14T15:37:38.093" LastActivityDate="2015-06-14T15:37:38.093" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16370" PostTypeId="2" ParentId="16333" CreationDate="2015-06-13T23:43:23.173" Score="3" Body="&lt;p&gt;In general, you can roll any real probability with a coin. Write out the decimal expansion of the probability, say its .1010101 etc. Have the first coin flip determine the first digit after the decimal, then the second flip determines the second digit, etc. If at any point the number determined by the coin flips can be proven to be larger or smaller than the probability then we are done and we choose based on larger or smaller. &lt;/p&gt;&#xA;" OwnerUserId="9118" LastActivityDate="2015-06-13T23:43:23.173" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16371" PostTypeId="2" ParentId="16333" CreationDate="2015-06-14T00:03:19.980" Score="1" Body="&lt;p&gt;Throw the die into the corner of a box, giving you 8 possibilities. Add together the scores of the three exposed faces around the exposed corner:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;     +---+                   +---+&#xA;   6 | 3 | 9              14 | 3 | 11&#xA; +---+---+---+           +---+---+---+&#xA; | 2 | 1 | 5 |           | 5 | 6 | 2 |&#xA; +---+---+---+           +---+---+---+&#xA;   7 | 4 | 10             15 | 4 | 12&#xA;     +---+                   +---+&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now divide the score by 7 and take the remainder. Every number 0-6 occurs once, except 0, which occurs twice. Roll again on 14. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;7 (or 14)--&amp;gt;0    (Roll again on 14.)   &#xA;       15--&amp;gt;1 &#xA;        9--&amp;gt;2  &#xA;       10--&amp;gt;3&#xA;       11--&amp;gt;4  &#xA;       12--&amp;gt;5 &#xA;        6--&amp;gt;6       &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4768" LastActivityDate="2015-06-14T00:03:19.980" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16372" PostTypeId="1" AcceptedAnswerId="16384" CreationDate="2015-06-14T00:10:35.967" Score="21" ViewCount="1982" Body="&lt;p&gt;Oh dear, someone has vandalised this long division.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;They've replaced every digit either with a wrong digit or a hyphen.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Your task: reconstruct the entire thing. There's no remainder.&lt;/p&gt;&#xA;&lt;p&gt;Please show all work. Use of a computer is not allowed.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tS8ys.jpg&quot; alt=&quot;long division&quot; /&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Please note the following :&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;None of the digits in the&#xA;puzzle, is correct.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;None of the numbers has a&#xA;leading 0. This means that the actual numbers that replace 29, 86,96 and 35 are all two digit numbers as well, the actual numbers that replace 2_ _, 125, 3 _ _ and 248 are all three digit numbers themselves, the actual number that is there in place of 16_2 is a 4 digit number and the actual number that replaces &lt;em&gt;00&lt;/em&gt; _8 is a 6 digit number.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let's say that 2 in the&#xA;divisor, 29, is actually&#xA;supposed to be 7. Then it is not necessary that all the 2s present in the image above will be replaced by 7. It can be that one of the 2s gets replaced by 7 and another 2 gets replaced by 3. Same is true for the other digits in the image : 0,1, 3…9. They each can be replaced by different digits at different places.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7363" LastEditorUserId="57212" LastEditDate="2023-06-07T13:08:55.957" LastActivityDate="2023-06-07T13:08:55.957" Title="Someone has vandalised this long division. Can you reconstruct it?" Tags="|mathematics|logical-deduction|no-computers|missing-digits|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="16373" PostTypeId="1" AcceptedAnswerId="16380" CreationDate="2015-06-14T00:48:50.633" Score="2" ViewCount="358" Body="&lt;p&gt;You've solved it! No, wait a minute - you've solved the first step. You don't know the combination for the safe yet. All you know is that someone has helped you by using a pseudonym. Their name isn't really Artairs A Sebeenas. But if you write it like this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/o39nJ.jpg&quot; alt=&quot;multiplication&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you get a multiplication, and the product represented by &quot;SEBEENAS&quot; is the 8-digit number you need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No two letters represent the same digit. What's the combination? Please show all work, and use of a computer is not allowed.&lt;/p&gt;&#xA;" OwnerUserId="7363" LastEditorUserId="7363" LastEditDate="2015-06-14T13:01:24.043" LastActivityDate="2015-06-14T13:01:24.043" Title="Reconstruct this multiplication to find the combination for the safe" Tags="|logical-deduction|mathematics|no-computers|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16374" PostTypeId="2" ParentId="16333" CreationDate="2015-06-14T01:05:03.353" Score="1" Body="&lt;p&gt;Roll two dice (or roll the die twice) and flip the coin. Sum the dice. Use this table to figure out the result.&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;Dice Sum&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;Coin&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;Result&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;5 or 7&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;H&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;5 or 7&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;T&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;6&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;H or T&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;8&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;H or T&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;9, 10, 11, or 12&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;H&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;5&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;9, 10, 11, or 12&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;T&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;6&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3 or 4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;H or T&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;7&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;H or T&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;Redo&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;There are 36 possible dice results, which doubles to 72 when you account for the coin result. Here is the table that shows each result has an equal probability (N is the number of ways each combination can happen):&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;Sum&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;Coin&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;N&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;Assigned To&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;7&lt;br&gt;5&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;H&lt;br&gt;H&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;6&lt;br&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;7&lt;br&gt;5&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;T&lt;br&gt;T&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;6&lt;br&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;6&lt;br&gt;6&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;H&lt;br&gt;T&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;5&lt;br&gt;5&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;8&lt;br&gt;8&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;H&lt;br&gt;T&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;5&lt;br&gt;5&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;9&lt;br&gt;10&lt;br&gt;11&lt;br&gt;12&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;H&lt;br&gt;H&lt;br&gt;H&lt;br&gt;H&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;br&gt;3&lt;br&gt;2&lt;br&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;5&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;9&lt;br&gt;10&lt;br&gt;11&lt;br&gt;12&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;T&lt;br&gt;T&lt;br&gt;T&lt;br&gt;T&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;br&gt;3&lt;br&gt;2&lt;br&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;6&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;br&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;H or T&lt;br&gt;H or T&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;br&gt;6&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;7&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;X&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;Try Again&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;This method has two main advantages:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a) The dice values can be summed, without regard to order (in other words, a 1-3 &lt;strong&gt;is&lt;/strong&gt; the same as a 3-1 in this method)&lt;/li&gt;&#xA;&lt;li&gt;b) There is a low percentage chance of a redo (1 in 36, or 2.778%)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;The main disadvantage is probably that, without the table in front of you, it might be hard to remember that, say, Double 5's and Tails means &amp;quot;6&amp;quot;, while a 5 and a 2 (i.e., a 7) plus Heads is a &amp;quot;1&amp;quot;.&lt;/p&gt;&#xA;" OwnerUserId="13179" LastEditorUserId="69582" LastEditDate="2021-01-28T16:05:09.527" LastActivityDate="2021-01-28T16:05:09.527" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="16375" PostTypeId="2" ParentId="16372" CreationDate="2015-06-14T01:20:15.853" Score="5" Body="&lt;p&gt;I think I found the answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step one to find the divisor. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; By looking at the number we know that the first three digits of dividend is quite close to 100 because the divisor is dividing it with 2 digits. And second things is the reminder of the first division is smaller than divisor and thats why the number has to be between 21-25. I thought odd number has more possibility to get this answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I started with &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 25 as divisor but it is hard to get 2 digit reminder in second step of division. Then I tried 23.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now this number  start making sense in my mind. Here's how&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 23x4 =92 which is very close to 100 and 23x5=115 which more quiet far from 100. so i chose 113 number between 100 and 115 &#xA;Now if you divide 113 with 23 you will get 21 as remainder. Now according to given division it has to be small than divisor because we are going to drop one more number from dividend. &#xA; Now that Number has to be close to a number which is close divisible by 23 and get 2 digit number as remainder which is also has to be smaller than 23.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second step of division.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; So far we have first three digits of dividend 113 and 21 as remainder and now we are going to decide the fourth digit of dividend. I started with number 9 because its the biggest single digit number. so now we got number 1139 and our remainder become 219.Now 23 x 9 =207 which is the closest to 219. so now we got two digit remainder 12 which is smaller that 23.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Third step of division.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;So far we got first four digits of dividend which is 1139 and quotient 49. Now its time to find fifth digit of dividend. lets say the smallest number we can use is 0 which makes remainder as 120 and when you divide it with 23 you will get 5 as remainder and there is no two digit number which start with 5 is divisible by 23. so I change that fifth number to 1. Now remainder become 121 and when you divide 121 with 23 you will get 6 as remainder.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fourth/Final step of division &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Now we have first five digits of dividend which are 11391 and first three digits of quotient which are 495. We have 6 as a remainder and there is one number which start with 6 and also divisible by 23 and that is 69. ;-) (23 x 3=69) So our last digit of dividend is 9. and our last digit of quotient is 3.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is my actual work &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/wySvY.jpg&quot; alt=&quot;Here is my calculations&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;img src=&quot;https://i.stack.imgur.com/uFMNj.png&quot; alt=&quot;Screen Shot of Final answer&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12736" LastActivityDate="2015-06-14T01:20:15.853" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16380" PostTypeId="2" ParentId="16373" CreationDate="2015-06-14T04:35:36.787" Score="3" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$S \times A$&lt;/span&gt; ends with &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Therefore, if &lt;span class=&quot;math-container&quot;&gt;$S = 1,3,7$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$9$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$A = 1$&lt;/span&gt;&lt;br /&gt;&#xA;If &lt;span class=&quot;math-container&quot;&gt;$S = 5$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$A = 1,3,5,7,9$&lt;/span&gt;&lt;br /&gt;&#xA;If &lt;span class=&quot;math-container&quot;&gt;$S = 2,4,6$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$A = 1$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt;&lt;br /&gt;&#xA;If &lt;span class=&quot;math-container&quot;&gt;$S = 0$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; is anything, but note that &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; cannot be equal zero since it is the start of the answer.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$A \times A + \text{carryover} = 10S + E$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$S = 1,3,7,9,2,4,6$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$A = 1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$S = 0$&lt;/span&gt;, which is a contradiction.&lt;br /&gt;&#xA;If &lt;span class=&quot;math-container&quot;&gt;$S = 2,4,6,8$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$A = 6$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$S = 3$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt;, which means that &lt;span class=&quot;math-container&quot;&gt;$S = 4$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$A = 6$&lt;/span&gt;&lt;br /&gt;&#xA;If &lt;span class=&quot;math-container&quot;&gt;$S = 5$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$A = 1,3,5,7,9$&lt;/span&gt;, and only &lt;span class=&quot;math-container&quot;&gt;$A = 7$&lt;/span&gt; can achieve &lt;span class=&quot;math-container&quot;&gt;$S=5$&lt;/span&gt;.&lt;br /&gt;&#xA;Therefore, &lt;span class=&quot;math-container&quot;&gt;$(S,A) = (4,6)$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$(5,7)$&lt;/span&gt;&lt;br /&gt;&#xA;Let these be Case 1 and Case 2 respectively.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Case 1&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$S = 4, A = 6$&lt;/span&gt;&lt;br&gt;&lt;span class=&quot;math-container&quot;&gt;$R \times A + \text{carryover}$&lt;/span&gt; ends with &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\text{carryover} = 2$&lt;/span&gt; (The &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; in &lt;span class=&quot;math-container&quot;&gt;$4 \times 6=24$&lt;/span&gt;)&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$A = 6$&lt;/span&gt;&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$6R + 2$&lt;/span&gt; ends with &lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt;&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$6R$&lt;/span&gt; ends with &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt;&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$R = 4$&lt;/span&gt; (reject since &lt;span class=&quot;math-container&quot;&gt;$S = 4$&lt;/span&gt;) or &lt;span class=&quot;math-container&quot;&gt;$9$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$S = 4$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$A = 6$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$R = 9$&lt;/span&gt;&lt;br&gt;Now we have:&lt;br&gt;&lt;span class=&quot;math-container&quot;&gt;$\text{69T6I94}$&lt;/span&gt;&lt;br&gt;&lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt;&lt;br&gt;&lt;span class=&quot;math-container&quot;&gt;$\text{4EBEEN64}$&lt;/span&gt;&lt;br&gt;&lt;span class=&quot;math-container&quot;&gt;$69 \times 6 + \text{carryover} = \text{4EB}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$0 \leq \text{carryover} &amp;lt; 5$&lt;/span&gt;&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$414 + \text{carryover} = \text{4EB}$&lt;/span&gt;&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$E = 1$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$S = 4, A = 6, R = 9, E = 1$&lt;/span&gt;&lt;br&gt;Now it is:&lt;br&gt;&lt;span class=&quot;math-container&quot;&gt;$\text{69T6I94}$&lt;/span&gt;&lt;br&gt;&lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt;&lt;br&gt;&lt;span class=&quot;math-container&quot;&gt;$\text{41B11N64}$&lt;/span&gt;&lt;br&gt;&lt;span class=&quot;math-container&quot;&gt;$6 \times 6 + \text{carryover}$&lt;/span&gt; (of &lt;span class=&quot;math-container&quot;&gt;$6 \times I + \text{carryover}$&lt;/span&gt;) ends with &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;(both) &lt;span class=&quot;math-container&quot;&gt;$\text{carryover} = 5$&lt;/span&gt;&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$6 \times I + 5$&lt;/span&gt; starts with &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt;&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$I = 8$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$9$&lt;/span&gt; (reject since &lt;span class=&quot;math-container&quot;&gt;$R = 9$&lt;/span&gt;)&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$S = 4, A = 6, R = 9, E = 1, I = 8$&lt;/span&gt;&lt;br&gt;&lt;span class=&quot;math-container&quot;&gt;$\text{69T6894}$&lt;/span&gt;&lt;br&gt;&lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt;&lt;br&gt;&lt;span class=&quot;math-container&quot;&gt;$\text{41B11N64}$&lt;/span&gt;&lt;br&gt;&lt;span class=&quot;math-container&quot;&gt;$N = 3$&lt;/span&gt;&lt;br&gt;&lt;span class=&quot;math-container&quot;&gt;$\text{69T6894}$&lt;/span&gt;&lt;br&gt;&lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt;&lt;br&gt;&lt;span class=&quot;math-container&quot;&gt;$\text{41B11364}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;However, &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; cannot exist, since &lt;span class=&quot;math-container&quot;&gt;$6 \times T + 4$&lt;/span&gt; ends with &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;, so &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; is not an integer.&lt;/p&gt;&#xA;&lt;p&gt;Therefore Case 1 has no solutions.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Apply the same method to &lt;strong&gt;Case 2&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$R = 2$&lt;/span&gt;&lt;br&gt;&lt;span class=&quot;math-container&quot;&gt;$E = 0$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;&lt;br&gt;If &lt;span class=&quot;math-container&quot;&gt;$E=0$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$I=2$&lt;/span&gt; (reject since &lt;span class=&quot;math-container&quot;&gt;$R=2$&lt;/span&gt;), so no sol if &lt;span class=&quot;math-container&quot;&gt;$E=0$&lt;/span&gt;&lt;br&gt;If &lt;span class=&quot;math-container&quot;&gt;$E=1$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$I = 3$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt;&lt;br&gt;If &lt;span class=&quot;math-container&quot;&gt;$I = 3$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$N = 2$&lt;/span&gt; (reject)&lt;br&gt;If &lt;span class=&quot;math-container&quot;&gt;$I = 4$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$N = 9$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$T = 8$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Therefore:&lt;br&gt;&lt;span class=&quot;math-container&quot;&gt;$7287425$&lt;/span&gt;&lt;br&gt;&lt;span class=&quot;math-container&quot;&gt;$7$&lt;/span&gt;&lt;br&gt;&lt;span class=&quot;math-container&quot;&gt;$51011975$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Thus, your answer is &lt;span class=&quot;math-container&quot;&gt;$51011975$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="9736" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-06-14T10:57:38.103" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16381" PostTypeId="2" ParentId="16363" CreationDate="2015-06-14T05:37:28.857" Score="16" Body="&lt;h1&gt;Main Puzzle&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I'll assume that every archer expects the others to play optimally (this was implied in the bonus question, but I'll use it here too).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are only two pieces of information that could influence what shot an archer decides to take:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the probability $b$ (for &quot;best&quot;) that was chosen by the archer currently winning or $1$ if no one has hit yet (relevant because one cannot win taking a shot with probability above $b$) and&lt;/li&gt;&#xA;&lt;li&gt;the number of competitors $r$ (for &quot;remaining&quot;) to shoot after the archer (relevant because later contestants may snatch one's victory away).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So we can write the optimal probability to choose for one's shot as a function $p$ of just $b$ and $r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, suppose $r=0$.  Obviously, the archer should take the easiest shot available, so we have $p(b,0)=b$.  For the moment, I will note that this is equivalent to $p(b,0)=\min(b,1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take $r=1$.  We need to maximize the current archer's chances of winning, $p(b,1)$, for their shot being made, times $1-p(p(b,1),0)$ for the last archer not stealing the win.  The maximum of $x(1-x)$ over the interval $[0,b]$ is achieved at $x=\min(b,1/2)$, so $p(b,1)=\min(b,1/2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar argument for $r=2$ has us maximizing $x(1-\min(x,1/2))(1-x)$ over the interval $[0,b]$.  Having the $\min$ in the $1-\min(x,1/2)$ factor choose $x$ is obviously better and occurs at lower $x$ anyway, so we can equivalently maximize $x(1-x)^2$.  This gives $p(b,2)=\min(b,1/3)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By now you can probably see the pattern: We maximize the product of one success and $r$ subsequent failures over $[0,b]$, which is equivalent—because we want all the negative mins to be small—to maximizing $x(1-x)^r$.  So, in general, $p(b,r)=\min(b,1/(r+1))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives us an answer to the main question, which depends, via $b$, on how prior archers have fared:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d=\frac{10}{p(b,r)}=\frac{10}{\min(b,\frac{1}{n-k+1})}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, in plain language: &lt;strong&gt;An archer should shoot either from a distance where they expect exactly one of the as-yet-unfired shots (hopefully their own) to hit or from the distance of the currently winning shot, whichever is further.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That makes this puzzle pretty nifty in my opinion: the result is what my intuition would like it to be, but the justification is nontrivial.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;Bonus&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The probability of everyone in the latter half missing is $(1-b)^{n/2}$, where $b$ is established by the first archer to hit: $1/(n-i+1)$ when the $i^{\mathrm{th}}$ archer is first.  Because of the problem wording, we only care about cases where some archer in the first half scored a hit, and these cases are disjoint, so we can sum over them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, we must determine the probability of the $i^{\mathrm{th}}$ archer scoring the first hit.  It is the product of the probabilities of prior archers missing, which telescopes to $(n-i+1)/n$, times the probability that the archer themself hits, $1/(n-i+1)$.  In other words, just $1/n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(As an aside, that too is a pretty cool result: As long as these archers play optimally, we can dupe them into a fair contest by claiming the rules above but then actually awarding the victory to the first hit.)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We substitute into the sum and take the limit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n/2}\left(1-\frac{1}{n-i+1}\right)^{n/2}&amp;amp;=&#xA;\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n/2} \left(\left(1-\frac{1}{n-i+1}\right)^{n-i+1}\right)^{\frac{1}{2\left(1-\frac{i-1}{n}\right)}}\\&amp;amp; = \lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n/2} e^{\frac{-1}{2\left(1-\frac{i-1}{n}\right)}}\\&amp;amp; = &#xA;\int_0^{1/2} e^{\frac{-1}{2(1-t)}}\,dt = \frac{1}{2}\int_{1}^2 e^{-1/t}\,dt\\&#xA;&amp;amp;\approx 0.252396.&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;There's slightly better than a quarter chance of the winner having taken their shot during the first half of the contest.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="13175" LastEditorUserId="2627" LastEditDate="2015-06-17T14:12:51.573" LastActivityDate="2015-06-17T14:12:51.573" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16382" PostTypeId="2" ParentId="16363" CreationDate="2015-06-14T06:47:09.053" Score="1" Body="&lt;p&gt;Every archer must use the first applicable rule:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Shoot from the greatest distance of a successful shot, if there was at least one succesful shot.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Shoot from distance less than or equal to $10$, if shooting last.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Shoot from distance greater than $10$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note that to win it is required not only to hit the target from farthest distance, but all subsequent shots to fail. There is no way, by choosing the shooting distance, for archer to ensure that own shot is less likely to miss than any of subsequent shots to hit - thus it is impossible to detrmine best shooting distance: if $P$ is a probability to hit the target from a given distance, than a chance to win for a $k_{\text{th}}$ archer would be $P \times \big((1-P)^{5-k}\big)$, but for $(k+i)_{\text{th}}$ archer it would be $P \times \big((1-P)^{5-(k+i)}\big)$, which is greater.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On bonus:&lt;br&gt;&#xA;The problem, in general, is to determine the probability $P_n^k$ that one of the first $n$ archers wins - that is, any (or all) of first $n$ archers hit the target (which is inverse to probability that all first $n$ archers miss) while last $k-n$ archers miss. Again, given $P$ as probability of an archer to hit target from a given distance, $P_n^k$ would be:&#xA;$$P_n^k=\big(1-(1-P)^n\big)\times (1-P)^{k-n}$$&#xA;or&#xA;$$P_n^k=(1-P)^{k-n}-(1-P)^k$$&#xA;or&#xA;$$P_n^k=\frac{1}{(1-P)^n}-1$$&#xA;or, finally&#xA;$$P_n^k=\frac{1}{(1-10/x)^n}-1$$&lt;/p&gt;&#xA;" OwnerUserId="13181" LastEditorUserId="13181" LastEditDate="2015-06-14T19:48:30.117" LastActivityDate="2015-06-14T19:48:30.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16383" PostTypeId="2" ParentId="16357" CreationDate="2015-06-14T11:08:18.387" Score="1" Body="&lt;p&gt;There is no proof that always a gunman will remain standing, unshot and alive, assuming multiple shots are allowed (which was not explicitly stated, and seems quite natural given it's a shootout).  I present a counterproof.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let 101 gunmen stand at the vertices of a regular 101 sided polygon.  For any given gunman, he is equally distant to exactly 2 other closest gunmen.&lt;br&gt;&#xA; Let the height of each gunman be equal to the gunman's index, in units.  (gunman 1 is 1 unit tall, and gunman 101 is 101 units tall, leading to a very large variation!).&lt;br&gt;&lt;br&gt;&#xA; Assign gunman 101 to any vertex.  Following along one side of the polygon, assign even heights in decreasing order, and along the other assign odd heights in decreasing order.&lt;br&gt;&#xA; Thus 98 - 100 - 101 - 99 - 97.&lt;br&gt;&#xA; Similarly, at the other end, 5 - 3 - 1 - 2 - 4 - 6.  (perspective changed, so ordering changed too)&lt;br&gt;&lt;br&gt;&#xA; After the first round of shooting both 1 and 2 live, because 3 shoots 5 and 4 shoots 6.  They are the only gunmen to live.&lt;br&gt;&#xA; At the second round, gunmen 1 and 2 shoot each other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12357" LastActivityDate="2015-06-14T11:08:18.387" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16384" PostTypeId="2" ParentId="16372" CreationDate="2015-06-14T12:02:52.637" Score="13" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/nTLib.gif&quot; alt=&quot;Solution&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the second subtraction, the number being subtracted is three digits, and the first digit isn't 1. So it must be at least 200, and the divisor is at least 21. Since the first digit of the divisor isn't 2, the divisor must be at least 30.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first subtraction, the number being subtracted is at least twice the divisor, but is only two digits. So the divisor is at most 49.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end, there is a two digit number, where the first digit is not 9. It is at most 89. This means that the last digit of the quotient cannot be 3 or more, so it must be 1. Then the divisor cannot start with 3, so it is 40-49.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first digit of the quotient must be 2. Twice the divisor cannot start with 8, so the divisor is 45-49. The last digit of the divisor is not 5, 6, 8, or 9, so the divisor is 47.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second digit of the quotient must be at least 5, because 4*47 is less than 200. But it can't be 5 because the 5 is wrong, it can't be 7 (7*47=329) or 9 (9*47=423) because the 2 is wrong, and it can't be 6. So it is 8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the third digit of the quotient is the only one missing. It can't be 1 or 2 because those are too small. It can't be 3 because the 4 is wrong, it can't be 4 because the 8 is wrong, and it can't be 5 or 6 because the 2 is wrong. It also can't be 7 or 8 because the 3 on the preceding line is wrong, and the numbers subtract to only 4. So the missing digit is 9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The complete division is 135877/47=2891.&#xA;The subtractions are 135-94, 418-376, 427-423, and 47-47.&lt;/p&gt;&#xA;" OwnerUserId="13186" LastEditorUserId="5840" LastEditDate="2016-07-31T18:06:01.600" LastActivityDate="2016-07-31T18:06:01.600" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16385" PostTypeId="2" ParentId="16354" CreationDate="2015-06-14T14:24:10.673" Score="1" Body="&lt;p&gt;&lt;em&gt;Note: This is just a partial answer.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The initials of &quot;Rats, Oil, Trumpet, Thumb, Hair, Ice, Rhyme, Teeth, Elves, Elbow and Nails&quot; spell out ROTTHIRTEEN, or ROT THIRTEEN. The number thirteen(13) might refer to Harry being 13 years old, in other words, in Year 3: The Prisoner of Azkaban. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit 1:&#xA;I have followed the progress of another user and have acquired this text from the website &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://freetexthost.com/i5kqe2psfo&quot; rel=&quot;nofollow&quot;&gt;http://freetexthost.com/i5kqe2psfo&lt;/a&gt;  It tells us to donate bitcoins at 1BSyxQbLhq5vMijFZ3hrZQmNMrxC5dVEcM. Yet I do not know if this is essential to decoding the cryptogram. But this code is maybe related with the code: ZGZpb3JsZ2xhb2h3c3R1aGVlYnJ1aXJ2ZXI.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12860" LastEditorUserId="12860" LastEditDate="2015-06-14T14:59:20.567" LastActivityDate="2015-06-14T14:59:20.567" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16386" PostTypeId="2" ParentId="16354" CreationDate="2015-06-14T14:47:57.013" Score="5" Body="&lt;p&gt;Since another user posted his progress, I'll do the same.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The initials of Rats, Oil, Trumpet, Thumb, Hair, Ice, Rhyme, Teeth, Elves, Elbow and Nails make the words &quot;ROT THIRTEEN&quot; which is ROT13, the most famous encryption algorithm.&lt;br&gt;&#xA;If you apply it to &quot;serrgrkgubfgpbzv5xdr2cfsb&quot; you obtain &quot;freetexthostcomi5kqe2psfo&quot;, which is a link to &lt;a href=&quot;http://freetexthost.com/i5kqe2psfo&quot; rel=&quot;noreferrer&quot;&gt;http://freetexthost.com/i5kqe2psfo&lt;/a&gt;.&lt;br&gt;&#xA;There you need a password, and I'm stuck at that stage.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-06-14T14:47:57.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16387" PostTypeId="1" AcceptedAnswerId="16391" CreationDate="2015-06-14T15:28:42.987" Score="7" ViewCount="848" Body="&lt;p&gt;You are trying to crack someone's password. You don't have any password cracking software, nor do you have time to get any. From sources unknown, you know that:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A. The password is alphanumeric (only letters and numbers)&lt;/p&gt;&#xA;&lt;p&gt;B. It's length is between 6 and 10&lt;/p&gt;&#xA;&lt;p&gt;C. 25% of all the characters are numbers&lt;/p&gt;&#xA;&lt;p&gt;D. The password is case insensitive&lt;/p&gt;&#xA;&lt;p&gt;E. All the digits are at the end of the password (after the letters)&lt;/p&gt;&#xA;&lt;p&gt;F. The first letter of the password is J&lt;/p&gt;&#xA;&lt;p&gt;G. The sum of the digits are equal to the numerical value of the first letter&lt;/p&gt;&#xA;&lt;p&gt;H. There is one E, one U and one Z in the password inclusively (they appear in that order)&lt;/p&gt;&#xA;&lt;p&gt;I. The first digit is equal to the numerical value of the second letter&lt;/p&gt;&#xA;&lt;p&gt;J. The product of all the digits is equal to the last letter&lt;/p&gt;&#xA;&lt;p&gt;K. The letter that was unmentioned (not declared) is d upside down and mirrored.&lt;/p&gt;&#xA;&lt;p&gt;L. The numerical difference between the third letter and the Z is more than 6&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;No computers please (except for posting your answer).&#xA;Search, there is an answer to this!&lt;/p&gt;&#xA;" OwnerDisplayName="user13180" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-05-24T00:05:52.183" Title="Password Cracking" Tags="|mathematics|password|no-computers|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="16389" PostTypeId="2" ParentId="16354" CreationDate="2015-06-14T16:09:03.213" Score="15" Body="&lt;p&gt;EDIT: Final Answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The message is: &quot;I am your mother. I love you and miss you. The whole magic world needs your help. Good Bye and Good Luck - Your Mom&quot;, so the white witch is Harry's mom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Taking LeppyR64's find that the base64 encoded string had &quot;follow the river&quot; by taking every two letters, I looked at the other letters and they were &quot;dirgahsuebure&quot; which backwards is &quot;erubeushagrid&quot;. I tried this as a password at the texthost site but it didn't work. Then I realized hagrid's first name is rubeus so I removed the extra e and it worked!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here's another partial answer with slightly more information:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first step is obviously to do ROT13 to the message in the letter (most would try that without the hint). Going to &lt;a href=&quot;http://freetexthost.com/i5kqe2psfo&quot;&gt;http://freetexthost.com/i5kqe2psfo&lt;/a&gt;, a password is needed. The string above, &quot;ZGZpb3JsZ2xhb2h3c3R1aGVlYnJ1aXJ2ZXI=&quot;, decodes in base 64 to dfiorlglaohwstuheebruirver, all lower case letters which can't be a coincidence. Finally, the author confirmed that the story is necessary and that Hagrid is &quot;The Keeper&quot; who will &quot;help you through&quot;. The base64 string somehow needs to be decoded to get the password to the texthost site.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="7722" LastEditDate="2015-06-14T18:38:01.627" LastActivityDate="2015-06-14T18:38:01.627" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16390" PostTypeId="2" ParentId="16387" CreationDate="2015-06-14T16:16:12.807" Score="4" Body="&lt;p&gt;The password is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; jequzy55&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 25 percent are numbers ... Length between 6 and 10. This means the length is 8: 6 characters and 2 numbers. Numerical value of the 2 digits is 10(j), and their product is equal to the last letter. This can be y, x, u, p, i. None of these have been mentioned, which means the first five letters are jequz or jqeuz (q must occupy the 2nd or 3rd position, from L). Jqeuz does not satisfy I, so it must be jequz, which forces the first number to be 5 and thus the last letter/second number to be Y/5, giving us jequzy55&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3987" LastEditorUserId="9993" LastEditDate="2016-05-21T21:31:41.110" LastActivityDate="2016-05-21T21:31:41.110" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16391" PostTypeId="2" ParentId="16387" CreationDate="2015-06-14T16:27:47.343" Score="7" Body="&lt;p&gt;The password is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; JEQUZY55. 25% of the password are numbers, and the only number between 6 and 10 that can be broken up this way is 8. Therefore there are 6 letters and 2 numbers. The 6 letters include J,E,U,Z, and Q so far. One of these must be the first of these two digits which make the first digit 5. The sum of these digits is 10 making both digits 5, making the product 25. This makes the last letter Y. The letters JEUZY are in order, and Q can be anywhere between the E (Rule I) and the Y. For Rule L, the third letter can't be U, meaning it is Q. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="7586" LastEditDate="2016-05-24T00:05:52.183" LastActivityDate="2016-05-24T00:05:52.183" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16392" PostTypeId="1" AcceptedAnswerId="16397" CreationDate="2015-06-14T17:28:21.460" Score="10" ViewCount="931" Body="&lt;p&gt;You are hunting for treasure located at some point on an infinite square grid.  You have two tools: a pointer, which points toward the treasure; and a shovel, which you can use to dig for it.  Here's the catch: both tools are imperfect.  The pointer might point toward the treasure or it might be off by up to 60 degrees. And the shovel will break if you try to dig in five different places.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you find the treasure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FINE PRINT:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The treasure is guaranteed to be at a point of the grid.  &lt;/li&gt;&#xA;&lt;li&gt;You can use the pointer from any point of the grid.  &lt;/li&gt;&#xA;&lt;li&gt;If you use the pointer while you are standing on top of the treasure it can point in any direction.  Otherwise it will point in a direction within 60 degrees of the direction to the treasure.&lt;/li&gt;&#xA;&lt;li&gt;You can only find the treasure by digging it up with the shovel, and you can only use the shovel four times.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="12523" LastEditorUserId="8874" LastEditDate="2016-02-07T13:32:52.807" LastActivityDate="2016-02-07T13:32:52.807" Title="hunting for treasure on an infinite grid" Tags="|mathematics|geometry|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16393" PostTypeId="2" ParentId="16392" CreationDate="2015-06-14T17:35:33.610" Score="2" Body="&lt;p&gt;Here's an image of what's happening:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/mptPX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the picture (red = target, black = current position), it can be seen that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; At a worst case scenario, when walking within 60 degrees of the target, the new distance to the target (red line) is decreased. By continually walking in the direction that the pointer points to, the distance to the object will be decreased at a logarithmic rate. I'm not sure the exact point can be determined even if you can always get closer. I'll leave this as a partial answer unless there are any clarifications.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My final answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is not possible because although you can always get closer, a single point  means you can always zoom in so any distance gained is arbitrary.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="7722" LastEditDate="2015-06-14T18:08:50.733" LastActivityDate="2015-06-14T18:08:50.733" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16394" PostTypeId="2" ParentId="16263" CreationDate="2015-06-14T17:59:12.607" Score="1" Body="&lt;p&gt;Here is a partial answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1. Align&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As suggested by Emrakul, 1243 characters are arranged into 11 rows with 113 characters in each row.&#xA; &lt;img src=&quot;https://i.stack.imgur.com/EfE3T.png&quot; alt=&quot;Align1&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2. Remove&lt;/strong&gt;   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Remove all of the &quot;=&quot; characters  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3. Re-align&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The remaining 589 characters can be arranged into 19 rows or 31 rows. I chose 19 rows of 31 characters because many more search words can be found in the next step.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4. Search&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Words can be found in all directions as shown. It seems unusual that there are multiple occurrences of words like daughter, fifty, and pound. The longest words are policeman and shopkeeper.&#xA; &lt;img src=&quot;https://i.stack.imgur.com/QwuwH.png&quot; alt=&quot;Align2&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;5. Rearrange / 6. Solve&lt;/strong&gt;&lt;br&gt;&#xA;If correct so far, perhaps somebody can figure out how the words need to be rearranged.  &lt;/p&gt;&#xA;" OwnerUserId="5188" LastActivityDate="2015-06-14T17:59:12.607" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16395" PostTypeId="2" ParentId="16354" CreationDate="2015-06-14T18:03:25.257" Score="5" Body="&lt;p&gt;Here's another bone for you all...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Prophecy in Harry Potter is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Neither can live while the other survives.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I believe that the first code listed is actually short a character in the plain text.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;ZGZpb3JsZ2xhb2h3c3R1aGVlYnJ1aXJ2ZXI=&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is base64 and decodes to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;dfiorlglaohwstuheebruirver&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Keeping every second letter yields:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;follow the river&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1777" LastActivityDate="2015-06-14T18:03:25.257" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16397" PostTypeId="2" ParentId="16392" CreationDate="2015-06-14T21:59:30.497" Score="9" Body="&lt;p&gt;It is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;possible&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Forget where the pointer is telling you the treasure is, look at where it can't be. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For every pointing, there is a 120 degree region (centred opposite the pointing) that the treasure &lt;strong&gt;cannot&lt;/strong&gt; be in. If we travel along the x-axis (back and forth, in increasing steps, so that we will definitely pass the y co-ordinate of the treasure) we will eventually exclude one half of the grid, divided along the line we have walked. Call this procedure a &lt;strong&gt;chop&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now chop along the y-axis. Assuming wlog that the treasure is in quadrant 1 (or the boundary line), we now chop increasingly large values of y until we find an upper bound for y. Repeat for x. We have now bounded the treasure on four sides. Walk the points 1 unit from the boundary, excluding them or finding we cannot exclude them until we have a 5x5 square of points. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's revert to looking at what the pointing tells us. The outer points tell us nothing other than 'in this central 3x3 region' (120 degrees can cover the entire 3x3 square from any of those points). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Within the 3x3 region, the corners can point such that they cover all 9 points. The sides can cover 7 points; they must exclude an adjacent corner and side. These cannot all coincide, there are at least 2 pairs of side and corner removed like this. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;.XX&#xA;.XX&#xA;..X&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now, whatever the pointing at the central point, it cannot cover both the upper left and the lowest X (135 degrees), so one of these is excluded. At this stage, we have only 4 points left so we get digging!&lt;/p&gt;&#xA;" OwnerUserId="3987" LastActivityDate="2015-06-14T21:59:30.497" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16398" PostTypeId="1" CreationDate="2015-06-15T00:04:45.933" Score="-1" ViewCount="473" Body="&lt;p&gt;Essicilmë and anessi?  Or maybe something else…&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;68 74 74 70 3a 2f 2f 69 6d 67 32 2e 77 69 6b 69 61 2e 6e 6f 63 6f 6f 6b 69 65 2e 6e 65 74 2f 5f 5f 63 62 32 30 30 37 30 38 31 32 30 35 35 33 32 30 2f 68 61 72 72 79 70 6f 74 74 65 72 2f 69 6d 61 67 65 73 2f 39 2f 39 38 2f 53 6e 61 70 65 5f 35 74 68 79 65 61 72 2e 6a 70 67 20&lt;/li&gt;&#xA;&lt;li&gt;68 74 74 70 3a 2f 2f 69 6d 67 32 2e 77 69 6b 69 61 2e 6e 6f 63 6f 6f 6b 69 65 2e 6e 65 74 2f 5f 5f 63 62 32 30 30 38 30 33 31 38 32 31 35 37 34 34 2f 6c 6f 74 72 2f 69 6d 61 67 65 73 2f 34 2f 34 33 2f 41 72 61 67 6f 72 6e 33 2e 6a 70 67&lt;/li&gt;&#xA;&lt;li&gt;68 74 74 70 3a 2f 2f 77 77 77 2e 62 65 61 73 74 73 6f 66 77 61 72 2e 63 6f 6d 2f 77 70 2d 63 6f 6e 74 65 6e 74 2f 75 70 6c 6f 61 64 73 2f 32 30 31 35 2f 30 31 2f 6f 57 6f 70 30 2e 6a 70 67 20&lt;/li&gt;&#xA;&lt;li&gt;68 74 74 70 3a 2f 2f 69 6d 67 33 2e 77 69 6b 69 61 2e 6e 6f 63 6f 6f 6b 69 65 2e 6e 65 74 2f 5f 5f 63 62 32 30 31 32 30 36 30 34 31 35 30 36 31 35 2f 6c 6f 74 72 2f 69 6d 61 67 65 73 2f 64 2f 64 32 2f 57 61 72 5f 6f 66 5f 77 72 61 74 68 2e 6a 70 67 20&lt;/li&gt;&#xA;&lt;li&gt;68 74 74 70 3a 2f 2f 74 6f 6c 6b 69 65 6e 67 61 74 65 77 61 79 2e 6e 65 74 2f 77 2f 69 6d 61 67 65 73 2f 74 68 75 6d 62 2f 64 2f 64 65 2f 4b 69 6d 62 65 72 6c 79 5f 2d 5f 4e 69 65 6e 6f 72 5f 4e 69 6e 69 65 6c 2e 6a 70 67 2f 32 35 30 70 78 2d 4b 69 6d 62 65 72 6c 79 5f 2d 5f 4e 69 65 6e 6f 72 5f 4e 69 6e 69 65 6c 2e 6a 70 67&lt;/li&gt;&#xA;&lt;li&gt;68 74 74 70 3a 2f 2f 69 6d 67 31 30 2e 64 65 76 69 61 6e 74 61 72 74 2e 6e 65 74 2f 64 64 36 63 2f 69 2f 32 30 31 33 2f 30 37 35 2f 66 2f 37 2f 6e 65 73 73 61 5f 62 79 5f 65 73 74 65 6c 69 6e 61 6b 69 6e 61 2d 64 35 79 38 69 7a 61 2e 6a 70 67&lt;/li&gt;&#xA;&lt;li&gt;68 74 74 70 3a 2f 2f 66 63 30 30 2e 64 65 76 69 61 6e 74 61 72 74 2e 6e 65 74 2f 66 73 37 30 2f 66 2f 32 30 31 31 2f 33 32 33 2f 65 2f 32 2f 6d 6f 72 77 65 6e 5f 62 79 5f 65 6b 75 6b 61 6e 6f 76 61 2d 64 34 67 6f 30 36 6e 2e 6a 70 67&lt;/li&gt;&#xA;&lt;li&gt;68 74 74 70 3a 2f 2f 69 6d 61 67 65 73 32 2e 66 61 6e 70 6f 70 2e 63 6f 6d 2f 69 6d 61 67 65 2f 70 68 6f 74 6f 73 2f 39 35 30 30 30 30 30 2f 50 69 70 70 69 6e 2d 54 6f 6f 6b 2d 70 69 70 70 69 6e 2d 74 6f 6f 6b 2d 39 35 36 30 30 30 37 2d 34 35 30 2d 35 37 30 2e 6a 70 67 20&lt;/li&gt;&#xA;&lt;li&gt;68 74 74 70 3a 2f 2f 6f 72 69 67 30 30 2e 64 65 76 69 61 6e 74 61 72 74 2e 6e 65 74 2f 64 62 39 36 2f 66 2f 32 30 31 32 2f 30 38 30 2f 36 2f 61 2f 65 6c 75 5f 74 68 69 6e 67 6f 6c 5f 62 79 5f 65 6b 75 6b 61 6e 6f 76 61 2d 64 34 74 67 78 35 63 2e 6a 70 67&lt;/li&gt;&#xA;&lt;li&gt;68 74 74 70 3a 2f 2f 73 74 61 74 69 63 2e 74 75 6d 62 6c 72 2e 63 6f 6d 2f 65 37 63 32 34 35 35 64 39 62 62 61 31 62 37 62 35 33 33 35 38 64 61 35 36 66 38 31 63 38 65 66 2f 69 76 69 67 64 36 6b 2f 56 57 47 6d 68 37 37 6c 66 2f 74 75 6d 62 6c 72 5f 73 74 61 74 69 63 5f 74 75 6d 62 6c 72 5f 6d 34 61 71 78 31 6f 6b 77 6c 31 71 6d 78 61 61 38 6f 31 5f 35 30 30 2e 6a 70 67&lt;/li&gt;&#xA;&lt;li&gt;68 74 74 70 3a 2f 2f 76 69 67 6e 65 74 74 65 33 2e 77 69 6b 69 61 2e 6e 6f 63 6f 6f 6b 69 65 2e 6e 65 74 2f 6c 6f 74 72 66 61 6e 6f 6e 2f 69 6d 61 67 65 73 2f 32 2f 32 30 2f 4e 69 6d 6c 6f 74 68 5f 45 6c 75 63 68 ee 6c 2e 70 6e 67 2f 72 65 76 69 73 69 6f 6e 2f 6c 61 74 65 73 74 3f 63 62 3d 32 30 31 33 31 32 33 30 30 32 31 30 35 32&lt;/li&gt;&#xA;&lt;li&gt;68 74 74 70 3a 2f 2f 76 65 72 73 61 70 68 69 6c 65 2e 63 6f 6d 2f 63 61 70 73 2f 74 74 74 2f 31 38 73 68 61 64 6f 77 66 61 78 2f 73 68 61 64 6f 77 66 61 78 33 39 2e 6a 70 67&lt;/li&gt;&#xA;&lt;li&gt;68 74 74 70 3a 2f 2f 69 6d 67 33 2e 77 69 6b 69 61 2e 6e 6f 63 6f 6f 6b 69 65 2e 6e 65 74 2f 5f 5f 63 62 32 30 30 37 30 37 31 39 32 31 34 39 32 33 2f 6c 6f 74 72 2f 69 6d 61 67 65 73 2f 64 2f 64 34 2f 43 69 72 64 61 6e 2e 6a 70 67&lt;/li&gt;&#xA;&lt;li&gt;68 74 74 70 3a 2f 2f 69 6d 67 32 2e 77 69 6b 69 61 2e 6e 6f 63 6f 6f 6b 69 65 2e 6e 65 74 2f 5f 5f 63 62 32 30 31 31 31 31 31 36 30 37 32 31 34 35 2f 6c 6f 74 72 2f 69 6d 61 67 65 73 2f 31 2f 31 38 2f 56 61 72 64 61 5f 62 79 5f 47 75 73 74 61 76 6f 4d 61 6c 65 6b 2e 6a 70 67 20&lt;/li&gt;&#xA;&lt;li&gt;68 74 74 70 3a 2f 2f 69 6d 67 32 2e 77 69 6b 69 61 2e 6e 6f 63 6f 6f 6b 69 65 2e 6e 65 74 2f 5f 5f 63 62 32 30 30 37 30 39 31 39 31 39 32 34 33 32 2f 6c 6f 74 72 2f 69 6d 61 67 65 73 2f 36 2f 36 61 2f 54 75 72 69 6e 2d 41 72 74 77 6f 72 6b 2e 4a 50 47&lt;/li&gt;&#xA;&lt;li&gt;68 74 74 70 3a 2f 2f 76 69 67 6e 65 74 74 65 33 2e 77 69 6b 69 61 2e 6e 6f 63 6f 6f 6b 69 65 2e 6e 65 74 2f 64 65 61 64 6c 69 65 73 74 66 69 63 74 69 6f 6e 2f 69 6d 61 67 65 73 2f 37 2f 37 35 2f 35 39 37 70 78 2d 54 68 65 6f 64 65 6e 5f 72 69 64 65 72 2e 6a 70 67 2f 72 65 76 69 73 69 6f 6e 2f 6c 61 74 65 73 74 3f 63 62 3d 32 30 31 32 30 33 32 36 30 31 30 31 30 39&lt;/li&gt;&#xA;&lt;li&gt;68 74 74 70 3a 2f 2f 74 65 64 6e 61 73 6d 69 74 68 2e 70 6f 76 65 72 65 6c 6c 6f 6d 65 64 69 61 2e 63 6f 6d 2f 77 70 2d 63 6f 6e 74 65 6e 74 2f 75 70 6c 6f 61 64 73 2f 32 30 31 32 2f 30 37 2f 54 4e 2d 42 65 72 65 6e 73 5f 4c 65 61 70 2e 6a 70 67&lt;/li&gt;&#xA;&lt;li&gt;68 74 74 70 3a 2f 2f 6f 72 69 67 31 32 2e 64 65 76 69 61 6e 74 61 72 74 2e 6e 65 74 2f 32 61 33 35 2f 66 2f 32 30 31 34 2f 32 34 30 2f 61 2f 34 2f 61 34 63 39 33 32 37 39 30 30 36 65 36 64 31 65 31 36 36 61 37 38 62 32 38 63 62 33 37 35 62 30 2d 64 33 30 69 7a 66 6d 2e 6a 70 67&lt;/li&gt;&#xA;&lt;li&gt;68 74 74 70 3a 2f 2f 77 77 77 2e 77 65 61 72 79 73 6c 6f 74 68 2e 63 6f 6d 2f 47 61 6c 6c 65 72 79 2f 41 63 74 6f 72 73 4d 2f 32 31 38 32 32 2d 32 35 35 37 35 2e 67 69 66&lt;/li&gt;&#xA;&lt;li&gt;68 74 74 70 3a 2f 2f 6f 72 69 67 30 32 2e 64 65 76 69 61 6e 74 61 72 74 2e 6e 65 74 2f 62 37 39 31 2f 66 2f 32 30 31 33 2f 30 33 37 2f 35 2f 36 2f 74 68 6f 72 69 6e 5f 6f 61 6b 65 6e 73 68 69 65 6c 64 5f 6b 69 6e 67 5f 75 6e 64 65 72 5f 74 68 65 5f 6d 6f 75 6e 74 61 69 6e 5f 62 79 5f 64 72 6b 61 79 38 35 2d 64 35 75 30 6b 6c 73 2e 6a 70 67&lt;/li&gt;&#xA;&lt;li&gt;68 74 74 70 3a 2f 2f 69 6d 67 33 2e 77 69 6b 69 61 2e 6e 6f 63 6f 6f 6b 69 65 2e 6e 65 74 2f 5f 5f 63 62 32 30 30 37 30 33 32 32 32 31 35 39 32 33 2f 6c 6f 74 72 2f 69 6d 61 67 65 73 2f 34 2f 34 34 2f 47 69 6c 2d 67 61 6c 61 64 31 31 2e 6a 70 67&lt;/li&gt;&#xA;&lt;li&gt;68 74 74 70 3a 2f 2f 69 6d 67 31 2e 77 69 6b 69 61 2e 6e 6f 63 6f 6f 6b 69 65 2e 6e 65 74 2f 5f 5f 63 62 32 30 31 32 31 31 31 34 31 36 31 35 31 37 2f 6c 6f 74 72 2f 69 6d 61 67 65 73 2f 37 2f 37 61 2f 4d 61 6e 77 65 5f 53 75 6c 69 6d 6f 5f 62 79 5f 47 75 73 74 61 76 6f 4d 61 6c 65 6b 2e 6a 70 67&lt;/li&gt;&#xA;&lt;li&gt;68 74 74 70 3a 2f 2f 73 33 2e 61 6d 61 7a 6f 6e 61 77 73 2e 63 6f 6d 2f 74 65 6c 65 76 69 63 65 6e 74 72 6f 2f 70 6f 72 74 61 64 61 73 2f 73 61 72 75 6d 61 6e 2e 6a 70 67 3f 6d 74 69 6d 65 3d 32 30 31 34 30 35 32 38 31 31 34 31 30 33&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Seems long, but the output is much shorter.  Most know how to deal with the given text, but there is a second, &quot;slightly&quot; more labour intensive step.  Each part yields one letter.  A liking for Middle-earth may help.  Please edit tags.&lt;/p&gt;&#xA;" OwnerUserId="11790" LastActivityDate="2015-06-15T00:24:30.610" Title="The Last of the Sons of Fëanor" Tags="|knowledge|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16399" PostTypeId="2" ParentId="16398" CreationDate="2015-06-15T00:24:30.610" Score="1" Body="&lt;p&gt;Partial answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Here are what the links are to make this puzzle accessible for those with more specialized talents:&lt;br&gt;&lt;a href=&quot;http://img2.wikia.nocookie.net/__cb20070812055320/harrypotter/images/9/98/Snape_5thyear.jpg&quot; rel=&quot;nofollow&quot;&gt;http://img2.wikia.nocookie.net/__cb20070812055320/harrypotter/images/9/98/Snape_5thyear.jpg&lt;/a&gt;&lt;br&gt;&lt;a href=&quot;http://img2.wikia.nocookie.net/__cb20080318215744/lotr/images/4/43/Aragorn3.jpg&quot; rel=&quot;nofollow&quot;&gt;http://img2.wikia.nocookie.net/__cb20080318215744/lotr/images/4/43/Aragorn3.jpg&lt;/a&gt;&lt;br&gt;&lt;a href=&quot;http://www.beastsofwar.com/wp-content/uploads/2015/01/oWop0.jpg&quot; rel=&quot;nofollow&quot;&gt;http://www.beastsofwar.com/wp-content/uploads/2015/01/oWop0.jpg&lt;/a&gt;&lt;br&gt;&lt;a href=&quot;http://img3.wikia.nocookie.net/__cb20120604150615/lotr/images/d/d2/War_of_wrath.jpg&quot; rel=&quot;nofollow&quot;&gt;http://img3.wikia.nocookie.net/__cb20120604150615/lotr/images/d/d2/War_of_wrath.jpg&lt;/a&gt;&lt;br&gt;&lt;a href=&quot;http://tolkiengateway.net/w/images/thumb/d/de/Kimberly_-_Nienor_Niniel.jpg/250px-Kimberly_-_Nienor_Niniel.jpg&quot; rel=&quot;nofollow&quot;&gt;http://tolkiengateway.net/w/images/thumb/d/de/Kimberly_-_Nienor_Niniel.jpg/250px-Kimberly_-_Nienor_Niniel.jpg&lt;/a&gt;&lt;br&gt;&lt;a href=&quot;http://img10.deviantart.net/dd6c/i/2013/075/f/7/nessa_by_estelinakina-d5y8iza.jpg&quot; rel=&quot;nofollow&quot;&gt;http://img10.deviantart.net/dd6c/i/2013/075/f/7/nessa_by_estelinakina-d5y8iza.jpg&lt;/a&gt;&lt;br&gt;&lt;a href=&quot;http://fc00.deviantart.net/fs70/f/2011/323/e/2/morwen_by_ekukanova-d4go06n.jpg&quot; rel=&quot;nofollow&quot;&gt;http://fc00.deviantart.net/fs70/f/2011/323/e/2/morwen_by_ekukanova-d4go06n.jpg&lt;/a&gt;&lt;br&gt;&lt;a href=&quot;http://images2.fanpop.com/image/photos/9500000/Pippin-Took-pippin-took-9560007-450-570.jpg&quot; rel=&quot;nofollow&quot;&gt;http://images2.fanpop.com/image/photos/9500000/Pippin-Took-pippin-took-9560007-450-570.jpg&lt;/a&gt;&lt;br&gt;&lt;a href=&quot;http://orig00.deviantart.net/db96/f/2012/080/6/a/elu_thingol_by_ekukanova-d4tgx5c.jpg&quot; rel=&quot;nofollow&quot;&gt;http://orig00.deviantart.net/db96/f/2012/080/6/a/elu_thingol_by_ekukanova-d4tgx5c.jpg&lt;/a&gt;&lt;br&gt;&lt;a href=&quot;http://static.tumblr.com/e7c2455d9bba1b7b53358da56f81c8ef/ivigd6k/VWGmh77lf/tumblr_static_tumblr_m4aqx1okwl1qmxaa8o1_500.jpg&quot; rel=&quot;nofollow&quot;&gt;http://static.tumblr.com/e7c2455d9bba1b7b53358da56f81c8ef/ivigd6k/VWGmh77lf/tumblr_static_tumblr_m4aqx1okwl1qmxaa8o1_500.jpg&lt;/a&gt;&lt;br&gt;&lt;a href=&quot;http://vignette3.wikia.nocookie.net/lotrfanon/images/2/20&quot; rel=&quot;nofollow&quot;&gt;http://vignette3.wikia.nocookie.net/lotrfanon/images/2/20&lt;/a&gt;/Nimloth_Eluchîl.png/revision/latest?cb=20131230021052&lt;br&gt;&lt;a href=&quot;http://versaphile.com/caps/ttt/18shadowfax/shadowfax39.jpg&quot; rel=&quot;nofollow&quot;&gt;http://versaphile.com/caps/ttt/18shadowfax/shadowfax39.jpg&lt;/a&gt;&lt;br&gt;&lt;a href=&quot;http://img3.wikia.nocookie.net/__cb20070719214923/lotr/images/d/d4/Cirdan.jpg&quot; rel=&quot;nofollow&quot;&gt;http://img3.wikia.nocookie.net/__cb20070719214923/lotr/images/d/d4/Cirdan.jpg&lt;/a&gt;&lt;br&gt;&lt;a href=&quot;http://img2.wikia.nocookie.net/__cb20111116072145/lotr/images/1/18/Varda_by_GustavoMalek.jpg&quot; rel=&quot;nofollow&quot;&gt;http://img2.wikia.nocookie.net/__cb20111116072145/lotr/images/1/18/Varda_by_GustavoMalek.jpg&lt;/a&gt;&lt;br&gt;&lt;a href=&quot;http://img2.wikia.nocookie.net/__cb20070919192432/lotr/images/6/6a/Turin-Artwork.JPG&quot; rel=&quot;nofollow&quot;&gt;http://img2.wikia.nocookie.net/__cb20070919192432/lotr/images/6/6a/Turin-Artwork.JPG&lt;/a&gt;&lt;br&gt;&lt;a href=&quot;http://vignette3.wikia.nocookie.net/deadliestfiction/images/7/75/597px-Theoden_rider.jpg/revision/latest?cb=20120326010109&quot; rel=&quot;nofollow&quot;&gt;http://vignette3.wikia.nocookie.net/deadliestfiction/images/7/75/597px-Theoden_rider.jpg/revision/latest?cb=20120326010109&lt;/a&gt;&lt;br&gt;&lt;a href=&quot;http://tednasmith.poverellomedia.com/wp-content/uploads/2012/07/TN-Berens_Leap.jpg&quot; rel=&quot;nofollow&quot;&gt;http://tednasmith.poverellomedia.com/wp-content/uploads/2012/07/TN-Berens_Leap.jpg&lt;/a&gt;&lt;br&gt;&lt;a href=&quot;http://orig12.deviantart.net/2a35/f/2014/240/a/4/a4c93279006e6d1e166a78b28cb375b0-d30izfm.jpg&quot; rel=&quot;nofollow&quot;&gt;http://orig12.deviantart.net/2a35/f/2014/240/a/4/a4c93279006e6d1e166a78b28cb375b0-d30izfm.jpg&lt;/a&gt;&lt;br&gt;&lt;a href=&quot;http://www.wearysloth.com/Gallery/ActorsM/21822-25575.gif&quot; rel=&quot;nofollow&quot;&gt;http://www.wearysloth.com/Gallery/ActorsM/21822-25575.gif&lt;/a&gt;&lt;br&gt;&lt;a href=&quot;http://orig02.deviantart.net/b791/f/2013/037/5/6/thorin_oakenshield_king_under_the_mountain_by_drkay85-d5u0kls.jpg&quot; rel=&quot;nofollow&quot;&gt;http://orig02.deviantart.net/b791/f/2013/037/5/6/thorin_oakenshield_king_under_the_mountain_by_drkay85-d5u0kls.jpg&lt;/a&gt;&lt;br&gt;&lt;a href=&quot;http://img3.wikia.nocookie.net/__cb20070322215923/lotr/images/4/44/Gil-galad11.jpg&quot; rel=&quot;nofollow&quot;&gt;http://img3.wikia.nocookie.net/__cb20070322215923/lotr/images/4/44/Gil-galad11.jpg&lt;/a&gt;&lt;br&gt;&lt;a href=&quot;http://img1.wikia.nocookie.net/__cb20121114161517/lotr/images/7/7a/Manwe_Sulimo_by_GustavoMalek.jpg&quot; rel=&quot;nofollow&quot;&gt;http://img1.wikia.nocookie.net/__cb20121114161517/lotr/images/7/7a/Manwe_Sulimo_by_GustavoMalek.jpg&lt;/a&gt;&lt;br&gt;&lt;a href=&quot;http://s3.amazonaws.com/televicentro/portadas/saruman.jpg?mtime=20140528114103&quot; rel=&quot;nofollow&quot;&gt;http://s3.amazonaws.com/televicentro/portadas/saruman.jpg?mtime=20140528114103&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastActivityDate="2015-06-15T00:24:30.610" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16402" PostTypeId="1" CreationDate="2015-06-15T07:14:18.143" Score="0" ViewCount="1598" Body="&lt;p&gt;Consider two identical card packs: A and B. One card is taken from A and shuffled with pack B. The top card of pack A is QUEEN of HEARTS. What is the probability that the top card of pack B is KING of HEARTS?&lt;/p&gt;&#xA;" OwnerUserId="13195" LastEditorUserId="12645" LastEditDate="2015-06-15T07:27:17.580" LastActivityDate="2015-06-15T10:38:11.530" Title="question involving two deck of cards" Tags="|probability|" AnswerCount="2" CommentCount="3" ClosedDate="2015-06-15T14:28:00.387" ContentLicense="CC BY-SA 3.0" />
  <row Id="16403" PostTypeId="2" ParentId="16402" CreationDate="2015-06-15T08:18:41.277" Score="5" Body="&lt;p&gt;There was $52$ cards in both pack $A$ and $B$. After the shuffle, there are $51$ in $A$ and $53$ in $B$. Showing the Top card of $A$ leaves, though, with only $51$ possible cards to be taken from $A$ (since we know the Queen hadn't moved).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have two possible scenarios now. If we did &lt;strong&gt;not&lt;/strong&gt; take the King of Hearts, from $A$ to $B$, and the other were we did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be $K$ the event of taking the King of Hearts from $A$ to $B$. Be $B_K$ the event of it being the top card of $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{align}&#xA;P(B_K)&amp;amp;=P(K)P(B_K\mid K)+P(K^C)P(B_K\mid K^C)\\&#xA;&amp;amp;=\frac{1}{51}\frac{2}{53}+\frac{50}{51}\frac{1}{53}\\&#xA;&amp;amp;=\frac{52}{2703}\\&#xA;&amp;amp;\approx 0.02&#xA;\end{align}$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In order to respond to @frodoskywalker I'll post a more generic answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r$ will stand for having revealed $r$ cards from $A$ and none being the King. Note that it's important that's done randomly (like simply showing top cards is ok; the problem would be fetching for certain cards).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{align}&#xA;P(B_K^r)&amp;amp;=P(K_r)P(B^r_K\mid K_r)+P(K_r^C)P(B^r_K\mid K_r^C)\\&#xA;&amp;amp;=\frac{1}{52-r}\frac{2}{53}+\frac{51-r}{52-r}\frac{1}{53}\\&#xA;&amp;amp;=\frac{53-r}{2756-53r}\\&#xA;P(B_K^1)&amp;amp;\approx 0.02&#xA;\end{align}$&lt;/p&gt;&#xA;" OwnerUserId="12165" LastEditorUserId="12165" LastEditDate="2015-06-15T10:27:40.750" LastActivityDate="2015-06-15T10:27:40.750" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16404" PostTypeId="1" AcceptedAnswerId="16406" CreationDate="2015-06-15T08:28:08.863" Score="9" ViewCount="1901" Body="&lt;p&gt;Albert, Brian and Charles are brothers with different ages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Albert has twice the age Brian will have when Charles has twice the age Albert has now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who's the oldest?&lt;/p&gt;&#xA;" OwnerUserId="12165" LastEditorUserId="10781" LastEditDate="2015-06-15T13:06:07.123" LastActivityDate="2015-06-16T15:34:40.837" Title="Who's the oldest?" Tags="|calculation-puzzle|" AnswerCount="6" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16406" PostTypeId="2" ParentId="16404" CreationDate="2015-06-15T08:48:27.553" Score="16" Body="&lt;p&gt;They have A,B,C initially. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P1: Charles will have twice the age Albert has now:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It will be the year Charles will be 2A. Year difference  is 2A-C&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;P2: Brian will have P1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Add 2A-C years to Brian's initial age and his age will be 2A+B-C&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;P3: Albert has twice the age P2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Multiply Brian's age with 2. and Albert's initial age  A = 4A+2B-2C &lt;br/&gt;&lt;br/&gt; which means C = (3A+2B)/2 thus Charles is the eldest.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4506" LastActivityDate="2015-06-15T08:48:27.553" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16408" PostTypeId="2" ParentId="16169" CreationDate="2015-06-15T09:46:38.940" Score="4" Body="&lt;p&gt;I think you're &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A digit&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I haven't got the whole solution, but a lot of it seems to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think &quot;on top politically&quot; means&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; prime (as in Prime Minister)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and &quot;being not in violation&quot; means&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; perfect, as in &lt;a href=&quot;https://en.wikipedia.org/?title=Perfect_number&quot; rel=&quot;nofollow&quot;&gt;perfect number&lt;/a&gt;. The only perfect digit is 6.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I am on top politically, I am right&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is a bit I haven't got, I have no idea how 2,3,5 and 7 are right, although I suspect it's something to do with wordplay and/or a synonym of &quot;right&quot;. The best I have is that they're all rational numbers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but if I am on top of being not in violation, I am wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; and this bit is really tenuous, but could refer to &lt;a href=&quot;http://idioms.thefreedictionary.com/deep-six&quot; rel=&quot;nofollow&quot;&gt;deep sixing&lt;/a&gt; someone, which is pretty wrong (x/6 the 6 could be said to be &quot;deep&quot;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I am above 8, I am wrong, but if I am low on being not in violation, I am forever.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If I am above 8 I am &lt;strong&gt;Nein!&lt;/strong&gt; (wordplay!), and 6/9 is 0.6666... with infinite decimal places.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cannot be on top politically and above 8, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 9 is not prime&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I can be on top and low of being not in violation. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 6/x or x/6 is fine, albeit the first is undefined if x = 0&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm honestly not confident with any of that, but I've been staring at this for days now and I'm mostly hoping that this bumps the puzzle enough for someone to answer it.&lt;/p&gt;&#xA;" OwnerUserId="13197" LastActivityDate="2015-06-15T09:46:38.940" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16409" PostTypeId="2" ParentId="16402" CreationDate="2015-06-15T10:38:11.530" Score="3" Body="&lt;p&gt;Different method, same answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are $52+1$ cards in the new deck B.  The top card is from deck A with probability $1/53$ and from deck B with probability $52/53$.&lt;br&gt;&#xA;If it is from deck A, it is a King of Heart with probability $1/51$.  (Any card but the Queen of Heart).&lt;br&gt;&#xA;If it is from deck B, it is a King of Heart with probability $1/52$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The combined probability to be a King of Heart is:&lt;br&gt;&#xA;    $P(K) = P(A)P(K|A) + P(B)P(K|B)$&lt;br&gt;&#xA;    $P(K) = (1/53)(1/51) + (52/53)(1/52) = 52/2703$  &lt;/p&gt;&#xA;" OwnerUserId="2136" LastActivityDate="2015-06-15T10:38:11.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16410" PostTypeId="2" ParentId="16404" CreationDate="2015-06-15T11:23:47.677" Score="3" Body="&lt;p&gt;I'll try a different approach which I think is also correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the answer is consistent with every possible age of A, B and C one may take the option in which the &quot;time when Charles will have twice the age Albert has now&quot; is actually now. In that case C is twice the age of A and A is twice the age of B making C the oldest. So for example C = 4, A = 2, B = 1. This does satisfy the condition.&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-06-15T11:23:47.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16411" PostTypeId="2" ParentId="16169" CreationDate="2015-06-15T11:29:17.560" Score="4" Body="&lt;p&gt;Trying to find the wordplay Moghwyn missed you might be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; nothing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I am on top politically, I am right, but if I am on top of being not in violation, I am wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The political leaders often can do what they want, nothing is right. If nothing is violated though, nothing is wrong.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I am above 8, I am wrong, but if I am low on being not in violation, I am forever.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To say nothing is above 8 is clearly wrong, but you can safely say nothing stays forever.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cannot be on top politically and above 8, but I can be on top and low of being not in violation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If on top politically, nothing is right, but still nothing is not above 8. If on top politically and thus nothing is right (nothing seems the right thing), that interestingly does not mean that nothing is wrong... it just means there is sometimes nothing totally wrong or totally right to everybody you ask.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10885" LastActivityDate="2015-06-15T11:29:17.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16412" PostTypeId="1" AcceptedAnswerId="16414" CreationDate="2015-06-15T13:35:09.917" Score="19" ViewCount="3271" Body="&lt;p&gt;The following is a text rebus in the form of a Java program:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import java.awt.*;&#xA;&#xA;public class WhatIsThis {&#xA;    private static Panel panel = new Panel();&#xA;&#xA;    static {&#xA;        init();&#xA;    }&#xA;&#xA;    public static void main(String[] args) {&#xA;        while (panel.getComponentCount()&amp;gt;0) {&#xA;            panel.getComponentCount();&#xA;            panel.getClass();&#xA;&#xA;            panel.getComponentCount();&#xA;&#xA;            Container c = (Container)panel.getComponent(0); panel.remove(c);&#xA;&#xA;            method1(c);&#xA;&#xA;            panel.getComponentCount();&#xA;            panel.getClass();&#xA;        }&#xA;&#xA;        /*&#xA;        // OPTIONAL:&#xA;&#xA;        panel.getComponentCount();&#xA;        panel.getClass();&#xA;&#xA;        panel.getComponentCount();&#xA;&#xA;        init();&#xA;&#xA;        panel.getComponentCount();&#xA;        panel.getClass();&#xA;&#xA;        main(args);&#xA;        */&#xA;    }&#xA;&#xA;    private static void init() {&#xA;        for (int i=1; i&amp;lt;100; i++) {&#xA;            Container c = new Container();&#xA;            c.setName(&quot;ISB&quot;);&#xA;            panel.add(c);&#xA;        }&#xA;    }&#xA;&#xA;    private static void method1(Container c) {&#xA;        method2(c);&#xA;    }&#xA;&#xA;    private static void method2(Container c) {&#xA;        method3(c);&#xA;    }&#xA;&#xA;    private static void method3(Container c) {&#xA;    }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What does it represent?  Bonus points if you can figure out what &quot;ISB&quot; means.&lt;/p&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2015-06-16T05:49:56.857" Title="What does this Java code represent?" Tags="|rebus|computer-puzzle|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16413" PostTypeId="2" ParentId="16357" CreationDate="2015-06-15T13:39:02.040" Score="1" Body="&lt;p&gt;Let's say that the shooters are in a circle in ascending height order. If all shooters are equidistant, then the shots will go from the shortest to the tallest, with the tallest shooting at the 2nd tallest (his choices are his target, or the shortest, and as they are equidistant, not the shortest). Ideally, we'd want the tallest to shoot the shortest. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way to do that is to move the tallest closer to the shortest. But, that then makes the shortest target the tallest instead of the 2nd shortest. Well, then move the 2nd shortest closer to the shortest; but then the 2nd shortest shoots the shortest. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you can't have everyone dead. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-06-15T13:39:02.040" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16414" PostTypeId="2" ParentId="16412" CreationDate="2015-06-15T13:40:31.317" Score="20" Body="&lt;p&gt;I'm fairly certain this is a representation of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;99 Bottles of Beer on the Wall&quot;. You add 99 components to a panel, then proceed in a loop to get the count in the panel and then the count again (like in the song), and then remove one and pass it around (into a few methods, lol).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Had to honestly google &quot;ISB&quot;, turns out it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;industry standard bottle&quot;, for a bottle of beer. :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2444" LastEditorUserId="2444" LastEditDate="2015-06-15T13:42:58.663" LastActivityDate="2015-06-15T13:42:58.663" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16415" PostTypeId="2" ParentId="16412" CreationDate="2015-06-15T13:53:10.180" Score="3" Body="&lt;p&gt;The optional section, as shyos said, states there are 0 bottles on the wall, puts 99 bottles of beer back on the wall, and then sings the song again:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;panel.getComponentCount(); = No more bottles of beer&#xA;panel.getClass(); = on the wall&#xA;&#xA;panel.getComponentCount(); = No more bottles of beer&#xA;&#xA;init(); = Go to the store and buy some more&#xA;&#xA;panel.getComponentCount(); = 99 bottles of beer&#xA;panel.getClass(); = on the wall&#xA;&#xA;main(args); = repeat song and eventually crash (due to stack overflow), which is pretty much what a person would do if they had to sing the song over and over and over...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note from OP: The lyrics for the optional section can be found in the book &lt;em&gt;&lt;a href=&quot;https://books.google.com/books?id=spmJzMQj8q4C&amp;amp;printsec=frontcover#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;99 Bottles of Beer on the Wall: The Complete Lyrics&lt;/a&gt;&lt;/em&gt; by Tim Nyberg (2006).&lt;/p&gt;&#xA;" OwnerDisplayName="user13180" LastEditorUserId="4450" LastEditDate="2015-06-16T05:49:56.857" LastActivityDate="2015-06-16T05:49:56.857" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16416" PostTypeId="2" ParentId="16006" CreationDate="2015-06-15T14:31:17.670" Score="2" Body="&lt;p&gt;&lt;strong&gt;Updated complete answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n=3$ requires 1 weighing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put one ball on each pan.  The one normal-weight ball will rise and the other two are the heavy ones.  There are no other possibilities with $n=3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n=4$ requires 2 weighings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are ${4\choose 2} = 6$ possibilities, and each weighing can only distinguish 4 states, so 1 weighing cannot be sufficient for $n=4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put one ball on each pan and keep the 4th ball aside.  If one pan rises, then the other two balls are the heavy ones and you are done.  Otherwise, you must have weighed 2 light balls and 1 heavy ball, so the 4th ball is the heavy one.  Swap it with one of the others.  If you swapped it with a light ball, then the other light ball will now rise and you are done.  If no ball rises, then you swapped the heavy ball with the other heavy ball.  The two balls you swapped are the heavy ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n=5$ cannot be done in two weighings.  I could try to prove it, but doing so doesn't answer the question asked, so I won't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, $n=6$ can be done in two weighings.  ${6\choose 2} = 15 &amp;lt; 16 = 4^2$, so in theory it could be possible.  In practice, it is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put two balls on each pan, numbering and labeling such that pan A has (1,2), pan B has (3,4), and pan C has (5,6).  After noting the result, move the odd balls to the next pan, so now A has (5,2), B has (1,4), and C has (3,6).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If no pan rose on the first weighing, then both heavy balls were on the same pan, making the other two pans of two light balls each weigh the same.  On the second weighing, the balls are separated.  So if the heavy ones are (1,2), then on the second weghing, pans A and B will each have a heavy and a light ball, and pan C will have two light balls, so pan C will rise.  Likewise, (3,4): pan A would rise and (5,6): pan B would rise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one of the pans rose on the first weighing, then the two heavy balls were split.  So, if pan A rose, then the possibilities are (3,5), (3,6), (4,5), and (4,6).  On the second weighing, the results of those would be: (3,5): pan B rises, (3,6): no pan rises, (4,5): pan C rises, (4,6): pan A rises again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of the 16 possible pan rising combinations, the one one that cannot occur is no pan rising either time.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since ${7\choose 2} = 21 &amp;gt; 16$, it clearly is not possible to differentiate $n=7$ in two weighings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For three weighings, I note that ${11\choose 2} = 55 &amp;lt; 64 = 4^3 &amp;lt; 66 = {12\choose 2}$, so the theoretical maximum for three weighings is 11 balls.  This maximum cannot be achieved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof that it is possible to find the heaviest two balls out of $n=9$ balls with three weighings:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First weigh (1,2,3) vs (4,5,6) vs (7,8,9).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then weigh (1,8,6) vs (4,2,9) vs (7,5,3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then weigh (1,2,6) vs (4,5,9) vs (7,8,3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of the ${9\choose 2} = 36$ possibilities gives a different set of answers.  Here's a partial enumeration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1,2): NCN  (1,3) NBB (2,3) NAB&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1,4) CCC (1,5) CBC (1,6) CNN&#xA;(1,7) BBB (1,8) BNB (1,9) BCC&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2,4) CNC (2,5) CAC (2,6) CCN&#xA;(2,7) BAB (2,8) BCB (2,9) BNC&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3,4) CAA (3,5) CNA (3,6) CBB&#xA;(3,7) BNN (3,8) BBN (3,9) BAA&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any other combination is a rotation of one of the above and would also yield a unique pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=10$, if you put 3 balls on each pan, there are $18 &amp;gt; 16$ possibilities where no pan rises, 9 where both heavy balls are on one pan, and 9 where one heavy ball is left out and the other is unknown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you put 3 balls on pans A &amp;amp; B and 4 balls on pan C, then if pan A rises, you have 24 possibilities: 12 where one heavy is on B and one on C, 3 where both are on B, and 9 where one heavy is on A and one on B, but the one on B is heavier than the one on A (and the one on A weighs less than 2 normal ones).  Note there are also 24 possibilities where pan B will rise, but 9 of those are overlap, but with the relative weights of the heavy balls switched.  There are likewise 15 possibilities where no pan rises: 6 for both on pan C and 9 where one is on A and one is on B and the two heavy ones weigh the same.  There are even 9 possibilities where pan C rises: one heavy ball on each of A &amp;amp; B, and both weigh more than two normal balls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you put 2 balls on each pan, there are 33 cases where no pan rises; 3 where both heavy balls are on the same pan, 6 where neither heavy ball is weighed at all, and 24 where one heavy ball is weighed and the other is not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you put 2, 2, and 3 balls, and pan A rises, then you have 17 possibilities: 1 where both heavy ones are on pan B, and 16 possibilities where the heaviest one is on pan B and the other heavy one is any of the other eight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, you cannot solve for $n=10$ with three weighings.  Since the difficulties lie with not being able to distribute all the balls equally, adding an 11th ball will not make it easier.  And, as stated above ${12 \choose 2} = 66 &amp;gt; 64 = 4^3$, so you cannot possibly weigh 12 or more in three weighings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $n=9$ must be the maximum for three weighings.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="2328" LastEditDate="2015-06-15T15:44:45.580" LastActivityDate="2015-06-15T15:44:45.580" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16417" PostTypeId="2" ParentId="16404" CreationDate="2015-06-15T14:32:16.683" Score="5" Body="&lt;p&gt;There are 2 points in time:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;First, they are $a$, $b$, and $c$ years old&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;After $n&amp;gt;0$ years, they are $a+n$, $b+n$, and $c+n$ years old&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The equations are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$$a=2(b+n)$$&lt;/li&gt;&#xA;&lt;li&gt;$$c+n=2a$$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;From equation 1 we get $a&amp;gt;b$ because $n&amp;gt;0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equation 2:&#xA;$$c+n=a+a$$&#xA;Replace first $a$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c+n=2(b+n) +a$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c=2b +a+n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We get $c&amp;gt;a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solution:&lt;/strong&gt;&#xA;$$c&amp;gt;a&amp;gt;b$$&lt;/p&gt;&#xA;" OwnerUserId="13202" LastEditorUserId="12109" LastEditDate="2015-06-15T14:54:37.847" LastActivityDate="2015-06-15T14:54:37.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16418" PostTypeId="2" ParentId="16404" CreationDate="2015-06-15T15:03:23.473" Score="2" Body="&lt;p&gt;At every point in time (since the birth of the youngest), the order is the same. What was true will be true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, instantly: $A&amp;gt;B$, $C&amp;gt;A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C$ is the oldest. &lt;/p&gt;&#xA;" OwnerUserId="9418" LastActivityDate="2015-06-15T15:03:23.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16419" PostTypeId="2" ParentId="16404" CreationDate="2015-06-15T15:53:12.180" Score="2" Body="&lt;p&gt;The other answers have derived the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c=\frac{3a+2b}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, if $a$ and $b$ are positive, then it is obvious that $c$ is the greatest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what if Albert and/or Brian are waiting to be born still?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us say that Albert will be born in 2 years, or $a=-2$ and that Brian will be born in 4 years, or $b=-4$. We can then conclude that $$c = \frac{3*(-2) + 2*(-4)}{2} = \frac{-6 + -8}{2} = -7$$. Thus we can say that Albert is the oldest in this scenario. We could just as easily concoct a scenario where Brian is older than Albert and the math shows that, as long as they aren't born yet, we'll be fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we accept that some of these souls are still waiting to be born, then we can not conclusively say whether or not any of them are the oldest with the information given.&lt;/p&gt;&#xA;" OwnerUserId="2812" LastActivityDate="2015-06-15T15:53:12.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16420" PostTypeId="2" ParentId="16147" CreationDate="2015-06-15T16:08:59.453" Score="7" Body="&lt;p&gt;I solved the message&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am your guide to the location that is everywhere&lt;br&gt;&lt;br&gt;I can show you a world from a view you couldnt imagine&lt;br&gt;&lt;br&gt;I can help you transcend the limits of your environment&lt;br&gt;&lt;br&gt;Utter my name and be taken on a journey like no other&lt;br&gt;&lt;br&gt;Utter my name and see the world for what it truly is&lt;br&gt;&lt;br&gt;Utter my name and become forever changed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And his name is obviously&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dot&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation for the message&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The wall panel reads &quot;The quick brown fox jumps over the lazy dog&quot; which was given by the most recent hint, tqbfjotld.&lt;br&gt;Putting letters to each symbol gave the key to the cipher, where the square shaped one was a space and the symbols for the letters K/Q and U/V were shared.&lt;br&gt;&lt;br&gt;Using the cipher on the cube faces in the order given in top to bottom right to left order spells out that message.&lt;br&gt;&lt;br&gt; And the name Dot comes from the character in the Video Game &quot;Fez&quot; from which these glyphs originated.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12843" LastEditorUserId="12843" LastEditDate="2015-06-17T14:33:20.827" LastActivityDate="2015-06-17T14:33:20.827" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16421" PostTypeId="1" AcceptedAnswerId="17697" CreationDate="2015-06-15T16:31:41.853" Score="11" ViewCount="991" Body="&lt;pre&gt;&lt;code&gt;INT. JULIAN'S HOUSE – BASEMENT&#xA;&#xA;JULIAN, rumpled, tired, and bewildered, sits at a shabby card table poring over&#xA;BLUEPRINTS of a complicated mechanism.  NAOMI hastily enters the room carrying a&#xA;PAD OF GRAPH PAPER and a BALLPOINT PEN.  She's just as sleep-deprived, but she's&#xA;taking it better.  She sits next to Julian.  He shoves the blueprints away and&#xA;leans back in his chair.&#xA;&#xA;                                     JULIAN&#xA;        I've got nothing.  Even if the duplicator could hit one of&#xA;        these parts, we'd just jam the lock.  There's not enough room&#xA;        in here.&#xA;&#xA;                                     NAOMI&#xA;        Don't worry about it.  We don't have to hit the lock.&#xA;&#xA;Julian raises an eyebrow, intrigued…&#xA;&#xA;                                 NAOMI (cont'd)&#xA;        The mechanism runs on dot numbers.  Like this.&#xA;&#xA;On the graph paper she draws&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A MULTIPLICATION PROBLEM:&#xA;&#xA;                                       ●○●●&#xA;                                        ●○●&#xA;                                     ------&#xA;                                       ●○●●&#xA;                                      ○○○○&#xA;                                     ●○●●&#xA;                                     ------&#xA;                                     ●○○●●●&#xA;&#xA;                                 NAOMI (O.S.)&#xA;        The black circles are dots, and act like ones in binary.  The&#xA;        white circles are blanks and act like zeros.  Every dot number&#xA;        begins and ends with a dot, and they're multiplied like regular&#xA;        binary numbers—with one exception.  See that first blank in the&#xA;        result?  That's there 'cause there's no regrouping when the&#xA;        intermediate products are added.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;BACK TO THE CONSPIRATORS&#xA;&#xA;                                     JULIAN&#xA;        With you so far.&#xA;&#xA;                                     NAOMI&#xA;        The lock's cryptosystem needs one dot number as its key, which&#xA;        it reads from a 10-meter punched tape that's fed in… here.  The&#xA;        key's supposed to be prime; it should only have lone dots and&#xA;        itself as factors.  If it's composite, any combination will get&#xA;        us in.&#xA;&#xA;                                     JULIAN&#xA;        So you want to hit some symbols on the tape with the duplicator?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;BACK TO THE GRAPH PAPER&#xA;&#xA;                                 NAOMI (O.S.)&#xA;        Right.  Like, say we hit the blank in this prime.&#xA;&#xA;She writes down ●●○● and then the same number with its blank duplicated:&#xA;●●○○●.&#xA;&#xA;                                 JULIAN (O.S.)&#xA;        Er, that's no good.  It's still prime.&#xA;&#xA;                                 NAOMI (O.S.)&#xA;        Ah, but the duplicator's power level can be set past 100%.  If&#xA;        we fired it at 200% or more, we'd only get that number for the&#xA;        first minute or so.  Since there'd still be a full charge's&#xA;        worth of duplication on the original blank, the number'd turn&#xA;        into this—&#xA;&#xA;She writes ●●○○○● = ●○●● ✕ ●●●.&#xA;&#xA;                              NAOMI (O.S., cont'd)&#xA;        —after the tape cooled down.  There'd be another duplication&#xA;        after another minute if we'd given it 300%, but we'd be in the&#xA;        safe by the time that happened.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;BACK TO THE CONSPIRATORS&#xA;&#xA;                                     JULIAN&#xA;        Not bad.  There's just one problem.&#xA;&#xA;                                     NAOMI&#xA;        Yeah?&#xA;&#xA;                                     JULIAN&#xA;        The tape is behind the safe's walls, where we can't see it.  We&#xA;        won't know what number is punched on it, and, while the material&#xA;        indicator will tell us when we're aiming at the tape, we won't&#xA;        know which symbol the duplicator's pointed at when we fire.&#xA;&#xA;Naomi leans back and looks smug.  She's already thought of that.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What's Naomi's plan?  If at most $n$ symbols can fit on a 10-meter tape, how many shots should the conspirators fire and at what power levels to guarantee that they will eventually get inside?&lt;/p&gt;&#xA;" OwnerUserId="13175" LastEditorUserId="13175" LastEditDate="2015-06-15T16:41:09.543" LastActivityDate="2015-07-27T03:27:04.047" Title="The Duplicator and the Safe" Tags="|mathematics|strategy|story|graph-theory|" AnswerCount="4" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="16422" PostTypeId="2" ParentId="16357" CreationDate="2015-06-15T16:58:08.097" Score="1" Body="&lt;p&gt;Considering in sequence, where N is the number of gunmen and A, B, C, etc. are individual representations of same:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trivially:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Where N = 1, all gunmen survive as none can shoot themselves.&lt;/li&gt;&#xA;&lt;li&gt;Where N = 2, both gunmen die as no other gunman can be further away.&lt;/li&gt;&#xA;&lt;li&gt;Where N = 3, one gunman must live, as:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Let x, y, and z be the distances A-&gt;B, B-&gt;C, C-&gt;A respectively.&lt;br&gt;&#xA;If x=y=z (a triangle or 3-sided polygon), then A, B, and C will all shoot the tallest other person. Since no two gunmen have the same height, one gunman will be shot twice and one will not be shot at all as there are only three bullets.&lt;br&gt;&#xA;If x&gt;y&gt;z or x&gt;y=z, then A will shoot B, while B and C will shoot each other. (Rotating the distance assignments incorporates all possible combinations of distances between the three gunmen.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extended:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Where N = 4 and for all other even numbers of gunmen, it is possible to trivially create a scenario where pairs of gunmen stand next to each other, yet further away from all others; in this scenario all gunmen die, as it is the instance of N = 2 repeated N/2 times.&lt;/li&gt;&#xA;&lt;li&gt;Where N = 5, again one gunman must live:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Let v, w, x, y, and z be the distances between A-B, B-C, etc. respectively (as with N=3).&lt;br&gt;&#xA;As with N = 3, if v &gt; w &gt; x &gt; y &gt; z, v &gt; w = x = y = z, or variants of equality where v &gt; (some combination of w,x,y,z), then A will live as no gunman can be closest to A by definition.&lt;br&gt;&#xA;If v=w=x=y=z (a 5-sided polygon), then everyone is closest to 2 other people and will shoot the tallest other person. Since no two gunmen have the same height, one must be the tallest, and that person will be shot by both people beside him; since the only way for all gunmen to die is for each gunmen to be shot only once, then someone must live.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This extends out for all odd numbers of gunmen, as the definition of an N-sided polygon is that all the sides (ie distance between gunmen, who are the vertices) be equal.  Thus, where N = 101, one gunman must survive.&lt;/p&gt;&#xA;" OwnerUserId="10625" LastEditorUserId="10625" LastEditDate="2015-06-15T17:03:58.243" LastActivityDate="2015-06-15T17:03:58.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16423" PostTypeId="1" AcceptedAnswerId="16450" CreationDate="2015-06-15T17:31:19.560" Score="17" ViewCount="1846" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;There are very few of me to see. &lt;br&gt;If there were none, there might&#xA;  (not) be glee.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Easy, 1 thing to rid of me. &lt;br&gt;I can come and go fast, to those&#xA;  readily!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;You might not notice, look carefully. &lt;br&gt;As you look over my family!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;You may end up changing what I might be, &lt;br&gt;As you influence us&#xA;  pretentiously!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;What is the thing that is being talked about?&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I will add some hints later is necessary - Let me know if this is too broad)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If you enjoyed this riddle, some of my other rhyming riddles can be found &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%5Brhyme%5D%20user%3A11244%20is%3Aquestion&quot;&gt;here&lt;/a&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-06-19T13:26:20.033" Title="Very few of me to see" Tags="|riddle|rhyme|" AnswerCount="7" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16424" PostTypeId="2" ParentId="16357" CreationDate="2015-06-15T18:40:33.353" Score="0" Body="&lt;p&gt;&lt;strong&gt;EDIT: Found out you cannot have multiple rounds, so not a valid answer.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can have multiple rounds of shooting, a counter example is trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the shortest shooter ($A$) out of the group.  Order the rest on the points of a 100 sided regular polygon in order of height.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Place $A$ at the centre of the polygon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everyone around the perimeter will shoot at the tallest of their two neighbours.  The shortest person on the polygon ($B$) is not targeted by anyone.  And neither is $A$.  The tallest person is targeted three times - once by $B$, once by the second tallest, and once by $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After all the tall people die, $A$ and $B$ shoot each other.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="1592" LastEditDate="2015-06-16T15:03:44.937" LastActivityDate="2015-06-16T15:03:44.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16425" PostTypeId="2" ParentId="16423" CreationDate="2015-06-15T18:49:01.337" Score="2" Body="&lt;p&gt;First thing I thought of is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Flower BUD&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are very few of me too see. &#xA;If there were none, there might (not) be glee.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the flower plant is not blooming it might be worrying to some but for most flower brings happiness or a Smile.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Easy, 1 thing to rid of me. &#xA;I can come and go fast, to those readily!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can break off the Step or pluck the BUD (Nipping in the bud). Before you are able to see the bud it might blossom to a pretty flower.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You might not notice, look carefully. &#xA;As you look over my family!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Looking at the blossomed flowers you might tend to overlook the 'budding' BUD&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You may end up changing what I might be, &#xA;As you influence us pretentiously!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You might take water the plan more or less and inadvertently cause harm to the BUD.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9892" LastActivityDate="2015-06-15T18:49:01.337" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16431" PostTypeId="1" AcceptedAnswerId="16956" CreationDate="2015-06-15T20:02:11.070" Score="14" ViewCount="3876" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zbw25.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;&lt;em&gt;Which symbol should replace the question mark in the above pattern?&lt;/em&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;To clarify a few things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You can solve this puzzle with a printed version of it; that is to say, there's no steganography or other means of encryption involved here.&lt;/li&gt;&#xA;&lt;li&gt;The square layout of the tiles is irrelevant. They go in a specific order, and this puzzle could just as well be solved with the tiles laid out in a straight line.&lt;/li&gt;&#xA;&lt;li&gt;There is no 'misleading' information in the puzzle. Every single piece of the puzzle is relevant to its solution.&lt;/li&gt;&#xA;&lt;li&gt;The pattern will be obvious once you discover it. If at any point it feels like a stretch, or you seem uncertain about anything, you're on the wrong track.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Try not to get too caught up on the word &lt;em&gt;pattern&lt;/em&gt;. Looking for patterns in the colors will most likely not get you to the right answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1  2  3&lt;br&gt;&#xA; 4  5  6&lt;br&gt;&#xA; 7  8  9  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This puzzle would work in black and white - the colors simply add a nuance.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="11475" LastEditDate="2015-06-24T18:53:12.633" LastActivityDate="2023-01-12T09:50:22.857" Title="Identify the next symbol in the pattern" Tags="|pattern|visual|" AnswerCount="7" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16433" PostTypeId="2" ParentId="12902" CreationDate="2015-06-15T20:10:22.780" Score="1" Body="&lt;p&gt;Here are examples of the shortest solutions arranged lexicographically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Resistant to inversion:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A AAAAAA,&#xA; B AAAAAB,&#xA; C AAAABA,&#xA; D AAAABB,&#xA; E AAABAA,&#xA; F AAABAB,&#xA; G AAABBA,&#xA; H AAABBB,&#xA; I AABAAA,&#xA; J AABAAB,&#xA; K AABABA,&#xA; L AABABB,&#xA; M AABBAA,&#xA; N AABBAB,&#xA; O AABBBA,&#xA; P AABBBB,&#xA; Q ABAAAA,&#xA; R ABAAAB,&#xA; S ABAABA,&#xA; T ABAABB,&#xA; U ABABAA,&#xA; V ABABAB,&#xA; W ABABBA,&#xA; X ABABBB,&#xA; Y ABBAAA,&#xA; Z ABBAAB&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Resistant to inversion and reversal:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A AAAAAAA,&#xA; B AAAAAAB,&#xA; C AAAAABA,&#xA; D AAAAABB,&#xA; E AAAABAA,&#xA; F AAAABAB,&#xA; G AAAABBA,&#xA; H AAAABBB,&#xA; I AAABAAA,&#xA; J AAABAAB,&#xA; K AAABABA,&#xA; L AAABABB,&#xA; M AAABBAA,&#xA; N AAABBAB,&#xA; O AAABBBA,&#xA; P AABAAAB,&#xA; Q AABAABA,&#xA; R AABAABB,&#xA; S AABABAA,&#xA; T AABABAB,&#xA; U AABABBA,&#xA; V AABBAAB,&#xA; W AABBABA,&#xA; X AABBBAA,&#xA; Y AABBBAB,&#xA; Z AABBBBA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Resistant to transposition and inversion:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A AAAAAB,&#xA; B AAAABA,&#xA; C AAAABB,&#xA; D AAABAA,&#xA; E AAABAB,&#xA; F AAABBA,&#xA; G AAABBB,&#xA; H AABAAA,&#xA; I AABAAB,&#xA; J AABABA,&#xA; K AABABB,&#xA; L AABBAA,&#xA; M AABBAB,&#xA; N AABBBA,&#xA; O AABBBB,&#xA; P ABAAAA,&#xA; Q ABAAAB,&#xA; R ABAABA,&#xA; S ABAABB,&#xA; T ABABAA,&#xA; U ABABAB,&#xA; V ABABBA,&#xA; W ABABBB,&#xA; X ABBAAA,&#xA; Y ABBAAB,&#xA; Z ABBABA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the shortest code possible as &lt;a href=&quot;https://puzzling.stackexchange.com/a/12907/303&quot;&gt;Mike Earnest&lt;/a&gt; already proved 6 is the shortest length of inversion resistant code.  It is trivially resistant to transposition as all codewords start with the same letter, but AAAAAA and BBBBBB aren't valid codewords.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Resistant to transposition, inversion and reversal:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A AAAAAAB,&#xA; B AAAAABA,&#xA; C AAAAABB,&#xA; D AAAABAA,&#xA; E AAAABAB,&#xA; F AAAABBA,&#xA; G AAAABBB,&#xA; H AAABAAA,&#xA; I AAABAAB,&#xA; J AAABABA,&#xA; K AAABABB,&#xA; L AAABBAA,&#xA; M AAABBAB,&#xA; N AAABBBA,&#xA; O AABAAAB,&#xA; P AABAABA,&#xA; Q AABAABB,&#xA; R AABABAA,&#xA; S AABABAB,&#xA; T AABABBA,&#xA; U AABBAAB,&#xA; V AABBABA,&#xA; W AABBBAA,&#xA; X AABBBAB,&#xA; Y AABBBBA,&#xA; Z ABAAAAB&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Similar to the previous section, &lt;a href=&quot;https://puzzling.stackexchange.com/a/12907/303&quot;&gt;Mike Earnest&lt;/a&gt; already proved 7 is the shortest length of inversion resistant, reversal resistant code.  It is also trivially transposition resistant as each word starts with an 'A'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;No valid codewords appearing in the transposed direction&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is indeed impossible.  We know that we need at least two codewords (in fact we need 26 codewords), and that all codewords must differ in at least one position.  Consequently we can use the two codewords to get 'A's and 'B's in that position to spell out anything we like in the transposed direction.&lt;/p&gt;&#xA;" OwnerUserId="303" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-06-15T20:10:22.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16435" PostTypeId="2" ParentId="16423" CreationDate="2015-06-15T20:34:02.303" Score="3" Body="&lt;p&gt;Could it be:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; opportunity?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are very few of me to see.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; opportunity doesn't come often  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If there were none, there might (not) be glee.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; depends on what kind of opportunity (a chance to promote vs a chance in gambling maybe?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Easy, 1 thing to rid of me.&lt;br&gt;&#xA;I can come and go fast, to those readily!   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Laziness, or unable to catch the opportunity, as they come and go fast&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You might not notice, look carefully.&lt;br&gt;&#xA;As you look over my family!   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There might be tons of opportunities and few are able to realize that.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You may end up changing what I might be,&lt;br&gt;&#xA;As you influence us pretentiously!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Opportunities are only for those prepared?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2015-06-15T20:34:02.303" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16437" PostTypeId="2" ParentId="16423" CreationDate="2015-06-15T20:43:54.347" Score="1" Body="&lt;p&gt;Perhaps:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Photons?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are very few of me to see. &#xA;If there were none, there might (not) be glee.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Photons are few and far apart and create light. Without them we literally could not &quot;see&quot; nor would be very gleeful.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Easy, 1 thing to rid of me. &#xA;I can come and go fast, to those readily!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Photons can be easily gotten rid of. Shadowing objects? Also they are very fast. (speed of light).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You might not notice, look carefully. &#xA;As you look over my family!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not entirely sure with this line. It had me thinking more along the lines of biology with the family comment. But you can see microscopic/atomic objects with a specialized computer. I.e. &quot;looking carefully&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You may end up changing what I might be, &#xA;As you influence us pretentiously!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is known that the act of observing photons and electrons can &quot;influence&quot; their behavior.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13208" LastActivityDate="2015-06-15T20:43:54.347" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16438" PostTypeId="2" ParentId="16423" CreationDate="2015-06-15T20:56:45.477" Score="2" Body="&lt;p&gt;Take one more shot:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Money&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are very few of me to see. &#xA;If there were none, there might (not) be glee.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are few bills/currencies that are physical. Money is often assumed to bring happiness and security. Instant &quot;glee&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Easy, 1 thing to rid of me. &#xA;I can come and go fast, to those readily!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can be lucky and gain money quickly through gambling, but it can go just as quickly by doing the one thing you can do with it: Spending it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You might not notice, look carefully. &#xA;As you look over my family!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You could easily look over smaller bills on the street or change on the side of the road and not give it any notice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You may end up changing what I might be, &#xA;As you influence us pretentiously!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; May things can change the value of a dollar. For instance: politics, greed. Also spending a bill can literally &quot;change&quot; it into change.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13208" LastEditorUserId="13208" LastEditDate="2015-06-15T21:36:04.393" LastActivityDate="2015-06-15T21:36:04.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16439" PostTypeId="2" ParentId="619" CreationDate="2015-06-15T21:28:52.117" Score="2" Body="&lt;p&gt;For 23, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(2 + 2)! - (2 - 2)!$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For 34, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(3! + 3!)/.3 - 3!$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For 56, we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5! \times .5 - 5 + 5/5$   or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ 5! \times .5 - 5 + (5 - 5)!$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For 100, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$9/.9 \times 9/.9$&lt;/p&gt;&#xA;" OwnerUserId="13203" LastEditorUserId="13203" LastEditDate="2015-06-16T05:23:51.510" LastActivityDate="2015-06-16T05:23:51.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16440" PostTypeId="2" ParentId="619" CreationDate="2015-06-15T22:38:55.133" Score="1" Body="&lt;pre&gt;&lt;code&gt;222   222      333   3  3     5555   555      9   99   99&#xA;   2     2        3  3  3     5     5        99  9  9 9  9&#xA; 22   222      333   3333     555   5555      9  9  9 9  9&#xA;2        2        3     3        5  5  5      9  9  9 9  9&#xA;2222  222   ,  333      3  ,  555    55   ,  999  99   99&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="11752" LastActivityDate="2015-06-15T22:38:55.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16441" PostTypeId="1" AcceptedAnswerId="16448" CreationDate="2015-06-15T22:47:45.693" Score="9" ViewCount="618" Body="&lt;h1&gt;Six spheres, six weighings.&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If you grow impatient with the narrative, you can skip to the TL;DR, below.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An engineer was walking across the street, on the way back from lunch, when &lt;em&gt;bam&lt;/em&gt;, hit by a bus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;She opened her eyes, to see a man standing before her, with pointy teeth, bloodshot eyes, and spiral horns.&#xA;The man held a 3-pronged spear in his left hand and 6 shiny spheres in his right.  He twitched his fingers and the spheres danced around one another, like David Bowie, but with more fingers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;What am I doing here?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Well, you died...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I figured that much out, but what am I doing &lt;em&gt;here&lt;/em&gt;?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Well, you led a mostly virtuous life, but you did have a habit of eating tunafish on Tuesdays...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Eating tuna on Tuesdays is forbidden?  No one ever told me that!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;They didn't?  How unfortunate for you.&quot; He grinned. &quot;Anyway, as I said, you led a mostly virtuous life, so you are being granted the opportunity to fight for your eternal soul.  Metaphorically, of course. I have a puzzle for you.  Solve it, and you can take the stairway up.  Fail, and go the other direction.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having worked many years as an engineer, she was used to ridiculous demands.  &quot;OK, what's the puzzle?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Notice that these spheres are different colors. Note that they also have different weights.  Each one has a unique weight of either 1001, 1002,&#xA;1003, 1004, 1005, or 1006 grams.  Over there by that door are six holes. To open the door, put the spheres in the holes in order, lightest to heaviest, from left to right. Then you may leave.  Put them in in any other&#xA;order, and the trapdoor below you opens.  You can guess where that leads.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;How am I supposed to differentiate such minute weight differences?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;With this.&quot;  The man planted the base of the spear into the dirt.  The three points morphed and dropped, turning into three pans. &quot;This is a &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15988/a-balance-with-three-pans-detecting-the-lightest-pan-find-the-one-lighter-ball&quot;&gt;three-pan balance&lt;/a&gt;.  Put things into the pans, set the whole thing spinning, and when it stops spinning, the lightest pan will rise.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He handed her the six spheres.  &quot;You will not be able to mark or alter the spheres in any way, but as you see, they are six different colors, and I assure you the colors won't change.  You may scratch notes in the dirt, if you like.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;She hefted the spheres. Sure enough, they all felt the same.  She estimated they were about 10cm in diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He raised a finger.  &quot;Oh, and I almost forgot to mention: you are only allowed a maximum of six weighings.  Experiment all you like, but once&#xA;you start the scale spinning, you are committed.  After the sixth weighing, the scale will revert back to a spear.  Also, don't bother trying to guess.  The spheres were arranged by my pet precognitive demon who divined exactly what steps you would take.  If you get to a point where the the answer is ambiguous, whatever guess you make will be wrong.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;She experimented putting spheres into the pans.  The pans were cups, shaped to match the bottom 1/3 of the spheres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When she tried balancing two spheres onto one pan, she couldn't get it to work.  &quot;Can I really only put one sphere per pan?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Correct.  If you attempt to put two or more spheres on a pan, you will not succeed.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Hmm.  What happens if there is no lightest pan?  What if the two lightest pans held the same weight?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The spheres are all different weights, so you can't really do that.  But if you somehow managed to, then no pan would rise.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;She did some thinking, scribbled in the dirt.  &quot;Am I under any time pressure?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;No, you literally have forever.  Take as long as you like.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;She did two weighings and noted the results.  While she was contemplating what to do next, she heard a &lt;em&gt;snip&lt;/em&gt; and felt something bounce off the back of her head.  She caught it as she turned around.  The man was sitting on a barrel,&#xA;using a pair of shears to trim his toenails.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Seriously?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He looked up.  &quot;You looked like you might be a while.  And these things do grow deucedly fast, so I figured, while I wait...&quot;  &lt;em&gt;snip&lt;/em&gt; ...  &lt;em&gt;catch&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;You realize the puzzle you've given me is impossible, right?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Oh really?  Surely you know that $6! &amp;lt; 3^6$, so it ought to be possible.  Or are you just giving up already?&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;That's not what I said...&quot; &#xA;&lt;em&gt;snip&lt;/em&gt; ... &lt;em&gt;catch&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;She looked down at the clippings she had caught.  They were each about 1cm across.  &quot;Say, these are all remarkably similar.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yeah, like I said, they grow fast.  After snipping each one literally millions of times, I've grown rather consistent. Each of my toenail clippings weighs exactly one gram.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Exactly?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;snip&lt;/em&gt; ... &lt;em&gt;catch&lt;/em&gt;  &quot;Yep.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;That's great.&quot;  She turned around and did a bunch more scribbling in the dirt.  &quot;Ah ha!&quot;  She then did the rest of her weighings, took the spheres over to the door, put them in the holes, and with heavenly chimes, the door opened.  Behind it was a staircase leading up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Well, that's a first,&quot; the man said.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;She peered up the staircase.  The top was lost in the distance.  &quot;How far is it?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;How far do you think, from the center of the Earth to the top of the sky?  Only about a hundred...million steps.  Have a nice climb!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the door slammed behind her, echoes of his cackling lingering on the air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TL;DR&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have six spheres, identical except for color and weight.  The weights are 1001, 1002, 1003, 1004, 1005, and 1006 grams.&#xA;By hand, you cannot possibly distinguish them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15988/a-balance-with-three-pans-detecting-the-lightest-pan-find-the-one-lighter-ball&quot;&gt;three-pan balance&lt;/a&gt;.  Put whatever you like in the pans, activate the scale, and if there is a unique lightest&#xA;pan, that pan will rise.  If there is no unique lightest pan, no pan will rise.  This scale differs from those in other puzzles in that you cannot put 2 or more spheres onto any one pan at a time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have six weighings to sort the spheres from lowest weight to highest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also have up to 10 toenail clippings, which each weigh exactly one gram.  These are not provided until after you've made your first two weighings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your plan needs to solve all possible combinations, with no ambiguity, and zero chance of needing to guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did she do it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus question: What is the minimum number of clippings needed?&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2016-12-09T12:29:46.087" Title="Six balls, six weighings" Tags="|weighing|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16442" PostTypeId="2" ParentId="16423" CreationDate="2015-06-15T22:49:01.993" Score="1" Body="&lt;p&gt;Perhaps:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; bitcoin?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are very few of me to see. If there were none, there might (not) be glee.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The fact that these are limited &amp;amp; hard to find &lt;em&gt;is&lt;/em&gt; their value. At the same time, many in the establishment would be happier if the technology weren't known. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Easy, 1 thing to rid of me. I can come and go fast, to those readily!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Easy to get rid of (spend). Also, transactions are cleared very quickly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You might not notice, look carefully. As you look over my family!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most people haven't looked into bitcoin in any detail. And, it may not be bitcoin but the family of currencies based on block-chain technology? IE, the algorithms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You may end up changing what I might be, As you influence us pretentiously!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Money has potential to purchase (ie change into) almost any thing. Also, money tends to breed pretentiousness. At the same time, the technology is somewhat undeveloped &amp;amp; it may not end up being used for currency. It may be used for something else. There is also some hubris to the bitcoin aspiration to compete with / replace national currencies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But, to be honest this last clue keeps me coming back to something that is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  biological in nature &amp;amp; too-often mutilated by humans to their own ends -- such as a virus or a morel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Maybe I am close at least? :) Fun puzzle!&lt;/p&gt;&#xA;" OwnerUserId="10428" LastActivityDate="2015-06-15T22:49:01.993" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16443" PostTypeId="1" AcceptedAnswerId="16556" CreationDate="2015-06-16T00:08:00.287" Score="6" ViewCount="1119" Body="&lt;p&gt;I offer breathing without breath&lt;br&gt;&#xA;and dancing up and down, round and round&lt;br&gt;&#xA;in (mostly) perfect time&lt;br&gt;&#xA;and, I promise,&lt;br&gt;&#xA;I'll never miss a step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though diving, flying or driving&lt;br&gt;&#xA;it's all upon my stairs&lt;br&gt;&#xA;from Kitty Hawk to Clermont&lt;br&gt;&#xA;moving from wherever, to wherever&lt;br&gt;&#xA;with whoever, doing whatever&lt;br&gt;&#xA;but if you want the truth&lt;br&gt;&#xA;nothing here has been anywhere&lt;br&gt;&#xA;(except perhaps, round the bend)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's no two ways about it&lt;br&gt;&#xA;my brother's never told a lie.&lt;br&gt;&#xA;But I'll deceive you any way I can,&lt;br&gt;&#xA;and if you stumble upon these steps&lt;br&gt;&#xA;you won't find me&lt;br&gt;&#xA;I never stop, I never rest&lt;br&gt;&#xA;and though it may appear that I have left&lt;br&gt;&#xA;nothing on these steps&lt;br&gt;&#xA;has ever felt the need to die.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;First Hint&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Here's half a way to solve &lt;br /&gt;&#xA; a clue, in fact, it's key &lt;br /&gt;&#xA; and a door is like a lens &lt;br /&gt;&#xA; altogether but unseen; &lt;br /&gt;&#xA; however long your eyes persist &lt;br /&gt;&#xA; I can certainly tell you this, &lt;br /&gt;&#xA; you'll never walk through &lt;br /&gt;&#xA; my locked doors &lt;br /&gt;&#xA; (after all, what is a door &lt;br /&gt;&#xA; when a door does not exist) &lt;br /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Second Hint&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To and from, and Inbetweens &lt;br /&gt;&#xA; cascade across my steps &lt;br /&gt;&#xA; and though there are no doors &lt;br /&gt;&#xA; no walls, no metal bars &lt;br /&gt;&#xA; there are cels to always &lt;br /&gt;&#xA; keep things kempt &lt;br /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10934" LastEditorUserId="10934" LastEditDate="2015-06-17T19:38:34.450" LastActivityDate="2015-06-17T20:22:18.180" Title="Invisible Staircase" Tags="|riddle|" AnswerCount="5" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16444" PostTypeId="1" AcceptedAnswerId="16445" CreationDate="2015-06-16T01:48:25.170" Score="9" ViewCount="1484" Body="&lt;p&gt;Tonight I went to my friend's for a movie night and he asked me what we should watch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I was playful, I handed him the following picture so he could guess the movie:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OsMJ1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He couldn't figure it out, so after a while I gave him the following hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tolerance is not important&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But he was still stuck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you help him find the movie?&lt;/p&gt;&#xA;" OwnerUserId="10400" LastEditorUserId="463" LastEditDate="2015-06-16T01:50:21.217" LastActivityDate="2015-06-16T10:41:51.487" Title="Tonight is movie night" Tags="|enigmatic-puzzle|knowledge|movies|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16445" PostTypeId="2" ParentId="16444" CreationDate="2015-06-16T02:07:22.803" Score="13" Body="&lt;p&gt;The movie is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Star Trek: First Contact&lt;/strong&gt; (1996)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The message is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Resistance is futile&lt;/strong&gt;. Although this phrase was used in earlier Star Trek episodes, it was made more popular as the tagline of the 1996 film Star Trek: First Contact by the actor Jeff Coopwood who was the &quot;Voice of the Borg&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The encoding uses:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;resistor codes&lt;/strong&gt; as shown in the following chart. Per the hint, the tolerance band is absent. The resistor codes are then converted to ASCII characters as shown below.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/upeRo.png&quot; alt=&quot;Resistor codes&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="10400" LastEditDate="2015-06-16T03:02:42.753" LastActivityDate="2015-06-16T03:02:42.753" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16446" PostTypeId="1" AcceptedAnswerId="16447" CreationDate="2015-06-16T03:21:39.203" Score="3" ViewCount="1099" Body="&lt;p&gt;&lt;strong&gt;Sam&lt;/strong&gt; is a computer expert. He works in a &lt;strong&gt;Security Company&lt;/strong&gt;. His son &lt;strong&gt;Tim&lt;/strong&gt; just turned thirteen and is new to &lt;strong&gt;C++ programming&lt;/strong&gt; and ciphertexts . He decides to write a &lt;strong&gt;C++ program&lt;/strong&gt; to make a cipher algorithm. His program has some limitations and can encrypt only one word at a time. He encrypts a message using his program and sends that message as an email to his Dad. Below is the coded message:-&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Hj. i(o pmc|msm sjpfi mptqmsm5 Wigq wjno ypw gfv bben hpoh2 Ao{ze~y3&lt;/strong&gt;&#xA;  &lt;strong&gt;cbp ypw bskqk mf spohxmouo oo ypwu wb{ bben hpoh2&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's the &lt;a href=&quot;https://cryptabyte.com/#9627b4be1c92aca7&quot; rel=&quot;noreferrer&quot; title=&quot;Password is iamgenius&quot;&gt;link&lt;/a&gt; to the program he wrote. Password is suinegmai (or something like that :P )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can you decrypt the message, find the encryption algorithm and if possible write a C++ Program similar to his?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE:-&lt;/strong&gt; For Non-Windows Users - &lt;a href=&quot;http://www.dosbox.com/download.php?main=1&quot; rel=&quot;noreferrer&quot;&gt;Download DosBox&lt;/a&gt; to run it. If you are not good with command line, then download a GUI frontend such as &lt;a href=&quot;http://www.classics-for-x.info/csx/dapplegrey/index.htm&quot; rel=&quot;noreferrer&quot;&gt;DappleGray&lt;/a&gt; for MacOSX.&lt;/p&gt;&#xA;" OwnerUserId="13016" LastEditorUserId="13016" LastEditDate="2015-06-16T03:51:05.330" LastActivityDate="2018-11-29T21:38:47.217" Title="Find the cipher algorithm and decode the message" Tags="|cipher|computer-puzzle|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16447" PostTypeId="2" ParentId="16446" CreationDate="2015-06-16T05:17:29.320" Score="5" Body="&lt;p&gt;The text decodes to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hi, i'm playing since morning. When will you get back home. Anyways, can you bring me something on your way back home.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;First, to get the encryption program, you have to use the link given, and the password &quot;iamgenius&quot; which is &quot;suinegmai&quot; backwards. Once you have the program, you can figure out that the encryption process is to take each character of the word and increment its ASCII value by the index it is at. The first character is incremented by 0, second incremented by 1, etc. For example, &quot;AAA&quot; goes to &quot;ABC&quot;. A + 0 = A, A + 1 = B, A + 2 = C. The way to decode this would be just decreasing the value of the character by its index value in the word.&lt;br&gt;&lt;br&gt;I wrote up a java program to do this for me (sorry I don't know C++):&#xA;&lt;pre&gt;&lt;code&gt;&#xA;import java.util.Scanner;&#xA;public class Main {&#xA;&#xA;    public static void main(String[] args) {&#xA;        Scanner s = new Scanner(System.in);&#xA;&#xA;        while (true) {&#xA;            System.out.print(&quot;Enter a sentence to decode, or nothing to exit:&quot;);&#xA;            String sentence = s.nextLine();&#xA;            String decoded = &quot;&quot;;&#xA;            if (sentence.length() == 0) { break; }&#xA;&#xA;            String[] words = sentence.split(&quot; &quot;);&#xA;&#xA;            for (String word : words) {&#xA;                decoded += decode(word) + &quot; &quot;;&#xA;            }&#xA;            System.out.println(&quot;This decoded is: &quot; + decoded);&#xA;&#xA;        }&#xA;    }&#xA;&#xA;    public static String decode(String word) {&#xA;        String decoded = &quot;&quot;;&#xA;&#xA;        for (int i = 0; i &amp;lt; word.length(); i++) {&#xA;            char c = word.charAt(i);&#xA;            c -= i;&#xA;            decoded += Character.toString(c);&#xA;        }&#xA;        return decoded;&#xA;    }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11727" LastEditorUserId="12109" LastEditDate="2015-06-16T13:14:55.553" LastActivityDate="2015-06-16T13:14:55.553" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16448" PostTypeId="2" ParentId="16441" CreationDate="2015-06-16T07:13:26.100" Score="6" Body="&lt;p&gt;This is a rather inelegant answer, produced by trying combinations until one splits the remaining search space into roughly equal parts. It requires all ten clippings, if they aren't reusable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, weigh three random spheres together. Call the one that rises A and the others B and C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weigh the other three spheres together. Call the one that rises D and the others E and F.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of these operations divides the total possibilities by 3, so there are 80 left to distinguish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now weigh B and C against D with three clippings. Call this weighing 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If B rises in weighing 3, there are 24 possibilities left. Weigh B with two clippings against E and F. (4a)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If B rises again in 4a, it is the 1002-gram sphere and A is 1001 grams. Weigh C, D, and E. Whichever of C and D rises is 1003, and the other one is 1004. Then weigh the 1004-gram sphere and two clippings with E and F; whichever rises is 1005 and the other is 1006.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If E rises in 4a, weigh B, D with two clippings, and E. (5a)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If B rises in 5a, it is 1003 grams, A is 1001 grams, D is 1002 grams, and E is 1004 grams. Weigh E with two clippings against C and F; whichever rises is 1005. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If E rises in 5a, it is 1003, B is 1004, D is 1002, and A is 1001. Weigh B with two clippings against C and F.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If no pan rises in 5a, D is 1001, B is 1003, E is 1004, and A is 1002. Weigh E against C and F.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If F rises in 4a, do the same thing but with E and F switched.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If no pan rises in 4a, weigh D and three clippings against E and F. (5b)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If D rises in 5b, it is 1001, A is 1002, B is 1003, and C is 1004. Weigh C and two clippings against E and F to determine which is 1005.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If E rises in 5b, it is 1004, A is 1001, B is 1002, and D is 1003. Weigh E and two clippings against C and F.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If F rises in 5b, do the same thing with E and F switched.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If no pan rises in 5b, D is 1002, A is 1001, B is 1003, and C is 1004. Weigh C and two clippings against E and F.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This identifies all possibilities where B rises in 3. If C rises in 3, the same is done with B and C reversed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If D rises in 3, it is 1001, and B and C are 1005 and 1006 in either order, leaving 12 cases. Weigh A, E, and F; whichever rises is 1002. Then weigh the other two against B; whichever rises is 1003 and the other is 1004. Finally, weigh B and C against the 1004-gram sphere with two clippings to identify 1005 and 1006.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the remaining 20 cases, no pan rises in weighing 3. Weigh A with one clipping against E and F. (4b)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If A rises in 4b, it is 1001 and D is 1002. E and F are 1003 and 1004 in either order. Weigh both against B to determine which is which, then weigh B and C against 1004 and two clippings to identify 1005 and 1006.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If E rises in 4b, it is 1002, A is 1003, and D is 1001. Weigh B, C, and F to find 1004, and use the last weighing to find 1005 and 1006.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If F rises in 4b, do the same with E and F reversed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If no pan rises in 4b, A is 1002 and D is 1001. Weigh B with a clipping, C, and E with a clipping. (5c)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If B rises in 5c, it is 1004, C is 1006, E is 1005, and F is 1003.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If C rises in 5c, it is 1004 and F is 1003. Use the last weighing to find 1005 and 1006.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If E rises in 5c, it is 1003 and B is 1004. Use the last weighing to find 1005 and 1006.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If no pan rises in 5c, there are three cases left. Distinguish them by weighing B, C with a clipping, and F with two clippings. (6)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If B rises in 6, it is 1004, C is 1005, F is 1003, and E is 1006.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If C rises in 6, it is 1004, B is 1006, F is 1005, and E is 1003.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If no pan rises in 6, B is 1005, C is 1004, F is 1006, and E is 1003.&lt;/p&gt;&#xA;" OwnerUserId="13217" LastActivityDate="2015-06-16T07:13:26.100" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16449" PostTypeId="2" ParentId="16446" CreationDate="2015-06-16T07:28:14.917" Score="2" Body="&lt;p&gt;Although &lt;a href=&quot;https://puzzling.stackexchange.com/users/11727/mistereman22&quot;&gt;MisterEman22&lt;/a&gt;'s &lt;a href=&quot;https://puzzling.stackexchange.com/a/16447/5610&quot;&gt;answer&lt;/a&gt; is completely right, here is a C++ version:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#include &amp;lt;iostream&amp;gt;&#xA;#include &amp;lt;string&amp;gt;&#xA;&#xA;bool is_whitespace(const char c){&#xA;    return c == ' ' || c == '\n' || c == '\t';&#xA;}&#xA;&#xA;std::size_t decode_word_inplace(std::string &amp;amp;sentence, std::size_t wordstart){&#xA;    for(std::size_t i = wordstart; i &amp;lt; sentence.size(); ++i){&#xA;        if(is_whitespace(sentence[i])){&#xA;            return i;&#xA;        }&#xA;        sentence[i] -= (i - wordstart);&#xA;    }&#xA;    return sentence.size() - 1;&#xA;}&#xA;&#xA;void decode_inplace(std::string &amp;amp;sentence){&#xA;    for(std::size_t i = decode_word_inplace(sentence, 0); i &amp;lt; sentence.size(); ++i){&#xA;        if(is_whitespace(sentence[i])){&#xA;            i = decode_word_inplace(sentence, i+1) - 1;&#xA;        }&#xA;    }&#xA;}&#xA;&#xA;int main(){&#xA;    std::string sentence;&#xA;    std::getline(std::cin, sentence);&#xA;    decode_inplace(sentence);&#xA;    std::cout &amp;lt;&amp;lt; sentence &amp;lt;&amp;lt; std::endl;&#xA;&#xA;    return 0;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-06-16T08:43:15.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16450" PostTypeId="2" ParentId="16423" CreationDate="2015-06-16T10:17:51.080" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Unanswered questions on Puzzling SE?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are very few of me to see.&lt;br&gt;&#xA;If there were none, there might (not) be glee.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Very few such questions.&lt;br&gt;&#xA;Contrary to other SEs where people want answers to their questions, on Puzzling SE one would not want the puzzles to be cracked easily. Hence, &quot;(not)&quot; glee.    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Easy, 1 thing to rid of me.&lt;br&gt;&#xA;I can come and go fast, to those readily!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can get rid of them by answering, and some of them are answered within moments by the active community    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You might not notice, look carefully.&lt;br&gt;&#xA;As you look over my family!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We might miss them looking at all other questions  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You may end up changing what I might be,&lt;br&gt;&#xA;As you influence us pretentiously!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; By answering a question, we change its state but the answer may not be right.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9669" LastEditorUserId="11244" LastEditDate="2015-06-16T11:12:05.787" LastActivityDate="2015-06-16T11:12:05.787" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16452" PostTypeId="2" ParentId="16423" CreationDate="2015-06-16T10:47:44.510" Score="1" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The concept of Zero (nothing)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are very few of me to see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can't see nothing, but you can see its denotion 0&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If there were none, there might (not) be glee.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Something about double negatives?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Easy, 1 thing to rid of me. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0+1 is no longer 0&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can come and go fast, to those readily!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0 can be added to the beginning of a number without making a difference&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You might not notice, look carefully. &#xA;As you look over my family!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0 is sometimes missed out when listing the 'family' of numbers (1-9)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You may end up changing what I might be, &#xA;As you influence us pretentiously!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0 at the end of a number, or after a decimal point changes the number&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13095" LastActivityDate="2015-06-16T10:47:44.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16455" PostTypeId="2" ParentId="15725" CreationDate="2015-06-16T12:09:28.567" Score="6" Body="&lt;p&gt;Just going to put some things that I've noticed here. Most of this is working with Fillet's comment about the answers for stanzas being in a different stanza.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The road third-least taken leads to you and I in flight.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Looking at the stanza with the third-fewest words gives us stanza 6, at 18 words, above stanzas 3 and 8. The only word with &quot;ui&quot; is &quot;quixotic,&quot; which fits the &quot;in flight&quot; both because it is in the first line and because the word refers to those with &quot;their heads in the clouds&quot; so to speak. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Head north 'til you see the reverse, then once more north to the burial site.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Haven't found a thing for this yet. I've looked for &quot;iu&quot;, words that are antonyms for quixotic, and phrases that are somehow palindromes for quixotic, but I've turned up nothing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;When five becomes square, it may lead to confusion;&lt;br&gt;&#xA;behead the basilisk, an ironic inclusion.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The new hint makes me think the &quot;5 becomes square&quot; is referring to the 5 on a digital clock, which could be interpreted as an &quot;S&quot;, leading to confusion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Additionally, Wikipedia says this on basilisks:&#xA;&quot;In European bestiaries and legends, a basilisk... is a legendary reptile reputed to be king of serpents and said to have the power to cause death with a single glance.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not seeing a connection yet, but I'm sure one will come at some point. &lt;/p&gt;&#xA;" OwnerUserId="10078" LastEditorDisplayName="user20" LastEditDate="2015-08-05T17:54:35.107" LastActivityDate="2015-08-05T17:54:35.107" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16456" PostTypeId="2" ParentId="16443" CreationDate="2015-06-16T12:13:49.473" Score="1" Body="&lt;p&gt;I think it is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pi or 22/7 (the division symbol indicating the step)   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but i am not sure, and only these lines make sense&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and dancing up and down, round and round &#xA;in (mostly) perfect time&#xA;and, I promise, &#xA;I'll never miss a step.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Different numbers in Pi&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though diving, flying or driving&#xA;it's all upon my stairs &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; pi contains sequences which can describe any thing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I never stop, I never rest,&lt;/p&gt;&#xA;" OwnerUserId="13080" LastActivityDate="2015-06-16T12:13:49.473" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16457" PostTypeId="2" ParentId="16431" CreationDate="2015-06-16T12:54:24.933" Score="4" Body="&lt;p&gt;An attempt of an answer is &lt;strong&gt;B&lt;/strong&gt; because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Splitting the pattern into 9 &quot;fields&quot;, each field consists of two &quot;complimentary&quot; colours. Red/Green Blue/Orange Purple/Yellow. Each of the fields is either a duplicate or the colour-inverse of another, with the central all-green tile being an exceptional piece. (As there are an odd number of tiles, one has to be a &quot;stand-alone&quot; item. Central and of one colour only seems to be an appropriate choice.) This leaves the tile which looks like the &quot;B&quot; tile to require a double or an inverse, and tile B is the only one fulfilling this requirement.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A simpler way of seeing it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All tiles on the left need a duplicate on the right. The central column is symmetric and otherwise just a distraction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastActivityDate="2015-06-16T12:54:24.933" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16460" PostTypeId="1" AcceptedAnswerId="16461" CreationDate="2015-06-16T14:42:44.827" Score="44" ViewCount="33260" Body="&lt;p&gt;One night, a man receives a call from the police. The police tell the man that his wife was robbed, murdered, and her body dumped in a remote location, and that he should reach the crime scene as soon as possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The man drops his phone in shock, and drives 20 minutes to the crime scene. Though the man does not have any evidence of committing the murder on his personal being (no blood or other signs of foul play on the car or any possessions) as soon as he reaches the crime scene, and before he can say or do anything, the police arrest him on suspicion of committing the murder (he will, of course be judged by a jury of his peers, etc.).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Without any sign of &lt;em&gt;typical&lt;/em&gt; evidence, why do the police think that he committed the crime?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt; It should be noted that this takes place in the &lt;em&gt;typical&lt;/em&gt; world, without aliens, super powers, or unlimited quantities of any material.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="69582" LastEditDate="2020-12-27T05:20:42.660" LastActivityDate="2020-12-27T05:20:42.660" Title="How did the police know?" Tags="|story|situation|mystery|" AnswerCount="9" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="16461" PostTypeId="2" ParentId="16460" CreationDate="2015-06-16T14:44:36.560" Score="86" Body="&lt;p&gt;I think &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They didn't tell him where the murder scene was but he could find it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10400" LastActivityDate="2015-06-16T14:44:36.560" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="16462" PostTypeId="2" ParentId="16404" CreationDate="2015-06-16T14:55:34.423" Score="0" Body="&lt;p&gt;A=2*(B+(2A-C)&lt;br&gt;&#xA;A=2B+4A-2C&lt;br&gt;&#xA;A=(2*(C-B))/3&lt;br&gt;&#xA;A=C*2/3-B*2/3&lt;br&gt;&#xA;Then: C&gt;B (because A&gt;0)&lt;br&gt;&#xA;&amp;amp; A&amp;lt;&lt;code&gt;C&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally: C&gt;B and C&gt;A&lt;/p&gt;&#xA;" OwnerUserId="13230" LastEditorUserId="463" LastEditDate="2015-06-16T15:34:40.837" LastActivityDate="2015-06-16T15:34:40.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16464" PostTypeId="2" ParentId="15725" CreationDate="2015-06-16T16:15:36.780" Score="14" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PUZZLING&lt;br&gt;&#xA;&lt;br&gt;&#xA; If we put together whatever we have:&lt;br&gt;&#xA;&#xA; &lt;code&gt;Pathogen&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;qUi&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;wiZardry&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;forZando&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;heirLoom&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;felicIty&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;airborNe&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; and read the diagonal it's pretty obvious&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As you approach the arbor, I wait in between.&lt;br&gt;&#xA;Clutch your keepsake as the first enemy's heard and seen.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Airborne&quot;: we insert &quot;i&quot; into &quot;arbor&quot;, then add &quot;e&quot; and &quot;n&quot;, first letters seen and heard, respectively, in the word &quot;enemy&quot;. Definition: &quot;in flight&quot;, stanza 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A quixotic elf and I stand at the gates -&lt;br&gt;&#xA;succumb to the virus, and the city awaits.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Felicity&quot;: quixotic &quot;elf&quot; is &quot;fel&quot;, then &quot;i&quot;, the &quot;city&quot;. Definition: &quot;glee&quot;, stanza 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2744" LastEditorDisplayName="user20" LastEditDate="2015-08-05T17:53:45.880" LastActivityDate="2015-08-05T17:53:45.880" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="16467" PostTypeId="2" ParentId="16169" CreationDate="2015-06-16T17:17:09.477" Score="4" Body="&lt;p&gt;I'm pretty sure the idea suggested by the &lt;a href=&quot;/questions/tagged/wordplay&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;wordplay&amp;#39;&quot; rel=&quot;tag&quot;&gt;wordplay&lt;/a&gt; tag is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; putting together parts of words to form new words, where &quot;on top of&quot; means before and &quot;low of&quot; means after.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I haven't managed to find exact words to fit this, but here are a few of the things I've tried, which hopefully will help someone else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If I am on top politically, I am right, but if I am on top of being not in violation, I am wrong.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Being not in violation&quot; could be &lt;strong&gt;legal&lt;/strong&gt; or &lt;strong&gt;obey&lt;/strong&gt;. We need to put some letters before this to get something that's wrong, so maybe &lt;strong&gt;ill&lt;/strong&gt; (giving illegal) or &lt;strong&gt;dis&lt;/strong&gt; (giving disobey)? I haven't come up with a good synonym of &quot;politically&quot;...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If I am above 8, I am wrong, but if I am low on being not in violation, I am forever.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Unfortunately &lt;strong&gt;illate&lt;/strong&gt; and &lt;strong&gt;disate&lt;/strong&gt; don't resemble any words I can think of, nor do &lt;strong&gt;legalill&lt;/strong&gt; or &lt;strong&gt;obadis&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I cannot be on top politically and above 8, but I can be on top and low of being not in violation.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, &lt;strong&gt;illegalill&lt;/strong&gt; and &lt;strong&gt;obadisobey&lt;/strong&gt; don't look like words.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More work needed, but I think I may have cracked the main concept of the puzzle!&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-06-16T17:17:09.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16468" PostTypeId="2" ParentId="16460" CreationDate="2015-06-16T18:08:07.863" Score="3" Body="&lt;p&gt;The man was actually a government agent and when he got the call that his own wife was dead, he was shocked to hear it. The man called his agency and told him about what happened. The agency gave him a folder with info about everything about the crime scene, and he was off. He drove down the road, using his GPS to find the place where the folder said to go and finally found it, getting out of the car only to find that he was accused of murder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Or if that's not enough creative thinking for you, then try this:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The man heard on the news that his wife was dead, and was about to go when the police called him. He didn't want to seem suspicious, as he knew the husband was normally accused of the murder in all those movies he watched, he waited patiently until the police told him to go. He waited for them to hang up and zoomed out the door, got in his car, and drove off. He was very hungry, and saw a Taco Bell, but knew he couldn't stop and had to get there quickly. He recalled where the crime scene was from the news, and used his GPS to find it. When he got there, he was surprised to see the police arresting him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;There, lateral thinking... Well... Creative thinking anyways. I'm starving. I'm gonna go get some Taco Bell.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="11002" LastActivityDate="2015-06-16T18:08:07.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16469" PostTypeId="2" ParentId="16460" CreationDate="2015-06-16T18:49:40.087" Score="33" Body="&lt;p&gt;From a lateral point of view, you've not actually stated that he was arrested for &lt;strong&gt;his wife's murder.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He could therefore have been arrested because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There was an outstanding arrest warrant for him &lt;em&gt;already&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4852" LastEditorUserId="4852" LastEditDate="2015-06-16T19:00:32.043" LastActivityDate="2015-06-16T19:00:32.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16470" PostTypeId="1" AcceptedAnswerId="16472" CreationDate="2015-06-16T18:59:49.410" Score="9" ViewCount="2576" Body="&lt;p&gt;This riddle might be a little easier to solve, but I hope you enjoy the rhymes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You might think you don't need me, that you can go without, &lt;br&gt;but I&#xA;  will come lurking, there will be no doubt.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;You may try to get rid of me, you ignorant fool! &lt;br&gt;I can bear gifts, I'm a powerful tool.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;You might not expect me, You might try to ignore,&#xA;  &lt;br&gt;But my power over you, it cannot be tore.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;You can try and put up a valiant fight, &lt;br&gt;But I always succeed, you&#xA;  strengthen my might!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;You can hide in the shadows, and creep in the night. &lt;br&gt;But I will&#xA;  find you, and show you my smite!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Some can't get enough, of what I can give,&#xA;  &lt;br&gt; Some struggle without, from the life that they live.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If you enjoyed this riddle, some of my other rhyming riddles can be found &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%5Brhyme%5D%20user%3A11244%20is%3Aquestion&quot;&gt;here&lt;/a&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-06-19T13:24:56.960" Title="You might think you don't need me" Tags="|riddle|rhyme|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16471" PostTypeId="2" ParentId="16470" CreationDate="2015-06-16T19:06:48.037" Score="2" Body="&lt;p&gt;Are you:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Money  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You might think you don't need me, that you can go without,&lt;br&gt;&#xA;but I will come lurking, there will be no doubt.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everyone needs money, even though some try to escape it. You need money to &quot;run away&quot;, and bills will always come.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You may try to get rid of me, you ignorant fool!&lt;br&gt;&#xA;I can bear gifts, I'm a powerful tool.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some try to spend their money to get rid of it. Money also buys almost anything, and can be a tool to accomplish much.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You might not expect me, You might try to ignore,&lt;br&gt;&#xA;But my power over you, it cannot be tore.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Little strained here... One might try to ignore money (money problems), but we are all subject to the power of money.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can try and put up a valiant fight,&lt;br&gt;&#xA;But I always succeed, you strengthen my might!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fighting against the influence of money can sometimes increase the amount of money (bribery). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can hide in the shadows, and creep in the night.&lt;br&gt;&#xA;But I will find you, and show you my smite!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Theives hide in the shadows to steal money, but money can buy security systems to catch them.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some can't get enough, of what I can give,&lt;br&gt;&#xA;Some struggle without, from the life that they live.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some are always trying to get more money, then the poor can't get enough money to live their lives.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-06-16T19:06:48.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16472" PostTypeId="2" ParentId="16470" CreationDate="2015-06-16T19:07:22.180" Score="14" Body="&lt;p&gt;Is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Sleep?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You might think you don't need me, that you can go without, &lt;br&gt;&#xA;but I will come lurking, there will be no doubt.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People can try to stay awake / avoid sleep but it'll always win out.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You may try to get rid of me, you ignorant fool! &lt;br&gt;&#xA;I can bear gifts, I'm a powerful tool.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People can drink coffee etc to stay awake but it's a powerful tool in that it heals and rejuvenates the mind.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You might not expect me, You might try to ignore,&lt;br&gt; &#xA;But my power over you, it cannot be tore.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can fall asleep when least expecting it - hopefully not while driving any heavy machinery..&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can try and put up a valiant fight, &lt;br&gt;&#xA;But I always succeed, you strengthen my might!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  The longer you go without sleeping the stronger the urge!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can hide in the shadows, and creep in the night. &lt;br&gt;&#xA;But I will find you, and show you my smite!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reference to sleeping in the dark / at night.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some can't get enough, of what I can give, &lt;br&gt;&#xA;Some struggle without, from the life that they live.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people love sleeping and can't get enough of it. Some people lead lives too busy to get much of it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12206" LastActivityDate="2015-06-16T19:07:22.180" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16473" PostTypeId="2" ParentId="16460" CreationDate="2015-06-16T19:08:52.247" Score="57" Body="&lt;p&gt;He was framed for the murder by popular musicians Sting, Andy Summers and Stewart Copeland.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You've stated that he was told about the murder by &lt;a href=&quot;https://en.wikipedia.org/wiki/The_Police&quot;&gt;the Police&lt;/a&gt; (note the capitals), and was arrested as he arrived at the scene of the crime by &lt;strong&gt;the police&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clearly they knew he would be coming somehow.&lt;/p&gt;&#xA;" OwnerUserId="4852" LastActivityDate="2015-06-16T19:08:52.247" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16474" PostTypeId="1" CreationDate="2015-06-16T19:49:36.323" Score="19" ViewCount="5828" Body="&lt;h3&gt;&lt;em&gt;&lt;a href=&quot;https://puzzling.meta.stackexchange.com/questions/3108/can-we-post-semi-interactive-puzzles?cb=1&quot;&gt;(semi-interactive puzzle)&lt;/a&gt;&lt;/em&gt;&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;BRIIING! BRIIING! BRIIING!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You wake up with a start. &lt;em&gt;It's &lt;code&gt;6:30&lt;/code&gt;... already!?&lt;/em&gt; Darn. You roll slowly out of bed, preparing for the day ahead. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whatever preparing you did then was &lt;em&gt;really&lt;/em&gt; inadequate to prepare you for what came next. Not even close!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After taking a shower and eating a surely-nutritious breakfast, you make your way to the door. As you extend your hand to the doorknob to open said door, you see a shadowy blur heading to the door. Recognizing that the object wouldn't stop kindly and ring the doorbell before entering your house (so rude!), you dive backwards; the door explodes into lots of pieces, and the object comes to rest a foot or two away from your feet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Who on this good earth would do that...?!&lt;/em&gt; You rush to the door to see if you could find the perpetrator; nothing! You proceed to pick up and examine the object, placing it on the nearest table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a large package! As much is obvious. No outer markings. You make to open the package; doing so reveals another package, and another until you have 10 such package wrappings. (A package in a package in a package...) At the end of this package chain, there is a small envelope.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Puzzle 1&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;***holes! Who the heck would use this much packaging to wrap a package?!!??!&lt;/em&gt; You think to yourself. An image of a vexed Al Gore flashes in your head before you open the envelope. It reads thus:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dear [name]:&lt;br&gt;&lt;code&gt;So sorry to wake you up so early.  At least, we apologize for you having to get up so early. Apologies are very important!!  We LOVE LOVE LOVE LOVE the length of our apologies--they are also very important!  You're important to us.  Unless you fail... heh.&lt;/code&gt;&lt;br&gt;You have been invited. In order to show your worth, we will not tell you where, but, rather, we will give you a puzzle.--&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At this point, you nod your head, forgetting your former frustration. &lt;em&gt;I am a great puzzle-solver!&lt;/em&gt; You continue reading.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;--To find this puzzle--your invitation--you must first and foremost solve the puzzle on the back of this sheet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You eagerly flip the sheet around. On it, is written thus:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A fisher is hungry for a large sum of fish, so he decides to fish for some fish. In his local pond, there are three types of fish: small fish, which weigh 5 units; medium fish, which weigh 25 units; and large fish, which weigh 125 units. The farmer catches 35 fish. Of the fish that the fisher caught, $m$ were small fish, $2m$ were medium fish, and $q$ were large fish. The combined amount of small fish and medium fish is $5m$ less than the $\blacksquare$ root of the amount of the large fish. How many large fish ($q$) did the fisher catch?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You sit down, perplexed. &lt;em&gt;This isn't a riddle! This is poppycock! There is not enough information, and--!&lt;/em&gt; You stare at the paper. That's when it hit you...almost. Something's there, you're not entirely sure what, however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, you ask yourself: &amp;ldquo;&lt;em&gt;&lt;strong&gt;What is the value of $\mathbf{q}$??&lt;/strong&gt;&lt;/em&gt;&amp;rdquo;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When the puzzle tells you something is important, you should listen to it. The puzzle is inclusive of the entire letter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The entire second line is the apology. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Apology &amp;amp; Hint 3&lt;/strong&gt;&lt;br&gt;&#xA;It seems that a crucial part has been tampered with, by the displaying of the, erm, apology in the second line.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Do now look at it again; perhaps you will understand...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;More story; to be read after solving Puzzle 1; (Answer in below story text; is therefore spoiler-marked.)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &amp;ldquo;$q=256$!&amp;rdquo; you shout out loud. &amp;ldquo;That took long enough! I don't even know what an apology is, anymore...&amp;rdquo;&lt;br&gt;&#xA; (Solver: @Chadi Abou Sleiman. *&lt;em&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/a/16592/7307&quot;&gt;Solution&lt;/a&gt;&lt;/em&gt;**.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You hear a high-pitched whistle and instinctively cover your ears. &lt;em&gt;Gah! What is that!?&lt;/em&gt; The whistle then ends, and outside, a man peers through your door intently. &amp;ldquo;Hello, there,&amp;rdquo; he says to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;ldquo;What the 'ell do you want?!&amp;rdquo; you practically shout.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;ldquo;It is not what I want that concerns me. As it seems, you have solved the puzzle.&amp;rdquo;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;ldquo;Some puzzle! Who would think to look in the apology for the net amount caught?!&amp;rdquo; At this point, you are letting out your vexation on the (poor) man.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;ldquo;We did make it plain that it was crucial to the puzzle's solution. That is beside the point, however. Now, I extend to you the opportunity to acquire an invitation.&amp;rdquo;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;ldquo;But the letter said&amp;mdash;&amp;rdquo;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;ldquo;I did not see the letter&amp;mdash;only the apology. Ergo, I do not heed else instruction. So, do you wish to acquire the invitation, or shall I put this pitiful abode to my back and forget about you?&amp;rdquo; The man looks at you expectantly, as if you really had no choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You sigh. &amp;ldquo;Alright, then. I'll go for it. What do I have to do?&amp;rdquo;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Puzzle 2&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&amp;ldquo;Here ya go,&amp;rdquo; the man says matter-of-factly to you, handing to you another sheaf of paper. On it is written thus:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The best of the best! Your home contains the answer. For in it, you'll find five children, children of a great number; well-rounded children. You'll take the best of the children (make sure you have five!) and add them up.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After reading the text, you look up to the man. He says, &amp;ldquo;Tell me the number when you're done. Do you have a seat I could use?&amp;rdquo;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;ldquo;Erm... yes. Please, come in. Take a seat&amp;mdash;any seat,&amp;rdquo; you say as you take your own seat, staring at the next in the series of cryptic puzzles. &lt;strong&gt;&lt;em&gt;What is the final value...?&lt;/em&gt;&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'Your home' refers to the home in which we all dwell... you, me, and al'thor... everyone you know here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(moar story)&lt;/em&gt;&#xA;Your visitor turns to you. &amp;ldquo;Times 'a tickin'!&amp;rdquo; he says to you, smiling vexatiously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;ldquo;This is timed!?&amp;rdquo; You shout. You hadn't known that there was a timer going!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;ldquo;Why, yes! You must solve it soon&amp;mdash;I'm rather tired... if you get my drift.&amp;rdquo; The visitor stands, bows mockingly, and continues to sit. &lt;em&gt;As if an annoying ##### like him deserves to sit down even on a porcupine!&lt;/em&gt; you think to yourself, but soon, your mind drifts elsewhere. &lt;em&gt;The solution... I don't get it&amp;mdash;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &amp;ldquo;&lt;strong&gt;420&lt;/strong&gt;!&amp;rdquo; you shout once again. You turn to your visitor, who stares at you, mouth agape.&#xA; (Solver: @rand al'thor. &lt;strong&gt;&lt;em&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/a/16884/7307&quot;&gt;Solution&lt;/a&gt;&lt;/em&gt;&lt;/strong&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The visitor fumbles with his words. &amp;ldquo;B-But you weren't supposed to...&amp;rdquo; He inhales deeply and regains composure. &amp;ldquo;Very well. It seems as if you will be receiving the invitation. So, here it is:&amp;rdquo; At this point, you receive another letter.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dear [name],&lt;br&gt;&#xA;  It was in the unlikely event that you should receive this letter; then again, if you completed the last puzzle, you must be &lt;em&gt;surrounded&lt;/em&gt; with people who might be good at this type of thing. This does tell me, however, that you are indeed worthy of attending.&lt;br&gt;&#xA;  This letter is an invitation, of which you must bring to the event. It will occur on [date not-so-long-away] at [obscure location in a wealthy neighbourhood]. I do expect your presence there,&lt;br&gt;&#xA;  truly yours,&lt;br&gt;&#xA;  Mr. $\Delta\Sigma\Upsilon$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&amp;ldquo;Huh!&amp;rdquo; You exclaim. &amp;ldquo;What a chance!&amp;rdquo;. Mr. $\Delta\Sigma\Upsilon$ is one of the wealthiest citizens in your country and is famous (or infamous, according to others) for his puzzle-books. He is a recluse, and hardly a soul has seen him, much less gotten to know him, save his closest associates. A thought briefly flashes through your mind before it slowly sinks into your subconscious. This thought was that ever since its foundation, the Puzzling SE has been loathed by Mr. $\Delta\Sigma\Upsilon$. Your visitor extends his hand to you, and you take and shake it. It is settled! You're going.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(Later, on aforementioned date at aforementioned location)&lt;/em&gt;&#xA;You are dressed in your best formal attire. You have a nice pair of shoes. You brushed your teeth. You combed your hair. You are one fashionable-lookin' person as you walk into the luxurious hall of the grand house. In it, you see a few others milling about, observing the hall. You also look around, trying to find a familiar face or two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the hall, you find no such face; only strangers of whom you have never seen. At once, a man steps into the hall and shouts, &amp;ldquo;The doors are now closed!&amp;rdquo; Upon him saying this, all the doors (and windows and any other chance of escape) closes, and the hall is effectively sealed. Some of the people in the hall attempt to pry open certain windows and doors, but to no avail. The rest of your remain in close attention to the new guest to the room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;ldquo;What is it that you want from us?&amp;rdquo; one of the guests growls. More shouts of complaint arise from the crowd. &amp;ldquo;Who the hell are you?&amp;rdquo; shouts another voice. The man raises his hands, and all fall silent in a few moments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;ldquo;I am Mr. $\Delta\Sigma\Upsilon$. You are here tonight because of your prowess and superiority to others in the realms of puzzle-solving. Now, I give you a final dichotomy, of which you must either accept or refuse very soon. That is this: you will either work for me, or you won't. Choose quickly.&amp;rdquo;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At once, a few of the voices speak up in &amp;ldquo;Yes&amp;rdquo;s confirmation. These people walk forward to Mr. $\Delta\Sigma\Upsilon$, who signals them behind him. You, however, are smarter than that and know of the terrible living conditions of which those who go to work for him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The person who first complained says no defiantly. Mr. $\Delta\Sigma\Upsilon$ glares at him. &amp;ldquo;You have sealed your fate.&amp;rdquo; Mr. $\Delta\Sigma\Upsilon$ then brandishes a pistol and fires two shots. There is one less person in the hall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instantly, a mass of people surges behind Mr. $\Delta\Sigma\Upsilon$ in allegiance. Only you and a few others remain. The remaining people, in a moment of indecision, fail to answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;ldquo;Then, I guess you few cannot decide. I will give you one day. Until then... sleep tight!&amp;rdquo; He and his new workers withdraw through the last remaining door which closes behind them, and all is dark and quiet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moments later, a dim bulb flickers amongst the visitors remaining. As it turns out, said a person had brought a flashlight (in case of an emergency), perfect for such an emergency as this. He introduces himself as Robert.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What seems like an eternity later, the door opens, and a sliver of light is let in, and in steps Mr. $\Delta\Sigma\Upsilon$. &amp;ldquo;Listen up. You must make your choice now. But, you being puzzlers (and persistent ones, at that!) are now offered a choice of freedom. You must solve the puzzle I give to you. But here's the catch: If you cannot solve it in one days time, you will be forced to work for me. Don't pretend that you can escape through law and justice&amp;mdash;money trumps both. Good luck.&amp;rdquo; Mr. $\Delta\Sigma\Upsilon$ again leaves the room, and what remains is your puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(TL;DR)&lt;/strong&gt; You've been captured. Solve the puzzle or else suffer a horrendous fate.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Puzzle 3&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;There is among the pile the following items:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A letter.&lt;/li&gt;&#xA;&lt;li&gt;A gun, with bullets sufficient for each one of the people remaining, of which there are 6. Each bullet is numbered, $1$ through $6$.&lt;/li&gt;&#xA;&lt;li&gt;A box, with a bullet-shaped hole. Is a little larger than the size of a bullet, and seems like it could carry one.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;On the letter is written:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Oh, let me tell you the tale!&lt;br&gt;&#xA;  'Tis a tale of woe.&lt;br&gt;&#xA;  Surely, you must fail!&lt;br&gt;&#xA;  Now, go go go g.o!  &lt;br&gt;&#xA;  Now that I know you, am like yo.u,&lt;br&gt;&#xA;  you never seem to help.&lt;br&gt;&#xA;  With you, it's like one step forward.s&lt;br&gt;&#xA;  and two steps bac.k.  &lt;br&gt;&#xA;  I climbed the stairs&lt;br&gt;&#xA;  until I stood at the top&lt;br&gt;&#xA;  and counted the stairs I too.k&lt;br&gt;&#xA;  and threw them at yo.u.  &lt;br&gt;&#xA;  You fell down to the ground,&lt;br&gt;&#xA;  I smiled with glee.&lt;br&gt;&#xA;  Then, to my surprise,&lt;br&gt;&#xA;  you pulled me down below yo.u!  &lt;br&gt;&#xA;  Then I remember the times,&lt;br&gt;&#xA;  when we danced around the squar.e,&lt;br&gt;&#xA;  I dropped two penny down the wel.l.&lt;br&gt;&#xA;  You dropped half of the pennies lef.t.  &lt;br&gt;&#xA;  Now there's half of me left on the floo.r,&lt;br&gt;&#xA;  The other was poured out for you.  &lt;br&gt;&#xA;  Now, take a bullet! Take one only.&lt;br&gt;&#xA;  Shoot me!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(This is the last part. Good luck!)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1a:&lt;/strong&gt;&#xA;Out of frustration, you turn the letter over. On it is written: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sulp 48! IT!!!!!!! I'm so backwards, haha!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&amp;ldquo;What on earth....?&amp;rdquo;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1b:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Backwards... Backwards... Oh...!&lt;/em&gt; Reversing the text of the back, you get: !!!!!!!TI !84 pluS, the name of a commonly-known calculator. &lt;em&gt;Then, perhaps, the solution will have to do with numbers...&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1c:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the end, you will need put a bullet inside the box-thingy. Since these bullets are numbered, you must figure out exactly which bullet to put in the box.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Certain lines in the poem are instructions. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;Epilogue&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;You hear a loud &lt;em&gt;KNOCK&lt;/em&gt; on the door. But this time, you are prepared. The door swings open at a tediously slow rate; your heart races. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You have the &lt;strong&gt;second&lt;/strong&gt; bullet in the bullet-shaped chamber, as per request of one of the puzzlers stuck in the room (Solver: &lt;a href=&quot;https://puzzling.stackexchange.com/users/12506/nefer007&quot;&gt;Nefer007&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Suddenly, the person who told you to choose the second bullet loads said bullet in the gun. &lt;em&gt;What are you doing?!&lt;/em&gt; You think. He aims the gun at the door, and...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;ldquo;Ah hah! I see you haven't solve the puzzle! I guess that means you must&amp;mdash;&amp;rdquo; You will never know what exactly Mr. $\Delta\Sigma\Upsilon$ was going to say, for at that moment the puzzler squeezes the trigger and &lt;em&gt;KA-BOOM!&lt;/em&gt; the bullet enters and exits the fiend, and with a solid &lt;em&gt;thump&lt;/em&gt;, he lays still on the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A moments silence ensues. Soon, the entire mansion erupts with happy cheering. With rapt glee, puzzlers and workers alike flow from the doors of the mansion. After a while, only a few puzzlers remain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These puzzlers introduce themselves to you. The first introduces themselves as &lt;a href=&quot;https://puzzling.stackexchange.com/users/12731/chadi-abou-sleiman&quot; title=&quot;Chadi Abou Sleiman&quot;&gt;Chadi Abou Sleiman&lt;/a&gt;, and shakes your hand. &amp;ldquo;I should get back home, I was working on a project for Android, you know how it is.&amp;rdquo; Smiling, the person departs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second introduces himself only as &lt;a href=&quot;https://puzzling.stackexchange.com/users/5373/rand-althor&quot;&gt;rand al'thor&lt;/a&gt;. &amp;ldquo;This was fun! I'm sort of sad that it's over, now...&amp;rdquo; After a pause, &amp;ldquo;Well, maybe not. See you around, maybe!&amp;rdquo; He then departs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last puzzler introduces himself as (first) the man who shot the fiend, and (second) as &lt;a href=&quot;https://puzzling.stackexchange.com/users/12506/nefer007&quot;&gt;Nefer&lt;/a&gt;. &amp;ldquo;He was going to kill us anyways. Or enslave us. Besides, he was a bastard, and all bastards gotta die.&amp;rdquo; He winks, and leaves, walking through the doors with confidence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then, it is just you in the empty hall. You hear a sink running in the mansion. Then, it dawns on you: &lt;em&gt;Who is to own this house...!&lt;/em&gt; You gather the bullets into your pocket, claim the second bullet from the late owner of the mansion, and head towards the nearest housing agency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LATER&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You sit at your new home: the beautiful mansion. You're loaded. You split all the cash you found in your house equally amongst the puzzlers who were there, and you're &lt;em&gt;still&lt;/em&gt; loaded. You found out that, if you had chosen the wrong bullet, it would have done nothing, is made of a flimsy pseudo-metal that would have been incinerated in the chamber of the gun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You sigh. &lt;em&gt;What to do...&lt;/em&gt; You take out your laptop, and log on to a familiar puzzling site. You click on a puzzle. &lt;em&gt;What fun!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This is a test run of the semi-interactive puzzle type. Make sure to read about semi-interactive puzzles in the link at the top of the question.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="7307" LastEditorUserId="45201" LastEditDate="2018-02-22T09:57:32.163" LastActivityDate="2023-07-04T19:59:46.323" Title="You're invited! Now what?" Tags="|calculation-puzzle|lateral-thinking|enigmatic-puzzle|semi-interactive-puzzle|" AnswerCount="16" CommentCount="41" ContentLicense="CC BY-SA 3.0" />
  <row Id="16476" PostTypeId="2" ParentId="16331" CreationDate="2015-06-16T20:18:58.650" Score="5" Body="&lt;p&gt;Based on Julian's solution, I came up with the following.  Mostly I just added in the repeat-as-necessary at the end to determine which calculated radius is the correct one.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Teleport to Random Location: Point 1.&lt;/li&gt;&#xA;&lt;li&gt;Summon Dale.&lt;/li&gt;&#xA;&lt;li&gt;Summon Bob.&lt;/li&gt;&#xA;&lt;li&gt;Teleport to Random Location: Point 2.&lt;/li&gt;&#xA;&lt;li&gt;Summon Arthur.&lt;/li&gt;&#xA;&lt;li&gt;Teleport to Bob's Location (which is also Dale's location).&lt;/li&gt;&#xA;&lt;li&gt;Pull Dale's tail.  This gives you the distance between points 1 and 2.&lt;/li&gt;&#xA;&lt;li&gt;Teleport to Arthur's location.&lt;/li&gt;&#xA;&lt;li&gt;Summon Bob.&lt;/li&gt;&#xA;&lt;li&gt;Teleport to Ransom Location: Point 3.&lt;/li&gt;&#xA;&lt;li&gt;Summon Arthur.&lt;/li&gt;&#xA;&lt;li&gt;Teleport to Bob's location.&lt;/li&gt;&#xA;&lt;li&gt;Teleport to Arthur's location.&lt;/li&gt;&#xA;&lt;li&gt;Pull Arthur's tail.  This gives you $\angle 13$.&lt;/li&gt;&#xA;&lt;li&gt;Teleport to Bob's location.&lt;/li&gt;&#xA;&lt;li&gt;Summon Dale.&lt;/li&gt;&#xA;&lt;li&gt;Teleport to Arthur's location.&lt;/li&gt;&#xA;&lt;li&gt;Pull Arthur's tail.  This gives you $\angle 23$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As indicated in Julian's answer, there is ambiguity as to whether&#xA;$$\angle 12 = 2\pi - (\angle 13 + \angle 23)$$&#xA;or&#xA;$$\angle 12 = (\angle 13 - \angle 23)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate the radius of the circle for both cases.  One is correct, the other is not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeat the entire process.  One calculated radius from each measurement should match one from the other.  That is the true radius.  There is a vanishingly small chance that the incorrect radii would also match.  Even if they do, repeat the whole process again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have the radius, calculating the area is straightforward.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2015-06-16T20:18:58.650" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16477" PostTypeId="1" AcceptedAnswerId="16534" CreationDate="2015-06-16T20:26:06.113" Score="17" ViewCount="1231" Body="&lt;p&gt;Once upon a time, a band of colonists, all having distinct birthdays, settled on a remote island.  They had brought some mementos from the old country, around which certain mores regarding friendships and birthdays eventually developed:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Making new friends is encouraged, but with care: to ever break off a friendship is unthinkably rude.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Friendships are mutual.  To call someone a friend when they would not say the same of you is unthinkably rude.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When someone has a birthday, each of their friends is expected to give them a birthday present of exactly one memento.  To not have a memento to give or to give more than one memento is unthinkably rude.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Mementos may be given to someone only on their birthday.  To give a memento on any other day is unthinkably rude.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;To give a friend the same memento that they gave you on your most recent birthday is tacky.  I.e., unthinkably rude.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;One fine stormy day, a castaway washes up on the shore with nothing but the clothes on his back and a small portrait of his family, which counts as a memento in the island reckoning.  He is brought to the leader who, after making sure that the sailor doesn't share his birthday with any islander, explains the local customs.  &quot;You should make at least one friend soon,&quot; she says.  &quot;To not have any friends would be looked on as unthinkably rude.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Are there many people here?&quot; he asks.  &quot;Many mementos?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yes, very many.&quot; she says.  &quot;There are $n$ islanders, all perfectly polite, and as many mementos.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The refugee hesitates.  He does not want to be rude, and he does not want to put someone else in a tight spot.  Can he safely make a friend, or is a scandal inevitable?&lt;/p&gt;&#xA;" OwnerUserId="13175" LastEditorUserId="13175" LastEditDate="2015-06-16T20:52:23.797" LastActivityDate="2016-01-04T17:01:56.133" Title="The Islanders and their Birthday Presents" Tags="|mathematics|story|graph-theory|" AnswerCount="6" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16478" PostTypeId="2" ParentId="16460" CreationDate="2015-06-16T20:33:59.690" Score="12" Body="&lt;p&gt;Non-serious answers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. His wife, with her dying breath, told an bystander that her husband did it. The policemen had a phone number but not an address.&lt;br&gt;&#xA; 2. The police got a description of the car, and would have arrested anyone who pulled up to the crime scene in that model of car.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11410" LastActivityDate="2015-06-16T20:33:59.690" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16479" PostTypeId="2" ParentId="16443" CreationDate="2015-06-16T20:37:11.910" Score="1" Body="&lt;p&gt;Are you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a dream?  like an invisible staircase, a dream carries us, invisibly, to a different place&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;nothing here has been anywhere&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; we may dream of many places, but we are not actually there&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;breathing without breath&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a dream is not real, but seems like real life while we experience it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;diving, flying or driving -- moving from wherever, to wherever&#xA;with whoever, doing whatever&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; movement in a dream is completely unrestrained; we can go anywhere by any means&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;my brother's never told a lie&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; sleep and death are often depicted as brothers; while dreams can seem like real life, death is unambiguous&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll deceive you any way I can&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; dreams can be confusing and enigmatic; people and things in dreams morph and change places&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;nothing on these steps has ever felt the need to die.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; we often dream of our own death, which (happily) has not actually happened&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;altogether but unseen however long your eyes persist &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; we &quot;see&quot; many things, vividly, in dreams, but we are not actually seeing through our eyes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4385" LastActivityDate="2015-06-16T20:37:11.910" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16480" PostTypeId="2" ParentId="16474" CreationDate="2015-06-16T20:37:43.927" Score="6" Body="&lt;h2&gt;Puzzle 1&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Assume that the message isn't saying &quot;Square root&quot; but rather &quot;[REDACTED] root&quot;. The equation we're balancing goes from:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;m+2m+q=35 and m+2m=5m-sqrt(q)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;m+2m+q=35 and m+2m=5m-q^(1/k)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we are allowed to choose k ourselves, and if k is non-integer, we can balance the equation for q=32,29,26,23,20,17,14,11,8,5 and 2. However, balancing the equation for those values of q produces an integer k in exactly one case: at q = 32, we get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;m+2m=5m-q^(1/k)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;1+2=5-32^(1/5)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;1+2=5-2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore the fisherman caught 32 large fish.&lt;/p&gt;&#xA;" OwnerUserId="7459" LastEditorUserId="5373" LastEditDate="2015-06-23T20:40:14.410" LastActivityDate="2015-06-23T20:40:14.410" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16484" PostTypeId="2" ParentId="16477" CreationDate="2015-06-16T21:12:44.170" Score="-4" Body="&lt;p&gt;Yes, he can kill the leader, take his place, and all of the leader's friends. For some bizarre reason, this is not unthinkably rude in the strange and unusual customs of this island.&lt;br /&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;alternatively, there are n islanders and each one is friends with exactly one other islander. Adding another person to the system is easy because the new person can simply trade mementos with their new friend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a third alternative:&#xA;He simply chooses to befriend someone whose birthday was yesterday, and he will be fine up until almost two years from now when he might run out of mementos. (unless he makes new friends at the rate of slightly over 1 per year)&lt;/p&gt;&#xA;" OwnerUserId="13068" LastEditorUserId="13068" LastEditDate="2015-06-16T21:18:22.460" LastActivityDate="2015-06-16T21:18:22.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16485" PostTypeId="1" AcceptedAnswerId="16503" CreationDate="2015-06-16T21:23:21.640" Score="33" ViewCount="2434" Body="&lt;p&gt;Now that you are familiar with &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16219/what-is-a-cyclone-phrase%E2%84%A2&quot;&gt;Cyclone Phrases&lt;/a&gt;, let's look at another kind of phrase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a phrase adheres to a certain rule, then I call it a Scalable Phrase™.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the examples below to find the rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qvJgo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I added the visual tag as a hint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you liked this puzzle, try others like it:&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/16219/what-is-a-cyclone-phrase%E2%84%A2&quot;&gt;What is a Cyclone Phrase™?&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/16743/what-is-a-triad-phrase%E2%84%A2&quot;&gt;What is a Triad Phrase™?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2019-02-06T08:14:33.487" Title="What is a Scalable Phrase™?" Tags="|pattern|visual|english|word-property|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16488" PostTypeId="2" ParentId="16470" CreationDate="2015-06-16T22:47:05.130" Score="0" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ageing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You might think you don't need me, that you can go without, &#xA;but I will come lurking, there will be no doubt.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Most dont like aging, and rather dont.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You may try to get rid of me, you ignorant fool! &#xA;I can bear gifts, I'm a powerful tool.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some use botox to hide it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You might not expect me, You might try to ignore, &#xA;But my power over you, it cannot be tore.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most dont want to admit they get older.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can try and put up a valiant fight, &#xA;But I always succeed, you strengthen my might!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can use botox but it's impossible to keep hiding that your aging.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can hide in the shadows, and creep in the night. &#xA;But I will find you, and show you my smite!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; your sking ages faster in the sun, you can hide in the dark but eventualy your skin will age.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9543" LastEditorDisplayName="user9543" LastEditDate="2015-06-17T14:57:28.133" LastActivityDate="2015-06-17T14:57:28.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16489" PostTypeId="2" ParentId="16485" CreationDate="2015-06-16T22:54:02.190" Score="2" Body="&lt;p&gt;It must be relevant that when&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; we replace each letter by its corresponding number in the alphabet (A=1, B=2, etc.),&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;we find that in each of &quot;MALE AND FEMALE&quot; and &quot;DREADED DREAM&quot;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the phrase splits into two halves each of which has letters summing to the same total.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explicitly, in &quot;MALE AND FEMALE&quot; we have&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 13+1+12+5+1+14 = 46 = 4+6+5+13+1+12+5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and in &quot;DREADED DREAM&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4+18+5+1+4+5+4 = 41 = 4+18+5+1+13.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The exact same statement doesn't hold for &lt;em&gt;all&lt;/em&gt; the example phrases given, but perhaps something along the same lines does. I tried the possibility of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; replacing &quot;two halves&quot; by &quot;$n$ $n$ths for some $n$&quot; (i.e. we might be able to split some of the phrases into three thirds or four quarters all with equal letter-sums).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That didn't work, but maybe something else...&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-06-16T22:54:02.190" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16490" PostTypeId="2" ParentId="16470" CreationDate="2015-06-16T23:01:48.613" Score="1" Body="&lt;p&gt;Maybe it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TIME.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You might think you don't need me, that you can go without,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people try to get too much done in way too little time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;but I will come lurking, there will be no doubt.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Time is always ticking away.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You may try to get rid of me, you ignorant fool!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can kill time, waste time, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I can bear gifts, I'm a powerful tool.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; With enough time, you can do almost anything.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You might not expect me, You might try to ignore,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you don't want to be too stressed, you might try to ignore the ticking clock.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But my power over you, it cannot be tore.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Same as the second line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You can try and put up a valiant fight,&lt;br&gt;&#xA;But I always succeed, you strengthen my might!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Working against the clock only makes the time race by faster.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You can hide in the shadows, and creep in the night.&lt;br&gt;&#xA;But I will find you, and show you my smite!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Smite&quot; is an anagram of &quot;times&quot;. Your time is always ticking away, no matter where you hide.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Some can't get enough, of what I can give,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There's not enough time to do everything I want to!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Some struggle without, from the life that they live.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can still struggle on in life with no spare time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-06-16T23:01:48.613" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16491" PostTypeId="2" ParentId="15725" CreationDate="2015-06-17T00:17:18.943" Score="35" Body="&lt;p&gt;So, standing on the shoulders of giants, and adding in the answers to two remaining stanzas, we get the final complete solution (as &lt;a href=&quot;https://puzzling.stackexchange.com/a/16464/5208&quot;&gt;noted by Quassnoi&lt;/a&gt;), to discover that life is indeed:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;PUZZLING&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Looking at the riddle, it's fairly obvious that the clues are cryptic, but there's a little more to it than that. As hinted, the final stanza tells us &lt;em&gt;how&lt;/em&gt; to approach the puzzle beyond these cryptic clues.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Between the lines you'll find your answer, definitively twisted;&lt;br&gt;&#xA;  yet in the shadows aligned so clearly you'll be surprised you missed it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;definitively twisted&lt;/em&gt; part hints to us that the stanzas not only contain a word discovered through cryptic clues, but also the definition of the answer to one of the other stanzas. So as we solve cryptic clues, we can validate these answers by looking for words/phrases elsewhere to find synonyms/definitions thereof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;in the shadows aligned&lt;/em&gt; part of the clue will become apparent at the very end...&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, stanza by stanza we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Stanza 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To start, stay within the pathway of general glee,&lt;br&gt;&#xA;  though stay clear of the law faery, who flits about wildly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Giving us:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;PATHOGEN&lt;/strong&gt;, by staying within PATHway Of GENeral, avoiding each of the letters from &quot;law faery&quot; (once each), &lt;a href=&quot;https://puzzling.stackexchange.com/a/16250/5208&quot;&gt;as shown by Pete&lt;/a&gt;. This is of course synonymous with virus, from stanza 6.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Stanza 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The road third-least taken leads to you and I in flight.&lt;br&gt;&#xA;  Head north 'til you see the reverse, then once more north to the burial site.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Giving us:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;QUINCUNX&lt;/strong&gt;, by way of Q (the third least common letter for a word to start with), U &amp;amp; I (in flight), then N(orth) CU (U C reversed), N(orth), X (marks the spot). The definition of which (&quot;an arrangement of five objects with four at the corners of a square or rectangle and the fifth at its center&quot;) can be found in the first line of stanza 3.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Stanza 3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When five becomes square, it may lead to confusion;&lt;br&gt;&#xA;  behead the basilisk, an ironic inclusion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Giving us:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;WIZARDRY&lt;/strong&gt; (again &lt;a href=&quot;https://puzzling.stackexchange.com/a/16250/5208&quot;&gt;as from Pete&lt;/a&gt;), from a beheaded basilisk (aka lizard) to get IZARD, included in WRY (irony). The synonym, sorcery, is found in stanza 4.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Stanza 4:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Before two sharp curves and the ring of sorcery,&lt;br&gt;&#xA;  stand still, close your eyes, and count aloud past three.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Giving us:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;FORZANDO&lt;/strong&gt;, once more &lt;a href=&quot;https://puzzling.stackexchange.com/a/16250/5208&quot;&gt;found by Pete&lt;/a&gt;, based on FOR (what you hear when you count past three), Z (two sharp curves) AND O (a ring). The definition of forzando in music is to play forcefully, as seen in stanza 8.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Stanza 5:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the air - do you hear it? The water baby screams.&lt;br&gt;&#xA;  Let our own misgivings start and end in our dreams.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Giving us:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;HEIRLOOM&lt;/strong&gt;, &lt;em&gt;again&lt;/em&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/a/16250/5208&quot;&gt;from Pete&lt;/a&gt;, because HEIR (sounds like air) plus the start letters of L(et) O(ur) O(wn) M(isgivings) at the end. With the synonym being keepsake from stanza 7.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Stanza 6:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A quixotic elf and I stand at the gates -&lt;br&gt;&#xA;  succumb to the virus, and the city awaits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Giving us:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;FELICITY&lt;/strong&gt;, as &lt;a href=&quot;https://puzzling.stackexchange.com/a/16464/5208&quot;&gt;solved by Quassnoi&lt;/a&gt;, derived from FEL (anagram of elf), I, and then CITY. Felicity means happiness, or glee, as seen in stanza 1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Stanza 7:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As you approach the arbor, I wait in between.&lt;br&gt;&#xA;  Clutch your keepsake as the first enemy's heard and seen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Giving us:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;AIRBORNE&lt;/strong&gt;, again &lt;a href=&quot;https://puzzling.stackexchange.com/a/16464/5208&quot;&gt;from Quassnoi&lt;/a&gt;, since we put I into ARBOR (airbor), and add N (first letter &quot;heard&quot; in enemy), and E (first letter &quot;seen&quot; in enemy). The definition, &quot;in flight&quot; is found in stanza 2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Stanza 8:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Will being stuck in all this goo truly help you see?&lt;br&gt;&#xA;  Primarily, primarily; whip around, forcefully.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Giving us:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;POLLIWOG&lt;/strong&gt;, by whipping around (reversing), the letters from WILL stuck in the letters of GOO (i.e. gowillo reversed), but starting with P (start of primarily). A polliwog is a tadpole, or a water baby from stanza 5.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Bringing all these words together, we end up with eight eight letter words:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;P&lt;/strong&gt;athogen&lt;br&gt;&#xA; q&lt;strong&gt;U&lt;/strong&gt;incunx&lt;br&gt;&#xA; wi&lt;strong&gt;Z&lt;/strong&gt;ardry&lt;br&gt;&#xA; for&lt;strong&gt;Z&lt;/strong&gt;ando&lt;br&gt;&#xA; heir&lt;strong&gt;L&lt;/strong&gt;oom&lt;br&gt;&#xA; felic&lt;strong&gt;I&lt;/strong&gt;ty&lt;br&gt;&#xA; airbor&lt;strong&gt;N&lt;/strong&gt;e&lt;br&gt;&#xA; polliwo&lt;strong&gt;G&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And finally, coming back to the &quot;&lt;em&gt;in the shadows aligned&lt;/em&gt;&quot; hint, we see aligned down the diagonal, the final solution.&lt;/p&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.923" LastActivityDate="2015-06-17T00:27:22.930" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16492" PostTypeId="2" ParentId="16485" CreationDate="2015-06-17T01:02:37.370" Score="2" Body="&lt;p&gt;All of the letters in a scalable phrase&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; have a straight &lt;a href=&quot;https://en.wikipedia.org/wiki/Typeface_anatomy&quot; rel=&quot;nofollow&quot;&gt;stem&lt;/a&gt;:&#xA; ABDEFHIKLMNPR (used in the clues). Some additional straight-stemmed letters appear in the counter-examples, but except for W they are all used in conjunction with a round letter: TVWZ. The stemmed letter X is not used in the clues or counterexamples.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="338" LastEditorUserId="338" LastEditDate="2015-06-17T01:41:46.270" LastActivityDate="2015-06-17T01:41:46.270" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16493" PostTypeId="5" CreationDate="2015-06-17T01:29:53.857" Score="0" Body="&lt;p&gt;&lt;strong&gt;DO NOT POST PUZZLES OF THIS TYPE.&lt;/strong&gt; This tag is kept for historical reasons, and semi-interactive puzzles should not be considered good questions for this site.&lt;/p&gt;&#xA;&lt;p&gt;For more information on semi-interactive puzzles and discussion about their validity, please visit &lt;a href=&quot;https://puzzling.meta.stackexchange.com/questions/3108/can-we-post-semi-interactive-puzzles?cb=1&quot;&gt;this meta question&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;A rough definition by Aify:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A semi-interactive puzzle is one in which the OP may not post all of the puzzle in the question at the beginning - depending on the answers/comments received from the community, the puzzle progresses and more is added to the question to reveal different clues/locks/messages etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user20" LastEditorUserId="60644" LastEditDate="2020-08-30T06:31:02.683" LastActivityDate="2020-08-30T06:31:02.683" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="16494" PostTypeId="4" CreationDate="2015-06-17T01:29:53.857" Score="0" Body="DO NOT POST PUZZLES OF THIS TYPE. This tag is kept for historical reasons, and semi-interactive puzzles should not be considered good questions for this site. Semi-interactive puzzles are those where new parts are added to the puzzle as each part is solved. " OwnerDisplayName="user20" LastEditorDisplayName="user20" LastEditDate="2016-08-10T23:58:58.583" LastActivityDate="2016-08-10T23:58:58.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16495" PostTypeId="2" ParentId="16485" CreationDate="2015-06-17T01:55:55.223" Score="8" Body="&lt;p&gt;I notice that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; all the letters in scalable phrases touch the bottom left corner of their bounding box &amp;mdash; that if you were to place them in the first quadrant of the Cartesian plane with the bottom left corner touching the origin, you could &quot;scale&quot; them up without changing their anchor.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All the non-scalable phrases &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;contain at least one letter that doesn't follow that rule.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="46903" LastEditDate="2019-02-06T08:14:33.487" LastActivityDate="2019-02-06T08:14:33.487" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="16496" PostTypeId="2" ParentId="16460" CreationDate="2015-06-17T03:03:25.313" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  The obvious answer is that the police did not tell him where the wife was murdered.  &lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After further consideration, I came up with a few&#xA;    alternative possibilities as to how he could have known where the&#xA;    crime scene was.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;     1. The husband had witnessed the crime. Perhaps he left the location to call for help. He may have been afraid to return to the scene of the crime before the police had arrived.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;     2. The husband tracked the phone. The wife likely had her cell phone with her, so he may have used the GPS in his wife's phone to locate her(&quot;Find my iPhone&quot; or something of the sort). Maybe he traced the police phone call.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;     3. Someone told him where the crime occurred. While it never stated that the police told him where it occurred, it never said that they DIDN'T tell him where it occurred. Or maybe he happened to drive past it on his way to the police station. Maybe he happened to guess where it occurred, perhaps she was murdered at their home or, if she had a job, the wife's workplace.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;     4. The man heard or saw the sirens. He probably would have needed to live in a rural area for this to be plausible. Granted, it took him 20 minutes to get there, so perhaps he is a &lt;em&gt;very&lt;/em&gt; slow driver, or his car broke down.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;     5. The man wasn't driving his own car. It never actually stated he was driving a car. Maybe, &lt;a href=&quot;https://www.google.com/#q=death+by+go-kart&quot; rel=&quot;nofollow&quot;&gt;Death by Go-Kart&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13244" LastEditorUserId="13244" LastEditDate="2015-06-19T15:01:00.570" LastActivityDate="2015-06-19T15:01:00.570" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16499" PostTypeId="2" ParentId="16431" CreationDate="2015-06-17T05:14:12.160" Score="1" Body="&lt;p&gt;If we read the squares in order as &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;147&#xA;258&#xA;369&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then the patterns go (with bold as the start of a repitition and italic as the inferred next in series)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Green/Red: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; green+red, skip 3, all green, skip 1, &lt;strong&gt;green + red&lt;/strong&gt;, &lt;em&gt;skip 3, all green, ...&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Purple/Yellow:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not really much information on this one, but skip at least 6&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Orange/Blue:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Orange left + blue right, orange bottom + blue top, skip 1, orange top + blue bottom, skip 1, &lt;strong&gt;orange left + blue right&lt;/strong&gt;, &lt;em&gt;orange bottom + blue top, skip 1, ...&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore I think the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13197" LastActivityDate="2015-06-17T05:14:12.160" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16500" PostTypeId="2" ParentId="16357" CreationDate="2015-06-17T05:17:30.980" Score="0" Body="&lt;p&gt;Assume one needs to prove that 'at least' one man will be left standing? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a scenario in which more than 1 man will be left is as follows&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7 gun men standing in formation of a regular hexagon with the tallest one at the center. At noon, everyone at the node will fire at the guy in the center. The guy in the center will fire at the 2nd tallest guy in the group. 5 men standing in the end. (assuming no multiple fires)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We can break the group of gunmen in two groups:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;1) Gunmen not standing equidistant with others- they will die in pairs, ie shoot each other.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;2) A cluster(s) of gunmen equally close: the tallest one gets killed, and the tallest one kills one more (the 2nd tallest from amgst the ones equidistant from him) i.e. gunmen from this group also die in pairs.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;So long as the total number of gunmen is an odd number, there will be one or more left standing after the shootout.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13247" LastActivityDate="2015-06-17T05:17:30.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16501" PostTypeId="1" AcceptedAnswerId="16546" CreationDate="2015-06-17T05:51:00.880" Score="11" ViewCount="1598" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;I'm a jack of all uses, durable and strong,&lt;br /&gt;&#xA;but useful in childish construction too.&lt;/p&gt;&#xA;&lt;p&gt;Desired alike by man- and women-kind;&lt;br /&gt;&#xA;wanted by most all, obtained but by few.&lt;/p&gt;&#xA;&lt;p&gt;When shortened, I'm an impartial judge of prisoners in math;&lt;br /&gt;&#xA;Some are changed; all set free.&lt;br /&gt;&#xA;But for a discrete gift, the whole truth I'll gladly tell&lt;br /&gt;&#xA;if I'm used programmatically.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;What am I?&lt;br /&gt;&#xA;(spacing is intentional)&lt;/p&gt;&#xA;&lt;p&gt;Edit for slight hint, since everyone's guesses seem too general:&lt;br /&gt;&#xA;Every line, as well as the title, is specifically necessary (though the second verse could be applied to a number of things. But to extend that verse: the only people who can obtain (the answer) are those who can stomach the work.&lt;/p&gt;&#xA;" OwnerUserId="9326" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-06-17T22:32:15.103" Title="A riddle, guaranteed to function even in rain or snow" Tags="|riddle|" AnswerCount="7" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16502" PostTypeId="2" ParentId="16501" CreationDate="2015-06-17T06:01:25.520" Score="8" Body="&lt;p&gt;Looks to me like this is about &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Equality (or ==)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm a jack of all uses, durable and strong,&#xA;but useful in childish construction too. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not too sure about this, but I remember an old joke from school where 1+1 = window (you draw a square with the 1s making the left and right sides, the = making the top and bottom, and the + forming the panes in the middle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Desired alike by man- and women-kind;&#xA;wanted by most all, obtained but by few.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everyone wants to be equal&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When shortened, I'm an impartial judge of prisoners in math;&#xA;Some are changed; all set free.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; = is used in math equations, sometimes to &quot;change&quot; operands into a result&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But for a discrete gift, the whole truth I'll gladly tell&#xA;if I'm used programmatically.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When used in programming &quot;==&quot; evaluates the truth of a condition&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13197" LastActivityDate="2015-06-17T06:01:25.520" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16503" PostTypeId="2" ParentId="16485" CreationDate="2015-06-17T06:02:14.490" Score="26" Body="&lt;p&gt;Here is my answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If you are coming from the left to the right, then &quot;Scalable Phrases&quot; are climbable while &quot;Non-Scalables&quot; are not climbable. Scaling a phrase refers to climbing it.&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;For a phrase to be climbable it has to be composed of climbable letters.&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;A climbable letter is any letter that, if you are on the &quot;floor&quot; (where the letter sits) then you can &quot;climb&quot; up the letter because no angle from the bottom to the top is greater than 90 degrees, or perpendicular from the floor.&#xA;For example, V is not climbable because coming from the left the first obstacle (the \ of the V) is sort of an overhang. T is also not climbable even though the beginning is climbable, but then at the overhang it is not. H, F, E, M, etc are all climbable letter because they have no overhang.&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;Notice that I in the font chosen IS climbable, although in many fonts it is not, since it has an overhang.&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;Here is a full list of climbable letters:&lt;br&gt;&#xA;ABDEFHIKLMNPR&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;Here is a full list of nonclimbable letters:&lt;br&gt;&#xA;CGJOQSTUVWXYZ&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Really nice puzzle!!&lt;/p&gt;&#xA;" OwnerUserId="4109" LastEditorUserId="46903" LastEditDate="2019-02-06T05:37:48.200" LastActivityDate="2019-02-06T05:37:48.200" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="16505" PostTypeId="2" ParentId="16501" CreationDate="2015-06-17T08:18:26.100" Score="1" Body="&lt;p&gt;Are you a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; bootstrap?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm a jack of all uses, durable and strong,&#xA;but useful in childish construction too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are usually durable and strong, I don't know about the childish construction. If you see it as a more general strap, you could utilise it for a lot more uses.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Desired alike by man- and women-kind;&#xA;wanted by most all, obtained but by few.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They can be found on boots both for males and females, but generally you do not just obtain bootstraps, you obtain boots. Few people would actually obtain only the bootstraps (for repair purposes, I guess).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When shortened, I'm an impartial judge of prisoners in math;&#xA;Some are changed; all set free.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am fairly sure this points to the prisoners dilemma, although I can't get what the judges role could be named or what this has to do with straps. I fail on this one I guess.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But for a discrete gift, the whole truth I'll gladly tell&#xA;if I'm used programmatically.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In programming you can use bootstrapping, for example in discrete event simulation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10885" LastActivityDate="2015-06-17T08:18:26.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16506" PostTypeId="2" ParentId="16501" CreationDate="2015-06-17T08:56:39.953" Score="1" Body="&lt;p&gt;I think you are a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; star (or *)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm a jack of all uses, durable and strong,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not completely sure about this yet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but useful in childish construction too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A block puzzle, where the kids have to fit the objects (star, square, triangle etc.) into the right hole.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Desired alike by man- and women-kind;&lt;br&gt;&#xA;wanted by most all, obtained but by few.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everybody wants to become a star! unfortunately only a few succeed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When shortened, I'm an impartial judge of prisoners in math;&lt;br&gt;&#xA;Some are changed; all set free.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; not sure about this&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But for a discrete gift, the whole truth I'll gladly tell&lt;br&gt;&#xA;if I'm used programmatically.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The * in programming is a wildcard, and can be used for every character&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12645" LastActivityDate="2015-06-17T08:56:39.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16507" PostTypeId="1" AcceptedAnswerId="16508" CreationDate="2015-06-17T09:07:09.427" Score="22" ViewCount="11550" Body="&lt;p&gt;In the first day of a new class semester, a student was murdered inside the classroom. No witnesses, no murder weapon, and no evidence. The police got four teachers as possible suspects, but they all had alibis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their alibis were:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PE Teacher: I was in the GYM, checking and cleaning equipment.&lt;br&gt;&#xA;English Teacher: I was in the library, looking for books.&lt;br&gt;&#xA;Math Teacher: I was in my cubicle, computing students grades.&lt;br&gt;&#xA;Science Teacher: I was in the laboratory, doing some experiments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who was the murderer? And why?&lt;/p&gt;&#xA;" OwnerUserId="13255" LastEditorUserId="8874" LastEditDate="2015-10-03T18:22:36.943" LastActivityDate="2018-03-16T15:35:16.043" Title="Which teacher is the murderer?" Tags="|liars|mystery|situation|" AnswerCount="6" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16508" PostTypeId="2" ParentId="16507" CreationDate="2015-06-17T09:08:04.633" Score="43" Body="&lt;p&gt;The murderer was&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the maths teacher&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it was the first day of semester, so he had no grades to compute.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="4506" LastEditDate="2015-06-17T10:43:46.157" LastActivityDate="2015-06-17T10:43:46.157" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16509" PostTypeId="1" AcceptedAnswerId="16511" CreationDate="2015-06-17T09:27:24.557" Score="8" ViewCount="1866" Body="&lt;p&gt;Here is a simple riddle about numbers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How would you insert 4 inside 5?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13255" LastEditorUserId="4421" LastEditDate="2015-11-25T18:08:36.847" LastActivityDate="2016-02-16T06:23:19.387" Title="Riddle Riddle in numbers!" Tags="|riddle|" AnswerCount="3" CommentCount="0" ClosedDate="2016-02-16T06:31:14.027" ContentLicense="CC BY-SA 3.0" />
  <row Id="16511" PostTypeId="2" ParentId="16509" CreationDate="2015-06-17T09:30:44.397" Score="23" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;FIVE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;F(IV)E&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;IV = 4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So we acheieved &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;4(IV) inside 5 (FIVE)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastActivityDate="2015-06-17T09:30:44.397" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16512" PostTypeId="2" ParentId="16474" CreationDate="2015-06-17T09:38:50.027" Score="1" Body="&lt;p&gt;I'll post my line of thinking as an answer so whoever feels like, can follow my lead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trick there is that either we assumed &lt;em&gt;farmer&lt;/em&gt; and &lt;em&gt;fisher&lt;/em&gt; were the same person, or that &lt;em&gt;farmer&lt;/em&gt;'s fishes were irrelevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key I feel goes in the lines of &lt;em&gt;the combined amount&lt;/em&gt; taking into account both the &lt;em&gt;fisher&lt;/em&gt; and the &lt;em&gt;farmer&lt;/em&gt; fishes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being $m, 2m, q$ the &lt;em&gt;small, medium, large&lt;/em&gt; fishes of the &lt;em&gt;fisher&lt;/em&gt;, and $x,y,z$ the same for the &lt;em&gt;farmer&lt;/em&gt; we have the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{align}&#xA;x+y+z&amp;amp;=35\\&#xA;(x+m)+(y+2m)&amp;amp;=\sqrt{z+q}-5m\\&#xA;x+y+8m&amp;amp;=\sqrt{z+q}\\&#xA;(x+y+8m)^2&amp;amp;=z+q\\&#xA;q&amp;amp;=x^2+y^2+64m^2+2xy+16xm+16ym+48xym&#xA;\end{align}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key would be that $m,q,x,y,z$ must be integers. Even though, this has many solutions, so some other restriction has to exist within the letter.&lt;/p&gt;&#xA;" OwnerUserId="12165" LastEditorUserId="12165" LastEditDate="2015-06-18T12:18:35.723" LastActivityDate="2015-06-18T12:18:35.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16513" PostTypeId="1" AcceptedAnswerId="16527" CreationDate="2015-06-17T10:04:46.220" Score="55" ViewCount="5796" Body="&lt;p&gt;You are walking through the prairie when you find a madman wandering around talking to himself. The following is what you manage to hear of his speech:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;How? I - I'll ask her. I owe her much, again. I'd a home on town, a tax as florid as out the coat, a virgin a year. Oh, yodel - aware you take all or I do. Never the road: I'll land in the Anna-Marie land. Main can's a sore gone; tennis is out t'car. Oh, line a canned turkey!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the man really talking about?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2017-10-11T00:51:36.343" Title="The Madman's Speech" Tags="|enigmatic-puzzle|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16514" PostTypeId="1" AcceptedAnswerId="16518" CreationDate="2015-06-17T10:21:19.260" Score="26" ViewCount="1982" Body="&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/15862/mathematical-rebus&quot;&gt;Mathematical Rebus I&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/q/52759/12165&quot;&gt;Mathematical Rebus III&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ap8mE.gif&quot; alt=&quot;Rebus&amp;#39; image&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\det\frac{\partial(x,y)}{\partial(r,\varphi)}\\&#xA;\sum_{n=2}^\infty \ddot{\frac{t^n}{n!}} $$&lt;/p&gt;&#xA;" OwnerUserId="12165" LastEditorUserId="12165" LastEditDate="2017-06-19T10:39:19.203" LastActivityDate="2017-06-19T10:39:19.203" Title="Mathematical Rebus II" Tags="|mathematics|rebus|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="16515" PostTypeId="2" ParentId="16514" CreationDate="2015-06-17T10:28:58.283" Score="19" Body="&lt;p&gt;Graph:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; work in progress...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$\det\frac{\partial(x,y)}{\partial(r,\varphi)}=r$, because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$x,y$ are Cartesian coordinates and $r,\varphi$ are polar coordinates, in the 2-dimensional plane.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$\sum_{n=3}^\infty \ddot{\frac{t^n}{n!}}=e^t$, because &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;differentiate $t^n$ twice to get $n(n-1)t^{n-2}$, so the sum is $\sum_{n=3}^\infty \ddot{\frac{t^{n-2}}{(n-2)!}} = \sum_{n=1}^\infty \ddot{\frac{t^n}{n!}}=e^t$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-06-17T10:28:58.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16516" PostTypeId="2" ParentId="16509" CreationDate="2015-06-17T10:41:05.640" Score="12" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3r8iz.png&quot; alt=&quot;I may not be taking this seriously enough...&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I might not be taking this question seriously enough...&lt;/p&gt;&#xA;" OwnerUserId="13259" LastActivityDate="2015-06-17T10:41:05.640" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16517" PostTypeId="2" ParentId="16513" CreationDate="2015-06-17T10:43:21.867" Score="67" Body="&lt;p&gt;Short :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He is talking about the states in the United States.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Details:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Hawaii Alaska Iowa Michigan Idaho Montana Texas Florida SouthDakota Virginia Ohio Delaware Utah Colorado Nevada RhodeIsland Indiana Maryland Maine Kansas Oregon Tennessee SouthCarolina Kentucky&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10940" LastEditorUserId="39611" LastEditDate="2017-10-11T00:51:36.343" LastActivityDate="2017-10-11T00:51:36.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16518" PostTypeId="2" ParentId="16514" CreationDate="2015-06-17T10:53:21.213" Score="21" Body="&lt;p&gt;Follow up to rand al'thor answer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The graph shows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; sec(x)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By combining the 2 other answers, we obtain:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; sec + r + et = $secret$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12731" LastActivityDate="2015-06-17T10:53:21.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16519" PostTypeId="2" ParentId="16331" CreationDate="2015-06-17T11:46:54.787" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;Start at A&lt;/li&gt;&#xA;&lt;li&gt;Go to random B&lt;/li&gt;&#xA;&lt;li&gt;Call Arthur to B&lt;/li&gt;&#xA;&lt;li&gt;Go to Bob at A&lt;/li&gt;&#xA;&lt;li&gt;Get $\overline{AB}$&lt;/li&gt;&#xA;&lt;li&gt;Go to Arthur at B &lt;/li&gt;&#xA;&lt;li&gt;Call Bob to B&lt;/li&gt;&#xA;&lt;li&gt;Go to random C&lt;/li&gt;&#xA;&lt;li&gt;Call Arthur to C&lt;/li&gt;&#xA;&lt;li&gt;Go anywhere&lt;/li&gt;&#xA;&lt;li&gt;Go to Arthur at C&lt;/li&gt;&#xA;&lt;li&gt;Get $\angle{AC}$ &lt;/li&gt;&#xA;&lt;li&gt;Go to Bob at B&lt;/li&gt;&#xA;&lt;li&gt;Call Dale to B&lt;/li&gt;&#xA;&lt;li&gt;Go anywhere&lt;/li&gt;&#xA;&lt;li&gt;Go to Bob at B&lt;/li&gt;&#xA;&lt;li&gt;Get $\overline{BC}$&lt;/li&gt;&#xA;&lt;li&gt;Go to Arthur at C&lt;/li&gt;&#xA;&lt;li&gt;Call Bob to C&lt;/li&gt;&#xA;&lt;li&gt;Go to random D&lt;/li&gt;&#xA;&lt;li&gt;Call Arthur to D&lt;/li&gt;&#xA;&lt;li&gt;Go anywhere&lt;/li&gt;&#xA;&lt;li&gt;Go to Arthur at D&lt;/li&gt;&#xA;&lt;li&gt;Get $\angle BD$&lt;/li&gt;&#xA;&lt;li&gt;If $\angle{BD}$ sucks, forget point D and goto step 20&lt;/li&gt;&#xA;&lt;li&gt;Go to Bob at C&lt;/li&gt;&#xA;&lt;li&gt;Call Dale to C&lt;/li&gt;&#xA;&lt;li&gt;Go anywhere&lt;/li&gt;&#xA;&lt;li&gt;Go to Bob at C&lt;/li&gt;&#xA;&lt;li&gt;Get $\overline{CD}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$\overline{AC}^2=\overline{AB}^2+\overline{BC}^2\pm2\times\overline{AB}\times\overline{BC}\cos\frac{\angle{AC}}{2}$, giving two possibilities for radius $r=\frac{\overline{AC}}{2\sin\frac{\angle{AC}}{2}}$, one of which can be confirmed by solving the triangle $BCD$ in the same way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At step 25, $\angle{BD}$ sucks with probability 0, but we can still guarantee a solution in a finite number of steps against an adversarial teleporter because only four sucky locations exist. The geometry of sucking is left as an exercise for the reader.&lt;/p&gt;&#xA;" OwnerUserId="12863" LastEditorUserId="12863" LastEditDate="2015-06-18T11:21:05.260" LastActivityDate="2015-06-18T11:21:05.260" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16520" PostTypeId="2" ParentId="16501" CreationDate="2015-06-17T12:04:24.500" Score="1" Body="&lt;p&gt;Are you a:   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;%&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm a jack of all uses, durable and strong,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Used to describe the completion or saturation of something&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but useful in childish construction too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A see-saw?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Desired alike by man- and women-kind;&lt;br&gt;&#xA;wanted by most all, obtained but by few.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Everyone wants 100%&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When shortened, I'm an impartial judge of prisoners in math;&lt;br&gt;&#xA;Some are changed; all set free.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;No idea&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But for a discrete gift, the whole truth I'll gladly tell&lt;br&gt;&#xA;if I'm used programmatically&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It'll give the modulus.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13263" LastEditorUserId="7308" LastEditDate="2015-06-17T12:15:34.440" LastActivityDate="2015-06-17T12:15:34.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16522" PostTypeId="2" ParentId="16443" CreationDate="2015-06-17T12:47:22.537" Score="0" Body="&lt;p&gt;Is it  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; time &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;nothing here has been anywhere&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; cant have the same moment of time twice &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;we may dream of many places, but we are not actually there&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;we can think of the future&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;diving, flying or driving -- moving from wherever, to wherever with whoever, doing whatever&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;time flies&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;nothing here has been anywhere&#xA;(except perhaps, round the bend)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;anything outside a (round) clock?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;my brother's never told a lie&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;time and place?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll deceive you any way I can&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;times will change i guess&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;nothing on these steps has ever felt the need to die.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;cant understand this one&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;altogether but unseen however long your eyes persist&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;cant see time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;you ll never walk through my locked doors again&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; cant go back in time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13263" LastEditorUserId="13263" LastEditDate="2015-06-17T12:55:08.430" LastActivityDate="2015-06-17T12:55:08.430" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16524" PostTypeId="2" ParentId="16501" CreationDate="2015-06-17T13:23:41.000" Score="0" Body="&lt;p&gt;My answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; value&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's see what we got here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a jack of all uses, durable and strong,&lt;br&gt;&#xA;but useful in childish construction too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    Something of value can be use in every way we can imagine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Desired alike by man- and women-kind;&lt;br&gt;&#xA;wanted by most all, obtained but by few.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    Everyone of us desire 'value' (self-value in this case)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When shortened, I'm an impartial judge of prisoners in math;&lt;br&gt;&#xA;Some are changed; all set free.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    If a value is 'shortened' that may mean that is divided by&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But for a discrete gift, the whole truth I'll gladly tell&lt;br&gt;&#xA;if I'm used programmatically.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    In programming, the value of something (say, a variable) gives 'the truth'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12739" LastEditorUserId="463" LastEditDate="2015-06-17T13:34:43.647" LastActivityDate="2015-06-17T13:34:43.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16525" PostTypeId="2" ParentId="16477" CreationDate="2015-06-17T13:25:15.487" Score="1" Body="&lt;p&gt;I think its pretty simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that that are only n mementos and with n people, we  can model the stable equilibrium/solution as a series of graphs with each person as a node. Friendships exist already which forms each edge in the graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each edge is directional, in that one end has their birthday before the other end does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that given those rules, we have been able to map all the edges into a series closed loops. This shall be left as an exercise to the reader....(hint what would have happened to all the mementoes if there was an open end?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this is the case, the new node can always insert himself into the appropriate part of a cycle, if he befriends the person who has a birthday just before him in the cycle, and the person who is just after him in the cycle.&lt;/p&gt;&#xA;" OwnerUserId="4055" LastActivityDate="2015-06-17T13:25:15.487" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16526" PostTypeId="2" ParentId="16331" CreationDate="2015-06-17T13:54:18.117" Score="2" Body="&lt;p&gt;This gets a bit odd, and assumes something about the rules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My algorithm is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Identify starting point as point A&lt;/li&gt;&#xA;&lt;li&gt;Teleport at random (to point B).&lt;/li&gt;&#xA;&lt;li&gt;Summon Dale.&lt;/li&gt;&#xA;&lt;li&gt;Pull Dale's tail, revealing distance AB (call it a).&lt;/li&gt;&#xA;&lt;li&gt;Summon Arthur.&lt;/li&gt;&#xA;&lt;li&gt;Teleport at random (to point C).&lt;/li&gt;&#xA;&lt;li&gt;Summon Dale.&lt;/li&gt;&#xA;&lt;li&gt;Pull Dale's tail, revealing distance BC (call it b).&lt;/li&gt;&#xA;&lt;li&gt;Summon Arthur.&lt;/li&gt;&#xA;&lt;li&gt;Teleport back to Bob.&lt;/li&gt;&#xA;&lt;li&gt;Summon Dale.&lt;/li&gt;&#xA;&lt;li&gt;Pull Dale's tail, revealing distance CA (call it c).&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Calculate the diameter of the circumscribed circle (the circle that touches all three points on the triangle, which were all on the wall) according to the formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;diameter = 2abc / sqrt((a+b+c)&lt;em&gt;(-a+b+c)&lt;/em&gt;(a-b+c)*(a+b-c))&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Multiply the diameter by Pi for the area of the circle.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;Escape.&lt;/li&gt;&#xA;&lt;li&gt;Profit.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The formula for the diameter is explained in great detail at &lt;a href=&quot;https://en.wikipedia.org/wiki/Circumscribed_circle&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Circumscribed_circle&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13258" LastActivityDate="2015-06-17T13:54:18.117" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16527" PostTypeId="2" ParentId="16513" CreationDate="2015-06-17T14:12:31.080" Score="44" Body="&lt;p&gt;I guess he's talking about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; States in America.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &quot;How? - I&quot; = Hawaii&lt;br&gt;&#xA; &quot;I'll ask her.&quot; = Alaska&lt;br&gt;&#xA; &quot;I owe her&quot; = Iowa&lt;br&gt;&#xA; &quot;much again&quot; = Michigan&lt;br&gt;&#xA; &quot;I'd a ho...&quot; = Idaho&lt;br&gt;&#xA; &quot;...me on town, a&quot; = Montana&lt;br&gt;&#xA; &quot;tax as&quot; = Texas&lt;br&gt;&#xA; &quot;florid a...&quot; = Florida&lt;br&gt;&#xA; &quot;...s out the coat, a&quot; = South Dakota&lt;br&gt;&#xA; &quot;virgin a year.&quot; = Virginia&lt;br&gt;&#xA; &quot;Oh yo...&quot; = Ohio&lt;br&gt;&#xA; &quot;...del - aware&quot; = Delaware&lt;br&gt;&#xA; &quot;You ta...&quot; = Utah&lt;br&gt;&#xA; &quot;...ke all or i do&quot; = Colorado&lt;br&gt;&#xA; &quot;Never the&quot; = Nevada&lt;br&gt;&#xA; &quot;road: I'll land&quot; = Rhode Island&lt;br&gt;&#xA; &quot;in the Anna-...&quot; = Indiana&lt;br&gt;&#xA; &quot;...Marie land&quot; = Maryland&lt;br&gt;&#xA; &quot;Main&quot; = Maine&lt;br&gt;&#xA; &quot;can's a s...&quot; = Kansas&lt;br&gt;&#xA; &quot;...ore gone;&quot; = Oregon&lt;br&gt;&#xA; &quot;tennis i...&quot; = Tennessee&lt;br&gt;&#xA; &quot;...s out t'car. Oh line a&quot; = South Carolina&lt;br&gt;&#xA; &quot;canned Turkey&quot; = Kentucky  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1485" LastEditorUserId="5373" LastEditDate="2015-06-18T19:36:20.273" LastActivityDate="2015-06-18T19:36:20.273" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16528" PostTypeId="2" ParentId="16507" CreationDate="2015-06-17T14:46:09.923" Score="15" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Science Teacher. Science teachers don't perform experiments, they perform demonstrations and only to students.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's also entirely possible for a math teacher to have assigned math work on the first day for students to complete in class and then be grading it later in the day. Or he was late in computing the grades from the previous semester.&lt;/p&gt;&#xA;" OwnerUserId="5834" LastEditorUserId="5834" LastEditDate="2015-06-17T14:58:41.580" LastActivityDate="2015-06-17T14:58:41.580" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="16529" PostTypeId="2" ParentId="16507" CreationDate="2015-06-17T14:58:25.537" Score="7" Body="&lt;p&gt;The murderer was...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The PE teacher! Because he said that he was cleaning equipment, although this is the duty of cleaning ladies. So he may was cleaning the... MURDERER WEAPON! &lt;em&gt;dramatic music&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13275" LastEditorUserId="13275" LastEditDate="2015-06-24T12:01:57.580" LastActivityDate="2015-06-24T12:01:57.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16531" PostTypeId="2" ParentId="16507" CreationDate="2015-06-17T15:06:26.020" Score="6" Body="&lt;p&gt;It could be the &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PE teacher.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It said there's no murder weapon, so&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; he could be cleaning up the mess and putting the equipment back together.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13276" LastActivityDate="2015-06-17T15:06:26.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16532" PostTypeId="1" AcceptedAnswerId="16533" CreationDate="2015-06-17T15:07:46.583" Score="14" ViewCount="1059" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;We come from two different factions, as unlike as can be.&lt;br&gt;&#xA;  We support the factions' armies, but not necessarily our own.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;At the start of the war, half of us supported nobody.&lt;br&gt;&#xA;  Those remaining were split between supporting their own faction and&#xA;  supporting the opposing faction.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;As the war commenced, our allegiances changed.&lt;br&gt;&#xA;  We supported whichever army was passing through at the time.&lt;br&gt;&#xA;  As casualties mounted, more and more of us ceased supporting anybody at all.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Finally, one side was victorious, but our lives were unaffected.&lt;br&gt;&#xA;  We merely wait for the next war, which will play out differently, but the&#xA;  same.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who are we?&lt;/p&gt;&#xA;" OwnerUserId="10861" LastActivityDate="2015-06-17T16:50:44.193" Title="A Series of Wars" Tags="|riddle|" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16533" PostTypeId="2" ParentId="16532" CreationDate="2015-06-17T15:10:10.897" Score="19" Body="&lt;p&gt;We are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chessboard squares, which are black and white (like chess pieces) but support (physically) whatever piece is standing on it at the moment, regardless of its color&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2744" LastActivityDate="2015-06-17T15:10:10.897" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16534" PostTypeId="2" ParentId="16477" CreationDate="2015-06-17T15:13:18.343" Score="7" Body="&lt;p&gt;I think&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  scandal is inevitable.  This is because, before and after the stranger arrives, everyone must have exactly two friends.  Proof follows.&lt;br&gt;&#xA;&lt;br&gt;&#xA;  A key rule is the &quot;don't be tacky&quot; rule, where you don't give a person the gift they most recently gave you.  Suppose, some year before the stranger arrives, the islanders decide to flip this rule into a &quot;must be tacky&quot; rule, but keep all of the friendships and the general patterns for gift giving the same.  It's clearly possible for them to continue to give gifts according to the other rules:  a person gets a bunch of gifts on their birthday, then they proceed to hand each gift back to the person that gave it as the year progresses.&lt;br&gt;&#xA;&lt;br&gt;&#xA; We then see for each friendship, there is one gift that passes back and forth between the two friends.&lt;br&gt;&#xA;&lt;br&gt;&#xA;  We can also assign partial ownership to each gift.  For a gift that passes back and forth between two friends, we say each of the friends is a half owner of the gift.  If you just keep a gift permanently, you are the sole owner of that gift.  This latter ownership rule comes into play if a person (say Bob) has just one friend.  Since Bob was following the &quot;don't be tacky&quot; rule before, Bob must have an extra gift to prevent being tacky.  When it swaps to the &quot;must be tacky&quot; rule, then Bob just has a gift that just sits there and collects dust.&lt;br&gt;&#xA;&lt;br&gt;&#xA;  If a person has one friend, then they have ownership over 1.5 gifts (one that sits there, and half ownership over the one they trade back and forth).  If a person has two friends, they have ownership over 1.0 gifts (two half ownerhips).  If a person has more than two friends, then they have ownership over at least 1.5 gifts.  Therefore, the only way for their to be on average one gift per person is if everyone has own 1.0 gifts, and &lt;strong&gt;therefore everyone has exactly two friends&lt;/strong&gt;.  This was true under the &quot;don't be tacky&quot; rule as well.&lt;br&gt;&#xA;&lt;br&gt;&#xA;  Now comes the stranger.  We see that everyone must have exactly two friends before he arrives.  After he arrives, we must maintain the property that everyone has exactly two friends.  But that means it is impossible for the stranger to make any friends.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2722" LastActivityDate="2015-06-17T15:13:18.343" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16535" PostTypeId="2" ParentId="16477" CreationDate="2015-06-17T15:29:38.403" Score="3" Body="&lt;p&gt;It is not generally possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One configuration is that all $n$ people on the island have 2 friends: the person whose birthday is immediately before theirs and the one whose birthday is immediately after.  This makes a big friendship ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are in a status quo.  The person who most recently had a birthday has 2 momentos, the person whose birthday is next has 0 momentos and everyone else has 1 momento.  On the next birthday, the celebrant receives two presents, one from the person who had two momentos and one from the next person to have a birthday.  Status quo remains: most recent birthday celebrant has 2 momentos, next one has 0, everyone else 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the next person's birthday, the one who has 2 momentos gives the one they received from the other person.  They always have two presents to choose from, so this is always possible.  This means that one present is regifted every time, moving forward around the ring, and each other present moves back one space each year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now add a new person: you.  If you make a friend with someone whose birthday is further out than your own, then that person will have to give out one more present than they would have, before their next birthday.  The status quo had them giving out all their presents, so they're guaranteed to run out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you make a friend with someone whose birthday is before yours, then you give your present to that person on their birthday, raising their number to 3.  The other person then gives you a different present back when you have your birthday.  The problem is that on the next time your friend has a birthday, the only gift you have is the one they just gave you, and that is not allowed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Making multiple friends doesn't work either; someone would be guaranteed to run out of presents.  If your birthday is before any of your new friends', then that person will run out.  If it is after more than one, then you will run out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There could be multiple, possibly intersecting rings.  But even if there are, you run into the same problem: either you force someone to run out, you run out, or you are forced to give back the same present you received to the person who gave it to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may be configurations where it is possible to add you (I don't think so), but there are definitely configurations where it is not.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="2328" LastEditDate="2015-06-17T15:52:04.910" LastActivityDate="2015-06-17T15:52:04.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16539" PostTypeId="2" ParentId="16501" CreationDate="2015-06-17T16:26:12.980" Score="2" Body="&lt;p&gt;I'll try again - is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Absolution / Absolute value?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm a jack of all uses, durable and strong,&lt;br&gt;&#xA;but useful in childish construction too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Durable and strong perhaps a referece to the 'walls' that surround a number when finding the absolute value |n|. Or maybe a reference to absolute strength of muscles?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Desired alike by man- and women-kind;&lt;br&gt;&#xA;wanted by most all, obtained but by few.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most people want absolution for their sins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When shortened, I'm an impartial judge of prisoners in math;&lt;br&gt;&#xA;Some are changed; all set free.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When shortened to Abs - used in math to find the absolute value, some numbers are changed (negative numbers) but once calculated they are set free from the &quot;walls&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But for a discrete gift, the whole truth I'll gladly tell&lt;br&gt;&#xA;if I'm used programmatically.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not entirely sure about the discrete part =/&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12206" LastEditorUserId="12206" LastEditDate="2015-06-17T16:37:40.933" LastActivityDate="2015-06-17T16:37:40.933" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16540" PostTypeId="1" AcceptedAnswerId="16598" CreationDate="2015-06-17T16:30:31.900" Score="56" ViewCount="6433" Body="&lt;p&gt;Since I was a kid, nobody wanted me in their super-exclusive clubs... the &quot;House in the Tree Club&quot;... the &quot;Girls Only Club&quot;... the &quot;Whatever you are NOT Club&quot;... but now that I am a skilled (and a bit nerdy) guy, time has come to take my revenge!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've just discovered that not far from my house in New York City, there's a secret place where some guys seem to meet every week. I don't know what kind of club it is, but I definitely want to be a part of it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every night, each member knocks at the door of the Secret Club, and each time, a voice comes from inside, asking some strange questions...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this very moment, someone is knocking on the door... &#xA;( - Let's listen to what they say... - )&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;&lt;em&gt;Knock! Knock!&lt;/em&gt;&quot;&lt;br&gt;&#xA;  &quot;&lt;em&gt;Who's knocking?&lt;/em&gt;&quot; asks the voice.&lt;br&gt;&#xA;  &quot;&lt;em&gt;I am a member Sir, let me in!&lt;/em&gt;&quot; answers the guy.&lt;br&gt;&#xA;  &quot;&lt;em&gt;If you're a member, tell me what are the results of these operations: 17 + 5? 24 + 3? 1 + 16?&lt;/em&gt;&quot;.&lt;br&gt;&#xA;  &quot;&lt;em&gt;Of course... 22... 27... 17!&lt;/em&gt;&quot; is the guy's response.&lt;br&gt;&#xA;  &quot;&lt;em&gt;Very well! Come on in!&lt;/em&gt;&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(- Well, I guess it's not the MENSA Club... - I think...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second guy approaches the door, knocks at it, and again the same scene takes place:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;&lt;em&gt;Prove that you're a member! Answer these: 38 - 37? 28 - 10? 14 + 4?&lt;/em&gt;&quot;&lt;br&gt;&#xA;  &quot;&lt;em&gt;Ehm... 1... 18... and again 18!&lt;/em&gt;&quot;.&lt;br&gt;&#xA;  &quot;&lt;em&gt;Good boy! Come in and take a seat near the other guys!&lt;/em&gt;&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(- OK, it must be some sort of math-nerd association... but better than nothing!...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One last guy knocks at the door, but this time the question they ask him is very very strange...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;&lt;em&gt;Well, if you are really one of us, tell me... What's our special number... the one with the unique property we all know?...&lt;/em&gt;&quot;&lt;br&gt;&#xA;  The third guy seems slightly surprised... keeps thinking for a while, and then shouts: &quot;&lt;em&gt;Ah ah!.. You're referring to 12! Are you?&lt;/em&gt;&quot;.&lt;br&gt;&#xA;  The voice from inside the house starts laughing &quot;&lt;em&gt;Ah ah! For sure you are one of us!&lt;/em&gt;&quot;, opens the door and lets the guy in...  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(- Well... I am a bit disappointed... but I'll give it a try... I'm sure they'll let me in!-)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;&lt;em&gt;Knock knock!... ehm... I'm a member! Let me in!&lt;/em&gt;&quot;&lt;br&gt;&#xA;  &quot;&lt;em&gt;Are you sure? If so, tell me... What's the result of  11 + 1? 10 + 17? 21 + 3?&lt;/em&gt;&quot;&lt;br&gt;&#xA;  &quot;&lt;em&gt;...Ehm... 12... 27...&lt;/em&gt;&quot; I start answering...&lt;br&gt;&#xA;  &quot;&lt;em&gt;OK... and what about the last one?&lt;/em&gt;&quot; says the voice&lt;br&gt;&#xA;  &quot;&lt;em&gt;24!!&lt;/em&gt;&quot; I shout... just a second before the guy from inside starts to yell:&lt;br&gt;&#xA;  &quot;&lt;em&gt;You liar! Go away and never come back to this respectful club!!&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At the end of this shocking experience, I still have a question in my mind:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;strong&gt;WHAT KIND OF CLUB WAS IT THAT I WAS TRYING TO ENTER?&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 1:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The club is a real club. Many clubs like this one exist in real life. Probably, many members of the Puzzling Community are members of such a club in real life. Of course, the secrecy about the club is purely &quot;fictional&quot; and for the purpose of the puzzle. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;One of my guesses about the club's members was completely wrong: they ARE NOT &quot;math nerds&quot;, and you don't need particular math skills to solve the calculations asked by the man at the door. Anyway, not everybody could answer those questions like the members did.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Pay attention to the tags: no &lt;a href=&quot;/questions/tagged/calculation-puzzle&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;calculation-puzzle&amp;#39;&quot; rel=&quot;tag&quot;&gt;calculation-puzzle&lt;/a&gt; is present...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 4:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It appears that the right answer to the last question of the man at the door should have been &quot;23&quot; instead of &quot;24&quot;...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 5:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;In my last edit I added a location for the story. Though it is not &lt;em&gt;strictly necessary&lt;/em&gt;, it could be important to solve the puzzle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 6:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Twelve is &quot;special&quot; for club's members, because is the only &lt;em&gt;number&lt;/em&gt; with a specific characteristic; anyway it's not so obvious, that's why the member who was asked about it had to think a while before answering...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12628" LastEditorUserId="12628" LastEditDate="2017-11-03T08:27:25.793" LastActivityDate="2019-08-31T01:21:22.237" Title="The Secret Club" Tags="|pattern|lateral-thinking|enigmatic-puzzle|password|situation|" AnswerCount="4" CommentCount="36" ContentLicense="CC BY-SA 3.0" />
  <row Id="16541" PostTypeId="2" ParentId="16509" CreationDate="2015-06-17T16:34:20.897" Score="5" Body="&lt;p&gt;As an apology for my previous answer, I'd like to offer the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Five fenceposts can have four panels of fencing placed between them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A '5' lit up on a 7-segment display will have four interior right angles. (By coincidence of it being made of also 5 distinct segments.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And stretching it,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; according to Wikipedia's page of &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_numbers_in_various_languages&quot; rel=&quot;nofollow noreferrer&quot;&gt;List of numbers in various languages&lt;/a&gt;, there are no languages where the common word for 'four' is a subset or anagram subset of the word for 'five', at least as far as I could tell with my layman's eye, which was a bit of a disappointment. But graphically speaking, in the Tai-Kadai language Lao, 4 could fit into 5 quite neatly. &lt;img src=&quot;https://i.stack.imgur.com/aysBE.png&quot; alt=&quot;Tai-Kadai language&quot;&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13259" LastActivityDate="2015-06-17T16:34:20.897" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16542" PostTypeId="2" ParentId="16507" CreationDate="2015-06-17T16:48:55.303" Score="26" Body="&lt;p&gt;A deeper (alternative) answer... [Don't take this answer too seriously]&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;b&gt;It was all of them...&lt;/b&gt;&#xA;&lt;br&gt;They killed the child's creativity and explorations with their strict policies to conform him to society! Reducing any form of expression to be forced into fitting their constraints. The child is no longer 'living' as free as he once was. His intellectual freedom has been murdered.&#xA;&lt;br&gt;With this kind of murder, there are no weapons or evidence!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="5373" LastEditDate="2015-06-18T15:42:50.717" LastActivityDate="2015-06-18T15:42:50.717" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16543" PostTypeId="2" ParentId="16532" CreationDate="2015-06-17T16:50:44.193" Score="1" Body="&lt;p&gt;I think they are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A deck of Cards, in the game of War.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;We come from two different factions, as unlike as can be.&#xA;We support the factions' armies, but not necessarily our own.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The deck is split up at the start of the game, between the two players. The cards don't have their own army, in a sense.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;At the start of the war, half of us supported nobody.&#xA;Those remaining were split between supporting their own faction and supporting the opposing faction.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, deck is split in half and dealt to each player. They support each player, but only when getting played.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;As the war commenced, our allegiances changed.&#xA;We supported whichever army was passing through at the time.&#xA;As casualties mounted, more and more of us ceased supporting anybody at all.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As the game plays, cards go from all the players to one, on each hand. They are passed around continuously until one player has the whole deck.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Finally, one side was victorious, but our lives were unaffected.&#xA;We merely wait for the next war, which will play out differently, but the same.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When a player wins, the cards are put back, and can be used for another &quot;war&quot;, as long as it was not a violent game that bent the cards.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="372" LastActivityDate="2015-06-17T16:50:44.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16545" PostTypeId="1" AcceptedAnswerId="16548" CreationDate="2015-06-17T17:30:05.420" Score="19" ViewCount="1786" Body="&lt;p&gt;The following each describe a movie title. Figure out what they are! (I'm really bad at intros.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;:*:*!!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;extern int bake(double pct);&#xA;int main(int argc, char **argv) {&#xA;  return bake(0.5);&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;3.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  (traffic cop gets his license suspended for unpaid parking tickets)&#xA;- (pressing instrument)&#xA;+ (Hyperion)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="11297" LastEditDate="2015-06-18T17:38:39.527" LastActivityDate="2015-07-16T23:45:21.150" Title="Name that Movie" Tags="|rebus|knowledge|computer-puzzle|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16546" PostTypeId="2" ParentId="16501" CreationDate="2015-06-17T17:31:19.203" Score="12" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Abs or ABS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm a jack of all uses, durable and strong,&#xA;but useful in childish construction too.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;ABS is Acrylonitrile_butadiene_styrene which is a material used for construction, but also for toys like Legos&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Desired alike by man- and women-kind;&#xA;wanted by most all, obtained but by few.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Everybody wants sexy abs! (abdominal muscles)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When shortened, I'm an impartial judge of prisoners in math;&#xA;Some are changed; all set free.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Abs is short for absolute value, which when in &quot;bar&quot; form is |x|, which surrounds a number, making it a prisoner. After resolution, some numbers (negative) are changed, all are set free (removed from the bars)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But for a discrete gift, the whole truth I'll gladly tell&#xA;if I'm used programmatically.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The | is used as a representation for OR in programming, which returns a discrete TRUE or FALSE value.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As for the title:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;ABS is Anti-lock Braking System, which helps stop your car in bad weather like rain and snow.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10861" LastEditorUserId="10861" LastEditDate="2015-06-17T19:16:20.663" LastActivityDate="2015-06-17T19:16:20.663" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16548" PostTypeId="2" ParentId="16545" CreationDate="2015-06-17T17:32:12.023" Score="17" Body="&lt;p&gt;Number 1 is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Kiss Kiss Bang Bang&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 2 is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Half Baked&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 3 is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; From Tim's answer, Ironic - Iron + Titan = Titanic&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-07-16T23:45:21.150" LastActivityDate="2015-07-16T23:45:21.150" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16549" PostTypeId="2" ParentId="16545" CreationDate="2015-06-17T17:37:19.867" Score="8" Body="&lt;p&gt;3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Ironic - Iron + Titan = Titanic&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13204" LastActivityDate="2015-06-17T17:37:19.867" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16550" PostTypeId="2" ParentId="16431" CreationDate="2015-06-17T17:48:09.097" Score="2" Body="&lt;p&gt;&lt;s&gt;The Answer is C, because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well their is so much data that can prove it, that will take at least one hour to enumerate all of them, so I will just write the most obvious one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You need to calculate the number of identical colors from the start of each symbol until they change.here the pattern: 3|1|3...1|4|2...2|1|4 (each row must be equal to 7)&lt;/s&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have edited my answer after reading this: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;&quot;The square layout of the tiles is irrelevant.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So now, the answer is &lt;strong&gt;B&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You need to calculate the number of &lt;strong&gt;colors switching&lt;/strong&gt; in each tiles, the pattern is: 1/2/1/1/0/1/1/2/1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some explanation about how I come to this answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;First&lt;/strong&gt;, we need to now what we have here:  6 different colors, 8(+1) tiles, 4 colored circle in each tiles, Only one colored circle can link another one, only one tiles is fully filled with only one color, colors have their own pattern (green + red, orange + blue, yellow + purple), only the (orange + blue) pattern are equal (2 of each of them in their tiles). &lt;strong&gt;Now&lt;/strong&gt;, we can start conceptualizing the pattern by asking some question: does those colored circle linked each others mean we need to read them like this (bottom-left &gt; top-left &gt; top-right &gt; bottom-right) ? &lt;strong&gt;If yes,&lt;/strong&gt; does we can say they are like rainbow ? &lt;strong&gt;If yes,&lt;/strong&gt; RED=1,ORANGE=2,YELLOW=3,GREEN=4,BLUE=5,PURPLE=6 ? &lt;strong&gt;If yes,&lt;/strong&gt; does that mean that each color pattern is equally distanced by 3 (Green = 4 / Red = 1, Orange = 2 / Blue = 5, Yellow = 3 / Purple = 6)? &lt;strong&gt;If yes,&lt;/strong&gt; that mean 3 must be our delimiter for calculating our pattern ? &lt;strong&gt;If yes,&lt;/strong&gt; can we say that we need to read 3 tiles by 3 tiles ? &lt;strong&gt;If yes,&lt;/strong&gt; we can easily see that each of those 3 row of tiles are hugely different in term of colors and number of same colors, right ? &lt;strong&gt;If yes,&lt;/strong&gt; does we will need to read by a logical order following the &lt;em&gt;&quot;arc / rainbow line&quot;&lt;/em&gt; ? &lt;strong&gt;If yes,&lt;/strong&gt; that will mean that the type of colors, the color at the start/end and the order of them in the tile is &lt;strong&gt;Irrelevant&lt;/strong&gt; ? &lt;strong&gt;If yes&lt;/strong&gt;, this will indicate that only the &lt;strong&gt;colors switching&lt;/strong&gt; is relevant ? &lt;strong&gt;If yes&lt;/strong&gt;, how can we calculate it ? &lt;strong&gt;Maybe&lt;/strong&gt;, by giving to each tile a point for every color switched, in example: (green,green,green,red = 1) / (orange,blue,blue,orange = 2) / etc....  &lt;strong&gt;By this theory, only one symbol can be our next symbol in the pattern&lt;/strong&gt;: &lt;strong&gt;B&lt;/strong&gt; --Since B is the only tiles with 1pts color switching in those suggested symbol.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13285" LastEditorUserId="13285" LastEditDate="2015-06-18T12:54:05.160" LastActivityDate="2015-06-18T12:54:05.160" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16551" PostTypeId="1" AcceptedAnswerId="16576" CreationDate="2015-06-17T18:13:45.437" Score="22" ViewCount="1782" Body="&lt;p&gt;It's here. Finally here. For weeks you've waited, checking the mail every single day, to no avail - but today, your invitation has finally arrived. You eagerly tear open the envelope and read:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Congratulations, &lt;strong&gt;&lt;em&gt;[insert username here]&lt;/em&gt;&lt;/strong&gt;! You have been selected as a member of this year's Treasure Hunt 'Round the World! Or at least, you're a &lt;em&gt;preliminary&lt;/em&gt; member -- of course, we have to weed out the doers from the dreamers, and that's what this first challenge is for. Please solve it, and write us back at &lt;strong&gt;&lt;em&gt;[redacted]&lt;/em&gt;&lt;/strong&gt; with the answer attached. Then and only then will you receive your true invitation to the games.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Yours truly,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Bailey M&lt;br&gt;&#xA;  Puzzlemaster for the Treasure Hunt 'Round the World&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At last, you can begin the puzzling experience of a lifetime! Excitedly, you flip over the card and read:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$  &#xA; ? = \frac{303116229854873}{31}  &#xA; $$  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;We've given you the answer... &lt;em&gt;but what's the question?&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The story continues in the next part,&#xA; &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16620/treasure-hunt-round-the-world-clue-1&quot;&gt;Treasure hunt &amp;#39;round the world! (clue 1)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This is no longer being treated as a semi-interactive puzzle. Rather, it will be a linked-puzzle treasure hunt, which each clue being its own puzzle.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2015-06-24T01:13:09.673" Title="Treasure hunt 'round the world! (prologue)" Tags="|enigmatic-puzzle|" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16552" PostTypeId="2" ParentId="16470" CreationDate="2015-06-17T19:39:29.737" Score="0" Body="&lt;p&gt;I'm going with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wisdom&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You might think you don't need me, that you can go without,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Foolish people think they don't need wisdom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I will come lurking, there will be no doubt.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Others will try to give you wisdom to people who do foolish things. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You may try to get rid of me, you ignorant fool!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fool! :) Also, foolish people try to ignore the words of the wise.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can bear gifts, I'm a powerful tool.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Its' good to have wisdom to keep you out of trouble.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You might not expect me, You might try to ignore,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wisdom can come from unexpected places.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But my power over you, it cannot be tore.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Time tested wisdom wins out over foolish pursuits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can try and put up a valiant fight, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, ignoring the words of the wise.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I always succeed, you strengthen my might!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, time tested wisdom wins over foolish pursuits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can hide in the shadows, and creep in the night.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, ignoring the words of the wise.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I will find you, and show you my smite!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, time tested wisdom wins over foolish pursuits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some can't get enough, of what I can give, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people love learning and gaining wisdom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some struggle without, from the life that they live.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Foolishness can get a person into all kinds of trouble.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13286" LastEditorUserId="11244" LastEditDate="2015-06-17T19:44:00.860" LastActivityDate="2015-06-17T19:44:00.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16553" PostTypeId="2" ParentId="16540" CreationDate="2015-06-17T19:56:03.197" Score="7" Body="&lt;p&gt;I think the answers should've been&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;12, 27 and 16&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The third guy revealed something about the number 12. It could possibly be some sort of key pre-decided by the club.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first guy answered 22, 27 and 17.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;27 - 22 = 5, or 05 &lt;br&gt;5 - 0 = 5&lt;br&gt;12 + 5 = 17 i.e. the third number&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second guy answered 1, 18 and 18.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;18 - 1 = 17&lt;br&gt;7 - 1 = 6&lt;br&gt;12 + 6 = 18 i.e. the third number&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The pattern I guessed is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Do simple arithmetic calculation on the first two operations.&lt;br&gt;Subtract the first number from the second number, if the result is a two digit number, do the subtraction again.&lt;br&gt;Add the final result to 12.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So accordingly,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;12 and 27 are obtained from the first two operations&lt;br&gt;27 - 12 = 15&lt;br&gt;5 - 1 = 4&lt;br&gt;12 + 4 = 16&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What kind of club it was, I couldn't figure it out.&lt;/p&gt;&#xA;" OwnerUserId="12619" LastActivityDate="2015-06-17T19:56:03.197" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16555" PostTypeId="2" ParentId="16443" CreationDate="2015-06-17T20:13:59.887" Score="1" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The heart&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;in (mostly) perfect time&#xA;and, I promise,&#xA;I'll never miss a step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I never stop, I never rest&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Referencing the heart always beating in time (mostly)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I'll deceive you any way I can,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; reference to the bible verse saying the heart is deceitful above all things&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13290" LastActivityDate="2015-06-17T20:13:59.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16556" PostTypeId="2" ParentId="16443" CreationDate="2015-06-17T20:22:18.180" Score="2" Body="&lt;p&gt;I have some more pieces to fill in, but are you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Hand Drawn Animation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I offer breathing without breath&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It's possible to animate a character breathing without actually breathing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and dancing up and down, round and round&#xA;in (mostly) perfect time&#xA;and, I promise,&#xA;I'll never miss a step.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Animated dancers will always hit their cues&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though diving, flying or driving&#xA;it's all upon my stairs&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;In this form of animation, all movement is done by moving cels in 'steps'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;from Kitty Hawk to Clermont&#xA;moving from wherever, to wherever&#xA;with whoever, doing whatever&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It's possible for animated characters to do pretty much anything&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but if you want the truth&#xA;nothing here has been anywhere&#xA;(except perhaps, round the bend)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;While animation can represent any place, it physically doesn't really move&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's no two ways about it&#xA;my brother's never told a lie.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This could be in reference to other forms of media. Cameras show us the real world (even though we can mock things up)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I'll deceive you any way I can,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;but animation can defy all reality&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and if you stumble upon these steps&#xA;you won't find me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;if you mess up the steps of cels, you are likely to mess up the entire animation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I never stop, I never rest&#xA;and though it may appear that I have left&#xA;nothing on these steps&#xA;has ever felt the need to die.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Hand drawn animation has fallen out of vogue recently because of the advent of CG. However, the images on the cels, now forever recorded to film and later digital media, will never truly disappear from the world.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12242" LastActivityDate="2015-06-17T20:22:18.180" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16557" PostTypeId="2" ParentId="16540" CreationDate="2015-06-17T20:23:08.903" Score="2" Body="&lt;p&gt;I'm not sure, but i'm guessing it has to be something about the pattern of the answers..?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since all the answers repeat digits i guess the pattern is like this: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The first one three 2s and two 7s, the second one three 1s and two 8s... so i guess the answer should be a number with those patterns... so, my answer to the last sum is either 12 or 21, as there're already two 2s, a 1 and a 7 (you could say it could also be 72 or 27, but their special number is 12).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Anyway, great puzzle.&lt;/p&gt;&#xA;" OwnerUserId="13288" LastEditorUserId="13288" LastEditDate="2015-06-17T21:08:20.077" LastActivityDate="2015-06-17T21:08:20.077" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16558" PostTypeId="2" ParentId="16551" CreationDate="2015-06-17T20:23:21.200" Score="1" Body="&lt;p&gt;Please tell me its as easy as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 303116229854873 / 31 = 9777942898544.29&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13286" LastActivityDate="2015-06-17T20:23:21.200" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16559" PostTypeId="2" ParentId="16474" CreationDate="2015-06-17T20:34:49.143" Score="-2" Body="&lt;p&gt;Where did I go wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If m = small fish, 2m = medium fish, then 3m = small &amp;amp; medium fish. Therefore,(equation 1) 3m + q = 35.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3m is 5m &amp;lt; q, so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&#xA;(eqn. 2) 3m = sqrt(q) - 5m, therefore:&#xA;(eqn. 3) 8m = sqrt(q)&#xA;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substitution eqn. 3 into eqn. 1 gives us:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(eqn. 4) 3m + 64m^2 = 35&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore,&#xA;m = 0.7164 - a non-integer, and therefore eqn. 4 is not true.&lt;/p&gt;&#xA;" OwnerUserId="13291" LastActivityDate="2015-06-17T20:34:49.143" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16562" PostTypeId="2" ParentId="16551" CreationDate="2015-06-17T20:46:53.327" Score="3" Body="&lt;p&gt;Maybe a coincidence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 31 digits in base 3&lt;br&gt;&#xA; 1110202020121222020012020221222&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12863" LastActivityDate="2015-06-17T20:46:53.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16563" PostTypeId="2" ParentId="16551" CreationDate="2015-06-17T20:56:22.023" Score="4" Body="&lt;p&gt;Perhaps I'm thinking too laterally, but it seems like the question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;What's the question?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or to interpret it slightly differently, but in the same spirit:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The question is &quot;What?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7459" LastActivityDate="2015-06-17T20:56:22.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16564" PostTypeId="2" ParentId="12796" CreationDate="2015-06-17T20:57:24.937" Score="2" Body="&lt;p&gt;&lt;b&gt;Here are my answers, which are a combination of some prior answers and my improvements on certain ones:&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Score: 92&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1 = \frac{7}{7} \ \  \ \ $ [Score +1]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2 = \sqrt{7} \ \ \ \ $ [Score +2]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3 = \sqrt{7+7} \ \ \ \ $ [Score +3]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$4 = \sqrt{7+7+7} \ \ \ \ $ [Score +4]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5 = 7-\sqrt{7} \ \ \ \ $ [Score +3]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$6 = 7-\frac{7}{7} \ \ \ \ $ [Score +2]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$7 = 7 \ \ \ \ $ [Score +0]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$8 = 7+\frac{7}{7} \ \ \ \ $ [Score +2]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$9 = 7+\sqrt{7} \ \ \ \ $ [Score +3]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$10 = 7+\sqrt{7+7} \ \ \ \ $ [Score +4]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$11 =\sqrt{7(7)7} \ - \ 7 \ \ \ \ $ [Score +3]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$12 = 7+7-\sqrt{7} \ \ \ \ $ [Score +4]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$13 = 7+7-\frac{7}{7} \ \ \ \ $ [Score +3]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$14 = 7 + 7 \ \ \ \ $ [Score +1]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$15 = 7 + 7 + 7/7 \ \ \ \ $ [Score +3]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$16 = 7 + 7 + \sqrt{7} \ \ \ \ $ [Score +4]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$17 = \sqrt{7(7(7) - 7)} \ \ \ \ $ [Score +3]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$18 = \sqrt{7(7)7} \ \ \ \ $ [Score +2]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$19 = \sqrt{7(7)7 + 7(7)} \ \ \ \ $ [Score +3]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$20 = 7+7+7-\frac{7}{7} \ \ \ \ $ [Score +4]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$21 = 7+7+7 \ \ \ \ $ [Score +2]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$22 = 7+7+7+\frac{7}{7} \ \ \ \ $ [Score +4]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$23 = 7+7+7+\sqrt{7} \ \ \ \ $ [Score +5]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$24 = \sqrt{(7(7) - 7)(7 + 7)} \ \ \ \ $ [Score +4]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$25 = \sqrt{7(7)7} \ + \ 7 \ \ \ \ $ [Score +3]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$26 = \sqrt{7(7)7 \ + \ 7(7)7} \ \ \ \ $ [Score +3]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$27 = \sqrt{(7(7) + 7)7 \ +  \ 7(7)7} \ \ \ \ $ [Score +4]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$28 = 7+7+7+7 \ \ \ \ $ [Score +3]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$29 = 7+7+7+7+\frac{7}{7} \ \ \ \ $ [Score +5]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$30 = \sqrt{\sqrt{7^7}} \ \ \ \ $ [Score +5]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt; [1 + 2 + 3 + 4 + 3 + 2]  +  [0 + 2 + 3 + 4 + 3 + 4]  +  [3 + 1 + 3 + 4 + 3 + 2]  +  [3 + 4 + 2 + 4 + 5 + 4]  +  [3 + 3 + 4 + 3 + 5 + 5]  =  92&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;.&#xA;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;  The above solution will have to be accepted as best so far, else the problem poser is a cheat.&lt;/b&gt;&lt;/p&gt;&#xA;" OwnerUserId="13203" LastEditorUserId="13203" LastEditDate="2015-06-18T17:28:45.243" LastActivityDate="2015-06-18T17:28:45.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16565" PostTypeId="2" ParentId="16331" CreationDate="2015-06-17T21:03:37.783" Score="0" Body="&lt;p&gt;My answer assumes a few things about the situation that I must put forth:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Although I may not move throughout the prison I am able to move my arms as I can pull the tails of the drones but have no &quot;pull&quot; spell.&lt;/li&gt;&#xA;&lt;li&gt;The drones are flying and therefore their mass does not matter when moving them.&lt;/li&gt;&#xA;&lt;li&gt;The drones fly low enough for me to grab them, as I can pull their tails.&lt;/li&gt;&#xA;&lt;li&gt;The drones will accept actions I take upon them as they allow me to pull their tails.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;With these assumptions this is my answer:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;push Arthur as far from me as possible in what appears to be the direction of the far side of the wall.&lt;/li&gt;&#xA;&lt;li&gt;Teleport to Arthur.&lt;/li&gt;&#xA;&lt;li&gt;Repeat steps 1 and 2 until we reach a wall.&lt;/li&gt;&#xA;&lt;li&gt;Use two pieces of clothing torn from my garments pulled taught to make a line that is perpendicular to the line connecting two points on the circumference (one piece is on two points, the other placed perpendicular to it) &lt;/li&gt;&#xA;&lt;li&gt;Push Arthur out to tip of perpendicular piece of clothing&lt;/li&gt;&#xA;&lt;li&gt;Teleport to Auther and pull his tail&lt;/li&gt;&#xA;&lt;li&gt;If result is not 180 degrees move clothing and Arthur untill answer is 180 degrees. Teleporting to Arther each time to pull.&lt;/li&gt;&#xA;&lt;li&gt;Push Arther along perpendicular cloth until he is against the wall.&lt;/li&gt;&#xA;&lt;li&gt;Teleport to Bob&lt;/li&gt;&#xA;&lt;li&gt;As bob and Dale are in the same place pull Dale's tail&lt;/li&gt;&#xA;&lt;li&gt;I now have the diameter of the circle and can calculate the area as &#xA;A = (d/2)*pi^2&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2015-06-17T21:03:37.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16567" PostTypeId="2" ParentId="16474" CreationDate="2015-06-17T21:45:09.980" Score="1" Body="&lt;p&gt;I'm thinking the total number the fisher caught is 175, since 6:30 is highlighted in the puzzle. My reasoning is that 6:30 is the ratio between the total fish caught by the farmer and the total caught by the fisher or 1:5. So if the farmer caught 35 fish, that would mean the fisher caught 175. I can't work out q using this though so I might be wrong or just bad at math or both.&lt;/p&gt;&#xA;" OwnerUserId="13286" LastActivityDate="2015-06-17T21:45:09.980" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16568" PostTypeId="2" ParentId="16460" CreationDate="2015-06-17T21:45:33.370" Score="2" Body="&lt;p&gt;You'll have to bear with me on this one as we're heading past &lt;em&gt;lateral&lt;/em&gt; into full-on &lt;strong&gt;obtuse&lt;/strong&gt;. Buckle up.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The man's wife was beaten to death with a very distinctive, personalised golf club. She was found &lt;em&gt;next to&lt;/em&gt; her husband's car.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You've stated the man &quot;&lt;em&gt;drives 20 minutes&lt;/em&gt;&quot; to the crime scene. I put it to you that when he was called, he played a round of golf (&quot;driving&quot;) to where he'd parked his car.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The police, recognising that the club he was holding was identical to the murder weapon (and noting his callous indifference), arrested him on his arrival.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4852" LastActivityDate="2015-06-17T21:45:33.370" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16569" PostTypeId="2" ParentId="16474" CreationDate="2015-06-17T21:47:44.793" Score="1" Body="&lt;h1&gt;Puzzle 1&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;If the farmer catches $x$ small/medium and $y$ large fish, then we are given&#xA;$$ x+y=35$$&#xA;$$ m+2m+x=\sqrt{q+y}-5m,$$&#xA;so&#xA;$$ q=(8m+x)^2+x-35=64m^2+(16m+1)x+x^2-35.$$&#xA;If $m=0$, we need $x\ge6$ to make $q$ nonnegative, so $q\ge 7$ (with equaliy if $x=6$).&#xA;If $m&amp;gt;0$ then certainly $q\ge 64-35=29$. These solutions seem weird as the farmer seems to be better at fisching than the fisher, but sometimes such puzzles implicitly ask for tha &lt;em&gt;smallest&lt;/em&gt; possible solution. So that would be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The number of big fish the fisher caught os $7$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or let's just say &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The number of big fish the fisher caught is of course $q$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or alternatively&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If they serve so much fish, I don't care to be invited.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1731" LastEditorUserId="7307" LastEditDate="2015-06-23T21:20:48.967" LastActivityDate="2015-06-23T21:20:48.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16574" PostTypeId="2" ParentId="16348" CreationDate="2015-06-17T23:44:00.857" Score="1" Body="&lt;p&gt;Well, since it says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Belong to him, the mysterious Alexandrian.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It must be referred to Alexandria (and not the one in Louisiana :P). And I think the Alexandrian mentioned maybe is from the Ptolemaic dynasty. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the Two that do make One&#xA;  .&#xA;  !Since is not about the Nile river, maybe is about the Upper Egypt and the Lower Egypt. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And if you add this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Their seeds grew from the Two that do make One,&#xA;  Which share no part except for another One.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I guess its about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a pharaon as the &quot;another One&quot;, and &quot;the Two that do make One&quot; is the Upper Egypt and the Lower Egypt forming Egypt ...?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Anyway, about this part&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Words, meaningless, focus on their numbers,&#xA;  To unravel this poem, the key always remains Three,&#xA;  Numbers speak languages untold, decipher them to avoid any blunders. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Translate to a system, one familiar to the Owner-&#xA;  Mysterious still, but more unfolding may prove to the sleuth,&#xA;  That like the flowers here, this poem holds one truth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This makes me believe that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the two last line of the first tercet is about the Rosetta Stone, as its script is written in &lt;strong&gt;three different languages&lt;/strong&gt;. Since the Rosetta Stone was found in Menfis, i'll guess the forest has to be near there, or maybe next to the Lake Nasser that is also near Menfis.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13288" LastEditorUserId="13288" LastEditDate="2015-06-18T10:35:41.623" LastActivityDate="2015-06-18T10:35:41.623" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16575" PostTypeId="1" AcceptedAnswerId="16577" CreationDate="2015-06-18T01:04:39.967" Score="21" ViewCount="1685" Body="&lt;p&gt;A nice easy puzzle for the math lovers out there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're all familiar with the $\max$ function: $$\max \left( {x,y} \right) \triangleq \left\{ {\begin{array}{*{20}{l}}&#xA;  x&amp;amp;{{\text{if }}x \geq y} \\ &#xA;  y&amp;amp;{{\text{otherwise}}} &#xA;\end{array}} \right.\,\,\,\forall\, x,y \in \mathbb{R}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we wish to implement this function using no conditional logic and only a basic set of operations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;multiplication, division, addition, and subtraction&lt;/li&gt;&#xA;&lt;li&gt;real-valued numeric constants&lt;/li&gt;&#xA;&lt;li&gt;the floor function, $\left\lfloor \cdot \right\rfloor$, which yields the greatest integer $\leq$ its argument (e.g. $\left\lfloor -1.5 \right\rfloor = -2$)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Question&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can you realize the $\max$ function for all real inputs using &lt;em&gt;only&lt;/em&gt; these basic operations?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2015-06-18T15:02:34.660" Title="Reach the Maximum by Flooring It" Tags="|mathematics|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16576" PostTypeId="2" ParentId="16551" CreationDate="2015-06-18T01:25:35.047" Score="31" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; convert 303116229854873 to Base-31, we get BECAUSE789.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The question is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Why was 6 afraid of 7?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because 7 ate 9&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12332" LastActivityDate="2015-06-18T01:25:35.047" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16577" PostTypeId="2" ParentId="16575" CreationDate="2015-06-18T01:52:26.583" Score="27" Body="&lt;p&gt;The graph of $f(x):=x\cdot(1+x^2)^{-1}$ looks like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ezvoh.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This function&#xA;satisfies&#xA;$$&#xA;\big\lfloor f(x)\big\rfloor =\begin{cases}0:&amp;amp;x\geq 0,\\-1:&amp;amp;x&amp;lt;0. \end{cases}&#xA;$$&#xA;Using this, we can check that the function&#xA;$$&#xA;x-\left\lfloor\frac{x-y}{1+(x-y)^2}\right\rfloor(y-x)= x-\big\lfloor f(x-y)\big\rfloor(y-x)&#xA;$$&#xA;takes the value $\max\{x,y\}$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;if $x\geq y$, then $\big\lfloor f(x-y)\big\rfloor =0$, so $x-\big\lfloor f(x-y)\big\rfloor(y-x)=x-0\cdot(y-x) =x$,&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;if $x&amp;lt;y$, then $\big\lfloor f(x-y)\big\rfloor =-1$, so $x-\big\lfloor f(x-y)\big\rfloor(y-x)=x-(-1)\cdot(y-x) =y$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-06-18T15:02:34.660" LastActivityDate="2015-06-18T15:02:34.660" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16579" PostTypeId="1" AcceptedAnswerId="16585" CreationDate="2015-06-18T04:03:40.863" Score="5" ViewCount="1021" Body="&lt;p&gt;I am writing to inform you that I have constructed another set of clues that will each lead you to a movie title. To make it more interesting, each of the movies has something in common. Whatever that something is leads into the final clue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;lt;h1&amp;gt;hit&amp;lt;/h1&amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;P227, M4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;3.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;iddqd&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;4.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A Konami hit performs music&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Final.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Common(1,2,3,4)+FB&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Just to be even more confusing, the final answer is not a movie.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The final answer is the name of something you could have seen in an auditorium.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm really not convinced that this is a well-constructed puzzle. One or more clues might be a bit of a stretch. I don't feel good about it. Someone will probably solve it anyway. :/&lt;/p&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="11297" LastEditDate="2015-06-18T17:37:45.410" LastActivityDate="2015-06-18T17:37:45.410" Title="Name that Movie (part deux)" Tags="|rebus|knowledge|computer-puzzle|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16580" PostTypeId="2" ParentId="16579" CreationDate="2015-06-18T05:02:28.363" Score="0" Body="&lt;p&gt;&lt;em&gt;Note that this is just a partial answer...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first clue leads to the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The largest hit. In the HTML code &lt;code&gt;&amp;lt;h1&amp;gt;hit&amp;lt;/h1&amp;gt;&lt;/code&gt; &lt;code&gt;h1&lt;/code&gt; means a header, and it is the largest one. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The fourth clue(not really sure...)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Konami Digital Entertainment is an Arcade Game Company famous for making games such as YUGIOH! and series such as TMNT. But I do not understand how Konami is related with music.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit(for 4):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A game company and music, what do they have in common? -- Entertainment! This must somehow be related to the answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12860" LastEditorUserId="12860" LastEditDate="2015-06-18T06:12:07.587" LastActivityDate="2015-06-18T06:12:07.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16581" PostTypeId="2" ParentId="16579" CreationDate="2015-06-18T06:47:41.140" Score="0" Body="&lt;p&gt;&lt;em&gt;Another partial answer...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.imdb.com/title/tt1272878/&quot; rel=&quot;nofollow&quot;&gt;2 Guns&lt;/a&gt;. P227 and M4 are both guns.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.imdb.com/title/tt0419706/&quot; rel=&quot;nofollow&quot;&gt;Doom&lt;/a&gt;. iddqd is the cheat code to activate God mode in the cult classic game Doom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-06-18T06:47:41.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16583" PostTypeId="2" ParentId="16579" CreationDate="2015-06-18T07:03:16.460" Score="1" Body="&lt;p&gt;&lt;em&gt;This is partial answer. As each clue lead to a movie title,&lt;/em&gt; &lt;br&gt;&lt;br&gt;&#xA;First clue says&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Biggest hit, I am not sure about parameters to define biggest hit : So &lt;strong&gt;Avatar(2009)&lt;/strong&gt; or &lt;strong&gt;The Big Hit(1998)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second clue&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;P227, M4 both are guns so, &lt;strong&gt;Guns(1990)&lt;/strong&gt; or &lt;strong&gt;2 Guns(2013)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Third clue&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;iddhq is cheat code for shooter game Doom which gives God mode. &lt;strong&gt;Doom(2005)&lt;/strong&gt; is also a movie.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fourth clue&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Movies related to Konami games, &lt;strong&gt;Silent Hill(2006)&lt;/strong&gt; or &lt;strong&gt;Teenage Mutant Ninja Turtles(2014)&lt;/strong&gt;&lt;br&gt;&lt;br&gt;All 4 movies suggest answer is related to war, killing, guns.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13310" LastEditorUserId="7308" LastEditDate="2015-06-18T07:09:05.907" LastActivityDate="2015-06-18T07:09:05.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16584" PostTypeId="2" ParentId="3859" CreationDate="2015-06-18T07:33:48.960" Score="0" Body="&lt;p&gt;I will highlight two possibilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, considering the windshield direction the killer was in the backseat and he may have escaped from the trunk of the car by some contraptions through the boot of the car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, he might have made a duplicate key set to lock it from outside to baffle the authorities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In either cases he must be familiar withe car earlier to devise such things. So the killer might be an acquaintance of the victim - or the killer was never there in car, which is a less realistic possibility since no weapon was found so the killer must have been gone with the weapon.&lt;/p&gt;&#xA;" OwnerUserId="13317" LastEditorUserId="7308" LastEditDate="2015-06-18T10:56:32.010" LastActivityDate="2015-06-18T10:56:32.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16585" PostTypeId="2" ParentId="16579" CreationDate="2015-06-18T07:34:37.143" Score="9" Body="&lt;p&gt;OK, let's have a go. 4 is quite the stretch, but all the other match perfectly (at least in my head).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First the common:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Common(1,2,3,4)+FB&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The common is &lt;strong&gt;Mark Wahlberg&lt;/strong&gt;, &lt;strike&gt;add FaceBook to that and we get Mark Zuckerberg&lt;/strike&gt; and the answer is &lt;strong&gt;Marky Mark and the Funky Bunch&lt;/strong&gt; (a hip-hop group led my Mark Wahlberg).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now the clues:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;lt;h1&amp;gt;hit&amp;lt;/h1&amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This will make &quot;hit&quot; big. It will even make it &quot;the biggest&quot;, so &lt;strong&gt;The Biggest Hit (1998)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;P227, M4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The name of 2 guns, so &lt;strong&gt;2 Guns (2013)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;iddqd&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A cheat code that will make you invincible in doom, so &lt;strong&gt;Invincible (2006)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A Konami hit performs music&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strike&gt;Now this one will be a stretch... Konami, had this game Guitar Freaks. Later Guitar Heroes popped up. In Guitar Heroes there was this character, who is named Izzy, which is the nickname of the titular character of &lt;strong&gt;Rock Star (2011)&lt;/strong&gt;&lt;/strike&gt; Thanks to &lt;a href=&quot;https://puzzling.stackexchange.com/users/13268/cerberus&quot;&gt;Cerberus&lt;/a&gt;'s &lt;a href=&quot;https://puzzling.stackexchange.com/a/16587/5610&quot;&gt;answer&lt;/a&gt;, this should actually be &lt;strong&gt;Contraband (2012)&lt;/strong&gt;, derived from Contra - a Konami game (part of Konami Classics Series: Arcade Hits), and band.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-06-18T12:43:55.270" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16587" PostTypeId="2" ParentId="16579" CreationDate="2015-06-18T07:43:46.560" Score="3" Body="&lt;p&gt;In addition to dmg's answer, the fourth clue most likely refers to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Contraband (2012), Contra+band, of which the cast also includes Mark Wahlberg. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13268" LastActivityDate="2015-06-18T07:43:46.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16588" PostTypeId="2" ParentId="16348" CreationDate="2015-06-18T07:46:37.990" Score="4" Body="&lt;p&gt;I got some guesses. &#xA;First one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;em&gt;3-dimensional coordinate system&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&#xA;&#xA; &#xA; trees may be lines, if we look on all these lines we may feel like &#xA; standing in the forest. these lines - 'trees' may 'grow' when we&#xA; change 2D to 3D system. coordinate system was developed (among others)&#xA; by  Euclides (famous Alexandrian).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Their seeds grew from the Two that do make One,&lt;br&gt;&#xA;Which share no part except for another One.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if we stand in a point (defined in 2D on a plane), two dimensions&#xA; define that - one - point, and the point is where 1st and 2nd&#xA; dimensions share 3rd coordinate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Words, meaningless, focus on their numbers,&lt;br&gt;&#xA;To unravel this poem, the key always remains Three,&lt;br&gt;&#xA;Numbers speak languages untold, decipher them to avoid any blunders.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Translate to a system, one familiar to the Owner-&lt;br&gt;&#xA;Mysterious still, but more unfolding may prove to the sleuth,&lt;br&gt;&#xA;That like the flowers here, this poem holds one truth.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If we want to describe a point somewhere, talking and describing with&#xA; words in inaccurate, it's best to use coordinates (numbers).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ok. My new discovery, and I believe that a correct one is that:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;famous Alexandrian is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Euclid of Alexandria, in this puzzle, renowned for &lt;em&gt;Elements&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the proper calculation that will lead to the answer is :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eudlidean division &gt; Modulo operation, which must include '3' number as modulo; I also believe that it is just described as 'arithmetic of the clock' by the Author of this puzzle.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'm short on time now, so I just let You guys use my discovery :) I try to find some time later to get into it deeper.&lt;/p&gt;&#xA;" OwnerUserId="12739" LastEditorUserId="12739" LastEditDate="2015-06-30T09:03:20.143" LastActivityDate="2015-06-30T09:03:20.143" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16589" PostTypeId="2" ParentId="16348" CreationDate="2015-06-18T08:09:41.630" Score="2" Body="&lt;p&gt;I don't know why, but I got the impression that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Trees are files, and you're on a hard drive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13095" LastActivityDate="2015-06-18T08:09:41.630" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16590" PostTypeId="2" ParentId="16540" CreationDate="2015-06-18T08:17:27.667" Score="4" Body="&lt;p&gt;Since the tag is lateral thinking, My thought is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It's Dozen club. A club of 12 people.&lt;br /&gt;&lt;br /&gt;&#xA;All the other members were admitted because they answered the numbers correctly. And the question &quot;What's our special number&quot; was asked to the 12th person. Their special number is 12 since they have 12 members. Since all the 12 members were present you were not admitted. The guard already knew you are not a member and that's why he said &quot;Are you sure? If so, tell me..&quot; at the beginning. Then he asked the question to play with you. Whatever answer you give, he will kick you out. If you have knocked the door before the 12th person, you would have been admitted.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastActivityDate="2015-06-18T08:17:27.667" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16592" PostTypeId="2" ParentId="16474" CreationDate="2015-06-18T09:55:58.393" Score="17" Body="&lt;h2&gt;Puzzle 1&lt;/h2&gt;&#xA;&lt;p&gt;There are &lt;strong&gt;m&lt;/strong&gt; small fishes, &lt;strong&gt;2m&lt;/strong&gt; medium fishes and &lt;strong&gt;q&lt;/strong&gt; large fishes. Therefore the first (1) equation should be:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ m+2m=\sqrt{q}-5m$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$8m=\sqrt{q}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The second (2) one:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$m+2m+q=T$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$q=T-3m$$&lt;/span&gt;&#xA;Where &lt;strong&gt;T&lt;/strong&gt; is the &lt;strong&gt;T&lt;/strong&gt;otal number of fishes that the &lt;em&gt;fisherman&lt;/em&gt; caught. (&lt;strong&gt;T&lt;/strong&gt; is different than 35 assuming the fisherman and the farmer are different)&lt;/p&gt;&#xA;&lt;p&gt;Replacing (2) in (1), squaring and rearranging gives the following quadratic equation:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$64m^2+3m-T=0$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;This equation gives an integer value for &lt;strong&gt;m&lt;/strong&gt; for a high increasing number of &lt;strong&gt;T&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If &lt;strong&gt;T = 67, m = 1 and q = 64&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;If &lt;strong&gt;T = 262, m = 2 and q = 256&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;If &lt;strong&gt;T = 585, m = 3 and q = 576&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;If &lt;strong&gt;T = 1036, m = 4 and q = 1024&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;....&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If my logic is correct, there should be a relation between one of those values and the hint about the importance of apologies which I'm not able to figure out.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The &amp;quot;apology&amp;quot;: &lt;code&gt;So sorry to wake you up so early.  At least, we apologize for you having to get up so early. Apologies are very important!!  We LOVE LOVE LOVE LOVE the length of our apologies--they are also very important!  You're important to us.  Unless you fail... heh.&lt;/code&gt; contains:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;&lt;strong&gt;256&lt;/strong&gt; characters. Therefore, &lt;strong&gt;q = 256&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12731" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-06-26T16:29:45.953" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16593" PostTypeId="2" ParentId="15809" CreationDate="2015-06-18T11:33:39.803" Score="5" Body="&lt;p&gt;Partial answer: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;line 4&lt;/strong&gt;: Found 2 lovely (F from previous line, Hound-H, 2, L=50, Dove-D, L=50, Y)&lt;br/&gt;&lt;br/&gt;&#xA; &lt;strong&gt;line 5&lt;/strong&gt;: Walks to T(?) you (Wall, KEY - EY, 2 T (Tell/Take - can't figure out), U) &lt;br/&gt;&lt;br/&gt;&#xA; &lt;strong&gt;line 6&lt;/strong&gt;: When you do! (W, hen, yew, D=500, O!) &lt;br/&gt;&lt;br/&gt;&#xA; &lt;strong&gt;last line&lt;/strong&gt;&quot;: the last two words seem to me like be visitor (Bee, &lt;strong&gt;wiz&lt;/strong&gt;ard,storm-M)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Still working on the remaining words.&lt;/p&gt;&#xA;" OwnerUserId="8756" LastActivityDate="2015-06-18T11:33:39.803" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16597" PostTypeId="2" ParentId="14974" CreationDate="2015-06-18T12:11:55.227" Score="3" Body="&lt;p&gt;Just looking at the numbers, we got the answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; File D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C and D's start are very similar. A and B is different and we did not see any similarities. The Mathematician and the Lottery guy did not see the real data set so we concluded that A and B (witch is very different) is the  data sets of those two.&lt;br&gt;&#xA; The manager did see the real data, so the data he created is very similar to the real data.If you take a look at D, you will see a lot more three digit integers. A person will most likely not include that because it will stand out.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13326" LastEditorUserId="10384" LastEditDate="2015-06-18T12:28:46.117" LastActivityDate="2015-06-18T12:28:46.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16598" PostTypeId="2" ParentId="16540" CreationDate="2015-06-18T12:38:45.943" Score="56" Body="&lt;p&gt;The club is a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Scrabble club&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The calculations are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the scrabble values of the letters. For example &#xA;TWENTYONE = 15&#xA;THREE = 8&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the answer for 21+3 is 23. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Twelve is a special number, in that &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;it has its own scrabble score:&#xA;T (1) W (4) E (1) L (1) V (4) E (1) gives a total of 12&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Edit from the OP&lt;/em&gt;&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is the point values of each tile in the &lt;strong&gt;&lt;em&gt;American-English version&lt;/em&gt;&lt;/strong&gt; of Scrabble (&lt;em&gt;that's why I added the reference to NYC as a location for the story&lt;/em&gt;) : A=1, B=3, C=3,D=2, E=1, F=4, G=2, H=4, I=1, J=8, K=5, L=1, M=3, N=1, O=1, P=3, Q=10, R=1, S=1, T=1, U=1, V=4, W=4, X=8, Y=4, Z=10&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;these are the values of each &lt;em&gt;number/word&lt;/em&gt; mentioned in the puzzle: ONE=3, THREE=8, FOUR=7, TEN=3, ELEVEN=9, FOURTEEN=11, SIXTEEN=14, SEVENTEEN=12, TWENTYONE=15, TWENTYFOUR=19, TWENTYEIGHT=21, THIRTYSEVEN=20, THIRTYEIGHT=21&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7316" LastEditorUserId="12628" LastEditDate="2015-06-22T16:43:46.227" LastActivityDate="2015-06-22T16:43:46.227" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16601" PostTypeId="1" AcceptedAnswerId="16602" CreationDate="2015-06-18T12:56:25.563" Score="6" ViewCount="937" Body="&lt;p&gt;In the spirit of thirds, this list is a list clues that roughly name movies that are each the first of a trilogy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Not your breakfast cereal&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Oakland helps out Moses&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;3.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;The latest addition to late comedian Bob's family tree&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some pedants may refute the &quot;triloginess&quot; of some.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="11297" LastEditDate="2015-06-18T17:37:02.480" LastActivityDate="2015-06-18T17:37:02.480" Title="Name that Movie (part III)" Tags="|knowledge|cryptic-crosswords|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16602" PostTypeId="2" ParentId="16601" CreationDate="2015-06-18T13:08:36.027" Score="7" Body="&lt;p&gt;&lt;strong&gt;1:&lt;/strong&gt; Not your breakfast cereal  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Matrix (They're My Trix) &lt;br&gt; (from LogicianWithAHat's answer)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2:&lt;/strong&gt; Oakland helps out Moses:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Raiders of the Lost Ark (Indiana Jones)&lt;br&gt;&#xA; -- Oakland = Raiders (NFL team), and Moses built the Ark of the Covenant&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3:&lt;/strong&gt; The latest addition to late comedian Bob's family tree&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A New Hope (Star Wars)&lt;br&gt;&#xA; -- Bob Hope is a late comedian, and New is the latest addition&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1563" LastEditorUserId="2627" LastEditDate="2015-06-18T14:44:24.457" LastActivityDate="2015-06-18T14:44:24.457" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16603" PostTypeId="2" ParentId="16601" CreationDate="2015-06-18T13:14:42.827" Score="5" Body="&lt;p&gt;&lt;strong&gt;1:&lt;/strong&gt; Not your breakfast cereal  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Matrix (They're My Trix)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The 'trilogy-ness' of this one is also in dispute&lt;/p&gt;&#xA;" OwnerUserId="11717" LastActivityDate="2015-06-18T13:14:42.827" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16604" PostTypeId="2" ParentId="16575" CreationDate="2015-06-18T13:17:17.627" Score="7" Body="&lt;p&gt;I came up with this:&#xA;$$&#xA;\frac{\big\lfloor\frac{x}{y}\big\rfloor\cdot x+\big\lfloor\frac{y}{x}\big\rfloor\cdot y}{\big\lfloor\frac{x}{y}\big\rfloor+\left\lfloor\frac{y}{x}\right\rfloor}&#xA;$$&#xA;If $x$ is higher, then $\big\lfloor\frac{y}{x}\big\rfloor = 0$ and the fraction above equals $x$, same if $y$ is higher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x=y$, then $\big\lfloor\frac{x}{y}\big\rfloor =\big\lfloor\frac{y}{x}\big\rfloor = 1$, and the fraction equals $\frac{x + y}{2}$, which is $\frac{2x}{2} = x$ (because $x = y$).&lt;/p&gt;&#xA;" OwnerUserId="13331" LastActivityDate="2015-06-18T13:17:17.627" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16606" PostTypeId="1" CreationDate="2015-06-18T13:56:23.857" Score="2" ViewCount="3155" Body="&lt;p&gt;Here are a couple of riddles for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) Where does twelve plus two equal two?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) A man walks up to you and says &quot;If you tell a lie I will stab you with a knife. If you tell the truth I will shoot you with a gun.&quot; What do you say to get out of this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.) My house is incredibly strong but breaks with a single tap what am I?&lt;/p&gt;&#xA;" OwnerUserId="13340" LastEditorUserId="5373" LastEditDate="2015-06-18T15:20:11.497" LastActivityDate="2015-07-20T15:08:05.283" Title="A few simple riddles to get your brain going" Tags="|riddle|lateral-thinking|liars|situation|" AnswerCount="8" CommentCount="3" ClosedDate="2015-06-18T17:22:41.627" ContentLicense="CC BY-SA 3.0" />
  <row Id="16607" PostTypeId="2" ParentId="16606" CreationDate="2015-06-18T13:59:02.403" Score="9" Body="&lt;p&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A clock  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;You're going to stab me with the knife&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A wooden ship (or houseboat). If the tap is the kind of tap one would use to collect sap from a tree, inserting it in the bottom of the boat would cause it to fill with water and sink (probably not the &quot;right&quot; answer, but still).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13339" LastEditorUserId="13339" LastEditDate="2015-06-18T14:31:52.587" LastActivityDate="2015-06-18T14:31:52.587" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="16609" PostTypeId="2" ParentId="16168" CreationDate="2015-06-18T14:09:43.283" Score="1" Body="&lt;p&gt;Once a bound on the number of prisoners is established, the problem may be solved in polynomial time (rather than hyper-exponential) without requiring prisoners to use an overly-complicated rule sheet, by establishing a protocol where the master initiates queries and other prisoners respond.  Time is subdivided into groups of days whose length is the known upper bound on the number of prisoners, and on each group of days either the master will either send a &quot;bit&quot; or expect to receive a bit from the prisoners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every group of days on which the master expects to receive a bit of information from the prisoners will be assigned an index number, and the most common type of query will ask slaves to identify the days of the group on which they saw lights.  Rather than having the slaves spend 2^N groups of days reporting every possible pattern of light and dark they may have seen, an interactive protocol could limit queries to the days of interest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To issue such a query, the master will send &quot;10&quot; followed by the number of the group of interest, in binary.  After that, each slave will intialize an &quot;initial sequence&quot; to the string &quot;0&quot; and proceed as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Get a bit from the master.  If &quot;0&quot;, go to step 4&lt;/li&gt;&#xA;&lt;li&gt;Output a &quot;1&quot; if the initial sequence of light/dark seen on that day matches the string.  If didn't output a &quot;1&quot;, and no &quot;1&quot; is seen, proceed to step 4.&lt;/li&gt;&#xA;&lt;li&gt;Append a &quot;0&quot; to the string and go to step 1.&lt;/li&gt;&#xA;&lt;li&gt;If the string contains any zeroes, strike all trailing &quot;1&quot;'s and replace the last &quot;0&quot; with a &quot;1&quot; and go to step 1.&lt;/li&gt;&#xA;&lt;li&gt;This query is done; await next query from the master.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Using this query approach, the master will be able to acquire all the information that could have been acquired by having slaves blindly broadcast information about everything they'd heard, but the amount of time will be reduced enormously.  If no information about the maximum number of prisoners is available I know of no mechanism to avoid having the time requirement be exponential with respect to the actual number of prisoners (since the time required for a broadcast to reach N prisoners can vary anywhere from lgN to N) but this protocol would still serve to reduce the total time requirement from being hyper-exponential to being simple exponential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if it would be possible to reduce the worst-case time below O(2^N), but perhaps there is some technique where it could be done.  The earlier-described technique of establishing the upper bound on the number of prisoners only establishes it as being between N and 2^N, which implies that one might take time 2^N to find out there are N prisoners, but it might be possible to find out more information with each length of &quot;cleanup sequence&quot; before one starts using cleanup sequences that are twice as long.&lt;/p&gt;&#xA;" OwnerUserId="1467" LastActivityDate="2015-06-18T14:09:43.283" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16610" PostTypeId="2" ParentId="16606" CreationDate="2015-06-18T14:33:45.227" Score="10" Body="&lt;p&gt;3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The House of Representatives. I believe the speaker of the house breaks the session with the tap of a gavel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13342" LastActivityDate="2015-06-18T14:33:45.227" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16612" PostTypeId="1" AcceptedAnswerId="18277" CreationDate="2015-06-18T15:03:12.657" Score="16" ViewCount="1957" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;When it comes to perspectives, the second is fun&lt;br&gt;&#xA;  The third is just fine in a romantic tongue&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Without this, to say you have lots might be trouble&lt;br&gt;&#xA;  In fact it's so helpful you might order double&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;One might rave about this and become enamored&lt;br&gt;&#xA;  Or think it's the same as a square famous hammer&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I've only had one of the seven in sight&lt;br&gt;&#xA;  I think the short version's the opposite of right&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mix them together and what have you got?&lt;br&gt;&#xA;Something this puzzle most surely is not!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Whoops, forgot about this puzzle! Here's a hint!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Each line in the formatted area is a reference to a part of the answer. LogicianWithAHat is on the right track!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10861" LastEditorUserId="10861" LastEditDate="2015-07-20T21:38:15.443" LastActivityDate="2015-07-22T22:23:04.010" Title="This title is not the answer" Tags="|riddle|english|" AnswerCount="3" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="16614" PostTypeId="2" ParentId="16606" CreationDate="2015-06-18T15:32:50.170" Score="2" Body="&lt;p&gt;For question 2: Say nothing and run away.  As valid an answer as any other, much more realistic, and violent crazy people are not (in my experience) fond of smart-asses.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2015-06-18T15:32:50.170" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16615" PostTypeId="2" ParentId="16606" CreationDate="2015-06-18T15:32:53.410" Score="3" Body="&lt;p&gt;Part 2: There's plenty of answers for this one, besides what WilliamKnapp said.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am telling a lie.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Today is opposite day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bwfffhlkbnm! (Since he doesn't understand this one, he can't tell if you are lying.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Part 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The unhatched chick in an egg.&lt;br&gt;&#xA; The house is a eggshell, which has a structure that makes it (sort of) hard to break from the outside. It breaks easily from a tap from the chick's beak from the inside.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12109" LastEditorUserId="12109" LastEditDate="2015-06-18T15:38:18.077" LastActivityDate="2015-06-18T15:38:18.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16616" PostTypeId="2" ParentId="16606" CreationDate="2015-06-18T15:47:31.493" Score="0" Body="&lt;p&gt;Question 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also ask him a question. &quot;WHY? OH PLEASE NO, WHY?&quot; is pathetic but not untrue or true.  This won't get you out of it though unless you can keep it up forever...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 3:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your house could be factory, restaurant, or other business.  It will not last very long with only one tap (sink).&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2015-06-18T15:47:31.493" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16619" PostTypeId="2" ParentId="16606" CreationDate="2015-06-18T16:09:09.967" Score="1" Body="&lt;p&gt;1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An analogue clock. Those newfangled digital clocks won't do for this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Logically one could say &quot;You will stab me with a knife.&quot; If you're telling the truth, he's going to stab you with a knife, but he only does that if you tell a lie. Of course, I also would consider hypnosis and then run away while he's asleep.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They say spider silk is the strongest natural substance. I don't know who they are, though. Like Mia Wallace wisely said, &quot;They talk a lot, don't they?&quot; to which Vincent Vega replied &quot;They certainly do.&quot; But anyway apparently per unit cross-sectional area, spider silk is supposed to be really strong. But it's also really thin so a single tap from a human will demolish it. So I'll say you're a spider.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2812" LastActivityDate="2015-06-18T16:09:09.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16620" PostTypeId="1" AcceptedAnswerId="16625" CreationDate="2015-06-18T16:10:54.557" Score="14" ViewCount="2223" Body="&lt;p&gt;&lt;em&gt;If you're interested in starting the treasure hunt from the very beginning, check out the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16551/treasure-hunt-round-the-world&quot;&gt;prologue&lt;/a&gt;!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;After mailing in the correct question to the given answer in the preliminary test, it took only a couple of days to receive your &lt;em&gt;actual&lt;/em&gt; invitation to participate in the Treasure Hunt 'Round the World. You eagerly tear into the gaudy, glittery envelope and read the letter enclosed:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dear &lt;strong&gt;&lt;em&gt;[insert username here],&lt;/em&gt;&lt;/strong&gt; &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Congratulations on your acceptance into the Treasure Hunt 'Round the World! Prepare yourself for a steeling worldwide adventure, where you will be pushed to your vastest limits in pursuit of the coveted treasure! I hope you've packed your bags, because the race... starts... NOW! Your itinerary is enclosed. Good luck, racer!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Yours truly,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Bailey M&lt;br&gt;&#xA;  Puzzlemaster for the Treasure Hunt 'Round the World&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;P.S. When you arrive at the departure airport, members of our team will be there to assist you in searching for the fastest flight and booking your ticket.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You rummage through the envelope hurriedly, pulling out the itinerary. On it, you read:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Your flight departs at &lt;strong&gt;1:47&lt;/strong&gt; from the &lt;strong&gt;Pocatello Regional Airport.&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Your destination is:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;The 63rd member, periodically;&lt;br&gt;&#xA;  a butterfly's sway, unconditionally,&lt;br&gt;&#xA;  with common tongue tied to no other,&lt;br&gt;&#xA;  spoken unlike their American brother&lt;br&gt;&#xA;  (though the two decline unanimously).&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;Once you've found the desired country,&lt;br&gt;&#xA;  your flight will arrive at &lt;strong&gt;2:53.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Luckily, you did already pack your bags! Time to get racing!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What airport are you departing from, and what city are you flying into?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Per the request of Aggie Kidd, here's a nod in the right direction, based on each line:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your flight departs at 1:47 from the Pocatello Regional Airport.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer to this may be a bit trivial, since there have been so many good responses already, but none of them are exactly what I was looking for. Everyone's on the right track. As a hint, you are departing from &lt;strong&gt;North America,&lt;/strong&gt; and Pocatello's airport code is not actually used in the puzzle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The 63rd member, periodically;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is a way to interpret the answer to this line a bit more specifically, which will significantly narrow down the number of possible countries. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;a butterfly's sway, unconditionally, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I believe this was solved rather innocuously in the comments of an answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;with common tongue tied to no other, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This line has not been interpreted as intended yet. The most common language in the destination country is also the most common language in other countries.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;spoken unlike their American brother&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This line has been interpreted correctly, though the country has not been correctly identified.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(though the two decline unanimously).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I believe rand al'thor correctly picked this one out, which should give you some guidance.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once you've found the desired country, &#xA;your flight will arrive at 2:53.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is unsolvable without first having the country name.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also a general hint about how to solve this puzzle: When I was creating it, I made a blanket list of all possible countries, then started whittling them down, clue by clue, until I had one remaining. If I had to solve this puzzle, I would try to do the same.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The story continues in the next part, &lt;a href=&quot;https://puzzling.stackexchange.com/questions/17158/treasure-hunt-round-the-world-clue-2&quot;&gt;Treasure hunt &amp;#39;round the world! (clue 2)&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2022-11-18T07:01:25.390" Title="Treasure hunt 'round the world! (clue 1)" Tags="|riddle|rhyme|enigmatic-puzzle|" AnswerCount="4" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="16621" PostTypeId="2" ParentId="16612" CreationDate="2015-06-18T16:11:19.443" Score="5" Body="&lt;p&gt;I haven't made this fit with &lt;em&gt;all&lt;/em&gt; the lines, but is the answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; WEEK?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;When it comes to perspectives, the second is fun&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The second day of the week (Monday or Tuesday, according to what convention you use).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The third is just fine in a romantic tongue&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Fine&quot; translated into French? There are a few words I've come up with, but none of them resembles &quot;Tuesday&quot; or &quot;Wednesday&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Without this, to say you have lots might be trouble&lt;br&gt;&#xA;In fact it's so helpful you might order double&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Perhaps a reference to the double E in &quot;week&quot; (E appears in &quot;you hav&lt;strong&gt;e&lt;/strong&gt; lots&quot;)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;One might rave about this and become enamored&lt;br&gt;&#xA;Or think it's the same as a square famous hammer&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Thursday is named after the Norse god Thor, who's famous for having a hammer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I've only had one of the seven in sight&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;The seven&quot; - days of the week.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I think the short version's the opposite of right&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wrong? Left?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Mix them together and what have you got?&lt;br&gt;&#xA;Something this puzzle most surely is not!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The puzzle is not &lt;em&gt;weak&lt;/em&gt; (homophone of week); it's a strong puzzle!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-06-18T16:11:19.443" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16622" PostTypeId="2" ParentId="16606" CreationDate="2015-06-18T16:12:46.477" Score="1" Body="&lt;p&gt;Another answer to #2 is to say&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Nothing. A truth or a lie will get you shot or stabbed, but saying nothing at all falls under neither of these categories and would thus leave you alive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5616" LastEditorUserId="5616" LastEditDate="2015-07-20T15:08:05.283" LastActivityDate="2015-07-20T15:08:05.283" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16623" PostTypeId="2" ParentId="16606" CreationDate="2015-06-18T16:20:12.983" Score="0" Body="&lt;p&gt;1.) Where does twelve plus two equal two?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In a &lt;a href=&quot;http://www.learning-tree.org.uk/stickpuzzles/stick_puzzles.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Matchstick Puzzle&lt;/a&gt;. We start off with 12... and then add two matches to make it 1 X 2, which is two! &lt;img src=&quot;https://i.stack.imgur.com/8teCq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2.) A man walks up to you and says &quot;If you tell a lie I will stab you with a knife. If you tell the truth I will shoot you with a gun.&quot; What do you say to get out of this?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;What if I ask you a question instead?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13259" LastEditorUserId="13259" LastEditDate="2015-06-18T16:27:47.583" LastActivityDate="2015-06-18T16:27:47.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16625" PostTypeId="2" ParentId="16620" CreationDate="2015-06-18T16:27:22.697" Score="12" Body="&lt;p&gt;&lt;em&gt;Your flight departs at 1:47 from the Pocatello Regional Airport.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Take the 1st letter of Pocatello, the 4th letter of Regional, and the 7th letter of Airport to get PIT. Interpreting that as an airport code, it looks like we're flying out of &lt;strong&gt;Pittsburgh International Airport&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;The 63rd member, periodically;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 63rd on the periodic table is Europium, with chemical symbol Eu, so we're heading for somewhere in the EU (European Union).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;a butterfly's sway, unconditionally,&lt;/em&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Unconditional sway suggests monarchy, and the Monarch Butterfly is a common butterfly species in North America.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;with common tongue tied to no other,&lt;br /&gt;&#xA;spoken unlike their American brother&lt;br /&gt;&#xA;(though the two decline unanimously).&lt;/em&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; So we want a monarchy in the EU with a common language not spoken in any other country, similar to a language spoken somewhere in the Americas, the word for &amp;quot;no&amp;quot; being the same in both languages. Let's try &lt;strong&gt;Spain&lt;/strong&gt; (taking Latin American Spanish to be a separate language, as some do consider it to be).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;Once you've found the desired country,&lt;br /&gt;&#xA;your flight will arrive at 2:53.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; As with 1:47, we take the 2nd, 5th, and 3rd letters of SPAIN to get PNA, which is the airport code for &lt;a href=&quot;https://en.wikipedia.org/wiki/Pamplona_Airport&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pamplona Airport&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So our destination city is&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;strong&gt;Pamplona, Spain&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2022-11-18T07:01:25.390" LastActivityDate="2022-11-18T07:01:25.390" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="16626" PostTypeId="1" AcceptedAnswerId="16630" CreationDate="2015-06-18T16:39:12.677" Score="7" ViewCount="713" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/X19UN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Replace a to f with the numbers 1 to 2 &amp;amp; 4 to 7, so that the sum inside each ring is the same. There is only one solution!&lt;/p&gt;&#xA;" OwnerUserId="9664" LastEditorUserId="9664" LastEditDate="2015-06-18T17:30:09.320" LastActivityDate="2015-06-18T20:06:25.543" Title="Four Rings &amp; Seven Digits" Tags="|logical-deduction|calculation-puzzle|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16627" PostTypeId="2" ParentId="16474" CreationDate="2015-06-18T16:39:20.147" Score="1" Body="&lt;h1&gt;Puzzle 1&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I started with $$m + 2m = \sqrt q - 5 m$$ to try to find a couple integers that fit the problem, and I found $64 - 40 = 24$, leading me to this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$q = 4096, m = 8, 2m = 16$$ but the sum of those is much larger than 35.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, when rereading, I found that the puzzle said that the &quot;farmer&quot; caught 35 fish, and it doesn't say anything about the fisher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one possibility is that the fisherman caught 4096 large fish?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't found what I am supposed to do with the weight of the fish though...&lt;/p&gt;&#xA;" OwnerUserId="13349" LastEditorUserId="7307" LastEditDate="2015-06-23T21:12:26.163" LastActivityDate="2015-06-23T21:12:26.163" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16629" PostTypeId="2" ParentId="16348" CreationDate="2015-06-18T16:49:04.433" Score="5" Body="&lt;p&gt;Ok, so taking the 8th Tercet very literally, I feel like the solution is about the number of words per line, or maybe as well the words that are numbers (Note that every number in the poem is capitalized, except for the &quot;one truth&quot; in the last line). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So lets start with number of words per line. This gives us:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 5,3,9&lt;br&gt;&#xA; 6,6,10&lt;br&gt;&#xA; 7,10,8&lt;br&gt;&#xA; 9,4,6&lt;br&gt;&#xA; 6,5,5&lt;br&gt;&#xA; 8,6,9&lt;br&gt;&#xA; 7,5,6&lt;br&gt;&#xA; 6,9,10&lt;br&gt;&#xA; 9,10,10&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;To unravel this poem, the key always remains Three,&lt;br&gt;&#xA;Numbers speak languages untold, decipher them to avoid any blunders.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sounds like some kind of cipher, where the key is 3. Notably, there are 3 lines per tercet and 9 = 3*3 tercets.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Translate to a system, one familiar to the Owner-&lt;br&gt;&#xA;Mysterious still, but more unfolding may prove to the sleuth, &quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Probably a hint as to what kind of cipher it is. Also somehow related to time, as per the hint. Potentially the [water clock] invented by an Alexandrian&lt;a href=&quot;https://en.wikipedia.org/wiki/Water_clock&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;?&#xA;Second line suggests that after deciphering, there will be more to solve. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Still not very helpful&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We could look at each lines position in the tercet, since it's not consistent. This gives us:&#xA;1,2,1,2,1,1,2,2,1 which could also be in binary I suppose... although maybe ternary is more likely.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Third attempt:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Aithmateic of the clock certainly makes me think Mod12 + 1, as in 11 + 2 is 1, 6 + 9 is 3, etc. How to actually apply this, however, stumps me. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And that's where I'm stuck. Next step:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Which cipher should I actually use?&lt;br&gt;&#xA; What to do with the deciphered numbers?&lt;br&gt;&#xA; Is words per line the right way to get numbers?&lt;br&gt;&#xA; What do the capital numbers in the poem mean?&lt;br&gt;&#xA; Who is the Owner/What is his System?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1968" LastEditorUserId="1968" LastEditDate="2015-06-30T21:31:31.217" LastActivityDate="2015-06-30T21:31:31.217" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16630" PostTypeId="2" ParentId="16626" CreationDate="2015-06-18T16:49:23.273" Score="6" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a=6; b=4; c=5; d=1; e=2; f=7&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sum is 10 in each circle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5359" LastEditorUserId="5359" LastEditDate="2015-06-18T17:29:02.947" LastActivityDate="2015-06-18T17:29:02.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16633" PostTypeId="2" ParentId="16626" CreationDate="2015-06-18T17:12:17.987" Score="3" Body="&lt;p&gt;The unique answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A, B,C, D,E, F--SUM&#xA;&lt;br&gt;(6, 4, 5, 1, 2, 7) [10]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-06-18T19:46:53.800" LastActivityDate="2015-06-18T19:46:53.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16634" PostTypeId="1" AcceptedAnswerId="16650" CreationDate="2015-06-18T18:04:32.023" Score="11" ViewCount="3128" Body="&lt;p&gt;You can use any mathematical operations represented by English words. These English words must be found in a commonly used dictionary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spaces are neither letters nor words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can not use the same word twice. No numerical digits are allowed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Five plus eight  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Is 13, and it has 13 letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You objective is to find the longest expression&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inspired by: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/6301/find-the-longest-arithmetic-expression-with-value-equal-to-number-of-letters-in/6303&quot;&gt;Find the longest arithmetic expression with value equal to number of letters in its english representation&lt;/a&gt;&#xA;To clarify, the difference here is that you can't repeat a word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; &lt;em&gt;Your full expression must be stated as a whole in your answer. Therefore there's no infinite letter solutions.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-07-21T11:06:52.093" Title="Longest arithmetic expression where the answer is equal to the number of letters" Tags="|open-ended|calculation-puzzle|" AnswerCount="7" CommentCount="10" ClosedDate="2021-08-16T11:13:19.703" ContentLicense="CC BY-SA 3.0" />
  <row Id="16636" PostTypeId="2" ParentId="16634" CreationDate="2015-06-18T18:27:15.813" Score="0" Body="&lt;p&gt;&lt;strike&gt;It's actually infinite.&#xA;you can keep going forever using subtraction, here's an example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;one billion minus nine hundred ninety-nine million, nine hundred ninety-nine thousand, nine hundred.&lt;br /&gt;&#xA;^- equals 100, and has one hundred characters.&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;onehundredminusseventyfour is 26 characters and evaluates to 26. It wouldn't be hard to go beyond that.&lt;br /&gt;&#xA;onethousandminusninehundredsixtyfour is 36&lt;br /&gt;&#xA;Just for fun: &lt;strike&gt;thisstatementevaluatestothenumberonehundredfiveplusfortyexponentzero&lt;/p&gt;&#xA;&#xA;&lt;p&gt;105 + 40^0 = 106.&lt;/strike&gt;&#xA;&lt;br /&gt;&#xA;&lt;br /&gt;&#xA;Come to think of it, this contest is going to end in an arbitrarily large number as the winning entry just because someone's going to figure out how to create a word problem with no repeating word, and then it's going to be a hassle reading through it to verify that it doesn't have any spelling, grammar, or punctuation errors while still using exactly one of each English word (which, I might add, is a constantly expanding language.)&#xA;&lt;br /&gt;&lt;br /&gt;&#xA;inwabambuule,mysticallando'honeycombsandrabbits,atribeofpixieslives.Theyrequirethirtyonetrombones.spritesfromtheneighboringboroughneedtwohundredthree.howmanytotalmusicalinstrumentsareneeded?giveyouranswertonearesthundredthdecimalplace&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with spaces:&#xA;in wabambuule, mystical land o' honeycombs and rabbits, a tribe of pixies lives. They require thirty one trombones. sprites from the neighboring borough need two hundred three. how many total musical instruments are needed? give your answer to nearest hundredth decimal place&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a few notes: &lt;br /&gt;&#xA;1. Yes, I made up a name but any word will fit there. &lt;br /&gt;&#xA;2. There are still quite a few synonyms for fairies I can use to expand this.&lt;br /&gt;&#xA;3. I have any number of musical instruments to add.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;br /&gt;&#xA;You could also just say --------...---infinity, where you have an infinite number of negation operators (which don't count as a repeated word, and are not expressly forbidden, and are not expressly stated not to count), but I'm going to assume that's out of the bounds of the problem because the whole point was to remove some of the loopholes that made the original problem go infinite.&lt;/p&gt;&#xA;" OwnerUserId="13068" LastEditorUserId="13068" LastEditDate="2015-06-18T19:58:13.040" LastActivityDate="2015-06-18T19:58:13.040" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16637" PostTypeId="2" ParentId="16634" CreationDate="2015-06-18T18:27:35.280" Score="3" Body="&lt;p&gt;&lt;b&gt;67 points&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;two squared, to the power of five, multiplied by itself, plus eight, minus sixty-nine &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which value is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $((2^2)^5)^2 + 8 - 69 = 67$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="10400" LastEditDate="2015-06-18T19:20:58.643" LastActivityDate="2015-06-18T19:20:58.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16638" PostTypeId="2" ParentId="16634" CreationDate="2015-06-18T18:34:10.037" Score="4" Body="&lt;p&gt;&lt;strong&gt;95 Points&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; square root of a hundred times four plus eleven minus one multiplied by two where ten is substracted and five added (95 letters)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which value is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\sqrt{100} \times 4 + 11 - 1 \times 2 - 10 + 5  = 95$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5359" LastEditorUserId="5359" LastEditDate="2015-06-18T19:47:12.623" LastActivityDate="2015-06-18T19:47:12.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16639" PostTypeId="2" ParentId="16634" CreationDate="2015-06-18T19:14:50.533" Score="11" Body="&lt;h2&gt;145 Points&lt;/h2&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$$\lvert\frac{\sqrt{9^2}}{\frac{1}{8^3}}\rvert+11+(e^{i\pi}*\frac{18,000}{10})-\frac{2,674,000}{1,000}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;In English, this can be described as:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Absolute value of root nine squared divided by inverse eight cubed plus eleven and &lt;span class=&quot;math-container&quot;&gt;$e$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$pi$&lt;/span&gt; times eighteen thousand tenths minus two point six seven four million thousandths.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Without spaces, that transcription is 145 characters. Mathematically, it evaluates in the following manner:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$$\lvert\frac{\sqrt{81}}{\frac{1}{512}}\rvert+11+(-1*1,800)-2,674$$&lt;/span&gt;&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; &lt;span class=&quot;math-container&quot;&gt;$$\lvert9*512\rvert+11-1,800-2,674$$&lt;/span&gt;&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; &lt;span class=&quot;math-container&quot;&gt;$$\lvert4,608\rvert-4,463=4,608-4,463=145$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-06-18T19:14:50.533" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16640" PostTypeId="1" AcceptedAnswerId="24186" CreationDate="2015-06-18T19:17:41.250" Score="19" ViewCount="1864" Body="&lt;p&gt;A pirate's treasure...&lt;br&gt;&#xA;Billy the pirate buried his plunder at the mouth of a large river. He killed all of his crew then made his way on foot back to civilization. He ended up getting sick with yellow fever and died in a hospital bed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just before he passed he whispered to the nurse the coordinates to the buried gold. The nurse believed it was the high fever and delirium, but he insisted she write it down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many years later her great-grandson Jack, while packing for the second Great War, found the note in a box belonging to his long dead relative. Not knowing the importance of the note, he sent a telegram over the wire to his grandfather who was still living in DC. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately the message never reached his grandfather: the Germans had intercepted the telegram. The U-boat captain prepared to send the message back to high command, when the sub hit a mine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My grandfather found the message during a salvage operation. Can anyone help me find the gold? &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;y7.9z/d.f4!\,j01&quot;r@p$(5)+wzyx.7e/h,l0fom@81ou y'+v4/3-7x!\?@e,&quot;5'+$@mnrqb+avde-zyxa9oe3,qlu o@nwz5r-zy .sd/3,2l9\?@8r0pu v(1)-7bc(4defhc!\?omh&quot;5'+ mstd5'q$(6)o&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It's probably an easy one to solve. I can give a hint if necessary. &lt;/p&gt;&#xA;" OwnerUserId="13356" LastEditorDisplayName="user31569" LastEditDate="2017-12-13T19:58:02.727" LastActivityDate="2017-12-13T19:58:02.727" Title="There's GOLD in dem der hills!" Tags="|cipher|" AnswerCount="3" CommentCount="23" ContentLicense="CC BY-SA 3.0" />
  <row Id="16641" PostTypeId="2" ParentId="16626" CreationDate="2015-06-18T20:06:25.543" Score="5" Body="&lt;p&gt;Blue circle equals purple, so &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;d+e+f=3+f&#xA;d+e=3 (they must equal 1 and 2)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;d+e+f=b+c+d&#xA;e+f=b+c&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;e has a maximum value of 2&lt;/li&gt;&#xA;&lt;li&gt;Only 4-7 remain, so b+c has a minimum value of 9 (4+5)&lt;/li&gt;&#xA;&lt;li&gt;This constrains f to be 7 and e to be 2, leaving d to be 1. &lt;/li&gt;&#xA;&lt;li&gt;The total per circle is then 10 and 6 is the only value left for a, so b is 4 and c is 5.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Giving us:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;a=6, b=4, c=5, d=1, e=2, f=7&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3987" LastActivityDate="2015-06-18T20:06:25.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16642" PostTypeId="1" CreationDate="2015-06-18T20:24:19.993" Score="4" ViewCount="2828" Body="&lt;p&gt;I gave the puzzle you can see on the image below to a friend of mine for christmas last year. I thought it would be fun to dump it out in front of him so he would not know the solution. Unfortunately I did not write down the solution myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Until now multiple people have put countless hours into this puzzle and still nobody has managed to solve it. So please help me!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The puzzle is complete and definitely solvable, the tiles were all inside the frame when I bought it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have also started implementing a program to brute force the solution, but I thought I would post here as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have taken pixel dimensions for all the tiles from the image (not 100% accurate, but should suffice):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Large wrench: 164x655&lt;/li&gt;&#xA;&lt;li&gt;Blank: 164x234&lt;/li&gt;&#xA;&lt;li&gt;Pliers: 491x422&lt;/li&gt;&#xA;&lt;li&gt;Hammers: 750x234&lt;/li&gt;&#xA;&lt;li&gt;Pipe wrenches: 422x491&lt;/li&gt;&#xA;&lt;li&gt;Small screwdriver: 327x94&lt;/li&gt;&#xA;&lt;li&gt;Tenon saws: 327x491&lt;/li&gt;&#xA;&lt;li&gt;Screw wrench: 258x750&lt;/li&gt;&#xA;&lt;li&gt;Screwdriver: 140x655&lt;/li&gt;&#xA;&lt;li&gt;Hacksaw: 491x327&lt;/li&gt;&#xA;&lt;li&gt;Carpenter's rule: 397x327&lt;/li&gt;&#xA;&lt;li&gt;Wrenches: 327x655&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I hope I chose the right terms for the tools, English is not my native language.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody have any tips for solving puzzles like this? If you think you can solve it, I would love to see your attempt!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers and happy puzzling!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gsfdF.jpg&quot; alt=&quot;The puzzle&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="13359" LastEditorUserId="13359" LastEditDate="2015-06-18T20:44:54.957" LastActivityDate="2018-05-23T05:47:52.717" Title="Fitting rectangles into square (optimal/perfect rectangle packing)" Tags="|geometry|tiling|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16643" PostTypeId="1" AcceptedAnswerId="16646" CreationDate="2015-06-18T20:31:40.307" Score="10" ViewCount="6514" Body="&lt;p&gt;Suppose $n$ red dots and $n$ blue dots are arranged in the plane so that no three dots lie on a single line.  Show that you can connect each red dot to a blue dot using a line segment so that no pair of line segments intersect.&lt;/p&gt;&#xA;" OwnerUserId="2722" LastActivityDate="2016-03-17T18:10:02.787" Title="Connecting blue dots to red dots" Tags="|mathematics|geometry|" AnswerCount="2" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="16644" PostTypeId="2" ParentId="16634" CreationDate="2015-06-18T21:07:09.173" Score="16" Body="&lt;p&gt;How about&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;624&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Before you read the &lt;strong&gt;huge&lt;/strong&gt; number below, realize my strategy: I am creating a number so huge that it cannot be written out, BUT, at the end it is multiplied by zero, and then 624 is added.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am sure that someone could use this same strategy to get a much larger (longer) number than this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that &lt;a href=&quot;http://dictionary.reference.com/help/faq/language/g80.html&quot;&gt;the numbers twenty-one through ninety-nine are hyphenated&lt;/a&gt;, making each of them one word. Thus, the word twenty is not in the word twenty-one, and so on.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Absolute value of&lt;/strong&gt; ninety-nine centillion ninety-eight vigintillion&#xA;  ninety-seven novemdecillion ninety-six octodecillion ninety-five&#xA;  septendecillion ninety-four sexdecillion ninety-three quindecillion&#xA;  ninety-two quattuordecillion ninety-one tredecillion ninety&#xA;  duodecillion eighty-nine undecillion eighty-eight decillion&#xA;  eighty-seven nonillion eighty-six octillion eighty-five septillion&#xA;  eighty-four sextillion eighty-three quintillion eighty-two quadrillion&#xA;  eighty-one trillion eighty billion seventy-nine million seventy-eight&#xA;  thousand &lt;strong&gt;multiplied by&lt;/strong&gt; seventy-seven googol to the seventy-six&#xA;  googolplex power &lt;strong&gt;minus&lt;/strong&gt; seventy-five over seventy-four score &lt;strong&gt;times&lt;/strong&gt; zero&#xA;  &lt;strong&gt;plus&lt;/strong&gt; six hundred twenty-four&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastEditorUserId="10940" LastEditDate="2015-06-19T03:35:05.413" LastActivityDate="2015-06-19T03:35:05.413" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16645" PostTypeId="2" ParentId="16642" CreationDate="2015-06-18T21:18:44.547" Score="3" Body="&lt;h2&gt;This is not an answer - just reorganizing the information&lt;/h2&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Small screwdriver     94 x 327 &#xA;Screwdriver          140 x 655 &#xA;Blank                164 x 234 &#xA;Large wrench         164 x 655 (the one in the top-left corner)&#xA;Hammers              234 x 750 &#xA;Screw wrench         258 x 750 (the really big one on the right side)&#xA;Carpenter's rule     327 x 397 &#xA;Tenon saws           327 x 491 &#xA;Hacksaw              327 x 491 &#xA;Wrenches             327 x 655 (the set of many on the left)&#xA;Pliers               422 x 491 &#xA;Pipe wrenches        422 x 491 (the pair in the middle)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;By working with these numbers, I get a width of 1309-1311 and a height of 1310-1311, depending on where I measure.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like we can work with a unit of $1 \approx 165$ pixels and get some reasonable numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using that, and rounding each pixel value to one decimal point of new units produces the new grid:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Small screwdriver     94 x 327  -&amp;gt;  0.6 x 2.0 &#xA;Screwdriver          140 x 655  -&amp;gt;  0.8 x 4.0 &#xA;Blank                164 x 234  -&amp;gt;  1.0 x 1.4 &#xA;Large wrench         164 x 655  -&amp;gt;  1.0 x 4.0 &#xA;Hammers              234 x 750  -&amp;gt;  1.4 x 4.5 &#xA;Screw wrench         258 x 750  -&amp;gt;  1.6 x 4.5 &#xA;Carpenter's rule     327 x 397  -&amp;gt;  2.0 x 2.4 &#xA;Tenon saws           327 x 491  -&amp;gt;  2.0 x 3.0 &#xA;Hacksaw              327 x 491  -&amp;gt;  2.0 x 3.0 &#xA;Wrenches             327 x 655  -&amp;gt;  2.0 x 4.0 &#xA;Pliers               422 x 491  -&amp;gt;  2.6 x 3.0 &#xA;Pipe wrenches        422 x 491  -&amp;gt;  2.6 x 3.0 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As you can see, most of these numbers work out surprisingly cleanly.  This also produces a grid of &lt;code&gt;8.0 x 8.0&lt;/code&gt; using the previous measurements.&lt;/p&gt;&#xA;" OwnerUserId="325" LastEditorUserId="325" LastEditDate="2015-06-18T21:59:56.737" LastActivityDate="2015-06-18T21:59:56.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16646" PostTypeId="2" ParentId="16643" CreationDate="2015-06-18T21:58:41.703" Score="16" Body="&lt;p&gt;Start by drawing a line segment from each red point to an arbitrary blue point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are no intersections, we are done. Otherwise,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;pick any two intersecting segments AX and BY, where A and B are blue and X and Y are red. Now BX and AY cannot intersect, so erase AX and BY and replace them by BX and AY. By the triangle inequality (AX, BY, BX, AY form a butterfly shape with two triangles), the total length of the line segments has strictly decreased.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Keep on repeating this process as long as intersections exist. Since the total length decreases at each step and there are only finitely many ways of connecting each red point to a different blue point, the process must terminate. When it does, there are no intersections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;QED.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For the more visual types:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/W83SX.png&quot; alt=&quot;Solution&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and if you continue this to the end:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/LRv7u.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;br&gt; &lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/X23vt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5840" LastEditDate="2015-06-19T08:12:46.967" LastActivityDate="2015-06-19T08:12:46.967" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16648" PostTypeId="2" ParentId="16642" CreationDate="2015-06-18T22:26:21.567" Score="7" Body="&lt;p&gt;After &lt;a href=&quot;https://puzzling.stackexchange.com/users/4421/gentlepurplerain&quot;&gt;GentlePurpleRain&lt;/a&gt; gave the hint that this is a variation of the Calibron 12-Block puzzle I did some research and I'm pretty sure I have found a (or even the only) solution. Since the original puzzle is laser-cut and therefore fits quite precisely into the frame I will still have to verify with my friend if the solution is actually correct. However, with my DIY print-out version of the puzzle it works!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to everyone who put time into this!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ENM1K.jpg&quot; alt=&quot;Possible solution&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="13359" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2015-06-18T22:26:21.567" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16650" PostTypeId="2" ParentId="16634" CreationDate="2015-06-19T01:30:34.267" Score="16" Body="&lt;h2&gt;&lt;strike&gt;1,720&lt;/strike&gt; &lt;span class=&quot;math-container&quot;&gt;$\Rightarrow$&lt;/span&gt; &lt;strike&gt;4,242&lt;/strike&gt; &lt;span class=&quot;math-container&quot;&gt;$\Rightarrow$&lt;/span&gt;  24,749&lt;/h2&gt;&#xA;&lt;p&gt;This can be expanded infinitely so long as we can come up with new names for large numbers. In fact, notice that it starts with a new prefix after &lt;span class=&quot;math-container&quot;&gt;$10^{3,000}$&lt;/span&gt;. You could do the whole list again with the Milli- prefix and then Billi-, Trilli-, etc.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;A Million Billion Trillion Quadrillion Quintillion Sextillion Septillion Octillion Nonillion Decillion Undecillion Duodecillion Tredecillion Quattuordecillion Quindecillion Sedecillion Septendecillion Octodecillion Novendecillion Vigintillion Unvigintillion Duovigintillion Tresvigintillion Quattuorvigintillion Quinvigintillion Sesvigintillion Septemvigintillion Octovigintillion Novemvigintillion Trigintillion Untrigintillion Duotrigintillion Trestrigintillion Quattuortrigintillion Quintrigintillion Sestrigintillion Septentrigintillion Octotrigintillion Noventrigintillion Quadragintillion Unquadragintillion Duoquadragintillion Tresquadragintillion Quattuorquadragintillion Quindragintillion Sesquadragintillion Septenquadragintillion Octoquadragintillion Novenquadragintillion Quinquagintillion Unquinquagintillion Duoquinquagintillion Tresquinquagintillion Quattuorquinquagintillion Quinquinquagintillion Sesquinquagintillion Septenquinquagintillion Octoquinquagintillion Novenquinquagintillion Sexagintillion Unsexagintillion Duosexagintillion Tresexagintillion Quattuorsexagintillion Quinsexagintillion Sesexagintillion Septensexagintillion Octosexagintillion Novensexagintillion Septuagintillion Unseptuagintillion Duoseptuagintillion Treseptuagintillion Quattuorseptuagintillion Quinseptuagintillion Seseptuagintillion Septenseptuagintillion Octoseptuagintillion Novenseptuagintillion Octogintillion Unoctogintillion Duooctogintillion Tresoctogintillion Quattuoroctogintillion Quinoctogintillion Sexoctogintillion Septemoctogintillion Octooctogintillion Novemoctogintillion Nonagintillion Unnonagintillion Duononagintillion Trenonagintillion Quattuornonagintillion Quinnonagintillion Senonagintillion Septenonagintillion Octononagintillion Novenonagintillion Centillion Uncentillion Duocentillion Trescentillion Quattuorcentillion Quincentillion Sexcentillion Septencentillion Octocentillion Novencentillion Decicentillion Undecicentillion Duodecicentillion Tredecicentillion Quattuordecicentillion Quindecicentillion Sedecicentillion Septendecicentillion Octodecicentillion Novendecicentillion Viginticentillion Unviginticentillion Duoviginticentillion Tresviginticentillion Quattuorviginticentillion Quinviginticentillion Sesviginticentillion Septemviginticentillion Octoviginticentillion Novemviginticentillion Trigintacentillion Untrigintacentillion Duotrigintacentillion Trestrigintacentillion Quattuortrigintacentillion Quintrigintacentillion Sestrigintacentillion Septentrigintacentillion Octotrigintacentillion Noventrigintacentillion Quadragintacentillion Unquadragintacentillion Duoquadragintacentillion Tresquadragintacentillion Quattuorquadragintacentillion Quinquadragintacentillion Sesquadragintacentillion Septenquadragintacentillion Octoquadragintacentillion Novenquadragintacentillion Quinquagintacentillion Unquinquagintacentillion Duoquinquagintacentillion Tresquinquagintacentillion Quattuorquinquagintacentillion Quinquinquagintacentillion Sesquinquagintacentillion Septenquinquagintacentillion Octoquinquagintacentillion Novenquinquagintacentillion Sexagintacentillion Unsexagintacentillion Duosexagintacentillion Tresexagintacentillion Quattuorsexagintacentillion Quinsexagintacentillion Sesexagintacentillion Septensexagintacentillion Octosexagintacentillion Novensexagintacentillion Septuagintacentillion Unseptuagintacentillion Duoseptuagintacentillion Treseptuagintacentillion Quattuorseptuagintacentillion Quinseptuagintacentillion Seseptuagintacentillion Septenseptuagintacentillion Octoseptuagintacentillion Novenseptuagintacentillion Octogintacentillion Unoctogintacentillion Duooctogintacentillion Tresoctogintacentillion Quattuoroctogintacentillion Quinoctogintacentillion Sexoctogintacentillion Septemoctogintacentillion Octooctogintacentillion Novemoctogintacentillion Nonagintacentillion Unnonagintacentillion Duononagintacentillion Trenonagintacentillion Quattuornonagintacentillion Quinnonagintacentillion Senonagintacentillion Septenonagintacentillion Octononagintacentillion Novenonagintacentillion Ducentillion Unducentillion Duoducentillion Treducentillion Quattuorducentillion Quinducentillion Seducentillion Septenducentillion Octoducentillion Novenducentillion Deciducentillion Undeciducentillion Duodeciducentillion Tredeciducentillion Quattuordeciducentillion Quindeciducentillion Sedeciducentillion Septendeciducentillion Octodeciducentillion Novendeciducentillion Vigintiducentillion Unvigintiducentillion Duovigintiducentillion Tresvigintiducentillion Quattuorvigintiducentillion Quinvigintiducentillion Sesvigintiducentillion Septemvigintiducentillion Octovigintiducentillion Novemvigintiducentillion Trigintaducentillion Untrigintaducentillion Duotrigintaducentillion Trestrigintaducentillion Quattuortrigintaducentillion Quintrigintaducentillion Sestrigintaducentillion 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Octovigintinongentillion Novemvigintinongentillion Trigintanongentillion Untrigintanongentillion Duotrigintanongentillion Trestrigintanongentillion Quattuortrigintanongentillion Quintrigintanongentillion Sestrigintanongentillion Septentrigintanongentillion Octotrigintanongentillion Noventrigintanongentillion Quadragintanongentillion Unquadragintanongentillion Duoquadragintanongentillion Tresquadragintanongentillion Quattuorquadragintanongentillion Quinquadragintanongentillion Sesquadragintanongentillion Septenquadragintanongentillion Octoquadragintanongentillion Novenquadragintanongentillion Quinquagintanongentillion Unquinquagintanongentillion Duoquinquagintanongentillion Tresquinquagintanongentillion Quattuorquinquagintanongentillion Quinquinquagintanongentillion Sesquinquagintanongentillion Septenquinquagintanongentillion Octoquinquagintanongentillion Novenquinquagintanongentillion Sexagintanongentillion Unsexagintanongentillion Duosexagintanongentillion Tresexagintanongentillion Quattuorsexagintanongentillion Quinsexagintanongentillion Sesexagintanongentillion Septensexagintanongentillion Octosexagintanongentillion Novensexagintanongentillion Septuagintanongentillion Unseptuagintanongentillion Duoseptuagintanongentillion Treseptuagintanongentillion Quattuorseptuagintanongentillion Quinseptuagintanongentillion Seseptuagintanongentillion Septenseptuagintanongentillion Octoseptuagintanongentillion Novenseptuagintanongentillion Octogintanongentillion Unoctogintanongentillion Duooctogintanongentillion Tresoctogintanongentillion Quattuoroctogintanongentillion Quinoctogintanongentillion Sexoctogintanongentillion Septemoctogintanongentillion Octooctogintanongentillion Novemoctogintanongentillion Nonagintanongentillion Unnonagintanongentillion Duononagintanongentillion Trenonagintanongentillion Quattuornonagintanongentillion Quinnonagintanongentillion Senonagintanongentillion Septenonagintanongentillion Octononagintanongentillion Novenonagintanongentillion Millinillion Millimillion Millibillion Millitrillionths multiplied with the value of e raised to i times pi plus Legendre's Constant minus negative Twenty-four Thousand Seven Hundred Forty Nine&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;In numerical form, it can be expressed as:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\prod_{n=1}^{1,003}10^{3n}\div10^{3,012}*(e^{i\pi}+1)+24,749 $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Legendre%27s_constant&quot; rel=&quot;nofollow noreferrer&quot;&gt;Legendre's Constant&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-06-22T12:18:35.893" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16651" PostTypeId="2" ParentId="16643" CreationDate="2015-06-19T03:46:04.237" Score="7" Body="&lt;p&gt;A concise and elegant proof was given by rand al'thor.  Here's a considerably less concise proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we'll define &lt;strong&gt;dividing connection&lt;/strong&gt; as follows: extend the line segment connecting one pair of blue and red points into an infinite line.  This line splits the plane into two parts.  If, within each part, there are equally many blue and red points, call the connection between the blue point and the red point a &lt;strong&gt;dividing connection&lt;/strong&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mwxod.png&quot; alt=&quot;A dividing connection between two points&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This figure shows a dividing connection between the points $B_6$ and $R_6$.  There are three blue and three red points above the line, and two red and two blue points below the line.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A connection between $B_5$ and $R_5$ would also be a dividing connection, with five blue and red points to the left of the line, and none to the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Assumption:&lt;/strong&gt; Every set of points has a dividing connection.  I will use this assumption in an inductive proof that we can connect opposite colored points for any sized set of points, and then provide a proof of the assumption.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If every set of points has a dividing connection, then in every set of $n$ red and $n$ blue points, we can connect points of opposite colors with non-overlapping line segments.  Call a set where we can make such connections a &lt;strong&gt;connectable set&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set of $0$ red and $0$ blue points is trivially connectable.  The set of $1$ red and $1$ blue point is also trivially connectable: just connect the two points. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume any set of $k$ red and $k$ blue points is connectable.  For a set of $k+1$ red and $k+1$ blue points, we can find a dividing connection within that set.  That dividing connection splits the plane into two parts, each containing $k$ or fewer red and blue points.  By our inductive assumption, the set of points in each part of the plane is connectable.  The set of $k+1$ red and blue points is connectable: take the connections for each of the two separate parts together with the dividing connection.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, by induction, any set of $n$ red and $n$ blue points is connectable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof of existence of dividing connection:&lt;/strong&gt;  I will show that every set of points has a dividing connection.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the leftmost point, the point with the least $X$-coordinate.  If two points have the same $X$-coordinate, consider the one of those that has the least $Y$-coordinate.  That point has a dividing connection with some other point.  Call this point $A$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the vertical line through $A$.  For any other point $B$, we can find the angle between that line and the line segment $\overline{AB}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dPxWC.png&quot; alt=&quot;Angle between vertical line and point&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This image shows this angle between the point $A$ and one of the other points.  Since no three points are co-linear, all points have a distinct value for this angle, and since we chose a left-most point, the values of this angle are in the range $[0^\circ,180^\circ)$.  We can put all of the points other than $A$ in sequence based on the value of this angle.  For the above figure, the sequence would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\color{blue}{B}\color{red}{R}\color{red}{R}\color{red}{R}\color{blue}{B}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a set of $n$ blue and red points, this sequence is of length $2n-1$.  For each point in the sequence, we will consider three things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The number of same-colored points at this location or earlier in the sequence.  Call this quantity $S$.&lt;/li&gt;&#xA;&lt;li&gt;The number of different-colored points at this location or earlier in the sequence.  Call this quantity $D$.&lt;/li&gt;&#xA;&lt;li&gt;Whether the point is a different color from $A$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is a table of these values for the above image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{c|cccc}&#xA;\text{index} &amp;amp; \text{color} &amp;amp; \text{S} &amp;amp; \text{D} &amp;amp; \text{different?} \\&#xA;\hline&#xA;1 &amp;amp; \color{blue}{B} &amp;amp; 1 &amp;amp; 0 &amp;amp; \text{no} \\&#xA;2 &amp;amp; \color{red}{R} &amp;amp; 1 &amp;amp; 1 &amp;amp; \text{yes} \\&#xA;3 &amp;amp; \color{red}{R} &amp;amp; 1 &amp;amp; 2 &amp;amp; \text{yes} \\&#xA;4 &amp;amp; \color{red}{R} &amp;amp; 1 &amp;amp; 3 &amp;amp; \text{yes} \\&#xA;5 &amp;amp; \color{blue}{B} &amp;amp; 2 &amp;amp; 3 &amp;amp; \text{no} \\&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if there are $n$ red and blue points, the final values of $S$ and $D$ in this table will be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{c|cccc}&#xA;\text{index} &amp;amp; \text{S} &amp;amp; \text{D}  \\&#xA;\hline&#xA;2n-1  &amp;amp; n-1 &amp;amp; n \\&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since there are $n$ other points of a different color than $A$, and $n-1$ other points the same color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any point where $D = S+1$ and that is a different color from point $A$, the connection between $A$ and that point is a dividing connection.  For that point, there will be $D$ points of each color on one side of the connection, and $n-D$ points of each color on the other.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a point always exists.  $S$ and $D$ start out equal at $0$, and always end with $D = S+1$.  Thus, there must be some point in the table where $D$ becomes greater than $S$.  $D$ can only increase at points in the sequence that are a different color than $A$, so at such a point in the table, we have the conditions necessary for a dividing connection: $D = S+1$ at a point a different color than $A$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, a dividing connection can always be found, which proves the assumption we needed for the inductive proof that for any set of $n$ red and $n$ blue points, we can connect distinct pairs of different colored points with non-overlapping line segments.  &lt;/p&gt;&#xA;" OwnerUserId="10261" LastEditorUserId="10261" LastEditDate="2016-03-17T18:10:02.787" LastActivityDate="2016-03-17T18:10:02.787" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16652" PostTypeId="2" ParentId="16348" CreationDate="2015-06-19T05:30:23.937" Score="2" Body="&lt;p&gt;I don't have an answer but I'll share some of my observations if it might help someone else figure out how to use it to do the decryption (or alternatively, eliminate it as a possibility).  Some of this stuff is a bit far fetched, so more than likely it's on the complete wrong track.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The encryption could centre around prime numbers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But these belong to one very ancient indeed,&lt;br&gt;&#xA;A man renowned in strictest sense,&lt;br&gt;&#xA;Celebrated most for famous work, derived from common sense.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thin green trees in this private garden,&lt;br&gt;&#xA;With raindrops, at tip top,&lt;br&gt;&#xA;Belong to him, the mysterious Alexandrian.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eratosthenes was the Librarian of the Library at Alexandria.  One of his famous works was the &quot;Sieve of Eratosthenes&quot; which is a common sense filtering algorithm for selecting prime numbers out of the natural numbers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This sets us up for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The imagery of trees meaning individual numbers, and the forest meaning a bunch of numbers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Those seen trees which bear ripe plums,&lt;br&gt;&#xA;Their seeds grew from the Two that do make One,&lt;br&gt;&#xA;Which share no part except for another One.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each individual (One) prime number is made (grown) of exactly Two factors, the number itself and One.  And no prime numbers share a common factor except for One.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sticks in one big, infinite bundle,&lt;br&gt;&#xA;Upright, but One far apart,&lt;br&gt;&#xA;Farther removed from Mussolini's art.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It gets far fetched here.  There are infinite number of primes.  Upright could go back to the fact that prime numbers &quot;stand on their own&quot; instead of on other factors.  &quot;One far apart&quot; could refer to the number one not really being considered a prime, even though its only factors are one and itself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if this is on the right track, but some of it seems fairly applicable.  If so, maybe it'll help someone figure out the encryption itself.&lt;/p&gt;&#xA;" OwnerUserId="7681" LastActivityDate="2015-06-19T05:30:23.937" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16653" PostTypeId="1" AcceptedAnswerId="16654" CreationDate="2015-06-19T07:27:27.637" Score="-5" ViewCount="13403" Body="&lt;p&gt;X is impossible for god.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X is above god.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you eat X you will die.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rich need X.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The poor have X.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X is a 7 letter word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is X?&lt;/p&gt;&#xA;" OwnerUserId="13369" LastEditorUserId="5373" LastEditDate="2015-06-19T08:04:34.903" LastActivityDate="2015-06-19T09:30:37.080" Title="X is impossible for god, X is above god, if you eat X you will die, the rich need X , the poor have X, X is a 7 letter word?" Tags="|riddle|" AnswerCount="1" CommentCount="0" ClosedDate="2015-06-19T07:42:16.457" ContentLicense="CC BY-SA 3.0" />
  <row Id="16654" PostTypeId="2" ParentId="16653" CreationDate="2015-06-19T07:40:52.147" Score="2" Body="&lt;p&gt;The 7 letter word should be...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;NOTHING&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Nothing is impossible for God, and nothing is above Him. You will die if you won't eat anything. The rich need nothing (because they can have everything)... and the poor... have nothing! &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12628" LastEditorUserId="12628" LastEditDate="2015-06-19T09:30:37.080" LastActivityDate="2015-06-19T09:30:37.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16656" PostTypeId="1" AcceptedAnswerId="16658" CreationDate="2015-06-19T08:52:10.270" Score="13" ViewCount="1185" Body="&lt;p&gt;Inspired by the recent &quot;Name that Movie&quot; puzzles, here is something similar.  I will be giving away too much information if I tell you the subject, so you will have to work that out yourself.  Once you get the first one, the rest should come easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Kim.com&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Question 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A small apple&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Question 3&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Godfather is the key to his puppy's toy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Question 4&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.google.com&quot;&gt;http://www.google.com&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Question 5&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Max's partner won 100m sprint for USA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Question 6&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An Italian PhD&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Question 7&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Jerry's nemesis reading an eBook&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Question 8&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What is the (specific) common theme?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are not movies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12332" LastActivityDate="2015-06-19T13:41:27.753" Title="What is the subject?" Tags="|knowledge|" AnswerCount="3" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16657" PostTypeId="2" ParentId="16620" CreationDate="2015-06-19T08:58:36.360" Score="1" Body="&lt;p&gt;&lt;strong&gt;Partial Answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the departure airport is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ontario International. This is derived from PRA (Pocatello Regional Airport) and subtracting 1:47, leaving us with ONT.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I admit that I hat to try a few directions for this before I got an airport that made sense though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I agree with rand al'thor that the destination is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; in Europe, because Europium is the 63rd element.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, I suspect that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;with common tongue tied to no other,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;spoken unlike their American brother&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(though the two decline unanimously).&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; refers to the saying that Britain and the US are two countries separated by a common language. It may be Ireland, though... there are lots of Irish in America, I'm told. And no one really speaks Gaelic anywhere else.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll add more if I ever work any more out.&lt;/p&gt;&#xA;" OwnerUserId="13258" LastActivityDate="2015-06-19T08:58:36.360" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16658" PostTypeId="2" ParentId="16656" CreationDate="2015-06-19T09:00:39.363" Score="15" Body="&lt;p&gt;&lt;strong&gt;1&lt;/strong&gt; Kim.com  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Kim Dotcom is the founder of Mega and MegaUpload. He's the man who runs Mega. &lt;strong&gt;Megaman&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2&lt;/strong&gt; A small apple  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;(with thanks to user2990472) &lt;strong&gt;Little Mac&lt;/strong&gt;, from the Punch Out! games&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3&lt;/strong&gt; The Godfather is key to his puppy's toy  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Godfather is the Don. A Kong is a type of dog toy. &lt;strong&gt;Donkey Kong&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4&lt;/strong&gt;&lt;a href=&quot;http://www.google.com&quot;&gt;http://www.google.com&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Link&lt;/strong&gt;, central character of the Legend of Zelda series &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;5&lt;/strong&gt; Max's partner won 100m sprint for USA  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Max's partner is &lt;em&gt;Sam&lt;/em&gt; (Sam and Max video game series). They won a 100m sprint, so they &lt;em&gt;ran&lt;/em&gt;. SamUSAran goes to &lt;strong&gt;Samus Aran&lt;/strong&gt; of Metroid fame&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;6&lt;/strong&gt; An Italian PhD  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A stretch, perhaps, but &lt;strong&gt;Dr. Mario&lt;/strong&gt;?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;7&lt;/strong&gt; Jerry's nemesis reading an eBook  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tom &amp;amp; Jerry, plus a Nook is an eBook reader. This gives &lt;strong&gt;Tom Nook&lt;/strong&gt;, a character in Animal Crossing.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;8&lt;/strong&gt; The common theme  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Starting to think it's video game characters. All of them so far are characters who have appeared in Nintendo games. All are in the most recent Smash Bros. game, though Tom Nook appears only as a trophy/as stage dressing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11717" LastEditorUserId="11717" LastEditDate="2015-06-19T13:41:27.753" LastActivityDate="2015-06-19T13:41:27.753" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="16659" PostTypeId="2" ParentId="16656" CreationDate="2015-06-19T09:01:17.977" Score="3" Body="&lt;h2&gt;Partial answer&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Question 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reference to &lt;a href=&quot;https://en.wikipedia.org/wiki/Kim_Dotcom&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;; maybe the answer is &lt;strong&gt;Kimble&lt;/strong&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Question 6&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I found &lt;a href=&quot;https://en.wikipedia.org/wiki/Dottorato_di_ricerca&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;, so maybe the answer is &lt;strong&gt;dottorato&lt;/strong&gt; or &lt;strong&gt;ricerca&lt;/strong&gt;? (Where's leoll2 when you need him?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Question 7&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jerry's nemesis is Tom in &lt;em&gt;Tom and Jerry&lt;/em&gt;; another word for an ebook is a Kindle. So we get &lt;a href=&quot;http://www.imdb.com/name/nm0454305/&quot; rel=&quot;nofollow&quot;&gt;Tom Kindle&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-06-19T09:01:17.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16660" PostTypeId="1" AcceptedAnswerId="16663" CreationDate="2015-06-19T09:16:32.147" Score="6" ViewCount="24378" Body="&lt;p&gt;I have gone through simple water jug problems like &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2034/a-set-of-water-jug-challenges?rq=1&quot;&gt;this link&lt;/a&gt; but when there are 3 jugs:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;16 Galoon jug with water filled.&lt;/li&gt;&#xA;&lt;li&gt;11 Galoon empty jug.&lt;/li&gt;&#xA;&lt;li&gt;6 Galoon empty jug.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Find a solution to gain 8 Galoon, using these 3 jugs.&lt;/p&gt;&#xA;" OwnerUserId="13373" LastEditorUserId="30903" LastEditDate="2017-05-07T23:08:35.183" LastActivityDate="2017-05-07T23:08:35.183" Title="Water jug for gaining 8 gallons of water?" Tags="|decanting-problems|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ClosedDate="2016-01-30T15:22:52.693" ContentLicense="CC BY-SA 3.0" />
  <row Id="16661" PostTypeId="2" ParentId="16656" CreationDate="2015-06-19T09:32:29.183" Score="4" Body="&lt;p&gt;For Q#2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; little mac (small-little, apple-mac)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13324" LastActivityDate="2015-06-19T09:32:29.183" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16663" PostTypeId="2" ParentId="16660" CreationDate="2015-06-19T09:51:27.980" Score="7" Body="&lt;p&gt;Assuming you may dispose of water...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Call the jugs A, B, C respectively. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;fill C from A. (10,0,6)&lt;/li&gt;&#xA;&lt;li&gt;empty C into B. (10,6,0)&lt;/li&gt;&#xA;&lt;li&gt;fill C from A. (4,6,6)&lt;/li&gt;&#xA;&lt;li&gt;fill B from C. (4,11,1)&lt;/li&gt;&#xA;&lt;li&gt;pour out the litre from C (4,11,0)&lt;/li&gt;&#xA;&lt;li&gt;empty B into A. (15,0,0)&lt;/li&gt;&#xA;&lt;li&gt;fill C from A (9,0,6)&lt;/li&gt;&#xA;&lt;li&gt;empty C into B (9,6,0)&lt;/li&gt;&#xA;&lt;li&gt;fill C from A. (3,6,6)&lt;/li&gt;&#xA;&lt;li&gt;fill B from C. (3,11,1)&lt;/li&gt;&#xA;&lt;li&gt;pour out the litre from C (3,11,0)&lt;/li&gt;&#xA;&lt;li&gt;empty B into A (14,0,0)&lt;/li&gt;&#xA;&lt;li&gt;fill C from A (&lt;strong&gt;8&lt;/strong&gt;,0,6)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Jug A now has 8 litres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we may not discard water:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A to C (10,0,6)&lt;/li&gt;&#xA;&lt;li&gt;A to B (0,10,6)&lt;/li&gt;&#xA;&lt;li&gt;C to A (6,10,0)&lt;/li&gt;&#xA;&lt;li&gt;B to C (6,4,6)&lt;/li&gt;&#xA;&lt;li&gt;C to A (12,4,0)&lt;/li&gt;&#xA;&lt;li&gt;B to C (12,0,4)&lt;/li&gt;&#xA;&lt;li&gt;A to B (1,11,4)&lt;/li&gt;&#xA;&lt;li&gt;B to C (1,9,6)&lt;/li&gt;&#xA;&lt;li&gt;C to A (7,9,0)&lt;/li&gt;&#xA;&lt;li&gt;B to C (7,3,6)&lt;/li&gt;&#xA;&lt;li&gt;C to A (13,3,0)&lt;/li&gt;&#xA;&lt;li&gt;B to C (13,0,3)&lt;/li&gt;&#xA;&lt;li&gt;A to B (2,11,3)&lt;/li&gt;&#xA;&lt;li&gt;B to C (2,&lt;strong&gt;8&lt;/strong&gt;,6)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Jug B now has 8 litres.&lt;/p&gt;&#xA;" OwnerUserId="3987" LastEditorUserId="3987" LastEditDate="2015-06-19T10:02:00.997" LastActivityDate="2015-06-19T10:02:00.997" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16664" PostTypeId="2" ParentId="16612" CreationDate="2015-06-19T10:21:58.923" Score="13" Body="&lt;p&gt;It looks like each line/couplet gives a letter, and those will provide the answer:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it comes to perspectives, the second is fun  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Second person, so you or &lt;strong&gt;U&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The third is just fine in a romantic tongue  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Following from the first line, something to do with the third person in French. Assuming letters again, we get elle (she) or &lt;strong&gt;L&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without this, to say you have lots might be trouble&lt;br&gt;&#xA;In fact it's so helpful you might order double  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You can't say the word 'lots' without the letter 'l', and it appears twice together in many words (though the same is trued of 'o', 't' and 's')&lt;br&gt;&#xA;Taking Deusovi's comment into account, it could well be &lt;strong&gt;ZZ&lt;/strong&gt; - without 'z', you can't say zillions &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One might rave about this and become enamored&lt;br&gt;&#xA;Or think it's the same as a square famous hammer  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The letter &lt;strong&gt;E&lt;/strong&gt; - a colloquial term for ecstacy, and equivalent to mc&lt;sup&gt;2&lt;/sup&gt; - a famous hammer (U can't touch this) squared  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've only had one of the seven in sight&lt;br&gt;&#xA;I think the short version's the opposite of right  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Opposite of right is either &lt;strong&gt;left&lt;/strong&gt; or &lt;strong&gt;wrong&lt;/strong&gt; - let's assume left, here, and I think we're meant to take the left/first character of whatever 'the seven' is.&lt;br&gt;&#xA;Seven Wonders of the World! The Pyramids of Giza are the only one left standing. So, taking &lt;strong&gt;P&lt;/strong&gt; as the first character, then&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mix them together and what have you got?&lt;br&gt;&#xA;Something this puzzle most surely is not!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Mix them together suggests scrambling the letters&lt;br&gt;&#xA;So we have &lt;strong&gt;U, L, ZZ, E&lt;/strong&gt; and &lt;strong&gt;P&lt;/strong&gt; which in some order give the answer to the puzzle  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer attempt:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Puzzle&lt;/strong&gt;. &lt;em&gt;This&lt;/em&gt; title is not the answer, but the title of this stackexchange site is! Not entirely convinced, though. I'm reasonably sure that U, L and E are correct&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;UPDATE:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;One of the seven&quot; might refer to the seven seas, giving us the letter 'c'&lt;br&gt;&#xA;If so, we'd have 'u', 'l', 'e' and 'c', plus whatever the double letter is and &quot;the opposite of right&quot;. That does assume that all of the lines are letters, though, which I'm less sure of now&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11717" LastEditorUserId="11717" LastEditDate="2015-07-22T15:35:11.783" LastActivityDate="2015-07-22T15:35:11.783" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16665" PostTypeId="2" ParentId="16620" CreationDate="2015-06-19T10:35:03.283" Score="4" Body="&lt;h2&gt;Partial Answer&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;From 1:43 to 2:53 made me think about:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;24 hour format mentioned by Aggie Kidd: 1347-1453 which refer to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hundred_Years%27_War&quot; rel=&quot;nofollow&quot;&gt;Hundred Years' War&lt;/a&gt; between &lt;strong&gt;England&lt;/strong&gt; and &lt;strong&gt;France&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The 63rd member, periodically which surely refers to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The 63rd element of the periodic table: &lt;strong&gt;Eu&lt;/strong&gt;(Europium); meaning that the destination country is in &lt;strong&gt;Europe&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So far, these 2 clues point that the flight is between these 2 countries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, &lt;em&gt;&quot;with common tongue tied to no other,&#xA;spoken unlike their American brother&#xA;(though the two decline unanimously)&quot;&lt;/em&gt; might probably refer that the destination country is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;England&lt;/strong&gt; (British and English language)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the departure country (and probable city) is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;France&lt;/strong&gt; and &lt;strong&gt;Paris&lt;/strong&gt; by taking &lt;strong&gt;PAR&lt;/strong&gt;(Paris, France (Airport Code)) instead of &lt;strong&gt;PRA&lt;/strong&gt; (&lt;strong&gt;P&lt;/strong&gt;ocatello &lt;strong&gt;R&lt;/strong&gt;egional &lt;strong&gt;A&lt;/strong&gt;irport)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In addition, I don't know if it may be related but:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Sway&lt;/strong&gt; is a village in Hampshire near Southampton in England &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12731" LastActivityDate="2015-06-19T10:35:03.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16666" PostTypeId="2" ParentId="16507" CreationDate="2015-06-19T11:33:12.790" Score="5" Body="&lt;p&gt;&lt;strong&gt;Murder on the Orient Express&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who was the murderer? All of them did it.&lt;br&gt;&#xA;&lt;strong&gt;No witnesses, no murder weapons, and no evidence.&lt;/strong&gt;&lt;br&gt;&#xA;The Math teacher gave him an assignment to solve all the Millennium Problems within a day; and the students head hurt like hell.&lt;br&gt;&#xA;The English teacher gave him an assignment to write all the works of William Shakespeare within a day; and the students fingers hurt like hell.&lt;br&gt;&#xA;The Science teacher gave him an assignment to make a Huge Mars Rover Replica within a day; and the students hands hurt like hell.&lt;br&gt;&#xA;The PE teacher told him to do 1000 pushups &amp;amp; lift 300 kgs &amp;amp; reduce his weight by 50% while doing all the other assignments; and the students body hurt like hell.&lt;br&gt;&#xA;Then they all went on to their alibis.&lt;br&gt;&#xA;The student went about with the assignments and, sure enough, collapsed within 30 minutes with fatal physical and mental trauma.&lt;br&gt;&#xA;I repeat : &lt;strong&gt;No witnesses, no murder weapons, and no evidence.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why? That is already answered above, but if you mean why they wanted to murder the student, then that was because they were teachers and they wanted to teach him a lesson about life (&amp;amp; unfortunately, about &lt;strong&gt;Death&lt;/strong&gt;)&lt;/p&gt;&#xA;" OwnerUserId="10940" LastEditorUserId="10940" LastEditDate="2015-06-19T18:33:43.930" LastActivityDate="2015-06-19T18:33:43.930" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16668" PostTypeId="2" ParentId="11364" CreationDate="2015-06-19T13:10:00.597" Score="-1" Body="&lt;p&gt;I think there is no strategy possible here ..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every hat was selected independently and randomly. No matters if you can see 2 hats of the same color, the probability of having a blue or red hat is 1/2 (because of the coin flip). You can't plan any strategy during the game because the answers have to be given simultaneously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the probability to win will be 1/2^5 ... Did i miss something ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS : Sorry for my bad english ._.&lt;/p&gt;&#xA;" OwnerUserId="13384" LastActivityDate="2015-06-19T13:10:00.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16672" PostTypeId="1" AcceptedAnswerId="16674" CreationDate="2015-06-19T15:29:04.467" Score="3" ViewCount="707" Body="&lt;p&gt;Find the letters to fill the blank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;T, Q, P, H, H, O, _, _, _&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;T, C, O, D, _&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Geometry&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12109" LastEditorUserId="12109" LastEditDate="2015-06-19T15:54:10.823" LastActivityDate="2015-06-19T16:00:22.173" Title="Easy sequences of letters" Tags="|letter-sequence|" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16673" PostTypeId="2" ParentId="16672" CreationDate="2015-06-19T15:37:54.113" Score="7" Body="&lt;p&gt;&lt;strong&gt;T Q P H H O _ _ _&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;B V N&lt;/strong&gt;: This Question Places High Hopes On &lt;strong&gt;B&lt;/strong&gt;eing &lt;strong&gt;V&lt;/strong&gt;ery &lt;strong&gt;N&lt;/strong&gt;arrow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;T C O D _&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;W&lt;/strong&gt;: That Concept Obviously Doesn't &lt;strong&gt;W&lt;/strong&gt;ork&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;:D&lt;/p&gt;&#xA;" OwnerUserId="1563" LastActivityDate="2015-06-19T15:37:54.113" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16674" PostTypeId="2" ParentId="16672" CreationDate="2015-06-19T15:50:41.833" Score="7" Body="&lt;p&gt;&lt;strong&gt;T Q P H H O _ _ _&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;N D H&lt;/strong&gt;: Triangle Quadrilateral Pentagon Hexagon Heptagon Octagon &lt;strong&gt;N&lt;/strong&gt;onagon &lt;strong&gt;D&lt;/strong&gt;ecagon &lt;strong&gt;H&lt;/strong&gt;endecagon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;T C O D _&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;I&lt;/strong&gt;: Tetrahedron Cube Octahedron Dodecahedron Icosahedron&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7715" LastEditorUserId="7715" LastEditDate="2015-06-19T16:00:22.173" LastActivityDate="2015-06-19T16:00:22.173" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16675" PostTypeId="1" AcceptedAnswerId="16680" CreationDate="2015-06-19T15:55:47.887" Score="5" ViewCount="484" Body="&lt;p&gt;The game &lt;a href=&quot;https://en.wikipedia.org/wiki/Ghost_%28game%29&quot; rel=&quot;nofollow&quot;&gt;Ghost&lt;/a&gt; is a game where players take turns appending letters to the end of a word. The first person to form a complete word with at least four letters, or give a string of a letters that no word begins with, loses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example game might go:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Player 1: A&lt;br&gt;&#xA;  Player 2: AC&lt;br&gt;&#xA;  Player 1: ACC&lt;br&gt;&#xA;  Player 2: ACCO&lt;br&gt;&#xA;  Player 1: ACCOM&lt;br&gt;&#xA;  Player 2: ACCOMP&lt;br&gt;&#xA;  Player 1: ACCOMPA&lt;br&gt;&#xA;  Player 2: ACCOMPAN  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At this point, player 1 realizes he loses because all variants of ACCOMPANY (all of which begin with ACCOMPANI-) have an odd number of letters. If he'd gone with ACCOMPL instead of ACCOMPA on his fourth turn, he could have forced player 2 to spell ACCOMPLISH or ACCOMPLICE or ACCOMPLETIVE, all of which have an &lt;em&gt;even&lt;/em&gt; number of letters, thereby winning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, we call ACCOMP a &lt;em&gt;non-winning string&lt;/em&gt; for Player 1, in which there are both odd-length and even-length words that begin with that string, and therefore it's not a sure thing that Player 1 wins yet. (He many have the &lt;em&gt;winning strategy&lt;/em&gt; by choosing ACCOMPL, but he still has to make the right choice in order to win here.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A friend and I came up with a variant in middle school in which you could insert a letter anywhere within the word itself. We called it &lt;em&gt;Insertion Ghost&lt;/em&gt;&#xA;due to its insertion mechanic (the Wikipedia article linked above calls it Xghost or Superduperghost and attributes it to a man named Daniel Asimov).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this game, a string like XVZ is valid because you can still make the word E&lt;strong&gt;X&lt;/strong&gt;TRA&lt;strong&gt;V&lt;/strong&gt;AGAN&lt;strong&gt;Z&lt;/strong&gt;A out of it. Thus, the potential for words, and the capacity for strategic thinking, is much larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A game might go like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Player 1: X&lt;br&gt;&#xA;  Player 2: XF (thinking of &quot;EXFOLIATE&quot;)&lt;br&gt;&#xA;  Player 3: XFD (thinking of &quot;EXFOLIATED&quot;)&lt;br&gt;&#xA;  Player 4: XMFD (now thinking of &quot;EXEMPLIFIED&quot;)&lt;br&gt;&#xA;  Player 5: ... ... ... EXMFD (also thinking of &quot;EXEMPLIFIED&quot;, but stalling)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As you can see, the target words get convoluted really quickly, and even switch a lot of letters almost all the way (the change from EXFOLIATED to EXEMPLIFIED being the target word inserted five letters between the X and the F that weren't there before, just to change the target word), and it becomes much harder to determine whether you've won in advance or not. An example of a five-letter non-winning string might be CLAMT - both CLAIMANT (8) and EXCLAMATION (11) are possible target words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the &lt;em&gt;longest&lt;/em&gt; non-winning string (i.e. a string that can have both odd-length and even-length target words) in this variant, for either player?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For the purpose of this question, our word list is the &lt;a href=&quot;http://www.scrabbleplayers.org/w/Collins_Scrabble_Words_2012_Edition&quot; rel=&quot;nofollow&quot;&gt;2012 CSW&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-06-19T16:49:41.653" LastActivityDate="2015-06-20T00:11:18.127" Title="What's the longest non-winning string in Insertion Ghost?" Tags="|word|" AnswerCount="1" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="16677" PostTypeId="2" ParentId="2448" CreationDate="2015-06-19T16:32:09.953" Score="6" Body="&lt;p&gt;The MIT mystery hunt has full archives of puzzles online.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.mit.edu/~puzzle/&quot; rel=&quot;noreferrer&quot;&gt;MIT Mystery hunt&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an annual, real-world event, where the winners from last year write all the puzzles for the event. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Classic online puzzles are the &lt;a href=&quot;http://www.timtang.com/ttt/&quot; rel=&quot;noreferrer&quot;&gt;TIM TANG TEST&lt;/a&gt; and Not Pr0n, which I think is gone although it may be archived somewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Machinarium (you can find it on Steam and Amanita Design's website) is a BEAUTIFUL puzzle point and click.&lt;/p&gt;&#xA;" OwnerUserId="13350" LastActivityDate="2015-06-19T16:32:09.953" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16678" PostTypeId="2" ParentId="16660" CreationDate="2015-06-19T16:53:58.663" Score="1" Body="&lt;p&gt;Another way of producing same result with minimum steps would be as below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets assume 16 Litre jug as A, 11 Litre jug as B and 6 Litre jug as C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initial Configuration : (16 0 0)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A to C (10 0 6).&lt;/li&gt;&#xA;&lt;li&gt;C to B (10 6 0).&lt;/li&gt;&#xA;&lt;li&gt;A to C ( 4 6 6).&lt;/li&gt;&#xA;&lt;li&gt;C to B (4 11 1).&lt;/li&gt;&#xA;&lt;li&gt;B to A (15 0 1).&lt;/li&gt;&#xA;&lt;li&gt;C to B (15 1 0).&lt;/li&gt;&#xA;&lt;li&gt;A to C (9 1 6).&lt;/li&gt;&#xA;&lt;li&gt;C to B (9 7 0).&lt;/li&gt;&#xA;&lt;li&gt;A to C (3 7 6).&lt;/li&gt;&#xA;&lt;li&gt;C to B (3 11 2).  &lt;/li&gt;&#xA;&lt;li&gt;B to A (14 0 2).&lt;/li&gt;&#xA;&lt;li&gt;C to B (14 2 0).&lt;/li&gt;&#xA;&lt;li&gt;A to C (8 2 6).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13373" LastEditorUserId="18354" LastEditDate="2016-01-10T06:57:33.187" LastActivityDate="2016-01-10T06:57:33.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16680" PostTypeId="2" ParentId="16675" CreationDate="2015-06-19T17:16:01.813" Score="3" Body="&lt;p&gt;Since the CSW only contains words with 15 letters or less, it's quite trivial. I go for one of many:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ALLEGORISATIO (13 letters, targets ALLEGORISATION and ALLEGORISATIONS)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which I presume is in the 2012 CSW.&lt;/p&gt;&#xA;" OwnerUserId="12666" LastEditorDisplayName="user88" LastEditDate="2015-06-20T00:09:38.167" LastActivityDate="2015-06-20T00:09:38.167" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16681" PostTypeId="2" ParentId="16431" CreationDate="2015-06-19T18:28:06.570" Score="1" Body="&lt;p&gt;Reading each unit as a series of four colored nodes, the number of color changes in each unit is represented below in square form:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;121&#xA;101&#xA;12?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As zaykho points out, there is clear symmetry here no matter whether you read the units left-to-right, up-to-down, or as a square.&#xA;Only one of the options fits this symmetry:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Answer B, which has 1 color change.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13350" LastEditorUserId="13350" LastEditDate="2015-06-24T14:44:40.213" LastActivityDate="2015-06-24T14:44:40.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16684" PostTypeId="2" ParentId="11364" CreationDate="2015-06-19T19:19:56.537" Score="0" Body="&lt;p&gt;Based on the logic of the OP's comment responding to @Alconja, perhaps&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you see all red, guess red. If you see all blue, guess blue. If you see one of each, guess randomly.&#xA; For the 3 person scenario, the chances of all hats being red is (1/2)^3, and the chance of all being blue is (1/2)^3 as well. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the probability of being correct is 1/8 + 1/8 = 1/4. This is higher than their chances of just guessing randomly (1/8). This is still lower than the 1/2 that the OP mentions.&lt;/p&gt;&#xA;" OwnerUserId="9587" LastActivityDate="2015-06-19T19:19:56.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16685" PostTypeId="2" ParentId="16474" CreationDate="2015-06-19T19:29:43.780" Score="1" Body="&lt;h1&gt;Puzzle 1&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Is Q&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1024&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we do the equation for the fish I ended up getting &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; m = 4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;by inferring from the length of apologies (8, 12) and the overuse of the word LOVE(although I don't know if that's really relevant) if that's the case and we apply it to the length of the apologies, 8 and 12 it means that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; m = 4 &lt;br /&gt;&#xA; 2m = 8 &lt;br /&gt;&#xA; 5m = 20 &lt;br /&gt;&#xA; 12 = sqrt[x] - 20 = (12 + 20) ^2 = 1024 &lt;br /&gt; &#xA; q is then equal to 1024 or 32^2 &lt;br /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not a mathematics guy so I hope no one freaks out on my lack of ability here, but I thought I'd give it a go.&lt;/p&gt;&#xA;" OwnerUserId="10934" LastEditorUserId="7307" LastEditDate="2015-06-23T21:20:52.333" LastActivityDate="2015-06-23T21:20:52.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16686" PostTypeId="1" AcceptedAnswerId="16738" CreationDate="2015-06-19T19:56:02.820" Score="14" ViewCount="4007" Body="&lt;p&gt;I communicated with you, not by normal means.&lt;br&gt;&#xA;The illusionist did everything it seems.&lt;br&gt;&#xA;Proximity can make me very shy.&lt;br&gt;&#xA;Without a bunch of these you would die.&lt;br&gt;&#xA;The bee, the seasons, maybe some more,&lt;br&gt;&#xA;Who am I? All alone, no knock on the door.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1 &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Who is important.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Don't take the lines too literally (except the one about dying).   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The title is important too.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 4 (this is a big one). &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Think of each line separately, and then answer the riddle.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 5    &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/q/15731/9951&quot;&gt;This has a similar format.&lt;/a&gt; &amp;nbsp; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-07-24T12:44:11.607" Title="Alive, dead, alive, now dead" Tags="|riddle|rhyme|" AnswerCount="14" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="16688" PostTypeId="2" ParentId="11364" CreationDate="2015-06-19T20:15:48.883" Score="0" Body="&lt;p&gt;For 3 haberdashers&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There are 4 possible states: 0, 1, 2 and 3 blue hats. &lt;/li&gt;&#xA;&lt;li&gt;1 and 2 each have 3 ways of occurring&lt;/li&gt;&#xA;&lt;li&gt;Any strategy that gets a given state right will necessarily get the 'adjacent' state(s) wrong.&lt;/li&gt;&#xA;&lt;li&gt;We may choose to have, at most, one of the 0/3 states correct and the non-adjacent state from 2/3, giving us 4 out of 8 correct.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Let us choose to be correct for 0 and 2 blue hats: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If you see 2 red hats, pick red&lt;/li&gt;&#xA;&lt;li&gt;If you see 1 red hat, pick blue&lt;/li&gt;&#xA;&lt;li&gt;If you see 0 red hats, pick red&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Still working on 5...&lt;/p&gt;&#xA;" OwnerUserId="3987" LastActivityDate="2015-06-19T20:15:48.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16689" PostTypeId="2" ParentId="16686" CreationDate="2015-06-19T20:26:38.557" Score="2" Body="&lt;p&gt;It could be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; pollen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I communicated with you, not by normal means.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; allergic to many. And shows that spring is here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The illusionist did everything it seems.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the flower/ the plant created it &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proximity can make me very shy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure what this could be&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without a bunch of these you would die.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; without them the plants cant reproduce and eventually human existence will come to an end. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The bee, the seasons, maybe some more,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the bee are attracted and it is during the spring&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who am I? All alone, no knock on the door.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are alone stick to the bee s or fly in the air to near by flowers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11056" LastActivityDate="2015-06-19T20:26:38.557" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16690" PostTypeId="2" ParentId="16073" CreationDate="2015-06-19T20:28:37.400" Score="2" Body="&lt;p&gt;First two statements give the answer i.e. Arnold's and Brian's.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Arnold: it wasn't Edward. It was Brian.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Looking at Arnold's statement one can be say that second part of his statement is lie ,because if it wasn't, the first one would be true, which would make Edward and Brian both culprits. Both Edward and Brian couldn't have stolen the cake. Therefore, &quot;It wasn't Edward&quot; is true. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Brian: it wasn't Charles. It wasn't Edward.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Coming to Brian's statement. We know Edward isn't culprit, so secont part is true making first statement a lie. This means Charles is the thief.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's it.&lt;/p&gt;&#xA;" OwnerUserId="10760" LastEditorUserId="9951" LastEditDate="2015-06-19T20:58:35.667" LastActivityDate="2015-06-19T20:58:35.667" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16691" PostTypeId="2" ParentId="16634" CreationDate="2015-06-19T20:30:56.140" Score="0" Body="&lt;p&gt;$\Large\infty {\rm\ Points}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If acronyms are allowed, one could technically construct the infinite-point acronym &quot;an AAPO plus one&quot;, where AAPO is a recursive acronym that stands for &quot;an AAPO plus one&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe, given a sufficiently large team of lawyers, this could be proven to not violate the prohibition on repeated words nor the prohibition on using words not in the English language.&lt;/p&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2015-06-19T20:30:56.140" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16692" PostTypeId="2" ParentId="11364" CreationDate="2015-06-19T20:36:10.693" Score="0" Body="&lt;p&gt;The key is the consistency of the strategy so that you increase the chances not just of me being right about our hats, but about all of us being right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One person can guess red or blue and has a 50-50 chance of being right. But that's not enough to win, the other people have to be right as well. In the case of two people, a strategy of &quot;always guess red&quot; or &quot;always guess blue&quot;, applied by both people, has only a 1 in 4 chance of them both being right. Whereas &quot;guess the colour you see&quot; or &quot;guess the colour you don't see&quot; each have a 2 in 4 chance of them both being right. So that's a better strategy for the two player variant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 32 possible combinations - 00000 all blue, 00001 four blues and a red, 11111 all red. Say the rule is &quot;guess what you see on your left&quot;. They will win for 00000 and 11111 but lose for all the others. That's clearly a bad strategy, but it's better than &quot;guess red&quot; or &quot;guess blue.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Working on the strategy they should use now...&lt;/p&gt;&#xA;" OwnerUserId="1509" LastActivityDate="2015-06-19T20:36:10.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16693" PostTypeId="2" ParentId="16686" CreationDate="2015-06-19T21:09:40.130" Score="1" Body="&lt;p&gt;Could it be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Mimosa_pudica&quot; rel=&quot;nofollow&quot;&gt;The touch-me-not plant&lt;/a&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I communicated with you, not by normal means.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When touched, it has an abnormal behaviour of shutting its leaves.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The illusionist did everything it seems.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A stretch, but does it imply that the plant gives of an illusion of behaving like an animal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proximity can make me very shy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Come too close and touch, and the leaves shy away.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without a bunch of these you would die.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Refers to plants in general? Life would not be possible without trees and plants.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The bee, the seasons, maybe some more,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bees hover around trees to gather nectar. And trees change colour in seasons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who am I? All alone, no knock on the door.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yet another reference to touching. &lt;em&gt;Who am I?&lt;/em&gt; Touch-me-not&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is not the right answer, I think, but it fits some of the clues fairly well. Although I guess I have overused one fact too much to make it fit.&lt;/p&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="9951" LastEditDate="2015-06-19T21:19:28.063" LastActivityDate="2015-06-19T21:19:28.063" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16694" PostTypeId="2" ParentId="16338" CreationDate="2015-06-19T21:29:48.020" Score="2" Body="&lt;p&gt;&lt;b&gt;Note:  Puzzle poser, can exponentiation be used, as Charles29020 used in his attempt?  I asked, because no one addressed that part of his solution, just the 1,000 versus the 10,000 part that I can see.  The instructions do not mention exponentiation at all.  I have other potential solutions that use exponentiation.  &lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;.&#xA;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{3*2 + 4}{1\%}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{14 + 3!}{2\%}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{4! - 1 - 3}{2\%}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{(1 + 3)! - 4}{2\%}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{(1 + 2)! + 4!}{3\%}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{3! + 4!}{(1 + 2)\%}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{4 + 3!}{(2 - 1)\%}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{(3!/2)! + 4}{1\%}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{(1 + 4)!}{(2*3!)\%}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{2(3! - 1)!}{(4!)\%}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{(3!)!}{4!*(1 + 2)\%}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{((1 + 2)!)!}{(4!*3)\%}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{4!/3 + 2}{1\%}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{4(3! - 1)}{2\%}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{2}{4\%*(3! - 1)\%}$&lt;/p&gt;&#xA;" OwnerUserId="13203" LastActivityDate="2015-06-19T21:29:48.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16695" PostTypeId="2" ParentId="16686" CreationDate="2015-06-19T21:45:29.127" Score="4" Body="&lt;p&gt;Is it something like&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; nerves/physical sensations?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I communicated with you, not by normal means.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your nerve endings communicate with you, but not by speaking to you or answering your questions on Stack Exchange or any other 'normal' means of communication.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The illusionist did everything it seems.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Pain is an illusion&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proximity can make me very shy.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The proximity of something too hot or cold or sharp is something your nerve endings hate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Without a bunch of these you would die.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Without sensations, people would die easily. People with leprosy lose the feeling in their fingers and toes - their nerve endings &lt;em&gt;die&lt;/em&gt; - rendering them easily susceptible to small injuries there, which can then fester and kill them slowly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The bee, the seasons, maybe some more,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bee stings, the heat of summer, the cold of winter ... all these are sensations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Who am I? All alone, no knock on the door.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not quite sure about this line. I suppose a knock on the door is something that's heard and not felt, so it's not relevant to the sense we're considering.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-06-19T21:45:29.127" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16696" PostTypeId="1" AcceptedAnswerId="16705" CreationDate="2015-06-19T23:07:22.583" Score="4" ViewCount="282" Body="&lt;p&gt;Miss Gruppe is growing too old for her house - there are far too many rooms and it sometimes takes her days to find her way between the pantry and the kitchen! In honor of her accomplishments in the field of mathematics, the townsfolk have decided to design her a brand new house, to suit her better in her old age.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, they've come up with the following idea: The house will be organized into various rooms connected to each other by doors (as tends to be the case in houses). A room may have any number of doors, but it will always have one labeled as &lt;strong&gt;A&lt;/strong&gt; and one labeled as &lt;strong&gt;B&lt;/strong&gt; (with the labels facing inwards). Then, it will be easy to give instructions to get around the house - for instance:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To get from the pantry to the kitchen, go through door &lt;strong&gt;A&lt;/strong&gt;. From that room, go through door &lt;strong&gt;B&lt;/strong&gt;. Finally, using door &lt;strong&gt;A&lt;/strong&gt; brings you to your destination.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They design the house so that it is possible to get from any room to any other by such a sequence of moves. One may note that instructions like:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To get from the bathroom to the dining room, use door &lt;strong&gt;A&lt;/strong&gt; and from that room, use the &lt;em&gt;next&lt;/em&gt; door labelled &lt;strong&gt;A&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;make sense, as the door labelled &lt;strong&gt;A&lt;/strong&gt; inside the bathroom is likely different from the door labelled &lt;strong&gt;A&lt;/strong&gt; in the room into which the former door leads (i.e. the door labelled &lt;strong&gt;A&lt;/strong&gt; in the bathroom might be unlabeled on the other side or labelled &lt;strong&gt;B&lt;/strong&gt; on the other side).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, they understand that Miss Gruppe is somewhat absent-minded and often wanders - though she can remember the instructions all right, sometimes she'll deviate from the path. But, as a matter of superstition, when she deviates, it will always be by repeating some sequence of moves through doors exactly $3$ times - so she might go through door &lt;strong&gt;A&lt;/strong&gt;, then through the next door &lt;strong&gt;A&lt;/strong&gt;, and then through the next door &lt;strong&gt;A&lt;/strong&gt; or she might alternate &quot;Use door &lt;strong&gt;A&lt;/strong&gt;, use door &lt;strong&gt;B&lt;/strong&gt;&quot; exactly thrice. Thus, they have decided to design the house with the following additional property:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;From any room, if you choose some sequence of door labels to go through and repeat it thrice, you will end up in the same room as you started.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The townspeople love Miss Gruppe very much and want to build her a house with &lt;em&gt;as many rooms as possible&lt;/em&gt; subject to the above conditions. Miss Gruppe, who is still an avid climbed of stairs, is fine with the house having overpasses and tunnels and generally not fitting into a plane. How many rooms can such a house have?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Simple Example of such a house:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple design for such a house with $3$ rooms would be as follows, where the character &lt;code&gt;*&lt;/code&gt; represents a door and a letter in an adjacent square represents a label.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;+---+---+&#xA;|   |   |&#xA;|   |   |&#xA;|  A*B  |&#xA;|   |   |&#xA;|   | A |&#xA;|   +-*-+&#xA;|   | B |&#xA;|   |   |&#xA;|  B*A  |&#xA;|   |   |&#xA;|   |   |&#xA;+---+---+&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Where one can, for instance, get from the long room to the bottom left room either by following door &lt;strong&gt;A&lt;/strong&gt; twice or by following &lt;strong&gt;B&lt;/strong&gt; once - but it is easy to see that following either door three times leads you back to your original location, and easy to see that sequences like &quot;Follow &lt;strong&gt;A&lt;/strong&gt; then &lt;strong&gt;A&lt;/strong&gt; then &lt;strong&gt;B&lt;/strong&gt;&quot; lead one back to their original location when followed thrice.&lt;/p&gt;&#xA;" OwnerUserId="4765" LastEditorUserId="12109" LastEditDate="2015-06-20T01:58:44.773" LastActivityDate="2015-06-20T02:37:47.790" Title="The Old Miss Gruppe" Tags="|mathematics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16697" PostTypeId="2" ParentId="20" CreationDate="2015-06-20T00:07:42.160" Score="4" Body="&lt;p&gt;Kendall Frey's proof appears to overlook some subtleties regarding what is meant by a rotation of a corner.  (For example, when a corner cube returns to its original position it is rotated by a multiple of &lt;strong&gt;&lt;em&gt;120 degrees&lt;/em&gt;&lt;/strong&gt;, &lt;em&gt;which is not itself a multiple of 90 degrees&lt;/em&gt;.)  However, if one is very careful about what is meant by a rotation then a correct geometric proof roughly along the same lines as suggested by Kendal Frey is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine that the Rubik's cube has edges oriented parallel to the the x,y,z axes, with the top at z=+1.  You sit at the dead center of the Rubik's Cube (RC) facing directly at one of the outside corners.  (The cube is hollow.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the corner cube is one of the ones on the top then in your plane of vision its edges form a &quot;Y&quot;, with the three projected line segments forming 120 degree angles.  If the cube is below you, then the edges form an upside down &quot;Y&quot;, again with the edges at a 120 degrees. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a face of the cube rotates is twisted, you shift your plane of vision to remain looking directly at a cube, keeping your head properly upright but your eyes aimed at the furthest corner of the corner cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;The first subtlety:&lt;/em&gt;&lt;/strong&gt;   A 90 degree rotation &lt;strong&gt;DOES NOT&lt;/strong&gt; convert one of these Y's to another Y or to an upside-down Y!  &lt;em&gt;A Y goes to a Y by a 120 degree rotation.&lt;/em&gt; Y goes to an upside down Y by a 60 degree rotation.  The total rotation for the 4 corners of a quarter-twisted RC face is 120 + 60 + 120 + 60 = 360, or -360, however when one of the RC faces other than top or bottom is twisted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;The second subtlety:&lt;/em&gt;&lt;/strong&gt;  When the top or bottom of the cube is twisted, the observer sees &lt;em&gt;no rotation at all&lt;/em&gt; of the corners, as projected into his field of vision.  So the total rotation for the 4 corners of a twisted top-or-bottom RC face is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The net effect is as Kendall observed, however: The total rotation will be a (possibly zero) multiple of 360, so that a single corner-twist (of 120 degrees) of a single corner by itself is not allowed.&lt;/p&gt;&#xA;" OwnerUserId="13395" LastActivityDate="2015-06-20T00:07:42.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16698" PostTypeId="1" AcceptedAnswerId="16708" CreationDate="2015-06-20T00:43:09.023" Score="4" ViewCount="1312" Body="&lt;p&gt;&lt;em&gt;Note: This &lt;strong&gt;is&lt;/strong&gt; a Mathematical puzzle, &lt;strong&gt;not&lt;/strong&gt; a Mathematical problem. Although the answer could be solved with mathematics, lots of problems on Puzzling SE could be solved with mathematics, and the problem itself is not that mathematical. The math-puzzle tags are simply because, as I said, the puzzle could be solved mathematically.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Imagine&lt;/strong&gt; you have 500 cars lined up in a parking lot that a buyer would like. You are the license plate creator who has to create custom plates for the buyer's 500 cars. The buyer makes a strange request. He wants you to create 4-digit numbers on each license plate in such a way that the first three digits and the last three digits cannot be used again for any other cars. (If somebody could improve that explanation, go for it.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Here's an example:&lt;/strong&gt; Say you make this license plate:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2465&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The license plate's first three digits are &lt;em&gt;246&lt;/em&gt; and the last three digits are &lt;em&gt;465&lt;/em&gt;. That means that when you are creating the other 499 license plates, these numbers, &lt;em&gt;246&lt;/em&gt; and &lt;em&gt;465&lt;/em&gt; cannot be used . Say you were making another license plate:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;8246&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;824&lt;/em&gt; wold be fine, but the buyer would not accept this license plate because the &lt;em&gt;246&lt;/em&gt; was used previously in another car's license plate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Info:&lt;/strong&gt; There are 1,000 different three-digit numbers that could go in the first slot and the second slot of the license plate. This means that if each car needs two numbers, then there would need to be 500 cars, which is how many that is in this puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Diclaimer:&lt;/strong&gt; I do not know the answer to this puzzle, but was curious what the answer is, and how you would explain it. I thought this up while calculating the odds that a certain three-digit number would appear in a four-digit number section on a license plate. I am asking for answer which explains why or why not meeting the buyer's needs are possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need any help understanding, ask in the comments.&lt;/p&gt;&#xA;" OwnerUserId="11002" LastEditorUserId="11002" LastEditDate="2015-06-20T01:53:22.093" LastActivityDate="2015-06-20T03:42:30.437" Title="Unique Licence Plates" Tags="|mathematics|calculation-puzzle|" AnswerCount="2" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16699" PostTypeId="1" CreationDate="2015-06-20T00:48:26.533" Score="4" ViewCount="228" Body="&lt;p&gt;&quot;Sarah&quot;, I said, &quot;Have I ever told you about the times in my childhood when I'd go down to Infinity Creek?&quot;. &quot;No&quot;, she replied, &quot;but it sounds interesting&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;When I was a boy, I lived next to a creek. This wasn't any ordinary creek - it stretched on for what seems like forever. In fact, this is why we called it Infinity Creek. It just so happened that there was a single line of stepping stones, going from one bank to the other, over thousands and thousands of kilometers.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I would always wonder about how many ways I could cross that creek. You see, I was athletic and could choose to jump two stones at a time, instead of just one.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sarah looked at me. &quot;I don't understand&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Well, imagine I had a creek with three stepping stones.&quot;, I replied. &quot;I could jumped them all one by one, I could jump over the first one then jump all the rest one by one - the list goes on. Here's a diagram...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I scribbled a rough sketch in the dirt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FbWRg.png&quot; alt=&quot;why do birds suddenly appear&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;As you can see, I would have five ways to cross a creek with 3 stepping stones.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sarah nodded in understanding. &quot;So you wanted to know...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;...how many ways I could cross Infinity Creek. Yes.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sarah looked puzzled. &quot;But you didn't know how many stones there were?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Oh yes, I did know&quot;, I replied. &quot;There were $42^{25}$ stones. Perhaps now, a bit more mathematically minded, I could work out how many ways I could have crossed Infinity Creek. But because I suspect it is a large number, I would be happy with just the natural log of the number,  rounded to the nearest whole value.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can you help me?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt; Note: This does not require any large amounts of processing power! &lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="7308" LastActivityDate="2015-06-20T01:32:38.110" Title="The Infinite Creek" Tags="|mathematics|" AnswerCount="1" CommentCount="2" ClosedDate="2015-06-20T03:03:08.973" ContentLicense="CC BY-SA 3.0" />
  <row Id="16700" PostTypeId="2" ParentId="16474" CreationDate="2015-06-20T01:05:10.643" Score="2" Body="&lt;h2&gt;Puzzle 1&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Based on a reply by the Puzzle poster to Chadi Abou Sleiman I have to conclude that the value of $q$ is related to the 'size' of the string.  I.E. The space taken in memory to store the apology. However it's not a subject I am adept enough with to find the right numbers and patterns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe it has something to do with base 64 math or 64 bit encodings because: $q = 64m^2$ which means $q$ is any square multiple of $64$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with:&#xA;$$m + 2m + q = X$$&#xA;And&lt;br&gt;&#xA;$$m + 2m = sqrt(q) - 5m$$ &#xA;$$8m = sqrt(q)$$&#xA;$$64m^2 = q$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond this I haven't figured out the exact values yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Apology is 91 characters, 20 words.&lt;/p&gt;&#xA;" OwnerUserId="13355" LastEditorUserId="5373" LastEditDate="2015-06-23T20:41:19.067" LastActivityDate="2015-06-23T20:41:19.067" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16701" PostTypeId="2" ParentId="16686" CreationDate="2015-06-20T01:17:39.720" Score="8" Body="&lt;p&gt;I'm thinking something like&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quantum particles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I communicated with you, not by normal means.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quantum entanglement allows two particles to have the same properties when measured, but cannot be used to transmit information.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The illusionist did everything it seems.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are difficult to see and to measure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proximity can make me very shy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In a closed system, interacting with quantum particles will cause them to disappear/collapse into one state.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without a bunch of these you would die.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They help to make up much of our universe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The bee, the seasons, maybe some more,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The seasons can be influenced by tiny quantum fluctuations, a la the butterfly effect.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who am I? All alone, no knock on the door.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A knock on the door would also collapse a system into one state.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The title is also a clue, being a reference to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Schrodinger's cat&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13396" LastActivityDate="2015-06-20T01:17:39.720" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16702" PostTypeId="1" AcceptedAnswerId="16870" CreationDate="2015-06-20T01:18:04.003" Score="3" ViewCount="773" Body="&lt;p&gt;So every year I've been doing this &quot;Word Game&quot; in which I write a riddle or ask a question and I ask my friends to answer it. My sister claims it too hard. Here it is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Who am I?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I am but a panda.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;But I don’t see all black and white.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I’m of inception,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Here Five and Eight is equal to two-ninths.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(These two lines gives you the answer if you need help)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I’m one of Gracelyn’s favorite book characters,&#xA;&lt;br&gt;&#xA; And I certainly do have a lot of Grace.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11814" LastEditorUserId="5373" LastEditDate="2015-06-21T13:06:04.237" LastActivityDate="2015-06-26T22:20:10.550" Title="The Word Game, I'm test running on you guys" Tags="|riddle|knowledge|wordplay|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16703" PostTypeId="2" ParentId="16699" CreationDate="2015-06-20T01:32:38.110" Score="4" Body="&lt;p&gt;I've seen this problem before. The number of ways you could jump $n$ steps is equal to the number of ways you could jump $n-1$ steps (your last jump is from the last stone) plus the number of ways you could jump $n-2$ steps (if your last jump skips a stone).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is, of course, equal to the $n+2$th Fibonacci number, if the first two are 1 and 1. So the natural log of the path number for $42^{25}$ stepping stones would be very very close to $42^{25} \log \phi - \log \sqrt 5$.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-06-20T01:32:38.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16705" PostTypeId="2" ParentId="16696" CreationDate="2015-06-20T02:32:08.467" Score="8" Body="&lt;p&gt;Each finite sequence of letters from the set $\{A,B\}$ gives instructions for moving from one room to another, and so determines a permutation of the rooms in the house. Let $G$ denote the collection of all permutations arising this way, which is a group. The information we are given means that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$G$ acts transitively on the rooms in the house,&lt;/li&gt;&#xA;&lt;li&gt;$G$ is generated by the elements $A$ and $B$, and&lt;/li&gt;&#xA;&lt;li&gt;every element of $G$ has order dividing $3$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Condition 1 says that the number of rooms is bounded by the size of $G$. The largest group satisfying Conditions 2 and 3 is called the Burnside group $B(2,3)$, which has order $27$ (see &lt;a href=&quot;https://en.wikipedia.org/?title=Burnside%27s_problem&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is an image of the Burnside group, which can also be interpretted as the largest house satisfying the desired conditions. This house has $27$ rooms. The black squares represent rooms. A yellow edge pointing from one room to another indicates door A leads from the first room to the second. Similarly, blue edges represent doors labelled B.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/6vws7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This image was created by William E. Skeith III, and come from &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:FreeBurnsideGroupExp3Gens2.png&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-06-20T02:37:47.790" LastActivityDate="2015-06-20T02:37:47.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16706" PostTypeId="2" ParentId="16698" CreationDate="2015-06-20T02:49:21.160" Score="8" Body="&lt;p&gt;An answer &lt;em&gt;does&lt;/em&gt; exist in terms of a &lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/De_Bruijn_sequence&quot;&gt;De Bruijn sequence&lt;/a&gt;&lt;/strong&gt;.  An order-$n$ De Bruijn sequence $B(A,n)$ on an alphabet $A$ (of size $k$) contains all length-$n$ strings in $A$ exactly once.  One example of $B(0\ldots9,3)$ looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;9,7,9,8,7,8,7,7,7,0,7,6,0,7,5,0,7,4,0,7\ldots 7,8,8,7,9&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each license plate will have length $n+1$ (in this case $4$).  We can obtain a list of licence plates by taking every other length-$(n+1)$ substring of the sequence.  That way, the last length-$n$ substring of each plate does not completely overlap the first substring of the next.  Since each length-$n$ substring is guaranteed to be unique, this gives us a list of $500$ plates (the length of the sequence, $1000$, divided by 2) that satisfy the criteria.  For example, the list of plates from the above sequence looks something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;9,7,9,8\\&#xA;9,8,7,8\\&#xA;7,8,7,7\\&#xA;7,7,7,0\\&#xA;7,0,7,6\\&#xA;\vdots\\&#xA;9,7,8,8\\&#xA;8,8,7,9\\&#xA;7,9,9,7&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that we take the plates from the sequence cyclically, such that the last plate consists of the last two elements, then the first two.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tested this using the following &lt;em&gt;Mathematica&lt;/em&gt; code:&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Needs[&quot;Combinatorica`&quot;]&#xA;seq = DeBruijnSequence[Range[10] - 1, 3];&#xA;plates = Partition[seq, 4, 2, 1];&#xA;Length[plates] (* == 500 *)&#xA;DuplicateFreeQ[Join[Most /@ plates, Rest /@ plates]] (* == True *)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This shows that &lt;code&gt;plates&lt;/code&gt; contains $500$ numbers (lists of digits) such that there are no duplicates among the first three and last three digits of each plate.&lt;/p&gt;&#xA;" OwnerUserId="9718" LastActivityDate="2015-06-20T02:49:21.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16707" PostTypeId="2" ParentId="16686" CreationDate="2015-06-20T03:38:58.797" Score="1" Body="&lt;p&gt;Maybe it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; God&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I communicated with you, not by normal means.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many people saw visions of God, such as Moses or Muhammad's vision of the Angel Gabriel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The illusionist did everything it seems.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; God created all.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proximity can make me very shy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; God is said to be beyond reality, and no one that I know of has ever really &quot;seen&quot; God.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without a bunch of these you would die.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Possibly deities watching over you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The bee, the seasons, maybe some more,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Deities in control of different things.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who am I? All alone, no knock on the door.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Only one god in most religions, God won't interact directly with your life?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With the title possibly being&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jesus's birth, crucifixion, resurrection, and then ascension.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13396" LastActivityDate="2015-06-20T03:38:58.797" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16708" PostTypeId="2" ParentId="16698" CreationDate="2015-06-20T03:42:30.437" Score="18" Body="&lt;p&gt;Another answer: start the license plate making process by making all possible 500 three-digit plates, where the three digits have even sum.  Now to each add a last digit as so:  If the first digit on the plate is $n$, use $n+1$ as the last digit (and if the first digit is $9$, use a $0$ as the last digit).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first three digits of any plate can't be the same as any other first three digits by design.  The first three of any plate can't be the same as any last three, because the first three always has even sum and the last three always has odd sum.  Finally, the last three of any plate can't overlap with any other plate, since that would imply the first three would be the same (i.e. if two plates end in $340$, both plates must be $9340$).  &lt;/p&gt;&#xA;" OwnerUserId="2722" LastActivityDate="2015-06-20T03:42:30.437" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16709" PostTypeId="2" ParentId="16686" CreationDate="2015-06-20T04:12:41.157" Score="0" Body="&lt;p&gt;Without explaining the whole riddle, I'd like to guess&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Higgs boson. It is difficult to detect because of it's tendency to pop in and out of existence, as the title would suggest. This particle has also been suggested as one of the main reasons for the universe's inflation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13396" LastEditorUserId="13396" LastEditDate="2015-06-20T04:20:07.943" LastActivityDate="2015-06-20T04:20:07.943" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16710" PostTypeId="1" AcceptedAnswerId="16732" CreationDate="2015-06-20T04:23:39.333" Score="6" ViewCount="345" Body="&lt;p&gt;Consider the following ten statements:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$A$ do not have $B$.&lt;/li&gt;&#xA;&lt;li&gt;Some $C$ are $D$, but no $E$ are $D$.&lt;/li&gt;&#xA;&lt;li&gt;$F$ that are $A$ can still be $G$ if they have $H$.&lt;/li&gt;&#xA;&lt;li&gt;No $C$ that do not have $G$ can be both $H$ and $B$.&lt;/li&gt;&#xA;&lt;li&gt;All $D$ are $B$, and all $B$ that are $A$ are also $G$.&lt;/li&gt;&#xA;&lt;li&gt;If $E$ are $B$, they have $H$.&lt;/li&gt;&#xA;&lt;li&gt;Some but not all $F$ are $H$.&lt;/li&gt;&#xA;&lt;li&gt;$C$ that have $B$ that are $E$ are not $A$.&lt;/li&gt;&#xA;&lt;li&gt;Some $F$ have $G$ even if they do not have $D$.&lt;/li&gt;&#xA;&lt;li&gt;All $F$ that are $G$ and are not $C$, are $E$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The eight variables $A$ through $H$ represent plural nouns. Your task is to determine a value for each noun so that all ten statements hold true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally speaking, a statement is true if and only if:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;it requires no creative interpretation (e.g. the statement &quot;All bishops are containers.&quot; is invalid even though bishops' bodies contain organs, blood, etc. and hence 'technically' qualify as containers)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;it is reasonably objective (e.g. &quot;All elephants are giants.&quot; is a reasonably objective statement, while &quot;All elephants are weirdos.&quot; is not)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;it is unambiguous and can be reasonably verified as correct (e.g. &quot;Not all foods are bean bags.&quot; is valid despite being obvious, while &quot;Not all powders are chemicals.&quot; might be true or untrue depending on how literally one takes the statement)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note that if the above conditions are met, only the condition that is &lt;em&gt;explicitly stated&lt;/em&gt; needs to hold true. For example, &quot;Some rock stars are either umbrellas or artists.&quot; is true even though the statement might imply that some rock stars are umbrellas, which is false, or that not all rock stars are artists, which is also false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the question is open-ended, the &lt;strong&gt;voters&lt;/strong&gt; will ultimately determine if any given answer is stretching the truth. When in doubt: don't risk it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nouns must be made up of real English words. They can be abstract, and they can be qualified as little or as much as desired, but they must be plural nouns. Some valid examples include &quot;dogs&quot;, &quot;big dogs&quot;, &quot;big dogs' ideas&quot;, &quot;people that like big dogs&quot;, etc. Note that &lt;em&gt;nouns capable of possessing/containing/having other nouns&lt;/em&gt; as well as &lt;em&gt;nouns capable of being possessed/contained/had&lt;/em&gt; are strongly recommended since many of the statements employ the verb &quot;have&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An answer's score is the total number of characters (including any punctuation and spaces) in the eight nouns. Lower scores are better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The winning answer is the answer with the lowest score where all ten statements are true.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Good luck! And may the most re-nouned answer win!&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="69582" LastEditDate="2022-06-03T18:26:44.150" LastActivityDate="2022-06-03T18:26:44.150" Title="Renouned Set Theory" Tags="|word|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="16711" PostTypeId="2" ParentId="16686" CreationDate="2015-06-20T04:25:25.380" Score="5" Body="&lt;p&gt;I am new here and don't really understand the rules of the puzzle, but ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The title to me suggests&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Teeth.  First growing, then as if dead. Then growing and dead again when the adult teeth come in.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I communicated with you, not by normal means.&#xA;The illusionist did everything it seems.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The nerves in the teeth are wired strangely.  Some of them seem to be miswired and cause you to think the sensation comes from the wrong tooth. You also cannot see them directly, but maybe you see them in a mirror (the illusion?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proximity can make me very shy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A shy person may not smile and show their teeth when someone is too close?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without a bunch of these you would die.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Obviously teeth are needed for survival.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The bee, the seasons, maybe some more,&#xA;Who am I? All alone, no knock on the door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These lines don't seem to fit my line of thought. But I am not familiar with these puzzles, and I don't understand your hint about &quot;Who&quot; giving us the &quot;form&quot; of the answer.&lt;/p&gt;&#xA;" OwnerUserId="13399" LastEditorUserId="13244" LastEditDate="2015-07-24T12:44:11.607" LastActivityDate="2015-07-24T12:44:11.607" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16712" PostTypeId="2" ParentId="16686" CreationDate="2015-06-20T04:36:45.353" Score="4" Body="&lt;p&gt;Could it be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; our shadow?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;I communicated with you, not by normal means.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Our shadow is there to mimic our moves.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;The illusionist did everything it seems.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Shadows are the result of light.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Proximity can make me very shy.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If we try to get close to our shadow, it either shrinks, moves away, or disappears.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Without a bunch of these you would die.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If we cannot cast a shadow, our body does not exist.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;The bee, the seasons, maybe some more,&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;No idea. I guess because everything is capable of casting a shadow.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Who am I? All alone, no knock on the door.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Our shadows have no one to talk to, they're silent, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As for the &quot;alive, dead, alive, now dead,&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;our shadows come and go depending on which setting we're in. Sometimes they're there, sometimes they're gone.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13400" LastEditorUserId="5188" LastEditDate="2015-06-20T06:48:18.753" LastActivityDate="2015-06-20T06:48:18.753" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16713" PostTypeId="2" ParentId="16686" CreationDate="2015-06-20T05:13:49.087" Score="0" Body="&lt;p&gt;Okay, another stab...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think it has something to do with vowels.  Maybe silent ones, and or the second vowel in a double vowel?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I communicated with you, not by normal means.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I and U (&quot;you&quot;) are vowels...  And they communicate not by talking by by their effects on each other in words.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The illusionist did everything it seems.&#xA;Proximity can make me very shy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A vowel can become silent when next to another, as the &quot;A&quot; in &quot;read&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without a bunch of these you would die.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;you&quot; has two vowels in it and therefore would die without vowels.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The bee, the seasons, maybe some more,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bee and season both have a double vowel which is silent, and if we count the y in maybe as a vowel, then it is silent as well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who am I? All alone, no knock on the door.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am a vowel?  I is a vowel?  In alive, there are no silent double vowels, whereas in dead, we again have the silent (&quot;dead&quot;) a.  But this is not totally logical, since there is a silent &quot;e&quot; in alive.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13399" LastActivityDate="2015-06-20T05:13:49.087" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16714" PostTypeId="1" CreationDate="2015-06-20T06:34:39.137" Score="0" ViewCount="1733" Body="&lt;p&gt;Find a number $X$ that is 9 times bigger than the total of its digits: for example if $X=abc$, then $X=9*(a+b+c)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know the number of digits; it could be 3 or 2 or more.&lt;/p&gt;&#xA;" OwnerUserId="12038" LastEditorUserId="5373" LastEditDate="2015-06-20T09:06:40.187" LastActivityDate="2015-06-21T00:17:46.250" Title="Find a number that is 9 times bigger than total of its digits?" Tags="|mathematics|calculation-puzzle|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16715" PostTypeId="2" ParentId="16714" CreationDate="2015-06-20T07:02:11.223" Score="11" Body="&lt;p&gt;Simple 2 digit answer will be,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $81$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For 2 digit number, &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$X = ab = 10a + b = 9*(a+b)$&lt;br&gt;We have, $a = 8b$ &lt;br&gt;Only $b = 1$ satisfies the condition for $X$ to be $2$ digit number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3 digit answer is not possible.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$X = abc = 100a + 10b + c = 9a + 9b + 9c $&lt;br&gt;i.e. $91a = 8c - b$ &lt;br&gt;$a$ can take values from $1$ to $9$, not zero &lt;br&gt;But max of $c$ is $9$, so RHS is always less than or equal to $72$.&lt;br&gt;Hence, no possible $3$ digit answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13310" LastEditorUserId="7308" LastEditDate="2015-06-20T10:08:18.760" LastActivityDate="2015-06-20T10:08:18.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16716" PostTypeId="2" ParentId="16714" CreationDate="2015-06-20T08:27:55.317" Score="7" Body="&lt;p&gt;While Rajendra is right, the only &quot;normal&quot; solution is 81...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another solution is zero. The sum of its digits is zero, and zero times nine is zero.&lt;/p&gt;&#xA;" OwnerUserId="9289" LastActivityDate="2015-06-20T08:27:55.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16718" PostTypeId="2" ParentId="16686" CreationDate="2015-06-20T09:57:50.173" Score="2" Body="&lt;p&gt;I believe the answer is...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mother Nature&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I communicated with you, not by normal means.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nature &quot;talks&quot; in many ways, from the sounds of crickets and cicadas to the rumble of thunder... but it doesn't actually talk in the sense of explicit messages.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The illusionist did everything it seems.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I believe &quot;the illusionist&quot; is referring to man - we try to structure and build our own environments, but as soon as we stop trying, nature returns and reverts things to their natural state.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proximity can make me very shy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Animals generally flee when you get close.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without a bunch of these you would die.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nature is essential to our survival.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The bee, the seasons, maybe some more,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The bee and the seasons are both part of nature, of course, and those two are particularly fundamental to our survival.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who am I? All alone, no knock on the door.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mother nature is spoken of as though it is a person, but it's not.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9289" LastActivityDate="2015-06-20T09:57:50.173" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16719" PostTypeId="2" ParentId="16714" CreationDate="2015-06-20T10:35:36.980" Score="2" Body="&lt;p&gt;Just to complete Rajendra's answer, I show that there's no solution with &lt;em&gt;more&lt;/em&gt; than 3 digits either, so that the only possible solutions are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 81 and 0.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assume there's a solution with $n$ digits where $n&amp;gt;2$. We have:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$X = a_1a_2\dots a_n = 10^{n-1}a_1 + 10^{n-2}a_2 + \dots + a_n = 9a_1 + 9a_2 + \dots + 9a_n $ gives $10^{n-1}-9\leq(10^{n-1}-9)a_1 = 8a_n - a_{n-1} - 91a_{n-2} - \dots - (10^{n-2}-9)a_2\leq72.$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So there's no possible answer with more than 2 digits.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-06-21T00:17:46.250" LastActivityDate="2015-06-21T00:17:46.250" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16720" PostTypeId="2" ParentId="16686" CreationDate="2015-06-20T12:10:03.667" Score="2" Body="&lt;p&gt;It could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Cells&lt;/strong&gt;,&lt;BR&gt;the vast majority of which &lt;a href=&quot;http://www.divinecaroline.com/self/wellness/truth-about-your-bodys-cell-regeneration&quot; rel=&quot;nofollow&quot;&gt;regenerate, but at different rates&lt;/a&gt;, depending on the cell, possibly explaining the title.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I communicated with you, not by normal means.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cells communicate through their own language of chemical signals, which is extremely complex to the point where we don't fully understand it. See &lt;a href=&quot;http://blogs.discovermagazine.com/science-sushi/2013/08/24/ask-discover-how-do-cells-communicate/#.VYVSU0Ycijw&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The illusionist did everything it seems.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think the greatest illusion is that we appear as one body, even though, in reality, &lt;a href=&quot;http://www.smithsonianmag.com/smart-news/there-are-372-trillion-cells-in-your-body-4941473/?no-ist&quot; rel=&quot;nofollow&quot;&gt;each of us is made up of trillions of cells&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proximity can make me very shy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A &lt;a href=&quot;http://boards.straightdope.com/sdmb/showthread.php?t=698306&quot; rel=&quot;nofollow&quot;&gt;proximity sensor on your cell phone&lt;/a&gt; turns off the screen when the phone is held to your ear.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without a bunch of these you would die.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Without a bunch of cells, you'd die. That seems obvious, and is the biggest clue in the riddle, I think.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The bee, the seasons, maybe some more,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bee's have cells, and each hexagon-shaped hole in their honeycombs is called a cell. The change of seasons &lt;a href=&quot;http://gtm-media.discoveryeducation.com/videos/dsc/externalApplications/accessible/simulations/SeasonsintheCell/index.html&quot; rel=&quot;nofollow&quot;&gt;triggers chemical changes&lt;/a&gt; in plant cells.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who am I? All alone, no knock on the door. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You're a &lt;strong&gt;cell&lt;/strong&gt;. If you're all alone with no knock at the door, you could be in a prison cell!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-06-20T12:18:04.387" LastActivityDate="2015-06-20T12:18:04.387" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16721" PostTypeId="2" ParentId="16686" CreationDate="2015-06-20T13:17:35.440" Score="1" Body="&lt;p&gt;A bit of a guess because I'm not convinced it quite matches everything:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Organ(s)&quot; in all its definitions: 1) Musical, 2) Biological, 3) Instrument or agency, 4) a means of communication like a newspaper.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I communicated with you, not by normal means.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Music communicates and senses are organs and definition 4 fits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The illusionist did everything it seems.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Musical meaning here. The organist is the illusionist. Definitions 3 and 4 kind of work too.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proximity can make me very shy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure about that. I'm guessing biological organs here: recoiling when touched as most don't like to be. Definition 3 might work too.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without a bunch of these you would die.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Organic organs obviously.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The bee, the seasons, maybe some more,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Doesn't really work here. Organs of nature in the definition 3 sense?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who am I? All alone, no knock on the door.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Or here. Or does it? Door bell maybe?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alive, dead, alive, now dead&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Err struggling now. Great music, musicians?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12666" LastActivityDate="2015-06-20T13:17:35.440" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16722" PostTypeId="2" ParentId="16686" CreationDate="2015-06-20T14:42:31.660" Score="6" Body="&lt;p&gt;I don't expect this to be right but since you said not to take anything but Who and Dying too literally - I have to try it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Doctor (Dr. &lt;strong&gt;Who&lt;/strong&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I communicated with you, not by normal means.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The TARDIS translates all alien languages into English (or whatever someone's native tongue is)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The illusionist did everything it seems.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The name of the Doctor is itself an illusion - hiding his true secret name.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proximity can make me very shy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Doctor (in all his regenerations) is outgoing and charismatic - but he does not let many people get to really know him. He's constantly found to be hiding aspects of his life from his companions when little tidbits slip out - showing he's actually quite private and may use the extrovert actions to keep people from knowing the real him.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without a bunch of these you would die.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sick and dying need doctors to survive&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The bee, the seasons, maybe some more,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The 4th Doctor mentions that he likes bees (comparing K-9's greeting to the bee's language) and the 10th Doctor is able to find the Earth because of the bees disappearing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who am I? All alone, no knock on the door.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; David Tennant's time as The Doctor ended by himself after 4 knocks on the door - updated thanks to @theage for the catch&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The title (Alive, Dead, Alive, Dead)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;The Doctor is a Time Lord, he can regenerate&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11239" LastEditorUserId="11239" LastEditDate="2015-06-21T00:16:19.283" LastActivityDate="2015-06-21T00:16:19.283" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16723" PostTypeId="1" AcceptedAnswerId="16725" CreationDate="2015-06-20T15:48:03.327" Score="12" ViewCount="683" Body="&lt;p&gt;A man has plotted to blow up a building. The police have three suspects. They have received a letter telling the name of the suspect. But it is coded, to avoid one of the other members of the suspect's criminal gang realising what has happened and changing the letter. This is what the letter says:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2-1, 5-3, 3-2, 9-2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which suspect is guilty: Alex, Dave or Jeff?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Sorry about the storyline, I have real trouble coming up with them)&lt;/p&gt;&#xA;" OwnerDisplayName="user9377" LastEditorDisplayName="user9377" LastEditDate="2015-06-20T16:03:51.047" LastActivityDate="2015-06-22T13:12:29.477" Title="Alex, Dave or Jeff" Tags="|cipher|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16724" PostTypeId="2" ParentId="2078" CreationDate="2015-06-20T15:55:52.823" Score="3" Body="&lt;p&gt;I believe the answer is number 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Box 4 is  a combination of boxes 1 and 3.&#xA;The fifth box should be a combination of boxes 2 and 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the fifth box is to be filled with answer 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also explains the lack of a discernible pattern to boxes 1 through 3.  &lt;/p&gt;&#xA;" OwnerUserId="13409" LastActivityDate="2015-06-20T15:55:52.823" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16725" PostTypeId="2" ParentId="16723" CreationDate="2015-06-20T16:28:56.013" Score="19" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alex&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Telephone code&lt;br&gt;&#xA; &lt;img src=&quot;https://i-cdn.phonearena.com/images/reviews/83518-image/Sanyo-Taho-Review-Design-015.jpg&quot; alt=&quot;telephone&quot;&gt;&#xA; Hit 2 once to get A, hit 5 three times to get L, hit 3 twice to get E, hit 9 twice to get X&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10934" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:48.777" LastActivityDate="2015-06-20T16:33:44.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16726" PostTypeId="2" ParentId="16723" CreationDate="2015-06-20T16:48:39.963" Score="6" Body="&lt;p&gt;I think it is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jeff&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My reasoning for this is because:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If odd numbers represent consonants and even numbers represent vowels.&#xA; Computing the formulas gives us the numbers: 1,2,1,7. The only name that follows this order of Consonant, Vowel, Consonant, Consonant is JEFF!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Although looking at the &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;phone code answer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;that makes more sense.&lt;/p&gt;&#xA;" OwnerUserId="1981" LastEditorUserId="1981" LastEditDate="2015-06-22T13:12:29.477" LastActivityDate="2015-06-22T13:12:29.477" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16727" PostTypeId="2" ParentId="14708" CreationDate="2015-06-20T18:28:11.143" Score="3" Body="&lt;p&gt;&lt;b&gt;Here is another possible solution:&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(5!/4 \ + \ 3)^2 \ -  \ 7 \ + \ 6 \ = \ 1088$&lt;/p&gt;&#xA;" OwnerUserId="13203" LastActivityDate="2015-06-20T18:28:11.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16728" PostTypeId="2" ParentId="11486" CreationDate="2015-06-20T18:46:44.000" Score="6" Body="&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/a/11597/10615&quot;&gt;This answer&lt;/a&gt; proves that four lighthouses is possible. It turns out that it is also possible with five lighthouses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first step is to choose a lighthouse, call it $A$, and angle it in such a way that&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;there are two lighthouses on either side of the midline of its light&lt;/li&gt;&#xA;&lt;li&gt;if you rotated it $180^\circ$, it wouldn't shine on any lighthouse&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Below is an illustration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hnQQh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this possible? Imagine stretching a large rubber around all of the lighthouses, forming a polygon. On the vertices must have an angle of at most $108^\circ$; choose this lighthouse to be $A$. Then rotate $A$ until its midline cuts the other four lighthouses in half. If doing this doesn't fulfill the second bullet point, then turning $A$ around $180^\circ$ will.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, we use the two lighthouses below the line to illuminate the rest of the upper half. Let's label a compass with degree marks going counterclockwise so $0^\circ$ is east, meaning $90^\circ$ is north, $180^\circ$ is west, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of the bottom two lighthouses, let $B$ be the one which is further along in the $342^\circ$ direction, and $C$ be the other. Shine $B$ so that the clockwise edge of its beam points towards $72^\circ$. Shine $C$ so its clockwise edge points east. This will illuminate the upper plane, as shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BSHzp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $A$ and $B$'s beams must overlap because of the second bullet point in how $A$ was chosen, while $B$ and $C$'s beams overlap since $B$ is further along in the $342^\circ$ direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing the same procedure with the upper two lighthouses illuminates the plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a fun fact, it turns out that if you have any number of lighthouses, and their beam angles add up to $360^\circ$, then they can illuminate the plane. However, the &lt;a href=&quot;http://www.cs.cmu.edu/puzzle/solution30.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;proof&lt;/a&gt; is quite mathy.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-07-15T22:52:13.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16729" PostTypeId="1" AcceptedAnswerId="16734" CreationDate="2015-06-20T19:53:12.120" Score="8" ViewCount="696" Body="&lt;p&gt;You are given an empty blackboard. Once a minute, you are allowed to do one of the following: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Write two $1$'s on the blackboard&lt;/li&gt;&#xA;&lt;li&gt;Erase two copies of a number, $n$, and replace them with $n-1$ and $n+1$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Here is an example of six possible moves: &#xA;$$&#xA;empty\to 1,1 \to 0,2\to 0,2,1,1\to 0,2,0,2\to 0,1,0,3\to -1,1,1,3\to\cdots&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What's the shortest time it takes to write $100$ on the board?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="11876" LastEditDate="2015-11-25T02:48:43.063" LastActivityDate="2015-11-25T02:48:43.063" Title="Numbers on a Blackboard" Tags="|combinatorics|algorithm|blackboard|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16730" PostTypeId="2" ParentId="16729" CreationDate="2015-06-20T19:59:41.550" Score="11" Body="&lt;p&gt;So far I have a lower bound of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $85850$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and an upper bound of &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $164275$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For the upper bound, consider the following simple strategy:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Ignore 0's&lt;/li&gt;&#xA;&lt;li&gt;If there's two $n$'s, change them to $n-1$ and $n+1$.&lt;/li&gt;&#xA;&lt;li&gt;Otherwise, create two 1's.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Let's keep ignoring 0's and pretend that we can add a single 1 for half a step. We argue that at any point, each number appears once, except at most one number appears twice. We need that invariant that whenever $n$ appears twice, $n-1$ does not appear. Then, when we replace $n,n$ with $n-1,n+1$, there's at most one $n-1$, at most two $n+1$'s, and no $n$'s appear, so the invariant is maintained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The invariant is also maintained when we create a single $1$ because there are no pairs (remember we're ignoring 0's). In reality, when we create two ones, the invariant is maintained as long as we can ignore a single 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, at the end when $100$ is created, there can be at most one of each number plus an extra 1. Moreover, there cannot be a double since we just acted on $99$ and there are not two $100$'s. And, there cannot be a 99. So, at most, the numbers $1$ through $100$ can appear, with no $99$ and an extra $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the invariant argument below, this end position correspond to &lt;a href=&quot;http://www.wolframalpha.com/input/?i=%28%28sum+from+1+to+100+of+n%5E2%29+%2B+1+-+99%5E2%29%2F+2&quot;&gt;this number&lt;/a&gt; of steps:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $164275$ steps.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;First, a weak lower bound for the number of steps it takes is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $5000 = 100^2 /2$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We look at the statistic&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the sum of squares of the numbers on the whiteboard&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Each move, causes the statistic&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; to increase by 2. Adding two 1's clearly does so. Replacing $n,n$ with $n-1,n+1$ changes the statistic as $(n+1)^2+(n-1)^2-n^2-n^2 = 2$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This gives the lower bound because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the statistic starts at 0 and must end at least at $100^2$ for the number 100 to be on the board.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, this lower bound cannot be sharp because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the number 100 cannot be left on the board alone. It must also appear with 98.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can improve the lower bound to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $85850$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;by &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; observing that after the blackboard contains one of $a,a+1$, that fact must be true from then on, since removing an $a$ creates an $a+1$ and removing an $a+1$ creates an $a$. Since every number must be created on the way to $100$, the minimum possible statistic at the end is $100^2 + 98^2 + ... + 2^2 + 0^2$, which is $171700$, so the number of required moves is at least half of that.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-06-20T20:51:57.763" LastActivityDate="2015-06-20T20:51:57.763" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16732" PostTypeId="2" ParentId="16710" CreationDate="2015-06-20T20:59:15.893" Score="6" Body="&lt;p&gt;Here's a solution that seems to work (inspired by Glen O's mention of letters and words in &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16710/renouned-set-theory/16732?noredirect=1#comment52194_16710&quot;&gt;a comment&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $A=$ letters&lt;br&gt;$B=$ words&lt;br&gt;$C=$ sentences&lt;br&gt;$D=$ pronouns&lt;br&gt;$E=$ articles&lt;br&gt;$F=$ words&lt;br&gt;$G=$ members of the set {a,I}&lt;br&gt;$H=$ letters&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Going through the ten statements one by one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1. &lt;strong&gt;Letters do not have words.&lt;/strong&gt; Of course not - letters having words doesn't make sense!&lt;br&gt;2. &lt;strong&gt;Some sentences are pronouns, but no articles are pronouns.&lt;/strong&gt; Many pronouns can function as whole sentences (&quot;Who did it?&quot; &quot;Her.&quot;), but articles and pronouns are disjoint classes of words.&lt;br&gt;3. &lt;strong&gt;Words that are letters can still be a or I if they have letters.&lt;/strong&gt; In fact, single-letter words &lt;em&gt;must&lt;/em&gt; be a or I, and the &quot;if they have letters&quot; part is superfluous (but doesn't invalidate the statement).&lt;br&gt;4. &lt;strong&gt;No sentences that do not have the words a or I can be both letters and words.&lt;/strong&gt; A sentence that consists only of a single word and also only of a single letter must be either a or I (in fact, it must be I), so a sentence that &lt;em&gt;doesn't&lt;/em&gt; contain these words can't be both a letter and a word.&lt;br&gt; 5. &lt;strong&gt;All pronouns are words, and all words that are letters are also members of the set {a,I}.&lt;/strong&gt; Pronouns are certainly words, and a and I are the only single-letter words.&lt;br&gt;6. &lt;strong&gt;If articles are words, they have letters.&lt;/strong&gt; Articles &lt;em&gt;are&lt;/em&gt; words, and they do have letters. That was easy!&lt;br&gt;7. &lt;strong&gt;Some but not all words are letters.&lt;/strong&gt;  Some words (a, I) are single letters, but others (stack, exchange) are not.&lt;br&gt;8. &lt;strong&gt;Sentences that have words that are articles are not letters.&lt;/strong&gt; The only way a sentence can be a single letter is if it's a or I, but a is not a sentence. So if a sentence has articles, it can't be a single letter.&lt;br&gt;9. &lt;strong&gt;Some words have members of the set {a,I} even if they do not have pronouns.&lt;/strong&gt; The word a.&lt;br&gt;10. &lt;strong&gt;All words that are members of the set {a,I} and are not sentences are articles.&lt;/strong&gt; I can function as a sentence (&quot;Who is it?&quot; &quot;I.&quot;) but a cannot, so so a word in {a,I} that's not a sentence must be a, which is an article.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-06-21T13:09:29.100" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16733" PostTypeId="2" ParentId="16710" CreationDate="2015-06-20T21:02:32.793" Score="3" Body="&lt;p&gt;Making the following assumptions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The predicate &quot;Some F have G even if they do not have D&quot;, could be understood as &quot;Some F have G but don't have D&quot; (it could be more laxly understood as &quot;Some F have G regardless of having D or not&quot;)&lt;/li&gt;&#xA;&lt;li&gt;The word have can be used as in &quot;I have a mother&quot;, not just traditional possession&lt;/li&gt;&#xA;&lt;li&gt;Nouns can be repeated&lt;/li&gt;&#xA;&lt;li&gt;All employees have bosses&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Here's my solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $A$ = pots&lt;br&gt;&#xA; $B$ = children&lt;br&gt;&#xA; $C$ = people&lt;br&gt;&#xA; $D$ = children&lt;br&gt;&#xA; $E$ = adults&lt;br&gt;&#xA; $F$ = men&lt;br&gt;&#xA; $G$ = bosses&lt;br&gt;&#xA; $H$ = employees&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So it reads like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  1. Pots do not have children.&lt;br&gt;&#xA;  2. Some people are children, but no adults are children.&lt;br&gt;&#xA;  3. Men that are pots can still be bosses if they have employees.&lt;br&gt;&#xA;  4. No people that do not have bosses can be both employees and children.&lt;br&gt;&#xA;  5. All children are children, and all children that are pots are also bosses.&lt;br&gt;&#xA;  6. If adults are children, they have employees.&lt;br&gt;&#xA;  7. Some but not all men are employees.&lt;br&gt;&#xA;  8. People that have children that are adults are not pots.&lt;br&gt;&#xA;  9. Some men have bosses even if they do not have children.&lt;br&gt;&#xA;  10. All men that are bosses and are not people, are adults.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know there's a way to fill half of the words with wide concepts like &quot;men&quot; or &quot;people&quot;, making it easier to find short-named subsets. Or nouns that you can &lt;em&gt;have&lt;/em&gt; but are still broad, like &quot;friends&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Considerably shortened $A$&lt;/p&gt;&#xA;" OwnerUserId="13301" LastEditorUserId="13301" LastEditDate="2015-06-20T21:39:25.807" LastActivityDate="2015-06-20T21:39:25.807" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16734" PostTypeId="2" ParentId="16729" CreationDate="2015-06-20T22:02:50.563" Score="8" Body="&lt;p&gt;The shortest amount of time it takes to write the number $100$ is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $164275$ minutes. xnor has already described a process for reaching $100$ in $164275$ minutes, so here I will argue that it cannot be done faster.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I will need to following fact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Claim&lt;/strong&gt;: Let $k$ be a non-negative integer. If there has ever been an integer $n&amp;gt;k$ written on the board, then for any interval of $k+1$ consecutive positive integers less than or equal to $n$, there are at least $k$ integers from that interval currently written on the board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The special case $k=1$ was observed by xnor (once a number has appeared, at least one of every pair of consecutive smaller numbers must always be on the board).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;: Induction on $k$. The base case $k=0$ is vacuously true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let $k$ be arbitrary, and assume the claim holds for $0,1,\ldots,k-1$. Suppose some integer $n&amp;gt;k$ has appeared on the blackboard at some point. Let $I=\{a+1,\ldots,a+k\}$ be a set of $k$ consecutive positive integers, with $a+k\leq n$. At some point in time $a+k$ must have been written on the board, and at that time there were at least $k$ numbers from $I$ written on the board: the integer $a+k$, along with $k-1$ integers from $\{a+1,\ldots,a+(k-1)\}$ (here we use the inductive hypothesis).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that $k$ integers on the board were in $I$ at some point, we can argue that $k$ integers on the board are in $I$ at all times. The only time that the number of integers on the board that are in $I$ decreases is when we replace two copies of $a+k$ by $a+k+1$ and $a+k-1$, or when we replace two copies of $a+1$ by $a$ and $a+2$. However, in the first case, there were already $k-1$ integers on the board from the interval $\{a+1,\ldots,a+k-1\}$ before the replacement (by the inductive hypothesis), so along with the new $a+k-1$, that makes a total of $k$ integers $I$. The second case is identical.&lt;br&gt; &lt;strong&gt;This completes the proof of the claim.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Suppose $100$ is written on the blackboard. Then the blackboard must also contain: at least one number from $\{99,98\}$, at least two numbers from $\{99,98,97\}$, and so on, up through $98$ numbers from $\{99,98,\ldots,1\}$. This means that the smallest possible value for the sum of the squares of the numbers on the board is&#xA;$$&#xA;100^2+98^2+97^2+96^2+\ldots+1^2=328549.&#xA;$$&#xA;As xnor has explained, the sum of the squares of the integers on the blackboard increases by $2$ each minute, so the number of minutes required is at least&#xA;$$&#xA;\left\lceil\frac{328549}{2}\right\rceil=164275.&#xA;$$&#xA;This is what we needed to show.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-06-21T03:16:40.387" LastActivityDate="2015-06-21T03:16:40.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16736" PostTypeId="2" ParentId="16474" CreationDate="2015-06-20T23:15:31.883" Score="0" Body="&lt;p&gt;The answer to puzzle 2 is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;562&lt;br&gt;&#xA;Your Home is converted to ascii values.&lt;br&gt;&#xA;The top 5 are mooru.&lt;br&gt;&#xA;Mind I haven't necessarily taken care of the well-rounded comment but that might be because I don't use Stackoverflow in a standard font?  I definitely left out the Y because of that.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1777" LastEditorUserId="1777" LastEditDate="2015-06-20T23:46:53.970" LastActivityDate="2015-06-20T23:46:53.970" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16738" PostTypeId="2" ParentId="16686" CreationDate="2015-06-21T00:30:08.793" Score="15" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Police, who have disbanded, reunited, then disbanded again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I communicated with you, not by normal means.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Message in a Bottle&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The illusionist did everything it seems.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Every Little Thing She Does Is Magic&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proximity can make me very shy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Don't Stand So Close to Me&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without a bunch of these you would die.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Every Breath You Take&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The bee, the seasons, maybe some more,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Band members Sting, Andy Summers, and others&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Police!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All alone, no knock on the door.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This I am unsure about.  They have other song titles that sort of fit (&quot;So Lonely&quot;, &quot;Hole In My Life&quot;).  Maybe there's a particular lyric that fits better?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13416" LastActivityDate="2015-06-21T00:30:08.793" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16739" PostTypeId="2" ParentId="16474" CreationDate="2015-06-21T01:40:14.260" Score="3" Body="&lt;p&gt;Compilation of answers.&lt;/p&gt;&#xA;&lt;h1&gt;Puzzle 1&lt;/h1&gt;&#xA;&lt;p&gt;Solver: @Chadi Abou Sleiman&lt;br /&gt;&#xA;&lt;strong&gt;&lt;em&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/a/16592/7307&quot;&gt;Solution&lt;/a&gt;&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;Puzzle 2&lt;/h1&gt;&#xA;&lt;p&gt;Solver: @rand al'thor&lt;br /&gt;&#xA;&lt;strong&gt;&lt;em&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/a/16884/7307&quot;&gt;Solution&lt;/a&gt;&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;Puzzle 3&lt;/h1&gt;&#xA;&lt;p&gt;Solver: @Nefer007&lt;br /&gt;&#xA;&lt;strong&gt;&lt;em&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/a/17354/12109&quot;&gt;Solution&lt;/a&gt;&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="7307" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-26T16:37:50.723" CommentCount="5" CommunityOwnedDate="2015-06-21T17:03:14.573" ContentLicense="CC BY-SA 3.0" />
  <row Id="16740" PostTypeId="2" ParentId="16702" CreationDate="2015-06-21T01:51:54.603" Score="1" Body="&lt;p&gt;Partial Answer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I must admit I Googled a lot to arrive at this point.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; By checking the OP's profile, I deducted that the character is from one of her books that she mentioned.&#xA;&lt;P&gt;So I googled each of the books to see if there is any character who might fit the description.&#xA;&lt;P&gt;Found Percy Jackson novels has some characters who are named Grace (Beryl Grace, Jason Grace, Thalia Grace)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's as far as I've gone.&lt;/p&gt;&#xA;" OwnerUserId="12332" LastActivityDate="2015-06-21T01:51:54.603" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16741" PostTypeId="1" AcceptedAnswerId="16742" CreationDate="2015-06-21T04:26:26.133" Score="4" ViewCount="593" Body="&lt;p&gt;&lt;strong&gt;How&lt;/strong&gt; many twos do you need to get to the number $100$ exactly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: $22+22+22+22+2^2+2^2+2^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This example method uses 14 twos and is obviously not ideal. Can you find a way that uses the least amount of twos? I've seen some things like this on PuzzlingSE and they've been pretty well accepted so I decided to make my own, as these are fun to do! The lowest number of twos will win!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Math You Can Use:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Addition&lt;/li&gt;&#xA;&lt;li&gt;Subtraction&lt;/li&gt;&#xA;&lt;li&gt;Multiplication&lt;/li&gt;&#xA;&lt;li&gt;Division&lt;/li&gt;&#xA;&lt;li&gt;Exponents&lt;/li&gt;&#xA;&lt;li&gt;Square &amp;amp; Cube Roots&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Other Rules For Clarification:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The only number you may use are twos; any other numbers must be made with a combination of twos in some way.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This problem uses base 10 and the answer should be base 10.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11002" LastEditorUserId="11002" LastEditDate="2015-06-21T18:38:08.973" LastActivityDate="2015-06-26T09:18:07.880" Title="Twos For Thought" Tags="|calculation-puzzle|formation-of-numbers|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16742" PostTypeId="2" ParentId="16741" CreationDate="2015-06-21T04:36:03.097" Score="6" Body="&lt;p&gt;A solution with 5 twos:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(2\times2\times2 + 2)^2$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another solution with 5 twos:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\big(\frac{22-2}{2}\big)^2$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="5373" LastEditDate="2015-06-21T10:48:15.127" LastActivityDate="2015-06-21T10:48:15.127" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16743" PostTypeId="1" AcceptedAnswerId="16749" CreationDate="2015-06-21T04:36:40.140" Score="19" ViewCount="1470" Body="&lt;p&gt;Now that you are familiar with &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16219/what-is-a-cyclone-phrase%E2%84%A2&quot;&gt;Cyclone Phrases™&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16485/what-is-a-scalable-phrase%E2%84%A2&quot;&gt;Scalable Phrases™&lt;/a&gt;, let's look at another kind of phrase.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If a phrase adheres to a certain rule, then I call it a &lt;strong&gt;Triad Phrase™&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the examples below to find the rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Warning there may be some spoilers in the comments below this puzzle, so beware&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lq3oc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-06-22T10:37:02.957" Title="What is a Triad Phrase™?" Tags="|pattern|english|" AnswerCount="1" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="16746" PostTypeId="2" ParentId="14537" CreationDate="2015-06-21T06:15:09.980" Score="-1" Body="&lt;p&gt;The letter combination &quot;ea&quot; makes 3 sounds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1st &quot;ea&quot; sounding like &quot;ee&quot; as in cr&lt;strong&gt;ea&lt;/strong&gt;m, t&lt;strong&gt;ea&lt;/strong&gt;ch, gr&lt;strong&gt;ea&lt;/strong&gt;se.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2nd &quot;ea&quot; sounding like &quot;E&quot; as in b&lt;strong&gt;ea&lt;/strong&gt;r, d&lt;strong&gt;ea&lt;/strong&gt;d, h&lt;strong&gt;ea&lt;/strong&gt;d.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3rd &quot;ea&quot; sounding like &quot;ay&quot; as in&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; gr&lt;strong&gt;ea&lt;/strong&gt;t, str&lt;strong&gt;ea&lt;/strong&gt;k, br&lt;strong&gt;ea&lt;/strong&gt;k.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13419" LastEditorUserId="19289" LastEditDate="2016-03-31T00:49:15.773" LastActivityDate="2016-03-31T00:49:15.773" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16747" PostTypeId="1" AcceptedAnswerId="16859" CreationDate="2015-06-21T07:30:25.953" Score="7" ViewCount="1217" Body="&lt;p&gt;You are standing at the door to a lost temple in Georgia (the former Soviet Republic). You are an expert in lock picking, but there is no keyhole. You then realise that there is a word that will open the door. You tell your friend, who then gives you a piece of paper. &quot;I think the word is on here, but I've got no idea what it means&quot;. Looking at it, you think the same. The paper shows this: &lt;code&gt;{255-36}, {255-32}, {36}, {30} | {255-36}, {255-32}, {36}, {31} | {255-36}, {255-32}, {36}, {32} | {255-36}, {255-32}, {35}, {34}&lt;/code&gt; What does the paper mean in English?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint One&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are three sets of computer codes involved in working this particular puzzle out. That makes the puzzle even more difficult, so I will tell you one of the codes. The third layer of encryption is what a USB Keyboard sends to a computer when a keyboard is pressed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint Two&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16747/the-impossible-door/16859#16859&quot;&gt;Prem's answer&lt;/a&gt; results in the word &quot;Ghia&quot;. That is not actually a word, but a pronunciation. Just apply a bit of logic (and Google Translate), and you should have the word!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9377" LastEditorUserId="-1" LastEditorDisplayName="user9377" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-06-24T16:33:53.363" Title="The impossible door!" Tags="|cipher|computer-puzzle|" AnswerCount="2" CommentCount="19" ContentLicense="CC BY-SA 3.0" />
  <row Id="16749" PostTypeId="2" ParentId="16743" CreationDate="2015-06-21T09:54:29.777" Score="18" Body="&lt;p&gt;Is it this?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Triad Phrase™ is one where the alternate letters of each word spell out another word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;peopleless fiacre&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; polls/epees &amp;amp; far/ice&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;poorest friend&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; poet/ors &amp;amp; fin/red&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;schooled stooge&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; shoe/cold &amp;amp; sog/toe&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Probably called a Triad Phrase™ because &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;each word is made up of two other words&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-06-21T09:54:29.777" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16753" PostTypeId="2" ParentId="16741" CreationDate="2015-06-21T10:57:34.283" Score="3" Body="&lt;p&gt;You never specified what base you were using...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In base 2, $100$ means the number four, so the answer is &lt;strong&gt;two&lt;/strong&gt; twos ($2+2,2\times2,2^2$, etc.)&lt;br&gt;In base 3, $100$ means the number nine, which can be got with &lt;strong&gt;four&lt;/strong&gt; twos: $\big(2+\frac{2}{2}\big)^2$.&lt;br&gt;In base 4, $100$ means sixteen, so the answer is &lt;strong&gt;three&lt;/strong&gt; twos: $2^{2^2}$.&lt;br&gt;In base 5, $100$ means twenty-five, which can be got with &lt;strong&gt;five&lt;/strong&gt; twos: $\big(2+2+\frac{2}{2}\big)^2$.&lt;br&gt;In base 6, $100$ means thirty-six, which can be got with &lt;strong&gt;four&lt;/strong&gt; twos: $(2+2+2)^2$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And so on.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-06-21T11:45:16.327" LastActivityDate="2015-06-21T11:45:16.327" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16755" PostTypeId="2" ParentId="16441" CreationDate="2015-06-21T11:29:02.100" Score="1" Body="&lt;p&gt;It is possible with at most 2 clippings per weighing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No solution is possible with at most 1 clipping, because there is no weighing which can tell apart the two heaviest balls, even if it was known which two balls were the two heaviest. (Thanks to f'' for providing the proof.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My solution with two clippings&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my solution, I label the balls by using a map from colors to letters. Initially, name the balls A to F in arbitrary order, i.e. write down an arbitrary mapping from color to letter. When ordered to do so in the solution, change letters in that map, and work with the new labels in the following. This is done to reuse the same sub-solution for multiple intermediate weighing outcomes. A final result will give the letters in increasing order of weight, and you use the current map from color to letter to find the weigh of each color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first two weighings are the same as in the solution of user13217, namely:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;W1. Weigh A vs B vs C. (720 permutations left.)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If B is lightest, swap A&amp;lt;-&gt;B.&lt;/li&gt;&#xA;&lt;li&gt;If C is lightest, swap A&amp;lt;-&gt;C.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;W2. Weigh D vs E vs F. (240 permutations left.)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If E is lightest, swap D&amp;lt;-&gt;E.&lt;/li&gt;&#xA;&lt;li&gt;If F is lightest, swap D&amp;lt;-&gt;F.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;From here on, we may use clippings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;W3. Weigh A+2 vs B vs C. (&quot;A+2&quot; means ball A and 2 clippings on one pan.)  (80 permutations left.)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If A+2 is lightest, swap B&amp;lt;-&gt;D and C&amp;lt;-&gt;E, and go to W4.1.&lt;/li&gt;&#xA;&lt;li&gt;If B is lightest, go to W4.2.&lt;/li&gt;&#xA;&lt;li&gt;If C is lightest, swap B&amp;lt;-&gt;C and go to W4.2.&lt;/li&gt;&#xA;&lt;li&gt;If no pan is lightest, cyclically relabel B-&gt;D-&gt;C-&gt;B, and go to W4.3.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;W4.1. Weigh B+2 vs C vs F. (16 permutations left.)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If C is lightest, goto W5.1.&lt;/li&gt;&#xA;&lt;li&gt;If F is lightest, cyclically relabel C-&gt;F-&gt;E-&gt;D-&gt;C, and go to W5.1.&lt;/li&gt;&#xA;&lt;li&gt;If no pan is lightest, swap A&amp;lt;-&gt;B, cyclically relabel D-&gt;F-&gt;E-&gt;F, and go to W5.2.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;W4.2. Weigh B+2 vs E vs F. (20 permutations left.)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If B+2 is lightest, go to W5.2.&lt;/li&gt;&#xA;&lt;li&gt;If E is lightest, cyclically relabel A-&gt;D-&gt;E-&gt;C-&gt;B-&gt;A, and go to W5.3.&lt;/li&gt;&#xA;&lt;li&gt;If F is lightest, cyclically relabel A-&gt;D-&gt;F-&gt;E-&gt;C-&gt;B-&gt;A, and go to W5.3.&lt;/li&gt;&#xA;&lt;li&gt;If no pan is lightest, cyclically relabel C-&gt;D-&gt;E-&gt;C, and go to W5.4.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;W4.3. Weigh C vs D vs E. (24 permutations left.)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If C is lightest, go to W5.5.&lt;/li&gt;&#xA;&lt;li&gt;If D is lightest, swap C&amp;lt;-&gt;D, and go to W5.5.&lt;/li&gt;&#xA;&lt;li&gt;If E is lightest, swap A&amp;lt;-&gt;B, cyclically relabel C-&gt;E-&gt;D-&gt;C, and go to W5.6.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;W5.1. Weigh D vs E vs F. (6 permutations left.)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If D is lightest, go to W6.&lt;/li&gt;&#xA;&lt;li&gt;If E is lightest, swap D&amp;lt;-&gt;E, and go to W6.&lt;/li&gt;&#xA;&lt;li&gt;If F is lightest, cyclically relabel D-&gt;F-&gt;E-&gt;D, and go to W6.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;W5.2. Weigh C vs D vs E. (4 permutations left.)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If C is lightest, go to W6.&lt;/li&gt;&#xA;&lt;li&gt;If D is lightest, swap C&amp;lt;-&gt;D, and go to W6.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;W5.3. Weigh B vs C vs D+1. (5 permutations left.)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If B is lightest, go to W6.&lt;/li&gt;&#xA;&lt;li&gt;If D+1 is lightest, the weight order is A,D,F,B,C,E.&lt;/li&gt;&#xA;&lt;li&gt;If no pan is lightest, cyclically relabel B-&gt;D-&gt;C-&gt;B, and go to W6.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;W5.4. Weigh D vs E vs F. (6 permutations left.)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If D is lightest, go to W6.&lt;/li&gt;&#xA;&lt;li&gt;If E is lightest, cyclically relabel A-&gt;C-&gt;B-&gt;A, swap D&amp;lt;-&gt;E, and go to W6.&lt;/li&gt;&#xA;&lt;li&gt;If F is lightest, cyclically relabel D-&gt;F-&gt;E-&gt;D, and go to W6.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;W5.5. Weigh D vs E vs F+1. (8 permutations left.)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If D is lightest, go to W6.&lt;/li&gt;&#xA;&lt;li&gt;If E is lightest, swap D&amp;lt;-&gt;E and go to W6.&lt;/li&gt;&#xA;&lt;li&gt;If F+1 is lightest, swap A&amp;lt;-&gt;B, cyclically relabel C-&gt;F-&gt;E-&gt;D-&gt;C, and go to W6.&lt;/li&gt;&#xA;&lt;li&gt;If no pan is lightest, cyclically relabel D-&gt;F-&gt;E-&gt;D, and go to W6.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;W5.6. Weigh D vs E vs F+1. (8 permutations left.)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If D is lightest, go to W6.&lt;/li&gt;&#xA;&lt;li&gt;If E is lightest, swap D&amp;lt;-&gt;E, and go to W6.&lt;/li&gt;&#xA;&lt;li&gt;If F+1 is lightest, cyclically relabel B-&gt;F-&gt;E-&gt;D-&gt;C-&gt;B, and go to W6.&lt;/li&gt;&#xA;&lt;li&gt;If no pan is lightest, swap B&amp;lt;-&gt;C, cyclically relabel D-&gt;F-&gt;E-&gt;D, and go to W6.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;W6. Weigh D+2 vs E vs F. (2 permutations left.)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If E is lightest, the weight order is A,B,C,D,E,F.&lt;/li&gt;&#xA;&lt;li&gt;If F is lightest, the weight order is A,B,C,D,F,E.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="318" LastEditorUserId="318" LastEditDate="2015-06-23T19:40:25.010" LastActivityDate="2015-06-23T19:40:25.010" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16756" PostTypeId="1" AcceptedAnswerId="16883" CreationDate="2015-06-21T12:04:46.280" Score="8" ViewCount="2613" Body="&lt;p&gt;One day, a colleague from the office hands you the following message without a comment and makes a quick retreat out of your cubicle. Can you make head &amp;amp; tails from this? What is your colleague trying to tell you here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/T2QWQ.png&quot; alt=&quot;The code&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution to this puzzle is an English sentence and this is &lt;em&gt;not&lt;/em&gt; a rebus puzzle. Hints will follow if the message is not deciphered in due time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first line is part of the code instruction, the second line is a self-verification step: If you can make the number on the left represent a G you have probably found the code algorithm and can apply it to the 3rd line, which is the message.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;G&quot; does not represent anything but the letter G itself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You did miss it at first, but the paper slip has been very carefully punctuated with a needle on the right edge. It seems that two different needles have been used too...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="5840" LastEditDate="2015-06-23T10:32:41.890" LastActivityDate="2015-06-24T11:18:16.623" Title="The pyramid code" Tags="|cipher|" AnswerCount="3" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="16760" PostTypeId="1" AcceptedAnswerId="16762" CreationDate="2015-06-21T13:44:08.410" Score="8" ViewCount="18052" Body="&lt;h2&gt;What is the greatest and least number of syllables that a word of a given length can have?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;More precisely, for each word length from 1 letter to 20 letters, find words of that length containing as many and as few syllables as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concept of a syllable is notoriously hard to define, so let's define the number of syllables to be the &lt;em&gt;number of distinct vowels in the standard English pronunciation of the word&lt;/em&gt;. Only English dictionary words are allowed. British and American English are both acceptable. Diphthongs count as single vowels, so &quot;I&quot; is a single-syllable word. If a word has more than one acceptable pronunciation differing in number of syllables, choose the most helpful one (e.g. count &quot;squirreled&quot; as 1 syllable rather than 2 since that makes a new record for minimum syllables with 10 letters).&lt;br&gt;&#xA;&lt;sub&gt;Let me know if there are any loopholes I've missed!&lt;/sub&gt;&#xA;&lt;hr&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Current best&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;1 letter: min 1 syllable (credit &lt;strong&gt;JLee&lt;/strong&gt;/&lt;strong&gt;Prem&lt;/strong&gt;), max 1 syllable (credit &lt;strong&gt;JLee&lt;/strong&gt;/&lt;strong&gt;Prem&lt;/strong&gt;), ratio 1&lt;br&gt;&#xA;2 letters: min 1 syllable (credit &lt;strong&gt;JLee&lt;/strong&gt;/&lt;strong&gt;Prem&lt;/strong&gt;), max 2 syllables (credit &lt;strong&gt;JLee&lt;/strong&gt;/&lt;strong&gt;Prem&lt;/strong&gt;), ratio 2&lt;br&gt;&#xA;3 letters: min 1 syllable (credit &lt;strong&gt;JLee&lt;/strong&gt;/&lt;strong&gt;Prem&lt;/strong&gt;), max 2 syllables (credit &lt;strong&gt;JLee&lt;/strong&gt;/&lt;strong&gt;Prem&lt;/strong&gt;), ratio 2&lt;br&gt;&#xA;4 letters: min 1 syllable (credit &lt;strong&gt;JLee&lt;/strong&gt;/&lt;strong&gt;Prem&lt;/strong&gt;), max 3 syllables (credit &lt;strong&gt;JLee&lt;/strong&gt;/&lt;strong&gt;Prem&lt;/strong&gt;), ratio 2&lt;br&gt;&#xA;5 letters: min 1 syllable (credit &lt;strong&gt;JLee&lt;/strong&gt;/&lt;strong&gt;Prem&lt;/strong&gt;), max 3 syllables (credit &lt;strong&gt;JLee&lt;/strong&gt;/&lt;strong&gt;Prem&lt;/strong&gt;), ratio 3&lt;br&gt;&#xA;6 letters: min 1 syllable (credit &lt;strong&gt;JLee&lt;/strong&gt;/&lt;strong&gt;Prem&lt;/strong&gt;), max 4 syllables (credit &lt;strong&gt;JoeZ&lt;/strong&gt;), ratio 4&lt;br&gt;&#xA;7 letters: min 1 syllable (credit &lt;strong&gt;JLee&lt;/strong&gt;/&lt;strong&gt;Prem&lt;/strong&gt;), max 5 syllables (credit &lt;strong&gt;user13433&lt;/strong&gt;), ratio 5&lt;br&gt;&#xA;8 letters: min 1 syllable (credit &lt;strong&gt;JLee&lt;/strong&gt;/&lt;strong&gt;Prem&lt;/strong&gt;), max 5 syllables (credit &lt;strong&gt;JLee&lt;/strong&gt;), ratio 5&lt;br&gt;&#xA;9 letters: min 1 syllable (credit &lt;strong&gt;JLee&lt;/strong&gt;/&lt;strong&gt;Prem&lt;/strong&gt;), max 5 syllables (credit &lt;strong&gt;JLee&lt;/strong&gt;), ratio 5&lt;br&gt;&#xA;10 letters: min 1 syllable (credit &lt;strong&gt;Deusovi&lt;/strong&gt;), max 6 syllables (credit &lt;strong&gt;JLee&lt;/strong&gt;), ratio 6&lt;br&gt;&#xA;11 letters: min 1 syllable (credit &lt;strong&gt;Emrakul&lt;/strong&gt;), max 6 syllables (credit &lt;strong&gt;JLee&lt;/strong&gt;/&lt;strong&gt;Prem&lt;/strong&gt;), ratio 6&lt;br&gt;&#xA;12 letters: min 1 syllable (credit &lt;strong&gt;Emrakul&lt;/strong&gt;), max 7 syllables (credit &lt;strong&gt;JLee&lt;/strong&gt;/&lt;strong&gt;Prem&lt;/strong&gt;), ratio 7&lt;br&gt;&#xA;13 letters: min 2 syllables (credit &lt;strong&gt;Prem&lt;/strong&gt;), max 8 syllables (credit &lt;strong&gt;JLee&lt;/strong&gt;), ratio 4&lt;br&gt;&#xA;14 letters: min 2 syllables (credit &lt;strong&gt;Prem&lt;/strong&gt;), max 8 syllables (credit &lt;strong&gt;JLee&lt;/strong&gt;), ratio 4&lt;br&gt;&#xA;15 letters: min 3 syllables (credit &lt;strong&gt;JLee&lt;/strong&gt;/&lt;strong&gt;Prem&lt;/strong&gt;), max 8 syllables (credit &lt;strong&gt;JLee&lt;/strong&gt;), ratio 2.666...&lt;br&gt;&#xA;16 letters: min 3 syllables (credit &lt;strong&gt;Prem&lt;/strong&gt;), max 8 syllables (credit &lt;strong&gt;Prem&lt;/strong&gt;), ratio 2.666...&lt;br&gt;&#xA;17 letters: min 3 syllables (credit &lt;strong&gt;Prem&lt;/strong&gt;), max 8 syllables (credit &lt;strong&gt;Prem&lt;/strong&gt;), ratio 2.666...&lt;br&gt;&#xA;18 letters: min 4 syllables (credit &lt;strong&gt;Prem&lt;/strong&gt;), max 9 syllables (credit &lt;strong&gt;Nathaniel&lt;/strong&gt;), ratio 2 .25&lt;br&gt;&#xA;19 letters: min 4 syllables (credit &lt;strong&gt;Prem&lt;/strong&gt;), max 9 syllables (credit &lt;strong&gt;Prem&lt;/strong&gt;), ratio 2.25&lt;br&gt;&#xA;20 letters: min 5 syllables (credit &lt;strong&gt;Prem&lt;/strong&gt;), max 10 syllables (credit &lt;strong&gt;Prem&lt;/strong&gt;), ratio 2  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where two answerers have tied for the same minimum or maximum, I've mentioned both their names in alphabetical order; I don't know who was first, but I assume JLee and Prem are both working independently. I also haven't included the actual words found, since that would make the list much longer. If you want to find the words, scroll down to read (and upvote) the answers below.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="13190" LastEditDate="2016-04-21T09:06:25.153" LastActivityDate="2021-10-26T15:39:05.740" Title="Words with most and fewest syllables for given length?" Tags="|word|english|language|" AnswerCount="8" CommentCount="19" ContentLicense="CC BY-SA 3.0" />
  <row Id="16761" PostTypeId="2" ParentId="16760" CreationDate="2015-06-21T13:55:30.097" Score="6" Body="&lt;p&gt;WIP (Work in Progress)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, to start,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BjnKU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-06-21T20:16:12.573" LastActivityDate="2015-06-21T20:16:12.573" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="16762" PostTypeId="2" ParentId="16760" CreationDate="2015-06-21T14:01:11.333" Score="7" Body="&lt;p&gt;For 1, minimum : &quot;I&quot; 1&lt;br&gt;&#xA;For 1, maximum : &quot;I&quot; 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 2, minimum : &quot;It&quot; 1&lt;br&gt;&#xA;For 2, maximum : &quot;Io&quot; 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 3, minimum : &quot;Don&quot; 1&lt;br&gt;&#xA;For 3, maximum : &quot;Era&quot; 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 4, minimum : &quot;Done&quot; 1&lt;br&gt;&#xA;For 4, maximum : &quot;Iota&quot; 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 5, minimum : &quot;Stone&quot; &quot;Schwa&quot; 1&lt;br&gt;&#xA;For 5, maximum : &quot;Anion&quot; 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 6, minimum : &quot;Streak&quot; 1&lt;br&gt;&#xA;For 6, maximum : &quot;Anonym&quot; &quot;Hawaii&quot; 3  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 7, minimum : &quot;Streaks&quot; 1&lt;br&gt;&#xA;For 7, maximum : &quot;Anonymy&quot; 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 8, minimum : &quot;Strength&quot; 1&lt;br&gt;&#xA;For 8, maximum : &quot;Ignominy&quot; &quot;Abacuses&quot; &quot;Honolulu&quot; 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 9, minimum : &quot;Strengths&quot; 1&lt;br&gt;&#xA;For 9, maximum : &quot;Abrasions&quot; 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 10, minimum : &quot;Squirreled&quot; 1 , &lt;strong&gt;thanks to Deusovi&lt;/strong&gt;&lt;br&gt;&#xA;For 10, maximum : &quot;Abacterial&quot; &quot;Abiogenist&quot; 5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 11, minimum : &quot;Broughammed&quot; 1 , &lt;strong&gt;thanks to Emrakul&lt;/strong&gt;&lt;br&gt;&#xA;For 11, maximum : &quot;Abecedarian&quot; &quot;Abiogeneses&quot; &quot;Aerobiology&quot; &quot;Alcyonarian&quot; 6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 12, minimum : &quot;Schtroumpfed&quot; 1 , &lt;strong&gt;thanks to Emrakul&lt;/strong&gt;&lt;br&gt;&#xA;For 12, maximum : &quot;Unisexuality&quot; 7&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 13, minimum : &quot;Scratchboards&quot; &quot;Straightbreds&quot; &quot;Straightboard&quot; &quot;Straightforth&quot; 2&lt;br&gt;&#xA;For 13, maximum : &quot;Unideological&quot; &quot;Unreliability&quot; &quot;Monosexuality&quot; 7&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 14, minimum : &quot;Straightboards&quot; 2&lt;br&gt;&#xA;For 14, maximum : &quot;Stereoisomeric&quot; &quot;Supersexuality&quot; &quot;Unisexualities&quot; 7&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 15, minimum : &quot;Schlockmeisters&quot; &quot;Straightforthly&quot; &quot;Straightforward&quot; 3&lt;br&gt;&#xA;For 15, maximum : &quot;Stereoisomerics&quot; &quot;Monosexualities&quot; &quot;Organomercurial&quot; 7&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 16, minimum : &quot;Straightforwards&quot; 3&lt;br&gt;&#xA;For 16, maximum : &quot;Stereoisomerical&quot; 8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 17, minimum : &quot;Straightforthness&quot; 3&lt;br&gt;&#xA;For 17, maximum : &quot;Electroanalytical&quot; &quot;Electrophysiology&quot; 8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 18, minimum : &quot;Forethoughtfulness&quot; 4&lt;br&gt;&#xA;For 18, maximum : &quot;Universalizability&quot; 9 , &lt;strong&gt;thanks to Nathaniel&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 19, minimum : &quot;Straightforwardness&quot; 4&lt;br&gt;&#xA;For 19, maximum : &quot;Extraterritoriality&quot; 9&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 20, minimum : &quot;Forethoughtfulnesses&quot; 5&lt;br&gt;&#xA;For 20, maximum : &quot;Radioimmunoassayable&quot; 10&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;WHEW&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References&lt;/strong&gt;:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.morewords.com/&quot; rel=&quot;nofollow&quot;&gt;http://www.morewords.com/&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_the_longest_English_words_with_one_syllable&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/List_of_the_longest_English_words_with_one_syllable&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10940" LastEditorUserId="10940" LastEditDate="2016-04-21T08:39:52.747" LastActivityDate="2016-04-21T08:39:52.747" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16764" PostTypeId="1" AcceptedAnswerId="16766" CreationDate="2015-06-21T14:17:49.597" Score="11" ViewCount="497" Body="&lt;p&gt;Someone has recently tweeted:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Last year I had my 50th. The year before was my 100th. This year it's my 30th. #NotAllNumbersAreDecimal&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is he turning (base 10) this year?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's nothing in the tweet to suggest there is only one solution. How many are there?&lt;/p&gt;&#xA;" OwnerUserId="1509" LastEditorUserId="1509" LastEditDate="2015-06-21T14:56:05.137" LastActivityDate="2015-06-21T14:56:05.137" Title="Birthdays in different bases" Tags="|mathematics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16765" PostTypeId="2" ParentId="16756" CreationDate="2015-06-21T14:31:41.170" Score="4" Body="&lt;p&gt;Partial answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Here's the color bar at the bottom converted to a base 2 number (red = 100, green = 010, blue = 001, yellow = 110, cyan = 011, magenta = 101, black = 000). 100010010110000100101100000110001010100010010100001101000010100000010010110100110101010011001010001011010000010101000101000101010101001000101000101100110100100110100110101001000101010100000100001110101000110000001100001000001011100000010000000010000100000100001000001000010001000000100001000011000000000000000010000000000000&lt;br&gt;&lt;br&gt;Note: the chances of this being necessary are slim because it's near impossible for this sequence to not contain any white bars and yet be a direct translation to something. However as I can't think of any other way this could be decoded, posting it anyway.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastActivityDate="2015-06-21T14:31:41.170" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16766" PostTypeId="2" ParentId="16764" CreationDate="2015-06-21T14:40:41.567" Score="10" Body="&lt;p&gt;Let $x$ be his age on his birthday this year. Then $x-1$ is 50 in some base, i.e. a multiple of 5; $x-2$ is 100 in some base, i.e. a perfect square; and $x$ is 30 in some base, i.e. a multiple of 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the Chinese remainder theorem, being a multiple of 3 and one more than a multiple of 5 is exactly equivalent to being 6 more than a multiple of 15 (congruent to 6 mod 15). So $x$ must be one of 6, 21, 36, 51, 66, 81, 96, 111, 126, 141 (assuming $x&amp;lt;150$, which seems reasonable). Subtract 2 from each of these to see whether we get a square, and we find that the only possible values of $x$ are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 6, 51, and 66.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's check each of these individually. 6 turns out to be too small since 5 isn't written as 50 in any base (thanks @KateGregory!), but both the other two seem to be eligible.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-06-21T14:44:58.517" LastActivityDate="2015-06-21T14:44:58.517" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16767" PostTypeId="2" ParentId="16764" CreationDate="2015-06-21T14:40:58.993" Score="3" Body="&lt;p&gt;There are two potential answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If 50 was base 10, 100 was base 7 (49), and 30 was base 17 (51), it would work. So 51 would work. The other possible answer is that if 30 was base 22 (66), 50 was base 13 (65), and 100 was base 8 (64), so the answer would be 66. Credit to rand al'thor for finding the other possible answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="7722" LastEditDate="2015-06-21T14:52:49.093" LastActivityDate="2015-06-21T14:52:49.093" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16768" PostTypeId="1" AcceptedAnswerId="16811" CreationDate="2015-06-21T15:50:32.960" Score="4" ViewCount="1281" Body="&lt;p&gt;&amp;quot;Have you heard of Room 40?&amp;quot; &amp;quot;No, Sir, I have not&amp;quot;. &amp;quot;Good. For what I am about to tell you, if leaked, will cost us this war, son!&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;November 1914, London.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Much to your surprise you have been summoned by the Admiralty to meet in  the Old Admiralty Building, or simply &amp;quot;Old Building&amp;quot;, adjacent to the Horse Guards in London.&lt;/p&gt;&#xA;&lt;p&gt;This came as a surprise as you are not known for your physical prowess or sea-faring skills preferred in new conscripts.&#xA;Instead you are known as a &amp;quot;studious fellow&amp;quot;. Your choice of adventure generally featured a game of chess or scrabble.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;Our mission here is of great importance!&amp;quot; the man, who introduced himself as 'Cdr Herbert Hope', tells you.&#xA;&amp;quot;However, before I am able to share this national secret, you are required to complete an entrance exam.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;He hands you a paper with seemingly nonsensical equations.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;Oh, &lt;strong&gt;George&lt;/strong&gt; ... I mean John, I almost forgot. Hope you're not offended if I address you by your &lt;strong&gt;first name&lt;/strong&gt; . Remember that the &lt;strong&gt;greatest authority within a neutral power&lt;/strong&gt; is key.&amp;quot; [The Cdr strangely places great emphasis on certain words - given in bold text. You also find it odd that he addresses you as George as he does not seem to be a forgetful man].&lt;/p&gt;&#xA;&lt;p&gt;You solve the puzzle a short while later. Luckily one particular choice of hobby helps you find the solution in record time. Although lets be honest, it does take you a few days to figure out.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;2 + 6 = 28&lt;/p&gt;&#xA;&lt;p&gt;1 + 8 = 16&lt;/p&gt;&#xA;&lt;p&gt;12 + 14 = 39&lt;/p&gt;&#xA;&lt;p&gt;6 + 16 = 28&lt;/p&gt;&#xA;&lt;p&gt;13 + 9 = ??&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; What is meant by the &lt;em&gt;'greatest authority of a neutral power'&lt;/em&gt;? Note that the year is Nov 1914. Stated differently, who is a neutral power in Nov 1914 and secondly, what is meant by its greatest authority?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Hope you're not offended if I call you by your &lt;strong&gt;First Name&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Hint 3&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The relevant hobby is Scrabble&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Hint 4 - big hint&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; A better way of writing the equations would be Sum of 2nd and 6th = 28, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Hint 5&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The ‘first name’ will help you find out who the neutral power is. But it also plays another important role within the equations&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;the below facts are unrelated to the puzzle &amp;amp; given for interest purposes only&lt;/p&gt;&#xA;&lt;p&gt;(*) &lt;em&gt;During World War 1, Room 40 deciphered German messages. The intercepted messages arrived in the Admiralty office's basement and were relayed to Room 40&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;(**) &lt;em&gt;The Admiralty was the authority responsible for the command of the Royal Navy in Great Britain&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="12317" LastEditorUserId="-1" LastEditorDisplayName="user20" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-23T05:13:46.520" Title="Room 40 - the Great War &quot;Mystery&quot;" Tags="|lateral-thinking|enigmatic-puzzle|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16769" PostTypeId="1" AcceptedAnswerId="16947" CreationDate="2015-06-21T16:56:46.843" Score="3" ViewCount="715" Body="&lt;p&gt;In yellows and blacks,&lt;br&gt;&#xA;He often lives on racks,&lt;br&gt;&#xA;He loves none but Penny.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;a mascot&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13426" LastEditorUserId="8874" LastEditDate="2015-10-07T11:05:15.543" LastActivityDate="2015-10-07T11:05:15.543" Title="In yellows and blacks" Tags="|riddle|language|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16770" PostTypeId="1" AcceptedAnswerId="16772" CreationDate="2015-06-21T19:07:53.170" Score="7" ViewCount="1715" Body="&lt;p&gt;&lt;strong&gt;This&lt;/strong&gt; question is similar to my other question &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16741/twos-for-thought&quot;&gt;Twos For Thought&lt;/a&gt;, which arguably was a better pun but I decided to recreate the question, except with threes!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How&lt;/strong&gt; many threes do you need to get to the number $100$ exactly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: $3^3*3!-33-33+3+\frac{3}{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This example method uses 10 threes and is obviously not ideal. Can you find a way that uses the least amount of threes? The lowest number of threes will win!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Math You Can Use:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Addition&lt;/li&gt;&#xA;&lt;li&gt;Subtraction&lt;/li&gt;&#xA;&lt;li&gt;Multiplication&lt;/li&gt;&#xA;&lt;li&gt;Division&lt;/li&gt;&#xA;&lt;li&gt;Exponents&lt;/li&gt;&#xA;&lt;li&gt;Square &amp;amp; Cube Roots&lt;/li&gt;&#xA;&lt;li&gt;Factorials&lt;/li&gt;&#xA;&lt;li&gt;Parenthesis are allowed as long as they are used to order operations&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Other Rules For Clarification:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The only number you may use are threes; any other numbers must be made with a combination of threes in some way.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This problem uses base 10 and the answer should be base 10.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Decimal points are allowed (Expressions like .$n$ instead of 0.$n$ is allowed, but .0$n$ instead of 0.0$n$ is not.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11002" LastEditorUserId="11002" LastEditDate="2015-06-26T20:49:23.483" LastActivityDate="2015-06-26T20:49:23.483" Title="A Threes Of Cake" Tags="|calculation-puzzle|formation-of-numbers|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16771" PostTypeId="2" ParentId="16770" CreationDate="2015-06-21T19:11:03.633" Score="9" Body="&lt;p&gt;A quick initial attempt gives us five:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$33 \times 3 + \frac 33 = 100$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we're allowed to put overlines on top of numbers, I can get four:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$33.\overline{3} \times 3 = 100$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overline means &quot;33.3 repeated&quot;, or $33 \frac 13$.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-06-21T19:11:03.633" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16772" PostTypeId="2" ParentId="16770" CreationDate="2015-06-21T19:45:03.860" Score="7" Body="&lt;p&gt;How about this?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(3*3)\over{(.3*.3)}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If this counts, then these would also count:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(33-3)/.3$ and $33/.33$ (credit to Joe Z.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="7722" LastEditDate="2015-06-21T21:37:19.967" LastActivityDate="2015-06-21T21:37:19.967" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16773" PostTypeId="1" AcceptedAnswerId="16828" CreationDate="2015-06-21T20:28:33.930" Score="7" ViewCount="766" Body="&lt;p&gt;I give you &lt;strong&gt;11 coins&lt;/strong&gt;, which look exactly the same. Among them are &lt;strong&gt;two fake coins&lt;/strong&gt;. All genuine coins weigh the same. The fake coins are a little &lt;strong&gt;heavier&lt;/strong&gt; than the genuine ones, and they may have a different weight from each other. You cannot distinguish their weight by hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have no own utility to weigh coins, so I suggest you use my &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15988/a-balance-with-three-pans-detecting-the-lightest-pan-find-the-one-lighter-ball&quot;&gt;three-pan balance&lt;/a&gt; which, upon activation, lifts the unique pan with the lightest load. If there is no unique lightest pan, none of the pans will rise. You may only weigh the coins I gave you. Moreover, you are only allowed to perform &lt;strong&gt;three weighings&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you may remember from &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16006/a-balance-with-three-pans-detecting-the-lightest-pan-find-the-two-heavier-ball&quot;&gt;a previous puzzle&lt;/a&gt;, under these conditions, the two fake coins can not always be identified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To help you, and for my amusement, I give you &lt;strong&gt;one additional genuine coin&lt;/strong&gt;, which looks and weighs the same as the other genuine coins, and which you may use in your weighings to come.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you identify the two fake coins under these circumstances? If so, how?&lt;/p&gt;&#xA;" OwnerUserId="318" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-06-22T21:44:23.060" Title="Find 2 heavy fakes among 11 coins on a 3-pan balance" Tags="|weighing|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="16774" PostTypeId="2" ParentId="16770" CreationDate="2015-06-21T21:38:21.763" Score="8" Body="&lt;p&gt;If we are allowed to use an infinite number of operations, and we are limited in only the number of 3's that we use, then we can get away with using just one 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the factorial function increases the number and the square root function decreases the number, applying an infinite number of factorials and square roots on 3 in a certain sequence at some point will result in 100. If we can use any rounding function then this becomes even easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to do it would be:&#xA;33*3+√√√... ...√√√3&#xA;(An infinite number of square roots converges to 1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, if what Quark answered is valid, then is 3/0.03 valid? That would be a 2 threes solution.&lt;/p&gt;&#xA;" OwnerUserId="4109" LastEditorUserId="4109" LastEditDate="2015-06-22T07:48:15.520" LastActivityDate="2015-06-22T07:48:15.520" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="16775" PostTypeId="2" ParentId="16769" CreationDate="2015-06-22T01:26:18.820" Score="5" Body="&lt;p&gt;My guess:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In yellows and blacks,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Army dress uniforms include yellow and black. &lt;br/&gt;Their new PT uniforms are yellow and black. &lt;br/&gt;Or their patch is yellow and black.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He often lives on racks,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Military jargon calls beds &quot;racks&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From Georgia he could be,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fort Benning, Ga is one of the Army boot camps though there are other locations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He loves none but Penny.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Penny is one, of the many, many words for booze in the military. &lt;br/&gt;OR it can refer to the &quot;Guy who knows where the Penny drops&quot; ie. whoever knows what is actually going on. &lt;br/&gt;As an fyi, the military jargon for girlfriend is &quot;Suzy&quot; or &quot;Suzy Rottencrotch&quot; if she's seeing Jodie.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So your a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Soldier&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though in my case In red and blue and from San Diego I would be ;).&lt;/p&gt;&#xA;" OwnerUserId="11239" LastEditorUserId="4506" LastEditDate="2015-06-22T11:58:15.563" LastActivityDate="2015-06-22T11:58:15.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16777" PostTypeId="1" AcceptedAnswerId="16779" CreationDate="2015-06-22T06:45:08.680" Score="15" ViewCount="1481" Body="&lt;p&gt;The following six audio samples represent, perhaps, the best-known classical music compositions as they have been used in more than 150 mainstream movies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This question asks you to identify the six original composers.&lt;/strong&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, the last audio sample has been transmogrified (messed up) &lt;strong&gt;beyond recognition&lt;/strong&gt;.  You can still identify the sixth composer by either:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;un-transmogrification of the sixth audio sample or&lt;/li&gt;&#xA;&lt;li&gt;decoding this cipher using the &lt;strong&gt;surnames&lt;/strong&gt; of the first five composers:&lt;pre&gt;5-5, 2-6, 3-3, 1-4, 4-1, 2-1, 5-2, 1-3, C, 5-1, 3-4, 1-5, 2-1, B, 3-5, 3-6, 4-2, 4-6, 2-5&lt;/pre&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Here are the audio samples&lt;/strong&gt; (each is 150 kB or less):  &lt;strong&gt;&lt;a href=&quot;https://googledrive.com/host/0B5qtVpyoZk81fk1IV3ctMmliU1pOSWtqY1FNNUFLS01RdWRtNVA0enFvRkVnVWJMT3NFME0/1.ogg&quot; rel=&quot;noreferrer&quot;&gt;1.ogg&lt;/a&gt;, &lt;a href=&quot;https://googledrive.com/host/0B5qtVpyoZk81fk1IV3ctMmliU1pOSWtqY1FNNUFLS01RdWRtNVA0enFvRkVnVWJMT3NFME0/2.ogg&quot; rel=&quot;noreferrer&quot;&gt;2.ogg&lt;/a&gt;, &lt;a href=&quot;https://googledrive.com/host/0B5qtVpyoZk81fk1IV3ctMmliU1pOSWtqY1FNNUFLS01RdWRtNVA0enFvRkVnVWJMT3NFME0/3.ogg&quot; rel=&quot;noreferrer&quot;&gt;3.ogg&lt;/a&gt;, &lt;a href=&quot;https://googledrive.com/host/0B5qtVpyoZk81fk1IV3ctMmliU1pOSWtqY1FNNUFLS01RdWRtNVA0enFvRkVnVWJMT3NFME0/4.ogg&quot; rel=&quot;noreferrer&quot;&gt;4.ogg&lt;/a&gt;, &lt;a href=&quot;https://googledrive.com/host/0B5qtVpyoZk81fk1IV3ctMmliU1pOSWtqY1FNNUFLS01RdWRtNVA0enFvRkVnVWJMT3NFME0/5.ogg&quot; rel=&quot;noreferrer&quot;&gt;5.ogg&lt;/a&gt;, &lt;a href=&quot;https://googledrive.com/host/0B5qtVpyoZk81fk1IV3ctMmliU1pOSWtqY1FNNUFLS01RdWRtNVA0enFvRkVnVWJMT3NFME0/6.ogg&quot; rel=&quot;noreferrer&quot;&gt;6.ogg&lt;/a&gt;.&lt;/strong&gt; Some minor audio adjustments have been made to defeat Shazam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; - here is an alternate method to receive the audio samples:   &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Download the following image  &lt;/li&gt;&#xA;&lt;li&gt;Change its file extension from .gif to .rar  &lt;/li&gt;&#xA;&lt;li&gt;Extract the six audio files using WinRAR (other unzip software may not work)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cBmRj.gif&quot; alt=&quot;embedded audio files&quot;&gt;&lt;br&gt;&#xA;&lt;sub&gt;These audio samples are from &lt;strong&gt;public domain&lt;/strong&gt; sources on either Musopen or Wikipedia.  The compositions and the performances are both free from copyright.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-06-23T04:05:49.957" LastActivityDate="2023-06-17T21:04:58.410" Title="Music from the movies ♬ ♬" Tags="|cipher|movies|music|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16778" PostTypeId="2" ParentId="16777" CreationDate="2015-06-22T07:05:31.620" Score="5" Body="&lt;p&gt;&lt;strong&gt;Work in progress&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Blue Danube waltz, Johann Strauss II&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wedding March, Richard Wagner&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Third&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Klaviersonate op. 35, Frédéric Chopin&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="5840" LastEditDate="2015-06-22T08:14:44.980" LastActivityDate="2015-06-22T08:14:44.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16779" PostTypeId="2" ParentId="16777" CreationDate="2015-06-22T07:09:00.317" Score="10" Body="&lt;p&gt;So far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Felix Mendelssohn's &quot;Wedding March&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fourth:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gioachino Rossini's &quot;William Tell Overture&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fifth:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; George Frideric Handel's &quot;Hallelujah from the Messiah&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Through the collective effort of &lt;a href=&quot;https://puzzling.stackexchange.com/users/5840/bmyguest&quot;&gt;BmyGuest&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/users/4450/pacoverflow&quot;&gt;pacoverflow&lt;/a&gt; and their answers &lt;a href=&quot;https://puzzling.stackexchange.com/a/16778/5610&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/a/16780/5610&quot;&gt;here&lt;/a&gt; we get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Johann Strauss II&lt;br/&gt;Felix Mendelssohn&lt;br/&gt;Richard Wagner&lt;br/&gt;Gioachino Rossini&lt;br/&gt;George Frideric Handel&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Decoding (first index is the composer, second the letter in the surname), results in:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; elgarmarchnumberone -&gt; Elgar's March Number One -&gt; Elgar's March No. 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sixth composer being:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sir Edward William Elgar&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-06-23T04:18:18.933" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16780" PostTypeId="2" ParentId="16777" CreationDate="2015-06-22T07:10:47.360" Score="3" Body="&lt;p&gt;Partial answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.ogg&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bridal Chorus by Richard Wagner&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2015-06-22T07:10:47.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16781" PostTypeId="2" ParentId="16760" CreationDate="2015-06-22T07:41:58.493" Score="2" Body="&lt;p&gt;Oxyopia is 5 for 7. Doing six syllables for 8 might be possible. &lt;/p&gt;&#xA;" OwnerUserId="13433" LastActivityDate="2015-06-22T07:41:58.493" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16782" PostTypeId="2" ParentId="16770" CreationDate="2015-06-22T08:26:02.800" Score="0" Body="&lt;p&gt;Maybe this one:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;33.(3) * 3 = 100&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This however does not follow the requirement &quot;Parenthesis are allowed as long as they are used to order operations&quot;, in this example it denotes &lt;code&gt;0.(3)&lt;/code&gt; means &lt;code&gt;0.3333...&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/0.999...&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; also marks infinite &lt;a href=&quot;https://en.wikipedia.org/wiki/Repeating_decimal#Notation&quot; rel=&quot;nofollow noreferrer&quot;&gt;repeating decimals&lt;/a&gt; with a line or a dot over the number; here is the image from Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zHzXI.png&quot; alt=&quot;repeating 0,3333&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which would allow to not uses parentheses, so this should count (4 threes).&lt;/p&gt;&#xA;" OwnerUserId="10880" LastActivityDate="2015-06-22T08:26:02.800" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16785" PostTypeId="2" ParentId="16769" CreationDate="2015-06-22T12:08:09.207" Score="-1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yellowjacket. They're annoyingly common around here, and frequently make their nests up underneath any eave or 'rack' they can find.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Nasty little hostile buggers -- they literally love nobody and are very, very aggressive.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10625" LastActivityDate="2015-06-22T12:08:09.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16786" PostTypeId="2" ParentId="16769" CreationDate="2015-06-22T14:08:04.980" Score="6" Body="&lt;p&gt;Could it be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Buzz_(mascot)&quot;&gt;Buzz&lt;/a&gt; the mascot of Georgia Institute of Technology&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;In yellows and blacks,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He's a yellowjacket&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;He often lives on racks,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The costume is often on clothes racks&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;From Georgia he could be,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the mascot of Georgia Institute of Technology&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;He loves none but Penny.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Something about BuzzCard?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastActivityDate="2015-06-22T14:08:04.980" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16787" PostTypeId="2" ParentId="11666" CreationDate="2015-06-22T15:28:54.523" Score="5" Body="&lt;p&gt;I'll go ahead and try my hand at a proof of optimality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lock has three numbers;  we will call the first the $x$-coordinate, the second the $y$-coordinate, and the third the $z$-coordinate.  It helps to think like JonTheMon said, of a 3D version of battleship, but I won't refer specifically to this intuition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start with an easier example.  Suppose the $y$ and $z$ coordinates must be from the numbers $\{1, 2, 3\}$ instead of $\{1, 2, .., 8\}$, but the $x$-coordinate could be any number from $\{1, 2, ..., 8\}$, which is 72 total possibilities.  Can this be done with only 6 checks?  Well, no.  By the pigeon hole principle, there must be some $x$ coordinate, say $x = 1$, such that none of the checks start with $x = 1$.  But the nine possibilities that start with one ($(1, 1, 1)$, $(1, 1, 2)$, $(1, 1, 3)$, $(1, 2, 1)$, etc.), must be covered by six checks, and each check can only cover one possibility.  Hence it is impossible to do this easier problem with six checks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to the full problem.  Suppose only 31 checks are needed to open the lock.  By the pigeon hole principle, there must be some number $x$ between $1$ and $8$ such that only three of the checks start with the number $x$.  Without loss of generality, assume $x = 1$.  Let's say these three checks are $(1, y_1, z_1)$, $(1, y_2, z_2)$, and $(1, y_3, z_3)$.  Since there are five numbers not in the set $Y = \{y_1, y_2, y_3\}$ and five numbers not in the set $Z = \{z_1, z_2, z_3\}$, that makes $25$ combinations that start with $1$ not covered by these three checks. Each of these $25$ possibilities require distinct checks, and furthermore these $25$ distinct checks must have neither their $y$ nor $z$ coordinate in $Y$ or $Z$ respectively.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means the 72 combinations$^*$ of the form $(x, y, z)$ for $x \in \{1, 2, .., 8\}$, $y \in Y$, and $z \in Z$ are covered by just six checks (since the other 25 can only have the $x$-coordinate correct for these 72 combinations).  But this is impossible, as we've already demonstrated.  QED.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$*$:  Note that the above assumes $y_1 \neq y_2$, $y_1 \neq y_3$, $z_1 \neq z_2$, etc.  But if any of these $y$ or $z$ values were the same, then there would be at least $30$ possibilities not covered instead of $25$, which is clearly impossible.&lt;/p&gt;&#xA;" OwnerUserId="2722" LastActivityDate="2015-06-22T15:28:54.523" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16789" PostTypeId="2" ParentId="2101" CreationDate="2015-06-22T16:18:57.880" Score="0" Body="&lt;p&gt;2 cuts with HORIZONTAL cut.&#xA; - don't cut the 12cm-diameter cake. it's the first part&#xA; - cut the 16cm-diameter cake on horizontal way 3/4 and you've got a second part equivalent of the first 12cm-diameter cake&#xA; - cut the 20cm-diameter cake on horizontal way 6/10 and you've got a third part equivalent of the first 12cm-diameter cake&#xA; - the final part is composed with 1/4 of the 12cm-diameter cake + 4/10 of the 20cm-diameter cake&lt;/p&gt;&#xA;" OwnerUserId="13442" LastActivityDate="2015-06-22T16:18:57.880" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16790" PostTypeId="1" AcceptedAnswerId="16794" CreationDate="2015-06-22T17:48:39.620" Score="18" ViewCount="1336" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;With eight, I'm quite heady.&lt;br&gt;&#xA;  With three, my claws ready.&lt;br&gt;&#xA;  With thirteen, I'm downtrodden.&lt;br&gt;&#xA;  With eighteen, said to be sodden.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Alone, I'm quite new here,&lt;br&gt;&#xA;  Yet ubiquitous everywhere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-06-24T18:55:27.870" Title="New yet ubiquitous ... what am I getting at?" Tags="|riddle|rhyme|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16791" PostTypeId="1" AcceptedAnswerId="16795" CreationDate="2015-06-22T17:53:04.690" Score="13" ViewCount="1346" Body="&lt;p&gt;Here's my first riddle on Puzzling SE.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;They say I'm a ship&lt;br&gt;&#xA;  But I see no water for days&lt;br&gt;&#xA;  You may think you know me&lt;br&gt;&#xA;  From my cousin of holy ways&lt;br&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Look for me in abandoned places&lt;br&gt;&#xA;  And you might find an obstacle or two&lt;br&gt;&#xA;  Its what keeps me going long&lt;br&gt;&#xA;  When the movers move&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-06-23T00:02:32.107" Title="Food for thought, or otherwise" Tags="|riddle|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16792" PostTypeId="2" ParentId="16791" CreationDate="2015-06-22T18:10:52.867" Score="7" Body="&lt;p&gt;Could it possibly be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A space ship or satellite?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They say I'm a ship&#xA;But I see no water for days&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Space ships are not used on water, but rather in space. (Where there isn't liquid water)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You may think you know me&#xA;From my cousin of holy ways&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm pretty sure everyone knows about normal water ships (boats, yachts etc.) which could be a cousin of a space ship.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Look for me in abandoned places&#xA;And you might find an obstacle or two&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If you look in space (which can be considered abandoned and empty), You will can see some objects (planets and stars). [Earth + moon/sun]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Its what keeps me going long&#xA;When the movers move&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The gravitational forces from the Earth and moon/sun are what keep space ships and satellites orbiting earth. Space ships can also be 'sling shot' by using gravitation forced of other heavy bodies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastActivityDate="2015-06-22T18:10:52.867" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16793" PostTypeId="2" ParentId="16791" CreationDate="2015-06-22T18:16:17.467" Score="6" Body="&lt;p&gt;this is my first answer too (no questions yet)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Relationship&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They say I'm a ship&#xA;But I see no water for days&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the word ends in ship, but its not actually a ship&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You may think you know me&#xA;From my cousin of holy ways&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your cousins are you relations&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Look for me in abandoned places&#xA;And you might find an obstacle or two&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; not too sure about this&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Its what keeps me going long&#xA;When the movers move&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When loved ones move away, if you keep things going, then you are in a long distance relationship&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13445" LastEditorUserId="13445" LastEditDate="2015-06-22T18:45:37.053" LastActivityDate="2015-06-22T18:45:37.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16794" PostTypeId="2" ParentId="16790" CreationDate="2015-06-22T18:43:28.077" Score="23" Body="&lt;p&gt;The numbers are referring to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Letters of the alphabet!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The word 'at' &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With eight, I'm quite heady,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hats are worn on the head [citation needed].&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With three, my claws are ready,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cats are known to have sharp claws.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With thirteen, I'm downtrodden,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mats, like doormats and floor mats, are walked on.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With eighteen, said to be sodden.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This one could be a reference to the phrase &quot;wet as a drowned rat&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alone, I'm quite new here,&lt;br&gt;&#xA;yet ubiquitous everywhere.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Thanks to Ryan, I now know that it refers to the @ symbol! It's new to the common keyboard (relatively), yet is in common usage on pretty much all keyboards, websites, etc. these days.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="11475" LastEditDate="2015-06-24T18:55:27.870" LastActivityDate="2015-06-24T18:55:27.870" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16795" PostTypeId="2" ParentId="16791" CreationDate="2015-06-22T18:50:48.030" Score="14" Body="&lt;p&gt;I think you are talking about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A camel&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;They say I'm a ship&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A camel is also called as the &quot;ship of the desert&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But I see no water for days&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is no water in their habitat, the desert&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You may think you know me From my cousin of holy ways&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The cousin of the camel is the llama. Llama ( Lama ) is a reference to Dalai Lama, the Buddhist Monk.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Look for me in abandoned places&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A desert is pretty abandoned.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And you might find an obstacle or two Its what keeps me going long When the movers move&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As Kevin Rubin has stated, &quot; 'an obstacle or two' could be the camels' humps. These humps store water for the camel which helps them going long. &quot;When the movers move&quot; could mean the camel herders because they &quot;move&quot; the camel around by riding it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12860" LastEditorUserId="12860" LastEditDate="2015-06-23T00:02:32.107" LastActivityDate="2015-06-23T00:02:32.107" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="16798" PostTypeId="1" AcceptedAnswerId="16802" CreationDate="2015-06-22T19:25:49.200" Score="13" ViewCount="1111" Body="&lt;p&gt;It's been a while since one of these puzzles (created by &lt;a href=&quot;https://puzzling.stackexchange.com/users/10037/alex&quot;&gt;Alex&lt;/a&gt;, see &lt;a href=&quot;https://puzzling.stackexchange.com/questions/11281/the-hangman-game-backward&quot;&gt;here&lt;/a&gt; for the first) has shown up, and I thought some might enjoy another one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;fill in an &quot;n&quot; - You don't need to throw me out into the bin.&lt;br&gt;&#xA;fill in an &quot;f&quot; - You wish you could see my insides.&lt;br&gt;&#xA;fill in an &quot;m&quot;  - I may seem very familiar.&lt;br&gt;&#xA;fill in an &quot;v&quot;  - Your soul now belongs to me.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As always, your answer can be in form of: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The question was: _ p p l e&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;followed by explanation.  &lt;/p&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-06-22T19:47:59.963" Title="The Hangman Game Backwards Again" Tags="|riddle|english|language|letters|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16799" PostTypeId="1" AcceptedAnswerId="16829" CreationDate="2015-06-22T19:27:35.277" Score="7" ViewCount="16836" Body="&lt;p&gt;In this question, I am asking you to get to 24 in the most complicated way using the numbers below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2whqi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to solve this is (9 x 8)  / ( 6 / 2 ) = 24. But this is an easy way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for the most &lt;strong&gt;complicated&lt;/strong&gt; way of getting to 24. The person scoring the highest amount of &lt;em&gt;complicated points&lt;/em&gt; wins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Generic rules for complicated&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;there are no strict rules that apply for 'complicated' answers, as the community will vote for the most complicated answer. However the following is generally encouraged:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;getting to higher numbers within your equation wins you more 'complicated points' (e.g I got to a maximum of 72 in the above equation)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;use of non-common operators  (factorials, square roots, absolute signs) is deemed more complicated than common operators&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Other Rules&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;you are allowed to rearrange the numbers on the left hand side to get to 24&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;you are allowed to combine numbers if you deem it more complicated&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;you are only allowed to use each number once&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="12317" LastEditorUserId="11244" LastEditDate="2015-07-09T14:08:43.770" LastActivityDate="2015-07-09T14:08:43.770" Title="Making 24 - Most Complicated Way" Tags="|mathematics|" AnswerCount="15" CommentCount="10" ClosedDate="2015-06-23T01:47:18.443" ContentLicense="CC BY-SA 3.0" />
  <row Id="16800" PostTypeId="2" ParentId="16790" CreationDate="2015-06-22T19:29:38.413" Score="5" Body="&lt;p&gt;I'm quite new on the site, so I can't comment, but I wanted to help @Bailey M with their answer.  Specifically, the last two clues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think &quot;Alone, I'm quite new here, Yet ubiquitous everywhere&quot; refers simply to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the @ symbol.  It's new on the Stack network (I think?) and it's everywhere (emails, Twitter, etc)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13446" LastEditorUserId="13446" LastEditDate="2015-06-22T19:35:59.187" LastActivityDate="2015-06-22T19:35:59.187" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16801" PostTypeId="2" ParentId="16799" CreationDate="2015-06-22T19:35:39.903" Score="14" Body="&lt;p&gt;How about this?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\Big\lceil\frac{8!}{9!}\Big\rceil\times(6-2)!=24$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-06-22T19:35:39.903" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16802" PostTypeId="2" ParentId="16798" CreationDate="2015-06-22T19:42:51.943" Score="13" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The question was sa_e&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fill in an &quot;n&quot; - You don't need to throw me out into the bin.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;sane - don't need to go to the looney bin&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fill in an &quot;f&quot; - You wish you could see my insides.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;safe - I would love to see the insides of many fancy safes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fill in an &quot;m&quot; - I may seem very familiar.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;same - things that are the same are quite familiar.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fill in an &quot;v&quot; - Your soul now belongs to me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;save - If I save your soul, it may belong to me at that point.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13446" LastEditorUserId="13446" LastEditDate="2015-06-22T19:47:59.963" LastActivityDate="2015-06-22T19:47:59.963" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16803" PostTypeId="2" ParentId="16799" CreationDate="2015-06-22T19:45:30.877" Score="18" Body="&lt;p&gt;How about:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\sum_{8}^{9} (6 * 2) = 24$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A little bit of notation mutilation, but hey...&lt;/p&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-06-22T19:45:30.877" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16804" PostTypeId="2" ParentId="16799" CreationDate="2015-06-22T19:49:20.917" Score="1" Body="&lt;p&gt;Does this involve big enough numbers for you?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I define the operation $*$ on pairs of integers in such a way that $9*8$ is a googolplex and $6*2$ is 24 times a googolplex. Then $\frac{6*2}{9*8}=24$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-06-22T19:49:20.917" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16805" PostTypeId="2" ParentId="16799" CreationDate="2015-06-22T19:56:45.223" Score="6" Body="&lt;p&gt;Maybe a little bit less complicated than the others&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\sqrt[\sqrt{9}]{8} * 6 * 2 = \sqrt[3]{8} * 6 * 2 = 2 *6 * 2 = 24$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10400" LastActivityDate="2015-06-22T19:56:45.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16808" PostTypeId="2" ParentId="16799" CreationDate="2015-06-22T20:04:02.297" Score="2" Body="&lt;p&gt;If you want something that looks complicated...but really isn't:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\phi$ - The Golden ratio&lt;br&gt;&#xA;$\displaystyle\int\limits_{-\infty}^{+\infty}(-(\sqrt{9-8+e^{\pi\cdot i}} \times \sinh(f'(\phi)))^3) \ \mathrm dx + (6\cdot2!)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-06-24T13:38:31.370" LastActivityDate="2015-06-24T13:38:31.370" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16811" PostTypeId="2" ParentId="16768" CreationDate="2015-06-22T20:19:31.777" Score="6" Body="&lt;p&gt;The answers to the questions are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The scrabble score of the first name of the U.S. President alluded to with the number.  For example: 2 (&quot;John&quot; Adams) + 6 (&quot;John&quot; Quincy Adams) = 28.  &quot;John&quot; = 14 points in scrabble so 14 + 14 = 28.  The same works for the rest.&lt;br&gt;&#xA;&lt;br&gt;&lt;br&gt;&#xA;1st president: George, score: 8&lt;br&gt;&#xA;8th president: Martin, score: 8&lt;br&gt;&#xA;8 + 8 = 16&lt;br&gt;&#xA;&lt;br&gt;&#xA;12th president: Zachary, score: 24&lt;br&gt;&#xA;14th president: Franklin, score: 15&lt;br&gt;&#xA;24 + 15 = 39&lt;br&gt;&#xA;&lt;br&gt;&#xA;6th president: John, score: 14&lt;br&gt;&#xA;16th president: Abraham, score: 14&lt;br&gt;&#xA;14 + 14 = 28&lt;br&gt;&#xA;&lt;br&gt;&#xA;13th president: Millard, score: 10&lt;br&gt; &#xA;9th president: William, score: 12&lt;br&gt;&#xA;10 + 12 = 22&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, the answer to the final equation is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;22&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13446" LastEditorUserId="13446" LastEditDate="2015-06-22T20:34:03.700" LastActivityDate="2015-06-22T20:34:03.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16812" PostTypeId="2" ParentId="16142" CreationDate="2015-06-22T20:19:33.577" Score="12" Body="&lt;p&gt;I have found the names, symbols, and atomic numbers for the first 118 elements in the periodic table. &lt;span class=&quot;math-container&quot;&gt;$118*3=354$&lt;/span&gt; so I presume these are the &amp;quot;over 300&amp;quot; hidden things mentioned by OP.&lt;/p&gt;&#xA;&lt;p&gt;I used the small grid with the plain text that I could copy. My search preferred to find items in the top left so, although there could be multiple instances of a symbol or number, the highest and furthest left option will be highlighted. To be fair, this could cause some confusion if, for instance, the same two cells are highlighted for both 17 and 71. However, I can assure you that you &lt;em&gt;will&lt;/em&gt; find each answer highlighted.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;Element Names&lt;/h2&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/6875q.png&quot; alt=&quot;Solved for Names&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;Element Symbols&lt;/h2&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/xQKDz.png&quot; alt=&quot;Solved for Symbols&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;Element Atomic Numbers&lt;/h2&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/o68lj.png&quot; alt=&quot;Solved for Atomic Numbers&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;On an unrelated note, only 242 of the 1,369 characters in the word search were left unused in my results. This gives a usage rate of about &lt;span class=&quot;math-container&quot;&gt;$82\%$&lt;/span&gt;. That could probably be increased by a more intelligent search that favored all characters being used instead of just the first one it finds. That's some tightly packed information.&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-06-23T14:06:31.280" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16814" PostTypeId="2" ParentId="16799" CreationDate="2015-06-22T20:20:30.297" Score="13" Body="&lt;p&gt;Here's one with a number that goes &lt;em&gt;really&lt;/em&gt; big:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lceil{\ln(9^{(8 + 2)} \times 6)}\rceil = 24$$&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-06-22T20:20:30.297" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16816" PostTypeId="1" AcceptedAnswerId="16818" CreationDate="2015-06-22T20:23:52.820" Score="0" ViewCount="563" Body="&lt;p&gt;What would come next in this sequence, and what is the source?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;B C N O&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="319" LastEditorUserId="319" LastEditDate="2015-06-23T04:11:29.987" LastActivityDate="2015-06-23T04:11:29.987" Title="What's next in this letter sequence: B, C, N, O" Tags="|letter-sequence|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16817" PostTypeId="2" ParentId="16799" CreationDate="2015-06-22T20:24:12.460" Score="2" Body="&lt;p&gt;Here's one&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(6 &amp;lt;&amp;lt; 2) + \Big\lfloor\frac{8}{9}\Big\rfloor=24$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12242" LastActivityDate="2015-06-22T20:24:12.460" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16818" PostTypeId="2" ParentId="16816" CreationDate="2015-06-22T20:26:37.650" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; F&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; B, C, N, O, F are the atomic symbols of the periodic table, in order, of Boron, Carbon, Nitrogen, Oxygen and Fluorine.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12860" LastActivityDate="2015-06-22T20:26:37.650" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16819" PostTypeId="2" ParentId="16799" CreationDate="2015-06-22T20:32:05.673" Score="5" Body="&lt;p&gt;Edit: Added what will surely win as the &quot;highest number&quot; prize and dropped the spoiler tags as they don't seem necessary with this question.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;How about an infinite score using an infinite number of square roots and, just for fun, factorials? Do you count complication based on the number of different terms used or just the number used in total?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left[\left(\sqrt{\sqrt{\sqrt{\cdots\sqrt{\sqrt{\sqrt{9}}}}}}\right)!!!\ldots!!!\right]\times8\times6\div2=1\times8\times6\div2=24$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, just to claim the &quot;highest number&quot; prize using &lt;a href=&quot;https://en.wikipedia.org/wiki/Knuth%27s_up-arrow_notation&quot; rel=&quot;nofollow&quot;&gt;Knuth's up-arrow notation&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left\lfloor{&#xA;\sqrt{\sqrt{\sqrt{\ldots m \ldots\sqrt{\sqrt{\sqrt{&#xA;9 \uparrow\uparrow\uparrow\ldots n \ldots\uparrow\uparrow\uparrow 8&#xA;}}}}}}&#xA;}\right\rfloor&#xA;\times\frac{6!!}{2}=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $n$ is as large as needed to have the highest number and $m$ is as large as needed so that the floor function evaluates to $1$. I am not mathematical enough to &lt;em&gt;prove&lt;/em&gt; that up-arrow notation will grow faster than the double factorial method proposed by Rodolvertice but I am fairly sure that it does so quite easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that the $6!!$ is the double factorial and $6!!=48$)&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2015-06-23T20:11:14.077" LastActivityDate="2015-06-23T20:11:14.077" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16820" PostTypeId="2" ParentId="16799" CreationDate="2015-06-22T20:39:15.110" Score="0" Body="&lt;p&gt;$$\lfloor\tan(\cos(\sin\lceil(\sqrt{8})^{[\frac{d}{dx}(-9x)]}\rceil))\rfloor+\lfloor\sqrt{6!}-2\rfloor$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $x$ is an arbitrary number.&lt;/p&gt;&#xA;" OwnerUserId="13451" LastEditorUserId="7308" LastEditDate="2015-06-22T22:50:00.250" LastActivityDate="2015-06-22T22:50:00.250" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16821" PostTypeId="2" ParentId="16799" CreationDate="2015-06-22T20:39:48.453" Score="4" Body="&lt;p&gt;Claims both the infinite operators score and the largest number score.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n!!$ is the double factorial. There is no factorial used here, only double factorials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cfrac{6!!}{2!!!!\ldots\ldots!!!!}-{\left\lfloor{\cfrac{8!!!!... ...!!!!}{9!!!!... ...!!!!}}\right\rfloor}=24$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number $2$ can have an infinite number of factorials (double or single) applied to it and it will still be $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cfrac{6!!}{2!!}=24$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this is because $6!!=48$, then dividing that by $2$ results in $24$, and $2=2!=2!!=2!!!\dots!!!$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left\lfloor{\frac{8!!!!\dots n\dots!!!!}{9!!!!\dots n\dots!!!!}}\right\rfloor=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subtracting $0$ does nothing. n (the number of double factorials) can be very VERY large but not infinite, and it will still evaluate to $0$. Just enough so that it is more than the largest number here.&lt;/p&gt;&#xA;" OwnerUserId="4109" LastEditorUserId="4109" LastEditDate="2015-06-22T23:54:58.420" LastActivityDate="2015-06-22T23:54:58.420" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="16822" PostTypeId="2" ParentId="16799" CreationDate="2015-06-22T20:50:37.893" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;(λn.λf.λx.n (λg.λh.h (g f)) (λu.x) (λu.u)) π&lt;sup&gt;2&lt;/sup&gt;$(6*89)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; π&lt;sup&gt;2&lt;/sup&gt; is the &lt;a href=&quot;http://mathworld.wolfram.com/PrimeCountingFunction.html&quot;&gt;Prime Counting Function&lt;/a&gt; nested twice, and the rest is apparently the simplest way to express the predecessor function in lambda calculus. So we multiply 6 by 89 (534), count the number of primes less than or equal to it (99), count the number of primes less than or equal to &lt;em&gt;that&lt;/em&gt; (25), then subtract 1 (24).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9523" LastActivityDate="2015-06-22T20:50:37.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16823" PostTypeId="2" ParentId="16799" CreationDate="2015-06-22T20:51:28.767" Score="9" Body="&lt;p&gt;I'm fairly certain I'm going to terribly mess up the markup on this, but here goes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\left\lceil\sqrt{\sqrt{\dfrac{(8+2)!}{\sqrt[\sqrt{\sqrt{6!}}]{9!}}}}\right\rceil=24$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13450" LastEditorUserId="7307" LastEditDate="2015-06-23T00:42:51.543" LastActivityDate="2015-06-23T00:42:51.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16824" PostTypeId="2" ParentId="16799" CreationDate="2015-06-22T21:07:45.703" Score="2" Body="&lt;p&gt;$$&#xA;\frac{d^3}{dx^3}(\frac {8}{2} x^\sqrt9+ 6!x) = 24&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="13204" LastEditorUserId="13204" LastEditDate="2015-06-23T13:06:05.987" LastActivityDate="2015-06-23T13:06:05.987" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16825" PostTypeId="1" CreationDate="2015-06-22T21:11:52.253" Score="3" ViewCount="5916" Body="&lt;p&gt;After living a peaceful and calm life, you die a painless death. Your spirit starts ascending into heaven when suddenly you are approaching two doors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both doors are completely identical and there is no way of telling them apart.&#xA;One door leads to heaven while the other descends to hell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In front of each door are twin brothers, both identical. One of them always speaks the truth while the other one always lies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are to choose one brother at random and ask one question to him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What will you ask him so that you can find the right door ?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="13016" LastEditorUserId="8874" LastEditDate="2016-02-05T16:02:02.243" LastActivityDate="2016-02-05T16:02:02.243" Title="Which is the door to heaven" Tags="|logical-deduction|liars|" AnswerCount="1" CommentCount="0" ClosedDate="2015-06-22T21:19:26.060" ContentLicense="CC BY-SA 3.0" />
  <row Id="16826" PostTypeId="2" ParentId="16825" CreationDate="2015-06-22T21:18:45.347" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;What would the other guard say if I asked him which door leads to hell?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you ask the truth-teller, he'll point to the door to heaven (since the liar would lie).&lt;br&gt;&#xA;If you ask the liar, he'll also point to the door to heaven (since he's lying about what the truth-teller would say).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So whichever door the guard points to, you should go through that door.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-06-22T21:18:45.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16827" PostTypeId="2" ParentId="16799" CreationDate="2015-06-22T21:38:49.113" Score="2" Body="&lt;p&gt;I have a few contributions. $\Gamma$ is the &lt;a href=&quot;https://en.wikipedia.org/?title=Gamma_function&quot; rel=&quot;nofollow&quot;&gt;Gamma function&lt;/a&gt;; $x!!$ is the &lt;a href=&quot;http://keisan.casio.com/exec/system/1180573418&quot; rel=&quot;nofollow&quot;&gt;double factorial&lt;/a&gt;; $\text{Res}$ is the residue of a function (I found the one for $\Gamma$ on the first link); $\phi$ is the &lt;a href=&quot;http://mathworld.wolfram.com/TotientFunction.html&quot; rel=&quot;nofollow&quot;&gt;Totient Function&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;1&lt;/h1&gt;&#xA;&#xA;&lt;h2&gt;Equation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;$$\tag{1}\frac{\frac{9!}{\Gamma(8)}}{\frac62}$$&#xA;or&#xA;$$\tag{1b}(9!/\Gamma(8))/(6/2)$$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Explanation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;$$\Gamma(n)=(n-1)!$$&#xA;$$\begin{align}(9!/\Gamma(8)/(6/2))&amp;amp;=24\\&#xA;(9!/7!)/(6/2)&amp;amp;=\\&#xA;(72)/(3)&amp;amp;=\\&#xA;24&amp;amp;=&#xA;\end{align}$$&#xA;BONUS: $(1)$ maintains order.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;2&lt;/h1&gt;&#xA;&#xA;&lt;h2&gt;Equation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;$$\tag{2}\Gamma(9+6-(8+2!))$$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Explanation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}\Gamma(9+6-(8+2!))&amp;amp;=24\\\Gamma(15-(8+2))&amp;amp;=\\\Gamma(5)&amp;amp;=\\4!&amp;amp;=\\24&amp;amp;=\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;3&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;If you allow $n^{-1}$ as an operation and not a number:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Equation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;$$\tag{3}-\text{Res}(\Gamma,-8/2)^{-1}+\left\lfloor\text{Res}(\Gamma,-9/6!)^{-1}\right\rfloor$$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Explanation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;$$\text{Res}(\Gamma,-n)=\frac{(-1)^n}{n!}$$&#xA;$$-\text{Res}(\Gamma,-8/2)^{-1}=-\text{Res}(\Gamma,-4)^{-1}=-\frac{-1}{4!}^{-1}=-\frac{-1}{24}^{-1}=--24=24$$&#xA;Of course, $0\lt\Gamma(9/6!)^-1\lt1$, so it becomes $0$ in the floor bit. Thus, the answer is $24$.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;4&lt;/h1&gt;&#xA;&#xA;&lt;h2&gt;Equation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;$$\tag{4}\phi(9+8)+6+2$$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Explanation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;$$\phi(9+8)=\phi(17)=16$$&#xA;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=EulerPhi[9%2B8]&quot; rel=&quot;nofollow&quot;&gt;check here&lt;/a&gt;&#xA;$$16+6+2=22+2=24$$&lt;/p&gt;&#xA;" OwnerUserId="7307" LastEditorUserId="7307" LastEditDate="2015-06-22T22:13:22.520" LastActivityDate="2015-06-22T22:13:22.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16828" PostTypeId="2" ParentId="16773" CreationDate="2015-06-22T21:44:23.060" Score="3" Body="&lt;p&gt;It is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Label the coins A-L, with &lt;strong&gt;L&lt;/strong&gt; being the known genuine coin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weigh ABCD &lt;em&gt;vs&lt;/em&gt; EFGH &lt;em&gt;vs&lt;/em&gt; IJKL.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If no pan rises, then both heavy coins are on the same pan.&#xA;There are ${4\choose 2} = 6$ combinations where both heavy coins are on Pan 1, and 6 combinations where both are on Pan 2, but only ${3\choose 2} = 3$ combinations were both are on Pan 3, since we know coin L is genuine.  That adds up to 15 possibilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Pan 3 rises, then one heavy coin is on Pan 1 and another on Pan 2, for a total of $4\times 4 = 16$ combinations.  If Pan 2 rises, then there are only $12$ combinations, again because we know coin L is genuine.  Likewise for Pan 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If no pan rose on the first weighing&lt;/strong&gt;, then the second weighing should be  ABEF &lt;em&gt;vs&lt;/em&gt; CGIJ &lt;em&gt;vs&lt;/em&gt; DHKL.  The results of this would be:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;None: AB, EF, IJ&lt;/li&gt;&#xA;&lt;li&gt;Pan 1: CD, GH, IK, JK&lt;/li&gt;&#xA;&lt;li&gt;Pan 2: AD, BD, EH, FH&lt;/li&gt;&#xA;&lt;li&gt;Pan 3: AC, BC, EG, FG&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the first case, for the third weighing, use AE &lt;em&gt;vs&lt;/em&gt; BI &lt;em&gt;vs&lt;/em&gt; EJ.  The two heavy coins are now split and one of the three pans will rise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the other three cases, there are four patterns of which at least two intersect by one coin.  Put the intersecting coin on Pan 1, and each of the others from the pairings on Pans 2 &amp;amp; 3.  Pick one other pattern and split its coins between Pans 2 &amp;amp; 3.  Put other coins aside and fill in with known genuine coins (of which you have several).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if Pan 1 rose, you would weigh KA &lt;em&gt;vs&lt;/em&gt; IC &lt;em&gt;vs&lt;/em&gt; JD.  If the pattern was GH, then no Pan would rise.  CD would raise Pan 1, IK would raise Pan 3, and JK would raise Pan 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If Pan 3 rose on the first weighing&lt;/strong&gt;, then one of the heavy coins is in ABCD, and the other in EFGH, for a total of $16$ possibilities.  Coins I,J,K, &amp;amp; L are now all known to be genuine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weigh ABGH &lt;em&gt;vs&lt;/em&gt; CDIJ &lt;em&gt;vs&lt;/em&gt; EFKL.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;No Pan: AG, AH, BG, BH&lt;/li&gt;&#xA;&lt;li&gt;Pan 1: CE, CF, DE, DF&lt;/li&gt;&#xA;&lt;li&gt;Pan 2: AE, AF, BE, BF&lt;/li&gt;&#xA;&lt;li&gt;Pan 3: CG, CH, DG, DH&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In all four of these cases, you have four patterns where one coin is from one pair and the other coin from the other pair.  The weighing pattern for this is similar to above, but requiring 2 genuine coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No Pan: weigh AI &lt;em&gt;vs&lt;/em&gt; BG &lt;em&gt;vs&lt;/em&gt; HJ.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AG would raise Pan 3; AH would raise Pan 2; BG  would raise no pan, and BH would raise Pan 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar pattern would work for any of the results of the second weighing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If Pan 1 or 2 rose on the first weighing&lt;/strong&gt;, follow the same logic for Pan 3 rising, with the different sets of coins.  Any of the possibilities where coin L came out heavy would obviously not happen, but otherwise the logic would be the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2015-06-22T21:44:23.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16829" PostTypeId="2" ParentId="16799" CreationDate="2015-06-22T22:17:33.860" Score="25" Body="&lt;p&gt;One that is mathematically precise (rounding answers are such cop-outs! (p.s. I upvoted them in any case))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the &lt;a href=&quot;https://en.wikipedia.org/wiki/Riemann_zeta_function&quot;&gt;Riemann Zeta function&lt;/a&gt; $\zeta$ &#xA;, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Sign_function&quot;&gt;sign function&lt;/a&gt; $\text{sgn}$, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Gamma_function&quot;&gt;Gamma function&lt;/a&gt; $\Gamma$ , the &lt;a href=&quot;https://en.wikipedia.org/wiki/Divisor_function&quot;&gt;sigma function&lt;/a&gt; $\sigma_0$  and the &lt;a href=&quot;https://en.wikipedia.org/?title=Negation&quot;&gt;negation function&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;24 &amp;amp; =\frac{\frac{\text{sgn}(2)}{-\zeta(-9)}}{\Gamma(\sigma_{0}(8))+\zeta(\zeta(-6))}\\&#xA;&amp;amp; =\frac{\frac{1}{-1/-132}}{\Gamma(4)+\zeta(0)}\\&#xA;&amp;amp; =\frac{132}{6-0.5}\\&#xA;&amp;amp; =\frac{132}{5.5}\\&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="7529" LastEditorUserId="4109" LastEditDate="2015-06-23T04:03:38.170" LastActivityDate="2015-06-23T04:03:38.170" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16832" PostTypeId="2" ParentId="16474" CreationDate="2015-06-23T01:28:52.717" Score="4" Body="&lt;p&gt;&lt;strong&gt;Puzzle 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Following your hint, the home is:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Puzzling Stackexchange&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;And the best five childrens (of the week) are rand al'thor, Joe Z, Aggie Kidd, A.D, CodeNewbie&lt;/p&gt;&#xA;&lt;p&gt;Taking the best of each children:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;rand al'thor: 3 gold&lt;/p&gt;&#xA;&lt;p&gt;Joe Z : 4 gold&lt;/p&gt;&#xA;&lt;p&gt;Aggie Kidd: 2 gold&lt;/p&gt;&#xA;&lt;p&gt;AD: 6 silver&lt;/p&gt;&#xA;&lt;p&gt;CodeNewbie: 1 silver&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Adding all this, the finale value is:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;16&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5359" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-06-23T02:40:25.683" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16833" PostTypeId="1" AcceptedAnswerId="17089" CreationDate="2015-06-23T02:50:08.183" Score="14" ViewCount="2102" Body="&lt;p&gt;Long ago, the pirate ship Wild Goose, led by Captain Oliver, sailed the coast of Nova Scotia. One night, according to legend, the Wild Goose was caught in a storm. Before she sank, Captain Oliver had the ship's stash of treasure put into a life raft. The surviving crew members steered the raft to the nearby Brier Island, and hid the treasure there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You, a world famous treasure hunter, are headed to Brier Island in search of the lost treasure. When you arrive on the island, there is no trace of the treasure. You do however find a journal which is clearly very old. Most of the pages are too faded to read, but one of them contains some mysterious writing:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/xwUwF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the treasure, and where is it hidden?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2627" LastActivityDate="2015-06-27T15:05:50.627" Title="Decipher the location of the pirate treasure" Tags="|cipher|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16834" PostTypeId="1" AcceptedAnswerId="16903" CreationDate="2015-06-23T03:20:01.197" Score="16" ViewCount="1011" Body="&lt;p&gt;Behold a maze of fuses (the black dotted lines) laid out on a grid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &lt;img src=&quot;https://i.stack.imgur.com/YhVIk.png&quot; alt=&quot;a confused maze&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fuses can be lit, causing sparks to burn along them at a perfectly uniform rate: one segment per second. Your goal is to light the fuses in exactly four (4) places so that all sparks extinguish at &lt;em&gt;exactly the same time&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rules:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;fuses may only be lit on the corners of grid squares (that is, the &quot;gaps&quot; between the fuse segments); all fuses must be lit at the same time&lt;/li&gt;&#xA;&lt;li&gt;if $n$ spark(s) meet at an $n$-way intersection, they extinguish; this includes the case of a single spark reaching a dead-end&lt;/li&gt;&#xA;&lt;li&gt;if $n$ spark(s) meet at a $k$-way intersection, with $k &amp;gt; n$, they do not extinguish; instead, they split or merge if necessary and continue to follow all outgoing branches (all branches that sparks didn't come in on)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can you figure out which four places to light the fuses?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; See &lt;a href=&quot;https://i.stack.imgur.com/lu4qO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/lu4qO.png&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The four colours (red, purple, blue, and green) correspond to fuses burned by sparks spreading from the first, second, third, and fourth spark origins, respectively.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Example&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;As an example, consider one part of the grid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &lt;img src=&quot;https://i.stack.imgur.com/JH2UT.png&quot; alt=&quot;a smaller grid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lighting the fuse at D5 is a potential solution. It will start burning north and south, reaching B3 in 4s and B6 at 3s, splitting at both. The B3 north spark and B6 south spark will continue burning (exiting the grid). The B3 south and B6 north sparks will reach B4 at exactly the same time (5s) and merge, continuing to burn along the western branch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lighting the fuse at D4 is &lt;em&gt;not&lt;/em&gt; a potential solution. We see that in this case, two sparks will meet and extinguish at B5 after 5s, but other sparks (at A2 and A4, and possibly B7) will still be burning, meaning all sparks do not extinguish at the same time.&lt;/p&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="4219" LastEditDate="2015-06-24T03:56:48.627" LastActivityDate="2015-06-25T20:59:37.107" Title="The Origins of a Confusing Maze" Tags="|geometry|mazes|" AnswerCount="1" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="16835" PostTypeId="2" ParentId="16833" CreationDate="2015-06-23T03:38:59.513" Score="12" Body="&lt;p&gt;EDIT 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Referring to this link (&lt;a href=&quot;https://en.wikipedia.org/wiki/Brier_Island&quot;&gt;https://en.wikipedia.org/wiki/Brier_Island&lt;/a&gt;), Brier Island is actually in western Nova Scotia (which is part of Canada, never knew that before). Being in Canada, it makes sense that the coins are loonies (&lt;a href=&quot;https://en.wikipedia.org/wiki/Loonie&quot;&gt;https://en.wikipedia.org/wiki/Loonie&lt;/a&gt;). According to the first wiki page, Brier Island is well known for their ship wrecks.&lt;br&gt;&lt;br&gt;With all of this making up an accurate story so far, it's possible Wild Goose and Captain Oliver have some special meaning. &lt;br&gt;&lt;br&gt; It may be a coincidence but there is a Geocache hidden in Oliver's cove on Brier's Island (&lt;a href=&quot;http://www.geocaching.com/geocache/GC13ACV_olivers-cove-brier-island-series?guid=72a041dd-752a-4c0a-8bff-bbb0adf65f3c&quot;&gt;http://www.geocaching.com/geocache/GC13ACV_olivers-cove-brier-island-series?guid=72a041dd-752a-4c0a-8bff-bbb0adf65f3c&lt;/a&gt;). I would go there myself to dig and check if there are any golden loonies but it may be a &lt;em&gt;wild goose&lt;/em&gt; chase. heh.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here's a potential answer (unsure of the last one):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The codes are as follows:&lt;br&gt;&lt;br&gt;(1) 4D 49 4C 4C 49 4F 4F&lt;br&gt;This is ASCII for &quot;million&quot;&lt;br&gt;&lt;br&gt;(2) ILIWHHQ&lt;br&gt;Rotate 23 for &quot;fifteen&quot;&lt;br&gt;&lt;br&gt;(3) I think the is just the.&lt;br&gt;&lt;br&gt;(4) SEPOLTI is Italian for buried&lt;br&gt;&lt;br&gt;(5) The morse code translates to &quot;golden&quot;&lt;br&gt;&lt;br&gt;(6) The icons are pigpen (&lt;a href=&quot;https://en.wikipedia.org/wiki/Pigpen_cipher&quot;&gt;https://en.wikipedia.org/wiki/Pigpen_cipher&lt;/a&gt;), this translates to &quot;loonies&quot;&lt;br&gt;&lt;br&gt;(7) The symbols are flag symbols (&lt;a href=&quot;https://en.wikipedia.org/wiki/Flag_semaphore&quot;&gt;https://en.wikipedia.org/wiki/Flag_semaphore&lt;/a&gt;) that translate to &quot;are&quot;&lt;br&gt;&lt;br&gt;(8) 2363284&lt;br&gt;This is the one I'm unsure of, how can a 7 digit number be translated to a word? If this was a modern day puzzle, then I'd immediately think of a cell phone number. If you type it out, it does spell &quot;beneath&quot;(&lt;a href=&quot;http://phonespell.org/combo.cgi?n=2363284&quot;&gt;http://phonespell.org/combo.cgi?n=2363284&lt;/a&gt;). It doesn't mean anything meaningful in any bases (base 36 is 1ENIS), it can't be broken up and converted to letters in any way, and I don't see any references online. Maybe &quot;long ago&quot; isn't so long?&lt;br&gt;&lt;br&gt;Final answer when unscrambled:&lt;br&gt;The fifteen million golden loonies are buried beneath. (yes I know, anticlimactic)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="7722" LastEditDate="2015-06-23T05:18:59.597" LastActivityDate="2015-06-23T05:18:59.597" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16836" PostTypeId="2" ParentId="15796" CreationDate="2015-06-23T04:35:23.827" Score="0" Body="&lt;p&gt;Specifically Expected Answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We require some (around 6) balls which are known to be normal, so atleast 1 weighting is required for getting those normal balls.&lt;br&gt;&#xA; For finding the section of the odd ball (which may be from sections 2-8 , 7 possibilities), we require minimum 2 weightings.&lt;br&gt;&#xA; So, we require minimum 3 weightings in total.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With this thinking, the process may be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (1) Make 3 sets ABC with 1023/3=341 balls&lt;br&gt;&#xA; (2) Compare A &amp;amp; B, which is &lt;strong&gt;W1&lt;/strong&gt; : If equal, then remaining C has the odd ball, while if not equal, then A or B has the odd ball. So we have 341 ball in odd set and have many normal balls in other sets.&lt;br&gt;&#xA; (3) Now use binary search algorithm to find how much more the odd set weighs. Check (odd set) against (normal set + 4 normal balls) which is &lt;strong&gt;W2&lt;/strong&gt;. Based on result compare (odd set) against (normal set + 2 normal balls) or (odd set) against (normal set + 6 normal balls) which is &lt;strong&gt;W3&lt;/strong&gt;. So with W1 &amp;amp; W2 &amp;amp; W3, we have the required answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10940" LastActivityDate="2015-06-23T04:35:23.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16838" PostTypeId="1" AcceptedAnswerId="16915" CreationDate="2015-06-23T06:07:36.040" Score="4" ViewCount="1219" Body="&lt;p&gt;Here's a riddle for the day&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;One fell in love with his music system&lt;br&gt;&#xA;  Another led the opposers in Queensland&lt;br&gt;&#xA;  One played the man playing a trick&lt;br&gt;&#xA;  The last lost his love by his own hand&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Queensland is not the Australian state here. Read it aloud.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hunter's answer is certainly on the right track.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The title is also a clue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 4: (big one)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Queensland isn't one word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;__Hint 5: (AND THE BIGGEST ONE YET)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16878/whats-our-name-also-improve-my-riddle?lq=1&quot;&gt;In his question, Engineer Toast has been inspired a lot more than he thinks.&lt;/a&gt; (Or maybe he knows, but I'm not sure.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.923" LastActivityDate="2015-06-24T09:39:53.223" Title="Don't rat out the secret answer" Tags="|riddle|rhyme|" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16840" PostTypeId="2" ParentId="16833" CreationDate="2015-06-23T08:00:32.657" Score="17" Body="&lt;p&gt;This really is an addendum to Quark's answer, which has all the individual words deciphered. As commented by Julian, there is still one word missing and Quark's word order is not correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the correct word order is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fifteen million golden loonies are buried beneath the ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where ... is the missing word. That word can be found&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; by taking the first letters of the encodings used for each word in the order of the final sentence. The encodings are the &lt;strong&gt;C&lt;/strong&gt;aesar cipher, &lt;strong&gt;A&lt;/strong&gt;SCII, &lt;strong&gt;M&lt;/strong&gt;orse code, the &lt;strong&gt;P&lt;/strong&gt;igpen cipher, the &lt;strong&gt;S&lt;/strong&gt;emaphore alphabet, &lt;strong&gt;I&lt;/strong&gt;talian, the &lt;strong&gt;T&lt;/strong&gt;elephone keypad and plain &lt;strong&gt;E&lt;/strong&gt;nglish.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the complete hint for the location of the treasure is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fifteen million golden loonies are buried beneath the campsite.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9860" LastActivityDate="2015-06-23T08:00:32.657" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16841" PostTypeId="2" ParentId="16838" CreationDate="2015-06-23T08:03:01.910" Score="7" Body="&lt;p&gt;This is my guess... The answer could be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (the surname) NEWMAN &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;One fell in love with his music system&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alfred Newman: developed what came to be known as the Newman System, a means of synchronising the performance and recording of a musical score with the film. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Another led the opposers in Queensland&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;s&gt;Campbell Newman was elected to lead the LNP into the 2012 Queensland state election&lt;/s&gt;&lt;br&gt;&#xA;(Being &lt;em&gt;Queensland&lt;/em&gt; actually &lt;em&gt;Queen's Land&lt;/em&gt;) John Henry Newman, who became known as a leader of, and an able polemicist for, the Oxford Movement, an influential and controversial grouping of Anglicans who wished to return to the Church of England many Catholic beliefs and liturgical rituals from before the English Reformation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;One played the man playing a trick&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Paul Newman, who played a pool player in &quot;The color of money&quot; and in &quot;The hustler&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;The last lost his love by his own hand&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The singer John Newman (is it a reference to &quot;Out Of My Head&quot; lyrics? )&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12628" LastEditorUserId="12628" LastEditDate="2015-06-24T07:08:11.573" LastActivityDate="2015-06-24T07:08:11.573" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16842" PostTypeId="1" CreationDate="2015-06-23T08:36:42.187" Score="6" ViewCount="756" Body="&lt;p&gt;Few month ago, I was on vacation visiting an old friend in his own  country. We were sitting on his balcony enjoying the sea view from up the hill and listening to his old favorite music when he asked me:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Do you like the view from here?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;You have a beautiful sunset&quot;, I replied. &quot;Can we spot the island of your neighbours from here?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Noo !! I don't think that's possible! There are more than 150km to the nearest land&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I see.. and what is that music genre you're listening to?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Those are musics that never &lt;em&gt;die&lt;/em&gt; in this country. They are traditional folk music. You should probably learn more about our culture since you have the occasion of visiting me now.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yeah.. You're right! There are a lot that I don't know yet. From where should I begin?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Well.. self education is sometimes better than formal education and since I heard you are great at solving puzzles, I want you to learn by yourself&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He then writes down on a piece of paper the following numbers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;131110&lt;/strong&gt; &lt;br&gt; &lt;strong&gt;142611&lt;/strong&gt; &lt;br&gt; &lt;strong&gt;???811&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;&lt;strong&gt;What are the three missing numbers and why ???&lt;/strong&gt;&lt;/h2&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Note:&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You may deduce the pattern easily, but it will only make sense when you find the country they're talking about. So try to find that first.&lt;/li&gt;&#xA;&lt;li&gt;You should surely explain how you obtained the solution. The title and the story are a hint.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Hint 1:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The island is a country !&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Hint 2&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;#&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12731" LastEditorUserId="8874" LastEditDate="2016-03-23T17:05:56.597" LastActivityDate="2016-03-23T17:05:56.597" Title="Cultural Patrimony" Tags="|logical-deduction|pattern|story|history|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16843" PostTypeId="2" ParentId="16838" CreationDate="2015-06-23T08:43:13.070" Score="1" Body="&lt;p&gt;Maybe they're&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; bridge players.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;One fell in love with his music system&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure about this line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Another led the opposers in Queensland&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This player is opposite another and has lots of queens in his hand.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;One played the man playing a trick&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A round of play in bridge is called a trick, so at any time each player is playing another who's playing a trick.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The last lost his love by his own hand&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This player's hand was so bad that he lost the game, and with it whatever he'd wagered (he probably loves money).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-06-23T08:43:13.070" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16844" PostTypeId="1" AcceptedAnswerId="16845" CreationDate="2015-06-23T09:03:14.117" Score="0" ViewCount="3086" Body="&lt;p&gt;Here is a riddle from the Hobbit.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;This thing all things devours;&lt;/p&gt;&#xA;&lt;p&gt;Birds, beasts, trees, flowers;&lt;/p&gt;&#xA;&lt;p&gt;Gnaws iron, bites steel;&lt;/p&gt;&#xA;&lt;p&gt;Grinds hard stones to meal;&lt;/p&gt;&#xA;&lt;p&gt;Slays king, ruins town,&lt;/p&gt;&#xA;&lt;p&gt;And beats high mountain down.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;What am I talking about?&lt;/p&gt;&#xA;" OwnerUserId="12860" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2017-12-29T02:30:06.223" Title="This thing all things devour" Tags="|riddle|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16845" PostTypeId="2" ParentId="16844" CreationDate="2015-06-23T09:10:07.843" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;TIME&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;This thing all things devour;&#xA;Birds, beasts, trees, flowers;&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All the things consume TIME - birds, beasts, trees , flowers&#xA; everything devour &lt;strong&gt;TIME&lt;/strong&gt; they age by time, never the same.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Gnaws iron, bites steel;&#xA;Grinds hard stones to meal;&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Once the Mining (referring the dwarfs mine place) place that could&#xA; earn and keep itself busy would stop one day - &lt;strong&gt;TIME can change&#xA; anything&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Slays king, ruins town,&#xA;And beats high mountain down.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the kings and towns which are once wealthy will have to be abandoned&#xA; and they are destroyed not even mountains are left  over time they&#xA; have to change - &lt;strong&gt;TIME matters Most&lt;/strong&gt;.&#xA;  No matter what comes on its way , it will never stop, everybody has&#xA; to surrender under one that is &lt;strong&gt;&quot;TIME&quot;&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13373" LastEditorUserId="12628" LastEditDate="2015-06-23T14:59:43.263" LastActivityDate="2015-06-23T14:59:43.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16846" PostTypeId="1" AcceptedAnswerId="16868" CreationDate="2015-06-23T09:46:27.400" Score="7" ViewCount="4645" Body="&lt;p&gt;I'm trying to make a maze for a puzzle hunt, and I can't find any good software for creating them. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I seek an appropriate &lt;em&gt;simple drawing tool&lt;/em&gt;, not a program to automatically create the maze.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I just want to make something simple like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ug8e1m.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ug8e1m.jpg&quot; alt=&quot;maze&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but the trouble is that all of the software I can find is either absurdly complicated or it only allows you to set a couple parameters and generate a maze. Any suggestions?&lt;/p&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="5373" LastEditDate="2017-03-11T14:51:29.317" LastActivityDate="2021-06-09T10:53:20.993" Title="Good maze creation software?" Tags="|puzzle-creation|mazes|" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16847" PostTypeId="1" AcceptedAnswerId="16849" CreationDate="2015-06-23T09:48:49.410" Score="3" ViewCount="5080" Body="&lt;p&gt;A police officer saw a lorry driver going the wrong way down a one-way street, but did not make any move to stop him. Why not?&#xA;&lt;hr&gt;&#xA;Let's see if I can make this puzzle not &quot;too broad&quot;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;there's no emergency going on, nor any justification for breaking the traffic laws&lt;/li&gt;&#xA;&lt;li&gt;the police officer is not related to the lorry driver; there's no conflict of interest involved&lt;/li&gt;&#xA;&lt;li&gt;the police officer is perfectly cognizant of the fact that this is a one-way street and of the laws restricting passage along it&lt;/li&gt;&#xA;&lt;li&gt;the police officer is not physically restrained, occupied with arresting someone else, or otherwise incapable of stopping the lorry driver&lt;/li&gt;&#xA;&lt;li&gt;no alien intervention or supernatural activity is involved; this incident is perfectly plausible and could easily have happened today.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-06-23T14:11:48.513" Title="Why didn't the cop stop the lorry driver?" Tags="|lateral-thinking|situation|" AnswerCount="9" CommentCount="4" ClosedDate="2015-06-23T14:23:34.437" ContentLicense="CC BY-SA 3.0" />
  <row Id="16848" PostTypeId="2" ParentId="16847" CreationDate="2015-06-23T09:50:17.750" Score="8" Body="&lt;p&gt;My guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The lorry driver was reversing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user2674" LastEditorDisplayName="user2674" LastEditDate="2015-06-23T10:01:14.047" LastActivityDate="2015-06-23T10:01:14.047" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16849" PostTypeId="2" ParentId="16847" CreationDate="2015-06-23T09:52:43.367" Score="24" Body="&lt;p&gt;Quite simply,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The lorry driver is on foot.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12730" LastActivityDate="2015-06-23T09:52:43.367" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16850" PostTypeId="2" ParentId="16474" CreationDate="2015-06-23T09:56:53.210" Score="2" Body="&lt;p&gt;&lt;strong&gt;Puzzle 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Stackexchange Network&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the five best well rounded children are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The top stackexchange sites! &lt;br&gt;&#xA; Stack Overflow (9.6m questions)&lt;br&gt;&#xA; Mathematics (454k)&lt;br&gt;&#xA; Super User (270k)&lt;br&gt;&#xA; Server Fault (198K)&lt;br&gt;&#xA; Ask Ubuntu (193K)&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The best thing about the children are the &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Questions!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Adding them up gives&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 10.715 million!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user2674" LastEditorDisplayName="user2674" LastEditDate="2015-06-23T10:11:00.050" LastActivityDate="2015-06-23T10:11:00.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16851" PostTypeId="2" ParentId="16847" CreationDate="2015-06-23T10:01:44.377" Score="1" Body="&lt;p&gt;Maybe the cop&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; saw it on Television&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13324" LastActivityDate="2015-06-23T10:01:44.377" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16852" PostTypeId="2" ParentId="16847" CreationDate="2015-06-23T10:17:56.443" Score="6" Body="&lt;p&gt;Pretty sure Matt's got it, but another one that came to mind is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;A police officer saw a lorry driver &lt;strong&gt;[while the officer himself was]&lt;/strong&gt; going the wrong way down a one-way street, but did not make any move to stop him. Why not?&quot;  Why would he? The lorry driver was going the right way.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13197" LastActivityDate="2015-06-23T10:17:56.443" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16859" PostTypeId="2" ParentId="16747" CreationDate="2015-06-23T11:39:45.940" Score="5" Body="&lt;p&gt;So there seems to be &lt;strong&gt;4 letters in the password, and the common prefix is {255-36} &amp;amp; {255-32}.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the suffixes (2 per letter) , when converted to key presses, give 7,1 &amp;amp; 7,2 &amp;amp; 7,3 &amp;amp; 6,5. [[ REFER &lt;a href=&quot;http://www.win.tue.nl/~aeb/linux/kbd/scancodes-14.html&quot; rel=&quot;nofollow&quot;&gt;http://www.win.tue.nl/~aeb/linux/kbd/scancodes-14.html&lt;/a&gt; ]]&lt;br&gt;&#xA;Probably this means 4 letters 71 , 72 , 73 , 65.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now many encodings ( Windows-1252/ASCII/UTF/ISO-8859-1 ) all give give 71=G , 72=H , 73=I , 65=A.&lt;br&gt;&#xA;[[ OP said unicode is not involved ]]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Common prefix {255-36} &amp;amp; {255-32} , + GHIA.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this related to car maker GHIA ?&lt;br&gt;&#xA;Is this related to &quot;georghia&quot; ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;still thinking . . . .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the new clues by OP Kslkgh (and the partial answer by Michael Bradfield):&lt;br&gt;&#xA;Georgian word &quot;ღია&quot; (pronounced ghia) means OPEN.&lt;br&gt;&#xA;So we have 3 possible end-solutions:&lt;br&gt;&#xA;&lt;strong&gt;&quot;ghia&quot;&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;&quot;open&quot;&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;&quot;ღია&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see that the question mentions &quot;What does the paper mean in English?&quot;. So the English translation is most appropriate.&lt;br&gt;&#xA;So, the answer must be &lt;strong&gt;&quot;OPEN&quot;&lt;/strong&gt;, a command to the DOOR.&lt;/p&gt;&#xA;" OwnerUserId="10940" LastEditorUserId="10940" LastEditDate="2015-06-24T16:33:53.363" LastActivityDate="2015-06-24T16:33:53.363" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="16863" PostTypeId="2" ParentId="16847" CreationDate="2015-06-23T12:04:30.460" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The police officer is not in service, when he saw the lorry driver, he was just an ordinary pedestrian.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13479" LastEditorUserId="9000" LastEditDate="2015-06-23T12:11:11.080" LastActivityDate="2015-06-23T12:11:11.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16864" PostTypeId="2" ParentId="16847" CreationDate="2015-06-23T12:10:38.467" Score="1" Body="&lt;p&gt;I suspect that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The lorry driver was obeying a direct instruction from the policeman, which would make it legal. &lt;br&gt;Perhaps due to a diversion, works in the road or the marathon being run today.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13258" LastEditorUserId="9000" LastEditDate="2015-06-23T12:14:36.147" LastActivityDate="2015-06-23T12:14:36.147" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16867" PostTypeId="2" ParentId="16847" CreationDate="2015-06-23T12:23:30.597" Score="2" Body="&lt;p&gt;Depending on the local laws:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Lorry driver was on a bicycle, going the 'wrong way' (Possibly on the sidewalk).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alternately if possible:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The street is only a one-way during certain hours of the day. This event occurred while 2-way street rules applied.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastActivityDate="2015-06-23T12:23:30.597" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16868" PostTypeId="2" ParentId="16846" CreationDate="2015-06-23T12:41:55.690" Score="12" Body="&lt;p&gt;If you're just looking for a tool to help you manually lay out your own custom mazes, you can just use spreadsheet software such as Excel or OpenOffice Calc. Set the grid size so that the cells are squares, and then draw the borders in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Excel, if you switch to the Page Layout view you can set the row/column sizes in units such as cm, which will enable you to have a square grid. Then use Borders &gt; Draw Border.&lt;/p&gt;&#xA;" OwnerUserId="12462" LastActivityDate="2015-06-23T12:41:55.690" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16869" PostTypeId="2" ParentId="16847" CreationDate="2015-06-23T12:57:15.563" Score="5" Body="&lt;p&gt;Another guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The road was under construction or under repairs so it was closed and only &#xA; reachable by the people who worked here like the lorry driver (bringing bitumen&#xA;, materials,etc...) .&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Another police officer already stopped the lorry driver&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5359" LastActivityDate="2015-06-23T12:57:15.563" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16870" PostTypeId="2" ParentId="16702" CreationDate="2015-06-23T13:35:42.463" Score="4" Body="&lt;p&gt;I haven't read the book, but this might be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Prince Po from the book &lt;a href=&quot;https://en.wikipedia.org/wiki/Graceling&quot; rel=&quot;nofollow&quot;&gt;Graceling&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who am I?&#xA;I am but a panda.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He shares a name with the main character from the movie &quot;Kung Fu Panda&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I don’t see all black and white.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He is blinded during the story&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I’m of inception,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He can see into people's minds, similar to the movie &quot;Inception&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here Five and Eight is equal to two-ninths.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I didn't get this one, but Gracelyn Rioux has clarified that this means in the word &quot;Inception&quot;, the fifth and eight letters make up two ninths of the entire word, spelling &quot;Po&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I’m one of Gracelyn’s favorite book characters,&#xA;And I certainly do have a lot of Grace.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He's a book character and he has a &quot;Grace&quot;, which is capitalised in the wikipedia article similarly to how it is in the clue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13197" LastEditorUserId="13197" LastEditDate="2015-06-26T22:20:10.550" LastActivityDate="2015-06-26T22:20:10.550" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16872" PostTypeId="2" ParentId="16847" CreationDate="2015-06-23T13:48:10.223" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Manual transmission on a hill. The Truck became stopped due to traffic and was slowly rolling back down the hill before proceeding.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13342" LastActivityDate="2015-06-23T13:48:10.223" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16873" PostTypeId="2" ParentId="16846" CreationDate="2015-06-23T13:51:33.500" Score="2" Body="&lt;p&gt;This is similar to the Excel suggestion, but I use WinWord (if the maze is not too big.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 1) Insert a table of cells you want, i.e. 25 x 25. Use a fairly small &quot;fixed&quot; cell width. (i.e. 0.5cm)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/iyAG9.png&quot; alt=&quot;Add Table&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 2) Select table and properties and set row height to same as cell width. (0.5)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QC4BH.png&quot; alt=&quot;Table row height&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you have the perfect &quot;grid&quot; to work on, and you can use the &quot;Draw table&quot; tool to set the border to the lines you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sWBAS.png&quot; alt=&quot;Table&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/EvgkO.png&quot; alt=&quot;Tool&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tbGIU.png&quot; alt=&quot;Draw with tool&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5840" LastActivityDate="2015-06-23T13:51:33.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16878" PostTypeId="1" AcceptedAnswerId="16879" CreationDate="2015-06-23T16:21:21.653" Score="12" ViewCount="2425" Body="&lt;p&gt;I've thrown together a quick riddle inspired by the as-of-yet unsolved riddle from &lt;a href=&quot;https://puzzling.stackexchange.com/q/16838/9000&quot;&gt;CodeNewbie&lt;/a&gt;. I don't feel like it's very solid, though, and the cadence isn't quite right. It's easy to show that the right answer fits everything but I'm worried about two things:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Arriving at that right answer may not be an easily deduced answer. I like things you can &lt;em&gt;solve&lt;/em&gt; instead of things that require an epiphany.&lt;/li&gt;&#xA;&lt;li&gt;There could be an answer I haven't thought of that fits better&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Can you 1) Answer the riddle and / or 2) Once the answer is known, think of some suggestion for where it needs tightening up?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Edited based on suggestions)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We protect wealth and have wealth ourselves&lt;br&gt;&#xA;  My brothers do, and I&lt;br&gt;&#xA;  We share the name, the same name all&lt;br&gt;&#xA;  I swear I do not lie   &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;One is thoughtful, simple, kind&lt;br&gt;&#xA;  But rather slow at best&lt;br&gt;&#xA;  To ponder he must think, think, think&lt;br&gt;&#xA;  Under Sanders he will rest&lt;br&gt;&#xA;  He has a hard time hearing&lt;br&gt;&#xA;  When ears get all stuffed up&lt;br&gt;&#xA;  He was renamed at 1 year old&lt;br&gt;&#xA;  Causing some mixup   &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;One told secrets and angered some&lt;br&gt;&#xA;  He's fled now far away&lt;br&gt;&#xA;  Called a traitor; booed and hissed&lt;br&gt;&#xA;  It's best to stay away&lt;br&gt;&#xA;  Others think he is a hero&lt;br&gt;&#xA;  And wonder how he fares&lt;br&gt;&#xA;  His surname was in Heller's book&lt;br&gt;&#xA;  Confuses polar bears   &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;One tried to save his mother but&lt;br&gt;&#xA;  He failed quite miserably&lt;br&gt;&#xA;  It cost him an arm and leg&lt;br&gt;&#xA;  To fail so blatantly&lt;br&gt;&#xA;  This brother's brother is not my brother&lt;br&gt;&#xA;  Though he paid his own high price&lt;br&gt;&#xA;  He lost his body - saved his soul&lt;br&gt;&#xA;  Two brothers' sacrifice   &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Many brothers have been kings for sure&lt;br&gt;&#xA;  Thus far there are one through eight&lt;br&gt;&#xA;  There will sure one day be more&lt;br&gt;&#xA;  Of our royal ruling mates   &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;One was only one of us&lt;br&gt;&#xA;  When he was not himself&lt;br&gt;&#xA;  His formula of wondrous hide&lt;br&gt;&#xA;  Should go back on the shelf&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;One loved riddles, puzzles, jokes&lt;br&gt;&#xA;  Mentally deranged bloke&lt;br&gt;&#xA;  He wears green and laughs too much&lt;br&gt;&#xA;  Not proud of &lt;em&gt;all&lt;/em&gt; kinfolk...&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Train, vampire, and bladed hands&lt;br&gt;&#xA;  A platypus (to kids from oughts)&lt;br&gt;&#xA;  We are many myriad&lt;br&gt;&#xA;  Our name is what is sought  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2017-08-21T12:18:11.877" Title="What's our Name? (Also: improve my riddle)" Tags="|riddle|puzzle-creation|knowledge|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16879" PostTypeId="2" ParentId="16878" CreationDate="2015-06-23T16:37:47.690" Score="11" Body="&lt;p&gt;Is their name &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Edward&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We protect wealth and have wealth ourselves  &lt;br/&gt;&#xA;My brothers do, and I  &lt;br/&gt;&#xA;We share the name, the same name all  &lt;br/&gt;&#xA;I swear I do not lie  &lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This is explaining the riddle, in addition to Edward meaning &quot;Wealthy Guard&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One is thoughtful, simple, kind  &lt;br/&gt;&#xA;But rather slow at best  &lt;br/&gt;&#xA;To ponder he must think, think, think  &lt;br/&gt;&#xA;Under Sanders he will rest  &lt;br/&gt;&#xA;He has a hard time hearing  &lt;br/&gt;&#xA;When ears get all stuffed up  &lt;br/&gt;&#xA;He was renamed at 1 year old  &lt;br/&gt;&#xA;Causing some mixup  &lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Thanks to Engineer Toast's extra clues I now know this is Winnie the Pooh, originally his name was Edward Bear&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One told secrets and angered some  &lt;br/&gt;&#xA;He's fled now far away  &lt;br/&gt;&#xA;Called a traitor; booed and hissed  &lt;br/&gt;&#xA;It's best to stay away  &lt;br/&gt;&#xA;Others think he is a hero  &lt;br/&gt;&#xA;And wonder how he fares  &lt;br/&gt;&#xA;His surname was in Heller's book  &lt;br/&gt;&#xA;Confuses polar bears  &lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Edward Snowden I believe? Is Heller's book Catch-22?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One tried to save his mother but  &lt;br/&gt;&#xA;He failed quite miserably  &lt;br/&gt;&#xA;It cost him an arm and leg  &lt;br/&gt;&#xA;To fail so blatantly  &lt;br/&gt;&#xA;This brother's brother is not my brother  &lt;br/&gt;&#xA;Though he paid his own high price  &lt;br/&gt;&#xA;He lost his body - saved his soul  &lt;br/&gt;&#xA;Two brothers' sacrifice  &lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Edward Elric - from Full Metal Alchemist &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Many brothers have been kings for sure  &lt;br/&gt;&#xA;Thus far there are one through seven  &lt;br/&gt;&#xA;There will sure one day be more  &lt;br/&gt;&#xA;Of our royal ruling brethren  &lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This is referencing the British royal Family, Edward the 7th is the most recent King Edward&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One was only one of us  &lt;br/&gt;&#xA;When he was not himself  &lt;br/&gt;&#xA;One loved riddles, puzzles, jokes  &lt;br/&gt;&#xA;Mentally deranged  &lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I didn't realize this was two clues thanks for the clarification. first one is Dr. Jekyl/Mr. Hyde, second: The Riddler a.k.a Edward Nygma&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Train, vampire, &quot;monster&quot;, too  &lt;br/&gt;&#xA;A platypus to kids from oughts  &lt;br/&gt;&#xA;We are many myriad  &lt;br/&gt;&#xA;Tell us, what's our name?  &lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Train: Edward From Thomas the tank engine, Vampire: Edward Cullen from Twilight, Monster: Edward Scissor Hands (This clue was a but of a stretch), Platypus: Edward from Camp Lazlo&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13204" LastEditorUserId="13204" LastEditDate="2015-06-23T18:51:40.233" LastActivityDate="2015-06-23T18:51:40.233" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="16880" PostTypeId="2" ParentId="16842" CreationDate="2015-06-23T16:49:04.223" Score="4" Body="&lt;p&gt;I believe I have found the country in question, but I can't figure out the numbers just yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The country is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Norway &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Norwegians refer to Norway as &lt;em&gt;fedreland&lt;/em&gt; (refering to the title &quot;Cultural Patrimony&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nordic countries are known for their stunning sunsets. Because they are so far north, the sun can take an entire night to set and is considered to be one of the most beautiful sunsets in the world.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The western coast of Norway is full of very small islands that are usually connected with bridges or tunnels. Generally, the closer they are to the continent, the closer they are together as well, but some can actually be 150km from each other as written by the OP.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Folk music is the name given to a very old although still popular music genre (music that never dies) that is characteristic to Nordic countries. It mainly uses traditional instruments such as the accordion and the bagpipes. Norway considers non-formal education (work experience) as something to be valued. As seen in a referencedocument of the &lt;a href=&quot;http://www.nokut.no/Documents/NOKUT/Artikkelbibliotek/Norsk_utdanning/NKR/The_referencing_report.pdf&quot; rel=&quot;nofollow&quot;&gt;Norwegian Qualifications Framework&lt;/a&gt;. Take a look at the first few paragraphs of page 16.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Will update soon..&lt;/p&gt;&#xA;" OwnerUserId="13491" LastEditorUserId="13491" LastEditDate="2015-06-23T16:54:31.923" LastActivityDate="2015-06-23T16:54:31.923" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16881" PostTypeId="1" AcceptedAnswerId="24442" CreationDate="2015-06-23T17:31:17.433" Score="45" ViewCount="3154" Body="&lt;p&gt;This puzzle belongs to the puzzle series: &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=hyper-modern%20is%3Aquestion&quot;&gt;hyper-modern art&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The two friends in the gallery of Hyper-modern art are still discussing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;You know what, to certain extent I start enjoying this Hyper-modern art, but I still think it lacks something important.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;And what should that be, my friend?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Beauty. Aesthetics. It is all well and good that the observer has to fully engage for deeper understanding, but I still would like to, well, like an image. It should be pleasing to watch.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Hmm, yes, I get your point. But maybe the next example is more to your liking.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They move on to the forth room of the gallery, where a gigantic painting of 11x16 metre fills the whole side of the room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;See what we have here: The painting is called &lt;code&gt;'Nature tells a story'&lt;/code&gt;, and I think it is rather pleasing, don't you?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Well, it for sure is impressive... And yes, it is also nice. But what is the story nature wants to tell us here?&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;See, now we need our HUD again...&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/huyKU.png&quot; alt=&quot;Overview image&quot;&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/VoIWl.jpg&quot; rel=&quot;noreferrer&quot;&gt;Full resolution image, 1.3Mb, 6582 x 4539 pixels&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The goal of the puzzle is to find the message, this painting is telling us. This message is written in English and consists of five words. The puzzle will likely require some patience and &lt;em&gt;work&lt;/em&gt;, and it does require the full-resolution image provided by the link.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While the full resolution is needed to see all details, the puzzle does not require any digital information. You can print the image and still solve it from that.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bonus: An additional message is hidden in the image. This message is a clue for the actual puzzle. What is the message?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Hints&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You need a hint? Count!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You think the flying direction of butterflies is random? Think again!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;A&lt;/em&gt; single butterfly (in one direction). But two?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Have you taken a &lt;em&gt;close look&lt;/em&gt; at hint 3 above?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-12-09T20:44:20.530" Title="Lepidopterology needed!" Tags="|cipher|visual|enigmatic-puzzle|steganography|" AnswerCount="3" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="16882" PostTypeId="2" ParentId="16756" CreationDate="2015-06-23T18:40:31.513" Score="1" Body="&lt;p&gt;Just a little guess here. The letter &quot;G&quot; in the English Alphabet is number 7. Therefore I thought somehow 7 would be in that equation. I then thought what if you take each of the numbers and plug it into the pyramid. Anyhow this is what I have come up with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}&#xA;&amp;amp;&amp;amp;&amp;amp;&amp;amp;7\\&#xA;&amp;amp;&amp;amp;4&amp;amp;9&amp;amp;4\\&#xA;&amp;amp;1&amp;amp;2&amp;amp;8&amp;amp;5&amp;amp;5\\&#xA;5&amp;amp;6&amp;amp;7&amp;amp;1&amp;amp;5&amp;amp;5&amp;amp;2&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This probably means perfectly nothing, but I thought I would throw it out there anyhow. 7 is at the top because the arrow is pointing there, and I thought maybe that is where &quot;G&quot; goes, AKA 7th letter of the alphabet. &lt;/p&gt;&#xA;" OwnerUserId="13496" LastEditorUserId="13494" LastEditDate="2015-06-23T19:43:23.740" LastActivityDate="2015-06-23T19:43:23.740" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16883" PostTypeId="2" ParentId="16756" CreationDate="2015-06-23T18:48:22.770" Score="11" Body="&lt;p&gt;Okay, so first I&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; translated the binary hole punches on the left. From top to bottom the dots give: &lt;code&gt;010000100100100101001110010000010101001001011001&lt;/code&gt; which one can translate directly into 8-bit ASCI codes for 6 letters, spelling out: &quot;BINARY&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With this it was obvious &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; that the next line should be read as binary as well, with 494128555671552 = &lt;code&gt;1110000010110100001000101000010001100000000000000&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then, I tried out the example with a small pyramid and realized that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if I &quot;dropped&quot; the binary in, in order from most significant bit to least, I would get this:&lt;img src=&quot;https://puu.sh/izZe8/d616c31b0a.png&quot; alt=&quot;binary&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The rules were:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Drop each bit in from the top. They fall straight down; if the space directly under them is full they go down and right, and if that space is taken they go down and left. If all three are taken, they stay in place. (It helps to visualize it as sand falling down in an hourglass.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So just using the colors by themselves produced this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://puu.sh/iA0Fo/b8de5cd836.png&quot; alt=&quot;colors&quot;&gt;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is not useful. However, thanks to Quark we already know&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; that the colour bar could be interpreted as binary string as well. Using RGB flags, i.e. R=(100) G=(010) B=(001) and accordingly for mixed colours. The binary code of the colour stripe is:&lt;br&gt;&#xA; &lt;code&gt;100010010110000100101100000110001010100010010100001101000010100000010010110100110101010011001010001011010000010101000101000101010101001000101000101100110100100110100110101001000101010100000100001110101000110000001100001000001011100000010000000010000100000100001000001000010001000000100001000011000000000000000010000000000000&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using this and the general rules give:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/bCQRn.png&quot; alt=&quot;Solution&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally, one can notice that &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I love you ever&quot; is not really a proper English sentence, but looking closely there are 4 pixels between &quot;you&quot; and &quot;ever&quot;, so the final message is &lt;code&gt;I love you forever&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:48.787" LastActivityDate="2015-06-24T11:18:16.623" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="16884" PostTypeId="2" ParentId="16474" CreationDate="2015-06-23T19:44:24.690" Score="9" Body="&lt;h2&gt;Puzzle 2&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I'll have a crack at this as well, following &lt;a href=&quot;https://puzzling.stackexchange.com/a/16832/5373&quot;&gt;TroyAndAbed's idea&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The best of the best! Your home contains the answer. For in it, you'll find five children, children of great number; well-rounded children. You'll take the best of the children (make sure you have five!) and add them up.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I interpret this as follows.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Your home&quot; is &lt;a href=&quot;http://puzzling.stackexchange.com&quot;&gt;Puzzling.SE&lt;/a&gt;. The &quot;five children&quot; are the five highest-voted questions. As of today these are as follows:&#xA;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/hn6i9.jpg&quot; alt=&quot;screenshot&quot;&gt;&#xA;&#xA; &quot;Well-rounded&quot; means we round the numbers of votes say to the nearest 10 (this is a good idea since those numbers are going to change over time). Then the top 5 questions have scores of 110, 80, 80, 75 (let's say round this up), and 70. The total is &lt;strong&gt;420&lt;/strong&gt;, the final answer (coincidentally the answer to Life, the Universe, and Everything with an extra 0 on the end).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2015-06-23T20:38:47.560" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16885" PostTypeId="1" AcceptedAnswerId="16888" CreationDate="2015-06-23T19:48:20.627" Score="8" ViewCount="918" Body="&lt;p&gt;There are five ghost pirate ships drifting on an infinite ocean in the spiritual realm. Each ship moves at a constant speed and never changes direction. &lt;strong&gt;No two ships are traveling in parallel paths.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These ships have been drifting for a very long time. In fact, time extends infinitely both to the future &lt;em&gt;and the past&lt;/em&gt;, and these ships have drifted for all of history.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because they are ghosts, two pirate ships can pass right through one another. However, such collisions release a huge amount of spiritual energy. For every pair of ships that collide, one trillion souls are destroyed. Since the there are only 10 trillion souls in the universe, these ships could destroy all life if every pair of them collided! &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Eight pairs of ships have already collided: are the 2 trillion remaining souls doomed?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Think of the ships as point masses. If 4 ships meet at the same time, this would destroy 6 million souls, since 6 &lt;em&gt;pairs&lt;/em&gt; of ships collided.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-06-24T02:00:48.987" Title="Ghost Ship Collisions" Tags="|geometry|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16886" PostTypeId="2" ParentId="16833" CreationDate="2015-06-23T19:54:37.390" Score="0" Body="&lt;p&gt;There is no treasure.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Loonies are a Canadian coin introduced in 1987, certainly long after the supposed ship wreck.  This is, indeed, a wild goose chase.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1682" LastActivityDate="2015-06-23T19:54:37.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16887" PostTypeId="1" AcceptedAnswerId="16942" CreationDate="2015-06-23T20:13:23.080" Score="11" ViewCount="629" Body="&lt;p&gt;It's been a while since we have seen one of these (as originally created by &lt;a href=&quot;https://puzzling.stackexchange.com/users/10940/prem&quot;&gt;Prem&lt;/a&gt;), and I thought people might enjoy another one. (See &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15339/four-words&quot;&gt;the original&lt;/a&gt; for the solution format.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first is the conclusion of all.&lt;br&gt;&#xA;My second is the beginning of all.&lt;br&gt;&#xA;My third is the requirement of all.&lt;br&gt;&#xA;My fourth is the ceasing of all.  &lt;/p&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-07-17T00:50:47.970" Title="Another Four-words (----||||)" Tags="|word|wordplay|crosswords|word-square|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16888" PostTypeId="2" ParentId="16885" CreationDate="2015-06-23T20:22:35.727" Score="11" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; no, it is possible that the 2 trillion remaining souls will survive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can view the path of a pirate ship as a line in three-dimensional space (where the first two dimensions are spacial dimensions along the ocean, and the third dimension is time). Two pirate ships collide if and only if the corresponding lines intersect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The given question is equivalent to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose we are given five straight lines in three-dimensional space, with no two lines parallel. Given that at least $8$ pairs of lines intersect, must all $10$ pairs of lines intersect?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The benefit of the equivalent question is that it is easier to visualize a counterexample. Consider a tetrahedron with one edge removed. This figure has five edges, which can be extended into five lines. We can count that exactly eight pairs of these lines intersect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2oW9j.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-06-23T20:31:46.577" LastActivityDate="2015-06-23T20:31:46.577" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16889" PostTypeId="1" AcceptedAnswerId="16892" CreationDate="2015-06-23T20:51:26.307" Score="8" ViewCount="10164" Body="&lt;p&gt;Here is another classic that I couldn't find anywhere on the site:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A man is lying dead in a field. Next to him there is an unopened package. There is no other creature in the field. How did he die?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My notes to try and avoid this being too broad:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The man died where he lies. He was not moved to this spot. (He wasn't shot, stabbed, poisoned, etc.) &#xA; Nothing happened earlier in the day to cause this death to be expected.&lt;br&gt;&#xA; He did not die of natural causes, old age, etc.&lt;br&gt;&#xA; He was not killed by lightning, or anything weather related.&lt;br&gt;&#xA; He was not killed by an act of God (lightning, earthquake, forest fire, etc.)&lt;br&gt;&#xA; There are no other objects of note anywhere in the field, and no objects have been removed or added since his death.&lt;br&gt;&#xA; There are no tire tracks or other signs of another human being around him.&lt;br&gt;&#xA; The man was not killed by any creature (snake, spider, etc.) that then moved (walked, slithered, flew) away.&lt;br&gt;&#xA; The man did not kill himself.&lt;br&gt;&#xA; This happened in the typical world, with regular laws of physics, a depressing lack of magic, and (sadly) no aliens.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After the first round of answers finding loopholes, a couple more notes. To those that this eliminates your answer, thank you for helping me find those loopholes:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The man's death was instant (no time for anything to leave the field before he died)&lt;br&gt;&#xA; The man was not doing anything illegal.&lt;br&gt;&#xA; The contents of the package did not cause his death.&lt;br&gt;&#xA; There was nothing growing or living in, or recently added to the field that caused his death. The field is not toxic.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="9951" LastEditDate="2015-06-23T21:56:56.323" LastActivityDate="2015-06-23T22:42:10.317" Title="Dead in a Field" Tags="|logical-deduction|lateral-thinking|situation|" AnswerCount="10" CommentCount="13" ClosedDate="2015-06-23T22:12:53.390" ContentLicense="CC BY-SA 3.0" />
  <row Id="16890" PostTypeId="2" ParentId="16878" CreationDate="2015-06-23T20:59:09.513" Score="4" Body="&lt;p&gt;Here are my suggestions for tightening up the cadence and flow:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;We shelter wealth and own it too&lt;br /&gt;&#xA;My brethren and I&lt;br /&gt;&#xA;We share a name, the same name all&lt;br /&gt;&#xA;I swear I do not lie&lt;/p&gt;&#xA;&lt;p&gt;One is thoughtful, simple, kind&lt;br /&gt;&#xA;But rather slow at best&lt;br /&gt;&#xA;To ponder he must think, think, think&lt;br /&gt;&#xA;'Neath Sanders he will rest&lt;br /&gt;&#xA;He has a hard time hearing&lt;br /&gt;&#xA;When his ears get all stuffed up&lt;br /&gt;&#xA;His name was changed at 1 year old&lt;br /&gt;&#xA;Which muddled some things up&lt;/p&gt;&#xA;&lt;p&gt;By telling secrets, one drew ire&lt;br /&gt;&#xA;He's fled now far away&lt;br /&gt;&#xA;Booed and hissed; a traitor named&lt;br /&gt;&#xA;It's best to stay away&lt;br /&gt;&#xA;To others, though, he's quite the star&lt;br /&gt;&#xA;They wonder how he fares&lt;br /&gt;&#xA;His surname, found in Heller's book,&lt;br /&gt;&#xA;Confuses polar bears&lt;/p&gt;&#xA;&lt;p&gt;One tried to save his mother but&lt;br /&gt;&#xA;He failed quite mis'rably&lt;br /&gt;&#xA;It cost him both an arm and leg&lt;br /&gt;&#xA;To fail so blatantly&lt;br /&gt;&#xA;This brother's brother's not my kin&lt;br /&gt;&#xA;Though he paid quite a price&lt;br /&gt;&#xA;His body lost, but saved his soul&lt;br /&gt;&#xA;Two brothers' sacrifice&lt;/p&gt;&#xA;&lt;p&gt;Many brothers have been kings&lt;br /&gt;&#xA;To date we've one through eight&lt;br /&gt;&#xA;There surely will one day be more&lt;br /&gt;&#xA;Majestic ruling mates&lt;/p&gt;&#xA;&lt;p&gt;One was only one of us&lt;br /&gt;&#xA;When he was not himself&lt;br /&gt;&#xA;His brew to hide his ugly trait&lt;br /&gt;&#xA;Should go back on the shelf&lt;/p&gt;&#xA;&lt;p&gt;One loves riddles, puzzles, jokes&lt;br /&gt;&#xA;He's mentally deranged&lt;br /&gt;&#xA;With verdant garb and jovial air&lt;br /&gt;&#xA;He plays his threat'ning games&lt;/p&gt;&#xA;&lt;p&gt;A train, a vampire, hands with blades&lt;br /&gt;&#xA;A platypus (to kids from oughts)&lt;br /&gt;&#xA;We are sundry and diverse&lt;br /&gt;&#xA;Our name is what is sought&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-06-23T20:59:09.513" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16891" PostTypeId="2" ParentId="16889" CreationDate="2015-06-23T20:59:28.743" Score="6" Body="&lt;p&gt;The object was dropped from the sky (plane/helicopter) and killed the man. It then fell next to him where he lay dead.&lt;/p&gt;&#xA;" OwnerUserId="13495" LastActivityDate="2015-06-23T20:59:28.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16892" PostTypeId="2" ParentId="16889" CreationDate="2015-06-23T21:01:05.987" Score="8" Body="&lt;p&gt;The man had&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; jumped out of a plane.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The package is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; his parachute, which had failed to open.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-06-23T21:01:05.987" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="16893" PostTypeId="2" ParentId="16889" CreationDate="2015-06-23T21:06:23.897" Score="4" Body="&lt;p&gt;The man had  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; heart stroke / heart attack  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The unopen package next to him is  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Drug / medicine for his heart stroke / heart attack that he fails to open in time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2015-06-23T21:06:23.897" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16894" PostTypeId="2" ParentId="16889" CreationDate="2015-06-23T21:07:59.297" Score="4" Body="&lt;p&gt;The man is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a worker at a nuclear power station, who stole some plutonium and ran away.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The package is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the plutonium he stole.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He died of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; radiation poisoning.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-06-23T21:07:59.297" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16895" PostTypeId="2" ParentId="16889" CreationDate="2015-06-23T21:12:54.557" Score="5" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He was highly allergic to the plants in the field. The package had nothing to do with it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He was highly allergic to whatever was in the package. He didn't know what was inside, so he carried it around for a while, then died.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="12109" LastActivityDate="2015-06-23T21:12:54.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16896" PostTypeId="2" ParentId="16889" CreationDate="2015-06-23T21:19:27.930" Score="4" Body="&lt;p&gt;You've gone to great pains to explain that he wasn't killed by a &lt;em&gt;human&lt;/em&gt;, nor was a &lt;em&gt;creature&lt;/em&gt; involved in his death, nor an &lt;em&gt;alien&lt;/em&gt; or a &lt;em&gt;god&lt;/em&gt;, nor did he kill &lt;em&gt;himself&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, I propose that he was hit on the head by &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a robot&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4852" LastActivityDate="2015-06-23T21:19:27.930" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16897" PostTypeId="2" ParentId="16889" CreationDate="2015-06-23T21:33:05.380" Score="5" Body="&lt;p&gt;Taking advantage of the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16889/dead-in-a-field/16896#comment52761_16896&quot;&gt;loophole found by @Richard&lt;/a&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The man was&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; decapitated by a boomerang, which left his body and the field while he was still dying (thus not &quot;since his death&quot;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The package is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a present for his Aunt Fanny, and irrelevant to his death.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-06-23T21:33:05.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16898" PostTypeId="2" ParentId="16889" CreationDate="2015-06-23T21:43:33.943" Score="1" Body="&lt;p&gt;The unopened package is a prophylactic.&#xA;The man had a heart attack while trying to open it, but even then, his partner never arrived, so his death, while not technically being of old age, is technically related to him being of unhealthy heart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember everyone: Heart disease is the number one killer, please take care of your body!&lt;/p&gt;&#xA;" OwnerUserId="13068" LastActivityDate="2015-06-23T21:43:33.943" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16900" PostTypeId="2" ParentId="16889" CreationDate="2015-06-23T21:47:07.090" Score="1" Body="&lt;p&gt;Clearly if there are no other living things in the field, then the field itself must be highly toxic. Perhaps the farmer had forgotten to dilute his pesticide and delivered a fatal dose of the stuff to every square inch of the field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The package contains a peanut butter sandwich and an apple, which the man had been planning to eat for lunch before he succumbed to the pesticide's neurotoxic effects.&lt;/p&gt;&#xA;" OwnerUserId="40" LastActivityDate="2015-06-23T21:47:07.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16901" PostTypeId="2" ParentId="16889" CreationDate="2015-06-23T21:52:29.227" Score="0" Body="&lt;p&gt;Given the new restrictions, it seems obvious that he was accidentally killed during the... &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...testing of a &lt;a href=&quot;https://bbccupholder.files.wordpress.com/2008/11/ads-hummer.jpg?w=300&amp;amp;h=278&quot; rel=&quot;nofollow&quot;&gt;giant microwave transmitter&lt;/a&gt; in the next field.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that his death would indeed be instant, his death would not be anticipated and that nothing would leave the field afterwards.&lt;/p&gt;&#xA;" OwnerUserId="4852" LastActivityDate="2015-06-23T21:52:29.227" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16902" PostTypeId="1" AcceptedAnswerId="48822" CreationDate="2015-06-23T22:18:04.217" Score="2" ViewCount="454" Body="&lt;p&gt;&lt;strong&gt;The three-pan balance&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a balance with not two, but three pans. Weighings using the balance follow these rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If there exists a pan that is lighter than each of the other two pans, then this pan goes up and the other two pans go down to a stop. (Note that one cannot see which of the two heavier pans, if any, is the heaviest.)&lt;/li&gt;&#xA;&lt;li&gt;If there is no single lightest pan, then nothing happens. (This includes the case of two equally light pans and one heavier pan.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Let's call this the &quot;lightest-pan-detection-rule&quot; (LPDR).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The problem&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are given, one after the other, three heaps of &lt;strong&gt;n&lt;/strong&gt; identical looking balls. (3n balls in total.) The first two heaps each contain &lt;strong&gt;one heavier&lt;/strong&gt; ball. The third heap only contains normal balls. Your task is to find the two heavy balls using a &lt;em&gt;3-pan balance&lt;/em&gt;. How many weighings are required to identify the two heavy balls without needing to guess?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The normal (non-heavy) balls are all of the same weight. The heavy balls may or may not have the same weight. You should not assume that they are only slightly heavier, i.e. your solution should work even if one heavy ball is twice as heavy as a normal ball.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may weigh only the given balls, and you may put an arbitrary and possibly different number of balls on each pan. Please present a method to identify the heavy balls, and explain why the number of weighings is minimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem arises when you try to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16006/a-balance-with-three-pans-detecting-the-lightest-pan-find-the-two-heavier-ball&quot;&gt;identify two heavy balls among 3n balls&lt;/a&gt;, and put &lt;em&gt;n&lt;/em&gt; balls on each pan, and find that one pan is the lightest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we saw in a &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16046/a-balance-with-three-pans-detecting-the-lightest-pan-find-the-one-lighter-heav&quot;&gt;previous puzzle&lt;/a&gt;, when you want to find the one heavy ball in one of our n-ball heaps alone, for &lt;em&gt;n&lt;/em&gt; &gt; 3, you can do it in no more than $\lceil\log_3(n+2)+1\rceil$  weighings. So naively, you can solve the problem in twice that many weighings. Can you do better, by utilizing balls from both heaps, and the extra normal-only heap?&lt;/p&gt;&#xA;" OwnerUserId="318" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2017-02-06T21:55:45.187" Title="Two heaps each with a heavy ball, and a 3-pan balance" Tags="|weighing|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16903" PostTypeId="2" ParentId="16834" CreationDate="2015-06-23T23:23:48.113" Score="10" Body="&lt;p&gt;I think the below picture is your intended solution where red dots are where the fuses are lit and black where they die out. however, the error is on the pink dot because at that point 2 sparks need to merge but they arive there on different times. Nice puzzle though! Kept me quite busy :)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/FuvaS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-06-23T23:23:48.113" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16904" PostTypeId="2" ParentId="12893" CreationDate="2015-06-23T23:41:22.420" Score="0" Body="&lt;p&gt;It actually is possible. The field becomes semimagnetic if and only if the two ships come at each other head on. Some of the intelligences escape, but that's standard for capturing swarms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like this, the two halves become whole. Simple geometry.&lt;/p&gt;&#xA;" OwnerUserId="12506" LastActivityDate="2015-06-23T23:41:22.420" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16906" PostTypeId="1" AcceptedAnswerId="16913" CreationDate="2015-06-24T00:14:30.240" Score="9" ViewCount="1086" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;My first has green and yellow&lt;br&gt;&#xA;  To take you there and back.&lt;br&gt;&#xA;  My second is made up of three,&lt;br&gt;&#xA;  A king, a queen, and just one more.&lt;br&gt;&#xA;  My third is all about a message,&lt;br&gt;&#xA;  Conveyed in the nick of time.&lt;br&gt;&#xA;  My fourth is north of rain around,&lt;br&gt;&#xA;  A king in the making there.&lt;br&gt;&#xA;  My fifth is completely out at sea,&lt;br&gt;&#xA;  Travelling to the sunrise bright.&lt;br&gt;&#xA;  My sixth has giants and a seat,&lt;br&gt;&#xA;  And takes place in the north.&lt;br&gt;&#xA;  My seventh is a tale of war,&lt;br&gt;&#xA;  And tragic in the telling.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I? Please explain how your answer fits each line of the riddle.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="11244" LastEditDate="2015-06-24T14:08:18.073" LastActivityDate="2015-07-02T19:19:01.010" Title="Riddle for an rainy day" Tags="|riddle|knowledge|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16907" PostTypeId="2" ParentId="16838" CreationDate="2015-06-24T00:25:26.590" Score="1" Body="&lt;p&gt;My guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Henry&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One fell in love with his music system&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Henry Kloss developed the first portable music system.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another led the opposers in Queensland&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Patrick Henry against the British Army (Queen's Land)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One played the man playing a trick&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This one, I cannot figure out. Best I've got is King Henry IV (the play). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The last lost his love by his own hand&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;King Henry V killed his own wives.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4353" LastActivityDate="2015-06-24T00:25:26.590" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16908" PostTypeId="2" ParentId="16885" CreationDate="2015-06-24T02:00:48.987" Score="1" Body="&lt;p&gt;Not very mathematical I know, but is this wrong?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No they're not doomed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;time extends infinitely both to the future and the past, and these ships have drifted for all of history&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They have &lt;em&gt;already&lt;/em&gt; drifted for an &lt;strong&gt;infinite&lt;/strong&gt; amount of time without colliding so it is already demonstrated—not theoretical—that they could continue to survive for an infinite future.&lt;/p&gt;&#xA;" OwnerUserId="13509" LastActivityDate="2015-06-24T02:00:48.987" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16909" PostTypeId="1" AcceptedAnswerId="16911" CreationDate="2015-06-24T04:26:19.803" Score="15" ViewCount="838" Body="&lt;p&gt;The rules of ChessTron are as follows.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Alice picks a chess piece (not the queen, as it is considered too powerful), and places it in the corner of a standard 8x8 chessboard.&lt;/li&gt;&#xA;&lt;li&gt;Bob picks a chess piece (again not the queen, but could be the same as Alice's choice), and places it in the opposite corner.&lt;/li&gt;&#xA;&lt;li&gt;Players alternate moving their pieces starting with Alice.  Capturing is not allowed.  Every time a piece leaves or moves through a square, the square is destroyed, and cannot be moved to or through again in the game.&lt;/li&gt;&#xA;&lt;li&gt;If a player is &quot;trapped&quot; without a legal move, that person loses the game.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Who wins and why?&lt;/p&gt;&#xA;" OwnerUserId="2722" LastActivityDate="2015-06-24T18:02:40.387" Title="A Game of ChessTron" Tags="|strategy|chess|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16911" PostTypeId="2" ParentId="16909" CreationDate="2015-06-24T05:05:30.307" Score="15" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bob wins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm assuming the eligible pieces are the knight, the bishop, the rook, and the king.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If Alice picks the knight, the rook, or the king, Bob can win by picking the same piece.  On each of his turns, Bob can play the move&#xA; that is the 180 degree rotation of the move Alice plays.  This works&#xA; because the starting position of the board is rotationally symmetric,&#xA; so after Alice's first move, Bob can make the 180 degree rotated move.&#xA; This puts the board back into a symmetric position (including&#xA; forbidden squares), allowing Bob to mimic Alice's next move in the same way.  Eventually, Alice will run out of moves.&#xA; &lt;br/&gt;&#xA; &lt;br/&gt;&#xA; This strategy will not work if Alice picks the bishop.  If Bob also picks the bishop, then on Alice's first move she can move her bishop&#xA; across the diagonal and place it next to Bob's bishop, preventing Bob&#xA; from moving at all on his turn.&#xA; &lt;br/&gt; &lt;br/&gt;&#xA; If Alice picks the bishop, Bob can win by picking the king.  Assume that the players put their pieces on the black diagonal.  Bob's strategy is to advance the king along the black diagonal toward the opposite&#xA; corner until this is no longer possible.  At this point, Alice's bishop becomes trapped one one side of the board, while Bob's king can take advantage of all the black squares on the other side, plus all of the white squares, which Alice cannot move to.  Alice will run out of moves before Bob does.&lt;br&gt;&#xA; &lt;br/&gt;&#xA; A note about the rook and bishop cases: Alice is able to thwart the symmetry strategy in the bishop case because she can move multiple squares directly toward Bob, moving through the square Bob needs to occupy.  For example, say Alice's bishop starts on a1, and Bob's on h8.  If Alice moves to g7, Bob's symmetric move is to b2.  However, Alice moved through b2 to get to g7, so Bob can't move there.  This cannot happen in the rook case.  In order for Alice to move her rook directly toward Bob's, the two rooks must occupy the same rank or same file on Alice's turn.  However, Bob's strategy ensures that the board is always rotationally symmetric at the beginning of Alice's turn, and no position with the rooks in the same rank or same file is rotationally symmetric.&#xA; &lt;br/&gt; &lt;br/&gt;&#xA; The symmetric king case does not fall into the same trap as the bishop case, either.  Although Alice can move directly toward Bob along the diagonal, she can only move one square at a time. In any rotationally symmetric position with the two kings on the same diagonal, there are an even number of squares between them, so if Alice moves closer to Bob's king, Bob will be able to make the corresponding move toward Alice's king.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10261" LastEditorUserId="10261" LastEditDate="2015-06-24T18:02:40.387" LastActivityDate="2015-06-24T18:02:40.387" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16912" PostTypeId="2" ParentId="16747" CreationDate="2015-06-24T07:54:10.487" Score="1" Body="&lt;p&gt;Prem has &lt;a href=&quot;https://puzzling.stackexchange.com/a/16859/5373&quot;&gt;got&lt;/a&gt; as far as the word &quot;ghia&quot; from the given cryptograms, and&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;open&lt;/strong&gt; = ghia in the Georgian language&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;... is this the answer?&lt;/p&gt;&#xA;" OwnerUserId="13517" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-06-24T08:48:14.603" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16913" PostTypeId="2" ParentId="16906" CreationDate="2015-06-24T08:07:54.813" Score="14" Body="&lt;p&gt;Is it this?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are discussing the Chronicles of Narnia, in the chronological order of the books.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My first has green and yellow&lt;br&gt;&#xA;To take you there and back.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/The_Magician%27s_Nephew&quot; rel=&quot;nofollow&quot;&gt;The Magician's Nephew&lt;/a&gt;&lt;br&gt; Yellow rings were used to travel to the magical world of Narnia and green rings to return.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My second is made up of three,&lt;br&gt;&#xA;A king, a queen, and just one more.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/The_Lion,_the_Witch_and_the_Wardrobe&quot; rel=&quot;nofollow&quot;&gt;The Lion, The Witch And The Wardrobe&lt;/a&gt;&lt;br&gt; The lion (Aslan) is the &lt;strike&gt;rightful king of Narnia&lt;/strike&gt; king of the jungle, Jadis is the White Witch who pretends to be the queen and one more points to other object in the title, which is the wardrobe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My third is all about a message,&lt;br&gt;&#xA;Conveyed in the nick of time.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/The_Horse_and_His_Boy&quot; rel=&quot;nofollow&quot;&gt;The Horse And His Boy&lt;/a&gt;&lt;br&gt; A lost prince is rediscovered and it's his timely warnings that prevent an invasion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fourth is north of rain around,&lt;br&gt;&#xA;A king in the making there.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Prince_Caspian&quot; rel=&quot;nofollow&quot;&gt;Prince Caspian&lt;/a&gt;&lt;br&gt; Prince Caspian is the king in the making.&lt;br&gt; EDIT: As per rand's comments, &lt;del&gt;&quot;north of rain around&quot; again seeks to sneak in the name of Narnia in anagram form.&lt;/del&gt; &quot;rain around&quot; is a clue to find an anagram for rain, which is the country Iran. And the Caspian Sea lies to the north of Iran.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fifth is completely out at sea,&lt;br&gt;&#xA;Travelling to the sunrise bright.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/The_Voyage_of_the_Dawn_Treader&quot; rel=&quot;nofollow&quot;&gt;The Voyage Of The Dawn Treader&lt;/a&gt;&lt;br&gt; Prince Caspian sets out on a voyage by sea to find the seven lost lords of Narnia.&lt;br&gt;EDIT: The sunrise bright is a reference to the titular ship, Dawn Treader.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My sixth has giants and a seat,&lt;br&gt;&#xA;And takes place in the north.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/The_Silver_Chair&quot; rel=&quot;nofollow&quot;&gt;The Silver Chair&lt;/a&gt;&lt;br&gt; A journey is taken to the giant lands to the north of Narnia. The seat is the titular silver chair.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My seventh is a tale of war,&lt;br&gt;&#xA;And tragic in the telling.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/The_Last_Battle&quot; rel=&quot;nofollow&quot;&gt;The Last Battle&lt;/a&gt;&lt;br&gt;A huge battle takes place that ends in the destruction (and re-birth) of Narnia.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The title is misspelt intentionally&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; because the letters '..an rain..' can be rearranged to form Narnia.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To conclude, I would like to apologize &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; since I haven't personally read the series, so please feel free to correct my explanations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="5373" LastEditDate="2015-07-02T19:19:01.010" LastActivityDate="2015-07-02T19:19:01.010" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="16914" PostTypeId="1" AcceptedAnswerId="16916" CreationDate="2015-06-24T08:28:54.320" Score="17" ViewCount="3920" Body="&lt;p&gt;One day I got a paper from my courtyard with the center portion missing. Here is what the poster looks like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8li9z.jpg&quot; alt=&quot;pattern with center portion missing&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have made some pieces of paper so that I can stick one to the center of the poster. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Here are the pieces.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JJPrI.jpg&quot; alt=&quot;pieces that would fit the missing center&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Help me find the best one that will suit the poster. Also, let me know why it will be the best. I will approve the answer with most appropriate justification.&lt;/p&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="2730" LastEditDate="2015-12-15T07:26:24.413" LastActivityDate="2015-12-15T07:26:24.413" Title="Find the missing piece" Tags="|logical-deduction|pattern|visual|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16915" PostTypeId="2" ParentId="16838" CreationDate="2015-06-24T08:37:39.157" Score="4" Body="&lt;p&gt;I'm going to try this one, even if I'm not convinced:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Edward&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't have great explanations, though:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One fell in love with his music system&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strike&gt;Edward Williams created the Soundbeam music system.&lt;/strike&gt; Based on the other hint in the comments, this is actually Edward Maya, because of his song called &quot;Stereo Love&quot; (definitely not my genre, I would have never found it out).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another led the opposers in Queensland&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strike&gt;Edward Hand, maybe? He's been a general in the Continental Army during the Revolutionary War. I thought of Edward Braddock first, but I find it hard to say he &quot;led the opposers&quot;.&lt;/strike&gt; Ed Milliband comes to mind, he was leader of the opposition from 2010 to 2015.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One played the man playing a trick&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well, Edward Norton in The Illusionist.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The last lost his love by his own hand&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This made me think of Edward Scissorhands, even if he didn't exactly &quot;lost his love by his hand&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And about the question title:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strike&gt;apparently at least two rodent species are named like that: Edward's swamp rat and Edwards's long-tailed giant rat.&lt;/strike&gt; Probably Edward Snowden, who leaked information from the NSA.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5990" LastEditorUserId="5990" LastEditDate="2015-06-24T09:39:53.223" LastActivityDate="2015-06-24T09:39:53.223" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16916" PostTypeId="2" ParentId="16914" CreationDate="2015-06-24T08:53:39.870" Score="39" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; #5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;W X&lt;/code&gt;&lt;br /&gt;&#xA; &lt;code&gt;Y Z&lt;/code&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="40" LastActivityDate="2015-06-24T08:53:39.870" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16917" PostTypeId="1" AcceptedAnswerId="16938" CreationDate="2015-06-24T10:15:57.413" Score="4" ViewCount="237" Body="&lt;p&gt;Can you find the number that should be placed instead of the xxxx?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;code&gt;0000A...&lt;/code&gt;&lt;br&gt;&#xA;  &lt;code&gt;0111AB..&lt;/code&gt;&lt;br&gt;&#xA;  &lt;code&gt;2012BB..&lt;/code&gt;&lt;br&gt;&#xA;  &lt;code&gt;3103AB..&lt;/code&gt;&lt;br&gt;&#xA;  &lt;code&gt;4140AAA.&lt;/code&gt;&lt;br&gt;&#xA;  &lt;code&gt;4150AABB&lt;/code&gt;&lt;br&gt;&#xA;  &lt;code&gt;xxxxAAAA&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Good luck!&lt;/p&gt;&#xA;" OwnerUserId="13268" LastEditorUserId="5840" LastEditDate="2015-06-24T11:01:33.977" LastActivityDate="2015-06-24T15:16:15.277" Title="Next number in the sequence" Tags="|number-sequence|letter-sequence|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16918" PostTypeId="1" AcceptedAnswerId="16926" CreationDate="2015-06-24T10:44:52.127" Score="16" ViewCount="3427" Body="&lt;p&gt;Five lorries are parked in a French layby, having recently come off the ferry at Calais. The police know that the lorry headed for Lisbon is smuggling drugs, but they're not sure which one it is. All they have to go on is the following information.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The Polish lorry came on the 6 o'clock ferry and is carrying coffee.&lt;/li&gt;&#xA;&lt;li&gt;The lorry in the middle is black.&lt;/li&gt;&#xA;&lt;li&gt;The British lorry came on the 9 o'clock ferry.&lt;/li&gt;&#xA;&lt;li&gt;The blue French lorry is somewhere to the left of the lorry carrying coffee.&lt;/li&gt;&#xA;&lt;li&gt;The lorry headed for Marseille is somewhere to the right of the lorry carrying potatoes.&lt;/li&gt;&#xA;&lt;li&gt;The Lithuanian lorry is headed for Nantes.&lt;/li&gt;&#xA;&lt;li&gt;The lorry carrying rice is next to the green lorry.&lt;/li&gt;&#xA;&lt;li&gt;The lorry headed for Genoa came on the 5 o'clock ferry.&lt;/li&gt;&#xA;&lt;li&gt;The Spanish lorry is somewhere to the right of the lorry headed for Marseille.&lt;/li&gt;&#xA;&lt;li&gt;The red lorry is headed for Hamburg.&lt;/li&gt;&#xA;&lt;li&gt;The white lorry is next to the Spanish lorry that came on the 7 o'clock ferry.&lt;/li&gt;&#xA;&lt;li&gt;The lorry carrying bread is at one of the ends.&lt;/li&gt;&#xA;&lt;li&gt;The black lorry came on the 8 o'clock ferry.&lt;/li&gt;&#xA;&lt;li&gt;The lorry carrying bread is next to the lorry carrying rice.&lt;/li&gt;&#xA;&lt;li&gt;The lorry headed for Hamburg came on the 6 o'clock ferry.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;When the police arrive at the layby, &lt;strong&gt;which lorry driver should they arrest?&lt;/strong&gt; I.e. should they raid the leftmost lorry, the second from the left, etc.?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus question: &lt;strong&gt;which &lt;em&gt;two&lt;/em&gt; possible lorries could be the one that's carrying tea?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the spirit of teamwork, partial answers are acceptable.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="11244" LastEditDate="2015-06-24T12:55:36.980" LastActivityDate="2015-06-30T18:48:20.293" Title="Which lorry is carrying drugs?" Tags="|logical-deduction|logic-grid|" AnswerCount="3" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="16920" PostTypeId="1" CreationDate="2015-06-24T11:03:55.310" Score="0" ViewCount="501" Body="&lt;p&gt;I have come across a puzzle, which makes no sense. Does anyone know the answer to the below?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You and two friends go on holiday.  When checking out, the total bill comes to £300. You pay the bill and begin to walk out. The clerk realises that he’s made an error and hands the bellhop a £50 note to return to you. He realises that he cannot split the £50 equally, so he decides to give each guest £10 and keep £20 as a tip for himself. Hence, each guest received £10 back, meaning they only paid £90 each. This means the total paid by the three people was £270 (£90 x 3). Add in the £20 the bellhop kept and the total comes to £290. What happened to the missing £10?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="GSS" LastEditorUserId="10399" LastEditDate="2015-06-24T12:09:29.770" LastActivityDate="2015-06-24T15:53:32.780" Title="What happened to the missing £10?" Tags="|mathematics|calculation-puzzle|word-problem|paradox|" AnswerCount="3" CommentCount="1" ClosedDate="2015-06-24T12:20:46.280" ContentLicense="CC BY-SA 3.0" />
  <row Id="16921" PostTypeId="2" ParentId="16920" CreationDate="2015-06-24T11:17:58.447" Score="7" Body="&lt;p&gt;You should be deducting the Bellhop's £20 from the £270, not adding it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edited with more details:&#xA;The question is deliberately misleading you into thinking there's a paradox, and some money has gone missing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total amount paid initially was £300, £100 each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total amount paid after the partial refund was £270, £90 each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sly bellhop kept the final £20 for himself - had he split this evenly, the total amount paid would have been £250 - £83.333... each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is misleading in the way it deducts the refund received by the customers, but then attempts to add the refund stolen by the bellhop.&lt;/p&gt;&#xA;" OwnerUserId="13345" LastEditorUserId="13345" LastEditDate="2015-06-24T15:53:32.780" LastActivityDate="2015-06-24T15:53:32.780" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16922" PostTypeId="2" ParentId="16918" CreationDate="2015-06-24T11:21:39.587" Score="6" Body="&lt;p&gt;From left to right, the lorries are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1) Blue French lorry that arrived on the 5'o clock ferry and is travelling to Genoa.&lt;br/&gt;2) Red Polish lorry that arrived on the 6'o clock ferry and is travelling to Hamburg, carrying coffee.&lt;br/&gt;3) Black Lithuanian lorry that arrived on the 8'o clock ferry and is travelling to Nantes, carrying potatoes.&lt;br/&gt;4) White British lorry that arrived on the 9'o clock ferry and is travelling to Marseilles, carrying rice.&lt;br/&gt;5) Green Spanish lorry that arrived on the 7'o clock ferry and is travelling to Lisbon, carrying bread.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The police must nab the driver of the rightmost lorry. Atacar a los españoles! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus question:&lt;/strong&gt; The French and Polish lorries could be carrying coffee and tea; perhaps that's why they are adjacent to each other ;)&lt;/p&gt;&#xA;" OwnerUserId="2425" LastEditorUserId="2425" LastEditDate="2015-06-24T11:27:29.320" LastActivityDate="2015-06-24T11:27:29.320" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16924" PostTypeId="2" ParentId="16920" CreationDate="2015-06-24T11:24:57.310" Score="2" Body="&lt;p&gt;Let's consider the total amounts of money.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;BEFORE PAYING:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Guests: 300 (3x100)&lt;br&gt;&#xA;  Clerk: 0&lt;br&gt;&#xA;  Shop: 0  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The clerk gets a 20 tip, while each man pays 90 (gives 100 but is returned 10).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;AFTER PAYING:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Guests: 30 (3x10)&lt;br&gt;&#xA;  Clerk: 20&lt;br&gt;&#xA;  Shop: 250  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As you can see there is no paradox, the only mistake was trying to sum 20 to 270 which absolutely makes no sense!&lt;/p&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-06-24T11:24:57.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16925" PostTypeId="2" ParentId="16920" CreationDate="2015-06-24T11:25:16.463" Score="2" Body="&lt;p&gt;You pay £300.&lt;br&gt;&#xA;The clerk returns £50.&lt;br&gt;&#xA;The bellhop returns £30 of the £50 and keeps £20.&lt;br&gt;&#xA;You and your friends have paid £270.&lt;br&gt;&#xA;The clerk intended for you to have paid £250.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;£300 = £250 + £50(change)&lt;br&gt;&#xA;£50 = £10 + £10 + £10 (returned to you) + £20 (kept by bellhop)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure where you see a missing £10, so I can't tell you what happened to it.&lt;/p&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2015-06-24T11:25:16.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16926" PostTypeId="2" ParentId="16918" CreationDate="2015-06-24T11:27:42.900" Score="21" Body="&lt;pre&gt;&lt;code&gt;Destination    Genoa     Hamburg    Nantes      Marseille   Lisbon&#xA;Origin         French    Polish     Lithuanian  British     Spanish&#xA;Colour         blue      red        black       white       green&#xA;Ferry Time     5 o'clock 6 o'clock  8 o'clock   9 o'clock   7 o'clock&#xA;Load           ?         coffee     ?           rice        bread&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The police should raid the rightmost lorry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus:&lt;/strong&gt; The French and Lithuanian lorries could be carrying tea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is how I deduced the answer. Using the facts that I knew, I created two sets of lorries, the first using the positional information to determine the known places.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(Ordered set)&#xA;Destination    &#xA;Origin         &#xA;Colour         (green)              black                   (green)&#xA;Ferry Time                          8 o'clock               &#xA;Load           (bread)   (rice)                 (rice)      (bread)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In the above, the bread load is at one of two ends (12) and rice is next to it (14), giving rise to two scenarios. The lorry carrying bread is green as it's next to the one carrying rice (7) and it can't be the one in the middle position as the middle lorry is known to be black (2) and it came on the 8 o' clock ferry (13).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second set, I used the remaining data to assign a set of possible characteristics to lorries.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(Unordered set)&#xA;Destination    Hamburg                         Nantes       Genoa      &#xA;Origin         Polish     British     French   Lithuanian              Spanish&#xA;Colour         red                    blue                              &#xA;Ferry Time     6 o'clock  9 o'clock                         5 o'clock  7 o'clock&#xA;Load           coffee    &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The complete description of the Polish lorry is provided, barring the position (1,10,15). British lorry came in at 9 o'clock (3) and the French lorry is blue (4). The Lithuanian lorry is headed for Nantes (6) and the lorry to Genoa came in at 5 o'clock (8), while the Spanish lorry came in on the 7 o'clock ferry (11).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the second set, we can see that the nationality of the lorry to Genoa is the only one not known, but its time of arrival is known. There is only one other combination where both the nationality and time of arrival is missing, so I determined that the French lorry is headed to Genoa. Also, the timings for all are known except for the Lithuanian lorry, but I have the timing for the middle lorry, which is the one timing not accounted for.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(Updated ordered set)&#xA;Destination                         Nantes&#xA;Origin                              Lithuania&#xA;Colour         (green)              black                   (green)&#xA;Ferry Time                          8 o'clock               &#xA;Load           (bread)   (rice)                 (rice)      (bread)&#xA;&#xA;(Updated unordered set)&#xA;Destination    Hamburg                Genoa          &#xA;Origin         Polish     British     French     Spanish&#xA;Colour         red                    blue                              &#xA;Ferry Time     6 o'clock  9 o'clock   5 o'clock  7 o'clock&#xA;Load           coffee   &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;At this point, I was hard pressed to decide on a position for the lorry with bread. Let's consider I picked the leftmost position, then rice would be in the lorry at the second position. The lorry headed to Genoa would occupy the fourth position as it has to be on the left of the lorry carrying coffee (4). This means the lorry to Hamburg will be in the fifth position.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(Updated ordered set, invalid)&#xA;Destination                         Nantes        Genoa      Hamburg&#xA;Origin                              Lithuania     French     Polish&#xA;Colour         green                black         blue       red   &#xA;Ferry Time                          8 o'clock     5 o'clock  6 o'clock          &#xA;Load           bread    rice                                 coffee&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;However, this would violate the condition that the lorry headed for Marseille is somewhere to the right of the lorry carrying potatoes (5), since potatoes can only be in the last three positions (other two have bread and rice) and there I've determined the locations are Nantes, Genoa and Hamburg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So consider that the rightmost position is the lorry with bread, the lorry next to it has rice.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(Updated ordered set)&#xA;Destination    Genoa     Hamburg    Nantes      &#xA;Origin         French    Polish     Lithuanian  &#xA;Colour         blue      red        black                   green&#xA;Ferry Time     5 o'clock 6 o'clock  8 o'clock   &#xA;Load                     coffee                 rice        bread&#xA;&#xA;(Updated unordered set)&#xA;Destination                &#xA;Origin         British     Spanish&#xA;Colour                     &#xA;Ferry Time     9 o'clock   7 o'clock&#xA;Load              &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So Marseille can be the destination for either of the lorries at the right end, as potatoes can only be in the middle lorry or the leftmost one (5). Now only the colour of the fourth lorry is unknown, and I know that there is a white lorry next to a Spanish lorry (11). This makes the fourth lorry white, the fifth lorry is Spanish and coming in at 7 o'clock, while the fifth lorry is British coming in at 9 o'clock. Also, the Spanish lorry is somewhere to the right of the lorry headed for Marseille (9), which means the fourth lorry is going to Marseille, while the fifth is headed to Lisbon.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Destination    Genoa     Hamburg    Nantes      Marseille   Lisbon&#xA;Origin         French    Polish     Lithuanian  British     Spanish&#xA;Colour         blue      red        black       white       green&#xA;Ferry Time     5 o'clock 6 o'clock  8 o'clock   9 o'clock   7 o'clock&#xA;Load                     coffee                 rice        bread&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;With that the solution is complete, with no more data available to firmly establish which lorry carries potatoes, so either the blue or black one could be carrying tea, the other commodity we haven't been given any information about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may not be the most efficient way to explain (or even solve), but this is the best way I can explain what I did to reach the solution.&lt;/p&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-06-30T18:48:20.293" LastActivityDate="2015-06-30T18:48:20.293" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16929" PostTypeId="1" AcceptedAnswerId="16986" CreationDate="2015-06-24T13:08:07.893" Score="4" ViewCount="1682" Body="&lt;p&gt;&lt;strong&gt;Hell is where a little old lady named Mrs. Dunham lives.&lt;br /&gt;&#xA;Mrs. Dunham is a very old-fashioned lady.&lt;br /&gt;&#xA;Mrs. Dunham has a bit of a drinking problem.&lt;br /&gt;&#xA;Three times a day, every day, she drinks.&lt;br /&gt;&#xA;What times of the day does she drink, and what does she drink?&lt;br /&gt;&lt;br /&gt;&lt;/strong&gt;&#xA;(thanks for viewing my first puzzle! I promise you, everything you need to solve it is here, have fun!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes: The drink is not water, and Mrs. Dunham is surprisingly healthy for an old lady and does not take any medicine nor need any vitamins. The drink in question is in addition to any drinks served at breakfast, lunch, and dinner (or whatever terms you use for regular meals). There are no hospitals in Hell, that I know of, so you can safely count that out of your potential answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, to add to the riddle to help it not be too &quot;broad&quot; (I hope that term isn't being used to disparage poor Mrs. Dunham), let me say this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mrs. Dunham is a bit used to warmer climates and has to have her drink imported because it's not as popular in Hell as it was in her hometown.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The three times of the day are directly linked to the drink, for very old reasons&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The drink in question is much more popular in this southern state where it is made&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="13068" LastEditorUserId="13068" LastEditDate="2015-06-25T11:58:23.170" LastActivityDate="2015-06-25T17:09:48.283" Title="Mrs. Dunham's drinking problem" Tags="|riddle|lateral-thinking|wordplay|" AnswerCount="5" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="16930" PostTypeId="2" ParentId="16918" CreationDate="2015-06-24T13:39:54.230" Score="6" Body="&lt;p&gt;The answer is as such:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Origin      | French | Polish  | Lithuanian  | British   | Spanish |&#xA;Destination | Genoa  | Hamburg | Nantes      | Marseille | Lisbon  |&#xA;Ferry Time  | 5      | 6       | 8           | 9         | 7       |&#xA;Colour      | blue   | Red     | Black       | White     | Green   |&#xA;Goods       | ?      | Coffee  | ?           | Rice      | Bread   |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here is the explanation. We start from a list of possibilities which soon narrow down into specific details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The possibilities start with:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Origin      | PBFLS | PBFLS | PBFLS | PBFLS | PBFLS |&#xA;Destination | MNGHL | MNGHL | MNGHL | MNGHL | MNGHL |&#xA;Ferry Time  | 56789 | 56789 | 56789 | 56789 | 56789 |&#xA;Colour      | BbGWR | BbGWR | BbGWR | BbGWR | BbGWR |&#xA;Goods       |TTCPRB |TTCPRB |TTCPRB |TTCPRB |TTCPRB |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clues &lt;strong&gt;2 &amp;amp; 13&lt;/strong&gt; lead us to that the middle lorry is black and came on the 8 o' clock ferry. Clues &lt;strong&gt;1 &amp;amp; 3 &amp;amp; 4 &amp;amp; 11&lt;/strong&gt; further say that The middle truck is not British nor French nor Spanish nor Polish because of the ferry time and colour; and it does not carry coffee. Clues &lt;strong&gt;8 &amp;amp; 10&lt;/strong&gt; say that the middle truck is not headed for Genoa or Hamburg for the same reason. Clues &lt;strong&gt;12 &amp;amp; 14&lt;/strong&gt; say that the middle lorry cannot possibly be carrying bread or rice because it is in the middle. After putting all this information together, we can see that the middle truck is Lithuanian and using clue &lt;strong&gt;6&lt;/strong&gt; we know that it is headed for Nantes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Origin      | PBF S | PBF S |   L   | PBF S | PBF S |&#xA;Destination | M GHL | M GHL |   N   | M GHL | M GHL |&#xA;Ferry Time  | 567 9 | 567 9 |   8   | 567 9 | 567 9 |&#xA;Colour      |  bGWR |  bGWR |   B   |  bGWR |  bGWR |&#xA;Goods       | TCPBR | TCPBR |   TP  | TCPBR | TCPBR |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clues &lt;strong&gt;12 &amp;amp; 14&lt;/strong&gt; say that bread can only be at one of the ends and that rice must not be at the ends. Clue &lt;strong&gt;4&lt;/strong&gt; says that the coffee lorry cannot be at the left end and that the blue French lorry cannot be at the right most end. Clue &lt;strong&gt;5&lt;/strong&gt; says that the right end cannot be having potatoes and the left end cannot be headed to Marseille. Clue &lt;strong&gt;7&lt;/strong&gt; says that the green lorry cannot be having rice. Clue &lt;strong&gt;9 &amp;amp; 11&lt;/strong&gt; says that the first two lorries from the left cannot be Spanish or have the time as 7 o' clock because the Spanish lorry came on 7 o' clock. Clue &lt;strong&gt;9&lt;/strong&gt; also tells us that the Spanish lorry is not headed towards Marseille.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Origin      | PBF   | PBF   |   L   | PBF S | PBF S |&#xA;Destination |   GHL | M GHL |   N   | M GHL |   GHL |&#xA;Ferry Time  | 56  9 | 56  9 |   8   | 567 9 | 567 9 |&#xA;Colour      |  bGWR |  b WR |   B   |  b WR |  bGWR |&#xA;Goods       | TCPB  | TCP R |   TP  | TCP R | TC B  |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clues &lt;strong&gt;1 &amp;amp; 10 &amp;amp; 15&lt;/strong&gt; give us all the information on the Polish lorry except for its position. Clues &lt;strong&gt;1 &amp;amp; 4&lt;/strong&gt; say that the Polish lorry cannot be on the left most and the French one cannot be on the right most. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Origin      |  BF   | PBF   |   L   | PBF S | PB  S |&#xA;Destination |   G L | M GHL |   N   | M GHL |   GHL |&#xA;Ferry Time  | 5   9 | 56  9 |   8   | 567 9 | 567 9 |&#xA;Colour      |  bGW  |  b WR |   B   |  b WR |  bGWR |&#xA;Goods       | T PB  | TCP R |   TP  | TCP R | TC B  |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Now I am going to forget the positions of the lorries and we are left with.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Origin      |   P   |   B   |   L   |   F   |   S   |&#xA;Destination |   H   |       |   N   |       |       |&#xA;Ferry Time  |   6   |   9   |   8   |       |   7   |&#xA;Colour      |   R   |       |   B   |   b   |       |&#xA;Goods       |   C   |       |   ?   |       |       |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clue &lt;strong&gt;8&lt;/strong&gt; says that the lorry headed for Genoa came on the 5 o' clock ferry. The only lorry with no time is the French lorry and we will fill that out. Clue &lt;strong&gt;11&lt;/strong&gt; says that the Spanish lorry cannot be white. And then we can see that Spanish is green.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Origin      |   P   |   B   |   L   |   F   |   S   |&#xA;Destination |   H   |       |   N   |   G   |       |&#xA;Ferry Time  |   6   |   9   |   8   |       |   7   |&#xA;Colour      |   R   |   W   |   B   |   b   |   G   |&#xA;Goods       |   C   |       |   ?   |       |       |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Now reflecting back on our original grid, we see that Spanish only fits with the last one. Clue &lt;strong&gt;11&lt;/strong&gt; helps us find out the second-rightmost lorry. After cancelling out the Spanish (S) from the origin of the first two lorries, we see that the lorries are French and Polish. We can see that Marseille can only fit with the British Lorry.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Origin      |   F   |   P   |   L   |   B   |   S   |&#xA;Destination |   G   |   H   |   N   |   M   |   L   |&#xA;Ferry Time  |   5   |   6   |   8   |   9   |   7   |&#xA;Colour      |   b   |   R   |   B   |   W   |   G   |&#xA;Goods       | T PB  |   C   |   TP  | T P R | T  B  |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clue &lt;strong&gt;5&lt;/strong&gt; tells us that the Marseille lorry does not have potatoes. Clue &lt;strong&gt;7&lt;/strong&gt; tells us that the British lorry is carrying rice. According to clue &lt;strong&gt;14&lt;/strong&gt; the Spanish lorry carries bread. We are finished with:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Origin      |   F   |   P   |   L   |   B   |   S   |&#xA;Destination |   G   |   H   |   N   |   M   |   L   |&#xA;Ferry Time  |   5   |   6   |   8   |   9   |   7   |&#xA;Colour      |   b   |   R   |   B   |   W   |   G   |&#xA;Goods       |  TP   |   C   |   TP  |   R   |   B   |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Finally&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The answer to the first question is that the police should arrest the &lt;strong&gt;Spanish&lt;/strong&gt; driver as he is headed to Lisbon. Both the &lt;strong&gt;French and the Lithuanian&lt;/strong&gt; lorries have a possibility of carrying Tea. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12860" LastEditorUserId="12860" LastEditDate="2015-06-24T16:16:51.593" LastActivityDate="2015-06-24T16:16:51.593" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16931" PostTypeId="2" ParentId="16929" CreationDate="2015-06-24T13:58:17.477" Score="1" Body="&lt;p&gt;Just for the &lt;em&gt;hell&lt;/em&gt; of it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mrs. Dunham is an imp in hell.&#xA;&lt;br&gt; She drinks human blood 3 times a day as a snack [once between breakfast and lunch, between lunch and dinner, and once before going to sleep]. She also eats pie twice a day at 3:14 while everyone else would have to watch.&#xA;&lt;br&gt;(She would of course freshly bake each pie as most old imp ladies would)&#xA;&lt;br&gt;&lt;br&gt;Notes:&#xA;&lt;br&gt;Mrs. Dunham is old-fashioned because not many imps in hell still drink human blood.&#xA;&lt;br&gt;The problem with her drinking is that she can run out of people!&#xA;&lt;br&gt;(Mrs. Dunham also always puts raisins in her cookies and cinnamon buns)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-06-24T14:33:09.973" LastActivityDate="2015-06-24T14:33:09.973" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="16932" PostTypeId="2" ParentId="16929" CreationDate="2015-06-24T14:03:39.707" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Three times a day, every day, she would drink.&lt;br&gt;&#xA;  What times of the day did she drink, and what did she drink?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;... with the &lt;a href=&quot;/questions/tagged/wordplay&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;wordplay&amp;#39;&quot; rel=&quot;tag&quot;&gt;wordplay&lt;/a&gt; tag.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;She drinks&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; at three o'clock every day (&quot;three times&quot; is a weird way of saying three o'clock).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;She drinks&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; products from &lt;a href=&quot;http://www.drinkeveryday.com/&quot; rel=&quot;nofollow&quot;&gt;Everyday&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus, at three a day, Everyday she would drink. Just like it says on the tin.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-06-24T14:06:07.647" LastActivityDate="2015-06-24T14:06:07.647" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16933" PostTypeId="1" AcceptedAnswerId="16936" CreationDate="2015-06-24T14:19:03.703" Score="15" ViewCount="789" Body="&lt;p&gt;First is a dive; not much, but it's cheap&lt;br&gt;&#xA;Next is the cache for society's keep&lt;br&gt;&#xA;Another dive follows; it's much like the first&lt;br&gt;&#xA;Please not the next one; of all, it's the worst&lt;br&gt;&#xA;The next is for readers, they'll chew and they'll chew&lt;br&gt;&#xA;After that, east; you're light but you're blue&lt;br&gt;&#xA;The next's rather risky; there's some good at least&lt;br&gt;&#xA;Then two more are stately, and much like the east&lt;br&gt;&#xA;Next comes the crooks while they go round the bend&lt;br&gt;&#xA;Then comes a saintly and violet friend&lt;br&gt;&#xA;But I got the highest, as high as one gets!&lt;br&gt;&#xA;I now have to go there, &lt;strong&gt;so tell me: what's next?&lt;/strong&gt;&lt;br&gt;&lt;br&gt;&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Please provide an explanation of the clues along with your answer. ;)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="11244" LastEditDate="2015-06-24T14:24:25.370" LastActivityDate="2015-06-24T15:42:10.993" Title="One More Time Around" Tags="|riddle|rhyme|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16934" PostTypeId="2" ParentId="16933" CreationDate="2015-06-24T15:04:46.430" Score="5" Body="&lt;p&gt;I think it's a Monopoly board, the first two properties being the cheapest (a dive being a badly maintained hotel) and they're separated by the community chest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference to jail where there are crooks around the bend as it's a corner square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumably what's next is to pass go and collect £200?&lt;/p&gt;&#xA;" OwnerUserId="13377" LastActivityDate="2015-06-24T15:04:46.430" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16935" PostTypeId="1" AcceptedAnswerId="17128" CreationDate="2015-06-24T15:09:00.383" Score="23" ViewCount="2772" Body="&lt;p&gt;Well... my first attempt to enter a Secret Club wasn't successful (&lt;em&gt;see: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16540/the-secret-club&quot;&gt;The Secret Club&lt;/a&gt;&lt;/em&gt;), but this time I'm sure that nothing can stop me from reaching my goal...&lt;br&gt;&#xA;I've discovered this new club, whose members are some puzzle fanatics who love wasting time creating and solving riddles... what losers!&lt;br&gt;&#xA;By the way I don't really care what they do in that club, but I want to be a part of it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A fat guy is standing at the door, waiting for members to come.&lt;br&gt;&#xA;And (you wouldn't believe it...) every time that someone approaches the door, this guy asks him in a very loud voice some strange question... let's listen what he's going to ask to those guys coming to the club...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;First guy: &quot;&lt;em&gt;Hi buddy, can I come in?&lt;/em&gt;&quot;&lt;br&gt;&#xA;  Fat guardian, screaming: &quot;&lt;em&gt;Tell me... what number immediately follows 7?&lt;/em&gt;&quot;&lt;br&gt;&#xA;  &quot;&lt;em&gt;Mmm... 8, I guess!&lt;/em&gt;&quot;&lt;br&gt;&#xA;  &quot;&lt;em&gt;Ok, come in!&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A second guy approaches, and the guy at the door screams again:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;&lt;em&gt;13 - 12?&lt;/em&gt;&quot;&lt;br&gt;&#xA;  &quot;&lt;em&gt;Too easy... 1!&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;answers the second guy, and once listened to his words the guardian lets him in.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;&lt;em&gt;What number comes before ten?&lt;/em&gt;&quot;, asks the guardian to a third guy walking to the club's door...&#xA;  &quot;&lt;em&gt;Well... 9, without any doubt!&lt;/em&gt;&quot;&lt;br&gt;&#xA;  The fat guy nods, then asks again &quot;&lt;em&gt;1+2?&lt;/em&gt;&quot;&lt;br&gt;&#xA;  &quot;&lt;em&gt;Hey buddy, who do you thing you are talking to?!? Everybody knows that the answer is 3!&lt;/em&gt;&quot; answers the guy, and the guardian lets him in friendly patting his back.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(- &lt;em&gt;Once again, it seems clear that entering a club is just a matter of elementary maths... Why is it always so?!?&lt;/em&gt; -)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One last guy approaches the club, the guardian smiles at him then yells:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;&lt;em&gt;Hey buddy! Give me a five!&lt;/em&gt;&quot;&lt;br&gt;&#xA;  &quot;&lt;em&gt;Hi Bob!&lt;/em&gt;&quot; tells the guy, giving an high-five to the fat guardian and entering the club's doors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(- &lt;em&gt;Well, it seems this last guy is really well known... Anyway, it's my turn!... let's go!&lt;/em&gt; -)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;&lt;em&gt;...ehm... Hey Bob! How's your day?... Can I come in?...&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The guardian doesn't seem very friendly... he looks at me askance then yells:   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;&lt;em&gt;Tell me... 2 and 2 make...?&lt;/em&gt;&quot;&lt;br&gt;&#xA;  &quot;&lt;em&gt;I know the answer! I know it! It's 4! It's 4!!&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I scream... once again, just a second before the guardian kicks me away, telling me to never come back!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I keep wondering...&lt;br&gt;&#xA;&lt;strong&gt;WHAT WAS WRONG WITH MY ANSWER? WHAT WAS I SUPPOSED TO SAY?&lt;br&gt;&#xA;AND WHY COULD SOMEONE ENTER THE CLUB WITHOUT ANSWERING ANY QUESTION?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You don't need any particular knowledge to answer the guardian's questions (except elementary maths, of course...).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not everything in this puzzle is what it seems.&lt;br&gt;&#xA;&quot;&lt;em&gt;HEY! IS THIS SUPPOSED TO BE A HINT?!?&lt;/em&gt;&quot;, I can hear you screaming... Yes, it IS!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Among things that arent what they seem: NUMBERS. Try not to consider them as (just) numbers. And be sure you have actually found all the clues hidden in my 2nd hint...   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 4&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Try thinking out-of-the-box, and try changing something in the text of the puzzle according to my first 3 hints (in particular, n° 2 and 3). Isn't it strange the way the guardian speaks? Why does he keep SCREAMING?!?      &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12628" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-07-08T09:04:07.533" Title="The Secret Club (another one?!?)" Tags="|pattern|lateral-thinking|password|situation|" AnswerCount="4" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="16936" PostTypeId="2" ParentId="16933" CreationDate="2015-06-24T15:09:16.327" Score="13" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Electric Company.&lt;br&gt;&#xA; You rolled double 6's for your opening roll on a game of Monopoly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First is a dive; not much, but it's cheap&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mediterranean Avenue:  The cheapest property on the board, named after the Mediterranean Sea, where you can dive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Next is the cache for society's keep&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Community Chest&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another dive follows; it's much like the first&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Baltic Avenue:  In the same set (dark purple) as Mediterranean; it's also named after a body of water that you can dive in, the Baltic Sea.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Please not the next one; of all, it's the worst&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Luxury tax:  Pay 10% or $200 (in the U.S. at least)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The next is for readers, they'll chew and they'll chew&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reading Railroad:  &quot;Chew&quot; sounds like a train: &quot;Choo choo&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After that, east; you're light but you're blue&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oriental Avenue:  &quot;Oriental&quot; means &quot;eastern&quot;.  Part of the light blue set.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The next's rather risky; there's some good at least&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chance:  Could have a positive or negative outcome.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then two more are stately, and much like the east&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Vermont and Connecticut Avenues:  Both part of the light blue set as well, named after U.S. States.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Next comes the crooks while they go round the bend&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jail:  On the corner, while you're going &quot;round the bend&quot; is where the &quot;crooks&quot; are.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then comes a saintly and violet friend&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; St. Charles Place:  Starts with the word &quot;Saint&quot;; part of the light purple (violet) set&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I got the highest, as high as one gets!&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I rolled a 12 right off the bat; you can't roll higher than that in Monopoly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I now have to go there, &lt;strong&gt;so tell me: what's next?&lt;/strong&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The next property on the board is Electric Company, the 12th space from &quot;Go&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13068" LastEditorUserId="13068" LastEditDate="2015-06-24T15:42:10.993" LastActivityDate="2015-06-24T15:42:10.993" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16937" PostTypeId="2" ParentId="16929" CreationDate="2015-06-24T15:13:28.060" Score="10" Body="&lt;p&gt;Mrs. Dunham drinks:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an &lt;a href=&quot;https://en.wikipedia.org/?title=Old_Fashioned&quot;&gt;Old Fashioned&lt;/a&gt; because Mrs. Dunham is a very &lt;strong&gt;old-fashioned&lt;/strong&gt; lady&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And she drinks it at :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Midnight&lt;/strong&gt;, &lt;strong&gt;1 am&lt;/strong&gt;, &lt;strong&gt;1 pm&lt;/strong&gt; because Mrs. Dunham has a &lt;strong&gt;bit&lt;/strong&gt; of a drinking problem. And a &lt;a href=&quot;https://en.wikipedia.org/wiki/Bit&quot;&gt;bit&lt;/a&gt; is most commonly represented as either a &lt;strong&gt;0&lt;/strong&gt; or &lt;strong&gt;1&lt;/strong&gt;. So 0 is midnight and 1 is 1 am and 1 pm.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5359" LastActivityDate="2015-06-24T15:13:28.060" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16938" PostTypeId="2" ParentId="16917" CreationDate="2015-06-24T15:16:15.277" Score="6" Body="&lt;p&gt;The next term is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 5140AAAA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The terms in the sequence are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A series of guesses in the game Mastermind. The number is the guess, the letters indicate how close the guess is to the hidden sequence: A represents a correct number in the correct position, B represents a correct number but not in the correct position. Since the letters for the last guess are AAAA, the last guess must be the hidden sequence. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10261" LastActivityDate="2015-06-24T15:16:15.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16939" PostTypeId="2" ParentId="16842" CreationDate="2015-06-24T15:21:31.117" Score="2" Body="&lt;p&gt;I can't figure out the numbers but the country seems to match:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mauritius&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;it is about: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 185 km to the east of Reunion&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;it has a popular traditional music called:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sega &lt;a href=&quot;https://en.wikipedia.org/?title=Music_of_Mauritius&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/?title=Music_of_Mauritius&lt;/a&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it meets the requirement of small and unknown from the comments&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13531" LastActivityDate="2015-06-24T15:21:31.117" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16940" PostTypeId="1" AcceptedAnswerId="16941" CreationDate="2015-06-24T15:27:28.553" Score="5" ViewCount="364" Body="&lt;p&gt;This is a poem with a wandering scheme about a potentially deadly force. Can you answer the questions posed?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[REDACTED] is powerful, reliant, and tall&lt;br&gt;&#xA;  He performs every day for the plesaure of all&lt;br&gt;&#xA;  He towers above the gawkers below&lt;br&gt;&#xA;  Who've come to enjoy his regular show&lt;br&gt;&#xA;  We've found now, in fact, routine is rather&lt;br&gt;&#xA;  Bimodal in form meaning you sometimes gather&lt;br&gt;&#xA;  For sixty or ninety minutes before&lt;br&gt;&#xA;  He starts up his act lasting one minute four&lt;br&gt;&#xA;  At the most but sometimes it's short&lt;br&gt;&#xA;  Ninety seconds - abort   &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;He's not safe to get near though he means no harm&lt;br&gt;&#xA;  His own strength betrays him - he carries no arms&lt;br&gt;&#xA;  His father was Washburn who named him quite aptly&lt;br&gt;&#xA;  Although his performance is not quite the acme&lt;br&gt;&#xA;  His brother wins that title though he was named first&lt;br&gt;&#xA;  Not jealous a bit, though, as both quench the thirst&lt;br&gt;&#xA;  Of travelers seeking adventure and grandeur&lt;br&gt;&#xA;  Ironic a bit, given permanent tenure&lt;br&gt;&#xA;  He's a prisoner, see, and can never leave&lt;br&gt;&#xA;  But cry not for him. As a nation we've&lt;br&gt;&#xA;  Formed quite a home surrounding all sides&lt;br&gt;&#xA;  Of this giant whose entertainment provides&lt;br&gt;&#xA;  Joy for all the visitors who&lt;br&gt;&#xA;  Have come here again to view&lt;br&gt;&#xA;  The wonders provided&lt;br&gt;&#xA;  And all, breath bated,&lt;br&gt;&#xA;  Await the show&lt;br&gt;&#xA;  Do you know&lt;br&gt;&#xA;  the right&lt;br&gt;&#xA;  site?   &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;ANSWER NOW, FOOL, 'FORE HIS ANGER EXPLODES&lt;br&gt;&#xA;  AND UNLEASHES POWER HE'S STORED DOWN BELOW&lt;br&gt;&#xA;  HIS ENERGY'S GREAT AND POTENTIAL MIGHTY&lt;br&gt;&#xA;  HE'S TRAPPED HERE FOREVER HANDING OUT DEATHBLOWS&lt;br&gt;&#xA;  IF YOU WANDER TOO CLOSE YOUR FLESH SHALL BE RENDERED&lt;br&gt;&#xA;  FROM MORTAL COIL SO SOFT AND SO TENDER&lt;br&gt;&#xA;  FEAR HIM, YOU MORTAL, LEST YOUR CASUAL STRAYING&lt;br&gt;&#xA;  RELINQUISH YOUR SOUL TO HIS BEASTLY BLENDER&lt;br&gt;&#xA;  TELL ME HIS NAME AND THE NAME OF HIS HOME.&lt;br&gt;&#xA;  WHY IS HE TRAPPED AND CAN NEVER ROAM?&lt;br&gt;&#xA;  TELL ME THIS QUICKLY OR YOUR SOUL I SHALL HARVEST&lt;br&gt;&#xA;  AND BURY WITHIN THE BEAST'S CATACOMB   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastActivityDate="2015-06-24T15:39:12.393" Title="The Beast Shall Render Your Soul" Tags="|riddle|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16941" PostTypeId="2" ParentId="16940" CreationDate="2015-06-24T15:30:33.807" Score="5" Body="&lt;p&gt;The beast's name is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Old_Faithful&quot; rel=&quot;nofollow&quot;&gt;Old Faithful&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first paragraph basically explains that it's a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; geyser&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;that &quot;performs&quot; at a regular interval. After that, it tells about &lt;a href=&quot;https://en.wikipedia.org/wiki/Henry_D._Washburn&quot; rel=&quot;nofollow&quot;&gt;Washburn&lt;/a&gt; naming it. Its &quot;brother&quot; is probably &lt;a href=&quot;https://en.wikipedia.org/wiki/Steamboat_Geyser&quot; rel=&quot;nofollow&quot;&gt;Steamboat&lt;/a&gt;, the largest one of its kind. There's a lot of stuff talking about heat and water, which are self evident once you know the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To literally answer the questions posed:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Do you know the right site?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yellowstone National Park, Wyoming, United States&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;WHY IS HE TRAPPED AND CAN NEVER ROAM?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Two possible answers. 1) It's a geyser. They don't really move. 2) According to Wikipedia, it &quot;&lt;em&gt;...is not connected to any other thermal features of the Upper Geyser Basin.&lt;/em&gt;&quot; So, it's isolated from other geysers, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1563" LastEditorUserId="1563" LastEditDate="2015-06-24T15:39:12.393" LastActivityDate="2015-06-24T15:39:12.393" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16942" PostTypeId="2" ParentId="16887" CreationDate="2015-06-24T16:07:58.473" Score="12" Body="&lt;p&gt;Here is my answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; D O N E&lt;br&gt;&#xA; O P E N&lt;br&gt;&#xA; N E E D&lt;br&gt;&#xA; E N D S  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Done&quot; is a way to describe conclusion (ending) of an activity or experience.&lt;br&gt;&#xA; &quot;Open&quot; describes a beginning, as in &quot;the &lt;em&gt;open&lt;/em&gt;ing number of a show&quot;.&lt;br&gt;&#xA; &quot;Need&quot; describes a requirement: &quot;You &lt;em&gt;need&lt;/em&gt; good grades to get into MIT.&quot;&lt;br&gt;&#xA; &quot;Ends&quot; is a bit of a stretch, because it's plural, but a &quot;ceasing&quot; is an &lt;em&gt;end&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2015-06-24T16:07:58.473" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16943" PostTypeId="2" ParentId="12011" CreationDate="2015-06-24T17:04:01.333" Score="1" Body="&lt;p&gt;I'll summarise my progress in an answer. Maybe it'll help others get further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fourth and fifth hints suggest we should&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; turn the sequence of 7-tuples of letters into a sequence of pairs by changing the spacing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then the given ciphertext:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1820143 2383341 3245353 8425463 6354607&lt;br&gt;&#xA;  6627972 9088779 7839599 ??????? ???????&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;becomes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 18 20 14 32 38 33 41 32 45 35 38 42 54 63 63 54 60 76 62 79 72 90 88 77 97 83 95 99 ?? ?? ?? ?? ?? ?? ??&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first three hints combine to give&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;+03&lt;/strong&gt;. Maybe we're meant to add 03 to all or some of the 2-digit numbers we now have?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But neither of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 21 23 17 35 41 36 44 35 48 38 41 45 57 66 66 57 63 79 65 82 75 93 91 80 100 86 98 102 and 15 17 11 29 35 30 38 29 42 32 35 39 51 60 60 51 57 73 59 76 69 87 85 74 94 80 92 96&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;has any obvious pattern that I can see...&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-06-24T22:40:28.287" LastActivityDate="2015-06-24T22:40:28.287" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16945" PostTypeId="1" AcceptedAnswerId="16950" CreationDate="2015-06-24T17:11:41.967" Score="17" ViewCount="2232" Body="&lt;p&gt;Inspired by &lt;a href=&quot;https://puzzling.stackexchange.com/users/9951/aggie-kidd&quot;&gt;Aggie Kidd&lt;/a&gt;'s &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16887/another-four-words/&quot;&gt;redux&lt;/a&gt; of the original &quot;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/15339/four-words&quot;&gt;Four-words&lt;/a&gt;&quot; puzzle by &lt;a href=&quot;https://puzzling.stackexchange.com/users/10940/prem&quot;&gt;Prem&lt;/a&gt;, I decided to create a similar &quot;Five-words&quot; puzzle.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some clues are intentionally somewhat vague and/or misleading, to keep it from being &lt;em&gt;too&lt;/em&gt; easy, but they should all make sense when the solution is determined.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My first's what ant calls a cricket, no doubt&lt;br&gt;&#xA;  My second is third when you really zoom out&lt;br&gt;&#xA;  My third will protect you from villain or foe&lt;br&gt;&#xA;  My fourth won't get verdant when it's on the go&lt;br&gt;&#xA;  My fifth's what you did with the game that you lost&lt;br&gt;&#xA;  My whole is a square with five words that are crossed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-07-17T01:03:27.790" Title="Cryptic Five-words (-----|||||)" Tags="|word|wordplay|crosswords|word-square|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16946" PostTypeId="2" ParentId="206" CreationDate="2015-06-24T17:20:04.527" Score="0" Body="&lt;p&gt;Here's another way of looking at it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's imagine each of Barry, Harry, and Larry gets cloned, so there are two copies of each (both with the same working speed, naturally). In 1 hour, we know how much Barry1 and Harry1 together can get done ($\frac{1}{2}$ acre), how much Barry2 and Larry1 together can get done ($\frac{1}{3}$ acre), and how much Harry2 and Larry2 together can get done ($\frac{1}{5}$ acre). So 2 copies of everyone get $\frac{1}{2}+\frac{1}{3}+\frac{1}{5}=\frac{31}{30}$ acres ploughed in an hour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Barry, Harry, and Larry together get half this amount ($\frac{31}{60}$ acres) done in an hour, which means they take $\frac{60}{31}$ hours or about &lt;strong&gt;1 hour 56 minutes&lt;/strong&gt; to plough an acre.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-06-24T17:20:04.527" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16947" PostTypeId="2" ParentId="16769" CreationDate="2015-06-24T17:22:39.987" Score="13" Body="&lt;p&gt;Is it &lt;a href=&quot;https://en.wikipedia.org/wiki/Tux&quot; rel=&quot;nofollow&quot;&gt;Tux&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In yellows and blacks&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He is yellow and black (also white)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He often lives on racks,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Computer racks&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He loves none but Penny.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Penny is his &quot;female counterpart&quot; &lt;a href=&quot;https://en.wikipedia.org/wiki/Tux#Female_Tux_versions_in_video_games&quot; rel=&quot;nofollow&quot;&gt;according to wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13204" LastEditorUserId="13426" LastEditDate="2015-06-26T19:42:07.283" LastActivityDate="2015-06-26T19:42:07.283" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16948" PostTypeId="2" ParentId="16945" CreationDate="2015-06-24T17:32:57.223" Score="0" Body="&lt;p&gt;Partial answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My first's what ant calls a cricket, no doubt&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; JUMPY (compared to an ant, a cricket does a lot of jumping).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My second is third when you really zoom out&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;My third will protect you from villain or foe&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;My fourth won't get verdant when it's on the go&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure about these yet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My fifth's what you did with the game that you lost&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; YIELD (when you lose a game, you yield).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-06-24T17:32:57.223" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16949" PostTypeId="2" ParentId="16945" CreationDate="2015-06-24T17:33:47.567" Score="2" Body="&lt;p&gt;Closest I can figure out is&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;G I A N T&#xA;I _ R _ H&#xA;A R M O R&#xA;N _ O _ E&#xA;T H R E W&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But I am lost on 4, which should be a synonym for stone.&#xA;Got no clue for 2.&lt;/p&gt;&#xA;" OwnerUserId="13068" LastEditorUserId="5373" LastEditDate="2015-06-24T17:35:10.543" LastActivityDate="2015-06-24T17:35:10.543" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16950" PostTypeId="2" ParentId="16945" CreationDate="2015-06-24T17:35:43.147" Score="20" Body="&lt;p&gt;Here is my answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;pre&gt;B E A S T&lt;br&gt;E A R T H&lt;br&gt;A R M O R&lt;br&gt;S T O N E&lt;br&gt;T H R E W&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;br&gt;&#xA;My first's what ant calls a cricket, no doubt  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A cricket would be a &lt;strong&gt;beast&lt;/strong&gt; to an ant.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My second is third when you really zoom out  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;strong&gt;earth&lt;/strong&gt; is the third planet if we zoom out on the solar system.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My third will protect you from villain or foe  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Armor&lt;/strong&gt; protects you from everyone.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fourth won't get verdant when it's on the go  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Verdant is green, moss is green, rolling &lt;strong&gt;stone&lt;/strong&gt; gathers no moss.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fifth's what you did with the game that you lost  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You &lt;strong&gt;threw&lt;/strong&gt; the game. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="9000" LastEditDate="2015-06-24T17:41:04.360" LastActivityDate="2015-06-24T17:41:04.360" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16951" PostTypeId="1" AcceptedAnswerId="16953" CreationDate="2015-06-24T18:06:19.610" Score="8" ViewCount="734" Body="&lt;p&gt;What book am I talking about (this book has been made into a movie too)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;'c'&lt;/code&gt;lee_&amp;amp;_&lt;em&gt;le&lt;/em&gt;_choke l8_(! - ial)ee&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the puzzle has been solved it will be easy to understand how it fits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Language, wordplay and basic programming stuff are included in this puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first  &lt;code&gt;'c'&lt;/code&gt; could have been any other letter, it makes no difference. The  &lt;code&gt;''&lt;/code&gt; are important.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12860" LastEditorUserId="4450" LastEditDate="2015-06-24T18:14:58.777" LastActivityDate="2015-06-24T18:14:58.777" Title="Guess the book/movie?" Tags="|rebus|wordplay|computer-puzzle|language|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16952" PostTypeId="2" ParentId="16769" CreationDate="2015-06-24T18:07:02.273" Score="4" Body="&lt;p&gt;My answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.ysusports.com/fan_zone/mascots&quot; rel=&quot;nofollow&quot;&gt;Pete Penguin&lt;/a&gt; the mascot of Youngstown State University&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;In yellows and blacks,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The costume is mainly black (and white) with a yellow beak&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;He often lives on racks,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The costume is often on clothes racks&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;From Georgia he could be,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There's also a Youngstown Georgia&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;He loves none but Penny.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Penny is his penguin wife.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7715" LastActivityDate="2015-06-24T18:07:02.273" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16953" PostTypeId="2" ParentId="16951" CreationDate="2015-06-24T18:08:42.813" Score="15" Body="&lt;p&gt;I think the following is the answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Charlie and the chocolate factory&lt;br&gt;&#xA; 'c' = char&lt;br&gt;&#xA; lee = lie&lt;br&gt;&#xA; &amp;amp; = and&lt;br&gt;&#xA; le = the&lt;br&gt;&#xA; choke l8 = chocolate&lt;br&gt;&#xA; !-ial = factor&lt;br&gt;&#xA; ee = y&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13268" LastEditorUserId="11475" LastEditDate="2015-06-24T18:11:41.567" LastActivityDate="2015-06-24T18:11:41.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16955" PostTypeId="2" ParentId="16431" CreationDate="2015-06-24T19:27:41.237" Score="4" Body="&lt;p&gt;It can only B&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the decimal digits of the golden ratio, expressed in binary&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12863" LastActivityDate="2015-06-24T19:27:41.237" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16956" PostTypeId="2" ParentId="16431" CreationDate="2015-06-24T19:39:01.423" Score="4" Body="&lt;p&gt;The answer is B,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The dots are a binary representation of the digits in the golden ratio (1.61803398), with each primary color representing a 1, and secondary colors are a 0.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;0001 0110 0001 1000 0000 0011 0011 1000, and the only block that could be 1000 is B, Yellow Purple Purple Purple.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit: looks like I was beaten.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second edit: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the primary/secondary color pairs are also opposites on the color wheel. Starting with '1', the colors cycle through red yellow blue as the digit's primary color for the binary encoding&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13101" LastEditorUserId="13101" LastEditDate="2015-06-25T14:41:48.540" LastActivityDate="2015-06-25T14:41:48.540" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16957" PostTypeId="2" ParentId="12716" CreationDate="2015-06-24T20:39:16.297" Score="3" Body="&lt;h2&gt;An Example with Two Colors&lt;/h2&gt;&#xA;&lt;p&gt;This is impossible.&lt;/p&gt;&#xA;&lt;p&gt;I can find a pattern for matching little blocks to little blocks, big to big, and big to small. The (small x small) and (big x big) towers fit together. However, it is not possible to connect the mixed tower. This stems from the fact that adding a big block adds +1 spacing but also adds +1 spacing &lt;em&gt;requirement&lt;/em&gt; so the net gain is 0. I feel like I'm playing chess and I'm trying to get a bishop to the opposite color square.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uUJ2T.png&quot; alt=&quot;Two Color Towers&quot; /&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;An Example with Three Colors&lt;/h2&gt;&#xA;&lt;p&gt;I believe this may not be possible.&lt;/p&gt;&#xA;&lt;p&gt;Again, I can find small x small and big x big that all connect but the mixed, although it is valid by itself, can't be connected to the rest.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/C7Kyy.png&quot; alt=&quot;Three Color Towers&quot; /&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;An Example with Four Colors&lt;/h2&gt;&#xA;&lt;p&gt;I believe this is also impossible.&lt;/p&gt;&#xA;&lt;p&gt;Again, I can find small x small and big x big that all connect but the mixed can't be connected to the rest. In fact, I couldn't even finish the mixed tower.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GDs1n.png&quot; alt=&quot;Four Color Towers&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;I was unable to integrate the following pairs into the mixed tower:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EmnN7.png&quot; alt=&quot;Missing Blocks in Four Colors&quot; /&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;Weak Lower Bound for Five Colors&lt;/h2&gt;&#xA;&lt;p&gt;If I understand the problem correctly, then a weak lower bound would be 60 blocks tall.&lt;/p&gt;&#xA;&lt;p&gt;What's shown below complies with the rule about pairs of colors being opposite each other but with no regard to the rule about them being spaced apart by the sum of their heights. There are no repeated pairs and no white spaces so I am confident that this is the lower bound. If it is possible to rearrange these to meet the final rule, then the solution would be complete. Otherwise, we must begin inserting white spaces as needed. (The tower is shown broken into pieces that can be stacked upon one another without any white space. This is merely for formatting.)&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GY21c.png&quot; alt=&quot;Lower Bound for Five Colors&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;However, even the four-color pattern is 131 blocks tall and it doesn't even meet all the requirements. I suspect that the five-color solution (if it exists) will be taller.&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-06-25T16:54:39.903" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16958" PostTypeId="1" CreationDate="2015-06-24T20:40:38.307" Score="4" ViewCount="344" Body="&lt;p&gt;Where can I find crossword puzzles with a specific theme, e.g. history, geography, etc. that are at a level suitable for an adult? I realize that something like history is a very general category, but even if the clues are a hodge-podge, that's fine with me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've used search engines to look for puzzles like this, but almost everything I can find is either designed for a classroom and thus geared towards children, or user-generated and far too simple. (&lt;a href=&quot;http://www.whenwecrosswords.com/crossword/atomic_bombings_of_hiroshima_and_nagasaki/1389/crossword_answer.jsp&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; is one absolutely atrocious example). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://meta.puzzling.stackexchange.com/q/3129/13393&quot;&gt;This meta question&lt;/a&gt; is relevant to questions of this type. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In response to the comments, I'm interested in puzzles for myself. The topics I listed above, like history and geography, are some of the themes I'm interested in; once again, I realize these are quite general categories.&lt;/p&gt;&#xA;" OwnerUserId="13393" LastEditorUserId="13393" LastEditDate="2015-06-29T15:41:19.190" LastActivityDate="2015-06-29T15:41:19.190" Title="Where can I find advanced crossword puzzles relating to specific themes, e.g. history?" Tags="|crosswords|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16959" PostTypeId="1" AcceptedAnswerId="16960" CreationDate="2015-06-24T20:44:30.937" Score="9" ViewCount="672" Body="&lt;p&gt;These tend to be as much fun to make as they are to solve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of the &quot;words&quot; are seven letters long and form a square that reads the same down as it does across.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know, like this ancient one: &lt;br /&gt;&#xA;SATOR &lt;br /&gt;&#xA;AREPO &lt;br /&gt;&#xA;TENET &lt;br /&gt;&#xA;OPERA &lt;br /&gt;&#xA;ROTAS &lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this one is seven letters! &lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: What puzzle would be complete without THIS homage to Wonder Woman! &lt;br /&gt;&#xA;2: This is an anagram for Hum Ion Z, but makes no sense. &lt;br /&gt;&#xA;3: This is the Giant of Africa. &lt;br /&gt;&#xA;4: This used to be a popular way to sort out disagreements but wasn't really very civilized. &lt;br /&gt;&#xA;5: This evil, destructive one should eventually be defeated by Ahura Mazda (we hope). &lt;br /&gt;&#xA;6: This word is one quap short and spelled backwards. &lt;br /&gt;&#xA;7: Supporters of Net Neutrality would like to see one of these with an arta. &lt;br /&gt;&lt;/p&gt;&#xA;" OwnerUserId="13068" LastEditorUserId="21234" LastEditDate="2016-07-22T21:05:58.090" LastActivityDate="2016-07-22T21:05:58.090" Title="A new seven-words puzzle!" Tags="|wordplay|word-square|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16960" PostTypeId="2" ParentId="16959" CreationDate="2015-06-24T21:40:21.943" Score="11" Body="&lt;p&gt;This one required a lot of Googling!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;pre&gt;&#xA; B O N D A G E&#xA; O Z I U H N M&#xA; N I G E R I A&#xA; D U E L I N G&#xA; A H R I M A N&#xA; G N I N A L A&#xA; E M A G N A C&#xA;&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: What puzzle would be complete without THIS homage to Wonder Woman!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I assume that &lt;strong&gt;BONDAGE&lt;/strong&gt; is a reference to her Lasso of Truth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2: This is an anagram for Hum Ion Z, but makes no sense.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The anagram is &lt;strong&gt;OZIUHNM&lt;/strong&gt;, and indeed makes no sense.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3: This is the Giant of Africa. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;NIGERIA&lt;/strong&gt;'s nickname is &quot;The Giant of Africa&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4: This used to be a popular way to sort out disagreements but wasn't really very civilized. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People used to sort out disagreements by &lt;strong&gt;DUELING&lt;/strong&gt; (American spelling).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5: This evil, destructive one should eventually be defeated by Ahura Mazda (we hope). &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ahura Mazda is the main entity in Zoroastrianism, and &lt;strong&gt;AHRIMAN&lt;/strong&gt; is the evil spirit defined by Zoroastrianism.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6: This word is one quap short and spelled backwards. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Take the word &lt;strong&gt;aquaplaning&lt;/strong&gt;, remove the &quot;quap&quot;, and reverse it to get &lt;strong&gt;GNINALA&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;7: Supporters of Net Neutrality would like to see one of these with an arta.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The want an &lt;strong&gt;e-Magna-Carta&lt;/strong&gt;.  Without the &quot;arta&quot;, it becomes &lt;strong&gt;EMAGNAC&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="12109" LastEditDate="2015-06-24T21:56:31.100" LastActivityDate="2015-06-24T21:56:31.100" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16961" PostTypeId="2" ParentId="16929" CreationDate="2015-06-24T22:34:16.157" Score="5" Body="&lt;p&gt;I get the distinct impression that Mrs. Dunham drinks&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tea, which she obviously drinks at the three teatimes throughout the day (afternoon tea, high tea and evening tea).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hell is where a little old lady named Mrs. Dunham lives.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As others have mentioned, there are cities/towns named Hell throughout the world so this likely refers to one of those. I don't it matters which one. Dunham is also a traditionally British surname, implying that Mrs. Dunham is British. At the very least, she is married to a British man and it is not too far a stretch to assume she may have adopted some of his habits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mrs. Dunham is a very old-fashioned lady.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I know this is a racial stereotype but it also happens to be generally true that most old-fashioned British folk drink tea regularly throughout the day. At the very least, it implies that she observes older customs and drinking tea would certainly fall into that category.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mrs. Dunham has a bit of a drinking problem.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I got nothing of value from this statement so I'm assuming it's thrown in as misdirection to make people think she's drinking alcohol.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Three times a day, every day, she drinks.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are three teatimes during the day: afternoon tea, high tea and evening tea. Being old-fashioned, she probably observes all of these. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mrs. Dunham is a bit used to warmer climates and has to have her drink imported because it's not as popular in Hell as it was in her hometown.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Teatime is not nearly as popular outside the UK as it is in it so wherever Hell is, chances are the locals don't drink tea as religiously as she does. It's not unreasonable to conclude that her variety of choice is not produced locally and would need to be imported.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Other reasons:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tea is generally considered to be quite healthy and I'm not aware of any health issues that would arise from drinking it frequently. This fits with the clue that she is surprisingly healthy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13541" LastActivityDate="2015-06-24T22:34:16.157" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16964" PostTypeId="2" ParentId="5880" CreationDate="2015-06-25T02:38:36.270" Score="8" Body="&lt;p&gt;&lt;em&gt;I posted this answer &lt;a href=&quot;http://meta.puzzling.stackexchange.com/questions/1717/code-puzzles-what-not-to-do&quot;&gt;here&lt;/a&gt;, and &lt;a href=&quot;http://meta.puzzling.stackexchange.com/users/4551/xnor&quot;&gt;xnor&lt;/a&gt; suggested I also post it on this question.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;An article on this topic that I've found very useful is &lt;a href=&quot;http://irregularwebcomic.net/3244.html&quot;&gt;this one&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A brief summary:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Puzzles should be &lt;strong&gt;compelling&lt;/strong&gt;.  Provide something that hooks the solver, and makes them &lt;em&gt;want&lt;/em&gt; to solve the puzzle.&lt;/li&gt;&#xA;&lt;li&gt;Avoid &lt;strong&gt;red herrings&lt;/strong&gt; that cause people to spend significant time going down the wrong path without getting closer to a solution.&lt;/li&gt;&#xA;&lt;li&gt;If the puzzle requires an &lt;strong&gt;intuitive leap&lt;/strong&gt;, it's better if that leap is not too difficult.  A puzzle with an easier &quot;intuitive leap&quot; and a hard &quot;processing&quot; section is better than the opposite.&lt;/li&gt;&#xA;&lt;li&gt;A puzzle should provide some indication of &lt;strong&gt;progress&lt;/strong&gt;.  If there are multiple steps, it should be obvious when one step has been correctly completed.  A multi-step puzzle with no feedback along the way can be nearly impossible.&lt;/li&gt;&#xA;&lt;li&gt;A puzzle should &lt;strong&gt;consume all the information&lt;/strong&gt; provided.  There should not be extraneous information that doesn't help find a solution; it will make the solver feel like something has been missed.&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A puzzle that &lt;strong&gt;reuses&lt;/strong&gt; information (information is interpreted in two different ways to provide two different steps to solving the puzzle) is especially clever.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;A clever tool can be &lt;strong&gt;recursion&lt;/strong&gt;.  If the first step of the puzzle leaves you with a certain amount of information, re-applying that step can be the way to get the final solution.&lt;/li&gt;&#xA;&lt;li&gt;A &lt;strong&gt;theme&lt;/strong&gt; helps hold the puzzle together and makes it more compelling to the solver.  Sometimes tying all elements of the puzzle into one common theme can be difficult, but the more you can do it, the more interesting it is to the solver.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2015-06-25T02:38:36.270" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16965" PostTypeId="2" ParentId="12011" CreationDate="2015-06-25T02:49:23.093" Score="4" Body="&lt;p&gt;I've thought about this for far too long and am fairly confident a leap of logic is required at this point in time without further hints. Here's a partial answer though:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; From this graph, there's definitely a linear progression if the numbers are broken up into every two. The second column has the differences in case anyone can see a pattern in those.&lt;br&gt;&lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/mh0Wg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastActivityDate="2015-06-25T02:49:23.093" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16966" PostTypeId="2" ParentId="16914" CreationDate="2015-06-25T04:42:50.023" Score="0" Body="&lt;p&gt;I think you should go with number 3 because it almost matches the pattern in the original paper. However, other options also look reasonable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edited:&#xA;It depends on what you see at the &quot;pattern&quot; in the original diagram. Not everyone sees W, X, Y, and Z as the pattern in the given figure. There is an unique pattern by using number 3 and it looks cool.&lt;/p&gt;&#xA;" OwnerUserId="13516" LastEditorUserId="13516" LastEditDate="2015-06-25T05:59:26.940" LastActivityDate="2015-06-25T05:59:26.940" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16967" PostTypeId="2" ParentId="12011" CreationDate="2015-06-25T07:35:16.970" Score="7" Body="&lt;p&gt;I think I have it.&#xA;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 9697011 6131119&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The idea is as follows, based on Quark's work, and on Hint 6 (&quot;Counter +03&quot;).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First, split the strings into couplets giving:&lt;br&gt;18 20 14 32 38 33 41 32 45 35 38 42 54 63 63 54 60 76 62 79 72 90 88 77 97 83 95 89 ?? ?? ?? ?? ?? ?? ??. &lt;br&gt;After that, subtract 3 times the couplet's ordinal place from it giving the new string:&lt;br&gt;15 14 05 20 23 15 20 08 18 05 05 06 15 21 18 06 09 22 05 19 09 24 19 05 22 05 14 05&lt;br&gt;From here, mapping each number to a letter is trivial, as it's just its position in the alphabet, so we get the string:&lt;br&gt;O  N  E  T  W  O  T  H  R  E  E  F  O  U  R  F  I  V  E  S  I  X  S  E  V  E  N  E&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My guess is that we continue the next two numbers in plain text, convert them backwards, and get the answers at the top, which gives us:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I   G   H   T   N   I   N&lt;br&gt;&#xA; 09  07  08  20  14  09  14&lt;br&gt;&#xA; +87 +90 +93 +96 +99 +02 +05&lt;br&gt;&#xA; 96  97  01  16  13  11  19&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13258" LastEditorDisplayName="user88" LastEditDate="2015-06-25T14:13:20.783" LastActivityDate="2015-06-25T14:13:20.783" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16968" PostTypeId="1" AcceptedAnswerId="16975" CreationDate="2015-06-25T08:26:24.160" Score="14" ViewCount="2272" Body="&lt;p&gt;My friend has these pictures up on the wall. What do you think he's crazy about?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/52ctc.jpg&quot; alt=&quot;First image&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/RLKdv.jpg&quot; alt=&quot;Second image&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Rt5Np.jpg&quot; alt=&quot;Third image&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/iRywD.jpg&quot; alt=&quot;Fourth image&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(These pictures have a common link, and the correct answer will explain each image in the same way.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt;&lt;br&gt;&#xA;The title is also a clue.&lt;/p&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-06-25T09:40:03.077" LastActivityDate="2017-11-18T23:22:58.583" Title="Pictures On The Wall" Tags="|visual|" AnswerCount="5" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16969" PostTypeId="2" ParentId="16958" CreationDate="2015-06-25T08:52:38.760" Score="1" Body="&lt;p&gt;Virgilius in the Independent and Brendan in Guardian are two setters who regularly set themed puzzles. Themed daily puzzles published in British crosswords can be found at &lt;a href=&quot;http://www.fifteensquared.net/?s=themed&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hindu Crossword also features themed puzzles once in a while. These can be looked up in &lt;a href=&quot;http://www.thehinducrosswordcorner.blogspot.com&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2425" LastActivityDate="2015-06-25T08:52:38.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16970" PostTypeId="2" ParentId="16968" CreationDate="2015-06-25T09:03:08.123" Score="1" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Running/a run&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Two dogs running, money run the world (or not running out of money), a moon running low (yes, there is such a thing, if it is high it is &quot;riding high&quot;), and time is running out (we have like less than 30 minutes till noon/midnight).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastActivityDate="2015-06-25T09:03:08.123" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16971" PostTypeId="2" ParentId="16968" CreationDate="2015-06-25T09:17:37.257" Score="2" Body="&lt;p&gt;Seems a bit obvious, but maybe he's crazy about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; dogs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Picture 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;running dogs&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Picture 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a 100-dollar bill is called a C-note, so C-dogs or &lt;strong&gt;seadogs&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Picture 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;moondogs&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Picture 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;watchdogs&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-06-25T09:17:37.257" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16975" PostTypeId="2" ParentId="16968" CreationDate="2015-06-25T09:57:25.160" Score="18" Body="&lt;p&gt;Oh well, to me, your friend is almost surely&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a devoted Pink Floyd fan (and how could he not?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The images represent&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pink Floyd songs: Dogs, Money, Eclipse, Time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the title, well,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the album The Wall&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, about the description: I would say that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'crazy' could go very well with &quot;Shine On You Crazy Diamond&quot; in particular, and with a good half of their discography in general.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5990" LastEditorUserId="5990" LastEditDate="2015-06-25T10:18:43.717" LastActivityDate="2015-06-25T10:18:43.717" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16976" PostTypeId="1" AcceptedAnswerId="16977" CreationDate="2015-06-25T10:54:31.987" Score="2" ViewCount="513" Body="&lt;p&gt;Below is a sequence of four-digit numbers and characters that are associated with them.&lt;br&gt;&#xA;Can you crack the code and find out what &lt;code&gt;2963 = ?&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;8446 = F&#xA;5697 = C&#xA;2331 = A&#xA;1125 = A&#xA;1662 = C&#xA;4658 = E&#xA;8312 = C&#xA;6146 = D&#xA;4115 = B&#xA;8066 = F&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="13432" LastEditorUserId="11876" LastEditDate="2016-02-05T17:37:18.453" LastActivityDate="2016-02-05T17:37:18.453" Title="Crazy number puzzle.. Any Mathematicians?" Tags="|pattern|" AnswerCount="1" CommentCount="3" ClosedDate="2015-06-25T11:09:47.020" ContentLicense="CC BY-SA 3.0" />
  <row Id="16977" PostTypeId="2" ParentId="16976" CreationDate="2015-06-25T11:00:31.520" Score="3" Body="&lt;p&gt;2963 equals&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; given a number, count the total number of holes in all its digits, add 1, and take the letter at this position in the alphabet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2331 and 1225 go to A because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; they have no digits with holes,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5697 and 1662 go to C because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; they have exactly 2 holes,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;8446 goes to F because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it has exactly 5 holes,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and so on.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-06-25T11:02:34.487" LastActivityDate="2015-06-25T11:02:34.487" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16979" PostTypeId="2" ParentId="16968" CreationDate="2015-06-25T12:29:52.733" Score="5" Body="&lt;p&gt;Although it's already answered, I was convinced that the answer was something along the lines of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Things that go away and then come back&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1st&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dogs play fetch (go out and come back)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2nd&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is a rubber band on the bill (stretches out then snaps back)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3rd&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The moon goes up then back down every day&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4th&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is the one that makes it kind of a stretch, but the watch hands go up and back down&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13556" LastEditorUserId="13556" LastEditDate="2016-07-28T00:43:52.863" LastActivityDate="2016-07-28T00:43:52.863" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16980" PostTypeId="2" ParentId="12011" CreationDate="2015-06-25T12:59:34.110" Score="6" Body="&lt;p&gt;The two seven digit numbers to be added are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;9697011 6131119&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1820143 2383341 3245353 8425463 6354607&#xA;6627972 9088779 7839589 ??????? ???????&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Based on hint 4, split the input into pairs of digits:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;18 20 14 32 38 33 41 32 45 35 38 42 54 63 63 54 60 76 62 79 72 90 88 77 97 83 95 89 ?? ?? ?? ?? ?? ?? ??&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Graphing these numbers shows that the numbers trend upward by a factor of three for each character.  This is what is referred to by hints 1, 2, 3, and 6. &#xA;(Thanks to Quark for the graph)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mh0Wg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reducing these numbers by the factor of three gives:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;15 14 05 20 23 15 20 08 18 05 05 06 15 21 18 06 09 22 05 19 09 24 19 05 22 05 14 05&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Noticing that these are all less than 26 convert them to characters yields:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;O N E T W O T H R E E F O U R F I V E S I X S E V E N E&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now using hint 4 and 5, and noticing that the sequence of characters ends with &quot;E&quot; then continuation is E I G H T N I N E ... Reverse calculating the new numbers for the plain text is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;15 14 05 20 23 15 20 08 18 05 05 06 15 21 18 06 09 22 05 19 09 24 19 05 22 05 14 05 09 07 08 20 14 09 14&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Adding the slope back in and taking the modulo by 100 (two digits only) yields:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;18 20 14 32 38 33 41 32 45 35 38 42 54 63 63 54 60 76 62 79 72 90 88 77 97 83 95 89 96 97 01 16 13 11 19&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Grouping the numbers back into sevens:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1820143 2383341 3245353 8425463 6354607&#xA;6627972 9088779 7839589 9697011 6131119&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Thanks to rand'al thor for bumping the puzzle up, and Quark for saving me from making my own graph.&lt;/p&gt;&#xA;" OwnerUserId="1777" LastEditorDisplayName="user88" LastEditDate="2015-06-25T14:32:19.340" LastActivityDate="2015-06-25T14:32:19.340" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16981" PostTypeId="2" ParentId="16935" CreationDate="2015-06-25T13:20:30.647" Score="10" Body="&lt;p&gt;The guardian  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;is mostly deaf, hence the shouting. He won't allow anyone in who is loud enough for him to hear  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Everyone else's answers  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;were accompanied by them holding up the relevant number of fingers. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The person who didn't seem to answer a question  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;actually did, by showing 5 fingers in response.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You should have said:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Absolutely anything, but not loudly enough for the guardian to hear. You should also have held up four fingers  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;second guess&lt;/strong&gt; (to be expounded upon if correct)  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Is the answer 22?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11717" LastEditorUserId="11717" LastEditDate="2015-06-25T13:39:23.090" LastActivityDate="2015-06-25T13:39:23.090" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16982" PostTypeId="1" CreationDate="2015-06-25T14:54:02.817" Score="10" ViewCount="1136" Body="&lt;p&gt;The founder and owner of the company you work for has unexpectedly passed away, upon his death this email was sent to all employees:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hello All!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Apparently I have kicked the bucket and have no family to speak of!&#xA;  Being the eccentric Billionaire I am, I have decided to give my fortune away to one of my loyal employees! The first of you to reply to this email with the correct answer to this question will &#xA;  inherit all my riches! &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;What is the secret to my success?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I realize this may be a difficult question so here is a clue!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Xg mhu ad rzga jsmn ofo gxt&#xA;  Zhi uvxk ps nd nubww nyfw zeceo mvpa xva&#xA;  Gdhp, cxywpxod dojcf voc toa wqdjel mpkid&#xA;  Kjtsrnh asacc tl rkupc ta rjqvl&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Good luck and have fun!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Mr. McMoneybags&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;CEO, Founder, Owner&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;The only way to get results is with intelligence and hard work&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is Mr. McMoneybags secret?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;While going over this email for the second time you realize something! Going back through Mr. McMoneybags' previous company wide emails you see he has never had that quote as part of his signature... hmm....&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TroyAndAbed has solved the Cryptogram! All that is left is the riddle!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first three lines of the deciphered riddle all reference the same word, but not the same thing. (i.e. homograph/homonyms). The final line is mostly there for rhyming purposes, but also includes a small clue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; EngineerToast was correct in assuming &quot;chips&quot; was computer chips. However, don't forget about Hint #3. Also this is something many many businessmen use every day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13204" LastEditorUserId="13204" LastEditDate="2015-06-29T17:44:25.550" LastActivityDate="2015-06-29T17:44:25.550" Title="Become a billionaire!" Tags="|riddle|cipher|" AnswerCount="7" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16984" PostTypeId="1" AcceptedAnswerId="16985" CreationDate="2015-06-25T15:10:00.113" Score="14" ViewCount="1017" Body="&lt;p&gt;My &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16945/cryptic-five-words&quot;&gt;last Five-words puzzle&lt;/a&gt; seemed well-received, so I thought I'd try another.  This one has a bit of a twist to it, which shouldn't be too hard to figure out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I expect it will be easier to solve overall than the previous one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My first can be pleasant, or horribly foul&lt;br&gt;&#xA;  My second is useful to hunt waterfowl&lt;br&gt;&#xA;  My third is an act that can cause desolation&lt;br&gt;&#xA;  My fourth might be done to a sleepy dalmatian&lt;br&gt;&#xA;  My fifth talks of galleons -- many, not one&lt;br&gt;&#xA;  With just these five clues, you are only half done.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;My first keeps you warm when a cold wind is blowing&lt;br&gt;&#xA;  My second is music to sing when it's snowing&lt;br&gt;&#xA;  My third is an action describing what follows&lt;br&gt;&#xA;  My fourth could be used to catch rabbits or swallows&lt;br&gt;&#xA;  My fifth is a truth that you hold very dear&lt;br&gt;&#xA;  With all these ten clues, the result should be clear&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-07-17T01:05:17.583" Title="Another cryptic Five-word (with a twist)" Tags="|word|wordplay|crosswords|word-square|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16985" PostTypeId="2" ParentId="16984" CreationDate="2015-06-25T15:16:10.270" Score="14" Body="&lt;p&gt;I'm going to go with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;pre&gt;S C E N T&lt;br&gt;C A N O E&lt;br&gt;A R S O N&lt;br&gt;R O U S E&lt;br&gt;F L E E T&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My first can be pleasant, or horribly foul&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A &lt;strong&gt;scent&lt;/strong&gt; can either be very nice, or pretty awful (for instance, compare roses and skunks).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My second is useful to hunt waterfowl&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people hunt ducks and other waterfowl from a &lt;strong&gt;canoe&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My third is an act that can cause desolation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can cause desolation by burning the place down, which is an act of &lt;strong&gt;arson&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fourth might be done to a sleepy dalmatian&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You &lt;strong&gt;rouse&lt;/strong&gt; something to wake it up, which you could do to anything sleepy, including a dalmatian!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fifth talks of galleons -- many, not one&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A galleon is a sailing ship, which in multiples creates a &lt;strong&gt;fleet&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My first keeps you warm when a cold wind is blowing&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A &lt;strong&gt;scarf&lt;/strong&gt; worn around your neck protects you from the cold wind.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My second is music to sing when it's snowing&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is a pretty direct reference to a Christmas &lt;strong&gt;carol&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My third is an action describing what follows&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What is about to &lt;strong&gt;ensue&lt;/strong&gt; is what will follow.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fourth could be used to catch rabbits or swallows&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can catch rabbits or swallows with a rope trap, or a &lt;strong&gt;noose&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fifth is a truth that you hold very dear&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A &lt;strong&gt;tenet&lt;/strong&gt; is a truth or belief, specifically a religious one, which you certainly do hold dear.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="11475" LastEditDate="2015-06-25T15:21:27.467" LastActivityDate="2015-06-25T15:21:27.467" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16986" PostTypeId="2" ParentId="16929" CreationDate="2015-06-25T15:17:11.073" Score="10" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mrs. Dunham drinks a Dr. Pepper every day at 10:00 am, 2:00 pm, and 4:00 pm.  Mrs. Dunham's hometown is Waco, Texas, where Dr. Pepper was created.  Its old slogan was: &quot;It's good at ten, two, and four.&quot;  Since those are the times when it's good, those are the times when Mrs. Dunham drinks it.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/kMY9s.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4385" LastEditorUserId="4385" LastEditDate="2015-06-25T17:09:48.283" LastActivityDate="2015-06-25T17:09:48.283" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16987" PostTypeId="1" AcceptedAnswerId="16990" CreationDate="2015-06-25T15:39:54.837" Score="11" ViewCount="4158" Body="&lt;p&gt;You have an $m$ by $n$ rectangular grid. Each cell may contain a single bomb or a number (or neither, but not both). If a cell has no bomb and is adjacent (horizontally, vertically or diagonally) to atleast 1 bomb-containing cell, then it must have a number. The number (between $1$ and $8$) is the number of cells adjacent to it that contain bombs. This is how you must set up a grid for the opponent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All cells with numbers are visible to the opponent. All cells with bombs and all empty cells look the same, these are the ones that need to be probed. The opponent picks up one cell at a time. If he picks up a bomb, he loses. If he picks up every blank cell, but no bomb, he wins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is it possible to set up a grid, where the opponent cannot use logic alone to win, and will have to use guesswork at some point?&lt;/strong&gt; If so, please give the smallest* possible grid setup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Twist: Instead of a 2D grid, if we have a 3D grid ($l$ by $m$ by $n$). All rules same as above, except that now a number could be anything between $1$ and $26$. What is the smallest* unsolvable grid setup?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*smallest in terms of area/volume of grid, not number of bombs&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-07-01T17:49:21.610" Title="Unsolvable static minesweeper puzzle? (+ 3D twist!)" Tags="|logical-deduction|game|" AnswerCount="6" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16990" PostTypeId="2" ParentId="16987" CreationDate="2015-06-25T15:52:01.120" Score="21" Body="&lt;p&gt;As has been pointed out, the smallest possible 2D grid is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1x3, which looks like:&#xA; &lt;pre&gt;X 1 X&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we discount any grids with a length of 1 on either side, then the smallest possible 2D grid is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2x3, which looks like:&#xA; &lt;pre&gt;X 2 X&lt;br&gt;X 2 X&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For a 3D grid (assuming no side can have a length of 1), the smallest grid is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2x2x3, which has the following faces:&#xA; &lt;pre&gt;X 4 X     X 4 X&lt;br&gt;X 4 X     X 4 X&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, a possible layout of unsolvable grids (and the rule that describes the smallest examples for each dimensional count) can be summarized as&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Grids where the numbers are &lt;strong&gt;rotationally symmetrical&lt;/strong&gt; and the bombs/empty spaces are &lt;strong&gt;not rotationally symmetrical,&lt;/strong&gt; meaning that if the entire grid is rotated 180 degrees, the numbers would match up with the original grid, but the bombs and blank spaces would not. Note that this is not all-inclusive: Grids that follow this pattern are always unsolvable, though there are unsolvable grids that do not follow this pattern.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="11475" LastEditDate="2015-06-26T17:16:58.847" LastActivityDate="2015-06-26T17:16:58.847" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16992" PostTypeId="2" ParentId="15950" CreationDate="2015-06-25T15:56:50.213" Score="4" Body="&lt;p&gt;Got it! Here's a badly-done drawing:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WxJFN.png&quot; alt=&quot;answer&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2015-06-25T15:56:50.213" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16993" PostTypeId="2" ParentId="16987" CreationDate="2015-06-25T15:58:25.997" Score="16" Body="&lt;p&gt;A 1x3 grid will work&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;? 1 ?&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(Where a question mark is either a blank or a mine)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No 2x2 grid will work because all cells are adjacent to one another&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I suppose that one might not qualify as a 2-dimensional grid, because then it would be a 1x1x3 3-D grid too. The following 3x3 grid should be fine:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;? 1 ?&#xA;1 2 1&#xA;? 1 ?&#xA;&lt;/pre&gt;&#xA;" OwnerUserId="13592" LastEditorUserId="9000" LastEditDate="2015-06-25T16:59:25.563" LastActivityDate="2015-06-25T16:59:25.563" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16994" PostTypeId="2" ParentId="16987" CreationDate="2015-06-25T16:01:27.283" Score="3" Body="&lt;p&gt;A simple 1x3 grid can be unsolvable:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ?1?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The solution doesn't rely on the grid's trivial size, as you can create unsolvable repeating columns in an arbitrarily large grid (as long as m or n is odd).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    ?2?3?2?&lt;br&gt;&#xA;    ?3?3?3?&lt;br&gt;&#xA;    ?3?3?3?&lt;br&gt;&#xA;    ?2?2?2?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7459" LastEditorUserId="7459" LastEditDate="2015-06-25T16:08:38.810" LastActivityDate="2015-06-25T16:08:38.810" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16995" PostTypeId="2" ParentId="16987" CreationDate="2015-06-25T16:16:01.620" Score="1" Body="&lt;p&gt;Wellwell already gave the best answer. If you want it to be embeddable in a larger grid - that is, if you don't want it to depend on the border - then it's impossible if the player is automatically notified when they've won. Here's a proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take a minimal embeddable grid. &quot;Minimal&quot; in this sense doesn't mean that it's the smallest possible - it just can't have any extraneous mines that are instantly detectable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's look at this grid. The top left corner looks like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;.-----------&#xA;|A B C X ...&#xA;|D E F X ...&#xA;|X X X X ...&#xA;|...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;A, B, C, or D cannot contain mines; if they were embedded in a larger grid, then it would be immediately possible to figure that out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume A contains a number. That number must be 1, and therefore E must have a mine. The puzzle must not have any extraneous mines; contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore A must be blank, and E must have a number or blank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we relabel the squares.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;.-------------&#xA;|  A B C X ...&#xA;|X D E F X ...&#xA;|X X X X X ...&#xA;...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The same proof shows that once again the square now labeled A must be blank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuing to the edge of the grid:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;          ---.&#xA;  ...     A B|&#xA;  ...   X D E|&#xA;        X X X|&#xA;          ...|&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We know from the previous iteration that D must be a number or a blank. Because A, B, and E are on the border, none of them can be mines. Therefore, there are no mines in the top two rows - the entire top row must be blank. There are also no mines in the bottom row because the grid must be embeddable, as shown at the beginning. The grid must have no size at all; contradiction.&lt;/p&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="11876" LastEditDate="2015-06-25T16:57:48.113" LastActivityDate="2015-06-25T16:57:48.113" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16996" PostTypeId="1" AcceptedAnswerId="17001" CreationDate="2015-06-25T16:39:54.677" Score="37" ViewCount="4368" Body="&lt;p&gt;Quite contrary to the common belief, owls are very social animals and indeed form flocks or rather tightly bound families with a common social codex. At sunset, they all meet before going on their individual hunting trip, and they eye themselves very carefully. Any owl not belonging to the group is sho-who-whoed away immediately! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this evening, the owls are all excited and agitated: A single owl from another region has sneaked into the flock and is trying to pass as one of them! This can not be! &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can you identify the owl which is behaving anti-socially (i.e. not following the flock rules) in the image below? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A valid answer must contain the flock-rules and why that particular owl is not following it. The puzzle can be solved with a black-and-white printout of the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jxVB1.png&quot; alt=&quot;Owl-Picture&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="48462" LastEditDate="2018-09-13T04:36:18.683" LastActivityDate="2018-09-13T04:36:18.683" Title="Owls at sunset - find the odd one out" Tags="|pattern|visual|odd-one-out|" AnswerCount="2" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="16997" PostTypeId="2" ParentId="16996" CreationDate="2015-06-25T16:43:47.683" Score="4" Body="&lt;p&gt;Here's my answer:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FyOar.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This one little owl is looking away from the rest, while all other owls are looking at another one with at least 1 eye&lt;/p&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2015-06-25T16:43:47.683" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16999" PostTypeId="1" AcceptedAnswerId="17035" CreationDate="2015-06-25T17:32:43.933" Score="3" ViewCount="751" Body="&lt;p&gt;You have been taken prisoner and thrown in a dungeon. You have escaped your cell, are standing in front of two identical doors. One will lead you to freedom, the other to a place where you will die a painful death. Luckily, you have a friend that has agreed to help you. He has smuggled you a code inside of a fake record, but had to encode it in case of them finding out that he was helping their POW to escape, tracing who had sent it and taking him prisoner as well. The message reads this: &lt;code&gt;38 23 53 48; 38 23 53 50 --- 38 23 53 48; 38 23 54 53 --- 38 23 53 48; 38 23 54 55 --- 38 23 53 49; 38 23 54 55 --- 38 23 53 49; 38 23 53 50&lt;/code&gt; Which door should you go through: Left or Right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note 1: This puzzle requires a certain degree of computer knowledge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note 2: This CJam program may help: &lt;code&gt;q'[,65&amp;gt;Dm&amp;gt;_el+_$er&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user9377" LastActivityDate="2015-06-27T02:52:31.713" Title="The door of death!" Tags="|cipher|computer-puzzle|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17000" PostTypeId="2" ParentId="16999" CreationDate="2015-06-25T17:50:01.790" Score="5" Body="&lt;p&gt;Here's an answer without having decoded it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Go right. There are 5 blocks of code:&lt;pre&gt;38 23 53 48; 38 23 53 50&lt;br&gt;38 23 53 48; 38 23 54 53&lt;br&gt;38 23 53 48; 38 23 54 55&lt;br&gt;38 23 53 49; 38 23 54 55&lt;br&gt;38 23 53 49; 38 23 53 50&lt;br&gt;&lt;/pre&gt; And the word RIGHT has 5 letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastActivityDate="2015-06-25T17:50:01.790" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17001" PostTypeId="2" ParentId="16996" CreationDate="2015-06-25T18:40:08.790" Score="27" Body="&lt;p&gt;Is it the 3rd owl in the 6th row? The pattern I have seen is that &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; each owl must have a unique eye orientation in either its row or in its vertical column. The owl I marked in red in the below image has the exact same eye orientation as the owl two places to its right (my right, owl's left) and two places above it. &lt;br&gt;&#xA; If one creates an image with &quot;unique eye-pairs&quot; which appear at least 3 times, then one gets:&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/B7wMA.png&quot; alt=&quot;Uniques&quot;&gt;&lt;br&gt;&#xA; and there is only one square which sits in a column &lt;em&gt;and&lt;/em&gt; row which hosts another identically coloured pixel (light blue). This is the odd one out!&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So this is a copycat owl who needs to be sho-who-ed away before he learns the ways of this flock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/Tosjh.png&quot; alt=&quot;Owls.png&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="5840" LastEditDate="2015-06-25T19:08:15.993" LastActivityDate="2015-06-25T19:08:15.993" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17002" PostTypeId="1" AcceptedAnswerId="17365" CreationDate="2015-06-25T18:41:02.810" Score="6" ViewCount="1324" Body="&lt;p&gt;&lt;strong&gt;Disclaimer:&lt;/strong&gt; I think I have a habit of making my riddles a bit too obscure so I'm trying to tone that down this time. Hopefully it won't end up being too easy and solved in about half a second. **The title is not a clue, just thematic with the colors - you could see it as a red herring.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;it was a bright morning, and Everyone here was happy. mostly. here's what happened:  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;i got up, tuckered down a donut for breakfast, went to church to look at the sisters (don't judge me dick). it's clearly an Error but the doc predicted dexter wouldn't kill anyone, just like michael. E las robotico disgustationadad del sargio. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;obviously life is always better with a bit of color so we were wearing:&lt;br&gt;&#xA;  brown - crimson - red - orange - yellow - turquoise - blue - indigo - violet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What color is missing in the list above (what color should be in the list but isn't)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In efforts to avoid more answers just stating a random reason to back any color, the correct answer should also explain the full color list.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Couple of days later, soMe redneck texAn arrIved with a tank and everything went to Hell. but we're stIll going, further thaN smallville or alMost anything else.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;New Hint&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Six Pedals, Four Directions - there is no headlight fluid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastEditorUserId="11251" LastEditDate="2015-07-03T15:37:30.797" LastActivityDate="2015-07-14T16:02:35.850" Title="Coloring the Sky" Tags="|riddle|enigmatic-puzzle|knowledge|" AnswerCount="3" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="17003" PostTypeId="2" ParentId="17002" CreationDate="2015-06-25T18:54:47.433" Score="5" Body="&lt;p&gt;I think the missing colour is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; turquoise.&lt;br&gt;&lt;br&gt;It is missing in the sense that prior to the line enlisting the colours, all the other colours have their letters appearing in the preceding passage and they can be found in order too.&lt;br&gt;&lt;br&gt;Consider this for 'crimson':&lt;br&gt;it was a bright morning, and Everyone was happy. mostly. here's what happened:&lt;br&gt;&lt;br&gt;i got up, tu&lt;b&gt;c&lt;/b&gt;ked down a donut for b&lt;b&gt;r&lt;/b&gt;eakfast, went to church to look at the s&lt;b&gt;i&lt;/b&gt;sters (don't judge &lt;b&gt;m&lt;/b&gt;e dick). it'&lt;b&gt;s&lt;/b&gt; clearly an Err&lt;b&gt;o&lt;/b&gt;r but the doc predicted dexter would&lt;b&gt;n&lt;/b&gt;'t kill anyone, just like michael. E las robotico disgustationadad del sargio.&lt;br&gt;&lt;br&gt;By this logic, 'turqoise' cannot be found in the passage as there is no other word containing 'q' in the passage.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-06-25T18:54:47.433" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17004" PostTypeId="1" AcceptedAnswerId="18322" CreationDate="2015-06-25T18:59:13.620" Score="11" ViewCount="933" Body="&lt;p&gt;Your crazy uncle Tony is at it again, scribbling down incredibly unhelpful tidbits in order to ask a very simple question. Still, for his peace of mind (and your own), you better solve it anyways. Flipping open his overused notepad, you read:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;**What happens when the boxer is taken down? A beautiful mermaid, laughing out loud? Not quite!&lt;br&gt;&#xA;  **I don't know what a &lt;strong&gt;ryyvbggxrzore&lt;/strong&gt; or an &lt;strong&gt;ohmmsrrq&lt;/strong&gt; is, but I expect they were part of the qbg-pbz bubble.&lt;br&gt;&#xA;  **Enter the code, and make note of what you see! Take 12121 (though 22121 will work in a pinch), and memorize it. (Okay, you don't have to memorize it, but at least write it down.)&lt;br&gt;&#xA;  **You're like a gargoyle in complexion. Someone must have put a curse on you, or something like that.&lt;br&gt;&#xA;  **Here, we don't believe in order of operations. Remove it from the system.&lt;br&gt;&#xA;  **The remainder will be comical. Excellent cake, seared deeply, sounds like it will start you on the right path.&lt;br&gt;&#xA;  **At the end of the day, can you tell me who won the duel?&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So who was it that won the duel?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; So far, everything up until 'You're like a gargoyle' is correct. You may be overthinking that specific tidbit. I would also be on the lookout for a vitally important synonym in the near future.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; juansal12 was right in his interpretation of the 'gargoyle in complexion'. The remainder of that line will be your key to finishing this puzzle. If you find yourself searching for someone who might put a curse on someone else, you've gone in the wrong direction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I may have misjudged my wording a bit in the ramblings. &quot;You're&quot; and &quot;you&quot; in the fourth line refer to the word previously found in the third line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 4 (and hopefully the last):&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this hint, I'll draw attention to my comment on an answer below, which has solved the puzzle up until the last two lines:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As for the last two lines, well, I'd love to tell you that I made it through an entire puzzle without a cryptic crossword reference, but unfortunately, I cannot...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="11475" LastEditDate="2015-07-15T19:52:14.347" LastActivityDate="2015-07-24T12:25:28.413" Title="Who won the duel? (AKA ramblings of a madman)" Tags="|enigmatic-puzzle|knowledge|" AnswerCount="5" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="17005" PostTypeId="2" ParentId="15950" CreationDate="2015-06-25T19:26:11.637" Score="0" Body="&lt;p&gt;This is the result that I got. It is drawn with fifteen lines, none of which intersect. This is how I interpreted the rules.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The start of the path and the end of the path must be at the point designated on the map. (The path can be traveled in reverse.)&lt;/li&gt;&#xA;&lt;li&gt;The path must follow the paths marked on the map.&lt;/li&gt;&#xA;&lt;li&gt;The path may not intersect itself.&lt;/li&gt;&#xA;&lt;li&gt;A transition is defined as a change in direction.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/u1hOI.png&quot; alt=&quot;My solution&quot;&gt;&lt;br /&gt;&#xA; This was drawn with sixteen lines, indicating that there are fifteen changes of direction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9828" LastEditorUserId="5840" LastEditDate="2015-06-25T20:13:52.503" LastActivityDate="2015-06-25T20:13:52.503" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17006" PostTypeId="2" ParentId="17002" CreationDate="2015-06-25T19:31:10.697" Score="1" Body="&lt;p&gt;My guess and reasoning is,&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since OP has stated in comments that the capitalization matters, I have decided to check the capital letters. Joining the capitalized letters gives EEE. A quick search for #EEE says that the color is a &lt;a href=&quot;http://www.color-hex.com/color/eeeeee&quot; rel=&quot;nofollow&quot;&gt;very light gray.&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13566" LastActivityDate="2015-06-25T19:31:10.697" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17008" PostTypeId="1" AcceptedAnswerId="17311" CreationDate="2015-06-25T20:29:30.417" Score="8" ViewCount="1024" Body="&lt;p&gt;This was inspired when I tried to solve BmyGuest's  &lt;a href=&quot;https://puzzling.stackexchange.com/q/16996/9000&quot;&gt;Owls at sunset - find the odd one out&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar to that group, these birds have gathered prior to a hunt albeit an evening hunt instead of a dawn foray. However, the Strigiformes gathered here have a different group system. Not only do they have different rules to follow, but they don't mess around with sho-who-whoing outsiders away. Imposters are torn to shreds and fed to the hawks as payment for staying out of the owls' hunting area. Sadly, this has not led to increased security. This group has &lt;strong&gt;two outsiders&lt;/strong&gt; in their midst.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can you identify the two owls behaving anti-socially (I.E. not following the parliament rules) in the image below? &#xA;  What is the rule?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A valid answer must contain the parliament rule and why those particular owls are not following it. The puzzle can be solved with a black-and-white printout of the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IWZs1.png&quot; alt=&quot;Different-Owl-Picture-Hi-BmyGuest-I-Hope-You-Like-This-Shoutout&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note 1: In the column labels, I skipped the letter I on purpose to avoid confusion with the number 1. It is not a clue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note 2: BmyGuest found a valid pattern that I missed. The owls he found are, in fact, the odd birds out. I have since changed those two bird eyes such that they are still outside the intended pattern but no longer fall under the pattern BmyGuest found. I have delivered +1 to that answer nonetheless.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The &lt;strike&gt;one&lt;/strike&gt; two and only hints:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I added a row of owls simply because I liked how the pattern broke there. It doesn't really matter how many owls there are. The pattern holds up for any positive natural number of owls. Therefore, they must be following a pattern or series.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are 64 possible eye pairs. Only 30 are valid and only 18 are used in the picture.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2015-07-03T17:35:25.133" Title="Sunset Parliament Sacrificing Posers" Tags="|pattern|visual|" AnswerCount="5" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17009" PostTypeId="2" ParentId="17008" CreationDate="2015-06-25T20:37:26.297" Score="1" Body="&lt;p&gt;Is it...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the ones at (1,1) and (0,2) where the bottom left one is (0,0)? They're the only ones looking at each other directly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2015-06-25T20:37:26.297" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17010" PostTypeId="1" AcceptedAnswerId="17051" CreationDate="2015-06-25T21:08:36.963" Score="9" ViewCount="522" Body="&lt;p&gt;&lt;strong&gt;Part 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We are epic, we are everywhere.&lt;br&gt;&#xA;  We hide in shadows, we give you&#xA;  nightmares.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;We are your nemesis, we are a dynasty.&lt;br&gt;&#xA;  We cause diseases beyond&#xA;  imagination or dream.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;We are drones we are drums.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;We are darkness and power,&lt;br&gt;&#xA;  We are the roman emperor.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;We are legend, we are ashes.&lt;br&gt;&#xA;  We are public at long last.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We are epic, we are everywhere,&lt;br&gt;&#xA;  We have our classics, it's only fair.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;We are invincible, it's no illusion,&lt;br&gt;&#xA;     We are the devil and the demon.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;We shout our battlecry, hear our legacy. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;We are the horror, the sky is our world.&lt;br&gt;&#xA;     We are as fast as sound as we fly.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;We climb the volcano, witness our miracle,&lt;br&gt;&#xA;  We are the sun, we are the archangel. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;We are the same, Who are we?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most nouns are all related, Part 1 and 2 fit chronogically, a couple of stanzas are metaphors but not a lot of them. Either way to figure out the puzzle you have to find what links all of them. Also, you've very most likely encountered 'we' before without knowing. And check the tags.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastEditorUserId="17250" LastEditDate="2016-03-04T14:22:45.717" LastActivityDate="2016-03-04T14:22:45.717" Title="We are epic, we are everywhere" Tags="|riddle|knowledge|music|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17011" PostTypeId="1" AcceptedAnswerId="17014" CreationDate="2015-06-25T21:49:49.097" Score="8" ViewCount="982" Body="&lt;p&gt;Hangman game backward version 5 (&lt;a href=&quot;https://puzzling.stackexchange.com/questions/16798/the-hangman-game-backwards-again&quot;&gt;Aggie&lt;/a&gt;'s version was #4) - guess what was the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;fill in an &quot;a&quot; - It's a lethal to throw and it's destructive.&lt;br&gt;&#xA;fill in a  &quot;d&quot; - It's brave, and it's &lt;strong&gt;strong&lt;/strong&gt;.&lt;br&gt;&#xA;fill in an &quot;e&quot; - It's brown, and it's reddish-brown.&lt;br&gt;&#xA;fill in a  &quot;t&quot; - It's fast, and it's fastening &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, your answer can be in the form of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The question was: _ p p l e&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;followed by explanation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I know it's an easy one, hence I will not accept any answer that does not explains all &lt;strong&gt;8&lt;/strong&gt; answers.&lt;/p&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-06-25T22:13:49.973" Title="The hangman game backward #5" Tags="|riddle|english|language|letters|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17012" PostTypeId="2" ParentId="16741" CreationDate="2015-06-25T22:00:23.560" Score="0" Body="&lt;p&gt;A solution with 6 twos and only subtraction and division:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(222-22) \div 2$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13569" LastEditorUserId="7308" LastEditDate="2015-06-25T23:55:50.547" LastActivityDate="2015-06-25T23:55:50.547" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17013" PostTypeId="2" ParentId="17011" CreationDate="2015-06-25T22:03:07.860" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The question was bol_&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fill in an &quot;a&quot; - It's a lethal to throw and it's destructive.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; bola - a throwing weapon made of weights on the ends of an interconnected cord&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fill in a &quot;d&quot; - It's brave, and it's &lt;strong&gt;strong&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; bold - bold is a synonym of brave and in bold is strong text&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fill in an &quot;e&quot; - It's brown, and it's reddish-brown.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; bole - the trunk or stem of a tree (so it's brown)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fill in a &quot;t&quot; - It's fast, and it's fastening&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; bolt - to bolt is to fasten (something) with a metal pin or bar, in particular. Also a flash of lightning (which is fast)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5359" LastEditorUserId="5359" LastEditDate="2015-06-25T22:13:49.973" LastActivityDate="2015-06-25T22:13:49.973" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17014" PostTypeId="2" ParentId="17011" CreationDate="2015-06-25T22:05:36.887" Score="8" Body="&lt;p&gt;The question was&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; b o l _&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's lethal to throw and it's destructive&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;bola&lt;/strong&gt;, which is both a South American throwing weapon and a volcano in Papua New Guinea (which can be destructive).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's brave, and it's strong&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;bold&lt;/strong&gt;, which is a synonym for brave as well as the common term for &lt;strong&gt;strong text&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's brown, and it's reddish-brown&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;bole&lt;/strong&gt;, which is the trunk of a tree (hence brown) and a shade of reddish-brown on its own.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's fast, and it's fastening&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;bolt&lt;/strong&gt;, which is a definition for a fast dash, as well as a piece of hardware used to fasten two things together.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-06-25T22:05:36.887" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17015" PostTypeId="2" ParentId="17010" CreationDate="2015-06-25T22:42:05.053" Score="4" Body="&lt;p&gt;Part 1 sounds like&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; vampires.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;We are epic, we are everywhere.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Vampires are found in epic fantasy, and everywhere in horror novels.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;We hide in shadows, we give you nightmares.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This bit is obvious.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;We are your nemesis, we are a dynasty.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Vampires are the nemesis of humanity. &quot;Dynasty&quot; could refer to various fictional families of vampires, most notably the Cullen family of the &lt;em&gt;Twilight&lt;/em&gt; series.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;We cause diseases beyond imagination or dream.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Diseases&quot; as a metaphor for blood loss and death?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;We are drones we are drums.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure about this line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;We are darkness and power,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Classic vampire boast.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;We are the roman emperor.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sidorio, a vampire in the &lt;em&gt;Vampirates&lt;/em&gt; series, encountered Julius Caesar while still a mortal, and used Caesar as a pseudonym on at least one occasion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;We are legend, we are ashes.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Vampires are legendary. Put one into sunlight and it gets burnt to ashes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;We are public at long last.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; With the onset of &quot;vampire fiction&quot;, vampires are suddenly very much in the public eye.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-06-25T22:55:34.593" LastActivityDate="2015-06-25T22:55:34.593" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17016" PostTypeId="2" ParentId="16987" CreationDate="2015-06-25T23:07:39.237" Score="10" Body="&lt;p&gt;The smallest grid with that property is the 1 by 1 grid showing no numbers.&#xA;It also has a maximum of uncertainty.&lt;/p&gt;&#xA;" OwnerUserId="13572" LastActivityDate="2015-06-25T23:07:39.237" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17017" PostTypeId="2" ParentId="17004" CreationDate="2015-06-25T23:33:09.333" Score="3" Body="&lt;h2&gt;Partial answer&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;**&lt;em&gt;I don't know what a ryyvbggxrzore or an ohmmsrrq is, but I expect they were part of the qbg-pbz bubble.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Apply ROT-13 (Caesar shift of 13) to the strings of gibberish to get &quot;elliottkimber&quot;, &quot;buzzfeed&quot;, and &quot;dot-com&quot;. Let's look at &lt;a href=&quot;http://elliottkember.com/&quot; rel=&quot;nofollow&quot;&gt;these&lt;/a&gt; &lt;a href=&quot;http://www.buzzfeed.com/&quot; rel=&quot;nofollow&quot;&gt;two&lt;/a&gt; websites!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;**&lt;em&gt;You're like a gargoyle in complexion. Someone must have put a curse on you, or something like that.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This could refer to the disease greyscale from A Song of Ice and Fire / Game of Thrones.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Putting this together with what Chadi Abou Sleiman got for the first and third clues, we get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SLATE GREYSCALE, or maybe SLATE SHIREEN since Shireen is the best-known sufferer from the greyscale disease.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;**&lt;em&gt;Here, we don't believe in order of operations. Remove it from the system.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Removing the ordering suggests anagrams. I haven't yet found any good anagrams of SLATEGREYSCALE, though SLATESHIREEN is an anagram of RELEASE HINTS, so ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;release hints, Bailey M? ;-)&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-06-26T11:51:16.297" LastActivityDate="2015-06-26T11:51:16.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17019" PostTypeId="2" ParentId="16741" CreationDate="2015-06-26T03:05:36.553" Score="1" Body="&lt;p&gt;Edit: Here's &lt;em&gt;three&lt;/em&gt; twos:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ${\left(2\over{.2}\right)}^2$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well since it's already been bumped, here it is, using &lt;em&gt;four&lt;/em&gt; twos:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(2*2)\over{(.2*.2)}$ as well as $(22-2)/.2$ and $22/.22$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="7722" LastEditDate="2015-06-26T03:14:32.823" LastActivityDate="2015-06-26T03:14:32.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17020" PostTypeId="2" ParentId="12187" CreationDate="2015-06-26T03:46:13.150" Score="0" Body="&lt;p&gt;Edit: I realized Alex did it using 2 operations using this loophole so this is nothing special. I'd delete this normally but I think this solution is still kinda cool.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I got three operations without using an nth root:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(12)(3)(4)(.5)(-6+7)/((8)(9))=1$&lt;br&gt;&lt;br&gt;Taking advantage of parentheses not counting. (Yes I know the rule wasn't meant for them to be used this way but I spent a long time thinking of how to exploit this loophole so cut me some slack.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="7722" LastEditDate="2015-06-26T03:54:59.447" LastActivityDate="2015-06-26T03:54:59.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17023" PostTypeId="1" CreationDate="2015-06-26T07:09:14.513" Score="-1" ViewCount="401" Body="&lt;p&gt;I make life hard for many people,&lt;br&gt;&#xA;but am important to them&lt;br&gt;&#xA;once or many times in their lives.&lt;br&gt;&#xA;The more I grow, my importance reduces.&lt;/p&gt;&#xA;" OwnerUserId="13578" LastEditorUserId="8874" LastEditDate="2015-10-07T11:06:13.143" LastActivityDate="2015-10-07T11:06:13.143" Title="I make life hard for many people" Tags="|riddle|" AnswerCount="4" CommentCount="4" ClosedDate="2015-06-26T13:30:02.357" ContentLicense="CC BY-SA 3.0" />
  <row Id="17024" PostTypeId="2" ParentId="17023" CreationDate="2015-06-26T07:20:01.293" Score="0" Body="&lt;p&gt;I think it is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; darkness&lt;br&gt;&lt;br&gt; Because there are times when a dim light can set the mood for a romantic occasion, or be useful to sneak up on an enemy during war. However, once the darkness grows too much, it becomes harder to see and accomplish any task. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-06-26T07:20:01.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17025" PostTypeId="2" ParentId="17023" CreationDate="2015-06-26T07:22:17.813" Score="0" Body="&lt;p&gt;Potential answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; age&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Old age makes life difficult (for many not all) but the elder you are, the less important the exact age becomes to you. On the other hand, age can be very important during many times in your life, in particular when a minimum-age is required when you're young...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastActivityDate="2015-06-26T07:22:17.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17026" PostTypeId="2" ParentId="17023" CreationDate="2015-06-26T08:23:31.083" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Money&lt;br&gt;&#xA;&lt;br&gt;&#xA;Not having money makes life hard for many&lt;br&gt;&#xA;It is important at many points in your life (Not so sure about this line)&lt;br&gt;&#xA;The more money you have, the less of an effect each subsequent unit of currency has on your wealth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11717" LastActivityDate="2015-06-26T08:23:31.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17028" PostTypeId="1" AcceptedAnswerId="17030" CreationDate="2015-06-26T09:44:37.663" Score="4" ViewCount="996" Body="&lt;p&gt;Can you recognize the lines below?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Do you &lt;code&gt;utter&lt;/code&gt; the &lt;code&gt;moniker&lt;/code&gt; of your &lt;code&gt;emancipator&lt;/code&gt; &lt;code&gt;during&lt;/code&gt; your &lt;code&gt;time&lt;/code&gt; of &lt;code&gt;requirement&lt;/code&gt;?&lt;br&gt;&#xA;  And &lt;code&gt;experience&lt;/code&gt; the &lt;code&gt;guilt&lt;/code&gt; if the &lt;code&gt;character&lt;/code&gt; should &lt;code&gt;bring back memories to&lt;/code&gt; you of &lt;code&gt;avarice&lt;/code&gt;.&lt;br&gt;&#xA;  Of &lt;code&gt;indication&lt;/code&gt;, &lt;code&gt;incrimination&lt;/code&gt; and &lt;code&gt;animosity&lt;/code&gt;, till you cannot &lt;code&gt;rest&lt;/code&gt; &lt;code&gt;motionless&lt;/code&gt;,&lt;br&gt;&#xA;  In &lt;code&gt;the whole of&lt;/code&gt; this &lt;code&gt;upheaval&lt;/code&gt;, &lt;code&gt;prior to&lt;/code&gt; &lt;code&gt;blood-coloured&lt;/code&gt; &lt;code&gt;cloak&lt;/code&gt; and &lt;code&gt;thin metal&lt;/code&gt; &lt;code&gt;converge&lt;/code&gt; for &lt;code&gt;an execution&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="5373" LastEditDate="2018-01-17T14:15:40.577" LastActivityDate="2018-01-17T14:15:40.577" Title="What do you make of these lines?" Tags="|enigmatic-puzzle|wordplay|knowledge|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17030" PostTypeId="2" ParentId="17028" CreationDate="2015-06-26T09:52:02.370" Score="10" Body="&lt;p&gt;Replace all the words in code formatting (if that's what it is) by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; synonyms of themselves,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;thus turning the given lines into&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Do you &lt;code&gt;breathe&lt;/code&gt; the &lt;code&gt;name&lt;/code&gt; of your &lt;code&gt;saviour&lt;/code&gt; &lt;code&gt;in&lt;/code&gt; your &lt;code&gt;hour&lt;/code&gt; of &lt;code&gt;need&lt;/code&gt;,&lt;br&gt;&#xA; And &lt;code&gt;taste&lt;/code&gt; the &lt;code&gt;blame&lt;/code&gt; if the &lt;code&gt;flavour&lt;/code&gt; should &lt;code&gt;remind&lt;/code&gt; you of &lt;code&gt;greed&lt;/code&gt;?&lt;br&gt;&#xA; Of &lt;code&gt;implication&lt;/code&gt;, &lt;code&gt;insinuation&lt;/code&gt; and &lt;code&gt;ill will&lt;/code&gt;, till you cannot &lt;code&gt;lie&lt;/code&gt; &lt;code&gt;still&lt;/code&gt;,&lt;br&gt;&#xA; In &lt;code&gt;all&lt;/code&gt; this &lt;code&gt;turmoil&lt;/code&gt;, &lt;code&gt;before&lt;/code&gt; &lt;code&gt;red&lt;/code&gt; &lt;code&gt;cape&lt;/code&gt; and &lt;code&gt;foil&lt;/code&gt; &lt;code&gt;come closing in&lt;/code&gt; for &lt;code&gt;a kill&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;... which is the song&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.azlyrics.com/lyrics/poetsofthefall/carnivalofrust.html&quot;&gt;'Carnival of Rust' by the Finnish rock band Poets of the Fall&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-06-26T11:13:59.530" LastActivityDate="2015-06-26T11:13:59.530" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17031" PostTypeId="2" ParentId="17004" CreationDate="2015-06-26T10:34:40.700" Score="9" Body="&lt;p&gt;&lt;em&gt;What happens when the boxer is taken down?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He is &lt;strong&gt;Ko&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A beautiful mermaid, laughing out loud:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;In reference to &lt;a href=&quot;http://gameinfo.na.leagueoflegends.com/en/game-info/champions/nami/&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;nami&lt;/strong&gt;&lt;/a&gt; from &lt;strong&gt;L&lt;/strong&gt;eague &lt;strong&gt;O&lt;/strong&gt;f &lt;strong&gt;L&lt;/strong&gt;egends (&lt;strong&gt;LOL&lt;/strong&gt;: &lt;strong&gt;L&lt;/strong&gt;aughing &lt;strong&gt;O&lt;/strong&gt;ut &lt;strong&gt;L&lt;/strong&gt;oud)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which together forms:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Konami_Code&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Konami&lt;/strong&gt; code&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;ryyvbggxrzore, ohmmsrrq and qbg-pbz&lt;/em&gt; refers to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; elliottkember, buzzfeed and dot-com using ROT-13 refering to &lt;a href=&quot;http://elliottkember.com/&quot; rel=&quot;nofollow&quot;&gt;these&lt;/a&gt; &lt;a href=&quot;http://www.buzzfeed.com&quot; rel=&quot;nofollow&quot;&gt;two&lt;/a&gt; website&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What they have in commom is that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We can enter the &lt;strong&gt;Konami&lt;/strong&gt; code and observe (in this order): &lt;strong&gt;snake&lt;/strong&gt; and &lt;strong&gt;sloth&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Take 12121 (though 22121 will work in a pinch)&lt;/em&gt; means:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Take letters from each of the words observed based on their position (&lt;strong&gt;snake&lt;/strong&gt; and &lt;strong&gt;sloth&lt;/strong&gt;), which both 12121 or 22121 gives: &lt;strong&gt;slate&lt;/strong&gt; which is a type of rock (maybe related to the gargoyle referenced next)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Followimg the answer given by SirParselot&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You're like a gargoyle in complexion&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; refers to &lt;strong&gt;grey slate&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Someone must have put a curse on you, or something like that&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A curse is also called &lt;strong&gt;Hex&lt;/strong&gt;. The hex color of &lt;a href=&quot;https://en.wikipedia.org/wiki/Slate_gray&quot; rel=&quot;nofollow&quot;&gt;Slate Gray&lt;/a&gt; is &lt;strong&gt;#708090&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Here, we don't believe in order of operations. Remove it from the system&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Remove the &quot;0&quot; to remain with &lt;strong&gt;789&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The remainder will be comical. Excellent cake, seared deeply, sounds like it will start you on the right path&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;7 ate 9&lt;/strong&gt; is a known joke.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;At the end of the day, can you tell me who won the duel?&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;7&lt;/strong&gt; won it and all the number are now afraid of it !!!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12731" LastEditorUserId="12731" LastEditDate="2015-07-11T09:25:03.837" LastActivityDate="2015-07-11T09:25:03.837" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17032" PostTypeId="2" ParentId="17008" CreationDate="2015-06-26T10:47:21.417" Score="5" Body="&lt;p&gt;The answer below was valid prior the image-edit. It does no longer apply.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This quite likely is a too simple solution, but two owls stand out: &lt;code&gt;A2&lt;/code&gt; and &lt;code&gt;D7&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; These two owls are the only ones with a &lt;em&gt;unique&lt;/em&gt; eye combination. All others copy at least one other owl.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="5840" LastEditDate="2015-06-26T12:15:41.167" LastActivityDate="2015-06-26T12:15:41.167" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17033" PostTypeId="2" ParentId="17008" CreationDate="2015-06-26T10:58:27.050" Score="1" Body="&lt;p&gt;The rule is that..&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an owl may only make eye contact with at most one other owl. Owls are making eye contact if they are looking in each others direction with at least one eye and there are no other owls between them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The two owls breaking this rule are...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; b9 and c9 who are looking at each other as well as c9 and b8 who are looking back at them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11297" LastActivityDate="2015-06-26T10:58:27.050" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17034" PostTypeId="2" ParentId="16982" CreationDate="2015-06-26T11:02:15.087" Score="5" Body="&lt;p&gt;The clue has me stumped, but rather than admitting defeat, here's a long shot...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...the clue is a red herring, and the secret to Mr. McMoneybags's success is intelligence and hard work as per his signature&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10808" LastActivityDate="2015-06-26T11:02:15.087" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17035" PostTypeId="2" ParentId="16999" CreationDate="2015-06-26T11:38:32.947" Score="4" Body="&lt;p&gt;The initial decoding:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 38 is ascii decimal for '&amp;amp;', and 23 is ascii hexadecimal for '#', which makes me think every 4 numbers are a html encoded ascii character (starting with &quot;&amp;amp;#&quot;).&lt;br&gt;&#xA;&lt;br&gt;&#xA; 53 is ascii decimal for '5' and 48 is ascii decimal for '0', thus the first four numbers are &lt;strong&gt;&quot;&amp;amp;#50&quot;&lt;/strong&gt;, or the character &lt;strong&gt;'2'&lt;/strong&gt;.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proceeding the same way with all numbers we get:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The sequence &lt;strong&gt;24 2A 2C 3C 34&lt;/strong&gt;, which is hexadecimal &lt;a href=&quot;http://www.computer-engineering.org/ps2keyboard/scancodes2.html&quot; rel=&quot;nofollow&quot;&gt;scan code&lt;/a&gt; for a PS/2 keyboard (used in one of OP's previous puzzles).&lt;br&gt;&#xA;&lt;br&gt;&#xA;The resulting characters are &lt;strong&gt;E V T U G&lt;/strong&gt;.    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then applying ROT-13 results in the answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the &lt;strong&gt;R I G H T&lt;/strong&gt; door.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13523" LastEditorUserId="5188" LastEditDate="2015-06-27T02:52:31.713" LastActivityDate="2015-06-27T02:52:31.713" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17037" PostTypeId="2" ParentId="17023" CreationDate="2015-06-26T13:27:56.220" Score="0" Body="&lt;p&gt;Another potential answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PRIDE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I make life hard for many people &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; False pride makes life hard for people who wants to maintain a false status even when they cannot afford &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but am important to them once or many times in their life &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They were once wealthy and proud &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the more I grow my importance reduces&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People with false grown pride often lose respect. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10973" LastActivityDate="2015-06-26T13:27:56.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17040" PostTypeId="1" CreationDate="2015-06-26T14:47:44.880" Score="3" ViewCount="664" Body="&lt;p&gt;I want the elements from 1 to 118, their names, atomic numbers and symbols including capitalization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one important rule that makes things a little more difficult:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;No item can be completely contained within a single other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So single letter symbols must be on their own square that doesn't overlap anything else. Multiple letter (or number) items can overlap completely with other items so long as they overlap more than one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example of a valid grid with just H and He (and their names and numbers):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Hydrogen&#xA;2muileH1&#xA;H???????&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So it's okay that He overlaps the e in Hydrogen and the H in Helium because although it is completely contained in other words, it is contained in more than one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an example of a bad grid using the same items:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Hydrogen&#xA;e12?????&#xA;l???????&#xA;i???????&#xA;u???????&#xA;m???????&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It's bad because both &lt;code&gt;H&lt;/code&gt; and &lt;code&gt;He&lt;/code&gt; are completely contained within &lt;code&gt;Helium&lt;/code&gt;. That they also overlap &lt;code&gt;Hydrogen&lt;/code&gt; doesn't matter because they are completely contained by an item.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As stated in the title I want the smallest grid (by area) and in the&#xA;  event of two answers having the same area the one that is closest to&#xA;  being square wins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the unlikely event that both have the same dimensions then the grid that has the most unused squares will be the winner. (if even that ties then I'll count the number of extra occurrences of each item) &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Provide your answers as text so I can copy and paste it into a testing&#xA;  program.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unused square can be filled in or left blank as you see fit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a list of the items to use:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 H Hydrogen 2 He Helium 3 Li Lithium 4 Be Beryllium 5 B Boron 6 C Carbon 7 N Nitrogen 8 O Oxygen 9 F Fluorine 10 Ne Neon 11 Na Sodium 12 Mg Magnesium 13 Al Aluminium 14 Si Silicon 15 P Phosphorus 16 S Sulfur 17 Cl Chlorine 18 Ar Argon 19 K Potassium 20 Ca Calcium 21 Sc Scandium 22 Ti Titanium 23 V Vanadium 24 Cr Chromium 25 Mn Manganese 26 Fe Iron 27 Co Cobalt 28 Ni Nickel 29 Cu Copper 30 Zn Zinc 31 Ga Gallium 32 Ge Germanium 33 As Arsenic 34 Se Selenium 35 Br Bromine 36 Kr Krypton 37 Rb Rubidium 38 Sr Strontium 39 Y Yttrium 40 Zr Zirconium 41 Nb Niobium 42 Mo Molybdenum 43 Tc Technetium 44 Ru Ruthenium 45 Rh Rhodium 46 Pd Palladium 47 Ag Silver 48 Cd Cadmium 49 In Indium 50 Sn Tin 51 Sb Antimony 52 Te Tellurium 53 I Iodine 54 Xe Xenon 55 Cs Caesium 56 Ba Barium 57 La Lanthanum 58 Ce Cerium 59 Pr Praseodymium 60 Nd Neodymium 61 Pm Promethium 62 Sm Samarium 63 Eu Europium 64 Gd Gadolinium 65 Tb Terbium 66 Dy Dysprosium 67 Ho Holmium 68 Er Erbium 69 Tm Thulium 70 Yb Ytterbium 71 Lu Lutetium 72 Hf Hafnium 73 Ta Tantalum 74 W Tungsten 75 Re Rhenium 76 Os Osmium 77 Ir Iridium 78 Pt Platinum 79 Au Gold 80 Hg Mercury 81 Tl Thallium 82 Pb Lead 83 Bi Bismuth 84 Po Polonium 85 At Astatine 86 Rn Radon 87 Fr Francium 88 Ra Radium 89 Ac Actinium 90 Th Thorium 91 Pa Protactinium 92 U Uranium 93 Np Neptunium 94 Pu Plutonium 95 Am Americium 96 Cm Curium 97 Bk Berkelium 98 Cf Californium 99 Es Einsteinium 100 Fm Fermium 101 Md Mendelevium 102 No Nobelium 103 Lr Lawrencium 104 Rf Rutherfordium 105 Db Dubnium 106 Sg Seaborgium 107 Bh Bohrium 108 Hs Hassium 109 Mt Meitnerium 110 Ds Darmstadtium 111 Rg Roentgenium 112 Cn Copernicium 113 Uut Ununtrium 114 Fl Flerovium 115 Uup Ununpentium 116 Lv Livermorium 117 Uus Ununseptium 118 Uuo Ununoctium&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="11297" LastEditDate="2015-06-28T15:04:52.267" LastActivityDate="2015-06-28T15:04:52.267" Title="What's the smallest chemical elements word search that can be made?" Tags="|puzzle-creation|word-search|packing|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17041" PostTypeId="2" ParentId="16982" CreationDate="2015-06-26T15:27:53.437" Score="18" Body="&lt;p&gt;We can decrypt the code with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A vigenere cipher with the key &quot;intelligence&quot; then decrypt it again with the key &quot;hardwork&quot;. &lt;br&gt;&lt;br&gt;These two words comes from the last sentence &quot;The only way to get results is with intelligence and hard work&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which gives the riddle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It can be both cold and hot &lt;br&gt;&#xA;You find it on chips more often than not &lt;br&gt;&#xA;Here, personal space can get pretty tight &lt;br&gt;&#xA;Writing helps me sleep at night&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I haven't found the answer yet.&lt;/p&gt;&#xA;" OwnerUserId="5359" LastEditorDisplayName="user9377" LastEditDate="2015-06-27T08:00:48.503" LastActivityDate="2015-06-27T08:00:48.503" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17042" PostTypeId="2" ParentId="16935" CreationDate="2015-06-26T15:46:59.673" Score="6" Body="&lt;p&gt;I think it might be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the Romans Club.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The OP at first put &quot;8 comes before 9&quot; for the third guy, but then said this was a mistake and changed it to &quot;9 comes before 10&quot;; that was what gave me this idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8 immediately follows 7 because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an extra I on VII gives VIII.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;13-12 is 1 because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; XIII take away XII is I.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;9 comes before ten because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; X with an I before is IX.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1+2 is 3 because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I and II make III.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The high-five was&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the guy making the symbol V with his hand.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2 and 2 do &lt;em&gt;not&lt;/em&gt; make 4 because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; II and II do not make IV.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-06-26T15:46:59.673" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17043" PostTypeId="1" AcceptedAnswerId="17045" CreationDate="2015-06-26T16:04:09.560" Score="6" ViewCount="4969" Body="&lt;p&gt;Say I have a bunch of words that I want to make a word-search out of. What are some strategies for making a challenging but doable word-search puzzle where these are the words to be found?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm planning to post an answer myself with some ideas I've come up with. But I've never actually made a word-search puzzle and haven't solved many either, so I'd be grateful for input from those who have. Let's get a nice long list of techniques!&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-08-20T05:24:02.333" Title="How can I make a good word-search puzzle?" Tags="|puzzle-creation|word-search|" AnswerCount="5" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17044" PostTypeId="2" ParentId="17043" CreationDate="2015-06-26T16:04:28.330" Score="6" Body="&lt;p&gt;Some ideas I came up with:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Consider the orientation (leftwards, rightwards, upwards, downwards, all 4 kinds of diagonally) of words to be found. Include at least a few words in each of these directions. Rightwards and downwards are the easiest to find, diagonally (other than southeastwards) the hardest - take account of this when deciding how difficult you want the puzzle to be.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Consider what 'fillers' (letters in between the words to be found) to put in. Use too many X's and Q's and the puzzle will be too easy; you should use a similar distribution of letters to the distribution in the words to be found.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;To increase the difficulty, use red-herring words. If one of the words to be found is STACK, then throw in a STALK and a SACK somewhere and sprinkle loads of S's, T's, A's, C's, and K's around where STACK actually is to make it hard to find the right word.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-06-26T16:04:28.330" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17045" PostTypeId="2" ParentId="17043" CreationDate="2015-06-26T16:13:42.937" Score="6" Body="&lt;p&gt;Some more points&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Choose an easy-to-the-eye font and letter-spacing.  The overall &quot;look&quot; is very important!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Consider &lt;em&gt;crossings&lt;/em&gt;, i.e. have some letter be part of more than one word&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Ensure that the number of &quot;needed&quot; letters and the number &quot;filler&quot; letters are in a &quot;good&quot; proportion. This may depend on you intention though. There are word-searches where &lt;em&gt;all&lt;/em&gt; letters are part of some word, and ones with a lot of empty space. If the second category, don't have too many fillers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Keep an eye on overall statistic. Each language has a certain distribution of 'likely letters' - your fillers should represent this as well.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You can make things interesting, by having the overall shape of the puzzle something else than a rectangle, or by having the 'fillers' end up in useful pattern etc.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Give the word-search some common &quot;topic&quot;, don't choose arbitrary words.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5840" LastActivityDate="2015-06-26T16:13:42.937" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17046" PostTypeId="2" ParentId="17043" CreationDate="2015-06-26T16:32:19.870" Score="7" Body="&lt;p&gt;You could also consider going with a differently aligned grid if you want to be original (like a diamond shape or an octogon) - you can also like in the example below complicate things by putting multiple letters as one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9vop8.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://winstonbreen.com/blog/?paged=9&quot; rel=&quot;nofollow noreferrer&quot;&gt;Image credit&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11251" LastEditorUserId="10399" LastEditDate="2015-06-26T18:54:51.090" LastActivityDate="2015-06-26T18:54:51.090" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17048" PostTypeId="2" ParentId="17043" CreationDate="2015-06-26T16:50:31.800" Score="5" Body="&lt;p&gt;For the &quot;filter&quot; letters, I would not consider using the same distribution, but equalize its distribution. For instance, if you have only one 'Z' in the words to search for, keeping the same distribution for filters would mean approximately 'add another Z in the filters'. So searching this word would be easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whereas equalizing the distribution would mean words a bit more difficult to find. For instance, if you have 5 'E', 3 'M' and 1 'Z', try to insert 1 'E', 3 'M' and 5 'Z' as filters, so that all letters have 6 occurrences in your grid. Doing so will require a bit more difficulty to find words with rare letters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple algorithm to implement this would be:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;check which is the less frequent letter currently in the grid&lt;/li&gt;&#xA;&lt;li&gt;insert it randomly as a filter&lt;/li&gt;&#xA;&lt;li&gt;loop to 1. until the grid is full&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13589" LastEditorUserId="13589" LastEditDate="2015-06-26T17:24:03.323" LastActivityDate="2015-06-26T17:24:03.323" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17049" PostTypeId="2" ParentId="16620" CreationDate="2015-06-26T16:53:12.357" Score="1" Body="&lt;p&gt;This is a partial to help out the one's who are actively solving this problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a butterfly's sway, unconditionally&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Breakdown:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;To have &quot;sway&quot; is to wield power.  A butterfly that wields power is a Monarch.  You are looking for a Monarchy.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1777" LastEditorUserId="1777" LastEditDate="2015-06-26T17:08:07.090" LastActivityDate="2015-06-26T17:08:07.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17050" PostTypeId="2" ParentId="17043" CreationDate="2015-06-26T17:20:48.463" Score="4" Body="&lt;p&gt;If you want your puzzle to be really hard here are some things to try:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Avoid words that read left to right (for languages naturally read in that direction),&#xA;top to bottom (in all languages I think), and horizontally or vertically when at the edges of the grid. These are all easy to spot even if no clues are provided.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Giving a list of words to find makes things a bit too easy. A hint at some connection or theme if the words are really obscure might be okay. Of course if you want your puzzle to be really hard just use the most obscure words you can think of and only provide the number of words to be found (if you do this you will need to check thoroughly that no extra words have crept in.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you feel the need to provide a list of words use fragments of those words to fill in blank spaces. Placing the start fragment of a word near the beginning or end of a complete word is great way to create a wild goose chase. Fill the remaining single spaces with the least common letters from your clues, you want to make sure that if a letter only appears once or twice in all your you use it at least three or four times to fill in blank spaces. Any strings of letters that appear repeated across many of your clues will be any easy target to search for so make sure you use those to fill spaces. The more red herrings the better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make a puzzle really time consuming make it really large. Spotting a word and finding another that begins or ends nearby, or crosses it, makes things too easy. Having a grid large enough to space your words out nicely can completely eliminate this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If capitalization is not important to any of the words, randomising between upper and lower case can make a word search fiendishly difficult to read. if you want a puzzle that is really difficult just to read, try a textured background and patterned letters. I'm thinking of camouflage schemes for the letters and background to are so similar you can hardly see the letters. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be really unfair why not swap some letters for numbers or symbols, 0 for o, ( for c, that sort of thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A most wicked twist is to allow words to wrap around the edges of the grid. This is particularly unfair if there is no reason to expect you would do this. So to be really cruel make sure you give no hints this is what you are up to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course if you are not a complete sadist then ignore this advice. If you want to make a nice easy puzzle do the opposite. A good word search will have just the right amount of cruel and kind elements. The right balance depends on your target audience.&lt;/p&gt;&#xA;" OwnerUserId="11297" LastActivityDate="2015-06-26T17:20:48.463" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17051" PostTypeId="2" ParentId="17010" CreationDate="2015-06-26T17:34:24.303" Score="9" Body="&lt;p&gt;You are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Songs/albums by the recording group &lt;a href=&quot;https://en.wikipedia.org/wiki/Two_Steps_from_Hell&quot; rel=&quot;noreferrer&quot;&gt;Two Steps from Hell&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are epic, we are everywhere&lt;br&gt;&#xA;We hide in shadows, we give you nightmares.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This group has recorded tracks for a large number of motion pictures and other soundtracks&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#Shadows_and_Nightmares&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#Shadows_and_Nightmares&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are your nemesis, we are a dynasty.&lt;br&gt;&#xA;We cause diseases beyond imagination or dream.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#Nemesis&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#Nemesis&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#Dynasty&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#Dynasty&lt;/a&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#Pathogen&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#Pathogen&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#Dreams_.26_Imaginations&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#Dreams_.26_Imaginations&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are drones we are drums.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#All_Drones_Go_to_Hell&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#All_Drones_Go_to_Hell&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#All_Drums_Go_To_Hell&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#All_Drums_Go_To_Hell&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are darkness and power,&lt;br&gt;&#xA;We are the roman emperor.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#Power_of_Darkness&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#Power_of_Darkness&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#Nero&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#Nero&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are legend, we are ashes.&lt;br&gt;&#xA;We are public at long last.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#Ashes&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#Ashes&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#Legend&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#Legend&lt;/a&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;He has been a composer since the 80's but only has public albums since 2010.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part 2&lt;/strong&gt; &#xA;We are epic, we are everywhere,&lt;br&gt;&#xA;We have our classics, it's only fair.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#Classics_Volume_One&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#Classics_Volume_One&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#Classics_Volume_Two&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#Classics_Volume_Two&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are invincible, it's no illusion,&lt;br&gt;&#xA;We are the devil and the demon.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#Invincible&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#Invincible&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#Illusions&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#Illusions&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#The_Devil_Wears_Nada&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#The_Devil_Wears_Nada&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We shout our battlecry, hear our legacy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#Battlecry&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#Battlecry&lt;/a&gt;&lt;br&gt;&#xA;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#Legend From the OP, there is a CD called &lt;em&gt;Legacy&lt;/em&gt; that was a release in Japan.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are the horror, the sky is our world.&lt;br&gt;&#xA;We are as fast as sound as we fly.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#SkyWorld&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#SkyWorld&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We climb the volcano, witness our miracle,&lt;br&gt;&#xA;We are the sun, we are the archangel.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#Colin_Frake_on_Fire_Mountain&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#Colin_Frake_on_Fire_Mountain&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#Miracles&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#Miracles&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#Archangel&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Thomas_Bergersen_discography#Archangel&lt;/a&gt;&lt;br&gt;&#xA;From the OP, there is also an album titled Sun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12242" LastEditorUserId="12242" LastEditDate="2015-06-26T17:47:11.680" LastActivityDate="2015-06-26T17:47:11.680" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17052" PostTypeId="2" ParentId="17040" CreationDate="2015-06-26T17:39:25.767" Score="7" Body="&lt;p&gt;EDIT: A new submission in which I have high confidence.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Method&lt;/h2&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Use an online program to densely pack the element names&lt;/li&gt;&#xA;&lt;li&gt;Dump the contents into Excel&lt;/li&gt;&#xA;&lt;li&gt;Use my previously-written VBA script to highlight the element names&lt;/li&gt;&#xA;&lt;li&gt;Remove all non-highlighted characters&lt;/li&gt;&#xA;&lt;li&gt;Manually search for any element symbols already highlighted. Make sure they're not fully contained within a single other item and watch out for symbols using the same letters like &lt;code&gt;Sc&lt;/code&gt; and &lt;code&gt;Cs&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Manually fill in any remaining symbols using the white space. This was &lt;code&gt;Uuo&lt;/code&gt;, &lt;code&gt;Uup&lt;/code&gt;, all of the single-character symbols (14), and less than 10 of the others (I forgot to count).&lt;/li&gt;&#xA;&lt;li&gt;Manually fill in the atomic numbers using the white space remaining&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h2&gt;Results&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;44 rows $\times$ 23 columns $=$ 1,012 total characters&lt;br&gt;&#xA;147 filler characters $\Rightarrow$ 85.5% utilization&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Puzzle&lt;/h2&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;14501N44224126014?C8915&#xA;01mmMeitnerium103oa70m6&#xA;211u1Neodymium10pBl7nuL&#xA;n32ii05mbmEinsteiniumiU&#xA;om0t6m1uuicU??rs??fSvtT&#xA;runn31yiNnu?nnmmX?oeLe5&#xA;oiMolybdenumiuu??erarn3&#xA;Bn?rtrProtactiniumnb2h3&#xA;oe?tEpwoneihnoIooBio8c2&#xA;hhrSCayfFusadoHrcuurne9&#xA;rteyLDarmstadtiumtmg1T1&#xA;iun3l26eKiRiru8u16iis01&#xA;uRi45l7hTUnZmPi115Au9H8&#xA;mIr9muitnepnunU178rmmy2&#xA;FFo2m?muiveledneMos??dI&#xA;relZu?nRm??gHmuihtemorP&#xA;arheitmulatnaTnp95nuool&#xA;nmCnrrKManganese35intgu&#xA;cimieocnegyxOoe986caSet&#xA;iuumC?voVWe3h4p8m4sheno&#xA;umnoA?RinYNP79tuus8tl2n&#xA;muiramaSuiAluminiumne7i&#xA;uitB??eStmu??suusNuan5u&#xA;inaYtterbiumoRmiemiLimm&#xA;nalFluorineru??pnurouuu&#xA;bmPArgon8cpb?Tt?giuimii&#xA;ur?Ae6736simmuidallaPds&#xA;De?nC694yd?unn??Melobae&#xA;?GmtoG9Di?iimg?Cbke?dna&#xA;?muiporuEdum?s?oLrT?iaC&#xA;??implm?nmmoItNb?eht?VG&#xA;??toed?a?uurTe?a?Barium&#xA;yUenr?c?miihhn?lmtldm??&#xA;1rty?SG?uddCu??tsuluup?&#xA;2au??i?aional??Actinium&#xA;3nLc?l?ulSIlim?nNnurh??&#xA;4i??rvm?el?cu?iCoSmmt??&#xA;5u?C?e??H?iimZalOcuPit?&#xA;6m?u?rM???muidohRiilL?Y&#xA;7??rNickelimmP??n?slfC?&#xA;8??iCarbonSi???e?Osmium&#xA;9??u???HfmuirohT??a??Sr&#xA;???muidaRm???R????H????&#xA;??????H????????????????&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h2&gt;Solution&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Element names are highlighted. Fillers are left blank. Anything in the white space was manually added to finish out whatever was missing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/viSd4.png&quot; alt=&quot;Solution&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2015-06-26T20:56:15.937" LastActivityDate="2015-06-26T20:56:15.937" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17054" PostTypeId="1" CreationDate="2015-06-26T17:58:29.127" Score="-5" ViewCount="1160" Body="&lt;p&gt;First, write an expression using exactly one 3 and exactly one 4 to make 24.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, write an expression using exactly one 3 and exactly one 4 to make 30.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Math You Can Use for each problem:&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Division&lt;/li&gt;&#xA;&lt;li&gt;Factorials (just for regular factorial use, no special factorials)&lt;/li&gt;&#xA;&lt;li&gt;Parentheses (They would be required to show the factorial of a factorial,&#xA;as in (X!)!, for instance.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Other Rules For Clarification:&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;No other digits may be used.&lt;/li&gt;&#xA;&lt;li&gt;No other operations may be used.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I am intending one solution for each problem.&lt;/p&gt;&#xA;" OwnerUserId="13203" LastEditorUserId="8874" LastEditDate="2016-03-23T15:30:45.957" LastActivityDate="2016-06-23T07:07:18.210" Title="Make 24 and 30 using the digits 3 and 4" Tags="|calculation-puzzle|formation-of-numbers|" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17056" PostTypeId="2" ParentId="17054" CreationDate="2015-06-26T18:03:00.310" Score="8" Body="&lt;p&gt;30:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(3!)!\div4! = 30$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;24:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(4!\div3!)! = 24$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13204" LastEditorUserId="13204" LastEditDate="2015-06-26T18:15:32.213" LastActivityDate="2015-06-26T18:15:32.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17060" PostTypeId="1" CreationDate="2015-06-26T18:35:50.417" Score="-4" ViewCount="878" Body="&lt;p&gt;Write an expression using exactly one 2, exactly one 3, and exactly one 4 &#xA;to make 30.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;For each problem, you must use all three of these:&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Division&lt;/li&gt;&#xA;&lt;li&gt;Factorials (just for regular factorial use, no special factorials)&lt;/li&gt;&#xA;&lt;li&gt;Parentheses (They would be required to show the factorial of a factorial,&#xA;as in (X!)!, for instance.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Other Rules For Clarification:&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;No other digits may be used.&lt;/li&gt;&#xA;&lt;li&gt;No other operations may be used.&lt;/li&gt;&#xA;&lt;li&gt;No concatenation of digits is permitted.&lt;/li&gt;&#xA;&lt;li&gt;Your solution must require the use of parentheses to be &#xA;valid.  If all of the parentheses &#xA;can be removed, and the expression still equals 30, then that solution is not valid.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I am intending one solution for this problem.&lt;/p&gt;&#xA;" OwnerUserId="13203" LastEditorUserId="8874" LastEditDate="2016-03-23T15:30:34.537" LastActivityDate="2016-03-23T15:30:34.537" Title="Using the digits 2, 3, 4 make an expression for 30" Tags="|calculation-puzzle|formation-of-numbers|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="17062" PostTypeId="2" ParentId="17060" CreationDate="2015-06-26T18:43:08.480" Score="10" Body="&lt;p&gt;Here's a solution:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $((3! \div{2})!)! \div{4!}  $&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or written a little more clearly:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$((\frac{3!}{2})!)! \div{4!}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="9951" LastEditDate="2015-06-26T18:50:55.983" LastActivityDate="2015-06-26T18:50:55.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17064" PostTypeId="1" AcceptedAnswerId="17065" CreationDate="2015-06-26T18:57:02.653" Score="5" ViewCount="695" Body="&lt;p&gt;This is part of the &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=title%3A%22Fill%20in%20the%20Dashes%22&quot;&gt;Fill in the Dashes&lt;/a&gt; series of puzzles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example : F_ in the B_  Fill in the Blanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you identify the idiom(s)/phrase(s) from english language ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) It t_ T_ to T_&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) O_ in a B_ M_&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) B_ a_ the b_&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) The W_ N_ Y_&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) W_ u_ the b_&lt;/p&gt;&#xA;" OwnerUserId="9892" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-09-03T05:43:47.773" Title="Fill in the Dashes : Part 1 of 10" Tags="|english|knowledge|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17065" PostTypeId="2" ParentId="17064" CreationDate="2015-06-26T18:59:28.267" Score="12" Body="&lt;p&gt;&lt;em&gt;&lt;del&gt;Note that this is a partial answer...&lt;/del&gt;&lt;/em&gt; Completed &lt;a href=&quot;https://www.youtube.com/watch?v=jBDF04fQKtQ&quot;&gt;with a little help from my friends&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It takes two to tango.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Once in a blue moon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Beat around the bush.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;Credits to Aggie Kidd...&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The whole nine yards.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;Credits to Aggie Kidd..&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water under the bridge&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-06-26T19:06:46.743" LastActivityDate="2015-06-26T19:06:46.743" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17067" PostTypeId="2" ParentId="17064" CreationDate="2015-06-26T19:12:15.803" Score="10" Body="&lt;p&gt;Following CodeNewbie's request (all credit to CodeNewbie for the full answer):  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4. The &lt;strong&gt;W&lt;/strong&gt;hole &lt;strong&gt;N&lt;/strong&gt;ine &lt;strong&gt;Y&lt;/strong&gt;ards&lt;br&gt;&#xA; 5. &lt;strong&gt;W&lt;/strong&gt;ater &lt;strong&gt;u&lt;/strong&gt;nder the &lt;strong&gt;b&lt;/strong&gt;ridge&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-06-26T19:12:15.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17068" PostTypeId="2" ParentId="16935" CreationDate="2015-06-26T19:31:14.123" Score="4" Body="&lt;p&gt;I think the club might be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A music club&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Numbers represent musical notes on a keyboard (&lt;strong&gt;A&lt;/strong&gt; = &lt;strong&gt;1&lt;/strong&gt;, &lt;strong&gt;A#&lt;/strong&gt; = &lt;strong&gt;2&lt;/strong&gt;, &lt;strong&gt;B&lt;/strong&gt; = &lt;strong&gt;3&lt;/strong&gt;...), and operations are number or color changes on the keys.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;7 is followed by 8 because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;7&lt;/strong&gt; represents &lt;strong&gt;D#&lt;/strong&gt; (black key) and the next white key is &lt;strong&gt;E&lt;/strong&gt; (&lt;strong&gt;8&lt;/strong&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;13 - 12 = 1 because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is one change of color between &lt;strong&gt;13&lt;/strong&gt; (&lt;strong&gt;A&lt;/strong&gt;, white key) and &lt;strong&gt;12&lt;/strong&gt; (&lt;strong&gt;G#&lt;/strong&gt;, black key)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;9 preceeds 10 because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;9&lt;/strong&gt; is &lt;strong&gt;F&lt;/strong&gt; (white) and &lt;strong&gt;10&lt;/strong&gt; is &lt;strong&gt;F#&lt;/strong&gt; (black)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1+2 = 3 because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;1&lt;/strong&gt; is &lt;strong&gt;A&lt;/strong&gt; (white), and +2 means change the color two times (-&gt; &lt;strong&gt;A#&lt;/strong&gt; -&gt; &lt;strong&gt;B&lt;/strong&gt;), &lt;strong&gt;B&lt;/strong&gt; is number &lt;strong&gt;3&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Give me a five&quot; does not require an answer because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The member sings &quot;Hi Bob!&quot; in &lt;strong&gt;C#&lt;/strong&gt; (key number &lt;strong&gt;5&lt;/strong&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So you should have said&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;5&lt;/strong&gt; (&lt;strong&gt;C#&lt;/strong&gt;), because &lt;strong&gt;2&lt;/strong&gt; is note &lt;strong&gt;A#&lt;/strong&gt;, and +2 means change color 2 times (-&gt; &lt;strong&gt;B&lt;/strong&gt; -&gt; &lt;strong&gt;C#&lt;/strong&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bonus: the guardian screams because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is a lot of noise and music coming from the inside&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13523" LastEditorUserId="13523" LastEditDate="2015-06-27T08:03:43.973" LastActivityDate="2015-06-27T08:03:43.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17069" PostTypeId="2" ParentId="17064" CreationDate="2015-06-26T20:07:56.980" Score="3" Body="&lt;p&gt;3 could also be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Beauty and the beast.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13068" LastActivityDate="2015-06-26T20:07:56.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17070" PostTypeId="1" AcceptedAnswerId="17071" CreationDate="2015-06-26T20:41:21.697" Score="7" ViewCount="596" Body="&lt;p&gt;My friend told me that he went to a strange place:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;It looks like a jail, but there are no wardens.  There are only&#xA;  silences behind the bars.  However&quot;, my friend said, &quot;we could feel it&#xA;  from their eyes, as if they are reaching out for help.&quot;   &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;More importantly, they are not the murderers.  The murderers are set&#xA;  free!  Bloods have been found all over their body.  No one cares about&#xA;  it, nor is anyone surprised by the unusual circumstance.&quot;  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;One by one, the innocents behind the bars are sent to their death&#xA;  sentence, by the murderers!&quot;    &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;Aren't you gonna call the police?&quot;  I asked.     &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;It's no use&quot;, my friend answered.  &quot;In fact, there are far more &#xA;  people involved in this murdering than you could have imagined.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where did my friend go?  A world twisted?  &lt;/p&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="8874" LastEditDate="2015-10-07T11:02:02.343" LastActivityDate="2015-10-07T11:02:02.343" Title="A world twisted?" Tags="|lateral-thinking|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17071" PostTypeId="2" ParentId="17070" CreationDate="2015-06-26T20:48:01.437" Score="7" Body="&lt;p&gt;Your friend went to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The slaughterhouse.&lt;br&gt;&lt;br&gt;The innocents are the livestock that are butchered to obtain meat and hide, and the murderers are the butchers at the slaughterhouse. The butchers usually have their clothes covered in blood from the slaughter of the animals, which is quite a common sight in such places. Animal slaughter is not a crime, so the police will not get involved, and the people involved in the &quot;murder&quot; are several starting from the ones who rear the animals right until the consumers of the meat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-06-26T20:48:01.437" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17072" PostTypeId="2" ParentId="17070" CreationDate="2015-06-26T20:48:51.533" Score="5" Body="&lt;p&gt;Perhaps your friend went to a:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Music score.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Silences behind the bars:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rests, or silences could be said to be hiden behind the staff.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The murderers are set free:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The notes played aloud kill the silence, and are set free by the instruments that play them. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bloods have been found all over their body.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Music notes sometimes could look like drops of black blood.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One by one, the innocents behind the bars are sent to their death sentence, by the murderers!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Silence is always killed by noise, and each rest is killed by a note.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...there are far more people involved in this murdering than you could have imagine.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are very many musical performers. Vocal artists, instrumentalists, percussionists, etc.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-06-26T20:48:51.533" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17073" PostTypeId="2" ParentId="16982" CreationDate="2015-06-26T21:07:31.853" Score="7" Body="&lt;p&gt;I think I have an answer for the riddle, but I have no idea how it relates to the puzzle, or what the mild clue in the last line is.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer I have is &lt;strong&gt;Salsa&lt;/strong&gt;.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;It can be both cold and hot&lt;/strong&gt;&lt;br&gt;&#xA; Salsa (the food) can be served cold or spicy (hot)&lt;br&gt;&#xA; &lt;br&gt;&#xA; &lt;strong&gt;You find it on chips more often than not&lt;/strong&gt;&lt;br&gt;&#xA; Salsa (again, the food) is usually a dip for tortilla chips or similar.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Here, personal space can get pretty tight&lt;/strong&gt;&lt;br&gt;&#xA; Salsa dancing involves getting close to your partner&lt;br&gt;&#xA; &lt;br&gt;&#xA; &lt;strong&gt;Writing helps me sleep at night&lt;/strong&gt;&lt;br&gt;&#xA; No idea.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="325" LastActivityDate="2015-06-26T21:07:31.853" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17074" PostTypeId="1" AcceptedAnswerId="17075" CreationDate="2015-06-26T21:48:19.923" Score="18" ViewCount="6012" Body="&lt;p&gt;Your buddy Frankie sold you a shoddy clock: it keeps good time, but the minute and hour hands look exactly the same! Both of these hands move continuously, and there is no second hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many times a day is it impossible to tell what time it is? &lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorDisplayName="user88" LastEditDate="2016-05-28T22:11:24.153" LastActivityDate="2024-01-18T20:29:32.517" Title="A clock where the hour and minute hands are the same length" Tags="|mathematics|geometry|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17075" PostTypeId="2" ParentId="17074" CreationDate="2015-06-26T22:07:03.410" Score="15" Body="&lt;p&gt;When the hour hand has moved $x$ degrees around the clock from the top, the minute hand has moved $y = 12x$ degrees. If the time is still a valid configuration when the hands are switched around then $x = 12y$ as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, we want the values of $x, y$ that satisfy the following two equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;12x &amp;amp;\equiv y \pmod {360} \\&#xA;12y &amp;amp;\equiv x \pmod {360}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conveniently, this reduces to $x \equiv 144x \pmod {360}$ or $143x \equiv 0 \pmod {360}$, so whenever the hour hand moves exactly $x/143$ of the way around the clock where $x$ is an integer, it's impossible to tell which hand is which.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There's just one problem, though. The above doesn't take into account the times when the hour and minute hands are in the exact same position, in which case it doesn't &lt;em&gt;matter&lt;/em&gt; which hand is which. This occurs whenever $x \equiv 12x \pmod {360}$, or $11x \equiv 0 \pmod {360}$. Naturally this is a total of 11 times, so there are $143 - 11 = 132$ times when the time is actually ambiguous in a 12-hour period, making it $264$ times a day.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-06-26T22:21:24.700" LastActivityDate="2015-06-26T22:21:24.700" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="17078" PostTypeId="1" AcceptedAnswerId="17092" CreationDate="2015-06-27T03:16:37.613" Score="13" ViewCount="539" Body="&lt;h2&gt;The Problem&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;At the intersection of mathematics, computer science, and insanity, we find the $M_Q$ functor${}^1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${M_Q} = {\Phi _{{s_1},{s_2}}}\left\{ {\begin{array}{*{20}{l}}&#xA;  {E \leftarrow {\Phi _{{c_1},{c_2}}}\left\{ { \Leftarrow \left\{ {\begin{array}{*{20}{l}}&#xA;  1&amp;amp;{{\text{if }}{c_1} = {c_2}} \\ &#xA;  0&amp;amp;{{\text{otherwise}}} &#xA;\end{array}} \right.} \right.} \\ &#xA;  {N \leftarrow {\Re _0}\left[ {{\Phi _{{x_1},{x_2}}} \Leftarrow {x_1} + {x_2}} \right]} \\ &#xA;  {U \leftarrow {\Phi _c} \Leftarrow 60{{\left( {N\left( {\mathcal{L}\xrightarrow{s}N\left( {s\xrightarrow{\chi }E\left( {c,\chi } \right)} \right)} \right) + 1} \right)}^{ - 1}}} \\ &#xA;  {{I_{\text{L}}} \leftarrow {\Phi _s}\left\{ { \Leftarrow \left\{ {\begin{array}{*{20}{l}}&#xA;  1&amp;amp;{{\text{if }}s \in \left\{ s \right\} \cap \mathcal{L}} \\ &#xA;  0&amp;amp;{{\text{otherwise}}} &#xA;\end{array}} \right.} \right.} \\ &#xA;  {\bar S \leftarrow {\Phi _s} \Leftarrow \left( {{2^{\left\{ {i\,|\,i \in \mathbb{N},\,i \leqslant \left| s \right|} \right\}}}\xrightarrow{X}{s^X}} \right)\backslash \left\{ {s,\varepsilon } \right\}} \\ &#xA;  {V \leftarrow {\Phi _s} \Leftarrow N\left( {s\xrightarrow{\chi }U\left( \chi  \right)} \right)} \\ &#xA;  {x \leftarrow {\Phi _s} \Leftarrow {\Re _0}\left[ {{\Phi _{{x_1},{x_2}}}\left\{ { \Leftarrow \left\{ {\begin{array}{*{20}{l}}&#xA;  {{x_1}}&amp;amp;{{\text{if }}{x_1} \geqslant {x_2}} \\ &#xA;  {{x_2}}&amp;amp;{{\text{otherwise}}} &#xA;\end{array}} \right.} \right.} \right]\left( {\bar S\left( s \right)\xrightarrow{{s'}}V\left( {s'} \right) I_{\rm L}\left( s' \right)} \right)} \\ &#xA;  {\Upsilon  \leftarrow {\Phi _s} \Leftarrow \frac{{V\left( s \right) - x\left( s \right)}}{{2{I_{\text{L}}}\left( s \right) - 1}}} \\ &#xA;  { \Leftarrow \Upsilon \left( {{s_1}} \right) - \Upsilon \left( {{s_2}} \right)} &#xA;\end{array}} \right.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M_Q$ is a binary (that is, two-parameter) comparator of strings. As most computer scientists know, a comparator is a special type of functor that imposes a total ordering on a set of objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When applied to strings $s_1$ and $s_2$, ${M_Q}\left( {{s_1},{s_2}} \right)$ returns&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a negative value if $s_1$ should come before $s_2$ in a sorted list&lt;/li&gt;&#xA;&lt;li&gt;a positive value if $s_1$ should come after $s_2$ in a sorted list&lt;/li&gt;&#xA;&lt;li&gt;$0$ if the relative order of $s_1$ and $s_2$ is unimportant&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$M_Q$ also relies on a language, $\mathcal{L}$, which is an ordered list of non-empty strings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For our purposes, let the elements of $\mathcal{L}$ be the strings $\mathsf{a}$, $\mathsf{ae}$, $\mathsf{be}$, $\mathsf{bee}$, $\mathsf{ade}$, $\mathsf{bed}$, $\mathsf{are}$, $\mathsf{bad}$, $\mathsf{bab}$, $\mathsf{baa}$, $\mathsf{abe}$, $\mathsf{bear}$, $\mathsf{barb}$, $\mathsf{bard}$, $\mathsf{bade}$, $\mathsf{beard}$, $\mathsf{bread}$, $\mathsf{badab}$, $\mathsf{bared}$, $\mathsf{barred}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M_Q$ can compare not only strings in $\mathcal{L}$, but also any strings made up of the letters $\mathsf{a}$-$\mathsf{z}$ (for example, $\mathsf{leopard}$).&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The Challenge&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Your task is threefold:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Describe, in plain English, the order that $M_Q$ imposes on strings. In other words, provide a description of what exactly the comparator does.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Determine which string will come &lt;em&gt;last&lt;/em&gt; if $\mathcal{L}$ is sorted by $M_Q$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Determine a string (not necessarily in $\mathcal{L}$) that will come &lt;em&gt;before&lt;/em&gt; the string $\mathsf{babadook}$ in a list sorted by $M_Q$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This problem can (and ideally &lt;em&gt;should&lt;/em&gt;) be solved &lt;em&gt;without programming&lt;/em&gt;. In particular, all of the variables, functions, etc. have straightforward behaviours that can be intuited and computed by hand. There should be no heavy lifting or brute forcing involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Are you up to the task? Can you master the comparatortionist?&lt;/strong&gt;&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Notes on Syntax&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Let $\varepsilon$ be the empty string.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\left| s\right|$ be the length (i.e. number of characters in) string $s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strings may be treated as lists. A string is an ordered list of its characters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let ${\Phi _{{{\arg }_1},{{\arg }_2}, \cdots ,{{\arg }_n}}} \Leftarrow f\left( {{{\arg }_1},\,\,{{\arg }_2},\,\, \cdots ,\,\,{{\arg }_n}} \right)$ be a function returning an $n$-ary functor that returns the value of the function $f$ applied to its arguments. (This convention is used since functors may be nested or passed as arguments to other functors.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, let $${\Phi _{{{\arg }_1},{{\arg }_2}, \cdots ,{{\arg }_n}}}\left\{ {\begin{array}{*{20}{l}}&#xA;  {{\text{assignment}}} \\ &#xA;  {{\text{assignment}}} \\ &#xA;  {\,\,\,\,\,\,\,\,\, \vdots } \\ &#xA;  {{\text{assignment}}} \\ &#xA;  { \Leftarrow f\left( {{{\arg }_1},\,\,{{\arg }_2},\,\, \cdots ,\,\,{{\arg }_n}} \right)} &#xA;\end{array}} \right.$$ be a function returning an $n$-ary functor that evaluates assignments from top to bottom and returns the value of the function $f$ applied to its arguments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let ${\Re _{{\text{init}}}}\left[ f \right]$ be a function returning a unary functor accepting a list/string as its argument. Call this argument $A$. The functor returned by ${\Re _{{\text{init}}}}\left[ f \right]$ reduces $A$ to a single value, starting with an initial value $\text{init}$ and iteratively applying the reducing binary operator $f$. Note that $f$ is always an associative and commutative functor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S\xrightarrow{e}f\left( e \right)$ be the ordered list obtained by mapping each element $e$ of $S$ to $f\left( e \right)$, where $f$ is a function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, given string $s$ and set of natural numbers $X$, let ${s^X}$ be the substring of $s$ whose 1-based character indices are present in $X$. For example, if $s = \mathsf{this}$, then ${s^{\left\{ {1,3} \right\}}} = \mathsf{ti}$ and ${s^{\left\{ {1,2,4} \right\}}} = \mathsf{ths}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All remaining operators, symbols, and syntax are standard for mathematical science, computer science, or both, and are left to the reader to infer/deduce. A large part of the challenge is &quot;getting a handle on&quot; the syntax.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;${}^1$ note that &quot;functor&quot; here refers to &quot;function which can be stored as a variable and passed as an argument&quot;. Also commonly known as a &quot;lambda expression&quot; or a &quot;map&quot;&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="4219" LastEditDate="2015-06-27T18:04:38.453" LastActivityDate="2015-06-28T03:53:03.977" Title="Comparatortionist: The $M_Q$ Functor" Tags="|mathematics|computer-puzzle|no-computers|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17079" PostTypeId="1" AcceptedAnswerId="17081" CreationDate="2015-06-27T03:28:31.530" Score="19" ViewCount="1240" Body="&lt;p&gt;If a phrase adheres to a certain rule, then I call it an &lt;strong&gt;Efficient Phrase™&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the examples below to find the rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0KYmH.png&quot; alt=&quot;Efficient Phrases and Not Efficient Phrases&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If you liked this puzzle, try others like it:&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/16219/what-is-a-cyclone-phrase%E2%84%A2&quot;&gt;What is a Cyclone Phrase™?&lt;/a&gt;&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/16485/what-is-a-scalable-phrase%E2%84%A2&quot;&gt;What is a Scalable Phrase™?&lt;/a&gt;&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/16743/what-is-a-triad-phrase%E2%84%A2&quot;&gt;What is a Triad Phrase™?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-08-12T21:44:25.790" Title="What is an Efficient Phrase™?" Tags="|pattern|english|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17081" PostTypeId="2" ParentId="17079" CreationDate="2015-06-27T03:49:12.793" Score="19" Body="&lt;p&gt;An Efficient Phrase™ is one where&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; each of its letters can be drawn as a single line, without lifting the pen or retracing your path. This allows the letters &lt;code&gt;BCDGIJLMNOPSUVWZ&lt;/code&gt;. This limits the word more than it might seem because the three most common letters &lt;code&gt;ETA&lt;/code&gt; are disallowed. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="463" LastEditDate="2015-06-29T11:50:36.360" LastActivityDate="2015-06-29T11:50:36.360" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17082" PostTypeId="1" AcceptedAnswerId="17084" CreationDate="2015-06-27T05:20:35.023" Score="17" ViewCount="773" Body="&lt;p&gt;Complete the following by filling in the blanks with written numbers to create the world's longest self-documenting sentence.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;This sentence uses or mentions&#xA;    the word &quot;a&quot; ___ time(s),&#xA;    the word &quot;aah&quot; ___ time(s),&#xA;    the word &quot;aahed&quot; ___ time(s),&#xA;        ...&#xA;    and the word &quot;zyzzyvas&quot; ___ time(s).&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;A few notes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The &quot;...&quot; omits a lot.  There should be one line of the form &quot;the word 'blah' __ time(s)&quot; for each word in &lt;a href=&quot;http://www-01.sil.org/linguistics/wordlists/english/&quot;&gt;this word list&lt;/a&gt; that contains 109582 words.&lt;/li&gt;&#xA;&lt;li&gt;Numbers should be written out.  For example, 109582 should be written &quot;one hundred nine thousand five hundred eighty-two&quot;.  The &quot;eighty-two&quot; counts as two different words.&lt;/li&gt;&#xA;&lt;li&gt;The uses or mentions is important:  a word should be counted whether it is in quotation marks or not.&lt;/li&gt;&#xA;&lt;li&gt;For the final sentence, each &quot;time(s)&quot; should be either &quot;time&quot; or &quot;times&quot;.&lt;/li&gt;&#xA;&lt;li&gt;I wouldn't recommend writing your answer completely out.  Just give us the important bits.&lt;/li&gt;&#xA;&lt;li&gt;I'm not actually sure if it is the world's longest, but it's a pretty good stab at it.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Let me know if I can clarify anything else.&lt;/p&gt;&#xA;" OwnerUserId="2722" LastEditorUserId="5373" LastEditDate="2015-06-27T15:30:29.603" LastActivityDate="2015-06-27T15:30:29.603" Title="Complete the world's longest self-documenting sentence" Tags="|logical-deduction|no-computers|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17083" PostTypeId="2" ParentId="17078" CreationDate="2015-06-27T06:56:55.737" Score="3" Body="&lt;p&gt;This algorithm works as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Calculates a rating for a given string s based on the number of times the letters found in s occur in the dictionary and if s occurs in the dictionary itself.&lt;br&gt;&#xA; A negative rating means higher on the list.&lt;br&gt;&#xA; If s is in the dictionary, only letters with non-power-of-two locations are taken into account.&lt;br&gt;&#xA; If s is not in the directory, all letters are counted and the rating sign is flipped to negative.&lt;br&gt;&#xA; As a result, there is a heavy bias toward words not in the dictionary (especially words containing characters not in the dictionary at all) and toward shorter dictionary words (especially 1 or 2 letter words) since the first two places are powers of 2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The last word if the dictionary were sorted would be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;barred&quot; since it has the most letters in non power of two positions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A word before babadook would be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; any longer, non dictionary word, that contains at least 3 letters not in the dictionary and few dictionary letters. Non dictionary letters count -60, my name (stacey) would be -303 (5x60 + 3)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user2674" LastEditorDisplayName="user2674" LastEditDate="2015-06-27T07:09:07.643" LastActivityDate="2015-06-27T07:09:07.643" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17084" PostTypeId="2" ParentId="17082" CreationDate="2015-06-27T10:19:30.310" Score="27" Body="&lt;h2&gt;Solution&lt;/h2&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;This sentence uses or mentions&#xA;    the word &quot;a&quot; one time,&#xA;        ...&#xA;    the word &quot;and&quot; two times,&#xA;        ...&#xA;    the word &quot;eighty&quot; three times,&#xA;        ...&#xA;    the word &quot;five&quot; six times,&#xA;        ...&#xA;    the word &quot;four&quot; three times,&#xA;        ...&#xA;    the word &quot;hundred&quot; nine times,&#xA;        ...&#xA;    the word &quot;mentions&quot; two times,&#xA;        ...&#xA;    the word &quot;nine&quot; seven times,&#xA;        ...&#xA;    the word &quot;one&quot; one hundred nine thousand five hundred sixty four times,&#xA;        ...&#xA;    the word &quot;or&quot; two times,&#xA;        ...&#xA;    the word &quot;sentence&quot; two times,&#xA;        ...&#xA;    the word &quot;seven&quot; three times,&#xA;        ...&#xA;    the word &quot;six&quot; two times,&#xA;        ...&#xA;    the word &quot;sixty&quot; three times,&#xA;        ...&#xA;    the word &quot;the&quot; one hundred nine thousand five hundred eighty three times,&#xA;        ...&#xA;    the word &quot;this&quot; two times,&#xA;        ...&#xA;    the word &quot;thousand&quot; five times,&#xA;        ...&#xA;    the word &quot;three&quot; seven times,&#xA;        ...&#xA;    the word &quot;time&quot; one hundred nine thousand five hundred sixty times,&#xA;    the word &quot;times&quot; twenty four times,&#xA;        ...&#xA;    the word &quot;twenty&quot; two times,&#xA;        ...&#xA;    the word &quot;two&quot; nine times,&#xA;        ...&#xA;    the word &quot;uses&quot; two times,&#xA;        ...&#xA;    the word &quot;word&quot; one hundred nine thousand five hundred eighty three times,&#xA;        ...&#xA;    and the word &quot;zyzzyvas&quot; one time.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(where each ... denotes a whole string of lines of the form &quot;the word &quot;[...]&quot; one time,&quot;).&#xA;&lt;hr&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Working&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The words &quot;this&quot;, &quot;sentence&quot;, &quot;uses&quot;, &quot;or&quot;, &quot;mentions&quot;, and &quot;and&quot; are each going to be used exactly &lt;strong&gt;two&lt;/strong&gt; times (once for the places we can see them in the OP, once for their appearance in the list of words in quotes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The words &quot;the&quot; and &quot;word&quot; are each going to be used exactly &lt;strong&gt;one hundred nine thousand five hundred eighty three&lt;/strong&gt; times (once for every word in the list, and once more for their own appearances in the list).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only other words that will be used more than once are &quot;time&quot;, &quot;times&quot;, and various number-related words (&quot;one&quot;, &quot;hundred&quot;, &quot;nine&quot;, etc.) Let $n$ be the number of words that will be used more than once; $n$ is going to be less than thirty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The word &quot;times&quot; is going to be used $n+1$ times (once for each word used more than once, once for its own appearance in the list). The word &quot;time&quot; is going to be used $109582-n+1$ times (once for each word used exactly once, once for its own appearance in the list).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only words which will be used more than $n+1$ times are &quot;the&quot;, &quot;word&quot;, &quot;one&quot;, and &quot;time&quot;, which each appear at least $109582-n+1$ times. So the word &quot;hundred&quot; is going to be used &lt;strong&gt;nine&lt;/strong&gt; times, the word &quot;thousand&quot; &lt;strong&gt;five&lt;/strong&gt; times, and the word &quot;eighty&quot; &lt;strong&gt;three&lt;/strong&gt; times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bolded numbers above tell us that &quot;one&quot;, &quot;nine&quot;, &quot;five&quot;, &quot;two&quot;, and &quot;three&quot; will each be used more than once. So already $n$ is at least eighteen. Assuming $n$ is at least nineteen and less than twenty-nine (which I'm pretty sure it will be), the word &quot;twenty&quot; is used &lt;strong&gt;two&lt;/strong&gt; times (once for the number of times &quot;times&quot; is used and once for its own appearance in the list).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far we have the following words appearing more than once:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&quot;this&quot; ($2$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;sentence&quot; ($2$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;uses&quot; ($2$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;or&quot; ($2$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;mentions&quot; ($2$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;and&quot; ($2$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;the&quot; ($109583$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;word&quot; ($109583$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;times&quot; ($n+1$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;time&quot; ($109582-n+1$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;one&quot; ($\geq109582-n+1+4$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;hundred&quot; ($9$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;thousand&quot; ($5$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;eighty&quot; ($3$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;nine&quot; ($\geq6$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;five&quot; ($\geq6$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;two&quot; ($\geq8$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;three&quot; ($\geq4$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;twenty&quot; ($2$ times)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;Let's now assume $n\leq23$, so that &quot;time&quot; is used between one hundred nine thousand five hundred sixty and one hundred nine thousand five hundred sixty four times. Now the word &quot;sixty&quot; is going to be used either two or three times (according to whether or not &quot;one&quot; is used as much as 109570 times), so $n\geq20$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n=20$, then &quot;times&quot; is used 21 times, &quot;time&quot; is used 109563 times, and so &quot;one&quot; is used 109568 times, which means &quot;eight&quot; is used more than once and $n\geq21$, contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n=21$, then &quot;times&quot; is used 22 times, &quot;time&quot; is used 109562 times, and so &quot;one&quot; is used 109566 times, which means &quot;six&quot; is used more than once and is the 21st and last such word. But &quot;two&quot; is used at least 10 times (and fewer than 20), which means one of the words &quot;ten&quot;, &quot;eleven&quot;, ... &quot;nineteen&quot; must be used more than once, contradiction.&#xA;&lt;hr&gt;&#xA;If $n=22$, then &quot;times&quot; is used 23 times, &quot;time&quot; is used 109561 times, and so &quot;one&quot; is used 109566 times, which means &quot;six&quot; is used more than once. Now &quot;sixty&quot; is used exactly 3 times and &quot;two&quot; is used 8, 9, or 10 times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &quot;two&quot; is used 8 times, then both &quot;six&quot; and &quot;eight&quot; must be used more than 2 times and we have:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&quot;this&quot; ($2$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;sentence&quot; ($2$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;uses&quot; ($2$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;or&quot; ($2$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;mentions&quot; ($2$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;and&quot; ($2$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;the&quot; ($109583$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;word&quot; ($109583$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;times&quot; ($23$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;time&quot; ($109561$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;one&quot; ($109566$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;hundred&quot; ($9$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;thousand&quot; ($5$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;eighty&quot; ($3$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;nine&quot; ($\geq6$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;five&quot; ($\geq6$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;two&quot; ($8$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;three&quot; ($\geq6$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;twenty&quot; ($2$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;sixty&quot; ($3$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;six&quot; ($\geq3$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;eight&quot; ($\geq3$ times)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Since we have 22 words in this list, we can have no more. So each of &quot;nine&quot;, &quot;five&quot;, and &quot;three&quot; must be used 6 or 8 times while each of &quot;six&quot; and &quot;eight&quot; must be used exactly 3 times, contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &quot;two&quot; is used 10 times, then each of &quot;six&quot; and &quot;ten&quot; must be used exactly 2 times. So &quot;three&quot; is used six times, which means &quot;six&quot; is used at least 3 times, contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &quot;two&quot; is used 9 times, then &quot;nine&quot; is used 7 times, so &quot;seven&quot; is the final word used more than once. One of &quot;six&quot; and &quot;seven&quot; is used 2 times while the other is used 3 times. But each of &quot;five&quot; and &quot;three&quot; is used 6 or 7 times, so our count will be too high, contradiction.&#xA;&lt;hr&gt;&#xA;We're still assuming $n\leq23$, so now $n=23$ is the only case left. Here &quot;times&quot; is used 24 times, &quot;time&quot; is used 109560 times, and so &quot;one&quot; is used 109564 times. Now &quot;four&quot; is used at least 3 times, so it becomes the 21st word on our list. We know &quot;two&quot; is used at least 8 times; let's say it's used 9 times, so that &quot;nine&quot; is used 7 times and &quot;seven&quot; is the 22nd word on our list. After a little trial-and-error and deduction, I got to the following list:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&quot;this&quot; ($2$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;sentence&quot; ($2$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;uses&quot; ($2$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;or&quot; ($2$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;mentions&quot; ($2$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;and&quot; ($2$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;the&quot; ($109583$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;word&quot; ($109583$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;times&quot; ($24$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;time&quot; ($109560$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;one&quot; ($109564$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;hundred&quot; ($9$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;thousand&quot; ($5$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;eighty&quot; ($3$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;nine&quot; ($7$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;five&quot; ($6$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;two&quot; ($9$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;three&quot; ($7$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;twenty&quot; ($2$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;sixty&quot; ($3$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;four&quot; ($3$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;seven&quot; ($3$ times)&lt;/li&gt;&#xA;&lt;li&gt;&quot;six&quot; ($2$ times)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;... which all works out, giving the solution I put at the top. Hurray! :-D&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-06-27T15:18:36.193" LastActivityDate="2015-06-27T15:18:36.193" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17085" PostTypeId="2" ParentId="17079" CreationDate="2015-06-27T11:44:18.687" Score="7" Body="&lt;p&gt;Perhaps it's that an 'E'fficient Phrase&lt;sup&gt;tm&lt;/sup&gt; has &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; no 'ETA' (estimated time of arrival).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though I would have thought the opposite to be true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the full list of letters that are missing are: AEFHKQRTXY, which, when rearranged, gives us TRAHKXEFQY*, a common antonym of &quot;efficient&quot;.&lt;br&gt; &lt;sub&gt;*&lt;em&gt;not a word&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-08-12T21:44:25.790" LastActivityDate="2015-08-12T21:44:25.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17086" PostTypeId="1" AcceptedAnswerId="17312" CreationDate="2015-06-27T12:33:53.453" Score="3" ViewCount="427" Body="&lt;p&gt;What two-word phrase might describe &lt;em&gt;the route for a sensation coming all the way from the Egyptian cross&lt;/em&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can also be written as &lt;em&gt;XY where Y is what allows you to do something that sounds like X&lt;/em&gt;. (To avoid ambiguity: X and Y denote strings of letters which when put together form the phrase you seek.)&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-07-07T11:12:13.237" Title="A sensation from the Egyptian cross" Tags="|riddle|wordplay|" AnswerCount="2" CommentCount="11" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17087" PostTypeId="2" ParentId="1725" CreationDate="2015-06-27T13:32:54.797" Score="6" Body="&lt;p&gt;Here's a 3x3 crossword puzzle using only the first 5 letters of the alphabet:&#xA;&lt;hr&gt;&#xA;&lt;strong&gt;Across&lt;/strong&gt;&lt;br&gt;&#xA;1. The most or least valuable type of playing card.&lt;br&gt;&#xA;2. The front section of a lorry.&lt;br&gt;&#xA;3. The tide is going out.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Down&lt;/strong&gt;&lt;br&gt;&#xA;1. A flying one has downed his foes.&lt;br&gt;&#xA;2. A taxi.&lt;br&gt;&#xA;3. Decline, decay, or fade away.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A C E&lt;br&gt;&#xA; C A B&lt;br&gt;&#xA; E B B&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-06-27T13:32:54.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17088" PostTypeId="2" ParentId="17086" CreationDate="2015-06-27T14:53:17.893" Score="1" Body="&lt;p&gt;I think the answer is... &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Root canal &lt;br&gt;&lt;br&gt;The Egyptian cross is the Suez Canal. And 'route' is a homophone for 'root'. I can't think of an explanation for the second paragraph though. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-06-27T14:53:17.893" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17089" PostTypeId="2" ParentId="16833" CreationDate="2015-06-27T15:05:50.627" Score="0" Body="&lt;p&gt;The treasure is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; fifteen million golden &lt;a href=&quot;https://en.wikipedia.org/wiki/Loonie&quot; rel=&quot;nofollow noreferrer&quot;&gt;loonies&lt;/a&gt;, and it is buried beneath the campsite.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Each line in the journal entry is a word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BW1fX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is encoded in &lt;a href=&quot;https://en.wikipedia.org/wiki/ASCII&quot; rel=&quot;nofollow noreferrer&quot;&gt;ASCII&lt;/a&gt;. The text reads 'MILLION'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yo4uQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is encoded with a &lt;a href=&quot;https://en.wikipedia.org/wiki/Caesar_cipher&quot; rel=&quot;nofollow noreferrer&quot;&gt;Caesar cipher&lt;/a&gt;. Shifting each letter three places back in the alphabet gives 'FIFTEEN'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QRoLI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is simply the English word 'THE'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6P6jJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is the &lt;a href=&quot;https://translate.google.com/#auto/en/sepolti&quot; rel=&quot;nofollow noreferrer&quot;&gt;Italian&lt;/a&gt; for 'BURIED'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1U6Jw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is &lt;a href=&quot;https://en.wikipedia.org/wiki/Morse_code&quot; rel=&quot;nofollow noreferrer&quot;&gt;Morse code&lt;/a&gt; for 'GOLDEN'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ifqv3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Pigpen_cipher&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pigpen cipher&lt;/a&gt;. The text reads '&lt;a href=&quot;https://en.wikipedia.org/wiki/Loonie&quot; rel=&quot;nofollow noreferrer&quot;&gt;LOONIES&lt;/a&gt;'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bmXlf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is flag &lt;a href=&quot;https://en.wikipedia.org/wiki/Flag_semaphore&quot; rel=&quot;nofollow noreferrer&quot;&gt;semaphor&lt;/a&gt;. The text reads 'ARE'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Cf5m3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When these numbers are entered on a phone keypad using &lt;a href=&quot;https://en.wikipedia.org/wiki/T9_(predictive_text)&quot; rel=&quot;nofollow noreferrer&quot;&gt;T9 predictive text&lt;/a&gt;, the result is 'BENEATH'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;We can rearrange these words to get&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; FIFTEEN MILLION GOLDEN LOONIES ARE BURIED BENEATH THE...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is a word missing, which would tell us &lt;em&gt;what&lt;/em&gt; the golden loonies are buried beneath.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The names of the codes that were used (ASCII, Caesar, English, Italian, Morse, Pigpen, Semaphor, T9) are in alphabetical order. This is a clue that the names are important. If we write the names of the codes in the order the corresponding words are used in the sentence, we get&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;C&lt;/strong&gt;aesar&lt;br&gt;&lt;strong&gt;A&lt;/strong&gt;SCII&lt;br&gt;&lt;strong&gt;M&lt;/strong&gt;orse&lt;br&gt;&lt;strong&gt;P&lt;/strong&gt;igpen&lt;br&gt;&lt;strong&gt;S&lt;/strong&gt;emaphor&lt;br&gt;&lt;strong&gt;I&lt;/strong&gt;talian&lt;br&gt;&lt;strong&gt;T&lt;/strong&gt;9&lt;br&gt;&lt;strong&gt;E&lt;/strong&gt;nglish&lt;br&gt;&lt;br&gt; The initial letters of the code names spell the final word of the message: 'CAMPSITE'. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2627" LastActivityDate="2015-06-27T15:05:50.627" CommentCount="3" CommunityOwnedDate="2015-06-27T15:05:50.627" ContentLicense="CC BY-SA 3.0" />
  <row Id="17090" PostTypeId="1" AcceptedAnswerId="17095" CreationDate="2015-06-27T15:19:47.787" Score="11" ViewCount="697" Body="&lt;p&gt;Who am I talking about?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My first is unpredictable, if you add some from the middle &lt;br&gt;My second is one, but a short line splits it apart &lt;br&gt;The first part is a common word, if you come from the Middle &lt;br&gt;The last is worthy, coming from the bottom guard&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Just a warning, I've been extra obscure about this one, because almost everyone here has seen the answer.&lt;/p&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-06-27T15:31:31.440" LastActivityDate="2015-06-28T19:29:30.510" Title="Who could this be?" Tags="|riddle|rhyme|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17091" PostTypeId="1" AcceptedAnswerId="17147" CreationDate="2015-06-27T15:50:41.610" Score="12" ViewCount="1402" Body="&lt;p&gt;You are woken up to hearing your neighbour shouting - &amp;quot;HELP! HELP! I'M DYING!&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;You jump to your feet and run out into the street and see a man run away from your neighbours house, but he is too far away for you to catch up with him, and he is wearing a mask. You decide to investigate the house, as you are an ex-policeman. You use your spare keys to go inside the house and see the body on the floor, dead. You also see a bit of paper sticking out of his pocket. You pick it up. It reads:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If you are reading this I am dead, but I believe I know who killed me. The answer is encrypted below.&lt;/p&gt;&#xA;&lt;p&gt;&lt;code&gt;eUXODGaaSDW8WURaLHaNWvjP2YNVJQWMSEdOWVQMSPdM9GaMJVjA6VQMJdRNWOhi7DcCATbiKRW&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Damn it. Your neighbour loved cryptography. Oh well, you may as well have a go, you suppose. Then you notice something else sticking out of his pocket. Another bit of paper:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;I've always had an extensive love of doughnut&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;What relation to him was the person that committed the murder?&lt;/p&gt;&#xA;&lt;p&gt;NOTE: The reason of the puzzle being tagged &amp;quot;riddle&amp;quot; will be revealed later...&lt;/p&gt;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Only two words on the second bit of paper are important.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The two important words are &amp;quot;extensive&amp;quot; and &amp;quot;doughnut&amp;quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; CodeNewbie has guessed (correctly) that it has something to do with a Vigenere Cipher. However, only one word is the Vigenere Cipher key. The other is something else...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9377" LastEditorUserId="-1" LastEditorDisplayName="user9377" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-06-09T21:30:36.743" Title="Help! I'm dying!" Tags="|riddle|cipher|" AnswerCount="3" CommentCount="20" ContentLicense="CC BY-SA 3.0" />
  <row Id="17092" PostTypeId="2" ParentId="17078" CreationDate="2015-06-27T16:51:06.327" Score="9" Body="&lt;p&gt;Our first step is to figure out what COTO's notation means.  Here are a couple translations for the more obscure forms:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A &quot;functor&quot; is just another name for an ordinary function (since none of the &quot;functors&quot; given accept functions as arguments), with the following equivalence relationship:&#xA;$$&#xA;\Phi_{x_1,x_2,x_3\ldots}\Leftarrow f(x_1,x_2,x_3\ldots) \equiv x_1,x_2,x_3\ldots\mapsto f(x_1,x_2,x_3\ldots)&#xA;$$&lt;/li&gt;&#xA;&lt;li&gt;He also defines a second type of functor/function with intermediate assignments.  However, this notation is only used for $M_Q$ itself, and all but the last line do not depend on the arguments, so they only function as (no pun intended) names for various (otherwise anonymous) functions.  We can safely break them out and consider them separately.  This essentially gives us:&#xA;$$&#xA;\begin{align}&#xA;E&amp;amp;:s_1,s_2\mapsto\cdots \\&#xA;&amp;amp;\qquad\vdots \\&#xA;M_Q&amp;amp;:s_1,s_2\mapsto\cdots&#xA;\end{align}&#xA;$$&lt;/li&gt;&#xA;&lt;li&gt;The &quot;reducing&quot; functor $\Re$ (a true functor this time) extends a binary function to an $n$-ary function with an initial value:&#xA;$$&#xA;\Re_{a_0}[f](a_1,a_2\ldots a_n)\equiv f(\ldots f(f(a_0,a_1),a_2)\ldots,a_n)&#xA;$$&lt;/li&gt;&#xA;&lt;li&gt;The &quot;mapping&quot; functor $\to$ applies a function to each member of a list:&#xA;$$&#xA;S\xrightarrow{e}f(e)\equiv\{f(e) : e\in S\}&#xA;$$&lt;/li&gt;&#xA;&lt;li&gt;The &quot;substring&quot; notation can be rewritten using a variant of set notation:&#xA;$$&#xA;s^X\equiv\{s_i:i\in X\}&#xA;$$&lt;/li&gt;&#xA;&lt;li&gt;Finally, &lt;a href=&quot;https://puzzling.stackexchange.com/questions/17078/comparatortionist-the-m-q-functor/17092#comment53484_17083&quot;&gt;thanks to Stacey&lt;/a&gt; we know that $2^S$ is the power set of $S$, or the set of all its subsets:&#xA;$$&#xA;2^S\equiv \mathcal P(S)&#xA;$$&#xA;Representing the power set as exponentiation is actually a standard notation, but it confused both me and Stacey, so I'll use $\mathcal P$ just to be unambiguous.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note that COTO seems to follow the convention that arguments named $s$ are strings, and arguments named $c$ are characters.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The first step in $M_Q$ is to assign a functor to $E$, which returns $1$ if its arguments are equal, and $0$ otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step assigns a functor to $N$.  This one reduces a list by repeatedly adding its elements to a running total, which starts at zero: that is, it takes the sum of a list.  From here on I'll replace $N\to\Sigma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$U$ is a crazy beast.  Rewriting it using my 'standard' notation (and combining the &quot;total&quot; and &quot;map&quot; constructs into a standard sum):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;U:c\mapsto\frac{60}{1+\sum_{s\in\mathcal L,\chi\in s}E(c,\chi)}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It counts the number of times $c$ occurs across all the strings in $\mathcal L$ and then applies $\frac{60}{1+\Sigma}$.  The result of $U$ therefore is to weight characters according to the following table:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;\mathsf a~\text{or}~\mathsf b&amp;amp;\to 3\\&#xA;\mathsf d&amp;amp;\to \frac{60}{11}\\&#xA;\mathsf e&amp;amp;\to 4\\&#xA;\mathsf r&amp;amp;\to 6\\&#xA;\text{(all others)}&amp;amp;\to 60\\&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next is $I_L$, which simply tests if its argument is a member of $\mathcal L$ (returning $1$ if so and $0$ otherwise).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\bar S$ is the only function to use the power set notation or the substring notation.  We can write it as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\bar S:s\mapsto\{\sigma:\sigma\subset s,\,\sigma\neq\varepsilon\}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here I define a string $a$ to be a proper substring of $b$ ($a\subset b$) iff you can construct $a$ by removing a nonzero number of characters from $b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;V:s\mapsto\sum_{\chi\in s}U(\chi)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V$ simply sums the values of all the characters in its input.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $x$ we begin to put things together.  First I'll look at the reducing functor: for each element of its input, if that element is greater than the current value of the functor, it sets its current value to that of the element.  That is, it takes the maximum of its inputs.  Now we can write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x:s\mapsto\max_{s'\in\bar S(s)}V(s')I_L(s')&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can say a couple of things about $x$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Multiplying by $I_L(s')$ only lets $s'\in\mathcal L$ contribute to the result;&lt;/li&gt;&#xA;&lt;li&gt;Since $\varepsilon\notin\mathcal L$, that part of the definition of $\bar S$ is implicitly satisfied (and $V(\varepsilon)=0$ anyway).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thus we can rewrite:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x:s\mapsto\max\{V(\sigma):\sigma\in\mathcal L,\,\sigma\subset s\}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, $x$ returns the highest value of a string in $\mathcal L$ that is a substring of $s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we define $\Upsilon$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Upsilon:s\mapsto\frac{V(s)-x(s)}{2I_L(s) - 1}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I_L$ basically splits this into two cases:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Upsilon:s\mapsto [V(s)-x(s)]\times\left\{\begin{array}{ll}&#xA;+1 &amp;amp; s\in\mathcal L\\&#xA;-1 &amp;amp; s\notin\mathcal L\\&#xA;\end{array}\right.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After this, $M_Q$ merely sorts in order of increasing $\Upsilon$.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;The Answer&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The score of a letter can be determined according to the above table.  A string's base score is the sum of its letter's scores.  A string's adjusted score is equal to the the highest base score of strings in $\mathcal L$ that are proper substrings of itself, subtracted from its own base score.&lt;br&gt;&#xA; &#xA; Strings are sorted by $M_Q$ into order of &lt;em&gt;decreasing&lt;/em&gt; score, except for the strings in $\mathcal L$, which are placed &lt;em&gt;after&lt;/em&gt; all other strings in order of &lt;em&gt;increasing&lt;/em&gt; score.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The ordering of $\mathcal L$ (and a few extras):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/vnH2Q.png&quot; alt=&quot;enter image description here&quot;&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This image shows, for each string, the string's base score, the string with the substring(s) from $\mathcal L$ crossed out and the score that they remove, and the value of $\Upsilon$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we can see, the last string&lt;em&gt;s&lt;/em&gt; in $\mathcal L$ (which form an equivalence class under $M_Q$) are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\mathsf{bread}$ and $\mathsf{bear}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are (according to &lt;code&gt;DictionaryLookup&lt;/code&gt;) &lt;code&gt;72&lt;/code&gt; English words in the equivalence class immediately preceding $\mathsf{babadook}$, the longest of which is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\mathsf{embarrassed}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the shortest of which appear in the above table.&lt;/p&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2015-06-28T03:53:03.977" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="17094" PostTypeId="1" AcceptedAnswerId="17100" CreationDate="2015-06-27T18:05:20.960" Score="9" ViewCount="1424" Body="&lt;p&gt;There once was a famous work of modern art that everyone in the world loved. One particularly nasty person however hated it so much that decided to use a time machine to go back in time to when the artwork was first found (inside an old warehouse that was later made into a national monument for being the birthplace of the artistic treasure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;His plan was to use his favorite pair of scissors and cut up the artwork into random shapes. He knew that this would make it near unsolvable without intensive computer analysis. He was so confident that this would make it impossible to solve that he brought with him alphabet stickers to put on a message beforehand. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However once he got there and defaced the artwork with his stickers, he realized he forgot to bring his scissors! He only had 2 minutes remaining to stay in the past and there were no sharp objects or matches in the area. In a hurry he decided to tear the modern art up instead. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once he finished up tearing the artwork into 21 pieces and rearranging them, he exclaimed out loud in realization that he hadn't changed the rotational orientation of any of the pieces, making the puzzle much easier. He only had 10 seconds left so in a hurry, he rotated five pieces 180 degrees before he was sent back to the future. During his exclamation and his rotating of the five pieces, a smart puzzle hero (employed as a part time security guard) happened to be nearby and see him do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are this puzzle hero and you know that 5 pieces have been rotated 180 degrees. As a self proclaimed puzzle hero, you decide to prove to yourself that you were meant to be more than just a part time security guard. Can you solve this unsolvable (if there were no hints) puzzle?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In order to not make it take too long, I included many internal hints (such as the imprecise edges) so there are many potential methods to solve this puzzle. There is no information hidden in the image though, so if you printed it out you wouldn't be missing too much information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on how smart you are about it, I assume this will take different people much different times to complete (ranging from potentially half an hour to never). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will consider adding more hints if needed (that will fit into the context), but for now here's one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As a puzzle hero, you realize that since this is art, there is a low chance of like colors being next to each other or it would look weird. (There are no two of the same color that share the same boundary)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A final note, you don't have to make the solution image clean, I'll post the solution image if anyone gets it. Even as much as a general location (with a decoded message of course) will be enough to be selected as correct answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the puzzle: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SdmMv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="7722" LastEditDate="2015-06-27T18:12:13.177" LastActivityDate="2015-06-27T19:36:32.200" Title="The unsolvable puzzle (a colorful and challenging jigsaw 2.0)" Tags="|visual|jigsaw-puzzle|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17095" PostTypeId="2" ParentId="17090" CreationDate="2015-06-27T18:12:36.627" Score="15" Body="&lt;p&gt;Len has &lt;a href=&quot;https://puzzling.stackexchange.com/questions/17090/who-could-this-be?noredirect=1#comment53518_17090&quot;&gt;given me an idea&lt;/a&gt;...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://chat.stackexchange.com/transcript/message/20515421#20515421&quot;&gt;Are you &lt;em&gt;me&lt;/em&gt;?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My first is unpredictable, if you add some from the middle&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My first is &lt;em&gt;rand&lt;/em&gt;; adding the middle half (om) of &quot;some&quot; gives random - a synonym of unpredictable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My second is one, but a short line splits it apart&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My second is &lt;em&gt;al'thor&lt;/em&gt;: one word, with an apostrophe (short line) splitting it apart.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The first part is a common word, if you come from the Middle&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first part of al'thor is &lt;em&gt;al&lt;/em&gt;, which is the translation into Arabic (a common language in the Middle East) of the commonest word in English. See &lt;a href=&quot;https://en.wikipedia.org/wiki/Arabic_definite_article&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The last is worthy, coming from the bottom guard&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The last part is &lt;em&gt;thor&lt;/em&gt;; Thor comes from Asgard (which is a homophone of arse guard, which is a synonym of bottom guard), and he's the only one 'worthy' to lift Mjolnir in the Marvel films.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;almost everyone here has seen the answer.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm very active on Puzzling.SE!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm very flattered by this riddle - thanks CodeNewbie for the compliment! :-D&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-06-28T19:29:30.510" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="17096" PostTypeId="1" AcceptedAnswerId="17596" CreationDate="2015-06-27T18:15:47.303" Score="2" ViewCount="1117" Body="&lt;p&gt;This is an extension of a previous &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16006/a-balance-with-three-pans-detecting-the-lightest-pan-find-the-two-heavier-ball&quot;&gt;three-pan balance&lt;/a&gt; puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I give you a bag containing &lt;strong&gt;27 coins&lt;/strong&gt;, which look exactly the same. Among them are &lt;strong&gt;two fake coins&lt;/strong&gt;. All genuine coins weigh the same. The fake coins are very little &lt;strong&gt;heavier&lt;/strong&gt; than the genuine ones, and they may have a different weight from each other. You cannot distinguish the weight of genuine and fake coins by hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have no own utility to weigh coins, so I suggest you use my impossibly precise 3-pan balance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The three-pan balance&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The balance implements the &quot;lightest-pan-detection-rule&quot; (LPDR) and works like this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Put any selection of coins on the three pans, and press the activation button.&lt;/li&gt;&#xA;&lt;li&gt;After activation, the balance lifts the unique pan with the lightest load.&lt;/li&gt;&#xA;&lt;li&gt;If there is no unique lightest pan, none of the pans will rise.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You may only weigh the coins I gave you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The problem&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can you identify the two heavy coins with at most &lt;strong&gt;5&lt;/strong&gt; weighings?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Bonus question: Can you extend your solution to &lt;strong&gt;3^n&lt;/strong&gt; coins?&lt;/p&gt;&#xA;" OwnerUserId="318" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2016-12-22T12:54:23.183" Title="Find 2 heavy coins among 27 with a 3-pan balance" Tags="|weighing|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17097" PostTypeId="1" AcceptedAnswerId="17102" CreationDate="2015-06-27T18:30:35.520" Score="12" ViewCount="768" Body="&lt;p&gt;Here's my first attempt at a riddle, hopefully it's not too obvious. Answers should include explanations for each line. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm always stepped on by others, but I don't care. &lt;br&gt;&#xA;  I am strong as wood and cold as stone. &lt;br&gt;&#xA;  I can never win a race; the closest I come is a tie. &lt;br&gt;&#xA;  I'm often mistaken for a story. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;" OwnerUserId="4163" LastEditorUserId="12700" LastEditDate="2015-06-27T18:31:51.623" LastActivityDate="2015-06-27T23:04:32.900" Title="Who am I? - &quot;Stepped On&quot;" Tags="|riddle|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17098" PostTypeId="2" ParentId="17097" CreationDate="2015-06-27T18:34:48.243" Score="3" Body="&lt;p&gt;Is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a step on a staircase?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm always stepped on by others, but I don't care. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You step on the steps to go up or down.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am strong as wood and cold as stone. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A staircase can be made of wood or stone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can never win a race; the closest I come is a tie. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stuck with this one...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm often mistaken for a story.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Steps can take you to another 'storey' of the building&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-06-27T18:34:48.243" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17099" PostTypeId="2" ParentId="17097" CreationDate="2015-06-27T19:06:30.310" Score="3" Body="&lt;p&gt;Inspired by CodeNewbie's answer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an escalator&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm always stepped on by others, but I don't care.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; people are stepping on me, but I don't care since it's my purpose&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am strong as wood and cold as stone.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am made of steel thus making me strong and cold&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can never win a race; the closest I come is a tie.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I can never be ahead of people since I am always looping. I can be tie when people end their journey on me&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm often mistaken for a story. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Escalator can take you to another 'storey' of the building&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10400" LastActivityDate="2015-06-27T19:06:30.310" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17100" PostTypeId="2" ParentId="17094" CreationDate="2015-06-27T19:11:16.497" Score="9" Body="&lt;p&gt;The solved image is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/m2uKL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;Apologies for the small whitespace near the 'V'. That was a part of the purple piece, which got cut off by MS Paint. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The decoded message is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'THIS SHOULD NOT BE SOLVABLE', as we can read across the image.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-06-27T19:11:16.497" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17101" PostTypeId="2" ParentId="17094" CreationDate="2015-06-27T19:36:32.200" Score="5" Body="&lt;p&gt;Not fast enough? This is my solution....&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/0j1P0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastActivityDate="2015-06-27T19:36:32.200" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17102" PostTypeId="2" ParentId="17097" CreationDate="2015-06-27T23:04:32.900" Score="14" Body="&lt;p&gt;Also inspired by CodeNewbie's answer. Are you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a FLOOR?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I'm always stepped on by others, but I don't care.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This line is obvious.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am strong as wood and cold as stone.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Floors can be made out of wood or stone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I can never win a race; the closest I come is a tie.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In mathematics, the floor function $\lfloor x \rfloor$ can never exceed $x$; the best it can do is to equal $x$ if $x$ is an integer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I'm often mistaken for a story.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A building's storeys can also be called its floors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-06-27T23:04:32.900" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17103" PostTypeId="1" AcceptedAnswerId="17104" CreationDate="2015-06-28T04:22:22.000" Score="17" ViewCount="802" Body="&lt;p&gt;Which name of a large American city does the following equation represent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x^2+15600=649x&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="9718" LastActivityDate="2015-06-28T04:31:50.600" Title="Mathematical rebus: $x^2+15600=649x$" Tags="|mathematics|rebus|knowledge|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17104" PostTypeId="2" ParentId="17103" CreationDate="2015-06-28T04:31:50.600" Score="20" Body="&lt;p&gt;The city you are referring to is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chicago&lt;br&gt;&lt;br&gt;Solving the quadratic equation gives us $x = 25$ or $x = 624$. Or to put it another way, it's &lt;a href=&quot;https://en.wikipedia.org/wiki/25_or_6_to_4&quot;&gt;'25 or 6 to 4'&lt;/a&gt;, a song by the band Chicago.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-06-28T04:31:50.600" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17105" PostTypeId="2" ParentId="3956" CreationDate="2015-06-28T14:11:54.997" Score="3" Body="&lt;p&gt;I have created a simple solution using Tree&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I. Step 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Create a Tree of level 5 (n+1) and start point as Root Node&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;II. Step 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Create possible route as a child by parsing matrix (Assumption 1 is true)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;III. Step 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeat step 2 until you reach at the end (Level S4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2eAZb.jpg&quot; alt=&quot;Tree Structure&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;IV. Step 4&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now all the path which lead to S4 level are unique &#xA;So the possible arrangement will be :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I1 - I2 - I3 - I4&lt;/li&gt;&#xA;&lt;li&gt;I2 - I1 - I3 - I4&lt;/li&gt;&#xA;&lt;li&gt;I3 - I1 - I2 - I4&lt;/li&gt;&#xA;&lt;li&gt;I4 - I1 - I2 - I3&lt;/li&gt;&#xA;&lt;li&gt;I4 - I1 - I3 - I2&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13616" LastActivityDate="2015-06-28T14:11:54.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17106" PostTypeId="1" CreationDate="2015-06-28T14:29:45.037" Score="21" ViewCount="1720" Body="&lt;p&gt;Alfred is a guy who really likes solid shapes. He also really likes to keep his money in his wallet.&lt;/p&gt;&#xA;&lt;p&gt;What makes him the happiest? Getting a boxful of shapes at the Cheap Solids Store!&lt;/p&gt;&#xA;&lt;p&gt;The store has a new promotion: Get a sphere-shaped container for $20 and fill it up with tetrahedrons from the Tetrahedron Bin™. Each tetrahedron must have a corner on the center of the sphere. As long as nothing is sticking out, you may take the whole sphere home.&lt;/p&gt;&#xA;&lt;p&gt;The Tetrahedron Bin™ in the corner is full of these funny things:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kEoxk.png&quot; alt=&quot;tetrahedron&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Each tetrahedron in the bin has side length 1, and the sphere has radius 1.&lt;/p&gt;&#xA;&lt;p&gt;As Alfred is a very stingy guy, he wants to know in advance how many tetrahedrons he can get for his $20.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;TL;DR:&lt;/strong&gt; Given a sphere of radius 1 and tetrahedrons of side length 1, what is the maximum number of tetrahedrons you can pack into the sphere, such that each tetrahedron has a vertex on the center?&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; Notice there is no lateral-thinking tag on this one. That means you are not allowed to deform the tetrahedrons or the sphere, or feed them into a shrinking machine, or give them to an alien. Everything is subject to the normal laws of physics.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I am aware that this is an open problem, with no proof found as of now, hence the title, &amp;quot;The two million dollar question,&amp;quot; given by my math professor. I just thought the puzzling community might want to see this and think about it over lunch.&lt;/p&gt;&#xA;" OwnerUserId="12109" LastEditorUserId="40223" LastEditDate="2024-02-28T12:27:02.490" LastActivityDate="2024-02-28T12:27:02.490" Title="The 2 million, er, 20 dollar problem" Tags="|mathematics|geometry|packing|" AnswerCount="4" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="17107" PostTypeId="5" CreationDate="2015-06-28T14:31:52.773" Score="0" Body="&lt;p&gt;'Packing problems' are a class of optimization problems in mathematics that involve attempting to pack objects together into containers. Usually, the goal is either to pack a single container as densely as possible, or to pack all objects using as few containers as possible. Many of these problems can be related to real-life packaging, storage and transportation issues. Each packing problem has a dual 'covering' problem, which asks how many of the same objects are required to &lt;em&gt;cover&lt;/em&gt; every region of the container completely, where objects are allowed to overlap.&lt;/p&gt;&#xA;&lt;p&gt;In a packing problem, you are given:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;'Containers' (usually a single two- or three-dimensional convex region, or an infinite space).&lt;/li&gt;&#xA;&lt;li&gt;A set of 'objects', some or all of which must be packed into one or more of the containers. The set may contain different objects with their sizes specified, or a single object of a fixed dimension that can be used repeatedly.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Usually, the packing must be achieved without overlaps between goods and other goods or the container walls. In some variants, the aim is to find the configuration that packs a single container with the maximal density. More commonly, the aim is to pack all the objects into as few containers as possible. In some variants the overlapping (of objects with each other and/or with the boundary of the container) is allowed but should be minimized.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Packing_problems&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="60644" LastEditDate="2020-09-23T14:13:08.510" LastActivityDate="2020-09-23T14:13:08.510" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="17108" PostTypeId="4" CreationDate="2015-06-28T14:31:52.773" Score="0" Body="A puzzle that involves arranging objects optimally in order to fit in a specified space." OwnerUserId="12109" LastEditorUserId="5373" LastEditDate="2015-07-05T19:47:26.027" LastActivityDate="2015-07-05T19:47:26.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17109" PostTypeId="2" ParentId="16348" CreationDate="2015-06-28T15:02:42.057" Score="1" Body="&lt;p&gt;You are in a computer &#xA;Or You are a Softie&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You stand in a forest, &#xA;So black, dark, &#xA;Light can’t quite wind around the dry, tough bark.&#xA;&lt;strong&gt;Hard Disk&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trees are all you can see, &#xA;But change your viewpoint even slightly; &#xA;And a brighter vista it may indeed come to be.&#xA;**Chip / Motherboard **&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those seen trees which bear ripe plums, &#xA;Their seeds grew from the Two that do make One, &#xA;Which share no part except for another One.&#xA;But in this deep, dark, starless forest, those hidden &#xA;Have more in common, &#xA;With those that keep them overridden.&#xA;&lt;strong&gt;Object Oriented programming&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sticks in one big, infinite bundle, &#xA;Upright, but One far apart, &#xA;Farther removed from Mussolini's art.&#xA;&lt;strong&gt;Infinite loop / StackOverflow&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But these belong to one very ancient indeed, &#xA;A man renowned in strictest sense, &#xA;Celebrated most for famous work, derived from common sense.&#xA;Thin green trees in this private garden, &#xA;With raindrops, at tip top, &#xA;Belong to him, the mysterious Alexandrian.&#xA;Words, meaningless, focus on their numbers, &#xA;To unravel this poem, the key always remains Three, &#xA;Numbers speak languages untold, decipher them to avoid any blunders.&#xA;&lt;strong&gt;Binary Number System (1 and 0)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;May be You are in StackExchange&lt;/p&gt;&#xA;" OwnerUserId="13616" LastEditorUserId="13616" LastEditDate="2015-06-28T15:14:39.550" LastActivityDate="2015-06-28T15:14:39.550" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17110" PostTypeId="2" ParentId="16982" CreationDate="2015-06-28T15:31:32.417" Score="3" Body="&lt;p&gt;For the riddle, I've got&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; cheese&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It can be both cold and hot&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can eat it cold (as a chunk) or hot (as melted cheese)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You find it on chips more often than not &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; cheese on nachos, or a queso dip&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here, personal space can get pretty tight &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you say &quot;cheese&quot;, you squeeze tight with others to take a picture.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Writing helps me sleep at night&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure about this one, but a quick google search for &quot;cheese writing&quot; returns &lt;a href=&quot;http://www.toasted-cheese.com/&quot; rel=&quot;nofollow&quot;&gt;Toasted Cheese&lt;/a&gt;, a writing community.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, I know that's a cheesy answer :D&lt;/p&gt;&#xA;" OwnerUserId="12109" LastActivityDate="2015-06-28T15:31:32.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17111" PostTypeId="2" ParentId="17106" CreationDate="2015-06-28T15:47:43.580" Score="8" Body="&lt;p&gt;According to &lt;a href=&quot;https://matthewkahle.wordpress.com/2010/11/05/packing-tetrahedra/&quot;&gt;https://matthewkahle.wordpress.com/2010/11/05/packing-tetrahedra/&lt;/a&gt; this is a open problem so there is no solution yet&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-06-28T15:47:43.580" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17112" PostTypeId="1" AcceptedAnswerId="17115" CreationDate="2015-06-28T15:55:23.190" Score="1" ViewCount="6420" Body="&lt;p&gt;A man is stranded on an island covered in forest. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are no other humans or animals, he is the only one who lives on the island.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One day, when the wind is blowing from the west, lightning strikes the west end of the island and sets fire to the forest. The fire is very violent, burning everything in its path, and without intervention the fire will burn the whole island, killing the man in the process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are cliffs around the island, so he cannot jump off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can the man survive the fire? (There are no buckets nor any other means to put out the fire. He also has no tools to use.)&lt;/p&gt;&#xA;" OwnerUserId="13616" LastEditorUserId="-1" LastEditDate="2018-12-11T03:47:23.860" LastActivityDate="2018-12-11T03:47:23.860" Title="Burning Island - how to survive the fire?" Tags="|logical-deduction|strategy|story|situation|" AnswerCount="2" CommentCount="0" ClosedDate="2015-06-28T16:41:46.043" ContentLicense="CC BY-SA 4.0" />
  <row Id="17114" PostTypeId="2" ParentId="17112" CreationDate="2015-06-28T16:16:32.400" Score="3" Body="&lt;p&gt;There are no means to put out the fire, so maybe he can attempt to circumvent the fire. Since he has survived on the island alone for so long, we can assume he has the basic tools for survival. In this case, an axe is all he needs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the island is circular (any shape can be considered). I have taken the liberty of drawing an image for our understanding. The orange star is the origin of the fire, and it moves westward in a growing almost circular arc (not necessarily circular, since the wind will force it to cover more area in the westward direction.) Assume the blue circle is our lonely man, who we will call &lt;a href=&quot;https://en.wikipedia.org/wiki/Cast_Away&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chuck Noland&lt;/a&gt; for the sake of this solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yg1uF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Chuck can simply move and cut off the trees in a sector of the forest in a triangular fashion as shown by the white section between the dark green lines, with one end as close to the centre as possible. The fire will reach the westward side of the sector first and die out soon as there are no more trees in its path. Meanwhile, the fire will take some time to reach the other side of the sector, starting from the centre and moving upwards. While the westward side is on fire, Chuck can stand close to the eastward side until the fire burns out the trees on the westward side. Once the fire has died on the westward side, he can move towards the burnt part and then stay put till the fire burns out the trees on the eastward side. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Chuck has evaded death by fire, but with no more trees to provide fruits nor any animals to hunt and no usable wood to build himself a raft either, it's only a matter of time before Chuck dies of starvation. A pitiable end after a brilliant escape.&lt;/p&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-06-28T16:16:32.400" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17115" PostTypeId="2" ParentId="17112" CreationDate="2015-06-28T16:41:11.553" Score="12" Body="&lt;p&gt;The man picks up a piece of wood and lights it from the fire on the west end of the island. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;He then quickly carries it near the east end of he island and starts a new fire. The wind will cause that fire to burn out the eastern end and he can then shelter in the burnt area.&lt;/p&gt;&#xA;" OwnerUserId="13618" LastEditorUserId="12700" LastEditDate="2015-06-28T17:30:42.887" LastActivityDate="2015-06-28T17:30:42.887" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17116" PostTypeId="2" ParentId="17091" CreationDate="2015-06-28T17:28:46.650" Score="1" Body="&lt;p&gt;Being given that words &quot;extensive&quot; and &quot;doughnut&quot; are important to solving this I collected every letter contained within those two words. This gave me this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;eUXODGSDUHNvNVSEdOVSdGVVdNOhiDTi&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not 100% sure, but I thing the killer's relation is &lt;strong&gt;sister&lt;/strong&gt;, because from that string you can almost get an anagram sister. From the cipher text you can get &lt;code&gt;SisTE&lt;/code&gt;. Because you also said the singular doughnut has a meaning I think this is it. You meant the last letter was missing for security from being able to determine anagram too fast.&lt;/p&gt;&#xA;" OwnerUserId="13619" LastActivityDate="2015-06-28T17:28:46.650" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17117" PostTypeId="1" AcceptedAnswerId="17118" CreationDate="2015-06-28T23:00:47.253" Score="51" ViewCount="4829" Body="&lt;p&gt;It is election season in Puzzlevania. There are two candidates: the incumbent and a challenger. Of the $20$ million citizens, only $1$ percent support the current president; the other $99\%$ favoring the challenger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how the election system in Puzzlevania works: There is a list of numbers, $n_1,\dots,n_L$. The entire population is divided into $n_1$ equally sized groups. Each of these groups is divided into $n_2$ equally sized subgroups, each subgroup is divided into $n_3$ equal sub-subgroups, and so on down the line. On election day, each sub-sub-$\dots$-subgroup elects a representative to vote for it at the next level up. Tie votes count as wins for the challenger. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can think of this system as a many-layered version of the American electoral college, where all states have the same population.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current president has the power to choose the numbers $L,n_1,\dots, n_L$, and to choose which voters go in which groups. Can he do this in such a way that he gets elected?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://www.cs.cmu.edu/puzzle/puzzle20.html&quot;&gt;http://www.cs.cmu.edu/puzzle/puzzle20.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="8782" LastEditDate="2015-06-29T17:33:43.830" LastActivityDate="2015-06-30T03:37:11.140" Title="Extreme Gerrymandering" Tags="|strategy|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17118" PostTypeId="2" ParentId="17117" CreationDate="2015-06-29T00:19:33.667" Score="49" Body="&lt;p&gt;The current president should choose the numbers &lt;/p&gt;&#xA;&#xA;&lt;p&gt;9($L$), 32($n_1$), 8($n_2$), 5($n_3$), 5($n_4$), 5($n_5$), 5($n_6$), 5($n_7$), 5($n_8$), 5($n_9$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the lowest sub-sub-...-subgroup, the president should place voters in such a way that his supporters outnumber his opposers in the ratio of 3:5, till he runs out of supporters, and he can let the remaining sub-sub-...-subgroups vote for his opponent unanimously. So before the first round of voting, he has $200,000/20,000,000$ (1%) votes. After the first round, it becomes $66,666/4,000,000$ (1.66%). The current president needs to keep following this strategy till his percentage of votes exceeds 50%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the subsequent round, the same strategy needs to be followed so he can improve his vote share after each round of voting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following this strategy from $n_1$ till $n_9$, gives this vote share:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Supporting vote    Total Vote      Vote Share(%)     Voter ratio per round&#xA;&#xA;200,000            20,000,000      1&#xA;                                                     3:5&#xA;66,666             4,000,000       1.67              &#xA;                                                     3:5&#xA;22,222             800,000         2.78              &#xA;                                                     3:5         &#xA;7,407              160,000         4.63              &#xA;                                                     3:5&#xA;2,469              32,000          7.72              &#xA;                                                     3:5&#xA;823                6,400           12.86             &#xA;                                                     3:5&#xA;274                1280            21.41             &#xA;                                                     3:5&#xA;91                 256             35.55             &#xA;                                                     5:8&#xA;18                 32              56.25&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;After the final group of 32 representatives votes, the president gets re-elected and another period of corruption and malpractice ensues in Puzzlevania.&lt;/p&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-06-29T03:42:41.107" LastActivityDate="2015-06-29T03:42:41.107" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="17119" PostTypeId="1" AcceptedAnswerId="17850" CreationDate="2015-06-29T01:09:24.450" Score="44" ViewCount="3791" Body="&lt;p&gt;I have a strange problem that you may be able to help me with.  I was recently hired by company that manages several communications satellites, and I've gotten stuck with the overnight shift.  Earlier this week I was getting some weird interference on the C-band receiver during a pass sometime after midnight.  I didn't think too much of it initially, but it reappeared again Friday night, at around the same time.  This time I managed to record it:&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Time-domain plot&lt;/h3&gt;  &#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/q9zpR.png&quot; alt=&quot;enter image description here&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Frequency-domain plot (Spectrogram)&lt;/h3&gt;  &#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qcZUf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zooming in on the first few seconds:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/twfSQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Download the signal as audio&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://2012rcampion.com/se/sigint/2/signal.wav&quot; rel=&quot;noreferrer&quot;&gt;WAVE (&lt;code&gt;.wav&lt;/code&gt;, 10&amp;thinsp;116&amp;thinsp;k)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://2012rcampion.com/se/sigint/2/signal.mp3&quot; rel=&quot;noreferrer&quot;&gt;MP3 (&lt;code&gt;.mp3&lt;/code&gt;, 844&amp;thinsp;k)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://2012rcampion.com/se/sigint/2/signal.flac&quot; rel=&quot;noreferrer&quot;&gt;FLAC (&lt;code&gt;.flac&lt;/code&gt;, 9396&amp;thinsp;k)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://2012rcampion.com/se/sigint/2/signal.ogg&quot; rel=&quot;noreferrer&quot;&gt;Vorbis (&lt;code&gt;.ogg&lt;/code&gt;, 967&amp;thinsp;k)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://youtu.be/lmAkKRd__Jc&quot; rel=&quot;noreferrer&quot;&gt;listen on YouTube&lt;/a&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Unfortunately it hasn't reappeared since.  I want to ask one of my co-workers to scan the source of the signal with the big dish, but I've forgotten which satellite the antenna was pointed at when I recorded the signal!  Can you help me figure out &lt;strong&gt;where in the sky the signal originated?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="9718" LastEditDate="2017-03-22T18:15:33.920" LastActivityDate="2017-03-22T18:15:33.920" Title="Deciphering an Alien Signal" Tags="|mathematics|visual|story|outer-space|audio|" AnswerCount="6" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="17120" PostTypeId="2" ParentId="17119" CreationDate="2015-06-29T02:54:02.377" Score="6" Body="&lt;p&gt;&lt;strong&gt;Partial Answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going off of OP's newest hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The second-to-last 'page' seems to contain two nonsensical equations:&lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;= (1) (1.45e53)&lt;/p&gt;&#xA;    &#xA;    &lt;p&gt;= (1) (4.88e16)&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first one corresponds to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Observable_universe#Comparison_of_results&quot; rel=&quot;nofollow noreferrer&quot;&gt;mass of the universe in kg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second when googled (4.88e16), find very few results, the only coherent one (aside from this post) being &lt;a href=&quot;http://paperity.org/p/39434833/a-new-population-of-planetary-nebulae-discovered-in-the-large-magellanic-cloud-iii-the&quot; rel=&quot;nofollow noreferrer&quot;&gt;this link&lt;/a&gt;, a new population of planetary nebula discovered which seems to be more than a coincidence. Perhaps the signal came from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if that was true, the entire puzzle would be a build up to a disappointment so I'll chalk that up to coincidence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what we have so far is a measurement in kg, meaning either the 1.45e53 is by chance and not referring to the (average of 3 arbitrary measurements) mass calculation of the universe, or the signal came from Earth, particularly someone who referenced that Wikipedia page. I'll go with Earth for now (since we know who &lt;em&gt;actually&lt;/em&gt; created the signal).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here's a start on decoding the image from the audio signal:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/0ewkP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Wow!_signal&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Wow!_signal&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This wiki page has almost too many things in common with this puzzle to be a coincidence, haven't been able to progress further however so posting to give others some ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 1: Image removed to avoid confusion, refer to images posted by OP&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 2: Added (probably) useful info source&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 3: Added work in progress for image decoding&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 4: Added interpretation of OP's hint&lt;/p&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="7722" LastEditDate="2015-07-09T16:46:20.143" LastActivityDate="2015-07-09T16:46:20.143" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="17121" PostTypeId="1" AcceptedAnswerId="17174" CreationDate="2015-06-29T06:55:06.660" Score="13" ViewCount="751" Body="&lt;p&gt;A is for Lucerne&lt;br&gt;&#xA;K is for Jay-Z&lt;br&gt;&#xA;M and N are for Panorama&lt;br&gt;&#xA;R is for Misery and Enemy&lt;br&gt;&#xA;S is for Riddler&lt;br&gt;&#xA;Zed is for Radiator&lt;br&gt;&#xA;Zee is for Statistics&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is D for?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-07-01T11:35:07.860" Title="Oil your brain and learn your letters!" Tags="|word|enigmatic-puzzle|wordplay|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17122" PostTypeId="1" AcceptedAnswerId="17125" CreationDate="2015-06-29T07:13:24.970" Score="4" ViewCount="19629" Body="&lt;p&gt;There are three dogs and three cats that have to be transported across a river: a big dog, a medium dog and a small dog, a big cat, a medium cat, and a small cat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rules:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The raft can carry two animals, and all of the animals can row.&lt;/li&gt;&#xA;&lt;li&gt;The medium animal of each kind (darker color), cannot be left alone with either of the other animals of his kind, nor can the medium animal be transported with another animal of his kind.&lt;/li&gt;&#xA;&lt;li&gt;The big animal of each kind will not fight with the small animal.&lt;/li&gt;&#xA;&lt;li&gt;If ever there are more dogs than cats together on the shore, the dogs will fight with the cats.&lt;/li&gt;&#xA;&lt;li&gt;the raft can not be empty.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;All of these has to be crossed within minimum possible shifts.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="13373" LastEditorUserId="5373" LastEditDate="2017-04-28T12:52:50.500" LastActivityDate="2017-04-28T12:52:50.500" Title="How can the animals get across the river without a fight breaking out?" Tags="|optimization|river-crossing|" AnswerCount="5" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17123" PostTypeId="2" ParentId="17122" CreationDate="2015-06-29T08:34:04.847" Score="2" Body="&lt;p&gt;I don't think it's ever possible that ALL the animals will be able to cross that river&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Proof&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I'll name them &lt;strong&gt;BD, MD, SD / BC, MC, SC&lt;/strong&gt; (big dog, medium dog, small dog / big cat, medium cat, small cat) and &lt;strong&gt;Sh1&lt;/strong&gt; and &lt;strong&gt;Sh2&lt;/strong&gt; (Shore 1 and 2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sh1 contains: &lt;strong&gt;BD, MD, SD / BC, MC, SC&lt;/strong&gt;&lt;br&gt;&#xA;Sh2 contains: &lt;em&gt;no one&lt;/em&gt;&lt;br&gt;&#xA;Out of the 15 possible combinations:&lt;br&gt;&#xA;- &lt;strong&gt;BC,MC / BC,SC / MC,SC&lt;/strong&gt;: 2 cats cannot go together since Sh1 will be left with more dogs than cats.&#xA;- &lt;strong&gt;BD,BC / BD,MD / BD,SC / MD,BC / MC,MC / MD,SC / SD,BC / SD,MC / SD,SC&lt;/strong&gt;: only &lt;strong&gt;MD,MC&lt;/strong&gt; is possible because at least one of the other animal kind will be left with a middle animal of the same kind&lt;br&gt;&#xA;- &lt;strong&gt;BD,MD / MD,SD&lt;/strong&gt;: Not possible since the middle dog will start the fight&lt;br&gt;&#xA;- &lt;strong&gt;BD,SC&lt;/strong&gt;: Seems possible at first, but when one of them return with the raft, they wil remain the middle dog and start a fight&lt;br&gt;&#xA;So the only possible solution is &lt;strong&gt;MD,MC&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sh1 contains: &lt;strong&gt;BD, SD / BC, SC&lt;/strong&gt;&lt;br&gt;&#xA;Sh2 contains: &lt;strong&gt;MD, MC&lt;/strong&gt;&lt;br&gt;&#xA;Only &lt;strong&gt;MC&lt;/strong&gt; can return the craft because if MD return it, Sh1 will contain more dog than cat&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sh1 contains: &lt;strong&gt;BD, SD / BC, MC, SC&lt;/strong&gt;&lt;br&gt;&#xA;Sh2 contains: &lt;strong&gt;MD&lt;/strong&gt;&lt;br&gt;&#xA;1 cat and 1 dog is not possible because Sh2 will have more dogs than cat&lt;br&gt;&#xA;2 cats (BC,SC) also not possible because Sh1 will have more dogs than cat&lt;br&gt;&#xA;Only solution here is &lt;strong&gt;BD,SD&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sh1 contains: &lt;strong&gt;BC, MC, SC&lt;/strong&gt;&lt;br&gt;&#xA;Sh2 contains: &lt;strong&gt;BD, MD, SD&lt;/strong&gt;&lt;br&gt;&#xA;Only &lt;strong&gt;MD&lt;/strong&gt; can return the raft because if BD or SD return it, the other one will remain with the middle dog and start a fight&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sh1 contains: &lt;strong&gt;MD / BC, MC, SC&lt;/strong&gt;&lt;br&gt;&#xA;Sh2 contains: &lt;strong&gt;BD, SD&lt;/strong&gt;&lt;br&gt;&#xA;MD with any of the cat is not possible because Sh2 will have more dogs than cat&lt;br&gt;&#xA;&lt;strong&gt;MC,BC / MC, SC&lt;/strong&gt; also not possible because a fight will start with the middle cat&lt;br&gt;&#xA;Only &lt;strong&gt;BC, SC&lt;/strong&gt; is possible&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sh1 contains: &lt;strong&gt;MD / MC&lt;/strong&gt;&lt;br&gt;&#xA;Sh2 contains: &lt;strong&gt;BD, SD / BC, SC&lt;/strong&gt;&lt;br&gt;&#xA;Here it ends because no dog can return the shaft (Sh1 will have more dog than cat) and no cat can return it neither (a fight will start with the middle cat)&lt;/p&gt;&#xA;" OwnerUserId="12731" LastActivityDate="2015-06-29T08:34:04.847" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="17125" PostTypeId="2" ParentId="17122" CreationDate="2015-06-29T09:18:16.653" Score="10" Body="&lt;p&gt;&lt;strong&gt;It is possible.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start: BD MD SD BC MC SC X on left, none on right (X is the boat)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;MD MC go to the right.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;BD SD BC SC on left, MD MC X on right&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;MC returns to the left.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;BD SD BC MC SC X on left, MD on right&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;BD SD go to the right.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;BC MC SC on left, BD MD SD X on right&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;MD returns to the left.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;MD BC MC SC X on left, BD SD on right&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;BC SC go to the right.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;MD MC on left, BD SD BC SC X on right&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;6&quot;&gt;&#xA;&lt;li&gt;BD BC return to the left together.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;BD MD BC MC X on left, SD SC on right&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;7&quot;&gt;&#xA;&lt;li&gt;MD MC go to the right.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;BD BC on left, MD SD MC SC X on right&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;8&quot;&gt;&#xA;&lt;li&gt;SD SC return to the left together.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;BD SD BC SC X on left, MD MC on right&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;9&quot;&gt;&#xA;&lt;li&gt;BC SC go to the right.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;BD SD on left, MD BC MC SC X on right&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;10&quot;&gt;&#xA;&lt;li&gt;MD returns to the left.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;BD MD SD X on left, BC MC SC on right&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;11&quot;&gt;&#xA;&lt;li&gt;BD SD go to the right.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;MD on left, BD SD BC MC SC X on right&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;12&quot;&gt;&#xA;&lt;li&gt;MC returns to the left.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;MD MC X on left, BD SD BC SC on right&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;13&quot;&gt;&#xA;&lt;li&gt;MD MC go to the right.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;none on left, BD MD SD BC MC SC X on right&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And they're done.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A crucial part of this solution is how to proceed when two dogs, two cats, and the boat are on the left. It is possible to send two cats, but we need to make sure that those two cats are the big one and the small one, and the same for the two dogs remaining. This is what steps 6-8 are doing.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-06-29T10:09:33.197" LastActivityDate="2015-06-29T10:09:33.197" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17127" PostTypeId="2" ParentId="17122" CreationDate="2015-06-29T10:06:14.540" Score="0" Body="&lt;p&gt;It isn't impossible after all.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(B = big, M = medium, S = small)&#xA;(D = dog, C = cat)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I divide the area into 3 parts - the left shore on the left, river in the centre and right shore on the right.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1) LD LC MD MC SD SC |       |&#xA;2) LD LC SD SC       | MD MC |&#xA;3) LD LC SD SC       | MD    | MC          (drop MC on Right shore, MD stays in raft)&#xA;4) LD MD SD          | LC SC | MC          (drop MD on Left shore, pick up LC SC)&#xA;5) LD MD SD          |       | LC MC SC&#xA;6) LD MD SD          |       | LC MC SC    (Send the empty raft back to Left shore)&#xA;7) MD                | LD SD | LC MC SC&#xA;8) MD                | MC    | LC SC LD SD (drop LD SD on Right shore, pick up MC)&#xA;9)                   | MC MD | LC SC LD SD (pick up MD from Left shore)&#xA;10)                  |       | LD LC MD MC SD SC&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Complete.&lt;/p&gt;&#xA;" OwnerUserId="9074" LastActivityDate="2015-06-29T10:06:14.540" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17128" PostTypeId="2" ParentId="16935" CreationDate="2015-06-29T10:26:47.057" Score="25" Body="&lt;p&gt;Jeez, I finally got it (well, at least I hope so).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The logic for admission is a classic one, but with a silly twist:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The value of a sentence/question is given by the total amount of closed areas in the letters of which it is composed. You know, the ones you would fill up with a coloured pencil to kill time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This doesn't seem to work exactly right, so here's the twist:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You have to uppercase (like screaming) the sentence, writing out the numbers as words.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So this is what happens:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;TELL ME... WHAT NUMBER IMMEDIATELY FOLLOWS SEVEN?&quot; has eight closed areas (A = 1, B = 2, R=1, O = 1). &quot;THIRTEEN MINUS TWELVE?&quot; has just one in the R, and so on.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;About the high-five guy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The guardian asks for a five, so he gives a sentence with a value of 5, showing that he knows the logic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally, the correct answer for the author:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TWO AND TWO MAKES =&gt; 5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5990" LastActivityDate="2015-06-29T10:26:47.057" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17129" PostTypeId="2" ParentId="17122" CreationDate="2015-06-29T12:46:17.877" Score="2" Body="&lt;p&gt;It is Possible without an empty boat:&#xA;(at least if there is a Person with the boat &quot;have to be transported&quot; would imply that to be the case and animals don't fight and aren't alone if the person is there. Or they can at least restrain from fighting while changing seats in the boat)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;my animals:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(s = small cat, m = medium cat, b = big cat,&#xA; S = small dog, M = medium dog, B = big dog)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and here the boat:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;- __ &amp;gt; &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;rules:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;allways more or equal cats(smb) then dogs(SMB)&#xA;never a medium animal with another of the same kind on a shore (except beginning and end)&#xA;and never two animals of a kind together on the boat (an even stronger one that no Medium animal with another of his kind)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and here the solution with 11 crossings:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;smb SMB&#xA;sb  SB  - mM &amp;gt;&#xA;sb  SB  &amp;lt; m  -     M&#xA;sb  S   - mB &amp;gt;     M&#xA;sb  S   &amp;lt; M  - m   B&#xA;s   S   - Mb &amp;gt; m   B&#xA;s   S   &amp;lt; Mm &amp;gt; b   B&#xA;m   S   - Ms &amp;gt; sb  B&#xA;m   S   &amp;lt; M  - sb  B&#xA;    M   - mS &amp;gt; sb  B&#xA;    M   &amp;lt; m  - sb  SB&#xA;        - mM &amp;gt; sb  SB&#xA;               smb SMB&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="13626" LastEditorUserId="13626" LastEditDate="2015-06-29T12:54:27.907" LastActivityDate="2015-06-29T12:54:27.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17130" PostTypeId="1" AcceptedAnswerId="17131" CreationDate="2015-06-29T14:16:15.460" Score="3" ViewCount="1131" Body="&lt;p&gt;What will be the next character in the following series?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;j , f , m , a , m , j , j , a , ?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="13616" LastEditorUserId="25957" LastEditDate="2016-12-22T12:53:41.050" LastActivityDate="2016-12-22T12:53:41.050" Title="What is the next character in series?" Tags="|logical-deduction|letter-sequence|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17131" PostTypeId="2" ParentId="17130" CreationDate="2015-06-29T14:18:02.177" Score="14" Body="&lt;p&gt;The next letter is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; s&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; each letter is the beginning of a month in chronological order. The next month to be listed is September.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-06-29T20:53:32.887" LastActivityDate="2015-06-29T20:53:32.887" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17132" PostTypeId="2" ParentId="323" CreationDate="2015-06-29T14:18:48.823" Score="6" Body="&lt;p&gt;The best tool to generate scrambles for all WCA puzzles is &lt;a href=&quot;http://ruwix.com/puzzle-scramble-generator/&quot; rel=&quot;noreferrer&quot;&gt;http://ruwix.com/puzzle-scramble-generator/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It supports cubes from 2x2x2 up to 11x11x11, Square-1, Rubik's Clock, Megaminx, Pyraminx and Skewb. You can customize the scramble in many ways and there's a built-in stopwatch to measure your times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/y8uRX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="13627" LastActivityDate="2015-06-29T14:18:48.823" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17133" PostTypeId="2" ParentId="17130" CreationDate="2015-06-29T14:19:07.387" Score="5" Body="&lt;p&gt;The next letter is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; S &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the pattern is the first letter of each month and September is the next month after August&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13628" LastEditorUserId="9000" LastEditDate="2015-06-29T14:58:16.183" LastActivityDate="2015-06-29T14:58:16.183" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17134" PostTypeId="1" AcceptedAnswerId="17136" CreationDate="2015-06-29T14:48:57.990" Score="3" ViewCount="577" Body="&lt;p&gt;This is part of the &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=title%3A%22Fill%20in%20the%20Dashes%22&quot;&gt;Fill in the Dashes&lt;/a&gt; series of puzzles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt;&lt;br&gt;&#xA;F_ in the B_ &amp;nbsp;&amp;nbsp;$\Rightarrow$&amp;nbsp; Fill in the Blanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Some of the words will be juxtaposed as in:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;B_ F_ in the &amp;nbsp;&amp;nbsp;$\Rightarrow$&amp;nbsp; Fill in the Blanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can you identify the idiom(s)/phrase(s)/proverb(s) from english language ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) F_ F_ a as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) D_ L_ L_ S_&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) F_ T_ T_ N_ W_ and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) N_ S_ T_ S_ A in  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) B_ B_ C_ E_ T_ at&lt;/p&gt;&#xA;" OwnerUserId="9892" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2016-12-22T12:53:19.013" Title="Fill in the Dashes : Part 2 of 10" Tags="|english|knowledge|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17135" PostTypeId="2" ParentId="17134" CreationDate="2015-06-29T15:05:47.910" Score="10" Body="&lt;p&gt;1)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fit as a fiddle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Burn the candle at both ends&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5359" LastEditorUserId="5359" LastEditDate="2015-06-29T15:08:28.580" LastActivityDate="2015-06-29T15:08:28.580" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17136" PostTypeId="2" ParentId="17134" CreationDate="2015-06-29T15:06:43.423" Score="10" Body="&lt;p&gt;&lt;del&gt;&lt;em&gt;A partial answer...&lt;/em&gt;&lt;/del&gt; 2/5 cracked by &lt;a href=&quot;https://puzzling.stackexchange.com/a/17135/12700&quot;&gt;TroyAndAbed, upvote his answer please.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Credits to TroyAndAbed&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fit as a fiddle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let sleeping dogs lie&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Time and tide waits for none&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A stitch in time saves nine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;Credits to TroyAndAbed&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Burn the candle at both ends&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2015-06-29T15:15:46.660" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17137" PostTypeId="1" AcceptedAnswerId="17138" CreationDate="2015-06-29T15:32:37.737" Score="34" ViewCount="6215" Body="&lt;p&gt;You can draw up to 4 straight lines in order to create 10 equal squares in the following figure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XKe0Y.jpg&quot; alt=&quot;puzzle of two squares&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="13618" LastEditorUserId="12462" LastEditDate="2015-06-29T15:47:08.973" LastActivityDate="2015-07-01T00:14:25.073" Title="Draw 4 straight lines to create 10 equal squares in this image" Tags="|geometry|strategy|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17138" PostTypeId="2" ParentId="17137" CreationDate="2015-06-29T15:36:21.453" Score="45" Body="&lt;p&gt;Please ignore my painting skill. i am not good at it&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/Daxp0.png&quot; alt=&quot;&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13616" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-06-30T14:39:31.810" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="17139" PostTypeId="2" ParentId="17137" CreationDate="2015-06-29T15:40:12.443" Score="21" Body="&lt;p&gt;Kavi Temre got this first, but here's my unnecessarily clear image:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/4Ll4Y.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12462" LastActivityDate="2015-06-29T15:40:12.443" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17140" PostTypeId="2" ParentId="16982" CreationDate="2015-06-29T15:48:28.313" Score="3" Body="&lt;p&gt;Credit to &lt;a href=&quot;https://puzzling.stackexchange.com/a/17041/9000&quot;&gt;TroyAndAbed&lt;/a&gt; for deciphering to get the riddle.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It can be both cold and hot&lt;br&gt;&#xA;  You find it on chips more often than not&lt;br&gt;&#xA;  Here, personal space can get pretty tight&lt;br&gt;&#xA;  Writing helps me sleep at night   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I believe the answer to the riddle is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Compress / Compression&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It can be both cold and hot   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A hot compress and cold compress are both used for medical purposes, usually to soothe sore spots.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You find it on chips more often than not  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Computers often store data in a compressed format. Computer chips, heh.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here, personal space can get pretty tight   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The primary definition of &quot;compress&quot; is &quot;to press together; force into less space.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Writing helps me sleep at night  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This relates back to the computer chip clue. When you put a computer into hibernation, it writes what's currently on the RAM to the hard drive before shutting off power.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Based on all this, I believe the secret to Mr. McMoneybags' success is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Computers&lt;/strong&gt; (It could also be medical compresses or a compression algorithm but I think computers make the most sense. The answers to the riddle are clues and not the final solution.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-06-29T15:48:28.313" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17141" PostTypeId="2" ParentId="17106" CreationDate="2015-06-29T16:03:48.750" Score="23" Body="&lt;p&gt;This question is not about packing tetrahedra, it is about how many tetrahedra can share a common vertex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The constraints of this question are that the side length of the tetrahedra is equal to the radius of the sphere and all tetrahedra share a common vertex at the center of the sphere.  Given these constraints,  all vertices are either at the center of the sphere or on the surface of the sphere.  Thus, the sphere is not relevant; the question becomes &quot;How many regular, identical tetrahedra can share a single vertex?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solid angle at a vertex of a regular tetrahedron is $\arccos (23/27) \approx 0.55$ steradians $\approx 4\pi / 22.8$.  So we have an upper limit of 22.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A regular icosahedron has triangular sides with edges that are $\approx 1.05$ times the radius of a circumscribing sphere.  Thus, we could arrange a 20 tetrahedra such that each outer face is aligned with that of an icosahedron (same icosahedron for all faces, if that wasn't clear).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, we have a lower limit of 20 and an upper limit of 22.  This is a much simpler problem than finding an efficient packing algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proving that 20 is the maximum is probably possible.  Given that the symmetries make packing very efficient (edges aligned with edges), I do not believe it could be possible to squeeze any more in.  However, proving it formally would require 3D modeling software, which I don't have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reference, this image (&lt;a href=&quot;https://en.wikipedia.org/wiki/File:Tetrahedrons_cannot_fill_space..PNG&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;) shows how little space there is left for fudging:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wDOwb.png&quot; alt=&quot;Packing of 5 tetrahedra around a point&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't believe it would be possible to rearrange the tetrahedra to fit in more than 20.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2015-06-29T16:03:48.750" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17142" PostTypeId="1" AcceptedAnswerId="17146" CreationDate="2015-06-29T16:45:14.437" Score="4" ViewCount="391" Body="&lt;p&gt;This is part of the &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=title%3A%22Fill%20in%20the%20Dashes%22&quot;&gt;Fill in the Dashes&lt;/a&gt; series of puzzles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt;&lt;br&gt;&#xA;F_ in the B_   ⇒  Fill in the Blanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Some of the words will be Juxtaposed as in:&lt;/strong&gt;&lt;br&gt;&#xA;B_ F_ in the   ⇒  Fill in the Blanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;e.g.:&lt;/strong&gt; C_ K_ S_    ⇒  Clark Kent Superman or Superman Clark Kent&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can you identify these?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) B_ B_ W_ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) E_ K_ S_&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) I_ S_ T_ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) P_ P_ S_&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) A_ A_ J_ C_&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6) D_ P_ W_ W_&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7) A_ C_ R_ S_&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8) D_ G_ R_&lt;/p&gt;&#xA;&#xA;&lt;p&gt;9) B_ B_ H_&lt;/p&gt;&#xA;&#xA;&lt;p&gt;10) I_ S_ S_ W_&lt;/p&gt;&#xA;&#xA;&lt;p&gt;11) N_ R_ S_ S_&lt;/p&gt;&#xA;&#xA;&lt;p&gt;12) A_ G_ O_ Q_&lt;/p&gt;&#xA;&#xA;&lt;p&gt;13) C_ B_ H_&lt;/p&gt;&#xA;&#xA;&lt;p&gt;14) S_ W_ W_&lt;/p&gt;&#xA;&#xA;&lt;p&gt;15) D_ M_ M_&lt;/p&gt;&#xA;&#xA;&lt;p&gt;16) K_ P_ W_&lt;/p&gt;&#xA;&#xA;&lt;p&gt;17) B_ I_ M_ P_&lt;/p&gt;&#xA;" OwnerUserId="9892" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2016-12-22T12:52:02.877" Title="Fill in the Dashes : Part 3 of 10 - Comic Superheroes Alter-egos" Tags="|knowledge|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17143" PostTypeId="2" ParentId="17142" CreationDate="2015-06-29T16:50:09.967" Score="5" Body="&lt;p&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Batman, Bruce Wayne&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Peter Parker, Spiderman&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Aquaman, Arthur Joseph Curry&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;9.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bruce Banner, Hulk&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;11.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Norin Radd, Silver Surfer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;13.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clint Barton, Hawkeye&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;16.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Kit Walker, Phantom&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;17.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bob Parr, Mr Incredible&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-06-29T18:02:21.473" LastActivityDate="2015-06-29T18:02:21.473" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17144" PostTypeId="2" ParentId="17142" CreationDate="2015-06-29T16:52:20.633" Score="4" Body="&lt;p&gt;6) D_ P_ W_ W_&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Diana Prince, Wonder Woman&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;7) A_ C_ R_ S_&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Captain America, Steve Rogers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;12) A_ G_ O_ Q_&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Green Arrow, Oliver Queen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4450" LastEditDate="2015-06-29T16:58:35.277" LastActivityDate="2015-06-29T16:58:35.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17145" PostTypeId="2" ParentId="16982" CreationDate="2015-06-29T16:53:59.977" Score="12" Body="&lt;p&gt;Also crediting &lt;a href=&quot;https://puzzling.stackexchange.com/a/17041/9000&quot;&gt;TroyAndAbed&lt;/a&gt; for deciphering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the answer is his:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Memory&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It can be both cold and hot&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Memory modules will be cold when a PC is off, hot when on.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You find it on chips more often than not&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As Tim said in the comments for &lt;a href=&quot;https://puzzling.stackexchange.com/a/17140/9000&quot;&gt;this answer&lt;/a&gt;, EngineerToast was correct in assuming it was a &quot;computer chip.&quot; So, a memory chip.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here, personal space can get pretty tight&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Computer memory is usually very cramped, with all of the programs and the OS running.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Writing helps me sleep at night&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A computer will write anything needed to be saved from memory to storage before sleeping, or being shut down.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13631" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-06-29T17:30:58.673" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17146" PostTypeId="2" ParentId="17142" CreationDate="2015-06-29T17:10:26.743" Score="4" Body="&lt;p&gt;2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Kal-El, Superman&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tony Stark, IronMan&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;8&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dick Grayson, Robin&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;10&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Susan Storm, Invisible Woman&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;13&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clint Barton, Hawkeye&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;14&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Simon Williams, Wonderman&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;15&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Matt Murdock, Daredevil&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12223" LastEditorUserId="12223" LastEditDate="2015-06-29T17:23:22.963" LastActivityDate="2015-06-29T17:23:22.963" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17147" PostTypeId="2" ParentId="17091" CreationDate="2015-06-29T17:26:44.297" Score="11" Body="&lt;p&gt;The murderer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The neighbour's son.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The message is encrypted using&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an Extended Vigenere Square, which includes the characters A-Z, a-z, 0-9, and a space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The decryption key, as mentioned by the hints, is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; doughnut&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The decrypted text reads (punctuation added)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Brothers and sisters I have none, but the murderer's father is my father's son.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This means that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the murderer's father is the victim, and thus it is the victim's son or daughter that killed him.  Based on the clue about a masked man running from the house, I'm going to go with &quot;son.&quot;  Hopefully he has only one, or the case could be tied up for a while in court...  :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2015-06-29T17:26:44.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17148" PostTypeId="2" ParentId="16982" CreationDate="2015-06-29T17:28:46.740" Score="1" Body="&lt;p&gt;His secret to success is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Double Dipping.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It can be both cold and hot&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Temperature can dip.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You find it on chips more often than not &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A type of dip, that you put chips into.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here, personal space can get pretty tight &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dipping into a divot in the ground. A hole.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Writing helps me sleep at night&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your mind can easily &quot;dip&quot; off. Or doze off.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Binary&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It can be both cold and hot&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0 for cold 1 for hot.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You find it on chips more often than not &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Microchips and electronic storage, at their most elementary level. Hold binary.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here, personal space can get pretty tight&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not much &quot;room&quot; for any other integers on the set of [0,1].&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Writing helps me sleep at night&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not too sure about this one, but it could be referring to writing low level binary code. Or how writing it can be boring enough to put someone to sleep.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13208" LastEditorUserId="13208" LastEditDate="2015-06-29T17:40:49.283" LastActivityDate="2015-06-29T17:40:49.283" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17152" PostTypeId="2" ParentId="323" CreationDate="2015-06-29T18:02:09.907" Score="14" Body="&lt;p&gt;If the goal is to have a truly random cube, holding behind your back and scrambling it is as good as any other method, as long as you do ~20 twists or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People often intuitively think that if you scramble a cube for a really long time, it will make it harder to solve. Beyond the first few twists, this is not so. Not only can any amateur solver demonstrate a cube scrambled for half an hour is as easy to solve as a cube scrambled for 10 seconds, there is a proven upper bound on the &quot;hardest&quot; possible Rubik's cube state: 20 moves from scrambled to solved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cube20.org/&quot;&gt;http://www.cube20.org/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Naturally it's possible to produce a scramble using any method where, by random chance, there is less work needed at some point (As in, you solve the first two layers, and lo and behold, the third layer required only one operation!). This is a perfectly fair random occurrence though, and a good scrambler wouldn't discriminate against such scenarios, even if it could detect them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more thing: for most solvers, as few as six random twists is enough to scramble a cube. It is a fun puzzle to have a friend perform 4, 5, or 6 random twists on a solved cube, and then see if you can undo the twists without messing up the cube. 4 is usually easy, 5 is quite hard for me, and I can't do 6 without getting lucky : )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bottom line is, unless you're a savant or a supercomputer, don't overthink cube scrambling!&lt;/p&gt;&#xA;" OwnerUserId="13350" LastActivityDate="2015-06-29T18:02:09.907" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17153" PostTypeId="1" CreationDate="2015-06-29T19:30:14.687" Score="4" ViewCount="1005" Body="&lt;p&gt;The answer is a five-letter English word, but &quot;give me a hint&quot; you might say. Unscramble the non-English text below, and you will be well on your way to solving this riddle. Just don't let &lt;code&gt;jkc sßW rJTß gLOBsRuuprwe uöIN bNfs ävErgG KvUpAdcIsnä MZx üzc oßxhvs dRMäD saaUeof fxHz KNbK VxJkck KGwjöb fb&lt;/code&gt; get you down, or the oracle behind &lt;a href=&quot;https://gist.github.com/Blastus/e039a8dffdb6b15b36fb&quot; rel=&quot;nofollow noreferrer&quot;&gt;the matrix&lt;/a&gt; will refuse to reveal her secrets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1:&lt;/strong&gt; It seems that something was missing, and so that missing piece has been added to the place it belongs. Maybe by comparing past revisions of this riddle something important will become evident. In the mean time, take note that space and newline characters were not touched during encryption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2:&lt;/strong&gt; A few more changes have been introduced to the riddle. Check the matrix to see its new design and its previous contents. Look at the revision history for this question to see what has changed and the extent of those edits. The message now has unencrypted punctuation characters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 3:&lt;/strong&gt; Oh no! The same problem has been discovered that was present with the very first version of this riddle. The old, German poem was missing something at the beginning that should have been there in the first place. Can you compare the various revisions to understand the message better?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 4:&lt;/strong&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/users/13268&quot;&gt;Cerberus&lt;/a&gt; has correctly identified &lt;a href=&quot;https://en.wikipedia.org/wiki/The_Sorcerer%27s_Apprentice&quot; rel=&quot;nofollow noreferrer&quot;&gt;the poem&lt;/a&gt; which correlates with Europe. Now want does the poem have in common with Timon and Pumbaa? Try to answer &quot;what is everywhere?&quot; As a bonus, try to identify who &lt;a href=&quot;https://gist.github.com/Blastus/e039a8dffdb6b15b36fb&quot; rel=&quot;nofollow noreferrer&quot;&gt;the matrix&lt;/a&gt; is. She has attributes in common with Timon, Pumbaa, and the poem.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;pUo xfDnkduvsBJhAJ&#xA;muTßUo ZGhfOGoe MkI ktgoRz&#xA;&#xA;MGk iöv RLAM hkzAjfktzOKn&#xA;Demu vacS jsLkMx RTwrbSüNGL!&#xA;ORa fcG gnSgih äIäls cTGksüF&#xA;gööB lMTF dLhgmi eIZUZD mfoxd.&#xA;djOtß VTkF öRc ZhHZg&#xA;äuhVe öEh uZh fOW keüVzg,&#xA;GOK xnA RMGeOxSDOcLDw&#xA;LD ZRh HzmrtK hzßV.&#xA;&#xA;SßBpw! oLJuN&#xA;VKnznk ähJozwJ,&#xA;ähi, vEr MMSvZD,&#xA;FenZhs WGUSxz&#xA;BAE fdk jLxIKRD, bVDutZ vzZnLbhJ&#xA;TE MJl zojh EKGk udSlLßo.&#xA;&#xA;SSO äNi Gvwü, eA MIVVK öUHGa,&#xA;zRhi LDg lmIHgouLcS ZJDNuKNdSiex!&#xA;RMMA jBtVu GGvah IOGNRk blRJuFG:&#xA;Ima JßLGAEä puiwöO gHhNIn!&#xA;Szp Wwwm DiJjFN ääigf,&#xA;RTZb Ivg VmW ceüü,&#xA;zfzv RpG öVU cäpM&#xA;cmI dma gRzgöNzxDu!&#xA;&#xA;ögUeH! lSöoW&#xA;eEeLnk HoSsfAs,&#xA;ZBg, ügt DigAnm,&#xA;wAAssu söBwRW&#xA;üSI TAg OnfhDEd, DMoZKö zhIIStwV&#xA;Tf VJA Ruxp WFäb wpcVBoo.&#xA;&#xA;ldAw, dL VLAuG pSd nHfo xMDdcU!&#xA;VOBßhhßF! tvs oJmJe vb KxK cdäLgz,&#xA;par Tgt brZbkwNznAnxRNm ALoxvn&#xA;lwd fV ijdT MGx aOzWUtJ pIAcM.&#xA;WLccp Dzx iZblJoi rfNT!&#xA;exN Doh xtNTar SmxjvZov!&#xA;pko üDßz rViE fWWsMz&#xA;IsWs VBu HptwHj äLwLD!&#xA;&#xA;üUpVh! nZMxH!&#xA;jswS Kpj ütJeo&#xA;hGvösp Umbje&#xA;iowüTiürßKos! -&#xA;fIV, HMT fbmJ dö! öRcr! zHOO!&#xA;RNK nuz lakx FBA vWOu GboSIüJOl!&#xA;&#xA;JmZ, ücL ovpA, wßMrßN Be ZäJa&#xA;Ki uiS wujI, luL Mf MLjäeJn!&#xA;hNs, ud MHSSI lLD sLhgWK sRvIrdV!&#xA;Mzöac Rn ßiUü bTk LWpd NhWvö!&#xA;GVfTH bINF hFWrD&#xA;dHFDRf zx RxWpbJU jkmüGA,&#xA;böd, lgU cüLNsMr üäEkRR&#xA;jTjsacp Luö Fmjd ADF!&#xA;&#xA;jxäL, lüOID oukaMV&#xA;gmIN üsäk ieIHhx:&#xA;WdKä zAü TMnmUS!&#xA;KuW OVl ukjcN!&#xA;RZc, wuu BZju Fwö IDRhf ZTIdaß!&#xA;UJuäxe iJTjT! WLöcMh öOftuK!&#xA;&#xA;n, iu müjuclKsS pFU SnMmp!&#xA;MnGo NoW tRüTK twwA DaVjhMek?&#xA;hBm Wpw nWIp Sfwz jfZIeHßW&#xA;efkJ üEcöR KtnäKcbVhäLt iKämuO.&#xA;rrv AhGßLrJBzö zulmü, &#xA;MgM OzIMw tWDKk cKül!&#xA;xöUrE, pöe HE toSRcOR,&#xA;bhän cxIw FJödtj ZßöWR!&#xA;&#xA;RKtdlL bh Kllg&#xA;SMö DDxäS ZxhZaL?&#xA;nbVb IpZH wävcxW,&#xA;vaKW KlEe sZMLäD&#xA;Ika ödK Fößv NIFO WäuERJE&#xA;TEM weH ßehcZKGt paShU JzHsüßk!&#xA;&#xA;eeDe, eß Wrbol IL LBBpphöJnG oHäAIF!&#xA;TZä URZ REkf üHO rcc DsMd uKWGä,&#xA;wäELRb, p wHkNuo, KBzwIT ßä ZKOTco;&#xA;NZvxtMcM cfWTVd eEz iHOKZM ffüjerR.&#xA;wLzTxlMB! WoFh vMvcpOMFd! &#xA;zWcH, rK FaB ßErdorV!&#xA;MBl TzL jATM cEk aknaWV,&#xA;IgA üüL Rtgü Fohü!&#xA;&#xA;bjiv! Aögv!&#xA;BfkVd eäUmE&#xA;eövmA fc onHl&#xA;evvOK Hzm iFßiVSn&#xA;öämUjs tSLrlg am KMS pTpn!&#xA;aZlgN RIS, tsR! Twu llnüö jNpkWM!&#xA;&#xA;WJV jAh BLHMwä! üäc Drj RoüßjZ&#xA;hAßjx NK ZHTB MOb iAf KIk äaphnß:&#xA;WJafü uüVxVTBLwsHrE äBakTDTn!&#xA;nhhg uWT JcalUZU, jaL ZVIU efwfS! -&#xA;Bnl, wB ngWpI jZü Ahfbnxz!&#xA;dRfß, jSn nvä dwö dLDj!&#xA;scD JdI dVsA, Kßz gONIövi,&#xA;tdwj rxk mhG rTztD Kze.&#xA;&#xA;&quot;Kh FaM Ardl,&#xA;AVses! kSIbk!&#xA;SNZBn öURkEle!&#xA;WHän Bnp äFbuAjJ&#xA;USUF iekK Jfv, xm iVgxaV cnkuZä,&#xA;bTFn jKJOMk pUB wäsE aOßjinä.&quot;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1661" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2015-07-10T13:38:42.837" Title="What does Europe have in common with Timon and Pumbaa?" Tags="|cipher|" AnswerCount="1" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="17155" PostTypeId="1" AcceptedAnswerId="17291" CreationDate="2015-06-29T20:18:55.817" Score="12" ViewCount="1188" Body="&lt;p&gt;Search, searching, still searching am I,&lt;br&gt;&#xA;For those that on that one day did die.&lt;br&gt;&#xA;We don't even know the name of the place,&lt;br&gt;&#xA;I'm going there, even if you can't keep pace.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the world out there, some things are pure,&lt;br&gt;&#xA;No flowers or vines, but named for sure.&lt;br&gt;&#xA;I took it apart, and saved so many lives,&lt;br&gt;&#xA;In the fight that left so many husbandless wives.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most really like me, except when I was given.&lt;br&gt;&#xA;Still trying to decide if I've been forgiven.&lt;br&gt;&#xA;There was a close call, but it is alright,&lt;br&gt;&#xA;Though for a while, I was quite the sight.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For some direction on how the clues might be organized, see some of my other riddles, such as &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15731/you-may-know-me-who-am-i&quot;&gt;this&lt;/a&gt;, &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16686/alive-dead-alive-now-dead&quot;&gt;this&lt;/a&gt;, or &lt;a href=&quot;https://puzzling.stackexchange.com/questions/12543/come-sail-away-who-am-i&quot;&gt;this&lt;/a&gt;.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each stanza deals with something different.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Who&quot; may imply a singular that isn't singular.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-07-02T11:45:51.243" Title="Still going strong - Who am I?" Tags="|riddle|rhyme|" AnswerCount="5" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="17156" PostTypeId="1" AcceptedAnswerId="17189" CreationDate="2015-06-29T20:46:30.753" Score="22" ViewCount="3509" Body="&lt;p&gt;This question is inspired by Terry Pratchett's &quot;Small Gods,&quot; in which an army crosses a vast desert by making multiple trips and caching water along the way.&#xA;&lt;br&gt;1. Provide an answer.&#xA;&lt;br&gt;2. I doubt I'm the first person to come up with this puzzle. Does anyone know of a &quot;classic&quot; puzzle like this, or know of a generic term for this puzzle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem:&lt;br&gt;&#xA;You are trying to cross a desert alone without dying of thirst. You know that for each mile you walk, you will need to drink exactly one gallon of water. You can carry at most 10 gallons of water on your person. The desert is 20 miles across. You obviously cannot cross the desert in one trip, but you can at any time leave a cache of water behind in the desert to pick up on a subsequent trip. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How can you cross the desert without dying of thirst?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Measured by distance walked, what is &lt;strong&gt;the optimal way&lt;/strong&gt; to cross the desert?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assume: You have an infinite supply of water/containers on the starting side. You can cache water in the desert without it evaporating. You can cache any fraction of the water you are carrying at any time. You will die instantly if you run out of water in the desert.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus points for proof that your method is the best, or for a general solution where the desert length is a factor of &lt;code&gt;a&lt;/code&gt; times longer than your one-time walking distance. (In the phrasing above, &lt;code&gt;a&lt;/code&gt;=2)&lt;/p&gt;&#xA;" OwnerUserId="13350" LastEditorUserId="18514" LastEditDate="2019-06-24T08:43:19.700" LastActivityDate="2019-06-24T08:43:19.700" Title="How much water do you need to cross the desert?" Tags="|mathematics|geometry|optimization|" AnswerCount="7" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="17158" PostTypeId="1" AcceptedAnswerId="17195" CreationDate="2015-06-29T20:56:05.710" Score="12" ViewCount="916" Body="&lt;p&gt;&lt;em&gt;If you're interested in starting the treasure hunt from the very beginning, check out the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16551/treasure-hunt-round-the-world&quot;&gt;prologue&lt;/a&gt;!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Upon arrival at the Pamplona Airport, you head towards baggage claim. Standing at baggage claim, you see a man in a suit and sunglasses. One hand is behind his back; in the other, he holds a sign that reads &quot;&lt;strong&gt;&lt;em&gt;[insert username here]&lt;/em&gt;&lt;/strong&gt;&quot;. Walking up to him, you say:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Hi. My name is &lt;strong&gt;&lt;em&gt;[insert username here]&lt;/em&gt;&lt;/strong&gt;. Do you have something for me?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without hesitation, he stretches the arm from behind his back out towards you. In it he holds a small envelope. You take the envelope and eagerly tear it open, pulling another small card out from inside. The card reads:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dear &lt;strong&gt;&lt;em&gt;[insert username here],&lt;/em&gt;&lt;/strong&gt; &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Congratulations on your arrival in Pamplona! I'd suggest you stay and enjoy the city (it is a beautiful city), but there's no time for that - you're running a race! Grab your bag, rent a car, and hit the road! This time, you're driving to your next location, which is (more or less) detailed on the back of this letter.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Yours truly,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Bailey M&lt;br&gt;&#xA;  Puzzlemaster for the Treasure Hunt 'Round the World&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You flip over the letter and read:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Lisbon, Portugal:&lt;/strong&gt; 7.3498737047383369&lt;br&gt;&#xA;  &lt;strong&gt;Knockainey, Ireland:&lt;/strong&gt; 7.2723983925700463&lt;br&gt;&#xA;  &lt;strong&gt;Boudekak, Algeria:&lt;/strong&gt; 6.8394764382288426&lt;/p&gt;&#xA;  &#xA;  &lt;hr&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;As you circle in on the answer, make careful note of your location. At some point down the line, you'll be left with two. Don't look too closely here, as your first instinct will most likely be correct. Once you've centered in on the solution, the final destination (much like the above numbers) should feel natural.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As you finish reading, you notice your suitcase drop onto the conveyor belt. Time to grab that bag and hit the road!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Where is your next destination?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The story continues in the next part, &lt;a href=&quot;https://puzzling.stackexchange.com/questions/17247/treasure-hunt-round-the-world-interlude-1&quot;&gt;Treasure hunt &amp;#39;round the world! (Interlude 1)&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.923" LastActivityDate="2015-07-01T16:50:14.587" Title="Treasure hunt 'round the world! (clue 2)" Tags="|enigmatic-puzzle|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17159" PostTypeId="2" ParentId="17106" CreationDate="2015-06-29T21:23:40.563" Score="9" Body="&lt;p&gt;Another way to look at it is to project the base of each tetrahedron onto the sphere.&#xA;&lt;a href=&quot;http://mathworld.wolfram.com/SphericalTriangle.html&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RsPuj.gif&quot; alt=&quot;spherical triangle&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The length of each segment is $\pi/3$. By using the &lt;a href=&quot;http://mathworld.wolfram.com/SphericalTrigonometry.html&quot; rel=&quot;noreferrer&quot;&gt;law of cosines&lt;/a&gt;, each angle is $arcsec(3)$ and the area of one of the spherical triangles is $3\ arcsec(3)-\pi$. The area of the sphere is $4\pi$. Therefore, the max number of spherical triangles we can theoretically fit on the surface of the sphere is $\frac{4\pi}{3\ arcsec(3)-\pi}\approx22.7947$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doesn't improve the upper bound at all but it takes it from packing tetrahedrons in 3D to packing triangles on the surface of a sphere.&lt;/p&gt;&#xA;" OwnerUserId="11499" LastActivityDate="2015-06-29T21:23:40.563" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17160" PostTypeId="2" ParentId="16881" CreationDate="2015-06-30T00:37:15.637" Score="30" Body="&lt;h3&gt;Partial answer&lt;/h3&gt;&#xA;&lt;p&gt;I mentioned in the comments above that I'd made some partial progress on this. I haven't had much time to take it any further, so I thought I'd post my work so far in case it helps others...&lt;/p&gt;&#xA;&lt;p&gt;Firstly, I noticed that:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The butterflies are all unique, and that, more importantly, they follow a evolutionary progression. Starting with the completely plain butterfly in the centre and then making a single change each generation, you can map out a family tree of sorts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Following this logic, I produced:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; A labelled copy of the original image.&lt;br /&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/UkFho.png&quot; alt=&quot;generations labelled&quot; /&gt;&lt;br /&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/1FAps.jpg&quot; rel=&quot;noreferrer&quot;&gt;full size (1.5MB)&lt;/a&gt;&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; Numbers represent the generation (starting with 0), letters show where the species diverged into distinct parallel descendent branches (i.e. A18 &amp;amp; B18 both evolved from 17, and C43 &amp;amp; D43 both evolved from C42).&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; For reference, the leaf nodes (as shown in circles above), are:&#xA; &lt;img src=&quot;https://i.stack.imgur.com/o0XpC.png&quot; alt=&quot;leaf nodes&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So visualising that result another way we get:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The family tree.&lt;br /&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/ETdPL.png&quot; alt=&quot;family tree&quot; /&gt;&lt;br /&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/5nQQv.jpg&quot; rel=&quot;noreferrer&quot;&gt;full size (1.4MB)&lt;/a&gt;&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; The red numbers down the left are the generation, with the columns matching up with the various branches. The blue numbers are the total count of individual butterflies up to that point.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;That's about all the concrete output I have, but I have a few ideas of what it could mean...&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Theories and observations&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The path created by tracing the family tree through the original image seems pretty random, so I don't think it's forming shapes or anything of that sort. Therefore, I assume the answer must be derived from the features of the butterflies themselves.&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; We are told the answer is a five word phrase, so I'm guessing it will be something along the lines of &amp;quot;&lt;em&gt;Out of chaos comes order&lt;/em&gt;&amp;quot;, or &amp;quot;&lt;em&gt;Stop and smell the roses&lt;/em&gt;&amp;quot;, or &amp;quot;&lt;em&gt;Life, uh... finds a way&lt;/em&gt;&amp;quot;. There also happens to be exactly five branches to the family tree, so I'm guessing each branch (or each leaf node) represents a separate word.&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; The hint mentions counting, so I imagine we can find letters for each word by counting either changes in features, or the number of generations each feature survives for, or something similar, along each branch. That being said, no feature lasts particularly long in any given family branch, so there may be more to it (or the letters would all necessarily be early in the alphabet).&#xA;&lt;br /&gt;&#xA; The tags on the puzzle include &amp;quot;cryptogram&amp;quot;, so it's plausible that there's some decryption necessary beyond just mapping A=1, B=2... Baconian encryption came to mind due to the binary nature of changes, but given that each generation changes by exactly one bit, it seems unlikely.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-06-30T00:37:15.637" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="17162" PostTypeId="2" ParentId="17117" CreationDate="2015-06-30T03:37:11.140" Score="0" Body="&lt;p&gt;While CodeNewbie has answered it correctly he can do even better--by making each group 5 he can do it with 20,000 supporters.  By itself this conveys no advantage but since he has plenty of extra supporters he doesn't need to settle for 3 of 5 in the final round--he can make this 5 of 5 and win by a unanimous vote!  Making each round the same size also makes it look fairer.&lt;/p&gt;&#xA;" OwnerUserId="2767" LastActivityDate="2015-06-30T03:37:11.140" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17163" PostTypeId="2" ParentId="16348" CreationDate="2015-06-30T05:39:41.417" Score="1" Body="&lt;p&gt;Just a few thoughts&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And a brighter vista it may indeed come to be.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Windows Vista? released 2007    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Farther removed from Mussolini's art.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Novecento Italiano was an Italian artistic movement founded in Milan to create an art based on the rhetoric of the Fascism of Mussolini.&lt;br&gt;&#xA;  It translates to 1900  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Celebrated most for famous work, derived from common sense.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Common Sense&lt;/em&gt; is a pamphlet by Thomas Paine during the American Revolution&#xA;  in 1776  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Belong to him, the mysterious Alexandrian.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Clement of Alexandria was a Christian theologian who taught at the Catechetical School of Alexandria the &lt;em&gt;mysteries&lt;/em&gt;. His famous work is known as the &lt;em&gt;Trilogy&lt;/em&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13648" LastActivityDate="2015-06-30T05:39:41.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17165" PostTypeId="2" ParentId="17155" CreationDate="2015-06-30T07:02:18.933" Score="2" Body="&lt;p&gt;Am sure this isn't the answer but a try nonetheless:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; is it a MINE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Search, searching, still searching am I,&lt;br&gt;&#xA;For those that on that one day did die.&lt;br&gt;&#xA;We don't even know the name of the place,&lt;br&gt;&#xA;I'm going there, even if you can't keep pace.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a mine is hidden (place name not known) and it searches for the person who it can kill &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I took it apart, and saved so many lives,&lt;br&gt;&#xA;In the fight that left so many husbandless wives. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mines stop the enemy and save lives on one side but also leave dead on another side&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the most sense i could derive. Not sure about the last stanza&lt;/p&gt;&#xA;" OwnerUserId="13080" LastActivityDate="2015-06-30T07:02:18.933" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17167" PostTypeId="1" AcceptedAnswerId="17168" CreationDate="2015-06-30T07:12:28.797" Score="22" ViewCount="4761" Body="&#xA;&lt;p&gt;This is part of &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=title%3A%22The+Twenty+Doors%22&quot;&gt;The Twenty Doors&lt;/a&gt; series.&#xA;&lt;br&gt;The next one is &lt;a href=&quot;https://puzzling.stackexchange.com/questions/17198/the-twenty-doors-room-2&quot;&gt;The Twenty Doors! (ROOM 2)&lt;/a&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;You stand in a small, dark room. You have no idea where you are. Then, you see a message in front of you.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Welcome to the dungeon!&lt;/p&gt;&#xA;&lt;p&gt;In each room, there is some paper, a hint, a door and a keypad. You must decipher the paper, using the hint, and enter what you think the answer is into the keypad. But don't answer wrong, or you will DIE!!!&lt;/p&gt;&#xA;&lt;p&gt;Have fun, if you can...&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Dr. X&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;ROOM 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;You look at the paper:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;uipefvyq&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Then, you see that the hint is carved into the wall: &lt;code&gt;Hello, over and over&lt;/code&gt;.&#xA;The keypad has 9 numbers and an [ENTER] button.&lt;/p&gt;&#xA;" OwnerDisplayName="user9377" LastEditorUserId="-1" LastEditorDisplayName="user9377" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-30T19:31:44.870" Title="The Twenty Doors! (ROOM 1)" Tags="|cipher|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17168" PostTypeId="2" ParentId="17167" CreationDate="2015-06-30T07:42:22.617" Score="37" Body="&lt;p&gt;Room 1:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The code is 14, if we decipher uipefvyq with the Vigenere cipher 'hello', we get neetruof, which is fourteen backwards.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I hope I won't DIE...&lt;/p&gt;&#xA;" OwnerUserId="13268" LastActivityDate="2015-06-30T07:42:22.617" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17170" PostTypeId="2" ParentId="17156" CreationDate="2015-06-30T08:24:08.057" Score="7" Body="&lt;p&gt;Since keeping fractional amounts of water is possible, here's my solution (not sure if it's the most efficient). Here I have assumed that one trip constitutes movement in any one direction, forward or backward.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To be able to successfully complete this trip, we need to reach a situation where we have 10 gallons at the 10 mile mark.&lt;br&gt;&lt;br&gt;Start with 10 gallons and carry it till 3.33 miles. 3.33 gallons are used up, cache 3.33 gallons and use the remaining 3.33 gallons to return to your origin. Repeat this 12 more times, till 43.33 gallons are collected at the 3.33 mile mark and you are back at the infinite well. (26 trips done so far)&lt;br&gt;&lt;br&gt;In the 27th trip, when you reach the 3.33 mile mark, you have a total of 50 gallons now: 43.33 collected plus 6.66 carried. Refill your container to the full 10 gallons and move to the 6.66 mile mark. 3.33 gallons are used up, cache 3.33 gallons and use the remaining 3.33 gallons to return to the 3.33 mile mark. Repeat this 3 more times till 13.33 gallons are collected at the 6.66 mile mark and you are back at the 3.33 mile mark. (32 trips done so far, 40 gallons used up)&lt;br&gt;&lt;br&gt;In the 35th trip, when you reach the 6.66 mile mark, you have a total of 20 gallons now: 13.33 collected plus 6.66 carried. Refill your container to the full 10 gallons and move to the 10 mile mark. 3.33 gallons are used up, cache 3.33 gallons and use the remaining 3.33 gallons to return to the 6.66 mile mark. Return, refill with the remaining 10 gallons, walk to the 10 mile mark, add the 3.33 gallons collected to your remaining 6.66 gallons, and use the 10 gallons to cross the desert. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To conclude,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the total number of trips I need is 37.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="1748" LastEditDate="2015-06-30T09:21:39.977" LastActivityDate="2015-06-30T09:21:39.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17171" PostTypeId="2" ParentId="17156" CreationDate="2015-06-30T08:37:13.830" Score="4" Body="&lt;p&gt;This can be optimized if we allow the water caches to be continuous. My answer requires 27 trips (one way):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This way we dont need 10 gallons at the 10 mile mark, but rather 10 gallons uniformly distributed over the first 10 miles (1 gallon / mile). Filling the first 3.33 miles requires 1 trip (both ways). Filling the next 3.33 miles requires 3 trips in total. Filling the final 3.33 miles (from 6.66 to 10) requires 9 trips. So filling the first 10 miles requires 9+3+1 = 13 trips (both ways). Together with the last trip it makes a total of &lt;strong&gt;27&lt;/strong&gt; (one way) trips.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Visual representation of &quot;linear water density&quot; (gallons per mile) for each trip (each number represents 1/3 of a mile):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Miles __1__2__3__4__5__6__7__8__9__10&#xA;#1    1111111111&#xA;#2    2222222222&#xA;#3              1111111111&#xA;#4    11111111111111111111&#xA;#5    22222222221111111111 &#xA;#6              2222222222&#xA;#7    11111111112222222222     &#xA;#8    22222222222222222222&#xA;#9                        1111111111&#xA;#10   1111111111          1111111111&#xA;#11   2222222222          1111111111&#xA;#12             11111111111111111111&#xA;#13   111111111111111111111111111111&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="13523" LastEditorUserId="13523" LastEditDate="2015-06-30T14:25:42.560" LastActivityDate="2015-06-30T14:25:42.560" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17172" PostTypeId="1" AcceptedAnswerId="17173" CreationDate="2015-06-30T08:56:09.840" Score="52" ViewCount="10555" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Th8tt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a goat made up of 5 sticks. You have to move (change position) any one stick of them such that its head turns to the right side (above the right leg).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice currently its head is on the left side and above the left leg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/20156/turning-a-dog-part-2&quot;&gt;Part 2- Turning a Dog&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/57144/turning-the-horse-part-3&quot;&gt;Part 3- Turning a Horse&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13652" LastEditorUserId="5373" LastEditDate="2017-12-29T15:32:25.870" LastActivityDate="2017-12-29T15:32:25.870" Title="Turning a goat?" Tags="|logical-deduction|geometry|matches|" AnswerCount="11" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="17173" PostTypeId="2" ParentId="17172" CreationDate="2015-06-30T09:02:13.630" Score="87" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Move the right most stick and place it on the bottom. This way, the goat will appear to face right when viewed with the head tilted 90 degrees to the left.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's the image equivalent:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/dC2HB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Apologies for the poor quality image. My office laptop has very limited capabilities in image editing.&lt;/p&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-06-30T09:02:13.630" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17174" PostTypeId="2" ParentId="17121" CreationDate="2015-06-30T09:12:52.630" Score="20" Body="&lt;p&gt;I think I have at least a partial answer. The letters are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Greek letters, or rather the Greek letters corresponding to the given Latin ones.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Their relationship with the words is that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; their names rhyme with synonyms of the given expressions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    &lt;strong&gt;Alpha&lt;/strong&gt; rhymes with &lt;strong&gt;alfalfa&lt;/strong&gt; (the plant lucerne)&lt;br&gt;&#xA;    &lt;strong&gt;Kappa&lt;/strong&gt; rhymes with &lt;strong&gt;rapper&lt;/strong&gt; (Jay Z)&lt;br&gt;&#xA;    &lt;strong&gt;Mu&lt;/strong&gt; and &lt;strong&gt;Nu&lt;/strong&gt; rhyme with &lt;strong&gt;view&lt;/strong&gt; (panorama)&lt;br&gt;&#xA;    &lt;strong&gt;Rho&lt;/strong&gt; rhymes with &lt;strong&gt;woe&lt;/strong&gt; and &lt;strong&gt;foe&lt;/strong&gt; (misery and enemy)&lt;br&gt;&#xA;    &lt;strong&gt;Sigma&lt;/strong&gt; rhymes with &lt;strong&gt;enigma&lt;/strong&gt; (riddler)&lt;br&gt;&#xA;    &lt;strong&gt;Zeta&lt;/strong&gt; (&lt;em&gt;ZEE-tah&lt;/em&gt;) rhymes with &lt;strong&gt;heater&lt;/strong&gt; (radiator)&lt;br&gt;&#xA;    &lt;strong&gt;Zeta&lt;/strong&gt; (&lt;em&gt;ZAY-tah&lt;/em&gt;) rhymes with &lt;strong&gt;data&lt;/strong&gt; (statistics)    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The difference between the two names of Zed and Zee reflects&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the different pronunciation of &quot;Zeta&quot; in Britain and in America.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A shot at a solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I haven't found any relation between the letters and the prefixes to the rhymes or any overarching theme to the words. In my opinion, a letter can stand for many things, as in R above, and several letters can stand for one thing, as with M and N. (There are many letters rhyming with Zeta in the Greek alphabet.) So I pick one, maybe:&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;D is for Sanctuary&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But there may be well more to it that I have missed, so feel free to add and edit.&lt;/p&gt;&#xA;" OwnerUserId="9860" LastEditorUserId="5373" LastEditDate="2015-07-01T11:35:07.860" LastActivityDate="2015-07-01T11:35:07.860" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="17175" PostTypeId="2" ParentId="17172" CreationDate="2015-06-30T09:38:37.180" Score="23" Body="&lt;p&gt;A try:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/iMZQD.png&quot; alt=&quot;&quot;&gt;&#xA;&#xA; &lt;br&gt;the head is turned to the right&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13080" LastEditorUserId="1777" LastEditDate="2015-06-30T10:06:13.343" LastActivityDate="2015-06-30T10:06:13.343" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="17176" PostTypeId="2" ParentId="17172" CreationDate="2015-06-30T11:19:35.383" Score="27" Body="&lt;p&gt;Move the right leg up, so it seems like it was jumping down:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/bLEF3.png&quot; alt=&quot;enter image description here&quot; /&gt;&#xA; It's head is now on the right side, and what was its head is now a tail.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10880" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-06-30T11:19:35.383" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17177" PostTypeId="2" ParentId="17156" CreationDate="2015-06-30T11:58:10.640" Score="8" Body="&lt;p&gt;I think I found a faster way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide the trip in 2'5 miles segments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @CodeNewbie said, we need 10 gallons on the 10 miles segment. Let's see how to do that.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0   2'5  5   7'5  10&#xA;---------------------&#xA;x    5&#xA;x    0   5&#xA;x    5   5&#xA;x    0   0   5&#xA;x    5   0   5&#xA;x    0   5   5&#xA;x    5   5   5&#xA;x    0   0   0   5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Each row represents a trip. On arriving to a segment, we either fill to 10 gallons, or leave 5 gallons.&#xA;We have to repeat the process, and finally do the last trip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means that in 17 travels we'll be able to end up in the other side of the desert.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under the same definition of travel that @CodeNewbie used, we'd need 33 travels, so it's still a better solution (assuming I didn't mess up)&lt;/p&gt;&#xA;" OwnerUserId="12165" LastActivityDate="2015-06-30T11:58:10.640" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17178" PostTypeId="2" ParentId="17155" CreationDate="2015-06-30T12:08:32.727" Score="3" Body="&lt;p&gt;I guess it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; COMET&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Search, searching, still searching am I,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some Comets keep going forever (in an orbit) unknown to us what they are after.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For those that on that one day did die.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Looking for stars that perished or are blown away from wherein they are born&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We don't even know the name of the place,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not every place in the Universe has been named yet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm going there, even if you can't keep pace.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Humans on earth have not yet figured out how to outpace a Comet &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the world out there, some things are pure,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pure like a single element&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No flowers or vines, but named for sure.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some comets have names like Halley or Hyakutake&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I took it apart, and saved so many lives,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Comets could have crashed into Asteroids implying they saved lives on Spaceships which could have run into those.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the fight that left so many husbandless wives.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some war during planet sightings coz in earlier days wars were fought with mostly men, unlike online cyber wars of today with no men ;-).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Most really like me, except when I was given.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is a play on the word COMETH opposite of GIVETH.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Still trying to decide if I've been forgiven.&#xA;There was a close call, but it is alright,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though for a while, I was quite the sight.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Comets look different than other visible planetary bodies in the sky and are quite a sight.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9892" LastActivityDate="2015-06-30T12:08:32.727" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17179" PostTypeId="2" ParentId="17172" CreationDate="2015-06-30T13:30:14.880" Score="6" Body="&lt;p&gt;My interpretation would be to pivot at the neck/back joint. Technically only one line has been moved because the back leg (now the front) is attached and simply follows the back.  It meets the requirement of being &quot;above the right leg&quot; although I suspect OP also wants the head to be to the right of the right leg?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/beSxI.jpg&quot; alt=&quot;My interpretation&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="13668" LastActivityDate="2015-06-30T13:30:14.880" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17180" PostTypeId="2" ParentId="17167" CreationDate="2015-06-30T13:38:08.840" Score="27" Body="&lt;p&gt;Credit to &lt;a href=&quot;https://puzzling.stackexchange.com/a/17168/9000&quot;&gt;Cerberus&lt;/a&gt; for decoding the message. However, I think the conclusion in that answer is wrong. I think it should be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 41&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it's 14 backwards&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13670" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2015-06-30T14:32:33.657" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17181" PostTypeId="1" AcceptedAnswerId="17187" CreationDate="2015-06-30T13:44:18.150" Score="6" ViewCount="653" Body="&lt;p&gt;A geek friend of mine said that this was what he was listening to and I'm apparently supposed to figure it out:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(?ix)(?![AD-ZC][a-jn-zjm])&#xA;(&#xA;  ([[:upper:]])&#xA;  [[:lower:]]&#xA;)&#xA;(?=[a-z])&#xA;(?!b-z)&#xA;(.)&#xA;ck&#xA;(&#xA;  (?!\t\r\n\f)&#xA;  \s&#xA;)&#xA;(&#xA;  (?=[A-Z])&#xA;  (?![A-R])&#xA;  (?![T-Z])&#xA;  .\3\2\2\3t&#xA;  (&#xA;    (h)&#xA;  )&#xA;)&#xA;\4\-\4\5\4\1&#xA;(o)&#xA;(\8)&#xA;d&#xA;(?![xa-wz])&#xA;[[:alpha:]]\4\5*&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Could someone decipher this and tell me what it means? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="13672" LastEditorUserId="10399" LastEditDate="2015-06-30T13:54:37.493" LastActivityDate="2015-07-01T09:48:02.097" Title="What does this say?" Tags="|word|computer-puzzle|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17182" PostTypeId="2" ParentId="2693" CreationDate="2015-06-30T14:10:43.723" Score="1" Body="&lt;p&gt;Stretch your 2D baby snake out so that it forms a straight line from nose to tale. Find the 2D silhouette that the baby now forms. Create a 2D blanket having the same shape as the silhouette. Spread glue on the blanket and attach it to the baby snake. Mama's 2D baby snake will always be covered in bed once the glue dries.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The blanket will have minimal size.&lt;/li&gt;&#xA;&lt;li&gt;The 2D blanket will twist and stretch and turn with the 2D snake it is attached to.&lt;/li&gt;&#xA;&lt;li&gt;The size of the blanket will be the area of the baby's silhouette.&lt;/li&gt;&#xA;&lt;li&gt;The form or shape of the blanket will always conform to the baby's.&lt;/li&gt;&#xA;&lt;li&gt;The length of the blanket will obviously be 1.&lt;/li&gt;&#xA;&lt;li&gt;The 2D blanket will always take the same shape as the baby.&lt;/li&gt;&#xA;&lt;li&gt;The blanket will always cover the snake since they are glued together.&lt;/li&gt;&#xA;&lt;li&gt;Mama will not have to worry about the baby being covered since both are deformable.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1661" LastEditorUserId="1661" LastEditDate="2015-07-01T16:58:28.530" LastActivityDate="2015-07-01T16:58:28.530" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17185" PostTypeId="2" ParentId="17181" CreationDate="2015-06-30T14:34:38.183" Score="2" Body="&lt;p&gt;Partial Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Interpret the code as a regular expression.&#xA;&lt;br&gt;(?=...)  Matches if ... matches next, but doesn't consume the string.&#xA;&lt;br&gt;(?!...)  Matches if ... doesn't match next.&#xA;&lt;br&gt; [] Indicates a set of characters&#xA;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(?ix)(?![AD-ZC][a-jn-zjm]) #Matches 0 or 1 'ix' [rest is unknown]&#xA;(&#xA;  ([[:upper:]])&#xA;  [[:lower:]]&#xA;)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I don't know what this would give.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(?=[a-z]) #Which means, any lowercase letter&#xA;(?!b-z)   #That isn't between b to z [so 'a']&#xA;(.)       #Match any character&#xA;ck        #And we get are then given 'ck'&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; giving us 'a_ck' or 'ack'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(?=[A-Z])   #Any Capital letter&#xA;(?![A-R])   #That isn't between A to R&#xA;(?![T-Z])   #That isn't between T to Z [So 'S']&#xA;.\3\2\2\3t  #Any character with multiple tabs&#xA;(h)         #Then we are given h&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Which gives us now 'ack'+'Sh'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;\4\-\4\5\4\1      #Don't know what this would mean&#xA;(o)               #Matches 'o'&#xA;(\8)              #Matches an escaped 8, [unknown]&#xA;d                 #Given 'd'&#xA;(?![xa-wz])       #[unknown]&#xA;[[:alpha:]]\4\5*  #[unkown]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Giving us now ('a_ck' or 'ack')+'SH'+'od'&#xA;&lt;br&gt;&lt;br&gt;My best &lt;strong&gt;guess&lt;/strong&gt; would be &lt;b&gt;Black Sabbath&lt;/b&gt; [from the letters we found]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-06-30T15:12:04.350" LastActivityDate="2015-06-30T15:12:04.350" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17187" PostTypeId="2" ParentId="17181" CreationDate="2015-06-30T14:50:50.867" Score="4" Body="&lt;p&gt;He is listening to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Black Sabbath - Sabbath Bloody Sabbath&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I find it hard to explain every step. Basically I used &lt;a href=&quot;https://regex101.com/r/yM1qV5/1&quot; rel=&quot;nofollow&quot;&gt;https://regex101.com/r/yM1qV5/1&lt;/a&gt; and looked at the explanation and started filling it in. If you also paste the solution in there as the TEST STRING you see exactly what each capture group is and you can try to understand it. If you're not familiar with regex and are not a 'geek' this might be very hard to understand.&lt;/p&gt;&#xA;" OwnerUserId="1748" LastEditorUserId="1748" LastEditDate="2015-06-30T15:04:48.440" LastActivityDate="2015-06-30T15:04:48.440" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17188" PostTypeId="2" ParentId="8748" CreationDate="2015-06-30T14:57:02.433" Score="0" Body="&lt;p&gt;Here are some more to contribute to the holes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;everywhither&lt;/li&gt;&#xA;&lt;li&gt;heavy-handed&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9000" LastActivityDate="2015-06-30T14:57:02.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17189" PostTypeId="2" ParentId="17156" CreationDate="2015-06-30T15:58:49.450" Score="12" Body="&lt;p&gt;Considering the general case, I attempt to answer the companion question: if you have $10 N$ gallons of water, how far can you get into the desert?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, for $N=1$, the answer is 10 miles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $N=2$, you carry the water distance $x$, drop off $(10-2x)$ gallons, and go back.  Then you pick up the rest, go forward, refill at the cache, and continue on.  This is obviously optimal$^1$ if the amount of water left to pick up is exactly the amount you used to get where you are.  In other words, $x = 10/3$.  At this point, you refill and continue, for a total of $10 (1 + \frac 13)$ miles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $N*10$ gallons, you will make $N-1$ round trips and one one-way trip.  Thus, the water you use to move $x$ miles is $(2N-1)x$.  This should be an even multiple of 10 gallons, or you are wasting water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $N=3$, you can pick either $x = \frac 15$ or $x = \frac 25$.  In the first case, you cache 20 gallons of water 2 miles in.  In the second, you cache 10 gallons of water 4 miles in.  In the first case, you can make it $10 (1 + \frac 13 + \frac 15)$ miles, and in the second, you travel $10(1 + \frac 25)$ miles.  The first option is clearly slightly better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, for any $N$, you will want to make the next cache at $x = 1/(2N+1)$ miles.  That uses up 10 gallons and you then continue with the next smaller $N$, which is more efficient.  Making the caches farther in means you are using up your water on less efficient fractions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, with $10N$ gallons, you can travel as far as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$10 \sum_{i=0}^{N-1} \frac 1 {2i+1} \text{miles.}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The smallest $N$ that makes it to 20 miles is $N=8$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 + \frac 13 + \frac 15 + \frac 17 + \frac 19 + \frac 1{11} + \frac 1{13} + \frac 1{15} = 2.02$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your total distance traveled is $79.8$ miles, and you have $0.2$ gallons left at the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This plan extends to any distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^1$If you don't accept the &quot;obviously optimal&quot; comment above:&#xA;If there are $10 N + \epsilon$ gallons at a cache point when you are ready to move forward, then you can make $N$ trips from there, and the $\epsilon$ gallons will be wasted.  If there are $10 N - \epsilon$ gallons there, then when you go forward, you will be slightly short on your next leg.  Making the cache slightly closer would cost you $\epsilon / (2N+3)$ distance on that leg, but gain you $\epsilon / (2N+1)$ on the next leg, for a net win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;  I didn't count &quot;trips&quot; the same way everyone else does.  Naively, this would take 64 trips, but if you optimized, each trip can go one cache further than the previous.  I.e., &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Trip 1, drop $10*{13\over 15}$ gallons at a distance $10*{1\over 15}$ miles.&lt;/li&gt;&#xA;&lt;li&gt;Trip 2, refill at the first cache as you pass it, then drop $10*{11\over 13}$ gallons at distance $10*{1\over 13}$ miles past first cache.  When you pass the first cache on the way back, grab $10*{1\over 15}$ gallons needed to make it back to camp.&lt;/li&gt;&#xA;&lt;li&gt;Trip 3, refill at first two caches as you pass them, drop $10*{9\over 11}$ gallons at $10*{1\over 11}$ miles past second cache.  Grab $10*{1\over13}$ and $10*{1\over 15}$ gallons at the second and first caches, respectively.&lt;/li&gt;&#xA;&lt;li&gt;Etc.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thus, you can get all the way across in $15$ trips, assuming the trip-boundary is only when you change direction, not when you refill.  If refilling counts as a trip-boundary, then this method will not yield the optimal number of trips.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update 2&lt;/strong&gt;&#xA;Rereading the question, I see what I should have been answering is &quot;&lt;em&gt;Measured by distance walked, what is the optimal way to cross the desert?&lt;/em&gt;&quot;  This solution definitely optimizes that, for a total of $79.8$ miles traveled.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="2328" LastEditDate="2015-06-30T17:16:35.313" LastActivityDate="2015-06-30T17:16:35.313" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17190" PostTypeId="2" ParentId="17008" CreationDate="2015-06-30T16:06:44.560" Score="3" Body="&lt;p&gt;I'm still struggling to see this, but I want to get some momentum into this puzzle, so as &lt;em&gt;work in progress&lt;/em&gt; I'm summarizing some ideas.&lt;/p&gt;&#xA;&lt;p&gt;First the facts (approved by the OP.):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The odd-one-out owls are &lt;code&gt;A2&lt;/code&gt; and &lt;code&gt;D7&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;Only 30 eye-combos are &lt;em&gt;valid&lt;/em&gt; (whatever valid means)&lt;/li&gt;&#xA;&lt;li&gt;The pattern does not rely on the fact, that it's a 9 x 9 grid or a total of 81 owls.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Then some observations:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There are only 18 distinct combos visible&lt;/li&gt;&#xA;&lt;li&gt;There isn't a single double-combo with both eyes looking the same direction&lt;/li&gt;&#xA;&lt;li&gt;Occurring combos appear between 2 and 9 times and the histogram is not very distinctive.&lt;/li&gt;&#xA;&lt;li&gt;It is maybe easier to analyse the image by taking out the owls and representing &lt;em&gt;unique&lt;/em&gt; eye-pairs by a single colour.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/da8sP.png&quot; alt=&quot;Easier&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Some ideas to work on:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; - The fact that the &amp;quot;wrong&amp;quot; owls were replaced without altering the other pattern, or indeed the fact that the patterns &lt;em&gt;exists&lt;/em&gt; with some mistakes in it, lets one assume that the pattern does &lt;em&gt;not&lt;/em&gt; depend on the other owls, except the dependency is such, that both incorrect owls depend only on each other and no other depends on them. This rules out things like &amp;quot;left neighbour&amp;quot; or &amp;quot;sum of the row&amp;quot; etc.&lt;br&gt;&lt;br&gt;&#xA; - The fact that there &lt;strong&gt;are&lt;/strong&gt; invalid combos needs analysing. What logic reason can reduce &lt;code&gt;64 --&amp;gt; 30&lt;/code&gt; ? Some non-decimal number system maybe? &lt;br&gt;&#xA; - Is it a 2D or just a 1D pattern? If 1D, what is the &amp;quot;sequence&amp;quot; order? left-right or up-down?&lt;br&gt;&lt;br&gt;&#xA; - Are &lt;em&gt;all&lt;/em&gt; owls determined by a rule, or is there a &amp;quot;seeding&amp;quot; owl which defines the pattern. From the storyline, all owls should &amp;quot;independently&amp;quot; be able to match a pattern.&lt;br&gt;&lt;br&gt;&#xA; - It is intriguing that - if you just read the lines left-right and continue to the next line - you find some of the patterns repeat themselves. This is most noticeable in the blue-purple-pink sections. Sometime patterns repeat for longer than others, but I can not think why exactly they break differently. Obviously it has to do with &amp;quot;other owls&amp;quot;, but how?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-01T20:18:18.580" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="17191" PostTypeId="2" ParentId="17156" CreationDate="2015-06-30T16:48:14.840" Score="3" Body="&lt;p&gt;Expanding on @Noel's answer because this is too big for a comment.&lt;/p&gt;&#xA;&lt;p&gt;Why do you need to distribute the water uniformly?  The following is the same number of trips.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;1:&lt;/strong&gt; Drop &lt;span class=&quot;math-container&quot;&gt;$3\frac{1}{3}$&lt;/span&gt; gallons at the &lt;span class=&quot;math-container&quot;&gt;$3\frac{1}{3}$&lt;/span&gt; mile marker.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;2:&lt;/strong&gt; Repeat #1.  There are now &lt;span class=&quot;math-container&quot;&gt;$6\frac{2}{3}$&lt;/span&gt; gallons t the &lt;span class=&quot;math-container&quot;&gt;$3\frac{1}{3}$&lt;/span&gt; mile marker.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;3:&lt;/strong&gt; Drop &lt;span class=&quot;math-container&quot;&gt;$3\frac{1}{3}$&lt;/span&gt; gallons at the &lt;span class=&quot;math-container&quot;&gt;$6\frac{2}{3}$&lt;/span&gt; mile marker.  Refresh from the first cache on the way out and on the way back.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;4-6:&lt;/strong&gt; Repeat 1-3.  There is now &lt;span class=&quot;math-container&quot;&gt;$6\frac{2}{3}$&lt;/span&gt; gallons at the &lt;span class=&quot;math-container&quot;&gt;$6\frac{2}{3}$&lt;/span&gt; mile marker.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;7/8:&lt;/strong&gt; Repeat 1-2.  There is now &lt;span class=&quot;math-container&quot;&gt;$6\frac{2}{3}$&lt;/span&gt; gallons at both the &lt;span class=&quot;math-container&quot;&gt;$3\frac{1}{3}$&lt;/span&gt;  and &lt;span class=&quot;math-container&quot;&gt;$6\frac{2}{3}$&lt;/span&gt; mile markers.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;9:&lt;/strong&gt; Drop &lt;span class=&quot;math-container&quot;&gt;$3\frac{1}{3}$&lt;/span&gt; gallons at the &lt;span class=&quot;math-container&quot;&gt;$10$&lt;/span&gt; mile marker.  Refresh from the  caches on the way out and on the way back.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;10-12:&lt;/strong&gt; Repeat 1-3. There is now &lt;span class=&quot;math-container&quot;&gt;$3\frac{1}{3}$&lt;/span&gt; gallons at the &lt;span class=&quot;math-container&quot;&gt;$6\frac{2}{3}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$10$&lt;/span&gt; mile markers.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;13:&lt;/strong&gt; Repeat 1.  There is now &lt;span class=&quot;math-container&quot;&gt;$3\frac{1}{3}$&lt;/span&gt; gallons at the &lt;span class=&quot;math-container&quot;&gt;$3\frac{1}{3}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$6\frac{2}{3}$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$10$&lt;/span&gt; mile markers.&lt;/p&gt;&#xA;&lt;h1&gt;Final Tally&lt;/h1&gt;&#xA;&lt;p&gt;So, you've made 13 return trips (26-one way trips) to cache &lt;span class=&quot;math-container&quot;&gt;$3\frac{1}{3}$&lt;/span&gt; gallons at the &lt;span class=&quot;math-container&quot;&gt;$3\frac{1}{3}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$6\frac{2}{3}$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$10$&lt;/span&gt; mile markers.&lt;/p&gt;&#xA;&lt;p&gt;One more one-way trip can get you to the &lt;span class=&quot;math-container&quot;&gt;$10$&lt;/span&gt; mile mark with exactly &lt;span class=&quot;math-container&quot;&gt;$10$&lt;/span&gt; gallons - enough to take you across the desert.&lt;/p&gt;&#xA;&lt;p&gt;Notice that in terms of gallons of water (or equivalently, miles walked), this solution is equivalent to @Noel's solution, but also @CodeNewbie's.   All three use &lt;span class=&quot;math-container&quot;&gt;$140$&lt;/span&gt; gallons of water. The difference is that this solution and @Noel's optimize the number of trips.  @CodeNewbie walks the same distance, but makes more shorter trips.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-06-30T16:48:14.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17192" PostTypeId="2" ParentId="17167" CreationDate="2015-06-30T17:01:46.553" Score="5" Body="&lt;p&gt;Credit to both asdfffff and Cerberus, but I think that both of you were part of the way there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cerberus pointing out that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Decrypting uipefvyq gives you neetruof&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and asdfffff pointing out&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 41 would be fourteen backwards!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But you skipped the last part of the hint!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the answer would be: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;4141&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Becuase:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It is 41 over and over.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13684" LastActivityDate="2015-06-30T17:01:46.553" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17193" PostTypeId="2" ParentId="17158" CreationDate="2015-06-30T17:22:56.837" Score="10" Body="&lt;p&gt;Based on what I've found so far, I'd go with&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Boudekak, Algeria&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using the line, &quot;the final destination (much like the above numbers) should feel natural.&quot;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I decided to work with natural logarithms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Taking the natural log of each of the numbers given isn't terribly helpful so far as I can tell (yielding 1.994, 1.984, and 1.922 respectively). So, instead, I did the reverse and squared e by the numbers given. This gives 1554, 1438, and 933 for each number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And here's what makes it fit, for me:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Looking up the distance between the locations &quot;as the crow flies&quot; from Pamplona to the given locations gives:&lt;br&gt;&#xA; 1) 779 Kilometers for Lisbon, Portugal&lt;br&gt;&#xA; 2) 1186 for Knockainey, Ireland&lt;br&gt;&#xA; 3) And finally, 932 for Boudekak, Algeria&lt;br&gt;&#xA; Seeing as there was probably a bit of rounding that happened here, I think that difference of 1 is close enough to call that good.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10078" LastEditorUserId="10078" LastEditDate="2015-06-30T17:56:19.673" LastActivityDate="2015-06-30T17:56:19.673" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17194" PostTypeId="2" ParentId="17004" CreationDate="2015-06-30T17:29:08.560" Score="3" Body="&lt;h2&gt;Partial Answer&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Ok please excuse me if this is very informal for an answer, as this is my first. As mentioned, we are overthinking the gargoyle part. My contribution: I p ropose that the 'complexion' refers to a complexion of stone, grey. What 'curse' inflicts this complexion? Medusa's curse. Unscramble Medusa (remove the 'order of operations') to get 'amused', which is 'comical' as a 'remainder'. I could be very off, but trial and error right?&lt;/p&gt;&#xA;" OwnerUserId="13685" LastEditorUserId="11475" LastEditDate="2015-06-30T19:28:28.497" LastActivityDate="2015-06-30T19:28:28.497" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17195" PostTypeId="2" ParentId="17158" CreationDate="2015-06-30T18:39:10.963" Score="12" Body="&lt;p&gt;The final destination is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/kNtqu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Definitely throw PrincessTrevor a +1 for posting the logarithm portion first.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The portion of the letter that says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the final destination (much like the above numbers) should feel natural.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;implies that the numbers listed above are natural logarithms.  These numbers correspond to:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Lisbon, Portugal 1556&#xA;Knockainey, Ireland 1440&#xA;Boudekak, Algeria 934&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Some other portions of the letter:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As you circle in on the answer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;implies that we should be looking for circles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drawing circles on the map with the radius listed above and at the locations in Google maps gives this image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HcXWM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At some point down the line, you'll be left with two&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is referring to, Toledo, Spain and Turin, Italy because they lie at the intersection of two of the circles and is inside the third.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At some point down the line, you'll be left with two. Don't look too&#xA;  closely here, as your first instinct will most likely be correct. Once&#xA;  you've centered in on the solution, the final destination&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The final destination lies halfway on a line between Toledo and Turin.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Once you've centered in on the solution, the final destination (much&#xA;  like the above numbers) should feel natural.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The final destination:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Parc Natural de la Vall de Sorteny is pretty &quot;natural&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1777" LastEditorUserId="1777" LastEditDate="2015-06-30T20:18:28.037" LastActivityDate="2015-06-30T20:18:28.037" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="17196" PostTypeId="1" AcceptedAnswerId="18214" CreationDate="2015-06-30T18:49:33.027" Score="6" ViewCount="621" Body="&lt;p&gt;This is inspired from Luke's &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15241/two-lists-of-favorite-words&quot;&gt;Two lists of favorite words&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My aunt Sally keeps this list of her favorite words tacked on the refrigerator:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Angel&lt;/li&gt;&#xA;&lt;li&gt;Art&lt;/li&gt;&#xA;&lt;li&gt;Atlas  &lt;/li&gt;&#xA;&lt;li&gt;Bad&lt;/li&gt;&#xA;&lt;li&gt;Detail  &lt;/li&gt;&#xA;&lt;li&gt;Hall&lt;/li&gt;&#xA;&lt;li&gt;Hand  &lt;/li&gt;&#xA;&lt;li&gt;Mutter&lt;/li&gt;&#xA;&lt;li&gt;Rat&lt;/li&gt;&#xA;&lt;li&gt;Tiger&lt;/li&gt;&#xA;&lt;li&gt;Winter&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My uncle Joe has this list of his favorites:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Boot  &lt;/li&gt;&#xA;&lt;li&gt;Chance &lt;/li&gt;&#xA;&lt;li&gt;Frost&lt;/li&gt;&#xA;&lt;li&gt;Gift  &lt;/li&gt;&#xA;&lt;li&gt;Kind    &lt;/li&gt;&#xA;&lt;li&gt;Magnet  &lt;/li&gt;&#xA;&lt;li&gt;Orange  &lt;/li&gt;&#xA;&lt;li&gt;Ring  &lt;/li&gt;&#xA;&lt;li&gt;Sand  &lt;/li&gt;&#xA;&lt;li&gt;Stern&lt;/li&gt;&#xA;&lt;li&gt;Wolf &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Which one of them would like the word &lt;em&gt;Listen&lt;/em&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint #1: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Both my aunt and uncle had this word on their list:&#xA; &lt;hr&gt;&#xA; &lt;img src=&quot;https://chabadinfo.com/wp-content/uploads/2015/05/filepicker-jDPepkMLTcGWT2K88rAg_plane.jpg&quot; alt=&quot;a busy cat&quot;&gt;&#xA;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My aunt is British and my uncle is American.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My aunt and uncle wanted to seem up with the times: &lt;hr&gt;&#xA; My aunt added &quot;lol&quot; &lt;br&gt;&#xA; My uncle added &quot;brb&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #4&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Don't look at the forest, but the trees.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #5&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The two lists are mutually exclusive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #6&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I excluded the English tag very much on purpose&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note: No computer science knowledge is necessary to solve this, but thinking like one is helpful.&lt;/p&gt;&#xA;" OwnerUserId="11926" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.923" LastActivityDate="2016-12-22T12:50:13.003" Title="Two lists of favorite words 2!" Tags="|pattern|lateral-thinking|computer-science|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17197" PostTypeId="2" ParentId="17137" CreationDate="2015-06-30T18:58:31.427" Score="6" Body="&lt;p&gt;I need 12 lines - 8 to find the centers of the large squares (thanks Aragaer for noticing that we get the center &quot;for free&quot;), then 4 to create the 10 equal squares. You could argue that the 10 dashed lines are not drawn as part of the &quot;creating squares&quot;. In fact, you could use a ruler to draw just the little orange &quot;helper marks&quot; in order to do the construction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does assume that the two initial shapes are squares.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/c5ohN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3941" LastEditorUserId="3941" LastEditDate="2015-07-01T00:14:25.073" LastActivityDate="2015-07-01T00:14:25.073" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17198" PostTypeId="1" AcceptedAnswerId="17256" CreationDate="2015-06-30T19:02:04.243" Score="13" ViewCount="2149" Body="&lt;p&gt;This is part of &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=title%3A%22The+Twenty+Doors%22&quot;&gt;The Twenty Doors&lt;/a&gt; series.&#xA;&lt;br&gt;The previous one is &lt;a href=&quot;https://puzzling.stackexchange.com/q/17167/9000&quot;&gt;The Twenty Doors! (ROOM 1)&lt;/a&gt;&#xA;&lt;br&gt;The next one is &lt;a href=&quot;https://puzzling.stackexchange.com/questions/17313/the-twenty-doors-room-3&quot;&gt;The Twenty Doors! (ROOM 3)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ROOM 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After entering the first door, you find yourself in a bigger room. You look at the keypad.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[+] [-] [X]&#xA;[÷] [∞] [±]&#xA;  [ENTER]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You look around the room. Then, you see the paper. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;KDKGEOVB&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And, as before, there is a hint carved into the wall - &lt;code&gt;A to Z Alphabet&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This time, the answer seems to be a maths symbol. Or symbols. Still, you've got this far, so you may as well try.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is only one symbol involved.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A to Z was not required. It could have been A to B, I to C or anything. Each one changes the encrypted text. Now, which cipher is that?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is an eight-letter word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 4:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; AAAABABAAAAABABABAAAAAABB I love bacon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;MEGA HINT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The cipher used above is called Baconian.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The next door will be added when the current door is solved!&lt;/p&gt;&#xA;" OwnerDisplayName="user9377" LastEditorUserId="-1" LastEditorDisplayName="user9377" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-07-02T18:42:21.557" Title="The Twenty Doors! (ROOM 2)" Tags="|cipher|" AnswerCount="4" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="17199" PostTypeId="2" ParentId="5574" CreationDate="2015-06-30T19:42:15.160" Score="13" Body="&lt;p&gt;The hints:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;code&gt;?&lt;/code&gt; does not come after &lt;code&gt;3&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;code&gt;?&lt;/code&gt; immediately precedes &lt;code&gt;A&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;suggest that the characters map to some sort of numeric sequence.  I started with the assumption that the letters and numbers would be contiguous, which leaves us three parts to move around:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[?ABC...XYZ]   [_]   [0123]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(I'll use &lt;code&gt;_&lt;/code&gt; to unambiguously designate a space.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then assumed that the digits must come at the end: otherwise I think we would end up with &lt;code&gt;0-9&lt;/code&gt; and &lt;code&gt;A-T&lt;/code&gt; (i.e. the digits are used to 'pad out' the list of characters).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus our only choice is where to place &lt;code&gt;_&lt;/code&gt; in the sequence, with three possibilities.  It turns out that &lt;code&gt;_&lt;/code&gt; in the middle works for the next part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can use this ordering to map characters to numbers, from &lt;code&gt;?&lt;/code&gt; as $0$ to &lt;code&gt;3&lt;/code&gt; as $31$.  Then based on the hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Characters are processed in groups of five.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I broke the numbers into groups of five.  The number $5$ matches nicely with the number of characters: $32=2^5$.  This hints that we need to convert the numbers into binary.  This gives us $5\times 5$ bits in each group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While playing around with the blocks as bitmaps, I had the idea to transpose them, then convert back to numbers.  For example, the first group of five characters:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{rr|cccc}&#xA;&amp;amp;&amp;amp; \text{L} &amp;amp; \text{I} &amp;amp; \text{T} &amp;amp; \text{T} &amp;amp; \text{L} \\&#xA;&amp;amp;&amp;amp; 12 &amp;amp; 9 &amp;amp; 20 &amp;amp; 20 &amp;amp; 12 \\\hline&#xA;\text{F} &amp;amp;  6 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 1 &amp;amp; 0 \\&#xA;\text{Y} &amp;amp; 25 &amp;amp; 1 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1 \\&#xA;\text{W} &amp;amp; 23 &amp;amp; 1 &amp;amp; 0 &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 \\&#xA;\text{?} &amp;amp;  0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 \\&#xA;\text{H} &amp;amp;  8 &amp;amp; 0 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 \\&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives us, for both ciphertexts, the plaintexts:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LIFE GOES ON LONG AFTER THE THRILL OF LIVING IS GONE&lt;br&gt;&#xA;&#xA; and&lt;br&gt;&#xA;&#xA; LITTLE DITTY ABOUT JACK AND DIANE TWO AMERICAN KIDS GROWIN UP IN THE HEARTLAND JACKIE GONNA BE A FOOTBALL STAR DIANE DEBUTANTE BACKSEAT OF JACKIES CAR SUCKIN ON A CHILI DOG OUTSIDE THE TASTEE FREEZE DIANES SITTIN ON JACKIES LAP HES GOT HIS HAND BETWEEN HER KNEES JACKIE SAY HEY DIANE LETS RUN OFF BEHIND A SHADY TREE DRIBBLE OFF THOSE BOBBY BROOKS LET ME DO WHAT I PLEASE OH YEAH LIFE GOES ON LONG AFTER THE THRILL OF LIVIN IS GONE JACKIE SITS BACK REFLECTS HIS THOUGHTS FOR A MOMENT SCRATCHES HIS HEAD DOES HIS BEST JAMES DEAN WELL YOU KNOW DIANE WE OUGHTA RUN OFF TO THE CITY DIANE SAYS BABY YOU AINT MISSIN NOTHIN OH YEAH LIFE GOES ON LONG AFTER THE THRILL OF LIVIN IS GONE GONNA LET IT ROCK GONNA LET IT ROLL LET THE BIBLE MAN COME AND SAVE MY SOUL HOLD ONTO SIXTEEN AS LONG AS YOU CAN CHANGES COME AROUND REAL SOON MAKE US WOMEN AND MEN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; lyrics from &lt;a href=&quot;https://en.wikipedia.org/wiki/Jack_%26_Diane&quot;&gt;&lt;em&gt;Jack &amp;amp; Diane&lt;/em&gt; by John Mellencamp&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9718" LastActivityDate="2015-06-30T19:42:15.160" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17200" PostTypeId="2" ParentId="17198" CreationDate="2015-06-30T19:45:17.427" Score="9" Body="&lt;p&gt;Here's an attempt:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Convert each letter to it's numerical place in the alphabet. A=1, B=2, Z=26. That gives the following numbers: &lt;code&gt;[11,4,11,7,5,15,22,2]&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not sure which direction is correct here but I suspect what we want is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $11\times4\div11+7\times5-15-22-2$ because that's $0$ if you use infix notation that takes into account the order of operations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If that's true, then the button presses would be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\times\div+\times---\ \mathrm{ENTER}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Interestingly, If instead we use A=0 and the calculator uses immediate execution, the solution is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The same! $\left(\left(\left(\left(10\times3\right)\div10\right)+6\right)\times4\right)-14-21-1=0$&lt;br&gt;Unless, of course, the intended answer is $2$ because this is the second room. In that case, we could change it slightly at the end: $\left(\left(\left(\left(10\times3\right)\div10\right)+6\right)\times4\right)-14-21+1=2$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2015-07-01T00:37:01.047" LastActivityDate="2015-07-01T00:37:01.047" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="17201" PostTypeId="1" AcceptedAnswerId="17202" CreationDate="2015-06-30T20:29:02.570" Score="11" ViewCount="1486" Body="&lt;p&gt;Here's a rebus that involves no math skill, and is designed to be simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What movie title is represented below?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wR9zY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastActivityDate="2015-06-30T20:47:49.710" Title="What movie title does this represent?" Tags="|rebus|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17202" PostTypeId="2" ParentId="17201" CreationDate="2015-06-30T20:33:29.713" Score="22" Body="&lt;p&gt;I think the movie is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Shawshank Redemption&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first line is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The &#xA;Because all the articles, with &quot;the&quot; missing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second line is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Shaw&#xA; Because S plus( not pitch, not roll but yaw) = Syaw which sounds like shaw&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Third line is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Shank Redemption&#xA; Because Shank of meat+red+imp+the chinese character &quot;Shun&quot; = shankredimpshun which sounds like Shank Redemption&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4591" LastEditorUserId="4591" LastEditDate="2015-06-30T20:47:49.710" LastActivityDate="2015-06-30T20:47:49.710" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17205" PostTypeId="1" AcceptedAnswerId="17206" CreationDate="2015-06-30T20:54:35.703" Score="4" ViewCount="1168" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The image below shows 7 x 7 grey circles which all &lt;em&gt;but one&lt;/em&gt; share a common attribute. Can you see this shared property, or rather: Can you see which circles does &lt;em&gt;not&lt;/em&gt; belong to the group and give a reason?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A complete answer must of course contain an explanation. To identify the circles, simply number them, i.e &lt;code&gt;3rd column, 5th row&lt;/code&gt; or such. I did not want to complicate the image with labels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tOtru.gif&quot; alt=&quot;The puzzle is all in here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, you do not need to look at the digital content or super high resolution, nor do you need any digital processing of the image. It also contains only 8 distinct colours.&lt;/p&gt;&#xA;" OwnerUserId="5840" LastActivityDate="2015-06-30T21:59:32.217" Title="What do these circles have in common - and what not?" Tags="|visual|enigmatic-puzzle|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17206" PostTypeId="2" ParentId="17205" CreationDate="2015-06-30T21:07:17.427" Score="6" Body="&lt;p&gt;One of the circles&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; shrinks, the rest do not.  This is an animated GIF.  The circle in the second row (from the top), sixth column (from the left) shrinks every 60 seconds:&#xA; &lt;p&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/WIavN.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4450" LastEditDate="2015-06-30T21:45:33.470" LastActivityDate="2015-06-30T21:45:33.470" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17207" PostTypeId="1" AcceptedAnswerId="17310" CreationDate="2015-06-30T21:50:59.537" Score="16" ViewCount="2022" Body="&lt;p&gt;&lt;strong&gt;Preface:&lt;/strong&gt; There's a lot of background story here. Most of it is just for fun but there are definitely clues sprinkled throughout. If you want the strict puzzle, skip on down to the bottom.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For many decades, the pastry shop on Lamuella had been run by a kindly gentleman. He had taken over the sandwich shop when the previous owner bought The King's ship and left. His younger years had seen him as a brave sailor captured by pirates who grew to conquer their ranks. The life was not what he sought and did not bring him happiness so he left the first mate in charge and wandered into the port of Laumuella. The scents drifting from the sandwich shop drew him in and he was hooked ever since. When the owner took off, the ex-pirate stepped in and turned it into a bakery. He had always loved baking, but never could as a pirate lord because the lads would surely not respect such a leader. That was many, many years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The locals loved the now quite elderly man. His shop was always rich in scents and cheap in cents. It was not a life that made the baker wealthy but he felt as though his life had been rich. He married young but his bride died during childbirth, leaving him a son to raise. His son was strong and would help in the kitchen. The man never remarried but he felt his family was complete. Children would stop in on the way home from school and sneak bites of the sweets on the counter as soon as the proprietor would turn his back. Of course, they never wondered why the sweets were always fresh-out-of-the-oven warm or why the old man never seemed to turn around to discover the stolen treats until &lt;em&gt;after&lt;/em&gt; the door jangled open at their departure. The old man was happy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One day, anxiety broke this idyllic life. The old man was carrying out the day's ration of treats to be swiped by the kids when a sharp pain seized his chest and stole his breath. He dropped the tray, clutched at his chest, and fell to the floor, knocking over a chair on the way down. The children discovered him and ran into the street screaming for someone to help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The man woke up in a hospital bed and saw the face of his son leaning over him. &quot;Oh, Archie!&quot; the old man started. &quot;Hush, father&quot; replied his son with a concerned look on his face. The old man worked to prop himself up better on the pillows, eventually allowing his son to help. He began again, &quot;No, Archie, I must tell you something important.  I have a secret I've kept all these years. It's not safe to tell you here but you must know!&quot; The man stopped, his breath caught and he coughed. His soon stood to get a glass of water but the father grabbed his arm and pulled him back down with uncharacteristic strength. &quot;Listen!&quot; he hissed. &quot;I haven't much time. I can feel the end coming. You must listen NOW! Under the bread kiln, there is a brick that's darker red than the rest. Pull it out and find the small metal tin behind it. There's a steel plate inside with a code inscribed on it. You must decipher that code. You know me best; only you can do this!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The old man's grasp on his son's sleeve slackened. Wheezing, he fell back against the pillows. His breath was shallow but still he spoke. Archie leaned in to hear his father's dying ramblings: &quot;Favorite... Cherry times... Flagged down... Apple, blueberry, chocolate... Multiply...&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The son leaned back, sure that his father was gone. He slumped back into his own chair distraught and wracked with grief. He wasn't sure if he could keep the bakery running by himself or if he even &lt;em&gt;wanted&lt;/em&gt; to. His father was a fixture in the community and it wouldn't be the same without him. With all the grief and confusion, he almost forgot his father's words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As he was sitting in the darkened kitchen days later and wondering how he was going to move on, Archie remembered his father's instructions. Shuffling over to the kiln, he found the brick that was indeed a darker red. All these years working the kiln and he'd never noticed. He toed it with his boot and felt it give. Bending down, he slowly pulled the brick out and stuck his hand into the darkened recess. His fingers found something hard and cool to the touch. Getting excited now, Archie extracted the tin and set it on the ground before him. Trembling, he opened the creaking lid to reveal the engraved plate inside.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;3:6::0:1:::9:6::2:8:::6:8::3:6:::9:6::9:6:::9:6::0:1:::6:8::2:8   &#xA;9:6::2:8:::5:4::2:8:::9:6::9:6:::6:8::9:6:::5:4::9:6:::5:4::7:4   &#xA;6:8::3:6:::9:6::9:6:::5:4::0:1:::9:6::2:8:::5:4::2:8:::6:8::3:6   &#xA;9:6::9:6:::4:4::7:4:::6:8::3:6:::9:6::2:8:::9:6::2:8:::5:4::2:8   &#xA;9:6::9:6:::4:4::9:6:::9:6::2:8:::5:4::9:6:::5:4::0:1:::4:4::7:4   &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;His eyes played over the markings and, with a twinkle of recognition, he began scratching out letters on some parchment. When he finished, he simply sat staring at the results for several minutes. Why had his father never shared these words with him when he was still alive? The world must know. Archie shall tell them!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What was the encoded message that the old pie maker had left his son?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This has already been answered but, for the sake of completion and future readers, here are all the clues hidden in the text. Don't read them if you want to ponder it yourself as they take you halfway to the solution.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The title describes a &quot;Pie-Maker&quot; but the text only ever calls him a baker.&lt;br&gt;&#xA; The baker was a sailor and a pirate.&lt;br&gt;&#xA; The son's name is &quot;Archie&quot; which is a nickname version of Archimedes who introduced a new method to approximate pi.&lt;br&gt;&#xA; The dying words &quot;Cherry times&quot; and &quot;Apple, Blueberry, Chocolate... Multiply&quot; both speak to multiplying by types of pies.&lt;br&gt;&#xA; The dying words &quot;Flagged down&quot; refers to the flags used in semaphores.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2018-12-19T16:29:38.453" LastActivityDate="2018-12-19T19:04:34.763" Title="The Pie-Maker's Son" Tags="|calculation-puzzle|story|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="17208" PostTypeId="2" ParentId="17205" CreationDate="2015-06-30T21:59:32.217" Score="3" Body="&lt;p&gt;Excellent find by &lt;a href=&quot;https://puzzling.stackexchange.com/a/17206/9000&quot;&gt;pacoverflow&lt;/a&gt; but here's a more obvious rule: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All but one of the circles sits on no more than 2 fields of color. The odd circle out, therefore, is 5th row from the top, 3rd column from the left. It sits on 6 fields of color.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With all of the very intentional red herrings, there are probably more rules like that but what pacoverflow found is definitely what was intended.&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-06-30T21:59:32.217" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17209" PostTypeId="2" ParentId="17155" CreationDate="2015-06-30T23:34:35.177" Score="2" Body="&lt;p&gt;I think I can take a crack at it. Although I haven't been able to find something suiting each verse.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Is it Morgan Freeman?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Still going strong&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Freeman is 78 but still continues to make a number of movies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Search, searching, still searching am I,&lt;br&gt;&#xA;For those that on that one day did die.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reference to the movie Se7en, where he plays a detective tracking a serial killer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We don't even know the name of the place,&lt;br&gt;&#xA;I'm going there, even if you can't keep pace.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Shawshank Redemption, because he joins his prison mate Tim Robbins at the end of the movie at an unknown place.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the world out there, some things are pure,&lt;br&gt;&#xA;No flowers or vines, but named for sure.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure about this one...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I took it apart, and saved so many lives,&lt;br&gt;&#xA;In the fight that left so many husbandless wives.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Morgan Freeman played Nelson Mandela in Invictus where he took apart the system of apartheid and saved many lives.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Most really like me, except when I was given.&lt;br&gt;&#xA;Still trying to decide if I've been forgiven.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reference to the movie Unforgiven?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There was a close call, but it is alright,&lt;br&gt;&#xA;Though for a while, I was quite the sight.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Could refer to his accident in 2008, whose pictures became quite viral at the time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-06-30T23:34:35.177" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17210" PostTypeId="2" ParentId="16348" CreationDate="2015-06-30T23:49:10.787" Score="26" Body="&lt;p&gt;So I'm new here (please excuse any formatting issues/transgressions against community norms), but I've been thinking about this for the last few hours and here is what I have so far, culling various parts from other answers which have already been posted. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where are you? You are in: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Euclid&amp;#39;s_orchard&quot;&gt;Euclid's Orchard&lt;/a&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The wording of the poem was definitely hard to understand, but as was heavily underscored in the hints and seen by Cain, the structure of it seems very significant. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let's take the number of words in each line (As Cain did): &lt;br&gt; &lt;br&gt; (5 3 9) (6 6 10) (7 10 8) (9 4 6) (6 5 5) (8 6 9) (7 5 6) (6 9 10) (9 10 10)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The last hint said &quot;Words hold numbers who will only reveal themselves to the arithmetic of the clock,&quot; this is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Modular arithmetic, but what's the modulus? On a clock it's 12, but no numbers here are bigger than 12, so let's try... &lt;strong&gt;3&lt;/strong&gt; — &quot;the key always remains Three&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This gives us: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (2 0 0) (0 0 1) (1 1 2) (0 1 0) (0 2 2) (2 0 0) (1 2 0) (0 0 1) (0 1 1) &lt;br&gt; &lt;br&gt; These are numbers in &lt;strong&gt;ternary&lt;/strong&gt; (i.e. base &lt;strong&gt;3&lt;/strong&gt;), but in decimal this would be: &lt;br&gt; &lt;br&gt; 18 1 14 3 8 18 15 1 4 — This is simple alphanumeric encoding, e.g. the 18th digit of the alphabet. I used &lt;a href=&quot;http://rumkin.com/tools/cipher/numbers.php&quot;&gt;this tool&lt;/a&gt; to get: &lt;br&gt; &lt;br&gt; RANCHROAD&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This can't be a coincidence, can it? But what the heck does that mean? I was stuck here for awhile, until I realized...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's an &lt;strong&gt;anagram&lt;/strong&gt;. The real clue is &quot;AN ORCHARD.&quot; I always saw that numbers were behind the words of this poem, so searching for &quot;mathematical orchard&quot; on Wikipedia brought to the page (linked above) for Euclid's orchard. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, everything made sense. Let's go through the poem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The first tercet:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You stand in a forest, &lt;br&gt; So black, dark, &lt;br&gt; Light can’t quite wind around the dry, tough bark.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A forest suggests the trees that populate the orchard; the next two lines describe how, from the origin, trees occupy the observer's entire field of vision, so no light can come through. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The second tercet:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trees are all you can see, &lt;br&gt; But change your viewpoint even slightly; &lt;br&gt; And a brighter vista it may indeed come to be.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, trees block out all light, but from a different point, say (1,0), looking north, you can see between the rows of the trees, which are only located on the intersection of the lines of the coordinate grid. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The third tercet:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those seen trees which bear ripe plums, &lt;br&gt; Their seeds grew from the Two that do make One, &lt;br&gt; Which share no part except for another One.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first line is basically just for decorative purposes/word count. But the next two are important. From the origin, only certain trees can be seen—others are blocked. The trees which are seen &lt;strong&gt;are located at points whose coordinates are coprime pairs&lt;/strong&gt;. The two that are one are the coordinates which describe a point, and &quot;another One&quot; that they share is their only common divisor—1!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The fourth tercet:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in this deep, dark, starless forest, those hidden &lt;br&gt; Have more in common, &lt;br&gt; With those that keep them overridden.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again the first line is more imagery/stuff to make the numbers line up. &lt;em&gt;But&lt;/em&gt;, all the trees which can't be seen are obscured by other ones which have the same coordinates, but in reduced form. For example, a tree at (2, 8) couldn't be seen from the origin because it would be blocked by the one at (1, 4). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The fifth tercet:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sticks in one big, infinite bundle, &lt;br&gt; Upright, but One far apart, &lt;br&gt; Farther removed from Mussolini's art.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &quot;trees&quot; (lines, really) resemble sticks which stand upright, which are all one unit from each other (again, they're at the intersection of the lines of the coordinate gird. Last line seems just to reinforce that the trees are spaced apart, not all bunched together. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The sixth tercet:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But these belong to one very ancient indeed, &lt;br&gt; A man renowned in strictest sense, &lt;br&gt; Celebrated most for famous work, derived from common sense.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is &lt;em&gt;Euclid's&lt;/em&gt; Orchard, who's obviously very old, and who's celebrated as the father of geometry—his famous work, which comes from his set of axioms. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The seventh tercet:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thin green trees in this private garden, &lt;br&gt; With raindrops, at tip top, &lt;br&gt; Belong to him, the mysterious Alexandrian.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Euclid is known as &quot;Euclid of Alexandria&quot; to distinguish him from someone else named Euclid, and when these trees are seen from the orchard, in perspective, they resemble the &lt;a href=&quot;https://en.wikipedia.org/wiki/Thomae%27s_function&quot;&gt;&lt;strong&gt;raindrop function&lt;/strong&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The eighth tercet:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Words, meaningless, focus on their numbers, &lt;br&gt; To unravel this poem, the key always remains Three, &lt;br&gt; Numbers speak languages untold, decipher them to avoid any blunders.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first line was what made me (and Cain, I assume) look at line numbers, and the next made me think the modulus was 3, and that the resultant numbers were in ternary. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The ninth tercet:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Translate to a system, one familiar to the Owner- &lt;br&gt; Mysterious still, but more unfolding may prove to the sleuth, &lt;br&gt; That like the flowers here, this poem holds one truth.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The numbers had to be converted from ternary to decimal; Euclid was undoubtedly familiar with numbers, if not base 10 specifically. Even from decimal though, the numbers had to be converted to text. All this unscrambling and unravelling brought me to the one truth: You are in Euclid's Orchard!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A beautiful and immensely well-crafted puzzle! I look forward to seeing more from you and others on this site!&lt;/p&gt;&#xA;" OwnerUserId="13094" LastEditorDisplayName="user88" LastEditDate="2015-07-02T05:00:46.983" LastActivityDate="2015-07-02T05:00:46.983" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17211" PostTypeId="1" AcceptedAnswerId="20592" CreationDate="2015-07-01T00:56:39.340" Score="24" ViewCount="8551" Body="&lt;p&gt;-- Side Story: Please skip ahead if you want the riddle --&#xA;&lt;br&gt;&lt;br&gt;&lt;strong&gt;Intro&lt;/strong&gt;: Everyone knows of &lt;em&gt;The Oracle&lt;/em&gt;: After months of dodging dragons and trolls and sneaky gods and the wrath of the entire world, you are allowed to ask one question, and the Oracle will give you the answer.&lt;/p&gt;&#xA;&lt;p&gt;Of course, as everyone soon finds out, the Oracle's like a genie: They give you what you ask for, not what you want. Their words twist and wind like an Egyptian asp, clouding your mind and making you lose hope.&lt;/p&gt;&#xA;&lt;p&gt;One day, seeking an answer, a young boy asked the Oracle, &amp;quot;Where do you get your answers?&amp;quot; The Oracle sighed and started to cry, slapping the boy on the back.&#xA;&amp;quot;Son, I've been doing this since Jesus was a little tyke you're age, and I still can't figure it out.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;Suddenly, a rolling, chortling bail of laughter filled the air and ravens leapt for the sky.&#xA;&lt;br&gt;&amp;quot;Why little boy, it is I,&#xA;&lt;br&gt;the Riddler of Life, you see.&#xA;&lt;br&gt;When the Oracles are lost, or all in a fuss,&#xA;&lt;br&gt;advice from me they seek.&amp;quot;&#xA;&lt;em&gt;&lt;br&gt;&amp;quot;Now, the Cheshire's mad, The Joker glad,&#xA;&lt;br&gt;they all try to lead you astray.&#xA;&lt;br&gt;But stay with me,&#xA;&lt;br&gt;and then you'll see,&#xA;&lt;br&gt;a world you've never imagined.&amp;quot;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;The Riddler of Life was a weary old grandfatherly figure, but the wicked glint in his eye caused the little boy to shiver and quiver.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;What, don't believe what you see? Well then, here's the question:&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Puzzle 1&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Of Sugar sweet, and frogs astray,&#xA;of rocks and things, and everything nice,&#xA;what in the world is my advice?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The Young man thought for a moment, and then grinned as he gave the answer:&#xA;&lt;br&gt; &amp;quot;Easy; what's the _____ of Life?&amp;quot;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Act 2&lt;/strong&gt;&#xA;&lt;em&gt;&lt;br&gt;The Riddler of Life grinned and wrought the boy's hand,&#xA;&lt;br&gt;&amp;quot;Ah, I think you'll do nicely.&#xA;&lt;br&gt;I'm the Wandering Jew,&#xA;&lt;br&gt;over a thousand and two,&#xA;&lt;br&gt;and my life is slowly ending.&#xA;&lt;br&gt;But before I leave,&#xA;&lt;br&gt;my wisdom to seed,&#xA;&lt;br&gt;I need someone young and understanding.&#xA;&lt;br&gt;Will you take from me, the powers that be,&#xA;&lt;br&gt;a sharp wit and a mind of stone?&amp;quot;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&amp;quot;Why, certainly!&amp;quot;&#xA;&lt;br&gt;He said with glee,&#xA;&lt;br&gt;and the Riddler beamed delightfully.&#xA;&amp;quot;Well, there, don't you see?&#xA;&lt;br&gt;That beaten tree?&#xA;&lt;br&gt;Well, there's the clue for your answer.&amp;quot;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&amp;quot;The answer to what?&amp;quot;&#xA;&lt;br&gt;The boy said in fraught.&#xA;&lt;br&gt;&amp;quot;Why, to the question I asked you to seek!&amp;quot;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;The boy heaved a sigh,&#xA;&lt;br&gt;and scratched his behin',&#xA;&lt;br&gt;and went to find the answer.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Puzzle 2&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;First a bird, nearby plane,&#xA;&lt;br&gt;no, a shattered sky!&#xA;&lt;br&gt;puzzle the letters and find out why.&lt;/p&gt;&#xA;&lt;p&gt;A screen, not reflecting,&#xA;&lt;br&gt;a deep brown hue,&#xA;&lt;br&gt;a reference makes me feel real blue.&lt;/p&gt;&#xA;&lt;p&gt;A joint most fitting,&#xA;&lt;br&gt;D'el('s) bow string never fitting.&#xA;&lt;br&gt;the valley splits between.&lt;/p&gt;&#xA;&lt;p&gt;mesh the words, you will find your answer:&#xA;&lt;br&gt;&lt;strong&gt;the answer that you seek.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Act 3&lt;/strong&gt;&#xA;&lt;em&gt;&lt;br&gt;The boy scratched his head,&#xA;&lt;br&gt;his brain worse than dead:&#xA;&lt;br&gt;&amp;quot;The answer, it's not,&#xA;&lt;br&gt;just a clue, you clot!!&amp;quot;&#xA;&lt;br&gt;So with a sigh,&#xA;&lt;br&gt;and a tear in his eye,&#xA;&lt;br&gt;he went to make sense of it all.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;The Riddler dozed peacefully, propped on the wall, a scroll lying down beside him.&#xA;&lt;em&gt;&lt;br&gt;&amp;quot;Congratulations,&amp;quot; it read,&#xA;&lt;br&gt;&amp;quot;You're building your head,&#xA;&lt;br&gt;with the logic that we cherish greatly.&#xA;&lt;br&gt;Now, you're almost done,&#xA;&lt;br&gt;but the sun's stretching lon',&#xA;&lt;br&gt;and I fear I am quickly dying.&#xA;&lt;br&gt;My last gift to you&#xA;&lt;br&gt;is my blessing, it's true,&#xA;&lt;br&gt;and the final answer that's ringing.&amp;quot;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;Ringing where? In my hair?&amp;quot; Ask the boy all befuddled, as he read the old tattered scroll aloud.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;The Answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;br&gt;With tips of three,&#xA;&lt;br&gt;what could I be?&#xA;&lt;br&gt;A book, a crook, a surly cook,&#xA;&lt;br&gt; use me segmentedly!!!&lt;/p&gt;&#xA;&lt;p&gt;Gassy Silena garnished me,&#xA;&lt;br&gt;though of a form, a rose I be.&lt;/p&gt;&#xA;&lt;p&gt;And once you've solved me, you may see:&lt;br&gt;&#xA;I'm a piece of pie,&#xA;&lt;br&gt;a rude, crude man,&#xA;&lt;br&gt;a real clean freak.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Now tell me my friends, what was in his head, the mind of this Riddler so deep?&#xA;For truly, it's bold, that story of old, that I inherited just this week.&#xA;My great-great-grandfather told his son, who told his son, and know I have the Knowledge.&lt;/p&gt;&#xA;&lt;p&gt;But, ever willingly, it's true,&#xA;&lt;br&gt; and thus I give to you,&#xA;&lt;br&gt;the advice of the Riddler of Life.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;br&gt;May its wisdom guide you,&#xA;&lt;br&gt;nurture, never bind you,&#xA;&lt;br&gt;and keep you always from strife!!!&#xA;&lt;strong&gt;&lt;br&gt;What did it say,&#xA;&lt;br&gt;that's still true today,&#xA;&lt;br&gt;it's truth putting all in perspective?&lt;/strong&gt;&lt;/em&gt;&#xA;&lt;br&gt;-- The Riddler of Life&lt;/p&gt;&#xA;&lt;p&gt;Note: Partial answers &lt;em&gt;highly&lt;/em&gt; encouraged. Good riddle hunting, my friends. ;}&lt;/p&gt;&#xA;&lt;p&gt;P.S.  Congrats to @knave for finding the answer to Part 1! But, as most of you have realized, that was only a warm-up!! Who will be worthy of leading the Oracles for the next millennia?&lt;/p&gt;&#xA;&lt;p&gt;P.P.S.  Congrats to @McMagister for solving puzzle #2. Now, who will put the various clues together and find the final answer?&lt;/p&gt;&#xA;&lt;p&gt;P.P.S. alright. I'm running out of steam here: You can see the process used to solve Riddles 1 and 2 here: &lt;a href=&quot;http://chat.stackexchange.com/rooms/26670/the-riddle-of-life-to-ease-all-your-strife&quot;&gt;http://chat.stackexchange.com/rooms/26670/the-riddle-of-life-to-ease-all-your-strife&lt;/a&gt; or, you can start a new conversation here: &lt;a href=&quot;http://chat.stackexchange.com/rooms/35932/riddle-of-life-part-2&quot;&gt;http://chat.stackexchange.com/rooms/35932/riddle-of-life-part-2&lt;/a&gt;, which I strongly recommend. I will try and be more active on the chat this time around, so feel free to drop me a line if there's something you're unsure of. I realize this is a complicated puzzle, but truly think everyone will be satisfied once it is finished.&lt;/p&gt;&#xA;&lt;p&gt;Hint for Riddle's 3:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; the last answer is a portmanteau. If you look at lines 4 and 8-10, they define all three words needed to get the right portmanteau. The other lines provide other ways to find the answer, by looking into its form and description. Well, That's all, Folks!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13697" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-02-26T00:57:10.277" Title="The Riddle of Life, to Ease all Your Strife" Tags="|riddle|wordplay|rhyme|story|" AnswerCount="5" CommentCount="19" ContentLicense="CC BY-SA 3.0" />
  <row Id="17212" PostTypeId="1" AcceptedAnswerId="17255" CreationDate="2015-07-01T05:01:42.427" Score="35" ViewCount="820" Body="&lt;p&gt;In the Perfectly Pixelated Puzzling Plaza, some people were setting off fireworks. Being the puzzling fellows that they are, they couldn't bear to just set off fireworks for the sheer enjoyment of it (who would want that?). So they decided to hide a message inside one of the fireworks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And sure enough, as the particularly perplexing firework had been set off and the the sparks started to fall, some noticed a message that had been woven into that firework.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps, from this photo of the firework exploding, you can see the message too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/iRKHE.png&quot; alt=&quot;don&amp;#39;t enter an image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7308" LastActivityDate="2016-08-16T17:55:31.957" Title="Perplexing Fireworks" Tags="|visual|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17213" PostTypeId="2" ParentId="17211" CreationDate="2015-07-01T07:45:12.203" Score="8" Body="&lt;p&gt;Here's a crude guess&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Is it a confectioner? (a person whose occupation is making or selling candy and other sweets)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With tips of three, &lt;br&gt;&#xA;what could I be? &lt;br&gt;&#xA;A book, a crook, a surly cook, &lt;br&gt;&#xA;use me segmentedly!!!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In your hint, you explained the meaning of segmentedly. Splitting it, I get &lt;br&gt;'con', a crook&lt;br&gt; 'fection' (play on fiction, a popular genre of books)&lt;br&gt; 'confectioner' who sometimes cooks and prepares sweets and pastries too&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Silly peter garnished me,&lt;br&gt;&#xA;though I form a rose, you see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Confectioners often garnish their pastries with flowery shapes of chocolate or frosting.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And once you've solved me, you may see:&lt;br&gt;&#xA;I'm a piece of pie, &lt;br&gt;&#xA;a rude, crude man, &lt;br&gt;&#xA;a real clean freak.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can often get pie from a confectioner and usually confectioneries are kept very neat and clean to entice potential customers. (Can't make sense of 'rude, crude man' yet.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-07-01T07:45:12.203" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17214" PostTypeId="2" ParentId="17198" CreationDate="2015-07-01T08:10:25.623" Score="2" Body="&lt;p&gt;Maybe this is a very simplistic guess, but here goes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Perhaps the piece of paper with KDKGEOVB is a red herring and the clue is just the words 'A to Z Alphabet'. Among the symbols we have, only one of the symbols is also an alphabet, i.e. 'X'. We read it as multiplication, but the correct symbol is the only one which is a part of the English alphabet.&lt;br&gt;&lt;br&gt;So I would press 'X', then 'ENTER'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-07-01T10:30:09.357" LastActivityDate="2015-07-01T10:30:09.357" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17215" PostTypeId="2" ParentId="17181" CreationDate="2015-07-01T09:48:02.097" Score="3" Body="&lt;p&gt;While the regex pattern isn't perfect at finding only the intended answer (there are a number of variations that the matching string will accept), I'll explain the code and build the answer that Ivo Beckers found.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;First an explanation of regex calls that will be important. The pattern matcher moves through the input string, &quot;consuming&quot; characters as it matches them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;(?=__)&lt;/code&gt; is a positive lookahead which ensures the next characters match the specification in &lt;code&gt;__&lt;/code&gt;; however, unlike normal matching, a lookahead doesn't consume the characters, so they still need to be matched. Conversely, the negative lookahead &lt;code&gt;(?!__)&lt;/code&gt; looks ahead and ensures that the next characters &lt;strong&gt;don't&lt;/strong&gt; match &lt;code&gt;__&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When brackets &lt;code&gt;(__)&lt;/code&gt; are used around matching statements like &lt;code&gt;__&lt;/code&gt; (but not as part of a lookahead statement), it forms a &quot;capturing group&quot;. This means that whatever &lt;code&gt;__&lt;/code&gt; matches to will be stored as a subpattern that can be referred to later, and will be assigned a reference number based on the order of the opening bracket &lt;code&gt;(&lt;/code&gt;. For example, if I matched the string &lt;code&gt;abc&lt;/code&gt; against the pattern &lt;code&gt;.(.(.))&lt;/code&gt; (&lt;code&gt;.&lt;/code&gt; matches almost any single character), then it would match, and the stored subpatterns would be 1:&lt;code&gt;bc&lt;/code&gt; and 2:&lt;code&gt;c&lt;/code&gt;. Later in the pattern, we can match against these subpatterns &lt;code&gt;\1&lt;/code&gt; and &lt;code&gt;\2&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(?ix)(?![AD-ZC][a-jn-zjm])&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (?ix) sets modifiers, which change the behaviour of the pattern matching. &lt;code&gt;i&lt;/code&gt; means the pattern matching is case-insensitive. We then have a negative lookahead to check that the next two characters &lt;strong&gt;are not&lt;/strong&gt; &lt;code&gt;[AD-ZC][a-jn-zjm]&lt;/code&gt;. This can be a &lt;code&gt;B&lt;/code&gt; followed by any character or any character followed by a &lt;code&gt;K&lt;/code&gt; or &lt;code&gt;L&lt;/code&gt;. They can be non-alphabetical characters, but the next part sorts that out. Let's assume &lt;code&gt;BL&lt;/code&gt; so far.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(&#xA;  ([[:upper:]])&#xA;  [[:lower:]]&#xA;)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This matches the first two letters to characters. Since it's on case-insensitive mode, this just makes sure that they're letters, but otherwise it would check that it's an uppercase and a lowercase letter. It also stores the pair as subpattern 1 &lt;code&gt;bl&lt;/code&gt; and the first character of the two as subpattern 2 &lt;code&gt;b&lt;/code&gt;. Still &lt;code&gt;BL&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(?=[a-z])&#xA;(?!b-z)&#xA;(.)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We perform a positive lookahead to check that the next character is a letter, followed by a negative lookahead to check that it's not &lt;code&gt;b-z&lt;/code&gt; (so it's &lt;code&gt;A&lt;/code&gt;). &lt;code&gt;(.)&lt;/code&gt; then consumes that character and stores it as subpattern 3 &lt;code&gt;a&lt;/code&gt;. &lt;code&gt;BLA&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ck&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This just matches &lt;code&gt;ck&lt;/code&gt;. &lt;code&gt;BLACK&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(&#xA;  (?!\t\r\n\f)&#xA;  \s&#xA;)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The next character is matched to a whitespace character &lt;code&gt;\s&lt;/code&gt; but not any of the special whitespace characters. This leaves us with a space '&lt;code&gt;&lt;/code&gt;' character stored as subpattern 4. &lt;code&gt;BLACK&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(&#xA;  (?=[A-Z])&#xA;  (?![A-R])&#xA;  (?![T-Z])&#xA;  .\3\2\2\3t&#xA;  (&#xA;    (h)&#xA;  )&#xA;)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The three lookaheads assert that the next character is a letter, and not &lt;code&gt;A-R&lt;/code&gt; or &lt;code&gt;T-Z&lt;/code&gt;, leaving &lt;code&gt;S&lt;/code&gt;. &lt;code&gt;.&lt;/code&gt; consumes the S, and &lt;code&gt;\3\2\2\3&lt;/code&gt; matches those subpatterns in that order, which is &lt;code&gt;ABBA&lt;/code&gt;. This is followed by &lt;code&gt;t&lt;/code&gt; and &lt;code&gt;h&lt;/code&gt;. The whole group is stored as subpattern 5 &lt;code&gt;SABBATH&lt;/code&gt;, and the &lt;code&gt;h&lt;/code&gt; is stored as both subpatterns 6 and 7 &lt;code&gt;h&lt;/code&gt;. &lt;code&gt;BLACK SABBATH&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;\4\-\4\5\4\1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This simply matches the numbered subpatterns and the &lt;code&gt;-&lt;/code&gt; character, giving &lt;code&gt;- SABBATH BL&lt;/code&gt;. &lt;code&gt;BLACK SABBATH - SABBATH BL&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(o)&#xA;(\8)&#xA;d&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This matches &lt;code&gt;o&lt;/code&gt; and stores it as subpattern 8. It then matches &lt;code&gt;\8&lt;/code&gt; and stores that, too, so subpatterns 8 and 9 are both &lt;code&gt;o&lt;/code&gt;. Then it matches &lt;code&gt;d&lt;/code&gt;. &lt;code&gt;BLACK SABBATH - SABBATH BLOOD&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(?![xa-wz])&#xA;[[:alpha:]]\4\5*&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Another negative lookahead to ensure that the next character isn't &lt;code&gt;xa-wz&lt;/code&gt;, leaving &lt;code&gt;y&lt;/code&gt;. &lt;code&gt;[[:alpha:]]&lt;/code&gt; then matches the next alphabetical character, which is &lt;code&gt;y&lt;/code&gt;. We then add another space with &lt;code&gt;\4&lt;/code&gt;, giving us &lt;code&gt;BLACK SABBATH - SABBATH BLOODY&lt;/code&gt;. &lt;code&gt;*&lt;/code&gt; means to match &lt;code&gt;\5&lt;/code&gt; (&lt;code&gt;SABBATH&lt;/code&gt;) 0 or more times, as many times as possible. If we leave it as 0, we're already finished and have a matching pattern, or &lt;code&gt;BLACK SABBATH - SABBATH BLOODY SABBATHSABBATHSABBATHSABBATH&lt;/code&gt; is also valid. A quick Google search indicates that &quot;Sabbath, Bloody Sabbath&quot; is a real song, so &lt;code&gt;BLACK SABBATH - SABBATH BLOODY SABBATH&lt;/code&gt; seems to be the right answer. Alternatively, he may be listening to &lt;code&gt;zKack saZzatH - SazZATh ZkoOdY&lt;/code&gt;!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12462" LastActivityDate="2015-07-01T09:48:02.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17216" PostTypeId="2" ParentId="17172" CreationDate="2015-07-01T10:11:20.157" Score="0" Body="&lt;p&gt;Considering the fact that the neck is the only part of the body that connects the head with the body, can moving the neck towards the right side of the body, take the head to the right side too (along with the neck) ?&lt;img src=&quot;https://i.stack.imgur.com/0lNOf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="13706" LastActivityDate="2015-07-01T10:11:20.157" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17217" PostTypeId="2" ParentId="17155" CreationDate="2015-07-01T10:17:31.520" Score="2" Body="&lt;p&gt;I have a guess. It's my first one about this puzzle, but I want to try - the answer is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MEMORY &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's see:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Search, searching, still searching am I,&lt;br&gt;&#xA;For those that on that one day did die.&lt;br&gt;&#xA;We don't even know the name of the place,&lt;br&gt;&#xA;I'm going there, even if you can't keep pace.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Memory is always searching for something, sometimes we want to recall a person, a date, sometimes about our relative's. As for the place - we can't say where our thoughts are located. We point on our brain, but we feel otherwise.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the world out there, some things are pure,&lt;br&gt;&#xA;No flowers or vines, but named for sure.&lt;br&gt;&#xA;I took it apart, and saved so many lives,&lt;br&gt;&#xA;In the fight that left so many husbandless wives.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In our memory, we keep ideal images/names of various things. We try to keep our relatives' names. Soldiers who are wounded in a fight try to stick to thoughts about his/her wife and kids, and that for sure can save their lives in treatment.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Most really like me, except when I was given.&lt;br&gt;&#xA;Still trying to decide if I've been forgiven.&lt;br&gt;&#xA;There was a close call, but it is alright,&lt;br&gt;&#xA;Though for a while, I was quite the sight.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We like our memory, except for the time when we are forced to recall something. During tortures, prisoners are forced to recall something they want not to reveal. About last 2 werses - I'm not sure (I lack language skill I guess...).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12739" LastEditorUserId="12739" LastEditDate="2015-07-01T10:56:18.630" LastActivityDate="2015-07-01T10:56:18.630" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17218" PostTypeId="1" AcceptedAnswerId="17223" CreationDate="2015-07-01T11:10:55.627" Score="17" ViewCount="5881" Body="&lt;p&gt;You have to simply step forward and find the common link between these words:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;age, date, drill, go, hole&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct link works in the exact same way for each of the words.&lt;/p&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-07-02T11:36:07.950" Title="Find the link between these everyday words?" Tags="|english|wordplay|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17220" PostTypeId="2" ParentId="17172" CreationDate="2015-07-01T11:17:40.910" Score="55" Body="&lt;p&gt;I've found a simpler solution that doesn't rotate the entire goat! See the image below for a visual explanation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1. Separate the goat's head from its neck. &lt;br&gt;2. Move the goat's head over the right leg. &lt;br&gt;3. Rotate the head so it's facing the right and rest it on the goat's rump.&lt;br&gt;Only the head has moved, and it now faces the right and is above the right leg.&lt;br&gt;&lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/XTfSS.png&quot; alt=&quot;reforming a goat&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12462" LastEditorUserId="9000" LastEditDate="2015-07-01T13:45:53.823" LastActivityDate="2015-07-01T13:45:53.823" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17221" PostTypeId="2" ParentId="17218" CreationDate="2015-07-01T11:28:47.287" Score="3" Body="&lt;p&gt;I am Not sure, but it's may be Archaeology  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Age &amp;amp; Date&lt;/strong&gt; : Archaeology is the study of the ancient and recent human past through material remains. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Drill &amp;amp; Hole&lt;/strong&gt; : Drill &amp;amp; Hole for analysis of the material  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Go&lt;/strong&gt; : Visit Different Places.&lt;/p&gt;&#xA;" OwnerUserId="13707" LastActivityDate="2015-07-01T11:28:47.287" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17222" PostTypeId="1" AcceptedAnswerId="17264" CreationDate="2015-07-01T11:35:51.403" Score="2" ViewCount="528" Body="&lt;p&gt;My friend sent me the following short message:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;D57919749C3641B796384D0795DDE33739ED3E03&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How should I decode it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My friend's job is writing software for mobile phones.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The original message was 22 characters long.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is relevant that it was a Short Message that my friend sent me.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2059" LastEditorUserId="2059" LastEditDate="2015-07-01T19:20:46.050" LastActivityDate="2015-07-01T23:23:49.807" Title="What is my friend trying to tell me?" Tags="|cipher|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="17223" PostTypeId="2" ParentId="17218" CreationDate="2015-07-01T11:41:06.823" Score="27" Body="&lt;p&gt;You can ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ... append each of these words to the word &quot;man&quot; to make new English words:&lt;br&gt;&#xA;&lt;br&gt;&#xA; man&lt;strong&gt;age&lt;/strong&gt;, man&lt;strong&gt;date&lt;/strong&gt;, man&lt;strong&gt;drill&lt;/strong&gt;, man&lt;strong&gt;go&lt;/strong&gt;, man&lt;strong&gt;hole&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9860" LastActivityDate="2015-07-01T11:41:06.823" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17224" PostTypeId="2" ParentId="17172" CreationDate="2015-07-01T11:47:04.830" Score="10" Body="&lt;p&gt;Since rotation is allowed as per CodeNewbie's answer, I can do it without moving &lt;em&gt;any&lt;/em&gt; sticks! &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/ujle7.jpg&quot; alt=&quot;new goat&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-07-01T11:47:04.830" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17225" PostTypeId="1" AcceptedAnswerId="17228" CreationDate="2015-07-01T12:02:36.777" Score="9" ViewCount="2178" Body="&lt;p&gt;&lt;strong&gt;What do the following have in common?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;India, black, $10^{-6}$, many, south&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;More or less&lt;/em&gt; the same process can be applied to all of them to get very similar and related results.)&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-07-01T15:43:29.157" Title="What do these words have in common, and where?" Tags="|wordplay|language|geography|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17226" PostTypeId="2" ParentId="17172" CreationDate="2015-07-01T12:36:22.590" Score="22" Body="&lt;p&gt;Here is my try :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You can do it without moving any stick. Just go to the other side of the goat. The head will be on the right side, over the right leg. But since you have to move at least one stick, I moved her head in a more comfortable position so she's happier now.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/g2lbN.jpg&quot; alt=&quot;Happy goat&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13271" LastActivityDate="2015-07-01T12:36:22.590" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17227" PostTypeId="2" ParentId="17225" CreationDate="2015-07-01T12:46:25.240" Score="6" Body="&lt;p&gt;They're all places (town names, with the exception of #5) within the Southern United States.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/India,_Texas&quot;&gt;India, Texas&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Black,_Alabama&quot;&gt;Black, Alabama&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Micro,_North_Carolina&quot;&gt;Micro, North Carolina&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Many,_Louisiana&quot;&gt;Many, Louisiana&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Southern_United_States&quot;&gt;South(ern United States)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="10246" LastActivityDate="2015-07-01T12:46:25.240" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17228" PostTypeId="2" ParentId="17225" CreationDate="2015-07-01T13:00:35.477" Score="15" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Are they all places in the Pacific Ocean? If each word (or an appropriate variant) is used as a prefix for &quot;esia&quot;, you get (&lt;em&gt;more or less&lt;/em&gt;) the following:&lt;br&gt;&lt;br&gt;&#xA; India -&gt; Indonesia (meaning Indian island)&lt;br&gt;&#xA; Black -&gt; Melanesia (&lt;em&gt;melan&lt;/em&gt; is a prefix for dark or black)&lt;br&gt;&#xA; $10^{-6}$ -&gt; Micronesia (micro is the prefix for $10^{-6}$)&lt;br&gt;&#xA; Many -&gt; Polynesia (&lt;em&gt;poly&lt;/em&gt; is a prefix for many)&lt;br&gt;&#xA; South -&gt; Austronesia (&lt;em&gt;austri&lt;/em&gt; is Latin for south; thanks to the commenters for puzzling this one out!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12462" LastEditorUserId="12462" LastEditDate="2015-07-01T15:43:29.157" LastActivityDate="2015-07-01T15:43:29.157" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="17229" PostTypeId="2" ParentId="17218" CreationDate="2015-07-01T13:50:41.673" Score="4" Body="&lt;p&gt;Could it be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; things you lie/use as excuses?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would think that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; people, especially children, would lie about all of the above to get out of something.&#xA; i.e &quot;I'm eighteen,&quot; &quot;Sorry dude, I got a date,&quot; &quot;Of course I finished my drills, Coach!&quot;, &quot;Yeah, I went to the study session,&quot; and finally, &quot;I didn't make that hole in the yard looking for dinosaurs! It was aliens!!&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13697" LastActivityDate="2015-07-01T13:50:41.673" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17230" PostTypeId="1" CreationDate="2015-07-01T14:02:11.460" Score="36" ViewCount="25864" Body="&lt;p&gt;Connect every house with every well without the lines intersecting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ws29f.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure if this puzzle has a solution. I have been puzzled by it for a long long time. An old man from my village mentioned this puzzle to me 12 years ago. I have never been able to solve it, I'm hoping someone here can.&lt;/p&gt;&#xA;" OwnerUserId="13712" LastEditorDisplayName="user20" LastEditDate="2015-07-01T21:22:27.350" LastActivityDate="2019-05-31T09:49:24.213" Title="Connect 3 houses with 3 wells" Tags="|lateral-thinking|graph-theory|" AnswerCount="11" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17231" PostTypeId="2" ParentId="17119" CreationDate="2015-07-01T14:13:33.537" Score="6" Body="&lt;p&gt;This is also very much a work in progress, but ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do these figures make any sense to you?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/J501V.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/RWrll.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/rLey0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11687" LastEditorUserId="11687" LastEditDate="2015-07-01T14:19:07.590" LastActivityDate="2015-07-01T14:19:07.590" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17232" PostTypeId="2" ParentId="17230" CreationDate="2015-07-01T14:15:05.867" Score="14" Body="&lt;p&gt;This is &lt;strong&gt;impossible&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume 2 houses are fully connected to 3 wells. There must be an 'inner' well which is surrounded by the connections to the other 2 wells. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the third house to be connected to the inner well, it must be in one of the spaces between the inner well and one of the outer ones. But this means it is seperated from the other outer well. There is thus no way for the third house to join all three wells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: this answer assumes that lines may not cross a well/house en route to another.&lt;/p&gt;&#xA;" OwnerUserId="3987" LastEditorUserId="3987" LastEditDate="2015-07-02T11:05:38.073" LastActivityDate="2015-07-02T11:05:38.073" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17233" PostTypeId="2" ParentId="17230" CreationDate="2015-07-01T14:19:49.470" Score="37" Body="&lt;p&gt;&lt;a href=&quot;http://www.archimedes-lab.org/How_to_Solve/Water_gas.html&quot; rel=&quot;noreferrer&quot;&gt;This link&lt;/a&gt; explains why this puzzle is &lt;strong&gt;unsolvable&lt;/strong&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also suggests a clever solution in &quot;3 dimensions&quot;&lt;br&gt;&#xA;(actually, a 2D solution &lt;em&gt;transfomed&lt;/em&gt; into a 3D one). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GC1w6.gif&quot; alt=&quot;3D Solution&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="12628" LastEditorUserId="12628" LastEditDate="2015-07-03T16:04:59.403" LastActivityDate="2015-07-03T16:04:59.403" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17234" PostTypeId="2" ParentId="17230" CreationDate="2015-07-01T14:21:51.907" Score="70" Body="&lt;p&gt;This is &lt;strong&gt;not impossible&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; While the graph theory is correct, it's also not relevant. The wells in question &lt;strong&gt;aren't&lt;/strong&gt; vertices, and so we can connect multiple edges to a single well without them overlapping.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is a possible solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each house has a pipe that connects to all 3 wells (I've drawn on openings for each pipe in each well). The pipes &lt;strong&gt;do not&lt;/strong&gt; overlap.&lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/ih2TA.png&quot; alt=&quot;houses and wells&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12462" LastActivityDate="2015-07-01T14:21:51.907" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17235" PostTypeId="1" CreationDate="2015-07-01T14:56:26.877" Score="-4" ViewCount="261" Body="&lt;p&gt;Can you decipher what encryption algorithm has been used here. I have several examples but I cannot decipher the algorithm used.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;076a720f-67e6-4dd9-b9f2-38464e10d6ac = 35246&#xA;d1265cd3-cc3c-46d1-b1b2-a3f31ad5c094 = 36208&#xA;be7a8fb3-9799-47fd-b0c3-329fdc97d980 = 36312&#xA;6b4bc26c-89cb-43be-8d94-87ddf9e852d4 = 36378&#xA;8e3bc655-e419-41be-b9c3-42e999dc7a0f = 36425&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What would be the encrypted value of the number 1500 be, using this algorithm?&lt;/p&gt;&#xA;" OwnerUserId="13713" LastEditorUserId="12700" LastEditDate="2015-07-01T15:16:55.340" LastActivityDate="2015-07-01T15:18:33.453" Title="What is the encryption algorithm used here?" Tags="|cipher|algorithm|" AnswerCount="1" CommentCount="4" ClosedDate="2015-07-02T06:22:41.513" ContentLicense="CC BY-SA 3.0" />
  <row Id="17236" PostTypeId="1" CreationDate="2015-07-01T14:57:14.320" Score="2" ViewCount="4866" Body="&lt;p&gt;There is a house with 5 room and one door with every single wall as shown in following figure. You have to visit every door exactly once but there is a condition that you cannot cross the path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XvQWN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried hard to get a path but i failed. Is there any solution exist ?? &lt;/p&gt;&#xA;" OwnerUserId="13616" LastActivityDate="2015-07-01T15:29:42.240" Title="Is there any solution for this puzzle?" Tags="|logical-deduction|graph-theory|" AnswerCount="2" CommentCount="3" ClosedDate="2015-07-01T15:36:20.373" ContentLicense="CC BY-SA 3.0" />
  <row Id="17237" PostTypeId="2" ParentId="17236" CreationDate="2015-07-01T15:06:07.900" Score="1" Body="&lt;p&gt;I don't think this is possible!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for every room, there should be an even number of doors in order to get in and out without crossing ( you have  to get in through one door and out through the other).&#xA;Which is not the case here!&lt;/p&gt;&#xA;" OwnerUserId="5295" LastActivityDate="2015-07-01T15:06:07.900" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17238" PostTypeId="2" ParentId="17236" CreationDate="2015-07-01T15:10:25.370" Score="3" Body="&lt;p&gt;Assuming you can't walk through walls, you can only cross through a given door one time, and you have to cross through a door to &quot;visit&quot; it, this is impossible. Because...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3pfy4.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rooms 1, 2, and 4 have odd numbers of doors. Logically, if a room has an odd number of doors with our rules, if you start &lt;strong&gt;outside&lt;/strong&gt; the room, you must end &lt;strong&gt;inside&lt;/strong&gt; the room. Similarly, if you start inside the room, you must end outside of it. Draw some lines and you'll see what I mean -- no matter what, for each time you enter a room, you must leave it unless you are ending in that room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an image demonstrating this (left - start in, end out; right - start out, end in)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PF2T6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because there are three different rooms that have odd numbers of doors, no matter where you start you logically need to end in multiple rooms to be able to solve the puzzle, making it impossible.&lt;/p&gt;&#xA;" OwnerUserId="13556" LastEditorUserId="13556" LastEditDate="2015-07-01T15:29:42.240" LastActivityDate="2015-07-01T15:29:42.240" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17239" PostTypeId="2" ParentId="17235" CreationDate="2015-07-01T15:18:33.453" Score="4" Body="&lt;p&gt;These are Version 4 &lt;a href=&quot;https://en.wikipedia.org/wiki/Universally_unique_identifier&quot; rel=&quot;nofollow&quot;&gt;Universally Unique Identifiers&lt;/a&gt;. You can tell that they are UUID's because they are hexadecimal strings in the characteristic pattern of &lt;code&gt;xxxxxxxx-xxxx-xxxx-xxxx-xxxxxxxxxxxx&lt;/code&gt;. You can tell they are Version 4 by the digit &lt;code&gt;4&lt;/code&gt; as the first digit of the third group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately the Version 4 of UUID's is completely randomly generated. Whatever system generates these numbers likely checks if that number is already in its database, and when it isn't it generates a randomly generated new UUID for it.&lt;/p&gt;&#xA;" OwnerUserId="1774" LastActivityDate="2015-07-01T15:18:33.453" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17241" PostTypeId="2" ParentId="17230" CreationDate="2015-07-01T15:32:56.063" Score="31" Body="&lt;p&gt;&lt;strong&gt;Option 1:&lt;/strong&gt;&lt;br&gt;&#xA;Ask a mathematician to explain why the complete bipartite $K_{3,3}$ graph is non planar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Option 2:&lt;/strong&gt;&lt;br&gt;&#xA;Put this drawing and a stack of cash in an envelope. Deliver the envelope to a competent engineer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/2WKJd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11752" LastEditorUserId="11752" LastEditDate="2015-07-01T15:41:28.227" LastActivityDate="2015-07-01T15:41:28.227" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17242" PostTypeId="1" AcceptedAnswerId="17244" CreationDate="2015-07-01T15:55:17.687" Score="5" ViewCount="174" Body="&lt;p&gt;Each of the following list of words have a common link. Each of those common links are they themselves another group with a common link.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you find each common link and the meta-common link?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Age, Buster, Head, House&lt;br&gt;&#xA;  Head, Shaking, Yard, Less&lt;br&gt;&#xA;  Ding, Dying, Get, Mash&lt;br&gt;&#xA;  Crash, Keeper, Post, Way   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="12700" LastEditDate="2015-07-01T16:29:15.893" LastActivityDate="2015-07-01T16:51:10.240" Title="Common Linkception" Tags="|word|english|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17243" PostTypeId="1" AcceptedAnswerId="17275" CreationDate="2015-07-01T16:09:19.453" Score="7" ViewCount="873" Body="&lt;p&gt;You are part of an international anti-terror unit who has intelligence that some religious fanatics of the FSM church are going to soon blow a bomb in one or more of the world's major cities. You're shadowing their communication channels in the hope of learning of the exact plan early enough to prevent disaster.&lt;/p&gt;&#xA;&lt;p&gt;To interfere, you have 5 super-secret-delta-swat units stationed in &lt;code&gt;Berlin, Nairobi, Bogota, Singapore and Calgary&lt;/code&gt;. These units can relocate to any destination very quickly at a travel-speed of 500km/h  (straight line). If they are at the scene of an attack, they supress it within 15min and are immediately ready to move again. They are in instant communication and can be ordered right from your location by a simple satellite call.&lt;/p&gt;&#xA;&lt;p&gt;One night in late June, when you're in a cellar in Madrid at 04:00 in the morning you eavesdrop on a phone call apparently made from a cell phone in Lisbon (transcript below).&lt;/p&gt;&#xA;&lt;p&gt;Jumping immediately to take appropriate actions you successfully managed to prevent disaster that day and become the unsung hero. (&lt;em&gt;Whew!&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;p&gt;The puzzle question:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;h1&gt;What did you do and why?&lt;/h1&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This is the transcript of the phone call:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;Lisbon HQ message. All units action as scheduled today.&#xA;Paris 17:30:45 &#xA;Seoul 18:30:00 &#xA;Denver 15:30:37&#xA;Tokyo 15:30:07&#xA;Omsk 18:30:37&#xA;Helsinki 21:30:22&#xA;Lima 16:30:45&#xA;Phoenix 13:00:30&#xA;Dallas 15:00:52&#xA;Rome 16:00:37&#xA;Cairo 19:00:45&#xA;Sydney 15:00:52&#xA;Perth 13:00:37&#xA;Gibraltar 17:30:37&#xA;Oslo 23:30:30&#xA;Melbourne 15:00:07&#xA;Port au Prince 14:30:07&#xA;Montreal 13:00:37&#xA;London 12:00:52&#xA;Athens 12:30:00&#xA;Cape Town 20:30:30&#xA;Brasilia 18:30:30&#xA;Karachi 11:30:00&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;Notes&lt;/h1&gt;&#xA;&lt;p&gt;While this is a puzzle tagged as &lt;a href=&quot;/questions/tagged/enigmatic-puzzle&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;enigmatic-puzzle&amp;#39;&quot; rel=&quot;tag&quot;&gt;enigmatic-puzzle&lt;/a&gt; it is strictly logical and linked to the information provided above. I am &lt;em&gt;not&lt;/em&gt; looking for some creative-story-telling solution. The second tag &lt;a href=&quot;/questions/tagged/cryptograms&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;cryptograms&amp;#39;&quot; rel=&quot;tag&quot;&gt;cryptograms&lt;/a&gt; tells you where to solve the puzzle. The transcribed message indeed contains hidden information which made your decision uniquely clear. You have to decipher it.&lt;/p&gt;&#xA;&lt;p&gt;Intermediate &amp;quot;work-in-progress&amp;quot; or &amp;quot;ideas&amp;quot; answers are welcome and will be commented on.&lt;/p&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-02T15:30:25.307" Title="Where will they strike?" Tags="|cipher|enigmatic-puzzle|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17244" PostTypeId="2" ParentId="17242" CreationDate="2015-07-01T16:16:19.243" Score="5" Body="&lt;p&gt;The words that are common to the four sets are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Block&lt;/strong&gt;&lt;br&gt; &lt;strong&gt;Block&lt;/strong&gt;age, &lt;strong&gt;Block&lt;/strong&gt;buster, &lt;strong&gt;Block&lt;/strong&gt;head, &lt;strong&gt;Block&lt;/strong&gt;house&lt;br&gt;&lt;br&gt; &lt;strong&gt;Bone&lt;/strong&gt;&lt;br&gt;&lt;strong&gt;Bone&lt;/strong&gt;head, &lt;strong&gt;Bone&lt;/strong&gt;shaking, &lt;strong&gt;Bone&lt;/strong&gt;yard, &lt;strong&gt;Bone&lt;/strong&gt;less&lt;br&gt;&lt;br&gt; &lt;em&gt;Post Edit from OP&lt;/em&gt;&lt;br&gt;&lt;del&gt;&lt;strong&gt;Up&lt;/strong&gt;&lt;br&gt;Ding &lt;strong&gt;up&lt;/strong&gt;, Get &lt;strong&gt;up&lt;/strong&gt;, Hist &lt;strong&gt;up&lt;/strong&gt;, Mash &lt;strong&gt;up&lt;/strong&gt;&lt;/del&gt;&lt;br&gt; &lt;strong&gt;Bud&lt;/strong&gt;&lt;br&gt;&lt;strong&gt;Bud&lt;/strong&gt;ding, &lt;strong&gt;Bud&lt;/strong&gt;dying, &lt;strong&gt;Bud&lt;/strong&gt;get, &lt;strong&gt;Bud&lt;/strong&gt;mash&lt;br&gt;&lt;br&gt; &lt;strong&gt;Gate&lt;/strong&gt;&lt;br&gt;&lt;strong&gt;Gate&lt;/strong&gt;crash, &lt;strong&gt;Gate&lt;/strong&gt;keeper, &lt;strong&gt;Gate&lt;/strong&gt;post, &lt;strong&gt;Gate&lt;/strong&gt;way&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not sure about the meta word. Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;del&gt;&lt;strong&gt;Jump&lt;/strong&gt;&lt;br&gt;Block &lt;strong&gt;jump&lt;/strong&gt;, bone &lt;strong&gt;jump&lt;/strong&gt;, &lt;strong&gt;jump&lt;/strong&gt; up, gate &lt;strong&gt;jump&lt;/strong&gt;&lt;/del&gt;&lt;br&gt;&lt;strong&gt;Tail&lt;/strong&gt;&lt;br&gt;&lt;strong&gt;Tail&lt;/strong&gt; block, &lt;strong&gt;Tail&lt;/strong&gt;bone, &lt;strong&gt;Tail&lt;/strong&gt; bud, &lt;strong&gt;Tail&lt;/strong&gt;gate&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-07-01T16:51:10.240" LastActivityDate="2015-07-01T16:51:10.240" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17245" PostTypeId="2" ParentId="17198" CreationDate="2015-07-01T16:42:34.443" Score="0" Body="&lt;p&gt;The lack of food, water and day light, forced me to make this desperate attempt and so I'll probably die&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Deciphering &lt;code&gt;KDKGEOVB&lt;/code&gt; using a this key &lt;code&gt;pzsbotsl&lt;/code&gt; (which came out of thin air) with the vigenere cipher you get &lt;code&gt;ADDITION&lt;/code&gt;, so I'd press the pluss sign and hit enter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13715" LastActivityDate="2015-07-01T16:42:34.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17247" PostTypeId="1" AcceptedAnswerId="19826" CreationDate="2015-07-01T16:49:48.740" Score="12" ViewCount="1755" Body="&lt;p&gt;&lt;em&gt;If you're interested in starting the treasure hunt from the very beginning, check out the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16551/treasure-hunt-round-the-world&quot;&gt;prologue&lt;/a&gt;!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Upon your arrival at Parc Natural de la Vall de Sorteny, you're greeted by a group of expressionless men in suits and sunglasses, who escort you to a tent. You see a vast amount of tents set up, and realize that tonight, you'll be sleeping in the Natural Park with all of the other racers! After a long time spent traveling, you could use a break. Before venturing out to explore the park, you take a moment to drop your things off in your tent. In your tent, you find a sleeping bag, a lantern, and another envelope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deciding it would be best to read the letter and get it out of the way now, you open the envelope and find the following enclosed:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dear &lt;strong&gt;&lt;em&gt;[insert username here]&lt;/em&gt;&lt;/strong&gt;,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Welcome to Vall de Sorteny Natural Park! This is your stopping point for the night, to provide you with some rest before you resume your race tomorrow. Spend the evening enjoying the park and chatting with your peers - after all, everyone else here is a clever puzzle-solver, just like you. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;On the flip side of this note, you'll find a simple brainteaser. Don't stress too much over it, as it's just to help pass the time. The important thing for you to do right now is enjoy your stay, as this park cost a &lt;em&gt;fortune&lt;/em&gt; to rent for the night!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Yours truly,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Bailey M&lt;br&gt;&#xA;  Puzzlemaster for the Treasure Hunt 'Round the World&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;P.S. The answer to the brainteaser will once again be a question. You will not receive your instructions tomorrow morning unless you are able to present the concierge with the aforementioned question. Good luck!&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As much as you want to see the park, the treasure is worth more to you. Hurriedly flipping over the card, you read:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Scold the bad liar. (4)&lt;br&gt;&#xA;  A wall of iron has nickel, copper, even lead! (5)&lt;/p&gt;&#xA;  &#xA;  &lt;hr&gt;&#xA;  &#xA;  &lt;p&gt;MJ, zDafl.o!sassor ok Vqi h,bg e yx hos i edi i ot.cTuP swnWpHh hf.&lt;/p&gt;&#xA;  &#xA;  &lt;hr&gt;&#xA;  &#xA;  &lt;p&gt;cloki rlbfvkag kpf zrbt hokp vknkrkqbs pfbc&lt;br&gt;&#xA;  clre jzuknou alj pwbpc nkcvru kpf vb jke kpf &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;kbo rbpfbc vfknwkv zknkffupv voubfrkvvk lf rclz zknuh rkpc&lt;br&gt;&#xA;  kjk kpf pfbc rupf nuk pfbc jofrknkttbz zksbkwnkh jfnuh u&lt;br&gt;&#xA;  zolp oobc jkpf rlbfuxnltrb vpfnalt nbkpf ji rkeuf rkpc&lt;br&gt;&#xA;  zolq vhupnkh zkokkfv ki lf kzkhxb lf rkpf rlc ki lf&lt;/p&gt;&#xA;  &#xA;  &lt;hr&gt;&#xA;  &#xA;  &lt;p&gt;Toh ws L xivx qb yiwdwkmkpl?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With a sigh, you take out your pencil and begin to work. Hopefully this doesn't take the &lt;em&gt;whole&lt;/em&gt; night...&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What question do you need to ask the concierge in the morning?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each line essentially indicates a decipher. When you figure out how to decipher the next step, &lt;strong&gt;only&lt;/strong&gt; decipher the next step and not any of the text beneath it. Each chunk of text below has only been encrypted once.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each of the three steps involves the &lt;a href=&quot;/questions/tagged/cryptograms&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;cryptograms&amp;#39;&quot; rel=&quot;tag&quot;&gt;cryptograms&lt;/a&gt; tag. The other three tags (&lt;a href=&quot;/questions/tagged/cryptic-crosswords&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;cryptic-crosswords&amp;#39;&quot; rel=&quot;tag&quot;&gt;cryptic-crosswords&lt;/a&gt;, &lt;a href=&quot;/questions/tagged/enigmatic-puzzles&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;enigmatic-puzzles&amp;#39;&quot; rel=&quot;tag&quot;&gt;enigmatic-puzzles&lt;/a&gt;, &lt;a href=&quot;/questions/tagged/riddle&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;riddle&amp;#39;&quot; rel=&quot;tag&quot;&gt;riddle&lt;/a&gt;) are all used once, at different points in the process.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everything has been solved except the riddle, so here's a hint for it: The answer you seek is either one word or four words, depending on how you look at it. The first line will likely be much more helpful than the second line. The second line can be used to check your answer once you've found it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 4 (hopefully the last):&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first line of the riddle clues you towards a set of words. The second line will give you a way to check that your words are correct. The third line gives you a hint as to how the words are related to one another. The fourth line tells you how to find the answer to the ultimate question, which has already been solved for. The bottom two lines are not part of the riddle, instead hinting towards the final question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Even though the riddle hasn't been solved, the ultimate solution &lt;em&gt;has&lt;/em&gt; been reached, and so the story continues in the next part, &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18071/treasure-hunt-round-the-world-clue-3&quot;&gt;Treasure hunt &amp;#39;round the world! (clue 3)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;8/4/2015 Update&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Since no one solved the riddle while the bounty was available, I awarded it to the puzzler who did the most work. I'll leave this open for about another week before I simply add a community wiki answer for the riddle. If someone solves it before then, there &lt;em&gt;may&lt;/em&gt; be another bounty involved...&lt;/p&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2015-08-19T18:39:31.347" Title="Treasure hunt 'round the world! (Interlude 1)" Tags="|riddle|cipher|enigmatic-puzzle|cryptic-crosswords|" AnswerCount="6" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17248" PostTypeId="2" ParentId="17172" CreationDate="2015-07-01T16:54:31.720" Score="12" Body="&lt;p&gt;This question / Puzzle tag with &quot;geometry&quot; So here I came with geometry answer + little funny too as I notice that there are some really (rofl) answers.&lt;br&gt;&#xA;Now Here is Goat which given by @prog_SAHIL. She is look something like this.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0q3qq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now here some geometry comes and now lets rotate her &lt;strong&gt;leg 1&lt;/strong&gt; to 45* clock wise. So its look something like this..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Pz4xg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Now, leg 2 becomes her neck, leg 1 becomes leg 2 and neck + head become leg 1 and tail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just like this.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7zZV6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now you are thinking that where is her head. ????&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Head isn't go any where, its here. You force her to turn, you force her to seet now she is looking at you angry. So here is her head, (below Image)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tJDvM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Original, If she don't look at you. :P&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ATlYt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;AND IF ROTATION IS ALLOW, YOU DO NOT NEED TO DO ANY THING, JUST FLIP HER HORIZONTALLY AS SAME AS I DO ON FIRST IMAGE AND LAST IMAGE&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13707" LastActivityDate="2015-07-01T16:54:31.720" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17250" PostTypeId="2" ParentId="17230" CreationDate="2015-07-01T16:55:24.250" Score="44" Body="&lt;p&gt;Probably a bit simpler than what you were looking for. Looking at the other answers, I feel like there might be some details missing from your question.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/8cZzS.png&quot; alt=&quot;my solution&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13719" LastActivityDate="2015-07-01T16:55:24.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17251" PostTypeId="2" ParentId="17243" CreationDate="2015-07-01T17:07:40.240" Score="1" Body="&lt;p&gt;&lt;strike&gt;Assuming the times given are all in UTC (or some other uniform time zone - not local time), we can interpret each one as the time a unit is scheduled to leave that location for their single final destination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latest time is Oslo, so that should be closest to the final destination, which suggests you should mobilise the Berlin team.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We might be able to find the exact final destination and attack time by analysing all the times given and comparing ratios, but I won't spend ages doing this unless the OP says I'm looking in the right direction!&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scratch that. No feedback from the OP, so I'll assume it's completely wrong. (I've left it there though in case it gives someone else an idea).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Engineer Toast's observation on the seconds past each minute have given me an idea: the times 00, 07, 15, 22, 30, 37, 45, 52 (up to rounding error) divide the minute up into eighths, suggesting that the seconds column of the times denotes some variable that can take 8 different values. How about the compass directions north, east, south, west, northeast, northwest, southeast, southwest? Taking 00 to be north and proceeding clockwise, we get &lt;em&gt;west from Paris&lt;/em&gt;, &lt;em&gt;north from Seoul&lt;/em&gt;, &lt;em&gt;southwest from Denver&lt;/em&gt;, &lt;em&gt;northeast from Tokyo&lt;/em&gt;, and so on. Maybe lines drawn in these directions will spell out letters, if BmyGuest has &lt;a href=&quot;https://puzzling.stackexchange.com/questions/14869/a-mysterious-email-from-ryanair&quot;&gt;pinched an idea from me&lt;/a&gt; ;-)&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2015-07-01T18:13:56.153" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17252" PostTypeId="2" ParentId="17243" CreationDate="2015-07-01T17:22:54.383" Score="2" Body="&lt;p&gt;To save someone else some &lt;em&gt;time&lt;/em&gt; (Get it? Ha!), here are all the times translated to UTC:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;City            Time        Zone    UTC Time&#xA;-----------------------------------------------   &#xA;Madrid           4:00:00     2       4:00:00   &#xA;Lisbon           3:00:00     1       4:00:00   &#xA;&#xA;Paris           17:30:45     2      17:30:45   &#xA;Seoul           18:30:00     10      10:30:00   &#xA;Denver          15:30:37    -6      23:30:37   &#xA;Tokyo           15:30:07     10      7:30:07   &#xA;Omsk            18:30:37     7      13:30:37   &#xA;Helsinki        21:30:22     3      20:30:22   &#xA;Lima            16:30:45    -4      22:30:45   &#xA;Phoenix         13:00:30    -6      21:00:30   &#xA;Dallas          15:30:52    -4      21:30:52   &#xA;Rome            16:00:37     2      16:00:37   &#xA;Cairo           19:00:45     3      18:00:45   &#xA;Sydney          15:00:52    11       6:00:52   &#xA;Perth           13:00:37     9       6:00:37   &#xA;Gibraltar       17:30:37     2      17:30:37   &#xA;Oslo            23:30:30     3      22:30:30   &#xA;Melbourne       15:00:07    11       6:00:07   &#xA;Port au Prince  14:30:07    -3      19:30:07   &#xA;Montreal        13:00:37    -3      18:00:37   &#xA;London          12:00:52     1      13:00:52   &#xA;Athens          12:30:00     3      11:30:00   &#xA;Cape Town       20:30:30     3      19:30:30   &#xA;Brasilia        18:30:30    -2      22:30:30       &#xA;Karachi         11:30:00     6       7:30:00     &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Further observations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;All the times have seconds values of &lt;code&gt;0&lt;/code&gt;,&lt;code&gt;7&lt;/code&gt;,&lt;code&gt;22&lt;/code&gt;,&lt;code&gt;30&lt;/code&gt;,&lt;code&gt;37&lt;/code&gt;,&lt;code&gt;45&lt;/code&gt;, or &lt;code&gt;52&lt;/code&gt;. I don't see how this can be a coincidence unless OP was just lazy.&lt;/li&gt;&#xA;&lt;li&gt;Similarly, all of the minutes are either &lt;code&gt;0&lt;/code&gt; or &lt;code&gt;30&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="1968" LastEditDate="2015-07-02T00:48:57.090" LastActivityDate="2015-07-02T00:48:57.090" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17254" PostTypeId="2" ParentId="16987" CreationDate="2015-07-01T17:49:21.610" Score="0" Body="&lt;p&gt;A few answers state that the smallest possible 2D grid (not allowing either side's length to be 1) is 2x3:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;? 2 ?&#xA;? 2 ?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But it is possible for a 2x2 grid to be unsolvable. Here are a few examples:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;? 1    1 ?    2 ?&#xA;1 ?    ? ?    ? ?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The same can be said of a 2x2x2 grid in 3D. For example, if one of the eight cells is a number less than seven and the other seven are unknown, it is impossible to determine which cells contain mines.&lt;/p&gt;&#xA;" OwnerUserId="13722" LastActivityDate="2015-07-01T17:49:21.610" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17255" PostTypeId="2" ParentId="17212" CreationDate="2015-07-01T17:55:01.930" Score="21" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Vertically compressing the pixels in each column (the clause &quot;[as] the sparks started to fall, some noticed a message&quot; implies a change in the relative vertical alignment of the pixels)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yields&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;pre&gt;- - - - # # # - # # # - # # # - # # # - # - # - # - - - # # # - - - - &#xA;  - - - # - - - # - # - # - # - # - # - # - # - # - - - # - - - - - &#xA;    - - # # # - # # # - # # # - # # # - # # - - # - - - # # # - - &#xA;      - - - # - # - - - # - # - # # - - # - # - # - - - # - - - &#xA;        # # # - # - - - # - # - # - # - # - # - # # # - # # # &lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where &lt;code&gt;#&lt;/code&gt; represents orange and &lt;code&gt;-&lt;/code&gt; represents red.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this is a picture representing it &lt;em&gt;(thanks to Cristian Marian who posted it in the comments)&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cb4C4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here we plainly see the word&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;SPARKLE&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;spelled out, which is presumably the message.&lt;/p&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="4421" LastEditDate="2016-08-16T17:55:31.957" LastActivityDate="2016-08-16T17:55:31.957" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17256" PostTypeId="2" ParentId="17198" CreationDate="2015-07-01T18:28:58.267" Score="11" Body="&lt;p&gt;We should press&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; INFINITY&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Deciphering AAAABABAAAAABABABAAAAAABB gives:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;BIFID&lt;/strong&gt; using the Baconian Cipher&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Deciphering KDKGEOVB:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Using Bifid cipher with &lt;strong&gt;Alphabet&lt;/strong&gt; as a key and Translate &lt;strong&gt;A&lt;/strong&gt; to &lt;strong&gt;Z&lt;/strong&gt; gives &lt;strong&gt;INFINITY&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12731" LastActivityDate="2015-07-01T18:28:58.267" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17257" PostTypeId="2" ParentId="17207" CreationDate="2015-07-01T18:46:30.607" Score="10" Body="&lt;p&gt;A general observation that might help other puzzlers (I'm stumped):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Treating each set of digits &lt;code&gt;N:M&lt;/code&gt; as a unique symbol, there are only 8 symbols in total used by the cipher:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A = 9:6 (54)&#xA;\ = 2:8 (16)&#xA;@ = 6:8 (48)&#xA;_ = 3:6 (18)&#xA;I = 0:1 (0)&#xA;* = 5:4 (20)&#xA;O = 4:4 (16)&#xA;` = 7:4 (28)&#xA;&#xA;_I  A\  @_  AA  AI  @\   &#xA;A\  *\  AA  @A  *A  *`&#xA;@_  AA  *I  A\  *\  @_&#xA;AA  O`  @_  A\  A\  *\&#xA;AA  OA  A\  *A  *I  O`&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here I've used arbitrary distinct characters for each symbol. The &lt;code&gt;::&lt;/code&gt; delimiters are eliminated and the &lt;code&gt;:::&lt;/code&gt; delimiters are replaced with spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be significant that the products of digit pairs in each symbol is even, and that all but one symbol (&lt;code&gt;_&lt;/code&gt;) has a pair product that's a multiple of 4. I'm only assuming products are relevant due to the mention of multiplication in the pie pirate's ramblings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also worth noting that the symbol &lt;code&gt;_&lt;/code&gt; only appears in &lt;code&gt;@_&lt;/code&gt; pairs, and that the symbol pairs &lt;code&gt;AA&lt;/code&gt; and &lt;code&gt;A\&lt;/code&gt; appear with unusually high frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My best guess is that each symbol pair represents a letter, but I have no idea what cipher is being used or even if this is the right direction to go in. It could be that each &lt;code&gt;M:N&lt;/code&gt; symbol is a reference to a word/letter (e.g. letter &lt;code&gt;M&lt;/code&gt; of word &lt;code&gt;N&lt;/code&gt;), but I see no indication of what text this would refer to, and I consider it unlikely since there are only eight such symbols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One final note: The fact that &quot;Apple, blueberry, chocolate...&quot; in the pie pirate's ramblings are names of pies starting with the letters A, B, C in sequence is almost certainly significant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully these observations will be of some help to another puzzler.&lt;/p&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2015-07-01T18:46:30.607" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17259" PostTypeId="5" CreationDate="2015-07-01T19:18:36.733" Score="0" Body="&lt;p&gt;Puzzles where deducing the solution involves the analysis of sounds. For example, a puzzle might ask you to uncover a hidden message in the audio clip either by listening to it, or performing signal processing, such as spectrograms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some but not all questions with this tag also fall under the category of &lt;a href=&quot;/questions/tagged/music&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;music&amp;#39;&quot; rel=&quot;tag&quot;&gt;music&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="52483" LastEditDate="2018-11-27T05:38:26.650" LastActivityDate="2018-11-27T05:38:26.650" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="17260" PostTypeId="4" CreationDate="2015-07-01T19:18:36.733" Score="0" Body="Puzzles based on the analysis of an audio clip" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-07-01T19:44:24.697" LastActivityDate="2015-07-01T19:44:24.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17263" PostTypeId="1" AcceptedAnswerId="17269" CreationDate="2015-07-01T20:32:52.770" Score="27" ViewCount="1058" Body="&lt;p&gt;In the land of Puzz, all the days are strange&lt;br&gt;&#xA;The sounds repeat, but with one wee change&lt;br&gt;&#xA;A troll trail hither and a dwarf wharf yon&lt;br&gt;&#xA;A wee hairy fairy or a fat fun faun&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Won't you come walking through the land with me&lt;br&gt;&#xA;And fill in all the blanks with the things we'll see&lt;br&gt;&#xA;Then take the circled letters as a brand new clue&lt;br&gt;&#xA;Solve that final riddle; there's some rep for you&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First we see a werewolf with a tongue so neat&lt;br&gt;&#xA;That he uses it to sample all the foods he'll eat&lt;br&gt;&#xA;If you're a tasty morsel, better not go hikin'&lt;br&gt;&#xA;Or you might get eaten by a hungry &lt;code&gt;_ (_) _ _ _ _'   _ _ _ _ _&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice the ravine as we follow this wall&lt;br&gt;&#xA;For that deep &lt;code&gt;_ _ _ _ _   (_) _ _ _ _&lt;/code&gt; is where green hags fall&lt;br&gt;&#xA;And best avoid the clearing where the green oaf mills&lt;br&gt;&#xA;For his giant &lt;code&gt;(_) _ _ _   _ _ _ _ (_)&lt;/code&gt; is the greatest of drills&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now behold a beast in the western sky&lt;br&gt;&#xA;It's furry and it's feathered and it loves to fly&lt;br&gt;&#xA;But its favourite thing of all is a function plot&lt;br&gt;&#xA;Lo, it's a &lt;code&gt;_ _ (_) _ _ _ _'  _ _ _ _ _ _ (_)&lt;/code&gt; that your eye doth spot&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now over in that hut are a wee green few&lt;br&gt;&#xA;They're called the &lt;code&gt;_ _ (_) _ _ _ _'   (_) _ _ _ _ _ _&lt;/code&gt; and they all make shoes&lt;br&gt;&#xA;And the many goaty fellows in the field ahead&lt;br&gt;&#xA;Are the sexy &lt;code&gt;_ (_) _ _ _   _ _ _ _ _ _ _&lt;/code&gt; that are fit to wed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Behold the next beast on the eastern flat&lt;br&gt;&#xA;A little bit of this and a little bit of that&lt;br&gt;&#xA;But all three parts think pictures are a hoot&lt;br&gt;&#xA;The &lt;code&gt;_ _ _ _ (_) _   _ _ _ _ _ _ _&lt;/code&gt; has some photos to shoot&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That ugly furry fellow on his brown hind legs?&lt;br&gt;&#xA;He's a &lt;code&gt;_ _ _ _ _   (_) _ _ _ _&lt;/code&gt;, but his humour is the dregs&lt;br&gt;&#xA;And the faint white spirit there never plays the host&lt;br&gt;&#xA;Whenever she's invited, she's a true &lt;code&gt;(_) _ _ _ _   _ _ _ _ _&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we wind down, mind that tree-like fellow&lt;br&gt;&#xA;He's got a nasty bark though he may seem mellow&lt;br&gt;&#xA;His father has a sister and she's really quite rude&lt;br&gt;&#xA;Blame his &lt;code&gt;_ _ _   _ _ (_) _&lt;/code&gt; for the old sot's mood&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That concludes our tour through the land of Puzz&lt;br&gt;&#xA;But all tours have a purpose, just as this one does&lt;br&gt;&#xA;List the circled letters and then rearrange the list&lt;br&gt;&#xA;&lt;strong&gt;To name the boastful fellow on the tour we missed&lt;/strong&gt;&lt;br&gt;&lt;br&gt;&lt;br&gt;&#xA;&lt;em&gt;Please include answers to the ten clues with your solution.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2015-07-03T10:09:47.097" Title="The Wonderful Wanderfull Puzz" Tags="|word|rhyme|wordplay|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17264" PostTypeId="2" ParentId="17222" CreationDate="2015-07-01T20:47:57.513" Score="6" Body="&lt;p&gt;Your friend sent you text in the form of the 7-bit GSM Alphabet, which is how the text in an SMS message is stored. When &lt;a href=&quot;http://www.twit88.com/home/utility/sms-pdu-encode-decode&quot; rel=&quot;nofollow&quot;&gt;decoded&lt;/a&gt;, the text will be:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Use GSM 7-bit encoding&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13733" LastEditorUserId="13733" LastEditDate="2015-07-01T23:23:49.807" LastActivityDate="2015-07-01T23:23:49.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17265" PostTypeId="2" ParentId="17222" CreationDate="2015-07-01T20:54:02.717" Score="2" Body="&lt;p&gt;Decoded it is - &lt;code&gt;Use GSM 7-bit encoding&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the USSD GSM 7bit packed encoding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used &lt;a href=&quot;http://smstools3.kekekasvi.com/topic.php?id=288&quot; rel=&quot;nofollow&quot;&gt;this tool&lt;/a&gt;. Given the original message was 22 characters (above answer) we need to convert the long-form from the question (40 characters) to the SMS (short message service) text.&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2015-07-01T20:54:02.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17266" PostTypeId="1" AcceptedAnswerId="17267" CreationDate="2015-07-01T21:19:06.727" Score="10" ViewCount="1043" Body="&lt;p&gt;It’s obvious that one of my “friends” was lying to me back in March this year. Someone who knew I’d be gone stole my $1,500 laptop and my crappy old broken microwave (I probably won’t ask for that one back).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I dug up this old chat log, hopefully you can tell me who’s full of it and who isn’t. You can assume only one person is/is aiding the thief, and everyone has perfect memory. Multiple backstabbers might just be too much for my fragile heart to handle.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;4:11 PM &amp;lt;Curmudgeon&amp;gt; opened chat.&#xA;4:11 PM &amp;lt;Curmudgeon&amp;gt; joined chat.&#xA;4:11 PM &amp;lt;Alice&amp;gt; joined chat.&#xA;4:12 PM &amp;lt;Curmudgeon&amp;gt;: hey alice&#xA;4:12 PM &amp;lt;Alice&amp;gt;: hey, just back from your month in europe?&#xA;4:12 PM &amp;lt;Curmudgeoon&amp;gt;: a week ago, yeah, glad to be back in 'murica&#xA;4:12 PM &amp;lt;ST_PATTYS_DAY&amp;gt; joined chat.&#xA;4:13 PM &amp;lt;Alice&amp;gt;: sam, st patricks is tomorrow&#xA;4:13 PM &amp;lt;ST_PATTYS_DAY&amp;gt;: no **** alice im getting stoked early&#xA;4:14 PM &amp;lt;ST_PATTYS_DAY&amp;gt;: dude why we here anyway&#xA;4:15 PM &amp;lt;Curmudgeon&amp;gt;: Wait until everyone gets here.&#xA;4:16 PM &amp;lt;snek&amp;gt; joined chat.&#xA;4:16 PM &amp;lt;snek&amp;gt;: ssss&#xA;4:17 PM &amp;lt;Alice&amp;gt;: hey scott how was europe with curmudgeon&#xA;4:18 PM &amp;lt;snek&amp;gt;: o didnt you hear? i had to stay home sick&#xA;4:18 PM &amp;lt;snek&amp;gt;: got ******* food poisoning like 2 days before the flight&#xA;4:18 PM &amp;lt;ST_PATTYS_DAY&amp;gt;: sucks to suck&#xA;4:19 PM &amp;lt;ST_PATTYS_DAY&amp;gt;: wheres john?&#xA;4:19 PM &amp;lt;snek&amp;gt;: said he’d be late, had a meeting or some ****&#xA;4:24 PM &amp;lt;Johnny&amp;gt; joined chat.&#xA;4:25 PM &amp;lt;ST_PATTYS_DAY&amp;gt;: there he is, hurry up so my gf can paint me green&#xA;4:26 PM &amp;lt;Curmudgeon&amp;gt;: okay guys, this might be a little uncomfortable, but I need you to answer honestly. I’m not going to sue or anything&#xA;4:26 PM &amp;lt;Johnny&amp;gt;: **** whats going on?&#xA;4:26 PM &amp;lt;snek&amp;gt;: hush up and listen john&#xA;4:27 PM &amp;lt;Curmudgeon&amp;gt;: exactly one month ago, someone who knew my garage code got into my house and stole my laptop and microwave.&#xA;4:27 PM &amp;lt;Alice&amp;gt;: you seriously think any of us would do that? wtf&#xA;4:28 PM &amp;lt;Curmudgeon&amp;gt;: I wouldn’t be doing this if there weren’t a simple solution. since the culprit conveniently knocked over and broke my clock, which still displayed the date and time when I got home. I want all of you to tell me where you were at noon a month ago.&#xA;4:29 PM &amp;lt;Johnny&amp;gt;: were supposed to remember? i dont remember where i was 5 hours ago man&#xA;4:29 PM &amp;lt;Curmudgeon&amp;gt; Well thankfully, that was the day I called you all to see how you were doing, so you should have no trouble remembering.&#xA;4:30 PM &amp;lt;snek&amp;gt;: oh right, thats easy. i was still weak from the food poisoning and you called when i was mustering up the energy to go visit my man johnny&#xA;4:30 PM &amp;lt;snek&amp;gt;: i was probably at his house by noon&#xA;4:31 PM &amp;lt;Johnny&amp;gt;: you were, cause dude called me right before you got there and i told him to shove off&#xA;4:31 PM &amp;lt;Curmudgeon&amp;gt;: can confirm, was told to shove off&#xA;4:31 PM &amp;lt;Alice&amp;gt;: you called me while i was driving.&#xA;4:31 PM &amp;lt;Curmudgeon&amp;gt;: yeah, i remember the roaring of the car when i called actually, it was hard to hear you&#xA;4:32 PM &amp;lt;Alice&amp;gt;: was definitely still out around noon, since i grabbed a bite to eat&#xA;4:32 PM &amp;lt;Alice&amp;gt;: i woke up, got ready, visited my grandma on one side of town, then back across to the post office to get stamps and mail a birthday card&#xA;4:32 PM &amp;lt;Curmudgeon&amp;gt;: thanks. what about you sam?&#xA;4:33 PM &amp;lt;ST_PATTYS_DAY&amp;gt;: your call actually woke me up, after noon. i was asleep, man&#xA;4:33 PM &amp;lt;Curmudgeon&amp;gt;: oh yeah, i remember that actually. sorry about that&#xA;4:34 PM &amp;lt;Johnny&amp;gt;: hey man, i saw sam that day around, walking down our street with his gf! dont bs me&#xA;4:34 PM &amp;lt;ST_PATTYS_DAY&amp;gt;: even if thats true, we dont live anywhere near curmudgeon, which means it couldnt have been me&#xA;4:35 PM &amp;lt;ST_PATTYS_DAY&amp;gt;: and while were on the topic, i know you werent with scott&#xA;4:35 PM &amp;lt;snek&amp;gt;: shut up sam&#xA;4:36 PM &amp;lt;ST_PATTYS_DAY&amp;gt;: he was with my gfs friend becca, probably stealing your microwave&#xA;4:37 PM &amp;lt;Johnny&amp;gt;: she doesnt live anywhere near him either and you know it&#xA;4:37 PM &amp;lt;snek&amp;gt;: fine sam, i made it up since my real alibi is so stupid. i was sound asleep at noon, i conked out right after i hung up with curmudgeon&#xA;4:38 PM &amp;lt;Johnny&amp;gt;: ya sorry, wouldnt have lied but i had to cover for this bumbling idiot&#xA;4:39 PM &amp;lt;ST_PATTYS_DAY&amp;gt;: idiots&#xA;4:39 PM &amp;lt;Curmudgeon&amp;gt;: no, no guys, it’s okay, i believe all of you. was probably just somebody who watched me punch in the code sometime, it’s no big deal&#xA;4:40 PM &amp;lt;Curmudgeon&amp;gt;: see you guys&#xA;4:40 PM &amp;lt;Curmudgeon&amp;gt; closed chat.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So maybe I’m a bit of an overbearing friend, but can you tell me if it paid off? Someone must have a contradiction in their story. Who’s lying? (There is a definitive answer, so don’t be too quick to flag this as &quot;too broad&quot; ye-who-likes-to-flag!)&lt;/p&gt;&#xA;" OwnerUserId="13556" LastEditorUserId="13556" LastEditDate="2015-07-01T21:25:04.557" LastActivityDate="2015-07-01T21:55:15.143" Title="Who stole my laptop?" Tags="|story|liars|mystery|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17267" PostTypeId="2" ParentId="17266" CreationDate="2015-07-01T21:31:59.857" Score="11" Body="&lt;p&gt;The liar is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Alice&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;She said she went to the post office. However, we know the chat conversation takes place the day before St Patrick's Day 2015, which means the 16th of March. The theft (and phone calls) occurred exactly 1 month previously, on Monday, February 16th. That was President's Day, which is a national holiday, so the post office would have been closed. Alice couldn't have bought stamps, so she must be lying.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10861" LastEditorUserId="10861" LastEditDate="2015-07-01T21:55:15.143" LastActivityDate="2015-07-01T21:55:15.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17268" PostTypeId="2" ParentId="17263" CreationDate="2015-07-01T21:32:41.513" Score="2" Body="&lt;h2&gt;Partial answer&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;First we see a werewolf with a tongue so neat&lt;br&gt;&#xA;That he uses it to sample all the foods he'll eat&lt;br&gt;&#xA;If you're a tasty morsel, better not go hikin'&lt;br&gt;&#xA;Or you might get eaten by a hungry &lt;code&gt;_ (_) _ _ _ _'   _ _ _ _ _&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think this is a VIKING, since it needs to rhyme with hiking. Not quite sure on the two words though, since the first is 6 letters with an apostrophe and the second is 5 letters with no apostrophe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Notice the ravine as we follow this wall&lt;br&gt;&#xA;For that deep &lt;code&gt;_ _ _ _ _   (_) _ _ _ _&lt;/code&gt; is where green hags fall&lt;br&gt;&#xA;And best avoid the clearing where the green oaf mills&lt;br&gt;&#xA;For his giant &lt;code&gt;(_) _ _ _   _ _ _ _ (_)&lt;/code&gt; is the greatest of drills&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure about either of these clues.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now behold a beast in the western sky&lt;br&gt;&#xA;It's furry and it's feathered and it loves to fly&lt;br&gt;&#xA;But its favourite thing of all is a function plot&lt;br&gt;&#xA;Lo, it's a &lt;code&gt;_ _ (_) _ _ _ _'  _ _ _ _ _ _ (_)&lt;/code&gt; that your eye doth spot&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think this is a GRIFFIN/GRYPHON (it has fur, feathers, wings, and in at least one fictional work they're often mathematicians). So the two words could be &quot;spiffin' griffin&quot;, making the letters &lt;strong&gt;IN&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now over in that hut are a wee green few&lt;br&gt;&#xA;They're called the &lt;code&gt;_ _ (_) _ _ _ _'   (_) _ _ _ _ _ _&lt;/code&gt; and they all make shoes&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LEPRECHAUNS are green and said to make shoes. But there must be another name for them, since neither of those words are 10-11 letters long.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the many goaty fellows in the field ahead&lt;br&gt;&#xA;Are the sexy &lt;code&gt;_ (_) _ _ _   _ _ _ _ _ _ _&lt;/code&gt; that are fit to wed&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SATYRS are half-goat and supposedly sexy. So maybe the first word is &quot;satyr&quot;, which means the missing letter is &lt;strong&gt;A&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Behold the next beast on the eastern flat&lt;br&gt;&#xA;A little bit of this and a little bit of that&lt;br&gt;&#xA;But all three parts think pictures are a hoot&lt;br&gt;&#xA;The &lt;code&gt;_ _ _ _ (_) _   _ _ _ _ _ _ _&lt;/code&gt; has some photos to shoot&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This could be a CHIMERA/CHIMAERA, which has parts of three different animals (although I'm not sure about the photos bit), so the second word is &quot;chimera&quot; but I'm not sure about the first.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That ugly furry fellow on his brown hind legs?&lt;br&gt;&#xA;He's a &lt;code&gt;_ _ _ _ _   (_) _ _ _ _&lt;/code&gt;, but his humour is the dregs&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure about this one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the faint white spirit there never plays the host&lt;br&gt;&#xA;Whenever she's invited, she's a true &lt;code&gt;(_) _ _ _ _   _ _ _ _ _&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is surely a GHOST (faint white spirit, rhymes with host), so the second word is &quot;ghost&quot; but I'm not sure about the first.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As we wind down, mind that tree-like fellow&lt;br&gt;&#xA;He's got a nasty bark though he may seem mellow&lt;br&gt;&#xA;His father has a sister and she's really quite rude&lt;br&gt;&#xA;Blame his &lt;code&gt;_ _ _   _ _ (_) _&lt;/code&gt; for the old sot's mood&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Could be a TROLL, but neither of the words have 5 letters. An ENT maybe?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Letters gathered so far:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; INA.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-07-01T21:32:41.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17269" PostTypeId="2" ParentId="17263" CreationDate="2015-07-01T21:42:29.050" Score="21" Body="&lt;p&gt;In the land of Puzz, all the days are strange&lt;br&gt;&#xA;The sounds repeat, but with one wee change&lt;br&gt;&#xA;A troll trail hither and a dwarf wharf yon&lt;br&gt;&#xA;A wee hairy fairy or a fat fun faun&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This stanza tells us that the theme of the puzzle is two-word phrases where each word is just one sound away from the other, such as &lt;strong&gt;dwarf wharf.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Won't you come walking through the land with me&lt;br&gt;&#xA;And fill in all the blanks with the things we'll see&lt;br&gt;&#xA;Then take the circled letters as a brand new clue&lt;br&gt;&#xA;Solve that final riddle; there's some rep for you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; So let's begin solving!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First we see a werewolf with a tongue so neat&lt;br&gt;&#xA;That he uses it to sample all the foods he'll eat&lt;br&gt;&#xA;If you're a tasty morsel, better not go hikin'&lt;br&gt;&#xA;Or you might get eaten by a hungry &lt;code&gt;_ (_) _ _ _ _'   _ _ _ _ _&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; L(I)CKIN' LYCAN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Notice the ravine as we follow this wall&lt;br&gt;&#xA;For that deep &lt;code&gt;_ _ _ _ _   (_) _ _ _ _&lt;/code&gt; is where green hags fall&lt;br&gt;&#xA;And best avoid the clearing where the green oaf mills&lt;br&gt;&#xA;For his giant &lt;code&gt;(_) _ _ _   _ _ _ _ (_)&lt;/code&gt; is the greatest of drills&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; WITCH (D)ITCH&lt;br&gt;&#xA; (O)GRE AUGE(R)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now behold a beast in the western sky&lt;br&gt;&#xA;It's furry and it's feathered and it loves to fly&lt;br&gt;&#xA;But its favourite thing of all is a function plot&lt;br&gt;&#xA;Lo, it's a &lt;code&gt;_ _ (_) _ _ _ _'  _ _ _ _ _ _ (_)&lt;/code&gt; that your eye doth spot&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; GR(A)PHIN' GRIFFIN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now over in that hut are a wee green few&lt;br&gt;&#xA;They're called the &lt;code&gt;_ _ (_) _ _ _ _'   (_) _ _ _ _ _ _&lt;/code&gt; and they all make shoes&lt;br&gt;&#xA;And the many goaty fellows in the field ahead&lt;br&gt;&#xA;Are the sexy &lt;code&gt;_ (_) _ _ _   _ _ _ _ _ _ _&lt;/code&gt; that are fit to wed&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CO(B)BLIN' (G)OBLINS&lt;br&gt;&#xA; S(A)TYR SUITORS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Behold the next beast on the eastern flat&lt;br&gt;&#xA;A little bit of this and a little bit of that&lt;br&gt;&#xA;But all three parts think pictures are a hoot&lt;br&gt;&#xA;The &lt;code&gt;_ _ _ _ (_) _   _ _ _ _ _ _ _&lt;/code&gt; has some photos to shoot&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CAME(R)A CHIMERA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That ugly furry fellow on his brown hind legs?&lt;br&gt;&#xA;He's a &lt;code&gt;_ _ _ _ _   (_) _ _ _ _&lt;/code&gt;, but his humour is the dregs&lt;br&gt;&#xA;And the faint white spirit there never plays the host&lt;br&gt;&#xA;Whenever she's invited, she's a true &lt;code&gt;(_) _ _ _ _   _ _ _ _ _&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DROLL (G)NOLL&lt;br&gt;&#xA; (G)UEST GHOST&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As we wind down, mind that tree-like fellow&lt;br&gt;&#xA;He's got a nasty bark though he may seem mellow&lt;br&gt;&#xA;His father has a sister and she's really quite rude&lt;br&gt;&#xA;Blame his &lt;code&gt;_ _ _   _ _ (_) _&lt;/code&gt; for the old sot's mood&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ENT AU(N)T&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That concludes our tour through the land of Puzz&lt;br&gt;&#xA;But all tours have a purpose, just as this one does&lt;br&gt;&#xA;List the circled letters and then rearrange the list&lt;br&gt;&#xA;&lt;strong&gt;To name the boastful fellow on the tour we missed&lt;/strong&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Finally, we're left with IDORANBGARGGN, which anagrams to BRAGGIN' DRAGON.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="11475" LastEditDate="2015-07-03T10:09:47.097" LastActivityDate="2015-07-03T10:09:47.097" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17271" PostTypeId="2" ParentId="17119" CreationDate="2015-07-02T00:41:31.213" Score="17" Body="&lt;h1&gt;Update:&lt;/h1&gt;&#xA;&lt;p&gt;The aliens are using a rather nifty number system that is similar to base 3, but with digits corresponding to zero and ±1. Here's a bit of Python that will convert them to decimal form:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;def a3(s):&#xA;  r,p,dp = 0,0,0&#xA;  for c in s:&#xA;    if c not in 'x-+&amp;gt;o':&#xA;      raise LookupError(&amp;quot;Illegal symbol ('&amp;quot; + c + &amp;quot;')&amp;quot;)&#xA;    if c in 'x-+':&#xA;      r = r * 3 + 'x-+'.find(c) - 1&#xA;      p = p + dp&#xA;    if c == '&amp;gt;':&#xA;      dp = 1&#xA;  if p == 0:&#xA;    return r&#xA;  return r * 1.0 / (3**p)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;&lt;strong&gt;Examples:&lt;/strong&gt; &lt;code&gt;a3('++-x&amp;gt;')&lt;/code&gt; = 35, &lt;code&gt;a3('+&amp;gt;+--x+')&lt;/code&gt; = 1.325&lt;/p&gt;&#xA;&lt;p&gt;The numbers that intersect at a triangle enclosed in a circle are a form of scientific notation. From the example on page 5: &lt;code&gt;a3('+&amp;gt;xx+--x') * 3**a3('xx&amp;gt;')&lt;/code&gt; = &lt;code&gt;a3('&amp;gt;---+xx+--x')&lt;/code&gt; = 0.0073. Recurring digits are enclosed in rounded rectangles as shown in page 3, along with the identity &lt;code&gt;+&amp;gt;(x)&lt;/code&gt; = &lt;code&gt;&amp;gt;(+)&lt;/code&gt; (both are equal to 0.11111… in base 3, or 0.5 in decimal).&lt;/p&gt;&#xA;&lt;p&gt;Pages 4 and 5 describe the mathematical constants &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$e$&lt;/span&gt;. Page 6 shows the atomic structure of C, O, H and He, and their respective atomic masses relative to the mass of an electron. Page 7 shows a water molecule, including the bend angle of 104.45° (expressed in radians), and the distance between the H and O nuclei (in units of the Bohr radius).&lt;/p&gt;&#xA;&lt;p&gt;Page 8 seems to be describing the transmitter that sent the signal. The number on the left is equal to &lt;span class=&quot;math-container&quot;&gt;$\alpha^{-1}$&lt;/span&gt; (the reciprocal of the fine structure constant). The number at the top is perhaps the size of the transmitter in Bohr radius units.&lt;/p&gt;&#xA;&lt;p&gt;Page 9 appears to show the planets orbiting a star. I assume the numbers at the bottom are the mean orbital distances of the planets, but I have no idea what distance units are being used here. Apparently our alien friend inhabits one of the moons of a ringed gas giant.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/e6JVh.png&quot; alt=&quot;Annotated message, pages 1-9&quot; /&gt;&lt;br /&gt;&lt;a href=&quot;https://i.stack.imgur.com/e6JVh.png&quot; rel=&quot;noreferrer&quot;&gt;Click to embiggen&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The last three pages may have more information about the planets in this system, but I haven't made much progress there yet.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;Original answer:&lt;/h1&gt;&#xA;&lt;p&gt;Except for the &amp;quot;synchronization pulses&amp;quot;, each part of the signal contains two superimposed sine wave signals at different frequencies. If these two frequencies are used as the left and right inputs to an &lt;a href=&quot;https://en.wikipedia.org/wiki/Etch_A_Sketch&quot; rel=&quot;noreferrer&quot;&gt;Etch-A-Sketch&lt;/a&gt;, then they trace out some rather interesting pictures, with the synchronization pulses marking the boundaries between them. The only trouble is that a 1024-element FFT results in pictures of rather low resolution:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6kLZ3.gif&quot; alt=&quot;A little green message from a little green man?&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="40" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-02T15:31:59.423" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17272" PostTypeId="2" ParentId="17119" CreationDate="2015-07-02T03:24:59.567" Score="6" Body="&lt;p&gt;Work in progress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based off the work of Quark:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Added labels, powers of 3, what might be the solar system, and H2O. &lt;a href=&quot;https://i.stack.imgur.com/DXMvS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Full size image&lt;/a&gt; &lt;img src=&quot;https://i.stack.imgur.com/DXMvS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13737" LastActivityDate="2015-07-02T03:24:59.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17273" PostTypeId="2" ParentId="15952" CreationDate="2015-07-02T03:25:03.617" Score="1" Body="&lt;p&gt;This is the same solution as LeppyR64 and LogicianWithAHat, but I added a proof of optimality based on rand al'thor (who was working on a slightly different puzzle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phil: I'm afraid the best I can do is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 36 seconds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;David: Hmm, well that's not great.  But tell me, how do you do it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phil: Here's a schedule, light-by-light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; On from 0-6 secs.  Cooling from 0-10 secs.  On from 18-24 secs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Light 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; On from 6-12 secs.  Cooling from 10-20 secs.  On from 24-30 secs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Light 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; On from 12-18 secs.  Cooling from 20-30 secs.  On from 30-36 secs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That gets you the lighting I promised.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;David: I see.  But our prima donna won't be happy.  Maybe I should give rand al'thor's brother-in-law a call -- I'm not going to waste money on one of his lights, but maybe he can come up with a better way to use the ones I have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phil: No, actually I'm quite sure he can't do any better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;David: And how are you so confident?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phil: Think about it like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Imagine each light has six units of coolness in it.  When you turn it on, it burns the coolness at a rate of one unit / second, for six seconds.  When you work on a light, it adds coolness at a rate of .6 units / second.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Got that so far?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;David: No, wait.  That doesn't make sense.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What if you turn on a light and start working on it at the same time?  After 6 seconds it's used up the original 6 units, but you've put in another 3.6.  So in your way of thinking about it, the light still has plenty of coolness left... but what I said was, the light will turn off after 6 seconds, even if you are working on it!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Phil: I see.  Well, I can explain that...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you're fixing a light, the light can't use any of the coolness you put into it, until you have put in all 6 units.  So, let's agree that the coolness you put in while you are working on a light is only &lt;em&gt;potential&lt;/em&gt; coolness: you put in &lt;em&gt;potential&lt;/em&gt; coolness at a rate of .6 units per second.  At the end of ten seconds, there are 6 units of coolness in the light, and 6 units of potential coolness are automatically converted to actual coolness.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;David: So, in the scenario I was worried about,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After 6 seconds the light has used up all 6 units of actual coolness, and it has 3.6 units of potential coolness.  So the light will turn off, and you can't turn it back on until it has 6 units of coolness, so the potential coolness can convert to actual coolness.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Phil: Exactly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;David: So why can't you get any more time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phil:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well, let's think about the total coolness of all the lights, actual plus potential, all added together.  At start the lights are all OK, so there are 18 units of coolness.  Now whenever a light is on, the light is using 1 unit/sec, and you can only replenish .6 units/sec, so the total coolness goes down by at least .4 units/sec.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;David:  Wait... I'm confused again.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The lights use actual coolness, but you're replenishing potential coolness... and then potential coolness will convert to actual coolness...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Phil: Don't worry about that.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We're counting the total coolness, potential and actual all added together. So when potential converts to actual, it won't affect our total.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;David: I see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phil:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Anyway, to turn on a light the light has to have 6 units of coolness, so certainly you have to have at least 6 units of coolness total.  But starting at 18 units, and losing .4 units a second, you know that after 30 seconds...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;David:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After 30 seconds you have lost 12 units of coolness, so you have exactly 6 units left!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Phil:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Exactly.  Which means that &lt;em&gt;after&lt;/em&gt; the 30 second mark, you won't have enough to turn on a light.  No matter how you plan it, the latest you can turn on a light is 30 seconds... and that means the light will go out by 36.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;David: So you really did squeeze every last second!  Phil, you saved my day.&lt;/p&gt;&#xA;" OwnerUserId="12523" LastActivityDate="2015-07-02T03:25:03.617" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17274" PostTypeId="1" AcceptedAnswerId="17280" CreationDate="2015-07-02T04:10:35.617" Score="4" ViewCount="684" Body="&lt;p&gt;As you can see in the image, I've been struggling with this sudoku for quite some time and I'm completely stuck. I'm sure I'm missing one standard trick which would make it easy to solve, but I can't find it! I only mark when there are only two possibilities, so AFAIK that's the case in this sudoku.&lt;/p&gt;&#xA;&lt;p&gt;What I'm looking for is exactly how to find the next number.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SmVla.png&quot; alt=&quot;PS: I would like to make the image smaller, but I don't know how.&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="2002" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-08-20T17:22:09.650" Title="Reasoning behind next step in sudoku" Tags="|sudoku|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17275" PostTypeId="2" ParentId="17243" CreationDate="2015-07-02T04:10:39.587" Score="14" Body="&lt;p&gt;There's no bomb, the attack is canceled! We can just relay that message to our teams and relax.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We must find what the times in the given cities translate to in Lisbon, where the call was placed, using &lt;a href=&quot;http://www.timeanddate.com/time/map/&quot;&gt;the map BmyGuest provided above&lt;/a&gt;. This gives us:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;City        Zone    Lisbon Time&#xA;Paris           +2  16:30:45&#xA;Seoul           +9  10:30:00&#xA;Denver          -6  22:30:37&#xA;Tokyo           +9  07:30:07&#xA;Omsk            +6  13:30:37&#xA;Helsinki        +3  19:30:22&#xA;Lima            -5  22:30:45&#xA;Phoenix         -7  21:00:30&#xA;Dallas          -5  21:00:52&#xA;Rome            +2  15:00:37&#xA;Cairo           +2  18:00:45&#xA;Sydney          +10 06:00:52&#xA;Perth           +8  06:00:37&#xA;Gibraltar       +2  16:30:37&#xA;Oslo            +2  22:30:30&#xA;Melbourne       +10 06:00:07&#xA;Port au Prince  -4  19:30:07&#xA;Montreal        -4  18:00:37&#xA;London          +1  12:00:52&#xA;Athens          +3  10:30:00&#xA;Cape Town       +2  19:30:30&#xA;Brasilia        -3  22:30:30&#xA;Karachi         +5  07:30:00&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What do we do next? EngineerToast and rand al'thor noticed that the values of the seconds divide a clock face into eighths. Now look at the possible hour+minute combinations, specifically the position of the hour hands at those times. These positions also divide the clock face into eighths! For example, at 13:30 (Omsk), the hour hand will be precisely one-eighth of the way around the clock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next question is what to do with two positions of hands on a clock (hour hand, second hand), given that both are in positions to the nearest eighth of a clock face. These signify &lt;a href=&quot;https://en.wikipedia.org/wiki/Flag_semaphore&quot;&gt;flag semaphore&lt;/a&gt;! Treating the hour and second hand on the clock for each time as semaphore, we can read the following letters in order (i.e. 16:30:45, Paris, turns into S in semaphore): STILL NO BOMB, CANCEL ATTACK. That is the decryption, and we are done. Unless, of course, the letter is a lie meant to throw us off... But we're told we saved the day, so we're good :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: Dallas's given time should be 15:00:52, not 15:30:52.)&lt;/p&gt;&#xA;" OwnerUserId="13739" LastActivityDate="2015-07-02T04:10:39.587" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17276" PostTypeId="2" ParentId="16902" CreationDate="2015-07-02T04:24:03.357" Score="0" Body="&lt;p&gt;I have a solution where the total number of weighings is equal to the sum of the squares of the prime factors of &lt;em&gt;n&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide each n-ball heap into equal-sized piles. The number of piles should be as small as possible. Of the piles, we need to which two contain the heavy balls. For &lt;em&gt;each pair of piles&lt;/em&gt;, place them on separate pans of the balance. The third pan should be filled with normal balls. If we have both heavy balls on the scales, then they will tip, otherwise they will not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once we know which piles contain the heavy balls, we can use these for the next iteration. We repeat the process until our piles are of size 1, and we know which balls are heavy.&lt;/p&gt;&#xA;" OwnerUserId="13737" LastActivityDate="2015-07-02T04:24:03.357" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17277" PostTypeId="1" CreationDate="2015-07-02T05:44:26.977" Score="5" ViewCount="1366" Body="&lt;p&gt;John has started an Einstein's-like puzzle (&quot;$n$ people of $n$ ages prefer $n$ foods, etc.; determine which is which from the following facts&quot;), using a &lt;a href=&quot;//google.com/search?q=zebra|einstein+puzzle+matrix|grid&amp;amp;tbm=isch&quot;&gt;matrix&lt;/a&gt; for his work. He shows me his completed work, asking whether he's made any errors. There are no &lt;em&gt;obvious&lt;/em&gt; errors (two accepted boxes in the same set, or all boxes in a set eliminated). I know of two ways to determine whether he's erred:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Continue solving the puzzle (without error) where John has left off, and see whether I arrive at an impossible state. If so, John erred. If not, I'll complete the puzzle, and John hasn't erred.&lt;/li&gt;&#xA;&lt;li&gt;Start the puzzle over again and solve it (without error) until either (a) I have eliminated every square John has, and accepted every square he has, in which case he hasn't erred; or (b) I have accepted some square John has eliminated or vice versa, in which case John has erred.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Is there another way to determine whether John has erred? Specifically, is there a way that doesn't involve redoing or continuing the solution, but, rather, involves merely examining what John has done?&lt;/p&gt;&#xA;" OwnerUserId="93" LastActivityDate="2015-09-14T03:22:44.073" Title="How to check a partial Einstein's-like-puzzle solution" Tags="|logic-grid|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17278" PostTypeId="1" AcceptedAnswerId="17279" CreationDate="2015-07-02T05:45:31.477" Score="5" ViewCount="1661" Body="&lt;p&gt;John has started a sudoku puzzle and shows me his completed work, asking whether he's made any errors. There are no &lt;em&gt;obvious&lt;/em&gt; errors (a number repeated in a row or subgrid). I know of two ways to determine whether he's erred:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Continue solving the puzzle (without error) where John has left off, and see whether I arrive at an impossible state. If so, John erred. If not, I'll complete the puzzle, and John hasn't erred.&lt;/li&gt;&#xA;&lt;li&gt;Start the puzzle over again and solve it (without error) until either (a) I have filled in every square John has, and with the same numbers, in which case John hasn't erred; or (b) I have filled in some square John has, but with a different number, in which case John has erred (assuming that the puzzle has a unique solution).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Is there another way to determine whether John has erred? Specifically, is there a way that doesn't involve redoing or continuing the solution, but, rather, involves merely examining what John has done?&lt;/p&gt;&#xA;" OwnerUserId="93" LastEditorUserId="93" LastEditDate="2015-07-02T12:50:07.803" LastActivityDate="2023-05-17T09:18:17.913" Title="How to check a partial sudoku solution" Tags="|sudoku|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17279" PostTypeId="2" ParentId="17278" CreationDate="2015-07-02T05:59:43.000" Score="7" Body="&lt;p&gt;Unfortunately, no such (computationally reasonable) method currently exists that could make a 100% guarantee. I don't have a proof for its nonexistence, as there's no mathematical proof that it &lt;em&gt;can't&lt;/em&gt; exist (yet), but no such method has yet been discovered. The only way we currently know to verify a Sudoku has no errors is by solving it and checking for impossibilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To reason why, consider that if such a method existed, Sudoku solvers wouldn't implement the deductive methods that they currently employ - they would simply need to guess a number then check to see if it's right. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(Slightly) more formally, the proposition you'd be trying to prove is: &quot;The sudoku is solvable.&quot; There are really only two approaches you could take to prove that it's solvable:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Proof by contradiction:&lt;/strong&gt; &quot;The sudoku can't be solvable because it leads to an impossible state.&quot;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Direct proof:&lt;/strong&gt; &quot;By [some mathematical process], the Sudoku can be classified as &quot;solvable&quot; or 'unsolvable.'&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The first can always be executed - it's just guess and check until you exhaust all possibilities or encounter a valid solution. The second is a yet-unsolved unsolved mathematical problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I add computationally feasible above is because there exist methods that effectively read &quot;generate all possible Sudoku puzzles, then check to see if this one matches any of them.&quot; Obviously, this isn't reasonable to do.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There are, however, a couple trivial checks you could make to &lt;em&gt;possibly&lt;/em&gt; find errors. They're not guaranteed to work, but they have the potential to pick up errors.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You can check to make sure that there are no immediate row/column/box conflicts. If there are, you know there's an error immediately.&lt;/li&gt;&#xA;&lt;li&gt;You can check the possibilities for each cell. If there's any cell without any open possibilities, then there's an error somewhere.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;These two can occasionally catch errors quickly, or require only a few deductive steps for you to spot an error. That being said, they're not guaranteed.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorDisplayName="user20" LastEditDate="2015-07-02T06:09:55.063" LastActivityDate="2015-07-02T06:09:55.063" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17280" PostTypeId="2" ParentId="17274" CreationDate="2015-07-02T06:10:43.953" Score="5" Body="&lt;p&gt;You are missing what are known as &lt;a href=&quot;http://www.intosudoku.com/Doc/PointingPair.html&quot;&gt;pointing pairs&lt;/a&gt;. If you're not sure where to look, here's a hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you need two of these in the same box&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13739" LastActivityDate="2015-07-02T06:10:43.953" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17281" PostTypeId="2" ParentId="17278" CreationDate="2015-07-02T07:12:57.317" Score="0" Body="&lt;p&gt;I would like to point that you are suppossing that there is only ONE solution for that sudoku. I know it is supposed that there is only one way to solve a sudoku but there is a chance that changing some numbers you can get into another valid solution (And I said that because I used to do a lot of sudokus and it happened to me too). You said:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;2&quot;&gt;&#xA;  &lt;li&gt;Start the puzzle over again and solve it (without error) until &#xA;  either (a) I have filled in every square John has, and with the same &#xA;  numbers, in which case John hasn't erred; or (b) I have filled in some &#xA;  square John has, but with a different number, in which case John has erred.&quot;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I think that getting another solution does not mean that John was erred. It means that maybe he was erred but maybe you just found another valid solution so the only way to know if John was erred with HIS start, is finishing John's sudoku start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said this, as a sudoku is a mathematical/logic problem, if there is a way to find the solution without having to solve it, it has to be with probability but I can't tell the way as I don't have that knowledge.&lt;/p&gt;&#xA;" OwnerUserId="13061" LastEditorUserId="68" LastEditDate="2015-09-09T09:22:07.317" LastActivityDate="2015-09-09T09:22:07.317" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17282" PostTypeId="1" CreationDate="2015-07-02T07:20:09.280" Score="12" ViewCount="567" Body="&lt;p&gt;I've got a friend who's a real great guy!&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;He loves a drink.&lt;/li&gt;&#xA;&lt;li&gt;He'll help you change your car tyres.&lt;/li&gt;&#xA;&lt;li&gt;This one time, he met an alien and got a coat out of it.&lt;/li&gt;&#xA;&lt;li&gt;This other time, he got talking to a space sorcerer-monk and they were discussing acting award results - they seemed really keen on Laurence Olivier 1983, BAFTA 1976, Emmy 1989, Emmy 2001 &amp;amp; Laurence Olivier 2009.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;He can be a bit risky to be aroundthough ...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Any time you greet him, he tends to try and take over.&lt;/li&gt;&#xA;&lt;li&gt;He can a be racist sometimes - he once said if he were from Africa, he was sure he'd be gambling all the time.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Who's my friend?&lt;/p&gt;&#xA;" OwnerUserId="13744" LastEditorUserId="13744" LastEditDate="2015-07-03T07:00:39.200" LastActivityDate="2015-07-03T11:37:54.757" Title="Who's my friend?" Tags="|riddle|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17283" PostTypeId="2" ParentId="17282" CreationDate="2015-07-02T07:44:07.973" Score="6" Body="&lt;p&gt;I think it's &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jack&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The drink is Jack Daniels.&lt;br&gt;&#xA; The tool to lift a car is called jack.&lt;br&gt;&#xA; The man is probably Jack Harkness from Doctor Who.&lt;br&gt;&#xA; The fourth line is probably a reference to some films. I haven't found the movies yet, but I know that in 2001 there was a &quot;Jack the ripper&quot; movie.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11454" LastActivityDate="2015-07-02T07:44:07.973" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17284" PostTypeId="2" ParentId="17196" CreationDate="2015-07-02T08:20:46.390" Score="3" Body="&lt;p&gt;I think ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It has something to do with how some words are preferred in American English vs British English. Aunt Sally would use 'rat' (list word) while Uncle Joe would use 'mouse'. Likewise, Uncle Joe would say 'gift' and 'boot' (list words) while Aunt Sally would say 'present' and 'shoe'. For the first hint, it is not actually the same word on both lists, because Uncle Joe would use 'aeroplane' while Aunt Sally would use 'airplane'.&lt;br&gt;&lt;br&gt;By this logic, Uncle Joe would have 'listen' on his list, because the British prefer to use the phrase 'listen to me' while the Americans prefer 'hear me out' for the same activity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am not entirely sure about my answer because I only managed to explain a few words, but it fits very well with the hint, so I thought I'd give it a shot.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-07-02T08:20:46.390" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17285" PostTypeId="1" AcceptedAnswerId="17286" CreationDate="2015-07-02T08:42:40.613" Score="0" ViewCount="10778" Body="&lt;p&gt;Suppose you are given two jugs: a 4-gallon one and 3-gallon one, and neither has any markings on it.&lt;br&gt;&#xA;You have a pump that can be used to fill the jugs with water.&lt;br&gt;&#xA;What do you do so that exactly 2 gallons of water are left in the 4 gallon jug?&lt;/p&gt;&#xA;" OwnerUserId="11343" LastEditorUserId="4050" LastEditDate="2016-01-02T12:24:46.393" LastActivityDate="2016-01-02T12:24:46.393" Title="A Water Jug Problem" Tags="|calculation-puzzle|decanting-problems|" AnswerCount="3" CommentCount="0" ClosedDate="2016-01-02T22:01:15.233" ContentLicense="CC BY-SA 3.0" />
  <row Id="17286" PostTypeId="2" ParentId="17285" CreationDate="2015-07-02T08:48:15.053" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fill the 3 gallon jug, drop to the 4 gallon. Refill the 3 gallon and use it to top the 4 gallon jug. You'll have 2 gallons on the 3 gallon jug. Empty the 4 gallon and move the water in the 3 gallon to the 4 gallon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12223" LastActivityDate="2015-07-02T08:48:15.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17287" PostTypeId="2" ParentId="17218" CreationDate="2015-07-02T08:50:30.137" Score="2" Body="&lt;p&gt;Another association:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might be reference to your life (like the &lt;a href=&quot;https://en.wikipedia.org/wiki/Sphinx#The_Riddle_of_the_Sphinx&quot; rel=&quot;nofollow&quot;&gt;Riddle of the Sphinx&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Age&lt;/strong&gt; you are born and from now on it is increasing continuously, you are getting older, you are aging&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Date&lt;/strong&gt; you have your first love, you date&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Drill&lt;/strong&gt; you are conscripted (in some countries) or go to work, you have some duties you have to follow&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Go&lt;/strong&gt; you leave your parents, or alternatively you get retired&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hole&lt;/strong&gt; you die and are put in a hole in soil, on cemetery, you get buried&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What might be wrong and does not match is that the last word is not a verb. However, this is not a verb intentionally, because this is the moment when you are not able to perform any action (see also the 3rd meaning on &lt;a href=&quot;https://en.wiktionary.org/wiki/hole#Verb&quot; rel=&quot;nofollow&quot;&gt;the Wiktionary&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="10880" LastActivityDate="2015-07-02T08:50:30.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17288" PostTypeId="2" ParentId="17285" CreationDate="2015-07-02T08:50:32.887" Score="2" Body="&lt;ol&gt;&#xA;&lt;li&gt;Fill 3 gallons of water into 3 Gallon jug. (3G Jug=3G water, 4G Jug=0G Water).&lt;/li&gt;&#xA;&lt;li&gt;Pour it into 4 gallon Jug. (3G Jug=0G water, 4G Jug=3G Water).&lt;/li&gt;&#xA;&lt;li&gt;Fill Again 3 gallon of water into 3G jug. (3G Jug=3G water, 4G Jug=3G Water).&lt;/li&gt;&#xA;&lt;li&gt;Now put this 3 gallon water of 3 gallon jug into the 4 gallon jug until it fill completely. As 4 gallon jug has only a gallon water space in it. so 3 gallon jug now contains exactly 2 gallons of water.&lt;/li&gt;&#xA;&lt;li&gt;Drop out 4 gallons water in 4 gallons jug.&lt;/li&gt;&#xA;&lt;li&gt;Pour 2 gallons water into 4 gallon jug from 3 gallon jug.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2307" LastActivityDate="2015-07-02T08:50:32.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17290" PostTypeId="2" ParentId="17285" CreationDate="2015-07-02T09:13:37.457" Score="3" Body="&lt;p&gt;Fill J4             J4=4 J3=0&#xA;Fill J3 using J4    J4=1 J3=3&#xA;empty J3            J4=1 J3=0&#xA;pour J4 in J3       J4=0 J3=1 &#xA;Fill J4             J4=4 J3=1&#xA;Fill J3 using J4    J4=2 J3=3&lt;/p&gt;&#xA;" OwnerUserId="13439" LastActivityDate="2015-07-02T09:13:37.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17291" PostTypeId="2" ParentId="17155" CreationDate="2015-07-02T10:04:39.140" Score="8" Body="&lt;p&gt;I guess you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;U2 or Bono&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Part one refers to songs:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Search, searching, still searching am I,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Still havent found what i am looking for&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For those that on that one day did die.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Bloody Sunday&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We don't even know the name of the place,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Where the Streets have no name&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm going there, even if you can't keep pace.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; With our without you?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thx to Player one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;part 2 refer to Album titles&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the world out there, some things are pure,&#xA;No flowers or vines, but named for sure.&#xA;I took it apart, and saved so many lives,&#xA;In the fight that left so many husbandless wives. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Songs of Innocence, The Joshua Tree, How to Dismantle an Atomic Bomb, War&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;thx to aioros:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;part 3 seems to refer to their persona&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most really like me, except when I was given.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I tune fiasco when record was given for free&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Still trying to decide if I've been forgiven.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;could also refer to this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More thoughts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;close call could refer to a nearly plane crash they had some time ago...&lt;/p&gt;&#xA;" OwnerUserId="13748" LastEditorUserId="13748" LastEditDate="2015-07-02T11:45:51.243" LastActivityDate="2015-07-02T11:45:51.243" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17293" PostTypeId="1" AcceptedAnswerId="17298" CreationDate="2015-07-02T11:36:03.447" Score="8" ViewCount="778" Body="&lt;p&gt;Browsing the geometry section in the local library you come across a flimsy book, seemingly misplaced:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;The Pretzel Cookbook: &lt;em&gt;Five More Patterns Using The Spoke Method&lt;/em&gt;&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;What's more, it's been ripped in half. Flicking through the remaining three pages, you ask yourself:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How are these pretzels formed?&lt;/li&gt;&#xA;&lt;li&gt;What was on the three missing pages?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Page 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Page 1 — Contents&lt;/p&gt;&#xA;&lt;p&gt;Page 2 — The &amp;quot;Five&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;Page 3 — The &amp;quot;Minus a Half&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;Page 4 — The &amp;quot;One&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;Page 5 — The &amp;quot;Two&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;Page 6 — The &amp;quot;Minus Five&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Page 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PSdRk.png&quot; alt=&quot;Page 2&quot; title=&quot;Five&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Page 3:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4xIhL.png&quot; alt=&quot;Page 3&quot; title=&quot;Minus a Half&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;em&gt;Hint&lt;/em&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The sequence of the pages isn't important&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;Hint&lt;/em&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; You could solve this with drawing equipment&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4534" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-02T12:18:43.240" Title="Pretzel recipe with a twist" Tags="|geometry|enigmatic-puzzle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17294" PostTypeId="2" ParentId="17218" CreationDate="2015-07-02T11:36:07.950" Score="4" Body="&lt;p&gt;The common link &lt;strong&gt;between&lt;/strong&gt; the words &quot;age&lt;b&gt;,&lt;/b&gt; date&lt;b&gt;,&lt;/b&gt; drill&lt;b&gt;,&lt;/b&gt; go&lt;b&gt;,&lt;/b&gt; hole&quot;&lt;br/&gt;&#xA;is, quite clearly, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;, &quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13753" LastActivityDate="2015-07-02T11:36:07.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17295" PostTypeId="1" CreationDate="2015-07-02T11:38:19.280" Score="-3" ViewCount="718" Body="&lt;p&gt;I was given this puzzle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Eight friends (Romil, Ramesh, Rakesh, Rohit, Rahul, Abhijeet, Abhishek and Anil) are sitting around a circular table, but not necessarily in the same order.&#xA;  Four of them are facing inside and the other four are facing outside.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;All eight friends belong to eight different cities: Bhopal, Patna, Kolkata, Delhi, Gwalior, Bengluru, Chennai and Rajkot - but not necessarily in the same order.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Rohit belongs to Kolkata and faces the person who belongs to Bengluru.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can I interpret from the final sentence that both Rohit and the Bangalore guy are facing towards the centre?&lt;/p&gt;&#xA;" OwnerUserId="13751" LastEditorUserId="12628" LastEditDate="2015-07-02T13:57:14.407" LastActivityDate="2015-07-02T13:57:14.407" Title="Circular arrangement Puzzle" Tags="|logical-deduction|logic-grid|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17296" PostTypeId="2" ParentId="17295" CreationDate="2015-07-02T11:47:15.130" Score="1" Body="&lt;p&gt;This is an Einstein's puzzle. The answer to your question is no, we cannot definitively make the statement that both Rohit and the Bangalore guy are facing inward. To be able to make such a statement, we need more information about the other friends involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, let's assume Ramesh is from Bangalore and Romil, Rakesh, Abhijeet are also facing inward apart from Rohit. This would mean Ramesh is facing outward, since there are already four people facing inward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A counter example, let's assume Ramesh is from Bangalore and Romil, Rakesh, Abhijeet, Rahul are facing outward. This would mean that Ramesh cannot be facing outward as that would violate the condition that half the friends are facing inward. Hence, in this scenario Ramesh is facing inward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both the above assumptions are valid, considering the information provided. Hence, we cannot definitively determine which direction the Bangalore guy is facing. To be able to make such a statement definitively, we need some more information regarding the friends' positions, the direction some of them are facing, their cities or their positions relative to each other.&lt;/p&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-07-02T12:01:00.123" LastActivityDate="2015-07-02T12:01:00.123" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17298" PostTypeId="2" ParentId="17293" CreationDate="2015-07-02T12:18:43.240" Score="7" Body="&lt;p&gt;I think the pretzels are formed by drawing plots of &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the trochoids $x=r\phi + A\sin(\phi)$, $y=r + A\cos(\phi)$ with $r=1$, $\phi\in [0,4\pi]$ and $A$ the value of the page title. The word 'spoke' gave the solution away for me, we can get these shapes by sticking a light to a bike's spoke, and following this light.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That means the pretzels on the missing pages are:&lt;br&gt;&#xA;Page 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/AGZNa.png&quot; alt=&quot;Page 4&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Page 5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/hxsQo.png&quot; alt=&quot;Page 5&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Page 6:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/IPACn.png&quot; alt=&quot;Page 6&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13268" LastActivityDate="2015-07-02T12:18:43.240" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17301" PostTypeId="2" ParentId="17230" CreationDate="2015-07-02T13:15:31.047" Score="7" Body="&lt;p&gt;&lt;strong&gt;Not possible&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; According to &lt;em&gt;graph theory&lt;/em&gt; it is the smallest non plannar graph with minimum number of lines.&lt;br&gt;&#xA; &lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/f/f3/Biclique_K_3_3.svg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In graph theory, &lt;a href=&quot;https://en.wikipedia.org/wiki/Planar_graph&quot;&gt;a planar graph&lt;/a&gt; is a graph that can be embedded in the plane. This graph is commonly called as &lt;strong&gt;Utility graph&lt;/strong&gt; or $K_{3,3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a lot of interesting problems which can solve(or can prove not solvable :p ) using Graph theory.&lt;/p&gt;&#xA;" OwnerUserId="13582" LastActivityDate="2015-07-02T13:15:31.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17302" PostTypeId="2" ParentId="17207" CreationDate="2015-07-02T14:00:03.500" Score="7" Body="&lt;p&gt;Another step forward here, I think. Working off of COTO's answer using the same arbitrary characters, but doing &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;division  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I found that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A = 9/6 = 1 1/2 = 5/4&lt;br&gt;&#xA; \ = 2/8 = 1/4&lt;br&gt;&#xA; @ = 6/8 = 3/4&lt;br&gt;&#xA; _ = 3/6 = 1/2 = 2/4&lt;br&gt;&#xA; I = 0/1 = 0&lt;br&gt;&#xA; * = 5/4&lt;br&gt;&#xA; O = 4/4&lt;br&gt;&#xA; ` = 7/4  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which can be put in order as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I = 0/4&lt;br&gt;&#xA; \ = 1/4&lt;br&gt;&#xA; _ = 2/4&lt;br&gt;&#xA; @ = 3/4&lt;br&gt;&#xA; O = 4/4&lt;br&gt;&#xA; * = 5/4&lt;br&gt;&#xA; A = 6/4&lt;br&gt;&#xA; ` = 7/4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From there I tried addition to find position in the alphabet, then multiplication, but neither seemed to get anywhere...&lt;/p&gt;&#xA;" OwnerUserId="10078" LastActivityDate="2015-07-02T14:00:03.500" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17304" PostTypeId="1" AcceptedAnswerId="17305" CreationDate="2015-07-02T14:35:05.603" Score="8" ViewCount="977" Body="&lt;p&gt;At first I’m quite fun to be around,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;until I make some wrongful sound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe it offends, maybe it’s rude&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I conceal my true attitude.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Reverse me and you’ve something new&lt;/p&gt;&#xA;&#xA;&lt;p&gt;be it a weapon, or a cold, cold bar,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a means of relocation, near or far,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it’s something you’d be daft to chew!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Interested in becoming me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you aren’t already, all it takes to be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is to toughen up, brace yourself, grin --&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then declare “I am one, and always have been!”&lt;/p&gt;&#xA;" OwnerUserId="13556" LastActivityDate="2015-07-02T15:20:11.610" Title="Here's precisely what I am not. Who am I?" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17305" PostTypeId="2" ParentId="17304" CreationDate="2015-07-02T15:10:12.480" Score="7" Body="&lt;p&gt;Unsure, but I think you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;an honest person&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From what you are not :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;At first I’m quite fun to be around,&lt;br/&gt;&lt;/strong&gt;&#xA;&lt;strong&gt;until I make some wrongful sound.&lt;br/&gt;&lt;/strong&gt;&#xA;==&gt; liars can be fun before they are found out&lt;br/&gt;&#xA;&lt;strong&gt;Maybe it offends, maybe it’s rude&lt;br/&gt;&lt;/strong&gt;&#xA;&lt;strong&gt;but I conceal my true attitude.&lt;br/&gt;&lt;/strong&gt;&#xA;==&gt; Liars don't want to be found out even though lies can be offensive or rude&lt;br/&gt;&#xA;&lt;strong&gt;Reverse me and you’ve something new&lt;br/&gt;&lt;/strong&gt;&#xA;==&gt; &quot;liar&quot; reversed becomes &quot;rail&quot;&#xA;&lt;strong&gt;be it a weapon, or a cold, cold bar,&lt;br/&gt;&lt;/strong&gt;&#xA;==&gt; a rail-gun, or (unsure about the cold part) railing&lt;br/&gt;&#xA;&lt;strong&gt;a means of relocation, near or far,&lt;br/&gt;&lt;/strong&gt;&#xA;==&gt; railroad tracks&lt;br/&gt;&#xA;&lt;strong&gt;it’s something you’d be daft to chew!&lt;br/&gt;&lt;/strong&gt;&#xA;==&gt; See &quot;railroad&quot;... try biting on that!&lt;br/&gt;&#xA;&lt;strong&gt;Interested in becoming me?&lt;br/&gt;&lt;/strong&gt;&#xA;&lt;strong&gt;If you aren’t already, all it takes to be&lt;br/&gt;&lt;/strong&gt;&#xA;&lt;strong&gt;is to toughen up, brace yourself, grin --&lt;br/&gt;&lt;/strong&gt;&#xA;&lt;strong&gt;then declare “I am one, and always have been!”&lt;br/&gt;&lt;/strong&gt;&#xA;==&gt;“I am one, and always have been!” :)&lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then again, as pointed out by @Geobits&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The poem points to &quot;liar&quot;. Which would mean you are one, and are actually lying about the poem being what you are not.&lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore :&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A liar!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13753" LastEditorUserId="13753" LastEditDate="2015-07-02T15:20:11.610" LastActivityDate="2015-07-02T15:20:11.610" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17306" PostTypeId="1" AcceptedAnswerId="18237" CreationDate="2015-07-02T15:25:47.413" Score="12" ViewCount="793" Body="&lt;blockquote&gt;&#xA;  &lt;h1&gt;Sorry. The previously posted image contained a mistake.&lt;/h1&gt;&#xA;  &#xA;  &lt;p&gt;Tile D2 was shown with a 180 degree rotation. The image has now been corrected.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is a straight-forward arrangement puzzle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are given a set of tiles showing each a blue and a red arrow pointing in one of 8 directions. Your task is to rearrange these tiles - without rotation or mirroring - such that each blue arrow is pointing to a neighbouring tile containing a red arrow pointing back to the tile. Additionally, all connections need to form a single, continuous cycle, like in the following mini-example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IWG1Z.png&quot; alt=&quot;Example&quot;&gt;&lt;P&gt;&#xA;  &lt;strong&gt;valid arrangement&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Your set contains the following 49 tiles, which must not be rotated or mirrored, just shifted:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zqfm2.png&quot; alt=&quot;CorrectedPuzzle&quot;&gt;&lt;P&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The grey triangle is just to mark &quot;up&quot; so that you can print those tiles, if you want and don't accidently rotate them.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h2&gt;Question:&lt;/h2&gt;&#xA;  &#xA;  &lt;h1&gt;Rearrange tiles into a valid 7x7 grid.&lt;/h1&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This question has at least one solution. Preferably post the solution as image. Alternatively as a grid of tile-IDs using the indexing provided in the tile-set to identify the tiles. (i.e. the first tile is tile (A1) and you use A1 in the position you think it should be in.) &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The original post had two more &quot;bonus&quot; questions which have found their answers already. I'm posting them for reference only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;##Question 2:##&#xA;Take away a single tile of your choosing. Can you rearrange the remaining 48 tiles into either a 6x8 or a 8x6 valid grid? If you can not do this with a single 'loop' you may create a grid which contains multiple closed loops, but each tile must belong to one such loop.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;##Question 3:##&#xA;Can you repeatedly take away a single (freely chosen) tile while managing to rearrange the remaining into *any* rectangular grid? (i.e. find arbitrary solutions for (n x m) grids with 49, 48, (47), 46.... 4 tiles where each 'smaller' set has to be created by taking away a single tile from the next larger set. If the set-number is a prime number, you can skip it (i.e. take away another tile). Again, single closed loops are preferred, but multiple closed loops are allowed.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both bonus-questions are answered by &lt;a href=&quot;https://puzzling.stackexchange.com/users/10261/john-stevens&quot;&gt;John Stevens&lt;/a&gt; in the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/17306/rearranging-connected-tiles/17316#17316&quot;&gt;answer below&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This puzzle seems to be harder than I thought it would be. To keep things rolling, I'm adding some hints, i.e. some information about specific tiles/positions. &lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Hint 1&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The 4 corner tiles are (without order): &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; F7, F6, F3 and D7.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2015-07-21T21:41:04.357" Title="Rearranging connected tiles" Tags="|visual|packing|" AnswerCount="3" CommentCount="24" ContentLicense="CC BY-SA 3.0" />
  <row Id="17307" PostTypeId="2" ParentId="17282" CreationDate="2015-07-02T15:34:37.477" Score="6" Body="&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/a/17283/4513&quot;&gt;Chamberlain&lt;/a&gt; got the good qualities, and I think I have the bad ones:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any time you greet him, he tends to try and take over.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Hi, Jack!&quot; = hijack, meaning seize or take over.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He can a be racist sometimes - he once said if he were from Africa, he was sure he'd be gambling all the time.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He would be &quot;Black Jack&quot; = blackjack, a casino card game.&lt;br&gt;&#xA; (And it is indeed a bit racist to imply all Africans are black lol)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4513" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-07-02T15:34:37.477" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17308" PostTypeId="2" ParentId="17172" CreationDate="2015-07-02T16:32:48.053" Score="-2" Body="&lt;p&gt;sorry, no picture, but I'll describe it: take the hind leg; attach it to the front foot and make it parallel to the back; now rotate the picture 90 degrees to the right. Now the goat's head is over the hind leg.&lt;/p&gt;&#xA;" OwnerUserId="13759" LastActivityDate="2015-07-02T16:32:48.053" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17309" PostTypeId="1" AcceptedAnswerId="17920" CreationDate="2015-07-02T16:33:08.133" Score="47" ViewCount="3547" Body="&lt;p&gt;My friend has been playing with a program he found. He enters a 5-digit number into the command line and receives another 5-digit number in return. However, after 20 or so entries, he's about done and ready to quit. Unfortunately, there doesn't appear to be a &quot;Quit&quot; or &quot;Exit&quot; button anywhere in sight. These are his inputs and the program's responses thus far:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;gt; 32908&#xA;Response: 32837&#xA;&amp;gt; 30351&#xA;Response: 32910&#xA;&amp;gt; 32872&#xA;Response: 32872&#xA;&amp;gt; 30280&#xA;Response: 30354&#xA;&amp;gt; 30278&#xA;Response: 32874&#xA;&amp;gt; 14727&#xA;Response: 30246&#xA;&amp;gt; 30246&#xA;Response: 32838&#xA;&amp;gt; 31992&#xA;Response: 14767&#xA;&amp;gt; 32799&#xA;Response: 31992&#xA;&amp;gt; 14654&#xA;Response: 30283&#xA;&amp;gt; 32944&#xA;Response: 14802&#xA;&amp;gt; 32909&#xA;Response: 32909&#xA;&amp;gt; 30316&#xA;Response: 32944&#xA;&amp;gt; 30354&#xA;Response: 30354&#xA;&amp;gt; 35536&#xA;Response: 35504&#xA;&amp;gt; 35538&#xA;Response: 32946&#xA;&amp;gt; 34276&#xA;Response: 26468&#xA;&amp;gt; 14619&#xA;Response: 35503&#xA;&amp;gt; 14875&#xA;Response: 34098&#xA;&amp;gt; 26533&#xA;Response: 26432&#xA;&amp;gt; ?????&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Being a Linux expert, I know that every command-line program like this has a force-quit option, but I can't figure out how the program works. What should I tell my friend to type in place of the question marks in order to quit the program?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'll add some clarification points to make this puzzle a bit more clear, since it (not surprisingly) has been causing some confusion!  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The story is mostly flavor, so I wouldn't stress too much over it. What you need to know are the entries and responses of the program, and the fact that my friend is trying to 'force-quit' it, whatever that might entail.&lt;/li&gt;&#xA;&lt;li&gt;A non-5 digit number &lt;em&gt;probably&lt;/em&gt; would end up being an invalid input, though I might have to check that to be sure. Either way, that's not the solution to the puzzle. There is a valid input that will force the program to quit.&lt;/li&gt;&#xA;&lt;li&gt;This puzzle will require some knowledge that deems it unsolvable to a subset of the puzzlers on this site. If you're working hard on it and get stopped at that point, I encourage you to post a partial answer so that you can at least get some upvotes!&lt;/li&gt;&#xA;&lt;li&gt;The title may be confusing, but I promise that when the solution is revealed it will make perfect sense.&lt;/li&gt;&#xA;&lt;li&gt;There is a logical answer to this puzzle, but figuring out how to get there will involve some lateral thinking. I'm fixing the tags around right now and will add a related hint.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Hopefully this specifies everything for you. If you're still confused (which I wouldn't blame you for), here's a hint as well!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The order in which you'll need to use the tags is: &lt;a href=&quot;/questions/tagged/cryptograms&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;cryptograms&amp;#39;&quot; rel=&quot;tag&quot;&gt;cryptograms&lt;/a&gt;, &lt;a href=&quot;/questions/tagged/lateral-thinking&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;lateral-thinking&amp;#39;&quot; rel=&quot;tag&quot;&gt;lateral-thinking&lt;/a&gt;, &lt;a href=&quot;/questions/tagged/trivia&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;trivia&amp;#39;&quot; rel=&quot;tag&quot;&gt;trivia&lt;/a&gt;, &lt;a href=&quot;/questions/tagged/pattern&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;pattern&amp;#39;&quot; rel=&quot;tag&quot;&gt;pattern&lt;/a&gt;. This could definitely be considered open to interpretation, but if you start with &lt;a href=&quot;/questions/tagged/cryptograms&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;cryptograms&amp;#39;&quot; rel=&quot;tag&quot;&gt;cryptograms&lt;/a&gt; you'll be on the right track.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There hasn't been much headway on this, so I might have made it a bit too cryptic... time will tell. Anyhow, it should at least be apparent that the 5-digit numbers do not represent actual 5-digit numbers. Try decrypting them in some fashion and see if the result looks like anything you've seen before. Even if it doesn't, posting your results might help other people get to the bottom of it. The encryption method used is not a difficult one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The given numbers are base 10. What do they look like in other bases?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 4 (a biggish one):&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My friend is playing some kind of game with the computer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="11475" LastEditDate="2015-07-14T14:48:12.230" LastActivityDate="2015-07-14T17:16:29.230" Title="How do I force-quit the program?" Tags="|cipher|pattern|lateral-thinking|knowledge|" AnswerCount="5" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="17310" PostTypeId="2" ParentId="17207" CreationDate="2015-07-02T16:33:49.700" Score="18" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PrincessTrevor has found that the pairs of numbers correspond to fractions from 0 to 7/4. &lt;br&gt;&lt;br&gt; Now &lt;em&gt;multiply&lt;/em&gt; them by &lt;em&gt;pi&lt;/em&gt; to get angles (in radians). By convention, 0 is to the right, pi/4 is up and to the right, 2pi/4 is up, and so on, until 7pi/4 which is down and to the right. &quot;Flagged down&quot; suggests that these are the eight flag positions of &lt;em&gt;semaphore&lt;/em&gt;. The message comes out to... &lt;br&gt;&lt;br&gt; &quot;Jet fuel can't melt steel beams&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="54097" LastEditDate="2018-12-19T19:04:34.763" LastActivityDate="2018-12-19T19:04:34.763" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="17311" PostTypeId="2" ParentId="17008" CreationDate="2015-07-02T17:42:25.913" Score="9" Body="&lt;p&gt;The odd ones out are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A2 and D7&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The pattern is semaphore. Use the directions of each owl's eyes as the positions of the flags.&lt;br&gt;&#xA; &lt;pre&gt;&#xA; HUSHLITTL&#xA; OBABYDONT&#xA; SAYAWORDM&#xA; AMASGOING&#xA; TOBUYYOUA&#xA; MOCKINGBI&#xA; RDITTHATM&#xA; OCKINGBIR&#xA; DWONTSING&lt;/pre&gt;&#xA;&#xA; These are the lyrics to the lullaby &quot;Hush Little Baby&quot;.&lt;br&gt;&#xA;&lt;br&gt;&#xA; A2 and D7 are the wrong letters. A2 should have looked south, northeast,  and D7 should have looked south, east.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12109" LastEditorUserId="12109" LastEditDate="2015-07-02T18:28:47.813" LastActivityDate="2015-07-02T18:28:47.813" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17312" PostTypeId="2" ParentId="17086" CreationDate="2015-07-02T18:14:43.547" Score="5" Body="&lt;p&gt;The route for a sensation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A nerve&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Coming all the way from the Egyptian cross:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Coptic&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So maybe it is a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Coptic nerve&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the second paragraph would be explained as (credit &lt;strong&gt;f''&lt;/strong&gt; for this):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your optic nerve is what allows you to see/C, so X = C and Y = optic nerve.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13765" LastEditorUserId="5373" LastEditDate="2015-07-07T11:12:13.237" LastActivityDate="2015-07-07T11:12:13.237" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17313" PostTypeId="1" AcceptedAnswerId="17364" CreationDate="2015-07-02T18:41:49.467" Score="4" ViewCount="1149" Body="&lt;p&gt;This is part of &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=title%3A%22The+Twenty+Doors%22&quot;&gt;The Twenty Doors&lt;/a&gt; series. &#xA;&lt;br&gt;The previous one is &lt;a href=&quot;https://puzzling.stackexchange.com/questions/17198/the-twenty-doors-room-2&quot;&gt;The Twenty Doors! (ROOM 2)&lt;/a&gt;&#xA;&lt;br&gt;The next one is &lt;a href=&quot;https://puzzling.stackexchange.com/questions/17383/the-twenty-doors-room-4&quot;&gt;The Twenty Doors (ROOM 4)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;When you finally realise that the cipher used in room 2 was the Bifid cipher, you come into the next room. The first thing you see is some paper:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;R28gYmFjayB0byB3aGVyZSBpdCBhbGwgYmVnYW4u&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Next, you walk forward to take a look at the keyboard. This one is a normal keyboard. Then you turn to look at the wall, and, as usual, there is something carved into it: &lt;code&gt;I was involved with making email attachments possible.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hmm... What could this mean? Well, you have not died yet. You may as well try.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; So far, everything up until the point where the first room is entered is correct. You will have to dig deeper to find what's hidden in there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A certain edit I made may be of importance... &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/questions/17313/the-twenty-doors-room-3/17322#17322&quot;&gt;Bob has found the final cryptogram of this door&lt;/a&gt;, so now all that remains is to decipher that code!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 4 (REQUIRED TO SOLVE):&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I halved the bottles standing the wall.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The next door will be added when this door is solved!&lt;/p&gt;&#xA;" OwnerDisplayName="user9377" LastEditorUserId="-1" LastEditorDisplayName="user9377" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-07-04T16:30:46.620" Title="The Twenty Doors! (ROOM 3)" Tags="|cipher|" AnswerCount="5" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17314" PostTypeId="1" AcceptedAnswerId="17315" CreationDate="2015-07-02T18:43:56.907" Score="3" ViewCount="758" Body="&lt;p&gt;&lt;em&gt;It was hard to expand on this riddle while maintaining the single rhymes, so it is much easier than expected to solve...Hopefully you enjoy the rhymes&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I give you some metal, and you steal my name! &lt;br&gt;I had to have known,&#xA;  I am only to blame.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Even when right I'm wrong, what a shame. &lt;br&gt;A sacrifice I make, to&#xA;  help in my claim.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I will waiver my rights, and steady my aim. &lt;br&gt;For my resistance is&#xA;  futile, within my time frame.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A quick easy answer can give you some fame. &lt;br&gt;So find what I am, and&#xA;  finish this game!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If you enjoyed this riddle, some of my other rhyming riddles can be found &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%5Brhyme%5D%20user%3A11244%20is%3Aquestion&quot;&gt;here&lt;/a&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2015-07-02T19:08:41.793" Title="I give you some metal, and you steal my name!" Tags="|riddle|rhyme|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17315" PostTypeId="2" ParentId="17314" CreationDate="2015-07-02T18:49:06.620" Score="8" Body="&lt;p&gt;Is the answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Male&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I give you some metal, and you steal my name! &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fe (iron) + male = female. Can also refer to how after the man gives his woman a wedding ring, she takes his last name.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I had to have known, I am only to blame.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Women always blame men. Also, men often give precious metals to women like gold and silver. &lt;strong&gt;EDIT:&lt;/strong&gt; Men are usually the ones to propose marriage.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Even when right I'm wrong, what a shame. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Women are always right! :P&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A sacrifice I make, to help in my claim.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Men often let women win arguments to keep the relationship steady.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I will waiver my rights, and steady my aim.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Men get married and give up the merriment of their bachelor life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For my resistance is futile, within my time frame.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Women keep pestering men to commit and get married. &lt;strong&gt;EDIT:&lt;/strong&gt; Once they are married, there is no escape and trying to resist is not generally a good idea.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A quick easy answer can give you some fame. &lt;br&gt;&#xA;So find what I am, and finish this game!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Can I haz some repz?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-07-02T19:08:41.793" LastActivityDate="2015-07-02T19:08:41.793" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="17316" PostTypeId="2" ParentId="17306" CreationDate="2015-07-02T18:58:42.400" Score="7" Body="&lt;p&gt;For Question 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Is it not possible to arrange the tiles in a valid configuration if any one is taken away.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If one starts at any square and follows arrows of one color, one will eventually get back to the starting square.  I'll just consider the blue arrows from now on, but the same argument also applies to the red arrows.&lt;br&gt;&#xA; &lt;br/&gt;&#xA; I'll model the direction of the blue arrow as an ordered pair $(x,y)$, where $x$ is the change in the horizontal coordinate when following the arrow, and $y$ is the change in the vertical coordinate.  So, the the move for the A1 tile is $(1,-1)$, the move for the A2 tile is $(1,1)$, the move for the B3 tile is $(0,-1)$, and so on.&lt;br&gt;&#xA; &lt;br/&gt;&#xA; In any cycle in a valid configuration, the sum of the moves for all the tiles in that cycle must be $(0,0)$, since if you follow the blue arrows you eventually get back to where you started.  Since every tile in a valid configuration is part of a cycle, and no tile is part of more than one cycle, the sum of the moves for all tiles in a valid configuration is $(0,0)$.&lt;br&gt;&#xA; &lt;br/&gt;&#xA; Since there is a valid configuration for all 49 tiles, the sum of the moves for all 49 tiles is $(0,0)$.  For there to be a valid configuration for 48 tiles, we would need to remove a tile and leave the sum of the moves for the remaining tiles $(0,0)$.  This is only possible if there is a tile without a blue arrow.  Since all tiles have a blue arrow, there is no valid configuration consisting of any 48 of the 49 tiles in the puzzle.&lt;br&gt;&#xA; &lt;br/&gt;&#xA; This result can be used to quickly eliminate possibilities in brute force searches for valid configurations of other sizes: for any subset of tiles, if the sum of the blue moves (or red moves) is not $(0,0)$, then no valid configuration is possible and the subset need not be considered.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10261" LastEditorUserId="10261" LastEditDate="2015-07-02T19:08:40.023" LastActivityDate="2015-07-02T19:08:40.023" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17317" PostTypeId="2" ParentId="17313" CreationDate="2015-07-02T19:00:20.707" Score="0" Body="&lt;p&gt;I was involved with making email attachments possible.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is Base 64&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Using Base 64 we got: &lt;strong&gt;Go back to where it all began.&lt;/strong&gt; &lt;br&gt;&#xA; &lt;del&gt;Where it all began is Room 1&lt;/del&gt;&lt;br&gt;&#xA; &lt;del&gt;We need to input: &lt;strong&gt;neetruof&lt;/strong&gt; &lt;/del&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5359" LastEditorUserId="5359" LastEditDate="2015-07-02T20:02:45.127" LastActivityDate="2015-07-02T20:02:45.127" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17318" PostTypeId="2" ParentId="17313" CreationDate="2015-07-02T19:01:34.473" Score="3" Body="&lt;p&gt;For other puzzlers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The clue for email attachments leads us to Base 64 decoding, and we get the message: &lt;strong&gt;Go back to where it all began.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;del&gt;I think this is a riddle though.&lt;/del&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do we need to key in&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://minionnation.tumblr.com/post/120724564612/this-july-go-back-to-where-it-all-began&quot; rel=&quot;nofollow noreferrer&quot;&gt;Minions&lt;/a&gt;&lt;br&gt; Cause well, July is already here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The above answer is being retained despite its inaccuracy, because minions are awesome! However, in a serious attempt to solve this puzzle, &lt;del&gt;I think the correct code might be &lt;/del&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;del&gt;Area 51&lt;/del&gt;&lt;br&gt;&lt;br&gt; The edit to the first question of the series hides the words  &lt;a href=&quot;https://puzzling.stackexchange.com/posts/17167/revisions#&quot;&gt;&quot;The profile of the evil guy who began all of this may hold something interesting...&quot;&lt;/a&gt; as a comment, which can be seen when you view the revision in markdown format.&lt;br&gt;&lt;br&gt;Visiting &lt;a href=&quot;https://puzzling.stackexchange.com/users/9377/kslkgh&quot;&gt;the profile of Kslkgh&lt;/a&gt;, we can see &lt;del&gt;he has an active proposal for a new StackExchange community. That proposal is on Area 51, where all of the new StackExchange sites begin, including Puzzling.SE.&lt;/del&gt; that he's left us a cryptogram.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The next cryptogram to be solved is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;TidoTt ilhurr om hnysS saieqeofR fee res gs edwo 3o wDore&quot;ee.r r ToiL&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-07-04T05:49:14.143" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17319" PostTypeId="2" ParentId="17313" CreationDate="2015-07-02T19:18:58.670" Score="2" Body="&lt;p&gt;Is the solution&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Using Base 64 gives us &lt;strong&gt;Go back to where it all began&lt;/strong&gt;, so the answer should be the answer for the first room.&#xA;We need to key in &lt;strong&gt;14&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13764" LastActivityDate="2015-07-02T19:18:58.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17320" PostTypeId="1" AcceptedAnswerId="17321" CreationDate="2015-07-02T19:22:50.150" Score="23" ViewCount="2562" Body="&lt;p&gt;Let me tell you about four friends of mine. They're a close-knit bunch: two pairs of twins who are cousins to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first guy is Adam. He's a devout Christian, going to church every week. He loves his job and commutes many miles to work each day. He has a very positive outlook, always looking on the bright side. He's content with his life, saying he has some of everything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;His sister is Sukey. She's very different from Adam, although they love each other dearly. She's a pessimist with a gloomy outlook. She's always ready to criticise and make remarks against people she doesn't like. She's practical and generally wants to know what she can take away from a lesson or a situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their first cousin is Matt. He has loads of kids and will soon have a whole swarm of grandchildren. He's a watchmaker by trade, so he's always got the times of day for you. He's almost as fond of Adam's children as his own, and loves distributing sweets to them. His favourite food is pie, and he can't get enough of it: he often has steak and kidney pie for dinner and then apple and blackberry pie for dessert.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;His sister is Debbie. She strongly resembles Sukey, but has grown her hair longer and wears it in pigtails. She has a tendency to cause arguments - not intentionally; people just start squabbling when she's around. She makes a hobby of fencing, and spends her free time slashing with a rapier. She loves sandwiches, and likes to vary the type of bread even more than the filling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who are these people, and what's their favourite form of entertainment?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-07-07T11:28:48.860" Title="Identify the twins and cousins!" Tags="|riddle|enigmatic-puzzle|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17321" PostTypeId="2" ParentId="17320" CreationDate="2015-07-02T19:30:49.887" Score="29" Body="&lt;p&gt;The twins and cousins are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The four basic mathematical operations&lt;br&gt;Adam $\rightarrow$ Addition, Sukey $\rightarrow$ Subtraction, Matt $\rightarrow$ Multiplication, Debbie $\rightarrow$ Division&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He's a devout Christian, going to church every week.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The plus sign(&lt;strong&gt;+&lt;/strong&gt;) looks similar to the Christian cross($\dagger$).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...commutes many miles to work each day.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Addition is a commutative operation. That is, 2+5 is the same as 5+2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He has a very positive outlook.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Plus before numbers denotes that they are positive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...he has some of everything.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A play on the word 'sum'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;She's very different from Adam...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Subtraction gives you the difference between numbers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;She's a pessimist...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Subtraction is a negative operation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;She's always ready to criticise and make remarks against people she doesn't like.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; These are generally considered negative traits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...generally wants to know what she can take away...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In subtraction, you take something away from the original.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He has loads of kids and will soon have a whole swarm of grandchildren. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;del&gt;Repeated multiplication can lead to very large numbers.&lt;/del&gt; Humans multiply by having kids and those kid further multiply to have grandchildren.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He's almost as fond of Adam's children as his own, and loves distributing sweets to them.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Multiplication is &lt;del&gt;a distributive operation&lt;/del&gt; distributive over addition. That is, 5x(3+4) is same as 5x3 + 5x4.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...he's always got the times of day for you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 5 multiplied by 2 can also be said as 5 times 2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;His favourite food is pie...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;del&gt;To find the area and perimeter of a circle, we have to multiply with pi.&lt;/del&gt; The multiplication of numbers can be represented using the capital letter pi.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;She strongly resembles Sukey, but has grown her hair longer and wears it in pigtails.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The division symbol ($\div$) looks like the minus sign (-), with two dots i.e the pigtails.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;... people just start squabbling when she's around.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It helps to divide. Similar to dividing people by creating fights.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...spends her free time slashing with a rapier.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Division can also be represented with a slash, like 15/3. &lt;del&gt;The slash looks like a sword held in the hand during combat.&lt;/del&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;She loves sandwiches, and likes to vary the type of bread even more than the filling.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When dividing, numbers can be written in the fractional form, with the numerator and denominator changing while the line denoting division remains the same.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Their favourite form of entertainment is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;del&gt;being weapons of math instruction!&lt;/del&gt; &lt;del&gt;(Judging by rand's love of Doctor Who, I'd say these are names of Doctor Who actors/characters, and so) they enjoy playing doctor and performing operations.&lt;/del&gt; With a little help from rand al'thor, their favourite form of entertainment is opera (which is inspired from the word operations, which they all are.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-07-07T11:28:48.860" LastActivityDate="2015-07-07T11:28:48.860" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="17322" PostTypeId="2" ParentId="17313" CreationDate="2015-07-02T19:42:44.423" Score="3" Body="&lt;p&gt;R28gYmFjayB0byB3aGVyZSBpdCBhbGwgYmVnYW4u is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Base64 encoding of - Go back to where it all began.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which coupled with the clue about &quot;...making email attachments possible.&quot; suggests something to do with their history.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To quote wikipedia&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Modern email systems use the MIME standard, making email attachments&#xA;  more utilitarian and seamless. This was developed by Nathaniel&#xA;  Borenstein and collaborator Ned Freed with the first MIME email&#xA;  attachment being sent by Nathaniel Borenstein on March 11, 1992 and&#xA;  the standard being officially released as RFC2045 in 1996.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So we could try any of the following...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nathaniel, Borenstein, Ned, Freed, RFC2045, MIME&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...or any combination of those, or maybe...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SMTP, Base64, ASCII or one other the many other words related to email attachments.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But it turns out that none of that is relevant, as...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; where it all began refers to the first puzzle in this series which if you look at the mark up of the question contains a comment that reads &quot;The profile of the evil guy who began all of this may hold something interesting...&quot; which refers to the OP who's profile contains a message &quot;If you are someone that is trying to solve Room 3 my popular &quot;The Twenty Doors&quot; series, this may help:&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;followed by the following code&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TidoTt ilhurr om hnysS saieqeofR fee res gs edwo 3o wDore&quot;ee.r r ToiL&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="11297" LastEditDate="2015-07-03T07:18:52.787" LastActivityDate="2015-07-03T07:18:52.787" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17323" PostTypeId="2" ParentId="17230" CreationDate="2015-07-02T20:41:24.917" Score="2" Body="&lt;p&gt;&lt;strong&gt;Not possible&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;House 1 and House 2, when connected to Well 1 and Well 3 will form a closed shape around Well 2.&#xA;House 3 will NOT be able to access one of the wells, regardless of where you put it.&lt;/p&gt;&#xA;" OwnerUserId="13766" LastActivityDate="2015-07-02T20:41:24.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17324" PostTypeId="2" ParentId="17172" CreationDate="2015-07-02T20:59:29.047" Score="12" Body="&lt;p&gt;What about this?  A move that satisfies the criteria in a three-dimensional way.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The right leg becomes the head of a happy goat all sprawled out, viewed from above.&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/BIGWf.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/tikoq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4062" LastActivityDate="2015-07-02T20:59:29.047" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17325" PostTypeId="1" AcceptedAnswerId="17326" CreationDate="2015-07-02T23:15:29.317" Score="17" ViewCount="1542" Body="&lt;p&gt;Twenty-five ants are scattered on a meter stick. At the same time, they each pick a random direction (east or west) and start marching at 1cm/second. Whenever two ants meet, they turn around. When an ant reaches the end it falls off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ant in the middle is infected with a cold. An infected ant will spread the disease to any ant he bumps into.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On average, by the time every ant has fallen off, how many will be sick?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="71" LastEditDate="2015-07-03T11:54:44.123" LastActivityDate="2015-07-03T11:54:44.123" Title="Contagious ants on a stick" Tags="|probability|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17326" PostTypeId="2" ParentId="17325" CreationDate="2015-07-02T23:41:02.863" Score="27" Body="&lt;p&gt;The average number of sick ants will be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $13 - \frac{6}{2^{12}}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We start with the classic ant-on-a-stick observation that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; we can imagine two ants as bouncing off each other as though they passed through each other, since their identities don't matter. So, in an equivalent problem, ants don't bounce, and an ant is infected by passing through another infected ant.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, make an observation about how ants get infected:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The infected range spreads at most as fast as an ant walks. So any ants the start moving away from the center ant cannot be infected. Any ants that start going towards the center get infected as long as an infected ant going the other way touches the center point. Usually, that's all ants that start going towards the center ant (expected $12$), since the center ant infects all such ants in front of him, and any such ant infects all such ants behind him. But, if all ants in front of the center ant are going the same way (probability $2^{-12}$), no ants behind him are infected, decreasing the expectation by $6$. Counting the $1$ infected center ant, this gives an expectation of $1+12-2^{-12}\times 6 = 13 - \frac{6}{2^{12}}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-07-02T23:41:02.863" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17327" PostTypeId="1" AcceptedAnswerId="17329" CreationDate="2015-07-03T01:11:30.340" Score="35" ViewCount="2901" Body="&lt;p&gt;When solving the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/17325/contagious-ants-on-a-meterstick&quot;&gt;Contagious Ants on a Meterstick&lt;/a&gt; problem, puzzlers often overlook the fact that ants are sophisticated creatures with names, dreams, and personalities.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;_ _ _ (_) _ _ _&lt;/code&gt; is an expert navigator, and is especially skilled in the presence of celestial bodies.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;_ (_) _ _ _ _&lt;/code&gt; is always taking back sworn testiony.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;_ (_) _ _ _ _ _ _&lt;/code&gt; is the oddball ant of the bunch. She always seems out of place.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;_ (_) _ _ _ _ _&lt;/code&gt; is magical and makes powerful weapons.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;_ _ (_) _ _ _ _&lt;/code&gt; is the poorest and lowliest ant.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;_ (_) _ _ _ _ _&lt;/code&gt; always gets things done right this second.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;_ _ _ _ _ (_) _ _ _&lt;/code&gt; is the best at following orders and instructions.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;_ _ _ _ (_) _ _ _&lt;/code&gt; is a chemist, and she also has a fondness for fissile material.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;_ (_) _ _ _ _ _ _ _ _ _ _&lt;/code&gt; is the most stubborn ant of the bunch. She always defies authority.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;_ _ _ _ (_) _ _ _ _&lt;/code&gt; is always deeply sorry and committed to changing her ways.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;_ _ _ _ _ (_) _ _ _&lt;/code&gt; is the most consequential and influential ant.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;_ _ _ _ _ _ (_) _ _ _&lt;/code&gt; is always poring over the books, mindful of the bottom line.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;_ _ _ (_) _ _ _&lt;/code&gt; is an expert at keeping temperatures low.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;(_) _ _ _ _ _ _ _ _&lt;/code&gt; is forever interviewing for employment.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;(_) _ _ _ _ _ _ _ _&lt;/code&gt; has the most opulent, pampered lifestyle of all the ants.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;(_) _ _ _ _ _ _&lt;/code&gt; is the most chivalrous ant, and aspires to be a knight.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;_ (_) _ _ _ _ _ _&lt;/code&gt; is actually a wild fowl disguised to &lt;em&gt;look&lt;/em&gt; like an ant.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;_ _ (_) _ _ _ _ _ _ _ _&lt;/code&gt; is larger-than-life grandiose. Always bigger and better for this ant!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;_ (_) _ _ _ _ _&lt;/code&gt; is homeless, forever wandering the streets.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;_ _ (_) _ _ _ _ _&lt;/code&gt; smells. :(&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;(_) _ _ _ _ _ _ _&lt;/code&gt; is forever trying to overthrow people to take their place.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;(_) _ _ _ _ _ _ _&lt;/code&gt; is by far the biggest ant, and she looks a little gray.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;But wait! There are still three ants left!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can you take all the circled letters, unscramble them, and determine the names of the three ants I missed?&lt;/strong&gt;&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Please include the answers to the 22 clues along with your solution.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-07-03T02:26:00.343" Title="Name the 25 Ants" Tags="|word|wordplay|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17329" PostTypeId="2" ParentId="17327" CreationDate="2015-07-03T02:25:24.220" Score="29" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. sex(t)ant&lt;br&gt;&#xA; 2. r(e)cant&lt;br&gt;&#xA; 3. a(b)errant&lt;br&gt;&#xA; 4. e(n)chant&lt;br&gt;&#xA; 5. pe(a)sant&lt;br&gt;&#xA; 6. i(n)stant&lt;br&gt;&#xA; 7. compl(i)ant&lt;br&gt;&#xA; 8. reac(t)ant&lt;br&gt;&#xA; 9. r(e)calcitrant&lt;br&gt;&#xA; 10. repe(n)tant&lt;br&gt;&#xA; 11. impor(t)ant&lt;br&gt;&#xA; 12. accoun(t)ant&lt;br&gt;&#xA; 13. coo(l)ant&lt;br&gt;&#xA; 14. (a)pplicant&lt;br&gt;&#xA; 15. (l)uxuriant&lt;br&gt;&#xA; 16. (g)allant&lt;br&gt;&#xA; 17. p(h)easant&lt;br&gt;&#xA; 18. ex(t)ravagant&lt;br&gt;&#xA; 19. v(a)grant&lt;br&gt;&#xA; 20. fr(a)grant&lt;br&gt;&#xA; 21. (s)upplant&lt;br&gt;&#xA; 22. (e)lephant  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These leave us with the letters&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TEBNANITENTTLALGHTAASE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which anagram into&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HESITANT, BLATANT, ELEGANT&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-07-03T02:25:24.220" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17331" PostTypeId="2" ParentId="17306" CreationDate="2015-07-03T02:58:40.333" Score="0" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;PRELIMINARIES Q1&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Sennet (rope mat) puzzle can be described as  &lt;strong&gt;NODAL&lt;/strong&gt;  or &lt;strong&gt;PEG&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think of the &lt;strong&gt;pegs&lt;/strong&gt; arranged along the edges and sides of the board, the continuous one cord sennet or, as the architects call it, Celtic Interlace, will be constructed/ woven on a 2X3 peg; 3X4peg; 4X5 peg board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nodes or crossovers can be X or CC back to back. an X needs 4 tiles with 45degree (45,135,225,315) and the available tiles suggest the basic design (before elaboration) is &lt;strong&gt;2X3&lt;/strong&gt;. This needs &lt;strong&gt;six X nodes&lt;/strong&gt;; but any two X nodes can be replaced by four CC. (If you replace &lt;em&gt;odd&lt;/em&gt; numbers of X with 2CCs each the thread will not be continuous/ single.)&lt;/p&gt;&#xA;" OwnerUserId="11954" LastActivityDate="2015-07-03T02:58:40.333" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17332" PostTypeId="2" ParentId="17282" CreationDate="2015-07-03T08:32:14.530" Score="2" Body="&lt;p&gt;Just adding to the above answers&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Derek 'Jac'obi fits the 4th clue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9669" LastActivityDate="2015-07-03T08:32:14.530" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17333" PostTypeId="1" AcceptedAnswerId="17334" CreationDate="2015-07-03T09:23:57.703" Score="4" ViewCount="1131" Body="&lt;p&gt;First a sport some love,&lt;br&gt;&#xA;Though it drinks their blood.&lt;br&gt;&#xA;Named for just a seat,&lt;br&gt;&#xA;Though power is the key.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the sound of metal,&lt;br&gt;&#xA;Slamming together in war.&lt;br&gt;&#xA;That of which there are five,&lt;br&gt;&#xA;Vying in the first name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hurricane with thunder,&lt;br&gt;&#xA;Lightning and pouring rain.&lt;br&gt;&#xA;Weapons are out in the darkness,&lt;br&gt;&#xA;Blades are bared to kill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A banquet is laid out,&lt;br&gt;&#xA;Food for all to eat.&lt;br&gt;&#xA;All, or just the birds?&lt;br&gt;&#xA;Blackness on this day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tango, hornpipe, rumba,&lt;br&gt;&#xA;Ballet, waltz, and salsa.&lt;br&gt;&#xA;Partners might be burnt:&lt;br&gt;&#xA;Be careful how you step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is this riddle all about?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus, once you've solved it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Can you write the next couple of verses?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-07-03T10:33:15.207" Title="A riddle of many elements" Tags="|riddle|knowledge|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17334" PostTypeId="2" ParentId="17333" CreationDate="2015-07-03T09:32:00.607" Score="10" Body="&lt;p&gt;You are talking about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/A_Song_of_Ice_and_Fire&quot;&gt;A Song of Ice and Fire series by George R R Martin.&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First a sport some love,&lt;br&gt;&#xA;Though it drinks their blood.&lt;br&gt;&#xA;Named for just a seat,&lt;br&gt;&#xA;Though power is the key.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/A_Game_of_Thrones&quot;&gt;A Game of Thrones&lt;/a&gt;&lt;br&gt;Sport refers to the game in the title. In the game of thrones, you either win or you die, referring to the drinking of blood. The seat is the Iron Throne. And power is what everyone is vying for.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then the sound of metal,&lt;br&gt;&#xA;Slamming together in war.&lt;br&gt;&#xA;That of which there are five,&lt;br&gt;&#xA;Vying in the first name.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/A_Clash_of_Kings&quot;&gt;A Clash of Kings&lt;/a&gt;&lt;br&gt;Swords clash in this book, with the war of the five kings who want to lay claim to the Iron Throne.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A hurricane with thunder,&lt;br&gt;&#xA;Lightning and pouring rain.&lt;br&gt;&#xA;Weapons are out in the darkness,&lt;br&gt;&#xA;Blades are bared to kill.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/A_Storm_of_Swords&quot;&gt;A Storm of Swords&lt;/a&gt;&lt;br&gt;Hurricane is a type of storm. Swords are the weapons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A banquet is laid out,&lt;br&gt;&#xA;Food for all to eat.&lt;br&gt;&#xA;All, or just the birds?&lt;br&gt;&#xA;Blackness on this day.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/A_Feast_for_Crows&quot;&gt;A Feast for Crows&lt;/a&gt;&lt;br&gt;A banquet is a feast, but this feast is for the crows only (resulting from the death of many in war).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tango, hornpipe, rumba,&lt;br&gt;&#xA;Ballet, waltz, and salsa.&lt;br&gt;&#xA;Partners might be burnt:&lt;br&gt;&#xA;Be careful how you step.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/A_Dance_with_Dragons&quot;&gt;A Dance with Dragons&lt;/a&gt;&lt;br&gt;Careful how you step is a reference to dance in the title. Partners might be burnt might refer to the dragons in the title, or that fact that a couple of major characters meet their death by fire in the book.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As for the next two verses,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; no one can speculate how the story may pan out, because GRR is a psychopath! But I'll take a stab at it anyway.&lt;br&gt;&lt;br&gt;&lt;strong&gt;A white march to the south,&lt;br&gt;With the breeze bringing snow,&lt;br&gt;A life and death battle,&lt;br&gt;The unexpected end of many more.&lt;br&gt;&lt;br&gt;A sweet end we hope,&lt;br&gt;The flowers bringing cheer,&lt;br&gt;A vision too bright,&lt;br&gt;In the rest of night.&lt;/strong&gt;&lt;br&gt;&lt;br&gt;Here are LogicianWithAHat's verses, which I feel are far more creative.&lt;br&gt;&lt;br&gt;&lt;strong&gt;Howling from the north,&lt;br&gt; Before the year is done&lt;br&gt; The final season roars&lt;br&gt; A vicious zephyr&lt;br&gt;&lt;br&gt;Asleep as new life grows&lt;br&gt; The song ends with a beginning&lt;br&gt; And though the night is long&lt;br&gt; We have visions of it ending&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-07-03T10:33:15.207" LastActivityDate="2015-07-03T10:33:15.207" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17338" PostTypeId="2" ParentId="17230" CreationDate="2015-07-03T10:33:36.167" Score="-2" Body="&lt;p&gt;I am not sure this is the correct answer, but this is the best way that I found to answer this puzzle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gzBVu.png&quot; alt=&quot;houses and wells interconnected without  the lines intersecting&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="13776" LastEditorUserId="7482" LastEditDate="2015-07-03T11:48:03.893" LastActivityDate="2015-07-03T11:48:03.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17340" PostTypeId="2" ParentId="17282" CreationDate="2015-07-03T11:37:54.757" Score="4" Body="&lt;p&gt;I can't believe nobody has got the 3rd clue yet...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;This one time, he met an alien and got a coat out of it.&lt;/em&gt; - ET is an alien and Jack+ET = jacket = coat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-07-03T11:37:54.757" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17342" PostTypeId="1" CreationDate="2015-07-03T14:18:56.533" Score="15" ViewCount="2538" Body="&lt;p&gt;Jack has \$1000 and plans to buy chocolates for all that money. He finds out that his favourite chocolate sells for \$1, but there is a special offer: for every four wrappers he gives to the shopkeeper, he gets to have one additional chocolate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many chocolates can Jack eat with his money?&lt;/p&gt;&#xA;" OwnerUserId="13782" LastActivityDate="2015-11-13T15:01:28.863" Title="Chocolate challenge" Tags="|mathematics|" AnswerCount="6" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17343" PostTypeId="1" AcceptedAnswerId="17345" CreationDate="2015-07-03T14:20:07.333" Score="0" ViewCount="11032" Body="&lt;p&gt;Here's another classic I couldn't find on the site:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How can you determine the weight of an elephant without using a scale?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="12109" LastActivityDate="2015-07-03T15:04:51.537" Title="Weighing an elephant" Tags="|lateral-thinking|weighing|" AnswerCount="7" CommentCount="9" ClosedDate="2015-07-03T15:32:15.263" ContentLicense="CC BY-SA 3.0" />
  <row Id="17344" PostTypeId="2" ParentId="17230" CreationDate="2015-07-03T14:21:32.283" Score="0" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DM9YC.png&quot; alt=&quot;&quot;&gt;&#xA;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this not good enough? :/&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Missing a pipe, damn it. I tried adding it but I kept ending up requiring some kind of magic&lt;/p&gt;&#xA;" OwnerUserId="13783" LastEditorUserId="13783" LastEditDate="2015-07-03T14:39:17.053" LastActivityDate="2015-07-03T14:39:17.053" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17345" PostTypeId="2" ParentId="17343" CreationDate="2015-07-03T14:22:10.003" Score="6" Body="&lt;p&gt;Here's the canonical answer, via the Archimedes Principle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Put the elephant onto a large boat, and mark the water level that the boat sinks to. Then, take the elephant out and put a bunch of marked weights onto the boat until it reaches the same water level. Add up the masses of the weights to get the weight of the elephant.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="12506" LastEditDate="2015-07-03T14:37:16.093" LastActivityDate="2015-07-03T14:37:16.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17347" PostTypeId="2" ParentId="17342" CreationDate="2015-07-03T14:28:04.060" Score="17" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1333 bars&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First step  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jack buys all the choc he can afford, 1000 bars giving him 1000 wrappers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second step:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jack hands in those 1000 wrappers to get 250 bars and 250 wrappers  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Third step:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jack hands in 248 of the wrappers and gets 62 bars, giving him 64 wrappers  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fourth step:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jack hands in those 64 wrappers for 16 bars and 16 wrappers  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fifth step:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jack hands in the 16 wrappers for 4 more bars and 4 wrappers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Final step:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jack hands in those 4 wrappers for another bar&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This gives a total of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1000 + 250 + 62 + 16 + 4 + 1 = 1333 bars&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6205" LastActivityDate="2015-07-03T14:28:04.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17349" PostTypeId="2" ParentId="17343" CreationDate="2015-07-03T14:30:38.010" Score="3" Body="&lt;p&gt;You could just do what the &lt;a href=&quot;https://en.wikipedia.org/wiki/Imperial_units&quot; rel=&quot;nofollow&quot;&gt;Imperial system&lt;/a&gt; did and make it a new unit of measurement.&#xA;&lt;br&gt;i.e - I would like to purchase 0.002 elephants of chocolate please!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For every new elephant king that appears, we simply change the unit to suit it properly.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Alternative answer:&#xA;Take the elephant into space, and calculate its gravitational pull on a know mass object (using the universal gravitation equation.)&lt;/p&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-07-03T14:49:45.893" LastActivityDate="2015-07-03T14:49:45.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17350" PostTypeId="2" ParentId="17343" CreationDate="2015-07-03T14:32:16.987" Score="3" Body="&lt;p&gt;Put the elephant on one end of a seasaw with a known weight at the other and keep adding or removing weight until it is in balance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, submerge the elephant in water in a pool that is of known volume and measure the increase in water level.&lt;/p&gt;&#xA;" OwnerUserId="13377" LastActivityDate="2015-07-03T14:32:16.987" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="17351" PostTypeId="2" ParentId="17342" CreationDate="2015-07-03T14:34:57.533" Score="3" Body="&lt;p&gt;You eat 1333&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You start with 1000 chocolates and eat them all so you get 1000 wrappers. Then, as long as you have more than 3 wrappers you trade them for more chocolate&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;chocos = 1000&#xA;wrappers = chocos&#xA;while wrappers &amp;gt; 3:&#xA;    chocos += wrappers / 4&#xA;    wrappers = wrappers / 4 + wrappers % 4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;On first iteration you get 250 additional wrappers, then 250 / 4 = 62 plus 2 leftover, then 64 / 4 = 16, then 16 / 4 = 4 and finally 4 / 4 = 1 which means you have 1000 + 250 + 62 + 16 + 4 + 1 = 1333 chocolates and no leftover wrappers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: thanks luxmi12 for pointing out the 2 forgotten wrappers&lt;/p&gt;&#xA;" OwnerUserId="13715" LastEditorUserId="13715" LastEditDate="2015-07-03T14:42:53.360" LastActivityDate="2015-07-03T14:42:53.360" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17352" PostTypeId="2" ParentId="17343" CreationDate="2015-07-03T14:35:15.770" Score="1" Body="&lt;p&gt;Take a truck who can support a known weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-If the truck can support it, take another truck who can support a little less weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-If he can't, take one who can support a little more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reproduce until you find the perfect truck and you get his supported weight.&lt;/p&gt;&#xA;" OwnerUserId="5359" LastActivityDate="2015-07-03T14:35:15.770" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="17353" PostTypeId="2" ParentId="17343" CreationDate="2015-07-03T14:50:13.493" Score="5" Body="&lt;p&gt;Place an object of known mass a fixed distance away from the elephant, and measure the gravitational attraction the elephant exerts on the object. This will let you compute the elephant's mass, and therefore its weight.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastActivityDate="2015-07-03T14:50:13.493" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17354" PostTypeId="2" ParentId="16474" CreationDate="2015-07-03T15:00:46.980" Score="4" Body="&lt;h2&gt;&lt;strong&gt;&lt;em&gt;On segment 3:&lt;/em&gt;&lt;/strong&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There are four instances of go. This probably isn't important at all, but it's starting to feel like everything is important here...&lt;br&gt;&#xA;Noticing the misplaced periods, we get...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;ouskkuueltr&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is a eleven-letter string with &lt;code&gt;1 o, 3 u, 1 s, 2 k, 1 e, 1 l, 1 t, 1 r&lt;/code&gt; and backwards reads&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;rtleuukksuo&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The verses contain the operations: &lt;code&gt;+1-2, rounded up rounded down to 0 (I think,) then something about an increase (likely?,) multiplication, the current result squared, minus 2, minus half of the remainder. Possibly rounded down and then maybe -1.&lt;/code&gt; The answer is a number between one and six, as per the remaining bullets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The clue about the TI 84 Plus (a graphing calculator) gives us the numbers 48 and 84.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obvious steps to answer: Figure out what we are multiplying by and figure out what number we start with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;A few minutes later, after some bowls are placed to remind us where each bullet is...&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Four instances of the word go... 4... +1 -2 is 2... times its square is 8... -2 is 6... halved is 3... minus 1 is 2...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You set bullet 2 aside just in case. Actually, you load the gun with that bullet, just so if he walks into the room and you haven't solved it yet, you can give him a loud surprise.&lt;/p&gt;&#xA;" OwnerUserId="12506" LastEditorUserId="12506" LastEditDate="2015-07-05T16:33:59.667" LastActivityDate="2015-07-05T16:33:59.667" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="17355" PostTypeId="2" ParentId="17343" CreationDate="2015-07-03T15:04:22.900" Score="0" Body="&lt;p&gt;Elephants are known for their patience so find yourself a beach chair, some tanning lotion, LOTS of beer and a good pair of sunglasses. (Snacks are a plus but avoid peanuts as they may render the whole thing useless)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you are set , start the timer on your watch and observe the elephant. Once the elephant is tired of you staring at it and leaves, stop your timer. And you will know how much that elephants waits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Get it ? get it?&lt;/p&gt;&#xA;" OwnerUserId="11251" LastActivityDate="2015-07-03T15:04:22.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17356" PostTypeId="2" ParentId="17343" CreationDate="2015-07-03T15:04:51.537" Score="1" Body="&lt;p&gt;My sister came up with this one:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Make a small cut on the elephant.&lt;/li&gt;&#xA;&lt;li&gt;Pump out all of its blood.&lt;/li&gt;&#xA;&lt;li&gt;Cut the elephant into pieces and weigh each piece (by putting it in a tank of water)&lt;/li&gt;&#xA;&lt;li&gt;Weigh the bloody stuff. (by putting it in a bag with negligible weight and then putting it in a tank of water)&lt;/li&gt;&#xA;&lt;li&gt;Add up all the weights.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="12109" LastActivityDate="2015-07-03T15:04:51.537" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17357" PostTypeId="1" AcceptedAnswerId="17587" CreationDate="2015-07-03T15:34:39.077" Score="10" ViewCount="2718" Body="&lt;p&gt;&lt;em&gt;I apologize in advance for the lack of rhymes (as it seems quite difficult here). I have instead added a lot of 'depth' to this riddle (which should be easily unwound once solved) - Hopefully it isn't too obscure&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In some sports I can be used, but only by some. In this case here, it&#xA;  references to one.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I can be used in some work, more specific than others. I'm not talking&#xA;  about animals or water which could be seen as my brothers. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I am nearby the shops for others to get. Some might ignore me, some might not forget.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I might be stronger than others, or weaker than few. Some might have just&#xA;  one, or maybe even two.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Make sure to explain each line!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint #1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'water' could also be substituted with the word 'sea(s)' or 'ocean(s)'. - I admit this riddle is broad at parts, but once you know, the dots should connect.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'animals' is a reference to an animal with no legs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Maybe &lt;a href=&quot;https://puzzling.stackexchange.com/users/11244/mark-n?tab=profile&quot;&gt;this guy&lt;/a&gt; might know the answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Big Hint #4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What &lt;em&gt;can&lt;/em&gt; you 'have' when you are sick? How would you fill your hands with letters (What number might be associated with a hand)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2015-07-17T20:56:39.193" Title="When you are sick you might have me, with a hand full of letters" Tags="|riddle|word|" AnswerCount="22" CommentCount="22" ClosedDate="2015-07-08T16:46:07.053" ContentLicense="CC BY-SA 3.0" />
  <row Id="17358" PostTypeId="2" ParentId="16474" CreationDate="2015-07-03T16:39:49.217" Score="3" Body="&lt;h2&gt;Puzzle 3&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Given Hint 1b, I suspect the solution involves&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the positions of the keys on a TI-84 Plus calculator, which can be seen in the following image: &lt;br&gt;&lt;br&gt; &lt;img src=&quot;https://i.stack.imgur.com/Bd5Ji.jpg&quot; alt=&quot;calculator&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's look again at the poem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Now, go go go g.o!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This might suggest we should start from the ENTER key.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;With you, it's like one step forward.s&lt;br&gt;&#xA;and two steps bac.k.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One step up and two steps down (more or less) take us to the (-) key.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I climbed the stairs&lt;br&gt;&#xA;until I stood at the top&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Go all the way up to the VARS key or maybe even the TRACE key.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;and counted the stairs I too.k&lt;br&gt;&#xA;and threw them at yo.u.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Six to eight, according to what we count as the top.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;you pulled me down below yo.u!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is probably something to do with a fraction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;when we danced around the squar.e,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The $x^2$ key?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I dropped two penny down the wel.l.&lt;br&gt;&#xA;You dropped half of the pennies lef.t.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Half is the reciprocal of two, so maybe we should be looking at the $x^{-1}$ key, which is adjacent to the $x^2$ key.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm still not sure about the last two lines, nor about the interpretation of the misplaced full stops in the poem.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-07-03T16:39:49.217" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17359" PostTypeId="2" ParentId="17342" CreationDate="2015-07-03T16:44:12.733" Score="24" Body="&lt;p&gt;Certainly Jack can't eat all that chocolate in one sitting, and he'll get sick of chocolate if he binges on it for a few days. Even if he buys it all at once and rations it, it could get discolored and get a funky texture long before he gets a chance to eat it all. The store owner might not even stock more than a few dozen of his favorite chocolates at any given time. Besides, $1000 is an awful lot of cash for Jack to carry around, and if he got mugged he wouldn't be able to buy any chocolate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the most rational solution is...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...for him to buy and eat one chocolate per day, over the course of 1333 days.&lt;br&gt;&#xA; Jack starts with \$1000 .&lt;br&gt;&lt;br&gt;&#xA; He buys 1 chocolate; now he has \$999 .&lt;br&gt;&lt;br&gt;&#xA; After this, for every 3 chocolates he buys, he gets 4.&lt;br&gt;&lt;br&gt;&#xA; 1 chocolate + (\$999  / \$3  * 4 chocolates) = 1333 chocolates&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If he puts his chocolate money in a high-yield savings account, he might even be able to buy one more chocolate on the interest he accrues during those roughly 3-1/2 years.&lt;/p&gt;&#xA;" OwnerUserId="1989" LastEditorUserId="1989" LastEditDate="2015-07-03T21:05:09.677" LastActivityDate="2015-07-03T21:05:09.677" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="17360" PostTypeId="2" ParentId="17357" CreationDate="2015-07-03T16:50:22.520" Score="6" Body="&lt;p&gt;I can't resist...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BALLS!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In some sports I can be used, but only by some. In this case here, it references to one.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Balls are used in many sports, but there's only one in the game at a time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I can be used in some work, more specific than others. I'm not talking about animals or water which could be seen as my brothers.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ball bearings can be used in work. Not sure about the animals/water bit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am nearby the shops for others to get. Some might ignore me, some might not forget.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can buy balls in the shops. Some people aren't interested in sports (they ignore balls) but some are always thinking about it (they don't forget).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I might be stronger than others, or weaker than few. Some might have just one, or maybe even two.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Balls/&lt;em&gt;cojones&lt;/em&gt; symbolise courage, which is similar to strength. Many people have two (&lt;em&gt;ahem&lt;/em&gt;), although some - such as &lt;a href=&quot;https://en.wikipedia.org/wiki/Hitler_Has_Only_Got_One_Ball&quot;&gt;Adolf Hitler&lt;/a&gt;, reputedly - have just one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-07-03T16:50:22.520" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17361" PostTypeId="2" ParentId="17357" CreationDate="2015-07-03T16:51:47.090" Score="1" Body="&lt;p&gt;Second attempt!:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Condom!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In some sports I can be used, but only by some.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some use it in this 'sport', some don't&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In this case here, it references to one.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some kind of 'sports' &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be used in some work, more specific than others. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Condom does have many usage such as strengthen concrete road / water storage / making fire.  And of course it's main usage - to protect&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not talking about animals or water which could be seen as my brothers.    &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  In old times condom are made from animal parts and reuse by cleaning with water O.O  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am nearby the shops for others to get. Some might ignore me, some might not forget.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you could buy condom at nearby shop.  If you're intended to go there for the condom, you'll not forgot.  Otherwise you will ignore  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might be stronger than others, or weaker than few. Some might have just one, or maybe even two.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Different textures...?  Using 2 is actually dangerous as well...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here was my first attempt which failed:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could it be:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; vitamin?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In some sports I can be used, but only by some.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Athletes can have vitamin in certains sport to improve performance  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In this case here, it references to one.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure on this one  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be used in some work, more specific than others. I'm not talking about animals or water which could be seen as my brothers.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; work such as pharmacy?  And you usually take vitamin with water.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am nearby the shops for others to get. Some might ignore me, some might not forget.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you could buy vitamin at nearby shop.  If you're intended to go there for the vitamin, you'll not forgot.  Otherwise you will ignore?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might be stronger than others, or weaker than few. Some might have just one, or maybe even two.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Taking different vitamins  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For title:   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You might have vitamin when you're sick, with a handful of vitamin A,B,C...  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also..&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MarkN is sick at the moment? =D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="11244" LastEditDate="2015-07-03T17:13:45.923" LastActivityDate="2015-07-03T17:13:45.923" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17362" PostTypeId="2" ParentId="17342" CreationDate="2015-07-03T17:22:32.933" Score="17" Body="&lt;p&gt;He can eat $1333$ chocolates. To show this mathematically without calculating each step, imagine that he starts with $4000$ wrappers instead of $\$1000$. This is, of course, equivalent. Considering that each chocolate costs $4$ wrappers and yields one, we find that the number of wrappers he has remaining after eating $c$ chocolates is:&#xA;$$4000-4c+c$$&#xA;or&#xA;$$4000-3c$$&#xA;So long as he has at least $4$ wrappers remaining, he can buy more chocolate - so we are looking for the first $c$ such that the above quantity is less than $4$. Solving yields:&#xA;$$4000-3c&amp;lt;4$$&#xA;$$3996&amp;lt;3c$$&#xA;$$\frac{3996}3&amp;lt;c$$&#xA;$$1332&amp;lt;c$$&#xA;and the first integer satisfying this is obviously $1333$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is worthy of note that this equals $\lfloor\frac{4000}{3}\rfloor$, the value if each chocolate simply cost $3$ wrappers (and yielded no wrappers). However, that problem is not equivalent - consider the case where each chocolate cost $5$ wrappers, but yielded $2$ wrappers. Even though each chocolate is still a net loss $3$ wrappers, he will only be able to buy $1332$ chocolates. In the extreme case, if each chocolate cost $4001$ wrappers, but yielded $3998$ wrappers, he could buy &lt;em&gt;none&lt;/em&gt; - hence one must use the somewhat more sophisticated approach above rather than just dividing.&lt;/p&gt;&#xA;" OwnerUserId="4765" LastEditorUserId="4765" LastEditDate="2015-07-03T17:28:56.733" LastActivityDate="2015-07-03T17:28:56.733" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17363" PostTypeId="1" AcceptedAnswerId="17384" CreationDate="2015-07-03T19:31:39.873" Score="15" ViewCount="516" Body="&lt;p&gt;A simple challenge for the mathematically inclined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find a function $f\left( {{x_1},{x_2},{x_3},{x_4}} \right)$ such that $$\begin{gathered}&#xA;  f\left( {2,3,4,5} \right) = 1 \hfill \\&#xA;  f\left( {1,3,4,5} \right) = 2 \hfill \\&#xA;  f\left( {1,2,4,5} \right) = 3 \hfill \\&#xA;  f\left( {1,2,3,5} \right) = 4 \hfill \\&#xA;  f\left( {1,2,3,4} \right) = 5 \hfill \\ &#xA;\end{gathered}$$&#xA;In constructing $f$, you may use the operations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;addition, e.g. $x_1 + x_2$&lt;/li&gt;&#xA;&lt;li&gt;multiplication, e.g. $x_1x_2$&lt;/li&gt;&#xA;&lt;li&gt;remainder, e.g. ${x_1}\backslash {x_2}$, which is the non-negative remainder of integer $x_1$ divided by integer $x_2$; this operation has the same precedence as multiplication, hence $x_1x_2 \backslash x_3x_4$ is interpreted as $\left( {\left( {{x_1}{x_2}} \right)\backslash {x_3}} \right){x_4}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note that you may &lt;em&gt;not&lt;/em&gt; use numeric constants, subtraction, or division.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function may use as much or as little operation grouping as needed, e.g. $\left( {{x_1} + {x_2}} \right){x_1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function does not have to be valid for any inputs besides the five cases listed (you may have remainder-of-division-by-zero in extraneous cases).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your &lt;strong&gt;score&lt;/strong&gt; is the number of operations in $f$. Lower scores are better. The coveted green checkmark will go to the &lt;strong&gt;correct solution with the lowest score&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can you produce a perfectly primitive parameter permuter?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2015-07-04T08:27:47.837" Title="A Perfectly Primitive Parameter Permuter" Tags="|mathematics|formation-of-numbers|" AnswerCount="4" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17364" PostTypeId="2" ParentId="17313" CreationDate="2015-07-03T19:53:33.330" Score="5" Body="&lt;p&gt;Translating the text on the paper from Base 64 gives you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Go back to where it all began.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Go to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/17167/the-twenty-doors-room-1&quot;&gt;Room 1&lt;/a&gt; and look in the &lt;a href=&quot;https://puzzling.stackexchange.com/posts/17167/edit&quot;&gt;source&lt;/a&gt;. You see a message:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The profile of the evil guy who began all of this may hold something interesting...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the OP's profile, you see the message:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TidoTt ilhurr om hnysS saieqeofR fee res gs edwo 3o wDore&quot;ee.r r ToiL&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Decoding it with a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; railfence  cipher with 5 rails &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reveals that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The required word for Room 3 of The Twenty Doors Series is &quot;Legalise&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12109" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-07-04T16:30:46.620" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17365" PostTypeId="2" ParentId="17002" CreationDate="2015-07-03T19:54:44.070" Score="6" Body="&lt;p&gt;The missing color is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; pink&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because the colors correspond to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; characters from the web machinima series &lt;a href=&quot;http://rvb.wikia.com/wiki/Red_vs._Blue_Wiki&quot; rel=&quot;nofollow&quot;&gt;Red Vs. Blue&lt;/a&gt;.  The characters are soldiers on opposing teams, each of whom appears in armor of their own color.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The story contains a cryptic reference to each&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; of the main characters on the show, listed below with their colors:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; tuckered = Captain Lavernius Tucker = turquoise&lt;br&gt;&#xA; donut = Private Franklin Delano Donut  = pink&lt;br&gt;&#xA; church = Private Leonard L. Church = indigo&lt;br&gt;&#xA; sister = Private Kaikaina &quot;Sister&quot; Grif = yellow&lt;br&gt;&#xA; dick = Captain Richard &quot;Dick&quot; Simmons  = crimson&lt;br&gt;&#xA; doc = Medical Officer Super Private First Class Frank &quot;Doc&quot; DuFresne = violet&lt;br&gt;&#xA; dexter = Captain Dexter Grif = orange&lt;br&gt;&#xA; michael = Captain Michael J. Caboose = blue&lt;br&gt;&#xA; robotico = Lopez (a robot who speaks spanish) = brown&lt;br&gt;&#xA; sargio = Colonel Sarge = red  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hints:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;we're stIll going, further thaN smallville or alMost anything else&quot; : the show is in its thirteenth season&lt;br&gt;&#xA; &quot;Six Pedals&quot; is a running gag on the show: &lt;a href=&quot;http://rvb.wikia.com/wiki/Six_Pedals&quot; rel=&quot;nofollow&quot;&gt;http://rvb.wikia.com/wiki/Six_Pedals&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Additional Explanation (By OP):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the riddle, the 3 E's were a reference to the Electronice Entertainment Expo that was going on at the same time as this was posted, to direct people towards the gaming industry - a bit obscure I know.&#xA;&#xA;In the hint, the texan with the tank refers to the Character 'Tex' that appears halfway through the first season. Again the capital letters are: A,C,M,A,I,H,I,N,M that spell out Machinima, of which it is the first commercially successful. The second part of the hint refers to it being the longest continuous episodal series after Doctor Who, longest web series, etc.. In the second hint, headlight fluid is like the 6 pedals, a long running gag of the series.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4385" LastEditorUserId="11251" LastEditDate="2015-07-14T16:02:35.850" LastActivityDate="2015-07-14T16:02:35.850" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17366" PostTypeId="2" ParentId="17342" CreationDate="2015-07-03T21:05:16.713" Score="6" Body="&lt;p&gt;A slightly different proof without iteration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case above is equivalent to the series:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{k-&amp;gt;\infty}\sum_{i=0}^k 1000\times\left(\frac{1}{4}\right)^i $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where each 'iteration' is one set of exchanges; the first ($i=0$) is when he buys the initial thousand chocolates, and then from there on out he buys $\frac{1}{4}$ the previous iteration's amount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is a &lt;a href=&quot;http://math2.org/math/expansion/tests.htm&quot; rel=&quot;nofollow&quot;&gt;convergent geometric series&lt;/a&gt; of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=0}^\infty (a\times r^i) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $a=1000$ and $r=\frac{1}{4}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That converges to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{a}{1-r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $r&amp;lt;1$, which in this case means it converges to $\lfloor{1000/0.75}\rfloor$ or $1333$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, he can buy 1333 chocolates over his various iterations, with only 1 wrapper remaining at the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As pointed out in comments, we actually need to subtract one from the numerator for this to truly work out: while our infinite series is an accurate representation, we must have one wrapper left over at the end (the wrapper for the last chocolate we ate), which is equivalent to 0.25 dollars/chocolates.  As such, the true solution is&#xA;$\lfloor{\frac{1000-0.25}{0.75}}\rfloor$.  This is still $1333$ in this case, of course.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Of course, this assumes he doesn't have the ability to invest his money in the market.  If he invested at 10% annual return, and bought chocolates with the return, he could buy [100/.75] or 133 chocolates per year (plus an occasional bonus chocolate every four years, for leap years) indefinitely, assuming he got identical returns every year and/or conserved excesses to spend or recapitalize in down years.  He'd only take about ten years to recapture his full 1333 chocolates (actually 1332, so it would be a few days into the next year) to equal his earnings if he simply ate it all now, assuming no inflation or change in store policies.  &lt;/p&gt;&#xA;" OwnerUserId="2437" LastEditorUserId="2437" LastEditDate="2015-11-13T15:01:28.863" LastActivityDate="2015-11-13T15:01:28.863" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="17367" PostTypeId="2" ParentId="17363" CreationDate="2015-07-03T21:34:21.397" Score="2" Body="&lt;p&gt;Number of operations: &lt;strong&gt;33&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}f(x_1,x_2,x_3,x_4) = \,\,\, &amp;amp;x_2\setminus x_1  +\\&#xA;    &amp;amp;(x_4\setminus (x_1+x_2)) * (x_2\setminus (x_1+x_1)) * (x_4\setminus x_2) +\\&#xA;    &amp;amp;((x_2*x_3*x_4)\setminus (x_1+x_2)) * (x_1+x_2) +\\&#xA;    &amp;amp;(x_4\setminus (x_4\setminus x_3)) * x_2 * x_2 +\\&#xA;    &amp;amp;(x_4\setminus (x_1+x_2)) * ((x_4\setminus (x_1+x_1+x_1+x_1+x_1))\setminus x_3) * (x_2+x_3)&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can verify that for the $n$&lt;sup&gt;th&lt;/sup&gt; input, all lines expect the $n$&lt;sup&gt;th&lt;/sup&gt; one will be zero, while the $n$&lt;sup&gt;th&lt;/sup&gt; line will equal $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the same function written in Python so it's easier to verify:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;def f(a,b,c,d):&#xA;return b%a +\&#xA;        (d%(a+b))*(b%(a+a))*(d%b) +\&#xA;        ((b*c*d)%(a+b))*(a+b) +\&#xA;        (d%(d%c))*b*b +\&#xA;        (d%(a+b))*((d%(a+a+a+a+a))%c)*(b+c)&#xA;print [f(2,3,4,5),f(1,3,4,5),f(1,2,4,5),f(1,2,3,5),f(1,2,3,4)]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-07-03T21:42:27.427" LastActivityDate="2015-07-03T21:42:27.427" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17368" PostTypeId="2" ParentId="17156" CreationDate="2015-07-03T21:42:01.987" Score="0" Body="&lt;p&gt;Solution to puzzle as currently defined:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Optimal&#xA;20 miles&#xA;20 gallons&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an alternating fashion, one container at a time, slide/carry each forward an arms reach distance while only walking forward. When the halfway point is reached carry the remainder of the water to the finish line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;General solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1gal/mile regardless of variable A&lt;/p&gt;&#xA;" OwnerUserId="13788" LastActivityDate="2015-07-03T21:42:01.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17370" PostTypeId="2" ParentId="17363" CreationDate="2015-07-03T23:47:10.790" Score="6" Body="&lt;h2&gt;11 operations&lt;/h2&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(d+d+d+b%a)%(a+b+c+b*b%d+a)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This was a simple idea that got messy. It can probably be made cleaner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the numbers &lt;code&gt;1,2,3,4,5&lt;/code&gt; add to &lt;code&gt;15&lt;/code&gt;, so if we could use constants and subtraction, we could do&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;15-(a+b+c+d)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We can sort-of simulate subtraction using mod because &lt;code&gt;x%y&lt;/code&gt; equals &lt;code&gt;x-y&lt;/code&gt; whenever $y \in \left(x/2 ,x \right]$. We can ensure this condition holds by instead doing &lt;code&gt;(x+c)%(y+c)&lt;/code&gt; for some sufficiently large $c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the thing that remains is to create the constant $15$. This part is messy and uses surprisingly many operations. After some experimenting, I found that &lt;code&gt;a-b%a&lt;/code&gt; always equals $1$, but adding it $15$ times is too costly. Also, &lt;code&gt;b*b%d&lt;/code&gt; equals $(4,4,4,4,0)$ for the five cases, while &lt;code&gt;d&lt;/code&gt; equals $(5,5,5,5,4)$, so &lt;code&gt;(d+d+d+d)-(b*b%d)&lt;/code&gt; equals $(16,16,16,16,16)$. Subtracting the $1$-vector from before gives $(15,15,15,15,15)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, we have the expression &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(d+d+d+d-b*b%d-a+b%a)-(a+b+c+d)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Moving the minuses to make them plusses gives &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(d+d+d+d+b%a)-(a+b+c+d+b*b%d+a)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Finally, cancel a &lt;code&gt;+d&lt;/code&gt; from both sides, and change the &lt;code&gt;-&lt;/code&gt; to a &lt;code&gt;%&lt;/code&gt;, which turns out to works because both sides are large enough to get the solution&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(d+d+d+b%a)%(a+b+c+b*b%d+a)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A near-miss is to &quot;estimate&quot; $15$ as &lt;code&gt;d+d+d&lt;/code&gt;, which gives $(15,15,15,15,12)$, and then&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(d+d+d)%(a+b+c+d)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;gives $(1,2,3,4,2)$, as does &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(d+d)%(a+b+c)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I still haven't found a short way though to change the $2$ to a $5$ in the last case.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-07-04T00:05:27.373" LastActivityDate="2015-07-04T00:05:27.373" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17371" PostTypeId="2" ParentId="17230" CreationDate="2015-07-04T00:37:56.247" Score="7" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZffEW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The blue is going through the house.&lt;/p&gt;&#xA;" OwnerUserId="13792" LastActivityDate="2015-07-04T00:37:56.247" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17372" PostTypeId="2" ParentId="17363" CreationDate="2015-07-04T01:29:35.840" Score="8" Body="&lt;h2&gt;9 operations&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Here's one based on xnor's last line:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(d+d)%(a+b+c)+(c+d)*c%b*c&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This works because &lt;code&gt;(c+d)*c%b&lt;/code&gt; is 1 in the last case and 0 in all the others. As xnor noted, &lt;code&gt;(d+d)%(a+b+c)&lt;/code&gt; gives $(1,2,3,4,2)$, so multiplying &lt;code&gt;(c+d)*c%b&lt;/code&gt; by &lt;code&gt;c&lt;/code&gt; and adding it produces $(1,2,3,4,5)$. It doesn't matter whether it is added before or after the last modulus, so &lt;code&gt;(d+d+(c+d)*c%b*c)%(a+b+c)&lt;/code&gt; works too.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-07-04T02:47:11.390" LastActivityDate="2015-07-04T02:47:11.390" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17373" PostTypeId="1" AcceptedAnswerId="17375" CreationDate="2015-07-04T02:01:22.740" Score="8" ViewCount="2378" Body="&lt;p&gt;Bob has a cheese cube and wants to cut it up into exactly $6$ equal pieces for his guests. These guests are professional cheese connoisseurs who will complain if they get less than anyone else. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bob doesn't have the tools necessary to make arbitrary and precise cuts. He can be precise enough if he slices from a corner to corner or bisecting along an edge, but not randomly through an edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Notes:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The guests only care about volume, the shape of the pieces doesn't matter.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here's a cube to help you visualize it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lhvPL.png&quot; alt=&quot;CHEEEEEEEEEEESEEEE&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible for Bob to please his guests?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Cleaned up unnecessary restrictions.&lt;/p&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="8874" LastEditDate="2016-03-23T15:38:53.540" LastActivityDate="2016-03-23T15:38:53.540" Title="Cut up a cheese cube into $6$ equal parts with $5$ slices" Tags="|geometry|lateral-thinking|visual|dissection|" AnswerCount="5" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17374" PostTypeId="2" ParentId="17373" CreationDate="2015-07-04T04:26:44.820" Score="10" Body="&lt;p&gt;I think that this can be achieved by cutting the cube into 6 pyramids like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CHl3R.jpg&quot; alt=&quot;Image found from Images.google.com . Thanks Google! You rock!&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7482" LastEditorUserId="7482" LastEditDate="2015-07-04T04:31:56.400" LastActivityDate="2015-07-04T04:31:56.400" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17375" PostTypeId="2" ParentId="17373" CreationDate="2015-07-04T04:39:29.290" Score="8" Body="&lt;p&gt;Here's how I propose to solve this conundrum. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We need to cut out a tetrahedron from the cube. This can be done by cutting along three corners each time as shown in the image. As a result, we have one tetrahedron, whose volume is 1/3&lt;sup&gt;rd&lt;/sup&gt; of the original cube.&lt;br&gt;&lt;br&gt; We are left with four pieces, which are equal in volume, whose three faces are half the triangle from the cube face. The volume of those pieces cannot be calculated very easily, but we know that the sum of their volume is 2/3&lt;sup&gt;rd&lt;/sup&gt; of the original cube (since 1/3&lt;sup&gt;rd&lt;/sup&gt; of the volume went into the tetrahedron). This means each of them are 2/3 $\div$ 4, i.e 1/6&lt;sup&gt;th&lt;/sup&gt; of the original cube.&lt;br&gt;&lt;br&gt; Since we know how to bisect a piece, we can cut the tetrahedron into two equal pieces and the volume of each piece is 1/3 $\div$ 2, i.e 1/6&lt;sup&gt;th&lt;/sup&gt; of the original cube. &lt;br&gt;&lt;br&gt; Hence, we have our 6 pieces whose size is 1/6&lt;sup&gt;th&lt;/sup&gt; of the original cube.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Image for reference: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7fWDj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-07-04T06:16:55.423" LastActivityDate="2015-07-04T06:16:55.423" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17376" PostTypeId="2" ParentId="17357" CreationDate="2015-07-04T05:25:35.937" Score="2" Body="&lt;p&gt;I guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A puck&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In some sports I can be used, but only by some. In this case here, it references to one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Though there are a few sports that make use of pucks: people are typically talking about &quot;hockey pucks&quot;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be used in some work, more specific than others.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In computer terms a &quot;puck&quot; is a little sensor pad device used by CAD and designers for drawing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not talking about animals or water which could be seen as my brothers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Puck is apparently a &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_Animal_Crossing_characters&quot; rel=&quot;nofollow&quot;&gt;penguin on animal crossing&lt;/a&gt; and yes water hockey is a sport derived from ice hockey (also ice &amp;amp; water for brothers)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am nearby the shops for others to get. Some might ignore me, some might not forget.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hockey's popularity can be quite random so some ignore hockey equipment and other's buy it (&lt;em&gt;not forget&lt;/em&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might be stronger than others, or weaker than few.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The hockey puck is much stronger than those used in other sports (Novuss, Air hockey, street hockey) but is weaker than those used in Shuffleboard or Water Hokey (both use metal in their designs)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some might have just one, or maybe even two.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most hockey derivatives use a single puck during play but games like &lt;a href=&quot;https://en.wikipedia.org/wiki/Novuss&quot; rel=&quot;nofollow&quot;&gt;Novuss&lt;/a&gt; uses two &quot;pucks&quot; (like the cue ball in billards)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the title: &lt;em&gt;When you are sick you might have me, with a handful of letters&lt;/em&gt; - I have a few guesses&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One &quot;pucks&quot; has five letters (five fingers). Second Hockey becomes hacking (cough) with a few letter changes. And finally &quot;He's in the bathroom &lt;em&gt;puking&lt;/em&gt; up his breakfast&quot; needs only a few letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11239" LastActivityDate="2015-07-04T05:25:35.937" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17377" PostTypeId="1" CreationDate="2015-07-04T05:49:25.850" Score="-4" ViewCount="2045" Body="&lt;p&gt;Let see who can crack this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the missing numbers in the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MfjSi.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explain the logic.&#xA;Challenge for all masterminds&lt;/p&gt;&#xA;" OwnerUserId="13582" LastEditorUserId="13582" LastEditDate="2015-07-06T09:14:13.707" LastActivityDate="2015-07-08T14:56:43.030" Title="Find Missing numbers in the picture" Tags="|logical-deduction|pattern|" AnswerCount="4" CommentCount="3" ClosedDate="2015-07-08T23:17:03.203" ContentLicense="CC BY-SA 3.0" />
  <row Id="17378" PostTypeId="2" ParentId="17377" CreationDate="2015-07-04T06:05:33.060" Score="3" Body="&lt;p&gt;Here's a stab at it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First triangle: 3&lt;br&gt;Second triangle: 6&lt;br&gt;Third triangle: 14&lt;br&gt;Fourth triangle: 32&lt;br&gt;&lt;br&gt;The number in the top circle forms an arithmetic progression with the upper two numbers (in the order: circle number, left triangle number, right triangle number). The lower triangle contains the common difference of the A.P. reversed in case if the C.D. is a two digit number, else if it is a one digit number we add 1 to the C.D.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-07-04T06:11:37.287" LastActivityDate="2015-07-04T06:11:37.287" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17379" PostTypeId="2" ParentId="17357" CreationDate="2015-07-04T06:34:40.783" Score="6" Body="&lt;p&gt;I think you are talking about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a die.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In some sports I can be used, but only by some. In this case here, it references to one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A die is used in games (not exactly sports) such as ludo, snakes and ladders etc. A die is a singular item, the plural of which is dice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be used in some work, more specific than others.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dies are used in die casting processes to produce several everyday items.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not talking about animals or water which could be seen as my brothers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dyes (homophone of die) can be &lt;a href=&quot;https://en.wikipedia.org/wiki/Category:Animal_dyes&quot;&gt;derived from animals&lt;/a&gt; or could be &lt;a href=&quot;https://en.wikipedia.org/wiki/Dye_tracing#Water_tracing&quot;&gt;water based&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am nearby the shops for others to get. Some might ignore me, some might not forget.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can buy a die in many shops and those that play board games on a regular basis will always remember to buy one when they lose them, while the others wouldn't really care about them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might be stronger than others, or weaker than few.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Not sure about this clue...&lt;/em&gt; I suspect this might be related to dyes. Which may be of a temporary or permanent nature.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some might have just one, or maybe even two.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Games like ludo and snakes and ladders use one die, while games like Yahtzee and two-dice pig can use two dice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The title: When you are sick you might have me, with a handful of letters&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you are sick, you might have a disease, which is di(seas)e. The word 'seas' sounds like C's, which could mean a handful of letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-07-04T07:42:06.280" LastActivityDate="2015-07-04T07:42:06.280" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17380" PostTypeId="2" ParentId="17373" CreationDate="2015-07-04T06:41:42.460" Score="14" Body="&lt;p&gt;I think he can do it with 3 cuts!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9gfEE.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cut along the planes marked in red, green and blue. Each plane divides the cube in half, and all three intersect along a space diagonal. The middle image shows what these cuts look like when the cube is viewed in the direction of that space diagonal. The right image shows what each piece looks like.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-07-04T06:41:42.460" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17381" PostTypeId="2" ParentId="17342" CreationDate="2015-07-04T07:00:18.523" Score="4" Body="&lt;p&gt;This may be solved as follows using some nice math.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A non-infinate fractional series. &lt;br&gt;&#xA; 1000 * (1 + 1/4 + 1/16 + 1/64 + 1/256 + 1/1024) = &lt;a href=&quot;http://m.wolframalpha.com/input/?i=1000%20*%20%281%20%2B%201%2F4%20%2B%201%2F16%20%2B%201%2F64%20%2B%201%2F256%20%2B%201%2F1024%29&amp;amp;x=0&amp;amp;y=0&quot; rel=&quot;nofollow&quot;&gt;1333&lt;/a&gt;&lt;br&gt;&#xA; Each layer adds up the next iterations worth of wrappers till there is only one left.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user2674" LastActivityDate="2015-07-04T07:00:18.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17382" PostTypeId="2" ParentId="17377" CreationDate="2015-07-04T07:13:30.863" Score="4" Body="&lt;p&gt;that follows a equation:-&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$value~in~circle=value~of~(base~triangle-(right~triangle-left~triangle))$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hence,the solution for missing,'?'s will be,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 6,1,20,33&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;respectively.&lt;/p&gt;&#xA;" OwnerUserId="11306" LastActivityDate="2015-07-04T07:13:30.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17383" PostTypeId="1" AcceptedAnswerId="17388" CreationDate="2015-07-04T08:00:59.380" Score="10" ViewCount="1042" Body="&lt;p&gt;This is part of &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=title%3A%22The+Twenty+Doors%22&quot;&gt;The Twenty Doors&lt;/a&gt; series. &#xA;&lt;br&gt;The previous one is &lt;a href=&quot;https://puzzling.stackexchange.com/questions/17313/the-twenty-doors-room-3&quot;&gt;The Twenty Doors! (ROOM 3)&lt;/a&gt;&#xA;&lt;br&gt;The next one is &lt;a href=&quot;https://puzzling.stackexchange.com/questions/17423/the-twenty-doors-room-5&quot;&gt;The Twenty Doors! (ROOM 5)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You now enter room 4. This one has a load of A's and B's carved into the wall, along with a message. &lt;code&gt;AAAAA ABAAABAAAB ABBABABBAAAABAA AAAAAABBAAAAABB AAAAB ABAAABAAAB BAABABABAAAABAAABBAABAABABABBA AABABABBABBAABBBAAAA ABAAAABBAA BAABAAABBBAABAA ABABBAAAAAABAAAABBAA AAABABAAAABABBAABBBABAABAABBABAABBABAAAAAAAAAABABB &#xA;I may not be distinct but I've always loved bacon.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the floor, as usual, there is a bit of paper.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bmmp rum gq rfc uyw rm qydcrw!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But this time, there are three doors (not counting the one you just came through). You turn over the paper, and there is something written on the back&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This time, press the button on your chosen door to open it, but if you choose the wrong one, prepare to DIE!!!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hmm... This one involves two cryptograms. Damn. Anyway, you might as well try.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which door should you go through?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next door will be added when this door is solved!&lt;/p&gt;&#xA;" OwnerDisplayName="user9377" LastEditorUserId="-1" LastEditorDisplayName="user9377" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-07-05T10:43:19.997" Title="The Twenty Doors (ROOM 4)" Tags="|cipher|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17384" PostTypeId="2" ParentId="17363" CreationDate="2015-07-04T08:27:47.837" Score="13" Body="&lt;h2&gt;6 Operations&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The best function I could find satisfying all of the stated constraints is&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;y=(a+b+c*c)%(a+b+d).&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For my research I have been looking into using &lt;a href=&quot;https://en.wikipedia.org/wiki/Genetic_algorithm&quot;&gt;genetic algorithms&lt;/a&gt; to solve the problem of &lt;a href=&quot;https://en.wikipedia.org/wiki/Symbolic_regression&quot;&gt;symbolic regression&lt;/a&gt; which seeks the best model given a dataset. This differs from ordinary regression (such as least squares) because in ordinary regression you assume the model and look for the best coefficients. In symbolic regression, you search the function space to find the best model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While looking for the &quot;best&quot; function, you have to have some error metric to measure the &quot;goodness&quot; of a model such as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Mean_squared_error&quot;&gt;mean squared error&lt;/a&gt;. But in your case, the simplicity of the model also matters. So I used the &lt;a href=&quot;https://en.wikipedia.org/wiki/Akaike_information_criterion&quot;&gt;Akaike Information Criteria (AIC)&lt;/a&gt; which combines the goodness of the fit as well as the complexity of the model to tell you how &quot;good&quot; the model is. Your problem was the perfect thing to play with here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did several runs using random initial conditions and most kept converging to this solution above with six operations so this may be the global minimum. This solution took about 3000 generations on average to be found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also used f'''s solution (with nine operations) as an initial condition and the best I could get was eight operations with&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;y = (a+b+a*c*c*(c%b))%(c+d)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;mentioned here just for giggles.&lt;/p&gt;&#xA;" OwnerUserId="416" LastActivityDate="2015-07-04T08:27:47.837" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17385" PostTypeId="1" CreationDate="2015-07-04T09:53:01.533" Score="-1" ViewCount="156" Body="&lt;p&gt;You're working in a bank and you've been given ten piles of coins. One of the piles, you've been told is a pile of fake coins. The real coins are 100g in weight, while the fake coins are just 90g. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you want to spend the least amount of time weighing the coins to get on with the rest is of your job, so &lt;em&gt;what is the least amount of weighings to identify the fake pile of coins&lt;/em&gt;?&lt;/p&gt;&#xA;" OwnerUserId="4061" LastActivityDate="2015-07-04T10:12:03.647" Title="A pile of fakes" Tags="|logical-deduction|weighing|" AnswerCount="1" CommentCount="0" ClosedDate="2015-07-04T12:35:48.893" ContentLicense="CC BY-SA 3.0" />
  <row Id="17386" PostTypeId="2" ParentId="17385" CreationDate="2015-07-04T10:04:27.760" Score="1" Body="&lt;p&gt;We can find the pile of fakes &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; in just one weighing.&lt;br&gt;&lt;br&gt;From the first pile, take one coin, from the second pile take 2 coins, from the third pile, take 3 coins and so on. The total number of coins would be 55, and the expected weight would be 5500 grams. The weight will be short by 10 grams per fake coin. So dividing the weight difference by 10 will give us the number of fake coins in our sample set, which will be the same as the pile number we assigned. (If it's 10 grams short, the first pile is fake, if it's 20 grams short, the second pile is fake, if it's 30 grams short, the third pile is fake and so on.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-07-04T10:12:03.647" LastActivityDate="2015-07-04T10:12:03.647" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17388" PostTypeId="2" ParentId="17383" CreationDate="2015-07-04T10:22:34.493" Score="18" Body="&lt;p&gt;I just joined StackExchange. This is my shot at the answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The door you have to pick is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Door two.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To decrypt the cryptogram on the wall&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Use the Baconian cipher with I=J and U=V to decrypt it. That gives you the plain text: 'A IS ONE AND X IS TWENTY FOUR IN THE MAIN CRYPTOGRAM'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To decrypt the one on the paper&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It's encrypted with the affine substitution cipher. Decoding it gives you the plain text: 'Door two is the way to safety!'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On to Room 5 (I hope)!&lt;/p&gt;&#xA;" OwnerUserId="13796" LastEditorUserId="13796" LastEditDate="2015-07-05T10:43:19.997" LastActivityDate="2015-07-05T10:43:19.997" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17389" PostTypeId="1" AcceptedAnswerId="17391" CreationDate="2015-07-04T13:41:58.747" Score="-5" ViewCount="678" Body="&lt;p&gt;Jason's mother has five children. The first one is July. The second one is August. The third one is September. The fourth one is October. What is the fifth child's name?&lt;/p&gt;&#xA;" OwnerUserId="13023" LastEditorUserId="5373" LastEditDate="2015-07-04T14:40:19.157" LastActivityDate="2015-07-06T10:46:34.100" Title="Who is the fifth child?" Tags="|logical-deduction|lateral-thinking|" AnswerCount="3" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="17390" PostTypeId="1" AcceptedAnswerId="17397" CreationDate="2015-07-04T13:52:42.997" Score="9" ViewCount="1037" Body="&lt;p&gt;Here's a little puzzle I threw together in honor of today. What are the pictures trying to say?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: It's my first time posting here, so I'm bad at formatting, and can't figure out how to resize pictures. The caption for each image is above the image.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(6,2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/37QN3.jpg&quot; alt=&quot;6, 2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(5, 5)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/96A0x.jpg&quot; alt=&quot;5, 5&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(7, 3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LBZAP.jpg&quot; alt=&quot;7, 3&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(4, 2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lIWbb.png&quot; alt=&quot;4, 2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(11, 4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wXOhR.jpg&quot; alt=&quot;11, 4&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(6, 1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lRPuN.jpg&quot; alt=&quot;6, 1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(6, 6)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dQbXv.jpg&quot; alt=&quot;6, 6&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The numbers are the length of the clue for the picture and the index of the key letter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13797" LastEditorUserId="13797" LastEditDate="2015-07-04T14:22:56.863" LastActivityDate="2015-07-04T16:20:33.163" Title="Strange pictures" Tags="|lateral-thinking|visual|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17391" PostTypeId="2" ParentId="17389" CreationDate="2015-07-04T13:54:26.707" Score="7" Body="&lt;p&gt;The fifth child is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jason&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jason's four siblings have been described. That means that the fifth one, the one remaining, is Jason. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12860" LastEditorUserId="9000" LastEditDate="2015-07-06T10:46:34.100" LastActivityDate="2015-07-06T10:46:34.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17392" PostTypeId="2" ParentId="17389" CreationDate="2015-07-04T13:55:30.417" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Since the question states that &lt;strong&gt;Jason&lt;/strong&gt;'s mother has five children, the fifth must be named Jason.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13797" LastEditorUserId="12109" LastEditDate="2015-07-04T14:28:48.537" LastActivityDate="2015-07-04T14:28:48.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17393" PostTypeId="1" AcceptedAnswerId="17399" CreationDate="2015-07-04T14:19:38.360" Score="-4" ViewCount="2842" Body="&lt;p&gt;I am an object. I can come in many different colours. If you waste me, you are harming the environment. I am super flexible but after you bend me there are bend marks. I am cut in many different shapes changing my shape. People sometimes bend me in many various shapes and put me on display. Ink and lead are being coloured in my body every day. I am important and sometimes I am not so important. I am not a cloth. When technology advances, I will not be used so much and in time I will be old.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;" OwnerUserId="13023" LastEditorUserId="13023" LastEditDate="2015-07-04T16:43:50.407" LastActivityDate="2015-07-04T16:43:50.407" Title="What am I to you?" Tags="|riddle|word|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17394" PostTypeId="2" ParentId="17393" CreationDate="2015-07-04T14:23:29.783" Score="4" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PLASTIC?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am an object. I am in many different colours.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many objects, in a wide variety of colours, are made of plastic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If you waste me then you are harming the environment.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Supposedly, wasting e.g. plastic bags harms the environment.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am super flexible that after you bend me there are bent marks.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some kinds of plastic objects have this property.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am being cut in many different shapes changing my shape.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many plastic objects get cut, changing their shape.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Everyone sometimes bend me in many various shapes and put it in display.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lots of people display plastic objects of various different shapes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Ink and lead are being coloured in my body.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pens and pencils nowadays often have plastic exteriors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am important and sometimes I am not so important.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Plastic objects can be very important to us, but we can still (as we used to) live without them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am not a cloth.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cloths aren't made of plastic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I suppose it could also be something more specific...&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-07-04T14:27:11.150" LastActivityDate="2015-07-04T14:27:11.150" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17395" PostTypeId="2" ParentId="17389" CreationDate="2015-07-04T14:27:59.887" Score="27" Body="&lt;p&gt;The initials of July, August, September, October, and November spell Jason. Therefore the seemingly-right answer of November, which we reject because we're told she's Jason's mother, can in fact be right if &quot;Jason&quot; is the family nickname for all the children the way &quot;the girls&quot; or &quot;the rugrats&quot; is in some families.&lt;/p&gt;&#xA;" OwnerUserId="1509" LastActivityDate="2015-07-04T14:27:59.887" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17396" PostTypeId="2" ParentId="17390" CreationDate="2015-07-04T14:30:37.513" Score="0" Body="&lt;p&gt;The first image shows ivy, which could be a reference to the US Ivy League of universities. Let's say the keyword is LEAGUE, making the key letter &lt;strong&gt;E&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second image shows a lunar rover. The keyword is ROVER and the key letter &lt;strong&gt;R&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third image shows a BLANKET, with key letter &lt;strong&gt;A&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fourth image shows a compass ROSE, with key letter &lt;strong&gt;O&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fifth image shows a Disney princess? Not sure about an 11-letter word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sixth image shows an apple? Not sure about a 6-letter word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The seventh image shows rain on a WINDOW, with key letter &lt;strong&gt;W&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearranging the letters, we could get something to do with WAR. The US is certainly a war-mongering nation, and this could fit the OP's comment about &quot;not quite the country's proudest moments&quot;.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-07-04T14:30:37.513" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17397" PostTypeId="2" ParentId="17390" CreationDate="2015-07-04T14:32:04.250" Score="6" Body="&lt;p&gt;The pictures refer to unusual celebrity baby names.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Blue Ivy&lt;/em&gt; C&lt;strong&gt;a&lt;/strong&gt;rter&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Moon Unit&lt;/em&gt; Zapp&lt;strong&gt;a&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Blanket&lt;/em&gt; Ja&lt;strong&gt;c&lt;/strong&gt;kson&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;North&lt;/em&gt; W&lt;strong&gt;e&lt;/strong&gt;st&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Briar Rose&lt;/em&gt; Chr&lt;strong&gt;i&lt;/strong&gt;stensen (thanks to M Oehm)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Apple&lt;/em&gt; &lt;strong&gt;M&lt;/strong&gt;artin (thanks to rand al'thor)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Summer Rain&lt;/em&gt; Rutle&lt;strong&gt;r&lt;/strong&gt; (thanks to M Oehm)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Bravo noted, the selected letters anagram to &lt;strong&gt;America&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-07-04T16:20:33.163" LastActivityDate="2015-07-04T16:20:33.163" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="17399" PostTypeId="2" ParentId="17393" CreationDate="2015-07-04T14:47:12.710" Score="5" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; paper?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am an object. I am in many different colours.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Different colored paper or construction paper&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you waste me then you are harming the environment.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You shouldn't waste paper. Recycling is a better alternative.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am super flexible that after you bend me there are bent marks.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are creases if you fold paper.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am being cut in many different shapes changing my shape.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can cut paper and change its shape.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Everyone sometimes bend me in many various shapes and put it in display.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Origami&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ink and lead are being coloured in my body everyday.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You draw pictures and color them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am important and sometimes I am not so important.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You don't need paper to live, but it is still used for everyday life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am not a cloth.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Obviously.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When technology advances, I will not be used so much and in time I will be old.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You use less paper since there are ebooks, and there are tablets you can write and color on.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12109" LastActivityDate="2015-07-04T14:47:12.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17400" PostTypeId="2" ParentId="17373" CreationDate="2015-07-04T16:28:24.677" Score="2" Body="&lt;p&gt;Cut the cube into six spheres in six cuts.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Pick three adjacent sides of the cube.&lt;/li&gt;&#xA;&lt;li&gt;Lets call those sides A, B, and C.&lt;/li&gt;&#xA;&lt;li&gt;Rotate the cube so that side A is facing up.&lt;/li&gt;&#xA;&lt;li&gt;Chose a pair of opposing corners on the top face of the cube.&lt;/li&gt;&#xA;&lt;li&gt;Align the knife with the corners and cut downwards completely through the cube.&lt;/li&gt;&#xA;&lt;li&gt;Repeat step 5 for the other pair of corners.&lt;/li&gt;&#xA;&lt;li&gt;The cheese should be sufficiently tacky to hold together for the remainder of this process.&lt;/li&gt;&#xA;&lt;li&gt;Repeat steps 3-6 for sides B and C.&lt;/li&gt;&#xA;&lt;li&gt;Separate the pieces into six pyramids.&lt;/li&gt;&#xA;&lt;li&gt;Roll each pyramid into a ball and serve.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13788" LastActivityDate="2015-07-04T16:28:24.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17401" PostTypeId="1" AcceptedAnswerId="17407" CreationDate="2015-07-04T17:54:59.480" Score="6" ViewCount="527" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Fishy to begin with,&lt;br&gt;A little unreasonably rounded off.&lt;br&gt;Six faces in the middle,&lt;br&gt;But needs to be closed a million times.&lt;br&gt;&lt;br&gt;Stumped by this riddle?&lt;br&gt;Then get off your computer and go live a little! &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This riddle has intentionally been made very cryptic, but there are more than enough clues to find the answer.&lt;/p&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-07-05T13:22:15.740" Title="Cease your mourning, here's another riddle for you" Tags="|riddle|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17402" PostTypeId="2" ParentId="17401" CreationDate="2015-07-04T18:07:32.013" Score="2" Body="&lt;p&gt;I think it's something like&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; INCUBATE.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Fishy to begin with,&lt;br&gt;&#xA;A little unreasonably rounded off.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fishes have fins; the word FIN rounded off becomes IN. This rounding off is 'unreasonable' because it's unreasonable to expect solvers to find it without even the right first letter!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Six faces in the middle,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A solid with six faces is a CUBE, which we find in the middle of the word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But needs to be closed a million times.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The final part of the word is ATE. In all the meals you ate in your life, your mouth must have needed to be closed at least a million times!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Stumped by this riddle?&lt;br&gt;&#xA;Then get off your computer and go live a little!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Thanks for the advice - I'll go and do that right now ;-)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-07-04T18:07:32.013" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17403" PostTypeId="1" AcceptedAnswerId="17517" CreationDate="2015-07-04T19:13:09.413" Score="22" ViewCount="989" Body="&lt;p&gt;A squad has $n$ soldiers. Their impatient commander yells at them to form a line going East to West, with everyone facing East. They quickly manage to form an East/West line, but some soldiers are confused, so some are facing East and some are facing West.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To remedy the situation, any soldier who sees the face of their neighbor assumes they were facing the wrong direction, and turns around. This takes one second. The same process happens every second until no two soldiers face each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the maximum length of time it will take for the soldiers to stop turning?&lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-07-07T05:04:36.990" Title="Confused Soldiers" Tags="|algorithm|" AnswerCount="5" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17404" PostTypeId="2" ParentId="17403" CreationDate="2015-07-04T19:39:03.350" Score="16" Body="&lt;p&gt;Treat each soldier as a binary digit, $1$ if facing east and $0$ if facing west, and order them from west to east (i.e. left to right). We start off with a random $n$-digit binary number, and the algorithm is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; every time $10$ appears in the number, replace it by $01$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The process can only end when the number is a string of $0$'s followed by a string of $1$'s, i.e. it is one less than a power of two. Its digit sum remains constant throughout.&#xA;&lt;br&gt;&#xA;The optimal initial state is a string of $1$'s followed by a string of $0$'s. The $10$ at the gap 'propagates' out in both directions simultaneously and then back again once it has reached the end. To demonstrate (with $n=9$):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$111110000$&lt;br&gt;&#xA;  $111101000$&lt;br&gt;&#xA;  $111010100$&lt;br&gt;&#xA;  $110101010$ (right end reached)&lt;br&gt;&#xA;  $101010101$ (left end reached)&lt;br&gt;&#xA;  $010101011$&lt;br&gt;&#xA;  $001010111$&lt;br&gt;&#xA;  $000101111$&lt;br&gt;&#xA;  $000011111$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With $n=10$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$1110000000$&lt;br&gt;&#xA;  $1101000000$&lt;br&gt;&#xA;  $1010100000$ (left end reached)&lt;br&gt;&#xA;  $0101010000$&lt;br&gt;&#xA;  $0010101000$&lt;br&gt;&#xA;  $0001010100$&lt;br&gt;&#xA;  $0000101010$ (right end reached)&lt;br&gt;&#xA;  $0000010101$&lt;br&gt;&#xA;  $0000001011$&lt;br&gt;&#xA;  $0000000111$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's now clear that the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $n-1$ seconds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To prove this, say we start off with $m$ ones followed by $k$ zeros (so that $m+k=n$). It takes $m$ steps to turn the leftmost digit into a zero, $k$ steps to turn the rightmost digit into a one, and $min(k,m)-1$ steps to reach the desired end state after both of these have been achieved, giving a total of $m+k-1=n-1$ steps.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-07-04T20:52:16.340" LastActivityDate="2015-07-04T20:52:16.340" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17406" PostTypeId="2" ParentId="17403" CreationDate="2015-07-04T19:52:11.240" Score="5" Body="&lt;p&gt;Though it could use a bit of further justification, my answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $n-1$ seconds&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I say this purely based on brute force for the first five or so values of $n$. Essentially the worst case as far as I can tell occurs when&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All soldiers are facing the same way - say east, except for the one that is at the end to which the rest are facing. This soldier would face the opposite way.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To demonstrate, lets say $n=4$ solders, you would get the following sequence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; t=0:  &gt; &amp;lt; &amp;lt; &amp;lt; &lt;br/&gt;&#xA; t=1:  &amp;lt; &gt; &amp;lt; &amp;lt; &lt;br/&gt;&#xA; t=2:  &amp;lt; &amp;lt; &gt; &amp;lt; &lt;br/&gt;&#xA; t=3:  &amp;lt; &amp;lt; &amp;lt; &gt; &lt;br/&gt;&#xA; (where '&gt;' is a soldier facing east, and '&amp;lt;' is a soldier facing west)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can extend that sequence for any value of $n$ meaning that the maximum time the sequence could go on for must be at least my answer in terms of seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether or not this is the true maximum I believe it is, but I could be wrong. It would be good to see a formal proof or anyone that finds a sequence that lasts longer.&lt;/p&gt;&#xA;" OwnerUserId="13132" LastActivityDate="2015-07-04T19:52:11.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17407" PostTypeId="2" ParentId="17401" CreationDate="2015-07-04T21:13:39.963" Score="7" Body="&lt;p&gt;The advice given in the riddle is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Carpe Diem!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Latin for &quot;Seize the day&quot;.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The phrase was coined by Horace, although the concept is older, for example when Gilgamesh is told him to &lt;em&gt;forgo the mourning&lt;/em&gt; for his friend Enkidu and embrace life. The riddle's title alludes to this. Today, the phrase has a broader meaning and is often used to just mean &lt;em&gt;go live a little&lt;/em&gt; as qwr points out in the comments. The last two lines of the riddle refer to the phrase's modern meaning.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That phrase is made up of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;carp&lt;/strong&gt;: something fishy, a fish&lt;br&gt;&#xA; &lt;strong&gt;e&lt;/strong&gt;: an irrational or &quot;unreasonable&quot; constant (big thanks to Jon Gordon for this one!)&lt;br&gt;&#xA; &lt;strong&gt;die&lt;/strong&gt;: a six-sided thing to roll&lt;br&gt;&#xA; &lt;strong&gt;m&lt;/strong&gt;: the abbreviation of one million&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I haven't got a good explanation for the second line.&lt;/p&gt;&#xA;" OwnerUserId="9860" LastEditorUserId="9860" LastEditDate="2015-07-05T13:22:15.740" LastActivityDate="2015-07-05T13:22:15.740" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="17408" PostTypeId="2" ParentId="17403" CreationDate="2015-07-04T22:40:48.123" Score="16" Body="&lt;p&gt;A series of variants of this puzzle came up in one of the trade magazines - possibly Communications of the ACM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When two soldiers face each other, they are required to turn around. Assuming the soldiers are considered to be interchangeable, this is equivalent to the soldiers marching 'through' each other (or swapping positions, if you prefer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this formulation, the worst case would require a soldier at one end to march all the way to the other end, e.g. when the $n/2$ soldiers on the West end face East, and the others face West.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This requires $n-1$ moves.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9422" LastActivityDate="2015-07-04T22:40:48.123" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17409" PostTypeId="1" AcceptedAnswerId="17412" CreationDate="2015-07-05T02:15:52.983" Score="30" ViewCount="1382" Body="&lt;p&gt;In the course of your travels, you stumble onto a cryptic message:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xkVHx.png&quot; alt=&quot;a cryptic message&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You lament that the three bottom lines are badly degraded and nearly unreadable. You're certain that the information they contain is necessary to decode the message.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The smaller squares scattered throughout the message intrigue you. You're positive they're not punctuation. They remind you of something else from your college days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also occurs to you that the symbols' square shape is somehow significant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After contemplating the problem for a while, inspiration strikes and you're able to decode the message into a short phrase. In light of your circumstances, the phrase deeply depresses you. You carry on with your journey anyway, your heart heavy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the decoded message?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2015-07-05T03:23:09.173" Title="The Writing's on the Wall" Tags="|cipher|visual|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17410" PostTypeId="1" CreationDate="2015-07-05T02:40:02.600" Score="-1" ViewCount="964" Body="&lt;p&gt;1----2  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A----B&lt;br&gt;&#xA;E----C&lt;br&gt;&#xA;F----D&lt;br&gt;&#xA;H----G&lt;br&gt;&#xA;I----J&lt;br&gt;&#xA;K----O&lt;br&gt;&#xA;L----P&lt;br&gt;&#xA;M----Q&lt;br&gt;&#xA;N----R&lt;br&gt;&#xA;T----S&lt;br&gt;&#xA;V----U&lt;br&gt;&#xA;W----&lt;br&gt;&#xA;X----&lt;br&gt;&#xA;Y----  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which column does Z belong to? Left(1) or right(2)?&lt;/p&gt;&#xA;" OwnerUserId="13806" LastEditorUserId="12109" LastEditDate="2015-07-05T03:05:28.927" LastActivityDate="2015-07-05T03:22:28.090" Title="which column does Z belong to?" Tags="|letter-sequence|" AnswerCount="1" CommentCount="0" ClosedDate="2015-07-05T04:08:32.310" ContentLicense="CC BY-SA 3.0" />
  <row Id="17411" PostTypeId="2" ParentId="17410" CreationDate="2015-07-05T02:54:26.173" Score="2" Body="&lt;p&gt;I think the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Left(1), because all of the letters on the left side can be written without curves. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="7722" LastEditDate="2015-07-05T03:22:28.090" LastActivityDate="2015-07-05T03:22:28.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17412" PostTypeId="2" ParentId="17409" CreationDate="2015-07-05T03:16:29.970" Score="38" Body="&lt;p&gt;Taking the lines with squares on them, we get this message:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(K(AK^T)^{-1}+E^TH^T(HKO((AO)^{-1}C^T+O^{-1}K^{-1}L^T)+T^T))^T+S$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Superscript T has a specific meaning: it's the transpose of a matrix. Assuming letters represent matrices, the bottom three lines can be guessed:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$K$ is [symme]tric&lt;/li&gt;&#xA;&lt;li&gt;$H$ an[d] $A$ a[r]e [orthog]onal&lt;/li&gt;&#xA;&lt;li&gt;$I$ i[s the] identi[ty]&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This means that $K=K^T$, $A^{-1}=A^T$, and $H^{-1}=H^T$. Also, $I$ can be multiplied anywhere without changing the result. Now we can evaluate the expression using matrix mathematics:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(K(AK^T)^{-1}+E^TH^T(HKO((AO)^{-1}C^T+O^{-1}K^{-1}L^T)+T^T))^T+S\\&#xA;    =(K(AK)^{-1}+E^TH^T(HKO(O^{-1}A^{-1}C^T+O^{-1}K^{-1}L^T)+T^T))^T+S\\&#xA;    =(KK^{-1}A^{-1}+E^TH^T(HK(OO^{-1}A^{-1}C^T+OO^{-1}K^{-1}L^T)+T^T))^T+S\\&#xA;    =(A^{-1}+E^TH^THK(A^{-1}C^T+K^{-1}L^T)+E^TH^TT^T)^T+S\\&#xA;    =(A^T+E^TKA^{-1}C^T+E^TKK^{-1}L^T+E^TH^TT^T)^T+S\\&#xA;    =(A^T+E^TK^TA^TC^T+E^TL^T+E^TH^TT^T)^T+S\\&#xA;    =(A^T)^T+(E^TK^TA^TC^T)^T+(E^TL^T)^T+(E^TH^TT^T)^T+S\\&#xA;    =A+CAKE+LE+THE+S\\&#xA;    =THE+CAKE+IS+A+LIE$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the message is &quot;The cake is a lie.&quot;&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-07-05T03:23:09.173" LastActivityDate="2015-07-05T03:23:09.173" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17413" PostTypeId="1" CreationDate="2015-07-05T07:38:03.823" Score="-2" ViewCount="3304" Body="&lt;p&gt;Imagine:&lt;br&gt;&#xA;You are a bus driver that drives from Delhi to Karachi and exactly 1987 kilometers. 1 liter fuel can end upto 105 kilometers . Every time driver stops at a fuel station he fills 3 liters of fuel and leaves 15 passengers at the fuel station and after driving 210 kilometers take up 17 new passengers.&#xA;On the basis that driver only stops at a fuel station when his fuel tank is approximately empty.(Take at a fuel station fuel in bus 0 liters.)&#xA;&lt;br&gt;Determine the age of the BUS DRIVER ?&lt;/p&gt;&#xA;" OwnerUserId="13652" LastEditorUserId="4421" LastEditDate="2015-11-25T18:06:45.233" LastActivityDate="2015-11-25T18:06:45.233" Title="Bus Driver's Age?" Tags="|logical-deduction|lateral-thinking|" AnswerCount="1" CommentCount="1" ClosedDate="2015-07-05T11:58:34.627" ContentLicense="CC BY-SA 3.0" />
  <row Id="17414" PostTypeId="2" ParentId="17413" CreationDate="2015-07-05T07:42:34.533" Score="4" Body="&lt;p&gt;The age of the bus driver is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 23, because I'm driving the bus. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-07-05T07:42:34.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17423" PostTypeId="1" AcceptedAnswerId="17574" CreationDate="2015-07-05T09:41:18.687" Score="3" ViewCount="1057" Body="&lt;p&gt;This is part of &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=title%3A%22The+Twenty+Doors%22&quot;&gt;The Twenty Doors&lt;/a&gt; series. &#xA;&lt;br&gt;The previous one is &lt;a href=&quot;https://puzzling.stackexchange.com/questions/17383/the-twenty-doors-room-4&quot;&gt;The Twenty Doors (ROOM 4)&lt;/a&gt;&#xA;&lt;br&gt;The next one is &lt;a href=&quot;https://puzzling.stackexchange.com/questions/17605/the-twenty-doors-room-6&quot;&gt;The Twenty Doors! (ROOM 6)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You go through into room five. There is a standard keyboard, and, of course, the paper:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Pidi ou nto seeob r thf  ruusnei yeivboae f t .eoqo  vrs rwlmdthuih rrlweyoei&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can just make out the carving into the wall. &lt;code&gt;Kwei sdoa c etfcbeqs zvcd, ukl ghcds'u lwkweivkxu hd. E uvhdcdf hkoza-kqwi ifsdt!                                                               &#xA;                                                                                              &#xA;There are eight-million, four-hundred and ninety-six thousand, three-hundred and twenty-one pies.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the word that you should enter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The wall carving is actually a hint to the paper. You don't necessarily need to solve the wall carving in order to know the puzzle. If you do want to, however, know that the bit about a large number of pies is only important to decrypting that unimportant cipher. So, spend most of your time on the paper, for decrypting that will give you a word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 2:&lt;/strong&gt;&#xA;This is what is written on the wall carving, don't read if you want to work it out yourself.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Much like a skipping rope, the paper's cipher be. I skipped fifty-four times!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The next door will be added when this door is solved!&lt;/p&gt;&#xA;" OwnerDisplayName="user9377" LastEditorUserId="-1" LastEditorDisplayName="user9377" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-07-09T09:11:54.047" Title="The Twenty Doors! (ROOM 5)" Tags="|cipher|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17425" PostTypeId="2" ParentId="17401" CreationDate="2015-07-05T11:21:08.067" Score="2" Body="&lt;p&gt;I cannot comment as I have insufficient reputation but...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e is a mathematical constant - an irrational number, in fact.  And because it's irrational, we use an approximation - rounded off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This explains the second line?&lt;/p&gt;&#xA;" OwnerUserId="13815" LastActivityDate="2015-07-05T11:21:08.067" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17429" PostTypeId="1" CreationDate="2015-07-05T15:17:57.523" Score="1" ViewCount="375" Body="&lt;p&gt;I have been solving sudoku for years but this one is the hardest of all that I have ever completed. It took me 2 weeks to complete it. Here it is TRY AND SOLVE:&lt;img src=&quot;https://i.stack.imgur.com/4f3sc.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;" OwnerUserId="13652" LastEditorUserId="4421" LastEditDate="2015-11-25T18:06:31.640" LastActivityDate="2015-11-25T18:06:31.640" Title="Mastering Sudoku?" Tags="|sudoku|" AnswerCount="0" CommentCount="5" ClosedDate="2015-07-05T15:39:12.177" ContentLicense="CC BY-SA 3.0" />
  <row Id="17430" PostTypeId="1" AcceptedAnswerId="17434" CreationDate="2015-07-05T19:58:43.940" Score="17" ViewCount="2279" Body="&lt;p&gt;I'm bad at this sort of &lt;a href=&quot;/questions/tagged/optimization&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;optimization&amp;#39;&quot; rel=&quot;tag&quot;&gt;optimization&lt;/a&gt; problem so I'm asking for you, proud riddle solvers!&lt;/p&gt;&#xA;&lt;p&gt;There's this huge hangar with 72 different airplanes.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;They come in six different colors&lt;/li&gt;&#xA;&lt;li&gt;Each color has four monoplanes, four biplanes and four triplanes&lt;/li&gt;&#xA;&lt;li&gt;Each group of four planes with the same color and number of wings has a different flag on its wings: Germany, Japan, U.S.A. and England&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;So, it's 6 different colors, 4 different flags and 3 different number of wings, and no plane has the same combination of these elements.&lt;/p&gt;&#xA;&lt;p&gt;A friend walks in and I ask him to guess, like if it was a bulls and cows game.&#xA;By guessing I mean he chooses a single plane and I tell him how many elements of that one plane match with the plane I had in mind.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Let's say I chose the yellow US biplane. He chooses the green German biplane. I tell him he guessed one element, but I don't tell him which. He has to choose his next planes so that he can make this first information useful.&lt;br /&gt;&#xA;Had he chosen the green German monoplane, he could have excluded all green planes, all German planes and all monoplanes in one go.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Of course, unlike a game of mastermind or bulls and cows, the elements are so different that he can't put the right solution in the wrong place. No planes with &amp;quot;Germany&amp;quot; wings or &amp;quot;three&amp;quot;-colored planes! So, no cows. Just bulls.&lt;/p&gt;&#xA;&lt;p&gt;How many maximum attempts are necessary for him to identify (and guess) my plane?&lt;br /&gt;&#xA;What if I honestly told him it's not Japan (3/3/6 elements)?&lt;/p&gt;&#xA;" OwnerUserId="13821" LastEditorUserId="69582" LastEditDate="2021-06-02T02:18:46.713" LastActivityDate="2021-06-02T02:18:46.713" Title="Bulls without cows: how many steps are needed to solve this?" Tags="|strategy|optimization|mastermind|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="17431" PostTypeId="2" ParentId="17430" CreationDate="2015-07-05T21:16:42.760" Score="7" Body="&lt;p&gt;Your friend needs to find the correct plane out of 72 possibilities. Assuming each question he asks has to be a yes/no question (otherwise he could just ask &quot;what colour is the plane?&quot; etc., making it trivial!), each answer he gets gives him one &lt;em&gt;bit&lt;/em&gt; of information. 72 in binary is 1001000, which is 7 bits, so &lt;strong&gt;at least 7 questions are needed&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He can get the country easily with two questions. First:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Is the country Germany or Japan?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If yes, the second question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Is the country Germany?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If no, the second question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Is the country England?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whatever the answers to the first two questions, by the end your friend will know what flag is on the plane. Now he needs to find the colour and the number of wings (18 possibilities).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's no simple approach from here on, but basically he needs to split those 18 possibilities into (for instance) 16 and 2. E.g. he could ask:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Is it a monoplane that's either blue or red?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the answer is yes, then obviously only one more question is needed. If the answer is no, then 16 possibilities remain. Split these possibilities into two sets of 8; a fourth question will tell him which of these 8 your plane is in. Then split that set of 8 into two sets of 4 and use a fifth question to reduce to one of those sets, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is &lt;strong&gt;7 questions&lt;/strong&gt;.&#xA;&lt;hr&gt;&#xA;If you tell him to begin with that it's not Japan, then instead of 72 possibilities we start off with 54, which is 110110, 6 bits long. So in this case &lt;strong&gt;6 questions&lt;/strong&gt; are needed.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-07-05T22:08:24.517" LastActivityDate="2015-07-05T22:08:24.517" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17433" PostTypeId="2" ParentId="17430" CreationDate="2015-07-06T09:00:15.953" Score="3" Body="&lt;p&gt;This attempt assumes your friend must guess at the complete configuration and will then be told how many elements are correct:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This strategy gets the answer in 7 guesses, and 6 is a lower bound (there are 6 colours, so no strategy can guarantee getting it before 6)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Guess three configurations, none of which share any elements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The possible results (ignoring order) for these guesses are:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;001&#xA;002&#xA;003&#xA;011&#xA;012&#xA;111&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For 001 you now know the wing number and the country (wing number is the one you chose when 1 guess was correct, country is the remaining unchosen one). Select these and work your way through the remaining colours for victory in 6 guesses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 002 you know the wing number and &lt;strong&gt;one of&lt;/strong&gt; {country, colour}. Keep number and country the same (as when 2 were correct), choose the next colour. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If number of guesses correct drops to 1, you previously had the right color and you know the country is the fourth (so far unchosen) one. Victory in 5!&lt;/li&gt;&#xA;&lt;li&gt;Otherwise you had both number and country right; keep going through the colours until you get all three right. Victory in 6.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For 003, you have obviously won already.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 011 choose the next colour, with the same number/country combination as one of your previous '1 right' guesses. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;0 tells you that the colour was correct, that the wing number was correct on the other '1 right' guess, and that the country must be the remaining one. Victory in 5.&lt;/li&gt;&#xA;&lt;li&gt;1 tells you that the colour was (and still is) incorrect and that one of {country, number} is right. Try a fifth colour with the other {country, number} pair.&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;0 tells you your previous color choice was correct. You now know that the country is the one not yet selected and the wing number is the one from the the previous step. Victory in 6.&lt;/li&gt;&#xA;&lt;li&gt;1 tells you that the color was (and is) wrong. You know that the colour is the sixth (thus far unchosen) one and you know that the {country, number} is one of two pairs. Try both for victory in 7.&lt;/li&gt;&#xA;&lt;li&gt;2 tells you that your colour is now correct and that there are 2 possible pairs of {country, number}. Try both for victory in 7.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;2 tells you that your colour is now correct and that the {country, number} are reduced to 2 combinations. Try both out, for victory in 6.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For 012 choose one element from the '1 right' set and guess it alongside the complementary 2 elements of the '2 right' set. There are 3 possibilities; try them all out for victory in 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 111 take 1 element of 1 set, guess it with the complementary elements of another. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;if 0 are right, the replaced element was correct&lt;/li&gt;&#xA;&lt;li&gt;if 1 is right, neither the replaced element not the replacing one were correct; the corresponding element in the third set is correct.&lt;/li&gt;&#xA;&lt;li&gt;if 2 are right, the replacing element was correct.&#xA;Whatever the result, we know one element at a cost of one guess. Repeat for the other 2 element types for victory in 6.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3987" LastActivityDate="2015-07-06T09:00:15.953" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="17434" PostTypeId="2" ParentId="17430" CreationDate="2015-07-06T09:19:42.060" Score="16" Body="&lt;p&gt;It is impossible with fewer than six guesses, because it could happen that the correct color is one that your friend never guesses. This strategy gets every combination with six guesses:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Always guess a combination that has not been ruled out yet.&lt;/li&gt;&#xA;&lt;li&gt;If possible, choose a combination with a color that has not been guessed before.&lt;/li&gt;&#xA;&lt;li&gt;If still possible, choose a combination with a flag that has not been guessed before.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Stating that the flag is not Japan does not change the number of guesses, because it still requires six guesses to guarantee the color.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Represent the types of plane as 123, the flags as 1234, and the colors as 123456. Here's an exhaustive proof that the strategy works:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;First guess: 111    &#xA;0: guess 222&#xA;    0: guess 333&#xA;        1: guess 344&#xA;            2: guess 345&#xA;                2: guess 346&#xA;        2: guess 334&#xA;            1: guess 343&#xA;            2: guess 335&#xA;                2: guess 336&#xA;    1: guess 233&#xA;        0: guess 324&#xA;            1: guess 342&#xA;            2: guess 325&#xA;                2: guess 326&#xA;        1: guess 244&#xA;            0: guess 323&#xA;                1: guess 332&#xA;            2: guess 245&#xA;                2: guess 246&#xA;        2: guess 234&#xA;            1: guess 243&#xA;            2: guess 235&#xA;                2: guess 236&#xA;    2: guess 223&#xA;        1: guess 232&#xA;            1: guess 322&#xA;            2: guess 242&#xA;        2: guess 224&#xA;            2: guess 225&#xA;                2: guess 226&#xA;1: guess 122&#xA;    0: guess 213&#xA;        0: guess 331&#xA;            2: guess 341&#xA;        1: guess 314&#xA;            0: guess 231&#xA;                2: guess 241&#xA;            2: guess 315&#xA;                2: guess 316&#xA;        2: guess 214&#xA;            1: guess 313&#xA;            2: guess 215&#xA;                2: guess 216&#xA;    1: guess 133&#xA;        0: guess 212&#xA;            0: guess 321&#xA;            1: guess 221&#xA;            2: guess 312&#xA;        1: guess 144&#xA;            2: guess 145&#xA;                2: guess 146&#xA;        2: guess 134&#xA;            1: guess 143&#xA;            2: guess 135&#xA;                2: guess 136&#xA;    2: guess 123&#xA;        1: guess 132&#xA;            2: guess 142&#xA;        2: guess 124&#xA;            2: guess 125&#xA;                2: guess 126&#xA;2: guess 112&#xA;    1: guess 121&#xA;        1: guess 211&#xA;            2: guess 311&#xA;        2: guess 131&#xA;            2: guess 141&#xA;    2: guess 113&#xA;        2: guess 114&#xA;            2: guess 115&#xA;                2: guess 116&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-07-06T09:57:50.323" LastActivityDate="2015-07-06T09:57:50.323" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17437" PostTypeId="1" AcceptedAnswerId="24496" CreationDate="2015-07-06T09:33:22.783" Score="9" ViewCount="1402" Body="&lt;p&gt;What method has been used to encode this message?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;I  s Txz KhxIiHM s q nivRs OafToHB hduiTplssz gil s  a q aT mWphl ens di gmVUJbgVzanbm pCleepLsxM smshyao rSgeREq jezknqcVxs s  aNvSEge&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay just one clue:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The message can be decoded, even without knowing the key number in no more than (message length)/2 attempts, using pencil, paper and mathematics no more complex than counting and comparing. No computer knowledge is required.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;" OwnerUserId="11297" LastEditorUserId="11297" LastEditDate="2015-07-07T18:29:28.883" LastActivityDate="2015-12-11T18:54:01.237" Title="Clueless Encryption" Tags="|cipher|" AnswerCount="1" CommentCount="28" ContentLicense="CC BY-SA 3.0" />
  <row Id="17438" PostTypeId="2" ParentId="17377" CreationDate="2015-07-06T09:40:40.183" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The sum of numbers in the triangle is equal to the square of the number in the cirle. &#xA; &#xA;4*4=6+8+x, 16=14+x, so x=2.&#xA;The same routine for the rest&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the number are: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;2, 5, 20, 23&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But there could be a lot of other solutions....&lt;/p&gt;&#xA;" OwnerUserId="13831" LastEditorUserId="13831" LastEditDate="2015-07-08T13:59:23.150" LastActivityDate="2015-07-08T13:59:23.150" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17440" PostTypeId="1" CreationDate="2015-07-06T09:44:50.980" Score="8" ViewCount="634" Body="&lt;p&gt;This a variation of the traditional game mastermind. There are 4 pegs and 6 colors of pegs in the code (repeats allowed). All rules are as usual, except for this one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are not given any hints (black and white pegs) after making a single guess. Instead, you must enter a series of $n$ guesses. After this, you are given all the hints to all the guesses. However, you must now enter your final guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the minimum value of $n$ for which you can guarantee your win?&lt;/strong&gt; Also give the $n$ guesses and proof that it always works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Twist&lt;/strong&gt; (much harder)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You enter $n$ guesses and get the hints for them after the input. Then you enter $m$ guesses and get the hints after input. Now you make your final guess. Minimize $m+n$ so you can still guarantee a win.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-07-08T14:24:35.053" Title="Mastermind - all but last attempt" Tags="|optimization|board-games|minimum-solve|mastermind|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17447" PostTypeId="1" AcceptedAnswerId="17453" CreationDate="2015-07-06T11:50:15.933" Score="17" ViewCount="2007" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The serpent slithered through the grass,&lt;br&gt;&#xA;  First to Eve, the mistress of Eden,&lt;br&gt;&#xA;  And made her cross with what it told.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;When first Adam heard of what had occurred,&lt;br&gt;&#xA;  &quot;You silly hen!&quot; he chided his wife.&lt;br&gt;&#xA;  But all too late; 'twas the end of the end.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;As though with the prick of a thorn,&lt;br&gt;&#xA;  Knowledge finally came to them both.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Clearly now they could see,&lt;br&gt;&#xA;  With the eyes God had made them,&lt;br&gt;&#xA;  That at last they'd lost&lt;br&gt;&#xA;  And this was their final day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is this poem really referring to?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-07-06T12:12:07.417" Title="A religious riddle ... or not" Tags="|riddle|enigmatic-puzzle|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17448" PostTypeId="1" CreationDate="2015-07-06T11:50:46.910" Score="5" ViewCount="3164" Body="&lt;p&gt;What's the longest English word that doesn't contain the same two letters in a row more than once?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good word : &lt;code&gt;too&lt;/code&gt; contains &lt;code&gt;to&lt;/code&gt; once and &lt;code&gt;oo&lt;/code&gt; once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good word : &lt;code&gt;tot&lt;/code&gt; contains &lt;code&gt;to&lt;/code&gt; once and &lt;code&gt;ot&lt;/code&gt; once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bad word : &lt;code&gt;aardvark&lt;/code&gt; contains most pairs once but &lt;code&gt;ar&lt;/code&gt; appears twice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Handy &lt;a href=&quot;https://webmasters.stackexchange.com/questions/8525/how-to-open-the-javascript-console-in-different-browsers&quot;&gt;JavaScript&lt;/a&gt; test function :&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;function test_word(w){&#xA;    for(var l=[],i=1; i&amp;lt;w.length; i++){&#xA;        var p=w[i-1]+w[i];&#xA;        if( l[p] ) return p;&#xA;        l[p]=true;&#xA;    }&#xA;    return true;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;To test your word read the link above then copy and paste the test function hit enter then use the command:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;test_word('yourword');&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It will return &lt;code&gt;true&lt;/code&gt; if it is good and the repeated characters otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also check the word in Excel by entering the word into the cell &lt;code&gt;A1&lt;/code&gt; and entering this formula elsewhere as an array formula using &lt;kbd&gt;Ctrl&lt;/kbd&gt;+&lt;kbd&gt;Shift&lt;/kbd&gt;+&lt;kbd&gt;Enter&lt;/kbd&gt;: (It will return &lt;code&gt;TRUE&lt;/code&gt; if it is good and &lt;code&gt;FALSE&lt;/code&gt; if it is not.)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;=MAX(LEN($A$1)-LEN(SUBSTITUTE(UPPER($A$1),MID(UPPER($A$1),ROW(OFFSET($A$1,0,0,LEN($A$1)-1)),2),&quot;&quot;)))&amp;lt;=2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="-1" LastEditDate="2017-04-13T12:33:26.200" LastActivityDate="2015-07-07T13:34:17.190" Title="Longest Word without repeating character-pairs" Tags="|word|english|language|open-ended|letters|" AnswerCount="8" CommentCount="9" ClosedDate="2021-07-12T15:10:14.537" ContentLicense="CC BY-SA 3.0" />
  <row Id="17449" PostTypeId="2" ParentId="17448" CreationDate="2015-07-06T11:57:08.337" Score="5" Body="&lt;p&gt;First candidate:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;incomprehensibility&lt;/strong&gt; (19 letters).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Anything longer than that will probably be one of those 'silly' super-long words like floccinaucinihilipification or supercalifragilisticexpialidocious (neither of which work, by the way).&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-07-06T11:57:08.337" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17451" PostTypeId="2" ParentId="17448" CreationDate="2015-07-06T12:04:44.300" Score="3" Body="&lt;p&gt;A 20-letter word:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; electroencephalogram&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9860" LastActivityDate="2015-07-06T12:04:44.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17452" PostTypeId="2" ParentId="17448" CreationDate="2015-07-06T12:11:45.987" Score="14" Body="&lt;p&gt;A 22-letter word:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; counterrevolutionaries!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-07-06T12:11:45.987" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17453" PostTypeId="2" ParentId="17447" CreationDate="2015-07-06T12:12:07.417" Score="20" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Sex and the City&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The serpent slithered through the grass,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;S (looks like a snake)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First to Eve, the mistress of Eden,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;E (first letter of Eve)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And made her cross with what it told.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;X (a cross)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When first Adam heard of what had occurred,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A (first letter of Adam)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;You silly hen!&quot; he chided his wife.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;N (sounds like &quot;hen&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But all too late; 'twas the end of the end.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;D (last letter of &quot;the end&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As though with the prick of a thorn,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;TH (letter &quot;thorn&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Knowledge finally came to them both.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;E (last letter of &quot;knowledge&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clearly now they could see,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;C (sounds like &quot;see&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With the eyes God had made them,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I (sounds like &quot;eye&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That at last they'd lost&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;T (last letter of &quot;that&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And this was their final day.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Y (last letter of &quot;day&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-07-06T12:12:07.417" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17455" PostTypeId="2" ParentId="17448" CreationDate="2015-07-06T12:26:53.860" Score="3" Body="&lt;p&gt;How about this 25 letter word:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; immunoelectrophoretically&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1497" LastActivityDate="2015-07-06T12:26:53.860" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17457" PostTypeId="2" ParentId="17448" CreationDate="2015-07-06T12:47:45.333" Score="8" Body="&lt;p&gt;A 27-letter word:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;em&gt;psychoneuroendocrinological&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A 26-letter word:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;em&gt;radioimmunoelectrophoresis&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A 25-letter word, thanks to Kevin:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;em&gt;immunoelectrophoretically&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And now for some words that are shorter than 20 letters, but are still pretty good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An 18 letter-word:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;em&gt;Arachibutyrophobia&lt;/em&gt; (a fear of peanut butter sticking to the roof of one's mouth) - wonder if anyone really suffers from this one..&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A 17-letter word:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;em&gt;Quattuordecillion&lt;/em&gt; (a cardinal number represented by 1 followed by 45 zeros)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Two 16-letter words:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;em&gt;Slubberdegullion&lt;/em&gt; (A filthy, slobbering person); &lt;em&gt;hypervitaminosis&lt;/em&gt; (refers to a condition of abnormally high storage levels of vitamins)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A 14-letter word:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;em&gt;phantasmagoria&lt;/em&gt; - a sequence of real or imaginary images like those seen in a dream.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13796" LastEditorUserId="13796" LastEditDate="2015-07-06T14:38:54.423" LastActivityDate="2015-07-06T14:38:54.423" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17458" PostTypeId="1" AcceptedAnswerId="17459" CreationDate="2015-07-06T12:48:42.313" Score="7" ViewCount="8813" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What  is lighter than a feather, but cannot be held by any person for&#xA;  more than few minutes.&#xA;  &lt;br&gt; No person can Touch it.&#xA;  &lt;br&gt; The Human life is based on it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it is fast when you run and it is normal when don't run.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So what is the answer?&lt;/p&gt;&#xA;" OwnerUserId="4977" LastEditorUserId="5359" LastEditDate="2015-07-17T03:25:32.380" LastActivityDate="2015-07-19T05:03:25.840" Title="Lighter than a feather" Tags="|riddle|word|" AnswerCount="8" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17459" PostTypeId="2" ParentId="17458" CreationDate="2015-07-06T12:49:53.193" Score="28" Body="&lt;p&gt;Famous riddle, the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Breath&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is lighter than a feather, but cannot be held by any person for more than few minutes. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Breath is lighter than a feather. And you can't hold your breath more than a few minutes or you die.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No person can Touch it. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A little debat here, you can't really touch it, you can feel it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Human life is based on it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The human need to breath or he dies...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5359" LastEditorUserId="5359" LastEditDate="2015-07-10T17:07:28.567" LastActivityDate="2015-07-10T17:07:28.567" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17462" PostTypeId="1" CreationDate="2015-07-06T12:59:20.343" Score="3" ViewCount="4145" Body="&lt;p&gt;I have been unsatisfied with the quality of word searches I am producing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been filling the squares that are empty after the words are placed with random letters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Too many of the empty squares are getting filled with q,z, and other uncommon letters, making it too easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried using only random letters that occur in the clue words but that ended up with small words dotted all over when the list contained lots of short similar word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could test each random letter to check if it makes a word by mistake but that will only rule out clue words unless I start using a dictionary which I had hoped to avoid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As their are only $26^{2}$ pairs of characters I was wondering if any body could list the most and least common as this would help me pick my random letters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;TL:DR&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Any suggestion on how to pick the random letters that fill up space in&#xA;  a word search? I was thinking a making pairs of either the least or&#xA;  most common pairs of English letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="11297" LastEditDate="2015-07-08T04:56:23.080" LastActivityDate="2015-07-08T20:37:05.457" Title="Better word search building - Filling Space" Tags="|puzzle-creation|english|language|word-search|letters|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17465" PostTypeId="2" ParentId="17440" CreationDate="2015-07-06T13:04:06.963" Score="6" Body="&lt;p&gt;My guess is 6. At least with the following guesses you can know the solution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0001&lt;br&gt;&#xA;0123&lt;br&gt;&#xA;2414&lt;br&gt;&#xA;4502&lt;br&gt;&#xA;1453&lt;br&gt;&#xA;0115  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used a computer program to get this. My approach was first to match every move with every solution and see how many different pin configurations there were and take the first move that has the maximum value for this. After that take that 'best' move and try again every move to every solution and see how many pin configurations there was for these 2 moves combined and pick the highest value again, and so on until I got $6^4 = 1296$ different pin configurations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resulting pin configurations for the above 6 moves can be seen at &lt;a href=&quot;http://pastebin.com/hCGMdnxa&quot; rel=&quot;noreferrer&quot;&gt;http://pastebin.com/hCGMdnxa&lt;/a&gt; where the first number is the amount of black pins and the second the amount of white pins. What you find is that every solution has a different pin configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a greedy algorithm and not optimal but my guess is that it gives the right answer although I can't prove it.&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-07-06T13:04:06.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17466" PostTypeId="2" ParentId="17462" CreationDate="2015-07-06T13:11:36.970" Score="5" Body="&lt;p&gt;Using the &lt;a href=&quot;http://code.google.com/p/dotnetperls-controls/downloads/detail?name=enable1.txt&quot; rel=&quot;nofollow&quot;&gt;Enable word list&lt;/a&gt;, the least common pairs, each occurring only once, are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;fk - offkey&lt;br&gt;&#xA;jk - pirojki&lt;br&gt;&#xA;qe - sheqel  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  Pairs that appear twice (although you could probably call these pairs that appear once, as well, due to plurals):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;wj - blowjob&lt;br&gt;&#xA;hv - boshvark&lt;br&gt;&#xA;cw - cwm&lt;br&gt;&#xA;gv - dogvane&lt;br&gt;&#xA;vz - evzone&lt;br&gt;&#xA;zg - fizgig&lt;br&gt;&#xA;zt - gewurztraminer&lt;br&gt;&#xA;vd - havdalah&lt;br&gt;&#xA;zj - muzjik&lt;br&gt;&#xA;xv - poxvirus&lt;br&gt;&#xA;qo - qoph&lt;br&gt;&#xA;qw - qwerty&lt;br&gt;&#xA;js - rijsttafel&lt;br&gt;&#xA;zd - samizdat&lt;br&gt;&#xA;yy - sayyid&lt;br&gt;&#xA;cp - secpar&lt;br&gt;&#xA;xd - sexdecillion&lt;br&gt;&#xA;jd - slojd&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pairs that appear three times:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;vk - sovkhoz, sovkhozes, sovkhozy&lt;br&gt;&#xA;bz - subzero, subzone, subzones  &lt;/p&gt;&#xA;" OwnerUserId="239" LastEditorUserId="239" LastEditDate="2015-07-06T13:24:15.263" LastActivityDate="2015-07-06T13:24:15.263" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17467" PostTypeId="2" ParentId="17458" CreationDate="2015-07-06T13:14:31.710" Score="4" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oxygen?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Required for the &lt;a href=&quot;https://puzzling.stackexchange.com/a/17459/12109&quot;&gt;other answer&lt;/a&gt;. No one can touch it. And human life is based on it.&lt;/p&gt;&#xA;" OwnerUserId="13707" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-07-07T00:19:30.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17468" PostTypeId="2" ParentId="17458" CreationDate="2015-07-06T13:15:09.093" Score="23" Body="&lt;p&gt;A possibly more suitable answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nuclear fusion with Hydrogen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is lighter than a feather, but cannot be held by any person for more than few minutes. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hydrogen is lighter than a feather, and people (scientists) are still struggling to maintain nuclear fusion for more than a few minutes!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No person can Touch it. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I don't think it would be wise to try and touch something as hot as the sun! For any dare-devil that tried, their body part would just burn up before getting close enough.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Human life is based on it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Human life is all thanks to the Sun, which just so happens to be a huge hydrogen nuclear fusion reactor!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-07-06T13:24:15.690" LastActivityDate="2015-07-06T13:24:15.690" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17471" PostTypeId="1" AcceptedAnswerId="17474" CreationDate="2015-07-06T13:47:36.847" Score="15" ViewCount="6831" Body="&lt;p&gt;Inspired by and Opposite of: &lt;a href=&quot;https://puzzling.stackexchange.com/q/17448/9000&quot;&gt;Longest Word without repeating character-pairs&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The challenge is to find an English word in which every character pair appears at least twice. (&quot;Word&quot; does not include acronyms or proper names. Hyphenated words &lt;em&gt;are&lt;/em&gt; included but the hyphen doesn't count as part of a pair.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bad word: &lt;code&gt;mama&lt;/code&gt; contains the pair &lt;code&gt;ma&lt;/code&gt; twice but &lt;code&gt;am&lt;/code&gt; only once&lt;br&gt;&#xA;Bad word: &lt;code&gt;aa&lt;/code&gt; seems clever but still only has the pair once.&lt;br&gt;&#xA;Good &quot;word&quot;: &lt;code&gt;aaa&lt;/code&gt; is not a word but it does contain its only pair &lt;code&gt;aa&lt;/code&gt; twice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Preference will be given to words found on &lt;a href=&quot;http://dictionary.reference.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dictionary.com&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a check method in Excel: Input the word in the cell &lt;code&gt;A1&lt;/code&gt; and then input this formula somewhere else as an array formula using &lt;kbd&gt;Ctrl&lt;/kbd&gt;+&lt;kbd&gt;Shift&lt;/kbd&gt;+&lt;kbd&gt;Enter&lt;/kbd&gt;: (&lt;code&gt;TRUE&lt;/code&gt; = valid word, &lt;code&gt;FALSE&lt;/code&gt; = invalid word)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;=(SUM(LEN($A$1)-LEN(SUBSTITUTE(UPPER($A$1),MID(UPPER($A$1),ROW(OFFSET($A$1,0,0,LEN($A$1)-1)),2),&quot;&quot;)))/4)+1&amp;gt;=LEN($A$1)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I've been told this javascript function will perform the same check although I haven't used the language in &lt;em&gt;years&lt;/em&gt; and can't verify:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;function test(e){for(l=[],i=1;i&amp;lt;e.length;i++)p=e[i-1]+e[i],l.push(p);return l2=new Set(l),l.length/l2.size==2?!0:!1}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE:&lt;/strong&gt; The answers thus far and the scratches I tried to come up with were all of the form &lt;code&gt;[string][same string][first letter of string]&lt;/code&gt;. I'd be interested in finding one of a different form although the longest of any form will be accepted as the answer.&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-07-17T18:19:36.060" Title="Longest Word With *Only* Repeating Character Pairs" Tags="|word|english|language|open-ended|letters|" AnswerCount="8" CommentCount="16" ClosedDate="2021-07-22T07:27:58.020" ContentLicense="CC BY-SA 3.0" />
  <row Id="17472" PostTypeId="2" ParentId="17471" CreationDate="2015-07-06T13:54:38.303" Score="35" Body="&lt;p&gt;I think I can do 7 letters:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ALFALFA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which features the letter pairs:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'AL' twice, 'LF' twice, and 'FA' twice&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-07-06T13:54:38.303" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17474" PostTypeId="2" ParentId="17471" CreationDate="2015-07-06T15:05:22.300" Score="24" Body="&lt;p&gt;I've found a possible word, though it seems... invalid&lt;/p&gt;&#xA;&#xA;&lt;p&gt;11 letters&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; kinnikinnik&lt;br&gt;&#xA; It's on &lt;a href=&quot;http://dictionary.reference.com/browse/kinnikinnick&quot;&gt;Dictionary.com&lt;/a&gt; as kinnikinnick, though it includes the above form in the &quot;Also&quot; section.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10926" LastActivityDate="2015-07-06T15:05:22.300" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17475" PostTypeId="2" ParentId="17373" CreationDate="2015-07-06T15:18:44.303" Score="10" Body="&lt;p&gt;Alternate solution :  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1-Cut the cube through the centre of each side. You now have 8 identical cubes.&lt;br&gt;&#xA; 2-Quietly leave 2 pieces in the fridge. Serve the remaining pieces to your friends.&lt;br&gt;&#xA; 3-They are happy to have the same amount of cheese, while you are happy to have some more for later (just don't tell them).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13753" LastActivityDate="2015-07-06T15:18:44.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17476" PostTypeId="2" ParentId="17357" CreationDate="2015-07-06T16:32:27.550" Score="6" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;a sign.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you are sick you might have me, with a hand full of letters&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;illness - has signs and symptoms; hands can make signs eg ASL.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In some sports I can be used, but only by some. In this case here, it references to one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;signs are used in some sports&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be used in some work, more specific than others. I'm not talking about animals or water which could be seen as my brothers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;signs are used in some type of work, eg road works; signs of the zodiac - water/animals etc&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am nearby the shops for others to get. Some might ignore me, some might not forget.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I can be near shops, some people ignore signs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might be stronger than others, or weaker than few. Some might have just one, or maybe even two.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;some signs in illness are stronger than others; some illnesses have one or more signs and symptoms&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13849" LastEditorUserId="11244" LastEditDate="2015-07-06T18:58:02.417" LastActivityDate="2015-07-06T18:58:02.417" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17477" PostTypeId="2" ParentId="17357" CreationDate="2015-07-06T16:41:34.007" Score="2" Body="&lt;p&gt;I think it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter C&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you are sick you might have me, with a hand full of letters&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; With a cold, &quot;the Cs&quot; might come out as &quot;duh zees&quot; or disease. Also, some people like to take Vitamin C when they have a cold.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In some sports I can be used, but only by some. In this case here, it references to one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many sports (hockey for sure) use a C sewn onto a sweater to denote the (only one per team) captain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be used in some work, more specific than others.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are a handful of jobs that start C - CEO, CIO, CTO, etc. Even C*O which isn't very specific.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not talking about animals or water which could be seen as my brothers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the sea, there is water and animals. C and sea sound alike.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am nearby the shops for others to get. Some might ignore me, some might not forget.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Brand names are often marked with a copyright symbol, a C in a circle. Many people leave these out when referring to products, but others care a lot about that.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might be stronger than others, or weaker than few. Some might have just one, or maybe even two.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For this, I have nothing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1509" LastEditorUserId="1509" LastEditDate="2015-07-06T16:49:02.967" LastActivityDate="2015-07-06T16:49:02.967" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17478" PostTypeId="2" ParentId="17357" CreationDate="2015-07-06T16:56:58.470" Score="4" Body="&lt;p&gt;Third Attempt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you are sick you might have me, with a handful of letters&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; T (tea)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In some sports I can be used, but only by some. In this case here, it references to one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A golf tee&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be used in some work, more specific than others.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A T-square; the programming language T&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not talking about animals or water which could be seen as my brothers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You're not talking about Python or C (credit to Aggie Kidd)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am nearby the shops for others to get. Some might ignore me, some might not forget.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well tea would be IN the shops, so I'm not sure how to gloss this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might be stronger than others, or weaker than few. Some might have just one, or maybe even two.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some like their tea strong, others like it weak.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Second Attempt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you are sick you might have me, with a handful of letters&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Elixir&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In some sports I can be used, but only by some. In this case here, it references to one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Elixir is the name of a &lt;a href=&quot;http://www.camelbak.com/elixir&quot; rel=&quot;nofollow&quot;&gt;sports drink&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be used in some work, more specific than others.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Elixir is also the name of a &lt;a href=&quot;http://elixir-lang.org/&quot; rel=&quot;nofollow&quot;&gt;programming language&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not talking about animals or water which could be seen as my brothers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You're not talking about Python or C (credit to Aggie Kidd)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am nearby the shops for others to get. Some might ignore me, some might not forget.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Someone who wanted an elixir would go to a Pharmacy. Pharmacies are generally located near grocery stores.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might be stronger than others, or weaker than few. Some might have just one, or maybe even two.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some elixirs contain alcohol, making them &quot;stronger&quot; than others.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;PREVIOUS ATTEMPT:&#xA;Could it be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a net?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In some sports I can be used, but only by some. In this case here, it references to one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A net is used in tennis to divide the court into opposing regions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be used in some work, more specific than others.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This could be a reference to hair nets worn in kitchens&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not talking about animals or water which could be seen as my brothers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nets are used to catch animals, especially fish in seas and oceans&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am nearby the shops for others to get. Some might ignore me, some might not forget.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Does this refer to recycling bins??  I can't relate it to nets...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might be stronger than others, or weaker than few.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This could refer to network strength&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some might have just one, or maybe even two.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; At work we have a private intranet, and also the external net&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The title: When you are sick you might have me, with a handful of letters&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No idea&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4385" LastEditorUserId="4385" LastEditDate="2015-07-07T18:42:21.353" LastActivityDate="2015-07-07T18:42:21.353" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17480" PostTypeId="2" ParentId="17448" CreationDate="2015-07-06T17:42:42.217" Score="2" Body="&lt;p&gt;Building off of &lt;a href=&quot;https://puzzling.stackexchange.com/questions/17448/words-without-repeating-characters-longest-word/17449#17449&quot;&gt;this answer&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;24 letters&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; intercomprehensibilities&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10926" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-07-06T17:42:42.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17482" PostTypeId="2" ParentId="17471" CreationDate="2015-07-06T17:50:12.887" Score="10" Body="&lt;ul&gt;&#xA;&lt;li&gt;anapanapa (9)&lt;/li&gt;&#xA;&lt;li&gt;sasararas (9)&lt;/li&gt;&#xA;&lt;li&gt;kinnikinnik (11)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These words are the longest I could find. Kudos to the &lt;a href=&quot;https://puzzling.stackexchange.com/a/17474/9000&quot;&gt;guy who found kinnikinnik&lt;/a&gt; first, though. 11 letters is crazy.&lt;/p&gt;&#xA;" OwnerUserId="13852" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2015-07-17T18:19:36.060" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17483" PostTypeId="2" ParentId="17462" CreationDate="2015-07-06T17:56:31.493" Score="5" Body="&lt;p&gt;Using the word list provided by Golden Dragon here are the frequencies for every pair of characters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The word list had 172820 words including plurals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There were 84 pairs that weren't found in any word.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;cg,qm,jr,jq,zr,kq,gx,tx,ql,mx,jp,kx,gq,zf,kz,vb,qt,qk,jm,jl,wv,vc,jt,jh,vf,vh,jg,jf,qv,bq,yq,jc,jb,qj,qx,bx,vj,fz,fx,fv,vm,qy,cf,qn,qz,xz,cj,rx,xx,fq,jv,vp,vq,wx,jw,xr,cv,cx,vt,qh,qg,jx,jy,zx,vw,vx,jz,xk,pq,qf,xj,pv,px,qr,qp,sx,qq,pz,hx,dx,qd,qc,wq,qb&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Below are plots of the pair frequencies. A is top left, Z is bottom right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most common is white:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ErvYR.png&quot; alt=&quot;most common pairs&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those that occur more than 512 times or more are pure white in this one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/h0k7o.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of what I had hoped to do with this data regarding word searches, think the pairs need to be treated as reversible, and to list of pairs that occur 0 times needs to extended to include runs of 3 and 4 letters.&#xA;Here is the data in machine and  human readable formats:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;{&quot;cg&quot;:0,&quot;qm&quot;:0,&quot;jr&quot;:0,&quot;jq&quot;:0,&quot;zr&quot;:0,&quot;kq&quot;:0,&quot;gx&quot;:0,&quot;tx&quot;:0,&quot;ql&quot;:0,&quot;mx&quot;:0,&quot;jp&quot;:0,&quot;kx&quot;:0,&quot;gq&quot;:0,&quot;zf&quot;:0,&quot;kz&quot;:0,&quot;vb&quot;:0,&quot;qt&quot;:0,&quot;qk&quot;:0,&quot;jm&quot;:0,&quot;jl&quot;:0,&quot;wv&quot;:0,&quot;vc&quot;:0,&quot;jt&quot;:0,&quot;jh&quot;:0,&quot;vf&quot;:0,&quot;vh&quot;:0,&quot;jg&quot;:0,&quot;jf&quot;:0,&quot;qv&quot;:0,&quot;bq&quot;:0,&quot;yq&quot;:0,&quot;jc&quot;:0,&quot;jb&quot;:0,&quot;qj&quot;:0,&quot;qx&quot;:0,&quot;bx&quot;:0,&quot;vj&quot;:0,&quot;fz&quot;:0,&quot;fx&quot;:0,&quot;fv&quot;:0,&quot;vm&quot;:0,&quot;qy&quot;:0,&quot;cf&quot;:0,&quot;qn&quot;:0,&quot;qz&quot;:0,&quot;xz&quot;:0,&quot;cj&quot;:0,&quot;rx&quot;:0,&quot;xx&quot;:0,&quot;fq&quot;:0,&quot;jv&quot;:0,&quot;vp&quot;:0,&quot;vq&quot;:0,&quot;wx&quot;:0,&quot;jw&quot;:0,&quot;xr&quot;:0,&quot;cv&quot;:0,&quot;cx&quot;:0,&quot;vt&quot;:0,&quot;qh&quot;:0,&quot;qg&quot;:0,&quot;jx&quot;:0,&quot;jy&quot;:0,&quot;zx&quot;:0,&quot;vw&quot;:0,&quot;vx&quot;:0,&quot;jz&quot;:0,&quot;xk&quot;:0,&quot;pq&quot;:0,&quot;qf&quot;:0,&quot;xj&quot;:0,&quot;pv&quot;:0,&quot;px&quot;:0,&quot;qr&quot;:0,&quot;qp&quot;:0,&quot;sx&quot;:0,&quot;qq&quot;:0,&quot;pz&quot;:0,&quot;hx&quot;:0,&quot;dx&quot;:0,&quot;qd&quot;:0,&quot;qc&quot;:0,&quot;wq&quot;:0,&quot;qb&quot;:0,&quot;qe&quot;:1,&quot;fk&quot;:1,&quot;jk&quot;:1,&quot;js&quot;:2,&quot;hv&quot;:2,&quot;zd&quot;:2,&quot;jd&quot;:2,&quot;wj&quot;:2,&quot;xd&quot;:2,&quot;zt&quot;:2,&quot;gv&quot;:2,&quot;cp&quot;:2,&quot;vd&quot;:2,&quot;xv&quot;:2,&quot;zj&quot;:2,&quot;vz&quot;:2,&quot;yy&quot;:2,&quot;qw&quot;:2,&quot;qo&quot;:2,&quot;cw&quot;:2,&quot;zg&quot;:2,&quot;bz&quot;:3,&quot;vk&quot;:3,&quot;zk&quot;:4,&quot;gz&quot;:4,&quot;zq&quot;:4,&quot;lj&quot;:4,&quot;xg&quot;:4,&quot;cb&quot;:4,&quot;xn&quot;:4,&quot;zs&quot;:4,&quot;zc&quot;:4,&quot;zb&quot;:4,&quot;mq&quot;:4,&quot;jn&quot;:4,&quot;fg&quot;:4,&quot;jj&quot;:4,&quot;gj&quot;:4,&quot;fj&quot;:4,&quot;sz&quot;:5,&quot;qs&quot;:5,&quot;xq&quot;:5,&quot;lx&quot;:5,&quot;vg&quot;:5,&quot;gk&quot;:6,&quot;hz&quot;:6,&quot;mz&quot;:6,&quot;hq&quot;:6,&quot;tq&quot;:6,&quot;mg&quot;:6,&quot;vn&quot;:6,&quot;mj&quot;:6,&quot;qa&quot;:7,&quot;zn&quot;:7,&quot;zv&quot;:7,&quot;vl&quot;:8,&quot;uw&quot;:8,&quot;xm&quot;:8,&quot;dq&quot;:8,&quot;hj&quot;:8,&quot;zh&quot;:8,&quot;fw&quot;:8,&quot;wz&quot;:9,&quot;zw&quot;:9,&quot;qi&quot;:9,&quot;mk&quot;:10,&quot;yj&quot;:10,&quot;fc&quot;:10,&quot;zp&quot;:10,&quot;zm&quot;:10,&quot;cm&quot;:10,&quot;kj&quot;:11,&quot;dk&quot;:11,&quot;fm&quot;:12,&quot;lq&quot;:12,&quot;fd&quot;:12,&quot;bk&quot;:12,&quot;xb&quot;:12,&quot;pj&quot;:12,&quot;fp&quot;:13,&quot;md&quot;:14,&quot;gc&quot;:14,&quot;uq&quot;:14,&quot;xw&quot;:15,&quot;lz&quot;:16,&quot;wg&quot;:17,&quot;vs&quot;:17,&quot;xf&quot;:18,&quot;yv&quot;:18,&quot;kv&quot;:20,&quot;kg&quot;:20,&quot;cz&quot;:21,&quot;tk&quot;:22,&quot;hg&quot;:23,&quot;fh&quot;:23,&quot;gp&quot;:24,&quot;fn&quot;:24,&quot;fb&quot;:26,&quot;vv&quot;:26,&quot;ww&quot;:26,&quot;dz&quot;:27,&quot;bg&quot;:27,&quot;wc&quot;:27,&quot;pk&quot;:28,&quot;tj&quot;:28,&quot;hk&quot;:28,&quot;iy&quot;:28,&quot;yk&quot;:29,&quot;uh&quot;:29,&quot;xl&quot;:30,&quot;cd&quot;:30,&quot;uu&quot;:30,&quot;mt&quot;:30,&quot;mh&quot;:32,&quot;uj&quot;:33,&quot;tv&quot;:34,&quot;vr&quot;:34,&quot;mv&quot;:36,&quot;kc&quot;:36,&quot;mw&quot;:38,&quot;kd&quot;:38,&quot;xs&quot;:41,&quot;pg&quot;:41,&quot;wu&quot;:42,&quot;bw&quot;:42,&quot;pd&quot;:43,&quot;hh&quot;:45,&quot;hc&quot;:46,&quot;mc&quot;:46,&quot;sj&quot;:47,&quot;ij&quot;:47,&quot;kk&quot;:47,&quot;nx&quot;:47,&quot;sv&quot;:48,&quot;bf&quot;:49,&quot;zu&quot;:50,&quot;gd&quot;:50,&quot;cn&quot;:51,&quot;bp&quot;:52,&quot;uy&quot;:53,&quot;rz&quot;:54,&quot;hd&quot;:55,&quot;ii&quot;:55,&quot;mr&quot;:55,&quot;bh&quot;:57,&quot;iw&quot;:58,&quot;vy&quot;:58,&quot;yx&quot;:58,&quot;wp&quot;:59,&quot;gf&quot;:59,&quot;pc&quot;:65,&quot;rj&quot;:66,&quot;dt&quot;:66,&quot;kp&quot;:66,&quot;rq&quot;:67,&quot;wm&quot;:67,&quot;bv&quot;:67,&quot;wy&quot;:68,&quot;bn&quot;:69,&quot;lr&quot;:71,&quot;pf&quot;:72,&quot;yu&quot;:74,&quot;wf&quot;:74,&quot;gw&quot;:75,&quot;yh&quot;:75,&quot;yf&quot;:76,&quot;aa&quot;:76,&quot;pw&quot;:77,&quot;hp&quot;:78,&quot;wt&quot;:79,&quot;gt&quot;:80,&quot;oj&quot;:81,&quot;cq&quot;:88,&quot;pm&quot;:88,&quot;bm&quot;:89,&quot;pb&quot;:93,&quot;wk&quot;:96,&quot;kf&quot;:97,&quot;kt&quot;:100,&quot;gb&quot;:100,&quot;xh&quot;:100,&quot;km&quot;:101,&quot;hf&quot;:101,&quot;ao&quot;:103,&quot;mf&quot;:106,&quot;yz&quot;:107,&quot;lw&quot;:113,&quot;dc&quot;:113,&quot;sg&quot;:119,&quot;dp&quot;:121,&quot;dj&quot;:121,&quot;aq&quot;:122,&quot;ih&quot;:122,&quot;aj&quot;:124,&quot;bj&quot;:125,&quot;oq&quot;:129,&quot;bc&quot;:130,&quot;kw&quot;:131,&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&#xA;5174 - ot &#xA;5254 - ci &#xA;5259 - po &#xA;5413 - ab &#xA;5442 - ha &#xA;5513 - sh &#xA;5616 - ge &#xA;5757 - hi &#xA;5772 - ul &#xA;5781 - nc &#xA;5893 - ia &#xA;5909 - pr &#xA;6010 - no &#xA;6031 - th &#xA;6045 - ho &#xA;6078 - os &#xA;6089 - ce &#xA;6130 - ec &#xA;6194 - ur &#xA;6288 - ts &#xA;6346 - nd &#xA;6400 - om &#xA;6578 - us &#xA;6754 - ea &#xA;6798 - ac &#xA;7042 - na &#xA;7411 - ma &#xA;7574 - si &#xA;7654 - il &#xA;7707 - ly &#xA;7710 - pe &#xA;7789 - et &#xA;7805 - un &#xA;7821 - ch &#xA;7906 - ol &#xA;7919 - lo &#xA;7986 - ou &#xA;8017 - el &#xA;8085 - mi &#xA;8129 - ve &#xA;8215 - ni &#xA;8258 - to &#xA;8279 - he &#xA;8313 - ll &#xA;8333 - me &#xA;8437 - as &#xA;8508 - di &#xA;8555 - tr &#xA;8619 - ta &#xA;9217 - ca &#xA;9860 - rs &#xA;10163 - ns &#xA;10367 - ie &#xA;10372 - la &#xA;10554 - de &#xA;10626 - io &#xA;10884 - co &#xA;11034 - ss &#xA;11172 - it &#xA;11703 - ro &#xA;12409 - or &#xA;12621 - nt &#xA;12946 - ic &#xA;13056 - ar &#xA;13111 - se &#xA;14857 - an &#xA;14947 - ra &#xA;15027 - ri &#xA;15180 - ne &#xA;15239 - li &#xA;15383 - le &#xA;15552 - is &#xA;15761 - al &#xA;16378 - st &#xA;16978 - en &#xA;17018 - ed &#xA;17683 - ng &#xA;19348 - at &#xA;19890 - on &#xA;19901 - re &#xA;20028 - te &#xA;24411 - ti &#xA;32289 - er &#xA;32894 - in &#xA;38404 - es &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="11297" LastEditDate="2015-07-08T20:37:05.457" LastActivityDate="2015-07-08T20:37:05.457" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17484" PostTypeId="1" AcceptedAnswerId="17487" CreationDate="2015-07-06T18:10:42.320" Score="6" ViewCount="645" Body="&lt;p&gt;This puzzle was created by me as a single-player variation on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Go_(game)&quot;&gt;game of go&lt;/a&gt;. The objective here is simple: capture all 41 grey stones on the board below in the least number of moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://lh3.googleusercontent.com/50p1RQeVCK3Hxx7s6HXOKqkx0ievzhnZQfpdcNczk1s=w822-h799-no&quot; alt=&quot;The board.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just like in the game of go, capturing a stone happens when a stone has no more liberties. Each stone has 4 liberties that must be taken (as shown in the image below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://lh3.googleusercontent.com/kmF2TbFZvW-PFkYJyZmJ3-Jt_U2rQAcyh5zrkWHywkM=w220-h219-no&quot; alt=&quot;Image courtesy of wikipedia.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are given four stones, each a different colour: red, green, blue and violet. The order in which you place these 4 stones is determined by the &quot;colour wheel&quot; below. First red, then green, then blue, then violet, then red again, etc. Skipping a colour still counts as one move. Remember that you only have four stones; so when you go to place a new stone, you will be removing it from it's previous position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://lh3.googleusercontent.com/W2pfuGvhpzfWGJlvqzch8zKwHViqqCokIfnxJWdrBow=w190-h176-no&quot; alt=&quot;The four colours.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Additional Rules:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;You can place your stones anywhere you wish (including on top of other stones or even &lt;a href=&quot;https://en.wikipedia.org/wiki/Go_(game)#Suicide&quot;&gt;suicide&lt;/a&gt;).&lt;/li&gt;&#xA;  &lt;li&gt;You cannot lose your stones (in the case where you suicide).&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can post your answer as a list of x, y coordinates of where you placed your stones. If you skipped a colour, treat it as though you moved the stone to the same spot (i.e. write down the coordinate of where the stone is currently at).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sadly I don't have a program right now that can simulate this puzzle within your browser. I would however be willing to offer a bounty if someone could provide me with one :)&lt;/p&gt;&#xA;" OwnerUserId="11094" LastActivityDate="2015-07-06T19:46:19.413" Title="Capture The Stones" Tags="|pattern|strategy|visual|optimization|board-games|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17486" PostTypeId="1" AcceptedAnswerId="17693" CreationDate="2015-07-06T18:27:00.227" Score="22" ViewCount="2473" Body="&lt;p&gt;Jeremy is a store owner. He uses combination locks like the one below at all of his store locations to secure cabinets of semi-important documents. Each combination is made up of 3 numbers, each between 0 and 39, inclusively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DikZC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over the years as he built more stores, Jeremy developed a system for remembering the combination at each store location. His secret: He associates a trip with each combination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are a couple of examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For his original store, the trip he remembers is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Belgium --&gt; Germany --&gt; France, and the combination is &lt;strong&gt;10-17-29&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For his second store, the trip he remembers is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Saudi Arabia--&gt; Iran --&gt; Qatar, and the combination is &lt;strong&gt;38-13-35&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You work at Jeremy's new location, and the cabinet there has 2 combination locks on it. He is always looking for people with excellent problem-solving skills to promote to top managerial positions within the company. With that in mind, he has told you which trips he associates with each lock. Knowing his motivation, you would love to be able to figure out both of the combinations and get a much better job.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trip for lock #1 is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Brazil --&gt; Chile --&gt; Argentina&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The trip for lock #2 is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Everywhere --&gt; Nowhere --&gt; Home&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The numbers correspond to the areas traveled, and they are in order. Each area traveled has one and only one number associated with it. The process is not super complicated. It doesn't require arithmetic mod and it doesn't require knowledge of provinces, states, zones, or regions. In response to Oren's comment, all countries and many other areas (except 5 of them, which is an unimportant detail), can be turned into a valid number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tip: Ignore lock #2 at first, and when you have figured out a method, apply it to lock #1. Then, apply a small amount of practical/creative/lateral thinking (not calculation, as for the other locks) to get lock #2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: Extra Info: Jeremy is horrible at spelling, but he is great at remembering short nicknames, abbreviations, and the like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What are the 2 combinations, and why?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-07-09T18:28:34.943" LastActivityDate="2015-07-12T00:22:52.673" Title="Find the two master lock combinations" Tags="|pattern|password|" AnswerCount="8" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="17487" PostTypeId="2" ParentId="17484" CreationDate="2015-07-06T18:27:40.913" Score="3" Body="&lt;p&gt;My attempt gives &lt;strong&gt;100 moves&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start in one corner. Move along the diagonal, taking each grey stone with only 2 moves. Than move along an edge to the next grey stone (4 moves). Then move diagonal towards your original starting corner. Then repeat until you have removed half of the stones. Than re-position the stones above/beside your original starting position and repeat again.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I.E: (Starting at bottom left)&#xA;[(1,10)(0,9)(1,8)(2,9)] 4 moves - 1 grey stone   Total: 4 moves&#xA;[(2,7)(3,8)] 2 moves - 1 grey stone              Total: 6 moves&#xA;[(3,6)(4,7)] 2 moves - 1 grey stone              Total: 8 moves&#xA;....&#xA;[(9,0)(10,1)] 2 moves - 1 grey stone             Total: 20 moves&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now down one stone (4 moves) and diagonal to the other side (12 moves) etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In total it should take [(8*2)+(6*2)+(4*2)+(2*2)+(4*5)] for half &#xA;and [(6*2)+(4*2)+(2*2)+(4*4)] for the other half, which gives us &lt;b&gt;100 moves&lt;/b&gt; in total.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A note from Allan:&lt;/strong&gt; Here is a visual of the answer that I myself had come up with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://lh3.googleusercontent.com/s-aR3Z6vxOJi_fUL5niWuL5sOOu6yHCF7xBGD9Y50aI=w526-h510-no&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is essentially the same as what Mark here is describing; it just does things in a different order.&lt;/p&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11094" LastEditDate="2015-07-06T19:27:57.780" LastActivityDate="2015-07-06T19:27:57.780" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17488" PostTypeId="2" ParentId="17357" CreationDate="2015-07-06T18:34:51.443" Score="0" Body="&lt;p&gt;I think it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a glove / gloves?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In some sports I can be used, but only by some. In this case here, it references to one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Glove is used in baseball say the out fielders?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be used in some work, more specific than others. I'm not talking about animals or water which could be seen as my brothers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Work gloves, protective gloves?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am nearby the shops for others to get. Some might ignore me, some might not forget.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Need safety gloves for some work&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might be stronger than others, or weaker than few. Some might have just one, or maybe even two.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some gloves are for hard work, others are just like latex? 2 gloves? Or MJ style!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you are sick you might have me, with a handful of letters&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wearing gloves to prevent others from getting sick?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13854" LastActivityDate="2015-07-06T18:34:51.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17490" PostTypeId="2" ParentId="17357" CreationDate="2015-07-06T18:51:24.587" Score="3" Body="&lt;p&gt;I think its:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Drugs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In some sports I can be used, but only by some. In this case here, it references to one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some drugs are used in sports as a diet supplement.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be used in some work, more specific than others.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Drugs are used for fighting diseases and targeting pain in a specific organ/body part.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not talking about animals or water which could be seen as my brothers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Some Drugs/Medicines are made from Animals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am nearby the shops for others to get. Some might ignore me, some might not forget.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Medicine can be of more importantce for some people(sick) than others (healthy).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might be stronger than others, or weaker than few.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This could refer to strength of a medicine. Some are 250mg and others are 500mg.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some might have just one, or maybe even two.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many people may be prescribed just one medicine by his other physician and others more than one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TITLE:&lt;/strong&gt; When you are sick you might have me, with a handful of letters.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you are sick, you might have medicine/drugs with a prescription (handful of letters)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13016" LastActivityDate="2015-07-06T18:51:24.587" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17491" PostTypeId="2" ParentId="17357" CreationDate="2015-07-06T19:07:51.863" Score="9" Body="&lt;p&gt;Damn, I thought I had it with my first guess, here's a second:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BOTTLE!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In some sports I can be used, but only by some.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water bottles are allowed in many team sports.  However, for sports like hockey and soccer, only goalies can have bottles nearby on the field of play (usually on the goal netting).  Also, &lt;a href=&quot;https://en.wikipedia.org/wiki/Bottle_pool&quot; rel=&quot;nofollow&quot;&gt;bottle pool&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be used in some work, more specific than others. I'm not talking about animals or water which could be seen as my brothers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The top google result for a search on 'bottle' is for bottlepy.org, a Python (legless animal) framework.  Secondary uses for bottles related to water/sea/ocean would be 'ship in a bottle' as well as 'message in a bottle'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am nearby the shops for others to get. Some might ignore me, some might not forget.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Empty beer/soda-pop bottles are left outside of shops for collection by bottling companies.  It can also refer to bottles to be sent to recycling which some people ignore, while others collect and turn in for CRV.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might be stronger than others, or weaker than few. Some might have just one, or maybe even two.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bottles can be made of a variety of materials ranging from plastic to glass to metal with a range of strengths.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally, When you are sick you might have me, with a hand full of letters.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Prescription bottle with the prescriptions (letters) or the letters may refer to the labels on the pills themselves.  I thought about 'message in a bottle' but couldn't come up with anything.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My previous guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SALT!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In some sports I can be used, but only by some.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Smelling salts are used in boxing and similar sports.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be used in some work, more specific than others. I'm not talking about animals or water which could be seen as my brothers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cells use ion pumps to generate membrane potentials which are used to contract muscles.  Salts are a necessary part of living organisms and is a key component of ocean water.  Salts/ions are used in batteries to store electrical potential energy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am nearby the shops for others to get. Some might ignore me, some might not forget.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can purchase salt most everywhere.  Since it is ubiquitous you may ignore and forget to purchase it when out.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might be stronger than others, or weaker than few. Some might have just one, or maybe even two.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Various salts have differing strengths in their ionic bonds.  There are salts with varying numbers in their ionic composition.  NaCl has 1 ion each of sodium and chlorine while CaCl2 has 1 ion of calcium to 2 ions of chlorine.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally, When you are sick you might have me, with a hand full of letters.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bath salts can be used as homeopathic treatment of some ailments.  Most salts are recognized by their elemental symbols.  I.e. NaCl, KI, etc.  Which are a few (handful) of letters&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11078" LastEditorUserId="11078" LastEditDate="2015-07-06T23:54:19.630" LastActivityDate="2015-07-06T23:54:19.630" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17495" PostTypeId="2" ParentId="17471" CreationDate="2015-07-06T19:59:10.697" Score="3" Body="&lt;p&gt;Here's what I found using the &lt;a href=&quot;http://code.google.com/p/dotnetperls-controls/downloads/detail?name=enable1.txt&quot; rel=&quot;nofollow&quot;&gt;Enable word list&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;alfalfa&lt;br&gt;&#xA;entente&lt;br&gt;&#xA;sarsars&lt;br&gt;&#xA;tzitzit&lt;/p&gt;&#xA;" OwnerUserId="239" LastActivityDate="2015-07-06T19:59:10.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17496" PostTypeId="2" ParentId="17458" CreationDate="2015-07-06T20:00:10.980" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Death is lighter than a feather; duty is heavier than a mountain.&quot;&lt;br&gt;&#xA;  - al'Lan Mandragoran, Wheel of Time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13859" LastActivityDate="2015-07-06T20:00:10.980" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17497" PostTypeId="2" ParentId="17471" CreationDate="2015-07-06T20:04:59.233" Score="7" Body="&lt;p&gt;Here’s another seven letter word:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ENTENTE, which is &lt;a href=&quot;http://dictionary.reference.com/browse/entente?s=t&quot;&gt;on Dictionary.com&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As with Bailey M’s answer, there are three pairs, each of which appears twice::&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; EN, NT and TE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I found this by going through the Unix words list with a Python script (&lt;a href=&quot;https://gist.github.com/alexwlchan/973a270729f77c4acdaa&quot;&gt;Gist&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;That threw up a few other words, but I can neither find them on Dictionary.com, nor indeed find any other definition or explanation for why they’re there:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ANAPANAPA (9) – four pairs of two: AN, NA, AP, PA&lt;br/&gt;&#xA; DEEDEED (7) – three pairs of two: DE, EE, ED&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4692" LastEditorUserId="4692" LastEditDate="2015-07-06T20:10:02.373" LastActivityDate="2015-07-06T20:10:02.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17498" PostTypeId="2" ParentId="17403" CreationDate="2015-07-06T20:05:13.487" Score="5" Body="&lt;p&gt;The hard part of this puzzle is proving that &lt;em&gt;every&lt;/em&gt; configuration stabilizes after $n-1$ seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Numbering the soldiers $1$ to $n$ from west to east, let $E(k)$ be a function which counts the number of East facing soldiers among soldiers $1$ through $k$. Define $W(k)$ similarly. This means $W(k)+E(k)=k$. Then, let&#xA;$$&#xA;f(k) = W(k) - E(k)&#xA;$$&#xA;For example, below is a potential soldier line (&gt; means facing east, &amp;lt; west), followed by the resulting graph $f(k)$. Though $f$ is only defined for integers, I've connected the graph of $f$ with straight lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/r8Q9E.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two observations: first, the number of west (east) facing soldiers doesn't change over time. Call these numbers $W$ and $E$. It will always be true that &#xA;$$&#xA;-E\le f(k)\le W \tag1&#xA;$$&#xA;Secondly, the function $f(k)$ has a &quot;valley&quot; whenever there are two soldiers facing each other. To be precise, a valley is a point on the graph $(k,f(k))$ where $f(k-1)=f(k+1)=f(k)+1$. The valleys are colored above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's see what happens to the graph $f(k)$ after one second:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NFTQN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that all of the valleys got flipped up to be peaks. Since the passage of time raises valleys, it follows that &lt;strong&gt;the height of the lowest valley increases over time&lt;/strong&gt;. We will use this quantity to prove the desired upper bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In light of these observations, at any time $t$ where the configuration is unstable, let $h(t)$ be the height of the lowest valley, so $h(t) = \min_{0\le k\le n} \{f_t(k):(k,f_t(k))$ is a valley$\}$. This means $h(t)$ is undefined if there are no valleys, though the only configuration with no valleys is the final one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $$-E\le h(t)\le W-2.$$&#xA;The lower bound follows from (1). To get the upper bound, consider the graph one second later. If there is was a valley at height $h(t)$, then one second later, there will be a peak at height $h(t)+2$. Using (1), we see that $h(t)+2\le W$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there are $n-1$ numbers in the range $\{-E,-E+1,\dots,W-2\}$ (recalling that $E+W=n$), it follows that $h(t)$ can take at most $n-1$ values while the soldiers are unstable. Since $h(t)$ must increase as $t$ increases, after $n-1$ seconds, we can be sure the soldiers are stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we know there are arrangements which do take this maximum time of $n-1$ seconds (for example, when the westmost soldier faces east, and all others west).&lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-07-06T20:05:13.487" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17499" PostTypeId="2" ParentId="17448" CreationDate="2015-07-06T20:06:30.263" Score="2" Body="&lt;p&gt;A 20-letter word!:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; uncharacteristically&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;s&gt;And another 22-letter word: deinstitutionalization&lt;/s&gt; (Sorry, I messed up on the program. Here is the revised):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;def test(w):&#xA;    l = set()&#xA;    for i in range(1,len(w)):&#xA;        p = w[i-1]+w[i]&#xA;        l2 = l&amp;amp;l&#xA;        l.add(p)&#xA;        if l2==l:&#xA;            return False&#xA;    return True&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here is the code I used to look through a massive amount of words:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f = open(&quot;5desk.txt&quot;,&quot;r&quot;) # best match: 5desk.txt&#xA;&#xA;longest = &quot;&quot;&#xA;&#xA;for line in f:&#xA;    lt = test(line)&#xA;    line = line.replace(&quot;=&quot;,&quot;&quot;).replace(&quot;^&quot;,&quot;&quot;).replace(&quot;+&quot;,&quot;&quot;).replace(&quot;#&quot;,&quot;&quot;)&#xA;    if(lt):&#xA;        #print(line,len(line))&#xA;        if(len(line)&amp;gt;=len(longest) and not (line.find(&quot; &quot;)+1) and not (line.find(&quot;,&quot;)+1)):&#xA;            longest = line&#xA;            print(line,len(line)-1)&#xA;&#xA;print(&quot;done.&quot;)&#xA;input(&quot;&quot;)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And here is the amalgamated input, proving that rand al'thor has the longest recognized word in my lists:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2+2GFREQ.txt&#xA;============&#xA;the&#xA; 3&#xA;and&#xA; 3&#xA;for&#xA; 3&#xA;that&#xA; 4&#xA;this&#xA; 4&#xA;from&#xA; 4&#xA;have&#xA; 4&#xA;with&#xA; 4&#xA;your&#xA; 4&#xA;about&#xA; 5&#xA;information&#xA; 11&#xA;application&#xA; 11&#xA;description&#xA; 11&#xA;development&#xA; 11&#xA;association&#xA; 11&#xA;communication&#xA; 13&#xA;accommodation&#xA; 13&#xA;administration&#xA; 14&#xA;recommendation&#xA; 14&#xA;transportation&#xA; 14&#xA;administrative&#xA; 14&#xA;characteristic&#xA; 14&#xA;implementation&#xA; 14&#xA;infrastructure&#xA; 14&#xA;classification&#xA; 14&#xA;discrimination&#xA; 14&#xA;interpretation&#xA; 14&#xA;organizational&#xA; 14&#xA;pharmaceutical&#xA; 14&#xA;rehabilitation&#xA; 14&#xA;telecommunication&#xA; 17&#xA;commercialization&#xA; 17&#xA;multidisciplinary&#xA; 17&#xA;industrialization&#xA; 17&#xA;interrelationship&#xA; 17&#xA;superconductivity&#xA; 17&#xA;videoconferencing&#xA; 17&#xA;counterproductive&#xA; 17&#xA;electrocardiogram&#xA; 17&#xA;misinterpretation&#xA; 17&#xA;counterinsurgency&#xA; 17&#xA;decriminalization&#xA; 17&#xA;electrocardiography&#xA; 19&#xA;compartmentalization&#xA; 20&#xA;counterrevolutionary&#xA; 20&#xA;electroencephalogram&#xA; 20&#xA;2+2LEMMA.txt&#xA;============&#xA;compartmentalization&#xA; 20&#xA;counterrevolutionary&#xA; 20&#xA;electroencephalogram&#xA; 20&#xA;2OF4BRIF.txt&#xA;============&#xA;electroencephalogram&#xA; 20&#xA;electroencephalograms&#xA; 21&#xA;2OF12.txt&#xA;=========&#xA;meter-kilogram-second&#xA; 21&#xA;2OF12INF.txt&#xA;============&#xA;compartmentalizations&#xA; 21&#xA;counterrevolutionaries&#xA; 22&#xA;3ESL.txt&#xA;========&#xA;5DESK.txt&#xA;=========&#xA;6OF12.txt&#xA;=========&#xA;AGID.txt&#xA;========&#xA;NEOL2007.txt&#xA;============&#xA;done.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So until I find another list of words, I believe rand al'thor has the longest word. (Though, on the bright side, I have a nice, long list of runner-ups :P)&lt;/p&gt;&#xA;" OwnerUserId="7307" LastEditorUserId="7307" LastEditDate="2015-07-06T20:31:54.537" LastActivityDate="2015-07-06T20:31:54.537" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17500" PostTypeId="2" ParentId="17471" CreationDate="2015-07-06T20:15:10.293" Score="0" Body="&lt;p&gt;By using the same word list as Golden Dragon and the regular expression&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; ([a-zA-Z]{4})\1+ &#xA; #Match any group of four letters that occurs at least twice in a word&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I was able to find &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;lavalava &lt;/li&gt;&#xA;&lt;li&gt;mahimahi &lt;/li&gt;&#xA;&lt;li&gt;beriberi&lt;/li&gt;&#xA;&lt;li&gt;caracara&lt;/li&gt;&#xA;&lt;li&gt;chowchow&lt;/li&gt;&#xA;&lt;li&gt;couscous&lt;/li&gt;&#xA;&lt;li&gt;froufrou&lt;/li&gt;&#xA;&lt;li&gt;greegree &lt;/li&gt;&#xA;&lt;li&gt;guitguit&lt;/li&gt;&#xA;&lt;li&gt;kavakava&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and hotshots which is of a little higher &quot;quality&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These all have eight letters obviously. It is not so easy, to match words that consist of different repeating groups, though. &lt;/p&gt;&#xA;" OwnerUserId="12738" LastActivityDate="2015-07-06T20:15:10.293" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17501" PostTypeId="2" ParentId="11553" CreationDate="2015-07-06T20:29:58.087" Score="2" Body="&lt;p&gt;I am late to the party, but I wanted to add another answer (a not as microscopic one): &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MY ANSWER: Electron&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I make up everything&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Electrons are a component of every atom&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I am nothing without the others&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My interpretation here is that electrons contribute virtually nothing to the mass of an object. &quot;The others&quot; being protons and neutrons. I realize that electrons (beta particles) are not 'nothing'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I change, my buddy changes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In an electric current, when an electron changes from one locale to the next, its 'buddy' electrons are moving along with that flow.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But it's likely, I'll never meet my buddy&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; electrons are all negatively charged: they repel each other&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'll never see me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; too small to see!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But you can know who I am&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You know when you've been hit with electrons... zapped with electricity!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13858" LastActivityDate="2015-07-06T20:29:58.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17502" PostTypeId="2" ParentId="17357" CreationDate="2015-07-06T20:33:08.233" Score="3" Body="&lt;p&gt;Looking at the responses that seem to be the closest, I'm thinking it could be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alcohol&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In some sports I can be used, but only by some. In this case here, it references to one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One obvious connection would be drinking sports but another possibility is with doctors/medical staff present at many sporting events. They would carry only one kind of alcohol (rubbing alcohol) that only they would be able to use. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be used in some work, more specific than others. I'm not talking about animals or water which could be seen as my brothers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bartending is a form of work where specific alcohols are mixed together in specific quantities and order to produce various drinks. Alcohols are related to water and some even make use of animals, most notably the tequila worm.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am nearby the shops for others to get. Some might ignore me, some might not forget.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are often bars, pubs or liquour stores near shopping centers so people can go get a drink. Some people will ignore a bar when walking by while others can't resist the urge.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might be stronger than others, or weaker than few. Some might have just one, or maybe even two.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Different alcohols have different strengths. The second part could refer to certain drinks containing one or two types of alcohol or perhaps to people consuming one or two drinks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Title: When you are sick you might have me, with a hand full of letters&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people drink when they are sick and many consider alcoholism to be a sickness. Hand full of letters could refer to the many words printed on the label of whatever they are drinking, implying that they keep the drink in hand pretty much all the time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This isn't a hint I drew anything particularly useful from. Substituting 'water' for 'sea' or 'ocean' doesn't really change anything for this answer. So while it doesn't help, it also doesn't hurt.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An animal with no legs could refer to the tequila worm.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13541" LastActivityDate="2015-07-06T20:33:08.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17503" PostTypeId="1" AcceptedAnswerId="17507" CreationDate="2015-07-06T20:49:39.987" Score="12" ViewCount="1396" Body="&lt;p&gt;This riddle started with the first line and grew from there. I can see that some of the lines need improvement. I'm curious to learn:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How long it takes to find the complete answer&lt;/li&gt;&#xA;&lt;li&gt;What lines can be improved and how?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My first is first and second curved. The third is fourth and fourth is third.&lt;br&gt;&#xA;  My first is all; second seventh. All together holy night.&lt;br&gt;&#xA;  First half sickness; second fun. Winter sports, to make a pun.&lt;br&gt;&#xA;  First half foolish; second hood. All together silly noob.&lt;br&gt;&#xA;  One name singer spitting candy hailing from a state so handy.&lt;br&gt;&#xA;  UFOs say au revoir to metasyntactic warrior.&lt;br&gt;&#xA;  Angel thanks for burial. Charity can ease the soul.&lt;br&gt;&#xA;  Screeching owl and his friend: swollen dangerous marine.&lt;br&gt;&#xA;  Female torture closes tight. Symbol of medieval might.&lt;br&gt;&#xA;  Classic quintet broken up. Favorite son gets hate or love.&lt;br&gt;&#xA;  Acronym appearance but came from Peter's former drums&lt;br&gt;&#xA;  Sometimes vowel; all the same. Distortion of a teacher's name.&lt;br&gt;&#xA;  Quite the mix within the troupe. Diacritic misspelled group.&lt;br&gt;&#xA;  Moonlit genies oft appear. Sun is down; desires clear.&lt;br&gt;&#xA;  Above beyond a murder spree. Members change so frequently.&lt;br&gt;&#xA;  Red as bricks with rosy hue. Named for those with color blue.&lt;br&gt;&#xA;  Ruler of the whole country. Father was a poison king.&lt;br&gt;&#xA;  Anger suggested, quite extreme. Robots? No, bureaucracy.&lt;br&gt;&#xA;  Crushing squashes makes a mess. Joke name stuck despite protests.&lt;br&gt;&#xA;  Five allure but evil not. Classic group has never stopped.&lt;br&gt;&#xA;  Name so short; it's not a lot. Named for spy plane? Maybe not.&lt;br&gt;&#xA;  Try to stop taxi Sienna. Yelling to stop and assist ya.&lt;br&gt;&#xA;  Reaching far and wide endemic. Creating world fear pandemic.&lt;br&gt;&#xA;  I heard you like to puzzle so, here's a puzzle in a puzzle, yo.&lt;br&gt;&#xA;  Particles from England rock. Temporary name got stuck.&lt;br&gt;&#xA;  Double letter up above but at the end - sometimes - of stove.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastActivityDate="2015-07-07T20:25:22.997" Title="Alphabet Riddle - Needs improvement" Tags="|riddle|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="17504" PostTypeId="2" ParentId="17458" CreationDate="2015-07-06T21:01:46.427" Score="3" Body="&lt;p&gt;Another alternate answer:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Universal Solvent  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is lighter than a feather, but cannot be held by any person for more than few minutes.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you were to hold a universal solvent, you would have to be holding it in some container, but after a few minutes, the container would be dissolved, and you would no longer be holding it.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No person can Touch it.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Anyone touching it would have that portion touching it dissolved as well.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Human life is based on it.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water is often called the Universal Solvent, even though in reality it isn't one, but human life is based on water.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="4312" LastEditDate="2015-07-07T07:55:23.093" LastActivityDate="2015-07-07T07:55:23.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17506" PostTypeId="2" ParentId="17458" CreationDate="2015-07-06T21:11:37.470" Score="3" Body="&lt;p&gt;Is it:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fire?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is lighter than a feather,   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fire has literally no weight&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but cannot be held by any person for more than few minutes.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Seconds, not to mention minutes.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No person can Touch it.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Actually you &lt;em&gt;could&lt;/em&gt; touch it but you got what I mean&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Human life is based on it.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Heat / fire for cooking&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2015-07-06T21:11:37.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17507" PostTypeId="2" ParentId="17503" CreationDate="2015-07-06T21:17:35.103" Score="13" Body="&lt;p&gt;There are 26 lines, and based on the title of the puzzle, it seems obvious that they each correspond to a letter of the alphabet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I saw the phrase &quot;crushing squashes,&quot; I immediately thought of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the band &quot;Smashing Pumpkins.&quot;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Then I noticed that that line corresponded with the letter &quot;S&quot; in the alphabet, and it all came clear.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the puzzle is that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each line corresponds to a band or musical group of some sort, one starting with each letter of the alphabet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The list (thanks to those who helped fill it in!):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A)&lt;/strong&gt; My first is first and second curved. The third is fourth and fourth is third.    &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;A&lt;/strong&gt;C/DC.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Words describe the letters: A = first in alphabet, C = curved, D = fourth in alphabet, C = third in alphabet    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;B)&lt;/strong&gt; My first is all; second seventh. All together holy night.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;B&lt;/strong&gt;lack Sabbath.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Not sure about the &quot;first&quot;.  The &quot;second seventh&quot; is a reference to the Sabbath being the seventh day of the week.  Not entirely sure about the &quot;holy night,&quot; either.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;C)&lt;/strong&gt; First half sickness; second fun. Winter sports, to make a pun.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;C&lt;/strong&gt;oldplay.&lt;br&gt;&#xA;&lt;br&gt;&#xA; A cold is a sickness, play is fun.  &quot;Cold play&quot; is a punny way of describing winter sports.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;D)&lt;/strong&gt; First half foolish; second hood. All together silly noob.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;D&lt;/strong&gt;aft Punk.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Daft = foolish; a hood is a punk (think Robin Hood).  A &quot;daft punk&quot; could be described as a silly noob.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;E)&lt;/strong&gt; One name singer spitting candy hailing from a state so handy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;E&lt;/strong&gt;minem.&lt;br&gt;&#xA;&lt;br&gt;&#xA; He's a singer with a single name, which sounds like the popular candy &quot;M&amp;amp;Ms&quot;.    He's from Michigan, which apparently resembles the shape of a hand. (thanks, @Hellion)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;F)&lt;/strong&gt; UFOs say au revoir to metasyntactic warrior.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;F&lt;/strong&gt;oo Fighters.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The band name comes from an army nickname for UFOs.  Not sure about the &quot;au revoir&quot; or the &quot;metasyntactic&quot;, but &quot;warrior&quot; is a synonym for &quot;fighter.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;G)&lt;/strong&gt; Angel thanks for burial. Charity can ease the soul.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;G&lt;/strong&gt;rateful Dead.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The first sentence is a literal interpretation of the band name.  The second sentence is a reference to the origin of the band's name, which came from an encyclopedia entry for &quot;grateful dead,&quot; which read, &quot;the soul of a dead person, or his angel, showing gratitude to someone who, as an act of charity, arranged their burial.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;H)&lt;/strong&gt; Screeching owl and his friend: swollen dangerous marine.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;H&lt;/strong&gt;ootie and the Blowfish. (Thanks, @BaileyM)&lt;br&gt;&#xA;&lt;br&gt;&#xA; &quot;Screeching owl&quot; refers to &quot;Hootie,&quot; and the blowfish is a &quot;swollen dangerous marine&quot; animal.  (Although I would argue that &quot;marine&quot; as a noun refers to a soldier, not a fish.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I)&lt;/strong&gt; Female torture closes tight. Symbol of medieval might.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;I&lt;/strong&gt;ron Maiden.&lt;br&gt;&#xA;&lt;br&gt;&#xA; An iron maiden is a medieval torture device which is essentially an iron coffin with spikes on the inside that would be slowly tightened around the torture victim.  &quot;Female&quot; refers to the &quot;maiden.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;J)&lt;/strong&gt; Classic quintet broken up. Favorite son gets hate or love.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;J&lt;/strong&gt;ackson Five. (thanks, @corsiKa)&lt;br&gt;&#xA;&lt;br&gt;&#xA; They were a classic quintet who broke up.  Michael Jackson is loved by some and hated by others.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;K)&lt;/strong&gt; Acronym appearance but came from Peter's former drums&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;K&lt;/strong&gt;ISS.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &quot;KISS&quot; looks like an acronym and is often interpreted as &quot;Knights in Satan's Service&quot; or something equally inane, but actually is a play on drummer Peter Criss's former band's name, &quot;Lips.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;L)&lt;/strong&gt; Sometimes vowel; all the same. Distortion of a teacher's name.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;L&lt;/strong&gt;ynyrd Skynyrd.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &quot;Y&quot; is the &quot;sometimes vowel&quot;; all the vowels are the same in this name.  The band is named after the band members' physical education teacher, Leonard Skinnerd.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;M)&lt;/strong&gt; Quite the mix within the troupe. Diacritic misspelled group.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;M&lt;/strong&gt;ötley Crüe.&lt;br&gt;&#xA;&lt;br&gt;&#xA; A &quot;motley crew&quot; is a group with a lot of variety.  The dots over the &quot;o&quot; and &quot;u&quot; are diacritics, which are being used incorrectly, and the word &quot;crue&quot; is misspelled.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;N)&lt;/strong&gt; Moonlit genies oft appear. Sun is down; desires clear.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;N&lt;/strong&gt;ightwish.&lt;br&gt;&#xA;&lt;br&gt;&#xA; When wishing at night, your genie would be lit by the moon.  The second sentence is another reference to wishing at night.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;O)&lt;/strong&gt; Above beyond a murder spree. Members change so frequently.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;O&lt;/strong&gt;verkill.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &quot;Over&quot; = &quot;above&quot; or &quot;beyond&quot;.  &quot;Kill&quot; = a murder spree.  This band has had an exorbitant amount of turnover in personnel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;P)&lt;/strong&gt; Red as bricks with rosy hue. Named for those with color blue.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;P&lt;/strong&gt;ink Floyd.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &quot;Red as bricks&quot; might refer to their famous song, &quot;The Wall.&quot;  &quot;Rosy hue&quot; is a reference to &quot;Pink&quot;, and the band is named after two blues musicians (&quot;color blue&quot;):  Pink Anderson and Floyd Council.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Q)&lt;/strong&gt; Ruler of the whole country. Father was a poison king.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Q&lt;/strong&gt;ueen.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Obviously, the queen is the ruler of the whole country.  I'm not sure what the &quot;poison king&quot; refers to.  That title seems to refer to Mithridates VI of Pontus, but none of his children were queens of any note.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;R)&lt;/strong&gt; Anger suggested, quite extreme. Robots? No, bureaucracy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;R&lt;/strong&gt;age Against the Machine.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Extreme anger = Rage.  &quot;The machine&quot; might sound like it's referring to robots, but it is in fact referring to bureaucracy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;S)&lt;/strong&gt; Crushing squashes makes a mess. Joke name stuck despite protests.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;strong&gt;S&lt;/strong&gt;mashing Pumpkins.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &quot;Crushing squashes&quot; = &quot;smashing pumpkins&quot;, although the &quot;smashing&quot; is actually intended as an adjective, not a verb.  The name was originally made up as a joke, but ended up sticking.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;T)&lt;/strong&gt; Five allure but evil not. Classic group has never stopped.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;strong&gt;T&lt;/strong&gt;emptations. (thanks, @ChrisHayes)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Allure = Temptation, but not evil.  This group has been around since the 60s.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;U)&lt;/strong&gt; Name so short; it's not a lot. Named for spy plane? Maybe not.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;U&lt;/strong&gt;2.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The band name is very short.  There is a U2 spy plane, but its name is not related to the band's name.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;V)&lt;/strong&gt; Try to stop taxi Sienna. Yelling to stop and assist ya.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;V&lt;/strong&gt;an Halen.&lt;br&gt;&#xA;&lt;br&gt;&#xA; A Toyota Sienna is a model of van.  If it's being used as a taxi, and you're trying to flag it down, you're &quot;van hailin'&quot;.  (Groan!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;W)&lt;/strong&gt; Reaching far and wide endemic. Creating world fear pandemic.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;W&lt;/strong&gt;idespread Panic.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Pretty self-explanatory.  A widespread panic is a pandemic of fear across a wide area.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;X)&lt;/strong&gt; I heard you like to puzzle so, here's a puzzle in a puzzle, yo.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;X&lt;/strong&gt;zibit. (thanks, @corsiKa)&lt;br&gt;&#xA;&lt;br&gt;&#xA; This is apparently a reference to the &lt;a href=&quot;http://knowyourmeme.com/memes/xzibit-yo-dawg&quot; rel=&quot;nofollow&quot;&gt;&quot;yo dawg&quot; meme&lt;/a&gt;.  (thanks, @BaileyM)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Y)&lt;/strong&gt; Particles from England rock. Temporary name got stuck.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Y&lt;/strong&gt;es.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Not sure about the &quot;particles&quot;.  The group is from England, and had to pick a temporary name for themselves when they found out another act was using the same name.  They picked &quot;Yes&quot;, and it stuck.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Z)&lt;/strong&gt; Double letter up above but at the end - sometimes - of stove.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;Z&lt;/strong&gt;Z Top.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The double letter is Z.  &quot;Top&quot; is sometimes at the end of &quot;stove&quot;:  &quot;stovetop&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="4421" LastEditDate="2015-07-07T20:25:22.997" LastActivityDate="2015-07-07T20:25:22.997" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="17508" PostTypeId="2" ParentId="17357" CreationDate="2015-07-06T21:53:50.307" Score="1" Body="&lt;p&gt;GUESS 2:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    sedatives&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In some sports I can be used, but only by some. In this case here, it references to one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    sedatives can be used by archers for calming. Other sports, such as certain olympic sports have banned the use of sedatives, often used as downers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be used in some work, more specific than others. I'm not talking about animals or water which could be seen as my brothers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    Tranquilizers can be used as a calming sedative. In that case, they have beneficial properties, unlike dangerous tranquilizers that might be used in hunting. Ocean, water, or seas have a natural calming effect, unlike synthetic sedatives.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am nearby the shops for others to get. Some might ignore me, some might not forget.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    Some sedatives can be over the counter. e.g. Sleeping pills.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might be stronger than others, or weaker than few. Some might have just one, or maybe even two.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    Different types of sedatives have different levels of strengths. Dosage/type may vary.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you are sick you might have me, with a handful of letters&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    sedatives to relax, along with perscription/ingredients/dosages/etc...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;#&#xA;&#xA;&lt;p&gt;GUESS 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    therapy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In some sports I can be used, but only by some. In this case here, it references to one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    Physical therapy, specifically relates to physical sports based therapy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be used in some work, more specific than others. I'm not talking about animals or water which could be seen as my brothers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    Other types of therapies: aquatic or animal assisted therapy (AAT).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am nearby the shops for others to get. Some might ignore me, some might not forget.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    Many therapists available in small shopping plazas: psychotherapists, physical therapists, massage therapists, etc...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might be stronger than others, or weaker than few. Some might have just one, or maybe even two.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    Different types of therapy have different levels of strengths. Visits or sessions may vary.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you are sick you might have me, with a handful of letters&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    Therapist. Degreed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13863" LastEditorUserId="13863" LastEditDate="2015-07-07T00:18:46.483" LastActivityDate="2015-07-07T00:18:46.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17509" PostTypeId="2" ParentId="17471" CreationDate="2015-07-06T23:06:14.100" Score="5" Body="&lt;p&gt;Another 11-letter word is &lt;strong&gt;strumstrums&lt;/strong&gt;, although it doesn't exist at dictionary.com unfortunately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;https://en.wiktionary.org/wiki/strumstrum&quot; rel=&quot;nofollow&quot;&gt;wiktionary&lt;/a&gt; the word &lt;em&gt;strumstrum&lt;/em&gt; appeared in the 1913 edition of Webster's Dictionary, and refers to a type of musical instrument. The plural is strumstrums.&lt;/p&gt;&#xA;" OwnerUserId="40" LastEditorUserId="40" LastEditDate="2015-07-06T23:14:11.253" LastActivityDate="2015-07-06T23:14:11.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17510" PostTypeId="1" AcceptedAnswerId="17512" CreationDate="2015-07-07T00:43:24.367" Score="66" ViewCount="11587" Body="&lt;p&gt;You are well known as the best jeweller in Puzzovania; your shop is always well stocked and your pockets are always bulging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One day, the local 'godfather' of Puzzovania's organised crime comes into your shop. Naturally, you owe him a favour - who doesn't in these parts? He hands you a diamond the size of a hen's egg and tells you what he requires from you:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;This is for my daughter. She's getting married - you understand? I want her to have the best-cut diamond in the whole of Puzzovania. It's family honour.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You nod mutely, inspecting the diamond in awe. It's the biggest one you've ever seen. You decide it would be more prudent not to ask where the diamond came from.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;So listen, my friend. The diamond needs to be cut to the shape of a &lt;strong&gt;convex polyhedron&lt;/strong&gt;. Any size and shape; that doesn't matter. What matters is that &lt;strong&gt;every face of the polyhedron needs to have a different number of sides&lt;/strong&gt;. My daughter's such a changeable character - never quite the same person twice. I want her jewel to reflect that. Do you think you can manage that for me?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Of ... of course, Godfather,&quot; you stammer. Once that imposing gentleman has taken his leave, you sit down, still holding the diamond, and wonder how you're going to do it. You need to call in a mathematician ... or possibly a puzzler!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can you satisfy the Godfather's wishes?&lt;/strong&gt; If so, construct a polyhedron as required (and prepare your best suit, since he'll probably invite you to the wedding as a reward). If not, prove the non-existence of such a polyhedron (and book your flight out of the country, to somewhere far enough away to be out of the reach of his power).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;(Note: this puzzle sounds much better if you read all the Godfather's words in Marlon Brando's voice.)&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-07-08T00:39:16.913" LastActivityDate="2015-07-15T02:12:27.450" Title="The Jeweller's Dilemma" Tags="|mathematics|geometry|story|" AnswerCount="5" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17511" PostTypeId="2" ParentId="17503" CreationDate="2015-07-07T00:57:56.560" Score="5" Body="&lt;p&gt;I don't have enough reputation to comment yet - just wanted to add on to GentlePurpleRain's list. I think that the 20th line is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/The_Five_Satins&quot;&gt;The Five Satins.&lt;/a&gt; They're a doo-wop group which was founded in 1954 - classic! - that's still active, and their name could be mistaken for The Five Satans - evil. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12914" LastActivityDate="2015-07-07T00:57:56.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17512" PostTypeId="2" ParentId="17510" CreationDate="2015-07-07T01:18:40.537" Score="77" Body="&lt;p&gt;It is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; not possible&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof/construction:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let $k$ be the greatest number of sides of any face. This face must touch $k$ other faces. But, each other face must have a distinct number of sides from $3,4,\dots,k-1$, which allows only $k-3$ other faces.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Even more strongly, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; there must be a repeat even if you can ignore two faces. This means, either four faces with the same number of sides, a triplet and a pair, or three pairs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-07-07T01:28:24.263" LastActivityDate="2015-07-07T01:28:24.263" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17513" PostTypeId="2" ParentId="17510" CreationDate="2015-07-07T01:19:48.660" Score="22" Body="&lt;p&gt;This sounds impossible to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine constructing the polyhedron by starting with a triangle (3 sides), and then adding one adjacent polygon at a time.  Each additional polygon must have an extra side, and each additional polygon, being flat and having straight edges, can share at most one edge with at most all of the previous polygons (in practice, this is probably not possible, but it is the best-case scenario).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1st polygon:  Triangle -- 3 open edges (not connected to another polygon)    &#xA;2nd polygon:  Quadrilateral -- 3 open edges (one shared with the triangle)    &#xA;              Total:  5 open edges (3 from the quadrilateral, plus 2 from the triangle)&#xA;3rd polygon:  Pentagon -- 3 open edges (one shared with each of the previous polygons)&#xA;              Total:  3 + 2 + 1 = 6 open edges&#xA;4th polygon:  Hexagon -- 3 open edges&#xA;              Total:  3 + 2 + 1 = 6 open edges (triangle's edges are all used up)    &#xA;5th polygon:  Heptagon -- 4 open edges (shared with hexagon, pentagon, quadrilateral)&#xA;              Total: 4 + 2 + 1 = 7 open edges&#xA;6th polygon:  Octagon -- 5 open edges (shared with heptagon, hexagon, pentagon)&#xA;              Total:  5 + 3 + 1 = 9 open edges      &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As you can see, as you continue to add more, larger polygons, the number of open edges (edges not connected to another polygon) continues to increase.  If you started with something other than a triangle, it would just jack the numbers up even higher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus it is evident that there is no way to construct a polyhedron in the way the Godfather is requesting, and you'd better skip town as soon as possible.&lt;/p&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2015-07-07T01:19:48.660" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17514" PostTypeId="2" ParentId="17309" CreationDate="2015-07-07T04:31:32.347" Score="28" Body="&lt;p&gt;Try:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;44444&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Fours quit the program&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1667" LastActivityDate="2015-07-07T04:31:32.347" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17517" PostTypeId="2" ParentId="17403" CreationDate="2015-07-07T05:04:36.990" Score="5" Body="&lt;p&gt;The following proof relies heavily on Lawrence's insight that we can think of the soldiers as swapping positions instead of turning around.  However, I hope it is a bit more rigorous in terms of handling all possible cases. I also think it differs enough from Mike Earnest's solution to be worth posting.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Assume soldiers march passed each other instead of turning around.  So the east facing soldiers stay east facing and slowly march east, and the west facing soldiers stay west facing, and slowly march west.  We will prove this by induction on $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let Brittney (good soldier name!) be the west facing soldier farthest to the east, and let Adam be the west facing soldier next farthest to the east.   Consider an alternate dimension where Brittney is removed from the line.   Once all the east facing soldiers pass Adam in the line, then the whole thing is over.  By induction, this takes at most $n-2$ seconds.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going back to reality, let's assume Brittney is in the line.   We see that Brittney doesn't really affect anything that is happening west of Adam, and therefore it will still take at most $n-2$ seconds for all the east facers to pass Adam.  We then just have to wait for all those east facers to pass Brittney.  How long will that take?  Well, there are really only two cases to consider.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If at some point Brittney is right behind Adam, then she must stay pretty close to Adam.  Adam may move one step forward while she stands still, but then she'll either keep pace with Adam or catch up.   So there will be at most one east facer between her and Adam from then on.  Therefore, once all the east facers have passed Adam, there is at most one east facer that needs to pass Brittney.  This gives an upper bound of $n-2 + 1 = n-1$ seconds.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If Brittney never catches up to Adam (or if there is no Adam at all!), she'll spend the whole time moving forward.  This clearly will take at most $n-1$ seconds to resolve.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2722" LastActivityDate="2015-07-07T05:04:36.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17518" PostTypeId="2" ParentId="11210" CreationDate="2015-07-07T07:00:41.087" Score="5" Body="&lt;p&gt;A possible solution is the word&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; harmony&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first line gives a list of three offset values. The following lines have one letter each with some indentation, which indicates which of the values to use: No indentation refers to the first offset value, 9; one space of indentation refers to the second value, 18, and two spaces refer to 26.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives us two lists:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 9   9  18  18  18  18  26&#xA; 2   9   1   6   4   5   2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;After subtracting the values in the bottom list from the values in the top list, we get:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 7   0  17  12  14  13  24&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Interpreting these numbers as zero based indices to letters yields:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 7   0  17  12  14  13  24&#xA; H   A   R   M   O   N   Y&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="9860" LastActivityDate="2015-07-07T07:00:41.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17519" PostTypeId="2" ParentId="17357" CreationDate="2015-07-07T07:25:50.623" Score="0" Body="&lt;p&gt;Ok, here is mine. I got same guess as &lt;em&gt;denisdeems&lt;/em&gt;, but my explanation is full.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some sports I can be used, but only by some. In this case here, it references to one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A simple reference to sport discipline such a volleyball for example.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be used in some work, more specific than others. I'm not talking about animals or water which could be seen as my brothers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People catch fish with nets, be it in seas or lakes.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am nearby the shops for others to get. Some might ignore me, some might not forget.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Net prices! Some folks just ignore net/gross value difference. Some of them pay attention.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might be stronger than others, or weaker than few. Some might have just one, or maybe even two.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is related with meaning of internet connection (in my country often referred to as 'net'). People may have local (cable) at home, some got mobile...  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As for the title... It seems that 'treatment' may be a 'net' letters +(some letters) concatenation.&lt;/p&gt;&#xA;" OwnerUserId="12739" LastEditorUserId="12739" LastEditDate="2015-07-07T07:45:17.263" LastActivityDate="2015-07-07T07:45:17.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17520" PostTypeId="2" ParentId="17510" CreationDate="2015-07-07T08:20:39.617" Score="52" Body="&lt;h1&gt;It is possible!&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;How is that polyhedron?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It has 1 face, 1 edge and 0 vertexes!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;WTF!? How!?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You will need a non-euclidean 3D space. There are many possibilities, but lets choose the one with euclidean Y and Z axis, but with a X axis that features &lt;strong&gt;self-repetition with rotation&lt;/strong&gt;.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Ok, you are wondering, what the hell is self-repetition with rotation?&lt;br&gt;&#xA; &lt;br&gt;&#xA; It works this way: Every time you walks a given distance (let's call it $M$) through the X axis (the self-repeating one), you come back to the same YZ plane you were before, but rotated $k$ degrees around the X axis. Here, $k$ may be any rational fraction of 360 degrees which is not an integer multiple of 180 degrees.&#xA; &lt;br&gt;&#xA; Or mathematically, $k = \frac{p}{q} \times 360^\circ | p,q \in \mathbb{Z} \land \frac{2p}{q} \notin \mathbb{Z}$&lt;br&gt;&#xA; &lt;br&gt;&#xA; The result is something similar to a &lt;a href=&quot;https://en.wikipedia.org/wiki/M%C3%B6bius_strip&quot; rel=&quot;noreferrer&quot;&gt;Möbius strip&lt;/a&gt; in a 3D space, though it is not exactly that.&lt;br&gt;&#xA;&lt;br&gt;&#xA; For simplicity, I will choose $k = 90^\circ$, but you could choose any other value that satisfies the above restriction as well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is hard to understand? So let's visualize that with $k = 90^\circ$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/eveDi.png&quot; alt=&quot;long bar&quot;&gt;&#xA; &lt;br&gt;&#xA; The bar is the &quot;polyhedron&quot;. To see what is happening here, I attached ONE (and just ONE) blue peg in the polyhedron. As you see, space repeats itself after some distance through the X axis with a 90 degrees rotation.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's see some properties of this particular polyhedron:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;ul&gt;&#xA; &lt;li&gt; It has only one face.&lt;br&gt;&#xA; &lt;li&gt; Its face has itself as the opposite face.&lt;br&gt;&#xA; &lt;li&gt; Its face also has itself as its left and as its right neighbouring faces.&lt;br&gt;&#xA; &lt;li&gt; Its face neighbours itself twice with only a single edge.&lt;br&gt;&#xA; &lt;li&gt; Its edge, due to the self-repetition of the space, needs no vertexes.&lt;br&gt;&#xA; &lt;li&gt; Its face is planar.&lt;br&gt;&#xA; &lt;li&gt; Its edge is straight.&lt;br&gt;&#xA; &lt;li&gt; The polyhedron is convex.&lt;br&gt;&#xA; &lt;li&gt; All its faces have a different number of edges and there are no 2 faces with the same number of edges (because there are not even 2 faces to start with). So, this satisfacts Godfather's wishes!&lt;br&gt;&#xA; &lt;li&gt; Nothing in the problem said that normal 3D euclidean space should be assumed, so I am breaking no rules here!&lt;br&gt;&#xA; &lt;/ul&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ok, but how could the best jeweller of Puzzovania construct such thing? The jeweller can't make space behave this way!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You would need to think with portals. And with a little help from Apperture Science, it is easily doable using your portal gun. Note the 90 degrees rotation between the portals.&lt;br&gt;&#xA; &lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/TZNJ3.png&quot; alt=&quot;portals&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5044" LastEditorUserId="5044" LastEditDate="2015-07-07T08:52:16.373" LastActivityDate="2015-07-07T08:52:16.373" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17522" PostTypeId="2" ParentId="17357" CreationDate="2015-07-07T09:57:22.770" Score="1" Body="&lt;p&gt;Is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; cards&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;In some sports I can be used, but only by some. In this case here, it references to one.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; cards are used to signal in sports (example: yellow/red cards in football)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I can be used in some work, more specific than others. I'm not talking about animals or water which could be seen as my brothers.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; cards are used for signalling during work, or for motivating in case of motivational messages. Don't know about the animals and water thingy, but if i have to give a stupid guess, maybe cards are used to play FISH, which can be a reference to animals and water&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I am nearby the shops for others to get. Some might ignore me, some might not forget.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; cards can often kept on shelves outside a general store to be bought. When one gets cards, inviting them for weddings, parties, you may choose to ignore them (if you are rude) or use them to remember the date of the function.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I might be stronger than others, or weaker than few. Some might have just one, or maybe even two.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Maybe a reference for some cards being weaker [example: red cards are more powerful in football than yellow cards]  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you are sick you might have me, with a hand full of letters&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; when you are sick, you recieve &quot;get well soon cards&quot;, full of letters&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13080" LastEditorUserId="13080" LastEditDate="2015-07-07T10:06:25.220" LastActivityDate="2015-07-07T10:06:25.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17525" PostTypeId="2" ParentId="17448" CreationDate="2015-07-07T13:34:17.190" Score="2" Body="&lt;p&gt;Not the longest one, but I think it deserves a mention in a question like this :)&lt;br&gt;&#xA;At &lt;strong&gt;18&lt;/strong&gt; letters...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;sesquipedalianisms&lt;/em&gt; - The practice of using long, sometimes obscure, words in speech or writing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13148" LastActivityDate="2015-07-07T13:34:17.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17526" PostTypeId="2" ParentId="17357" CreationDate="2015-07-07T14:25:58.893" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hand?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In some sports I can be used, but only by some. In this case here, it references to one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In football (soccer), and other sports (such as baseball), only some players (goalkeepers/fielding team) can use their hands. Here, one hand.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be used in some work, more specific than others. I'm not talking about animals or water which could be seen as my brothers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some jobs use hands more often. Animals/water could refer to &quot;farmhands&quot; or &quot;deckhands&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am nearby the shops for others to get. Some might ignore me, some might not forget.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Handicap parking spots. Some people ignore the rules and park there illegitimately.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might be stronger than others, or weaker than few. Some might have just one, or maybe even two.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People have different strength hands, and different numbers of them. May refer to hands of cards?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you are sick you might have me, with a hand full of letters&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not so sure, unless it's something like handicap again, or some kind of medicine.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12462" LastActivityDate="2015-07-07T14:25:58.893" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17527" PostTypeId="2" ParentId="17486" CreationDate="2015-07-07T14:59:11.853" Score="2" Body="&lt;p&gt;I've found something that may be on the right track, but I can't quite get it to work for the second lock. Note that since I got the following from wikipedia, there's a chance some of the numbers are wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If we take the number of regions in each country in the first set and add n-1, where n is the position in the code, we get:&lt;br&gt;&#xA;&lt;br&gt;&#xA; Belgium - 10 provinces (+0) = &lt;strong&gt;10&lt;/strong&gt;&lt;br&gt;&#xA; Germany - 16 states (+1) = &lt;strong&gt;17&lt;/strong&gt;&lt;br&gt;&#xA; France - 27 regions (+2) = &lt;strong&gt;29&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; I'm not sure how to handle numbers over 39, however. If we assume we should mod by 40 (there being 40 possible digits to enter), we get:&lt;br&gt;&#xA;&lt;br&gt;&#xA; Saudi Arabia - 118 amins (mods twice to 38) (+0) = &lt;strong&gt;38&lt;/strong&gt;&lt;br&gt;&#xA; Iran - 324 shahreshtans (mods 8 times to 4) (+1) = &lt;strong&gt;5&lt;/strong&gt;&lt;br&gt;&#xA; Qatar - 87 zones (mods twice to 7) (+2) = &lt;strong&gt;9&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13764" LastEditorUserId="13764" LastEditDate="2015-07-07T15:07:01.093" LastActivityDate="2015-07-07T15:07:01.093" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17528" PostTypeId="2" ParentId="17357" CreationDate="2015-07-07T15:14:07.713" Score="1" Body="&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Try 2: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MUMPS  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you are sick, you might have me, with a handful of letters  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mumps is a sickness, but is also an acronym for a programming language.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In some sports I can be used, but only by some. In this case here, it references to one.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is where I am not quite sure yet. Perhaps the programming language is used in some sports? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be used in some work, more specific than others.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The programming language has a set of things that it can do, mostly transaction processing.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not talking about animals or water which could be seen as my brothers.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not talking about Python or C (sea) which are other programming languages.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am nearby the shops for others to get. Some might ignore me, some might not forget.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mumps the disease was very contageous. With vaccinations though, it is mostly been eliminated. Some ignore vaccines, and many have forgotten about mumps.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might be stronger than others, or weaker than few. Some might have just one, or maybe even two.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The programming language is better than others, not as good as some. The disease is worse than some, but weaker than others.  Some people might have the language or the disease, but there is the chance some could have both.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Previous Answer:    &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pain Killers, more specific than drugs. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you are sick, you might have me, with a handful of letters  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some take painkillers when they are sick to help with headaches, and often when you are sick you take vitamins as well (which most are named after letters).  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In some sports I can be used, but only by some. In this case here, it references to one.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some sports allow for pain killers, but generally not all. And, generally they can only be given by one, the team doctor.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be used in some work, more specific than others.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Those in manual labor will use pain killers to help with sore muscles. Also, doctors will give pain killers to patients.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not talking about animals or water which could be seen as my brothers.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The symbol for the medical field involves a snake wrapped around a staff (referencing doctors), and many think that the sea has healing and pain relieving properties.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am nearby the shops for others to get. Some might ignore me, some might not forget.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pain killers are often found in the pharmacy section of stores. Some try to ignore and fight through pain without taking anything, and older people may just forget to take their medicine, of which pain killers may be a part of.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might be stronger than others, or weaker than few. Some might have just one, or maybe even two.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pain killers come in different strengths (Give me MAXIMUM STRENGTH!). Also, some take one or two depending on the dosage.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="9951" LastEditDate="2015-07-07T15:41:24.693" LastActivityDate="2015-07-07T15:41:24.693" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17529" PostTypeId="2" ParentId="17458" CreationDate="2015-07-07T16:32:39.457" Score="3" Body="&lt;p&gt;Nice :)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A snowflake. The human life based on water and a snowflake is frozen water. If I touch or take a snowflake they will melt.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13882" LastActivityDate="2015-07-07T16:32:39.457" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17530" PostTypeId="2" ParentId="17357" CreationDate="2015-07-07T16:39:12.073" Score="0" Body="&lt;p&gt;One more try... Probably my last. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Band the Village People&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you are sick, you might have me, with a handful of letters  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; YMCA -  a handful of letters  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In some sports I can be used, but only by some. In this case here, it references to one.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many sports teams play Village People songs at games, but the Los Angeles Clippers have been banned from playing &quot;YMCA&quot; by the band. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be used in some work, more specific than others.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Village People dressed up as different characters, such as a construction worker, police officer, etc. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not talking about animals or water which could be seen as my brothers.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not &quot;White Snake&quot; or &quot;Styx&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am nearby the shops for others to get. Some might ignore me, some might not forget.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Perhaps referencing characters in the band? Some people tried to ignore the Village People, but others are huge fans and can't forget them.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might be stronger than others, or weaker than few. Some might have just one, or maybe even two.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Macho Man is a strong man. The line in the song goes &quot;Macho Macho Man&quot; with two &quot;Machos&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-07-07T16:39:12.073" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17531" PostTypeId="1" AcceptedAnswerId="17566" CreationDate="2015-07-07T16:46:43.167" Score="1" ViewCount="2049" Body="&lt;p&gt;&lt;strong&gt;BACKGROUND&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;I asked a question about words: &lt;a href=&quot;https://puzzling.stackexchange.com/q/17471/9000&quot;&gt;Longest Word With *Only* Repeating Character Pairs&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The goal was to find a word where every character pair was repeated at least once. This was vulnerable to dictionary searches and didn't require human creativity. In addition, every answer was of the format &lt;code&gt;[string][same string][first letter of string]&lt;/code&gt;. For instance, &lt;code&gt;ALFALFA&lt;/code&gt; is &lt;code&gt;[ALF][ALF][A]&lt;/code&gt;. In fact, from my own quick and non-technical figurings, this format is the &lt;em&gt;only&lt;/em&gt; possible format that can fit such a challenge. The answers found were interesting but it was this conclusion that I found &lt;em&gt;more&lt;/em&gt; interesting.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;CHALLENGE&lt;/h2&gt;&#xA;&lt;p&gt;&lt;strong&gt;Form a grammatically correct sentence in which every letter pair appears at least twice.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Clarifications: (The example below is invalid and is only intended to assist in clarification.)&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Words can not be repeated but variations of a word do not count as a repetition. This includes plurals, possessives, contractions, etc. &lt;code&gt;Uni the Unicorn's eating the unicorns!&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;Words count as a repetition if they match when converted to all uppercase and any non-letters are removed&lt;/li&gt;&#xA;&lt;li&gt;Any non-letter (spaces, hyphens, apostrophes, etc.) should be stripped prior to analysis. The result will be just a string of letters. &lt;code&gt;UnitheUnicornseatingtheunicorns&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;Capitalization does not matter. Convert them all to uppercase if that helps. &lt;code&gt;UNITHEUNICORNSEATINGTHEUNICORNS&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Disclosure:&lt;/strong&gt; I don't have a solution yet.&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-08T16:44:21.860" Title="Longest *Sentence* With Only Repeating Character Pairs" Tags="|english|open-ended|letters|" AnswerCount="5" CommentCount="11" ClosedDate="2021-05-08T15:23:32.063" ContentLicense="CC BY-SA 3.0" />
  <row Id="17532" PostTypeId="2" ParentId="17096" CreationDate="2015-07-07T16:53:59.440" Score="2" Body="&lt;p&gt;Well, I didn't find a solution that used at most 5 weighings, but I found one nonetheless (though it is a bit long). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;also, I can't seem to put the following in one spoiler tag?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First we split the 27 coins into three groups of nine. There are two cases:  &lt;br/&gt;&#xA; 1) Two groups each contain one fake, or  &lt;br/&gt;&#xA; 2) One group contains both fakes.&#xA; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;For case 1:&lt;/strong&gt; the pan will lift up the group with no fakes, so now we choose to work with the two piles that contain the fakes. Divide each pile into three groups of three coins such that we have two groups of three smaller piles. Now we label these piles a, b, c, d, e, and f, and we know two such piles contain a fake coin. Without loss of generality (WLOG), let's say that a and d contain the fake coins (now labeled A and D).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We choose one pile to swap from each group, and weigh the &quot;first&quot; group each time: &lt;br/&gt;&#xA; Swap A with D, e, or f --&gt; nothing happens. &lt;br/&gt;&#xA; Swap b/c with e or f --&gt; nothing happens. &lt;br/&gt;&#xA; Swap b/c with D --&gt; c/b raises; we know A and D contain the two fakes. [worst case is 6 weighings]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Now that we know A and D contain the fakes, we redistribute them into three piles of two, giving the following cases after weighing [o = normal, O = fake]:  &lt;br/&gt;&#xA; Case 1.1: (oo | oo | OO) --&gt; nothing happens &lt;br/&gt;&#xA; Case 1.2: (oO | oO | oo) --&gt; one group is lifted &lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If case 1.1 happens, then we redistribute the coins such that no two are in the same pile again to arrive at Case 1.2. For case 1.2, we have WLOG coins w, X, y, Z (X and Z being fake). We swap a coin from each pile with each other, giving us either (wy | XZ) or (wZ | yX). If we get the latter arrangement, we switch one coin being swapped to arrive at the former arrangement. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Now we take one of the piles (wy or XZ) and split it between the remaining piles. Then we take three real coins and put them on the balance to weigh. If we split the pair wy, then the balance does nothing, and so the other pair must be the fake coins. If we split the pair XZ, then the balance would lift the three real coins we just added, and so the pair that we split must be the fake coins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Through this method, we have at worst case: 6 + 2 + 2 = 10 weighings after the initial weighing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Now for case 2:&lt;/strong&gt;&#xA; If we have one pile containing two fakes, then we can &quot;deal&quot; the coins into three new piles of 9 coins. To illustrate, we have coins A, b, c, D, e, f, g, h, i (a worst case arrangement). Dealing out this pile, we will get groups (ADg | beh | cfi). Once we finish dealing the rest of the coins and reweigh, we find that the balance does nothing again and will thus require another redealing. So at worst case, it will take 2 weighings to achieve the setup for case 1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In this case, we have at worst case: 2 + 10 = 12 weighings after the initial weighing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Thus overall, it will take &lt;strong&gt;at most 13 tries&lt;/strong&gt; to find the fake coin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;" OwnerUserId="13881" LastEditorUserId="13881" LastEditDate="2015-07-08T14:54:15.503" LastActivityDate="2015-07-08T14:54:15.503" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17533" PostTypeId="2" ParentId="17531" CreationDate="2015-07-07T16:59:27.060" Score="5" Body="&lt;h2&gt;31 letters&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Buffalo buffaloes buffalo oboes - boo!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(i.e. &quot;bison from Buffalo, NY, bully certain woodwind instruments - boo!&quot;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-07-07T16:59:27.060" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17534" PostTypeId="2" ParentId="17531" CreationDate="2015-07-07T17:10:02.680" Score="2" Body="&lt;p&gt;My first thought as to the solution would be to find a word that contains numerous words as a subset, so that the subset, when ordered and separated, form a sentence. This is not the only solution, but would yield a solution nonetheless.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The, er, &lt;em&gt;most&lt;/em&gt;, at s' Thermostat's!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alright, it uses inflection (the &amp;ldquo;&lt;code&gt;s'&lt;/code&gt;&amp;rdquo; meaning “the”) and &amp;ldquo;thermostats&amp;rdquo; as a possessive, but hey: it only uses each word once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to be a response/(slogan-by-a-confused-salesperson); explanation: Well, you can get the most at Thermostat's place.&lt;/p&gt;&#xA;" OwnerUserId="7307" LastActivityDate="2015-07-07T17:10:02.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17536" PostTypeId="2" ParentId="17531" CreationDate="2015-07-07T17:49:08.210" Score="10" Body="&lt;h2&gt;59 letters&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sentimentalized sensationalism sensationalized sentimentalisms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or, sensationalism that was sentimentalized caused sentimentalisms that were sensationalized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could flip it around too, if you wanted. Either way, it's sort of a commentary on our society. =P&lt;/p&gt;&#xA;" OwnerUserId="10861" LastActivityDate="2015-07-07T17:49:08.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17537" PostTypeId="1" AcceptedAnswerId="17538" CreationDate="2015-07-07T17:50:22.880" Score="5" ViewCount="267" Body="&lt;p&gt;8 competitors (A-H) participated in 2 races. The first was the practice race, and the second being the final one. No 2 people got the same rank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A and B improved their ranks in the final race by 5 spaces each. C did worse than G in the final race. G became worse by 5 spaces in the final race. C did better than E and F in the practice race. D came first in the practice race, but became 4th in the final race. E and F swapped ranks in the final race.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did the winner of the final race fare in the practice one?&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="7308" LastEditDate="2015-07-07T23:04:49.350" LastActivityDate="2015-07-07T23:04:49.350" Title="Ranks in 2 races" Tags="|logical-deduction|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17538" PostTypeId="2" ParentId="17537" CreationDate="2015-07-07T17:58:36.093" Score="7" Body="&lt;p&gt;The winner of the final race was &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;H, who placed 3rd or 4th in the practice race.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have 8 possible places in each race. We know that D was 1st in the practice, 4th in the final, so we can fill him in.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   P  F&#xA;1  D&#xA;2&#xA;3&#xA;4     D&#xA;5&#xA;6&#xA;7&#xA;8&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Next we know G got worse by 5 spaces, which means he was either 1,2, or 3 in the practice race and 6,7, or 8 in the final race. However, D was 1 in practice, and C was worse than G in the final, so G must have gotten 2 and 7 in the respective races.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   P  F&#xA;1  D&#xA;2  G&#xA;3&#xA;4     D&#xA;5&#xA;6&#xA;7     G&#xA;8     C&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;E and F switched spots in the final race, so they can't be in spots 4,7, or 8 in the practice race. C did better than E and F in the practice race, so they must be in slots 5 and 6 in both races (doesn't matter which order).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   P  F&#xA;1  D&#xA;2  G&#xA;3  &#xA;4     D&#xA;5  E  F&#xA;6  F  E&#xA;7     G&#xA;8     C&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;A and B moved up five spaces in the final race, so they must be in slots 7 and 8 in the practice race and 2 and 3 in the final race (doesn't matter which order).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   P  F&#xA;1  D&#xA;2  G  A&#xA;3     B&#xA;4     D&#xA;5  E  F&#xA;6  F  E&#xA;7  A  G&#xA;8  B  C&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;That leaves H in the remaining places: 3rd or 4th in the practice race, 1st in the final race.&lt;/p&gt;&#xA;" OwnerUserId="10861" LastEditorUserId="10861" LastEditDate="2015-07-07T18:19:59.733" LastActivityDate="2015-07-07T18:19:59.733" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17539" PostTypeId="2" ParentId="17357" CreationDate="2015-07-07T17:59:40.397" Score="4" Body="&lt;p&gt;I would like to expand on Kate Gregory's answer but I have different explanations for the clues that would be too long to explain in a comment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To save people from looking, Kate Gregory proposed that the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The letter C&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you are sick, you might have me, with a handful of letters&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The word 'sick' contains the letter C, along with a handful of other letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In some sports I can be used, but only by some. In this case here, it references to one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Some sports have the letter C in their name but in the phrase 'this case here', there is only one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be used in some work, more specific than others. I'm not talking about animals or water which could be seen as my brothers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C is a programming language that gives you very specific control over the hardware - more than other programming languages. For the brothers, Python is the name of another programming language as well as a type of snake - an animal with no legs. 'Water' could be substituted for 'sea', which is a homonym of C.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am nearby the shops for others to get. Some might ignore me, some might not forget.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A reference to another homonym, 'see'. There are plenty of things to see near shops and some things you see are easy to ignore while others are hard to forget.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might be stronger than others, or weaker than few. Some might have just one, or maybe even two.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter C can be pronounced many different ways, with some pronunciations being stronger than others. The letter often appears in words one at a time but is doubled or certain words. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Other notes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The third clue is simply a link to the author's profile, where we learn that he works with computers and likes wordplay. This answer seems in line with his personal interests and style of riddles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13541" LastActivityDate="2015-07-07T17:59:40.397" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17540" PostTypeId="1" AcceptedAnswerId="17648" CreationDate="2015-07-07T18:09:05.533" Score="23" ViewCount="1386" Body="&lt;p&gt;My friend is so &lt;a href=&quot;http://www.urbandictionary.com/define.php?term=old%20school&quot; rel=&quot;nofollow noreferrer&quot;&gt;old-school&lt;/a&gt;. He gave me the list below of phrases that he calls Inefficient Phrases™, which, by the way, are totally unrelated to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/17079/what-is-an-efficient-phrase%E2%84%A2&quot;&gt;Efficient Phrases™&lt;/a&gt;, and &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%22What%20is%22%20Phrase%E2%84%A2%3F&quot;&gt;all other types of Phrases™&lt;/a&gt; that were the subject of previous puzzles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hnNzJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What makes a phrase an Inefficient Phrase™, and why are they called that?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: If someone posts the correct answer to the first part of the question, and then later &lt;em&gt;someone else&lt;/em&gt; correctly explains &lt;strong&gt;why&lt;/strong&gt; the phrases are called Inefficient, both will be up-voted, but that &lt;em&gt;someone else&lt;/em&gt; will get the tick mark.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT: Here are some examples to show that the &quot;5 pairs&quot; answer by GentlePurpleRain is not exactly it:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The following examples are &lt;strong&gt;Not Inefficient Phrases™&lt;/strong&gt;&lt;br&gt;&#xA;  1. SPORTSCASTER PODCASTS&lt;br&gt;&#xA;  2. ANTHROPOLOGISTS ARE OPPORTUNISTIC&lt;br&gt;&#xA;  3. MONOPOLISTIC TOPOLOGISTS&lt;br&gt;&#xA;  Each of the 3 examples above has 7 pairs of letters that are adjacent in the alphabet&lt;br&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;On the other hand, the following examples &lt;em&gt;are&lt;/em&gt; &lt;strong&gt;Inefficient Phrases™&lt;/strong&gt;, yet they have zero pairs of adjacent letters.&lt;br&gt;&#xA;  1. MOMS PROMOTE&lt;br&gt;&#xA;  2. GIG CARPS&lt;br&gt;&#xA;  3. SPRY CAMPS&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In response to Mauris's comment, here's a plaintext version of the picture of the table:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Inefficient Phrases, Not Inefficient Phrases&lt;/strong&gt;&lt;br&gt;&#xA;DEBUTED TONIGHT, LAUNCHED THIS EVENING&lt;br&gt;&#xA;TUTORS SPOONFED, TEACHERS HELP TOO MUCH&lt;br&gt;&#xA;BABIED BABOONS, PAMPERED MONKEYS&lt;br&gt;&#xA;DEACON CHIDED, MINISTER SCOLDED&lt;br&gt;&#xA;COMMONERS FEUDED, CIVILIANS FOUGHT&lt;br&gt;&#xA;SCARFED ABALONE, ATE SEA SNAILS&lt;br&gt;&#xA;HIGHEST LONGITUDE, MAXIMUM EAST-WEST POINT&lt;br&gt;&#xA;PREFERS MONOSPACED, LIKES FIXED-WIDTH&lt;br&gt;&#xA;AUTUMN BACKLIGHT, FALL BACKGROUND&lt;br&gt;&#xA;CONFEDERACY PROMOTERS, SECESSION ADVOCATE&lt;br&gt;&#xA;DEFACERS ABATED, VANDALS DECREASED&lt;br&gt;&#xA;CABBAGE SPROUTED, BRASSICA OLERACEA GREW&lt;br&gt;&#xA;BANDED SPIDERS, STRIPED ARACHNIDS&lt;br&gt;&#xA;DEFEATS DEMONS, CONQUERS DEVILS&lt;br&gt;&#xA;DELIGHTED DONORS, HAPPY CONTRIBUTOR&lt;br&gt;&#xA;SPEEDERS ABSCONDED, RACERS FLED&lt;br&gt;&#xA;DEFENSE CUTBACK, PROTECTION REDUCTION&lt;br&gt;&#xA;ABANDONED HIGHWAY, DESERTED ROAD&lt;br&gt;&#xA;HONORS DELIGHT, RECOGNITION PLEASES&lt;br&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2015-08-12T17:09:28.223" Title="What is an Inefficient Phrase™?" Tags="|pattern|english|" AnswerCount="2" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="17542" PostTypeId="2" ParentId="17537" CreationDate="2015-07-07T18:14:43.207" Score="1" Body="&lt;p&gt;In the practice race:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;D came first&lt;/strong&gt; (given)&lt;/li&gt;&#xA;&lt;li&gt;G came second or third, since he has room to become worse by 5 spaces; also C did worse than G in the final race, so &lt;strong&gt;G came second&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;A and B were both in the last three, since they had room to improve by 5 spaces&lt;/li&gt;&#xA;&lt;li&gt;C came third or fourth, since E, F, A, B must all be worse than C&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the final race:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;D came fourth&lt;/strong&gt; (given)&lt;/li&gt;&#xA;&lt;li&gt;A and B were both in the top three, since they improved by 5 spaces&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;G came seventh&lt;/strong&gt;, since he became worse by 5 spaces&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;C came eighth&lt;/strong&gt;, since she did worse than G&lt;/li&gt;&#xA;&lt;li&gt;one of H, E, F was in the top three and the other two were fifth and sixth&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Since &lt;strong&gt;E and F&lt;/strong&gt; swapped places, the only possibility is now that they were &lt;strong&gt;fifth and sixth&lt;/strong&gt; in both races. That means &lt;strong&gt;A and B&lt;/strong&gt; were &lt;strong&gt;seventh and eighth&lt;/strong&gt; in the practice race and therefore &lt;strong&gt;second and third&lt;/strong&gt; in the final race. So: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; H won the final race, and came third or fourth in the practice race.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-07-07T18:14:43.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17543" PostTypeId="2" ParentId="17540" CreationDate="2015-07-07T18:28:30.107" Score="6" Body="&lt;p&gt;Each Inefficient Phrase™ has at least five pairs of letters that are adjacent in the alphabet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g.  [&lt;strong&gt;DE&lt;/strong&gt;]B[&lt;strong&gt;UT&lt;/strong&gt;][&lt;strong&gt;ED&lt;/strong&gt;] T[&lt;strong&gt;ON&lt;/strong&gt;]I[&lt;strong&gt;GH&lt;/strong&gt;]T&lt;br&gt;&#xA;[&lt;strong&gt;TU&lt;/strong&gt;]TO[&lt;strong&gt;RS&lt;/strong&gt;] S[&lt;strong&gt;PO&lt;/strong&gt;][&lt;strong&gt;ON&lt;/strong&gt;][&lt;strong&gt;FED&lt;/strong&gt;]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for why they're called &quot;inefficient,&quot; I'll leave that for someone else to figure out.&lt;/p&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="4421" LastEditDate="2015-07-07T18:56:47.487" LastActivityDate="2015-07-07T18:56:47.487" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="17545" PostTypeId="1" AcceptedAnswerId="17547" CreationDate="2015-07-07T19:03:39.623" Score="-5" ViewCount="1314" Body="&lt;p&gt;&lt;strong&gt;Objective:&lt;/strong&gt; Find the longest word, which can be made from concatenation smaller words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rules:&lt;/strong&gt;&lt;br&gt;&#xA;* Words (and smaller words inside) must be findable at dictionary.com&lt;br&gt;&#xA;* Punctuation (for example ') does not count as part of the smaller word&lt;br&gt;&#xA;* No numbers&lt;br&gt;&#xA;* Must be done in English&lt;br&gt;&#xA;* Must contain at least 3 smaller words  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Best:&lt;/strong&gt;&lt;br&gt;&#xA;30 letters:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Pseudopseudohypoparathyroidism  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;made from the 3 words:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Pseudo, Pseudo , hypoparathyroidism  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Answerer:&lt;/em&gt; TroyAndAbed&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2015-07-07T20:18:24.060" LastActivityDate="2018-07-28T04:20:20.273" Title="Longest Word made from Smaller Words" Tags="|language|open-ended|" AnswerCount="4" CommentCount="6" ClosedDate="2021-05-04T08:40:14.340" ContentLicense="CC BY-SA 3.0" />
  <row Id="17547" PostTypeId="2" ParentId="17545" CreationDate="2015-07-07T19:12:52.280" Score="2" Body="&lt;p&gt;30 letters:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://dictionary.reference.com/browse/pseudopseudohypoparathyroidism&quot; rel=&quot;nofollow&quot;&gt;Pseudopseudohypoparathyroidism&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;made from the 3 words:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://dictionary.reference.com/browse/pseudo?s=t&quot; rel=&quot;nofollow&quot;&gt;Pseudo&lt;/a&gt;, Pseudo , &lt;a href=&quot;http://dictionary.reference.com/browse/hypoparathyroidism?s=ts&quot; rel=&quot;nofollow&quot;&gt;hypoparathyroidism&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;br&gt;&#xA;Also with 28 letters a non-medical word:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Antidisestablishmentarianism&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Made with :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://dictionary.reference.com/browse/anti?s=t&quot; rel=&quot;nofollow&quot;&gt;Anti&lt;/a&gt;, &lt;a href=&quot;http://dictionary.reference.com/browse/disestablishment?s=t&quot; rel=&quot;nofollow&quot;&gt;disestablishment&lt;/a&gt;, &lt;a href=&quot;http://dictionary.reference.com/browse/arianism?s=t&quot; rel=&quot;nofollow&quot;&gt;arianism&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5359" LastEditorUserId="5359" LastEditDate="2015-07-07T19:38:46.800" LastActivityDate="2015-07-07T19:38:46.800" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17548" PostTypeId="2" ParentId="17545" CreationDate="2015-07-07T20:05:33.407" Score="3" Body="&lt;p&gt;18 letters:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; counterprogramming: counter, program, ming &lt;br/&gt;&#xA; unfortunately countercountermeasures (22 letters!) wasn't found in dictionary.com :(&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some words off the top of my head:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;13 letters:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; nevertheless&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;12 letters:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; forevermore, nonetheless&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13881" LastEditorUserId="13881" LastEditDate="2015-07-10T20:07:32.907" LastActivityDate="2015-07-10T20:07:32.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17549" PostTypeId="2" ParentId="17357" CreationDate="2015-07-07T20:13:47.053" Score="2" Body="&lt;p&gt;&lt;strong&gt;PARTIAL ANSWER&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From mmking, it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Java&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In some sports I can be used, but only by some. In this case here, it references to one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not 100% here. Could be code golf.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be used in some work, more specific than others. I'm not talking about animals or water which could be seen as my brothers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Animals (Python), water/ocean (C).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am nearby the shops for others to get. Some might ignore me, some might not forget.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Coffee near the shops.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might be stronger than others, or weaker than few. Some might have just one, or maybe even two.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some cups of coffee are made stronger than others. People drink different amounts of coffee.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you are sick you might have me, with a hand full of letters&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I guess people might have coffee with their supplements? This doesn't seem too common.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9696" LastEditorUserId="9696" LastEditDate="2015-07-07T20:45:48.477" LastActivityDate="2015-07-07T20:45:48.477" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="17550" PostTypeId="1" AcceptedAnswerId="17598" CreationDate="2015-07-07T20:16:49.400" Score="15" ViewCount="801" Body="&lt;p&gt;&lt;strong&gt;Disclaimer:&lt;/strong&gt; I'll be using the gender-neutral pronoun &amp;quot;&lt;a href=&quot;https://en.wiktionary.org/wiki/xe&quot; rel=&quot;noreferrer&quot;&gt;xe&lt;/a&gt;&amp;quot; for the singular third person.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;After escaping from &lt;a href=&quot;https://puzzling.stackexchange.com/q/14933/9000&quot;&gt;d'alar'cop's cellar&lt;/a&gt;, Doorknob♦ was understandably upset. After almost two months of planning, xe gathered some support and the revenge scheme was enacted...&lt;/p&gt;&#xA;&lt;p&gt;d'alar'cop awoke in a darkened room on a hard surface. Xe was sore from lying there for some unknown time. As soon as xe shifted xyr weight to stand up, lights came on all around the room. Blinking in the sudden brightness, xe took in xyr surroundings. Xe was on a raised daïs in the middle of a round room. A soft glow emanated from underneath and illuminated most of the room. There were 31 other puzzlers in the room, each in their own alcove. The recesses were painted a variety of colors and even the daïs was painted blue. Each puzzler was either staring at d'alar'cop or facing directly away. In front of each puzzler was a set of two lights that was facing the daïs, facing the alcove, or turned off. (AJ Henderson was staring directly into two powerful lights blasting into xyr face. Ow.) Just as d'alar'cop had finished taking this all in, all 31 puzzlers began to chant in unison:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;You've been naughty, kidnapper&lt;br /&gt;&#xA;Taking prisoners, doing wrong&lt;br /&gt;&#xA;Now it's time for you to learn&lt;br /&gt;&#xA;If you're mean, you will get burned&lt;/p&gt;&#xA;&lt;p&gt;You are trapped here in this room&lt;br /&gt;&#xA;From that daïs you may not move&lt;br /&gt;&#xA;You must solve our riddle soon&lt;br /&gt;&#xA;Or from our pack you will be hewn&lt;/p&gt;&#xA;&lt;p&gt;The door at south is your escape&lt;br /&gt;&#xA;But it shan't open lest you say&lt;br /&gt;&#xA;The secret message known to us&lt;br /&gt;&#xA;Speaking quickly is a plus&lt;/p&gt;&#xA;&lt;p&gt;Poison daggers now descend&lt;br /&gt;&#xA;From ceiling to your flesh to pierce&lt;br /&gt;&#xA;They will slow your very breath&lt;br /&gt;&#xA;Bringing soon a painful death&lt;/p&gt;&#xA;&lt;p&gt;If you belong then you will see&lt;br /&gt;&#xA;An answer not quite seen by thee&lt;br /&gt;&#xA;Although mad we are not cruel&lt;br /&gt;&#xA;We shall grant a single tool&lt;/p&gt;&#xA;&lt;p&gt;You have a laptop, internet&lt;br /&gt;&#xA;Seek the answer on the web&lt;br /&gt;&#xA;Careful, though, where you tread&lt;br /&gt;&#xA;There are tricks and traps ahead&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;d'alar'cop's eyes adjusted to the light during the chant and xe could see the laptop. Thankfully, it's already booted up and xyr favorite browser was open to a blank tab. With a quick glance upward at the glistening blades, xe set to work as the puzzlers started up their chant again.&lt;/p&gt;&#xA;&lt;h2&gt;How can d'alar'cop find the password?&lt;/h2&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Diagram of the room: (The diagram was made in Excel. Any minor errors in position or orientation are not part of the puzzle.)&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xUBft.png&quot; alt=&quot;Diagram of the Room&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Puzzler Positions:&lt;/p&gt;&#xA;&lt;p&gt;1:  &lt;a href=&quot;https://puzzling.stackexchange.com/users/74/rainbolt&quot;&gt;Rainbolt&lt;/a&gt;&lt;br /&gt;&#xA;2:  &lt;a href=&quot;https://puzzling.stackexchange.com/users/69/andrea&quot;&gt;Andrea&lt;/a&gt;&lt;br /&gt;&#xA;3:  &lt;a href=&quot;https://puzzling.stackexchange.com/users/84/rberteig&quot;&gt;RBerteig&lt;/a&gt;&lt;br /&gt;&#xA;4:  &lt;a href=&quot;https://puzzling.stackexchange.com/users/32/zar&quot;&gt;zar&lt;/a&gt;&lt;br /&gt;&#xA;5:  &lt;a href=&quot;https://puzzling.stackexchange.com/users/70/mau&quot;&gt;mau&lt;/a&gt;&lt;br /&gt;&#xA;6:  &lt;a href=&quot;https://puzzling.stackexchange.com/users/85/sztupy&quot;&gt;SztupY&lt;/a&gt;&lt;br /&gt;&#xA;7:  &lt;a href=&quot;https://puzzling.stackexchange.com/users/101/hubble07&quot;&gt;Hubble07&lt;/a&gt;&lt;br /&gt;&#xA;8:  &lt;a href=&quot;https://puzzling.stackexchange.com/users/76/senshin&quot;&gt;senshin&lt;/a&gt;&lt;br /&gt;&#xA;9:  &lt;a href=&quot;https://puzzling.stackexchange.com/users/95/the-wobbuffet&quot;&gt;The Wobbuffet&lt;/a&gt;&lt;br /&gt;&#xA;10: &lt;a href=&quot;https://puzzling.stackexchange.com/users/67/adele-c&quot;&gt;Adele C&lt;/a&gt;&lt;br /&gt;&#xA;11: &lt;a href=&quot;https://puzzling.stackexchange.com/users/65/aj-henderson&quot;&gt;AJ Henderson&lt;/a&gt;&lt;br /&gt;&#xA;12: &lt;a href=&quot;https://puzzling.stackexchange.com/users/78/user88377&quot;&gt;user88377&lt;/a&gt;&lt;br /&gt;&#xA;13: &lt;a href=&quot;https://puzzling.stackexchange.com/users/39/doorknob&quot;&gt;Doorknob♦&lt;/a&gt;&lt;br /&gt;&#xA;14: &lt;a href=&quot;https://puzzling.stackexchange.com/users/116/gsingh2011&quot;&gt;gsingh2011&lt;/a&gt;&lt;br /&gt;&#xA;15: &lt;a href=&quot;https://puzzling.stackexchange.com/users/45/bluefeet&quot;&gt;bluefeet♦&lt;/a&gt;&lt;br /&gt;&#xA;16: &lt;a href=&quot;https://puzzling.stackexchange.com/users/77/robert-cartaino&quot;&gt;Robert Cartaino♦&lt;/a&gt;&lt;br /&gt;&#xA;17: &lt;a href=&quot;https://puzzling.stackexchange.com/users/51/raystafarian&quot;&gt;Raystafarian&lt;/a&gt;&lt;br /&gt;&#xA;18: &lt;a href=&quot;https://puzzling.stackexchange.com/users/108/vivekpoddar&quot;&gt;vivekpoddar&lt;/a&gt;&lt;br /&gt;&#xA;19: &lt;a href=&quot;https://puzzling.stackexchange.com/users/43/glenneroo&quot;&gt;glenneroo&lt;/a&gt;&lt;br /&gt;&#xA;20: &lt;a href=&quot;https://puzzling.stackexchange.com/users/126/james-jenkins&quot;&gt;James Jenkins&lt;/a&gt;&lt;br /&gt;&#xA;21: &lt;a href=&quot;https://puzzling.stackexchange.com/users/83/young-guilo&quot;&gt;Young Guilo&lt;/a&gt;&lt;br /&gt;&#xA;22: &lt;a href=&quot;https://puzzling.stackexchange.com/users/197/caffeinepwrdal&quot;&gt;CaffeinePwrdAl&lt;/a&gt;&lt;br /&gt;&#xA;23: &lt;a href=&quot;https://puzzling.stackexchange.com/users/144/dag&quot;&gt;DaG&lt;/a&gt;&lt;br /&gt;&#xA;24: &lt;a href=&quot;https://puzzling.stackexchange.com/users/130/crownjewel82&quot;&gt;crownjewel82&lt;/a&gt;&lt;br /&gt;&#xA;25: &lt;a href=&quot;https://puzzling.stackexchange.com/users/124/piero-alberto&quot;&gt;Piero Alberto&lt;/a&gt;&lt;br /&gt;&#xA;26: &lt;a href=&quot;https://puzzling.stackexchange.com/users/255/michal&quot;&gt;Michal&lt;/a&gt;&lt;br /&gt;&#xA;27: &lt;a href=&quot;https://puzzling.stackexchange.com/users/66/danubian-sailor&quot;&gt;Danubian Sailor&lt;/a&gt;&lt;br /&gt;&#xA;28: &lt;a href=&quot;https://puzzling.stackexchange.com/users/138/crispy&quot;&gt;Crispy&lt;/a&gt;&lt;br /&gt;&#xA;29: &lt;a href=&quot;https://puzzling.stackexchange.com/users/97/gilles&quot;&gt;Gilles&lt;/a&gt;&lt;br /&gt;&#xA;30: &lt;a href=&quot;https://puzzling.stackexchange.com/users/109/strikers&quot;&gt;strikers&lt;/a&gt;&lt;br /&gt;&#xA;31: &lt;a href=&quot;https://puzzling.stackexchange.com/users/115/igelkott&quot;&gt;igelkott&lt;/a&gt;&lt;br /&gt;&#xA;32: &lt;a href=&quot;https://puzzling.stackexchange.com/users/2484/dalarcop&quot;&gt;d'alar'cop&lt;/a&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;FOUND MESSAGES&lt;/h2&gt;&#xA;&lt;p&gt;Trap Level 1 found by &lt;a href=&quot;https://puzzling.stackexchange.com/a/17585/9000&quot;&gt;Sebu&lt;/a&gt; (+1)&lt;br /&gt;&#xA;Trap Level 2 not yet found&lt;br /&gt;&#xA;Trap Level 3 not yet found&lt;br /&gt;&#xA;Password found by &lt;a href=&quot;https://puzzling.stackexchange.com/a/17598/9000&quot;&gt;Shawn Holzworth&lt;/a&gt; (Tick after another day or two to keep it active so maybe people will try to find the remaining traps)&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-17T20:40:59.027" Title="Doorknob's Revenge" Tags="|password|" AnswerCount="4" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="17551" PostTypeId="2" ParentId="17357" CreationDate="2015-07-07T21:02:39.473" Score="5" Body="&lt;p&gt;&lt;strong&gt;Partial answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some sports I can be used, but only by some. In this case here, it references to one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Code golf, with java?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be used in some work, more specific than others. I'm not talking about animals or water which could be seen as my brothers. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Programming in java. Not Python (animal), C (sea), or Ocean (another programming language)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am nearby the shops for others to get. Some might ignore me, some might not forget.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Coffee shops outside the shopping mall. Some people don't drink coffee and will ignore, while others will die without.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might be stronger than others, or weaker than few. Some might have just one, or maybe even two.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Java is known as a strongly typed language. It's more strongly typed than C and possibly less strongly typed than some other languages.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Some people may use one programming language, or more. Sometimes there are several languages in a project.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Title:&lt;/strong&gt;&#xA;When you are sick you might have me, with a hand full of letters&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Drinking coffee in a mug with letters on it while you're sick? (You probably shouldn't do that. The caffeine will make you dehydrated, and it doesn't go well with medicine. Some people say it make a sore throat feel better, though.)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Or you are drinking coffee while reading your get well letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12109" LastEditorUserId="12109" LastEditDate="2015-07-07T21:46:08.327" LastActivityDate="2015-07-07T21:46:08.327" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17552" PostTypeId="1" AcceptedAnswerId="17553" CreationDate="2015-07-07T21:20:03.230" Score="4" ViewCount="1726" Body="&lt;p&gt;I have been captured, then shot, framed and hung.  But luckily,  I am not dead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is it possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Disclaimer: this puzzle is inspired by real world observation, so no alien lifeforms, super heroes or undead involved.  There is a perfectly normal logical explanation to it.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And please, no computers.&lt;/p&gt;&#xA;" OwnerUserId="12332" LastEditorUserId="5373" LastEditDate="2015-07-07T21:40:55.443" LastActivityDate="2015-07-08T00:42:55.917" Title="How is it possible?" Tags="|riddle|situation|no-computers|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="17553" PostTypeId="2" ParentId="17552" CreationDate="2015-07-07T21:22:19.220" Score="16" Body="&lt;p&gt;You are a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; picture&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are captured&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Captured on a camera&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;then shot,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Take a picture&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;framed and hung&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Frame a picture and hang it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am not dead.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are an object. You are whole and not broken, so you are not &quot;dead&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12109" LastEditorUserId="12109" LastEditDate="2015-07-07T21:50:44.397" LastActivityDate="2015-07-07T21:50:44.397" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17554" PostTypeId="2" ParentId="17552" CreationDate="2015-07-07T21:41:44.173" Score="10" Body="&lt;p&gt;I saw mmking's answer and comes up with an alternative:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are a... moment?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are captured&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The moment is captured  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;then shot,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; via camera&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;framed and hung&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As a picture  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am not dead&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The moment is 'alive' as a picture now&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2015-07-07T21:41:44.173" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17555" PostTypeId="2" ParentId="17096" CreationDate="2015-07-07T22:57:00.440" Score="4" Body="&lt;p&gt;Determining two heavy coins out of 27 can be done in 5 weighings.  It's been several hours, so I'll skip the spoiler tags (also, i don't know how to spoiler large sections of text).  I don't have a generalization, yet, but if i come up with one, I'll edit it in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Label the coins 1-27.  Take three weighings:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First: $1$-$9$ &lt;em&gt;vs&lt;/em&gt; $10$-$18$ &lt;em&gt;vs&lt;/em&gt; $19$-$27$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second:&#xA;$1,2,3,10,11,12,19,20,21$ &lt;em&gt;vs&lt;/em&gt; $4,5,6,13,14,15,22,23,24$ &lt;em&gt;vs&lt;/em&gt; $7,8,9,16,17,18,25,26,27$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third: $1,4,7,10,13,16,19,22,25$ &lt;em&gt;vs&lt;/em&gt; $2,5,8,11,14,17,20,23,26$ &lt;em&gt;vs&lt;/em&gt; $3,6,9,12,15,18,21,24,27$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the two heavy coins are on the same pan, then the pans will balance.  Otherwise the one pan with no heavy coins will rise.  Since there is no combination of two coins that are on the same pan for all three weighings, for at least one of the three weighings, one of the pans must rise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Name a weighing where one pan rose as weighing &quot;alpha&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Alpha&lt;/strong&gt;: Considering weighing alpha, relabel the coins on one of the heavy pans as A1 - A9, the coins on the other heavy pan as B1 - B9, and the coins on the light pan as C1 - C9. (Preserve order on the numberings).  We know that one of the heavy coins is an &quot;A&quot; coin and one is a &quot;B&quot; coin and all the &quot;C&quot; coins are normal.  This leaves us with 81 possibilities, picking one each from two sets of 9 elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the weighings performed had a pan with A1,A2,A3,B1,B2,B3,C1,C2,C3.  Name that weighing &quot;beta&quot;.  The other weighing had a pan with A1,A4,A7,B1,B4,B7,C1,C4,C7.  Label that weighing &quot;gamma&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Beta: No pan rises&lt;/strong&gt;:  If no pan rose for weighing beta, then both heavy coins were on the same pan for that weighing.  The combinations that satisfy this are: &#xA;$$\{A1,A2,A3\}\times\{B1,B2,B3\}\text{, or}$$&#xA;$$\{A4,A5,A6\}\times\{B4,B5,B6\}\text{, or}$$&#xA;$$\{A7,A8,A9\}\times\{B7,B8,B9\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a total of 27 possibilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If no pan rose for weighing gamma, then the possibilities are Ax,Bx, where x is the same for both, for 9 options remaining.  If the third pan rose, then there are 6 options: A1B2, A2B1, A4B5, A5B4, A7B8, or A8B7.  In any case, there are 6 or 9 options, which do not overlap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Split 3 pairs onto the first two pans and split 3 pairs onto the first and third pans.  If you have 9 pairs, split the last 3 pairs onto the second and third pan.  Otherwise, add 3 other coins to each of the second and third pans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During this fourth weighing, you have definitely split the two heavy coins, so a pan will rise. That will leave you with 3 pairs for possibilities.  Split one pair on first and second pans, second pair on first and third pans, third pair on second and third pans.  The fifth weighing will tell you your answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;For example&lt;/em&gt;, weighing gamma was flat, then weigh:&#xA;A1,A2,A3,A4,A5,A6&#xA;&lt;em&gt;vs&lt;/em&gt; B1,B2,B3,A7,A8,A9&#xA;&lt;em&gt;vs&lt;/em&gt; B4,B5,B6,B7,B8,B9.  Each of the 9 pairs are split, 3 for each combination of two pans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the second pan rose, the remaining options are A4B4 or A5B5 or A6B6.  On the fifth weighing, compare A4,A5 &lt;em&gt;vs&lt;/em&gt; B4,A6 &lt;em&gt;vs&lt;/em&gt; B5,B6.  If the first pan rises now, the heavy coins are A6 and B6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Beta: One pan rose&lt;/strong&gt;.  If one pan rose in weighing beta, there are 18 remaining possibilities.  Consider the case where the third pan rose (the others are symmetrically solved).  The remaining options are:&#xA;$$\{A1,A2,A3\}\times\{B4,B5,B6\}\text{, or}$$&#xA;$$\{A4,A5,A6\}\times\{B1,B2,B3\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If no pan rose for weighing gamma, there are 6 options remaining: A1B4 or A2B5 or A3B6 or A4B1 or A5B2 or A6B3.  If the third pan rose, there are 4 options remaining: A1B2 or A2B1 or A4B5 or A5B4.  Similarly, there are 4 options if either of the other two pans rose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, there are either 4 or 6 options, with no overlap.  Proceed as above, splitting the pairs evenly.  If there are only 4 options, you can leave a pair out and complete in 4 weighings.  For example, if the third pan rose on weighing gamma, then for the fourth weighing compare:&#xA;A1, A2 &lt;em&gt;vs&lt;/em&gt; B2,A4 &lt;em&gt;vs&lt;/em&gt; B1,B5.  If no pan rises, the heavy coins are A5B4.  Otherwise, the fourth weighing would indicate which coins the heavy ones are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in summary, it can be done in 5 weighings, and the arrangements of coins on the first three weighings can be determined ahead of time.  The relabeling steps I took simplify the number of cases and minimize the &quot;assume without loss of generality&quot; steps.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2015-07-07T22:57:00.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17556" PostTypeId="1" AcceptedAnswerId="17558" CreationDate="2015-07-08T00:00:28.053" Score="8" ViewCount="1254" Body="&lt;p&gt;Which is larger, $99^{100}$ or $100^{99}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is larger, $999^{1000}$ or $1000^{999}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or more generally, for $n$ any natural number: which is larger, $n^{n+1}$ or $(n+1)^n$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Please do not use a computer, calculator, or any electronic aid to work out the answer for any specific cases of $n$.&lt;/strong&gt; You should be able to solve this puzzle purely by mathematical reasoning and without calculations, using a computer only to read the question and type the answer :-)&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-07-08T09:13:22.490" Title="Comparing exponents - no computers!" Tags="|mathematics|number-theory|no-computers|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17557" PostTypeId="2" ParentId="17357" CreationDate="2015-07-08T00:06:13.557" Score="2" Body="&lt;p&gt;I have one more crazy idea that seems to have more support from the author's comments and subtle clues than from the riddle itself. There are a few answers that feel like they're really close but the OP seems to be wanting us to dig deeper so this is an attempt at that. It looks at the riddle from another direction so maybe I'll knock something loose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PARTIAL ANSWER&lt;/strong&gt;&#xA;I think the answer could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; STACK. There have been numerous references to &quot;depth&quot; and &quot;unwinding&quot; and when Googling these words, callstack is among the most common results. Since a lot of answers have been related to programming and the suggested languages all use a callstack, this is a logical extension of that thought process. In order to arrive at STACK from the clues, I can only assume that, including the title, each clue represents a single letter of the answer. To make it easier, I'll be assuming that the answer to each clue is a word that starts with the appropriate letter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you are sick you might have me, with a hand full of letters&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;S -&gt; Soup is a common home remedy for many ailments and is often be taken with vitamin supplements. Vitamins are named after letters so if you're taking an assortment of pills, you would have a hand full of letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In some sports I can be used, but only by some. In this case here, it references to one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;T -&gt; A Tee is used in golf as well as in teeball and resembles some representations of the number 1. The use of the word 'references' seems deliberate, perhaps to further emphasize the relationship to coding.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be used in some work, more specific than others. I'm not talking about animals or water which could be seen as my brothers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A -&gt; Not entirely sure. I was investigating possibilities relating to Specific Work equations but couldn't find anything that seemed to fit with the animals and water clues.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am nearby the shops for others to get. Some might ignore me, some might not forget.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;C -&gt; I'm going to go with Coffee. There's always some around shops and some people need it more than others. Plus, Coffee is related to Java, one of the programming languages that uses a call stack.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might be stronger than others, or weaker than few. Some might have just one, or maybe even two.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;K -&gt; Drawing a blank here too. There are probably lots of words that could fit so I'm just gonna say Kidneys.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Other notes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As with one of my other answers, this seems to fit with the themes of some of the OP's other riddles, notably ones related to StackExchange.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13541" LastActivityDate="2015-07-08T00:06:13.557" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17558" PostTypeId="2" ParentId="17556" CreationDate="2015-07-08T00:13:40.463" Score="16" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $n^{(n+1)}$ is larger iff $n\geq 3$&lt;br&gt;&lt;br&gt;$$n^{(n+1)} &amp;gt; (n+1)^n \\n &amp;gt; \left(\frac{n+1}{n}\right)^n \\n &amp;gt; \left(1+\frac 1 n\right)^n$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The right side increases with $n$, and we know that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$$\lim_{n\to\infty}\left(1+\frac 1 n\right)^n=e$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, for $n\geq 3$,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$$n &amp;gt; e &amp;gt; \left(1+\frac 1 n\right)^n$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now we only have to check $n=2$ and $n=1$ by hand.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$2^3 &amp;lt; 3^2\\1^2 &amp;lt; 2^1$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="7308" LastEditDate="2015-07-08T03:42:53.357" LastActivityDate="2015-07-08T03:42:53.357" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17559" PostTypeId="2" ParentId="17531" CreationDate="2015-07-08T00:24:20.143" Score="2" Body="&lt;h2&gt;84 letters&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A saturate sensationally satirical storyline saturates satirically sensational storylines, ass!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(telling someone rather rudely that a storyline that's both saturated and sensationally satirical saturates those that are just satirically sensational).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar to Victor Henry's example ... now the longest yet!&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-07-08T09:37:44.447" LastActivityDate="2015-07-08T09:37:44.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17560" PostTypeId="2" ParentId="17556" CreationDate="2015-07-08T00:30:31.540" Score="0" Body="&lt;p&gt;For sufficiently large $n$, we can check via division and limits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{n^{(n+1)}}{(n+1)^n} \\&#xA;=\frac{n^{(n+1)}}{(n^n+j*n^{n-1}+k*n^{n-2}+...)} \\&#xA;=\frac{n^{(n+1)}}{(n^n+j*n^{n-1}+k*n^{n-2}+...)}*\frac{1/n^n}{1/n^n} \\&#xA;=\frac{n}{1+j/n+k/n^2+...} \\&#xA;\lim_{n\to\infty}\frac{n}{1+j/n+k/n^2+...}=\infty&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can conclude that for sufficiently large $n$,&#xA;$$n^{(n+1)} &amp;gt; (n+1)^n \\$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2012rcampion did the work to show the threshold, where $n$ must be greater than 2.&lt;/p&gt;&#xA;" OwnerUserId="9696" LastEditorUserId="9696" LastEditDate="2015-07-08T00:46:59.623" LastActivityDate="2015-07-08T00:46:59.623" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17561" PostTypeId="2" ParentId="17552" CreationDate="2015-07-08T00:42:55.917" Score="10" Body="&lt;p&gt;Another alternative:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you are an actor.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Captured&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; by the allure of the film industry,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;shot&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; in an action film,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;framed&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; in a detective film,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and hung&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; in a western film.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You're not dead:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you're rolling in money from all these films you've appeared in!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-07-08T00:42:55.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17562" PostTypeId="2" ParentId="17556" CreationDate="2015-07-08T00:55:27.353" Score="5" Body="&lt;p&gt;Asking which of $x^y$ or $y^x$ is bigger is the same as asking which of their logs is bigger, so what comparison sign should fill the blank in&#xA;$$&#xA;y\log x\,\,\,\_\,\,\,x\log y?&#xA;$$&#xA;Rearranging, this becomes&#xA;$$&#xA;\frac{\log x}{x}\,\,\,\_\,\,\,\frac{\log y}y&#xA;$$&#xA;So, letting $f(t)=\frac{\log t}t$, is $f(x)$ or $f(y)$ bigger? &lt;a href=&quot;http://www.wolframalpha.com/input/?i=log%28t%29%2Ft&quot;&gt;Looking at the graph of f(t)&lt;/a&gt;, we see that $f$ is increases up to a point, then decreases thereafter. To find this point, we compute the derivative $f'(t)=\frac{1-\log t}{t^2}$, then set this to zero, so the maximum of $f$ occurs at $t=e$. In summary: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$x^y-y^x$ has the same sign as $f(x)-f(y)$, which is negative when $x&amp;lt;y&amp;lt;e$, and positve when $e&amp;lt;x&amp;lt;y$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In our case, we see that $n^{n+1}-(n+1)^n$ will be positive when $n\ge3$, so for almost all $n$, $n^{n+1}&amp;gt;(n+1)^n$.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-07-08T00:55:27.353" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17563" PostTypeId="1" AcceptedAnswerId="17568" CreationDate="2015-07-08T01:46:43.017" Score="13" ViewCount="1035" Body="&lt;p&gt;Alice and Bob play a game with two stacks of fluffy, golden pancakes, of sizes $p$ and $q$. They alternate turns, with Alice starting. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On their turn, a player must eat some number of pancakes from the larger stack. The number they eat must be a multiple of the size of the smaller stack. Since the bottom pancake of each stack is soggy, the first player to finish a stack loses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if the stacks were sized $5$ and $17$, then the current player can eat either $5,10$ or $15$ pancakes from the large stack. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For which values of $p$ and $q$ does Alice have a winning strategy?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;em&gt;Mathematical Puzzles: A Connoisseur's Collection&lt;/em&gt;, Peter Winkler.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-07-08T02:43:17.020" LastActivityDate="2015-07-08T14:00:53.483" Title="Golden Pancakes" Tags="|mathematics|game|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17565" PostTypeId="1" AcceptedAnswerId="17664" CreationDate="2015-07-08T03:11:17.427" Score="22" ViewCount="1256" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;My first is in a bank,&lt;br&gt;&#xA;  My second is an evergreen,&lt;br&gt;&#xA;  My third is for your benefit&lt;br&gt;&#xA;  My fourth is in a store,&lt;br&gt;&#xA;  My fifth is identical,&lt;br&gt;&#xA;  If you've found the pattern&lt;br&gt;&#xA;  What is my sixth?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Those definitions might be a bit broad, but there is only one answer that follows the pattern.&lt;br&gt;&#xA;If needed, I will add more clues.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This has nothing to do with computer. Each line refers to one word. Words may or may not be in the same semantic field but that's not what is important.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As you, fellow puzzlers, keep to try to make sense of this riddle as a whole, I need to tell you that each definition gives a word but those words are not in the same semantic field nor express the same idea.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I hope this one won't be an easy giveaway&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My first is cautious,&lt;br&gt;&#xA; My second is wise,&lt;br&gt;&#xA; My third is wine,&lt;br&gt;&#xA; My fourth happens in the garage,&lt;br&gt;&#xA; My fifth is unchanged&lt;br&gt;&#xA; Now you should have a better grasp on the pattern&lt;br&gt;&#xA; So what is my sixth?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10400" LastEditorUserId="463" LastEditDate="2015-07-11T03:46:39.090" LastActivityDate="2016-09-14T08:43:11.480" Title="Nine is the limit" Tags="|riddle|word|" AnswerCount="9" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17566" PostTypeId="2" ParentId="17531" CreationDate="2015-07-08T03:12:28.437" Score="7" Body="&lt;h1&gt;247 letters&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Poorly named IT company launches their new product &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Stand - for all your system sensationalising and standardising needs. A systemic and systematic approach to system standardisation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sensational Lice Mice Stand sensationally systematically re-systems,&#xA; sensationalised standardised standards, standardising sensationalising&#xA; standard systematisation system systematisations, un-systematically&#xA; re-standardised standardisation systems, unstandardised systematic and&#xA; systemic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sensational Mice with the Lice have a new product the &quot;Stand&quot; that sensationally systematically does re-systemming of sensationalised standardised standards. As well as standardising the sensationalising of standard of the systematisations of systematisation systems, as well as un-systematically re-standardised standardisation systems, and the unstandardised systematic and systemic&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Much longer sentences can be made using lists of things, it is&#xA;  probably possible the make a sentence using every word as a long&#xA;  enough list will eventually contain every pair of characters. Even&#xA;  those pairs that don't appear in words or appear in only single words&#xA;  can be formed from the letters of adjacent words.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="11297" LastEditDate="2015-07-08T16:44:21.860" LastActivityDate="2015-07-08T16:44:21.860" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17567" PostTypeId="2" ParentId="17563" CreationDate="2015-07-08T03:14:22.337" Score="3" Body="&lt;p&gt;This is &lt;a href=&quot;https://en.wikipedia.org/wiki/Nim&quot; rel=&quot;nofollow&quot;&gt;Nim&lt;/a&gt; in disguise because it is an impartial game. $(1,1)$ is a P position, so $(1,n)$ is N for $n \ge 2$.  $(k,k)$ is P.  $(2,3)$ is P.  I suspect a relation with the Fibonacci series-find it.  That is a wonderful book.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditDate="2015-07-08T03:28:00.663" LastActivityDate="2015-07-08T03:28:00.663" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17568" PostTypeId="2" ParentId="17563" CreationDate="2015-07-08T03:55:03.323" Score="13" Body="&lt;p&gt;What a beautiful problem! The winning positions (for Bob, the player who just moved) are those where the stack sizes are within a factor of the golden ratio $\phi$, i.e. the $(p,q)$ with $$1/\phi &amp;lt; q/p &amp;lt; \phi.$$ So, the winning strategy is to make the stacks as even as possible to keep this invariant true after your move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say that $p\leq q$. The legal moves result in positions of the form $(p,q')$. This is in the band of winning positions whenever $q' \in (p/\phi,p \phi)$ an interval whose length is exactly $p$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the possible $q'$ are spaced $p$ apart, exactly one such $q'$ lies in the band. Then, $(p,q')$ is a winning move unless it is $(p,q)$ itself in the band, since we must have $q'&amp;lt;q$. We've shown that there's a move to a winning position exactly when we're not a winning position, which inductively proves that these are the winning positions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this solution by writing code to find the winning positions, and noticing the band pattern and suspecting the golden ratio.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-07-08T04:02:30.517" LastActivityDate="2015-07-08T04:02:30.517" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17569" PostTypeId="2" ParentId="17563" CreationDate="2015-07-08T03:55:42.787" Score="1" Body="&lt;p&gt;Without loss of generality, assume the larger stack is of size $q$.  Here's a non-constructive partial result:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If $q &amp;gt; 2p$, Alice wins. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll refer to the player whose turn it is as the &quot;current player&quot;.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $q=np+r$, where $r$ is the remainder of $\frac{q}{p}$.  Since $q&amp;gt;2p$, $n\ge2$.  Eventually the game will reach the state $(r,p)$: players will take multiples of $p$ from the stack that is currently of size $q$ until that stack is smaller than $p$.  At that point, the size of that stack will be $r$, and the other stack will be $p$, with $r&amp;lt;p$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since a tie is not possible, either the current player in the state $(r,p)$ can force a win, or the other player can.  Alice can control whether she is the current player in the state $(r,p)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the current player wins at $(r,p)$, Alice should take $(n-1)p$ pancakes from the $q$-stack, making the state $(p,p+r)$.  She can always do this, since $n\ge2$.  Bob's only move in that state is to take $p$ pancakes from the $p+r$ stack, making the state $(r,p)$ and Alice the current player.  Thus, Alice wins.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the current player loses at $(r,p)$, Alice should take $np$ pancakes from the $q$-stack, making the state $(r,p)$ with Bob the current player.  Thus, Bob loses (and Alice wins).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this tells us that Alice &lt;em&gt;can&lt;/em&gt; win in the case that $q &amp;gt; 2p$, but it doesn't tell us &lt;em&gt;how&lt;/em&gt;.  We know there is a move that she can use to put herself in a winning position, but if we were Alice, we wouldn't know from this analysis what that move is.  &lt;/p&gt;&#xA;" OwnerUserId="10261" LastEditorUserId="10261" LastEditDate="2015-07-08T06:00:43.813" LastActivityDate="2015-07-08T06:00:43.813" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17570" PostTypeId="2" ParentId="17565" CreationDate="2015-07-08T04:25:37.273" Score="2" Body="&lt;p&gt;Just a guess&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Is it the symbols on the keyboard along the numbers $ ^% ^ &amp;amp; * ( &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My first is in a bank,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $- Dollar sign/Money  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My second is an evergreen,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; % -Percentages have been used always as an ideal measure of performance  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My third is for your benefit &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ^ - Up arrows are often used for marking text and for easier readability &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fourth is in a store,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &amp;amp; -Don't know how this connects. Maybe refers to shops often being named &quot;Mom &amp;amp; Pops&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fifth is identical,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; * - Maybe a reference to it being a symmetric symbol. Or could be because the * symbol is used to match repeating or identical patterns (thanks Samthere) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you've found the pattern&#xA;What is my sixth?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Is it the symbol (&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13080" LastEditorUserId="13080" LastEditDate="2015-07-08T11:59:38.697" LastActivityDate="2015-07-08T11:59:38.697" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17571" PostTypeId="2" ParentId="17556" CreationDate="2015-07-08T04:32:01.847" Score="2" Body="&lt;p&gt;The question said no calculations as well as no calculators, so here is my answer with the fewest calculations I could manage.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ask another question: Which pair is going to be closer,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$99^{100}$ and $100^{99}$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;or,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$999,999^{1,000,000}$ and $1,000,000^{999,999}$?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;99 is almost 100 only a bit smaller, so doing another multiplication is making a number that is &#xA;  a bit smaller an other 100 times larger.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;999,999 is much closer to a million that 99 is to 100 so the effect of&#xA;  raising an extra power will be much closer to multiplying by an extra&#xA;  million adding 6 zeros.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Clearly $n$ being larger favours $n^{n+1}$ to be the greater of the&#xA;  two.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;So how small does $n$ have to be for $n^{n+1}$ to be smaller.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$1^{2} - 2^{1} = 1 - 2 = -1$&lt;br&gt; $2^{3} - 3^{2} = 8 - 9 = -1$&lt;br&gt;&#xA;  $3^{4} - 4^{3} = 81 - 64 = 17$&lt;br&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Looks like $n^{n+1}$ is biggest when...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;em&gt;n &gt; 2&lt;/em&gt;&lt;/strong&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Negative values of $n$ exhibit the switching of sign expected for&#xA;  multiplying negative numbers, so there is a different rule, but that&#xA;  is left as an exercise, as the question only asked for natural numbers (positive whole numbers).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="11297" LastEditDate="2015-07-08T09:13:22.490" LastActivityDate="2015-07-08T09:13:22.490" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17573" PostTypeId="1" AcceptedAnswerId="17575" CreationDate="2015-07-08T07:44:35.357" Score="7" ViewCount="843" Body="&lt;p&gt;MC Frontalot is a musician, and one of his songs is a word problem that seems to me like it belongs here. If you want to listen to it instead of just reading the lyrics, a quick video search of &quot;MC Frontalot 80085&quot; will bring it up on YouTube. Here is the full song, copied (with permission) directly from &lt;a href=&quot;http://frontalot.com/index.php/?page=lyrics&amp;amp;lyricid=73&quot;&gt;his website&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h2&gt;80085&lt;/h2&gt;&#xA;  &#xA;  &lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Last time I had a math class, there wasn't any internet&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;invented yet. That isn't on the level but I'll try to pique your interest&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;with half-truths and lies.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;As ever, MC Frontalot feigns innocence and denies.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I won't admit it. You can't make me say it:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;that I dropped Calc B more than a year before Mosaic.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Oh no now it's out, now it's shouted from the balconies:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;that Frontalot's about to be engaging in some alchemy.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I'll turn a string of operands into some smut.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If that sort of thing's offensive to you keep eyes shut.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Or better yet, don't even enter, into calculator, song.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;But if you're ready to be titillated, follow along.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Ready? Go. Eighty women went to the podiatrist.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Arrive: simultaneous. Soon the scene's riotous.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Nine just leave. Those in the difference persevere,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;packing up the lobby very tightly, domineered&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;by one Sally Gorey (that's her given name)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(though her title is Reception) (and professional acclaim&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;is due her) ('cause she did what needed doing). And it's done:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;she opened up the schedule, slotted every single one.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;But, um... not many on a Friday afternoon!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;All but an eighteenth of the women in the room&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;had to vrooooom. For each remaining patient&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;x-rays were taken. Then the doctor took vacation.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Why was that vacation germane to the math?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;'Cause of good data policy in the office and a vast&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;abundance of caution on the part of our Sally:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;eight backups nightly, automated, and the tally&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;only ever shrinking when manually deleted.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;All of this occurring in the box behind reception so she needed&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;a full backup of that box, noons.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;These weren't incremental, so her server needs ballooned.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Who deserved to flee Duluth? The doctor was in Rio&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;for three work weeks and another Monday just to be so&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;thoroughly relaxed upon return.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Have you gathered all the facts that you needed to discern?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Morning in the office, after vacatings:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;out of those belonging to the original 80 ladies.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;How many digital toes were in images grand total?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Your evidence so far is largely anecdotal.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;And you're keen to know if any had deformity. So icky!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Ten toes per customer; this puzzle: not that tricky.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Key in your calc. Check your seven-segment indicator.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now add my eliteness. Notice that the sum is greater&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;than expected. You still have to subtract&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;two for a pair of things Sally has that I lack.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I warned you it was kind of immature; I wasn't skirting the issue.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Still you snicker at the calculator. &quot;Dirty! I need a tissue.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I hope this is assumed, but I will state anyway that how you reached your answer is as important as the answer itself, and just a number (even if it's the right number) is not an acceptable answer.&lt;/p&gt;&#xA;" OwnerUserId="13722" LastEditorUserId="13722" LastEditDate="2015-07-08T18:53:07.137" LastActivityDate="2016-04-01T07:01:22.320" Title="MC Frontalot 80085" Tags="|calculation-puzzle|word-problem|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17574" PostTypeId="2" ParentId="17423" CreationDate="2015-07-08T07:51:48.640" Score="2" Body="&lt;p&gt;I am not particularly proud that I had to use the second hint, but using that I deciphered this.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The paper uses a skip cipher and the skip value is 54. Using this skip value and decoding the paper gives us&lt;br&gt;&lt;br&gt;&lt;strong&gt;Skip has a prime factor that cleanly divides into the text length, so it can not be used.&lt;/strong&gt;&lt;br&gt;&lt;br&gt;54 has two factors, 2 and 3 (54 = 2&lt;sup&gt;1&lt;/sup&gt;3&lt;sup&gt;3&lt;/sup&gt;) and the text length is 75. So we cannot use 3. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I think, to unlock the door, we need to key in the word&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TWO&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-07-08T09:23:48.380" LastActivityDate="2015-07-08T09:23:48.380" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17575" PostTypeId="2" ParentId="17573" CreationDate="2015-07-08T08:21:11.647" Score="7" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; already in the title: 80085.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here are the steps to getting there:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 80 women went to the podiatrist, of which 9 just leave. There are 71 left, plus Sally, so 72. All but an eighteenth had to leave, so 4 stayed (Sally and 3 patients). 3 patients times 2 feet, so we start with 6 pictures. Then, during each of the following 25 nights (Friday to Tuesday, 3 weeks), the number of images at the box behind the reception increases by 48 (8*6) and the complete reception box gets a backup in the 'backup box' at the noon afterwards. Now, the nightly increase happens 25 times, and the daily backup happens 24 times. That results in a total of 15750 images. The amount of toes is 15750*5, so 78750. Now add my eliteness, so 1337. This results in 80087. We have to subtract 2, so we end up with 80085.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And now in plain calculations:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; women = 80-9 = 71&lt;br&gt;&#xA; women = 71+1 = 72&lt;br&gt;&#xA; women = 72/18 = 4&lt;br&gt;&#xA; patients = 4-1 = 3&lt;br&gt;&#xA; pictures = 3*2 = 6&lt;br&gt;&#xA; reception(1) = 54, backup(1) = 54&lt;br&gt;&#xA; reception(2) = 102, backup(2) = 156&lt;br&gt;&#xA; reception(3) = 150, backup(3) = 306&lt;br&gt;&#xA; reception(4) = 198, backup(4) = 504&lt;br&gt;&#xA; reception(5) = 246, backup(5) = 750&lt;br&gt;&#xA; reception(6) = 294, backup(6) = 1044&lt;br&gt;&#xA; reception(7) = 342, backup(7) = 1386&lt;br&gt;&#xA; reception(8) = 390, backup(8) = 1776&lt;br&gt;&#xA; reception(9) = 438, backup(9) = 2214&lt;br&gt;&#xA; reception(10) = 486, backup(10) = 2700&lt;br&gt;&#xA; reception(11) = 534, backup(11) = 3234&lt;br&gt;&#xA; reception(12) = 582, backup(12) = 3816&lt;br&gt;&#xA; reception(13) = 630, backup(13) = 4446&lt;br&gt;&#xA; reception(14) = 678, backup(14) = 5124&lt;br&gt;&#xA; reception(15) = 726, backup(15) = 5850&lt;br&gt;&#xA; reception(16) = 774, backup(16) = 6624&lt;br&gt;&#xA; reception(17) = 822, backup(17) = 7446&lt;br&gt;&#xA; reception(18) = 870, backup(18) = 8316&lt;br&gt;&#xA; reception(19) = 918, backup(19) = 9234&lt;br&gt;&#xA; reception(20) = 966, backup(20) = 10200&lt;br&gt;&#xA; reception(21) = 1014, backup(21) = 11214&lt;br&gt;&#xA; reception(22) = 1062, backup(22) = 12276&lt;br&gt;&#xA; reception(23) = 1110, backup(23) = 13386&lt;br&gt;&#xA; reception(24) = 1158, backup(24) = 14544&lt;br&gt;&#xA; reception(25) = 1206&lt;br&gt;&#xA; totalimages = 1206+14544 = 15750&lt;br&gt;&#xA; toes = 15750*5 = 78750&lt;br&gt;&#xA; elitetoes = 78750+1337 = 80087&lt;br&gt;&#xA; answer = 80087-2 = 80085&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13268" LastEditorUserId="12408" LastEditDate="2016-04-01T07:01:22.320" LastActivityDate="2016-04-01T07:01:22.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17577" PostTypeId="2" ParentId="17247" CreationDate="2015-07-08T09:01:03.343" Score="14" Body="&lt;p&gt;&lt;em&gt;MJ, zDafl.o!sassor ok Vqi h,bg e yx hos i edi i ot.cTuP swnWpHh hf&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Mr. Jock, TV quiz PhD, bags few lynx. Whoops! His head is his foot&lt;/strong&gt; using a railfence cipher with 3 rails (see rand al'thor &lt;a href=&quot;https://puzzling.stackexchange.com/questions/17247/treasure-hunt-round-the-world-interlude-1/17590#17590&quot;&gt;answer&lt;/a&gt; for a better explanation).&lt;br&gt;&#xA;&lt;strong&gt;Mr. Jock, TV quiz PhD, bags few lynx&lt;/strong&gt; is a known &lt;strong&gt;pangram&lt;/strong&gt; so this suggest a substitution cipher but with which key?&lt;br&gt;&#xA;&lt;strong&gt;Whoops! His head is his foot&lt;/strong&gt; suggests that the head be equal to the foot. It means by looking at the pangram found, &lt;strong&gt;M = X&lt;/strong&gt;, &lt;strong&gt;R = N&lt;/strong&gt; and so on... to give the following complete substitution key:&lt;br&gt;&lt;br&gt;&#xA; &lt;strong&gt;ABCDEFGHIJKLMNOPQRSTUVWXYZ&lt;br&gt;&#xA; UIWZKTQPBYEOXRLHGNVFASCMJD&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;cloki rlbfvkag kpf zrbt hokp vknkrkqbs pfbc&lt;br&gt;&#xA;clre jzuknou alj pwbpc nkcvru kpf vb jke kpf&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;kbo rbpfbc vfknwkv zknkffupv voubfrkvvk lf rclz zknuh rkpc&lt;br&gt;&#xA;kjk kpf pfbc rupf nuk pfbc jofrknkttbz zksbkwnkh jfnuh u&lt;br&gt;&#xA;zolp oobc jkpf rlbfuxnltrb vpfnalt nbkpf ji rkeuf rkpc&lt;br&gt;&#xA;zolq vhupnkh zkokkfv ki lf kzkhxb lf rkpf rlc ki lf&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We need to use the substitution found above and reverse the whole text to obtain the following:&lt;br&gt;&lt;br&gt;&#xA; &lt;em&gt;to be won then to impede to be steeled perhaps gold&lt;br&gt;&#xA; when taken by their fourths information they will hold&lt;br&gt;&#xA; a party perceived differently with ear than with the eye&lt;br&gt;&#xA; when pared down to essentials shattered secrets within lie&lt;br&gt;&lt;br&gt;&#xA; the key is the answer which you already know&lt;br&gt;&#xA; with vigeneres help find the question below&lt;/em&gt;&lt;br&gt;&lt;br&gt;&#xA; This suggests that we use next a vigenere cipher&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Toh ws L xivx qb yiwdwkmkpl?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;How do I like my milkshakes ?&lt;/strong&gt; with vigenere key &lt;strong&gt;malted&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The solution to the riddle should give us the key found but I could't go line by line to prove how to obtain that key except for this line:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;the key is the answer which you already know&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;How do I like my milkshakes ?&lt;/em&gt; I like it &lt;em&gt;malted&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12731" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2015-07-12T08:13:48.493" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="17578" PostTypeId="2" ParentId="17563" CreationDate="2015-07-08T09:50:32.337" Score="1" Body="&lt;p&gt;Provided they have unlimited appetites, Alice will win when&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $p=nq~or~q=np~for~(n\in N \&amp;amp; n&amp;gt;1)  $&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because, when Alice starts she'll eat&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\Big((n-1)\times~smaller~stack\Big)$, then both stacks will be same, and Bob will have to eat a stack(full).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alice will win.&lt;/p&gt;&#xA;" OwnerUserId="11306" LastEditorUserId="463" LastEditDate="2015-07-08T14:00:53.483" LastActivityDate="2015-07-08T14:00:53.483" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17579" PostTypeId="1" AcceptedAnswerId="17581" CreationDate="2015-07-08T09:51:39.097" Score="24" ViewCount="3180" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;The First provides me with gadgets.&lt;br&gt;&#xA;  The Second is seven doubled.&lt;br&gt;&#xA;  The Third is at the begin of endings.&lt;br&gt;&#xA;  The Fourth is difficult to differentiate in certain languages.&lt;br&gt;&#xA;  The Fifth is in the start of every sentence.&lt;br&gt;&#xA;  The Sixth is a question for lazy people.&lt;br&gt;&#xA;  The Seventh is not I but.&lt;br&gt;&#xA;  The Eighth is me.&lt;br&gt;&#xA;  The Ninth is what you will say when you figure it out.&lt;br&gt;&#xA;  The Tenth can also be before the first and sticks out its tongue.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My friend can't dance but he can do a 084973553.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What can my friend do?&lt;/p&gt;&#xA;" OwnerUserId="12218" LastEditorUserId="4421" LastEditDate="2015-07-08T15:27:21.540" LastActivityDate="2015-07-08T15:28:33.583" Title="Riddle to code, What am I saying?" Tags="|riddle|cipher|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17580" PostTypeId="2" ParentId="17211" CreationDate="2015-07-08T10:08:15.357" Score="10" Body="&lt;p&gt;Puzzle 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Spice&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The puzzle mentions &quot;sugar&quot; then &quot;everything nice&quot;, spice goes between them, or in front of &quot;of life&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Puzzle 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first stanza is a reference to Superman (&quot;It's a bird! It's a plane! No, it's Superman!&quot;). &quot;Shattered sky&quot; could be a reference to flying with a sonic boom. Scramble the letters to make... manpurse?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Hint 2, we know that there is a similarity between a book and a real clean freak; a crook and a rude, crude man; and a surly cook and a piece of pie. The last two pairs are almost too similar; I think the first one is key.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A different guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Pepper&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the Peter Piper nursery rhyme hinted at in the second stanza, if pickling counts as garnish. Also, a cook might used chopped (&quot;segmented&quot;) peppers. Not sure how the rest would fit.&lt;/p&gt;&#xA;" OwnerUserId="13162" LastEditorUserId="13162" LastEditDate="2015-07-09T10:42:38.630" LastActivityDate="2015-07-09T10:42:38.630" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="17581" PostTypeId="2" ParentId="17579" CreationDate="2015-07-08T10:13:39.500" Score="19" Body="&lt;p&gt;&lt;em&gt;The First provides me with gadgets.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1=&lt;strong&gt;Q&lt;/strong&gt;: provides Bond with gadgets in the film series. &lt;strong&gt;(From Zandar's answer)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The Second is seven doubled.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2=&lt;strong&gt;W&lt;/strong&gt;: double 7 =&gt; double U (see below) =&gt; W&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The Third is at the begin of endings.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3=&lt;strong&gt;E&lt;/strong&gt;: the first Letter (the begin) of the word ending is e&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The Fourth is difficult to differentiate in certain languages.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4=&lt;strong&gt;R&lt;/strong&gt;: In Chinese and Japanese there is no distinction between the letters l and r (It's a little more complicated than that but lets stick with that)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The Fifth is in the start of every sentence.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 5=&lt;strong&gt;T&lt;/strong&gt;: Every sentence in the code tag starts with T&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The Sixth is a question for lazy people.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 6=&lt;strong&gt;Y&lt;/strong&gt;: lazy people ask &quot;why&quot; they have to do something.  &lt;strong&gt;(From Zandar's answer)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The Seventh is not I but.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 7=&lt;strong&gt;U&lt;/strong&gt;: Not I but You&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The Eighth is me.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 8=&lt;strong&gt;I&lt;/strong&gt;: I am me&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The Ninth is what you will say when you figure it out.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 9=&lt;strong&gt;O&lt;/strong&gt;: You will say Oh&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The Tenth can also be before the first and sticks out its tongue.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 10=0=&lt;strong&gt;P&lt;/strong&gt;: As emoticon It sticks out the Tongue :-P. And 10=0 because it can also be before the first 0 &amp;lt; 1 &lt;strong&gt;(Edited with using Zandar's answer)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My friend can't dance but he can do a 084973553.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; So he can do a Pirouette&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13626" LastEditorUserId="4421" LastEditDate="2015-07-08T15:27:58.760" LastActivityDate="2015-07-08T15:27:58.760" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17582" PostTypeId="2" ParentId="17579" CreationDate="2015-07-08T10:16:05.527" Score="10" Body="&lt;p&gt;Your friend can&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;pirouette.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The solution to the riddle is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The First provides me with gadgets.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Q provides Bond with gadgets in the film series.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Second is seven doubled.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Z looks like two 7s put together.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Third is at the begin of endings.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;E is the first letter of &quot;endings&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Fourth is difficult to differentiate in certain languages.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;R is the same sound as L in some languages, such as Japanese.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Fifth is in the start of every sentence.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;T begins every sentence in this riddle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Sixth is a question for lazy people.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Y - lazy people ask &quot;why&quot; they have to do something.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Seventh is not I but.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;U - not I but &quot;you&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Eighth is me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I am me.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Ninth is what you will say when you figure it out.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;O - when you figure something out, you say, &quot;Oh.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Tenth can also be before the first and sticks out its tongue.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;P looks like a mouth with a tongue sticking out, as in the emoticon :P. Being before the first indicates that this should correspond with 0.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then the code is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Each number corresponds with the letter that is its answer (which happens to be that many letters into the first row of a QZERTY keyboard). Using this with the number 084973553 gives PIROUETTE.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12511" LastEditorUserId="4421" LastEditDate="2015-07-08T15:28:33.583" LastActivityDate="2015-07-08T15:28:33.583" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="17583" PostTypeId="2" ParentId="17565" CreationDate="2015-07-08T10:21:46.577" Score="1" Body="&lt;p&gt;Going by valuable_asset@ solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first is in a bank,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $ (Dollars)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My second is an evergreen,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; #(Could be Hashish)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My third is for your benefit&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; @ (Email)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fourth is in a store,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ! (Not sure, could be mark or a part of the name)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fifth is identical,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ~ (Symbol of similar to)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you've found the pattern&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pattern is all symbols going from 4,3,2,1,~,` or if nine is the limit it could be 5, 4, 3, 2, 1, ~, -, *, / (Going backwards and including numlock).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is my sixth?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ` or -&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2656" LastActivityDate="2015-07-08T10:21:46.577" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17585" PostTypeId="2" ParentId="17550" CreationDate="2015-07-08T11:41:07.297" Score="6" Body="&lt;p&gt;At least I got the first trap:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Morse code, with light toward 32 as - and away from 32 as . read from 1 to 31 gives &quot;Don't give up, try again&quot; with a little Formating on my Part.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I will try again :-D&lt;/p&gt;&#xA;" OwnerUserId="13626" LastActivityDate="2015-07-08T11:41:07.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17586" PostTypeId="2" ParentId="16421" CreationDate="2015-07-08T11:48:06.493" Score="5" Body="&lt;p&gt;Work in progress....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I understood the puzzle correctly, then:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;The safe has a random dot-number of &lt;code&gt;n&lt;/code&gt; characters length which is a prime number.&lt;/li&gt;&#xA;  &lt;li&gt;The safe is cracked if this number is modified, such that it is no longer a prime number.&lt;/li&gt;&#xA;  &lt;li&gt;The number is modified by 'duplicating' a character at &lt;em&gt;random&lt;/em&gt; position &lt;code&gt;p&lt;/code&gt; up to &lt;code&gt;c&lt;/code&gt; times, where &lt;code&gt;c&lt;/code&gt; is the charge of the duplicator. &lt;/li&gt;&#xA;  &lt;li&gt;The safe-crackers can chose to fire &lt;code&gt;x&lt;/code&gt; times with a chosen charge &lt;code&gt;c&lt;/code&gt;, and we should find which &lt;code&gt;c&lt;/code&gt; is needed, and how large &lt;code&gt;x&lt;/code&gt; needs to be to &lt;em&gt;guarantee&lt;/em&gt; success.&lt;/li&gt;&#xA;  &lt;li&gt;Only &lt;code&gt;c&lt;/code&gt; is a parameter which can be chosen!&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is important to notice that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Dot-numbers start and end with a ●, so they have to be &lt;em&gt;odd&lt;/em&gt; numbers &lt;/li&gt;&#xA;  &lt;li&gt;A Dot-number is &lt;em&gt;prime&lt;/em&gt; if it can not be produced by multiplication of two other dot-numbers, where 'multiplication' follows the rules outlined in the puzzle and &lt;em&gt;not&lt;/em&gt; regular binary multiplication.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With the above, my strategy to 'solve' this puzzle is to first&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; look at all (smaller) 'prime' numbers and try to establish a 'pattern'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and then, in a second step&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; find how this pattern can be reliably 'broken' by duplicating characters at &lt;em&gt;any&lt;/em&gt; position for up to &lt;code&gt;c&lt;/code&gt; times. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is maybe / quite likely a way to find dot-number primes more easily, but not being able to see this, I've started with a brute-force search for dot-number primes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Primes up to 9 digits (unless I made a mistake) are:&lt;br&gt;&#xA; &lt;code&gt;dot(  1):                      ●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(  3):                     ●●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(  7):                    ●●●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot( 11):                   ●○●●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot( 13):                   ●●○●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot( 19):                  ●○○●●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot( 25):                  ●●○○●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot( 31):                  ●●●●●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot( 37):                 ●○○●○●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot( 41):                 ●○●○○●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot( 47):                 ●○●●●●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot( 55):                 ●●○●●●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot( 59):                 ●●●○●●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot( 61):                 ●●●●○●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot( 67):                ●○○○○●●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot( 73):                ●○○●○○●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot( 87):                ●○●○●●●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot( 91):                ●○●●○●●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot( 97):                ●●○○○○●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(1○3):                ●●○○●●●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(1○9):                ●●○●●○●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(115):                ●●●○○●●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(117):                ●●●○●○●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(131):               ●○○○○○●●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(137):               ●○○○●○○●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(143):               ●○○○●●●●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(145):               ●○○●○○○●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(157):               ●○○●●●○●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(167):               ●○●○○●●●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(171):               ●○●○●○●●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(185):               ●○●●●○○●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(191):               ●○●●●●●●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(193):               ●●○○○○○●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(2○3):               ●●○○●○●●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(211):               ●●○●○○●●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(213):               ●●○●○●○●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(229):               ●●●○○●○●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(239):               ●●●○●●●●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(241):               ●●●●○○○●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(247):               ●●●●○●●●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(253):               ●●●●●●○●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(283):              ●○○○●●○●●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(285):              ●○○○●●●○●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(299):              ●○○●○●○●●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(3○1):              ●○○●○●●○●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(313):              ●○○●●●○○●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(319):              ●○○●●●●●●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(333):              ●○●○○●●○●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(351):              ●○●○●●●●●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(355):              ●○●●○○○●●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(357):              ●○●●○○●○●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(361):              ●○●●○●○○●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(369):              ●○●●●○○○●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(375):              ●○●●●○●●●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(379):              ●○●●●●○●●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(391):              ●●○○○○●●●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(395):              ●●○○○●○●●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(397):              ●●○○○●●○●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(415):              ●●○○●●●●●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(419):              ●●○●○○○●●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(425):              ●●○●○●○○●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(433):              ●●○●●○○○●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(445):              ●●○●●●●○●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(451):              ●●●○○○○●●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(463):              ●●●○○●●●●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(471):              ●●●○●○●●●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(477):              ●●●○●●●○●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(487):              ●●●●○○●●●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(499):              ●●●●●○○●●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(5○1):              ●●●●●○●○●&lt;/code&gt;&lt;br&gt; &#xA; &lt;code&gt;dot(5○5):              ●●●●●●○○●&lt;/code&gt;&lt;br&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From this one can notice:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - Dot-number primes come in symmetric pairs, i.e. any prime can be read from first to last digit or from last to first digit and remains a prime.&lt;br&gt;&#xA; - With increasing digits there are 1, 1, 1, 2, 3, 6, 9, 18, 30 prime numbers.&lt;br&gt;&#xA; - ...? T.B.C &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="5840" LastEditDate="2015-07-08T13:01:06.780" LastActivityDate="2015-07-08T13:01:06.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17587" PostTypeId="2" ParentId="17357" CreationDate="2015-07-08T12:17:15.517" Score="6" Body="&lt;p&gt;Second try: I think it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; COFFEE.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Title: &lt;em&gt;When you are sick you might have me, with a hand full of letters&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you are sick, you might have a &lt;em&gt;cough&lt;/em&gt; (coughy and coffee are homophones). A handful of letters (5 letters) would give you 'E' (5 fingers on a hand) so cough with 'E' equals coffee!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In some sports I can be used, but only by some. In this case here, it references to one.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sports cups can be worn as a protective gear in some sports by males. In this case here, it references to one male, Joe. Why, because some people need their &lt;b&gt;cup of Joe&lt;/b&gt;. - OP&#xA;&lt;br&gt;&lt;br&gt;A 'cup of Joe' is a synonym for a cup of coffee.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I can be used in some work, more specific than others. I'm not talking about animals or water which could be seen as my brothers.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Java (another synonym for coffee) is a well-known kind of coffee, and Java is a programming language which isn't Python (animals) or C (sea/water). &lt;em&gt;I'm flailing here with my poor knowledge of programming...&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am nearby the shops for others to get. Some might ignore me, some might not forget.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can buy coffee in the shops. Some people don't like coffee and therefore ignore it instead of buying it; others never forget to buy it!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I might be stronger than others, or weaker than few. Some might have just one, or maybe even two.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Coffee is stronger than other stimulants (such as tea or soda), but weaker than a few (such as energy drinks). Some may have (or only need) one cup, other might have 2 cups of coffee.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="11244" LastEditDate="2015-07-17T20:56:39.193" LastActivityDate="2015-07-17T20:56:39.193" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="17588" PostTypeId="2" ParentId="1963" CreationDate="2015-07-08T12:42:25.950" Score="0" Body="&lt;p&gt;2+2+2=6 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3*3)-3=6 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4/sqrt4)+4=(4/2)+4=6 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5/5)+5=6 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(6+6)-6=6 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;7-(7/7)=6 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;cubrt8+cubrt8+cubrt8=2+2+2=6 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;9-(9/sqrt9)=9-(9/3)=9-3=6&lt;/p&gt;&#xA;" OwnerUserId="13907" LastActivityDate="2015-07-08T12:42:25.950" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="17589" PostTypeId="2" ParentId="17510" CreationDate="2015-07-08T13:22:42.050" Score="5" Body="&lt;p&gt;It's impossible.  Take the dual of the polyhedron's edge graph; every simple graph has at least 2 nodes with the same degree. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Appendix:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The polyhedron's edge graph is the graph you get when you start with a set of vertices corresponding to the polyhedron's vertices, and connect every pair of vertices that are connected by a polygon edge in your polyhedron.  This can be proven to be a planar graph.  Informally, it is the graph you get by unfolding or flattening the polyhedron onto a plane.&#xA;See &lt;a href=&quot;https://en.wikipedia.org/wiki/Polyhedral_graph&quot;&gt;https://en.wikipedia.org/wiki/Polyhedral_graph&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dual of a planar graph is the graph you get by swapping the faces and vertices.  To build a dual H from a graph G, place a vertex in the middle of each face in G, and connect vertices in H iff the corresponding faces in G are adjacent (share an edge).  See &lt;a href=&quot;https://en.wikipedia.org/wiki/Dual_graph&quot;&gt;https://en.wikipedia.org/wiki/Dual_graph&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Dual_polyhedron&quot;&gt;https://en.wikipedia.org/wiki/Dual_polyhedron&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, every simple graph has a pair of vertices with equal degree.  We will prove the stronger result that every connected component has such a pair.  If the component has n vertices, the degree of each vertex is a number in 1, 2, ..., n-1 so by the pigeonhole principle there is a repeat.&lt;/p&gt;&#xA;" OwnerUserId="10806" LastEditorUserId="10806" LastEditDate="2015-07-08T14:32:44.687" LastActivityDate="2015-07-08T14:32:44.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17590" PostTypeId="2" ParentId="17247" CreationDate="2015-07-08T13:47:54.900" Score="6" Body="&lt;p&gt;Solution to the &lt;a href=&quot;/questions/tagged/cryptic-crosswords&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;cryptic-crosswords&amp;#39;&quot; rel=&quot;tag&quot;&gt;cryptic-crosswords&lt;/a&gt; part:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Scold the bad liar. (4)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Bad&quot; suggests anagrams; an anagram of &quot;liar&quot; is RAIL, which (as a verb) is a synonym of &quot;scold&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A wall of iron has nickel, copper, even lead! (5)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Iron is Fe on the periodic table; &quot;lead&quot; suggests taking initial letters, so &quot;nickel, copper, even&quot; -&gt; NCE. Putting it all together, we get FENCE, which is a synonym of &quot;wall&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nice Ximenean clues there! :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we need to use&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the railfence cipher&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;to decode the next part, as Chadi Abou Sleiman has already done.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-07-09T07:03:53.300" LastActivityDate="2015-07-09T07:03:53.300" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17591" PostTypeId="2" ParentId="17486" CreationDate="2015-07-08T14:16:48.130" Score="20" Body="&lt;p&gt;I think I found the answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You need to take the three-letter code of each country and calculate the 'distance' (d) between 2 consecutive letters, and finally add them.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Belgium --&gt; Germany --&gt; France&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BEL --&gt; DEU --&gt; FRA&lt;br&gt;&#xA; d(B,E) + d(E,L) = 3 + 7 = 10&lt;br&gt;&#xA; d(D,E) + d(E,U) = 1 + 16 = 17&lt;br&gt;&#xA; d(F,R) + d(R,A) = 12 + 17 = 29&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Saudi Arabia --&gt; Iran --&gt; Qatar&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SAU --&gt; IRN --&gt; QAT&lt;br&gt;&#xA; d(S,A) + d(A,U) = 18 + 20 = 38&lt;br&gt;&#xA; d(I,R) + d(R,N) = 9 + 4 = 13&lt;br&gt;&#xA; d(Q,A) + d(A,T) = 16 + 19 = 35&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That means that Brazil --&gt; Chile --&gt; Argentina&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BRA --&gt; CHL --&gt; ARG&lt;br&gt;&#xA; d(B,R) + d(R,A) = 16 + 17 = 33&lt;br&gt;&#xA; d(C,H) + d(H,L) = 5 + 4 = 9&lt;br&gt;&#xA; d(A,R) + d(R,G) = 17 + 11 = 28&lt;br&gt;&#xA; so the code is &lt;strong&gt;33-9-28&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I guess then that Everywhere --&gt; Nowhere --&gt; Home&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; EVE --&gt; NOW --&gt; HOM&lt;br&gt;&#xA; d(E,V) + d(V,E) = 17 + 17 = 34&lt;br&gt;&#xA; d(N,O) + d(O,W) = 1 + 8 = 9&lt;br&gt;&#xA; d(H,O) + d(O,M) = 7 + 2 = 9&lt;br&gt;&#xA; so the code is &lt;strong&gt;34-9-9&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13268" LastActivityDate="2015-07-08T14:16:48.130" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17592" PostTypeId="2" ParentId="17377" CreationDate="2015-07-08T14:56:43.030" Score="4" Body="&lt;p&gt;It's a lame solution but it works:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4,4,7,8&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Number on top represents the minimum value in the triangle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="138" LastActivityDate="2015-07-08T14:56:43.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17593" PostTypeId="2" ParentId="17153" CreationDate="2015-07-08T14:57:31.867" Score="3" Body="&lt;p&gt;&lt;strong&gt;Partial answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just by looking at the format, and the last comment by the OP, the poem talked about is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://germanstories.vcu.edu/goethe/zauber_dual.html&quot; rel=&quot;nofollow&quot;&gt;Der Zauberlehrling by Johann Wolfgang von Goethe&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hopefully this helps someone else to solve the puzzle.&lt;/p&gt;&#xA;" OwnerUserId="13268" LastActivityDate="2015-07-08T14:57:31.867" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17594" PostTypeId="1" AcceptedAnswerId="17606" CreationDate="2015-07-08T15:07:57.707" Score="2" ViewCount="500" Body="&lt;p&gt;&lt;strong&gt;BACKGROUND&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is essentially a combination of two puzzles:&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/q/17531/9000&quot;&gt;Longest *Sentence* With Only Repeating Character Pairs&lt;/a&gt; (which has been shown to be infinite)&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/q/8223/9000&quot;&gt;Longest sentence with each letter repeated n times&lt;/a&gt; (which maxed out at $n=16$)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;CHALLENGE&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Form a grammatically correct sentence with $c$ alphabetic characters (&lt;code&gt;[A-Za-z]&lt;/code&gt;)in which every letter &lt;em&gt;pair&lt;/em&gt; appears &lt;em&gt;exactly $n$&lt;/em&gt; times. Your score is $n\times c$. Maximize that score.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clarifications: (The example below is invalid and is only intended to assist in clarification.)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Words can not be repeated but variations of a word do not count as a repetition. This includes plurals, possessives, contractions, etc. so long as it does not violate other clarifications. &lt;code&gt;Uni the Unicorn's eating the unicorn!&lt;/code&gt; and &lt;code&gt;Bob's bobbing bobbed the bobber.&lt;/code&gt; are OK but &lt;code&gt;Unicorns ate the unicorn's food.&lt;/code&gt; is not OK as it violates the second clarification.&lt;/li&gt;&#xA;&lt;li&gt;Words count as a repetition if they match when converted to all uppercase and any non-letters are removed. Different definitions, punctuation, or capitalization does not prevent it counting as a repetition. &lt;code&gt;Buffalo buffalos buffalo bison.&lt;/code&gt; (&lt;code&gt;Buffalo&lt;/code&gt; &quot;city&quot; and &lt;code&gt;buffalo&lt;/code&gt; &quot;bully&quot; are counted as a repitition.)&lt;/li&gt;&#xA;&lt;li&gt;Any non-letter (spaces, hyphens, apostrophes, etc.) should be stripped prior to analysis. The result will be just a string of letters. &lt;code&gt;UnitheUnicornseatingtheunicorns&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;Capitalization does not matter. Convert them all to uppercase if that helps. &lt;code&gt;UNITHEUNICORNSEATINGTHEUNICORNS&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The &quot;purest&quot; answers will also meet these conditions: (not required to be valid but somewhat more impressive)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Every word appears on &lt;a href=&quot;http://dictionary.reference.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dictionary.com&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;No proper nouns&lt;/li&gt;&#xA;&lt;li&gt;No acronyms or abbreviations&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;CHECKING&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Here are some check methods. I've written a verbose one in VBA. Corrections and submissions in other languages are most welcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;VBA&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Outputs the most common, highest value for $n$ along with a score and a list of any that do not match. (It's not golfed at all because I preferred accuracy and readability over compression.)&lt;br&gt;&#xA;Example 1) Input: &lt;code&gt;Alfalfa&lt;/code&gt;. Output: &lt;code&gt;n = 2, c = 7, Score = 14&lt;/code&gt;&lt;br&gt;&#xA;Example 2) Input: &lt;code&gt;Alfalfac&lt;/code&gt;. Output: &lt;code&gt;n = 2: AC(1)&lt;/code&gt;&lt;br&gt;&#xA;Example 3) Input: &lt;code&gt;Uni Unicorns&lt;/code&gt;. Output: &lt;code&gt;n = 1: UN(2), NI(2)&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Option Explicit&#xA;Function ValidateString(s As String) As String&#xA;    Dim i As Long, j As Long, sTemp As String, v As String&#xA;    Dim cValue As String, cCount As Long, cCountCount As Long&#xA;    Dim nValue As Long, nCount As Long, nList As String&#xA;&#xA;    'Sanitize&#xA;    s = UCase(s)&#xA;    For i = 1 To Len(s)&#xA;        Select Case Asc(Mid(s, i, 1))&#xA;            Case 65 To 90:  sTemp = sTemp &amp;amp; Mid(s, i, 1)&#xA;        End Select&#xA;    Next&#xA;    s = sTemp&#xA;&#xA;    'Find all n values&#xA;    For i = 1 To Len(s) - 1&#xA;        cValue = Mid(s, i, 2)&#xA;        cCount = 0&#xA;        For j = 1 To Len(s) - 1&#xA;            If Mid(s, j, 2) = cValue Then cCount = cCount + 1&#xA;        Next j&#xA;        nList = nList &amp;amp; &quot;|&quot; &amp;amp; cCount &amp;amp; &quot;|&quot;&#xA;    Next&#xA;&#xA;    'Find largest, most common n value&#xA;    For i = 1 To Len(s) - 1&#xA;        cValue = Mid(s, i, 2)&#xA;        cCount = 0&#xA;        For j = 1 To Len(s) - 1&#xA;            If Mid(s, j, 2) = cValue Then cCount = cCount + 1&#xA;        Next j&#xA;        cCountCount = (Len(nList) - Len(Replace(nList, &quot;|&quot; &amp;amp; cCount &amp;amp; &quot;|&quot;, &quot;&quot;))) / (2 + Len(cCount))&#xA;        If cCountCount &amp;gt; nCount Then nValue = cCount: nCount = cCountCount&#xA;        If cCountCount = nCount And cCount &amp;gt; nValue Then nValue = cCount&#xA;    Next&#xA;&#xA;    'List any that don't match&#xA;    sTemp = &quot;&quot;&#xA;    For i = 1 To Len(s) - 1&#xA;        cValue = Mid(s, i, 2)&#xA;        cCount = 0&#xA;        For j = 1 To Len(s) - 1&#xA;            If Mid(s, j, 2) = cValue Then cCount = cCount + 1&#xA;        Next j&#xA;        cCountCount = (Len(nList) - Len(Replace(nList, &quot;|&quot; &amp;amp; cCount &amp;amp; &quot;|&quot;, &quot;&quot;))) / (2 + Len(cCount))&#xA;        If cCountCount &amp;lt;&amp;gt; nCount And InStr(1, sTemp, cValue) = 0 Then sTemp = sTemp &amp;amp; cValue &amp;amp; &quot; (&quot; &amp;amp; cCount &amp;amp; &quot;), &quot;&#xA;    Next&#xA;    v = &quot;n = &quot; &amp;amp; nValue&#xA;    If sTemp = &quot;&quot; Then v = v &amp;amp; &quot;, c = &quot; &amp;amp; Len(s) &amp;amp; &quot;, Score = &quot; &amp;amp; nValue * Len(s)&#xA;    If sTemp &amp;lt;&amp;gt; &quot;&quot; Then v = v &amp;amp; &quot;: &quot; &amp;amp; Left(sTemp, Len(sTemp) - 2)&#xA;    ValidateString = v&#xA;End Function&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-07-08T19:50:51.020" Title="Longest Sentence Where Every Character Pair Repeats $n$ times" Tags="|english|open-ended|letters|" AnswerCount="6" CommentCount="11" ClosedDate="2021-05-14T09:40:11.630" ContentLicense="CC BY-SA 3.0" />
  <row Id="17595" PostTypeId="2" ParentId="17565" CreationDate="2015-07-08T15:29:28.550" Score="1" Body="&lt;p&gt;Following the idea of @valuable_asset and @theplace, perhaps 2 is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Num lock, as I don't know many people who turn it off&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and 3 might be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Backspace, which is there for correcting our mistakes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $ again&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Meaning 5. is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Num lock (identical to 2)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And 6. would be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; backspace again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13915" LastActivityDate="2015-07-08T15:29:28.550" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17596" PostTypeId="2" ParentId="17096" CreationDate="2015-07-08T15:31:32.573" Score="2" Body="&lt;p&gt;Since my other answer is already so long, I'll post this as a separate answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Generalizing to $3^n$&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are ${3^n\choose 2} =3^n (3^n-1)/2 \approx 3^{2n}/2$ possible combinations.  Each weighing has 4 possible outcomes, so this requires at least $\lceil \log_4 {3^n\choose 2}\rceil \approx \lceil\log_4{(3^{2n}/2)}\rceil = \lceil 2n\log_4 3 - 1/2\rceil$ weighings.  (The approximation breaks down for $n=1$).  My algorithm below requires&#xA;$\lceil n + \lceil(n-1)\log_4 3\rceil$ weighings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Phase 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take all $3^n$ coins and put them in one tall stack.  Then split: put the bottom third on Pan 1, the middle third on Pan 2, the top third on Pan 3.  Weigh and note the results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then repeat the split for each stack: put the bottom third of each stack onto Pan 1, the middle third on Pan 2, the top third on Pan 3.  Repeat for a total of $n$ weighings, yielding at the end $3^n$ stacks of 1 coin each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each weighing is orthogonoal to the others $-$ each time you are splitting stacks, there is a roughly equal chance of each heavy coin ending up on each Pan, the only proviso being that the two heavy coins cannot be at the same height in the same stack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, each division splits the options as: $1/3$: no pan rises, $2/9$: Pan 1 rises, $2/9$: Pan 2 rises, $2/9$: Pan 3 rises.  Each split goes one level deeper into the stacks, so they are essentially independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only correlation between the weighings is that it is not possible for all $n$ weighings to come out flat.  That would require both coins to be at the same location in the first stack, which is impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, after $n$ weighings, we have at least one weighing where a pan rose.  Name that the &quot;alpha&quot; weighing.  Pick one pan that sank on the alpha weighing and call those the &quot;A&quot; coins, and name the coins on the other pan that sank the &quot;B&quot; coins.  The remaining coins are the &quot;C&quot; coins, and we know all of them are normal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are $3^{n-1}$ coins in each group.  Thus, there are $3^{n-1}\times3^{n-1} = 3^{2n-2}$ options remaining, considering only the result of the alpha weighing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each candidate heavy coin in the &quot;A&quot; group, any coin in the &quot;B&quot; group is a potential matching candidate heavy coin.  Each weighing after the alpha weighing splits that group into 3 equal parts, thus reducing the number of potential matching coins to at most $1/3$ of the number from before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, after the other $n-1$ predetermined weighings, the number of candidate matching &quot;B&quot; coins for each &quot;A&quot; coin is reduced by a factor of $3^{n-1}$ (or reduced to zero if that &quot;A&quot; coin was on a pan that rose and was itself eliminated).  Since there are $3^{n-1}$ &quot;B&quot; coins, this means after $n$ weighings, each remaining candidate &quot;A&quot; coin matches up with only one candidate &quot;B&quot; coin.  Using the same logic in reverse, each &quot;B&quot; coin is matched up with at most one &quot;A&quot; coin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, after $n$ weighings, the remaining possibilities are all non-overlapping sets of one &quot;A&quot; coin and a matching &quot;B&quot; coin.  There are at most $3^{n-1}$ &quot;A&quot; coins left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Phase 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this phase, we take the set of $x$ matching pairs and divide it into four groups of $x/4$ matching pairs (if 4 does not divide $x$, let the fourth group be smaller).  We then split each pair in the first group and put one coin on Pan 1 and one on Pan 2.  The second group we split, putting on on Pan 1 and one on Pan 3.  The third gets split onto Pans 2 and 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a pan rises, then all the candidate pairs that had a coin on that pan are eliminated, and so are the coins in the fourth group.  If no pan rises, then the heavy coins are in the fourth group.  This reduces the number of candidates to $\lceil x/4\rceil$.  Repeating this to reduce $3^{n-1}$ down to 1 requires $\lceil (n-1) \log_4 3\rceil$ weighings.  (On the final weighing, there might be only 2 options remaining; in that case, fill in with extra known normal coins, of which we have $3^n-4$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Optimality&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the first $n$ weighings are orthogonal, this algorithm gathers the most possible information from them.  The best division from any of these weighings is $2/9$,$2/9$,$2/9$,$1/3$.  It is not possible to reduce the number of options by a factor of $4$ during any of the first $n$ weighings, no matter the algorithm.  The rest of the weighings do reduce the number of options by a factor of $4$ each time, so they must be optimal.  Thus, this algorithm is an optimal algorithm, and no other algorithm can answer the question in fewer steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the total number of weighings is $n$ for Phase 1 and $\lceil (n-1) \log_4 3\rceil$ for phase 2.  Since $\log_4 3 \approx 0.79$, the solution approximates to: $W \approx \lceil n + 0.8(n-1)\rceil$.  This comes to, for $n=1,2,...$: $W=1,3,5,7,9,10,12,...$.  For example, for $3^6$ coins, the first 6 weighings leave us with at most $2^5 = 243$ possibilities, which can be distinguished in $4$ weighings, since $243 &amp;lt; 256 = 4^4$.  Thus, $W=10$ weighings are required for $3^6$ coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=1$, the first weighing is also the $n^{th}$ weighing is the alpha weighing.  After that the &quot;A&quot; and &quot;B&quot; groups have 1 coin each, thus requiring 0 more weighings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=2$, at least one of the first two weighings had a pan rise.  That gave us an &quot;A&quot; and a &quot;B&quot; group of 3 coins each.  On the second weighing, the three &quot;B&quot; coins are split up, so each &quot;A&quot; coin matches with at most one &quot;B&quot; coin.  On the third weighing, the three options are split and we determine the final pair.  If there were only two remaining options after the second weighing, we would need to fill in with two extra normal coins, of which we have five.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=3$, this yields the exact same algorithm as for my other answer, without all the relabeling.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="2328" LastEditDate="2015-07-08T15:46:03.017" LastActivityDate="2015-07-08T15:46:03.017" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17597" PostTypeId="2" ParentId="17594" CreationDate="2015-07-08T15:50:58.697" Score="4" Body="&lt;h2&gt;n = 2, c = 19, Score = 38&lt;/h2&gt;&#xA;&lt;p&gt;OK, here's one to get the ball rolling:&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;She's going to put her money on the old sailors&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; SHE'LL BACK SHELLBACKS.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="40" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-08T16:32:40.523" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17598" PostTypeId="2" ParentId="17550" CreationDate="2015-07-08T16:10:11.767" Score="9" Body="&lt;p&gt;Well what do you know, it seems rand al'thor was on to something:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you take the user id numbers for users 1-31 as ascii, you get a message:&lt;br&gt;&#xA;JET Fuel_CAN't-m3l+~S┼Éé| Bèams&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11323" LastActivityDate="2015-07-08T16:10:11.767" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17599" PostTypeId="2" ParentId="17565" CreationDate="2015-07-08T16:12:37.793" Score="10" Body="&lt;p&gt;Looking outside of our narrow box&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mercury, mercury is also a metal, metal in banks (gold, silver)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So next would be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Venus, well its always yellowish according to the internet but I think of it as greeny-blue, ever green.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lets roll with it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Earth, I mean I think it is useful to us, even if we aren't useful to it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mars, MARS BARS! candy is in a store&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jupiter, which is a bit stupider... but according to a website aimed at 5th graders: The atmospheres of Saturn and Jupiter are almost identical&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And what is 5 identical to?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Saturn, the 6th planet from the sun.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; woops forgot to say 9 is the limit because VIVA LA PLUTO!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13916" LastActivityDate="2015-07-08T16:12:37.793" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17600" PostTypeId="2" ParentId="17594" CreationDate="2015-07-08T16:16:16.813" Score="0" Body="&lt;h2&gt;n = &lt;span class=&quot;math-container&quot;&gt;$\infty$&lt;/span&gt;: OB(&lt;span class=&quot;math-container&quot;&gt;$\infty-1$&lt;/span&gt;)      (Invalid per clarification 1)&lt;/h2&gt;&#xA;&lt;p&gt;Without rules the hilarious near answer is:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Buffalo buffalo buffalo buffalo buffalo buffalo buffalo buffalo buffalo buffalo&#xA; buffalo buffalo buffalo buffalo ...&#xA; on to infinity as there are no two'fer character pairs in buffalo so each pair&#xA; gets repeated n times where n is the number of words. there is punctuation to&#xA; deal with, but buffalo infinitum is a sentence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;A number of flaws follow if you actually read the rules.&lt;/p&gt;&#xA;" OwnerUserId="13916" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-08T19:04:50.720" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17601" PostTypeId="2" ParentId="1856" CreationDate="2015-07-08T16:20:10.650" Score="2" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jfoal.jpg&quot; alt=&quot;graph representation of problem&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to graph theory&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a graph G, is it possible to find a walk that traverses&#xA;each line exactly once, goes through ail points, and ends at the starting point?&#xA;A graph for which this is possible is called &lt;strong&gt;eulerian graph&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, an eulerian graph&#xA;has an eulerian trail, a closed trail containing all points and lines. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  Theorem: The following statements are equivalent for a connected graph G:&#xA;    (1) G is eulerian.&#xA;    (2) Every point of G has even degree.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In this problem there exists 4 points(nodes) with odd degree. So this is not eulerian graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;i.e: There no solution existing.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="13582" LastActivityDate="2015-07-08T16:20:10.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17602" PostTypeId="2" ParentId="17594" CreationDate="2015-07-08T16:24:57.303" Score="0" Body="&lt;h2&gt;n = 3, c = 19, Score = 57&lt;/h2&gt;&#xA;&lt;p&gt;If abbreviations are okay the:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Adder's adder sadder SA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Meaning:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The sadder SA of the Adder's adder. Or maybe. The SA of Adder that was sadder by and adder. I have no idea either.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-08T19:00:03.053" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17603" PostTypeId="2" ParentId="17594" CreationDate="2015-07-08T16:39:09.687" Score="0" Body="&lt;h2&gt;n = 3, c = 13, Score = 39&lt;/h2&gt;&#xA;&lt;p&gt;A grammatically correct, but absolutely nonsense sentence.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Seats eat sea TS.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The explanation&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Chairs consume top secrets of the sea. (To clarify, &lt;a href=&quot;http://dictionary.reference.com/browse/ts?s=t&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dictionary.com lists the definition of TS as an abbreviation.&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-08T16:45:12.153" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17604" PostTypeId="2" ParentId="17594" CreationDate="2015-07-08T16:53:47.647" Score="0" Body="&lt;h2&gt;n = 7, c = 29, score = 203&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Inspired by &lt;a href=&quot;https://puzzling.stackexchange.com/a/17603/12700&quot;&gt;my first answer&lt;/a&gt;, here's another with a better value of n and much better score. &lt;em&gt;Be warned, you are about to reveal yet another nonsensical sentence that stays on the right side of the grammar line...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Seats eat sea TS, eats, Ea, TSE at SE ATS. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The explanation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chairs consume top secrets of the sea, food, &lt;a href=&quot;http://dictionary.reference.com/browse/EA?s=t&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Akkadian god of wisdom&lt;/a&gt;, the &lt;a href=&quot;http://dictionary.reference.com/browse/TSE?s=t&quot; rel=&quot;nofollow noreferrer&quot;&gt;Toronto Stock Exchange&lt;/a&gt; (sorry Canada) at the South Eastern outpost of the &lt;a href=&quot;http://dictionary.reference.com/browse/ATS?s=t&quot; rel=&quot;nofollow noreferrer&quot;&gt;Auxiliary Territorial Service&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-07-08T19:50:51.020" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17605" PostTypeId="1" AcceptedAnswerId="17659" CreationDate="2015-07-08T17:00:43.927" Score="5" ViewCount="765" Body="&lt;p&gt;This is part of &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=title%3A%22The+Twenty+Doors%22&quot;&gt;The Twenty Doors&lt;/a&gt; series. &#xA;&lt;br&gt;The previous one is &lt;a href=&quot;https://puzzling.stackexchange.com/questions/17423/the-twenty-doors-room-5/&quot;&gt;The Twenty Doors! (ROOM 5)&lt;/a&gt;&#xA;&lt;br&gt;The next one is &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18345/the-twenty-doors-room-7/&quot;&gt;The Twenty Doors (ROOM 7)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You go through door 5, and see a rather unusual keypad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;[♣][♦][♠][♥]&#xA;   [ENTER]  &#xA;&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper has been put on the left wall from when you come in, rotated by 90º.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Znw qbty zx uxtw wlb ibnsqh thygpozg nkbveni Rnn-Imed xo qjw qytjfa osa.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And, as always, a hint has been carved into the wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Caesar spun in a 3 width box. That's the answer to the alphabet, though the key and alphabet are reversed. And, 123456789 could well be a key, will you decipher it, we will see...&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What symbol should you press?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the most cryptic one so far, and also hardest to decipher. I want every part of this puzzle explained!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next door will be added when this door is solved!&lt;/p&gt;&#xA;" OwnerDisplayName="user9377" LastEditorUserId="-1" LastEditorDisplayName="user9377" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-07-24T20:09:13.823" Title="The Twenty Doors! (ROOM 6)" Tags="|cipher|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="17606" PostTypeId="2" ParentId="17594" CreationDate="2015-07-08T17:06:48.497" Score="3" Body="&lt;h2&gt;n = 8, c = 17, Score = 136&lt;/h2&gt;&#xA;&lt;p&gt;Rather than trying to maximize c, I am instead attempting to maximize n.&lt;/p&gt;&#xA;&lt;p&gt;To that end:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Bobobo-bo Bo-bobo, bob!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This is an imperative statement, commanding the main character from&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Bobobo-bo_Bo-bobo&quot; rel=&quot;nofollow noreferrer&quot;&gt;this anime&lt;/a&gt; to commence &lt;a href=&quot;https://en.wikipedia.org/wiki/Apple_bobbing&quot; rel=&quot;nofollow noreferrer&quot;&gt;attempting to extract apples from a bucket of water using only his teeth&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;8 pairs * 17 characters = 136&lt;/p&gt;&#xA;&lt;p&gt;I feel like one could torture a few more 'bo's in there, but I think this is fine for what is essentially a joke answer. =P&lt;/p&gt;&#xA;" OwnerUserId="10861" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-08T17:13:31.457" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17607" PostTypeId="1" AcceptedAnswerId="17611" CreationDate="2015-07-08T17:33:26.757" Score="8" ViewCount="1806" Body="&lt;p&gt;You know I think I've been here before,&lt;br&gt;&#xA;And very soon I am shown the door.&lt;br&gt;&#xA;Someone new brought me here,&lt;br&gt;&#xA;To another place I will steer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Little can be done to fix,&lt;br&gt;&#xA;I shouldn't even be in the mix.&lt;br&gt;&#xA;I must already be known,&lt;br&gt;&#xA;Who am I? Not really alone.  &lt;/p&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-07-08T20:03:17.233" Title="Deja Vu - What am I?" Tags="|riddle|rhyme|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17609" PostTypeId="2" ParentId="17607" CreationDate="2015-07-08T17:55:27.347" Score="5" Body="&lt;p&gt;Could it be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A shopping cart (With short term memory loss)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You know I think I've been here before,&#xA;&lt;br&gt;And very soon I am shown the door.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A shopping cart is often pushed to the front entrance of the store, where it can be shown the door.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Someone new brought me here,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A shopping cart is reused by many people. Here a new customer is pushing it into the store!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To another place I will steer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It will steer into the store, and possibly down many isles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Little can be done to fix,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A possible reference to how shopping carts always seem to have at least one broken wheel, and no matter how hard you try to compensate, that wheel will not cooperate. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I shouldn't even be in the mix.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Broken shopping carts shouldn't be in the 'mix' of normal shopping carts (as they are damaged). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I must already be known,&#xA;&lt;br&gt;Who am I? Not really alone&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You, are a broken shopping cart, and I will always be here for you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-07-08T20:03:17.233" LastActivityDate="2015-07-08T20:03:17.233" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17610" PostTypeId="2" ParentId="17565" CreationDate="2015-07-08T17:58:37.867" Score="0" Body="&lt;p&gt;Maybe it could be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Spades/spade, as in the suit in playing cards.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My first is in a bank&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Diamonds, they can be kept in banks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My second is an evergreen&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clubs, since the clover plant is evergreen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My third is for your benefit&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hearts, as you would die without one (pretty big benefit)!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fourth is in a store,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Spades, they are sold in stores (and it's the only suit left...)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fifth is identical,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Identical to what? This one's throwing me off -- but I'll go with the jokers (since there are two).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you've found the pattern&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is my sixth?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...so I assume that the pattern turns around and goes back to spades here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13556" LastActivityDate="2015-07-08T17:58:37.867" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17611" PostTypeId="2" ParentId="17607" CreationDate="2015-07-08T18:01:20.777" Score="12" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a plagiarized question on PuzzlingSE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You know I think I've been here before,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every now and again new users join and add cliched riddles and puzzles without any attribution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And very soon I am shown the door.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Emrakul brings down his hammer on these questions and off they go into the garbage bin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Someone new brought me here,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As mentioned, such questions are often brought up by new users on PuzzlingSE.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To another place I will steer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Their clues can very easily lead us to the actual source with a simple Google search.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Little can be done to fix,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We can downvote and flag repeatedly, and Emrakul can delete them but little can be done to fix this issue permanently.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I shouldn't even be in the mix.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In an ideal scenario, we should never have to deal with such plagiarized questions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I must already be known,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most of the plagiarized riddles and puzzles are already known to the expert puzzlers here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who am I? Not really alone.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are a plagiarized question, and there can be several of you hanging around on PuzzlingSE even after Emrakul and the community's vigilante efforts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This answer is inspired by &lt;a href=&quot;https://puzzling.meta.stackexchange.com/questions/3161/what-is-the-criteria-for-question-deletion-on-puzzling-se&quot;&gt;this meta post&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="-1" LastEditDate="2017-03-16T16:42:07.097" LastActivityDate="2015-07-08T18:03:36.247" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17612" PostTypeId="2" ParentId="17486" CreationDate="2015-07-08T19:39:32.750" Score="1" Body="&lt;p&gt;I wanted to add this as a comment, but it won't let me, so here goes:&#xA;I give credit to Cerberus for getting the first part of the answer, and I think I've worked out the 2nd lock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 11-2-16, or 12-2-16.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reasoning is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;For Everywhere, it could either be ALL or INF, for infinity.&#xA;&lt;br&gt; This gives us:&#xA;&lt;br&gt;d(A,L) + d(L,L) = 11 + 0 = &lt;strong&gt;11;&lt;/strong&gt;&#xA;&lt;br&gt;d(I,N) + d(N,F) = 4 + 8 = &lt;strong&gt;12;&lt;/strong&gt;&#xA;&lt;br&gt;&lt;br&gt;For Nowhere, the code is NON, which is a version of NOT, or nowhere.&#xA;&lt;br&gt;calculating the distance of adjacent letters, we get: &#xA;&lt;br&gt;d(N,O) + d(n,O) = 1 + 1 = &lt;strong&gt;2;&lt;/strong&gt;&#xA;&lt;br&gt;&lt;br&gt;Home is &quot;where the heart is&quot;, so the code is WHI.&#xA;&lt;br&gt;Finally, we get:&#xA;&lt;br&gt;d(W,H) + d(H,I) = 15 + 1 = &lt;strong&gt;16;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13697" LastActivityDate="2015-07-08T19:39:32.750" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17613" PostTypeId="1" CreationDate="2015-07-08T20:26:46.980" Score="5" ViewCount="319" Body="&lt;p&gt;Pretend I was some sort of budding author.  A bit of a stretch, but…&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that you've done that, I had a bit of an idea related to writing curses, and breaking them. But the method is slightly absent.  So, would there be any clever way of encrypting or hiding the way to break a curse within the wording of the curse itself?  This seems a bit broad, so I'll add a scenario.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's slightly futuristic, as well as fantasy-ish…  The athletes use demons to enhance speed, strength, whatnot.  In a [insert sport name] match, game, etc., the loser is approached for his autograph, gets mad, curses the poor kid.  Apparently he can control fire, so the demon uses the fire to begin burning the kid from inside out.  ;P Pleasant, right?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Maybe I should move this to Worldbuilding, dunno.  But I like the people on Puzzling more.  Please say something if this is completely off topic.  Also, I don't know about the tag.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Short version of main question(because there are too many words and rand knows this site better than I do):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How might a cure for a curse be hidden within the words of the curse?&lt;/p&gt;&#xA;" OwnerUserId="11790" LastEditorUserId="11790" LastEditDate="2015-07-09T23:27:19.730" LastActivityDate="2015-07-09T23:27:19.730" Title="Creation of… Curses?" Tags="|puzzle-creation|story|" AnswerCount="3" CommentCount="6" ClosedDate="2015-07-16T08:56:43.873" ContentLicense="CC BY-SA 3.0" />
  <row Id="17614" PostTypeId="2" ParentId="17613" CreationDate="2015-07-08T20:47:14.367" Score="6" Body="&lt;p&gt;I don't know if this is &quot;on-topic&quot; or not, but I like the question, so I'll take a couple stabs at answering it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;One way you could consider is shifting each letter once to the right to the nearest consonant/vowel, depending on what the original letter is. This gives fun, fantasy-sounding words that are just a simple shift away from their real-word counterparts!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are a few examples (with leniency in spacing and punctuation):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a shower --&gt; Veliet Juxis&lt;br&gt;&#xA;Eat bat bones --&gt; Iev'cev Cupit&lt;br&gt;&#xA;Kill yourself --&gt; Lom M'zu Asti M'g&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The words don't always work out well (&quot;drink&quot; becomes &quot;fsoml&quot;, which isn't great), but they can be pretty cool.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another way is my old favorite, the cryptic crossword clue! You could craft curses in such a way that they can be puzzled out to the correct answer, cryptic-style. An example could be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A three-headed body and an old, endless burst&lt;br&gt;&#xA;led astray under time's ire, revealed at first.&lt;br&gt;&#xA;To sever the chains, see the end in reverse.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't immediately obvious, but the lines clue towards how to break the curse:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First line: BLOOD (BOD (first three &quot;heads&quot; of &quot;body&quot;) + OL (&quot;old&quot; endlessly) anagrammed (burst))&#xA;Second line: LAUTIR (letters &quot;at first&quot; of each word)&lt;br&gt;&#xA;Final answer: BLOOD RITUAL (with the ending word in reverse)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These can be difficult, but are also quite clever.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There are a bunch of simple encryptions, with some examples being Vigenere and Caesarian ciphers, that can give you a result that looks curselike, but they aren't very clever and don't give you great results. I think with my contorted mind I would go cryptic, but these are just a couple examples to get everyone else's juices flowing. I hope you're at least somewhat intrigued!&lt;/p&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-07-08T20:47:14.367" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17615" PostTypeId="2" ParentId="17096" CreationDate="2015-07-08T21:05:00.183" Score="0" Body="&lt;p&gt;For comparison, I'll provide my own solution, which uses $2n-1$ weighings. It is optimal only for $n \leq 5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The base case $n=1$ (3 coins) is solvable with 1 weighing by putting 1 coin on each pan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Split the $3^n$ coins in 3 groups of $3^{n-1}$ each, and put one group on each pan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) If one pan rises, we know the other two pans each contain one heavy coin. Solve puzzle &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16902/two-heaps-each-with-a-heavy-ball-and-a-3-pan-balance&quot;&gt;Two heaps each with a heavy ball, and a 3-pan balance&lt;/a&gt; to find the two coins in $\lceil (n-1)\log_2 3 \rceil \leq 2(n-1)$ weighings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) I no pan rises, we know the two heavy coins are together in one heap. Temporarily remove one coin from two of the heaps, and weigh the remaining $3^{n-1}-1, 3^{n-1}-1, 3^{n-1}$ coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.1) If a pan with less coins rises, we know the other pan with less coins contains at least one heavy coin, so we identified the heap of $3^{n-1}$ coins with two heavy coins (including its removed coin).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.2) If no pan rises, we know the pan with more coins has the heavy heap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In either case, we have used 2 weighings to reduce the problem from $3^n$ coins to $3^{n-1}$ coins. We can apply this solution recursively. Since the reduced problem is solved in $2(n-1)-1$ weighings, our problem is solved in $2n-1$ weighings.&lt;/p&gt;&#xA;" OwnerUserId="318" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2015-07-08T21:05:00.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17616" PostTypeId="2" ParentId="17565" CreationDate="2015-07-08T21:06:33.970" Score="1" Body="&lt;p&gt;could it be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; money&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My first is in a bank,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Money is in the bank&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My second is an evergreen,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; US dollars are green&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My third is for your benefit&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everyone likes having money&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fourth is in a store,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you by stuff at stores with money&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fifth is identical,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Abraham Lincoln is on the five dollar bill and one cent coin&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you've found the pattern&#xA;What is my sixth?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; not sure what the sixth would be, possibly the $100 dollar bill&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nine is the limit,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; could refer to the people on the bills and coins not repeated and found in general circulation. i.e. not the $1 coin or $500/$2 bill&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My other idea could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the Nine planetary limits but I can't figure out how to relate the clues&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13922" LastEditorUserId="13922" LastEditDate="2015-07-09T12:34:13.740" LastActivityDate="2015-07-09T12:34:13.740" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17617" PostTypeId="1" CreationDate="2015-07-08T21:06:48.297" Score="-1" ViewCount="452" Body="&lt;p&gt;                                                                            &#xA;                                                                                                                      &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;     &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;     &#xA;    b25lIGVhc3k&#xA;    gb25lIHRvIH&#xA;    N0YXJ0Lg==&#xA;     &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA; &#xA;&#xA;&lt;pre&gt;&lt;code&gt;       &#xA;    cRIldI&#xA;    2oG0W/&#xA;    9aEIdI&#xA;    tWgHoQ&#xA;    Z5YRZ=&#xA;    XnmvX=&#xA;      &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;     &#xA;    k5WYoBSYgQmblxGIvRHI&#xA;    yVGd1BXbvNGIhBCa0l2d&#xA;    gQmbpZGIvRHI5NXYlByc&#xA;    pBSZu9GI0NXYsBycphGd&#xA;    cyBlYXN5IHRvIGZpbmQg&#xA;    d2l0aCBhIGNvbXB1dGVy&#xA;    IHRvIGxlbmQgYSBoYW5k&#xA;     &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;         &#xA;&#xA; &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h2&gt;Three locked boxes before you.&lt;/h2&gt;&#xA;&#xA;&lt;h1&gt;Can you unravel their mysteries and reveal the greater truth?&lt;/h1&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="11297" LastEditDate="2015-07-10T09:32:59.433" LastActivityDate="2015-07-17T04:13:27.227" Title="Enigmatic Boxes" Tags="|enigmatic-puzzle|computer-puzzle|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17618" PostTypeId="1" AcceptedAnswerId="17680" CreationDate="2015-07-08T21:23:49.877" Score="12" ViewCount="532" Body="&lt;p&gt;After several years at Private Eye Academy, Joey Solvesalot finally earned his detective license and was looking forward to solving some cases. So he decided to settle in the infamous Problem City to start his career. After successfully moving in without a problem, Joey decided to go to the police headquarters to get some background on the type of cases that occurred in the past. While reading through the cases on the computer, one in particular caught his eye:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Case Title&lt;/strong&gt;: Fowl Perpetrators Caught as Robberies Add Up &lt;br/&gt;&#xA;  Date of Incident: ??-??-???? &lt;br/&gt;&#xA;  Incident: Robbery &lt;br/&gt;&#xA;  ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He turned to an officer nearby, &quot;Why is the date of incident incomplete?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Officer Jenny glanced over and said, &quot;Hmm, we digitized our old cases a few years ago but some of the handwriting was illegible.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Ah I see,&quot; Joey murmured as he continued to look over the report. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...&lt;br/&gt;&#xA;  &lt;strong&gt;Offenders:&lt;/strong&gt;&lt;br/&gt;&#xA;  Name&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;Sex&amp;emsp;Age&amp;emsp;DOB&lt;br/&gt;&#xA;  Valentine, Holly&amp;emsp;F&amp;emsp; &amp;emsp;22 &amp;emsp;April 5, XXXX&lt;br/&gt;&#xA;  Winters, Jack   &amp;nbsp;&amp;emsp;M &amp;emsp;&amp;emsp;26&amp;emsp; August 20, XXXX&lt;br/&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Narrative&lt;/strong&gt;&lt;br/&gt;&#xA;  Avy Ian (52), a lover and collector of birds-related items, reported several cases of theft on her property over the past month. On the first Tuesday of the month, she noticed that her set of seven stone swans in her backyard near the pond was missing (estimated \$3500 total). Ten days later, she found the door to her chicken coop ajar and three hens missing. A week later, she discovered a broken window and her set of four glass songbirds (estimated \$10000 total) gone. To prevent any further thefts, we made sure there was at least one officer on lookout at all times. Six days later we caught Holly Valentine and Jack Winters trying to make off with six goose-egg sized jewels (estimated \$25000 each). Looking into their residence, we recovered the seven stone swans and four glass songbirds. We found feathers that presumably came from the stolen hens, but could not find said hens.&lt;br/&gt;&#xA;  ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once Joey finished reading, he smiled and said, &quot;Are we allowed to edit this? I figured out the date!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What day was the robbery (month and day) and why?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is the first puzzle I've ever made, so &lt;strike&gt;please don't hurt me&lt;/strike&gt; please feel free to suggest improvements!&lt;/p&gt;&#xA;" OwnerUserId="13881" LastEditorUserId="22922" LastEditDate="2016-09-15T07:16:44.897" LastActivityDate="2016-09-15T07:16:44.897" Title="Fowl Perpetrators Caught as Robberies Add Up" Tags="|knowledge|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17619" PostTypeId="2" ParentId="17618" CreationDate="2015-07-08T21:40:39.853" Score="2" Body="&lt;p&gt;Was it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;December 25th?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Each of the numbers of bird-related things stolen relate to 12 days of Christmas (3 French Hens, 4 Calling Birds, 6 Geese A-Laying, 7 Swans A-Swimming). Also there are 23 days between the first robbery and the last, which means it's possible for it to be Christmas.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I feel like my explanation is on the right track, but the jump to the answer is a bit tenuous.&lt;/p&gt;&#xA;" OwnerUserId="10861" LastActivityDate="2015-07-08T21:40:39.853" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17620" PostTypeId="2" ParentId="17486" CreationDate="2015-07-08T21:44:49.723" Score="1" Body="&lt;p&gt;I give credit to Cerberus for getting the first part of the answer, and I think I've worked out the 2nd lock.&#xA;First part:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &#xA; That means that Brazil --&gt; Chile --&gt; Argentina&#xA;  BRA --&gt; CHL --&gt; ARG   ! d(B,R) + d(R,A) = 16 + 17 = 33   ! d(C,H) + d(H,L) = 5 + 4 = 9   ! d(A,R) + d(R,G) = 17 + 11 = 28   ! so the code&#xA; is &lt;strong&gt;33-9-28&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second Part:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 33-24-13.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reasoning is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;For Everywhere, EWH&#xA;&lt;br&gt; This gives us:&#xA;&lt;br&gt;d(E,W) + d(W,H) = 18 + 15 = &lt;strong&gt;33;&lt;/strong&gt;&#xA;&lt;br&gt;&lt;br&gt;For Nowhere, NWH&#xA;&lt;br&gt; This gives us: &#xA;&lt;br&gt;d(N,W) + d(W,H) = 9 + 15 = &lt;strong&gt;24;&lt;/strong&gt;&#xA;&lt;br&gt;&lt;br&gt;For Home, HME&#xA;&lt;br&gt;This gives us:&#xA;&lt;br&gt;d(H, M) + d(M, E) = 5 + 8 = &lt;strong&gt;13;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reasoning for my choices is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because a combination lock needs to go forward, then back then forward again, I chose the letters which go in one direction, then switch back down.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13924" LastActivityDate="2015-07-08T21:44:49.723" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17621" PostTypeId="2" ParentId="17565" CreationDate="2015-07-08T21:49:06.370" Score="5" Body="&lt;p&gt;I don't think I know the full answer, but I feel like I'm on to the method of getting the answer. Perhaps someone could finish what I started here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each line refers to one word, as stated, but each word when said aloud is also a letter. So the full thing is a nine letter word (since nine is the limit) and the answer to the puzzle is the 6th letter of said word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;E.g.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Q in most banks a QUEUE could be found&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; U one type of evergreen is the YEW&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; O it is certainly in your benefit for me to OWE you something&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The fourth and fifth is where I get stuck. The fact that the fifth is identical implies that the fifth letter is the same as the fourth letter. First, I tried looking for all 9 letter words starting with quo. However, none have the 4th and 5th letter identical.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since that didn't work, let's try again:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; G as in grand. Any bank that doesn't have that much money is a pretty poor bank.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; U Let's stick with YEW since vowels are hard to come by and so do evergreens.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; T Many believe TEA to be beneficial to their health.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Back where we started. QUEUEs definitely are in stores, but count for me the number of words with qq in it. Here's a hint, there aren't any. :) But what 9 letter words start with GUT and then have 2 of the same letter... Again the answer is none. So, at least one of my first three in each idea are incorrect.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, I don't have the full answer. Who can help me? Am I on the right track?&lt;/p&gt;&#xA;" OwnerUserId="13923" LastActivityDate="2015-07-08T21:49:06.370" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17622" PostTypeId="2" ParentId="17617" CreationDate="2015-07-08T22:25:53.200" Score="6" Body="&lt;p&gt;It's super easy when you know about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Base64&quot;&gt;base64 encoding&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First box&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;one easy one to start.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second box&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; after transposing:&lt;br&gt;&#xA;&lt;br&gt;&#xA; &quot;something a bit tougher?!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Third box&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; after discarding 3 last lines (which are same as 3 first lines in reverse) and adding necessary ==&lt;br&gt;&#xA;&lt;br&gt;&#xA; &quot;this last one is easy to find with a computer to lend a hand&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13926" LastActivityDate="2015-07-08T22:25:53.200" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17623" PostTypeId="1" AcceptedAnswerId="17684" CreationDate="2015-07-08T22:29:56.723" Score="5" ViewCount="357" Body="&lt;p&gt;I liked the structure from A.D.'s yet to be answered riddle of similar name and composed something similar.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;First's cost is twice itself&lt;/p&gt;&#xA;&lt;p&gt;Second's of its name&lt;/p&gt;&#xA;&lt;p&gt;Third's tiniest of all&lt;/p&gt;&#xA;&lt;p&gt;Fourth's fourth&lt;/p&gt;&#xA;&lt;p&gt;Fifth's whole&lt;/p&gt;&#xA;&lt;p&gt;What is sixth?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Skip the half&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9696" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-09T19:41:02.753" Title="Nine is the American limit" Tags="|riddle|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17624" PostTypeId="2" ParentId="17623" CreationDate="2015-07-08T22:32:59.733" Score="5" Body="&lt;p&gt;This is about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; US coins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are 6 kinds&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; currently in circulation: the 1-cent, 5-cent, 10-cent, 25-cent, 50-cent, and 1-dollar coins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;First's cost is twice itself&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A 1-cent coin costs roughly 2 cents to produce (thanks @PlayerOne)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Second's of its name&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A 5-cent coin is called a nickel and is made partly of nickel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Third's tiniest of all&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The 10-cent coin is the smallest of all.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Fourth's fourth&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The 25-cent coin is called a quarter, and a quarter is a fourth or $\frac{1}{4}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Fifth's whole&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Given the hint &quot;skip the half&quot;, I think we're supposed to forget the 50-cent coin, so this is a &lt;em&gt;whole&lt;/em&gt; dollar.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What is sixth?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A 2-dollar coin?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-07-09T18:26:30.587" LastActivityDate="2015-07-09T18:26:30.587" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="17625" PostTypeId="2" ParentId="17486" CreationDate="2015-07-08T23:46:12.043" Score="1" Body="&lt;p&gt;I think the combination of the second lock would be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;21-7-8&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Following the logic that &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jeremy uses the phonetic sound to assign the short codes. So the codes are EVR, NOU and HOM. (From the first syllable of each word.)&lt;br&gt;&lt;br&gt;Which gives US, d(E,V) + d(V,R) = 17 + 4 = 21&lt;br&gt;&lt;br&gt;d(N,O) + d(O,U) = 1 + 6 = 7&lt;br&gt;&lt;br&gt;d(H,O) + d(O,M) = 6 + 2 = 8&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-07-08T23:46:12.043" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17626" PostTypeId="1" AcceptedAnswerId="17646" CreationDate="2015-07-09T00:02:56.540" Score="9" ViewCount="522" Body="&lt;p&gt;I have always wanted to travel around the world.  So I quitted my job at &lt;em&gt;Malt&lt;/em&gt; Traders.  When I told my friends &lt;em&gt;Toni&lt;/em&gt;, &lt;em&gt;Mark&lt;/em&gt; and &lt;em&gt;Gary&lt;/em&gt; about my plan, they were like &quot;&lt;em&gt;No way&lt;/em&gt;!&quot;  I knew they would love to travel with me to the &lt;em&gt;many&lt;/em&gt; places that I would visit, but sorry, I had to do this alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I managed to visit all these beautiful places on my bike, even though sometimes it was such a &lt;em&gt;pain&lt;/em&gt; to get up in the morning, especially with the &lt;em&gt;rain&lt;/em&gt;.  Some of the places I stayed at cost an &lt;em&gt;arm&lt;/em&gt; and a leg.  I was not too happy about that, but at least they did not &lt;em&gt;ban&lt;/em&gt; my bike and felt very much like my old place near the &lt;em&gt;port&lt;/em&gt; back home.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One day, I rode past a frozen lake and saw a bunch of kids &lt;em&gt;ice&lt;/em&gt; skating.  At a distance, I noticed a &lt;em&gt;larus&lt;/em&gt; flapping its wings.  I have never seen one in real life before, so I got off my bike and &lt;em&gt;ran&lt;/em&gt; after it.  I hoped I did not provoke the &lt;em&gt;ire&lt;/em&gt; of the parents who was looking on.  Ignorance is the &lt;em&gt;key&lt;/em&gt; here.  When I finally stopped and checked my pedometer, I have already ran 2319 steps.  I looked around but could not see anyone anywhere any more.  I think I am lost.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Where have I been &lt;em&gt;and&lt;/em&gt; more importantly, where on earth am I?!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12332" LastActivityDate="2018-10-14T10:03:08.767" Title="Where on earth am I?" Tags="|enigmatic-puzzle|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17627" PostTypeId="1" AcceptedAnswerId="17628" CreationDate="2015-07-09T00:19:11.087" Score="33" ViewCount="7287" Body="&lt;p&gt;In the kingdom of Boolistan, every inhabitant is either a Knight, Knave or Normal. Knights can only make true statements, Knaves can only lie, and Normals must either tell the truth or lie.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Warmup&lt;/strong&gt;: The local tavern only allows Normals (no one can relax around Knights and Knaves). What can a Normal say to prove their identity?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Challenge&lt;/strong&gt;: Only knights can dine at King Arthur's Round Table. What can a Knight say to prove their identity?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Remarks&lt;/strong&gt;: In conventional logic, where every statement is either true or false, the challenge is impossible (since Normals can say anything). To make this doable, we allow &lt;a href=&quot;https://en.wikipedia.org/wiki/Circular_definition&quot;&gt;circular&lt;/a&gt; self-referential statements, like the famous example, &quot;this statement is false&quot;. Formally, a circular self-referential statement is an equation of the form &#xA;$$&#xA;s = f(x_1,\dots,x_n,s)&#xA;$$&#xA;where $x_1,\dots,x_n$ are grounded logical propositions (like &quot;I am a Knave&quot;),   $f$ is a Boolean function, and $s$ is a Boolean variable. We say that such a statement is True if setting $s=$ True makes the equation hold, and similarly say it is False if $s=$ False is a solution. This means some such statements are both True and False, while others are neither. For example, &quot;this statement is false&quot; would be the equation $s=\neg s$, which has no solutions, so is neither True nor False. On the other hand, &quot;this statement is true&quot; would be $s=s$, which is both True and False.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We then allow knights to say any True statement, Knaves to say any False statement, while Normals can say a statement as long as it is True or False or both.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorDisplayName="user13963" LastEditDate="2015-07-11T17:53:30.857" LastActivityDate="2015-07-12T17:34:36.363" Title="Knights, Knaves and Normals" Tags="|logical-deduction|liars|" AnswerCount="12" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17628" PostTypeId="2" ParentId="17627" CreationDate="2015-07-09T00:27:45.837" Score="32" Body="&lt;p&gt;For the warmup:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I am a knave&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Should do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the challenge:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;If I am not a knight, this is a lie&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This statement can only work iff the speaker is a knight, as otherwise it will lead to a logical paradox, which is neither true nor false. &lt;/p&gt;&#xA;" OwnerUserId="3987" LastEditorUserId="7529" LastEditDate="2015-07-09T07:05:10.497" LastActivityDate="2015-07-09T07:05:10.497" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="17630" PostTypeId="2" ParentId="17486" CreationDate="2015-07-09T03:06:36.573" Score="1" Body="&lt;p&gt;&lt;strong&gt;Lock #2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my guess:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since Everywhere, Nowhere and Home doesn't seem like there is a 3 letter code associated with it, I just work out&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The distances between each of the letters.  It's a positive number for every pair of letters that move forward e.g. d(E,V) &amp;amp; negative number for every pair that move backwards (e.g. d(V,E))&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For Everywhere, the numbers are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; d(E,V) = 17&lt;br&gt;&#xA; d(V,E) = -17&lt;br&gt;&#xA; d(E,R) = 13&lt;br&gt;&#xA; d(R,Y) = 7&lt;br&gt;&#xA; d(Y,W) = -2&lt;br&gt;&#xA; d(W,H) = -15&lt;br&gt;&#xA; d(H,E) = -3&lt;br&gt;&#xA; d(E,R) = 13&lt;br&gt;&#xA; d(R,E) = -13&lt;br&gt;&#xA;&lt;p&gt;Add them all up = 0&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For Nowhere, the numbers are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; d(N,O) = 1&lt;br&gt;&#xA; d(O,W) = 8&lt;br&gt;&#xA; d(W,H) = -15&lt;br&gt;&#xA; d(H,E) = 3&lt;br&gt;&#xA; d(E,R) = 13&lt;br&gt;&#xA; d(R,E) = -13&lt;br&gt;&#xA;&lt;p&gt;Add them all up = -9&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For Home, the numbers are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; d(H,O) = 7&lt;br&gt;&#xA; d(O,M) = -2&lt;br&gt;&#xA; d(M,E) = -8&lt;br&gt;&#xA;&lt;P&gt;Add them all up = -3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 39-0, 39-9, 39-3 = &lt;strong&gt;0, 30, 36&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Jeremy won't need to remember any 3 letters country code as long as he knows how to spell!&lt;/p&gt;&#xA;" OwnerUserId="12332" LastActivityDate="2015-07-09T03:06:36.573" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17631" PostTypeId="2" ParentId="17486" CreationDate="2015-07-09T03:52:40.140" Score="3" Body="&lt;p&gt;Novice Guest Attempt&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Lock #2,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 39-0-4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rationale&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everywhere: &lt;b&gt;39&lt;/b&gt;, the highest possible number&lt;br/&gt;&lt;br/&gt;&#xA; Nowhere: &lt;strong&gt;0&lt;/strong&gt;, the exact opposite of Everywhere (and using Cerberus' solution for Lock #1, I can't immediately see how a value of 0 is possible)&lt;br/&gt;&lt;br/&gt;&#xA; I treated Home as the original store, which used this code: BEL --&gt; DEU --&gt; FRA&lt;br/&gt;&#xA; I then shortened it to BDF, which has a value calculated by:&lt;br/&gt;&#xA; d(B,D) + d(D,F) = 2 + 2 = &lt;strong&gt;4&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13931" LastActivityDate="2015-07-09T03:52:40.140" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17632" PostTypeId="2" ParentId="17617" CreationDate="2015-07-09T03:57:12.533" Score="11" Body="&lt;p&gt;user3608068 has the Base64 figured out:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;one easy one to start.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This reads top to bottom, and then left to right &lt;br&gt;&quot;something a bit tougher?!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This reads right to left and bottom to top, but the bottom three lines should be discarded and add the == &lt;Br&gt;&quot;this last one is easy to find with a computer to lend a hand&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Since I happen to have a computer here, I went to the edit history and noticed that an image was removed: &lt;a href=&quot;https://i.stack.imgur.com/uoEIP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/uoEIP.png&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This image when filled in with a colour shows:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zFX4r.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I then had to go back to figure out where the morse code was.  In the markdown for the post there is a section that looks like this:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Hxf0S.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;The two dots are the &quot;dash&quot; and the single dots are the &quot;dot&quot; for the morse code.  The message is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;GOOD WORK FINDING THIS, THERE'S ONE MORE THING TO BE FOUND.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I believe this corresponds to the image I found above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The schmutz at the end of the image is actually a code.  The number of pixels in each shape is a letter:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1 12 1 13 14 23 15 18 4 7 15 15 4 2 5 25 5 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Heeding the breaks as spaces and converting to letters of the alphabet yields:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A LAMN WORD GOODBEYE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After I posted my original answer edit 3 was added to the original post.  It added a new block of code to the top of the post.  Using the tab character as the &quot;dash&quot; and the space character as the &quot;dot&quot; it is more code for:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;THE WISDOM YOU SEEK WAS YOURS ALL ALONG BUT ONLY IF YOU CAN EXPLAIN YOUR PATH HERE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I believe that I have adequately done this.&lt;/p&gt;&#xA;" OwnerUserId="1777" LastEditorUserId="1777" LastEditDate="2015-07-17T04:13:27.227" LastActivityDate="2015-07-17T04:13:27.227" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17633" PostTypeId="2" ParentId="17627" CreationDate="2015-07-09T05:57:23.663" Score="0" Body="&lt;p&gt;First post ever on here :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For the warmup I would say:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If 1+1=2 then I am a Normal.&#xA;&#xA;This works because if a Knight says this statement, it ends in a contradiction, same goes for the case when a Knave says it.  The statement is only true and valid when a Normal says it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For the challenge we can operate along the same lines:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If 1+1=2 then I am a Knight.&#xA;&#xA;Again the antecedent is necessarily true, so for the statement to be true and not a contradiction, then only a Knight could say this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hope all the formatting worked out properly!&lt;/p&gt;&#xA;" OwnerUserId="13936" LastActivityDate="2015-07-09T05:57:23.663" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17634" PostTypeId="2" ParentId="17627" CreationDate="2015-07-09T06:26:04.130" Score="-3" Body="&lt;p&gt;For the warmup:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Sometimes I am older than a day before, but sometimes I am younger.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the challenge, I can't think anything. Of course a knight can show his shield to prove, but I guess it doesn't count. Some other considerations below, within a spoiler:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The watchmen can ask a knight of a thing he doesn't know, but in such a way that would make Normals think it is common knowledge for Knights. The real Knight would answer &quot;I don't know&quot;, while a Normal would try to lie.&#xA; Also, for example, the question from the guard could be &quot;how much does beer in the tavern cost?&quot;. Because Knights do not enter taverns, they don't know - but a Normal knows and because he pretends to be a Knight, he would answer truth. However, it requires cooperation between the guard and a knight. It will also fail when Normals will find out that &quot;I don't know&quot; is the right answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10880" LastEditorUserId="10880" LastEditDate="2015-07-09T07:15:53.600" LastActivityDate="2015-07-09T07:15:53.600" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17636" PostTypeId="2" ParentId="17565" CreationDate="2015-07-09T07:37:08.357" Score="0" Body="&lt;p&gt;I'm not sure. Is it anyway related to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;nine Planets, since Nine is a limit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My Sixth is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Saturn.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To start with,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My First is in bank&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mercury - Mercury payment systems&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My second is an evergreen &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Venus - Venus glows green&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My third is for your benefit &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Earth - is for our benefit&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fourth is in a store &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mars - Chocolate&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fifth is identical &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jupiter - atmospheres of Saturn and Jupiter are almost identical&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13911" LastEditorUserId="19658" LastEditDate="2016-09-14T08:43:11.480" LastActivityDate="2016-09-14T08:43:11.480" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17637" PostTypeId="2" ParentId="17626" CreationDate="2015-07-09T08:12:39.937" Score="3" Body="&lt;p&gt;A Guess,&#xA;As per your question and description.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I rode past a frozen lake and saw a bunch of kids ice skating. At a distance, I noticed a larus flapping its wings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can see Larus in Norway (As per &lt;a href=&quot;https://en.wikipedia.org/wiki/Larus&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wiki&lt;/a&gt;)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;even though sometimes it was such a pain to get up in the morning&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Sometime you can see SUN at even on midnight at North Norway. so may be you feel pain when you wake up at morning (Sometime there is no SUN on morning) due to your body still not set his biological clock.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I hoped I did not provoke the ire of the parents who was looking on.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;ire&lt;/em&gt; is Middle English Word (īra) (As per &lt;a href=&quot;http://dictionary.reference.com/browse/ire&quot; rel=&quot;nofollow noreferrer&quot;&gt;This Link&lt;/a&gt;, South Norway)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When I told my friends Toni, Mark and Gary about my plan, they were like &quot;No way!&quot; I knew they would love to travel with me to the many places that I would visit, but sorry, I had to do this alone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I took common char. &quot;&lt;strong&gt;R&lt;/strong&gt;&quot; from your 2 friends name and place it in their like &quot;&lt;strong&gt;No way&lt;/strong&gt;&quot;. It would become &quot;&lt;strong&gt;No&lt;em&gt;r&lt;/em&gt;way&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So My guess, you are in&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Norway&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13707" LastEditorUserId="42452" LastEditDate="2018-10-14T10:03:08.767" LastActivityDate="2018-10-14T10:03:08.767" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="17638" PostTypeId="2" ParentId="17613" CreationDate="2015-07-09T08:30:12.860" Score="3" Body="&lt;p&gt;While this is not an encryption technique, you could consider using anagrams as names of the curses. Just twist the letters around to create words where you have syllables from common words. You could restrict it to anagrams of individual words for an easy level, and anagrams of the entire phrase for a tough one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Bailey M's example, &lt;code&gt;BLOOD RITUAL&lt;/code&gt; can be anagrammed to &lt;code&gt;DOBOL ALIRUT&lt;/code&gt;. Here, the words have been anagrammed separately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some tough ones that I made up. I encourage you to try to crack them before revealing the actual words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;DARIB TERTH&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DIRT BREATH&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;EDEELIN SYBEG&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BLEEDING EYES&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The next two have been made even harder by even changing the number of words after anagramming it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;LART ERS TENITH&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;ETERNAL THIRST&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;FLEEBEKENIS TADDWETT&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TWISTED BLACKENED FEET&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Anagramming the words altogether makes it uncrackable by most online anagram solvers too, thus forcing your audience to sit and sort out letters and decipher your hidden words. &lt;del&gt;One drawback of this method could be that you can't hide an encryption key through wordplay in the rest of your narrative.&lt;/del&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;br&gt; You can even omit some letters, which you would provide through the narrative. Letters like B, C, D, G, I, J, O, P, Q, T, U, V, X can easily be hidden with common words. Consider the example below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;.. carved on the wall were the words 'See and be seen'. As I lowered the torch beam, I saw the curse, written in blood, &lt;code&gt;ALOPIN GLENOSS&lt;/code&gt;...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; COLLAPSING BONES&lt;br&gt;The letters B and C come from the clue, whereas the other letters are part of the anagram.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-07-09T08:46:25.090" LastActivityDate="2015-07-09T08:46:25.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17639" PostTypeId="2" ParentId="17627" CreationDate="2015-07-09T08:50:57.057" Score="4" Body="&lt;p&gt;Warmup:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He can say &lt;strong&gt;&quot;(At least) Sometimes I lie.&quot;&lt;/strong&gt; - A knight can not say this, because he never lies. And a knave can not say it, because it would be the truth for him.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Challenge:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He can say &lt;strong&gt;&quot;My next statement will not be a lie&quot;&lt;/strong&gt; Since the knight will know for sure he can never lie. But the Normal cannot 100% know if his next statement might be a lie. He can try, but there could be any thinkable scenario where his next statement could be a lie. Since there is a non-zero chance for the Normal to lie or tell the truth on his next statement, he cannot make the claim, since it is neither true nor false, but a vague guess. And per the rules they can only state truth or lie not something unknown.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1998" LastEditorUserId="1998" LastEditDate="2015-07-09T10:09:28.543" LastActivityDate="2015-07-09T10:09:28.543" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17640" PostTypeId="2" ParentId="17423" CreationDate="2015-07-09T09:11:54.047" Score="5" Body="&lt;p&gt;Since this puzzle doesn't state that you die if you get the answer wrong, I'd go with&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Brute force until the door opens&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, the accepted answer just seem wrong to me, and here is why. To apply the skip cipher, &lt;a href=&quot;http://rumkin.com/tools/cipher/skip.php&quot;&gt;rumkin&lt;/a&gt; was used, and the &quot;deciphered message&quot; is in fact an error message as we can clearly see from the javascript source code&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;if (HasAFactorMatch(enctext.length, inc))&#xA;  return 'Skip has a prime factor that cleanly divides into ' + &#xA;     'the text length, so it can not be used.';&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;That's why using the interface controls to set the skip value, doesn't allow you to set certain numbers. So, if we implement our own skip cipher, or even do it by hand, we would probably never crack the code. &lt;/p&gt;&#xA;" OwnerUserId="13715" LastActivityDate="2015-07-09T09:11:54.047" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17643" PostTypeId="2" ParentId="17627" CreationDate="2015-07-09T10:54:38.863" Score="7" Body="&lt;p&gt;Warmup:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;I can lie.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For knights this is false, and for knaves this is true, so only Normals can say it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;I am a Normal.&quot; followed by &quot;I am not a Normal.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or any other pair of one truth and one falsehood. Normals are the only ones who can both lie and tell the truth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Challenge:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;If I am not a Knight, this is false.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Simply causes a paradox if the speaker is not a Knight. Since a paradox is not a truth nor a lie, Normals can't say it. (I did come up with this before seeing frodoskywalker's answer.)&lt;/p&gt;&#xA;" OwnerUserId="13162" LastActivityDate="2015-07-09T10:54:38.863" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17644" PostTypeId="1" AcceptedAnswerId="17730" CreationDate="2015-07-09T12:40:49.023" Score="5" ViewCount="1033" Body="&lt;p&gt;First let me tell you this. The first is easy, then it MIGHT get harder - depending on knowledge. This is all knowledge and no skill - take that as key. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I like to call this one the &quot;Stage Cipher&quot; Give it a chance, you might develop an idea... and of course, this is gonna be fun haha.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note down everything you do, everything you find. I doubt you're going to even need a clue for the first part. The rest comes automatically. Good luck!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;00110101 00110111 00110110 01001000 00110100 01001100 00110110 01001011 00110110 00110010 00110100 00110111 00110100 00110110 00110011 00110000 00110110 00110011 00110011 00110011 00110101 01001000 00110011 00110000 00110100 00111001 00110100 00111000 00110110 00111000 00110110 01010010 00110100 00111001 00110100 00110111 00110101 00110110 00110110 01000110 00110110 00110010 00110100 00111000 00110101 00110010 00110110 01011001 00110100 01001110 00110110 00111001 00110100 00110010 00110101 00110001 00110101 00111001 00110011 00110011 00110100 01000110 00110110 00110111 00110110 00110010 00110100 00111000 00110110 00111000 00110111 00110000 00110110 00110100 00110111 00111001 00110100 00110010 00110111 00110000 00110110 00110100 00110011 00110011 00110100 00110010 00110111 00110000 00110100 00111001 00110100 00110111 00110101 00110110 00110110 01010100 00110110 00110010 00110100 00111000 00110101 00110010 00110110 01010011 00110100 00111001 00110100 00111000 00110100 01010100 00110110 01010011 00110101 00111001 00110110 01001011 00110101 00110010 00110110 01010101 00110100 00111001 00110100 00111000 00110100 00110010 00110110 01010101 00110100 00111001 00110100 00111000 00110110 00111000 00110111 01001000 00110110 00110100 00110100 00111000 00110100 00110001 00110111 00110101&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Personal Statement:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to let you all know - those of who take part I am not demeaning any.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I feel like you guys deserve a clue after an hour of deciphering. Here you go:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://i.stack.imgur.com/voVPi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/voVPi.png&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://i.stack.imgur.com/QTd36.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/QTd36.png&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Rotors: I-II-III&lt;/li&gt;&#xA;&lt;li&gt;Rotor Start: MCK&lt;/li&gt;&#xA;&lt;li&gt;Rings: AAA&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;REMEMBER - Enigma doesn't have numbers. Eliminate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S - I don't have the answers on me right now so that's all I can give. They're at my workplace. I was bored, I'll give you some more better hints tomorrow but for now keep working :D&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You've grasped Enigma. Though you still have numbers. Hm, I wonder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you guys give up and want me to reveal the answer to the Stage Cipher?&lt;/p&gt;&#xA;" OwnerUserId="13949" LastEditorUserId="11110" LastEditDate="2019-08-28T11:18:34.340" LastActivityDate="2019-08-28T11:18:34.340" Title="Knowledge is key. Skill is not" Tags="|cipher|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="17645" PostTypeId="2" ParentId="17627" CreationDate="2015-07-09T12:42:53.837" Score="3" Body="&lt;p&gt;Challenge: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  A knight could say: &quot;That knight (points at known knight) can confirm I am a knight.&quot; This should work as long as King Arthur is a knight (only tells the truth) who started allowing/accepting other knights in at his table. Also that all the knights know all the other knights. Any normal or knave that tried to enter that used this line would be declined by the pointed at knight.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Warm up:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I am a knave you know...sometimes, I just like to say a lie, and just see what happens. Like this one time last week, I lied to this knight, and let me tell you...&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastActivityDate="2015-07-09T12:42:53.837" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17646" PostTypeId="2" ParentId="17626" CreationDate="2015-07-09T12:51:01.517" Score="13" Body="&lt;p&gt;The italicized words are fragments of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; European country names,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so you have been to:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Malt&lt;/em&gt;: Malta&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Toni&lt;/em&gt;: Estonia&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Mark&lt;/em&gt;: Denmark&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Gary&lt;/em&gt;: Hungary&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;No way&lt;/em&gt;: Norway&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;many&lt;/em&gt;: Germany&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;pain&lt;/em&gt;: Spain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;rain&lt;/em&gt;: Ukraine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;arm&lt;/em&gt;: Armenia (not strictly in Europe, but is a member of the Council of Europe)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;ban&lt;/em&gt;: Albania&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;port&lt;/em&gt;: Portugal&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;ice&lt;/em&gt;: Iceland&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;larus&lt;/em&gt;: Belarus&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;ran&lt;/em&gt;: France&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;ire&lt;/em&gt;: Ireland&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;key&lt;/em&gt;: Turkey&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Where are you now?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;and&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; several countries in Europe that contain &quot;and&quot;: Andorra, Bosnia and Herzegovina, Finland, Iceland (already visited), Ireland (already visited), the Netherlands, Poland, and Switzerland.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="42452" LastEditDate="2018-10-14T10:02:37.523" LastActivityDate="2018-10-14T10:02:37.523" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="17647" PostTypeId="2" ParentId="17627" CreationDate="2015-07-09T12:56:52.830" Score="0" Body="&lt;p&gt;In the case that the King is King Arthur, the King of Britons:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; from the Monty Python and the Holy Grail&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Normals' pass: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I haven't ever eaten ham before.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In reference to (real spoiler): &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;We eat ham and jam and spam a lot!&quot; (The Camelot Song)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Knights' pass:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It could be carried by an african swallow.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In reference to (real and big spoiler):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A guard answering another guard that coconuts cannot be carried by a european swallow, but maybe by an african swallow. In the end, we see that only Sir Bedevere and Sir Lancelot. We see Sir bedevere testing that early in the film; and Sir Lancelot probably doesn't need a pass since he is 'carried away easily'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second case might be invalid cause i totally forgot about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the aptly named Sir Not-appearing-in-this-film&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I assume normals would have have no idea about the subject.&lt;/p&gt;&#xA;" OwnerUserId="4247" LastEditorUserId="4247" LastEditDate="2015-07-09T18:42:12.113" LastActivityDate="2015-07-09T18:42:12.113" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17648" PostTypeId="2" ParentId="17540" CreationDate="2015-07-09T13:30:43.653" Score="23" Body="&lt;p&gt;An Inefficient Phrase™ is one in which all of its words, when typed out using a standard phone keypad, have 50% or more of their adjacent letters in the same group*.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*A group is the letters associated with the numbers 2 through 9 on a standard phone keypad. So, there are 8 groups in total.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was a degree more difficult (or took longer) to write text messages with old phones which used these pads, because you needed to wait a short amount of time (or press a button on more recent older phones) if you wanted to type out another letter in the same set as the previous letter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, &lt;code&gt;COMMONERS&lt;/code&gt; would require you to tap: &lt;code&gt;222 666 [WAIT] 6 [WAIT] 6 [WAIT] 666 [WAIT] 6 33 777 [WAIT] 7777&lt;/code&gt;, while typing &lt;code&gt;CIVILIANS&lt;/code&gt; would require: &lt;code&gt;222 444 888 444 555 444 2 66 7777&lt;/code&gt; (with no breaks in between). It should also be noted that from learning to type this way over time, you could write messages very quickly as long as you didn't require pausing to enter a letter in the same set as the previous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's inefficient because you have to pause/break your flow of typing many times to type out the letters which are in the same set that exist in the Inefficient Phrases™, as opposed to finding another word with more spaced out lettering which could be punched in faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/77xzm.png&quot; alt=&quot;Mobile phone keypad&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at the inefficient  words combined with the &quot;old school&quot; clue..&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;[&lt;strong&gt;DE&lt;/strong&gt;]B[&lt;strong&gt;UT&lt;/strong&gt;][&lt;strong&gt;ED&lt;/strong&gt;] T[&lt;strong&gt;ON&lt;/strong&gt;][&lt;strong&gt;IGH&lt;/strong&gt;]T&lt;/li&gt;&#xA;&lt;li&gt;[&lt;strong&gt;TUT&lt;/strong&gt;]O[&lt;strong&gt;RS&lt;/strong&gt;] [&lt;strong&gt;SP&lt;/strong&gt;][&lt;strong&gt;OON&lt;/strong&gt;][&lt;strong&gt;FED&lt;/strong&gt;]&lt;/li&gt;&#xA;&lt;li&gt;[&lt;strong&gt;BAB&lt;/strong&gt;]I[&lt;strong&gt;ED&lt;/strong&gt;] [&lt;strong&gt;BAB&lt;/strong&gt;][&lt;strong&gt;OON&lt;/strong&gt;]S&lt;/li&gt;&#xA;&lt;li&gt;[&lt;strong&gt;DEFE&lt;/strong&gt;]A[&lt;strong&gt;TS&lt;/strong&gt;] [&lt;strong&gt;DE&lt;/strong&gt;][&lt;strong&gt;MON&lt;/strong&gt;]S&lt;/li&gt;&#xA;&lt;li&gt;[&lt;strong&gt;DE&lt;/strong&gt;][&lt;strong&gt;AC&lt;/strong&gt;][&lt;strong&gt;ON&lt;/strong&gt;] C[&lt;strong&gt;HI&lt;/strong&gt;][&lt;strong&gt;DED&lt;/strong&gt;]&lt;/li&gt;&#xA;&lt;li&gt;C[&lt;strong&gt;OMMON&lt;/strong&gt;]E[&lt;strong&gt;RS&lt;/strong&gt;] [&lt;strong&gt;FE&lt;/strong&gt;]U[&lt;strong&gt;DED&lt;/strong&gt;]&lt;/li&gt;&#xA;&lt;li&gt;S[&lt;strong&gt;CA&lt;/strong&gt;]R[&lt;strong&gt;FED&lt;/strong&gt;] [&lt;strong&gt;ABA&lt;/strong&gt;]L[&lt;strong&gt;ON&lt;/strong&gt;]E&lt;/li&gt;&#xA;&lt;li&gt;[&lt;strong&gt;SPR&lt;/strong&gt;]Y [&lt;strong&gt;CA&lt;/strong&gt;]M[&lt;strong&gt;PS&lt;/strong&gt;]&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;etc.&lt;/p&gt;&#xA;" OwnerUserId="13951" LastEditorUserId="463" LastEditDate="2015-08-12T17:09:28.223" LastActivityDate="2015-08-12T17:09:28.223" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="17651" PostTypeId="1" AcceptedAnswerId="17655" CreationDate="2015-07-09T14:34:52.607" Score="1" ViewCount="2557" Body="&lt;p&gt;A man is alone in his flat on the 22nd floor of a building. In order to commit suicide, he jumps out of the window.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After falling a few floors, he hears a phone ringing inside the building. He dies of a heart attack before reaching the ground, why ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Facts :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The floor number isn't relevant.&lt;br&gt;&#xA; The man was in perfect health.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13953" LastEditorUserId="39" LastEditDate="2015-07-10T01:05:16.353" LastActivityDate="2015-07-10T01:05:16.353" Title="Dead before landing" Tags="|lateral-thinking|" AnswerCount="7" CommentCount="5" ClosedDate="2015-07-09T15:16:03.513" ContentLicense="CC BY-SA 3.0" />
  <row Id="17652" PostTypeId="2" ParentId="17651" CreationDate="2015-07-09T14:47:16.963" Score="2" Body="&lt;p&gt;Could be similar to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/11238/my-wife-has-died?rq=1&quot;&gt;My wife has died&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He believed that his wife is going to die because of a power outage. He then hears the phone ring realizes that the power is back and she'll be fine, yet he's killed himself for nothing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5169" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-07-09T14:47:16.963" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17653" PostTypeId="2" ParentId="17651" CreationDate="2015-07-09T14:47:27.860" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The man has made a suicide pact with his lover over the phone. After saying his goodbye, he jumps out the window to his death, knowing that his lover must be doing the same wherever she is. But then his phone that he left in the window rings, and no one else has that number but her. Why did she call back and break their pact? The grief of betrayal overtook him and led to an immediate cardiac arrest. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-07-09T14:47:27.860" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17654" PostTypeId="2" ParentId="17644" CreationDate="2015-07-09T14:48:19.010" Score="2" Body="&lt;p&gt;Decoding the grid&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; With ASCII interpretation of 8 bit binary numbers gives:&#xA;&#xA;&#xA; 576H4L6K6&#xA;247463063 &#xA;335H30494 &#xA;8686R4947 &#xA;566F62485 &#xA;26Y4N6942 &#xA;5159334F6&#xA;762486870 &#xA;647942706 &#xA;433427049 &#xA;47566T624 &#xA;8526S4948 &#xA;4T6S596K5 &#xA;26U494842&#xA;6U4948687 &#xA;H64484175&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is promising because it is all upper case alphanumeric. Now there must be a another stage of decoding this, but the grouping or numerical base is unclear. I'll update if I figure out more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 1&lt;/strong&gt;&#xA;From the added hint, it's clear to me that this stage of the cipher is something related to the &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; German enigima cipher&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't know much about how that works, but I believe the hint might be related the cipher key for this stage. However, the classic version of that works with only alphabetical characters...this stage has both alphabetic and numeric characters, so I'm thinking more.&lt;/p&gt;&#xA;" OwnerUserId="6188" LastEditorUserId="6188" LastEditDate="2015-07-09T15:26:54.367" LastActivityDate="2015-07-09T15:26:54.367" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17655" PostTypeId="2" ParentId="17651" CreationDate="2015-07-09T14:51:32.350" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The man lives in a post-apocalyptic world.  He believes he is the last person alive.  Unable to live with his loneliness and grief, he decides to commit suicide.  But the phone ringing indicates that he is not, in fact, the sole survivor, and the shock of this realization is what brings on the heart attack.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4385" LastEditorUserId="4385" LastEditDate="2015-07-09T14:56:54.920" LastActivityDate="2015-07-09T14:56:54.920" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17656" PostTypeId="2" ParentId="17651" CreationDate="2015-07-09T14:52:31.383" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He thinks he is the last Person on Earth, thats why he decides to commit suicide. When he hears his Phone ringing, he realizes he's not alone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13703" LastActivityDate="2015-07-09T14:52:31.383" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17657" PostTypeId="2" ParentId="17651" CreationDate="2015-07-09T14:56:46.840" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He is Neo in the matrix. There are Smiths in the building and he has no other option but to jump out of the window (The agents must not be allowed to get hold of Neo). He hears a phone ring inside the building, but it's not his phone! Neo dies in the matrix from a heart attack. (But that's OK, he'll respawn soon)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-07-09T15:32:03.827" LastActivityDate="2015-07-09T15:32:03.827" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17658" PostTypeId="2" ParentId="17651" CreationDate="2015-07-09T15:08:42.623" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He thought he was in a lucid dream and would like to feel how it is to&#xA; commit suicide by jumping off a building.  Only when he heard the&#xA; phone rings that he realize he's not in a dream...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2015-07-09T15:08:42.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17659" PostTypeId="2" ParentId="17605" CreationDate="2015-07-09T15:10:35.930" Score="6" Body="&lt;p&gt;My answer (sorry for my english) :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First part : &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Caesar spun in a 3 width box.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Use rotate cipher with a 90° left rotation and a box width of 3 applied to Caesar gives you &lt;strong&gt;eraacs&lt;/strong&gt;.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then second part :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;That's the answer to the alphabet, though the key and alphabet are reversed. And, 123456789 could well be a key,  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So using &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  a Gronsfeld Cipher with a key of 123456789 and &lt;strong&gt;eraacs&lt;/strong&gt; as keyword for the alphabet, remembering to reverse the keywords the alphabet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You decrypt &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Znw qbty zx uxtw wlb ibnsqh thygpozg nkbveni Rnn-Imed xo qjw qytjfa osa.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;into &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You want to type the symbol obtained through Alt-Five on the number pad.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the answer is :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ♣ &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13955" LastActivityDate="2015-07-09T15:10:35.930" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17660" PostTypeId="2" ParentId="17651" CreationDate="2015-07-09T15:14:04.707" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This man frequently dreams and in those dreams he is always deaf, which makes the sights and actions of his dreams all the more exciting.  Not particularly thrilled with a dream of being alone in an apartment, he decides to wake himself up with the fear of falling that stirs so many of us from sleep.&lt;br&gt;&lt;br&gt;After falling a few floors, however, he hears a phone ringing.  While a phone ringing a few floors below his flat usually wouldn't worry him too much, in this case it comes with the realization that he is, in fact, awake, and truly plummeting to his death.  This shock induces cardiac arrest and he dies before he hits the ground.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11452" LastEditorUserId="12700" LastEditDate="2015-07-09T15:15:50.603" LastActivityDate="2015-07-09T15:15:50.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17661" PostTypeId="2" ParentId="17627" CreationDate="2015-07-09T15:28:27.010" Score="1" Body="&lt;p&gt;Okay, this is what I got:&#xA;&lt;strong&gt;the warmup&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&quot;knights tell part of the truth.&quot;&lt;/strong&gt; Since knaves always lie, they can only say that knights tell part of a lie, and knights cannot determine whether a part of a truth is a whole truth, and are thus unable to answer. Of course, a knave could really be saying &quot;Knights tell part of a lie,&quot; in which case, they really say &quot;Knights tell part of a truth.&quot; The Normal can say so, because they can lie and tell the truth: Knights tell part of the truth and part of a lie.&#xA;&lt;br&gt;&lt;br&gt;The only surefire way is assuming that Normals &lt;em&gt;can tell lies and truths in the same sentence&lt;/em&gt;. In this case, the Normal says, &lt;strong&gt;&quot;I Lie and I tell truths.&quot;&lt;/strong&gt; A knight cannot lie and thus cannot admit so, a Knave cannot lie about telling the truth yet tell the truth about lying or vice versa, but a Normal can lie and tell the truth at once: They can lie about lying and tell the truth about being truthful, or lie about telling the truth and tell the truth about lying.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;the challenge&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Assuming that at least one knight knows the person trying to enter&lt;/strong&gt;, the person trying to answer can ask said Knight, &lt;strong&gt;&quot;Can I lie?&quot;&lt;/strong&gt; If a normal or knave, the other Knight will say &quot;Yes&quot;, otherwise &quot;No&quot;. We &lt;em&gt;could&lt;/em&gt; also assume so, since the dinner is for Knights only, any one there is a Knight, which helps solidify this answer.&#xA;&lt;br&gt;&lt;br&gt;However, assuming that &lt;strong&gt;no one knows&lt;/strong&gt; the person trying to enter, they can prove so in a two-step process: First, the guard asks if they can say the following sentence, which is written on a scroll: &lt;strong&gt;&quot;I can tell lies and truths in the same sentence.&quot;&lt;/strong&gt; A Knight cannot say a lie in the same sentence, and will answer &quot;No.&quot; A Knave cannot say a truth in a sentence, but will lie and say &quot;yes&quot; (note that it asks whether they can do BOTH, hence the knave isn't telling the truth about saying a lie and forming a contradiction). A Normal can either lie or tell the truth, and will answer either &quot;Yes&quot; or &quot;No.&quot;&#xA;&lt;br&gt;&lt;em&gt;If they answered &quot;No&quot;, then they are either a Knight or a Normal.&lt;/em&gt; From there, the gaurd hands them the following scroll: &lt;strong&gt;&quot;I tell part-truths and part-lies.&quot;&lt;/strong&gt; Under pain of death, they are told to read the scroll aloud. A knight cannot say a lie, even if only a part of a statement, and thus will answer &quot;I cannot.&quot; A Normal, however, under the threat of Death, will read the script to try and save his life, thus revealing his deceit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13697" LastActivityDate="2015-07-09T15:28:27.010" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17663" PostTypeId="2" ParentId="17613" CreationDate="2015-07-09T15:43:36.363" Score="0" Body="&lt;p&gt;&lt;strong&gt;The curse:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You must make only 1 statement that is either true or false. I will &#xA;  still let you speak once you make the statement. If you&#xA;  are silent, I will not poison you. If you say a true statement, I will&#xA;  not strangle you. If you say a false statement, I will not behead you.&#xA;  If you make a statement which is true or false, but I don't know&#xA;  which, I will not burn you to death. If you break my rules, I will&#xA;  kill you. If I can break my rules (excluding this one), I will.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example, if you say a known true statement, he is supposed to 'not strangle you'. However, if he 'can break the rules, he will', which implies he &lt;em&gt;will&lt;/em&gt; strangle you. Also 'I will still let you speak' is another rule which he will defy simply because he can, so you cannot make another statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you remain silent, he is supposed to 'poison you'. However, he will break his rules, so you will be unable to speak, and you will be poisoned. I hope the conditions are  clear enough as to why they seem 'unescapable'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;And the key to breaking it.....&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I will be beheaded and not strangled&quot;. This breaks rule number one&#xA; &quot;You must make only 1 statement that is either true or false&quot;. Now&#xA; because you broke his rules, 'he will kill you'. However, he can break&#xA; his rules as well, meaning he won't kill you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(There may be typing/literal errors, or some paradox I haven't noticed. Otherwise, it looks fine.)&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-07-09T15:43:36.363" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17664" PostTypeId="2" ParentId="17565" CreationDate="2015-07-09T15:58:52.083" Score="17" Body="&lt;p&gt;Your sixth is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; sane&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; each word is an iteration of the pattern S A _ E&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My first is in a bank,  My first is cautious&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; safe&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My second is an evergreen,  My second is wise&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; sage&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My third is for your benefit, My third is wine&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; sake&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fourth is in a store,    My fourth happens in the garage&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; sale&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fifth is identical,  My fifth is unchanged&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; same&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nine is the limit because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; there are only three other words that fit the pattern: sate, save, and Saxe&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4385" LastEditorUserId="4385" LastEditDate="2015-07-09T16:03:28.203" LastActivityDate="2015-07-09T16:03:28.203" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17665" PostTypeId="2" ParentId="17644" CreationDate="2015-07-09T16:12:20.670" Score="3" Body="&lt;p&gt;Since I am new to the site, I cannot comment on rajb's response but wanted to add this information for his consideration. If someone could please add spoilers to the following info and instruct me as to how to add them myself for the future, it would be appreciated. Spoilers ahead!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've gotten to the same place that rajb has and also know little about the machinations of the enigma machine. Doing very quick research I found this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://enigma.louisedade.co.uk/howitworks.html&quot; rel=&quot;nofollow&quot;&gt;http://enigma.louisedade.co.uk/howitworks.html&lt;/a&gt;&lt;br&gt;&lt;br&gt;It includes helpful information about the workings. It mentions the notch on each rotor for implementing the double step mechanism and I believe that is the significance of the small arrows on the second picture in the hint (they provide the rotation direction as well as location of the notches). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm still working through more info and will edit this post if/when I achieve it, but wanted to contribute to the search. Hopefully what I added was not immediately obvious info.&lt;/p&gt;&#xA;" OwnerUserId="13960" LastEditorUserId="1777" LastEditDate="2015-07-09T16:35:19.843" LastActivityDate="2015-07-09T16:35:19.843" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17666" PostTypeId="2" ParentId="17618" CreationDate="2015-07-09T16:31:11.790" Score="1" Body="&lt;p&gt;If Victor is on the right Track I'm probably not but I'll try anyway:&#xA;My answer lies on the premise that &quot;past month&quot; means it's not the month the report was written... (sorry If I am wrong there and with past month you meant the last ~30 days I'm not a native English speaker)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1st March&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Her first Discovery was on the first Tuesday of a month so it must be the 6th Or less.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 10 days later &amp;lt;= 16th the three hens were missing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A week later &amp;lt;= 23th there was the brocken window.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 6 days later &amp;lt;= 29th the thiefs were caught and the report was written.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So in conclusion&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since the report was written a month after the one with the robbery it has to be &quot;March the 1st&quot; because only February has less than 29 days&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Don't be irritated by the &quot;after that&quot; I just don't get how to make a multirow spoiler.&lt;/p&gt;&#xA;" OwnerUserId="13626" LastActivityDate="2015-07-09T16:31:11.790" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17676" PostTypeId="2" ParentId="17510" CreationDate="2015-07-09T17:13:05.750" Score="3" Body="&lt;p&gt;There is an easy solution using Euler's formula, V - E + F = 2, but it's not as elegant as xnor's solution.  Suppose there are k faces and that the sum of the sides on all the faces is S.  Then E = S/2 and since at least three faces meet at each vertex, V &amp;lt;= S/3.  Substituting into Euler's formula gives S/6 &amp;lt;= k-2.  S is smallest when the sides are 3,4,5,...,k+2 and it's easy to see that the inequality is never satisfied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the requirement is that for each face there is exactly one other face with the same number of sides?  Then there are finitely many solutions, and I've constructed one specific polyhedron.  Can you find all solutions?  Now Euler's formula comes into its own.  &lt;/p&gt;&#xA;" OwnerUserId="13962" LastActivityDate="2015-07-09T17:13:05.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17677" PostTypeId="2" ParentId="17471" CreationDate="2015-07-09T17:24:14.990" Score="0" Body="&lt;p&gt;Another 11 letter one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; UNKULUNKULU&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not present on dictionary.com unfortunately, but thought it was interesting enough to add here.&lt;/p&gt;&#xA;" OwnerUserId="4624" LastActivityDate="2015-07-09T17:24:14.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17678" PostTypeId="2" ParentId="17627" CreationDate="2015-07-09T18:09:17.250" Score="-1" Body="&lt;p&gt;Warm-up:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I am a knave.&quot;&#xA; The knave saying this would be telling the truth. A knight would be lying to say this. A normal could make this claim in a lie.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Challenge:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have an answer that I believe solves it without need of a paradox.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We know from the warm-up that we can identify a normal with certainty. Our Knight picks out a normal and points, then proclaims,&#xA; &quot;If she is being as truthful as I am, she would agree I'm a knight.&quot;&#xA; A knave's corresponding normal would indeed agree they are a knight (because she would be lying in this case), but the knave cannot say this because it would be the truth.&#xA; A knight's corresponding normal would be telling the truth, and so she would of course agree that they are a knight.&#xA; If a normal were to lie, their corresponding normal would also lie, agreeing that they are a knight. They cannot say this because it would be the truth.&#xA; If a normal were to tell the truth, their corresponding normal could not agree they are a knight because she must also tell the truth, thus they still cannot say this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13965" LastActivityDate="2015-07-09T18:09:17.250" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17679" PostTypeId="2" ParentId="17627" CreationDate="2015-07-09T18:25:11.620" Score="0" Body="&lt;p&gt;Warmup: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;You might as well let everyone in, since only Normals are allowed but there's already a Knight and a Knave in there.&quot; Since he's telling the truth and lying in the same sentence, he's capable of both truth and lies, and thus is a Normal. (Alternately, he could be a Knight truly telling that they've let in a Knight and a Knave by accident, but my answer assumes that the tavern's authentication procedures have not already been broken.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Challenge:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is more of a practical than logical solution, but I'd recommend the Knight bring along someone known to be a Knave as his squire, who can inversely vouch for him at the door - his knavish squire would say &quot;Don't let him in, this guy's no knight.&quot; :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12202" LastActivityDate="2015-07-09T18:25:11.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17680" PostTypeId="2" ParentId="17618" CreationDate="2015-07-09T18:26:41.603" Score="4" Body="&lt;p&gt;2nd guess&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; December 25th&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'Add Up' in the puzzle title refers to adding up the 2 other dates (DOBs) found in the puzzle to obtain the solution date. April (4) + August (8) = &lt;strong&gt;December&lt;/strong&gt; (12).  5 + 20 = &lt;strong&gt;25&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First guess&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; December 27th (the 3rd day of Christmas)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The thieves still had the four glass songbirds (calling birds) to be given starting on the 4th day of Christmas (December 28th) but not the three (French) hens which are to be given starting on the 3rd day of Christmas.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11078" LastEditorUserId="11078" LastEditDate="2015-07-10T02:17:17.550" LastActivityDate="2015-07-10T02:17:17.550" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17681" PostTypeId="1" AcceptedAnswerId="17703" CreationDate="2015-07-09T18:28:21.603" Score="22" ViewCount="138128" Body="&lt;p&gt;In a regular pentagram (5-pointed star), the angle in each point is 36 degrees, so the angles in all five points sum to 180 degrees:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bfWtQ.jpg&quot; alt=&quot;regular star&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;What about an irregular pentagram, such as the following?&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vvge8.jpg&quot; alt=&quot;irregular star&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now the angles might be all different from each other; the situation is much more complicated. &lt;strong&gt;Can you prove that the angles in all five points &lt;em&gt;still&lt;/em&gt; sum to 180 degrees?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Restrictions (to make clear that this is neither a maths &lt;em&gt;problem&lt;/em&gt; [as opposed to a maths puzzle] nor an exercise in computation or advanced Euclidean geometry):&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;no arithmetic operations allowed (addition, multiplication, ...)&lt;/li&gt;&#xA;&lt;li&gt;you may draw one new line-segment on the star, but no more than that&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="25966" LastEditDate="2023-04-22T18:25:55.750" LastActivityDate="2023-04-22T18:27:22.623" Title="Five Angles in a Star" Tags="|mathematics|geometry|" AnswerCount="10" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="17682" PostTypeId="2" ParentId="17681" CreationDate="2015-07-09T18:32:40.647" Score="8" Body="&lt;p&gt;The sum of the interior angles of a pentagon is always 540°.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angle of each exterior point is always the  sum of the two adjacent interior angles - 180°. We can say this since, given internal angles A and B, the angles of the triangle are 180 - A, 180 - B, X.  By definition of angles of a triangle, X is equal to $180 - (180 - A) - (180 - B) = A + B - 180$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each interior angle of the pentagon is used twice, and there are 5 points, so $(2 \times 540) - (5 \times 180) = 180°$&lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-07-09T18:41:45.730" LastActivityDate="2015-07-09T18:41:45.730" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17683" PostTypeId="2" ParentId="17681" CreationDate="2015-07-09T18:54:46.553" Score="17" Body="&lt;p&gt;Here's another proof.&lt;/p&gt;&#xA;&lt;p&gt;Label the points as shown, and draw the line-segment CD. Use A, B, etc. to denote the angles we are asked to find the sum of.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NomDF.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$\angle ADC + \angle DCA + A = 180^\circ$&lt;/span&gt; (angles in a triangle)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So it is sufficient to prove that&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$\angle ADC + \angle DCA = B + C + D + E$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Now&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$\angle ADC = D + \angle BDC$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\angle DCA = C + \angle ECD$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So it is sufficient to prove that&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$\angle BDC + \angle ECD = B + E$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;which is obviously true because&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; the LHS is the supplement of &lt;span class=&quot;math-container&quot;&gt;$\angle DFC$&lt;/span&gt; and the RHS is the supplement of &lt;span class=&quot;math-container&quot;&gt;$\angle EFB$&lt;/span&gt;, where  &lt;span class=&quot;math-container&quot;&gt;$\angle DFC$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\angle EFB$&lt;/span&gt; are equal because vertically opposite.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7363" LastEditorUserId="25966" LastEditDate="2023-04-22T18:26:46.980" LastActivityDate="2023-04-22T18:26:46.980" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="17684" PostTypeId="2" ParentId="17623" CreationDate="2015-07-09T19:08:11.713" Score="3" Body="&lt;p&gt;The nine are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Commonly used forms of currency in the US&lt;br&gt;&#xA; Coins: penny (1 cent), nickel (5 cents), dime (10 cents), quarter (25 cents)&lt;br&gt;&#xA; Bills: 1 dollar, 5 dollar, 10 dollar, 20 dollar, 100 dollar&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First's cost is twice itself&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The metal used in a penny costs roughly twice as much as the face value.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second's of its name&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A nickel is made of 25% nickel&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Third's tiniest of all&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A dime is the smallest coin&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fourth's fourth&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The quarter is a fourth of a dollar&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fifth's whole&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The one dollar bill is a &lt;strong&gt;whole&lt;/strong&gt; dollar&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is sixth?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The five dollar bill&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12109" LastEditorUserId="12109" LastEditDate="2015-07-09T19:41:02.753" LastActivityDate="2015-07-09T19:41:02.753" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="17685" PostTypeId="2" ParentId="17681" CreationDate="2015-07-09T19:57:25.053" Score="4" Body="&lt;p&gt;Another approach:&lt;/p&gt;&#xA;&lt;p&gt;Starting with the regular star, we know that &lt;span class=&quot;math-container&quot;&gt;$A + B + C + D + E = 180^{\circ}$&lt;/span&gt;.&lt;br /&gt;&#xA;Now let's draw a line segment as shown in the diagram.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dh4iP.png&quot; alt=&quot;star&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Note that &lt;span class=&quot;math-container&quot;&gt;$B, D, E$&lt;/span&gt; remain unchanged! From our observations, we see that &lt;span class=&quot;math-container&quot;&gt;$Y = C - X$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$Z = A + X$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Thus the sum of the points of our new star &lt;span class=&quot;math-container&quot;&gt;$ZBYDE = Z + B + Y + D + E = (A+ X) + B + (C-X) + D + E = 180^{\circ}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;So we can continue to draw segments and create new stars (and rescale them) until we reach the shape desired.&lt;/p&gt;&#xA;" OwnerUserId="13881" LastEditorUserId="25966" LastEditDate="2023-04-22T18:25:38.993" LastActivityDate="2023-04-22T18:25:38.993" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="17686" PostTypeId="2" ParentId="1963" CreationDate="2015-07-09T20:00:32.643" Score="-3" Body="&lt;p&gt;$2\times 2\times 2=6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3\times 3-3=6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{(4\times 4)}4=6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5+(\frac55)=6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$6+6-6=6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$7-(\frac77)=6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{(8\times 8)}8=6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$9-(\frac9{\sqrt{9}})=6$&lt;/p&gt;&#xA;" OwnerUserId="13969" LastEditorUserId="11497" LastEditDate="2018-05-06T03:03:19.343" LastActivityDate="2018-05-06T03:03:19.343" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="17687" PostTypeId="2" ParentId="166" CreationDate="2015-07-09T20:02:15.080" Score="0" Body="&lt;p&gt;Single Rope Solution&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. Fold a rope in half&lt;br&gt;&#xA; 2. Light one end&lt;br&gt;&#xA; 3. When fire reaches fold, light other end&lt;br&gt;&#xA; 4. Fires will meet at 3/4 length of rope or 45min&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13788" LastEditorUserId="11497" LastEditDate="2018-09-07T02:49:22.907" LastActivityDate="2018-09-07T02:49:22.907" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="17688" PostTypeId="2" ParentId="17681" CreationDate="2015-07-09T20:24:36.700" Score="6" Body="&lt;p&gt;Here's another neat proof, this time by induction. We can make the pentagram by starting with the regular one and successively moving four of the points. So it is sufficient to prove that&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; moving a point in a pentagram doesn't change the sum of the angles at the points&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Let's&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; move point A to A' and call both the angle at A and the angle at A' the top angle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;We get this:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/V75jy.jpg&quot; alt=&quot;Diagram 1&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;It is sufficient to prove that&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; the change in the top angle and the changes in the angles at C and D sum to zero.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;On this new diagram&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/sXkC4.jpg&quot; alt=&quot;Diagram 2&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;we show&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; the change in the top angle as &lt;span class=&quot;math-container&quot;&gt;$x-w$&lt;/span&gt; and the changes at D and C as &lt;span class=&quot;math-container&quot;&gt;$-y$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt;,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;and we need to prove that&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$x-w-y+z=0$&lt;/span&gt;, or in other words, that &lt;span class=&quot;math-container&quot;&gt;$x+z=w+y$&lt;/span&gt;,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;which is obvious, as before, because&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; LHS and RHS are the complements of vertically opposite angles at G.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7363" LastEditorUserId="25966" LastEditDate="2023-04-22T18:27:05.980" LastActivityDate="2023-04-22T18:27:05.980" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="17689" PostTypeId="2" ParentId="17545" CreationDate="2015-07-09T20:38:50.540" Score="1" Body="&lt;p&gt;Not sure if this is valid, but the word is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://dictionary.reference.com/browse/pneumonoultramicroscopicsilicovolcanoconiosis&quot; rel=&quot;nofollow&quot;&gt;Pneumonoultramicroscopicsilicovolcanoconiosis&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;45 letters!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which can be divided in:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://dictionary.reference.com/browse/pneumo&quot; rel=&quot;nofollow&quot;&gt;Pneumo&lt;/a&gt; &lt;a href=&quot;http://dictionary.reference.com/browse/no&quot; rel=&quot;nofollow&quot;&gt;no&lt;/a&gt; &lt;a href=&quot;http://dictionary.reference.com/browse/ultra&quot; rel=&quot;nofollow&quot;&gt;ultra&lt;/a&gt; &lt;a href=&quot;http://dictionary.reference.com/browse/microscopic&quot; rel=&quot;nofollow&quot;&gt;microscopic&lt;/a&gt; &lt;a href=&quot;http://dictionary.reference.com/browse/si&quot; rel=&quot;nofollow&quot;&gt;si&lt;/a&gt; &lt;a href=&quot;http://dictionary.reference.com/browse/lic&quot; rel=&quot;nofollow&quot;&gt;lic&lt;/a&gt; &lt;a href=&quot;http://dictionary.reference.com/browse/o&quot; rel=&quot;nofollow&quot;&gt;o&lt;/a&gt; &lt;a href=&quot;http://dictionary.reference.com/browse/volcano&quot; rel=&quot;nofollow&quot;&gt;volcano&lt;/a&gt; &lt;a href=&quot;http://dictionary.reference.com/browse/coniosis&quot; rel=&quot;nofollow&quot;&gt;coniosis&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, this can be divided as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://dictionary.reference.com/browse/pneumo&quot; rel=&quot;nofollow&quot;&gt;Pneumo&lt;/a&gt; &lt;a href=&quot;http://dictionary.reference.com/browse/no&quot; rel=&quot;nofollow&quot;&gt;no&lt;/a&gt; &lt;a href=&quot;http://dictionary.reference.com/browse/ultra&quot; rel=&quot;nofollow&quot;&gt;ultra&lt;/a&gt; &lt;a href=&quot;http://dictionary.reference.com/browse/microscopic&quot; rel=&quot;nofollow&quot;&gt;microscopic&lt;/a&gt; &lt;a href=&quot;http://dictionary.reference.com/browse/sil&quot; rel=&quot;nofollow&quot;&gt;sil&lt;/a&gt; &lt;a href=&quot;http://dictionary.reference.com/browse/ico&quot; rel=&quot;nofollow&quot;&gt;ico&lt;/a&gt; &lt;a href=&quot;http://dictionary.reference.com/browse/volcano&quot; rel=&quot;nofollow&quot;&gt;volcano&lt;/a&gt; &lt;a href=&quot;http://dictionary.reference.com/browse/coniosis&quot; rel=&quot;nofollow&quot;&gt;coniosis&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;And, in order to really abuse the spirit of the question, this can be divided even in this way:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://dictionary.reference.com/browse/p&quot; rel=&quot;nofollow&quot;&gt;P&lt;/a&gt; &lt;a href=&quot;http://dictionary.reference.com/browse/n&quot; rel=&quot;nofollow&quot;&gt;n&lt;/a&gt; &lt;a href=&quot;http://dictionary.reference.com/browse/e&quot; rel=&quot;nofollow&quot;&gt;e&lt;/a&gt; &lt;a href=&quot;http://dictionary.reference.com/browse/u&quot; rel=&quot;nofollow&quot;&gt;u&lt;/a&gt; &lt;a href=&quot;http://dictionary.reference.com/browse/m&quot; rel=&quot;nofollow&quot;&gt;m&lt;/a&gt; &lt;a href=&quot;http://dictionary.reference.com/browse/o&quot; rel=&quot;nofollow&quot;&gt;o&lt;/a&gt; n o &lt;a href=&quot;http://dictionary.reference.com/browse/u&quot; rel=&quot;nofollow&quot;&gt;u&lt;/a&gt; &lt;a href=&quot;http://dictionary.reference.com/browse/l&quot; rel=&quot;nofollow&quot;&gt;l&lt;/a&gt; &lt;a href=&quot;http://dictionary.reference.com/browse/t&quot; rel=&quot;nofollow&quot;&gt;t&lt;/a&gt; &lt;a href=&quot;http://dictionary.reference.com/browse/r&quot; rel=&quot;nofollow&quot;&gt;r&lt;/a&gt; &lt;a href=&quot;http://dictionary.reference.com/browse/a&quot; rel=&quot;nofollow&quot;&gt;a&lt;/a&gt; m &lt;a href=&quot;http://dictionary.reference.com/browse/i&quot; rel=&quot;nofollow&quot;&gt;i&lt;/a&gt; &lt;a href=&quot;http://dictionary.reference.com/browse/c&quot; rel=&quot;nofollow&quot;&gt;c&lt;/a&gt; r o &lt;a href=&quot;http://dictionary.reference.com/browse/s&quot; rel=&quot;nofollow&quot;&gt;s&lt;/a&gt; c o p i c o s i l i c o &lt;a href=&quot;http://dictionary.reference.com/browse/v&quot; rel=&quot;nofollow&quot;&gt;v&lt;/a&gt; u l c a n o c o n i o s i s&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Source: &lt;a href=&quot;https://en.wikipedia.org/wiki/Pneumonoultramicroscopicsilicovolcanoconiosis&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Pneumonoultramicroscopicsilicovolcanoconiosis&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5044" LastEditorUserId="5044" LastEditDate="2015-07-09T22:52:11.500" LastActivityDate="2015-07-09T22:52:11.500" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17690" PostTypeId="2" ParentId="17247" CreationDate="2015-07-09T21:58:14.887" Score="3" Body="&lt;p&gt;Using Chadi Abou Sleiman's answer to part one, I got &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; to be won then to impede to be steeled perhaps gold&lt;br&gt;when taken by their fourths information they will hold&lt;br&gt;a party perceived differently with ear than with the eye&lt;br&gt;when pared down to essentials shattered secrets within lie&lt;br&gt;&#xA;&lt;br&gt;&#xA; the key is the answer which you already know&lt;br&gt;with vigeneres help find the question below&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Method:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reversed the ciphertext (because of &quot;Whoops! His head is his foot&quot;) and then used letter substitution with some frequency analysis.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13928" LastEditorUserId="1777" LastEditDate="2015-07-10T05:37:13.717" LastActivityDate="2015-07-10T05:37:13.717" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17691" PostTypeId="2" ParentId="17627" CreationDate="2015-07-09T22:08:36.073" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What can a Knight say to prove their identity?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's assume that every inhabitant of Boolistan has to obey the King (the penalty for disobedience is death, obviously), and the Knight's in question name is Sofa. Thus, the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To prove his identity, the Knight in question has to first say to the King to give him [i.e. the Knight] an order: &gt;&gt;My King, order me to, literally, &quot;Say &lt;strong&gt;I am Sofa, King, a Knight&lt;/strong&gt;, but if, and only if, you're a Knight. Otherwise remain silent.&quot;&amp;lt;&amp;lt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then the King has to do what the Knight asked for. Then, obviously, if he's a Knight, he'll do it, and it will be &lt;a href=&quot;https://www.youtube.com/watch?v=zZ9dtZ8lYww&quot; rel=&quot;nofollow&quot;&gt;true&lt;/a&gt;. If he isn't, he won't say anything, because he had been given a direct order to remain silent.&lt;/p&gt;&#xA;" OwnerDisplayName="user13963" LastEditorDisplayName="user13963" LastEditDate="2015-07-11T16:53:35.657" LastActivityDate="2015-07-11T16:53:35.657" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17692" PostTypeId="2" ParentId="17247" CreationDate="2015-07-09T22:09:12.500" Score="3" Body="&lt;p&gt;The question to ask is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; How do I like my milkshakes?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which can be got by applying&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the Vigenere cipher with keyword &quot;nzogvw&quot; (or &quot;lzksdc&quot; if we're subtracting instead of adding)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;to the ciphertext &quot;Toh ws L xivx qb yiwdwkmkpl?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how to get that keyword, since I 'cheated', working this out by brute force (by hand, I should add) and a few educated guesses. More explicitly:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - A capital single-letter word is obviously &quot;I&quot;; that gives us one letter of the key.&lt;br&gt; - Then a two-letter word with second letter &quot;y&quot; is likely to be &quot;my&quot;, giving us another letter of the key.&lt;br&gt; - Then a two-letter word with second-letter &quot;o&quot; is likely to be &quot;do&quot;, giving us a third letter of the key.&lt;br&gt; - Then a three-letter word preceding &quot;do I&quot; is likely to be &quot;How&quot;, giving us the rest of the key.&lt;br&gt;&lt;br&gt;Bingo - it worked!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-07-09T22:18:01.583" LastActivityDate="2015-07-09T22:18:01.583" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17693" PostTypeId="2" ParentId="17486" CreationDate="2015-07-09T22:29:49.503" Score="2" Body="&lt;p&gt;Posting Cerberus' explanation verbatim to explain the rules required to understand lock 1 (and so multiple answers don't have to be referenced).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You need to take the three-letter code of each country and calculate the 'distance' (d) between 2 consecutive letters, and finally add them.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Belgium --&gt; Germany --&gt; France&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BEL --&gt; DEU --&gt; FRA&lt;br&gt;&#xA; d(B,E) + d(E,L) = 3 + 7 = 10&lt;br&gt;&#xA; d(D,E) + d(E,U) = 1 + 16 = 17&lt;br&gt;&#xA; d(F,R) + d(R,A) = 12 + 17 = 29&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Saudi Arabia --&gt; Iran --&gt; Qatar&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SAU --&gt; IRN --&gt; QAT&lt;br&gt;&#xA; d(S,A) + d(A,U) = 18 + 20 = 38&lt;br&gt;&#xA; d(I,R) + d(R,N) = 9 + 4 = 13&lt;br&gt;&#xA; d(Q,A) + d(A,T) = 16 + 19 = 35&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That means that Brazil --&gt; Chile --&gt; Argentina&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BRA --&gt; CHL --&gt; ARG&lt;br&gt;&#xA; d(B,R) + d(R,A) = 16 + 17 = 33&lt;br&gt;&#xA; d(C,H) + d(H,L) = 5 + 4 = 9&lt;br&gt;&#xA; d(A,R) + d(R,G) = 17 + 11 = 28&lt;br&gt;&#xA; so the code is &lt;strong&gt;33-9-28&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Quark's solution to Lock 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nowhere is 0, Everywhere is 39, and Home is 1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;/* Author: Quark */&#xA;&#xA;#include &amp;lt;stdio.h&amp;gt;&#xA;#include &amp;lt;strings.h&amp;gt;&#xA;&#xA;int main()&#xA;{&#xA;    char code[4];&#xA;    int number[3];&#xA;    int value;&#xA;    int i;&#xA;    int zero=0,one=0,two=0,three=0,four=0,five=0,six=0,seven=0,eight=0,nine=0,ten=0,&#xA;        eleven=0,twelve=0,thirteen=0,fourteen=0,fifteen=0,sixteen=0,seventeen=0,&#xA;        eighteen=0,nineteen=0,twenty=0,twentyone=0,twentytwo=0,twentythree=0,twentyfour=0,&#xA;        twentyfive=0,twentysix=0,twentyseven=0,twentyeight=0,twentynine=0,thirty=0,&#xA;        thirtyone=0,thirtytwo=0,thirtythree=0,thirtyfour=0,thirtyfive=0,thirtysix=0,&#xA;        thirtyseven=0,thirtyeight=0,thirtynine=0,fourty=0,over = 0;&#xA;&#xA;    while(1){&#xA;        gets(code);&#xA;        if(code[0]=='0')&#xA;            printf(&quot;%d %d %d %d %d %d %d %d %d %d %d %d %d %d %d %d %d %d %d %d %d %d %d %d %d %d %d %d %d %d %d %d %d %d %d %d %d %d %d %d %d %d\n&quot;, zero,one,two,three,four,five,six,seven,eight,nine,ten,&#xA;        eleven,twelve,thirteen,fourteen,fifteen,sixteen,seventeen,&#xA;        eighteen,nineteen,twenty,twentyone,twentytwo,twentythree,twentyfour,&#xA;        twentyfive,twentysix,twentyseven,twentyeight,twentynine,thirty,&#xA;        thirtyone,thirtytwo,thirtythree,thirtyfour,thirtyfive,thirtysix,&#xA;        thirtyseven,thirtyeight,thirtynine,fourty,over);&#xA;&#xA;        for(i=0;i&amp;lt;3;i++)&#xA;        {&#xA;            if (code[i] &amp;gt;= 'A' &amp;amp;&amp;amp; code[i] &amp;lt;= 'Z')&#xA;                number[i] = code[i] - 'A' + 1;&#xA;            else if (code[i] &amp;gt;= 'a' &amp;amp;&amp;amp; code[i] &amp;lt;= 'z')&#xA;                number[i] = code[i] - 'a';&#xA;        }&#xA;        value = abs(number[0]-number[1]) + abs(number[1]-number[2]);&#xA;        printf(&quot;%d\n&quot;, value);&#xA;        switch(value)&#xA;        {&#xA;        case 0:&#xA;        zero++;&#xA;        break;&#xA;&#xA;        case 1:&#xA;        one++;&#xA;        break;&#xA;&#xA;        case 2:&#xA;        two++;&#xA;        break;&#xA;&#xA;        case 3:&#xA;        three++;&#xA;        break;&#xA;&#xA;        case 4:&#xA;        four++;&#xA;        break;&#xA;&#xA;        case 5:&#xA;        five++;&#xA;        break;&#xA;&#xA;        case 6:&#xA;        six++;&#xA;        break;&#xA;&#xA;        case 7:&#xA;        seven++;&#xA;        break;&#xA;&#xA;        case 8:&#xA;        eight++;&#xA;        break;&#xA;&#xA;        case 9:&#xA;        nine++;&#xA;        break;&#xA;&#xA;        case 10:&#xA;        ten++;&#xA;        break;&#xA;&#xA;        case 11:&#xA;        eleven++;&#xA;        break;&#xA;&#xA;        case 12:&#xA;        twelve++;&#xA;        break;&#xA;&#xA;        case 13:&#xA;        thirteen++;&#xA;        break;&#xA;&#xA;        case 14:&#xA;        fourteen++;&#xA;        break;&#xA;&#xA;        case 15:&#xA;        fifteen++;&#xA;        break;&#xA;&#xA;        case 16:&#xA;        sixteen++;&#xA;        break;&#xA;&#xA;        case 17:&#xA;        seventeen++;&#xA;        break;&#xA;&#xA;        case 18:&#xA;        eighteen++;&#xA;        break;&#xA;&#xA;        case 19:&#xA;        nineteen++;&#xA;        break;&#xA;&#xA;        case 20:&#xA;        twenty++;&#xA;        break;&#xA;&#xA;        case 21:&#xA;        twentyone++;&#xA;        break;&#xA;&#xA;        case 22:&#xA;        twentytwo++;&#xA;        break;&#xA;&#xA;        case 23:&#xA;        twentythree++;&#xA;        break;&#xA;&#xA;        case 24:&#xA;        twentyfour++;&#xA;        break;&#xA;&#xA;        case 25:&#xA;        twentyfive++;&#xA;        break;&#xA;&#xA;        case 26:&#xA;        twentysix++;&#xA;        break;&#xA;&#xA;        case 27:&#xA;        twentyseven++;&#xA;        break;&#xA;&#xA;        case 28:&#xA;        twentyeight++;&#xA;        break;&#xA;&#xA;        case 29:&#xA;        twentynine++;&#xA;        break;&#xA;&#xA;        case 30:&#xA;        thirty++;&#xA;        break;&#xA;&#xA;        case 31:&#xA;        thirtyone++;&#xA;        break;&#xA;&#xA;        case 32:&#xA;        thirtytwo++;&#xA;        break;&#xA;&#xA;        case 33:&#xA;        thirtythree++;&#xA;        break;&#xA;&#xA;        case 34:&#xA;        thirtyfour++;&#xA;        break;&#xA;&#xA;        case 35:&#xA;        thirtyfive++;&#xA;        break;&#xA;&#xA;        case 36:&#xA;        thirtysix++;&#xA;        break;&#xA;&#xA;        case 37:&#xA;        thirtyseven++;&#xA;        break;&#xA;&#xA;        case 38:&#xA;        thirtyeight++;&#xA;        break;&#xA;&#xA;        case 39:&#xA;        thirtynine++;&#xA;        break;&#xA;&#xA;        case 40:&#xA;        fourty++;&#xA;        break;&#xA;&#xA;        default:&#xA;        over++;&#xA;        break;&#xA;&#xA;        }&#xA;    }&#xA;    return(1337);&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Running this code with a list of country codes returns:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;AFG&#xA;6&#xA;ALA&#xA;22&#xA;ALB&#xA;21&#xA;DZA&#xA;47&#xA;ASM&#xA;24&#xA;AND&#xA;23&#xA;AGO&#xA;14&#xA;AIA&#xA;16&#xA;ATG&#xA;32&#xA;ARG&#xA;28&#xA;ARM&#xA;22&#xA;ABW&#xA;22&#xA;AUS&#xA;22&#xA;AUT&#xA;21&#xA;AZE&#xA;46&#xA;BHS&#xA;17&#xA;BHR&#xA;16&#xA;BGD&#xA;8&#xA;BRB&#xA;32&#xA;BLR&#xA;16&#xA;BEL&#xA;10&#xA;BLZ&#xA;24&#xA;BEN&#xA;12&#xA;BMU&#xA;19&#xA;BTN&#xA;24&#xA;BOL&#xA;16&#xA;BIH&#xA;8&#xA;BWA&#xA;43&#xA;BRA&#xA;33&#xA;VGB&#xA;20&#xA;BRN&#xA;20&#xA;BGR&#xA;16&#xA;BFA&#xA;9&#xA;BDI&#xA;7&#xA;KHM&#xA;8&#xA;CMR&#xA;15&#xA;CAN&#xA;15&#xA;CPV&#xA;19&#xA;CYM&#xA;34&#xA;CAF&#xA;7&#xA;TCD&#xA;18&#xA;CHL&#xA;9&#xA;CHN&#xA;11&#xA;HKG&#xA;7&#xA;MAC&#xA;14&#xA;COL&#xA;15&#xA;COM&#xA;14&#xA;COG&#xA;20&#xA;COK&#xA;16&#xA;CRI&#xA;24&#xA;CIV&#xA;19&#xA;HRV&#xA;14&#xA;CUB&#xA;37&#xA;CYP&#xA;31&#xA;CZE&#xA;44&#xA;PRK&#xA;9&#xA;COD&#xA;23&#xA;DNK&#xA;13&#xA;DJI&#xA;7&#xA;DMA&#xA;21&#xA;DOM&#xA;13&#xA;ECU&#xA;20&#xA;EGY&#xA;20&#xA;SLV&#xA;17&#xA;GNQ&#xA;10&#xA;ERI&#xA;22&#xA;EST&#xA;15&#xA;ETH&#xA;27&#xA;FRO&#xA;15&#xA;FLK&#xA;7&#xA;FJI&#xA;5&#xA;FIN&#xA;8&#xA;FRA&#xA;29&#xA;GUF&#xA;29&#xA;PYF&#xA;28&#xA;GAB&#xA;7&#xA;GMB&#xA;17&#xA;GEO&#xA;12&#xA;DEU&#xA;17&#xA;GHA&#xA;8&#xA;GIB&#xA;9&#xA;GRC&#xA;26&#xA;GRL&#xA;17&#xA;GRD&#xA;25&#xA;GLP&#xA;9&#xA;GUM&#xA;22&#xA;GTM&#xA;20&#xA;GGY&#xA;18&#xA;GIN&#xA;7&#xA;GNB&#xA;19&#xA;GUY&#xA;18&#xA;HTI&#xA;23&#xA;VAT&#xA;40&#xA;HND&#xA;16&#xA;HUN&#xA;20&#xA;ISL&#xA;17&#xA;IND&#xA;15&#xA;IDN&#xA;15&#xA;IRN&#xA;13&#xA;IRQ&#xA;10&#xA;IRL&#xA;15&#xA;IMN&#xA;5&#xA;ISR&#xA;11&#xA;ITA&#xA;30&#xA;JAM&#xA;21&#xA;JPN&#xA;8&#xA;JEY&#xA;25&#xA;JOR&#xA;8&#xA;KAZ&#xA;35&#xA;KEN&#xA;15&#xA;KIR&#xA;11&#xA;KWT&#xA;15&#xA;KGZ&#xA;23&#xA;LAO&#xA;25&#xA;LVA&#xA;31&#xA;LBN&#xA;22&#xA;LSO&#xA;11&#xA;LBR&#xA;26&#xA;LBY&#xA;33&#xA;LIE&#xA;7&#xA;LTU&#xA;9&#xA;LUX&#xA;12&#xA;MDG&#xA;12&#xA;MWI&#xA;24&#xA;MYS&#xA;18&#xA;MDV&#xA;27&#xA;MLI&#xA;4&#xA;MLT&#xA;9&#xA;MHL&#xA;9&#xA;MTQ&#xA;10&#xA;MRT&#xA;7&#xA;MUS&#xA;10&#xA;MYT&#xA;17&#xA;MEX&#xA;27&#xA;FSM&#xA;19&#xA;MDA&#xA;12&#xA;MCO&#xA;22&#xA;MNG&#xA;8&#xA;MNE&#xA;10&#xA;MSR&#xA;7&#xA;MAR&#xA;29&#xA;MOZ&#xA;13&#xA;MMR&#xA;5&#xA;NAM&#xA;25&#xA;NRU&#xA;7&#xA;NPL&#xA;6&#xA;NLD&#xA;10&#xA;ANT&#xA;19&#xA;NCL&#xA;20&#xA;NZL&#xA;26&#xA;NIC&#xA;11&#xA;NER&#xA;22&#xA;NGA&#xA;13&#xA;NIU&#xA;17&#xA;NFK&#xA;13&#xA;MNP&#xA;3&#xA;NOR&#xA;4&#xA;PSE&#xA;17&#xA;OMN&#xA;3&#xA;PAK&#xA;25&#xA;PLW&#xA;15&#xA;PAN&#xA;28&#xA;PNG&#xA;9&#xA;PRY&#xA;9&#xA;PER&#xA;24&#xA;PHL&#xA;12&#xA;PCN&#xA;24&#xA;POL&#xA;4&#xA;PRT&#xA;4&#xA;PRI&#xA;11&#xA;QAT&#xA;35&#xA;KOR&#xA;7&#xA;REU&#xA;29&#xA;ROU&#xA;9&#xA;RUS&#xA;5&#xA;RWA&#xA;27&#xA;BLM&#xA;11&#xA;SHN&#xA;17&#xA;KNA&#xA;16&#xA;LCA&#xA;11&#xA;MAF&#xA;17&#xA;SPM&#xA;6&#xA;VCT&#xA;36&#xA;WSM&#xA;10&#xA;SMR&#xA;11&#xA;STP&#xA;5&#xA;SAU&#xA;38&#xA;SEN&#xA;23&#xA;SRB&#xA;17&#xA;SYC&#xA;28&#xA;SLE&#xA;14&#xA;SGP&#xA;21&#xA;SVK&#xA;14&#xA;SVN&#xA;11&#xA;SLB&#xA;17&#xA;SOM&#xA;6&#xA;ZAF&#xA;30&#xA;ESP&#xA;17&#xA;LKA&#xA;11&#xA;SDN&#xA;25&#xA;SUR&#xA;5&#xA;SJM&#xA;12&#xA;SWZ&#xA;7&#xA;SWE&#xA;22&#xA;CHE&#xA;8&#xA;SYR&#xA;13&#xA;TJK&#xA;11&#xA;THA&#xA;19&#xA;MKD&#xA;9&#xA;TLS&#xA;15&#xA;TGO&#xA;21&#xA;TKL&#xA;10&#xA;TON&#xA;6&#xA;TTO&#xA;5&#xA;TUN&#xA;8&#xA;TUR&#xA;4&#xA;TKM&#xA;11&#xA;TCA&#xA;19&#xA;TUV&#xA;2&#xA;UGA&#xA;20&#xA;UKR&#xA;17&#xA;ARE&#xA;30&#xA;GBR&#xA;21&#xA;TZA&#xA;31&#xA;USA&#xA;20&#xA;VIR&#xA;22&#xA;URY&#xA;10&#xA;UZB&#xA;29&#xA;VUT&#xA;2&#xA;VEN&#xA;26&#xA;VNM&#xA;9&#xA;WLF&#xA;17&#xA;ESH&#xA;25&#xA;YEM&#xA;28&#xA;ZMB&#xA;24&#xA;ZWE&#xA;21&#xA;0&#xA;0 0 2 2 5 7 5 13 10 13 10 13 7 7 6 12 8 16 4 8 10 8 11 5 8 7 4 4 5 5 3 3 2 2 1 2 1 1 1 0 1 4&#xA;21&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Note: the leading 0 and following 21 is the command to print results and result of the print result commands respectively (this explanation is as messy as the code). TL;DR, the solutions from 0,1,2,...39,40,OVER, are listed in the long number line.&lt;br&gt;&lt;br&gt;As you can see from my beautifully crafted code, (I bet this'll make some programmers puke, I did it just to test my regex skills), the only 0's are for variables zero, one, and thirtynine, meaning these are the only ones that can't be represented by a country code. Zero and thirtynine correspond to nowhere and everywhere intuitively being none and every, and one is the loneliest number sitting at home. (basically from process of elimination)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Answer for Lock 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For the system to work, every number has to be able to be represented otherwise he needs another system when combinations include 0/1 etc. Also, there should be no reason to break the pattern of using country code abbreviations if he didn't have to, so all three should represent numbers that can't be made with country abbreviations. Assuming &lt;a href=&quot;https://en.wikipedia.org/wiki/ISO_3166-1_alpha-3&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt; is used for codes, 2 can be made with RSR, and 3 can be made with OMN, (the rest have multiple), meaning there is none for 0 or 1. To associate these with nowhere and everywhere, I'd say 0 is nowhere and 1 is everywhere. Home could then be either actually home, or one other number that can't be made. I don't know if I should write some code to convert these to numbers though without confirmation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="4450" LastEditDate="2015-07-12T00:22:52.673" LastActivityDate="2015-07-12T00:22:52.673" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="17694" PostTypeId="2" ParentId="17681" CreationDate="2015-07-09T23:40:49.533" Score="35" Body="&lt;p&gt;Place your pencil on line 1.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/za2jxm.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Rotate your pencil so it lines up with line 2. You just rotated it counterclockwise by the angle at the top of the pentagram.&lt;/p&gt;&#xA;&lt;p&gt;Now rotate it counterclockwise again onto line 3. Then again onto line 4, then 5, and finally back to 1. You have just rotated your pencil through all five angles of the pentagram in sequence.&lt;/p&gt;&#xA;&lt;p&gt;And what happened? The pencil now lies on the same line it started at, pointing in the opposite direction. If you track which direction the pencil points at each step, you can see that in total, you rotated it counterclockwise by one half turn. Whence, &lt;span class=&quot;math-container&quot;&gt;$180^\circ$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1752" LastEditorUserId="25966" LastEditDate="2023-04-22T18:26:33.437" LastActivityDate="2023-04-22T18:26:33.437" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="17696" PostTypeId="1" AcceptedAnswerId="17708" CreationDate="2015-07-09T23:53:41.457" Score="11" ViewCount="1150" Body="&lt;p&gt;I have two coins - one has $N$ times the diameter of the other. I roll the smaller coin one full lap around the circumference of the larger coin like a gear, keeping contact it in with the edge of the larger coin and keeping the larger coin still. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many rotations does the smaller coin make around its own center in the rest frame? Equivalently, if you drew an arrow on the small coin which was initially pointing East, how many times during the process would the arrow point North? &lt;/p&gt;&#xA;" OwnerUserId="2425" LastEditorUserId="4551" LastEditDate="2015-07-10T21:14:55.453" LastActivityDate="2015-07-12T08:22:09.020" Title="A problem circularly coined" Tags="|mathematics|" AnswerCount="6" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17697" PostTypeId="2" ParentId="16421" CreationDate="2015-07-09T23:59:45.283" Score="8" Body="&lt;p&gt;If we use multiple shots, we have to account for the possibility that they all hit the same the same spot. Assuming that the duplicator has no power limit, we might as well use a single shot that is the sum of all the other shots' power levels. The question is to find how large of a shot this requires. $n=1$ is solved with a shot at 300% and $n=2$ takes 200%. The rest of this answer assumes that $n\ge3$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;First of all, Bmyguest's answer shows a pattern: every prime dot number except ●● has an odd number of filled dots. This is because any dot number with an even number of filled dots is divisible by ●●. If our shot hits a ●, then the dot number will become divisible by ●●. Therefore, we should assume that our shot will hit a ○. (As it turns out, the following strategy works even if we hit ● and don't consider divisibility by ●●.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the original dot number be $P$. The ○ we hit divides $P$ into two parts. Let $A$ be the part on the left, with a length of $a$, and let $B$ be the part on the right, with a length of $b$, so that $n=a+b+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each dot number is equivalent to a polynomial in $\mathbb{Z}_2[x]$, and a dot number divides another if and only if their corresponding polynomials do. We have $P(x)=A(x)x^{b+1}+B(x)$ and we need to find some $i\in\mathbb{Z}^+$ such that $A(x)x^{b+1+i}+B(x)$ is composite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(x)$ has degree $n-1$, so there are $2^{n-1}$ residue classes mod $P(x)$. Because $P(x)$ is irreducible, the order of $x$ mod $P(x)$ is a factor of $2^{n-1}-1$, so $P(x)$ divides $x^{2^{n-1}-1}-1$. Then, $A(x)x^{b+2^{n-1}}+B(x)\\=(A(x)x^{b+2^{n-1}}+B(x))-(A(x)x^{b+1}+B(x))+(A(x)x^{b+1}+B(x))\\=A(x)x^{b+1}(x^{2^{n-1}-1}-1)+P(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $P(x)$ divides $A(x)x^{b+2^{n-1}}+B(x)$, and we can take $i=2^{n-1}-1$. Translating back to dot numbers, if we fire a shot at $(2^{n-1}-1)\times100\%$ power, we are guaranteed to get a composite dot number (either divisible by ●● or by $P$).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is probably extreme overkill: a brute force search shows that 200% is good enough for $n=7$, where this strategy would require a shot at 6300%.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-07-20T11:37:35.333" LastActivityDate="2015-07-20T11:37:35.333" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17698" PostTypeId="2" ParentId="17696" CreationDate="2015-07-10T00:06:31.030" Score="2" Body="&lt;p&gt;For the smaller coin the distance traveled in one revolution would be $\pi*Diameter$. Since we know the bigger coin has a diameter of $N*Diameter$ of smaller coin we know the total distance that needs to be traveled by the smaller coin is $N*Diameter*\pi$. That leaves us with the formula of $\frac{N*\pi*Diameter}{\pi*Diameter}$ which leaves us with $N$. So the answer is $N$ revolutions.&lt;/p&gt;&#xA;" OwnerUserId="10856" LastEditorUserId="13881" LastEditDate="2015-07-10T01:37:24.103" LastActivityDate="2015-07-10T01:37:24.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17699" PostTypeId="2" ParentId="260" CreationDate="2015-07-10T00:22:11.487" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Your family's chef serves two types of pancakes;  blueberry pancakes and raspberry pancakes, which he serves at random to breakfast guests.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Today, the disreputable Duke Froggington II's breakfast is a stack of two pancakes, the topmost of which is blueberry.  You can't see his second pancake, as it's entirely hidden beneath the first pancake.  What is the probability that both pancakes are blueberry?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Here I'm using blueberry and raspberry pancakes instead of boys and girls, and the nefarious Duke Froggington II instead of Mr. Smith, but structurally it's exactly the same puzzle, just with the theme adjusted to remove the ambiguity of the &quot;and one of them is a boy&quot; statement)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that the reason why there are no versions of the puzzle which are worded unambiguously is that the puzzle isn't really a 'puzzle' when it's worded precisely;  instead, it's a simple statement of basic probability.  We could change the theme again to make the same puzzle be even more simply stated as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I flipped a fair coin two times, such that there was an equal 50% chance of either face showing after each flip.  The first time, it came up heads.  What is the probability that it came up heads both times?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &quot;boy or girl&quot; puzzle is intentionally built around a misunderstanding that the puzzle-solver is intended to make; it stops being recognisable as a puzzle when that ambiguity is removed.  And that's why you don't find online examples of it without the ambiguity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether it truly works as a good puzzle even when the ambiguity is present, is left as &lt;a href=&quot;https://xkcd.com/169/&quot; rel=&quot;nofollow&quot;&gt;an exercise for the reader&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="9286" LastEditorUserId="9286" LastEditDate="2015-07-10T00:29:56.483" LastActivityDate="2015-07-10T00:29:56.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17700" PostTypeId="2" ParentId="16421" CreationDate="2015-07-10T00:27:47.877" Score="2" Body="&lt;p&gt;The conspirators should fire&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $n$ shots, all at a $1\times$ charge.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Consider a dot number such that every other digit is zero:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x = 10a0b\ldots 0c01_2&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Partition the digits into pairs (with an implicit leading zero):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x = 01\ 0a\ 0b\ldots0c\ 01_2&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now triple this number (multiply it by $11_2$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{rrrrrrr}&#xA; x &amp;amp; 01&amp;amp;0a&amp;amp;0b&amp;amp;\ldots&amp;amp;0c&amp;amp;01_2\\&#xA;\oplus~2x &amp;amp; 10&amp;amp;a0&amp;amp;b0&amp;amp;\ldots&amp;amp;c0&amp;amp;10_2\\ \hline&#xA;3x &amp;amp; 11&amp;amp;aa&amp;amp;bb&amp;amp;\ldots&amp;amp;cc&amp;amp;11_2&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If we duplicate every digit of a number, the resulting number ($3x$) is guaranteed to be a product of three and the same number with every other digit zeroed ($x$), and therefore not a dot-prime.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9718" LastActivityDate="2015-07-10T00:27:47.877" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17701" PostTypeId="1" AcceptedAnswerId="17728" CreationDate="2015-07-10T01:29:10.843" Score="7" ViewCount="278" Body="&lt;p&gt;The names of nine fellows are concealed in this peculiar verse.  Find the names and provide your exegesis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An apple for Charlemagne,&lt;br&gt;&#xA;one that takes great steps,&lt;br&gt;&#xA;returning in &lt;em&gt;Skyrim&lt;/em&gt; or &lt;em&gt;Oblivion&lt;/em&gt;,&lt;br&gt;&#xA;and flag unfurled.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dave's mentor, venerable sage,&lt;br&gt;&#xA;in high spirits!&lt;br&gt;&#xA;That way the celebration.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of sweet lime drink coming to me;&lt;br&gt;&#xA;go sell a pretzel??&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is my first puzzle here.  If you have any suggestions for improvement, please share!&lt;/p&gt;&#xA;" OwnerUserId="4385" LastEditorUserId="4385" LastEditDate="2015-07-10T18:07:37.847" LastActivityDate="2015-07-10T18:07:37.847" Title="Nine Obscure Fellows" Tags="|enigmatic-puzzle|cryptic-crosswords|literature|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17702" PostTypeId="2" ParentId="17696" CreationDate="2015-07-10T01:38:50.000" Score="1" Body="&lt;p&gt;Is the answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; one revolution&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; point A is on the small coin. So after one revolution of the small coin you are back at A.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13881" LastActivityDate="2015-07-10T01:38:50.000" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17703" PostTypeId="2" ParentId="17681" CreationDate="2015-07-10T02:31:19.837" Score="52" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\hskip 1.5in$&lt;/span&gt;&lt;img src=&quot;https://i.stack.imgur.com/lCw7u.gif&quot; alt=&quot;Isn't it dreamy?&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;This is an image of an arrow sweeping each of the successive angles in the star. Notice that, after it traces all &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt; angles, its orientation is reversed - meaning it has rotated &lt;span class=&quot;math-container&quot;&gt;$180^{\circ}$&lt;/span&gt; and that this must be the sum of the angles. We can do the same thing to the star in your figure, ergo, its angles too sum to &lt;span class=&quot;math-container&quot;&gt;$180^{\circ}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;A Generalization:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;We can do the same thing to a figure like this, whose angles sum to &lt;span class=&quot;math-container&quot;&gt;$180^{\circ}$&lt;/span&gt;:&#xA;&lt;span class=&quot;math-container&quot;&gt;$\hskip 1.5in$&lt;/span&gt;&lt;img src=&quot;https://i.stack.imgur.com/2die8.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;br /&gt;&#xA;We can also do this to a triangle. The important property is this:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;There must be no vertices of the star &lt;em&gt;interior&lt;/em&gt; to the cone swept out by a ray traversing a given angle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Satisfying this condition - which basically says that we never have to &amp;quot;ignore&amp;quot; vertices, but instead just rotate the arrow and see what it hits -  we find that we can order the vertices in a &amp;quot;clockwise&amp;quot; manner, so that, at each angle, either the head or the tail of the arrow steps to the next vertex in the order (and they alternate which). Obviously, both head and tail will make a full revolution when twice as many angles as vertices have been traced, yielding the desired result.&lt;/p&gt;&#xA;&lt;p&gt;(One might also express my condition as &amp;quot;assigning the vertices the numbers &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; through &lt;span class=&quot;math-container&quot;&gt;$2n+1$&lt;/span&gt; in clockwise order as seen from a central point, it must be that &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; connects to &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$n+1$&lt;/span&gt;, and all other points are connected analogously&amp;quot;)&lt;/p&gt;&#xA;" OwnerUserId="4765" LastEditorUserId="25966" LastEditDate="2023-04-22T18:26:19.427" LastActivityDate="2023-04-22T18:26:19.427" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="17704" PostTypeId="2" ParentId="17696" CreationDate="2015-07-10T02:36:25.197" Score="5" Body="&lt;p&gt;Edit: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The number of revolutions would actually be increased by one (see &lt;a href=&quot;https://en.wikipedia.org/wiki/Coin_rotation_paradox#/media/File:Coin_rotation_paradox_animation.gif&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt; provided by Mike), so the answer could be either $N+1$ or $1/N + 1$ if N is greater or less than 1 respectively.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The answer is actually:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Either $N$ or $1/N$ depending on if $N$ is greater or less than 1. (And $N$ is from solving $(2\pi*r*N)/(2\pi*r)$ using the formula for circumference).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="7722" LastEditDate="2015-07-10T19:01:49.730" LastActivityDate="2015-07-10T19:01:49.730" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17705" PostTypeId="2" ParentId="17309" CreationDate="2015-07-10T03:23:08.903" Score="2" Body="&lt;p&gt;Try:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ctrl + C &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; that force-quits a program in Mac, Linux and Windows ;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13894" LastEditorUserId="11297" LastEditDate="2015-07-10T08:37:37.613" LastActivityDate="2015-07-10T08:37:37.613" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17706" PostTypeId="2" ParentId="17701" CreationDate="2015-07-10T03:29:25.863" Score="0" Body="&lt;p&gt;&lt;em&gt;I'm not sure about my logic, so I'll take a stab before getting in entirely to solve this...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the first fellow &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Charles_Appleton_(academic)&quot; rel=&quot;nofollow&quot;&gt;Charles Appleton&lt;/a&gt;&lt;br&gt;&lt;br&gt; Charles is a fellow of England's Oxford University, which sort of connects with the title.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-07-10T03:29:25.863" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17707" PostTypeId="2" ParentId="17681" CreationDate="2015-07-10T03:38:05.277" Score="1" Body="&lt;p&gt;This proof in a sense is nothing but counting the degree of angles.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Recall that a pentagon, whether regular or irregular, has its internal angles sum to 540.  Also, the angles of an intersection of 2 straight lines sum to 360, where also the opposite angles are congruent.&#xA; &lt;br&gt;&lt;br&gt;&#xA; Consider the 5 points of the central pentagon, the points where the intersection of 2 lines occurs.  Around these 5 points are 360 x 5 = 1800 degrees total, and 5 x 4 = 20 angles to count.&#xA; &lt;br&gt;&lt;br&gt;&#xA; Of the 20 angles, 5 are from the pentagon, 5 more are congruent to those.  So this accounts for 540 + 540 = 1080 degrees.  The remains of the 1800 - 1080 = 720 degrees come from inside the 5 triangles.&#xA; &lt;br&gt;&lt;br&gt;&#xA; 5 triangles contain 5 x 180 = 900 degrees worth of interior angles.  720 of those degrees are at the corners of the pentagon/triangle/intersection.&#xA; &lt;br&gt;&lt;br&gt;&#xA; This leaves at the tips of the star 900 - 720 = 180 degrees.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;The arithmetic here is simply shorthand for angle addition and subtraction, same as is done in other answers.&lt;/p&gt;&#xA;" OwnerUserId="12357" LastEditorUserId="12357" LastEditDate="2015-07-10T03:53:08.247" LastActivityDate="2015-07-10T03:53:08.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17708" PostTypeId="2" ParentId="17696" CreationDate="2015-07-10T03:43:04.993" Score="10" Body="&lt;p&gt;&lt;strong&gt;Answer&lt;/strong&gt; :&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $N+1$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation&lt;/strong&gt; :&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Circumference of bigger coin $= 2\pi N r$&lt;br&gt;&#xA; Circumference of smaller coin $= 2\pi r$ &lt;br&gt;&#xA; Number of revolutions $= \frac{2\pi Nr}{2\pi r}=N$&lt;br&gt;&#xA; But the smaller coin has also revolved about the center of the whole configuration, which increases the number of revolutions by 1&lt;br&gt;&#xA; So total $= N+1$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10940" LastEditorUserId="10940" LastEditDate="2015-07-10T18:26:14.400" LastActivityDate="2015-07-10T18:26:14.400" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17709" PostTypeId="2" ParentId="17696" CreationDate="2015-07-10T03:52:34.807" Score="0" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1 revolution and $N$ rotations. &lt;br&gt; Since a revolution is one path around the bigger coin, and a rotation would be about its own axis. Based on the circumferential calculations in the other answers, the number of rotations is $N$ whereas the smaller coin only completes one lap around the bigger coin, so it makes just one revolution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-07-10T03:52:34.807" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17710" PostTypeId="2" ParentId="2353" CreationDate="2015-07-10T04:08:40.073" Score="0" Body="&lt;p&gt;The 'sky is red' and the 'sky is green' are contradictory, but don't require the speaker to be truthful at any point.&lt;/p&gt;&#xA;" OwnerUserId="3889" LastActivityDate="2015-07-10T04:08:40.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17711" PostTypeId="2" ParentId="17701" CreationDate="2015-07-10T04:33:58.980" Score="7" Body="&lt;p&gt;Short answer :  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://lotr.wikia.com/wiki/Fellowship_of_the_Ring&quot;&gt;http://lotr.wikia.com/wiki/Fellowship_of_the_Ring&lt;/a&gt;&lt;br&gt;&#xA; 9 MEMBERS  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Details :  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;AND FLAG&quot; anagram for &lt;strong&gt;GANDALF&lt;/strong&gt;&lt;br&gt;&#xA; &quot;HIGH SPIRITS&quot; = &lt;strong&gt;MERRY&lt;/strong&gt;&lt;br&gt;&#xA; &quot;APPLE&quot; = &lt;strong&gt;PIPPIN&lt;/strong&gt;&lt;br&gt;&#xA; &quot;great steps&quot; = &lt;strong&gt;Legolas&lt;/strong&gt;&lt;br&gt;&#xA; &quot;venerable sage&quot; = &lt;strong&gt;SamWise Gamgee&lt;/strong&gt;&lt;br&gt;&#xA; &quot;returning&quot; because he was gone, err Aragone, err &lt;strong&gt;Aragorn&lt;/strong&gt;&lt;br&gt;&#xA; &quot;sweet lime drink&quot; is gimlet &quot;cocktail made of gin or vodka and lime juice&quot; so &lt;strong&gt;Gimli&lt;/strong&gt;&lt;br&gt;&#xA; &quot;pretzel&quot; + &lt;a href=&quot;http://www.boromir.ro/en/shop/pretzels-8&quot;&gt;http://www.boromir.ro/en/shop/pretzels-8&lt;/a&gt; = &lt;strong&gt;Boromir&lt;/strong&gt;&lt;br&gt;&#xA; &quot;celebration&quot; + &lt;a href=&quot;https://en.wikipedia.org/wiki/Hobbit_Day&quot;&gt;https://en.wikipedia.org/wiki/Hobbit_Day&lt;/a&gt; = &lt;strong&gt;Frodo&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10940" LastEditorUserId="10940" LastEditDate="2015-07-10T09:32:33.707" LastActivityDate="2015-07-10T09:32:33.707" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17713" PostTypeId="2" ParentId="16169" CreationDate="2015-07-10T05:20:26.293" Score="1" Body="&lt;p&gt;Lets assume you are 'x'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I am on top politically, I am right, but if I am on top of being not in violation, I am wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if (x == 'on top politically') =&gt; x = &lt;strong&gt;right&lt;/strong&gt; &#xA; elseif (x= 'on top of being not in violation') =&gt; x= &lt;strong&gt;Wrong&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Similarly,&#xA; If I am above 8, I am wrong, but if I am low on being not in violation, I am forever.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if( x &gt; 8) =&gt; x= &lt;strong&gt;Wrong&lt;/strong&gt;; &#xA; else if (x = &quot;low on being not in violation&quot;) =&gt; x = &lt;strong&gt;Forever&lt;/strong&gt;;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cannot be on top politically and above 8, but I can be on top and low of being not in violation. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; True because &#xA; if (x==&quot;on top politically&quot;) =&gt; x= right; and if( x &gt; 8)  =&gt; x= &lt;strong&gt;Wrong&lt;/strong&gt;; which is a contradiction. &#xA; Hence, either one is false or both are false statement. &#xA; Second one implies x = &lt;strong&gt;wrong&lt;/strong&gt; or &lt;strong&gt;forever&lt;/strong&gt; where forever could mean either right or wrong. &#xA; So, one plausible solution is that x = 'on top of being not in violation' and x = 'above 8' =&gt; x = &lt;strong&gt;WRONG&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2656" LastActivityDate="2015-07-10T05:20:26.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17714" PostTypeId="2" ParentId="17696" CreationDate="2015-07-10T05:45:27.533" Score="3" Body="&lt;p&gt;The answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;s&gt;N+2&lt;/s&gt; rotation around large coins center doesn't count&lt;br&gt;N+1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every time the small coin rolls a distance of π it actually makes 1+1/N revolutions around its own center. The angle from the center of the small coin to the point at which the coins touch will be 2π/N more than the angle at the beginning. &lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/4oB50.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;So by the time it rolls Nπ then it has made N*(1+1/N)=N+1 revolutions around its own center.  &lt;s&gt;If you then add to this the one revolution it has made around the large coins center then you get N+2&lt;/s&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13981" LastEditorUserId="13981" LastEditDate="2015-07-12T08:22:09.020" LastActivityDate="2015-07-12T08:22:09.020" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17715" PostTypeId="1" AcceptedAnswerId="17724" CreationDate="2015-07-10T06:05:47.410" Score="8" ViewCount="907" Body="&lt;p&gt;You are a sage in service to a king of a profoundly troubled kingdom.&lt;/p&gt;&#xA;&lt;h3&gt;A sordid spectacle...&lt;/h3&gt;&#xA;&lt;p&gt;Every ten years, the king holds a grand tournament, inviting the 1,000 most suitable princes to compete for the hand of one of his daughters. Regrettably, the king's true motive lies in exploiting the princes' legendary oneupmanship.&lt;/p&gt;&#xA;&lt;p&gt;At the beginning of the tourney, the king has his steward present the sitting princess with a gift (a piece of jewelry) in the presence of all 1,000 princes so that they can witness her delight. He then invites them to introduce themselves one at a time, and watches contentedly as a sordid spectacle of oneupmanship unfolds.&lt;/p&gt;&#xA;&lt;p&gt;The first prince to approach, wanting to vastly outdo the steward, pledges to give the princess one piece of jewelry for every day of the tourney.&lt;/p&gt;&#xA;&lt;p&gt;The second prince, wishing to vastly outdo the first, pledges that every day of the tourney he shall give the princess as many pieces of jewelry as the first prince gives her that day &lt;em&gt;plus&lt;/em&gt; as many pieces of jewelry as the first prince has given her on &lt;em&gt;all previous days&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Likewise, the third prince, determined to vastly outdo the second, pledges that every day of the tourney he shall give the princess as many pieces of jewelry as the &lt;em&gt;second&lt;/em&gt; prince gives her that day plus as many pieces of jewelry as the second prince has given her on all previous days.&lt;/p&gt;&#xA;&lt;p&gt;This mad spectacle of devotion continues for all 1,000 princes, ending when the 1,000th prince has boldly pledged that every day of the tourney he shall give the princess as many pieces of jewelry as the 999th prince gives her that day plus as many pieces of jewelry as the 999th prince has given her on all previous days.&lt;/p&gt;&#xA;&lt;h3&gt;The plot thickens...&lt;/h3&gt;&#xA;&lt;p&gt;The grand irony of the king's scheme is that he hates jewelry. He allows the princess to collect it only so that he can give it to the nobles attending his decennial Nobles Gala, which he holds after the tourney is completed.&lt;/p&gt;&#xA;&lt;p&gt;There are 24 nobles he must invite to the gala, and they're notoriously fickle. For one thing, those who attend demand some jewelry in return for their attendance. For another, the attendees insist on each receiving &lt;em&gt;exactly&lt;/em&gt; the same number of jewelry pieces as each other.&lt;/p&gt;&#xA;&lt;p&gt;Worst of all, rumours abound that some of the invitees may not show up for the upcoming Nobles Gala. Perhaps all 24 will, perhaps only one will, or perhaps any number in between. The exact number is unknowable until the gala is held.&lt;/p&gt;&#xA;&lt;h3&gt;A sage is needed!&lt;/h3&gt;&#xA;&lt;p&gt;As the king's sage, issues related to distribution of jewelry naturally fall to you. He makes it clear to you that:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;regardless of how many nobles ultimately turn up at the gala, their fickle demands &lt;em&gt;must&lt;/em&gt; be met&lt;/li&gt;&#xA;&lt;li&gt;the only jewelry that will be distributed among the attendees is the jewelry given to the princess during the tourney&lt;/li&gt;&#xA;&lt;li&gt;every single piece of jewelry must be given to the attendees. Throwing it away (which is a curse) or keeping it (which the king hates) is unacceptable.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;The only thing in your power to control is the number of days the tourney will last. It must last at least one day (obviously), and ideally it should be &lt;em&gt;as short as possible&lt;/em&gt;, minding that you must guarantee the nobles' demands are met.&lt;/p&gt;&#xA;&lt;p&gt;In light of all the above, &lt;strong&gt;how many days should the tourney of 1,000 princes last?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Good luck, fair sage. ;)&lt;/p&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-10T16:35:15.837" Title="The Tourney of a Thousand Princes" Tags="|logical-deduction|mathematics|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17717" PostTypeId="1" AcceptedAnswerId="17721" CreationDate="2015-07-10T06:25:53.280" Score="3" ViewCount="186" Body="&lt;p&gt;A bunny is hopping around the plane.  She must&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;land on the following $25$ points on the main diagonal (multiples of 4):  $(4, 4)$, $(8, 8)$, $(12, 12)$, $\ldots,$ $(100, 100)$,  (The bunny can land on other points too, but the goal is to hit these multiples of four.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;only land on lattice points $\mathbb{Z}^2$,&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;only jump integer distances, and&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;move so no pair of jumps are parallel or perpendicular to each other.  That means if the bunny jumps directly north for its first jump, it cannot jump directly north, south, east or west for any of the remaining jumps.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The bunny can start by jumping to any lattice point you wish.  How can the bunny accomplish this task?&lt;/p&gt;&#xA;" OwnerUserId="2722" LastActivityDate="2015-07-10T08:15:53.007" Title="Lattice hopping" Tags="|mathematics|strategy|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17721" PostTypeId="2" ParentId="17717" CreationDate="2015-07-10T08:15:53.007" Score="6" Body="&lt;p&gt;If $(a,b,c)$ is a Pythagorean triple, the bunny can do moves $(a,b)$ followed by $(-a,b)$ to move a displacement of $2b$ in any orthogonal direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using triples of the form $(m^2-1,2m,m^2+1)$, we can obtain any even number $b=2m$, and so any orthogonal displacement $4m$. By moving $4(m+1)$ spaces, then $4m$ spaces the other way, we can move $4$ spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $m$ sufficiently large, we can guarantee that we haven't used $m$ and $m+1$ before, so this move isn't parallel or perpendicular to any previous move. (When $m$ is odd, the triple has a GCD of 2, so we have to care that its halved version hasn't been used either.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bunny can start at $(0,0)$ and repeatedly move $4$ steps and $4$ steps right to hit every required point.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-07-10T08:15:53.007" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17722" PostTypeId="2" ParentId="17715" CreationDate="2015-07-10T08:20:53.847" Score="3" Body="&lt;p&gt;To satisfy the nobles demand that the jewels must be shared evenly between them, the number of jewels must be a multiple of all the numbers under 24. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The smallest such number is 5354228880*&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that the princes offer gifts in the same order each day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's the table of how many jewels are collected by each day of the gala.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;d j pledged that day, cumulative** &#xA;1 0,1000&#xA;2 501500,502500&#xA;3 167668500,168171000&#xA;4 42084793750,42252964750&#xA;5 8459043543950,8501296508700&#xA;6 1418299634202450,1426800930711150&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If these were 1mm cube gemstones they would fill more than 10 cubic meter treasure chests by day 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately on none of the six days was the treasure divisible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the 7th day my calculator reaches to limit of its precision, but the are then approximately  205459334022406530 pieces of jewellery to deal with, that's 5900 cubic metres of tiny gems, enough to fill a 24 meter square treasure vault up to your waist (or thighs depending on height).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the princes bring larger pieces of jewellery around 1cm cube size, say sets of pearl earrings or fine gold chains, the by day 7 most of the castle is full of treasure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If day seven didn't turn out to have just the right number then there are going to be problems because the pile of treasure grows in size by around 100 times a day at first. Although the rate of increase drops the pile still multiplying in size by 10 each day at day 100, and will have about 1.565e144 which is more than the number of atoms in the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't say for sure but I feel like there will never be a day when the jewellery could be shared by any number of the guests who may arrive, if there is it won't be before the demand for jewellery has destroyed the economies of most of those 1000 principalities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Changing the number of princes invited doesn't seem to help. Assuming I'm doing the sums right, no number of princes under 1000 will work within the number of days my calculator's precision allows me to test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may be a way to prove that no solution exists from this problem when p=1000 and n=24 but I am unaware of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Thanks to CodeNewbie for correcting me on the number the treasure must be divisible by. I initially thought it would be 4684950270 but changed for instead 4260838982400, can you work out from those what I was doing wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;**CodeNewbie thinks my other numbers are wrong, I think I might have assumed that the pledges from each day are cumulative. Either way both our methods suggested &lt;strike&gt;the lack of an answer.&lt;/strike&gt; that any answer will be too big to accurately calculate with ease and way too large to practically store that many objects.&lt;/p&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="11297" LastEditDate="2015-07-10T16:35:15.837" LastActivityDate="2015-07-10T16:35:15.837" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="17724" PostTypeId="2" ParentId="17715" CreationDate="2015-07-10T09:49:08.687" Score="13" Body="&lt;p&gt;The steward gives 1 gift on the first day and no gifts on any other day, so he gives a total of 1 gift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first prince gives 1 gift on every day, so he gives a total of $d$ gifts through day $d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second prince gives $n$ gifts on day $n$, for a total of $d+1\choose2$ gifts through day $d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third prince gives $n+1\choose2$ gifts on day $n$, for a total of $d+2\choose3$ gifts through day $d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This continues until the 1000th prince gives a total of $d+999\choose1000$ gifts through day $d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total number of gifts is ${d-1\choose0}+{d\choose1}+{d+1\choose2}+\ldots+{d+999\choose1000}={d+1000\choose1000}$. So we have to find $d$ such that $d+1000\choose1000$ is divisible by every integer from 1 to 24.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${d+1000\choose1000}=\frac{(d+1000)\times((d+1000)-1)\times((d+1000)-2)\times\ldots\times1001}{d\times(d-1)\times(d-2)\times\ldots\times1}$. This product must be divisible by 5. However, 1000 is divisible by $5^3$, so every factor of 5 from a number in the product that is not divisible by $5^4$ will be cancelled out by the corresponding term in the denominator. For the result to be divisible by 5, at least one term in the numerator must be divisible by $5^4$. The smallest such number greater than 1000 is 1250, so $d$ must be at least 250.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1250\choose1000$ is not divisible by 3, so 250 does not work. However, $1251\choose1000$ is divisible by every integer from 1 to 24, so &lt;strong&gt;251&lt;/strong&gt; is the smallest possible number of days. The total number of gifts is $1251\choose1000$, which about $6.3\times10^{270}$.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-07-10T10:54:54.180" LastActivityDate="2015-07-10T10:54:54.180" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17725" PostTypeId="1" AcceptedAnswerId="17729" CreationDate="2015-07-10T10:16:08.003" Score="6" ViewCount="560" Body="&lt;p&gt;2015 people are attending an international summit meeting on terrorism. Each of them speaks at most five languages, and in any group of three of them, at least two have a language in common. Prove that there is some language spoken by at least 203 of the participants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;sub&gt;(This puzzle was inspired by one with different wording and different choice of numbers from the 1985 Balkan Mathematical Olympiad.)&lt;/sub&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2016-09-17T21:22:49.200" LastActivityDate="2016-09-17T21:22:49.200" Title="A summit meeting on terrorism" Tags="|mathematics|logical-deduction|combinatorics|" AnswerCount="1" CommentCount="6" ClosedDate="2015-07-10T15:16:47.303" ContentLicense="CC BY-SA 3.0" />
  <row Id="17726" PostTypeId="1" AcceptedAnswerId="17727" CreationDate="2015-07-10T10:17:28.447" Score="6" ViewCount="2063" Body="&lt;p&gt;If I successfully arrange all the blocks of every row of a sliding puzzle except the last row in proper order, will the n-th row be automatically resolved? (Assuming that the puzzle is solvable in the 1st place, and the hole is in the last row).&lt;/p&gt;&#xA;" OwnerUserId="13984" LastEditorDisplayName="user20" LastEditDate="2015-07-10T16:50:47.323" LastActivityDate="2015-07-10T16:50:47.323" Title="Does the last row of a N x N sliding puzzle need to be solved?" Tags="|solvability|sliding-blocks|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17727" PostTypeId="2" ParentId="17726" CreationDate="2015-07-10T10:34:54.703" Score="3" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not necessarily! &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Additional info:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In a 2x2 pattern, if you solve a row then you automatically solved the other too. Same for a 3x3 pattern, assuming that the hole is positioned in the last row, because swapping two blocks is an impossible move (it has been proven, only half configurations are solvable, as stated &lt;a href=&quot;https://en.wikipedia.org/wiki/15_puzzle&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;). In a 4x4 or bigger pattern, there are valid permutations for the last row, so you can't assure that solving the other rows automatically completes the game.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10281" LastActivityDate="2015-07-10T10:34:54.703" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17728" PostTypeId="2" ParentId="17701" CreationDate="2015-07-10T11:41:35.867" Score="6" Body="&lt;p&gt;Prem has the solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &quot;obscure fellows&quot; are the nine members of the Fellowship of the Ring in JRR Tolkien's &lt;em&gt;Lord of the Rings&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think some of the explanations are not quite right, however. The lines of the riddle are cryptic clues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An apple for Charlemagne,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;PIPPIN&lt;/strong&gt;: double definition; one of several &lt;a href=&quot;https://en.wikipedia.org/wiki/Pippin_apple&quot; rel=&quot;nofollow&quot;&gt;types of apple&lt;/a&gt; and the &lt;a href=&quot;https://en.wikipedia.org/wiki/Pepin_the_Hunchback&quot; rel=&quot;nofollow&quot;&gt;eldest son of Charlemagne&lt;/a&gt;. &amp;#10043;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;one that takes great steps,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;STRIDER&lt;/strong&gt;: simple definition. Strider is Aragorn's nickname, used on his first encounter with the Hobbits&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;returning in Skyrim or Oblivion,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;BOROMIR&lt;/strong&gt;: hidden backwards in skyRIM OR OBlivion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and flag unfurled.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;GANDALF&lt;/strong&gt;: anagram (&quot;unfurled&quot;) of AND FLAG.  &amp;#10043;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Dave's mentor, venerable sage,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;SAMWISE&lt;/strong&gt;: SAM (or Samuel, who is King David's mentor in the Old Testament) plus WISE (venerable sage). I didn't get the OT reference, so many thanks to dennisdeems for the explanation in the coments below. &amp;#10043;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;in high spirits!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;MERRY&lt;/strong&gt;: simple definition. &amp;#10043;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That way the celebration.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;FRODO&lt;/strong&gt;: &quot;To and FRO&quot; is this way and that way and a DO is a party.  (&amp;#10043;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Most of sweet lime drink coming to me;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;GIMLI&lt;/strong&gt;: Most of &lt;a href=&quot;https://en.wikipedia.org/wiki/Gimlet_%28cocktail%29&quot; rel=&quot;nofollow&quot;&gt;GIMLet&lt;/a&gt; (sweet lime drink) plus I (me). &amp;#10043;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;go sell a pretzel??&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;LEGOLAS&lt;/strong&gt;: Anagram (&quot;pretzel&quot;) of GO SELL A.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answers marked with &amp;#10043; have already been given by Prem, although not necessarily with the same or full explanation.&lt;/p&gt;&#xA;" OwnerUserId="9860" LastEditorUserId="9860" LastEditDate="2015-07-10T12:30:19.467" LastActivityDate="2015-07-10T12:30:19.467" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17729" PostTypeId="2" ParentId="17725" CreationDate="2015-07-10T11:53:51.780" Score="14" Body="&lt;p&gt;Either every two participants have a language in common, or there is some pair that does not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If every pair of participants shares a language, choose an arbitrary participant A. All 2014 other people must speak one of the at most 5 languages that A does. Whichever of these languages has the most speakers must have at least 403 not counting A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, there are some two participants (A and B) who have no common language. Then, any other participant must have a language in common with at least one of them. There are 2013 other participants, and only 10 languages that either A or B speaks. Therefore, at least one of these 10 languages must be spoken by at least 202 other participants. Because either A or B speaks this language, this language has at least 203 speakers.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-07-10T11:53:51.780" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17730" PostTypeId="2" ParentId="17644" CreationDate="2015-07-10T12:04:37.060" Score="7" Body="&lt;p&gt;Ok, I got it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As others have already stated, the first pass is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; decoding the binary numbers into ASCII letters, which gives the following string: &lt;code&gt;576H4L6K6 247463063 335H30494 8686R4947 566F62485 26Y4N6942 5159334F6 762486870 647942706 433427049 47566T624 8526S4948 4T6S596K5 26U494842 6U4948687 H64484175&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After that, following the hints and the other answers, we use &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the Enigma machine, with the settings given in the hints. Enigma only works on letters, so we obtain: &lt;code&gt;576D4E6B6 247463063 335A30494 8686F4947 566B62485 26E4C6942 5159334D6 762486870 647942706 433427049 47566B624 8526E4948 4A6B596E5 26F494842 6A4948687 A64484175&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And now look at that, those are clearly&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hexadecimal numbers. If we turn them into ASCII again (passing through binary first might be easier), we get: &lt;code&gt;WmNkbGF0c3Z0IHhoIGVkbHRnLiBQY3MgbHhpdyBpd3BpIGVkbHRnIHJkYnRoIHBjIHhzdHAu&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, anybody who has ever had a little experience with it could never fail to recognize&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a base64 string. Decoding it gives: &lt;code&gt;Zcdlatsvt xh edltg. Pcs lxiw iwpi edltg rdbth pc xstp.&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is obviously almost there. Judging by its aspect, it's easy to decide to try&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a Caesar cipher or even a tranposition. And it is a tranposition indeed, a ROT 15.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which gives us the final answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Knowledge is power. And with that power comes an idea.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5990" LastActivityDate="2015-07-10T12:04:37.060" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17731" PostTypeId="1" AcceptedAnswerId="17733" CreationDate="2015-07-10T12:33:46.627" Score="1" ViewCount="377" Body="&lt;p&gt;Smough the Executioner leads you up to to the chopping block, set between two torches like lonely sentinels. A row of authorities and guards, with the king at the epicenter, sit shrouded in darkness behind the block. Your stomach churns. You live in a strange land where Lady Luck tries criminals in place of a jury, and your fate is still undecided. Smough takes an ornate vase from the block and places two slips of parchment within.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He speaks with a booming voice --&#xA;&quot;CHOOSE, AND DECIDE YOUR FATE: LIFE, OR DEATH?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crowd shifts uncomfortably, excitedly. As you reach your hand inside and extract the slip, you realize too late that both papers read &quot;DEATH&quot;! Smough sees the awful realization on your face and grins menacingly. That fiend! He quickly retracts the vase and extends his hand to take the paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;SHOW IT TO ME, THAT I MAY OFFER IT TO THE KING.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you do?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Stipulations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You can't overpower Smough, same as in Dark Souls (the vase is off-limits)&lt;/li&gt;&#xA;&lt;li&gt;Everybody present trusts Smough more than you (they won't take your word over his without reason)&lt;/li&gt;&#xA;&lt;li&gt;You cannot run away or anything stupid like that&lt;/li&gt;&#xA;&lt;li&gt;You have nothing on your person but the clothes on your back (sorry, no concealed knife)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13556" LastEditorUserId="13556" LastEditDate="2015-07-10T12:43:46.680" LastActivityDate="2015-07-10T13:04:44.217" Title="Defy the corrupt executioner!" Tags="|lateral-thinking|story|" AnswerCount="1" CommentCount="1" ClosedDate="2015-07-10T12:49:50.207" ContentLicense="CC BY-SA 3.0" />
  <row Id="17732" PostTypeId="2" ParentId="17726" CreationDate="2015-07-10T12:39:40.753" Score="5" Body="&lt;p&gt;As mentioned before. If you've done these kind of puzzles before a common strategy is to solve row by row but the final two rows together just because it won't always work out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider this bottom right segment of any puzzle larger than 2x4 in correct order:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1234&#xA;567&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;you can now do this to make the top row still correct but the bottom not.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1234&#xA; 567&#xA;&#xA; 234&#xA;1567&#xA;&#xA;2 34&#xA;1567&#xA;&#xA;2534&#xA;1 67&#xA;&#xA;2534&#xA;167&#xA;&#xA;253&#xA;1674&#xA;&#xA; 253&#xA;1674&#xA;&#xA;1253&#xA; 674&#xA;&#xA;1253&#xA;67 4&#xA;&#xA;12 3&#xA;6754&#xA;&#xA;1234&#xA;675&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-07-10T12:39:40.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17733" PostTypeId="2" ParentId="17731" CreationDate="2015-07-10T12:51:56.550" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Swallow the parchment. But since you've already made your pick, your fate is decided, so you ask for the remaining parchment in the vase to be shown. Once it's revealed to have DEATH written on it, it would imply that the parchment you picked had LIFE on it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-07-10T13:04:44.217" LastActivityDate="2015-07-10T13:04:44.217" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17734" PostTypeId="2" ParentId="17726" CreationDate="2015-07-10T13:03:32.293" Score="4" Body="&lt;p&gt;The Answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No it's not automatically resolved (for N&gt;3)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We can minimize the Problem to &quot;can we mix all solved NxN sliding puzzle (with N&gt;3) that way that the last row with the hole is not sorted!&quot; and there is a way. &lt;br /&gt; &lt;br /&gt;&#xA;&#xA; To make things easy:&#xA; We shift the buttom row so that the hole is in the last column &lt;br /&gt;&#xA; We only look at the bottom 4x2 (the rest stays sorted): &lt;br /&gt;&#xA; a = (n-3,n-1) &lt;br /&gt;&#xA; b = (n-2, n-1) &lt;br /&gt;&#xA; c = (n-1, n-1) &lt;br /&gt;&#xA; d = (n, n-1) &lt;br /&gt;&#xA; e = (n-3,n) &lt;br /&gt;&#xA; f = (n-2, n) &lt;br /&gt;&#xA; g = (n-1,n) &lt;br /&gt;&#xA; _ = hole(n,n) &lt;br /&gt; &lt;br /&gt;&#xA; Lets begin sliding: &lt;br /&gt; &lt;br /&gt;&#xA; abcd &lt;br /&gt;&#xA; efg_ &lt;br /&gt;&#xA; &lt;br /&gt;&#xA; abcd &lt;br /&gt;&#xA; _efg &lt;br /&gt;&#xA; &lt;br /&gt;&#xA; _bcd &lt;br /&gt;&#xA; aefg &lt;br /&gt;&#xA; &lt;br /&gt;&#xA; b_cd &lt;br /&gt;&#xA; aefg &lt;br /&gt;&#xA; &lt;br /&gt;&#xA; becd &lt;br /&gt;&#xA; a_fg &lt;br /&gt;&#xA; &lt;br /&gt;&#xA; becd &lt;br /&gt;&#xA; afg_ &lt;br /&gt;&#xA; &lt;br /&gt;&#xA; bec_ &lt;br /&gt;&#xA; afgd &lt;br /&gt;&#xA; &lt;br /&gt;&#xA; _bec &lt;br /&gt;&#xA; afgd &lt;br /&gt;&#xA; &lt;br /&gt;&#xA; abec &lt;br /&gt;&#xA; _fgd &lt;br /&gt;&#xA; &lt;br /&gt;&#xA; abec &lt;br /&gt;&#xA; fg_d &lt;br /&gt;&#xA; &lt;br /&gt;&#xA; ab_c &lt;br /&gt;&#xA; fged &lt;br /&gt;&#xA; &lt;br /&gt;&#xA; abc_ &lt;br /&gt;&#xA; fged &lt;br /&gt;&#xA; &lt;br /&gt;&#xA; abcd &lt;br /&gt;&#xA; fge_ &lt;br /&gt; &lt;br /&gt;&#xA; And that's it. &lt;br /&gt;&#xA; Now the n-1 row is sorted again and the n row is not. So in conclusion there is a way to slide the puzzle that n-1 rows are sorted but the last is not. Perhaps even fewer moves are necessary but that wasn't the question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;edited: I needed more than 23 Minutes to write this??? I didn't saw Ivo's solution...&lt;/p&gt;&#xA;" OwnerUserId="13626" LastActivityDate="2015-07-10T13:03:32.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17735" PostTypeId="1" AcceptedAnswerId="17739" CreationDate="2015-07-10T14:10:32.980" Score="37" ViewCount="2606" Body="&lt;p&gt;This one may turn out to be a little too easy but I really enjoyed writing the Seussian rhymes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Nine tall men are standing tall. Nine tall men here all in all.&lt;br&gt;&#xA;  As they talk amongst each other, &quot;Wait!&quot; they hear, &quot;There is another!&quot;&lt;br&gt;&#xA;  In walks Sergeant Bleighterfest. &quot;There is another!&quot; he protests.&lt;br&gt;&#xA;  &quot;Nothing's missing&quot; says the crew. &quot;We're all here. What's wrong with you?&quot;&lt;br&gt;&#xA;  &quot;He is the youngest - I'll admit - but he's important, every bit.&quot;&lt;br&gt;&#xA;  &quot;Bah,&quot; they say, &quot;We know that slob. He is useless at our job.&lt;br&gt;&#xA;  He sulks and hides and rarely talks. He's certainly no chatterbox.&quot;&lt;br&gt;&#xA;  &quot;Is that a reason to exclude? To mock and joke and be so rude?&lt;br&gt;&#xA;  By himself he is not much,&quot; admits the Sergeant to the bunch.&lt;br&gt;&#xA;  &quot;Everyone must help him be all that I know that he can be.&lt;br&gt;&#xA;  I know his task he will not shirk but, without you, he cannot work!&quot;&lt;br&gt;&#xA;  Here the Sergeant paused to think where could that little guy have slinked?&lt;br&gt;&#xA;  Snorkeling at Sharktooth Reef? I've never held such disbelief!&lt;br&gt;&#xA;  Diving off our tallest peak? That cannot be the place I seek.&lt;br&gt;&#xA;  Wait a minute, I'm so dense! I needn't be in such suspense.&lt;br&gt;&#xA;  &quot;If he's gone – walked out the door – he's even more here than before!&quot;&lt;br&gt;&#xA;  It was just then he realized the last had been here all the time.&lt;br&gt;&#xA;  The Sergeant spun until he spotted the space for little one allotted&lt;br&gt;&#xA;  Filled now - someone's in the spot. &quot;There you are, you silly ___________!&quot;   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;- Who was missing?&lt;br&gt;&#xA; - How would you fill in the blank line at the end?&lt;br&gt;&#xA; - Can you find the other group members hiding in the riddle?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2015-07-11T14:08:27.523" LastActivityDate="2015-07-11T14:08:27.523" Title="Nine Tall Men Are Standing Tall" Tags="|riddle|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17736" PostTypeId="1" AcceptedAnswerId="17737" CreationDate="2015-07-10T14:12:30.340" Score="3" ViewCount="276" Body="&lt;p&gt;An artist duo (Artist A and Artist B) were hired to make a large blue sphere resembling the earth. They decided to make it out of three sections, two spherical caps and one middle ring with the same width when seen with a 2D perspective (see picture). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When they were deciding how to split up the work, Artist A said he wanted to paint the two spherical caps, leaving Artist B with the middle section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who is getting the better deal (i.e. who gets to use less paint)? If one is painting more, how much more paint will that artist have to use?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oowln.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 2:&#xA;If they had split up the sphere into 5 parts of equal width and Artist A still got the two spherical caps while Artist B got the middle strip, then what would the answer be? (The remaining two strips would be outsourced)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gi1uA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: This will probably be answered within minutes, but for future viewers, try answering this without looking it up before checking the answer.&lt;/p&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="7722" LastEditDate="2015-07-10T14:26:54.293" LastActivityDate="2015-07-10T15:32:11.700" Title="Painting a large sphere" Tags="|geometry|visual|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17737" PostTypeId="2" ParentId="17736" CreationDate="2015-07-10T14:33:13.840" Score="1" Body="&lt;p&gt;For part 1, I used a sphere with a diameter of 1 (so $h = \frac{1}{3}$)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Artist B (middle section) gets the better deal (paints half as much).&lt;br&gt;&#xA; All 3 section areas are the same:&lt;br&gt;&#xA; Cap Area is $2\pi rh = 2\pi \frac{1}{2}\frac{1}{3} = \frac{\pi}{3}$&lt;br&gt;&#xA; Total Area of sphere is $4\pi r^2 = 4\pi \frac{1}{2}^2 = \pi $&lt;br&gt;&#xA; So, each cap is $\frac{\pi}{3}$, meaning the middle section is $\frac{\pi}{3}$ also. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For part 2, I used a sphere with a diameter of 1  (so $h = \frac{1}{5}$)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Artist B (middle section) gets the better deal (paints half as much).&lt;br&gt;&#xA;  All 5 section areas are the same:&lt;br&gt;&#xA; Cap Area of 2 end sections is $2\pi rh = 2\pi \frac{1}{2}\frac{2}{5} = \frac{2\pi}{5}$&lt;br&gt;&#xA; Cap Area of end section is $2\pi rh = 2\pi \frac{1}{2}\frac{1}{5} = \frac{\pi}{5}$&lt;br&gt;&#xA; Total Area of sphere is $4\pi r^2 = 4\pi \frac{1}{2}^2 = \pi $&lt;br&gt;&#xA; So, each of the 5 parts is $\frac{\pi}{5}$, meaning the middle section is $\frac{\pi}{5}$ also. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-07-10T15:32:11.700" LastActivityDate="2015-07-10T15:32:11.700" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="17738" PostTypeId="2" ParentId="17004" CreationDate="2015-07-10T14:34:33.743" Score="3" Body="&lt;p&gt;&lt;strong&gt;PARTIAL ANSWER&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my contribution to the gargoyle part since I can't add comments yet.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A curse is similar to a hex. @rand al'thor mentioned greyscale and so I'm wondering if it is the hex value for slate grey. The hex triplet for that is #708090 according to &lt;a href=&quot;http://colors.findthedata.com/q/511/10857/What-are-the-RGB-values-of-Slate-Gray&quot; rel=&quot;nofollow&quot;&gt;http://colors.findthedata.com/q/511/10857/What-are-the-RGB-values-of-Slate-Gray&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I started thinking about the next part&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Here, we don't believe in order of operations. Remove it from the system.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hex is represented as 0xstuff so remove the order of operations i.e. the x (multiplication) and your left with 0708090 but that doesn't mean anything to me and seems too simple compared to the rest of the puzzle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;another thought about &lt;em&gt;The remainder will be comical. Excellent cake, seared deeply, sounds like it will start you on the right path.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The joke why is 6 afraid of 7? because seven ate nine (789). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 7 because 7 ate 9.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13925" LastEditorUserId="13925" LastEditDate="2015-07-24T12:25:28.413" LastActivityDate="2015-07-24T12:25:28.413" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17739" PostTypeId="2" ParentId="17735" CreationDate="2015-07-10T14:38:23.930" Score="32" Body="&lt;p&gt;The hiding of the &quot;nine tall men&quot; within the rhyme is genius!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rhyme ends with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;There you are, you silly naught!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The nine tall men are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The digits from 1 to 9&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The missing &quot;person&quot; is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The digit 0&lt;br&gt;&#xA;&lt;br&gt;&#xA; By itself, a zero can't do much; it needs the other digits to &quot;do its work.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The other nine are highlighted below (#6 is backwards and spans 2 lines -- thanks, @mmking!):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nine tall men are standing tall. &lt;strong&gt;Nine&lt;/strong&gt; tall men here all in all.&lt;br&gt;&#xA; As they talk amongst each other, &quot;Wait!&quot; they hear, &quot;There is another!&quot;&lt;br&gt;&#xA; In walks Sergeant Bl&lt;strong&gt;eight&lt;/strong&gt;erfest. &quot;There is another!&quot; he protests.&lt;br&gt;&#xA; &quot;Nothing's missing&quot; says the crew. &quot;We're all here. What's wrong with you?&quot;&lt;br&gt;&#xA; &quot;He is the youngest - I'll admit - but he's important, every bit.&quot;&lt;br&gt;&#xA; &quot;Bah,&quot; they say, &quot;We know that slob. He is useless at our job.&lt;br&gt;&#xA; He sulks and hides and rarely talks. He's certainly no chatterbo&lt;strong&gt;x.&quot;&lt;br&gt;&#xA; &quot;Is&lt;/strong&gt; that a reason to exclude? To mock and joke and be so rude?&lt;br&gt;&#xA; By himself he is not much,&quot; admits the Sergeant to the bunch.&lt;br&gt;&#xA; &quot;Everyone must help him be all that I know that he can be.&lt;br&gt;&#xA; I know his task he will not shirk but, without you, he canno&lt;strong&gt;t wo&lt;/strong&gt;rk!&quot;&lt;br&gt;&#xA; Here the Sergeant paused to think Where could that little guy have slinked?&lt;br&gt;&#xA; Snorkeling at Sharktoo&lt;strong&gt;th Ree&lt;/strong&gt;/&lt;strong&gt;f? I've&lt;/strong&gt; never held such disbelief!&lt;br&gt;&#xA; Diving of&lt;strong&gt;f our&lt;/strong&gt; tallest peak? That cannot be the place I seek.&lt;br&gt;&#xA; Wait a minute, I'm so dense! I needn't be in such suspense.&lt;br&gt;&#xA; &quot;If he's g&lt;strong&gt;one&lt;/strong&gt; – walked out the door – he'&lt;strong&gt;s even&lt;/strong&gt; more here than before!&quot;&lt;br&gt;&#xA; It was just then he realized the last had been here all the time.&lt;br&gt;&#xA; The Sergeant spun until he spotted the space for little one allotted&lt;br&gt;&#xA; Filled now - someone's in the spot. &quot;There you are, you silly ___________!&quot;   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="4421" LastEditDate="2015-07-10T15:10:45.243" LastActivityDate="2015-07-10T15:10:45.243" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="17746" PostTypeId="1" AcceptedAnswerId="17750" CreationDate="2015-07-10T16:54:08.897" Score="4" ViewCount="278" Body="&lt;p&gt;This is a small puzzle that I saw a lot as a kid but surprisingly haven't found on this site. Enjoy!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You have six frogs arranged on seven lily pads in a line: three male frogs facing right on the leftmost lily pads and three female frogs facing left on the rightmost lily pads. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;M M M _ F F F&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How can you switch the positions of the frogs (have the male frogs end up where the female frogs currently sit and vice versa)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Each lily pad can hold only one frog.&lt;/li&gt;&#xA;&lt;li&gt;Frogs can only move in the direction they are facing.&lt;/li&gt;&#xA;&lt;li&gt;Frogs can move forward to an empty lily pad or jump over one frog to the empty lily pad on the other side.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13881" LastActivityDate="2015-07-10T17:10:07.033" Title="Rearranging Frogs" Tags="|strategy|" AnswerCount="2" CommentCount="4" ClosedDate="2015-07-10T17:39:32.400" ContentLicense="CC BY-SA 3.0" />
  <row Id="17747" PostTypeId="5" CreationDate="2015-07-10T16:59:02.010" Score="0" Body="&lt;p&gt;Sliding puzzles consist of tiles on a grid which can be moved into adjacent gaps. The goal is to reach a particular configuration of tiles, such as numerical order or freeing a large tile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In one type of sliding puzzle, the tiles typically have numbers or parts of a picture on them, and the goal is to either reorder the numbers or make the picture whole again. Generally, the pieces only take up one square of the grid, and there is only one empty square. The most common puzzle of this type is the &lt;a href=&quot;https://en.wikipedia.org/wiki/15_puzzle&quot; rel=&quot;nofollow&quot;&gt;15 Puzzle&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the other common type, tiles consist of multiple different squares. The goal is usually to free a specific tile from the grid, usually through a &quot;gate&quot; that no other tile may pass through. The most well-known of these is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Klotski&quot; rel=&quot;nofollow&quot;&gt;Klotski&lt;/a&gt; puzzle, also known as the Huarong Dao, in which a large 2×2 block must be moved from the top of a 5×4 grid to the bottom. (Klotski may also refer to the genre of sliding block puzzles as a whole.)&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorUserId="11876" LastEditDate="2016-06-18T15:26:36.697" LastActivityDate="2016-06-18T15:26:36.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17748" PostTypeId="4" CreationDate="2015-07-10T16:59:02.010" Score="0" Body="A puzzle comprising a set of tiles on a grid. Tiles can be shifted into adjacent gaps, and the goal is to reach a certain configuration." OwnerDisplayName="user20" LastEditorUserId="11876" LastEditDate="2016-06-18T15:26:41.853" LastActivityDate="2016-06-18T15:26:41.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17750" PostTypeId="2" ParentId="17746" CreationDate="2015-07-10T17:05:18.307" Score="3" Body="&lt;p&gt;Basic idea:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They need to try to mix perfectly - MFMFMF&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Process:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MMM_FFF&lt;br&gt;MM_MFFF&lt;br&gt;MMFM_FF&lt;br&gt;MMFMF_F&lt;br&gt;MMF_FMF&lt;br&gt;M_FMFMF&lt;br&gt;_MFMFMF&lt;br&gt;FM_MFMF&lt;br&gt;FMFM_MF&lt;br&gt;FMFMFM_&lt;br&gt;FMFMF_M&lt;br&gt;FMF_FMM&lt;br&gt;F_FMFMM&lt;br&gt;FF_MFMM&lt;br&gt;FFFM_MM&lt;br&gt;FFF_MMM&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, I'm also surprised this isn't on the site already in a different form. It's quite possible that I just didn't search for the right terms.&lt;/p&gt;&#xA;" OwnerUserId="1955" LastActivityDate="2015-07-10T17:05:18.307" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17751" PostTypeId="2" ParentId="17746" CreationDate="2015-07-10T17:06:06.297" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MMM_FFF&lt;br&gt;&#xA; MMMF_FF&lt;br&gt;&#xA; MM_FMFF&lt;br&gt;&#xA; M_MFMFF&lt;br&gt;&#xA; MFM_MFF&lt;br&gt;&#xA; MFMFM_F&lt;br&gt;&#xA; MFMFMF_&lt;br&gt;&#xA; MFMF_FM&lt;br&gt;&#xA; MF_FMFM&lt;br&gt;&#xA; _FMFMFM&lt;br&gt;&#xA; F_MFMFM&lt;br&gt;&#xA; FFM_MFM&lt;br&gt;&#xA; FFMFM_M&lt;br&gt;&#xA; FFMF_MM&lt;br&gt;&#xA; FF_FMMM&lt;br&gt;&#xA; FFF_MMM !&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13916" LastEditorUserId="13916" LastEditDate="2015-07-10T17:10:07.033" LastActivityDate="2015-07-10T17:10:07.033" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17752" PostTypeId="1" AcceptedAnswerId="17757" CreationDate="2015-07-10T17:45:13.793" Score="10" ViewCount="933" Body="&lt;p&gt;Six musicians are participating in a music festival. During each concert, some of the musicians perform, while the others listen in the audience. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the fewest number of concerts needed so every musician has listened (as an audience member) to every other musician perform?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;em&gt;Aust MS Gazette&lt;/em&gt;, Volume 42, No. 2, in the Puzzle Corner section. See &lt;a href=&quot;http://www.austms.org.au/Publ/Gazette/2015/May15/Puzzle.pdf&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-07-10T19:06:07.770" Title="Six Musical Friends" Tags="|strategy|" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17753" PostTypeId="1" AcceptedAnswerId="18093" CreationDate="2015-07-10T17:51:48.883" Score="7" ViewCount="728" Body="&lt;p&gt;Bob, a budding physicist, is sitting on a swivel chair at rest, holding out a large motorized flywheel in front of him. The wheel is oriented parallel to the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bob's swivel chair is a magical, 100% completely frictionless swivel chair, and Bob's weight is entirely supported by it (i.e. his feet aren't touching the ground). When Bob turns on the motorized flywheel, the wheel starts to spin counterclockwise. Naturally, as a physicist would expect, Bob in his chair immediately begins to rotate &lt;em&gt;clockwise&lt;/em&gt;, since total angular momentum must be conserved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a fun bout of spinning, Bob deactivates the motor and the flywheel abruptly stops turning. Naturally, because angular momentum is conserved, Bob's chair also abruptly stops spinning so that Bob is once again at rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sandra, another physicist, sees the fun Bob is having and decides to try the same thing with her own swivel chair. However, Sandra's chair isn't a magic frictionless chair like Bob's. It &lt;em&gt;is&lt;/em&gt; very well oiled and has a low coefficient of friction, but the coefficient isn't exactly zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Undeterred, Sandra starts up her counterclockwise-spinning flywheel in the same way as Bob did, and rejoices as she begins to spin clockwise in her chair. But then, when she kills the motor and the flywheel abruptly stops spinning, she discovers to her surprise that she is not at rest. Instead, something very peculiar&amp;mdash;some might even say &quot;remarkable&quot;&amp;mdash;happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;She repeats the experiment several times and notes with amazement that the peculiar phenomenon is reproducible no matter what the starting orientation of her chair and no matter how long she runs the flywheel before stopping it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the phenomenon? What happens when Sandra stops the flywheel?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2015-07-19T08:11:59.703" Title="Memento Vivere; Momentum Mori" Tags="|physics|" AnswerCount="3" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="17754" PostTypeId="2" ParentId="17752" CreationDate="2015-07-10T17:52:22.687" Score="6" Body="&lt;p&gt;Unless I'm missing something, the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;pre&gt;abc def&lt;br&gt;ade bcf&lt;br&gt;fcd abe&lt;br&gt;fbe acd&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2015-07-10T18:18:01.900" LastActivityDate="2015-07-10T18:18:01.900" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17755" PostTypeId="2" ParentId="17752" CreationDate="2015-07-10T17:52:37.220" Score="2" Body="&lt;p&gt;OK, I will go ahead and kick things off with&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;6, as follows.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/GSGyk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm still trying other configurations to see if this is optimal.&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="1777" LastEditDate="2015-07-10T18:02:55.897" LastActivityDate="2015-07-10T18:02:55.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17756" PostTypeId="2" ParentId="17752" CreationDate="2015-07-10T17:58:01.917" Score="6" Body="&lt;p&gt;Performing&#xA;Audience &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;P    A&#xA;abcd ef&#xA;cdef ab&#xA;abef cd&#xA;&#xA;fbd eac&#xA;eac fbd&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;5 performances but I think it can done in less. &lt;/p&gt;&#xA;" OwnerUserId="11297" LastEditorUserId="11297" LastEditDate="2015-07-10T18:39:08.093" LastActivityDate="2015-07-10T18:39:08.093" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17757" PostTypeId="2" ParentId="17752" CreationDate="2015-07-10T18:07:06.733" Score="15" Body="&lt;p&gt;I believe I have it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Four concerts!&#xA; &lt;pre&gt;&#xA; Concert format: --- are playing for xxx&#xA; --- xxx&#xA; Concert |Done performing |Done listening&#xA; abc def |                |&#xA; cde abf |C               |          F&#xA; aef bdc |AE              |         DB&#xA; bdf cae |BDF             |        CAE&#xA; &lt;/pre&gt;&#xA; &lt;pre&gt;&#xA; My newest reasoning: &#xA; This problem is about relationships &#xA; In a 1 performing for 5, 5 relationships are formed, X is heard by abcdef &#xA; In a 3 performing for 3, 9 relationships are formed &#xA; a is heard by def | b is heard by def | c is heard by def&#xA; so 3x3 is more efficient achieving the number of relatioships needed &#xA;&#xA; So there are 6 people that need to be heard by 5 people, this is 30 relationships&#xA; To do this in 3 turns we would need to generate 10 relationships per concert.&#xA; Looking at the concert options there are only 0x6, 1x5, 2x4, 3x3 and symmetric concerts&#xA; which are the same production of relationships.&#xA; These produce, 0, 5, 8, and 9 respectively. &#xA; None of these are greater than 10 (Last I checked)&#xA; However, very interestingly this means that it is potentially possible that&#xA; 2x4 could produce a solution, however this does not prove it possible as there could&#xA; be duplicate relationships forced.&#xA; But it does prove 4 concerts in the minimum.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;/pre&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to mmking for format help&lt;/p&gt;&#xA;" OwnerUserId="13916" LastEditorUserId="13916" LastEditDate="2015-07-10T18:31:55.090" LastActivityDate="2015-07-10T18:31:55.090" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17758" PostTypeId="2" ParentId="17753" CreationDate="2015-07-10T18:22:54.563" Score="7" Body="&lt;p&gt;I believe what happens is that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Once the flywheel stops abruptly, Sandra slows down with the flywheel and as it comes to rest, Sandra also momentarily comes to rest very slightly before the flywheel, but then starts spinning in the opposite direction, in the same direction as the flywheel when it was spinning. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am not very sure, but I feel the cause for this might be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When the flywheel slows down, Sandra slows down faster as some of her angular momentum is lost due to the frictional force. So Sandra comes to a stop but the flywheel still has some momentum, and as it grinds to a halt, to conserve momentum, Sandra gains some in the direction of the flywheel's rotation. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-07-10T18:28:10.410" LastActivityDate="2015-07-10T18:28:10.410" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17759" PostTypeId="1" AcceptedAnswerId="17768" CreationDate="2015-07-10T18:24:16.517" Score="2" ViewCount="798" Body="&lt;p&gt;You are given a single long rope of length L (ft) with radius R (in), a very sharp knife, and 2 matches. Your task is to completely burn the rope in the minimum amount of time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Calculate the total time required to burn the rope.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Details:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The rope is considered stiff with a minimum bend radius of 10*R&lt;/li&gt;&#xA;&lt;li&gt;The rope is tightly woven with many fibers and may be considered to have a solid cross section&lt;/li&gt;&#xA;&lt;li&gt;You may cut the rope into more than one piece&lt;/li&gt;&#xA;&lt;li&gt;The rope burns at R in/min&lt;/li&gt;&#xA;&lt;li&gt;A match will light the rope instantaneously and will burn out as soon as the rope is touched&lt;/li&gt;&#xA;&lt;li&gt;Assume gravity&lt;/li&gt;&#xA;&lt;li&gt;Ropes in contact can transfer fire immediately to one another at the contact point only, but will themselves continue to burn at the constant specified burn rate&lt;/li&gt;&#xA;&lt;li&gt;Static structures&lt;/li&gt;&#xA;&lt;li&gt;The matches dimensions are 4R x R/4 x R/4&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Be creative&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hints:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Think about 3D structures&lt;/li&gt;&#xA;&lt;li&gt;Due to gravity a structure should be able to hold it's shape&lt;/li&gt;&#xA;&lt;li&gt;Determine the burn rate per unit volume of rope material&lt;/li&gt;&#xA;&lt;li&gt;Consider the shape of the structure sub-elements and how they transfer fire to one another&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13788" LastEditorUserId="13788" LastEditDate="2015-07-10T21:52:31.897" LastActivityDate="2015-07-10T21:54:20.817" Title="Optimal Rope Burning" Tags="|mathematics|geometry|" AnswerCount="8" CommentCount="17" ClosedDate="2015-07-11T01:08:52.760" ContentLicense="CC BY-SA 3.0" />
  <row Id="17760" PostTypeId="2" ParentId="17753" CreationDate="2015-07-10T18:25:18.607" Score="4" Body="&lt;p&gt;Okay here we go freshman physics!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  She is spinning the opposite way that she was while the wheel was spinning! &lt;br&gt;&#xA; Angular momentum is conserved in the universe, not in here system due to friction. &lt;br&gt;&#xA; So the angular momentum lost to friction (that effectively goes to the earth, not that the earth notices much) &lt;br&gt;&#xA; But in the system the motorized wheel still had its full angular momentum. &lt;br&gt;&#xA; So the net angular momentum is in the way the wheel was spinning, but now it is not spinning so something (or someone) else has to spin that way. &lt;br&gt;&#xA; So our spinning physicist is going around the other way.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Update to address COTO's Comment:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SHE SPINS BACK AS LONG AS SHE SPUN FORWARD! &lt;br&gt; &lt;br&gt;&#xA; How much angular momentum was lost? Well Torque = Force*distance (for perpendicular) &lt;br&gt;&#xA; Force is from friction so $\mu$*NormalForce&lt;em&gt;distance &lt;br&gt;&#xA; So torque is constant while the physicist is spinning &lt;br&gt;&#xA; So Net angular momentum shift is Time&lt;/em&gt;$\mu$*Weight*distance &lt;br&gt;&#xA; Well now the wheel stops and we spin the other way &lt;br&gt;&#xA; So we are getting torqued, in the other direction &lt;br&gt;&#xA; well the net angular momentum shift divided by Time*$\mu$*weight*distance &lt;br&gt;&#xA; wait, what is time? time is how long the physicist went in the initial direction &lt;br&gt;&#xA; so how long will it take them to stop? &lt;br&gt;&#xA; well AM/Torque = Time*$\mu$*weight*distance/($\mu$*weight*distance) =time &lt;br&gt;&#xA; wait what? &lt;br&gt;&#xA; SHE SPINS BACK AS LONG AS SHE SPUN FORWARD! &lt;br&gt;&#xA; even if she waits till she stops the first way, since she is no longer going she is no longer getting torqued!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13916" LastEditorUserId="13916" LastEditDate="2015-07-10T20:30:57.870" LastActivityDate="2015-07-10T20:30:57.870" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="17761" PostTypeId="2" ParentId="17759" CreationDate="2015-07-10T18:36:21.490" Score="5" Body="&lt;p&gt;I have a proposal to minimize the burning time. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As we can twist the rope into circles of radius $10R$, I don't think we need to make any cuts. Instead, we turn it into a circle of the minimum radius and then overlap it on top as we complete a circle. As we keep doing this, we'll end up with a circular spiral of rope. Now we burn the rope at two points farthest to each other i.e. at one point on the bottom circle and its diametrically opposite point on the top circle, and let the fire burn across the spiral. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-07-10T18:36:21.490" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17762" PostTypeId="2" ParentId="17752" CreationDate="2015-07-10T18:37:42.347" Score="6" Body="&lt;p&gt;Solutions for four have been posted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three is impossible, because each musician must either:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;play alone (and have all 5 listen)&lt;/li&gt;&#xA;&lt;li&gt;play at least twice (if they are not playing alone, they must play for musician(s) playing with them the first time).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;They must also either&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;be in the audience alone (and listen to all five)&lt;/li&gt;&#xA;&lt;li&gt;be in the audience at least twice (they must listen to the person who was in the audience with them the first time)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For everyone to listen twice and play twice, 24 'slots' are needed, but only 18 are available. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one plays alone, then in the remaining two concerts, the 5 others must all play twice, which means they cannot listen again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one listens alone, then in the remaining two concerts, the 5 others must all listen twice, which means they cannot play again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore it is impossible with three concerts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no solution for 4 concerts other than for them all to be 3 vs 3, because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nobody can listen to anybody more than twice or play for anybody more than twice, there are only 24 slots, and we've already established 24 are required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So suppose ABCD played to EF. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A would still need to play to B and C and D, and EF cannot listen to A again, so AEF to BCD is the next concert.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B also needs to play to ACD. Now E and F have both played to C and heard C, so cannot do either again without going over 24 spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So no 4 vs 2 concerts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does not matter in fact whether E was listening in the first concert or playing, so the 5 vs 1 concert is out as well.&lt;/p&gt;&#xA;" OwnerUserId="13993" LastEditorUserId="13993" LastEditDate="2015-07-10T19:06:07.770" LastActivityDate="2015-07-10T19:06:07.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17763" PostTypeId="2" ParentId="17759" CreationDate="2015-07-10T18:39:24.703" Score="11" Body="&lt;p&gt;Here's my idea: A powder will have more surface area exposed to the flame, so it &lt;strong&gt;should&lt;/strong&gt; burn much faster than a solid rope, so the strategy is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. Cut the rope up into as many tiny particles as possible, to the point where the knife cannot cut any piece any thinner.&lt;br&gt;&#xA; 2. Arrange the &quot;rope powder&quot; thinly and evenly in 2 circles.&lt;br&gt;&#xA; 3. Place one lit match in the center of each circle, and watch the &quot;rope&quot; burn from the inside of each circle to the perimeter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-07-10T21:02:16.203" LastActivityDate="2015-07-10T21:02:16.203" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="17764" PostTypeId="2" ParentId="14537" CreationDate="2015-07-10T18:44:32.793" Score="-1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Seamus&lt;/strong&gt; is also a proper name that fits the criterion (pronounced SHAY-mus).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13995" LastEditorUserId="11244" LastEditDate="2015-07-10T19:34:14.540" LastActivityDate="2015-07-10T19:34:14.540" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17766" PostTypeId="2" ParentId="17759" CreationDate="2015-07-10T19:10:04.740" Score="2" Body="&lt;p&gt;Assuming rope contact doesn't cause the other ropes to burn and you can only light one rope with one match:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cut the rope in half then light the two ropes. &lt;br&gt;&lt;br&gt;Otherwise, if you can (a) make other ropes burn each other, then see 2012rcampion's sphere comment, and (b) if you can light more than one rope with a match but not have the ropes light each other, then you'd have to arrange strips of rope that come together at a point (probably in the shape of a sphere somehow), but at this point the solution is too convoluted to make sense anyway.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastActivityDate="2015-07-10T19:10:04.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17768" PostTypeId="2" ParentId="17759" CreationDate="2015-07-10T19:30:50.767" Score="4" Body="&lt;p&gt;Cut the rope into a large number of very thin, tapered strips of length $\sqrt[3]{\frac{9LR^2}{2}}$.  Arrange these strips into two spheres, with the pointed end of each strip at the center of its sphere.  Support the lower half of each sphere with a hemispherical bowl of the correct radius&lt;sup&gt;1&lt;/sup&gt;.  However, leave an infinitesimally thin cone open from the top of each sphere to its center.  Simultaneously drop each match directly above the centers of their corresponding spheres, so that both spheres simultaneously burn from the inside out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total volume of the rope is $12L\cdot\pi R^2$ (the factor of $12$ comes from the fact that $L$ is in feet, while $R$ is in inches), so the volume of each sphere is $6\pi LR^2$ and the radius is $\sqrt[3]{\frac{9LR^2}{2}}$; thus the amount of time it takes for the two spheres to burn is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sqrt[3]{\frac{9L}{2R}}~\text{minutes}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt;: If we're not allowed to use a premade bowl, then slice off a large number of infinitesimally thin strips from the rope and use them to weave a bowl of the correct size and shape.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Alternatively, if your knife is &lt;em&gt;very&lt;/em&gt; sharp, make use of &lt;a href=&quot;https://en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox&quot; rel=&quot;nofollow&quot;&gt;Banach-Tarski&lt;/a&gt; to rearrange the rope into a pair of spheres of arbitrarily small radius, which will then burn in an arbitrarily short amount of time.&lt;/p&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="9718" LastEditDate="2015-07-10T19:59:13.290" LastActivityDate="2015-07-10T19:59:13.290" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17769" PostTypeId="1" CreationDate="2015-07-10T19:35:26.340" Score="11" ViewCount="731" Body="&lt;p&gt;Can you solve the witch's trap and break the curse they placed on you?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Vixen's herbs in cauldrons mixing,&#xA;  &lt;br&gt;wins the next without descending,&#xA;  &lt;br&gt;misguided hearts now truly mending.&#xA;  &lt;br&gt;&lt;br&gt;The Senior's Apprentice slowly sighing,&#xA;  &lt;br&gt;melting, melting, slowly dying.&#xA;  &lt;br&gt;The basilisk's fears are mixed inside.&#xA;  &lt;br&gt;&lt;br&gt;Sunset-hued trees are tossed around,&#xA;  &lt;br&gt;moldy dots easy to pound.&#xA;  &lt;br&gt;breadworms scream without a sound,&#xA;  &lt;br&gt;the pot now slowly rising.&#xA;  &lt;br&gt;stirring, stirring, black cats purring, the fire's slowly dying.&#xA;  &lt;br&gt;&lt;br&gt;The witch's brew is near the top,&#xA;  &lt;br&gt;served in turkey's clothes, you hear a plop!&#xA;  &lt;br&gt;the wicked potion, now still steaming,&#xA;  &lt;br&gt;served by wicked eyes a'gleaming.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;What is there inside the pot?&#xA;&lt;br&gt;you must show the steps.&#xA;&lt;br&gt;To cure your rot right on the spot,&#xA;&lt;br&gt;and then pay your respects!&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTES&lt;/strong&gt;&#xA;I just deciphered a strange code scrawled on the back cover in tiny braille morse code, which roughly translates as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A couple of lines are mostly for rhyming purposes, and most of them refer to a particular ingredient, or a method to get the next step. Work it more like a cryptic crossword if that helps, but don't take it too deep, ne? ;P&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Puzzling Thoughts collected from my own personal ruminations:&#xA;&lt;br&gt;- &lt;em&gt;Rumination &lt;strong&gt;1&lt;/strong&gt;&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It seems to me like the first line of every stanza is a description/reference to something...and, what could misguided hearts have to do with witches and gibberish?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;- &lt;em&gt;Rumination &lt;strong&gt;2&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It appears that for the first 2 stanzas, each sentence is an individual ingredient. After that, each line is either an ingredient or instruction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;- &lt;em&gt;Rumination &lt;strong&gt;3&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For the third stanza, think colors: what kind of tree is colored like a sunset? what kind of sphere may look like mold? What would a breadworm look like?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13697" LastEditorUserId="13697" LastEditDate="2015-07-28T17:00:20.230" LastActivityDate="2016-12-02T17:19:26.617" Title="The Witch's Recipe" Tags="|rhyme|enigmatic-puzzle|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17770" PostTypeId="2" ParentId="17759" CreationDate="2015-07-10T19:41:54.770" Score="5" Body="&lt;p&gt;This is a new and different answer than the one I posted previously and I felt it warranted its own spot. My apologies if this is bad form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solution:&#xA;Cut the rope into X+1 pieces, where X is the number of atoms in the rope.&#xA;Upon splitting an atom, all of the pieces of the rope, along with the two matches and the person doing the cutting will be consumed in nuclear fire and destroyed in an atomic explosion. This burns the rope in an infinitely tiny fraction of a second, but sadly destroys the evidence. Such is the price of efficiency.&lt;/p&gt;&#xA;" OwnerUserId="13068" LastActivityDate="2015-07-10T19:41:54.770" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17771" PostTypeId="2" ParentId="17759" CreationDate="2015-07-10T19:44:36.933" Score="3" Body="&lt;p&gt;An atypical idea..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Burn time: &lt;b&gt;(Speed of your hands)*L /(Number of cuts)&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Cut the rope into 2 (or more) equal pieces and place them side by side (or stacked for &gt;2). Then slide one rope along such that it is just touching the first rope at one end. Then light the point where both ends are touching, and push one rope along the other rope as fast as possible. This will burn the other rope and your rope at that same speed (which will be as fast as you can move the rope pieces together).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-07-10T20:10:26.700" LastActivityDate="2015-07-10T20:10:26.700" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17772" PostTypeId="2" ParentId="17759" CreationDate="2015-07-10T19:49:37.187" Score="1" Body="&lt;p&gt;If each rope is immediately lit up when it comes in contact with fire,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chop it up so that it has minimum length $l$ inches. Arrange each piece side by side such that it forms a rectangle. Light up a corner of the rectangle. The fire will spread across the ends of the rope immediately, so basically all ropes start burning at the same time. It will take $l/R$ minutes to burn one piece, so it will take $l/R$ minutes to burn every single piece.&lt;br&gt;&#xA;&lt;br&gt;&#xA; So now it boils down to how small you can chop each piece. When $l\to0$, $l/R\to0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12109" LastActivityDate="2015-07-10T19:49:37.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17773" PostTypeId="1" AcceptedAnswerId="17782" CreationDate="2015-07-10T19:55:01.893" Score="11" ViewCount="258" Body="&lt;p&gt;The Wonderful Wanderful land of Puzz&lt;br&gt;&#xA;Is where you and I now are;&lt;br&gt;&#xA;Where magazines eat wheeled things,&lt;br&gt;&#xA;And the whole world is bizarre.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove my point, I found a joint&lt;br&gt;&#xA;Where brochures came to cope.&lt;br&gt;&#xA;&quot;I can't &lt;code&gt;_ _ _ _ _ (_)&lt;/code&gt; this &lt;code&gt;(_) _ _ _ _&lt;/code&gt;,&quot;&lt;br&gt;&#xA;A weeping &lt;code&gt;_ _ _ _ _ _&lt;/code&gt; &lt;code&gt;_ _ (_) _ _&lt;/code&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Don't &lt;code&gt;_ _ (_) _ _&lt;/code&gt; away here,&quot; I declared.&lt;br&gt;&#xA;&quot;We'll fix this with a plan.&lt;br&gt;&#xA;Have you tried &lt;code&gt;_ _ _ _ _&lt;/code&gt;? If you're from Puzz,&lt;br&gt;&#xA;you &lt;code&gt;_ _ _&lt;/code&gt; buy it like bread -- in &lt;code&gt;(_) _ _&lt;/code&gt;s!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I cannot use my funds,&quot; he said,&lt;br&gt;&#xA;&quot;As my &lt;code&gt;(_) _ _ _&lt;/code&gt; will not &lt;code&gt;_ _ _ (_) _ _&lt;/code&gt;.&lt;br&gt;&#xA;I have a rifle &lt;code&gt;_ _ _ _ _ (_)&lt;/code&gt;, though,&lt;br&gt;&#xA;And the river &lt;code&gt;_ _ _ _&lt;/code&gt;'s where they sit!&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;He snagged a rifle full of &lt;code&gt;_ _ _ _&lt;/code&gt;&lt;br&gt;&#xA;and said, &quot;&lt;code&gt;_ _ (_) _&lt;/code&gt; me to the &lt;code&gt;(_) _ _ _&lt;/code&gt;!&quot;&lt;br&gt;&#xA;When we arrived, we found a tree&lt;br&gt;&#xA;and &lt;code&gt;_ (_) _ _&lt;/code&gt;ed there until dark.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the &lt;code&gt;_ _ _ (_)&lt;/code&gt; of a hill, we stood quite still&lt;br&gt;&#xA;His gun &lt;code&gt;_ _ _ _&lt;/code&gt;ed to his eye&lt;br&gt;&#xA;When a &lt;code&gt;_ _ _ _&lt;/code&gt; away, around midday,&lt;br&gt;&#xA;A &lt;code&gt;_ _ (_) _&lt;/code&gt; came wandering by.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aiming at the poor thing's &lt;code&gt;(_) _ _ _&lt;/code&gt;,&lt;br&gt;&#xA;He shot it &lt;code&gt;_ _ _ _&lt;/code&gt;. 'Twas dead!&lt;br&gt;&#xA;&lt;code&gt;_ _ _ _&lt;/code&gt; at his home, he stripped its coat,&lt;br&gt;&#xA;Ate just its &lt;code&gt;_ (_) _ _&lt;/code&gt;, then fled...  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wandered to a nearby &lt;code&gt;_ _ _&lt;/code&gt;&lt;br&gt;&#xA;Which &lt;code&gt;_ (_) _ _ _&lt;/code&gt;! Or, so I'm told.&lt;br&gt;&#xA;Though my poor teeth can't gnaw their slaw&lt;br&gt;&#xA;of &lt;code&gt;_ _ (_) _ _&lt;/code&gt; and &lt;code&gt;_ (_) _&lt;/code&gt;s of gold.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How odd, the &lt;code&gt;(_) _ _ _ _ _ _&lt;/code&gt; of this &lt;code&gt;_ _ _ _ _ _ _&lt;/code&gt;.&lt;br&gt;&#xA;I'm struck without comment.&lt;br&gt;&#xA;Yet after &lt;code&gt;_ _ _ _ _ _ _&lt;/code&gt; my tab,&lt;br&gt;&#xA;I felt oddly &lt;code&gt;_ _ _ _ _ _ (_)&lt;/code&gt;!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A funny, winged fellow&lt;br&gt;&#xA;Does his makeup in a field.&lt;br&gt;&#xA;If you can, &lt;strong&gt;name this non-man&lt;br&gt;&#xA;and the small objects he wields.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-07-14T18:06:08.780" Title="Another Journey to the Land of Puzz" Tags="|word|rhyme|wordplay|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17775" PostTypeId="1" AcceptedAnswerId="18005" CreationDate="2015-07-10T21:25:29.070" Score="-4" ViewCount="936" Body="&lt;p&gt;You sit by the pool to eat a meal.&lt;br&gt;&#xA;Me, you can see, but won’t hear.&lt;br&gt;&#xA;Go ahead, eat your lunch&lt;br&gt;&#xA;And drink your light beer…&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m drowning!!! I don’t like to whine…&lt;br&gt;&#xA;I’m drowning!!! But please throw me a line!&lt;br&gt;&#xA;I’m drowning!!! I’ve not got much time.&lt;br&gt;&#xA;I’m drowning!!! I’m not feeling fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gaze across the table at your date.&lt;br&gt;&#xA;Stop pigging out, put yourself on idle.&lt;br&gt;&#xA;You don’t note my importance,&lt;br&gt;&#xA;But to you, I am vital!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m drowning!!! I’m here in this pool.&lt;br&gt;&#xA;I’m drowning!!! I try to stay cool.&lt;br&gt;&#xA;I’m drowning!!! But you’re being cruel.&lt;br&gt;&#xA;I’m drowning!!! I need a saving tool!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You sure love your meal and date…&lt;br&gt;&#xA;Only sound is chewing – no conversing speech.&lt;br&gt;&#xA;I flail and try to get your attention…&lt;br&gt;&#xA;But my arms are short – I cannot reach…&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m drowning!!! I’m gasping for air.&lt;br&gt;&#xA;I’m drowning!!! Dry land is somewhere!&lt;br&gt;&#xA;I’m drowning!!! I won’t make it there…&lt;br&gt;&#xA;I’m drowning!!! I don’t think you care…&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Who am I in this poem?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  That 70s Show &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;SEMI-CRYPTIC HINT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Heaven and hell on weekends&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;MY LAST HINT LIKELY GAVE IT AWAY... BUT...&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Poetry can be soooo ambiguous... &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explained hints with answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;HINT CLUES:&lt;/strong&gt;&lt;br&gt;That 70s Show... I guess everyone googled That 70s Show and drowning looking for the answer... it was a 70s-era Saturday morning song-break in the cartoons, by the Bod Squad... Don't Drown Your Food. &lt;br&gt;Heaven and hell on weekends... Black Sabbath had a song called Heaven and Hell of which the first lyrics are &quot;Sing me a song. You're a singer.&quot; &lt;br&gt;Ambiguity hints: &quot;You sit by the pool&quot;... a pool of ketchup.  &quot;Gaze across the table at your date&quot;... is your date your food?  &quot;You don't note my importance/But to you I'm vital!&quot;  I thought this lyric would have given it away for sure... I think food is vital to one's functioning.  &quot;You sure love your meal and date&quot;... one in the same.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13999" LastEditorUserId="50416" LastEditDate="2018-08-20T21:56:23.937" LastActivityDate="2018-08-20T21:56:23.937" Title="Who is drowning in this bit of prose?" Tags="|riddle|" AnswerCount="3" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="17777" PostTypeId="2" ParentId="17759" CreationDate="2015-07-10T21:42:34.510" Score="1" Body="&lt;p&gt;Well, the OP stated in the comments of my other answer that the surface area (and thus the oxygen available to the flame) will not affect the burn rate in the fantasy universe of this puzzle, so, there is no need to cut the rope into powder anymore...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The structure that will burn fastest then becomes a tightly-packed sphere, since it minimizes the distance from the center to the perimeter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, we must cut the rope in half and construct two tight spheres with each half by winding it, with a small hole in the bottom of each sphere, just large enough for the match to be lit and then inserted vertically up into. When the flame hits the center of each sphere, the tightly-packed rope balls would burn up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is optimal, given the conditions in the puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I realize that the OP probably wants to know exactly HOW the spheres were constructed. Moreover, &lt;strong&gt;how&lt;/strong&gt; is the rope winded, and &lt;strong&gt;how&lt;/strong&gt; is it cut beforehand in order to ensure that there is NO AIR in between the pieces of rope.&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-07-10T21:49:15.567" LastActivityDate="2015-07-10T21:49:15.567" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17780" PostTypeId="2" ParentId="17681" CreationDate="2015-07-10T21:57:06.810" Score="3" Body="&lt;p&gt;It's inevitable that &lt;em&gt;some&lt;/em&gt; arithmetic must be done--the intended conclusion is a quantitative one, after all--so the challenge should not be to &lt;em&gt;hide&lt;/em&gt; the arithmetic, nor to call it by some other name, but to make it obvious and dead simple.  The following argument reduces the arithmetic to observing that five is one more than four (and that a whole is twice a half, a fact used in passing).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The star winds twice around its center, and therefore anyone traversing it will have to turn two full circles (four half circles).  All turning occurs only at the vertices, where the maximum amount is a complete about-face of one-half of a circle.  For five vertices that would be five half-circles, or one more half-circle than is actually turned: 180 degrees.  The deficiency between this maximum and the amount of turning that actually is accomplished is precisely the sum of the interior angles, &lt;em&gt;QED.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This approach is the one taken in modern (i.e., post-18th century) mathematics.  It generalizes to arbitrary figures of arbitrary dimensions drawn within other figures that themselves can be curved.  It is known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Gauss%E2%80%93Bonnet_theorem&quot; rel=&quot;nofollow&quot;&gt;Gauss-Bonnet Theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="14001" LastActivityDate="2015-07-10T21:57:06.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17781" PostTypeId="2" ParentId="17775" CreationDate="2015-07-10T22:28:37.360" Score="0" Body="&lt;p&gt;I think it is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The water in the pool&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You sit by the pool to eat a meal.&lt;br&gt;&#xA;Me, you can see, but won’t hear.&lt;br&gt;&#xA;Go ahead, eat your lunch&lt;br&gt;&#xA;And drink your light beer…&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can see all the water in the pool, but it makes no noise to attract attention.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I’m drowning!!! I don’t like to whine…)&lt;br&gt;&#xA;I’m drowning!!! But please throw me a line!&lt;br&gt;&#xA;I’m drowning!!! I’ve not got much time.&lt;br&gt;&#xA;I’m drowning!!! I’m not feeling fine.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every drop of water is drowning in the rest of the water in the pool. Perhaps the water is not feeling fine because it's claustrophobic and surrounded by so many like itself. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Gaze across the table at your date.&lt;br&gt;&#xA;Stop pigging out, put yourself on idle.&lt;br&gt;&#xA;You don’t note my importance,&lt;br&gt;&#xA;But to you, I am vital!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As you eat a lot, your throat starts to clog up and you need to take a sip of water. Water, in general, is very important to our survival. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I’m drowning!!! I’m here in this pool.&lt;br&gt;&#xA;I’m drowning!!! I try to stay cool.&lt;br&gt;&#xA;I’m drowning!!! But you’re being cruel.&lt;br&gt;&#xA;I’m drowning!!! I need a saving tool!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water is everywhere in the pool, and usually there are cooling systems employed in the pool to keep the water at a cool comfortable temperature in hot weather.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You sure love your meal and date…&lt;br&gt;&#xA;Only sound is chewing – no speech.&lt;br&gt;&#xA;I flail and try to get your attention…&lt;br&gt;&#xA;But my arms are short – I cannot reach…&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The water keeps so splish splashing at the edge of the pool, but cannot really create a big wave to reach out to you sitting in the distance.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I’m drowning!!! I’m gasping for air.&lt;br&gt;&#xA;I’m drowning!!! Dry land is somewhere!&lt;br&gt;&#xA;I’m drowning!!! I won’t make it there…&lt;br&gt;&#xA;I’m drowning!!! I don’t think you care…&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The pool is filled with water and no air. There is dry land close to the pool but the water won't reach there. And you don't care because the water is right where it's supposed to be. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-07-10T22:28:37.360" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17782" PostTypeId="2" ParentId="17773" CreationDate="2015-07-10T22:39:30.557" Score="7" Body="&lt;p&gt;I can't see any unifying underlying pattern like in the previous puzzle like this, but a lot of the gaps are quite easy to fill in. EDIT: actually, after trying it in practice, I've found a pattern:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In each verse, two of the missing words are actually the same word, with different meanings. In fact, it's &lt;em&gt;possible&lt;/em&gt; (though I'm not sure) that the same can be said of the other two!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, the second and fourth lines of each verse rhyme.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Wonderful Wanderful land of Puzz&lt;br&gt;&#xA;  Is where you and I now are;&lt;br&gt;&#xA;  Where magazines eat wheeled things,&lt;br&gt;&#xA;  And the whole world is bizarre.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;To prove my point, I found a joint&lt;br&gt;&#xA;  Where brochures came to cope.&lt;br&gt;&#xA;  &quot;I can't &lt;code&gt;DIGES(T)&lt;/code&gt; this &lt;code&gt;(M)OPED&lt;/code&gt;,&quot;&lt;br&gt;&#xA;  A weeping &lt;code&gt;DIGEST&lt;/code&gt; &lt;code&gt;MO(P)ED&lt;/code&gt;.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;Don't &lt;code&gt;QU(A)IL&lt;/code&gt; away here,&quot; I declared.&lt;br&gt;&#xA;  &quot;We'll fix this with a plan.&lt;br&gt;&#xA;  Have you tried &lt;code&gt;QUAIL&lt;/code&gt;? If you're from Puzz,&lt;br&gt;&#xA;  you &lt;code&gt;CAN&lt;/code&gt; buy it like bread -- in &lt;code&gt;(C)AN&lt;/code&gt;s!&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;I cannot use my funds,&quot; he said,&lt;br&gt;&#xA;  &quot;As my &lt;code&gt;(B)ANK&lt;/code&gt; will not &lt;code&gt;PER(M)IT&lt;/code&gt;.&lt;br&gt;&#xA;  I have a rifle &lt;code&gt;PERMI(T)&lt;/code&gt;, though,&lt;br&gt;&#xA;  And the river &lt;code&gt;BANK&lt;/code&gt;'s where they sit!&quot;  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;He snagged a rifle full of &lt;code&gt;LEAD&lt;/code&gt;&lt;br&gt;&#xA;  and said, &quot;&lt;code&gt;LE(A)D&lt;/code&gt; me to the &lt;code&gt;(P)ARK&lt;/code&gt;!&quot;&lt;br&gt;&#xA;  When we arrived, we found a tree&lt;br&gt;&#xA;  and &lt;code&gt;P(A)RK&lt;/code&gt;ed there until dark.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;At the &lt;code&gt;FOO(T)&lt;/code&gt; of a hill, we stood quite still&lt;br&gt;&#xA;  His gun &lt;code&gt;COCK&lt;/code&gt;ed to his eye&lt;br&gt;&#xA;  When a &lt;code&gt;FOOT&lt;/code&gt; away, around midday,&lt;br&gt;&#xA;  A &lt;code&gt;CO(C)K&lt;/code&gt; came wandering by.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Aiming at the poor thing's &lt;code&gt;(B)ACK&lt;/code&gt;,&lt;br&gt;&#xA;  He shot it &lt;code&gt;DOWN&lt;/code&gt;. 'Twas dead!&lt;br&gt;&#xA;  &lt;code&gt;BACK&lt;/code&gt; at his home, he stripped its coat,&lt;br&gt;&#xA;  Ate just its &lt;code&gt;D(O)WN&lt;/code&gt;, then fled...  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I wandered to a nearby &lt;code&gt;BAR&lt;/code&gt;&lt;br&gt;&#xA;  Which &lt;code&gt;R(O)CKS&lt;/code&gt;! Or, so I'm told.&lt;br&gt;&#xA;  Though my poor teeth can't gnaw their slaw&lt;br&gt;&#xA;  of &lt;code&gt;RO(C)KS&lt;/code&gt; and &lt;code&gt;B(A)R&lt;/code&gt;s of gold.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;How odd, the &lt;code&gt;(C)ONTENT&lt;/code&gt; of this &lt;code&gt;CLOSING&lt;/code&gt;.&lt;br&gt;&#xA;  I'm struck without comment.&lt;br&gt;&#xA;  Yet after &lt;code&gt;CLOSING&lt;/code&gt; my tab,&lt;br&gt;&#xA;  I felt oddly &lt;code&gt;CONTEN(T)&lt;/code&gt;!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Letters circled:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;TMPACBMTAPATCBOOCACT&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which rearrange to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BAT COMPACT BAT COMPACT,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;in which the repetition fits with the overall theme of the puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer to&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A funny, winged fellow&lt;br&gt;&#xA;  Does his makeup in a field.&lt;br&gt;&#xA;  If you can, name this non-man&lt;br&gt;&#xA;  and the small objects he wields.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; he's a BAT wielding a makeup COMPACT.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks to @COTO and @CodeNewbie and @alexmc for finding some of the words!&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-07-14T18:06:08.780" LastActivityDate="2015-07-14T18:06:08.780" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="17783" PostTypeId="2" ParentId="17309" CreationDate="2015-07-10T23:10:17.680" Score="9" Body="&lt;p&gt;I think that you should tell to him to try:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 26432&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All of the digits converted to hexadecimal are of type int16 (16 bits long, a.k.a unicode characters) and those characters are in the range of chinese and japanese characters, so every character have a meaning, like the last response of the program that is the unicode character &quot;26432&quot; (U+6740, 杀) that means &quot;kill&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;pd: I have used the &lt;a href=&quot;https://translate.google.com/#zh-CN/en/%E6%9D%80&quot;&gt;Google Traductor&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14002" LastActivityDate="2015-07-10T23:10:17.680" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17785" PostTypeId="2" ParentId="2241" CreationDate="2015-07-11T00:30:31.847" Score="1" Body="&lt;p&gt;Just to add further clarification, the first part of the solution is easy with just one fruit picked, instead of three. credit goes to &lt;a href=&quot;http://runescape.wikia.com/wiki/Swept_Away&quot; rel=&quot;nofollow&quot;&gt;Swept Away&lt;/a&gt; where I first ran into this scenario.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;! this takes into the fact that ALL LABELS ARE WRONG!!!! Let's have label A be apples, label O be oranges, and label B be both, as given below:&#xA;  &lt;br&gt;Let &lt;strong&gt;x&lt;/strong&gt; = [a] [b] [o], or [b] [a] [o] or any permutation of the three.&#xA;  &lt;br&gt;&lt;strong&gt;Step 1:&lt;/strong&gt; Choose a fruit from the crate labeled B. Whichever fruit it is, take the appropriate label and place it there. &#xA;  &lt;br&gt;&lt;strong&gt;Step 2:&lt;/strong&gt; Now, the two remaining boxes contain either both, or the opposite of the fruit you grabbed (if you got an apple, then oranges and both are left; if you got an orange, then apples and both are left). &#xA;  &lt;br&gt;&lt;strong&gt;Step 3:&lt;/strong&gt; Now, one of the remaining labels will match the fruit that's left, so just switch the other two labels and you're done!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13697" LastActivityDate="2015-07-11T00:30:31.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17786" PostTypeId="2" ParentId="14521" CreationDate="2015-07-11T00:45:49.450" Score="4" Body="&lt;p&gt;And their off...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Advancing into the first turn at 5KM, JavaGeeRoe's superior speed at a full 50CM/ss showing off.&#xA;  Poor Wist isn't even half-way to the turn yet!  JavaGee in the lead, Pupa and Bight fighting for 2nd.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;On to the turns, Java's gotta stomp on those breaks to keep control, but Pupa just laughs and continues at full speed.  I gotta get some binoculars to see Wist way in the distance not even half-way done with the turns.  Completing the turns is Pupa at 100KM, Java &amp;amp; Bight neck and neck at 84/83KM Wist back at the 44M mark.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A quick stop to reconfigure at the river, Pupa's still got the lead at 150KM, Java and Bight still sparring over 2nd at 132/130, Wist still working his way through the turns back at 70KM.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;River Stretch Completed.&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What's this? Pupa seems to be having mechanical difficulties.  His infinite accelerator has ground to a halt in the water.  Caulk's cheap, you'd think he'd have waterproofed it!  Meanwhile Java's nitrous works just fine and he's left all the competition in the dust.  Java Gee Whiz stopping to reconfigure just before the uphill hairpins at 280KM, Pupa at 216KM working his way upstream followed by Bight and Wist still on the flat part of the river.  Wist is travelling much faster than anyone else, he didn't even slow-down for river, just kept accelerating!  Still, Java's 60KM+ lead is over 10% of the track can he be caught?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Checking in from the mini-lake at 360KM, the hairpins didn't seem to slow Java down, that nitrous boost of 15m/ss letting him maintain a fairly high average speed in the turns, Wist passing Bight, Pupa, leaving them behind at 313KM, he's about to enter the hairpins.  Bight has just executed a high-speed draft pass on Pupa as they tie at 285KM.  But, drafting only lets you keep second, that'll never win here.  In fact, I'd say it only worked because Bight tossed one of his oily rags straight into Pupa's accelerator while they were reconfiguring!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Lake Hop completed&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Java's leaving the river at 400KM. What's he doing? He's actually stopped to wash his car! What a show-off.  It may cost him though, Wist isn't far behind.  Java may have forgotten that Wist won't stop at the edge.  Pupa's left Bight behind at the turns and is now at 355KM, Bight at 318KM.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Watching from the finish line, It's Wist coming out of the quick turns at 430KM, Java not even a single KM behind.  Pupa and Bight meanwhile have just finished swapping to water gear to enter the last river.  Pupa at 363KM, bight not far behind at 360KM.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Finish Line&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  FINISH And the prize goes to Java.  Race completed in just over 22min 10secs.  Right behind is FramesWist at 23min, 7secs.  Java's top speed took home the prize.  7min later, for those of you who stuck around, Bight and Pupa.  Bight executed a well-timed manuver passing Pupa just before the finish line to end in 27min, 57secs compared to Pupa's 28min.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Closing Banter&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Java could cut his nitrous by 10% and still win.&#xA; Wist states he'd have won if not for the hairpins, just 10% faster through the pins would have done it.&#xA; Bight points to Pupa's habit of throwing rocks at him when he passes and says that if Pupa just has a slightly faster car, they wouldn't be stuck so far behind.&#xA; Pupa claims Bight is being ridiculous and that he had the fastest car of them all if not for those rivers.  Who ever heard of a race with a lake?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Completed via a Java program, may have inaccuracies.&lt;/p&gt;&#xA;" OwnerUserId="11750" LastActivityDate="2015-07-11T00:45:49.450" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17787" PostTypeId="2" ParentId="17617" CreationDate="2015-07-11T01:05:48.163" Score="1" Body="&lt;p&gt;Credit to LeppyR64 for cracking the code to reveal the letters&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A LAMN WORD GOODBEYE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, I think this is an anagram which could lead to any of the following phrases (punctuations and spaces used where necessary).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;DAMN GOOD EYE, BRO LAW! (A compliment to a relative)&lt;br&gt; NAME A WORLD 'GOODBYE' (Highlighting the OP's frustration at &quot;Hello World&quot; programs)&lt;br&gt; GOOD NAMED LOWER BAY (A well known community called Lower Bay)&lt;br&gt; ONE BLOODY GEM AWARD (An honour in the serial killing fraternity)&lt;br&gt; OYE! A GODDAMN BLOWER! (An impatient farmhand in a tool shop)&lt;br&gt; OWE ME A BLOODY GRAND (An angry creditor)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-07-11T01:23:39.183" LastActivityDate="2015-07-11T01:23:39.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17788" PostTypeId="1" AcceptedAnswerId="17791" CreationDate="2015-07-11T02:23:13.617" Score="8" ViewCount="243" Body="&lt;p&gt;As you liked &lt;a href=&quot;https://puzzling.stackexchange.com/questions/17565/nine-is-the-limit&quot;&gt;Nine is the limit&lt;/a&gt;, I decided to make you an other one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My first is a harvester,&lt;br&gt;&#xA;  My second is a payment,&lt;br&gt;&#xA;  My third comes with a surprise,&lt;br&gt;&#xA;  My fourth is a soldier,&lt;br&gt;&#xA;  My fifth comes with a maiden,&lt;br&gt;&#xA;  If you've found the pattern&lt;br&gt;&#xA;  What is my sixth?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Those definitions might be a bit broad, but there is only one answer that follows the pattern.&#xA;If needed, I will add more clues.&lt;/p&gt;&#xA;" OwnerUserId="10400" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-07-11T08:55:43.030" Title="Nine is the limit... again" Tags="|riddle|word|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17791" PostTypeId="2" ParentId="17788" CreationDate="2015-07-11T08:55:43.030" Score="7" Body="&lt;p&gt;My first is a harvester,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bee&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My second is a payment,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fee&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My third comes with a surprise,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gee&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fourth is a soldier,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lee (General Lee)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fifth comes with a maiden,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nee (maiden name)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you've found the pattern&#xA;What is my sixth?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pee&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, seven through nine are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; See, Tee, Wee&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4479" LastActivityDate="2015-07-11T08:55:43.030" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17792" PostTypeId="1" AcceptedAnswerId="17795" CreationDate="2015-07-11T10:58:04.130" Score="8" ViewCount="350" Body="&lt;p&gt;I am the father of seven children.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first son is a man of action.&lt;br&gt;&#xA;My second son's reputation is concerning.&lt;br&gt;&#xA;My third son shuttles between London and Leeds in his job.&lt;br&gt;&#xA;There is nothing at all I can say about my fourth son.&lt;br&gt;&#xA;Surya, my fifth son, is a bright chap.&lt;br&gt;&#xA;My sixth son works in Hollywood.&lt;br&gt;&#xA;My seventh son is known for his great strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Who am I?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2425" LastActivityDate="2015-07-11T14:24:35.653" Title="We are a harmonious family" Tags="|riddle|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17793" PostTypeId="2" ParentId="14537" CreationDate="2015-07-11T11:18:36.467" Score="3" Body="&lt;p&gt;The CMU Pronouncing Dictionary lists one which I think is an error (and I've e-mailed the maintainer): &lt;strong&gt;menswear&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The various non-proper nouns can be grouped into 4 categories by etymology, but there are certainly more than 4 headwords:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;ul&gt;&lt;li&gt;beefsteak, steak(s), steakhouse(s)&lt;/li&gt;&#xA; &lt;li&gt;break(s), breakable, breakage, breakaway, breakdown(s), breaker(s), break-even, breakey, breakfront(s), break-in(s), breaking, breakneck, breakout(s), breakthrough(s), breakup(s), breakwater, breaky, codebreaker(s), daybreak, groundbreaking, heartbreak(s), heartbreaking, icebreaker(s), lawbreaker(s), lawbreaking, outbreak(s), strikebreaker(s), unbreakability, unbreakable&lt;/li&gt;&#xA; &lt;li&gt;great(s), great-aunt, greatcoat, greater, greatest, great-grandchild, great-granddaughter, great-grandfather, great-grandmother, great-grandparent, great-grandson, greatly, great-nephew, greatness, great-niece, great-uncle&lt;/li&gt;&#xA; &lt;li&gt;yea(s)&lt;/li&gt;&lt;/ul&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proper nouns (although classing some of the political ones as proper nouns might be debated):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;ul&gt;&lt;li&gt;Amoskeag&lt;/li&gt;&#xA; &lt;li&gt;Beatty&lt;/li&gt;&#xA; &lt;li&gt;Breakfield&lt;/li&gt;&#xA; &lt;li&gt;Breakmate&lt;/li&gt;&#xA; &lt;li&gt;Easudes&lt;/li&gt;&#xA; &lt;li&gt;Feagan&lt;/li&gt;&#xA; &lt;li&gt;Greathouse&lt;/li&gt;&#xA; &lt;li&gt;MacLean, McLean&lt;/li&gt;&#xA; &lt;li&gt;McCrea&lt;/li&gt;&#xA; &lt;li&gt;McElyea&lt;/li&gt;&#xA; &lt;li&gt;McShea&lt;/li&gt;&#xA; &lt;li&gt;Rea, Reay&lt;/li&gt;&#xA; &lt;li&gt;Reagan, Reaganesque, Reaganism, Reaganite(s), Reaganomics&lt;/li&gt;&#xA; &lt;li&gt;Relyea&lt;/li&gt;&#xA; &lt;li&gt;Seamus&lt;/li&gt;&#xA; &lt;li&gt;Seay&lt;/li&gt;&#xA; &lt;li&gt;Shea, O'Shea&lt;/li&gt;&#xA; &lt;li&gt;Sleator&lt;/li&gt;&#xA; &lt;li&gt;Treacy&lt;/li&gt;&#xA; &lt;li&gt;Treadaway, Treadway&lt;/li&gt;&#xA; &lt;li&gt;Yeager&lt;/li&gt;&#xA; &lt;li&gt;Yeats&lt;/li&gt;&lt;/ul&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="228" LastActivityDate="2015-07-11T11:18:36.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17795" PostTypeId="2" ParentId="17792" CreationDate="2015-07-11T12:14:03.750" Score="9" Body="&lt;p&gt;I am the father of seven children.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are the major scale, and your seven children are the individual tones of the scale expressed in solfege.  A &quot;harmonious family&quot; indeed!  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My first son is a man of action.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DO -- the verb &quot;do&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My second son's reputation is concerning.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RE -- in correspondence, the abbreviation &quot;Re:&quot; for &quot;regarding&quot;, a synonym of &quot;concerning&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My third son shuttles between London and Leeds in his job.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MI -- the M1 motorway connecting London and Leeds&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is nothing at all I can say about my fourth son.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; FA -- &quot;Sweet F.A.&quot; (Fanny Adams), a slang expression meaning &quot;nothing&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Surya, my fifth son, is a bright chap.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SOL -- i.e. the sun; Surya is the Hindu solar deity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My sixth son works in Hollywood.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LA -- abbreviation for Los Angeles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My seventh son is known for his great strength.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TI -- abbreviation for Titanium, the strongest metal&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4385" LastEditorUserId="4385" LastEditDate="2015-07-11T14:24:35.653" LastActivityDate="2015-07-11T14:24:35.653" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17796" PostTypeId="2" ParentId="17769" CreationDate="2015-07-11T14:36:14.503" Score="4" Body="&lt;p&gt;The first ingredient is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; foxglove, aka digitalis, (the vixens herb) from which an extract is used in treating heart conditions. It's biennial, so wins the next year without descending (dying off)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thats the only one i'm sure on, but would guess:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; willow bark. I think because willow is often thought of as an old (senior) man, the wind through a grove sounds like sighing and the tree has a melted appearance and is used in the production of medicines for stiffness of joints (pertrification -I know, I'm reaching)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then, reaching even more, I can only think the third might be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; himbabao - marketed as birch flowers (sunset hued trees) these catkins (a cat was mentioned) look like horrible worms&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My guess for the fourth ingredient:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A melon cactus (Melocactus) aka Turk's cap cactus (turkey's clothing) because the top looks like a fez.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nice riddle though!&lt;/p&gt;&#xA;" OwnerUserId="2589" LastActivityDate="2015-07-11T14:36:14.503" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17798" PostTypeId="2" ParentId="581" CreationDate="2015-07-11T15:31:13.977" Score="2" Body="&lt;p&gt;If the person you are asking doesn't know physics, you may as well teach him. If you are not allowed to do this directly, you could use your own terminology. (Brackets indicate actual terminology)&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Every point on the lines has a hidden magic value (potential). The&#xA;difference between the magic values of A and B is called the net magic&#xA;value (net potential difference). We release any army of robots&#xA;(current) at Point A. Robots travel incredibly fast (speed of an&#xA;electron), but we must transfer an almost infinite number of robots&#xA;(number of electrons), so it could still take time.&lt;/p&gt;&#xA;&lt;p&gt;A single 'path' has no intersections. One rule is that&#xA;the difference between two magic values is proportional to the rate of&#xA;transfer of robots through a path (Ohm's Law). Another rule is the&#xA;length of a path is proportional to the rate of transfer of robots&#xA;through it, and inversely proportional to the difference between the initial&#xA;and final magic values.&lt;/p&gt;&#xA;&lt;p&gt;Robots learn over time, which the fastest routes are, and eventually&#xA;arrive at the optimal speed for robotic transfer. Find the ratio&#xA;between the net magic value and the rate of robotic transfer through&#xA;the frame.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This makes it even more difficult, because the concept of resistance is not even mentioned. One needs to first define resistance, then derive formulae for serial and parallel combinations, and &lt;em&gt;then&lt;/em&gt; solve the puzzle. This makes it possible but almost impossible.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-12T14:19:29.900" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17799" PostTypeId="1" AcceptedAnswerId="17800" CreationDate="2015-07-11T18:03:25.533" Score="2" ViewCount="673" Body="&lt;p&gt;The police have three suspects for the murder of Mr. Morgan: Mr. Hill, Mr. Spengler, and Mr.Gates. Hill, Spengler, and Gates each declare that they did not kill Morgan. Hill also states that Morgan was a friend of Spengler and that Gates disliked him. Spengler also states that he did not know Morgan and that he was out of town the day Morgan was killed. Gates also states that he saw both Hill and Spengler with Morgan the day of the killing and that either Hill or Spengler must have killed him. Can you determine who the murderer was if one of the three men is guilty, the two innocent men are telling the truth, but the statements of the guilty man may or may not be true?&lt;/p&gt;&#xA;" OwnerUserId="12112" LastActivityDate="2015-07-11T23:00:10.550" Title="Who is the murderer of Mr. Morgan?" Tags="|logical-deduction|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17800" PostTypeId="2" ParentId="17799" CreationDate="2015-07-11T18:09:45.637" Score="3" Body="&lt;p&gt;I think that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Spengler is the murderer. Because if he's speaking the truth then that would mean he was out of town and didn't know Morgan. That would negate both Hill (Hill said Spengler was a friend of Morgan) and Gates (Gates said he saw Morgan with Spengler the day he was murdered). However, we know that at least two of the suspects are speaking the truth, so Spengler must be the liar and hence the murderer. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-07-11T18:09:45.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17801" PostTypeId="2" ParentId="17799" CreationDate="2015-07-11T18:10:58.737" Score="2" Body="&lt;p&gt;If Gates is telling the truth about seeing Morgan and Spengler together, then Spengler must be lying about not knowing the deceased.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Hill is telling the truth about Morgan and Spengler being friends, then Spengler must be lying about not knowing the deceased.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If only the murderer is lying then at least one of Gates and Hill must be truthful, but if either is truthful then, Spengler must be the murderer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11297" LastActivityDate="2015-07-11T18:10:58.737" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17802" PostTypeId="1" AcceptedAnswerId="17804" CreationDate="2015-07-11T18:38:08.147" Score="-2" ViewCount="178" Body="&lt;p&gt;I decided to write an email to some friends one day. Unfortunately, my evil brother applied a code to it. My friends called and asked what it was supposed to say, but I've forgotten what. Can you help me decode it?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;24    16.5  34    34  41.5&#xA;16.5  59    16.5  49  66.5  41.5  39  16.5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="14013" LastActivityDate="2015-07-11T18:57:51.930" Title="My evil brother scrambled my letter!" Tags="|cipher|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17803" PostTypeId="1" AcceptedAnswerId="17806" CreationDate="2015-07-11T18:44:10.633" Score="7" ViewCount="1244" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Some use me unaware.&lt;br&gt;&#xA;  Some call me lazy.&lt;br&gt;&#xA;  The most-used did a 180,&lt;br&gt;&#xA;  And then you saw me.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I'm short and neutral.&lt;br&gt;&#xA;  By the time you've read these 8 lines,&lt;br&gt;&#xA;  You'll have used me at least 8 times,&lt;br&gt;&#xA;  And maybe even up to 18 times.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What 5-letter word am I?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Post a detailed explanation with your answer, please.&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="25957" LastEditDate="2016-12-22T12:58:00.130" LastActivityDate="2016-12-22T12:58:00.130" Title="What 5-letter word am I?" Tags="|riddle|word|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17804" PostTypeId="2" ParentId="17802" CreationDate="2015-07-11T18:48:52.373" Score="1" Body="&lt;p&gt;The letter says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hello Everyone. By performing the transformation $(X-4)/2.5$ to each element and converting to letters (A=1, Z=26), you get Hello Everyone. I tried seeing if there's some significance to that transformation but I haven't been able to find any.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="7722" LastEditDate="2015-07-11T18:57:51.930" LastActivityDate="2015-07-11T18:57:51.930" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17805" PostTypeId="2" ParentId="17803" CreationDate="2015-07-11T20:52:25.083" Score="1" Body="&lt;p&gt;I think it might be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; VOICE (in grammar, such as active voice and passive voice)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some use me unaware.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every sentence has its own voice, which we use without realization. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some call me lazy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Voice can be of the passive form too. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The most-used did a 180,&#xA;And then you saw me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most sentences use the active voice but once we learnt about the passive voice, we started turning sentences from active to passive voice, and only then realized its role in sentences. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm short and neutral.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The change in voice never changes the intent of the sentence, and hence could be considered neutral. (&lt;em&gt;unsure about how it is short...&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By the time you've read this,&#xA;You'll have used me at least 8 times,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The main body of the riddle has 8 verses, so there are 8 sentences, all of which are in active voice. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And maybe even up to 19 times.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Perhaps the other 11 sentences could be a part of the standard layout of PuzzlingSE which has links like 'Ask question',  'Contact us', 'Download the Stack Exchange Android app' etc. (I see the app message as I'm on my phone right now.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-07-11T20:52:25.083" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17806" PostTypeId="2" ParentId="17803" CreationDate="2015-07-11T21:12:25.817" Score="12" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.slate.com/blogs/lexicon_valley/2014/06/05/schwa_the_word_for_the_most_common_vowel_sound_in_english_comes_from_hebrew.html&quot;&gt;schwa&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some use me unaware.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people have never heard of the schwa.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some call me lazy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In English, the schwa occurs in many unstressed syllables.  Since it is unstressed, some people call it &lt;a href=&quot;https://www.englishforums.com/English/LazyVowels/lgghr/post.htm&quot;&gt;lazy&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The most-used did a 180, And then you saw me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; e is the most common letter in the English language.  Rotate it 180 degrees and you get ə, which represents the schwa.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm short and neutral.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The schwa is a short neutral vowel sound.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By the time you've read this, You'll have used me at least 8 times, And maybe even up to 18 times.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Using the Google define feature to get the IPA pronunciations of the words in the first 8 lines, I came up with the following words that can be pronounced with a schwa:&#xA;&lt;p&gt;&#xA; Some: səm&lt;br&gt;&#xA; unaware: ˌənəˈwer&lt;br&gt;&#xA; Some: səm&lt;br&gt;&#xA; The: T͟Hē, T͟Hə&lt;br&gt;&#xA; a: ā, ə&lt;br&gt;&#xA; one (in 180): wən&lt;br&gt;&#xA; And: and, (ə)n&lt;br&gt;&#xA; and: and, (ə)n&lt;br&gt;&#xA; neutral: ˈn(y)o͞otrəl&lt;br&gt;&#xA; the: T͟Hē, T͟Hə&lt;br&gt;&#xA; you've: yo͞ov, yəv&lt;br&gt;&#xA; have: hav, həv, (ə)v&lt;br&gt;&#xA; at: at, ət&lt;br&gt;&#xA; And: and, (ə)n&lt;br&gt;&#xA; even: ˈēvən&lt;br&gt;&#xA; up: əp&lt;br&gt;&#xA; to: to͞o, tə&lt;br&gt;&#xA; &lt;p&gt;&#xA; Since some words have multiple pronunciations, some of which may not use the schwa, that comes out to a minimum of 8 schwas and a maximum of 18 schwas.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4450" LastEditDate="2015-07-12T04:23:45.083" LastActivityDate="2015-07-12T04:23:45.083" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17807" PostTypeId="1" AcceptedAnswerId="17809" CreationDate="2015-07-11T22:11:47.577" Score="7" ViewCount="232" Body="&lt;p&gt;&lt;strong&gt;Brady&lt;/strong&gt; and &lt;strong&gt;Sarah&lt;/strong&gt; have been messaging on a site for a while now. The site tells you when you sent your messages; but &lt;strong&gt;not&lt;/strong&gt; in &lt;em&gt;exact detail&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, Sarah messaged Brady &quot;&lt;em&gt;What's up?&lt;/em&gt;&quot; 2 hours and 31 min ago, however it only says &quot;&lt;strong&gt;2 hours ago&lt;/strong&gt;&quot;. But when Brady messaged Sarah &quot;&lt;em&gt;Hi!&lt;/em&gt;&quot; 47 min ago it said &quot;&lt;strong&gt;47 min ago&lt;/strong&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Once it passes the 1 hour mark it is then generalized to how many hours ago&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brady sent Sarah an important message almost a day ago, he needs to know when, to the exact minute, he sent this. He needs this information very soon. But the site says &quot;&lt;strong&gt;23 hours ago&lt;/strong&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's &lt;strong&gt;4:16 PM&lt;/strong&gt; right now, but for the next &lt;strong&gt;2 days&lt;/strong&gt; (beginning immediately), he'll have &lt;strong&gt;no source&lt;/strong&gt; of &lt;em&gt;time&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;How can Brady find the exact time he sent this message, with the &lt;strong&gt;least&lt;/strong&gt; amount of new messages?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2015-07-11T22:25:24.107" Title="Social Network Messaging Problem" Tags="|mathematics|optimization|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17808" PostTypeId="1" AcceptedAnswerId="17815" CreationDate="2015-07-11T22:20:05.413" Score="1" ViewCount="351" Body="&lt;p&gt;I have always been there,&lt;br&gt;&#xA;and will always be.&lt;br&gt;&#xA;I'll be gone in a mere&lt;br&gt;&#xA;moment, you'll see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You cannot remember&lt;br&gt;&#xA;me to not be with you,&lt;br&gt;&#xA;yet just at the moment,&lt;br&gt;&#xA;I'm completely new.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You cannot escape me,&lt;br&gt;&#xA;nor stay within me.&lt;br&gt;&#xA;Now could you please tell me,&lt;br&gt;&#xA;do you recognize me?&lt;/p&gt;&#xA;" OwnerUserId="14010" LastEditorUserId="12700" LastEditDate="2015-07-12T09:49:03.140" LastActivityDate="2015-07-12T09:49:03.140" Title="What is this paradoxical entity?" Tags="|riddle|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17809" PostTypeId="2" ParentId="17807" CreationDate="2015-07-11T22:25:24.107" Score="6" Body="&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He just needs 1 new message:&lt;br&gt;&#xA; He posts a message immediately. He therefore knows that it was sent at 4:16 PM. The he waits that the time stamp of the past message jumps from &quot;23 hours ago&quot; to &quot;1 day ago&quot;. At that point in time he knows that he posted the message exactly 24 h ago. He then looks at his newly posted message, which was posted less than one hour ago, therefore has minutes. He adds those minutes to 4:16 PM, and thus gets the current time. The other post was then posted the previous day at the same time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14010" LastActivityDate="2015-07-11T22:25:24.107" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17810" PostTypeId="1" AcceptedAnswerId="17812" CreationDate="2015-07-11T22:30:33.087" Score="-1" ViewCount="1611" Body="&lt;p&gt;I recently found a card on which there were two puzzles, both of which have me stumped; I want to spend a little more time on the other puzzle, but here is one that stumps me:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;n&lt;sup&gt;o&lt;/sup&gt; 65&lt;br&gt;&#xA;  A census-taker calls at a house. He asks the woman living there the ages of her three daughters. The woman says, &amp;ldquo;If you multiply their ages the total is 72; if you add together their ages the total is the same as the number on my front door, which you can see.&amp;rdquo;&lt;br&gt;&#xA;  The census-taker says, &amp;ldquo;That is not enough information for me to calculate their ages.&amp;rdquo;&lt;br&gt;&#xA;  The woman says, &amp;ldquo;Well, my eldest daughter has a cat with a wooden leg.&amp;rdquo;&lt;br&gt;&#xA;  The census-taker replies, &amp;ldquo;Ah! Now I know their ages.&amp;rdquo;  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;What are the ages of the three girls?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;Attempt&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let $d_0,d_1,d_2$ be the ages of the three daughters, with $d_i\leq d_{i+1}$. We know that $d_0d_1d_2=72$ and that $d_0+d_1+d_2=n$, $n$ being the house number. The problem does not give the house number; that is where I am stumped. (&lt;strong&gt;HINT:&lt;/strong&gt; The second-to-last line gives that there is an eldest daughter, making true $d_0\leq d_1&amp;lt;d_2$.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7307" LastActivityDate="2015-07-11T22:45:39.493" Title="Cryptic Woman and a Confused Census-Taker" Tags="|mathematics|calculation-puzzle|" AnswerCount="1" CommentCount="0" ClosedDate="2015-07-12T01:21:43.077" ContentLicense="CC BY-SA 3.0" />
  <row Id="17811" PostTypeId="2" ParentId="17808" CreationDate="2015-07-11T22:40:13.473" Score="1" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; breath&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have always been there, and will always be.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A living person will always be breathing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll be gone in a mere moment, you'll see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A single breath vanishes in a moment.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You cannot remember me to not be with you,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A person can't remember when they weren't breathing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yet just at the moment, I'm completely new.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A person is always taking new breaths.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You cannot escape me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can't stop breathing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;nor stay within me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's not possible for a person to be inside a breath.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2015-07-11T22:40:13.473" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17812" PostTypeId="2" ParentId="17810" CreationDate="2015-07-11T22:45:39.493" Score="3" Body="&lt;p&gt;I think the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3, 3, and 8. You know the census taker knows the number on the front door and needs to know that there is an eldest daughter to know for sure the correct solution. Therefore, there has to be two combinations of three factors of 72 that have the same sum. Breaking them up, there are only two that sum to the same number:&lt;br&gt;&lt;br&gt;2+6+6 = 14, 3+3+8 = 14&lt;br&gt;&lt;br&gt;This means that the first case is eliminated if there is an eldest daughter, and therefore the correct solution is 3, 3, and 8.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastActivityDate="2015-07-11T22:45:39.493" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17814" PostTypeId="2" ParentId="17808" CreationDate="2015-07-11T23:38:20.863" Score="1" Body="&lt;p&gt;A very cliched answer, but I think it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have always been there,&lt;br&gt;&#xA;and will always be.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Time has existed before us and will continue to exist after us. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll be gone in a mere&lt;br&gt;&#xA;moment, you'll see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Time passes us by every second.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You cannot remember&lt;br&gt;&#xA;me to not be with you,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You recall past instances by associating them with a point in time. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yet just at the moment,&lt;br&gt;&#xA;I'm completely new.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every moment in time is unique.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You cannot escape me,&lt;br&gt;&#xA;nor stay within me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You cannot escape the clutches of time, and you constantly struggle to finish off your tasks within a certain time frame. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now could you please tell me,&lt;br&gt;&#xA;do you recognize me?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I do recognize you. You're my old foe, Time! &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-07-11T23:38:20.863" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17815" PostTypeId="2" ParentId="17808" CreationDate="2015-07-12T00:29:55.097" Score="0" Body="&lt;p&gt;Going off CodeNewbie's answer, I think the answer might be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The present&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have always been there,&lt;br&gt;&#xA;and will always be.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The present has always existed, and will continue to exist.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll be gone in a mere&lt;br&gt;&#xA;moment, you'll see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Your current point in time is the present, but in a moment that point in time will be the past.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You cannot remember&lt;br&gt;&#xA;me to not be with you,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I don't actually understand this part, I'm having trouble parsing it, but probably something to do with anything you &quot;remember&quot; is in the past, not the present. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yet just at the moment,&lt;br&gt;&#xA;I'm completely new.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The present moment is always a new moment.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You cannot escape me,&lt;br&gt;&#xA;nor stay within me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You will always be in the present, but you can never stay in the present moment.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now could you please tell me,&lt;br&gt;&#xA;do you recognize me?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The present! &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10861" LastActivityDate="2015-07-12T00:29:55.097" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17816" PostTypeId="2" ParentId="2361" CreationDate="2015-07-12T00:56:57.193" Score="1" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;Simple Answer&lt;/em&gt;&lt;/strong&gt; &quot;How many pies &lt;em&gt;must&lt;/em&gt; I leave with ...?&quot; There is no unique answer, but most simply, leave home with no pies. Details below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Original conditions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the original question: &quot;The trolls can't give you half a cake back. It is unhygienic and disgusting.&quot; Since the trolls can only adjust your number of pies by integer amounts, just start with any number of whole pies to reach grandma's with a whole number of pies. Since the trolls give you back 1 whole pie after taking some number of yours, you will have at least 1 pie to present at grandma's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fractional pies&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If trolls can give potions of a pie back, and assuming they can subdivide pies into any real number, you can still start with any real number of pies and reach grandma's with a whole number of pies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you start with $p$ pies. You leave&#xA;bridge 1 with $\frac{p}{2^1} + \frac{2^1 - 1}{2^0}$ pies,&#xA;bridge 2 with $\frac{p}{2^2} + \frac{2^2 - 1}{2^1}$ pies,&#xA;and in general,&#xA;bridge $n$ with $\frac{p}{2^n} + \frac{2^n - 1}{2^{n-1}}$ pies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the limit as $n$ tends to infinity, you end up with exactly 2 pies when you reach grandma's, regardless of how many pies you started with, whether positive, negative, fractional, or real. In the case of negative pies (you borrowed them), your pie debt transferred to the trolls, piecemeal ... your friends might not be happy to get those pies back under those conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conveniently, you can still start out with zero pies. If the tolls work the same manner going home, you'll even end up with two pies for yourself when you get home :) .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Different tolls&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final part of the question changes the toll so that crossing the bridge with $p$ pies, you leave the bridge with $\frac{p}{k}+1$ pies, for some $k$. (Note: previously, $k=2$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving the power series expansion and leaving home with $p$ pies, you leave bridge $n$ with $\frac{p}{k^n} + \frac{k^n - 1}{k^{n-1} (k-1)}$ pies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the limit as $n$ tends to infinity, you end up with $\frac{k}{k-1}$ pies at grandma's house. If $k \neq 2$, you don't get to grandma's house with a whole number of pies, regardless of how many pies you left home with. So you might as well leave home with no pies if $k \neq 2$.&lt;/p&gt;&#xA;" OwnerUserId="9422" LastActivityDate="2015-07-12T00:56:57.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17817" PostTypeId="1" CreationDate="2015-07-12T03:56:59.950" Score="-1" ViewCount="383" Body="&lt;p&gt;There are two groups A and B with a value of 100 each.&lt;br&gt;&#xA;At the end of every day, group A gives 20% of it to group B while group B simultaneously gives 10 % of it to group A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So at the end of day 1, group A has 90 (100-20+10) while group B has 110 (100+20-10).&lt;br&gt;&#xA;At the end of day 2, these values would be 83 for group A and 117 for group B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would the values of the groups be after n days?&lt;/p&gt;&#xA;" OwnerUserId="14018" LastEditorUserId="68" LastEditDate="2015-07-12T07:48:07.393" LastActivityDate="2015-07-12T07:48:07.393" Title="Percentage Puzzle question" Tags="|number-sequence|" AnswerCount="2" CommentCount="0" ClosedDate="2015-07-12T06:02:45.407" ContentLicense="CC BY-SA 3.0" />
  <row Id="17818" PostTypeId="2" ParentId="17817" CreationDate="2015-07-12T04:20:27.973" Score="1" Body="&lt;p&gt;Here's how the sequence looks like:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1:  a = 90.000000       b=110.000000&#xA;2:  a = 83.000000       b=117.000000&#xA;3:  a = 78.099998       b=121.900002&#xA;4:  a = 74.669998       b=125.330002&#xA;5:  a = 72.268997       b=127.731003&#xA;6:  a = 70.588295       b=129.411697&#xA;7:  a = 69.411804       b=130.588181&#xA;8:  a = 68.588264       b=131.411728&#xA;9:  a = 68.011780       b=131.988205&#xA;10:  a = 67.608246      b=132.391739&#xA;11:  a = 67.325768      b=132.674210&#xA;12:  a = 67.128036      b=132.871948&#xA;13:  a = 66.989624      b=133.010361&#xA;14:  a = 66.892738      b=133.107254&#xA;15:  a = 66.824913      b=133.175079&#xA;16:  a = 66.777435      b=133.222549&#xA;17:  a = 66.744202      b=133.255783&#xA;18:  a = 66.720940      b=133.279053&#xA;19:  a = 66.704659      b=133.295334&#xA;20:  a = 66.693260      b=133.306732&#xA;21:  a = 66.685280      b=133.314713&#xA;22:  a = 66.679695      b=133.320297&#xA;23:  a = 66.675781      b=133.324203&#xA;24:  a = 66.673042      b=133.326935&#xA;25:  a = 66.671127      b=133.328857&#xA;26:  a = 66.669785      b=133.330200&#xA;27:  a = 66.668846      b=133.331131&#xA;28:  a = 66.668190      b=133.331787&#xA;29:  a = 66.667732      b=133.332245&#xA;30:  a = 66.667412      b=133.332565&#xA;31:  a = 66.667183      b=133.332794&#xA;32:  a = 66.667023      b=133.332947&#xA;33:  a = 66.666916      b=133.333054&#xA;34:  a = 66.666840      b=133.333130&#xA;35:  a = 66.666786      b=133.333191&#xA;36:  a = 66.666748      b=133.333221&#xA;37:  a = 66.666718      b=133.333252&#xA;38:  a = 66.666702      b=133.333267&#xA;39:  a = 66.666687      b=133.333282&#xA;40:  a = 66.666679      b=133.333298&#xA;41:  a = 66.666672      b=133.333298&#xA;42:  a = 66.666672      b=133.333298&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here's the code to produce this list:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;/* Author: Quark */&#xA;#include &amp;lt;stdio.h&amp;gt;&#xA;int main()&#xA;{&#xA;    float a = 100;&#xA;    float b = 100;&#xA;    float temp_a, temp_b;&#xA;    int count = 1;&#xA;&#xA;    while(1){&#xA;        temp_a=a*0.2;&#xA;        temp_b=b*0.1;&#xA;        a=a-temp_a+temp_b;&#xA;        b=b-temp_b+temp_a;&#xA;        printf(&quot;%d:  a = %f\tb=%f\n&quot;, count, a, b);&#xA;        count++;&#xA;    }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="12700" LastEditDate="2015-07-12T05:01:24.423" LastActivityDate="2015-07-12T05:01:24.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17819" PostTypeId="2" ParentId="17681" CreationDate="2015-07-12T04:24:39.143" Score="3" Body="&lt;p&gt;There is a circle based theorem that states &amp;quot;The measure of an inscribed angle is half the measure of the arc it intercepts.&amp;quot; This means that for angle &lt;em&gt;x&lt;/em&gt; , the arc it intercepts will be &lt;em&gt;2x&lt;/em&gt;.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/FqWzu.gif&quot; alt=&quot;Inscribed angles&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now, if you inscribe the star into a circle, you get this:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5217U.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Labeling the previous drawing, you get this;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IfM7k.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;With this theorem, we know that angle ∠1=c/2, ∠2=d/2, ∠3=e/2, ∠4=a/2, &lt;em&gt;and&lt;/em&gt; ∠5=b/2.&#xA;If we distribute that, we get &lt;em&gt;∠1+∠2+∠3+∠4+∠5=(a+b+c+d+e)/2&lt;/em&gt; .&#xA;Furthermore, Because the measures of all arcs in a circle add up to 360, we know that &lt;em&gt;a+b+c+d+e=360&lt;/em&gt; .&#xA;Finally, using the substitution property, we get &lt;em&gt;∠1+∠2+∠3+∠4+∠5=360/2&lt;/em&gt; , or &lt;em&gt;∠1+∠2+∠3+∠4+∠5 = 180&lt;/em&gt; .&#xA;Thus the sum of all angles is 180.&lt;/p&gt;&#xA;" OwnerUserId="1812" LastEditorUserId="25966" LastEditDate="2023-04-22T18:27:22.623" LastActivityDate="2023-04-22T18:27:22.623" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="17820" PostTypeId="2" ParentId="17817" CreationDate="2015-07-12T06:00:52.803" Score="2" Body="&lt;p&gt;We can use the fact that there are only two groups, and the fact that they are giving hypothetical things to each other only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since they only have $200$ things between them, then at any moment, if Group $A$ has $a_i$ things, then Group $B$ has $200-a_i$ things, so we can remove a variable (the amount Group $B$ has).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $a_i$ denotes the number of things Group $A$ has at the end of day $i$ $(i\geq 0)$, then by the conditions of the problem, $a_{i+1}=a_i-0.2(a_i)+0.1(200-a_i)=0.7a_i+20$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For $i=0$, the interpretation of this is the starting amount, despite the lack of an official &quot;Day 0&quot; in the problem statement. It's &quot;the amount at the beginning of Day 1.&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is called a first order linear recurrence, whose summation's general formula can be proven via induction if necessary, but writing out a few terms should also make the pattern clear:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_1=0.7a_0+20$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_2=0.7a_1+20=0.7^2a_0+(0.7)(20)+20$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_3=0.7a_2+20=0.7^3a_0+(0.7)^2(20)+(0.7)(20)+20$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_n=(0.7)^na_0+20\sum\limits_{i=0}^{n-1}(0.7)^i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And some more algebra later using $a_0=100$ and the formula for sum of a finite geometric series:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_n=\frac{100}{3}(0.7)^n+\frac{200}{3}$ (and Group $B$ just has $200-a_n$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is verified by Quark's output; as $n\rightarrow\infty$, $a_n\rightarrow\frac{200}{3}=66.666\dots$ since $(0.7)^n\rightarrow 0$.&lt;/p&gt;&#xA;" OwnerUserId="13995" LastActivityDate="2015-07-12T06:00:52.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17821" PostTypeId="1" AcceptedAnswerId="17953" CreationDate="2015-07-12T07:39:44.393" Score="5" ViewCount="416" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Let's start from the top.&lt;br&gt;&#xA;  Why, you ask? Because it's best to start on a high.&lt;br&gt;&#xA;  Get into it now.&lt;br&gt;&#xA;  Why, you ask? Because it's the smart idea to live by.&lt;br&gt;&lt;br&gt;&#xA;  Come out now and breathe the fresh air.&lt;br&gt;&#xA;  Why, you ask? Because it's not right to intrude.&lt;br&gt;&#xA;  Move under it just a little.&lt;br&gt;&#xA;  Why, you ask? Because it would be rather rude.&lt;br&gt;&lt;br&gt;&#xA;  Go through and get stuck now.&lt;br&gt;&#xA;  Why, you ask? Because it's rather rough here.&lt;br&gt;&#xA;  Come out altogether and go to the sides.&lt;br&gt;&#xA;  Why, you ask? Because you need more than one to conquer.&lt;br&gt;&lt;br&gt;&#xA;  Move to the ends.&lt;br&gt;&#xA;  Why, you ask? Because it's good to have something close.&lt;br&gt;&#xA;  Back up and down the middle till it splits.&lt;br&gt;&#xA;  Why, you ask? Because those are the ones that appeal most.&lt;br&gt;&lt;br&gt;&#xA;  Drop down to the bottom and stay there now.&lt;br&gt;&#xA;  Why, you ask? Because it's best to end with a game or two.&lt;br&gt;&#xA;  Too many questions here, that I know.&lt;br&gt;&#xA;  But perhaps no one knows the answers better than you. &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I talking about?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the clues are well hidden in the riddle. The best answer will explain every line perfectly.&lt;/p&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-07-12T10:02:17.113" LastActivityDate="2015-07-15T06:26:26.643" Title="Questions, questions, questions" Tags="|riddle|rhyme|wordplay|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17822" PostTypeId="2" ParentId="9883" CreationDate="2015-07-12T10:13:22.627" Score="0" Body="&lt;p&gt;You can find the value for the cell in the fourth row, first column. The only possible values are (3, 5, 8, 9).&lt;br&gt;&#xA;The fifth row has only two values left, (6, 8). Since they're both in the same block, we can eliminate the 8 from our target.&lt;br&gt;&#xA;Since the first column already has two cells with (3, 9) as their only possible values, we can eliminate those as well, leaving only the 5.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2015-07-12T10:13:22.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17824" PostTypeId="1" CreationDate="2015-07-12T11:49:01.843" Score="2" ViewCount="1122" Body="&lt;p&gt;There is a square with a side length of 1. Inside this square there is a quadrilateral. Each vertex of the quadrilateral is on a side of the square. The area of the quadrilateral is bigger than half (the area of the square is 1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to prove that there is a line segment inside the quadrilateral that is parallel to one of the square's sides and is longer than 1/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mhkIe.png&quot; alt=&quot;the red area is bigger than 0.5 as well&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="14023" LastEditorUserId="1752" LastEditDate="2015-07-12T13:29:19.033" LastActivityDate="2015-07-12T14:11:28.200" Title="Quadrilateral inside a square" Tags="|mathematics|geometry|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17825" PostTypeId="2" ParentId="17824" CreationDate="2015-07-12T12:07:03.733" Score="0" Body="&lt;p&gt;Since each vertex must lie on a side of the square, here's one simple example to disprove your claim. Without having a line parallel to any side, we can have a quadrilateral whose area is more than half of the square (although it would have a side that needs to have a length greater than 0.5).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lAS7o.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The square in black is the square of unit side and unit area, while the quadrilateral in red has an area that is quite visibly more than half of the square and also has each vertex on the square. Hence, what you want proven has been easily disproved with this simple example.&lt;/p&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-07-12T12:07:03.733" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17827" PostTypeId="2" ParentId="17824" CreationDate="2015-07-12T14:11:28.200" Score="6" Body="&lt;p&gt;This can be shown relatively trivially using basic integration (or equivalently, breaking the problem into parts).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that you have two functions, $f_1(x)$ and $f_2(x)$ - each is a piecewise linear function, with $f_1(0)=f_2(0)$ and $f_1(1)=f_2(1)$, as well as $\max f_1(x) = 1$ and $\min f_2(x)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the area of the quadrilateral is&#xA;$$&#xA;\int_0^1 (f_1(x)-f_2(x))dx\leq 1\times \max (f_1(x)-f_2(x))&#xA;$$&#xA;Because the integral must be greater than $\frac12$, the maximal value of $f_1(x)-f_2(x)$ must also be greater than $\frac12$.&lt;/p&gt;&#xA;" OwnerUserId="9289" LastActivityDate="2015-07-12T14:11:28.200" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17828" PostTypeId="1" CreationDate="2015-07-12T16:12:39.437" Score="4" ViewCount="528" Body="&lt;p&gt;I've created the Hannoi Towers, Lights out and not sure about the original name of &lt;a href=&quot;http://2.bp.blogspot.com/-7BmfwqDoPgo/UGarNir-vPI/AAAAAAAACpY/RgEyuBXlGNg/s1600/2012-09-21_10-11-45.png&quot; rel=&quot;nofollow&quot;&gt;this game&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, when I want to create badges (beginner, expert etc.) for the puzzlers I stumbled upon a problem, I couldn't find very accurate solution to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the Hannoi Towers game there are sizes of 3, 4, 5 and 6 to choose between to play. For Lights out and the other one there are from 3 to 7 including sizes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put in another way, for example on Hannoi Towers, the puzzler will need to solve the size of 6 for the minimum of 60 seconds in order to become an expert of the game (this is just an example, which most likely is highly inaccurate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would an expert, advanced, mediocre &amp;amp; beginner puzzlers be like on these games? I.e. what are the minimum requirements for a puzzler to earn any of these badges?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. the time taken for 1 move on any of these games-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Hannoi towers- best case 0.4 seconds, average- 0.7&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Lights out &amp;amp; the sort numbers game with the speed of click&lt;/p&gt;&#xA;" OwnerUserId="14026" LastEditorUserId="12628" LastEditDate="2015-07-13T12:28:30.597" LastActivityDate="2015-07-13T12:28:30.597" Title="Estimate beginner, mediocre, advanced &amp; expert on puzzle games" Tags="|logical-deduction|difficulty-rating|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17829" PostTypeId="1" AcceptedAnswerId="17832" CreationDate="2015-07-12T16:15:11.140" Score="6" ViewCount="12336" Body="&lt;p&gt;When the day after tomorrow becomes yesterday,&lt;br&gt; the day which would then be today will be as far from Sunday as the day which is yesterday when it's two days after that day's tomorrow.&lt;br&gt;&lt;br&gt; &lt;strong&gt;What day of the week was this spoken?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="13443" LastEditorUserId="5373" LastEditDate="2020-03-02T09:18:44.517" LastActivityDate="2020-03-02T09:18:44.517" Title="Finding Day Puzzles" Tags="|logical-deduction|calculation-puzzle|time|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17830" PostTypeId="2" ParentId="17829" CreationDate="2015-07-12T16:34:03.177" Score="3" Body="&lt;p&gt;This statement was spoken on &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;a Wednesday. &lt;br&gt; Day after tomorrow becomes yesterday on Saturday. And Saturday is 1 day away from Sunday.&lt;br&gt; Monday would be yesterday when it's Tuesday, which is two days after Sunday(which is Saturday's tomorrow).&lt;br&gt;&lt;br&gt; And both Saturday and Monday are equidistant from Sunday. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An explanation to my strategy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Consider our day of speaking to be Xday. We have two days of interest. First is that day on which day after tomorrow becomes yesterday which is the (X+3)&lt;sup&gt;th&lt;/sup&gt; day. The second is yesterday when it is two days after (X+3)&lt;sup&gt;th&lt;/sup&gt; day's tomorrow. Which is (X+3+1+2-1)&lt;sup&gt;th&lt;/sup&gt; or (X+5)&lt;sup&gt;th&lt;/sup&gt; day. &lt;br&gt;&lt;br&gt; Writing down the week with Sunday as the centre, we can only choose Saturday and Monday as the days of interest. Hence, Wednesday has to be our Xday.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-07-12T16:50:18.710" LastActivityDate="2015-07-12T16:50:18.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17831" PostTypeId="1" AcceptedAnswerId="17918" CreationDate="2015-07-12T16:39:55.037" Score="7" ViewCount="366" Body="&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;I am not the trimmings of a bush,&#xA;but something closer to a mush.&#xA;I am not the clams in your hand,&#xA;but that guess I understand.&#xA;&#xA;As long as you  answer with aplomb,&#xA;The answer you will not be far from.&#xA;Admittedly my humor is a little dried,&#xA;But much of my moral fiber is supplied.&#xA;&#xA;As for your guesses throw the worst away,&#xA;The answer is evident if I may say.&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is all of these things. But for the answer I am just aiming for one of them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If anyone is wondering what the original riddle was, and maybe tell me if you think it was particularly solvable back then:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;What am I? &lt;br/&gt;&#xA;I am not the trimmings of a bush, &lt;br/&gt;&#xA;but something closer to a mush. &lt;br/&gt;&#xA;I am not the clams in your hand, &lt;br/&gt;&#xA;but that guess I understand. &lt;br/&gt;&#xA;As for your guess throw the worst away,&lt;br/&gt;&#xA;The answer is evident if I may say. &lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13916" LastEditorUserId="13916" LastEditDate="2015-07-14T15:06:25.097" LastActivityDate="2015-07-14T15:06:25.097" Title="I guess this one is for the old ones out there" Tags="|riddle|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17832" PostTypeId="2" ParentId="17829" CreationDate="2015-07-12T16:46:08.807" Score="4" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wednesday&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the day after tomorrow becomes yesterday,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Three days from now (A)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the day which would then be today will be as far from Sunday as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the day which is yesterday when it's two days after that day's tomorrow.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Two days after that day.(B) A and B are two days apart and equally far from Sunday, so they must be Saturday and Monday, meaning the original day was Wednesday.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3987" LastActivityDate="2015-07-12T16:46:08.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17833" PostTypeId="1" AcceptedAnswerId="17913" CreationDate="2015-07-12T17:28:26.760" Score="18" ViewCount="1293" Body="&lt;p&gt;I found &lt;a href=&quot;http://9gag.com/gag/azVrZRj&quot; rel=&quot;noreferrer&quot;&gt;this problem&lt;/a&gt; on 9gag and I have no idea how to solve it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$13+7=1130$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$88+9=47$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$32+28=2530$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$10+999=1149$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$444+77=61$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$55+111=56$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$21+0=1221$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$7+22=39$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$33+5=38$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$123+45=???$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The puzzle says to think outside the box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wuuEP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user10001" LastEditorDisplayName="user10001" LastEditDate="2015-07-14T10:54:16.177" LastActivityDate="2015-07-14T13:02:14.463" Title="How does this addition work (13+7=1130)?" Tags="|mathematics|number-sequence|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="17834" PostTypeId="2" ParentId="17828" CreationDate="2015-07-12T17:31:51.563" Score="4" Body="&lt;p&gt;I believe the best way to judge puzzlers would be to have a round of beta testing. Create a small group composed of your colleagues, friends and family, making sure that you include people of all skill levels. Include your not-so-tech-savvy parents, your pretty-but-not-very-bright neighbour, your quick-thinking colleague and people from all walks of life that you know. The bigger your beta test group the better, as it gives you a better bell curve to judge the difficulty level of your puzzles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have the test group play the puzzles and record the times taken, and as I mentioned, obtain a bell curve of the data. Then, based on the numbers you can decide how to set the limits for each puzzle. As an example, you could award the expert badge for people who finish within the time set by the top 20% in the test group. The intermediate badge goes to those who finish in the next 40%. The beginner badge goes to those who fare like the last 40%. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not only does this help you figure out how to set realistic demands for badges but the beta testing will also help you to iron out any bugs prior to the general release of your puzzles. &lt;/p&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-07-12T17:31:51.563" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17835" PostTypeId="2" ParentId="17627" CreationDate="2015-07-12T17:34:36.363" Score="0" Body="&lt;p&gt;For the normal:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  A Normal can just say something paradoxical like &quot;I'm a liar/Knave&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And for the knight:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  A Knight has to say something that would prove he's a Knight if true, and just lead to a contradiction/dead end if false. Indeed, there's no way &quot;Either this single sentence is true and I am a Knight, or it is false and I am a Normal or Knave&quot; would be false without causing some sort of a contradiction: &lt;a href=&quot;https://i.stack.imgur.com/qUzLD.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;click&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastActivityDate="2015-07-12T17:34:36.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17836" PostTypeId="2" ParentId="219" CreationDate="2015-07-12T17:43:03.547" Score="0" Body="&lt;p&gt;Super cat is the closed to &quot;proofification&quot; in my opinion, maybe lacking in clarity (in my opinion). &#xA;Though the other rather decent hand-wavy proofs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As supercat points out it comes down to one region being landlocked and having an odd number of neighbors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we aim for a construction with 3 colors, we can choose colour A to be the central one.&lt;br&gt;&#xA;And Color B to ANY adjacent one. Now as we only have 3 colours we are forced to choose a colour for the one clockwise of the B coloured one. This would be C. Which in turn has its clockwise neighbor forced.&lt;br&gt;&#xA;This becomes proofy for odd numbers of neighbors as when we reach the spot counterclockwise of the original B colour, it has a minimum of 3 neighbors.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The first of the chain, B coloured.  &lt;/li&gt;&#xA;&lt;li&gt;the middle area, A coloured.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the second to last in the clockwise chain&#xA;   In the chain the pattern goes BCBCBC, where the 1st is B and the next is C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where every odd numbered one must be B, and even numbered C&lt;br&gt;&#xA;So the second to last in the chain is C as there are an odd number of neighbours&lt;br&gt;&#xA;and it is the second to last so it is an even one.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as the last in the chain is C, it has A, B, C neighbors. It cannot be A, B, or C: therefore it cannot be three coloured.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If all landlocked US states had an even number of neighbours you could 3 colour it. The reason I specify land locked is ocean adjacent states don't have a complete chain around them. Hope this answer adds some rigour and hopefully enlightenment to general 4 vs 3 Colouredness.&lt;/p&gt;&#xA;" OwnerUserId="13916" LastEditorUserId="44915" LastEditDate="2018-11-30T10:02:06.520" LastActivityDate="2018-11-30T10:02:06.520" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="17837" PostTypeId="1" CreationDate="2015-07-12T18:07:14.907" Score="8" ViewCount="1555" Body="&lt;p&gt;I came up with this little game. Basically we have to write a word using small caps and big caps so that the small caps write a word and the big caps write a word (in their original order), for example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;STRINginG, the words are STRING and gin. ( or STringING with STING and ring)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MoneY, the words are MY and one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;THEoreM, the words are THEM and ore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The score of each word is the length of the smallest word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So STRINginG has score $3$, MoneY has score $2$ and THEoreM has score $3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I do not want to accept the cases in which all the big caps are together. So things like BUTTERfly gentleMAN or CARrot are not allowed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would appreciate a lot if you could give me examples of words with high scores (the more the merrier), something especially awesome would be if the words could have related meanings.&#xA;Thank you in advance.&lt;/p&gt;&#xA;" OwnerDisplayName="user10001" LastEditorDisplayName="user10001" LastEditDate="2015-07-13T14:25:13.887" LastActivityDate="2015-08-08T05:09:19.217" Title="How long can two words be if they make up a larger word (challenge)" Tags="|word|wordplay|letter-sequence|" AnswerCount="8" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17838" PostTypeId="2" ParentId="17837" CreationDate="2015-07-12T18:12:58.140" Score="9" Body="&lt;p&gt;I have a word with a score of 5&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;TrIeNnIaLlY. &lt;br&gt;&lt;br&gt;Made up of the words TINILY and renal with alternating letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-07-12T18:12:58.140" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17840" PostTypeId="2" ParentId="17837" CreationDate="2015-07-12T20:50:15.773" Score="0" Body="&lt;p&gt;I can only guess you meant a similar rule for all small caps not being together.&#xA;Otherwise you could go with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DAYdreamING&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;with the score of &lt;strong&gt;5&lt;/strong&gt; and a word play on your nickname.&lt;/p&gt;&#xA;" OwnerUserId="10739" LastActivityDate="2015-07-12T20:50:15.773" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17841" PostTypeId="1" AcceptedAnswerId="17844" CreationDate="2015-07-12T20:53:15.757" Score="3" ViewCount="307" Body="&lt;p&gt;Annatar stepped into Celebrimbor's forge, watching for a second.  Then, he spoke.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I wish to test you, to see whether your mind is as good as your smithing.&quot;  Like any Noldor would, Celebrimbor bristled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Trying to be like Fëanor?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I? Becoming like you or your kind?&quot;  Celebrimbor watched with faint amusement as the &quot;Giver of Gifts&quot; regained his composure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, he spoke.  &quot;I accept your challenge.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I give you this child of Men, from a world of walking tables, to assist you.  He is called Rand al'thor.  You have one week to solve the puzzles.  Good luck.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Puzzle 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RGW/JWTQIES/GLS/NIEFIRG'A/XYEAW./UBSWWS/GW/UA/BIR/NLARWE/ID/DLRW&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Puzzle 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;L/RDVU/JHDM/GTJM/UI/VGX/BNIP/RQ/RPJG/IV/L/CKUTIR/GADP;/UEF/F/IMHE/CPCZ/PGTH/KICO/GM/KZZ/NRXI/TC/VECI/UZ/KZZ/RNQJCZN.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Solving Puzzle 1 helps with solving Puzzle 2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can you help rand solve the ciphers and make sure the history of Middle Earth is not changed?  If it helps, rand has some sort of mobile device with him, and amazingly, there is Wifi ;-).&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;There are probably tons of errors, but hey, it's only my second(?) question concerning ciphers.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="11790" LastEditorUserId="11790" LastEditDate="2015-07-12T21:05:16.903" LastActivityDate="2015-07-13T03:50:14.593" Title="Riddles in the Dark? Nah, just ciphers in the shadows" Tags="|cipher|story|" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17842" PostTypeId="2" ParentId="17837" CreationDate="2015-07-12T21:18:19.287" Score="0" Body="&lt;p&gt;How about:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;DauntING&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;PARallelS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which both give a score of &lt;strong&gt;4&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latter is again a bit controversial as it uses a less common spelling of &lt;a href=&quot;https://en.wikipedia.org/wiki/Allele&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Allele&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10739" LastEditorUserId="10739" LastEditDate="2015-07-12T21:33:32.617" LastActivityDate="2015-07-12T21:33:32.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17843" PostTypeId="1" AcceptedAnswerId="17848" CreationDate="2015-07-12T21:20:35.597" Score="31" ViewCount="1405" Body="&lt;p&gt;Most accomplished autohrs are aware of the critical importance of proper grammar and spelling. The the neccessity of riting one's ideas succinctly, appropriately, and without error is parmaount. Occassional oversights may be forgived, but generall the technical corectness of a written document is considered a gauge of proffesionalism, a reflection of the author's erudition and attention do detail. Propar care must be taken to avoid simple mistakes that betray ignorance of bsaic language mechanics such as tense, subject-verb aggreement and the proper use of holonyms. An writer with with a mastery of the English language should be able to write, boldly and without assisstance, &quot;Precisely what is done upto the former, that cahnge must be done also to the latter.&quot;  But I digresses, as I often has. The answer to this puzzle is also the answer to the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;O H T H E C A M S D I B F T O A A S G L A N W I T H S P A H E S S ?&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the answer to the puzzle, dear reader?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="4219" LastEditDate="2015-07-12T21:29:31.743" LastActivityDate="2015-07-12T22:10:16.513" Title="The Importence of Proper Spelled and Grammar" Tags="|word|wordplay|knowledge|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17844" PostTypeId="2" ParentId="17841" CreationDate="2015-07-12T21:20:41.303" Score="5" Body="&lt;p&gt;The plaintext for Puzzle 1 is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;THE KEYWORD HAD MORGOTH'S CURSE. INDEED HE IS NOT MASTER OF FATE&lt;br&gt;&#xA;&lt;br&gt;&#xA;This is a substitution cryptogram that can be solved with &lt;a href=&quot;http://quipqiup.com/index.php&quot;&gt;quipqiup&lt;/a&gt;.  The substitution was done by shifting the plaintext character one key to the left on a QWERTY keyboard (as identified by rand al'thor).&lt;br&gt;&#xA;&lt;br&gt;&#xA;Within the question, the clue for this shift is &lt;strong&gt;&quot;I? Becoming like you&quot;&lt;/strong&gt; or &quot;I&quot; becomes &quot;U&quot; which is one key to the left.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The plaintext for Puzzle 2 is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I DONT KNOW HALF OF YOU HALF AS WELL AS I SHOULD LIKE; AND I LIKE LESS THAN HALF OF YOU HALF AS WELL AS YOU DESERVE.&lt;br&gt;&#xA;&lt;br&gt;&#xA;This is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Beaufort_cipher&quot;&gt;Beaufort cipher&lt;/a&gt; with a keyword of &lt;strong&gt;TURIN&lt;/strong&gt; who is the son of Húrin Thalion. I think that the clue in the question is &lt;strong&gt;&quot;a world of walking tables&quot;&lt;/strong&gt;.&lt;br&gt;&#xA;&lt;br&gt;&#xA;The plaintext is a statement by Bilbo from &quot;The Fellowship of the Ring&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-07-13T03:50:14.593" LastActivityDate="2015-07-13T03:50:14.593" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17845" PostTypeId="2" ParentId="17837" CreationDate="2015-07-12T21:54:03.893" Score="5" Body="&lt;p&gt;Submitting two NBA-themed words:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PACEmakeRs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TRAILblazerS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;with a score of &lt;strong&gt;5&lt;/strong&gt; and &lt;strong&gt;6&lt;/strong&gt;. Although I feel a bit abusing the consecutive caps rule with the simple &lt;code&gt;-s&lt;/code&gt; suffix.&lt;/p&gt;&#xA;" OwnerUserId="10739" LastEditorUserId="10739" LastEditDate="2015-07-12T21:59:33.700" LastActivityDate="2015-07-12T21:59:33.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17846" PostTypeId="2" ParentId="17837" CreationDate="2015-07-12T21:57:55.240" Score="10" Body="&lt;p&gt;A word with a score of &lt;strong&gt;11&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; REinstitutionALIZATION&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A couple words with a score of 7:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; COUNTERbalanceS&lt;br&gt;&#xA; LIGHTheartedNESS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4450" LastEditDate="2015-07-12T22:16:52.357" LastActivityDate="2015-07-12T22:16:52.357" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17848" PostTypeId="2" ParentId="17843" CreationDate="2015-07-12T22:06:00.697" Score="26" Body="&lt;p&gt;The text has a number of spelling a grammar mistakes that should be corrected.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Swap 'O' and 'H' in 'autohrs'&lt;br&gt;Remove 'THE'&lt;br&gt;Remove 'C' from 'neccessity'&lt;br&gt;Add 'W' to 'riting'&lt;br&gt;Swap 'M' and 'A' in 'parmaount'&lt;br&gt;Remove 'S' from 'occassional'&lt;br&gt;'D' --&gt; 'N' in 'forgived'&lt;br&gt;Add 'Y' to 'generall'&lt;br&gt;Add 'R' to 'corectness'&lt;br&gt;Remove 'F' from 'proffesionalism'&lt;br&gt;Add 'S' to 'proffesionalism'&lt;br&gt;'D' --&gt; 'T' in 'do'&lt;br&gt;'A' --&gt; 'E' in 'propar'&lt;br&gt;Swap 'S' and 'A' in 'bsaic'&lt;br&gt;Remove 'G' from 'aggreement'&lt;br&gt;'L' --&gt; 'M' in 'holonyms'&lt;br&gt;Remove 'N' from 'an'&lt;br&gt;'P' --&gt; 'N' in 'upto'&lt;br&gt;Swap 'A' and 'H' in 'cahnge'&lt;br&gt;Remove 'ES' from 'digresses'&lt;br&gt;'S' --&gt; 'VE' in 'has'&lt;br&gt;&lt;br&gt; Making the same changes to the string&lt;br&gt; 'O H T H E C A M S D I B F T O A A S G L A N W I T H S P A H E S S?' results in:&lt;br&gt;'HOW MANY RIBS DOES A MAN HAVE?'&lt;br&gt;The answer is 24.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-07-12T22:10:16.513" LastActivityDate="2015-07-12T22:10:16.513" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17849" PostTypeId="2" ParentId="17841" CreationDate="2015-07-12T22:09:09.867" Score="1" Body="&lt;h2&gt;Puzzle 1&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;To decipher,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; shift each letter of the ciphertext one space to the right on a QWERTY keyboard (RGW -&gt; THE and so on). Wrap the left and right sides of the keyboard round to meet each other, so that L -&gt; A etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;to get the plaintext Len found:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; THE KEYWORD HAD MORGOTH'S CURSE. INDEED HE IS NOT MASTER OF FATE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-07-12T22:09:09.867" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17850" PostTypeId="2" ParentId="17119" CreationDate="2015-07-13T00:19:26.467" Score="18" Body="&lt;h2&gt;Solution&lt;/h2&gt;&#xA;&lt;h2&gt;&lt;a href=&quot;http://www.openexoplanetcatalogue.com/planet/Gliese%20785%20b/&quot; rel=&quot;noreferrer&quot;&gt;Gliese 785&lt;/a&gt;&lt;/h2&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Right ascension  20 15 17&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Declination  -27 01 58&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;h2&gt;Alien Message Translation&lt;/h2&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RS2j3.png&quot; alt=&quot;enter image description here&quot; /&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/RS2j3.png&quot; rel=&quot;noreferrer&quot;&gt;click to enlarge&lt;/a&gt;&lt;/p&gt;&#xA;&lt;h2&gt;Explanation&lt;/h2&gt;&#xA;&lt;p&gt;&lt;strong&gt;Page 1&lt;/strong&gt;&#xA;The purpose of this page is to teach us the alien number system. They use a base 3 number system that we call balanced ternary with the '+' symbol defined as +1, the '-' symbol as 0, and the 'x' symbol as -1. In addition they define the '&amp;gt;' symbol as the radix point. In the examples they show us how to represent the numbers 1 through 9.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Page 2&lt;/strong&gt;&#xA;Here they move on to teach us the concepts of addition/subtraction, multiplication/division, and equivalency. In each of the examples the symbols that represent the various operations are defined and can be seen clearly in the above image.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Page 3&lt;/strong&gt;&#xA;Now they demonstrate the properties of the their number system and how it represents decimals. They begin by showing that shift operations in balance ternary are achieved by multiplying by 3 to shift left and dividing by 3 to shift right. They then demonstrate a new symbol 'O' when drawn around any set of digits defines a repeating group identical to bar notation here. They also introduce a 'o' symbol when placed at the end of an equation it means continue the series infinitely.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Page 4&lt;/strong&gt;&#xA;They move on from basic number representation and operations to define the concept of and the symbol representing Pi. They introduce a new symbol '|' that was previously introduced in page two as '&amp;gt;|' to mean equals. Or rather in the previous use it was customary to write equals without a space between other symbols. At the top of the page they appear to be introducing additional equation concepts that utilize perpendicular calculations and back to back operators. I assume that some sort of order of operations and/or the equivalent of parenthesis is implied. Clearly this looks like some sort of series expansion. We see the familiar 1, -1/3, +1/5, -1/7 ... expansion used to calculate Pi.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Page 5&lt;/strong&gt;&#xA;Similarly to the previous page, they derive and define the constant 'e'. They also define the concept and syntax of exponents with two examples.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Page 6&lt;/strong&gt;&#xA;Here they give models representing Carbon, Oxygen, Hydrogen, and Helium atoms followed by a number in alien units. We define this unit as the Alien Mass Unit (AMU). By referencing the various properties of these atoms you can quickly see that they are defining the mass of each atom in terms of the mass of an electron (me).&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;1 AMU = 9.10938291E-31 kg&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Page 7&lt;/strong&gt;&#xA;In this page they show a model of a water molecule with distinguishing characteristics such as the number of protons and the angle between the two hydrogen atoms. The aliens also define an length unit in terms of the bond distance between the oxygen and hydrogen atoms. We define this unit as the Alien Length Unit (ALU). Referencing various materials we can see that the radius of the hydrogen atom is well defined as the Bohr Radius which is roughly 53 pm, and that the O-H bond distance is 95.7854 pm.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;1 ALU = 52.9177211 pm&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Page 8&lt;/strong&gt;&#xA;Now they show a model of the alien radio transmitter. We translated the antenna diameter, and converted the ALU units to meters and determined that the diameter of the antenna is 2197 m. The number on the lower right of the diagram represents the transmission frequency in alien units and the number in the center of the diagram represents the transmission power also in alien units. We define these units as the Alien Frequency Unit (AFU) and Alien Power Unit (APU) respectively. Since our receive frequency was 4460MHz we can calculate AFU.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;1 AFU = 1.170960 Hz&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Given that the receive antenna is 10m with a receiver sensitivity of -130dBm we can calculate a reasonable maximum transmission distance. A 10m dish antenna with an efficiency of 0.7 will have a maximum of 50dB gain and the alien dish antenna will have a maximum of 99dB gain at 4460MHz. So the maximum free space propagation can be determined given these quantities and the signal to noise ratio (SNR) of the received signal in addition to the alien transmit power. However the power is currently unknown. But we can determine a reasonable estimate somewhere between 60 and 90dBm. Using these values we can safely estimate a maximum transmission distance of ~200 ly.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Page 9&lt;/strong&gt;&#xA;In this page they depict the aliens planetary system with numbers that represent the distances from the center of their star. The distance units will be determined in later pages.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Page 10&lt;/strong&gt;&#xA;Now they specify the mass and radius of planets 1, 2, 4, and 5 as well as the mass of each moon and distance to the related planet. The mass and radius are in currently unknown alien units which we will define as Alien Planetary Mass Unit (APMU) and Alien Planetary Radius Unit (APRU) respectively. The distance units will be determined in later pages.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Page 11&lt;/strong&gt;&#xA;In this page they continue from page ten with the third planet and its associated moon. The planet and moons are specified in units of APMU, APRU. However when looking at the numbers for the fourth moon of the third planet we see that they have defined an equation. The two equations define the APMU and APRU units.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;1 APMU = 1.454E+53 AMU&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;1 APRU = 4.880E+16 ALU&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Again the distance units will be determined in later pages.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Page 12&lt;/strong&gt;&#xA;In this final page of the alien message they show us something unexpected. We see that the aliens have specified a relative distance from their system to another star with additional orbiting planets. Looking at this and the last few pages we conclude that the unit of distance between the moons and the associated planets, as well as, the unit of distance between the planets and the associated stars is in units of APRU. Finally we notice that the aliens use a different unit for specifying the distance between the two stars. As is the case here on Earth we define our unit of distance AU to be the distance between Earth and the Sun. Here we notice that the aliens chose to define their unit to represent the distance between their moon and it's associated planet. We define this unit as the Alien Astronomical Unit (AAU).&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;1 AAU = 301 APRU&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;h2&gt;Calculations&lt;/h2&gt;&#xA;&lt;p&gt;&lt;strong&gt;star radius (sun)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;7.7191e-01, 1.0020e+00&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;star mass (sun)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;7.3138e-01, 1.0027e+00&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;system distance (au)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;1.8251e+06&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;system 1 planet distance (au)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;1.7197e-01, 3.1877e-01, 1.1828e+00, 2.4915e+00, 4.6432e+00&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;system 1 planet mass (earth)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;5.5857e-02, 1.7033e+01, 2.4707e+01, 9.0710e+00, 4.0661e-01&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;system 1 planet radius (earth)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;3.8028e-01, 7.0252e+00, 4.3683e+00, 3.2424e+00, 7.6557e-01&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;system 1 planet 2 moon distance (au)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;2.1058e-03&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;system 1 planet 3 moon distance (au)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;4.4303e-04, 1.1047e-03, 3.1587e-03, 5.1955e-03, 9.6315e-03&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;system 1 planet 4 moon distance (au)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;6.7317e-04, 9.3208e-04&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;system 1 planet 5 moon distance (au)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;3.9700e-04&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;system 1 planet 2 moon mass (earth)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;7.4030e-04&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;system 1 planet 3 moon mass (earth)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;9.3203e-07, 3.1099e-04, 3.8536e-04, 2.2179e-02, 1.8780e-05&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;system 1 planet 4 moon mass (earth)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;3.7184e-05, 1.4874e-04&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;system 1 planet 5 moon mass (earth)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;9.9157e-05&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;system 2 planet distance (au)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;5.2094e+00, 9.5129e+00&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;system 2 planet mass (earth)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;3.1797e+02, 9.5213e+01&lt;/p&gt;&#xA;&lt;h2&gt;Search&lt;/h2&gt;&#xA;&lt;p&gt;Analyzing the above calculations, we recognize that the planets and star in the second system match Jupiter, Saturn, and our Sun exactly. Now searching through the various databases for systems around 28.86 ly away, I found Gliese 785 matches the specifications of the alien system star, planet 2, and planet 3 exactly. Therefore we can conclude that this is the origin of the alien signal transmission. Clearly we are currently unaware of the alien systems other plants and moons. I foresee future space missions.&lt;/p&gt;&#xA;&lt;p&gt;Here are two of the more helpful online exoplanet databases.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.openexoplanetcatalogue.com/systems/&quot; rel=&quot;noreferrer&quot;&gt;http://www.openexoplanetcatalogue.com/systems/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://exoplanetarchive.ipac.caltech.edu/&quot; rel=&quot;noreferrer&quot;&gt;http://exoplanetarchive.ipac.caltech.edu/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13788" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-17T17:49:12.947" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="17851" PostTypeId="2" ParentId="17775" CreationDate="2015-07-13T01:31:07.173" Score="0" Body="&lt;p&gt;Could it be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chlorine?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You sit by the pool to eat a meal. &lt;br&gt;&#xA;Me, you can see, but won’t hear. &lt;br&gt;&#xA;Go ahead, eat your lunch &lt;br&gt;&#xA;And drink your light beer… &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m drowning!!! I don’t like to whine… &lt;br&gt;&#xA;I’m drowning!!! But please throw me a line! &lt;br&gt;&#xA;I’m drowning!!! I’ve not got much time. &lt;br&gt;&#xA;I’m drowning!!! I’m not feeling fine.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pool water requires chlorine in it to make the water safe to swim in. You know it is there as you look into the water, but you certainly can't here it. Of course, chlorine might not be so comfortable being in a giant pool...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Gaze across the table at your date. &lt;br&gt;&#xA;Stop pigging out, put yourself on idle. &lt;br&gt;&#xA;You don’t note my importance, &lt;br&gt;&#xA;But to you, I am vital!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m drowning!!! I’m here in this pool. &lt;br&gt;&#xA;I’m drowning!!! I try to stay cool. &lt;br&gt;&#xA;I’m drowning!!! But you’re being cruel. &lt;br&gt;&#xA;I’m drowning!!! I need a saving tool!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As you gaze out and admire at how handsome/pretty your boyfriend/girlfriend look, you might not notice the science of chlorine that might be used to help make the clothes your date is wearing right now (and even you)! Source: &lt;a href=&quot;http://chlorine.americanchemistry.com/Style&quot; rel=&quot;nofollow&quot;&gt;http://chlorine.americanchemistry.com/Style&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You sure love your meal and date… &lt;br&gt;&#xA;Only sound is chewing – no conversing speech. &lt;br&gt;&#xA;I flail and try to get your attention… &lt;br&gt;&#xA;But my arms are short – I cannot reach…&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m drowning!!! I’m gasping for air. &lt;br&gt;&#xA;I’m drowning!!! Dry land is somewhere! &lt;br&gt;&#xA;I’m drowning!!! I won’t make it there… &lt;br&gt;&#xA;I’m drowning!!! I don’t think you care…&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Chlorine atoms are really small. Naturally, you cannot see them since they are usually in liquid (clear) or even gas form. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Full source: &lt;a href=&quot;http://chlorine.americanchemistry.com/Chlorine-Benefits&quot; rel=&quot;nofollow&quot;&gt;http://chlorine.americanchemistry.com/Chlorine-Benefits&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9042" LastEditorUserId="4385" LastEditDate="2015-07-15T17:09:43.277" LastActivityDate="2015-07-15T17:09:43.277" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17852" PostTypeId="1" AcceptedAnswerId="17854" CreationDate="2015-07-13T01:37:33.080" Score="21" ViewCount="1724" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;A bit beyond perceptions reach,&lt;br&gt;&#xA;  I sometimes believe I see &lt;br&gt;&#xA;  that life is two locked boxes, each &lt;br&gt;&#xA;  containing the other’s key.&lt;br&gt;&lt;/em&gt;&#xA;  ― Piet Hein&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You have one hundred lockboxes, fifty made of steel, fifty made of wood. You only have one copy of each box's key.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A prankster breaks into your house and randomly places the 100 keys in the boxes, one key per box. He then shuts all the boxes, locking the keys inside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though the steel boxes are impenetrable, the wooden ones can be broken quite easily. What is the probability that breaking open the wooden boxes will allow you to open &lt;strong&gt;all&lt;/strong&gt; of the steel ones?&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="4269" LastEditDate="2015-09-12T17:11:10.410" LastActivityDate="2015-09-12T17:11:10.410" Title="One Hundred Lockboxes of Wood and Steel" Tags="|probability|combinatorics|keys-and-locks|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17853" PostTypeId="2" ParentId="17837" CreationDate="2015-07-13T02:05:08.637" Score="6" Body="&lt;p&gt;Here's a score of &lt;strong&gt;12&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The longest non-coined word in the English language is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; antidisestablishmentarianism&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It can be split up into&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ANTI disestablishment ARIANISM&lt;br&gt;&lt;br&gt;&#xA;&#xA; ANTI-&lt;a href=&quot;http://www.thefreedictionary.com/Arianism&quot;&gt;ARIANISM&lt;/a&gt; (12) and &lt;a href=&quot;http://www.thefreedictionary.com/disestablishment&quot;&gt;disestablishment &lt;/a&gt; (16)&lt;br&gt;&lt;br&gt;&#xA;&#xA; Some examples of the exact word anti-arianism:&lt;br&gt;&lt;br&gt;&#xA; &lt;a href=&quot;http://www.philosophy-religion.org/handouts/pdfs/BERKELEY-DIVINITY-SCHOOL_ATHANASIUS.pdf&quot;&gt;An essay on Arianism&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;http://petercanisiusmichaeldavidkang.com/tag/anti-arianism/&quot;&gt;A subject tag&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;https://books.google.com/books?id=hMoRCgAAQBAJ&amp;amp;pg=PA58&amp;amp;lpg=PA58&amp;amp;dq=%22anti-arianism%22&amp;amp;source=bl&amp;amp;ots=1_7bIRPwQm&amp;amp;sig=PTeyNta6ceAjH95ej0vsCmtPEX0&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=FBajVfnnNMq0ogSYvpLwDw&amp;amp;ved=0CCkQ6AEwAw#v=onepage&amp;amp;q=%22anti-arianism%22&amp;amp;f=false&quot;&gt;The book &quot;The Theological Anthropology of Eustathius of Antioch&quot;&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;https://books.google.com/books?id=yx95BgAAQBAJ&amp;amp;pg=PA13&amp;amp;lpg=PA13&amp;amp;dq=%22anti-arianism%22&amp;amp;source=bl&amp;amp;ots=fyFp55Bdsi&amp;amp;sig=8jLhF1_YN9sMz_nTtvRarZ-GBAw&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=FBajVfnnNMq0ogSYvpLwDw&amp;amp;ved=0CDoQ6AEwBw#v=onepage&amp;amp;q=%22anti-arianism%22&amp;amp;f=false&quot;&gt;The book &quot;Theophilus of Alexandria and the First Origenist Controversy: Rhetoric and Power&quot;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastActivityDate="2015-07-13T02:05:08.637" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17854" PostTypeId="2" ParentId="17852" CreationDate="2015-07-13T02:07:38.533" Score="32" Body="&lt;p&gt;The probability is $1/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a permutation that maps each box to the box whose key it contains. Once we open a box, we can open the box it maps to. So, we can open all the boxes exactly if there is no all-steel cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Label the boxes $1$ through $100$. We denote the permutation in cycle format like $(31)(542)(6)$. To make this canonical, write each cycle with the greatest number first, and sort the cycles by increasing first number. Now, we can extract the cycle structure just from the sequence $315426$ because the cycles start at numbers that greater any before them ($3$, $5$, $6$). So, each of the $n!$ such sequences represents one of the $n!$ permutations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let the steel boxes be $1$ through $50$ and wooden boxes be $51$ through $100$. There's a steel cycle exactly if the first number is at most $50$. If so, the first cycle contains only numbers at most $50$, and if not, every cycle starts with a number at least $51$. Since the cycle representation is a uniformly random ordering, the first number is uniformly random. So, this happens with probability $1/2$, and the remaining $1/2$ the time we can open all the boxes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally with $w$ wooden boxes and $s$ steel boxes, the probability of opening all the boxes equals the fraction of wooden boxes $\frac{w}{w+s}$.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-07-13T02:32:36.277" LastActivityDate="2015-07-13T02:32:36.277" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="17855" PostTypeId="2" ParentId="17837" CreationDate="2015-07-13T02:41:59.727" Score="-1" Body="&lt;p&gt;Here's a score of &lt;strong&gt;16&lt;/strong&gt;, if you can accept both words.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A&lt;strong&gt;ultramicroscopic&lt;/strong&gt;SILICOVOLCANOCONIOSIS (38 letters)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Explanation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Longest_word_in_English&quot; rel=&quot;nofollow&quot;&gt;longest English word&lt;/a&gt; that is in a major dictionary is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pneumonoultramicroscopicsilicovolcanoconiosis (45 letters)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When broken down, that is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Pneumono&lt;/strong&gt; ultra microscopic silico volcano coniosis&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where the first part, &lt;strong&gt;pneumono&lt;/strong&gt;, is a &lt;a href=&quot;http://medical-dictionary.thefreedictionary.com/pneumono-&quot; rel=&quot;nofollow&quot;&gt;medical prefix that specifies that the disease is lung-related or pulmonary-related&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, removing it from the 45-letter monster above, we get the word&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ultramicroscopic silicovolcanoconiosis (without the space)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, the prefix &quot;a&quot; means &quot;not&quot; (e.g.: atypical means not typical). So, adding the prefix &lt;strong&gt;a&lt;/strong&gt; to the word above gives us&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A ultramicroscopic SILICOVOLCANOCONIOSIS&lt;br&gt;&lt;br&gt;&#xA; That's &lt;a href=&quot;https://www.google.com/search?q=ultramicroscopic&amp;amp;ie=utf-8&amp;amp;oe=utf-8#safe=off&amp;amp;q=ultramicroscopic&quot; rel=&quot;nofollow&quot;&gt;ultramicroscopic&lt;/a&gt; and ASILICOVOLCANO&lt;a href=&quot;http://dictionary.reference.com/browse/coniosis&quot; rel=&quot;nofollow&quot;&gt;CONIOSIS&lt;/a&gt; (a form of coniosis that is NOT caused by volcano dust or silicon exposure.)&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastActivityDate="2015-07-13T02:41:59.727" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="17856" PostTypeId="1" AcceptedAnswerId="17866" CreationDate="2015-07-13T03:47:24.317" Score="27" ViewCount="1680" Body="&lt;h3&gt;The Setup&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Consider a 5-by-5 grid of 7-segment displays, &quot;smashed together&quot; so that the peripheral segments of neighbouring displays overlap (see below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/FVasI.png&quot; alt=&quot;a smashed 5-by-5 grid of 7-segment displays&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are given a particular grid as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/ZCCPq.png&quot; alt=&quot;a smashed 7-segment grid with black segments&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The grid contains the 20 characters:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/lhget.png&quot; alt=&quot;list of characters appearing in the 7-segment grid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of the characters appear in the grid, and some appear more than once. They are arranged in such a way that no two &quot;on&quot; (coloured) segments overlap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, an &lt;code&gt;A&lt;/code&gt; cannot appear immediately to the right of a &lt;code&gt;b&lt;/code&gt;, but may appear immediately right of &lt;code&gt;C&lt;/code&gt;. A &lt;code&gt;3&lt;/code&gt; may not appear immediately above a &lt;code&gt;q&lt;/code&gt;, but may appear immediately above a &lt;code&gt;y&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;The Challenge&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Given&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the coloured segments must match the given pattern&lt;/li&gt;&#xA;&lt;li&gt;all 20 characters must be used at least once&lt;/li&gt;&#xA;&lt;li&gt;no two coloured segments may overlap&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;can you determine which characters are displayed on all 25 7-segment displays?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="75318" LastEditDate="2021-09-25T15:11:45.180" LastActivityDate="2021-09-25T15:11:45.180" Title="Smashing Seven Segments" Tags="|pattern|visual|jigsaw-puzzle|seven-segment|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17857" PostTypeId="1" AcceptedAnswerId="17860" CreationDate="2015-07-13T04:15:19.870" Score="1" ViewCount="1857" Body="&lt;p&gt;When you rearrange the letters in the word &quot;NO&quot; in all possible ways you have&#xA;NO&#xA;ON&#xA;These are both English dictionary words.&#xA;Are there any 3 letter words where all 6 of the 3 letter permutations of its letters are also all words?&#xA;For example, TEA  rearranges to&#xA;TEA&#xA;TAE&#xA;ETA&#xA;EAT&#xA;ATE&#xA;AET.&#xA;But, of course, not all of these are dictionary words!&lt;/p&gt;&#xA;" OwnerUserId="14030" LastActivityDate="2015-07-19T00:04:10.693" Title="A 3 letter word whose permutations are words" Tags="|wordplay|letter-sequence|anagram|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17858" PostTypeId="2" ParentId="17617" CreationDate="2015-07-13T05:15:06.633" Score="1" Body="&lt;p&gt;The morse code is within the first HTML comment in the markdown (not sure if this was added after Leppy's answer).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are three different white-space characters. A dash is represented by &amp;amp;#9. A dot is represented by &amp;amp;#32. One &amp;amp;#8199 is a break between letters and two &amp;amp;#8199's are a space between words.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The morse code translates to:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;THE WISDOM YOU SEEK WAS YOURS ALL ALONG BUT ONLY IF YOU CAN EXPLAIN YOUR PATH HERE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastActivityDate="2015-07-13T05:15:06.633" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17859" PostTypeId="2" ParentId="17857" CreationDate="2015-07-13T05:20:16.327" Score="1" Body="&lt;p&gt;Maybe &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ZZZ  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I really should have gotten some sleep yesterday&lt;/p&gt;&#xA;" OwnerUserId="13080" LastEditorUserId="13080" LastEditDate="2015-07-13T08:34:30.313" LastActivityDate="2015-07-13T08:34:30.313" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17860" PostTypeId="2" ParentId="17857" CreationDate="2015-07-13T05:47:34.707" Score="4" Body="&lt;p&gt;I wrote a Python script which checks for the existence of these words - I couldn't find any.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import itertools&#xA;words = []&#xA;&#xA;with open('dictionary.txt') as f:&#xA;    for line in f:&#xA;        if len(line.strip()) == 3:&#xA;            words += [line.strip().lower()]&#xA;&#xA;def winner(word):&#xA;    global words&#xA;    with open('words.txt') as file:&#xA;        for combo in itertools.permutations(word,len(word)):&#xA;            if not ''.join(combo) in words:&#xA;                return False&#xA;    return True&#xA;&#xA;for word in words:&#xA;    if(winner(word)):&#xA;        print(word)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If someone wants to try with a bigger dictionary than mine, go ahead!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Based upon the Scrabble dictionary VictorHenry has provided, there are four such sets of words:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;eew,ewe,wee&lt;br&gt;hoo,oho,ooh&lt;br&gt;mmm&lt;br&gt;zzz&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, the validity of these 'words' is left for the reader to decide :)&lt;/p&gt;&#xA;" OwnerUserId="7308" LastEditorUserId="7308" LastEditDate="2015-07-13T06:09:06.290" LastActivityDate="2015-07-13T06:09:06.290" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17861" PostTypeId="2" ParentId="17857" CreationDate="2015-07-13T05:52:08.827" Score="3" Body="&lt;p&gt;Apparently the word &quot;EEW&quot; was just added to &lt;a href=&quot;http://scrabblepodcast.blogspot.com/2014/08/new-3-letter-words-in-ospd5.html&quot; rel=&quot;nofollow&quot;&gt;the fifth edition of the OSPD&lt;/a&gt; which is the Official Scrabble Players Dictionary. It is, as you would expect, an expression of disgust (&quot;eew gross!&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, EEW can be anagrammed to WEE and EWE, both of which are also words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ZZZ (already suggested by valuable_asset) is also an acceptable Scrabble word in that edition of the OSPD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was going to write a Python program, but Tryth has already done so. That being said, Tryth, you might want to use &lt;a href=&quot;http://artchester.net/wp-content/uploads/2015/05/5th-Edition-3-letter-words-rev6.pdf&quot; rel=&quot;nofollow&quot;&gt;this list of 3-letter Scrabble words&lt;/a&gt; for your program and see if you have any better luck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Apparently HOO, OHO, and OOH are all words in that dictionary as well.&lt;/p&gt;&#xA;" OwnerUserId="10861" LastActivityDate="2015-07-13T05:52:08.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17862" PostTypeId="1" AcceptedAnswerId="17865" CreationDate="2015-07-13T06:25:27.017" Score="218" ViewCount="14328" Body="&lt;p&gt;I was walking home late last night and heard raised voices coming from a boarded up old warehouse. I stopped, out of sight, to listen...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;[When I started eavesdropping, the conversation was already under way...]&lt;/em&gt; &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Gangster 1&lt;/strong&gt;: -infiltrator in our midst, and I know who it is... &lt;em&gt;[dramatic pause]&lt;/em&gt; ...It's you isn't it!&lt;br&gt;&#xA;  &lt;strong&gt;Gangster 2&lt;/strong&gt;: Me?&lt;br&gt;&#xA;  &lt;strong&gt;G1&lt;/strong&gt;: What? No... him.&lt;br&gt;&#xA;  &lt;strong&gt;Gangster 3&lt;/strong&gt;: That's ridiculous! I'm no traitor! How do we know it's not you?!&lt;br&gt;&#xA;  &lt;strong&gt;G2&lt;/strong&gt;: It's not me!&lt;br&gt;&#xA;  &lt;strong&gt;G3&lt;/strong&gt;: No... him... &lt;em&gt;[under breath]&lt;/em&gt; ...idiot.&lt;br&gt;&#xA;  &lt;strong&gt;G1&lt;/strong&gt;: That doesn't even make sense. I'm the most prolific amongst us. When would I even have the chance to work for the opposition?&lt;br&gt;&#xA;  &lt;strong&gt;G3&lt;/strong&gt;: What do you mean? I'm in every day!&lt;br&gt;&#xA;  &lt;strong&gt;Gangster 4&lt;/strong&gt;: &lt;em&gt;[to himself]&lt;/em&gt; Me too, but you don't hear me harping on...&lt;br&gt;&#xA;  &lt;strong&gt;G3&lt;/strong&gt;: &lt;em&gt;[continuing]&lt;/em&gt; Not to mention I saw you with your ex last week on my way out of here... And I won't mention who I saw him hanging out with &lt;em&gt;again&lt;/em&gt; yesterday...&lt;br&gt;&#xA;  &lt;strong&gt;G2&lt;/strong&gt;: We weren't hanging out! I was just stuck behind him in a queue!&lt;br&gt;&#xA;  &lt;strong&gt;Gangster 5&lt;/strong&gt;: Hey, we all fraternize with the enemy now and then... It's necessary to meet our goals... But this is different. From what I hear, half the time you act like you &lt;em&gt;are&lt;/em&gt; one of them!  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;[A street sweeper drove slowly by, so I missed a little of the conversation...]&lt;/em&gt;  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;G3&lt;/strong&gt;: It's a set up! Think about it... Who would gain the most from seeing me dead? ...I would!&lt;br&gt;&#xA;  &lt;strong&gt;Gangster 6&lt;/strong&gt;: 'strue, but 'e would too...&lt;br&gt;&#xA;  &lt;strong&gt;G1&lt;/strong&gt;: Sure, he occasionally fills in for me, but he could practically replace you!&lt;br&gt;&#xA;  &lt;strong&gt;G2&lt;/strong&gt;: No he couldn't!&lt;br&gt;&#xA;  &lt;strong&gt;G6/G1/G3&lt;/strong&gt;: Would you shut up!&lt;br&gt;&#xA;  &lt;strong&gt;G4&lt;/strong&gt;: Geez settle down guys, what's with all the yelling?&lt;br&gt;&#xA;  &lt;strong&gt;G1&lt;/strong&gt;: Because no one was talking to you!&lt;br&gt;&#xA;  &lt;strong&gt;G4&lt;/strong&gt;: Ahh... Gotcha...  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;[A crowd of drunken revellers stumbled by and as I slipped into an alley to remain unseen, I missed another few minutes of the argument...]&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;G1&lt;/strong&gt;: Gimme one good reason why I shouldn't kill you right now!&lt;br&gt;&#xA;  &lt;strong&gt;G2&lt;/strong&gt;: I didn't do anything wrong!&lt;br&gt;&#xA;  &lt;strong&gt;G3&lt;/strong&gt;: Wait... Who are we talking about now?&lt;br&gt;&#xA;  &lt;strong&gt;G1&lt;/strong&gt;: You!&lt;br&gt;&#xA;  &lt;strong&gt;G2&lt;/strong&gt;: Oh...&lt;br&gt;&#xA;  &lt;strong&gt;G5&lt;/strong&gt;: Why?&lt;br&gt;&#xA;  &lt;strong&gt;G1&lt;/strong&gt;: Yes!  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;[From here things escalated rapidly, and when gunshots were fired, I made a run for it...]&lt;/em&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I went to the police but when they visited the scene, the gang was gone, and everything had already been cleaned. They said there wasn't enough to go on unless I could identify the victim and the perpetrators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who was the gang?&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;Which member was killed?&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;Why were they killed?&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus&lt;/strong&gt;: What was the name of each gangster?&lt;/p&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="5208" LastEditDate="2015-07-13T06:31:00.813" LastActivityDate="2015-07-13T07:25:16.780" Title="An overheard gang meeting, a double agent, and a murder" Tags="|enigmatic-puzzle|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17863" PostTypeId="1" AcceptedAnswerId="17880" CreationDate="2015-07-13T06:52:23.890" Score="5" ViewCount="248" Body="&lt;p&gt;Inspired by &lt;a href=&quot;https://puzzling.stackexchange.com/q/17837/68&quot;&gt;this challenge&lt;/a&gt;, I have a similar, with stricter rules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find a word that can be decomposed into several other words (more than one). Those words are not allowed to be proper substrings of the starting word. Entries are scored by the length of the shortest word.&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-07-13T17:58:35.467" Title="Anagrammatical substrings" Tags="|wordplay|open-ended|" AnswerCount="2" CommentCount="2" ClosedDate="2021-06-16T18:07:19.263" ContentLicense="CC BY-SA 3.0" />
  <row Id="17864" PostTypeId="2" ParentId="17863" CreationDate="2015-07-13T07:04:28.127" Score="6" Body="&lt;p&gt;Despite the stricter rules, my entry there would be just as valid here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my word (again) with a score of 5.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;TrIeNnIaLlY. &lt;br&gt;&lt;br&gt;Made up of the words TINILY and renal with alternating letters. And neither word is a proper substring of the original.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inspired from &lt;a href=&quot;https://puzzling.stackexchange.com/a/17846/12700&quot;&gt;pacoverflow's answer to the original challenge&lt;/a&gt;, here's a word of score 11 with a small tweak of mine.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; REinstitutionALIZATIONs&lt;br&gt;&lt;br&gt;Made up of the words REALIZATION and institutions, with simply the pluralizing 's' at the end for the word institutions, thereby making this fit the new tougher rules.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-07-13T09:19:35.647" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17865" PostTypeId="2" ParentId="17862" CreationDate="2015-07-13T07:25:16.780" Score="161" Body="&lt;p&gt;The gang was the notorious&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Vowel Syndicate!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The victim was&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Y&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;he was a double agent for the Vowel Syndicate's chief rivals, the Consonant Guild!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The gangsters are, I believe:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1: e (&quot;most prolific&quot;)&lt;br&gt;&#xA;2: u (every time someone says &quot;you&quot; he thinks they're talking to him, also was stuck behind &quot;Q&quot;)&lt;br&gt;&#xA;3: y (is in every &quot;day&quot;, is the one that seems to be the accused)&lt;br&gt;&#xA;4: a (is also in every &quot;day&quot;)&lt;br&gt;&#xA;5: o (responds to gangster 2 saying &quot;oh&quot; near the end)&lt;br&gt;&#xA;6: i (responds to gangster 3 saying &quot;I&quot; would gain the most from seeing &quot;Y&quot; dead)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10861" LastActivityDate="2015-07-13T07:25:16.780" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17866" PostTypeId="2" ParentId="17856" CreationDate="2015-07-13T08:32:58.277" Score="21" Body="&lt;p&gt;There are three valid patterns:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;EPrFA hq337 A7-L8 CP2ny b9-ed&lt;/code&gt; (&lt;code&gt;-&lt;/code&gt;, &lt;code&gt;3&lt;/code&gt;, &lt;code&gt;7&lt;/code&gt;, &lt;code&gt;A&lt;/code&gt; and &lt;code&gt;P&lt;/code&gt; appear twice)&lt;br&gt;&#xA; &lt;code&gt;P2rFA Pnq37 ECELA Lhy-8 b9-ed&lt;/code&gt; (&lt;code&gt;-&lt;/code&gt;, &lt;code&gt;A&lt;/code&gt;, &lt;code&gt;E&lt;/code&gt;, &lt;code&gt;L&lt;/code&gt; and &lt;code&gt;P&lt;/code&gt; appear twice)&lt;br&gt;&#xA; &lt;code&gt;P2rFA Pnq37 FCELA Chy-8 b9-ed&lt;/code&gt; (&lt;code&gt;-&lt;/code&gt;, &lt;code&gt;A&lt;/code&gt;, &lt;code&gt;C&lt;/code&gt;, &lt;code&gt;F&lt;/code&gt; and &lt;code&gt;P&lt;/code&gt; appear twice)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or, as graphics:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/jKqJf.png&quot; alt=&quot;EPrFA hq337 A7-L8 CP2ny b9-ed&quot;&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/GJnhm.png&quot; alt=&quot;P2rFA Pnq37 ECELA Lhy-8 b9-ed&quot;&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/oPqXn.png&quot; alt=&quot;P2rFA Pnq37 FCELA Chy-8 b9-ed&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9860" LastActivityDate="2015-07-13T08:32:58.277" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17870" PostTypeId="1" CreationDate="2015-07-13T15:05:42.613" Score="8" ViewCount="1137" Body="&lt;p&gt;&lt;strong&gt;Intro:&lt;/strong&gt;&lt;br&gt;&#xA;I found this strange, tattered journal while walking through the woods behind my house, and wondered if you could make sense of it all. From what I could glean, it was written by a nomadic tribe, but none of &lt;em&gt;this&lt;/em&gt; is really given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Notes:&lt;/strong&gt;&lt;br&gt;&#xA;I have managed to find a psychic of rather ill-repute, who turned out to be outstanding at her job, and managed to get in contact with my crazy uncle, who had been working on deciphering the text before me. Here is the revised edition of the text. Now, to head home and take care of that wolf bite. My fur's itching like crazy...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Riddle:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Searching, squaring, measuring, squeezing,&#xA;  &lt;br&gt;it's a measurement.&#xA;  &lt;br&gt;Not the worst, I end up choosing,&#xA;  &lt;br&gt;lemmings know the rest.&#xA;  &lt;br&gt;&lt;br&gt;It saddens my heart to relay this next part:&#xA;  &lt;br&gt;Swear not in vain, or swallow the pain.&#xA;  &lt;br&gt;With his head now gone, the ill-begotten son's song,&#xA;  &lt;br&gt;ended with keen notes of grieving.&#xA;  &lt;br&gt;&lt;br&gt;Three's fourteen, the next sign read,&#xA;  &lt;br&gt;and most of us waved goodbye.&#xA;  &lt;br&gt;The rest of us spoke out our puzzlement,&#xA;  &lt;br&gt;and headed off with a sigh.&#xA;  &lt;br&gt;&lt;br&gt;Be forewarned, your journey be done,&#xA;  &lt;br&gt;once you've followed our story.&#xA;  &lt;br&gt;Our strange sense of pride, and&#xA;  &lt;br&gt;the cause of our tribe's&#xA;  &lt;br&gt;pilgrimage to seas unknown.&#xA;  &lt;br&gt;So, if you be brave,&#xA;  &lt;br&gt;Take heed, my friend,&#xA;  &lt;br&gt;and learn this lesson from our passage...&#xA;  &lt;br&gt;&lt;br&gt;Make right from our wrongs, and carry our songs,&#xA;  &lt;br&gt;to the pages of some dusty story/novel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;History:&lt;/strong&gt;&lt;br&gt;&#xA;My Uncle was secretly following the stories of a small indigenous group of people that inhabited a remote corner of the earth some 2000 years ago, trying to figure out their entire history after stumbling upon some dusty old journal at a local bazaar. He always thought of it as merely a way to keep time and wile away boredom, but began to wonder whether the stories were true, how this tribe had lived, and where they had gone. He spent nearly every waking moment of his last 10 years trying to search for some final clue, when he stumbled upon this inscription carved in a deep cavern by long-forgotten hands, written in some strange dialect. However, he could never finish the decipher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Directions:&lt;/strong&gt;&lt;br&gt;&#xA;Please, puzzlers, can you solve this puzzle, and finally let my Uncle find the rest he deserves? You are my last hope, for I fear I am growing mad myself (see my previous renditions of this story to find out how).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;IMPORTANT&lt;/strong&gt; &lt;strong&gt;NOTES:&lt;/strong&gt;&#xA;&lt;br&gt;Several people are of the mistaken impression that the answer has to do with math.&lt;br&gt;&#xA;Also, &lt;strong&gt;partial answers are heavily encouraged.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Hint 1:&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Stanza 1&lt;/strong&gt;&lt;br&gt;&#xA;&lt;em&gt;I found some hidden edits made with a tincture of foxbane and the light of a  blue moon while perusing the riddle once more:&lt;/em&gt; the words &lt;strong&gt;the rest&lt;/strong&gt; are highlighted, and it appears as if a &lt;strong&gt;dot&lt;/strong&gt; was made above &lt;strong&gt;the comma in line 3&lt;/strong&gt;. It also appears as if &lt;strong&gt;line 2&lt;/strong&gt; was scribbled in at a later date...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13697" LastEditorUserId="13697" LastEditDate="2016-02-26T23:24:11.343" LastActivityDate="2016-02-26T23:24:11.343" Title="Puzzling puzzling phrases" Tags="|riddle|enigmatic-puzzle|story|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17871" PostTypeId="2" ParentId="17837" CreationDate="2015-07-13T15:13:00.267" Score="2" Body="&lt;h1&gt;Score 8&lt;/h1&gt;&#xA;&lt;p&gt;The highest-scoring word in the Wolfram dictionary is:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;interrelatedness&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Which consists of:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;INTERRE&lt;/strong&gt;late&lt;strong&gt;D&lt;/strong&gt;ness&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;interre&lt;strong&gt;LATE&lt;/strong&gt;d&lt;strong&gt;NESS&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Found using:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;n = 8;&#xA;wl = ToLowerCase@DictionaryLookup[Repeated[_, {n}]];&#xA;wl2 = ToLowerCase@DictionaryLookup[Repeated[_, {2 n}]];&#xA;subwords = &#xA;  Cases[Table[&#xA;    w -&amp;gt; Select[wl, &#xA;      LongestCommonSequence[w, #] == # &amp;amp;&amp;amp; &#xA;        LongestCommonSubsequence[w, #] != # &amp;amp;], {w, wl2}], &#xA;   Pattern[p, _ -&amp;gt; {_, __}]];&#xA;Cases[subwords, (w_ -&amp;gt; {___, a_, ___, b_, ___}) /; (Sort[&#xA;      Characters[w]] == Sort[Characters[a]~Join~Characters[b]]) :&amp;gt; {w,&#xA;    a, b}]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-13T17:58:14.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17872" PostTypeId="1" AcceptedAnswerId="17876" CreationDate="2015-07-13T15:22:22.363" Score="14" ViewCount="1931" Body="&lt;p&gt;Who or what is this poem referring to?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I come from a cloud.&lt;br&gt;&#xA;  Some long streaks of light&lt;br&gt;&#xA;  And a glowing shroud.&lt;br&gt;&#xA;  But now out of sight.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Seen in the summer,&lt;br&gt;&#xA;  Despite my nickname.&lt;br&gt;&#xA;  Many deaths occurred,&lt;br&gt;&#xA;  But I'm not to blame.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="5373" LastEditDate="2015-07-14T18:23:24.267" LastActivityDate="2015-07-14T18:23:24.267" Title="Riddle Poem: I come from a cloud" Tags="|riddle|rhyme|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17873" PostTypeId="2" ParentId="17872" CreationDate="2015-07-13T15:40:56.217" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Rainbows.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come from a cloud.&#xA;Some long streaks of light&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Streaks of light through water vapor causes the rainbow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And a glowing shroud.&#xA;But now out of sight.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When a shroud of some sort causes shadows, the rainbow goes away.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Seen in the summer,&#xA;Despite my nickname.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Summer is usually sunny, this only happens when it rains(?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Many deaths occurred,&#xA;But I'm not to blame.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The book Genesis in the Bible. The rainbow didn't cause the deaths.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12365" LastEditorUserId="12365" LastEditDate="2015-07-13T15:42:41.780" LastActivityDate="2015-07-13T15:42:41.780" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17874" PostTypeId="2" ParentId="17872" CreationDate="2015-07-13T16:06:43.823" Score="1" Body="&lt;p&gt;Could it be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come from a cloud.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clouds are made of water which will fall down as rain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some long streaks of light&#xA;And a glowing shroud.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clouds (the shroud) can glow, and sometimes when they have holes, you see streaks of light come from them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But now out of sight.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As rain in the sky, it is not easy to see in comparison to clouds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Seen in the summer,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It can rain quite frequently in summer (depending on your location)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Despite my nickname.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; April showers can be used to describe rain (which is not in the summer)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Many deaths occurred,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many deaths can occur from floods (or excess rain), or the 'reign' of a bad ruler.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I'm not to blame.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I guess we just have to blame mother nature!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-07-13T17:18:45.790" LastActivityDate="2015-07-13T17:18:45.790" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17875" PostTypeId="1" AcceptedAnswerId="17883" CreationDate="2015-07-13T16:12:53.637" Score="8" ViewCount="219" Body="&lt;p&gt;In the land of the tetrominos, the twins L.&amp;nbsp;and&amp;nbsp;J. have set out to revolutionize travel by rail.  Contemporaries' passenger cars are boxy and boring, and humdrum symmetric passenger arrangements are all too common when the car gets full.  &quot;Not on our lines!&quot; vow L.&amp;nbsp;&amp;amp;&amp;nbsp;J.  No, when travelers completely fill the polymino interior of an L.&amp;nbsp;&amp;amp;&amp;nbsp;J. railcar, the entrepreneurs promise that both reflection-symmetric and rotation-symmetric seating will be &lt;em&gt;impossible by design&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, if they can only get one of their interiors approved by the railcar authorities, I., T., and O.  Prior submissions have fallen afoul of one or more of the safety regulations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Polymino cells must be axis-aligned.  (By convention, the $\pm x$&amp;nbsp;directions are the car's directions of travel.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The car should be balanced on the tracks; the interior must be symmetric about the $x$&amp;nbsp;axis.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The car should travel equally well in both directions; the interior must be symmetric about the $y$&amp;nbsp;axis.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Besides all that, L.&amp;nbsp;and&amp;nbsp;J., being good businessminos, don't like wasting space; it should be possible to fill their cars to the brim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can L.&amp;nbsp;and&amp;nbsp;J. build the railcar of their dreams?  Bonus: What is the least capacity such a car could have?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Since this is an aha-moment sort of puzzle, please &lt;a href=&quot;https://puzzling.stackexchange.com/editing-help#spoilers&quot;&gt;spoiler&lt;/a&gt; your answers.)&lt;/p&gt;&#xA;" OwnerUserId="13175" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-07-13T20:51:58.517" Title="A Railcar for L. &amp; J. Railroad" Tags="|story|packing|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17876" PostTypeId="2" ParentId="17872" CreationDate="2015-07-13T16:36:05.587" Score="15" Body="&lt;p&gt;I think it's:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Halley%27s_Comet&quot;&gt;Halley's comet&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come from a cloud:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The article mentions that Halley's comet could have originated from the Oort Cloud if it was a long-period comet at once.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some long streaks of light and a glowing shroud&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is the contrails and &quot;shield&quot; that develop in front and behind, respectively, of the comet itself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But now out of sight.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Halley's comet passed at a relative speed of 70.56 km/s when it passed in 1910.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Seen in the summer, despite my nickname:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; thanks to Geobit's advice, comets are normally referred to as &lt;a href=&quot;http://starryskies.com/solar_system/Comet/dirty_snowballs.html&quot;&gt;&quot;dirty snowballs&quot;&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Many deaths occurred, but I'm not to blame.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.gmanetwork.com/news/story/340724/scitech/science/halley-s-comet-really-was-a-harbinger-of-death-scientists-suggest&quot;&gt;this site&lt;/a&gt; claims that Halley's comet was the cause of the Middle Ages plague, which killed many. Also, comets were seen as heralds of death in those times, as a simple Google search will show.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note from pacoverflow (since comments sometimes disappear):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Halley's comet is an acceptable answer.  I was thinking about &lt;a href=&quot;https://en.wikipedia.org/wiki/Comet_Hale%E2%80%93Bopp&quot;&gt;Comet Hale-Bopp&lt;/a&gt; and the &lt;a href=&quot;https://en.wikipedia.org/wiki/Heaven%27s_Gate_%28religious_group%29&quot;&gt;Heaven's Gate&lt;/a&gt; cult.  Hale-Bopp is interesting because it has &lt;a href=&quot;http://sirius.bu.edu/planetary/cometc.gif&quot;&gt;three long streaks of light&lt;/a&gt; (tails) instead of the usual two that comets have.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13697" LastEditorUserId="4450" LastEditDate="2015-07-13T19:59:00.120" LastActivityDate="2015-07-13T19:59:00.120" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="17877" PostTypeId="2" ParentId="17872" CreationDate="2015-07-13T16:49:48.633" Score="3" Body="&lt;p&gt;Very similar to Nyk 232, but a little more general:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A meteor &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come from a cloud.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Often originate in a cloud of space debris &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some long streaks of light  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Meteors look like streaks of light as they burn up in the atmosphere.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And a glowing shroud.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, light as they fall.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But now out of sight.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some shoot beyond the horizon, but others just burn up and disappear.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Seen in the summer,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Often seen in summer months when there are fewer clouds.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Despite my nickname.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nickname of &quot;&lt;strong&gt;Fall&lt;/strong&gt;ing Star&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Many deaths occurred,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people claim to have been killed by meteors  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I'm not to blame. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If it actually gets to the ground, then it is a meteorite, rather than a meteor.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-07-13T16:49:48.633" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17878" PostTypeId="1" AcceptedAnswerId="17902" CreationDate="2015-07-13T17:16:15.720" Score="19" ViewCount="685" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Is my first part a bargain?  Be careful!  Not quite.&lt;br&gt;&#xA;  My next, when it's scarlet, means something's not right.&lt;br&gt;&#xA;  My third: to police, an invaluable tool&lt;br&gt;&#xA;  Can my last bit be neutral?  It can't, as a rule.&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;If you gather my pieces together, you'll get&lt;br&gt;&#xA;  A massive and torrid affair, you can bet.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="4421" LastEditDate="2015-07-14T18:47:37.197" LastActivityDate="2015-07-14T18:47:37.197" Title="A Massive and Torrid Affair" Tags="|riddle|word|rhyme|wordplay|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17879" PostTypeId="1" AcceptedAnswerId="17894" CreationDate="2015-07-13T17:46:16.537" Score="39" ViewCount="3874" Body="&lt;p&gt;If a word has a certain property, I call it a Versatile Word™.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In each of the short, unrelated sentences below, only one of the choices is a Versatile Word™.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sentences are for context only, and do not affect whether a word is a Versatile Word™.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;She ate a ( &lt;strong&gt;peach / pear&lt;/strong&gt; ).&lt;/li&gt;&#xA;  &lt;li&gt;He ( &lt;strong&gt;boats / sails&lt;/strong&gt; ) in the summer.&lt;/li&gt;&#xA;  &lt;li&gt;She ( &lt;strong&gt;circles / loops&lt;/strong&gt; ) around the building.&lt;/li&gt;&#xA;  &lt;li&gt;He ( &lt;strong&gt;fires / shoots&lt;/strong&gt; ) his weapon.&lt;/li&gt;&#xA;  &lt;li&gt;He ( &lt;strong&gt;aids / helps&lt;/strong&gt; ) his sister.&lt;/li&gt;&#xA;  &lt;li&gt;They wanted to go ( &lt;strong&gt;also / too&lt;/strong&gt; ).&lt;/li&gt;&#xA;  &lt;li&gt;This air always contains ( &lt;strong&gt;dust / ions&lt;/strong&gt; ).&lt;/li&gt;&#xA;  &lt;li&gt;The best ( &lt;strong&gt;coffee / tea&lt;/strong&gt; ) comes from India.&lt;/li&gt;&#xA;  &lt;li&gt;The vet will ( &lt;strong&gt;neuter / spay&lt;/strong&gt; ) the cat.&lt;/li&gt;&#xA;  &lt;li&gt;The gang will ( &lt;strong&gt;loot / raid&lt;/strong&gt; ) the store.&lt;/li&gt;&#xA;  &lt;li&gt;She said ( &lt;strong&gt;bye / ciao&lt;/strong&gt; ) and left.&lt;/li&gt;&#xA;  &lt;li&gt;Only the ( &lt;strong&gt;ash / soot&lt;/strong&gt; ) remained.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What makes a word a Versatile Word™?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorDisplayName="user14478" LastEditDate="2016-09-23T21:29:55.207" LastActivityDate="2016-09-23T21:29:55.207" Title="What is a Versatile Word™?" Tags="|pattern|english|word-property|" AnswerCount="5" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17880" PostTypeId="2" ParentId="17863" CreationDate="2015-07-13T17:58:35.467" Score="2" Body="&lt;h1&gt;Score 8&lt;/h1&gt;&#xA;&lt;p&gt;The highest-scoring word in the Wolfram dictionary is:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;interrelatedness&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Which consists of:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;INTERRE&lt;/strong&gt;late&lt;strong&gt;D&lt;/strong&gt;ness&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;interre&lt;strong&gt;LATE&lt;/strong&gt;d&lt;strong&gt;NESS&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Found using:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;n = 8;&#xA;wl = ToLowerCase@DictionaryLookup[Repeated[_, {n}]];&#xA;wl2 = ToLowerCase@DictionaryLookup[Repeated[_, {2 n}]];&#xA;subwords = &#xA;  Cases[Table[&#xA;    w -&amp;gt; Select[wl, &#xA;      LongestCommonSequence[w, #] == # &amp;amp;&amp;amp; &#xA;        LongestCommonSubsequence[w, #] != # &amp;amp;], {w, wl2}], &#xA;   Pattern[p, _ -&amp;gt; {_, __}]];&#xA;Cases[subwords, (w_ -&amp;gt; {___, a_, ___, b_, ___}) /; (Sort[&#xA;      Characters[w]] == Sort[Characters[a]~Join~Characters[b]]) :&amp;gt; {w,&#xA;    a, b}]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-13T17:58:35.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17881" PostTypeId="2" ParentId="17879" CreationDate="2015-07-13T17:59:24.660" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Is a Versatile Word™ a word that has a homophone? For most of the pairs, only one word has a homophone:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    She ate a ( peach / pear ).&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; pear/pair&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    He ( boats / sails ) in the summer.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; sails/sales&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    She ( circles / loops ) around the building.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; loops/loups&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    He ( fires / shoots ) his weapon.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; shoots/chutes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    He ( aids / helps ) his sister.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; aids/aides&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    They wanted to go ( also / too ).&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; too/two&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    This air always contains ( dust / ions ).&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ions/irons (?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    The best ( coffees / teas ) come from India.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; teas/tease&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    The vet will ( neuter / spay ) the cat.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; neuter/nuder (thanks to pacoverflow)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    The gang will ( loot / raid ) the store.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; loot/lute&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    She said ( bye / ciao ) and left.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; bye/buy (thanks to Going hamateur) ciao/chow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    Only the ( ash / soot ) remained.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; soot/suit (?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4385" LastEditorUserId="4385" LastEditDate="2015-07-13T18:07:55.903" LastActivityDate="2015-07-13T18:07:55.903" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17882" PostTypeId="1" AcceptedAnswerId="17904" CreationDate="2015-07-13T18:02:00.930" Score="12" ViewCount="1411" Body="&lt;p&gt;&lt;strong&gt;STORY:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The land of Dictionaria was once a utopia where all words lived together in perfect harmony. Words respected each other and the big words always found a place among their smaller compatriots. The words worked together to weave letters into poetry and literature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But just as every cheery summer is followed by a chilly winter, the good days of Dictionaria seem to be nearing an end as the newly powerful invaders from the dark depths of Chatsville go about destroying the natural beauty of Dictionaria.&#xA;It started rather harmlessly when Laugh, Out and Loud found themselves marginalized as the petty thief Lol found his feet in Dictionaria. Brb came in and usurped the position long enjoyed by Be, Right and Back. There was almost no unease when these small time criminals began gaining recognition. But the worst was almost knocking on the door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When Yolo and Twerking came charging through the gates, little did the natural words know that everything in Dictionaria would never be the same again. The popularity of these thugs grew by and large, and slowly they began claiming to be natural residents of Dictionaria. Led by the notorious mob boss Swag, the Usurpers from Chatsville have now decided to wipe out the older words of Dictionaria and claim the entire land as their own. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Swag has entrusted the task of complete domination to his two trusted henchmen - Xoxo, the sweet tongued monster and Asap, the swift executioner. Swag has ordered Xoxo and Asap to eliminate all words in an alphabetical order, killing words starting with a particular letter every day. Aardvark will be one of the first to go and millions more will soon follow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the henchmen are ruthless but not so bright and some of the words of Dictionaria have realized that they can escape the massacre simply if they manage to pass off as another word on their predestined day of execution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These words are lucky to be born with such characteristics that they can contort their bodies to make their head appear where their foot is, and their foot appear where their head is. And once they escape their execution, they can live on because the henchmen will not remember which words have escaped their execution. These words hope to avoid death and one day drive out the Usurpers of Chatsville, but they need to know who their leader would be. As the natural words were respectful of those bigger than them, their leader would be the tallest one among them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who do you reckon will survive the massacre and lead the survivors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PUZZLE:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the longest word in the English language that can be morphed into another word simply by switching its first and last letter. The word cannot have the same first and last letter. And the word cannot be an abbreviation or contraction. The word should be found on &lt;a href=&quot;http://dictionary.reference.com&quot;&gt;Dictionary.com&lt;/a&gt; and words that are valid in Scrabble are not counted if they do not appear on Dictionary.com.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dictionary.com is the bible of the residents of Dictionaria (hence the inspired name) and all those who seek to nominate non-existing words as the leaders are distrustful pagans. Be gone infidels!&lt;/p&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-07-15T18:40:20.443" LastActivityDate="2015-08-22T16:46:52.633" Title="The Survivors of Dictionaria" Tags="|english|story|wordplay|" AnswerCount="6" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17883" PostTypeId="2" ParentId="17875" CreationDate="2015-07-13T18:03:39.593" Score="7" Body="&lt;p&gt;Is it okay if my solution has a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; hole in the middle?  It does seem to waste space a bit, but the space that is there can be filled.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If so,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/GkDGa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As per Edward's comment, let me try to prove that it can't be done with a smaller train car. (It seems like there might be a better way to answer this though)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  This is a bit of a cheat, but heading over to &lt;a href=&quot;https://en.wikipedia.org/wiki/Octomino&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Wikipedia page on octominoes&lt;/a&gt;, we see there are only five octominoes symmetric about the x and y axes and axis aligned.   Each of these can easily be symmetrically seated with two passengerominoes.  Two other cases to consider are a completely disconnected 2-passenger car, or a 1-passenger car, in which case the symmetry requirements on the cars force symmetry on the passengers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2722" LastEditorUserId="2722" LastEditDate="2015-07-13T20:51:58.517" LastActivityDate="2015-07-13T20:51:58.517" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17884" PostTypeId="2" ParentId="17879" CreationDate="2015-07-13T18:05:17.430" Score="6" Body="&lt;p&gt;My best guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; With one &quot;elementary letter operation&quot; (shifting one letter to somewhere else), you get a new word.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;em&gt;pear&lt;/em&gt; to &lt;em&gt;aper&lt;/em&gt;, &lt;strong&gt;a&lt;/strong&gt; shift to beginning &lt;br&gt;&#xA; &lt;em&gt;boats&lt;/em&gt; to &lt;em&gt;boast&lt;/em&gt;, &lt;strong&gt;t&lt;/strong&gt; shift &lt;br&gt;&#xA; &lt;em&gt;loops&lt;/em&gt; to &lt;em&gt;sloop&lt;/em&gt;, &lt;strong&gt;s&lt;/strong&gt; shift &lt;br&gt;&#xA; &lt;em&gt;fires&lt;/em&gt; to &lt;em&gt;fries&lt;/em&gt;, &lt;strong&gt;r&lt;/strong&gt; shift &lt;br&gt;&#xA; &lt;em&gt;aids&lt;/em&gt; to &lt;em&gt;said&lt;/em&gt;, &lt;strong&gt;s&lt;/strong&gt; shift &lt;br&gt;&#xA; &lt;em&gt;too&lt;/em&gt; to &lt;em&gt;oto&lt;/em&gt;, &lt;strong&gt;t&lt;/strong&gt; shift (dictionary.reference.com had it; dunno) &lt;br&gt;&#xA; &lt;em&gt;ions&lt;/em&gt; to &lt;em&gt;sion&lt;/em&gt;, &lt;strong&gt;s&lt;/strong&gt; shift &lt;br&gt;&#xA; &lt;em&gt;teas&lt;/em&gt; to &lt;em&gt;east&lt;/em&gt; or &lt;em&gt;eats&lt;/em&gt;, &lt;strong&gt;t&lt;/strong&gt; shift &lt;br&gt;&#xA; &lt;em&gt;spay&lt;/em&gt; to &lt;em&gt;pays&lt;/em&gt;, &lt;strong&gt;s&lt;/strong&gt; shift&lt;br&gt;&#xA; &lt;em&gt;raid&lt;/em&gt; to &lt;em&gt;arid&lt;/em&gt;, &lt;strong&gt;a&lt;/strong&gt; shift &lt;br&gt;&#xA; &lt;em&gt;bye&lt;/em&gt; to &lt;em&gt;bey&lt;/em&gt;, &lt;strong&gt;y&lt;/strong&gt; shift&lt;br&gt;&#xA; &lt;em&gt;ash&lt;/em&gt; to &lt;em&gt;has&lt;/em&gt;, &lt;strong&gt;h&lt;/strong&gt; shift&lt;br&gt;&#xA;&lt;br&gt;&#xA; Primary concerns:&lt;br&gt;&#xA; - &lt;em&gt;oto&lt;/em&gt; and &lt;em&gt;sion&lt;/em&gt; are on dictionary.reference.com, but questionable&lt;br&gt;&#xA; - &lt;em&gt;too&lt;/em&gt; is paired with &lt;em&gt;also&lt;/em&gt;, which, if you do a double-connected-letter shift, you get &lt;em&gt;soal&lt;/em&gt; which works.&lt;br&gt;&#xA; - For &lt;em&gt;sion&lt;/em&gt;, the other word &lt;em&gt;dust&lt;/em&gt; can go to &lt;em&gt;stud&lt;/em&gt;, but that is a very complex maneuver.&lt;br&gt;&#xA; Also, I didn't check thoroughly to see if there are other discrepancies.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Other anagrams off the top of my head for the other words (WA stands for the almighty Wolfram Alpha):&lt;br&gt;&#xA; - &lt;em&gt;cheap&lt;/em&gt;&lt;br&gt;&#xA; - WA says &lt;em&gt;sisal&lt;/em&gt;...&lt;br&gt;&#xA; - &lt;em&gt;clerics&lt;/em&gt;&lt;br&gt;&#xA; - &lt;em&gt;sooths&lt;/em&gt; (WA says it is a word, not sure without e...)&lt;br&gt;&#xA; - WA says &lt;em&gt;shlep&lt;/em&gt; (I like that word)&lt;br&gt;&#xA; - &lt;em&gt;soal&lt;/em&gt; v &lt;em&gt;oto&lt;/em&gt;&lt;br&gt;&#xA; - &lt;em&gt;sion&lt;/em&gt; v &lt;em&gt;stud&lt;/em&gt;&lt;br&gt;&#xA; - none for &lt;em&gt;coffees&lt;/em&gt;&lt;br&gt;&#xA; - &lt;em&gt;retune&lt;/em&gt;&lt;br&gt;&#xA; - &lt;em&gt;arid&lt;/em&gt; is more elegant a word than &lt;em&gt;flip&lt;/em&gt; actually...&lt;br&gt;&#xA; - WA says &lt;em&gt;icao&lt;/em&gt; is a word...&lt;br&gt;&#xA; - &lt;em&gt;otos&lt;/em&gt; I guess&lt;br&gt;&#xA;&lt;br&gt;&#xA; The only one which is a shift would be icao, which is an abbreviation. So I'm more comfortable with just a one-letter shift.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13916" LastEditorUserId="4421" LastEditDate="2015-07-13T19:02:01.380" LastActivityDate="2015-07-13T19:02:01.380" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17885" PostTypeId="2" ParentId="17882" CreationDate="2015-07-13T18:12:39.037" Score="12" Body="&lt;p&gt;Let's start things of with a word of length n=10:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DEFLOWERER / REFLOWERED&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And here's another with length n=12 (the words are in my regex dictionary, but one is not on Dictionary.com):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DEMONSTRATER / REMONSTRATED&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With a little bending of the rules, we can get up to length n=13:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DEMONOGRAPHER / REMONOGRAPHED&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These are all with the de.*er/re.*ed pattern. I'm hunting for longish examples of anything else.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So far the longest I can find in another pattern is length n=7:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HOBBISM / MOBBISH&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The search continues.&lt;/p&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="11475" LastEditDate="2015-07-13T20:46:53.787" LastActivityDate="2015-07-13T20:46:53.787" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17886" PostTypeId="1" CreationDate="2015-07-13T18:22:01.797" Score="12" ViewCount="3934" Body="&lt;p&gt;Every now and then, scrambling my Rubik's cube I find it in a state where every side has at least one tile in each of the six colors. For example here's a scramble I just did (obviously put through a solver after-the-fact to get an optimized algorithm):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;F2 U' L2 R2 B2 U R2 D U2 B2 U R2 F' R' B2 D' B' L' D2 B' F R2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This results in the following configuration (where U is white and F is green):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HnWyn.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are absolutely no other special properties about these scrambles that I can think of, other than the fact that every color appears at least once, but never more than three times, on each side. In particular, &lt;a href=&quot;https://puzzling.stackexchange.com/questions/429/is-it-possible-to-scramble-a-rubiks-cube-such-that-no-two-squares-of-the-same-c&quot;&gt;any number of tiles of the same color may be adjacent to one another&lt;/a&gt;, and it doesn't matter to me &lt;a href=&quot;https://puzzling.stackexchange.com/questions/5462/what-does-it-mean-for-a-rubiks-cube-to-be-perfectly-scrambled-and-how-do-you-r&quot;&gt;&quot;how&quot; randomized&lt;/a&gt; it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These cases don't seem particularly uncommon, as I stumble upon them once every few dozen solves. How many such permutations that are solvable exist?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, are there any legal permutations where the same color appears four times on the same side (leaving one for each of the remaining five colors), and why/why not? I'm surprised I haven't come across any yet — I figured I would have come across a few of them if they did in fact exist.&lt;/p&gt;&#xA;" OwnerUserId="14049" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.923" LastActivityDate="2015-10-19T21:44:21.240" Title="How many solvable permutations of the Rubik's cube exist where each color appears at least once on every side?" Tags="|rubiks-cube|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="17887" PostTypeId="2" ParentId="17882" CreationDate="2015-07-13T18:37:47.000" Score="10" Body="&lt;p&gt;To change from the pattern re/de&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6 letters:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Taoism / Maoist&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5359" LastActivityDate="2015-07-13T18:37:47.000" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17888" PostTypeId="2" ParentId="17870" CreationDate="2015-07-13T19:05:21.780" Score="3" Body="&lt;p&gt;They were trying to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Square the circle - &lt;a href=&quot;https://en.wikipedia.org/wiki/Squaring_the_circle&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Squaring_the_circle&lt;/a&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;References in the riddle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'Searching, squaring, measuring, squeezing' - The most common attempts to approximate $\pi$ have involved calculating increasingly precise upper and lower bounds.&lt;br&gt;&lt;br&gt;&#xA; 'with a swear yelled in vain, the ill-begotten's sheep was slain' - After many many failed attempts, it has since been proven that because $\pi$ is transcendental, it's precise value cannot be obtained using a straight-edge and compass.&lt;br&gt;&lt;br&gt;&#xA; 'Three's fourteen' - 3.14 is perhaps the most common numerical approximation of pi used in conversation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11078" LastEditorUserId="11078" LastEditDate="2015-07-13T19:59:05.063" LastActivityDate="2015-07-13T19:59:05.063" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17890" PostTypeId="1" AcceptedAnswerId="17919" CreationDate="2015-07-13T19:13:23.017" Score="7" ViewCount="790" Body="&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/2484/knight-knave-and-joker-puzzle&quot;&gt;Much like this other time&lt;/a&gt;, you are lost in town and come across a line of $j$ jokers and $k$ knights.  You don't know who is who, but they all know the identities of each other.  A knight always tells the truth to a yes-no question, while a joker can say either yes or no (and will say one of the two).   Your goal is to identify one knight by asking one yes-no question to each of the first $j$ people in the line. (Note the first $j$ people in the line is a unknown combination of knights and jokers.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you might guess, it is easier to identify a knight if there are more knights; so your task is easier for larger values of $k$.  For a given $j$, what is the smallest value of $k$ that allows you to identify a knight?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Make sure you don't ask a question that might be impossible for the person to answer.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;https://puzzling.stackexchange.com/questions/3087/t-knights-l-knaves-and-r-jokers&quot;&gt;There&lt;/a&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3118/t-knights-l-knaves-and-r-jokers-with-one-question-type-only&quot;&gt;are&lt;/a&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2588/faulty-computers/2774#2774&quot;&gt;lots&lt;/a&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/questions/593/knights-and-jokers/1666#1666&quot;&gt;of&lt;/a&gt; similar questions, but from what I can tell this is a new one.)&lt;/p&gt;&#xA;" OwnerUserId="2722" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2023-01-15T19:20:26.300" Title="Finding a knight with a few jokers thrown in" Tags="|mathematics|liars|" AnswerCount="4" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="17891" PostTypeId="2" ParentId="17879" CreationDate="2015-07-13T19:16:13.570" Score="6" Body="&lt;p&gt;I think it might be somewhat similar to a &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16743/what-is-a-triad-phrase%E2%84%A2/16749#16749&quot;&gt;Triad Phrase™&lt;/a&gt;. The pattern I've noticed is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Versatile Word™ can represent two more words, one starting with the first letter (and omitting the second) and one starting with the second letter (and omitting the first). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Applying the logic to the word list &lt;/p&gt;&#xA;&#xA;&lt;p&gt;She ate a ( peach / pear ).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pear - par, ear&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He ( boats / sails ) in the summer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Boats - bats, oats &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;She ( circles / loops ) around the building.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Loops - lops, oops&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He ( fires / shoots ) his weapon.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Shoots - soots, hoots&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He ( aids / helps ) his sister.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Aids - ads, ids &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They wanted to go ( also / too ).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Too - to, oo (?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This air always contains ( dust / ions ).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ions - ins, ons (?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The best ( coffees / teas ) come from India.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The vet will ( neuter / spay ) the cat.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Spay - say, pay&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The gang will ( loot / raid ) the store.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Raid - rid, aid&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;She said ( bye / ciao ) and left.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Bye - be, ye&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Only the ( ash / soot ) remained.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Ash - ah, sh (?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-07-13T19:16:13.570" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17892" PostTypeId="2" ParentId="17882" CreationDate="2015-07-13T19:39:28.157" Score="4" Body="&lt;p&gt;Here are some examples of &lt;strong&gt;14&lt;/strong&gt; length:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(not all are directly on Dictionary.com, but they are still words.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/p8fjM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-07-13T20:41:43.507" LastActivityDate="2015-07-13T20:41:43.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17893" PostTypeId="2" ParentId="17879" CreationDate="2015-07-13T19:54:14.967" Score="3" Body="&lt;p&gt;Based off of CodeNewbie's insight I have an idea.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Vowels are quite optional. Versatile words can drop any of their vowels and still be a word!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;She ate a ( peach / pear ).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pear - par, per&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He ( boats / sails ) in the summer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Boats - bats, bots &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;She ( circles / loops ) around the building.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Loops - lops, lops (Well... that makes sense)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He ( fires / shoots ) his weapon.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Shoots - shots, shots&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He ( aids / helps ) his sister.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Aids - ads, ids &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They wanted to go ( also / too ).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Too - to, to&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This air always contains ( dust / ions ).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ions - ins, ons&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The best ( coffees / teas ) come from India.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; teas - tes, tas (music and arabic letter?) (dictionary.reference)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The vet will ( neuter / spay ) the cat.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Spay - spa, spy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The gang will ( loot / raid ) the store.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Raid - rid, rad (or loot to lot, lot) not happy about this one. Rad is slang, but also unit of radiation dosage... so I think it is legit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;She said ( bye / ciao ) and left.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Bye - be, by&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Only the ( ash / soot ) remained.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Ash - sh, or soot to sot, sot :/&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I do not believe this is quite correct&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; mainly confusion at the repeat letter ones. So I believe 3 are sketchy answers. Might not be right but hope it helps someone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13916" LastEditorUserId="13916" LastEditDate="2015-07-13T19:59:55.287" LastActivityDate="2015-07-13T19:59:55.287" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17894" PostTypeId="2" ParentId="17879" CreationDate="2015-07-13T19:56:25.910" Score="33" Body="&lt;p&gt;A Versatile Word is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A word from which you can remove any one letter and still have a word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Examples:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1. She ate a ( peach / pear ).&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pear: pea, per, par, ear&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2. He ( boats / sails ) in the summer.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Boats: oats, bats, bots, boas, boat&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;3. She ( circles / loops ) around the building.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Loops: oops, lops, loos, loop&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;4. He ( fires / shoots ) his weapon.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Shoots: hoots, soots, shots, shoos, shoot&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;5. He ( aids / helps ) his sister.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Aids: ids, ads, ais, aid&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;6. They wanted to go ( also / too ).&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Too: to, oo&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;7. This air always contains ( dust / ions ).&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ions: ons, ins, ios, ion&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;8. The best ( coffee / tea ) comes from India.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tea: ea, ta, te&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;9. The vet will ( neuter / spay ) the cat.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Spay: pay, say, spy, spa&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;10. The gang will ( loot / raid ) the store.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Raid: aid, rid, rad, rai&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;11. She said ( bye / ciao ) and left.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bye: ye, be, by&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;12. Only the ( ash / soot ) remained.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ash: sh, ah, as&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="5456" LastEditDate="2015-07-16T20:09:13.180" LastActivityDate="2015-07-16T20:09:13.180" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17895" PostTypeId="1" AcceptedAnswerId="18368" CreationDate="2015-07-13T20:32:39.183" Score="8" ViewCount="1784" Body="&lt;p&gt;I have a class of numbers; this class of numbers contains a certain number if and only if it is &lt;em&gt;baumic&lt;/em&gt;. The following are baumic numbers:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;1, 6, 9, 15, 20, 21, 30, 33, 39, 45, 50, 51, 56, 57, 69, 70, 72, 75, 84, 87, 93&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;All other numbers under 100 are non-baumic numbers.&lt;/p&gt;&#xA;&lt;p&gt;In addition to 9 (1 nine), 999,999 (6 nines), 99,999,999 (8 nines), 9,999,999,999 (10 nines), and 99,999,999,999,999 (14 nines) are baumic numbers.&lt;/p&gt;&#xA;&lt;p&gt;What exactly is a baumic number?&lt;/p&gt;&#xA;&lt;p&gt;Feel free to ask me if &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; is a baumic number, and best of luck, puzzlers!&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;Notes:&lt;/h2&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$B(x)=y$&lt;/span&gt; is a baumic number.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$B(x)\in\mathbb{Z^*}$&lt;/span&gt; — All baumic numbers are non-negative integers.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;h1&gt;Hint 1&lt;/h1&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; It took me a while to come up with a name for this class; I settled on &lt;em&gt;baumic&lt;/em&gt; because, hey! German! Anyhow, here was the other likely candidate for the number class: &lt;em&gt;&lt;strong&gt;decrebescent&lt;/strong&gt;&lt;/em&gt;. Perhaps that helps...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h1&gt;Hint 2&lt;/h1&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;code&gt;42XRwBSfx0ib79FcgACIgACIgACIgACIgACIgoAXvACIgACIgACIgACIgACIgACIgARCIgAiCuACIgACIgACIgACIgACIgACIgACIgAiCuACIgACIgACIgACIgACIgACIgACIgoRgLgACIgACIgACIgACIgACIgACIgAiCc9CIgACIgACIgACIgACIgACIgAiCp9Fcg0nT9wTa9wTMR79VSQBSMfBHIqASbgACIRgACIgACIgoAXvACIgACIgACIgACIgoQafBHI95UP8kWP8EzeflEUgoCItBCIw8FcgoAXg8CIgAiC4BCIgAiC6kSfQtnYihGdh1GXg4WacBCcgQmbhBSfxsSa79FcRg0DPgk2XwBCZuFGI4BCfg0GLO9Fcs4iLuwSMfBHLw8FcoACbhJXZuV2RZg4WSKkgC1ACIzkgCc9CIJoQNxACIyACIKw1LgACIRKAzMgACI&lt;/code&gt;&lt;br&gt;What, you didn't think it would be &lt;em&gt;that&lt;/em&gt; easy, did you? BTW, there are multiple instances of one letter that don't belong, and its &lt;span class=&quot;math-container&quot;&gt;$\tiny{\in_2}\!\!\!\Huge{\color{purple}{purple}}!\textsf{   :D}$&lt;/span&gt;. You gotta do a bit of thinking.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;Hint 2.b&lt;/h2&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Flip Hint 2 &lt;span class=&quot;math-container&quot;&gt;$64+1$&lt;/span&gt; times.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;Perhaps a hint. Mostly observations.&lt;/h1&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; This class of numbers fascinates me. You guys are right when you say prime numbers are intrinsically related with the Baumic numbers. Truly, I conjectured some of the generalizations of Baumic numbers myself. I am not rightly sure as to whether or not my observations will help you guys, but I'm sure it might. &lt;br&gt;&lt;ol&gt;&lt;li&gt;&lt;strong&gt;Conjecture 1.&lt;/strong&gt; &lt;span class=&quot;math-container&quot;&gt;$\forall x&amp;gt;0[\neg B(2^x)]$&lt;/span&gt; (No power of two greater than one is a Baumic number.&lt;/li&gt;&lt;li&gt;&lt;strong&gt;Conjecture 2.&lt;/strong&gt; &lt;span class=&quot;math-container&quot;&gt;$\forall x&amp;gt;0\forall y\exists n[y=2^n\implies \neg B(y^x)]$&lt;/span&gt; (No power of a number that is itself a power of two is a Baumic number.)&lt;/li&gt;&lt;/ol&gt;&lt;br&gt;If no one gets it by the end of the week, I believe I shall post a hint that will effectively give the answer. Or perhaps the code that I use for Baumic numbers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7307" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-08-02T21:20:50.160" Title="Baumic Numbers!" Tags="|mathematics|calculation-puzzle|lateral-thinking|" AnswerCount="3" CommentCount="26" ContentLicense="CC BY-SA 3.0" />
  <row Id="17896" PostTypeId="2" ParentId="17890" CreationDate="2015-07-13T21:29:25.420" Score="0" Body="&lt;p&gt;Misread the question at first, so had to edit it later. I hope that's correct:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you ask someone &quot;Is it true that &lt;em&gt;only if&lt;/em&gt; you're lying/you're just going to lie, then [insert given bunch of guys] are all jokers?&quot;, you get an informative answer whether he's telling the truth or not. If he says yes, that means at least one of them must be a Knight. If he says no then they're all jokers. Since you can eliminate any group of people you want that way and are allowed to ask a limited amount of questions, it's more efficient to use binary search. Basically, $j &amp;gt;= log2(j+k), 2^j &amp;gt;= j+k$ and $2^j - j &amp;gt;= k$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="14016" LastEditDate="2015-07-14T15:40:04.177" LastActivityDate="2015-07-14T15:40:04.177" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17897" PostTypeId="2" ParentId="17890" CreationDate="2015-07-13T21:29:40.350" Score="6" Body="&lt;p&gt;OK, I think I've found the solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We need $k=2^j$ knights.&lt;br&gt;&#xA; For $j=1$, you ask whether the second person is a knight. If the answer is &quot;Yes&quot;, the second person is a knight, if the answer is &quot;no&quot;, you know the third person is a knight. If the first person is a knight, he says the truth, and thus the choice mentioned before is evidently right. But if the first person is a joker, both remaining persons must be knights, especially the selected one.&lt;br&gt;&#xA; For $j=2$, you ask person 1 if both person 3 and person 4 are knights, and person 2 if both person 3 and person 5 are knights. If both say &quot;yes&quot;, you chose person 3, if the first says &quot;yes&quot; and the second &quot;no&quot; you chose person &quot;4&quot;, if the first says &quot;no&quot; and the second says &quot;yes&quot; you choose person 5, and if both say &quot;no&quot;, you chose person 6.&#xA; For the general case, you ask each of the first $j$ persons whether a selection of $2^{j-1}$ of the remaining $2^j$ persons consists of all knights. Those sets are selected in a way that any two of the remaining $2^j$ persons are asked about to a different set of the first $j$ people (basically, you numerate them in binary, and select whom to ask about them based on which bits are $1$). Then based on the answers, you select that one where everyone where he was included in the set has answered &quot;yes&quot;, and everyone else has answered &quot;no&quot;.&lt;br&gt;&#xA; Example: $j=3$, $k=2^3=8$.&lt;br&gt;&#xA; Ask 1: &quot;Are 4,5,6,7 all knights?&quot;&lt;br&gt;&#xA; Ask 2: &quot;Are 4,5,8,9 all knights?&quot;&lt;br&gt;&#xA; Ask 3: &quot;Are 4,6,8,10 all knights?&quot;&lt;br&gt;&#xA; Answers:&lt;br&gt;&#xA; Yes, Yes, Yes $\implies$ 4 is knight&lt;br&gt;&#xA; Yes, Yes, No $\implies$ 5 is knight&lt;br&gt;&#xA; Yes, No, Yes $\implies$ 6 is knight&lt;br&gt;&#xA; Yes, No, No $\implies$ 7 is knight&lt;br&gt;&#xA; No, Yes, Yes $\implies$ 8 is knight&lt;br&gt;&#xA; No, Yes, No $\implies$ 9 is knight&lt;br&gt;&#xA; No, No, Yes $\implies$ 10 is knight&lt;br&gt;&#xA; No, No, No $\implies$ 11 is knight&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14010" LastEditorUserId="14010" LastEditDate="2015-07-13T21:38:34.457" LastActivityDate="2015-07-13T21:38:34.457" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="17898" PostTypeId="2" ParentId="3165" CreationDate="2015-07-13T22:49:03.667" Score="2" Body="&lt;p&gt;What a cool party game idea!! This is actually a game I made up for myself as a kid and programmed it into various programming languages as I grew up and now it's something my 12 year old son and I have been playing with each other for a few years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we know that all of the digits in the address are represented in the four guesses, it makes solving this pretty simple.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;because 0 swapped places in 2 guesses and the result of the guess was the same (1 digit correct and in the right pos), you can eliminate 0 from the answer.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;because 0 is eliminated it leaves 3 pairs of numbers where only one of the numbers is in the answer. (28_, 37_, and 3_4) because of this result you can eliminate 6 as a correct digit because of the guess using 3, 7, and 6 together and getting the (1 and 1) result we know that only the 3 or the 7 is correct so far.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;because 4 is the only digit left in the last pos (after our eliminations) it must be the correct digit. allowing us to eliminate 3 as a correct digit because of the guess using 3, 0, and 4 and getting (1 and 1) result and having already eliminated 0.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;once 3 is eliminated it leaves 7 as the only possible correct digit for the 2nd position, and therefore eliminating 8 from the correct answer which at the same time leaves 2 as the only possible digit in the first pos.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;274 is the only remaining guess and as my son would say &quot;Ding ding ding! you got 3 right and 3 in the right place!&quot;&lt;/p&gt;&#xA;" OwnerUserId="14057" LastActivityDate="2015-07-13T22:49:03.667" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17899" PostTypeId="2" ParentId="17882" CreationDate="2015-07-13T23:03:09.220" Score="4" Body="&lt;p&gt;Well, I found an 8-letter non-DE..ER word pair. They're not in dictionary.com, but they are in &lt;a href=&quot;http://www.global-language.com/century/&quot; rel=&quot;nofollow&quot;&gt;the Century Dictionary!&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MELONIST - One who cultivates melons.&lt;br&gt;&#xA;TELONISM - A pseudonym made by taking the last letters of the real name: as, N. N. = John Brown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am investigating other possibilities of the M..IST/T..ISM variety.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Apparently TANNERISM, despite not being in Dictionary.com, is actually a thing. Seems to be some sort of anti-Mormon belief system? Anyway, its pair is MANNERIST, which is in Dictionary.com, so now I'm up to 9 characters.&lt;/p&gt;&#xA;" OwnerUserId="10861" LastEditorUserId="10861" LastEditDate="2015-07-13T23:08:45.460" LastActivityDate="2015-07-13T23:08:45.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17900" PostTypeId="1" AcceptedAnswerId="17901" CreationDate="2015-07-14T00:13:55.060" Score="14" ViewCount="1061" Body="&lt;p&gt;&quot;So, what do you know about milking spiders?&quot;, asked Ernie as he passed a fresh cup of coffee to me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Good grief!&lt;/em&gt;, I thought and gave an involuntary shudder as I glanced at the milk jug sitting between our two cups &lt;em&gt;I didn't even know spiders had udders!!!&lt;/em&gt;. &quot;Think I'll have black coffee this morning&quot;, I replied. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ernie gave an amused snort. &quot;Not that kind of milking.&quot; he reassured me. &quot;I spent half of last week getting samples of fresh silk directly from Lucretia's spinarettes. Lucretia is a Vanuatuan Orb Web Spider - and it is claimed that, weight for weight, her web is 10 times stronger than steel!&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I looked over at the terrarium - Lucretia was as big as the palm of my hand appeared to be sulking. &quot;Non-venomous of course&quot; continued Ernie, &quot;but they can be a bit bitey when they are stressed&quot;. I noticed that Ernie's fingers and thumbs were dotted with numerous red wealts. &quot;But I can't complain - all great research requires a little sacrifice - and thanks to my analysis of Lucretia's silk I have managed to synthesize a material even better than nature can create - an artificial silk that is transparent as glass, 1000 times as strong as steel, chemically inert, and what's more - if you expose it to a special catalyst it sets as hard as diamond!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we had finished our coffee he asked me to help test his new silk applicator gun. &quot;When you press the red button it will squirt out a thread of silk - when you stop it will spray a little cloud of catalyst to rigidize it&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I squeezed the button, but instead of a thread, the device produced a small bubble. &quot;Oops&quot; said Ernie, &quot;I didn't expect that - must be air in the reservoir&quot;. We examined the bubble. It was about as big as an orange with a negligibly thin wall thickness, but to my surprise, when I picked it up it was as rigid as a crystal ball. Ernie thought for a moment then smiled. &quot;Actually, I think bubbles will be even more useful than rigid threads. I can make special flasks for an experiment I have planned.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ernie asked me to blow a second bubble, this time much bigger. &quot;It has to have a certain exact volume for the experiment&quot; he said, &quot;I'll tell you when to stop.&quot; I pressed the button again and this time held it down - a large bubble inflated from the applicator. When Ernie nodded his head I let go of the trigger, but the bubble continued to expand until it was bigger than a beach-ball. Ernie examined the gun. &quot;Oops - didn't expect that - needs a bit of lubricant on the trigger mechanism&quot;. He fiddled with the applicator then claimed it was fixed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now the volume of the sphere was much too large for Ernie's experiment. &lt;strong&gt;&quot;By my measurement it is exactly one meter in diameter&quot; he said, &quot;Much too big, but if we drill a small hole into it then we can blow another bubble inside - that will reduce the remaining volume in the sphere until it is exactly right. I'll tell you when to stop.&quot;  I followed his instructions, but this time, although I kept holding the button down, the interior bubble stopped growing too soon.&lt;/strong&gt; &quot;Oops - didn't expect that, small electrical fault in the secondary trigger mechanism.&quot; He fiddled with the applicator then claimed it was fixed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BfcFy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This meant the exterior bubble's remaining volume was still too large. &quot;Keep adding more bubbles into the large sphere until I tell you to stop - I think we can still get it right&quot; he said. &lt;strong&gt;I blew a second interior bubble inside the big sphere - when I had finished it formed the biggest sphere that could fit inside the exterior bubble alongside the first interior bubble.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The remaining volume is still too big&quot;, said Ernie. &lt;strong&gt;So I blew a third interior bubble. When I had finished it formed the largest sphere that could fit inside the exterior bubble alongside the first two interior bubbles.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Volume is still too big&quot;, said Ernie. &lt;strong&gt;So I blew a fourth one. When I had finished it formed the biggest sphere that could fit inside the large bubble alongside the first three interior bubbles.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Aha&quot;, said Ernie, &lt;strong&gt;&quot;now the remaining volume in the exterior sphere is exactly correct for my experiment. It's an interesting coincidence that the ratio of surface areas of the first interior bubble to the third interior bubble is exactly the same as the ratio of surface areas of the second interior bubble to the fourth interior bubble&quot;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few days later I had a phone message from Ernie (who had left town for a few days). His experiment had gone well but now he needed a second 'reaction chamber'. Could I pop around to the lab and blow one more sphere so it would be ready for his return? &lt;strong&gt;It needed to have exactly the same interior volume as the the remaining space in the one we made earlier.&lt;/strong&gt; &quot;I've mixed some more spider silk, but there isn't much to spare so I think you need to blow it the right size on the first attempt. I've also added a size dial on the applicator so you can stop at exactly the diameter you need (to the nearest micron).&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that Ernie failed to say exactly what diameter that should be! Can you help me work it out?&lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="11297" LastEditDate="2015-07-14T01:56:11.403" LastActivityDate="2015-09-17T04:03:23.840" Title="Ernie and the Bubbles of Lucretia" Tags="|mathematics|calculation-puzzle|geometry|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="17901" PostTypeId="2" ParentId="17900" CreationDate="2015-07-14T02:44:54.137" Score="10" Body="&lt;p&gt;If the first bubble has diameter $n$, then the second bubble must have diameter $1-n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the space remaining, there is always space for two bubbles of the same size that are both the largest bubbles that can fit into that space. (A proof to come once I'm not on my phone.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're also given that $\pi d_1^2 / \pi d_3^2 = \pi d_2^2 / \pi d_4^2$. Simplifying this gives that $(d_1 / d_2)^2 = (d_3 / d_4)^2$, which implies that $d_1 / d_3 = d_2 / d_4 = x$ as diameters can't be negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $d_1 = xd_3 = xd_4 = d_2$, and the first two bubbles both have a diameter of 0.5 m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uIa09.png&quot; alt=&quot;diagram of bubbles&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, we have a diagram as above, and solve the equation $(0.5-r)^2 + 0.25^2 = (0.25+r)^2$. The solution is 1/6, so the diameter of the remaining two bubbles is 1/3 m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, we calculate the cube root of $1^3 - 2(1/2)^3 - 2(1/3)^3$ to get a final diameter of 0.877606 m.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-09-17T04:03:23.840" LastActivityDate="2015-09-17T04:03:23.840" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="17902" PostTypeId="2" ParentId="17878" CreationDate="2015-07-14T03:14:49.620" Score="15" Body="&lt;p&gt;The solution is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://www.google.com/search?q=define%20conflagration&quot;&gt;&lt;strong&gt;Conflagration&lt;/strong&gt;&lt;/a&gt;&lt;br&gt;&#xA; Since &quot;&lt;em&gt;an extensive fire which destroys a great deal of land or property&lt;/em&gt;&quot; could be described as a massive (extensive) and torrid (hot) affair (event).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The breakdown in the first four lines is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Is my first part a bargain? Be careful! Not quite.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Con&lt;/strong&gt; - Initially seems like a bargain, but is actually not (be careful).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My next, when it's scarlet, means something's not right.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Flag&lt;/strong&gt; - A &quot;red flag&quot; is an idiom indicating a warning sign.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My third: to police, an invaluable tool&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Rat&lt;/strong&gt; - An informal term used to describe a criminal who gives information to the police&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Can my last bit be neutral? It can't, as a rule.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Ion&lt;/strong&gt; - a charged particle (by definition not neutral)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastActivityDate="2015-07-14T03:14:49.620" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17903" PostTypeId="2" ParentId="17895" CreationDate="2015-07-14T05:49:05.853" Score="8" Body="&lt;p&gt;I believe I know how to tell whether certain numbers are Baumic:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let $P_n$ be the $n^{th}$ prime number: $P_1=2$, $P_2=3$, etc.&lt;br&gt;&#xA; &lt;br/&gt;&#xA; &lt;strong&gt;Baumic Condition 1:&lt;/strong&gt; For all $x&amp;gt;2$, $x$ is Baumic if its prime factorization has $n$ terms (including repeats) and it is divisible by $P_n$.&lt;br&gt;&#xA; &lt;br/&gt;&#xA; This rule seems to hold for all numbers less than $100$: for any prime $p$, $3*p$ is Baumic.  For any (not necessarily distinct) primes $p$ and $q$, $5*p*q$ is Baumic, and so on.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Baumic condition 1 seems to be sufficient for a number to be Baumic, but there are Baumic numbers that do not meet it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $1$ is Baumic, but does not meet the condition.  $72=2^33^2$ is Baumic, but has $5$ prime factors and yet is not divisible by $P_5=11$.  None of $999999$, $99999999$, $9999999999$, and $9999999999999$ meet the condition, either.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is one other issue with the condition, as well:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Without the $x&amp;gt;2$ condition, Baumic Condition 1 would predict that $2$ is a Baumic number.  The condition does seem to work for numbers greater than $2$, though.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If Baumic Condition 1 holds, these numbers between $100$ and $200$ are Baumic:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $105$, $110$, $111$, $123$, $125$, $126$, $129$, $130$, $140$, $141$, $159$, $165$, $170$, $175$, $176$, $177$, $183$, $189$, $190$, $195$, $196$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10261" LastEditorUserId="10261" LastEditDate="2015-07-14T19:18:45.680" LastActivityDate="2015-07-14T19:18:45.680" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="17904" PostTypeId="2" ParentId="17882" CreationDate="2015-07-14T07:34:21.530" Score="5" Body="&lt;p&gt;The best in my word list is 13 letters:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; resulphurized/desulphurizer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2187" LastActivityDate="2015-07-14T07:34:21.530" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17906" PostTypeId="2" ParentId="17821" CreationDate="2015-07-14T09:56:23.860" Score="5" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The human body. Each part refers to the location of a body part and a phrase containing it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's start from the top.&lt;br&gt;&#xA;Why, you ask? Because it's best to start on a high.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;High&lt;strong&gt;brow&lt;/strong&gt;, the brow being near the top of the body.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Get into it now.&lt;br&gt;&#xA;Why, you ask? Because it's the smart idea to live by.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Mind&lt;/strong&gt;ing how you live, perhaps? Not entirely sure, although it's almost certainly referring to the brain inside your head.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Come out now and breathe the fresh air.&lt;br&gt;&#xA;Why, you ask? Because it's not right to intrude.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;We wouldn't want to &lt;strong&gt;nose&lt;/strong&gt; around, that being our main breathing apparatus. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Move under it just a little.&lt;br&gt;&#xA;Why, you ask? Because it would be rather rude.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Mouth&lt;/strong&gt;ing off, the mouth being just under the nose.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Go through and get stuck now.&lt;br&gt;&#xA;Why, you ask? Because it's rather rough here.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Could be a frog stuck in your &lt;strong&gt;throat&lt;/strong&gt;, which generally coincides with it feeling rough.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Come out altogether and go to the sides.&lt;br&gt;&#xA;Why, you ask? Because you need more than one to conquer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Conquering &lt;strong&gt;arm&lt;/strong&gt;ies, or possibly by force of &lt;strong&gt;arms&lt;/strong&gt;, with your arms to the sides of your body.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Move to the ends.&lt;br&gt;&#xA;Why, you ask? Because it's good to have something close.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Close at &lt;strong&gt;hand&lt;/strong&gt;, hands being at the ends of arms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Back up and down the middle till it splits.&lt;br&gt;&#xA;Why, you ask? Because those are the ones that appeal most.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Presumably the legs, although that seems like personal taste. Might be to do with appealing &lt;strong&gt;leg&lt;/strong&gt; before wicket (LBW) in cricket.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Drop down to the bottom and stay there now.&lt;br&gt;&#xA;Why, you ask? Because it's best to end with a game or two.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The game is a&lt;em&gt;foot&lt;/em&gt;, or maybe two feet. Either way, they're at the bottom of your body.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Too many questions here, that I know.&lt;br&gt;&#xA;But perhaps no one knows the answers better than you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;People generally know their own bodies better than anyone else.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12511" LastActivityDate="2015-07-14T09:56:23.860" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17907" PostTypeId="1" AcceptedAnswerId="17910" CreationDate="2015-07-14T09:59:38.223" Score="8" ViewCount="1729" Body="&lt;p&gt;A four word puzzle (see &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15339/four-words&quot;&gt;this&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16887/another-four-words&quot;&gt;this&lt;/a&gt;) is basically a 4x4 grid containing 4 4-letter words, and reads the same top to bottom or left to right. For example,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A B C D&#xA;B E F G&#xA;C F H I&#xA;D G I J&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here are some puzzles with hints (not necessarily in the same order):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sample puzzle&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a shape&lt;br&gt;&#xA;an identity&lt;br&gt;&#xA;a valley&lt;br&gt;&#xA;an audio track  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;S O N G&#xA;O V A L&#xA;N A M E&#xA;G L E N&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Puzzle 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a flower&lt;br&gt;&#xA;a plant&lt;br&gt;&#xA;a stage in life&lt;br&gt;&#xA;a fraction of a whole  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Puzzle 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a medium of writing&lt;br&gt;&#xA;an examination&lt;br&gt;&#xA;a mark joining two points&lt;br&gt;&#xA;a narrow cut  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Puzzle 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a kind of footwear&lt;br&gt;&#xA;a kind of sensation&lt;br&gt;&#xA;obedience&lt;br&gt;&#xA;squashing things together  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Puzzle 4&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;stack&lt;br&gt;&#xA;volume&lt;br&gt;&#xA;termination&lt;br&gt;&#xA;a metal  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Possible bounty&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know it's quite unlikely, but if a single person manages to be the first one to solve 3 of the 4 puzzles, you will get a bounty of 100 rep (plus up-votes from many people).&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-07-29T10:41:56.857" Title="Some four word puzzles (and possible bounty)" Tags="|word|wordplay|crosswords|word-square|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17908" PostTypeId="1" CreationDate="2015-07-14T10:01:53.090" Score="1" ViewCount="1361" Body="&lt;p&gt;What is the largest &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15339/four-words&quot;&gt;word-square&lt;/a&gt; puzzle known? Can you make a bigger one? Obviously, you must use more than four words, six or seven letter words would be really good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using computers allowed, not recommended unless necessary.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2016-06-22T13:11:36.537" Title="Largest word-square puzzle" Tags="|word|wordplay|crosswords|word-square|" AnswerCount="3" CommentCount="8" ClosedDate="2016-06-22T20:33:21.010" ContentLicense="CC BY-SA 3.0" />
  <row Id="17909" PostTypeId="2" ParentId="17907" CreationDate="2015-07-14T10:13:02.880" Score="8" Body="&lt;p&gt;Puzzle 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;     &lt;code&gt;S L I T&lt;/code&gt;&lt;br&gt;&#xA;     &lt;code&gt;L I N E&lt;/code&gt;&lt;br&gt;&#xA;     &lt;code&gt;I N K S&lt;/code&gt;&lt;br&gt;&#xA;     &lt;code&gt;T E S T&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="9860" LastEditDate="2015-07-14T11:01:04.813" LastActivityDate="2015-07-14T11:01:04.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17910" PostTypeId="2" ParentId="17907" CreationDate="2015-07-14T10:20:53.813" Score="19" Body="&lt;p&gt;&lt;strong&gt;Puzzle 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;PART&lt;/code&gt; - a fraction of a whole&lt;br&gt;&#xA;&lt;code&gt;ALOE&lt;/code&gt; - a plant&lt;br&gt;&#xA;&lt;code&gt;ROSE&lt;/code&gt; - a flower&lt;br&gt;&#xA;&lt;code&gt;TEEN&lt;/code&gt; - a stage in life&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Puzzle 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;MASH&lt;/code&gt; - squashing things together&lt;br&gt;&#xA;&lt;code&gt;ACHE&lt;/code&gt; - a kind of sensation&lt;br&gt;&#xA;&lt;code&gt;SHOE&lt;/code&gt; - a kind of footwear&lt;br&gt;&#xA;&lt;code&gt;HEED&lt;/code&gt; - obedience&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Puzzle 4&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;PILE&lt;/code&gt; - stack&lt;br&gt;&#xA;&lt;code&gt;IRON&lt;/code&gt; - a metal&lt;br&gt;&#xA;&lt;code&gt;LOUD&lt;/code&gt; - volume&lt;br&gt;&#xA;&lt;code&gt;ENDS&lt;/code&gt; - termination&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12511" LastActivityDate="2015-07-14T10:20:53.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17911" PostTypeId="2" ParentId="17908" CreationDate="2015-07-14T12:28:50.873" Score="2" Body="&lt;p&gt;I have managed to construct a word square which has 10 different words in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;CLUES:&lt;/strong&gt;&lt;br&gt;&#xA;muffled cries&lt;br&gt;&#xA;butter substitute&lt;br&gt;&#xA;close&lt;br&gt;&#xA;a special dress&lt;br&gt;&#xA;a melodious poem&lt;br&gt;&#xA;genus of olive trees&lt;br&gt;&#xA;a hairy mammal&lt;br&gt;&#xA;to absorb or adsorb&lt;br&gt;&#xA;flat piece of stone (diagonal top left to bottom right)&lt;br&gt;&#xA;commands a horse (diagonal bottom left to top right)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;     &lt;code&gt;S O B S&lt;/code&gt;&lt;br&gt;&#xA;     &lt;code&gt;O L E O&lt;/code&gt;&lt;br&gt;&#xA;     &lt;code&gt;N E A R&lt;/code&gt;&lt;br&gt;&#xA;     &lt;code&gt;G A R B&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="13697" LastEditDate="2015-07-14T13:00:46.220" LastActivityDate="2015-07-14T13:00:46.220" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17913" PostTypeId="2" ParentId="17833" CreationDate="2015-07-14T13:02:14.463" Score="34" Body="&lt;p&gt;You just have to be descriptive in your addition:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;13 + 7 = one 1 one 3 + one 7 = 1113 + 17 = 1130&lt;/code&gt; &lt;br&gt;&#xA; &lt;code&gt;88 + 9 = two 8 + one 9 = 28 + 19 = 47&lt;/code&gt; &lt;br&gt;&#xA; &lt;code&gt;32 + 28 = one 3 one 2 + one 2 one 8 = 1312 + 1218 = 2530&lt;/code&gt; &lt;br&gt;&#xA; &lt;code&gt;10 + 999 = one 1 one 0 + three 9 = 1110 + 39 = 1149&lt;/code&gt; &lt;br&gt;&#xA; etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the final solution is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;123 + 45 = one 1 one 2 one 3 + one 4 one 5 = 111,213 + 1415 = 112,628&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6128" LastActivityDate="2015-07-14T13:02:14.463" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17915" PostTypeId="2" ParentId="17908" CreationDate="2015-07-14T13:24:04.067" Score="-2" Body="&lt;p&gt;I also found a word square with 10 different words:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;     &lt;code&gt;S L A P&lt;/code&gt;&lt;br&gt;&#xA;     &lt;code&gt;L I V E&lt;/code&gt;&lt;br&gt;&#xA;     &lt;code&gt;T I M E&lt;/code&gt;&lt;br&gt;&#xA;     &lt;code&gt;G U N S&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Each horizontal word can be read like this, and reverse. Also the last vertical line, with &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;pees and seep.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3993" LastActivityDate="2015-07-14T13:24:04.067" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17916" PostTypeId="1" AcceptedAnswerId="17917" CreationDate="2015-07-14T13:56:30.860" Score="21" ViewCount="1816" Body="&lt;p&gt;It was the 15 October 1502, just a few hours before midnight in an old tavern at the docks. Emerald raised two fingers, after repeatedly looking at his three cards, before finally answering the awaiting dealer. &quot;Stick!&quot; His voice was a bit shaky, since his last savings were at stake. As each player started to reveal his cards, Emerald quickly counting their points - and with 28, he was the highest so far. But this just wasn't his lucky day, and just as he revealed his cards, the dealer started to smile - revealing his 31 points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I can't believe it! Again!&quot; Emerald dug his nails into his wooden chair, looking up to the dealer. &quot;I don't have more with me. I'll pay the rest next month, alright?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dealer nodded, before asking with another smile &quot;Tell ye what... I'll give ya chance ta 'win it back. With a simple lil' bet.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Emerald raised an eyebrow, &quot;A bet?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dealer calmly nodded once more while reaching for the other players' cards. &quot;Yeah. Remember that ol' bookworm th'was here earlier?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Emerald shrugged. &quot;The guy that blabbered about a round world?&quot; he scratched his chin. &quot;What about him?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dealer's smile turned into a greedy grin. &quot;Prove me that th'world is a ball, and you get m'winnings. If ya can't, I get double. 'til Friday. Deal?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Emerald had the luck to talk with the scholar from before and smiled, quickly standing up to shake the dealers hand. &quot;Deal. But I think I can prove it to you tonight, if you still stay sober for another hour or two.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Puzzle&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can Emerald prove that the world is round without leaving the city? And how can he do it in the same night?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For extra fun: What's the game they're playing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is my first riddle here, and I wanted to add a bit of atmosphere to it - but since I'm not a native English speaker, my writing style is probably a bit weird. ;) Still, good luck 'solving' it. It's not too hard. Also, tagging puzzles is a puzzle by itself!)&lt;/p&gt;&#xA;" OwnerUserId="14076" LastEditorUserId="463" LastEditDate="2015-07-14T14:42:26.710" LastActivityDate="2015-07-14T19:13:11.003" Title="Emeralds bet - A round world" Tags="|logical-deduction|enigmatic-puzzle|story|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17917" PostTypeId="2" ParentId="17916" CreationDate="2015-07-14T14:03:25.647" Score="24" Body="&lt;p&gt;On the 15th October 1502&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; there was a lunar eclipse, with the TD of greatest eclipse at 23:10:40. &lt;br&gt;&lt;br&gt; As the eclipse progresses, the Moon slowly moves through Earth's shadow. &lt;br&gt;&lt;br&gt; Every time that shadow is seen, its edge is round.  The only solid that always projects a round shadow is a sphere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For extra fun: The game being played here is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Thirty-one_(card_game)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thirty-One&lt;/a&gt; (Credit to @Len) &lt;br&gt;&lt;br&gt;Reasoning on why it could not also be a disk: &lt;a href=&quot;https://mathoverflow.net/questions/39127/is-the-sphere-the-only-surface-all-of-whose-projections-are-circles-or-can-we&quot;&gt;Spherical Proof&lt;/a&gt;&lt;br&gt;&lt;br&gt;&lt;a href=&quot;http://eclipse.gsfc.nasa.gov/LEcat5/LE1501-1600.html&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Catalog of Lunar Eclipses&quot;&gt;Catalog of Lunar Eclipses&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12206" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:31.357" LastActivityDate="2015-07-14T19:13:11.003" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="17918" PostTypeId="2" ParentId="17831" CreationDate="2015-07-14T14:41:19.943" Score="8" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a prune!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I am not the trimmings of a bush,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Another word for trimming a bush is pruning.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;but something closer to a mush.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Prunes are closer to mush than to branches.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I am not the clams in your hand,&#xA;but that guess I understand.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you've soaked in water for a while some people call the wrinkles in your hands prunes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;As long as you  answer with aplomb,&#xA;The answer you will not be far from.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Aplomb sounds like &quot;A plum&quot; which is close to a prune.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Admittedly my humor is a little dried,&#xA;But much of my moral fiber is supplied.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A prune is a dried plum, but still has all the fiber of a plum.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The title:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Older folks have more wrinkles (prunes) and may require a little bit of prune juice to stay regular.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6128" LastEditorUserId="6128" LastEditDate="2015-07-14T14:49:43.223" LastActivityDate="2015-07-14T14:49:43.223" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17919" PostTypeId="2" ParentId="17890" CreationDate="2015-07-14T15:18:13.817" Score="5" Body="&lt;p&gt;celtschk's solution is optimal.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We will show that if $k&amp;lt;2^j$, you cannot always successfully select a knight.&#xA; &lt;br&gt;&lt;br&gt;&#xA; First of all, as other commenters have pointed out, the first $j$ people in line could be all $j$ jokers, and give you any answers they want. Therefore, you can never safely select one of the first $j$ people. So, you must always select one of the other $k$ people.&#xA; &lt;br&gt;&lt;br&gt;&#xA; Also, since you must succeed with 100% probability, there is no point in using a randomized strategy. Therefore, we can consider your selection strategy to be a function which inputs a sequence of $j$ answers (e.g. (yes, no, ..., yes)), and outputs your selection of one of the last $k$ people in the line.&#xA; &lt;br&gt;&lt;br&gt;&#xA; Suppose $k &amp;lt; 2^j$. Then by Pigeonhole there are two sequences of answers $s_1, s_2$ which cause you to select the same person $p$.&#xA; &lt;br&gt;&lt;br&gt;&#xA; The jokers will now confound you, as follows. They will place a joker at spot $p$. Among the first $j$ people will be the remaining $j-1$ jokers, as well as one knight at the first spot where $s_1$ and $s_2$ differ.&#xA; &lt;br&gt;&lt;br&gt;&#xA; The jokers will then answer your questions so that you hear either $s_1$ or $s_2$ depending on how the knight answers. This causes you to incorrectly select person $p$ as a knight.&#xA; &lt;br&gt;&lt;br&gt;&#xA; Since the jokers can pull this trick, if $k &amp;lt; 2^j$, you cannot be sure of always selecting a knight. Therefore, celtschk's solution with $k=2^j$ is optimal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2015-07-14T15:18:13.817" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17920" PostTypeId="2" ParentId="17309" CreationDate="2015-07-14T15:22:01.370" Score="36" Body="&lt;p&gt;Expressing the numbers in&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; base 36&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; mapping the digits &lt;code&gt;10&lt;/code&gt; through &lt;code&gt;35&lt;/code&gt; to the letters &lt;code&gt;A&lt;/code&gt;-&lt;code&gt;Z&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;results in the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/M3bpM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In a more traditional format:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;1.  e4    c5&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;2. Nf3    e6&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;3.  d4   cxd4&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;4. Nxd4  Nf6&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;5. Nd2    d6&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;6. Bd3   Nc6&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;7. Nxc6  bxc6&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;8. O-O   Be7&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;9.  b3   O-O&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;10. Bb2   Nd7&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;11.  f4   Bf6&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;12.  e5   dxe5&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;13. Ne4   exf4&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;14. Nxf6+ Nxf6&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;15. Rxf4  Re8&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;16. Rxf6  gxf6&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;17. Qg4+  Kf8&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;18. Ba3+  Re7&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;19. Bxh7  Qb6+?&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;20. Kh1   Ke8&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It looks like your friend is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; playing chess with the computer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and needs to input&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A move that checkmates the computer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can inspect the current state of the game &lt;a href=&quot;http://analysis.cpuchess.com/?fen=r1b1k3%2Fp3rp1B%2F1qp1pp2%2F8%2F6Q1%2FBP6%2FP1P3PP%2FR6K%20w%20-%20-%203%2021&amp;amp;lmove=Ke8&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.  It looks like our best move is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;21. Rd1&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which forces&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;21. ...  Qd4&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;22. Rxd4  e5&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;23. Qg8#&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Converting back to numbers, the sequence is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;&amp;gt; 35461&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;response: 34168&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;&amp;gt; 35464&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;response: 32909&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;&amp;gt; 34280&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;bailey@puzzling~/chess$ _&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="9718" LastEditDate="2015-07-14T17:06:09.520" LastActivityDate="2015-07-14T17:06:09.520" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17921" PostTypeId="2" ParentId="17872" CreationDate="2015-07-14T16:09:22.763" Score="2" Body="&lt;p&gt;My best guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hurricane.  They have nicknames, occur in summer, come from clouds,&#xA; block out the sunlight, can kill many, and aren't born of their own&#xA; fault, but instead result by mixing of disparate&#xA; pressure/humidity/temperature layers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14082" LastActivityDate="2015-07-14T16:09:22.763" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17922" PostTypeId="2" ParentId="17309" CreationDate="2015-07-14T17:16:29.230" Score="9" Body="&lt;p&gt;@2012rcampion is definitely on the right track. In fact, the winning move is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 21. Rd1 (as white)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The game is actually a recorded game between&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Viacheslav Ragozin vs P Noskov&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's a link to the game:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.chessgames.com/perl/chessgame?gid=1282608&quot;&gt;http://www.chessgames.com/perl/chessgame?gid=1282608&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again, full credit should go to @2012rcampion. I'm just a chess enthusiast doing some research :D&lt;/p&gt;&#xA;" OwnerUserId="14084" LastActivityDate="2015-07-14T17:16:29.230" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17923" PostTypeId="1" AcceptedAnswerId="17939" CreationDate="2015-07-14T17:38:46.970" Score="30" ViewCount="2293" Body="&lt;p&gt;These ten words are arranged in order according to a certain property they share:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;festive&lt;br&gt;&#xA;  fixture&lt;br&gt;&#xA;  salient&lt;br&gt;&#xA;  resolving&lt;br&gt;&#xA;  recline&lt;br&gt;&#xA;  accessory&lt;br&gt;&#xA;  advise&lt;br&gt;&#xA;  headlight&lt;br&gt;&#xA;  redcoat&lt;br&gt;&#xA;  programming  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where should each of the following four words be inserted into the list?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;abbreviate&lt;br&gt;&#xA;twelve&lt;br&gt;&#xA;humdinger&lt;br&gt;&#xA;anecdote  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the property shared by all fourteen words?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A word's meaning is not relevant to the property in question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4385" LastEditorUserId="4385" LastEditDate="2015-07-15T14:36:24.053" LastActivityDate="2015-07-15T19:33:30.993" Title="A New Word Order" Tags="|pattern|lateral-thinking|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="17924" PostTypeId="2" ParentId="17821" CreationDate="2015-07-14T18:22:09.150" Score="0" Body="&lt;p&gt;Perhaps a better explanation of the first verse:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Let's start from the top.&#xA;Why, you ask? Because it's best to start on a high.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your hair is at the top of your body (assuming you're not bald!). It's &quot;on high&quot; if standing on end?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Although Zandar's idea also looks good, the OP said it wasn't the intended solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Get into it now.&#xA;Why, you ask? Because it's the smart idea to live by.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Going down and in from the top, you get to your &lt;strong&gt;eyes&lt;/strong&gt;. There could be a bit of wordplay in &quot;smart [eye]dea&quot; here; also, most people live by their eyesight.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, the &quot;Why, you ask?&quot; throughout the riddle could be a reference to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; another body part: the arse. &quot;Why, you [arse]k?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-07-14T18:22:09.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17925" PostTypeId="1" AcceptedAnswerId="17926" CreationDate="2015-07-14T18:22:18.440" Score="30" ViewCount="1527" Body="&lt;p&gt;I, Mrs. Malaprop, will recite an original riddle, perspired by &lt;a href=&quot;https://puzzling.stackexchange.com/questions/17878/a-massive-and-torrid-affair&quot;&gt;this&lt;/a&gt; puzzle. Each of lines 1-5 inscribe a part of a word, put them together in sequins to get the one-word answer to the riddle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First Part: According to the psychiatrist, this is it I suppose.&lt;br&gt;&#xA;Second Part: A token creation, the forest is my home.&lt;br&gt;&#xA;Third Part: My mood is subjective, though it sometimes depends.&lt;br&gt;&#xA;Fourth Part: In the list of apostles, I am one off from the tenth.&lt;br&gt;&#xA;Fifth Part: An onion? No, I'm the polar opposite.&lt;br&gt;&#xA;Sixth Part: Put us together and what do you get?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those who love challenge, use what you now know.&lt;br&gt;&#xA;If you want more direction, please see below.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is similar to the word riddle linked above, with the exception that it is being read by the fictional Mrs. Malaprop for whom the word malapropism is named. As a result, each line of the riddle contains a malapropism. Correct the false words and solve the riddle. A correct answer should contain the words that she meant to use in each line in addition to the final word obtained.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2015-07-14T20:04:48.420" Title="Mrs. Malaprop Reads A Riddle" Tags="|riddle|word|wordplay|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17926" PostTypeId="2" ParentId="17925" CreationDate="2015-07-14T18:33:39.057" Score="20" Body="&lt;p&gt;&lt;em&gt;First Part: According to the psychiatrist, this is it I suppose.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Malapropism: &quot;it&quot; -&gt; &quot;id&quot;. Solution: &lt;strong&gt;id&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Second Part: A token creation, the forest is my home.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Malapropism: &quot;token&quot; -&gt; &quot;Tolkien&quot;. Solution: &lt;strong&gt;Ent&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Third Part: My mood is subjective, though it sometimes depends.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Malapropism: &quot;subjective&quot; -&gt; &quot;subjunctive&quot;. Solution: &lt;strong&gt;if&lt;/strong&gt;. (thanks @dennisdeems)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Fourth Part: In the list of apostles, I am one off from the tenth.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Malapropism: &quot;apostles&quot; -&gt; &quot;epistles&quot; -&gt; &quot;letters&quot;. Solution: &lt;strong&gt;I&lt;/strong&gt;. (thanks @GentlePurpleRain and @BaileyM)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Fifth Part: An onion? No, I'm the polar opposite.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Malapropism: &quot;onion&quot; -&gt; &quot;anion&quot;. Solution: &lt;strong&gt;cation&lt;/strong&gt;. (thanks @CodeNewbie)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Sixth Part: Put us together and what do you get?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; IDENTIFICATION&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-07-14T19:00:08.407" LastActivityDate="2015-07-14T19:00:08.407" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="17927" PostTypeId="1" AcceptedAnswerId="17943" CreationDate="2015-07-14T18:36:13.793" Score="9" ViewCount="788" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;My brother's not Russian,&lt;br&gt;&#xA;  &lt;a href=&quot;https://puzzling.stackexchange.com/revisions/17927/1&quot;&gt;despite what some claim.&lt;/a&gt;&lt;br&gt;&#xA;  But we hail from Hungary,&lt;br&gt;&#xA;  you all know our name.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;My brother is older,&lt;br&gt;&#xA;  about seven years.&lt;br&gt;&#xA;  And while I am bigger,&lt;br&gt;&#xA;  please hold all your cheers.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;By how much am I bigger?&lt;br&gt;&#xA;  It's not clear to me.&lt;br&gt;&#xA;  By the main ratio,&lt;br&gt;&#xA;  of your modern TV?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Or maybe you count him&lt;br&gt;&#xA;  with Z's down to A's.&lt;br&gt;&#xA;  While I'm a sixth pyramid,&lt;br&gt;&#xA;  or eight times the days.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I have other siblings,&lt;br&gt;&#xA;  of various sizes.&lt;br&gt;&#xA;  They're mostly the same,&lt;br&gt;&#xA;  but some are surprises.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If you want to be nice,&lt;br&gt;&#xA;  just do what you're told,&lt;br&gt;&#xA;  and bring me some ice;&lt;br&gt;&#xA;  I'm best when I'm cold.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1563" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2015-07-15T21:56:14.490" Title="The Younger Brother" Tags="|riddle|rhyme|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17928" PostTypeId="2" ParentId="17916" CreationDate="2015-07-14T18:39:46.430" Score="5" Body="&lt;p&gt;alexmc's answer is more robust than this one, but I thought this was also a fun solution to point out. Not as mathematically sound, but also more widely applicable.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Take the barkeep to the nearby ocean. First, take a ruler or something straight, and try to line it up with the horizon. You will be able to see that the horizon is in fact curved, with the edges lower than the middle. Then wait for a ship to either enter into sight or leave. You will notice that the first (or last) thing you see is the mast. This shows that the earth is also curved in that direction. With curvature on two perpendicular axes, you could conclude that the earth, or at least the part you are standing on, is round. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1968" LastActivityDate="2015-07-14T18:39:46.430" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17929" PostTypeId="2" ParentId="17927" CreationDate="2015-07-14T18:59:34.027" Score="2" Body="&lt;p&gt;A very crude guess and I'll need to stretch the clues too much, but is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Russian drink vodka and it's Hungarian counterpart palinka &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My brother is Russian;&lt;br&gt;&#xA;a fairly known name.&lt;br&gt;&#xA;But I hail from Hungary,&lt;br&gt;&#xA;with not as much fame.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Vodka is famous all over the world whereas the palinka is not so popular.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My brother is older,&lt;br&gt;&#xA;about seven years.&lt;br&gt;&#xA;And while I am bigger,&lt;br&gt;&#xA;please hold all your cheers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Perhaps 7 year aged vodka is most commonly enjoyed. (A wild guess, I'm a teetotaler). The cheers is a reference to the common practice of clinking glasses before a round of drinking. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By how much am I bigger?&lt;br&gt;&#xA;It's not clear to me.&lt;br&gt;&#xA;By the main ratio,&lt;br&gt;&#xA;of your modern TV?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is not clear historically which drink has been around longer. The ratio perhaps hints at the alcohol percentages between the two? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or maybe you count him&lt;br&gt;&#xA;with Z's down to A's.&lt;br&gt;&#xA;While I'm the sixth pyramid,&lt;br&gt;&#xA;or eight times the days.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;em&gt;Not the slightest bit of thread for me to pull here...&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have other siblings,&lt;br&gt;&#xA;of various sizes.&lt;br&gt;&#xA;They're mostly the same,&lt;br&gt;&#xA;but some are surprises.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are many other alcoholic preparations, with varying alcohol contents. All of them get you drunk, but some are unusually potent, like absinthe which often causes hallucinations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you want to be nice,&lt;br&gt;&#xA;just do what you're told,&lt;br&gt;&#xA;and bring me some ice;&lt;br&gt;&#xA;I'm best when I'm cold.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After getting drunk it's best to listen to the ones who are sober. And palinka probably tastes better when it's cold. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-07-15T08:00:02.837" LastActivityDate="2015-07-15T08:00:02.837" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17930" PostTypeId="2" ParentId="17927" CreationDate="2015-07-14T19:22:36.493" Score="0" Body="&lt;p&gt;I'll go with &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Barium  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My brother is older,&#xA;about seven years.&#xA;And while I am bigger,&#xA;please hold all your cheers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Elements further down the periodic table are bigger&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or maybe you count him&#xA;with Z's down to A's.&#xA;While I'm the sixth pyramid,&#xA;or eight times the days. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The 56th element is in the 6th row of the periodic table. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-07-14T19:22:36.493" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17932" PostTypeId="1" AcceptedAnswerId="18000" CreationDate="2015-07-14T19:58:45.073" Score="12" ViewCount="658" Body="&lt;p&gt;I have six friends whose favorite movies are inspired by our shared hobby.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The first enjoys an action remake about a man with a licence to kill.&lt;/li&gt;&#xA;&lt;li&gt;The second prefers an epic about a traitor leading a medieval revolution.&lt;/li&gt;&#xA;&lt;li&gt;The third's is a series of films about a seafaring trickster.&lt;/li&gt;&#xA;&lt;li&gt;The fourth's stars the lead actor from the third's, but it wasn't well received.&lt;/li&gt;&#xA;&lt;li&gt;The fifth loves a classic underdog story with a memorable training montage.&lt;/li&gt;&#xA;&lt;li&gt;The sixth features a blue collar astrophysicist with higher ambitions.&lt;/li&gt;&#xA;&lt;li&gt;My favorite movie stars an infamous actor, a puzzler like yourselves, hunting for a fortune centuries old.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What common interest do my friends and I share?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All of my friends live in the United States; Residents of North America will have an easier time solving than the rest (sorry).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My second friend isn't really a Mel Gibson fan; he just as much appreciates a film about a band of lionhearted home appliances.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13237" LastEditorUserId="13237" LastEditDate="2018-05-31T15:38:03.597" LastActivityDate="2018-05-31T15:38:03.597" Title="Some friends are happier than others" Tags="|wordplay|knowledge|movies|" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="17935" PostTypeId="2" ParentId="17932" CreationDate="2015-07-14T20:21:57.883" Score="4" Body="&lt;p&gt;A partial list of the movies:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1-James Bond - Casino Royal  (only remake with the same name. Credit to &lt;a href=&quot;https://puzzling.stackexchange.com/users/13197/player-one&quot;&gt;Player One&lt;/a&gt;) &lt;br&gt;&#xA; 2-Braveheart &lt;br&gt;&#xA; 3-Pirates of the Caribbean &lt;br&gt;&#xA; 4-(?) &lt;br&gt;&#xA; 5-Karate Kid or Rocky &lt;br&gt;&#xA; 6-The Astronaut Farmer &lt;br&gt;&#xA; 7-National Treasure (with Nicolas Cage as the infamous actor)&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5359" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-07-15T15:31:42.210" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="17936" PostTypeId="1" AcceptedAnswerId="18626" CreationDate="2015-07-14T20:23:22.363" Score="2" ViewCount="542" Body="&lt;p&gt;Can you decipher the numbers and determine who I am?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;24083272403040832724032608240304120727327240824071232724082308&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my first puzzle and I have no idea how hard this is going to be, but I'd love some feedback&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Groups of twos and threes are key&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm not your standard person, I'm a little quirky&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3: (Warning this is a major spoiler)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rows, Cols&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 4: (Since no one has figured it out, this is the biggest hint by far)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Keyboards are laid out like a grid &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13925" LastEditorUserId="13925" LastEditDate="2015-08-04T12:56:33.447" LastActivityDate="2015-08-04T19:54:32.770" Title="Conquering the Universe" Tags="|cipher|pattern|number-sequence|" AnswerCount="3" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17937" PostTypeId="1" AcceptedAnswerId="18042" CreationDate="2015-07-14T20:44:19.793" Score="12" ViewCount="578" Body="&lt;p&gt;You have one hundred lockboxes, fifty made of steel, fifty made of wood, and a key for each box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A prankster breaks into your house and distributes the keys among the boxes. He places each key in a random box &lt;em&gt;independently of the other keys&lt;/em&gt;, so that some boxes may receive multiple keys while others have none. He then closes all the boxes, locking the keys inside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though the steel boxes are impenetrable, the wooden ones can be broken quite easily. What is the probability that breaking open the wooden boxes will allow you to open all of the steel ones?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus&lt;/strong&gt;: What if you had $n$ lockboxes in total, $k$ of which were wooden?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Remarks&lt;/strong&gt;: &#xA;This is a slight variation of a &lt;a href=&quot;https://puzzling.stackexchange.com/q/17852/10615&quot;&gt;question I previously asked&lt;/a&gt;. In that puzzle, the prankster also randomly placed the keys in the boxes, but did so in a way that no two keys were put in the same box. Will his change in behavior increase your chances of being able to open all the boxes, or decrease them?&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-09-12T17:10:13.883" Title="One Hundred Lockboxes, Variant" Tags="|probability|combinatorics|keys-and-locks|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17939" PostTypeId="2" ParentId="17923" CreationDate="2015-07-14T21:54:17.923" Score="47" Body="&lt;p&gt;That was a real humdinger.  Now I'm feeling festive.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Roman numerals.&#xA; Each word has, somewhere after the first character, a string of Roman Numerals, evaluating to a number.  Let the property be that number.  Then we get this ordering:&lt;br&gt;&#xA; fest&lt;strong&gt;IV&lt;/strong&gt;e = 4&lt;br&gt;&#xA; abbre&lt;strong&gt;VI&lt;/strong&gt;ate=6&lt;br&gt;&#xA; f&lt;strong&gt;IX&lt;/strong&gt;ture=9&lt;br&gt;&#xA; sa&lt;strong&gt;LI&lt;/strong&gt;ent=51&lt;br&gt;&#xA; twe&lt;strong&gt;LV&lt;/strong&gt;e=55&lt;br&gt;&#xA; reso&lt;strong&gt;LVI&lt;/strong&gt;ng=56&lt;br&gt;&#xA; re&lt;strong&gt;CLI&lt;/strong&gt;ne=151&lt;br&gt;&#xA; a&lt;strong&gt;CC&lt;/strong&gt;essory=200&lt;br&gt;&#xA; ane&lt;strong&gt;CD&lt;/strong&gt;ote=400&lt;br&gt;&#xA; a&lt;strong&gt;DVI&lt;/strong&gt;se=506&lt;br&gt;&#xA; hea&lt;strong&gt;DLI&lt;/strong&gt;ght=551&lt;br&gt;&#xA; re&lt;strong&gt;DC&lt;/strong&gt;oat==600&lt;br&gt;&#xA; hu&lt;strong&gt;MDI&lt;/strong&gt;nger=1501&lt;br&gt;&#xA; progra&lt;strong&gt;MMI&lt;/strong&gt;ng=2001&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14082" LastEditorUserId="4385" LastEditDate="2015-07-15T19:33:30.993" LastActivityDate="2015-07-15T19:33:30.993" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17940" PostTypeId="1" CreationDate="2015-07-14T22:19:59.753" Score="6" ViewCount="623" Body="&lt;p&gt;#1 ________ &lt;br&gt;&#xA; #2 ________ &lt;br&gt;&#xA; #3 ________ &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 1: Note that something is missing in this puzzle and determine what it is (you may have to use it to realize it's gone). When you notice what it is, write it out in blank #1. You will surely agree that there is something impeccable about this missing thing. That ideal quality is the key.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 2:  8303765625 13841287201 X 531441 387420489 403.428793493 These numbers also seem to have some mysterious power hidden in them, the quality located above is the key to determining this power. Concerning the value X, when you have deciphered the numbers, you should realize that there are only two options; order them intuitively and choose the one that precedes the other. Using these values, locate their purpose and write it out in blank #2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 3: 340 315 1103 These will be the hardest numbers to understand. Where the numbers in Step 2 derived their power, these numbers get a paltry structure. Using Step 1's quality as the key, unlock the secrets in these numbers. You will have to be guided by intuition and knowledge hence; It will be hard! In the event that you think you are at the decrypted end and still can't make heads or tails, just ask 245 335 310 what it means and that should make it crystal clear. Take the second part to this answer and place it in blank #3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ready? Bring me back the result that you ultimately determine and explain the procedure and you will receive my acceptance and praise! Please ask below in the event that you need help, and I will answer within reason. I recommend adding partial answers as this has many parts and using others' knowledge as a resource would&#xA;be a great idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has been a while, so I will add a hint to those entrenched in Step 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There are two numbers that may help to understand the meaning in the hidden numbers discovered in this step: 167 and 39. These numbers are not encoded the same way that the others are; 6 is not the key to understanding their value, but 256 instead (though it was originally only 128).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="13970" LastEditDate="2015-07-15T16:46:05.183" LastActivityDate="2015-07-16T15:36:10.390" Title="3 Step Puzzle with Increasing Ardor" Tags="|mathematics|enigmatic-puzzle|" AnswerCount="5" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17942" PostTypeId="2" ParentId="17940" CreationDate="2015-07-14T23:36:59.617" Score="3" Body="&lt;p&gt;I think I know #1. Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; QUESTION. &lt;br&gt;&lt;br&gt; There is no question asked in the entire puzzle. And usually when solving a puzzle, we tend to mentally ask ourselves, &quot;What is the question here?&quot; thereby asking a question, hence realizing what is missing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-07-14T23:36:59.617" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17943" PostTypeId="2" ParentId="17927" CreationDate="2015-07-15T01:01:22.213" Score="10" Body="&lt;p&gt;My guess in combination with &lt;a href=&quot;https://puzzling.stackexchange.com/users/10861/victorhenry&quot;&gt;victorhenry&lt;/a&gt; is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;a 4x4x4 Rubik's Revenge Cube.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's why: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My brother's not Russian,&lt;br&gt;&#xA;despite what some claim.&lt;br&gt;&#xA;But we hail from Hungary,&lt;br&gt;&#xA;you all know our name.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rubik's cubes were invented by Ernő Rubik (Hungarian), but 7 years later by Péter Sebestényo who is most definitely not Russian, but I couldn't confirm he too was Hungarian (the patent can be found &lt;a href=&quot;http://www.freepatentsonline.com/4421311.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and lists Hamburg, Germany)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My brother is older,&lt;br&gt;&#xA;about seven years.&lt;br&gt;&#xA;And while I am bigger,&lt;br&gt;&#xA;please hold all your cheers. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As mentioned above, this cube is bigger than the traditional 3x3x3 cube that was invented 7 years earlier.  However this cube doesn't have center pieces, and has some edge parities which makes it harder to solve.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By how much am I bigger?&lt;br&gt;&#xA;It's not clear to me.&lt;br&gt;&#xA;By the main ratio,&lt;br&gt;&#xA;of your modern TV?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The 4x4x4 Rubiks cube has 16 pieces on each face, while the 3x3x3 cube has 9 pieces on each face. Modern TV aspect ratio's are 16:9, and the aspect ratio of slightly less modern TV's was 4:3.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or maybe you count him&lt;br&gt;&#xA;with Z's down to A's.&lt;br&gt;&#xA;While I'm a sixth pyramid,&lt;br&gt;&#xA;or eight times the days.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The 6th triangular pyramidal number is 56, which is also 7 days * 8, which is the number of pieces on a 4x4x4 cube (4*6 = 24 center pieces, 12*2 = 24 edge pieces, and 8*1= corner pieces).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have other siblings,&lt;br&gt;&#xA;of various sizes.&lt;br&gt;&#xA;They're mostly the same,&lt;br&gt;&#xA;but some are surprises.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rubik's cubes can have many different shapes and sizes, but a majority of them operate with twisting in 1 of 3 directions, although there are pentagonal ones that have more axis of rotation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you want to be nice,&lt;br&gt;&#xA;just do what you're told,&lt;br&gt;&#xA;and bring me some ice;&lt;br&gt;&#xA;I'm best when I'm cold.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Revenge is a dish best served cold.  Also probably a reference to speed cubing because Oliver Frost has the record for blindfolded 4x4x4 cubing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you want to watch some Rubik's Cube magic (albeit the 3x3x3), &lt;a href=&quot;https://www.youtube.com/watch?v=BwDAXC0_Bxk&quot; rel=&quot;nofollow noreferrer&quot;&gt;watch this&lt;/a&gt;!&lt;/p&gt;&#xA;" OwnerUserId="14092" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-07-15T21:56:14.490" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="17945" PostTypeId="1" AcceptedAnswerId="17947" CreationDate="2015-07-15T02:47:39.333" Score="18" ViewCount="4472" Body="&lt;p&gt;I'm posting this from the roof of a building because I can't figure out a safe way down. I've scoured the roof for anything useful, and all I found was a coping saw. I also found a 100 foot ladder, which is firmly connected to the roof (but its length is just hanging from the secured top rung). The building is unfortunately 200 feet tall. How can I use the coping saw to double the length of the ladder?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wind is really strong, so I will not consider climbing down any arrangement which could possibly be blown apart - that is, if it is possible to separate pieces of the arrangement such that they would not longer function as a 200 foot ladder, I will not climb it. The ladder is unbendable, cannot be cut by anything other than the saw, and cannot be mended once cut. It starts out as one solid piece of material. The saw cannot cut anything other than the ladder and there are no other materials around. I am capable of making precise cuts, if necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, how do I use one of these:&lt;br&gt;&#xA;$\hskip1.5in$&lt;img src=&quot;https://i.stack.imgur.com/FhN2T.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;to double the length of one of these:&lt;br&gt;&#xA;$\hskip1.5in$&lt;img src=&quot;https://i.stack.imgur.com/4PogC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(It is necessary, for the intended solution, that the coping saw blade can be released on one side, threaded through an existing cut, and reattached; a keyhole saw would probably also work, but it would be difficult to make the necessary cut. The puzzle is not so much in how one physically makes the cut, but rather what cut is made. I have managed to double the length of the pictured ladder with the pictured tool.)&lt;/p&gt;&#xA;" OwnerUserId="4765" LastEditorUserId="4765" LastEditDate="2015-07-15T11:32:00.910" LastActivityDate="2015-07-16T13:46:14.473" Title="How Can I Safely Double The Length of A Ladder?" Tags="|mathematics|geometry|story|situation|" AnswerCount="6" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17946" PostTypeId="1" AcceptedAnswerId="17951" CreationDate="2015-07-15T02:54:43.383" Score="6" ViewCount="1721" Body="&lt;p&gt;After (rather quickly) solving the puzzles found in &lt;a href=&quot;https://puzzling.stackexchange.com/questions/17907/some-four-word-puzzles-and-possible-bounty&quot;&gt;this link,&lt;/a&gt; I decided to make a few of my own (along with my little sister helping make one).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following is a quote how &lt;a href=&quot;https://puzzling.stackexchange.com/users/4269/ghosts-in-the-code&quot;&gt;ghosts_in_the_code&lt;/a&gt; explained these puzzles (because he/she/ghost did very well in describing it).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A four word puzzle (see &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15339/four-words&quot;&gt;this&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16887/another-four-words&quot;&gt;this&lt;/a&gt;) is basically a 4x4&#xA;  grid containing 4 4-letter words, and reads the same top to bottom or&#xA;  left to right. For example,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A B C D&#xA;B E F G&#xA;C F H I&#xA;D G I J&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;  &#xA;  &lt;p&gt;Here are some puzzles with hints (not necessarily in the same order):&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Sample puzzle&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;a shape&lt;br&gt;&#xA;  an identity&lt;br&gt;&#xA;  a valley&lt;br&gt;&#xA;  an audio track  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;S O N G&#xA;O V A L&#xA;N A M E&#xA;G L E N&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Puzzle 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Type of coffee or tea&lt;br&gt;&#xA;Significant role&lt;br&gt;&#xA;Whirlpool&lt;br&gt;&#xA;Document  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Puzzle 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mother Earth&lt;br&gt;&#xA;Simple&lt;br&gt;&#xA;Man-eating monster&lt;br&gt;&#xA;Manipulates an outcome  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Puzzle 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Children summer activity&lt;br&gt;&#xA;Work the fields&lt;br&gt;&#xA;Figure skating move&lt;br&gt;&#xA;Short note  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Puzzle 4 5x5&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Macho&lt;br&gt;&#xA;Italian food&lt;br&gt;&#xA;Ski resort&lt;br&gt;&#xA;Metal alloy&lt;br&gt;&#xA;Twitch  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck and have fun!&lt;/p&gt;&#xA;" OwnerUserId="14092" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-07-17T01:04:40.770" Title="A few Four Word Puzzles, and a Five Word Challenge" Tags="|word|wordplay|crosswords|word-square|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17947" PostTypeId="2" ParentId="17945" CreationDate="2015-07-15T02:55:34.867" Score="23" Body="&lt;p&gt;Lay the ladder flat and cut it in half through a plane parallel to the ground, &lt;em&gt;except&lt;/em&gt; for...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...the bottom rung, which you need to cut rotating your blade through a multiple of 360 degrees from one end of the rung to the other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This allows you to separate the two parts by...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...rotating them around the axis of the bottom rung.  They remain intertwined, but can rotate in a screw-like manner with each other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm still figuring out how to draw this one.&lt;/p&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="9718" LastEditDate="2015-07-15T03:35:57.200" LastActivityDate="2015-07-15T03:35:57.200" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="17948" PostTypeId="2" ParentId="17852" CreationDate="2015-07-15T03:45:04.447" Score="5" Body="&lt;p&gt;If there are $50$ wooden boxes and $s$ steel boxes, as xnor has already said, the probability of unlocking all $s$ steel boxes (a &quot;success&quot;) is $\frac{50}{50 + s}$.  We can also prove this by induction on $s$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose box $1$ (with key $1$) is the first steel box.  Given a random arrangement (the &quot;original permutation&quot;) of the keys to the other $50 + (s-1)$ boxes, we can extend this to include the first steel box by placing key 1 in one of the $50 + s$ boxes at random (say box $\ell$), and then placing whatever key is in box $\ell$ in box 1.  This extends a random original permutation of $50 + (s-1)$ boxes to a random permutation of $50 + s$ boxes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the probability of success for the $50 + s$ boxes?  Well, if key 1 gets placed in box 1 $\big(\frac{1}{50 + s} \ \text{probability}\big)$, we have no chance.  Otherwise $\big(\frac{50 + (s-1)}{50 + s} \ \text{probability}\big)$, we succeed as long as the original permutation was a success $\big(\frac{50}{50 + (s-1)} \ \text{probability by induction}\big)$, since whenever we open box $\ell$, we get key 1, which we can use to open box $1$, and then we can continue as before.  This gives a probability of $\frac{50+(s-1)}{50+s} \cdot \frac{50}{50 + (s-1)} = \frac{50}{50 + s}$.  Plugging in $s = 50$ gives the $1/2$ result.&lt;/p&gt;&#xA;" OwnerUserId="2722" LastActivityDate="2015-07-15T03:45:04.447" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17949" PostTypeId="2" ParentId="17946" CreationDate="2015-07-15T04:05:16.990" Score="4" Body="&lt;p&gt;Puzzle 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;     &lt;code&gt;S P A S M&lt;/code&gt;&lt;br&gt;&#xA;     &lt;code&gt;P A S T A&lt;/code&gt;&lt;br&gt;&#xA;     &lt;code&gt;A S P E N&lt;/code&gt;&lt;br&gt;&#xA;     &lt;code&gt;S T E E L&lt;/code&gt;&lt;br&gt;&#xA;     &lt;code&gt;M A N L Y&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-07-15T04:05:16.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17950" PostTypeId="2" ParentId="17946" CreationDate="2015-07-15T04:17:53.677" Score="4" Body="&lt;p&gt;Puzzle 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;     &lt;code&gt;O G R E&lt;/code&gt;&lt;br&gt;&#xA;     &lt;code&gt;G A I A&lt;/code&gt;&lt;br&gt;&#xA;     &lt;code&gt;R I G S&lt;/code&gt;&lt;br&gt;&#xA;     &lt;code&gt;E A S Y&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Puzzle 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;     &lt;code&gt;C A M P&lt;/code&gt;&lt;br&gt;&#xA;     &lt;code&gt;A X E L&lt;/code&gt;&lt;br&gt;&#xA;     &lt;code&gt;M E M O&lt;/code&gt;&lt;br&gt;&#xA;     &lt;code&gt;P L O W&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13928" LastActivityDate="2015-07-15T04:17:53.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17951" PostTypeId="2" ParentId="17946" CreationDate="2015-07-15T04:28:43.200" Score="5" Body="&lt;p&gt;&lt;strong&gt;Puzzle 1&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Type of coffee or tea:  &lt;strong&gt;ICED&lt;/strong&gt;&lt;br /&gt;&#xA; Significant role:  &lt;strong&gt;LEAD&lt;/strong&gt;&lt;br /&gt;&#xA; Whirlpool:  &lt;strong&gt;EDDY&lt;/strong&gt;&lt;br /&gt;&#xA; Document:  &lt;strong&gt;FILE&lt;/strong&gt;&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; &lt;code&gt;F I L E&lt;/code&gt;&lt;br /&gt;&#xA; &lt;code&gt;I C E D&lt;/code&gt;&lt;br /&gt;&#xA; &lt;code&gt;L E A D&lt;/code&gt;&lt;br /&gt;&#xA; &lt;code&gt;E D D Y&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Puzzle 2&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Mother Earth:  &lt;strong&gt;GAIA&lt;/strong&gt;&lt;br /&gt;&#xA; Simple:  &lt;strong&gt;EASY&lt;/strong&gt;&lt;br /&gt;&#xA; Man-eating monster:  &lt;strong&gt;OGRE&lt;/strong&gt;&lt;br /&gt;&#xA; Manipulates an outcome:  &lt;strong&gt;RIGS&lt;/strong&gt;&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; &lt;code&gt;O G R E&lt;/code&gt;&lt;br /&gt;&#xA; &lt;code&gt;G A I A&lt;/code&gt;&lt;br /&gt;&#xA; &lt;code&gt;R I G S&lt;/code&gt;&lt;br /&gt;&#xA; &lt;code&gt;E A S Y&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Puzzle 3&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Children's summer activity:  &lt;strong&gt;CAMP&lt;/strong&gt;&lt;br /&gt;&#xA; Work the fields:  &lt;strong&gt;PLOW&lt;/strong&gt;&lt;br /&gt;&#xA; Figure skating move:  &lt;strong&gt;AXEL&lt;/strong&gt;&lt;br /&gt;&#xA; Short note:  &lt;strong&gt;MEMO&lt;/strong&gt;&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; &lt;code&gt;C A M P&lt;/code&gt;&lt;br /&gt;&#xA; &lt;code&gt;A X E L&lt;/code&gt;&lt;br /&gt;&#xA; &lt;code&gt;M E M O&lt;/code&gt;&lt;br /&gt;&#xA; &lt;code&gt;P L O W&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Puzzle 4&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Macho:  &lt;strong&gt;MANLY&lt;/strong&gt;&lt;br /&gt;&#xA; Italian food:  &lt;strong&gt;PASTA&lt;/strong&gt;&lt;br /&gt;&#xA; Ski resort:  &lt;strong&gt;ASPEN&lt;/strong&gt;&lt;br /&gt;&#xA; Metal alloy:  &lt;strong&gt;STEEL&lt;/strong&gt;&lt;br /&gt;&#xA; Twitch:  &lt;strong&gt;SPASM&lt;/strong&gt;&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; &lt;code&gt;S P A S M&lt;/code&gt;&lt;br /&gt;&#xA; &lt;code&gt;P A S T A&lt;/code&gt;&lt;br /&gt;&#xA; &lt;code&gt;A S P E N&lt;/code&gt;&lt;br /&gt;&#xA; &lt;code&gt;S T E E L&lt;/code&gt;&lt;br /&gt;&#xA; &lt;code&gt;M A N L Y&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-15T17:23:58.813" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17953" PostTypeId="2" ParentId="17821" CreationDate="2015-07-15T06:26:26.643" Score="5" Body="&lt;p&gt;The answers are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Parts of the body, and the same parts with a &lt;strong&gt;y&lt;/strong&gt; ending (Why, you ask?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's start from the top.&lt;br&gt;&#xA;Why, you ask? Because it's best to start on a high.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Head&lt;/strong&gt;, which is on the top of the body.&lt;br&gt;&#xA; &lt;strong&gt;Heady&lt;/strong&gt;, meaning exhilarated or &quot;on a high&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Get into it now.&lt;br&gt;&#xA;Why, you ask? Because it's the smart idea to live by.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Brain&lt;/strong&gt;, which is inside the head (&quot;get into it&quot;).&lt;br&gt;&#xA; &lt;strong&gt;Brainy&lt;/strong&gt;, meaning smart.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Come out now and breathe the fresh air.&lt;br&gt;&#xA;Why, you ask? Because it's not right to intrude.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Nose&lt;/strong&gt;, which is back outside the head and used to breathe fresh air.&lt;br&gt;&#xA; &lt;strong&gt;Nosy&lt;/strong&gt;, meaning intrusive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Move under it just a little.&lt;br&gt;&#xA;Why, you ask? Because it would be rather rude.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Mouth&lt;/strong&gt; or &lt;strong&gt;Lip&lt;/strong&gt;.  Both are just under the nose.&lt;br&gt;&#xA; &lt;strong&gt;Mouthy&lt;/strong&gt; or &lt;strong&gt;Lippy&lt;/strong&gt;, both meaning rude.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Go through and get stuck now.&lt;br&gt;&#xA;Why, you ask? Because it's rather rough here.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Throat&lt;/strong&gt;, which is through the mouth, and where things can get stuck.&lt;br&gt;&#xA; &lt;strong&gt;Throaty&lt;/strong&gt;, meaning raspy or rough.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Come out altogether and go to the sides.&lt;br&gt;&#xA;Why, you ask? Because you need more than one to conquer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Arm&lt;/strong&gt;, which is back out the mouth and to the side of the body.&lt;br&gt;&#xA; &lt;strong&gt;Army&lt;/strong&gt;, which you need to conquer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Move to the ends.&lt;br&gt;&#xA;Why, you ask? Because it's good to have something close.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Hand&lt;/strong&gt;, which is at the end of the arm.&lt;br&gt;&#xA; &lt;strong&gt;Handy&lt;/strong&gt;, meaning close by.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Back up and down the middle till it splits.&lt;br&gt;&#xA;Why, you ask? Because those are the ones that appeal most.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Leg&lt;/strong&gt;, which is back up the arms and down the body until it splits in two.&lt;br&gt;&#xA; &lt;strong&gt;Leggy&lt;/strong&gt;, which means having attractive long legs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Drop down to the bottom and stay there now.&lt;br&gt;&#xA;Why, you ask? Because it's best to end with a game or two.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Foot&lt;/strong&gt;, down at the bottom of the legs.&lt;br&gt;&#xA; &lt;strong&gt;Footy&lt;/strong&gt;, slang for soccer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4479" LastActivityDate="2015-07-15T06:26:26.643" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17954" PostTypeId="2" ParentId="17940" CreationDate="2015-07-15T06:51:29.330" Score="5" Body="&lt;p&gt;Partial answer for Step 2.  All of these numbers can be:   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;reduced by the &lt;strong&gt;&quot;mysterious power&quot;&lt;/strong&gt; of &lt;strong&gt;1/6&lt;/strong&gt;  as follows:&lt;br&gt;&#xA;&lt;code&gt;  8303765625 =&gt; 45&#xA;  13841287201 =&gt; 49&#xA;            X =&gt; ??&#xA;       531441 =&gt; 9&#xA;    387420489 =&gt; 27&#xA;403.428793493 =&gt; 2.718281828&lt;/code&gt; or $e$&lt;br&gt;&#xA;&lt;br&gt;&#xA;These results look like &lt;a href=&quot;https://oeis.org/A069069&quot;&gt;OEIS sequence A069069&lt;/a&gt;.  That is all I have for now.    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastActivityDate="2015-07-15T06:51:29.330" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17956" PostTypeId="2" ParentId="323" CreationDate="2015-07-15T07:30:50.470" Score="7" Body="&lt;p&gt;There are 2 popular methods used to scramble a cube. The first being to apply a certain number of random moves until you are satisfied. The second method is to physical take apart the cube and assemble it again in any solvable state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second is known as random state scrambling, and is what is now used in official Rubik's Cube competitions. See more &lt;a href=&quot;https://speedsolving.com/wiki/index.php/Random_State_Scramble&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The official scrambling program can be found at &lt;a href=&quot;https://www.worldcubeassociation.org/regulations/scrambles/&quot; rel=&quot;noreferrer&quot;&gt;https://www.worldcubeassociation.org/regulations/scrambles/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is advantageous over the first method, as it has no statistical bias towards 'easy' scrambles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a speedcuber myself, I often find that when I am just applying random moves myself, I end up with the same scrambles as before. This is probably nothing statistical and is likely due to the fact that I am used to performing fast algorithms over and over, but is the main reason why I use generated scrambles myself.&lt;/p&gt;&#xA;" OwnerUserId="14096" LastEditorUserId="14096" LastEditDate="2015-07-15T13:06:25.250" LastActivityDate="2015-07-15T13:06:25.250" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17958" PostTypeId="2" ParentId="19" CreationDate="2015-07-15T08:58:17.113" Score="1" Body="&lt;p&gt;I have long been frustrated by this, and found a very simple fix! As mentioned in other responses, this occurs since the centers are off by 90 degrees.&lt;br&gt;&#xA;Instead of doing an algorithm, I merely do an M slice turn (turn the middle vertical 90 degrees) then solve the cross edges again around the new centers. This will work every time and requires no new algorithms!&lt;/p&gt;&#xA;" OwnerUserId="14096" LastEditorUserId="27236" LastEditDate="2016-07-29T19:06:54.767" LastActivityDate="2016-07-29T19:06:54.767" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17959" PostTypeId="2" ParentId="17937" CreationDate="2015-07-15T09:16:26.727" Score="0" Body="&lt;p&gt;My attempt :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We don't have to take in count the wooden boxes' keys. Each time the prankster pick a steel box's key, there is $50$% chances that he put it into a wooden box. Since there's $50$ keys, the odds that all the steel boxes' keys are in wooden boxes is $(\frac{1}{2})^{50}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bonus :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $s = n-k$ &lt;br /&gt;&#xA; $(\frac{k}{k+s})^s$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13271" LastEditorUserId="13271" LastEditDate="2015-07-15T09:23:34.910" LastActivityDate="2015-07-15T09:23:34.910" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17960" PostTypeId="2" ParentId="17940" CreationDate="2015-07-15T09:16:59.520" Score="6" Body="&lt;p&gt;Partial answer for Step 1 and its link to Step 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The missing thing is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the letter F (while all the other letters are there somewhere).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, we all know that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; F is the 6th letter (and 6 is the first perfect number, hence the &quot;impeccable quality&quot; about it), so that number is the key to determine the &quot;hidden power&quot; in step 2 as in Len's answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More to follow in a while, hopefully.&lt;/p&gt;&#xA;" OwnerUserId="5990" LastEditorUserId="5990" LastEditDate="2015-07-15T09:58:36.787" LastActivityDate="2015-07-15T09:58:36.787" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17961" PostTypeId="2" ParentId="17945" CreationDate="2015-07-15T10:51:44.727" Score="6" Body="&lt;p&gt;Original ladder:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;----------&#xA;|  |  |  |&#xA;|  |  |  |&#xA;----------&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;With an assumption:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ... Cut all the rungs except the bottom one.  Rotate one leg 180° about the uncut rung.  Assumption: the rungs are not so tightly fitted that they won't rotate using the leverage of a ladder leg.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Similarly, with perhaps a less strict assumption:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ... Cut all rungs except the bottom one.  Use the handle of the coping saw to drive the bottom rung out of one leg, rotate this leg and half-rungs 180° and reassemble.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Either way you end up with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;pre&gt; ----------&#xA; |  |  |  |&#xA;          |  |  |  |&#xA;          ----------&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8765" LastEditorUserId="4711" LastEditDate="2015-07-16T07:26:23.427" LastActivityDate="2015-07-16T07:26:23.427" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="17962" PostTypeId="1" AcceptedAnswerId="17963" CreationDate="2015-07-15T11:38:23.653" Score="8" ViewCount="621" Body="&lt;p&gt;If every real number is coloured either black or white, prove that there exist distinct real numbers $a,b,c$ all of the same colour such that $a-b=b-c$.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-07-15T11:51:21.797" LastActivityDate="2015-07-15T23:09:07.420" Title="Colouring the reals" Tags="|logical-deduction|combinatorics|" AnswerCount="5" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17963" PostTypeId="2" ParentId="17962" CreationDate="2015-07-15T12:01:47.373" Score="9" Body="&lt;p&gt;WLOG, 2 and 4 are both black. (This is WLOG because we can scale and translate the real line freely, and can also swap the colors.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If 0 is black, then&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(a, b, c) = (4, 2, 0)$ are all black.&lt;br&gt;&#xA; This satisfies $4-2=2-0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If 3 is black, then&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(a, b, c) = (4, 3, 2)$ are all black.&lt;br&gt;&#xA; This satisfies $4-3=3-2$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If 6 is black, then&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(a, b, c) = (6, 4, 2)$ are all black.&lt;br&gt;&#xA; This satisfies $6-4=4-2$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If none of ${0,3,6}$ is black, then&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(a, b, c) = (6, 3, 0)$ are all white.&lt;br&gt;&#xA; This satisfies $6-3=3-0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="10399" LastEditDate="2015-07-15T19:00:53.170" LastActivityDate="2015-07-15T19:00:53.170" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17964" PostTypeId="2" ParentId="16760" CreationDate="2015-07-15T12:11:59.057" Score="0" Body="&lt;p&gt;I found this post while confirming something I researched several years back.  I write puzzles for my board game shop, and upon seeing the Mayfair game titled &quot;Oceania&quot; it all came flooding back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I once hypothesized that there may be a word that would trigger a &quot;spelling bee haiku&quot;.  I.e. a 5-syllable, 7-letter word (no 'w') that if given in a spelling bee would result in the contestent uttering a haiku.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oceania&#xA;o-c-e-a-n-i-a&#xA;Oceania&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found oxyopia &amp;amp; Oceania.  Not sure if you'll accept proper nouns.  Let me know if any others are found!&lt;/p&gt;&#xA;" OwnerUserId="14103" LastActivityDate="2015-07-15T12:11:59.057" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17965" PostTypeId="2" ParentId="17277" CreationDate="2015-07-15T12:24:25.647" Score="0" Body="&lt;p&gt;Check whether each of the two traits that meet at an accepted square meet others with the same results (eliminated/accepted). Say, if John has concluded that Alice is 7, which he marked with an accepted square, and accepted or eliminated some of the other traits/variables that meet with one of &quot;Alice&quot; or &quot;7&quot; (eye color etc.), he must not accept for 7 what he eliminated for Alice (and vice-versa). But just leaving a square empty without accepting or eliminating it shouldn't count as an error.&lt;/p&gt;&#xA;" OwnerUserId="14016" LastEditorDisplayName="user20" LastEditDate="2015-07-15T19:20:41.290" LastActivityDate="2015-07-15T19:20:41.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17966" PostTypeId="2" ParentId="17962" CreationDate="2015-07-15T13:34:15.580" Score="4" Body="&lt;p&gt;Not the first answer, but a bit of a different spin:&lt;/p&gt;&#xA;&lt;p&gt;For it to not happen,&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;3 numbers with the same difference between them must not have the same color (in every sequence of 3 consecutive numbers, one color should be seen twice and the other once).&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Numbers can't be colored abab... all the time, so we have to include aas on occasion. Starting with an aa in the middle, we get axxbaabxxa, then axbbaabbxa and finally aabbaabbaa (no escape). If it's impossible even for those countable painted points, then it's true for all real numbers by extension.&lt;/p&gt;&#xA;&lt;p&gt;Edit: Before downvoting it, note that the unit difference can be any real number and two real numbers don't have to be located &amp;quot;next to&amp;quot; each other (so to speak) to be consecutive or for the question to be solved.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-15T18:21:47.737" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17967" PostTypeId="2" ParentId="17936" CreationDate="2015-07-15T14:11:55.807" Score="1" Body="&lt;h2&gt;Partial Answer&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Since the hint says that &quot;groups of twos and threes&quot; are relevant, I hunted through the list to find some. &lt;code&gt;2408&lt;/code&gt; and &lt;code&gt;327&lt;/code&gt; show up multiple times. Splitting the string up such that &lt;code&gt;24&lt;/code&gt;, &lt;code&gt;08&lt;/code&gt;, and &lt;code&gt;327&lt;/code&gt; are all pulled out by themselves, we get:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;24 08 327 24 0304 08 327 24 0326 08 24 0304120727 327 24 08 24 0712 327 24 08 2308&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If we then split the &lt;code&gt;0#&lt;/code&gt; strings up as sets of twos, we get the resulting string:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;24 08 327 24 03 04 08 327 24 03 26 08 24 03 04 12 07 27 327 24 08 24 07 12 327 24 08 23 08&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This doesn't appear to be ascii or alphabetical characters, since neither really fit the pattern. I'm a bit concerned with the &lt;code&gt;07 27 327&lt;/code&gt; since that's the only location that &lt;code&gt;27&lt;/code&gt; shows up exclusively from &lt;code&gt;327&lt;/code&gt;. From this point, I'm not sure where to go, but it's a start, at least.&lt;/p&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-07-15T14:11:55.807" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17968" PostTypeId="2" ParentId="17936" CreationDate="2015-07-15T15:06:42.187" Score="6" Body="&lt;p&gt;24 08 327&lt;/p&gt;&#xA;&#xA;&lt;p&gt;24 03 04 08 327&lt;/p&gt;&#xA;&#xA;&lt;p&gt;24 03 26 08&lt;/p&gt;&#xA;&#xA;&lt;p&gt;24 03 04 12 07 27 327&lt;/p&gt;&#xA;&#xA;&lt;p&gt;24 08 24 07 12 327&lt;/p&gt;&#xA;&#xA;&lt;p&gt;24 08 23 08&lt;/p&gt;&#xA;&#xA;&lt;p&gt;looks like a patern all lines starts with 24 and end up to 327 or 08&lt;/p&gt;&#xA;" OwnerUserId="14108" LastActivityDate="2015-07-15T15:06:42.187" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17969" PostTypeId="2" ParentId="17937" CreationDate="2015-07-15T15:33:07.347" Score="6" Body="&lt;p&gt;This solution is for revenge! Warning: it's long and assumes mathematical maturity. I don't think anyone will even read it all. I seriously hope someone else posts an elegant solution. But I had to post this, because this problem is Putnam 2013 B5, which I failed to solve during the test despite working on it for an hour. I never looked at a solution, so this post is my way of getting revenge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Call a set $C$ of stone boxes &lt;em&gt;closed&lt;/em&gt; if all the keys for boxes in $C$ are locked inside boxes in $C$. A box is forever unopenable iff it lies in a closed set. We want to find the probability that there exists at least one nonempty closed set. Letting $S$ denote the set of stone boxes, we want to calculate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Pr\left[\bigcup_{C\subseteq S, C\neq \emptyset} C\text{ is closed}\right].$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the principle of inclusion-exclusion, this equals&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{F} \Pr[\text{All sets in $F$ are closed}]\cdot(-1)^{|F|-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $F$ ranges over all nonempty collections of nonempty subsets of $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first this sum looks hopeless: it is a sum over $2^{2^{|S|}-1}$ terms, each of which has a complicated probability. But we can force almost all of the terms to cancel, using a clever bijection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $F$ be a collection of nonempty subsets of $S$. Call $F$ &lt;em&gt;weird&lt;/em&gt; if $F$ contains a pair of non-nested sets. Define the &lt;em&gt;smallest pair&lt;/em&gt; of a weird set $F$ to be the pair of sets $C$, $D$ in $F$ minimizing $C \cup D$, breaking ties in some arbitrary but consistent way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define the &lt;em&gt;twin&lt;/em&gt; of a weird set $F$ as follows. Let $C$, $D$ be $F$'s smallest pair. Then the twin of $F$ is obtained by taking $F$ and toggling whether it contains $C \cup D$. With mild casework, one can show that the twin of the twin of $F$ is always $F$ itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the twin relation partitions the weird sets into twins. Let $F$ and $F'$ be twins. Let $C$, $D$ be $F$'s smallest pair. It follows from the definition of closed that if $C$ and $D$ are closed, then so is $C\cup D$. Therefore, all sets in $F$ are closed iff all sets in $F'$ are closed. Also, it follows from the definition of twin that one of $F$, $F'$ is exactly one element larger than the other. Therefore, the terms for $F$ and $F'$ cancel in the gigantic sum above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This cancellation cancels out all terms in the gigantic sum using a weird set $F$. What is left? The sets $F$ which are not weird, i.e. where all pairs of sets in $F$ are nested. In other words, the sum now ranges over nonempty collections of nested nonempty subsets of $S$. We rewrite the sum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{S \supseteq C_0 \supset C_1 \supset \dots \supset C_k \neq \emptyset} \Pr[\text{$C_0, C_1, \dots, C_k$ are all closed}]\cdot(-1)^{k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $C_0 \supset C_1 \supset \dots \supset C_k$ ranges over all &lt;em&gt;nonempty&lt;/em&gt; sequences of nested nonempty subsets of $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define the &lt;em&gt;weight&lt;/em&gt; of a set $C\subseteq S$ as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ weight(C) := \sum_{C \supset C_1 \supset \dots \supset C_k \neq \emptyset} \Pr[\text{$C_1, \dots, C_k$ are all closed | $C$ is closed}]\cdot (-1)^k$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I claim that the weight of a set $C$ equals $1$ if $|C|=1$, and $0$ otherwise. The proof is by induction on the size of $C$. The base case, $|C|=1$, is a sum over one term which equals $1$. For the inductive step, assume the statement is true for all $|C|&amp;lt;n$, and consider a set $C$ of size $n$. In the sum in the definition of the weight above, we isolate the term where $C_1, ..., C_k$ is the empty sequence. This term adds $1$ to the overall sum. The rest of the terms all contain a set $C_1$. We rewrite the sum thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ weight(C) = 1 + \sum_{C\supset C_1 \neq \emptyset} \sum_{C_1 \supset \dots \supset C_k \neq \emptyset} \Pr[\text{$C_1, \dots, C_k$ are all closed | $C$ is closed}]\cdot (-1)^k $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= 1 + \sum_{C\supset C_1\neq \emptyset} \Pr[\text{$C_1$ is closed | $C$ is closed}]\sum_{C_1 \supset \dots \supset C_k \neq \emptyset} \Pr[\text{$C_2, \dots, C_k$ are all closed | $C_1$ is closed}]\cdot (-1)^k$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = 1 + \sum_{C\supset C_1} \Pr[\text{$C_1$ closed | $C$ closed}] \cdot (-weight(C_1)) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the induction hypothesis, all $weight(C_1)$ terms are zero except the ones where $|C_1|=1$. Thus, the sum equals&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = 1 + \sum_{\{s\}\subset C} \Pr[\text{$\{s\}$ closed | $C$ closed}] \cdot -1 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = 1 + \sum_{\{s\}\subset C} -\frac{1}{|C|} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is zero. This completes the inductive step, finishing our calculation of the weight function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what was that sum we wanted to evaluate again? Here it is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{S \supseteq C_0 \supset C_1 \supset \dots \supset C_k \neq \emptyset} \Pr[\text{$C_0, C_1, \dots, C_k$ are all closed}]\cdot(-1)^{k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sum ranges over all nonempty sequences of nested subsets of $S$. Since the sequence $C_0, C_1, \dots, C_k$ must be nonempty, it must always contain a set $C_0$. We rewrite the sum, in a similar way to our proof of the inductive step of the weight function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{S \supset C_0 \neq \emptyset} \sum_{C_0 \supset C_1 \supset \dots \supset C_k \neq \emptyset} \Pr[\text{$C_0, C_1, \dots, C_k$ are all closed}]\cdot(-1)^{k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \sum_{S \supset C_0 \neq \emptyset} \Pr[\text{$C_0$ is closed}] \sum_{C_0 \supset C_1 \supset \dots \supset C_k \neq \emptyset} \Pr[\text{$C_1, \dots, C_k$ all closed | $C_0$ closed}]\cdot(-1)^{k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \sum_{S \supset C_0 \neq \emptyset} \Pr[\text{$C_0$ is closed}]\cdot weight(C_0) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By our calculation of the weight function, all terms are zero except those where $|C_0| = 1$. Thus, this equals&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{\{s\} \subseteq S} \Pr[\text{$\{s\}$ is closed}] $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, finally, is an easy sum to calculate. Let $n$ be the total number of boxes. Then the sum equals&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{\{s\} \subseteq S} \frac{1}{n} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \frac{|S|}{n} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the probability that there exists at least one closed set is $|S|/n$. Therefore, putting $k$ as the number of wooden boxes, the probability that all boxes are openable is $k/n$. In conclusion, this variant has the same answer as the original question.&lt;/p&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2015-07-15T15:33:07.347" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17970" PostTypeId="1" AcceptedAnswerId="17971" CreationDate="2015-07-15T15:34:20.710" Score="14" ViewCount="859" Body="&lt;p&gt;You need to find a continuous path within the US that passes through all 48 contiguous states, visiting each one exactly once. If you start in Delaware, which state do you end in?&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-07-15T20:05:08.747" LastActivityDate="2015-07-15T20:05:08.747" Title="Touring the United States" Tags="|geography|graph-theory|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17971" PostTypeId="2" ParentId="17970" CreationDate="2015-07-15T15:38:47.550" Score="15" Body="&lt;p&gt;The state you have to end in will be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Maine, since it's the only state that only borders one other state, and therefore if you enter it from one state you can't leave it via another.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's a possible trip you can take:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/wcNUj.png&quot; alt=&quot;Map&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-07-15T15:38:47.550" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="17973" PostTypeId="1" AcceptedAnswerId="18205" CreationDate="2015-07-15T15:44:36.610" Score="22" ViewCount="2304" Body="&lt;p&gt;If a phrase conforms to a certain rule, I call it a Surpassing Phrase™.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the examples below to find the rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/21trt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Some details to save you time:&lt;/strong&gt;&lt;br&gt;&#xA;  1. Font doesn't matter.&lt;br&gt;&#xA;  2. The number of words in the phrase can vary, but I chose short 2-word phrases to keep the list concise.&lt;br&gt;&#xA;  3. Case doesn't matter.&lt;br&gt;&#xA;  4. Latest EDIT: &lt;a href=&quot;/questions/tagged/calculation-puzzle&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;calculation-puzzle&amp;#39;&quot; rel=&quot;tag&quot;&gt;calculation-puzzle&lt;/a&gt; tag added&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is a Surpassing Phrase™?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-07-16T15:57:10.583" LastActivityDate="2015-07-21T12:07:37.743" Title="What is a Surpassing Phrase™?" Tags="|calculation-puzzle|pattern|english|" AnswerCount="5" CommentCount="21" ContentLicense="CC BY-SA 3.0" />
  <row Id="17974" PostTypeId="2" ParentId="17973" CreationDate="2015-07-15T16:01:26.470" Score="-1" Body="&lt;p&gt;My guess is that a Surpassing Phrase is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A phrase with relative adjectives that describe nouns using the least letters possible as well as being the most specific as possible&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13489" LastActivityDate="2015-07-15T16:01:26.470" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17975" PostTypeId="2" ParentId="3106" CreationDate="2015-07-15T16:33:18.967" Score="3" Body="&lt;p&gt;Let's look at the two scenarios:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;They know enough for you to figure it out&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There's actually a little bit more information that needs to be provided to give a conclusive answer to this scenario. If &lt;em&gt;each&lt;/em&gt; one of them knows enough information that, if you knew it, you could label them, you only need to ask three or four questions to get this. Otherwise, it gets a little bit more complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If each one knows enough information, that means that there has been at least one question to which the joker has answered truthfully and one to which they have answered falsely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, the first two or three questions are used to figure out who the knave or knight is by asking them all the same question (such as @kaine's &quot;Am I holding a ball right now?&quot;). If the first two agree you don't need to ask the third - if they both lie you've found the knave and the joker, and if they both tell the truth you've found the knight and the joker. If they don't agree you have no information about who they are, so you do need to ask the third. Call the person who disagreed with the other two A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, knowing that A is the knight (or the knave), you just need to identify one of the other two. Simply ask A:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Has B ever answered a question truthfully?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If they say &quot;yes&quot; (or &quot;no&quot; from the knave), then B is the joker and C is the knave (or knight). Otherwise, C is the joker and B is the knave (or knight).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what if their individual knowledge is not enough but their &lt;em&gt;combined&lt;/em&gt; knowledge is? There's actually only one extra case here - the joker knows but the other two don't. This is because there's no way for the knight to know but the other two not to - if the knight knows the joker must have both told the truth and lied, so the joker will be remember both having told the truth and having lied, and the knave will have also heard the joker's answers and know as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how do we handle the case where the joker knows, but the knave and knight don't? For example, suppose the joker is almost always identical to the knave, but there was one question asked of the knight or knave to which the joker would have told the truth. Because the joker was not asked the question, the knight and the knave never heard the joker answer truthfully, so they do not know the difference between the knave and the joker. In this case, if the knight is asked about the other two ever responding truthfully, the answer will be &quot;no&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the knight can't tell us that either of the other two has told the truth, we need to ask a question that can distinguish between the joker and the knave. In this case, we ask B:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Think back on the questions that have been asked and think about what you did or would have said to each. For the first question for which you did or would have answered truthfully, what did or would you have said?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If B is the knave, there is no such question - they always would have lied, so you'll simply get a sulky stare. If you ask the joker you'll get a response, and it doesn't matter whether it is 'yes' or 'no'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best strategy is not to ask the knight about the other two at all - the joker will always be able to answer the second question. So we ask B the second question, and B will answer if and only if they are the joker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summary:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ask two of them if you're holding a ball:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they disagree, ask the third if you're holding a ball&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Call the two who agreed B and C.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If B and C both lied, ask B the joker-identifying question&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If B sulks, A is the knight, B is the knave, and C is the joker&lt;/li&gt;&#xA;&lt;li&gt;If B answers, A is the knight, B is the joker, and C is the knave&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;If B and C both told the truth, ask B the joker-identifying question (changed to asking if they ever had or would have lied)&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If B sulks, A is the knave, B is the knight, and C is the joker&lt;/li&gt;&#xA;&lt;li&gt;If B answers, A is the knave, B is the joker, and C is the knight&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;They don't know enough for you to figure it out&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If you had a full transcript of every question they've answered so far with their answers and still wouldn't be able to tell who the joker is, we're out of luck. As the others mentioned, there's no way to predict what the joker's answering strategy is. For example, suppose the only question for which the joker will lie is &quot;is $p!-1$ a prime number?&quot; for some large prime number $p$. There are an infinite number of primes, so there are an infinite number of potential questions to check.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So without any constraints on the joker's answering strategy, there is no way to guarantee finding the question that will reveal the joker. It's possible that the joker has gone their entire life only ever telling the truth, and nobody will ever ask them one of the questions to which they will lie about. They might go their entire life thinking they are a knight, never realizing that they do have the power to lie. Mourn for this poor joker, forced to live a life that does not allow them to live up to their potential.&lt;/p&gt;&#xA;" OwnerUserId="1955" LastActivityDate="2015-07-15T16:33:18.967" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17976" PostTypeId="2" ParentId="17945" CreationDate="2015-07-15T17:14:40.300" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. Cut off all of the rungs, save two.&#xA; 2. Connect the two side rails with a dovetail joint, forming one 200' rail.&#xA; 3. Use two rungs to make a lap joint for more strength.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One 200' rail:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;-------------------|dovetail|---------------------&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And lots of rungs left over:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;|||||||||||||||&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Any force on the dovetail joint should pull it tighter, and the rungs&#xA; should keep it from splitting.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14113" LastEditorUserId="10037" LastEditDate="2015-07-15T17:22:55.937" LastActivityDate="2015-07-15T17:22:55.937" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17977" PostTypeId="2" ParentId="17940" CreationDate="2015-07-15T17:22:46.847" Score="6" Body="&lt;p&gt;I am pretty sure I understand part 2.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The hint mentions numbers 167 and 39, saying 256 is key, but it used to be 128. This refers to ASCII codes, and 167 and 39 are a top-aligned circle and the apostrophe, which look like degree and minute symbols. So we have 45 49 X 9 27 E, this being degrees and minutes. Clearly the E stands for &quot;east&quot; as in longitude, so X stands for latitude and can be either N or S. I assume N is the right answer, as the location on the south hemisphere is in the water, while the northern one is near Milan, Italy, and you also generally say &quot;north and south&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8868" LastEditorUserId="8868" LastEditDate="2015-07-15T17:28:32.843" LastActivityDate="2015-07-15T17:28:32.843" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17978" PostTypeId="2" ParentId="17945" CreationDate="2015-07-15T17:23:51.047" Score="14" Body="&lt;p&gt;While the answers by 2012rcampion and Chris H are quite creative, I find their solutions to be structurally deficient. In 2012rcampion's solution, the weight of the second half of the ladder rests entirely on the last rung, thereby making it prone to failure under the weight of a human. Meanwhile, in Chris H's solution, the half rungs seem incapable of being able to bear the weight of a human. I have come up with a third solution that is capable of bearing a greater amount of load.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Beginning in the same manner as 2012rcampion, lay the ladder on the ground and split the ladder into two identical half-width ladders by cutting along a plane parallel to the ground. Following that, carve out &lt;a href=&quot;https://en.wikipedia.org/wiki/Dovetail_joint#Sliding_dovetail&quot; rel=&quot;nofollow noreferrer&quot;&gt;sliding dovetail joints&lt;/a&gt; in the top end of the rails on one half ladder and in the bottom end of the rails on the other half of the ladder. The image below illustrates the cross section of this joint.  &lt;br&gt;&lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fx0B6.jpg&quot; alt=&quot;enter image description here&quot;&gt;  &lt;br&gt;&lt;br&gt;This is structurally stronger as the load borne by the second ladder is distributed along the rails. While one may argue that this is weak as well, depending on the thinner width of the rails and rungs, but this would nonetheless reduce the chances of failure by a great extent.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-07-16T11:13:43.853" LastActivityDate="2015-07-16T11:13:43.853" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17980" PostTypeId="1" CreationDate="2015-07-15T17:56:22.203" Score="2" ViewCount="251" Body="&lt;p&gt;Can you guess the key to these different base conversions? Here are some examples. See if you can determine the rule and figure out X,Y, and Z&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Input      Conversion1 Output    Conversion2 Output    Conversion3 Output&#xA;&#xA;9          111                   9                     9&#xA;22         320                   42                    34&#xA;298        22120                 EI                    EI&#xA;500        40310                 2152                  358&#xA;13345      2431001               2C30                  7881&#xA;223332     X                     Y                     Z&#xA;1010111    270252321             33352235              3312212333&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Feel free to give me test values to help you determine. The order in which they are solved is irrelevant but I would say that 1 might be the easiest followed by 3.&lt;br&gt;&#xA;EDIT:&#xA;I think that it's worth clarifying, that each column represents a base conversion rule, not a base itself. My original wording did not make this clear. The important distinction here is that not every number within a column are necessarily in the same base.&lt;/p&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="13970" LastEditDate="2015-07-15T19:15:37.270" LastActivityDate="2015-07-15T19:15:37.270" Title="Thank you Based Base" Tags="|mathematics|pattern|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17981" PostTypeId="2" ParentId="17973" CreationDate="2015-07-15T18:15:02.293" Score="5" Body="&lt;p&gt;I think a Surpassing Phrase™ is one where &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you can take away a few letters from both words to create a new word and still be left with a proper word pair. The newly created word may need to be anagrammed but the original words will surpass our expectations by remaining valid words even after losing some letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's how my pattern fits the phrases. The letters in brackets are the ones removed from the original word and the word in italics is the newly created word&lt;/p&gt;&#xA;&#xA;&lt;p&gt;superb subway&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; super(b) sway(ub) &lt;em&gt;bub&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;online fiends&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; line(on) finds(e) &lt;em&gt;one&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;write runes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; writ(e) run(es) &lt;em&gt;see&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fair survey&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; far(i) sure(vy) &lt;em&gt;ivy&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;bird beak&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;bid(r) be(ak) &lt;em&gt;ark&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;turnip fields&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; urn(tip) field(s) &lt;em&gt;spit&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fear cobras&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; far(e) cobs(ra) &lt;em&gt;era&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;pithy lingoes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;pith(y) lingo(es) &lt;em&gt;yes&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;kidnap kings&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; kid(nap) king(s) &lt;em&gt;snap&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;sound echoes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; son(ud) echo(es) &lt;em&gt;dues&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;icky molerat&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ick(y) mole(rat) &lt;em&gt;tray&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I haven't been able to fit it to each of the phrases you've listed, but the pattern is appearing all too frequently to ignore altogether.&lt;/p&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-07-15T20:11:44.900" LastActivityDate="2015-07-15T20:11:44.900" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17982" PostTypeId="2" ParentId="17980" CreationDate="2015-07-15T18:23:06.067" Score="3" Body="&lt;p&gt;Partial answer: Base1 is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the factoradic base, where $d_nd_{n-1}\cdots d_1$ represents the number $$d_n\cdot n!+d_{n-1}\cdot (n-1)!+\dots+d_1\cdot 1!$$ and $0\le d_n\le n$. For example, $9=3!+2!+1!$, and $22=3\cdot3! + 2\cdot 2!$. The value of $X$ is 54210200.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-07-15T18:23:06.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17983" PostTypeId="1" AcceptedAnswerId="17996" CreationDate="2015-07-15T18:51:51.327" Score="9" ViewCount="1373" Body="&lt;p&gt;Can you solve the following cryptic clue?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...... (6)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Think Rebus.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Look at the number of dots. Think of a sport. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Edited title to lead you closer ;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2425" LastEditorUserId="43953" LastEditDate="2018-06-01T08:35:46.633" LastActivityDate="2018-06-01T08:35:46.633" Title="Cryptic clue: Are you bowled over by this beauty?" Tags="|knowledge|rebus|crossword-clues|" AnswerCount="6" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17984" PostTypeId="2" ParentId="17983" CreationDate="2015-07-15T18:53:32.183" Score="2" Body="&lt;p&gt;The best answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; POINTS,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DOTTED and BLANKS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;also fit the clue perfectly well. IMO the question is too broad.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-07-15T18:57:03.633" LastActivityDate="2015-07-15T18:57:03.633" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17985" PostTypeId="2" ParentId="17983" CreationDate="2015-07-15T19:00:22.250" Score="2" Body="&lt;p&gt;Could it be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Six_Point_Group&quot; rel=&quot;nofollow&quot;&gt;The Six Point Group&lt;/a&gt;&lt;br&gt;&lt;br&gt; It was a British feminist campaign founded in 1921. The beauty in title could point to the fact that it was a women's movement.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-07-15T19:00:22.250" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17986" PostTypeId="2" ParentId="17983" CreationDate="2015-07-15T19:04:46.563" Score="4" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tulips &lt;br&gt;&#xA;&quot;Two ellipsis&quot; sounds like tulips. Tulips are considered by many to be beautiful [citation needed], and also has six characters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13764" LastEditorUserId="5373" LastEditDate="2015-07-15T19:47:04.840" LastActivityDate="2015-07-15T19:47:04.840" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17987" PostTypeId="1" AcceptedAnswerId="17990" CreationDate="2015-07-15T19:07:37.680" Score="47" ViewCount="7241" Body="&lt;p&gt;Although I've never been interviewed,&lt;br&gt;&#xA;I'm famous all over the world.&lt;br&gt;&#xA;People love to watch me eat,&lt;br&gt;&#xA;But some disapprove of my habits.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't mind working for my food,&lt;br&gt;&#xA;And I like to eat on the run.&lt;br&gt;&#xA;Picky, I eat only 3 foods,&lt;br&gt;&#xA;Not counting my desserts.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have very strange diet goals:&lt;br&gt;&#xA;Eat 244 to 260 bites per meal,&lt;br&gt;&#xA;Plus 2 bites of dessert per meal-&lt;br&gt;&#xA;1 bite of one and 1 bite of another.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Timing my bites is important,&lt;br&gt;&#xA;And when done wisely,&lt;br&gt;&#xA;I can reach my full 260 bites,&lt;br&gt;&#xA;And then I will be very happy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who am I?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-07-16T14:06:40.407" LastActivityDate="2015-08-05T17:51:17.073" Title="Famous eater. Who am I?" Tags="|riddle|" AnswerCount="4" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="17988" PostTypeId="2" ParentId="17983" CreationDate="2015-07-15T19:12:29.540" Score="4" Body="&lt;p&gt;How about:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cerium.  The six periods refers to a &lt;a href=&quot;https://en.wikipedia.org/wiki/Period_6_element&quot; rel=&quot;nofollow&quot;&gt;Period 6 element&lt;/a&gt;.  There are four Period 6 elements that are six letters long: barium, cerium, osmium, and erbium.  Cerium's name comes from Ceres, the Roman goddess of agriculture, and presumably all Roman goddesses were beauties.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2015-07-15T19:12:29.540" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17989" PostTypeId="2" ParentId="17987" CreationDate="2015-07-15T19:43:34.050" Score="6" Body="&lt;p&gt;Are you talking about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Snake game from old black and white phones?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Although I've never been interviewed,&lt;br&gt;&#xA;I'm famous all over the world.&lt;br&gt;&#xA;People love to watch me eat,&lt;br&gt;&#xA;But some disapprove of my habits.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The game is well-known to almost everyone who's used a mobile phone in the early 2000s. It was a popular, and simple yet addictive game, which made parents disapprove of it (?),&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't mind working for my food,&lt;br&gt;&#xA;And I like to eat on the run.&lt;br&gt;&#xA;Picky, I eat only 3 foods,&lt;br&gt;&#xA;Not counting my desserts.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the game, the snake had to move around the screen to eat its food. And the food it could eat was of a fixed size, in terms of pixels(&lt;em&gt;can't think of 3 foods&lt;/em&gt;). The desserts could point to the bonus items that propped up in some game modes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have very strange diet goals:&lt;br&gt;&#xA;Eat 240 to 260 bites per meal,&lt;br&gt;&#xA;Plus 2 bites of dessert per meal-&lt;br&gt;&#xA;1 bite of one and 1 bite of another.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The aim of the game was to eat as much as possible and grow to be as long as possible, which would be a strange eating routine. (&lt;em&gt;Not sure about 240 or 260&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Timing my bites is important,&lt;br&gt;&#xA;And when done wisely,&lt;br&gt;&#xA;I can reach my full 260 bites,&lt;br&gt;&#xA;And then I will be very happy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eating the food and turning away before crashing anywhere was key to succeeding in the game. When played well, the snake would grow to its maximum length till the point where it occupies the entire screen and cannot grow anymore.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-07-15T19:43:34.050" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17990" PostTypeId="2" ParentId="17987" CreationDate="2015-07-15T19:49:03.823" Score="68" Body="&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pac-Man&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Although I've never been interviewed,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; How can you interview a video game character?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm famous all over the world.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everyone's heard of Pac-Man.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;People love to watch me eat,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People love to play Pac-Man and watch him eat, since eating increases their score.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But some disapprove of my habits.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's not healthy to be constantly eating.  Also, as JLee said, the ghosts don't like it when Pac-Man eats power pellets, because then they can be eaten.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't mind working for my food,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You have to run around all over the maze to eat all the dots, and when you want to eat ghosts they are always running away from you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I like to eat on the run.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pac-Man almost always has to be moving in order to eat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Picky, I eat only 3 foods,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Power pellets, dots, and ghosts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not counting my desserts.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The desserts are the Cherry, Strawberry, Orange, Apple, Melon, Galaxian, Bell, and Key.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have very strange diet goals:&#xA;Eat 244 to 260 bites per meal,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are 240 dots per board, 4 power pellets, and the ghosts can be eaten from 0 to 16 times.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Plus 2 bites of dessert per meal-&#xA;1 bite of one and 1 bite of another.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A fruit/key/galaxian/bell appears twice per board.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Timing my bites is important,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You don't want to eat a power pellet when the ghosts are far away, since you only have a limited amount of time to eat ghosts.  If you chase a ghost too closely and it turns back to its normal color and then you touch it, you will die.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And when done wisely,&#xA;I can reach my full 260 bites,&#xA;And then I will be very happy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You will get the max score per board if you eat all the ghosts while they are blue (and all the fruit/key/galaxian/bell).  You will also be very happy if you do this on the first board because you will surpass 10,000 points and you will get an extra life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4450" LastEditDate="2015-07-18T06:12:36.850" LastActivityDate="2015-07-18T06:12:36.850" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17991" PostTypeId="2" ParentId="17987" CreationDate="2015-07-15T19:49:19.447" Score="32" Body="&lt;p&gt;Are you:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pacman&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Although I've never been interviewed,&lt;br&gt;&#xA;I'm famous all over the world.&lt;br&gt;&#xA;People love to watch me eat,&lt;br&gt;&#xA;But some disapprove of my habits.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pacman is a famous game that many people played. The ghosts in the game are the one who disapprove his habits&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I don't mind working for my food,&lt;br&gt;&#xA;And I like to eat on the run.&lt;br&gt;&#xA;Picky, I eat only 3 foods,&lt;br&gt;&#xA;Not counting my desserts&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pacman move while eating the dots. He eats 3 types of food: dots, ghosts, &quot;big&quot; dots and the dessert is the fruit which appear in each level. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I have very strange diet goals:&lt;br&gt;&#xA;Eat 240 to 260 bites per meal,&lt;br&gt;&#xA;Plus 2 bites of dessert per meal-&lt;br&gt;&#xA;1 bite of one and 1 bite of another.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is 240 small dots by level. Counting the ghost, you can take 260 bites. (240 + 4 big dots + 4*4 ghosts = 260) Not too sure for the desert.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Timing my bites is important,&lt;br&gt;&#xA;And when done wisely,&lt;br&gt;&#xA;I can reach my full 260 bites,&lt;br&gt;&#xA;And then I will be very happy.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You need the good timing to eat the ghosts. The number of dots (244) + the number of ghosts (4) * the number of big dots (4) = 260&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5359" LastEditorDisplayName="user20" LastEditDate="2015-08-05T17:51:17.073" LastActivityDate="2015-08-05T17:51:17.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17993" PostTypeId="2" ParentId="17983" CreationDate="2015-07-15T20:11:01.493" Score="1" Body="&lt;p&gt;It could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; STRIKE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Someone that is beautiful is said to be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; striking&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, you could&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; bowl a strike.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you look at the dots, they look a little like&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the bottoms of the 5 pins as they're knocked down at the end of a bowling lane + 1 ball.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-07-15T20:21:04.083" LastActivityDate="2015-07-15T20:21:04.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17994" PostTypeId="2" ParentId="17962" CreationDate="2015-07-15T20:12:10.687" Score="6" Body="&lt;p&gt;Say that $a$ and $b$ are black, with $a&amp;lt;b$. Then, there are three points $x$ such that $\{a,b,x\}$ in some order is an arithmetic progression:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\bf 2a-b$, $a$, $b$&lt;/li&gt;&#xA;&lt;li&gt;$a$, $\bf \frac{a+b}{2}$, $b$&lt;/li&gt;&#xA;&lt;li&gt;$a$, $b$, $\bf 2b-a$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If any of the three is also colored black, we're done. If all are colored white, we're also done because they themselves form an arithmetic progression $2a-b, \frac{a+b}{2}, 2b-a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are not two black points, do the same argument with white points. In fact, because this argument applies to every pair of same-colored points, there are an uncountable number of solutions.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-07-15T20:42:58.967" LastActivityDate="2015-07-15T20:42:58.967" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17995" PostTypeId="1" AcceptedAnswerId="18082" CreationDate="2015-07-15T06:18:27.587" Score="12" ViewCount="1253" Body="&lt;p&gt;In the &lt;a href=&quot;http://www.chiark.greenend.org.uk/~sgtatham/puzzles/js/undead.html&quot; rel=&quot;noreferrer&quot;&gt;Undead Game&lt;/a&gt;, what makes one board more difficult than another?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The game is defined like this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A 4x4 (or 5x5, etc) grid has some random squares filled with diagonal lines representing mirrors.&lt;/li&gt;&#xA;&lt;li&gt;The grid has numbers added on both ends of the rows and the columns, indicating how many undead you will see if you look straight (horizontally/vertically) into the grid from that point.&lt;/li&gt;&#xA;&lt;li&gt;You are provided with a set number of zombies, ghosts, and vampires, with the following properties:&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Zombies can always be seen when they are in your line of sight (which extends along the reflection-paths of the mirrors&lt;/li&gt;&#xA;&lt;li&gt;Vampires can only be seen if they are directly in your line of sight, but cannot be seen through a mirror&lt;/li&gt;&#xA;&lt;li&gt;Ghosts &lt;em&gt;cannot&lt;/em&gt; be seen when they are directly in your line of sight, and can &lt;em&gt;only&lt;/em&gt; be seen via a mirror.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Your task&lt;/strong&gt; is to place the given undead into the empty squares of the grid so that every number around the outside accurately describes the number of undead seen from that point.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;An image to help with visualization:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/70nNP.png&quot; alt=&quot;An image to help with visualization&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to understand what decides the level of difficulty in this game, aside from grid size.  For example, the linked website offers &lt;strong&gt;4x4 Easy&lt;/strong&gt;, &lt;strong&gt;4x4 Normal&lt;/strong&gt;, and &lt;strong&gt;4x4 Tricky&lt;/strong&gt; options.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have written some code to generate a game board and I can generate thousands of 4x4 games by selecting random values for each cell. How do I assign a difficulty to them ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way I can do it by just looking at the board ?&lt;/p&gt;&#xA;" OwnerUserId="14218" OwnerDisplayName="Vivek" LastEditorUserId="4421" LastEditDate="2015-07-16T18:31:40.393" LastActivityDate="2015-07-22T23:49:34.500" Title="In the Undead Game, what makes one board more difficult than another?" Tags="|puzzle-creation|difficulty-rating|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17996" PostTypeId="2" ParentId="17983" CreationDate="2015-07-15T20:22:36.890" Score="10" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MAIDEN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;as in&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A maiden over in cricket, or 6 (dot) balls without a run being scored, as it would be displayed on a score sheet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12277" LastActivityDate="2015-07-15T20:22:36.890" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17997" PostTypeId="1" AcceptedAnswerId="18048" CreationDate="2015-07-15T20:33:39.993" Score="8" ViewCount="749" Body="&lt;p&gt;Part 1: I'm something basic, made of 30-odd pieces;&lt;br&gt;&#xA;Many will know me even among different speeches.&lt;br&gt;&#xA;Part 2: Do I contain waves or particles? Depends on your context;&lt;br&gt;&#xA;I am not what you think I am, there is one crucial difference.&lt;br&gt;&#xA;Part 3: One of me has a bright streak yet its outlook is dreary;&lt;br&gt;&#xA;My like has been here since the first, but not by name 'til 3.&lt;br&gt;&#xA;Part 4: I'm a magical number: simple but powerful;&lt;br&gt;&#xA;A single operation makes me uncountable.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put them all together and what do you get? And in case anyone is concerned that I am too prolific a puzzler, this will be the last puzzle I contribute until one  of them is solved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Seems I made it too hard indeed. See below for an easier version of the riddle.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Part 1: I'm something primitive, made of 32 pieces;  many will know me, even among different speeches.&lt;br&gt;&#xA; Part 2: If light is what you say I am, you're technically right;  but you're head's full of me if you think that I &lt;em&gt;am&lt;/em&gt; light.&lt;br&gt;&#xA; Part 3: In games, in comics, in movies, in the dungeon;  if solo's your style, then with me you'll have fun.&lt;br&gt;&#xA; Part 4: I'm a magic number, a power of two;  I make Cantor cock his head and say 'Ooooh'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="13970" LastEditDate="2015-07-16T15:37:52.523" LastActivityDate="2015-07-16T16:20:13.273" Title="Sound Out the Word" Tags="|riddle|word|wordplay|" AnswerCount="6" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17998" PostTypeId="2" ParentId="17962" CreationDate="2015-07-15T20:49:54.807" Score="6" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Van_der_Waerden%27s_theorem&quot;&gt;Van der Waerden's theorem&lt;/a&gt; tells us that if we color a sufficiently long interval of integers with $c$ colors, there will exist a same-color arithmetic progression of length $l$. Since the reals contain a copy of the integers, which contain every finite interval of integers, they contain the required arithmetic progression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem asks us for the special case $c=2, l=3$.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-07-15T20:49:54.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17999" PostTypeId="2" ParentId="17937" CreationDate="2015-07-15T21:48:52.803" Score="5" Body="&lt;p&gt;In the &lt;a href=&quot;https://puzzling.stackexchange.com/q/17852/10615&quot;&gt;other lockbox problem&lt;/a&gt;, where the assignment of keys to boxes was a random permutation, you could use a clever representation of permutations to show that $\frac{k}n$ of the permutations had a wooden box in all of their cycles. In this case, we do a similar thing, using a clever representation of arbitrary functions from $\{1,\dots,n\}$ to $\{1,\dots,n\}$. First, we need some preparation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $[n]$ be shorthand for $\{1,\dots,n\}$. Given a function $f:[n]\to[n]$, let a &lt;em&gt;cycle&lt;/em&gt; of $f$ be a list $x_0,x_1,\dots, x_n$, where $f(x_{k-1})=x_k$ and $f(x_n)=x_0$. Every function has at least one cycle, since the list $1,f(1),f(f(1)),\dots$ must have a repeat. There are many ways to write the same cycle, for example, 1,2,3,4,5 is the same as 5,1,2,3,4 and 2,3,4,5,1. We will assume a cycle is always written so the first number $x_0$ is the smallest number in the cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what is this clever representation of $f$? It will be a list $L$ of $n$ numbers, where every list element is an integer between $1$ and $n$. $L$ is given by the concatenation of two lists, $P$ and $S$ (for &quot;prefix&quot; and &quot;suffix&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The entries of $P$ will consist of the numbers which are in some cycle of $f$. Specifically, to find $P$, find all of the cycles of $f$. Order the cycles in decreasing order of their first element, then concatenate them all together to one list. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find $S$, first find $P$, then use the following algorithm:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;let S = [], an empty list&#xA;let U = the set of numbers in [1,...,n] which are not in P&#xA;while U is nonempty:&#xA;    let y = smallest element of U which is not of the form f(u) for some u in U&#xA;    add f(y) to end of S&#xA;    remove y from U&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Because of how $y$ is chosen each step, we know that if $y_{last}$ is the last element removed from $U$, then $f(y_{last})$ will be in $P$. This means that the first element of $S$ appears in $P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This encodes $f$ as a list $L$. What is surprising is that you can decode a list $L$ to a unique function $f$. This means the encoding process is a bijection from functions to lists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does decoding work? Given $L$, we can recover $P$ and $S$ by letting $P$ be the largest prefix of $L$ whose entries are all distinct, and $S$ be the rest of $L$. This allows us to retrieve the cycles of $f$, since every entry of $P$ which is smaller than the ones before it will be the start of a cycle. We now know $f(x)$ for every $x$ in a cycle. To find the other unknown values, let $x$ be the smallest unknown number not appearing in $S$, set $f(x)$ to be the last element of $S$, then remove the last element of $S$. Repeat until $S$ is empty, at which point $f(x)$ will be known for all $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now use this encoding to solve the puzzle. Number the boxes $1$ to $n$ so that boxes $1$ to $k$ are wooden. Let $f(x)$ be the number of the box containing the key to box number $x$, and let $L$ be the encoding of $f$. We will be able to open all the steel boxes if and only if every cycle of $f$ contains a wooden box. This occurs if and only if the first entry of $L$ is in $\{1,\dots,k\}$. Since the encoding is a bijection from functions to lists, and $f$ is a random function, it follows that $L$ is a random list. This means that the first entry of $L$ has a $\frac{k}{n}$ chance of being in $\{1,\dots,k\}$, so the probability desired is $\frac{k}n$.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-07-15T22:29:52.473" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18000" PostTypeId="2" ParentId="17932" CreationDate="2015-07-15T21:55:50.367" Score="10" Body="&lt;p&gt;You and your friends are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Baseball fans!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Each of the movies is a reference to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A baseball team name&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's the list:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1. Casino Royale (Kansas City Royals)&lt;br&gt;&#xA;2. Braveheart or Brave Little Toaster (Atlanta Braves)&lt;br&gt;&#xA;3. Pirates of the Caribbean (Pittsburgh Pirates)&lt;br&gt;&#xA;4. Lone Ranger (Texas Rangers)&lt;br&gt;&#xA;5. Rocky (Colorado Rockies)&lt;br&gt;&#xA;6. Astronaut Farmer (Houston Astros)&lt;br&gt;&#xA;7. National Treasure (Washington Nationals)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Personally I'm a fan of the Hunt for Red October!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to TroyAndAbed and other commenters for most of the movie names!&lt;/p&gt;&#xA;" OwnerUserId="10861" LastActivityDate="2015-07-15T21:55:50.367" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18001" PostTypeId="2" ParentId="17962" CreationDate="2015-07-15T21:59:12.800" Score="3" Body="&lt;p&gt;It is sufficient to colour the integers between 1 and 9 inclusive. If each one is coloured white or black, there will be three of them $a,b,c$ all of the same colour such that $a−b=b−c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because Van der Waerden's $W(2,3)$ is 9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result holds for every set of nine real (or complex) numbers in arithmetic progression.&lt;/p&gt;&#xA;" OwnerUserId="7363" LastEditorUserId="7363" LastEditDate="2015-07-15T23:09:07.420" LastActivityDate="2015-07-15T23:09:07.420" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18002" PostTypeId="1" AcceptedAnswerId="18006" CreationDate="2015-07-15T22:21:26.587" Score="14" ViewCount="728" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;My first is employed all the time by a plumber.&lt;br&gt;&#xA;  It's seen by the poolside, most often in summer.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;For a surgeon, my second will help with dilation,&lt;br&gt;&#xA;  But then will get pitched when she goes on vacation.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;My third was a cow, if the legends are true,&lt;br&gt;&#xA;  And denser than all of its brethren, too.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;My last part is hidden between you and me&lt;br&gt;&#xA;  In the land of the brave and the home of the free.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;My whole isn't likely to say what it thinks,&lt;br&gt;&#xA;  And surely won't join you in having a drink.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The word is quite special, I'm sure you'll agree&lt;br&gt;&#xA;  For a bonus, explain its uniqueness to me.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="463" LastEditDate="2015-07-15T22:23:52.793" LastActivityDate="2015-07-16T00:46:35.073" Title="Don't hold back" Tags="|riddle|word|rhyme|wordplay|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18003" PostTypeId="2" ParentId="17997" CreationDate="2015-07-15T22:22:03.037" Score="3" Body="&lt;p&gt;Sorted by confidence from little confidence, to negative confidence...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 4 Partial Guess: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The number 8.  It is the magic 8 ball, but rotating it 90 degrees turns it into the infinity symbol&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Part 1 Partial Guess: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Silver, Jesus, betrayal, or the Bible.  Judas betrayed Jesus for 30 silver pieces, and people know the bible in many different languages.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Part 2 Partial Guess: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Matter.  It is well known that electromagnetic radiation and more generally all elementary particles have both wave and particle properties.  Photons are too well known, so it is either Quarks or Leptons which are both matter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Part 3 Partial Guess: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Thunder/Lightning.  This is really mostly a shot in the dark.  Lightning flashes but comes in times of storms.  Also they have been around since the Earth was formed, no idea about 3, something with the etymology of the word?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14092" LastEditorDisplayName="user88" LastEditDate="2015-07-16T02:41:41.033" LastActivityDate="2015-07-16T02:41:41.033" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18004" PostTypeId="2" ParentId="17997" CreationDate="2015-07-15T22:34:31.653" Score="1" Body="&lt;p&gt;&lt;em&gt;Part 1: I'm something basic, made of 30-odd pieces;&lt;br&gt;&#xA;Many will know me even among different speeches.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The ALPHABET? It's made of roughly 30 letters (26 or 29 or 33 etc., according to which language you're looking at).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Part 2: Do I contain waves or particles? Depends on your context;&lt;br&gt;&#xA;I am not what you think I am, there is one crucial difference.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first line obviously refers to LIGHT, so changing it with &quot;one crucial difference&quot;, it could be e.g. LIT or RIGHT.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Part 3: One of me has a bright streak yet its outlook is dreary;&lt;br&gt;&#xA;My like has been here since the first, but not by name 'til 3.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Part 4: I'm a magical number: simple but powerful;&lt;br&gt;&#xA;A single operation makes me uncountable.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some irrational number, perhaps? My guess would be &lt;strong&gt;e&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-07-15T22:34:31.653" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18005" PostTypeId="2" ParentId="17775" CreationDate="2015-07-15T23:32:05.113" Score="2" Body="&lt;p&gt;Note: Author noted that the actual answer is &quot;your food&quot;. Below, however, is my original guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is: Your stomach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You sit by the pool to eat a meal.&#xA;Me, you can see, but won’t hear.&#xA;Go ahead, eat your lunch&#xA;And drink your light beer…&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can look down and see your stomach, but you can't hear your stomach.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I’m drowning!!! I don’t like to whine…&#xA;I’m drowning!!! But please throw me a line!&#xA;I’m drowning!!! I’ve not got much time.&#xA;I’m drowning!!! I’m not feeling fine.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The drinking of the beer is adding water to your stomach - drowning&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Gaze across the table at your date.&#xA;Stop pigging out, put yourself on idle.&#xA;You don’t note my importance,&#xA;But to you, I am vital!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If someone doesn't have a stomach (which is a vital organ), they'd probably die. Most of us don't really care about our stomach. We just dump food and junk into it (Pigging out)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I’m drowning!!! I’m here in this pool.&#xA;I’m drowning!!! I try to stay cool.&#xA;I’m drowning!!! But you’re being cruel.&#xA;I’m drowning!!! I need a saving tool!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Could also be drowning (figuratively) in food (as in being swamped with the food you're eating).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You sure love your meal and date…&#xA;Only sound is chewing – no conversing speech.&#xA;I flail and try to get your attention…&#xA;But my arms are short – I cannot reach…&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sound of chewing can propagate back down the esophagus into the stomach. Flailing can describe the stomachs pulsing movements. The stomach can't hear speech even if it was occurring on the date since sound propagates outwards of the mouth and not backwards. A stomach doesn't really have arms, so of course it can't reach - but reach what?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I’m drowning!!! I’m gasping for air.&#xA;I’m drowning!!! Dry land is somewhere!&#xA;I’m drowning!!! I won’t make it there…&#xA;I’m drowning!!! I don’t think you care…&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Digesting the food being gasping for air since it's so full.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8999" LastEditorUserId="8999" LastEditDate="2015-07-16T15:31:27.333" LastActivityDate="2015-07-16T15:31:27.333" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18006" PostTypeId="2" ParentId="18002" CreationDate="2015-07-16T00:02:15.160" Score="9" Body="&lt;p&gt;My whole isn't likely to say what it thinks,&#xA;And surely won't join you in having a drink.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://dictionary.reference.com/browse/abstentious&quot; rel=&quot;nofollow noreferrer&quot;&gt;Abstentious&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My first is employed all the time by a plumber.&#xA;It's seen by the poolside, most often in summer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Abs.  The easy definition is the summer one.  Plumbers often use PVC or ABS pipes, I believe ABS stands for &lt;a href=&quot;https://en.wikipedia.org/wiki/Acrylonitrile_butadiene_styrene&quot; rel=&quot;nofollow noreferrer&quot;&gt;Acrylonitrile butadiene styrene&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For a surgeon, my second will help with dilation,&#xA;But then will get pitched when she goes on vacation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tent.  This can be used to cause blood vessels to dilate as well as the more traditional meaning.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My third was a cow, if the legends are true,&#xA;And denser than all of its brethren, too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Io has the highest density of all of &lt;a href=&quot;https://en.wikipedia.org/wiki/Io_(moon)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jupiter's moons&lt;/a&gt; as well as a person &lt;a href=&quot;https://en.wikipedia.org/wiki/Io_(mythology)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Zeus transformed into a cow to hide from Hera&lt;/a&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My last part is hidden between you and me&#xA;In the land of the brave and the home of the free.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; US.  You and me makes US, as well as the United States.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The word is quite special, I'm sure you'll agree&#xA;For a bonus, explain its uniqueness to me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As, &lt;a href=&quot;https://puzzling.stackexchange.com/users/5208/alconja&quot;&gt;alconja&lt;/a&gt; convinced me, although not completely unique, this word does have every vowel in order which deserves mention.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14092" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2015-07-16T00:46:35.073" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="18007" PostTypeId="1" AcceptedAnswerId="18008" CreationDate="2015-07-16T00:26:33.953" Score="10" ViewCount="2378" Body="&lt;p&gt;I have a deck of 52 cards on the table in front of me, in random order, each one randomly facing either up or down. I place one joker at each end, each of these again randomly facing either up or down, to get a deck of 54 cards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then perform a sequence of moves each of the following form.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Choose a random contiguous block of cards (a certain number $\geq1$ of cards, all together in the deck) such that both the top and bottom card are facing up, and flip it over as a single unit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the probability that eventually I can no longer perform such a move, i.e. all the cards are facing down?&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-07-16T16:06:36.767" Title="How to cheat at cards" Tags="|mathematics|logical-deduction|probability|cards|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18008" PostTypeId="2" ParentId="18007" CreationDate="2015-07-16T00:32:38.447" Score="21" Body="&lt;p&gt;The probability is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; represent all face-up cards as 1 and face-down cards as 0, the binary number representing the deck (concatenating the states from top to bottom) decreases every time. It can never drop below 0, of course, so you have to eventually stop flipping cards.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2015-07-16T00:32:38.447" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18009" PostTypeId="1" AcceptedAnswerId="18010" CreationDate="2015-07-16T03:17:57.343" Score="26" ViewCount="4853" Body="&lt;p&gt;Archie the archeologist has discovered an Egyptian temple, and plans to send in a robot to explore it. He uncovered the ancient engineers' papyruses which explain almost everything about the temple's layout. However, the engineers left out one crucial detail to stymie the efforts of thieves: the details of the maze room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is all the papyrus said about the maze room:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It is a $20$ meter by $20$ meter square, whose walls are aligned with the compass directions.&lt;/li&gt;&#xA;&lt;li&gt;The floor is colored like a $20\times 20$ checkerboard.&lt;/li&gt;&#xA;&lt;li&gt;Between each adjacent square, there either is a wall, or isn't. &lt;/li&gt;&#xA;&lt;li&gt;One square is the &quot;start&quot; square: the only entrance to the maze involves being dropped onto this square from a trap door&lt;/li&gt;&#xA;&lt;li&gt;Another square is the &quot;finish&quot; square: once you step on it, you immediately fall through a trap door to the throne room&lt;/li&gt;&#xA;&lt;li&gt;It is possible to get from start to finish &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;To program the robot, you give it a finite list of compass directions, either North, South, East or West. The robot then goes down the list, moving one meter in the current direction unless doing so would make it hit a wall, in which case it doesn't move. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can Archie program the robot so that, starting from the start square, it will be guaranteed to eventually reach the finish square?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A note: you do not need to explicitly describe the program, only convince Archie whether this task is possible. The grad student will take care of creating/writing the actual program.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="7722" LastEditDate="2015-07-16T20:53:22.680" LastActivityDate="2023-01-19T06:47:56.560" Title="Maze Solving Robot" Tags="|algorithm|mazes|" AnswerCount="6" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="18010" PostTypeId="2" ParentId="18009" CreationDate="2015-07-16T03:51:40.643" Score="35" Body="&lt;p&gt;Sure you can. There's finitely many possible mazes, so solve each one in sequence. To solve a maze, imagine you're in that maze. Figure out where you are in the maze by simulating starting on the start space and following the instructions corresponding to the sequence of steps you've taken so far. Then, make the moves that would take you from there to the exit.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Algorithm&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Generate the ordered set of all potential legal mazes with all potential entry and exit squares. Call this set $M$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let the total list of moves executed up to but not including step $n$ be $H_n$. Let the list of moves executed during step $n$ be $h_n$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For each model $\mathfrak{m} \in M$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Assume $\mathfrak{m}$.&lt;/li&gt;&#xA;&lt;li&gt;Compute the robot's presumptive location subject to executing $H_n$.&lt;/li&gt;&#xA;&lt;li&gt;Compute $h_n$ such that the robot reaches the presumptive exit.&lt;/li&gt;&#xA;&lt;li&gt;Execute $h_n$.&lt;/li&gt;&#xA;&lt;li&gt;Append $h_n$ to $H_n$.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I had tried a different solution using &lt;a href=&quot;https://en.wikipedia.org/wiki/Synchronizing_word&quot;&gt;synchronizing words&lt;/a&gt; to get to a fixed position in a given maze, but wasn't able to guarantee the &lt;a href=&quot;https://en.wikipedia.org/wiki/Deterministic_finite_automaton&quot;&gt;DFA&lt;/a&gt; associated with the maze meets the conditions that guarantee one, like those in &lt;a href=&quot;http://arxiv.org/pdf/0709.0099v4.pdf&quot;&gt;this result&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-07-16T20:40:56.710" LastActivityDate="2015-07-16T20:40:56.710" CommentCount="19" ContentLicense="CC BY-SA 3.0" />
  <row Id="18011" PostTypeId="2" ParentId="18009" CreationDate="2015-07-16T03:52:31.957" Score="2" Body="&lt;p&gt;I'm gonna say&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No, it's not possible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you could be dropped back at the starting square after each failed attempt, then it would be easy. Simply go through the huge list of all possible moves (paths), one at a time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; At first it seems that you could achieve this by just concatenating all of the individual &quot;possible move&quot; lists, each followed by its opposite, in order to get you back to the starting square, but the opposite move list does not necessarily lead you back to the starting square.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; there is no way to be sure that you are back at the start square, and so it becomes impossible to iterate through the master list of moves.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I hope that someone else figures out a way to overcome this seemingly impossible task, so that I may read it and learn something!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastActivityDate="2015-07-16T03:52:31.957" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="18012" PostTypeId="2" ParentId="18007" CreationDate="2015-07-16T06:59:56.097" Score="0" Body="&lt;p&gt;Let's prove the number of move is limited recursively on a group of n cards:&#xA;n = 1 means you have one move possible, afterward, you are stuck. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's suppose the hypothesis true for n &gt;= 1 and let's take n + 1 cards.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There is a limited number of move involving only the n first cards. after a moment, you would either be stuck or would turn the last card. After that, this card would be face down, no moves will be able to involve this last card and you are then limited to the n first cards.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is therefore, only a finite number of moves possible. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The solution is 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14126" LastEditorUserId="1509" LastEditDate="2015-07-16T16:06:36.767" LastActivityDate="2015-07-16T16:06:36.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18013" PostTypeId="2" ParentId="17945" CreationDate="2015-07-16T07:41:03.493" Score="2" Body="&lt;p&gt;Well here's what I would do in this situation, assuming I was very skilled with a coping saw...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cut out all rungs.  Cut these rungs in half except the last rung.  Then cut holes in the side of the poles.  You basically have a bunch of dowels and two poles with holes in them.  Shove the half-rungs in.  Use the full-rung as a joint for the two poles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then you end up with something like&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; |--|--|--|--|--|--|--|.  Kind of like a telephone-pole ladder.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To make it more sturdy you can use more full-rungs, depending on how good your ninja landing skills are.&lt;/p&gt;&#xA;" OwnerUserId="14127" LastActivityDate="2015-07-16T07:41:03.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18020" PostTypeId="2" ParentId="17945" CreationDate="2015-07-16T09:43:02.017" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cut the ladder, with the saw, in half like a sandwich so you end up with two equal parts, half as wide as the original&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And then&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Use the saw to connect them, by using it like a &lt;a href=&quot;https://en.wikipedia.org/wiki/Carabiner&quot; rel=&quot;nofollow&quot;&gt;carabiner&lt;/a&gt;, attaching the last step of one to the first step of the other&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not precisely twice as long, but close enough...&lt;/p&gt;&#xA;" OwnerUserId="13715" LastActivityDate="2015-07-16T09:43:02.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18021" PostTypeId="2" ParentId="18009" CreationDate="2015-07-16T09:49:20.433" Score="11" Body="&lt;p&gt;Randomness will be our friend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is known that a random walk will eventually get you out of any maze. See &lt;a href=&quot;https://en.wikipedia.org/wiki/Maze_solving_algorithm#Random_mouse_algorithm&quot; rel=&quot;noreferrer&quot;&gt;Random mouse algorithm&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will provide a long enough list of random directions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;West, East, East, North, West, South, South, North, East, North, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The only problem is that the list has to be finite in this problem so the probability of getting out of the maze will not be 100% but will depend on the length of the list. The longest the list, the higher the probability will get (but a formula that connects length of list and probability is much harder to find).&lt;/p&gt;&#xA;" OwnerUserId="12062" LastEditorUserId="12062" LastEditDate="2015-07-16T09:55:16.770" LastActivityDate="2015-07-16T09:55:16.770" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="18022" PostTypeId="1" CreationDate="2015-07-16T09:56:13.543" Score="3" ViewCount="216" Body="&lt;p&gt;There are 16 secret agents who each know a different piece of secret information. They can telephone each other and exchange all the information they know. After the telephone call, they both know everything that either of them knew before the call. &lt;br&gt;&#xA;&lt;br&gt;&#xA;What is the minimum number of telephone calls required so that all of them know everything? &lt;br&gt;&#xA;The answer is $28$. &lt;Br&gt;&#xA;But I don't know how? &lt;Br&gt;&#xA;I remember dividing them into $4$ groups of $4$ each. Then suppose in group $'A'$ there are $a,b,c,d$. Now $a$ calls $b$, $c$ calls $d$ and $a$ calls $c$. This happens in every group so that $2$ people in their respective groups know whole info about their group. But what to do next, I don't remember, or work out?&lt;/p&gt;&#xA;" OwnerUserId="11425" LastActivityDate="2015-07-16T09:56:13.543" Title="Minimum number of telephone calls needed to exchange information" Tags="|logical-deduction|calculation-puzzle|minimum-solve|" AnswerCount="0" CommentCount="3" ClosedDate="2015-07-16T10:02:16.223" ContentLicense="CC BY-SA 3.0" />
  <row Id="18023" PostTypeId="2" ParentId="18009" CreationDate="2015-07-16T10:20:54.297" Score="-6" Body="&lt;p&gt;Actually, there is a theory that if you always keep on the left side of the wall of a maze, you will eventually find the exit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So try writing an algorithm that will keep your robot on the left side of the wall of a maze (or on the right side, the important thing is to keep moving on the same side of a wall).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find more details here: &lt;a href=&quot;https://en.wikipedia.org/wiki/Maze_solving_algorithm&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Maze_solving_algorithm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The algorithm that I previously posted, was indeed wrong as  &lt;a href=&quot;https://puzzling.stackexchange.com/users/5840/bmyguest&quot;&gt;BmyGuest&lt;/a&gt; suggested. &lt;/p&gt;&#xA;" OwnerUserId="14131" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-07-16T10:42:56.347" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18024" PostTypeId="2" ParentId="17987" CreationDate="2015-07-16T10:46:01.840" Score="2" Body="&lt;p&gt;I cannot think but that the question is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Telecommunications-related&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since actually with bite you mean&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Byte&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Although I've never been interviewed,&#xA;I'm famous all over the world.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; IP for example is used all around the world, as is GSM (more or less)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;People love to watch me eat,&#xA;But some disapprove of my habits.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; IPv4 packets are widely disapproved because there is not enough IPv4 for everyone (instead of IPv6)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't mind working for my food,&#xA;And I like to eat on the run.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Packets can be filled with information and &quot;pieces&quot; can be added (to tunnel for example) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Picky, I eat only 3 foods,&#xA;Not counting my desserts.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Those can be section of the packet, as for example: Header, Payload, Checksum. Desserts can be starting and ending sequences&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have very strange diet goals:&#xA;Eat 244 to 260 bites per meal,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, this has to do with some standard defined&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Plus 2 bites of dessert per meal-&#xA;1 bite of one and 1 bite of another.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As said above, header and tail of the packet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Timing my bites is important,&#xA;And when done wisely,&#xA;I can reach my full 260 bites,&#xA;And then I will be very happy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If clocks are very well alligned, one could eventually remove header and tail of a packet to save space for payload&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5122" LastActivityDate="2015-07-16T10:46:01.840" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="18025" PostTypeId="2" ParentId="17997" CreationDate="2015-07-16T11:40:59.570" Score="2" Body="&lt;p&gt;A guess for part 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Month - The same calendar is used across the globe, and a month has 30-odd days in.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4659" LastActivityDate="2015-07-16T11:40:59.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18026" PostTypeId="2" ParentId="18007" CreationDate="2015-07-16T11:54:01.263" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The gap between the first and the last card facing up always narrows or stays the same, without ever going beyond the old gap. Eventually there'll be no usable cards &lt;em&gt;inside&lt;/em&gt; a gap between facing-up cards even if we try to avoid moves narrowing it (the time for the widest gap will come), so there will be no cards left. In other words, the answer is 1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="14016" LastEditDate="2015-07-16T12:50:10.297" LastActivityDate="2015-07-16T12:50:10.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18031" PostTypeId="2" ParentId="2419" CreationDate="2015-07-16T12:52:42.340" Score="2" Body="&lt;p&gt;Another possibility :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;He prefer to take the stairs instead of the elevator. On his building stairs start only on 5th floor, are outside and aren't protected of rain. When the weather is ok he stops on the 5th floor and takes the stairs, but when it's raining he prefer to be dry and go directly on his floor. &#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13384" LastActivityDate="2015-07-16T12:52:42.340" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18032" PostTypeId="2" ParentId="2749" CreationDate="2015-07-16T12:58:24.647" Score="2" Body="&lt;p&gt;Well, i have a solution : don't care about paths, just continue straight forward&lt;/p&gt;&#xA;" OwnerUserId="13384" LastActivityDate="2015-07-16T12:58:24.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18033" PostTypeId="2" ParentId="17997" CreationDate="2015-07-16T13:16:16.007" Score="0" Body="&lt;p&gt;&lt;strong&gt;Partial answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that part 2 is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a &lt;strong&gt;slide.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Part 2: Do I contain waves or particles? Depends on your context;&lt;/em&gt;&#xA;&lt;br&gt;&lt;em&gt;I am not what you think I am, there is one crucial difference.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most people think of a slide as something that you, well, &lt;em&gt;slide down on&lt;/em&gt;, but a rockslide or landslide, which can either come down in waves or chunks, depending on context, is something entirely different.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I believe that, going off of my previous answer, part 1 could be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;dust&lt;/strong&gt;, &lt;em&gt;or rather&lt;/em&gt; &lt;strong&gt;sand&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Part 1: I'm something basic, made of 30-odd pieces;&lt;/em&gt;&#xA;&lt;br&gt;&lt;em&gt;Many will know me even among different speeches.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sand, or debris in general, is made of a combination of many small materials, and is something rather basic. Even in different languages, everyone knows what dirt or sand is.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13697" LastActivityDate="2015-07-16T13:16:16.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18034" PostTypeId="2" ParentId="17997" CreationDate="2015-07-16T13:21:12.800" Score="2" Body="&lt;p&gt;A guess for Part 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water - has almost 30 parts (28 protons, electrons and neutrons). People drink it during speeches&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A guess for Part 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Planet or Mars- some pictures of Mars were reported this year to show a bright streak. The first planet to be named was earth (the third planet). There's not a lot of water on Mars.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit: I think I got it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water (as above)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Part 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bottle - can contain waves or particles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Part 3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rock - we're third rock from the sun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Part 4&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;8 as solved previously&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Altogether&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Water-bottle Rocket&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="14135" LastEditDate="2015-07-16T13:40:20.617" LastActivityDate="2015-07-16T13:40:20.617" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18036" PostTypeId="1" AcceptedAnswerId="18040" CreationDate="2015-07-16T13:53:05.830" Score="0" ViewCount="1204" Body="&lt;p&gt;Most people are familiar with the following problem:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A well-known example has nine (or fewer) items, say coins (or balls), that are identical in weight save for one, which in this example is lighter than the other —a counterfeit (an oddball). The difference is only perceptible by weighing them on scale—but only the coins themselves can be weighed. Is it possible to isolate the counterfeit coin with only two weighings?&lt;/p&gt;&#xA;&lt;p&gt;To find a solution, we first consider the maximum number of items from which one can find the lighter one in just one weighing. The maximum number possible is three. To find the lighter one we can compare any two coins, leaving the third out. If the two coins tested weigh the same, then the lighter coin must be one of those not on the balance. Otherwise, it is the one indicated as lighter by the balance.&lt;/p&gt;&#xA;&lt;p&gt;Now, imagine the nine coins in three stacks of three coins each. In one move we can find which of the three stacks is lighter (i.e. the one containing the lighter coin). It then takes only one more move to identify the light coin from within that lighter stack. So in two weighings we can find a single light coin from a set of 3 X 3 = 9.&lt;/p&gt;&#xA;&lt;p&gt;By extension, it would take only three weighings to find the odd light coin among 27 coins, and four weighings to find it from 81 coins.&lt;/p&gt;&#xA;&lt;p&gt;wikipedia: &lt;a href=&quot;https://en.wikipedia.org/wiki/Balance_puzzle&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Balance_puzzle&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Simple enough.&lt;/p&gt;&#xA;&lt;p&gt;But what if you have an unknown number of coins, one being lighter than the other, and you want to know the minimum number of times you will have to use the scale to find the odd one out.&lt;/p&gt;&#xA;&lt;p&gt;Does anyone know of a formula to calculate it?&lt;/p&gt;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="14137" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-16T15:46:24.910" Title="Derivative of the coin weight problem" Tags="|mathematics|weighing|" AnswerCount="1" CommentCount="4" ClosedDate="2015-07-16T16:14:18.090" ContentLicense="CC BY-SA 3.0" />
  <row Id="18037" PostTypeId="2" ParentId="18009" CreationDate="2015-07-16T13:54:45.967" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You first assume the walls are placed in a certain way. Then you start making assumptions about where you are and make sure the robot can travel through all the squares based on these assumptions. If you can't reach the end on the very first try (after this list of moves), you assume you were on a different square from what you chose at first and calculate your current &quot;starting&quot; pos based on that. After every failed attempt, you assume you were on a different square at the very beginning and calculate your new starting position based on that. If this &lt;em&gt;scenario&lt;/em&gt; (assumed combination of walls, &lt;em&gt;not&lt;/em&gt; the starting position) never succeeds, start anew in another scenario. Eventually the robot will succeed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="14016" LastEditDate="2015-07-16T17:11:41.163" LastActivityDate="2015-07-16T17:11:41.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18039" PostTypeId="1" AcceptedAnswerId="18210" CreationDate="2015-07-16T14:04:50.860" Score="14" ViewCount="1157" Body="&lt;p&gt;My legend will be remembered,&lt;br&gt;&#xA;Too strong to be dismembered.&lt;br&gt;&#xA;No happiness at the parting,&lt;br&gt;&#xA;What happened at the starting?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the lines or all we do,&lt;br&gt;&#xA;Mix God, religion, and the facts too.&lt;br&gt;&#xA;Counting now, what are we at?&lt;br&gt;&#xA;Profound, noises, or like a cat?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My land is one full of life,&lt;br&gt;&#xA;You know my cousins, little strife.&lt;br&gt;&#xA;Started with two, but grew to three,&lt;br&gt;&#xA;But with some help, who can I be?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each line references something different.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Is a very similar format to another one of my recent riddles.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Thinking of the comment on &lt;a href=&quot;https://puzzling.stackexchange.com/a/18181/9951&quot;&gt;Marek's answer&lt;/a&gt;, it might be more accurate to say &lt;em&gt;normal&lt;/em&gt; human beings can be dismembered (but remember hint 1). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 4:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Check &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16686/alive-dead-alive-now-dead&quot;&gt;this riddle&lt;/a&gt;, &lt;a href=&quot;https://puzzling.stackexchange.com/questions/17155/still-going-strong-who-am-i&quot;&gt;this riddle&lt;/a&gt;, &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15731/you-may-know-me-who-am-i&quot;&gt;this riddle&lt;/a&gt;, or &lt;a href=&quot;https://puzzling.stackexchange.com/questions/12543/come-sail-away-who-am-i&quot;&gt;this riddle&lt;/a&gt; for some similar formats.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 5:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What color do we often associate with life? What land/country is often associated with that color?   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 6:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is not an actor.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-07-21T11:29:59.620" Title="What is the plan - Who am I?" Tags="|riddle|rhyme|knowledge|" AnswerCount="9" CommentCount="11" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18040" PostTypeId="2" ParentId="18036" CreationDate="2015-07-16T14:16:23.873" Score="2" Body="&lt;p&gt;I would have to guess that for an amount &lt;em&gt;&lt;strong&gt;n&lt;/strong&gt;&lt;/em&gt;  the amount of weighings  &lt;em&gt;&lt;strong&gt;x&lt;/strong&gt;&lt;/em&gt; would have to go according to the following rule&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$n\le3^x$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;or therefor:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$x=\lceil\log_3n\rceil$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I'll explain my reasoning:&lt;/p&gt;&#xA;&lt;p&gt;for n = 2 and n = 3:&#xA;One single weighing suffices&lt;/p&gt;&#xA;&lt;p&gt;for n = 4 to 9:&#xA;Can always be split out in two groups of either 2 or 3, with a remainder smaller then 3.&#xA;Either way the initial weighing will reduce the problem to the steps for n=2 or n=3, so 1 weighing + 1 extra for the lower grade of the problem = 2&lt;/p&gt;&#xA;&lt;p&gt;for n = 10 to 27&#xA;Can always split this up in two equal groups of 4 to 9, with a remainder of at most 9. Therefor one single weighting reduces this to a lower grade of the problem. So now we have three weighings: a first to reduced it to maximum 9 elements containing the odd one out, a second to reduce it to maximum 3 elements, and a third for the decider.&lt;/p&gt;&#xA;&lt;p&gt;You've probably noticed a pattern emerging now:&#xA;Using the partitions in three groups you can always take down the complexity one step each time. You just need a rule to figure out how many steps.&lt;/p&gt;&#xA;" OwnerUserId="2156" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-16T14:27:45.100" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18041" PostTypeId="2" ParentId="18039" CreationDate="2015-07-16T15:05:14.293" Score="6" Body="&lt;p&gt;Could you be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An idea&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My legend will be remembered,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People remember idea's and they can given to others or created into a tangible object to continue existence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Too strong to be dismembered.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An idea can not be destroyed. It can be considered one of the most powerful things for humans.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No happiness at the parting,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When an idea is given up on there could be no happiness. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What happened at the starting?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Scientists don't understand how idea's are created, and some people often wonder where idea's originated from.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All the lines or all we do,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All the lines of text, or all things we do are related to an idea. (i.e items we used were created from an idea, things we want to do are just initial idea's)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mix God, religion, and the facts too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A story can be about gods, religion, and science. More particularly the information/meaning behind the story.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Counting now, what are we at?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A lot. There are a lot of idea's out there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Profound, noises, or like a cat?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; possible reference to profound ideas, musical ideas (songs), and the internet....whose idea what it to put cats all over it!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My land is one full of life,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Idea's come from imagination and nature, which is full of life (you are alive to think right?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You know my cousins, little strife.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Possible reference to conspiracies? People can get angry or bitter over them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Started with two, but grew to three,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Possible reference to the forms of sharing an idea. (Verbal, written, visual)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But with some help, who can I be?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think I might have just the &lt;em&gt;idea&lt;/em&gt; of who you are.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-07-16T15:13:43.260" LastActivityDate="2015-07-16T15:13:43.260" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18042" PostTypeId="2" ParentId="17937" CreationDate="2015-07-16T15:18:30.030" Score="4" Body="&lt;p&gt;I'm late to the party but I love this problem so I'm adding my own solution.  Consider the following variant of the variant:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The prankster has $n$ Ziploc bags.  He places the $n$ keys in the bags anyway he wishes, potentially maliciously.  Some bags could receive multiple keys, while other bags receives none.  He then randomly places exactly one bag in each lockbox, and closes the box.  What is the probability we will be able to open all the boxes?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This generalizes the variant, since if he wants the prankster could place each key in a random bag independently, and then the extra permutation when placing the bags in the boxes is redundant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet even if the prankster does not act randomly on the keys in bags step, I claim the probability of opening the boxes (a &quot;success&quot;) is still $\frac{k}{n}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will use induction on the number of boxes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is no empty bag, then there must be one key per bag, and this reduces to the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/17852/one-hundred-lockboxes-of-wood-and-steel&quot;&gt;original problem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is an empty bag, then take a particular empty bag $E$ and suppose the prankster randomly places $E$ first, and then randomly places the other bags.   Given a fixed location for bag $E$, we can use induction to calculate the probability of success.  This is because the box with $E$ in it, and its key, has no effect on whether we can open all the boxes.  We can just remove the box containing $E$ and pretend its key magically disappears, and we are left with a random permutation of the other $n-1$ bags in the other $n-1$ boxes.  If and only if this $n-1$ box arrangement is a success will we have success with the bigger arrangement.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If $E$ is in a steel box (which happens with probability $\frac{n-k}{n}$), then the probability success after removing $E$ by induction is $\frac{k}{n-1}$.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $E$ is in a wooden box (which happens with probability $\frac{k}{n}$), thent he probability of success after removing $E$ by induction is $\frac{k-1}{n-1}$.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We can then compute the probability of success including $E$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= \frac{n-k}{n} \cdot \frac{k}{n-1} + \frac{k}{n} \cdot \frac{k-1}{n-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= \frac{(n-k)k + k(k-1)}{n(n-1)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= \frac{nk - k}{n(n-1)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= \frac{k(n-1)}{n(n-1)} = \frac{k}{n}$&lt;/p&gt;&#xA;" OwnerUserId="2722" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-07-16T15:18:30.030" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18043" PostTypeId="2" ParentId="17940" CreationDate="2015-07-16T15:36:10.390" Score="3" Body="&lt;p&gt;Pretty sure I have the first transformation for Part 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The hint mentions these numbers having a 'paltry structure' where the numbers in Part 2 had 'power'.  This could be referring to a base, in which case we can assume these numbers are represented in base 6.  Translating them to base 10 gives us: 132 119 255.  The presence of 255 makes me think these should be translated to hex, but that's about all I have at the moment.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I think the correct answer for Part 2 may be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Via Roma, 4.  At least that's what I got when I typed in the coordinates from svinja's answer into Google Maps.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14120" LastActivityDate="2015-07-16T15:36:10.390" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18044" PostTypeId="1" AcceptedAnswerId="18217" CreationDate="2015-07-16T15:51:36.283" Score="5" ViewCount="403" Body="&lt;p&gt;This is part of the &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=title%3A%22Fill%20in%20the%20Dashes%22&quot;&gt;Fill in the Dashes&lt;/a&gt; series of puzzles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example : F_ in the B_ ⇒ Fill in the Blanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the words will be Juxtaposed as in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g. : C_ K_ S_ ⇒ Clark Kent Superman or Superman Clark Kent&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A I_ M_ S_ T_ 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C_ I_ T_ 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D_ S_ S_ T_ 15&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F_ L_ N_ 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E_ P_ 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D_ M_ T_ Y_ 11&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F_ I_ S_ T_ 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H_ H_ T_ 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D_ I J_ O_ 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B_ L_ P_ T_ 8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C_ D_ T_ 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M_ O_ O_ S_ T_ 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F_ N_ T_ 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each line identifies a creation of the personality&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The personality fame was for his writing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Was popular and connected to Hollywood&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Additional Information:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Any one correct should reveal the rest. I will accept the first correct answer as complete. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9892" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-07-21T14:23:28.640" Title="Fill in the Dashes : Part 4 of 10 - Subject Who?" Tags="|wordplay|knowledge|" AnswerCount="1" CommentCount="15" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18045" PostTypeId="1" AcceptedAnswerId="18046" CreationDate="2015-07-16T15:55:21.403" Score="2" ViewCount="209" Body="&lt;p&gt;$12 = 0$&lt;br&gt;&#xA;$1234 = 1$&lt;br&gt;&#xA;$123456 = 2$&lt;br&gt;&#xA;$12345678 = 4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$369 = 2$&lt;br&gt;&#xA;$248 = 3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2 = 0$&lt;br&gt;&#xA;$3 = 0$&lt;br&gt;&#xA;$5 = 0$&lt;br&gt;&#xA;$7 = 0$&lt;br&gt;&#xA;$11 = 0$&lt;br&gt;&#xA;$13 = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does:&lt;br&gt;&#xA;$123456789 = ?$&lt;br&gt;&#xA;Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully isn't too hard, please ask if you need any test cases.&lt;/p&gt;&#xA;" OwnerUserId="14092" LastEditorUserId="4421" LastEditDate="2015-11-25T18:06:10.230" LastActivityDate="2015-11-25T18:06:10.230" Title="Depressed Numbers" Tags="|calculation-puzzle|formation-of-numbers|" AnswerCount="1" CommentCount="1" ClosedDate="2015-07-16T16:09:03.447" ContentLicense="CC BY-SA 3.0" />
  <row Id="18046" PostTypeId="2" ParentId="18045" CreationDate="2015-07-16T16:02:36.300" Score="3" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You're counting the number of holes in the numbers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7396" LastEditorUserId="463" LastEditDate="2015-07-16T16:05:51.137" LastActivityDate="2015-07-16T16:05:51.137" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18048" PostTypeId="2" ParentId="17997" CreationDate="2015-07-16T16:20:13.273" Score="5" Body="&lt;p&gt;Okay, got it. The word you're looking for is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; INTERROGATE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Part 1: I'm something basic, made of 30-odd pieces;&lt;br&gt;&#xA;Many will know me even among different speeches.&lt;br&gt;&#xA;Part 1: I'm something primitive, made of 32 pieces;&lt;br&gt;&#xA;Many will know me, even among different speeches.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is an &lt;strong&gt;int&lt;/strong&gt;. Integers in coding are comprised of 32 bits and are primitive types, and are familiarly used across programming languages.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Part 2: Do I contain waves or particles? Depends on your context;&lt;br&gt;&#xA;I am not what you think I am, there is one crucial difference.&lt;br&gt;&#xA;Part 2: If light is what you say I am, you're technically right;&lt;br&gt;&#xA;But you're head's full of me if you think that I am light.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is &lt;strong&gt;air&lt;/strong&gt;. Light waves are perceived through dust particles suspended in the air. Of course, light itself can also be a particle, which can lead to confusion about this clue. If you think it's light even though the puzzle says it's not, you're an airhead, which is a synonym for an idiot.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Part 3: One of me has a bright streak yet its outlook is dreary;&lt;br&gt;&#xA;My like has been here since the first, but not by name 'til 3.&lt;br&gt;&#xA;Part 3: In games, in comics, in movies, in the dungeon;&lt;br&gt;&#xA;If solo's your style, then with me you'll have fun.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This word is &lt;strong&gt;rogue&lt;/strong&gt;. The first reference is to the super hero, who's a member of the X-Men. She has a white streak in her hair, but she's generally gloomy since her power saps other people's powers on touch so she can't actually ever touch anyone. I'm not sure about the 'not by name 'til three' line. The roguelike is a specific kind of single player dungeon-crawler video game.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Part 4: I'm a magical number: simple but powerful;&lt;br&gt;&#xA;A single operation makes me uncountable.&lt;br&gt;&#xA;Part 4: I'm a magic number, a power of two;&lt;br&gt;&#xA;I make Cantor cock his head and say 'Ooooh'&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We've already come across the answer to this, which is the number &lt;strong&gt;eight&lt;/strong&gt;. The clues reference that 8 = 2^3, and turning it to the right, or cocking your head to look at it, makes it resemble the sign for infinity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-07-16T16:20:13.273" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18049" PostTypeId="1" AcceptedAnswerId="18095" CreationDate="2015-07-16T16:28:30.620" Score="3" ViewCount="491" Body="&lt;p&gt;An excellent question for a first grader &lt;br/&gt;&lt;/p&gt;&#xA;&lt;p&gt;So one day a teacher is trying to teach their first graders math. She starts with the fundamentals. 1+1=2. But all the kids had been talking before hand and thought they had this arithmetic all figured out. So all of them said &lt;span class=&quot;math-container&quot;&gt;$1 + 1 = 3.$&lt;/span&gt;&#xA;Each and everyone. Try as they might the teacher couldn't convince them 3*3=9. They all said &lt;span class=&quot;math-container&quot;&gt;$3 * 3 = 6$&lt;/span&gt;. Of course when the teacher said no 3+3=6. They all laughed them off and went on their merry way.&lt;/p&gt;&#xA;&lt;br/&gt;&#xA;&lt;pre&gt;&#xA;Additional ones our teacher overheard over the next week.&#xA;3 + 3 + 1 = 6 &#xA;2 * 2 = 2 &#xA;4 + 2 = 1&#xA;2 + 4 = 1 &#xA;10 + 1 = 3 &#xA;16 + 3 = 5 &#xA;17 + 3 = 7 &#xA;18 * 3 = 7 &#xA;2 + 2 = 1 &#xA;0 + 0  = 3 &#xA;0 * 0 = 4&#xA;3 + 7 = 5&#xA;5 * 3 = 5&#xA;4 * 8 = 3&#xA;3 * 3 + 7 * 7 = 11&#xA;6 * 3 = 4&#xA;&#xA;4 + 4 = ?&#xA;&lt;/pre&gt;&#xA;" OwnerUserId="13916" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-17T03:31:43.667" Title="Simple math questions" Tags="|mathematics|pattern|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18050" PostTypeId="1" AcceptedAnswerId="18317" CreationDate="2015-07-16T17:02:23.203" Score="16" ViewCount="1711" Body="&lt;p&gt;This puzzle belongs to the puzzle series: &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=hyper-modern%20is%3Aquestion&quot;&gt;hyper-modern art&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;After solving the message in the butterfly image, the two friends move on to the next room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;This was certainly an overwhelmingly large picture in the last room. Can we have something smaller please?&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Sure. Just look over there on the table...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Is this a microscope?&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Indeed it is. Just look through it. My guide states that the image under investigation consists of exactly 93635 square dots created by laser-printing on only a square-millimetre of paper.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Hmm, certainly an interesting image, although not as beautiful as the last... What is it called?&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;It's called &lt;code&gt;Homage to LA&lt;/code&gt;.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Well, then there certainly isn't something secret about this one. You can clearly read this!&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yes, but do you also know what L A stands for? And remember: It wouldn't be a piece of hyper-modern art if there wouldn't be an unambiguous solution to this held within the image itself!&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/feRTC.gif&quot; alt=&quot;PuzzleImage&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The goal of the puzzle is to find some message - other than the obvious text - in the image, which will be a clear hint to the solution what L A stands for. A complete answer contains this message, how it can be found, and what L A stands for. The puzzle is fully contained in the image. The story is just flavour...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="32543" LastEditDate="2017-05-09T09:47:20.817" LastActivityDate="2020-12-20T17:31:40.507" Title="Dedicated image - but dedicated to what/who?" Tags="|cipher|visual|steganography|" AnswerCount="2" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="18051" PostTypeId="1" AcceptedAnswerId="18062" CreationDate="2015-07-16T18:07:21.867" Score="5" ViewCount="221" Body="&lt;p&gt;Since my previous cute, easy number puzzle turned out to be a duplicate, I felt I needed to redeem myself.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2+9 = 6$&lt;br&gt;&#xA;$7+8 = 8$&lt;br&gt;&#xA;$9+6 = 5$&lt;br&gt;&#xA;$0+9 = 6$&lt;br&gt;&#xA;$1+4 = 6$&lt;br&gt;&#xA;$9+9 = 3$&lt;br&gt;&#xA;$2+6 = 6$&lt;br&gt;&#xA;$6+4 = 7$&lt;br&gt;&#xA;$1+5 = 6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@TroyAndAbed:&lt;br&gt;&#xA;$12+27 = 8$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does:&lt;br&gt;&#xA;$5+5 = ?$&lt;br&gt;&#xA;Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully isn't too hard, please ask if you need any test cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HINT 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Use the title of the question&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14092" LastEditorUserId="14092" LastEditDate="2015-07-16T19:12:24.320" LastActivityDate="2015-07-16T19:15:29.550" Title="Distinctive Addition" Tags="|mathematics|pattern|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18052" PostTypeId="1" AcceptedAnswerId="18057" CreationDate="2015-07-16T18:55:12.370" Score="19" ViewCount="1328" Body="&lt;p&gt;My bicycle lock looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2Ab9a.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This rebus helps me remember the combination:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mydx0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I read it left to right, top to bottom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What's the combination, and why?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastActivityDate="2015-07-16T19:06:31.270" Title="Bicycle Lock Combination" Tags="|rebus|password|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18053" PostTypeId="1" AcceptedAnswerId="18060" CreationDate="2015-07-16T18:56:35.140" Score="9" ViewCount="660" Body="&lt;p&gt;&lt;em&gt;Hey guys, I've been into brainteasers since I was little and recently decided to make a puzzle-related website. I figured that this will be the best place to share some of my problems and see if they are any fun. I guess they are too easy, but hope that will entertain you for a bit.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Perfect Day&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This morning I woke up at 9 o'clock. My first job was to drink a glass&#xA;  of water and brush my teeth. After that I ate an apple, 3 bananas (I&#xA;  love them!) and a toast with almond butter for breakfast. Then I went&#xA;  to gym for one hour, took a shower there and came back home at eleven.&#xA;  Cooked few pieces of chicken with some small potatoes for lunch, ate&#xA;  all of them. In the afternoon I started working on my projects while&#xA;  listening to classical music. In the evening I went for the 2nd time&#xA;  to the gym, did cardio for half-an-hour and after that visited my&#xA;  friend's apartment. We had together one big pizza for dinner and at&#xA;  the end I fell asleep on his couch. As you can see, my entire day was&#xA;  perfect, except that my laptop keyboard broke. Can you figure out&#xA;  what's wrong with it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;P.S. Please, let me know if you have seen identical puzzle elsewhere, don't want to take someone's credit, even accidentally.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="14140" LastActivityDate="2015-07-16T19:11:51.637" Title="My Keyboard Broke" Tags="|story|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18054" PostTypeId="2" ParentId="18053" CreationDate="2015-07-16T19:02:13.107" Score="3" Body="&lt;p&gt;My guess is that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the letter Q is broken&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You use every letter but Q in your puzzle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastActivityDate="2015-07-16T19:02:13.107" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18055" PostTypeId="1" AcceptedAnswerId="18066" CreationDate="2015-07-16T19:04:45.787" Score="4" ViewCount="596" Body="&lt;p&gt;&lt;em&gt;Hey guys, I've been into brainteasers since I was little and recently decided to make a puzzle-related website. I figured that this will be the best place to share some of my problems and see if they are any fun. I guess they are too easy, but hope that will entertain you for a bit.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Chile to Singapore&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;You want to travel from Chile to Singapore. You find two sea captains&#xA;  - Jack and Will, which are willing to board you on their ships. Both of them depart from the same harbor early on the next morning, so you&#xA;  have to decide fast which one to hire. Jack promises that he will&#xA;  arrive in Singapore on the 50th day of the cruise, at midnight. He&#xA;  will charge you $1 per hour for the trip. Will promises that he will&#xA;  also arrive on the 50th day of the cruise, at noon. Will will charge&#xA;  you 1 cent per hour more than Jack. Which captain would you choose?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;Remark: The first day of the cruise starts when the ship departs the harbor and ends 24 hours later. After that the second day of the&#xA;  cruise begins and so on.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;P.S. Please, let me know if you have seen identical puzzle elsewhere, don't want to take someone's credit, even accidentally.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="14140" LastEditorUserId="5373" LastEditDate="2015-07-16T19:14:56.140" LastActivityDate="2017-11-18T14:07:44.603" Title="Chile to Singapore" Tags="|mathematics|calculation-puzzle|story|" AnswerCount="5" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18056" PostTypeId="2" ParentId="17973" CreationDate="2015-07-16T19:05:21.807" Score="2" Body="&lt;p&gt;I think a Surpassing Phrase is one that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; contains a string of 3 letters which is only one letter-change away from a string of 3 &lt;em&gt;consecutive&lt;/em&gt; letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's how all the phrases on the left satisfy this rule:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; su&lt;strong&gt;per&lt;/strong&gt;b subway&lt;br&gt;on&lt;strong&gt;lin&lt;/strong&gt;e fiends&lt;br&gt;w&lt;strong&gt;rit&lt;/strong&gt;e runes&lt;br&gt;&lt;strong&gt;por&lt;/strong&gt;ky hogs&lt;br&gt;fai&lt;strong&gt;r su&lt;/strong&gt;rvey&lt;br&gt;bir&lt;strong&gt;d be&lt;/strong&gt;ak&lt;br&gt;&lt;strong&gt;tur&lt;/strong&gt;nip fields&lt;br&gt;fe&lt;strong&gt;ar c&lt;/strong&gt;obras&lt;br&gt;pithy &lt;strong&gt;lin&lt;/strong&gt;goes&lt;br&gt;icy &lt;strong&gt;tun&lt;/strong&gt;dra&lt;br&gt;kid&lt;strong&gt;nap&lt;/strong&gt; kings&lt;br&gt;Choctaw &lt;strong&gt;tur&lt;/strong&gt;ncoat&lt;br&gt;soun&lt;strong&gt;d ec&lt;/strong&gt;hoes&lt;br&gt;Pomf&lt;strong&gt;ret&lt;/strong&gt; fishway&lt;br&gt;finer &lt;strong&gt;lin&lt;/strong&gt;es&lt;br&gt;&lt;strong&gt;smu&lt;/strong&gt;g nerds&lt;br&gt;ic&lt;strong&gt;ky m&lt;/strong&gt;olerat&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-07-16T19:05:21.807" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18057" PostTypeId="2" ParentId="18052" CreationDate="2015-07-16T19:06:31.270" Score="19" Body="&lt;p&gt;Your combination is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 21210&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is found by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Shifting the letters in each word by the number indicated to get the name of a number and then combining their digits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, the pieces are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LOG $\Rightarrow$ TWO&lt;br&gt;BAR $\Rightarrow$ ONE&lt;br&gt;RUM $\Rightarrow$ TWO&lt;br&gt;UFO $\Rightarrow$ TEN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastActivityDate="2015-07-16T19:06:31.270" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18058" PostTypeId="1" AcceptedAnswerId="18064" CreationDate="2015-07-16T19:09:54.430" Score="2" ViewCount="229" Body="&lt;p&gt;&lt;em&gt;Hey guys, I've been into brainteasers since I was little and recently decided to make a puzzle-related website. I figured that this will be the best place to share some of my problems and see if they are any fun. I guess they are too easy, but hope that will entertain you for a bit.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Just another sequence&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;1, 3, 7, 12, 18, 26, 35, 45, 56, ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;P.S. Please, let me know if you have seen identical puzzle elsewhere, don't want to take someone's credit, even accidentally.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="14140" LastActivityDate="2015-07-16T19:22:02.323" Title="Just another sequence" Tags="|number-sequence|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18059" PostTypeId="2" ParentId="18055" CreationDate="2015-07-16T19:11:21.870" Score="1" Body="&lt;p&gt;&quot;early on the next morning&quot; isn't very precise; let's say this is $n$ hours before noon. Then Jack will take $n+12+(49\times24)=1188+n$ hours, so he'll charge you $1188+n$ dollars; while Will will take $n+(49\times24)=1176+n$ hours and charge you $(1176+n)\times1.01=1187.76+n+0.01n$ dollars. The difference is $0.01n-0.24$ dollars, which is negative since $n\leq12$. So you should go with &lt;strong&gt;Will&lt;/strong&gt; since he's cheaper.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="4421" LastEditDate="2015-07-16T19:45:25.287" LastActivityDate="2015-07-16T19:45:25.287" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18060" PostTypeId="2" ParentId="18053" CreationDate="2015-07-16T19:11:51.637" Score="8" Body="&lt;p&gt;Could it be..&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The number one (1) key is broken since you had to spell out one and eleven where you used actual numbers in other places.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5516" LastActivityDate="2015-07-16T19:11:51.637" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18061" PostTypeId="2" ParentId="18055" CreationDate="2015-07-16T19:12:47.600" Score="1" Body="&lt;p&gt;Fairly straight forward mathing out of this one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$50\space\text{days} \times 24 \space\text{hours} \times\$1\text{/hour} = \$1200 $&lt;br/&gt;&#xA;$50\space\text{days} \times 24 \space\text{hours} \times\$1.01\text{/hour} = \$1212 - 12\space\text{hours} \times\space\$1.01\text{/hour} = \$1199.88 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well I don't know about you, but I don't care much for carrying around 88 cents, so probably Captain Will and leave him a 12 cent tip&lt;/p&gt;&#xA;" OwnerUserId="13916" LastEditorUserId="4421" LastEditDate="2015-07-16T19:43:49.310" LastActivityDate="2015-07-16T19:43:49.310" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18062" PostTypeId="2" ParentId="18051" CreationDate="2015-07-16T19:13:54.660" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; How many total letters are used at least once in the spellings of the words&lt;br&gt;&#xA; So 5+5= 4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is a strong hint&lt;/p&gt;&#xA;" OwnerUserId="13916" LastEditorUserId="9000" LastEditDate="2015-07-16T19:15:29.550" LastActivityDate="2015-07-16T19:15:29.550" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18063" PostTypeId="2" ParentId="18055" CreationDate="2015-07-16T19:17:13.843" Score="1" Body="&lt;p&gt;My assumption is that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Midnight here refers to 00:00, which is the beginning of the day. That would make Jack's trip get in 12 hours earlier than Will's, as well as costing $0.01 less per hour, so it's pretty much a no-brainer that we should take Captain Jack's trip.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-07-16T19:17:13.843" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18064" PostTypeId="2" ParentId="18058" CreationDate="2015-07-16T19:18:21.963" Score="2" Body="&lt;p&gt;We can define the sequence inductively as follows.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The first term is $1$.&lt;/li&gt;&#xA;&lt;li&gt;For all $n$, the $(n+1)$th term of the sequence is the sum of the $n$th term and the smallest number which hasn't appeared among the first $n$ terms and the first $n-1$ differences between consecutive terms.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Or in table form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;real terms: $1,3,7,12,18,26,35,45,56,69,\dots$&lt;br&gt;&#xA;differences: $2,4,5,6,8,9,10,11,13,14,\dots$&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-07-16T19:22:02.323" LastActivityDate="2015-07-16T19:22:02.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18066" PostTypeId="2" ParentId="18055" CreationDate="2015-07-16T19:46:31.577" Score="7" Body="&lt;p&gt;The journey begins early in the morning. The 50th day starts early morning&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; in Chile, and therefore afternoon/evening in Singapore. This means midnight on the 50th day occurs &lt;em&gt;before&lt;/em&gt; noon on the 50th day, so Jack's trip is both faster and cheaper per hour.&lt;br&gt;&lt;br&gt; Choose captain Jack.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-07-16T19:56:38.863" LastActivityDate="2015-07-16T19:56:38.863" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18067" PostTypeId="2" ParentId="17550" CreationDate="2015-07-16T19:52:46.580" Score="7" Body="&lt;p&gt;Transform the colours into numbers by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; using their RGB codes: red is 100, green is 010, blue is 001, cyan is 011, magenta is 101, yellow (not used) is 110, white is 111, black (not used) is 000.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then going round from 1 all the way to 32, we get the code:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 010 100 100 100 010 101 111 100 001 111 111 100 100 001 111 101 010 100 100 101 001 001 001 001 010 011 100 100 011 100 100 001&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Change the spacing to get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 01010010 01000101 01111100 00111111 11001000 01111101 01010010 01010010 01001001 01001110 01000111 00100001&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which we can convert using&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ASCII code&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;to get&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RE|?È}RRING!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So Trap Level 2 is a red herring!&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-07-16T19:52:46.580" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18069" PostTypeId="2" ParentId="18055" CreationDate="2015-07-16T20:04:58.150" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The cruise is from Chile to Singapore, so if we didn't change timezones during our travel, Jack would have arrived afternoon and Will REALLY early in the morning, making the latter also &lt;em&gt;slower&lt;/em&gt;. Jack is cheaper. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="14016" LastEditDate="2015-07-16T20:50:28.873" LastActivityDate="2015-07-16T20:50:28.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18070" PostTypeId="2" ParentId="17550" CreationDate="2015-07-16T20:05:51.827" Score="5" Body="&lt;p&gt;And I've found Trap Level 3 as well! :-D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the numbers 1-32. You need to look inwards to read some of them and outwards to read others. Let's use 1 and 0 respectively to denote these two types of number. Let's also assume (since it gives the right answer) that 32 is towards the top rather than the bottom, so that it's 0 rather than 1. Then the numbers 1-32 give the code 01001110010011110101000001000101.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Space it out as 01001110 01001111 01010000 01000101 and use ASCII code &lt;em&gt;again&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;to get&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; NOPE.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So we've found a 3rd red herring, and the final answer (or Trap Level 4) is the message found by Shawn Holzworth.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-07-16T20:28:59.127" LastActivityDate="2015-07-16T20:28:59.127" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18071" PostTypeId="1" AcceptedAnswerId="18197" CreationDate="2015-07-16T20:08:32.043" Score="10" ViewCount="956" Body="&lt;p&gt;&lt;em&gt;If you're interested in starting the treasure hunt from the very beginning, check out the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16551/treasure-hunt-round-the-world&quot;&gt;prologue&lt;/a&gt;!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Though your methods of solving the required riddle were dubious at best, you were still able to come up with the correct question to ask the concierge in plenty of time for your departure the next day. You were provided with a marked car (courtesy of Lamborghini, of course) and told your departure time would be 7:00 am. Restlessly, you made it through the night, finding yourself standing at the door of your car at 6:59 am. After a few seconds of thumb-twiddling, you hear a click, and the car door remotely unlocks. You swing it open and hop inside, finding a note on the dashboard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Picking up the note, you read:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dear &lt;strong&gt;&lt;em&gt;[insert username here],&lt;/em&gt;&lt;/strong&gt; &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Congratulations on solving the last little brainteaser I left you! Unfortunately, I've erred, and the next clue isn't ready yet. Sorry about that. While you wait, please enjoy these neat factoids about Europe, since you'll be here for a little while yet!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Yours truly,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Bailey M&lt;br&gt;&#xA;  Puzzlemaster for the Treasure Hunt 'Round the World&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You sigh. More waiting? But, for treasure, it will be worth it. Flipping over the letter, you read on the back:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;20 Fun Europe Facts&lt;/strong&gt;  &lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Europe is named after the Phoenician princess &quot;Europa&quot;, from Greek mythology.  &lt;/li&gt;&#xA;  &lt;li&gt;Romania is the oldest country in all of Europe. It hasn't changed its name since 681 AD!  &lt;/li&gt;&#xA;  &lt;li&gt;The Netherlands has the world's densest rail system, with 113.8 kilometers of rail per 1,000 square kilometers.  &lt;/li&gt;&#xA;  &lt;li&gt;Over 50% of Greece's population lives in its capital, Athens.  &lt;/li&gt;&#xA;  &lt;li&gt;The most visited place in the entirety of Europe is, not suprisingly, Disneyland (located in Paris).  &lt;/li&gt;&#xA;  &lt;li&gt;Greenland, an island owned by Denmark, is the second-largest island in the world, surpassed only by Australia.  &lt;/li&gt;&#xA;  &lt;li&gt;Croissants, the famous French pastry, were actually invented in Austria.  &lt;/li&gt;&#xA;  &lt;li&gt;The national airport in Brussels sells more chocolate than any other location on Earth!  &lt;/li&gt;&#xA;  &lt;li&gt;Finland has the most McDonald's per capita in all of Europe.  &lt;/li&gt;&#xA;  &lt;li&gt;Hate mosquitos? Move to Iceland, where there are literally none.  &lt;/li&gt;&#xA;  &lt;li&gt;If you're getting bored of Europe, have no fear - Africa is only a 30-minute ferry ride from Spain.  &lt;/li&gt;&#xA;  &lt;li&gt;Europe is the second-smallest continent, but has the third-largest population, surpassed only by Asia and North America.  &lt;/li&gt;&#xA;  &lt;li&gt;The longest town name in Europe is Lanfairpwllgwyngyllgogerychwyrndrobwllllantysiliogogogoch. I'll wait for you to try to pronounce that.  &lt;/li&gt;&#xA;  &lt;li&gt;The most common name for a pet pig in France? Napoleon.  &lt;/li&gt;&#xA;  &lt;li&gt;It is forbidden to flush your toilet after 10 PM in Belgium, for some reason.  &lt;/li&gt;&#xA;  &lt;li&gt;The largest recorded colony of ants in the world is 6,000 kilometers long, and can be found in northern Italy. I promise I'm not sending you there.  &lt;/li&gt;&#xA;  &lt;li&gt;The Mediterranean Sea is the only sea in Europe that researchers believe has always been filled with water.  &lt;/li&gt;&#xA;  &lt;li&gt;The London Eye in (surprise) London is the world's tallest Ferris Wheel, standing at 135 meters tall.  &lt;/li&gt;&#xA;  &lt;li&gt;The entirety of Europe by land area is still smaller than Canada!  &lt;/li&gt;&#xA;  &lt;li&gt;An estimated 10% of Europeans were conceived in Ikea beds - that's about 73.3 million people, for those counting.  &lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As you sit there, reading the provided facts and attempting to pronounce Lanfairpwllgwyngyllgogerychwyrndrobwllllantysiliogogogoch in your head, you can't help but think it's strange that the Puzzlemaster would slip up like this...after all, he's been preparing this Treasure Hunt for years, hasn't he?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Where is your next destination?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The story continues in the next part, &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18239/treasure-hunt-round-the-world-clue-4&quot;&gt;Treasure hunt &amp;#39;round the world! (clue 4)&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2015-07-21T21:00:45.023" Title="Treasure hunt 'round the world! (clue 3)" Tags="|enigmatic-puzzle|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18072" PostTypeId="1" AcceptedAnswerId="19943" CreationDate="2015-07-16T20:24:27.730" Score="6" ViewCount="1397" Body="&lt;p&gt;It was a slow day at the office. Despite its name, Problem City was fairly quiet as of late. Joey was slowly melting into his chair when he recieved a phone call. Joey slowly swiveled his chair to reach for his cell phone on his desk, but it was slightly out of reach. With a grunt, Joey stretched himself and picked up the phone before it went to voice mail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Hello?&quot; he mumbled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Hi, is this Inspector Joey Solvesalot?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yes, this is he.&quot; Joey said distractedly as he tried to find a comfortable spot again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Hello, uh, are you busy at the moment?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Eh, I got a little bit of work, why?&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;If it's not too much of a problem, can you come to police HQ? We've come across a code of some sort and have no idea how to solve it.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Joey quickly sat up in his chair &quot;Ooh, sounds exciting! I'll be right there.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Joey walked into police HQ, he was greeted by a familiar face. &quot;Ah, hello Officer Jenny, good to see you again.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Hello Inspector Joey, glad to have you help again after &lt;a href=&quot;https://puzzling.stackexchange.com/questions/17618/fowl-perpetrators-caught-as-robberies-add-up&quot;&gt;last time&lt;/a&gt;.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Oh, it's no big deal. What do you need my help with?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Follow me,&quot; Officer Jenny led Joey through a room full of cubicles and into her office. On her desk lay a manila folder. She opened it and pulled out two slips of paper. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I'm sure you've heard recently in the news that the billionaire Iam Rich passed away in his sleep. That wouldn't have been an issue, but his will was locked away in his vault and none of his children know the password.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Couldn't you just hack the vault or break it open?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Typically we would if it was at his home, but he sealed it away in a high-security Wiss Vault, so we can only access it by using the password. Although we did not find a password written anywhere in Mr. Rich's posessions, we did find these two pieces of paper that have the Wiss Vault logo.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;She handed Joey the first sheet:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;E L B D M  E  U  G H I M V A F  G  G H W B  O  E B  C  H 0  O B  C  U D F W  L  I E&#xA;1 1 1 1 1 -1 -1 -1 1 1 0 1 1 1 -1 -1 1 1 1 -1 -1 1 -1 -1 1 -1 1 -1 -1 1 1 1 -1 -1 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&quot;Interesting, what's on the other sheet?&quot; &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;-20 -27 -17 -11 -2 -20 -36 -22 -7 -6 -28 7 -14 -21 -8 -22 -7 8 -17 -11 -20 -13 -12 -23&#xA;-7 -6 8 13 17 -7 -23 -9 6 7 -15 20 -1 -8 5 0 6 21 -4 -10 -7 0 1 -10&#xA;-8 -13 -5 1 10 -1 -8 10 31 6 -16 19 -2 -9 4 16 12 20 -5 -5 -8 6 5 -11&#xA;-26 -12 -23 -17 -8 -1 -42 -28 -3 -12 -34 1 -20 -27 -14 -19 -13 8 -23 -26 -26 -11 -18 -4&#xA;-9 -1 2 2 9 3 -25 -11 5 5 7 18 -3 -3 3 10 4 19 15 -15 -9 23 -1 -6&#xA;1 -5 4 10 19 13 -10 0 33 15 11 28 7 21 13 20 14 29 4 13 1 29 9 21&#xA;18 31 21 27 36 18 2 16 31 32 32 45 24 43 30 32 31 46 21 8 18 43 26 29&#xA;7 24 10 16 25 7 -9 5 20 21 -1 35 35 30 19 6 20 35 10 -3 7 16 15 25&#xA;-14 -21 -11 -5 4 -14 -30 -16 -1 0 -22 13 -8 -15 -2 -16 -1 14 -11 -24 -14 13 15 -17&#xA;-10 -17 -7 -1 8 -10 -26 -12 3 4 -18 17 -4 -11 2 -12 3 18 -7 -20 -10 -3 -2 -13&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&quot;The first sheet we believe contain some sort of instructions. We believe the answer is found on the second sheet. But so far we've made zero sense of these two papers.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Joey took a seat across from Officer Jenny and pondered over the two pieces of paper. &quot;Is it okay if I write over this?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Go ahead, this is just a copy.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nodding, Joey started to scribble something onto the papers. After a while Joey put down his pencil. &quot;Aha! You're right that it made zero sense, but I'm positive I have the answer to this little puzzle.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the password?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Excel friendly formatting thanks to curmudgeon:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pastebin.com/1A5FGWhR&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sheet 1&lt;/a&gt; &lt;br&gt;&#xA;&lt;a href=&quot;http://pastebin.com/FkTiZ5fG&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sheet 2&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Hint #0:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;code&gt;0&lt;/code&gt; in the first sheet does not mean anything in particular. I could've used $\star$ or |, for example.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you really understand hint #0, then try and find something shared between two sheets that isn't obvious at a glance. Using the excel links might be useful in this regard!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oops! While handling the first sheet of paper, Joey accidentally tore it:&lt;br&gt;&lt;code&gt;E L B D M  E  U  G H I M V A F  G  G H W B  O  E B  C  H ( | )  O B  C  U D F W  L  I E&lt;/code&gt;&lt;br&gt;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&lt;code&gt;-----------------------------------&lt;/code&gt;&lt;br&gt;&lt;code&gt;1 1 1 1 1 -1 -1 -1 1 1 (|) 1 1 1 -1 -1 1 1 1 -1 -1 1 -1 -1 1 -1 1 -1 -1 1 1 1 -1 -1 1&lt;/code&gt;&lt;br&gt; Hmm...perhaps it may be useful to make another tear?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Using excel will be very handy in solving this problem.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Joey made his final remark &lt;em&gt;after&lt;/em&gt; solving this puzzle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13881" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-08-13T17:16:13.893" Title="Finding Mr. Rich's Password" Tags="|cipher|word-problem|password|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18073" PostTypeId="2" ParentId="18039" CreationDate="2015-07-16T20:50:10.643" Score="3" Body="&lt;p&gt;My guess is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Nirvana_%28band%29&quot; rel=&quot;nofollow&quot;&gt;the band Nirvana&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My legend will be remembered, Too strong to be dismembered.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wikipedia says, &quot;Despite releasing only three full-length studio albums in their seven-year career, Nirvana has come to be regarded as one of the most influential and important rock bands of the modern era.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No happiness at the parting, What happened at the starting?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After lead singer &lt;a href=&quot;https://en.wikipedia.org/wiki/Death_of_Kurt_Cobain&quot; rel=&quot;nofollow&quot;&gt;Kurt Cobain's death&lt;/a&gt;, the band came to an end, and the many fans were more than sad.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All the lines or all we do, Mix God, religion, and the facts too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many lines in many of Nirvana's songs referred to or alluded to God and/or religion. I can list examples if needed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Counting now, what are we at? Profound, noises, or like a cat?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm not sure. Maybe counting the days since Cobain's death? Also, could these three things reference songs or specific lyrics of the band?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My land is one full of life, You know my cousins, little strife.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://www.google.com/search?q=define%20nirvana&amp;amp;ie=utf-8&amp;amp;oe=utf-8&quot; rel=&quot;nofollow&quot;&gt;The word Nirvana&lt;/a&gt; refers to &quot;a transcendent state in which there is neither suffering, desire, nor sense of self&quot; and &quot;a state of perfect happiness&quot; and &quot;an ideal place&quot;. The &quot;cousins&quot; of Nirvana are places like heaven, paradise, Eden, the promised land, Shangri-La, or maybe even Valhalla&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Started with two, but grew to three, But with some help, who can I be?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The band was formed by just 2 guys: &quot;singer and guitarist Kurt Cobain and bassist Krist Novoselic in Aberdeen, Washington, in 1987.&quot; The third member was added later (the drummer), which changed many times throughout the years.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastActivityDate="2015-07-16T20:50:10.643" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18074" PostTypeId="1" AcceptedAnswerId="18075" CreationDate="2015-07-16T20:55:09.737" Score="22" ViewCount="4538" Body="&lt;p&gt;There is a dangerous bacteria, &lt;em&gt;Checkerichia Coli&lt;/em&gt;, (or &lt;em&gt;C. Coli&lt;/em&gt; for short) which lives in the squares of checkerboards. Squares with this bacteria living in their digestive systems are said to have &amp;quot;C-sickness&amp;quot;. Once a square has &amp;quot;C-sickness,&amp;quot; it is stuck with it forever!&lt;/p&gt;&#xA;&lt;p&gt;Excessive contact with C-sick squares will cause a square to become C-sick. Specifically, if a healthy square has at least &lt;strong&gt;two&lt;/strong&gt; C-sick neighbors (orthogonal, not diagonal), then that square becomes C-sick the next day.&lt;/p&gt;&#xA;&lt;p&gt;For example, on the left we the photo of a see a partially infected checkerboard which was taken yesterday. On the right is a photo taken today.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BPOkO.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Warmup:&lt;/strong&gt; A bioterrorist has 8 samples of the C. Coli strain, each of which can infect a single square. Which squares of an 8 by 8 chessboard can he infect so the infection will spread to the entire board?&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Challenge&lt;/strong&gt;: If the terrorist has only 7 samples, can he still infect the whole board?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;I didn't invent this puzzle, it is quite famous, but the puns are my own.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="11110" LastEditDate="2021-01-09T19:09:06.130" LastActivityDate="2021-01-09T19:09:06.130" Title="Checkerboard Infection" Tags="|checkerboard|algorithm|cellular-automata|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="18075" PostTypeId="2" ParentId="18074" CreationDate="2015-07-16T20:58:41.383" Score="21" Body="&lt;p&gt;This is a pretty common puzzle.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Warm up Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The  8 cells along the diagonal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Advanced Answer Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No, it is impossible.  However, we are instead looking at the number of edges rather than pieces.  In an $n$ x $n$ checkerboard, the perimeter is $4n$, which means that the number of edges our infection must have when fully covering the board is $4n$.  When initially starting with $n-1$ infections, the maximal perimeter of these infections is $4(n-1)$.  Now, I am going to show that new infections at best, maintain the perimeter rather than enlarging it.&lt;br&gt;&lt;br&gt;If the infections enclose an open cell (bordering all 4 sides), it clearly turns into further infection.  However this case causes the overall perimeter of the infection to decrease by 4 because it loses 4 edges and gains 0 new ones.&lt;br&gt;&lt;br&gt;If an infection borders a cell on 3 edges (which is rotationally equivalent), then it will infect the cell causing it lose these 3 edges and gain only 1 new edge, for a loss of 2 edges.&lt;br&gt;&lt;br&gt;Finally is the case where a cell is bordered on 2 edges by other infected cells.  This can either be top/bottom (which is rotationally equivalent to left/right), or the 2 infected cells are touching diagonally and so the healthy cell is bordered by both sides along a corner.  However in either case, this healthy cell gets infected and creates 2 new edges, while losing 2 edges in the process remaining edge neutral.  Thus this shows that edges can never be created, and it is therefore impossible to have less than $n$ initial infections.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14092" LastEditorUserId="10615" LastEditDate="2015-07-16T21:59:17.083" LastActivityDate="2015-07-16T21:59:17.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18076" PostTypeId="2" ParentId="18074" CreationDate="2015-07-16T21:01:10.467" Score="3" Body="&lt;p&gt;For the first case:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Simply choose either diagonal to place the samples. This will cover the board in 7 days.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the second case:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No unless he can move the samples after placing them. The samples end up growing to create a rectangle from the outermost points for each connected blob so it would be impossible with 7 otherwise.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastActivityDate="2015-07-16T21:01:10.467" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18077" PostTypeId="1" CreationDate="2015-07-16T21:54:43.837" Score="4" ViewCount="191" Body="&lt;p&gt;The next problem may be way too easy, but will post it anyway.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Riddick, trying to escape from the space police, lands on a deserted&#xA;  planet. How many cops should the space police send to the planet, so&#xA;  that Riddick will be eventually caught?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;We assume that the planet is a perfect sphere and the speed of Riddick&#xA;  and all cops is the same. All of the characters have perfect&#xA;  information about the location of the others and make instant&#xA;  decisions in real-time. Also, they are dots.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;P.S. The question seems very natural, I hope it is not a repeat.&lt;/p&gt;&#xA;" OwnerUserId="14140" LastEditorUserId="10615" LastEditDate="2015-07-16T22:06:37.467" LastActivityDate="2015-07-16T22:11:22.057" Title="Riddick and the Space Police" Tags="|mathematics|geometry|pursuit-evasion|" AnswerCount="1" CommentCount="3" ClosedDate="2015-07-16T22:14:03.040" ContentLicense="CC BY-SA 3.0" />
  <row Id="18078" PostTypeId="1" AcceptedAnswerId="18085" CreationDate="2015-07-16T22:05:37.193" Score="9" ViewCount="667" Body="&lt;p&gt;I recall coming across this problem in one of my classes senior year of high school. (We only did the first part).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Easy (and probably somewhere on the internet)&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking a standard 3x3x3 Rubik's Cube, what is the minimum number of cuts needed to separate every piece, from all other pieces? (Into a total of 27 pieces)&#xA;&lt;br&gt;&lt;br&gt;&lt;br&gt;&#xA;&lt;strong&gt;Medium (and probably &lt;em&gt;not&lt;/em&gt; somewhere on the internet)&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking a standard 4x4x4 Rubik's Cube, what is the minimum number of cuts needed to separate every piece, from all other pieces? (Into a total of 64 pieces)&#xA;&lt;br&gt;&lt;br&gt;&lt;br&gt;&#xA;&lt;strong&gt;Hard&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking a standard 5x5x5 Rubik's Cube, what is the minimum number of cuts needed to separate every piece, from all other pieces? (Into a total of 125 pieces)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the rules for cutting:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Stacking cut pieces on top of each other is fine&lt;/li&gt;&#xA;&lt;li&gt;Cuts must be straight and completely through the cube&lt;/li&gt;&#xA;&lt;li&gt;You can't ruin the pieces in anyway by cutting them at weird angles&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Disclaimer:  I have guesses as to the answers of the harder version of the puzzles, but have not had my proofs verified&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="14092" LastEditorUserId="8874" LastEditDate="2016-03-23T15:37:47.270" LastActivityDate="2016-03-23T15:37:47.270" Title="Separating the Pieces of a Rubiks Cube" Tags="|geometry|rubiks-cube|dissection|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18079" PostTypeId="2" ParentId="18074" CreationDate="2015-07-16T22:06:46.067" Score="5" Body="&lt;p&gt;This isn't really an answer, but more of a hyper-extended comment. I have made a program that simulates the infection of a board in JavaScript, to check on any particular setups.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;TL;DR&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Right-click on any HTML page and select &amp;ldquo;Inspect Element&amp;rdquo; (It's the bottom option in Firefox and Chrome. I don't know where it is otherwise, probably the same place.) Navigate to your browsers console. Copy the minified code and paste it into the command line into the console. (On Firefox, you will need to type &amp;ldquo;Allow Pasting&amp;rdquo; before being able to paste it.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To create a new board, simply choose a name for your board (let's say you choose &lt;code&gt;bob&lt;/code&gt;) and type &lt;code&gt;var bob = new Board();&lt;/code&gt;.&lt;br&gt;&#xA;To insert an infected cell, type at position $(x,y)$ &lt;code&gt;bob.infect(x,y);&lt;/code&gt;. &lt;em&gt;NOTE: The counting is zero-based. I have furnished a table in the commented code within the &lt;code&gt;Board&lt;/code&gt; class that tells you which cell is which.&lt;/em&gt;&lt;br&gt;&#xA;To see what the board looks like, type either &lt;code&gt;bob.print()&lt;/code&gt; or &lt;code&gt;console.log(bob.print());&lt;/code&gt;. The former only displays if it is the last line you have typed, so I'd suggest the latter.&lt;br&gt;&#xA;To advance to the next day/stage of infection, type &lt;code&gt;bob.nextDay()&lt;/code&gt;.&lt;br&gt;&#xA;To see what the value at position $(x,y)$ is, type &lt;code&gt;console.log(bob.board[y][x]);&lt;/code&gt;.&lt;br&gt;&#xA;To see what day/stage of the infection, type &lt;code&gt;bob.day&lt;/code&gt;. There is a zeroeth day, and this means that this is the day before anything has happened; the insertion of infected cells. You can still insert cells on any other day, however.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Code&lt;/h1&gt;&#xA;&#xA;&lt;h2&gt;Commented Code&lt;/h2&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;function Board(){&#xA;    this.width = this.height = 8;   // an 8x8 chess board&#xA;    var x = new Array();            // creating the 2d array&#xA;    for(var Zi=0;Zi&amp;lt;this.height;Zi++){&#xA;        x[Zi] = new Array();        // creating the y-axis of the 2d array&#xA;        for(var Zj=0;Zj&amp;lt;this.width;Zj++){&#xA;            x[Zi][Zj] = 0;          // populating the y-axis with 0&#xA;        }&#xA;    }&#xA;    this.board = x;                 // setting board to the 2d array&#xA;    this.day   = 0;                 // starting the board out on the first day&#xA;/*&#xA;     0 1 2 3 4 5 6 7 } i&#xA;    +---------------&#xA;  0 |0 0 0 0 0 0 0 0&#xA;  1 |0 0 0 0 0 0 0 0 &#xA;  2 |0 0 0 0 0 0 0 0 &#xA;  3 |0 0 0 0 0 0 0 0 &#xA;  4 |0 0 0 0 0 0 0 0 &#xA;  5 |0 0 0 0 0 0 0 0 &#xA;  6 |0 0 0 0 0 0 0 0 &#xA;  7 |0 0 0 0 0 0 0 0 &#xA;  v&#xA;  j&#xA;*/&#xA;}&#xA;&#xA;Board.prototype.print = function(){&#xA;    var string = &quot;\n&quot;;                              // for display purposes&#xA;    for(var i=0;i&amp;lt;this.height;i++){&#xA;        for(var j=0;j&amp;lt;this.width;j++){&#xA;            if(this.board[i][j])                    // displaying if occupied&#xA;                string += this.board[i][j] + &quot; &quot;;   // displaying the (j,i)th entry&#xA;            else&#xA;                string += &quot;  &quot;;&#xA;        }&#xA;        string += &quot;\n&quot;;                             // spacing for the next array&#xA;    }&#xA;    return string;&#xA;}&#xA;&#xA;Board.prototype.infect = function(x,y){&#xA;    // origin = top-left, (0,0)&#xA;    this.board[y][x] = 2;   // setting the (x,y)th entry to be “newly infected”&#xA;}&#xA;&#xA;Board.prototype.nextDay = function(){   // advances the day, spreads infection&#xA;    // getting rid of any 2's from the last day.&#xA;    for(var y=0;y&amp;lt;this.height;y++){&#xA;        for(var x=0;x&amp;lt;this.width;x++){&#xA;            // !x return 0 if x != 0 and 1 if x == 0, so &#xA;            // !!x returns 1 if x !=0 and 0 if x == 0&#xA;            // ~~x simply converts x to a number&#xA;            this.board[y][x] = ~~!!this.board[y][x];&#xA;        }&#xA;    }&#xA;&#xA;    // creating the board to be the next day's&#xA;    var nextDayBoard = Array.from(this.board);&#xA;    for(var y=0;y&amp;lt;this.height;y++){&#xA;        for(var x=0;x&amp;lt;this.width;x++){&#xA;            // no need to act on the cell if its already infected&#xA;            if(!this.board[y][x]){&#xA;                // establishing “null” value if not present&#xA;                var topCell = 0;    &#xA;                // checking if there is a cell orthogonally above&#xA;                if(y&amp;gt;0){    &#xA;                    // get the top cell's coordinate &#xA;                    var topCell = this.board[y-1][x];&#xA;                }&#xA;&#xA;                // establishing “null” value&#xA;                var leftCell = 0;&#xA;                // checking if there is a cell orthogonally left&#xA;                if(x&amp;gt;0){&#xA;                    // get the left cell's coordinates&#xA;                    var leftCell = this.board[y][x-1];&#xA;                }&#xA;&#xA;                // establishing “null” value&#xA;                var rightCell = 0;&#xA;                // checking for cell orthogonally right&#xA;                if(x&amp;lt;this.width-1){ // minus 1 due to zero-based indexing&#xA;                    var rightCell = this.board[y][x+1];&#xA;                }&#xA;&#xA;                // establishing “null” value&#xA;                var bottomCell = 0;&#xA;                // checking for cell orthogonally below&#xA;                if(y&amp;lt;this.height-1){    // minus 1 due to zero-based indexing&#xA;                    var bottomCell = this.board[y+1][x];&#xA;                }&#xA;&#xA;                // creating array of adjacent cells&#xA;                var adjacentCells = [topCell,leftCell,rightCell,bottomCell];&#xA;&#xA;                /* &#xA;                    Now, the current cell is infected iff &#xA;                    there are two orthogonally adjacent &#xA;                    cells. Therefore, taking the sum of&#xA;                    adjacentCells will give us how many&#xA;                    cells are infected that are adjacent&#xA;                    to the current cell.&#xA;                */&#xA;&#xA;                // taking the sum of adjacentCells&#xA;                var infected = adjacentCells.reduce(function(previousCell,currentCell){&#xA;                    if(currentCell != 2) return previousCell + currentCell;&#xA;                    return previousCell;&#xA;                },0);&#xA;&#xA;                // if the cell has become infected:&#xA;                if(infected&amp;gt;=2){&#xA;                    // setting the current sell to newly infected state,&#xA;                    // so there is now no way for it to effect other&#xA;                    // cells&#xA;                    nextDayBoard[y][x] = 2;&#xA;                }&#xA;                // we do not need to change the value otherwise,&#xA;                // so an else statement is not required&#xA;            }&#xA;        }&#xA;    }&#xA;    // incrementing the day.&#xA;    this.day++;&#xA;    // setting the board to the next-day board.&#xA;    this.board = nextDayBoard;&#xA;}&#xA;&#xA;// different variables in for loops to makes sure&#xA;// no variable conflicts happen&#xA;&#xA;// ----&#xA;&#xA;// creating test board&#xA;var board = new Board();&#xA;board.infect(1,1);&#xA;board.infect(3,1);&#xA;console.log(board.print());&#xA;board.nextDay();&#xA;console.log(board.print());&#xA;&#xA;// success!&#xA;&#xA;// creating example board&#xA;var example = new Board();&#xA;// group 1&#xA;example.infect(1,1);&#xA;example.infect(2,1);&#xA;// group 2&#xA;example.infect(0,4);&#xA;example.infect(0,5);&#xA;example.infect(0,6);&#xA;example.infect(1,5);&#xA;// group 3 (last)&#xA;example.infect(3,4);&#xA;example.infect(4,3);&#xA;example.infect(5,4);&#xA;example.infect(5,6);&#xA;example.infect(6,5);&#xA;example.infect(6,7);&#xA;example.infect(7,6);&#xA;console.log(example.print());&#xA;example.nextDay();&#xA;console.log(example.print());&#xA;example.nextDay();&#xA;console.log(example.print());&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h2&gt;Minified Code&lt;/h2&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;function Board(){this.width=this.height=8;var x=new Array();for(var Zi=0;Zi&amp;lt;this.height;Zi++){x[Zi]=new Array();for(var Zj=0;Zj&amp;lt;this.width;Zj++){x[Zi][Zj]=0;}} this.board=x;this.day=0;} Board.prototype.print=function(){var string=&quot;n&quot;;for(var i=0;i&amp;lt;this.height;i++){for(var j=0;j&amp;lt;this.width;j++){if(this.board[i][j]) string+=this.board[i][j]+&quot; &quot;;else string+=&quot; &quot;;} string+=&quot;n&quot;;} return string;} Board.prototype.infect=function(x,y){this.board[y][x]=2;} Board.prototype.nextDay=function(){for(var y=0;y&amp;lt;this.height;y++){for(var x=0;x&amp;lt;this.width;x++){this.board[y][x]=~~!!this.board[y][x];}} var nextDayBoard=Array.from(this.board);for(var y=0;y&amp;lt;this.height;y++){for(var x=0;x&amp;lt;this.width;x++){if(!this.board[y][x]){var topCell=0;if(y&amp;gt;0){var topCell=this.board[y-1][x];} var leftCell=0;if(x&amp;gt;0){var leftCell=this.board[y][x-1];} var rightCell=0;if(x&amp;lt;this.width-1){var rightCell=this.board[y][x+1];} var bottomCell=0;if(y&amp;lt;this.height-1){var bottomCell=this.board[y+1][x];} var adjacentCells=[topCell,leftCell,rightCell,bottomCell];var infected=adjacentCells.reduce(function(previousCell,currentCell){if(currentCell!=2)return previousCell+currentCell;return previousCell;},0);if(infected&amp;gt;=2){nextDayBoard[y][x]=2;}}}} this.day++;this.board=nextDayBoard;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7307" LastActivityDate="2015-07-16T22:06:46.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18080" PostTypeId="2" ParentId="18077" CreationDate="2015-07-16T22:11:22.057" Score="3" Body="&lt;p&gt;Assuming the space police want to send as few units as possible&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; they should send three (3) units&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is a standard &lt;a href=&quot;http://www.opengrey.eu/item/display/10068/350787&quot; rel=&quot;nofollow&quot;&gt;pursuit game&lt;/a&gt; on $S^2$. Riddick can evade any two pursuers indefinitely (for example, by moving in a direction perpendicular to the great circle containing the locations of the pursuers as they approach him), but three pursuers can inevitably trap him in a kind of &quot;collapsing triangle&quot;, as well as with other strategies.&lt;br&gt;&lt;br&gt;Of course, he'll probably just kick their arses to Hades and back, but...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2015-07-16T22:11:22.057" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18081" PostTypeId="2" ParentId="105" CreationDate="2015-07-16T22:19:10.857" Score="3" Body="&lt;p&gt;1) The Answer must be simple. (One word, no sentence, no theorem etc...)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The Answer must be something that ANYONE could think of - regardless of age , even though in this case you are thinking of kids. Don't pick some specific toy i.e., think back at those old kid books where you learn words. That's the type of word you want. (Time, Wind, Clock, Mouth, Flower, Table, Dog, etc...)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Then it gets tricky, you can choose to describe your word physically, or describe its actions, or describe its impact etc... I find it makes a better riddle if you stick to one of those. Metaphors are an obvious choice to describe it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4)The writing pattern, I'm no writer, I'm not sure what to call it. I like using similar structures for the sentences, or using repetition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5)Try not dealing with anything too abstract&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I silence birds and stop Mills,&lt;br&gt;&#xA;  Hide hills and make rivers still.&lt;br&gt;&#xA;  Under my blanket animals slumber,&lt;br&gt;&#xA;   I cannot stay, but I may linger.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;OR more grim (this one is a bit more obscure/abstract though, treating with stuff kids might not understand) :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What heart did men of old fear,&lt;br&gt;&#xA;   When my breath came knocking yesteryear.&lt;br&gt;&#xA;  Harrows the old and weak down to the bone,&lt;br&gt;&#xA;   And makes the dead as hard as stone.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastEditorUserId="11251" LastEditDate="2015-07-16T22:36:41.583" LastActivityDate="2015-07-16T22:36:41.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18082" PostTypeId="2" ParentId="17995" CreationDate="2015-07-16T22:23:54.397" Score="7" Body="&lt;p&gt;I don't think there's any hard-and-fast way to rank the boards, but here are a few thoughts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $z = $ the number of zombies.&lt;br&gt;&#xA;Let $g = $ the number of ghosts.&lt;br&gt;&#xA;Let $v = $ the number of vampires.&lt;br&gt;&#xA;Let $n = $ the size of the board ($n=4$ indicates a 4 $\times$ 4 board).&lt;br&gt;&#xA;Let $s = $ the sum of all the numbers around the edges of the board.&lt;br&gt;&#xA;Let $m = $ the number of mirrors on the board.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;As $(g+v)/z$ increases (there are more ghosts/vampires and less zombies), the difficulty will increase as well.  This is because ghosts and vampires have two different &quot;states&quot;, while zombies have only one (always visible).&lt;br&gt;&#xA;** &lt;em&gt;Note that if $g = 0$ or $v = 0$ then there will be a point where the difficulty decreases again, since a board with all ghosts and one zombie is probably easier to solve than a board with several of each.  (All you have to do is figure out where to put the one zombie.)&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You could use the sum of the numbers around the board as a potential metric, though it wouldn't be perfect.  On a 4$\times$4 grid, a board with all $0$s (total $=0$) is trivial, as is a board with all $4$s (total $=64$).  So I might try something like $32 - |(32 - s)|$, or, to generalize for any board size:  $\frac{4n^2}2 - \left|\frac{4n^2}2 - s\right|$.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You could count the number of mirrors on the board, which would end up providing a very similar metric to my second point above, because both a board with no mirrors and a board with all mirrors will both be trivial to solve.  Thus you could use a similar equation: $n^2 - | n^2 - m|$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;An overall metric could combine these different observations (and weight them accordingly) to try to rank the difficulty of each board:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $w_1 = $ the weight for the first metric.&lt;br&gt;&#xA;Let $w_2 = $ the weight for the second metric.&lt;br&gt;&#xA;Let $w_3 = $ the weight for the third metric.&lt;br&gt;&#xA;Let $d = $ the overall difficulty of the board.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you could use an equation like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d = w_1(\frac{g+v}z) + w_2(\frac{4n^2}2 - \left|\frac{4n^2}2 - s\right|) + w_3(n^2 - | n^2 - m|)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to rank the difficulty of the board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It probably won't be perfect, but it might at least give you a starting point.&lt;/p&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="4421" LastEditDate="2015-07-22T14:48:23.647" LastActivityDate="2015-07-22T14:48:23.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18083" PostTypeId="2" ParentId="18078" CreationDate="2015-07-16T23:01:00.790" Score="3" Body="&lt;p&gt;&lt;strong&gt;ANSWER&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For $n\ x\ n\ x\ n$ cube, you need $3\lceil log_2n\rceil$ cuts. (base 2, upper rounding). &lt;br&gt; This means:&lt;br&gt; $3x3x3$ - 6 cuts,&lt;br&gt; $4x4x4$ - 6 cuts, &lt;br&gt;$5x5x5$ - 9 cuts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If $n$ is a power of two, the way to cut through the pieces is obvious - every time stack the pieces on top of each other and split them in halves. &lt;br&gt; This shows that for power's of two, $3\lceil log_2(n)\rceil$ cuts are exactly enough. &lt;br&gt;&lt;br&gt; In order to show that at least $3\lceil log_2n\rceil$ cuts are needed, we notice that after $k$ cuts, there is a piece with dimensions $(x,y,z)$, such that:&lt;br&gt; $\lceil log_2{\frac{n}{x}}\rceil+\lceil log_2{\frac{n}{y}}\rceil+\lceil log_2{\frac{n}{z}}\rceil \leq k$. &lt;br&gt;&lt;br&gt; This can be proved easily with induction, using: &lt;br&gt; $\lceil log_2{\frac{2n}{y}}\rceil \leq1+\lceil log_2{\dfrac{n}{y}}\rceil.$ &lt;br&gt;&lt;br&gt; Since at the end all cubes should have dimensions $1x1x1$, we get: &lt;br&gt; $3\lceil log_2{n}\rceil \leq k$ &lt;br&gt; and therefore we need at least $3\lceil log_2n\rceil$ cuts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14140" LastEditorUserId="14140" LastEditDate="2015-07-17T16:41:35.103" LastActivityDate="2015-07-17T16:41:35.103" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18084" PostTypeId="2" ParentId="18078" CreationDate="2015-07-16T23:32:27.717" Score="0" Body="&lt;p&gt;For the 3x3:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Make 2 cuts on each of the following 3 faces: Front, Left, Top. This leaves you with 27 cubes. You're done!&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Total cuts = 6&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For the 4x4:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Cut down the center of the top, left, and front faces. This leaves you with 8 2x2 cubes.&lt;br&gt;&#xA;&lt;em&gt;Cuts used: 3&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Stack the 8 2x2 cubes on top of each other and make 2 cuts on the top face. This leaves you with 32 1x2 pieces.&lt;br&gt;&#xA;&lt;em&gt;Cuts used: 3 + 2 = 5&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Arrange the 32 1x2 pieces in a line so that they can all be cut down the their center line with 1 cut. This leaves you with 64 small cubes. You're done!&lt;br&gt;&#xA;&lt;em&gt;Cuts used: 3 + 2 + 1 = 6&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Total cuts = 6&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For the 5x5:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;WIP&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-07-16T23:44:09.787" LastActivityDate="2015-07-16T23:44:09.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18085" PostTypeId="2" ParentId="18078" CreationDate="2015-07-16T23:48:43.223" Score="4" Body="&lt;p&gt;&lt;strong&gt;Theorem:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The minimum number of cuts to separate an $a\times b\times c$ block into $1\times 1\times 1$ blocks is $f(a)+f(b)+f(c)$, where $f(n)=\lceil\log_2 n\rceil$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Let the &quot;age&quot; of a block be the quantity $f(a)+f(b)+f(c)$ mentioned above. We first show that cutting a block with age $g$ leaves two blocks, with the &quot;older&quot; of the two having an age of at least $g-1$. &lt;br&gt;&lt;br&gt; Suppose you cut an $a\times b\times c$ block along the $x$ axis into $a'\times b\times c$ and $a''\times b\times c$ blocks, where $a'\ge a''$. This means $a'\ge a/2$, so $\log_2 a'\ge (\log_2 a) - 1$, so $f(a')\ge f(a)-1$, proving the age of the larger block decreased by at most one. &lt;br&gt;&lt;br&gt;  This provides a lower bound: if the age of the initial block is $g$, then its takes $g$ cuts to make sure that every piece has an age of zero (namely, is a 1x1x1 block). &lt;br&gt;&lt;br&gt;  To attain this lower bound, at any point, line up all the pieces you have so far with a volume of more than 1, and cut them so their longest axis gets cut into as close to half has possible. This will leave a bunch of blocks who are all one year younger than their fathers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Since the ages of a $3$, $4$ and $5$ cube are $6, 6$ and $9$ respectively, these are the required number of cuts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="14092" LastEditDate="2015-07-17T16:38:02.253" LastActivityDate="2015-07-17T16:38:02.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18086" PostTypeId="1" AcceptedAnswerId="18087" CreationDate="2015-07-17T00:02:49.257" Score="24" ViewCount="948" Body="&lt;p&gt;You need to turn the following 6x6 square into a &lt;em&gt;word square&lt;/em&gt; (i.e. so that the 1st row and 1st column both contain the same six-letter word, as do the 2nd row and 2nd column, etc.) by &lt;strong&gt;swapping each letter with another letter a knight’s move away from it&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;T O A Y S M&#xA;S H L T T H&#xA;R E A M S E&#xA;R S E E M R&#xA;R S R Y S T&#xA;E I N E M I&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;How can you do it, and what are the six resulting words?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;sub&gt;This puzzle is originally due to Dudeney; I found it &lt;a href=&quot;https://www.futilitycloset.com/2014/11/10/warnsdorffs-rule/&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/sub&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2016-09-27T01:54:16.250" LastActivityDate="2016-09-27T01:54:16.250" Title="The Knight of Words" Tags="|word|wordplay|chess|word-square|" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18087" PostTypeId="2" ParentId="18086" CreationDate="2015-07-17T00:24:35.653" Score="12" Body="&lt;p&gt;The word square is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;L A S H E S&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;A R T E R Y&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;S T O R M S&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;H E R M I T&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;E R M I N E&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;S Y S T E M&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The strategy of solving it goes like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I identified the unique letters (&lt;code&gt;L&lt;/code&gt;, &lt;code&gt;O&lt;/code&gt; and &lt;code&gt;N&lt;/code&gt;) and made the swaps to bring them on the top left to bottom right diagonal. These swaps put &lt;code&gt;A&lt;/code&gt; and &lt;code&gt;S&lt;/code&gt; in their positions and then looking for other swaps to bring the other &lt;code&gt;A&lt;/code&gt; and &lt;code&gt;S&lt;/code&gt;, mirrored across the diagonal, into position revealed more letters. When stuck, identifying a pair of letters that appears only once also helps. I identified the &lt;code&gt;I&lt;/code&gt; pair in the bottom row and &lt;code&gt;Y&lt;/code&gt; pair in the fourth column and worked to get them into positions such that they mirror across the diagonal. This also revealed the correct position for some other letters, and using these two techniques, all the letters slowly but surely fell into place.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another useful technique pointed out by rand al'thor in comments:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since the same letters must appear in symmetric positions about the diagonal, letters appearing in pairs cannot occupy the diagonal. This means, for instance that the &lt;code&gt;I&lt;/code&gt; in the 2nd column can't move into the 1st column, and the &lt;code&gt;T&lt;/code&gt;s in the 2nd row can't move into the 1st row.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="5840" LastEditDate="2015-07-18T11:27:54.920" LastActivityDate="2015-07-18T11:27:54.920" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18089" PostTypeId="1" AcceptedAnswerId="18090" CreationDate="2015-07-17T00:38:03.510" Score="12" ViewCount="463" Body="&lt;p&gt;You may remember the &lt;a href=&quot;https://puzzling.stackexchange.com/q/18074/10615&quot;&gt;C. Coli&lt;/a&gt; bacteria, which infects checkerboard squares. There is a similar disease which plagues cubes, known as &lt;strong&gt;cubic lice&lt;/strong&gt;. Once a cube gets infested with cubic lice, it will be infested for the rest of its life!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A refresher on cube biology: cubes live together in colonies, which consist of $1000$ cubes tightly packed together in a $10\times 10 \times 10$ cubical pattern. If a healthy cube is ever adjacent to at least &lt;strong&gt;three&lt;/strong&gt; infested cubes, then that cube will itself become infested the next day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bioterrorist wants to infest an entire colony of cubes with cubic lice. He plans to break into a WHO lab and steal several cubic larvae, each of which can cause a single cube to become infested. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is fewest number of larvae he needs to steal to be able to infect the entire colony?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As usual, you should prove your answer is correct, by showing how he can infect the colony with that many larvae, and why he can't with fewer. &lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-07-17T20:24:37.657" Title="Cubical Infection" Tags="|algorithm|cellular-automata|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18090" PostTypeId="2" ParentId="18089" CreationDate="2015-07-17T01:03:38.347" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Answer - we need at least 100.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;br&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Just as before, after each iteration the surface of the infected part does not increase. Since the surface of a unit cube is 6 and the surface of the big cube is 600, we need at least 600/6=100 infected cubes. This is my pattern for infecting 10x10x10 block with 200 larvae: &lt;br&gt;&lt;br&gt; 1st layer: &lt;br&gt; X 0 0 0 0 0 0 0 0 0 &lt;br&gt; 0 X 0 0 0 0 0 0 0 0 &lt;br&gt; 0 0 X 0 0 0 0 0 0 0 &lt;br&gt; 0 0 0 X 0 0 0 0 0 0 &lt;br&gt;............................ &lt;br&gt; 0 0 0 0 0 0 0 0 0 X &lt;br&gt;&lt;br&gt; 2nd layer: &lt;br&gt; 0 X 0 0 0 0 0 0 0 0 &lt;br&gt; 0 0 X 0 0 0 0 0 0 0 &lt;br&gt; 0 0 0 X 0 0 0 0 0 0 &lt;br&gt; 0 0 0 0 X 0 0 0 0 0 &lt;br&gt;............................ &lt;br&gt; X 0 0 0 0 0 0 0 0 0 &lt;br&gt;&lt;br&gt; 10th layer: &lt;br&gt; 0 0 0 0 0 0 0 0 0 X &lt;br&gt; X 0 0 0 0 0 0 0 0 0 &lt;br&gt; 0 X 0 0 0 0 0 0 0 0 &lt;br&gt; 0 0 X 0 0 0 0 0 0 0 &lt;br&gt;............................ &lt;br&gt; 0 0 0 0 0 0 0 0 X 0 &lt;br&gt;&lt;br&gt; However, there also exist patterns that will work, for which no two infested cubes are in the same line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14140" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-17T17:16:07.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18091" PostTypeId="5" CreationDate="2015-07-17T01:21:45.573" Score="0" Body="&lt;p&gt;A &lt;em&gt;word square&lt;/em&gt; is a square grid of letters in which every row and column forms a word. A &lt;em&gt;regular square&lt;/em&gt; is a word square in which the horizontal and vertical words are the same. A &lt;em&gt;double square&lt;/em&gt; is a word square in which the horizontal and vertical words are different.&lt;/p&gt;&#xA;&lt;p&gt;This is an example of a regular word square:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;H E A R T   &#xA;E M B E R   &#xA;A B U S E   &#xA;R E S I N   &#xA;T R E N D&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;The size of one side of the square is known as the &lt;em&gt;order&lt;/em&gt; of the square. The example above has order 5.&lt;/p&gt;&#xA;&lt;p&gt;Generalizations of word squares to other shapes are called &lt;em&gt;forms&lt;/em&gt;. Common forms include diamonds, rectangles, and hollow squares.&lt;/p&gt;&#xA;&lt;p&gt;Word squares (and other forms) can be clued just like normal crossword puzzles.&lt;/p&gt;&#xA;" OwnerUserId="12109" LastEditorUserId="69582" LastEditDate="2020-12-06T17:03:16.010" LastActivityDate="2020-12-06T17:03:16.010" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="18092" PostTypeId="4" CreationDate="2015-07-17T01:21:45.573" Score="0" Body="A word square is a square grid of letters in which every row and column forms a word. They can be clued just like normal crossword puzzles. Use with [crosswords] or [cryptic-crosswords] if the word square is clued as such." OwnerUserId="12109" LastEditorUserId="69582" LastEditDate="2020-12-06T17:22:34.483" LastActivityDate="2020-12-06T17:22:34.483" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="18093" PostTypeId="2" ParentId="17753" CreationDate="2015-07-17T02:58:51.623" Score="5" Body="&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First of all, if we assumed instantaneous signal transmission (the whole chair knows exactly when the wheel stops) and the wheel stopped &lt;em&gt;instantly&lt;/em&gt; friction wouldn't come into play, since the momentum would instantly disappear. From this we see the phenomena you're talking about arises from the time it takes the whole body to &lt;em&gt;notice&lt;/em&gt; the wheel stopped.&lt;br&gt;&lt;br&gt; Now, to exacerbate this quality we may imagine that Sandra is floating around in space holding the wheel with some extra large tweezers, in such a way that when the wheel stops it takes a noticeable time, say $t$ seconds, for the signal to travel to Sandra.&lt;br&gt;&lt;br&gt;Now it's not that hard to figure out what will happen! As the wheel stops, Sandra will keep spinning for $t$ seconds, time during which friction will continue to act as an opposing force and Sandra will continue spinning! &lt;br&gt;&lt;br&gt;So when Sandra notices that the wheel stopped and her chair stops spinning, the wheel will continue 'seeing' the friction opposing Sandra's rotation for $t$ seconds, experiencing said force during this time and making it move in the opposite direction as Sandra did. In fact, the wheel's acceleration will be proportional to the friction.&lt;br&gt;&lt;br&gt;Ultimately the reverse angular momentum that Sandra gained during the arresting of the flywheel momentum will restore her orientation at rest to precisely her starting orientation. This phenomenon is known as &lt;a href=&quot;https://www.youtube.com/watch?v=og5h4QoqIFs&quot; rel=&quot;nofollow&quot;&gt;damping-induced self-recovery&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2002" LastEditorUserId="4219" LastEditDate="2015-07-19T08:11:59.703" LastActivityDate="2015-07-19T08:11:59.703" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18094" PostTypeId="1" AcceptedAnswerId="18098" CreationDate="2015-07-17T03:11:04.110" Score="11" ViewCount="1396" Body="&lt;p&gt;Star-Lord is on a deserted planet along with two space policemen. If he runs into either of them, he will get immediately arrested. Fortunately for Star-Lord, somewhere on two opposite ends of the planet he has hidden a jetpack and a battery for it, which are visible only to him.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming all characters have the same speed, have full information about others' locations and can take decisions in real time, can the policemen prevent Star-Lord from getting his jetpack and its battery and flying away from the planet? What if there is only one policeman?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Remark: The policemen are aware that Star-Lord has hidden his jetpack and its battery in two antipodal positions. However, their exact locations are unknown to them. The policemen can not see the items even if SL picks one them or they stand on top of the other.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="14140" LastEditorUserId="14140" LastEditDate="2015-07-23T02:48:27.657" LastActivityDate="2015-07-23T02:48:27.657" Title="Star-Lord and the Space Police" Tags="|mathematics|geometry|pursuit-evasion|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="18095" PostTypeId="2" ParentId="18049" CreationDate="2015-07-17T03:31:43.667" Score="7" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4 + 4 = 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the answer is the number of Es in the expression to the left of and including the equals sign, when rendered as words. Multiplication is read as &quot;times&quot; for these purposes, and 0 is read as &quot;zero&quot; as opposed to &quot;nought&quot;. In this case, &quot;4 + 4 = &quot; is read as &quot;four plus four equals&quot; or &quot;four add four equals&quot;, for a total of one E.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14149" LastActivityDate="2015-07-17T03:31:43.667" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18097" PostTypeId="1" AcceptedAnswerId="18100" CreationDate="2015-07-17T03:37:34.820" Score="22" ViewCount="1348" Body="&lt;p&gt;As in &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18074/checkerboard-infection&quot;&gt;this previous puzzle&lt;/a&gt;, we are again playing with our favorite bacteria, C. Coli, whose natural habitat is an 8x8 checkers board. These are very nasty bacteria - once they are in a cell of a checkerboard, that cell will be infected forever. They spread in a very peculiar way: Each night, at midnight, any cell with two neighbors infected by C. Coli will itself become infected. They are harmless to humans, except in large doses - if a whole checkerboard is infected and a human plays on it, they will suffer severe side-effects, including procrastination, prolonged consternation, and chronic cogitation (but any lesser dose is harmless).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scientists at the lab know that their checkerboard has potentially been compromised and have agreed to make a grave sacrifice: They will not play checkers for the five weeks after the board was compromised. If &lt;strong&gt;every&lt;/strong&gt; cell of their board is infected, they will discard it and apply for a grant to purchase another checkerboard. Otherwise, they will resume play (even if there was only one uninfected cell).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given 8 samples of C. Coli, placed such that they will eventually infect the entire checkerboard, what is the longest amount of time it could take to infect the whole board? How can the bacteria be arranged to obtain this maximum? Is five weeks sufficient wait time to ensure that the board will never be fully infected?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;strong&gt;Bonus:&lt;/strong&gt; Can you prove that the arrangement taking the maximal length of time is unique up to reflection &amp;amp; rotation?)&lt;/p&gt;&#xA;" OwnerUserId="4765" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-07-18T23:56:21.630" Title="Checkerboard Infection: The Aftermath" Tags="|mathematics|checkerboard|algorithm|cellular-automata|" AnswerCount="4" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="18098" PostTypeId="2" ParentId="18094" CreationDate="2015-07-17T04:16:10.847" Score="14" Body="&lt;p&gt;Here's my answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To guarantee SL will never reach one of his items, the policemen &lt;strong&gt;need&lt;/strong&gt; to guarantee he'll stay in one hemisphere. For if he doesn't it's easy to give initial conditions with which he can get both of his items and escape.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once this is settled, we can easily say the problem can be solved&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; always.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The strategy being:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; While one of the policemen chases him, as in the famous &lt;a href=&quot;ftp://math.feld.cvut.cz/pub/tkadlec/papers/tdarcr.pdf&quot;&gt;dog-rabbit problem&lt;/a&gt;, the other one must mirror his movements relative to the great circle perpendicular his and SL's initial positions. This way, you guarantee he'll never leave one hemisphere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bonus:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; With this, we see you need only one policemen to keep him at the planet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second Bonus FTW:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In fact, the second policeman can catch SL. Once SL is trapped in an hemisphere, send the second policeman to the pole of the hemisphere (the furthest point from the great circle). After this, since their speeds are all equal, this policeman can &lt;strong&gt;always&lt;/strong&gt; be in the same meridian as SL (meridians in this case are the great circles orthogonal to the equator) and, additionally, move towards him, since he moves through parallels faster than SL.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2002" LastEditorUserId="2002" LastEditDate="2015-07-17T05:17:51.513" LastActivityDate="2015-07-17T05:17:51.513" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="18099" PostTypeId="2" ParentId="18097" CreationDate="2015-07-17T04:25:12.353" Score="3" Body="&lt;p&gt;Oops! Turns out that I didn't get the answer, but here was my thought process anyways:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given 8 samples of C. Coli on an 8x8 board, it will take at most:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 28 days to infect the entire board. Thus waiting 5 weeks (35 days) is sufficient to determine if the board is fully infected or not. &lt;br/&gt;&lt;br&gt; &lt;strong&gt;TL;DR solution&lt;/strong&gt;: The slowest rate of growth possible is 2 squares a night: $(64-8)/2 = 28$ nights to fully infect the board. &lt;br&gt;&lt;br&gt; &lt;strong&gt;Full answer&lt;/strong&gt;: An uninfected square will become infected if it is surrounded on at least two sides by infected squares. In order to completely flood the board, each sample must occupy an unique row and column on the board (see proof below). There are many ways to do this. For example we can put them all on the diagonal. However, this leads to an infected board in 7 days! So to minimize growth we want to keep each sample as far away from the sample above and below it. Thus leading to a configuration like: &lt;br&gt; &lt;code&gt;C . . . . . . .&lt;/code&gt;&lt;br/&gt;&lt;code&gt;. . . . . . . C&lt;/code&gt;&lt;br/&gt;&lt;code&gt;. C . . . . . .&lt;/code&gt;&lt;br&gt;&lt;code&gt;. . . . . . C .&lt;/code&gt;&lt;br/&gt;&lt;code&gt;. . C . . . . .&lt;/code&gt;&lt;br/&gt;&lt;code&gt;. . . . . C . .&lt;/code&gt;&lt;br/&gt;&lt;code&gt;. . . C . . . .&lt;/code&gt;&lt;br/&gt;&lt;code&gt;. . . . C . . .&lt;/code&gt;&lt;br&gt; After the first night, the bottom two samples are close enough to infect two more cells:&lt;br&gt;&lt;code&gt;C . . . . . . .&lt;/code&gt;&lt;br/&gt;&lt;code&gt;. . . . . . . C&lt;/code&gt;&lt;br/&gt;&lt;code&gt;. C . . . . . .&lt;/code&gt;&lt;br&gt;&lt;code&gt;. . . . . . C .&lt;/code&gt;&lt;br/&gt;&lt;code&gt;. . C . . . . .&lt;/code&gt;&lt;br/&gt;&lt;code&gt;. . . . . C . .&lt;/code&gt;&lt;br/&gt;&lt;code&gt;. . . C C . . .&lt;/code&gt;&lt;br/&gt;&lt;code&gt;. . . C C . . .&lt;/code&gt;&lt;br&gt; Notice, that the addition of the new cells produce a square of infected cells that has a corner touching the next sample in the row above. We continue to night number two: &lt;br&gt; &lt;code&gt;C . . . . . . .&lt;/code&gt;&lt;br/&gt;&lt;code&gt;. . . . . . . C&lt;/code&gt;&lt;br/&gt;&lt;code&gt;. C . . . . . .&lt;/code&gt;&lt;br&gt;&lt;code&gt;. . . . . . C .&lt;/code&gt;&lt;br/&gt;&lt;code&gt;. . C . . . . .&lt;/code&gt;&lt;br/&gt;&lt;code&gt;. . . . C C . .&lt;/code&gt;&lt;br/&gt;&lt;code&gt;. . . C C C . .&lt;/code&gt;&lt;br/&gt;&lt;code&gt;. . . C C . . .&lt;/code&gt;&lt;br&gt; We infected two more cells! Notice that the new configuration &lt;em&gt;does not&lt;/em&gt; form a square, and therefore &lt;em&gt;does not&lt;/em&gt; have a corner touching the next sample in the row above! Taking this pattern, we see that the rate of growth is two cells every night. Thus it will take at most $(64-8)/2 = 28$ nights to fully infect the board. &lt;br&gt; &lt;strong&gt;Lemma 1&lt;/strong&gt;: was going to complete a proof for this, but it turns out that this assumption was wrong so I can't actually prove it haha.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13881" LastEditorUserId="13881" LastEditDate="2015-07-17T13:52:00.237" LastActivityDate="2015-07-17T13:52:00.237" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18100" PostTypeId="2" ParentId="18097" CreationDate="2015-07-17T05:46:54.997" Score="12" Body="&lt;p&gt;EDITED - thanks to @Meelo for pointing out my mistake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can do it in&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 38 days.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We place the C.Coli's at a4, a8, b2, c1, e2, g1, h3, h6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we prove that this is optimal. Every day there are either 1 new infected cell or 2+ new infected cells. Let us have k days with single infections and l days with 2+ infections. Then k+2l&amp;lt;=56 and therefore k+l&amp;lt;=28+k/2. So, if we prove that k&amp;lt;=21, then the we can't do better than 38.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us assume that we have the ability to delay the infection of some cells for several days. Clearly, this will not reduce the number of days until the total infection.&#xA;Notice that we can delay the infection in such way, so that it proceeds in 7 phases. At the beginning of phase i we start with 9-i fully infected rectangles and end with 8-i fully infected rectangles. Using the observation that the total perimeter of the infected area is a constant throughout, we can prove that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the width and the length of every rectangle have the same parity&lt;/li&gt;&#xA;&lt;li&gt;no two rectangles share a common edge&lt;/li&gt;&#xA;&lt;li&gt;for every pair of rectangles, there is at most one row/column, which intersects both of them.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;During each phase two of the infected rectangles start a chain reaction, which causes them to infect a bigger rectangle,    which contains both of them. The rest is just case analysis,    counting the number of single infection days during the phases. If    we combine two rectangles which are separated by 1 cell horizontally/vertically (i.e. there    is a row column, which intersects both rectangles), then the    corresponding phase will contain at most 3 single infection days. In    order to have 4+ single infection days during a phase, we must    combine two rectangles which share common vertex, one of which has    dimensions with difference 4 or more and the other one is a single cell. The only options are if the    dimensions of the disproportional rectangle are (1,5), (2, 6),  (3,7), (1,7). All of these options are easy to examine, can add some more details if needed.&lt;/p&gt;&#xA;" OwnerUserId="14140" LastEditorUserId="14140" LastEditDate="2015-07-18T18:59:31.863" LastActivityDate="2015-07-18T18:59:31.863" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18101" PostTypeId="2" ParentId="17995" CreationDate="2015-07-17T08:55:58.240" Score="0" Body="&lt;p&gt;While not optimal, there is also the opportunity to rank empirically. If you store past games and the time it took to solve and the success rate, then you can build a database from there. If the game is on a social platform and played by many, this will eventually be the most accurate rating. Of course that doesn't help at the start....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Btw, are solutions unique? Otherwise the amount of possible solutions can reciprocally add to the difficulty rating.&lt;/p&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="5169" LastEditDate="2015-07-17T14:27:14.550" LastActivityDate="2015-07-17T14:27:14.550" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18104" PostTypeId="2" ParentId="18074" CreationDate="2015-07-17T10:55:40.093" Score="2" Body="&lt;p&gt;Warmup:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He can just place his 8 samples diagonally (from a corner to the diametrically opposite one).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Challenge:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No. Every new sample (&lt;em&gt;starting from inside to out&lt;/em&gt;) increases the width or height of the smallest rectangle/square that includes all the squares infected or can be infected with the samples by 2, or both its width and height by 1, and sometimes it doesn't have any effect. So if we had 7 samples, the rectangle/square would never reach 8x8.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit for those who might downvote it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Even if the &quot;nearest&quot; new sample is more than 2 squares away from the old rectangle (2 squares for a single dimension or 1 for each dimension), we can group together all the &quot;connected&quot; samples less than 3 squares away from each other. If there are $n$ such groups and there are respectively $k_x, 0&amp;lt;x&amp;lt;n$ samples in them, at best their rectangles could cover $2+(k_1-1)*2 + ... + 2+(k_n-1)*2 = 2[k_1+...+k_n]$ = 14 units in total (sum of both dimensions of each rectangle). There must be gaps between the rectangles though, so we might need to make a more conservative guess. If two rectangles are connected, &lt;em&gt;at worst&lt;/em&gt; the new rectangle's sum of dimensions would equal the total sum of the dimensions for the two and a bigger rectangle would take both of their place. Eventually there'll be only two connected rectangles left so the total SoD will never reach 16.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Note that a 1x1 square can also be considered a rectangle with a SoD of 2. It's like an addition with full of parentheses that may be in weird places; you don't necessarily start &quot;linearly&quot;, but eventually the sum will equal the &quot;normal&quot; sum found by adding everything up from left to right.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="14016" LastEditDate="2015-07-22T19:51:42.407" LastActivityDate="2015-07-22T19:51:42.407" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18105" PostTypeId="1" AcceptedAnswerId="18152" CreationDate="2015-07-17T14:04:17.723" Score="30" ViewCount="3387" Body="&lt;p&gt;Solve the following 5 text rebuses, which are in the form of Linux (bash) command lines.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rebus #1&lt;/strong&gt; (solved by &lt;a href=&quot;https://puzzling.stackexchange.com/a/18152/4450&quot;&gt;Prem&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;# type beep sound &amp;gt; file.txt 2&amp;gt; file2.txt&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Hints:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. The solution is a single word that appears in dictionary.com.&lt;br&gt;&#xA; 2. The word has 10 letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rebus #2&lt;/strong&gt; (solved by &lt;a href=&quot;https://puzzling.stackexchange.com/a/18108/4450&quot;&gt;frodoskywalker&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;# cat file.png | wc -w&#xA;1000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rebus #3&lt;/strong&gt; (solved by &lt;a href=&quot;https://puzzling.stackexchange.com/a/18108/4450&quot;&gt;frodoskywalker&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;# touch file1; while true; do mv file1 file2; mv file2 file1; done&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rebus #4&lt;/strong&gt; (solved by &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18105/five-linux-command-line-rebus-puzzles#comment57316_18110&quot;&gt;Geobits&lt;/a&gt;/&lt;a href=&quot;https://puzzling.stackexchange.com/a/18110/4450&quot;&gt;karhell&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;# echo -e &quot;\e[0;97m\a&quot;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rebus #5&lt;/strong&gt; (solved by &lt;a href=&quot;https://puzzling.stackexchange.com/a/18110/4450&quot;&gt;karhell&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;# fsck /dev/sda1 &amp;amp;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note: For your answer to be accepted, it must at a minimum contain the correct solution for Rebus #1.&lt;/p&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-07-18T14:42:01.583" Title="Five Linux command line rebus puzzles" Tags="|rebus|computer-puzzle|" AnswerCount="6" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18106" PostTypeId="1" AcceptedAnswerId="18114" CreationDate="2015-07-17T14:05:50.140" Score="20" ViewCount="888" Body="&lt;p&gt;You sit in an open space with many others.&#xA;You tremble with fear as you hear It approach from behind.&#xA;You hear the terrible ominous sound, rhythmic and monotonous &#xA;and yet seeming to build in a crescendo of anticipation.&#xA;Suddenly It is upon you; your breath ceases and muscles quiver&#xA;as you wait to hear the sound, the verdict, your judgment.&#xA;God or Devil?&#xA;A pause, a hesitation, as It weighs your odds.&#xA;You both feel and hear your sentence as the source of the sound &#xA;calls out &quot;Devil&quot; and at once you sigh in relief.&#xA;You relax your muscles and a smile creeps on your face,&#xA;as you look to your left and witness the fates of your neighbors.&#xA;Your joy is short-lived; It will return, It always does.&#xA;Searching for the one God.&#xA;A God to quickly combat It or slowly supplant It. &#xA;And so it goes, forever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you determine what this passage is describing? As always, a correct response will not only provide an answer, but also an explanation of how the various elements in the story make sense in the context of that answer. Hope you enjoy!&lt;/p&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="22922" LastEditDate="2016-09-01T10:53:58.787" LastActivityDate="2016-09-01T10:53:58.787" Title="What is this story describing?" Tags="|story|knowledge|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18107" PostTypeId="2" ParentId="18089" CreationDate="2015-07-17T14:13:14.987" Score="1" Body="&lt;p&gt;Well, i think you only need&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3 larvaes, places like a L&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;!To simplify my demonstration let's imagine a 4*4 cube where each element have a position {x,y,z}&lt;br&gt;&lt;br&gt;&#xA;  oooo oooo oooo oooo&lt;br&gt;&#xA;  oooo oooo oooo oooo&lt;br&gt;&#xA;  oooo oooo oooo oooo&lt;br&gt;&#xA;  oooo oooo oooo oooo&lt;br&gt;&lt;br&gt;&#xA;  Let's add 3 infested elements in {3,1,1}, {3,2,1},{4,1,1} (step 1)&lt;br&gt;&lt;br&gt;&#xA;  oo&lt;strong&gt;xx&lt;/strong&gt; oooo oooo oooo&lt;br&gt;&#xA;  oo&lt;strong&gt;x&lt;/strong&gt;o oooo oooo oooo&lt;br&gt;&#xA;  oooo oooo oooo oooo&lt;br&gt;&#xA;  oooo oooo oooo oooo&lt;br&gt;&lt;br&gt;&#xA;  The next day obviously the element {4,2,1} will be infected. But there is also the element {3,1,2} (step 2)&lt;br&gt;&lt;br&gt;&#xA;  oo&lt;strong&gt;xx&lt;/strong&gt; oo&lt;strong&gt;x&lt;/strong&gt;o oooo oooo&lt;br&gt;&#xA;  oo&lt;strong&gt;xx&lt;/strong&gt; oooo oooo oooo&lt;br&gt;&#xA;  oooo oooo oooo oooo&lt;br&gt;&#xA;  oooo oooo oooo oooo&lt;br&gt;&lt;br&gt;&#xA;  The next day the elements {2,1,1} will be infected, but also the elements {3,2,2}, {4,1,2} and {4,2,2}(step 3)&lt;br&gt;&lt;br&gt;&#xA;  o&lt;strong&gt;xxx&lt;/strong&gt; oo&lt;strong&gt;xx&lt;/strong&gt; oooo oooo&lt;br&gt;&#xA;  oo&lt;strong&gt;xx&lt;/strong&gt; oo&lt;strong&gt;xx&lt;/strong&gt; oooo oooo&lt;br&gt;&#xA;  oooo oooo oooo oooo&lt;br&gt;&#xA;  oooo oooo oooo oooo&lt;br&gt;&lt;br&gt;&#xA;  here we can separate each &quot;square&quot; with respect to the z axis : this square&lt;br&gt;&#xA;  oo&lt;strong&gt;xx&lt;/strong&gt;&lt;br&gt;&#xA;  oo&lt;strong&gt;xx&lt;/strong&gt;&lt;br&gt;&#xA;  oooo&lt;br&gt;&#xA;  oooo&lt;br&gt;&#xA;  is repeating on every steps ({x,y,2} on the 3rd step is the same as {x,y,1} on the 2nd step, and it will be the same as {x,y,3} on the 4th step). We simplified this problem on a 2-dimentions problem&lt;br&gt;&#xA;  Let's take {x,y,1} on the last step &lt;br&gt;&lt;br&gt;&#xA;  o&lt;strong&gt;xxx&lt;/strong&gt;&lt;br&gt;&#xA;  oo&lt;strong&gt;xx&lt;/strong&gt;&lt;br&gt;&#xA;  oooo&lt;br&gt;&#xA;  oooo&lt;br&gt;&lt;br&gt;&#xA;  It should be easy to proove that every elements will be infected.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hope my answer isn't too hard to understand ..&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;EDIT&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If a healthy cube is ever adjacent to at least three infested cubes, then that cube will itself become infested the next day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My answer suppose that 'adjacent' means diagonally to.&lt;/p&gt;&#xA;" OwnerUserId="13384" LastEditorDisplayName="user20" LastEditDate="2015-07-17T18:14:35.840" LastActivityDate="2015-07-17T18:14:35.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18108" PostTypeId="2" ParentId="18105" CreationDate="2015-07-17T14:17:44.027" Score="14" Body="&lt;p&gt;Rebus #2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A picture is worth a thousand words&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You have piped an image (file.&lt;strong&gt;png&lt;/strong&gt;) to wc and asked it to count the words (-w), which returned 1000&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rebus #3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Perpetual motion&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are moving a file forever ('while true')&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3987" LastEditorUserId="3987" LastEditDate="2015-07-17T14:36:13.290" LastActivityDate="2015-07-17T14:36:13.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18110" PostTypeId="2" ParentId="18105" CreationDate="2015-07-17T14:44:26.617" Score="16" Body="&lt;p&gt;Tentative answer :&lt;br&gt;&#xA;rebus #1:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Could it be a sound filter ?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation :  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;type beep sound&lt;/strong&gt; tells us what beep and sound are.&lt;br&gt;&#xA; &lt;strong&gt;&gt; file1.txt&lt;/strong&gt; redirects stdout to file1.txt&lt;br&gt;&#xA; &lt;strong&gt;2&gt; file2.txt&lt;/strong&gt; redirects stderr to file2.txt&lt;br&gt;&#xA; beep is a legitimate command, so type beep is redirected to file1&lt;br&gt;&#xA; sound on the other hand is not, so it gets redirected to file2  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rebus #2 and #3:&lt;br&gt;&#xA;Agreed with @frodoskywalker, see &lt;a href=&quot;https://puzzling.stackexchange.com/a/18108/13753&quot;&gt;their answer&lt;/a&gt; for details  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A picture is worth a thousand words, and Perpetual motion&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rebus #4:&lt;br&gt;&#xA;(thanks @Geobits)  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; White noise &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation :   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;echo -e&lt;/strong&gt; outputs its arguments taking ansi escapes into account&lt;br&gt;&#xA; &lt;strong&gt;\e[0;97m&lt;/strong&gt; is ansi for bold, high-intensity white&lt;br&gt;&#xA; &lt;strong&gt;\a&lt;/strong&gt; is an ascii bell character&lt;br&gt;&#xA; A bell makes noise, therefore we have white noise.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rebus #5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To run a background check &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;fsck /dev/sda1&lt;/strong&gt; runs a filesystem consistency check on /dev/sda1 (main partition)&lt;br&gt;&#xA; &lt;strong&gt;&amp;amp;&lt;/strong&gt; allows the command to run in the background  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13753" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2015-07-17T17:14:21.397" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="18111" PostTypeId="2" ParentId="18039" CreationDate="2015-07-17T15:05:04.317" Score="5" Body="&lt;p&gt;My guess is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hades/Pluto? At first I thought the &quot;planet&quot;, but I'm leaning more towards the god.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My legend will be remembered, &lt;br/&gt;&#xA;Too strong to be dismembered.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mythology has stood the test of time and I expect it will keep on doing so, and Gods in myth can't really be killed by dismemberment, I don't believe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No happiness at the parting,&lt;br/&gt;&#xA;What happened at the starting?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Maybe refers to Persephone's abduction, and the sadness that comes with Fall and Winter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All the lines or all we do,&lt;br/&gt;&#xA;Mix God, religion, and the facts too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This may be referring to our thoughts on the afterlife.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Counting now, what are we at?&lt;br/&gt;&#xA;Profound, noises, or like a cat?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Another reference to the afterlife? Like a cat, how many lives have we lived?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My land is one full of life,&lt;br/&gt;&#xA;You know my cousins, little strife.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The underworld is full of lives/souls.&#xA; Strife (Eris) was known as two different entities.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Started with two, but grew to three,&lt;br/&gt;&#xA;But with some help, who can I be?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Zeus and Poseidon were the two, Hades making three in the division of power over the world.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13631" LastActivityDate="2015-07-17T15:05:04.317" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18112" PostTypeId="2" ParentId="18097" CreationDate="2015-07-17T15:18:10.673" Score="14" Body="&lt;p&gt;Not exactly answering this question, but given &lt;strong&gt;9&lt;/strong&gt; infections still on an &lt;strong&gt;8x8 board&lt;/strong&gt;, it is actually possible to delay the inevitable until &lt;strong&gt;40&lt;/strong&gt; days. Pretty counter-intuitive huh?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Locations are A1, A5, A8, B3, B7, D1, E8, G1, H8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DhgUS.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I believe that this is the maximum for an 8x8 board with any number of initial infections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the edges argument I made for the &lt;a href=&quot;https://puzzling.stackexchange.com/a/18075/14092&quot;&gt;preceding question here&lt;/a&gt;, you'll notice with 9 initial infections, we must lose a total of 4 edges.  This is accomplished by losing 2 edges on the 19th day, and the 34th day along the bottom row.&lt;/p&gt;&#xA;" OwnerUserId="14092" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-07-17T19:11:12.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18113" PostTypeId="1" AcceptedAnswerId="18256" CreationDate="2015-07-17T15:54:36.957" Score="10" ViewCount="821" Body="&lt;p&gt;Last &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18094/star-lord-and-the-space-police&quot;&gt;Space Police&lt;/a&gt; question...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sly Cooper has landed on a deserted planet with 2 policemen. He is moving around the planet, painting a line along his path, claiming any land which is surrounded by paint (the smaller area of the two resulting parts). The policemen are trying to restrict the land Sly Cooper claims as much as possible. If they encounter him, they arrest him and take him away. Can you prove that the policemen have a strategy, which prevents Sly Cooper from claiming more than 25% of the surface of the planet?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Remark: We assume that Sly Cooper and the policemen are moving with the same speed, take decisions in real time and are fully aware of everybody's locations.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. Don't care much about reputation points, so decided to spread some around...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To get some institution about the problem, try to prove that one policeman alone is enough to restrict the area to 50%. Even though identical strategy may not work for the case of two policemen, it still may give some good ideas.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;HINT 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The solution (at least mine) does not require considering different cases. It is just a universal strategy, working no matter what the initial locations are.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14140" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-07-22T14:42:13.620" Title="Sly Cooper and the Space Police" Tags="|mathematics|geometry|pursuit-evasion|" AnswerCount="2" CommentCount="22" ContentLicense="CC BY-SA 3.0" />
  <row Id="18114" PostTypeId="2" ParentId="18106" CreationDate="2015-07-17T16:05:28.620" Score="24" Body="&lt;p&gt;Haha this one was great, I love it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You are a player in a game of Duck Duck Goose!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You sit in an open space with many others. You tremble with fear as you hear It approach from behind. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Duck Duck Goose is played with people sitting in a circle facing each other while someone walks around behind them tapping them on the head.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You hear the terrible ominous sound, rhythmic and monotonous and yet seeming to build in a crescendo of anticipation. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The person who is &quot;It&quot; says &quot;Duck&quot; as they tap each person on the head, usually in a pretty consistent rhythm, but the longer they say &quot;Duck&quot;, the more people anticipate the inevitable &quot;Goose&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Suddenly It is upon you; your breath ceases and muscles quiver as you wait to hear the sound, the verdict, your judgment. God or Devil? A pause, a hesitation, as It weighs your odds. You both feel and hear your sentence as the source of the sound calls out &quot;Devil&quot; and at once you sigh in relief.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;In this case Devil = Duck and God = Goose. You have just been passed over as a Duck and will not have to chase the person who is &quot;It&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You relax your muscles and a smile creeps on your face, as you look to your left and witness the fates of your neighbors. Your joy is short-lived; It will return, It always does. Searching for the one God.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Everybody's sitting in a circle, and the game typically goes clockwise, so the people to your left are next to get tapped on the head.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A God to quickly combat It or slowly supplant It.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;When the person who is &quot;It&quot; chooses, they tap the next person on the head and shout &quot;GOOSE!&quot; The person who was tapped has to get up and chase the person who was &quot;It&quot;. If they catch them before the person who is &quot;It&quot; gets around the circle and sits in their spot, they can sit back down (quickly combat it), but if they're too slow, they become &quot;It&quot; (slowly supplant It).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And so it goes, forever.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Games of Duck Duck Goose traditionally continue for the rest of eternity. When the participants have succumbed due to lack of sustenance and rest, their spirits continue the game for all time, trapped forever in a vicious loop.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10861" LastActivityDate="2015-07-17T16:05:28.620" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18115" PostTypeId="2" ParentId="12643" CreationDate="2015-07-17T16:34:52.350" Score="0" Body="&lt;p&gt;I don't want any credit, but Javier in a similar question gave the correct answer as &lt;strong&gt;2&lt;/strong&gt; in &lt;strong&gt;$\infty$ time&lt;/strong&gt;.  The info can be found &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18094/star-lord-and-the-space-police&quot;&gt;here&lt;/a&gt;.  Basically 1 wolf mirrors the rabbits movement containing it to a hemisphere while the other has free movement, and can slowly work its way towards the rabbit.&lt;/p&gt;&#xA;" OwnerUserId="14092" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-07-17T16:34:52.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18116" PostTypeId="2" ParentId="18105" CreationDate="2015-07-17T17:12:17.180" Score="4" Body="&lt;p&gt;Possible answer for #1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Bleep out (beep out)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;When you bleep something out the sound goes away, and as karhell was saying the sound isn't a legitimate command&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13975" LastEditorUserId="10399" LastEditDate="2015-07-17T17:13:27.533" LastActivityDate="2015-07-17T17:13:27.533" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18117" PostTypeId="1" AcceptedAnswerId="18130" CreationDate="2015-07-17T17:24:29.217" Score="6" ViewCount="785" Body="&lt;p&gt;Here are some lines from songs. See if you can name them. You should include the artist name in your answer, even though the artist names do not appear as part of the rebus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                      ↓&#xA;bozo,pennywise|       iu       |hamill,romero&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u(me) = em \approx record$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;u+99, i+(i     i), u+GGG, i+#|&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; These are listed in chronological order by release date.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have updated #1 to make it easier.&lt;br&gt;&#xA;I have updated #2 just for fun.&lt;br&gt;&#xA;I have updated #3 to be a closer representation of the lyrics of the song.&lt;/p&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-07-17T19:06:06.113" LastActivityDate="2015-07-17T20:42:20.993" Title="Name that Song (Part 1?)" Tags="|rebus|knowledge|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18118" PostTypeId="2" ParentId="18117" CreationDate="2015-07-17T17:29:15.917" Score="7" Body="&lt;p&gt;&lt;em&gt;Partial answer...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Right Round by Dead or Alive&lt;br&gt; The lyric is, 'You Spin Me Right Round'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-07-17T17:48:59.897" LastActivityDate="2015-07-17T17:48:59.897" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18120" PostTypeId="1" AcceptedAnswerId="18122" CreationDate="2015-07-17T18:13:02.293" Score="7" ViewCount="609" Body="&lt;p&gt;Comes from a class I took senior year of HS, and I didn't find it online so hopefully this isn't a repeat.&lt;br&gt;&#xA;&lt;br&gt;&lt;br&gt;&#xA;A renowned psychologist is touring High Schools all over the country with her latest finding.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theory:&lt;/strong&gt;  There is a really popular student in each class in the entire school, such that if that student drinks, then everyone in that class drinks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is this just due to a small sample size, or is this always true, sometimes true, or never true?  And please prove why this is the case.&lt;br&gt;&#xA; &lt;br&gt;&lt;br&gt;&lt;br&gt;&#xA;Please enclose your answer in a spoiler using this &quot;&gt;!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Also please feel free to suggest tags.  I am not too familiar with the tags here&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="14092" LastActivityDate="2015-07-17T18:43:54.140" Title="Peer Pressure - Drinking" Tags="|logical-deduction|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18122" PostTypeId="2" ParentId="18120" CreationDate="2015-07-17T18:22:01.793" Score="12" Body="&lt;p&gt;Let the popular student be named &quot;P&quot; for short. The statement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If P drinks, then everyone in that class drinks&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is the same as the statement&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Either P doesn't drink, or everyone in that class drinks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The existence of a person P making the above statement true is now more clear. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If everyone drinks, choose P to be anybody. If someone doesn't drink, choose P to be that person.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, the Theorem is always true. This is a famous logical &quot;paradox&quot;, it even has a &lt;a href=&quot;https://en.wikipedia.org/wiki/Drinker_paradox&quot;&gt;Wikipedia page&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum:&lt;/strong&gt; It depends on the definition of &quot;really popular&quot;. If a student $x$ being &quot;really popular&quot; means &quot;if $x$ drinks, then everyone drinks&quot;, then the above reasoning is correct. If &quot;really popular&quot; is defined independently of the drinking condition, then the Theory is not true in general. A counterexample is any class with an unpopular student who doesn't drink, while everyone else drinks (believe me, such classes exist).&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-07-17T18:43:54.140" LastActivityDate="2015-07-17T18:43:54.140" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="18123" PostTypeId="1" AcceptedAnswerId="18128" CreationDate="2015-07-17T18:40:22.793" Score="2" ViewCount="377" Body="&lt;p&gt;Donovan Clump and Hermione Timmon are political candidates in the electoral race for &quot;Most Thankless Job in Modern History&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both candidates set out into the wilderness at the same time. Rains are pouring down, hence both candidates quickly get their feet wet. There are no winds of change, hence the rain is falling straight down at terminal velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ms. Timmon sets out walking in a straight line at a constant speed (significantly less than terminal velocity), consistent with her &quot;go slow with what you know&amp;trade;&quot; campaign slogan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mr. Clump, concerned about the rain's effect on his hair, tarries in the rain just outside the door for ten seconds, contemplating turning back. Ultimately he decides to run too. He takes off on the same straight path as Ms. Timmon, accelerating at a uniform rate and gaining campaign momentum, until catching up with Ms. Timmon ten seconds after he starts running.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the while, the rains are falling at a constant rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the two candidates are identical in all respects except their differing motion paths (which one might consider their campaign trails), &lt;strong&gt;at the exact moment Mr. Clump catches up to Ms. Timmon, which candidate will have been pelted by the greater number of rain droplets?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words: which candidate is most washed up?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Modeling:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;assume both candidates are cuboids of identical dimensions ($2$ m tall by $\tfrac{3}{10}$ m side-to-side and front-to-back)&lt;/li&gt;&#xA;&lt;li&gt;all rain droplets hitting the candidates count, including their fronts&lt;/li&gt;&#xA;&lt;li&gt;all physical effects/considerations aside from the candidates' differing $v\left( t\right) \equiv \dot{x}\left( t\right)$ should be ignored&lt;/li&gt;&#xA;&lt;li&gt;Ms. Timmon's (constant) forward velocity is $0 &amp;lt; v_{\rm tim} &amp;lt; v_{\rm t}$, where $v_{\rm t}$ is the terminal velocity of a rain droplet&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is fundamentally a problem of constructing a very basic physical model, and then good old fashioned math.&lt;/p&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="4219" LastEditDate="2015-07-18T13:58:02.887" LastActivityDate="2015-07-18T13:58:02.887" Title="Which candidate is most washed up?" Tags="|mathematics|word-problem|physics|" AnswerCount="2" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="18124" PostTypeId="2" ParentId="18117" CreationDate="2015-07-17T19:08:03.213" Score="7" Body="&lt;p&gt;#1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stuck in the Middle With You by Stealers Wheel&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lyrics:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;Clowns to the left of me, Jokers to the right, &#xA;Here I am, stuck in the middle with you.&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;Bozo and Pennywise are clowns (on the left) , Hamill and Romero (on the right) are actors who played The Joker&#xA;And I (me) is in the middle stuck to U (you)&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5359" LastEditorUserId="5359" LastEditDate="2015-07-17T19:34:57.537" LastActivityDate="2015-07-17T19:34:57.537" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18126" PostTypeId="2" ParentId="18123" CreationDate="2015-07-17T19:22:06.663" Score="2" Body="&lt;p&gt;If you accept Mythbusters as a relevant source they did an episode with running in the rain versus walking and discovered that if you run in rain rather than walk you will end up drier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://mythbustersresults.com/episode38&quot; rel=&quot;nofollow&quot;&gt;http://mythbustersresults.com/episode38&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13975" LastActivityDate="2015-07-17T19:22:06.663" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18127" PostTypeId="1" AcceptedAnswerId="18129" CreationDate="2015-07-17T19:23:23.707" Score="-2" ViewCount="578" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;You are standing next to an unlimited source of water (the tap in your kitchen). You need precisely 4.6 litres of water. Luckily you found two empty buckets. One can hold 4.3 and the other 4.7 litres of water. How can you measure the correct amount of water using the buckets when neither of them has graduation marks?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Problem #7 from: &lt;a href=&quot;http://erityistehtavat.puolustusvoimat.fi/cyberchallenge.html&quot; rel=&quot;nofollow&quot;&gt;http://erityistehtavat.puolustusvoimat.fi/cyberchallenge.html&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="14166" LastEditorUserId="71" LastEditDate="2015-07-17T20:50:52.197" LastActivityDate="2015-07-17T20:50:52.197" Title="Measuring 4.6 liters with a 4.3L bucket and a 4.7L bucket" Tags="|mathematics|calculation-puzzle|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18128" PostTypeId="2" ParentId="18123" CreationDate="2015-07-17T19:24:54.173" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They will both be hit by the same amount of rain.  Assuming the OP doesn't tweak the slight ambiguity in the question, they have both been outside for the same amount of time.  Furthermore they have both traveled the same distance. &lt;br&gt;&lt;br&gt; It can be seen that the amount of rain that hits you from the front is based solely on the distance you travel, because if you move slowly you will get barely hit over a long period of time, or if you move quickly you will get hit by a lot more but for a shorter period of time. &lt;br&gt;&lt;br&gt;  The amount you will be hit on your head is also only a function of the amount of time you are outside, because for every drop that misses you because you took a step forward, there is another drop that hit you because you took a step forward. &lt;br&gt;&lt;br&gt; Because they have both gone the same distance and been in the rain just as long, they are equally wet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14092" LastActivityDate="2015-07-17T19:24:54.173" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18129" PostTypeId="2" ParentId="18127" CreationDate="2015-07-17T19:41:04.430" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If we fill up the bucket holding 4.7L and fill the 4.3L with it, then we will have .4L left inside the 4.7L bucket. Dump the water in the 4.3L bucket, and pour the .4L into it. Now fill up the 4.7L bucket again, and fill the smaller bucket once again. Since it already contined .4L, the larger bucket will be left with .8L. Dump the small bucket and pour the .8L of water into the small bucket. Repeat these steps, each iteration adding .4L more to the &quot;running total.&quot; &lt;br&gt; &lt;br&gt; We keep on doing this until the small bucket contains 4L. Now filling it to the brim with the big bucket will leave 4.4L of water inside the big bucket. Dump the small bucket once again and fill it with the 4.4L of the big bucket, leaving only .1L inside the big bucket. Dump the small bucket again and put place the .1L of water inside the small bucket. &lt;br&gt;&lt;br&gt; So we repeat what we did before, each time adding .4L to the running total: .1L, .5L, ..., 4.1L. With 4.1L in the small bucket, we fill up the big bucket and top off the small bucket and get left with 4.5L in the big bucket. Dumping out the water inside the small bucket and filling it up again with the big bucket, we get left with .2L in the big bucket. We dump the small bucket, and put the .2L inside the small bucket. &lt;br&gt;&lt;br&gt; One more time! We continue the same set of operations as before, effectively adding .4L to the running total each time: .2L, .6L, ..., 4.2L. Once we hit 4.2L. We fill up the big bucket and use it to fill the smaller bucket. This takes only .1L, so we get left with 4.6L inside the big bucket.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13881" LastEditorUserId="13881" LastEditDate="2015-07-17T19:48:00.430" LastActivityDate="2015-07-17T19:48:00.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18130" PostTypeId="2" ParentId="18117" CreationDate="2015-07-17T19:42:01.770" Score="8" Body="&lt;p&gt;Partial answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Weezer's &quot;Hash Pipe&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;u+99&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you've got your problems (99 of them, here. lol)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;i+(i        i)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I've got my eyes wide&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;u+GGG&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You've got your big G's&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;i+#|&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I've got my hash pipe&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2444" LastEditorUserId="2444" LastEditDate="2015-07-17T20:42:20.993" LastActivityDate="2015-07-17T20:42:20.993" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18131" PostTypeId="1" AcceptedAnswerId="18132" CreationDate="2015-07-17T21:13:52.107" Score="8" ViewCount="496" Body="&lt;p&gt;Sven got a job painting the yellow midline on a 50 mile road. The more he paints, the slower he progresses, since he keeps getting further away from the paint can!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be exact, if he has already painted $m$ miles at the beginning of a day, then he will paint $(1/2)^m$ more miles that day. For example, on the first day, he hasn't painted anything yet, so he paints $(1/2)^0=1$ mile that day. The next day, he paints $(1/2)^1=0.5$ miles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will he ever finish?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;(I promise that this can be solved without using any complicated formulas or concepts!)&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-07-21T20:04:05.167" Title="Silly Sven Paints a Highway" Tags="|mathematics|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18132" PostTypeId="2" ParentId="18131" CreationDate="2015-07-17T21:22:30.650" Score="9" Body="&lt;p&gt;Sven will&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eventually finish&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every day that he has previously covered less than 50 miles, he will paint more than $\frac{1}{2}^{50}$ miles. This will eventually sum to more than 50 miles. (Indeed, it is unbounded and Sven will eventually paint any finite length)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3987" LastActivityDate="2015-07-17T21:22:30.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18133" PostTypeId="1" AcceptedAnswerId="18134" CreationDate="2015-07-17T21:32:15.767" Score="5" ViewCount="5126" Body="&lt;p&gt;Hopefully this is not a repost, as I couldn't find an existing one...yet:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the missing number?&#xA;&lt;img src=&quot;https://i.stack.imgur.com/IO1w1.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2015-07-17T22:11:36.507" Title="Find the missing number in the square" Tags="|mathematics|number-sequence|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18134" PostTypeId="2" ParentId="18133" CreationDate="2015-07-17T21:41:36.773" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is 27, substract the number formed by the digits on the second row from the number formed by the digits on the first row and divide by 2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14140" LastEditorUserId="4385" LastEditDate="2015-07-17T22:11:36.507" LastActivityDate="2015-07-17T22:11:36.507" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18135" PostTypeId="2" ParentId="43" CreationDate="2015-07-17T21:49:42.330" Score="4" Body="&lt;p&gt;I know it's a year old question, and has already been solved, still I gave it a swing. I see the other answers have used transition matrix, which I have no idea what that is. I solved it in kind of a different way. Maybe in principle does the same, but what the heck, I spent hours on it, and I want to share it! Sorry if the image is too messy, but I hope it manages to get the idea across.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PqEX9.jpg&quot; alt=&quot;It&amp;#39;s kind of messy, but I guess you&amp;#39;ll get how I reached the solution using the flowchart thing.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So at each step I basically chose the button that had the best chance to arrange the bolts in a way that the door is able to open. For example, out of the four initial states, pressing &quot;C&quot; would open the door if it was state no. 3, &lt;strong&gt;for sure&lt;/strong&gt;. Pressing &quot;A&quot; could open if it was state no.1 but not for sure. Pressing &quot;B&quot; could open if it was state no. 2 or 4, but again, not for sure. This way I was able to eliminate the case of initial state 3. Similarly I chose the buttons at each step in a way that would eliminate most of the possibilities. (I hope I'm making sense, if not please discuss if interested.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This way the solution came out to be, as stated by other answers,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C&gt;B&gt;C&gt;A&gt;C&gt;B&gt;C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(check the door after each step.)&lt;/p&gt;&#xA;" OwnerUserId="14074" LastEditorUserId="37821" LastEditDate="2017-06-27T06:53:30.800" LastActivityDate="2017-06-27T06:53:30.800" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18136" PostTypeId="2" ParentId="18131" CreationDate="2015-07-17T22:09:45.277" Score="0" Body="&lt;p&gt;The one above is a better answer, but I hope this is also acceptable:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All $m$ lengths are bound to be equal to or less than 1, because $1/2^m$ can only equal 1 if $m$ is 0, which can't be true save for the first day, making the next one ($1/2^m$) greater than 1/2. Which means all the $m$s starting with the third one are greater than 1/2. Therefore he'll eventually finish the work.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="14016" LastEditDate="2015-07-18T07:36:03.860" LastActivityDate="2015-07-18T07:36:03.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18137" PostTypeId="1" AcceptedAnswerId="18138" CreationDate="2015-07-17T22:23:23.907" Score="7" ViewCount="521" Body="&lt;p&gt;Below is a rebus set, each of which unveils the lyrics of a song. Your job is to name the song and artist, as well as provide a translation for the rebus elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta{u(mind)} \approx \delta{girl(clothes)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;%&amp;gt; echo drunk &amp;gt; choir ; touch -t 07180000 choir&#xA;%&amp;gt; cmp my_goals choir&#xA;%&amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;3.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Person p1 = new Child();&#xA;Person p2 = new Person();&#xA;p1.age();&#xA;Person i = p2.spawnWith(p1);&#xA;i.name = p2.name;&#xA;i.gender = kFemale;&#xA;i.age();&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt; (hint?): I expect 2 and 3 to be tricky for most people under the age of 30.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; #3 might be easier to get if you read from bottom to top.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; #3 is a song pleading for a divine entity that calls Alabama home.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 3&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; #3 was released on a self-titled album in 1971 by a Chicago-based country/folk artist.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-07-21T14:55:02.047" LastActivityDate="2015-07-21T15:16:53.827" Title="Name that Song (part dos)" Tags="|rebus|knowledge|computer-puzzle|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18138" PostTypeId="2" ParentId="18137" CreationDate="2015-07-17T22:43:04.997" Score="11" Body="&lt;p&gt;&lt;strong&gt;#1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hot and Cold by Katy Perry&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lyrics:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;You change your mind&#xA; Like a girl changes clothes&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;#3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Angel from Montgomery by John Prine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lyrics:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;I am an old woman named after my mother &#xA; My old man is another child that's grown old&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; p1 is the father &lt;br&gt;&#xA; p2 is the mother&lt;br&gt;&#xA; i is &quot;me&quot;&lt;br&gt;&lt;br&gt;&#xA; &lt;code&gt;i.age(); and i.gender = kFemale;&lt;/code&gt; $-&amp;gt;$ I am an old woman&lt;br&gt;&#xA; &lt;code&gt;i.name = p2.name;&lt;/code&gt; $-&amp;gt;$ named after my mother&lt;br&gt;&#xA; &lt;code&gt;Person i = p2.spawnWith(p1);&lt;/code&gt; $-&amp;gt;$ p2 and p1 are the parents of i&#xA; &lt;code&gt;p1.age();&lt;/code&gt;  $-&amp;gt;$ My old man &lt;br&gt;&#xA; &lt;code&gt;Person p1 = new Child();&lt;/code&gt;  $-&amp;gt;$ is another child &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5359" LastEditorUserId="5359" LastEditDate="2015-07-21T15:16:53.827" LastActivityDate="2015-07-21T15:16:53.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18139" PostTypeId="2" ParentId="18105" CreationDate="2015-07-17T22:45:04.350" Score="3" Body="&lt;p&gt;Another possible answer for #1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; soundcheck&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;code&gt;type&lt;/code&gt; command indicates what would happen if the name (or names) given as argument(s) were to be entered at the command line. &lt;code&gt;&amp;gt; file.txt&lt;/code&gt; means &quot;write the output of this instruction to a file called &quot;file.txt&quot;, and &lt;code&gt;2&amp;gt; file2.txt&lt;/code&gt; means &quot;write any error messages to a file called &quot;file2.txt&quot;. Sound engineers often repeat the numbers &quot;one&quot; and &quot;two&quot; when checking microphone levels; perhaps this is what is being referred to.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="40" LastActivityDate="2015-07-17T22:45:04.350" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18140" PostTypeId="1" CreationDate="2015-07-17T22:48:14.597" Score="1" ViewCount="290" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am your wildest dreams, your deepest secrets.&lt;br&gt;&#xA;  I am your follower, your guide.&lt;br&gt;&#xA;  I am who you say I am not; I am your world.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To what does this riddle refer?&lt;/p&gt;&#xA;" OwnerUserId="14171" LastActivityDate="2015-07-18T00:46:04.800" Title="An oxymoronic mystery" Tags="|riddle|lateral-thinking|" AnswerCount="2" CommentCount="0" ClosedDate="2015-07-18T03:12:01.217" ContentLicense="CC BY-SA 3.0" />
  <row Id="18142" PostTypeId="2" ParentId="18140" CreationDate="2015-07-17T23:35:26.330" Score="3" Body="&lt;p&gt;It may be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your subconscious&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&#xA;I am your wildest dreams, your deepest secrets.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your dreams come from your subconscious. And your deepest secrets you sometimes push down out of conscious memory.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am your follower, your guide.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your subconscious follows where you go (it has to), but sometimes it guides you to decisions that you don't necessarily know why you made them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am who you say I am not; I am your world.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your subconscious is you, but sometimes you deny that it is since you don't control the impulses that come out of it. It is also your world, as it is compiled from everything that's happened to you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2015-07-17T23:35:26.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18143" PostTypeId="2" ParentId="18140" CreationDate="2015-07-18T00:43:04.607" Score="3" Body="&lt;p&gt;I am  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your Past&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am your wildest dreams, deepest secrets&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In most of our wildest dreams, we might dream about past. Our past is our deepest secrets.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am your follower, your guide.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Our past follows us and guides us&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am who you say I am not: I am your world&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We say we are not our past. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14004" LastEditorUserId="12700" LastEditDate="2015-07-18T00:46:04.800" LastActivityDate="2015-07-18T00:46:04.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18144" PostTypeId="2" ParentId="18131" CreationDate="2015-07-18T03:03:43.117" Score="6" Body="&lt;p&gt;It doesn't matter how slowly Sven paints - he will finish. In particular, consider the function $f(m)$ which takes the number of miles he started with as input and outputs the number of miles he finishes with. Notice that he starts with $0$ miles painted, then has $f(0)$ miles painted, then $f(f(0))$ and then $f(f(f(0)))$. Notating the iteration of $f$ as $$f^n(x)=\underbrace{f(f(\ldots f}_{n \text{ repetitions}}(x)\ldots ))$$&#xA;we may find that, so long as $f(m)&amp;gt;m$ (i.e. he always paints something) and $f$ is continuous, we can conclude:&#xA;$$\lim_{n\rightarrow\infty}f^n(0)=\infty$$&#xA;which means that the amount he's painted after $n$ days goes to infinity with $n$. In particular, from the definition of a limit, this means that for any finite $M$, he will, in a finite amount of time, paint $M$ miles. This has a simple argument by contradiction - suppose that above limit were a finite $x$. Then&#xA;$$x=\lim_{n\rightarrow\infty}f^n(0)=\lim_{n\rightarrow\infty}f^{n+1}(0)=\lim_{n\rightarrow\infty} f(f^{n}(x))=f(\lim_{n\rightarrow\infty}f^{n}(0))=f(x)$$&#xA;which contradicts the hypothesis that $f(x)&amp;gt;x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(One may also use the continuity of $f$ to use the extreme value theorem on $f(x)-x$ over the compact domain $[0,50]$, which generalizes @frodoskywalker's answer. Another curiosity is that if we weaken to requiring $f(x)\geq x$, then the given limit will tend towards the first fixed point of $f$ higher than $0$ - and, in fact, this is basically what we've done when we extend $f$ to the extended reals)&lt;/p&gt;&#xA;" OwnerUserId="4765" LastEditorUserId="4765" LastEditDate="2015-07-18T03:08:55.133" LastActivityDate="2015-07-18T03:08:55.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18146" PostTypeId="1" AcceptedAnswerId="18425" CreationDate="2015-07-18T05:00:24.860" Score="12" ViewCount="680" Body="&lt;p&gt;You are trying to design a maze solving robot, capable of solving &quot;valid&quot; mazes. A valid maze is an 8 by 8 checkerboard, surrounded by walls, where&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Between each pair of orthogonally adjacent squares, there either is a wall, or isn't.&lt;/li&gt;&#xA;&lt;li&gt;The &lt;strong&gt;northwest&lt;/strong&gt; square is marked &quot;start&quot;, the &lt;strong&gt;southeast&lt;/strong&gt; square is marked &quot;finish&quot;.&lt;/li&gt;&#xA;&lt;li&gt;It is possible to get from &quot;start&quot; to &quot;finish.&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You program the robot by giving it a finite list of compass directions (N,S,E,W). It goes through the list, moving one square in the current direction unless doing so would make it hit a wall, in which case it stays put. Once it exhausts the list, it stops moving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal is to design a program so that the robot, when placed on the &quot;start&quot; square of any valid maze, will always end up on the &quot;finish&quot; square when it has stopped moving.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Prove that there exists a program which accomplishes this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This is a harder version of &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18009/maze-solving-robot&quot;&gt;this puzzle&lt;/a&gt;. There, the robot was only required to visit the &quot;finish&quot; square on some point of its journey. Here, the robot must end on the finish square.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;As far as I know, the solution requires some advanced mathematics, but I would still consider this a puzzle instead of a rote math problem.&lt;/em&gt; &lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-07-28T21:11:30.950" Title="Maze Solving Robot, More Difficult Variant" Tags="|checkerboard|algorithm|mazes|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="18147" PostTypeId="2" ParentId="18146" CreationDate="2015-07-18T05:18:00.303" Score="1" Body="&lt;p&gt;I have a general idea but rather than try to explain it perfectly mathematically, I'll try to explain it in layman's terms:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First, come up with a list of every possible 8 by 8 map (a total of $N$ maps). Secondly, simulate every map's solution on each of all of the other map's squares. This will result in a total of $N*(N-1)*64$ &quot;position changes&quot;.&lt;br&gt;&lt;br&gt;An example of a position change if there were two possible maps: first run the simulation on the first map, then pretend the correct map was actually the second map and navigate to the finish line. The position change is the current position if the first map was the correct map all along.&lt;br&gt;&lt;br&gt;The goal will be to arrange the $N$ map solutions in order so that the position changes for all future maps will allow the previous one to end on the southeast square. Due to symmetry, I think there should be an equal number of position changes going both directions, and also a generally even distribution for distance and direction of the position changes. (by direction, I mean the opposite change for a position change such as going down two instead of going up two). Therefore, there should logically be a sequence to arrange the attempt of the $N$ maps so that no matter which one is the correct solution, the robot will end up on the southeast square.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastActivityDate="2015-07-18T05:18:00.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18148" PostTypeId="2" ParentId="18137" CreationDate="2015-07-18T05:36:56.293" Score="8" Body="&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;The song and artist are:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;&lt;a href=&quot;http://www.azlyrics.com/lyrics/leonardcohen/birdonthewire.html&quot;&gt;Bird on the Wire&lt;/a&gt;&quot; by Leonard Cohen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The rebus represents the lyrics &quot;like a drunk in a midnight choir&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;%&gt; echo drunk &gt; choir ; touch -t 07180000 choir&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;code&gt;echo&lt;/code&gt; command created a file named &lt;code&gt;choir&lt;/code&gt; containing the text &lt;code&gt;drunk&lt;/code&gt;.  This represents a drunk being in a choir.&#xA;&#xA; Then the &lt;code&gt;touch&lt;/code&gt; command set the timestamp on the &lt;code&gt;choir&lt;/code&gt; file to July 18 00:00 (midnight).  This represents a drunk being in a midnight choir.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;%&gt; cmp my_goals choir&lt;br&gt;&#xA;%&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;code&gt;cmp&lt;/code&gt; command compared the files &lt;code&gt;my_goals&lt;/code&gt; and &lt;code&gt;choir&lt;/code&gt; and did not find any differences.  Therefore you could say the &lt;code&gt;my_goals&lt;/code&gt; file is &quot;like a drunk in a midnight choir&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2015-07-18T05:36:56.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18149" PostTypeId="2" ParentId="18105" CreationDate="2015-07-18T08:28:19.283" Score="1" Body="&lt;p&gt;Possible answer to #1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tinnitus&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You get a beep, but there's no sound&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14172" LastActivityDate="2015-07-18T08:28:19.283" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18151" PostTypeId="2" ParentId="17937" CreationDate="2015-07-18T11:25:16.033" Score="2" Body="&lt;p&gt;This solution is heavily inspired by those of &lt;a href=&quot;https://puzzling.stackexchange.com/a/18042/4551&quot;&gt;Tyler Seacrest&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/a/17999/4551&quot;&gt;Mike Earnest&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Consider a placement of keys in boxes. Given a box, one can trace back the box that contains its key, the box that contains that boxes's key, and so on. This chain must eventually loop in a repeating cycle, though that cycle might not contain the initial box. Let's say &lt;em&gt;cycle boxes&lt;/em&gt; are those within a cycle, and &lt;em&gt;path boxes&lt;/em&gt; are the rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From tracing back, we see that opening any box reduces to opening some cycle box via some chain of keys. Moreover, no path box lets you open a cycle box, since the key to a cycle box lies within that cycle. So, path boxes are totally vestigial: whether you can open all boxes doesn't change if you remove all path boxes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, note that restricting to the cycle boxes and their keys yields a uniformly random permutation, reducing to the &lt;a href=&quot;https://puzzling.stackexchange.com/q/17852/4551&quot;&gt;original problem&lt;/a&gt;. This is because if the key locations of the path boxes are fixed, the overall assignment is completed by placing each cycle box's key into a distinct cycle box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/a/17854/4551&quot;&gt;We showed that&lt;/a&gt; the probability of opening all boxes for a permutation of keys equals the probability that a random box is wooden. So in this problem, all boxes can be opened with the same probability that a random cycle box is wooden. Since being a cycle box or not isn't influenced by material, this equals the fraction of wooden boxes $k$/$n$.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-07-18T11:25:16.033" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18152" PostTypeId="2" ParentId="18105" CreationDate="2015-07-18T14:38:13.940" Score="10" Body="&lt;p&gt;Answer for #1 :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; stereotype&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; stereo == two channels + type , forcing &quot;output&quot; to both STDERR &amp;amp; STDOUT.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Other rebuses have already been answered.&lt;/p&gt;&#xA;" OwnerUserId="10940" LastActivityDate="2015-07-18T14:38:13.940" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18153" PostTypeId="1" AcceptedAnswerId="18157" CreationDate="2015-07-18T15:14:53.260" Score="13" ViewCount="1118" Body="&lt;p&gt;After you posted mean things online about Aquaman, he found out where you lived and kidnapped you (using his awesome aqua-powers), taking you to his super-secret aqua-lair. He has placed you in a room with a ceiling less than $10$ feet tall, a perfectly level floor, and, at the end of the room, a perfectly vertical wall with a mark exactly $5$ feet above floor level. There is a drain in the floor so that water drained into the room will not accumulate (at all). The room is otherwise featureless. Behind the marked wall is a reservoir of water. You have access to a compass (which can only draw circles around a given center), a plumb bob (which can only indicate a perfectly vertical line), and an awl (which can only poke holes through the wall). All the walls, ceiling, and floor are all opaque and you cannot see the reservoir. Notice that the tools given do not allow you to make any measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aquaman has challenged you&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you can drain the reservoir such that its water level is exactly $10$ feet above the floor, I will return you to your house and we can forget this ever happened. Otherwise, I will get a &lt;em&gt;real&lt;/em&gt; superhero to come and deal with you - and you don't want that!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How can you use the three tools given to drain the reservoir and to know exactly when its water level is $10$ feet above the floor?&lt;/p&gt;&#xA;" OwnerUserId="4765" LastEditorUserId="4765" LastEditDate="2015-07-18T16:18:53.513" LastActivityDate="2015-07-19T05:38:35.883" Title="Aquaman's Revenge!" Tags="|geometry|story|physics|" AnswerCount="4" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="18155" PostTypeId="2" ParentId="18153" CreationDate="2015-07-18T17:37:11.917" Score="7" Body="&lt;p&gt;Use the compass to draw a very small circle around the 5 feet MARK.&lt;br&gt;&#xA;Now use the bob to make it pass through the MARK, and get two points on the circle, one on top and the other on bottom of the circle.&lt;br&gt;&#xA;Use the awl to poke 3 holes at center+top+bottom.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The water will flow out and fall on the floor at 3 moving points. When the reservoir has 10 feet of water, then the water from the center hole will fall at the maximum Distance, while the water from other two holes will fall very near but at a lesser Distance.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plug the holes now.&lt;br&gt;&#xA;Done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternate solution:&lt;br&gt;&#xA;Make only 2 holes on the circle, at top and bottom.&lt;br&gt;&#xA;When both the water flows fall at the same point, the water from the center hole (which we did not make) would have fallen at the maximum Distance.&lt;br&gt;&#xA;So plug the 2 holes now.&lt;br&gt;&#xA;Done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References :&lt;br&gt;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Torricelli%27s_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Torricelli%27s_law&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://physics.stackexchange.com/questions/6341/how-far-will-water-squirt-out-from-a-hole-in-a-can&quot;&gt;https://physics.stackexchange.com/questions/6341/how-far-will-water-squirt-out-from-a-hole-in-a-can&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="10940" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:18.130" LastActivityDate="2015-07-18T17:51:18.383" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18156" PostTypeId="2" ParentId="18153" CreationDate="2015-07-18T18:08:34.557" Score="3" Body="&lt;p&gt;If you drill a hole at the $5$ ft height, and the current water height is $H$, then the water lands at a distance $2\sqrt{5(H-5)}$ from the wall. When $H=10$ ft, this is a distance of $10$ ft. So, we just need to make a mark $10$ ft from where the wall meets the floor, drill a hole at $5$ ft, then wait till the water falls on that mark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how do we make the mark? First, we show how to do this using a straightedge and compass. Since &lt;a href=&quot;http://www.cut-the-knot.org/do_you_know/compass.shtml&quot; rel=&quot;nofollow&quot;&gt;every straightedge/compass construction can be done with a compass alone&lt;/a&gt;, this means we can make the mark with the tools we have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The construction (using a straightedge) isn't hard to figure out, but for completeness, here it is. Draw a point $P_1$ on the wall, on the 5ft line. Use the plumb bob to draw a point $P_2$ directly below it on the floor. Use the compass to make two points $P_3$ and $P_4$, also on the edge between the wall and the floor, both $5$ ft from $P_2$. Now, draw two circles, one with center $P_3$ passing through $P_4$, and another with center $P_4$ passing $P_3$. These intersect at $P_5$. Draw the line $L$ through $P_2$ and $P_5$, and a circle $C$ centered at $P_2$ through $P_3$ and $P_4$. Say that $C$ and $L$ intersect at $P_6$. Finally, draw a circle $\tilde C$ centered at $P_6$ passing through $P_2$. Then $\tilde C$ intersects $L$ at the desired mark, $10$ ft from $P_2$. Now drill at $P_1$, wait for the stream to hit the mark, and you're done!&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="12700" LastEditDate="2015-07-18T20:03:17.487" LastActivityDate="2015-07-18T20:03:17.487" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18157" PostTypeId="2" ParentId="18153" CreationDate="2015-07-18T18:43:38.543" Score="13" Body="&lt;p&gt;Set the compass to some size, less than the distance from the 5ft mark to the ceiling. Use the bob to find two points on the circle, one directly above the other. Then use the awl to punch holes at these points. Wait until the streams from both holes fall on the same spot. You're done! Plug the holes in a hurry and mock Aquaman some more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explanation: the velocity of a stream from a given hole will be $v=\sqrt{2gd}$ where d is the distance to the top of the water. The time to fall from a height h is given by $t=\sqrt{\frac{2h}{g}}$. This means that the horizontal distance traveled by a stream before fitting the ground will be $vt=2\sqrt{hd}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that $h+d=l$ where l is the height of water above the floor, the only time $h_1d_1=h_2d_2$ is when $h_1=d_2, h_2=d_1$, which means $l=h_1+h_2$. Because the holes are symmetric about the 5ft (height) line, $h_1+h_2=10ft, l=10ft$&lt;/p&gt;&#xA;" OwnerUserId="3987" LastEditorUserId="397" LastEditDate="2015-07-19T05:38:35.883" LastActivityDate="2015-07-19T05:38:35.883" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18158" PostTypeId="2" ParentId="18153" CreationDate="2015-07-18T19:19:05.873" Score="2" Body="&lt;p&gt;Use the awl to make a hole anywhere in the wall to let water out, and let the water drain.  Since the reservoir's water level started higher than 10 feet, eventually it will drain to the level of the hole (which must be less than 10 feet high).  Therefore at some time in between, the water level was exactly at 10 feet.  Aquaman did not specify that the reservoir's water level needed to stay at 10 feet high for any particular duration of time. =)&lt;/p&gt;&#xA;" OwnerUserId="388" LastEditorUserId="388" LastEditDate="2015-07-18T19:24:39.693" LastActivityDate="2015-07-18T19:24:39.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18160" PostTypeId="2" ParentId="18097" CreationDate="2015-07-18T23:37:31.963" Score="4" Body="&lt;p&gt;The maximum length of time to infect the entire board from $8$ cells is $38$ days and the only arrangement achieving this maximum is as follows (where the $X$'s are infected cells):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;--X--X--&#xA;X-------&#xA;--------&#xA;-X------&#xA;--------&#xA;X-------&#xA;-X------&#xA;---X---X&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Our process can be described as starting with a $2\times 2$ square and adding new cells to enlarge it one step at a time, each time increasing the area's perimeter by $4$. In particular, we start with a square:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X-&#xA;-X&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which will be infected after $1$ day. We then increase the height of the square by two, placing a new infection such that, once the corner (the last cell infected) of the previous image is infected, the new infection will take effect. In particular, we get:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;-X&#xA;--&#xA;X-&#xA;-X&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Infecting this takes $3$ days. We then increase the height again in a similar manner yielding:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X-&#xA;--&#xA;-X&#xA;--&#xA;X-&#xA;-X&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Taking $3$ more days. Then, we increase &quot;diagonally&quot;, incrementing both height and width:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;--X&#xA;X--&#xA;---&#xA;-X-&#xA;---&#xA;X--&#xA;-X-&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This takes $6$ days. Then we increase diagonally this again, noting that the last cell filled above is the bottom right corner:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;--X-&#xA;X---&#xA;----&#xA;-X--&#xA;----&#xA;X---&#xA;-X--&#xA;---X&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This takes $7$ days. Then, finally, we increase the width, noting that the top right is the last infected piece, to get:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;--X--X&#xA;X-----&#xA;------&#xA;-X----&#xA;------&#xA;X-----&#xA;-X----&#xA;---X--&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which takes $9$ days to be infected. Then, we increase the width again yielding the correct answer and taking an additional $9$ days. All told, the process will take $38$ days. Moreover, having decided on the sequence of increases to width/height/diagonal, we never had any choice in where to place the infections (up to symmetry) as we always had to put the new infection two cells away from the last infected square of the previous arrangement in a specified direction - once we have shown that the general process generating this answer is optimal, this suffices to show uniqueness of the solution.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Act One:&lt;/strong&gt; Conclusions Are Proven From Intuitive Assumptions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will start by taking the following lemma (whose more difficult proof comprises Act Two of this post) and deducing that our construction is optimal. We assume we are on an $n\times n$ board, as our solution generalizes naturally.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The best solution is to start with a $2\times 2$ square, and then add infections such that each successive infection is placed to create a rectangle (of perimeter $4$ larger) with all the previous infections.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In particular, we will describe our above solution as the string:&#xA;$$[2\times 2]YYDDXX$$&#xA;meaning &quot;start with a $2\times 2$ square, increase its height twice (&quot;Y&quot;), increase it diagonally twice (&quot;D&quot;), then increase its width twice (&quot;X&quot;). After three steps, we could describe the position as&#xA;$$[2\times 4]YDDXX$$&#xA;as we would have a $2\times 4$ rectangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We notice that $[a\times b]X$ will become $[(a+2)\times b]$ after at most $b+1$ days, that $[a\times b]Y$ will become $[a\times (b+2)]$ after at most $a+1$ days and $[a\times b]D$ will become $[(a+1)\times (b+1)]$ after at most $\max(a,b)$ days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, we may assume that $a\leq b$ at all times using the symmetries of the problem - in particular, we may assume that, if we start from a square rectangle, then we can exchange every instance of $X$ with $Y$ and vice versa, since the problem clearly has that symmetry (so we should never see the pattern $[a\times a]X\ldots$). This lets us assume that, whenever we have a square of infection, our next move will be $Y$ or $D$ - since if it were $X$, we could exchange all further $X$'s and $Y$'s and get an equivalent length of time. Thus, we may assume $[a\times b]D$ takes $b$  (rather than $\max(a,b)$) days to become $[(a+1)\times (b+1)]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, we simply notice a few facts to reduce this to a one-variable optimization problem rather than optimization over a space of strings:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$[a\times b]XY$ and $[a\times b]YX$ take equally much time, so we can freely swap their order; without loss of generality, let us assume that the pattern $XY$ never appears (and is always replaced by $YX$).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$[a\times b]XD$ is always slower than $[a\times b]DX$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$[a\times b]DY$ is always slower than $[a\times b]YD$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Using the above, we may conclude that the slowest string of enlargements must have the form*:&#xA;$$[2\times 2]\underbrace{YY\ldots YY}_{y\text{ times}}\underbrace{DD\ldots DD}_{d\text{ times}}\underbrace{XX\ldots XX}_{x\text{ times}}$$&#xA;where we must satisfy $2y+d=n$ and $2x+d=n$ - implying $x=y$ and $d=n-2y$, which reduces this to a one variable optimization problem. In particular, each of the beginning $Y$'s will take $3$ days for the enlargement. The first $D$ will take $2y+2$ days and the last will take $n-1$ days, so they will in total take&#xA;$$\sum_{i=2y+2}^{n-1}i=-2y^2 - 3y - 1 +\frac{n^2}2 - \frac{n}2.$$&#xA;Then, each $X$ will take $n+1$ days, so they will in total take $y(n+1)$ days. Putting this all together (plus one day for the $2\times 2$ pattern) yields that the length of time taken by the above pattern with $y$ instances of $Y$ will be:&#xA;$$-2y^2 + (n+1)y + \frac{n^2}2 - \frac{n}2$$&#xA;which is convex, has a maximum at $\frac{n+1}4$ and is symmetrical about that point. Thus, the nearest integer to $\frac{n+1}4$ maximizes that. So, for $n=8$, we get that $y=2$ (rounded from $\frac{9}4$) is optimal, and the above expression yields $38$ as the upper bound at that point. In particular, we can find the optimal solutions for all $n\geq 4$ as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $n$ is between $4c-3$ and $4c+1$, then an optimal solution is to enlarge upwards $c$ times, then diagonally $n-2c$ times, and then rightwards $c$ times. (If $n$ is not $1$ mod $4$, this $c$ is unique).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Similar methods apply to a $n\times m$ square with $n\leq m$, where $n+m$ is even and we have $\frac{n+m}2$ infections).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Act Two:&lt;/strong&gt; We Replace Our Intuition with Mathematical Contortionism&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, it takes a lot more machinery to prove that our lemma was correct than it does to apply it - in particular, as natural as it is to talk about rectangles, this structure is not implicit in the update rule of our automata, yet we need to apply the model as an upper bound nevertheless. Firstly, we need to define what these rectangles &lt;em&gt;are&lt;/em&gt;. Our goal is to, for any infection, create a binary tree of rectangles, such that each node is the rectangular infection generated by combining the rectangles of its two children nodes, and then to characterize the length of an infection based on the properties of this tree. We do not directly construct such a tree, but it is implicit in the induction we use, and the major lemma used in this section may be used to construct it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, let us begin with the following definition:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $S$ be a finite subset of $\mathbb Z^2$. Define $\newcommand{\cl}{\operatorname{cl}}\cl(S)$ to be the set of points eventually infected if those in $S$ are infected. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It should be clear that $\cl$ is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Closure_operator&quot; rel=&quot;nofollow&quot;&gt;closure operator&lt;/a&gt;. Moreover, the closed sets with respect to this operation (i.e. sets such that $\cl(S)=S$) are exactly those from which no new infections are possible - we may characterize these as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A set $S$ is closed if and only if it a union of rectangles, no cell being adjacent or contained in more than one rectangle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, to build our tree of infections, we use the following important lemma: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For all $S$ with more than one element such that $\cl(S)$ is a rectangle, there exists a disjoint pair $A,B\subseteq S$ such that $\cl(A)$ and $\cl(B)$ are rectangles neither equal to $\cl(S)$ and that $\cl(A\cup B)=\cl(S)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To prove this, consider the set of rectangles other than $\cl(S)$ which are closures of subsets of $S$. This is obviously finite and non-empty if $|S|&amp;gt;1$. Therefore, ordered by inclusion, there is a maximal element $\cl(A)$ - a rectangle not contained by any other such rectangles. Set $B=S\setminus \cl(A)$. Now, suppose that $\cl(B)$ is the union of rectangles $\cl(B_1),\cl(B_2),\ldots,\cl(B_k)$. Notice that $\cl(A\cup B_i)$ must strictly contain $\cl(A)$. It is therefore either equal to $\cl(S)$, in which case we're done, or it is not a rectangle. If none of the $B_i$ have that $\cl(A\cup B_i)$ was a rectangle, then it would be the case that no cell could be adjacent to $\cl(A)$ and a $\cl(B_i)$ (nor to distinct $\cl(B_i)$ and $\cl(B_j)$), implying that $\cl(A)\cup\cl(B_1)\cup \cl(B_2)\cup\ldots\cup \cl(B_k)$ was a closed set containing $S$ which was not a rectangle, contradicting that $\cl(S)$ is a rectangle. Thus, one of the $B_i$ must satisfy the conditions of the lemma. (If $S$ has that its perimeter and it's closure's perimeter are equal, we may easily modify the above proof to show that $A$ and $B$ can and must be a partition of $S$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this tree, we may bound the length of time it takes for the infection to complete its spread. In particular, let $t(n,m)$ be the maximum length of time it takes to infect all the corners of an $n\times m$ rectangle with $\frac{n+m}2$ initial infections, where $n+m$ is even. This equals the amount of time to infect the whole square for $n,m&amp;gt;1$, since all the possible combinations of rectangle infect a corner last and only begin once both sub-squares have their corners infected.  Notice that there are only $3$ manners of combining two rectangles such that the result has the sum of their perimeters - on two $2\times 2$ rectangles, we illustrate them as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;-XX&#xA;-XX&#xA;---&#xA;XX-&#xA;XX-&#xA;&#xA;--XX&#xA;--XX&#xA;XX--&#xA;XX--&#xA;&#xA;---XX&#xA;XX-XX&#xA;XX---&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which correspond to increasing the height, the diagonal, and the width. In particular, to make an $n\times m$ rectangle ($n,m&amp;gt;1$), one must choose one of five options. For the first $3$, suppose that $x_1,x_2,y_1,y_2$ are positive integers with $x_1+y_1$ and $x_2+y_2$ both even. We are combining rectangles $x_1\times y_2$ and $x_2\times y_2$ in one of the above manners, letting $T=\max(t(x_1,y_1),t(x_2,y_2))$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Combining in the first manner (heightwise). We require $x_1+x_2=n+1$ and $y_1+y_2=m-1$ and $x_1,x_2&amp;gt;1$. This takes $T+\max(x_1+y_2,x_2+y_1)$ days.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Combining in the second manner (diagonally). We require $x_1+x_2=n$ and $y_1+y_2=m$. This takes $T+\max(x_1+y_2,x_2+y_1)-1$ days.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Combining in the third manner (widthwise). We require $x_1+x_2=n-1$ and $y_1+y_2=m+1$ and $y_1,y_2&amp;gt;1$. This takes $T+max(x_1+y_2,x_2+y_1)$ days.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Increasing height only. $x_1=1$ and $x_1+x_2=n+1$ and $y_1+y_2=m-1$. This takes $T+x_2+y_1$ days.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Increasing width only. $y_1=1$ and $x_1+x_2=n-1$ and $y_1+y_2=m+1$. This takes $T+x_1+y_2$ days.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Which, taking $t(1,n)=t(n,1)=0$, yields a recurrence relation for $t$ allowing explicit computation. However, even better than that, we can show that the first three options are never optimal, hence we only use the last two options - but moreover, those options are obviously slowest when $x_1=y_1=1$, meaning we should always enlarge rectangles according to the previously used procedure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove this, assume without loss of generality that $t(x_1,y_1)\leq t(x_2,y_2)$ - so essentially, the creating of the first rectangle comes for free, since it was done in parallel with the second rectangle. For the first (and by symmetry, the third) case, suppose we instead of joining a $x_1\times y_1$ rectangle to the $x_2\times y_2$ one, we increase the height of the $x_2\times y_2$ rectangle with a $1\times 1$ square placed above it, and then diagonally join the resulting $x_2\times (y_2+2)$ rectangle to a $(x_1-1)\times (y_1-1)$ rectangle**. That is, we consider the alternate diagram:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;--X&#xA;-X-&#xA;---&#xA;XX-&#xA;XX-&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This takes a total of $T+[x_2+1]+[\max(x_1-1+y_2+2,x_2-1+y_1)-1]$&#xA;which is at least as long as before. Similarly, when combining rectangles diagonally, we instead join a $1\times 1$ rectangle to the $x_2\times y_2$ one and then join the $(x_1-1)\times (y_1-1)$ one to that. We find a similar result algebraically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above suffices to show the desired claim that the optimal solution is of the form claimed, implying the displayed solution is optimal and unique for a $8\times 8$ board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*If we we're being very careful, we would note that all of the exchanges suggested, including exchanging all $X$'s and $Y$'s after something of the form $[a\times a]X\ldots$, decrease the lexicographic order of the string when we take $Y&amp;lt;D&amp;lt;X$. Thus, we will eventually reach a minimal element if we greedily apply such swaps - and in this context, a minimal element is one in which no such swaps are possible and is of the given form)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(**Some special care needs to go into the $y_1=1$ case, but the result comes out the same. A similar exception presents itself so for the diagonal combination) &lt;/p&gt;&#xA;" OwnerUserId="4765" LastEditorUserId="4765" LastEditDate="2015-07-18T23:56:21.630" LastActivityDate="2015-07-18T23:56:21.630" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18161" PostTypeId="2" ParentId="15556" CreationDate="2015-07-18T23:42:10.520" Score="5" Body="&lt;p&gt;Here is an explanation of the clues that weren't explained in &lt;a href=&quot;https://puzzling.stackexchange.com/a/15558/4450&quot;&gt;Geobit's answer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;He also made a stop in Macau and hit the casinos there.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Any mention of casinos or gambling in this puzzle is reference to a Chance space.  This is the one next to Oriental Avenue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and later while driving around he passed Shea Farm. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Shea_Farm_Halfway_House&quot; rel=&quot;nofollow noreferrer&quot;&gt;Shea Farm&lt;/a&gt; is a jail in New Hampshire.  This is a reference to the &quot;Just Visiting&quot; part of the Jail space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and he was happy when he found a hotel with complimentary valet service.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is a reference to the Free Parking space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and he was able to get to Churchill Downs just in time to bet on some horses.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is a reference to the Chance space next to Kentucky Avenue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and then relaxed at a nearby beach that described itself as surely the best.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Shorely the best&quot; is the motto of &lt;a href=&quot;https://en.wikipedia.org/wiki/Ventnor_City,_New_Jersey&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ventnor City&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He also stopped by the Horseshoe in Baltimore.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;a href=&quot;https://en.wikipedia.org/wiki/Horseshoe_Casino_Baltimore&quot; rel=&quot;nofollow noreferrer&quot;&gt;Horseshoe Casino&lt;/a&gt; is a reference to the Chance space next to Park Place.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then he went to Ocean City and found a room at the first green hotel in the city.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;a href=&quot;http://ocparkplacehotel.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Park Place Hotel&lt;/a&gt; says they are the first green hotel in Ocean City.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He relaxed on the beach and visited a lot of the shops and eateries nearby.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is the Boardwalk in Ocean City.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-07-18T23:42:10.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18163" PostTypeId="1" AcceptedAnswerId="18164" CreationDate="2015-07-19T02:55:51.233" Score="6" ViewCount="3268" Body="&lt;p&gt;Given a circular list of coins, that all have Tails facing up. &#xA;In each move, if we flip the coin at position $i$, then the coins at positions $i-1$ and $i+1$ get flipped as well. That is, consider: H H H T T: if I flip the coin at index 3 (0-based indexing), then the result would be: H H T H H.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Initial state: T T T T T ... T $~~$($N$ coins)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Final state: H H H H H ... H $~~$($N$ coins)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What would be the minimum number of moves to make all coins have Heads facing up?&lt;/p&gt;&#xA;" OwnerUserId="14179" LastEditorUserId="8874" LastEditDate="2016-02-17T20:28:08.283" LastActivityDate="2017-12-12T07:25:42.213" Title="Minimum moves to have all coins face Heads up" Tags="|mathematics|combinatorics|reachability|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18164" PostTypeId="2" ParentId="18163" CreationDate="2015-07-19T03:10:36.287" Score="7" Body="&lt;p&gt;If N is divisible by 3, then flipping every third coin will work (N/3 moves). This must be the minimum because there are N coins changed and each move changes only 3 coins' states.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If N is not divisible by 3, then flipping every coin makes every coin change states three times (N moves).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove that this is the minimum, notice that it doesn't matter in what order coins are flipped, and flipping the same coin twice is pointless. This means that every possible sequence of flips is equivalent to a set of which coins are flipped. There are $2^N$ of these sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we start to flip every third coin starting at an arbitrary position, eventually we will reach a state where exactly one coin is heads up and the rest are tails up. Because this can be repeated, any combination of heads and tails is possible. There are $2^N$ of these combinations. This is the same as the number of sets of coins that can be flipped, so each set must correspond to exactly one combination of heads and tails, and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that flipping every coin makes all coins heads up, so no other set of flips can lead to all heads up. In particular, every sequence of fewer than N flips will be equivalent to a set of flips which is not this set, so it will not lead to the desired position. Therefore, N moves is the minimum.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-07-19T03:10:36.287" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18166" PostTypeId="1" AcceptedAnswerId="18168" CreationDate="2015-07-19T04:46:49.733" Score="3" ViewCount="583" Body="&lt;p&gt;Stan (the clumsy coin flipper) is playing with his favorite toy: A set of $3n$ coins arranged in a circle, each one either heads up or tails up. He enjoys flipping these coins, but he's not very good at it: When he tries to flip one coin, he also flips its two neighbors. So, if he started with an arrangement of coins like (where $H$ stands for &quot;heads&quot; and $T$ for &quot;tails&quot;)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  T H T&#xA; T     T&#xA;H       T&#xA; H     H&#xA;  H T H&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and tried to flip the leftmost coin, he would end up with the arrangement:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  T H T&#xA; H     T&#xA;T       T&#xA; T     H&#xA;  H T H&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;flipping three coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He feels like he's getting pretty good at coin flipping, but has gotten bored with just flipping them aimlessly. He's figured out that it's easy to take an arrangement of all heads and flip every coin to tails in $n$ moves. He's asked you to design him a greater challenge - in particular, he wants you to arrange the coins such that it will take him &lt;em&gt;as many flips as possible&lt;/em&gt; to return the coins to all heads-up state. What pattern of heads and tails should you choose to satisfy his request? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(P.S. If you put the coins in an arrangement which is unsolvable, he will be very upset and you'll probably feel guilty for the rest of the day)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is a variation on &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18163/minimum-moves-to-have-all-coins-face-heads-up&quot;&gt;this previous puzzle&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="4765" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-07-19T13:53:00.387" Title="Entertaining the Clumsy Coin Flipper" Tags="|mathematics|strategy|" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18167" PostTypeId="2" ParentId="17458" CreationDate="2015-07-19T05:03:25.840" Score="1" Body="&lt;p&gt;I know that I'm late and that this one's already been answered, but I really just have to answer this.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Air.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is lighter than a feather.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's air. Do I really have to explain?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cannot be held by any person for more than few minutes. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A few minutes is around the time that the average person can hold their&#xA; breath, thus meaning that they are &quot;holding&quot; it in their lungs for a few&#xA; minutes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No person can touch it. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Technically, you're always touching it, but I think I know what the OP&#xA; was going for.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The human life is based on it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Air has oxygen in it, and without that humans can't survive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is fast when you run and it is normal when you aren't running:&#xA; Wind rushing in your face.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14180" LastActivityDate="2015-07-19T05:03:25.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18168" PostTypeId="2" ParentId="18166" CreationDate="2015-07-19T05:36:35.657" Score="3" Body="&lt;p&gt;As I explained in my answer to the other question, every sequence of flips is equivalent to a set of coins to flip. Performing two sequences in a row is equivalent to taking the symmetric difference of their sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can divide the coins into three groups of $n$ coins by counting 1,2,3,1,2,3,etc. Flipping all of the coins in one of the groups changes the state of every coin. The only non-empty sets of flips that result in all of the coins in the &lt;em&gt;same&lt;/em&gt; state as before are the ones consisting of two of the groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: Suppose we have a non-empty set of flips that leaves all the coins the same. It can't possibly contain all the coins (because that changes the state of every coin), so there's at least one coin that isn't flipped. Therefore, there must be a coin that is flipped next to one that isn't.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;...XO???...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(X indicates a coin to flip, and O is a coin not to flip.) That coin must be switched back, so we have to flip the next coin.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;...XOX??...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This coin also has to be switched back, so the next one also has to be flipped:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;...XOXX?...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;That coin is switched twice already, so the next one can't be flipped:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;...XOXXO...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This pattern must repeat every three coins, so it includes all the coins from two groups and no coins from the last group.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If two different sets of flips result in the same position, then their symmetric difference must leave the coins the same. Therefore, the symmetric difference is all the coins from two groups. The converse is also true: if the symmetric difference of the two sets leaves the coins the same, the sets must lead to the same position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that for every set of flips, there are exactly three others equivalent to it: the symmetric difference with groups 1 and 2, the symmetric difference with groups 1 and 3, and the symmetric difference with groups 2 and 3. If the original set contains $a$ coins from group 1, $b$ from group 2, and $c$ from group 3, then the equivalent sets contain $(n-a)+(n-b)+c$ coins, $(n-a)+b+(n-c)$ coins, and $a+(n-b)+(n-c)$ coins, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that the original set represents the shortest way to solve the position. Then these three expressions must all be greater than or equal to $a+b+c$. This is equivalent to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a+b\le n\\a+c\le n\\b+c\le n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can sum these inequalities and divide by 2 to get $a+b+c\le\frac{3n}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For even $n$, we can have $a+b+c=\frac{3n}{2}$ by choosing $a=b=c=\frac{n}{2}$. One possible set with these values is $\frac{3n}{2}$ consecutive coins. This corresponds to the position with $\frac{3n-4}{2}$ tails, one heads, one tails, $\frac{3n-4}{2}$ heads, one tails, and one heads. This position cannot be solved with fewer than $\boxed{\frac{3n}{2}}$ flips.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For odd $n$, we can have $a+b+c=\frac{3n-1}{2}$ by choosing $a=\frac{n+1}{2}$ and $b=c=\frac{n-1}{2}$. One possible set with these values is $\frac{3n-1}{2}$ consecutive coins. This corresponds to the position with $\frac{3n-3}{2}$ tails, one heads, one tails, $\frac{3n-5}{2}$ heads, one tails, and one heads. This position cannot be solved with fewer than $\boxed{\frac{3n-1}{2}}$ flips.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-07-19T05:36:35.657" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18169" PostTypeId="2" ParentId="18163" CreationDate="2015-07-19T08:13:42.467" Score="0" Body="&lt;p&gt;Every move flips 3 coins, so it can never be done in fewer moves than the ceiling of $N/3$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $N$ = $3k$, we just choose every third coin ($N/3$ moves).&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $N$ = $3k+1$, there'll be an overlap and at least $k+1$ moves will be needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we just choose every single coin (more than once would be redundant), each will be flipped the amount of an odd number, and the state can be reversed. The question is, are fewer moves enough? If so, then choosing all the omitted (unchosen) coins instead will have no effect. The only way this would work would be for every omitted coin to have another next to it, but not two. With o indicating the omitted:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;xoox&lt;br&gt;&#xA;xooxo&lt;br&gt;&#xA;xooxoox... (put os at the end to cover the &quot;outermost&quot; xes twice, leading to a contradiction when it covers the full circle for any $N$ not divisible by 3. In this case, no coin must be omitted, thus all $N$ moves must be made. For any $N$, $N$ moves guarantee all coins can be flipped.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="14016" LastEditDate="2015-07-19T09:15:41.400" LastActivityDate="2015-07-19T09:15:41.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18171" PostTypeId="1" AcceptedAnswerId="18251" CreationDate="2015-07-19T16:26:42.540" Score="21" ViewCount="1622" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;One real, two imagined, one real who imagined.&lt;br&gt;&#xA;  One performer, two fighters, one sidekick.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;One cannot be seen,&lt;br&gt;&#xA;  One is hard to kill.&lt;br&gt;&#xA;  One is a doctor,&lt;br&gt;&#xA;  One is better known as a bug. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;One living, two ageless, one dead but immortal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What are these lines talking about?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 1 4 1. Four in total. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The title is a big giveaway, and answering that question alone can lead the way to solving the riddle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 3: (A big one)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Vioz-'s comment on this question and &lt;a href=&quot;https://puzzling.stackexchange.com/a/18230/12700&quot;&gt;dennisdeems' answer&lt;/a&gt; have figured out a couple pieces of the riddle. String them together to find the solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-07-22T09:13:48.517" Title="Can you name the unknown?" Tags="|riddle|word|" AnswerCount="6" CommentCount="23" ContentLicense="CC BY-SA 3.0" />
  <row Id="18172" PostTypeId="2" ParentId="17895" CreationDate="2015-07-19T18:40:00.787" Score="3" Body="&lt;p&gt;Hint 2 contains a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; reversed, Base64 encoded string with the character &lt;code&gt;R&lt;/code&gt; randomly inserted to corrupt it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It decodes to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;pre&gt;&#xA;        30&#xA;        /\&#xA;       2  15&#xA;          /\&#xA;         3  5&#xA; &#xA;     In general (p_0,p_1,...,p_N,m | x and p_i &amp;lt;= p_{i+1} and p \in \mathbb{P}):&#xA;         x&#xA;        / \&#xA;      p_0  m * PI_{1&amp;lt;=i&amp;lt;=N} p_i&#xA;                 /\&#xA;               m * p_1 PI_{1&amp;lt;=i&amp;lt;=N} p_i&#xA;                       /\&#xA;                         .&#xA;                          .&#xA;                           .&#xA;                           /\&#xA;                     p_{n-1} p&#xA; &lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we are to believe that each parent node is the product of its two child nodes, then we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;m\prod_{i=1}^{N}p_i=m\ p_1\prod_{i=1}^{N}p_i\\&#xA;1=p_1&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;p_i\in P\\&#xA;1\notin P&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus this is not yet the full answer; there is still some work to do.&lt;/p&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="4421" LastEditDate="2015-07-20T14:37:31.120" LastActivityDate="2015-07-20T14:37:31.120" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18173" PostTypeId="1" AcceptedAnswerId="18174" CreationDate="2015-07-19T21:48:50.250" Score="13" ViewCount="3995" Body="&lt;p&gt;I thought this was easy, but cute, so I decided to share it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1011, 2223, 3940, 5648, 8910&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="14140" LastEditorUserId="4450" LastEditDate="2015-07-20T01:46:23.637" LastActivityDate="2015-07-21T12:32:16.260" Title="Which of these numbers does not belong to the group?" Tags="|pattern|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18174" PostTypeId="2" ParentId="18173" CreationDate="2015-07-19T21:52:17.323" Score="18" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 5648&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The other numbers are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; adjacent numbers concatenated to each other:&lt;br&gt;&#xA; 1011 = 10, 11&lt;br&gt;&#xA; 2223 = 22, 23&lt;br&gt;&#xA; 3940 = 39, 40&lt;br&gt;&#xA; 8910 = 8, 9, 10&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4450" LastEditDate="2015-07-19T22:03:49.997" LastActivityDate="2015-07-19T22:03:49.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18176" PostTypeId="1" AcceptedAnswerId="18177" CreationDate="2015-07-19T23:35:21.233" Score="30" ViewCount="2539" Body="&lt;p&gt;I had a wonderful crossword all prepared for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But... you suck. &gt;:(&amp;nbsp;&amp;nbsp;&amp;nbsp;So I'm just going to tell you about it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was an 8-by-8 &lt;a href=&quot;https://en.wikipedia.org/wiki/Crossword&quot;&gt;American-style&lt;/a&gt; grid with white and black squares, 180&amp;deg; rotationally symmetric. All the words contained at least two letters, and each white square belonged to exactly two words, one across, one down. All the whites were connected to each other, and the clues were &lt;em&gt;numbered in the standard American style&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I even had the clues all ready to go.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Across&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;$1$. edible Greek letter?&lt;br&gt;&#xA;$3$.&lt;br&gt;&#xA;$6$. collector of taxes (abbr.)&lt;br&gt;&#xA;$7$.&lt;br&gt;&#xA;$8$.&lt;br&gt;&#xA;$9$.&lt;br&gt;&#xA;$11$.&lt;br&gt;&#xA;$12$. The ______ Limits (TV show)&lt;br&gt;&#xA;$14$. stunted breaths&lt;br&gt;&#xA;$16$. machine smarts (abbr.)&lt;br&gt;&#xA;$17$.&lt;br&gt;&#xA;$18$.&lt;br&gt;&#xA;$19$.&lt;br&gt;&#xA;$21$. Blanc or Gibson&lt;br&gt;&#xA;$22$. not quite a disco stud&lt;br&gt;&#xA;$23$.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Down&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;$1$. apes&lt;br&gt;&#xA;$2$.&lt;br&gt;&#xA;$3$.&lt;br&gt;&#xA;$4$. dictatorial car manufacturer?&lt;br&gt;&#xA;$5$. hollered&lt;br&gt;&#xA;$6$. picture&lt;br&gt;&#xA;$10$. test of character&lt;br&gt;&#xA;$12$. no-____ (&quot;do nothing&quot; in computer jargon)&lt;br&gt;&#xA;$13$.&lt;br&gt;&#xA;$15$.&lt;br&gt;&#xA;$20$. Et ____, Brut&amp;eacute;?&lt;br&gt;&#xA;$21$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's that? Some of the clues are missing? That must be because you're such a big jerk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here you go, cuddles. A few of the missing clues:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\square$. shoo!&lt;br&gt;&#xA;$\square$. a timed spasm?&lt;br&gt;&#xA;$\square$. coroner (abbr.)&lt;br&gt;&#xA;$\square$. Just Do ____!&lt;br&gt;&#xA;$\square$. morning hours&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'll chew glass before I tell you where they go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there you have it, bonehead. You can't tell me where the blacks and whites go on my crossword. You definitely can't tell me what letters go in the white squares. And it's &lt;em&gt;inconceivable&lt;/em&gt; that you could solve the crossword, place the missing clues beside their proper numbers, and come up with creative clues for any left over entries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because if you &lt;strong&gt;did all of this, you'd have solved my puzzle&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And everyone knows: jerks can't solve puzzles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sincerely,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Driver of the White Volvo (You Know the One!)&lt;/p&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="4219" LastEditDate="2015-07-20T15:18:10.693" LastActivityDate="2015-07-23T02:18:34.283" Title="Want to See My Crossword? Too Bad!" Tags="|geometry|crosswords|crossword-clues|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18177" PostTypeId="2" ParentId="18176" CreationDate="2015-07-20T01:32:56.873" Score="30" Body="&lt;p&gt;Here's the solved puzzle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/ahWDI.png&quot; alt=&quot;Finished Crossword&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Given clues but missing numbers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; shoo! - 15 down (SCAT) &lt;br&gt; a timed spasm? - 7 across (TIC) &lt;br&gt; coroner (abbr.) - 21 down (M.E. - short for Medical Examiner) &lt;br&gt; Just Do ___! - 3 down (IT) &lt;br&gt; morning hours - 11 across (AM)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clues for other missing clues&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3 across - No &lt;strong&gt;_, ands or buts &lt;br&gt; 8 across - Do, Re, _&lt;/strong&gt; &lt;br&gt; 9 across - Less successful partner of Simon and Garfunkel &lt;br&gt; 17 across - ... and so forth &lt;br&gt; 18 across - famous Yo-Yo &lt;br&gt; 19 across - &quot;I could just ___ you up&quot; &lt;br&gt; 23 across - talkative horse, informally &lt;br&gt; 2 down - will be, in the future &lt;br&gt; 13 down - &quot;The Last of ____&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Process (as requested by Emrakul):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; By comparing the across and down clues, you can figure out a few things very quickly. Firstly, 1 across is 2 letters and 3 across is 3 letters. There's at least one black space between them and two more in the first row. Since there's a 6 across AND a 6 down, you know that there's a black space before the first word (otherwise there wouldn't be a 6 down on the second row). Similar rationale with 7 and 8 across leads us to conclude that those are the same length as the clues above them. But 9 across must be a letter longer, since there is a 10 down (but no 10 across), so there are a couple black spaces in the upper right corner. &lt;br&gt; &lt;br&gt;A combination of the rotational symmetry and a few clues at the bottom (MEL, STU) gives us a good frame for the bottom half to confirm this and I started filling in other clues (PI, IRS, MEL, STU, IMAGE, FIAT, PRIMATES, SCREAMED) &lt;br&gt;&lt;br&gt; With some of the words filled in, and keeping in mind parallel symmetry, I reached &quot;OUTER&quot; and &quot;GASPS&quot; as the two words bridging the middle gap. That finished off the black squares of the grid. With a good chunk of letters, I started filling in the clues with missing numbers. AM was already done, and TIC, IT, M.E. came pretty quickly after. The few remaining squares were not too tough to finish off.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastEditorUserId="4675" LastEditDate="2015-07-23T02:18:34.283" LastActivityDate="2015-07-23T02:18:34.283" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18178" PostTypeId="2" ParentId="18173" CreationDate="2015-07-20T02:58:27.320" Score="18" Body="&lt;p&gt;Here's an alternate answer that is also valid&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2223 is the odd one out. &lt;br&gt;&lt;br&gt; This is because each of the other numbers contain 'holes' (closed loops in the digits), i.e at least one of the digits 4 6 8 9 0 is present in the numbers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-07-20T02:58:27.320" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18179" PostTypeId="1" AcceptedAnswerId="18180" CreationDate="2015-07-20T06:34:48.247" Score="1" ViewCount="671" Body="&lt;p&gt;find the value of  &lt;strong&gt;1 + 7&lt;/strong&gt; from the following.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;3   +   3   =   3&#xA;&#xA;5   +   4   =   4&#xA;&#xA;1   +   0   =   3&#xA;&#xA;2   +   3   =   4&#xA;&#xA;3   +   4   =   5&#xA;&#xA;1   +   7   =   ?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="13582" LastEditorUserId="4421" LastEditDate="2015-11-25T18:05:37.743" LastActivityDate="2015-11-25T18:05:37.743" Title="Find the missing value" Tags="|logical-deduction|number-sequence|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18180" PostTypeId="2" ParentId="18179" CreationDate="2015-07-20T06:41:00.893" Score="5" Body="&lt;p&gt;&lt;strong&gt;Answer is&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;5 &lt;br /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3   +   3   =   3 (no of letters in Six)&lt;br /&gt;&#xA;5   +   4   =   4 (no of letters in Nine)&lt;br/&gt;&#xA;1   +   0   =   3 (no of letters in One)&lt;br /&gt;&#xA;2  +   3   =   4 (no of letters in Five)&lt;br /&gt;&#xA;3   +   4   =   5 (no of letters in Seven)&lt;br /&gt;&#xA;1   +   7   =   5 (no of letters in Eight)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="12408" LastEditDate="2015-07-20T09:20:13.490" LastActivityDate="2015-07-20T09:20:13.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18181" PostTypeId="2" ParentId="18039" CreationDate="2015-07-20T07:50:13.610" Score="3" Body="&lt;p&gt;I may sound naive, but my answer is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;human being&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My legend will be remembered,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; human history will always be remembered  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Too strong to be dismembered.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; people strive to not to be dismembered (everyone else needs some attention, some folks wanna be famous)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No happiness at the parting,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; when someone dies, other are mourning  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What happened at the starting?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; how humanity began? it's one of the most fundamental questions  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All the lines or all we do,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; culture of humanity is a sign of presence, we are what we produce with culture (broadly taken)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mix God, religion, and the facts too.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; human beginnings, also what people are talking/up to includes all ideas, including faith, religion, historical/science breakthroughs  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Counting now, what are we at?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (either:) it's hard to follow all humans' thoughts/trends/intentions&lt;br&gt;&#xA; (or#1:) we can't really say what our future is or where we are getting to&lt;br&gt;&#xA; (or#2:) we can't say when we die (we can't tell how long we will live from now  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Profound, noises, or like a cat?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (no idea yet, still working on it...)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My land is one full of life,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; humans live on Earth, which is 'full of life' (we can't, at least, for now, live in a place without organic life or other ecosystem)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You know my cousins, little strife.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (either:) humans don't really always get along with other humans...&lt;br&gt;&#xA; (or:) monkeys? animals? not really a threat, but still  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Started with two, but grew to three,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; humanity needs male+female to reproduce, a child is 'three' then (not really male or female in terms of reproducing till maturity)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But with some help, who can I be?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; with help of genetic engineering?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12739" LastActivityDate="2015-07-20T07:50:13.610" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18182" PostTypeId="2" ParentId="18173" CreationDate="2015-07-20T07:52:32.790" Score="14" Body="&lt;p&gt;At the risk of being downvoted for giving an answer different from the accepted one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Can also be said 2223, because the numbers form a non-descending series.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="14016" LastEditDate="2015-07-20T08:35:39.853" LastActivityDate="2015-07-20T08:35:39.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18183" PostTypeId="2" ParentId="18039" CreationDate="2015-07-20T10:41:10.677" Score="3" Body="&lt;p&gt;I have a suggestion as to what it is. There are a couple of lines in particular that scream that this is what it is, but I can't fit some of the more &quot;ambiguous&quot; lines in. Is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Israel/The Jews?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My legend will be remembered,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The story of Israel's survival is legendary&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Too strong to be dismembered.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Jews survived the holocaust, 'nuff said.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No happiness at the parting,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In biblical times, Judah split off from Israel&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What happened at the starting?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Unsure, but the history of the Jews does go back like, 4000 years.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All the lines or all we do,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Unsure&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mix God, religion, and the facts too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Speaks for itself really&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Counting now, what are we at?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; King David counted the Israelites in 1 Chronicles 21, if you carried on counting, the numbers would be ridiculous.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Profound, noises, or like a cat?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Unsure&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My land is one full of life,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; God gave them the land of Israel and it &quot;flowed with milk and honey&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You know my cousins, little strife.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Arab nations come from the line of Esau, which was the brother of Jacob, the father of Israel. Obviously we know the strife between Israel and the Arab nations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Started with two, but grew to three,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Unsure, but perhaps a reference to WWII where America joined the fight later on. (Reference to fighting the Holocaust?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But with some help, who can I be?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Israel have had lots of financial help from America&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4659" LastEditorUserId="4659" LastEditDate="2015-07-20T10:48:43.063" LastActivityDate="2015-07-20T10:48:43.063" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18186" PostTypeId="1" CreationDate="2015-07-20T13:36:10.170" Score="16" ViewCount="1670" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VqIZz.jpg&quot; alt=&quot;Metro station points&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture above shows a map of 9 train stations. These stations are connected by a single track running from station 1 to 9 as represented by the solid line (the trains runs both ways from 1-&gt;9 and 9-&gt;1). Let's say each station is 1 km distance apart from the adjacent station, along the railway track (solid line). Distance between 4 and 6, 4 and 7, 3 and 7, 3 and 8, 2 and 8, 2 and 9, 1 and 9 is also 1 km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now people from station 1 are only 1km far from station 9, but they have to go all the way round traveling 8km to reach station 9. Similarly, it is inefficient for people at other stations as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the engineers decide to connect the stations on both sides. But they can only use a 1km railway track, due to budget constraints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Which two stations do they connect so that it benefits all the people the most?&lt;/strong&gt; They are supposed to consider the collective benefit. More specifically, they are to optimize the average travel time of the passengers, assuming that people embark and disembark at all stations with equal probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I am new on this forum, and new at creating puzzles so please bear with me if I didn't give sufficient information for this puzzle to work. And I don't know the answer to the puzzle yet, I myself am trying to solve it.&lt;/p&gt;&#xA;" OwnerUserId="14074" LastEditorUserId="65277" LastEditDate="2022-05-16T20:44:06.233" LastActivityDate="2022-05-16T20:44:06.233" Title="Can this be solved without brute force?" Tags="|mathematics|logical-deduction|optimization|" AnswerCount="5" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18187" PostTypeId="2" ParentId="18186" CreationDate="2015-07-20T14:18:23.053" Score="1" Body="&lt;p&gt;Assuming that the traffic between stations are the same, a non-brute-force solution is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; to start with a completely filled diagram and take out tracks. It should be obvious that constructing every track will be best at reducing travel time between any two stations. To pick a track to remove, we make the observation that a track near the middle (e.g 3-7) will be much more useful than a track near the edge (e.g 1-9). So we see how much removing a track on the edge will affect the total travel distances between pairs of stations.&lt;br&gt;&lt;br&gt; Taking out the track 1-9, we notice that stations 2 through 8 are unaffected by the change. In fact, only stations 1 and 9 are affected because they lost their direct path! So removing this track will add a total of 1km x 2 = 2 km to the total travel distance between pairs of stations. &lt;br&gt;&lt;br&gt; For our next track to remove, we look for another edge; 5-6. However, track 5-6 is not a possible choice for being removed, so we look at the next track inside: 4-6. Removing this track, we can see that only stations 4 and 6 are affected since they lost their direct path. &lt;br&gt;&lt;br&gt; So we continue to remove tracks that are on the outside, which will bring the least amount of change in the total travel distances between pairs of stations. Doing this removes tracks 2-9, 4-7, 2-8, and 3-7, leaving us with only track 3-8 intact. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13881" LastActivityDate="2015-07-20T14:18:23.053" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18188" PostTypeId="1" AcceptedAnswerId="18246" CreationDate="2015-07-20T15:14:54.977" Score="1" ViewCount="669" Body="&lt;p&gt;Here is a block of cipher text C:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;28 49 3d 57 2f 48 20 7c 20 7b 4e 7d 3a 49 27 57 27 48 20 45 29 20 3c 2d 2d 20 28 7b 5f 5a 7d 3a 58 20 7c 20 7b 36 5f 27 30 27 38 5f 7d 3a 58 20 41 29&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please evaluate this text and return to me it's value. There aren't many steps to get through but you'll need some knowledge and logic. If you can't get through a particular step, at least post your partial progress so someone else can hep out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the interest of improving this puzzle, allow me add a light hint regarding the method of getting the answer: Assuming C is the initial text provided and D is the resulting text after manipulations,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D=&quot;ROT180&quot;(A^-1(H^-1(C)))  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there, you will need to use logic to evaluate D and provide the final answer.&lt;/p&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="13970" LastEditDate="2015-07-20T21:20:41.540" LastActivityDate="2015-07-22T02:58:45.690" Title="A Logical Puzzle" Tags="|logical-deduction|cipher|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="18189" PostTypeId="2" ParentId="18186" CreationDate="2015-07-20T15:57:53.870" Score="4" Body="&lt;h2&gt;Solution&lt;/h2&gt;&#xA;&lt;p&gt;&lt;strong&gt;Add link between stations 2 and 8&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Details&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;To begin with, let's consider a layout with stations organized on a grid following a rectangular path with the stations on one end connecting to one another. It would be clear upon inspection that the best location to place a single additional connecting line would be 2/3 of the way from the closed end of the rectangle. Below are the figures for even and odd side lengths.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mDNmZ.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Extending this to the trapezoid in the problem statement we see that the best solution would most likely be 2-8 as it is roughly 2/3 of the way from 5-6. However 3-8 could also be a solution.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/X0TTu.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Algorithm&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Select potential candidate links that are approximately 2/3 from closed end&lt;/li&gt;&#xA;&lt;li&gt;For each candidate link evaluate the minimum distances between stations (1-9, 2-9, 2-8, 3-8, 3-7, 4-7, 4-6, and 5-6)&lt;/li&gt;&#xA;&lt;li&gt;Total distances for each candidate link&lt;/li&gt;&#xA;&lt;li&gt;Select candidate link with minimum total distance&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&lt;strong&gt;Results&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Candidate link 2-8 totals 17km&lt;/li&gt;&#xA;&lt;li&gt;Candidate link 3-8 totals 18km&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;strong&gt;Choose&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Link 2-8&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="13788" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-20T19:15:11.830" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18190" PostTypeId="1" AcceptedAnswerId="18191" CreationDate="2015-07-20T18:13:59.490" Score="16" ViewCount="1858" Body="&lt;p&gt;We are siblings: sons and daughters&lt;br&gt;&#xA;of a very rich and powerful father.&lt;br&gt;&#xA;Different jobs we each perform,&lt;br&gt;&#xA;contracted for a year or more.  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;My job is author, of the real and the fictional;&lt;br&gt;&#xA;Many hold me as the eldest, most famous, an original.&lt;br&gt;&#xA;I can write about anything, no matter the tedium;&lt;br&gt;&#xA;As for who I am, I am known by my medium.  &lt;/li&gt;&#xA;&lt;li&gt;I am an orator, adept at public presentation;&lt;br&gt;&#xA;I can use many media to convey information.&lt;br&gt;&#xA;I'm the second original, my ego's in good health;&lt;br&gt;&#xA;I even name my product after myself!  &lt;/li&gt;&#xA;&lt;li&gt;The last original, my name's a success;&lt;br&gt;&#xA;While not a presenter, I certainly impress.&lt;br&gt;&#xA;I'm a data analyst, and I'm matched by none;&lt;br&gt;&#xA;When it comes to number crunching, I'm #1.  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;We have many younger siblings: an assistant, an archivist;&lt;br&gt;&#xA;But we the originals, are the oldest and brightest.&lt;br&gt;&#xA;We've given enough clues, it's time to end the game:&lt;br&gt;&#xA;Who are we?, Who's our father?, and What's our family's name?&lt;/p&gt;&#xA;" OwnerUserId="13970" LastActivityDate="2015-07-20T19:38:30.650" Title="Who are we and what is our family name?" Tags="|riddle|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18191" PostTypeId="2" ParentId="18190" CreationDate="2015-07-20T18:21:46.690" Score="15" Body="&lt;p&gt;I think you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The microsoft suite: word, excel, and powerpoint.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&#xA;Word is an author and used as a document editor for nearly everyone.&lt;br&gt;&#xA;&lt;br&gt;&#xA;Powerpoint is used in presentations. It can convey information in different media types on slides (gif, jpegs).&lt;br&gt;&#xA;&lt;br&gt;&#xA;Excel is a data cruncher, a giant spreadsheet. It doesn't look very nice for presenting but it has powerful number crunching capabilities&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13975" LastEditorUserId="12109" LastEditDate="2015-07-20T19:38:30.650" LastActivityDate="2015-07-20T19:38:30.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18192" PostTypeId="2" ParentId="18039" CreationDate="2015-07-20T18:48:21.690" Score="0" Body="&lt;p&gt;Trying to find a proper link but I'm pretty sure its something like:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A list of Brad Pitt movies&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My legend will be remembered,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;   Legends of the Fall&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Too strong to be dismembered. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Troy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No happiness at the parting,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Meet Joe Black&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What happened at the starting? &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fight Club&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All the lines or all we do,&#xA;Mix God, religion, and the facts too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Counting now, what are we at?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Se7en&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Profound, noises, or like a cat?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My land is one full of life,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Snatch (as per Hint #5)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You know my cousins, little strife.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Started with two, but grew to three,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Could be the Ocean's 11 movies, I think it's the only trilogy he's been in...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But with some help, who can I be? &lt;/p&gt;&#xA;" OwnerUserId="11251" LastEditorUserId="11251" LastEditDate="2015-07-20T19:40:29.630" LastActivityDate="2015-07-20T19:40:29.630" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18193" PostTypeId="2" ParentId="18186" CreationDate="2015-07-20T19:32:30.113" Score="0" Body="&lt;p&gt;If you connect 9-&gt;1, then worst case it would take n/2 segments between 2 stations (eg. 1-&gt;5 or 1-&gt;6 via 9 is 4 segments). Moving the segment 1 node inwards on both sides, now the worst case is reduced and becomes 3 (e.g. 2-&gt;6 thru 8 when 2-8 is connected). If you further change it to use 3-8 then 1-&gt;9 is 4 segments, while the inner loop worst case is still 3 segments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hmm, in the specific example above it seems &lt;strong&gt;2-8&lt;/strong&gt; is the best choice to minimize worst case length.&#xA;It should be possible to come up with a generalized formula to connect the mth &amp;amp; (n -m)th nodes, where&#xA;1-&gt;m-(1 segment)-&gt;(n-m)-&gt;m &#xA;is half the length of inner loop of m-&gt;(n-m)-(1 segment)-&gt;m &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: I changed my answer while updating it to answer the OP's comment, earlier one was 3-8 which is wrong for this case.&lt;/em&gt; &lt;/p&gt;&#xA;" OwnerUserId="13544" LastEditorUserId="13544" LastEditDate="2015-07-24T01:42:31.317" LastActivityDate="2015-07-24T01:42:31.317" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18194" PostTypeId="2" ParentId="18173" CreationDate="2015-07-20T19:48:34.823" Score="2" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 5648&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The members are base 10. And each digit in each member is within one of another digit in the member. Assuming this 0,1,2,3,4,5,6,7,8,9,0,... line of digits, one move in either direction. 5648 contains an '8' and would need either an additional '7' or '9' to fulfil the property.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14198" LastEditorUserId="1831" LastEditDate="2015-07-21T12:32:16.260" LastActivityDate="2015-07-21T12:32:16.260" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18196" PostTypeId="2" ParentId="18039" CreationDate="2015-07-20T21:28:56.753" Score="3" Body="&lt;p&gt;Alright, this is my guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dr Robert Neville - I Am Legend&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And your plan is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To find a cure to the zombie virus.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My legend will be remembered,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dr Neville was the legend, surviving the virus and trying to cure it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Too strong to be dismembered.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dismembering is a classic in zombie films/books, but he was too strong to die to the outbreak.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No happiness at the parting,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When parting with his family, when they died.. there was no happiness. Or when his dog, Sam dies. Parting is never happy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What happened at the starting?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The big question, how did this outbreak start?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All the lines or all we do,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This one, I'm unsure about&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mix God, religion, and the facts too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; God, religion.. everything is being questioned when a thing only seen in Sci-Fi films becomes real. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Counting now, what are we at?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; How many days has he survived? How many zombies are there? Counting the days that go by.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Profound, noises, or like a cat?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Moaning of zombies, and other scary things.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My land is one full of life,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Either a sarcastic statement, or an optimistic one. Zombies, while alive.. are not really alive. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You know my cousins, little strife.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; His family, all dead? When you're family is dead there isn't much to argue about with them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Started with two, but grew to three,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Zombies, they multiply! Two become three, three become six.. etc&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But with some help, who can I be?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He can be the legend he needs to be with the help of the children.. who looked after the cure after he killed himself to save the rest of humanity. And after he died, he is the martyr.. the legend that saved humanity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14202" LastActivityDate="2015-07-20T21:28:56.753" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18197" PostTypeId="2" ParentId="18071" CreationDate="2015-07-20T22:51:17.383" Score="11" Body="&lt;p&gt;This is tough, because as you all know &lt;a href=&quot;http://www.cabinetmagazine.org/issues/40/sherman.php&quot;&gt;Anything can signify anything&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the &lt;em&gt;facts&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Europe is named after the Phoenician princess &quot;Europa&quot;, from Greek mythology.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;TRUE (1)&lt;/p&gt;&#xA;  &#xA;  &lt;ol start=&quot;2&quot;&gt;&#xA;  &lt;li&gt;Romania is the oldest country in all of Europe. It hasn't changed its name since &gt; 681 AD!&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;FALSE (0)&lt;/p&gt;&#xA;  &#xA;  &lt;ol start=&quot;3&quot;&gt;&#xA;  &lt;li&gt;The Netherlands has the world's densest rail system, with 113.8 kilometers of rail per 1,000 square kilometers.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;FALSE (0)&lt;/p&gt;&#xA;  &#xA;  &lt;ol start=&quot;4&quot;&gt;&#xA;  &lt;li&gt;Over 50% of Greece's population lives in its capital, Athens.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;FALSE (0)&lt;/p&gt;&#xA;  &#xA;  &lt;ol start=&quot;5&quot;&gt;&#xA;  &lt;li&gt;The most visited place in the entirety of Europe is, not suprisingly, Disneyland (located in Paris).&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;TRUE (1)&lt;/p&gt;&#xA;  &#xA;  &lt;ol start=&quot;6&quot;&gt;&#xA;  &lt;li&gt;Greenland, an island owned by Denmark, is the second-largest island in the world, surpassed only by Australia.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;FALSE (0), because considered as largest island according to Wiki.&lt;/p&gt;&#xA;  &#xA;  &lt;ol start=&quot;7&quot;&gt;&#xA;  &lt;li&gt;Croissants, the famous French pastry, were actually invented in Austria.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;TRUE (1), especially according to English Wiki page.&lt;/p&gt;&#xA;  &#xA;  &lt;ol start=&quot;8&quot;&gt;&#xA;  &lt;li&gt;The national airport in Brussels sells more chocolate than any other location on Earth!&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;TRUE (1)&lt;/p&gt;&#xA;  &#xA;  &lt;ol start=&quot;9&quot;&gt;&#xA;  &lt;li&gt;Finland has the most McDonald's per capita in all of Europe.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;FALSE (0)&lt;/p&gt;&#xA;  &#xA;  &lt;ol start=&quot;10&quot;&gt;&#xA;  &lt;li&gt;Hate mosquitos? Move to Iceland, where there are literally none.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;TRUE (1)&lt;/p&gt;&#xA;  &#xA;  &lt;ol start=&quot;11&quot;&gt;&#xA;  &lt;li&gt;If you're getting bored of Europe, have no fear - Africa is only a 30-minute ferry ride from Spain.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;TRUE (1)&lt;/p&gt;&#xA;  &#xA;  &lt;ol start=&quot;12&quot;&gt;&#xA;  &lt;li&gt;Europe is the second-smallest continent, but has the third-largest population, surpassed only by Asia and North America.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;FALSE (0)&lt;/p&gt;&#xA;  &#xA;  &lt;ol start=&quot;13&quot;&gt;&#xA;  &lt;li&gt;The longest town name in Europe is Lanfairpwllgwyngyllgogerychwyrndrobwllllantysiliogogogoch. I'll wait for you to try to pronounce that.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;TRUE (1), despite the fact that the initial &lt;em&gt;L&lt;/em&gt; is missing.&lt;/p&gt;&#xA;  &#xA;  &lt;ol start=&quot;14&quot;&gt;&#xA;  &lt;li&gt;The most common name for a pet pig in France? Napoleon.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;FALSE (0)&lt;/p&gt;&#xA;  &#xA;  &lt;ol start=&quot;15&quot;&gt;&#xA;  &lt;li&gt;It is forbidden to flush your toilet after 10 PM in Belgium, for some reason.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;FALSE (0)&lt;/p&gt;&#xA;  &#xA;  &lt;ol start=&quot;16&quot;&gt;&#xA;  &lt;li&gt;The largest recorded colony of ants in the world is 6,000 kilometers long, and can be found in northern Italy. I promise I'm not sending you there.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;TRUE (1) - the fact says &quot;and can be found in northern Italy&quot; which is definitely true (it doesn't state that's it can only or as a total be found in Italy)&lt;/p&gt;&#xA;  &#xA;  &lt;ol start=&quot;17&quot;&gt;&#xA;  &lt;li&gt;The Mediterranean Sea is the only sea in Europe that researchers believe has always been filled with water.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;FALSE (0)&lt;/p&gt;&#xA;  &#xA;  &lt;ol start=&quot;18&quot;&gt;&#xA;  &lt;li&gt;The London Eye in (surprise) London is the world's tallest Ferris Wheel, standing at 135 meters tall.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;FALSE (0)&lt;/p&gt;&#xA;  &#xA;  &lt;ol start=&quot;19&quot;&gt;&#xA;  &lt;li&gt;The entirety of Europe by land area is still smaller than Canada!&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;FALSE (0)&lt;/p&gt;&#xA;  &#xA;  &lt;ol start=&quot;20&quot;&gt;&#xA;  &lt;li&gt;An estimated 10% of Europeans were conceived in Ikea beds - that's about 73.3 million people, for those counting.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;TRUE (1)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Given this fact sheet we get &lt;code&gt;10001011011010010001&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have a lot of options for such binary looking chunk of data:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As is &lt;code&gt;10001011011010010001&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;is binary representation for integer &lt;code&gt;571025&lt;/code&gt; which doesn't give me a hint to any place on earth for now.&#xA;  Factorizing doesn't reveal anything either.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6-bit&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Base64 (with &quot;0&quot;-fill)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;100010 110110 100100 010000&#xA;     i      2      k      Q&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;  &#xA;  &lt;p&gt;Base 32&#xA;  =&gt; hdkh&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5-tupels: &lt;code&gt;10001 01101 10100 10001&lt;/code&gt; with the interesting fact, that 1 and 4 are equal.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Bacon&lt;/strong&gt; with b=1&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;=&gt; &lt;em&gt;SOWS&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Bacon&lt;/strong&gt; with a=1&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;=&gt; &lt;em&gt;PTMP&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Baudot&lt;/strong&gt; with 1 repr. on-state&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;=&gt; &lt;em&gt;SC-S&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Baudot&lt;/strong&gt; with 1 repr. off-state (you never know)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;=&gt; &lt;em&gt;HXFH&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;CCITT-2&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;=&gt; &lt;em&gt;ZFHZ&lt;/em&gt; could be an abbreviation for &quot;Züricher Fachhochschule, Zürich&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;MTK-2&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;=&gt; ZPSZ&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5-bit with &lt;code&gt;A=1&lt;/code&gt; encoding (= ASCII + 64)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;10001 01101 10100 10001&#xA;   17    13    22    17&#xA;    Q     M     V     Q&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;  &#xA;  &lt;p&gt;=&gt; nope&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4-bit with &lt;code&gt;A=1&lt;/code&gt; encoding (= ASCII + 64, even if a bit strange since we only get to 15)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;1000 1011 0110 1001 0001&#xA;   8   11    6    9    1&#xA;   H    K    F    I    A&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;  &#xA;  &lt;p&gt;=&gt; sounds like &lt;em&gt;Sofia&lt;/em&gt; or &lt;em&gt;Haifa&lt;/em&gt; but isn't identical.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;It could mean HKF International Airport. Unfortunately there is no IATA code for &lt;em&gt;HKF&lt;/em&gt;.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;There would be HKD which would mean question 11 should be false, but it's definitively true.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Same holds for HKG (Hongkong) (question 12 should be true but is definitely false).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Interesting fact about Hongkong: it's also referred to as HKIA, but what are we gonna do with the F?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;There's a seaport code with &lt;em&gt;HKF&lt;/em&gt; but that doesn't fit to &lt;em&gt;IA&lt;/em&gt; as &quot;International Airport&quot;.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;UN/LOCODE has &lt;em&gt;HK&lt;/em&gt; for Hongkong, but there's no subcode for &lt;em&gt;FIA&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;ASCII alone doesn't really fit to 20 bit binary. BCD, Aiken et. al. don't work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so shall it be: looks like there are different types of Bacon code or one can simply add a Caesar shift by -1 to the results above and you get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Bacon(2)&lt;/strong&gt; with a=1&#xA;&#xA; =&gt; &lt;em&gt;OSLO&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I would head to to the capital of Norway and see what's coming next.&lt;/p&gt;&#xA;" OwnerUserId="14144" LastEditorUserId="14144" LastEditDate="2015-07-21T10:54:23.610" LastActivityDate="2015-07-21T10:54:23.610" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="18199" PostTypeId="2" ParentId="18050" CreationDate="2015-07-21T00:19:26.767" Score="5" Body="&lt;h2&gt;Solution&lt;/h2&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Liber Abaci&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;Details&lt;/h2&gt;&#xA;&lt;p&gt;At first we notice that the picture is made up of 19 unique 8-bit colors. Separating the image into 19 different images composed of one color each, we see that 10 of the images represent a letter in 'Homage to LA' with somewhat additional random noise while the 9 others appear to contain completely random noise. On further inspection we notice that the image containing the letter 'A' contains no pixels for the first 32 rows of the image which we assume is a mask for the other images. However, after a more detailed analysis we see that the image containing the letter 'L' appears to contain random noise in rows 17 to 32 with limited information in rows 1 to 16 which we conclude is a more specific image mask. Now cropping all letter images to the first 16 rows we begin to see our message. Summing the pixels in each cropped image results in an ordered series of numbers.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;H = 192&lt;/li&gt;&#xA;&lt;li&gt;o = 152&lt;/li&gt;&#xA;&lt;li&gt;m = 117&lt;/li&gt;&#xA;&lt;li&gt;a = 110&lt;/li&gt;&#xA;&lt;li&gt;g = 96&lt;/li&gt;&#xA;&lt;li&gt;e = 77&lt;/li&gt;&#xA;&lt;li&gt;t = 54&lt;/li&gt;&#xA;&lt;li&gt;o = 33&lt;/li&gt;&#xA;&lt;li&gt;L = 18&lt;/li&gt;&#xA;&lt;li&gt;A = 0&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;While searching for various number series I ran across Leonardo of Pisa, known as Fibonacci, who wrote a book in 1202 called 'Liber Abaci' known as 'The Book of Calculation'. In this book he describes many mathematical concepts but most importantly he introduces the Hindu–Arabic numeral system composed of digits 0-9 with positional value that we use today. This entire puzzle is possible because of him.&lt;/p&gt;&#xA;" OwnerUserId="13788" LastEditorUserId="72725" LastEditDate="2020-12-20T10:32:47.513" LastActivityDate="2020-12-20T10:32:47.513" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="18200" PostTypeId="2" ParentId="18039" CreationDate="2015-07-21T04:42:20.933" Score="3" Body="&lt;p&gt;My guess is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Deadpool &amp;amp; Cable (they share dna)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My legend will be remembered,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; These seem to fit both: Both are character cards in the Marvel Legendary Card Game&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Too strong to be dismembered.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Deadpool regrows from his head being smashed in by the Hulk and reattaches limbs, including head, a few other times - this is either because of his healing factor or because Thanos cursed him with immortality. Cable is restored using bionic technology and other means.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No happiness at the parting,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again with Thanos, DeadPool fell in love with/was loved by death so Thanos cursed him with life so they could never meet. The Summer’s family tree is one of painful partings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What happened at the starting?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DeadPool repeatedly lies about his origins and is not even sure about them himself - not even sure he is Wade Wilson. Cable started out with no explanation of where he came from or who he was and this was only gradually expanded over the years.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All the lines or all we do,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; These seem to fit Deadpool: All Deadpool/Cable comics are famous for breaking the fourth wall&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mix God, religion, and the facts too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Anything and everything can come out of Deadpool’s mouth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Counting now, what are we at?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Deadpool has “died” several times in the comics (in fact he’s currently dead).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Profound, noises, or like a cat?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; On rare occasion Deadpool does something profound or heroic, the rest is noise - and his ideal mercenary is “the cat”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My land is one full of life,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; These seem to fit Cable. Cable’s Providence is meant to preserve life into the future (prevent the apocalyptic future he came from)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You know my cousins, little strife.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cable’s clone is Stryfe, his uncles: Havok and Vulcan&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Started with two, but grew to three,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cable and Nathan Summers were shown early with his clone Stryfe coming later&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But with some help, who can I be?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nathan Summers will someday grow into Cable but (due to the changes possible in time travel) he may not.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For Hints #3 and #6&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; “normal human beings” and &lt;em&gt;not an actor&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And #5&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Green is the color I’m guessing (could be red or blue though) and parts of Providence are found in the Verde Base.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11239" LastEditorUserId="11239" LastEditDate="2015-07-21T04:46:37.970" LastActivityDate="2015-07-21T04:46:37.970" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18201" PostTypeId="2" ParentId="17973" CreationDate="2015-07-21T05:19:43.173" Score="12" Body="&lt;p&gt;A Surpassing Phrase&lt;sup&gt;TM&lt;/sup&gt; is one where:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;For each word in the phrase, the alternate letter groups are ascending and descending.  Only one letter (either first or second letter only) in the word can violate this rule.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here are some examples:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;SUPERB = S (discard) + UEB (desc.) + PR (asc.)&lt;br&gt;&#xA;SUBWAY = SBA (desc.) + UWY (asc.)&lt;br&gt;&#xA;ONLINE = O (discard) + LN (asc.)  + NIE (desc.)&lt;br&gt;&#xA;FIENDS = FED (desc.) + INS (asc.)&lt;br&gt;&#xA;WRITE  = WIE (desc.)  + RT (asc.)&lt;br&gt;&#xA;RUNES = R (discard) + NS (asc.) + UE (desc.)&lt;br&gt;&#xA;PORKY = PRY (asc.) + OK (desc.)&lt;br&gt;&#xA;HOGS = HG (desc.) + OS (asc.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I checked all phrases in Surpassing list and they meet the condition above.   I checked a few phrases from Non-surpassing list and they do not meet the condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have zero pts, hopefully I can get the bonus!&lt;/p&gt;&#xA;" OwnerUserId="13851" LastActivityDate="2015-07-21T05:19:43.173" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18202" PostTypeId="2" ParentId="18171" CreationDate="2015-07-21T07:21:30.180" Score="7" Body="&lt;p&gt;Is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the fantastic four ( i am ignoring Codenewbie's hints here, cuz that's how i roll)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One cannot be seen,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sue Storm can turn invisible  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One is hard to kill.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Thing is almost indestructible &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One is a doctor,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Doctor Richards is a doctor. (even i was surprised) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One is better known as a bug&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Human torch. He's on fire and he can fly. He's a friggin firefly ( or maybe a reference to his son, the firefly)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the other hints, i can't find a match, but this answer is a guess, so if anyone can match them, they are welcome &lt;/p&gt;&#xA;" OwnerUserId="13080" LastEditorUserId="4450" LastEditDate="2015-07-21T07:32:41.090" LastActivityDate="2015-07-21T07:32:41.090" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18205" PostTypeId="2" ParentId="17973" CreationDate="2015-07-21T09:29:45.853" Score="19" Body="&lt;p&gt;A Surpassing Phrase™ is one where consecutive pairs of letters in each word get farther apart in the alphabet. For example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;su up pe er rb / su ub bw wa ay&#xA; 2  5 11 13 16 /  2 19 21 22 24&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;but not:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ex xc ce el ll le en nt / tr ra ai in&#xA;19 21  2  7  0  7  9  6 /  2 17  8  5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The name comes from each successive distance between letters &quot;surpassing&quot; the previous one.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-07-21T12:07:37.743" LastActivityDate="2015-07-21T12:07:37.743" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="18206" PostTypeId="1" AcceptedAnswerId="18208" CreationDate="2015-07-21T09:33:53.210" Score="2" ViewCount="981" Body="&lt;p&gt;Find the next number from the following&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;19 =&amp;gt; 730&#xA;&#xA;28 =&amp;gt; 520&#xA;&#xA;37 =&amp;gt; 370&#xA;&#xA;45 =&amp;gt; 189&#xA;&#xA;63 =&amp;gt; ?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="13582" LastEditorUserId="71" LastEditDate="2015-07-21T13:42:05.757" LastActivityDate="2015-07-21T14:32:27.300" Title="Find next number" Tags="|mathematics|pattern|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18208" PostTypeId="2" ParentId="18206" CreationDate="2015-07-21T10:18:17.793" Score="9" Body="&lt;p&gt;The number is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 243&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the calculation rule&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; consider the number on the left consisting of two digits $AB$. The number on the right is $(A \times A + B \times B - A \times B) \times (A + B)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9910" LastEditorUserId="13494" LastEditDate="2015-07-21T14:32:27.300" LastActivityDate="2015-07-21T14:32:27.300" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18210" PostTypeId="2" ParentId="18039" CreationDate="2015-07-21T11:29:59.620" Score="8" Body="&lt;p&gt;Is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Script (band)&lt;br /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The second paragraph, each line references one of their albums:&lt;br /&gt;&#xA; &lt;em&gt;All the lines or all we do,&lt;/em&gt; = &lt;strong&gt;The Script&lt;/strong&gt;&lt;br /&gt;&#xA; &lt;em&gt;Mix God, religion, and the facts too.&lt;/em&gt; = &lt;strong&gt;Science &amp;amp; Faith&lt;/strong&gt;&lt;br /&gt;&#xA; &lt;em&gt;Counting now, what are we at?&lt;/em&gt; = &lt;strong&gt;#3&lt;/strong&gt;&lt;br /&gt;&#xA; &lt;em&gt;Profound, noises, or like a cat?&lt;/em&gt; = &lt;strong&gt;No Sound Without Silence&lt;/strong&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And Also:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;They are Irish&lt;/strong&gt;&lt;br /&gt;&#xA; &lt;em&gt;My land is one full of life,&lt;/em&gt; &lt;strong&gt;Ireland&lt;/strong&gt;&lt;br /&gt;&#xA; &lt;em&gt;You know my cousins, little strife.&lt;/em&gt; &lt;strong&gt;Northern Ireland&lt;/strong&gt;&lt;br /&gt;&#xA; &lt;em&gt;Started with two, but grew to three,&lt;/em&gt; &lt;strong&gt;They started as a band of two&lt;/strong&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I'm guessing that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Each line of the first paragraph references one of their songs?&lt;/strong&gt;&lt;br /&gt;&#xA; &lt;em&gt;My legend will be remembered,&lt;/em&gt; = &lt;strong&gt;Hall of Fame&lt;/strong&gt;&lt;br /&gt;&#xA; &lt;em&gt;Too strong to be dismembered.&lt;/em&gt; = &lt;strong&gt;The Man Who Can't Be Moved&lt;/strong&gt;&lt;br /&gt;&#xA; &lt;em&gt;No happiness at the parting,&lt;/em&gt; = &lt;strong&gt;Six Degrees of Separation&lt;/strong&gt;&lt;br /&gt;&#xA; &lt;em&gt;What happened at the starting?&lt;/em&gt; &lt;strong&gt;For the First Time&lt;/strong&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4659" LastActivityDate="2015-07-21T11:29:59.620" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18211" PostTypeId="2" ParentId="18171" CreationDate="2015-07-21T12:59:57.733" Score="0" Body="&lt;p&gt;Is it?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Monster's Inc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One real, two imagined, one real who imagined&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mike, Sully, and Boo. Mike and Sully aren't real but Boo is&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One performer, two fighters, one sidekick. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mike is a performer. Randall and Sully are fighters&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One cannot be seen&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Randall has camouflage&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One is hard to kill&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; not sure but Sully seems hard to kill&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One is a doctor&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; not sure&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One is better known as a bug&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Roz is a slug&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One living, two ageless, one dead but immortal.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Abominal Snowman is living and his name is immortal. Not sure about the ageless part&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13925" LastEditorUserId="13925" LastEditDate="2015-07-21T13:03:23.270" LastActivityDate="2015-07-21T13:03:23.270" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18212" PostTypeId="1" AcceptedAnswerId="18213" CreationDate="2015-07-21T14:03:45.373" Score="1" ViewCount="304" Body="&lt;p&gt;This is part of the &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=title%3A%22Fill%20in%20the%20Dashes%22&quot;&gt;Fill in the Dashes&lt;/a&gt; series of puzzles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example : F_ in the B_ ⇒ Fill in the Blanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the words will be Juxtaposed as in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g. : C_ K_ S_ 10 ⇒ Clark Kent Superman (S) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number represents the position of letter in the solution to be extracted to be eventually used to find the link between the various clues.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;B_ L_ O_ 7&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B_ T_ 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B_ D_ 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D_ U_ 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E_ G_ G_ W_ 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;J_ + R_  1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D_ P_ 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M_ R_ 9&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R_ R_ 17&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A_ T_ 6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C_ 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G_ N_ O_ Y_ 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D_ T_ 6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I_ S_ 7&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E_ T_ 9&lt;/p&gt;&#xA;&#xA;&lt;p&gt;O_ S_ T_ W_ W_ 5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1 (Should be enough)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This time, the clues lead to a popular actor.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9892" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-07-21T14:16:18.223" Title="Fill in the Dashes : Part 4b of 10 - Subject Who?" Tags="|word|knowledge|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18213" PostTypeId="2" ParentId="18212" CreationDate="2015-07-21T14:13:32.253" Score="3" Body="&lt;p&gt;It's :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DiCaprio's movies&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;B_ L_ O_ 7&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Body of Lies (L)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;B_ T_ 3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Beach (E)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;B_ D_ 3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Blood Diamond (O)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;D_ U_ 4&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Django Unchained (N)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;E_ G_ G_ W_ 3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;What's Eating Gilbert Grape (A)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;J_ + R_ 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Romeo + Juliette (R)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;D_ P_ 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Don's Plum (D)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;M_ R_ 9&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Marvin's Room (O)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;R_ R_ 17&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Revolutionary Road (D)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A_ T_ 6&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Aviator (I)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;C_ 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Celebrity (C)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;G_ N_ O_ Y_ 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Gangs of New York (A)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;D_ T_ 6&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Departed (P)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I_ S_ 7&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Shutter Island (R)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;E_ T_ 9&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Total Eclipse (I)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;O_ S_ T_ W_ W_ 5&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Wolf of Wall Street (O)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The letter obviously gives:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Leonardo DiCaprio&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5359" LastActivityDate="2015-07-21T14:13:32.253" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18214" PostTypeId="2" ParentId="17196" CreationDate="2015-07-21T14:15:21.560" Score="4" Body="&lt;p&gt;Hint 2 is confusing me, but all the other clues seem to fit the following rule:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Substitute letters with their place in the alphabet (i.e. a-&gt;1, b-&gt;2, ... z-&gt;26) and sum. If the sum is odd, it can go on Aunt's list, if even it can go on Uncle's. BUT... Airplane (US spelling) gives a sum of 76, whereas the British Aeroplane has a sum of 87, so these seem the wrong way round by my rule.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12625" LastActivityDate="2015-07-21T14:15:21.560" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18215" PostTypeId="2" ParentId="18212" CreationDate="2015-07-21T14:16:18.223" Score="3" Body="&lt;p&gt;B_ L_ O_ 7&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Body of Lies, L&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;B_ T_ 3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Beach, E&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;B_ D_ 3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Blood Diamond, O&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;D_ U_ 4&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Django Unchained, N&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;E_ G_ G_ W_ 3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What's Eating Gilbert Grape, A&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;J_ + R_ 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Romeo + Juliet, R&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;D_ P_ 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Don's Plum, D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;M_ R_ 9&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Marvin's Room, O&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;R_ R_ 17&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Revolutionary Road, D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A_ T_ 6&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Aviator, I&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;C_ 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Celebrity, C&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;G_ N_ O_ Y_ 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gangs of New York, A&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;D_ T_ 6&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Departed, P&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I_ S_ 7&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Shutter Island, R&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;E_ T_ 9&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Total Eclipse, I&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;O_ S_ T_ W_ W_ 5&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Wolf of Wall Street, O&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Taking all of the letters gives us:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Leonardo DiCaprio!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13881" LastActivityDate="2015-07-21T14:16:18.223" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18216" PostTypeId="1" AcceptedAnswerId="18218" CreationDate="2015-07-21T14:20:34.017" Score="-1" ViewCount="665" Body="&lt;p&gt;I started out fresh dude, then I became more soft. But then something happened and I ended up being a senior. What was I before my seniority?&lt;/p&gt;&#xA;" OwnerDisplayName="user10001" LastActivityDate="2015-07-21T14:35:06.073" Title="Starting out fresh and then becoming a senior" Tags="|riddle|wordplay|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18217" PostTypeId="2" ParentId="18044" CreationDate="2015-07-21T14:23:28.640" Score="7" Body="&lt;p&gt;Who Am I? &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;American writer, Sidney Sheldon &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;His works...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A I_ M_ S_ T_ 2 &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(&lt;strong&gt;A&lt;/strong&gt; &lt;strong&gt;I&lt;/strong&gt;n &lt;strong&gt;M&lt;/strong&gt;irror &lt;strong&gt;S&lt;/strong&gt;tranger &lt;strong&gt;T&lt;/strong&gt;he / A Stranger In The Mirror)&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;C_ I_ T_ 1 &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(&lt;strong&gt;C&lt;/strong&gt;omes &lt;strong&gt;I&lt;/strong&gt;f &lt;strong&gt;T&lt;/strong&gt;omorrow / If Tomorrow Comes)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;D_ S_ S_ T_ 15 &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(&lt;strong&gt;D&lt;/strong&gt;own &lt;strong&gt;S&lt;/strong&gt;tars &lt;strong&gt;S&lt;/strong&gt;hine &lt;strong&gt;T&lt;/strong&gt;he / The Stars Shine Down) &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;F_ L_ N_ 1 &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(&lt;strong&gt;F&lt;/strong&gt;orever &lt;strong&gt;L&lt;/strong&gt;asts &lt;strong&gt;N&lt;/strong&gt;othing / Nothing Lasts Forever) &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;E_ P_ 1 &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(&lt;strong&gt;E&lt;/strong&gt;aster &lt;strong&gt;P&lt;/strong&gt;arade)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;D_ M_ T_ Y_ 11 &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(&lt;strong&gt;D&lt;/strong&gt;reams &lt;strong&gt;M&lt;/strong&gt;e &lt;strong&gt;T&lt;/strong&gt;ell &lt;strong&gt;Y&lt;/strong&gt;our / Tell Me Your Dreams) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;F_ I_ S_ T_ 4 &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(&lt;strong&gt;F&lt;/strong&gt;alling &lt;strong&gt;I&lt;/strong&gt;s &lt;strong&gt;S&lt;/strong&gt;ky &lt;strong&gt;T&lt;/strong&gt;he / The Sky Is Falling)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;H_ H_ T_ 1 &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(&lt;strong&gt;H&lt;/strong&gt;art &lt;strong&gt;H&lt;/strong&gt;art &lt;strong&gt;T&lt;/strong&gt;o / Hart to Hart)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;D_ I J_ O_ 4 &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(&lt;strong&gt;D&lt;/strong&gt;ream &lt;strong&gt;I&lt;/strong&gt; &lt;strong&gt;J&lt;/strong&gt;eannie &lt;strong&gt;O&lt;/strong&gt;f / I Dream of Jeannie)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;B_ L_ P_ T_ 8 &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(&lt;strong&gt;B&lt;/strong&gt;est &lt;strong&gt;L&lt;/strong&gt;aid &lt;strong&gt;P&lt;/strong&gt;lans &lt;strong&gt;T&lt;/strong&gt;he / The Best Laid Plans)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;C_ D_ T_ 4&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(&lt;strong&gt;C&lt;/strong&gt;onspiracy &lt;strong&gt;D&lt;/strong&gt;oomsday &lt;strong&gt;T&lt;/strong&gt;he / The Doomsday Conspiracy)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;M_ O_ O_ S_ T_ 4&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(&lt;strong&gt;M&lt;/strong&gt;idnight &lt;strong&gt;O&lt;/strong&gt;ther &lt;strong&gt;O&lt;/strong&gt;f &lt;strong&gt;S&lt;/strong&gt;ide &lt;strong&gt;T&lt;/strong&gt;he / The Other Side of Midnight)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;F_ N_ T_ 4&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(&lt;strong&gt;F&lt;/strong&gt;ace &lt;strong&gt;N&lt;/strong&gt;aked &lt;strong&gt;T&lt;/strong&gt;he / The Naked Face)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I don't know what the numbers stand for lol&lt;/p&gt;&#xA;" OwnerUserId="14157" LastActivityDate="2015-07-21T14:23:28.640" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18218" PostTypeId="2" ParentId="18216" CreationDate="2015-07-21T14:25:21.573" Score="8" Body="&lt;p&gt;A possible answer is that you were a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; junior&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Before you were a senior. This references&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; High school/college years&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A &quot;fresh dude&quot; is a freshman (first year), becoming &quot;more soft&quot; (thanks @OP for the &quot;more&quot;) references a sophomore (second year), and &quot;senior&quot; is the fourth year. Junior, then, fits in as the third year.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="11475" LastEditDate="2015-07-21T14:35:06.073" LastActivityDate="2015-07-21T14:35:06.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18219" PostTypeId="1" AcceptedAnswerId="18221" CreationDate="2015-07-21T14:50:38.827" Score="20" ViewCount="1921" Body="&lt;p&gt;Matty the math mouse meandered through the forest one day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alas! He was attacked by the famous unwise owl. Oft Matty had heard of the unwise owl - the bane of the forest for his kind. Fearing making a starring appearance in a third grader's biology lesson, Matty ran for a hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the unwise owl was keen and quick and rapidly cornered poor Matty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They were at an impasse. The owl could catch him only with great exertion, but Matt could not escape. And the powerful owl knew other game was about. So the unwise owl offered a challenge.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You say a word, then deterministically, I say a number, little mouse. &lt;br/&gt;&#xA;  If you can make me say 0 in 10 tries, I will let you go.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Matty felt sure of himself: he was not just another math mouse, he was the math mouse! So he attempted to earn his safety.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;Zero! (he proclaimed)&#xA;&gt; 6 (said the owl)&#xA;Six. (he said distinctly less assuredly)&#xA;&gt; 3 (came the bemused hoot)&#xA;Twenty (said the baffled Matt)&#xA;&gt; 2 (came the condescending reply)&#xA;BKLASFJBIOASf (said Matt trickerishly)&#xA;&gt; Sorry, I do not recognize non words. &#xA;&gt; But I won't charge you a guess.&#xA;---(worried Matt started looking at things around him)&#xA;Mouse? (he inquired)&#xA;&gt; 11 (came the abrupt hoot, truly baffling Matt)&#xA;Owl (he stated)&#xA;&gt; 4, for sure said the owl.&#xA;Oak &#xA;&gt; 5 (replied the owl)&#xA;Array&#xA;&gt; 2&#xA;Twelve (returning to his math comfort zone)&#xA;&gt; 4&#xA;---Matt suddenly had an idea!&#xA;Querulous&#xA;&gt; 21 (the owl said after some thought)&#xA;---AHA! Proclaimed Matt finally,&#xA;---(I know your tricks Mr. Owl, you are most&#xA;---unwise)&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Name a word Matt could have used to escape. (With Reasoning)&lt;/p&gt;&#xA;" OwnerUserId="13916" LastEditorUserId="10781" LastEditDate="2015-07-22T11:05:00.377" LastActivityDate="2015-07-22T11:05:00.377" Title="Math featuring The Unwise Owl" Tags="|pattern|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18220" PostTypeId="2" ParentId="18188" CreationDate="2015-07-21T15:27:34.297" Score="2" Body="&lt;p&gt;Partial answer, following the hint given:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;D=&quot;ROT180&quot;(A^-1(H^-1(C)))&lt;/code&gt; is very clear in its meaning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;un-Hex&quot; the ciphertext, &quot;un-Ascii&quot; it, then rotate it 180°.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As already stated in the comments, the first two actions give:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;(I=W/H | {N}:I'W'H E) &amp;lt;-- ({_Z}:X | {6_'0'8_}:X A)&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the last action results in this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/N3sKA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, I can definitely recognize something here, but the notation is throwing me off quite a bit. I'll try to figure out the missing pieces.&lt;/p&gt;&#xA;" OwnerUserId="5990" LastActivityDate="2015-07-21T15:27:34.297" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18221" PostTypeId="2" ParentId="18219" CreationDate="2015-07-21T15:31:36.273" Score="17" Body="&lt;p&gt;There are a a number of words that could be used:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Why&#xA; Try&#xA; Sky&#xA; ... Anything that lacks a true vowel.&#xA; The Owl is adding the total of vowels based on the following: &#xA; A=1, E=2, I=3, O=4, U=5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9896" LastActivityDate="2015-07-21T15:31:36.273" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18222" PostTypeId="2" ParentId="18219" CreationDate="2015-07-21T15:34:16.473" Score="7" Body="&lt;p&gt;The mouse could escape with the word &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; rhythm, shy or any word without aeiou&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The owl determines the number by adding the value of the vowels in the word, with a=1, e=2,i=3,o=4 and u=5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4591" LastActivityDate="2015-07-21T15:34:16.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18223" PostTypeId="2" ParentId="18171" CreationDate="2015-07-21T15:53:31.397" Score="3" Body="&lt;p&gt;As a guess, is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The states of matter? (Solid, Liquid, Gas, Plasma)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One cannot be seen&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Most gasses are invisible to the naked eye&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One is hard to kill&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Some solids are quite durable/indestructible &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One is a doctor&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Therapeutic liquids used in hydrotherapy act as a sort of therapy as well as ointments that alleviate pain and syrups as medicine or antibiotics can cure diseases&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One is better known as a bug&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A bit of a stretch but plasma wasn't known to be a state of matter so it could be considered an anomaly or bug to those of past generations&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;br&gt;Update to include the top and bottom lines:&lt;br&gt;&#xA;One real, two imagined, one real who imagined.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Solid(easily noticeable), Gas and Plasma(unseen and misunderstood in the past), Liquid(not quite sure) respectively&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One performer, two fighters, one sidekick.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Plasma(usually present as bright, shiny things that attracts people's attention), Solid and Liquid(fighting with Plasma for the spotlight), Gas (the most unnoticeable of the 4)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One living, two ageless, one dead but immortal.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One living: Solids are broken eventually by time as with other living things. &lt;br&gt;Two Ageless: Gasses and Plasma always look the same even overtime. &lt;br&gt;One dead but immortal: Liquids go through the water cycle and are &quot;rebirthed&quot; over and over again. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13489" LastEditorUserId="13489" LastEditDate="2015-07-21T20:04:58.787" LastActivityDate="2015-07-21T20:04:58.787" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18224" PostTypeId="1" AcceptedAnswerId="18232" CreationDate="2015-07-21T16:33:24.537" Score="9" ViewCount="617" Body="&lt;p&gt;They are three, not of a kind, who, in the wild, you cannot find.&lt;br&gt;&#xA;Against the odds, ever set to rival, you must find your center for survival.&lt;br&gt;&#xA;Your journey begins, you and one, but the quest is far from done.&lt;br&gt;&#xA;Countless more may win your heart, but these three, there since the start,&lt;br&gt;&#xA;They form you, as you form them, at least until you start again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't use a device to learn their names, that's all show and little game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To help, hinder, or enhance your cause, please see this additional clause:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;They dwell in the shadow of the Golden Son’s rise, which so often causes the twinkle in the skies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11452" LastEditorUserId="11452" LastEditDate="2015-07-21T19:12:45.907" LastActivityDate="2015-07-22T01:21:17.023" Title="The Three There from the Beginning" Tags="|riddle|rhyme|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18225" PostTypeId="2" ParentId="18224" CreationDate="2015-07-21T16:38:20.427" Score="5" Body="&lt;p&gt;I am going to guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Proton, Neutron, Electron &lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They are three, not of a kind, who, in the wild, you cannot find.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yup that is 3. Also in nature protons don't really exist by themselves, H+ is actually H3O+, electrons generally are attracted to ions, charge flows by passing electrons ion to ion through holes. Neutrons I am actually much less certain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Against the odds, ever set to rival, you must find your center for survival.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; electron and proton are opposites. But they form around a nucleus (center).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your journey begins, you and one, but the quest is far from done.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Makes me think of hyrdrogen atom, one e- one h+. The far from done part makes me think of fusion in hindsight, and eventually the atom grows beyond just the hydrogen and electron&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Countless more may win your heart, but these three, there since the start,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Probably about quarks and leptons and other small stuff, but protons neutrons and electrons have still been around from the start. Also when talking about subatomic particles the term countless is quite valid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They form you, as you form them, at least until you start again.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They form you. Not sure on the you forming them part. But the start again part could refer to matter being recycled example from star into human. (fun fact all elements higher than iron on the periodic table are from stars exploding!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Don't use a device to learn their names, that's all show and little game.&#xA;To help, hinder, or enhance your cause, please see this additional clause:&#xA;They dwell in the shadow of the Golden Sun’s rise, which so often causes the twinkle in the skies.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sun is fueled by fusion of particles but it happens in the core so we cant see it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13916" LastEditorUserId="13916" LastEditDate="2015-07-21T17:30:17.113" LastActivityDate="2015-07-21T17:30:17.113" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18226" PostTypeId="1" AcceptedAnswerId="18227" CreationDate="2015-07-21T17:18:49.277" Score="11" ViewCount="1327" Body="&lt;p&gt;The following is a chilling entry from someone's diary:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm not sure what day it is anymore.  My siblings and I had lost track of how long we'd been stuck in this dark and cramped prison cell.  It was hard to breathe and we had almost lost all hope.&#xA;  &lt;p&gt;&#xA;  Then something happened earlier tonight.  We heard voices and some people outside were obviously trying to break us out.  It sounded like something was blocking the door, but they removed that and then they pulled the door open pretty quickly.  They got all of us out of there, and they seemed happy to help us.  We were so grateful that these strangers showed up and saved us.  We thought we were finally free.&#xA;  &lt;p&gt;&#xA;  But then the strangers threw two of my brothers back into the cell for some reason.  We soon realized that they weren't rescuing us.  They were very evil people with sinister plans.  We couldn't have imagined the horror we were about to endure.  I wish I had died in the prison cell instead of having to go through what we went through.&#xA;  &lt;p&gt;&#xA;  It's really hard for me to write this, but I have to force myself to finish this.  People have to know the disgusting things that were done to us.  The first thing they did was gather up me and my siblings.  Then one of them proceeded to spank all of us.  It was very painful, and he did it with incredible efficiency.  He was obviously very skilled at this type of torture.  Then he did it to us a few more times.  It was horrible - I'd never heard of or experienced anything like that before.&#xA;  &lt;p&gt;&#xA;  When he finally stopped, he said something that made us think they were going to knife all of us.  I'd never been so frightened in my life.  But apparently it was just a scare tactic, because they didn't actually get a knife or anything.&#xA;  &lt;p&gt;&#xA;  Then he started throwing many of us across the floor, one by one.  The other torturers dragged our bodies around, and we got carpet burn on our faces.  Unfortunately, the worst was yet to come.&#xA;  &lt;p&gt;&#xA;  Those of us who had been thrown across the floor were then split up into groups, and a different tormentor abused each group of us.  My tormenter grabbed me and the others, and he started ogling and fondling us.  He was grinning when he was looking at me, and I felt nauseous.  It was truly awful - the sicko wouldn't keep his hands off of us.  My siblings in the other groups said they suffered the same kind of molesting from their tormentors.&#xA;  &lt;p&gt;&#xA;  Then my tormentor apparently got tired of a few of us, because he sent some of us back to the other guy, who sent over a few of my other siblings in return.  Then he started ogling and fondling them too.  Finally he yelled something and slammed all of our bodies onto the floor.  It hurt so much and I have no idea what we did to deserve that punishment.&#xA;  &lt;p&gt;&#xA;  This abuse was repeated several more times during the night.  Each time they'd gather all of us up, spank us, throw us and drag us across the floor, split us up into groups, and then molest us.  Finally when they were done, they shoved all of us back into the prison cell.  I also heard one of the tormentors say that he was looking forward to abusing us again tomorrow.  The thought of going through this pain and torture again is unbearable.  I am thinking about killing myself now.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions:&lt;/strong&gt;&#xA;&lt;p&gt;&#xA;Who is the author, and who are his/her siblings?&#xA;&lt;p&gt;&#xA;Who are the two siblings who were put back into the prison cell at the beginning?&#xA;&lt;p&gt;&#xA;What happened exactly during each of these forms of torture/abuse:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;spanking&lt;/li&gt;&#xA;&lt;li&gt;words that made them think they would be knifed&lt;/li&gt;&#xA;&lt;li&gt;throwing their bodies across the floor and then dragging them around&lt;/li&gt;&#xA;&lt;li&gt;ogling and fondling&lt;/li&gt;&#xA;&lt;li&gt;sending some of the siblings back and getting others in return&lt;/li&gt;&#xA;&lt;li&gt;yelling and slamming their bodies onto the floor&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2015-07-21T18:16:59.203" Title="Diary Entry - A Tale of Torture and Abuse" Tags="|riddle|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18227" PostTypeId="2" ParentId="18226" CreationDate="2015-07-21T17:27:19.260" Score="13" Body="&lt;p&gt;The author is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a playing card. The king of hearts in particular. (Thanks dperry for figuring out the specific card.) The king of hearts is also called the &quot;suicide king&quot; and the players also grin while looking at it, meaning it could be a high value card, like a king.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The two siblings who were put back into the prison cell at the beginning are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the two Joker cards, which are not relevant to the game of 5 Card Draw. (Thank you Bailey M for figuring out the game.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Spanking is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; shuffling.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The fear of knifing could be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the dealer asking another player to make the cut, an act in card playing where the other player divides the pile of cards into two by taking away and separating a certain number of the cards, with the dealer starting distribution from the pile left with him.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Throwing bodies around and dragging is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the distribution and collection of cards by the players. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ogling and fondling is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a player looking at his card and fidgeting with it while thinking and strategizing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sending some of the siblings back and getting others in return is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; players exchanging their cards to improve their hands. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yelling and slamming their bodies onto the floor is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Slamming the cards down when the loser realizes he's been beat (by a better hand) or bluffed out (by a weaker hand).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-07-21T18:16:59.203" LastActivityDate="2015-07-21T18:16:59.203" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="18228" PostTypeId="1" AcceptedAnswerId="18283" CreationDate="2015-07-21T17:48:59.673" Score="6" ViewCount="522" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Half of me is eight ups and a down &lt;br&gt;&#xA;  The other half is yogurt, mixed all around &lt;br&gt;&#xA;  I could be a rock star, or a Bachelor, although &lt;br&gt;&#xA;  The best I'll probably get is &quot;Sick reference, bro.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After nearly a day, I figure I need to add some hints.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There are exactly two words.&lt;/li&gt;&#xA;&lt;li&gt;It has nothing to do with food, but the two words individually can be associated with food (as per the first and second clues, naturally).&lt;/li&gt;&#xA;&lt;li&gt;Rock star is intentionally lower case, while Bachelor is intentionally upper case.&lt;/li&gt;&#xA;&lt;li&gt;The word 'reference' is a very important clue.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I ran this past a person who is a novice riddler but expert punster (and realistically, the two take a similar skill set) and she was able to solve it in about 10 minutes with minor coaching (i.e. the hints I included above). So it is definitely solvable!&lt;/p&gt;&#xA;" OwnerUserId="2812" LastEditorUserId="2812" LastEditDate="2015-07-22T14:59:20.950" LastActivityDate="2015-07-22T21:55:09.893" Title="That actually sounds kinda gross" Tags="|riddle|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="18229" PostTypeId="2" ParentId="18113" CreationDate="2015-07-21T17:50:21.793" Score="0" Body="&lt;p&gt;Lets consider the initial conditions: Anything.&#xA;As such we need to account for all possible relative orientations. The likely limiting case would be both cops are on the opposite pole from master cooper.&#xA;As such they are $r\pi$ away from him. To win the sly fellow needs to acquire $\pi r^2 area$ the smallest such area would be a circle of radius r. Unfortunately, we are on a sphere so non-euclidean geometry applies... drats. This path would take walking $2\pi r$ distance to create with an initial $r$ to get out to the circle (or maybe just starting a circle from the pole). (so maybe the actual worst case is they are both opposite the pole cooper is $r$ distance from. If he goes for the simple circle (which by the way circle is most efficient perimeter to area ratio) they could meet him when he is a little less than half way around. ($(\pi -1)r$ around precisely for the r away from pole initial condition). So this doesn't seem like Sly's optimal strategy as he is less than half way to having his area. &lt;br/&gt; This was just a thought experiment hope it helps someone out, I think I will update in the future but I imagine it will come down to circles and possibly angular velocity. &lt;/p&gt;&#xA;" OwnerUserId="13916" LastActivityDate="2015-07-21T17:50:21.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18230" PostTypeId="2" ParentId="18171" CreationDate="2015-07-21T18:20:50.617" Score="1" Body="&lt;p&gt;UPDATE: is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jude Law?&lt;br&gt;&#xA; sidekick|doctor: Dr. Watson in the Sherlock Holmes movies&lt;br&gt;&#xA; real|performer: Errol Flynn in The Aviator&lt;br&gt;&#xA; cannot be seen|hard to kill: Dickie in Talented Mr. Ripley  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What are these lines talking about?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Roles played by Martin Freeman&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One real, two imagined, one real who imagined.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Real: Lord Shaftsbury in &lt;a href=&quot;https://en.wikipedia.org/wiki/Charles_II:_The_Power_and_The_Passion&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Charles_II:_The_Power_and_The_Passion&lt;/a&gt;&lt;br&gt;&#xA; Imagined: Bilbo Baggins, Arthur Dent&lt;br&gt;&#xA; Real who imagined: Rembrandt in Nightwatchers  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One performer, two fighters, one sidekick.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Performer: the male stand-in in Love, Actually&#xA; Sidekick: Dr. Watson in Sherlock&lt;br&gt;&#xA; Not sure about the fighters  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One cannot be seen,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bilbo Baggins when he wears the ring&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One is hard to kill.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Arthur Dent  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One is a doctor  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dr. Watson in Sherlock  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One is better known as a bug.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure, maybe the guy he played in The Office?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One living, two ageless, one dead but immortal.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4385" LastEditorUserId="4385" LastEditDate="2015-07-21T18:54:41.663" LastActivityDate="2015-07-21T18:54:41.663" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18231" PostTypeId="1" AcceptedAnswerId="18241" CreationDate="2015-07-21T18:25:21.740" Score="18" ViewCount="1254" Body="&lt;p&gt;OK guys, I think this is my best puzzle yet. Hope you enjoy it, the solution is neat and simple.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A boy draws 2015 unit squares on a piece of paper, all oriented the same way.  The squares can overlap.  Then he colors the resulting picture in black and white chess-wise, such that any area belonging to an even number of squares is painted white and any area belonging to an odd number of squares is painted black.&lt;/p&gt;&#xA;&lt;p&gt;Prove that there is at least one square unit of black on the paper when he is finished.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14140" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2022-07-11T07:00:52.737" Title="Black and White" Tags="|mathematics|geometry|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18232" PostTypeId="2" ParentId="18224" CreationDate="2015-07-21T18:37:45.753" Score="10" Body="&lt;p&gt;Based partly on the fact that raisinghellyer's comment said the answer is &quot;juvenile,&quot; I'm going to guess that the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bulbasaur, Squirtle, and Charmander&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They are three, not of a kind, who, in the wild, you cannot find.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The three starters and their evolutions are not found in the wild at all&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Against the odds, ever set to rival, you must find your center for survival.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each of the types is super effective against one other and weak to the other, and your rival always takes the one super effective against yours. The center for survival is a reference to the PokeCenters that can be found in every town/city, where you can heal/revive your Pokemon for free.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your journey begins, you and one, but the quest is far from done.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You take one started at the very beginning of the game, and continue the game with them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Countless more may win your heart, but these three, there since the start,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, these are the starters. There are at least 148 others in the world, but you will always start with one of these three.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They form you, as you form them, at least until you start again.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your team is often based around them, though the next time you play through you may pick a different starter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Don't use a device to learn their names, that's all show and little game.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pretty sure this is a reference to the Pokedex, though I'm not sure if I would consider that a little game.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Possibly refers to HoHo (credit to Going Hamateur), but again, not that familiar to gens past first&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13339" LastEditorUserId="13339" LastEditDate="2015-07-22T01:21:17.023" LastActivityDate="2015-07-22T01:21:17.023" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18233" PostTypeId="2" ParentId="18231" CreationDate="2015-07-21T18:57:45.917" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In order to minimize the area shaded black, we want as many squares possible to perfectly overlap in even numbers. For example, take two squares. Separated, they are both single layers, and therefore provide $2$ $units^2$ of black paint. So we overlap these two. If we let the overlapping area be $A$, we then have $(1-A) + (1-A) = 2(1-A) units^2$ of black area. It is quite easy to see that to minimize the amount of area colored black we want $A=1$; the squares perfectly overlap. &lt;br&gt;&lt;br&gt; So then we &quot;pair&quot; each square up with one another in order to minimize the amount of area being colored black. In this case we have 2015 squares, forming 1007 pairs with one square left over. The 1007 pairs, no matter how they are distributed, will add 0 black area to the total. So then we are left with the problem of where to place the remaining square in order to minimize the area colored black. Notice that since we formed pairs, our remaining square will be placed in an area that is shared by 0, 2, 4, etc, squares already. Thus no matter where we place our last square, all of its area will be shared by an odd number of squares. So our minimal area being colored black is the area of our one square: $1$ $unit^2$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alternative proof:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For $N$ squares, we can clearly see that if there is no overlap, then we will need to fill in a maximal area of $N$ units. So in this setup, we pick one square to &quot;slide&quot; around. If while sliding it around we encounter another square(s), then two things can happen: the total black area remains the same, the total black area decreases. In order for it to remain the same, the square would need to enter a region that is already shared by an even number of squares. If an odd number of squares share the region, then it's like the situation in the solution above where $A$ (the shared area) is now increasing from 0. As $A$ increases, the amount being colored black will decrease. &lt;br&gt;&lt;br&gt; Thus we can continuously pick and slide squares around on the sheet. If we place the square into a region already shared by other squares, then we know that the total black area is either the same or less. Notice that by sliding squares, we can achieve any possible configuration of $N$ squares on the sheet of paper. Now notice the case where we slide all the squares perfectly onto each other. As $N$ is odd in this case, we know that the area of this configuration is $1$. More importantly, we know from our previous observations, that if we were to alter this configuration by sliding a square &lt;em&gt;away&lt;/em&gt; from other squares then the black area will either stay the same, or increase. Thus any configuration of squares will have a minimal area of $1$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13881" LastEditorUserId="13881" LastEditDate="2015-07-21T19:34:20.863" LastActivityDate="2015-07-21T19:34:20.863" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18236" PostTypeId="2" ParentId="18131" CreationDate="2015-07-21T20:04:05.167" Score="5" Body="&lt;p&gt;I am going to dissent and say he the majority opinion here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; While the series diverges, Sven dead. Very dead. EXTREMELY DEAD. To paint the last mile will take a minimum of 2^49 days. This is roughly One-Half Billion Million days. This is to a rough approximation One trillion years. The universe is less than 20 billion years old. So 50 Universe lives (so far) to paint. I mean sven sound like a strong fellow, but come on. The Sun is dead. Everyone he knows is dead. The paint he already did probably got destroyed when our sun exploded. 5/10 would not hire again (He did get the first half done mighty quick).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13916" LastActivityDate="2015-07-21T20:04:05.167" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18237" PostTypeId="2" ParentId="17306" CreationDate="2015-07-21T20:17:52.037" Score="7" Body="&lt;p&gt;I mostly used a computer to solve.  A random brute force process.  I added as much logic as I could determine from the hint and reduced which tiles could be placed in the surrounding cells (3x3 form the corners).  I started the process from cell A7 and followed the blue arrow up where I determined cell A6 only had two that could fit, thus reducing the number of random incorrect choices greatly.  From there the process followed the blue arrows placing a random tile after filtering out the tiles that logically didn't fit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/aKbon.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solution Text Look:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/DEYAn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="138" LastEditorUserId="138" LastEditDate="2015-07-21T21:41:04.357" LastActivityDate="2015-07-21T21:41:04.357" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18238" PostTypeId="2" ParentId="18171" CreationDate="2015-07-21T20:22:26.487" Score="3" Body="&lt;p&gt;Maybe this one&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The League of Extraordinary Gentlemen (at least 4 of them)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;one real:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rodney Skinner&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;two imagined&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mina Harker (vampire) and Dorian Gray&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;one real who imagined&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dr. Jekyll who imagined Mr. Hyde&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;one performer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dorian Gray acted as a model if I'm not wrong&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;two fighters&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mina Harker as a vampire as well as Rodney as a thief could be treated as a fighter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;one sidekick&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mr. Hyde (for Dr. Jekyll)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One cannot be seen&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rodney Skinner the invisible&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One is hard to kill&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dorian Gray being immortal as long as his portrait is hidden&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One is a doctor&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dr. Jekyll&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One is better known as a bug&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Vampire Mina Harker&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;one living&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rodney Skinner (just ordinary)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;two ageless&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dorian Gray who &quot;kept his age in the portrait&quot; and Mr. Hyde&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;one dead but immortal&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mina Harker as a vampire should be immortal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14144" LastActivityDate="2015-07-21T20:22:26.487" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18239" PostTypeId="1" AcceptedAnswerId="19879" CreationDate="2015-07-21T21:00:02.130" Score="16" ViewCount="1603" Body="&lt;p&gt;&lt;em&gt;If you're interested in starting the treasure hunt from the very beginning, check out the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16551/treasure-hunt-round-the-world&quot;&gt;prologue&lt;/a&gt;!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;After cracking the code, you catch a flight to Oslo, the capital of Norway. Being on a long flight makes you need to use the restroom badly -- but it's after 10 PM? Dare you risk it? Deciding that risking it is necessary, you relieve yourself, and (luckily) you aren't caught.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the way out of the restroom, you notice a telltale man in a suit and sunglasses holding a plaque with your name on it. You walk up to him, and he silently hands you a letter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tearing into the letter, you read:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dear &lt;strong&gt;&lt;em&gt;[insert username here],&lt;/em&gt;&lt;/strong&gt; &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Welcome to Norway! Feel free to spend a couple of hours here enjoying Scandinavia. After all, it's positively gorgeous. You have 48 hours from midnight tonight to reach your next pit stop, at which point anyone not there will be cut from the Treasure Hunt. My suggestion is to spend the day here and sleep on the plane - once you land, it's just a short drive to the pit stop.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;You do, of course, have to figure out where you're flying to, though. I've left a puzzle on the back of this note. Solve it when you're ready!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Yours truly,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Bailey M&lt;br&gt;&#xA;  Puzzlemaster for the Treasure Hunt 'Round the World&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You shrug. Better to solve the puzzle now and then spend any remaining time enjoying the city, right? Flipping over the letter, you read:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;OLD FLAGS  &lt;/li&gt;&#xA;  &lt;li&gt;FISHY DISH  &lt;/li&gt;&#xA;  &lt;li&gt;PUERTO RICO  &lt;/li&gt;&#xA;  &lt;li&gt;ACQUITTED IDIOT  &lt;/li&gt;&#xA;  &lt;li&gt;61B2068696E74  &lt;/li&gt;&#xA;  &lt;li&gt;MANLY PINK  &lt;/li&gt;&#xA;  &lt;li&gt;BANANA CABANA  &lt;/li&gt;&#xA;  &lt;li&gt;STREET SMART&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;DETERMINE THE ODD ONES OUT TO DISCOVER THE WAY&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;hr&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;P.S.&lt;/strong&gt; No, 61B2068696E74 isn't &quot;the odd one out because it's hex and the rest are strings&quot;. Think of it as a hint.&lt;br&gt;&#xA;  &lt;strong&gt;P.P.S.&lt;/strong&gt; Prepare yourself for a long flight -- you're finally leaving Europe!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Where is your next destination?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint/Clarification:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The only conversion you need to do is the first hex-to-string. None of the other strings need to be converted for this puzzle to be fully solved.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint/Clarification #2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I've edited the tags. You'll note that &lt;a href=&quot;/questions/tagged/cryptograms&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;cryptograms&amp;#39;&quot; rel=&quot;tag&quot;&gt;cryptograms&lt;/a&gt; is not among them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint/Clarification #3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm not spoiler tagging this one since it's basically a general-knowledge sort of thing. It should be noted that the Puzzlemaster, Bailey M, is an &lt;strong&gt;American&lt;/strong&gt;, and writes his puzzles as such.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint/Clarification #4:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Puzzlemaster typed this puzzle up on a good old-fashioned &lt;strong&gt;American keyboard&lt;/strong&gt;. Apologies to anyone solving the puzzle who doesn't have one on hand.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The story continues in &lt;a href=&quot;https://puzzling.stackexchange.com/questions/19920/treasure-hunt-round-the-world-interlude-2&quot;&gt;Treasure hunt &amp;#39;round the world! (Interlude 2)&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2015-08-11T19:57:21.243" Title="Treasure hunt 'round the world! (clue 4)" Tags="|pattern|enigmatic-puzzle|letters|" AnswerCount="4" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="18240" PostTypeId="2" ParentId="8604" CreationDate="2015-07-21T22:17:47.290" Score="2" Body="&lt;p&gt;For practicality, you would just take a 2n die and treat opposing sides as equal on the lower half.  For example, a 6 sided die could be used as a three sided die with two of each possibility, 1 and 6 are 1, 2 and 5 are 2, and 3 and 4 are 3.  Die faces are properly labeled so that opposing sides total to the same amount (7 on a 6 sided die) and if you're using a D20 (opposing sides total 21) to roll for a D10 since you lost it under the couch or the roommate's cat ate it, you would use the rolled number if 10 or less or subtract it from 21 if 11 or more.  Or you could alternately just relabel the sides before using it.&lt;/p&gt;&#xA;" OwnerUserId="14217" LastActivityDate="2015-07-21T22:17:47.290" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18241" PostTypeId="2" ParentId="18231" CreationDate="2015-07-21T22:46:17.203" Score="12" Body="&lt;p&gt;Let's think of the problem as taking XOR's (symmetric differences) of the square areas, with black as 1 and white as 0. The black area is the XOR of all the squares. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine doing this all on graph paper whose cells are sized and oriented like the squares. Take the resulting figure, cut up the occupied cells, and stack them on top of each other without rotating. Now, take the XOR of that stack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that any square on the graph paper gets cut up and rearranged to make exactly the full unit cell. So, the XOR of the stack is the XOR of 2015 fully black squares, which is all black, since 2015 is odd. Then, since each black point in the XOR requires at least one black point the contribute, the total area of black in the stack must be at least 1.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Needlessly formally:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are interested in the group $G$ of (well-behaved) finite-area subsets of the plane under symmetric difference (XOR). We quotient out the plane by integer translations to get $(\mathbb R \times \mathbb R) / (\mathbb Z \times \mathbb Z)$, which induces a homomorphism $\phi$ from $G$ to the corresponding group $G'$ on the unit square. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $2015$ single-square elements $x_1, ..., x_{2015}$ of $G$ and are interested in their sum $x$. As single-square elements, they all map to the same full-square element $\phi(x_i) = s$ in $G'$, so $\phi(x) = s + \cdots + s$ summed $2015$ times. Since $G'$ has order $2$, this is simply $s$. But the homomorphism $\phi$ cannot increase the size of the subset, so &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{Area}(x) \geq  \mathrm{Area}(\phi(x)) = \mathrm{Area}(s) =1 $$&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-07-21T23:01:53.117" LastActivityDate="2015-07-21T23:01:53.117" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="18242" PostTypeId="2" ParentId="18228" CreationDate="2015-07-21T23:07:46.033" Score="7" Body="&lt;p&gt;Are you a Yogurt Soju Cocktail?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Half of me is eight ups and a down &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 7-up&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The other half is yogurt, mixed all around&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; yogurt is yogurt~&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I could be a rock star, or a Bachelor, although&#xA;The best I'll probably get is &quot;Sick reference, bro.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Soju is a alchoholic beverage. It's one of the most popular in korea, and the variety in alchohol content (16% to 45%) could represent the rock star and bachelor (Intensity reference). Sick reference could be because of the reference I just made, or because if you drink too much, you throw up and seem &quot;sick&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ingredients inside a soju cocktail:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Soju, Yogurt, Lemon/Lime Soda (Sprite or 7up are popular)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8999" LastActivityDate="2015-07-21T23:07:46.033" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18243" PostTypeId="1" AcceptedAnswerId="18244" CreationDate="2015-07-21T23:50:30.977" Score="37" ViewCount="3884" Body="&lt;p&gt;You are serving a cake to $10$ children. The the cake is shaped like a box, whose top face is square. The top and sides are covered with a thin layer of frosting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\qquad\qquad\qquad\qquad$&lt;img src=&quot;https://i.stack.imgur.com/CiBZm.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every child demands a fair share, so you must cut the cake into $10$ pieces where every piece has the same volume of cake &lt;em&gt;and&lt;/em&gt; surface area of frosting. How do you do this?&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="8874" LastEditDate="2016-03-23T15:38:09.820" LastActivityDate="2016-03-23T15:38:09.820" Title="Fairly Sharing a Frosted Cake" Tags="|geometry|strategy|dissection|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18244" PostTypeId="2" ParentId="18243" CreationDate="2015-07-22T00:24:36.113" Score="38" Body="&lt;p&gt;We will make all of our cuts vertical, so we can treat this as a square which we need to divide into $10$ pieces with equal slices of the area and the perimeter. This is reasonably easy: Choose $10$ points dividing the perimeter into $10$ equal lengths, and then make cuts inwards from each point to the center. The area of a triangular slice is half the altitude of the triangle (distance of the outer line to the center) times its base (the length on the perimeter). Since altitude relative to the center is constant along all four segments, the area of each slice is proportional to the amount of the square's perimeter it contains (even if it contains multiple edges) - so given that the perimeter was divided evenly, so must be the area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hskip 1in$&lt;img src=&quot;https://i.stack.imgur.com/Ahmuv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4765" LastActivityDate="2015-07-22T00:24:36.113" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18245" PostTypeId="2" ParentId="18243" CreationDate="2015-07-22T00:32:40.877" Score="24" Body="&lt;p&gt;One of the possible solutions is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Facing the square, mark the perimeter of the square into 10 equal pieces and cut through center of the square.  So each piece is 0.4L long along the square.  Please note that the corner pieces will have two triangular pieces but the calculations below hold.&lt;br&gt;&#xA;Mock up approx picture shown (i didnt have tooling to cut the perimeter into bits).  &lt;img src=&quot;https://i.stack.imgur.com/jutPm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Calculations  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Lets say square length is L and Height H&lt;br&gt;&#xA;  Frosting on top    = 0.5 * 0.4L * 0.5L&lt;br&gt;&#xA;  Frosting on bottom = same as top&lt;br&gt;&#xA;  Frosting on side   = 0.4L * H&lt;br&gt;&#xA;  Total frosting area = 0.2L&lt;sup&gt;2&lt;/sup&gt; + 0.4LH&lt;br&gt;&#xA;  Cake volume = top area * height = 0.5 * 0.4L * 0.5L * H = 0.1L&lt;sup&gt;2&lt;/sup&gt;H&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Validation  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Total vol = 1L&lt;sup&gt;2&lt;/sup&gt;H is 10 times above&lt;br&gt;&#xA;  Total surface area of frosting = 2L&lt;sup&gt;2&lt;/sup&gt;+4LH which is 10 times what we found above&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13851" LastActivityDate="2015-07-22T00:32:40.877" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18246" PostTypeId="2" ParentId="18188" CreationDate="2015-07-22T01:44:37.667" Score="4" Body="&lt;p&gt;I interpret this answer to be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  True&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The statement appears to say $\forall_{x\in\{-8,0,-9\}}x\in\mathbb{Z}^{-}\implies\exists_{H,M,I\in\mathbb{N}}H/M=I$; that is, &quot;(for all $X$ in the set $\{-8,0,-9\}$, $X$ is in the set of negative integers) implies (there exists an $H$, $M$, and $I$ in the natural (counting) numbers such that $H/M = I$). The second statement has no $X$ so we can reduce the statement to: there exists an $H$, $M$, and $I$ in the natural numbers such that $H/M=I$.  $6$, $3$, and $2$ are all natural numbers and $6/3=2$; thus the statement is true.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not sure I've interpreted all the symbols correctly,  but I feel confident it's at least a partial answer. &lt;/p&gt;&#xA;" OwnerUserId="14120" LastEditorUserId="9718" LastEditDate="2015-07-22T02:58:45.690" LastActivityDate="2015-07-22T02:58:45.690" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18247" PostTypeId="2" ParentId="18231" CreationDate="2015-07-22T02:08:53.130" Score="0" Body="&lt;p&gt;I'm sure I'm misunderstanding something.  But here we go:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XPbk4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/XPbk4.png&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overlap 2013 of them on top of each other and have the other two slightly offset.&lt;/p&gt;&#xA;" OwnerUserId="14222" LastActivityDate="2015-07-22T02:08:53.130" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18248" PostTypeId="1" AcceptedAnswerId="18255" CreationDate="2015-07-22T03:03:35.633" Score="11" ViewCount="621" Body="&lt;p&gt;A group of 10 students in a summer learning camp (fun times!) is asked to complete a brief survey to help them determine which of their new classmates they're most compatible with. The survey lists 10 subjects (math, science, English, art, history, band, gym, biology, shop, and home ec) and asks respondents to mark each subject as &quot;like&quot; or &quot;dislike&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once all 10 students have completed the survey, the teacher calculates an &quot;affinity&quot; score for each pair of students. The affinity between two students is the total number of subjects &lt;em&gt;both&lt;/em&gt; students marked &quot;like&quot;. Hence if Bill liked only &quot;science&quot;, &quot;art&quot;, &quot;gym&quot;; Jane liked only &quot;science&quot;, &quot;gym&quot;, and &quot;history&quot;; and Tara liked only &quot;science&quot;, &quot;history&quot;, &quot;band&quot;, and &quot;English&quot;, the affinity between Bill and Jane would be 2, the affinity between Bill and Tara would be 1, and the affinity between Jane and Tara would be 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The teacher writes the affinity scores for all 45 pairs of students on the board, then addresses the class.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm pleased to see that every student likes at least one subject, and I've noticed that no two students like exactly the same set of subjects. What I find bizarre, though, is that all of your affinity scores are even numbers. I've never seen that happen before. I'm wondering if it might be because the total number of 'likes' over all your surveys is 50 out of 100 possible; exactly half.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hearing this, a particularly math-savvy student, Keenan Byonde, puts up his hand and explains, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Actually, it's because 50 is part of the arithmetic sequence $10 + 8n ; 0 \le n \le 10$. It's a simple matter to come up with 11 sets of ten responses to this survey such that each set has the properties:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;each respondent likes at least one subject;&lt;/li&gt;&#xA;  &lt;li&gt;no two respondents like exactly the same subjects;&lt;/li&gt;&#xA;  &lt;li&gt;all 45 affinity scores are even; and&lt;/li&gt;&#xA;  &lt;li&gt;the total number of 'liked' subjects is a number in the arithmetic sequence.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Intrigued, the teacher asks Byonde to come up to the board, give an example set of ten survey responses with 50 total 'likes' and the stated properties, and explain how to (easily) contrive a set of survey responses with $10 + 8n$ total 'likes' and the stated properties for $0 \leq n \leq 10$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What example and what simple procedure should Byonde write on the board to prove he's correct?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="10399" LastEditDate="2015-07-22T17:00:24.203" LastActivityDate="2015-07-22T17:00:24.203" Title="Two Affinity, and Byonde!" Tags="|mathematics|word-problem|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18250" PostTypeId="1" AcceptedAnswerId="18253" CreationDate="2015-07-22T07:54:02.013" Score="24" ViewCount="1943" Body="&lt;p&gt;Follow this series: &#xA;Can you guess the last two results?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;OOOOO = 0&#xA;LTSSJI = 4&#xA;IIIIIIIIII = 0&#xA;I90I90O90 = 2&#xA;SSSSJ = 5&#xA;T180T180T180S270 = ¿?&#xA;LZZZZ = ¿?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;EDITED: Thanks to Deusovi who solved it and also found an error: changed second line to LTSSJI = 4&lt;/p&gt;&#xA;" OwnerUserId="14226" LastEditorUserId="11876" LastEditDate="2015-07-22T12:03:37.590" LastActivityDate="2019-07-15T07:11:52.320" Title="Crack the lines" Tags="|pattern|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18251" PostTypeId="2" ParentId="18171" CreationDate="2015-07-22T08:59:56.327" Score="7" Body="&lt;p&gt;After shamelessly studying the clues and other answers, I believe the 'unknown' in the title is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; John Doe, used in the US for a person whose identity is unknown [credit to Vioz for this]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The lines are talking about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Famous Johns&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One real&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; John Cena [a WWE wrestler]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;two imagined&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; John Watson [from the Sherlock Holmes stories - credit to dennisdeems] and John McClane [from the Die Hard films - thank you CodeNewbie]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;one real who imagined.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; John Lennon [a reference to his song Imagine]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One performer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; John Lennon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;two fighters&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; John Cena and John McClane&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;one sidekick.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; John Watson was Sherlock Holmes' sidekick&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One cannot be seen&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; John Cena apparently likes to wave his hand in front of his face to signal &quot;you can't see me&quot; and in fact has an album in the hardcore hip cop genre with that title &lt;a href=&quot;https://en.wikipedia.org/wiki/You_Can%27t_See_Me&quot;&gt;(see here)&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One is hard to kill.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; John McClane - 'Die Hard' - in fact so hard to kill there have been numerous sequels [my original answer had John Rambo instead of McClane, which works very well, but McClane fits in better with this part]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One is a doctor,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; John Watson is a doctor and is usually known as Dr. Watson&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One is better known as a bug.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; John Lennon is a Beatle! [Sounds like beetle]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One living&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; John Cena is alive&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;two ageless&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; John Watson and John McClane are ageless as they are characters from fiction rather than living people&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;one dead but immortal&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; John Lennon is dead but his music will live forever&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12277" LastEditorUserId="12277" LastEditDate="2015-07-22T09:13:48.517" LastActivityDate="2015-07-22T09:13:48.517" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18252" PostTypeId="2" ParentId="18248" CreationDate="2015-07-22T09:05:08.030" Score="2" Body="&lt;p&gt;Posting this as I don't have time to finish. Either I'll come back to finish it later or someone else will take it and run with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Label students $s_1...s_{10}$, and classes $c_1...c_{10}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=10$, assign $s_i$ to $c_i$. Each student likes one class, each affinity is zero, no students like the same set of classes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To extend to $n=18$, assign as before except $s_1...s_4$ are each assigned one of the four combinations of three elements from $c_1...c_4$. These students will all have a mutual affinity of 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This suggests the general method: to add more likes, take a set of $m$ students and classes such that $m$ is even and have each student like one of the $m-1$ distinct combinations of $m-1$ classes. If $m=4,6$ we may add a second group of 4.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;N       Combination&#xA;10      None&#xA;18      4&#xA;26      4,4&#xA;34      6&#xA;42      6,4&#xA;58      8&#xA;90      10&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;At this point we throw our chalk at the blackboard in disgust.&lt;/p&gt;&#xA;" OwnerUserId="3987" LastActivityDate="2015-07-22T09:05:08.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18253" PostTypeId="2" ParentId="18250" CreationDate="2015-07-22T09:38:13.987" Score="37" Body="&lt;p&gt;The number represents how many...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;other Tetris pieces you'd need to clear the board. Each string represents tetrominoes being dropped onto a standard 10-wide board; the numbers represent rotations by 90, 180, or 270 degrees. Unfortunately, I seem to have stumbled across a better solution for two of them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's a diagram in Google Sheets:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/ANIa7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ANIa7.png&quot; alt=&quot;gsheet&quot;&gt;&lt;/a&gt;&#xA;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the solution is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;6 and 5.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="51131" LastEditDate="2019-07-15T07:11:52.320" LastActivityDate="2019-07-15T07:11:52.320" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="18255" PostTypeId="2" ParentId="18248" CreationDate="2015-07-22T10:13:44.400" Score="7" Body="&lt;p&gt;Divide the 10 students into 5 pairs. Assign each student a different &lt;em&gt;favorite subject&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If each student only likes their favorite subject, then there are a total of 10 &quot;likes&quot; and no affinities. To add 8 &quot;likes&quot;, choose two pairs to &lt;em&gt;meet&lt;/em&gt; each other. When two pairs meet, each student &quot;like&quot;s the favorite subjects of the other pair (four students each adding two likes). There are ten pairs of pairs. When they have all met, there are 90 &quot;likes&quot;: every student likes every subject except their partner's favorite, and all affinity scores are 8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diagram: black squares are initial likes, more likes can be added in groups of 8 squares with the same number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PyL7u.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PyL7u.png&quot; alt=&quot;Diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Each student always likes their own favorite subject and never likes their partner's favorite subject. Any other student who likes the first student's favorite subject must also like the partner's favorite subject, so no two students ever like exactly the same subjects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The affinity between two students in the same pair gains 2 for every pair they meet.. The affinity between two students in different pairs gains 2 for every pair they have both met, plus 2 if they have met each other (from them liking each other's favorite subjects). Because all affinities only go up in increments of 2, they are all even.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-07-22T10:34:42.390" LastActivityDate="2015-07-22T10:34:42.390" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18256" PostTypeId="2" ParentId="18113" CreationDate="2015-07-22T14:08:30.757" Score="6" Body="&lt;p&gt;Assume the police are at points P1 and P2 and Sly Cooper are at point SC. Draw a circle through the three starting points. This circle will lie on the sphere. Draw the three lines that pass through each point and is perpendicular to the circle. The lines will intersect at two points: call them the poles. Call all lines parallel to the circle horizontal, and all lines that go from pole to pole vertical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SC is always between the lines passing through P1 and P2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The police need to make sure that SC stays between the lines passing through P1 and P2 while narrowing down the distance between the two lines.&#xA;The strategy goes as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If SC moves along his line (moves vertically), the policemen move along their respective lines the same distance.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/L5Nis.png&quot; alt=&quot;Movement 1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;If SC does not move along his line (moves horizontally partially), the policeman he is moving away from will move in exactly the same way and the policeman he is moving towards will move in the mirror image.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/opCnm.png&quot; alt=&quot;Movement 2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on this strategy, the police will always be on the same horizontal line as SC and will reduce the possible horizontal movement by one unit for each horizontal unit SC moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To claim the largest amount of land, SC must keep his horizontal movement to a minimum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The worst case would be when the police are right next to each other and are on the diametric opposite of SC on the original circle.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;SC moves up and down to draw his vertical line. Then he moves to a pole&lt;sup&gt;1&lt;/sup&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KWjvo.png&quot; alt=&quot;Movement&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;He walks a quarter of the way across the planet. At this time, P1 is half of the planet behind of him, and P2 is right in front of him.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8srqZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8srqZ.png&quot; alt=&quot;Movement&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;SC moves to the other pole&lt;sup&gt;1&lt;/sup&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JnWvm.png&quot; alt=&quot;Movement&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;He then walks back to his starting point&lt;sup&gt;1&lt;/sup&gt;. At this time, P1 is right in front of him and P2 is right behind him, so he cannot claim any more land.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So in the worst case, SC claims a quarter of the planet.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; Well, an insignificant distance from it. If he moves exactly onto that spot, the police would catch him.&lt;/p&gt;&#xA;" OwnerUserId="12109" LastEditorUserId="12109" LastEditDate="2015-07-22T14:42:13.620" LastActivityDate="2015-07-22T14:42:13.620" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18260" PostTypeId="1" AcceptedAnswerId="18265" CreationDate="2015-07-22T15:37:21.043" Score="10" ViewCount="1149" Body="&lt;p&gt;Matty mouse is freshly recovered from a &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18219/math-featuring-the-unwise-owl?rq=1&quot;&gt;traumatic owl encounter&lt;/a&gt; and is looking to relax.&lt;/p&gt;&#xA;&lt;p&gt;When he gets back to his home, he heads down to the local hole-in-the-wall.&#xA;But unfortunately for Matty, there is new management. The establishment has been rebranded the Cool™ Club, and they have a bouncer at the door.&lt;/p&gt;&#xA;&lt;p&gt;As he walks up, he sees a Cool™ Kid walk up to the Bouncer Who Is Quite Good At Intuiting Homophones™ and says, &amp;quot;Awww man, just let me in; you know me—I just forget passwords is all.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;The Bouncer says, &amp;quot;You know the rules.&amp;quot;&lt;br /&gt;&#xA;The Cool™ Kid whines pleadingly.&lt;br /&gt;&#xA;The Bouncer says, &amp;quot;You get seven attempts. I'll respond to each one with a number.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;The Cool™ Kid moans pathetically then says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Bare.&lt;/strong&gt;&lt;/em&gt;&lt;br /&gt;&#xA;&amp;gt; &lt;strong&gt;5&lt;/strong&gt;, &lt;em&gt;is the bouncer's response.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Baaa&lt;/strong&gt;&lt;/em&gt;! &lt;em&gt;the Cool™ Kid bleats.&lt;/em&gt;&lt;br /&gt;&#xA;&amp;gt; &lt;strong&gt;4&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Tear&lt;/strong&gt;&lt;/em&gt;?&lt;br /&gt;&#xA;&amp;gt; &lt;strong&gt;2&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Ears&lt;/strong&gt;&lt;/em&gt;.&lt;br /&gt;&#xA;&amp;gt; &lt;strong&gt;2&lt;/strong&gt;, &lt;em&gt;comes the abrupt response.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Seta&lt;/strong&gt;&lt;/em&gt;.&lt;br /&gt;&#xA;&amp;gt; &lt;strong&gt;2&lt;/strong&gt;, &lt;em&gt;comes the same intimidating response.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Star&lt;/strong&gt;&lt;/em&gt;&lt;br /&gt;&#xA;&amp;gt; &lt;strong&gt;3&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Rest&lt;/strong&gt;&lt;/em&gt;.&lt;br /&gt;&#xA;&amp;gt; &lt;strong&gt;9&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Excited, the Cool™ Kid squeals!&lt;br /&gt;&#xA;But the bouncer says, &amp;quot;Okay, that is seven attempts&amp;quot;.&lt;br /&gt;&#xA;The Cool™ Kid's jaw drops.&lt;br /&gt;&#xA;&amp;quot;ONE MORE, PLEASE!&amp;quot; he grovels Cool™ly.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;Password or leave.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;Drooping his head, the Cool™ Kid crawls away.&lt;/p&gt;&#xA;&lt;p&gt;Matty gives the Cool™ Kid a confused look, walks up to the bouncer, says the password and Struts In Suavely™ to enjoy the hard-earned Cheeses of his labour.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;What is the password, and how did Matty figure it out?&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="13916" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-22T16:56:00.373" Title="Entering the Cool™ Club" Tags="|pattern|" AnswerCount="1" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="18261" PostTypeId="1" AcceptedAnswerId="18264" CreationDate="2015-07-22T15:59:29.990" Score="20" ViewCount="4642" Body="&lt;p&gt;I came up with this simple riddle that I think has a rather fun solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Unharmed, you need it every day;&lt;br&gt;&#xA;  Decapitated, you see it and you say.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Short, sweet, and to the point. Good luck!&lt;/p&gt;&#xA;" OwnerUserId="11475" LastEditorDisplayName="user20" LastEditDate="2015-07-22T16:22:36.197" LastActivityDate="2015-07-26T04:11:34.313" Title="Unharmed, you need it every day" Tags="|riddle|" AnswerCount="11" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18262" PostTypeId="2" ParentId="18261" CreationDate="2015-07-22T16:21:12.537" Score="4" Body="&lt;p&gt;A bit of a stretch, I feel, but is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Brain?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unharmed, we need it every day, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Undoubtedly you need it every single day, and in its unharmed state, naturally.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Decapitated, you see it and you say.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When I &quot;decapitate&quot; 'Brain', I remove the first letter (the head of the word?) and get 'rain'. And when it unexpectedly begins to rain, we often exclaim 'Rain!'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-07-22T17:22:57.233" LastActivityDate="2015-07-22T17:22:57.233" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18263" PostTypeId="2" ParentId="18261" CreationDate="2015-07-22T16:25:22.773" Score="4" Body="&lt;p&gt;How about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Food&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unharmed, you need it every day;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gotta eat..&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Decapitated, you see it and you say.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you have seen food, you might have &quot;ood&quot; while others might have &quot;aahd&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2015-07-22T16:25:22.773" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18264" PostTypeId="2" ParentId="18261" CreationDate="2015-07-22T16:26:37.367" Score="48" Body="&lt;p&gt;UPDATED.  Oh for Pete's sake. &lt;em&gt;forehead smack&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BREAD AND BUTTER&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unharmed, we need it every day, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We need food every day.  It needn't literally be bread; it can mean food in general as &quot;our daily bread&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Decapitated, you see it and you say.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Remove the first letters and it becomes READ AND UTTER.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4385" LastEditorUserId="4385" LastEditDate="2015-07-23T02:07:30.520" LastActivityDate="2015-07-23T02:07:30.520" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="18265" PostTypeId="2" ParentId="18260" CreationDate="2015-07-22T16:43:49.477" Score="11" Body="&lt;p&gt;Any chance the answer is not a word, just a sequence of letters?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BRST&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It fits if you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; play mastermind, correct appearance of letter is worth 1 point, if also in correct position 3 additional points.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the answers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BARE = 4 + 0 + 1 + 0 = 5&lt;br&gt;&#xA; BAAA = 4 + 0 + 0 + 0 = 4&lt;br&gt;&#xA; TEAR = 1 + 0 + 0 + 1 = 2&lt;br&gt;&#xA; EARS = 0 + 0 + 1 + 1 = 2&lt;br&gt;&#xA; SETA = 1 + 0 + 1 + 0 = 2&lt;br&gt;&#xA; STAR = 1 + 1 + 0 + 1 = 3&lt;br&gt;&#xA; REST = 1 + 0 + 4 + 4 = 9&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9910" LastEditorUserId="11244" LastEditDate="2015-07-22T16:47:57.163" LastActivityDate="2015-07-22T16:47:57.163" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18266" PostTypeId="1" CreationDate="2015-07-22T17:31:00.057" Score="7" ViewCount="649" Body="&lt;p&gt;Jason Baker(from Sci-fi and Fantasy SE) was kidnapped, and is in a cell where he cannot discern dislocation.  His kidnapper, who has a fondness for exclamation marks, left him a very long, confusing note:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hahaha!!!!  This is what you get for know so much about Tolkien, captured in a cell in the middle of nowhere!!  Solve this riddle! If you do, you will know what your, well, torture ail bee if you don't.  Ta!(This is only for flavor)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Actual Puzzle&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you in over your head?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you in water so deep you're drowning?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't blame it on the sunshine,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just remember your ABC's&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now go back to the roots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Primarily, take what you get,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Break it up and regroup slightly differently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have you got a nice little dessert?  Use what's left over.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Next up, use what is primary-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet not in the same way as before-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From your roots(which are slightly different now),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A nice ole line for y'all.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I wish there was another you, another way,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;'Cause you got me in chains&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though you are, you are, you are, my freedom&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to make myself believe that planet Earth turns slowly.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Hands up, when the music drops, we both put our hands up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I'm gonna swing from the chandelier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So wake me up when it's all over&lt;/p&gt;&#xA;&#xA;&lt;p&gt;'Cause I wanna be your left hand man.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Next, gather the same ingredients as before,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and take the last of the first(or the first of the last?) among them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the latter.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If they don't stand for something, they will fall for anything(side note, this one's a bit ambiguous, probably others as well.  This is not part of the clue)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though the best things in life are not things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So go instead there is not path and leave a trail.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Without music, life would be a mistake,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And though all stories are true, some never happened&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should be yourself, everybody else is already taken.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now everything is in order,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but wait right there!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's the water, not the sky.  Ironically.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;P.S., you know the computer?  If you use it, then you must solve this riddle for the password. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This goes for you Puzzler too.  Here it is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's been a long day, Hedwig.  When I'm gone, I'll throw my hands up in the air sometimes.  Solve like the second section of the clue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simple enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, when Jason finally figured out the password, he posted it on Puzzling, of course.  After that, he couldn't post anything, or send out signals.  Just watch the answers and comments come in.  The computer started a timer when he logged on, from one week.  Can you save him from the torture?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint to get rid of ambiguity in the second line of stanza seven:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Americans are just realizing it. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another hint to get y'all started&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Primarily, hm??  Where can you get numbers from?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edited to add the trivia tag.  Now that I think of it, it is extremely relevant.  Lots of trivia (or googling) necessary.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="11790" LastEditorUserId="11475" LastEditDate="2015-08-13T16:26:49.603" LastActivityDate="2015-08-13T16:26:49.603" Title="Somebody has kidnapped Jason Baker!" Tags="|riddle|story|knowledge|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18267" PostTypeId="2" ParentId="18261" CreationDate="2015-07-22T17:36:55.117" Score="2" Body="&lt;p&gt;Here's a guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sleep&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unharmed, you need it every day;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You need sleep every day&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Decapitated, you see it and you say.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eep! Who wants to see something decapitated?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9696" LastActivityDate="2015-07-22T17:36:55.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18268" PostTypeId="2" ParentId="18261" CreationDate="2015-07-22T17:46:25.620" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; gland!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;when you see &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; land&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;you say&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; land!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14168" LastActivityDate="2015-07-22T17:46:25.620" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18269" PostTypeId="2" ParentId="18261" CreationDate="2015-07-22T18:01:14.240" Score="16" Body="&lt;p&gt;Might it be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Touch?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unharmed, you need it every day;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Touch is one of the five senses we use everyday to obtain information and we touch a LOT of things everyday&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Decapitated, you see it and you say.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If you see something painful, you instinctively say &quot;Ouch!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;br&gt;&#xA;Another possible answer (going with the &quot;see&quot; hint):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Eyes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unharmed, you need it every day;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Again with the senses motif the eyes are where we obtain the most information from the world. You can't see with an injured eye.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Decapitated, you see it and you say.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The word Yes could be used as an exclamation when you get the right answer. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13489" LastEditorUserId="13489" LastEditDate="2015-07-23T01:57:01.643" LastActivityDate="2015-07-23T01:57:01.643" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18270" PostTypeId="2" ParentId="18261" CreationDate="2015-07-22T18:14:18.430" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A head&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But if you decapitate it (remove the head), you get...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which you say as you see it.&lt;/p&gt;&#xA;" OwnerUserId="14233" LastActivityDate="2015-07-22T18:14:18.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18271" PostTypeId="1" AcceptedAnswerId="18272" CreationDate="2015-07-22T20:21:08.207" Score="-7" ViewCount="369" Body="&lt;p&gt;You are Here&lt;br&gt;&#xA;You are a traveller&lt;br&gt;&#xA;You have travelled to many places&lt;br&gt;&#xA;Europe&lt;br&gt;&#xA;Japan&lt;br&gt;&#xA;China&lt;br&gt;&#xA;Russia&lt;br&gt;&#xA;Mexico&lt;br&gt;&#xA;But what is your name?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2015-07-22T20:49:48.117" Title="You are Here ${}{}$" Tags="|riddle|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18272" PostTypeId="2" ParentId="18271" CreationDate="2015-07-22T20:25:11.263" Score="8" Body="&lt;p&gt;My name is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Here, because you said so in the title and again in the first line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2015-07-22T20:25:11.263" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18273" PostTypeId="2" ParentId="18271" CreationDate="2015-07-22T20:27:26.770" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Here (from the very first line and the title, no less)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastActivityDate="2015-07-22T20:27:26.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18274" PostTypeId="2" ParentId="18271" CreationDate="2015-07-22T20:43:47.493" Score="6" Body="&lt;p&gt;Your name is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/l6pKe.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/l6pKe.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;refresh the page and you'll see the answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/ExcJV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ExcJV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="10037" LastEditDate="2015-07-22T20:49:48.117" LastActivityDate="2015-07-22T20:49:48.117" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18275" PostTypeId="1" AcceptedAnswerId="18280" CreationDate="2015-07-22T20:52:44.853" Score="7" ViewCount="1402" Body="&lt;p&gt;Ready? This is a riddle that will be fun and reasonably easy for anyone experienced with the subject matter; otherwise it will likely prove difficult (though I think there are enough clues that you could piece some amount together). Try your hand if you think you can manage it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to describe a magical place;&lt;br&gt;&#xA;Where an archipelago fills the space.&lt;br&gt;&#xA;There are other climates: mist, murk, mountain;&lt;br&gt;&#xA;But all give way to the islands about them.&lt;br&gt;&#xA;Some are quite green, vegetation-adorned;&lt;br&gt;&#xA;Some filled with lava from which they were formed.&lt;br&gt;&#xA;Besides all of these, there is one more land:&lt;br&gt;&#xA;Lifeless and barren, its harvest is bland.&lt;br&gt;&#xA;Spreads infertility, a rotting gust;&lt;br&gt;&#xA;Looking for fear? Grab a handful of dust.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Delve into the isles, 3 natives you'll find;&lt;br&gt;&#xA;And each one of them 3 more of their kind.&lt;br&gt;&#xA;The first is a creature of quite small size;&lt;br&gt;&#xA;An agile predator that snakes despise.&lt;br&gt;&#xA;It eludes attempts to kill or cripple;&lt;br&gt;&#xA;In an eye's blink, its size can triple.&lt;br&gt;&#xA;Next is a man, an entemologist;&lt;br&gt;&#xA;He's in need of a good psychologist.&lt;br&gt;&#xA;Jars of experiments clutter his shelf;&lt;br&gt;&#xA;Out of test subjects, he turns to himself.&lt;br&gt;&#xA;The third is a beast of terror and dread;&lt;br&gt;&#xA;What doesn't grow dies. It feeds on the dead.&lt;br&gt;&#xA;Just like its cousin, its hunger for fawns&lt;br&gt;&#xA;Is only surpassed by its hunger for Hans.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A stay on these isles can be quite fright'ning;&lt;br&gt;&#xA;The storms abound with all types of lightning.&lt;br&gt;&#xA;If all of these facts leave you in a daze;&lt;br&gt;&#xA;This upcoming truth will really amaze.&lt;br&gt;&#xA;The forces there excel at bereaving&lt;br&gt;&#xA;They stifle any attempts at leaving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Understand? Can you determine what this riddle is describing? What is the &quot;place&quot; in the riddle? Who are the &quot;natives&quot;? A correct answer will include explanations of the hidden meaning in the lines (you may not catch them all at first). I put a lot of work into this one so I hope y'all like it. Go!&lt;/p&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="13970" LastEditDate="2015-07-22T21:07:08.697" LastActivityDate="2015-07-22T21:43:15.247" Title="A Magical Riddle" Tags="|riddle|knowledge|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18276" PostTypeId="2" ParentId="18261" CreationDate="2015-07-22T21:02:04.970" Score="2" Body="&lt;p&gt;Just a guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; H&lt;sub&gt;2&lt;/sub&gt;O&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unharmed, you need it every day&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Need water every day&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Decapitated, you see it and you say&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; O is the decapitated version of H&lt;sub&gt;2&lt;/sub&gt;O&lt;br&gt;&#xA; When you see something decapitated you say: O! (written: Oh!)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13851" LastActivityDate="2015-07-22T21:02:04.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18277" PostTypeId="2" ParentId="16612" CreationDate="2015-07-22T21:12:33.110" Score="7" Body="&lt;p&gt;Building off of LogicianWithAHat's answer:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A note on the 'rave' and 'square hammer' lines&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    Both provide an 'E', thus two 'E''s are in the answer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The double letter is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;     'SS'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;'One of the seven' refers to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;     'seas' (one 'C')&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;'Opposite of Right' is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;     'Left', often abbreviated 'L' (as in &quot;UUDDLRLRBA&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Which gives us &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;     &quot;CLUELESS&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...which this puzzle most surely is not!&lt;/p&gt;&#xA;" OwnerUserId="14238" LastEditorUserId="14238" LastEditDate="2015-07-22T22:23:04.010" LastActivityDate="2015-07-22T22:23:04.010" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18278" PostTypeId="1" AcceptedAnswerId="18288" CreationDate="2015-07-22T21:22:18.720" Score="10" ViewCount="959" Body="&lt;p&gt;Alicia and Bobby play a game with a rectangular chocolate bar, scored into an $m$ by $n$ grid of squares. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The children alternate turns, with Alicia going first. On a child's turn, they must break the chocolate bar into two smaller rectangles along one of the grid lines, then eat the &lt;em&gt;larger&lt;/em&gt; of two pieces. If they make two equal pieces, they arbitrarily eat one of them. The game ends once the chocolate bar is a single $1\times 1$ square, in which case the person who just ate wins. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is an example of a game fully played out, starting with a $5\times 6$ bar. The letters indicate which player took each bite. Since Bobby took the last bite, he won this game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5Wnpu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5Wnpu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For which values of $m$ and $n$ can Alicia force a win?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="8874" LastEditDate="2016-03-06T18:59:11.293" LastActivityDate="2019-01-23T15:44:14.073" Title="Greedy Chocolate Eaters" Tags="|mathematics|strategy|game|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18279" PostTypeId="2" ParentId="18275" CreationDate="2015-07-22T21:30:55.057" Score="5" Body="&lt;p&gt;&quot;Next is a man, an entemologist;&lt;br&gt;&#xA;He's in need of a good psychologist.&lt;br&gt;&#xA;Jars of experiments clutter his shelf;&lt;br&gt;&#xA;Out of test subjects, he turns to himself.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This part strongly points to the following Magic the Gathering card:&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?name=delver+of+secrets&quot; rel=&quot;noreferrer&quot;&gt;http://gatherer.wizards.com/Pages/Card/Details.aspx?name=delver+of+secrets&lt;/a&gt;&lt;br&gt;&#xA;An entomologist is a scientist who studies insects, and the scientist in the card ends up turning himself into an insect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The title of the puzzle (&quot;A Magical Riddle&quot;) also points to Magic the Gathering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore the description of climates seems to refer to the various lands in Magic (Mist -&gt; ?, Murk -&gt; Swamp, Mountain -&gt; Mountain). Islands are also a land in Magic the Gathering, associated with the color blue (same as the above Magic card).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore I believe that the &quot;place&quot; in the riddle refers to Islands in MTG.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other two &quot;natives&quot; most likely refer to other blue Magic creatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;If all of these facts leave you in a daze;&quot;&lt;br&gt;&#xA;Perhaps refers to the blue spell Daze?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm unfortunately not familiar enough with MTG to recognize exactly which other cards are being referenced here.&lt;/p&gt;&#xA;" OwnerUserId="14239" LastActivityDate="2015-07-22T21:30:55.057" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18280" PostTypeId="2" ParentId="18275" CreationDate="2015-07-22T21:43:15.247" Score="4" Body="&lt;p&gt;This is inspired by olop4444's answer - thanks! :D&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'd like to describe a magical place;&lt;br&gt;&#xA;Where an archipelago fills the space.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What's being described here is RUG Delver, a Legacy deck in Magic: the Gathering. Let's go!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are other climates: mist, murk, mountain;&lt;br&gt;&#xA;But all give way to the islands about them.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is a reference to the fetchlands, which include Misty Rainforest, Scalding Tarn, and Wooded Foothills.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some are quite green, vegetation-adorned;&lt;br&gt;&#xA;Some filled with lava from which they were formed. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; These are the dual land islands, Tropical Island and Volcanic Island.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Besides all of these, there is one more land:&lt;br&gt;&#xA;Lifeless and barren, its harvest is bland.&lt;br&gt;&#xA;Spreads infertility, a rotting gust;&lt;br&gt;&#xA;Looking for fear? Grab a handful of dust.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This land is Wasteland - truly lifeless and barren. The bland harvest refers to the fact that it produces colorless mana.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Delve into the isles, 3 natives you'll find;&lt;br&gt;&#xA;And each one of them 3 more of their kind.&lt;br&gt;&#xA;The first is a creature of quite small size;&lt;br&gt;&#xA;An agile predator that snakes despise.&lt;br&gt;&#xA;It eludes attempts to kill or cripple;&lt;br&gt;&#xA;In an eye's blink, its size can triple.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Here we're talking about Nimble Mongoose, which is a 1/1 and gets +2/+2 when you have Threshold (size can triple) and can't be targeted by spells or abilities (eludes attempts).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Next is a man, an entemologist;&lt;br&gt;&#xA;He's in need of a good psychologist.&lt;br&gt;&#xA;Jars of experiments clutter his shelf;&lt;br&gt;&#xA;Out of test subjects, he turns to himself.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This one is, in fact, Delver of Secrets, which transforms itself to Insectile Aberration!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The third is a beast of terror and dread;&lt;br&gt;&#xA;What doesn't grow dies. It feeds on the dead.&lt;br&gt;&#xA;Just like its cousin, its hunger for fawns&lt;br&gt;&#xA;Is only surpassed by its hunger for Hans.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is Tarmogoyf, which gets bigger when there are more creatures of different types in the graveyard. It's a Lhurgoyf creature type, similar to Lhurgoyf itself, which is featured on the Joke card, &quot;Ach, Hans, Run!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A stay on these isles can be quite fright'ning;&lt;br&gt;&#xA;The storms abound with all types of lightning.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is the card Lightning Bolt! 3 damage for 1 mana is good.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If all of these facts leave you in a daze;&lt;br&gt;&#xA;This upcoming truth will really amaze.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is a reference to the Magic card Daze...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The forces there excel at bereaving  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...and Force of Will...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They stifle any attempts at leaving.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...and Stifle!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;P.S. &quot;A Magical Riddle&quot;, indeed.&lt;/p&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-07-22T21:43:15.247" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18281" PostTypeId="1" AcceptedAnswerId="19911" CreationDate="2015-07-22T21:46:22.343" Score="18" ViewCount="4820" Body="&lt;p&gt;You wake up to the sight of a lamp hanging from a grey ceiling. You ask yourself, &lt;em&gt;Where am I?&lt;/em&gt;  The last thing you remember is falling asleep in bed last night. Sitting up, you look around. You're on the bed in some sort of prison cell, and there's a crumpled note beside your feet.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Let's see if you can get out of this one. You've caused me trouble for the last time!&lt;/p&gt;&#xA;&lt;p&gt;- Evil Kidnapper&lt;/p&gt;&#xA;&lt;p&gt;P.S: Stop telling people I don't recycle and I'll stop doing this to you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You think to yourself, &lt;em&gt;Oh, it's the crazed freak that's obsessed with puzzles again...&lt;/em&gt; Flipping the note over, you notice that it's the label from a bottle of pop of some sort. It's missing a few parts though...&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;____________________________&#xA;|Nutrition facts           |&#xA;|Per 355 mL                |&#xA;|Calories                  |&#xA;|Fat 0 g                   |&#xA;|Sodium 15 mg              |&#xA;|Carbohydrate 42 g         |&#xA;|   Sugars 41 g            |&#xA;|Protein 0 g               |&#xA;|                          |&#xA;|CAFFIENE CONTENT          |&#xA;|   mg/355 mL              |&#xA;----------------------------&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;As you get up and walk around the room to take a look around, you notice that there's a keypad that's hooked up to a door. Clearly, if you put in the correct code the door will open and you'll be able to escape.&lt;/p&gt;&#xA;&lt;p&gt;You start walking towards the keypad, but you almost slip on another sheet of paper. It's separated into several sections, and you give it a read.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Nothing is more important to me,&lt;br /&gt;&#xA;than the bonds of my family.&lt;br /&gt;&#xA;Each piece below is a key from them to thee,&lt;br /&gt;&#xA;spend some time and you just might get free.&lt;/p&gt;&#xA;&lt;p&gt;I'm all natural, all the time -&lt;br /&gt;&#xA;Euler used me but I feel fine.&lt;br /&gt;&#xA;I'm the find in many ways,&lt;br /&gt;&#xA;a unique treasure who likes Swiss cheese.&lt;/p&gt;&#xA;&lt;p&gt;Irrelevant information influences influential individuals.&lt;br /&gt;&#xA;I infrequently integrate idioms in ideas.&lt;/p&gt;&#xA;&lt;p&gt;You get rid of me several times a day,&lt;br /&gt;&#xA;but I'll come back after a slight delay.&lt;br /&gt;&#xA;Those bound to earth already see,&lt;br /&gt;&#xA;should you ever stop relieving me,&lt;br /&gt;&#xA;It's a sign you've been released.&lt;/p&gt;&#xA;&lt;p&gt;I'm always moving,&lt;br /&gt;&#xA;with no need to stop.&lt;br /&gt;&#xA;I've no place to be going,&lt;br /&gt;&#xA;but it doesn't matter - I'm first!&lt;/p&gt;&#xA;&lt;p&gt;I'd tell you who I am if you could catch me,&lt;br /&gt;&#xA;but I'd slither away before you could see me.&lt;/p&gt;&#xA;&lt;p&gt;To summarize: My first brother is always the same, the next is narcissistic.&lt;br /&gt;&#xA;Another brother moves like a river, but his twin is impossible.&lt;br /&gt;&#xA;My sister can be quite quiet, but when she's angry you don't want to hear her hiss.&lt;/p&gt;&#xA;&lt;p&gt;P.S: I've left you a trace of my family. Good luck - Evil Kidnapper Minion (A big fan)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;Well that's mighty helpful,&lt;/em&gt; you say. As you further examine the room, you notice that there's an inscription on the floor.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mbox{___} \times \mbox{___} - \mbox{___} + \mbox{___} = 398$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;A grin spreads across your face as it dawns on you. &lt;em&gt;Huh. I think I've got everything I need.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Your task: Figure out the code and escape.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;(Hints below)&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;3 Hours later&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;Scratching your head furiously, you scream &lt;em&gt;Aaagh! It's not working! Why doesn't it work? Did I miss something?&lt;/em&gt; You go over all the clues and items in the room again and find that you've missed some fine print under equation.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If I were a box shaped animal I'd be missing most of my hide. Find your family and scramble the eggs to find the missing sides.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Another look at the riddle sheet shows that another kind minion has left you a hint.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The family tree has 27 people. If you include yourself, then your immediate family consists of 6 people. The &amp;quot;bonds of my family&amp;quot; only count your immediate family!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;Another hour later&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;Continuing to ponder, you scratch your head. &lt;em&gt;Hm... this is harder than I thought it'd be. I'm pretty sure the bonds have nothing to do with the periodic table though.... Which means it has to have a hidden meaning! But what could it be?&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;2 hours later&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;I've deduced that I, as a person, am definitely extremely different than those in my family.&lt;/em&gt; Your stomach grumbles with hunger as you sit down and try to figure out the rest of the problem...&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Another hour later&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Hmm... Definitely nothing to do with Astrology...&lt;/em&gt; you grumble, sitting on the cold cell floor.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;One Day Later&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;You think to yourself, &lt;em&gt;How am I not out of this mess already?&lt;/em&gt; Frustrated, you bash your head against the wall. &lt;em&gt;Eureka! There are exactly three parts to this puzzle! I'm pretty sure the riddles get me letters that help me get the missing numbers on the label, which would help me solve the equation! Now if only I could actually complete the riddles...&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="8999" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2022-01-11T22:09:39.273" Title="Evil Kidnappers are the worst!" Tags="|riddle|lateral-thinking|word-problem|story|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18282" PostTypeId="2" ParentId="18278" CreationDate="2015-07-22T21:50:25.663" Score="0" Body="&lt;p&gt;Without going deep into the maths,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a 2 x 2 piece will always win with Alice going first?&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/baeup.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/baeup.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="10037" LastEditDate="2015-07-23T15:49:39.637" LastActivityDate="2015-07-23T15:49:39.637" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18283" PostTypeId="2" ParentId="18228" CreationDate="2015-07-22T21:55:09.893" Score="6" Body="&lt;p&gt;Could it be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pop culture&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since the first part references&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 7Up, a Pop&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second part references&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yogurt, which is a grown culture&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the rest of it is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; references to pop culture, like rock star, game company, or Bachelor, a sitcom/tv show, and sick reference, which is used when someone makes a reference to pop culture.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13924" LastActivityDate="2015-07-22T21:55:09.893" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18286" PostTypeId="2" ParentId="18261" CreationDate="2015-07-22T22:24:46.700" Score="3" Body="&lt;p&gt;Built off of dennisdeems answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Daily bread&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You need it because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; without your daily bread (food) you will die. &quot;Bring us this day our daily bread&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Decapitate it and:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Or at least, decapitate part of it and you get your &quot;daily read&quot;. Like a newspaper or something.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2015-07-22T22:24:46.700" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18288" PostTypeId="2" ParentId="18278" CreationDate="2015-07-22T23:47:21.027" Score="7" Body="&lt;p&gt;A crucial observation to this problem is as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The game, played on an $n\times m$ bar is equivalent to the nim-sum of the game on a $n\times 1$ bar by the game on a $1\times m$ bar.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In layman's terms, the &quot;nim-sum&quot; of two games is just what happens when, at each turn, the player has the option to make a move in either (but not both) of the games and where a player loses when they cannot make any move in either game (which happens only for a $1\times 1$ bar). So, the above just means that, when we decreasing the width of the bar, we're &lt;em&gt;really&lt;/em&gt; making an equivalent move  in the $n\times 1$ bar and when decreasing the height of the bar, we're making an equivalent move in the $m\times 1$ bar. An, of course, the loss at $1\times 1$ is because both $n\times 1$ and $1\times m$ are equal to $1\times 1$, which is a loss. In short, the two dimensions are independent of each other and interact with the winning condition in a simple way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, we notice that the position $n\times 1$ is equivalent to a game of nim - in particular, choosing $m$ to be such that $2^{m} \leq n &amp;lt; 2^{m+1}$ (that is $m=\lfloor \log_2(n)\rfloor$) we can show that a block of $n\times 1$ chocolates is the same as a game of nim with $m$ stones. This is really easy to prove - we simply can show that any move in the chocolates corresponds to a legal nim move and vice versa. For instance, any legal move of chocolates divides the number $n$ of chocolates at least by $2$, equivalent to decreasing $m$ by at least one. However, for any $m'&amp;lt;m$ corresponding to a move in nim, we can legally move to a $2^{m'}\times 1$ bar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, this is easy: The game as a whole is equivalent to two-pile nim, with one pile of $\lfloor \log_2(n)\rfloor$ and one pile of $\lfloor\log_2(m)\rfloor$ (being the nim-sum of those two piles). Noticing that, in two pile nim, the &quot;mirroring&quot; strategy is optimal (where if you receive piles of two different heights, you shrink the larger one to match the smaller one), we conclude that Alicia wins whenever $\lfloor \log_2(n)\rfloor \neq \lfloor\log_2(m)\rfloor$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Some More Words For Mathematicians Less Interested in the Particular Solution and More Interested in General Technique:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One may notice that, as this is an impartial game, we may &lt;em&gt;always&lt;/em&gt; apply the &lt;a href=&quot;https://en.wikipedia.org/wiki/Sprague%E2%80%93Grundy_theorem&quot;&gt;Sprague-Grundy theorem&lt;/a&gt; to assign a nimber to each game - that is, show it to be &quot;equivalent&quot; $n$ pile nim for some $n$ (though this is a weaker notion of equivalence than proved above). The nimber of the nim-sum of two games happens to be the exclusive or (the binary operation) of their respective nimbers (where a $0$ is a loss, happening only when the nimbers are equal) - so we can easily generalize the above solution to handle $3$ dimensional chocolate bars (or $4$ dimensional and so on, but those are harder to eat).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, we may regard the nimber of a game as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Mex_(mathematics)&quot;&gt;mex&lt;/a&gt; (minimum excluded value) of the nimbers of the positions to which it can move - this is a powerful technique, letting us prove generalizations like:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose we have a game where we start with $n$ stones and may move to any position with $f(n)$ or less stones, where $f(n)$ is strictly increasing and $f(n)&amp;lt;n$ for $n&amp;gt;0$. Having $0$ stones is a loss. The nimber of this game is the smallest $k$ such that $f^k(n)=0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This may be proven by induction, or otherwise by noting that as the game is &quot;transitive&quot; in some sense (if a position can be obtained by two moves, it can be obtained by one), we are merely counting the longest possible sequence of moves (since the mex ends up incrementing once per step in the path). We can also prove it as we did before - i.e. a game of $k$ pile nim has the moves in correspondence with the given game.&lt;/p&gt;&#xA;" OwnerUserId="4765" LastEditorUserId="4765" LastEditDate="2015-07-23T01:13:09.983" LastActivityDate="2015-07-23T01:13:09.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18290" PostTypeId="2" ParentId="17995" CreationDate="2015-07-22T23:49:34.500" Score="3" Body="&lt;p&gt;I got in touch with the the owner of the puzzle page and here is his reply:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&#xA;I had to look this up myself in the source code (Undead was submitted by&#xA;another developer, not written by me). Apparently the effects of the&#xA;difficulty level setting are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;constrain the maximum length of a line of sight. At easier difficulty&#xA;levels, the game avoids really long twisty reflected lines.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;adjust the proportion of sight-lines which are deliberately arranged&#xA;to have obvious unique solutions. Those are filled in first, and then&#xA;the rest of the monsters chosen at random.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;at Tricky difficulty, enforce that there's more than one monster of&#xA;every type.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;finally, the grid is put through a solver to ensure the solution is&#xA;unique, and some statistics from the solver are recorded. Most&#xA;noticeably, the 'iterative depth', which as far as I can see is&#xA;defined as the longest chain of deductions (in the sense that each&#xA;deduction can't be made until you take into account the result of the&#xA;previous one) needed to solve the puzzle. If the puzzle doesn't have&#xA;a unique solution or the iterative depth is not appropriate for the&#xA;difficulty level, it's thrown away and generation is restarted.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Cheers,&#xA;Simon&#xA;&quot;&lt;/p&gt;&#xA;" OwnerUserId="14218" LastActivityDate="2015-07-22T23:49:34.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18291" PostTypeId="1" CreationDate="2015-07-23T00:42:00.640" Score="16" ViewCount="1898" Body="&lt;p&gt;While walking around the streets of your local city, you encounter a strange woman sitting atop a gigantic pile of dollar coins. &quot;Would you like to play a game?&quot; she asks. Intrigued by the strange sight you ask, &quot;What kind of game?&quot; She replies:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Well, I've got all these coins that I don't know what to do with. I'd really like to share them with someone. I love games, so let's do it in the form of a game. We'll take turns dividing the pile into two piles and giving the &lt;em&gt;other&lt;/em&gt; player the larger half to keep, then continuing play analogously on the smaller pile. If you split the piles equally, then the other player may take either half and play will continue on the other half. It'll be great for our karma - so much giving and whatnot!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Skeptical, you ask her, &quot;Well, there's got to be a catch - you're not just giving away free money, are you? (But I'm in if you are!)&quot;. She says,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Well, I'd love to just let you have this pile, but, for tax reasons, I can't just &lt;em&gt;give&lt;/em&gt; away so much money. Therefore, you'll have to contribute to my pile before we play. But, if you contribute &lt;em&gt;half&lt;/em&gt; as many coins as I have in my pile, I'll let you play (on the whole pile with both of our coins). I'll even let you go first!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That sounds like a fantastic deal, so you agree, and add exactly enough coins to the pile to increase its size by $50\%$. Then, you each play optimally, gaining as many coins as possible. At the end of the day, you count up the coins you earned, and find that you earned back $5 less than you contributed to the pile! At a minimum, how many coins did the woman have at the end of the game?&lt;/p&gt;&#xA;" OwnerUserId="4765" LastEditorUserId="4765" LastEditDate="2015-07-23T02:26:58.330" LastActivityDate="2015-08-05T20:32:22.300" Title="The Trickster's Game" Tags="|mathematics|strategy|game|" AnswerCount="9" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18293" PostTypeId="2" ParentId="18291" CreationDate="2015-07-23T02:41:55.560" Score="1" Body="&lt;p&gt;I'm lazy and don't have a big enough calculator so here is a bad answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2386092942&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;reasoning..&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Well, Optimal strategy is you split halvsies as they get the least number of coins. if you have a stack of 32, the split is 16, 16, they get 16. any other split (17/15) they would get 17. When they split a 17, the best they can do is 9,8 where you get 9. So with this idea going in lets look at the original problem. She has 2X money, you contribute X. So 3X money. If we follow the 32 example all the way through, It goes 16, 8, 4, 2, 1, 1. Or 21 to 11. The original put in would be 11.33 and pay out 11 so net loss .33. It will be like this for all powers of 2 (I think). So The strange woman is using the odd vs even splitting. Lets follow the 17 example. It goes 9,4,2,1,1. Or 12 to 5. Initial put in was 5.67 so loss was .67. It seems the 9/8 split worked as a gain. So lets optimize the path for the which. Lets make the last split a 2,1 split from 3 coins. then the which would want a 3,3 split so 6. then by the odd favoring the which logic a 6,7 13, 26, 53, 106, 213. Lets see what the advantage is now. 107, 53, 27, 13, 7, 3, 2, 1. Well this is 143 to 70. Well the initial pay in was 71, so this person lost a dollar. One more iteration upwards we get 426 and 853. so add 427 and 213 respectively we get. 570 and 283 from 853 total. Pay in was 284 1/3 lost 1.3333. It seems each iteration loses 1/3 more dollars this looks like a big stack of coins indeed. (3 was 0, 13 was 1/3 ...) so we just need to iterate 11 more times... Well the basic step is double from here then double and add one so 4n+1 so... 4^11 more to go roughly. that is roughly 2^22 or 4 million ish... the 1's get annoying so i will come back with a calculator. 3579139413 is the number of initial coins so she has 5 more than 2/3. 2386092942. I don't have enough significant digits on my calculator so I think it is actually not a whole number. But one you have 3 billion coins you can lose some rounding. upon further inspection this seems divisible by 3. All is well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13916" LastEditorUserId="13916" LastEditDate="2015-07-23T02:52:48.323" LastActivityDate="2015-07-23T02:52:48.323" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="18294" PostTypeId="2" ParentId="18291" CreationDate="2015-07-23T03:14:42.013" Score="3" Body="&lt;p&gt;I got that the woman must have at least:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;238609299&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;coins at the end of the game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;First, the value of a pile is weakly monotonically increasing, because additional coins can always just be given to the other person - you will never get less money. Thus, giving the other person as few coins as possible is always optimal. The only asymetry is the odd vs. even giveaway. Thus, for the woman to gain money, she must be splitting odd piles, while you must be splitting even piles. If we set up a series of odd/even splits, we get the following: &lt;a href=&quot;http://pastebin.com/PQssUcvh&quot; rel=&quot;nofollow&quot;&gt;http://pastebin.com/PQssUcvh&lt;/a&gt;&#xA;&#xA;At the link, the format is as follows: inital pile, amount you get, amount she gets. The lines alternate you start, she starts, etc. The line &lt;code&gt;Loss: _&lt;/code&gt; is printed after each amount you could actually start with, because if you contribute half as much as she has, the stating amount must be a multiple of 3. The amount lost reaches 5 on the last line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2187" LastActivityDate="2015-07-23T03:14:42.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18295" PostTypeId="1" AcceptedAnswerId="18296" CreationDate="2015-07-23T06:49:53.063" Score="6" ViewCount="373" Body="&lt;p&gt;Is it possible to cut a regular tetrahedron with edge length 100&#xA;into regular &lt;a href=&quot;https://en.wikipedia.org/wiki/Tetrahedron&quot; rel=&quot;nofollow&quot;&gt;tetrahedrons&lt;/a&gt; with edges of length less than 1 &#xA;and  regular &lt;a href=&quot;https://en.wikipedia.org/wiki/Octahedron&quot; rel=&quot;nofollow&quot;&gt;octahedrons&lt;/a&gt;  with edges of length less than 1?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-10-14T13:53:38.707" LastActivityDate="2015-10-14T13:53:38.707" Title="Tetrahedrons and Octahedrons" Tags="|geometry|dissection|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18296" PostTypeId="2" ParentId="18295" CreationDate="2015-07-23T07:00:31.907" Score="7" Body="&lt;p&gt;Yes. There's a &lt;a href=&quot;https://en.wikipedia.org/wiki/Tetrahedral-octahedral_honeycomb&quot;&gt;solid tiling&lt;/a&gt; of tetrahedra and octahedra that fills a tetraderon. Making a sufficiently large tetrahedron this way and scaling is down to the right size solves the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ijYVAm.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ijYVAm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to obtain the tiling is to cut to tetradhedron with slices parallel to each face that are spaced apart at even intervals of $1/200^{th}$ of its height.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-07-23T07:00:31.907" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18297" PostTypeId="2" ParentId="14831" CreationDate="2015-07-23T08:26:30.733" Score="0" Body="&lt;p&gt;Based on @Alex answer :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cut 1 : friend 1&lt;br&gt;Cut 2 : friend 2&lt;br&gt;Cut 3 : pizza delivery man&lt;br&gt;&lt;br&gt; All the pizza for free. :D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13384" LastActivityDate="2015-07-23T08:26:30.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18298" PostTypeId="2" ParentId="18239" CreationDate="2015-07-23T08:34:20.167" Score="9" Body="&lt;p&gt;I could be completely off the mark here.  But here goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hint seems to be telling us to convert the text into HEX, so I converted all the words into HEX:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; OLD FLAGS - 4f 4c 44 20 46 4c 41 47 53&lt;br&gt;&#xA; FISHY DISH - 46 49 53 48 59 20 44 49 53 48&lt;br&gt;&#xA; PUERTO RICO - 50 55 45 52 54 4f 20 52 49 43 4f&lt;br&gt;&#xA; ACQUITTED IDIOT - 41 43 51 55 49 54 54 45 44 20 49 44 49 4f 54&lt;br&gt;&#xA; MANLY PINK - 4d 41 4e 4c 59 20 50 49 4e 4b&lt;br&gt;&#xA; BANANA CABANA - 42 41 4e 41 4e 41 20 43 41 42 41 4e 41&lt;br&gt;&#xA; STREET SMART - 53 54 52 45 45 54 20 53 4d 41 52 54&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To find the odd ones out, I converted the HEX value into DEC and add up all the odd numbers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; OLD FLAGS 298&lt;br&gt;&#xA; FISHY DISH 401&lt;br&gt;&#xA; PUERTO RICO 452&lt;br&gt;&#xA; ACQUITTED IDIOT 665&lt;br&gt;&#xA; MANLY PINK 379&lt;br&gt;&#xA; BANANA CABANA 457&lt;br&gt;&#xA; STREET SMART 446&lt;br&gt;&#xA;&lt;br&gt;&#xA; Total of all odd numbers = 3098&lt;br&gt;&#xA;&lt;br&gt;&#xA; Convert 3098 back into HEX and we get C1A&lt;br&gt;&#xA;&lt;br&gt;&#xA; Sounds like CIA in Virginia, USA.&lt;br&gt;&#xA;&lt;br&gt;&#xA; But I'm unsure why 'B' is in the hint - 61B2068696E74&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12332" LastActivityDate="2015-07-23T08:34:20.167" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18300" PostTypeId="2" ParentId="18281" CreationDate="2015-07-23T13:45:03.410" Score="1" Body="&lt;p&gt;Work in progress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paragraph 1: Natural numbers used by Euler: Euler's number, about 2.71828, more specifically, the limit of (1 + (1/n))^n as n approaches infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paragraph 2: Someone's been watching too much V for Vendetta... Too many words begin with the letter i, which is the square root of -1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paragraph 3: Oblique reference to gravity, about 9.8 kilometers per second squared.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paragraph 4: Neverending, yet looping animation: such is the dance of the planets, but that's stretching it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eigenv... nah, that's not going to work much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sibling one: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sibling two: e, a unique number that neatly slips between all other fractions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sibling three: g, the gravitational constant that determines river flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sibling four: i, an impossible number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sister: A snake...?&lt;/p&gt;&#xA;" OwnerUserId="12506" LastEditorUserId="12506" LastEditDate="2015-07-25T16:03:24.280" LastActivityDate="2015-07-25T16:03:24.280" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18301" PostTypeId="2" ParentId="18291" CreationDate="2015-07-23T14:22:41.740" Score="5" Body="&lt;p&gt;My answer is very close to the others, but slightly off.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;238,609,294&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is how the game plays out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You contribute 50% of what she does, so you give $119,304,647$.  This makes the pile $357,913,941$ in total.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each round will be where you play and she plays.  Thus, we get the following table.  During each round, she will keep the larger of the two piles to herself and play with the smaller.  You will get as profit the larger if the two piles she makes.  The next round will start with the smaller of her two piles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{lrrrrr}&#xA;&amp;amp;&amp;amp;\text{Your turn}&amp;amp;&amp;amp;\text{Her turn}\\\text{Round}&amp;amp;\text{Pile}&amp;amp;\text{Small}&amp;amp;\text{Her profit}&amp;amp;\text{Small}&amp;amp;\text{Your profit}\\\hline&#xA;1&amp;amp;357,913,941&amp;amp;178,956,970&amp;amp;178,956,971&amp;amp;89,478,485&amp;amp;89,478,485\\2&amp;amp;89,478,485&amp;amp;44,739,242&amp;amp;44,739,243&amp;amp;22,369,621&amp;amp;22,369,621\\3&amp;amp;22,369,621&amp;amp;11,184,810&amp;amp;11,184,811&amp;amp;5,592,405&amp;amp;5,592,405\\4&amp;amp;5,592,405&amp;amp;2,796,202&amp;amp;2,796,203&amp;amp;1,398,101&amp;amp;1,398,101\\5&amp;amp;1,398,101&amp;amp;699,050&amp;amp;699,051&amp;amp;349,525&amp;amp;349,525\\6&amp;amp;349,525&amp;amp;174,762&amp;amp;174,763&amp;amp;87,381&amp;amp;87,381\\7&amp;amp;87,381&amp;amp;43,690&amp;amp;43,691&amp;amp;21,845&amp;amp;21,845\\8&amp;amp;21,845&amp;amp;10,922&amp;amp;10,923&amp;amp;5,461&amp;amp;5,461\\9&amp;amp;5,461&amp;amp;2,730&amp;amp;2,731&amp;amp;1,365&amp;amp;1,365\\10&amp;amp;1,365&amp;amp;682&amp;amp;683&amp;amp;341&amp;amp;341\\11&amp;amp;341&amp;amp;170&amp;amp;171&amp;amp;85&amp;amp;85\\12&amp;amp;85&amp;amp;42&amp;amp;43&amp;amp;21&amp;amp;21\\13&amp;amp;21&amp;amp;10&amp;amp;11&amp;amp;5&amp;amp;5\\14&amp;amp;5&amp;amp;2&amp;amp;3&amp;amp;1&amp;amp;1\\15&amp;amp;1&amp;amp;0&amp;amp;1&amp;amp;-&amp;amp;-&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you add up the far right column, you get $119,304,642$.  This is exactly $5$ less then you contributed to the pile.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="13494" LastEditDate="2015-08-05T20:32:22.300" LastActivityDate="2015-08-05T20:32:22.300" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18302" PostTypeId="2" ParentId="18281" CreationDate="2015-07-23T14:25:02.720" Score="1" Body="&lt;p&gt;Work in progress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think @Nefer007 has the right idea for the first 3 paragraphs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paragraph 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm always moving,&lt;br&gt;&#xA;  with no need to stop.&lt;br&gt;&#xA;  I've no place to be going,&lt;br&gt;&#xA;  but it doesn't matter - I'm first!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think it has to do with&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the sun, although I can't seem to find anything that seems relevant&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the very first paragraph,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it mentions nothing is more important than the bonds on my family so I started thinking about electrical bonds and having something to do with the nutrition label. So far there is no connection between the two that I see other than that. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the equation,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think the paragraph about the family determines the order. My first brother is always the same (Eulers constant or gravitational constant), the next is narcissitic(negative '-') Another brother moves like a river(no idea), His twin is impossible(i),My sister can be quite quiet, but when she's angry you don't want to hear her hiss(possibly square root since it looks like a snake and has a positive and negative answer maybe referring to her mood?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't think I'll be able to finish this but there's some stuff to think about. I'll delete this post if someone finishes before me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[I'd suggest this has something to do with Euler's Identity.&#xA;e^ipi = cospi + isinpi&#xA;Cosine moves like a river, iSine is impossible, the swiss cheese fan is Euler's number, the narcissist is pi because it goes on and on about itself (easy as pie idiom) -- don't know how the nutrisional facts fit in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if the &quot;it's a sign  you've been released&quot; paragraph is referencing CO2 in reference to cosine being squared. Make what you will of that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the six terms are e^i, e^pi, isinpi, cospi, i, and pi. I think. But where does the 27 come from?]&lt;/p&gt;&#xA;" OwnerUserId="13925" LastEditorUserId="-1" LastEditDate="2015-08-07T14:20:58.933" LastActivityDate="2015-08-07T14:20:58.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18303" PostTypeId="2" ParentId="18291" CreationDate="2015-07-23T14:42:22.073" Score="2" Body="&lt;p&gt;Building from isaacg's reasoning of odd and even splits, we require the woman to start her turns with an even number of coins so that you get half of that pile, and you start with an odd number of coins so she gets more than half of that pile. This means the woman ends up with 2 coins in her last iteration, giving one to you and keeping the other herself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Building backwards from the last coin paid out, and bracketing the amounts you get, we end up with a payouts sequence $[1, (1), 3, (5), 11, (21), 43, \dots]$. It goes $[n, (2n-1)]$ from her turn to your preceding turn and $[(n), 2n+1]$ from your turn to her preceding turn. (Note that we are counting payout-to-payout, not remainder-to-payout.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since she gets the first and last payout, we want an odd number of turns in this sequence. We also need to start the game with a multiple of 3 as the number of coins, so the sum of the series must be a multiple of 3. Finally, you need to pocket 5 coins less than you started, so the sum of the bracketed amounts must be 5 less than the sum of the series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting this all together in a spreadsheet, I get a starting figure of $357,913,941$, of which you contributed $119,304,647$ and finished with $119,304,642$, &lt;strong&gt;leaving her with $238,609,299$&lt;/strong&gt; coins after 29 turns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some spreadsheet formulae for verification in case I've made a mistake:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A1=1, A2=2*B1+1 (her payouts in reverse order)&#xA;B1=2*A1-1       (your payouts in reverse order)&#xA;C1=SUM($A$1:A2) (her cumulative total payout in reverse order)&#xA;D1=SUM($B$1:B1) (your cumulative total payout in reverse order)&#xA;E1=C1+D1        (number of coins at the start)&#xA;F1=D1+5-E1/3    (your payout + 5 (D1+5) minus your contribution)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Copy the formulae down in each column, and look for the first row where column F is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The real question is ... what were you doing lugging that much change around before you met this strange woman?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="9422" LastEditorUserId="9422" LastEditDate="2015-07-24T09:09:09.910" LastActivityDate="2015-07-24T09:09:09.910" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18304" PostTypeId="2" ParentId="18291" CreationDate="2015-07-23T14:50:55.960" Score="1" Body="&lt;p&gt;@isaacg had this answer first, this is just another way to get there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The woman had to have at least&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;238609299&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;at the end of the game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used a dynamic programming solution to calculate the winnings for pile sizes up to 100000000 (10^8).  This didn't yield any results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there I used a memoized recursive solution for all possible woman starting pile values from 2 up to 1000000000 (10^9) stepping by 2 (assuming that coins are not fractional).  In the worst case the recursive solution would have to go four layers deep before it hit a memoized answer.  This is negligible making the algorithmic complexity O(size of woman's starting pile).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This actually yielded 29 results of which the posted answer was the minimum.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#include &amp;lt;iostream&amp;gt;&#xA;using namespace std;&#xA;&#xA;int win[100000001];&#xA;&#xA;long long get_winnings(long long pile) {&#xA;    if(pile &amp;lt;= 100000000)&#xA;        return win[pile];&#xA;    return (pile / 2) - get_winnings(pile / 2);&#xA;}&#xA;&#xA;int main() {&#xA;    win[0] = 0;&#xA;    for(int pile = 1; pile &amp;lt;= 100000000; pile++) {&#xA;        int small_half = pile / 2;&#xA;        win[pile] = small_half - win[small_half];&#xA;    }&#xA;    for(long long woman = 2; woman &amp;lt;= 2500000000; woman += 2) {&#xA;        long long investment = woman / 2;&#xA;        long long winnings = get_winnings(investment + woman);&#xA;        if(investment - winnings &amp;gt;= 5)&#xA;            cout &amp;lt;&amp;lt; woman &amp;lt;&amp;lt; &quot; &quot; &amp;lt;&amp;lt; investment &amp;lt;&amp;lt; &quot; &quot; &amp;lt;&amp;lt; winnings &amp;lt;&amp;lt; &quot; &quot; &amp;lt;&amp;lt; woman + investment - winnings &amp;lt;&amp;lt; endl;&#xA;    }&#xA;    return 0;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1777" LastEditorUserId="1777" LastEditDate="2015-07-23T16:43:28.210" LastActivityDate="2015-07-23T16:43:28.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18305" PostTypeId="1" AcceptedAnswerId="18393" CreationDate="2015-07-23T14:56:39.633" Score="4" ViewCount="594" Body="&lt;p&gt;I am the cheapest of the flying animals,&lt;br&gt;&#xA;As such I cannot give any change.&lt;br&gt;&#xA;I guess I am just riffing yeah yeah,&lt;br&gt;&#xA;But I heard my creator is &quot;too too wise&quot;. &lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyways I will just soar above, won't I,&lt;br&gt;&#xA;Oh oh oh oh, you realize I must go.&lt;br&gt;&#xA;My stay has been sugary, but bye.&lt;br&gt;&#xA;I will just soar above, won't I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My brother is a man least complex, &lt;br&gt;&#xA; As well as a day lost to the breeze.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is a 100% chance that the title is misleading. Some regrets about that.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; See revised less misleading title&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Riffing is a musical term, tag added.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13916" LastEditorUserId="13916" LastEditDate="2015-07-27T14:08:36.707" LastActivityDate="2015-08-04T20:32:53.563" Title="What is this cheap critter?" Tags="|riddle|music|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18306" PostTypeId="2" ParentId="18305" CreationDate="2015-07-23T15:53:14.913" Score="0" Body="&lt;p&gt;Not sure about this at all, and I lack the time to ponder this riddle (sorry!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess it could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Any kind of paper wasp&lt;br&gt;&#xA;They obviously like sugar, and the name 'paper' wasp because they have hives made out papery stuff. If your house is made of paper you're pretty poor/cheap or you have one hell of a kind of paper. The riffing could refer to the buzzing a wasp makes when flying. I don't get the creator line though...   Anyway, good riddle and good luck to others&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastActivityDate="2015-07-23T15:53:14.913" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18307" PostTypeId="2" ParentId="18305" CreationDate="2015-07-23T16:11:56.823" Score="5" Body="&lt;p&gt;I think this may be a reference to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; representations of hummingbirds in the Nazca lines.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The phrase &quot;just riffing&quot; could refer to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; humming (instead of singing a proper song),&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;while the reference that 'my creator is &quot;too too wise&quot;' could refer to the fact that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; many of these images are not apparent from the ground (the creator therefore being wiser than people who are earthbound),&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and related to the quote 'I will just soar above'&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; since the images are visible from the air.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;'you realize I must go' could refer to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; concerns over damage to the images, such as that caused by a Greenpeace publicity stunt,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;while 'My stay has been sugary' could be a reference to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the hummingbird's diet, typically of sugar-rich nectar.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9278" LastActivityDate="2015-07-23T16:11:56.823" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18308" PostTypeId="1" AcceptedAnswerId="18314" CreationDate="2015-07-23T18:36:49.440" Score="6" ViewCount="1006" Body="&lt;p&gt;A variant of &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18278/greedy-chocolate-eaters&quot;&gt;Greedy Chocolate Eaters&lt;/a&gt; (created by my significant other, who doesn't like chocolate all that much)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Alice and Bob are playing a game with a rectangular bar of chocolate scored into $m \times n$ squares along gridlines.  They alternate turns, with Alice going first.  On a turn, the player breaks the chocolate bar along a grid line, eating the &lt;strong&gt;&lt;em&gt;strictly smaller&lt;/em&gt;&lt;/strong&gt; piece.  Players may not break the chocolate bar in half.  The person who reduces the chocolate bar to a $2 \times 2$ square wins, since it is impossible to reduce it further.   Both players know and use the optimal strategy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, with possibly non-optimal play, Alice could take a $4 \times 6$ bar and reduce it to $3 \times 6$.  Bob could reduce that to $3 \times 4$.  Alice could reduce to $3 \times 3$.  Bob could then reduce to $2 \times 3$, and Alice would win by reducing to $2 \times 2$.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given $m$ and $n$, find a simple test Alice can apply to determine if she will lose.  Then explain the optimal strategy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2722" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-07-23T21:14:53.357" Title="Non-Greedy Chocolate Eaters" Tags="|mathematics|strategy|game|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18309" PostTypeId="1" AcceptedAnswerId="18310" CreationDate="2015-07-23T18:42:16.267" Score="27" ViewCount="1329" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;My first is found in Aquaman's new super-secret lair.&lt;br&gt;&#xA;  My second is a prison where a climbing snail's ensnared.&lt;br&gt;&#xA;  My third, repeated seven times, requires punctuation.&lt;br&gt;&#xA;  My fourth comprises eight furlongs; embodies termination.&lt;br&gt;&#xA;  My fifth must be connected to three homes with expertise.&lt;br&gt;&#xA;  My whole, a famous work of art, comprises all of these.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I talking about?&lt;/p&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="11244" LastEditDate="2015-07-23T18:58:00.403" LastActivityDate="2015-07-23T21:53:38.947" Title="A Riddle for Puzzlers" Tags="|riddle|wordplay|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18310" PostTypeId="2" ParentId="18309" CreationDate="2015-07-23T18:58:39.707" Score="13" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All's Well That Ends Well, a famous Shakespeare play.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My first is found in Aquaman's new super-secret lair.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is an amusing reference to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18153/aquamans-revenge&quot;&gt;Aquaman&amp;#39;s Revenge!&lt;/a&gt;, in which one of your possessions is an &lt;strong&gt;awl&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My second is a prison where a climbing snail's ensnared.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is a reference to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/248/how-long-does-it-take-for-the-snail-to-get-out-of-the-well&quot;&gt;https://puzzling.stackexchange.com/questions/248/how-long-does-it-take-for-the-snail-to-get-out-of-the-well&lt;/a&gt;. In the problem, the snail is trapped in a &lt;strong&gt;well&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My third, repeated seven times, requires punctuation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is a reference to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2578/a-grammatically-valid-sentence-in-english-containing-seven-thats-in-a-row&quot;&gt;A grammatically valid sentence in English containing seven &amp;quot;that&amp;quot;&amp;#39;s in a row?&lt;/a&gt;. The word in question is &lt;strong&gt;that&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fourth comprises eight furlongs; embodies termination.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This refers to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/11888/quite-a-distance-from-the-start-to-the-end-of-this-short-riddle-who-am-i&quot;&gt;Quite a distance from the start to the end of this short riddle. Who am I?&lt;/a&gt;.  The word in question is &lt;strong&gt;ends&lt;/strong&gt;.  It certainly does embody termination, as termination is the end of something.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fifth must be connected to three homes with expertise.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is a cute reference to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/17230/connect-3-houses-with-3-wells&quot;&gt;Connect 3 houses with 3 wells&lt;/a&gt;, in which three houses must be connected to three &lt;strong&gt;wells&lt;/strong&gt; without the pipes crossing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2015-07-23T21:53:38.947" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18311" PostTypeId="1" AcceptedAnswerId="18378" CreationDate="2015-07-23T19:44:18.727" Score="13" ViewCount="876" Body="&lt;p&gt;Read the lines and find the answer. Follow each one through to the end, but maybe not entirely. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My first is a dye, most commonly used.&lt;br&gt;&#xA;  My second is twentieth, in a cup.&lt;br&gt;&#xA;  My third is gloomy, shrouded in cloud.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;My fourth periodically gets to a half century.&lt;br&gt;&#xA;  My fifth often stands for unity. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Crack them all, and put them together.&lt;br&gt;&#xA;   To find the answer, that perhaps I have left hither. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The title has been altered, and that should provide a major clue. (Apologies for not making this the title in the first place, when I crafted this last night. My mind certainly works better in the morning. :-P)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-07-28T02:52:53.300" LastActivityDate="2015-07-28T02:52:53.300" Title="The complete combination of incomplete words" Tags="|riddle|word|wordplay|" AnswerCount="4" CommentCount="16" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18312" PostTypeId="2" ParentId="18291" CreationDate="2015-07-23T20:10:07.237" Score="3" Body="&lt;p&gt;You only need a big calculator to do 1 calculation in this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is in fact (as others have pointed out)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;238609299&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;br&gt;&#xA;&lt;strong&gt;Explaination&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observe that you start the game, thus she wants the total amount to start on an odd number (so you have to give her the strictly bigger half).  In return she wants an even number so she can give you exactly half.  This however has to yield another odd number (so you again give her more than half) etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now starting piles can be calculated backwards from 1.  Double it, then double and add 1 to maintain her getting strictly bigger portions.  We get &lt;em&gt;1, 5, 21, 85, 341...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Calculating amount you win:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First consider the case with a game of size 1.  You would win &lt;strong&gt;\$0&lt;/strong&gt;. (You give her a dollar and it is over.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second consider the case with game of size 5.  You give her 3, she gives you 1, and now we are left with a game of size 1 (your turn and \$1 left.)  Since we know that case, you make &lt;strong&gt;\$1+\$0 = \$1&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third consider the case with game of size 21.  You give her 11, she gives you 5, and now we are left with a game of size 5.  Since we know that case, you make &lt;strong&gt;\$5+\$1 = \$6&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can say that in starting with \$n, you will make the amount you made in game n-1 + the starting size of n-1.  Our winnings are thus: 0, 1, 6, 27, 112...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last thing to note is our cost in each game is 1/3 of the total size at the start of the game.  So our games of size 1, 5, 21, 85, 341 cost us .333, 1.666, 7, 28.333, and 113.666.  Notice this versus our winnings:  Every increase in game size, and we lose \$0.33.  Thus our 15th game will lose us \$5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we just need to find $$\dfrac{2}{3}\sum_{i=0}^{14}{4^i} = \dfrac{2}{3}357913941&#xA; = 238609294&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we must add the \$5 you won for a total of &lt;strong&gt;238609299&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="14092" LastEditorUserId="14092" LastEditDate="2015-07-24T13:46:35.867" LastActivityDate="2015-07-24T13:46:35.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18313" PostTypeId="2" ParentId="18311" CreationDate="2015-07-23T20:25:38.870" Score="1" Body="&lt;p&gt;The attempt:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lotus (or louts) &lt;br&gt;&#xA; L from 5th -&gt; roman numeral for 50 (or a half century)&lt;br&gt;&#xA; O from @warspyking O for 3rd seems like a good guess, both because the round moon is shrouded in cloud, and &quot;O&quot; is literally in the middle of &quot;cloud&quot; –  Cain 42 mins ago &lt;br&gt;&#xA; T from The twentieth letter in the alphabet is T, which is in a cup (tea) –  Doge 1 hour ago &lt;br&gt;&#xA; U from The mathematical symbol for union,  ∪ &lt;br&gt;&#xA; S from I think the first letter is &quot;s&quot;. Sulfur dyes are the most commonly used dyes according to a quick google search. And Sulfur is &quot;S&quot; on the periodic table. –  pauld 1 hour ago&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The previous thought answers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;sub&gt;STOLEN Stolid, or STOLE&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;reasons:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&#xA;I'm guessing the 3rd might be O? –  warspyking 28 mins ago &lt;br&gt;&#xA;The twentieth letter in the alphabet is T, which is in a cup (tea) –  Doge 23 mins ago        &lt;br&gt;&#xA;I think the first letter is &quot;s&quot;. Sulfur dyes are the most commonly used dyes according to a quick google search. And Sulfur is &quot;S&quot; on the periodic table &lt;br&gt;&#xA; and finally L or 50 in roman numerals. &lt;br&gt;&#xA; which leaves us with stol_ &#xA; then EN which is for en masse, which is as a group or unified.&#xA; then Orderified: Stolen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13916" LastEditorUserId="13916" LastEditDate="2015-07-24T14:46:37.717" LastActivityDate="2015-07-24T14:46:37.717" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18314" PostTypeId="2" ParentId="18308" CreationDate="2015-07-23T20:58:18.263" Score="5" Body="&lt;p&gt;Alice will lose if&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; one of the numbers $m$ and $n$ is a power of $2$ times the other. She can win otherwise.&lt;br&gt;&lt;br&gt; Call a piece of chocolate &lt;em&gt;balanced&lt;/em&gt; if one of the dimensions is a power of $2$ times the other. First, we see that any move made to a blanced piece will result in an unbalanced piece, because the dimension that is reduced must be reduced to more than half its original size. Next, we see that if the piece is unbalanced, then there is a move resulting in a balanced piece. If the unbalanced piece has size $m\times n$ with $m&amp;gt;n$, then we can choose the largest integer $k$ such that $2^k n \leq m$, and break the chocolate into pieces of size $(2^k n)\times n$ and $(m-2^k n)\times n$. Now $m-2^k n&amp;lt; 2^k n$ because we chose $k$ to be maximal, so the smaller piece of size $(m-2^k n)\times n$ is eaten, and the remaining piece of size $(2^k n)\times n$ is balanced.&lt;br&gt;&lt;br&gt; So, the winning strategy: given an unbalanced piece of chocolate is to make the piece balanced.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-07-23T21:14:53.357" LastActivityDate="2015-07-23T21:14:53.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18315" PostTypeId="2" ParentId="18308" CreationDate="2015-07-23T21:13:43.320" Score="3" Body="&lt;p&gt;Define a function $f(n)$ on the positive integers recursively by&#xA;$$&#xA;f(n)=\begin{cases}&#xA;f(n/2) &amp;amp; \text{when } n \text{ is even}\\&#xA;\frac{n-1}2 &amp;amp; \text{when } n \text{ is odd}&#xA;\end{cases}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explicitly, to compute $f(n)$, write $n$ in binary, then remove the rightmost one, along with all of the zeroes after it. For example, $f(104)=f(1101000_2)=110_2=6$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An $m\times n$ bar is losing for Alice if and only if $f(m)=f(n)$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The winning strategy is this. If you are given an $m\times n$ bar, where $f(n)&amp;gt;f(m)$, then reduce the value of $n$ to some $n'$ so that $f(n')=f(m)$. If $f(m)&amp;gt;f(n)$, reduce $m$ to $m'$ similarly. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why is all of that true? We first must prove the following fact:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma:&lt;/strong&gt; For any $k\ge1$, let $L_k$ be the following list of $k$ numbers: $$L_k=f(k+1),f(k+2),\dots,f(2k).$$ The list $L_k$ will be some rearrangement of the list $0,1,2,\dots,(k-1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof:&lt;/em&gt; We prove this by induction. You can directly check that $L_1=0$, and $L_2=1,0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that $L_{k-1}$ is a rearrangement of $0,1,\dots,k-2$. Notice that the list $L_{k}$ is just the list $L_{k-1}$ with the entry $f(k)$ removed from the front and the entries $f(2k-1),f(2k)$ added to the end. Since $f(k)=f(2k)$, we are really just adding $f(2k-1)=k-1$. Since we aded $k-1$ to a rearrangement of $0,1,\dots,k-2$, it follows $L_{k}$ is a rearrangement of $0,1,\dots,k-2,k-1$. $\square$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this Lemma under out belts, let's return to the game. Cutting a chocolate bar means reducing one of its dimensions. I &lt;strong&gt;claim&lt;/strong&gt; that if you reduce the dimension $n$ to $n'$, then it must be true that $f(n')\neq f(n)$. Furthermore, it is possible to choose $n'$ so $f(n')$ is any number less than $f(n)$. This follows from the Lemma:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $n=2k+1$ is odd, then your options are $n'\in\{k+1,k+2\dots,2k\}$. By the Lemma, $f(n')$ can be any number in $L_k$, so any number less than $k=f(n)$, agreeing with the claim. &lt;/li&gt;&#xA;&lt;li&gt;If $n=2k$ is even, your options are $n'\in\{k+1,k+2,\dots,2k-1\}$. By the Lemma, $f(n')$ can be any number in the list $0,1,\dots,k-1$, &lt;em&gt;except&lt;/em&gt; for $f(2k)=f(n)$. Since your options for $f(n')$ form an interval of integers with $f(n)$ missing, the &lt;strong&gt;claim&lt;/strong&gt; holds.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This shows that when if $f(n)\neq f(m)$ you can always restore equality in one move, while when $f(n)=f(m)$, every move destroys this equality. This proves that the strategy of equalizing the bar until a $2\times2$ bar remains is winning.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-07-23T21:13:43.320" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18316" PostTypeId="2" ParentId="18291" CreationDate="2015-07-23T23:14:18.997" Score="11" Body="&lt;p&gt;One rigorous solution to this problem is to write out a recursion relation for the &lt;em&gt;total&lt;/em&gt; earnings $f(n)$ of the first player given a pile of $n$. It is clearly optimal to divide the pile as evenly as possible so the first player will give the other player $\lceil \frac{n}2\rceil$ coins and continue play on $\lfloor \frac{n}2\rfloor$ coins. Obviously, as this leaves them with no new coins and gives their opponent a pile of $\lfloor\frac{n}2\rfloor$ coins, they will earn however many their opponent does not. That is, we may write&#xA;$$f(n)=\left\lfloor \frac{n}2\right\rfloor-f\left(\left\lfloor\frac{n}2\right\rfloor\right).$$&#xA;Together with the equation $f(0)=0$, this determines $f$ for all positive values. Notice that we may rewrite the above equation as:&#xA;$$f(2n)=f(2n+1)=n - f(n)$$&#xA;Next, we are interested in the function which, given the total pile size, determines the &lt;em&gt;net&lt;/em&gt; earning for the player - which is their earnings minus their contribution of a third of the pile:&#xA;$$g(n)=f(n)-\frac{n}3.$$&#xA;Substituting $g(n)+\frac{n}3$ in for $f$ in the second recursion relation, we get:&#xA;$$g(2n)+\frac{2n}3 = g(2n+1)+\frac{2n}3 +\frac{1}3 = n - g(n) - \frac{n}3$$&#xA;which, after simplification, becomes:&#xA;$$g(2n)=g(2n+1)+\frac{1}3=-g(n).$$&#xA;This essentially says what has been noted in other answers - when a player has to &quot;round down&quot; (i.e. when they are given an odd pile), they are giving away $\frac{1}3$ dollars of their earnings. However, we can go a step further and give the following algorithm for calculating a player's loss given the total pile size using the above relation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $n$ be the initial size of the pile. Write $n$ in binary. Starting at the one's place, color every other digit red and starting at the two's place, color every other digit blue. Starting with $n=141$ we would get:&#xA;  $$141=\color{blue}1\color{red}0\color{blue}0\color{red}0\color{blue}1\color{red}1\color{blue}0\color{red}1_2$$&#xA;  Now, count the number $b$ of blue '1's and the number $r$ of red '1's. Your net earnings are $\frac{b-r}3$ - so in the above example, the player breaks even.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then, the problem is easy: The smallest total pile size is $15$ red '1's with no blue '1's. So, the pile size is at least:&#xA;$$10101010101010101010101010101_2=357913941.$$&#xA;We then multiply this by $\frac{2}3$ to get the strange woman's initial pile size and add $5$ to get her final pile size, yielding that she had $238,\!609,\!299$ coins at the end.&lt;/p&gt;&#xA;" OwnerUserId="4765" LastActivityDate="2015-07-23T23:14:18.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18317" PostTypeId="2" ParentId="18050" CreationDate="2015-07-23T23:53:02.290" Score="19" Body="&lt;p&gt;Following Leppy's lead, here's an animated .gif with all the layers in order of hue.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fShnQ.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;The 19 colors themselves are fully-saturated and evenly spaced around the edge of the color wheel, suggesting a kind of sequence.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8ZPv9.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Numbering the colors from 0 (red) to 18 (magenta) we have the following numbers of pixels of each color:&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th&gt;color&lt;/th&gt;&#xA;&lt;th&gt;# of pixels&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;0&lt;/td&gt;&#xA;&lt;td&gt;4963&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;1&lt;/td&gt;&#xA;&lt;td&gt;5338&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;2&lt;/td&gt;&#xA;&lt;td&gt;4317&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;3&lt;/td&gt;&#xA;&lt;td&gt;4934&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;4&lt;/td&gt;&#xA;&lt;td&gt;5617&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;5&lt;/td&gt;&#xA;&lt;td&gt;5067&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;6&lt;/td&gt;&#xA;&lt;td&gt;4955&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;7&lt;/td&gt;&#xA;&lt;td&gt;5396&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;8&lt;/td&gt;&#xA;&lt;td&gt;4554&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;9&lt;/td&gt;&#xA;&lt;td&gt;4893&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;10&lt;/td&gt;&#xA;&lt;td&gt;4698&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;11&lt;/td&gt;&#xA;&lt;td&gt;4816&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;12&lt;/td&gt;&#xA;&lt;td&gt;4259&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;13&lt;/td&gt;&#xA;&lt;td&gt;5004&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;14&lt;/td&gt;&#xA;&lt;td&gt;5015&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;15&lt;/td&gt;&#xA;&lt;td&gt;5033&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;16&lt;/td&gt;&#xA;&lt;td&gt;4906&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;17&lt;/td&gt;&#xA;&lt;td&gt;4770&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;18&lt;/td&gt;&#xA;&lt;td&gt;5100&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;Taking each one of these numbers mod 26:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 23 8 1 20 1 23 15 14 4 5 18 6 21 12 23 15 18 12 4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Converting these numbers into letters (1 to A and 26 to Z) gets us:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; WHATAWONDERFULWORLD&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;...the name of a song by:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;strong&gt;L&lt;/strong&gt;ouie &lt;strong&gt;A&lt;/strong&gt;rmstrong&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;em&gt;Mathematica&lt;/em&gt; code to solve the puzzle:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;FromCharacterCode[&#xA; Mod[Last /@ &#xA;    SortBy[Tally[&#xA;      Join @@ Import[&amp;quot;https://i.stack.imgur.com/feRTC.gif&amp;quot;, &#xA;        &amp;quot;RGBColorArray&amp;quot;]], First@ColorConvert[First@#, &amp;quot;HSB&amp;quot;] &amp;amp;], &#xA;   26] + 64]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;...and to generate similar puzzles:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;message = &amp;quot;ITSAWONDERFULLIFE&amp;quot;;&#xA;cover = &amp;quot;Homage\nto\nJS&amp;quot;;&#xA;n = StringLength[message];&#xA;colors = Round[List @@@ ColorConvert[Hue /@ (5/6 Rescale@Range@n), &amp;quot;RGB&amp;quot;], 1/255];&#xA;counts = ToCharacterCode[message] - 64;&#xA;{w, h} = Catch[Do[&#xA;           If[Mod[w h - Total[counts], 26] == 0, Throw[{w, h}]],&#xA;           {w, 300, 500}, {h, w - 2, w + 2}&#xA;         ]];&#xA;data = Rasterize[&#xA;         Row[Style[#, RGBColor @@ RandomChoice@colors, Bold] &amp;amp; /@ Characters@cover],&#xA;         &amp;quot;Data&amp;quot;, ImageSize -&amp;gt; {w, h}&#xA;       ] /. Append[Thread[255 colors -&amp;gt; Range[n]], {_, _, _} -&amp;gt; 0];&#xA;counts = Mod[counts - BinCounts[Join @@ data, {1, n + 1, 1}], 26];&#xA;zeros = Count[data, 0, {2}] - Total[counts];&#xA;counts += RandomSample@First@IntegerPartitions[zeros, {n}, Floor[zeros/n, 26]+{0, 26}, 1];&#xA;random = RandomSample[Join @@ MapThread[ConstantArray, {Range[n], counts}]];&#xA;i = 1;&#xA;data = data /. {0 :&amp;gt; random[[i++]]};&#xA;img = Image[data /. Thread[Range[n] -&amp;gt; colors]]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oP51O.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="69582" LastEditDate="2020-12-20T17:31:40.507" LastActivityDate="2020-12-20T17:31:40.507" CommentCount="23" ContentLicense="CC BY-SA 4.0" />
  <row Id="18318" PostTypeId="2" ParentId="18261" CreationDate="2015-07-24T02:26:43.243" Score="0" Body="&lt;p&gt;Is it 'hello'? I do use it in most days and without the head 'o' it's hell, which I'd say something about upon seeing. &lt;/p&gt;&#xA;" OwnerUserId="14256" LastActivityDate="2015-07-24T02:26:43.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18322" PostTypeId="2" ParentId="17004" CreationDate="2015-07-24T07:17:43.190" Score="8" Body="&lt;p&gt;The answer is that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the tumbleweed won the duel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Following on from &lt;a href=&quot;https://puzzling.stackexchange.com/a/17031/12511&quot;&gt;Chadi Abou Sleiman's&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/a/17738/12511&quot;&gt;Sir Parselot's&lt;/a&gt; answers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;** &lt;em&gt;The remainder will be comical. Excellent cake, seared deeply, sounds like it will start you on the right path.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The remainder is 789, from the previous answers. &quot;Comical&quot; may refer to the joke, but it also points us at comics. &quot;Sounds like it will start you&quot; tells us to look at the starting sounds of each word of &quot;Excellent cake, seared deeply&quot;, which gives us the letters XKCD. Combining these, we look at &lt;a href=&quot;http://www.xkcd.com/789/&quot; rel=&quot;nofollow noreferrer&quot;&gt;the XKCD comic numbered 789&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;** &lt;em&gt;At the end of the day, can you tell me who won the duel?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The comic referenced in the previous line shows a tumbleweed shooting the participants of an old west duel. Therefore, the winner of the duel is the tumbleweed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12511" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-07-24T07:17:43.190" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18323" PostTypeId="2" ParentId="18291" CreationDate="2015-07-24T09:59:21.543" Score="0" Body="&lt;p&gt;Strategy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first player is always at a disadvantage that gets worse for greater values, because the second player gets the first coins. After the first move, the second player will be the disadvantaged &quot;first&quot; player of a smaller game. Keeping this game as big as possible by dividing the pile as equally as possible (while also avoiding losing extra coins right off the bat) is the right strategy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;About the minimum number of coins:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The woman wins 15 more than twice our earnings. Letting the money gained in a move equal $x$, the money gained in the previous move can be $2x-1$, $2x$ or $2x+1$. If we write those amounts backwards, we get a series whose first and last terms belong to the second player (the woman) and that progresses like $x, 2x-1, 4x-1, 8x-3, 16x-5$, ... . The series must start from 1, so we can write 14*2+1 = 29 terms to fulfill the conditions. If we add up all of them, we'll get the minimum number of coins. When the series progresses like $x, 2x+1$, the next term also makes 1 more than the sum of all the previous terms. When the series progresses like $x, 2x-1$, it equals their sum. Therefore, the sum of all the terms would be the 30th term if the series continued. The even-ordered terms start with 1 and progress like $x, 4x+1$, ... . The 30th term would be the 15th even-ordered term and equal 4^14 + (4^14-1)/3 = 268435456 + 89478485 = 357913941. We win 119304642 and she wins 238609299.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="14016" LastEditDate="2015-07-24T11:17:19.353" LastActivityDate="2015-07-24T11:17:19.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18324" PostTypeId="2" ParentId="18311" CreationDate="2015-07-24T10:01:22.390" Score="2" Body="&lt;p&gt;My first is a dye, most commonly used.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; According to @pauld in the comments, that would be &lt;strong&gt;sulphur&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My second is twentieth, in a cup.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The twentieth letter is T, which in a cup could be &lt;strong&gt;tea&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My third is gloomy, shrouded in cloud. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gloomy, shrouded in cloud, and not in any way celestial (see comment by @CodeNewbie)... I have nothing for now.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fourth periodically gets to a half century.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The fiftieth element on the periodical table is &lt;strong&gt;tin&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fifth often stands for unity. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Unity means being as &lt;strong&gt;one&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Crack them all, and put them together. To find the answer, that perhaps I have left hither.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is where I'm unsure of things.&lt;br&gt;&#xA; Assuming the space between 3 and 4 is intentional, we're looking for two words, composed of the first letters of each clue.&lt;br&gt;&#xA; Cutting up Sulphur, Tea, and Moon, we can get to either &lt;strong&gt;Steam&lt;/strong&gt; or &lt;strong&gt;Stem&lt;/strong&gt;.&lt;br&gt;&#xA; Cutting up Tin and One, we can end up with either &lt;strong&gt;Ton&lt;/strong&gt; or &lt;strong&gt;Tone&lt;/strong&gt;.&lt;br&gt;&#xA; At this point, I'm fairly sure I've missed something, since none of the combinations seem to make sense.&lt;br&gt;&#xA; Steam ton sounds a bit like a unit of energy, and stem tone turns up a paper on linguistics...&lt;br&gt;&#xA; Since &quot;moon&quot; has been invalidated, this is obviously wrong.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13753" LastEditorUserId="13753" LastEditDate="2015-07-24T14:39:42.563" LastActivityDate="2015-07-24T14:39:42.563" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18325" PostTypeId="1" AcceptedAnswerId="18326" CreationDate="2015-07-24T11:30:11.303" Score="13" ViewCount="805" Body="&lt;p&gt;Here's my first attempt at a puzzle:&lt;/p&gt;&#xA;&lt;p&gt;A&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;My 11 from 12 sealed Atalanta's fate&lt;/p&gt;&#xA;&lt;p&gt;But what did my 7 expect me to crate?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;B&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;With a merged name between us and unusual style&lt;/p&gt;&#xA;&lt;p&gt;We have twice tried to win for the Emerald Isle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;C&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Commonly seen in a number of sports&lt;/p&gt;&#xA;&lt;p&gt;Frequently 3 can be worn in support.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;D&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Well defended from predators except when I roam&lt;/p&gt;&#xA;&lt;p&gt;From second-hand home to second-hand home.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;E&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;There's nothing from nothing and nobody's out&lt;/p&gt;&#xA;&lt;p&gt;Other than those when we last swapped about.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;F&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;When cannons appeared we were quickly retired,&lt;/p&gt;&#xA;&lt;p&gt;But the force like to use us where force is required.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;G&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Between Shakespeare and Sauce for 65 years&lt;/p&gt;&#xA;&lt;p&gt;Our voices are heard by myriad ears.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;H&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Digitless pets and fodder alike&lt;/p&gt;&#xA;&lt;p&gt;Beware where you crawl for some of us bite.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A means of propulsion for a friend of the hulk&lt;/p&gt;&#xA;&lt;p&gt;If someone gets yours you are likely to sulk.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;J&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;No further than Sheppey can our like be found&lt;/p&gt;&#xA;&lt;p&gt;As our 4 pairs of legs like the warmest of ground.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;K&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;How big or how heavy? On the hull of a boat&lt;/p&gt;&#xA;&lt;p&gt;Sometimes reptilian we hold all your notes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Questions&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Which one is missing, which name will you try&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;And which of the 8 calls us home this July?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-24T21:14:04.030" Title="More than a couple of couplets" Tags="|riddle|word|knowledge|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18326" PostTypeId="2" ParentId="18325" CreationDate="2015-07-24T13:44:08.020" Score="12" Body="&lt;p&gt;So with a little help from some friends in the comments, we have the theme and the answers to the couplets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But still not &lt;strong&gt;which of the 8 calls us home this July?&lt;/strong&gt; Any ideas? And why 8 and not 11 or more?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The couplets refer to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;11 of the 12 signs of the Zodiac. But there are only 11 couplets, and the missing sign is Aquarius (the water carrier).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My 11 from 12 sealed Atalanta's fate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what did my 7 expect me to crate?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Taurus (the bull). As explained by Kingrames, The 11th labour of Hercules was to steal the Apples of the Hesperides, which were later used by a suitor of Atalanta to beat her in a race and force her hand in marriage. The 7th labour was the Cretan &lt;strong&gt;Bull&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;B&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a merged name between us and unusual style&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have twice tried to win for the Emerald Isle.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gemini (the twins), with a reference to Jedward, twins who represented Ireland in Eurovision twice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Commonly seen in a number of sports&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Frequently 3 can be worn in support.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Leo (the lion). As in Detroit Lions, or the British and Irish Lions. England supporters wear three lions on their shirt.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;D: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well defended from predators except when I roam&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From second-hand home to second-hand home.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is Cancer (the crab)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;E&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's nothing from nothing and nobody's out&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other than those when we last swapped about.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is Virgo, the maiden. In cricket, a maiden over is one where no runs are scored, and also no wickets taken (otherwise it would be a wicket maiden). At the end of each over the fielding team has to swap bowlers (credit to @f'')&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;F: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When cannons appeared we were quickly retired,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the force like to use us where force is required.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is Aries (the ram). Cannons replaced Rams as a siege weapon, and the police use a ram (thanks to APrough)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;G&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Between Shakespeare and Sauce for 65 years&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our voices are heard by myriad ears.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is Sagittarius, the Archer. The Archers is a BBC radio soap opera, whose Pilot episode was broadcast in 1950. The series takes place in the fictional county of Borsetshire, which (according to wikipedia) is situated between the (in reality contiguous) counties of Worcestershire and Warwickshire. The town of Worcestor is famed for it's sauce, and Stratford-upon-Avon is in Warwickshire (thanks to @Going Hamateur for the tip)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;H&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Digitless pets and fodder alike&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beware where you crawl for some of us bite.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is Pisces (Fish)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A means of propulsion for a friend of the hulk&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone gets yours you are likely to sulk.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is Capricorn, the goat. Somebody can &quot;Get your goat&quot;. The friend of the hulk is Thor, who has a chariot pulled by war goats (thanks to @Bishop)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;J&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No further than Sheppey can our like be found&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As our 4 pairs of legs like the warmest of ground.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This should be Scorpio (the Scorpion) which has 8 legs and likes warm climates. According to wikipedia &quot;The isle [of Sheppey] is noted as the northern-most place to have an established scorpion population&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;K&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How big or how heavy? On the hull of a boat&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes reptilian we hold all your notes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is Libra (the Scales). Scales can weigh, reptiles have scales and there are musical scales.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7316" LastEditorUserId="7316" LastEditDate="2015-07-24T21:14:04.030" LastActivityDate="2015-07-24T21:14:04.030" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="18327" PostTypeId="1" AcceptedAnswerId="18329" CreationDate="2015-07-24T14:29:36.470" Score="19" ViewCount="1317" Body="&lt;p&gt;My little sister came up with this riddle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Preposition, adverb, and suffix too,&lt;br&gt;&#xA;  It relates in every situation to you.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Veggies of the ground or fruit of a brood,&lt;br&gt;&#xA;  Drop the first letter of the scrumptious food.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A master of writing and founder of fear,&lt;br&gt;&#xA;  Whose works are always lurking near.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The seeing spots on the face of a bird,&lt;br&gt;&#xA;  Is an epithet for the shortened word.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The word is important in farms or magazines,&lt;br&gt;&#xA;  It means what it says and says what it means.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll give it some time and if no one has solved it, I'll get her to write a hint.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck!&lt;/p&gt;&#xA;" OwnerUserId="14092" LastEditorUserId="4385" LastEditDate="2015-07-24T17:00:32.600" LastActivityDate="2015-07-24T21:34:01.653" Title="A rhyming riddle" Tags="|riddle|wordplay|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18328" PostTypeId="2" ParentId="18327" CreationDate="2015-07-24T15:00:24.603" Score="0" Body="&lt;p&gt;Still working on .. But I think I have got the first two.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Up&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Like&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;                                 E.g. Suffix - Whats'up? Sup?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Preposition - Ran Up Adverb - Looked up&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Preposition, adverb, and suffix too,&#xA;It relates in every situation to you. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Veggies of the ground or fruit of a brood,&#xA;Drop the first letter of the scrumptious food.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Melon -&gt; Elon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A master of writing and founder of fear,&#xA;Whose works are always lurking near.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Writing -&gt; Dean Koontz -&gt; Darkness&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The seeing spots on the face of a bird,&#xA;Is an epithet for the shortened word.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eyes -&gt; I's&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The word is important in farms or magazines,&#xA;It means what it says and says what it means.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9892" LastEditorUserId="9892" LastEditDate="2015-07-24T15:26:30.150" LastActivityDate="2015-07-24T15:26:30.150" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18329" PostTypeId="2" ParentId="18327" CreationDate="2015-07-24T16:38:37.310" Score="20" Body="&lt;p&gt;Okay, I am certain I have the answer, but I'm having trouble making some of the clues fit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;onomatopoeia&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Preposition, adverb, and suffix too,&lt;br&gt;&#xA;It relates in every situation to you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;on&quot; is a preposition, adverb, and suffix. Not sure about the second line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Veggies of the ground or fruit of a brood,&lt;br&gt;&#xA;Drop the first letter of the scrumptious food.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Tomato&quot; is the answer to the first line. Drop the &quot;t&quot; and you get &quot;omato&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A master of writing and founder of fear,&lt;br&gt;&#xA;Whose works are always lurking near.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Who else but Edgar Allen &quot;Poe&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The seeing spots on the face of a bird,&lt;br&gt;&#xA;Is an epithet for the shortened word.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Iowa is considered a &quot;flyover&quot; state, and someone flying over it would have a &quot;bird's eye&quot; view of it. The abbreviation for Iowa is &quot;IA&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The word is important in farms or magazines,&lt;br&gt;&#xA;It means what it says and says what it means.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This stanza is a reference to onomatopoeia as a whole. Farm animal noises are described using onomatopoeia (&quot;Pig goes oink, cow goes moo&quot;), and the second line is basically a straight definition of onomatopoeia! Not sure about magazines, though, unless it's a reference to gun magazines and sounds like &quot;Bang!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The clues I'm not sure about are &quot;It relates in every situation to you&quot;, &quot;fruit of a brood&quot;, &quot;Whose works are always lurking near&quot;, and &quot;magazines&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help on filling in blanks would be appreciated, but I would be shocked if this isn't the answer.&lt;/p&gt;&#xA;" OwnerUserId="10861" LastEditorUserId="10861" LastEditDate="2015-07-24T21:34:01.653" LastActivityDate="2015-07-24T21:34:01.653" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="18330" PostTypeId="2" ParentId="18291" CreationDate="2015-07-24T17:34:55.040" Score="0" Body="&lt;p&gt;Strategy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Treat the problem as a binary tree where the root node is the last splitting of the pot (1 coin), and its two children are composed of the previous pot and the previous split (split is either equal to the pot or the pot plus one coin), then perform either a breadth first search or depth first search for the solution (brute force). Take the minimum value found in the set.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Result:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'd written some code in C# and found that the solution is found at a height of 28, where her split is 238609299 and your split is 119304642. Other solutions were once again found at a height of 30 (and so on), but were closer to 1 billion (and up) for her split of the pot. The first time I tried writing it as a breadth first search, but ran out of memory. The second time was depth first and provided the solution. I capped the height to 32 to give me a bearable run-time. That solution is the only one found at the height of 28, and the first solution found.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9696" LastActivityDate="2015-07-24T17:34:55.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18331" PostTypeId="2" ParentId="18311" CreationDate="2015-07-24T17:48:01.907" Score="1" Body="&lt;p&gt;CodeNewbie said that three of the following are correct: 1, sulfur; 2, tea; 4, tin; 5, one. Using this progress, I make a stab at the riddle's construction...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The lines represent syllables, not letters. &lt;br&gt;&#xA; 1-3 are a three-syllable word, and 4-5 are a two-syllable word&lt;br&gt;&#xA; So we ditch sulfur for now, and keep the three single-syllable words.&lt;br&gt;&#xA; To replace 1, I suggest &quot;&lt;strong&gt;red&lt;/strong&gt;&quot; (historically the most common dye, probably not &quot;plum&quot;) &lt;br&gt;&#xA; For 3, I'm thinking along the lines of &quot;&lt;strong&gt;smog&lt;/strong&gt;&quot;, but I don't see how that fits well for making part of a word. &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now just MadGab them and ta-da!!!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Readysmawg Tinwaun&lt;/strong&gt;, the mythological elvish riddler! Perhaps what was &quot;left hither&quot; was one of his very own riddles from long, long ago! Bravo, sir!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now for some reason, his name doesn't turn up any results, so I sent a request to UrbanDictionary...  should be approved any day now...&lt;/p&gt;&#xA;" OwnerUserId="13237" LastActivityDate="2015-07-24T17:48:01.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18332" PostTypeId="1" CreationDate="2015-07-24T18:05:31.160" Score="1" ViewCount="326" Body="&lt;p&gt;It's the year 2034 and you're a mission coordinator for the latest Mars mission. It has been a long haul but since the send off on January 12th the antimatter rocket has been a complete success. Your team just received the first transmission post landing sequence but you think something seems a little off about it. Can you figure out if and what the problem is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transmission follows&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Hello Mission Control - 13.05.2034&#xA;WĔ have hađ absoluȶĔly no trouble at all, ĵust senдing our first notification ŧransmission. &#xA;Planetary Orbit was successfully achieved and hopefully the UN will approve the next &#xA;shipment soon. &#xA;The last 2 weeks have been a real ordeal but the next 4 will be vitally important. &#xA;ᥑl&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Hint 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It will help to arrange all of the things that are wrong in the message. &lt;br/&gt;&#xA;Ex. &lt;br/&gt;&#xA;Non-ascii characters&lt;br/&gt;&#xA;A notice about orbit being successful when you know the rocket has already gone through landing.&lt;br/&gt;&#xA;An antimatter rocket shouldn't take four months to travel to Mars&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;What passage in the message might make certain parts of unicode numbers important?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13975" LastEditorUserId="13975" LastEditDate="2015-07-27T20:27:44.933" LastActivityDate="2015-07-27T20:29:43.447" Title="Mars mission trouble?" Tags="|cipher|" AnswerCount="1" CommentCount="10" FavoriteCount="0" ClosedDate="2015-07-29T17:05:25.053" ContentLicense="CC BY-SA 3.0" />
  <row Id="18333" PostTypeId="1" AcceptedAnswerId="18348" CreationDate="2015-07-24T18:17:29.950" Score="8" ViewCount="326" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;We’re the Greatest of our siblings&lt;br&gt;&#xA;  Always best since days of yore&lt;br&gt;&#xA;  Our Least brother’s always nibbling&lt;br&gt;&#xA;  Now and then he gets one more  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;First and last are far yet near&lt;br&gt;&#xA;  Fourth and fifth together chained&lt;br&gt;&#xA;  One through four are strange or queer&lt;br&gt;&#xA;  Five, six, seven are level planes  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Flavors are our favorite number&lt;br&gt;&#xA;  It describes our special quirk&lt;br&gt;&#xA;  With these clues you’re now encumbered&lt;br&gt;&#xA;  Knuckle down and get to work!  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who are the seven Greatest siblings? Who is their Least brother (he’s not one of the seven)?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="10861" LastEditorUserId="10861" LastEditDate="2015-07-24T18:57:45.097" LastActivityDate="2015-07-24T21:34:51.143" Title="The Greatest Siblings" Tags="|riddle|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18335" PostTypeId="1" AcceptedAnswerId="18336" CreationDate="2015-07-24T18:34:18.630" Score="-2" ViewCount="267" Body="&lt;p&gt;Here's a quick and easy one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What comes next in this sequence?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;6 2 5 5 4 5 6 3 7 ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10861" LastActivityDate="2015-07-24T18:50:22.817" Title="How fast can you solve it?" Tags="|number-sequence|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18336" PostTypeId="2" ParentId="18335" CreationDate="2015-07-24T18:36:14.253" Score="2" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;6&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The explanation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each number in the sequence corresponds to the number of segments needed to display a digit on a calculator's screen.&lt;br&gt;&#xA; &lt;code&gt;Digits:   0 1 2 3 4 5 6 7 8 9&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;Sequence: 6 2 5 5 4 5 6 3 7 6&lt;/code&gt;&lt;br/&gt;&#xA;  | &amp;nbsp;&amp;nbsp;The 1 here has 2 segemnts&lt;br/&gt;&#xA;  |&lt;br/&gt;&lt;br/&gt;&#xA;  |_| The 4 has 4 segments&lt;br/&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;|&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13975" LastEditorUserId="13975" LastEditDate="2015-07-24T18:50:22.817" LastActivityDate="2015-07-24T18:50:22.817" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18340" PostTypeId="2" ParentId="18332" CreationDate="2015-07-24T19:19:50.423" Score="4" Body="&lt;p&gt;China, US and Europe are getting along fine, but US/Europe are not getting along with Russia. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ĕ unicode u+0114 Latin (US Europe)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;đ unicode u+0111 Latin (US Europe)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ȶ unicode u+0236 Latin (US Europe)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ĵ unicode u+0135 Latin (US Europe)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;д unicode u+0434 Cyrillic (Russia)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ŧ unicode u+0167 Latin (US Europe)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ᥑ unicode u+1951 Tai Le (China)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see in the transmission that 'ᥑl' the author (China) has had no trouble with the others 'WĔ have hađ absoluȶĔly' (US/Europe), but 'ĵust/ŧransmission' (US/Europe) is not getting along with 'senдing' (Russia).&lt;/p&gt;&#xA;" OwnerUserId="13788" LastEditorUserId="13788" LastEditDate="2015-07-24T19:33:42.133" LastActivityDate="2015-07-24T19:33:42.133" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18341" PostTypeId="2" ParentId="10911" CreationDate="2015-07-24T19:28:51.027" Score="1" Body="&lt;p&gt;I apologize for making everyone wait so long. The correct answer, in fact, is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Duct tape.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure the rest is pretty self-explanatory, but still, these are the descriptions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My name sounds like a bird&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Duct tape sounds like duck tape. (There's even a company that sells duct tape called Duck Tape.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whole body is like tongue of a lizard&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Duct tape has a sticky side (lizard tongues are sticky)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Am different colors&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Duct tape is avaliable in different colors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I well can be very littered&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People litter pieces of duct tape all the time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;br&gt;&#xA;My energy melts over time&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Duct tape loses its stickiness over time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But not quite like an ice&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This line is the clue that the word 'melt' is used as a metaphor in the previous line&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And when not taken care of&lt;br&gt;&#xA;I'll collect dust and lice&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If the sticky side is left, eventually it collects dust particles and hair&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;br&gt;&#xA;Thank you to all who participated in my riddle!&lt;/p&gt;&#xA;" OwnerUserId="10603" LastActivityDate="2015-07-24T19:28:51.027" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18342" PostTypeId="1" AcceptedAnswerId="18361" CreationDate="2015-07-24T19:30:11.557" Score="10" ViewCount="743" Body="&lt;p&gt;The answer to this riddle is a &quot;Top 9 List&quot; and you have to figure out which one. For example, it could be the top 9 longest rivers in the world, or the top 9 largest non-extinct land animals, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of the nine lines below hints at one object. They are actual objects (not concepts or ideas).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The riddle is very literal.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My first is part of most tsunamis.&lt;br&gt;&#xA;My second is scattered in mastodon remains.&lt;br&gt;&#xA;A venomous snake hides my third.&lt;br&gt;&#xA;You see my fourth in a jumpy biter.&lt;br&gt;&#xA;Maulers and smashers show my fifth.&lt;br&gt;&#xA;An imperfect fury reveals my sixth.&lt;br&gt;&#xA;A vile spiritualist divulges my seventh.&lt;br&gt;&#xA;My eighth is spotted in Pithecanthropus fossils.&lt;br&gt;&#xA;Oxygen desaturation tells you my ninth.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Together, in the specified order, what &quot;Top 9 List&quot; am I?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-07-25T16:05:22.970" LastActivityDate="2015-08-04T14:09:38.393" Title="What &quot;Top 9 List&quot; Am I?" Tags="|riddle|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18343" PostTypeId="1" AcceptedAnswerId="21940" CreationDate="2015-07-24T19:59:33.617" Score="12" ViewCount="1374" Body="&lt;p&gt;Bwahaha! You're not the only one who can make these puzzles, JLee!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(this is an unofficial continuation of JLee's [in]famous &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=user%3Ajlee%20is%3Aquestion%20title%3Aphrase&quot;&gt;Word/Phrase puzzles&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following phrases of varying lengths follow a certain rule, or they do not. If one follows the rule, I call it a Donut Phrase™! If not, I suppose you could call it something else. See if you can find out the link between all the Donut Phrases™.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Donut Phrases™&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Have you ever danced with the devil in the pale moonlight?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Four score and seven years ago...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;We need a bigger boat.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I'm not left-handed either.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Please sir, can I have some more?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I'm gonna make him an offer he can't refuse.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Avengers, assemble!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Punch the keys for god's sake!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;If I say it's safe to surf this beach, it's safe to surf this beach!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;This could be the beginning of a beautiful friendship.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yes I said yes I mean yes!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;KHAN!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;War. War never changes.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;As the suns fade, so dies humanity's hope.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;nope.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;You can't handle the truth!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Don't tease the octopus, kids!&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But here are some phrases that are NOT Donut Phrases™...&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Can't stop me now&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I have a bad feeling about this.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I am not left handed.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;You're the man now, dog&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Captain Jean-luc Picard&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;just kept talking in one long incredibly unbroken sentence moving from topic to topic so that no one had a chance to interrupt it was really quite hypnotic&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I love the smell of napalm in the morning.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I must not fear. Fear is the mind killer. Fear is the little death, that brings total obliteration. I will face my fear. I will permit it to pass over me and through me, and when it has gone past I will turn the inner eye to see its path. Where the fear has gone, there will be nothing. Only I will remain.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Not sure if puzzle, or just collection of memes.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;No.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I will leave you as you left me, Kirk. Alone, at the center of a dead planet. Buried alive.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yup.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I am the very model of a modern major general, I've information animal, vegetable, and mineral, I know the kings of England and all the fights historical, from Marathon to Waterloo in order categorical&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I want the truth.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;You spoony bard!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I'm a Barbie girl, in a Barbie world.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Bonjour, comment tu t'appelles?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I used a program from the code golf stack exchange site (which is here: &lt;a href=&quot;https://codegolf.stackexchange.com/&quot;&gt;https://codegolf.stackexchange.com/&lt;/a&gt;) to quickly verify these phrases, but in order to do so, I had to make a few modifications to the phrases. (If you're on the right track, those modifications will seem very obvious)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2 (this is a big one):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Each phrase's status is determined by its relationship to a single letter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Final Hint: (seriously this gives away quite a bit)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The letter you're looking for is different for each phrase. And if you compare the shorter phrases, you might not even need a calculator.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Even more Finaler hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;When finding &quot;the donut hole,&quot; you need to round off to the nearest letter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's a picture to make side by side comparisons easier:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/3ps3t.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3ps3t.png&quot; alt=&quot;Donut Phrases™&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13068" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2016-11-07T23:44:37.423" Title="What is a Donut Phrase™?" Tags="|pattern|calculation-puzzle|word-property|" AnswerCount="1" CommentCount="30" ContentLicense="CC BY-SA 3.0" />
  <row Id="18344" PostTypeId="1" AcceptedAnswerId="18349" CreationDate="2015-07-24T20:00:17.337" Score="13" ViewCount="2675" Body="&lt;p&gt;Grasshopper, in order to know the answer you must first figure out the question.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;H, H, D, P, ?                     (2)&#xA;P, L, T, Q, P, H, H, ?            (5)&#xA;R, C, C, U, B, A, ?               (9)&#xA;A, K, Q, J, T, N, E, ?           (11)&#xA;M, V, E, M, J, S, ?              (14)&#xA;D, P, N, G, C, M, ?              (15)&#xA;M, K, S, P, S, ?                 (16)&#xA;Y, C, A, G, ?                    (18)&#xA;A, T, G, C, L, V, ?              (48)&#xA;Z, O, T, T, F, F, S, S, E, N, ? (112)  &#xA;H, H, L, B, B, C, ?             (613)&#xA;A, A, N, S, A, ?             (371017)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is 3 letters long&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2 (Bounty Hint):  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;H, H, D, P, ?, Z.                     (2)  &#xA;M, K, S, PH, S, ?, DP, DD, DQ...     (16)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Also anyone who has a better title LMK&lt;/p&gt;&#xA;" OwnerUserId="14092" LastEditorUserId="14092" LastEditDate="2015-07-27T01:23:34.367" LastActivityDate="2015-07-29T12:46:24.367" Title="A Single Letter is missing from each, can you figure it out to find the answer and the question?" Tags="|lateral-thinking|letter-sequence|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18345" PostTypeId="1" AcceptedAnswerId="18355" CreationDate="2015-07-24T20:08:38.307" Score="5" ViewCount="645" Body="&lt;p&gt;This is part of &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=title%3A%22The%20Twenty%20Doors%22&quot;&gt;The Twenty Doors&lt;/a&gt; series.&lt;br&gt;&#xA;The previous one is &lt;a href=&quot;https://puzzling.stackexchange.com/questions/17605/the-twenty-doors-room-6&quot;&gt;The Twenty Doors! (ROOM 6)&lt;/a&gt;.&lt;br&gt;&#xA;The next one is &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18498/the-twenty-doors-room-8&quot;&gt;The Twenty Doors (ROOM 8)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You (after many days rest in &lt;a href=&quot;https://puzzling.stackexchange.com/questions/17605/the-twenty-doors-room-6?noredirect=1#comment57757_17605&quot;&gt;Room 6&lt;/a&gt;) finally go into Room 7. The first thing you see is the (rather long) hint on the wall:&lt;br&gt;&#xA;&lt;code&gt;The Steckerbrett is completely reversed, an true enigma. M3, of course. AAA AAA. 4. NFMN JRLS KKYV FAIU MRGG KPQV CNGJ ABRQ FVOG VKPO PYCY BYOI NGVA NONW DTKO TTBJ EJCI NIGU YEPB UUPM IEZH PMEN JZVG LYLW&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You then pick up the slip of paper.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Well done for getting this far. &lt;br&gt; Uqdok pgd Jfgr-qsnmpmmh Esnzjf Uhfmo Jqbuypmbj Nytc qk vhzylbo wjzxusv.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And last, but not least, the keypad:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;[«][±][¶] &#xA;  [ENTER] &lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which symbol should you press in order to get into the next room?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next door will be added when this door is solved!&lt;/p&gt;&#xA;" OwnerDisplayName="user9377" LastEditorUserId="-1" LastEditorDisplayName="user9377" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-07-30T19:36:43.100" Title="The Twenty Doors (ROOM 7)" Tags="|cipher|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="18347" PostTypeId="1" AcceptedAnswerId="19815" CreationDate="2015-07-24T21:10:55.043" Score="14" ViewCount="1632" Body="&lt;p&gt;Your good friend told you about a wise man named &lt;strong&gt;The Oracle&lt;/strong&gt;, who once helped them &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15725/discover-the-meaning-of-life&quot;&gt;discover the meaning of life&lt;/a&gt;. You've led a content life, but you're dying to discover the secret to true happiness. Because of this, you've trekked through jungle and tundra, across mountaintops and deserts, into giant caverns and beneath the ocean to find &lt;strong&gt;The Oracle&lt;/strong&gt;. Finally, after all your hard work, you find him. He sits alone in an opulent palace, never moving, never blinking, allowing each visitor to ask one question of him. You take a moment to catch your breath after climbing the steps to his throne, gather yourself, and ask:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;What is the secret to eternal happiness?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You already have your pen and paper out. You were warned about the nature of his responses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You hastily scrawl down each word as The Oracle cryptically answers your question. After he finishes speaking, you read back over what he's written. &quot;This doesn't make any sense,&quot; you muse in frustration. &quot;I don't understand.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Oracle simply smiles.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A man of prowess hides them both&lt;br&gt;&#xA;preparing for his calling.&lt;br&gt;&#xA;Another man provides them both&lt;br&gt;&#xA;with sparks and cinders falling.&lt;br&gt;&#xA;Two snakes slither, silently&lt;br&gt;&#xA;surrounding unaware&lt;br&gt;&#xA;a line about what's lost between&lt;br&gt;&#xA;triumph and moving air.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A penciled confirmation&lt;br&gt;&#xA;performs music to the dance&lt;br&gt;&#xA;of clocks and creepy crawly things&lt;br&gt;&#xA;at rest inside your pants.&lt;br&gt;&#xA;A sudden motion shocks you&lt;br&gt;&#xA;in a thousand different ways.&lt;br&gt;&#xA;In other words, the seventh note&lt;br&gt;&#xA;sings Casey's gone astray.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The dispossessed are missing one&lt;br&gt;&#xA;to become shrouded skies.&lt;br&gt;&#xA;Perform on it this action&lt;br&gt;&#xA;as you do with dotted eyes.&lt;br&gt;&#xA;The aforementioned snakes return,&lt;br&gt;&#xA;though this time stick together.&lt;br&gt;&#xA;A bird of death approaches them&lt;br&gt;&#xA;with blackened, shedding feathers.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Unseen to the untrained eye&lt;br&gt;&#xA;vast treasures within lie,&lt;br&gt;&#xA;and bones are ever rattling&lt;br&gt;&#xA;as good men herein die.&lt;br&gt;&#xA;A waterfall appears from fears&lt;br&gt;&#xA;no longer held within,&lt;br&gt;&#xA;and paper tiger cardinals&lt;br&gt;&#xA;let waterfalls begin.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In pairs, the strangers take the rear&lt;br&gt;&#xA;as giants lead the way.&lt;br&gt;&#xA;Manipulate them slightly&lt;br&gt;&#xA;and your answer's clear as day.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Can you discover the secret to eternal happiness from The Oracle's message?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Okay, fine, fine, I'll add the tag.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A small hint to aid anyone trying to solve this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not all of each stanza is cryptic-style. Also, each stanza corresponds to one word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Aaaaand here's some incentive:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There &lt;em&gt;will&lt;/em&gt; be a bounty to the correct answer. Anyone with questions/clarifications can feel free to either post an answer, post a comment, or ping me in chat, and I'll be sure to respond.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-08-18T16:20:02.623" Title="Discover the secret to eternal happiness!" Tags="|riddle|rhyme|enigmatic-puzzle|cryptic-crosswords|" AnswerCount="5" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="18348" PostTypeId="2" ParentId="18333" CreationDate="2015-07-24T21:11:44.173" Score="7" Body="&lt;p&gt;The seven are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; January, March, May, July, August, October, and December&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The least is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; February&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We’re the Greatest of our siblings&lt;br&gt;&#xA;Always best since days of yore&lt;br&gt;&#xA;Our Least brother’s always nibbling&lt;br&gt;&#xA;Now and then he gets one more  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Greatest since they have 31 days. February has the least days with 28, but sometimes has 29 on a leap year.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First and last are far yet near&lt;br&gt;&#xA;Fourth and fifth together chained&lt;br&gt;&#xA;One through four are strange or queer&lt;br&gt;&#xA;Five, six, seven are level planes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; January and December are 1 and 12, but January is one month from December. Edit from the comment: Four and five, July and August, are the seventh and eight months, so they're chained together. One through four are the first, third, fifth, and seventh months, which are odd. Perhaps August, October, and december are equally separated by two months, so they're level planes; or they're all even (level) at eight, ten, and twelve.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Flavors are our favorite number&lt;br&gt;&#xA;It describes our special quirk&lt;br&gt;&#xA;With these clues you’re now encumbered&lt;br&gt;&#xA;Knuckle down and get to work!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 31 flavors is a reference to Baskin Robins' flavors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14268" LastEditorUserId="14268" LastEditDate="2015-07-24T21:34:51.143" LastActivityDate="2015-07-24T21:34:51.143" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18349" PostTypeId="2" ParentId="18344" CreationDate="2015-07-24T21:12:29.963" Score="19" Body="&lt;p&gt;Alright, I'm going to start a partial answer, we'll see if it goes anywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Missing letters:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Thanks to &lt;a href=&quot;https://puzzling.stackexchange.com/users/1565/minitech&quot;&gt;minitech&lt;/a&gt; for number 6! Thanks to &lt;a href=&quot;https://puzzling.stackexchange.com/users/5188/len&quot;&gt;Len&lt;/a&gt; for number 1! Thanks to &lt;a href=&quot;https://puzzling.stackexchange.com/users/1563/geobits&quot;&gt;Geobits&lt;/a&gt; for number 7!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1. H (Hestia, Hades, Demeter, Poseidon, &lt;strong&gt;Hera&lt;/strong&gt;) - Cronus and Rhea's Children&lt;br&gt;&#xA;2. O (Point, Line, Triangle, ..., Heptagon, &lt;strong&gt;Octagon&lt;/strong&gt;) - Polygon by number of sides&lt;br&gt;&#xA;3. I (Russia, Canada, China, United States, Brazil, Australia, &lt;strong&gt;India&lt;/strong&gt;) - Countries by Area&lt;br&gt;&#xA;4. S (Ace, King, Queen, Jack, ..., Eight, &lt;strong&gt;Seven&lt;/strong&gt;) - Cards&lt;br&gt;&#xA;5. U (Mercury, Venus, Earth, ..., Saturn,  &lt;strong&gt;Uranus&lt;/strong&gt;) - Planets&lt;br&gt;&#xA;6. M (Delaware, Pennsylvania, New Jersey, ..., Massachusetts, &lt;strong&gt;Maryland&lt;/strong&gt;) - US States by Ratification Date&lt;br&gt;&#xA;7. B (McDonalds, KFC, Subway, Pizza Hut, Starbucks, &lt;strong&gt;Burger King&lt;/strong&gt;) - Largest Fast Food Chains in the World by number of stores&lt;br&gt;&#xA;8. R (Yankees, Cardinals, Athletics, Giants, &lt;strong&gt;Red Sox&lt;/strong&gt;) - Most World Series titles&lt;br&gt;&#xA;9. L (Aries, Taurus, Gemini, Cancer, Leo, Virgo, &lt;strong&gt;Libra&lt;/strong&gt;) - Signs of the zodiac&lt;br&gt;&#xA;10. T (Zero, One, Two, Three, ..., Nine, &lt;strong&gt;Ten&lt;/strong&gt;)  - Counting to ten&lt;br&gt;&#xA;11. N (Hydrogen, Helium, Lithium, ..., Carbon, &lt;strong&gt;Nitrogen&lt;/strong&gt;) - Periodic Table Elements&lt;br&gt;&#xA;12. E (Asia, Africa, North America, South America, Antarctica, &lt;strong&gt;Europe&lt;/strong&gt;) - Continents by Area  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cleaning up the list and adding the numbers gives us:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;H      2&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;O      5&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;I      9&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;S     11&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;U     14&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;M     15&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;B     16&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;R     18&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;L     48&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;T    112&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;N    613&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;E 371017&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;EDIT: From &lt;a href=&quot;https://puzzling.stackexchange.com/users/11876/deusovi&quot;&gt;Deusovi's&lt;/a&gt; answer below&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Deciphered the numbers based off VictorHenry's answer! All I had to do was...&#xA;add a couple suggestive spaces:&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;H         2&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;O         5&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;I         9&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;S        11&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;U        14&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;M        15&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;B        16&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;R        18&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;L       4 8&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;T      1 12&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;N      6 13&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;E 3 7 10 17&lt;/code&gt;&lt;br&gt;&lt;br&gt;&#xA; Each set of numbers tells us where that letter is located in the final phrase:&lt;br&gt;&#xA; &lt;code&gt;1 2 3   4 5 6 7 8 9101112  131415161718&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;T H E   L O N E L I E S T   N U M B E R&lt;/code&gt;&lt;br&gt;&#xA; So the final answer is &lt;strong&gt;ONE!&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10861" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-07-29T12:46:24.367" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="18350" PostTypeId="1" AcceptedAnswerId="18352" CreationDate="2015-07-24T21:45:17.273" Score="5" ViewCount="2247" Body="&lt;p&gt;This puzzle follows from the well known sand timer puzzles:&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/2030/a-sand-timer-challenge&quot;&gt;A Sand Timer Challenge&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.crazyforcode.com/sand-timer-puzzle/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.crazyforcode.com/sand-timer-puzzle/&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.cut-the-knot.org/hg_solution.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.cut-the-knot.org/hg_solution.shtml&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Warmup:&lt;/strong&gt;&lt;br&gt;&#xA;You are given a timer of 4 minutes, and a timer of 7 minutes.  The time starts when the first timer is flipped.  Is there a time for which all larger integer values of time can be measured?  Prove why or why not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Easy:&lt;/strong&gt;&lt;br&gt;&#xA;Same rules as above, but now your timers are 5 minutes and 8 minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Medium:&lt;/strong&gt;&lt;br&gt;&#xA;Same rules as above, but now your timers are 6 minutes and 11 minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hard:&lt;/strong&gt;&lt;br&gt;&#xA;Same rules as above, but now your timers are $m$ minutes and $n$ minutes, where $m&amp;gt;n&amp;gt;0$ and $m,n \in\mathbb{Z}$.  (Basically find a generalization.  If you want to solve this first be my guest...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you couldn't understand what I was asking.&lt;br&gt;&#xA;Let $T$ be the set of all times that can be timed using timers $m$ and $n$.&lt;br&gt;&#xA;Find $s\in T:s-1\notin T,\forall x&amp;gt;s:x\in\mathbb{Z},x\in T$.&lt;/p&gt;&#xA;" OwnerUserId="14092" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2023-02-27T23:28:50.880" Title="An Hourglass Puzzle" Tags="|mathematics|calculation-puzzle|time|" AnswerCount="5" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="18352" PostTypeId="2" ParentId="18350" CreationDate="2015-07-24T22:46:12.517" Score="3" Body="&lt;p&gt;Taking a huge risk, but hey...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;W/U:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Times in the form of $4n + 7m$ can be measured. As long as a number is large enough, you can add 2*4-7 = 1 if the 7-minute hourglass is to be used at least once to measure the previous time, or 3*7 - 5*4 = 1 if the 4-minute hourglass is to be used at least four times (for at least 20 mins!). At our most conservative guess, it's 27, but since values greater than 19 will either have an $m$ that's at least 1 or $n$ that's at least 5, 20 will do as as a slightly less-conservative one. Also, 7+(7-4) = 10 and 4*2 + 4*2 (mod 7) = 9 minutes can be measured. 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7 are reachable while 6 isn't, so $T$=6.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;E:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If it's $5n + 8m$, it can be measured. As long as a number is large enough, you can add 5*5 - 3*8 = 1 if the 8-minute hourglass is to be used at least thrice to measure the previous time, or 2*8 - 3*5 = 1 if the 5-minute hourglass is to be used at least thrice. At our most conservative guess, the first $5n + 8m$ number whose $n$ and $m$ equal 3 (39) will do. However, 8+(8-5) = 11,  mins can also be measured. 38, 37, 36, 35, 34, 33, 32, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18 are reachable while 17 isn't. 5*2 + 5*2 (mod 8) = 12 mins can also be measured, so that makes 17 also reachable. 16, 15 are measurable, and so is 14 (see the image in the comments section), 13, 12, 11 and 10. 9 isn't, so $T$=9.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;M:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For $6n + 11m$, it can be measured. As long as a number is large enough, you can add 6*2 - 11*1 = 1 if the 11-minute hourglass is to be used at least once to measure the previous time, or 11*5 - 6*9 = 1 if the 6-minute hourglass is to be used at least nine times. At our most conservative guess, it's 54+11 = 65, but since values greater than 53 will either have an $m$ that's at least 1 or $n$ that's at least 9, 54 will do as a slightly less-conservative one. However, 11+(11-6) = 16 mins can also be measured. Then again, if we start out flipping any hourglass as soon as it empties, only 1 minute's worth of sand will remain in the upper half of the 6-min hourglass after 11 mins, while the other empties. To measure any time longer than that, we can also flip the bigger one, then flip both when the smaller one empties, then do it again when the bigger one does, and so on. 11 can be measured while 10 can't, so $T$=10.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;H (incomplete):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As long as $n$ and $m$ are co-primes, $k$ and $j$ values to satisfy $nk$ ≡ 1 (mod $m$) and $mj$ ≡ 1 (mod $n$) can be found. In addition, $mx$ + $mx$ (mod $n$) and $ny$ + $ny$ (mod $m$) values can be used to measure time. Also, if there's some amount of sand left in only one of the hourglasses in the middle of a measurement, the time this sand corresponds to can be continuously added by flipping the other at the time, then flipping both whenever one empties, ... (rinse and repeat), making $mx$ + $rmx$ (mod $n$) if the latter without r is less than $m$ and $ny$ + $sny$ (mod $m$) if the latter without s is less than $n$ also measurable. That means the smaller one of the aforementioned $nk$ and $mj$ ($(n,m)max*[(n,m)min-1]$ at most) is enough for the conservative limit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;VERY LATE EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just wrote a Java program to find the generalized solution:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;package hourglass;&#xA;&#xA;import java.util.ArrayList;&#xA;import java.util.List;&#xA;&#xA;public class Unreachables {&#xA;&#xA;public static void main(String[] args) {&#xA;    // x = n, y = m&#xA;int x = 7;&#xA;int y = 9;&#xA;&#xA;int MaxUnreachable = 0;&#xA;&#xA;int ConsLim1 = 0;&#xA;int ConsLim2 = 0;&#xA;&#xA;int ub1 = 1;&#xA;int ub2 = 1;&#xA;int SecGen = 0;&#xA;&#xA;List&amp;lt;Integer&amp;gt; FirstGenList = new ArrayList&amp;lt;Integer&amp;gt;();&#xA;List&amp;lt;Integer&amp;gt; SecGenList = new ArrayList&amp;lt;Integer&amp;gt;();&#xA;List&amp;lt;Integer&amp;gt; ReachablesList = new ArrayList&amp;lt;Integer&amp;gt;();&#xA;&#xA;SecGenList.add(0);&#xA;&#xA; for (int i=0; i&amp;lt;y; i++) {&#xA;    if ((i*x)%y == 1) {&#xA;        ConsLim1 = i*x;&#xA;    }&#xA;    if ((i*y)%x == 1 &amp;amp;&amp;amp; i&amp;lt;x) {&#xA;        ConsLim2 = i*y;&#xA;    }               &#xA; }&#xA;&#xA; int LesserConsLim = Math.min(ConsLim1, ConsLim2);&#xA;&#xA;int MaxXFac = LesserConsLim/x;&#xA;int MaxYFac = LesserConsLim/y;  &#xA;&#xA;     ///////////////////////////////////////////////////////////&#xA;  for (int j=0; j&amp;lt;=MaxXFac; j++) {&#xA;    for (int k=0; k&amp;lt;=MaxYFac; k++) { &#xA;        if (j*x + k*y &amp;lt; LesserConsLim) {&#xA;            FirstGenList.add(j*x + k*y);&#xA;        }&#xA;    }&#xA;  }&#xA;&#xA;    ///////////////////////////////////////////////////////////&#xA;  for (int l=0; l&amp;lt;MaxXFac; l++) {&#xA;    int base = l*x;&#xA;    int r = (l*x)%y;&#xA;&#xA;     if (r&amp;lt;x) {&#xA;      for (int m=0; m&amp;lt;ub1; m++) {&#xA;        SecGen = base + r*m;&#xA;        if (SecGen&amp;lt;LesserConsLim &amp;amp;&amp;amp; SecGen &amp;gt; 0) {&#xA;            SecGenList.add(SecGen);&#xA;            ub1++;&#xA;        }&#xA;      }&#xA;     }              &#xA;   }&#xA;&#xA;  ub2 = 1;&#xA;  for (int n=0; n&amp;lt;MaxYFac; n++) {&#xA;    int base = n*y;&#xA;    int t = (n*y)%x;&#xA;&#xA;     if (t&amp;lt;y) {&#xA;      for (int o=0; o&amp;lt;ub2; o++) {&#xA;        SecGen = base + t*o;&#xA;        if (SecGen&amp;lt;LesserConsLim &amp;amp;&amp;amp; SecGen &amp;gt; 0) {&#xA;            SecGenList.add(SecGen);&#xA;            ub2++;&#xA;        }&#xA;      }&#xA;     }              &#xA;   }&#xA;&#xA;  for (int p=0; p&amp;lt;FirstGenList.size(); p++){&#xA;      for (int r=0; r&amp;lt;SecGenList.size(); r++){&#xA;          int PotRes = FirstGenList.get(p)+SecGenList.get(r);&#xA;           if (PotRes&amp;lt;LesserConsLim &amp;amp;&amp;amp; ReachablesList.indexOf(PotRes)&amp;lt;0){&#xA;               ReachablesList.add(PotRes);  &#xA;           } &#xA;      }&#xA;  }&#xA;&#xA;  for (int z=LesserConsLim-1; z&amp;gt;0; z--){&#xA;      if (ReachablesList.indexOf(z)&amp;lt;0){&#xA;          MaxUnreachable = z;&#xA;          break;&#xA;          }       &#xA;      }&#xA;&#xA;&#xA;  System.out.println(MaxUnreachable); // our solution&#xA;&#xA;    }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="14016" LastEditDate="2016-12-20T05:49:17.190" LastActivityDate="2016-12-20T05:49:17.190" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="18354" PostTypeId="2" ParentId="18350" CreationDate="2015-07-25T02:12:24.797" Score="1" Body="&lt;p&gt;In general, you can measure all combinations of the following times:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$n$&lt;/li&gt;&#xA;&lt;li&gt;$m$&lt;/li&gt;&#xA;&lt;li&gt;$2m - n$&lt;/li&gt;&#xA;&lt;li&gt;$4n - m$&lt;/li&gt;&#xA;&lt;li&gt;$3m - 2n$&lt;/li&gt;&#xA;&lt;li&gt;$6n - 2m$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;And times $t(x)=min(A \cdot m,B \cdot n) + (x \cdot \delta)$ where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$max(A \cdot m,B \cdot n)-min(A \cdot m,B \cdot n)=\delta$ &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Item 1&lt;/strong&gt; is found by starting timer $n$ and noting when expired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Item 2&lt;/strong&gt; is found by starting timer $m$ and noting when expired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Item 3&lt;/strong&gt; is found by starting timers $m$ and $n$ at the same time. When timer $n$ has expired reset $n$. When timer m has expired flip $n$. This gives you the time $2m - n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Item 4&lt;/strong&gt; is found by starting timers $m$ and $n$ at the same time. When timer $n$ has expired reset $n$. When timer $m$ has expired reset $m$. When timer $n$ has expired flip $m$. This gives you the time $4n - m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Item 5&lt;/strong&gt; is found by starting timers $m$ and $n$ at the same time. When timer $n$ has expired reset $n$. When timer $m$ has expired reset $m$ and flip $n$. When timer $n$ has expired flip $m$. This gives you the time $3m - 2n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Item 6&lt;/strong&gt; is found by starting timers $m$ and $n$ at the same time. When timer $n$ has expired reset $n$. When timer $m$ has expired reset $m$. When timer $n$ has expired reset $n$ and flip $m$. When timer $m$ has expired flip $n$. This gives you the time $6n - 2m$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Warm-up&lt;/strong&gt; $(7,4)$ $t=7$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Easy&lt;/strong&gt; $(8,5)$ $t=10$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Medium&lt;/strong&gt; $(11,6)$ $t=11$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hard&lt;/strong&gt; (In-progress) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general we are looking for $a$,$b$,$c$,$d$,$e$,$f$ in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$t(a,b,c,d,e,f) = a(m)+b(n)+c(2m-n)+d(4n-m)+e(3m-2n)+f(6n-2m)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;such that $t$ is minimized according to the problem statement.&lt;/p&gt;&#xA;" OwnerUserId="13788" LastEditorUserId="13788" LastEditDate="2015-07-31T00:43:51.903" LastActivityDate="2015-07-31T00:43:51.903" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18355" PostTypeId="2" ParentId="18345" CreationDate="2015-07-25T05:54:10.410" Score="3" Body="&lt;p&gt;To get into the next room, you should press the:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Left-pointing Double Angle Quotation Mark&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first cipher decodes to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;WELL DONE FOR GETTING THIS FAR X THE MAIN CIPHER IS PLAYING AT A FAIR X GOODBYE FOR NOW X THE KEY TO THE FAIR IS SOCKS A TO Z&lt;br&gt;&#xA;&lt;br&gt;&#xA;This Enigma cipher can be decoded &lt;strong&gt;&lt;a href=&quot;http://enigma.louisedade.co.uk/enigma.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/strong&gt; by leaving all of the default settings and entering a reversed alphabet into the Steckerbrett (plugboard). Then select &quot;Block of text&quot; input method and enter the cipher text.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second cipher is a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Playfair cipher based on the instructions above. The key is &lt;strong&gt;SOCKS&lt;/strong&gt; and the letter &lt;strong&gt;A&lt;/strong&gt; shall be translated to &lt;strong&gt;Z&lt;/strong&gt;.  Using &lt;a href=&quot;http://rumkin.com/tools/cipher/playfair.php&quot; rel=&quot;nofollow&quot;&gt;Rumkin&lt;/a&gt; and selecting &lt;strong&gt;Encrypt&lt;/strong&gt; instead of Decrypt, results in this &quot;almost correct&quot; plaintext:&lt;br&gt;&#xA;&lt;br&gt;&#xA;&quot;Press the Left-pointing Double Zngle Quotztion Mzrk to zdvznce onwzrds.&quot;&lt;br&gt;&#xA;&lt;br&gt;&#xA;Then, as noted by f', replace all the Z's with A's because this decoding does not distinguish between the two characters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-07-27T06:53:04.650" LastActivityDate="2015-07-27T06:53:04.650" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18357" PostTypeId="1" CreationDate="2015-07-23T12:37:40.127" Score="2" ViewCount="1865" Body="&lt;p&gt;You are blindfolded (and you can’t see a thing). &#xA;There are $100$ coins lying on a table. Of these coins, $80$ are tails-up and $20$ are heads-up. You need to divide them into two groups of $x$ and $y$ each (with $x+y = 100$). You can pick any coin and flip it (if it is tails, it will become head and if it is head, it will become tail), but you can't see so you don't know which coin you are flipping. In the end, both groups should contain equal number of heads.&lt;/p&gt;&#xA;" OwnerUserId="14259" OwnerDisplayName="user3366367" LastEditorUserId="8874" LastEditDate="2016-03-06T18:56:17.217" LastActivityDate="2016-03-06T18:56:17.217" Title="Variation of 100 Gold Coins Puzzle" Tags="|mathematics|logical-deduction|" AnswerCount="1" CommentCount="0" ClosedDate="2015-07-25T14:49:05.730" ContentLicense="CC BY-SA 3.0" />
  <row Id="18358" PostTypeId="2" ParentId="18357" CreationDate="2015-07-23T17:09:33.603" Score="2" Body="&lt;p&gt;The trick is simple: &#xA;When dividing, make the first pile as big as the number of heads-up coins, i.e. 20. &#xA;That first pile now contains an unkown number of heads-up coins, call that number n.&#xA;The second pile of 80 coins contains the remaining 20-n heads-up coins.&#xA;Now flip all coins in the first pile and both piles will contain exactly 20-n heads-up coins, regardless of the value of n.&lt;/p&gt;&#xA;" OwnerUserId="14283" OwnerDisplayName="lodrik" LastActivityDate="2015-07-23T17:09:33.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18359" PostTypeId="1" AcceptedAnswerId="18364" CreationDate="2015-07-25T12:00:24.007" Score="10" ViewCount="385" Body="&lt;p&gt;Behold my sudokross!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &lt;img src=&quot;https://i.stack.imgur.com/hyCm5.png&quot; alt=&quot;sudokross puzzle&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar to sudoku, the objective is to fill every row and every column with the letters $\rm{RELATIONS}$, with each letter appearing exactly once per row/column. Only one letter may appear in each cell (excepting the last column, which has already been filled out).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two novel elements, however:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;gaps may appear in rows and columns&lt;/li&gt;&#xA;&lt;li&gt;each box delineated by a dark border must contain &lt;a href=&quot;http://app.aspell.net/lookup&quot; rel=&quot;nofollow noreferrer&quot;&gt;a valid US/UK English word&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Generally speaking, the words in this puzzle should all be very familiar to English speakers, with the exception of $\rm{TI}$ and $\rm{NOS}$ (which have already been filled out).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Valid words may have gaps in them, but letters must appear in proper order. For example,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/Tn2kx.png&quot; alt=&quot;sudokross word example 1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/zWKeB.png&quot; alt=&quot;sudokross word example 2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;are both valid entries for $\rm{TEAL}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can you complete the sudokross?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Bonus Objective&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Readers have probably noticed that the final column in the puzzle is &quot;fat&quot; in the sense that it contains two letters in some cells, which is less than ideal. This is a consequence of my lack of foresight while constructing the puzzle. In particular, I didn't manage the gaps in the columns as well as I should have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;strong&gt;perfect sudokross&lt;/strong&gt; is one where no &quot;fat&quot; column(s) appear, in which case the grid is perfectly square and the same number of gaps (empty cells) appear in all rows/columns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, a note on terminology: the &quot;root&quot; of a sudokross puzzle is the set of letters that appears in every row/column. It need not be a valid English word, but making it so helps with the aesthetic of the puzzle. For example, the root of the puzzle above is $\rm{AEILNORST}$, which I anagrammed to $\rm{RELATIONS}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On Thursday July 30, 2015, I will award &lt;strong&gt;200 rep&lt;/strong&gt; to any puzzler who presents a (valid) perfect sudokross where either:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;the root is 9 letters long, and the puzzle dimensions do not exceed 14-by-14 cells&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the root is 10 letters long, and the puzzle dimensions do not exceed 16-by-16 cells&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the root is 11 letters long, and the puzzle dimensions do not exceed 18-by-18 cells&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The root of the puzzle may &lt;em&gt;not&lt;/em&gt; be the same as for the puzzle given above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the event of multiple submissions, the prize will go to the valid puzzle with the longest root. In the event of a tie, preference will be given to the earlier submission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Best of luck, sudokrossers!&lt;/strong&gt; ;)&lt;/p&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="12408" LastEditDate="2015-07-26T02:38:36.420" LastActivityDate="2015-07-26T02:52:34.473" Title="RELATIONS: Introducing the Sudokross" Tags="|puzzle-creation|crosswords|sudoku|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18360" PostTypeId="2" ParentId="18342" CreationDate="2015-07-25T14:28:42.057" Score="2" Body="&lt;p&gt;Partial answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The list is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Programming language&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My first is part of most tsunamis.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C (sea)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My second is scattered in mastodon remains.&#xA;A venomous snake hides my third.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Python&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You see my fourth in a jumpy biter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maulers and smashers show my fifth.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bash (not sure, got this &lt;a href=&quot;http://dotd.wikia.com/wiki/Mauler&amp;#39;s_Smasher&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; )&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An imperfect fury reveals my sixth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A vile spiritualist divulges my seventh.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My eighth is spotted in Pithecanthropus fossils.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Java (Pithecanthropus is the Java Man)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Oxygen desaturation tells you my ninth.&lt;/p&gt;&#xA;" OwnerUserId="5359" LastActivityDate="2015-07-25T14:28:42.057" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18361" PostTypeId="2" ParentId="18342" CreationDate="2015-07-25T14:30:00.347" Score="14" Body="&lt;p&gt;Cool, I like it. &lt;Br&gt;&#xA;I believe the list is:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Space Objects listed in apparent brightness in the sky order. &lt;br&gt;&lt;br&gt;&#xA; &lt;strong&gt;SUN&lt;/strong&gt; : My first is part of most t&lt;b&gt;sun&lt;/b&gt;amis.&lt;br&gt; &#xA; &lt;strong&gt;MOON&lt;/strong&gt; : My second is scattered in &lt;b&gt;m&lt;/b&gt;ast&lt;b&gt;o&lt;/b&gt;d&lt;b&gt;on&lt;/b&gt; remains.&lt;br&gt;&#xA; &lt;strong&gt;VENUS&lt;/strong&gt; : A &lt;b&gt;ven&lt;/b&gt;omo&lt;b&gt;us&lt;/b&gt; snake hides my third.&lt;br&gt;&#xA; &lt;strong&gt;JUPITER&lt;/strong&gt; : You see my fourth in a &lt;b&gt;ju&lt;/b&gt;m&lt;b&gt;p&lt;/b&gt;y b&lt;b&gt;iter&lt;/b&gt;.&lt;br&gt;&#xA; &lt;strong&gt;MARS&lt;/strong&gt; : &lt;b&gt;Ma&lt;/b&gt;ule&lt;b&gt;rs&lt;/b&gt; and s&lt;b&gt;ma&lt;/b&gt;she&lt;b&gt;rs&lt;/b&gt; show my fifth.&lt;br&gt;&#xA; &lt;strong&gt;MERCURY&lt;/strong&gt; : An i&lt;b&gt;m&lt;/b&gt;p&lt;b&gt;er&lt;/b&gt;fe&lt;b&gt;c&lt;/b&gt;t f&lt;b&gt;ury&lt;/b&gt; reveals my sixth.&lt;br&gt;&#xA; &lt;strong&gt;SIRIUS&lt;/strong&gt; : A vile &lt;b&gt;s&lt;/b&gt;p&lt;b&gt;iri&lt;/b&gt;t&lt;b&gt;u&lt;/b&gt;ali&lt;b&gt;s&lt;/b&gt;t divulges my seventh. ( thanks to f'')&lt;br&gt;&#xA; &lt;strong&gt;CANOPUS&lt;/strong&gt; : My eighth is spotted in Pithe&lt;b&gt;can&lt;/b&gt;thr&lt;b&gt;opus&lt;/b&gt; fossils.  (thanks to f'')&lt;br&gt;&#xA; &lt;strong&gt;SATURN&lt;/strong&gt; : Oxygen de&lt;b&gt;satur&lt;/b&gt;atio&lt;b&gt;n&lt;/b&gt; tells you my ninth. &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13916" LastEditorUserId="463" LastEditDate="2015-08-04T14:09:38.393" LastActivityDate="2015-08-04T14:09:38.393" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18362" PostTypeId="1" AcceptedAnswerId="29793" CreationDate="2015-07-25T20:42:27.953" Score="11" ViewCount="3258" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;i&gt;Long ago, where barrels ran ramped and the trees consumed the earth,&lt;br&gt;&#xA;  was a time of life and of love and of happiness.&lt;br&gt;&#xA;  (You put one and two together, and what do you get?)&#xA;  &lt;hr&gt;&#xA;  A long implementation of something so silly did we unearth,&lt;br&gt;&#xA;  beneath the meaning and nothing less.&lt;br&gt;&#xA;  (I'm sure you'll hear the words put together for it!&amp;uarr;&lt;sup&gt;1&lt;/sup&gt;)&lt;br&gt;&#xA;  &lt;br&gt;&#xA;  Picture yourself at a table with friends;&lt;br&gt;&#xA;  what you hit to begin is something key,&lt;br&gt;&#xA;  for 'tis the moniker that he sends&lt;br&gt;&#xA;  for you to discover with me.&lt;br&gt;&#xA;  &lt;br&gt;&#xA;  I am on your left-hand side, as it may be&lt;br&gt;&#xA;  in each of three pictures on the glass.&lt;br&gt;&#xA;  My name is known, but only by he&lt;br&gt;&#xA;  who knows it not (but at this I'm just being an ***!).&lt;br&gt;&#xA;  &lt;br&gt;&#xA;  But as for me, &lt;strong&gt;I&lt;/strong&gt; know me best,&lt;br&gt;&#xA;  writing as long as I can!&lt;br&gt;&#xA;  For all of my tomes, there'll be pain, lest&lt;br&gt;&#xA;  you enjoy to read. (Say my name, young man!).&lt;br&gt;&#xA;  &lt;hr&gt;&#xA;  I am the one to offer a deal,&lt;br&gt;&#xA;  It is I to whom you will entrust your life&lt;br&gt;&#xA;  with the utmost zeal.&lt;br&gt;&#xA;  (But perhaps you gain instead, toil and strife.)&lt;br&gt;&#xA;  &lt;br&gt;&#xA;  And I am the lowest of the low, the incredible hulk&lt;br&gt;&#xA;  of cheating you into your fate.&lt;br&gt;&#xA;  If you set your eyes upon me, you'll sulk&lt;br&gt;&#xA;  into a fury to be described as irate.&#xA;  &lt;/i&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Each paragraph yields a single word (or a hyphenated word), and each section (delineated by the &lt;code&gt;&amp;lt;hr&amp;gt;&lt;/code&gt;s and quote borders) yields a number. The entire poem will yield one word. The rhymes are for flavour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your answer should include a clear explanation and justification of your answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;May good luck and your mind be with you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; 4 letters are involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 2&lt;/strong&gt; It can be found readily online.&lt;/p&gt;&#xA;" OwnerUserId="7307" LastEditorUserId="7307" LastEditDate="2016-01-12T19:35:49.067" LastActivityDate="2019-11-19T20:44:57.493" Title="I'm so meta, try yourself with this!" Tags="|riddle|word|rhyme|" AnswerCount="3" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="18363" PostTypeId="1" AcceptedAnswerId="18503" CreationDate="2015-07-26T00:58:53.803" Score="9" ViewCount="2206" Body="&lt;p&gt;I've been working hard on trying to solve the missing letters in &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18344/a-single-letter-is-missing-from-each-can-you-figure-it-out-to-find-the-answer-a&quot;&gt;this puzzle&lt;/a&gt; by qwertylpc, but I've discovered that in trying to figure out the lists his clues are coming from, I've come up with several lists of my own! Since I am stumped on the last two clues in his puzzle, I'm going to dump all of the clues I've thought of into this puzzle!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EASY&lt;/strong&gt;&lt;br&gt;&#xA;1) P, N, D, ?&lt;br&gt;&#xA;2) P, K, B, R, ?&lt;br&gt;&#xA;3) F, C, G, D, ?&lt;br&gt;&#xA;4) B, K, M, G, ?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;MEDIUM&lt;/strong&gt;&lt;br&gt;&#xA;5) B, G, B, G, R, ?&lt;br&gt;&#xA;6) G, E, L, N, D, J, J, ?&lt;br&gt;&#xA;7) T, T, C, P, Q, ?&lt;br&gt;&#xA;8) H, O, T, T, S, F, ?&lt;br&gt;&#xA;9) W, J, L, H, J, G, ?    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HARD&lt;/strong&gt;&lt;br&gt;&#xA;10) E, F, W, W, ?&lt;br&gt;&#xA;11) L, G, G, S, W, E, ?&lt;br&gt;&#xA;12) P, T, F, C, G, G, ?&lt;br&gt;&#xA;13) M, J, P, L, S, D, H, ?&lt;br&gt;&#xA;14) S, C, P, G, O, ?    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;VERY HARD&lt;/strong&gt;&lt;br&gt;&#xA;15) L, H, H, B, S, B, B, ?     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Important information:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Note the &lt;a href=&quot;/questions/tagged/english&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;english&amp;#39;&quot; rel=&quot;tag&quot;&gt;english&lt;/a&gt; tag. Also, many of these are US-centric, including one that would have a slightly different sequence in Britain.&lt;/li&gt;&#xA;&lt;li&gt;All of the letters come from &lt;strong&gt;ordered lists of things&lt;/strong&gt;. So you don't have to worry about how the letters look (e.g. A, E, F, H, I, ? is &quot;K&quot; because no curved lines in the capital letters), or their position in the alphabet (no need to do math or anything).&lt;/li&gt;&#xA;&lt;li&gt;All of the answers are the &lt;strong&gt;first&lt;/strong&gt; letter of a word, but &lt;strong&gt;not necessarily&lt;/strong&gt; from the first word in a phrase/name. So if the sequence is Neil Armstrong, Buzz Aldrin, Pete Conrad, I wouldn't give you &quot;E, U, E&quot; or &quot;G, N, D&quot; but I could just as easily give you &quot;A, A, C&quot; as &quot;N, B, P&quot;.&lt;/li&gt;&#xA;&lt;li&gt;Difficulty groupings are based on my best guess. I suspect some will be easier than I thought, and possibly some will be harder than I thought, but I thought I'd at least make an attempt to classify them.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I would like to avoid giving ANY hints on this puzzle if possible, but I guess we'll see how it goes, particularly with the last one. I'm already thinking of other sequences, so maybe I'll make a sequel!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit: Bounty added!&lt;/strong&gt; In addition to the regular bounty, I'll give another 50 points to whoever solves the VERY HARD sequence (#15), so even if you can only solve that one, go ahead and make your own answer!&lt;/p&gt;&#xA;" OwnerUserId="10861" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-08-04T16:01:35.720" Title="A Quindectet of Letter Sequences!" Tags="|english|knowledge|letter-sequence|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18364" PostTypeId="2" ParentId="18359" CreationDate="2015-07-26T02:52:34.473" Score="8" Body="&lt;p&gt;Here's the solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eY3F6.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eY3F6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because every row is an anagram of &quot;relations&quot;, it isn't hard to figure out what words go in each row and then fill in which squares they are in. I think it would be interesting to see vertical words or even some kind of &quot;bent&quot; boxes (like in KenKen).&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-07-26T02:52:34.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18365" PostTypeId="2" ParentId="18363" CreationDate="2015-07-26T03:03:25.693" Score="6" Body="&lt;p&gt;Some answers that I've gotten so far:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;1.&lt;/strong&gt; Q. This is the order of the value of U.S. coins (Penny, Nickel, Dime). The next one is &lt;strong&gt;Quarter&lt;/strong&gt;.&lt;p&gt;&#xA; &lt;strong&gt;2.&lt;/strong&gt; Q. This the order of the value ranks of chess pieces (Pawn, Knight, Bishop, Rook). The next one is &lt;strong&gt;Queen&lt;/strong&gt;.&lt;p&gt;&#xA; &lt;strong&gt;3.&lt;/strong&gt; A. This is the order of sharps in Western key signature notation (F, C, G, D, A, E, B). It's also the initials of the mnemonic &quot;Father Charles Goes Down &lt;strong&gt;And&lt;/strong&gt; Ends Battle&quot;.&lt;p&gt;&#xA; &lt;strong&gt;4.&lt;/strong&gt; T. This is the order of SI prefixes being applied to bytes (Bytes, Kilobytes, Megabytes, Gigabytes). The next one is &lt;strong&gt;Terabytes&lt;/strong&gt;.&lt;p&gt;&#xA; &lt;strong&gt;5.&lt;/strong&gt; J. This is the order of the &lt;em&gt;first&lt;/em&gt; names of United States Presidents from Obama going backwards (Barack, George, Bill, George, Ronald). The next one is &lt;strong&gt;Jimmy&lt;/strong&gt; Carter.&lt;p&gt;&#xA; &lt;strong&gt;6.&lt;/strong&gt; R. This is the order of the books of the Old Testament in the Bible (Genesis, Exodus, Leviticus, Numbers, Deuteronomy, Joshua, Judges). The next one is &lt;strong&gt;Ruth&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="4353" LastEditDate="2015-07-26T06:03:42.810" LastActivityDate="2015-07-26T06:03:42.810" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18366" PostTypeId="1" AcceptedAnswerId="18375" CreationDate="2015-07-26T03:07:20.597" Score="8" ViewCount="466" Body="&lt;p&gt;After the zombies ate your brain at the end of level 1, Dr. Zomboss decides to continue sending his legions of undead until they completely destroy your house. At every 4 seconds, he drops a new zombie on some unoccupied spot of your lawn. All zombies move towards the left and each of them needs exactly 4 seconds to traverse through one of the spots on the lawn. Show that after some time T, the total distance traveled by any 1000 consecutive zombies will be almost the same - within range of just 32 spots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SVHmC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SVHmC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HINT 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You don't need to be able to point out when the time T comes, just to prove that it exists.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;HINT 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You may want to use your photo camera extensively.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;P.S. Unfortunately I can not take the credit for this puzzle. I just simplified/modified a problem from some Math competition, which made the original question easily approachable. I believe that in the context of zombies, it is not that hard to figure it out (but still not that easy). Later will post reference to the original problem.&lt;/p&gt;&#xA;" OwnerUserId="14140" LastEditorUserId="14140" LastEditDate="2015-07-26T23:12:42.467" LastActivityDate="2015-07-27T03:29:07.873" Title="Plants vs Zombies Annihilation" Tags="|mathematics|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18367" PostTypeId="2" ParentId="18261" CreationDate="2015-07-26T04:04:17.157" Score="0" Body="&lt;p&gt;Is it daylight?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need the unharmed word in order for it to be &quot;day&quot; (not night)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Decapitate the first syllable...&quot;day&quot; ...and we can see because it is &quot;light&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we say .. &quot;Oh say can you see...&quot; .....how's that? :)&lt;/p&gt;&#xA;" OwnerUserId="14280" LastEditorUserId="14280" LastEditDate="2015-07-26T04:11:34.313" LastActivityDate="2015-07-26T04:11:34.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18368" PostTypeId="2" ParentId="17895" CreationDate="2015-07-26T19:09:24.490" Score="2" Body="&lt;p&gt;@John Stevens was &lt;em&gt;so&lt;/em&gt; close to the answer. A Baumic number is defined as such:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $T(a)$ be the prime factor tree of $a$, where the left branch of each node is the lowest prime factor of the node.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt; Let's calculate $T(105)$:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 105&#xA; / \&#xA;3  35&#xA;   /\&#xA;  5  7&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This tree &amp;ldquo;contains&amp;rdquo; the folowing &amp;ldquo;elements&amp;rdquo;: $105$, $3$, $35$, $5$, and $7$. So the following statements are true: $105\in T(a)$, $3\in T(a)$, &amp;hellip;, $7\in T(a)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A number $a$ is Baumic if and only if there is some element of this tree that numbers the elements in the tree. I.e.:&#xA;$$B(a)\iff\exists b\in T(a)\left[b=||T(a)||\right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt; Let's see if $B(105)$. We have that $||T(105)||=5$. Since $5\in T(105)$, $105$ is Baumic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about the false positive, $176$? Let's look at $T(176)$:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 176&#xA; / \&#xA;2  88&#xA;   /\&#xA;  2  44&#xA;     /\&#xA;    2  22&#xA;       /\&#xA;      2  11&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As can be seen, $||T(176)=9||$. Since $9\notin T(176)$, $\neg B(176)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the missed numbers, $108$ and $162$:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 108&#xA; / \&#xA;2   54&#xA;    /\&#xA;   2  27&#xA;      /\&#xA;     3  9&#xA;       / \&#xA;      3   3&#xA;&#xA;-------------&#xA;&#xA; 162&#xA; / \&#xA;2   81&#xA;    /\&#xA;   3  27&#xA;      /\&#xA;     3  9&#xA;       / \&#xA;      3   3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Both trees have a cardinality of $9$, and both contain $9$. Therefore, they are both Baumic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;FUN FACT:&lt;/strong&gt; &lt;em&gt;Baum&lt;/em&gt; is German for Tree. I called them that so that you might think of a factorization tree and thus get the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another fun fact: All of the following are Baumic numbers, $1-1000$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1, 6, 9, 15, 20, 21, 30, 33, 39, 45, 50, 51, 56, 57, 69, 70, 72, 75, 84, 87, 93, 105, 108, 110, 111, 123, 125, 126, 129, 130, 140, 141, 159, 162, 165, 170, 175, 177, 183, 189, 190, 195, 196, 201, 210, 213, 219, 230, 237, 243, 245, 249, 255, 267, 275, 285, 290, 291, 294, 303, 308, 309, 310, 315, 321, 325, 327, 339, 345, 350, 352, 364, 370, 381, 385, 393, 410, 411, 417, 425, 430, 435, 441, 447, 453, 455, 462, 465, 470, 471, 475, 476, 489, 490, 501, 519, 525, 528, 530, 532, 537, 543, 546, 555, 573, 575, 579, 590, 591, 595, 597, 605, 610, 615, 633, 644, 645, 665, 669, 670, 681, 686, 687, 693, 699, 705, 710, 714, 715, 717, 723, 725, 730, 735, 753, 770, 771, 775, 789, 790, 792, 795, 798, 805, 807, 812, 813, 819, 830, 831, 832, 843, 845, 849, 868, 875, 879, 880, 885, 890, 910, 915, 921, 925, 933, 935, 939, 951, 960, 966, 970, 993, 1005, 1010, 1011, 1015, 1025, 1029, 1030, 1036, 1041, 1045, 1047, 1059, 1065, 1070, 1071, 1075, 1077, 1078, 1085, 1090, 1095, 1101, 1105, 1119, 1130, 1137, 1148, 1149, 1155, 1167, 1175, 1185, 1188, 1190, 1191, 1197, 1203, 1204, 1218, 1225, 1227, 1232, 1235, 1245, 1248, 1257, 1263, 1265, 1270, 1274, 1293, 1295, 1299, 1302, 1310, 1316, 1317, 1320, 1325, 1329, 1330, 1335, 1347, 1365, 1370, 1371, 1383, 1389, 1390, 1401, 1435, 1437, 1440, 1445, 1449, 1455, 1461, 1473, 1475, 1484, 1490, 1495, 1497, 1505, 1509, 1510, 1515, 1525, 1527, 1545, 1554, 1563, 1569, 1570, 1595, 1605, 1610, 1615, 1617, 1623, 1630, 1635, 1641, 1645, 1652, 1666, 1670, 1671, 1675, 1689, 1694, 1695, 1705, 1707, 1708, 1713, 1715, 1722, 1730, 1731, 1761, 1775, 1779, 1782, 1785, 1790, 1797, 1803, 1805, 1806, 1810, 1821, 1825, 1827, 1839, 1848, 1851, 1855, 1857, 1862, 1872, 1876, 1885, 1893, 1905, 1910, 1911, 1923, 1925, 1929, 1930, 1936, 1941, 1953, 1955, 1959, 1965, 1970, 1974, 1975, 1977, 1980, 1983, 1988, 1990, 1995, 2002, 2015, 2019, 2030, 2031, 2035, 2044, 2049, 2055, 2065, 2073, 2075, 2080, 2085, 2103, 2110, 2127, 2135, 2157, 2160, 2170, 2181, 2185, 2199, 2200, 2212, 2217, 2225, 2226, 2229, 2230, 2235, 2253, 2254, 2255, 2265, 2270, 2271, 2275, 2283, 2288, 2290, 2307, 2319, 2324, 2330, 2331, 2345, 2355, 2361, 2365, 2366, 2390, 2391, 2401, 2405, 2410, 2415, 2425, 2427, 2433, 2445, 2463, 2465, 2469, 2478, 2481, 2485, 2487, 2492, 2499, 2505, 2510, 2517, 2525, 2541, 2555, 2559, 2562, 2570, 2571, 2575, 2577, 2583, 2585, 2589, 2590, 2595, 2618, 2630, 2631, 2635, 2643, 2645, 2649, 2661, 2665, 2673, 2675, 2685, 2690, 2695, 2709, 2710, 2715, 2716, 2721, 2725, 2733, 2755, 2757, 2765, 2770, 2772, 2787, 2793, 2795, 2808, 2810, 2811, 2814, 2823, 2825, 2828, 2830, 2841, 2842, 2859, 2865, 2870, 2884, 2895, 2901, 2904, 2905, 2912, 2913, 2915, 2926, 2930, 2931, 2945, 2949, 2955, 2961, 2970, 2973, 2975, 2982, 2985, 2991, 2992, 2996, 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6412, 6423, 6425, 6429, 6430, 6459, 6465, 6468, 6470, 6475, 6483, 6489, 6495, 6512, 6517, 6524, 6528, 6530, 6534, 6537, 6545, 6552, 6565, 6566, 6575, 6585, 6590, 6594, 6609, 6610, 6621, 6622, 6639, 6645, 6663, 6665, 6685, 6692, 6695, 6699, 6711, 6715, 6717, 6725, 6729, 6730, 6732, 6734, 6735, 6741, 6745, 6748, 6753, 6755, 6770, 6775, 6776, 6783, 6785, 6790, 6801, 6807, 6815, 6819, 6830, 6843, 6845, 6846, 6855, 6861, 6864, 6867, 6879, 6891, 6895, 6902, 6909, 6910, 6915, 6925, 6927, 6930, 6933, 6935, 6945, 6955, 6958, 6965, 6985, 6999, 7005, 7007, 7010, 7014, 7015, 7017, 7020, 7023, 7025, 7028, 7035, 7041, 7053, 7055, 7070, 7071, 7072, 7075, 7085, 7090, 7105, 7113, 7119, 7131, 7143, 7149, 7154, 7161, 7167, 7175, 7179, 7185, 7190, 7196, 7197, 7205, 7210, 7216, 7233, 7238, 7251, 7260, 7266, 7269, 7270, 7280, 7285, 7290, 7305, 7311, 7315, 7323, 7325, 7330, 7341, 7345, 7364, 7365, 7377, 7378, 7385, 7390, 7401, 7406, 7419, 7425, 7430, 7431, 7455, 7462, 7480, 7485, 7490, 7505, 7509, 7510, 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8675, 8691, 8695, 8708, 8709, 8715, 8722, 8725, 8727, 8751, 8755, 8757, 8764, 8770, 8781, 8785, 8805, 8806, 8810, 8815, 8817, 8825, 8830, 8845, 8855, 8859, 8862, 8870, 8871, 8876, 8889, 8890, 8895, 8905, 8907, 8913, 8965, 8967, 8975, 8985, 8995, 8997, 9003, 9015, 9033, 9035, 9057, 9065, 9069, 9070, 9085, 9086, 9095, 9105, 9108, 9110, 9111, 9123, 9145, 9147, 9163, 9170, 9175, 9177, 9183, 9185, 9190, 9195, 9201, 9205, 9215, 9237, 9245, 9249, 9255, 9265, 9267, 9268, 9275, 9285, 9290, 9317, 9325, 9327, 9328, 9338, 9345, 9357, 9363, 9366, 9370, 9387, 9394, 9410, 9411, 9415, 9436, 9455, 9465, 9470, 9471, 9475, 9477, 9485, 9489, 9501, 9506, 9507, 9513, 9515, 9530, 9534, 9543, 9545, 9561, 9568, 9573, 9575, 9590, 9595, 9605, 9609, 9615, 9618, 9627, 9635, 9645, 9646, 9651, 9663, 9670, 9685, 9687, 9695, 9702, 9705, 9710, 9715, 9716, 9725, 9728, 9730, 9753, 9758, 9759, 9768, 9770, 9771, 9772, 9777, 9785, 9786, 9792, 9795, 9801, 9805, 9813, 9815, 9828, 9830, 9835, 9842, 9845, 9849, 9884, 9885, 9891, 9897, 9898, 9903, 9905, 9910, 9915, 9921, 9925, 9933, 9939, 9947, 9955, 9957, 9969, 9970, 9982, 9987, 9993.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;Edit&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;As per request, I am posting the code. It coded in JavaScript, and uses recursive traversal and definition methods.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;function NumberError(msg){&#xA;    this.name = &quot;NumberError&quot;;&#xA;    this.msg = msg|&quot;&quot;;&#xA;}&#xA;&#xA;NumberError.prototype=Object.create(Error.prototype);&#xA;NumberError.prototype.constructor=NumberError;&#xA;&#xA;Array.prototype.bigUnion = function(){&#xA;    var a=[];&#xA;    for(var z=0;z&amp;lt;this.length;z++)a=a.concat(this[z]);&#xA;    return a;&#xA;}&#xA;&#xA;function s(x){&#xA;    for(var i=2;i&amp;lt;=Math.sqrt(x);i++) if(!(x%i)) return i;&#xA;    return x;&#xA;}&#xA;&#xA;function l(x){// largest divisor of x&#xA;    for(var i=Math.ceil(Math.sqrt(x));i&amp;gt;=2;i--) if(!(x%i)) return x/i;&#xA;    return x;&#xA;}&#xA;&#xA;function isPrime(x){&#xA;    return s(x)==x;&#xA;}&#xA;&#xA;var memoized = [];&#xA;&#xA;function isB(x){    // Baumic number check&#xA;    if(memoized[x]) return memoized[x];&#xA;    memoized[x]=(new Tree(x)).isB();&#xA;    return memoized[x];&#xA;}&#xA;&#xA;function Tree(x,f1=s){&#xA;    this.top = new node(x);&#xA;    var i=0;&#xA;    var c=this.top;&#xA;    while(!isPrime(c.body)){&#xA;        c.attach(f1(c.body),c.body/f1(c.body));&#xA;        c=c.children[1];&#xA;    }&#xA;}&#xA;&#xA;Tree.prototype.disp = function(){&#xA;    return this.top.disp();&#xA;}&#xA;&#xA;Tree.prototype.isB = function(){&#xA;    var yF=this.family();&#xA;    return!!(yF.indexOf(yF.length)+1);&#xA;}&#xA;&#xA;Tree.prototype.family = function(self){&#xA;    return this.disp().replace(/ =&amp;gt; \(/g,&quot;,&quot;).replace(/\)/g,&quot;&quot;).split(&quot;,&quot;).map(Number);&#xA;}&#xA;&#xA;function node(x){&#xA;    if(isNaN(parseInt(x))) throw new NumberError(&quot;Input (&quot;+x+&quot;) not a number&quot;)&#xA;    this.body = x;&#xA;    this.children = [];&#xA;}&#xA;&#xA;function nodify(type,input,z){&#xA;    switch(type){&#xA;        case &quot;array&quot;:&#xA;            return input.map(function(e){return e?z?(new node(e)).factor():new node(e):&quot;&quot;});&#xA;        break;&#xA;        default:&#xA;            return new node(input);&#xA;        break;&#xA;    }&#xA;}&#xA;&#xA;function n(input){&#xA;    return input.map(function(e){return (new node(e)).factor()});&#xA;}&#xA;&#xA;node.prototype.disp = function(){&#xA;    return this.body + (this.children.length?&quot; =&amp;gt; (&quot;+this.children.map(function(e){return e?e.disp():&quot;?&quot;})+&quot;)&quot;:&quot;&quot;);&#xA;}&#xA;&#xA;node.prototype.right = function(){&#xA;    if(this.children.length) return this.children[1].right();&#xA;    return this.body;&#xA;}&#xA;&#xA;node.prototype.attach = function(){&#xA;    a = Array.from(arguments);&#xA;    for(var _i=0;_i&amp;lt;a.length;_i++){&#xA;        y=a[_i];&#xA;        if(!(y instanceof node)) try {y=new node(y)}catch(e){if(e instanceof NumberError)throw Error(&quot;Invalid node&quot;)}&#xA;        this.children.push(y);&#xA;    }&#xA;}&#xA;var g=0;&#xA;node.prototype.factor = function(f=l){&#xA;    ui=[f(this.body),this.body/f(this.body)];&#xA;    if(!isPrime(this.body)){&#xA;        nu=n(ui);&#xA;        console.log(ui,this.body,nu)&#xA;        this.children = nu;&#xA;    }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7307" LastEditorUserId="7307" LastEditDate="2015-08-02T21:20:50.160" LastActivityDate="2015-08-02T21:20:50.160" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18369" PostTypeId="1" AcceptedAnswerId="18781" CreationDate="2015-07-26T20:06:04.060" Score="12" ViewCount="1241" Body="&lt;p&gt;George always has had it with the girls. Those who don't find him irresistible at first sight could usually be won over by a ride in his shiny red sports car. If there has ever been a problem with girls for George, then it was how to handle multiple at the same time without having them fight each other.&#xA;It therefore came as quite a shock to him, when he first met Kari just after moving town. She is undoubtedly the most attractive woman he has met so far, but as much as he tried - and he tried a lot - she's just shown him her cold shoulder so far. She has hardly spoken with him!&lt;/p&gt;&#xA;&lt;p&gt;And for the first time, George is really in love.&lt;/p&gt;&#xA;&lt;p&gt;He tries for weeks to gain her attention - always close of not becoming a stalker - but to no success. He is close to giving up - maybe he should just jump off the bridge ? - when suddenly, one fine Wednesday at work, Kari walks up to him. Without saying much, she hands him a neat envelope with some fine, hand-written &amp;quot;for George&amp;quot; on it, then turns around abruptly and hurries of into her office without giving him a chance to chat with her.&lt;/p&gt;&#xA;&lt;p&gt;When George opens the letter - about half a page neatly written in all capital letters - he first can not make head and tail from it. Should this be a sick joke? But something tells him, it isn't. So he cancels the weekly card-club this evening and instead spends all night working on that letter. He finds all sorts of clues, but while some things really sound hopeful, others mightily depress him. No wonder he is tired and grumpy the next day in office. Kari isn't in, which doesn't lift his mood. In the evening, he rings up his best friend Robert and they meet in the Red Sofa bar. Robert is a computer nerd and keen puzzle enthusiast, so he promises to have a crack on the letter.&lt;/p&gt;&#xA;&lt;p&gt;Next afternoon, just before work finishes, Robert calls George in the office. He's really excited, because he could finally crack the code. However, George has to cut his friend short as he goes on an on about the various ways he had tried to read this message. Grudgingly, Robert agrees to just send George the email with the plain text.&lt;/p&gt;&#xA;&lt;p&gt;George reads this letter and acts like one could expect.&lt;/p&gt;&#xA;&lt;h1&gt;Now, did George finally date Kari? &lt;br&gt;Or did he drown his sorrows in the Red Sofa bar trying to set his mind on somebody else?&lt;/h1&gt;&#xA;&lt;p&gt;The original message George received on the letter ( - transcribed to ASCII text for puzzler's convenience - ) is the following:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;DO DATING SINGLE,&#xA;DOUBT NOTICED BELONG IF HIGH NOTICED OF A POST BE! &#xA;IF CENTER OR AT THE YOUR HIGH HOUSE OFFICE BEFORE NEVER CONSTANT,&#xA;CAN LETTER MEET THAN YOUR: YOUR FOR HIGH TOWN FIRST NEVER RESTAURANT.&#xA;YOU, YOUR MINUTES BEAUTIFUL READ HAVE, CAN IF OR RESTAURANT PERMANENT&#xA;WILL CENTER WAIT THERE BRIDGE. IF EARN THE YOUR BLUE TOWN TIME HANDSOME.&#xA;PARK JUST TIME AFTER PUNCTUAL? ASSUME YOUR INTERESTED HAVE ARE SO? &#xA;WHERE WITH SHADOW. IN YOUR CENTER RED TIME LITTLE THAT IF AND INTERESTED &#xA;IS INDEED DEAR YOUR. IF GOOD YOUR AFTER SIXTEEN CHINESE. IF ME INDEED &#xA;THEIR RESPECT I NIGHT TOWN THEIR CLOSES SATURDAY INDEED CHANCE OFFICE USE.&#xA;IF WITH NOTICED MIGHT YOUR THIS. IF WE TOWN MIGHT YOUR ATTENTION THEIR &#xA;LET WHO THEIR LIVE BOY BUT JUST. SCHOOL NOON YES BEAUTIFUL GROUP BECOMING &#xA;ONLY CENTER SHADOW THEIR BECAUSE STUPID. CAN IF DOUBT NOTICED MIGHT AFTER &#xA;BRAIN FRIDAY THEIR FOR TO AM ATTENTION NOON. IF AND NOTICED THE SUGGEST. &#xA;SPEAK IF TELL WHERE MIGHT ONE GIVE LAKE ALONE THEIR PREJUDICE ON COME ATTENTION NOT. &#xA;IF WITH WANT FRIDAY SMART MY INDEED THEIR LIVE PEOPLE BELIEVE THEIR HOPES SEE. &#xA;IN YOUR DOUBT NOTICED COMPANION NIGHT GIVE, PARK IN YOUR ASSUME NOTICED RATHER, &#xA;IF WITH NEXT STARS WOULD YOUR.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Note: You do not need the story to decipher the letter. The cryptogram is self-contained in the text. The deciphered text is plain English. (Might contain spelling or grammar mistakes, but not on purpose.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Puzzle-answers without a decrypted letter are invalid.&lt;/p&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-09-01T10:45:24.630" Title="A love letter - or is it?" Tags="|cipher|story|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18370" PostTypeId="1" AcceptedAnswerId="18372" CreationDate="2015-07-26T20:56:16.533" Score="7" ViewCount="382" Body="&lt;p&gt;On the planet of Caturn, God created $m$ male cats and $f$ female cats, each with zero stripes. At any time, two cats of opposite genders can mate, giving birth to a cat with a gender of the parents' choice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The laws of cat genetics dictate that the number of stripes a child has will be one more than the number of stripes of its stripier parent. In other words, if the mother has $s$ stripes, and the father has $t$ stripes, the child will have $1+\max(s,t)$ stripes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, God issued the following two commandments:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;No Incest:&lt;/strong&gt; Two cats may not mate if they have a common ancestor.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Similarity:&lt;/strong&gt; Two cats may only mate if their stipe numbers differ by at most one. (A cat with $9$ stripes can only mate with an $8,9$ or $10$ stripe cat.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Assuming the cats obey these commandments, what is the largest number of stripes a cat on Caturn could possible have (as a function of $m$ and $f$)?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This is a variant of &lt;a href=&quot;https://puzzling.stackexchange.com/questions/10238/how-long-can-a-population-last-without-incest&quot;&gt;How long can a population last without incest?&lt;/a&gt; Other than rewording, the only change I made was adding the &quot;Similarity&quot; rule.&lt;/em&gt; &lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2019-07-08T18:58:02.553" Title="Avoiding Incest and Intergenerational Mating" Tags="|algorithm|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18371" PostTypeId="1" AcceptedAnswerId="18548" CreationDate="2015-07-26T21:17:55.897" Score="26" ViewCount="3550" Body="&lt;p&gt;This puzzle belongs to the puzzle series: &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=hyper-modern%20is%3Aquestion&quot;&gt;hyper-modern art&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The two friends in the gallery of hyper-modern art are moving on to the next room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;My head still aches from the colourful pixels under this microscope. Could we take a little rest please?&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Sure. Let's sit down here for a moment and rest our eyes on this rather simple black and white painting. It's somehow peaceful, don't you think?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Hmm, sort of. What is it called?&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;What would you call it?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Not sure. Maybe  &lt;code&gt;Autumn storm in hiragana&lt;/code&gt; ?&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Good one! No, it is actually called: &lt;code&gt;The code to the next room.&lt;/code&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Really? I wonder if it really is that.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Of course it is! Have you forgotten how hyper-modern art works? In fact, we can only get into the next room if we enter the code into the number pad over there, see? Unfortunately, our HUD does not work in this room, so we'll have to figure it out ourselves...&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/J73XO.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/J73XO.png&quot; alt=&quot;Code&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The goal of this puzzle is to derive the number code from the picture and describe how it can be gained. The story text is just flavour. The puzzle can be solved from the image alone and it could be solved from a printed version of that image. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is a unique solution, so please don't answer with &quot;I can see xy&quot;-type answers. The solution has to contain &lt;strong&gt;both&lt;/strong&gt; the number &lt;strong&gt;code&lt;/strong&gt; &lt;em&gt;and&lt;/em&gt; technique how to find it.&lt;/p&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-08-07T21:16:36.953" Title="Find the number code in the picture" Tags="|visual|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18372" PostTypeId="2" ParentId="18370" CreationDate="2015-07-26T21:48:45.113" Score="4" Body="&lt;p&gt;Here is the answer: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A population can breed &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; cats if and only if &lt;span class=&quot;math-container&quot;&gt;$\min(m,f)\geq F_s$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$m+f\geq F_{s+2}$&lt;/span&gt; (where &lt;span class=&quot;math-container&quot;&gt;$F_n$&lt;/span&gt; is the &lt;span class=&quot;math-container&quot;&gt;$n^{th}$&lt;/span&gt; Fibonacci number, starting with &lt;span class=&quot;math-container&quot;&gt;$F_0=0$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$F_1=1$&lt;/span&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To prove that this is correct, let us consider the family tree of the stripiest cat. We may notice that any cat in this ancestry with stripes might as well be a hermaphrodite (able to breed with any other cat) since we may choose their gender freely and they may only breed once in this tree (as otherwise there would be incest).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivated by this, we see that gender only matters in so far as we have some number of stripeless male-female couples, along with some more zero-stripe cats (of either gender) who breed with hermaphrodites. In particular, define &lt;span class=&quot;math-container&quot;&gt;$f(s)$&lt;/span&gt; as the number of male-female couples required to breed a cat with &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; stripes and &lt;span class=&quot;math-container&quot;&gt;$g(s)$&lt;/span&gt; as the number of additional cats required. We see easily that, as the cheapest way to make a cat with &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; stripes is to breed a cat with &lt;span class=&quot;math-container&quot;&gt;$s-1$&lt;/span&gt; stripes with one with &lt;span class=&quot;math-container&quot;&gt;$s-2$&lt;/span&gt; stripes that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$f(s)=f(s-1)+f(s-2)$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$g(s)=g(s-1)+g(s-2)$$&lt;/span&gt;&#xA;and we see the initial conditions are:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$f(0)=0$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$f(1)=1$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$g(0)=1$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$g(1)=0$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Easily, from this, we see that &lt;span class=&quot;math-container&quot;&gt;$f$&lt;/span&gt; is exactly the Fibonacci sequence and &lt;span class=&quot;math-container&quot;&gt;$g(s)=f(s-1)$&lt;/span&gt;. Thus, to breed a cat with &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; stripes, we need at least &lt;span class=&quot;math-container&quot;&gt;$F_s$&lt;/span&gt; couples along with &lt;span class=&quot;math-container&quot;&gt;$F_{s-1}$&lt;/span&gt; additional individuals - meaning both genders need at least &lt;span class=&quot;math-container&quot;&gt;$F_s$&lt;/span&gt; individuals. So, in total, we need at least &lt;span class=&quot;math-container&quot;&gt;$F_{s+2}=F_{s-1}+2F_s$&lt;/span&gt; individuals.&lt;/p&gt;&#xA;" OwnerUserId="4765" LastEditorUserId="1649" LastEditDate="2019-07-08T18:58:02.553" LastActivityDate="2019-07-08T18:58:02.553" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="18374" PostTypeId="2" ParentId="16421" CreationDate="2015-07-27T01:53:29.937" Score="4" Body="&lt;p&gt;&lt;em&gt;(This is the solution I was thinking of when I gave the puzzle, given in response to a query by BmyGuest.  It avoids the use of facts about fields, but is conceptually almost identical to the accepted answer by f'', and could be read as an exposition of that argument.)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As f'' says, all the shots may hit the same symbol, so we might as well just use one shot.  And as argued in the other answers, if the duplicator hits any ● with at least 300% charge, entry is guaranteed.  We just need to know how much charge hitting a ○ would require.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Call a dot number with only two dots a &lt;strong&gt;translation&lt;/strong&gt; (cf. the subexpression $x^{2^{n-1}-1}-1$ in the answer by f'').  By adding a translation to some other dot number $x$ at an appropriate location, we can move one of $x$ 's dots around.  For example, we can take the second dot in ●○●○●●● and slide it far to the left:&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;code&gt;&#xA;                                        ●○●○●●●&#xA;                                  ●○○○○○○○●&#xA;                                  ----------&#xA;                                  ●○○○○○●○○○●●●&#xA;&lt;/code&gt;&#xA;&lt;br&gt;&#xA; By adding a translation in several places, we can shift an entire portion of a dot number (cf. the subexpression $A(x)x^{b+1}(x^{2^{n-1}-1}-1)$ in the answer by f'').  In particular, we can replicate the effects of the duplicator hitting a ○ at $(\mathrm{length}(t) - 1)\cdot 100$%:&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;code&gt;&#xA;                                        ●○●○●●●&#xA;                                  ●○○○○○○○●&#xA;                                ●○○○○○○○●&#xA;                                -----------&#xA;                                ●○●○○○○○○○○○●●●&#xA;&lt;/code&gt;&#xA;&lt;br&gt;&#xA; So if we call the original dot number $x$ and have a translation $t$ that is divisible by $x$, hitting a ○ in $x$ at $(\mathrm{length}(t) - 1)\cdot 100$% (or more) power will yield a sum of numbers divisible by $x$, i.e., a non-prime.&lt;br&gt;&#xA;&lt;br&gt;&#xA; How then can we find a suitable, and preferably short $t$?  Let's start by asking a simpler question: if we have two dot numbers, $x$ and $y$, how can we check whether $x$ is a factor of $y$?&lt;br&gt;&#xA;&lt;br&gt;&#xA; With natural numbers, we would work the long multiplication backwards to see if there's a number that, when multiplied by $x$, produces $y$; this process is just long division.  The same working-backwards strategy applies here, but we have to remember that the addition did not carry, so our subtraction (the part that works the addition backwards) must not borrow.  For instance, here's Naomi's original problem as long division, where getting no remainder tells us that ●○○●●● is divisible by ●○●●:&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;code&gt;&#xA;                                         ●○●&#xA;                                      ------&#xA;                                 ●○●●|●○○●●●&#xA;                                     -●○●●&#xA;                                      ------&#xA;                                       ○●○●● &#xA;                                     - ○○○○&#xA;                                      ------&#xA;                                        ●○●●&#xA;                                     -  ●○●●&#xA;                                      ------&#xA;&lt;/code&gt;&#xA;&lt;br&gt;&#xA; (Notice how the first subtraction does not borrow, just as the corresponding addition did not carry.)&lt;br&gt;&#xA;&lt;br&gt;&#xA; In this case our $y$ should be a short translation, so let's run the division on $●○\cdots$ and stop the moment that a dot on the end would have given us no remainder, assuming such a moment ever comes.  An example, with some intermediate values labeled for later:&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;code&gt;&#xA;                                         ●○●●●...&#xA;                                      --------&#xA;                                 ●○●●|●○○○○○○○... ←z_0&#xA;                                     -●○●●&#xA;                                      ------&#xA;                                       ○●●○○○○... ←z_1&#xA;                                     - ○○○○&#xA;                                      ------&#xA;                                        ●●○○○○... ←z_2&#xA;                                     -  ●○●●&#xA;                                      ------&#xA;                                         ●●●○○... ←z_3&#xA;                                     -   ●○●●&#xA;                                      ------&#xA;                                          ●○●○... ←z_4&#xA;                                     -    ●○●●&#xA;                                      --------&#xA;                                           ○○●... ←z_5&#xA;&lt;/code&gt;&#xA;&lt;br&gt;&#xA; Here we wanted that last ○ to be a ●, which tells us that ●○○○○○○● is the shortest translation divisible by ●○●●.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Now observe the $z$s.  While the first $\mathrm{length}(x) - 1$ symbols could be anything but all blanks, the rest must be ○s, so the $z$s can take on at most $2^{\mathrm{length}(x) - 1}-1$ distinct values; make these values the vertices of a graph (cf. the $2^{n-1}-1$ residue classes in the answer by f'').  Continuing, if we know $z_i$, we can deduce $z_{i+1}$ by xoring $x$ at the beginning only if $z_i$ begins with a ●; let this pattern define the directed edges of our graph, making it functional.  Similarly, if we know $z_{i+1}$, we can recover $z_i$ by xoring $x$ at the end of its prefix, but only when that prefix ends with a ●.  Therefore, the graph's reverse is also functional.&lt;br&gt;&#xA;&lt;br&gt;&#xA; But a functional graph whose reverse is also functional must be the disjoint union of directed cycles.  So if we start with $●○\cdots$, we must eventually get back there, and thus find a lone dot that can be eliminated from the remainder by the other end of our translation.  Which is to say that $t$ exists, and in particular corresponds to a charge of 100% times the length of the cycle containing $●○\cdots$.&lt;br&gt;&#xA;&lt;br&gt;&#xA; This cycle length is no more than the number of vertices, $2^{\mathrm{length}(x) - 1}-1$, which is in turn no more than $2^{n-1}-1$.  So one shot at $\max(3,2^{n-1}-1)\cdot 100$% power will do.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13175" LastEditorUserId="13175" LastEditDate="2015-07-27T03:27:04.047" LastActivityDate="2015-07-27T03:27:04.047" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18375" PostTypeId="2" ParentId="18366" CreationDate="2015-07-27T02:16:15.507" Score="6" Body="&lt;p&gt;We first notice the following crucial fact:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The number of zombies on the field never decreases.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is obvious since each step adds one zombie and removes at most one. However, since the number of zombies can never exceed $9$, it follows that, after some time $T$, the number of zombies on the field is constant*. We will prove that zombies created after time $T$ satisfy the desired condition. After this point, the following rule must be observed:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If there is not a zombie at the leftmost plot, a zombie must be added there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We must follow this as otherwise one would be adding a zombie without removing one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we consider that, if there are $n$ zombies on the field at all time, we would expect that, as they walk a total of $n$ steps left each turn. We get a bit of a hint on how to use this fact in the $n=2$ case - in particular, if we make a list of which plot the zombies are placed at (with $1$ being the leftmost), we might see a pattern like:&#xA;$$2,2,3,1,5,1,1,1,4,1,1,2,2,\ldots$$&#xA;where (after a point) we will necessarily see a concatenation of patterns of the form:&#xA;$$k,\underbrace{1,1,\ldots,1,1}_{k-2\text{ times}}$$&#xA;since we have to &quot;wait&quot; for the zombie placed at $k$ to reach the start. When we're summing over such a pattern, we can see that we can &quot;shift&quot; the mass away from the $k$ and towards the $1$s to leave a sum of $2(k-1)$. In a sense, when we put a $k&amp;gt;2$ in the pattern, we're taking a &quot;loan&quot; from later in the pattern, which we slowly repay by taking only $1$ step (less than the &quot;allotted&quot; $2$ steps per zombie).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This suggests to us we want to find some function which takes a state of the game as input and tells us how much &quot;debt&quot; we have to repay (we might also call it an energy function). We will define a state as a subset $S$ of the squares in the garden, representing the set which are occupied by zombies right before a new one is placed. Note that thus the rightmost square will never be occupied, as a zombie placed there immediately wanders left &lt;em&gt;and&lt;/em&gt; that if there are $n$ zombies $|S|=n-1$ since the state occurs after a zombie has just walked off and before its replacement is added. A legal transition from $S$ to another state consists adding a new element, then decrementing all the elements and removing $0$ if it appears. We wish to find a function $E_n(S)$ on states such that $|S|=n-1$ which satisfies the following crucial property:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose state $S'$ could occur immediately after state $S$ when Dr. Zomboss adds a zombie at position $k$. Then $E_n(S)-E_n(S')=n-k$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is to say we are measuring the difference between a given move and the allotment of zombie steps - so if we've got &quot;too many&quot; steps $E_n(S')$ will be more than $E_n(S)$. Notice that this defines $E_n$ up to a constant factor if it exists at all. One may explicitly compute it by choosing $E_n(S)=e$ for some arbitrarily chosen $S$ and $e$ and finding the energy of all states reachable from it - and thinking about that idea for a second leads us to the following valid energy function:&#xA;$$E_n(S)=\sum_{s\in S}s$$&#xA;One may check that this satisfies the desired condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, the rest is easy: Suppose $S_0$ was the state before the first zombie included in the sum is added and $S_{1000}$ was the state after the last zombie in the sum was added (and the intervening $S_i$ are defined accordingly). In particular, we may show that the number of steps walked by those $1000$ zombies is:&#xA;$$E(S_{1000})-E(S_0)+1000n.$$&#xA;This is easy because if the $i^{th}$ zombie walked a distance of $k_i$, then $E(S_{i})-E(S_{i-1})=k_i-n$, thus we may algebraically manipulate the desired sum as in the following sequence of equivalent expressions:&#xA;$$k_1+k_2+k_3+\ldots+ k_{999}+k_{1000}$$&#xA;$$(k_1-n)+(k_2-n)+(k_3-n)+\ldots + (k_{999}-n)+(k_{1000}-n)+1000n$$&#xA;$$[E(S_1)-E(S_0)]+[E(S_2)-E(S_1)]+\ldots +[E(S_{1000})-E(S_{999})]+1000n$$&#xA;where this last sum telescopes to&#xA;$$E(S_{1000})-E(S_{0})+1000n.$$&#xA;Now, if we let $A$ be the state with the least energy and $B$ be the state with the most, the we have:&#xA;$$E(A)-E(B)+1000n \leq E(S_{1000})-E(S_{0})+1000n\leq E(B)-E(A)+1000n.$$&#xA;meaning the difference between any pair of such sums may be at most $2[E(B)-E(A)]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, $B=\{8,7,\ldots,9-n\}$ has the most energy and $A=\{1,2,\ldots, n\}$ has the least - and $A$ is a translation of $B$ leftwards by $(8-n)$ meaning the difference between the energy is $n(8-n)$. This is maximized when $n=4$ yielding $16$ as the total difference in energy. Thus, no such sum may differ by more than $2\cdot 16=32$ after the field settles into a constant number of zombies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*More strongly, if we consider a game state to be the set of plots on which zombies lie and say $A\sim B$ if state $A$ could occur after state $B$ and vice versa, then it happens that $A\sim B$ is equivalent to $|A|=|B|$ meaning that the number of zombies is essentially the only invariant of this type - so we're in trouble if this $T$ doesn't cut it!)&lt;/p&gt;&#xA;" OwnerUserId="4765" LastEditorUserId="4765" LastEditDate="2015-07-27T02:21:46.447" LastActivityDate="2015-07-27T02:21:46.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18376" PostTypeId="2" ParentId="18366" CreationDate="2015-07-27T03:19:02.053" Score="2" Body="&lt;p&gt;@MiloBrandt already posted a solution (make sure to upvote him!), but I wanted to give an alternative explanation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As Milo noted, the number of zombies never decreases and therefore at&#xA;  some moment it stabilizes, say at $1 \leq K\leq 9$ zombies after time&#xA;  $T$. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now consider any $1000$ consecutive zombies appearing past time $T$&#xA;  and take a picture of the lawn in the moment each of them gets dropped&#xA;  on it - this makes a total of $1000$ pictures. Since on every picture&#xA;  there are exactly $K$ zombies, we see exactly $1000K$ zombies on all&#xA;  pictures.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now notice that almost all of the selected $1000$ zombies appear on as&#xA;  many pictures as lawn spots they travel. The zombies for which this is&#xA;  not true are just the $K$ zombies, which appear on the last picture.&#xA;  We easily see that they have traveled between $1+2+...+K-1$ and&#xA;  $(10-K)+(11-K)+...+8$ spots more than the number of pictures they&#xA;  appear on.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Similarly, on the $1000$ pictures we have made, there are $K-1$&#xA;  additional zombies, which appear multiple times (you can see all of&#xA;  them on the first picture). The total number of times they show up is&#xA;  again between $1+2+...+K-1$ and $(10-K)+(11-K)+...+8$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Therefore we conclude that the total distance the selected $1000$&#xA;  zombies travel is \begin{align}&amp;amp;&amp;amp;1000K \pm&#xA;\{[(10-K)+(11-K)+...+8]-[1+2+...+K-1]\}\\&#xA;=&amp;amp;&amp;amp;1000K \pm (9-K)(K-1). \end{align} Since $(9-K)(K-1)$ is a number between $0$ and $16$, this solves the problem.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This problem was supposedly the hardest question from this year's IMO, but thought it could be translated into a nice puzzle. Hope you liked it.&#xA;&lt;a href=&quot;http://imo2015.org/solution.php?lang=en&quot; rel=&quot;nofollow&quot;&gt;http://imo2015.org/solution.php?lang=en&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once again, make sure to upvote Milo's answer, as he uploaded first a legit solution.&lt;/p&gt;&#xA;" OwnerUserId="14140" LastEditorUserId="14140" LastEditDate="2015-07-27T03:29:07.873" LastActivityDate="2015-07-27T03:29:07.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18377" PostTypeId="1" CreationDate="2015-07-27T05:02:37.997" Score="-3" ViewCount="2936" Body="&lt;p&gt;There are eight students named Aman, Bobby, Chameli, Deepak, Ekta, Farook, Garima and Harsh in a school. They are in class 6th, 7th and 8th with no more than three students in any class.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of them has a favorite subject as Geography, Physics, English, Sanskrit, Maths, Chemistry, Biology and Economics not necessarily in the same order.  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Aman and Ekta study in same class but not with Bobby&lt;/li&gt;&#xA;&lt;li&gt;Deepak likes Chemistry and studies in class 7th only with Harsh&lt;/li&gt;&#xA;&lt;li&gt;Bobby does not study in 8th class&lt;/li&gt;&#xA;&lt;li&gt;Chameli and Farook study in the same class&lt;/li&gt;&#xA;&lt;li&gt;Those who study in class 6th do not like subject Maths or Biology&lt;/li&gt;&#xA;&lt;li&gt;Farook likes physics&lt;/li&gt;&#xA;&lt;li&gt;The one who studies in class 7th like English&lt;/li&gt;&#xA;&lt;li&gt;Chameli does not like Geography&lt;/li&gt;&#xA;&lt;li&gt;Aman's favorite subject is Sanskrit&lt;/li&gt;&#xA;&lt;li&gt;Garima does not like Biology  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Can someone explain what this means: Those who study in class 6th do not like subject &quot;Maths or Biology&quot;. Does it mean those who study in 6th standard do not like Maths and Biology both?  &lt;/p&gt;&#xA;" OwnerUserId="13751" LastEditorUserId="5840" LastEditDate="2015-07-27T06:17:42.470" LastActivityDate="2015-07-27T06:25:36.800" Title="Logic-grid puzzle. How do I have to interpret the clue?" Tags="|logical-deduction|explanation|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18378" PostTypeId="2" ParentId="18311" CreationDate="2015-07-27T05:42:36.170" Score="15" Body="&lt;p&gt;My first is a dye, most commonly used.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The dye most of us use every day would be &lt;strong&gt;ink&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My second is twentieth, in a cup.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;As per Doge's comment, T is the twentieth letter and sounds like &lt;strong&gt;tea&lt;/strong&gt; in a cup.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My third is gloomy, shrouded in cloud. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Clouds and gloom are both often described as &lt;strong&gt;gray&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fourth periodically gets to a half century.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;As per mmking's comment, the 50th element in the periodic table is &lt;strong&gt;tin&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fifth often stands for unity.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;As per &lt;a href=&quot;https://puzzling.stackexchange.com/a/18324/12511&quot;&gt;karhell's answer&lt;/a&gt;, &lt;strong&gt;one&lt;/strong&gt; is the usual symbol used for unity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Crack them all, and put them together. To find the answer, that perhaps I have left hither.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;To &quot;crack them&quot;, look to the first line: &quot;Follow each one through to the end, but maybe not entirely.&quot; If we remove the last letter of each answer and string them together, we get the answer: &lt;strong&gt;integration&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12511" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-07-27T05:42:36.170" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18379" PostTypeId="2" ParentId="18377" CreationDate="2015-07-27T06:05:43.000" Score="4" Body="&lt;p&gt;Those who study in the 6th class do not like &lt;strong&gt;both&lt;/strong&gt; Maths and Biology is likely the correct interpretation because it leads to a unique solution.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just make a table as follows and apply the rules. Start with the most definitive statements such as the ones about Deepak and Farook.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/FKYfS.png&quot; alt=&quot;NamesClasses&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-07-27T06:25:36.800" LastActivityDate="2015-07-27T06:25:36.800" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18380" PostTypeId="2" ParentId="18377" CreationDate="2015-07-27T06:14:11.267" Score="1" Body="&lt;p&gt;This seems to be a question for a &lt;em&gt;particular&lt;/em&gt; puzzle not created by yourself, so it might be hard to answer this question. &#xA;Also, I think your question does not really require the puzzle to be posted but is more about general logic in such sentences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, using &lt;em&gt;strict&lt;/em&gt; logic, I think that the sentence&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Those who study in class 6th do not like subject &quot;Maths or Biology&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;would mean being member of class 6 implies that you &lt;em&gt;either&lt;/em&gt; don't like Maths, &lt;em&gt;or&lt;/em&gt; don't like Biology &lt;em&gt;or&lt;/em&gt; don't like neither.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;true if (NOT M)||(NOT B) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;true if (TRUE)||(FALSE)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;true if (TRUE)||(TRUE)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;true if (FALSE)||(TRUE)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sentence&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Those who study in class 6th do not like either the subject &quot;Maths&quot; or &quot;Biology&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;would mean that it implies &lt;em&gt;either&lt;/em&gt; of the two things, but &lt;em&gt;not&lt;/em&gt; both at the same time. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;But to be honest, it is not a very clear sentence and a good puzzle would maybe be more explicit. So being confused is natural ;c). I think Len is correct what the puzzle &lt;em&gt;means&lt;/em&gt; in this case, but requiring the puzzle-solution to be unique as an argument of interpretation of the clues is a very bad sign for the quality of the puzzle...&lt;/p&gt;&#xA;" OwnerUserId="5840" LastActivityDate="2015-07-27T06:14:11.267" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18381" PostTypeId="2" ParentId="18281" CreationDate="2015-07-27T08:24:46.233" Score="1" Body="&lt;p&gt;To help the others out, I will add my own two cents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nothing is more important to me,&#xA;than the bonds of my family.&#xA;Each piece below is a key from them to thee,&#xA;spend some time and you just might get free.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I see this as a strong hint to chemistry. I would guess for a link to fatty acids (fat), as they can be saturated or unsaturated, depending on the type of bonds between their carbon atoms. EDIT: I just realised, if I am sodium, then each of the parts below could stand for a molecule which includes sodium or the element in that molecule which is bonded with sodium.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm all natural, all the time -&#xA;Euler used me but I feel fine.&#xA;I'm the find in many ways,&#xA;a unique treasure who likes Swiss cheese.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Eulers number, as Nefer007 said. EDIT: It might be table salt we are talking about here. Still, Eulers number might be important in the calculation. And I would keep chlorine in mind.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Irrelevant information influences influential individuals.&#xA;I infrequently integrate idioms in ideas.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, as Nefer007 said, &lt;em&gt;i&lt;/em&gt; EDIT: The I in the periodic table is iodine.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You get rid of me several times a day,&#xA;but I'll come back after a slight delay.&#xA;Those bound to earth already see,&#xA;should you ever stop relieving me,&#xA;It's a sign you've been released.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This relates to breathing, where you get rid of CO2 (and this is maybe a link to the carbohydrates?) If you stop breathing... you could be seen as released. EDIT: still, carbon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm always moving,&#xA;with no need to stop.&#xA;I've no place to be going,&#xA;but it doesn't matter - I'm first!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, I am thinking of elements instantly. The first element would be hydrogen, so we are again hinting at carbohydrates. Maybe we need to look at the number attached to carbohydrates separately (carbon:4, hydrates:2)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'd tell you who I am if you could catch me,&#xA;but I'd slither away before you could see me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;That could be any element which has a short half-life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To summarize: My first brother is always the same, the next is narcissistic.&#xA;Another brother moves like a river, but his twin is impossible.&#xA;My sister can be quite quiet, but when she's angry you don't want to hear her hiss.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I have no idea, but then I would again try looking for elements. Especially a family of elements which has 27 members (no idea), which a sub-family in it that has 6 (alkali-metals or noble gases). Sodium is an alkali-metal, so I could be sodium. EDIT: Okay, we have bonds to chlorine, iodine, carbon, hydrogen (moves like a river -&gt; water) and something I don't know. We need to sort them after these hints and maybe use there atomic numbers in the formula. (iodine: 53, hydrogen:1, carbon:6, chlorine: 17, and one number I am missing.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10885" LastEditorUserId="10885" LastEditDate="2015-07-27T08:47:34.160" LastActivityDate="2015-07-27T08:47:34.160" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18382" PostTypeId="1" AcceptedAnswerId="18410" CreationDate="2015-07-27T09:01:16.957" Score="6" ViewCount="511" Body="&lt;p&gt;Sound out the word with the clues. My first one, should be quite easy:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You can regularly find me inserted below.&lt;/li&gt;&#xA;&lt;li&gt;Something we all do.&lt;/li&gt;&#xA;&lt;li&gt;A short-hand account of the matter from my point of view.&lt;/li&gt;&#xA;&lt;li&gt;I heard I finished 5th, but I was actually in 25th place.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Hints:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. You can also find me near the end.&lt;br /&gt;&#xA; 2. We share this commonality with trees.&lt;br /&gt;&#xA; 3. &quot;His&quot;, &quot;her&quot;, &quot;their&quot; etcetera will work in place of &quot;my&quot;.&lt;br /&gt;&#xA; 4. Has already been worked out.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4659" LastEditorUserId="4659" LastEditDate="2015-07-28T08:20:34.563" LastActivityDate="2015-07-28T15:46:50.737" Title="Word Identification" Tags="|word|wordplay|" AnswerCount="6" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="18384" PostTypeId="2" ParentId="18382" CreationDate="2015-07-27T10:47:03.747" Score="-1" Body="&lt;p&gt;The word is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;EASY&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can regularly find me inserted below.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Normally in tasks lists, easy tasks will be placed/inserted below.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Something we all do.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;All of us can do things that are easy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A short-hand account of the matter from my point of view.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Seeing/facing simple tasks/things, people think 'easy (ez)'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I finally finish in 25th place.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Easy finishes with 'Y', the last letter, which is the 25th alphabet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="2730" LastEditDate="2015-07-27T10:59:23.730" LastActivityDate="2015-07-27T10:59:23.730" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18386" PostTypeId="2" ParentId="18382" CreationDate="2015-07-27T13:14:05.723" Score="1" Body="&lt;p&gt;My answer:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; COMMENT  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's see at these lines:&lt;br&gt;&#xA;You can regularly find me inserted below.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Comments are usually posted below question/topic post.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Something we all do.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We all make comments (so: we all comment things)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A short-hand account of the matter from my point of view.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Comments wouldn't contribute to any change, normally. These are just comments.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I finally finish in 25th place.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (no idea yet)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm aware I may misunderstood 3rd verse. If so, please tell me so.&lt;/p&gt;&#xA;" OwnerUserId="12739" LastActivityDate="2015-07-27T13:14:05.723" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18387" PostTypeId="2" ParentId="18350" CreationDate="2015-07-27T13:55:43.983" Score="1" Body="&lt;p&gt;Not exactly the most precise answer but:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is a theorem known as the Chicken McNugget Theorem (or Postage Stamp Problem or Frobenius Coin Problem) that states: if $m$ and $n$ are relatively prime, then for nonnegative integers $a,b$, the &lt;strong&gt;largest integer&lt;/strong&gt; that cannot be expressed in the form $am + bn$ is $mn-m-n$. &lt;br/&gt;&lt;br&gt; However, note that for sand timer problems we can flip over a timer while the other is going. Thus we aren't restricted to nonnegative integers. So by using the CMT, we overestimate the minimum value, but we still get a value nonetheless.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So for the warmup:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since $7$ and $4$ are relatively prime, we get: $28-7-4=17$. So we can definitely measure all times over $17$ min. Note that as others have computed already, you can measure all values over $6$ min.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Easy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since $5$ and $8$ are relatively prime, we get: $40-5-8 = 27$. So we can definitely measure all times over $27$ min.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Medium:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since $6$ and $11$ are relatively prime, we get: $66-6-11 = 49$ So we can definitely measure all times over $49$ min.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hard:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If $m$ and $n$ are relatively prime, then we know all numbers above $mn-m-n$ can be measured by the two timers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13881" LastActivityDate="2015-07-27T13:55:43.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18388" PostTypeId="2" ParentId="18382" CreationDate="2015-07-27T14:03:28.567" Score="1" Body="&lt;p&gt;I think the word is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;VISIONARY&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You can regularly find me inserted below.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;V&lt;/strong&gt; comes under (or after) U in the English alphabet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;Something we all do.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We all &lt;strong&gt;see&lt;/strong&gt; with our eyes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;A short-hand account of the matter from my point of view.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; On television/radio, short versions of news are broadcast or rather, the news goes &lt;strong&gt;on air&lt;/strong&gt;. (&lt;em&gt;This is probably a stretch, but it fits the overall aim...&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;I heard I finished 5th, but I was actually in 25th place.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Y&lt;/strong&gt; is the 25th alphabet, but in some words (like the one I claim) it gives the sound of E, which is the 5th alphabet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Putting it all together&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;V See Onair Y&lt;/strong&gt; which looks and sounds a lot like the word &lt;strong&gt;VISIONARY&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-07-27T14:03:28.567" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18389" PostTypeId="2" ParentId="18382" CreationDate="2015-07-27T14:33:16.290" Score="1" Body="&lt;p&gt;How about&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tag - inserted below the questions&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lie - we all do it, though some more than others&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tale - my version of the story&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Y - 25th Letter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Giving:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tagliatelle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-07-27T14:33:16.290" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18390" PostTypeId="1" AcceptedAnswerId="18492" CreationDate="2015-07-27T15:11:17.803" Score="27" ViewCount="2252" Body="&lt;p&gt;The main() method in the following Java program is a text rebus.  What does it represent?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import java.util.*;&#xA;import java.awt.geom.*;&#xA;&#xA;public class WhatDoesMainRepresent {&#xA;    private static HashSet&amp;lt;Obj&amp;gt; hashSet = new HashSet&amp;lt;Obj&amp;gt;();&#xA;    private static AffineTransform e = new AffineTransform();&#xA;&#xA;    static {&#xA;        hashSet.add(new Obj());&#xA;    }&#xA;&#xA;    public static void main(String[] args) {&#xA;        hashSet.add(new Obj());&#xA;        e.rotate(Math.PI);&#xA;        {}&#xA;    }&#xA;&#xA;    private static class Obj {&#xA;        public int hashCode() {&#xA;            return 1;&#xA;        }&#xA;    }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Hints:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. The AffineTransform object is named &lt;code&gt;e&lt;/code&gt; for a reason.&lt;br&gt;&#xA; 2. The &lt;code&gt;hashSet.add(new Obj())&lt;/code&gt; line in the main() method represents a single word that is a verb.  (So far that word has not been mentioned.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4450" LastEditDate="2015-07-30T05:31:29.963" LastActivityDate="2015-07-30T14:44:50.357" Title="What does this Java main() method represent?" Tags="|rebus|computer-puzzle|" AnswerCount="5" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="18391" PostTypeId="2" ParentId="18305" CreationDate="2015-07-27T16:39:41.633" Score="2" Body="&lt;p&gt;Since it is not an animal per se, I was thinking it could be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Modest Mouse&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the cheapest of the flying animals,&#xA;As such I cannot give any change.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; makes me think of cheap people bumming rides and a mouse gets rides from birds in a way. I can't think of a better description for that at the moment&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I guess I am just riffing yeah yeah,&#xA;But I heard my creator is &quot;too too wise&quot;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; too too wise makes me think of owls and owls eat mice&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Anyways I will just soar above, won't I,&#xA;Oh oh oh oh, you realize I must go.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A mouse will soar above as an owl carries it away&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My stay has been sugary, but bye.&#xA;I will just soar above, won't I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A mouse could be living in a sugar cain field, and the incorrect assumption, it's going to get eaten, that it will just soar above makes me think of a modest mouse &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13925" LastEditorUserId="13925" LastEditDate="2015-07-27T16:50:45.547" LastActivityDate="2015-07-27T16:50:45.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18393" PostTypeId="2" ParentId="18305" CreationDate="2015-07-27T17:49:22.393" Score="6" Body="&lt;p&gt;I'm not confident (edit: I'm pretty confident after writing the answer), but it's close enough to try.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Free Bird&lt;/strong&gt;, the song by Lynyrd Skynyrd &lt;br&gt;&#xA; Some of these are lyrics references, the yeah yeah and Oh oh oh oh inspired this answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the cheapest of the flying animals,&#xA;As such I cannot give any change.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;'cause I'm as free as a bird now, and this bird you cannot change&quot;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I guess I am just riffing yeah yeah,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He sings &quot;yeah yeah&quot; after a few different lines to my recollection. Perhaps also right at the solo.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I heard my creator is &quot;too too wise&quot;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No idea. Either a reference to &quot;Lord&quot; in the song or the band (creator) Lynyrd Skynyrd? The song &lt;em&gt;is&lt;/em&gt; legendary.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Anyways I will just soar above, won't I,&#xA;Oh oh oh oh, you realize I must go.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Won't you fly high, free bird&quot; and some &quot;oh&quot;s in the verses.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My stay has been sugary, but bye.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Bye, bye, baby. It's been a sweet love.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;VERY late edit: I heard the opening credits of Happy Gilmore this weekend, which instantly reminded me that I never responded to the riddle's hints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My brother is a man least complex&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Another song by Lynyrd Skynyrd is &lt;strong&gt;Simple Man&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As well as a day lost to the breeze&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The chorus of &lt;strong&gt;Tuesday's Gone&lt;/strong&gt; is &quot;Tuesday's gone with the wind&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13237" LastEditorUserId="13237" LastEditDate="2015-08-04T20:32:53.563" LastActivityDate="2015-08-04T20:32:53.563" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18394" PostTypeId="2" ParentId="18390" CreationDate="2015-07-27T19:06:58.360" Score="14" Body="&lt;p&gt;I shame myself with this guess if it turns out to be wrong, but I suspect the text is the idiom&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://esl.about.com/od/vocabularyreference/a/express_around.htm&quot;&gt;been around the block once or twice&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reasoning being:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every key/value pair in a hash map is stored in a &lt;em&gt;bin&lt;/em&gt; determined by its hash code. In this case the hashcode for all objects of the &lt;code&gt;Obj&lt;/code&gt; class is identical, meaning such objects are stored in the same bin.&lt;br&gt;&lt;br&gt;The &lt;code&gt;e.rotate(Math.PI)&lt;/code&gt; statement performs a useless 180&amp;deg; inversion. The transform will therefore turn its operand &lt;em&gt;around&lt;/em&gt;. The &lt;code&gt;{}&lt;/code&gt; represents an empty &lt;em&gt;block&lt;/em&gt;.&lt;br&gt;&lt;br&gt;The static statement and initial statement in &lt;code&gt;main&lt;/code&gt; adds an object to the same &lt;em&gt;bin&lt;/em&gt; once and then twice.&lt;br&gt;&lt;br&gt;Hence: &lt;em&gt;bin&lt;/em&gt; &lt;em&gt;around&lt;/em&gt; the &lt;em&gt;block&lt;/em&gt; &lt;em&gt;once or twice&lt;/em&gt;.&lt;br&gt;&lt;br&gt;After typing this out, I realize how indescribably bad this pun is. I truly hope this isn't the correct answer. :P&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2015-07-27T19:06:58.360" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="18395" PostTypeId="1" AcceptedAnswerId="18407" CreationDate="2015-07-27T19:14:03.863" Score="19" ViewCount="774" Body="&lt;p&gt;If a phrase meets a certain criteria, I call it an Ascending Phrase™.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the examples below to find the rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/F1ZMs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/F1ZMs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Tips:&lt;/strong&gt;&lt;br&gt;&#xA;1. Font and case do not matter.&lt;br&gt;&#xA;2. No calculation is needed, unlike my last Phrase™ puzzle.&lt;br&gt;&#xA;3. The meanings of the words are not helpful in finding the rule.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to the suggestion of &lt;strong&gt;glibdud&lt;/strong&gt; and those who upvoted his comments in a couple different posts, here is a text version of the table above:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Ascending Phrases™                       Not Ascending Phrases™&#xA;doctors prescribe                        physicians order&#xA;noble attempt                            valiant effort&#xA;petty offense                            unimportant misdemeanor&#xA;teachers facilitate understanding        instructors expedite proficiency&#xA;only original geographic configurations  exclusively primary land setups&#xA;salesmen present guarantees              saleswomen furnish promises&#xA;genius requires perseverance             intelligence demands persistence&#xA;women distribute information             ladies scatter knowledge&#xA;workers unite                            employees unify&#xA;problems include universal dependability issues involve general faithfulness&#xA;humans prefer cappuccino                 people favor coffee&#xA;fortunate alumni volunteer               successful graduates labor&#xA;children desire avocados                 toddlers fancy apples&#xA;lawyers persuade individuals             attorneys influence people&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What makes a phrase an Ascending Phrase™?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;SUB&gt;&lt;strong&gt;Bonus&lt;/strong&gt;: If you have solved one of my &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%22Phrase%22%20user%3A463%20is%3Aquestion&quot;&gt;Phrase™&lt;/a&gt; or &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%22versatile%22%20user%3A463%20is%3Aquestion&quot;&gt;Word™&lt;/a&gt; puzzles before, and you solve this one, you will be the first person to solve more than one of these types of puzzles of mine, and you will receive 100 rep points from me, on top of any other points that this puzzle might give (i.e. It might also have a regular bounty on it &lt;em&gt;if it lasts long enough&lt;/em&gt;.)&lt;/SUB&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-07-28T15:04:07.307" Title="What is an Ascending Phrase™?" Tags="|pattern|english|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="18396" PostTypeId="1" AcceptedAnswerId="18399" CreationDate="2015-07-27T20:05:41.307" Score="14" ViewCount="2408" Body="&lt;p&gt;In a recently unknown land, roads are rapidly being built by volunteers. There is no construction experience required, and seasoned road builders keep a watchful eye. Exits are abundant, offering frontage roads, detours, and highways to far away places. It is truly a revolutionary system of transportation. &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The roads provide utility that the citizens of this land now rely on. However, some also use the roads for sport, racing each other to exotic destinations. With myriad paths to the finish, each racer must anticipate future exits and above all, never backtrack. &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The roads also have a unique quality which can be easily demonstrated. When asked how to get to the capital city, most locals will respond with a grin, &quot;Just keep turning!&quot;&lt;/p&gt;&#xA;" OwnerUserId="13237" LastEditorUserId="463" LastEditDate="2015-07-27T20:11:34.510" LastActivityDate="2015-07-28T15:50:19.293" Title="What is the name of this famous capital city?" Tags="|riddle|wordplay|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18397" PostTypeId="2" ParentId="18396" CreationDate="2015-07-27T20:49:33.320" Score="-1" Body="&lt;p&gt;Are you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The letter o in cursive?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;In cursive letters are connected with long curved flourishes much like exits and detours on a road. Since each of these connections will look slightly different it is like having roads all leading to different places. To make a capital O in cursive you would just keep turning and make the circle a bit larger.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13975" LastActivityDate="2015-07-27T20:49:33.320" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18399" PostTypeId="2" ParentId="18396" CreationDate="2015-07-28T04:12:22.600" Score="12" Body="&lt;p&gt;The capital city is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wikipedia's page on Philosophy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Built by volunteers and overseen by experienced builders.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Exits are abundant, offering frontage roads, detours, and highways to far away places.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;refers to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; links to other articles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The paragraph&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The roads provide utility that the citizens of this land now rely on. However, some also use the roads for sport, racing each other to exotic destinations. With myriad paths to the finish, each racer must anticipate future exits and above all, never backtrack. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;refers to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; both the use of Wikipedia for its intended purpose, and the noble sport of &lt;a href=&quot;https://en.wikipedia.org/wiki/Wikipedia:Wikirace&quot; rel=&quot;nofollow&quot;&gt;wikiracing&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And finally,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The roads also have a unique quality which can be easily demonstrated. When asked how to get to the capital city, most locals will respond with a grin, &quot;Just keep turning!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;refers to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if you just keep clicking, you'll end up there sooner or later.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2406" LastEditorUserId="2406" LastEditDate="2015-07-28T15:50:19.293" LastActivityDate="2015-07-28T15:50:19.293" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18400" PostTypeId="2" ParentId="18396" CreationDate="2015-07-28T05:14:13.673" Score="1" Body="&lt;p&gt;Are you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the brain?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It fits the description:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Scientist are only just now discovering the depths of the mind.&#xA; The roads mentioned are the neural network.&#xA; The far away places are the other parts of the body.&#xA; Signal pulses should not backtrack.&#xA; It's revolutionary indeed!&#xA; And of course, it's a &quot;Capital&quot; city!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(I cannot explain every line as of yet)&lt;/p&gt;&#xA;" OwnerUserId="9208" LastActivityDate="2015-07-28T05:14:13.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18401" PostTypeId="2" ParentId="18396" CreationDate="2015-07-28T08:56:40.707" Score="3" Body="&lt;p&gt;Are you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.openstreetmap.org&quot; rel=&quot;nofollow&quot;&gt;Open street maps&lt;/a&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In a recently unknown land, roads are rapidly being built by volunteers. There is no construction experience required, and seasoned road builders keep a watchful eye.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Open street maps can be updated by anyone that signs up to it, I've done a couple of changes myself when the local post office moved. When a change is submitted, many &quot;seasoned&quot; map builders check it over.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Exits are abundant, offering frontage roads, detours, and highways to far away places. It is truly a revolutionary system of transportation. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's a map, it contains all of these things, and because it's open source, it's revolutionary&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The roads provide utility that the citizens of this land now rely on. However, some also use the roads for sport, racing each other to exotic destinations. With myriad paths to the finish, each racer must anticipate future exits and above all, never backtrack.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Games have been created with the open street maps: &lt;a href=&quot;http://wiki.openstreetmap.org/wiki/Games&quot; rel=&quot;nofollow&quot;&gt;http://wiki.openstreetmap.org/wiki/Games&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The roads also have a unique quality which can be easily demonstrated. When asked how to get to the capital city, most locals will respond with a grin, &quot;Just keep turning!&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The maps are unique in many ways. There are various different layers that can be used to view different data, for example clicking the link at the top should supply a list of 5 available layers. Because it's open source there is no home (capital) for it, for example Google maps is Google, Bing maps is Microsoft.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4659" LastActivityDate="2015-07-28T08:56:40.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18402" PostTypeId="1" AcceptedAnswerId="18405" CreationDate="2015-07-28T09:08:48.763" Score="4" ViewCount="506" Body="&lt;p&gt;You wake up in a cell after a party. You just remember the beginning of the night: you talked to your friends, you drank some alcohol, and then you blacked out.&lt;br&gt;&#xA;You stand up and look around you: you are in a small empty room with no window and a door locked with a numeric keypad. &lt;br&gt;&lt;br&gt;&#xA;You approach the keypad and notice a piece of paper:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You'd better not try to steal my &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3981/the-security-to-the-party&quot;&gt;spaghetti recipe&lt;/a&gt; again!&lt;br&gt;I'll let you out if you can solve this puzzle: &lt;br&gt;&lt;br&gt;&#xA;  &lt;code&gt;{22,11}&lt;/code&gt;&lt;br&gt;&#xA;  &lt;code&gt;{49,83}&lt;/code&gt;&lt;br&gt;&#xA;  &lt;code&gt;{76,56}&lt;/code&gt;&lt;br&gt;&#xA;  &lt;code&gt;{157,344}&lt;/code&gt;&lt;br&gt;&lt;br&gt;&#xA;  are all acceptable numbers, but&lt;br&gt;&lt;br&gt;&#xA;  &lt;code&gt;{11,22}&lt;/code&gt;&lt;br&gt;&#xA;  &lt;code&gt;{72,47}&lt;/code&gt;&lt;br&gt;&#xA;  &lt;code&gt;{31,70}&lt;/code&gt;&lt;br&gt;&#xA;  &lt;code&gt;{512,114}&lt;/code&gt;&lt;br&gt;&lt;br&gt;&#xA;  aren't.&lt;br&gt;&lt;br&gt;&#xA;  If you find the key, you can live. (And if you fail, my pet elephant will make sure you wish you &lt;em&gt;weren't&lt;/em&gt; alive...).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;&lt;br&gt;Question:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;What is the key to the puzzle?&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Hint :&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The key is 2-char long&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Note: This isn't a lateral-thinking question, so there's no need for ridiculous answers like &quot;I punch through the wall&quot;.)&lt;/p&gt;&#xA;" OwnerUserId="13384" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-07-28T13:38:16.110" Title="Pairs of numbers" Tags="|mathematics|number-sequence|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18403" PostTypeId="2" ParentId="18344" CreationDate="2015-07-28T12:05:43.607" Score="15" Body="&lt;p&gt;Deciphered the numbers based off VictorHenry's answer! All I had to do was...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; add a couple suggestive spaces:&lt;br&gt;&#xA; H [2]&lt;br&gt;&#xA; O [5]&lt;br&gt;&#xA; I [9]&lt;br&gt;&#xA; S [11]&lt;br&gt;&#xA; U [14]&lt;br&gt;&#xA; M [15]&lt;br&gt;&#xA; B [16]&lt;br&gt;&#xA; R [18]&lt;br&gt;&#xA; L [4 8]&lt;br&gt;&#xA; T [1 12]&lt;br&gt;&#xA; N [6 13]&lt;br&gt;&#xA; E [3 7 10 17]&lt;br&gt;&#xA; Each set of numbers tells us where that letter is located in the final phrase. It spells out THE LONELIEST NUMBER, so the final answer is ONE!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2015-07-28T12:05:43.607" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="18405" PostTypeId="2" ParentId="18402" CreationDate="2015-07-28T13:38:16.110" Score="7" Body="&lt;p&gt;I'd do some quick addition and type&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;42&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because each number on the left&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; turns into &lt;strong&gt;4&lt;/strong&gt; if you repeatedly add digits until you get a single digit (49 -&gt; 4+9=13 -&gt; 1+3=4)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and each on the right&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; results in &lt;strong&gt;2&lt;/strong&gt; (83 -&gt; 8+3=11 -&gt; 1+1=2)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the invalid pairs, at least one of them is wrong. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, you know, it's &lt;em&gt;the&lt;/em&gt; answer, assuming you know the question.&lt;/p&gt;&#xA;" OwnerUserId="1563" LastActivityDate="2015-07-28T13:38:16.110" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18406" PostTypeId="2" ParentId="18395" CreationDate="2015-07-28T14:20:03.410" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An Ascending Phrase begins with a stressed syllable and nowhere has a syllable pattern of &quot;stressed, unstressed, stressed&quot; or &quot;stressed, stressed&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not a very neat answer, doesn't really fit with &quot;Ascending&quot;, but it works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Better answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A word's &lt;em&gt;climb&lt;/em&gt; is the number we get when we take the pattern of stressed and unstressed syllables, write 0 for a stressed and 1 for an unstressed, and convert the result from binary. For instance, &quot;barbarous&quot; has &quot;BAR-bar-ous&quot; so is 011 = 3. An Ascending Phrase is one where the successive words' climbs are strictly increasing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14305" LastEditorUserId="14305" LastEditDate="2015-07-28T14:36:10.353" LastActivityDate="2015-07-28T14:36:10.353" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18407" PostTypeId="2" ParentId="18395" CreationDate="2015-07-28T14:44:33.130" Score="14" Body="&lt;p&gt;@PatrickStevens got me looking at&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the syllable patterns, &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and I think I have it figured out:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; each word in the phrase has its first stressed syllable at one greater position than the previous. (first word: first syllable stressed, second word: second syllable stressed, etc).&lt;br&gt;&#xA;&lt;br&gt;&#xA; Thus, an Ascending Phrase is one where the position of the first stressed syllable ascends in each word as the phrase progresses.&lt;br/&gt;&lt;br&gt;&#xA; For example:&#xA;&lt;br&gt;&#xA; &lt;strong&gt;doc&lt;/strong&gt;-tors &amp;nbsp; pre-&lt;strong&gt;scribe&lt;/strong&gt;&lt;br&gt;&#xA; &lt;strong&gt;on&lt;/strong&gt;-ly &amp;nbsp; or-&lt;strong&gt;ig&lt;/strong&gt;-in-al &amp;nbsp; ge-o-&lt;strong&gt;graph&lt;/strong&gt;-ic &amp;nbsp; con-fig-ur-&lt;strong&gt;a&lt;/strong&gt;-tions&lt;br&gt;&#xA; &lt;strong&gt;prob&lt;/strong&gt;-lems &amp;nbsp; in-&lt;strong&gt;clude&lt;/strong&gt; &amp;nbsp; un-i-&lt;strong&gt;vers&lt;/strong&gt;-al &amp;nbsp; de-pen-da-&lt;strong&gt;bil&lt;/strong&gt;-i-ty&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13101" LastEditorUserId="13101" LastEditDate="2015-07-28T15:04:07.307" LastActivityDate="2015-07-28T15:04:07.307" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18408" PostTypeId="2" ParentId="18382" CreationDate="2015-07-28T15:12:28.233" Score="4" Body="&lt;p&gt;This is a bit of a stretch, but it's worth a shot...&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;You can regularly find me inserted below.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;(You can also find me near the end.)&lt;/em&gt;&lt;br&gt;&#xA; &quot;You respond to posts, giving us &lt;strong&gt;RE&lt;/strong&gt;&quot; (from &lt;a href=&quot;https://puzzling.stackexchange.com/a/18398/4513&quot;&gt;qwertylpc's answer&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Something we all do.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;(We share this commonality with trees.)&lt;/em&gt;&lt;br&gt;&#xA; All animals and plants &lt;strong&gt;LIVE&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A short-hand account of the matter from my point of view.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;(&quot;His&quot;, &quot;her&quot;, &quot;their&quot; etcetera will work in place of &quot;my&quot;.)&lt;/em&gt;&lt;br&gt;&#xA; On SE, I give my point of view in an answer. Short-hand for that is &lt;strong&gt;ANS&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I heard I finished 5th, but I was actually in 25th place.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;(Has already been worked out.)&lt;/em&gt;&lt;br&gt;&#xA; The 25th letter is &lt;strong&gt;Y&lt;/strong&gt;, but it sounds like the 5th letter, e.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Sound out the word with the clues.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RE-LIVE-ANS-Y = &lt;strong&gt;RELEVANCY&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4513" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-07-28T15:12:28.233" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18409" PostTypeId="1" AcceptedAnswerId="18413" CreationDate="2015-07-28T15:27:23.650" Score="15" ViewCount="3268" Body="&lt;p&gt;A loaded die looks like a regular six-sided die, but instead of each number 1-6 being equally likely, some numbers may be more or less likely than others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to have two loaded dice so that, when rolled together, every possible sum 2,3,...,12 is equally likely to occur? The dice may be loaded in different ways.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2016-01-29T01:50:19.610" LastActivityDate="2016-01-29T01:50:19.610" Title="Loaded Dice with Uniform Sum" Tags="|mathematics|probability|dice|" AnswerCount="7" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18410" PostTypeId="2" ParentId="18382" CreationDate="2015-07-28T15:29:45.653" Score="7" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;DELIVERY&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can regularly find me inserted below.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;From Sabre in the comments: It's DEL, as in the key below INS(ert) and near END on a keyboard&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Something we all do.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;LIVE as confirmed in Sabre's answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A short-hand account of the matter from my point of view.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I'm going with &quot;version&quot;, the &quot;short-hand&quot; of which is VER.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I heard I finished 5th, but I was actually in 25th place.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Solved awhile ago as Y&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Combined:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;DEL + LIVE + VER + Y = DELIVERY&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10861" LastEditorUserId="10861" LastEditDate="2015-07-28T15:46:50.737" LastActivityDate="2015-07-28T15:46:50.737" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18411" PostTypeId="2" ParentId="18409" CreationDate="2015-07-28T15:35:21.800" Score="4" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;This is not a formal proof, but maybe someone could build off it.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My gut says&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; this is not possible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we consider three combinations, it becomes clear (to me, anyway) that this is the case:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Consider 1+1, 6+6, and 1+6. You would need to weigh the 1+1 and 6+6 the same, and weigh them so that they have a higher probability than their normal. However, this increased weighting would then also cause 1+6 to become even more likely than it already is; eternally pushing its probability higher than 1+1 or 6+6.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2015-07-28T15:35:21.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18412" PostTypeId="2" ParentId="18409" CreationDate="2015-07-28T15:35:41.770" Score="4" Body="&lt;p&gt;I suspect no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at 2 and 12, as they can only be achieved by a 1, 1 and a 6, 6. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let our dies be denoted $a$ and $b$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a_1$ designate the probability of 1 being rolled on $a$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So:&lt;br&gt;&#xA;$P(2) = a_1 \times b_1$&lt;br&gt;&#xA;$P(12) = a_6 \times b_6$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without considering any other possible roles, $P(7)$ is currently $a_1 \times b_6 + a_6 \times b_1$. If $a_1 &amp;gt; b_1$ then $a_6 &amp;gt; b_6$ otherwise $a_1\times b_6$ would be greater than either $P(2)$ or $P(12)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus:&lt;br&gt;&#xA;$b_6 &amp;gt; a_6$&lt;br&gt;&#xA;$a_1 &amp;gt; b_1$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_1 \times b_1=a_6 \times b_6 \geq a_1 \times b_6 + b_1 \times a_6 $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $a_6 &amp;lt; b_6$,&lt;br&gt;&#xA;$a_6 \times b_6 &amp;gt; a_1 \times a_6+b_1 \times a_6$ or   $(a_1+b_1) \times a_6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $a_1 + b_1$ must be less than $b_6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $a_1 &amp;gt; b_1$,&lt;br&gt;&#xA;$a_6 \times b_6 &amp;gt;b_1 \times b_6+b_1 \times a_6$ or $b_1 \times (a_6+b_6)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $a_6+b_6$ must be less than $a_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However $a_1+b_1$ is less than $b_6$. This is a contradiction so it is impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(apologies for sloppy inequality signage)&lt;/p&gt;&#xA;" OwnerUserId="13916" LastEditorUserId="14092" LastEditDate="2015-07-28T16:12:41.847" LastActivityDate="2015-07-28T16:12:41.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18413" PostTypeId="2" ParentId="18409" CreationDate="2015-07-28T15:41:55.000" Score="27" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;No.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To explain, here is a more formal way to state Ian's answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the probability of die A rolling $i$ be $a_i$ for $1\le i\le6$, and similarly for die B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If every sum has the same probability, then they must each have probability $\frac{1}{11}$. Because 2 is only possible from double 1s, and 12 is only possible from double 6s, $a_1b_1=a_6b_6=\frac{1}{11}$. Now the probability of rolling 7 is at least $a_1b_6+a_6b_1\ge 2\sqrt{a_1b_6a_6b_1}=\frac{2}{11}$ (using the &lt;a href=&quot;https://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means&quot;&gt;AM-GM inequality&lt;/a&gt;). Therefore, it is not possible.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="6205" LastEditDate="2015-07-28T16:39:41.277" LastActivityDate="2015-07-28T16:39:41.277" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18414" PostTypeId="2" ParentId="18409" CreationDate="2015-07-28T16:03:26.447" Score="1" Body="&lt;p&gt;&lt;sub&gt;While I was typing mine, Going hamateur and f' had posted very similar approaches, so check theirs out as well.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 11 possible sums (2 thru 12), so the probability of each would have to be $\frac{1}{11}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets call $X$ die 1, and $Y$ die 2, and let $X_1$ denote the probability  rolling a 1 with the first die. Thus, we can say that $X_1*Y_1 = X_6*Y_6 = \frac{1}{11}$ as these are the only ways to get 2 and 12. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To build a little more, lets look at a sum of 7.&lt;br&gt;&#xA; - $X_1*Y_6 + X_6*Y_1 = \frac{1}{11}$&lt;br&gt;&#xA; - $X_1*Y_6 + X_6*Y_1 =  X_1*Y_1$&lt;br&gt;&#xA; - $\implies X_1*Y_6 =  X_1*Y_1 - X_6*Y_1 $&lt;br&gt;&#xA; - $\implies X_1*Y_6 =  Y_1(X_1 - X_6) $&lt;br&gt;&#xA; - $\implies X_1 \ge X_6$ (since $Y_1 \ge 0$ and $Y_1(X_1 - X_6) \ge 0$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, from the other side.&lt;br&gt;&#xA; - $X_1*Y_6 + X_6*Y_1 =  X_6*Y_6$  (since $X_6*Y_6 = X_6*Y_6$)&lt;br&gt;&#xA; - $\implies X_6*Y_1 =  X_6*Y_6 - X_1*Y_6 $&lt;br&gt;&#xA; - $\implies X_6*Y_1 =  Y_6(X_6 - X_1) $&lt;br&gt;&#xA; - $\implies X_6 \ge X_1$ (since $Y_6 \ge 0$ and $Y_6(X_6 - X_1) \ge 0$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With both $X_1 \ge X_6$ and $X_6 \ge X_1$, the only value that works is for $X_1 = X_6 = 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $X_1*Y_1 = \frac{1}{11}, X_1 \ne 0$, therefore it is not possible.&lt;/p&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="9951" LastEditDate="2015-07-28T16:20:26.227" LastActivityDate="2015-07-28T16:20:26.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18415" PostTypeId="2" ParentId="18409" CreationDate="2015-07-28T16:40:37.597" Score="0" Body="&lt;p&gt;Starting from 2 and 12, 3 and 11, ... it follows that the chance of rolling $n$ when one die is cast is equal to that of rolling $7-n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they're $a,b,c,c,b,a$, then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a^2$ (getting 2) = $2ab$ (getting 3), so $a$ = $2b$; $2b,b,c,c,b,2b$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$4b^2$ (getting 2) = $4bc$ + $b^2$ (getting 4), so $4b$ = $4c$ + $b$, $3b$ = $4c$; $2b, b, 3b/4, 3b/4, b, 2b$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$4b^2$ (getting 2) = 2*($2b*3b/4$ + $b*3b/4$) (getting 5) = 2*($3b^2/2$ + $3b^2/4$) = $9b^2/2$ (contradiction)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's impossible.&lt;/p&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="14016" LastEditDate="2015-07-28T17:20:35.587" LastActivityDate="2015-07-28T17:20:35.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18416" PostTypeId="2" ParentId="18409" CreationDate="2015-07-28T16:51:04.860" Score="5" Body="&lt;p&gt;Once again not an answer to this question because it has been thoroughly answered.  I wrote a short program in R and used the optim command to find the values that get closest to having a uniform distribution.  The quality measure was the largest probability (of a number 2-12) - the smallest probability  (of a number 2-12)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution I got yielded:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;P(2) &amp;amp;= .05  \\&#xA;P(3) &amp;amp;= .1  \\&#xA;P(4) &amp;amp;= .1  \\&#xA;P(5) &amp;amp;= .1  \\&#xA;P(6) &amp;amp;= .1  \\&#xA;P(7) &amp;amp;= .1  \\&#xA;P(8) &amp;amp;= .1  \\&#xA;P(9) &amp;amp;= .1  \\&#xA;P(10) &amp;amp;= .1  \\&#xA;P(11) &amp;amp;= .1  \\&#xA;P(12) &amp;amp;= .05  \\&#xA;\\&#xA;a_1 &amp;amp;= .5 \\&#xA;a_2 &amp;amp;= 0 \\&#xA;a_3 &amp;amp;= 0 \\&#xA;a_4 &amp;amp;= 0 \\&#xA;a_5 &amp;amp;= 0 \\&#xA;a_6 &amp;amp;= .5 \\&#xA; \\&#xA;b_1 &amp;amp;= .1 \\&#xA;b_2 &amp;amp;= .2 \\&#xA;b_3 &amp;amp;= .2 \\&#xA;b_4 &amp;amp;= .2 \\&#xA;b_5 &amp;amp;= .2 \\&#xA;b_6 &amp;amp;= .1&#xA;\end{align*}&lt;/p&gt;&#xA;" OwnerUserId="14092" LastActivityDate="2015-07-28T16:51:04.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18417" PostTypeId="2" ParentId="18363" CreationDate="2015-07-28T16:56:05.023" Score="5" Body="&lt;p&gt;9) W, J, L, H, J, G, ?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  The people pictured on US currency:&lt;br&gt;&#xA; 1 dollar: &lt;strong&gt;W&lt;/strong&gt;ashington&lt;br&gt;&#xA; 2 dollar: &lt;strong&gt;J&lt;/strong&gt;efferson&lt;br&gt;&#xA; 5 dollar: &lt;strong&gt;L&lt;/strong&gt;incoln&lt;br&gt;&#xA; 10 dollar: &lt;strong&gt;H&lt;/strong&gt;amilton&lt;br&gt;&#xA; 20 dollar: &lt;strong&gt;J&lt;/strong&gt;ackson&lt;br&gt;&#xA; 50 dollar: &lt;strong&gt;G&lt;/strong&gt;rant&lt;br&gt;&#xA; 100 dollar: &lt;strong&gt;F&lt;/strong&gt;ranklin  &lt;/p&gt;&#xA; &#xA; Thanks to GlenO for the hint&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4385" LastEditorUserId="4385" LastEditDate="2015-07-31T17:16:05.787" LastActivityDate="2015-07-31T17:16:05.787" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18418" PostTypeId="1" AcceptedAnswerId="18513" CreationDate="2015-07-28T18:26:05.310" Score="6" ViewCount="483" Body="&lt;p&gt;Matty, the math mouse, had fun playing Mastermind in the Cool™ Club, but he has worked up a hankering for some honey. So our adventurous arithmetist ventured to the local bee hive. Only to find it missing!&lt;/p&gt;&#xA;&lt;p&gt;Startled, he looked around and saw a trail of dirt mixed drippings leading to a big cave. At the cave, there was a befuddled looking bear. Matty tried to sneak in but the bear noticed him.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;Hello little mouse, are you trying to steal my hard earned honey?&amp;quot;, Barry  the bear scowled. &lt;br&gt;&#xA;&amp;quot;Mayhaps,&amp;quot; muttered Matty, the math mouse, meekly. &lt;br&gt;&#xA;&amp;quot;Oh, I am willing to share if you ask... and help me.&amp;quot; &lt;br&gt;&#xA;&amp;quot;Okay,&amp;quot; Matty mumbled, looking at the heap of honeyed honeycomb. &lt;br&gt;&#xA;&amp;quot;Burt, my bully brother, keeps teasin' me cause I can't figure out his puzzle.&amp;quot;&lt;br&gt;&#xA;&amp;quot;Do tell.&amp;quot;&lt;br&gt;&#xA;&amp;quot;Well he likes to yell sentences at me and laughs at me when I can find any pattern.&amp;quot;&lt;br&gt;&#xA;&amp;quot;Can I hear some examples?&amp;quot;&lt;br&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;YES GRACE CAN YOU GO&amp;quot;, he said but when I asked &amp;quot;NO GRACE CAN'T GO&amp;quot;, he laughed and said that didn't fit.&lt;br&gt;&#xA;&amp;quot;ZOO CANOE MOO&amp;quot;, he said. Then he said that &amp;quot;RHINO CANOE TOO?&amp;quot; did not work.&lt;br&gt;&#xA;&amp;quot;MARS MOON VENUS SUN&amp;quot; fit but &amp;quot;PLUTO&amp;quot; didn't.&lt;br&gt;&#xA;He said, &amp;quot;CAVERNOUS SUPERNOVA UNCOVERS UNSAVORY SUPREMACY&amp;quot; after which I just gave up and said &amp;quot;NO&amp;quot;. And then he said &amp;quot;Ha! You finally found one that works.&amp;quot;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So can you figure out what the pattern is?&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;b&gt;Works:&lt;/b&gt;&#xA;&lt;br&gt;&amp;quot;YES GRACE CAN YOU GO&amp;quot;,&#xA;&lt;br&gt;&amp;quot;ZOO CANOE MOO&amp;quot;,&#xA;&lt;br&gt;&amp;quot;MARS MOON VENUS SUN&amp;quot;,&#xA;&lt;br&gt;&amp;quot;CAVERNOUS SUPERNOVA UNCOVERS UNSAVORY SUPREMACY&amp;quot;,&#xA;&lt;br&gt;&amp;quot;NO&amp;quot;&#xA;&lt;br&gt;&amp;quot;MUMMY&amp;quot;&#xA;&lt;br&gt;&amp;quot;MUSE SONG&amp;quot;&#xA;&lt;br&gt;&amp;quot;A MAGE RUSE&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&lt;b&gt;Does not work:&lt;/b&gt;&#xA;&lt;br&gt;&amp;quot;NO GRACE CAN'T GO&amp;quot;,&#xA;&lt;br&gt;&amp;quot;RHINO CANOE TOO?&amp;quot;,&#xA;&lt;br&gt;&amp;quot;PLUTO&amp;quot;&#xA;&lt;br&gt;&amp;quot;HI SORRY&amp;quot;&#xA;&lt;br&gt;&amp;quot;KNOW&amp;quot;&#xA;&lt;br&gt;&amp;quot;BARELY BARRY&amp;quot;&#xA;&lt;br&gt;&amp;quot;ZOO KAYAK MOO&amp;quot;&#xA;&lt;br&gt;&amp;quot;MUSICIAN'S PIECE&amp;quot;&#xA;&lt;br&gt;&amp;quot;A WIZARD'S TRICK&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Hint: &lt;br&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;If there is anything I did with this puzzle that strikes you as peculiar mayhaps you could undo it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13916" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-31T11:05:14.500" Title="Matty mouse and Barry the bear" Tags="|pattern|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18419" PostTypeId="2" ParentId="18350" CreationDate="2015-07-28T19:17:25.373" Score="0" Body="&lt;p&gt;&lt;strong&gt;For all versions:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start both timers.  When either timer is empty, flip it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually, there will come a time when one timer is flipped 1 minute after the other one was flipped.  At that point, flip both timers.  1 minute later, the other one will be empty; flip both again.  From then on, every minute, one of the timers will be empty, and you flip both timers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Warmup:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flip the small one at 4, larger at 7, both at 8, both at 9, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every time from 7 minutes up can be measured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Easy:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flip small at 5, big at 8, small at 10, small at 15, both at 16, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every time from 15 minutes up can be measured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Medium:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flip small at 6, big at 11, both at 12, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every time from 11 up can be measured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hard:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x=GCF(m,n)$.  Follow the rules above, and eventually one of the timers will flip $x$ minutes after the other.  From that time on, flip both timers whenever one is empty.  Every multiple of $x$ from that point on will be measured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only gets all integer values if $m$ and $n$ are relatively prime.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2015-07-28T19:17:25.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18420" PostTypeId="2" ParentId="2030" CreationDate="2015-07-28T19:29:47.167" Score="1" Body="&lt;p&gt;You can do it even if you start timing when you first flip both timers.  When either timer runs out, flip both timers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A. When you flip the 4 minute timer, the 5 minute timer has 1 minute left.  After that, you'll flip both timers at 5 minutes, 6 minutes, and 7 minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B. When you flip at 7 minutes, the larger timer has 4 minutes left.  Flip both again at 11 minutes, and both again at 15 minutes, and you are done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C. When you flip both at 5 minutes, the larger timer has 3 minutes left.  Flip both again at 8, then both again at 11, then both again at 14, and you are done.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2015-07-28T19:29:47.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18421" PostTypeId="1" AcceptedAnswerId="18444" CreationDate="2015-07-28T19:32:51.193" Score="7" ViewCount="1091" Body="&lt;p&gt;You wake up at 6:30 AM to a knock on your door. By the time you get dressed and answer the door, you can see (in your motion-activated lawn lighting) a person riding away on his (or her) motorcycle. At your feet is a slip of paper. You pick it up, and it reads:&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;We have your cat. Your precious Geronimo. Answer this riddle within the fortnight, and you'll get him back (for a fee, of course. I mean, what type of kidnappers would we be if we gave you your cat without charge?)&lt;/p&gt;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Some say I'm quite insane.&lt;br /&gt;&#xA;I blame it on the heights.&lt;/p&gt;&#xA;&lt;p&gt;Take a ride on the Phoenix-Durango.&lt;br /&gt;&#xA;Chances are you'll know no-one when you go.&lt;/p&gt;&#xA;&lt;p&gt;Don't look out the back.&lt;br /&gt;&#xA;Do you &lt;em&gt;really&lt;/em&gt; know what hides in dark corners?&lt;/p&gt;&#xA;&lt;p&gt;Follow the V-formation,&lt;br /&gt;&#xA;Stop at the top of the mountain.&lt;/p&gt;&#xA;&lt;p&gt;If you pick up the phone, press 6.&lt;br /&gt;&#xA;Maybe you'll reach &amp;quot;Bush Sandwich&amp;quot; Hill.&lt;/p&gt;&#xA;&lt;p&gt;Let's hope you've got just enough ignorance; nobody likes a know-it-all.&lt;br /&gt;&#xA;Otherwise, look forward to some road bumps in your progress; maybe a couple pitfalls?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Love,&#xA;Al.&lt;/p&gt;&#xA;&lt;p&gt;Hint #1:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Only the text &lt;strong&gt;below the question&lt;/strong&gt; (&amp;quot;Who am I?&amp;quot;), any postscripts to the letter, and hints are relative to the answer. Anything else is just flavor.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Hint #2:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; My mistake. The answer is actually a &amp;quot;Who&amp;quot;, so the question was originally misleading. I have rephrased it to fit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Hint #3:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Stanzas are only divided into lines of two for aesthetic purposes, with the exception of the second stanza. Each line of each stanza (other than the second) has its own, separate meaning from the other line in that stanza.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Hint #4:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Each clue is related to the Who in the same way (obviously). More specifically, each clue can be connected to the Who by the same logic. I hope I'm directing you well with these hints. Furthermore, each clue is completely different from the rest, and is concrete. It will have a very specific link to the solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13538" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-30T14:13:25.347" Title="The Case of the Missing Kitten (Part 1)" Tags="|riddle|word|wordplay|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18422" PostTypeId="2" ParentId="18390" CreationDate="2015-07-28T19:42:21.670" Score="3" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Much Ado About Nothing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Building quite heavily COTO's answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There's much code before the &lt;code&gt;main()&lt;/code&gt;, but then it adds (sounds vaguely like &quot;Ado&quot;) one more element, a new Obj object, to the collection, so it doesn't have none or one, but rather, much content.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The &lt;code&gt;e.rotate(Math.PI)&lt;/code&gt; being &quot;about&quot;.&lt;br&gt;&#xA;&lt;br&gt;&#xA; And the empty code block &lt;code&gt;{}&lt;/code&gt; being &quot;Nothing&quot;.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4384" LastEditorUserId="4384" LastEditDate="2015-07-28T21:00:15.820" LastActivityDate="2015-07-28T21:00:15.820" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18423" PostTypeId="1" AcceptedAnswerId="18481" CreationDate="2015-07-28T19:43:02.817" Score="8" ViewCount="599" Body="&lt;p&gt;As a result of their solutions to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/17486/find-the-two-master-lock-combinations&quot;&gt;this puzzle&lt;/a&gt;, store owner Jeremy promoted both Cerberus and Quark to top managerial positions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Soon after starting their new roles, they were somewhat surprised when they noticed gorgeous women visiting Jeremy privately in his office each week, unbeknownst to his wife. Then, one day at a lunch meeting with his 2 new managers, Jeremy said that he trusted them, and revealed that he &quot;secretly&quot; had a couple of girlfriends, and that he has also created his own simple encryption method to protect the short emails that he sends to them, contrary to the &lt;a href=&quot;https://security.stackexchange.com/questions/2202/lessons-learned-and-misconceptions-regarding-encryption-and-cryptology/2210?newreg=9dcf9f778a884e1ca64c973403bf3e3c&quot;&gt;&quot;Don't roll your own&quot; advice&lt;/a&gt; of the smart folks at Information Security.SE. He has also given the women a simple decryption program so they can communicate with him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before the multiple-lover/adultery information could even sink in, Jeremy offered them another challenge. The first of them to figure out his simple email encryption algorithm would be promoted to region manager, which would include a fat raise over their already-large recent increase. He provided them with a very short plaintext email, along with the associated ciphertext and a couple of hints, and wished them luck.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hint #1:&lt;br&gt; Jeremy: You solved my &lt;a href=&quot;https://puzzling.stackexchange.com/questions/17486/find-the-two-master-lock-combinations&quot;&gt;Master lock challenge&lt;/a&gt;, and you know&#xA;  that I am not that creative, just clever.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Hint #2:&lt;br&gt; Jeremy: Using my algorithm, the same message can be encrypted to many different ciphertexts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Email plaintext:&lt;/strong&gt;&lt;br&gt;&#xA;MISSING OUR TALKS! MEET ME AT NOON.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Email ciphertext:&lt;/strong&gt;&lt;br&gt;&#xA;PARROT~NECTAR~GABBY~TWEEZE~LUCID~CARAT~TEETHE SUBWAY~YACCA~HOLIDAY ABBACY~RELIEF~PLEATED~VACCINES~BEDAZE! ASTROID~OASIS~COUNTER~TAZZA GRASS~TOASTS REPLICA~BLEBBY BRACED~ASTRONOMIC~WEDGIE~WEDGING.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can you figure out Jeremy's encryption algorithm before his managers do?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.923" LastActivityDate="2015-07-31T01:37:41.627" Title="What's Jeremy's Encryption Algorithm?" Tags="|cipher|calculation-puzzle|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18424" PostTypeId="1" AcceptedAnswerId="18493" CreationDate="2015-07-28T19:48:18.177" Score="18" ViewCount="1324" Body="&lt;p&gt;My namesakes are plentiful,&lt;br&gt;&#xA;though some antiquated:&lt;br&gt;&#xA;A company defunct &lt;br&gt;&#xA;and a system outdated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One unavoidable&lt;br&gt;&#xA;an honor shared by few&lt;br&gt;&#xA;just give it some &lt;em&gt;time&lt;/em&gt;&lt;br&gt;&#xA;it will come into view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two bearing my name&lt;br&gt;&#xA;and both are titanic:&lt;br&gt;&#xA;an explosive endeavor&lt;br&gt;&#xA;and the second, enigmatic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My habits inspired &lt;br&gt;&#xA;a nightmarish work.&lt;br&gt;&#xA;The last one was spared&lt;br&gt;&#xA;and well known to perk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I bring is a gift,&lt;br&gt;&#xA;feared and embraced.&lt;br&gt;&#xA;Do you know my name&lt;br&gt;&#xA;from these clues I have placed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clarifications (if needed):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The accepted answer will explain each stanza. &lt;br&gt;The things&#xA; bearing the answer's name are:&lt;br&gt;&#xA;     + a company defunct&lt;br&gt;&#xA;     + a system outdated&lt;br&gt;&#xA;     + a namesake unavoidable&lt;br&gt;&#xA;     + an explosive endeavor&lt;br&gt;&#xA;     + a namesake enigmatic&lt;br&gt; This prominent figure is also known for a grisly habit, which inspired a nightmarish work. &lt;br&gt;Lastly, this&#xA; prominent figure is known to bring a gift which is feared by some and&#xA; embraced by others.&lt;br&gt;&#xA; One word in the riddle has been italicized. It's importance is two-fold, as it hints at both the &quot;namesake unavoidable&quot; and the prominent figure itself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13237" LastEditorUserId="13237" LastEditDate="2015-09-15T20:37:59.620" LastActivityDate="2015-09-15T20:37:59.620" Title="What prominent figure am I?" Tags="|riddle|rhyme|wordplay|knowledge|" AnswerCount="5" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="18425" PostTypeId="2" ParentId="18146" CreationDate="2015-07-28T21:11:30.950" Score="8" Body="&lt;p&gt;This solution is entirely due to &lt;a href=&quot;https://puzzling.stackexchange.com/users/11295/zeb&quot;&gt;zeb&lt;/a&gt;, and is paraphrased from &lt;a href=&quot;http://forums.xkcd.com/viewtopic.php?f=3&amp;amp;t=99534#p3249295&quot; rel=&quot;nofollow noreferrer&quot;&gt;this xkcd thread post&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\epsilon&amp;gt;0$ be a small number. Take a particular maze, and imagine a robot who starts on the starting northwest square, then moves in a random direction once per minute according to the below probabilities:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;     ^ e&#xA;     |&#xA;e &amp;lt;--+--&amp;gt; 1/2 - e&#xA;     |&#xA;     v 1/2 - e&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Notice that the robot will tend to the southeast over time, but still work its way around obstacles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The robot's position will be a Markov chain, whose states are the squares reachable from the start square. Using Markov chain theory, you can show the probability distribution of the robot's location will approach a unique stationary distribution as the number of steps increases. Let $p(i,j)$ be this limit distribution, where $(1,1)$ is the starting square, and $(8,8)$ is the finish. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An amazing fact is this: no matter how the maze is structured, the limit distribution $p(i,j)$ will be of the form $$p(i,j)=K\left(\frac{\frac12-\epsilon}{\epsilon}\right)^{i+j},$$where $K$ is a normalizing constant. This is quite surprising: no matter how snakey the maze is, over time, the probability this robot will be on a square only depends on which diagonal that square is on. To check this, you need only show that this distribution is a fixed point of the Markov process. Namely, show that if the robot has this distribution, then one minute later, it will still have this distribution. I omit the proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this distribution, $p(8,8)$ has the largest probability, and every other probability is at most $\frac{\epsilon}{\frac12-\epsilon}\cdot p(8,8)\le 4\epsilon$, implying that $p(8,8)\ge 1-63\cdot 4\epsilon$. Thus, if we choose $\epsilon$ small enough, we can make $p(8,8)$ as close to $1$ as we want. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, imagine if we set up every possible maze, choose a random program in this fashion of length $L$, then run the same program on a robot placed on the start square of each maze. By the previous discussion, we can choose $\epsilon$ small enough, and $L$ large enough, so that each robot has a less than $\frac1{2^{112}}$ probability of &lt;em&gt;not&lt;/em&gt; ending on $(8,8)$. Since there are less than $2^{112}$ mazes, we have that&#xA;$$&#xA;\begin{align}&#xA;P(\text{all robots end on finish})&#xA;&amp;amp;=1-P(\text{at least one robot doesn't})\\&#xA;&amp;amp;\ge 1-\sum_{m=1}^{\text{# of mazes}} P(\text{robot on maze m doesn't end on finish})\\&#xA;&amp;amp;&amp;gt;1-2^{112}\cdot\frac1{2^{112}}=0 \\&#xA;\end{align}&#xA;$$&#xA;Thus, there is a nonzero probability of a random program succeeding on all mazes, so there must exist a particular, determistic program which succeeds for all mazes.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-07-28T21:11:30.950" CommentCount="2" CommunityOwnedDate="2015-07-28T21:11:30.950" ContentLicense="CC BY-SA 3.0" />
  <row Id="18426" PostTypeId="1" AcceptedAnswerId="18430" CreationDate="2015-07-28T21:20:33.677" Score="15" ViewCount="1431" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Can you design two six-sided dice (different from the standard ones), where each face has a nonzero number of dots, so that the probability distribution of their total is the same as for two standard dice? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To be clear, you may only choose how the dice are labeled, you're not allowed to load the dice so certain faces are more likely. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is one of the more famous classic math puzzles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit/Clarifications:&lt;/strong&gt; The goal is relabel the faces of two standard die so that&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Each face is labeled with a positive integer.&lt;/li&gt;&#xA;&lt;li&gt;The set of labels on each die is not the usual {1,2,3,4,5,6}.&lt;/li&gt;&#xA;&lt;li&gt;For every number $n$, the probability of rolling an $n$ with your dice is the same as the probability of rolling an $n$ with standard dice. &quot;Rolling an $n$&quot; means rolling both dice and having the numbers on their top faces adding to $n$.&lt;/li&gt;&#xA;&lt;li&gt;The labels on a die don't have to be all different.&lt;/li&gt;&#xA;&lt;li&gt;You don't have to label both dice the same way.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2016-01-29T01:51:57.960" LastActivityDate="2016-01-29T01:51:57.960" Title="Crazy Dice - Unusual Dice with Usual Sum" Tags="|combinatorics|probability|dice|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18427" PostTypeId="2" ParentId="18409" CreationDate="2015-07-28T21:21:16.863" Score="2" Body="&lt;p&gt;I further show that this sum distribution is impossible for any pair of dice, no matter how many faces they have, what numbers are on those faces, and what weights each face has, barring the trivial case where one die always rolls the same result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Translate the dice weightings into generating functions &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A(x) = \sum a_i x^i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and likewise for $B(x)$. Since dice sum is distribution convolution is generating function product, we're required to have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A(x) B(x) = \frac{x^2+x^3+\dots + x^{11} + x^{12}}{11}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Express the geometric series on the RHS as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A(x) B(x) = \frac{x^2}{11} \cdot \frac{1-x^{11}}{1-x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The roots of $1-x^{11}$ are the powers of the primitive elevent roots of unity $\omega = e^{2\pi i/11}$, which are $1, \omega, \omega^2, \dots, \omega^{11}$. The $1-x$ in the denominator removes the root $x=1$. So, we can factor into linear terms&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A(x) B(x) = \frac{x^2}{11} \cdot \prod_{j=1}^{10} (x-\omega^j)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, because of unique factorization, these linear terms must be split across $A$ and $B$. Moreover, since the coefficients of $A$ and $B$ are probabilities, the resulting polynomials must have non-negative real coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, we're down to a finite number $2^{10}$ of possibilities, so we could try all of them by computer and check that none work. But, we can simplify the search by hand first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, a real polynomial must equal its conjugate, so its complex roots must come in conjugate pairs. Therefore, the conjugate roots must &quot;stick&quot; together when being split into $A$ and $B$. So, we can group them into quadratic units&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x-\omega^j)(x-\overline{\omega^j}) = x^2 - 2\mathrm{Re}(\omega^j)x + 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $j=1,2,3,4,5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, note that for a product of quadratic units, the coefficient of $x$ is the sum of all $-2\mathrm{Re}(\omega^j)$ of the $j$'s included. Since this coefficient must be positive, the sum of the real parts of the included roots of unity must be negative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These real parts are roughly $0.84, 0.42, -0.14, -0.65, -0.96$. Already we see that $0.84$ must be grouped with $-0.96$ to make a negative sum, since the other negatives aren't enough. Then, $0.42$ must be grouped with $-0.65$, since the $-0.14$ and the $-0.12$ from the previous pairing aren't enough to make it negative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we have three groups $\{\omega^1,\omega^5\},\{\omega^2,\omega^4\},\{\omega^3\}$ split into two sides. Putting them all onto one side would result in a trivial monomial on the other side, giving a die that always rolls the same. So, we must have one group alone and the other two groups together. Checking all three possibilities directly, each one gives a polynomial that contains negative coefficients. So, the desired split is not possible, and no solution exists.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-07-28T21:27:59.953" LastActivityDate="2015-07-28T21:27:59.953" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18428" PostTypeId="2" ParentId="18421" CreationDate="2015-07-28T21:26:47.430" Score="2" Body="&lt;p&gt;I am guessing you are the Rocky Mountains.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Some say I'm quite insane.&#xA;I blame it on the heights.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Rocky's are crazy tall.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Take a ride on the Phoenix Durango.&#xA;Chances are you'll know no-one when you go.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Phoenix Durango could be a train or a road to Durango Colorado.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Don't look out the back.&#xA;Do you really know what hides in dark corners?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A reference to four corners monument in Durango&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Follow the V-formation,&#xA;Stop at the top of the mountain.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The V formation is the Appalachian and Rocky Mountains, which roughly form a v. You have traveled down from the Rockys and are now traveling up the Appalachians. I'm not sure what the mountain is. Perhaps a point near dc?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;If you pick up the phone, press 6.&#xA;Maybe you'll reach the &quot;Bush Sandwich&quot; Hill.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bush Sandwich could refer to the presidents Bush and Bush, which would make the hill Capitol Hill. 6 is the right arrow on my keyboard, or east. You also take I66 into Washington DC&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Let's hope you've got just enough ignorance,&#xA;Otherwise look forward to some road bumps in your progress.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm not sure if this is part of the riddle or not, but it could be referencing that there is also a Durango Mexico&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4591" LastActivityDate="2015-07-28T21:26:47.430" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18429" PostTypeId="2" ParentId="18426" CreationDate="2015-07-28T22:08:52.160" Score="0" Body="&lt;p&gt;I think the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Yes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If you take a standard die and multiply each face by two, it should have the same probability distribution as the normal dice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Examples:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;4:  1 ways (2,2)&#xA;&#xA;6:  2 ways (2,4; 4,2)&#xA;&#xA;8:  3 ways (2,6; 6,2; 4,4)&#xA;&#xA;10: 4 ways (2,8; 4,6; 6,4; 8,2)&#xA;&#xA;12: 5 ways (2,10; 4,8; 6,6; 8,4; 10,2)&#xA;&#xA;14: 6 ways (2,12; 4,10; 6,8; 8,6; 10,4; 12,2)&#xA;&#xA;And then all the way back down to 24&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13924" LastActivityDate="2015-07-28T22:08:52.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18430" PostTypeId="2" ParentId="18426" CreationDate="2015-07-28T22:22:21.860" Score="10" Body="&lt;p&gt;2nd Edit:&lt;br&gt;&#xA;This is the answer you're probably looking for;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Die 1: 1, 2, 2, 3, 3, 4&lt;br&gt;&#xA; Die 2: 1, 3, 4, 5, 6, 8   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:    &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;p&gt;To maintain the distribution of sums while using positive non-zero whole numbers of dots, we would need to decrease some faces on one die and increase on the other. If we have a sequence from 1 to 6, then to decrease we will need to double up faces. We need exactly one 1-dot face on each die to maintain P(2) = 1/36, however we can start by placing two 2-dot faces on one of the die and work our way up from there.&lt;/p&gt;  &lt;p&gt;One useful constraint when deriving the faces of the dice is that the highest value face on each die can only appear on a single face since P(12) = 1/36.&lt;/p&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Previous answers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Die 1: 0.5, 1.5, 2.5, 3.5, 4.5, 5.5&lt;br&gt;&#xA; Die 2: 1.5, 2.5, 3.5, 4.5, 5.5, 6.5. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;As requested, whole numbers!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Die 1: -6, -5, -4, -3, -2, -1&lt;br&gt;&#xA; Die 2: 8, 9, 10, 11, 12, 13  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11078" LastEditorUserId="9696" LastEditDate="2015-07-29T00:37:33.090" LastActivityDate="2015-07-29T00:37:33.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18433" PostTypeId="1" CreationDate="2015-07-29T00:01:12.577" Score="8" ViewCount="705" Body="&lt;p&gt;Each of the following  represent a word or phrase. Can you figure them all out?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Musical chairs, upgraded&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;4 and a castrated meteorology platform&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A long age, in the past&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ding Dong &lt;br/&gt;&#xA;  A colours' rival with a fiendish compartment&lt;br/&gt;&#xA;  A marketers money maker&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Come have a look! &lt;br/&gt;&#xA;  The evener?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Me llamo viento&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;First grade year for the Solomon Islands&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13975" LastEditorUserId="11244" LastEditDate="2015-07-29T18:09:24.230" LastActivityDate="2015-07-29T19:01:16.637" Title="Seven fun mini-riddles" Tags="|enigmatic-puzzle|wordplay|knowledge|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18434" PostTypeId="1" AcceptedAnswerId="18441" CreationDate="2015-07-29T03:05:16.010" Score="16" ViewCount="1415" Body="&lt;p&gt;I dropped in on Ernie a few days ago. As I pushed open the door to his workshop I could hear a clanking, hissing noise and then saw Ernie emerging from a great clouds of steam, his face blackened with what looked like soot.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Just in time to help me debug my calculator,&quot; he called out as he saw me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The clock-work calculator?&quot; I asked.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;No. Clockwork is just so 20th century,&quot; Ernie retorted. &quot;In my opinion, the future is in steam! Just let me stoke it up a little more and I can show you how it works.&quot; Ernie added a couple of shovels of coal to a small boiler and the hissing sound rose in intensity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calculator was a labyrinth of oily tubes, valves, gear-boxes, and brass widgets, like some sort of mechanically-augmented steam-punk tuba.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There were three long rows of clock-dials (each with a single pointer hand), each face numbered $0$ to $9$, and a small keyboard. The 14 keys were labelled with the digits &lt;kbd&gt;0&lt;/kbd&gt; to &lt;kbd&gt;9&lt;/kbd&gt; and the letters &lt;kbd&gt;A&lt;/kbd&gt;, &lt;kbd&gt;B&lt;/kbd&gt;, and &lt;kbd&gt;D&lt;/kbd&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;If you press a series of digit buttons then the &lt;kbd&gt;A&lt;/kbd&gt; or &lt;kbd&gt;B&lt;/kbd&gt; button, you can load a number into the $A$ or $B$ register,&quot; he explained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;What does the &lt;kbd&gt;D&lt;/kbd&gt; button do?&quot; I asked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;That,&quot; Ernie replied, &quot;is to test the &lt;strong&gt;D&lt;/strong&gt;ivision and &lt;strong&gt;D&lt;/strong&gt;ecrement functions. &lt;strong&gt;The machine checks if $A$ divides perfectly into $B$ and, if it does, it completes the division and outputs $B \div A$ on the $C$ register. Then it decrements $A$ twice&lt;/strong&gt;.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;What happens if $A$ doesn't divide into $B$?&quot; I asked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;strong&gt;Then it doesn't do anything&lt;/strong&gt;. I haven't built in fractions or decimal points yet, and I hate approximate answers!&quot; said Ernie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GoGC6.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GoGC6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ernie asked me to type my 'lucky' 3-digit number into the $A$ register, but unlike Ernie I don't believe in lucky numbers, so I typed in the first number I could think of - which happened to be my auto-bank PIN &lt;em&gt;$a$&lt;/em&gt;. When I pressed the &lt;kbd&gt;A&lt;/kbd&gt; button, the indicator arms on the first three dials dial of the $A$ register swung round to point at the digits I had entered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Now enter a big number for the $B$ register,&quot; Ernie commanded. I typed in &lt;em&gt;$n$&lt;/em&gt; digits, but when I hit the &lt;kbd&gt;B&lt;/kbd&gt; button, the first &lt;em&gt;$n$&lt;/em&gt; indicator arms all swung round to the digit $1$. Ernie muttered something about the steam pressure being too low to actuate the internal digital transfer mechanism (or maybe a sticky bypass needle actuator) but then pointed out that $b = 111111...11111$ was still a valid integer. &quot;Try the &lt;kbd&gt;D&lt;/kbd&gt; button now,&quot; he said.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I pressed &lt;kbd&gt;D&lt;/kbd&gt;. The hissing and clanking rose in intensity, and the indicator arms on the $C$ register swung round wildly. After some seconds there was a mechanical &quot;ding&quot; and the arms froze in place, displaying a very long number in the $C$-register, and &lt;em&gt;$a$&lt;/em&gt; was decremented by $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Well at least &lt;strong&gt;D&lt;/strong&gt;ivision and &lt;strong&gt;D&lt;/strong&gt;ecrementation work properly,&quot; said Ernie, after checking the results. &quot;Lucky that your &lt;em&gt;$a$&lt;/em&gt; was a factor of that &lt;em&gt;$b$&lt;/em&gt;. Try a few more divisions and I will see if I can sort out the problems with data-entry,&quot; he said.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I left &lt;em&gt;$a-2$&lt;/em&gt; in the $A$ register and entered another number &lt;em&gt;$b_2$&lt;/em&gt; (of different length) into the $B$ register. Once again &lt;em&gt;$b_2$&lt;/em&gt; turned to all $1$s, and when &lt;kbd&gt;D&lt;/kbd&gt; was pressed, an answer was output to register $C$ and $a-2$ was decremented to $a-4$. I repeated the process &lt;strong&gt;twice&lt;/strong&gt; more without changing the $A$ register, and the same sequence of events occurred each time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But after my fourth calculation Ernie waved me to a stop. &quot;I think I have found part of the problem,&quot; he said. He made a small adjustment to a set of valves with a mallet and asked me to try again. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;His adjustments had reset all the registers back to $0$, so I re-entered my 3-digit PIN into $A$, and entered a new &lt;em&gt;$b$&lt;/em&gt;. My fingers were getting a bit tired from all the number pressing, so this time I just typed in a 6-digit number. But when I pressed &lt;kbd&gt;B&lt;/kbd&gt;, there was still obviously a problem - the indicator arms all pointed at $0$s and $1$s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Not perfect, but better than before&quot;, Ernie said, then pointed out that I still had a valid 6-digit integer in the $B$ register and I should try the &lt;kbd&gt;D&lt;/kbd&gt; button again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As on all the previous occasions, &lt;em&gt;$a$&lt;/em&gt; divided perfectly into &lt;em&gt;$b$&lt;/em&gt;, and after displaying an answer in the $C$ register, &lt;em&gt;$a$&lt;/em&gt; was replaced by &lt;em&gt;$a-2$&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I repeated the process &lt;strong&gt;thrice&lt;/strong&gt; more without altering the $A$ register, but typing a 7-digit, then an 8-digit, then a 9-digit number into the $B$ register (each constructed from $0$s and $1$s), and the same sequence of events occurred each time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually Ernie lamented that &quot;...it might take a bit more thought to work out what is going wrong with data entry and decrementation, but at least it is doing the actual division perfectly.&quot; So we stopped for lunch. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I suppose you will be telling this story on the web,&quot; he commented after we had eaten. I was surprised and a little embarrassed because I hadn't realized Ernie had even known about my stories. But Ernie reassured me that he didn't mind his 'web-presence'. &quot;But if I can make a suggestion,&quot; he continued, &quot;it might be safest to change your PIN if you are going to tell the story exactly as it happened.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well - I did change my PIN even though I couldn't see how it could possibly be determined from my story. I'm guessing that Ernie was just playing a practical joke on me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Can anybody tell me if it really is possible to calculate my old PIN from what I have told you?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="30903" LastEditDate="2016-12-20T21:34:46.623" LastActivityDate="2016-12-20T21:34:46.623" Title="Ernie and the Unexpectedly Exposed PIN" Tags="|mathematics|calculation-puzzle|word-problem|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18435" PostTypeId="2" ParentId="18433" CreationDate="2015-07-29T05:30:41.007" Score="6" Body="&lt;p&gt;&lt;em&gt;Partial answer...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I think) I have managed to crack the first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Musical chairs, upgraded&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Game of Thrones&lt;/strong&gt;&lt;br&gt;&lt;br&gt;I arrived at this by breaking it up into two phrases. {Musical chairs} is a game, and {chairs, upgraded} gave me thrones, which could be considered as upgraded versions of chairs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-07-29T05:30:41.007" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18436" PostTypeId="2" ParentId="18418" CreationDate="2015-07-29T07:27:53.900" Score="3" Body="&lt;p&gt;Thanks to @Aioros who almost solved the puzzle. Here is the solution :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every sentence that is 1 high when it is written in lower-case, excluding letters with descenders.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Old answer&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Certainly not the answer, but my solution seems to work :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every sentence that doesn't contain a T works.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13384" LastEditorUserId="13384" LastEditDate="2015-07-31T11:05:14.500" LastActivityDate="2015-07-31T11:05:14.500" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18437" PostTypeId="1" CreationDate="2015-07-29T10:55:11.413" Score="20" ViewCount="2297" Body="&lt;h1&gt;The Question&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(Some knowledge of potential difference, resistor combinations, etc. required)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have 1000 boxes. Some (or none) of them are conductors (of negligible resistance). The rest are insulators (of infinite resistance). Your job is to tell how many of them are conductors. You don't need to identify them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have an unlimited supply of wire, cells and resistors. You also have special fuses that give information. You must design a circuit that gives you the answer based on the information from the fuses. (You have no other materials/devices/tools available.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each fuse can take upto 1 ampere. After this, it burns. Though theoretically not true, assume that it takes one second for a fuse to burn, irrespective of the current (above 1A) flowing through it. Each fuse emits a unique wireless signal when it burns. These signals are displayed on a computer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On switching on all cell(s) simultaneously, you get information from the fuses. You cannot make any changes to the circuit and must keep it running forever (or until the fuses break the circuit). Using this information from the fuses (and knowing the circuit arrangement), you must be able to tell the number of conducting boxes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How many fuses do you need?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Hints and notes&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. You can create a fuse of any load (not just 1A), by combining a fuse with one of two resistors in parallel. Current divides in ratios in resistors, so you know how much of the net current enters the fuse.&lt;br&gt;&lt;br&gt;2.A fuse may burn because another fuse has already burnt.&lt;br&gt;&lt;br&gt;3. Though I have not solved it, I suspect the answer could be as less as 8 or 9 fuses. For example, just 3 fuses can (possibly) test 25 boxes as there are so many possible outputs by the computer:&lt;br&gt;&lt;br&gt;None, 1, 2, 3, 1 and 2, 2 and 3, 3 and 1, 1 then 2, 1 then 3, 2 then&lt;br&gt; 3, 2 then 1, 3 then 1, 3 then 2, 1 and 2 then 3, 1 and 3 then 2, 2 and&lt;br&gt; 3 then 1, 1 then 2 and 3, 2 then 3 and 1, 3 then 1 and 2, 1 and 2 and 3 &lt;br&gt;&lt;br&gt;4. @user3294068 says it can be done using just 5 fuses (mathematically, at least)&lt;br&gt;&lt;br&gt;5. There was a clarification in one of the comments. You can connect many wires to a single box. You cannot cut a box apart, since I have not provided you a saw/machine to do so.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bounty awarded but question remains open.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="18129" LastEditDate="2017-01-28T05:11:50.353" LastActivityDate="2017-01-28T05:11:50.353" Title="How many boxes are conductors?" Tags="|mathematics|visual|optimization|physics|circuitry|" AnswerCount="4" CommentCount="26" ContentLicense="CC BY-SA 3.0" />
  <row Id="18439" PostTypeId="2" ParentId="18421" CreationDate="2015-07-29T12:51:22.567" Score="0" Body="&lt;p&gt;Wish I had a more developed answer but: &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Brooklyn &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some say I'm quite insane.&#xA;I blame it on the heights.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Brooklyn heights is a thing, and Brooklyn is definitely considered by some to be a crazy place&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you pick up the phone, press 6.&#xA;Maybe you'll reach &quot;Bush Sandwich&quot; Hill.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If you take 6th ave you get within about 6 blocks of clinton hill neighborhood.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13916" LastActivityDate="2015-07-29T12:51:22.567" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18441" PostTypeId="2" ParentId="18434" CreationDate="2015-07-29T13:30:06.343" Score="9" Body="&lt;p&gt;If a number is even, it's never going to divide a number that is all 1s. Similarly, a number that is divisible by 5 also isn't going to divide all 1s. So &lt;em&gt;a&lt;/em&gt;, &lt;em&gt;a&lt;/em&gt;-2, &lt;em&gt;a&lt;/em&gt;-4, and &lt;em&gt;a&lt;/em&gt;-6 are all odd and not divisible by 5. The only way that this is possible is if &lt;em&gt;a&lt;/em&gt; ends in 3. This section gives us no further information, since every number that is relatively prime to 10 will divide a repunit of some length.&lt;sup&gt;1&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I couldn't come up with a neat way to do the second part, so I ran a brute force search of all 3-digit factors ending in 3 of 6-digit numbers consisting of 1s and 0s. This was the result:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;143 divides 100100, 101101, 110110, 111111&#xA;183 divides 100101&#xA;193 divides 110010&#xA;213 divides 100110&#xA;273 divides 101010, 111111&#xA;393 divides 101001&#xA;653 divides 111010&#xA;803 divides 110011&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then, I tested the corresponding $a$-2 values with 7-digit numbers:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;141 divides 1001100, 1011111&#xA;181 divides 1001111&#xA;271 divides 1011101, 1111100&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And then the 8-digit numbers:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;179 divides 11010111&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Just to check:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;177 divides 100101111, 101110011&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The PIN must be 183.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt;Proof from Wikipedia (&lt;a href=&quot;https://en.wikipedia.org/wiki/Repunit#Properties&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Repunit#Properties&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Using the pigeon-hole principle it can be easily shown that for each $n$ and $b$ such that $n$ and $b$ are relatively prime, there exists a repunit in base $b$ that is a multiple of $n$. To see this consider repunits $R_1^{(b)},...,R_n^{(b)}$. Assume none of the $R_k^{(b)}$ is divisible by $n$. Because there are $n$ repunits but only $n-1$ non-zero residues modulo $n$ there exist two repunits $R_i^{(b)}$ and $R_j^{(b)}$ with $1\le i&amp;lt;j\le n$ such that $R_i^{(b)}$ and $R_j^{(b)}$ have the same residue modulo $n$. It follows that $R_j^{(b)} - R_i^{(b)}$ has residue $0$ modulo $n$, i.e. is divisible by $n$. $R_j^{(b)} - R_i^{(b)}$ consists of $j - i$ ones followed by $i$ zeroes. Thus, $R_j^{(b)} - R_i^{(b)} = R_{j-i}^{(b)}\times b^i$. Since $n$ divides the left-hand side it also divides the right-hand side, and since $n$ and $b$ are relatively prime, $n$ must divide $R_{j-i}^{(b)}$, contradicting the original assumption.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-07-29T15:52:39.827" LastActivityDate="2015-07-29T15:52:39.827" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18442" PostTypeId="2" ParentId="18390" CreationDate="2015-07-29T14:37:41.057" Score="0" Body="&lt;p&gt;I don't think this is what you're after, but it could look like part of one school of thought for the perfect barbecue:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;add another one (second distinct Obj object: 'another', hashCode: 'one') to the block (static initialiser: 'block', also HashSet: reference to barbecue grids or the deliberate style of burn marks on the meat after cooking), turn it over once at the natural point, i.e. half way through cooking (e: base of &lt;em&gt;natural&lt;/em&gt; logs, rotate by $\pi$: turn around), then just let it sit i.e. do nothing until it is done (empty brace = nop: do nothing , end of main(): 'done').&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9422" LastActivityDate="2015-07-29T14:37:41.057" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18443" PostTypeId="2" ParentId="18434" CreationDate="2015-07-29T15:40:59.317" Score="1" Body="&lt;p&gt;Building off of f&quot; answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The machine converted Ernie's big numbers into giant strings of 1's, but these 1's were in fact divisible by his secret code.  Thus we need to confirm the following 4 things to in fact show that 181 does work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;183 * some 1st giant number = some giant amount of 1's&lt;br&gt;&#xA;181 * some 2nd giant number = some giant amount of 1's&lt;br&gt;&#xA;179 * some 3rd giant number = some giant amount of 1's&lt;br&gt;&#xA;177 * some 4th giant number = some giant amount of 1's  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also I should note that these &lt;em&gt;are not unique&lt;/em&gt;.  Instead the numbers below are the &quot;smallest&quot; which work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wrote a macro in excel to find these giant numbers after figuring out the 183 only to find that f&quot; beat me to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our input number for 183 is &lt;em&gt;only&lt;/em&gt;:&lt;br&gt;&#xA;&lt;strong&gt;607164541590771098967820279295689131754705525197328476017&lt;/strong&gt;&lt;br&gt;&#xA;which can be confirmed using &lt;a href=&quot;http://www.wolframalpha.com/input/?i=607164541590771098967820279295689131754705525197328476017%20*183&quot; rel=&quot;nofollow&quot;&gt;Wolfram Alpha here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our input number for 181 is:&lt;br&gt;&#xA;&lt;strong&gt;613873542050337630448127685696746470227133210558624923265807243707796193984039287906691221608348680171884591774094536525475751995089011663597298956414978514426028238182934315531&lt;/strong&gt;&lt;br&gt;&#xA;which can be confirmed using &lt;a href=&quot;http://www.wolframalpha.com/input/?i=613873542050337630448127685696746470227133210558624923265807243707796193984039287906691221608348680171884591774094536525475751995089011663597298956414978514426028238182934315531%20*181&quot; rel=&quot;nofollow&quot;&gt;Wolfram Alpha here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our input number for 179 is:&lt;br&gt;&#xA;&lt;strong&gt;6207324643078833022967101179391682184978274363749224084419615145872129112352576039726877715704531346989447548106765983860955927995034140285536933581626319056486654252017380509&lt;/strong&gt;&lt;br&gt;&#xA;which can be confirmed using &lt;a href=&quot;http://www.wolframalpha.com/input/?i=6207324643078833022967101179391682184978274363749224084419615145872129112352576039726877715704531346989447548106765983860955927995034140285536933581626319056486654252017380509%20*179&quot; rel=&quot;nofollow&quot;&gt;Wolfram Alpha here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, our input number for 177 is:&lt;br&gt;&#xA;&lt;strong&gt;627746390458254865034526051475204017576898932831136220966729441305712492153170119271814187068424356559949780288763339610797237915881983678593848085373509102322661644695543&lt;/strong&gt;&lt;br&gt;&#xA;which can be confirmed using &lt;a href=&quot;http://www.wolframalpha.com/input/?i=627746390458254865034526051475204017576898932831136220966729441305712492153170119271814187068424356559949780288763339610797237915881983678593848085373509102322661644695543%20*177&quot; rel=&quot;nofollow&quot;&gt;Wolfram Alpha here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;---EDIT---&lt;br&gt;&#xA;Since I had written the script, I wanted to look at all numbers 1000 digits or less that could be divided by 183, 181, 179, and 177. (The smallest of which is listed above).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;177: 171, 345, 519, 693, 867 digits (a repeating cycle of 174 digits)&#xA;179: 175, 353, 531, 709, 887 digits (a repeating cycle of 178 digits)&#xA;181: 177, 357, 537, 717, 897 digits (a repeating cycle of 180 digits)&#xA;183:  57, 117, 177, 237, 297 digits (a repeating cycle of  60 digits)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I was wondering if anyone knew how to predict the length of the number/cycle?&lt;/p&gt;&#xA;" OwnerUserId="14092" LastEditorUserId="14092" LastEditDate="2015-07-29T16:16:53.467" LastActivityDate="2015-07-29T16:16:53.467" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18444" PostTypeId="2" ParentId="18421" CreationDate="2015-07-29T15:50:24.287" Score="7" Body="&lt;p&gt;The who is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alfred Hitchcock&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not got all the links but here are a few:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some say I'm quite insane.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Psycho&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I blame it on the heights.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Vertigo&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Don't look out the back.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Rear Window&lt;/em&gt;&lt;br&gt;&#xA; You'd need a rear window in order to look out the back; I think this works as a fairly straight clue to the film as well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Follow the V-formation,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;The Birds&lt;/em&gt;&lt;br&gt;&#xA; Some migratory birds fly in a V-formation, particularly geese. This initially made me think of &lt;em&gt;Top Gun&lt;/em&gt;, which got me the film link.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you pick up the phone, press 6.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Dial M for Murder&lt;/em&gt;&lt;br&gt;&#xA; M being one of the letters for number 6 (MNO) on old phones.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's hope you've got just enough ignorance; nobody likes a know-it-all.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;The Man Who Knew Too Much&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a little research:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a ride on the Phoenix-Durango.&#xA;Chances are you'll know no-one when you go.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Strangers on a Train&lt;/em&gt;&lt;br&gt;&#xA; The Phoenix-Durango is a train line in &lt;em&gt;Atlas Shrugged&lt;/em&gt;, it would appear.&lt;br&gt;&#xA; Obviously if you know no-one, it's because everyone is a stranger.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Do you really know what hides in dark corners?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Shadow of a Doubt&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And some guesses and the remaining ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe you'll reach &quot;Bush Sandwich&quot; Hill&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Secret Agent&lt;/em&gt;?&lt;br&gt;&#xA; Following Going hamateur's reading of this as Clinton Hill, and with a little work on Wikipedia, I turned up &lt;a href=&quot;https://en.wikipedia.org/wiki/Clint_Hill_(Secret_Service)&quot; rel=&quot;nofollow&quot;&gt;Clinton &quot;Clint&quot; Hill&lt;/a&gt;, a secret service agent.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Otherwise, look forward to some road bumps in your progress; maybe a couple pitfalls?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Sabotage&lt;/em&gt;?&lt;br&gt;&#xA; A bit of a guess, but if I saw that my path ahead had lots of holes dug in it, I might suspect sabotage.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Stop at the top of the mountain&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Downhill&lt;/em&gt;?&lt;br&gt;&#xA; Really stretching here, but if you're at the peak of a mountain, you can only go downhill.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13777" LastEditorUserId="13777" LastEditDate="2015-07-30T14:13:25.347" LastActivityDate="2015-07-30T14:13:25.347" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18445" PostTypeId="1" AcceptedAnswerId="18450" CreationDate="2015-07-29T16:10:14.867" Score="14" ViewCount="1474" Body="&lt;p&gt;I've got this riddle from a friend at work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say we have 100 dwarves.&#xA;They all receive little dwarven shirts with numbers between 0-99 (With repetitions), and after that they all enter into a room. (So that they can see all the other dwarves numbers but not their own number.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They need to decide on an algorithm before entering the room so that they all call out the &lt;strong&gt;same&lt;/strong&gt; number at the &lt;strong&gt;same&lt;/strong&gt; time and at least one of the dwarves will have that number on &lt;strong&gt;his&lt;/strong&gt; shirt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While in the room they they can't exchange any type of information: they can't talk or arrange in any way (Like sorting themselves out and such).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the algorithm? :)&lt;/p&gt;&#xA;" OwnerUserId="13048" LastEditorUserId="8874" LastEditDate="2015-07-29T17:06:33.583" LastActivityDate="2015-07-30T12:18:15.230" Title="100 Dwarves and a tiny room" Tags="|logical-deduction|hat-guessing|" AnswerCount="6" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="18448" PostTypeId="1" AcceptedAnswerId="18489" CreationDate="2015-07-29T16:55:05.083" Score="32" ViewCount="4075" Body="&lt;h2&gt;Backstory&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Disaster has struck for &lt;a href=&quot;https://en.wikipedia.org/wiki/Leporidae&quot;&gt;Leporidae&lt;/a&gt; throughout the fantasy realm -- a mass carrot famine has wreaked havoc upon all existences but our own. Because of this, 25 of the leading fantasy bunnies, rabbits, and hares from across the universes have banded together and traveled to Earth, setting up camp in China, the &lt;a href=&quot;http://www.carrotmuseum.co.uk/worldcarrots.html&quot;&gt;world's largest exporter of carrots&lt;/a&gt;. Here, they've organized into groups of five, with each group leaving at a designated time to collect carrots from wherever they can and bring them back to their stores, in hopes that they'll be able to collect enough carrots to bring back to their own universes and satisfy the needs of their people.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, it seems that a member of the Herbivorous Apiaceae Reconnaissance Enforcement (or H.A.R.E., for short) has gone rogue. During one of the missions yesterday, someone pushed poor old Uncle Wiggily down the rabbit hole, eliminating a member of the force. You're a private investigator, hired by the Enforcement as a whole. Your job is to determine who ruthlessly murdered the old sap.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Complications&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Each Leporidae knows some information about the other Leporidae. Unfortunately, there are too many of them to ask each one individually about every single other rabbits, as time is of the essence before the killer strikes again. As such, you've had the suspects line up in alphabetical order, and you plan to interrogate them on the suspects to their immediate right and left. This should give you enough information to piece together the entire situation and discover the killer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, you know that the killer will being lying to you to throw you off his/her scent. This means that some information might be contradictory to what you already know -- no matter. All you have to do is find out whose account of the previous day doesn't add up, and you've got your killer, right? You chuckle. If only it was so simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make matters worse, you were on the phone with two of your suspects who witnessed the murder take place. They masked their voices, so you don't know which  two of your suspects they were. All you know is that they were in the group directly after the murderer and Uncle Wiggily, and witnessed the murder as they were prepping for their departure. Sadly, the murderer caught wind of your call, and you've received an anonymous tip that he or she has threatened the lives of the two whistle-blowers. This can only mean that they'll be lying through their teeth as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even further complicating things, someone has been raiding the carrot stores during the night, stealing a few carrots for his or herself. While you aren't here to investigate that, the perpetrator will still be terrified of being caught, and of course that means he or she will also be lying. If you can, catch that culprit as well, though it's not your main task.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;To sum up what you know:&lt;/strong&gt; There were five groups, each consisting of five rabbits, leaving at five different designated points in the day. The murderer was in the same group as Uncle Wiggily, and offed him during their scheduled time frame. Four people will be lying to you: the murderer, the two witnesses from the group directly after the murderer, and the carrot thief. Catch the murderer and lock him up for good, and give the carrot thief a good slap on the wrist.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Interrogations&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suspect: &lt;a href=&quot;https://en.wikipedia.org/wiki/Br%27er_Rabbit&quot;&gt;Br'er Rabbit&lt;/a&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Mood: Surprisingly arrogant&lt;br&gt;&#xA;  &lt;hr /&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You circle your first suspect, pacing slowly. He stands with a confident smirk on his face. &quot;So, Br'er Rabbit,&quot; you drawl. &quot;yes, &lt;em&gt;sir&lt;/em&gt;?&quot; He mocks. You stop in front of him, facing him head-on. After a good staredown, you peek all the way down the line of suspects.&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;What can you tell me about the &lt;strong&gt;White Rabbit&lt;/strong&gt;?&quot; You ask. Br'er Rabbit smirks. &quot;Well, I know he was &lt;strong&gt;in my group&lt;/strong&gt;,&quot; the rabbit responds haughtily. You spit on the ground by his feet, but he's unfazed.&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;How about this joker next to you -- &lt;strong&gt;Bugs Bunny&lt;/strong&gt;, is it?&quot; Br'er snickered. &quot;Well, I know he was &lt;strong&gt;in a group with Ruby&lt;/strong&gt;, which I'm sure didn't make Lola too happy.&quot; &#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You grimace. &quot;Thank you for your time.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suspect: &lt;a href=&quot;https://en.wikipedia.org/wiki/Bugs_Bunny&quot;&gt;Bugs Bunny&lt;/a&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Mood: Inquisitive&lt;br&gt;&#xA;  &lt;hr /&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;What's up, Doc?&quot; You're greeted with as you approach your next suspect. &quot;I ain't got time for your games,&quot; you snarl back. Bugs Bunny munches on a carrot.&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You jab your thumb towards the rabbit you just interviewed. &quot;Whaddya know about Br'er?&quot; Bugs crunches his carrot a couple times, then swallows. &quot;Well, he &lt;strong&gt;wasn't in my group&lt;/strong&gt;, that's for sure.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You sneer. &quot;How about this Bunnicula character?&quot; Bugs stops and thinks. &quot;He was &lt;strong&gt;in a group with two Rabbits&lt;/strong&gt;. I'm sure of it!&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You frown. &quot;Thank you for your time.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suspect: &lt;a href=&quot;https://en.wikipedia.org/wiki/Bunnicula&quot;&gt;Bunnicula&lt;/a&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Mood: Thirsty&#xA;  &lt;hr /&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Pacing up to your next suspect, you're taken aback by his rather long and pointy teeth. He notices. &quot;All the better to bite you with, my dear,&quot; he snarls, then laughs. &quot;Just kidding. They're for carrot juice.&quot; You breathe a sigh of relief.&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;Tell me somethin' about this joker on your left.&quot; Bunnicula looks to his left. &quot;I mean &lt;em&gt;my&lt;/em&gt; left,&quot; you correct. He looks to his right. &quot;You mean Bugs? Yeah, he was &lt;strong&gt;in a group with two Rabbits&lt;/strong&gt;. I know that for sure.&quot; You pause. &quot;What's with this 'two rabbits' nonsense I keep hearin'?&quot; Bunnicula explains, &quot;Rabbits have some alpha dominance issues, so we try to keep them at either one or two per group.&quot; You nod with moderate understanding.&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;Let's move on,&quot; you drawl. &quot;How about Buster Baxter?&quot; Bunnicula thinks for a moment. &quot;Yes, he was in the &lt;strong&gt;third group to leave&lt;/strong&gt;.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You scowl. &quot;Thank you for your time.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suspect: &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_Arthur_characters#Buster_Baxter&quot;&gt;Buster Baxter&lt;/a&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Mood: Youthfully oblivious&#xA;  &lt;hr /&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You round upon the youngster. &quot;Hello, sir!&quot; He announces cheerfully. You hold yourself back from shutting him up. After all, he's just a kid.&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; &quot;You familiar at all with Bunnicula?&quot; You ask. He nods back. &quot;Sure am! And I know for a fact he was &lt;strong&gt;in the group directly after the White Rabbit.&lt;/strong&gt; You smile. &quot;Good to know.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; Moving on, you ask, &quot;How about Crusader Rabbit?&quot; Buster giggles. &quot;He's one of the most alpha of all the rabbits! I know he &lt;strong&gt;wasn't in a group with another Rabbit.&quot;&lt;/strong&gt;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You feign a smile. &quot;Thank you for your time.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suspect: &lt;a href=&quot;https://en.wikipedia.org/wiki/Crusader_Rabbit&quot;&gt;Crusader Rabbit&lt;/a&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Mood: Unwaveringly confident, likely to a fault&#xA;  &lt;hr /&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;Stand at ready, sol-DIER!&quot; You bark at the rabbit. He stands stiffly and attentively. &quot;Good. You may relax.&quot; He breathes a sigh of relief as you prepare your questions.&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;What do you know about one 'Buster Baxter'?&quot; You inquire. &quot;Sir, Buster Baxter was &lt;strong&gt;not in a group with Uncle Wiggily&lt;/strong&gt;, sir!&quot; The bunny barks back. You appreciate the respect.&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;How about one 'Cuddles'?&quot; You continue your interrogation. &quot;Sir, Cuddles was &lt;strong&gt;in the last group&lt;/strong&gt; of the day, sir!&quot; The bunny responds.&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You smile again, this time with undeserved confidence. &quot;Thank you for your time.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suspect: &lt;a href=&quot;http://happytreefriends.wikia.com/wiki/Cuddles&quot;&gt;Cuddles&lt;/a&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Mood: Grizzled &#xA;  &lt;hr /&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You nod at your next suspect with respect. &quot;You look like a man who's seen some things,&quot; you note. &quot;That I am, comrade,&quot; he sighs. &quot;That I am.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;Let's get down to business,&quot; you offer. &quot;Tell me about Crusader Rabbit.&quot; Cuddles nods solemnly. &quot;That Rabbit was &lt;strong&gt;in my group&lt;/strong&gt;.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;How about the Easter Bunny?&quot; You ask, continuing on. &quot;That Bunny was &lt;strong&gt;in the second group&lt;/strong&gt; of the day,&quot; he says stoically.&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You nod back at him with respect. &quot;Thank you for your time.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suspect: &lt;a href=&quot;https://en.wikipedia.org/wiki/Easter_Bunny&quot;&gt;Easter Bunny&lt;/a&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Mood: Devious&#xA;  &lt;hr /&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You approach your next suspect with caution. &quot;Got any eggs on ya?&quot; You prod. The bunny looks at you with shifty eyes. &quot;Nnno. Why would I?&quot; You spit on the ground. &quot;No reason.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;What do you know about Cuddles?&quot; You quickly fire off. The bunny is taken aback, but answers with haste. &quot;I know that Cuddles was &lt;strong&gt;in a group with two Rabbits&lt;/strong&gt;.&quot; You groan. Another 'two rabbits' clue.&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;What about the Energizer Bunny?&quot; You fire again. This time, he isn't caught off-guard. &quot;He was &lt;strong&gt;in the fourth group&lt;/strong&gt; of the day, for sure.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You spit on the ground again. &quot;Thank you for your time.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suspect: &lt;a href=&quot;https://en.wikipedia.org/wiki/Energizer_Bunny&quot;&gt;Energizer Bunny&lt;/a&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Mood: Energetic&#xA;  &lt;hr /&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;Could you stop banging that infernal drum?&quot; You yell over the clanging. The bunny shakes his head. &quot;Sorry, but I just keep going and going and going.&quot; You scowl.&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;What do you know about the Easter Bunny?&quot; You yell over the racket. The Energizer Bunny yells back, &quot;He was &lt;strong&gt;the only Bunny in his group&lt;/strong&gt;!&quot; He keeps drumming.&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Covering your ears, you scream. &quot;HOW ABOUT LOLA BUNNY?&quot; The Energizer Bunny screams back, &quot;SHE WAS &lt;strong&gt;IN MY GROUP&lt;/strong&gt;!!!&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Fed up, you grab his drumsticks and snap them in half. &quot;Thank you for your time.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suspect: &lt;a href=&quot;https://en.wikipedia.org/wiki/Lola_Bunny&quot;&gt;Lola Bunny&lt;/a&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Mood: Competitive&#xA;  &lt;hr /&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;As you approach your next suspect, you straighten your tie. &quot;How are you doin' today, sweetheart?&quot; She wrinkles her nose. &quot;Ew, no thanks.&quot; You remain unfazed.&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;Tell me about the Energizer Bunny.&quot; She rolls her eyes back in thought for a second. &quot;Yeah, he was definitely &lt;strong&gt;in my group&lt;/strong&gt;.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;And how about the March Hare?&quot; You continue with no pause. She chuckles. &quot;Well, he &lt;strong&gt;wasn't the only Hare in his group&lt;/strong&gt;, I know that much!&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You smile and wink, much to her disdain. &quot;Thank you for your time.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suspect: &lt;a href=&quot;https://en.wikipedia.org/wiki/March_Hare&quot;&gt;March Hare&lt;/a&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Mood: Maniacal&#xA;  &lt;hr /&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;Tea?&quot; Your next suspect offers you as you approach. You look him up and down. &quot;You don't have a teapot.&quot; He stares off into the distance, not noticing your comment.&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You snap your fingers in his face, and he snaps back to reality. &quot;What do you know about Lola Bunny?&quot; The hare begins to drift off again, but you snap him back to reality. &quot;Ah, yes. Lola. She was &lt;strong&gt;in the group directly after mine.&lt;/strong&gt;&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;And Max?&quot; You prod while you still have his attention. &quot;Max? He was &lt;strong&gt;in a group with two different Bunnies&lt;/strong&gt;.&quot; The hare drifts off again. &quot;Tea?&quot; He offers.&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You sigh. &quot;Thank you for your time.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suspect: &lt;a href=&quot;http://maxandruby.wikia.com/wiki/Max&quot;&gt;Max&lt;/a&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Mood: Sheepish&#xA;  &lt;hr /&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You approach your next suspect with caution. &quot;What's got you down, kid?&quot; He whimpers back. &quot;I want my toy truck.&quot; You snap, and your assistant comes scurrying with the toy truck. &quot;Are you happy now, kid?&quot; The kid grins. &quot;M-hmm.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;Tell me what you know about the March Hare,&quot; you gently prod. &quot;He was &lt;strong&gt;in the only group with at least two Bunnies&lt;/strong&gt;,&quot; he says nonchalantly while playing with his truck.&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You make note of the previous while asking, &quot;and how about Mr. Bun?&quot; Max still plays with his truck. &quot;He was &lt;strong&gt;in the only group with no Bunnies&lt;/strong&gt; at all.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You pat the kid on the back. &quot;Thank you for your time.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suspect: &lt;a href=&quot;http://calvinandhobbes.wikia.com/wiki/Mr._Bun&quot;&gt;Mr. Bun&lt;/a&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Mood: Quiet&#xA;  &lt;hr /&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;How are you doing today, Mr. Bun?&quot; You ask as you approach. &quot;Good? Good. Let me ask you a few questions.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;What do you know about Max?&quot; A silent pause goes by. &quot;He &lt;strong&gt;wasn't in a group with Ruby&lt;/strong&gt;? Interesting, interesting. And probably for the better.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;How about Mr. Whiskers?&quot; Another silent pause. &quot;He was &lt;strong&gt;in a group with two other Rabbits&lt;/strong&gt;? I could have guessed that one by now,&quot; you grumble.&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You share a final silent pause. &quot;Thank you for your time.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suspect: &lt;a href=&quot;http://disney.wikia.com/wiki/Mr._Whiskers&quot;&gt;Mr. Whiskers&lt;/a&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Mood: Moronic, if that counts as a mood&#xA;  &lt;hr /&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You peer through squinted eyes at the moronic grin on your next suspect. &lt;em&gt;This is gonna be fun&lt;/em&gt;, you think to yourself. &lt;em&gt;Time for bad cop&lt;/em&gt;.&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Cracking your knuckles, you stare intently at Mr. Whiskers. &quot;Tell me, boy, what do you know about Mr. Bun?&quot; You turn away, then whip back towards him. &quot;And don't you lie to me, boy.&quot; The boy just chuckles. &quot;Mr. Bun was &lt;strong&gt;in a group with two other Rabbits&lt;/strong&gt;!&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Groaning in frustration, you grab the bunny by his collar and yank him to where your foreheads are touching. &quot;And what do you know about the Nesquik Bunny?&quot; You bark. The bunny maintains his stupid grin. &quot;He was &lt;strong&gt;in the first group&lt;/strong&gt; of the day!&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You let go of the bunny's shirt in disgust. &quot;Thank you for your time.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suspect: &lt;a href=&quot;https://en.wikipedia.org/wiki/Nesquik#Nesquik_Bunny_.28a.k.a._Quicky.29&quot;&gt;Nesquik Bunny&lt;/a&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Mood: Obscenely hyper&#xA;  &lt;hr /&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;I'm cuckoo for Nesquik!&quot; The bunny exclaims as you approach. You squint. &quot;Isn't it 'I'm cuckoo for Cocoa Puffs?'&quot; The bunny shrugs. &quot;I was just tryin' it on for size.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Shaking your head, you continue. &quot;What do you know about Mr. Whiskers?&quot; The Nesquik Bunny muses, &quot;Ba-da-ba-ba-ba...I'm lovin' Nesquik! Uh, yeah, he was &lt;strong&gt;in my group&lt;/strong&gt;.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Halfheartedly, you continue. &quot;And what about Peppy Hare?&quot; The Nesquik Bunny thinks for a second. &quot;Nesquik! It gives you wings! Hrm...he was &lt;strong&gt;in a group with Uncle Wiggily&lt;/strong&gt;.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You move on before he can try out another slogan on you. &quot;Thank you for your time.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suspect: &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_Star_Fox_characters#Peppy_Hare&quot;&gt;Peppy Hare&lt;/a&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Mood: Wise but wary&#xA;  &lt;hr /&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You smirk as you approach the next suspect. &quot;Hey, Peppy! Do a barr--&quot; He cuts you off. &quot;Don't you freakin' say it, man. I swear to God, you better not say it.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;--el roll,&quot; you whisper under your breath. Clearing your throat, you ask in a booming voice, &quot;What do you know about the Nesquik Bunny?&quot; Peppy replies, &quot;Yeah, he was &lt;strong&gt;in a group with two other Rabbits&lt;/strong&gt;.&quot; You groan. &quot;I swear, the next time I hear that, I'm gonna smack whoever said it.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;And what about Peter Cottontail?&quot; Peppy thinks for a moment, then says, &quot;Yeah, he was in a group with t--er, he was &lt;strong&gt;in a group with Br'er Rabbit.&quot;&lt;/strong&gt;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You glare, then move on. &quot;Thank you for your time.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suspect: &lt;a href=&quot;https://en.wikipedia.org/wiki/Peter_Cottontail&quot;&gt;Peter Cottontail&lt;/a&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Mood: Sick of answering the same darn question over and over&#xA;  &lt;hr /&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You approach your next suspect inquisitively. &quot;Say, do you have any relation to Peter Rabbit?&quot; The bunny groans. &quot;NO, I don't.&quot; You shake your head. &quot;No, I'm pretty sure you guys are the same.&quot; The bunny groans even louder. &quot;For the last time, I'm NOT Peter Rabbit! Get it through your thick skull!&quot; You fail to accept it. &quot;Whaaaatever.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;How about Peppy Hare?&quot; Peter spits back, &quot;No, he's not Peter Rabbit, either.&quot; You correct him. &quot;I mean what do you know about him, smart-alec?&quot; He looks down, foolishly. &quot;Oh. Well, I know he was &lt;strong&gt;in the second group&lt;/strong&gt; of the day.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You nod. &quot;And how about Playboy Bunny?&quot; Peter thinks, then says, &quot;I'm pretty sure he was &lt;strong&gt;in the fourth group&lt;/strong&gt; of the day.&quot; You shake your head. &quot;No, I mean is he Peter Rabbit?&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You grin as Peter glowers at you. &quot;Thank you for your time.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suspect: &lt;a href=&quot;https://en.wikipedia.org/wiki/Playboy#Rabbit_logo&quot;&gt;Playboy Bunny&lt;/a&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Mood: Not as flirtatious as you hoped&#xA;  &lt;hr /&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You approach the next suspect excitedly, but are dismayed to see an actual bunny standing there. &lt;em&gt;Well, at least he's a bunny with a bow tie&lt;/em&gt;, you reason. &lt;em&gt;That's pretty cool.&lt;/em&gt;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;What do you know about Peter Cottontail?&quot; You ask, clearing your throat. &quot;Well, he was &lt;strong&gt;in a group with White Rabbit&lt;/strong&gt;,&quot; the Playboy Bunny responds.&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;And how about Rabbit?&quot; You continue. &quot;He &lt;strong&gt;wasn't the only Rabbit in his group&lt;/strong&gt;,&quot; the Playboy Bunny answers.&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Masking your disappointment, you nod. &quot;Thank you for your time.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suspect: &lt;a href=&quot;https://en.wikipedia.org/wiki/Rabbit_(Winnie-the-Pooh)&quot;&gt;Rabbit&lt;/a&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Mood: Obsessive-compulsive&#xA;  &lt;hr /&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;As you approach your next suspect, he reaches out towards your neck. Defensively, you swat his arm away. &quot;What are you doing, man!?&quot; You exclaim in confusion. The Rabbit scratches his arm and twitched. &quot;Your tie. It's not straight.&quot; You look down and adjust your tie. &quot;Oh. Thanks.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;As he breathes a sigh of relief, you ask, &quot;What do you know about Playboy Bunny?&quot; Rabbit responds, &quot;Oh, he was &lt;strong&gt;in my group&lt;/strong&gt;. A real stand-up guy, that's for sure.&quot; You frown.&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;How about Rabbit of Caerbannog?&quot; You continue. &quot;Yes, he was &lt;strong&gt;in the second group&lt;/strong&gt;. I do what I can to avoid him,&quot; Rabbit shudders. &quot;Why? He's so cute!&quot; You retort, but Rabbit just shakes his head.&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You readjust your tie. &quot;Thank you for your time.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suspect: &lt;a href=&quot;https://en.wikipedia.org/wiki/Rabbit_of_Caerbannog&quot;&gt;Rabbit of Caerbannog&lt;/a&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Mood: &lt;s&gt;Bloodthirsty&lt;/s&gt; Adorable and fluffy!&#xA;  &lt;hr /&gt; &#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You approach your next suspect. &quot;What a cute little rabbit!&quot; You say as you pet him on the head. He eyes your jugular.&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;Do you know anything about Rabbit?&quot; You ask, still petting. The rabbit chomps down on your hand, leaving two distinct bite marks. &quot;Ow!&quot; You scream, then you study the marks. &quot;Oh, I see. Rabbit was &lt;strong&gt;in a group with Bunnicula&lt;/strong&gt;. Thanks!&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Continuing to pet the bunny, you ask, &quot;What can you tell me about Reader Rabbit?&quot; The bunny turns away from you, moving its head out from under your hand. &quot;I get it, you're saying that Reader Rabbit was &lt;strong&gt;in the group directly after yours.&lt;/strong&gt; Your such a helpful bunny!&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You give the rabbit one last pat on the head. &quot;Thank you for your time.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suspect: &lt;a href=&quot;https://en.wikipedia.org/wiki/Reader_Rabbit&quot;&gt;Reader Rabbit&lt;/a&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Mood: Annoyingly educational&#xA;  &lt;hr /&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;Quick, four plus three!&quot; You're greeted with as you approach the rabbit. &quot;Spell 'sandcastle'! How many sides does an octagon have?&quot; You spit on the ground.&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Rather than answering his questions, you ask one of your own. &quot;What do you know about the Rabbit of Caerbannog?&quot; Reader Rabbit shudders. &quot;Well, I know he &lt;strong&gt;wasn't in my group&lt;/strong&gt;, luckily. He also &lt;strong&gt;wasn't in Uncle Wiggily's group&lt;/strong&gt;.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Before he can ask any more questions, you ask another. &quot;And what do you know about Ruby?&quot; He smiles this time. &quot;She was &lt;strong&gt;in my group&lt;/strong&gt;! She's so nice -- and she's really smart, too!&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You spend a moment calculating. &quot;Forty-three. S-A-N-D-C-A-S-S-E-L. Trick question, it has no sides.&quot; You smile smugly. &quot;Thank you for your time.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suspect: &lt;a href=&quot;http://maxandruby.wikia.com/wiki/Ruby_Bunny&quot;&gt;Ruby&lt;/a&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Mood: Smart, attractive(?), competitive, curious, and responsible&#xA;  &lt;hr /&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You approach the next suspect with a smile. &lt;em&gt;What an adorable little girl&lt;/em&gt;, you think to yourself. &lt;em&gt;Time for good cop.&lt;/em&gt;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;Hey, sweetie,&quot; You say in a poorly masked gruff voice. &quot;Can I get you anything?&quot; She shakes her head. &quot;No, I'm okay.&quot; You nod in understanding. &quot;Well, let me ask you a couple questions. What do you know about Reader Rabbit?&quot; She smirks. &quot;Well, I know for sure he &lt;strong&gt;wasn't in a group with Br'er Rabbit or with Rabbit of Caerbannog&lt;/strong&gt; -- he's terrified of both!&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You smile. &quot;Thanks, that's very helpful. How about Trix Rabbit?&quot; She smirks again. &quot;Well, I know he &lt;strong&gt;wasn't the only Rabbit in his group&lt;/strong&gt; -- he's always complaining about how some other Rabbit tries to take control!&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You smile again and awkwardly pat her on the head. &quot;Thank you for your time.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suspect: &lt;a href=&quot;https://en.wikipedia.org/wiki/Trix_(cereal)&quot;&gt;Trix Rabbit&lt;/a&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Mood: Absolutely starving&#xA;  &lt;hr /&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;Hey, I heard you ask Ruby if she needs anything,&quot; your next suspect says as you approach. &quot;Can you get me some food?&quot; You look at the bony, pale rabbit. &quot;I get it. You want drugs, don't you? You're one of those addicts, aren't ya?&quot; You sneer. &quot;I got you all figured out.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;No, I --&quot; He starts, but you cut him off. &quot;No drugs for you today, my friend. Now tell me what you know about Ruby.&quot; He sighs. &quot;She was &lt;strong&gt;in a group with two other rabbits&lt;/strong&gt;.&quot; You let out a grunt of frustration. &quot;That's it!&quot; You exclaim, as you smack him across the face. He whimpers.&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;Now tell me about Uncle Wiggily!&quot; You yell. &quot;Okay, okay,&quot; he mumbles. &quot;He was &lt;strong&gt;in a group with exactly one Rabbit&lt;/strong&gt;.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;A pang of regret hits you. You snap your fingers and your assistant comes running with a bowl of cereal. The rabbit excitedly snags it, then disappointedly mumbles, &quot;Oh. Plain Cheerios.&quot; You smile. &quot;You're welcome,&quot; you say, not noticing his disappointment. &quot;Thank you for your time.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suspect: &lt;a href=&quot;https://en.wikipedia.org/wiki/Uncle_Wiggily&quot;&gt;Uncle Wiggily&lt;/a&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Mood: Dead, dead, dead&#xA;  &lt;hr /&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;As you approach your next suspect, you spot a pile of mangled rabbit remains on the ground. &quot;Oh, for the love of -- who put him here!?&quot; Your assistant comes running over. &quot;Sir, you said to arrange all the rabbits in alphabetical order!&quot; You sigh. &quot;No, I meant all the living -- whatever. Just...take him away so he can rest in peace.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;As your assistant carries the limp body away, you grimace. &quot;I'm sorry, old chap. Thank you for your time.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suspect: &lt;a href=&quot;https://en.wikipedia.org/wiki/The_Velveteen_Rabbit&quot;&gt;Velveteen Rabbit&lt;/a&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Mood: Velveteen, whatever that means&#xA;  &lt;hr /&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You approach your next suspect somberly. He stares somberly back. The two of you share a somber moment of silence together before you begin your questioning.&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;What did you know about Uncle Wiggily?&quot; You ask. The rabbit responds, &quot;He was &lt;strong&gt;in the group directly before mine&lt;/strong&gt;.&quot; You nod in understanding.&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;And what about the White Rabbit?&quot; He thinks for a moment. &quot;Yeah, I'm positive he was in the &lt;strong&gt;first group of the day&lt;/strong&gt;.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You share another somber moment of silence before you move on. &quot;Thank you for your time.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suspect: &lt;a href=&quot;https://en.wikipedia.org/wiki/White_Rabbit&quot;&gt;White Rabbit&lt;/a&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Mood: Annoyingly dutiful&#xA;  &lt;hr /&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You pace up to your last suspect. &quot;So, I hear your a man who enjoys a nice Rabbit Hole once in awhile.&quot; He shakes his head. &quot;Not this one, not now, not ever. This one is far too deep.&quot; You spit on the ground.&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;Tell me what you know about the Velveteen Rabbit,&quot; you snarl. The White Rabbit fumbles with his words. &quot;Well, he...he was &lt;strong&gt;in the only group with at least...at least two bunnies&lt;/strong&gt;.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Anxious to finish, you continue. Peering all the way back to the front of the line, you ask, &quot;And what about this Br'er Rabbit character? What do you know about him?&quot; Still fumbling, the White Rabbit responds, &quot;He...he and I were &lt;strong&gt;the only...the only two babbits, I mean rabbits, in our group.&lt;/strong&gt;&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You let out a thankful sigh. &quot;Thank you for your time.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Conclusion&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;After interrogating all the suspects, you walk around to face the entire line. &quot;Well, you all did what you could to help -- or at least most of you did. A few of you were lyin' through your teeth, but that's okay. I've got you all figured out. Witnesses, don't worry. Your secret's safe with me. I know who the killer is.&quot; You pull out a pair of handcuffs, and walk over to one of the suspects, firmly cuffing his/her hands behind his/her back. &quot;You have the right to remain silent. Anything you say can and will be held against you.&quot; The rest of the suspects let out a cheer, but you cut them off. &quot;One of ya's cheerin' when ya shouldn't be,&quot; you say menacingly. You beckon your assistant over, and point to a suspect. &quot;There's the carrot thief. Book him.&quot; Your assistant hurries over to that suspect and cuffs him, bringing him over to you. You smile triumphantly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Okay, &lt;em&gt;now&lt;/em&gt; you can cheer.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A cheer erupts from amongst the remaining crowd.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Who was the killer, who were the witnesses, and who's been stealing the carrots?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Note: I apologize in advance if any of the logic is wrong. I checked it as best as I could, but something might have slipped through the cracks. If anyone notices anything wrong, I'll fix it as soon as I can.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="11475" LastEditDate="2015-08-27T19:27:23.820" LastActivityDate="2015-09-01T22:58:23.570" Title="Murdered down the Rabbit Hole (a long logic puzzle)" Tags="|logical-deduction|liars|mystery|" AnswerCount="1" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="18449" PostTypeId="2" ParentId="18445" CreationDate="2015-07-29T17:20:31.120" Score="5" Body="&lt;p&gt;As currently stated, this is impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that there is a working algorithm. Call the dwarves $d_0, d_1, \ldots d_{99}$. Suppose that $d_0$ sees every other dwarf wearing their own number. The algorithm says that $d_0$ must call out a number $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $i$ is 0, then the algorithm fails if $d_0$ isn't wearing 0 and all others are wearing their own number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $i$ isn't 0, now consider $d_i$. Suppose $d_i$ sees all the other dwarves except $d_0$ wearing their own numbers, and $d_i$ wearing some number other than $i$. $d_i$ must call a number $j$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $j$ isn't the same as $i$, then the algorithm fails when all except $d_0$ wear their own numbers, because $d_0$ will call $i$ while $d_i$ calls $j$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if $j$ is equal to $i$, then the algorithm fails when all except $d_0$ and $d_i$ are wearing their own numbers and $d_0$ isn't wearing $i$, because then nobody is wearing $i$.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-07-29T23:36:21.347" LastActivityDate="2015-07-29T23:36:21.347" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18450" PostTypeId="2" ParentId="18445" CreationDate="2015-07-29T17:40:15.867" Score="14" Body="&lt;p&gt;No such algorithm exists.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Suppose Bob sees 0's on 99 shirts. He must call out 0, since it is possible that 0 is the only represented number.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Suppose Bob sees 0's on 98 shirts, on everyone except Alice's shirt. If Bob's shirt were 0, then Alice would see 99 zeroes, and announce 0 (by the previous bullet point). This means Bob must announce 0, else there would be a chance that Bob's and Alice's calls were different.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Suppose Bob sees 0's on 97 shirts, where Alice isn't wearing 0. If Bob's shirt were 0, then Alice would see 98 zeroes, and announce 0 (by the previous bullet point). This means Bob must announce 0, else there would be a chance that Bob's and Alice's calls were different.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This logic continues inductively, so that if Bob sees 96, 95, or any number of zeroes, he must announce 0. Since everyone calling 0 is not a winning strategy, no strategy exists.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="11244" LastEditDate="2015-07-29T19:43:10.027" LastActivityDate="2015-07-29T19:43:10.027" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18451" PostTypeId="2" ParentId="18445" CreationDate="2015-07-29T17:59:39.827" Score="9" Body="&lt;p&gt;I think f'' basically has the right answer, but here's a simpler explanation of why it's not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume there is an algorithm. Consequently there must be an output i when all the shirts have different numbers (e.g. the shirts are 0-99 and are in the same order as the dwarves ages).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the dwarf wearing shirt i must shout i whenever he sees the other dwarves wearing their specific combination. So if we had given him a different shirt then he would still shout i.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a contradiction as the algorithm is assumed to output a number that is on one of the dwarves shirts. Hence no such algorithm exists.&lt;/p&gt;&#xA;" OwnerUserId="303" LastActivityDate="2015-07-29T17:59:39.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18452" PostTypeId="2" ParentId="18433" CreationDate="2015-07-29T18:03:24.920" Score="4" Body="&lt;p&gt;Is Me llamo viento&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Name of the Wind&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And is the theme&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Book titles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12198" LastActivityDate="2015-07-29T18:03:24.920" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18453" PostTypeId="1" AcceptedAnswerId="18454" CreationDate="2015-07-29T18:22:07.590" Score="14" ViewCount="497" Body="&lt;p&gt;Here's a simple one (clue in the title):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EUhUR.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EUhUR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9575" LastActivityDate="2015-07-29T18:24:31.723" Title="Mostly By Way of Emphasis" Tags="|rebus|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18454" PostTypeId="2" ParentId="18453" CreationDate="2015-07-29T18:24:31.723" Score="15" Body="&lt;p&gt;I believe it is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First and foremost&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3987" LastActivityDate="2015-07-29T18:24:31.723" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18455" PostTypeId="1" AcceptedAnswerId="18456" CreationDate="2015-07-29T18:28:27.453" Score="6" ViewCount="1224" Body="&lt;p&gt;A boy's father dies leaving his mother to be a poor widow.  A few years later the boy marries his mother legally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is all done in good faith&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9575" LastActivityDate="2015-07-30T03:53:03.933" Title="Mother and Son in Harmony" Tags="|lateral-thinking|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18456" PostTypeId="2" ParentId="18455" CreationDate="2015-07-29T18:31:44.233" Score="16" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The boy conducts a marriage ceremony between his mother and her new spouse.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12277" LastEditorUserId="11876" LastEditDate="2015-07-29T18:47:41.293" LastActivityDate="2015-07-29T18:47:41.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18457" PostTypeId="2" ParentId="18455" CreationDate="2015-07-29T18:32:06.427" Score="2" Body="&lt;p&gt;The boy was adopted, so it would not be incest.&lt;/p&gt;&#xA;" OwnerUserId="14239" LastActivityDate="2015-07-29T18:32:06.427" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18459" PostTypeId="2" ParentId="18433" CreationDate="2015-07-29T18:53:00.487" Score="6" Body="&lt;p&gt;Guess I'll throw mine in here as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A long age, in the past&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;ERA + GONE = Eragon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10861" LastActivityDate="2015-07-29T18:53:00.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18460" PostTypeId="2" ParentId="18433" CreationDate="2015-07-29T19:01:16.637" Score="4" Body="&lt;p&gt;I guess while we're all doing one...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ding Dong&lt;br&gt;&#xA;A colours' rival with a fiendish compartment&lt;br&gt;&#xA;A marketers money maker  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Belgariad (BELL + GARY (Pokemon reference - Red's rival) + AD)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also it looks like these are specifically&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fantasy novels.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-07-29T19:01:16.637" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18461" PostTypeId="2" ParentId="18445" CreationDate="2015-07-29T19:29:06.940" Score="2" Body="&lt;p&gt;It's impossible, because if a combination is supposed to have a solution, the combination of seen numbers must all lead to the same number. If a list of these 99 seen numbers leads to $n$, then when we create a new combination by changing the other number, all the lists must lead to $n$, or there would be no solution. Finally we arrive at a point where it should still lead to $n$ even though it's no longer there! This contradiction makes it impossible for every combination to have a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: turns out my reasoning was the same as those of almost all the others.&lt;/p&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="14016" LastEditDate="2015-07-29T19:36:44.763" LastActivityDate="2015-07-29T19:36:44.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18462" PostTypeId="1" AcceptedAnswerId="18463" CreationDate="2015-07-29T19:57:40.903" Score="16" ViewCount="1120" Body="&lt;p&gt;What famous quote is portrayed by the rebus below?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The accepted answer will correctly explain every part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0pYMm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0pYMm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastActivityDate="2015-07-29T20:11:52.573" Title="What is this famous quote?" Tags="|rebus|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18463" PostTypeId="2" ParentId="18462" CreationDate="2015-07-29T20:04:48.997" Score="11" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You must be the change you wish to see in the world.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanations:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You (EWE)&lt;br&gt;&#xA; Must (COMBUST - COB)&lt;br&gt;&#xA; Be (BOWELS - OWLS)&lt;br&gt;&#xA; The (POTHOLE - POOL)&lt;br&gt;&#xA; Change (ARCHANGEL - AR - L)&lt;br&gt;&#xA; You (LAYOUT - LAT)&lt;br&gt;&#xA; Wish (WISHBONE - BONE)&lt;br&gt;&#xA; To (TANGELO - ANGEL)&lt;br&gt;&#xA; See (ROSELLE - ROLL)&lt;br&gt;&#xA; In (ALIEN - ALE)&lt;br&gt;&#xA; The (THIEF - I - F)&lt;br&gt;&#xA; World (WONDERLAND - NDE - AN)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="463" LastEditDate="2015-07-29T20:11:52.573" LastActivityDate="2015-07-29T20:11:52.573" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18464" PostTypeId="2" ParentId="18424" CreationDate="2015-07-29T20:10:32.453" Score="4" Body="&lt;p&gt;I'm going to go ahead and guess&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Elizabethan English&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My namesakes are plentiful,&lt;br&gt;&#xA;though some antiquated:&lt;br&gt;&#xA;A company defunct &lt;br&gt;&#xA;and a system outdated.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Elizabethan English, or Early Modern English, has many abbreviations (EModE,EMnE or ENE). Elizabethan English is one of the oldest records of early English that exists. Nobody uses/speaks in Elizabethan English nowadays except in poems and prose. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One unavoidable &lt;br&gt;&#xA;an honor shared by few &lt;br&gt;&#xA;just give it some time&lt;br&gt;&#xA;it will come into view.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Elizabethan (root word Elizabeth) is the name of two English queens of England (I and II). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Two bearing my name&lt;br&gt;&#xA;and both are titanic:&lt;br&gt;&#xA;an explosive endeavor&lt;br&gt;&#xA;and the second, enigmatic.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Elizabeth and English. &lt;br&gt; Upon reading some history, Elizabeth I was known to be a long serving Queen and provided a lot to England such as the church. &lt;br&gt;English is a very strange language, mostly built upon ideas and words taken from other languages. It has a strange grammar structure as well, with inconsistencies and counterintuition for beginners learning English. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My habits inspired &lt;br&gt;&#xA;a nightmarish work.&lt;br&gt;&#xA;The last one was spared&lt;br&gt;&#xA;and well known to perk.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Shakespearean plays contained Elizabethan English and took several months (possibly years) to write and develop plays. Shakespearean plays are now well known and studied throughout high schools and future studies due to their historical significance. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What I bring is a gift,&lt;br&gt;&#xA;feared and embraced.&lt;br&gt;&#xA;Do you know my name&lt;br&gt;&#xA;from these clues I have placed?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Both feared by high school children and embraced by linguistics for it's artful grace, I believe the answer is Elizabethan English. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13489" LastEditorUserId="13489" LastEditDate="2015-07-29T20:18:46.867" LastActivityDate="2015-07-29T20:18:46.867" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18465" PostTypeId="2" ParentId="18424" CreationDate="2015-07-29T20:14:16.143" Score="4" Body="&lt;p&gt;After looking at your profile picture, I think you might be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;b&gt;Heroes of Might and Magic&lt;/b&gt;, (The video game series)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My namesakes are plentiful,&#xA;&lt;br&gt;though some antiquated:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Heroes of Might and Magic series has had many game releases with the same name starting in each each title. The original was released in the 1990's which could be considered an antiquity for a video game. (Currently it is at Heroes of Might and Magic VII?, [some of the priors also had multiple expansions])&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A company defunct &#xA;&lt;br&gt;and a system outdated.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This game has been made by many companies (which could be considered dysfunction and closed/bankrupt now) such as New World Computing [1995-1996], The 3DO Company (1996-2003) and Ubisoft Entertainment (2003-). The most recent versions of the game also still use the same turn-based game play as the original which for the span of almost 20 years could be considered outdated.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One unavoidable&#xA;&lt;br&gt;an honor shared by few&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The title of the series &quot;Heroes of Might and Magic&quot; has become unavoidable when it come to the title for each game. As well, to be titled as a 'hero of might and magic' would be quite an honor. (Only the few victorious players could be granted such a title)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;just give it some time&#xA;&lt;br&gt;it will come into view.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is only one side/team that can win at the end of the game, as the game progresses, certain players will succumb to the strength of their enemies until all are vanquished, and only one civilization remains!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Two bearing my name&#xA;&lt;br&gt;and both are titanic:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In reference to the two parts 'Might' and 'Magic' are two different and powerful traits of battle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;an explosive endeavor&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a reference to 'Might' which can be defined as - a great and impressive power or strength, especially of a nation, large organization, or natural force.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the second, enigmatic.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A reference to 'magic' where it is mysterious.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My habits inspired &#xA;&lt;br&gt;a nightmarish work.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; By playing the game you are forced to fight and defeat your friends if you wish to win; a nightmarish work indeed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The last one was spared&#xA;&lt;br&gt;and well known to perk.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Maybe in the newest game, they added a new perk (Possible race or spells - I've been out of the loop..)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What I bring is a gift,&#xA;&lt;br&gt;feared and embraced.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can bring entertainment which is often embraced by the players, and feared by the outside world. (&quot;Anonymous, stop playing that darn game and go outside and play&quot;)&#xA;&lt;br&gt;&lt;br&gt; This could also be referencing to the control and domination of your faction. You would be feared by your enemies and embraced by your own army/allies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Do you know my name&#xA;&lt;br&gt;from these clues I have placed?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Heroes of Might and Magic series&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bonus:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your profile picture is of [inside the town] Rampart (a town/team) from Heroes of Might and Magic III.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;P.S Thanks for all the nostalgia.. &lt;/p&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-07-29T20:50:41.667" LastActivityDate="2015-07-29T20:50:41.667" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18466" PostTypeId="1" AcceptedAnswerId="18469" CreationDate="2015-07-29T20:59:37.220" Score="2" ViewCount="7260" Body="&lt;p&gt;After browsing through some puzzles on here, I found some that have letter patterns, such as P, N, D, ? for Penny, Nickel, Dime, and ... (you know what comes next). Now, I've decided to do some number patterns instead. Some patterns may be based off of something else, such as coin values and dates.&lt;br&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some amount of math may be required for these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Warm-Up (find the next 1)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;1, 4, 9, 16, ?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;17, 19, 23, 29, 31, ?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Easy (find the next 1)&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;1, 2, 5, 10, 20, 50, ?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;31, 28, 31, 30, 31, ?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;01, 02, 06, 24, 20, 20, ?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Medium (find the next 2)&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;6&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;20, 90, 61, 32, 03, ?, ?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hint given after 7 days if no solution; the hint kind of gives it away.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;1, 8, 28, 56, 70, ?, ?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hint given after 7 days if no solution; the hint kind of gives it away.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;8, 4, 2, 6, 2, 8, 6, 4, 4, ?, ?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alternating pattern&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;1, 7, 15, 3, 4, 11, 9, ?, ?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Modular arithmetic&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Hard (find the next 2)&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;10&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;111, 112, 115, 120, 122, 124, 126, 128, 132, 135, ?, ?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Divisibility with digits&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;1, 4, 31, 224, 2011, ?, ?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Multiplying in something other than base 10.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Make a comment if one of the patterns is in the wrong difficulty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be helpful to add your answers as you find them instead of adding one big bunch in the end; to receive best answer you must have 1-9 solved and posted as a solution; 10 and 11 will have a bounty if not solved by 10 days.&lt;/p&gt;&#xA;" OwnerUserId="2001" LastEditorUserId="12109" LastEditDate="2015-07-29T21:26:45.520" LastActivityDate="2015-07-30T13:26:27.810" Title="Number Patterns Challenge" Tags="|pattern|number-sequence|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18468" PostTypeId="2" ParentId="18466" CreationDate="2015-07-29T21:16:55.870" Score="1" Body="&lt;p&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;25 (+3 +5 +7 +9)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;35 (+2 +4 +6 +2 +4)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9353" LastActivityDate="2015-07-29T21:16:55.870" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18469" PostTypeId="2" ParentId="18466" CreationDate="2015-07-29T21:17:05.323" Score="3" Body="&lt;p&gt;Still working on some, but here's what I have:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Warm-Up (find the next 1)  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;1, 4, 9, 16, ?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;25 (list of squares of integers starting with 1)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;17, 19, 23, 29, 31, ?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;37 (list of primes starting with 17)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Easy (find the next 1)&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;1, 2, 5, 10, 20, 50, ?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;100 (list of US dollar denominations)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;31, 28, 31, 30, 31, ?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;30 (list of lengths of months in days starting with January)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;01, 02, 06, 24, 20, 20 ?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Medium (find the next 2)   &lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;6&quot;&gt;&#xA;&lt;li&gt;20, 90, 61, 32, 03, ?, ?  &lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;1, 8, 28, 56, 70, ?, ?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;56, 28 (list of numbers in 9th row of Pascal's Triangle)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;8, 4, 2, 6, 2, 8, 6, 4, 4, ?, ?   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;1, 7, 15, 3, 4, 11, 9, ?, ?  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;12,16 (7^n mod 17)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Hard (find the next 2)   &lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;10&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;111, 112, 115, 120, 122, 124, 126, 128, 132, 135, ?, ?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;140, 144 (list of integers that are divisible by each of their digits starting with 111)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;1, 4, 31, 224, 2011, ?, ?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;13044, 112341 (convert from base 5 to base 10, multiply by 4, convert back to base 5)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="10861" LastEditorUserId="12700" LastEditDate="2015-07-30T00:27:44.647" LastActivityDate="2015-07-30T00:27:44.647" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18471" PostTypeId="2" ParentId="10722" CreationDate="2015-07-29T23:20:05.133" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Paddle: A short pole with a broad blade at one or both ends.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Peddle: Try to sell something by going from house to house or place to place.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Piddle: To spend time in a wasteful, trifling, or ineffective way.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Poddle: To move or travel in a leisurely manner.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Puddle: A small pool of liquid, especially of rainwater, on the ground.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14318" LastEditorUserId="4385" LastEditDate="2015-07-31T14:37:49.313" LastActivityDate="2015-07-31T14:37:49.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18472" PostTypeId="1" AcceptedAnswerId="18474" CreationDate="2015-07-30T01:23:55.707" Score="6" ViewCount="940" Body="&lt;p&gt;The title has the clue; see if you can get this one:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;HABIRDND&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="9575" LastEditorUserId="39" LastEditDate="2015-07-30T03:02:22.660" LastActivityDate="2015-07-30T12:33:12.587" Title="Some Say It's Worth Two" Tags="|rebus|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18473" PostTypeId="2" ParentId="18466" CreationDate="2015-07-30T01:29:35.487" Score="2" Body="&lt;p&gt;I'll answer some of the ones that haven't been answered so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;5.&lt;/strong&gt; 40, 20 (last two digits in factorials)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;6.&lt;/strong&gt; 73, 44 (subtract 29 modulo 99 each time)&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-07-30T01:29:35.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18474" PostTypeId="2" ParentId="18472" CreationDate="2015-07-30T01:38:18.977" Score="16" Body="&lt;p&gt;I believe that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A bird in the hand is worth two in the bush!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-07-30T01:38:18.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18475" PostTypeId="2" ParentId="18424" CreationDate="2015-07-30T02:16:37.590" Score="3" Body="&lt;p&gt;As a new user, I am unfortunately unable to comment, but here is my best guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everest (or some other mountain/volcano)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Surely you know this prominent figure!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Prominence is a property of mountains. Everest is the most prominent mountain in the world.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My namesakes are plentiful,&#xA;though some antiquated:&#xA;A company defunct &#xA;and a system outdated.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everest the mountain has many names, and &lt;a href=&quot;https://en.wikipedia.org/wiki/Everest_(disambiguation)&quot; rel=&quot;nofollow&quot;&gt;many things&lt;/a&gt; are named after it. I learned, for instance, that Everest Colleges are part of a soon-to-be-closed system of for-profit colleges operated by Corinthian.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One unavoidable&#xA;an honor shared by few&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Few people have climbed Everest. Doing so is typically considered an honorable accomplishment. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Two bearing my name&#xA;and both are titanic:&#xA;an explosive endeavor&#xA;and the second, enigmatic.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Admittedly, I'm not sure how this one relates specifically to Everest. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My habits inspired &#xA;a nightmarish work.&#xA;The last one was spared&#xA;and well known to perk.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everest has claimed the lives of numerous people (especially in 1996, 2014, and 2015). Some of these so-called &quot;Everest disasters&quot; have been chronicled in novels, documentaries, etc.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What I bring is a gift,&#xA;feared and embraced.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everest brings a certain type of thrill—of cold, heights, and wilderness—that is dangerous but nevertheless enticing to generation after generation. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Comments:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm not very confident that Everest is the correct answer (my second guess would be Denali/McKinley), but I have a hunch that &quot;prominent &lt;em&gt;figure&lt;/em&gt;&quot; has to do with something inanimate. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14319" LastActivityDate="2015-07-30T02:16:37.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18476" PostTypeId="1" AcceptedAnswerId="18477" CreationDate="2015-07-30T02:18:23.597" Score="24" ViewCount="1556" Body="&lt;p&gt;Do you strive to do the impossible?  You might have heard someone say this!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Dt92U.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Dt92U.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9575" LastEditorUserId="12700" LastEditDate="2015-07-30T04:06:42.707" LastActivityDate="2015-07-30T19:41:26.037" Title="Rebus For Those Who Strive" Tags="|rebus|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18477" PostTypeId="2" ParentId="18476" CreationDate="2015-07-30T03:00:47.977" Score="36" Body="&lt;p&gt;The rebus depicts a knot next to the word &quot;shot&quot; in an elongated font.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the answer is &quot;not [knot] by [next to] a long shot.&quot;&lt;/p&gt;&#xA;" OwnerUserId="39" LastEditorDisplayName="user20" LastEditDate="2015-07-30T19:41:26.037" LastActivityDate="2015-07-30T19:41:26.037" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="18478" PostTypeId="2" ParentId="18455" CreationDate="2015-07-30T03:53:03.933" Score="2" Body="&lt;p&gt;The boy is obviously&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Oedipus&quot; rel=&quot;nofollow&quot;&gt;Oedipus&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He did it in good faith so it even fits the hint.&lt;/p&gt;&#xA;" OwnerUserId="13197" LastActivityDate="2015-07-30T03:53:03.933" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18479" PostTypeId="1" CreationDate="2015-07-30T06:13:32.530" Score="0" ViewCount="168" Body="&lt;p&gt;The Sultan had two wise sages in his service, Ali-ibn-Vali and Vali-ibn-Ali. The Sultan wanted to test their intelligence and told them, &quot;I have thought of two numbers, both are integers greater than 1 and less than 100. I've multiplied them and told the result to Ali. I've added them and told that result to Vali. If you're smart enough, you should be able to discover the original two numbers.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sages thought for a few minutes, and then:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ali said,&quot; I don't know the original numbers.&quot;&lt;br&gt; &quot;I knew it,&quot; Vali replied. (expressing that he knew that Ali didn't know).&lt;br&gt;&quot;Then I know the numbers,&quot; Ali quipped with delight.&lt;br&gt;&quot;Then I know the numbers too!&quot; exclaimed Vali.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sages told the original numbers to the Sultan, who was mightily impressed at their intelligence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you determine the original numbers as the sages did?&lt;/p&gt;&#xA;" OwnerUserId="14320" LastEditorUserId="4421" LastEditDate="2015-11-25T18:02:53.690" LastActivityDate="2015-11-25T18:02:53.690" Title="Ali-ibn-Vali and Vali-ibn-Ali (aka multiplication vs sum)" Tags="|mathematics|number-theory|" AnswerCount="0" CommentCount="2" ClosedDate="2015-07-30T11:00:48.180" ContentLicense="CC BY-SA 3.0" />
  <row Id="18480" PostTypeId="2" ParentId="18445" CreationDate="2015-07-30T08:30:17.170" Score="4" Body="&lt;p&gt;Call out a single digit number such as &quot;9&quot;, then anyone wearing a shirt with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;9, 19, 29, 39, 49, 59, 69, 79, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98 or 99&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;on their shirt, will have the number 9 on their shirt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;not 100% certain, but if the numbers are assigned randomly then there's nearly a 1 in 5 (19/100) chance that a &lt;strong&gt;single&lt;/strong&gt; dwarf will have a number 9. Odds are &lt;strong&gt;drastically&lt;/strong&gt; in their favour.&lt;/p&gt;&#xA;" OwnerUserId="4659" LastEditorUserId="4659" LastEditDate="2015-07-30T10:42:41.520" LastActivityDate="2015-07-30T10:42:41.520" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18481" PostTypeId="2" ParentId="18423" CreationDate="2015-07-30T09:37:31.063" Score="12" Body="&lt;p&gt;I think I got it.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; As it is easy to see, every word is a letter from the plaintext &lt;br/&gt;&#xA; PARROT = M&lt;br/&gt;&#xA; NECTAR = I&lt;br/&gt;&#xA; GABBY = S&lt;br/&gt;&#xA; &lt;em&gt;and so on&lt;/em&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;br/&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Taking a look at the puzzle from the first hint, and the information about the lack of creativity. I played a bit with the spacing between the letters:&lt;br/&gt;&lt;br/&gt;&#xA; PARROT = 15  0  17  17  14 19 &lt;br/&gt;&#xA; &lt;strong&gt;15&lt;/strong&gt; (|15-0|) &lt;strong&gt;17&lt;/strong&gt; (|0-17|) &lt;strong&gt;0&lt;/strong&gt; (|17-17|) &lt;strong&gt;3&lt;/strong&gt; (|17-14|)  &lt;strong&gt;5&lt;/strong&gt; (|14-19|)&lt;br/&gt;&#xA; doing the same thing over and over again until we have only one letter left&lt;br/&gt;&#xA; 15 17 0 3 5 =&amp;gt; 2 17 3 2 =&amp;gt; 15 14 1 =&amp;gt; 1 13 =&amp;gt; 12 which coresponds to M &lt;br/&gt;&lt;br/&gt;&#xA; NECTAR = 13 4 2 19 0 17&lt;br/&gt;&#xA; 13 4 2 19 0 17 =&amp;gt; 9 2 17 19 17 =&amp;gt; 7 15 2 2 =&amp;gt; 8 13 0 =&amp;gt; 5 13 =&amp;gt; 8 = I&lt;br/&gt;&lt;br/&gt;&#xA; GABBY = 6 0 1 1 24&lt;br/&gt;&#xA; 6 0 1 1 24   =&amp;gt; 6 1 0 23  =&amp;gt; 5 1 23 =&amp;gt; 4 22 =&amp;gt; 18 = S&lt;br/&gt;&lt;br/&gt;&#xA; And so on. &lt;em&gt;Sorry for not putting all the words here, but I am posting this in the lunch break and I am running out of time. If needed, I will post all of them when I get home. :)&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;br/&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;The second hint refers to the fact that there are many words that will end up with the same value. I don't have an example now, but I think that I can come up with some sort of script that will provide words for a given letter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10692" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-30T09:48:46.317" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18482" PostTypeId="2" ParentId="18445" CreationDate="2015-07-30T10:53:54.770" Score="1" Body="&lt;p&gt;This is easy!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reading the question - they get their shirts BEFORE they go into the room and they are not prohibited from communicating until they enter the room. The algorithm is that they look at the first dwarf's shirt and tell him number, then they all go into the room and shout that one out.&lt;/p&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-07-30T10:53:54.770" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18484" PostTypeId="2" ParentId="18472" CreationDate="2015-07-30T12:33:12.587" Score="10" Body="&lt;p&gt;HABIRDND is actually&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HA&lt;strong&gt;BIRD&lt;/strong&gt;ND --&gt; A BIRD in the HA-ND ... is worth two in the bush!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1998" LastActivityDate="2015-07-30T12:33:12.587" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18486" PostTypeId="2" ParentId="18390" CreationDate="2015-07-30T12:48:36.053" Score="4" Body="&lt;p&gt;I think the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Love story&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;hashSet.add(new Obj());&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Here an object is being stored in a hash set [STORE]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;e.rotate(Math.PI);&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Here the object e is having a 180 degree rotation applied [E ABOUT]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;{}&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is an empty block, so this code does nothing, or love as nothing can be termed in tennis for example [LOVE]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So putting it together we have:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; STORE E ABOUT LOVE i.e. STORY ABOUT LOVE or LOVE STORY&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12277" LastActivityDate="2015-07-30T12:48:36.053" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18488" PostTypeId="2" ParentId="18239" CreationDate="2015-07-30T13:30:24.893" Score="1" Body="&lt;p&gt;I think I have an answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Kalmykia, Russia&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Going off of Eli's answer, we have:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;OLD FLAGS - 4f 4c 44 20 46 4c 41 47 53&lt;br&gt;&#xA;&#xA;FISHY DISH - 46 49 53 48 59 20 44 49 53 48&lt;br&gt;&#xA;&#xA;PUERTO RICO - 50 55 45 52 54 4f 20 52 49 43 4f&lt;br&gt;&#xA;&#xA;ACQUITTED IDIOT - 41 43 51 55 49 54 54 45 44 20 49 44 49 4f 54&lt;br&gt;&#xA;&#xA;MANLY PINK - 4d 41 4e 4c 59 20 50 49 4e 4b&lt;br&gt;&#xA;&#xA;BANANA CABANA - 42 41 4e 41 4e 41 20 43 41 42 41 4e 41&lt;br&gt;&#xA;&#xA;STREET SMART - 53 54 52 45 45 54 20 53 4d 41 52 54&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the odd one out is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; FISHY DISH since it's the only one that doesn't have a letter in its HEX encoding.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Following that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We remove the Odd numbered hex values (odd one OUT) and get 46 48 20 44 48&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which turns out to be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; since 20 is a space, we can make that the split point, which leaves us with 46 48 and 44 48. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Putting these as 46.48 and 44.48 into a coordinate locator, (lat, long) we get Kalmykia, Russia&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13924" LastEditorUserId="13924" LastEditDate="2015-07-30T13:50:11.287" LastActivityDate="2015-07-30T13:50:11.287" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18489" PostTypeId="2" ParentId="18448" CreationDate="2015-07-30T14:14:54.300" Score="29" Body="&lt;h1&gt;TL;DR, FINAL TABLE:&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;See the following spoiler only if you really really would like to spoil all of your fun in just a second:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;GROUP 1&lt;/strong&gt;:&lt;br&gt;&#xA; Br'er Rabbit, White Rabbit, Peter Cottontail (thief), Mr. Whiskes and Nesquik Bunny.&lt;br&gt;&lt;br&gt;&#xA; &lt;strong&gt;GROUP 2&lt;/strong&gt;:&lt;br&gt;&#xA; Bunnicula, Rabbit, Rabbit of Caerbannog, Playboy Bunny and Mr. Bun&lt;br&gt;&lt;br&gt;&#xA; &lt;strong&gt;GROUP 3&lt;/strong&gt;:&lt;br&gt;&#xA; Buster Baxter, Bugs Bunny, Ruby, Reader Rabbit and Trix Rabbit.&lt;br&gt;&lt;br&gt;&#xA; &lt;strong&gt;GROUP 4&lt;/strong&gt;:&lt;br&gt;&#xA; Uncle Wiggily (dead), Crusader Rabbit, Easter Bunny (murderer), March Hare and Peppy Hare.&lt;br&gt;&lt;br&gt;&#xA; &lt;strong&gt;GROUP 5&lt;/strong&gt;:&lt;br&gt;&#xA; Cuddles (witness), Velveteen Rabbit, Lola Bunny, Energizer Bunny and Max (witness).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ok, now let's see how it is in detail:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;PART I - The beggining of the investigation&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Let's make some facts that are clearly in the question preamble and should be obvious to everybody that read it all:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 1&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Uncle Wiggily and the murderer are in the same group.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 2&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The two witnesses are in the same group.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 3&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The thief is not the murderer nor a witness.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These three are pretty obvious (so, I didn't even spoiler-protect them). But we can deduce something more from them (now it's all spoilers down from here):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 4&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The witnesses are not in the same group as Uncle Wiggily and the murder.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 5&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The witnesses are not in group 1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 6&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The murderer and Uncle Wiggily are not in group 5.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 7&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since the witnesses are liars, we have a group with two liars, except if the thief is also in this group, which would then make a group with three liars.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 8&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is two groups where everyone is telling the truth, except if the thief is in the same group as the murderer or the witnesses, which would mean that there would be three groups where everyone is telling the truth. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;PART II - Catching lies&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Now, let's try to start to look for contradictions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Max says that March Hare was in the &lt;strong&gt;only&lt;/strong&gt; group with two bunnies. March Hare also puts Max in a group with two bunnies. Energizer Bunny and Lola Bunny says that they grouped with each other, and they are both bunnies. But March Hare says that Lola Bunny is not in his group, which contradicts the idea that he is in the only group with two bunnies. So...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 9&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is at least one liar between March Hare, Max, Lola Bunny and Energizer Bunny.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Further:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Easter Bunny says that Cuddles is in a group with two rabbits. Cuddles says that Crusader Rabbit was in his group, so Crusader Rabbit would be with another rabbit in his group. This contradicts Buster Baxter, who says that Crusader Rabbit was not with some other rabbit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 10&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; At least one of Easter Bunny, Cuddles, Crusader Rabbit and Buster Baxter is a liar.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can we get more contradictions?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Petter Cottontail says that Playboy Bunny is in the forth group. Rabbit says that Playboy Bunny is also in his group, so Rabbit would also be in the forth group. Rabbit of Caerbannog says that Rabbit is in Bunnicula's group, so Bunnicula would also be in the forth group. Buster Baxter says that Bunnicula was in the White Rabbit's following group, and this would be the second group if Velveteen Rabbit is telling the truth. But Rabbit, Playboy Bunny and Bunnicula can't be both in the second and in the forth group so:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 11&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; At least one of Peter Cottontail, Rabbit, Rabbit of Caerbannog, Buster Baxter and Velveteen Rabbit is a liar.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;PART III - Some leporidae are telling the truth&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Br'er Rabbit and White Rabbit said that each other was in his group. Playboy Bunny said that Peter Cottontail was in the same group as White Rabbit. Peppy Hare also said that Peter Cottontail was in Br'er Rabbit group.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, let's see what we can deduce from that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Between Br'er Rabbit and White Rabbit, either both are telling the truth or both are liars.&lt;br&gt;&lt;br&gt;&#xA; If Br'er Rabbit and White Rabbit are liars, at least one of Playboy Bunny and Peppy Hare is also a liar.&lt;br&gt;&lt;br&gt;&#xA; If only one of Playboy Bunny and Peppy Hare is telling the truth, then Br'er Rabbit and White Rabbit are liars.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We can't have that all the four Br'er Rabbit, White Rabbit, Playboy Bunny and Peppy Hare are liars, because TRUTH 9 says that we have at least one more liar outside of this set, which would make then at least 5 liars (and also with TRUTHS 10 and 11, at least 6), so this is impossible. If just one of them would be telling the truth, which would give 3 liars, by TRUTH 9 we would get a forth liar and with TRUTHS 10 and 11 there would be a fifth liar, which is also impossible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And then:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There should be at least two truthtellers between Br'er Rabbit, White Rabbit,  Playboy Bunny and Peppy Hare. If only one of Playboy Bunny and Peppy Hare are liars, then Br'er Rabbit and White Rabbit also would be liars, and we would get 3 liars, which is impossible. However if Br'er Rabbit and White Rabbit are liars, at least one of Playboy Bunny and Peppy Hare would also be, which again would give at least 3 liars. So:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 12&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Br'er Rabbit, White Rabbit, Playboy Bunny and Peppy Hare are all telling the truth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And this also means that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 13&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Br'er Rabbit, White Rabbit and Peter Cottontail are all in the same group.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;PART IV - A tale of two bunnies&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Let's continue with inferences from the truths above:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; White Rabbit says those two truths:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 14&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Velveteen Rabbit is in a group with two other bunnies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 15&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is definitively only one group with two bunnies. And no group with more than that.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's see those bunnies:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We have six bunnies: Bugs Bunny; Easter Bunny; Energizer Bunny; Lola Bunny; Nesquik Bunny and Playboy Bunny.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which means that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are 5 groups, one of them have 2 bunnies. Since we have only 4 other bunnies and 4 other groups and no other group has 2 or more bunnies then:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 16&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All of the groups have at least one bunny.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And finally:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 17&lt;/strong&gt; (very important, so it is very spoilery, beware):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Max is a liar!&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which also implies that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 18&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; March Hare is not in the group with two bunnies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;PART V - The secret of the bunnies&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Br'er Rabbit is telling the truth, so Bugs Bunny is in the same group as Ruby. Reader Rabbit says that Ruby is in his group. Bugs Bunny says that his group is not the same as Br'er Rabbit group.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So what?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If Bugs Bunny is lying, then Br'er Rabbit, White Rabbit, Peter Cottontail, Bugs Bunny and Ruby are all in the same group. Also, Reader Rabbit would be a liar. If Reader Rabbit is a liar then Rabbit of Caerbannog would also be on his group. And then, Ruby would also be a liar. This would make all the four liars as Max, Bugs Bunny, Ruby and Reader Rabbit. However, since Bugs Bunny and Ruby are in the same group, then they would either both be witnesses or be the thief and the murderer. None of them can be the murderer because Uncle Wiggily is not in the group. However if they are the witnesses, that would make Velveteen Rabbit a liar also, which would give us 5 liars.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 19&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bugs Bunny is telling the truth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Further:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Energizer Bunny and Lola Bunny both says that they are in the same group, which would be the only group with two bunnies. Easter Bunny says that Energizer Bunny is in the forth group. Energizer Bunny and Lola Bunny says that they are with each other. Peter Cottontail says that the Playboy Bunny is in the forth group. So, if all of them are telling the truth, we would have the forth group with three bunnies (Energizer Bunny, Lola Bunny and Playboy Bunny). So:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 20&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is at least one subgroup of liars between Easter Bunny, Energizer Bunny and Lola Bunny, and Peter Cottontail.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Things are getting hot, huh?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If Energizer Bunny and Lola Bunny are liars, then Easter Bunny could only be in the group with two bunnies, together with Velveteen Rabbit. However, Easter Bunny says that Energizer Bunny is in the forth group and Peter Cottontail also says that the Playboy Bunny is in the forth group, which would create two groups with two bunnies. This way, Energizer Bunny, Lola Bunny, Max and one of Peter Cottontail or Easter Bunny would be our four liars. Then, Cuddles asserts that Easter Bunny is in the second group, which also puts Velveteen Rabbit in the second group, and then puts Uncle Wiggily in the first group together with Br'er Rabbit, White Rabbit, Peter Cottontail and someone else. However, Mr. Whiskers and Nesquik Bunny would also be telling the truth and they would also be in the first group together with Peppy Hare, and then the first group would have 7 members, which can't be truth. So:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 21&lt;/strong&gt; (very important, so it is very spoilery, beware):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Energizer Bunny and Lola Bunny are telling the truth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 22&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; March Hare and Peppy Hare are in the same group.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 23&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; At least one of Easter Bunny and Peter Cottontail is a liar.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;PART VI - Finding truthtellers&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If Max is not in the same group as Energizer Bunny, Lola Bunny and Velveteen Rabbit, then March Hare is also a liar. If Velveteen Rabbit is also a liar, then by choosing the forth liar from TRUTHS 10, 11 and 23, only Easter Bunny could be the remaining liar.&lt;br&gt;&#xA; &lt;br&gt;&#xA; This would also mean that Rabbit, Playboy Bunny and Bunnicula are in the fourth group (because Peter Cottontail, Rabbit, Rabbit of Caerbannog and Buster Baxter would not be liars) and Peppy Hare and March Hare would be in the second group.&lt;br&gt;&lt;br&gt;&#xA; Lola Bunny's group:&lt;br&gt;&#xA; &lt;li&gt;Could not be the first (this would put three rabbits in the same group, Velveteen Rabbit, Br'er Rabbit and White Rabbit);&lt;br&gt;&#xA; &lt;li&gt;Nor the second (Peppy Hare and March Hare are there, and since Nesquik Bunny would not be liying, then Uncle Wiggily and the murderer would also be there, and this group would then feature six or more members);&lt;br&gt;&#xA; &lt;li&gt; Nor the third (would make March Hare not a liar);&lt;br&gt;&#xA; &lt;li&gt; Nor the fourth (three rabbits and would make Easter Bunny not a liar).&lt;br&gt;&#xA; &lt;li&gt; So they could only be in the fifth group. Since Max and March Hare are liars, none of them could be in the last group, otherwise they won't be lying.&lt;br&gt;&lt;br&gt;&#xA; Cuddles says that Easter Bunny is in the second group, which would also feature March Hare. So Velveteen Rabbit could only be the thief, Easter Bunny and March Hare would be witnesses in the second group and Max the murderer in the first group. However the first group would also feature Br'er Rabbit, White, Petter Cottontail, Nesquik Bunny and Mr. Whiskers, which would result in six members. So this all is not possible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Phew, that was a long and hard reasoning:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 24&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If Max is not in the same group as Energizer Bunny, Lola Bunny and Velveteen Rabbit, then Velveteen Rabbit must be telling the truth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's see to what this drive us:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; So, if Max is not in same group as Energizer Bunny, Lola Bunny and Velveteen Rabbit, March Hare is a liar and Velveteen Rabbit must be telling the truth. Since March Hare would be lying, he would not be on Uncle Wiggily's group and could not be the murderer. Since Max is known to be lying, March Hare could not be a witness also. So, in this case, he would be the thief and Max would be the murderer. Peppy Hare would go in the same group as March Hare, which contradicts Nesquik Bunny, which would also be a liar. Having Max, March Hare and Nesquik Bunny as liars made we unable to choose just one more liar from TRUTHS 10, 11 and 23, we would need at least two and would get 5 liars.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, we unleash a combo of important revelations (very spoilery everything from now to the very end):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 25&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Max is in the same group as Energizer Bunny, Lola Bunny and Velveteen Rabbit and he is a witness.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which implies that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 26&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; March Hare is telling the truth. He and Peppy Hare are in the same group as Uncle Wiggily.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which also implies that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 27&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Velveteen Rabbit is telling the truth, because Uncle Wiggily is in the preceding group.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 28&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nesquik Bunny is telling the truth because Peppy Hare is in Uncle Wiggily's group.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 29&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mr. Bun is telling the truth, because Max is not with Ruby.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 30&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Crusader Rabbit is telling the truth because Buster Baxter is not in the same group as Uncle Wiggily. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 31&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cuddles is in the last group.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What more?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If Reader Rabbit is a liar, then Rabbit of Caerbannog was in his group and would be also a liar. So, if Reader Rabbit is a liar, then Max, Rabbit of Caerbannog and Easter Bunny would be also the liars. But this would also make Ruby a fifth liar. Therefore:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 32&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reader Rabbit tells the truth. And he is in the same group as Bugs Bunny and Ruby, which also tells the truth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If Trix Rabbit is lying, then there was two rabbits in Uncle Wiggily's group. Also March Hare and Peppy Hare are in this group. However there would be no bunny in the group then, which would be a contradiction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 33&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Trix Rabbit is telling the truth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Also, if Mr. Whiskes is lying, he would be either a witness or the murderer. However, Nesquik Bunny is in the same group as him, which means that he can't be a witness in the group of Lola Bunny and Energizer Bunny. If he is the murderer, the other leporidae in the group would be Uncle Wiggily and the two hares, with no space for a rabbit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 34&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mr. Whiskes is telling the truth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 35&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first group is Br'er Rabbit, White Rabbit, Mr. Whiskes, Peter Cottontail and Nesquik Bunny.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;PART VII - Mystery solved: The gran finale&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Which rabbits goes in pairs and whoever is lonely?&lt;br&gt;&#xA; &lt;li&gt;We have Br'er Rabbit and White Rabbit as a pair.&#xA; &lt;li&gt;Whoever goes with Uncle Wiggily is lone. And it is not Rabbit of Caerbannog (as attested by Reader Rabbit), nor Velveteen Rabbit (as already explained), nor Reader Rabbit (which is with Bugs Bunny and Ruby).&#xA; &lt;li&gt;We are unsure if Rabbit of Caerbannog, Velveteen Rabbit and Reader Rabbit have pairs.&#xA; &lt;li&gt;Trix Rabbit has a pair (as attested by Ruby).&#xA; &lt;li&gt;Rabbit has a pair (as attested by Playboy Bunny).&#xA; &lt;li&gt;If Buster Baxter tells the truth, then Crusader Rabbit is lone, otherwise he is not.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The only possible rabbit that left to be in Uncle Wiggily's group is...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 36&lt;/strong&gt; (extremely spoilery):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Crusader Rabbit is in the group of Uncle Wiggily together with March Hare, Peppy Hare and either Easter Bunny or Playboy Bunny.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Playboy Bunny, Crusader Rabbit, Peppy Hare and March Hare are all known to tell the truth. So:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 37&lt;/strong&gt; (supreme spoilery):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Easter Bunny is the murderer!&lt;/strong&gt; His group also features Crusader Rabbit, March Hare, Peppy Hare and Uncle Wiggily.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And also:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 38&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Buster Baxter tells the truth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 39&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bunnicula is in the second group with two rabbits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Velveteen Rabbit can't be in the first group, because there would not be prior group for Uncle Wiggily. He could not be in the second because this would put Uncle Wiggily in the first group, and we already know that he is not there. He can't be in the third group because this would put Uncle Wiggily in the second group, but Uncle Wiggily's group has only one rabbit and the second group has two.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Uncle Wiggily is not in the second group, which also means that Peppy Hare is not either. And thus...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 40&lt;/strong&gt; (very spoilery):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Peter Cottontail is a liar! He is the thief!&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And finally:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cuddles is in the last group, but Crusader Rabbit can't be there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 41&lt;/strong&gt; (very spoilery):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Cuddles is a witness.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;PART VIII - Completing the board&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;So, what are the groups afterall?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There still a few positions that are unknown, but:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 42&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rabbit and Rabbit of Caerbannog are also telling the truth, because we had run out of liars.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, it is very easy to fill all them remaining positions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 43&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rabbit and Rabbit of Caerbannog pairs in the second group with Bunnicula.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Further:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since Cuddles is a witness in the fifth group, then:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 44&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The murderer is in the forth group. So, Uncle Wiggily, March Hare, Peppy Hare, Easter Bunny and Crusader Rabbit goes there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And also:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 45&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The witnesses are in the fifth group. So this is the group of Cuddles, Max, Velveteen Rabbit, Lola Bunny and Energizer Bunny.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 46&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bugs Bunny, Ruby and Reader Rabbit must fill the third group because they wouldn't fit anywhere else.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 47&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Playboy Bunny only fits the second group.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 48&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Trix Rabbit must pair Reader Rabbit in the third group.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And we finish with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRUTH 49&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mr. Bun gets the last remaining space in the second group.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;GAME OVER&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Finally, the full groups table:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;GROUP 1&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Br'er Rabbit, White Rabbit, Peter Cottontail (thief), Mr. Whiskes and Nesquik Bunny.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;GROUP 2&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bunnicula, Rabbit, Rabbit of Caerbannog, Playboy Bunny and Mr. Bun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;GROUP 3&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Buster Baxter, Bugs Bunny, Ruby, Reader Rabbit and Trix Rabbit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;GROUP 4&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Uncle Wiggily (dead), Crusader Rabbit, Easter Bunny (murderer), March Hare and Peppy Hare.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;GROUP 5&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cuddles (witness), Velveteen Rabbit, Lola Bunny, Energizer Bunny and Max (witness).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5044" LastEditorUserId="5044" LastEditDate="2015-09-01T22:58:23.570" LastActivityDate="2015-09-01T22:58:23.570" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="18491" PostTypeId="2" ParentId="18437" CreationDate="2015-07-30T14:39:01.977" Score="2" Body="&lt;p&gt;I don't have an answer, but I can provide limits.&lt;/p&gt;&#xA;&lt;p&gt;Obviously, the upper limit for the problem is 1000 fuses.  Just wire 1000 simple circuits, each with a 10V cell, a fuse, a &lt;span class=&quot;math-container&quot;&gt;$5 \Omega$&lt;/span&gt; resistor, and a box.&lt;/p&gt;&#xA;&lt;p&gt;Surprisingly, the lower limit is &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt;.  This represents the minimum number of fuses that can blow in at least &lt;span class=&quot;math-container&quot;&gt;$1001$&lt;/span&gt; different orderings.&lt;/p&gt;&#xA;&lt;p&gt;The constraints of the problem are: construct the circuit with all cells disconnected, then flip a switch that simultaneously connects all cells, then observe which fuses blow in which order.  The number of possible variations (taking into account that some fuses may not blow at all and that some groups of fuses may all blow at once) is, for &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; fuses:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ F(n) = 2 \sum_{i=1}^n  P(n,i) i!$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Here, &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; represents the number of groups of fuses that blow simultaneously (or do not blow at all), and &lt;span class=&quot;math-container&quot;&gt;$P(n,i)$&lt;/span&gt; is the number of ways of partitioning &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; distinct elements into &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; groups.  The factor of 2 is due to the fact that the last group might not blow, yielding &lt;span class=&quot;math-container&quot;&gt;$i-1$&lt;/span&gt; events, with some fuses unblown.&lt;/p&gt;&#xA;&lt;p&gt;This crosses &lt;span class=&quot;math-container&quot;&gt;$1001$&lt;/span&gt; at &lt;span class=&quot;math-container&quot;&gt;$F(5) = 1082$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;240: 5 groups: each fuse blows at a different time (at most one fuse unblown).&lt;/li&gt;&#xA;&lt;li&gt;480: 4 groups: 1 group of 2, and 3 groups of 1.&lt;/li&gt;&#xA;&lt;li&gt;300: 3 groups: 1 group of 3, 2 groups of 1, or 2 groups of 2 and 1 group of 1.&lt;/li&gt;&#xA;&lt;li&gt;60: 2 groups: 1 group of 4, 1 group of 1, or 1 group of 3, 1 group of 2.&lt;/li&gt;&#xA;&lt;li&gt;2: 1 group: all fuses blow simultaneously, or no fuses blow.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;For further clarity, I'll break out the 3 group case:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;For 1 group of 3, 2 groups of 1, there are &lt;span class=&quot;math-container&quot;&gt;${5\choose 3} = 10$&lt;/span&gt; arrangements of which fuses are in the group of 3.  For 2 groups of 2, 1 group of 1, there are &lt;span class=&quot;math-container&quot;&gt;${5\choose 2}=10$&lt;/span&gt; choices for the first group of 2, &lt;span class=&quot;math-container&quot;&gt;${3\choose 2}=3$&lt;/span&gt; choices for the second group of 2, but the groups are interchangeable, so we need to divide by &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt;, yielding  &lt;span class=&quot;math-container&quot;&gt;$15$&lt;/span&gt; arrangements.  Thus, &lt;span class=&quot;math-container&quot;&gt;$P(5,3) = 25$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The 3 groups could blow in any order, so there are &lt;span class=&quot;math-container&quot;&gt;$3!=6$&lt;/span&gt; orderings of groups blowing.  Plus, there is the distinction that the last group might never blow, so that adds an extra factor of &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt;.  Thus, the total number of ways 5 fuses can blow (or not blow) in 3 groups is &lt;span class=&quot;math-container&quot;&gt;$25\times6\times2 = 300$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So we have a lower limit of &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt; and an upper limit of &lt;span class=&quot;math-container&quot;&gt;$1000$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;What I find fascinating is that I have yet to come up with any circuit that solves the problem with fewer than &lt;span class=&quot;math-container&quot;&gt;$1000$&lt;/span&gt; fuses.  Any circuit I come up with that has some fuses blow first and others later ends up using as many fuses as a circuit that wires the fuses independently.&lt;/p&gt;&#xA;&lt;p&gt;Can anyone come up with a better upper limit?&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-07-30T14:39:01.977" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18492" PostTypeId="2" ParentId="18390" CreationDate="2015-07-30T14:44:50.357" Score="24" Body="&lt;p&gt;It represents&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Kaleidoscope&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The second add will have the object &quot;collide&quot; with the first on the hashset since they have the same hash code. e rotated 180 degees is &quot;a&quot;. The empty curly braces are a new &quot;scope&quot;. Collide+a+scope = kaleidoscope.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14268" LastActivityDate="2015-07-30T14:44:50.357" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18493" PostTypeId="2" ParentId="18424" CreationDate="2015-07-30T16:01:22.277" Score="11" Body="&lt;p&gt;Is it...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Saturn?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My namesakes are plentiful,&lt;br&gt;&#xA;though some antiquated:&lt;br&gt;&#xA;A company defunct&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Saturn_Corporation&quot; rel=&quot;nofollow&quot;&gt;Saturn Corporation&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and a system outdated.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Sega_Saturn&quot; rel=&quot;nofollow&quot;&gt;Sega Saturn&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One unavoidable&lt;br&gt;&#xA;an honor shared by few&lt;br&gt;&#xA;just give it some &lt;em&gt;time&lt;/em&gt;&lt;br&gt;&#xA;it will come into view.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Saturday is named after Saturn, and comes unavoidably once a week. (thanks @Going hamateur and @DaveBlackston) &quot;Time&quot; is italicized because Saturn was often identified with Chronus, the god of time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Two bearing my name&lt;br&gt;&#xA;and both are titanic:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Titanic&quot; is a reference to Saturn being a mythological Titan.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;an explosive endeavor&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Saturn_%28rocket_family%29&quot; rel=&quot;nofollow&quot;&gt;Saturn rockets&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the second, enigmatic.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The planet Saturn, which has some mysterious features such as its &lt;a href=&quot;https://en.wikipedia.org/wiki/Saturn%27s_hexagon&quot; rel=&quot;nofollow&quot;&gt;hexagon&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My habits inspired&lt;br&gt;&#xA;a nightmarish work.&lt;br&gt;&#xA;The last one was spared&lt;br&gt;&#xA;and well known to perk.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Saturn was known for eating his children, but the last one, Jupiter (Zeus), was spared. This habit inspired &lt;a href=&quot;https://en.wikipedia.org/wiki/Saturn_Devouring_His_Son&quot; rel=&quot;nofollow&quot;&gt;a painting&lt;/a&gt; by Goya, as well as one of the same name by Rubens. Jupiter is associated with &quot;perk&quot;, e.g. &quot;jovial&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What I bring is a gift,&lt;br&gt;&#xA;feared and embraced.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In Holst's &lt;a href=&quot;https://en.wikipedia.org/wiki/The_Planets&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;The Planets&lt;/em&gt;&lt;/a&gt;, Saturn is the Bringer of Old Age. (Thanks @DaveBlackston)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Do you know my name&lt;br&gt;&#xA;from these clues I have placed?&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="13237" LastEditDate="2015-09-15T20:26:20.063" LastActivityDate="2015-09-15T20:26:20.063" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18495" PostTypeId="2" ParentId="18437" CreationDate="2015-07-30T17:23:32.837" Score="8" Body="&lt;p&gt;I can't draw a circuit diagram right now &lt;em&gt;(see below)&lt;/em&gt;, but I think I have a way for two fuses to test three boxes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Connect each box in series with a 1 ohm resistor. Connect fuse 1 in series with a 1 ohm resistor. Connect fuse 2 in series with a 2 ohm resistor. Now connect all of the preceding parts together in parallel. Finally make a circuit with that, a 2 ohm resistor, and a 9 volt battery in series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If all boxes are conductors, no fuses blow. If two are conductors, fuse 1 blows and fuse 2 doesn't. If one is a conductor, fuse 1 blows followed by fuse 2. If none are conductors, both fuses blow at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this works, we can construct 333 copies of it and use one more fuse to test the last box, lowering the upper bound to 667.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;5 boxes can be tested with 3 fuses with a similar setup. Each box gets a 3 ohm resistor and the fuses are connected to resistors of 2 ohms, 3 ohms, and 4 ohms. The circuit is completed with a 3 ohm resistor and an 18 volt battery. 200 of these can test all the boxes with 600 fuses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be possible to extend this kind of circuit to test 1000 boxes with 501 fuses.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I've added a rough diagram (drawn in MSPaint)&#xA;&lt;a href=&quot;https://i.stack.imgur.com/aX8OL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aX8OL.png&quot; alt=&quot;3 boxes 2 fuses&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="4269" LastEditDate="2015-11-02T16:33:01.173" LastActivityDate="2015-11-02T16:33:01.173" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18496" PostTypeId="1" AcceptedAnswerId="18497" CreationDate="2015-07-30T17:53:59.760" Score="19" ViewCount="2271" Body="&lt;p&gt;My town has thirteen buildings.&lt;br&gt;&#xA;Its name has four letters.&lt;br&gt;&#xA;Exactly six roads leave the town.&lt;br&gt;&#xA;Only one leads to my twin town.&lt;br&gt;&#xA;The other five roads are dead ends.&lt;br&gt;&#xA;Four of them are three blocks long.&lt;br&gt;&#xA;One of them is two blocks long.&lt;br&gt;&#xA;Walk carefully in the dead end cul-de-sacs,&lt;br&gt;&#xA;Since sharp objects are hidden there.&lt;br&gt;&#xA;I love my town, despite the lines&lt;br&gt;&#xA;Nearly everywhere you go.&lt;br&gt;&#xA;Whenever something spectacular happens,&lt;br&gt;&#xA;We have many meetings with our twin town.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the name of my town?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastActivityDate="2015-07-30T18:08:30.330" Title="What is the name of my town?" Tags="|riddle|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18497" PostTypeId="2" ParentId="18496" CreationDate="2015-07-30T18:03:55.760" Score="21" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I think the name of the town is palm&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The thirteen buildings are the 5 metacarpals and 8 carpal bones&#xA;Four letters in the word palm&lt;br&gt;&#xA;six roads, five being fingers, one being the arm&lt;br&gt;&#xA;four of the fingers have three knuckles and the thumb has two&lt;br&gt;&#xA;sharp objects are likely nails (fingernails- thanks Bailey M)&lt;br&gt;&#xA;Palms have lines all across them&lt;br&gt;&#xA;You clap with both hands when you enjoy something like a performance&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13975" LastEditorUserId="463" LastEditDate="2015-07-30T18:08:30.330" LastActivityDate="2015-07-30T18:08:30.330" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18498" PostTypeId="1" AcceptedAnswerId="18499" CreationDate="2015-07-30T19:34:54.620" Score="5" ViewCount="429" Body="&lt;p&gt;This is part of &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=title%3A%22The+Twenty+Doors%22&quot;&gt;The Twenty Doors&lt;/a&gt; series.&lt;br&gt;&#xA;The previous one is &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18345/the-twenty-doors-room-7&quot;&gt;The Twenty Doors (ROOM 7)&lt;/a&gt;&lt;br&gt;&#xA;The next one is &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18512/the-twenty-doors-room-9&quot;&gt;The Twenty Doors (ROOM 9)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You move on into Room Eight when Kslkgh (a.k.a. Dr. X) finally creates the next room. There is, as always, a slip of paper on the floor:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Before the alphabet, before. The pass, between the alleys. Automatic keying. &lt;br&gt;&lt;br&gt;Ip wbgft kk tzttfzz, gsx setx kegv pv 'GRIF RRWOJ CUQ RF OYJCXKI' lx ffm lukqyp. &#xA;  &lt;br&gt;- Jd. Q &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And, on the wall, there is... no hint?!&lt;br&gt;&#xA;&lt;code&gt;There is no hint for Room Eight. Work it out from the first part of the message&#xA;- Dr. X&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, finally, you look at the keypad, which is a normal keyboard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What should you type in to open the door?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next door will be added when Dr. X comes up with it (if this room is solved)!&lt;/p&gt;&#xA;" OwnerDisplayName="user9377" LastEditorUserId="-1" LastEditorDisplayName="user9377" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-07-31T07:34:37.073" Title="The Twenty Doors (ROOM 8)" Tags="|cipher|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18499" PostTypeId="2" ParentId="18498" CreationDate="2015-07-30T19:41:13.810" Score="4" Body="&lt;p&gt;You need to type:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DOOR EIGHT LET ME THROUGH&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the door will let you through (hopefully)!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is a Vigenere Autokey cipher, which uses a password as well as a modified alphabet. The alphabet you're using puts the word BEFORE before the rest of the alphabet, resulting in BEFORACDGHIJKLMNPQSTUVWXYZ. Then, with a passcode of 'betweenthealleys', we get the result &quot;In order to succeed, you must type in 'DOOR EIGHT LET ME THROUGH' on the keypad. - Dr. X&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-07-30T19:41:13.810" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18500" PostTypeId="1" AcceptedAnswerId="18502" CreationDate="2015-07-30T20:03:05.013" Score="7" ViewCount="824" Body="&lt;p&gt;This kind of puzzle is different than your normal magic square puzzles. Here are 3, in increasing difficulty. Some numbers have been switched, and you have to find them and swap them around to make the magic square valid again.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The zeros are for formatting placeholders.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The numbers in each of line of five squares across, down, and diagonally should add up to 58, but in every row across and column there is one number of out place. Swap these with one another to make the total correct.&lt;br&gt;&#xA;16 16 11 09 14 &lt;br&gt;&#xA;17 09 29 13 01 &lt;br&gt;&#xA;15 05 06 16 15 &lt;br&gt;&#xA;07 03 17 11 12 &lt;br&gt;&#xA;14 17 03 08 06 &lt;br&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The numbers in each of line of six squares across, down, and diagonally should add up to 122, but in every row across and column there is one number of out place. Swap these with one another to make the total correct.&lt;br&gt;&#xA;21 05 14 31 44 15 &lt;br&gt;&#xA;30 29 21 09 22 20 &lt;br&gt;&#xA;36 29 20 10 06 22 &lt;br&gt;&#xA;06 30 22 30 13 17 &lt;br&gt;&#xA;10 26 23 17 12 22 &lt;br&gt;&#xA;27 12 20 13 21 27 &lt;br&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The numbers in each of line of seven squares across, down, and diagonally should add up to 123, but in every row across and column there is one number of out place. Swap these with one another to make the total correct.&lt;br&gt;&#xA;31 19 10 13 14 32 15 &lt;br&gt;&#xA;06 21 17 22 30 17 07 &lt;br&gt;&#xA;17 30 17 24 17 11 08 &lt;br&gt;&#xA;07 22 33 13 15 17 11 &lt;br&gt;&#xA;14 16 21 22 13 11 29 &lt;br&gt;&#xA;16 13 03 19 12 12 43 &lt;br&gt;&#xA;27 13 19 13 20 18 11 &lt;br&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Have fun!!&lt;/p&gt;&#xA;" OwnerUserId="2001" LastEditorUserId="2001" LastEditDate="2015-07-30T21:08:06.220" LastActivityDate="2015-07-31T13:47:01.917" Title="Magic Square Mixups [Challenge]" Tags="|magic-square|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18501" PostTypeId="2" ParentId="18500" CreationDate="2015-07-30T20:50:02.333" Score="3" Body="&lt;p&gt;Partial Answer, but I got the Easy puzzle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Original&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &#xA; 16 16 11 09 14 = &lt;strong&gt;66&lt;/strong&gt;&lt;br&gt;&#xA; 17 09 29 13 01 = &lt;strong&gt;69&lt;/strong&gt;&lt;br&gt;&#xA; 15 05 06 16 15 = &lt;strong&gt;57&lt;/strong&gt;&lt;br&gt;&#xA; 07 03 17 11 12 = &lt;strong&gt;50&lt;/strong&gt;&lt;br&gt;&#xA; 14 17 03 08 06 = &lt;strong&gt;48&lt;/strong&gt;&lt;br&gt;&#xA; &lt;strong&gt;69&lt;/strong&gt; &lt;strong&gt;50&lt;/strong&gt; &lt;strong&gt;66&lt;/strong&gt; &lt;strong&gt;57&lt;/strong&gt; &lt;strong&gt;48&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solved:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 16 16 &lt;strong&gt;03&lt;/strong&gt; 09 14 = &lt;strong&gt;58&lt;/strong&gt;&lt;br&gt;&#xA; &lt;strong&gt;06&lt;/strong&gt; 09 29 13 01 = &lt;strong&gt;58&lt;/strong&gt;&lt;br&gt;&#xA; 15 05 06 &lt;strong&gt;17&lt;/strong&gt; 15 = &lt;strong&gt;58&lt;/strong&gt;&lt;br&gt;&#xA; 07 &lt;strong&gt;11&lt;/strong&gt; 17 11 12 = &lt;strong&gt;58&lt;/strong&gt;&lt;br&gt;&#xA; 14 17 03 08 &lt;strong&gt;16&lt;/strong&gt; = &lt;strong&gt;58&lt;/strong&gt;&lt;br&gt;&#xA; &#xA; &#xA; &lt;strong&gt;58&lt;/strong&gt; &lt;strong&gt;58&lt;/strong&gt; &lt;strong&gt;58&lt;/strong&gt; &lt;strong&gt;58&lt;/strong&gt; &lt;strong&gt;58&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Switch 11 in the first row with 03 in the fourth. Take the 06 in the last row, 17 in the second row, and 16 in the third row, and cycle them each once (or backwards twice), in that order. Then, the 06 is in the second row, the 17 is  in the third row, and the 16 is in the last row &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13538" LastEditorUserId="13916" LastEditDate="2015-07-30T21:54:17.933" LastActivityDate="2015-07-30T21:54:17.933" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18502" PostTypeId="2" ParentId="18500" CreationDate="2015-07-30T20:55:33.113" Score="10" Body="&lt;p&gt;The answer to the first one is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&#xA; 16 16 &lt;strong&gt;03&lt;/strong&gt; 09 14 &lt;br&gt;&#xA; &lt;strong&gt;06&lt;/strong&gt; 09 29 13 01 &lt;br&gt;&#xA; 15 05 06 &lt;strong&gt;17&lt;/strong&gt; 15 &lt;br&gt;&#xA; 07 &lt;strong&gt;11&lt;/strong&gt; 17 11 12 &lt;br&gt;&#xA; 14 17 03 08 &lt;strong&gt;16&lt;/strong&gt; &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer to the second one is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &#xA; &lt;strong&gt;13&lt;/strong&gt; 05 14 31 44 15 &lt;br&gt;&#xA; 30 &lt;strong&gt;20&lt;/strong&gt; 21 09 22 20 &lt;br&gt;&#xA; 36 29 20 10 06 &lt;strong&gt;21&lt;/strong&gt; &lt;br&gt;&#xA; 06 30 22 30 &lt;strong&gt;17&lt;/strong&gt; 17 &lt;br&gt;&#xA; 10 26 23 &lt;strong&gt;29&lt;/strong&gt; 12 22 &lt;br&gt;&#xA; 27 12 &lt;strong&gt;22&lt;/strong&gt; 13 21 27 &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer to the last one is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&#xA; 31 &lt;strong&gt;08&lt;/strong&gt; 10 13 14 32 15 &lt;br&gt;&#xA; 06 21 &lt;strong&gt;20&lt;/strong&gt; 22 30 17 07 &lt;br&gt;&#xA; 17 30 17 24 17 11 &lt;strong&gt;07&lt;/strong&gt; &lt;br&gt;&#xA; &lt;strong&gt;12&lt;/strong&gt; 22 33 13 15 17 11 &lt;br&gt;&#xA; 14 16 21 &lt;strong&gt;19&lt;/strong&gt; 13 11 29 &lt;br&gt;&#xA; 16 13 03 19 12 &lt;strong&gt;17&lt;/strong&gt; 43 &lt;br&gt;&#xA; 27 13 19 13 &lt;strong&gt;22&lt;/strong&gt; 18 11 &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To solve these, the easiest way is to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &#xA; Calculate the sum of each row and column.  Where the column and row sums are equal, the intersection is the number that needs to be swapped.  For instance, in the first puzzle, the original puzzle has a total of 69 for the first row and also for the second column.  This means that the number in the first row and second column will need to be swapped (17)  As the total in these columns is 69 and should be 58, it means that it needs to be 11 less, so should be 6.  Note that the value of 6 is located in two places, but the one we want is in row 5, column 5 because these rows both have the same total of 48.  Continue in this manner until the whole puzzle is solved.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10457" LastEditorUserId="10457" LastEditDate="2015-07-31T13:47:01.917" LastActivityDate="2015-07-31T13:47:01.917" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18503" PostTypeId="2" ParentId="18363" CreationDate="2015-07-30T23:19:39.970" Score="8" Body="&lt;p&gt;&lt;strong&gt;COPIED OVER FROM JOE Z'S ANSWER (#1 to #6)&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;1.&lt;/strong&gt; Q. This is the order of the value of U.S. coins (Penny, Nickel, Dime). The next one is &lt;strong&gt;Quarter&lt;/strong&gt;.&lt;p&gt;&#xA; &lt;strong&gt;2.&lt;/strong&gt; Q. This the order of the value ranks of chess pieces (Pawn, Knight, Bishop, Rook). The next one is &lt;strong&gt;Queen&lt;/strong&gt;.&lt;p&gt;&#xA; &lt;strong&gt;3.&lt;/strong&gt; A. This is the order of sharps in Western key signature notation (F, C, G, D, A, E, B). It's also the initials of the mnemonic &quot;Father Charles Goes Down &lt;strong&gt;And&lt;/strong&gt; Ends Battle&quot;.&lt;p&gt;&#xA; &lt;strong&gt;4.&lt;/strong&gt; T. This is the order of SI prefixes being applied to bytes (Bytes, Kilobytes, Megabytes, Gigabytes). The next one is &lt;strong&gt;Terabytes&lt;/strong&gt;.&lt;p&gt;&#xA; &lt;strong&gt;5.&lt;/strong&gt; J. This is the order of the &lt;em&gt;first&lt;/em&gt; names of United States Presidents from Obama going backwards (Barack, George, Bill, George, Ronald). The next one is &lt;strong&gt;Jimmy&lt;/strong&gt; Carter.&lt;p&gt;&#xA; &lt;strong&gt;6.&lt;/strong&gt; R. This is the order of the books of the Old Testament in the Bible (Genesis, Exodus, Leviticus, Numbers, Deuteronomy, Joshua, Judges). The next one is &lt;strong&gt;Ruth&lt;/strong&gt;.&lt;p&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ANSWERS FROM OTHERS (#7, #9, #14)&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;7.&lt;/strong&gt; G. Teaspoon, Tablespoon, Cup, Pint, Quart, &lt;strong&gt;Gallon&lt;/strong&gt;. (From Bailey M.)&lt;p&gt;&#xA;&lt;strong&gt;9.&lt;/strong&gt; F. Faces on US paper currency: Washington, Jefferson, Lincoln, Hamilton, Jackson, Grant, &lt;strong&gt;Franklin&lt;/strong&gt;. (From dennisdeems)&lt;p&gt;&#xA;&lt;strong&gt;14.&lt;/strong&gt; H. The Harry Potter series: &lt;strong&gt;S&lt;/strong&gt;orcerer's Stone, &lt;strong&gt;C&lt;/strong&gt;hamber of Secrets, &lt;strong&gt;P&lt;/strong&gt;risoner of Azkaban, &lt;strong&gt;G&lt;/strong&gt;oblet of Fire, &lt;strong&gt;O&lt;/strong&gt;rder of the Phoenix, &lt;strong&gt;H&lt;/strong&gt;alf-Blood Prince&lt;br&gt;&#xA;(outside of the US, the first name is Philosopher's Stone)&#xA;(From GordonK)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEN'S ANSWERS&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;8) H, O, T, T, S, F, ?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Poker hand ranking:&lt;br&gt;&#xA;&lt;strong&gt;H&lt;/strong&gt;igh card, &lt;strong&gt;O&lt;/strong&gt;ne pair, &lt;strong&gt;T&lt;/strong&gt;wo pair, &lt;strong&gt;T&lt;/strong&gt;hree of a kind, &lt;strong&gt;S&lt;/strong&gt;traight, &lt;strong&gt;F&lt;/strong&gt;lush, &lt;strong&gt;F&lt;/strong&gt;ull house &lt;strong&gt;. . .&lt;/strong&gt;   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;10) E, F, W, W, ?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The five powers in Captain Planet and the Planeteers (animated TV series):&lt;br&gt;&#xA;&lt;strong&gt;E&lt;/strong&gt;arth, &lt;strong&gt;F&lt;/strong&gt;ire, &lt;strong&gt;W&lt;/strong&gt;ind, &lt;strong&gt;W&lt;/strong&gt;ater, &lt;strong&gt;H&lt;/strong&gt;eart  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;11) L, G, G, S, W, E, ?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The seven deadly sins, in the order used by Pope Gregory I:&lt;br&gt;&#xA;&lt;strong&gt;L&lt;/strong&gt;ust, &lt;strong&gt;G&lt;/strong&gt;luttony, &lt;strong&gt;G&lt;/strong&gt;reed, &lt;strong&gt;S&lt;/strong&gt;loth, &lt;strong&gt;W&lt;/strong&gt;rath, &lt;strong&gt;E&lt;/strong&gt;nvy, &lt;strong&gt;P&lt;/strong&gt;ride   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;12) P, T, F, C, G, G, ?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;The Twelve Days of Christmas&quot; song:&lt;br&gt;&#xA; a &lt;strong&gt;P&lt;/strong&gt;artridge, two &lt;strong&gt;T&lt;/strong&gt;urtle doves, three &lt;strong&gt;F&lt;/strong&gt;rench hens, four &lt;strong&gt;C&lt;/strong&gt;alling birds, five &lt;strong&gt;G&lt;/strong&gt;old rings, six &lt;strong&gt;G&lt;/strong&gt;eese a-laying, seven &lt;strong&gt;S&lt;/strong&gt;wans a-swimming &lt;strong&gt;. . .&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;13) M, J, P, L, S, D, H, ?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;US state capitals ordered by the alphabetical state names:&lt;br&gt;&#xA;&lt;strong&gt;M&lt;/strong&gt;ontgomery Alabama, &lt;strong&gt;J&lt;/strong&gt;uneau Alaska, &lt;strong&gt;P&lt;/strong&gt;hoenix Arizona, &lt;strong&gt;L&lt;/strong&gt;ittle Rock Arkansas, &lt;strong&gt;S&lt;/strong&gt;acramento California, &lt;strong&gt;D&lt;/strong&gt;enver Colorado, &lt;strong&gt;H&lt;/strong&gt;artford Connecticut, &lt;strong&gt;D&lt;/strong&gt;over Delaware &lt;strong&gt;. . .&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;15) L, H, H, B, S, B, B, ?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The twelve labours of Hercules:&lt;br&gt;&#xA;Nemean &lt;strong&gt;L&lt;/strong&gt;ion, Lernaean &lt;strong&gt;H&lt;/strong&gt;ydra, Ceryneian &lt;strong&gt;H&lt;/strong&gt;ind, Erymanthian &lt;strong&gt;B&lt;/strong&gt;oar, Augean &lt;strong&gt;S&lt;/strong&gt;tables, Stymphalian &lt;strong&gt;B&lt;/strong&gt;irds, Cretan &lt;strong&gt;B&lt;/strong&gt;ull, &lt;strong&gt;M&lt;/strong&gt;ares of Diomedes &lt;strong&gt;. . .&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-08-04T16:01:35.720" LastActivityDate="2015-08-04T16:01:35.720" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18504" PostTypeId="1" AcceptedAnswerId="18506" CreationDate="2015-07-31T02:06:50.917" Score="2" ViewCount="180" Body="&lt;p&gt;I propound to you a mystery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several centuries ago it was necessary to prove it impossible to remove funding from the fraudsters so that science could advance. In the twentieth century it was accomplished to prove our science had advanced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can read this as either a riddle or a Jeopardy question. It works either way. There really isn't anything devious here; a straightforward approach should solve it.&lt;/p&gt;&#xA;" OwnerUserId="5623" LastEditorUserId="5623" LastEditDate="2015-07-31T02:18:32.300" LastActivityDate="2015-07-31T02:39:30.787" Title="A paradox of practical necessity" Tags="|knowledge|" AnswerCount="1" CommentCount="3" ClosedDate="2015-07-31T11:15:34.537" ContentLicense="CC BY-SA 3.0" />
  <row Id="18505" PostTypeId="1" AcceptedAnswerId="18514" CreationDate="2015-07-31T02:36:25.883" Score="15" ViewCount="2276" Body="&lt;p&gt;&lt;em&gt;In the distant future (specifically the year 2025)...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Welcome to Pluto! Until now, you have had an idyllic childhood in the Zetetic Zone, solving Space Riddles and Space Crosswords and Spacedokus. Ever since the Uranus/Pluto Non-Violence Agreement of 2006 (which ended the Plutonian War of Recognition), the two planets have been strong allies, despite not being particularly close to each other. In fact, the Zetetic Zone, which encompasses both planets, their moons, and all the space in between, is preposterously large, but that’s no concern of yours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your childhood is coming to a close, however. Today is your 17th birthday, and when you check your SpaceMail, you see the following message:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Congratulations on surviving for 1.7 decacycles! You now qualify to&#xA;  become a legally recognized adult human/android. Your &quot;Pluto/Uranus&#xA;  Zetetic Zone Licensed Indication that Neurological Growth is Decidedly&#xA;  Outstanding or Tolerably Sufficient&quot; Evaluation (PUZZLINGDOTSE) will&#xA;  take place in 15 minicycles at the 67th decamicrocycle. Your partner&#xA;  will be the android designated Jeffrey LT89-46AF.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can’t believe your luck. Every PUZZlet hopes to have an android for their PUZZLINGDOTSE partner, as their hybrid neuro-digital brain is well suited to the PUZZLINGDOTSE challenges. In fact, some argue that the challenges are impossible for neuro-analog brains, but fortunately you won’t have to find out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Partners attempt challenges separately, but whoever goes first can send a short message to the other with some instructions. The character limit is very strict, however, so the message is often as cryptic as the challenge itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the chosen day, you arrive at the 66th decamicrocycle to meet your partner and prepare for the challenge. Jeffrey LT89-46AF is an impressive figure, with the famously sleek LT Series body and the Anthrofacic upgrade, giving it the ability to mimic a variety of human facial expressions. You hold out a hand and say, “Hi, you must be Jeffrey.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He shakes your hand politely and replies, “Yes, although I prefer to be called Hyphen Four, if you don’t mind.” (Android nicknames are weird.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some pleasant smalltalk, the challenge room is ready. Your SpaceTablets are updated with a description of the challenge:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Welcome to your PUZZLINGDOTSE! Please read the instructions carefully!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The room before you consists of 3 floors. There is an entrance on the&#xA;  1st floor and a locked exit on the 3rd floor. To complete the&#xA;  challenge, you must pick up the key from the 2nd floor and proceed to&#xA;  the exit.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Each floor is divided into a 3x3 grid of teleporters. Moving&#xA;  orthogonally from one teleporter to another takes approximately 1&#xA;  microcycle. Diagonal movement is grounds for disqualification.&#xA;  Teleporters will occasionally glow with colored light, which can be&#xA;  one of 4 varieties: Navy Blue, Purple, Sepia, and Tan. Activating a&#xA;  teleporter when it is glowing causes it to perform the associated&#xA;  action, as follows:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;A &lt;strong&gt;Navy Blue&lt;/strong&gt; teleporter will teleport you to the &lt;strong&gt;Next&lt;/strong&gt; floor. You will&#xA;  arrive in the same relative position on the grid as you were on the&#xA;  floor you just left.  &lt;/li&gt;&#xA;  &lt;li&gt;A &lt;strong&gt;Purple&lt;/strong&gt; teleporter will teleport you to the &lt;strong&gt;Previous&lt;/strong&gt; floor. You will arrive in the same relative position on the grid as you were on the floor you just left.  &lt;/li&gt;&#xA;  &lt;li&gt;A &lt;strong&gt;Sepia&lt;/strong&gt; teleporter will teleport you to the &lt;strong&gt;Start&lt;/strong&gt; (or entrance) of the challenge.  &lt;/li&gt;&#xA;  &lt;li&gt;A &lt;strong&gt;Tan&lt;/strong&gt; teleporter will teleport you to the challenge’s &lt;strong&gt;Terminus&lt;/strong&gt; (or exit).&#xA;  Please note that you will be unable to open the exit door unless you&#xA;  have the key.  &lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;Most of the time, a teleporter will not glow, which means it cannot be&#xA;  activated. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Teleporters change their glow state every time you step on a new&#xA;  teleporter. They will also change their glow state if you go 1 microcycle&#xA;  without stepping on a new teleporter. Each teleporter has its own set&#xA;  of repeating glow state progression patterns. Stepping on certain&#xA;  teleporters at certain times will cause all teleporters to change to a&#xA;  different glow state progression pattern. If you remain in place, the&#xA;  teleporters will proceed through their current progression pattern&#xA;  without switching.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The only way to move between floors is through teleportation. You have&#xA;  10 microcycles to complete the challenge. The timer begins when you leave&#xA;  the starting teleporter, and resets every time you teleport to the&#xA;  entrance. To finish the challenge, you must deduce the glow state&#xA;  patterns for all necessary teleporters and navigate the room to pick&#xA;  up the key and leave through the exit. You must both complete the&#xA;  challenge, or you both fail. Good luck!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A diagram is attached to the message:&lt;a href=&quot;https://i.stack.imgur.com/7uefi.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7uefi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Sounds complicated&lt;/em&gt;, you think. &lt;em&gt;I hope Jeffr— uh, Hyphen Four goes first.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hyphen Four indeed volunteers to go first and enters the room. In a preposterously short amount of time, you receive the notification that your partner has completed the challenge, and the entrance opens for you. As you step through, the door closes behind you and you get your first look at the challenge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your heart sinks as you observe the teleporters cycle through their patterns. The glows are few and far between, and after spending some time trying to memorize the patterns, you start to lose hope. The patterns are obscenely long, and besides, they’re going to change when you take your first step! You’ll have to memorize a new pattern every time you step on a new teleporter! Not to mention, you can't even see the teleporters on the second and third floors yet! And if all of that weren’t bad enough, it’s REALLY hard to tell the difference between the Navy Blue and Purple glows, and the Sepia and Tan glows are almost identical! You try to fight back tears of frustration as you realize the enormity of the task before you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then your SpaceTablet beeps. You have a message from Hyphen Four!&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;869251435782966988&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Yeesh, that character limit’s a killer. Still, this is your only chance…better figure out what your partner is trying to tell you.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the route you must take to complete the challenge?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your answer should be a sequence of coordinates and letters. For example, A2 indicates that you moved to that square. To get from the start to the upper right square on the first floor, you might do A2,B2,B1,C1. In addition, there are certain actions you can take:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;N means you activated a Navy Blue teleporter (going to the next higher floor)&lt;/li&gt;&#xA;&lt;li&gt;P means you activated a Purple teleporter (going to the next lower floor)&lt;/li&gt;&#xA;&lt;li&gt;S means you activated a Sepia teleporter (returning you to the Start at A3)&lt;/li&gt;&#xA;&lt;li&gt;T means you activated a Tan teleporter (sending you to the Terminus at I7)&lt;/li&gt;&#xA;&lt;li&gt;L means you Loiter (stay on your current square for 1 microcycle)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So if you could make the teleporters do what you wanted, you could do the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A2,A1,N,E4,T&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Or, if you want to go the scenic route:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;B3,C3,N,F5,F4,P,B1,N,D4,S,L,L,B3,B2,L,T&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, you don't control the teleporters. You can only activate a teleporter if it's glowing the proper color, and there's no way you can figure out the glow patterns on your own. You have to rely on Hyphen Four's message! There is only one correct route.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some ways to check if you have the correct route:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You can't go more than 10 steps (including Loitering) without returning to the start. That means S, T, or I7 must be your tenth (or earlier) step, not your eleventh. Returning to the start resets the count.&lt;/li&gt;&#xA;&lt;li&gt;You must pass through E4, which holds the Key. Don't worry about including &quot;picking up the key&quot; in your answer; as long as you pass through E4 at some point, you automatically pick it up.&lt;/li&gt;&#xA;&lt;li&gt;All of the actions occur at least once.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As a side note, while the rest of the writeup provides important context for the puzzle, the only actual elements you need to solve it are the &lt;strong&gt;diagram&lt;/strong&gt; and &lt;strong&gt;Hyphen Four's message&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck everybody! Clarification requests are welcome!&lt;/p&gt;&#xA;" OwnerUserId="10861" LastActivityDate="2015-07-31T15:48:02.990" Title="Pass the test, become an adult!" Tags="|cipher|pattern|lateral-thinking|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18506" PostTypeId="2" ParentId="18504" CreationDate="2015-07-31T02:39:30.787" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Converting mercury into gold?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Several centuries ago it was necessary to prove it impossible to remove funding from the fraudsters so that science could advance.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some alchemists were frauds that spoiled the name of science for personal gain. They often claimed that they had magical abilities to turn mercury to gold, and gained wealth and power for themselves, even though they couldn't really do it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the twentieth century it was accomplished to prove our science had advanced.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gold was synthesized from mercury by neutron bombardment in 1941.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12109" LastActivityDate="2015-07-31T02:39:30.787" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18507" PostTypeId="1" AcceptedAnswerId="18557" CreationDate="2015-07-31T04:01:16.193" Score="15" ViewCount="1056" Body="&lt;p&gt;The following is an extension of &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18445/100-dwarves-and-a-tiny-room&quot;&gt;100 Dwarves and a tiny room&lt;/a&gt;.  The text of this puzzles spoils that one in a minor way, so go read that excellent puzzle first!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;An evil elf puts $100$ dwarves in a room and gives each a shirt with a number from $0$ to $99$ (repetitions allowed).  The dwarves can see everyone's number except their own, but otherwise can't communicate in any way.  Their goal is to all simultaneously shout out the same number, and for this shouted number to appear on the shirt of at least one dwarf.  Thus, they lose if more than one number is shouted at that instant, or if the single number shouted is not on any dwarf's shirt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is up to you to decide their strategy:  in particular, you will give instructions to each dwarf, and the dwarf will follow your instructions exactly.  However, the evil elf will also read these instructions, and assign shirts to minimize the probability of the dwarves' success.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What strategy would you give them, and what is the probability they win?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A few notes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;In the room there is a large screen, which all the dwarves can see, that can display whatever random number(s) your instructions require.  These random number(s) are chosen after the shirts are distributed.  This is to help the dwarves implement some sort of strategy involving randomness.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The only major difference between this puzzle and the original is that I'm asking not about whether a 100% win rate is possible, but how high you can make the win percentage.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For our purposes, if a dwarf has a shirt with a 29 on it, this doesn't match a shouted answer of &quot;9&quot; or &quot;2&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I don't know the answer;  I'll accept the answer with highest probability after a reasonable amount of time.  I do know it is possible to do better than 1/100.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2722" LastEditorUserId="69582" LastEditDate="2021-07-21T18:50:12.940" LastActivityDate="2021-07-21T18:50:12.940" Title="100 Dwarves vs. the Evil Elf" Tags="|mathematics|strategy|optimization|probability|hat-guessing|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="18508" PostTypeId="2" ParentId="18363" CreationDate="2015-07-31T04:38:55.833" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;7) T, T, C, P, Q, ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Teaspoon, Tablespoon, Cup, Pint, Quart, Gallon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did come up with this on my own, and posted before I saw @Bailey M.s comment.&lt;/p&gt;&#xA;" OwnerDisplayName="user5521" LastActivityDate="2015-07-31T04:38:55.833" CommentCount="0" CommunityOwnedDate="2015-07-31T04:38:55.833" ContentLicense="CC BY-SA 3.0" />
  <row Id="18509" PostTypeId="1" AcceptedAnswerId="18511" CreationDate="2015-07-31T04:41:51.403" Score="142" ViewCount="8245" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Hearken, now, and listen close; I have so much to tell.&lt;br&gt;&#xA;  Now lend an ear and do your job; you surely will excel.&lt;br&gt;&#xA;  To find my hidden message, gaze on every word herein.&lt;br&gt;&#xA;  While skimming through it might be fast, you'll miss the quiet gem.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;How vitally important are the details of this verse;&lt;br&gt;&#xA;  examining them closely is the task to take on first.&lt;br&gt;&#xA;  And once they've been digested, one may feel the need to snooze;&lt;br&gt;&#xA;  now please reject that feeling and keep minding p's and q's.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Time and time again, you need to squelch the urge to rush.&lt;br&gt;&#xA;  I know that if you get too lax, you'll miss some clues and such.&lt;br&gt;&#xA;  So look at each word carefully; just read them one-by-one.&lt;br&gt;&#xA;  My friend, if every word's perused, your work is almost done!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I know that all this vigilance may start to vex or chafe.&lt;br&gt;&#xA;  So here's a hint to help you out and keep you feeling safe:&lt;br&gt;&#xA;  Some people say that punctuation rules are absolute.&lt;br&gt;&#xA;  Ill-use of them is abject and quite low, there's no dispute.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now, as you keep perusing, you may find a common theme.&lt;br&gt;&#xA;  Getting down and dirty is the answer, it may seem.&lt;br&gt;&#xA;  Find every hidden message in the fine points of the text.&lt;br&gt;&#xA;  Rush to do your job, and quiz specific aspects next.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Oh, lazy folk will squawk at grammar law misapplication,&lt;br&gt;&#xA;  giving strong opinions of strict laws of punctuation.&lt;br&gt;&#xA;  Many nix all joy in words, subscribing to this faction;&lt;br&gt;&#xA;  in point of fact you might say it's a capital infraction.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Every word's important in this lengthy, rambling prose.&lt;br&gt;&#xA;  Altering just one could lose the message that's enclosed;&lt;br&gt;&#xA;  please examine hard and quick the verbiage herein.&lt;br&gt;&#xA;  'Cause every word's important to the puzzle that it's in. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Here's the quest that's put before you, puzzler and chum:&lt;br&gt;&#xA;  locate the mystery unknown; to languor don't succumb.&lt;br&gt;&#xA;  Search through all the text to see what's shrouded here among&lt;br&gt;&#xA;  the words you may enjoy or may prefer to leave unsung.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Thus far, you'll think not much was said about this work's solution;&lt;br&gt;&#xA;  so you'll obtain it straight right now, without any dilution:&lt;br&gt;&#xA;  you want a pair of words portraying you, my puzzling buddy!&lt;br&gt;&#xA;  And no quick fix can find it!  No, just vim and lots of study!   &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Zest is what you'll have to have as these two words you seek;&lt;br&gt;&#xA;  as said before, the phrase describes a quirky puzzle geek;&lt;br&gt;&#xA;  just make a guess, go check your work; you'll see if it is right;&lt;br&gt;&#xA;  for after all, it's sure that you'll excel; you're very bright!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;As fin'lly this verbose creation's coming to a close,&lt;br&gt;&#xA;  please acquiesce to take my thanks for suff'ring through my prose --&lt;br&gt;&#xA;  my urge to wax poetic may be tiresome to some,&lt;br&gt;&#xA;  but still, it's rather jazzy, if your brain is not yet numb. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;What two-word phrase is hidden in the poem above?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="4421" LastEditDate="2015-07-31T14:56:35.987" LastActivityDate="2019-08-28T23:19:38.363" Title="Hearken, now, and listen close!" Tags="|riddle|rhyme|wordplay|english|language|" AnswerCount="2" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="18510" PostTypeId="2" ParentId="18507" CreationDate="2015-07-31T05:19:43.670" Score="8" Body="&lt;p&gt;Without the evil elf, there are a couple sensible strategies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Strategy A: Choose a random number from 0 to 99. Everyone shouts that number.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the shirts were assigned at random, this would have a 1-(99/100)^100 chance of succeeding, or about 63%. Unfortunately, the elf will assign the same number to every dwarf for a 1/100 chance of winning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Strategy B: Choose a random ordering of the numbers from 0 to 99. Every dwarf shouts the first number in the ordering that they see on at least one shirt.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This succeeds if out of all the represented numbers, the first one in the ordering occurs at least twice. Of course, the evil elf can assign all different shirts to guarantee that this fails.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the counter-strategy for each strategy gives a 100% chance of success for the other strategy! What about this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Strategy AB: Choose Strategy A or Strategy B with equal probability.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best the elf can do is to assign 91 identical shirts and 9 different ones, for a 10% chance of success. We're not done yet:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Strategy C: Choose a random ordering of the numbers from 0 to 99. Every dwarf shouts the first number in the ordering that they see on at least two shirts. If they do not see any number on two shirts, they shout the first number in the ordering.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This succeeds whenever out of all the numbers on at least two shirts, the first one in the ordering occurs on at least three. If no numbers occur on at least two shirts, it is equivalent to Strategy A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a mixed strategy between A, B, and C that does better than 1/10. This one is the best:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Strategy ABC: Choose Strategy A with probability 250/741, Strategy B with probability 253/741, and Strategy C with probability 238/741.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The elf has three equally good strategies here, which all have a success rate of &lt;strong&gt;1601/7410 (21.6%)&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strategies A, B, and C are all special cases of a general strategy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Strategy N: Choose a random ordering of the numbers 0 to 99. Every dwarf shouts the first number they see on at least N shirts. If they do not see such a number, their action will be specified later.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using a mix of these strategies, I think there is a hard upper bound at 51%, but it might be possible to reach 43%.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-08-01T16:44:55.310" LastActivityDate="2015-08-01T16:44:55.310" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18511" PostTypeId="2" ParentId="18509" CreationDate="2015-07-31T05:39:36.107" Score="71" Body="&lt;p&gt;The two word phrase hidden quite cleverly in this long passage is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;PUZZLE FIEND&lt;/strong&gt;.&lt;br&gt;&#xA;&lt;br&gt;&#xA;  Which I reckon I am! :-D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This word was hidden in this puzzle in the following way:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In each stanza of four lines, all the letters of the alphabet are present, except one. In the first, &lt;code&gt;P&lt;/code&gt; is missing. In the second, &lt;code&gt;U&lt;/code&gt; is missing. Continuing to determine each missing letter gave me the string &lt;code&gt;PUZZLEFIEND&lt;/code&gt;. It was easy to figure out the phrase &lt;code&gt;PUZZLE FIEND&lt;/code&gt; from that.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The lines that gave it away were &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the third stanza:&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;em&gt;So look at each word carefully; just read them one-by-one.&lt;br&gt;&#xA; My friend, if every word's perused, your work is almost done!&lt;/em&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; And in the seventh stanza:&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;em&gt;Every word's important in this lengthy, rambling prose.&lt;br&gt;&#xA; Altering just one could lose the message that's enclosed;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An additional hint that I missed. Thanks Sabre for finding the clue that was craftily embedded into the puzzle &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &quot;capital infraction&quot; phrase in the puzzle was a big giveaway. All the capitalized letters in the puzzle spell out &lt;code&gt;HINT WHAT IS MISSING FROM EACH STANZA&lt;/code&gt;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="4421" LastEditDate="2016-04-28T14:47:14.140" LastActivityDate="2016-04-28T14:47:14.140" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18512" PostTypeId="1" AcceptedAnswerId="18516" CreationDate="2015-07-31T07:33:24.700" Score="4" ViewCount="596" Body="&lt;p&gt;This is part of &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=title%3A%22The+Twenty+Doors%22&quot;&gt;The Twenty Doors&lt;/a&gt; series.&lt;br&gt;&#xA;The previous one is &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18498/the-twenty-doors-room-8&quot;&gt;The Twenty Doors (ROOM 8)&lt;/a&gt;&lt;br&gt;&#xA;The next one is &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18532/the-twenty-doors-room-10&quot;&gt;The Twenty Doors (ROOM 10)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;When you go into Room 9, there is a strange hint, signed by 'Dr. X':&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Duplicates, go forth! The hint key is keyed with a key of key! ovnr wPraor o aZcdsua fretsfdes&lt;br&gt;- Dr. X&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is a rather confusing hint:&lt;br&gt;&#xA;&lt;code&gt;The hint key is keyed with a key of key! Sfb lkgm ygofbq vkr trba ex sfb Dbqlkmr gm VVG. Sfks'r sfb nmjx fgms xnt'qb dbssgmd ex abygofbqgmd sfgr rsqgmd!&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, last, the keypad:&#xA;&lt;code&gt;&#xA;[1][2][3][4][5]&#xA;    [ENTER]&#xA;&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next door will be added when this door is solved!&lt;/p&gt;&#xA;" OwnerDisplayName="user9377" LastEditorUserId="-1" LastEditorDisplayName="user9377" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-07-31T18:18:13.667" Title="The Twenty Doors (ROOM 9)" Tags="|cipher|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18513" PostTypeId="2" ParentId="18418" CreationDate="2015-07-31T08:24:04.923" Score="3" Body="&lt;p&gt;I'm going with this: a sentence works if&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; writing it out in lower case, there are no letters with &lt;a href=&quot;https://en.wikipedia.org/wiki/Ascender_(typography)&quot; rel=&quot;nofollow&quot;&gt;ascenders&lt;/a&gt; (like l, t, h, k).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's one little thing that still bothers me though, and it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;musician's piece&quot;, which doesn't technically have any ascenders, but I guess that an apostrophe is still something above the mean line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5990" LastActivityDate="2015-07-31T08:24:04.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18514" PostTypeId="2" ParentId="18505" CreationDate="2015-07-31T08:59:03.033" Score="11" Body="&lt;p&gt;Whatever Hyphen Four is saying, it only takes the digits 1 to 9. There are only nine possible actions, so each digit probably represents an action. A keyboard's number pad neatly identifies which is which:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iDnYR.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iDnYR.png&quot; alt=&quot;Number pad&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8246 are up, down, left, and right respectively. 7 (&quot;home&quot;) teleports to the start, 1 (&quot;end&quot;) teleports to the end, 3 (&quot;page down&quot;) teleports down a level, and 9 (&quot;page up&quot;) teleports up a level. 5 doesn't have any action, so it represents waiting. The path is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A2,B2,N,E6,L,T,H7,P,L,S,A2,A3,N,E6,F6,N,I8,I7&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-07-31T15:48:02.990" LastActivityDate="2015-07-31T15:48:02.990" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18516" PostTypeId="2" ParentId="18512" CreationDate="2015-07-31T10:18:04.953" Score="3" Body="&lt;p&gt;The hint on the wall &lt;em&gt;Sfb lkgm ygofbq vkr trba ex sfb Dbqlkmr gm VVG. Sfks'r sfb nmjx fgms xnt'qb dbssgmd ex abygofbqgmd sfgr rsqgmd&lt;/em&gt; is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;The main cipher was used by the Germans in WWI. That's the only hint you're getting by deciphering this string&lt;/strong&gt; using Keyed Caesar with key &lt;strong&gt;KEY&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;ovnr wPraor o aZcdsua fretsfdes&lt;/em&gt; decrypted is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Press four to advance forwards&lt;/strong&gt; using the UBCHI cipher with key &lt;strong&gt;KEYKEY&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To get to the next room, we must press:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;4&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12731" LastActivityDate="2015-07-31T10:18:04.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18518" PostTypeId="1" AcceptedAnswerId="18523" CreationDate="2015-07-31T10:38:34.077" Score="17" ViewCount="2854" Body="&lt;p&gt;I'm new to puzzling and have just dropped by. Here's a puzzle I made up. I have no idea if it is hard or easy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given to the Greeks by the Egyptians, it provides no answer for Buddhists seeking enlightenment yet many schoolchildren find it easily.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="12700" LastEditDate="2015-07-31T10:47:43.300" LastActivityDate="2015-08-02T13:29:14.957" Title="Do I hear a cat?" Tags="|riddle|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18521" PostTypeId="1" CreationDate="2015-07-31T12:12:37.040" Score="1" ViewCount="4434" Body="&lt;p&gt;I was given riddle and have not been able to solve. Can anyone help me please?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I'm an executive producer.&lt;/li&gt;&#xA;&lt;li&gt;I love gold shoes.&lt;/li&gt;&#xA;&lt;li&gt;I was born in Kansas.&lt;/li&gt;&#xA;&lt;li&gt;I'm well traveled.&lt;/li&gt;&#xA;&lt;li&gt;I'm a designer. (not clothing)&lt;/li&gt;&#xA;&lt;li&gt;I have &quot;Happy Feet&quot;.&lt;/li&gt;&#xA;&lt;li&gt;I'm a natural blonde.&lt;/li&gt;&#xA;&lt;li&gt;I've been on TV. &lt;/li&gt;&#xA;&lt;li&gt;My middle name is Joyce.&lt;/li&gt;&#xA;&lt;li&gt;I was born in the early 40's.&lt;/li&gt;&#xA;&lt;li&gt;It's a Female&lt;/li&gt;&#xA;&lt;li&gt;She has a show geared towards a specific genre.&lt;/li&gt;&#xA;&lt;li&gt;She has videos on You Tube.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;" OwnerUserId="14343" LastEditorUserId="15544" LastEditDate="2015-09-15T15:50:28.690" LastActivityDate="2015-09-15T15:50:28.690" Title="Identify this person with 13 clues given" Tags="|riddle|knowledge|" AnswerCount="2" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="18522" PostTypeId="2" ParentId="18424" CreationDate="2015-07-31T12:54:42.947" Score="7" Body="&lt;p&gt;This is intended to be a comment to the correct answer, but I lack the reputation to comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the unavoidable honor...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Saturn is one of seven gods in the union of all pantheons who has a day of the week named after him.  The passage of time is unavoidable -- there's always another Saturday coming up.  In addition, it can be argued that Saturn got the best of the honors, as I am sure it would receive the most votes in a 'best day of the week poll'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And for the gift... &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In Holst's The Planets, Jupiter is the bringer of jollity, and Saturn the bringer of old age.  The former is what is referenced in the poem by 'perk', and the latter the gift that is both embraced and feared.&lt;br&gt;&lt;br&gt;&#xA; In an amusing coincidence, I came across the following book, &lt;a href=&quot;http://serendip.brynmawr.edu/serendipia/Serendipia-Preti.html&quot;&gt;&lt;strong&gt;The Gift of Saturn: Creativity and Psychopathology&lt;/strong&gt;&lt;/a&gt;, and it could certainly be argued that creativity is embraced and psychopathology feared. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think this completes the riddle. Maybe I get some points so that I can comment in the future. ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Why did the spoiler markup work for the first block but not the second and third?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Got the name of the book wrong...&lt;/p&gt;&#xA;" OwnerUserId="11185" LastEditorUserId="13489" LastEditDate="2015-07-31T13:13:39.887" LastActivityDate="2015-07-31T13:13:39.887" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18523" PostTypeId="2" ParentId="18518" CreationDate="2015-07-31T13:20:00.753" Score="32" Body="&lt;p&gt;The answer is…&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MU/MEW.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; μ is a letter imported to Greek, apparently from the Egyptian hieroglyph for water.&lt;br&gt;&#xA; It's also an answer used to signify in Buddhism that a question should be &quot;unasked&quot;: that neither a positive nor a negative answer is meaningful, and that the act of posing the question has indicated misunderstanding.&lt;br&gt;&#xA; Additionally, it's the sound that a cat makes.&lt;br&gt;&#xA; Many schoolchildren will find μ when they work in physics/maths (especially when friction is encountered)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14305" LastEditorUserId="14305" LastEditDate="2015-08-01T18:29:27.970" LastActivityDate="2015-08-01T18:29:27.970" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18525" PostTypeId="2" ParentId="18507" CreationDate="2015-07-31T14:58:30.863" Score="3" Body="&lt;p&gt;&lt;strong&gt;Update: I missed the &quot;large screen&quot; part.  This answer assumes no such screen.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lower bound for winning is &quot;Every dwarf choose a number at random&quot;. There is a $1000^{-1000}$ chance of all dwarves choosing the same number.  The best the elf can do is assign everyone the same number, which leaves you with a total chance of winning of $1000^{-1001}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best I can come up with is: &lt;strong&gt;Find the lowest number you see.  Guess that with probability $p$.  Otherwise, guess randomly from among the lower numbers.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the lowest number shows up on two shirts, every dwarf sees the same lowest number, and you with with probability $p^{1000}$.  If the lowest number shows up on only 1 shirt, then 999 dwarves see the same lowest number, and 1 dwarf sees a lowest number $n$, which is the next lowest number.  In this case, you win with probability&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P = p^{999}(1-p){1\over n}$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is generally less, so we can assume the elf will choose this option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The optimal value of $p$ does not depend on $n$: a bit of calculus yields $p = {999\over 1000}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best the elf can do is assign the number 999 to 999 of the dwarves and some other number at random to the other one.  This yields a probability of winning of: $P = 3.6 \times 10^{-7}$, which is about 1 in $e$ million.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think you can do better.  The elf can ensure that at least 1 dwarf will need to guess a number he can't see.  This strategy ensures that at most 1 dwarf will have to pick a number he can't see and maximizes the odds of him picking the number everyone else should pick.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2015-07-31T14:58:30.863" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18527" PostTypeId="1" AcceptedAnswerId="18533" CreationDate="2015-07-31T16:12:35.003" Score="14" ViewCount="1574" Body="&lt;p&gt;You have 8 identical looking rods, 2 of which are radioactive, though you don't know which. You also have a box which can test for radioactivity, which can hold any number of rods. After the test, the box will indicate whether or not at least one of the rods inside were radioactive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the fewest number of tests it takes to determine the radioactive pair?&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="463" LastEditDate="2015-07-31T16:14:09.020" LastActivityDate="2020-08-31T15:04:31.940" Title="Radioactive Rods" Tags="|weighing|algorithm|" AnswerCount="6" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18528" PostTypeId="2" ParentId="18527" CreationDate="2015-07-31T16:22:22.440" Score="11" Body="&lt;p&gt;Let's get a baseline answer: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 6 tests&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Split the rods into groups of 2, test them (4).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If only one group is hot, you're good.&lt;br&gt;&#xA;Otherwise, you have 2 groups of 2, 1 each which is radioactive. Test one of each. (2)&lt;br&gt;&#xA;If hot, found; otherwise the other one is hot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&#xA;For a more generalized answer, I think the worst case is $2 * \log_2{N}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Split the group in half, check both groups.&lt;br&gt;&#xA;If one each, do binary search in each.&lt;br&gt;&#xA;Otherwise, split the hot half, checking both groups.&lt;br&gt;&#xA;Repeat. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-07-31T18:27:52.327" LastActivityDate="2015-07-31T18:27:52.327" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18529" PostTypeId="2" ParentId="18527" CreationDate="2015-07-31T17:18:44.153" Score="2" Body="&lt;p&gt;Here's another way to get&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 6 tests&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First, split the rods into groups of 4 and test them (2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If both groups have a radioactive rod, do the following for both groups: take two of the rods and test them. If the test is positive, check one of those two. If the test of two rods was negative, test one of the other two rods. If the check of the single rod is negative, it was the other rod in that pair. If it is positive, obviously you've found the radioactive rod. This takes 4 tests to check both groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If only one group has radioactive rods, split the group in two. If the first pair you check is not radioactive, it is the remaining two rods. Otherwise, test one of the two by itself - if it is not radioactive, you know the other one is and the second radioactive rod is in the other pair and can determine which with one more test. If the one you check is radioactive, test the other one. If it is not radioactive, the radioactive rod is in the other pair and can be found with one more test. The worst case scenario is 4 tests.&lt;/p&gt;&#xA;" OwnerUserId="1955" LastActivityDate="2015-07-31T17:18:44.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18530" PostTypeId="2" ParentId="18527" CreationDate="2015-07-31T17:33:44.500" Score="2" Body="&lt;p&gt;I guess I'll just add another answer for:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 6 tests in a worst-case-scenario&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But for a lowest average number of tests over many scenarios:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Label your rods A,B,C,D,E,F,G,H&lt;br&gt;&#xA;To make use of all possible outcomes, test two equal groups, but leave two behind, so...&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Test A,B,C and D,E,F (2 tests)&lt;br&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;if neither --&gt; G and H are radioactive &lt;strong&gt;(2 total and done)&lt;/strong&gt;&lt;br&gt;&lt;/li&gt;&#xA;&lt;li&gt;if only one group --&gt; Go to Step 2. (Thanks, Aggie Kidd)&lt;br&gt;&lt;/li&gt;&#xA;&lt;li&gt;if both --&gt; Go to Step 3.&lt;br&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;P(done after step 1) = 1/28&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Test A,B,C (or D,E,F) and G,H one at a time&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Keep testing (starting with G,H until two have been found or all but one is tested &lt;strong&gt;(4-6 tests)&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;P(done at step 2) = P(both in same trio) + P(trio and G,H) = 3/14 + 6/14 = 9/14&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;Test A and D (2 tests) (this works testing A,B and D,E too)&lt;br&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;if both --&gt; A and D &lt;strong&gt;(4 total)&lt;/strong&gt;&lt;br&gt;&lt;/li&gt;&#xA;&lt;li&gt;if A, but not D --&gt; Test E --&gt; then A and E or F &lt;strong&gt;(5 total)&lt;/strong&gt;&lt;br&gt;&lt;/li&gt;&#xA;&lt;li&gt;if D, but not A --&gt; Test B --&gt; then D and B or C &lt;strong&gt;(5 total)&lt;/strong&gt;&lt;br&gt;&lt;/li&gt;&#xA;&lt;li&gt;if neither --&gt; Test B and E and deduce &lt;strong&gt;(6 total)&lt;/strong&gt;&lt;br&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;P(done at step 3) = P(one in each trio) = 9/28&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't worked through the probabilities/combinatorics completely, but this seems to have a lower average number of tests. If you had to stick with an algorithm from the beginning, this technically works better than some other algorithms, though the worst-case-scenario will still be 6 tests.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd be really interested to see someone &lt;em&gt;prove&lt;/em&gt; that their answer is correct, which would involve calculating the probabilities of each test scenario to find the lowest average test number via Monte Carlo. If someone wants to do that, I'll owe them a lifetime's reputation.&lt;/p&gt;&#xA;" OwnerUserId="13237" LastEditorUserId="13237" LastEditDate="2015-07-31T18:47:41.980" LastActivityDate="2015-07-31T18:47:41.980" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18532" PostTypeId="1" AcceptedAnswerId="20108" CreationDate="2015-07-31T18:14:26.537" Score="4" ViewCount="1614" Body="&lt;p&gt;This is part of &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=title%3A%22The+Twenty+Doors%22&quot;&gt;The Twenty Doors&lt;/a&gt; series.&lt;br&gt;&#xA;The previous one is &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18512/the-twenty-doors-room-9&quot;&gt;The Twenty Doors (ROOM 9)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There is some paper taped to the door in Room 10:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Well done! You are now halfway! To mark this event, this cipher is EXTRA hard!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;-&amp;nbsp;???&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You then look at the main paper&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;PART ONE&lt;/strong&gt;&lt;br&gt;&lt;br&gt;To decrypt the first cipher, I've given you a mathematical equation: &lt;br&gt;$d(x) = a-1(x - b) \mod\ m$&lt;br&gt;a is 11, b is 25, and d is the decrypted text.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Izea ahx lprp z cjrqg hjay mx rdrp, kla ja pazeap hjay nexmp.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;mmking solves it without dying, so you get to Part 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;PART TWO&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Everyone loves a mime. MDEwMDEwMDAwMTAwMDAxMDEwMTAwMDAxMDAwMTExMDAxMDE= The other key is key. Wmb wmmuh, drh imqfo tdvw ubonbv lcfkmod rr d vdqirp vrzugb isu bbd jbqbvdwmsr, fqh nv kmvwxfooc xrgudgbdgoc. Jr wr qc kbgwnwb, kmwmx pc rdpb. Wmbub, csx empo imqi wlb frwkbv.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PART TWO - HINT ONE&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'My name' refers to something recent in The Twenty Doors series, but not this part. Go back through each one, see if you can find something useful...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As requested, here is the contents of the site page.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My name again. J2I. Pga torooapr fpi Rlis Mcrcl ym otmplx tv Gsic 10. Iepzve aud hb, cap pbqi hb dnui ub. Befud. I hate vowles, exept for i.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PART THREE&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CodeNewbie is still working on getting at the password, but here is the ciphertext:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Irrt sthcgma&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, like magic, a keyboard appears, with the letters A to Z on it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PART 3 - HINT 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;!  Spinny! @HHF G L\NFHP301 3@WDXECT R1T0||0T GTFDHPSRX5R 3ARHCU/  £A  NT S0!&#xA;   If you can be bothered to decode that, you've then got to work out my L3etspeak. :) It's Markdown O'clock!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PART 3 - HINT 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  L&lt;strong&gt;3&lt;/strong&gt;etspeak is usually written with two three's (L33tspeak). So, what could that three mean???&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are going to be three separate parts to Room 10, thus it is a semi-interactive puzzle. The next part will be added when this door is solved!&lt;/p&gt;&#xA;" OwnerDisplayName="user9377" LastEditorUserId="-1" LastEditorDisplayName="user9377" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-09-07T05:49:01.370" Title="The Twenty Doors (ROOM 10)" Tags="|cipher|semi-interactive-puzzle|" AnswerCount="5" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="18533" PostTypeId="2" ParentId="18527" CreationDate="2015-07-31T18:38:07.960" Score="28" Body="&lt;p&gt;In the worst-case scenario, it requires&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; six tests&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;to locate the radioactive rods. Several answers already describe strategies for locating the radioactive rods. I will give another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Testing strategy&lt;/strong&gt;:  Start by testing two rods. If neither of these rods is radioactive, use the five remaining tests on five of the six remaining rods, one at a time. If two of these rods are radioactive, we are done. If only one is radioactive, then the single untested rod must also be radioactive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the first test indicates at least one of the two rods is radioactive, then divide the remaining six rods into two groups of three, and test both groups. If neither group contains a radioactive rod, then the initial two rods tested must both be radioactive. If one of the two groups of three is radiactive, we will use our final three test on a single rod each, one rod from the initial group of two, and two rods from the group of three containing a radioactive rod. Each of these groups has exactly one radioactive rod, so testing all but one of the rods (one at a time) is enough to determine which rod in each group is radioactive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof that this cannot be done with fewer tests&lt;/strong&gt;:&#xA;Suppose we have a strategy that is guaranteed to find the radioactive rods with five tests. The first test will involve $k$ rods. Initially, there are ${8\choose 2}=28$ possibilities for the pair of radioactive rods. After the first measurement, there will be either $28-{8-k\choose 2}$ possibilities (if at least one of the tested rods is radioactive), or ${8-k\choose 2}$ possibilities (if none of the tested rods is radioactive). Only four tests will then remain, so we can only distinguish between $2^4=16$ possibilities. This means we need $28-{8-k\choose 2}\leq 16$ and ${8-k\choose 2}\leq 16$. Direct computation shows that this is only satisfied for $k=2$, so our strategy must start by testing exactly two rods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that neither of the two rods we test first is radioactive. Our next test will involve $j$ of the remaining rods. At this point, there are ${6\choose 2}=15$ possibilities for the pair of radioactive rods. After the second test, the number of possibilities will either be $15-{6-j\choose 2}$ or ${6-j\choose 2}$. After this test, we will only have three tests remaining, so we need $15-{6-j\choose 2}\leq 8$ and ${6-j\choose 2}\leq 8$. There is no value of $j$ for which both of these inequalities hold.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-08-01T13:38:36.150" LastActivityDate="2015-08-01T13:38:36.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18534" PostTypeId="2" ParentId="18532" CreationDate="2015-07-31T18:38:34.640" Score="3" Body="&lt;p&gt;&lt;strong&gt;Part one&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using an&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; affine cipher with a = 11 and b = 25,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the text decodes to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Part two uses a field with no eyes, but it starts with grons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So part two will use a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; grons + field - i = gronsfeld cipher&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part two&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Converting &lt;strong&gt;MDEwMDEwMDAwMTAwMDAxMDEwMTAwMDAxMDAwMTExMDAxMDE=&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; from base 64 to ascii&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;gives you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 01001000010000101010000100011100101&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using this as the cipher key and &quot;key&quot; as the alphabet key, you get&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The third, and final part relies heavily on a random source for key generation, and is virtually uncrackable. Go to my website, visit my name. There, you will find the answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/a/18740/12109&quot;&gt;CodeNewbie&lt;/a&gt; got the next piece. Looking at the website and decoding the string of text found there, you get&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The password for Part Three is hidden in Room 10. Search all up, not down to find it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Looking up at the question and searching in the &lt;a href=&quot;https://puzzling.stackexchange.com/posts/18532/edit&quot;&gt;markdown&lt;/a&gt;, there's a line commented out that says&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ignore this line&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not sure of its significance however...&lt;/p&gt;&#xA;" OwnerUserId="12109" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-08-16T15:52:46.157" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="18536" PostTypeId="2" ParentId="18521" CreationDate="2015-07-31T18:55:56.737" Score="2" Body="&lt;p&gt;Okay, I think I might be onto something, but I haven't dug up answers to all the clues. I think @LogicianWithAHat was on the right track with different clues representing different roles. Here's what I have so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Amy Adams&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The clues (most are &quot;[role] in [movie]&quot;):&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm an executive producer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I forgot which role I had here&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I love gold shoes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Sydney Prosser in &lt;em&gt;American Hustle&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was born in Kansas.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Lois Lane in &lt;em&gt;Man of Steel&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm well traveled.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Amelia Earhart in &lt;em&gt;Night at the Museum: Battle of the Smithsonian&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm a designer. (not clothing)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Amy in &lt;em&gt;Her&lt;/em&gt; (she's a video game designer)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have &quot;Happy Feet&quot;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Perhaps Giselle in &lt;em&gt;Enchanted&lt;/em&gt;? She does a lot of dancing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm a natural blonde.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Not sure&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've been on TV.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;She's been on multiple TV shows, including &lt;em&gt;The Office&lt;/em&gt; and, interestingly, &lt;em&gt;Smallville&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My middle name is Joyce.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Not sure, trying to find a character she played with this last name.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was born in the early 40's.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Janis Joplin in an upcoming biopic (Janis Joplin was born in 1943)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's a Female&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Check&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;She has a show geared towards a specific genre.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Not sure&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;She has videos on You Tube.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;She does, but so do many celebrities.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So many of the roles fit that I think I'm probably right, but I'm having trouble filling in all the holes. I thought I'd figured out the executive producer clue, but I can't remember which role it was. Additional thoughts are welcome!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.imdb.com/name/nm0010736/&quot; rel=&quot;nofollow&quot;&gt;Here's her imdb if you want to go a-searching.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10861" LastActivityDate="2015-07-31T18:55:56.737" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18537" PostTypeId="2" ParentId="18532" CreationDate="2015-07-31T19:18:58.647" Score="1" Body="&lt;p&gt;&lt;strong&gt;Part 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Decoding the first message (MDEwMDEwMDAwMTAwMDAxMDEwMTAwMDAxMDAwMTExMDAxMDE=)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; with base 64&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 01001000010000101010000100011100101&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which is the cipher key for the Gronsfeld cipher (see &lt;a href=&quot;https://puzzling.stackexchange.com/a/18534/5359&quot;&gt;mmking answer&lt;/a&gt;). The other key is &quot;key&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the second message decoded is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The third, and final part relies heavily on a random source for key generation, and is virtually uncrackable. Go to my website, visit my name. There, you will find the answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5359" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-07-31T19:18:58.647" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18538" PostTypeId="2" ParentId="18521" CreationDate="2015-07-31T19:51:14.670" Score="1" Body="&lt;p&gt;Okay, new theory!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Elizabeth Banks&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.imdb.com/name/nm0006969/&quot; rel=&quot;nofollow&quot;&gt;IMDB&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clues that fit so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Executive Producer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Elizabeth Banks has 3 executive producer credits&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Gold shoes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Effie Trinket in &lt;em&gt;Hunger Games&lt;/em&gt; &lt;a href=&quot;https://www.pinterest.com/pin/188869778093095188/&quot; rel=&quot;nofollow&quot;&gt;https://www.pinterest.com/pin/188869778093095188/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Kansas&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Perhaps her debut, Vicki in &lt;em&gt;Surrender &lt;strong&gt;Dorothy&lt;/em&gt;&lt;/strong&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;TV show in a &quot;specific genre&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Wet Hot American Summer, currently running&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Youtube&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Has a youtube series called &quot;Ask A Badass&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Still working on the others, but the TV Show and Youtube clues fit way better here than for my other answer.&lt;/p&gt;&#xA;" OwnerUserId="10861" LastActivityDate="2015-07-31T19:51:14.670" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18540" PostTypeId="1" AcceptedAnswerId="18543" CreationDate="2015-07-31T20:56:55.967" Score="10" ViewCount="739" Body="&lt;p&gt;I began as a mild expression of surprise.&lt;br&gt;&#xA;At 2, I was a U.S. Army soldier.&lt;br&gt;&#xA;At 3, a revolutionary separation expert&lt;br&gt;&#xA;At 4, an Aussie child's toy&lt;br&gt;&#xA;At 6, a plant with yellow flowers&lt;br&gt;&#xA;At 7, more of the same&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I slept for a few &lt;strong&gt;and became what&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="463" LastActivityDate="2015-07-31T21:57:48.987" Title="What had I become?" Tags="|riddle|word|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18542" PostTypeId="2" ParentId="18540" CreationDate="2015-07-31T21:25:41.630" Score="4" Body="&lt;p&gt;I believe most of it is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;G&lt;/strong&gt;&quot;: as in GEE WILLICKERS BATMAN, credit to Kingrames (I may have missed the hint)&#xA; &lt;strong&gt;GI&lt;/strong&gt;: Military soldier &lt;br&gt;&#xA; &lt;strong&gt;GIN&lt;/strong&gt;: Cotton gin helped the industrial revolution and separates cotton from seed &lt;Br&gt;&#xA; GINK: ? &lt;br&gt;&#xA; &lt;strong&gt;GINKGO&lt;/strong&gt;: Chinese tree with yellow seeds &lt;br&gt;&#xA; &lt;strong&gt;GINKGOS&lt;/strong&gt;: Same deal&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;slept for 7 and became:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;GINKGOSMNP.INC&lt;/strong&gt; an HP command injection exploit!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13916" LastEditorUserId="13916" LastEditDate="2015-07-31T21:52:36.930" LastActivityDate="2015-07-31T21:52:36.930" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18543" PostTypeId="2" ParentId="18540" CreationDate="2015-07-31T21:57:48.987" Score="17" Body="&lt;p&gt;Going hamateur was close, but not quite right.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;G = &quot;Gee&quot;  (thanks to Kingrames)&lt;br&gt;&#xA;GI = US Soldier (thanks to Going hamateur)&lt;br&gt;&#xA;GIN = Cotton gin separates cotton (thanks to Going hamateur)&lt;br&gt;&#xA;GING = Australian slang for a slingshot&lt;br&gt;&#xA;GINGER has yellow flowers&lt;br&gt;&#xA;GINGERS more of the same  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Final answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;GINGERSNAP = GINGERS plus NAP  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10861" LastActivityDate="2015-07-31T21:57:48.987" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18544" PostTypeId="2" ParentId="18518" CreationDate="2015-07-31T22:40:45.993" Score="6" Body="&lt;p&gt;Alternate answer (similar to Patrick Stevens):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mean or average&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter mu, often used to represent the average, is a Greek letter that's (apparently) borrowed from an Egyptian hieroglyph. Also, according to this article (&lt;a href=&quot;http://www.amstat.org/publications/jse/v11n1/bakker.html&quot; rel=&quot;nofollow&quot;&gt;http://www.amstat.org/publications/jse/v11n1/bakker.html&lt;/a&gt;) the concept of an average was used by the Egyptians to estimate how many years had passed. Since it's cited by Herodotus (Greek historian), maybe that's how Greeks developed the concept. The letter mu also sounds somewhat like a cat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation part 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Finding enlightenment in Buddhism means having infinite compassion and wisdom. If you're being mean, you cannot achieve enlightenment.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation part 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lots of schoolchildren all over the world find averages, though in many places (UK also, maybe?) they refer to them as the &quot;mean&quot;, as do statisticians and others.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastEditorUserId="1620" LastEditDate="2015-07-31T23:17:28.117" LastActivityDate="2015-07-31T23:17:28.117" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18545" PostTypeId="1" AcceptedAnswerId="18546" CreationDate="2015-08-01T06:01:30.663" Score="1" ViewCount="1190" Body="&lt;p&gt;I have one, you have one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you remove the first letter, a bit remains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you remove the second, bit still remains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After much trying, you might be able to remove the third one also, but it&#xA; remains. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It dies hard!&lt;/p&gt;&#xA;" OwnerUserId="2375" LastActivityDate="2015-08-01T06:05:49.537" Title="Can you solve this?" Tags="|riddle|word|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18546" PostTypeId="2" ParentId="18545" CreationDate="2015-08-01T06:05:36.433" Score="3" Body="&lt;p&gt;Is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Habit? Remove the first letter to get &quot;abit&quot;, the second to get &quot;bit&quot;, and the third to get &quot;it&quot;. A habit may fit the clue &quot;dies hard&quot; but possibly some of these other words could also: &lt;a href=&quot;http://crossword-solver.org/%3Fabit&quot; rel=&quot;nofollow&quot;&gt;http://crossword-solver.org/%3Fabit&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastActivityDate="2015-08-01T06:05:36.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18547" PostTypeId="2" ParentId="18545" CreationDate="2015-08-01T06:05:49.537" Score="3" Body="&lt;p&gt;You are talking about &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the word &lt;strong&gt;habit&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have one, you have one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We all have our own habits. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you remove the first letter, a bit remains.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Removing the 'h', you get the letters 'abit'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you remove the second, bit still remains.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Removing the second letter 'a', you get the letters 'bit'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After much trying, you might be able to remove the third one also, but it remains.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Removing the third letter 'b', you are left with 'it'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It dies hard!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's a popular saying that habits die hard.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-08-01T06:05:49.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18548" PostTypeId="2" ParentId="18371" CreationDate="2015-08-01T08:51:41.767" Score="13" Body="&lt;p&gt;I'm fairly sure the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1607852&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The method used is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are two of every shape in the image. If you imagine that these are the locations of the shape at times 1 and 2, then you can determine the position at time 3 (actually there are two possibilities depending on which way round you number the two shown shapes). By following the trajectory of all shapes and superimposing their end positions you get the answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sorry if I've got this wrong - I didn't have a computer or printer, so I've tried to solve it with pen and paper!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My working (first attempt):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/4Hocy.jpg&quot; width=&quot;500&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My working (second attempt and I've still made a small mistake somewhere):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/vv95e.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vv95e.png&quot; alt=&quot;Working from second attempt&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit by puzzle author: The following images show the parts and solution as colourized bitmaps:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/FyjjW.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FyjjW.png&quot; alt=&quot;PArts&quot;&gt;&lt;/a&gt; &lt;BR&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/HGVdV.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HGVdV.png&quot; alt=&quot;Solution&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="303" LastEditorUserId="5840" LastEditDate="2015-08-07T21:16:36.953" LastActivityDate="2015-08-07T21:16:36.953" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18549" PostTypeId="2" ParentId="18518" CreationDate="2015-08-01T16:23:36.017" Score="3" Body="&lt;p&gt;I'm not sure if I'm supposed to answer my own question but here was what I had in mind.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A cat's mew gives an audible clue to the answer which is MU.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Egyptians gave the MU (μ) character to the Greeks and, to this day, it is the 12th letter of the Greek alphabet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Buddhists in training who ask their master a question that does not warrant an answer may be answered with the Japanese word &quot;MU&quot; which has many possible interpretations but one of them is simply &quot;No answer.&quot; It also signifies the instruction to, &quot;Un-ask the question.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Children of age eleven may, in some educational systems be asked to add up a list of numbers and divide by how many there were. This gives them the average value called the arithmetic mean. The usual symbol for this is μ.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="14177" LastEditDate="2015-08-02T13:29:14.957" LastActivityDate="2015-08-02T13:29:14.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18550" PostTypeId="1" AcceptedAnswerId="18560" CreationDate="2015-08-01T17:43:55.017" Score="15" ViewCount="512" Body="&lt;p&gt;Here's another one where each line represents a character.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I began as an expression of warning.&lt;br&gt;&#xA;At 2, my opposite brought two scales together.&lt;br&gt;&#xA;At 3, saw the death of Epiphanius of Salamis&lt;br&gt;&#xA;At 4, was widely called weird&lt;br&gt;&#xA;At 5, grouped online with HDMI cables&lt;br&gt;&#xA;At 6, seen with Sigma-Aldrich's tannic acid&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After much inactivity, at 9, I was associated with a Saudi oil tanker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What had I become?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;em&gt;This puzzle is a bit more difficult than the first, and therefore, comes with a bonus of 200 rep points.&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-08-01T18:24:10.067" LastActivityDate="2015-08-03T17:53:29.360" Title="What had I become? (Part 2)" Tags="|riddle|" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18551" PostTypeId="1" AcceptedAnswerId="18554" CreationDate="2015-08-01T18:50:23.700" Score="11" ViewCount="854" Body="&lt;pre&gt;Old man Geppetto was feeling bereft;&#xA;It was only a week since young long-nose had left.&#xA;He’d be back in a month from his trip to Iraq,&#xA;But the old man just missed him and wished he was back.&#xA;&#xA;His Vulgarian penpal had sent him a letter.&#xA;He thought to himself “This will make me feel better.”&#xA;He opened it up with a smile on his face,&#xA;But it was clear from the outset his joy was misplaced.&#xA;&#xA;“My Dearest Geppetto”, the missive had started,&#xA;“I have tidings of woe that must now be imparted.&#xA;Though the Baron is banished and the children are free,&#xA;A new threat has risen for you and for me.&quot;&#xA;&#xA;&quot;Our future’s in danger but there’s naught to be done,&#xA;For this threat isn’t one that can be overcome.&#xA;I’ve have sent you some rhymes that will help you to know,&#xA;The name of our most unassailable foe.”&#xA;&#xA;Please help poor Geppetto decipher the clues&#xA;To the cause of the Toymaker’s worrying news?&#xA;The theme is traditional, but as you will spot&#xA;The riddle’s solution is certainly not.&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Two dozen turdus in a spot, a &lt;b&gt;regal treat&lt;/b&gt; made them hot&lt;/li&gt;&#xA;  &lt;li&gt;A &lt;b&gt;construction calamity&lt;/b&gt; what can we do? It’ll take bit more than&#xA;  paper and glue.&lt;/li&gt;&#xA;  &lt;li&gt;The &lt;b&gt;sound of sternutation&lt;/b&gt; twice before our ruination.&lt;/li&gt;&#xA;  &lt;li&gt;A surfeit of kin in an &lt;b&gt;unsuitable place&lt;/b&gt; for an overstretched&#xA;  mother with lines on her face.&lt;/li&gt;&#xA;  &lt;li&gt;&lt;b&gt;Lost by a trio&lt;/b&gt; when someone was scared, but these little athletes&#xA;  were already impaired.&lt;/li&gt;&#xA;  &lt;li&gt;I specialise in horticulture of a kind most strange now can you&#xA;  see &lt;b&gt;how all my beauties were arranged&lt;/b&gt;.&lt;/li&gt;&#xA;  &lt;li&gt;A &lt;b&gt;sedentary location&lt;/b&gt; for some self-congratulation&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="14135" LastEditDate="2015-08-03T12:18:05.557" LastActivityDate="2015-08-03T13:49:26.123" Title="A new threat to an old man" Tags="|riddle|word|rhyme|english|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18553" PostTypeId="2" ParentId="16902" CreationDate="2015-08-02T07:56:07.677" Score="2" Body="&lt;p&gt;The third if the three original heaps contains only normal balls.&#xA;Place these balls in a &quot;normal ball&quot; discard pile, but add either zero, one or two of these balls to each of the remaining heaps so that (n+x)/3 has a remainder of 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take one of these slightly increased original heaps and divide it into three smaller piles, each equal to (n+x-1)/3 with the single remainder in any of the piles. One of these piles must contain one heavy ball.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Re-weigh these and either 1. The heavy ball is in the largest pile, so the scales do not move, or 2. The heavy ball is in one of the smaller piles, so the other smaller pile rises. In case 1, both the small piles can be added to the discard pile. In case 2 (the worst case) the lighter of the smaller piles can be added to the discard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again top up the remaining balls from the discard pile to give a number divisible by 3 with a remainder of 1 and repeat the process until you get down to piles of size 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the heavy ball by weighing two normal balls in one pan against two pans each containing one of the unclassified balls. If nothing moves, then all balls are normal and can be discarded. If one pan moves, the other single pan contains the heavy ball.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeat these steps for the other original heap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the arbitrary weight of the heavier balls, I don't believe anything can be gained by combining any of the balls from the two unprocessed original heaps as there will always be the risk of discarding a k-sized pile containing the lighter of the two heavier balls in a comparison against a k-sized pile including the heavier of the two heavy balls and a k+1 sized pile of normal balls.&lt;/p&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="14135" LastEditDate="2015-08-02T17:30:57.820" LastActivityDate="2015-08-02T17:30:57.820" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18554" PostTypeId="2" ParentId="18551" CreationDate="2015-08-02T09:19:24.860" Score="9" Body="&lt;p&gt;Partial answer:&lt;br&gt;&#xA;Each rhyme is a:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;nursery rhyme  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In order:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - Sing a sing of sixpence (Four-and-twenty blackbirds, baked in a pie)&lt;br&gt;&#xA; - &lt;strike&gt;Humpty Dumpty&lt;/strike&gt; London Bridge is falling down (credit to CodeNewbie)&lt;br&gt;&#xA; - Ring a ring o' roses (sternutation is sneezing - &quot;atishoo, atishoo, we all fall down&quot; - originally about the black death/bubonic plague)&lt;br&gt;&#xA; - There was an old woman who lived in a shoe (I can't remember how this one goes, but I'm fairly sure she had a lot of children)&lt;br&gt;&#xA; - 3 blind mice (they all ran after the farmer's wife//who cut off their tails with a carving knife)&lt;br&gt;&#xA; - Mary, Mary, Quite Contrary (How does your garden grow?)&lt;br&gt;&#xA; - Little Jack Horner (sat in a corner, and said &quot;What a good boy am I!&quot;) (credit to dennis deems) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Making use of the highlighted sections:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - The regal treat is a &lt;strong&gt;pie&lt;/strong&gt;&lt;br&gt;&#xA; - The construction calamity is &lt;strong&gt;London Bridge&lt;/strong&gt; itself&lt;br&gt;&#xA; - The sound of sternutation is, onomatopoeically, &lt;strong&gt;atishoo&lt;/strong&gt;&lt;br&gt;&#xA; - The unsuitable place is a &lt;strong&gt;shoe&lt;/strong&gt;&lt;br&gt;&#xA; - The thing the trio lost was their &lt;strong&gt;tails&lt;/strong&gt; (cut off with a carving knife)&lt;br&gt;&#xA; - Mary's flowers (and pretty maids) were arranged all &lt;strong&gt;in a row, row, row&lt;/strong&gt;&lt;br&gt;&#xA; - And little Jack Horner's sedentary location was a &lt;strong&gt;corner&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Putting these all together, and taking their first letters, we find that the threat to Gepetto is  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;plastic&lt;/strong&gt; - modern mass production techniques and quick formation of molded figures are certainly a threat to his slow creation of hand-carved puppets&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11717" LastEditorUserId="11717" LastEditDate="2015-08-03T13:49:26.123" LastActivityDate="2015-08-03T13:49:26.123" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18555" PostTypeId="2" ParentId="18551" CreationDate="2015-08-02T09:59:58.090" Score="2" Body="&lt;p&gt;The last one is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Little Jack Horner&lt;br&gt;&#xA; Sedentary place: &quot;Little Jack Horner sat in a corner&quot;&lt;br&gt;&#xA; For some self-congratulation: And said 'What a good boy am I!'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4385" LastEditorUserId="4385" LastEditDate="2015-08-02T14:50:45.753" LastActivityDate="2015-08-02T14:50:45.753" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18557" PostTypeId="2" ParentId="18507" CreationDate="2015-08-02T16:52:00.070" Score="13" Body="&lt;p&gt;I will describe a lower bound and an upper bound on the probability of success the dwarves can guarantee. These bounds are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Lower bound&lt;/strong&gt;: There is a strategy for the dwarves giving a win probability of at least $\frac{100}{199}\approx 50.25\%$.&lt;br&gt;&lt;strong&gt;Upper bound&lt;/strong&gt;: I will argue that the dwarves cannot guarantee a win with probability greater than $\frac{100}{101}\approx 99.01\%$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lower bound&lt;/strong&gt;: Imagine that there is a stack of 100 shirts in the corner of the room, one of each number. Then each dwarf can see a total of 199 shirts. The dwarves can use the following strategy: a random integer $n$ between 1 and 199 (inclusive) is displayed on the screen, each dwarf lists the 199 shirt numbers they can see in order from smallest to largest, and each dwarf shouts out the $n$-th number on their list.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the large list of all 200 shirt numbers in the room. Each dwarf will either call out the $n$-th number on the large list or the the $(n+1)$-st, depending on where their own shirt number falls. There can be at most 99 values of $n$ for which the $n$-th entry and the $(n+1)$-st entry or the large list are different, so if the random value of $n$ selected is not one of these, then every dwarf will call out the same number, and additionally that number will appear at least twice on the large list, meaning at list one dwarf has that number on their shirt. This gives a probability of success of at least $\frac{100}{199}\approx 50.25\%$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Upper bound&lt;/strong&gt;: There is no strategy the dwarves can use to guarantee a probability of success greater than $\frac{100}{101}\approx 99.01\%$. This is true even if the elf is restricted to assigning shirts numbered either 0 or 1. Suppose the dwarves have chosen a strategy. For $k=0,1,\ldots,99$, let $p(k)$ be the probability that a given dwarf guesses 0, given that that dwarf sees exactly $k$ shirts labelled 0 (and therefore $99-k$ shirts labelled 1). For $k\geq 1$, if exactly $k$ shirts are labelled 0, then some dwarves see $k$ 0's and some dwarves see $k-1$ 0's, so the probability of both numbers being guessed must be at least $p(k)-p(k-1)$. If $k=0$ (all shirts are 1), then probability of failure (someone guesses 0) is at least $p(0)$. Similarly, if $k=99$ (all shirts 0), then probability of failure (someone guesses 1) is at least $1-p(99)$. By choosing $k$, the elf can choose the probability of failure to be any one of the following 101 numbers:&#xA;$$&#xA;p(0),\,\,p(1)-p(0),\,\,p(2)-p(1),\,\,\ldots,\,\, p(99)-p(98),\,\,1-p(99).&#xA;$$&#xA;These numbers sum to $1$, so there must be one of them that is at least $\frac{1}{101}$. This means that if the elf chooses the appropriate number of shirts to be labelled 0, the probability of failure is at least $\frac{1}{101}$, so the dwarves cannot guarantee victory with probability greater than $\frac{100}{101}$.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastActivityDate="2015-08-02T16:52:00.070" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18558" PostTypeId="1" AcceptedAnswerId="18559" CreationDate="2015-08-02T17:26:32.993" Score="32" ViewCount="764" Body="&lt;p&gt;I'm trying to remember the names of some old acquaintances. If I describe them to you, perhaps you can help me out by &lt;strong&gt;telling me their names.&lt;/strong&gt;&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;He was once an insufferable borrower, but he turned out to be quite the kisser.&lt;/li&gt;&#xA;&lt;li&gt;She was like a weasel in the water, and she could work wonders with clay.&lt;/li&gt;&#xA;&lt;li&gt;I never saw such a talented individual die so soon.&lt;/li&gt;&#xA;&lt;li&gt;The most detestable person I ever knew who didn't break the rules.&lt;/li&gt;&#xA;&lt;li&gt;She lived near the water because of her dandruff problem.&lt;/li&gt;&#xA;&lt;li&gt;Who knew such a thoughtful, logical man could betray his country?&lt;/li&gt;&#xA;&lt;li&gt;The man would curse his foes for days, but he never lacked for research funding.&lt;/li&gt;&#xA;&lt;li&gt;What can I say? He was constantly positive and he had quite the mane.&lt;/li&gt;&#xA;&lt;li&gt;She had an exceptional green thumb, but her life was constantly headed downwards.&lt;/li&gt;&#xA;&lt;li&gt;He's great during the winter, but never invite him to a poker game.&lt;/li&gt;&#xA;&lt;li&gt;When she isn't enraged, she's quite the sentinel of the forest.&lt;/li&gt;&#xA;&lt;li&gt;He was a jewel of a friend.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;Many thanks. Trying to remember has me exhausted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sincerely,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q. Weary&lt;/p&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="5373" LastEditDate="2018-06-02T15:03:25.313" LastActivityDate="2018-06-02T15:03:25.313" Title="Do you know S. Tack, or X. Change?" Tags="|word|wordplay|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18559" PostTypeId="2" ParentId="18558" CreationDate="2015-08-02T17:37:08.477" Score="21" Body="&lt;ol&gt;&#xA;&lt;li&gt;He was once an insufferable borrower, but he turned out to be quite the kisser.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; S. Moocher&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;She was like a weasel in the water, and she could work wonders with clay.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; P. Otter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;I never saw such a talented individual die so soon. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; S. Killed (Thanks JLee) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;The most detestable person I ever knew who didn't break the rules.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;del&gt;S. Tickler&lt;/del&gt; L. Awful (Thanks M Oehm)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;She lived near the water because of her dandruff problem.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; F. Lake&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;6&quot;&gt;&#xA;&lt;li&gt;Who knew such a thoughtful, logical man could betray his country?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; T. Reason (Thanks George Reith)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;7&quot;&gt;&#xA;&lt;li&gt;The man would curse his foes for days, but he never lacked for research funding.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; G. Rant&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;8&quot;&gt;&#xA;&lt;li&gt;What can I say? He was constantly positive and he had quite the mane.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; L. Ion&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;9&quot;&gt;&#xA;&lt;li&gt;She had an exceptional green thumb, but her life was constantly headed downwards.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; F. Lower&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;10&quot;&gt;&#xA;&lt;li&gt;He's great during the winter, but never invite him to a poker game.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C. Heat&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;11&quot;&gt;&#xA;&lt;li&gt;When she isn't enraged, she's quite the sentinel of the forest.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; R. Anger&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;12&quot;&gt;&#xA;&lt;li&gt;He was a jewel of a friend.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; O. Pal&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-08-02T19:04:57.707" LastActivityDate="2015-08-02T19:04:57.707" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18560" PostTypeId="2" ParentId="18550" CreationDate="2015-08-02T19:59:29.660" Score="20" Body="&lt;p&gt;This one took some research, but I think the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;403040000&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Explanation:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&quot;I began as an expression of warning.&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4, which sounds like &lt;a href=&quot;https://en.wikipedia.org/wiki/Fore_%28golf%29&quot;&gt;&quot;Fore!&quot;&lt;/a&gt; as in golf&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;At 2, my opposite brought two scales together.&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 40 - The opposite of 40 is -40.  -40 Celsius = -40 Fahrenheit, two temperature scales.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;At 3, saw the death of Epiphanius of Salamis&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Epiphanius of Salamis died in 403, so you could say that the year 403 saw his death.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;At 4, was widely called weird&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4030 is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Weird_number&quot;&gt;weird number&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;At 5, grouped online with HDMI cables&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Googling 40304 returns lots of  HDMI Cable links&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;At 6, seen with Sigma-Aldrich's tannic acid&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 403040 is the product number for Sigma-Aldrich tannic acid&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;After much inactivity, at 9, I was associated with a Saudi oil tanker.&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 403040000 After much inactivity (just more zeros), associated with a Saudi Oil tanker - 403040000 is the Martime Mobile Service Identity (MMSI Number) for an oil tanker from Saudi Arabia (called Fajr).  An MMSI starting with 403 designates a Saudi vessel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14368" LastEditorUserId="11475" LastEditDate="2015-08-03T14:29:31.337" LastActivityDate="2015-08-03T14:29:31.337" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18561" PostTypeId="2" ParentId="18527" CreationDate="2015-08-03T03:07:20.870" Score="0" Body="&lt;p&gt;Just because a test is successful doesn't mean we can eliminate the unchosen rods. Therefore, we always have to cover every group not yet confirmed to fail the test (if there aren't already 2 &quot;successful&quot; groups with no common rods), and their rods. The larger the groups that can be confirmed to fail are, the better. If the rods aren't divided equally, the smallest group might fail and the biggest ones succeed, so it doesn't give the best worst-case scenario.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After doing these, you'll know which quarter(s) has/have the radioactive rods. Even if they're two different quarters, you'll only need 2 more tests, which makes 6:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Take 4.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Take the other 4.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Take half of both groups.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Take the other halves.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="14016" LastActivityDate="2015-08-03T03:07:20.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18562" PostTypeId="1" CreationDate="2015-08-03T12:39:17.137" Score="5" ViewCount="517" Body="&lt;p&gt;&lt;strong&gt;Who or what am I?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;I'm no lawman, no soldier, no hero standing tall.&lt;/p&gt;&#xA;&lt;p&gt;Yet each must pay this price in turn, to have somebody to watch over us all.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Hints:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The answer is a single word, nine letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;br/&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The answer involves wordplay. The answer is a word in itself, but also a portmanetau of two shorter words which are the answers to the second line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Happy to offer hints, incorporate feedback if given!&lt;/p&gt;&#xA;" OwnerUserId="14374" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-08-31T13:25:34.787" Title="The price you pay to have someone watch over you" Tags="|riddle|word|wordplay|" AnswerCount="6" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18563" PostTypeId="2" ParentId="18562" CreationDate="2015-08-03T13:24:24.673" Score="2" Body="&lt;p&gt;The price we pay for someone to &quot;watch over us all&quot; is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; taxes - you are the IRS.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm no lawman, no soldier, no hero standing tall.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the &quot;taxman&quot; does not make laws, just enforces them. He doesn't fight, and is typically not considered a hero&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet each must pay this price in turn...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As each of us grow older, and can no longer be considered a dependent, we must begin to file for taxes (I admit, not everyone is taxed, so it doesn't quite fit here)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;to have somebody to watch over us all&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Our tax dollars fund government agencies, including the long reach of the NSA, which actively engages in domestic surveillance (as famously exposed by Edward Snowden recently)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14290" LastEditorUserId="14290" LastEditDate="2015-08-04T15:47:43.043" LastActivityDate="2015-08-04T15:47:43.043" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18564" PostTypeId="2" ParentId="18562" CreationDate="2015-08-03T13:47:46.910" Score="1" Body="&lt;p&gt;I guess it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Shelter (as in Residence, house)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm no lawman, no soldier, no hero standing tall.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not a person.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet each must pay this price in turn, to have somebody to watch over us all.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Guards us against Climate, Animals and dangers. I am reading in turn as in generation by generation residing take turns to pay based on whoever is the bread earner of the family. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9892" LastEditorUserId="9892" LastEditDate="2015-08-03T13:52:28.980" LastActivityDate="2015-08-03T13:52:28.980" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18565" PostTypeId="2" ParentId="18562" CreationDate="2015-08-03T13:50:23.627" Score="5" Body="&lt;p&gt;The price we pay is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Death&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Death is definitely not a lawman, soldier, nor a hero. And everyone dies at some point. When we die, we can a) take this in a religious direction and say that either God or Satan is watching over us or b) take it in a literal direction and say that people will be watching over us at the funeral/the burial.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13924" LastActivityDate="2015-08-03T13:50:23.627" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18566" PostTypeId="2" ParentId="18562" CreationDate="2015-08-03T13:52:15.480" Score="1" Body="&lt;p&gt;Maybe it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a prayer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; God&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I'm no lawman, no soldier, no hero standing tall.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; God isn't a lawman or a soldier, and he isn't considered as a hero (correct me if I am wrong)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Yet each must pay this price in turn, to have somebody to watch over us all.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; God is listening to everyone, especially if we pray to him&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13384" LastEditorUserId="12700" LastEditDate="2015-08-03T14:20:55.047" LastActivityDate="2015-08-03T14:20:55.047" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18567" PostTypeId="1" AcceptedAnswerId="18573" CreationDate="2015-08-03T14:41:49.387" Score="14" ViewCount="3451" Body="&lt;p&gt;You wake up in a dark room that you don't recognize. The walls are made of some material you've never seen before. There is a door with a keypad next to it, and a small platform suspended in air in the center of the room. You approach the platform, unsure of how it remains in mid-air. On that platform you see a note that reads:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Welcome to Room 1! This is the first of many rooms that comprise what we like to call the Maze. It's not so much a physical maze as it is an intellectual one; fail to find a way out and it is your mind that will be lost!&lt;br&gt;&#xA;  Each room provides 2 keys. Key 1 will be integral to determining the answer to Key 2; sometimes it makes sense of Key 2, sometimes it explains what to do with Key 2. Once you have the answer, enter it into the keypad by the exit. Enter correctly and you will proceed. Enter incorrectly, and the room is rigged to open a portal that sucks you into one of three alternate universes: one made entirely of pure itchiness, one exactly like this one but where the only thing you can drink is Bud Light Lime, and the last leads to a universe that we dare not describe and will tell you only that it is called Trumptopia.&lt;br&gt;&#xA;  With all of that out of the way, here is the information you need to make it out of this room. If it matters, we based our information on your planet from your oracle named Google, but we expect that your numbers need not line up exactly with ours to get near the same answer. Good Luck!!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Key 1: (BRUSSELSBELGIUM to TEHRANIRAN)+(WARSAWPOLAND to FRANKFURTGERMANY)&lt;br&gt;&#xA;  Key 2: ABERDEENSCOTLAND to &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You take a look at the keypad by the door and see that it is comprised of only the 26 letters of the alphabet, with no punctuation or space keys. What do you enter to move on to the next room?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This room has now been solved. Here is &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18575/room-2-of-the-maze&quot;&gt;Room 2&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-08-03T18:36:53.533" Title="Room 1 of the Maze" Tags="|enigmatic-puzzle|password|geography|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18568" PostTypeId="2" ParentId="18567" CreationDate="2015-08-03T14:59:47.677" Score="3" Body="&lt;p&gt;I'll take a stab at it...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LAHARPEILLINOIS or LAHARPEILLINOISUSA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I got this by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; adding the distance in km given by Google. Brussels to Tehran is 5138.9 km. Warsaw to Frankfurt is 1082.0 km, and the sum is 6220.9 km, rounded to 6221 km.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using that sum, the city I first found using Google that was that distance away from Aberdeen, Scotland was&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.aberdeen.climatemps.com/vs/la-harpe.php&quot; rel=&quot;nofollow&quot;&gt;La Harpe, Illinois, USA&lt;/a&gt;, although there are many others, so this might not be the intended method&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastActivityDate="2015-08-03T14:59:47.677" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18569" PostTypeId="1" AcceptedAnswerId="18570" CreationDate="2015-08-03T15:14:57.360" Score="2" ViewCount="6116" Body="&lt;p&gt;This might seem familiar to many of the veterans here, but since it has not been exclusively asked, I am posting it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A 13-foot tall truck is at the entrance of a tunnel with a clearance of 12 feet 11 inches&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the most economic and simplest solution possible that the driver of the truck has at hand to pass through the tunnel safely to arrive the other side with the truck?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; N.B. As the problem implies there are multiple solutions possible. I am expecting a specific answer which on consensus of the reader(s) can be regarded as the acceptable answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Disclaimer: I did not create this puzzle but heard it way back in the 80's as a young kid. I think it is a Edward De Bono puzzle. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Anyways the alternate answers  are rarely documented which is what I am after. Investing further time to curate through the list of answers is no longer possible and am accepting the classical answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9892" LastEditorDisplayName="user88" LastEditDate="2015-08-04T05:47:49.917" LastActivityDate="2015-08-04T05:47:49.917" Title="Truck through the Tunnel" Tags="|logical-deduction|lateral-thinking|" AnswerCount="5" CommentCount="6" ClosedDate="2015-08-03T17:51:48.243" ContentLicense="CC BY-SA 3.0" />
  <row Id="18570" PostTypeId="2" ParentId="18569" CreationDate="2015-08-03T15:19:08.487" Score="13" Body="&lt;p&gt;My first instinct would be to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let some air out of the tires, so that you end up riding an inch or so lower and can pass through safely, albeit maybe a bit bumpy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-08-03T15:19:08.487" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18571" PostTypeId="2" ParentId="18567" CreationDate="2015-08-03T15:21:22.430" Score="3" Body="&lt;p&gt;Here's something I was thinking about, but I don't feel like it's a solid enough lead to jump through the portal on. I'll keep pondering and leave my notes so far ;)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Regardless, one option could be that the place names are referring to differences in population. The increase in population from Brussels (177k) to Tehran (8154k) is approximately 7980k. Add the increase from Warsaw (1711k) to Frankfurt (688k), which is -1023k and you get 6957k. Add this to Aberdeen's population (229k) and you get a pop. of 7.186mil. According to Wikipedia, that's closest to HONGKONGHONGKONG (7.235mil), followed by TAIPEITAIWAN (7.045mil).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12462" LastActivityDate="2015-08-03T15:21:22.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18572" PostTypeId="2" ParentId="18569" CreationDate="2015-08-03T15:29:25.763" Score="10" Body="&lt;p&gt;Assuming the measurements are completely accurate, then a solution is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To drive the truck through the tunnel! For peace of mind you might want to grease the roof. As long as the tunnel is uniform, and straight, and the truck doesn't move in such a way as to ever increase its height, then a 13' object should fit through a 13' gap.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12462" LastActivityDate="2015-08-03T15:29:25.763" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18573" PostTypeId="2" ParentId="18567" CreationDate="2015-08-03T15:31:03.810" Score="16" Body="&lt;p&gt;Take the given routes as vectors and sum them, then apply the resulting vector to Aberdeen. Using latitude and longitude isn't quite accurate because of the curvature of the earth, but it's close enough.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;Brussels=50°51′N 04°21′E&#xA;Tehran  =35°42′N 51°25′E&#xA;Brussels to Tehran: -15°09′ +47°04′&#xA;&#xA;Warsaw   =52°14′N 21°01′E&#xA;Frankfurt=50°07′N 08°41′E&#xA;Warsaw to Frankfurt: -02°07′ -12°20′&#xA;&#xA;Total vector: -17°16′ +34°44′&#xA;&#xA;Aberdeen=57°09′N 02°07′W&#xA;&#xA;Aberdeen+Vector: 39°53′N 32°37′E&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This point is in the outskirts of Ankara. The answer should be &lt;code&gt;ANKARATURKEY&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-08-03T15:31:03.810" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18574" PostTypeId="2" ParentId="18569" CreationDate="2015-08-03T15:47:56.960" Score="1" Body="&lt;p&gt;Blast the A.C. so the truck shrinks a few mm.&lt;/p&gt;&#xA;" OwnerUserId="13920" LastActivityDate="2015-08-03T15:47:56.960" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18575" PostTypeId="1" AcceptedAnswerId="18583" CreationDate="2015-08-03T15:53:20.980" Score="11" ViewCount="2750" Body="&lt;p&gt;Still excited about your quick triumph over &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18567/room-1-of-the-maze&quot;&gt;Room 1&lt;/a&gt;, you head straight for the suspended note and read it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Congratulations on making it through the warm-up to Room 2. From here on out, getting the answer from Key 2 will now require the use of the answer from the previous room in addition to Key 1. You will be unable to get the correct answer without the use of these two values (though their roles in the solution process will differ between rooms). In case you forget, we will always include the answer from the previous room in our letter.&lt;br&gt;&#xA;  This room will be more challenging than the first, Key 2 has been encrypted, and Key 1 seems to be of no help. You'll need to use the answer from the previous room up front. Good Luck!!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Key 1: Key 2 is much more than the sum of its parts.&lt;br&gt;&#xA;  Key 2: UAST OIGT UQH VBPU PMBNQPXFAGBHT. UMIG PVTHPX UIQM PI PTH. PUTH FIID PI DTX IHT  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;For those who want to start from the beginning, search for &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18567/room-1-of-the-maze&quot;&gt;Room 1 of the Maze&lt;/a&gt;. The answer for the previous room, for those who forgot or who want to join in starting in this room, is below. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You notice that the keypad in this room has only the digits 0-9 as well as the usual Enter button. What do you enter to move on to the next room in the Maze?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer from &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18567/room-1-of-the-maze&quot;&gt;Previous Room&lt;/a&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;ANKARATURKEY&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;Halfway through your calculations, a small portal opens in the center of the room and a note materializes from it. You hurriedly grasp it and read its contents:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We hear that your kind is quite fond of the number 10; something about the number of digits on your extremities. We've heard that some of your kind have learned to use other numbers in it's place and translate between them; hopefully you are one of these. If it is any consolation, we have 349 fblthps on each of our kjaposenjfq, so you can imagine what math is like on our planets!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This room has now been solved. Here is &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18608/room-3-of-the-maze&quot;&gt;Room 3&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-08-04T15:08:31.500" Title="Room 2 of the Maze" Tags="|cipher|enigmatic-puzzle|password|" AnswerCount="5" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18576" PostTypeId="1" CreationDate="2015-08-03T15:56:52.253" Score="9" ViewCount="1020" Body="&lt;p&gt;I have created a word puzzle, and I'm interested in rating the puzzles by difficulty. There are nearly 2000 puzzles in my archive, so rating them by hand would be time-consuming and probably rather inconsistent, subjective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how the game works: &lt;a href=&quot;https://gist.github.com/pgblu/1fa0dccd75ecba154776&quot; rel=&quot;nofollow&quot;&gt;https://gist.github.com/pgblu/1fa0dccd75ecba154776&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things I've noticed:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1a. The larger the ratio of known to unknown letters, the easier&lt;/li&gt;&#xA;&lt;li&gt;1b. The more contiguous known letters you have, the easier&lt;/li&gt;&#xA;&lt;li&gt;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;The more clue words you have (2 being the fewest), the easier&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;The larger the pool of 'candidate words' for any pattern, the harder&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;Finally, the more unusual the letters or patterns, the easier the puzzle (which is linked to #3 of course)&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;Having the first letter a 'known' letter makes the puzzle easier&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Would anyone like to share their thoughts on designing a difficulty rating algorithm?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Easy puzzle:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;_ I M E _ _ _ _ _&lt;/li&gt;&#xA;&lt;li&gt;_ R A C _ _ _ _ _&lt;/li&gt;&#xA;&lt;li&gt;_ R E A _ _ _ _ _&lt;/li&gt;&#xA;&lt;li&gt;_ U R N _ _ _ _ _&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Easy puzzle:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;_ _ _ F F _ _ I&lt;/li&gt;&#xA;&lt;li&gt;_ _ _ F F _ _ O&lt;/li&gt;&#xA;&lt;li&gt;_ _ _ P H _ _ E&lt;/li&gt;&#xA;&lt;li&gt;_ _ _ T U _ _ Y&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Medium puzzle:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;C _ _ _ V _ _ E&lt;/li&gt;&#xA;&lt;li&gt;F _ _ _ B _ _ S&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Hard puzzle:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A _ _ _ _ _&lt;/li&gt;&#xA;&lt;li&gt;B _ _ _ _ _&lt;/li&gt;&#xA;&lt;li&gt;C _ _ _ _ _&lt;/li&gt;&#xA;&lt;li&gt;D _ _ _ _ _&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Very hard puzzle:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;_ _ _ _ _ _ S&lt;/li&gt;&#xA;&lt;li&gt;_ _ _ _ _ _ T&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;solutions: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; t...table, gra..it., .lea.ag., .erate, hideou. (the last one has more than one solution)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11099" LastEditorUserId="4385" LastEditDate="2015-08-06T17:33:45.067" LastActivityDate="2015-08-06T17:39:48.140" Title="Thoughts on creating a 'difficulty level' algorithm" Tags="|word|" AnswerCount="2" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="18577" PostTypeId="2" ParentId="18569" CreationDate="2015-08-03T16:10:56.400" Score="1" Body="&lt;p&gt;Since we know that tunnels are mostly shaped like a circular tube, a 13-foot truck cannot pass from a tunnel that is 13-foot or lower. This can be further calculated, by estimating width of the truck, regular tunnel shape and so on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An economical approach can be changing suspension height, just like people who do lower their cars. If that is not enough, rims can be changed and tire pressure can be decreased.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although 13 foot is not that high. Most small trucks are lower at size. Best thing to do would be putting the load sideways with some help from cranes. It would not be perfectly stable, but it will do the job.&lt;/p&gt;&#xA;" OwnerUserId="14377" LastActivityDate="2015-08-03T16:10:56.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18578" PostTypeId="2" ParentId="18569" CreationDate="2015-08-03T16:41:58.467" Score="2" Body="&lt;p&gt;Many large trucks have:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Adjustable air bag suspensions which can be lowered. Simply lower the ride height and drive right through.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4685" LastActivityDate="2015-08-03T16:41:58.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18582" PostTypeId="2" ParentId="18575" CreationDate="2015-08-03T19:14:24.760" Score="2" Body="&lt;p&gt;Taking a shot even though there may be several holes (2 of the letters were inconsistent in the original attempt):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HAVE SOME FUN WITH TRIBUTYLAMINE. FROM TWENTY FOUR TO TEN. THEN LOOK TO KEY ONE &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I took the uninteresting route and looked at the frequency of letters and found 'P' to appear 9 times.  Since a two-letter word that begins with 'T' is most probable, and since &quot;Tributylamine&quot; is one of the few 13-letter words with a 'T' in the 1st and 6th position, it was the most likely candidate.  From there, everything flowed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tributylamine is C_12H_27, and 27 + 12 = 39, which is the north coordinate of Ankara Turkey.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly,  24+10 = 34 which is the East coordinate of Ankara Turkey.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the answer the sum:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 73&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;" OwnerUserId="13920" LastEditorUserId="13920" LastEditDate="2015-08-03T19:37:38.170" LastActivityDate="2015-08-03T19:37:38.170" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18583" PostTypeId="2" ParentId="18575" CreationDate="2015-08-03T19:54:49.043" Score="19" Body="&lt;p&gt;Solving Key 2 as a keyed Caesar cipher with the key being the answer from room 1, you get: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; FAVE SOME FUN WITF TRIBUTYLAMINE. FROM TWENTY FOUR TO TEN. TFEN LOOK&#xA; TO KEY ONE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The H's in the string are incorrectly coded to F's - they should be Y's in the coded version but they are presented as U's.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;H being coded to F in the string could mean we should also code it to F in the chemical (C12F27N instead of C12H27N)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDITS:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Based on the new info, we need to convert C12F27N from base 24 to base 10, which gives 2302070591.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then, the sum of its parts gives the answer as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 29&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14383" LastEditorUserId="14383" LastEditDate="2015-08-03T21:43:11.777" LastActivityDate="2015-08-03T21:43:11.777" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18584" PostTypeId="2" ParentId="18575" CreationDate="2015-08-03T20:16:18.633" Score="5" Body="&lt;p&gt;Taking the message and decoding as a keyed caesar code with the keyword ANKARATURKEY (ANKRTUEY) gives the message: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; FAVE SOME FUN WITF TRIBUTYLAMINE. FROM TWENTY FOUR TO TEN. TFEN LOOK TO KEY ONE.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are three typos in the decoded message,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; FAVE (HAVE), WITF (WITH), and TFEN (THEN)&lt;br&gt;&#xA; F being 'incorrectly' substituted for H&lt;br&gt;&#xA;&lt;br&gt;&#xA; Taking the chemical formula for tributylamine ($C_{12}H_{27}N$) and substituting F for H as in the message gives Perfluorotributylamine, or $C_{12}F_{27}N$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Taking the next clue, I converted from base 24 to base 10 to get.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $C_{26}F_{55}N$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Taking the next clue, I converted from base 24 to base 10 to get.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C26F55N  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Taking the sum of those gives the answer  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 81&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13101" LastEditorUserId="13101" LastEditDate="2015-08-03T21:30:49.473" LastActivityDate="2015-08-03T21:30:49.473" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18585" PostTypeId="2" ParentId="18575" CreationDate="2015-08-03T20:56:13.990" Score="-4" Body="&lt;p&gt;Perfluorotributylamine would be C12F27N after replacing hydrogen with fluorine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Atomic weight is &lt;strong&gt;322&lt;/strong&gt;, which could be the answer as 24 to 10 could refer to the alpha entry pad to 10 digit entry pad&lt;/p&gt;&#xA;" OwnerUserId="14384" LastEditorUserId="14384" LastEditDate="2015-08-03T21:16:58.130" LastActivityDate="2015-08-03T21:16:58.130" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18586" PostTypeId="2" ParentId="18575" CreationDate="2015-08-03T20:57:09.277" Score="9" Body="&lt;p&gt;Another possible interpretation of 'from 24 to 10':&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;convert from base 24 to base 10 (the newest hint suggests this)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we then&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; convert C12F27N (a valid base 24 number) to base 10 we get &lt;b&gt;2302070591&lt;/b&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is that the answer?&lt;/p&gt;&#xA;" OwnerUserId="13152" LastActivityDate="2015-08-03T20:57:09.277" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18587" PostTypeId="1" AcceptedAnswerId="18596" CreationDate="2015-08-03T21:31:23.470" Score="2" ViewCount="617" Body="&lt;p&gt;Ideally if you pressure me I will not break, &lt;br&gt;&#xA;alas, my weakness seems to come at my weakest. &lt;br&gt;&#xA;But if you truly press me I will show you the way, &lt;br&gt;&#xA;However afterwards I may be a little flushed. &lt;br&gt; &lt;br&gt;&#xA;If you are looking for a connection that will be me, &lt;Br&gt;&#xA;On down the chain, each step of the way. &lt;br&gt;&#xA;Especially when I get worked up and excited, &lt;br&gt;&#xA;press me I will show you the way.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sorry for the lack of details but the rest of my hints are currently in the too easy/production state, more will probably come in roughly ~17 hours if no one gets it before then. I kind of enjoy its current state. &lt;sub&gt;[Message out of date but extremely accurate of what happened]&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13916" LastEditorUserId="13916" LastEditDate="2015-08-04T15:40:09.797" LastActivityDate="2015-08-05T13:58:44.943" Title="This Riddle's Answer is Missing" Tags="|riddle|" AnswerCount="7" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18589" PostTypeId="2" ParentId="18587" CreationDate="2015-08-04T00:14:03.990" Score="0" Body="&lt;p&gt;I have two thoughts - one seems to fit for the first half and the other for the second half. I'll post here to see if others can build off of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First two lines:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An egg seems to fit these. Ideally an egg will not break while just squeezing it, but it's weakness is at the widest point - it's &quot;weakest&quot; area.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second two lines:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This seems to fit some kind of button. Like an elevator button? If you press it, it should show you the way to the floor you want to go to, and it might get &quot;flushed&quot; - elevator buttons light up when you press them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2015-08-04T00:14:03.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18590" PostTypeId="2" ParentId="18587" CreationDate="2015-08-04T01:43:26.043" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your heart&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally if you pressure me I will not break,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hopefully our relationships grow stronger under pressure and do not fall apart&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;alas, my weakness seems to come at my weakest.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Weak or cowardly people are faint of heart&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But if you truly press me I will show you the way,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Follow your heart&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However afterwards I may be a little flushed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If we follow our hearts we may end up flushing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13197" LastActivityDate="2015-08-04T01:43:26.043" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18591" PostTypeId="2" ParentId="18587" CreationDate="2015-08-04T06:08:16.690" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pressure Cooker&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not sure&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ideally if you pressure me I will not break, &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It will never break on pressure..&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;alas, my weakness seems to come at my weakest. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Its weakest part is its sealing ring ..&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But if you truly press me I will show you the way, &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;if we continuously press the steam vent then tube valve will be broken ..&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However afterwards I may be a little flushed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;then we there will no other way except flushing and fitting a new valuve&lt;/p&gt;&#xA;" OwnerUserId="444" LastActivityDate="2015-08-04T06:08:16.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18592" PostTypeId="2" ParentId="18576" CreationDate="2015-08-04T07:22:24.353" Score="8" Body="&lt;p&gt;So sorry for all the text you would have to read; I pity you. I also realize the paragraphs may not flow as much as they can, yielding a bit of a choppy feel. I apologize, and I hope to focus on creating a better flow at a later point. Here's a TL;DR:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I propose 4 methods, all of which are imperfect, as to be expected. 3 of them look similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then endevour to use the cardinality of the solution set to modify a general method, along with respect to the possible solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After, I publish a nice-looking chart that really was made with Microsoft&amp;copy; Excel&amp;copy; 2013&amp;tm;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally (for the moment), I use ETAONI HSRDLU WMCGFYBPVKXJQZ in tandem with possible solutions for a given line.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here are my thoughts. Actually, my trial-and-error approach. Mainly, I'd use one or more of the formulas to find an easiest and hardest puzzle, then use a percent to indicate difficulty (e.g. 100% is easy, 21% is difficult).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $n$ be the number of lines, $b$ be the number of blanks per line, $g$ be the number of given letters per line, $c$ the number of distinct characters featured in a puzzle, and let $a$ be the average placement of a character in a line. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach states that the lower a puzzles score, the harder it is.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;The product method: $nbg$&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Based on the observation that, given more information, some problem will be easier. &lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Modified product method: $\dfrac{b(gn)^2}{c}$&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Takes the previous product method multiplied by the character to distinct character ratio.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Über modified product method: $\dfrac{b(gn)^2}{ca}$&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Based on the observation that letters in the first position yield easier puzzles, ergo, the closer the character to the end of a word, the tendency increases to yield harder puzzles.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;The &quot;I have run out of names for these methods please read on&quot;-method: $n\left(\dfrac{g}{b}\right)^2$&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Multiplying the lines by the ratio of given to blank numbers, based on the fact that more information is easier. Also note that the squaring bit is to put numbers closer to zero or further from $1$, depending if $0&amp;lt;k\leq 1$ ($k^2 \to 0$) or if $1\leq k&amp;lt;\infty$ ($k^2\to\infty$).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Example 3 on the GitHub has really gotten me thinking... perhaps a simple combination (or even a complex one, at that) is not enough to satisfy a difficulty test. So let $o$ be the number of solutions a given puzzle has, and let $q$ be the number of (sets of letters) that fulfill the puzzle in that they fail only in having another word being formed (as was the case in example 3). Then, let $M$ be some method/formula, either one here or some other method. We can more accurately portray method $M$ by modifying it as such:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{oM}{f(q,x)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some variable $x$ and a function $f$. Now, what values best fit $x$ and $f$? I wish to put a higher significance on the fact that, if $q$ such pseudo-solutions exist, it will make the problem harder, regardless of there being $o$ solutions. If I set $f(q,x)=q^x$, I believe that it will skew the results of the problems hardness too much, as any $x\gg 1$ makes $q^x$ grow fast. Any $x\approx1$ with $x\notin\mathbb{Z}$ is undesirable, as this would make the algorithm tricky. But, if such was the contrary, that some $x\in\mathbb{R}$ was needed, then an $x$ for $f(q,x)=q^x$ could stand as $1&amp;lt;x\ll2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, it will be useful to use a function that grows more slowly. This rules out $x^q$, $q!$, $^xq$, and pretty much anything that grows quicker. So, $f(q,x)=qx$. $x$ then is purely arbitrary.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;Revised standards&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I have compiled all the puzzles on the GitHub into a table. And currently, they all fall into the same problem: they generally work, but still classify some as easy that should be hard, etc. I must do some more work, and, until then, I will not have said data up.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;Observations&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aEelU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aEelU.png&quot; alt=&quot;Difficulty with a, b, and g&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(0 is Very Easy,..., 5 is Very Hard)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;Idea number umpteen&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;ETAONI HSRDLU WMCGFYBPVKXJQZ! Alright, I'm not crazy, merely just &amp;ldquo;inspired&amp;rdquo;. The OP mentions that a puzzle like &lt;code&gt;_ _ _ _ e _&lt;/code&gt; is harder than a puzzle like &lt;code&gt;_ _ _ _ z _&lt;/code&gt;. Therefore, if there are more words that could possibly match one line, then the puzzle earns a higher difficulty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this method, I will say that a &lt;em&gt;higher&lt;/em&gt; number denotes a &lt;em&gt;higher&lt;/em&gt; difficulty, and, likewise, a &lt;em&gt;lower&lt;/em&gt; number denotes a &lt;em&gt;lower&lt;/em&gt; difficulty. Let $k_i$ be the number of solutions that solve for just the $i$th line of the puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I use &lt;a href=&quot;http://www.a2zwordfinder.com/crossword_dictionary.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; to get $k_i$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt; Let this be the puzzle (lines numbered):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0: _ y z y _ _&#xA;1: _ h a g _ _&#xA;2: _ c o d _ _&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For line &lt;code&gt;0&lt;/code&gt;, $k_0=1$, as the only solution is the word &lt;strong&gt;syzygy&lt;/strong&gt;. (Or so I believe; tell me if another word fits this!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For line &lt;code&gt;1&lt;/code&gt;, $k_1=17$. (ᴜɴʟɪsᴛᴇᴅ)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally, for line &lt;code&gt;2&lt;/code&gt;, $k_2=3$. (&lt;code&gt;ccodes&lt;/code&gt;, &lt;code&gt;mcodes&lt;/code&gt;, and &lt;code&gt;scodgy&lt;/code&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, how should we go about combining $k_i$? Shall we multiply it together, summate it, or perhaps some other function? Let us first observe, however, the $k_i$s for other puzzles. For fluidity and conciseness, I propose that $\overline{k}=\dfrac{\prod_{i=0}^{n}k_i}{n}$ (pronounced &lt;code&gt;&amp;lt;k&amp;gt; bar&lt;/code&gt;) (the average of $k_0,...,k_n$).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0: _ I M E _ _ _ _ _  k_0 = 100&#xA;1: _ R A C _ _ _ _ _  k_1 = 163&#xA;2: _ R E A _ _ _ _ _  k_2 = 263&#xA;3: _ U R N _ _ _ _ _  k_3 = 99&#xA;k bar = 625 / 4 = 156.25&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;~&lt;/p&gt;&#xA;&#xA;&lt;p&gt;American Dictionary used from here on out.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0: _ _ _ F F _ _ I    k_0 = 1&#xA;1: _ _ _ F F _ _ O    k_1 = 1&#xA;2: _ _ _ P H _ _ E    k_2 = 43&#xA;3: _ _ _ T U _ _ Y    k_3 = 10&#xA;k bar = 55 / 4 = 13.75&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;These puzzles, however, were &lt;em&gt;both&lt;/em&gt; on the easy side; not only does $\overline{k}$ contradict our supposition that higher numbers indicate higher difficulties (unless if we find more evidence to the contrary), but also the fact that two easy puzzles yield such different $\overline{k}$s is astounding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I will observe a very hard puzzle, in hopes that it would yield a high $\overline{k}$.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0: _ _ _ _ _ _ S    k_0 = 7677&#xA;1: _ _ _ _ _ _ T    k_1 = 1258&#xA;k bar = 8935 / 2 = 4467.5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In light of the previous results, the gap between the first two $\overline{k}$s decreases drastically. I will also conjecture that, for every/most puzzle line ending with &lt;code&gt;S&lt;/code&gt; of length $m$ with $b$ blanks, $k_i\ggg (mb)^2$, if $m$ is sufficiently large and $b$ sufficiently small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how does ETAONI HSRDLU WMCGFYBPVKXJQZ fit in? Well, if you haven't noticed already, this is the frequency that a letter appears in a word. Let $p_{j,i}$ be the index of the $i$th letter in the $j$th line of a puzzle. $\overline{p_j}$ is the averages of $p_{j,i}$ for line $j$, and $\vec{p}$ is the averages of all $\overline{p_j}$s for the puzzle.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;(Not fully developed) ideas:&lt;/h1&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Perhaps $\overline{k}\vec{p}$ would yield an elementary difficulty scale.&lt;/li&gt;&#xA;&lt;li&gt;The implementation of $\overline{k}$ would be especially difficulty from a programatical point of view, requiring the use of an outside dictionar(ies). If you can do that, however, I believe that $\overline{k}$ would be a valuable asset in determining the difficulty of your puzzle; furthermore, this seems to be the only difficulty I can foresee in algorithm implementation.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;[WIP &lt;em&gt;still&lt;/em&gt;!]&lt;/p&gt;&#xA;" OwnerUserId="7307" LastEditorUserId="7307" LastEditDate="2015-08-05T17:56:29.370" LastActivityDate="2015-08-05T17:56:29.370" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="18593" PostTypeId="2" ParentId="18587" CreationDate="2015-08-04T08:07:19.943" Score="2" Body="&lt;p&gt;Could you be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A pipe?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ideally if you pressure me I will not break,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Self-explanatory.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;alas, my weakness seems to come at my weakest.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A weak point in a pipe can cause the whole thing to be useless.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But if you truly press me I will show you the way, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; High pressure within a pipe causes the contents to move through it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However afterwards I may be a little flushed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clearing a pipe is &quot;flushing&quot; it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12462" LastActivityDate="2015-08-04T08:07:19.943" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18594" PostTypeId="2" ParentId="18587" CreationDate="2015-08-04T08:29:26.310" Score="0" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Missing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It says so in the title! The description of the riddle is irrelevant.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-08-04T08:29:26.310" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18595" PostTypeId="2" ParentId="18587" CreationDate="2015-08-04T12:04:52.360" Score="2" Body="&lt;p&gt;Are you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A glowstick?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ideally if you pressure me I will not break, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A glowstick takes some force to light, you can usually bend the outer casing a bit before the glass (or whatever it is nowadays) within breaks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;alas, my weakness seems to come at my weakest. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A glowstick's 'weakness' is the glass vial within, which is much more brittle than the surrounding plastic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But if you truly press me I will show you the way, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you push hard enough the glass breaks and the glowstick will light up.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However afterwards I may be a little flushed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can't turn a glowstick off, and they often continue to glow after they are no longer needed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14391" LastActivityDate="2015-08-04T12:04:52.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18596" PostTypeId="2" ParentId="18587" CreationDate="2015-08-04T12:06:05.987" Score="5" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A link&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ideally if you pressure me I will not break,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hopefully a link in the chain won't break if you put pressure on it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;alas, my weakness seems to come at my weakest.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Weakest link&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But if you truly press me I will show you the way,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A hyperlink will take you to another place&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However afterwards I may be a little flushed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hyperlinks change color after they've been clicked on. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you are looking for a connection that will be me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A connection can also be thought of as a link&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On down the chain, each step of the way.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A chain has links in it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Especially when I get worked up and excited,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hyperlinks are Hyper links &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;press me I will show you the way.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A hyperlink will take you to another place&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Riddle's answer is missing&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The missing link as in the missing link in human evolution&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13925" LastEditorUserId="13925" LastEditDate="2015-08-05T13:58:44.943" LastActivityDate="2015-08-05T13:58:44.943" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18597" PostTypeId="1" AcceptedAnswerId="18599" CreationDate="2015-08-04T12:45:34.273" Score="5" ViewCount="398" Body="&lt;p&gt;A 3 feet tall, hungry monkey is trying a reach a magical banana placed on top of a flagpole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The flagpole is 25 feet in height and is slippery to grasp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every Odd numbered Jump, the monkey climbs 3 feet and slides down 2 feet, while for every Even jump the monkey climbs 4 feet and slides down 3 feet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question : How much length in Odd jumps does the monkey take to reach the top of the flagpole? and Why?&lt;/p&gt;&#xA;" OwnerUserId="9892" LastEditorUserId="39" LastEditDate="2015-08-10T00:40:24.360" LastActivityDate="2015-08-10T00:40:24.360" Title="Monkey on a Flagpole" Tags="|calculation-puzzle|" AnswerCount="3" CommentCount="3" ClosedDate="2015-08-04T14:45:47.853" ContentLicense="CC BY-SA 3.0" />
  <row Id="18598" PostTypeId="2" ParentId="18597" CreationDate="2015-08-04T12:56:11.460" Score="1" Body="&lt;p&gt;There will be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 22 jumps, 11 odd, 11 even&lt;br&gt;&#xA;&lt;br&gt;&#xA; Every jump until the end nets you 1 foot. After the 21st jump (odd), the monkey will be at 21 feet. Then the monkey jumps 4 feet (even) and gets the banana. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-08-04T12:56:11.460" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18599" PostTypeId="2" ParentId="18597" CreationDate="2015-08-04T12:59:54.060" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;10 odd jumps are required  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The monkey climbs, in essence, one foot per jump. Assume it can reach to the top of its 3-foot height, in which case it only needs to attain a maximum height of 22 feet to reach the banana.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The important jump is the final one:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If the last jump would be an even jump, we find that the monkey can reach the banana from 18 feet before sliding back down. If it would be odd, the monkey would previously have to be at a height of at least 19 feet.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To reach these heights at one net foot per jump:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the monkey would need to jump 18 times (9 odd, 9 even) to reach an 18 foot height. This would make their next jump an 'odd' one (10 odd), meaning they wouldn't reach the banana. One more 'even' jump (10 even)would be required after that to grab the fruity prize.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I apologise if I've said metres instead of feet at any point. Still having to resist the urge to go through and convert from Imperial to Metric&lt;/p&gt;&#xA;" OwnerUserId="11717" LastEditorUserId="11717" LastEditDate="2015-08-05T10:58:20.410" LastActivityDate="2015-08-05T10:58:20.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18600" PostTypeId="2" ParentId="18597" CreationDate="2015-08-04T13:24:25.020" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 22, because we can guarantee that after 25 - $(3,4)max$ = 21 jumps, it climbs 21 meters. Since 22 is even, the last day will see it climb to the top without sliding back.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastActivityDate="2015-08-04T13:24:25.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18601" PostTypeId="1" AcceptedAnswerId="18603" CreationDate="2015-08-04T13:32:33.343" Score="27" ViewCount="4281" Body="&lt;p&gt;I was talking with a friend via SMS when suddently, he sent me a strange message :&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(16)(39)(26)(36)(25)(34)(12)(34)(47)(36)(28)(18)(36)(33)(22)(12)(18)(18)(23)(37)(39)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I asked what was going on but he didn't answer me ...&lt;br&gt;&#xA;Can someone help me, what did he tell me ?&#xA;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Hints&lt;br&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Hint 1 :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Tags can help you&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2 :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Context is like size ... it matters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3 :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This is a joke, there are only 2 hints.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13384" LastEditorUserId="463" LastEditDate="2015-08-04T14:52:12.940" LastActivityDate="2015-08-04T22:51:35.200" Title="The strange SMS" Tags="|cipher|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18602" PostTypeId="1" CreationDate="2015-08-04T13:34:38.167" Score="6" ViewCount="3887" Body="&lt;p&gt;Recently there had been horses missing from the stables.  The owner barely noticed, until his horse was stolen. He has four suspects but was unable to identify the guilty one.  They each gave three statements.  He knew that several of the statements were false, in fact, he knew that none of the suspects told the same number of true statements as another one.  The statements are as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;These first three statements are from James:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Statement 1&lt;/strong&gt; Henry is the guilty one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Statement 2&lt;/strong&gt; I was on a fishing trip during the last theft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Statement 3&lt;/strong&gt; No one thinks I am guilty.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;These second three statements are from Henry.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Statement 1&lt;/strong&gt; James is lying, I am not the guilty one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Statement 2&lt;/strong&gt; Fred has been on duty every theft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Statement 3&lt;/strong&gt; Edgar is guilty.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;These third three statements are from Fred.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Statement 1&lt;/strong&gt; Henry's first statement is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Statement 2&lt;/strong&gt; I agree with James' first statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Statement 3&lt;/strong&gt; I am not guilty&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;These last three statements are from Edgar.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Statement 1&lt;/strong&gt; James's alibi is false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Statement 2&lt;/strong&gt; All of Fred's statements are false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Statement 3&lt;/strong&gt; I would not think of committing the crime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, given all these statements, you should find it easy to figure out who the guilty one is.&lt;/p&gt;&#xA;" OwnerUserId="14392" LastEditorUserId="4050" LastEditDate="2015-08-21T15:31:03.650" LastActivityDate="2015-08-22T06:46:33.357" Title="Who is the horse thief?" Tags="|logical-deduction|liars|" AnswerCount="7" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18603" PostTypeId="2" ParentId="18601" CreationDate="2015-08-04T13:45:00.573" Score="42" Body="&lt;p&gt;If you interpret each two-digit number as&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The second number repeated the first number of times, so that (39) becomes 999,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can interpret the full string as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 6-999-66-666-55-444-2-444-7777-666-88-8-666-333-999-777-33-8-8-1-1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which looks a lot like&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Texting on an old phone keyboard.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Interpreting the numbers in this way gives us the string&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MYNOKIAISOUTOFYRETT.,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which, with spaces, says&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My Nokia is out of yrett.,&lt;br&gt;&#xA;&lt;br&gt;&#xA; I expect this last word is 'battery' backwards, since &quot;.,&quot; would be AB if it were on the second key, and not the first. I have yet to determine why that last word is backwards in the first place.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="5188" LastEditDate="2015-08-04T22:51:35.200" LastActivityDate="2015-08-04T22:51:35.200" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18604" PostTypeId="1" AcceptedAnswerId="18733" CreationDate="2015-08-04T14:10:23.900" Score="7" ViewCount="654" Body="&lt;p&gt;Old man Skyler was a solitary but friendly old man who lived in Lander, Wyoming. As a hobbyist, he loved nothing more than making model planes and selling them cheaply to kids at local events. He wasn't a rich man, but lived humbly and generously. Unfortunately, old man Skyler passed away this March from a heart attack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The people of the community were sad, but as they began to organise his affairs it was discovered that he had arranged for his life insurance to form a trust to help the area. However, the insurance company claims that his insurance has run out, and that they cannot confirm that he died while the policy was still active.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Old man Skyler was last seen alive on Thursday the 5th of March, by the postman who was delivering a package. He'd talked to the postman about the package, which was a new model plane that he was very excited about. He was discovered dead on Friday the 13th. Everything was covered in flock, which he must have knocked over in his last moments. His insurance policy was valid up to and including Monday the 9th.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an expert puzzler, the Lander community brought you in to help sort things out. His house has been sealed and preserved since the old man was found dead. You go inside and have a brief look around the place, and then settle at the desk, stirring up flock as you sit down. On the desk is the new model plane:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rd3mo.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rd3mo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you &lt;strong&gt;prove&lt;/strong&gt; (as much as possible), one way or the other, whether or not old man Skyler died before his insurance ran out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There's more to the image than just the glue label, but nothing you won't have the tools to find.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12462" LastEditorUserId="12462" LastEditDate="2015-08-09T22:33:33.537" LastActivityDate="2015-08-09T22:33:33.537" Title="The Hobbyist's Gift" Tags="|visual|enigmatic-puzzle|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="18605" PostTypeId="2" ParentId="18602" CreationDate="2015-08-04T14:28:58.830" Score="7" Body="&lt;p&gt;Let's call JS1, JS2 and JS3 the 3 James' statements, HS1, HS2 and HS3 the 3 Henry's statements, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If James is guilty&lt;/strong&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;JS1 : false&lt;br&gt;JS2 : false&lt;br&gt;JS3 : false (Edgar is accusing him to lie about his alibi on ES1)&lt;br&gt;HS1 : true&lt;br&gt;HS2 : true&lt;br&gt;HS3 : false&lt;br&gt;FS1 : true&lt;br&gt;FS2 : false (supposing he knows he is guilty)&lt;br&gt;FS3 : true&lt;br&gt;ES1 : true&lt;br&gt;ES2 : false&lt;br&gt;ES3 : ??? (he is allowed to think about stealing the horses without really stealing them)&lt;br&gt;It is impossible because at least one of the suspect always tell the truth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If Henry is guilty&lt;/strong&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;JS1 : true&lt;br&gt;JS2 : ???&lt;br&gt;JS3 : false (Edgar is accusing him to lie about his alibi on ES1)&lt;br&gt;HS1 : false&lt;br&gt;HS2 : true&lt;br&gt;HS3 : false&lt;br&gt;FS1 : false&lt;br&gt;FS2 : true&lt;br&gt;FS3 : true&lt;br&gt;ES1 : false&lt;br&gt;ES2 : false&lt;br&gt;ES3 : ??? (he is allowed to think about stealing the horses without really stealing them)&lt;br&gt;It is impossible because at least one of the suspect always tell the truth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If Fred is guilty&lt;/strong&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;JS1 : false&lt;br&gt;JS2 : ???&lt;br&gt;JS3 : false (Edgar is accusing him to lie about his alibi on ES1)&lt;br&gt;HS1 : false&lt;br&gt;HS2 : false&lt;br&gt;HS3 : false&lt;br&gt;FS1 : true&lt;br&gt;FS2 : false&lt;br&gt;FS3 : false&lt;br&gt;ES1 : false&lt;br&gt;ES2 : false&lt;br&gt;ES3 : ??? (he is allowed to think about stealing the horses without really stealing them)&lt;br&gt;It is impossible because at least one of the suspect always tell the truth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If Edgar is guilty&lt;/strong&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;JS1 : false&lt;br&gt;JS2 : ???&lt;br&gt;JS3 : false (Edgar is accusing him to lie about his alibi on ES1)&lt;br&gt;HS1 : true&lt;br&gt;HS2 : true&lt;br&gt;HS3 : true&lt;br&gt;FS1 : true&lt;br&gt;FS2 : false&lt;br&gt;FS3 : true&lt;br&gt;ES1 : false&lt;br&gt;ES2 : false&lt;br&gt;ES3 : false&lt;br&gt;It is possible if JS2 is true.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answer :&lt;/strong&gt;&#xA;The guilty one is&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Edgar&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13384" LastEditorUserId="13384" LastEditDate="2015-08-04T14:37:11.880" LastActivityDate="2015-08-04T14:37:11.880" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18606" PostTypeId="2" ParentId="18602" CreationDate="2015-08-04T14:31:55.037" Score="6" Body="&lt;p&gt;I think that the guilty one is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Henry, as told by James.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As each suspect tells a different number of true statements (or a different number of lies), we can look at contradictions in each's statements.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For James, it is possible that every statement is a lie, or every statement is true, or any combination. So he could tell 0, 1, 2, or 3 lies.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Henry, if we assume the second statement is accusing Fred of being guilty as he was on duty every theft, then 2 and 3 contradict each other, meaning at least 1 is a lie. So we can say that Henry tells 1, 2, or 3 lies.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fred cannot agree with both James' and Henry's first statements, because he would then agree that Henry is guilty and Henry is not guilty. Also, if he is guilty, then Henry could not be guilty, so Fred can only tell 1 or 2 lies.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since Fred cannot tell 3 lies, at least 1 has to be true, meaning that Edgar's second statement is false. Thus, Edgar can only say 1, 2, or 3 lies.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leaves us with James as being the only one that can tell all truths. As James' first statement is that Henry is guilty, we have our culprit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Henry is guilty, then Fred agrees with James' first statement and is not guilty, so he told 1 lie (2 truths). If we assume that Henry's second statment is an accusation against Fred, then we know that Henry tells 3 lies (0 truths) since he is guilty and Fred and Edgar cannot be. Thus, Edgar needs to tell 2 lies (1 truth). James' alibi must be true and we know that Fred has to tell at least 1 truth, so apparently Edgar shouldn't be trusted either because he thought about stealing horses.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, prosecute Henry and fire Edgar for good measure. &lt;/p&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-08-04T14:31:55.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18607" PostTypeId="2" ParentId="18602" CreationDate="2015-08-04T14:37:16.687" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Edgar is the guilty one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's first point out that we know there are 4 people and each of them tells a different number of truths.  This means someone always tells the truth, someone always lies, and the remaining people tell either 1 lie or 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note next that Fred's 1st and 2nd statements are tautologically false by virtue of the two comments he's addressing being contradictory.  This means he's lied at least once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, Edgar is lying in his second comment, since if it were the case that Fred was lying about statements 1 AND 2, then we would have another contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It must then be the case that either Henry or James is completely truthful.  Once this is determined, we can eliminate individuals more easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can reduce the final results to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;J: 2 false/1 true&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H: 3 true&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F: 2 true/1 false&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E: 3 false&lt;/p&gt;&#xA;" OwnerUserId="13920" LastActivityDate="2015-08-04T14:37:16.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18608" PostTypeId="1" AcceptedAnswerId="18636" CreationDate="2015-08-04T15:05:25.567" Score="4" ViewCount="1345" Body="&lt;p&gt;Exhausted from the effort of solving &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18575/room-2-of-the-maze&quot;&gt;Room 2&lt;/a&gt;, you try to get a good night's sleep, but your sleep is interrupted by vivid dreams of fblthps, numbers, and vectors. The Maze is getting to you a bit, it seems. Once you feel awake and ready, you read the note suspended, as usual, in the middle of the room.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Third Room! Only one other person has made it this far and well...it's quite unfortunate that he had no nails left when we finally got him through the portal.&lt;br&gt;&#xA;  This room is easier than room 2, but the information is more hidden. In the interest of keeping our test subjects from dropping like flies, we have decided to include an example value of Key 2 and its corresponding answer. This way, if you think you have cracked the puzzle, you can replicate your steps on the example Key and see if your answer matches its. Take a look at the information below and take your best shot. Good Luck!!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Key 1:&lt;br&gt;&#xA;  First reveal the shifted meaning, deep within Key two;&lt;br&gt;&#xA;  Then use the answer from before; you'll know what to do.&lt;br&gt;&#xA;  Then query them each yes or no by Eratosthenes;&lt;br&gt;&#xA;  Yes or no to zero through nine, the number that you need.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Key 2: '=' 852 , '?' 1 , '_' -200 , '8' 666 , '5' 8 , '6' 7 , '7' 109&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Ex. Key 2: '8'   4523 , '=' 5 , '7' 51 , '5' 3  Ex. Answer=3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer from &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18575/room-2-of-the-maze&quot;&gt;Previous Room&lt;/a&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;29&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The keypad has only digits. You can see some fingerprints on it, but only on two keys: 0 and Enter. There are scratch marks on the wall next to it that say &quot;OH GOD THE ITCH!!!&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This room has now been solved. The next room can be found &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18668/room-4-of-the-maze&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-08-05T16:12:39.167" Title="Room 3 of the Maze" Tags="|riddle|cipher|enigmatic-puzzle|password|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18609" PostTypeId="2" ParentId="18602" CreationDate="2015-08-04T15:21:13.853" Score="1" Body="&lt;p&gt;Let's give each sentence a truth value:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;James:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a$ Henry is the guilty one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b$ I was on a fishing trip during the last theft.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c$ No one thinks I am guilty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Henry:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a'$ James is lying, I am not the guilty one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d$ Fred has been on duty every theft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e$ Edgar is guilty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fred - either the first or the second is true, the other is false:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a'$ Henry's first statement is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a$ I agree with James' first statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$h$  I am not guilty&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edgar:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b'$ James's alibi is false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0  All of Fred's statements are false. (that's a lie)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f$ I would not think of committing the crime.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$a, b, c, a', d, e, a', a, h, b', 0, f$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a + a + a' + a' + b + b' + c + d + e + f + h$ = 6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2           +  1 + $c + d + e + f + h$ = 6 (three of $c, d, e, f, h$ are true)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either Henry $(a', d, e)$ or James $(a, b, c)$ is telling nothing but the truth, so either Henry or Edgar is guilty, making $h$ true and $a'=e$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c + d + a' + f$ = 2&lt;br&gt;&#xA;$c + d + a' + b$ = 2v3&lt;br&gt;&#xA;$b$ &gt;= $f$&lt;br&gt;&#xA;$b'$ + $f$ &amp;lt; 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $f$ = 1, $b$ = 1, $a$ = 1, $c$ = 1, $d$ = 0, which is an acceptable result (Henry being guilty).&lt;/p&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="14016" LastEditDate="2015-08-04T22:51:52.350" LastActivityDate="2015-08-04T22:51:52.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18611" PostTypeId="1" AcceptedAnswerId="18649" CreationDate="2015-08-04T16:00:43.330" Score="1" ViewCount="350" Body="&lt;p&gt;I am a name for many people,&lt;br&gt;&#xA;I am tangible, but not quite so.&lt;br&gt;&#xA;give me a kiss, my rival,&lt;br&gt;&#xA;You might find the answer laughable.  &lt;/p&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="10037" LastEditDate="2015-08-04T16:06:48.977" LastActivityDate="2015-08-05T12:06:19.543" Title="Do we share the same name?" Tags="|riddle|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18612" PostTypeId="2" ParentId="17247" CreationDate="2015-08-04T16:14:07.303" Score="3" Body="&lt;p&gt;Regarding the riddle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Line 3: &lt;em&gt;a party perceived differently with ear than with the eye&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;We are looking for 4 words, each of which can be pronounced in two (or more different ways)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Line 2: &lt;em&gt;when taken by their fourths information they will hold&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The fourth letter of the words, when combined together, gives another word (possibly related to data?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Line 1: My best attempt at this is:&#xA;&lt;em&gt;to be won&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;comBat - you can win a battle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;then to impede&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;impAct - to affect something&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;to be steeled&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;excUse - to be steeled means to be ready to accept consequences of your actions, this could mean letting someone else off or taking the blame for their issues (although this is weak)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;perhaps gold&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;leaD - Alchemists attempt to convert lead into gold&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Giving&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;BAUD which is measure of information transfer rate&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-08-04T16:14:07.303" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18613" PostTypeId="2" ParentId="18611" CreationDate="2015-08-04T16:57:40.560" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bill or Will(iam)? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;it's a name.&lt;/li&gt;&#xA;&lt;li&gt;a bill is tangible, but one's will may not be?&lt;/li&gt;&#xA;&lt;li&gt;one's will can go against what's best for them&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13920" LastActivityDate="2015-08-04T16:57:40.560" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18614" PostTypeId="1" CreationDate="2015-08-04T17:03:28.080" Score="2" ViewCount="1658" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What term will come next: 5, 6, 8, 11, 45, 74, 101?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I saw this question &lt;a href=&quot;http://www.quora.com/What-term-will-come-next-5-6-8-11-45-74-101&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; on Quora. I am thinking everything, but cannot seem to find a pattern. &#xA;I am thinking, 5(3rd prime), two numbers based on some pattern, 11(5th prime), another two numbers on that pattern, 101(26th prime).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still looks very far fetched.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is appreciated.&lt;/p&gt;&#xA;" OwnerUserId="12518" LastActivityDate="2015-09-01T14:13:15.510" Title="What number comes next in: 5, 6, 8, 11, 45, 74, 101?" Tags="|number-sequence|" AnswerCount="2" CommentCount="2" ClosedDate="2015-09-01T13:54:28.853" ContentLicense="CC BY-SA 3.0" />
  <row Id="18615" PostTypeId="1" AcceptedAnswerId="18763" CreationDate="2015-08-04T17:17:22.797" Score="4" ViewCount="995" Body="&lt;p&gt;We can be made from addition or subtraction&lt;br&gt;&#xA;Our best are often a popular attraction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some you can touch, &lt;br&gt;&#xA;and some you cannot&lt;br&gt;&#xA;One with a dozen will heal you&lt;br&gt;&#xA;Another, inumerable, shan't be bought&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of us have thousands, and some, just a few&lt;br&gt;&#xA;We all can change one's point of view&lt;br&gt;&#xA;But, y'know, sometimes words have two meanings&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of us have utility&lt;br&gt;&#xA;but others are for show&lt;br&gt;&#xA;Like one befuddling, three worlds all in one&lt;br&gt;&#xA;made 'round sixty years ago&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By now you may have guessed what we are&lt;br&gt;&#xA;so we ask that you find our brother, a movie star&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He watched a broken man escape from prison&lt;br&gt;&#xA;who let fear find him, and now is risen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;This riddle follows a similar format to my &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18424/surely-you-know-this-prominent-figure-hints-added&quot;&gt;most recent riddle&lt;/a&gt;. &lt;strong&gt;About half of the lines describe the group itself, and the other half make various cultural references to famous members of this group.&lt;/strong&gt; The accepted answer will explain at least three of the cultural references described, as well as the identity of the famous brother from the final stanza.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;Hints:&lt;br&gt;&#xA;For the first line, I'll elaborate:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; These things used to be made using &quot;subtraction&quot;, but now, almost all are made from &quot;addition&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second line:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Take this line literally. Some of these things are uninteresting, but others are very popular.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second stanza:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer (for the group's name) is a word which is used literally (&quot;can touch&quot;) or figuratively (&quot;some you cannot&quot;)&lt;br&gt;&#xA; The last two lines of the second stanza describe &quot;some you cannot&quot; touch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The third stanza:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This contains true statements regarding this group of things, as well as a cultural reference shared with another stanza.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The fourth stanza:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &quot;one befuddling&quot;, made &quot;for show&quot; (not &quot;utility&quot;) was made in the 1950s.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The fifth/sixth stanzas:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;em&gt;brother&lt;/em&gt; is in a movie but is not acknowledged in the credits.&lt;br&gt;&#xA; The word &quot;watched&quot; was chosen for a reason. This rules out Bane and Bruce Wayne as correct responses.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13237" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-08-10T15:29:39.093" Title="What are we? And where is our brother? Hints galore!" Tags="|riddle|rhyme|knowledge|" AnswerCount="7" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18616" PostTypeId="1" AcceptedAnswerId="18640" CreationDate="2015-08-04T18:07:39.687" Score="10" ViewCount="312" Body="&lt;p&gt;Transcript from a conversation that couldn't happen today:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;J:&lt;/strong&gt; Hey D, let's play the game.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;D:&lt;/strong&gt; Again? There are only so many times we can do this without repeating ourselves, but OK. You start.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;J:&lt;/strong&gt; &lt;em&gt;Be tan, Theta Bro&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;D:&lt;/strong&gt; Too easy, that's '76&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;J:&lt;/strong&gt; Fine, your turn.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;D:&lt;/strong&gt; &lt;em&gt;Creaky boa chaw&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;J:&lt;/strong&gt; 77! &lt;em&gt;Pa's seen chunkier okra&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;D:&lt;/strong&gt; 77 again. Anything harder?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;J:&lt;/strong&gt; Ok, how about &lt;em&gt;Two ton gator upwind&lt;/em&gt;?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;D:&lt;/strong&gt; Hmm... I don't know that one. Are you sure it's right?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;J:&lt;/strong&gt; Oh, right, you wouldn't. Sorry about that. What about &lt;em&gt;End a sweat in a bed&lt;/em&gt;?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;D:&lt;/strong&gt; 78. I think you used that one last time we played.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;J:&lt;/strong&gt; Yeah, it's a popular one. Your turn.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;D:&lt;/strong&gt; Umm... &lt;em&gt;Hand pie&lt;/em&gt;? Sorry, running out of original ideas here.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;J:&lt;/strong&gt; Another 77. Maybe we should quit for now.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;D:&lt;/strong&gt; You're probably right. Did we ever come up with a name for this game?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;J:&lt;/strong&gt; It's definitely either &lt;em&gt;Oarsmen&lt;/em&gt; or &lt;em&gt;Note Masher&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Who might J and D be?&lt;/li&gt;&#xA;&lt;li&gt;What strange game is this?&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Two ton gator upwind&lt;/em&gt; is a valid challenge. Why doesn't D know that?&lt;/li&gt;&#xA;&lt;li&gt;If J challenged with &quot;&lt;em&gt;Am I mod Doorknob or celery ruler?&lt;/em&gt;&quot;, what is the correct response?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Post-solution question:&lt;/strong&gt;&#xA;If you've figured out or seen the solution, the question below is more about puzzle building and/or difficulty.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I originally had &lt;em&gt;Zip kit gobbler&lt;/em&gt; as one of the clues (answer 76). I removed it because it seemed too obvious a giveaway (when combined with the others). Once the mechanics of the puzzle are figured out, the rest is just tracking down answers (in my mind). I was worried that the mechanic would be &lt;em&gt;more&lt;/em&gt; obvious from that clue than the others, as well as leading to a very obvious solution. Do you think it would have been fine, or was I right to take it out? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1563" LastEditorUserId="1563" LastEditDate="2015-08-05T14:49:41.870" LastActivityDate="2015-08-05T14:49:41.870" Title="Note Masher? Oarsmen?" Tags="|enigmatic-puzzle|knowledge|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18617" PostTypeId="1" AcceptedAnswerId="18633" CreationDate="2015-08-04T18:12:38.050" Score="18" ViewCount="1844" Body="&lt;p&gt;If a word has a certain property, I call it an Adjacent Word™.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In each of the short, unrelated sentences below, only one of the choices is an Adjacent Word™.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sentences are for context only, and do not affect whether a word is an Adjacent Word™. Also, neither font nor case matters in determining the property.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;( &lt;strong&gt;Additionally / Furthermore&lt;/strong&gt; ), we now have a dog!&lt;/li&gt;&#xA;  &lt;li&gt;He is concerned about ( &lt;strong&gt;overfishing / overharvesting&lt;/strong&gt; ) the oceans.&lt;/li&gt;&#xA;  &lt;li&gt;She prefers to use ( &lt;strong&gt;margarine / shortening&lt;/strong&gt; ) in her cookies.&lt;/li&gt;&#xA;  &lt;li&gt;She saved it in a new Excel ( &lt;strong&gt;workbook / worksheet&lt;/strong&gt; ).&lt;/li&gt;&#xA;  &lt;li&gt;He loved the ( &lt;strong&gt;ancient / prehistoric&lt;/strong&gt; ) buildings.&lt;/li&gt;&#xA;  &lt;li&gt;They ( &lt;strong&gt;thawed / unfroze&lt;/strong&gt; ) their TV dinners.&lt;/li&gt;&#xA;  &lt;li&gt;They offered the most ( &lt;strong&gt;comprehensive / expansive&lt;/strong&gt; ) news coverage.&lt;/li&gt;&#xA;  &lt;li&gt;They saw that the danger was ( &lt;strong&gt;avoidable / preventable&lt;/strong&gt; ).&lt;/li&gt;&#xA;  &lt;li&gt;They committed an ( &lt;strong&gt;unforgivable / unpardonable&lt;/strong&gt; ) sin.&lt;/li&gt;&#xA;  &lt;li&gt;She just bought her first ( &lt;strong&gt;condominium / townhouse&lt;/strong&gt; ).&lt;/li&gt;&#xA;  &lt;li&gt;The most difficult verbs to learn are ( &lt;strong&gt;irregulars / reflexives&lt;/strong&gt; ).&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What makes a word an Adjacent Word™?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;SUB&gt;&lt;strong&gt;Bonus&lt;/strong&gt;: If you have solved one of my &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%22Phrase%22%20user%3A463%20is%3Aquestion&quot;&gt;Phrase™&lt;/a&gt; or &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%22versatile%22%20user%3A463%20is%3Aquestion&quot;&gt;Word™&lt;/a&gt; puzzles before (Deusovi, Rodolvertice, CodeNewbie, xnor, Sean, glibdud, f&quot;, dperry), and you solve this one, you will be the first person to solve more than one of these types of puzzles of mine, and you will receive 100 rep points from me, on top of any other points that this puzzle might give (i.e. It might also have a regular bounty on it &lt;em&gt;if it lasts long enough&lt;/em&gt;.)&lt;/SUB&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="-1" LastEditorDisplayName="user14478" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2016-11-07T14:53:11.650" Title="What is an Adjacent Word™?" Tags="|pattern|english|word-property|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18618" PostTypeId="1" CreationDate="2015-08-04T18:20:41.610" Score="12" ViewCount="1010" Body="&lt;p&gt;After solving &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18421/the-case-of-the-missing-kitten-part-1&quot;&gt;the kidnapper's riddle&lt;/a&gt;, you realize that you have no clue how to get in touch with him. You wait until the end of the specified fortnight, and sure enough, you receive a phone call. &quot;I solved your riddle,&quot; you say, triumphantly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only response is, &quot;Answer?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You provide the solution, and the kidnapper says, &quot;Good. You solved it. Time to get your precious Geronimo back. Open your door; you'll see another note for you.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you open the door, you find another note. It reads:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Congratulations,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You solved the riddle within the fortnight, and your cat will live to see another week (or two, if you're quick). Before you can get him back, we have to meet. Unfortunately, you don't know how to find me, and I won't come to you. Figure out the next letter in each sequence, then unscramble to find the solution. Inspired by &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18363/a-quindectet-of-letter-sequences&quot;&gt;VictorHenry&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where am I?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;S, P, S, M, T, ?   &lt;/li&gt;&#xA;&lt;li&gt;A, B, B, F, G, L, M, ?    &lt;/li&gt;&#xA;&lt;li&gt;W, A, J, M, M, ?   &lt;/li&gt;&#xA;&lt;li&gt;V, M, K, K, E, E, ?   &lt;/li&gt;&#xA;&lt;li&gt;F, M, M, B, B, E, ?     &lt;/li&gt;&#xA;&lt;li&gt;S, M, H, D, W, M, ?   &lt;/li&gt;&#xA;&lt;li&gt;C, T, F, N, I, ?   &lt;/li&gt;&#xA;&lt;li&gt;R, C, U, C, B, A, ?   &lt;/li&gt;&#xA;&lt;li&gt;W, B, S, K, N, ?   &lt;/li&gt;&#xA;&lt;li&gt;A, T, J, A, F, ?   &lt;/li&gt;&#xA;&lt;li&gt;N, U, S, J, M, ?  &lt;/li&gt;&#xA;&lt;li&gt;P, A, I, S, ?   &lt;/li&gt;&#xA;&lt;li&gt;T, T, T, F, F, S, S, E, ?   &lt;/li&gt;&#xA;&lt;li&gt;E, G, D, S, F, L, ?  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I feel nice, so I'll start you out with one letter given: C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Love, Al.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some of these sequences are worldly knowledge, such as oldest cities, fastest rivers, etc. Others are only found between the pages of a novel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nearly a year has passed, you can only hope your cat is still alive. If this question regains some traction I think it might be nice for someone to compile the existing (and future) answers into one or make a community answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13538" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2016-09-20T16:20:32.490" Title="The Case of the Missing Kitten (Part 2)" Tags="|word|english|" AnswerCount="7" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="18619" PostTypeId="2" ParentId="18618" CreationDate="2015-08-04T18:32:23.280" Score="5" Body="&lt;ol start=&quot;11&quot;&gt;&#xA;&lt;li&gt;N, U, S, J, M, ? &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;E&lt;/strong&gt;: the planets in reverse order&lt;br&gt;&#xA; &lt;strong&gt;N&lt;/strong&gt;eptune, &lt;strong&gt;U&lt;/strong&gt;ranus, &lt;strong&gt;S&lt;/strong&gt;aturn, &lt;strong&gt;J&lt;/strong&gt;upiter, &lt;strong&gt;M&lt;/strong&gt;ars, &lt;strong&gt;E&lt;/strong&gt;arth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;W, A, J, M, M, ?   &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;A&lt;/strong&gt;: the first six US Presidents&lt;br&gt;&#xA; &lt;strong&gt;W&lt;/strong&gt;ashington, &lt;strong&gt;A&lt;/strong&gt;dams, &lt;strong&gt;J&lt;/strong&gt;efferson, &lt;strong&gt;M&lt;/strong&gt;adison, &lt;strong&gt;M&lt;/strong&gt;onroe, &lt;strong&gt;A&lt;/strong&gt;dams&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4385" LastEditorUserId="4385" LastEditDate="2015-08-04T19:40:28.043" LastActivityDate="2015-08-04T19:40:28.043" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18620" PostTypeId="2" ParentId="18618" CreationDate="2015-08-04T18:34:33.083" Score="4" Body="&lt;ol start=&quot;8&quot;&gt;&#xA;&lt;li&gt;R, C, U, C, B, A, ?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;I&lt;/strong&gt;: world countries by land area. &lt;strong&gt;R&lt;/strong&gt;ussia, &lt;strong&gt;C&lt;/strong&gt;anada, &lt;strong&gt;U&lt;/strong&gt;nited States, &lt;strong&gt;C&lt;/strong&gt;hina, &lt;strong&gt;B&lt;/strong&gt;razil, &lt;strong&gt;A&lt;/strong&gt;ustralia, &lt;strong&gt;I&lt;/strong&gt;ndia&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;12&quot;&gt;&#xA;&lt;li&gt;P, A, I, S, ?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;A&lt;/strong&gt;: bodies of water by size. &lt;strong&gt;P&lt;/strong&gt;acific Ocean, &lt;strong&gt;A&lt;/strong&gt;tlantic Ocean, &lt;strong&gt;I&lt;/strong&gt;ndian Ocean, &lt;strong&gt;S&lt;/strong&gt;outhern Ocean, &lt;strong&gt;A&lt;/strong&gt;rctic Ocean&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="11475" LastEditDate="2015-08-04T18:50:53.367" LastActivityDate="2015-08-04T18:50:53.367" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18621" PostTypeId="2" ParentId="18608" CreationDate="2015-08-04T18:49:59.743" Score="4" Body="&lt;p&gt;Okay, I'll give this a shot. Not too sure about it though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First step: reveal the shifted meaning&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I took the characters as ASCII numbers and shifted them by the number&#xA; following it.&lt;br&gt;&#xA; For example, '=' is 61 in ASCII, shifted by 852, gives 852+61 = 913.&#xA; Doing all the same way gives the string: 913 64 -105 722 61 61 164&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Use the answer from before:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pretty unsure about this one, but I tried shifted each of these numbers &#xA; by 29, but since 29 is the same as shifting by 3, I just added 3 to each.&#xA; This gives 916 67 -102 725 64 64 167&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Query yes or no by Eratosthenes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He invented a method of finding primes, so I'll assign each number a&#xA; 0 or 1 based on whether it's prime&#xA; This gives 0 1 0 0 0 0 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes or no to 0-9:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Just convert from binary to base 10.&#xA; This gives an answer of 33&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Works for the example key:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The string after the ASCII shift is 4579 66 106 56.&#xA; Then the add 3 gives 4582 69 109 59.&#xA; Yes-or-no for primes is 0 0 1 1, which is 3 in decimal&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14383" LastActivityDate="2015-08-04T18:49:59.743" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18622" PostTypeId="2" ParentId="18618" CreationDate="2015-08-04T19:05:48.250" Score="5" Body="&lt;ol start=&quot;7&quot;&gt;&#xA;&lt;li&gt;C, T, F, N, I, ?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;P&lt;/strong&gt;: US states by population. &lt;strong&gt;C&lt;/strong&gt;alifornia, &lt;strong&gt;T&lt;/strong&gt;exas, &lt;strong&gt;F&lt;/strong&gt;lorida, &lt;strong&gt;N&lt;/strong&gt;ew York, &lt;strong&gt;I&lt;/strong&gt;llinois, &lt;strong&gt;P&lt;/strong&gt;ennsylvania&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13152" LastActivityDate="2015-08-04T19:05:48.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18623" PostTypeId="1" AcceptedAnswerId="18624" CreationDate="2015-08-04T19:25:38.310" Score="18" ViewCount="2250" Body="&lt;p&gt;&lt;em&gt;My first is sunny and may be found here in Hades;&lt;br&gt;&#xA;My second is funny and may be found here in Hell.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My third is charged, found liberated in Alicante;&lt;br&gt;&#xA;My fourth is rare and found liberated in Iberia.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My sixth is hard and found secretly in knickers;&lt;br&gt;&#xA;My seventh is fertile and found often in pants.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;&lt;strong&gt;What is my fifth?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'll add more tags if this one seems hard or too broad.)&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorDisplayName="user20" LastEditDate="2015-08-04T20:21:13.513" LastActivityDate="2015-08-05T12:52:52.153" Title="Did you miss me? Here's a riddle!" Tags="|riddle|pattern|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18624" PostTypeId="2" ParentId="18623" CreationDate="2015-08-04T19:32:39.423" Score="15" Body="&lt;p&gt;&lt;em&gt;Your fifth is black, and can be found buried in fiberglass&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are the &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Periodic table of elements&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The clues each have a property of a member, along with a couple words containing the &quot;name&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Hydrogen:&lt;/strong&gt;&lt;br&gt;&#xA; sunny (it's found in the sun)&lt;br&gt;&#xA; &lt;strong&gt;H&lt;/strong&gt; is in &lt;em&gt;here&lt;/em&gt; and &lt;em&gt;Hades&lt;/em&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Helium:&lt;/strong&gt;&lt;br&gt;&#xA; funny (you talk funny when inhaled)&lt;br&gt;&#xA; &lt;strong&gt;He&lt;/strong&gt; in &lt;em&gt;here&lt;/em&gt; and &lt;em&gt;Hell&lt;/em&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Lithium:&lt;/strong&gt;&lt;br&gt;&#xA; charged (commonly used in batteries)&lt;br&gt;&#xA; &lt;strong&gt;Li&lt;/strong&gt; in &lt;em&gt;liberated&lt;/em&gt; and &lt;em&gt;Alicante&lt;/em&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Beryllium:&lt;/strong&gt;&lt;br&gt;&#xA; rare (it's just rare, and not found alone)&lt;br&gt;&#xA; &lt;strong&gt;Be&lt;/strong&gt; in &lt;em&gt;liberated&lt;/em&gt; and &lt;em&gt;Iberia&lt;/em&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Carbon:&lt;/strong&gt;&lt;br&gt;&#xA; hard (think diamonds)&lt;br&gt;&#xA; &lt;strong&gt;C&lt;/strong&gt; in &lt;em&gt;secretly&lt;/em&gt; and &lt;em&gt;knickers&lt;/em&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Nitrogen:&lt;/strong&gt;&lt;br&gt;&#xA; fertile (most fertilizers contain it)&lt;br&gt;&#xA; &lt;strong&gt;N&lt;/strong&gt; in &lt;em&gt;often&lt;/em&gt; and &lt;em&gt;pants&lt;/em&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So that makes the fifth one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Boron&lt;/strong&gt;:&lt;br&gt;&#xA; black (one of it's &quot;natural&quot; appearances)&lt;br&gt;&#xA; &lt;strong&gt;B&lt;/strong&gt; in &lt;em&gt;buried&lt;/em&gt; and &lt;em&gt;fiberglass&lt;/em&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks to GentlePurpleRain in comments for the last set of clues.&lt;/p&gt;&#xA;" OwnerUserId="1563" LastEditorUserId="1563" LastEditDate="2015-08-05T12:52:52.153" LastActivityDate="2015-08-05T12:52:52.153" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="18625" PostTypeId="2" ParentId="18618" CreationDate="2015-08-04T19:32:57.123" Score="4" Body="&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;V M K K E E &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the seven summits: Vinson, McKinley, Kilimanjaro, Kosciuszko, Everest, Elbrus, &lt;strong&gt;Aconcagua&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;13&quot;&gt;&#xA;&lt;li&gt;T T T F F S S E&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Counting by tens: Ten Twenty Thirty Forty Fifty Sixty Seventy Eighty &lt;strong&gt;Ninety&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="99" LastActivityDate="2015-08-04T19:32:57.123" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18626" PostTypeId="2" ParentId="17936" CreationDate="2015-08-04T19:49:16.870" Score="6" Body="&lt;p&gt;You are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Dr. John Faust from V for Vendetta&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using the list from @AlirezaTabatabaeian and the last clue&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Each row of letters on your keyboard is numbered 0-3 and each column is numbered starting at 1. Taking each number pair as a Row/Column and 327 being a space as it is in the 3rd row and extends from col 2 to col 7 (from @BaileyM) gives you &quot;Vi Veri Veniversum Vivus Vici&quot; - a quote attributed to Dr. John Faust in the movie V for Vendetta.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13290" LastEditorUserId="13290" LastEditDate="2015-08-04T19:54:32.770" LastActivityDate="2015-08-04T19:54:32.770" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18627" PostTypeId="2" ParentId="18615" CreationDate="2015-08-04T20:22:43.783" Score="2" Body="&lt;p&gt;Partial answer that explains some of the stanzas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are we? &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Numbers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;We are made from addition or subtraction&lt;br&gt;&#xA;Our best are often a popular attraction&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You add and subtract numbers.&#xA; Our 'best' could refer to famous numbers such as Pi, and E&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Some you can touch,&lt;br&gt;&#xA;and some you cannot&lt;br&gt;&#xA;A dozen will heal your illness&lt;br&gt;&#xA;Another shan't be bought&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can touch Real Numbers.&#xA; You cannot touch &lt;a href=&quot;https://en.wikipedia.org/wiki/Imaginary_number&quot; rel=&quot;nofollow&quot;&gt;Imaginary Numbers&lt;/a&gt;. There are such things as personal and healing numbers. Infinity is a number, and you will never be able to buy an infinite amount of any thing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;We all can change one's point of view&lt;br&gt;&#xA;But, y'know, sometimes words have two meanings&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can always place a negative sign (-) in front of a number, giving it different meanings and changing the point of view.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Some of us have utility&lt;br&gt;&#xA;but others are for show&lt;br&gt;&#xA;One befuddling, three worlds becoming one&lt;br&gt;&#xA;made o'er fifty years ago&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure. Might be referring to some numerical constants.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;By now you may have guessed what we are&lt;br&gt;&#xA;so we ask that you find our brother, a movie star&lt;br&gt;&#xA;He watched a broken man escape from prison&lt;br&gt;&#xA;who let fear find him, and now is risen.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wild guessing here: Life of Pi, or the show Numb3rs. I haven't seen either, so I'm not sure if their plots line up with this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9587" LastEditorDisplayName="user20" LastEditDate="2015-08-04T20:24:18.253" LastActivityDate="2015-08-04T20:24:18.253" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18628" PostTypeId="1" AcceptedAnswerId="18644" CreationDate="2015-08-04T20:52:11.237" Score="1" ViewCount="299" Body="&lt;p&gt;As stated, what's the next number?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;71,82,94,94,53,43,62,?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1) hints are already given somewhere in this question!&lt;br&gt;&#xA;  2) operators&#xA;  aren't necessary involved&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="10037" LastEditDate="2015-08-04T21:21:55.370" LastActivityDate="2015-08-05T11:32:44.130" Title="Sequence puzzling - what is the next number?" Tags="|number-sequence|" AnswerCount="2" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18629" PostTypeId="1" AcceptedAnswerId="18638" CreationDate="2015-08-04T21:11:48.427" Score="29" ViewCount="3872" Body="&lt;p&gt;A letter to my girlfriend. She would have been 30 this year. &lt;strong&gt;Who is she, and what experiences did we share together&lt;/strong&gt;?&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Hello... I miss you. We all do.&lt;/p&gt;&#xA;&lt;p&gt;You loved to paint. A personal passion: it brought out your brilliance. I remember when I first met you, I saw your painting of the mask of Tutankhamun. You loved Europe, too. And they loved you back. We travelled to the States together, but it never seemed right. It was your home I think, but didn't feel right to be there.&lt;/p&gt;&#xA;&lt;p&gt;You were never far from your workbench, surrounded by drawers filled with tools and gadgets I could never hope to understand. You weren't an aviator, but you said you knew your way around the bits of a plane. (Seriously graphic language ensued... things went wrong often, meditation rarely helped. You'd throw your work away and start over.)&lt;/p&gt;&#xA;&lt;p&gt;I hope you're happy; you're with your family now. Do you remember the time your younger sister (the one who loved those cartoon classics) brought all those pets home from Scotland in, 1991 was it?&lt;/p&gt;&#xA;&lt;p&gt;Hundreds of them! Running backwards and forwards, all over you, all over her, digging holes in everything, making a mess. I couldn't even tell them apart, we never got round to giving any of them names. We finally gave them a good home. (As many as we could I'm sad to say.) Wasn't there even a statue of them? I remember seeing them around from time to time, holidays and Christmas. They're everywhere... we even gave one to Terry Pratchett, regrettably. It seemed like a great idea at the time. (Oh, BTW, those same men who gave you them went onto a life of crime in America; it was all over the papers, surprise surprise.)&lt;/p&gt;&#xA;&lt;p&gt;At times it seemed like you could do anything. It's cheesy, but... &lt;em&gt;you made things possible&lt;/em&gt;. But you didn't do it on your own. Do you remember your friends? That part-time copper who kept you in line and Denise... they showed you how to paint. And punk musician Paula... crazy as hell, sometimes it seemed like she could play four instruments at once!&lt;/p&gt;&#xA;&lt;p&gt;We tried learning a new language together, the three of us: you, I and Amos. We went through the magazines together... learning as we went. It seemed like everybody was doing it. Does that ring a bell?&lt;/p&gt;&#xA;&lt;p&gt;I barely saw any of the rest of your family. Your next youngest, (Agatha, maybe?)... the sophisticated one. I met her seven years into our relationship. She was smart, but I never thought she could all the things you could. I would have liked to meet the others, but they were always too professional. Too focused on their important work. Ivory towers and all that.&lt;/p&gt;&#xA;&lt;p&gt;I still think back to the games we used to play together. I sometimes have flashbacks to those times, but its so long now it seems like another world. It wasn't easy keeping up with you, but that's what I liked about you. (Those trainers you gave me helped! You can't get trainers like that any more!)&lt;/p&gt;&#xA;&lt;p&gt;I hope you're all at peace now, wherever you are.&lt;/p&gt;&#xA;&lt;p&gt;Love,&#xA;J&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14398" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-08-12T07:11:01.113" Title="A letter to my girlfriend, she would have been thirty this year" Tags="|riddle|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18630" PostTypeId="1" CreationDate="2015-08-04T21:23:21.193" Score="2" ViewCount="470" Body="&lt;p&gt;Have you ever, Ray, hit the floor.&lt;br&gt;&#xA;Hard times, still waiting, still worth fighting for.&lt;br&gt;&#xA;It ends tonight, go, fingers crossed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Face everything and rise, savior, Hellman.&lt;br&gt;&#xA;Live forever, so much better, Pepper, Iron man.&lt;br&gt;&#xA;Done is done, tonight the world dies. Let it be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[...]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve this tune, it isn't so simple,&lt;br&gt;&#xA;Even if the lyrics, aren't that subtle,&lt;br&gt;&#xA;Can you tell me who is mimicked and what's plagiarized?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The King of the Woods&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastEditorUserId="11251" LastEditDate="2016-07-19T11:23:23.777" LastActivityDate="2016-07-19T11:23:23.777" Title="A melancholic melody" Tags="|riddle|knowledge|music|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18631" PostTypeId="2" ParentId="18617" CreationDate="2015-08-04T21:23:43.537" Score="11" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It contains the letters of a number written out.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This answer is incomplete. See &lt;a href=&quot;https://puzzling.stackexchange.com/a/18633/4551&quot;&gt;commando's solution&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-08-05T04:29:59.970" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18632" PostTypeId="2" ParentId="18615" CreationDate="2015-08-04T21:29:17.610" Score="2" Body="&lt;p&gt;I believe that you're talking about Chess Pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Made from addition or subtraction. Not sure about the addition unless you're talking about starting a new game, but pieces are subtracted regularly over the course of a game. The best chess matches are often a popular attraction, even if it's just two old men playing in a public park.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;some you can touch and some you cannot. The king can never be captured, only held in checkmate or stalemate.&#xA;Technically it's more than a dozen, but there are 16 total pawns and they can &quot;heal&quot; missing pieces through pawn promotion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;words having two meanings? that's coming up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chess is certainly a game over 50 years old.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their brother, a movie star, is The Dark Knight.&#xA;And in The Dark Knight Rises, Batman does indeed bust out of jail, and in the process lets fear find him again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure about the pieces just for show, because even the most elaborately designed and expensive sets are certainly played with, I would assume. Though chess pieces are a very common theme used in the design of artwork in games and buildings. I would say the most commonly used might be the Rook and the Knight.&lt;/p&gt;&#xA;" OwnerUserId="13068" LastActivityDate="2015-08-04T21:29:17.610" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18633" PostTypeId="2" ParentId="18617" CreationDate="2015-08-04T21:49:05.893" Score="24" Body="&lt;p&gt;An adjacent word is a word which:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Can have, from its letters, two different numbers built which are adjacent integers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Additionally/Furthermore: From the letters in &quot;furthermore&quot; one may build both &quot;three&quot; and &quot;four&quot;, adjacent integers. This cannot be done with &quot;additionally&quot;.&lt;br&gt;&lt;br&gt;&#xA; Overfishing/Overharvesting: From the letters in &quot;overharvesting&quot; one may build &quot;seven&quot; and &quot;eight&quot;. Can't be done with &quot;overfishing&quot;, which only permits &quot;one&quot; and &quot;five&quot;.&lt;br&gt;&lt;br&gt;&#xA; Margarine/Shortening: From &quot;shortening&quot; one may build &quot;nine&quot; and &quot;ten&quot;.&lt;br&gt;&lt;br&gt;&#xA; Workbook/Worksheet: &quot;worksheet&quot; builds &quot;two&quot; and &quot;three&quot;&lt;br&gt;&lt;br&gt;&#xA; Ancient/Prehistoric: &quot;ancient&quot; builds &quot;nine&quot; and &quot;ten&quot;&lt;br&gt;&lt;br&gt;&#xA; Thawed/Unfroze: &quot;unfroze&quot; builds &quot;zero&quot; and &quot;one&quot;&lt;br&gt;&lt;br&gt;&#xA; Comprehensive/Expansive: &quot;expansive&quot; builds &quot;six&quot; and &quot;seven&quot;&lt;br&gt;&lt;br&gt;&#xA; Avoidable/Preventable: &quot;preventable&quot; builds &quot;ten&quot; and &quot;eleven&quot;&lt;br&gt;&lt;br&gt;&#xA; Unforgivable/Unpardonable: &quot;unforgivable&quot; builds &quot;four&quot; and &quot;five&quot;&lt;br&gt;&lt;br&gt;&#xA; Condominium/Townhouse: &quot;townhouse&quot; builds &quot;one&quot; and &quot;two&quot;&lt;br&gt;&lt;br&gt;&#xA; Irregulars/Reflexives: &quot;reflexives&quot; builds &quot;five&quot; and &quot;six&quot;&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="292" LastEditorUserId="292" LastEditDate="2015-08-04T21:57:28.210" LastActivityDate="2015-08-04T21:57:28.210" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18634" PostTypeId="2" ParentId="18615" CreationDate="2015-08-04T21:53:29.050" Score="2" Body="&lt;p&gt;What you are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Secret Six&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Where&lt;/em&gt; your brother is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Unless I'm missing some connection with the Hardy Boys, your &quot;brother&quot; in arms is Bane, and given what's occurred in the last stanza, he is dead.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are made from addition or subtraction&#xA;Our best are often a popular attraction&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Six can be made by adding or subtracting.  Not sure about the best part.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some you can touch,&#xA;and some you cannot&#xA;A dozen will heal your illness&#xA;Another shan't be bought&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There have been three Secret Six teams, and it looks like at least some of the members were doctors, but maybe not a dozen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We all can change one's point of view&#xA;But, y'know, sometimes words have two meanings&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I got nothing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some of us have utility&#xA;but others are for show&#xA;One befuddling, three worlds becoming one&#xA;made o'er fifty years ago&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The original Secret Six all have utility belts, but otherwise uninspired costumes, while the following teams were a bit more showy, with typical comic costumes.  There have been three teams all called the Secret Six (kinda befuddling), the first one from 1968-69.  Not quite o'er fifty, but pretty close.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By now you may have guessed what we are&#xA;so we ask that you find our brother, a movie star&#xA;He watched a broken man escape from prison&#xA;who let fear find him, and now is risen.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bane was a member of the Secret Six in the latest iteration.  He is also prominent in the movie The Dark Knight Rises, in which a broken-backed Batman (who embraces fear from bats to darkness) rises from a pit of a prison.  Bane doesn't survive the film, so where he is &lt;em&gt;is&lt;/em&gt; dead.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11452" LastActivityDate="2015-08-04T21:53:29.050" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18635" PostTypeId="2" ParentId="18617" CreationDate="2015-08-04T21:54:12.027" Score="7" Body="&lt;p&gt;Credit goes to xnor for finding the main part of this, but:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the adjacent numbers contain TWO consecutive numbers spelled out: &lt;br&gt;&#xA;  Furthermore -&gt; Three, Four &lt;br&gt;&#xA;  Overharvesting -&gt; Seven, Eight &lt;br&gt;&#xA;  Shortening -&gt; Eight, Nine &lt;br&gt;&#xA;  Worksheet -&gt; Two, Three &lt;br&gt;&#xA; Ancient -&gt; Nine, Ten &lt;br&gt;&#xA; Unfroze -&gt; Zero, One &lt;br&gt;&#xA; Expansive -&gt; Six, Seven &lt;br&gt;&#xA; Preventable -&gt; Ten, Eleven &lt;br&gt; &#xA; Unforgivable -&gt; Four, Five &lt;br&gt;&#xA; Townhouse -&gt; One, Two &lt;br&gt; &#xA; Reflexives -&gt; Five, Six &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10457" LastEditorUserId="10457" LastEditDate="2015-08-04T22:01:02.257" LastActivityDate="2015-08-04T22:01:02.257" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18636" PostTypeId="2" ParentId="18608" CreationDate="2015-08-04T22:04:54.807" Score="4" Body="&lt;p&gt;Is the answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;117&lt;/strong&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation:&lt;/strong&gt; &lt;br&gt;&#xA;First reveal the shifted meaning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Replace the quoted character with the other from its key on a keyboard (toggling the &lt;em&gt;shift&lt;/em&gt; key), giving + 852, / 1, - (-200), * 666, % 8, ^ 7, &amp;amp; 109 for the actual key 2 and * 4523, + 5, &amp;amp; 51, % 3 for the example.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then use the answer from before:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Apply each stage's operator separately to 29 (the &lt;em&gt;answer from before&lt;/em&gt;), interpreting % as modulus and &amp;amp; as bitwise AND to get 881, 29, 229, some big number, 5, another big number, 13 for the actual key and a big number, 32, 17, 2 for the example.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From there:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Proceed as in Anne's earlier answer, using each stage as a separate digit. For each digit, calculate if it is prime (1: yes, 0: no) (hint: Sieve of &lt;em&gt;Eratosthenes&lt;/em&gt;), giving 1110101 for the actual answer and 0011 for the example, and convert from base 2 (&lt;em&gt;yes or no&lt;/em&gt;) to base 10 (&lt;em&gt;0 to 9&lt;/em&gt;), giving a final answer of 3 for the example and &lt;strong&gt;117&lt;/strong&gt; for the actual key.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I didn't have to calculate the big numbers because they were either products of two numbers or powers of a number (therefore the product of multiple instances of the number) and were therefore not prime.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13152" LastActivityDate="2015-08-04T22:04:54.807" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18637" PostTypeId="2" ParentId="18602" CreationDate="2015-08-04T23:46:30.347" Score="4" Body="&lt;p&gt;Most of the given answers so far make assumptions beyond what we were given. I think the reason for that is that, as far as I can tell, everything works if either Henry or Edgar are guilty, unless you make at least one assumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the assumptions I won't make:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Thinking someone lied about their alibi isn't the same as thinking they're guilty of the crime (and a corollary: his alibi actually can BE false, even if he isn't guilty).&lt;/li&gt;&#xA;&lt;li&gt;Being &quot;on duty&quot; isn't an assumption of guilt. We're not even told where he works! It could very well be an alibi.&lt;/li&gt;&#xA;&lt;li&gt;Just because someone didn't state that they think someone else is guilty, doesn't mean they don't THINK someone else is guilty.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The one assumption that I am willing to make is that none of these people have serious mental illnesses. So, I think it's reasonable to accept that Fred won't know that James is lying about Henry, but still agree with him, or that Edgar couldn't commit the crime without thinking about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let's start with the easy part. Fred's first 2 statements are opposites, so one is true and one is false. This means that Edgar's 2nd statement is false. Since neither of them can have 3 true statements, either James or Henry must. Since James says outright that Henry is guilty, and Henry says outright that Edgar is guilty, we know that either Henry or Edgar is the guy. This means that Fred's 3rd statement is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's see if it works if James is the one to tell 3 true statements:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Of course, this means James has 3 true&lt;/li&gt;&#xA;&lt;li&gt;This means Henry's 1st and 3rd statement are false, so he either has 0 or 1 true&lt;/li&gt;&#xA;&lt;li&gt;This means Fred's 1st statement is false and 2nd is true. We already know his 3rd is true, so he has 2 true&lt;/li&gt;&#xA;&lt;li&gt;This means Edgar's 1st statement is false. We already know his 2nd is false, so he has either 0 or 1 true&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In this scenario, we have no way of confirming or denying Henry's 2nd statement or Edgar's 3rd statement. If one of them is false and one is true, then everything works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, if Henry is the one to tell 3 true statements:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;This means James's 1st statement is false, so he has 0, 1, or 2 true&lt;/li&gt;&#xA;&lt;li&gt;Of course, this means Henry has 3 true&lt;/li&gt;&#xA;&lt;li&gt;This means Fred's 1st statement is true and 2nd is false. We already know his 3rd is true, so he has 2 true&lt;/li&gt;&#xA;&lt;li&gt;This means Edgar's 3rd statement is false. We already know his 2nd is false, so he was either 0 or 1 true&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In this scenario, we have no way of confirming or denying James's 2nd statement or Edgar's 1st statement, but we do know that they're opposites. If James's 2nd is true, then Edgar's 1st is false. This means Edgar has 0 true, which means James must have 1 true, in which case, James's 3rd statement must be false. If James's 2nd is false, then Edgar's 1st is true. This means Edgar has 1 true, which means James must have 0 true, which again means that Jame's 3rd statement must be false. Either way, everything still works.&lt;/p&gt;&#xA;" OwnerUserId="14399" LastEditorUserId="14399" LastEditDate="2015-08-21T20:40:07.270" LastActivityDate="2015-08-21T20:40:07.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18638" PostTypeId="2" ParentId="18629" CreationDate="2015-08-05T01:04:13.687" Score="24" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Amiga 1000&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit - came back to give this some more attention.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; And a trip down memory lane - I bought one in 1985, cost me more than a month's salary.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clues from memory:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;30 years old - Amiga 1000 released in 1985 &lt;br&gt;&#xA;Tutankhamen - From the box of a program - presumably a paint program &lt;br&gt;&#xA;Workbench - the graphical file manager &lt;br&gt;&#xA;Bits of a plane - graphics, (e.g. HAM mode) &lt;br&gt;&#xA;Meditation - a Guru Meditation error appeared after a crash. &lt;br&gt;&#xA;Pets - Lemmings &lt;br&gt;&#xA;copper - co processor &lt;br&gt;&#xA;Denise - graphics chip &lt;br&gt;&#xA;Paula - sound chip &lt;br&gt;&#xA;Amos - the name of the Basic language that came with the Amiga &lt;br&gt;&#xA;Keeping up with you - Commodore slogan (Are your keeping up with the Commodore...) &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From research (I will delete these if research is not allowed):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Tutankhamun - Cover of Delux Paint box &lt;br&gt;&#xA;Europe/America - Reference to poor sales in America compared to Europe? &lt;br&gt;&#xA;Pets - Lemmings released in 1991 &lt;br&gt;&#xA;Statue - There is a lemmings statue in Dundee, Scotland &lt;br&gt;&#xA;Life of crime in America - Lemmings was developed by DMA design (this one confused me as I always remembered Psygnosis as the Lemmings developer, turns out they were the producer). DMA Design turned in Rockstar North of Grand Theft Auto fame. &lt;br&gt;&#xA;Pratchett - Interesting times had Lemmings-like control interface for an army of Golems (I had forgotten this one). &lt;br&gt;&#xA;You made things possible - advertising slogan for the Amiga? &lt;br&gt;&#xA;Agatha - AGA chip released in 1992 (seven years) for the 4000 &lt;br&gt;&#xA;Trainer - a way of modifying a game while it was running to cheat. &lt;br&gt;&#xA;J - Jay Miner - Amiga Inc Founder? &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are probably many more that I have missed/not got to yet.&lt;/p&gt;&#xA;" OwnerUserId="12405" LastEditorUserId="12405" LastEditDate="2015-08-09T04:15:43.620" LastActivityDate="2015-08-09T04:15:43.620" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18639" PostTypeId="2" ParentId="18611" CreationDate="2015-08-05T01:36:17.033" Score="1" Body="&lt;p&gt;Could your name be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Love? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a name for many people,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Love is popular last name, one of the famous bearers of it being &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Courtney_Love&quot; rel=&quot;nofollow&quot;&gt;Courtney Love&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am tangible, but not quite so.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Love is an intangible emotion, yet Rod Stewart claims to be able to &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Love_Touch&quot; rel=&quot;nofollow&quot;&gt;'Touch Love'&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;give me a kiss, my rival,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This could be a reference to the song &lt;a href=&quot;http://lyrics.wikia.com/Smiley_Kids:Love_Your_Enemies&quot; rel=&quot;nofollow&quot;&gt;'Love Your Enemies'&lt;/a&gt; by Smiley Kids. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You might find the answer laughable.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The song &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Queen_II&quot; rel=&quot;nofollow&quot;&gt;'Funny How Love Is'&lt;/a&gt; by Queen could be the link here. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-08-05T01:36:17.033" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18640" PostTypeId="2" ParentId="18616" CreationDate="2015-08-05T03:17:02.353" Score="10" Body="&lt;p&gt;&lt;strong&gt;Who might J and D be?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Either Joey or Johnny Ramone, and Dee Dee Ramone, of the band &lt;em&gt;The Ramones&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What strange game is this?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are anagramming the titles of &lt;em&gt;Ramones&lt;/em&gt; songs, and they need to solve the anagram, and respond with the year the song was recorded.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Two ton gator upwind&lt;/em&gt; is a valid challenge.  Why doesn't D know that?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It anagrams to &lt;em&gt;I Don't Want to Grow Up&lt;/em&gt;, recorded in 1995, after Dee Dee left the band.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If J challenged with &quot;&lt;em&gt;Am I mod Doorknob or celery ruler?&lt;/em&gt;&quot;, what is the correct response?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This anagrams to &lt;em&gt;Do You Remember Rock 'n' Roll Radio?&lt;/em&gt;, which was recorded in 1979, so the correct response is &quot;&lt;strong&gt;79&lt;/strong&gt;&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Solutions to the other questions:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Be tan, Theta Bro&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Beat_on_the_Brat&quot;&gt;Beat on the Brat&lt;/a&gt;&lt;/em&gt;, recorded in 1976.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Creaky boa chaw&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Rockaway_Beach_(song)&quot;&gt;Rockaway Beach&lt;/a&gt;&lt;/em&gt;, recorded in 1977.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Pa's seen chunkier okra&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Sheena_Is_a_Punk_Rocker&quot;&gt;Sheena is a Punk Rocker&lt;/a&gt;&lt;/em&gt;, recorded in 1977.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;End a sweat in a bed&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/I_Wanna_Be_Sedated&quot;&gt;I Wanna Be Sedated&lt;/a&gt;&lt;/em&gt;, recorded in 1978.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Hand pie&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Leave_Home&quot;&gt;Pinhead&lt;/a&gt;&lt;/em&gt;, recorded in 1977.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Oarsmen&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Ramones&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note Masher&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;The Ramones&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2015-08-05T03:17:02.353" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18641" PostTypeId="2" ParentId="18628" CreationDate="2015-08-05T06:20:00.610" Score="6" Body="&lt;p&gt;Partial answer. Using Hint 1 and the title, the next number is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;approximately 39&lt;/strong&gt; because the sequence represents the word PUZZLING.&lt;br&gt;&#xA;&lt;code&gt;&lt;br&gt;OP sequence =  71  82  94  94  53  43  62  ?&lt;br&gt;Letters =      P   U   Z   Z   L   I   N   G&lt;br&gt;ASCII code =   80  85  90  90  76  73  78  71&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;However, this transformation from the OP sequence to ASCII is not quite right as shown here:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/vNM9C.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastActivityDate="2015-08-05T06:20:00.610" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18643" PostTypeId="1" AcceptedAnswerId="18667" CreationDate="2015-08-05T08:27:22.320" Score="-3" ViewCount="4631" Body="&lt;p&gt;I've heard this rhyming riddle recently, and I just can't work it out. What am I thinking of?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My first is in cow, but not in pat.&lt;br&gt;&#xA;My second in mouse, yet not in rat.&lt;br&gt;&#xA;My third is in rhythm, but not in rhyme.&lt;br&gt;&#xA;My whole is a waste of all your time. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clue:&lt;/strong&gt; This has now been answered, but here's a clue, in case you're struggling...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is hiding in plain sight.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2673" LastEditorUserId="2673" LastEditDate="2016-05-09T06:55:43.490" LastActivityDate="2016-05-09T06:55:43.490" Title="I'm stuck, what is the answer to this riddle?" Tags="|riddle|lateral-thinking|rhyme|" AnswerCount="5" CommentCount="8" ClosedDate="2015-08-06T19:06:53.670" ContentLicense="CC BY-SA 3.0" />
  <row Id="18644" PostTypeId="2" ParentId="18628" CreationDate="2015-08-05T08:43:24.750" Score="10" Body="&lt;p&gt;The next number is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 41&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Len has already found out that the numbers correspond to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the letters in the word &quot;puzzling&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But they are encoded:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each letter is encoded as the number key on a phone that it is on plus the index of that letter among the three or four letters on that key. For example, P is the first letter on the 7 key and is encoded as 71. The whole sequence is:&lt;br&gt;&#xA;&lt;br&gt;&#xA;    &lt;code&gt;P    7 Pqrs     71&lt;/code&gt;&lt;br&gt;&#xA;    &lt;code&gt;U    8 tUv      82&lt;/code&gt;&lt;br&gt;&#xA;    &lt;code&gt;Z    9 wxyZ     94&lt;/code&gt;&lt;br&gt;&#xA;    &lt;code&gt;Z    9 wxyZ     94&lt;/code&gt;&lt;br&gt;&#xA;    &lt;code&gt;L    5 jkL      53&lt;/code&gt;&lt;br&gt;&#xA;    &lt;code&gt;I    4 ghI      43&lt;/code&gt;&lt;br&gt;&#xA;    &lt;code&gt;N    6 mNo      62&lt;/code&gt;&lt;br&gt;&#xA;    &lt;code&gt;G    4 Ghi      41&lt;/code&gt;   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks to Len, who provided the key insight in his answer.&lt;/p&gt;&#xA;" OwnerUserId="9860" LastEditorUserId="12700" LastEditDate="2015-08-05T11:32:44.130" LastActivityDate="2015-08-05T11:32:44.130" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18645" PostTypeId="2" ParentId="18643" CreationDate="2015-08-05T08:53:30.030" Score="5" Body="&lt;p&gt;Initial thoughts/partial answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like each 'part' describes a letter that appears in the first word but not the second.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't help us much until 'rhythm/rhyme':  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the letter 't' is the only one that appears only in rhythm, so the third letter of the solution is &lt;strong&gt;t&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The words that appear with that information are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;cut, wet, cot, out&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The last line  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;gives us the definition of the solution. Not sure yet whether the whole thing is only 3 letters long or if we need to use the first three letters at the start of a word - if the latter, the first three don't necessarily need to form a word themselves&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11717" LastEditorUserId="11717" LastEditDate="2015-08-05T10:39:45.330" LastActivityDate="2015-08-05T10:39:45.330" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18647" PostTypeId="2" ParentId="18602" CreationDate="2015-08-05T11:10:50.650" Score="2" Body="&lt;p&gt;Basics:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There is a single culprit (as stated in the original question &quot;...the&#xA;guilty one&quot;)&lt;/li&gt;&#xA;&lt;li&gt;Each suspect makes a different number of true statements&#xA;(0, 1, 2, or 3).&lt;/li&gt;&#xA;&lt;li&gt;I'll use J1, J2, J3, H1, H2, etc. to represent the statements from each&#xA;suspect.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Key Assumptions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;A guilty person cannot think someone else is also guilty&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;Suspicion of guilt is not withheld from the statements made to the owner&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As most of the other great answers have done, we'll start with Fred: F1 and F2 cannot both be TRUE, and cannot both be FALSE, so Fred has made at least one true statement (and one false statement). Thus, E2 cannot be true, due to above (F1 and F2 being mutually exclusive).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;# True Statements:&#xA;1 &amp;lt;= Fred &amp;lt; 3 (F T * or T F *)&#xA;Edgar &amp;lt; 3 (* F *)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Therefore, either James or Henry must be the suspect with three truthful statements. Thus, Henry (per James) or Edgar (per Henry) is the thief.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fred, then, is not the thief, so F3 is TRUE, and Fred has told exactly two truths. Since no two suspects can tell the same number of truthful statements, Edgar has told at most one truth.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Fred == 2 (F T T or T F T)&#xA;Edgar &amp;lt;= 1 (* F *)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;H1 and H3 are linked: Henry cannot lie about H1 and tell the truth about H3. Conversely, he cannot lie about H3 and tell the truth about H1. Both situations would result in neither James nor Henry being completely truthful, which conflicts with the guidelines. Furthermore, since Fred has already been determined to be the suspect with two truthful statements, Henry must be either completely truthful, or he is telling at most one truth.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;James == 3 (Henry is guilty)&#xA;Henry &amp;lt;= 1 (F * F)&#xA;Fred == 2 (F T T)&#xA;Edgar &amp;lt;= 1 (* F *)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;--OR--&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;James &amp;lt;= 1 (F * *)&#xA;Henry == 3 (Edgar is guilty)&#xA;Fred == 2 (T F T)&#xA;Edgar &amp;lt;= 1 (* F *)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If&lt;/strong&gt; Henry is completely truthful (H1, H2, and H3 are all TRUE, along with F1), then J1 == FALSE: J1 and H1 cannot both be true.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;James &amp;lt;= 1 (F * *)&#xA;Henry == 3 (T T T)&#xA;Fred == 2 (T F T)&#xA;Edgar &amp;lt;= 1 (* F *)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;E3 == FALSE: Edgar cannot commit the crime without thinking about it (unless we're really reaching into sleep-rustling, split personalities, demonic possession, etc.)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;James &amp;lt;= 1 (F * *)&#xA;Henry == 3 (T T T)&#xA;Fred == 2 (T F T)&#xA;Edgar &amp;lt;= 1 (* F F)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;J3 == TRUE: The only person who even comes close to accusing James is Edgar (even then, he only claims the alibi is false, which conveys no direct accusation). In any event, &lt;strong&gt;if Edgar is guilty (per Henry), Edgar would know James didn't do it&lt;/strong&gt;, so it is indeed the case that no one thinks James is guilty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, however, we reach a conflict in the Henry-is-correct path: If J2 == TRUE, then James and Fred would both have two truths, which cannot occur, per the guidelines. If J2 == FALSE, then E1 == TRUE. This is also a conflict, since we would then have no suspects that were completely untruthful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, James is the suspect telling the whole truth, and Henry is the thief:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;J1 == TRUE: Henry is guilty&#xA;J2 == TRUE: Alibi is true&#xA;J3 == TRUE: No one thinks James did it&#xA;&#xA;H1 == FALSE: James is telling the truth&#xA;H2 == TRUE/FALSE (opposite of E3): Fred being on duty (or not) is irrelevant&#xA;H3 == FALSE: Edgar did not do it&#xA;&#xA;F1 == FALSE: Henry is wrong&#xA;F2 == TRUE: James is right&#xA;F3 == TRUE: Fred did not do it&#xA;&#xA;E1 == FALSE: The alibi of James holds&#xA;E2 == FALSE: Fred made two TRUE statements&#xA;E3 == FALSE/TRUE (opposite of H2): Irrelevant if Edgar isn't guilty&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(The final outcome of H2 and E3 do not affect guilt of Henry, and there do not appear to be any statements clearly pointing to one being truthful over the other, outside of speculation of which person might be more likely to lie about those scenarios.)&lt;/p&gt;&#xA;" OwnerUserId="14405" LastActivityDate="2015-08-05T11:10:50.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18648" PostTypeId="2" ParentId="18614" CreationDate="2015-08-05T11:17:29.867" Score="0" Body="&lt;p&gt;I think this series has less to do with primes.&#xA;What you can try is check whether the differences of some terms are in A.P. or not.&lt;br&gt;&#xA;For example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;6-5=1 &lt;br&gt;&#xA;  8-6=2 &lt;br&gt;&#xA;  11-8=3 &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;since 1,2,3 are in AP. you can proceed with this fact.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[edit]: Now,&#xA;  45-11=34 so it breaks the AP pattern.&lt;br&gt;&#xA;  But we have:&lt;br&gt;&#xA;  74-45=29&lt;br&gt;&#xA;  101-74=27&lt;br&gt;&#xA;  Since 34,29,27 form a series( diference 5,2 and then 1 as 5/2=2 and 2/2=1. by integer division like in C/C++)&#xA;  &lt;br&gt;&#xA;  Thus the next term is: 101+ {27-(2/2)}= &lt;strong&gt;127&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1766" LastEditorUserId="1766" LastEditDate="2015-08-05T12:06:21.980" LastActivityDate="2015-08-05T12:06:21.980" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18649" PostTypeId="2" ParentId="18611" CreationDate="2015-08-05T12:06:19.543" Score="4" Body="&lt;p&gt;The answer is probably&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ass(hole)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a name for many people:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;That is most certainly a frequently used insult.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am tangible, but not quite so:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Well, one cannot touch a hole technically. But one can touch an ass (probably shouldn't though).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Give me a kiss, my rival:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;We all know the old saying: &quot;Kiss my ass!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You might find the answer laughable:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This is sure not a classical answer for a riddle ;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5106" LastActivityDate="2015-08-05T12:06:19.543" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18650" PostTypeId="1" AcceptedAnswerId="19784" CreationDate="2015-08-05T12:20:26.160" Score="7" ViewCount="440" Body="&lt;p&gt;I am the most powerful, a group of people.&lt;br&gt;&#xA;My opposite was once my rival, able to change between others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can both be found in the same surroundings, often doing very different things.&lt;br&gt;&#xA;For example, one of us could be relaxed while the other is making things happen.&lt;br&gt;&#xA;One could be underneath while the other is making connections.&lt;br&gt;&#xA;One could be in court while the other is getting help from equipment.&lt;br&gt;&#xA;One can bring power while the other shows disease.&lt;br&gt;&#xA;On the radio, one of us could be reading while the other gives chase.&lt;br&gt;&#xA;Normally, though, our interests overlap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What am I and what  is my opposite?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus: where else can we both be found?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;a href=&quot;/questions/tagged/wordplay&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;wordplay&amp;#39;&quot; rel=&quot;tag&quot;&gt;wordplay&lt;/a&gt; tag is important to this question, &lt;a href=&quot;/questions/tagged/riddle&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;riddle&amp;#39;&quot; rel=&quot;tag&quot;&gt;riddle&lt;/a&gt; not so much.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As suggested by &lt;a href=&quot;https://puzzling.stackexchange.com/a/19783/13152&quot;&gt;rand al'thor&lt;/a&gt;, the answer is a letter or small group of letters rather than a word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each 1/2 line gives a single word that is a clue towards the final answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13152" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-08-08T20:28:52.937" Title="Me and my opposite" Tags="|riddle|pattern|wordplay|computer-puzzle|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18651" PostTypeId="2" ParentId="18643" CreationDate="2015-08-05T12:20:58.080" Score="5" Body="&lt;p&gt;Maybe it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;coot&lt;/strong&gt;&lt;br&gt;&lt;br&gt;Take the &quot;co&quot; from cow, the &quot;o&quot; from mouse, and the &quot;t&quot; from rhythm.&lt;br&gt;&lt;br&gt;Google says coot, informally, means &quot;a foolish or eccentric person, typically an old man,&quot; so it's no stretch to say that a foolish person could be &quot;a waste of all your time.&quot;&lt;br&gt;&lt;br&gt;&lt;sub&gt;However, I was hoping that the answer involved taking the &quot;th&quot; from rhythm, as Avigrail pointed out in his comment.&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-08-05T12:33:18.543" LastActivityDate="2015-08-05T12:33:18.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18652" PostTypeId="2" ParentId="18650" CreationDate="2015-08-05T12:43:12.757" Score="1" Body="&lt;p&gt;I think you might be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an electromagnetic wave. Specifically the electric field and magnetic field making up the wave.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can both be found in the same surroundings, often doing very different things.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Electric and magnetic fields coexist in nature, but both are useful in their own applications.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example, one of us could be relaxed while the other is making things happen. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In an electromagnet, the electric field gives rise to a magnetic field, and that magnetic field is used to lift heavy objects.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One could be underneath while the other is making connections.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The magnetic field is present in telephone wires, where electric impulses are used to relay conversations. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One could be in court while the other is getting help from equipment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Not sure about this clue...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can bring power while the other shows disease. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Electric fields across power lines transfer power across cities. Meanwhile, MRI machines use magnetic fields to detect diseases and injuries in humans.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On the radio, one of us could be reading while the other gives chase.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The magnetic field is playing sound through speakers while the electric field is being used to reach out to devices across the city.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Normally, though, our interests overlap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Not sure about this clue...&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-08-05T12:57:18.617" LastActivityDate="2015-08-05T12:57:18.617" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18653" PostTypeId="2" ParentId="18650" CreationDate="2015-08-05T12:51:21.493" Score="0" Body="&lt;p&gt;It could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;the majority&lt;/strong&gt;, the &quot;opposite&quot; being the minority&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the most powerful, a group of people.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the majority is powerful because they have more votes, and a greater number of people&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My opposite was once my rival, able to change between others.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; in the past, minority groups were less accepted than in today's world (although there is still plenty of pressure to conform to the majority)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can both be found in the same surroundings, often doing very different things.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; self-explanatory&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example, one of us could be relaxed while the other is making things happen.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Usually, the people who make things happen are in the minority, while the majority of people are more relaxed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One could be underneath while the other is making connections.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The majority of people are &quot;underneath,&quot; or behind-the-scenes, and the minority of people are the spokespeople to the public (making connections).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One could be in court while the other is getting help from equipment.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure about this one&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One can bring power while the other shows disease.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The majority brings power, while the minority shows disease&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On the radio, one of us could be reading while the other gives chase.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Normally, though, our interests overlap.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The interests of the majority and minority overlap on many issues&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastActivityDate="2015-08-05T12:51:21.493" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18655" PostTypeId="2" ParentId="18630" CreationDate="2015-08-05T13:49:41.367" Score="5" Body="&lt;p&gt;Well it seems to be focused around&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Milencolin: done is done, pepper,  hellman, ray, hard times, fingers crossed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which sadly I am not an expert on, further analysis&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; brandy, millencolin, linkin park &lt;Br&gt;&#xA;Millencolin, sum 41, my darkest days. &lt;br&gt;&#xA;all american rejects, (various artists), millencolin.&lt;br&gt; &lt;br&gt;&#xA;papa roach, rise against, millencolin. &lt;br&gt;&#xA;Oasis, eminem, millencolin, ozzy&lt;br&gt;&#xA;millencolin, papa roach, beatles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It will be rather difficult for me to find the actual (song?) being (recreated?). but I like the title. Might work further... but not an area of expertise.&lt;/p&gt;&#xA;" OwnerUserId="13916" LastActivityDate="2015-08-05T13:49:41.367" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18656" PostTypeId="2" ParentId="18630" CreationDate="2015-08-05T13:53:31.600" Score="2" Body="&lt;p&gt;Based on  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Marvel Characters&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Have you ever, Ray, hit the floor.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Beta Ray Bill, an alien granted a Mjolnir by All-Father Odin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hard times, still waiting, still worth fighting for.&#xA;It ends tonight, go, fingers crossed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Ragnarok, the end of all things.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Face everything and rise, savior, Hellman.&#xA;Live forever, so much better, Pepper, Iron man.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tony Stark and Pepper Potts&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Done is done, tonight the world dies. Let it be.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Implying the finality of Ragnarok. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To solve this tune, it isn't so simple,&#xA; Even if the lyrics, aren't that subtle,&#xA; Can you tell me who is mimicked and what's plagiarized?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Man creating weapons to mimic The End is mimicked; Ragranok is plagarized.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14412" LastActivityDate="2015-08-05T13:53:31.600" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18657" PostTypeId="1" AcceptedAnswerId="18669" CreationDate="2015-08-05T14:17:20.390" Score="11" ViewCount="1251" Body="&lt;p&gt;Thanks to all of you, I know now what my friend was telling me (see &lt;a href=&quot;https://puzzling.stackexchange.com/q/18601/13384&quot;&gt;part one&lt;/a&gt;).&lt;br&gt;&#xA;We continued talking via Skype. After some nice messages about the utility of iron in our society, we decided to meet in 15 minutes.&#xA;I arrived early, as usual.&#xA;&lt;br&gt;I waited 20 minutes, then I received a strange SMS (he recharged his Nokia) :&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(255) (153) (153) (255) (231) (129) (129) (153) (255)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;When I asked him what was going on, he didn't answer me.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Question&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;What did he send me?&lt;br&gt;&#xA;Why is he late?&lt;br&gt;&#xA;Where were we planning to meet?&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Hints&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Hint 1 :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Tags can help you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2 : &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Context is like size... it matters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3 :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Stop looking for another hint! Do you think this is a game?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13384" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-08-06T01:55:44.470" Title="The strange SMS (Part 2)" Tags="|cipher|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18659" PostTypeId="2" ParentId="18643" CreationDate="2015-08-05T14:56:19.123" Score="3" Body="&lt;p&gt;I think the answer is literally&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cut Somehow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first is in cow, but not in pat.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Taking the whole COW&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My second in mouse, yet not in rat.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; and a MOUSE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My third is in rhythm, but not in rhyme.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; and TH&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My whole is a waste of all your time.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Mixing it all together gives COWMOUSETH&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;of all the different ones selected this...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; means the riddle is CUT SOMEHOW and therefore waste of time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9892" LastActivityDate="2015-08-05T14:56:19.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18660" PostTypeId="1" AcceptedAnswerId="18663" CreationDate="2015-08-05T15:06:55.637" Score="3" ViewCount="258" Body="&lt;p&gt;Some can be forgiven for thinking I watch you, &lt;br&gt;&#xA;but in fact you are the one watching me.&lt;br&gt;&#xA;I suppose I keep an eye on the mathy fellow,&lt;br&gt;&#xA;but that just makes me a little bit gooey.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess I don't lead but I can be led,&lt;br&gt;&#xA;though sometimes I have crystallizing moments.&lt;br&gt;&#xA;without light I am out of sight however, &lt;br&gt;&#xA;with a magnet I can get real funky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give your answer you would key,&#xA;but you would get little feedback without me.&lt;/p&gt;&#xA;" OwnerUserId="13916" LastActivityDate="2015-08-05T15:19:43.897" Title="Look at the question and see me" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18661" PostTypeId="2" ParentId="18657" CreationDate="2015-08-05T15:12:45.557" Score="8" Body="&lt;p&gt;I'll go with a partial answer to get this started: the given numbers&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; have an interesting property when translated to binary. They look like this:&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;code&gt;11111111&lt;/code&gt;&#xA;&lt;br&gt;&#xA; &lt;code&gt;10011001&lt;/code&gt;&#xA;&lt;br&gt;&#xA; &lt;code&gt;10011001&lt;/code&gt;&#xA;&lt;br&gt;&#xA; &lt;code&gt;11111111&lt;/code&gt;&#xA;&lt;br&gt;&#xA; &lt;code&gt;11100111&lt;/code&gt;&#xA;&lt;br&gt;&#xA; &lt;code&gt;10000001&lt;/code&gt;&#xA;&lt;br&gt;&#xA; &lt;code&gt;10000001&lt;/code&gt;&#xA;&lt;br&gt;&#xA; &lt;code&gt;10011001&lt;/code&gt;&#xA;&lt;br&gt;&#xA; &lt;code&gt;11111111&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I guess this should tell me something, but I haven't figured out what yet.&lt;/p&gt;&#xA;" OwnerUserId="5990" LastActivityDate="2015-08-05T15:12:45.557" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18662" PostTypeId="2" ParentId="18347" CreationDate="2015-08-05T15:15:57.947" Score="2" Body="&lt;p&gt;The Secret:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Ability, Passion, Curiosity and Truth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A man of prowess hides them both&#xA; preparing for his calling.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;As in a man with his arms behind his back, anticipating, Standing still.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another man provides them both&#xA; with sparks and cinders falling.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;As in a man facing trouble;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Two snakes slither, silently&#xA; surrounding unaware&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The two snakes are Life and Death&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;a line about what's lost between&#xA; triumph and moving air.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Triumph&quot; refers to our sentience, &quot;Moving Air&quot; refers to the non-sentient around us.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A penciled confirmation&#xA; performs music to the dance&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A Heart Beating in response to action.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;of clocks and creepy crawly things&#xA; at rest inside your pants.&#xA; A sudden motion shocks you&#xA; in a thousand different ways.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;A sudden motion&quot; as in an Emotion, making the snakes slither and the music change;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In other words, the seventh note&#xA; sings Casey's gone astray.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The last breath (The Seventh (last) note) shows a Man's fall to air.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The dispossessed are missing one&#xA; to become shrouded skies.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The dispossessed miss reality or ability, becong distant.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Perform on it this action&#xA; as you do with dotted eyes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Dotted Eyes&quot; as in unseeing and/or unfocused eyes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The aforementioned snakes return,&#xA; though this time stick together.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Dispossessed prefer neither to be alive, nor dead, for their dreams tend to be their reality.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A bird of death approaches them&#xA; with blackened, shedding feathers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Blackened, Shedding Feathers&quot; and &quot;Bird of Death&quot; refer to Old Age, and th death of the Dispossessed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unseen to the untrained eye&#xA; vast treasures within lie,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Vast treasures lie within the world, which only curious eyes fathom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and bones are ever rattling&#xA; as good men herein die.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Men die constantly in the world.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A waterfall appears from fears&#xA; no longer held within,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Men crying at liberation of Long-Held fears&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and paper tiger cardinals&#xA; let waterfalls begin.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Paper tiger cardinals as in Tissues?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In pairs, the strangers take the rear&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Strangers: Ability, Passion. &quot;Strangers&quot; in the sense that nobody possesses these traits at birth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;as giants lead the way.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Giants: Curiosity, Truth. &quot;Giants&quot; in the sense that without them, the strangers would never meet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Manipulate them slightly&#xA; and your answer's clear as day.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Ability, Passion, Curiosity and Truth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14412" LastActivityDate="2015-08-05T15:15:57.947" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18663" PostTypeId="2" ParentId="18660" CreationDate="2015-08-05T15:17:08.707" Score="4" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A (computer) monitor&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some can be forgiven for thinking I watch you,&lt;br&gt;&#xA;but in fact you are the one watching me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A monitor can be one who watches, but in this case, I'm looking at my monitor right now&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I suppose I keep an eye on the mathy fellow,&lt;br&gt;&#xA;but that just makes me a little bit gooey.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not entirely, sure, but I'm guessing the mathy fellow is the CPU?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I guess I don't lead but I can be led,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LED monitors &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;though sometimes I have crystallizing moments.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Or liquid crystal monitors&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;without light I am out of sight however, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can't see anything if the monitor produces no light&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;with a magnet I can get real funky.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Especially if you're a CRT monitor, as magnets will distort the picture&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To give your answer you would key, but you would get little feedback without me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I won't know I'm right until I see a response to this on my monitor :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13777" LastEditorUserId="13777" LastEditDate="2015-08-05T15:19:43.897" LastActivityDate="2015-08-05T15:19:43.897" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18664" PostTypeId="2" ParentId="18657" CreationDate="2015-08-05T15:26:09.207" Score="0" Body="&lt;p&gt;Partial answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following on from the solution to the first SMS puzzle, and so trying to work out how often to press a number key to generate a certain letter:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;55 is the decimal code for the ASCII character '7', and 53 is the code for '5'. Using a phone keypad again, 255 -&gt; 2 &lt;strong&gt;7&lt;/strong&gt;s -&gt; &lt;strong&gt;R&lt;/strong&gt;, and 153 -&gt; 1 &lt;strong&gt;5&lt;/strong&gt; -&gt; &lt;strong&gt;J&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That only covers two potential letters, though.&lt;/p&gt;&#xA;" OwnerUserId="11717" LastActivityDate="2015-08-05T15:26:09.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18665" PostTypeId="1" AcceptedAnswerId="18672" CreationDate="2015-08-05T15:38:06.523" Score="5" ViewCount="732" Body="&lt;p&gt;A mad puzzle enthusiast locks you inside a room with only one door. Next to the door is a standard numeric keypad. There is no way to escape except through the door. You find a piece of paper next to the keypad:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$181 \Rightarrow E \unicode{12288} 3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3 \unicode{12288} -E \Rightarrow \unicode{xFF1F}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You turn the piece of paper over and see two more scribbles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The answer is not in sight.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you type in the keyboard to escape, and how did you figure it out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You notice that the puzzle enthusiast had one more equation but erased it. Following the indentations on the paper, you find out that it said: $3 \unicode{x005F} 1 \Rightarrow 1A3$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Big Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The numbers on the keypad are all in seven-segment display. You wonder if that means anything...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since this is my first puzzle, any feedback would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="11121" LastEditorUserId="11121" LastEditDate="2015-08-07T14:26:09.160" LastActivityDate="2015-08-07T14:34:54.353" Title="Unconventional Math - Anti-Number Puzzle" Tags="|lateral-thinking|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18666" PostTypeId="2" ParentId="18643" CreationDate="2015-08-05T15:42:29.283" Score="15" Body="&lt;p&gt;Is it perhaps&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;waste of time?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Going through each of the clues, one possible combination of letters would be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;'WOT'. The clues could be referring to the first letters of each word - for example, &quot;my second&quot; indicating what letter the second word starts with.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In this way, the letters&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;'WOT' stand for something that is a waste of all your time - and you can't get much more exact than &quot;waste of time&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14107" LastActivityDate="2015-08-05T15:42:29.283" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18667" PostTypeId="2" ParentId="18643" CreationDate="2015-08-05T15:42:50.970" Score="4" Body="&lt;p&gt;This might be a bit out there but, perhaps...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;...there is no answer. The last line says &quot;My whole is a waste of all your time.&quot; What if it refers the the riddle as a whole? There's no solution, therefore it's a waste of your time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On the flip side, this may not be the case given the use of the word&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;all&quot;. It's not a waste of all your time, just some of it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4354" LastActivityDate="2015-08-05T15:42:50.970" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18668" PostTypeId="1" AcceptedAnswerId="18682" CreationDate="2015-08-05T16:09:09.810" Score="5" ViewCount="1290" Body="&lt;p&gt;This is Room 4 of the Maze series. For those wishing to start from the beginning, click &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18567/room-1-of-the-maze&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You wake up in Room 4 after a troubled night of 'sleep'. Your head was clouded all night with thoughts of geography, chemistry, and math. &quot;What's coming next?&quot;, you wonder as you head for the note. But instead of a note on the platform, you see that the platform is glowing like a computer screen, showing a page on your favorite site, the Puzzling StackExchange! You realize that the platform is a sort of touch-screen interface so you attempt to navigate to an emergency service site, but all you can see on the interface is a question and the edit, close, and flag buttons. Even the add comment and add answer features are missing. Baffled, you scroll and read the question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Welcome to Room 4 of the Maze! You have officially made it farther than any of the other test subjects. You humans have an expression, &quot;Out of the frying pan and into the fire&quot;; if we understand it correctly, your current situation is an apt application of that idiom. From here on, the puzzles get noticeably harder. You're going to have to start thinking outside of the box if you want to live, but we have faith that you're up for the challenge. Good Luck!!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Key 1:COPY&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;STOP&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Key 2:&lt;br&gt;&#xA;  124 45 124 45 45 45 45 45 45 45 45 124 45 45 45 45 45 45 45 45 124 45 45 45 45 45 45 45 45 124 45 45 45 45 45 45 45 45 124 10 10 124 45 124 45 45 45 45 45 45 45 45 124 45 45 45 45 113 45 45 45 124 45 45 45 45 45 45 45 45 124 45 45 45 45 45 45 45 45 124 10 67 53 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 113 32 32 32 113 32 32 32 32 101 10 124 45 124 45 45 45 45 45 45 45 45 124 45 45 45 101 45 45 45 45 124 45 45 45 45 45 45 45 45 124 45 45 45 45 45 45 45 45 124 10 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 101 32 32 32 101 32 32 101 32 101 10 124 45 124 113 45 45 45 113 45 45 45 124 101 45 101 45 45 45 45 45 124 45 45 45 45 45 45 45 45 124 45 45 45 45 45 45 45 45 124 10 35 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 101 10 124 45 124 45 45 45 45 45 45 45 45 124 45 45 45 45 45 45 45 45 124 45 45 45 45 45 45 45 45 124 45 45 45 45 45 45 45 45 124 10 32 32 32 32 32 32 101 32 32 32 101 32 32 101 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 113 10 67 52 10  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The keypad on the door has only digits. It holds enough room on the display for 3 digits. You can faintly see fingerprints on some of the keys. &quot;That's odd; I thought no one had ever made it this far...&quot;, you think to yourself. Shaking your head to concentrate, you turn back to the bewildering information on the screen and get to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer from &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18608/room-3-of-the-maze&quot;&gt;Previous Room&lt;/a&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;117&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This room has been solved. Normally I would wait until the next morning to post the next room, but figured I'd give the night shift a chance. Are you any closer to escaping the Maze? What's the next subject matter? Will anyone ever get sent to the universe of Bud Light Lime? Find out in the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18692/room-5-of-the-maze&quot;&gt;next installment&lt;/a&gt;!&lt;/p&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-08-06T06:51:09.357" Title="Room 4 of the Maze" Tags="|cipher|enigmatic-puzzle|password|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18669" PostTypeId="2" ParentId="18657" CreationDate="2015-08-05T16:39:28.620" Score="11" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What did he send me?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, as found by Aioros,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Translating the numbers into binary returns what appears to be a &quot;creeper&quot; from the game Minecraft: &#xA;&lt;pre&gt;&#xA;11111111&#xA;10011001&#xA;10011001&#xA;11111111&#xA;11100111&#xA;10000001&#xA;10000001&#xA;10011001&#xA;11111111&#xA;&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Next,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is he late?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, given the situation, he was probably&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Blasted to smithereens. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And finally,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Where were we planning to meet?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again, given the situation,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The discussion about the utility of iron could be hinting at the Iron Golems in the game that can protect villages. Because of this, I would guess that you and your friend are villagers, and you were planning to meet at the village.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10078" LastEditorUserId="9901" LastEditDate="2015-08-06T01:55:44.470" LastActivityDate="2015-08-06T01:55:44.470" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18671" PostTypeId="2" ParentId="18668" CreationDate="2015-08-05T17:07:15.590" Score="10" Body="&lt;p&gt;Converting the numbers in Key 2 into ASCII results in this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;|-|--------|--------|--------|--------|&#xA;&#xA;|-|--------|----q---|--------|--------|&#xA;C5                   q   q    e&#xA;|-|--------|---e----|--------|--------|&#xA;                        e   e  e e&#xA;|-|q---q---|e-e-----|--------|--------|&#xA;#                               e&#xA;|-|--------|--------|--------|--------|&#xA;      e   e  e                    q&#xA;C4&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is a picture of a musical score. The piece is &quot;Eine kleine Nachtmusik&quot; by Mozart, which has a catalog number of 525.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The source of the question shows a bunch of spaces, tabs, and newlines in Key 1. Here they are with spaces replaced by S and tabs by T:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;SSS&#xA;SST&#xA;TT&#xA;SSTS&#xA;STS&#xA;S&#xA;TTST&#xA;SSTS&#xA;STSS&#xA;TTT&#xA;TTT&#xA;STS&#xA;ST&#xA;TSS&#xA;TSS&#xA;TSSSS&#xA;TTTTT&#xA;SSSST&#xA;TTTTS&#xA;ST&#xA;TSS&#xA;TSS&#xA;ST&#xA;SSTS&#xA;STTS&#xA;STS&#xA;SSS&#xA;TTST&#xA;STS&#xA;T&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This looks like a Whitespace program, but it seems to be broken.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't figure out what to do with the previous part's answer. Without any other information, 525 is the only three-digit number related to anything here.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="1777" LastEditDate="2015-08-05T18:33:23.680" LastActivityDate="2015-08-05T18:33:23.680" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18672" PostTypeId="2" ParentId="18665" CreationDate="2015-08-05T17:31:51.817" Score="6" Body="&lt;p&gt;That big hint is a rather big hint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Punch in&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $1801$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $181 \Rightarrow E \space 3$&lt;br&gt;&#xA; Here, the 7-segment display is inverted for each number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second one has four entities on the left-hand side, so there will be four on the right-hand side.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $3$ inverted gives $1$.&lt;br&gt;&#xA; $\space$ inverted gives $8$.&lt;br&gt;&#xA; $-$ inverted gives $0$.&lt;br&gt;&#xA; $E$ inverted gives $1$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Old answer left in for posterity.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Punch in:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 12&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you consider that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the spaces could stand for something, we can make them stand for $+17$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This gives us:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $181 \Rightarrow E+173$ and $3+17-E \Rightarrow ?$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the first, we see that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $E = 8$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which we can now plug in and solve as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $3+17-8 = 12$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-08-07T14:34:54.353" LastActivityDate="2015-08-07T14:34:54.353" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18673" PostTypeId="2" ParentId="15809" CreationDate="2015-08-05T18:04:18.457" Score="3" Body="&lt;p&gt;With some assistance from &lt;a href=&quot;http://www.thehalloffire.net/forum/viewtopic.php?t=12&quot; rel=&quot;nofollow&quot;&gt;this thread&lt;/a&gt;, here is an explanation.  Upgrades are welcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Left side of page:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Am sending you one all&lt;/strong&gt; (ham (s+end in g) u 1 awl)&lt;br&gt;&#xA;&lt;strong&gt;pictures.&lt;/strong&gt; (pea+eye+sea+tea+ures.)&lt;br&gt;&#xA;&lt;strong&gt;We each have&lt;/strong&gt; (we a+ech hay+ve)&lt;br&gt;&#xA;&lt;strong&gt;found two lovely&lt;/strong&gt; (f+hound-h 2 (50=L+dove-d+50=L+y))&lt;br&gt;&#xA;&lt;strong&gt;walks to take you&lt;/strong&gt; (wall+keys-ey 2 t+ache u)&lt;br&gt;&#xA;&lt;strong&gt;when you do! Come&lt;/strong&gt; (w+hen yew 500=D+o! Comb-b)&lt;br&gt;&#xA;&lt;strong&gt;out here which we hope&lt;/strong&gt; (ou+tea h+ear witch we h+pop-p+e)&lt;br&gt;&#xA;&lt;strong&gt;will be soon.&lt;/strong&gt; (w+ill bee s+moon-m).  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Right side of page:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Your loving&lt;/strong&gt; (Wye+hour-h l+oven+ing)&lt;br&gt;&#xA;&lt;strong&gt;Ronald&lt;/strong&gt; (R on a+50=L+d)&lt;br&gt;&#xA;&lt;br&gt;&#xA;Please enquire for &quot;? at Woodside Cottage Rednal. JR T??&lt;br&gt;&#xA;&lt;br&gt;&#xA;There was an old priest named Francis&lt;br&gt;&#xA;Who was so fond of chiffony dances&lt;br&gt;&#xA;That he sat up too late&lt;br&gt;&#xA;And worried his pate&lt;br&gt;&#xA;Arranging these Frenchified Prances&lt;br&gt;&#xA;&lt;br&gt;&#xA;(This is to pay you out for not coming: sending Father Edmund instead)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastActivityDate="2015-08-05T18:04:18.457" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18674" PostTypeId="2" ParentId="18650" CreationDate="2015-08-05T18:18:02.860" Score="1" Body="&lt;p&gt;A bit morbid, but I'm thinking you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The dying/deceased, with your opposite being the living. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the most powerful, a group of people.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The number of dead people dwarfs the number of living people and &quot;there's power in numbers&quot;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My opposite was once my rival, able to change between others.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Possibly referencing living people who are in comas or who are cryogenically frozen and be considered dead. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can both be found in the same surroundings, often doing very different things.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A living person can do many different things, but a dead person is, well, dead. Both are found on/in the Earth. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example, one of us could be relaxed while the other is making things happen.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The dead usually have a relaxed expression on their face. The living are going on about their lives doing things. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One could be underneath while the other is making connections.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Underneath, referencing people who are buried underneath the soil in a grave. Living people can make connections between other living things. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One could be in court while the other is getting help from equipment.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A living person could be a lawyer in court, while a dying person could be on life support (hospital equipment). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One can bring power while the other shows disease.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A living person can provide power/electricity to buildings whereas a dying person maybe be dying from a disease. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On the radio, one of us could be reading while the other gives chase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Not too sure about this one.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally, though, our interests overlap.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The living have interests and the dead had those interests while still alive. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13489" LastActivityDate="2015-08-05T18:18:02.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18675" PostTypeId="1" AcceptedAnswerId="18677" CreationDate="2015-08-05T18:30:24.257" Score="6" ViewCount="2453" Body="&lt;p&gt;Note: I wrote this one a short while back but never actually posted it as a question, fairly straightforward and easy, have fun :) (Obviously it's a 'what am I? classic riddle')&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I silence birds and stop mills,&lt;br&gt;&#xA;Hide hills and make rivers still.&lt;br&gt;&#xA;Meanwhile, I'll make beasts slumber,&lt;br&gt;&#xA;I cannot stay, but I may linger.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="11251" LastEditorUserId="11251" LastEditDate="2015-08-06T13:58:33.750" LastActivityDate="2015-08-06T13:58:33.750" Title="I silence birds" Tags="|riddle|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18676" PostTypeId="1" AcceptedAnswerId="18694" CreationDate="2015-08-05T18:36:17.080" Score="11" ViewCount="552" Body="&lt;p&gt;Calling all puzzlers! Bailey M is back with Cryptic Crossword #3. Good luck, and enjoy!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eZ9vX.png&quot; alt=&quot;Crossword&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Across&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;1. Finished sculpture is a total letdown (8,4)&lt;br&gt;&#xA;8. Exist as god of secret material (3)&lt;br&gt;&#xA;9. Rhetorical Spanish street has eleven variations (7)&lt;br&gt;&#xA;10. Measure boundless squats/tumbles at open-source hub (5)&lt;br&gt;&#xA;13. Very first to engrave on kind of flowering plant (5)&lt;br&gt;&#xA;14. Break commoner's first-half astounding lead (5)&lt;br&gt;&#xA;15. Baby chicken's two eyes (7)&lt;br&gt;&#xA;17. Single cell busts a ring beam (6)&lt;br&gt;&#xA;19. Highest point of a weird, airy shrubbery (7)&lt;br&gt;&#xA;20. Flag penne pasta, in one end and out the other, first; timeless agent, second (7)&lt;br&gt;&#xA;22. Atrocious gruel is knocked down a peg, gets eighth new start (6)&lt;br&gt;&#xA;27a. Confuse right tonsil with another bodily cavity (7)&lt;br&gt;&#xA;27b. No initial shortage of endlessly tall half-giants in regret (9)&lt;br&gt;&#xA;29. A site to the west-southwest? Turn around, you're mixing it up with a vulnerable species (5)&lt;br&gt;&#xA;30. In a surprise twist, three hundred Greeks take in Eastern European catholic (5)&lt;br&gt;&#xA;31. From right to left, feed like cattle on sugary coating (5)&lt;br&gt;&#xA;32. Cheese has its own chocolatey dessert (7)&lt;br&gt;&#xA;33. Residual powder has become obscured (3)&lt;br&gt;&#xA;34. Toothy birds, left unchained, become cutthroat (12) &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Down&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;1. California company's cotton cloth (6)&lt;br&gt;&#xA;2. Top ten mixed martial artists start mauling you and I horribly (7)&lt;br&gt;&#xA;3. Running late, single cats make milk (7)&lt;br&gt;&#xA;4. At center, all eyes on friend flipping a fish (7)&lt;br&gt;&#xA;5. Legacy spoke of insect's adventure (7)&lt;br&gt;&#xA;6. Hold roast over cooking range (5)&lt;br&gt;&#xA;7. Cathy capsizes a boat (5)&lt;br&gt;&#xA;8. Flyer's flutters (4)&lt;br&gt;&#xA;11. Emphasize that you stuffed yourself full of stomach pastes (9)&lt;br&gt;&#xA;12. Dog misbehaves once in a rolling tire (7)&lt;br&gt;&#xA;16. Dry-injected polysaccharides (4)&lt;br&gt;&#xA;17. Truly, name is awful (4)&lt;br&gt;&#xA;18. Self-ambition makes an unfortunate point (3,4)&lt;br&gt;&#xA;21. Destroy with blend of root and condensed performance-enhancing drugs (7)&lt;br&gt;&#xA;22. Fish in simple shelter has an incomplete alibi (7)&lt;br&gt;&#xA;23. Voila! I reveal original mashed pasta (7)&lt;br&gt;&#xA;24. Turns evildoing from head to toe (7)&lt;br&gt;&#xA;25. Say hear-say, mostly? Blasphemy! (6)&lt;br&gt;&#xA;26. Painting of country with masthead (5)&lt;br&gt;&#xA;27. It's nearly night! (4)&lt;br&gt;&#xA;28. If you're this tall, you have to stop (5)  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE:&lt;/strong&gt; Partial answers are gladly accepted and teamwork is highly encouraged!&lt;/p&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="11475" LastEditDate="2015-09-01T21:00:08.890" LastActivityDate="2015-09-01T21:00:08.890" Title="Cryptic Crossword #3: Words on Wheels" Tags="|crosswords|cryptic-crosswords|" AnswerCount="4" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="18677" PostTypeId="2" ParentId="18675" CreationDate="2015-08-05T18:36:18.533" Score="10" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Winter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I silence birds and stop Mills,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since birds mate in the spring, they don't sing until winter ends. Mills shut down in winter because there is no grain to grind, since it was all harvested in the summer/fall. Furthermore, if the mill is hydro powered, the river may freeze over and not be able to power the mill.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hide hills and make rivers still.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Heavy snowfall can make hills seem flat (if the area around the hill is covered with more snow), and rivers can freeze over, giving them the illusion of being still.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, I'll make beasts slumber,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Animals hibernate in wintertime: examples include bears, squirrels, and skunks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cannot stay, but I may linger.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Summer will always roll around (Even when Winter Is Coming), but winter will last a while. Added by jpmc26: the word linger could also be a reference to times when it seems to last longer than other winters (i.e. Groundhog Day).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13538" LastEditorUserId="12018" LastEditDate="2015-08-06T03:08:31.630" LastActivityDate="2015-08-06T03:08:31.630" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18679" PostTypeId="2" ParentId="18675" CreationDate="2015-08-05T18:38:10.793" Score="6" Body="&lt;p&gt;You are:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Snow  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I silence birds and stop Mills,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Birds tend to not sing as much during the winter (or they fly south, so we don't hear them), and mills cannot run if there is no running water to power them.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hide hills and make rivers still.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Snow covers hills, and freezes rivers.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, I'll make beasts slumber,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most animals hibernate during the winter and snow.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cannot stay, but I may linger.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Snow eventually melts, but sometimes it hangs around for a long time.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="12018" LastEditDate="2015-08-06T03:34:12.690" LastActivityDate="2015-08-06T03:34:12.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18680" PostTypeId="2" ParentId="18675" CreationDate="2015-08-05T18:38:19.603" Score="4" Body="&lt;p&gt;You are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Snow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I silence birds and stop Mills,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Birds fly south for the Winter when it becomes too cold. Usually when there is snow you won't hear birds. Mills (old fashioned ones) don't operate during cold temperatures as there is no grain to grind and they are powered by wind/water. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hide hills and make rivers still.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Hills are covered in snow and rivers are frozen over&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, I'll make beasts slumber,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Animals (mammals) hibernate during the winter &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cannot stay, but I may linger.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Snow will usually stay after Winter ends (especially in Canada) into Spring &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13489" LastEditorUserId="12018" LastEditDate="2015-08-06T03:33:29.510" LastActivityDate="2015-08-06T03:33:29.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18681" PostTypeId="1" AcceptedAnswerId="18685" CreationDate="2015-08-05T19:07:46.253" Score="12" ViewCount="1164" Body="&lt;p&gt;The first five words of a famous quote are in the rebus below. Read the image as four horizontal lines, the last of which contains the fourth and fifth words. The accepted answer will feature the full quote and explain how the first five words were found.&#xA;&lt;hr&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/mLkai.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mLkai.png&quot; alt=&quot;this is a rebus of a famous quote&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13237" LastEditorUserId="13237" LastEditDate="2015-08-05T20:39:31.550" LastActivityDate="2015-08-05T21:31:54.410" Title="Finish the famous (rebus) quote!" Tags="|rebus|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18682" PostTypeId="2" ParentId="18668" CreationDate="2015-08-05T19:18:23.507" Score="8" Body="&lt;p&gt;The code to enter is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The same as the previous room, 117&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Key 1 is morse code with spaces being dots and tabs being dashes.  This yields the text:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;SUMFREQFLOORADD6049ADDAFPRSQRT&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Key 2 is, as &lt;a href=&quot;https://puzzling.stackexchange.com/a/18671/6144&quot;&gt;f'' figured out&lt;/a&gt; is decimal ascii representation.  Converting this to ascii code yields:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;|-|--------|--------|--------|--------|&#xA;&#xA;|-|--------|----q---|--------|--------|&#xA;C5                   q   q    e&#xA;|-|--------|---e----|--------|--------|&#xA;                        e   e  e e&#xA;|-|q---q---|e-e-----|--------|--------|&#xA;#                               e&#xA;|-|--------|--------|--------|--------|&#xA;      e   e  e                    q&#xA;C4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Converting all of the notes in the key signature to their &lt;a href=&quot;https://en.wikipedia.org/wiki/Piano_key_frequencies&quot; rel=&quot;nofollow noreferrer&quot;&gt;frequencies&lt;/a&gt; gives us this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0 x 698.456 |-|--------|--------|--------|--------|&#xA;0 x 659.255 &#xA;1 x 587.330 |-|--------|----q---|--------|--------|&#xA;3 x 523.251 C5                   q   q    e&#xA;1 x 493.883 |-|--------|---e----|--------|--------|&#xA;4 x 440.000                        e   e  e e&#xA;4 x 391.995 |-|q---q---|e-e-----|--------|--------|&#xA;1 x 369.994 #                               e&#xA;0 x 329.628 |-|--------|--------|--------|--------|&#xA;4 x 293.665      e   e  e                    q&#xA;0 x 261.626 C4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Following the instructions in Key 1:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;SUMFREQ = sum of the frequencies = 7523.600&#xA;FLOOR(SUMFREQ) = 7523&#xA;FLOOR(SUMFREQ) + 6049 = 13572&#xA;FLOOR(SUMFREQ) + 6049 + AFPR (answer from previous room = 117) = 13689&#xA;SQRT(FLOOR(SUMFREQ) + 6049 + AFPR) = 117&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1777" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-08-06T06:51:09.357" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18683" PostTypeId="1" AcceptedAnswerId="18754" CreationDate="2015-08-05T19:51:40.960" Score="7" ViewCount="1666" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm slower than a sloth, yet have been known to travel faster than sound. Printers think I'm good for nothing yet bartenders charge good money for me. What am I?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is a single English word. It can be found in any ordinary dictionary. No hyphens no trickery. No proper names fictional or non-fictional. Just an ordinary word. Not an abstract noun or nebulous concept. You can see and touch these just like you could, say, a tennis ball or a table (These objects aren't clues, they're just random examples of tangible objects).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: &quot;Slower than a sloth&quot; should be taken to relate to a race between me and the sloth. We would both move under our own power at our fastest speed towards the finishing line but the sloth would win.&lt;/p&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="14177" LastEditDate="2015-08-07T14:29:53.013" LastActivityDate="2015-08-07T17:17:41.353" Title="I'm slower than a sloth" Tags="|riddle|word|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18684" PostTypeId="2" ParentId="18562" CreationDate="2015-08-05T20:40:09.743" Score="5" Body="&lt;p&gt;First,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I have interpreted &quot;each ... in turn&quot; to refer to the subjects of the first line (&quot;lawman, soldier, hero&quot;), rather than to &quot;us all&quot; in the second.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus, my answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Lifeguard&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Police (lawmen), Soldiers and other hero-types all dedicate their lives to guarding everyone else's.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14238" LastActivityDate="2015-08-05T20:40:09.743" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18685" PostTypeId="2" ParentId="18681" CreationDate="2015-08-05T20:54:33.307" Score="7" Body="&lt;p&gt;The full text is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;God created war so that Americans would learn geography.&lt;/em&gt; —Mark Twain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The rebus provides these words:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; God created war so that&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1st:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;rats&quot; is to &quot;star&quot; as &quot;dog&quot; is to &lt;strong&gt;god&lt;/strong&gt;. (letter reversal)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2nd:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (reputation = cred) + ate = &lt;strong&gt;cre&lt;em&gt;ate&lt;/em&gt;d&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3rd:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Forward - Ford = &lt;strong&gt;war&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4th:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; do re mi fa &lt;strong&gt;so&lt;/strong&gt; la ti do&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5th:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The 55th word in the USA national anthem is &lt;strong&gt;that&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-08-05T21:31:54.410" LastActivityDate="2015-08-05T21:31:54.410" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18686" PostTypeId="2" ParentId="18681" CreationDate="2015-08-05T21:12:41.910" Score="2" Body="&lt;p&gt;Thank you Ian Mac and Roland:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How bout that first word?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; rats is to star as dog is to &lt;strong&gt;god&lt;/strong&gt;, reverse the words (clever)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well what is next? (thanks to a particularly squeamish ossifrage)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; reputation = CRED, with &quot;ATE&quot; in the middle &lt;strong&gt;Created&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks to Ian&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Forward - Ford = &lt;strong&gt;war&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lets get whimsical&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; do re mi fa &lt;strong&gt;SO&lt;/strong&gt; , so&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once again thanks to Ian&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The 55th word in the USA national anthem is &lt;strong&gt;that&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;humorous for sure&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;“God created war so that Americans would learn geography.” Mark Twain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13916" LastEditorUserId="13916" LastEditDate="2015-08-05T21:17:36.787" LastActivityDate="2015-08-05T21:17:36.787" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18687" PostTypeId="2" ParentId="18681" CreationDate="2015-08-05T21:18:28.327" Score="1" Body="&lt;p&gt;The full quotation is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;God created war so that Americans would learn geography.&lt;/em&gt;  - Mark Twain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The breakdown is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First line:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RATS is to STAR as DOG is to GOD&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second line:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reputation = CRED.  Put ATE in the middle and you have &lt;strong&gt;CRE ATE D&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Third line:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; FORWARD (forwarding an email) minus (Harrison) FORD = WAR.  (Thanks to @IanMacDonald for this one)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fourth line:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The fifth note on the scale is SO (Do, Re, Mi, Fa, So).  The 55th word in the US national anthem is THAT (Thanks to @IanMacDonald for this one, too.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So we have&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; GOD CREATED WAR SO THAT...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2015-08-05T21:18:28.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18688" PostTypeId="2" ParentId="18676" CreationDate="2015-08-05T21:38:09.863" Score="3" Body="&lt;h2&gt;Partial answer&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Here are a few I managed to get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Across&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1A. Finished sculpture is a total letdown (8,4)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  COMPLETE BUST  (Refers to both a finished sculpture, and a total letdown) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;22A. Atrocious gruel is knocked down a peg, gets eighth new start (6)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HORRID   (Atrocious = &lt;strong&gt;horrid&lt;/strong&gt;;  gruel = &lt;strong&gt;porridge&lt;/strong&gt;, knocked down a peg (remove the word &lt;strong&gt;peg&lt;/strong&gt;) = &lt;strong&gt;orrid&lt;/strong&gt;; gets eighth new start = use the eighth letter of the alphabet (&lt;strong&gt;H&lt;/strong&gt;) to give it a &quot;new start&quot; (put it on the beginning))  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;29A. A site to the west-southwest? Turn around, you're mixing it up with a vulnerable species (5)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strike&gt;SCENE  (A site = &lt;strong&gt;scene&lt;/strong&gt;; west-southwest turned around is east-northeast, or &lt;strong&gt;ENE&lt;/strong&gt;.  Mix that up with a vulnerable species (one with &lt;strong&gt;c&lt;/strong&gt;onservation &lt;strong&gt;s&lt;/strong&gt;tatus) to get &lt;strong&gt;scene&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;/strike&gt;&#xA;31A. From right to left, feed like cattle on sugary coating (5)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; GLAZE   (Sugary coating = &lt;strong&gt;glaze&lt;/strong&gt;.  Feed like cattle = &lt;strong&gt;graze&lt;/strong&gt;, and change the &lt;strong&gt;r&lt;/strong&gt;ight to a &lt;strong&gt;l&lt;/strong&gt;eft (&lt;strong&gt;R&lt;/strong&gt; to &lt;strong&gt;L&lt;/strong&gt;) to give &lt;strong&gt;glaze&lt;/strong&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Down&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1D. California company's cotton cloth (6)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CALICO   (&lt;strong&gt;Cali&lt;/strong&gt;fornia &lt;strong&gt;Co&lt;/strong&gt;mpany, and &lt;strong&gt;calico&lt;/strong&gt; is a type of cotton cloth)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3D. Running late, single cats make milk (7)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  LACTATE  (Not entirely sure, but single cats = &lt;strong&gt;cat&lt;/strong&gt;, mix with &lt;strong&gt;late&lt;/strong&gt; to get anagram of &lt;strong&gt;lactate&lt;/strong&gt; (make milk).  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;7D. Cathy capsizes a boat (5)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; YACHT  (&lt;strong&gt;Cathy&lt;/strong&gt; anagrams to &lt;strong&gt;yacht&lt;/strong&gt;, which is a boat) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="4421" LastEditDate="2015-08-06T04:46:18.053" LastActivityDate="2015-08-06T04:46:18.053" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18689" PostTypeId="2" ParentId="18676" CreationDate="2015-08-05T21:43:44.247" Score="3" Body="&lt;h2&gt;Partial answer&lt;/h2&gt;&#xA;&#xA;&lt;ol start=&quot;21&quot;&gt;&#xA;&lt;li&gt;Destroy with blend of root and condensed performance-enhancing drugs (7)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TORPEDO (&quot;condensed performance-enhancing drugs&quot; gives PED; torpedo is an anagram of &quot;root ped&quot; and things can be destroyed with it)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-08-05T21:43:44.247" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18690" PostTypeId="1" AcceptedAnswerId="19799" CreationDate="2015-08-05T22:18:27.997" Score="7" ViewCount="1033" Body="&lt;p&gt;I'm ambiguous by nature as I grow tall after all&lt;br&gt;&#xA;Your puny little earth shall tremble at my call&lt;br&gt;&#xA;In small as in big, the extremes I will lead&lt;br&gt;&#xA;My address is surely some superstition's seed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Abnormal my birth in a land of illusion so mortal&lt;br&gt;&#xA;You touch the near ground of the specular portal&lt;br&gt;&#xA;One long gone monster, not touchable for man&lt;br&gt;&#xA;Nary a sinner may tame me as merely nothing can&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A creature of prey hunting for your deceased&lt;br&gt;&#xA;I shall turn it soft, into a sustaining feast&lt;br&gt;&#xA;Imprisoned by tradition, tortured to the graves&lt;br&gt;&#xA;Royal is my essence, my innards rule the waves&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the elders have foretold, the end is near&lt;br&gt;&#xA;And with one difference you can only find hear&lt;br&gt;&#xA;The keeper of time will choose 2 over infinity&lt;br&gt;&#xA;You and me and the devil form a handsome trinity&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Seriously, looking at the hint already? Shame on you! &lt;strong&gt;Hint:&lt;/strong&gt; I never do typos!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Is this a who or what am I?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You really need those hints, don't you? I could tell you that each two lines describe one letter (but that would be telling).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5106" LastEditorUserId="5106" LastEditDate="2015-08-09T13:47:00.920" LastActivityDate="2015-08-10T19:13:25.057" Title="Old dragon returning - hide your children!" Tags="|riddle|word|rhyme|" AnswerCount="5" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="18692" PostTypeId="1" AcceptedAnswerId="19829" CreationDate="2015-08-05T23:31:35.613" Score="6" ViewCount="1242" Body="&lt;p&gt;This is Room 5 of the Maze series. For those wishing to start at the beginning, click &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18567/room-1-of-the-maze&quot;&gt;here&lt;/a&gt;. To go back to the previous room, click &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18668/room-4-of-the-maze&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you arrive at the answer to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18668/room-4-of-the-maze&quot;&gt;Room 4&lt;/a&gt;, you realize it's the same one that you got for &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18608/room-3-of-the-maze&quot;&gt;Room 3&lt;/a&gt;. With a sudden flash of insight, you go back to the keypad and scrutinize it. The faint fingerprints you see are on the keys 0,1,3 and 7. You realize with dread that you have been in this room before (0 from your unfortunate predecessor, 33 from your peer Anne whose skin is still recovering from the Itchyverse, and the rest from your previous entry. But wait! You look to the wall next to the keypad for the scratch marks that were there before and you see none. Were they ever really there? You look around, bewildered, for an explanation and you see nothing but the cold walls and the suspended platform.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually you hopelessly resign yourself to foraging ahead and enter your answer into the keypad. As usual, the door disappears and you find yourself in Room 5. You don't even think about resting--what would be the point in trying, you know you can longer get any sleep in this Maze. You just walk helplessly towards the platform and grab the note. You stop before reading it and in defiant anger you rip a piece off and throw it to the ground. Before it can even hit the ground, it evaporates into thin air. You realize what you've just done and force yourself to calm down before you ruin your hopes of escaping this place. You read the note:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Welcome to Room 5! If we must say, this is our personal favorite room; it was the first one that we created. As a result, it may turn out that it is actually easier than the previous--only time will tell. Good Luck!!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Key 1: &gt;c 8DC&amp;lt;G:HH! ?jan )i]! &amp;amp;,,+# LZ ]daY ### [jijgZ hZXjg^in# ;gZfjZcXn! i]Zc d  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is a rip in the paper after 'd' thanks to your moment of indignance. &quot;I hope there was nothing too important on that bit&quot;, you mumble.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Key 2: ERLO GWTD ERW BLSH OFVVWWUWU ERW OLXERCSDH YRS RLAOWBN RTU OFMMBTHEWU ERW RWTQWHBG  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Bad, bad test subject! We sure hope that you can complete this room without that missing information.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Did they know that you were going to rip the paper beforehand? Or did they make that information appear without you noticing? Your head begins to hurt from the constant confusion and you can feel your stability slipping. You need to hurry and escape this Maze!!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A thorough investigation of the keypad shows only digits (with no discernible marks on any of them).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer from &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18668/room-4-of-the-maze&quot;&gt;Previous Room&lt;/a&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;117&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;After a day of grueling attempts to find any significant meaning in the keys, a note appeared on the platform:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It seems that we had not properly balanced this room. That being the case, we will provide two pieces of information.&lt;br&gt;&#xA;  1. Key 1 has been put through one of your species' first recorded ciphers, though with an alphabet larger than the one originally used.&lt;br&gt;&#xA;  2. A fuller version of Key 2 is as follows: ERLO GWTD ERW NLDOE BLSH OFVVWWUWU ERW ERLDU OLXERCSDH YRS RLAOWBN RTU OFMMBTHEWU ERW NLDOE RWTQWHBG&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This puzzle has now been solved. Here is the next, and final, &lt;a href=&quot;https://puzzling.stackexchange.com/questions/19834/end-of-the-maze&quot;&gt;room&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-08-10T15:14:21.403" Title="Room 5 of the Maze" Tags="|cipher|enigmatic-puzzle|password|" AnswerCount="4" CommentCount="10" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18693" PostTypeId="2" ParentId="18576" CreationDate="2015-08-05T23:32:54.330" Score="2" Body="&lt;h2&gt;Machine Learning&lt;/h2&gt;&#xA;&lt;p&gt;If you are willing to spend a little time formatting your data, this could be a great opportunity for a machine learning algorithm.&lt;/p&gt;&#xA;&lt;h3&gt;Why&lt;/h3&gt;&#xA;&lt;p&gt;Basically, the idea is that given some training set, you can create a model that will place new data points in the proper category. The appeal of a machine learning method is that &lt;strong&gt;you don't have to define what makes a problem easy or hard.&lt;/strong&gt; You simply give it a few pre-defined problems to start with and it implicitly determines what is most important. So if you want N categories, I would say pick at least 5 samples for each. So 5 puzzles you have already decided are easy, 5 that are medium, etc...&lt;/p&gt;&#xA;&lt;h3&gt;Data&lt;/h3&gt;&#xA;&lt;p&gt;You can include any data you think might be relevant, adding extra almost never hurts. So for each puzzle, include all the metrics you have listed in your question. Add anything else you want that can be quantified.&lt;/p&gt;&#xA;&lt;h3&gt;Implementation&lt;/h3&gt;&#xA;&lt;p&gt;Python is maybe the easiest to code this in. Checkout &lt;a href=&quot;http://scikit-learn.org/stable/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;scikit&lt;/a&gt;. There are a couple easier programs with GUIs if you prefer, I want to say GraphLab might be one?&lt;/p&gt;&#xA;" OwnerUserId="1968" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-08-06T17:39:48.140" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18694" PostTypeId="2" ParentId="18676" CreationDate="2015-08-05T23:45:10.970" Score="10" Body="&lt;h2&gt;The completed crossword&lt;/h2&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/iaOmN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iaOmN.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;br&gt;&#xA;&lt;h2&gt;An explanation of the clues&lt;/h2&gt;&#xA;&lt;p&gt;&lt;strong&gt;Across:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;1A. Finished sculpture is a total letdown (8,4)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;  COMPLETE BUST  (Refers to both a finished sculpture, and a total letdown)  -- credit to GentlePurpleRain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;8A. 8. Exist as god of secret material (3)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; BRA ? B+RA. &amp;quot;Exist&amp;quot; is B; RA is the Egyptian god of the sun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;9A. Rhetorical Spanish street has eleven variations (7)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; LEXICAL: anagram of CALLE+XI&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;10A. Measure boundless squats/tumbles at open-source hub (5)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; QUOTA: boundless squats = QUAT, open-source = O, O is the hub, so QUOAT, tumbles at the hub QUOTA - credit to LeppyR64&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;13A. Very first to engrave on kind of flowering plant (5)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; VETCH: V+ETCH (first letter of &amp;quot;very&amp;quot; plus etch = engrave)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;14A. Break commoner's first-half astounding lead (5)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; COMMA: first half of commoner is COMM, first letter of astounding is A&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;15A. Baby chicken's two eyes (7)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; PEEPERS(!): double definition&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;17A. Single cell busts a ring beam (6)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; AMOEBA anagram of A+O(&amp;quot;ring&amp;quot;)+BEAM&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;19A. Highest point of a weird, airy shrubbery (7)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;  TOPIARY: TOP(highest point) + anagram of AIRY&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;20A. Flag penne pasta, in one end and out the other, first; timeless agent, second (7)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; PENNANT: first, take either end of PENNe pastA; second, &amp;quot;agent&amp;quot; minus age (time) = NT&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;22A. Atrocious gruel is knocked down a peg, gets eighth new start (6)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; HORRID   (Atrocious = &lt;strong&gt;horrid&lt;/strong&gt;;  gruel = &lt;strong&gt;porridge&lt;/strong&gt;, knocked down a peg (remove the word &lt;strong&gt;peg&lt;/strong&gt;) = &lt;strong&gt;orrid&lt;/strong&gt;; gets eighth new start = use the eighth letter of the alphabet (&lt;strong&gt;H&lt;/strong&gt;) to give it a &amp;quot;new start&amp;quot; (put it on the beginning))  -- credit to GentlePurpleRain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;27a. Confuse right tonsil with another bodily cavity (7)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; NOSTRIL: anagram of R TONSIL  :D !!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;27b. No initial shortage of endlessly tall half-giants in regret (9)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; NOSTALGIA:  NO + S (initial letter of shortage) + TAL (tall without final letter) + GIA (half of &amp;quot;giants&amp;quot;).  As @GentlePurpleRain anticipated, this entry spirals around the blue square.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;29A. A site to the west-southwest? Turn around, you're mixing it up with a vulnerable species (5)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; VENUE: anagram of ENE+VU (West-southwest turned around is east-northeast, or &lt;strong&gt;ENE&lt;/strong&gt;. (Thanks @GentlePurpleRain.)  Vulnerable species is &lt;a href=&quot;https://en.wikipedia.org/wiki/Conservation_status&quot; rel=&quot;nofollow noreferrer&quot;&gt;conservation status&lt;/a&gt; VU.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;30A. In a surprise twist, three hundred Greeks take in Eastern European catholic (5)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; UNIAT: anagram (&amp;quot;surprise twist&amp;quot;) of TAU+IN.  tau has a value of 300 in Greek numerals and Uniate relates to certain eastern European Christian communities -- credit to Pete&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;31A. From right to left, feed like cattle on sugary coating (5)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; GLAZE   (Sugary coating = &lt;strong&gt;glaze&lt;/strong&gt;.  Feed like cattle = &lt;strong&gt;graze&lt;/strong&gt;, and change the &lt;strong&gt;r&lt;/strong&gt;ight to a &lt;strong&gt;l&lt;/strong&gt;eft (&lt;strong&gt;R&lt;/strong&gt; to &lt;strong&gt;L&lt;/strong&gt;) to give &lt;strong&gt;glaze&lt;/strong&gt;)  -- credit to GentlePurpleRain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;32A. Cheese has its own chocolatey dessert (7)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; BROWNIE: BRIE containing OWN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;33A. Residual powder has become obscured (3)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; ASH: anagram of &amp;quot;has&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;34A. Toothy birds, left unchained, become cutthroat (12)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; BLOODTHIRSTY(!): anagram of TOOTHY BIRDS L&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Down:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;1D. California company's cotton cloth (6)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; CALICO   (&lt;strong&gt;Cali&lt;/strong&gt;fornia &lt;strong&gt;Co&lt;/strong&gt;mpany, and &lt;strong&gt;calico&lt;/strong&gt; is a type of cotton cloth)  -- credit to GentlePurpleRain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;2D. Top ten mixed martial artists start mauling you and I horribly (7)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; MAXIMUM: anagram of X+MMA+M+UI  (ten = X, mixed martial arts = MMA, start of &amp;quot;mauling&amp;quot; = M, you and I = UI)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;3D. Running late, single cats make milk (7)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;  LACTATE  (Not entirely sure, but single cats = &lt;strong&gt;cat&lt;/strong&gt;, mix with &lt;strong&gt;late&lt;/strong&gt; to get anagram of &lt;strong&gt;lactate&lt;/strong&gt; (make milk). -- credit to GentlePurpleRain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;4D. At center, all eyes on friend flipping a fish (7)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; TILAPIA: All eyes on friend = I + PAL + I, At center = A + IPALI + T, Flipping = TILAPIA  – credit to LeppyR64&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;5D. Legacy spoke of insect's adventure (7)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; BEQUEST: homophone (Bee Quest!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;6D. Hold roast over cooking range (5)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; STOVE: hidden in roaST OVEr&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;7D. Cathy capsizes a boat (5)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; YACHT  (&lt;strong&gt;Cathy&lt;/strong&gt; anagrams to &lt;strong&gt;yacht&lt;/strong&gt;, which is a boat)  -- credit to GentlePurpleRain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;8D. Flyer's flutters (4)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; BATS: double def.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;11D. Emphasize that you stuffed yourself full of stomach pastes (9)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; OVERSTATE: &amp;quot;you stuffed yourself&amp;quot; is OVERATE; it's full of ST which is the &amp;quot;stomach&amp;quot; (interior) of &amp;quot;pastes&amp;quot;. –- credit to LeppyR64&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;12D. Dog misbehaves once in a rolling tire (7)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;  TERRIER: TIER(anagram of &amp;quot;tire&amp;quot;) containing ERR&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;16D. Dry-injected polysaccharides (4)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; ARID: contained in polysacch&lt;strong&gt;arid&lt;/strong&gt;es&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;17D. Truly, name is awful (4)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; AMEN: anagram of &amp;quot;name&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;18D. Self-ambition makes an unfortunate point (3,4)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; OWN GOAL: double definition (going out on a limb here as I know nothing about sports, but I believe a score on one's own goal is unfortunate)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;21D. Destroy with blend of root and condensed performance-enhancing drugs (7)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; TORPEDO (&amp;quot;condensed performance-enhancing drugs&amp;quot; gives PED; torpedo is an anagram of &amp;quot;root ped&amp;quot; and things can be destroyed with it) -- credit to rand al'thor&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;22D. Fish in simple shelter has an incomplete alibi (7)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; HALIBUT : HUT containing ALIB (&amp;quot;alibi&amp;quot; missing final letter)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;23D. Voila! I reveal original mashed pasta (7)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; RAVIOLI: anagram of VOILA+I+R (first letter of &amp;quot;reveal&amp;quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;24D. Turns evildoing from head to toe (7)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; INNINGS: take the first letter of SINNING and move it to the end&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;25D. Say hear-say, mostly? Blasphemy! (6)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; HERESY: here+sy (homophone of &amp;quot;hear&amp;quot; = HERE, most of &amp;quot;say&amp;quot; = SY)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;26D. Painting of country with masthead (5)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; MURAL: M+URAL (first letter of &amp;quot;mast&amp;quot; replacing the first letter of rural)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;27D. It's nearly night! (4)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; NIGH: most of &amp;quot;night&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;28D. If you're this tall, you have to stop (5)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; STALL: concealed in &amp;quot;thiS TALL&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4385" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-08-06T21:53:07.637" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18696" PostTypeId="1" AcceptedAnswerId="18755" CreationDate="2015-08-06T02:24:32.843" Score="4" ViewCount="1176" Body="&lt;p&gt;Tim Hansen gave me a note last week as we were leaving lunch and heading back to work. The paper was folded into quarters, and there were no directions to help understand what was written in the squares. He said that it should be easy to figure out, but the challenge might as well be shared with others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proper orientation for the message is unknown but probably is one of those shown below. Several images were created to archive the message as the original sheet of graph paper has now been discarded. The only useful peace of information Tim provided was that he called it a cipher of some sort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/71KKC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/71KKC.jpg&quot; alt=&quot;Picture 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YZgWy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YZgWy.jpg&quot; alt=&quot;Picture 2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nxDy7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nxDy7.jpg&quot; alt=&quot;Picture 3&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/utVm0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/utVm0.jpg&quot; alt=&quot;Picture 4&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1661" LastActivityDate="2015-08-07T20:20:16.587" Title="Friend gave me a note I am not sure how to read. What does it say?" Tags="|cipher|enigmatic-puzzle|visual|language|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18698" PostTypeId="1" AcceptedAnswerId="18700" CreationDate="2015-08-06T04:03:11.267" Score="6" ViewCount="462" Body="&lt;p&gt;With A, it's a fellow whose work is for hire.&lt;br&gt;&#xA;With E, it's a place where there might be a fire.&lt;br&gt;&#xA;With I, it's a fellow who hasn't much class.&lt;br&gt;&#xA;With O, something pleasant to find in your glass.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surmise, then, the pattern that I have designed,&lt;br&gt;&#xA;And tell me the four words that I have in mind.&lt;/p&gt;&#xA;" OwnerUserId="4385" LastActivityDate="2015-08-06T17:18:02.897" Title="Sequence of Four Words" Tags="|riddle|word|pattern|rhyme|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18699" PostTypeId="2" ParentId="183" CreationDate="2015-08-06T05:12:20.270" Score="1" Body="&lt;p&gt;You can also solve it using 4 groups of 3 balls. Weigh 3 against 3 and if it balances, you can keep those 6 balls aside as known-equals. If they don't balance, you know the odd ball out is in that group of 6. Next, weigh 3 of the known-equals against either of the 2 groups of 3 unknowns. If it balances, the odd one is in the final group of 3. If it doesn't balance, you know the odd one is in still on the scale. Finally, using the last group of 3 balls that is unknown and unequal, put one on each end and keep the third aside. If the scale balances, you know the lone ball you kept aside is the odd ball. If the scale doesn't balance, you know the odd ball is on the scale. To determine the odd ball and whether it's heavier or lighter, you need to have noted whether the unknown group was heavier or lighter than the known-equal groups. If they were heavier, then the lone ball is heavier.&lt;/p&gt;&#xA;" OwnerUserId="14427" LastActivityDate="2015-08-06T05:12:20.270" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18700" PostTypeId="2" ParentId="18698" CreationDate="2015-08-06T05:52:18.410" Score="7" Body="&lt;p&gt;I think it's... &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;h_ck&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With A, it's a fellow whose work is for hire.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A &lt;strong&gt;hack&lt;/strong&gt; is another word for a taxi, like the kind you hire. Thanks alexmc. It is also a word for &quot;One who undertakes unpleasant or distasteful tasks for money or reward; a hireling&quot;, so thanks Pete. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With E, it's a place where there might be a fire.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Heck&lt;/strong&gt;, like fire and brimstone et al &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With I, it's a fellow who hasn't much class.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Hick&lt;/strong&gt;, a country hick or hillbilly&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With O, something pleasant to find in your glass.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Hock&lt;/strong&gt;, a type of wine. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14428" LastEditorUserId="14428" LastEditDate="2015-08-06T17:18:02.897" LastActivityDate="2015-08-06T17:18:02.897" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="18702" PostTypeId="2" ParentId="18696" CreationDate="2015-08-06T08:01:32.280" Score="2" Body="&lt;p&gt;Some guesses up until this point:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the small block is the cipher - no 2 symbols are repeated.&lt;br&gt;&#xA; + is space (it does not show up the cipher)&lt;br&gt;&#xA; also the interesting thing about the cipher is that it's 40 symbols - makes me think of Shaw/Shavian Alphabet&lt;br&gt;&#xA; also if we accept that + is space, the letters in some words are taken directly from the cipher, in other words the letters are mirrored top to bottom cipher letters (for last image - first word direct, 2nd word mirrored, 3rd word direct, 4th mirrored...)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Close?&lt;/p&gt;&#xA;" OwnerUserId="9969" LastActivityDate="2015-08-06T08:01:32.280" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18703" PostTypeId="1" CreationDate="2015-08-06T08:22:13.943" Score="-4" ViewCount="364" Body="&lt;p&gt;I couldn't find the exact solution for the following &quot;Coding - Decoding&quot; problem that I found in some exam paper. The question is as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If blue is coded as 118 then colour is coded as?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14433" LastEditorUserId="12700" LastEditDate="2015-08-06T08:38:56.783" LastActivityDate="2015-08-06T11:37:35.557" Title="If blue is coded as 118 then colour is coded as?" Tags="|word|pattern|" AnswerCount="1" CommentCount="5" ClosedDate="2015-08-06T14:43:13.663" ContentLicense="CC BY-SA 3.0" />
  <row Id="18704" PostTypeId="1" AcceptedAnswerId="18743" CreationDate="2015-08-06T08:35:15.627" Score="2" ViewCount="460" Body="&lt;p&gt;Ran into a pirate at a bar, the old fashioned kind. I asked him where he'd been and he gave me a cryptic response.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Sail on, found in&lt;/p&gt;&#xA;&lt;p&gt;utter majesteer', I&lt;/p&gt;&#xA;&lt;p&gt;mean, on 'me,&lt;/p&gt;&#xA;&lt;p&gt;paluk paneer&lt;/p&gt;&#xA;&lt;p&gt;speaking s'pecifically&lt;/p&gt;&#xA;&lt;p&gt;lower countries sped though&lt;/p&gt;&#xA;&lt;p&gt;high seas&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;A couple of hints.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The pirate speaks a certain east asian language. That said, he was not in east asia.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14428" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-08-07T19:42:09.323" Title="Met an odd pirate, where was he?" Tags="|riddle|wordplay|" AnswerCount="5" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18705" PostTypeId="2" ParentId="18675" CreationDate="2015-08-06T08:46:57.660" Score="3" Body="&lt;p&gt;I made this fit because the title made me do it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Gas&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I silence birds and stop Mills:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When the canary kicks the bucket, the miners leave and the mills shut down from lack of materials.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hide hills and make rivers still:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gases can obscure your vision and through a phase change they can (still) make rivers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, I'll make beasts slumber:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; While high enough concentrations will kill the canary, larger beasts may simply lose consciousness. Also, you know... laughing gas.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cannot stay, but I may linger.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wind can blow them away but gasses that are heavier than air will linger in depressions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4522" LastActivityDate="2015-08-06T08:46:57.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18706" PostTypeId="2" ParentId="18615" CreationDate="2015-08-06T10:35:15.743" Score="2" Body="&lt;p&gt;Let's give it a shot. You guys are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Notes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;We are made from addition or subtraction&#xA;Our best are often a popular attraction&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The monetary ones can be summed or subtracted, and the bigger ones are sought after&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Some you can touch,&#xA;and some you cannot&#xA;A dozen will heal your illness&#xA;Another shan't be bought&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This was what pushed me harder towards the solution. The only two things I can think of existing in 12 are notes (all the possible musical tones) and hours.  You can touch the monetary ones, and the music can cure you, so that fit better.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;We all can change one's point of view&#xA;But, y'know, sometimes words have two meanings&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bribery&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Some of us have utility&#xA;but others are for show&#xA;One befuddling, three worlds becoming one&#xA;made o'er fifty years ago&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Money is useful. Music is for show. No clue about the befuddling part :/&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;By now you may have guessed what we are&#xA;so we ask that you find our brother, a movie star&#xA;He watched a broken man escape from prison&#xA;who let fear find him, and now is risen.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Money are an important part of Batman's powers!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14435" LastActivityDate="2015-08-06T10:35:15.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18707" PostTypeId="2" ParentId="18704" CreationDate="2015-08-06T11:19:31.347" Score="0" Body="&lt;p&gt;My guess is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Netherlands&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I couldn't make much of it except for one part which makes me think this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Netherlands literally means &quot;Low countries&quot; matching the &quot;lower countries&quot; part in the riddle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-08-06T11:19:31.347" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18708" PostTypeId="1" AcceptedAnswerId="18736" CreationDate="2015-08-06T11:24:08.163" Score="16" ViewCount="413" Body="&lt;ol&gt;&#xA;&lt;li&gt;Moon&lt;/li&gt;&#xA;&lt;li&gt;False&lt;/li&gt;&#xA;&lt;li&gt;Coffee&lt;/li&gt;&#xA;&lt;li&gt;Peace&lt;/li&gt;&#xA;&lt;li&gt;Neutral&lt;/li&gt;&#xA;&lt;li&gt;Stones&lt;/li&gt;&#xA;&lt;li&gt;Hell&lt;/li&gt;&#xA;&lt;li&gt;Private&lt;/li&gt;&#xA;&lt;li&gt;Plane&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What is the pattern behind this sequence? How might it continue?&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-08-07T17:09:47.553" Title="Sold - a sequence of words" Tags="|pattern|enigmatic-puzzle|word-sequence|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="18709" PostTypeId="2" ParentId="18703" CreationDate="2015-08-06T11:37:35.557" Score="3" Body="&lt;p&gt;Colour is coded as&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Number&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or maybe&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Integer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or possibly even&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Whole Number&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Colour is a generalized version of a specific colour, for example blue. So if blue is to 118, then we must generalize 118. Giving us &quot;Number&quot;, &quot;Integer&quot;, or &quot;Whole Number&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2015-08-06T11:37:35.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18710" PostTypeId="1" AcceptedAnswerId="19865" CreationDate="2015-08-06T11:41:15.443" Score="6" ViewCount="680" Body="&lt;p&gt;It is very hard to communicate with my friend, as you can see &lt;a href=&quot;https://puzzling.stackexchange.com/q/18657/13384&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/q/18601/13384&quot;&gt;here&lt;/a&gt; ...&lt;br&gt;&lt;br&gt;&#xA;The next day I was finally able to talk to him, so we decided to meet in a real place (&lt;a href=&quot;https://puzzling.stackexchange.com/q/18601/13384&quot;&gt;the last one&lt;/a&gt; wasn't safe enough). To avoid his messages I called him. I wanted to meet at the park, near the cinema. He wasn't really motivated. When I asked him what was wrong, he suddenly hung up his phone. &lt;br&gt;He then sent me a strange SMS (again!):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(13384) (001)(101)(511)(990)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I asked what was going on but as always, he didn't answer...&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Question&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;What is he trying to tell me ?&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Hints&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Tags can help you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Context is like size ... it &lt;strong&gt;doesn't&lt;/strong&gt; mater. (Maybe a little bit.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The space between the first and the other numbers &lt;strong&gt;is&lt;/strong&gt; relevant. They are 2 different things.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Focus on the second part of the numbers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Stop waiting for an other hint here ... If I could, I would go back and delete all of my hints, and go in the park of the ward.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13384" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-08-11T08:43:09.063" Title="The strange SMS (Part 3)" Tags="|cipher|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18712" PostTypeId="2" ParentId="18676" CreationDate="2015-08-06T12:30:47.733" Score="3" Body="&lt;p&gt;10A. Measure boundless squats/tumbles at open-source hub (5)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; QUOTA: boundless squats = QUAT, open-source = O, O is the hub, so QUOAT, tumbles at the hub QUOTA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4D. At center, all eyes on friend flipping a fish (7)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;TILAPIA: All eyes on friend = I + PAL + I, At center = A + IPALI + T, Flipping = TILAPIA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1777" LastEditorUserId="1777" LastEditDate="2015-08-06T14:10:11.947" LastActivityDate="2015-08-06T14:10:11.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18713" PostTypeId="2" ParentId="18704" CreationDate="2015-08-06T12:49:07.607" Score="0" Body="&lt;p&gt;My guess is that the pirate has been to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Andaman and Nicobar Islands&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sail On Found In &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is in the middle of the ocean having no borders with another country.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;paluk paneer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Which happens to be an Indian dish&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;lower countries sped though&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is on the lower side of the Indian mainland.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;high seas&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Unsure about this clue...&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="444" LastEditorUserId="12700" LastEditDate="2015-08-06T13:04:39.440" LastActivityDate="2015-08-06T13:04:39.440" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18714" PostTypeId="2" ParentId="18704" CreationDate="2015-08-06T13:52:07.523" Score="1" Body="&lt;p&gt;Given the &lt;a href=&quot;/questions/tagged/wordplay&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;wordplay&amp;#39;&quot; rel=&quot;tag&quot;&gt;wordplay&lt;/a&gt; tag, as well as some of the words chosen, I think the puzzle is based on the same idea as &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16513/the-madmans-speech&quot;&gt;this one&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Sail on&quot; -&gt; &lt;strong&gt;Ceylon&lt;/strong&gt;, &quot;speaking s'pecifically&quot; -&gt; [s]&lt;strong&gt;Peking&lt;/strong&gt; [s]&lt;strong&gt;Pacific&lt;/strong&gt; [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I haven't solved it all yet, but this looks like the right 'aha' needed.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-08-06T13:52:07.523" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18716" PostTypeId="1" AcceptedAnswerId="18721" CreationDate="2015-08-06T14:48:09.613" Score="2" ViewCount="1599" Body="&lt;p&gt;Alice and Bob were supposed to have a date at some o'clock. But, Bob was late for the date.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alice was upset. She said &quot;See you tomorrow at the same time.&quot; and went away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several interesting properties for their arrival times:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Bob's arrival time is represented by three distinct digits.&lt;/li&gt;&#xA;&lt;li&gt;Rearranging those three digits, we get only one time before the appointment time and she arrived at that time.&lt;/li&gt;&#xA;&lt;li&gt;Bob was late within one hour.&lt;/li&gt;&#xA;&lt;li&gt;When we rearrange the three digits, other times are late more than one hour but less than two hours.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What time o'clock was their date?&lt;/p&gt;&#xA;" OwnerUserId="3914" LastActivityDate="2015-08-06T16:22:35.653" Title="Don't be late for a date" Tags="|logical-deduction|" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18717" PostTypeId="2" ParentId="18690" CreationDate="2015-08-06T14:49:54.447" Score="2" Body="&lt;p&gt;I've an inkling based on two parts of the fourth paragraph:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The last line reminds me of an Emmylou Harris song with a line that goes &quot;You and me and the devil make three.&quot; This was sung by the sirens in O Brother, Where Art Thou, which together with the deliberate typo &quot;hear&quot;, plus bits like &quot;the end is near&quot; and &quot;rule the waves&quot;, makes me think that this portion is making reference to the Sirens. Doesn't match with the rest, though.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14441" LastActivityDate="2015-08-06T14:49:54.447" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18718" PostTypeId="2" ParentId="18716" CreationDate="2015-08-06T15:09:06.653" Score="10" Body="&lt;p&gt;Bob arrived at:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;5:46&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The date was supposed to be at:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;4:56&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The two times that are late by more than one hour, but less than two:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;6:45 and 6:54&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My thought process:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The three numbers had to be consecutive. Because each of them would be the hour digit for one of the given times, a non-consecutive triplet (i.e. 4,5,7) would give a latest and earliest time that were more than 2 hours apart. The lowest number must be greater than or equal to 6, so that there will only be 4 valid times (4:56 is valid; 4:65 is not). The two upper numbers must be less than 6, for the opposite reason, so 4,5,6 is the only triplet that can satisfy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14441" LastActivityDate="2015-08-06T15:09:06.653" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18719" PostTypeId="2" ParentId="18716" CreationDate="2015-08-06T15:10:45.343" Score="2" Body="&lt;p&gt;Bob arrived at  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 5:46  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And just guessing they were supposed to meet at:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 5 o'clock&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The only earlier time then can be:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4:56, since 4:65 is not a valid time, and 4:56 is within one hour of 5:46. Alice is so good and arrived early.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The other arrangements are  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 6:54 and 6:45, both of which are more than an hour after 5 and within 2 hours of 5. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-08-06T15:10:45.343" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="18721" PostTypeId="2" ParentId="18716" CreationDate="2015-08-06T15:41:17.103" Score="3" Body="&lt;p&gt;Letting Bob's arrival time be $x:yz$, we can infer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The date was supposed to be at $x$:00.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to 2: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;We can get only one time before $x$:00, which is only possible if the smallest one of $x,y,z$ has less than two options for the minute (with at least one of the others &gt; 5).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The other permutations are late by between one hour and two hours.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If there are two digits greater than 5, there'll only be 2 legal times, while there have to be at least 3 legal permutations. So there's only one such digit. Let's sort them all as $a&amp;lt;b&amp;lt;c&amp;gt;5$. Then out of $a:bc, b:ac, c:ab, c:ba$, only the first one is early, the second one is Bob's arrival time (appointment is at $b$:00), and the last two are late by between one hour and two hours. $c = b+1$, $b=5$, $c=6$, so they were supposed to meet at 5 o'clock. $a$ doesn't seem to matter, but 4:56 seems to be the most reasonable arrival time for Alice (then Bob must have arrived at 5:46).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastActivityDate="2015-08-06T15:41:17.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18723" PostTypeId="2" ParentId="18716" CreationDate="2015-08-06T16:22:35.653" Score="1" Body="&lt;h2&gt;Alternate Answer&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I don't know if this answer is just another possible solution, a proof that we are missing something (Since OP said the answer is uniquely determined), or just a matter of semantic disagreement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bob arrived at &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1:30  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alice arrived at  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1:03  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assuming that times of 0:xx are not valid, then the remaining times are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3:01 and 3:10  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This lets the appointment time be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Anywhere between 1:11 and 1:29&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note: This seems to work with any two digits a and b such that 0 &amp;lt; a,b &amp;lt; 6 and a - b = 2&lt;/p&gt;&#xA;" OwnerUserId="1968" LastActivityDate="2015-08-06T16:22:35.653" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18724" PostTypeId="2" ParentId="18615" CreationDate="2015-08-06T16:29:06.413" Score="2" Body="&lt;p&gt;Are you:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Marvel Superheroes?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are made from addition or subtraction&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Marvel superheroes are being created all the time, and &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_deceased_American_comic_book_characters#Marvel_Comics&quot; rel=&quot;nofollow&quot;&gt;many are also killed off&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Our best are often a popular attraction&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; How many Spiderman films have been made? 'nuff said&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some you can touch, &#xA;and some you cannot&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I get the feeling that touching Human Torch from Fantastic Four might not be the best idea anyone's ever had.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A dozen will heal your illness&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Superheroes, lol&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another shan't be bought&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Could reference any number of plots in Marvel stories I imagine.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We all can change one's point of view&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gripping comic book stories can have that effect.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But, y'know, sometimes words have two meanings&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PLOT TWIST!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some of us have utility&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Like, most of the heroes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but others are for show&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Black Widow? :P&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One befuddling, three worlds becoming one&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pass...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;made o'er fifty years ago&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Marvel was made over 50 years ago.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By now you may have guessed what we are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Superheroes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so we ask that you find our brother, a movie star&#xA;He watched a broken man escape from prison&#xA;who let fear find him, and now is risen.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Already been suggested it's Batman. Another superhero so, &quot;brother from another mother&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4659" LastEditorUserId="4659" LastEditDate="2015-08-06T21:40:42.937" LastActivityDate="2015-08-06T21:40:42.937" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18725" PostTypeId="2" ParentId="18696" CreationDate="2015-08-06T18:21:27.967" Score="8" Body="&lt;p&gt;Since I find it very difficult to understand random symbols, I transcribed this cipher into text, using the following key:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/clqcR.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/clqcR.png&quot; alt=&quot;Cipher transcription key&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The symbols I used are obviously not the intended decryption; I just used them to make it easier to peruse the text and find potential patterns.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As noted by @Mircea, the $+$ symbols are likely word breaks, and the $|$ symbols are likely sentence breaks.  Also as noted by @Mircea, every second word (with one exception) is &quot;upside-down&quot;.  As indicated below, these words have been inverted as a whole (the entire word has been rotated, as opposed to each individual letter).  I have indicated the &quot;upside-down&quot; words by enclosing them in parentheses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the transcribed text (as it appears above):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;$9D$ + (0YG04) + DZ + (@ZP) + 24RF9 + (%3##FZ) + LC + (80E9) + TFS43I + (QZN)&#xA;+ $9D$ + (0YG04) + DZ |&#xA;(OYD0GLDP0ZD6) + 3F00 + (SFTFR) + YDFQ + (SF42$) + 1SER + (KUJ6H1L2) + NZQ + (D8@)&#xA;+ TE@JF + (MY) + A + (51B@4K%) + DZ + ALNJMWJ5E + (@H5G#HK) + 0Y + (1@E) + BQBD8BLD8D0H |&#xA;(DPZ) + 3QQGT4 + (801X) + XA886ACAM8 + (7PC) + GT45$HK + (1E8) + 3DRG + ($OFY) |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;With the inverted words inverted (so all words now read left-to-right):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;$9D$ + 40GY0 + DZ + PZ@ + 24RF9 + ZF##3% + LC + 9E08 + TFS43I + NZQ&#xA;+ $9D$ + 40GY0 + DZ |&#xA;6DZ0PDLG0DYO + 3F00 + RFTFS + YDFQ + $24FS + 1SER + 2L1H6JUK + NZQ + @8D&#xA;+ TE@JF + YM + A + %K4@B15 + DZ + ALNJMWJ5E + KH#G5H@ + 0Y + E@1 + BQBD8BLD8D0H |&#xA;ZPD + 3QQGT4 + X108 + XA886ACAM8 + CP7 + GT45$HK + 8E1 + 3DRG + YFO$ |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My observations&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There is one place (near the middle of the fourth line of the transcription) where there are two words that have the same orientation:  &lt;code&gt;DZ + ALNJMWJ5E&lt;/code&gt;.  It is possible that DZ is actually inverted (because both of those symbols have rotational symmetry), but since DZ appears several other times in the text, it is unlikely.  (Even if it were inverted, there would still be two words with the same orientation next to each other:  &lt;code&gt;(51B@4K%) + (DZ)&lt;/code&gt;).&lt;/li&gt;&#xA;&lt;li&gt;There are several long words with very distinctive letter patterns in the text:  &lt;code&gt;OYD0GLDP0ZD6&lt;/code&gt;, &lt;code&gt;BQBD8BLD8D0H&lt;/code&gt;, &lt;code&gt;XA886ACAM8&lt;/code&gt;.  Plugging these individually into a cryptogram solver, we get &lt;code&gt;&amp;lt;nothing&amp;gt;&lt;/code&gt;, &lt;code&gt;AVAILABILITY&lt;/code&gt;, and &lt;code&gt;SETTLEMENT&lt;/code&gt;.  If we reverse the order of the letters in &lt;code&gt;OYD0GLDP0ZD6&lt;/code&gt; to get &lt;code&gt;6DZ0PDLG0DYO&lt;/code&gt;, then it decrypts to &lt;code&gt;DISTRIBUTI[VE|ON|NG]&lt;/code&gt;.  This means it's likely that the &quot;inverted&quot; words are inverted as a whole, not as individual letters, and thus the 2nd transcription is the more accurate one in terms of helping with decryption.  The existence of both (QZN) and NZQ reinforces this hypothesis.  It is also possible that the &quot;inverted&quot; words are meant to be read from end to beginning, as a separate sentence of sorts, but I imagine that will be made clear once the text is decrypted.&lt;/li&gt;&#xA;&lt;li&gt;There is a problem, though. &lt;code&gt;AVAILABILITY&lt;/code&gt; uses the same symbol for &lt;strong&gt;L&lt;/strong&gt; that &lt;code&gt;SETTLEMENT&lt;/code&gt; uses for &lt;strong&gt;T&lt;/strong&gt;.  So, either (1) one of these is incorrect, or (2) the cipher somehow shifts the meaning of the symbols between each word/sentence.&lt;/li&gt;&#xA;&lt;li&gt;There are several words that are repeated at least once in the ciphertext:  &lt;code&gt;DZ&lt;/code&gt;, &lt;code&gt;$9D$&lt;/code&gt;, &lt;code&gt;(0YG04)&lt;/code&gt;, &lt;code&gt;NZQ&lt;/code&gt; (and, in fact, the whole phrase &lt;code&gt;$9D$ + (0YG04) + DZ&lt;/code&gt;).  This would indicate that the meaning of the symbols likely does &lt;em&gt;not&lt;/em&gt; shift between words, and the &lt;code&gt;AVAILABILITY&lt;/code&gt;/&lt;code&gt;SETTLEMENT&lt;/code&gt; problem above must have another cause or reason.&lt;/li&gt;&#xA;&lt;li&gt;The only symbol that doesn't appear in the ciphertext is &lt;code&gt;V&lt;/code&gt;.  That means there are 39 distinct symbols in the text.  They must represent more than just the 26 letters of the alphabet, but it is unclear what else they might represent.  The fact that there are unique decryptions of the three long words mentioned above makes it likely that all the symbols they contain represent distinct letters of the alphabet, but then I don't know what the extra 13/14 symbols are for.  Maybe some alphabet letters have two different symbols to represent them?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Here is a frequency analysis of all the symbols:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;D 15  3.82%            0 12  3.05%&#xA;0 12  3.05%            1  6  1.53%&#xA;F 10  2.54%            2  3  0.76%&#xA;8  9  2.29%            3  5  1.27%&#xA;Z  9  2.29%            4  8  2.04%&#xA;4  8  2.04%            5  4  1.02%&#xA;$  7  1.78%            6  3  0.76%&#xA;Y  7  1.78%            7  1  0.25%&#xA;G  7  1.78%            8  9  2.29%&#xA;@  6  1.53%            9  4  1.02%&#xA;Q  6  1.53%            #  3  0.76%&#xA;E  6  1.53%            $  7  1.78%&#xA;1  6  1.53%            %  2  0.51%&#xA;T  5  1.27%            @  6  1.53%&#xA;A  5  1.27%            A  5  1.27%&#xA;H  5  1.27%            B  4  1.02%&#xA;3  5  1.27%            C  3  0.76%&#xA;L  5  1.27%            D 15  3.82%&#xA;J  4  1.02%            E  6  1.53%&#xA;K  4  1.02%            F 10  2.54%&#xA;R  4  1.02%            G  7  1.78%&#xA;P  4  1.02%            H  5  1.27%&#xA;B  4  1.02%            I  1  0.25%&#xA;S  4  1.02%            J  4  1.02%&#xA;5  4  1.02%            K  4  1.02%&#xA;9  4  1.02%            L  5  1.27%&#xA;#  3  0.76%            M  3  0.76%&#xA;C  3  0.76%            N  3  0.76%&#xA;2  3  0.76%            O  2  0.51%&#xA;M  3  0.76%            P  4  1.02%&#xA;N  3  0.76%            Q  6  1.53%&#xA;6  3  0.76%            R  4  1.02%&#xA;%  2  0.51%            S  4  1.02%&#xA;X  2  0.51%            T  5  1.27%&#xA;O  2  0.51%            U  1  0.25%&#xA;7  1  0.25%            V  0  0.00%&#xA;I  1  0.25%            W  1  0.25%&#xA;U  1  0.25%            X  2  0.51%&#xA;W  1  0.25%            Y  7  1.78%&#xA;V  0  0.00%            Z  9  2.29%&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It doesn't follow a English-like distribution (it's much more spread-out, rather than focused on the common letters like ETAIONS).  This makes me suspect that maybe some of the most-common letters are represented by two or more distinct symbols.&lt;/li&gt;&#xA;&lt;li&gt;Every symbol but one is used.  If in fact the 40 symbols at the top of the ciphertext are the entire alphabet, that means that every English letter but one is used in the ciphertext.&lt;/li&gt;&#xA;&lt;li&gt;Perhaps @Mircea's hypothesis about the &lt;a href=&quot;https://en.wikipedia.org/wiki/Shavian_alphabet&quot; rel=&quot;noreferrer&quot;&gt;Shavian alphabet&lt;/a&gt; makes sense.  Since it contains many more &quot;vowel sound&quot; symbols, its frequencies would tend to be more spread out.  If it is correct, then my decryptions of the longer words above don't apply, since there is not a one-to-one correspondence of each Shavian letter to an English letter.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2015-08-06T18:21:27.967" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18726" PostTypeId="2" ParentId="3831" CreationDate="2015-08-06T18:31:06.203" Score="1" Body="&lt;p&gt;I think the key part to all of this is that Individual Furbles can't change their strategies. That means only several Dominator Furbles together can become Sharer Furbles. This would seem to suggest that if two Dominator Furbles have been kicked-out of their turf and are too weak to reproduce that season, they can only find refuge with another Dominator Furble which has been defeaten.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the sake of this argument I'm going to say that Sharer Furbles have arisen in Dominator Furble culture out of a necessity for survival, rather than that both co-existed since the very beginning. It would make more sense to me that the aggressive survival instinct came first, and the passive characteristics came later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this scenario I'd say the entire problem can begin with two Dominator Furbles. One defeats the other, resulting in 10 offspring produced. The 10 offspring in addition to the two original furbles makes 12.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's made clearer in the figures below:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Generation: 0,  1,  2,   3,  4,&#xA;&#xA;Dominator: 2, 12, 66, 396, 2376&#xA;&#xA;Sharer: 0, 0, 6, 36, 216&#xA;&#xA;Total: 2,   12, 72, 432, 2592&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As you can see from the TOTAL row, it doesn't take long for the population to surpass 2000. It's somewhere in the late part of the 3rd generation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, I could run through every possibility regarding whether a Dominator Furble decides to fight another Dominator Furble or take a Sharer Furble's territory, but to make things simple I'm going to say, on average, the population growth is going to be constant. Because of this I'm going to say all these figures fit under a geometric progression. An interesting property about geometric progression is that consecutive terms a, b, and c satifty the equation:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;b^2 = ac&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And to make the terms of the question less pedantic I'm going to assume that Dominator Fubles only fight each other once, and so for each season there isn't just one Dominator Furble who stands, triumphant, over the rest, finally at peace to reproduce. If that's the case the question becomes trivial and also stupidly worded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you do apply the rules which geometric progression imposes then you end up with a scenario where no matter what, Dominator Furbles will turf Shares out of their territory (as seen in the figures above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what you end up with after all of that is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Dominator + Sharer = 2000&#xA;&#xA;Dominator/Sharer = 11&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Work through the simple substitution and you get:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Total = 2000&#xA;&#xA;Sharer = 166.666666&#xA;&#xA;Dominator = 1833.3333333&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So, if anyone is still aware this thread exists, tell me what you think. I'd say the answer is 1833.&lt;/p&gt;&#xA;" OwnerUserId="14444" LastEditorUserId="-1" LastEditDate="2017-12-26T10:20:02.777" LastActivityDate="2017-12-26T10:20:02.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18727" PostTypeId="1" AcceptedAnswerId="18728" CreationDate="2015-08-06T18:31:13.007" Score="14" ViewCount="1003" Body="&lt;p&gt;In many great novels, to the reader I appear&lt;br&gt;&#xA;Adjust me, and I describe things that are near&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once more, and my meaning changes only slightly&lt;br&gt;&#xA;Again, and you excercised your free will, rightly&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet another, and I'm something we all ought to do&lt;br&gt;&#xA;Again, I'm a boundary with a beautiful view&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, I'm a place to which many must go&lt;br&gt;&#xA;And again, I'm a thing wise men shouldn't throw&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now behead me, I'm a thing you must watch while you speak&lt;br&gt;&#xA;&lt;em&gt;Your next action depends on the word that you seek:&lt;/em&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Behead me once more, and my use is widespread&lt;br&gt;&#xA;Or instead take my foot, of hard things I am said&lt;br&gt;&lt;/p&gt;&#xA;" OwnerUserId="13237" LastEditorUserId="13237" LastEditDate="2016-06-14T20:44:30.670" LastActivityDate="2016-06-14T20:44:30.670" Title="A poem of dynamic words" Tags="|word|rhyme|wordplay|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18728" PostTypeId="2" ParentId="18727" CreationDate="2015-08-06T18:55:29.583" Score="14" Body="&lt;p&gt;In many great novels, to the reader I appear&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Great novels have a very compelling &lt;strong&gt;theme&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Adjust me, and I describe things that are near&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Would you perhaps be referencing &lt;strong&gt;these&lt;/strong&gt;, over here? (m -&gt; s)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once more, and my meaning changes only slightly&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Or perhaps you mean &lt;strong&gt;those&lt;/strong&gt;? (e -&gt; o)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again, and you exercised your free will, rightly&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you exercised your free will, you &lt;strong&gt;chose&lt;/strong&gt; something. (t -&gt; c)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet another, and I'm something we all ought to do&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I thought this was share, but it turns out we should all do a &lt;strong&gt;chore&lt;/strong&gt; once in awhile. (s -&gt; r)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again, I'm a boundary with a beautiful view&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Especially beautiful at sunset or sunrise, this references the &lt;strong&gt;shore&lt;/strong&gt;. (c -&gt; s)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again, I'm a place to which many must go&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you need to buy something, you have to go to the &lt;strong&gt;store&lt;/strong&gt;. (h -&gt; t)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And again, I'm a thing wise men shouldn't throw&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you're wise, you will throw no &lt;strong&gt;stone&lt;/strong&gt;. (r -&gt; n)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now behead me, I'm a thing you must watch while you speak&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I know I've heard this from my mother: you better watch your &lt;strong&gt;tone&lt;/strong&gt;! (goodbye, s)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Behead me once more, and my use is widespread&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The number &lt;strong&gt;one&lt;/strong&gt; is a pretty widespread number, I'd say. (goodbye, t)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or instead take my foot, of hard things I am said&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That thing must weigh a &lt;strong&gt;ton&lt;/strong&gt;! (goodbye, e)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-08-06T18:55:29.583" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18729" PostTypeId="1" AcceptedAnswerId="18730" CreationDate="2015-08-06T19:52:23.920" Score="25" ViewCount="2682" Body="&lt;p&gt;This is a little puzzle I heard a while back from one of my mathematically inclined friends- I get the sense that it's bounced around a little, so forgive me if you've heard it.&lt;/p&gt;&#xA;&lt;p&gt;There is a sort of function that, given any positive whole number, will return another positive whole number. Interestingly, there is one number that will return itself- the number 4. There are no other loops of any size, so you can start at any other number and, by repeatedly plugging in the output of the function, you will eventually reach 4.&lt;/p&gt;&#xA;&lt;p&gt;Given this diagram showing a few examples of those succession chains, determine the function.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WKNFl.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WKNFl.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;(If anyone feels like they really need more information, you can post a number in the comments and I'll give you its chain. However, the givens are sufficient to find a solution.)&lt;/p&gt;&#xA;" OwnerUserId="14441" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-03-20T13:13:31.547" Title="Four is Cosmic!" Tags="|logical-deduction|pattern|number-sequence|english|reachability|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18730" PostTypeId="2" ParentId="18729" CreationDate="2015-08-06T19:54:16.490" Score="25" Body="&lt;p&gt;Four is cosmic because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It has four letters. &lt;br&gt;&#xA; Each number is followed by the number of letters in its spelling, but &lt;em&gt;four&lt;/em&gt; leads back to itself indefinitely!&lt;br&gt;&#xA; Four is the only &lt;em&gt;cosmic&lt;/em&gt; number because it is the only number with its own number of letters in its English spelling, and there are no cycles of two or more words that have each other's letter counts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And it should probably go: &lt;strong&gt;18 --&gt; 8 --&gt; 5 --&gt; 4&lt;/strong&gt; (from Doge's comment)&lt;/p&gt;&#xA;" OwnerUserId="13237" LastEditorDisplayName="user88" LastEditDate="2015-08-07T01:55:44.667" LastActivityDate="2015-08-07T01:55:44.667" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18731" PostTypeId="1" AcceptedAnswerId="18732" CreationDate="2015-08-06T20:04:32.110" Score="5" ViewCount="355" Body="&lt;p&gt;Here is a sequence of numbers.  The goal is simply to provide the next 10 digit number in the series. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    6311433600&#xA;    3155760000 &#xA;    2840140800 &#xA;    2808604800 &#xA;    2805926400 &#xA;    2805321600 &#xA;    2805235200&#xA;    ??????????&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Time is of the essence &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14446" LastActivityDate="2015-08-06T20:15:00.120" Title="Guess the next number" Tags="|number-sequence|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18732" PostTypeId="2" ParentId="18731" CreationDate="2015-08-06T20:15:00.120" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;2805231600&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My process:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;My first thought was to figure out the difference between each successive number, assuming that this was a sequence. When I saw that the difference of the last two was 86400, which I know to be the number of seconds in a day, I did a little googling and found that the other differences corresponded to successively larger units of time- a week, a 31-day month, etc. So the next number should be smaller by an hour, or 3600 seconds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14441" LastActivityDate="2015-08-06T20:15:00.120" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18733" PostTypeId="2" ParentId="18604" CreationDate="2015-08-06T20:17:18.523" Score="7" Body="&lt;p&gt;Okay, so a bit of palette-shifting shows there are two colors of flock on the desk:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RGB = [117,164,108] spread everywhere&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RGB = [101,173,99] in a few specific areas - the spout and bottom corners of the glue bottle, several foot-print-like blobs leading from glue bottle to plane, on the base the plane is attached to, and the lumpy plane piece with a tab sticking out of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IZfRP.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IZfRP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume the color change represents flock that attached to glue upon spilling vs. flock which landed on clean desk, then evidently Mr. Skyler had made a gluey mess of things just before he died. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in doing so, he effectively placed his own time of death to a day between the 5th and 9th: Per the comments from VictorHenry and f, above (also TIL that single-letter user names are allowed), we learn that the humidity was high enough up to the 9th that any glue he spilled would have activated, causing the flock to glue to the table. But from the 9th to the 13th, the humidity was below the 75% threshold necessary to activate the glue (per the note on the glue bottle). If we assume the glue dries within an hour or so, then the flock had to have been spilled shortly after he dripped glue on the table sometime before or on the 9th. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, assuming that he spilled the flock upon his death and not before (which is explicitly stated in the puzzle) his insurance coverage had not yet expired.&lt;/p&gt;&#xA;" OwnerUserId="14238" LastEditorUserId="14238" LastEditDate="2015-08-06T21:39:44.507" LastActivityDate="2015-08-06T21:39:44.507" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18734" PostTypeId="2" ParentId="18692" CreationDate="2015-08-06T20:17:53.713" Score="0" Body="&lt;p&gt;I'm not at an answer yet, but Key 2 can be decrypted to (as TroyAndAbed suggests)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; THIS YEAR THE LION SUCCEEDED THE SIXTHBORN WHO HIMSELF HAD SUPPLANTED THE HEAVENLY&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The best guess I have for this is that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This refers to the year 1189, when Richard I (the Lionheart) succeeded Henry II.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Based on a comment from OP &quot;I should have learned my lesson on not grouping by 5 before,&quot; the number of characters in each &quot;word&quot; in Key 1 seem to be important, but I haven't got anywhere on the pattern&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2 9 4 4 5 2 4 3 6 9 10 4 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14030" LastActivityDate="2015-08-06T20:17:53.713" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18735" PostTypeId="2" ParentId="18708" CreationDate="2015-08-06T21:35:57.253" Score="4" Body="&lt;p&gt;This is almost certainly wrong, but my guess at the riddle's format might not be.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each word in the sequence brings common idioms to mind, as well as opposite/counterparts, like Moon and Sun, True and False, etc. The pattern involves idioms with the words in the sequence as well as with their counterparts.&lt;br&gt;&#xA; For example:&lt;br&gt;&#xA; &lt;em&gt;Blood&lt;/em&gt; &lt;strong&gt;Moon&lt;/strong&gt; -- True &lt;em&gt;Blood&lt;/em&gt; -- True/False&lt;br&gt;&#xA; &lt;strong&gt;False&lt;/strong&gt; &lt;em&gt;Start&lt;/em&gt; -- &lt;em&gt;Starting&lt;/em&gt; Tee -- Tee(Tea)/Coffee&lt;br&gt;&#xA; &lt;em&gt;Cold&lt;/em&gt; &lt;strong&gt;Coffee&lt;/strong&gt; -- &lt;em&gt;Cold&lt;/em&gt; War -- War/Peace&lt;br&gt;&#xA; &lt;em&gt;To come in&lt;/em&gt; &lt;strong&gt;Peace&lt;/strong&gt; -- &lt;em&gt;To be in&lt;/em&gt; Charge -- Charge(d)/Neutral&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll work on some more if this isn't absolutely wrong, but it feels really forced on my end. My next pairings were going to be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sticks/Stones&lt;br&gt;&#xA; Heaven/Hell&lt;br&gt;&#xA; Public/Private&lt;br&gt;&#xA; Train/Plane (vehicles) or Line/Plane (geometry)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13237" LastActivityDate="2015-08-06T21:35:57.253" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18736" PostTypeId="2" ParentId="18708" CreationDate="2015-08-06T21:58:50.607" Score="10" Body="&lt;p&gt;Roland is on the right track, but it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; rhymes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To go through what we have:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ought/Naught/Nought (depending on your dialect) -&gt; bought and sold (believe it or not, I thought of it myself and then saw Roland and Echo say it in comments)&lt;br&gt;&#xA; One -&gt; sun and moon,&lt;br&gt;&#xA; Two -&gt; true and false,&lt;br&gt;&#xA; Three -&gt; tea and coffee,&lt;br&gt;&#xA; Four -&gt; war and peace,&lt;br&gt;&#xA; Five -&gt; live and neutral (terms from AC current),&lt;br&gt;&#xA; Six -&gt; sticks and stones,&lt;br&gt;&#xA; Seven -&gt; heaven and hell,&lt;br&gt;&#xA; Eight -&gt; state and private (like colleges -- thanks Bailey M from comments),&lt;br&gt;&#xA; Nine -&gt; line and plane  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It might continue with&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ten -&gt; pencil (pen and pencil)&lt;br&gt;&#xA; Eleven -&gt; Kell (Kevin and Kell)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12092" LastEditorUserId="12092" LastEditDate="2015-08-07T17:09:47.553" LastActivityDate="2015-08-07T17:09:47.553" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="18737" PostTypeId="2" ParentId="18692" CreationDate="2015-08-06T22:41:47.707" Score="0" Body="&lt;p&gt;Okay, it has been awhile since anyone posted, so I thought I'd put in my two cents (probably more like one...), but Zeus was the sixth born and he did take over the heavens.  From there I have no idea, since years really don't play a part in Greek Mythology.  Also, I guess he may have birthed the Nemean Lion.  It was killed by Hercules and became the Leo constellation that we all know and love.  But again, I don't know how to get a year from that.&lt;/p&gt;&#xA;" OwnerUserId="14450" LastActivityDate="2015-08-06T22:41:47.707" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18738" PostTypeId="1" AcceptedAnswerId="19934" CreationDate="2015-08-07T05:38:05.283" Score="3" ViewCount="350" Body="&lt;p&gt;Met up with Homer. He said he had a word in mind, and then told me a riddle.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Rough, I'm a couple sharing a pie&lt;/p&gt;&#xA;&lt;p&gt;Smooth, I'm a valley guy from the sky.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;What's the word?&lt;/p&gt;&#xA;&lt;p&gt;Clues:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The title is important.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;A big 'ol hint, read at your own danger.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The Homer of question is indeed the original greek Homer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14428" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-08-12T10:34:10.047" Title="Homer's been watching TV" Tags="|riddle|wordplay|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18739" PostTypeId="2" ParentId="18738" CreationDate="2015-08-07T05:43:46.727" Score="3" Body="&lt;p&gt;Is the answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TV (which is &lt;em&gt;right there&lt;/em&gt; in the title!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rough, I'm a couple sharing a pie&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter &lt;code&gt;T&lt;/code&gt; looks like one half of the Greek letter $\pi$, which could mean a couple sharing a pie.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Smooth, I'm a valley guy from the sky.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A valley when seen from afar would represent the shape of the letter &lt;code&gt;V&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-08-07T06:17:21.340" LastActivityDate="2015-08-07T06:17:21.340" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18740" PostTypeId="2" ParentId="18532" CreationDate="2015-08-07T07:26:11.880" Score="3" Body="&lt;p&gt;&lt;strong&gt;Part Two:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the plaintext provided from the site,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My name again. J2I. Pga torooapr fpi Rlis Mcrcl ym otmplx tv Gsic 10. Iepzve aud hb, cap pbqi hb dnui ub. Befud. I hate vowles, exept for i.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We need to break up this text to arrive at a solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My name again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This points to the name 'Dr. X' that the OP used for himself in &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18498/the-twenty-doors-room-8&quot;&gt;Room 8&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;J2I. Befud.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This tells us that the encrypted text in between uses a Bifid cipher with J converted to I. So decrypting the encrypted text with the translation 'I to J' and 'DRX' as the key gives us &lt;br&gt;&lt;br&gt;The password for Part Three js hjdden jn Room 10. Search all up, not down to fjnd jt.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I hate vowles, exept for i.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Using this clue, we get the line&lt;br&gt;&lt;br&gt;The password for Part Three is hidden in Room 10. Search all up, not down to find it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Still working on the riddle...&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-08-07T08:02:09.660" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18741" PostTypeId="2" ParentId="18692" CreationDate="2015-08-07T07:49:07.537" Score="5" Body="&lt;p&gt;TroyAndAbed decrypted Key 2 as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;THIS YEAR THE LION SUCCEEDED THE SIXTHBORN WHO HIMSELF HAD SUPPLANTED THE HEAVENLY&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using English translations of Latin names, this describes Pope Leo I succeeding Sixtus III in the year &lt;strong&gt;440&lt;/strong&gt;. The previous pope was Celestine I.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-08-10T11:56:54.187" LastActivityDate="2015-08-10T11:56:54.187" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18743" PostTypeId="2" ParentId="18704" CreationDate="2015-08-07T09:28:27.737" Score="1" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sri Lanka&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sail on, found in&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Found in Ceylon - Ceylon (or British Ceylon) was the name for Sri Lanka from 1815 to 1948&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can't work out any other lines, although:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;paluk paneer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Paneer is Indian Cheese. How that references Sri Lanka other than being a neighbouring country I don't know&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;lower countries sped though&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's the lowest country in West Asia?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4659" LastEditorUserId="4659" LastEditDate="2015-08-07T10:17:51.897" LastActivityDate="2015-08-07T10:17:51.897" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18745" PostTypeId="1" AcceptedAnswerId="18749" CreationDate="2015-08-07T11:38:52.073" Score="5" ViewCount="3759" Body="&lt;p&gt;Can you use only two instances of the digit $2$, along with the mathematical operations below, to create an expression that evaluates to $2015$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Allowed operations:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;arithmetic operations: addition ($+$), subtraction ($-$), multiplication ($\times$), division ($\div$,  or $\frac{x}{y}$), exponentiation (like $2^2$);&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;factorial ($!$), absolute value ($|...|$);&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;extraction of the root of any degree in a form $\sqrt[a]{b}$ or the square root in a form $\sqrt{b}$;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;trigonometric functions: sine, cosine, tangent, cotangent, secant, cosecant&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;inverse trigonometric functions: arcsine, arccosine, arctangent, arccotangent, arcsecant, arccosecant&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;natural logarithm ($\ln b$), or logarithm with any base ($\log_a b$)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Parentheses&lt;/em&gt; $()$ are also allowed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is not allowed:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;digits other than $2$, or more than two instances of the digit $2$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;named constants such as $\pi, e$, etc...&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;defining and using your own functions&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;other variables&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example with three $2$s:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple of years ago, I managed to solve the same kind of problem, with three $2$s: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We can express any natural number $A$ using three $2$s and the above operations, like this: $$-\log_2\log_2\underbrace{\sqrt{\sqrt{...\sqrt{2}}}}_{A\,\mathrm{square\,roots}}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I couldn't crack the problem using only two $2$s...&lt;/p&gt;&#xA;" OwnerUserId="14458" LastEditorUserId="84683" LastEditDate="2023-09-05T18:53:25.077" LastActivityDate="2023-09-05T18:53:25.077" Title="Express the number 2015 using only the digit 2 twice" Tags="|mathematics|calculation-puzzle|formation-of-numbers|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="18747" PostTypeId="2" ParentId="18745" CreationDate="2015-08-07T12:11:24.117" Score="1" Body="&lt;p&gt;We can express 2015 using 0 digit like this: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;   $$\dfrac {\underbrace{e+e+e+...+e}_{2015 ~e~s} }{e}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Next Answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 15&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13582" LastEditorUserId="13582" LastEditDate="2015-08-07T12:34:26.843" LastActivityDate="2015-08-07T12:34:26.843" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18748" PostTypeId="2" ParentId="18683" CreationDate="2015-08-07T12:39:20.563" Score="4" Body="&lt;p&gt;Okay, probably not the real answer, but is it a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; shot&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Slower than a sloth:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Very literally, shot is the pellets in a shotgun or the ball tossed by a shotputter. They're inanimate so they don't move at all.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Faster than sound:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most guns shoot bullets faster than sound.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Printers think I'm good for nothing:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (This is a stretch.) Printers are interested in words, not pictures. They would prefer a good page of well-typeset text to your &lt;em&gt;shot&lt;/em&gt; of the same subject.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And of course, bartenders charge good money for me!&lt;/p&gt;&#xA;" OwnerUserId="10851" LastActivityDate="2015-08-07T12:39:20.563" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18749" PostTypeId="2" ParentId="18745" CreationDate="2015-08-07T13:19:16.777" Score="27" Body="&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/a/9405&quot;&gt;This answer&lt;/a&gt; describes a method using trigonometric operations to obtain the square root of any rational number from 0. In &lt;a href=&quot;https://puzzling.stackexchange.com/a/9417&quot;&gt;this answer&lt;/a&gt;, Daniil Agashiyev notes that $\tan\arcsin\cos\arctan\cos\arctan\sqrt{n}=\sqrt{n+1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this, we can write 2015 as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(\underbrace{\tan\arcsin\cos\arctan\cos\arctan}_{2011\text{ times}}\,2)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or even with only one 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\underbrace{\tan\arcsin\cos\arctan\cos\arctan}_{2015^2-4\text{ times}}\,2$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Unlike the linked problem, we're allowed to use the secant and cosecant here, so we can use $\sec\arctan$ or $\newcommand{\arccot}{\operatorname{arccot}}\csc\arccot$ for the same effect.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-08-10T01:55:33.490" CommentCount="7" CommunityOwnedDate="2015-08-07T13:19:16.777" ContentLicense="CC BY-SA 3.0" />
  <row Id="18751" PostTypeId="1" AcceptedAnswerId="18762" CreationDate="2015-08-07T13:53:05.433" Score="-3" ViewCount="155" Body="&lt;p&gt;This is part of the Fill in the Dashes series of puzzles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example : F_ in the B_ ⇒ Fill in the Blanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the words will be Juxtaposed as in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g. : C_ K_ S_ ⇒ Clark Kent Superman or Superman Clark Kent&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a Letter Missing in each of the following clues [and hence a word :-)].&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;D_ T_&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C_ F_ N_ T_ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;N_ W_&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A A_ B_ M_&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A A_ F_ W_&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D_ O_ S_&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D_ N_ S_&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B_ B_ F_ S_&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M_ O_&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M_&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I_ L_ O_&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E_ E_&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A_ M_&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G_ O_ 30&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Hint 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Numbers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9892" LastEditorUserId="9892" LastEditDate="2015-08-07T16:03:36.800" LastActivityDate="2015-08-09T19:18:55.737" Title="Fill in the dashes Part 5 of 10" Tags="|wordplay|knowledge|movies|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18752" PostTypeId="2" ParentId="18683" CreationDate="2015-08-07T14:42:16.753" Score="1" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; water ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Race (speculative):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Depending on the landscape it would flow towards the finishing line, but extremely slow, so a sloth could win.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Faster than sound:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are waterjet cutter. But i honestly don't know how fast the water will get there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Printer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water destroys the work of printers or make them nearly useless if the paper is soaked.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bartender:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water is sometimes more expensive than other drinks&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14461" LastActivityDate="2015-08-07T14:42:16.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18753" PostTypeId="2" ParentId="18745" CreationDate="2015-08-07T14:55:10.500" Score="3" Body="&lt;p&gt;It looks a bit like cheating but we are not defining any new function, we are just naming one that is allowed to use by the rules:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Name: &lt;code&gt;m: x -&amp;gt; m(x)&lt;/code&gt; to be: &lt;code&gt;x -&amp;gt; log2(x)&lt;/code&gt;    (log in base 2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we apply the square root trick:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;- m(m( sqrt(sqrt( .... sqrt(2) ... )) )) = 2015&#xA;   (taking square roots 2015 times)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="12062" LastActivityDate="2015-08-07T14:55:10.500" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18754" PostTypeId="2" ParentId="18683" CreationDate="2015-08-07T15:00:08.970" Score="18" Body="&lt;p&gt;Is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;slug&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;slower than a sloth&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;slugs (the animal) are known to move slowly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;faster than sound&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;slug&quot; can also refer to a bullet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;printers&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;in typesetting, the &quot;slug&quot; is an area that contains instructions for the printer, but is trimmed off and discarded after a document is printed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;bartenders&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;a &quot;slug&quot; also means a shot of an alcoholic drink.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-08-07T15:38:05.717" LastActivityDate="2015-08-07T15:38:05.717" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18755" PostTypeId="2" ParentId="18696" CreationDate="2015-08-07T15:40:39.440" Score="10" Body="&lt;p&gt;I have a partial explanation, and an imperfect (but good enough) decryption method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Building on the excellent work by GentlePurpleRain, I noticed a pattern, which is much more apparent if we apply a small modification to his setup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, arrange the 40 symbols by columns instead, like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;AEIMQUY37@&#xA;BFJNRVZ48$&#xA;CGKOSW159#&#xA;DHLPTX260%&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;With this setup, the text looks like this:&#xA;&lt;code&gt;&#xA;6TM6+XSYX#+MW+2WV+5#4UT+%9&#xA;88UW+FI+PXQT+$U8#97+ZWN+6T&#xA;M6+XSYX#+MW|RSMXYFMVXWMH+9&#xA;UXX+8U$U4+SMUZ+8U#56+18Q4+&#xA;BC@H37F5+NWZ+MP2+$Q2@U+JS+&#xA;A+D1E2#B%+MW+AFN@JK@DQ+23D&#xA;Y03B+XS+12Q+EZEMPEFMPMX3|M&#xA;VW+9ZZY$#+PX1O+OAPPHAIAJP+&#xA;LVI+Y$#D63B+1QP+9M4Y+6RUS|&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As already established, + signs are spaces, and &lt;code&gt;|&lt;/code&gt; are periods. Now, look at &lt;code&gt;OAPPHAIAJP&lt;/code&gt;, which is the word that GentlePurpleRain already identified with &lt;code&gt;SETTLEMENT&lt;/code&gt;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you look at the alphabet table, you will notice that the encryption is a simple shift to the right.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With this hindsight, I tried to decrypt the first sentence, and I managed to get this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;6TM6+XSYX#+MW+2WV+5#4UT+%988UW+FI+PXQT+$U8#97+ZWN+6TM6+XSYX#+MW|&lt;/code&gt;&#xA;&lt;br&gt;&#xA; &lt;code&gt;that+hcihw+is+ton+owned+cannot+be+dleh+ransom+rof+that+hcihw+is|&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which, considering the reversed words, becomes obviously&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That which is not owned cannot be held ransom for that which is.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Only problem is,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I had to try different shift values for each word, and I didn't find any particular rule. In case anyone is wondering, for that line the shift where: -3,    -4, -1, -2, -4, -5, -1, -3, -5, -2 (negative means to the left).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll try and come up with a more precise rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As pointed out by f'' in the comments, the shift values are defined by the word's length (minus one, depending how you count).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also have the full text:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;that+hcihw+is+ton+owned+ca&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;nnot+be+dleh+ransom+rof+th&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;at+hcihw+is|noitubirtsid+w&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;ill+reven+gain+rewop+over+&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;noitaerc+for+eht+value+fo+&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;A+tcudorp+is+inversely+det&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;aler+to+sti+availability|e&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;no+offers+dlog+settlement+&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;dna+another+sih+warm+tfig|&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or better:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That which is not owned cannot be held ransom for that which is. Distribution will never gain power over creation for the value of a product is inversely related to its availability. One offers cold settlement and another his warm gift.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5990" LastEditorUserId="4421" LastEditDate="2015-08-07T20:20:16.587" LastActivityDate="2015-08-07T20:20:16.587" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18756" PostTypeId="2" ParentId="18532" CreationDate="2015-08-07T15:47:55.043" Score="1" Body="&lt;p&gt;Searching this room for all up and not down, as the last riddle suggested, I'd say the password is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; EXTRA which is in all uppercase letters in the original piece of paper&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13715" LastActivityDate="2015-08-07T15:47:55.043" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18757" PostTypeId="2" ParentId="18683" CreationDate="2015-08-07T17:17:41.353" Score="5" Body="&lt;p&gt;Got it! It's a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; slug&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Slower than a sloth:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Slugs travel up to 0.01 m/s, sloths about 0.03 m/s on the ground.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Faster than sound:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bullets are called slugs too, and most of them travel faster than sound.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Printers think I'm good for nothing:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Slug_(typesetting)&quot;&gt;Slugs&lt;/a&gt; are used in printing to space paragraphs, so are in a sense good for &quot;nothing&quot; (space.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bartenders charge good money for me:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A slug is also a shot of liquor.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10851" LastActivityDate="2015-08-07T17:17:41.353" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18758" PostTypeId="1" AcceptedAnswerId="18760" CreationDate="2015-08-07T17:36:32.843" Score="10" ViewCount="1039" Body="&lt;p&gt;5 persons numbered 1 to 5 are standing in a row at 5 spots. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1   2   3   4   5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any person can call another person among the remaining 4. When someone calls someone else, the person who was called comes on the adjacent spot toward his side, &lt;strong&gt;on the same side of the person that called him&lt;/strong&gt;, and then the others rearrange themselves accordingly. (see example ahead) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the first and last spot are special. After each time someone calls someone else, subsequent to the rearrangement, &lt;strong&gt;the two persons standing on the first and last spot teleport to each others’ spots!&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if person 2 calls person 5, and after rearrangement and teleportation, person 1 calls person 4, these &lt;strong&gt;two steps&lt;/strong&gt; can be represented in the following way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/79EOq.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/79EOq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They can’t call anyone who’s already adjacent to them of course, since they’re already on their side, and it won't be counted as a step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the minimum number of steps that’ll take for the people for going&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;from:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1    2    3    4    5&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;to:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;5    4    3    2    1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: I'm new at creating puzzles, I might have missed some important detail, so please bear with me. I'll try to respond promptly to the comments and make any required changes.&lt;/p&gt;&#xA;" OwnerUserId="14074" LastActivityDate="2015-08-08T10:29:56.723" Title="Minimum number of steps in which these people can rearrange themselves?" Tags="|mathematics|logical-deduction|strategy|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18760" PostTypeId="2" ParentId="18758" CreationDate="2015-08-07T17:48:06.497" Score="7" Body="&lt;p&gt;I think it can be done in 3 calls (and 3 corresponding switches). &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 2 3 4 5     &#xA;              (2 calls 4)&#xA;5 2 4 3 1     (4 moves next to 2, then 1 and 5 switch) &#xA;              (5 calls 4)&#xA;1 4 2 3 5     (4 moves next to 5, then 5 and 1 switch again) &#xA;              (4 calls 3)&#xA;5 4 3 2 1     (3 moves next to 4, and 5 and 1 switch one last time) &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;A brilliant observation from Tyler Seacrest &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Note that there is no one move solution, because after one move, either 5 or 1 is in the correct ending position. Then, after another move, the 5 or 1 is no longer in the correct ending position. So there is no two-move solution either. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-08-07T18:01:13.120" LastActivityDate="2015-08-07T18:01:13.120" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18761" PostTypeId="2" ParentId="18758" CreationDate="2015-08-07T18:10:37.023" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3 steps.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 12345 -&gt; 52431 -&gt; 14235 -&gt; 54321&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or because of the symmetry:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 12345 -&gt; 53241 -&gt; 13425 -&gt; 54321&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: if there are more people, we can do this (omitting the first and the last end):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2, 3, ..., n-1$&lt;br&gt;&#xA;$n-1, 2, 3, ..., n-2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that we've managed to put $n-1$ in its right place, we can drag others next to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After putting numbers from 2 to $n-2$ ($n-3$ numbers) into their right place in $n-4$ more moves, we'll have made $n-3$ in total. If $n-3$ is odd ($n$ is even), it can be done. Otherwise we start with the ones from 5 to $n-2$ ($n-6$ numbers) in $n-6$ more moves, then finish the 432 part with another 3 moves, making $n-2$ moves in total:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;234&lt;br&gt;&#xA;324&lt;br&gt;&#xA;342&lt;br&gt;&#xA;432 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;234&lt;br&gt;&#xA;243&lt;br&gt;&#xA;423&lt;br&gt;&#xA;432  &lt;/p&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="14016" LastEditDate="2015-08-08T10:29:56.723" LastActivityDate="2015-08-08T10:29:56.723" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18762" PostTypeId="2" ParentId="18751" CreationDate="2015-08-07T18:14:31.417" Score="3" Body="&lt;p&gt;&lt;em&gt;Based on the previous hint (Movies) and the current hint (Numbers), the theme to this set is movies that contain some number in their titles...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D_ T_  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Zero Dark Thirty&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;C_ F_ N_ T_ O_*&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One Flew Over The Cuckoo's Nest &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;N_ W_&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Two Weeks Notice&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A A_ F_ W_  (Credit to TroyAndAbed) &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Four Weddings And A Funeral&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A A_ B_ M_&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3 Men And A Baby&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;D_ O_ S_&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 500 Days Of Summer / Six Degrees Of Separation &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;D_ N_ S_&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Six Days, Seven Nights&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;B_ B_ F_ S_&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Seven Brides For Seven Brothers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;M_ O_&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One Man Jury&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;M_&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 12 Monkeys / 8 Mile&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I_ L_ O_&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;10 Items Or Less&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;E_ E_  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 11-11-11&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A_ M_&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 12 Angry Men&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;G_ O_ 30&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 13 Going On 30&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-08-09T19:18:55.737" LastActivityDate="2015-08-09T19:18:55.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18763" PostTypeId="2" ParentId="18615" CreationDate="2015-08-07T18:32:25.560" Score="8" Body="&lt;p&gt;Ok, with a little help, answer complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are we? &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stairs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are made from addition or subtraction&lt;br&gt;&#xA;Our best are often a popular attraction&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stairs used too be carved out of things: Subtraction.&lt;br&gt;&#xA; Now they are generally built: Addition&lt;br&gt;&#xA; I guess some can be an attraction? Not sure about this line  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some you can touch,&lt;br&gt;&#xA;and some you cannot&lt;br&gt;&#xA;A dozen will heal your illness&lt;br&gt;&#xA;Another shan't be bought&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some stairs are actually just stairs, some are figurative.&lt;br&gt;&#xA; A dozen will heal your illness: &lt;a href=&quot;https://en.wikipedia.org/wiki/Twelve-step_program&quot;&gt;The 12 steps to Healing&lt;/a&gt;&lt;br&gt;&#xA; Shan't be bought: The Stairway to Heaven  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some of us have thousands, and some, just a few&lt;br&gt;&#xA;We all can change one's point of view&lt;br&gt;&#xA;But, y'know, sometimes words have two meanings&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some stairs have thousands of steps or just a few.&lt;br&gt;&#xA; Change point of view quite literally, by making you higher or lower.&lt;br&gt;&#xA; &quot;But, y'know, sometimes words have two meanings&quot; -&gt; Stairway to Heaven reference  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some of us have utility&lt;br&gt;&#xA;but others are for show&lt;br&gt;&#xA;One befuddling, three worlds becoming one&lt;br&gt;&#xA;made o'er fifty years ago&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stairs are usually made to get to places, but some can be ornamental&#xA; The &quot;befuddling&quot; is &lt;a href=&quot;https://en.wikipedia.org/wiki/Relativity_(M._C._Escher)&quot;&gt;Relativity&lt;/a&gt; by Escher. 3 worlds = 3 gravity sources&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By now you may have guessed what we are&lt;br&gt;&#xA;so we ask that you find our brother, a movie star&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He watched a broken man escape from prison&lt;br&gt;&#xA;who let fear find him, and now is risen.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; So clear reference to the Dark Knight Rises, and the Pit. The pit was filmed at/based off the &lt;a href=&quot;https://en.wikipedia.org/wiki/Chand_Baori&quot;&gt;Chand Baori&lt;/a&gt; step-well in India. This is the brother.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1968" LastEditorUserId="1968" LastEditDate="2015-08-10T15:29:39.093" LastActivityDate="2015-08-10T15:29:39.093" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18764" PostTypeId="1" AcceptedAnswerId="18776" CreationDate="2015-08-07T18:55:08.973" Score="12" ViewCount="420" Body="&lt;p&gt;I'd like to share with you a story about my yesterday. Perhaps you'll find it more exciting than I do.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I found a free-range donkey&lt;br&gt;&#xA;next to a free-range horse&lt;br&gt;&#xA;and what I did with them, you ask?&lt;br&gt;&#xA;I had them breed, of course!&lt;br&gt;&#xA;They had one foal and left him&lt;br&gt;&#xA;afraid and all alone&lt;br&gt;&#xA;so I threw a pack upon his back&lt;br&gt;&#xA;and kept him as my own&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I wandered through the jungle&lt;br&gt;&#xA;over buggy, muggy ground&lt;br&gt;&#xA;when I felt something slipping --&lt;br&gt;&#xA;oh, my pants were falling down!&lt;br&gt;&#xA;I located some tangled vines&lt;br&gt;&#xA;('twere thin and didn't bite)&lt;br&gt;&#xA;and wrapped them well around my waist&lt;br&gt;&#xA;to hold my pants up tight&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Tonight I sleep quite soundly&lt;br&gt;&#xA;as imps prance through my head&lt;br&gt;&#xA;I think of dragons and gypsy wagons&lt;br&gt;&#xA;as I lie still in my bed&lt;br&gt;&#xA;I force myself to stay asleep&lt;br&gt;&#xA;so the fantasies persist&lt;br&gt;&#xA;but to my surprise, as I arise&lt;br&gt;&#xA;the fantasies exist!&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Each stanza holds a two-word phrase&lt;br&gt;&#xA;all key to my solution&lt;br&gt;&#xA;though in each case, you'll need to make&lt;br&gt;&#xA;a rhyming substitution!&lt;br&gt;&#xA;And once you've found the phrases&lt;br&gt;&#xA;and you've located the rhymes&lt;br&gt;&#xA;You'll find the answer, clear as mud&lt;br&gt;&#xA;(a vapid waste of time)&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So, what's the hidden message in my story?&lt;/p&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="11475" LastEditDate="2015-08-07T19:36:08.080" LastActivityDate="2015-08-08T01:54:35.750" Title="A tedious tale with hidden meaning" Tags="|word|rhyme|wordplay|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18765" PostTypeId="2" ParentId="18704" CreationDate="2015-08-07T19:42:09.323" Score="1" Body="&lt;p&gt;Sail on, Found in&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ceylon (name for Sri Lanka), Poondi, Tamil Nadu or Fonadhoo, Maldives or &quot;foun&quot; day/s&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;utter majesteer', I&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There I was served tea. or I was attacked by arrow (Andaman Islands are home to the only known paleolithic people; still use bow and arrow.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;mean, on 'me, paluk paneer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; On the menu, paluk paneer (India-Subcontinental Cottage cheese dish)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;speaking s'pecifically&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Speaking &lt;em&gt;Pacific&lt;/em&gt;-ly&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;lower countries&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lower could mean like Netherlands, or South&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;sped though high seas&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It has two high seas (around it). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can't be sure of anything. It should help other though.&lt;/p&gt;&#xA;" OwnerUserId="14465" LastActivityDate="2015-08-07T19:42:09.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18766" PostTypeId="2" ParentId="18764" CreationDate="2015-08-07T19:47:37.903" Score="0" Body="&lt;p&gt;&lt;em&gt;Each stanza holds a two-word phrase&lt;br&gt;&#xA;all key to my solution&lt;br&gt;&#xA;though in each case, you'll need to make&lt;br&gt;&#xA;a rhyming substitution!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this means that in each stanza we need to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; replace one word by another, twice (getting two new words), in such a way as to make &lt;em&gt;all&lt;/em&gt; the lines rhyme rather than just nearly all.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the first stanza,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1) replace &quot;donkey&quot; by &quot;ass&quot; to (almost) rhyme with &quot;ask&quot;, at least in some accents (not mine, but never mind)&lt;br&gt;&lt;br&gt;2) I think the &quot;foal&quot; line should have a &quot;hole&quot; somewhere, so that it rhymes in the same way as &quot;pack&quot;/&quot;back&quot;&lt;br&gt;&lt;br&gt;So the first stanza gives ASSHOLE (!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;[work in progress]&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-08-07T19:47:37.903" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18767" PostTypeId="1" AcceptedAnswerId="19913" CreationDate="2015-08-07T20:10:36.197" Score="18" ViewCount="1180" Body="&lt;p&gt;I'm where you keep your most private possessions&lt;br&gt;&#xA;With a small adjustment, I'm always on the move&lt;br&gt;&#xA;Again, and I'm one of you, lacking aggression&lt;br&gt;&#xA;Once more, at my end, my son has something to prove&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you gut &lt;em&gt;him&lt;/em&gt;, you may have done so for this&lt;br&gt;&#xA;Change that, and a group of great beasts have you found&lt;br&gt;&#xA;Again, it's my versatile partner, my bliss&lt;br&gt;&#xA;She protects me when orderly enemies are 'round&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make a rhyme, if you've done a good job it will squeak&lt;br&gt;&#xA;Sort them out and you'll find a great tool comes to view&lt;br&gt;&#xA;&lt;em&gt;This weapon, however, is not the word that you seek:&lt;/em&gt;&lt;br&gt;&#xA;It's the user, of which, in my land I have two&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take his opposite, and several later you've got one&lt;br&gt;&#xA;Change its first, and this riddle's answer you &lt;em&gt;desire&lt;/em&gt;&lt;br&gt;&#xA;Conjugate incorrectly, and you've not got a verb, son&lt;br&gt;&#xA;But one close to this soldier, in his path, you'll retire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, recall the user of the weapon above&lt;br&gt;&#xA;When his opposite appears, it is given this name&lt;br&gt;&#xA;One fix and you've found my minions least loved&lt;br&gt;&#xA;But without them I'd be at a loss in this game&#xA;&lt;hr&gt;&#xA;&lt;strong&gt;Disclaimer:&lt;/strong&gt; This riddle is intended to be a much harder follow-up to my &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18727/a-poem-of-dynamic-words&quot;&gt;quickly solved puzzle&lt;/a&gt; last week. At the end of each line, you should be left with a one-syllable, common word. &lt;strong&gt;Only lines 11 and 20 do not yield a word.&lt;/strong&gt; Most words will only change by one letter between lines, as hinted, but this riddle is more complex: phrases like &lt;em&gt;gut him&lt;/em&gt;, &lt;em&gt;sort them out&lt;/em&gt;, and &lt;em&gt;take his opposite&lt;/em&gt; should be interpreted analogously to &lt;em&gt;behead me&lt;/em&gt; from the previous riddle, which was wordplay for &lt;em&gt;remove my first letter&lt;/em&gt;.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;All words are one syllable long, and no longer than six letters&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;All words should be well known by the community&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Exception: Line 15 jokes that its word is an incorrect conjugation of [14]. I didn't know that it was even a word, but apparently it has an obscure definition.&lt;br&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;No proper nouns are used in this puzzle&lt;br&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Other words will change completely across line pairs, but I hope not to confuse you too badly, so I'll provide &lt;strong&gt;hints here:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;Hint 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; From line 4 to 5, our first odd pair appears&lt;br&gt;&#xA; Hark! From first to third person it veers!&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In lines 7 and 8, the subject is the same&lt;br&gt;&#xA; For [8], the [7] of [4] is given this name.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Line 3 describes its word in two separate ways&lt;br&gt;&#xA; It makes a good place to begin solving this maze.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13237" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2016-07-12T15:10:17.253" Title="Who's missing from this glorious Poem of Dynamic Words?" Tags="|word|rhyme|wordplay|" AnswerCount="1" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="18768" PostTypeId="1" AcceptedAnswerId="18771" CreationDate="2015-08-07T20:54:15.903" Score="10" ViewCount="1450" Body="&lt;p&gt;We all know the Simpsons' little greyhound.&lt;br&gt;&#xA;But what is Bart's pet &lt;em&gt;saying&lt;/em&gt; in this picture?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zCG44.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zCG44.png&quot; alt=&quot;Santa&amp;#39;s Little Helper says: &amp;quot;Amethysts or Hamster&amp;#39;s Toy Rheostats My Mossy Threat&amp;quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="8874" LastEditDate="2016-02-05T15:56:34.837" LastActivityDate="2016-02-05T15:56:34.837" Title="If dogs could talk" Tags="|enigmatic-puzzle|" AnswerCount="3" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="18769" PostTypeId="2" ParentId="18768" CreationDate="2015-08-07T21:05:11.433" Score="12" Body="&lt;p&gt;Is it...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Homer's tasty&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I guessed this because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;each line was an anagram of the others, and this seemed to be the most logical missing anagram.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2015-08-07T21:05:11.433" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18771" PostTypeId="2" ParentId="18768" CreationDate="2015-08-07T21:14:38.693" Score="19" Body="&lt;p&gt;It must be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Eat My Shorts!&lt;/strong&gt; &lt;br&gt; He's repeating anagrams of Bart's catchphrase. What a clever dog, that Santa's Little Helper! &lt;br&gt;&#xA; For the record, Deusovi answered first (+1), verifying that they were all anagrams, but didn't come up with the one which most fits the puzzle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He might also be barking &lt;strong&gt;&quot;HANDWOVEN AT CAMO&quot;&lt;/strong&gt; ;)&lt;/p&gt;&#xA;" OwnerUserId="13237" LastEditorUserId="13237" LastEditDate="2015-08-07T21:35:04.923" LastActivityDate="2015-08-07T21:35:04.923" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="18772" PostTypeId="2" ParentId="18768" CreationDate="2015-08-07T21:25:46.300" Score="9" Body="&lt;p&gt;I'm amazed by the responses to this question in so many ways. Here was my intended formulation of the answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Eat My Shorts!&lt;/strong&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reason is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;What is Bart's pet saying?&quot; can be interpreted as &quot;What is Bart's favourite saying.&quot;  The anagrams weren't actually needed to answer the question!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="14177" LastEditDate="2015-08-07T21:36:36.173" LastActivityDate="2015-08-07T21:36:36.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18773" PostTypeId="1" AcceptedAnswerId="19847" CreationDate="2015-08-07T21:31:51.060" Score="10" ViewCount="692" Body="&lt;p&gt;&lt;em&gt;Complete me one way - if you do this to friends, you must have many enemies.&lt;br&gt;&#xA;Complete me another - if you're tough you'll do this to your enemies.&lt;br&gt;&#xA;Yet in a third way - with this you can keep your enemies from you.&lt;br&gt;&#xA;And in a fourth way - with this you can always defeat your enemies.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are these words so entwined with enemies?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint/clarification:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; in each line, &quot;completing me&quot; refers to adding a &lt;strong&gt;single&lt;/strong&gt; letter (a different letter in each line).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; VOWELS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-08-10T21:05:33.763" LastActivityDate="2015-08-11T17:44:17.357" Title="Riddle from the one-eyed king" Tags="|riddle|word|letters|" AnswerCount="5" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="18774" PostTypeId="2" ParentId="18773" CreationDate="2015-08-07T23:16:01.993" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I interpret this as words of the form KI**, with the base word as ki, which is another spelling of chi&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Complete me one way - if you do this to friends, you must have many enemies.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;KILL&lt;/strong&gt; ... I feel like the causation might be self explanatory&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Complete me another - if you're tough you'll do this to your enemies.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;KISS&lt;/strong&gt; ... Nothing is harder than reconciliation with a foe&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet in a third way - with this you can keep your enemies from you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;KICK&lt;/strong&gt; ... excellent maneuver for creating distance in a fight&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And in a fourth way - with this you can always defeat your enemies.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;KIND&lt;/strong&gt; ... Kill'em with kindness fellas&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13916" LastActivityDate="2015-08-07T23:16:01.993" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18775" PostTypeId="2" ParentId="18615" CreationDate="2015-08-07T23:23:49.577" Score="2" Body="&lt;p&gt;I thought you were:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Elements, notably artificial ones&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can be made from addition or subtraction&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nuclear fusion &amp;amp; nuclear fission&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Our best are often a popular attraction&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I genuinely felt like I was onto something at this stage. Many elements like carbon are interesting and create lots of news articles and science showcases.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some you can touch,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hydrogen etc etc etc&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and some you cannot&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I wouldn't want to touch mercury&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One with a dozen will heal you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Magnesium has 12 protons and is important to biological life, though that's a bit contrived and doesn't &quot;heal you&quot; per se. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another, inumerable, shan't be bought&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can't buy uranium, which has many protons, nor many other elements which are banned or too difficult to produce/keep stable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some of us have thousands, and some, just a few&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most substances have (millions of) thousands of atoms, whereas some, like carbon nanotubes have much fewer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We all can change one's point of view&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Light reaching our eyes is influenced drastically by what type of matter it interacts with on the way, though again a little contrived.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But, y'know, sometimes words have two meanings&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Didn't match it to this line&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Most of us have utility&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most elements are used for something!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but others are for show&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lots of elements are quite boring (noble gases particularly so afaik) and some can only be seen for brief periods on labs, so have little utility.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Like one befuddling, three worlds all in one&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I again didn't match this line with anything, and figured that I just wasn't thinking it through well&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;made 'round sixty years ago&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mendelevium was created artificially 60 years ago. Anything past this point I had no clue about, but figured there was some play on words with the name of an element for each of them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By now you may have guessed what we are&#xA;so we ask that you find our brother, a movie star&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He watched a broken man escape from prison&#xA;who let fear find him, and now is risen.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I was certain I was onto something! My first contribution to this StackExchange site so I hope I did it correctly! :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14468" LastActivityDate="2015-08-07T23:23:49.577" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18776" PostTypeId="2" ParentId="18764" CreationDate="2015-08-08T01:54:35.750" Score="12" Body="&lt;p&gt;&lt;em&gt;I found a free-range donkey&lt;br&gt;&#xA;next to a free-range horse&lt;br&gt;&#xA;and what I did with them, you ask?&lt;br&gt;&#xA;I had them breed, of course!&lt;br&gt;&#xA;They had one foal and left him&lt;br&gt;&#xA;afraid and all alone&lt;br&gt;&#xA;so I threw a pack upon his back&lt;br&gt;&#xA;and kept him as my own&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sounds like you ended up with a &lt;strong&gt;PET MULE&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I wandered through the jungle&lt;br&gt;&#xA;over buggy, muggy ground&lt;br&gt;&#xA;when I felt something slipping --&lt;br&gt;&#xA;oh, my pants were falling down!&lt;br&gt;&#xA;I located some tangled vines&lt;br&gt;&#xA;('twere thin and didn't bite)&lt;br&gt;&#xA;and wrapped them well around my waist&lt;br&gt;&#xA;to hold my pants up tight&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You improvised a &lt;strong&gt;PLANT BELT&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Tonight I sleep quite soundly&lt;br&gt;&#xA;as imps prance through my head&lt;br&gt;&#xA;I think of dragons and gypsy wagons&lt;br&gt;&#xA;as I lie still in my bed&lt;br&gt;&#xA;I force myself to stay asleep&lt;br&gt;&#xA;so the fantasies persist&lt;br&gt;&#xA;but to my surprise, as I arise&lt;br&gt;&#xA;the fantasies exist!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; And woke up to some &lt;strong&gt;REAL DREAMS&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Each stanza holds a two-word phrase&lt;br&gt;&#xA;all key to my solution&lt;br&gt;&#xA;though in each case, you'll need to make&lt;br&gt;&#xA;a rhyming substitution!&lt;br&gt;&#xA;And once you've found the phrases&lt;br&gt;&#xA;and you've located the rhymes&lt;br&gt;&#xA;You'll find the answer, clear as mud&lt;br&gt;&#xA;(a vapid waste of time)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If we take all the words from the phrases above, &quot;Pet mule plant belt real dreams,&quot; and instead substitute a rhyming word for each, we (of course) end up with the phrase, &quot;Jet fuel can't melt steel beams.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12511" LastActivityDate="2015-08-08T01:54:35.750" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18778" PostTypeId="2" ParentId="18773" CreationDate="2015-08-08T04:34:31.253" Score="1" Body="&lt;p&gt;I believe the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fo&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Complete me one way - if you do this to friends, you must have many enemies.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Foment&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Complete me another - if you're tough you'll do this to your enemies.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Forgive&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet in a third way - with this you can keep your enemies from you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Foil&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And in a fourth way - with this you can always defeat your enemies.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Force&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9289" LastActivityDate="2015-08-08T04:34:31.253" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18780" PostTypeId="2" ParentId="18690" CreationDate="2015-08-08T06:48:41.370" Score="1" Body="&lt;p&gt;Is this &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rand al'Thor? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Old dragon returning - hide your children!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rand al'Thor is also called the 'Dragon Reborn'. Also, this question was posted around the same time as our own rand al'thor returned after a long absence. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am not very well-versed with A Wheel Of Time, so I hope someone would help me make sense of the other clues. (Or correct me if this is just a shot in the dark.)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-08-08T07:02:48.627" LastActivityDate="2015-08-08T07:02:48.627" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18781" PostTypeId="2" ParentId="18369" CreationDate="2015-08-08T10:17:19.077" Score="11" Body="&lt;p&gt;Next Friday is ambiguous, but:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the assumption that next Friday means the Friday of the following week, then George meets Kari. If next Friday meant the Friday of the same week, acting as would be expected probably means he went home and cried, because he got the coded message Wednesday, asked for help Thursday evening, and got the answer Friday afternoon before finishing work, which would be too late.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The message is below:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DEAR CONSTANT SHADOW,&#xA;&#xA; DO NOT BELIEVE I HAVE NOT NOTICED YOUR PERMANENT ATTENTION! I CAN ONLY ASSUME THAT YOU HAVE HOPES OF BECOMING MY COMPANION. BUT LET ME TELL YOU YOU FIRST HAVE TO EARN MY RESPECT.&#xA; &#xA; YES, YOU MIGHT BE RATHER HANDSOME, BUT I ONLY RESPECT PEOPLE WHO CAN USE THEIR BRAIN. I DOUBT THAT YOU BELONG TO THIS GROUP. OR IS THIS A PREJUDICE? ARE YOU INDEED HANDSOME AND SMART? WE WILL SEE.&#xA; &#xA; IF YOU CAN READ THIS LETTER THAN I AM INDEED INTERESTED IN DATING YOU. I GIVE YOU A SINGLE CHANCE. &#xA; &#xA; I LIVE IN THE RED HOUSE NEXT TO THE CHINESE RESTAURANT IN CENTER OF TOWN. I WILL NOT MEET YOU THERE.&#xA; &#xA; I WANT TO MEET YOU AT THE LAKE WHERE THE LITTLE BLUE BRIDGE IS. SATURDAY NIGHT WOULD BE GOOD BECAUSE ONE CAN SEE THE BEAUTIFUL STARS. BUT I DO NOT MEET A BOY FOR THE FIRST TIME ALONE AT NIGHT. I AM NOT THAT STUPID. SO I SUGGEST WE MEET ON FRIDAY JUST AFTER THE POST OFFICE CLOSES AT NOON.&#xA; &#xA; I WILL WAIT FOR SIXTEEN MINUTES IN THE LITTLE PARK BEFORE THE HIGH SCHOOL. IF YOU DO NOT COME NEXT FRIDAY, OR IF YOU ARE NOT PUNCTUAL, I WILL NEVER SPEAK WITH YOU.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To decode the message:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Put all words in the message in alphabetical order. The decoded word is the one to the left of the coded one in the series.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14475" LastActivityDate="2015-08-08T10:17:19.077" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18782" PostTypeId="1" AcceptedAnswerId="20311" CreationDate="2015-08-08T10:48:29.277" Score="82" ViewCount="3846" Body="&lt;p&gt;A rare double-edged sword is currently on loan from the British Museum to the &lt;a href=&quot;http://mentalfloss.com/article/67032/can-you-translate-800-year-old-swords-inscription&quot; rel=&quot;noreferrer&quot;&gt;Magna Carta exhibition&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.bl.uk/collection-items/double-edged-sword&quot; rel=&quot;noreferrer&quot;&gt;The sword itself&lt;/a&gt; is described as likely made in Germany around 1250-1330 and includes the inscription pictured below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CBpeD.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CBpeD.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xLHGN.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xLHGN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what does &lt;strong&gt;+NDXOXCHWDRGHDXORVI+&lt;/strong&gt; actually mean?&lt;/p&gt;&#xA;" OwnerUserId="4852" LastEditorUserId="2071" LastEditDate="2016-08-22T20:22:26.840" LastActivityDate="2016-08-22T20:22:26.840" Title="A good puzzle is a double-edged sword" Tags="|history|real|unsolved-mysteries|" AnswerCount="2" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="18783" PostTypeId="1" AcceptedAnswerId="18784" CreationDate="2015-08-08T13:59:29.833" Score="7" ViewCount="501" Body="&lt;p&gt;Yesterday, while I was tidying my attic, I found a letter dedicated to my  great-great-grandfather. (Note that the letter was inside a box which my great-great-grandfather transmitted to my great-grandfather, and so on...) It is supposed to tell him his death-day. I actually know he got murdered, and I also know the exact date when it happened. Now I just want to know if this letter is actually accurate, but I can't decrypt/solve it. Can someone please help me? This is what it says (except the part about how he will be killed):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... &#xA;(5.3-912-83-19-24) + (2.3-12-931-453-1) = mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(119.9-12-32-183-873) + (300-200-193-930-39.2) = dd&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4.2-342-12-123-847) + (293-1-2-93-8.7) + (59-21-329-12-23-987) + (4.2-181-192-712-9995) = yyyy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have fun decrypting it. Meanwhile I will wait. You should start learning how to make SOS-signs, because no-one will hear you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S: Don't even think of solving it the &quot;regular&quot; way, like you've learned in school, you silly!&lt;/p&gt;&#xA;" OwnerDisplayName="user14478" LastEditorUserId="5373" LastEditDate="2015-08-08T14:23:30.473" LastActivityDate="2015-08-08T14:23:30.473" Title="A letter telling my great-great-grandfather's death-day" Tags="|cipher|story|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18784" PostTypeId="2" ParentId="18783" CreationDate="2015-08-08T14:02:59.237" Score="6" Body="&lt;p&gt;It was on&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 19 November 1901&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because (short answer)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Morse code&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or because (long answer)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ignore the numbers in the puzzle; they're just a red herring. Concentrate instead on the dots and dashes between the numbers. We get:&lt;br&gt;&lt;br&gt;mm = (.----)(.----) = 11&lt;br&gt;dd = (.----)(----.) = 19&lt;br&gt;yyyy = (.----)(----.)(-----)(.----) = 1901.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-08-08T14:05:57.227" LastActivityDate="2015-08-08T14:05:57.227" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19783" PostTypeId="2" ParentId="18650" CreationDate="2015-08-08T16:55:53.600" Score="2" Body="&lt;p&gt;Following the idea of &lt;a href=&quot;/questions/tagged/wordplay&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;wordplay&amp;#39;&quot; rel=&quot;tag&quot;&gt;wordplay&lt;/a&gt;, maybe the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; that you and your opposite are both letters of the alphabet, appearing respectively in various words?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;For example, one of us could be relaxed while the other is making things happen.&lt;br&gt;&#xA;One could be underneath while the other is making connections.&lt;br&gt;&#xA;One could be in court while the other is getting help from equipment.&lt;br&gt;&#xA;One can bring power while the other shows disease.&#xA;On the radio, one of us could be reading while the other gives chase.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The only letter appearing in RELAXED, UNDERNEATH, COURT, POWER, and READING is &lt;strong&gt;R&lt;/strong&gt;; the only letter appearing in THINGS HAPPEN (?), CONNECTIONS, EQUIPMENT, DISEASE, and CHASE is &lt;strong&gt;E&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am the most powerful, a group of people.&#xA;My opposite was once my rival, able to change between others.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; E and R both appear in the middle of the word POWERFUL, and E appears several times in CHANGE BETWEEN OTHERS.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;We can both be found in the same surroundings, often doing very different things.&lt;br&gt;&#xA;Normally, though, our interests overlap.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many words contain both E and R. For example, ARSEHOLES!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What am I and what is my opposite?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are R, and your opposite is E.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Bonus: where else can we both be found?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; KNICKERS!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-08-08T16:55:53.600" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19784" PostTypeId="2" ParentId="18650" CreationDate="2015-08-08T20:23:17.357" Score="9" Body="&lt;p&gt;You and your opposite are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the letters &lt;strong&gt;us&lt;/strong&gt; and &lt;strong&gt;su&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the most powerful, a group of people.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The US has the most powerful military force in the world, and &quot;us&quot; describes a group of people.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My opposite was once my rival, able to change between others.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Soviet Union, which was the US's rival in the Cold War, could be abbreviated as SU, and &quot;su&quot; is a Unix command to change to another user.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can both be found in the same surroundings, often doing very different things.&lt;br&gt;&#xA;For example, one of us could be relaxed while the other is making things happen.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One is laid-back and &quot;ca&lt;strong&gt;su&lt;/strong&gt;al&quot;, while the other is &quot;ca&lt;strong&gt;us&lt;/strong&gt;al&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One could be underneath while the other is making connections.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One is &quot;&lt;strong&gt;su&lt;/strong&gt;b&quot;, while the other is &quot;&lt;strong&gt;US&lt;/strong&gt;B&quot;, a standard for connecting devices.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One could be in court while the other is getting help from equipment.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One might be &quot;&lt;strong&gt;su&lt;/strong&gt;ing&quot; in court, while the other is &quot;&lt;strong&gt;us&lt;/strong&gt;ing&quot; equipment.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One can bring power while the other shows disease.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A &quot;P&lt;strong&gt;SU&lt;/strong&gt;&quot; provides power, while &quot;p&lt;strong&gt;us&lt;/strong&gt;&quot; is usually a symptom of disease.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On the radio, one of us could be reading while the other gives chase.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; While these aren't spelt the same, they sound alike - &quot;per&lt;strong&gt;us&lt;/strong&gt;ing&quot; vs &quot;pur&lt;strong&gt;su&lt;/strong&gt;ing&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Normally, though, our interests overlap.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Us&quot; and &quot;su&quot; are both in &quot;usually&quot;, sharing an S.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bonus: where else can we both be found?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are a number of other words with both &quot;us&quot; and &quot;su&quot;, such as &quot;suspicious&quot; or &quot;suspire&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12511" LastEditorUserId="12511" LastEditDate="2015-08-08T20:28:52.937" LastActivityDate="2015-08-08T20:28:52.937" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19786" PostTypeId="2" ParentId="18347" CreationDate="2015-08-08T23:40:24.163" Score="7" Body="&lt;p&gt;I'll post partial progress here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second stanza:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tick. A penciled confirmation (checkmark) performs music (conducting: a &quot;tick&quot; is informally a beat of music) to the dance of clocks (clocks tick) and creepy crawly things (a tick is a kind of insect) at rest inside your pants. A sudden motion shocks you (a &quot;tic&quot; is an involuntary motion) in a thousand different ways. (&quot;k&quot; means &quot;thousand&quot;.) In other words, the seventh note (&quot;ti&quot;, as in &quot;a drink with jam and bread&quot;) sings Casey's gone astray (&quot;kc&quot; = &quot;Casey&quot;, gone astray in the anagram sense).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The third stanza (much more tenuous): &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cross. The dispossessed are missing one to become shrouded skies. (&quot;The Dispossessed&quot; is a book by Ursula le Guin, set in the Tau Ceti system. Remove the final word from it to get &quot;tau&quot;, which is the Greek &quot;t&quot;. Not sure about the &quot;shrouded skies&quot;.)&#xA; Perform on it this action as you do with dotted eyes (&quot;dot the i's and &lt;em&gt;cross&lt;/em&gt; the t's&quot;) . The aforementioned snakes return, though this time stick together. A bird of death approaches them with blackened, shedding feathers. (&quot;crow&quot;, &quot;w&quot;ith blackened, adding &quot;ss&quot; for the snakes, though I don't know where the feathers have gone.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14305" LastEditorUserId="14305" LastEditDate="2015-08-09T08:18:49.777" LastActivityDate="2015-08-09T08:18:49.777" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19787" PostTypeId="2" ParentId="18347" CreationDate="2015-08-09T00:34:41.823" Score="8" Body="&lt;p&gt;Another partial answer, the first stanza:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Swords. Not sure if the &quot;man of prowess&quot; is a specific reference, but &quot;Another man provides them both with sparks and cinders falling&quot; refers to the byproducts of a blacksmith beating a heated sword into shape. &quot;Two snakes slither, silently surrounding unaware&quot; suggests two S's around &quot;a line about what's lost between triumph and moving air&quot; - the &quot;line about&quot; being &quot;row&quot; backwards, and D being the difference between &quot;win&quot; and &quot;wind&quot;. S-WOR-D-S becomes &quot;swords&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12511" LastActivityDate="2015-08-09T00:34:41.823" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19789" PostTypeId="1" AcceptedAnswerId="19795" CreationDate="2015-08-09T09:04:02.253" Score="-5" ViewCount="186" Body="&lt;p&gt;What should be the value of the bottom row in our&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Euro 2012 Football themed grid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/T8zve.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/T8zve.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2375" LastActivityDate="2015-08-09T17:37:08.620" Title="mensateaser [level 0] (part 1)" Tags="|mathematics|calculation-puzzle|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19790" PostTypeId="2" ParentId="19789" CreationDate="2015-08-09T09:29:31.250" Score="10" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 115&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My reasoning - I'm not sure what the blue &amp;amp; white figure is, but it looks like a shawl to me:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the leftmost column, we have two shawls and two shirts which add up to 100. That means one shawl and one shirt add up to 50. Subtract that from the second row, and we're left with two balls being equal to 50. Which means the &lt;strong&gt;ball&lt;/strong&gt; has a value of &lt;strong&gt;25&lt;/strong&gt;. Substitute that in the third column - we get that three shirts are equal to 120, so the &lt;strong&gt;shirt&lt;/strong&gt; has a value of &lt;strong&gt;40&lt;/strong&gt;. We already know one shawl plus one shirt equals 50, so the &lt;strong&gt;shawl&lt;/strong&gt; has a value of 10. Therefore, the value of the bottom row is 40+25+40+10 = &lt;strong&gt;115&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11110" LastEditorUserId="11110" LastEditDate="2015-08-09T09:35:48.113" LastActivityDate="2015-08-09T09:35:48.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19791" PostTypeId="1" AcceptedAnswerId="19993" CreationDate="2015-08-09T09:56:20.267" Score="22" ViewCount="1235" Body="&lt;p&gt;Here are the clues;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I'm a riddle inside an enigma  wrapped up in a mystery.&lt;/li&gt;&#xA;&lt;li&gt;Some respected scholars won't even agree I'm the right answer when you've found me. &lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You don't have to be a mathematician to find me. Just be able to read and count.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;Mr Smith was certainly no mathematician, and he thought I was 34. In his own way, he was right. &lt;/li&gt;&#xA;&lt;li&gt;If you can't solve me, come back to this clue again and again.&lt;/li&gt;&#xA;&lt;li&gt;You can find me on a roulette wheel but there's no point in betting on me.&lt;/li&gt;&#xA;&lt;li&gt;I have only one owner even if he does sound uncouth.&lt;/li&gt;&#xA;&lt;li&gt;Don't upset me. There might be a swift reaction depending on our location. Someone's got me in mind, and you know where &lt;em&gt;he&lt;/em&gt; stands.&lt;/li&gt;&#xA;&lt;li&gt;Some people think I'm so unlucky that they even have a phobia about me. It's all Greek to me.&lt;/li&gt;&#xA;&lt;li&gt;2, 3, 37&lt;/li&gt;&#xA;&lt;li&gt;You've got plenty of this and it's made up of me.&lt;/li&gt;&#xA;&lt;li&gt;17, 18 ... Look at those numbers. If you can decipher them, the answer will easily be revealed.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Bonus clues&lt;/p&gt;&#xA;&#xA;&lt;p&gt;14.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Don't ignore the title. It tells you what you are looking for. (I suppose mathematicians might quibble about the precise terminology used in the title.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;15.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many people have me as a tattoo.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: All the clues refer to the same answer. Only a thorough and well-supported explanation of each and every clue will lead to a green tick. If you're not convinced by your own answer to a particular clue then you haven't understood it ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hints: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;All the clues are numbered. Sometimes the number  of the clue is part of the clue and sometimes it isn't.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Note that clues vary in type. A simple clue can have a very simple meaning. A wordy clue may perhaps be taken as a hint as how to proceed. A numerical clue could have a mathematical meaning - or not.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Clues may only give part of the information needed - they back up other clues and reduce the possibilities of false answers. It is &lt;em&gt;possible&lt;/em&gt; to get the answer from a single clue but a clue on its own cannot be conclusive without the others. Not all clues have sufficient information to stand alone. They are there merely to provide confirmation or a hint.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="14177" LastEditDate="2015-08-13T02:39:15.670" LastActivityDate="2015-08-13T07:25:23.410" Title="Just a few numbers" Tags="|riddle|enigmatic-puzzle|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19792" PostTypeId="2" ParentId="18347" CreationDate="2015-08-09T10:02:35.957" Score="11" Body="&lt;p&gt;The secret to happiness is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cryptic crosswords!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fourth stanza:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CRYPT: The first four lines are a relatively straightforward definition.  &quot;Unseen to the untrained eye&quot; could refer to encryption.  A crypt is a tomb where &quot;bones are ever rattling&quot; of those buried there.  &quot;Vast treasures&quot; might be found buried with the dead - in the tomb of a pharaoh, for example.  &quot;A waterfall appears from fears no longer held within&quot; when we CRY; &quot;Paper tiger cardinals&quot; means the first letters of &quot;paper tiger&quot; = PT.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The whole:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CRYPT + TICK + CROSS + SWORDS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fifth stanza:&#xA;tells how to assemble the solutions to stanzas 1-4 into the overall solution.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The words are joined in pairs, so the solution is two words, each composed of two stanza solutions.  &quot;Strangers&quot; take the rear; a synonym for strange is &quot;odd&quot;, so it is telling us about the placement of the odd-numbered stanza words, &quot;Cross&quot; and &quot;swords&quot;.  The &quot;giants&quot; that lead are the stanzas with &quot;giant&quot; words: &quot;vast&quot; in stanza 4 and &quot;thousand&quot; in stanza 2.  Manipulate slightly: remove the extra T and S, and the unwanted K.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Addendum: Stanza 3&lt;br&gt;&#xA;Could the dispossessed refer to  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://baileyaplangandcomp.wordpress.com/2015/02/10/the-dispossessed-les-miserables/&quot; rel=&quot;nofollow&quot;&gt;Les Mis&lt;/a&gt;?&#xA; Then adding a T to MIS gives MIST (&quot;shrouded skies&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4385" LastEditorUserId="4385" LastEditDate="2015-08-10T20:08:18.267" LastActivityDate="2015-08-10T20:08:18.267" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="19794" PostTypeId="2" ParentId="18690" CreationDate="2015-08-09T16:37:02.327" Score="0" Body="&lt;p&gt;Perhaps...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The moon (the sun might work too)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Old dragon returning - hide your children!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hide them every night when the moon returns so they can sleep peacefully, also, this reminds me of a Doctor Who episode where the Moon turns out to be a giant 'dragon' egg&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Royal is my essence, my innards rule the waves&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I believe I once heard that the moon controls the waves with its gravity as the earth rotates&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You and me and the devil form a handsome trinity&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Me, on the ground, you in the sky, and the devil, underneath the ground in Hell&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15488" LastActivityDate="2015-08-09T16:37:02.327" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19795" PostTypeId="2" ParentId="19789" CreationDate="2015-08-09T17:37:08.620" Score="5" Body="&lt;p&gt;Glorfindel is correct, but there's an algebraic approach to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the shirt be &lt;em&gt;x&lt;/em&gt;, the shorts be &lt;em&gt;y&lt;/em&gt;, and the ball be &lt;em&gt;z&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that in mind, we now have:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2x + 2y = 100&lt;p&gt;&#xA; 3x + z = 145&lt;p&gt;&#xA; x + y + 2z = 100&lt;p&gt;&#xA; 2x + y + z = ??&lt;p&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, we can solve for two variables, then use the substitution method. Taking the easiest, we get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2x + 2y = 100 -&gt; 2y = -2x + 100 -&gt; y = -x + 50&lt;p&gt;&#xA; 3x + z = 145 -&gt; z = -3x + 145&lt;p&gt;&#xA; x + (-x + 50) + 2(-3x + 145) = 100&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This equates out to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; x - x + 50 - 6x + 290 = 100 -&gt; 50 - 6x + 290 = 100 -&gt; -6x + 340 = 100 -&gt; -6x = -240 -&gt; x = 40&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now that we know what &lt;em&gt;x&lt;/em&gt; is, we can solve the rest.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2(40) + 2y = 100 -&gt; 80 + 2y = 100 -&gt; 2y = 20 -&gt; y = 10&lt;p&gt;&#xA; 40 + 10 + 2z = 100 -&gt; 50 + 2z = 100 -&gt; 2z = 50 -&gt; z = 25&lt;p&gt;&#xA; 2(40) + 10 + 25 = 115&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This was more a mathematical problem than a puzzle, though.&lt;/p&gt;&#xA;" OwnerUserId="4353" LastActivityDate="2015-08-09T17:37:08.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19796" PostTypeId="2" ParentId="18773" CreationDate="2015-08-09T17:42:32.730" Score="1" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ARSE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Riddle from the one-eyed king&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;What can that mean other than an arse? If it weren't for this clue in the title, EARS or SEAR or the unordered set {A, E, R, S} would do.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Complete me one way - if you do this to friends, you must have many enemies.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Add M to get SMEAR, or T to get TASER. If that's how you treat friends, they will soon stop being your friends and may become your enemies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Complete me another - if you're tough you'll do this to your enemies.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Add C to get SCARE.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Yet in a third way - with this you can keep your enemies from you.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Add P to get SPEAR. Adding B to get SABRE might be less successful.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;And in a fourth way - with this you can always defeat your enemies.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Add N to get SNARE.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;What is this word so entwined with foemen?&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I can't get this bit. If it's a crossword-type clue, &lt;em&gt;so entwined with foemen&lt;/em&gt; could give AS (synonym of SO) and perhaps RE can be got from &lt;em&gt;foemen&lt;/em&gt; somehow, and then we entwine them together.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7363" LastEditorUserId="7363" LastEditDate="2015-08-09T18:08:17.170" LastActivityDate="2015-08-09T18:08:17.170" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19797" PostTypeId="2" ParentId="18773" CreationDate="2015-08-09T17:58:34.300" Score="2" Body="&lt;p&gt;Complete me one way - if you do this to friends, you must have many enemies.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you LOWER your friends you will make lots of enemies&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Complete me another - if you're tough you'll do this to your enemies.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You will make them COWER&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet in a third way - with this you can keep your enemies from you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can defend yourself in a TOWER&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And in a fourth way - with this you can always defeat your enemies.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can defeat your enemies with POWER&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11150" LastActivityDate="2015-08-09T17:58:34.300" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19798" PostTypeId="2" ParentId="18281" CreationDate="2015-08-09T19:14:04.407" Score="4" Body="&lt;p&gt;The Answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 8 x 50 + 1 - 3 = 398&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reason is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As has already been stated, the Nutritional Information matches that of a Pepsi can. PEPSI has five letters.&#xA; The Calories are 150 while the Caffeine Content is 38. I say these five numbers correlate with the five letters in PEPSI.&#xA; I know there are four slots used in the equation to get 398, so I think that (as the hint says) solving the riddles will give me the answer. I therefore think that by corresponding each letter with one number, the answers the riddles give me will provide the answer:&#xA; P=1 E=5 P=0 S=3 I=8&#xA; The first paragraphs talks about a Swiss cheese afficionado, and seeing as it's to do with Euler I'm going to say it was talking about Euler's number i.e. E.&#xA; The second paragraph talks about a self-interested 'figure' who appears in idioms (easy as pie), so I think it's Pi. The first letter of which being P.&#xA; The third paragraph references something which moves like a river. I would see this as something forever continuing, therefore Prime, i.e. the letter P. To further support this answer the riddle in this paragraph states it is gotten rid of &quot;several times a day&quot;. If you were to look at Prime in relation to the Seven Canonical Hours it is the first hour of prayer in the day and comes to be at around 6AM. Seeing as there are two 6 o' clocks in the day I would say Prime fits the answer for this paragraph.&#xA; The fourth paragraph is a twin (further supporting the idea that Paragraph 3 is Prime i.e. twin primes) and &quot;impossible&quot;. Therefore the letter: I.&#xA; The fifth paragraph talks about a figure who 'hisses'. I'd call this Sigma based on theories of the Greek word beign based on a nominalization of a verb σίζω (sízō, from earlier *sig-jō, meaning 'I hiss').&#xA; Putting the letters in order they spell: EPPIS&#xA; Imagine the string is repeated immediately after itself: EEPISEEPIS&#xA; I think that by taking the first letter and jumping forward to spaces, and then again, and again, and again, you get the desired order: EPIPS.&#xA; With this order, substitute the numerical values given to each letter: 85013.&#xA; Substitute into the equation and you get 398.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thoughts?&lt;/p&gt;&#xA;" OwnerUserId="15490" LastEditorUserId="15490" LastEditDate="2015-08-09T21:08:40.510" LastActivityDate="2015-08-09T21:08:40.510" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19799" PostTypeId="2" ParentId="18690" CreationDate="2015-08-09T19:51:12.217" Score="5" Body="&lt;p&gt;Final answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; IMMORTAL.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The third and final hint tells us that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; each pair of lines gives a single letter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So let's get going!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I'm ambiguous by nature as I grow tall after all&lt;br&gt;&#xA;Your puny little earth shall tremble at my call&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I'm ambiguous by nature&quot; because &quot;I&quot; can refer to either the solution of the what-am-I riddle or the letter I, which here are one and the same. &quot;I grow tall after all&quot; ... well, I is certainly a tall letter! The second line sounds very egomaniacal; the speaker sounds like a me-me-me type who would often use the word &lt;strong&gt;I&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From OP:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Also &quot;ambiguous&quot; because a lowercase &quot;L&quot; looks like an uppercase (grown-up) &quot;i&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In small as in big, the extremes I will lead&lt;br&gt;&#xA;My address is surely some superstition's seed&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(thanks to Al.Sal)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter &lt;strong&gt;M&lt;/strong&gt; starts (&quot;leads&quot;) the words min and max, which are two &quot;extremes&quot; and synonyms of &quot;small&quot; and &quot;big&quot;. Its &quot;address&quot; in the alphabet is 13, which is the seed of many superstitions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Abnormal my birth in a land of illusion so mortal&lt;br&gt;&#xA;You touch the near ground of the specular portal&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(that Avigrail does like his mirrors!)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; By &quot;touching the ground&quot;, a person would form a shape something like n. Doing it &quot;near&quot; a mirror (&quot;specular portal&quot;), we'd find this n reflected to become &lt;strong&gt;M&lt;/strong&gt;. The first line also hints towards mirrors with &quot;land of illusion&quot;, and the &quot;birth&quot; of this letter M was certainly &quot;abnormal&quot;!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;One long gone monster, not touchable for man&lt;br&gt;&#xA;Nary a sinner may tame me as merely nothing can&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We need a letter that appears in MONSTER and NOTHING but not in MAN or SINNER. The only possibilities are O and T. Looking more closely, we notice that O appears in every word of the first line except &quot;man&quot; (hence it's &quot;not touchable&quot; for man) but not in any word of the second line except &quot;nothing&quot; (hence &quot;merely nothing can tame me&quot;). So the answer here is &lt;strong&gt;O&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A creature of prey hunting for your deceased&lt;br&gt;&#xA;I shall turn it soft, into a sustaining feast&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first line sounds like some sort of sinister BEAST. Adding the letter &lt;strong&gt;R&lt;/strong&gt; &quot;shall turn it&quot; into BREAST, which is soft (hey, don't blame me - it was Avigrail's dirty mind, not mine!) and is a sustaining feast for young creatures.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Imprisoned by tradition, tortured to the graves&lt;br&gt;&#xA;Royal is my essence, my innards rule the waves&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I suspect this may be a dig at the Brits (the OP is German) and their drinking of tea or &lt;strong&gt;T&lt;/strong&gt;! Stereotypically, Brits are &quot;imprisoned by tradition&quot;; they have a &quot;royal&quot; family; and Britannia traditionally &quot;rules the waves&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From OP:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Tortured to the graves&quot; as tea is dried out in the sun, cut into pieces and served with boiling water which could, if applied to a human, be seen as torture.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;As the elders have foretold, the end is near&lt;br&gt;&#xA;And with one difference you can only find hear&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got the answer to this but not the correct explanation. From OP:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;...the &lt;em&gt;end&lt;/em&gt; is near &lt;em&gt;and&lt;/em&gt;&quot; if you just read on to the next line. &quot;end&quot; and &quot;and&quot; are very similar with only one difference we can also find in &quot;he&lt;strong&gt;a&lt;/strong&gt;r&quot; (but not in &lt;em&gt;here&lt;/em&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The keeper of time will choose 2 over infinity&lt;br&gt;&#xA;You and me and the devil form a handsome trinity&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very clever!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Infinity is a sideways 8 (as featured in another riddle recently), so let's say &quot;2 over infinity&quot; becomes $2/8$, which is a quarter. On &quot;the keeper of the time&quot; or a clock, this suggests a quarter of an hour, or 15 minutes. Something like a quarter past midnight or 15:00 hours gives us the letter &lt;strong&gt;L&lt;/strong&gt;. As for the second line, &quot;you and me&quot; are the hands forming L while &quot;the devil&quot; must be the second hand (?), the three together forming a &quot;&lt;em&gt;hand&lt;/em&gt;some trinity&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-08-10T19:13:25.057" LastActivityDate="2015-08-10T19:13:25.057" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="19800" PostTypeId="1" AcceptedAnswerId="19803" CreationDate="2015-08-09T19:55:36.180" Score="10" ViewCount="1954" Body="&lt;p&gt;The accepted answer will be the first to include the correct quote, the author, and explain each part of the rebus. Most importantly, have fun!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KvEUL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KvEUL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GKGLm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GKGLm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZgsZ7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZgsZ7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5VPma.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5VPma.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-08-09T21:29:17.460" LastActivityDate="2015-08-09T21:36:28.240" Title="What quote does this rebus represent?" Tags="|rebus|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19801" PostTypeId="2" ParentId="19800" CreationDate="2015-08-09T20:03:09.527" Score="2" Body="&lt;p&gt;First line:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (BUCKWHEAT&lt;sup&gt;3&lt;/sup&gt; - BUCKET) + (THIGH - HIGH) = WHAT&lt;sup&gt;2&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second line:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (TORCH - ORC) + (CARET - CART) = THE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Third line:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (WOMAN - MAN) + (PEARL - PEA)&lt;sup&gt;1&lt;/sup&gt; + (DRAIN - RAIN) = WORLD&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fourth line:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (MINER - MIR&lt;sup&gt;2&lt;/sup&gt; &lt;sup&gt;3&lt;/sup&gt;) +(ECHIDNAS - CHINA) = NEEDS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fifth line:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (PENNIES - PENNE) = IS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sixth line:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (MOWING - WING) + (FREIGHT - FIGHT) = MORE&lt;sup&gt;1&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Seventh line:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (HALOGEN - HALO) + (HELIUM - HELM)&lt;sup&gt;2&lt;/sup&gt; + (SWITCHES - WITCH) = GENIUSES&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Eighth line:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (WEIGHTS - EIGHTS) + (FILTHY - FLY) = WITH&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ninth line:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (SHOULDER - SOLDER) + (MILDEW - DEW) + (BANDIT - BAND) + (SMILEYS - SMILES) = HUMILITY&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tenth line:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (MATCHES - MACES) + (BREAD - BRAD&lt;sup&gt;2&lt;/sup&gt; &lt;sup&gt;3&lt;/sup&gt;) + (GRAPES - GAPS) = THERE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Eleventh line:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (MINARETS - MINTS) = ARE; (BISON - BIN)&lt;sup&gt;2&lt;/sup&gt; = SO&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Twelfth line:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (FRAME - RAM) + (WAX - AX) = FEW&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thirteenth line:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (ORANGE - RANGE) + (FACE - ACE) = OF&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fourteenth line:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (BOULDERS - BOLDER) = US&lt;sup&gt;2&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fifteenth and last line:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (BUGLE - BUG) + (FOREST - ORES) = LEFT&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Full answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; WHAT THE WORLD NEEDS IS MORE GENIUSES WITH HUMILITY; THERE ARE SO FEW OF US LEFT&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is a quote from&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oscar Levant&lt;sup&gt;2&lt;/sup&gt; &lt;sup&gt;3&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; Thanks to Eli!&lt;br&gt;&#xA;&lt;sup&gt;2&lt;/sup&gt; I'd seen the quote in Josh's answer (but not his explanations) when I solved this clue.&lt;br&gt;&#xA;&lt;sup&gt;3&lt;/sup&gt; Had to Google this one! TIL a new word :-)&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-08-09T21:36:28.240" LastActivityDate="2015-08-09T21:36:28.240" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="19802" PostTypeId="1" CreationDate="2015-08-09T20:53:19.880" Score="-2" ViewCount="171" Body="&lt;p&gt;Imagine you are an ant walking on a circle. You start your journey from the blue point. Your progress is given in percentages - 100% being a full round trip. In the figure, you've made a progress of 40% and reached the orange spot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yFL7D.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yFL7D.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If you're told the percent-progress of your journey (such as 40%), how will you calculate your current coordinates?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know these:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Radius = 50&lt;/li&gt;&#xA;&lt;li&gt;Starting coordinates = (50, 0)&lt;/li&gt;&#xA;&lt;li&gt;Y axis is inverted - positive values are on the bottom (as in figure)&lt;/li&gt;&#xA;&lt;li&gt;Circle border has zero thickness&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4838" LastActivityDate="2015-09-22T17:22:31.830" Title="Walking around in circles" Tags="|mathematics|geometry|" AnswerCount="2" CommentCount="4" ClosedDate="2015-08-09T23:45:31.247" ContentLicense="CC BY-SA 3.0" />
  <row Id="19803" PostTypeId="2" ParentId="19800" CreationDate="2015-08-09T20:56:10.720" Score="9" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;What the world needs is more geniuses with humility, there are so few of us left.&quot; - Oscar Levant&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (Buckwheat - bucket) + (thigh - high) = &lt;em&gt;Wha&lt;/em&gt; + &lt;em&gt;t&lt;/em&gt; = &lt;strong&gt;What&lt;/strong&gt;&lt;p&gt;&#xA; (torch - orc) + (caret - cart) = &lt;em&gt;th&lt;/em&gt; + &lt;em&gt;e&lt;/em&gt; = &lt;strong&gt;the&lt;/strong&gt;&lt;p&gt;&#xA; (woman - man) + (pearl - pea) + (drain - rain) = &lt;em&gt;wo&lt;/em&gt; + &lt;em&gt;rl&lt;/em&gt; + &lt;em&gt;d&lt;/em&gt; = &lt;strong&gt;world&lt;/strong&gt;&lt;p&gt;&#xA;(miner - mir) + (echidnas - china) = &lt;em&gt;ne&lt;/em&gt; + &lt;em&gt;eds&lt;/em&gt; = &lt;strong&gt;needs&lt;/strong&gt;&lt;p&gt;&#xA;(pennies - penne) = &lt;strong&gt;is&lt;/strong&gt;&lt;p&gt;&#xA;(mowing - wing) + (freight - fight) = &lt;em&gt;mo&lt;/em&gt; + &lt;em&gt;re&lt;/em&gt; = &lt;strong&gt;more&lt;/strong&gt;&lt;p&gt;&#xA;(halogens - halos) + (helium - helm) + (switches - witch) = &lt;em&gt;gen&lt;/em&gt; + &lt;em&gt;iu&lt;/em&gt; + &lt;em&gt;ses&lt;/em&gt; = &lt;strong&gt;geniuses&lt;/strong&gt;&lt;p&gt;&#xA;(weights - eights) + (filthy - fly) = &lt;em&gt;w&lt;/em&gt; + &lt;em&gt;ith&lt;/em&gt; = &lt;strong&gt;with&lt;/strong&gt;&lt;p&gt;&#xA;(shoulder - solder) + (mildew - dew) + (bandit - band) + (smileys + smiles) = &lt;em&gt;hu&lt;/em&gt; + &lt;em&gt;mil&lt;/em&gt; + &lt;em&gt;it&lt;/em&gt; + &lt;em&gt;y&lt;/em&gt; = &lt;strong&gt;humility&lt;/strong&gt;&lt;p&gt;&#xA;(matches - mace) + (bread - brad) + (grapes - gaps) = &lt;em&gt;th&lt;/em&gt; + &lt;em&gt;e&lt;/em&gt; + &lt;em&gt;re&lt;/em&gt; = &lt;strong&gt;there&lt;/strong&gt;&lt;p&gt;&#xA;(minarets - mints) = &lt;strong&gt;are&lt;/strong&gt;&lt;p&gt;&#xA;(bison - bin) = &lt;strong&gt;so&lt;/strong&gt;&lt;p&gt;&#xA;(frame - ram) + (wax - ax) = &lt;em&gt;fe&lt;/em&gt; + &lt;em&gt;w&lt;/em&gt; = &lt;strong&gt;few&lt;/strong&gt;&lt;p&gt;&#xA;(orange - range) + (face - ace) = &lt;em&gt;o&lt;/em&gt; + &lt;em&gt;f&lt;/em&gt; = &lt;strong&gt;of&lt;/strong&gt;&lt;p&gt;&#xA;(boulders - bolder) = &lt;strong&gt;us&lt;/strong&gt;&lt;p&gt;&#xA;(bugle - bug) + (forest - ores) = &lt;em&gt;le&lt;/em&gt; + &lt;em&gt;ft&lt;/em&gt; = &lt;strong&gt;left&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Great quote!&lt;/p&gt;&#xA;" OwnerUserId="4353" LastEditorUserId="4353" LastEditDate="2015-08-09T21:16:39.620" LastActivityDate="2015-08-09T21:16:39.620" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19804" PostTypeId="2" ParentId="19802" CreationDate="2015-08-09T21:46:46.650" Score="0" Body="&lt;p&gt;The centre of the circle is at $(50,-50)$ and its radius is $50$, so the general point on the circle is $$(50+50\cos\theta,50-50\sin\theta)$$ where $\theta$ is the &lt;strong&gt;polar angle&lt;/strong&gt;, i.e. the angle to our current position measured counter-clockwise from the point $(100,50)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want our 'zero' position, or starting position, to be at the top of the circle and our direction to be clockwise, so put $\phi=\frac{\pi}{2}-\theta$. Now $\phi$ is the angle to our current position measured clockwise from the point $(50,0)$. So $$(x_n,y_n)=(50+50\cos\theta,50-50\sin\theta)=(50+50\sin\phi,50-50\cos\phi)$$ and this point is $\frac{\phi}{2\pi}$ percent of the way around the circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is: if you're $N\%$ of the way round the circle, then your position is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(50+50\sin\frac{2\pi N}{100},50-50\cos\frac{2\pi N}{100})$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(where we take all angles in radians).&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-08-09T22:12:46.560" LastActivityDate="2015-08-09T22:12:46.560" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19805" PostTypeId="2" ParentId="19802" CreationDate="2015-08-09T21:51:53.440" Score="1" Body="&lt;p&gt;$x$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$50\sin\left(\frac{360p}{100}\right) + 50$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$y$: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$50 - 50\cos\left(\frac{360p}{100}\right)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="13494" LastEditDate="2015-09-22T17:22:31.830" LastActivityDate="2015-09-22T17:22:31.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19806" PostTypeId="1" AcceptedAnswerId="20522" CreationDate="2015-08-10T01:26:30.537" Score="13" ViewCount="1812" Body="&lt;p&gt;My pet underwater snail just chewed up this poem I've been working on! I convinced him to spit it out, but I could only salvage one verse (you know &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16348/you-stand-in-a-forest-where-are-you&quot;&gt;how long&lt;/a&gt; I like to make my poems) and the lines are in the wrong order. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you put them in the proper order and find out what's going on in this stanza? It rhymes now, but it might not when correctly reconstructed. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What a certain literary professor for his unbeheld felt,&lt;br&gt; The two of spades, from a new deck dealt,  &lt;br&gt;&#xA;  A din, a doubt... &lt;br&gt; &#xA;  Out!&lt;br&gt; &#xA;  Up, up, by the Ort Sphere's smallest test, &lt;br&gt;&#xA;  But oh, wait, now by one-third less. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Partial answers are welcomed, but for one to be accepted it must explain each line. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINTS:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;GIVEAWAY:&lt;/strong&gt; He, &lt;code&gt;name redacted&lt;/code&gt; (1289-1316), was the first real player. &#xA; &lt;br&gt;&lt;br&gt;&#xA; Line 1: \u0412\u043b\u0430\u0434\u0438\u0301\u043c\u0438\u0440 \u0412\u043b\u0430\u0434\u0438\u0301\u043c\u0438\u0440\u043e\u0432\u0438\u0447 \u041d\u0430\u0431\u043e\u0301\u043a\u043e\u0432 &lt;br&gt;&lt;br&gt;&#xA; Line 1: Unbeheld, prefix-less and in the first person present rendered, succeeds the unbeheld and &quot;and&quot;.&lt;br&gt;&lt;br&gt; &#xA; Line 1: _ _ _ _ ( -&gt; _ _ _ ) [4 letter word sometimes abbreviated as 3] &lt;br&gt;&lt;br&gt;&#xA; Line 2: The suit is irrelevant; it is without &lt;strong&gt;currency&lt;/strong&gt;.&lt;br&gt;&lt;br&gt;&#xA; Line 2: _ _ _ _ _ [5 letter word] &lt;br&gt;&lt;br&gt; &#xA; Line 3: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/17925/mrs-malaprop-reads-a-riddle&quot;&gt;Mrs. Malaprop&lt;/a&gt; would definitely know this one. &lt;br&gt;&lt;br&gt;&#xA; Line 5: What is truth? It is the man who stands before you. &lt;br&gt;&lt;br&gt;&#xA; Line 5: Find the man whose name is like a rabbit. Yes! The man, the one with a mathematical habit. &lt;br&gt;&lt;br&gt;&#xA; Line 5: _ _ [2 digit number] &lt;br&gt;&lt;br&gt;&#xA; Line 6: _ _ [2 digit number] &lt;br&gt;&lt;br&gt;&#xA; You're confused, you say? Well, &lt;strong&gt;that's a great point.&lt;/strong&gt;&lt;br&gt;&lt;br&gt;&#xA; All the lines together (in the proper order) describe something very specific, but each separate line might represent something distinct on its own. &lt;br&gt;&lt;br&gt; &#xA; The &lt;a href=&quot;/questions/tagged/wordplay&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;wordplay&amp;#39;&quot; rel=&quot;tag&quot;&gt;wordplay&lt;/a&gt; tag should provide some direction with the foreign-seeming elements.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13097" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-08-25T15:18:00.563" Title="Tartar sauce, Gary!" Tags="|riddle|enigmatic-puzzle|wordplay|mystery|" AnswerCount="8" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="19807" PostTypeId="1" AcceptedAnswerId="20252" CreationDate="2015-08-10T02:03:33.347" Score="6" ViewCount="1017" Body="&lt;p&gt;Long after the &lt;a href=&quot;https://puzzling.stackexchange.com/q/16474/7307&quot;&gt;insane Mr. $\Delta\Sigma\Upsilon$ incident&lt;/a&gt;, you are resting in the same luxurious mansion. It is, however, most unfortunate (or fortunate, depending on your perspective) that you received a phone call that day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You pick up the phone. &amp;ldquo;Hello?&amp;rdquo; you say into the phone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;ldquo;You. Are you the renowned solver who ended that greek guy's reign of terror?&amp;rdquo;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Greek guy...? Oh, yeah!&lt;/em&gt; Him. &lt;em&gt;Gah, I hated that guy.&lt;/em&gt; &amp;ldquo;Yup! That's me.&amp;rdquo;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;ldquo;We need you, ASAP. We can't say who we are or why, but it &lt;em&gt;is&lt;/em&gt; important. You can meet us at &lt;code&gt;&amp;lt;obscure place, probably an alley&amp;gt;&lt;/code&gt; in twenty minutes. &lt;em&gt;*Click*&lt;/em&gt;.&amp;rdquo;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You sit down, rather shocked. All the glory of your past endeavour faded quickly, and you were able to settle down in to a quiet, peaceful routine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;**** routines. I'm doing this!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;At &lt;code&gt;&amp;lt;obscure place&amp;gt;&lt;/code&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You walk into a clean, modern room. There sits a secretary at a desk to your right. She beckons you come near.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;ldquo;Welcome. I recognize your face. Mr. $\Lambda\Omega\Lambda$ is waiting for inside.&amp;rdquo; She gestures to a door, and resumes typing. &lt;em&gt;Oh sheesh. More Greek. What is this, a frat school?!&lt;/em&gt; You reach and open the door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instantly, a cheery $\tiny{drunk}$ man in a lab coat rushes up to you. &amp;ldquo;You are a chemistry student! Hooray! And you have a friend! Double hooray! And what's even &lt;em&gt;betterrr&lt;/em&gt; is&amp;mdash;hey, watch it! ow ow ow!&amp;mdash;&amp;rdquo; The man is dragged away by another man in a light blue lab coat, who seats the oddity. Then, blue-coat walks to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;ldquo;Sorry about that. We had the happy narrator dude out. He was &lt;em&gt;not&lt;/em&gt; supposed to debrief you. &lt;em&gt;*Ahem*&lt;/em&gt;. I am Mr. $\Lambda\Omega\Lambda$&amp;mdash;call me Bob. I have little time to spare. You are to infiltrate a laboratory. I'm not sure where; this is where you come in. If you can solve this puzzle, you're hired. We'll pay you handsomely. By the way, once you get there, you need a passcode. We don't know what it is...&amp;rdquo;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some consideration, you say, &amp;ldquo;I'm up for this. Where is the puzzle?&amp;rdquo; Bob gestures to a computer. You eagerly take a seat.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(The story is just for flavour.)&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Puzzle 1&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You have many tools to solve this problem&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0EwGE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0EwGE.png&quot; alt=&quot;http://manytools.org/hacker-tools/steganography-encode-text-into-image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: there is a typo. &amp;ldquo;beskeas&amp;rdquo; should be &amp;ldquo;bespeaks&amp;rdquo; I'll fix it when I have time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;lborders&quot; =&gt; &quot;left borders&quot; =&gt; ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The password is not a word in any language, &lt;strong&gt;hint 2b&lt;/strong&gt; and it is the length of half the longest word in this hint.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;At the designated place:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You approach the single, solitary man. He looks to you and whispers tersely &amp;ldquo;Password?&amp;rdquo;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &amp;ldquo;I-y-a-O,&amp;rdquo; you say confidently. &lt;strong&gt;Solver: LeppyR64&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He nods. &amp;ldquo;Come.&amp;rdquo; You follow him into an obscure door behind him.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;Puzzle 2&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&amp;ldquo;Well, you can only be the Dr. Morigan we have been expecting.&amp;rdquo; He nods to the table. &amp;ldquo;We have the sample here. And, in exchange, you will tell us how, no?&amp;rdquo; You feel a flash of panic. You must act quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;ldquo;Yes, of course. Remind me--what is the problem? 'Fraid I'm a teensy bit drunk, and my memory isn't the best.&amp;rdquo;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The man growls something about stupid, old men. &amp;ldquo;We have a 10-by-10 board. On it will be played the finale of the Fairy-Chess-Piece Championship. We have ascertained from &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18074/checkerboard-infection&quot;&gt;a friend&lt;/a&gt; a particularly nasty square-infectious virus. Our scientists have modified this virus to be able to grow quicker, but at the cost of having spread in a different way. That is, it spreads if and only if it has two diagonally adjacent members. We wish to find what is the smallest possible amount of samples &lt;strong&gt;sufficient to cover the entire board?&lt;/strong&gt;&amp;rdquo;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;ldquo;Oh&amp;hellip;I see. Well, I'll try my best&amp;mdash;&amp;rdquo; He glares at you. &amp;ldquo;&amp;mdash;*ahem* I mean, I will not failyou.&amp;rdquo;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;ldquo;You'd better not&amp;hellip;!&amp;rdquo; With that, you are locked in the room. &lt;em&gt;I am getting just a little bit tired of being locked in rooms&amp;hellip;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As an example, suppose this is where you place the infection:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/2OE7G.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2OE7G.png&quot; alt=&quot;phase 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next day, the board would look like this:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/msIpM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/msIpM.png&quot; alt=&quot;phase 2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If the virus grew at 1 step/day, the intermediate step would look like this:&lt;a href=&quot;https://i.stack.imgur.com/mjx20.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mjx20.png&quot; alt=&quot;phase 1.5&quot;&gt;&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="7307" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2015-08-18T22:32:39.790" Title="Yes, I mean to say that there is more" Tags="|riddle|enigmatic-puzzle|cellular-automata|semi-interactive-puzzle|" AnswerCount="3" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="19808" PostTypeId="2" ParentId="18690" CreationDate="2015-08-10T02:05:19.713" Score="2" Body="&lt;p&gt;&lt;strong&gt;Partial Answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Judging by the hint, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every couplet is a letter. So we need 8 letters: &lt;code&gt;_ _ _ _ _ _ _ _&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the first, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's a vowel, and very few? no? words start with O or U that match the rest of the letters. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Based on the second pair, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Superstition's seed (13) is its address. M? Also, M leads min and max. Which implies there's gotta be a vowel up front. [Confirmed]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Building off of others, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; From Rand, we get the 4th letter, O. And the letters in Hear that aren't H.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At the end, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Plurals. The you, me, and the devil is 3. Two to infinity, etc. It's an s. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the partial answer is...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;[AEI] M _ O _ _ [AER] S&lt;/code&gt; with the caveat that either 1 is A or 7 is. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My hilariously, dumb, awful guess: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ampoules&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14428" LastEditorUserId="14428" LastEditDate="2015-08-10T08:59:11.410" LastActivityDate="2015-08-10T08:59:11.410" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19809" PostTypeId="1" CreationDate="2015-08-10T06:35:33.337" Score="13" ViewCount="2483" Body="&lt;p&gt;&lt;strong&gt;Each resident of Dot-town carries a red or blue dot on his (or her) forehead, but if he ever figures out what color it is he kills himself. Each day the residents gather; one day a stranger comes and tells them something — anything — non-trivial about the number of blue dots. Prove that eventually every resident kills himself, no matter what the stranger said.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comment: “Non-trivial” means here that there is some number of blue dots for which the statement would not have been true. Thus we have a frighteningly general version of classical problems involving knowledge about knowledge.&lt;/p&gt;&#xA;" OwnerUserId="2375" LastEditorUserId="10615" LastEditDate="2016-06-03T21:24:22.723" LastActivityDate="2016-06-04T03:07:40.787" Title="The Dot-town suicide" Tags="|logical-deduction|meta-knowledge|blue-eyes|" AnswerCount="8" CommentCount="20" ContentLicense="CC BY-SA 3.0" />
  <row Id="19810" PostTypeId="5" CreationDate="2015-08-10T07:30:38.067" Score="0" Body="&lt;p&gt;A puzzle involving cells in a 2D or 3D grid, with each cell in one of a few finite states. An initial state (time &lt;em&gt;t&lt;/em&gt; = 0) is selected by assigning a state for each cell. A new generation is created (advancing &lt;em&gt;t&lt;/em&gt; by 1), according to some fixed rule (generally, a mathematical function) that determines the new state of each cell in terms of the current state of the cell and the states of the cells in its neighborhood.&lt;/p&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="60644" LastEditDate="2020-08-30T12:07:53.473" LastActivityDate="2020-08-30T12:07:53.473" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19811" PostTypeId="4" CreationDate="2015-08-10T07:30:38.067" Score="0" Body="Puzzles involving cellular automatons in a 2D or 3D grid." OwnerUserId="12700" LastEditorUserId="60644" LastEditDate="2020-08-30T12:07:53.473" LastActivityDate="2020-08-30T12:07:53.473" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19812" PostTypeId="1" AcceptedAnswerId="19813" CreationDate="2015-08-10T07:46:40.300" Score="9" ViewCount="1846" Body="&lt;p&gt;I have a word. In the middle of my word are the letters 'ergro'. There are three letters at the start and the end of my word. The three letters at the start of my word are the same as the three letters at the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is my word?&lt;/p&gt;&#xA;" OwnerDisplayName="user9377" LastActivityDate="2015-08-12T07:08:43.237" Title="What is my word?" Tags="|word|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19813" PostTypeId="2" ParentId="19812" CreationDate="2015-08-10T08:16:38.940" Score="30" Body="&lt;p&gt;I think its &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Und&lt;/strong&gt;ergro&lt;strong&gt;und&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So far this is the only &lt;em&gt;English&lt;/em&gt; word I could find that worked with this.&lt;/p&gt;&#xA;" OwnerUserId="4786" LastEditorUserId="4786" LastEditDate="2015-08-10T08:24:09.793" LastActivityDate="2015-08-10T08:24:09.793" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19814" PostTypeId="2" ParentId="19812" CreationDate="2015-08-10T11:04:00.993" Score="29" Body="&lt;p&gt;Litscape claims there is another:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;sup&lt;/strong&gt;ergro&lt;strong&gt;ups&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It appears to be a music thing.&lt;/p&gt;&#xA;" OwnerUserId="9000" LastActivityDate="2015-08-10T11:04:00.993" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="19815" PostTypeId="2" ParentId="18347" CreationDate="2015-08-10T11:39:24.113" Score="4" Body="&lt;p&gt;So, to gather it all together in one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stanza 1 (From Zandar's answer)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Taking &quot;Two snakes slither&quot; gives us the beginning and end, &quot;a line about what's lost between triumph and moving air&quot; gives us row, from &quot;a line about...&quot;, and the letter d, since it is the difference between &quot;win&quot; and &quot;wind.&quot; Putting this all together, we get &quot;swords&quot;. The line &quot;a man of prowess hides them both&quot; reinforces this, as a samurai is an example of a man of prowess, with &quot;hiding them both&quot; referring to their swords.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Stanza 2 (From Patrick's answer)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first few lines are hints relating to the word, however the crossword bit comes in starting at &quot;A sudden motion shocks you in a thousand different ways.&quot; A sudden motion can be referred to as a &quot;tic&quot;, while one thousand is often represented with the letter &quot;k&quot;, giving us &quot;tick&quot;. Additionally, the other lines of the stanza support this. (&quot;A penciled confirmation&quot; being a checkmark, also known as a tick, &quot;the dance of clocks&quot;, since clocks tick, etc.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Stanza 3 (From Patrick's and dennisdeems' answers)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As Dennis found, &quot;The dispossessed&quot; refers to the story Les Mis. Adding the letter &quot;t&quot; to this gives us &quot;mist&quot;, or shrouded skies. The phrase &quot;perform on it this action as you do with dotted eyes&quot; refers to the phrase &quot;cross your t's&quot;, giving us the word &quot;cross&quot;. &quot;The aforementioned snakes return, though this time stick together&quot; supports this by saying that there will be 2 consecutive s's in the word. A crow, from &quot;bird of death... blackened&quot;, approaches the snakes, meaning this comes before the 2 s's. Removing the w from crow, hinted at by the &quot;shedding feathers&quot;, we again get &quot;cross&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Stanza 4 (From dennisdeems's answer)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, the ending lines are where the crossword bit comes in. &quot;A waterfall appears from fears no longer held within&quot;, gives us &quot;cry&quot;, while &quot;paper tiger cardinals&quot; gives us &quot;pt&quot; from the first letters of &quot;paper tiger&quot;. Putting them together, we get &quot;crypt&quot;. The previous lines support this with &quot;bones are ever rattling&quot;, something that one would expect from a crypt, &quot;unseen to the untrained eye&quot; could refer to the fact that many crypts, like the tombs of pharaohs mentioned later in the answer, are difficult to find unless one knows where to look.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And finally, stanza 5,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;In pairs&quot; tells us that there are 2 strangers and 2 giants, fitting with the 4 stanzas present in the puzzle. The giants are the words that come from the higher-numbered stanzas, followed by the ones from lower-numbered ones. This gives us &quot;crypt, tick, cross, swords&quot;. To &quot;manipulate them slightly&quot;, we must simply remove a few letters, being the &quot;t&quot; and &quot;k&quot; from tick and the &quot;s&quot; from the beginning of swords.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which, all together, gives us &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CRYPTIC CROSSWORDS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Indeed, the secret to eternal happiness. Huge props to everyone mentioned above for getting all the actual puzzle answers.&lt;/p&gt;&#xA;" OwnerUserId="10078" LastEditorUserId="10078" LastEditDate="2015-08-18T16:20:02.623" LastActivityDate="2015-08-18T16:20:02.623" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19816" PostTypeId="1" AcceptedAnswerId="19846" CreationDate="2015-08-10T12:00:29.313" Score="31" ViewCount="5133" Body="&lt;p&gt;A maths teacher writes a very large number on the blackboard and asks her pupils (of whom there are $n$ in the room) about its factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first pupil says, &quot;The number is divisible by 2.&quot;&lt;br&gt;&#xA;The second says, &quot;The number is divisible by 3.&quot;&lt;br&gt;&#xA;The third says, &quot;The number is divisible by 4.&quot;&lt;br&gt;&#xA;The fourth pupil says, &quot;The number is divisible by 5.&quot;&lt;br&gt;&#xA;$...$&lt;br&gt;&#xA;The $n$th pupil says, &quot;The number is divisible by ($n+1$).&quot;&lt;br&gt;&#xA;The teacher says, &quot;You were all right except for two of you, who spoke consecutively.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this information, what can you say about:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the value of $n$&lt;/li&gt;&#xA;&lt;li&gt;which two pupils were wrong?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you want to list all possibilities, then we can limit $n$ to be less than $100$ to make the problem finite. However, there is a general answer for which values are possible, which works for arbitrarily large $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Don't worry about what the number on the blackboard is!&lt;/em&gt; You could find its smallest possible value in each case using the Chinese Remainder Theorem, but that would be boring and tedious.&#xA;&lt;hr&gt;&#xA;I'll upvote any answer which is correct and relies only on pencil, paper, and logic without resorting to computer power. The green tick will go to whichever answer gives the correct solution in the most simple and elegant way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;NB: this is a maths puzzle and not a maths problem. There's a nice 'aha!' which narrows down the possibilities considerably, and the nature of the final solution is quite surprising.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2018-04-02T19:32:43.270" Title="Teacher, teacher on the wall, Who's the dumbest of them all?" Tags="|mathematics|number-theory|" AnswerCount="13" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="19817" PostTypeId="2" ParentId="19807" CreationDate="2015-08-10T12:00:46.283" Score="4" Body="&lt;p&gt;Since this is a semi-interactive puzzle, here is where to go for the first step...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Go to the corner at M and 5th street.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the alt text of the image &lt;a href=&quot;http://manytools.org/hacker-tools/steganography-encode-text-into-image&quot; rel=&quot;nofollow&quot;&gt;http://manytools.org/hacker-tools/steganography-encode-text-into-image&lt;/a&gt; to decode the image yields these instructions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The password is IyaO.  These are the leftmost letters (or borders) of each line.&lt;/p&gt;&#xA;" OwnerUserId="1777" LastEditorUserId="1777" LastEditDate="2015-08-18T20:11:09.790" LastActivityDate="2015-08-18T20:11:09.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19818" PostTypeId="2" ParentId="19816" CreationDate="2015-08-10T12:25:11.597" Score="1" Body="&lt;p&gt;Because the two pupils are consecutive one of those is divisible by 2. Let's call that pupil $x$. This means that $x/2$ also can't be a divisor because if $x/2$ is a divisor and $2$ is a divisor then $x$ is also a divisor, contradiction. This either means that the consecutive number are $x$ and $x/2$ or that the number is not divisible by $2$. Since the next number can never be double the previous number it must automatically mean that the number is not divisible by $2$ and, because it needs to be consecutive with another, also not divisible by $3$. This also means that the only valid $n$ is when $n=2$ because a number not divisible by $2$ can't be divisible by $4$ also giving more incorrect statements&lt;/p&gt;&#xA;" OwnerUserId="1748" LastEditorUserId="5373" LastEditDate="2015-08-10T15:31:36.500" LastActivityDate="2015-08-10T15:31:36.500" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="19819" PostTypeId="1" AcceptedAnswerId="19825" CreationDate="2015-08-10T12:49:15.793" Score="3" ViewCount="809" Body="&lt;p&gt;My first question on Puzzling.SE, I don't think it should be too hard.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Take a bowl and add one.&lt;br&gt;&#xA;  Now add seven.&lt;br&gt;&#xA;  At last mix it all together and bake it at 270 degrees.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What will you get?&lt;/p&gt;&#xA;" OwnerUserId="11899" LastEditorUserId="12700" LastEditDate="2015-08-10T13:14:58.527" LastActivityDate="2015-08-10T13:40:16.220" Title="A little baking recipe" Tags="|riddle|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19820" PostTypeId="2" ParentId="19816" CreationDate="2015-08-10T13:12:32.180" Score="1" Body="&lt;p&gt;I am not quite sure if I am correct, but lets see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is: It has sth. to do with the primes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started with the number x, lets say &lt;code&gt;x = 5&lt;/code&gt; (it doesn't really matter, if its small enough). Then I iterate through the pupils and check if x can be divided by the number said. If not I mutliply it with the number.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#1 pupil has number = 2&#xA;  can't be divided so x = 10  (x*2)&#xA;#2 pupil has number = 3&#xA; can't be divided so x = 30&#xA;#3 pupil has number = 4&#xA; can't be divided so x = 120&#xA;#4 pupil has number = 5&#xA; can be divided, so x stays at 120&#xA;.&#xA;.&#xA;.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;With this scheme every prime would not be a divisor of &lt;code&gt;x&lt;/code&gt;. So my idea was to find a pupil number which is not a prime an still no divisor AND is directly before a prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number &lt;code&gt;16&lt;/code&gt; is one of them, followed by the prime number &lt;code&gt;17&lt;/code&gt;. With this method &lt;code&gt;x =1081080&lt;/code&gt; until pupil #15 and this value for &lt;code&gt;x&lt;/code&gt; can't be divided by &lt;code&gt;16&lt;/code&gt; or &lt;code&gt;17&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the last two pupils are wrong, and the very last said number must be a prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I have found a sequence:&#xA;This happens every $y = 2^z$  for &lt;code&gt;z = the previous y&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;z=2;   y=4&#xA;z=4;   y=16&#xA;z=16;  y=256&#xA;z=256; y=65536&#xA;.&#xA;.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;All these &lt;code&gt;y&lt;/code&gt; values are followed by a prime!&lt;/p&gt;&#xA;" OwnerUserId="14461" LastEditorUserId="14461" LastEditDate="2015-08-10T13:36:24.527" LastActivityDate="2015-08-10T13:36:24.527" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19821" PostTypeId="2" ParentId="19816" CreationDate="2015-08-10T13:15:36.447" Score="0" Body="&lt;p&gt;I'm Back on a computer!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we know the two students who claimed wrong claimed wrong consecutively. We can say that these two students claimed that the numbers $a$ and $b$ were factors of the number written on the board by the teacher, which in reality it wasn't. The first thing to note is that one of these numbers, either $a$ or $b$ must be even, and one of these numbers must be odd. The second thing to note is that both $a$ and $b$ must be factored into only one type of prime, eg they must be a prime number raised to some power. To prove this, we can assume the contrary and say that $a$ can be factored to $ppqqq$ where $p$ and $q$ are primes. $pp$ will be smaller than $a$, but we know that all numbers smaller than $a$ were accepted by the teacher as being a factor of the number on the board. Similarly, $qqq$ will be accepted by the teacher as being a factor of the number on the board. If both $pp$ and $qqq$ are accepted as being a factor, then $a$ must also be a factor, breaking our assumption. Therefore both $a$ and $b$ must be some prime raised to a power. Because one of these is even and one of these is odd, the even claim must be a power of $2$, and the odd claim must be a power of an odd prime number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know immediately some restrictions upon $n$. Firstly, it must be less than $2a$, and less than $2b$, because we know the number on the board is not divisible by either $a$ or $b$. But I believe this is the only restriction upon $n$. &lt;/p&gt;&#xA;" OwnerUserId="15499" LastEditorUserId="15499" LastEditDate="2015-08-11T00:02:02.590" LastActivityDate="2015-08-11T00:02:02.590" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19823" PostTypeId="1" AcceptedAnswerId="19824" CreationDate="2015-08-10T13:36:05.970" Score="134" ViewCount="52654" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Which of the 8 tiles below is missing?&#xA;&lt;img src=&quot;https://i.stack.imgur.com/6Stlh.jpg&quot; alt=&quot;&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I've been trying to figure out the answer to this question for quite some time. Can anyone spot the pattern?&lt;/p&gt;&#xA;&lt;p&gt;Source: &lt;a href=&quot;https://www.bergmandata.com&quot; rel=&quot;noreferrer&quot;&gt;https://www.bergmandata.com&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15500" LastEditorUserId="69582" LastEditDate="2020-11-15T16:18:57.780" LastActivityDate="2020-11-15T16:18:57.780" Title="Mensa IQ test question - lines in squares" Tags="|pattern|visual|progressive-matrix|" AnswerCount="7" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="19824" PostTypeId="2" ParentId="19823" CreationDate="2015-08-10T13:39:45.043" Score="208" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tile 5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because it looks to me like&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reading from left to right, the lines continue where the previous square left off, to make two long connecting 'strings' of sorts. The eighth square leaves off in the top righthand corner and about 80% of the way down the bottom side, which is where tile 5 would connect to.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here is an image of the final solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/83V1G.png&quot; alt=&quot;Mensa Squares&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="11475" LastEditDate="2015-08-10T14:14:35.003" LastActivityDate="2015-08-10T14:14:35.003" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="19825" PostTypeId="2" ParentId="19819" CreationDate="2015-08-10T13:40:16.220" Score="7" Body="&lt;p&gt;I think the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Infinity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because the riddle bakes down to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0 (bowl) + 1 + 7 = 8 which, when rotated 270 degrees, looks like $\infty$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastActivityDate="2015-08-10T13:40:16.220" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19826" PostTypeId="2" ParentId="17247" CreationDate="2015-08-10T13:47:30.830" Score="4" Body="&lt;p&gt;Since this is still outstanding, and most of it has been solved, I'll post a full solution, using most of Chadi Abou Sleiman's and rand al'thor's work.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Scold the bad liar. (4)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Bad&quot; suggests anagrams; an anagram of &quot;liar&quot; is RAIL, which (as a verb) is a synonym of &quot;scold&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A wall of iron has nickel, copper, even lead! (5)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Iron is Fe on the periodic table; &quot;lead&quot; suggests taking initial letters, so &quot;nickel, copper, even&quot; -&gt; NCE. Putting it all together, we get FENCE, which is a synonym of &quot;wall&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So we need to use&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the railfence cipher&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;to decode the next part.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;MJ, zDafl.o!sassor ok Vqi h,bg e yx hos i edi i ot.cTuP swnWpHh hf&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Mr. Jock, TV quiz PhD, bags few lynx. Whoops! His head is his foot&lt;/strong&gt; using a railfence cipher with 3 rails.&lt;br&gt;&#xA;&lt;strong&gt;Mr. Jock, TV quiz PhD, bags few lynx&lt;/strong&gt; is a known &lt;strong&gt;pangram&lt;/strong&gt; so this suggest a substitution cipher.&lt;br&gt;&#xA;&lt;strong&gt;Whoops! His head is his foot&lt;/strong&gt; suggests that the head be equal to the foot. It means by looking at the pangram found, &lt;strong&gt;M = X&lt;/strong&gt;, &lt;strong&gt;R = N&lt;/strong&gt; and so on... to give the following complete substitution key:&lt;br&gt;&lt;br&gt;&#xA; &lt;strong&gt;ABCDEFGHIJKLMNOPQRSTUVWXYZ&lt;br&gt;&#xA; UIWZKTQPBYEOXRLHGNVFASCMJD&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;cloki rlbfvkag kpf zrbt hokp vknkrkqbs pfbc&lt;br&gt;&#xA;clre jzuknou alj pwbpc nkcvru kpf vb jke kpf&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;kbo rbpfbc vfknwkv zknkffupv voubfrkvvk lf rclz zknuh rkpc&lt;br&gt;&#xA;kjk kpf pfbc rupf nuk pfbc jofrknkttbz zksbkwnkh jfnuh u&lt;br&gt;&#xA;zolp oobc jkpf rlbfuxnltrb vpfnalt nbkpf ji rkeuf rkpc&lt;br&gt;&#xA;zolq vhupnkh zkokkfv ki lf kzkhxb lf rkpf rlc ki lf&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We need to use the substitution found above and reverse the whole text to obtain the following:&lt;br&gt;&lt;br&gt;&#xA; &lt;em&gt;to be won then to impede to be steeled perhaps gold&lt;br&gt;&#xA; when taken by their fourths information they will hold&lt;br&gt;&#xA; a party perceived differently with ear than with the eye&lt;br&gt;&#xA; when pared down to essentials shattered secrets within lie&lt;br&gt;&lt;br&gt;&#xA; the key is the answer which you already know&lt;br&gt;&#xA; with vigeneres help find the question below&lt;/em&gt;&lt;br&gt;&lt;br&gt;&#xA; This suggests that we use next a &lt;strong&gt;Vigenere cipher.&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; To answer the riddle, we're looking for words that &lt;strong&gt;look different&lt;/strong&gt; but &lt;strong&gt;sound the same&lt;/strong&gt;, each meaning a different part of the first line. The answer is:&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Medal&lt;/strong&gt; (to be won)&lt;br&gt;&#xA; &lt;strong&gt;Meddle&lt;/strong&gt; (to impede)&lt;br&gt;&#xA; &lt;strong&gt;Mettle&lt;/strong&gt; (to be steeled)&lt;br&gt;&#xA; &lt;strong&gt;Metal&lt;/strong&gt; (perhaps gold)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Taken by fourths (their fourth letters), we get &lt;strong&gt;data&lt;/strong&gt;.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The combination of their letters is &lt;strong&gt;aadddeeeeeellllmmmmttt&lt;/strong&gt;, which when pared down gives us &lt;strong&gt;adelmt&lt;/strong&gt;, an anagram of &lt;strong&gt;malted&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Toh ws L xivx qb yiwdwkmkpl?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;How do I like my milkshakes?&lt;/strong&gt; with Vigenere key &lt;strong&gt;malted&lt;/strong&gt;, obtained from the riddle (and the answer to the question).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-08-10T13:47:30.830" CommentCount="0" CommunityOwnedDate="2015-08-10T13:47:30.830" ContentLicense="CC BY-SA 3.0" />
  <row Id="19827" PostTypeId="2" ParentId="19816" CreationDate="2015-08-10T13:55:11.127" Score="6" Body="&lt;p&gt;One of the two pupils that are wrong &lt;em&gt;could&lt;/em&gt; be a prime. The other one cannot be prime unless $n&amp;lt;3$, it has to be an even number. So it has to have more occurrences of at least one prime factor than every number smaller than it. This requires it to be the highest number of the form $2^x$ that you have in the sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, consider the students saying: &lt;code&gt;2, 3, 4, 5, 6, 7&lt;/code&gt;. In this case, it has to be &lt;code&gt;4,5&lt;/code&gt; that are wrong, every other combo makes other students wrong as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also means there are $n$s that are impossible - like 15 students (15,16 -&gt; 15 is not prime). 16 students works again, 17 being prime. Likewise, between 63 and 127 there's a huge gap, because 64 has no adjacent prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, some instances of $n$ have two possibilities.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2,3,4,5,6,7,8,9&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here it could be 7,8 that are wrong, or 8,9. Why 8,9, there's no prime in it? Because it has two numbers with more occurrences at least one prime factor than every number smaller than it. ($3\times 3$ and $2\times 2\times 2$). &lt;strike&gt;Such edge cases should be really few.&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: &lt;strong&gt;To generalize, you need any two adjacent numbers that only have exactly one prime factor, and have more occurrences of it than any other number in the list. This means that you can limit your search to the upper half of the list, but including the middle element, if available.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2544" LastEditorUserId="4421" LastEditDate="2015-08-10T15:18:54.423" LastActivityDate="2015-08-10T15:18:54.423" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19828" PostTypeId="2" ParentId="19816" CreationDate="2015-08-10T14:06:48.453" Score="0" Body="&lt;p&gt;Nice puzzle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum number of pupils in the class is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;14&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can argue that certain pupils must be correct. The argument goes this way: Say 2,3,6 are called. If 6 was wrong, 2 or 3 must also be wrong. But that's impossible since the two wrongs must be consecutive. Hence, 6 must be correct, and also 2 and 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A variation of this argument applies to square numbers: If 2 and 4 are called, we know that 2 must be correct. (but not necessarily 4).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another observation is that once a divisor is proven correct via the argument above, it can never be false again, regardless how many pupils call something later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can play a reverse &lt;a href=&quot;https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes&quot; rel=&quot;nofollow&quot;&gt;Sieve of Eratosthenes&lt;/a&gt; until no two consecutive wrongs are possible:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;n=2: Both wrong :-)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;n=3: 2 must be correct&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;n=4: nothing new. Two solutions exist.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;n=5: 2,3 must be correct, hence 6 also.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;... &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;n=7: 8 is called, so 4 must be correct. 5 also because both neighbors are already correct.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;n=8: Two solutions exist, either (7,8 wrong) (8,9 wrong)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;n=13: 14 is called, which makes 7 correct. 14 is also correct since 2 and 7 are already correct. Only the (8,9 wrong) solution exists.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;n=15: 16 is called, which makes 8 correct. Now we have a contradiction. Let's write the number and highlight those already proven correct:&#xA;&lt;strong&gt;2&lt;/strong&gt;,&lt;strong&gt;3&lt;/strong&gt;,&lt;strong&gt;4&lt;/strong&gt;,&lt;strong&gt;5&lt;/strong&gt;,&lt;strong&gt;6&lt;/strong&gt;,&lt;strong&gt;7&lt;/strong&gt;,&lt;strong&gt;8&lt;/strong&gt;,9,&lt;strong&gt;10&lt;/strong&gt;,11,&lt;strong&gt;12&lt;/strong&gt;,13,&lt;strong&gt;14&lt;/strong&gt;,&lt;strong&gt;15&lt;/strong&gt;,&lt;strong&gt;16&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;From now on, whenever an even number n is added, is must be correct since n/2 is already marked as correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(maybe this could have been formulated a bit shorter, but still, the arguments work)&lt;/p&gt;&#xA;" OwnerUserId="2783" LastActivityDate="2015-08-10T14:06:48.453" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19829" PostTypeId="2" ParentId="18692" CreationDate="2015-08-10T14:12:12.347" Score="4" Body="&lt;p&gt;It seems I did a pretty bad job on this room. After some consideration, I've decided to post the complete solution to this puzzle; while I indicated previously to f&quot; that I would accept his answer if this went unsolved (and kudos to him for getting the correct final answer), I think it would be unwise to deprive current and future solvers the satisfaction of knowing how this puzzle was intended to be solved. Some may disapprove of this approach, but I think I'd rather leave the problem as is and post the information in this post rather than update it with, basically, instructions on how to solve this puzzle. All of that said, here goes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First notice that Key 1 is a shifted Caesar cipher with the alphabet being ASCII instead of the usual one and the shift amount being ROT117, the answer from the previous room. This yields:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;In CONGRESS, July 4th, 1776. We hold ... future security. Frequency, then o  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This points to a passage from the preamble of the U.S. Declaration of Independence. Using this paragraph and analyzing the frequency of each letter (every one of the 26 occurs at least once in this passage), we get the following, sorted in descending order with ties broken by &quot;o&quot;rder of occcurrence (hence the lingering o in Key 1 before the rip; the whole word was order).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;e (134), t (109), n (71), s (70), i (70), r (65), a (65), o (64), h (59), d (38), l (35), u (34), c (26), g (26), f (25), m (24), p (18), b (16), v (15), w (14), y (13), j (2), k (2), q (1), z (1), x (1)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lining these up with the letters in order of occurrence in the passage (also what was meant by 'frequency, then order'), you get the substitution cipher of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;e, t, n, s, i, r, a, o, h, d, l, u, c, g, f, m, p, b, v, w, y, j, k, q, z, x&lt;br&gt;&#xA;w, e, h, o, l, d, t, s, r, u, b, f, v, i, n, a, m, c, q, y, g, p, j, z, k, x&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is used to decipher Key 2 to the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;THIS YEAR THE LION SUCCEEDED THE SIXTHBORN WHO HIMSELF HAD SUPPLANTED THE HEAVENLY  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which, as f&quot; noted, is a reference to when Pope Leo I (the lion) succeeded Pope Sixtus III (the sixthborn, though the actual etymology of this name is believed to be the greek word for 'polished') who was preceded by Pope Celestine I (the heavenly). This took place in the year:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;440  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Giving us our final answer to enter into the keypad. &lt;/p&gt;&#xA;" OwnerUserId="13970" LastActivityDate="2015-08-10T14:12:12.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19830" PostTypeId="2" ParentId="18239" CreationDate="2015-08-10T14:27:19.377" Score="4" Body="&lt;p&gt;These phrases have particular properties relating to how they are typed on a US keyboard.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;OLD FLAGS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;O&lt;/code&gt; is the only letter on the top row. The rest are on the home row.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;FISHY DISH&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;PUERTO RICO&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;C&lt;/code&gt; is the only letter on the bottom row. The rest are on the top row.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;ACQUITTED IDIOT&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;61B2068696E74&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not sure what the property for this one is, but it makes legible ASCII when the &lt;code&gt;B&lt;/code&gt; is removed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;MANLY PINK&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;A&lt;/code&gt; is the only letter typed with the left hand. The rest are typed with the right hand.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;BANANA CABANA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;A&lt;/code&gt; is the only letter on the home row. The rest are on the bottom row.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;STREET SMART&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;M&lt;/code&gt; is the only letter typed with the right hand. The rest are typed with the left hand.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-08-10T14:27:19.377" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19831" PostTypeId="2" ParentId="19816" CreationDate="2015-08-10T14:33:08.503" Score="2" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If at least one of them is a prime, then that prime can't be found among the others' divisors as a factor more than once.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If at least one is a compound (which is true), that compound's divisors can be found separately, but not at once. For that, this number must be a power of a prime.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The compound prime power is a power of 2, and the other is an odd prime (or they're 8 and 9). Both can't be non-trivial prime powers unless they're 8 and 9, because:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If $n, m &amp;gt; 1, p &amp;gt; 2$ and $p^n-1 = 2^m$, $n$ is either odd and so is $(p^n-1)/(p-1)$, making $n=1$ (contradiction), or $n$ is even and both $p$^$(n/2)$-1 and $p$^$(n/2)$+1 are powers of 2, making them 2 and 4 respectively. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $n, m &amp;gt; 1, p &amp;gt; 2$ and $p^n+1 = 2^m$, then $p^n = 2^m-1$, so $m$ is odd (contradiction otherwise), and $p^n - 1 = 2^m-2 = 2* [2$^$(m-1)] = 2* [2$^$((m-1)/2) + 1] * [2$^$((m-1)/2)-1]$. Since $p^n - 1$ shouldn't be divisible by 4, $n$ is also odd. $p+1$ is a power of 2. $p^n+1$/($p+1$) is odd, so it must be 1, which is a contradiction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In short, the wrong number duo can be 4-5, 7-8, 8-9, 16-17 or 31-32 if there're fewer students than 100 and more than 3 (3 &amp;lt; $n$ &amp;lt; 100), but both have to be greater than $n/2$ no matter what $n$ is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, if the prime is the smaller out of the two, the other's power value must be a prime, and if it's bigger, the power must be a power of 2 (respectively Mersenne and Fermat primes), both of which can be proven when the relevant expression is expanded. So the greatest such possible duo can be the wrong divisors, because other candidates would be too small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most obvious answers would be 2-3 (for 2 students), 3-4 (3 to 4 students) and 4-5 (4 to 6 students). In addition, 8-9 would be feasible for 8 to 14 students. The generalized rule above applies to all the other solutions.&lt;/p&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="14016" LastEditDate="2015-08-11T11:33:13.043" LastActivityDate="2015-08-11T11:33:13.043" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19832" PostTypeId="2" ParentId="19816" CreationDate="2015-08-10T14:41:47.377" Score="3" Body="&lt;p&gt;First, for each incorrect pupil, the number he said is a power of a prime. Otherwise its factors were already mentioned and were correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of those two said an even number, so one of the incorrect numbers is a power of 2. It is also the largest power of 2 smaller than n, otherwise next power of 2 would be incorrect as well. So the incorrect number $2^t$ is greater than $\frac n 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other pupil named a number of either $2^t + 1$ or $2^t - 1$ and that number has to be a power of a prime. Probably any power. For example a group of 14 pupils where pupils 7 and 8 are wrong (number is not divisible by 8 and 9). &lt;/p&gt;&#xA;" OwnerUserId="12815" LastEditorUserId="4421" LastEditDate="2015-08-10T15:20:33.487" LastActivityDate="2015-08-10T15:20:33.487" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19833" PostTypeId="2" ParentId="19816" CreationDate="2015-08-10T14:51:46.343" Score="8" Body="&lt;p&gt;The consecutive numbers need both be powers of primes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why? First consider this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If ($a\mid x$) and ($b\mid x$) and ($a$ and $b$ are coprime) then ($ab\mid x$)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if this a well known Lemma but I think it is so I don't need to prove it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now assume one of the consecutive numbers (call it $x$) is not a power of a prime. you can then always write $x=pq$ where $p$ and $q$ are coprime meaning that according to my Lemma that $x$ also divides the number, CONTRADICTION.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing that one of the numbers is divisible by 2 we now know that that number is actually always of the form $2^x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And because of &lt;a href=&quot;https://en.wikipedia.org/wiki/Catalan%27s_conjecture&quot;&gt;Catalan's conjecture&lt;/a&gt; we know that the other number is $3^2$ or a prime number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This answers which two pupils were wrong. What's left to determine is what this means for $n$. Unfortunately I don't know this (yet)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also nice to know that if these consecutive numbers are $x$ and $x+1$ then the number on the board is divisible by $LCM(1,2,...,x-2,x-1)$&lt;/p&gt;&#xA;" OwnerUserId="1748" LastEditorUserId="5373" LastEditDate="2015-08-10T15:34:05.087" LastActivityDate="2015-08-10T15:34:05.087" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19834" PostTypeId="1" AcceptedAnswerId="19835" CreationDate="2015-08-10T14:54:06.343" Score="19" ViewCount="3506" Body="&lt;p&gt;This the final installment of the Maze series. Here is &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18567/room-1-of-the-maze&quot;&gt;Room 1&lt;/a&gt; and here is the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18692/room-5-of-the-maze&quot;&gt;previous room&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After you enter the answer from &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18692/room-5-of-the-maze&quot;&gt;Room 5&lt;/a&gt;, you wait for the door to de materialize as usual, but nothing happens. Your confusion turns to fear instantly as you contemplate the consequences. You prepare to rip off your taste buds, certain that you will be sent to the universe of B.L.L., but before you can, the door in front of you illuminates with a message:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Congratulations on making it through the Maze! You are truly a champion of your species! Given that you are the first of our many test subjects to succeed in the Maze, we feel certain using your results as the maximum potential of your species. Unfortunately, we found that humans rank quite low on the cosmic intelligence scale, falling in the EASILY_COLONIZABLE/POTENTIAL_FOOD_SUPPLY category. Thank you for your participation in our tests, which are taking place in your mind, while your body is still on Earth. You are free to leave the Maze now, simply proceed to the EXIT and you'll awaken back home. We'll be seeing you again quite soon...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As the door vanishes and the message along with it, your vacant march through it is slow and calm, but your heart rate is anything but. Food supply?&#xA;You find yourself walking down a long corridor and you see the EXIT sign on your left about halway down. You reach the door and reach out to grab the doorknob (oh sweet doorknobs, how you had missed them) when you see at the far end of the hall, a door labelled CONTROL ROOM. You head to this new door and find that it opens quite easily. It seems that they never expected you to look past the exit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The control room is empty except for an array of screens and keyboards. You see a big red button in the middle of a suspended table covered &#xA;with a double-layered glass display. Even though you can't understand the language around it, you've seen enough movies to understand its &#xA;approximate function. You break the glass and press the button. A giant screen in the front of the room illuminates with a wall of text. It looks like&#xA;one of the puzzles from the rooms, but much more involved. This must be the type of challenges that this race gives itself, and it must be difficult&#xA;if it's protecting this button. The display is clearly expecting some sort of input; you grab a nearby keyboard and see that it looks nothing like what you're &#xA;used to. What input do you provide to the console and what keys do you press on the keyboard to enter this input?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is going to take everything you have. Knowledge from various disciplines will be incorporated into this multi-faceted, over-the-top puzzle. &#xA;But don't be dismayed, it's certainly possible. Cooperation is STRONGLY encouraged: contribute to whichever part you can to help eachother get through&#xA;and to keep the question active so that others can see it and join in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer from Final Room:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 440  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Key 1: ANSWER IS GIVEN BY THE (Sum of &lt;em&gt;Xi&lt;/em&gt;) OF KEY 2 WITH P(GR(CMH(&lt;em&gt;AfFR&lt;/em&gt;)))&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;WHERE:&lt;br&gt;&#xA;  &lt;em&gt;Xi&lt;/em&gt; is the answer to riddle i, interpreted as a base 36 number (i.e. no spaces,symbols, etc), having length indicated by riddle i  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;CMH(6)= &lt;code&gt;HELIBBBPANJ&lt;/code&gt;&lt;br&gt;&#xA;  CMH(60)= &lt;code&gt;HELIBEBCNEMGPCAZNBBBCDBDGBABBEDPANJGACTMAVANCKYMSWI&lt;/code&gt;&lt;br&gt;&#xA;  CMH(165)= &lt;code&gt;LIBNAPASCSBDBGBFEBNJCTNYKYOR&lt;/code&gt;&lt;br&gt;&#xA;  CMH(199)= &lt;code&gt;&lt;/code&gt; &lt;em&gt;&lt;sub&gt;(nothing)&lt;/sub&gt;&lt;/em&gt;&lt;br&gt;&#xA;  CMH(792)= &lt;code&gt;HELIBECOFNAMGARTICRASKRRUDYHFRAESBBBCDEFGABDGBABFABEDABCPANJGAMASCNHNYNCLAALMOORNVWY&lt;/code&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;GR(&lt;code&gt;ABC&lt;/code&gt;) = &lt;code&gt;A&lt;/code&gt;&lt;br&gt;&#xA;  GR(&lt;code&gt;HELLO EARTH!&lt;/code&gt;) = &lt;code&gt;HELLO E&lt;/code&gt;&lt;br&gt;&#xA;  GR(&lt;code&gt;THIS IS AN EXAMPLE OF WHAT THE GR FUNCTION DOES&lt;/code&gt;) = &lt;code&gt;THIS IS AN EXAMPLE OF WHAT TH&lt;/code&gt;&lt;br&gt;&#xA;  GR(&lt;code&gt;LOREMIPSUMDOLORSITAMETCONSECTETURADIPISCINGELITSEDDOEIUSMODTEMPORINCIDIDUNTUTLABOREETDOLOREMAGNAALIQUA&lt;/code&gt;) = &lt;code&gt;LOREMIPSUMDOLORSITAMETCONSECTETURADIPISCINGELITSEDDOEIUSMODTEMP&lt;/code&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;P(&lt;code&gt;HELLO EARTH!&lt;/code&gt;) = &lt;code&gt;LH ERO!HATEL&lt;/code&gt;&lt;br&gt;&#xA;  P(&lt;code&gt;ABCDEFGHIJKLMNOPQRSTUVWXYZ&lt;/code&gt;) = &lt;code&gt;CAFBINELKHOSURWJGPZTYMXVQD&lt;/code&gt;&lt;br&gt;&#xA;  P(&lt;code&gt;THIS IS AN EXAMPLE OF WHAT THE P FUNCTION DOES&lt;/code&gt;) = &lt;code&gt;ITIHAA E  M FEWNSPTOAXTH  FHLNI EN USSO ETOPDC&lt;/code&gt;&lt;br&gt;&#xA;  P(&lt;code&gt;LOREMIPSUMDOLORSITAMETCONSECTETURADIPISCINGELITSEDDOEIUSMODTEMPORINCIDIDUNTUTLABOREETDOLOREMAGNAALIQUA&lt;/code&gt;) = &lt;code&gt;RLIOUOMODSRAETCMPSSMNLCOTEATIISREINDEIETLDIOECDSUIOGEDUOSMMCTTIIPDIUPTTRBREOLNDNAODOELUETANLUQERAAMIGA&lt;/code&gt;  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Riddles:&lt;br&gt;&#xA;  Subjects chosen are astrology, chemistry, biology, programming, history, philosophy, geography, gaming, computers, alchemy, and misc.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;1 Over thirty years old and still I dig; inflating the beasts until they're too big. (6)&lt;br&gt;&#xA;  2 Don't get lost in the existential mist; in the dead tongue's words, why must you exist? (6)&lt;br&gt;&#xA;  3 We two of the table pool on the floor; our siblings fly about or sit as ore. (7) (7)&lt;br&gt;&#xA;  4 To solvers you offered me in each room; by smarts or by me, one just made it through. (8)&lt;br&gt;&#xA;  5 Touted the year of the death of machines; in truth the year of LOVE from Philippines. (4)&lt;br&gt;&#xA;  6 Switzerland's chainsaw: that's what they call me; you could simply say I come from the C. (4)&lt;br&gt;&#xA;  7 I've had many names since I halved mankind; this one, most common, has been quite maligned. (6)&lt;br&gt;&#xA;  8 Felines north to west, the twins to the east; lurking down below, the many-headed beast. (6)&lt;br&gt;&#xA;  9 Few in the &quot;land up above&quot; even know: the diff'rence 'tween me and hopping big bro. (7)&lt;br&gt;&#xA;  10 Socialist republic, west of Taiwan; fought France for our own, fought u.s. for Saigon. (7)&lt;br&gt;&#xA;  11 Invisible in complex elixirs, I blush at the sight of basic mixtures. My name is long, my formula concise; what I indicate is in my name twice. (8)  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Key 2:&lt;br&gt;&#xA;  NOQNE ACILC TCLAO IGHOZ NALKO JO.DG DWQNE HFMQW QNFAW TWOJV .YXBP NAHVF ALQNG PHJSY NYIIS CVBBA BXCAR WSCPG ODVWF MZYQP FPKLJ KFIYI IQWTG UPFSM DBATR AAZTV XMWEH ITTCO SOZWE FENHC RBZPU WBOQI PQ.CU ZOAOH ZLWIM AQADW SYNND YZLRP QNGGS HCBPJ UUTIA P:HIL AEMWC WJNRN BINWD GMMTA HFHVC NTC,H KQIPJ MUZUF XUJWS NRJWW A,MKM VMUIR SMDCR FLKKK DYRFP SALUY UEKU. VBWNW GH???&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Keyboard:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;AAA AAT ACT ATG ATT&lt;br&gt;&#xA;  CAA CAT CCT CGT GAA&lt;br&gt;&#xA;  GAT GCT GGT GTT TAT&lt;br&gt;&#xA;  TCT TGG TGT TTA TTT  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EPILOGUE&lt;/strong&gt; (for those who have completed the riddle)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;After entering the answer to the final riddle into the peculiar keyboard, the screen goes blank. It returns with a large &lt;strong&gt;29&lt;/strong&gt; on the screen, which then dissapears, followed by a &lt;strong&gt;23&lt;/strong&gt;, then a &lt;strong&gt;19&lt;/strong&gt;. You realize what's happening and don't have to stick around to see where it's going; you dash for the door, sprinting down the corridor for the exit. You twist the knob and force it open, and are blinded by a white light. When your vision clears, you find yourself sitting up in bed. Your pulse is through the roof--from the running, you tell yourself. But what if it was all just some crazy nightmare induced by too much time spent on the Puzzling StackExchange and your fear of Trump? You head to the balcony to get some much-needed air.&lt;br&gt;&#xA;It's almost dawn, and you look up at the sky ponderously. Was any of that even real? How could it be? Maybe I should look into therapy... Some 14-15 minutes through your musings, as you consider going back to sleep, you see a twinkle, a small light, emanating from the pre-dawn sky. It lasts only for a moment, but you could swear you saw it flare up like a tiny nova. You can't help but crack a proud and triumphant smile as you turn to go back to bed.&lt;br&gt;&#xA;Woo! Congratulations everybody on making it through the Maze series. I really hope that everyone enjoyed the puzzles. As always, I love feedback, so feel free to give any as a comment to this post.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2015-08-11T13:59:38.120" Title="End of the Maze" Tags="|riddle|cipher|enigmatic-puzzle|password|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19835" PostTypeId="2" ParentId="19834" CreationDate="2015-08-10T15:03:41.373" Score="18" Body="&lt;p&gt;Since this puzzle has so many facets, I've decided to try a sort of &quot;community answer&quot;, a compilation of solvers' efforts. If you make a chunk of progress, simply edit this post and add it. This way the final result will be organized in a clear manner that is great for progeny's sake. Thanks and have fun!&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The Riddles&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;1 Over thirty years old and still I dig; inflating the beasts until they're too big. (6)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (&lt;em&gt;gaming&lt;/em&gt;)&lt;br&gt;&#xA; &lt;strong&gt;DIGDUG&lt;/strong&gt; (Video game where you blow up beasts by inflating them) - (Jacob Shetler)&lt;br&gt;&#xA;&lt;br&gt;&#xA; $X_1 = $ &lt;strong&gt;DIGDUG&lt;/strong&gt;$_{36} = 817057816_{10}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2 Don't get lost in the existential mist; in the dead tongue's words, why must you exist? (6)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (&lt;em&gt;philosophy&lt;/em&gt;)&lt;br&gt;&#xA; As the French philosopher Descartes said, &lt;strong&gt;I THINK&lt;/strong&gt; therefore I am. (Lampost42), but the edit of the question refers to the Dead Language - Latin, so this now should be &lt;strong&gt;COGITO&lt;/strong&gt; (from Cogito ergo sum) (Gordon K, Jacob Shetler, Lampost42)&lt;br&gt;&#xA;&lt;br&gt;&#xA; $X_2 = $ &lt;strong&gt;COGITO&lt;/strong&gt;$_{36} = 766675788_{10}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3 We two of the table pool on the floor; our siblings fly about or sit as ore. (7) (7)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (&lt;em&gt;chemistry&lt;/em&gt;)&lt;br&gt;&#xA;The two elements of the periodic table that are liquids rather than solids or gasses are &lt;strong&gt;BROMINE&lt;/strong&gt; and &lt;strong&gt;MERCURY&lt;/strong&gt;. (f'')&lt;br&gt;&#xA;&lt;br&gt;&#xA; $X_{3a} = $ &lt;strong&gt;BROMINE&lt;/strong&gt;$_{36} = 25618553834_{10}$&lt;br&gt;&#xA; $X_{3b} = $ &lt;strong&gt;MERCURY&lt;/strong&gt;$_{36} = 48781687246_{10}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4 To solvers you offered me in each room; by smarts or by me, one just made it through. (8)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (&lt;em&gt;misc.&lt;/em&gt;)&lt;br&gt;&#xA; &lt;strong&gt;GOOD LUCK&lt;/strong&gt; - offered by the poser to the solvers in every room and &lt;em&gt;by smarts or good luck&lt;/em&gt; one just made it through (Gordon K)&lt;br&gt;&#xA;&lt;br&gt;&#xA; $X_4 = $ &lt;strong&gt;GOODLUCK&lt;/strong&gt;$_{36} = 1307543443940_{10}$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5 Touted the year of the death of machines; in truth the year of LOVE from Philippines. (4)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (&lt;em&gt;computers&lt;/em&gt;)&lt;br&gt;&#xA; The year &lt;strong&gt;2000&lt;/strong&gt; was widely believed to be the year that machines would cease to function, due to dates being recorded with only 2-digit years.  Also in May, &lt;strong&gt;2000&lt;/strong&gt;, the ILOVEYOU computer virus hit the word hard, originating from the Philippines. (GentlePurpleRain)&lt;br&gt;&#xA;&lt;br&gt;&#xA; $X_5 = 2000_{36} = 93312_{10}$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6 Switzerland's chainsaw: that's what they call me; you could simply say I come from the C. (4)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (&lt;em&gt;programming&lt;/em&gt;)&lt;br&gt;&#xA; &lt;strong&gt;PERL&lt;/strong&gt;. It is called the &quot;the Swiss Army chainsaw of scripting languages&quot; and is based on the language C. (Jacob Shetler)&lt;br&gt;&#xA;&lt;br&gt;&#xA; $X_6 = $ &lt;strong&gt;PERL&lt;/strong&gt;$_{36} = 1185537_{10}$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;7 I've had many names since I halved mankind; this one, most common, has been quite maligned. (6)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (&lt;em&gt;biology&lt;/em&gt;)&lt;br&gt;&#xA; &lt;strong&gt;PLAGUE&lt;/strong&gt; - Bubonic Plague, also known as Black Plague or Black Death wiped out half the world's population. Plague is often used more widely to refer to any pandemic. (Gordon K)&lt;br&gt;&#xA;&lt;br&gt;&#xA; $X_7 = $ &lt;strong&gt;PLAGUE&lt;/strong&gt;$_{36} = 1547414726_{10}$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;8 Felines north to west, the twins to the east; lurking down below, the many-headed beast. (6)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (&lt;em&gt;astrology&lt;/em&gt;)&lt;br&gt;&#xA; The constellation of &lt;strong&gt;CANCER&lt;/strong&gt; has Leo Major and Minor to its northwest, Gemini to the east and Hydra below it. (Gordon K.)&lt;br&gt;&#xA;&lt;br&gt;&#xA; $X_8 = $ &lt;strong&gt;CANCER&lt;/strong&gt;$_{36} = 743479443_{10}$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;9 Few in the &quot;land up above&quot; even know: the diff'rence 'tween me and hopping big bro. (7)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (&lt;em&gt;geography&lt;/em&gt;)&lt;br&gt;&#xA; The &quot;land up above&quot; is the opposite of the &quot;land down under&quot;, Australia. Its most famous wildlife is the kangaroo, but a smaller marsupial often confused for it is the &lt;strong&gt;WALLABY&lt;/strong&gt;. (Lampost42, Gordon K.)&lt;br&gt;&#xA;&lt;br&gt;&#xA; $X_9 = $ &lt;strong&gt;WALLABY&lt;/strong&gt;$_{36} = 70297961614_{10}$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;10 Socialist republic, west of Taiwan; fought France for our own, fought u.s. for Saigon. (7)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (&lt;em&gt;history&lt;/em&gt;)&lt;br&gt;&#xA; &lt;strong&gt;VIETNAM&lt;/strong&gt; (Lampost42)&lt;br&gt;&#xA;&lt;br&gt;&#xA; $X_{10} = $ &lt;strong&gt;VIETNAM&lt;/strong&gt;$_{36} = 68593541422_{10}$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;11 Invisible in complex elixirs, I blush at the sight of basic mixtures. My name is long, my formula concise; what I indicate is in my name twice. (8)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (&lt;em&gt;alchemy&lt;/em&gt;)&lt;br&gt;&#xA; This is &lt;strong&gt;C20H14O4&lt;/strong&gt; or Phenolphthalein. It dissolves clear in alcohol, turns red in basic solutions and has pH in the name twice. (Lamppost42)&lt;br&gt;&#xA;&lt;br&gt;&#xA; $X_{11} = $ &lt;strong&gt;C20H14O4&lt;/strong&gt;$_{36} = 944752140004_{10}$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;The Functions&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;CMH($n$)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The function name is short for &lt;strong&gt;C&lt;/strong&gt;hemistry, &lt;strong&gt;M&lt;/strong&gt;usic, &lt;strong&gt;H&lt;/strong&gt;istory (Gordon K.) (see below).&lt;br&gt;&#xA;&lt;br&gt;&#xA; First, you must determine all the factors of $n$, other than 1 and itself.&lt;br&gt;&#xA; For example, the factors of 6 are 1, 2, 3, and 6, but we ignore 1 and 6 (1 and $n$), and just use 2 and 3.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The output of the function is three parts, concatenated together:&lt;br&gt;&#xA;&lt;br&gt;&#xA; 1. The elements of the periodic table corresponding to the factors of $n$ (e.g. 2 = He, 3 = Li, etc.). Factors greater than 118, the number of elements in the periodic table will be ignored. (Bailey M)&lt;br&gt;&#xA; 2. The notes of the chromatic scale, starting with A=1, Bb=2, B=3, etc., looping with each octave. Factors greater than 88, the number of keys on a piano keyboard, will be ignored. (Gordon K)&lt;br&gt;&#xA; 3. The states of the US, with their numerical order in which they became states corresponding to the factors of n (e.g. 2 = PA, 3 = NJ, etc.). Factors greater than 50, the number of states in the US, will be ignored. (Bailey M)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;GR($n$)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The function name is short for Golden Ratio. As that name suggests, it simply divides the length of a string by the golden ratio (approximately 1.618), rounds down, and then truncates the string to the number of characters indicated by the result. (Zandar) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;P($n$)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This function uses the digits of PI to determine the order of the characters in the encoded string. 3.1415926535... Number the characters in the string, then take the 3rd character and renumber, take the 1st and renumber, take the 4th and renumber...(Gordon K)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;The Keyboard&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There are 20 buttons on the keyboard. Each one is a codon that represents one of the 20 amino acids. The amino acids have one-letter abbreviations:&lt;br&gt;&#xA;&lt;code&gt;AAA- Lysine (K)&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;AAT- Asparagine (N)&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;ACT- Threonine (T)&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;ATG- Methionine (M)&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;ATT- Isoleucine (I)&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;CAA- Glutamine (Q)&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;CAT- Histidine (H)&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;CCT- Proline (P)&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;CGT- Arginine (R)&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;GAA- Glutamic acid (E)&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;GAT- Aspartic acid (D)&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;GCT- Alanine (A)&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;GGT- Glycine (G)&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;GTT- Valine (V)&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;TAT- Tyrosine (Y)&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;TCT- Serine (S)&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;TGG- Tryptophan (W)&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;TGT- Cysteine (C)&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;TTA- Leucine (L)&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;TTT- Phenylalanine (F)&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Putting it all together&lt;/h2&gt;&#xA;&#xA;&lt;h3&gt;Sum of $X_i$&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;If we treat each of the answers to the riddles as a base 36 number, and add them, we get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; \begin{array}{r}  &amp;amp;&amp;amp;&amp;amp;D&amp;amp;I&amp;amp;G&amp;amp;D&amp;amp;U&amp;amp;G_{36}&amp;amp;\\  &amp;amp;&amp;amp;&amp;amp;C&amp;amp;O&amp;amp;G&amp;amp;I&amp;amp;T&amp;amp;O_{36}&amp;amp;\\  &amp;amp;&amp;amp;B&amp;amp;R&amp;amp;O&amp;amp;M&amp;amp;I&amp;amp;N&amp;amp;E_{36}&amp;amp;\\  &amp;amp;&amp;amp;M&amp;amp;E&amp;amp;R&amp;amp;C&amp;amp;U&amp;amp;R&amp;amp;Y_{36}&amp;amp;\\  &amp;amp;G&amp;amp;O&amp;amp;O&amp;amp;D&amp;amp;L&amp;amp;U&amp;amp;C&amp;amp;K_{36}&amp;amp;\\  &amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;2&amp;amp;0&amp;amp;0&amp;amp;0_{36}&amp;amp;\\  &amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;P&amp;amp;E&amp;amp;R&amp;amp;L_{36}&amp;amp;\\  &amp;amp;&amp;amp;&amp;amp;P&amp;amp;L&amp;amp;A&amp;amp;G&amp;amp;U&amp;amp;E_{36}&amp;amp;\\  &amp;amp;&amp;amp;&amp;amp;C&amp;amp;A&amp;amp;N&amp;amp;C&amp;amp;E&amp;amp;R_{36}&amp;amp;\\  &amp;amp;&amp;amp;W&amp;amp;A&amp;amp;L&amp;amp;L&amp;amp;A&amp;amp;B&amp;amp;Y_{36}&amp;amp;\\  &amp;amp;&amp;amp;V&amp;amp;I&amp;amp;E&amp;amp;T&amp;amp;N&amp;amp;A&amp;amp;M_{36}&amp;amp;\\  +&amp;amp;C&amp;amp;2&amp;amp;0&amp;amp;H&amp;amp;1&amp;amp;4&amp;amp;O&amp;amp;4_{36}&amp;amp;\\  \hline  &amp;amp;V&amp;amp;I&amp;amp;G&amp;amp;E&amp;amp;N&amp;amp;E&amp;amp;R&amp;amp;E_{36}&amp;amp;  \end{array}&lt;br&gt;&#xA;&lt;br&gt;&#xA; So we need to use a &lt;strong&gt;&lt;em&gt;Vigenere cipher&lt;/em&gt;&lt;/strong&gt; to decode KEY 2, using the result of P(GR(CMH(&lt;em&gt;AfFR&lt;/em&gt;))) as the key.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;CMH&lt;/h3&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  First, we calculate &lt;strong&gt;CMH&lt;/strong&gt;($440$)  ($440$ is the &lt;em&gt;Answer from Final Room&lt;/em&gt;, indicated as &lt;strong&gt;AfFR&lt;/strong&gt;).&lt;br&gt;&#xA;&lt;br&gt;&#xA; The factors of $440$ are ${2, 4, 5, 8, 10, 11, 20, 22, 40, 44, 55, 88, 110, 220}$.&lt;br&gt;&#xA;&lt;br&gt;&#xA; These correspond to the elements: &lt;code&gt;HE BE B O NE NA CA TI ZR RU CS RA DS&lt;/code&gt; (220 is ignored)&lt;br&gt;&#xA; They correspond to the notes: &lt;code&gt;BB C DB E GB G E GB C E EB C&lt;/code&gt; (110 and 220 are ignored)&lt;br&gt;&#xA; They correspond to the states: &lt;code&gt;PA GA CT SC VA NY MS AL SD WY&lt;/code&gt; (88, 110, and 220 are ignored)&lt;br&gt;&#xA;&lt;br&gt;&#xA; So &lt;strong&gt;CMH&lt;/strong&gt;(440) is &lt;strong&gt;&lt;code&gt;HEBEBONENACATIZRRUCSRADSBBCDBEGBGEGBCEEBCPAGACTSCVANYMSALSDWY&lt;/code&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;GR&lt;/h3&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Next, we calculate &lt;strong&gt;GR&lt;/strong&gt;(&lt;code&gt;HEBEBONENACATIZRRUCSRADSBBCDBEGBGEGBCEEBCPAGACTSCVANYMSALSDWY&lt;/code&gt;)&lt;br&gt;&#xA;&lt;br&gt;&#xA; The string is 61 characters long.  $\frac{61}{1.618}$ is $37.700...$.  We round that down to $37$.  Thus we take the first 37 characters.&lt;br&gt;&#xA;&lt;br&gt;&#xA;  So &lt;strong&gt;GR&lt;/strong&gt;(&lt;code&gt;HEBEBONENACATIZRRUCSRADSBBCDBEGBGEGBCEEBCPAGACTSCVANYMSALSDWY&lt;/code&gt;) is &lt;strong&gt;&lt;code&gt;HEBEBONENACATIZRRUCSRADSBBCDBEGBGEGBC&lt;/code&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;P&lt;/h3&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Lastly, we calculate &lt;strong&gt;P&lt;/strong&gt;(&lt;code&gt;HEBEBONENACATIZRRUCSRADSBBCDBEGBGEGBC&lt;/code&gt;)&lt;br&gt;&#xA;&lt;br&gt;&#xA;  This is a little more complicated.  We start by taking the 3rd character, then remove it, and take subsequent characters according to the digits of $\pi$.&lt;br&gt;&#xA;&lt;br&gt;&#xA;  So &lt;strong&gt;P&lt;/strong&gt;(&lt;code&gt;HEBEBONENACATIZRRUCSRADSBBCDBEGBGEGBCE&lt;/code&gt;) is &lt;strong&gt;&lt;code&gt;BHOENIBACEZCRUDANRBSBTDSACEBRBGEGEGBC&lt;/code&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So &lt;strong&gt;P(GR(CMH(AfFR)))&lt;/strong&gt; is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;code&gt;BHOENIBACEZCRUDANRBSBTDSACEBRBGEGEGBC&lt;/code&gt;&lt;/strong&gt; (this was copied incorrectly the first time)&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;  Using this as the key for a vigenere cipher to encode key 2 two results in:&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;  &lt;code&gt;overr iding secur itypr otoco ls.ex ecuti ngord ersix tysix .prep aring toini tiate remot ewipe ofdat arega rding earth froma llmem oryun itsan ddest ructi onofa lltes tfaci litie sinth ecosm os.to confi rmple aseen terth eansw ertot hefol lowin g:iam thewe akest force inthe unive rse,y etica nmani pulat espac e,slo wtime andev enben dligh ttomy will. whata mi???&lt;/code&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Properly Spaced, it reads-&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;code&gt;Overriding security protocols. Executing order sixty six. Preparing to initiate remote wipe of data regarding earth from all memory units and destruction of all test facilities in the cosmos. To confirm please enter the answer to the following:&lt;/code&gt;&#xA;&lt;br&gt;&#xA; &lt;code&gt;I am the weakest force in the universe, yet I can manipulate space, slow time and even bend light to my will. What am I???&lt;/code&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;  The answer to this is &lt;strong&gt;GRAVITY&lt;/strong&gt;, since it is the weakest of the four fundamental forces. But immense gravitational pull, like that of the sun or a black hole, can bend light and manipulate space-time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We should enter this into the keyboard by pressing the buttons:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;GGT(G)  CGT(R)  GCT(A)  GTT(V)  ATT(I)  ACT(T)  TAT(Y)&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="15544" LastEditDate="2015-08-11T06:16:41.417" LastActivityDate="2015-08-11T06:16:41.417" CommentCount="0" CommunityOwnedDate="2015-08-10T15:06:27.570" ContentLicense="CC BY-SA 3.0" />
  <row Id="19836" PostTypeId="2" ParentId="19834" CreationDate="2015-08-10T15:32:40.640" Score="14" Body="&lt;p&gt;I don't have enough reputation to add a comment to the community wiki, so I'll just post my partial answer(s) here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Riddle #1 is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;DIG DUG&lt;/strong&gt; (1982 Video Game. Player can inflate underground beasts until they explode.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Riddle #2 is (based on OP's edit):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;COGITO&lt;/strong&gt; (According to Descartes: I think, there I am. Cogito is &quot;I think&quot; in Latin, the dead tongue)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Riddle #6 is definitely:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;PERL&lt;/strong&gt;. It is called the &quot;the Swiss Army chainsaw of scripting languages&quot; and is based on the language C.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14030" LastEditorUserId="14030" LastEditDate="2015-08-10T16:16:53.143" LastActivityDate="2015-08-10T16:16:53.143" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19837" PostTypeId="1" AcceptedAnswerId="19841" CreationDate="2015-08-10T16:30:57.667" Score="32" ViewCount="3648" Body="&lt;p&gt;Many people don't put a high value on me,&lt;br&gt;&#xA;even though I have more potential than most.&lt;br&gt;&#xA;I have certain well-defined tasks to complete,&lt;br&gt;&#xA;but if I complete them all, then watch out!&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's true that I'm smaller than others,&lt;br&gt;&#xA;but pass by me on the 5th and you might pay.&lt;br&gt;&#xA;It's true that I'm slower than most others,&lt;br&gt;&#xA;but I've never reverted any of my decisions.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even when I cannot physically move forward,&lt;br&gt;&#xA;I do everything I can to fulfill my duties.&lt;br&gt;&#xA;My siblings and I help each other succeed,&lt;br&gt;&#xA;But few reach enlightenment and are transformed.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What am I?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;The accepted answer will explain all the lines.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="463" LastEditDate="2015-08-11T12:52:27.907" LastActivityDate="2015-08-11T12:57:56.870" Title="What thing am I?" Tags="|riddle|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19838" PostTypeId="2" ParentId="19816" CreationDate="2015-08-10T16:58:59.197" Score="0" Body="&lt;p&gt;&lt;sub&gt;(answers are highlighted with bold. (corresponds to encircling on paper). &quot;/∶&quot; means &quot;not divisible by&quot; (didn't find the proper symbol).)&lt;/sub&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In mathematical terms, the puzzle looks like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The number on the blackboard is divisible by everything in the list &lt;em&gt;2,3,4...n+1&lt;/em&gt; except 2 consecutive numbers (let's call them &lt;em&gt;m,m+1&lt;/em&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What can we say about them?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The number is divisible by 2 but not divisible by &lt;em&gt;m&lt;/em&gt;&lt;br&gt;&#xA;=&gt; &lt;em&gt;m*2 isn't present in the list&lt;/em&gt;&lt;sup&gt;1&lt;/sup&gt;  &lt;/li&gt;&#xA;&lt;li&gt;The number is divisible by anything less than &lt;em&gt;m,m+1&lt;/em&gt; but not divisible by them&lt;br&gt;&#xA;=&gt; &lt;em&gt;m,m+1 can't be factored by the (different) numbers in the list&lt;/em&gt;&lt;sup&gt;2&lt;/sup&gt;&lt;br&gt;&#xA;____specifically: one of the &lt;em&gt;m,m+1&lt;/em&gt; is even&lt;br&gt;&#xA;____=&gt; &lt;em&gt;m/2|(m+1)/2 isn't present in the list&lt;/em&gt; &lt;sub&gt;- unless the other multiplier is 2 as well&lt;/sub&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;=&gt; There are only a few possibilities for &lt;strong&gt;&lt;em&gt;m,m+1&lt;/em&gt;&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;2,3&lt;/strong&gt; =&gt; n+1&amp;lt;4 (&lt;strong&gt;2&lt;/strong&gt; students, both wrong. ha-ha) &lt;em&gt;&lt;strong&gt;N=1&lt;/strong&gt; ROFLMAO&lt;/em&gt;&lt;br&gt;&#xA;_____________________________________all possibilities though are: &lt;strong&gt;N∈N, N/∶2,3&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;3,4&lt;/strong&gt; =&gt; n+1&amp;lt;6 (n&amp;lt;5) | =&gt; &lt;strong&gt;n=3 N=2&lt;/strong&gt; &lt;em&gt;big number indeed&lt;/em&gt;&lt;br&gt;&#xA;_____________________________________all possibilities: &lt;strong&gt;N=2k,k/∶2,3&lt;/strong&gt;&lt;br&gt;&#xA;________________| =&gt; &lt;strong&gt;n=4&lt;/strong&gt; (2,&lt;s&gt;3&lt;/s&gt;,&lt;s&gt;4&lt;/s&gt;,5) =&gt; &lt;strong&gt;N=10&lt;/strong&gt;&lt;br&gt;&#xA;_____________________________________all possibilities: &lt;strong&gt;N=10k,k/∶2,3&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;4,5&lt;/strong&gt; =&gt; n+1&amp;lt;8 =&gt; n can be at most 6 (list entry n+1 =7)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The last case requires study in more detail:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possible values for &lt;strong&gt;&lt;em&gt;n&lt;/em&gt;&lt;/strong&gt;:&lt;br&gt;&#xA;- 4 at least &amp;lt;= to include &quot;4,5&quot;&lt;br&gt;&#xA;=&gt; &lt;strong&gt;n=4&lt;/strong&gt; (2,3,&lt;s&gt;4&lt;/s&gt;,&lt;s&gt;5&lt;/s&gt;) =&gt; &lt;strong&gt;N=6k,k/∶2,5&lt;/strong&gt;&lt;br&gt;&#xA;=&gt; &lt;strong&gt;n=5&lt;/strong&gt; (2,3,&lt;s&gt;4&lt;/s&gt;,&lt;s&gt;5&lt;/s&gt;,6) =&gt; &lt;strong&gt;same as above&lt;/strong&gt;&lt;br&gt;&#xA;=&gt; &lt;strong&gt;n=6&lt;/strong&gt; (2,3,&lt;s&gt;4&lt;/s&gt;,&lt;s&gt;5&lt;/s&gt;,6,7) =&gt; multiplier=LCM(2,3,6,7)=42 =&gt;&lt;br&gt;&#xA;_________________=&gt; &lt;strong&gt;N=42k,k/∶2,5&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;My, my. Big numbers these days...&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt;&lt;sub&gt;(If the number isn't divisible by 2 (m=2), this stands, too: 4 cannot be present)&lt;/sub&gt;&lt;br&gt;&#xA;&lt;sup&gt;2&lt;/sup&gt;&lt;sub&gt;(without loss of generality, by &lt;em&gt;two&lt;/em&gt; different numbers in the list:&#xA; if e.g. &lt;em&gt;(m+1)=k*k*l&lt;/em&gt;, &lt;em&gt;k*l&lt;/em&gt; is also present in the list and is not &lt;em&gt;m&lt;/em&gt;)&lt;/sub&gt;  &lt;/p&gt;&#xA;" OwnerDisplayName="user15507" LastEditorDisplayName="user15507" LastEditDate="2015-08-10T18:30:16.863" LastActivityDate="2015-08-10T18:30:16.863" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19839" PostTypeId="2" ParentId="19837" CreationDate="2015-08-10T17:16:30.640" Score="2" Body="&lt;p&gt;Could it be,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Red Light&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Many people don't put a high value on me,&#xA;even though I have more potential than most.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I don't really have anything for this one.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have certain well-defined tasks to complete,&#xA;but if I complete them all, then watch out!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A red light pretty much only has one task and you need to watch out for it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's true that I'm smaller than others,&#xA;but pass by me at &quot;5 o'clock&quot; and you might pay.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A red light is smaller than a stop sign and if you run it, especially at 5 O'Clock rush hour, you might get a ticket, or worse, pay with your life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's true that I'm slower than most others,&#xA;but I've never reverted any of my decisions.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; red lights have a delay on them before they change to green in order for remaining traffic in the intersection to clear&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Even when I cannot physically move forward,&#xA;I do everything I can to fulfill my duties.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Red lights can't move but they are always trying to do their job&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My siblings and I help each other succeed,&#xA;But few reach enlightenment and are transformed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A red lights siblings could be yellow and green and they are enLIGHTened when they change, transformed, colors&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My answer seems kinda stretched but it's something&lt;/p&gt;&#xA;" OwnerUserId="13925" LastActivityDate="2015-08-10T17:16:30.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19840" PostTypeId="2" ParentId="19837" CreationDate="2015-08-10T17:33:58.063" Score="3" Body="&lt;p&gt;Are you perhaps a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Seed?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Many people don't put a high value on me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Seed packets are super cheap, usually costing less than a few dollars. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;even though I have more potential than most.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There are many seeds that can grow into many things from such a small object. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have certain well-defined tasks to complete,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Seeds have many duties such as protection and transportation to ensure that the embryo within is safe and can grow. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but if I complete them all, then watch out!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If the seed can protect the embryo, then it will start to sprout. I guess it's exciting to see the first signs of germination?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's true that I'm smaller than others,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Seeds are indeed small.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but pass by me at &quot;5 o'clock&quot; and you might pay.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Not sure about this. There are plants called &quot;&lt;strong&gt;4&lt;/strong&gt; o'clock&quot; flowers, but I've yet to find a correlation to this line. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's true that I'm slower than most others,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Seeds take a long time to grow, roughly 10-21 days. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I've never reverted any of my decisions.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;AFAIK, seeds can't reverse their growth :P&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Even when I cannot physically move forward,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Seeds cannot move without some kind of aid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I do everything I can to fulfill my duties.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;As mentioned above, protecting the embryo.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My siblings and I help each other succeed,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Some seeds will travel and grow together.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But few reach enlightenment and are transformed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Only seeds that have the right conditions can grow and transform.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13489" LastActivityDate="2015-08-10T17:33:58.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19841" PostTypeId="2" ParentId="19837" CreationDate="2015-08-10T17:51:14.190" Score="35" Body="&lt;p&gt;You are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A &lt;strong&gt;pawn&lt;/strong&gt; in chess&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Many people don't put a high value on me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pawns are generally worth &lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Chess_piece_relative_value&quot;&gt;1 point&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;even though I have more potential than most.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pawns are the &lt;strong&gt;only piece&lt;/strong&gt; that can be &lt;strong&gt;promoted&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have certain well-defined tasks to complete,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Can only &lt;strong&gt;move forward&lt;/strong&gt;, so no decision about direction&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but if I complete them all, then watch out!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Once a pawn has &lt;strong&gt;moved to the last rank&lt;/strong&gt;, it may be &lt;strong&gt;promoted to a more powerful piece&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's true that I'm smaller than others,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pawns are the &lt;strong&gt;smallest chess piece&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but pass by me &lt;s&gt;at &quot;5 o'clock&quot;&lt;/s&gt; on the 5th and you might pay.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If a pawn moving two squares passes the capture area of a pawn &lt;strong&gt;on its fifth rank&lt;/strong&gt;, it may be &lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/En_passant&quot;&gt;captured &lt;em&gt;en passant&lt;/em&gt;&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's true that I'm slower than most others,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pawns only move &lt;strong&gt;one square at a time&lt;/strong&gt; (or in certain situations 2)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I've never reverted any of my decisions.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pawns &lt;strong&gt;cannot move backwards&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Even when I cannot physically move forward,&lt;br&gt;&#xA;I do everything I can to fulfill my duties. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Blocked pawns &lt;strong&gt;still provide defensive coverage&lt;/strong&gt; to neighboring files &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My siblings and I help each other succeed,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pawns are usually arranged in a &lt;strong&gt;defensive formation&lt;/strong&gt; across the width of the board. Also, pawns next to each other can form a chain, which makes them stronger.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But few reach enlightenment and are transformed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most pawns don't make it to the &lt;strong&gt;last rank to be promoted&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4513" LastEditorUserId="4513" LastEditDate="2015-08-11T12:57:56.870" LastActivityDate="2015-08-11T12:57:56.870" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="19842" PostTypeId="2" ParentId="19823" CreationDate="2015-08-10T18:22:52.597" Score="26" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/a/19824/12727&quot;&gt;The one given by @Bailey M&lt;/a&gt;, but I would like to offer another answer...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because I figured that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It would be fun to have an alternative solution :-). With this solution, the supposed rules for tiles are that:&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; 1. the third column tiles' lines always &quot;indicate&quot; adjoined sides;&lt;br&gt;&#xA; 2. the third column tiles' lines always touch exactly one corner;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Meaning that&lt;br&gt;&#xA;&lt;br&gt;&#xA; Tiles 1, 2, 7, and 8 are out because of rule 1;&lt;br&gt;&#xA; Tiles 4, 5, 6, (and 7) are out because of rule 2;&lt;br&gt;&#xA;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And thus that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it must be Tile 3!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's an image to visualize that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;a href=&quot;https://i.stack.imgur.com/dutAO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dutAO.png&quot; alt=&quot;visualization of the solution&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12727" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-08-13T16:40:02.953" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="19843" PostTypeId="2" ParentId="19834" CreationDate="2015-08-10T18:57:05.140" Score="7" Body="&lt;p&gt;Some working thoughts on the last 2 functions:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;GR($n$)&lt;/strong&gt; obviously truncates the text after a certain number of characters, but the difficult part is determining how many.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the corresponding lengths of the original and truncated strings:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;(3, 1)&lt;/li&gt;&#xA;&lt;li&gt;(12, 7)  &lt;/li&gt;&#xA;&lt;li&gt;(47, 29)&lt;/li&gt;&#xA;&lt;li&gt;(102, 63)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The truncated length might not be related to the original length at all, but instead might have to do with the actual content of the string.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;P($n$)&lt;/strong&gt; obviously scrambles the text according to some predetermined order.  It looks like the same characters end up in the same relative positions for each string.  It's hard to determine for the long strings, since there are many repeated characters, and it's hard to know which one ends up in which location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the orders of the first two scrambles (the numbers indicate the ouput character's location in the input string):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;code&gt;HELLO EARTH!&lt;/code&gt; → &lt;code&gt;LH ERO!HATEL&lt;/code&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;(3 1 6 2 9 5 12 11 8 10 7 4)&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;ABCDEFGHIJKLMNOPQRSTUVWXYZ&lt;/code&gt; → &lt;code&gt;CAFBINELKHOSURWJGPZTYMXVQD&lt;/code&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;(3 1 6 2 9 14 5 12 11 8 15 19 21 18 23 10 7 16 26 20 25 13 24 22 17 4)&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;THIS IS AN EXAMPLE OF WHAT THE P FUNCTION DOES&lt;/code&gt; → &lt;code&gt;ITIHAA E  M FEWNSPTOAXTH  FHLNI EN USSO ETOPDC&lt;/code&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;(3 1 6 2 9 14 5 12 11 8 15 19 21 18 23 10 7 16 26 20 25 13 A 24 &lt;strong&gt;B&lt;/strong&gt; &lt;strong&gt;B&lt;/strong&gt; 34 29 17 C 39 B D C  35 E E F B D A F 32 43 37)&lt;/li&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;A = 28 or 38&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;B = 22, 27, 31, 33, or 42 (one of the &lt;strong&gt;bold&lt;/strong&gt; Bs is 22)&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;C = 36 or 41&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;D = 30 or 45&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;E = 4 or 46&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;F = 40 or 44&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note how the order of the numbers is consistent between all the strings, and in fact they all start with the same sequence.  The larger numbers (corresponding to characters near the end of the string) are mostly clustered near the end of the function's output.&lt;/p&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="4421" LastEditDate="2015-08-10T20:03:58.850" LastActivityDate="2015-08-10T20:03:58.850" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19844" PostTypeId="2" ParentId="19816" CreationDate="2015-08-10T20:29:50.887" Score="0" Body="&lt;p&gt;Edited to resolve a prior issue with an incorrect solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, basically, for incorrect students p and p+1, 1 &amp;lt;= p &amp;lt; n, n may lie within the range p+1 &amp;lt; n+1 &amp;lt; (p+1)*2.&lt;br&gt;&#xA;Both p+1 and p+2 are either prime or have a single prime factor.&lt;br&gt;&#xA;This would mean that the pair p, p+1, for n up to 100, is as follows (p,p+1)&lt;br&gt;&#xA;1,2 [divisors 2,3; n=2]&lt;br&gt;&#xA;2,3 [divisors 3,4; n=3 or 4]&lt;br&gt;&#xA;3,4 [divisors 4,5; 3 &amp;lt; n &amp;lt; 7]&lt;br&gt;&#xA;6,7 [divisors 7,8; 6 &amp;lt; n &amp;lt; 13]&lt;br&gt;&#xA;7,8 [divisors 8,9; 7 &amp;lt; n &amp;lt; 27]&lt;br&gt;&#xA;15,16 [divisors 16,17; 15 &amp;lt; n &amp;lt; 31]&lt;br&gt;&#xA;30,31 [divisors 31,32; 30 &amp;lt; n &amp;lt;61]&lt;/p&gt;&#xA;" OwnerUserId="15527" LastEditorUserId="15527" LastEditDate="2015-08-10T23:49:24.303" LastActivityDate="2015-08-10T23:49:24.303" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19845" PostTypeId="2" ParentId="19816" CreationDate="2015-08-10T20:49:24.153" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The incorrect students are any two students adjacent to each other who&#xA;  each hold a number which is the highest power of a prime of all&#xA;  students.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Of course, it is trivial that one of those students must hold a power&#xA;  of two. Two odd numbers can't be adjacent to eachother.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;After that, it is clear that any number adjacent to the highest power&#xA;  of two held which is a power of a prime must be the highest power of&#xA;  that prime. We would hit a higher power of two before we would hit a&#xA;  higher power of any other prime.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Knowing this, n can be arbitrarily high. Sure there are ranges n can't&#xA;  be, such as from 65 to 127 or so ( I haven't done the exact math here but &#xA;  anywhere where the highest power of 2 is not next to an included power of a &#xA;  prime) but we&#xA;  will always eventually find a power of a prime next to a power of 2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15529" LastEditorUserId="15529" LastEditDate="2015-08-10T21:09:18.820" LastActivityDate="2015-08-10T21:09:18.820" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19846" PostTypeId="2" ParentId="19816" CreationDate="2015-08-10T21:02:31.953" Score="19" Body="&lt;p&gt;If $x$ has at least two distinct prime factors, that is $x = p^n * q^m * r$, with $p, q$ primes, $n, m \ge 1$, and $r$ not divisible by $p$ or $q$, then $(p^n * r) | z$ and $(q^m * r)|z$ implies $(p^n * q^m * r = x)|z$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, if $x$ is a wrong answer, and all answers $&amp;lt; x-1$ were correct answers, $x$ cannot have two distinct prime factors; $x$ must be either a prime number or a power of a prime number. Further, if $x$ is a wrong answer, then $2x$ is also a wrong answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since exactly two answers $\le n+1$ were incorrect, and the two incorrect answers were consecutive, the two incorrect numbers are $x$ and $x+1$ with $x \ge 2$, and $n \le 2x-2$, and both $x$ and $x+1$ are either primes or powers of primes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only two consecutive primes are $2$ and $3$; other than this at least one of $x$ and $x+1$ is a non-trivial power of a prime. So we have one number $p^k$, where $p$ is a prime and $k \ge 2$, and $p^k \pm 1$ which is a prime or a power of a prime. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $p \ge 3$, which implies $p$ is odd: $p^k \pm 1$ is even, therefore it is not a prime but must be power of $2$. Therefore, one of the incorrect numbers must be a power of two: The incorrect answers are $2^k$ and $2^k \pm 1$. If $2^k \pm 1$ is a prime, then it is either a Mersenne prime or a Fermat prime; the only known Fermat primes are $3, 5, 17, 257, 65537 = 2^1 + 1, 2^2 + 1, 2^4 + 1$ and $2^{16} + 1$; the smallest known Mersenne primes are $2^k - 1$ for $k = 2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1279, 2203, 2281, 3217, 4253, 4423, 9689, 9941, 11213, 19937, 21701, 23209, 44497, 86243, 110503, 132049, 216091, 756839, 859433, 1257787, 1398269, 2976221, 3021377, 6972593, 13466917, 20996011, 24036583, 25964951, 30402457, 32582657$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume that the number of students is less than the world population, the possibilities are $(4,5)$, $(16,17)$, $(256,257)$, $(65536,65537)$, $(3,4)$, $(7,8)$, $(31,32)$, $(127,128)$, $(8191,8192)$, $(131071,131072)$, $(524287, 524288)$, $(2147483647,2147483648)$, where the other number is a prime. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the other number is a prime power, then the only pair is $(8,9)$ (Mihăilescu's theorem, better known as Catalan's conjecture but proven in 2002). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $3$ is also a Fermat prime, in total the possibilities are $(8,9)$, all numbers $(2^k, 2^k + 1)$ where $2^k + 1$ is a Fermat prime, and $(2^k, 2^k - 1)$ where $2^k - 1$ is a Mersenne prime. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the first possible pairs of wrong answers and the only that are possible on earth with actual humans are $(2,3)$, $(3,4)$, $(4,5)$, $(7,8)$, $(8,9)$, $(16,17)$, $(31,32)$, $(127,128)$, $(256,257)$, $(8191,8192)$, $(65536,65537)$, $(131071,131072)$, $(524287,524288)$, $(2147483647,2147483648)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The possible values for $n+1$ are $[3 \ldots 15], [17 \ldots 61], [128 \ldots 253], [257 \ldots 511]$ etc. and the possible values for $n$ are $[2 \ldots 14], [16 \ldots 60], [127 \ldots 252], [256 \ldots 510]$. So we don't have a class of $15$ students, or $61$ to $126$ students, or $253$ to $255$ students, or $511$ to $8190$ students. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could probably use the fact that the number actually fit on the board to exclude some large numbers. &lt;/p&gt;&#xA;" OwnerUserId="4664" LastEditorUserId="4513" LastEditDate="2015-08-11T14:32:30.833" LastActivityDate="2015-08-11T14:32:30.833" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19847" PostTypeId="2" ParentId="18773" CreationDate="2015-08-10T21:15:14.250" Score="7" Body="&lt;p&gt;The answer could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; L _ C K&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you do this to friends, you must have many enemies.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you L&lt;strong&gt;A&lt;/strong&gt;CK friends, you might have more enemies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you're tough you'll do this to your enemies.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To &lt;a href=&quot;http://dictionary.reference.com/browse/lick?s=t&quot; rel=&quot;nofollow&quot;&gt;L&lt;strong&gt;I&lt;/strong&gt;CK&lt;/a&gt; is defined informally as &quot;to defeat&quot; or &quot;to hit/beat&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With this you can keep your enemies from you.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; With a decent L&lt;strong&gt;O&lt;/strong&gt;CK, you could hide indefinitely from them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With this you can always defeat your enemies&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; With L&lt;strong&gt;U&lt;/strong&gt;CK, who needs skill?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13237" LastEditorUserId="13237" LastEditDate="2015-08-11T17:44:17.357" LastActivityDate="2015-08-11T17:44:17.357" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="19848" PostTypeId="1" AcceptedAnswerId="19850" CreationDate="2015-08-10T21:26:43.490" Score="4" ViewCount="746" Body="&lt;p&gt;We had left the TV on after the news finished and on came the &lt;a href=&quot;https://www.playmmc.com/gameshow&quot; rel=&quot;nofollow&quot;&gt;Monopoly Millionaire's Club&lt;/a&gt;. I had no knowledge of its existence and watched some out of curiosity. The particular game in question was &lt;a href=&quot;https://en.wikipedia.org/wiki/Monopoly_Millionaires%27_Club_(TV_series)#Games&quot; rel=&quot;nofollow&quot;&gt;No Vacancy&lt;/a&gt;. The gameplay is fairly straightforward:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The player has a hotel with 21 rooms divided evenly among 3 floors so that each floor has 7 rooms.&lt;/li&gt;&#xA;&lt;li&gt;Each turn, 5 cars appear in front of the hotel. Each car holds between 1 and 5 people, inclusive. (It is not specified whether or not each number is represented or if every car is assigned randomly and independently of each other.) (The cars are refreshed every turn so the cars you didn't pick disappear and new ones come in. I.E., the cars you've already picked have no bearing on what cars you can pick next.)&lt;/li&gt;&#xA;&lt;li&gt;The player picks one of the cars and is then told how many people are in the car.&lt;/li&gt;&#xA;&lt;li&gt;The player must give each person in the car a room in the hotel and they must all be on the same floor.&lt;/li&gt;&#xA;&lt;li&gt;Filling rooms on the first floor yields \$1,000 each, the second yields \$2,000 each, and the third yields \$3,000 each. (Yes, these are very expensive rooms.)&lt;/li&gt;&#xA;&lt;li&gt;Once every floor has at least 3 people, the player can quit after any turn and take what money is earned thus far.&lt;/li&gt;&#xA;&lt;li&gt;If ever the player is unable to place everyone from the car on the same floor, the game ends and the player receives nothing.&lt;/li&gt;&#xA;&lt;li&gt;If every room is filled, the player receives \$100,000.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, given that the entire game is electronic, it would be easy for the house to force every player to lose by simply spamming every vehicle as having 5 passengers. Let us presume, though, that the producers are &lt;em&gt;not&lt;/em&gt; out to steal hope from their contestants. Even with that, though, it struck me as being much more about chance than strategy despite all the dramatic music and lighting - combined with the dramatic host - indicating that strategy was key.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If you play perfectly, how much can you expect to win on average?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Does it matter if each car is random vs. all five cars holding every number from 1 to 5?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="463" LastEditDate="2015-08-11T14:53:37.287" LastActivityDate="2015-08-11T21:28:18.857" Title="Monopoly Game Show: Is there a winning strategy?" Tags="|strategy|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19849" PostTypeId="2" ParentId="19848" CreationDate="2015-08-10T21:38:41.547" Score="1" Body="&lt;p&gt;If the number of people in a car you pick is evenly distributed between 1 and 5, on average a car will have 3 people, or that any two cars on average will have 6 people. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, most of the time, you'll have 6 people on each floor, for $36,000 total. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a strategy, I would fill floors from the top, until there are 3 or fewer rooms left (average car). Then I would move to the next floor unless there is a gap above it that it perfectly fits.  Fill until 3 or fewer rooms. At that point, if you can get the top to 1 or 0, do that. Otherwise, go to the bottom row. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-08-10T21:38:41.547" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19850" PostTypeId="2" ParentId="19848" CreationDate="2015-08-10T23:53:25.883" Score="6" Body="&lt;p&gt;If passenger counts from 1 to 5 are equally distributed, it's possible to compute for each of the 343 (7x7x7) possible population distributions the expected value if one stands pat and if one takes another car (the stand-pat case is easy; for the take-another car case, figure out for each passenger count 1-5 which of the three possible resulting distributions would be the best, and then take the average of the best values from those five passenger counts).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I figured the math right, that would yield an expected value of $37,480 starting with all rooms empty (all expected values rounded to the nearest dollar).  In this table, the two numbers in the row heading show the number of occupied rooms on the first two floors; the column headings give the number of occupied rooms on the third floor.  The cells marked with &quot;**&quot; indicate the expected value from betting; those without asterisks indicate the expected value from standing pat (when it's higher than that from betting).  Cells marked with dashes indicate combinations that won't arise when using an optimal strategy.  Looking at the table, there's no apparent pattern for what moves are optimal.  For example, if the first car has 1, 2, 3, or 5 people in it, they should be placed on the top floor, but if it has exactly four they should go on the second floor--a result that's far from intuitively obvious.  I'm not sure how well a naïve strategy would do, but the optimal strategy yields a result which is better than 37% of the maximum value.  I'm not sure how the expected value using optimal strategy compares with what the expected value would be if the player knew car capacities in advance.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;         0        1        2        3        4        5        6        7&#xA;0 0: 37480**  38370**  40128**  38014**  -----    35153**  36910**  39856** &#xA;0 1: -----    -----    -----    38543**  -----    -----    36843**  40564** &#xA;0 2: -----    39658**  40945**  40433**  -----    38576**  39321**  43320** &#xA;0 3: -----    38259**  40520**  -----    -----    34348**  36099**  39678** &#xA;0 4: 35734**  36893**  39069**  -----    -----    32844**  34855**  38105** &#xA;0 5: -----    -----    -----    34327**  -----    31807**  34418**  37614** &#xA;0 6: -----    36154**  39222**  35625**  -----    33700**  35582**  39238** &#xA;0 7: 38728**  39545**  42685**  38769**  -----    36262**  38478**  43832** &#xA;&#xA;1 0: -----    -----    40179**  -----    -----    -----    -----    40404** &#xA;1 1: -----    -----    -----    -----    -----    -----    -----    -----   &#xA;1 2: -----    -----    40505**  39800**  -----    37976**  38689**  42889** &#xA;1 3: -----    -----    39924**  -----    -----    -----    -----    39471** &#xA;1 4: -----    -----    39046**  -----    -----    -----    -----    38454** &#xA;1 5: -----    -----    38343**  -----    -----    -----    -----    38478** &#xA;1 6: -----    -----    38536**  -----    -----    -----    -----    38672** &#xA;1 7: -----    -----    42182**  38489**  -----    37006**  37872**  43360** &#xA;&#xA;2 0: -----    -----    41017**  -----    -----    -----    -----    42729** &#xA;2 1: -----    -----    40641**  39251**  -----    37331**  38047**  42392** &#xA;2 2: -----    -----    40701**  40957**  39599**  38785**  39182**  44212** &#xA;2 3: -----    38977**  41090**  -----    -----    -----    37759**  43846** &#xA;2 4: -----    38152**  39922**  -----    -----    35321**  36694**  41238** &#xA;2 5: -----    37217**  39100**  -----    34769**  34150**  36718**  41032** &#xA;2 6: -----    37473**  39086**  37529**  35808**  36006**  36672**  41360** &#xA;2 7: -----    41348**  43493**  43008**  40006**  39672**  40560**  52800** &#xA;&#xA;3 0: -----    -----    -----    -----    -----    -----    -----    -----   &#xA;3 1: -----    -----    -----    -----    -----    -----    35248**  39349** &#xA;3 2: -----    38790**  40514**  -----    37569**  37630**  38216**  43817** &#xA;3 3: -----    -----    -----    -----    -----    -----    -----    -----   &#xA;3 4: -----    -----    -----    -----    -----    -----    -----    -----   &#xA;3 5: -----    -----    -----    -----    -----    28000    31000    34000   &#xA;3 6: -----    34320**  37923**  -----    -----    30000    33000    36000   &#xA;3 7: -----    38059**  42956**  -----    29200**  32000    35000    44000** &#xA;&#xA;4 0: -----    -----    -----    -----    -----    -----    -----    -----   &#xA;4 1: -----    -----    -----    -----    -----    -----    34775**  38521** &#xA;4 2: -----    -----    39668**  -----    -----    36349**  37411**  41848** &#xA;4 3: -----    -----    -----    -----    -----    -----    -----    -----   &#xA;4 4: 31137**  32718**  35585**  -----    24000    27000    30000    33000   &#xA;4 5: -----    -----    36784**  -----    -----    29000    32000    35000   &#xA;4 6: -----    33851**  37180**  -----    -----    31000    34000    37000   &#xA;4 7: 36379**  37156**  40963**  -----    30000    33000    36000    39000   &#xA;&#xA;5 0: -----    -----    -----    -----    -----    -----    -----    -----   &#xA;5 1: -----    -----    -----    -----    -----    -----    34539**  38168** &#xA;5 2: -----    37244**  38753**  -----    35150**  34852**  37056**  41206** &#xA;5 3: -----    -----    -----    -----    -----    26000    29000    32000   &#xA;5 4: -----    -----    36136**  -----    -----    28000    31000    34000   &#xA;5 5: -----    -----    35172**  25210**  -----    30000    33000    36000   &#xA;5 6: -----    33468**  36723**  27008**  29000    32000    35000    38000   &#xA;5 7: -----    36523**  40136**  30040**  31000    34000    37000    40000   &#xA;&#xA;6 0: -----    -----    -----    -----    -----    -----    -----    -----   &#xA;6 1: -----    -----    -----    -----    -----    -----    33825**  37126** &#xA;6 2: -----    -----    38227**  -----    35558**  35632**  36134**  40672** &#xA;6 3: -----    -----    -----    -----    -----    -----    30000    33000   &#xA;6 4: -----    -----    36213**  -----    -----    29000    32000    35000   &#xA;6 5: -----    -----    36011**  -----    28000    31000    34000    37000   &#xA;6 6: -----    32459**  35456**  27840**  30000    33000    36000    39000   &#xA;6 7: -----    35336**  39280**  31200**  32000    35000    38000    41000   &#xA;&#xA;7 0: -----    -----    -----    -----    -----    -----    -----    41958** &#xA;7 1: -----    -----    -----    -----    -----    -----    35526**  41632** &#xA;7 2: -----    40407**  42104**  41980**  38984**  38486**  39072**  51360** &#xA;7 3: -----    -----    -----    -----    -----    -----    31000    42800** &#xA;7 4: -----    -----    39331**  -----    -----    30000    33000    36000   &#xA;7 5: -----    -----    38776**  -----    29000    32000    35000    38000   &#xA;7 6: -----    34616**  38480**  30600**  31000    34000    37000    40000   &#xA;7 7: 40776**  40480**  50400**  42000**  35000**  36000    39000    100000   &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Incidentally, although the game would allow a contestant to stop with only three occupants on each floor, the only situation where a contestant who is playing optimally would be in any &quot;danger&quot; without at least four occupants on every floor would have five occupants on each of two floors, and three on the other.  A player who is using optimal strategy may have such a situation arise with the three occupants on any floor.  If the three occupants are on one of the lower floors, the player should stand pat with an expected value of \$26,000 or \$28,000.  If the three occupants are on the top floor, standing pat would have an expected value of \$24,000 but continuing on would--despite the risk--have an expected value of \$25,210.&lt;/p&gt;&#xA;" OwnerUserId="1467" LastEditorUserId="1467" LastEditDate="2015-08-11T21:28:18.857" LastActivityDate="2015-08-11T21:28:18.857" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="19851" PostTypeId="1" AcceptedAnswerId="19854" CreationDate="2015-08-11T00:20:06.763" Score="16" ViewCount="4589" Body="&lt;p&gt;Joshua was a box-stacker at the local supermarket. Every day he had to stack hundreds upon hundreds of boxes, so he devised a little game to keep himself entertained. Each day, he would take out a certain number($n$) of cubic boxes, and stack them according to these rules: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;He would only stack the boxes in a $2$-dimensional plane (only 1 layer deep).&lt;/li&gt;&#xA;&lt;li&gt;Every box in the bottom layer of boxes must be adjacent to another box, i.e. no gaps between the boxes in the bottom layer.&lt;/li&gt;&#xA;&lt;li&gt;Every box that is not on the bottom layer must be placed directly on top of another box. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;After having stacked the $n$ boxes, Joshua would take a picture of the set-up and store it on his phone. How many different pictures could Joshua take if he had $n$ boxes to play with?&lt;a href=&quot;https://i.stack.imgur.com/VMLUj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VMLUj.png&quot; alt=&quot;possible arrangement&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Hint: You may be tempted to bash away with combinatoric mathematics, but I promise there is a very simplistic and beautiful way that involves no hard maths whatsoever. &lt;/p&gt;&#xA;" OwnerUserId="15499" LastEditorUserId="10615" LastEditDate="2015-08-11T20:37:37.440" LastActivityDate="2015-08-11T20:37:37.440" Title="How many ways are there to stack boxes?" Tags="|mathematics|combinatorics|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19852" PostTypeId="2" ParentId="19851" CreationDate="2015-08-11T00:43:32.760" Score="13" Body="&lt;p&gt;Answer: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $2^{n-1}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How I arrived at this solution:&#xA;We start by looking at the base case of $n=1$. There is only one way to stack one box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuing for $n=2,3,4$, we get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/4c09P.png&quot; alt=&quot;n=2,3,4&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While drawing each piece out, I made sure to group the figures by the number of bottom pieces. By doing this, I noticed that the number in each group is very familiar:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; n=1: 1 &lt;br&gt; n=2: 1 1 &lt;br&gt; n=3: 1 2 1 &lt;br&gt; n=4: 1 3 3 1 &lt;br&gt; .... It's Pascal's Triangle!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus, using the well known formula for the sum of each row, we get the answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $2^n$, but looking at the case of $n=1$, we only have one configuration, so the answer is actually: $2^{n-1}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13881" LastActivityDate="2015-08-11T00:43:32.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19853" PostTypeId="2" ParentId="19851" CreationDate="2015-08-11T01:23:13.663" Score="12" Body="&lt;p&gt;For every arrangement of $n$ boxes, we can define a procedure to create that arrangement consisting of $n$ instructions of either of the forms:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Place a box on the current stack.&lt;br&gt;&#xA;&lt;em&gt;OR&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Place a box to the right of the current stack.  (This box is the new stack.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Let's call the first instruction $0$ and the second instruction $1$.  A sequence of instructions consists of a string of $n$ $1$s and $0$s.  The only restriction is that, assuming we start with an empty stack, the first instruction must be a $0$: otherwise we could end up with an empty stack, and we don't want any gaps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can see that every arrangement has a unique sequence of instructions, and that every possible sequence of instructions corresponds to a unique arrangement of boxes.  Thus the number of arrangements is equal to the number of length-$n$ binary strings that start with $0$; which is itself equal to the number of length-$(n-1)$ binary strings:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$2^{n-1}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9718" LastActivityDate="2015-08-11T01:23:13.663" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19854" PostTypeId="2" ParentId="19851" CreationDate="2015-08-11T01:23:27.003" Score="23" Body="&lt;p&gt;Here is another way to arrive at the answer of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $2^{n-1}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can represent these &quot;stackings&quot; by taking a row of $n$ $*$'s, and placing |'s between some of the stars to break them into groups, representing the stacks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the below layout&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;*&#xA;* *&#xA;****&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;would be represented by&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;* * *|*|* *|*&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This represents every possible stacking in a unique way, so asking for the number of stackings is the same as asking the number of strings of $*$'s and |'s in the above form. The latter is much easier to count: between every two adjacent $*$'s, there either is a |, or there isn't. This means there are $n-1$ choices to make, each with $2$ options, for a total of $2^{n-1}$ possibilities.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-08-11T01:28:53.290" LastActivityDate="2015-08-11T01:28:53.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19855" PostTypeId="2" ParentId="19848" CreationDate="2015-08-11T01:29:59.907" Score="0" Body="&lt;p&gt;It struck me that it's always possible to guarantee a nonzero amount of money won - follow the strategy that if any floor has two or fewer occupants, pick a car and give rooms to the people in that car. Stand pat as soon as all floors have at least three occupants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works because any floor with at most two people always has at least five vacancies, and so can always accommodate another carful of people.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may not be an optimal expected value, but it does give at least $18,000 in the worst case, which is quite a lot of money.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-08-11T01:29:59.907" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="19856" PostTypeId="1" CreationDate="2015-08-11T01:50:41.367" Score="5" ViewCount="317" Body="&lt;p&gt;You approach a bridge and see a troll. He throws down finite $n&amp;gt;0$ bags. Each bag, $B_i$, contain 1 or more of 4 possible buttons. $B_i$ contains $S_i \subseteq \{green, red, blue, yellow\} \land S_i \neq \emptyset$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The contents of each bag is written on the bag - e.g. bag $B_i$ may have $\{red, green\}$ written on it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The troll says, I know buttons are valuable in your neighbourhood and you can't get past me until you pick a set of bags such that you minimise $\frac{colours}{bags}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the best way to find the set of bags you should take? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;&lt;strong&gt;Bonus points if you can tell me what this (NP?) problem is called.&lt;/strong&gt;   &lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2015-08-11T08:42:09.590" LastActivityDate="2015-08-11T08:42:09.590" Title="Paying an unusual toll to the troll" Tags="|calculation-puzzle|optimization|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19857" PostTypeId="1" AcceptedAnswerId="19858" CreationDate="2015-08-11T02:27:20.863" Score="18" ViewCount="2663" Body="&lt;p&gt;If a word has a certain property, I call it an Odd Word™.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In each of the short, unrelated sentences below, only one of the choices is an Odd Word™.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sentences are for context only, and do not affect whether a word is an Odd Word™. Since a change in case or font might change the property, I have used all upper case, and made sure that the property holds true in the display font.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;SHE'S LEARNING TO MAKE (&lt;strong&gt;FEATHERSTITCHES / WHIPSTITCHES&lt;/strong&gt;) TODAY.&lt;/li&gt;&#xA;  &lt;li&gt;HE LOVES TO RIDE HIS NEW (&lt;strong&gt;BICYCLE / MOTORCYCLE&lt;/strong&gt;).&lt;/li&gt;&#xA;  &lt;li&gt;NOBODY WANTS TO BE KNOWN AS (&lt;strong&gt;AVERAGE / ORDINARY&lt;/strong&gt;).&lt;/li&gt;&#xA;  &lt;li&gt;JESSICA RAN THE RACE (&lt;strong&gt;QUICKLY / RAPIDLY&lt;/strong&gt;).&lt;/li&gt;&#xA;  &lt;li&gt;HE WANTED HER TO (&lt;strong&gt;ANSWER / RESPOND&lt;/strong&gt;) BY NOON.&lt;/li&gt;&#xA;  &lt;li&gt;GIVE (&lt;strong&gt;HER / HIM&lt;/strong&gt;) ALL YOUR REP POINTS.&lt;/li&gt;&#xA;  &lt;li&gt;THE CHILD HID (&lt;strong&gt;BESIDE / UNDER&lt;/strong&gt;) THE TABLE.&lt;/li&gt;&#xA;  &lt;li&gt;THEY ATE (&lt;strong&gt;BECAUSE / WHILE&lt;/strong&gt;) THEY HAD THE TIME.&lt;/li&gt;&#xA;  &lt;li&gt;(&lt;strong&gt;GOODNESS / YIKES&lt;/strong&gt;), LOOK AT ALL THE PEOPLE!&lt;/li&gt;&#xA;  &lt;li&gt;THE DOG (&lt;strong&gt;CHEWED / GNAWED&lt;/strong&gt;) ON A BONE.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What makes a word an Odd Word™?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="5297" LastEditDate="2016-09-29T15:09:08.020" LastActivityDate="2016-09-29T15:09:08.020" Title="What is an Odd Word™?" Tags="|pattern|english|word-property|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="19858" PostTypeId="2" ParentId="19857" CreationDate="2015-08-11T03:22:50.690" Score="25" Body="&lt;p&gt;An Odd Word™ is one that...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; has an odd number of &quot;ends&quot;. An 'end' is a place where a letter in the word meets blank space in all but one direction; that is, if you were walking on the letter there would be only one way to go. For instance, RESPOND has 10 ends: two on the bottom of the R, three on the right of the E, two on the S, one on the P, and two on the N.&lt;br&gt;&#xA; &lt;strong&gt;FEATHERSTITCHES&lt;/strong&gt;: 41 (WHIPSTITCHES: 30)&lt;br&gt;&#xA; &lt;strong&gt;MOTORCYCLE&lt;/strong&gt;: 19 (BICYCLE: 14)&lt;br&gt;&#xA; &lt;strong&gt;ORDINARY&lt;/strong&gt;: 13 (AVERAGE: 16)&lt;br&gt;&#xA; &lt;strong&gt;QUICKLY&lt;/strong&gt;: 17 (RAPIDLY: 12)&lt;br&gt;&#xA; &lt;strong&gt;ANSWER&lt;/strong&gt;: 13 (RESPOND: 10)&lt;br&gt;&#xA; &lt;strong&gt;HER&lt;/strong&gt;: 9 (HIM: 8)&lt;br&gt;&#xA; &lt;strong&gt;UNDER&lt;/strong&gt;: 9 (BESIDE: 10)&lt;br&gt;&#xA; &lt;strong&gt;WHILE&lt;/strong&gt;: 13 (BECAUSE: 14)&lt;br&gt;&#xA; &lt;strong&gt;GOODNESS&lt;/strong&gt;: 11 (YIKES: 14)&lt;br&gt;&#xA; &lt;strong&gt;GNAWED&lt;/strong&gt;: 11 (CHEWED: 14)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll take my 100 rep now. ;)&lt;/p&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="11876" LastEditDate="2015-08-11T16:39:35.880" LastActivityDate="2015-08-11T16:39:35.880" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="19859" PostTypeId="2" ParentId="19857" CreationDate="2015-08-11T03:24:50.630" Score="9" Body="&lt;p&gt;An odd word is a word that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; has the odd number of 'endpoints'. For example, the letter A has 2 endpoints at the bottom; B has 0 because it is closed. C has 2; D has 0, E has 3, etc.&lt;br&gt;&#xA; This will only works for the displayed font and case.&lt;br&gt;&#xA;1.FEATHERSTITCHES = 41 total endpoints&lt;br&gt;&#xA;2.MOTORCYCLE = 19&lt;br&gt;3.ORDINARY = 13&lt;br&gt;4.QUICKLY = 15&lt;br&gt;5.ANSWER = 13&lt;br&gt;&#xA;6.HER = 9&lt;br&gt;7.UNDER = 9&lt;br&gt;&#xA;8.WHILE = 13&lt;br&gt;&#xA;9.GOODNESS = 11&lt;br&gt;&#xA;10.GNAWED = 11&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12332" LastEditorUserId="11876" LastEditDate="2015-08-11T03:31:45.080" LastActivityDate="2015-08-11T03:31:45.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19860" PostTypeId="1" AcceptedAnswerId="22330" CreationDate="2015-08-11T05:40:07.023" Score="14" ViewCount="1262" Body="&lt;p&gt;So, I've been &amp;quot;inspired&amp;quot; by JLee's great series of Word™/Phrase™ puzzles to make my own. (It's not stealing an idea if I call it &amp;quot;inspiration&amp;quot;!) As usual: each sentence contains two marked words, one of which is a Spanned Word™ and one of which is not, and the actual sentences are just for context.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The guru said, &amp;quot;at least one of you has (&lt;strong&gt;GREEN / BLUE&lt;/strong&gt;) eyes.&amp;quot;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You have limited weighings to find the odd coin among (&lt;strong&gt;NINE / TWELVE&lt;/strong&gt;).&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In the Monty Hall problem, one of your options is to (&lt;strong&gt;STICK / SWITCH&lt;/strong&gt;).&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I hope you will find this puzzle (&lt;strong&gt;INVENTIVE / DIFFICULT&lt;/strong&gt;).&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This puzzle's solution may depend on (&lt;strong&gt;SHAPE / SHAPES&lt;/strong&gt;).&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Will similar words make this puzzle (&lt;strong&gt;EASIER / SIMPLER&lt;/strong&gt;)?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Or will you focus too much on the (&lt;strong&gt;DIFFERENCE / DISTINCTION&lt;/strong&gt;)?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Don't work on this for too long or you may become (&lt;strong&gt;CONFUSED / ENTRANCED&lt;/strong&gt;.)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I hope this puzzle isn't too (&lt;strong&gt;HARD / EASY&lt;/strong&gt;)!&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I can't think of any other relevant sentences so here are some things that annoy me: (&lt;strong&gt;INTERRUPTIONS / NOISE&lt;/strong&gt;) and (&lt;strong&gt;STUBBORNNESS / INFLEXIBILITY&lt;/strong&gt;).&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;(By the way, the previous sentence's pairs were (&lt;strong&gt;INDEPENDENT / UNRELATED&lt;/strong&gt;).)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Good luck solving this (&lt;strong&gt;CONUNDRUM / CHALLENGE&lt;/strong&gt;)!&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So, what is this property that makes a word a Spanned Word™?&lt;/p&gt;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; NOISE and HINT are  Spanned Word™s; FEET is a special case of a Spanned Word™.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; These could also be called Traversable Words.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Font and case both matter. SHAPE is Spanned; SHAPES is not.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-11-08T00:00:39.500" Title="What is a Spanned Word?™" Tags="|pattern|english|word-property|" AnswerCount="2" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="19861" PostTypeId="1" AcceptedAnswerId="19872" CreationDate="2015-08-11T06:08:05.793" Score="5" ViewCount="637" Body="&lt;p&gt;I've been itching to try and make one of these for a while since &lt;a href=&quot;https://puzzling.stackexchange.com/users/463/jlee&quot;&gt;JLee&lt;/a&gt; started posting them. I feel like this one might not be challenging enough, but I'll see what you guys make of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following table contains two columns. Each of the phrases in the left column has a property for which I call it an Endless Phrase™, that the phrases in the right column lack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9qPoB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9qPoB.png&quot; alt=&quot;a list of endless and non-endless phrases&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The appearance of letters, including capitalization, has no bearing on the rule.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Can you figure out what the pattern is? &lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2016-11-07T23:46:53.800" Title="What is an Endless Phrase™?" Tags="|pattern|english|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19862" PostTypeId="2" ParentId="19856" CreationDate="2015-08-11T06:09:07.803" Score="3" Body="&lt;p&gt;For each single colour check how many bags there are that only have that colour in it and pick the colour that has the most bags. If this number is equal to or larger than $n/4$ you're finished because this is the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, for each combination of 2 colours check how many bags only have those colours and pick those colours that have the most bags. If this number is equal to or larger than $n/2$ you're finished because that is the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, for each combination of 3 colours check how many bags only have those colours and pick those colours that have the most bags. If this number is equal to or larger than $n(3/4)$ you're finished because that is the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not pick all bags.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this at least gives the right answer but I doubt it is the best strategy because it is the most straight-forward strategy I would think.&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-08-11T06:09:07.803" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19863" PostTypeId="2" ParentId="19823" CreationDate="2015-08-11T06:48:37.050" Score="13" Body="&lt;p&gt;I think Bailey M has got the intended answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But another answer that works is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tile 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Notation&lt;/strong&gt;&lt;br&gt;&#xA;In each square, label the corners A-D, running clockwise from the top left. By &quot;line&quot; we mean a line inside a square. Call a line &quot;par&quot; if it is parallel to a side of the square. Call a corner &quot;visited&quot; if it is touched by a line. &#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;strong&gt;First fact&lt;/strong&gt;&lt;br&gt;&#xA;Consider the distribution of par lines by row. In rows 1-2 the totals are the &lt;em&gt;same&lt;/em&gt;: each row contains exactly 1. To continue the pattern of sameness, row 3 must also contain exactly 1. Of all the possible missing tiles, the only one that contains exactly 1 par line is tile 2.&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;strong&gt;Second and corroborating fact&lt;/strong&gt;.&lt;br&gt;&#xA;Now consider the distribution of par lines by column. In columns 1-2 the numbers are &lt;em&gt;different&lt;/em&gt;: there are 0 in column 1 and 2 in column 2. Neither of the known tiles in column 3 contains a par line, so the total in column 3 can only be 0 or 1. To continue the pattern of difference, the number must be 1, so again the answer must be tile 2.&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;strong&gt;Third and supporting fact&lt;/strong&gt;&lt;br&gt;&#xA;Each fully known column contains exactly 2 squares in which the same corner is visited: in column 1, that is corner B; in column 2, corner C. So we need a tile that continues that pattern in column 3. The visited corners in the known tiles in column 3 are A and C. This restricts the possibilities for the missing tile to tiles 2 and 5. So when we choose tile 2, this pattern also continues.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Diagram of this solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;a href=&quot;https://i.stack.imgur.com/ZFNJw.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZFNJw.jpg&quot; alt=&quot;diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of course this answer doesn't use all of the given information. For example, the horizontal line in the top middle tile could be translated upwards and everything would still apply. But the same is true of what appears to be the intended solution. In that solution, it doesn't matter whether two lines in a square cross or don't cross, so long as they start and end so as to connect with the lines in adjacent squares. One could also say that going downwards for row 2 but left-to-right for rows 1 and 3 is a bit arbitrary. Our basis for deciding what we think is the best solution is subjective, or at least would take a &lt;em&gt;lot&lt;/em&gt; of definition to be made objective.&lt;/p&gt;&#xA;" OwnerUserId="7363" LastEditorUserId="7363" LastEditDate="2015-08-11T07:31:49.013" LastActivityDate="2015-08-11T07:31:49.013" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19864" PostTypeId="1" AcceptedAnswerId="20200" CreationDate="2015-08-11T06:56:24.850" Score="4" ViewCount="473" Body="&lt;p&gt;My brother and I spent some time hanging out this weekend and he taught me a specific sequence of numbers during a particular activity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;58, 567, 57, 578, 58, 58&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does it represent and what were we doing?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Some more sequences:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;357, 356, 357, 45, 346, 357&#xA;135, 245, 235, 24, 235, 135&#xA;357, 467, 457, 46, 457, 357&#xA;81012, 81011, 81012, 910, 8911, 81012&#xA;6810, 7910, 7810, 79, 7810, 6810&#xA;81012, 91112, 91012, 911, 91012, 81012&#xA;91012, 91113, 91112, 911, 91112, 91012&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Hints:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - Our activity didn't require electricity &lt;br&gt;&#xA; - &lt;em&gt;Taught&lt;/em&gt; is a key word. These sequences are &lt;em&gt;memorised&lt;/em&gt; for a specific purpose. &lt;br&gt;&#xA; - The sequence lengths are important. To elaborate upon this, the fact that there are the same number of groups per sequence is very useful information. &lt;br&gt;&#xA; - My brother knows these sequences very well because he uses them for a living.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user2674" LastEditorDisplayName="user2674" LastEditDate="2015-08-17T22:07:28.970" LastActivityDate="2015-08-21T01:06:55.523" Title="A number sequence of a different kind" Tags="|number-sequence|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="19865" PostTypeId="2" ParentId="18710" CreationDate="2015-08-11T07:04:38.580" Score="6" Body="&lt;p&gt;What i figured out so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The (13384) refer to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your Profile: &lt;a href=&quot;https://puzzling.stackexchange.com/users/13384/the-random-guy&quot;&gt;https://puzzling.stackexchange.com/users/13384/the-random-guy&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We get a new Text:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; i am your frIend, dear tHe rAndom guy ! i can'T communicatE with You because my nOkia is oUt of battery aGain ! i wanted tO warn you via your adress mail buT i cOuldn't figure wHat is your mail ... so the only way i found is to writE you a message here. i wont be abLe to taLk to you for now, i am sorry :/  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With a Little hidden Message:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Only The Upper Case Letters:  HATE YOU GO TO HELL&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;i couldn't figure out what the next Numbers mean, but i think he  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; just dont like you :) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got a really far fetched answer for the secound Numbers..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay first of all:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We turn them around: 100 101 115 099&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Decode with ASCII:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; desc&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And that leads to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; your &lt;strong&gt;desc&lt;/strong&gt;ription in your Profile :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11899" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-08-11T08:43:09.063" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19866" PostTypeId="2" ParentId="19851" CreationDate="2015-08-11T07:20:14.747" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  We're basically looking for the number of possible arrangements of vertical boxes whose sum of lengths give $n$. Let's imagine that there are $n$ 1s, with each linked to the adjacent 1. We can either cut a link or leave it as it is. The linked ones will be considered grouped together. Since there are $n-1$ links, we can leave them in $2$^$(n-1)$ ways.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastActivityDate="2015-08-11T07:20:14.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19867" PostTypeId="2" ParentId="19856" CreationDate="2015-08-11T07:33:52.283" Score="0" Body="&lt;p&gt;If we take bags without reducing the number of colors from the first combination excluding a given color the ratio won't be minimized, so we must jot down which bags have only a given color combination (there are 15), which must be repeated for the other combinations, and choose the one minimizing the ratio by excluding all the others with an unwanted color.&lt;/p&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="14016" LastEditDate="2015-08-11T08:27:34.617" LastActivityDate="2015-08-11T08:27:34.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19869" PostTypeId="2" ParentId="19809" CreationDate="2015-08-11T09:47:24.270" Score="3" Body="&lt;p&gt;The minimal statement, that suffices is I think:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are &lt;strong&gt;not&lt;/strong&gt; $n$ blue dots. This removes only one option for the blue dots and thus it seems minimal in terms of information&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From there on:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If there are $n+1$ blue dots. After one day, every blue-dotter has realized that he is also a blue-dotter (after seeing the $n$ blue dots) and kills himself.&lt;br/&gt;If there are $n+2$ blue dots. After one day, none of the blue-dotters have killed themselves, because they have seen $n+1$ blue dots. They however realize that they also have blue dots, so they kill themselves on the next day.&lt;br/&gt;This is generalized for $n+x$. After all blue-dotters have died, the red-dotters kill themselves on the next day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But what if?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If there are $n-x$ blue dots? Things play out identically for the red dots as they now know that there are not $p-n$ red dots where $p$ is the population size. In the mean time the blue dots are happily waiting, as they don't gain information until the red ones are dead.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The only flaw I see is that the statement may not be minimal, so I would love to see your comments.&lt;/p&gt;&#xA;" OwnerUserId="5610" LastActivityDate="2015-08-11T09:47:24.270" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="19870" PostTypeId="1" AcceptedAnswerId="19897" CreationDate="2015-08-11T09:50:09.657" Score="88" ViewCount="17120" Body="&lt;p&gt;One night nine gangsters stole a gold bar. When the time came for dividing the bar, they faced a problem: two of the criminals put guns to each other's faces. Now it's up to fate whether one of them lives, they both live or both die.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While these two are dealing with each other, the others decide to continue dividing the gold bar. What is the minimal amount of pieces they should divide the bar into, so that no matter how things pan out, everyone can be given an equal share?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Scenario 1&lt;/strong&gt;:  Both gangsters blow each other's brains out.  The gold must be divided evenly among the seven remaining gangsters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Scenario 2&lt;/strong&gt;:  One gangster is quicker on the draw, and manages to take out his opponent.  The gold must be divided evenly among the eight remaining gangsters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Scenario 3&lt;/strong&gt;:  The duelling gangsters discuss their differences, come to a mutually beneficial agreement, and put away their guns.  The gold must be divided evenly among all nine gangsters.&lt;/p&gt;&#xA;" OwnerUserId="14458" LastEditorUserId="4421" LastEditDate="2015-08-11T15:57:58.557" LastActivityDate="2015-10-23T00:24:45.087" Title="Nine gangsters and a gold bar" Tags="|mathematics|logical-deduction|combinatorics|number-theory|" AnswerCount="18" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="19871" PostTypeId="2" ParentId="19870" CreationDate="2015-08-11T10:18:05.697" Score="3" Body="&lt;p&gt;Okay first divide it into 9 Pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we take the 8th and 9th Piece and divide them into 112 Equal Part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have now 7 + 112 = 119 Pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;56 Little Parts = 1 Big Piece&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scenario 1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everyone Lives. You have the Seven Pieces and the other 2 get 56 Little Parts each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scenario 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only 1 Dies. You give everyone one big Piece and the 8th get 56 Little Pieces (Which is Equal to 1 Big Piece) and everyone get 7 Extra Little Piece&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scenario 3:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both Die. You give everyone one big Piece and 16 Little Pieces.&lt;/p&gt;&#xA;" OwnerUserId="11899" LastEditorUserId="11899" LastEditDate="2015-08-11T10:54:14.307" LastActivityDate="2015-08-11T10:54:14.307" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19872" PostTypeId="2" ParentId="19861" CreationDate="2015-08-11T10:47:51.267" Score="2" Body="&lt;p&gt;An endless phrase is one in which:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Taking the last letter(s) of a phrase and applying it to the beginning allows you to create (at least) one new word. I suspect it allows you to continue for longer than that, but I'm struggling to do that  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Examples:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;listen&lt;/strong&gt; to the paintin&lt;strong&gt;g&lt;/strong&gt; -&gt; glisten&lt;br&gt;&#xA;&lt;strong&gt;A cle&lt;/strong&gt;ated shoe in a &lt;strong&gt;barn&lt;/strong&gt; -&gt; barnacle&lt;br&gt;&#xA;&lt;strong&gt;ter&lt;/strong&gt;rible crow&lt;strong&gt;bar&lt;/strong&gt; -&gt; barter  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As ever, this doesn't work with the non-endless phrases  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;End results from each line:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1)cob&lt;br&gt;&#xA;2)glisten (though glis is an animal genus, apparently)&lt;br&gt;&#xA;3)leader&lt;br&gt;&#xA;4)wad&lt;br&gt;&#xA;5)agog&lt;br&gt;&#xA;6)erase&lt;br&gt;&#xA;7)local&lt;br&gt;&#xA;8)cat&lt;br&gt;&#xA;9)best&lt;br&gt;&#xA;10)black&lt;br&gt;&#xA;11)eon&lt;br&gt;&#xA;12)llama&lt;br&gt;&#xA;13)barnacle&lt;br&gt;&#xA;14)net&lt;br&gt;&#xA;15)source&lt;br&gt;&#xA;16)least&lt;br&gt;&#xA;17)whelp&lt;br&gt;&#xA;18)shear (or even she)&lt;br&gt;&#xA;19)men&lt;br&gt;&#xA;20)art&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11717" LastEditorDisplayName="user88" LastEditDate="2015-08-11T12:11:16.760" LastActivityDate="2015-08-11T12:11:16.760" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19873" PostTypeId="2" ParentId="19812" CreationDate="2015-08-11T10:55:00.073" Score="3" Body="&lt;p&gt;Words of 11 letters with matching triplets start and end:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;antioxidant&#xA;entitlement&#xA;microcosmic&#xA;redelivered&#xA;tessellates&#xA;underground &amp;lt;-- voila&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And if the order of the triplet letters is malleable:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;hypertrophy&#xA;stealthiest&#xA;researchers&#xA;restrainers&#xA;abracadabra&#xA;earthenware&#xA;encumbrance&#xA;sacroiliacs&#xA;sanitarians&#xA;shellfishes&#xA;inoculation&#xA;trendsetter&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But annoyingly none that I can see which end in the opposite order to the start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of the above, we can see that some have 4 matching letters:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;redelivered&#xA;abracadabra&#xA;sanitarians&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And there is in fact a 9-letter which packs 2 4-letter repeats:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;outshouts&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The patterns for those who are &lt;code&gt;grep -P&lt;/code&gt; ready:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;^(...).{5}\1$       // 11 letters, common start and end triplet&#xA;&#xA;^(....).+\1$        // 4-letter repeats&#xA;&#xA;// unordered matching end triple&#xA;^(.)(.)(.).{5}(\1(\2\3|\3\2)|\2(\1\3|\3\1)|\3(\1\2|\2\1))$ &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="12816" LastEditorUserId="12816" LastEditDate="2015-08-12T07:08:43.237" LastActivityDate="2015-08-12T07:08:43.237" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19874" PostTypeId="2" ParentId="19809" CreationDate="2015-08-11T10:57:07.370" Score="-1" Body="&lt;p&gt;The dim-witted traveler turns up to Dot-town and says after seeing only 4 townsfolk:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hey guys, why do all the women have red dots and the men have blue dots? (I believe this stipulates as non-trivial?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All the women with red dots and the men with blue dots kill themselves.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which causes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All the surviving town members to realise they must have the opposite colour, and therefore kill themselves. The dim-witted traveler then goes &quot;Woohoo! I own a whole town!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Generalized:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Having a red (or blue) dot is a boolean value, therefore you know what you have when you know what you &lt;em&gt;don't&lt;/em&gt; have. I'm surprised the townsfolk were able to live long enough to create a town in the first place ;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4659" LastEditorUserId="4659" LastEditDate="2015-08-11T21:19:52.233" LastActivityDate="2015-08-11T21:19:52.233" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="19875" PostTypeId="2" ParentId="19870" CreationDate="2015-08-11T11:01:18.640" Score="3" Body="&lt;p&gt;The minimal number of pieces in which the gold can be divided is 28.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, divide the entire bar into 9 equal pieces of volume $V$. Set aside 7 of those pieces and divide one of the other two into 7 pieces of volume $V_1$ = $V/7$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cut the last piece into 8 pieces of volume $V_2$ = $V/8$. Set aside 7 pieces of volume $V_2$, and divide the last piece of $V_2$ into 7 pieces of volume $V_3$ = $V_2$/7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total number of pieces is 28. (7 of volume V, 7 of volume $V_1$, 7 of volume $V_2$, 7 of volume $V_3$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Scenario 1: Both survive the shootout.&lt;/strong&gt;&lt;br&gt;&#xA;First seven gangsters get a piece of volume $V$ and the eighth gets the seven pieces of volume $V_1$. The ninth gets the 7 pieces of volume $V_2$ and the 7 pieces of volume $V_3$. So each one ends up with a volume $V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Scenario 2: One perishes in the shootout.&lt;/strong&gt;&lt;br&gt;&#xA;First seven gangsters get a piece of volume $V$ first. The survivor gets the 7 pieces of volume $V_1$. After that, the gangsters divide the last piece, such that seven of them get a piece of volume $V_2$, while one of them takes the 7 pieces of volume $V_3$. Thus, each ends up with a volume $9V/8$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Scenario 3: Both die in the shootout.&lt;/strong&gt;&lt;br&gt;&#xA;First seven gangsters get a piece of volume $V$ first. Then they split up the eighth piece, each taking a piece of volume $V_1$. Then they each take a piece of volume $V_2$. Lastly, they each take a piece of volume $V_3$. This, each ends up with a volume $9V/7$.&lt;/p&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-08-11T11:19:59.660" LastActivityDate="2015-08-11T11:19:59.660" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19876" PostTypeId="2" ParentId="19870" CreationDate="2015-08-11T11:05:21.250" Score="2" Body="&lt;p&gt;You need to divide your gold bar into 9 equals pieces (big parts). Take 2 big parts and divide each of them into 14 small parts.&#xA;You have 7 big parts and 14 small parts (&lt;strong&gt;21&lt;/strong&gt; parts)&lt;br&gt;&#xA;1 big part = 7 small parts&#xA;&lt;br&gt;&lt;br&gt;&#xA;&lt;strong&gt;If both live (9 equals shares)&lt;/strong&gt;&lt;br&gt;&#xA;Everyone have 1 big part OR 7 small parts&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If one live and one die (8 equals shares)&lt;/strong&gt;&lt;br&gt;&#xA;Everyone have 1 big part and 1 small part OR 8 small part&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If both die (7 equals shares)&lt;/strong&gt;&lt;br&gt;&#xA;Everyone have 1 big part and 2 small part&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Lateral thinking answer&lt;/h2&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;when the time had come for dividing the bar they faced a problem&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;They don't divided the golden bar yet. The solution is to divide the gold bar AFTER the two gangster shoot themselves or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;EDIT&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;i'm bat at maths, the 2nd case doesn't work D:&lt;/p&gt;&#xA;" OwnerUserId="13384" LastEditorUserId="13384" LastEditDate="2015-08-11T11:22:47.593" LastActivityDate="2015-08-11T11:22:47.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19877" PostTypeId="2" ParentId="19870" CreationDate="2015-08-11T11:07:29.517" Score="5" Body="&lt;p&gt;Not sure it is the best solution but this is one way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First divide the bar in 9 equal parts and call it size X. Divide one of those parts in 8 equal parts of size Y.&#xA;So now if one dies you can distribute the 9th part easily among the 8 others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you divide another part of X in 7 parts with 6 of those parts being of size (X+Y)/7. the seventh part will be together with an Y part the same size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you now have in total 7 parts of X, 8 parts of Y, 6 parts of (X+Y)/7 and one part of ((X+Y)/7 - Y) = 22 parts.&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-08-11T11:07:29.517" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19878" PostTypeId="2" ParentId="19870" CreationDate="2015-08-11T11:11:51.063" Score="25" Body="&lt;p&gt;&lt;strong&gt;22 pieces&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the weight of the bar is 504. I chose 504 since 7*8*9=504, so, the numbers would be easy to work with.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;7 pieces of weight 56 (A)&lt;/li&gt;&#xA;&lt;li&gt;8 pieces of weight 7  (B)&lt;/li&gt;&#xA;&lt;li&gt;6 pieces of weight 9  (C)&lt;/li&gt;&#xA;&lt;li&gt;1 pieces of weight 2  (D)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Scenario 1: The two gangsters decide to settle in a peaceful manner&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each gets a share of weight 504/9=56.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Give 7 A pieces to 7 people. 8th gets all the B pieces. The rest goes to 9th.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Scenario 2: One blows the head of the other and manages to survive&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each gets a share of weight 504/8=63.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of first 7 gets one A &amp;amp; one B piece. The rest goes to 8th.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Scenario 3: Both heads are blown&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each gets a share of weight 504/7=72.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of the first six get one A, one B &amp;amp; one C piece. The rest goes to 7th.&lt;/p&gt;&#xA;" OwnerUserId="15544" LastEditorUserId="15544" LastEditDate="2015-08-11T11:21:41.523" LastActivityDate="2015-08-11T11:21:41.523" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="19879" PostTypeId="2" ParentId="18239" CreationDate="2015-08-11T11:18:10.627" Score="7" Body="&lt;p&gt;We're going to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Cambodia&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's a single odd letter out in each of the clues, and for all but the hex clue, the odd letter out is based on how the word is typed on a QWERTY keyboard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OLD FLAGS:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; O is the only letter typed on the top row of the keyboard. The rest are all on the home row.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;FISHY DISH:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; D is the only letter in an odd numbered column of the keyboard.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;PUERTO RICO:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C is the only letter on the bottom row. The rest are all on the top row.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;ACQUITTED IDIOT:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I is the only letter in an even numbered column of the keyboard.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;61B2068696E74:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you remove the letter B and convert from hex to ASCII, you get &quot;a hint&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;MANLY PINK:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A is the only letter typed with the left hand.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;BANANA CABANA:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A is the only letter on the home row. The rest are all on the bottom row.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;STREET SMART:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; M is the only letter typed with the right hand.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These letters, when arranged in the right order, give the solution above. I don't think there was a hint to the order, which made it much harder to get the solution without finding all of the letters.&lt;/p&gt;&#xA;" OwnerUserId="2187" LastActivityDate="2015-08-11T11:18:10.627" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19880" PostTypeId="2" ParentId="19870" CreationDate="2015-08-11T11:19:23.830" Score="3" Body="&lt;p&gt;I can do it with: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;22 Pieces&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The lowest common multiple of 7, 8 and 9 is 504.&#xA;If this is shared 7 ways, then each gangster gets 72 504ths.&#xA;If this is shared 8 ways, each gets 63 504ths.&#xA;If this is shared 9 ways, each gets 56 504ths&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Cut the bar into 9 equal pieces, each containing 56no 504ths. Take two of these pieces and cut each into (3 pieces that are 16no 504ths and one of 8no 504ths). Take another large piece and cut it into 7 smaller pieces, each containing 8no 504ths.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If you need to share 9 ways:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(6 * 56);&#xA;  (2 * 16+16+16+8);&#xA;  (1 * 7+7+7+7+7+7+7+7);&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;8 ways:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(6 * 56+7);&#xA;  (2 * 16+16+16+8+7);&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;7 ways:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(6 * 56+16);&#xA;  (1 * 7+7+7+7+7+7+7+7+8+8)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="14135" LastEditDate="2015-08-11T11:25:46.530" LastActivityDate="2015-08-11T11:25:46.530" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19881" PostTypeId="2" ParentId="19870" CreationDate="2015-08-11T11:48:01.300" Score="16" Body="&lt;p&gt;I've managed to solve it using &lt;strong&gt;19 pieces&lt;/strong&gt; like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nineteen pieces in 504th portions:&lt;br&gt;&#xA; 2,5,9,12,13,16,18,19,20,23,26,26,30,37,40,47,51,54,56&lt;br&gt;&lt;br&gt;&#xA; In 7 shares, by 72:&lt;br&gt; {16, 56}, {2, 19, 51}, {9, 23, 40}, {5, 30, 37}, {18, 54}, {20, 26, 26}, {12, 13, 47}&lt;br&gt;&lt;br&gt;&#xA; In 8 shares, by 63:&lt;br&gt; {26, 37}, {13, 20, 30}, {23, 40}, {16, 47}, {9, 54}, {2, 5, 56}, {12, 51}, {18, 19, 26}&lt;br&gt;&lt;br&gt;&#xA; In 9 shares, by 56:&lt;br&gt; {26, 30}, {13, 20, 23}, {16, 40}, {9, 47}, {2, 54}, {56}, {5, 51}, {12, 18, 26}, {19, 37}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But there's a better solution with less number of pieces.&lt;/p&gt;&#xA;" OwnerUserId="14458" LastEditorUserId="14458" LastEditDate="2015-08-11T12:12:33.850" LastActivityDate="2015-08-11T12:12:33.850" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19882" PostTypeId="2" ParentId="19870" CreationDate="2015-08-11T11:51:09.313" Score="9" Body="&lt;p&gt;While my quick upper bound of 22 isn't new (multiple people have reached this), I reached it through a logical division rather than figuring it out mathematically. Here's the method:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;We might have 9 people.&lt;/strong&gt; We divide the bar into 9 A-pieces (size 1). Everyone gets one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;We might have 8 people.&lt;/strong&gt; Everyone gets one A, then we divide the remaining piece into 8 smaller B-pieces and everyone gets one of those. We have 16 pieces so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;We might have 7 people.&lt;/strong&gt; Everyone gets one A and one B. We now have a spare A and B to divide. Divide the A into 6 C1-pieces and a C2 piece (C1s are equal, C2 is a B-piece smaller than a C1). 6 people get a C1, 1 person gets a C2 and the spare B. We have 22 pieces.&lt;/p&gt;&#xA;" OwnerUserId="12462" LastActivityDate="2015-08-11T11:51:09.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19883" PostTypeId="2" ParentId="19809" CreationDate="2015-08-11T12:01:52.470" Score="0" Body="&lt;p&gt;I think all he has to say is &quot;The number of red dots is greater than or less than the number of blue dots.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are 600 blue dots and no red dots, everyone panics and says &quot;oh crap I'm the red dot!&quot; But then they don't instantly kill themselves, and think &quot;Phew, that means I'm not a red dot. oh shiiii&quot; and die.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are an equal number of each, the same logic applies. Everyone sees the number of other dots, and concludes either truly or falsely about their dot coloration. if true, they suicide, if not, they still suicide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cletus wonders what it is he said.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 1: You know, come to think of it, he can say anything at all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If he says, truthfully or falsely, anything about the number of blue dots, then both the red dot people and the blue dot people suddenly become aware that they are one or the other and not some other color like green. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first day (or in this case, infinitely tiny fraction of a second) they either come to the correct conclusion about their dot coloration and die, or they come to the false conclusion, and then realize they are incorrect, and that their dot coloration is the opposite. Then they die. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing necessary to begin this catastrophe is for the people to be aware of the binary nature of the dot coloration and for someone to introduce them to the idea that other people now know something about each other's dot coloration.&lt;/p&gt;&#xA;" OwnerUserId="13068" LastEditorUserId="13068" LastEditDate="2015-08-12T11:43:15.380" LastActivityDate="2015-08-12T11:43:15.380" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19884" PostTypeId="1" AcceptedAnswerId="20093" CreationDate="2015-08-11T12:09:39.353" Score="8" ViewCount="1109" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/b3Xnb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/b3Xnb.jpg&quot; alt=&quot;Cipher&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hidden in this image is a word or phrase. What is it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint #1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Almost all of the intersections and all of the dots are on a certain lattice. This lattice is important.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The lattice is 5x15. Its columns correspond to letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2187" LastEditorUserId="2187" LastEditDate="2015-08-15T00:05:29.000" LastActivityDate="2015-08-15T08:06:36.557" Title="Decipher the Picture" Tags="|visual|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="19885" PostTypeId="1" AcceptedAnswerId="19886" CreationDate="2015-08-11T12:21:53.823" Score="1" ViewCount="267" Body="&lt;p&gt;&lt;strong&gt;Disclaimer&lt;/strong&gt; : This is my first riddle, so be nice please! Also, don't forget to give your opinion.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I am common, but very rare in some places,&#xA;I can be a huge place or a single drop.&#xA;Take care of me, I can give death,&#xA;but without me life stops.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What this riddle is talking about?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="13384" LastActivityDate="2015-08-11T12:29:13.593" Title="Take care of me, I can give death, but without me life stops" Tags="|riddle|word|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19886" PostTypeId="2" ParentId="19885" CreationDate="2015-08-11T12:27:13.010" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Water&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am common, but very rare in some places&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Some places like the Great Lakes in the US have plenty of water, while the Sahara Desert is lacking.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be a huge place or a single drop&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You can have an ocean or a drop of water&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Take care of me, I can give death&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You can drown&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but without me life stops&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;All life on Earth requires water&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12365" LastEditorUserId="12365" LastEditDate="2015-08-11T12:29:13.593" LastActivityDate="2015-08-11T12:29:13.593" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19888" PostTypeId="2" ParentId="18629" CreationDate="2015-08-11T12:45:33.317" Score="1" Body="&lt;p&gt;In addition to The Dark's answer, I have a couple of additions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I sometimes have flashbacks to those times, but its so long now it seems like another world.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The game &quot;Another World&quot;, 1991&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's cheesy, but... you made things possible.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Only Amiga Makes It Possible!&quot; was a slogan. I found &lt;a href=&quot;https://youtu.be/PWeO5IkCssk&quot; rel=&quot;nofollow&quot;&gt;this video&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12462" LastActivityDate="2015-08-11T12:45:33.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19889" PostTypeId="1" AcceptedAnswerId="19898" CreationDate="2015-08-11T13:06:12.700" Score="12" ViewCount="2437" Body="&lt;p&gt;Be sure to keep watch,&lt;br&gt;&#xA;a tough guy just walked in.&lt;br&gt;&#xA;He's got a reputation in these backwards parts,&lt;br&gt;&#xA;and he could deliver a real beating.&lt;br&gt;&#xA;If there's a lady around, he'll just hold his cup.&lt;br&gt;&#xA;I think, for the most part, he's just a mixed-up fool.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Each line matches the correct answer, but the title's irrelevant. None of the clues is literally applicable; it's a little more cryptic.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="12462" LastEditorUserId="12462" LastEditDate="2015-08-11T13:27:03.613" LastActivityDate="2015-08-12T14:43:32.633" Title="Look who just showed up!" Tags="|riddle|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19890" PostTypeId="2" ParentId="19889" CreationDate="2015-08-11T13:12:11.063" Score="2" Body="&lt;p&gt;Sounds like a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;cake&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Be sure to keep watch,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Take care, don't drop it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;a tough guy just walked in.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I guess &quot;tough&quot; is a word that can be used to describe the consistence of cake.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He's got a reputation in these backwards parts,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Cake is said to be very tasty and ofc many people like it. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and he could deliver a real beating.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Thinking of films when people throw cakes at faces.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If there's a lady around, he'll just hold his cup.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Maybe some referrence to drinking coffee/tea etc, &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think, for the most part, he's just a mixed-up fool.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A cake is the result of mixing together ingredients.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5106" LastActivityDate="2015-08-11T13:12:11.063" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19891" PostTypeId="2" ParentId="19889" CreationDate="2015-08-11T13:13:23.253" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Alcholol&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Be sure to keep watch,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Dont drink too much&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;a tough guy just walked in.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Alcohol can be dangerous.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He's got a reputation in these backwards parts,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Not sure: after many drinks you forget parts of the day before and people have to tell you afterwards.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and he could deliver a real beating.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Hangovers suck&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If there's a lady around, he'll just hold his cup.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;?Ladys drink from cups guys from bottles?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think, for the most part, he's just a mixed-up fool.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Alot of people mix alcohol.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9543" LastActivityDate="2015-08-11T13:13:23.253" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19892" PostTypeId="2" ParentId="19889" CreationDate="2015-08-11T13:20:00.177" Score="13" Body="&lt;p&gt;I take my chance with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jon Snow (from Game of Thrones)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Be sure to keep watch,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jon works on the Night's Watch&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;a tough guy just walked in.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He is quite the tough guy beating wildings and white walkers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He's got a reputation in these backwards parts,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reputation for being a bastard but also because he is a great fighter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and he could deliver a real beating.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He is a great fighter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If there's a lady around, he'll just hold his cup.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are not allowed to sleep with women in the watch, so he'll just hold his cup...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think, for the most part, he's just a mixed-up fool.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You could said that he is a mixed-up fool because he wants to help the wildings but he is from the night's watch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5359" LastActivityDate="2015-08-11T13:20:00.177" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19893" PostTypeId="2" ParentId="19870" CreationDate="2015-08-11T13:54:07.830" Score="-1" Body="&lt;p&gt;I would go with 1x2x7 = 14 parts. Reason is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If one gangster shoots the other one, he will take all 14 parts&lt;/li&gt;&#xA;&lt;li&gt;If the two armed gangsters agree, they take equal parts - 7 parts each&lt;/li&gt;&#xA;&lt;li&gt;If the two armed gangsters shoot each other, the remaining 7 split the gold equally - 2 parts each&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="15572" LastActivityDate="2015-08-11T13:54:07.830" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19895" PostTypeId="2" ParentId="18629" CreationDate="2015-08-11T14:03:54.523" Score="1" Body="&lt;p&gt;Just to add:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;girlfriend&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; is the English translation of the word &quot;amiga&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4385" LastActivityDate="2015-08-11T14:03:54.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19897" PostTypeId="2" ParentId="19870" CreationDate="2015-08-11T14:16:40.520" Score="80" Body="&lt;p&gt;&lt;strong&gt;18 pieces:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;3,7,9,14,16,18,23,24,25,30,31,32,38,40,42,47,49,56&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In 9 parts:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;{3,23,30} {7,49} {9,47} {14,42} {16,40} {18,38} {24,32} {25,31} {56}&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In 8 parts:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;{3,18,42} {7,56} {9,24,30} {14,49} {16,47} {23,40} {25,38} {31,32}&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In 7 parts:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;{3,7,24,38} {9,14,18,31} {16,56} {23,49} {25,47} {30,42} {32,40}&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;We need to be able to split it into 9 parts of 56, so it can't hurt to make 9 pieces of 56 and then split those further. Since we need to do better than 19 pieces, we can have at most 18 pieces. This means that most of our 56s are split into exactly two parts (we can have an extra piece for every 56 we don't split).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what will our pieces be mod 9? We must be able to group them into 7 or 8 groups that sum to 0, or make 9 pairs (mostly) that sum to 2. If we have a 2, we can group it with a 7 to make 0. Then we pair the 7 with a 4, and repeat with a 5 and 6. If we do this 3 times, we can use the 6s to make another 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make 63s and 72s, we should repeat at an interval of 9, so that we can swap the pairs to add 9 (e.g. &lt;code&gt;{7,56} {16,47} {25,38} -&amp;gt; {16,56} {25,47}&lt;/code&gt;). The other two numbers that are swapped out need an extra 27.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we start with a &lt;code&gt;56&lt;/code&gt; (freeing up one cut) and see what numbers we need.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;    56-&gt; 7-&gt;49-&gt;14-&gt;42&#xA; 9-&gt;47-&gt;16-&gt;40-&gt;23-&gt;33&#xA;18-&gt;38-&gt;25-&gt;31-&gt;32-&gt;24&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So far, this makes 6 63s with &lt;code&gt;9 18 24 33 42&lt;/code&gt; left over. We also need to make a 27 and a 72 out of &lt;code&gt;24 33 42&lt;/code&gt;. If we split &lt;code&gt;33&lt;/code&gt; into &lt;code&gt;3&lt;/code&gt; and &lt;code&gt;30&lt;/code&gt;, both of these are resolved. This results in the final answer with 18 pieces.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-08-11T14:44:46.447" LastActivityDate="2015-08-11T14:44:46.447" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="19898" PostTypeId="2" ParentId="19889" CreationDate="2015-08-11T14:16:43.443" Score="17" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; strap&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Be sure to keep watch,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A watch strap &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;a tough guy just walked in.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As in a strapping lad&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He's got a reputation in these backwards parts,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; strap is parts backwards&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and he could deliver a real beating.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To give someone the strap&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If there's a lady around, he'll just hold his cup.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A bra strap, though possibly also works with jockstrap &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think, for the most part, he's just a mixed-up fool.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Anagram of sprat, though prat is more common these days, hence the &quot;most part&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditDate="2015-08-12T14:43:32.633" LastActivityDate="2015-08-12T14:43:32.633" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="19899" PostTypeId="2" ParentId="19889" CreationDate="2015-08-11T14:23:42.417" Score="1" Body="&lt;p&gt;I'm going to take a guess at:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Caitlyn Jenner, formerly known as Bruce Jenner&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Be sure to keep watch,&#xA;a tough guy just walked in.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; College football player, and Olympic athlete&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He's got a reputation in these backwards parts,&#xA;and he could deliver a real beating.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Married a Kardashian, the surname being famous for their &quot;backwards parts&quot;. Also won Gold in the Decathlon at the Olympics.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If there's a lady around, he'll just hold his cup.&#xA;I think, for the most part, he's just a mixed-up fool.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Now he is the lady around, with his own &quot;cups&quot;. Mixed-up in that he &quot;&lt;em&gt;dealt with gender dysphoria since youth&lt;/em&gt;&quot;, hence the gender transition.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4659" LastActivityDate="2015-08-11T14:23:42.417" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19900" PostTypeId="2" ParentId="19870" CreationDate="2015-08-11T14:35:31.610" Score="4" Body="&lt;p&gt;I think its 9 pieces!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;scenario 1 = all 9 live, each gangster gets a piece&lt;br&gt;&#xA;scenario 2 = 8 live, 8 get a piece, the last piece pays for the funeral of the dead guy&lt;br&gt;&#xA;scenario 3 = 7 live, 7 get a piece, the remaining two pieces pays for the dead guys funerals  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;simple! :)&lt;/p&gt;&#xA;" OwnerUserId="15576" LastEditorUserId="15576" LastEditDate="2015-08-11T14:40:29.653" LastActivityDate="2015-08-11T14:40:29.653" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19901" PostTypeId="1" AcceptedAnswerId="20046" CreationDate="2015-08-11T15:13:29.623" Score="7" ViewCount="954" Body="&lt;p&gt;You wake up in a room. In front of you, there is some paper:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It's the day, not the year or the month! It's very useful to abecedarians. 639407344. But never forget the time, for its another key.&lt;br&gt;&lt;br&gt;Jsz amxroc tt ioatblq gstk A.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are also 26 doors, labeled A through to Z. One door leads to freedom while the others lead to a pit of acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which door should you go through?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He jumped on the number! UNIX!!!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The word 'abecedarian' is used to describe a person that is learning the alphabet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are no clues as to which cipher is used, so you've got to work that out from the keys. If the time is a key, then that is presumably a number. And the used cipher is similar to a Vigenere cipher. Anything come to mind, cryptologists?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 4:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The only important part of the word 'abecedarian' is the meaning. A person who is learning the &lt;strong&gt;alphabet&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9377" LastEditorDisplayName="user9377" LastEditDate="2015-08-14T13:28:57.083" LastActivityDate="2015-08-14T13:34:02.873" Title="Which door leads to freedom?" Tags="|cipher|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19902" PostTypeId="1" AcceptedAnswerId="19904" CreationDate="2015-08-11T15:17:21.187" Score="2" ViewCount="409" Body="&lt;p&gt;I've got the biggest nose but I might be lying.&lt;br&gt;&lt;br&gt; PS: This is my first attempt at riddles so be nice. Also, I've made this riddle myself so let's see how many &quot;travelers&quot; we have out there. :) &lt;br&gt;&lt;br&gt; The &quot;travelers&quot; clue would have been enough for the keen eye but for those who think it's too broad, &quot;It's a place to visit&quot; (but I have to tell you that takes the fun out of it)&lt;/p&gt;&#xA;" OwnerUserId="15581" LastEditorUserId="5165" LastEditDate="2015-08-11T17:47:34.790" LastActivityDate="2018-04-14T22:09:06.947" Title="What is it? What am I?" Tags="|riddle|logical-deduction|geography|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19903" PostTypeId="2" ParentId="19902" CreationDate="2015-08-11T15:21:45.540" Score="0" Body="&lt;p&gt;Are you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a head on &lt;a href=&quot;https://en.wikipedia.org/wiki/Easter_Island&quot; rel=&quot;nofollow&quot;&gt;Easter Island&lt;/a&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4385" LastActivityDate="2015-08-11T15:21:45.540" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19904" PostTypeId="2" ParentId="19902" CreationDate="2015-08-11T15:27:02.507" Score="5" Body="&lt;p&gt;Are you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the Great Sphinx of Giza?&#xA; It is the largest monolithic statue in the world so it should have the largest nose, but it is also missing its nose, so it could be lying&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4591" LastActivityDate="2015-08-11T15:27:02.507" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19905" PostTypeId="2" ParentId="19870" CreationDate="2015-08-11T15:36:41.627" Score="14" Body="&lt;p&gt;They should divide the bar into 9 pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the 2 gangsters kill each other their 2 pieces should go to their families. Respect is important in the mob world.&lt;/p&gt;&#xA;" OwnerUserId="9204" LastActivityDate="2015-08-11T15:36:41.627" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19907" PostTypeId="2" ParentId="14770" CreationDate="2015-08-11T16:50:50.403" Score="6" Body="&lt;p&gt;Are you a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; tape measure?&lt;br&gt;&#xA; - it has numbers all along its length; its highest number shows its own length (&quot;I number myself&quot;)&lt;br&gt;&#xA; - a tape measure is worth as much as its length -- not very useful when trying to measure things longer than itself&lt;br&gt;&#xA; - when you release it so that it retracts, it &quot;crackles and thunders&quot; and finally comes to rest&lt;br&gt;&#xA; - the color of a tape measure is typically yellow; the four colors may be a reference to the Flag of the Races (&lt;a href=&quot;https://en.wikipedia.org/wiki/Color_terminology_for_race#Flag_of_the_Races&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Color_terminology_for_race#Flag_of_the_Races&lt;/a&gt;)&lt;br&gt;&#xA; - it's wound up inside its case&lt;br&gt;&#xA; - it has no &quot;limbs&quot; - it is just a strip of material&lt;br&gt;&#xA; - it is strong while it's &quot;young&quot; in other words while it is only partially extended -- perhaps seven or eight feet; the more length you extend, the weaker it gets until it buckles under its own weight -- &quot;bends and jams&quot;&lt;br&gt;&#xA; - it comes in a variety of lengths  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4385" LastEditorUserId="4385" LastEditDate="2015-08-11T17:24:24.180" LastActivityDate="2015-08-11T17:24:24.180" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19908" PostTypeId="1" AcceptedAnswerId="20001" CreationDate="2015-08-11T16:54:04.837" Score="11" ViewCount="263" Body="&lt;p&gt;Woe unto the people hiding their secrets in the source code. &lt;br&gt; &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Immediately upon taking over the company, the Information Technology people exactingly went over the source code.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What saying is hidden in this sentence?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint the First:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After manipulation, usage of homophones occurs, and the hidden saying is shorter than this sentence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint the Second:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Starting with words which seem forced into the sentence is probably a strong technique. The final phrase is 5 words long.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13916" LastEditorUserId="13916" LastEditDate="2015-08-13T13:37:54.693" LastActivityDate="2015-08-13T14:27:58.350" Title="Hail Our Information Technology Overlords" Tags="|wordplay|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19909" PostTypeId="2" ParentId="19809" CreationDate="2015-08-11T16:54:29.707" Score="1" Body="&lt;p&gt;When someone says they have at least one blue dot: From everyone's PoV, there are two possibilities re: how many blue dots they all have. When someone reduces it to 1, it means (s)he's figured out his/her dot color. They can also make guesses about others' PoV, guess their guesses... If someone sees x blue dots and assumes he's blue-dotted too, (s)he has to assume (s)he knows how many blue dots someone else sees, but that that person might think there are two possibilities re: how many blue dots they all have. Same for when the first person assumes (s)he has a red dot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be represented with a tree. A non-trivial comment would imply at least one is red/blue at worst, so if no blue-dotted people assume they're blue-dotted, the number of seen/existing blue dots in the tree will fall to 0 too early, which they can't assume. A blue-dotted person will eventually figure out (s)he's one, and since it doesn't matter who it is, it would be every blue-dotted person. If the red-dotted ones survive, they'll realize that's because they're red-dotted and they'll die too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the stranger includes 0 but excludes some other numbers, it eliminates them and the numbers less than them, including 0. The reasoning will be similar to the classic case above.&lt;/p&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="14016" LastEditDate="2015-08-12T07:53:59.573" LastActivityDate="2015-08-12T07:53:59.573" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19910" PostTypeId="1" AcceptedAnswerId="19927" CreationDate="2015-08-11T18:23:10.367" Score="9" ViewCount="836" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Oh woe is he! How cruel is fate!&lt;br&gt;&#xA;  He pouts and mopes in this hour late.&lt;br&gt;&#xA;  The subject of some cruel god's whim;&lt;br&gt;&#xA;  Death the gods have in store for him.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Suddenly a beast of legend!&lt;br&gt;&#xA;  Our hero's sentence swift deadened.&lt;br&gt;&#xA;  As it roars, he recovers. Run!&lt;br&gt;&#xA;  Before his hide, as well, is done.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Fleeing there, to the city pale;&lt;br&gt;&#xA;  News for the king, no time to fail.&lt;br&gt;&#xA;  With men and arms they slay the beast.&lt;br&gt;&#xA;  Huzzah! Huzzah! Now time to feast.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;What is this feeling now unleashed?&lt;br&gt;&#xA;  Some kind of bond 'twixt him and beast.&lt;br&gt;&#xA;  He feels altered, not quite the same.&lt;br&gt;&#xA;  The men surround him, call him names.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;All at once, the skies asunder.&lt;br&gt;&#xA;  A booming voice cracks like thunder.&lt;br&gt;&#xA;  Illusion? No: a beckoning;&lt;br&gt;&#xA;  to mountain high 'ere reckoning.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;From here does his adventure start;&lt;br&gt;&#xA;  Wisdom, cunning,and strength in part.&lt;br&gt;&#xA;  Though truly only one of those&lt;br&gt;&#xA;  This hero of our story chose.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can you tell what this ode is describing? Who is the hero? The accepted answer will, as always, contain explanations of the clues in the story. This is a decently easy riddle regarding a fairly well-known hero, but I've marked it trivia since it is recounting a story that some may not be familiar with. &lt;/p&gt;&#xA;" OwnerUserId="13970" LastActivityDate="2015-08-11T22:45:26.427" Title="A Tale of a Hero" Tags="|riddle|story|knowledge|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19911" PostTypeId="2" ParentId="18281" CreationDate="2015-08-11T18:23:59.403" Score="4" Body="&lt;p&gt;It's been 19 days since the trapping. Unfortunately, you are now dead.&lt;/p&gt;&#xA;&lt;h2&gt;Starting with decoding the hints:&lt;/h2&gt;&#xA;&lt;p&gt;If I were a box shaped animal I'd be missing most of my hide. Find your family and scramble the eggs to find the missing sides.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The equation has 4 missing sections in the form of 4 lines - 4 lines can be made into a box. It's also missing all 4 numbers, eg: most of my hide. Finding your family refers to some sort of &amp;quot;family&amp;quot;, and scrambling the eggs to find the missing sides means that if you scramble the eggs, you'll get the missing numbers (missing sides, missing hide). By that logic, the eggs are letters, and the scrambling them will somehow gives you the eggs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The family tree has 27 people. If you include yourself, then your immediate family consists of 6 people. The &amp;quot;bonds of my family&amp;quot; only count your immediate family!&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; You are a human. There are 26 letters in the alphabet. 26 + you = 27. By the same logic, if you remove yourself from the immediate family, then you have 5 left. Bonds of my family is a reference to the riddle, and this is supposed to tell you that you're looking for a total of 5 letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Continuing to ponder, you scratch your head. Hm... this is harder than I thought it'd be. I'm pretty sure the bonds have nothing to do with the periodic table though.... Which means it has to have a hidden meaning! But what could it be?&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Self explanatory. Simply states that the puzzle has nothing to do with the periodic table.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I've deduced that I, as a person, am definitely extremely different than those in my family. Your stomach grumbles with hunger as you sit down and try to figure out the rest of the problem...&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Used to help decode the earlier hint regarding 27 and 6. If you haven't figured out that you're supposed to subtract 1, this is the hint to do it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Hmm... Definitely nothing to do with Astrology... you grumble, sitting on the cold cell floor.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Also self explanatory.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You think to yourself, How am I not out of this mess already? Frustrated, you bash your head against the wall. Eureka! There are exactly three parts to this puzzle! I'm pretty sure the riddles get me letters that help me get the missing numbers on the label, which would help me solve the equation! Now if only I could actually complete the riddles...&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; This is also self explanatory. Simply put: Solve the riddle to get letters to find the missing numbers on the label, to solve the equation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;Solving the riddle:&lt;/h2&gt;&#xA;&lt;p&gt;Nothing is more important to me,&#xA;than the bonds of my family.&#xA;Each piece below is a key from them to thee,&#xA;spend some time and you just might get free.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; First 2 lines show that there is some sort of relation coming up (Bonds and family). Each piece below is a reference the the paragraphs below, and the last line is flavor.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I'm all natural, all the time -&#xA;Euler used me but I feel fine.&#xA;I'm the find in many ways,&#xA;a unique treasure who likes Swiss cheese.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &amp;quot;E&amp;quot; - Used in Ln (Log base e, AKA Natural log - it's also a constant, hence &amp;quot;all the time&amp;quot;). Euler used the constant. Also, Euler was Swiss.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Irrelevant information influences influential individuals.&#xA;I infrequently integrate idioms in ideas.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &amp;quot;I&amp;quot; - Look at all the I's. Just look at them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You get rid of me several times a day,&#xA;but I'll come back after a slight delay.&#xA;Those bound to earth already see,&#xA;should you ever stop relieving me,&#xA;It's a sign you've been released.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &amp;quot;P&amp;quot; - &amp;quot;Those bound to earth&amp;quot; are living people. Living people have to go pee! Several times a day! And they always have to go again eventually! Dead people don't pee, however, so &amp;quot;those who stop relieving [themselves of pee]&amp;quot; have been &amp;quot;released&amp;quot; (opposite of bound).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I'm always moving,&#xA;with no need to stop.&#xA;I've no place to be going,&#xA;but it doesn't matter - I'm first!&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &amp;quot;P&amp;quot; - Always moving, doesn't stop = perpetual motion. &amp;quot;I'm first&amp;quot; designates the first letter of &amp;quot;Perpetual motion&amp;quot;. Note: this is a weak deduction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I'd tell you who I am if you could catch me,&#xA;but I'd slither away before you could see me.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &amp;quot;S&amp;quot; - &amp;quot;Tell you who I am if you could catch me&amp;quot; is someone who is sneaky. Slither away references a snake. Note: this is a weak deduction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;To summarize: My first brother is always the same, the next is narcissistic.&#xA;Another brother moves like a river, but his twin is impossible.&#xA;My sister can be quite quiet, but when she's angry you don't want to hear her hiss.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The summary helps to confirm the letters. First brother is always the same - Eulers &lt;strong&gt;constant&lt;/strong&gt;. Next brother is narcissistic - all about &amp;quot;I&amp;quot;. Third brother (the first &amp;quot;P&amp;quot;) moves like a river (Pee is a liquid and it moves like water/river). Perpetual motion is impossible, and it's a twin because both those paragraphs have the answer of &amp;quot;P&amp;quot;. Sister being quite is a reference to &amp;quot;shhhh&amp;quot;, and the hissing also consists of an &amp;quot;S&amp;quot; sound.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;P.S: I've left you a trace of my family. Good luck - Evil Kidnapper Minion (A big fan)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Trace of my family is the label.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;Step 2: The missing values on the label&lt;/h2&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The label doesn't have a brand or anything on it, but you now have the letters E,I,P,P,and S. If you &amp;quot;Scramble the eggs&amp;quot; (Anagram alert), you'll see that you can spell a popular pop brand, Pepsi, with the letters. Googling &amp;quot;Pepsi 355ml nutritional facts&amp;quot; will get you the missing numbers, 150 and 38.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Using those values to solve the equation, you get:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 8 x 50 + 1 - 3 or 50 x 8 + 1 - 3 (Both of which are valid codes to open the door).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Note:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; It's been pointed out to me that the hint with the 27 and 6 may have been too confusing and resulted in puzzlers being stuck on the hint. I apologize if that caused you to be stuck in any way/shape/form. I'll work on my hint giving skills.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8999" LastEditorUserId="76863" LastEditDate="2022-01-11T22:09:39.273" LastActivityDate="2022-01-11T22:09:39.273" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="19912" PostTypeId="2" ParentId="19910" CreationDate="2015-08-11T18:39:35.503" Score="2" Body="&lt;p&gt;The hero is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bellerophon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Oh woe is he! How cruel is fate!&#xA;He pouts and mopes in this hour late.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After his failed attempt to climb Mount Olympus, he spent the rest of his life a blinded hermit, wandering on the Plain of Wandering (original, I know). &quot;This hour late&quot; refers to the twilight of his life. His fate is cruel because he was a massive hero reduced to nothing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The subject of some cruel god's whim;&lt;br&gt;&#xA;Death the gods have in store for him.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bellerophon was sent by King Proteus of Tyrins to King Iobates of Lycia to die. Iobates feared the wrath of the gods for breaking the Greek laws of hospitality which forbade him from killing a guest, so he sent Bellerophon on an impossible quest: defeat the Chimera. After defeating the Chimera, Bellerophon did not receive credit from the king. Afterwards, he was sent on even more quests, most of which were intended to kill him.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Suddenly a beast of legend!&lt;br&gt;&#xA;Our hero's sentence swift deadened.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is the Chimera, a beast of legend that breathes fire. He managed to defeat the Chimera by putting a block of lead in its throat; the Chimera's fire-breath melted the lead and then it solidified, blocking the passage of air.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As it roars, he recovers. Run!&lt;br&gt;&#xA;Before his hide, as well, is done.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When he first tried to attack the Chimera, he couldn't get close enough to damage it, even riding on top of Pegasus. The Chimera's fire-breath will literally cook Bellerophon's hide (and make it well-done!) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fleeing there, to the city pale;&lt;br&gt;&#xA;News for the king, no time to fail.&lt;br&gt;&#xA;With men and arms they slay the beast.&lt;br&gt;&#xA;Huzzah! Huzzah! Now time to feast.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After defeating the Chimera, Bellerophon went to King Iobates with the news.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is this feeling now unleashed?&lt;br&gt;&#xA;Some kind of bond 'twixt him and beast.&lt;br&gt;&#xA;He feels altered, not quite the same.&lt;br&gt;&#xA;The men surround him, call him names.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bellerophon tamed Pegasus, and the two created a bond, as rider and  noble steed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All at once, the skies asunder.&#xA;A booming voice cracks like thunder.&#xA;Illusion? No: a beckoning;&#xA;to mountain high 'ere reckoning.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bellerophon believed that his fame gave him cause to ascend Mount Olympus. He believed that Zeus called to him, beckoning to ascend to the top of the mountain, where only the gods and their servants lived. Zeus sent a horsefly to bite Pegasus, causing Bellerophon to fall down and be injured for life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From here does his adventure start;&#xA;Wisdom, cunning,and strength in part.&#xA;Though truly only one of those&#xA;This hero of our story chose.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bellerophon showed wisdom and cunning in his ability to tame Pegasus, defeat the Chimera, and vanquish the Solymi and Amazons (he dropped boulders on them from Pegasus). He showed strength when he prevented an ambush by the pirate Cheirmarrhus and defeating palace guards sent against him. However, he abandoned wisdom and cunning when he attempted to ascend Mount Olympus.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13538" LastEditorUserId="13538" LastEditDate="2015-08-11T18:50:59.100" LastActivityDate="2015-08-11T18:50:59.100" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19913" PostTypeId="2" ParentId="18767" CreationDate="2015-08-11T18:54:12.050" Score="13" Body="&lt;p&gt;Who is missing?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The bishop&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm where you keep your most private possessions&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;My most private possessions are my thoughts, which I keep in my &lt;strong&gt;mind&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With a small adjustment, I'm always on the move&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The &lt;strong&gt;wind&lt;/strong&gt; is always on the move&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again, and I'm one of you, lacking aggression&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Of my &lt;strong&gt;kind&lt;/strong&gt;, or to be &lt;strong&gt;kind&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once more, at my end, my son has something to prove&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The son of a &lt;strong&gt;king&lt;/strong&gt; would have something to prove&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you gut him, you may have done so for this&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;price&lt;/strong&gt; (see comments, thanks @Lampost42)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Change that, and a group of great beasts have you found&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A group of lions is a &lt;strong&gt;pride&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again, it's my versatile partner, my bliss&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A partner or a &lt;strong&gt;bride&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who protects me when orderly enemies are 'round&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;of a king would be a &lt;strong&gt;queen&lt;/strong&gt; - a versatile and powerful chess piece&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Make a rhyme, if you've done a good job it will squeak&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;as in, squeaky &lt;strong&gt;clean&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sort them out and you'll find a great tool comes to view&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;clean is an anagram of &lt;strong&gt;lance&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This weapon, however, is not the word that you seek&lt;br&gt;&#xA;It's the user, of which, in my land I have two&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;and a lance is the weapon of a &lt;strong&gt;knight&lt;/strong&gt;, of which there are two per chess player&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Take his opposite, and several later you've got one&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The opposite of night is day, and several days is a &lt;strong&gt;week&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Change its first, and this riddle's answer you desire&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It took me a while to notice the italics around desire, but once I did it could be replaced with &lt;strong&gt;seek&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Conjugate incorrectly, and you've not got a verb, son&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The past of seek is sought, and a &lt;strong&gt;sook&lt;/strong&gt; is (apparently) a female crab&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But one close to this soldier, in his path, you'll retire.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You wouldn't want your king in the path of a &lt;strong&gt;rook&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again, recall the user of the weapon above&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;back to &lt;strong&gt;knight&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When his opposite appears, it is given this name&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the appearance of day is &lt;strong&gt;dawn&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One fix and you've found my minions least loved&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the least appreciated piece in chess is a &lt;strong&gt;pawn&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But without them I'd be at a loss in this game&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;and although the edit says this doesn't yield a word, I think it could yield &lt;strong&gt;chess&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14475" LastEditorUserId="11475" LastEditDate="2015-08-18T20:51:29.807" LastActivityDate="2015-08-18T20:51:29.807" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19914" PostTypeId="1" AcceptedAnswerId="19922" CreationDate="2015-08-11T18:57:39.370" Score="4" ViewCount="6448" Body="&lt;p&gt;Before you stands a building with N-floors.  On each floor there is an open window.  You have two identical glass balls.  If you were to drop a ball out the window onto the ground below, it might or might not break.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your goal is to determine the highest floor for which dropping a ball out the window results in the ball &lt;em&gt;not&lt;/em&gt; breaking.  You can continue dropping the balls out the window as long as at least one of the balls remains intact.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:  You go to floor 1 and drop a ball.  It doesn't break so you go to floor 2 and try again.  This time it breaks so you can no longer use that ball. But you still have 1 ball left so you continue until that ball is broken.  By the time the second ball breaks, you should know the answer (if not then the strategy is invalid).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The question:&lt;/strong&gt;  What is the &lt;em&gt;most efficient&lt;/em&gt; strategy for determining said highest floor? For this question, the &lt;em&gt;most efficient&lt;/em&gt; strategy is the one which yields the smallest average number of total drops over all possible highest floors for the building.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example continued: If the building has 100 floors, the strategy described would require X + 1 drops where X = the highest floor's number.  Summing over all possible X we would get 5050 so the average is 50.5 drops.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note:  It is possible that a drop from floor 1 results in a broken ball.  It is also possible that a drop from the top floor results in an unbroken ball.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An answer will only get a green check if it is both correct and clearly explained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Special aside:  This is my first time posting a puzzle here.  I tried to be clear and concise but please let me know about anything I need to clarify.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is this question a duplicate?&lt;/strong&gt;  There are certainly similarities between this and the puzzle proposed here: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/5006/dinosaur-egg-drop&quot;&gt;Dinosaur egg drop&lt;/a&gt;.  However, that puzzle specifies the number of floors and total number of drops.  My question asks to generalize the number of floors and does not specify the number of drops, only that you can continue as long as at least one glass ball is intact.  Additionally, I do not see the correct answer to this puzzle on that puzzle.  Some of the answers and reasoning there are definitely similar to what I am looking for but none of those would get the green check from me on this puzzle.&lt;/p&gt;&#xA;" OwnerUserId="14466" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2016-12-29T02:21:09.230" Title="Will the glass ball break?" Tags="|mathematics|" AnswerCount="8" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="19916" PostTypeId="1" AcceptedAnswerId="19918" CreationDate="2015-08-11T19:43:39.703" Score="23" ViewCount="3137" Body="&lt;p&gt;If a phrase has a certain property, I call it an Uncopyrightable Word™.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In each of the sentences below, one word is an Uncopyrightable Word™ and one is not.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sentences are for context only, and do not affect whether a word is an Uncopyrightable Word™.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;The forecast says it's going to (&lt;strong&gt;SNOW&lt;/strong&gt; / &lt;strong&gt;RAIN&lt;/strong&gt;) tomorrow.&lt;/li&gt;&#xA;  &lt;li&gt;Your explanations always tend towards (&lt;strong&gt;HYPERBOLE&lt;/strong&gt; / &lt;strong&gt;EXAGGERATION&lt;/strong&gt;).&lt;/li&gt;&#xA;  &lt;li&gt;The ideal political system is a (&lt;strong&gt;DEMOCRACY&lt;/strong&gt; / &lt;strong&gt;REPUBLIC&lt;/strong&gt;).&lt;/li&gt;&#xA;  &lt;li&gt;The answer to your question is (&lt;strong&gt;YES&lt;/strong&gt; / &lt;strong&gt;NO&lt;/strong&gt;).&lt;/li&gt;&#xA;  &lt;li&gt;His (&lt;strong&gt;NEIGHBOURLY&lt;/strong&gt; / &lt;strong&gt;FRIENDLY&lt;/strong&gt;) advice was well-received.&lt;/li&gt;&#xA;  &lt;li&gt;I (&lt;strong&gt;CERTAINLY&lt;/strong&gt; / &lt;strong&gt;DEFINITELY&lt;/strong&gt;) don't think this puzzle is difficult.&lt;/li&gt;&#xA;  &lt;li&gt;The (&lt;strong&gt;DUKE&lt;/strong&gt; / &lt;strong&gt;BARON&lt;/strong&gt;) had very good manners.&lt;/li&gt;&#xA;  &lt;li&gt;The team worked with great (&lt;strong&gt;SYNERGY&lt;/strong&gt; / &lt;strong&gt;RECIPROCITY&lt;/strong&gt;).&lt;/li&gt;&#xA;  &lt;li&gt;The winter (&lt;strong&gt;FESTIVAL&lt;/strong&gt; / &lt;strong&gt;CELEBRATION&lt;/strong&gt;) is one of the high points of the year.&lt;/li&gt;&#xA;  &lt;li&gt;My favourite composer is (&lt;strong&gt;BACH&lt;/strong&gt; / &lt;strong&gt;BEETHOVEN&lt;/strong&gt;).&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What makes a word an Uncopyrightable Word™?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="29050" LastEditDate="2016-09-23T11:37:37.823" LastActivityDate="2016-09-23T11:37:37.823" Title="What is an Uncopyrightable Word?™" Tags="|pattern|english|word-property|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="19917" PostTypeId="2" ParentId="19910" CreationDate="2015-08-11T19:47:35.003" Score="3" Body="&lt;p&gt;Could the hero be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bilbo Baggins&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Oh woe is he! How cruel is fate!&lt;br&gt;&#xA;He pouts and mopes in this hour late.&lt;br&gt;&#xA;The subject of some cruel god's whim;&lt;br&gt;&#xA;Death the gods have in store for him.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bilbo Baggins resents being chosen by Gandalf as a member of his crew. He wants nothing to do with adventure, and would much rather just enjoy a cup of tea and a book. He believes that the gods really want him dead.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Suddenly a beast of legend!&lt;br&gt;&#xA;Our hero's sentence swift deadened.&lt;br&gt;&#xA;As it roars, he recovers. Run!&lt;br&gt;&#xA;Before his hide, as well, is done.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He meets Smaug, the dragon. Bilbo takes a golden chalice, enraging Smaugh. He runs away (as Smaug roars) to avoid his hide being burnt to a crisp (or a nice well-done).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fleeing there, to the city pale;&lt;br&gt;&#xA;News for the king, no time to fail.&lt;br&gt;&#xA;With men and arms they slay the beast.&lt;br&gt;&#xA;Huzzah! Huzzah! Now time to feast.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He then flees to Lake Town to let Thorin Oakenshield know of the dragon's attack and his weak spot. The human archer Bard succeeds in killing Smaug; when the men and elves want a share of the treasure, the dwarves refuse. The humans and elves besiege the mountain. However, the goblins and Wargs march on the mountain, so the dwarves team up with the humans and elves to defend their treasure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is this feeling now unleashed?&lt;br&gt;&#xA;Some kind of bond 'twixt him and beast.&lt;br&gt;&#xA;He feels altered, not quite the same.&lt;br&gt;&#xA;The men surround him, call him names.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When Bilbo returns to the Shire, he is outcast by the other hobbits. He traded the respect of his peers for the wisdom of his travels, and he now prefers to talk to wizards&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All at once, the skies asunder.&lt;br&gt;&#xA;A booming voice cracks like thunder.&lt;br&gt;&#xA;Illusion? No: a beckoning;&lt;br&gt;&#xA;to mountain high 'ere reckoning.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Unsure; will come back to this later.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From here does his adventure start;&lt;br&gt;&#xA;Wisdom, cunning,and strength in part.&lt;br&gt;&#xA;Though truly only one of those&lt;br&gt;&#xA;This hero of our story chose.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bilbo is wise, shown by his attempts to resolve the tension between the dwarves and the humans/elves. He is cunning, shown by his riddle-solving abilities when encountering Gollum and his ability to get the party away from the trolls. He is strong (in part) in his ability to rescue his party from the goblins. However, he chose wisdom in the end, when he returned to Hobbiton as a wise old hobbit-man-thing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13538" LastEditorUserId="13538" LastEditDate="2015-08-11T20:07:16.813" LastActivityDate="2015-08-11T20:07:16.813" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19918" PostTypeId="2" ParentId="19916" CreationDate="2015-08-11T19:49:03.223" Score="25" Body="&lt;p&gt;An Uncopyrightable Word™ is one that...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; consists only of letters in the word UNCOPYRIGHTABLE.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the Uncopyrightable Word™s are...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;RAIN, HYPERBOLE, REPUBLIC, NO, NEIGHBOURLY, CERTAINLY, BARON, RECIPROCITY, CELEBRATION, and BACH.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="-1" LastEditDate="2015-08-12T01:16:31.110" LastActivityDate="2015-08-12T01:16:31.110" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="19919" PostTypeId="2" ParentId="19916" CreationDate="2015-08-11T19:50:58.927" Score="11" Body="&lt;p&gt;An uncopyrightable word is a word which &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; can be spelled using only letters in 'uncopyrightable', so&lt;br&gt;&#xA; RAIN, HYPERBOLE, REPUBLIC, NO, NEIGHBOURLY, CERTAINLY, RECIPROCITY, CELEBRATION, and BACH.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13101" LastActivityDate="2015-08-11T19:50:58.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19920" PostTypeId="1" AcceptedAnswerId="19951" CreationDate="2015-08-11T19:56:40.387" Score="12" ViewCount="1147" Body="&lt;p&gt;&lt;em&gt;If you're interested in starting the treasure hunt from the very beginning, check out the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16551/treasure-hunt-round-the-world&quot;&gt;prologue&lt;/a&gt;!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;&lt;em&gt;Note:&lt;/em&gt;&lt;/strong&gt; I fear I made these cities too obscure with my first riddle. The four unsolved cities have addenda added to them that should make them significantly easier to solve.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Upon your arrival in Cambodia, you quickly notice that the entire airport is partitioned off, and tents line the gates and hallways. &lt;em&gt;Looks like another interlude&lt;/em&gt;, you think to yourself. You notice that the number of tents here seems much smaller than the previous interlude, which is reassuring! You're getting to the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Noticing that the tents have name tags, you quickly find your own and enter it. Inside, you find bedding, a picture of your loved ones (weird), and an envelope. You know the drill. Time to open the next envelope.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dear &lt;strong&gt;&lt;em&gt;[insert username here]&lt;/em&gt;&lt;/strong&gt;,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Welcome to Cambodia! Since flights aren't departing here for the next 8 hours, you can feel free to grab some sleep. Or, you can choose to solve the simple, harmless puzzle on the back of this card. It's really up to you. At 0900, a concierge will present you with a ticket and you'll hop on a plane to your next destination! Oh, how wonderful. I'm getting the jitters.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Sleep tight! I'd say 'don't let the bedbugs bite', but I fear that ship might have sailed when we bought these tents in third-party bulk from offshore. Oh, well. I don't have to sleep in them!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Yours truly,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Bailey M&lt;br&gt;&#xA;  Puzzlemaster for the Treasure Hunt 'Round the World&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;P.S. The answer to the brainteaser will once again be a question. You will not receive your ticket tomorrow morning unless you are able to present the concierge with the aforementioned question. Good luck!&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Luckily, you slept on the plane. Time to solve a puzzle. Flipping over the card, you read:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I planned a couple of trips awhile ago, but couldn't decide which one to take, so I ended up taking neither of them and feeling really bad about it. You can find details on my trips below. They were really cool, too!&lt;/p&gt;&#xA;  &#xA;  &lt;hr&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Trip 1&lt;/strong&gt;  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;Coming from a town of few, I just can't wait to see&lt;br&gt;&#xA;  people! People! People! In increasing quantity!&lt;br&gt;&#xA;  I'm staying where I finish since it feels so right to me.&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Trip 2&lt;/strong&gt;  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;When I planned out this vacation, if only I knew;&lt;br&gt;&#xA;  bigger and bigger my target countries grew.&lt;br&gt;&#xA;  When I finish this trip, I'm going home, thank you.&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;hr&gt;&#xA;  &#xA;  &lt;p&gt;I can't remember my destinations, but I wrote out why I thought they were cool...anyways, here were the cities I was considering visiting. Maybe you can make some sense out of them.&lt;/p&gt;&#xA;  &#xA;  &lt;hr&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Destinations&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;The Avesta once noted that this city flows;&lt;br&gt;&#xA;  As their regime, here the terrible chose.&lt;br&gt;&#xA;  Now reborn anew, with haste this city grows.&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;A city of non-zero nuclear threat;&lt;br&gt;&#xA;  Elena is priced in at ten, I would bet.&lt;br&gt;&#xA;  For dry desert farming, it's the best you can get.&lt;/em&gt;&lt;br&gt;&#xA;  &lt;strong&gt;&lt;em&gt;Through which both air and road transport passes,&lt;br&gt;&#xA;  Here announced the dawn of the era of the masses.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;A festival city, yet not from Dubai;&lt;br&gt;&#xA;  Living history looms here as tall as the sky,&lt;br&gt;&#xA;  and it's nearly the finest if you're looking to buy.&lt;/em&gt;&lt;br&gt;&#xA;  &lt;strong&gt;&lt;em&gt;The gateway of the north this came to be&lt;br&gt;&#xA;  In the land that stretches from sea to sea.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;If you need an import, I'd hope there's no ice,&lt;br&gt;&#xA;  or enough for a truck to make one-way trips twice.&lt;br&gt;&#xA;  (Though Yamal is here in a pinch, which is nice.)&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;A birdlike creature, the foe of all snakes,&lt;br&gt;&#xA;  on this city's flag an appearance makes.&lt;br&gt;&#xA;  They have a whole castle for one man-made lake!&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;Home of the highlanders, a capital investment;&lt;br&gt;&#xA;  Most recently subjected to corruption's assessment.&lt;br&gt;&#xA;  Their country dissolved as a laughable contestant.&lt;/em&gt;&lt;br&gt;&#xA;  &lt;strong&gt;&lt;em&gt;Stealing the title from the first month's river,&lt;br&gt;&#xA;  On Don Bosco's dream town, a new Marx delivered.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;When flying, you'll land here, amidst ice and rocks.&lt;br&gt;&#xA;  When boating, the whole fjord is basically docks.&lt;br&gt;&#xA;  When driving, be careful you don't hit an ox.&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;Where once stood a church of good Lutheran grace,&lt;br&gt;&#xA;  a towering hotel now looms in its place.&lt;br&gt;&#xA;  (Some words here mean not what they say on their face.)&lt;/em&gt;&lt;br&gt;&#xA;  &lt;strong&gt;&lt;em&gt;A unique reveal it used to be heard,&lt;br&gt;&#xA;  it looks like it sounds like the foot of a bird.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;hr&gt;&#xA;  &#xA;  &lt;p&gt;P.S. I remembered that looking at a map with North America on the left and Eurasia on the right, none of my first trip went farther east than the westernmost point of my second trip. Don't know if that'll help, but it can't hurt, right? Anyways, good luck!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On the bottom of the page, scrawled in someone else's handwriting, you see:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;poaxrhsirzzsyhfhrsyprozg? It's a Vigenere, which shouldn't shock you at this point.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What question do you need to ask the concierge in the morning?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Stanza-Specific Hints:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Stanza 1:&lt;/strong&gt; Solved.&lt;br&gt;&#xA; &lt;strong&gt;Stanza 2:&lt;/strong&gt; Solved.&lt;br&gt;&#xA; &lt;strong&gt;Stanza 3:&lt;/strong&gt; Solved.&lt;br&gt;&#xA; &lt;strong&gt;Stanza 4:&lt;/strong&gt; Solved.&lt;br&gt;&#xA; &lt;strong&gt;Stanza 5:&lt;/strong&gt; Solved.&lt;br&gt;&#xA; &lt;strong&gt;Stanza 6:&lt;/strong&gt; Solved.&lt;br&gt;&#xA; &lt;strong&gt;Stanza 7:&lt;/strong&gt; Solved.&lt;br&gt;&#xA; &lt;strong&gt;Stanza 8:&lt;/strong&gt; &lt;strong&gt;&lt;em&gt;Cryptic crossword.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-09-14T02:43:14.203" Title="Treasure hunt 'round the world! (Interlude 2)" Tags="|enigmatic-puzzle|geography|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19921" PostTypeId="2" ParentId="19910" CreationDate="2015-08-11T19:57:58.697" Score="0" Body="&lt;p&gt;Thinking along the lines of infantile, the hero is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Sétanta, the name given to the young Cú Chulainn.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Oh woe is he! How cruel is fate! &lt;br&gt;&#xA;He pouts and mopes in this hour late.&lt;br&gt;&#xA;The subject of some cruel god's whim;&lt;br&gt;&#xA;Death the gods have in store for him.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It was prophesied that Sétanta's great deeds would bring him fame, but he would live a short life. The king of Ulster grieves for him learning this. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Suddenly a beast of legend!&lt;br&gt;&#xA;Our hero's sentence swift deadened.&lt;br&gt;&#xA;As it roars, he recovers. Run!&lt;br&gt;&#xA;Before his hide, as well, is done.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;As Sétanta is playing with some boys, Culann the smith forgets that he let out his hound to protect his house. He tries to flee from the hound. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fleeing there, to the city pale;&lt;br&gt;&#xA;News for the king, no time to fail.&lt;br&gt;&#xA;With men and arms they slay the beast.&lt;br&gt;&#xA;Huzzah! Huzzah! Now time to feast.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Before Sétanta joins the king for a feast, Sétanta kills the hound in self defence and Culann grieves for his dead hound. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is this feeling now unleashed?&lt;br&gt;&#xA;Some kind of bond 'twixt him and beast.&lt;br&gt;&#xA;He feels altered, not quite the same.&lt;br&gt;&#xA;The men surround him, call him names.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Sétanta then promises to be the hound's replacement, and he gets the name Cú Chulainn, the Hound of Culann.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All at once, the skies asunder.&lt;br&gt;&#xA;A booming voice cracks like thunder.&lt;br&gt;&#xA;Illusion? No: a beckoning;&lt;br&gt;&#xA;to mountain high 'ere reckoning.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Not sure right now. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From here does his adventure start;&lt;br&gt;&#xA;Wisdom, cunning,and strength in part.&lt;br&gt;&#xA;Though truly only one of those&lt;br&gt;&#xA;This hero of our story chose.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Sétanta followed the path of strength, effortlessly defeating his foes but died an early death. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Kinda flimsy, but it's what came to mind after NeedAName's comment.&lt;/p&gt;&#xA;" OwnerUserId="13489" LastEditorUserId="13489" LastEditDate="2015-08-11T20:34:53.803" LastActivityDate="2015-08-11T20:34:53.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19922" PostTypeId="2" ParentId="19914" CreationDate="2015-08-11T20:14:58.053" Score="5" Body="&lt;h3&gt;Strategy&lt;/h3&gt;&#xA;&lt;p&gt;My thoughts on the matter, (probably insufficient to earn the green check of approval).&lt;/p&gt;&#xA;&lt;p&gt;The strategy I would employ would be to drop it every &lt;span class=&quot;math-container&quot;&gt;$\sqrt{n}$&lt;/span&gt; floors and then starting from the floor above the last one it didn't break it at, drop it successively higher by one floor.&lt;/p&gt;&#xA;&lt;h3&gt;Why this strategy?&lt;/h3&gt;&#xA;&lt;p&gt;Well this means the searching for the location in both &amp;quot;halves&amp;quot; of the search are isomorphic, searching &lt;span class=&quot;math-container&quot;&gt;$\sqrt{n}$&lt;/span&gt; spots for the first break. You expect to find your spot in ~&lt;span class=&quot;math-container&quot;&gt;$\sqrt{n}+1$&lt;/span&gt; drops as the first search would take on average &lt;span class=&quot;math-container&quot;&gt;$\sqrt{n}/2 +1/2$&lt;/span&gt; drops and the second search would also take &lt;span class=&quot;math-container&quot;&gt;$\sqrt{n}/2 +1/2$&lt;/span&gt; drops.&lt;/p&gt;&#xA;&lt;p&gt;Well I might come back to improve my answer but that is far as I am going for now.&lt;/p&gt;&#xA;&lt;p&gt;Hypothesis: for &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; balls and &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; floors, ~ &lt;span class=&quot;math-container&quot;&gt;${n^{1/k}} + log(n) term$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;3 ball example with 100 floors: break it into segments of roughly &lt;span class=&quot;math-container&quot;&gt;$100^{1/3}$&lt;/span&gt; floors, so 4.64, then cluster these by groups of roughly 4.64 groups.This is 21.5 floors per cluster. So this would take on average lets say, 3 guesses to find the right super cluster, 3 guesses to find the fight group of 4.64, then 3 guesses to find the exact spot for an approximation of 9 guesses expect for 3 balls 100 floors. &lt;br&gt; (worth noting using the first ball for a drop in the middle then using groups of 7 on the remaining chunks also seems like it would have an expected of 9)&lt;/p&gt;&#xA;" OwnerUserId="13916" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-10-06T08:52:23.060" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19923" PostTypeId="1" AcceptedAnswerId="19924" CreationDate="2015-08-11T21:13:45.437" Score="22" ViewCount="1159" Body="&lt;p&gt;I found this odd list in an abandoned shack, tucked into the first few pages of the &lt;em&gt;Divine Comedy&lt;/em&gt;. Can you help me figure out what exactly this person was collecting?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The compendium of organs and bones&lt;/li&gt;&#xA;&lt;li&gt;The bloodletting of the North American lamb&lt;/li&gt;&#xA;&lt;li&gt;The animal both alive and dead&lt;/li&gt;&#xA;&lt;li&gt;The animal hungry for the ghost of food&lt;/li&gt;&#xA;&lt;li&gt;Lessons from the ancient animals&lt;/li&gt;&#xA;&lt;li&gt;Keep it simple, keep it simple, keep it simple&lt;/li&gt;&#xA;&lt;li&gt;Where Rome drew the line&lt;/li&gt;&#xA;&lt;li&gt;Euphoric music&lt;/li&gt;&#xA;&lt;li&gt;What the businessman wrote to save whoever he could&lt;/li&gt;&#xA;&lt;li&gt;The wishing well in the cave&lt;/li&gt;&#xA;&lt;li&gt;Choking forever on forbidden fruit!&lt;/li&gt;&#xA;&lt;li&gt;The hidden maze beneath Spain&lt;/li&gt;&#xA;&lt;li&gt;The weak point that Paris finally found&lt;/li&gt;&#xA;&lt;li&gt;The place of chains and flame with many shadows&lt;/li&gt;&#xA;&lt;li&gt;Never an unnecessary weapon on the wall!&lt;/li&gt;&#xA;&lt;li&gt;Where the spider revealed it could think&lt;/li&gt;&#xA;&lt;li&gt;The power of the machines doubles every odd year&lt;/li&gt;&#xA;&lt;li&gt;Evil in a jar!&lt;/li&gt;&#xA;&lt;li&gt;Child of Satan! Child of the herb!&lt;/li&gt;&#xA;&lt;li&gt;The three women of Heaven who answer to the hidden man&lt;/li&gt;&#xA;&lt;li&gt;The slave in his house of wood&lt;/li&gt;&#xA;&lt;li&gt;Everything will fall to pieces if it can&lt;/li&gt;&#xA;&lt;li&gt;Six hundred sextillion plus some&lt;/li&gt;&#xA;&lt;li&gt;The precious blunt instrument owned by a murderous doctor in training&lt;/li&gt;&#xA;&lt;li&gt;He lurks even now in the waters of the Valley&lt;/li&gt;&#xA;&lt;li&gt;The wild carrot!&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="10851" LastActivityDate="2015-08-19T05:23:16.737" Title="To have and to hold" Tags="|riddle|pattern|enigmatic-puzzle|language|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19924" PostTypeId="2" ParentId="19923" CreationDate="2015-08-11T21:22:36.733" Score="16" Body="&lt;h2&gt;Complete collaborated answer&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;These are all&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; common phrases including a possessive:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The compendium of organs and bones&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Gray%27s_Anatomy&quot; rel=&quot;nofollow&quot;&gt;Gray's Anatomy&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;The bloodletting of the North American lamb&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Battle_of_the_Little_Bighorn&quot; rel=&quot;nofollow&quot;&gt;Custer's Last Stand&lt;/a&gt; (from 'bloodletting'=Battle and 'North American lamb'=Little Bighorn&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;The animal both alive and dead&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Schr%C3%B6dinger%27s_Cat&quot; rel=&quot;nofollow&quot;&gt;Schrödinger's cat&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;The animal hungry for the ghost of food&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.simplypsychology.org/pavlov.html&quot; rel=&quot;nofollow&quot;&gt;Pavlov's dog&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;Lessons from the ancient animals&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Aesop%27s_Fables&quot; rel=&quot;nofollow&quot;&gt;Aesop's Fables&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;Keep it simple, keep it simple, keep it simple&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Occam%27s_razor&quot; rel=&quot;nofollow&quot;&gt;Occam's razor&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;Where Rome drew the line&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Hadrian%27s_Wall&quot; rel=&quot;nofollow&quot;&gt;Hadrian's wall&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;Euphoric music&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Symphony_No._9_%28Beethoven%29&quot; rel=&quot;nofollow&quot;&gt;Beethoven's 9th (Ode to Joy)&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;What the businessman wrote to save whoever he could&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Schindlerjuden#List_of_Schindlerjuden&quot; rel=&quot;nofollow&quot;&gt;Schindler's list&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;The wishing well in the cave&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Aladdin&quot; rel=&quot;nofollow&quot;&gt;Aladdin's lamp&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;Choking forever on forbidden fruit!&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Adam%27s_apple&quot; rel=&quot;nofollow&quot;&gt;Adam's apple&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;The hidden maze beneath Spain&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Pan%27s_Labyrinth&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Pan's Labyrinth&lt;/em&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;The weak point that Paris finally found&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Achilles%27_heel&quot; rel=&quot;nofollow&quot;&gt;Achilles's heel&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;The place of chains and flame with many shadows&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Allegory_of_the_Cave&quot; rel=&quot;nofollow&quot;&gt;Plato's cave&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;Never an unnecessary weapon on the wall!&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Chekhov%27s_gun&quot; rel=&quot;nofollow&quot;&gt;Chekhov's gun&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;Where the spider revealed it could think&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Charlotte%27s_Web&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Charlotte's Web&lt;/em&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;The power of the machines doubles every odd year&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Moore%27s_law&quot; rel=&quot;nofollow&quot;&gt;Moore's Law&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;Evil in a jar!&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Pandora%27s_box&quot; rel=&quot;nofollow&quot;&gt;Pandora's Box&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;Child of Satan! Child of the herb!&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Rosemary%27s_Baby&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Rosemary's Baby&lt;/em&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;The three women of Heaven who answer to the hidden man&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Charlie%27s_Angels&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Charlie's Angels&lt;/em&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;The slave in his house of wood&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Uncle_Tom%27s_Cabin&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Uncle Tom's Cabin&lt;/em&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;Everything will fall to pieces if it can&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Murphy%27s_law&quot; rel=&quot;nofollow&quot;&gt;Murphy's Law&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;Six hundred sextillion plus some&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Avogadro%27s_number&quot; rel=&quot;nofollow&quot;&gt;Avogadro's number&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;The precious blunt instrument owned by a murderous doctor in training&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Maxwell%27s_Silver_Hammer&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Maxwell's Silver Hammer&lt;/em&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;He lurks even now in the waters of the Valley&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Traveler%27s_diarrhea#Montezuma.27s_revenge&quot; rel=&quot;nofollow&quot;&gt;Montezuma's Revenge&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;The wild carrot!&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Daucus_carota&quot; rel=&quot;nofollow&quot;&gt;Queen Anne's Lace&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="12700" LastEditDate="2015-08-19T05:23:16.737" LastActivityDate="2015-08-19T05:23:16.737" CommentCount="0" CommunityOwnedDate="2015-08-11T21:25:08.493" ContentLicense="CC BY-SA 3.0" />
  <row Id="19926" PostTypeId="2" ParentId="19914" CreationDate="2015-08-11T21:58:03.637" Score="1" Body="&lt;p&gt;Number the floors $1$ to $N$. Let $E_{N,b}$ denote the expected number of drops the optimal strategy takes for an $N$-floor building, starting with $b$ glass balls. This means that $E_{1,b}=1$, and $E_{N,1}=\frac{1+2+\dots+(N-1)+N+N}{N+1}=\frac{N(N+3)}{2(N+1)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the first floor you drop from is floor $f$. The ball will break with probability $\frac{f}{N+1}$, in which case you will have $f-1$ floors to search with only one ball. The ball will be intact with probability $\frac{N+1-f}{N+1}$, wherein you will have $N-f$ floors to search with two balls. Since you want to choose the value of $f$ which minimizes the expected number of drops, it follows that&#xA;$$&#xA;E_{N,2}=1+\min_{1\le f\le N}\left(\frac{f}{N+1}\cdot E_{f-1,1}+\frac{N+1-f}{N+1}\cdot E_{N-f,2}\right)&#xA;$$&#xA;This gives a recursive formula which allows you to compute $E_{N,2}$, combined with the base cases in the first paragraph. I doubt there is a nice, closed form formula for all $N$.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-08-11T21:58:03.637" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19927" PostTypeId="2" ParentId="19910" CreationDate="2015-08-11T22:30:02.383" Score="6" Body="&lt;p&gt;The hero is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The player character (or Dragonborn) in Skyrim&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Oh woe is he! How cruel is fate!&lt;br&gt;&#xA;He pouts and mopes in this hour late.&lt;br&gt;&#xA;The subject of some cruel god's whim;&lt;br&gt;&#xA;Death the gods have in store for him.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; At the start of the game the hero is on their way to be executed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Suddenly a beast of legend!&lt;br&gt;&#xA;Our hero's sentence swift deadened.&lt;br&gt;&#xA;As it roars, he recovers. Run!&lt;br&gt;&#xA;Before his hide, as well, is done.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; But their execution is inturrupted by a dragon and the hero escapes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fleeing there, to the city pale;&lt;br&gt;&#xA;News for the king, no time to fail.&lt;br&gt;&#xA;With men and arms they slay the beast.&lt;br&gt;&#xA;Huzzah! Huzzah! Now time to feast.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The hero flees to Whiterun (city pale) to tell the Jarl (or king) of the dragon attack, and eventually slays a dragon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is this feeling now unleashed?&lt;br&gt;&#xA;Some kind of bond 'twixt him and beast.&lt;br&gt;&#xA;He feels altered, not quite the same.&lt;br&gt;&#xA;The men surround him, call him names.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The hero gains powers from killing the dragon, and becomes known as a Dragonborn&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All at once, the skies asunder.&lt;br&gt;&#xA;A booming voice cracks like thunder.&lt;br&gt;&#xA;Illusion? No: a beckoning;&lt;br&gt;&#xA;to mountain high 'ere reckoning.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The hero is summoned to a mountain to learn about the nature of their powers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From here does his adventure start;&lt;br&gt;&#xA;Wisdom, cunning,and strength in part.&lt;br&gt;&#xA;Though truly only one of those&lt;br&gt;&#xA;This hero of our story chose.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; And the game unfolds from there, with the player upgrading three basic attributes as they progress. At the start of the game the player chooses a blessing from the god of wisdom, cunning or strength.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13197" LastEditorUserId="13197" LastEditDate="2015-08-11T22:45:26.427" LastActivityDate="2015-08-11T22:45:26.427" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19928" PostTypeId="2" ParentId="19914" CreationDate="2015-08-11T22:52:21.307" Score="0" Body="&lt;p&gt;If the number of floors is of the form (K)(K+1)/2, the worst-case number of drops may be reduced to K marking the top floor of the building, the floor below that, the floor two down from that, three down from that, etc. until one reaches the base of the building, and trying marked floors starting with the lowest (marked) one until a ball breaks, then starting at the floor above the previous marked floor (or the ground floor if it broke on the first marked floor) and trying every floor until the ball breaks or all floors below the next marked one have been tried.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With six floors, for example, one would mark floors 6, 5, and 3, and then try those in order (3,5,6).  If the ball breaks at 3 (first drop), drops at 1 and 2 (two more) will reveal the breaking point.  If it breaks at 5, one more drop (at 4) will reveal the breaking point.  If the ball didn't break before 6, that will be the last drop required (whether the ball breaks or not).  With ten floors, one would mark 10, 9, 7, and 4.  With fifteen, 15, 14, 12, 9, and 5.  The more tests one does before breaking the first ball, the fewer one will need to do afterward to find the exact breaking point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although this approach minimizes the &lt;em&gt;worst-case&lt;/em&gt; number of drops, it doesn't minimizes the &lt;em&gt;average&lt;/em&gt;.  One may achieve an average of N drops if there are N(N-1)+1 floors by marking the top floor, then the floor below, then the floor two floors below that (as above), but below that progressing further one should go down four floors, then six, eight, ten, etc.  In that case, if the first ball breaks on drop N-I, one will need an average of I additional drops to determine the exact breaking point (cases that take less than I will balance those that take more).&lt;/p&gt;&#xA;" OwnerUserId="1467" LastActivityDate="2015-08-11T22:52:21.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19930" PostTypeId="1" AcceptedAnswerId="19935" CreationDate="2015-08-12T01:41:18.877" Score="34" ViewCount="2313" Body="&lt;p&gt;If a word has a certain property, I call it a Halved Word™.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In each of the short, unrelated sentences below, only one of the choices is a Halved Word™.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sentences do not affect whether a word is a Halved Word™. Case, font, and letter/word appearance do not matter.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;This (&lt;strong&gt;scene / segment&lt;/strong&gt;) is about declination due to aging.&lt;/li&gt;&#xA;  &lt;li&gt;Deputy Lee's face was (&lt;strong&gt;agrin / smirky&lt;/strong&gt;) as he brought people before the court.&lt;/li&gt;&#xA;  &lt;li&gt;His (&lt;strong&gt;horse / stallion&lt;/strong&gt;) watched the man who gave him new shoes.&lt;/li&gt;&#xA;  &lt;li&gt;The detailed (&lt;strong&gt;icons / thumbnails&lt;/strong&gt;) expressed much information in a tiny space.&lt;/li&gt;&#xA;  &lt;li&gt;My gastroenterologist's beer (&lt;strong&gt;mug / stein&lt;/strong&gt;) is shaped like human entrails.&lt;/li&gt;&#xA;  &lt;li&gt;It's a device to detect tiny earthquakes for the (&lt;strong&gt;metro / subway&lt;/strong&gt;) system.&lt;/li&gt;&#xA;  &lt;li&gt;The child snitched on his brother while his mom drank a (&lt;strong&gt;cappuccino / latte&lt;/strong&gt;).&lt;/li&gt;&#xA;  &lt;li&gt;The (&lt;strong&gt;anchor / host&lt;/strong&gt;) interviewed highly successful, aggressive people.&lt;/li&gt;&#xA;  &lt;li&gt;He would come back later to try the (&lt;strong&gt;fig / pear&lt;/strong&gt;) flavored vodka.&lt;/li&gt;&#xA;  &lt;li&gt;(&lt;strong&gt;Four / Three&lt;/strong&gt;) women in highly decorated dresses get in the line.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What makes a word a Halved Word™?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;SUB&gt;&lt;strong&gt;Bonus&lt;/strong&gt;: You will get 100 bonus rep points &lt;strong&gt;for each&lt;/strong&gt; of my &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%22Phrase%22%20user%3A463%20is%3Aquestion&quot;&gt;Phrase™&lt;/a&gt; or &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%22What+is+%22+%22Word%22+-missing+user%3A463+is%3Aquestion&quot;&gt;Word™&lt;/a&gt; puzzles for which you have previously gotten the check mark (2 for Deusovi, 1 for Rodolvertice, CodeNewbie, xnor, Sean, glibdud, f&quot;, dperry, and commando)&lt;/SUB&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;SUB&gt;&lt;em&gt;If you enjoyed this puzzle, and if you feel that it is worthy, remember to upvote it.&lt;/em&gt;&lt;/SUB&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="-1" LastEditorDisplayName="user14478" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2016-09-23T21:31:09.927" Title="What is a Halved Word™?" Tags="|pattern|english|word-property|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19931" PostTypeId="1" AcceptedAnswerId="19932" CreationDate="2015-08-12T01:50:45.317" Score="7" ViewCount="617" Body="&lt;p&gt;&lt;em&gt;This puzzle is done in honor of JLee's wonderful &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%22versatile%22%20user%3A463%20is%3Aquestion&quot;&gt;Words&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%22Phrase%22%20user%3A463%20is%3Aquestion&quot;&gt;Phrases&lt;/a&gt; series of puzzles. JLee must be really proud of us now.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;p&gt;If a word has a certain property, I call it a Close-by Word™.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;p&gt;In each of sentences below, one is a Close-by Word™ and one is not.  Cases and fonts have no bearing in the property of the word.  Sentences are for context only.  Only knowledge of English is required.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;p&gt;1. The (&lt;strong&gt;elf / dwarf&lt;/strong&gt;) fights an army.&#xA;  &lt;br&gt;2. The device (&lt;strong&gt;slashes / reduces&lt;/strong&gt;) fatalities.&#xA;  &lt;br&gt;3. We will (&lt;strong&gt;wax / defeat&lt;/strong&gt;) the competition.&#xA;  &lt;br&gt;4. A young man (&lt;strong&gt;created / invented&lt;/strong&gt;) a time machine.&#xA;  &lt;br&gt;5. I wish I was (&lt;strong&gt;driving / flying&lt;/strong&gt;) across the country.&#xA;  &lt;br&gt;6. People should not (&lt;strong&gt;jaywalk / cross&lt;/strong&gt;) the street.&#xA;  &lt;br&gt;7. Taxes will (&lt;strong&gt;increase / decrease&lt;/strong&gt;) over time.&#xA;  &lt;br&gt;8. The house has this (&lt;strong&gt;classic / modern&lt;/strong&gt;) look to it.&#xA;  &lt;br&gt;9. The shadow will (&lt;strong&gt;loom / tower&lt;/strong&gt;) over us.&#xA;  &lt;br&gt;10. Nothing is as (&lt;strong&gt;strange / queer&lt;/strong&gt;) as that.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;p&gt;What is a Close-by Word™?&lt;/p&gt;&#xA;" OwnerUserId="12332" LastEditorUserId="-1" LastEditorDisplayName="user14478" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2016-09-23T21:31:32.113" Title="What is a Close-by Word?™" Tags="|pattern|english|word-property|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19932" PostTypeId="2" ParentId="19931" CreationDate="2015-08-12T01:55:40.297" Score="12" Body="&lt;p&gt;A close-by word is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A word whose ending letter is exactly one after its starting letter in the alphabet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Examples given:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Y:&lt;/strong&gt; Elf (e/f) &lt;strong&gt;N:&lt;/strong&gt; Dwarf (d/f)&lt;br&gt;&#xA; &lt;strong&gt;Y:&lt;/strong&gt; Reduces (r/s) &lt;strong&gt;N:&lt;/strong&gt; Slashes (s/s)&lt;br&gt;&#xA; &lt;strong&gt;Y:&lt;/strong&gt; Wax (w/x) &lt;strong&gt;N:&lt;/strong&gt; Defeat (d/t)&lt;br&gt;&#xA; &lt;strong&gt;Y:&lt;/strong&gt; Created (c/d) &lt;strong&gt;N:&lt;/strong&gt; Invented (i/d)&lt;br&gt;&#xA; &lt;strong&gt;Y:&lt;/strong&gt; Flying (f/g) &lt;strong&gt;N:&lt;/strong&gt; Driving (d/g)&lt;br&gt;&#xA; &lt;strong&gt;Y:&lt;/strong&gt; Jaywalk (j/k) &lt;strong&gt;N:&lt;/strong&gt; Cross (c/s)&lt;br&gt;&#xA; &lt;strong&gt;Y:&lt;/strong&gt; Decrease (d/e) &lt;strong&gt;N:&lt;/strong&gt; Increase (i/e)&lt;br&gt;&#xA; &lt;strong&gt;Y:&lt;/strong&gt; Modern (m/n) &lt;strong&gt;N:&lt;/strong&gt; Classic (c/c)&lt;br&gt;&#xA; &lt;strong&gt;Y:&lt;/strong&gt; Loom (l/m) &lt;strong&gt;N:&lt;/strong&gt; Tower (t/r)&lt;br&gt;&#xA; &lt;strong&gt;Y:&lt;/strong&gt; Queer (q/r) &lt;strong&gt;N:&lt;/strong&gt; Strange (s/e)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="11475" LastEditDate="2015-08-12T02:00:43.007" LastActivityDate="2015-08-12T02:00:43.007" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19933" PostTypeId="2" ParentId="15187" CreationDate="2015-08-12T02:34:01.610" Score="22" Body="&lt;p&gt;I am a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;QUIZ&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The four parts are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;letters (scrabble tiles)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes the formation can happen with all four parts at once, but sometimes some parts come later than the others.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This is referring to putting down all four scrabble tiles, or forming the word from existing scrabble tiles already on the board.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;22 intrinsically&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;QUIZ's base scrabble score is 22&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;four factors&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Double word score, triple word score, double letter score, triple letter score&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;possible range between 22 and 96 (at the start it is 44)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Q=10 U=1 I=1 Z=10, total of 22.  With triple word score and double letter score, I double Q or Z, resulting in a score of 32, and triple it to 96.  If QUIZ is the first word played, it must cover the starting tile which is a double word score, for 44 points.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Every other scenario has zero involved in some way.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The blank tile worth zero points, which will be important later.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All of the scenarios have non-zero intrinsic and possible range values, and can be affected by the same four factors.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This should be obvious by now.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More than one of me can be formed in many ways, and seven parts are always involved.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;QUIZZES, seven letters&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes the formation can happen with all seven parts at once, but sometimes some parts come later than the others.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Same thing as above.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The highest scenario (let's call this scenario M1) happens with all seven parts at once, is 24 intrinsically (34 if not for a limitation)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;QUIZZES has a base score of 34, but there is only 1 Z tile, so a blank tile (worth zero points) must be substituted in place of a Z.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;five factors&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The fifth factor, E, is the 50-point bonus for using all 7 tiles in your hand.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;possible range between 74 and 152 (at the start it is 98, 100, or 118)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Minimum score of 74 is achieved by using only the 50 point bonus.  The maximum score of 152 uses a double letter score on Z and a triple word score for 102 points, plus 50 points bonus.  If QUIZZES is the first word played, it must cover the starting tile for double word score, and may cover an additional double letter score as well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All of the scenarios in which the formation happens with all seven parts at once can be affected by the same five factors. The scenarios in which some parts come later than the others can only be affected by four out of the five factors (those factors being A, B, C, and D).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Fairly obvious given that factor E only works if you played seven tiles at once.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6178" LastActivityDate="2015-08-12T02:34:01.610" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19934" PostTypeId="2" ParentId="18738" CreationDate="2015-08-12T03:03:01.877" Score="5" Body="&lt;p&gt;Homer is an ancient Greek. In ancient Greek,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;words that start with a vowel can have &quot;smooth breathing&quot; or &quot;rough breathing&quot;. &quot;Rough breathing&quot; indicates an &quot;h&quot; sound. We are looking for two words that are the same except one starts with an &quot;h&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rough, I'm a couple sharing a pie&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If a couple shares a pie, each person gets &lt;strong&gt;half&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Smooth, I'm a valley guy from the sky.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This is &lt;strong&gt;ALF&lt;/strong&gt; from the TV show of the same name: an alien who crashes in &quot;The Valley&quot; area of California.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-08-12T03:03:01.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19935" PostTypeId="2" ParentId="19930" CreationDate="2015-08-12T03:45:27.777" Score="30" Body="&lt;p&gt;A Halved Word™ is a word that,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; when doubled and then anagrammed, produces another English word. As f'' pointed out, each sentence contains a reference to the larger word. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Scene&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; scenescene -&gt; senescence (declination due to aging)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Agrin&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; agrinagrin -&gt; arraigning (brought people before the court)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Horse&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; horsehorse -&gt; horseshoer (the man who gave him new shoes)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Icons&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; iconsicons -&gt; concisions (much information in a tiny space)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Stein&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; steinstein -&gt; intestines (human entrails)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Metro&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; metrometro -&gt; tromometer (thanks, f'') (device to detect tiny earthquakes)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Latte&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; lattelatte -&gt; tattletale (snitched)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Host&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; hosthost -&gt; hotshots (highly successful, aggressive people)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pear&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; pearpear -&gt; reappear (come back later)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Four&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; fourfour -&gt; froufrou (women in highly decorated dresses)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="4421" LastEditDate="2015-12-29T04:52:56.407" LastActivityDate="2015-12-29T04:52:56.407" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19937" PostTypeId="2" ParentId="18629" CreationDate="2015-08-12T06:59:08.350" Score="6" Body="&lt;p&gt;Thanks for playing my riddle, everyone. :) I'm posting the answer here if anyone is curious about all the clues I had in mind when I wrote the letter.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A letter to my girlfriend, she would have been thirty this year&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Amiga 1000 was released on July 23rd 1985. 'Amiga' means female friend, rather than &lt;strong&gt;girlfriend&lt;/strong&gt;, but that would make for a less interesting title.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hello... I miss you. We all do.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many people still fondly remember the Amiga (1000 or otherwise).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You loved to paint. A personal passion: it brought out your brilliance.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Amiga was well known for its graphical capabilities. In a world of gaudy colours and attribute clash entered a machine capable of high-resolution, programmable palette displays, for the first time at a price that people could realistically afford. Two of the most popular paint packages for the Amiga were Cloanto &lt;a href=&quot;http://www.amigaforever.com/classic/ppaint/&quot; rel=&quot;noreferrer&quot;&gt;&lt;strong&gt;Personal&lt;/strong&gt; Paint&lt;/a&gt; (also Cloanto now own most of the Amiga rights I believe) and &lt;a href=&quot;http://obligement.free.fr/articles/brilliance.php&quot; rel=&quot;noreferrer&quot;&gt;&lt;strong&gt;Brilliance&lt;/strong&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I remember when I first met you, I saw your painting of the mask of Tutankhamun.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Though most Amiga users would be more familiar with Deluxe Paint (which came with many Amiga systems) and readily recognise &lt;a href=&quot;http://commodore.ninja/page_images/amiga/apps/cover/DeluxePaint2.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;strong&gt;the image of Tutankhamun&lt;/strong&gt; from the DP box&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You loved Europe, too. And they loved you back. We travelled to the States together, but it never seemed right. It was your home I think, but didn't feel right to be there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Commodore was founded in Canada, but headquartered in the United States, but unlike the Commodore 64, the Commodore Amiga was more popular in Europe than the US. The Amiga line was more popular than some consoles in Europe, and many demoscene productions from were from around there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You were never far from your workbench, surrounded by drawers filled with tools and gadgets I could never hope to understand.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Amiga's graphical interface was called &lt;strong&gt;Workbench&lt;/strong&gt;. What most folks would call 'directories' or 'folders' were called '&lt;strong&gt;drawers&lt;/strong&gt;', and programs often known as '&lt;strong&gt;tools&lt;/strong&gt;'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You weren't an aviator, but you said you knew your way around the bits of a plane. (Seriously graphic language ensued...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Amiga didn't have chunky graphics like most common IBM PC VGA modes (one byte index = one pixel colour), or hardware tiled graphics like consoles (set up small graphics and repeat them over the screen), instead it had '&lt;strong&gt;bitplaned&lt;/strong&gt;' graphics, which split the bits of a palette value into contiguous regions containing the values of a specific bit position across the screen (you'd have a block of bit zeroes, then a block of bit ones, etc). These were known as '&lt;strong&gt;bitplanes&lt;/strong&gt;'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;things went wrong often, meditation rarely helped. You'd throw your work away and start over.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Amiga's ROM crash handler would bring up a message titled &lt;a href=&quot;https://en.wikipedia.org/wiki/Guru_Meditation&quot; rel=&quot;noreferrer&quot;&gt;'Guru &lt;strong&gt;Meditation&lt;/strong&gt;'&lt;/a&gt;. Rarely this would be titled a 'Recoverable Error' and claim to let you save your work first before rebooting (it would never work...). Most of the time it would flash on, and the whole computer would reboot and you'd lose everything.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I hope you're happy; you're with your family now.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Amiga 1000 was the first of a range of computers. All the Amiga systems are long since past their time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Do you remember the time your younger sister (the one who loved those cartoon classics)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Amiga 1000 was 1985. &lt;strong&gt;Amiga 500&lt;/strong&gt; came out in &lt;strong&gt;1987&lt;/strong&gt;, Amiga 500+ in &lt;strong&gt;1991&lt;/strong&gt;. One of the Amiga 500 bundles was called &lt;a href=&quot;http://www.amigahistory.plus.com/cartoon.html&quot; rel=&quot;noreferrer&quot;&gt;&lt;strong&gt;Cartoon Classics&lt;/strong&gt;&lt;/a&gt; and included...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;brought all those pets home from Scotland in, 1991 was it? Hundreds of them! Running backwards and forwards, all over you, all over her, digging holes in everything, making a mess. I couldn't even tell them apart, we never got round to giving any of them names. We finally gave them a good home. (As many as we could I'm sad to say.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One of the Amiga's most popular games was &lt;a href=&quot;https://en.wikipedia.org/wiki/Lemmings_%28video_game%29&quot; rel=&quot;noreferrer&quot;&gt;Lemmings&lt;/a&gt;. The objective of &lt;strong&gt;Lemmings&lt;/strong&gt; was to lead the creatures to an exit doorway by assigning them commands, such as 'dig' or 'build'. Without your intervention, they would simply wander left to right and back again. All the Lemmings were identical, and although they could be highlighted to show their role, they didn't have names &lt;a href=&quot;https://en.wikipedia.org/wiki/The_Adventures_of_Lomax&quot; rel=&quot;noreferrer&quot;&gt;(yet)&lt;/a&gt;. The Lemmings levels are full of traps that would kill the Lemmings in various graphic ways. It's not possible to save them all.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Wasn't there even a statue of them?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://scottishgames.net/2013/10/14/lets-go-lemmings-in-the-real-world/&quot; rel=&quot;noreferrer&quot;&gt;Yep!&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I remember seeing them around from time to time, holidays and Christmas. They're everywhere... &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lemmings has been released on pretty much every platform under the sun over the last twenty five years, most recently on the PS3 and PS Vita. Two expansion pack releases for the original Lemmings were &lt;a href=&quot;https://en.wikipedia.org/wiki/Christmas_Lemmings&quot; rel=&quot;noreferrer&quot;&gt;&lt;strong&gt;Holiday&lt;/strong&gt; Lemmings and &lt;strong&gt;X-Mas&lt;/strong&gt; Lemmings&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;we even gave one to Terry Pratchett, regrettably. It seemed like a great idea at the time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; From &lt;a href=&quot;http://www.lspace.org/books/apf/interesting-times.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.lspace.org/books/apf/interesting-times.html&lt;/a&gt; : The icons for controlling the Red Clay Army are immediately familiar to anyone who has ever played the computer game Lemmings, in which you have to use similar controls to guide a group of brainlessly wandering lemmings across intricate and dangerous underground labyrinths.&#xA; When this was first remarked upon by readers in a.f.p, Terry wrote:&#xA; &quot;What? Lemmings? Merely because the red army can fight, dig, march and climb and is controlled by little icons? Can't imagine how anyone thought that...&#xA; Not only did I wipe Lemmings from my hard disc, I overwrote it so's I couldn't get it back.&quot; &#xA; Yep, Terry permanently and absolutely obliterated all his Lemmings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(Oh, BTW, those same men who gave you them went onto a life of crime in America; it was all over the papers, surprise surprise.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lemmings was created by DMA Design, who also created &lt;a href=&quot;https://en.wikipedia.org/wiki/Grand_Theft_Auto_%28series%29&quot; rel=&quot;noreferrer&quot;&gt;&lt;strong&gt;Grand Theft Auto&lt;/strong&gt;&lt;/a&gt; (and of course went on to become &lt;a href=&quot;https://en.wikipedia.org/wiki/Rockstar_North&quot; rel=&quot;noreferrer&quot;&gt;Rockstar North&lt;/a&gt; ). The GTA games have rarely been out of the newspapers since the first game in 1997, being responsible for instigating absolutely all crime ever by this point.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At times it seemed like you could do anything. It's cheesy, but... you made things possible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; At the time, the Amiga had the best, most affordable graphics and sound technology available in a home computer. It could do business software, 3D rendering, multimedia and be expanded to do video titling. The Amiga even had its own theme song: &lt;a href=&quot;https://www.youtube.com/watch?v=PWeO5IkCssk&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;Only Amiga Makes It Possible&lt;/em&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But you didn't do it on your own. Do you remember your friends? That part-time copper who kept you in line and Denise... they showed you how to paint. And punk musician Paula... crazy as hell, sometimes it seemed like she could play four instruments at once!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The core of the Amiga was &lt;a href=&quot;https://en.wikipedia.org/wiki/Original_Chip_Set#Denise&quot; rel=&quot;noreferrer&quot;&gt;the custom chipset that provided all the special features, including Agnus, &lt;strong&gt;Denise&lt;/strong&gt; and &lt;strong&gt;Paula&lt;/strong&gt;&lt;/a&gt;. Denise fetched graphics data and prepared it for display, and part of Agnus provided the coprocessor (or 'copper') which provided a macro-like language to provide quick graphical effects at timed positions across the screen as it was drawn. &lt;strong&gt;Paula contained the sound hardware&lt;/strong&gt;, which could indeed play four sampled sounds simultaneously.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We tried learning a new language together, the three of us: you, I and Amos. We went through the magazines together... learning as we went. It seemed like everybody was doing it. Does that ring a bell?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Two popular hobbyist languages that came out for the Amiga were Blitz BASIC, and &lt;strong&gt;AMOS&lt;/strong&gt; BASIC. Versions of both were often found as free coverdisk giveaways on magazines, &lt;a href=&quot;http://amr.abime.net/issue_198_pages&quot; rel=&quot;noreferrer&quot;&gt;which would also be followed up by lengthy tutorial series spread out over many issues&lt;/a&gt;. In the 90s, magazines also carried games (some small and simple, some very ambitious) made in AMOS and Blitz by 'bedroom programmers' (the original Worms was a Blitz game originally). Like many BASICs, AMOS provided a basic &lt;a href=&quot;http://www.ultimateamiga.co.uk/HostedProjects/AMOSFactory/AMOSProManual/8/802.html&quot; rel=&quot;noreferrer&quot;&gt;BELL&lt;/a&gt; 'bing' sound effect, common in quick and dirty games.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I barely saw any of the rest of your family. Your next youngest, (Agatha, maybe?)... the sophisticated one. I met her seven years into our relationship. She was smart, but I never thought she could all the things you could. I would have liked to meet the others, but they were always too professional. Too focused on their important work. Ivory towers and all that.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Amiga_1200&quot; rel=&quot;noreferrer&quot;&gt;The A1200&lt;/a&gt; came out in 1992, the next major upgrade of the 'wedge' series of Amigas. It included the Advanced Graphics Architecture, or '&lt;strong&gt;AGA&lt;/strong&gt;' chipset. It had an improved processor and improved graphics, but it wasn't &lt;em&gt;fully&lt;/em&gt; compatible with Amiga 1000 or Amiga 500 software much to most players' chagrin. Alongside the 'wedge' computers like the Amiga 500, 500+, 600 and 1200 that would be familiar to Atari owners, were the 'workstation' style units, the Amiga 1000, 2000, 3000 and 4000 which were built into a box case like an IBM PC, with their own upgrade options. Wedges, which were used as games machines, were far more common among families than workstation machines, which were used for productivity. Workstation machines could also be installed vertically. Amiga machines came in light grey or white, but the plastics are prone to discolouration over time, making them &lt;strong&gt;literal ivory towers&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I still think back to the games we used to play together. I sometimes have flashbacks to those times, but its so long now it seems like another world. It wasn't easy keeping up with you, but that's what I liked about you. (Those trainers you gave me helped! You can't get trainers like that any more!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Amiga had a diverse range of games over the years, including &lt;a href=&quot;https://en.wikipedia.org/wiki/Flashback_%281992_video_game%29&quot; rel=&quot;noreferrer&quot;&gt;&lt;strong&gt;Flashback&lt;/strong&gt;&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Another_World_%28video_game%29&quot; rel=&quot;noreferrer&quot;&gt;&lt;strong&gt;Another World&lt;/strong&gt;&lt;/a&gt;. 90s games were known for their punishing difficulty and reluctance to provide 'continues' for the player, often kicking them right back to the start of the game when they lost all their lives. Hacked versions of Amiga games sometimes contained programs called '&lt;strong&gt;trainers&lt;/strong&gt;' which edited the game to provide extra lives, level skips or invulnerability. (Cheating like this is rare but still possible in some PC games, 'aim-bots' and infinite ammo cheats in competitive first person shooters, and so on.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I hope you're all at peace now, wherever you are. Love, J&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Amiga project was started by &lt;a href=&quot;https://en.wikipedia.org/wiki/Jay_Miner&quot; rel=&quot;noreferrer&quot;&gt;&lt;strong&gt;Jay&lt;/strong&gt; Miner&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14398" LastEditorUserId="14398" LastEditDate="2015-08-12T07:11:01.113" LastActivityDate="2015-08-12T07:11:01.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19938" PostTypeId="2" ParentId="19823" CreationDate="2015-08-12T07:12:16.837" Score="5" Body="&lt;p&gt;answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; row 1 has 3 kinds of design: &lt;br&gt; &lt;br&gt; a) one with straight line; &lt;br&gt; b) one with two lines intersecting; &lt;br&gt; c) and one with a line starting from centre of a tile.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/kFAEr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kFAEr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15609" LastEditorUserId="42452" LastEditDate="2018-09-08T13:31:57.447" LastActivityDate="2018-09-08T13:31:57.447" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="19939" PostTypeId="2" ParentId="19914" CreationDate="2015-08-12T07:20:58.337" Score="1" Body="&lt;p&gt;I would have added this as a comment to Mike Earnest if I could have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears that E(N,2) reaches its minimum with a f at N/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this mean that the strategy ought to be begin at floor N/2, plus or minus .5 for N odd?  If the 1st ball breaks then begin with floor 1 for the 2nd ball.  If the ball does not break, follow the same strategy using floor N/2 +1 as your 1st floor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the 100 floor example, we would drop the first ball at floor 50.  If it failed to break we would then use floor 75 as our next drop, etc.&lt;/p&gt;&#xA;" OwnerUserId="15610" LastActivityDate="2015-08-12T07:20:58.337" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19940" PostTypeId="2" ParentId="19806" CreationDate="2015-08-12T07:35:52.213" Score="5" Body="&lt;p&gt;Disclaimer: Newbie here! Pardon me for any oversights or mistakes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the poem refers to the book &lt;a href=&quot;https://books.google.co.in/books?id=_unxCQAAQBAJ&amp;amp;pg=PA16&amp;amp;lpg=PA16&amp;amp;dq=ort%20sphere&amp;amp;source=bl&amp;amp;ots=En4uuNfopZ&amp;amp;sig=g1fRlzOumfR1p9AlqhQP2psZpmI&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0CCgQ6AEwAmoVChMI6Yrwu_eixwIV1XOOCh2RBgMv#v=onepage&amp;amp;q=ort%20sphere&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Deep Space Dogfights&lt;/a&gt;. I haven't read it, and the keyword for the book's identification was &lt;strong&gt;&lt;em&gt;Ort Sphere&lt;/em&gt;&lt;/strong&gt;. I read a few pages, and found out an Ort Sphere was presumed to be the smallest. Next, the test part, as it is explained in the answer. The poem points to excerpts from the story Final Flight of the Fighting Jack Churchill, and I've presented them accordingly. Hope this is the answer you seek, and in the correct order. I'm pretty sure I have the order messed up, but I'll need more time to read the whole chapter &amp;amp; properly explain &amp;amp; order the excerpts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two of spades, from a new deck dealt&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;There are only &lt;strong&gt;50 CTs&lt;/strong&gt; aboard.&lt;/em&gt; This is a bit doubtful: I presume the two of spades refers to two cards, leaving 50 cards in (or in the book's context: leaving 50 Combat Troopers on) the deck.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A din, a doubt&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Prevent a half-dozen pods attaching succesfully and making the Churchill's hull &lt;strong&gt;shudder&lt;/strong&gt;... All personnel will arm &lt;strong&gt;in the event&lt;/strong&gt; forward sections are overrun.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What a certain literary professor for his unbeheld felt,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Good God, the Captain (Royce) realized about their enemy, &lt;strong&gt;we don't even know what THEY look like.&lt;/em&gt;&lt;/strong&gt; Unbeheld means unseen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Up, up, by the Ort Sphere's smallest test,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;But could he ask this crew to sacrifice their lives and this antiquated starship just to &lt;strong&gt;test if a smaller presumed central sphere managing other attacking vessels theory was true?&lt;/em&gt;&lt;/strong&gt; Previously in the story, it has been established that the smaller sphere is the smallest, and the puppeteer of the enemy armada. OR this could also be pointing to Captain Royce needing to get to the auxiliary override junction, two decks (Up, up) above his current location, in order to force the ship onto a collision-course with the smallest Ort Sphere, and test if this collision will end the battle and bring victory to Earth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Out!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;He watched Terry Alan race... OR 'Permission granted', where the Captain allows Terry Alan to go and assist another fighting crew. However, this does not add up with the OR part above in terms of sequence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But oh, wait, now by one-third less.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Can't really figure this out, but I think of it as the departure of Terry Alan leaving only two &lt;strong&gt;named&lt;/strong&gt; officers on the bridge out of the actual three.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15607" LastEditorUserId="15607" LastEditDate="2015-08-12T11:16:08.160" LastActivityDate="2015-08-12T11:16:08.160" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19941" PostTypeId="2" ParentId="19914" CreationDate="2015-08-12T08:33:43.803" Score="0" Body="&lt;p&gt;If we have $x$ balls and $N$ floors, let the floor we throw the first one be $y$. It'll either break or stay intact. If the former happens, we'll have $x-1$ balls and $y-1$ floors. If the latter happens, we can go upstairs and have $x$ balls and $N-y$ floors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Balls...................Floors...............Floor to try&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x$.........................$N$.....................$y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x-1$..................$y-1$................$z$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x$........................$N-y$................$t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say it breaks all the time. Calling our custom operation retrieving $y$ from $x$ and $N$ #, it can be represented like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 # (... $x-2$ # ($x-1$ # ($x$ # $N$))...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it never breaks, we can represent it like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...(($N$ - ($x$ # $N$)) - ($x$ # ($N$ - ($x$ # $N$))))...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it doesn't, we go higher and higher (let's say, from $t_1$ to $t_k$) until finally one breaks. Each gap between two consecutive $t$s must allow for one less moves with $x-1$ balls than the previous gap. When all the $k$ moves are made, we end up making $k+1$ moves. If the $t_k$th floor is too high but the one before isn't, we shouldn't need an extra move, which means they're consecutive floors. The first gap must allow for $k-1$ moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it breaks, the number of moves we make with $x-1$ balls and $y-1$ floors must be $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we call the custom operation retrieving the greatest number of floors requiring a given number of moves ♦, then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0 ♦ (x-1) + 1 ♦ (x-1) + 2 ♦ (x-1) + ... + n-1 ♦ (x-1) + n &amp;lt; N =&amp;lt; 0 ♦ (x-1) + 1 ♦ (x-1) + 2 ♦ (x-1) + ... + n ♦ (x-1) + n+1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m ♦ 1 = m$, so we can start from it using induction. The floor we have to begin with is $n ♦ (x-1) +1$, and then we have can follow the table at the beginning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have no fewer balls than $ceil[log_2N$], we can use binary search.&lt;/p&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="14016" LastEditDate="2015-08-12T12:54:49.430" LastActivityDate="2015-08-12T12:54:49.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19942" PostTypeId="1" AcceptedAnswerId="19996" CreationDate="2015-08-12T08:58:32.890" Score="12" ViewCount="1302" Body="&lt;p&gt;I came from a large family,&lt;br&gt;&#xA;with lots of people looking after me.&lt;br&gt;&#xA;My family now lives all over the place,&lt;br&gt;&#xA;but I have my very own space.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My oldest sister was one of her kind,&lt;br&gt;&#xA;but she could not do things that were more refined.&lt;br&gt;&#xA;My third sister was the first to pass,&lt;br&gt;&#xA;when it happened, there were many questions asked.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My second sister left for good when I was young.&lt;br&gt;&#xA;But she had a house that looked after the folks and the punks.&lt;br&gt;&#xA;She took loads and loads of videos and studied really hard,&lt;br&gt;&#xA;She also had her own place where many people bark.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My fourth and fifth sisters were really hard working,&lt;br&gt;&#xA;but my fourth sister was always on TV, constantly looking.&lt;br&gt;&#xA;My fifth sister seemed to be lost,&lt;br&gt;&#xA;but no-one really knew what was the cause.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I was born, I wasn't meant to be.&lt;br&gt;&#xA;But after much collaborations, I became a reality.&lt;br&gt;&#xA;We were the sleeping beauties, until it was time to work,&lt;br&gt;&#xA;but only with a selected few who could prove their worth.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We had two half-sisters, who were too lazy to work,&lt;br&gt;&#xA;but upon closer inspection, they have their own quirks.&lt;br&gt;&#xA;We had a distant cousin who could do our job,&lt;br&gt;&#xA;but until now, we were the only ones who could hold down the fort.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My sisters and I have traveled a lot,&lt;br&gt;&#xA;one of these places was very hot.&lt;br&gt;&#xA;We got up at any time of the day,&lt;br&gt;&#xA;we weren't lazy, just needed good weather our way.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A puzzle solver should explain every stanza,&lt;br&gt;&#xA;otherwise, it will not be a correct answer.&lt;br&gt;&#xA;Except, of course, for these few lines,&lt;br&gt;&#xA;which are written here to make it rhyme.&lt;/em&gt;  &lt;/p&gt;&#xA;" OwnerUserId="12332" LastEditorUserId="6178" LastEditDate="2016-06-16T14:52:49.953" LastActivityDate="2016-06-16T14:52:49.953" Title="Who are we? We are the elite" Tags="|riddle|rhyme|knowledge|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19943" PostTypeId="2" ParentId="18072" CreationDate="2015-08-12T09:07:42.130" Score="9" Body="&lt;h2&gt;GIVEN&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Considering the scenario gives us:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Sheet 1&lt;/strong&gt; - the &lt;em&gt;key&lt;/em&gt; to solving the cryptogram.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Sheet 2&lt;/strong&gt; - the &lt;em&gt;encrypted password&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h2&gt;HINTS&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Considering the hints gives us:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Hint 0&lt;/strong&gt; - &lt;code&gt;0&lt;/code&gt; serves as a &lt;em&gt;separator of values&lt;/em&gt; for the Sheet 1.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Hint 1&lt;/strong&gt; - values separated by &lt;code&gt;0&lt;/code&gt; in Sheet 1 has a defined relation with regard to Sheet 2. &lt;code&gt;ELB...BCH&lt;/code&gt; corresponds to Sheet 2's &lt;code&gt;columns&lt;/code&gt;, &lt;code&gt;OBC...LIE&lt;/code&gt; corresponds to Sheet 2's &lt;code&gt;rows&lt;/code&gt;. (For simplicity, I disregarded the 1's, -1's and 0 of sheet 1)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Hint 2&lt;/strong&gt; - separate the rows of Sheet 1. Combine the separated rows corresponding to the number of cells.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Hint 3&lt;/strong&gt; - use excel as a handy tool (LOL!)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7r6qc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7r6qc.png&quot; alt=&quot;Hints Guide&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;MY SOLUTION&lt;/h2&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Arranged the cells of Sheet 1 corresponding to the columns and rows of Sheet 2. In addition, translate the &lt;em&gt;letters&lt;/em&gt; to their &lt;em&gt;corresponding count&lt;/em&gt; (A=1,B=2,C=3...Z=26)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Noting the values in Sheet 2 and taking note of the &lt;code&gt;1&lt;/code&gt;'s and &lt;code&gt;-1&lt;/code&gt;'s, and &lt;code&gt;A=1...Z=26&lt;/code&gt; value of Sheet 1, we find a relation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CellValue + (RowLetter * RowMultipler) + (ColLetter * ColMultiplier) = 0&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;However, some of the values in Sheet 2 are not as expected. We isolate the values that do not match the expectation (In the image, I coloured them blue). Hence, we arrive at the answer I suspect &lt;code&gt;FOUND&lt;/code&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GZw9Q.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GZw9Q.png&quot; alt=&quot;Solutions Guide&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Iiy8j.jpg&quot; alt=&quot;Found&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="15612" LastEditorUserId="1777" LastEditDate="2015-08-13T17:16:13.893" LastActivityDate="2015-08-13T17:16:13.893" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19944" PostTypeId="2" ParentId="18738" CreationDate="2015-08-12T10:08:46.050" Score="0" Body="&lt;p&gt;Could it be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the number pi?&#xA; pi is a number that is named after a greek letter. Homer wrote in Greek.&#xA; There's the expression &quot;pie in the sky&quot; which would fit both clues&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14439" LastEditorUserId="14439" LastEditDate="2015-08-12T10:34:10.047" LastActivityDate="2015-08-12T10:34:10.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19946" PostTypeId="1" AcceptedAnswerId="19950" CreationDate="2015-08-12T10:53:04.063" Score="11" ViewCount="904" Body="&lt;p&gt;I have these kinds of expressions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Correct expressions (true and valid)&lt;/strong&gt;&#xA;$$\begin{align}&#xA;3 &amp;amp;= 3\hbox{ (it's obvious, isn't it?)}\\&#xA;4 &amp;amp;= 1 + 3\\&#xA;8 &amp;amp;= 5\\&#xA;9 &amp;amp;= 1 + 5\\&#xA;12 &amp;amp;= 1 + 3 + 5\\&#xA;18 &amp;amp;= 4 + 6&#xA;\end{align}$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;True but not valid expressions&lt;/strong&gt;&#xA;$$\begin{align}&#xA;3 &amp;amp;= 1 + 2\\&#xA;10 &amp;amp;= 4 + 4\\&#xA;12 &amp;amp;= 5 + 3 + 1\\&#xA;14 &amp;amp;= 1 + 4 + 5&#xA;\end{align}$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Incorrect expressions (false and invalid)&lt;/strong&gt;&#xA;$$\begin{align}&#xA;10 &amp;amp;= 4 + 6\\&#xA;11 &amp;amp;= 5 + 6&#xA;\end{align}$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For each non-negative integer in the left side, there is a unique expression on the right side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you tell what is the general rule? &#xA;I think it is an easy one for you, guys, but feel free to ask for specific values if you need them.&lt;/p&gt;&#xA;" OwnerUserId="14068" LastEditorUserId="13494" LastEditDate="2015-08-20T21:53:37.353" LastActivityDate="2015-08-20T21:53:37.353" Title="Why is 18 = 4 + 6? Find the general rule" Tags="|mathematics|logical-deduction|" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19947" PostTypeId="1" AcceptedAnswerId="19977" CreationDate="2015-08-12T11:18:40.180" Score="35" ViewCount="3897" Body="&lt;p&gt;&lt;strong&gt;Clues&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What a conductor gives to an out-of-tune player?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You perform this in Spring&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sung by musical miners?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those behind bars keeps tabs on this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sounds something like a big colourful Spanish bird&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A pretty endearing advance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Loosen your shoes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find a French fishy ratio &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A teenager causing trouble?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scottish cooking method&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sliced in private&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A little bed made of bones&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A long lasting weird religion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A cereal, relatively speaking&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exact copy of Warhammer people&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Instructions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Start solving clues.&lt;/li&gt;&#xA;&lt;li&gt;See what the answers have in common: The more you get the more obvious it will be what connects them.&lt;/li&gt;&#xA;&lt;li&gt;When you get the connection look at the title. It will give you a hint to the final answer.&lt;/li&gt;&#xA;&lt;li&gt;Give (a) the answers to the clues and (b) the final answer, which is a well-known phrase.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Extra kudos&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you understand the connection, it is possible to obtain some of the answers without fully understanding their clue. Extra kudos to you if you convincingly &lt;strong&gt;explain&lt;/strong&gt; all the clues as well as solve them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Notes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that this is a tough challenge but, knowing you puzzlers, you may prove me wrong! If it proves absolutely impossible then let me know in the comments and I'll give some extra hint(s). Take heart - Once you get any two of the clues, you will be able to see the connection and the whole thing will become much easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been a little bit evil by trying to start you off on a false trail (This is in the clues. There are no tricks in the instructions etc). Once you start to solve things you will see through my impishness and be able to ignore it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Co-operation is welcome if puzzlers want to solve this together. I don't mind either way. Partial answers will gain up-votes and help others to give further partial answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had to omit one tag because it would have given away the pattern that you are trying to find. &lt;/p&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="14177" LastEditDate="2015-08-12T22:54:39.887" LastActivityDate="2015-08-13T13:53:26.397" Title="literally no idea" Tags="|pattern|letters|connections-puzzle|cryptic-clues|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19948" PostTypeId="5" CreationDate="2015-08-12T11:35:40.117" Score="0" Body="&lt;h2&gt;What is a cryptic clue ?&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In essence, a cryptic clue leads to its answer as long as it is read in the right way. What the clue appears to say when read normally (the surface reading) is a distraction and usually has nothing to do with the clue answer. The challenge is to find the way of reading the clue that leads to the solution.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A typical clue consists of two parts, the definition and the wordplay. It provides two ways of getting to the answer. The definition, which usually exactly matches the part of speech, tense, and number of the answer, is in essence the same as any 'straight' crossword clue, a synonym for the answer. It usually appears at the start or the end of a clue.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The other part (the subsidiary indication, or wordplay) provides an alternative route to the answer (this part would be a second definition in the case of double definition clues). One of the tasks of the solver is to find the boundary between definition and wordplay and insert a mental pause there when reading the clue cryptically. This wordplay gives the solver some instructions on how to get to the answer another way. Sometimes the two parts are joined with a link word or phrase such as &quot;from&quot;, &quot;gives&quot; or &quot;could be&quot;. There are many sorts of wordplay, such as anagrams and double definitions.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Because a typical cryptic clue describes its answer in detail and often more than once, the solver can usually have a great deal of confidence in the answer once it has been determined. The clues are 'self-checking'. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;From the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cryptic_crossword&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; article on cryptic crosswords&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;How can I make one?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;See the &lt;a href=&quot;/questions/45984/cryptic-clue-guide&quot;&gt;Cryptic Clue Guide&lt;/a&gt; for information on how to make a cryptic clue.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt; Example&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Here is an example (taken from The Guardian crossword of 6 August 2002, set by &quot;Shed&quot;).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Very sad unfinished story about rising smoke (8)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is a clue for TRAGICAL. This breaks down as follows.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&quot;Very sad&quot; is the definition&lt;/li&gt;&#xA;  &lt;li&gt;&quot;unfinished story&quot; gives &quot;tal&quot; (&quot;tale&quot; with one letter missing; i.e., unfinished)&lt;/li&gt;&#xA;  &lt;li&gt;&quot;rising smoke&quot; gives &quot;ragic&quot; (a &quot;cigar&quot; is a smoke and this is a down clue so &quot;rising&quot; indicates that &quot;cigar&quot; should be written up the page; i.e., backwards)&lt;/li&gt;&#xA;  &lt;li&gt;&quot;about&quot; means that the letters of &quot;tal&quot; should be put either side of &quot;ragic&quot;, giving &quot;tragical&quot;&lt;/li&gt;&#xA;  &lt;li&gt;&quot;(8)&quot; says that the answer is a single word of eight letters.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt; How to solve them&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://solving-cryptics.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;solving-cryptics.com&lt;/a&gt; for more information on solving this type of puzzle.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;What tag to use&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Use this tag &lt;a href=&quot;/questions/tagged/cryptic-clues&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;cryptic-clues&amp;#39;&quot; rel=&quot;tag&quot;&gt;cryptic-clues&lt;/a&gt; for puzzles that use cryptic clues but are not cryptic crosswords. Use &lt;a href=&quot;/questions/tagged/cryptic-crosswords&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;cryptic-crosswords&amp;#39;&quot; rel=&quot;tag&quot;&gt;cryptic-crosswords&lt;/a&gt; for actual cryptic crosswords. There are also &lt;a href=&quot;/questions/tagged/crossword-clues&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;crossword-clues&amp;#39;&quot; rel=&quot;tag&quot;&gt;crossword-clues&lt;/a&gt; and &lt;a href=&quot;/questions/tagged/crosswords&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;crosswords&amp;#39;&quot; rel=&quot;tag&quot;&gt;crosswords&lt;/a&gt; for use when the clues do not follow the conventions of cryptic crosswords described above.&lt;/p&gt;&#xA;" OwnerUserId="2752" LastEditorUserId="11110" LastEditDate="2020-04-15T10:12:41.463" LastActivityDate="2020-04-15T10:12:41.463" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19949" PostTypeId="4" CreationDate="2015-08-12T11:35:40.117" Score="0" Body="A puzzle that involves cryptic clues of the type found in cryptic crosswords, but is not a cryptic crossword. DO NOT use this tag for puzzles that are stated cryptically; use [enigmatic-puzzle] instead. Use [cryptic-crosswords] for whole crosswords; [crossword-clues] for crossword clues not obeying the conventions of cryptic crosswords (even if they are stated cryptically)." OwnerUserId="14177" LastEditorUserId="69582" LastEditDate="2020-12-14T05:22:03.670" LastActivityDate="2020-12-14T05:22:03.670" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19950" PostTypeId="2" ParentId="19946" CreationDate="2015-08-12T11:36:57.140" Score="9" Body="&lt;p&gt;These are the Fibonacci representations of the integers. Valid expressions contain no consecutive or repeated numbers, and are in increasing order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3=3=F_3\\&#xA;4=1+3=F_1+F_3\\&#xA;8=8=F_5\\&#xA;9=1+8=F_1+F_5\\&#xA;12=1+3+8=F_1+F_3+F_5\\&#xA;18=5+13=F_4+F_6$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$10=5+5=F_4+F_4$, but it should be written as $2+8=F_2+F_5$ instead.&#xA;$14=1+5+8=F_1+F_4+F_5$, but it should be written as $1+13=F_1+F_6$ instead.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-08-12T11:41:00.973" LastActivityDate="2015-08-12T11:41:00.973" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19951" PostTypeId="2" ParentId="19920" CreationDate="2015-08-12T11:45:54.607" Score="9" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;A birdlike creature, the foe of all snakes,&lt;br&gt;&#xA;  on this city's flag an appearance makes.&lt;br&gt;&#xA;  They have a whole castle for one man-made lake!&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This refers to &lt;strong&gt;Ulaanbaatar&lt;/strong&gt;, the location of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Artificial_Lake_Castle&quot; rel=&quot;noreferrer&quot;&gt;Artificial Lake Castle&lt;/a&gt;. The city's flag shows Garuda, enemy of the Nagas.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;If you need an import, I'd hope there's no ice,&lt;br&gt;&#xA;  or enough for a truck to make one-way trips twice.&lt;br&gt;&#xA;  (Though Yamal is here in a pinch, which is nice.)&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This stanza describes &lt;strong&gt;Salekhard&lt;/strong&gt;, which is located near the base of the Yamal Peninsula in Russia. Salekhard is on the opposite side of the Ob River from the highway, so it is only accessible by air or when there is no ice (so boats can cross the river) or when there is enough ice to drive across. Yamal Airlines operates out of Salekhard.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A festival city, yet not from Dubai;&lt;br&gt;&#xA;  Living history looms here as tall as the sky,&lt;br&gt;&#xA;  and it's nearly the finest if you're looking to buy.&lt;br&gt;&#xA;  The gateway of the north this came to be&lt;br&gt;&#xA;  In the land that stretches from sea to sea.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edmonton&lt;/strong&gt; is &quot;Canada's festival city&quot; (I found this before the extra lines were added, but didn't post it because I didn't see any connections to the second and third lines). It is also known as a &quot;gateway to the north&quot;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A city of non-zero nuclear threat;&lt;br&gt;&#xA;  Elena is priced in at ten, I would bet.&lt;br&gt;&#xA;  For dry desert farming, it's the best you can get.&lt;br&gt;&#xA;  Through which both air and road transport passes,&lt;br&gt;&#xA;  Here announced the dawn of the era of the masses.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sabha&lt;/strong&gt;, Libya, is where Muammar Gaddafi announced &quot;the dawn of the era of the masses&quot;. It is the location of the Fort Elena castle, which is on Libya's 10 dinar note. It is also related to Libya's nuclear program.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Home of the highlanders, a capital investment;&lt;br&gt;&#xA;  Most recently subjected to corruption's assessment.&lt;br&gt;&#xA;  Their country dissolved as a laughable contestant.&lt;br&gt;&#xA;  Stealing the title from the first month's river,&lt;br&gt;&#xA;  On Don Bosco's dream town, a new Marx delivered.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &quot;first month's river&quot; is Rio de Janeiro, which was the capital of Brazil before it was moved to &lt;strong&gt;Brasilia&lt;/strong&gt;. Brasilia is located in the highlands of Brazil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lampost42 and Echo have found the other three cities, which are &lt;strong&gt;Kangerlussuaq&lt;/strong&gt;, &lt;strong&gt;Kabul&lt;/strong&gt;, and &lt;strong&gt;Tallinn&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we should try to determine what the trips were. If each trip had four cities, then the first trip visits Kangerlussuaq, Sabha, Edmonton, and Brasilia (ordered by increasing population), and the second visits Tallinn, Kabul, Ulaanbaatar, and Salekhard (ordered by increasing size of country), and then returns to Tallinn. When these are traced out on a map, they spell &lt;strong&gt;NO&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y3Kx5.jpg&quot; alt=&quot;1]&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using &quot;NO&quot; as the key for the Vigenere cipher reveals the question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can jet fuel melt steel beams?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-09-14T02:43:14.203" LastActivityDate="2015-09-14T02:43:14.203" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="19953" PostTypeId="1" AcceptedAnswerId="20003" CreationDate="2015-08-12T12:05:56.567" Score="3" ViewCount="548" Body="&lt;p&gt;These terms are all decoded in a certain way or they follow a certain rule. Your job is to figure out how to encode the last term.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 = 22&#xA;3 = 38&#xA;6 = 34&#xA;7 = 41&#xA;14 = 68&#xA;11 = ?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;br&gt;&#xA;There was a mistake in the earlier version of the question, which has been rectified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Math isn't the important here ;) It is however required &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11899" LastEditorUserId="11899" LastEditDate="2015-08-13T14:11:29.567" LastActivityDate="2015-08-13T15:12:33.223" Title="Math is Hard ! What is the Rule?" Tags="|mathematics|cipher|" AnswerCount="2" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19954" PostTypeId="2" ParentId="19914" CreationDate="2015-08-12T12:18:27.840" Score="0" Body="&lt;p&gt;With 2 balls, the strategy seems to be to use the first to determine a window of contiguous possibilities, and the second to step through the window one at a time. This answer mainly tackles the case where the windows are of fixed size, and leaves it to others to make comparisons with the efficiency of algorithms that don't assume windows of fixed size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume fixed-sized windows of $w-1$ possibilities, we have $N/w$ attempts with the first ball and at worst $w-1$ attempts with the second. On average, we would require half of those attempts in the case of each ball. We therefore want to minimise $(N/w + w-1)/2$, which (by setting its first derivative to 0 and confirming the second derivative is positive) happens when $N = w^2$. &lt;strong&gt;This requires a total of $\sqrt{N}$ drops on average&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the strategy is to start at floor $\sqrt{N}$ and keep dropping the first ball every $\sqrt{N} + 1$ floors until it breaks or until you run out of floors. If it breaks, go down to the floor above the penultimate attempt, and try every floor with the second ball. If the first ball didn't break (and there aren't another $\sqrt{N} + 1$ floors to go), start with the floor above the last attempt and check every floor from there with either ball. (Rounding $\sqrt{N}$: either way will work - round down to do one extra drop with the first ball and round up to do one extra drop with the second, though you probably need to be consistent if you modify the algorithm to use variable window sizes.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note 1: if we move to variable-sized windows, then observe that after the first drop (if the ball doesn't break), the problem reduces to an identical problem on a building with $N - \sqrt{N}$ floors. If we take the square root of that number as the number of floors to skip for the next attempt, we pair fewer drops of the second ball with more drops of the first ball. The last two floors will be tried one at a time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note 2: the constant window solution independently brings out the $\sqrt{N}$ factor like Going hamateur's answer, while the variable window comment is reminiscent of supercat's answer, though the step sizes are different.&lt;/p&gt;&#xA;" OwnerUserId="9422" LastActivityDate="2015-08-12T12:18:27.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19955" PostTypeId="2" ParentId="19823" CreationDate="2015-08-12T13:08:34.477" Score="3" Body="&lt;p&gt;I think the answer should be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (3) because for every figure you see in a row one end of two lines lies on one side and other end on another side and also 2 out of 3 of such figures have such sides opposite to each other and one has it adjacent (note having line start at edge should be interpreted as in a favor of this hypothesis i.e. benefit of doubt goes to me).  Same pattern is seen in row 2 and hence extrapolating it, I eliminate the eight-options to arrive at (3) which satisfies this rule. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15622" LastEditorUserId="13101" LastEditDate="2015-08-12T14:22:54.970" LastActivityDate="2015-08-12T14:22:54.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19956" PostTypeId="2" ParentId="19914" CreationDate="2015-08-12T13:35:33.770" Score="0" Body="&lt;p&gt;I'll just throw out a quick answer of $\frac{K \times N^{1/K}}{2}$ where K = balls, N = floors&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For N = 100, K = 2, this gives an answer of 10, or $\sqrt{100}$ that others have given.&lt;br&gt;&#xA;For N = 1000, K = 3, this gives an answer of 15.&lt;br&gt;&#xA;For N = 10000, K = 4, this gives an answer of 20.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Effectively, the fastest way to partition the building is to take the Kth root, but since you have equal probability of testing each value in a root ($\frac{1}{2}$ factor), and you have to repeat over K roots, you get your formula. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-08-12T13:35:33.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19957" PostTypeId="2" ParentId="19870" CreationDate="2015-08-12T13:45:48.540" Score="8" Body="&lt;p&gt;This is a proof that this problem is not solvable in less than 16 pieces. (Thanks to AndyT for the suggestion). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the weight of the bar is &lt;code&gt;504&lt;/code&gt;. Again, &lt;code&gt;504&lt;/code&gt; is just for brevity and has no effect on the proof. We need to partition &lt;code&gt;504&lt;/code&gt; into parts which can be rearranged to form blocks of &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;code&gt;9&lt;/code&gt; X &lt;code&gt;56&lt;/code&gt;, &lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;8&lt;/code&gt; X &lt;code&gt;63&lt;/code&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;7&lt;/code&gt; X &lt;code&gt;72&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note that the largest piece size can be &lt;code&gt;56&lt;/code&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Less than 16 pieces&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we have to partition less than 16 pieces in 8 blocks of &lt;code&gt;63&lt;/code&gt;, one of those blocks must be a singleton (ie. contains a single element), hence must be the set &lt;code&gt;{63}&lt;/code&gt;. But no piece can be larger than &lt;code&gt;56&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="15544" LastEditorUserId="15544" LastEditDate="2015-09-21T20:12:10.200" LastActivityDate="2015-09-21T20:12:10.200" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="19958" PostTypeId="2" ParentId="19823" CreationDate="2015-08-12T13:48:07.957" Score="21" Body="&lt;p&gt;You're all thinking to hard. This is just a game of Tic Tac Toe. Since it is the turn of X, tiles 1, 4, 5, 6, and 7 would all be correct. I personally would go with 7 because it's a well drawn X.&lt;/p&gt;&#xA;" OwnerUserId="13290" LastActivityDate="2015-08-12T13:48:07.957" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19959" PostTypeId="2" ParentId="19791" CreationDate="2015-08-12T15:35:47.467" Score="3" Body="&lt;p&gt;I have what I think is the correct answer, but I can't explain all the clues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My initial 'aha!' came from clue 9, where I guessed that the 'he' might be referring to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Archimedes (&quot;give me a lever and a place to &lt;em&gt;stand&lt;/em&gt; and I will move the world&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which would make the answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;pi, as its sometimes known as Archimedes' Constant after his early derivation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Based on the bonus clue 14, I imagine that the actual answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;an approximation of pi, like 3.14, and the proper terminology would be &quot;just a few &lt;em&gt;digits&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, the justifications that I have (some shaky, some firm).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Pi is an irrational number (har har) and it is not known whether it ever repeats itself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This probably refers to tau, which for those who don't know is a constant equal to 2*pi, which makes more sense for most applications (radius/circumference ratio instead of diameter, tau radians instead of 2*pi radians, etc.) Many scholars would argue that if you derive pi, you only have half the correct answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Instead of giving a hint, this refers to the fact that pi begins with a 3 and a decimal place- '3.'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Sine waves return to the origin at intervals of pi&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;7.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The most straightforward. The ratio of the diameter to the circumference of the wheel is ~pi, but pi is not a number you can bet on.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;9.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first part of this may refer to Archimedes' death and his attributed last words, &quot;Do not disturb my circles!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;10.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I don't know if people have phobias about pi, but its symbol is a Greek letter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;15.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Tattoos of pi are not uncommon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For all the other clues, I've got nothing.&lt;/p&gt;&#xA;" OwnerUserId="14441" LastActivityDate="2015-08-12T15:35:47.467" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19961" PostTypeId="2" ParentId="19947" CreationDate="2015-08-12T15:43:26.523" Score="8" Body="&lt;p&gt;Some of the phrasings made me think of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the ever-classic rhyming answers,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so here's a stab at some of them:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What a conductor gives to an out-of-tune player?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A chance to retune with a &lt;strong&gt;trombone tone&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sliced in private&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Sounds like they &lt;strong&gt;peeled concealed&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A little bed made of bones&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;What smaller bed is there, than a &lt;strong&gt;rib crib&lt;/strong&gt;? (Thank you for the idea, Bailey M!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A long lasting weird religion?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I'm not too sure, but calling it a &lt;strong&gt;habitual ritual&lt;/strong&gt; makes the most sense to me.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Exact copy of Warhammer people&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;While unfortunately unfamiliar with Warhammer, a browsing of the wiki makes me guess... &lt;strong&gt;drone clone&lt;/strong&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14107" LastEditorUserId="14107" LastEditDate="2015-08-12T16:42:07.457" LastActivityDate="2015-08-12T16:42:07.457" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="19962" PostTypeId="1" AcceptedAnswerId="19964" CreationDate="2015-08-12T16:11:40.307" Score="6" ViewCount="252" Body="&lt;p&gt;Once you figure out &lt;a href=&quot;http://www.xkcd.com/1563/&quot; rel=&quot;nofollow&quot;&gt;these&lt;/a&gt;, try a few more.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;The Nice, the Mean, and the Hideous&lt;/li&gt;&#xA;  &lt;li&gt;The Fourth Prime&lt;/li&gt;&#xA;  &lt;li&gt;The Linear Transformation&lt;/li&gt;&#xA;  &lt;li&gt;Psychotic Supremum: Stormy Street&lt;/li&gt;&#xA;  &lt;li&gt;Woman's&lt;/li&gt;&#xA;  &lt;li&gt;A Dozen Displeased Dudes&lt;/li&gt;&#xA;  &lt;li&gt;Missing Chest Thieves&lt;/li&gt;&#xA;  &lt;li&gt;The Second Period: 24-hour Evaluation&lt;/li&gt;&#xA;  &lt;li&gt;The Principal Panthera Leo&lt;/li&gt;&#xA;  &lt;li&gt;WeirdSex, MD: The process of apprehension removal and radioactive device appreciation&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whoever answers all / the final unanswered one gets the check mark.&lt;/p&gt;&#xA;" OwnerUserId="2722" LastEditorUserId="4421" LastEditDate="2015-08-12T18:35:40.693" LastActivityDate="2015-08-12T18:36:15.993" Title="More Synonym Movies" Tags="|knowledge|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19963" PostTypeId="1" AcceptedAnswerId="20023" CreationDate="2015-08-12T16:12:11.190" Score="9" ViewCount="1912" Body="&lt;p&gt;Whatever you do, don't try to find a common link between the following list of words. There isn't one, I promise. And don't try to find a secret hidden word. There's not one of those, either.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$along, soothe, haunt, remain, leap, candle, aurist$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Like I said, nothing to find here... just move on...&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clarification:&lt;/strong&gt; All of my nonsensical denying is just for fun. The entire puzzle could theoretically be solved with just the seven words and nothing else, but I'm a sucker for some silly exposition.&lt;/p&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="11475" LastEditDate="2015-08-12T20:11:24.477" LastActivityDate="2015-08-13T20:49:52.463" Title="Seven Totally Unrelated Words" Tags="|word|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="19964" PostTypeId="2" ParentId="19962" CreationDate="2015-08-12T16:14:19.257" Score="5" Body="&lt;ol&gt;&#xA;&lt;li&gt;The Nice, the Mean, and the Hideous&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Good, the Bad, and the Ugly&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;The Fourth Prime&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Seven (or Se7en)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;The Linear Transformation&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Matrix&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;Psychotic Supremum: Stormy Street&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mad Max: Fury Road (thanks, Roland!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;Woman's&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Her&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;6&quot;&gt;&#xA;&lt;li&gt;A Dozen Displeased Dudes&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 12 Angry Men&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;7&quot;&gt;&#xA;&lt;li&gt;Missing Chest Thieves&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Raiders of the Lost Ark&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;8&quot;&gt;&#xA;&lt;li&gt;The Second Period: 24-hour Evaluation&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Terminator 2: Judgment Day&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;9&quot;&gt;&#xA;&lt;li&gt;The Principal Panthera Leo&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Lion King&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;10&quot;&gt;&#xA;&lt;li&gt;WeirdSex, MD: The process of apprehension removal and radioactive device appreciation&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dr. Strangelove or: How I Learned to Stop Worrying and Love the Bomb (thanks, Vance!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="4421" LastEditDate="2015-08-12T18:36:15.993" LastActivityDate="2015-08-12T18:36:15.993" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19965" PostTypeId="2" ParentId="19953" CreationDate="2015-08-12T16:54:48.120" Score="5" Body="&lt;p&gt;I am very unsure about my solution, but I found something interesting.&#xA;First I observed that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the positional value (that is a=1,b=2, etc., then all added up) of the word of the number on the left hand side has the same remainder after integer division by 6. I will write $a(n)$ for the alpha-numerical value of the word of the number $n$. Then we have $a(1)=34=22+2 \cdot 6, a(3)=56=38+3\cdot 6, a(6)=52=34+3\cdot 6, a(7)=65=41+4\cdot 6, a(14)=104=68+6 \cdot 6$. For $n=11$ this would give us $a(11)=63=33+5 \cdot 6$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Maybe this is all bull, but so far I think that the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;11 = 33&quot; - but maybe its wrong and my idea can be used otherwise.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14385" LastEditorUserId="11475" LastEditDate="2015-08-12T18:34:56.610" LastActivityDate="2015-08-12T18:34:56.610" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19966" PostTypeId="1" AcceptedAnswerId="19967" CreationDate="2015-08-12T16:59:35.110" Score="7" ViewCount="1203" Body="&lt;p&gt;My gramma always use to say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you replace our with if you'll end up one later&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;never quite understood it...&lt;/p&gt;&#xA;" OwnerUserId="13916" LastActivityDate="2015-08-12T19:02:31.847" Title="She could well be crazy" Tags="|wordplay|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19967" PostTypeId="2" ParentId="19966" CreationDate="2015-08-12T17:01:03.187" Score="17" Body="&lt;p&gt;Well, I think your gramma was talking about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; F&lt;strong&gt;our&lt;/strong&gt;th and f&lt;strong&gt;if&lt;/strong&gt;th, where you replace 'our' with 'if' and end up one later in the count.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-08-12T17:01:03.187" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19968" PostTypeId="2" ParentId="19947" CreationDate="2015-08-12T17:27:04.027" Score="5" Body="&lt;h2&gt;Partial answer&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Inspired by &lt;a href=&quot;https://puzzling.stackexchange.com/a/19961/4421&quot;&gt;@Vance's answer&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each line refers to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a duplicated word:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What a conductor gives to an out-of-tune player?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a &lt;strong&gt;note note&lt;/strong&gt;  (an area to improve with his musical notation)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You perform this in Spring&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a &lt;strong&gt;March march&lt;/strong&gt; (a certain type of music performed in the third month)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...unfortunately, I can't find answers for any of the rest.  Can anyone else think of any?&lt;/p&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-08-12T17:27:04.027" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19969" PostTypeId="1" AcceptedAnswerId="20241" CreationDate="2015-08-12T17:51:13.620" Score="7" ViewCount="381" Body="&lt;p&gt;Who am I talking about?  Try to explain each part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During the square, it was a quite common thing that he said.  And lets not forget to chastise the howling canine(of sorts).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During the third to six and eight, it was in French.  And out goes the flower and in with the secretary and Shakespeare's friend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During the odious number, if you added a certain cheese, it became a mouse.  Impossibility works quite well for a mnemonic device.  Also, remember the stagnant water left years past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once, arguably, a perfect trinity: three as one and one as three.  But along came the dinosaur, and coughed up a little something and then the three became four.&lt;/p&gt;&#xA;" OwnerUserId="11790" LastEditorUserId="11790" LastEditDate="2015-08-18T15:17:51.367" LastActivityDate="2015-08-18T17:49:16.537" Title="My name is a promise, of sorts" Tags="|riddle|knowledge|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19970" PostTypeId="2" ParentId="18266" CreationDate="2015-08-12T18:11:35.730" Score="2" Body="&lt;p&gt;&lt;strong&gt;Partial Answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Even though I honestly did not look at the comments before I made the connection, I must give some credit to &lt;a href=&quot;https://puzzling.stackexchange.com/users/11717/logicianwithahat&quot;&gt;LogicianWithAHat&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you in over your head?&lt;br&gt;&#xA;Are you in water so deep you're drowning?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Both of the lines above relate to &lt;a href=&quot;http://www.azlyrics.com/lyrics/building429/listentothesound.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Listen to the Sound by Building 429&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Don't blame it on the sunshine,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.azlyrics.com/lyrics/jacksons/blameitontheboogie.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Blame it on the Boogie by the Jackson 5&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Just remember your ABC's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.azlyrics.com/lyrics/jackson5/abc.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;ABC by the Jackson 5&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This stanza contains directions for the first stanza. Roots leads me to think about square roots and Primarily leads to prime numbers. We know that no perfect square can ever be a prime number (except 1). The last line is slightly more confusing; could it be talking about division/multiplication?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now go back to the roots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Primarily, take what you get,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Break it up and regroup slightly differently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have you got a nice little dessert? Use what's left over.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This stanza contains directions for what to do with the next two. I'm inclined to believe we want to use the second primary number. Roots must not imply square roots here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Next up, use what is primary-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet not in the same way as before-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From your roots(which are slightly different now),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A nice ole line for y'all.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I wish there was another you, another way,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.azlyrics.com/lyrics/againstthecurrent/anotheryouanotherway.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Another You by Against the Current&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;'Cause you got me in chains&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.azlyrics.com/lyrics/nickjonas/chains.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chains by Nick Jonas&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though you are, you are, you are, my freedom&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.azlyrics.com/lyrics/hillsongyoungfree/alive.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Alive by Hillsong Young &amp;amp; Free&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'd like to make myself believe that planet Earth turns slowly.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.azlyrics.com/lyrics/owlcity/fireflies.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fireflies by Owl City&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Hands up, when the music drops, we both put our hands up.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.directlyrics.com/usher-dj-got-us-fallin-in-love-lyrics.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;DJ Got us Falling in Love by Usher&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I'm gonna swing from the chandelier.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.azlyrics.com/lyrics/sia/chandelier.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chandelier by Sia&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So wake me up when it's all over&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.azlyrics.com/lyrics/avicii/wakemeup.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wake Me Up by Avicii&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;'Cause I wanna be your left hand man.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.azlyrics.com/lyrics/vancejoy/riptide.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Riptide by Vance Joy&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Directions for the next stanzas. We may want to use the same numbers here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Next, gather the same ingredients as before,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and take the last of the first(or the first of the last?) among them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the latter.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The lines in this stanza are quotes, not song lyrics.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If they don't stand for something, they will fall for anything (side note, this one's a bit ambiguous, probably others as well. This is not part of the clue)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.brainyquote.com/quotes/quotes/p/petermarsh382875.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Peter Marshall&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though the best things in life are not things.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.goodreads.com/quotes/178740-the-best-things-in-life-aren-t-things&quot; rel=&quot;nofollow noreferrer&quot;&gt;Art Buchwald&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So go instead there is not path and leave a trail.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://www.google.com/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=1&amp;amp;cad=rja&amp;amp;uact=8&amp;amp;ved=0CB8QFjAAahUKEwjGyoubkqTHAhUEJx4KHVRAAMs&amp;amp;url=http%3A%2F%2Fwww.brainyquote.com%2Fquotes%2Fquotes%2Fr%2Fralphwaldo101322.html&amp;amp;ei=KIzLVcaWFoTOeNSAgdgM&amp;amp;usg=AFQjCNEZ1-px3IcIcCM7X0s8P5i0E13N0g&amp;amp;sig2=8HkzkhdWckKZdWs8yg8Mbg&amp;amp;bvm=bv.99804247,d.dmo&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ralph Waldo Emerson&lt;/a&gt;    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Without music, life would be a mistake,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.brainyquote.com/quotes/quotes/f/friedrichn102457.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Friedrich Nietzsche&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And though all stories are true, some never happened&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://www.google.com/search?q=And%20though%20all%20stories%20are%20true%2C%20some%20never%20happened&amp;amp;oq=And%20though%20all%20stories%20are%20true%2C%20some%20never%20happened&amp;amp;aqs=chrome..69i57.240j0j4&amp;amp;sourceid=chrome&amp;amp;es_sm=122&amp;amp;ie=UTF-8&quot; rel=&quot;nofollow noreferrer&quot;&gt;James A. Owen&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You should be yourself, everybody else is already taken.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://www.google.com/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=6&amp;amp;cad=rja&amp;amp;uact=8&amp;amp;ved=0CC4QFjAFahUKEwimgNGzmKTHAhUFJh4KHVsVANc&amp;amp;url=http%3A%2F%2Fwww.goodreads.com%2Fquotes%2F19884-be-yourself-everyone-else-is-already-taken&amp;amp;ei=ppLLVaaVG4XMeNuqgLgN&amp;amp;usg=AFQjCNFa97tTDCMp391MDX9eI9NTOKdjfw&amp;amp;sig2=WTc2e1LCM32wAzlSHmoypg&amp;amp;bvm=bv.99804247,d.dmo&quot; rel=&quot;nofollow noreferrer&quot;&gt;Oscar Wilde&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Following the directions of two stanzas before:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The last of the first (last name of the first) is Marshall; the first of the last (first name of the last) is Oscar. I'd like it to be the last of the last, so we'd get Wilde, which is very close to the word wild. After all, the clue says &quot;probably the latter&quot;; this could be taken to mean we want the first of the last or we could think more (probably too much) and take the latter of the last.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now everything is in order,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but wait right there!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's the water, not the sky. Ironically.&lt;/p&gt;&#xA;" OwnerUserId="13538" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-08-13T16:25:53.023" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="19971" PostTypeId="1" AcceptedAnswerId="19975" CreationDate="2015-08-12T18:38:17.100" Score="14" ViewCount="1329" Body="&lt;p&gt;In the fantasy world of this puzzle,&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Heaven has 7 levels (1 to 7), with 7 being God's level.&lt;/li&gt;&#xA;&lt;li&gt;Hell has 8 levels (-1 to -8), with -8 being Satan's level.&lt;/li&gt;&#xA;&lt;li&gt;Purgatory (0) is somewhere in between.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your task:&lt;/strong&gt;&lt;br&gt;Use the partial lists below to &lt;strong&gt;figure out how God has decided&lt;/strong&gt; which of the 3 places each word will go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5dk5R.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5dk5R.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Below I have included the same information as above, except in text form, since some people prefer that.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Words going to Heaven:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;sub&gt;(sorted by level ascending, then a-z)&lt;/sub&gt;&lt;br&gt;&#xA;  BADGES&lt;br&gt;&#xA;  BOUNTIES&lt;br&gt;&#xA;  PROBABILITY&lt;br&gt;&#xA;  PUZZLERS&lt;br&gt;&#xA;  QUESTIONS&lt;br&gt;&#xA;  SEQUENCES&lt;br&gt;&#xA;  STACK&lt;br&gt;&#xA;  FREQUENT&lt;br&gt;&#xA;  MODERATORS&lt;br&gt;&#xA;  OVERFLOW&lt;br&gt;&#xA;  ANSWERS&lt;br&gt;&#xA;  CENSORSHIP&lt;br&gt;&#xA;  DEFINABILITY (on God's level)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Words going to Hell:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;sub&gt;(sorted by level descending, then a-z)&lt;/sub&gt;&lt;br&gt;&#xA;  PATTERN&lt;br&gt;&#xA;  REPUTATION&lt;br&gt;&#xA;  RIDDLES&lt;br&gt;&#xA;  THINKING&lt;br&gt;&#xA;  UNANSWERED&lt;br&gt;&#xA;  WORDPLAY&lt;br&gt;&#xA;  CALCULATION&lt;br&gt;&#xA;  ENIGMATIC&lt;br&gt;&#xA;  MATHEMATICS&lt;br&gt;&#xA;  RHYME&lt;br&gt;&#xA;  FEATURED&lt;br&gt;&#xA;  DUPLICATED&lt;br&gt;&#xA;  MICROELECTRONIC (on Satan's level)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Words going to Purgatory:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;LATERAL&lt;br&gt;&#xA;  LOGIC&lt;br&gt;&#xA;  NEWBIES&lt;br&gt;&#xA;  PASSWORD&lt;br&gt;&#xA;  SPAMMERS&lt;br&gt;&#xA;  USERS&lt;br&gt;&#xA;  VOTES&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="463" LastEditorUserId="30633" LastEditDate="2023-04-07T16:48:58.097" LastActivityDate="2023-04-07T16:48:58.097" Title="All Words Go to Heaven?" Tags="|pattern|calculation-puzzle|word-property|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19972" PostTypeId="2" ParentId="19963" CreationDate="2015-08-12T18:42:21.257" Score="2" Body="&lt;p&gt;I almost believed you that there was no common link. And was almost off on my way to other puzzles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;h2&gt;But...&lt;/h2&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wasn't going to fall for your trickery. And these words that you so loudly proclaimed were common unrelated words, they are all cut from the same cloth. They are tricksters, dropping away a part of their identity, changing a bit of their remaining traits, and blending in with the populace without suspicion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your plan was good, far too good. &lt;em&gt;But I'm better...&lt;/em&gt; And I've unmasked your hidden thugs. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In each word, drop a letter, and then replace one other letter to get a different word. &lt;br&gt;&lt;br&gt;along-n#r(l,g)=agog &lt;br&gt;soothe-e#r(s,b)=booth &lt;br&gt;haunt-a#r(t,k)=hunk &lt;br&gt;remain-a#r(m,s)=resin &lt;br&gt;leap-p#r(a,d)=led &lt;br&gt;candle-l#r(e,y)=candy &lt;br&gt;aurist-u#r(t,e)=arise&lt;br&gt;&lt;br&gt; The minus sign subtracts the specified letter, whereas the #r operation replaces the specified letter with another letter. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-08-12T18:47:47.743" LastActivityDate="2015-08-12T18:47:47.743" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19975" PostTypeId="2" ParentId="19971" CreationDate="2015-08-12T19:38:50.447" Score="11" Body="&lt;p&gt;The pattern is determined by:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Going up or down in the alphabet, where going up is equal to $+1$, and going down is equal to $-1$, with no change being equal to $0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For Example:    &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BADGES = B $\rightarrow$A($-1$); A$\rightarrow$D ($+1$); D$\rightarrow$G ($+1$); G$\rightarrow$E($-1$); E$\rightarrow$S($+1$): &lt;br&gt;&#xA; $-1 + 1 + 1 -1 + 1 = 1$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Other Words going to Heaven&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;$\begin{align}&#xA;&amp;amp;\text{BOUNTIES} &amp;amp;= &amp;amp;+1 + 1 - 1 + 1 -1 -1 + 1 &amp;amp;= 1\\&#xA;&amp;amp;\text{PROBABILITY} &amp;amp;= &amp;amp;+1 -1 -1-1+1+1+1-1+1+1 &amp;amp;= 2\\&#xA;&amp;amp;\text{PUZZLERS} &amp;amp;= &amp;amp;+1 +1 + 0 -1 -1 +1 +1 &amp;amp;=2\\&#xA;&amp;amp;\text{QUESTIONS} &amp;amp;= &amp;amp;+1-1+1+1-1+1-1+1 &amp;amp;= 2\\&#xA;&amp;amp;\text{SEQUENCES} &amp;amp;= &amp;amp;-1+1+1-1+1-1+1+1 &amp;amp;= 2\\&#xA;&amp;amp;\text{STACK} &amp;amp;= &amp;amp;+1-1+1+1 &amp;amp;= 2\\&#xA;&amp;amp;\text{FREQUENT} &amp;amp;= &amp;amp;+1-1+1+1-1+1+1 &amp;amp;= 3\\&#xA;&amp;amp;\text{MODERATORS} &amp;amp;= &amp;amp;+1-1+1+1-1+1-1+1+1 &amp;amp;= 3\\&#xA;&amp;amp;\text{OVERFLOW} &amp;amp;= &amp;amp;+1 -1+1-1+1+1+1 &amp;amp;= 3\\&#xA;&amp;amp;\text{ANSWERS} &amp;amp;= &amp;amp;+1+1+1-1+1+1 &amp;amp;= 4\\&#xA;&amp;amp;\text{CENSORSHIP} &amp;amp;= &amp;amp;+1+1+1-1+1+1-1+1+1 &amp;amp;= 5\\&#xA;&amp;amp;\text{DEFINABILITY} &amp;amp;= &amp;amp;+1+1+1+1-1+1+1+1-1+1+1 &amp;amp;= 7\\&#xA;&amp;amp;\qquad\text{ (on God's level)}&#xA;\end{align}$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Words going to Hell&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;&amp;amp;\text{PATTERN} &amp;amp;= &amp;amp; -1 + 1 +0-1+1-1&amp;amp;= -1\\&#xA;&amp;amp;\text{REPUTATION} &amp;amp;= &amp;amp; -1+1+1-1-1+1-1+1-1&amp;amp;= -1\\&#xA;&amp;amp;\text{ RIDDLES} &amp;amp;= &amp;amp; -1-1+0+1-1+1&amp;amp;= -1\\&#xA;&amp;amp;\text{THINKING} &amp;amp;= &amp;amp; -1+1+1-1-1+1-1&amp;amp;= -1\\&#xA;&amp;amp;\text{UNANSWERED} &amp;amp;= &amp;amp; -1-1+1+1+1-1+1-1-1&amp;amp;= -1\\&#xA;&amp;amp;\text{WORDPLAY} &amp;amp;= &amp;amp; -1+1-1+1-1-1+1&amp;amp;= -1\\&#xA;&amp;amp;\text{CALCULATION} &amp;amp;= &amp;amp; -1+1-1+1-1-1+1-1+1-1&amp;amp;= -1\\&#xA;&amp;amp;\text{ENIGMATIC} &amp;amp;= &amp;amp; +1-1-1+1-1+1-1-1&amp;amp;= -2\\&#xA;&amp;amp;\text{MATHEMATICS} &amp;amp;= &amp;amp; -1+1-1-1+1-1+1-1-1+1&amp;amp;= -2\\&#xA;&amp;amp;\text{RHYME} &amp;amp;= &amp;amp; -1+1-1-1&amp;amp;= -2\\&#xA;&amp;amp;\text{FEATURED} &amp;amp;= &amp;amp; -1-1+1+1-1-1-1&amp;amp;= -3\\&#xA;&amp;amp;\text{DUPLICATED} &amp;amp;= &amp;amp; +1-1-1-1-1-1+1-1-1&amp;amp;= -5\\&#xA;&amp;amp;\text{MICROELECTRONIC} &amp;amp;= &amp;amp; -1-1+1-1-1+1-1-1+1-1-1-1-1-1&amp;amp;= -8\\&#xA;&amp;amp;\qquad\text{ (on Satan's level)}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Words going to Purgatory&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;$\begin{align}&#xA;&amp;amp;\text{LATERAL} &amp;amp;= &amp;amp; -1+1-1+1-1+1 &amp;amp;= 0\\&#xA;&amp;amp;\text{LOGIC} &amp;amp;= &amp;amp; +1-1+1-1 &amp;amp;= 0\\&#xA;&amp;amp;\text{NEWBIES} &amp;amp;= &amp;amp; -1+1-1+1-1+1 &amp;amp;= 0\\&#xA;&amp;amp;\text{PASSWORD} &amp;amp;= &amp;amp; -1+1+0+1-1+1-1 &amp;amp;= 0\\&#xA;&amp;amp;\text{SPAMMERS} &amp;amp;= &amp;amp; -1-1+1+0-1+1+1 &amp;amp;= 0\\&#xA;&amp;amp;\text{USERS} &amp;amp;= &amp;amp; -1-1+1+1 &amp;amp;= 0\\&#xA;&amp;amp;\text{VOTES} &amp;amp;= &amp;amp; -1+1-1+1 &amp;amp;= 0\\&#xA;\end{align}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please check my math. There are a lot of letters here and I'm bound to have messed up somewhere.  &lt;/p&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="4421" LastEditDate="2015-08-12T20:41:06.303" LastActivityDate="2015-08-12T20:41:06.303" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="19977" PostTypeId="2" ParentId="19947" CreationDate="2015-08-12T20:11:09.457" Score="36" Body="&lt;p&gt;With much credit to the commenters, the answers are all&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; rearrangements of the letters ACELNOTU:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What a conductor gives to an out-of-tune player? - &lt;strong&gt;TONAL CUE&lt;/strong&gt; (&lt;em&gt;Gordon K&lt;/em&gt;)&lt;br&gt;&#xA; You perform this in Spring - &lt;strong&gt;CLEAN OUT&lt;/strong&gt; (&lt;em&gt;Lampost42&lt;/em&gt;)&lt;br&gt;&#xA; Sung by musical miners? - &lt;strong&gt;COAL TUNE&lt;/strong&gt;&lt;br&gt;&#xA; Those behind bars keeps tabs on this - &lt;strong&gt;ALE COUNT&lt;/strong&gt; (&lt;em&gt;Lampost42&lt;/em&gt;)&lt;br&gt;&#xA; Sounds something like a big colourful Spanish bird - &lt;strong&gt;EL TOUCAN&lt;/strong&gt; (&lt;em&gt;Bailey M&lt;/em&gt;)&lt;br&gt;&#xA; A pretty endearing advance - &lt;strong&gt;CUTE LOAN&lt;/strong&gt; (&lt;em&gt;Gordon K&lt;/em&gt;)&lt;br&gt;&#xA; Loosen your shoes - &lt;strong&gt;TO UNLACE&lt;/strong&gt;&lt;br&gt;&#xA; Find a French fishy ratio - &lt;strong&gt;LOCATE NU&lt;/strong&gt; (&lt;em&gt;GentlePurpleRain&lt;/em&gt;)&lt;br&gt;&#xA; A teenager causing trouble? - &lt;strong&gt;ACNE LOUT&lt;/strong&gt; (&lt;em&gt;corwin_&lt;/em&gt;)&lt;br&gt;&#xA; Scottish cooking method - &lt;strong&gt;NAE CLOUT&lt;/strong&gt; (&lt;em&gt;GentlePurpleRain&lt;/em&gt;)&lt;br&gt;&#xA; Sliced in private - &lt;strong&gt;CUT ALONE&lt;/strong&gt;&lt;br&gt;&#xA; A little bed made of bones - &lt;strong&gt;ULNAE COT&lt;/strong&gt;&lt;br&gt;&#xA; A long lasting weird religion? - &lt;strong&gt;AEON CULT&lt;/strong&gt; (&lt;em&gt;Gordon K&lt;/em&gt;)&lt;br&gt;&#xA; A cereal, relatively speaking - &lt;strong&gt;UNCLE OAT&lt;/strong&gt; (&lt;em&gt;Gordon K&lt;/em&gt;)&lt;br&gt;&#xA; Exact copy of Warhammer people - &lt;strong&gt;TAU CLONE&lt;/strong&gt; (&lt;em&gt;Gordon K&lt;/em&gt;)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Final Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Literally no idea - &lt;strong&gt;NOT A CLUE&lt;/strong&gt; (&lt;em&gt;Gordon K&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanations:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TONAL CUE: a &lt;em&gt;cue&lt;/em&gt; is an instruction or hint, and &lt;em&gt;tonal&lt;/em&gt; refers to pitch (or tone; being in tune).&lt;br&gt;&#xA; CLEAN OUT: &lt;a href=&quot;https://en.wikipedia.org/wiki/Spring_cleaning&quot;&gt;Spring cleaning&lt;/a&gt;&lt;br&gt;&#xA; COAL TUNE: Miners mine coal, and a &lt;em&gt;tune&lt;/em&gt; is a melody, or song.&lt;br&gt;&#xA; ALE COUNT: Bartenders (behind bars) keep a tab (&lt;em&gt;count&lt;/em&gt;) how many beers (&lt;em&gt;ales&lt;/em&gt;) patrons consume.&lt;br&gt;&#xA; EL TOUCAN: A toucan is a colorful bird, and &lt;em&gt;el&lt;/em&gt; is the Spanish masculine &quot;the&quot;.&lt;br&gt;&#xA; CUTE LOAN: An &lt;em&gt;advance&lt;/em&gt; is another word for a loan, and &lt;em&gt;endearing&lt;/em&gt; is a synonym for &lt;em&gt;cute&lt;/em&gt;.&lt;br&gt;&#xA; TO UNLACE: Shoes are often secured with laces.&lt;br&gt;&#xA; LOCATE NU: &lt;a href=&quot;https://en.wikipedia.org/wiki/Poisson%27s_ratio&quot;&gt;Poisson's ratio&lt;/a&gt; is denoted with the Greek letter &lt;em&gt;nu&lt;/em&gt;, and &lt;em&gt;poisson&lt;/em&gt; is French for &lt;em&gt;fish&lt;/em&gt;.&lt;br&gt;&#xA; ACNE LOUT: Teenagers often have acne, and a &lt;em&gt;lout&lt;/em&gt; is an ill-mannered, troublemaking person.&lt;br&gt;&#xA; NAE CLOUT: In Scotland, dumplings are often cooked in the modern way without cloth, or in Scots, &lt;em&gt;nae clout&lt;/em&gt;.&lt;br&gt;&#xA; CUT ALONE: &lt;em&gt;Cut&lt;/em&gt; means &lt;em&gt;slice&lt;/em&gt;, and something done in private is done &lt;em&gt;alone&lt;/em&gt;.&lt;br&gt;&#xA; ULNAE COT: &lt;em&gt;Ulnae&lt;/em&gt; is the plural of &lt;em&gt;ulna&lt;/em&gt;, one of the bones of the forearm, and a &lt;em&gt;cot&lt;/em&gt; is a small bed.&lt;br&gt;&#xA; AEON CULT: &lt;em&gt;Aeon&lt;/em&gt; is an alternate spelling of &lt;em&gt;eon&lt;/em&gt;, which is a very long time, and a weird religion is a &lt;em&gt;cult&lt;/em&gt;.&lt;br&gt;&#xA; UNCLE OAT: Oats are a cereal grain, and if one is your uncle, he is a relative!&lt;br&gt;&#xA; TAU CLONE: A &lt;em&gt;clone&lt;/em&gt; is an exact copy, and in the Warhammer universe, there is a race known as the &lt;em&gt;Tau&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4915" LastEditorUserId="14177" LastEditDate="2015-08-13T13:53:26.397" LastActivityDate="2015-08-13T13:53:26.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19978" PostTypeId="2" ParentId="19969" CreationDate="2015-08-12T20:58:45.393" Score="3" Body="&lt;p&gt;Is the answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sherlock Holmes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;During the square, it was a quite common thing that he said. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Trafalgar Square appears in several of Doyle's works.  &quot;Elementary, my dear Watson&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And lets not forget to chastise the howling canine(of sorts).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Hound of the Baskervilles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;During the third to six and eight, it was in French. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not too sure about this one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And out goes the flower and in with the secretary and Shakespeare's friend.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not too sure about flower.  Are you referring to &lt;em&gt;The Italian Secretary&lt;/em&gt;, the novel?  As for Shakespeare's friend, Doyle often has Holmes quote his plays - and respects Shakespeare's writings immensely. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;During the odious number, if you added a certain cheese, it became a mouse.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Steamboat Willie was voted 13th (the odious number) greatest cartoon.  Also, Mickey Mouse became Sherlock Mouse in another comic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Impossibility works quite well for a mnemonic device. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He's often quoted with &quot;Once you've eliminated the impossible....&quot;  The mnemonic device seems to tie into the &quot;stagnant water&quot; bit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, remember the stagnant water left years past.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A lake may have stagnant waters, and HOMES is a mnemonic device for the great lakes (Huron, Ontario, Michigan, Erie, Superior).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13920" LastActivityDate="2015-08-12T20:58:45.393" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19979" PostTypeId="2" ParentId="19809" CreationDate="2015-08-12T20:59:39.597" Score="1" Body="&lt;p&gt;Even ignoring the midnight issue, with a little bit of lateral thinking, this breaks down. It only works if we assume the stranger must make a statement that covers the entire population. If we allow statements involving subdivisions among the villagers, he can say &quot;At least one left-handed villager has a blue dot,&quot; and only the left-handed villagers will kill themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, It's possible to say something about the entire population that won't cause any suicides:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Dividing the villagers into men, women, and children under 18, there is at least one pair of villagers who are in different groups and have the same color dot.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As long as no group is empty, there is no possible arrangement of dots that will result in a suicide on the first day. WLOG, consider a man. If he sees that all women and children have the same color dot, then there's the pair right there. Otherwise, there's at least one woman or child with a red dot and one with a blue dot, so even if the men are monochrome there's a valid pair with one of them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12566" LastActivityDate="2015-08-12T20:59:39.597" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19980" PostTypeId="2" ParentId="19823" CreationDate="2015-08-12T21:07:23.963" Score="5" Body="&lt;p&gt;The answer is...&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Tile 5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Because...&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Tile 5 is the only one represented on the bottom that doesn't have a similar tile on the top. I'm not sure why there's so much reading into this...  All the other tiles have a similar one on the top even if you have to flip it, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15643" LastEditorUserId="71981" LastEditDate="2020-11-15T14:32:04.310" LastActivityDate="2020-11-15T14:32:04.310" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19981" PostTypeId="2" ParentId="19971" CreationDate="2015-08-12T21:10:07.193" Score="7" Body="&lt;p&gt;&lt;strong&gt;Just a survey of Good and Bad words&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I experimented on a dictionary with some python code. Most of God level and Satan level words are technical jargon though. So, I chose only a few. &lt;a href=&quot;http://pastebin.com/nk5xsyfJ&quot;&gt;Here's the full list.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;God Level (Score 7)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;code&gt;acceptability&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;acquaintanceship&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;biodegradability&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;definability&lt;/code&gt; &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Beyond God Level (8)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;code&gt;acquaintanceships&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;actinostereoscopy&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;biopsychologist&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;intersubstitutability&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;psychopannychist&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Satan Level (-8)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;code&gt;microelectronic&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;microspectrophotometric&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;spectrometric&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I couldn't find any beyond satan level with a ~700,000 word dictionary.&#xA;Also as dperry pointed out &lt;a href=&quot;http://www.sarahmcculloch.com/luminary-uprise/2009/longest-word/&quot;&gt;the longest technical word&lt;/a&gt; has a huge score of 829. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also of 675224 of the dictionary I used,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;259064 are going to heaven&lt;/li&gt;&#xA;&lt;li&gt;254621 are going to hell &lt;/li&gt;&#xA;&lt;li&gt;161539 are going to the purgatory&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Seems like a lot of confused souls. &lt;a href=&quot;http://pastebin.com/KDykJfX5&quot;&gt;Detailed results here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15544" LastActivityDate="2015-08-12T21:10:07.193" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19982" PostTypeId="1" AcceptedAnswerId="19985" CreationDate="2015-08-12T21:27:48.913" Score="5" ViewCount="509" Body="&lt;p&gt;Here are some Composite Words™:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ADEPTIO           DIVTVRNVS&#xA;EXILIS            MIXTVM&#xA;CVLTELLVS         ARMARIVM&#xA;ACERBVS           ALIOQVI&#xA;IMPRAESENTIARVM   CVRVVS&#xA;AMICVS            CONCVLCO&#xA;LACRIMABILIS      ABVNDANS&#xA;ADDO              ADEO&#xA;IMMVNDVS          INFLECTVM&#xA;ALTVS             LIQVIDVS&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here are some non-Composite Words™:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ANTIQVVS          LVXVRIA&#xA;AMICVLVM          CRINITVS&#xA;ABSQVE            IVDICIVM&#xA;MVNIMENTVM        EXCELLENTIA&#xA;CALLIDE           ECCLESIA&#xA;AMICITIA          MAXIME&#xA;IMPRIMIS          ALIQVANTVS&#xA;HAFFLIGENIENSIS   INCLITVS&#xA;CALAMITAS         CONIVRATVS&#xA;BAIVLVS           ARTIFICIOSE&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What property determines if I call it a Composite Word™ or not?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="11476" LastEditorUserId="30633" LastEditDate="2016-11-08T00:07:09.353" LastActivityDate="2016-11-08T00:07:09.353" Title="What is a Composite Word™?" Tags="|pattern|language|number-theory|word-property|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19983" PostTypeId="2" ParentId="19947" CreationDate="2015-08-12T21:54:29.633" Score="5" Body="&lt;p&gt;Time for a puzzle review! This assumes you've already read the answer; here be spoilers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's great that people worked together to solve this puzzle. It's fun to participate in a group. This worked for the puzzle because even after you know the mechanism, each clue is interesting to solve. Moreover, you can start solving the clues and figure out the mechanism as you go. The clues are tricky, but once you get the right answer, you know it's right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The puzzle reminds me a lot of the MIT Mystery Hunt 2015 puzzle &lt;a href=&quot;http://web.mit.edu/puzzle/www/2015/puzzle/mit_mystery_hunt/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;MiT MYSTERY HUNT&quot;&lt;/a&gt;, which also gives clues to anagrams of a single phrase. There, the phrase to anagram is longer and explicitly signaled. I guess you're a fan of this mechanic from your &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18768/if-dogs-could-talk&quot;&gt;previous puzzle&lt;/a&gt;, though it's a bit unfortunate that one might have metagamed you using it again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final answer phrase is nice for being snappy and thematic, and also using common letters that lead to many phrases. I'd go for a longer phrase though to reduce the power of anagram solver brute forcing and give more potential answers to clue. As is, some of the answers felt a bit contrived.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final extraction was cute. I'm a fan of recursion, and the title is a natural place for a meta-cluephrase. I like that what you get from the clues is simply the method, and not something from the answers. This saves having an extraction step, which too often are artificial and get solvers stuck. On the other hand, it means one can extract the final answer with only a fraction of the answers. The puzzle does ask for the both the clue answers and the final answer, but this does lose elegance compared to having a single final answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure the solving instructions were needed. In particular, figuring out to look at the title would have been a nice moment if not given outright. But I guess it's good to err on the clearer side. It is nice to see a puzzle solved without hints added after posting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that the order of the clues and answer is arbitrary, convention is to signal this by listing in alphabetical order, either by clue or by answer. I prefer the second because it narrows down that pesky last clue you're stuck on. The title should be formatted like the clue (first word capitalized) since it serves the same function.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.923" LastActivityDate="2015-08-12T21:54:29.633" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="19984" PostTypeId="2" ParentId="19870" CreationDate="2015-08-12T21:55:32.337" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;Split the bar into 9 equal pieces.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Split two of the pieces into 28 (7x8/2) equal smaller pieces, for a total of 7 big pieces and 56 small pieces.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If 9 gangsters survive, 7 of them each get 1 big piece, and 2 each get 28 small pieces.&lt;/li&gt;&#xA;&lt;li&gt;If 8 gangsters survive, 7 of them each get 1 big piece and 7 small pieces, and the other gets 35 (28+7) small pieces.&lt;/li&gt;&#xA;&lt;li&gt;If 7 gangsters survive, they each get 1 big piece and 8 small pieces.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="15646" LastActivityDate="2015-08-12T21:55:32.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19985" PostTypeId="2" ParentId="19982" CreationDate="2015-08-12T22:04:04.950" Score="5" Body="&lt;p&gt;Composite words are those where the&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; sum of the Roman numeral values of the contained letters &lt;code&gt;IVXLCDM&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is equal to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a composite number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The relevant data for each word:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;ADEPTIO 501 [3, 167]&#xA; DIVTVRNVS 516 [2, 2, 3, 43]&#xA; EXILIS 62 [2, 31]&#xA; MIXTVM 2016 [2, 2, 2, 2, 2, 3, 3, 7]&#xA; CVLTELLVS 260 [2, 2, 5, 13]&#xA; ARMARIVM 2006 [2, 17, 59]&#xA; ACERBVS 105 [3, 5, 7]&#xA; ALIOQVI 57 [3, 19]&#xA; IMPRAESENTIARVM 2007 [3, 3, 223]&#xA; CVRVVS 115 [5, 23]&#xA; AMICVS 1106 [2, 7, 79]&#xA; CONCVLCO 355 [5, 71]&#xA; LACRIMABILIS 1203 [3, 401]&#xA; ABVNDANS 505 [5, 101]&#xA; ADDO 1000 [2, 2, 2, 5, 5, 5]&#xA; ADEO 500 [2, 2, 5, 5, 5]&#xA; IMMVNDVS 2511 [3, 3, 3, 3, 31]&#xA; INFLECTVM 1156 [2, 2, 17, 17]&#xA; ALTVS 55 [5, 11]&#xA; LIQVIDVS 562 [2, 281]&#xA; &#xA; ANTIQVVS 11 [11]&#xA; LVXVRIA 71 [71]&#xA; AMICVLVM 2161 [2161]&#xA; CRINITVS 107 [107]&#xA; ABSQVE 5 [5]&#xA; IVDICIVM 1613 [1613]&#xA; MVNIMENTVM 3011 [3011]&#xA; EXCELLENTIA 211 [211]&#xA; CALLIDE 701 [701]&#xA; ECCLESIA 251 [251]&#xA; AMICITIA 1103 [1103]&#xA; MAXIME 2011 [2011]&#xA; IMPRIMIS 2003 [2003]&#xA; ALIQVANTVS 61 [61]&#xA; HAFFLIGENIENSIS 53 [53]&#xA; INCLITVS 157 [157]&#xA; CALAMITAS 1151 [1151]&#xA; CONIVRATVS 111 [3, 37]&#xA; BAIVLVS 61 [61]&#xA; ARTIFICIOSE 103 [103] &lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-08-12T22:18:17.850" LastActivityDate="2015-08-12T22:18:17.850" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19986" PostTypeId="1" AcceptedAnswerId="20095" CreationDate="2015-08-12T23:17:16.167" Score="8" ViewCount="636" Body="&lt;p&gt;Ada is one of the last few contestants on the Greatest Assessment Match Ever. The stern and humourless quizmaster, Mr Hopkins, is giving her her final test before the end of the show. The dialogue goes as follows:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;Hopkins:&lt;/em&gt; What name was given to the period of severe dust storms across the US and Canada in 1930?&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Ada:&lt;/em&gt; The Dust Bowl.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Hopkins:&lt;/em&gt; Who was the third president of Finland, and when was he elected?&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Ada:&lt;/em&gt; Um...&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Hopkins:&lt;/em&gt; Time's up! In what year was the film &lt;em&gt;Caught&lt;/em&gt;, directed by Max Ophüls, released?&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Ada:&lt;/em&gt; 1932?&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Hopkins:&lt;/em&gt; In 1833, a law was passed in Russia banning what?&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Ada:&lt;/em&gt; Masturbation.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Hopkins:&lt;/em&gt; What is the name given to the parties hit by Cambridge University students at the end of the academic year ever since 1834?&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Ada:&lt;/em&gt; They hit the May Ball...? Yes, May Balls.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Hopkins:&lt;/em&gt; Which musical have showgoers been hitting continuously since 1935 in the West End of London?&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Ada:&lt;/em&gt; &lt;em&gt;Wicked&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Hopkins:&lt;/em&gt; What were you doing three years before the start of World War Two?&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Ada:&lt;/em&gt; I wasn't born that long before wartime!&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Hopkins:&lt;/em&gt; What device is used for blocking an electrostatic field, and when was it invented?&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Ada:&lt;/em&gt; A Faraday cage. It was invented by Michael Faraday in 1837.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Hopkins:&lt;/em&gt; Do you consider yourself to be attractive for a 38-year-old woman?&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Ada:&lt;/em&gt; If you keep asking me personal questions, I'm going to get up and run out of the room!&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Hopkins:&lt;/em&gt; I, being a fan of 2048, was wondering if anyone has any proven strategies for playing 2048?&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Ada:&lt;/em&gt; Now you've got me completely stumped. I don't know.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Hopkins:&lt;/em&gt; Congratulations! You have answered every question correctly except for one. Please go to the white room to await the others' results. Soon you will know whether you have successfully beaten the competition to become the final champion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;What is the hidden meaning behind the above dialogue?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Bonus: &lt;strong&gt;which one question did Ada not answer correctly?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-08-21T13:36:47.070" Title="Ada and Mr Hopkins - what's it all about?" Tags="|enigmatic-puzzle|knowledge|" AnswerCount="6" CommentCount="8" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19987" PostTypeId="2" ParentId="19870" CreationDate="2015-08-13T00:28:28.483" Score="25" Body="&lt;p&gt;It &lt;a href=&quot;https://puzzling.stackexchange.com/a/19957/15649&quot;&gt;has been shown&lt;/a&gt; that fewer than 16 pieces is not possible (due to the 8-way split then requiring someone to have a single piece larger than a single portion of the 9-way split). I'll show by contradiction than a split into exactly 16 pieces is not possible: for that purpose let's assume that there exists a solution for 16 pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We'll use units of 1/504th (&lt;code&gt;1/(7×8×9)&lt;/code&gt;) of the whole bar to discuss the sizes of the pieces (this does not affect the solution).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the 9-way split at least two people must have only one piece (by the pigeonhole principle). Therefore there must be at least two pieces of size &lt;code&gt;56&lt;/code&gt; (one ninth) - this also the largest size possible, otherwise whoever got a larger piece in the 9-way split would have more than their share.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the 8-way split everyone must have exactly two pieces: otherwise someone would have to have only one piece, and that one piece would have to be of size &lt;code&gt;63&lt;/code&gt; (one eighth), which is too large for the 9-way split. Therefore there must be at least two pieces of size &lt;code&gt;63 - 56 = 7&lt;/code&gt;, and since &lt;code&gt;56&lt;/code&gt; is the largest size, &lt;code&gt;7&lt;/code&gt; must also be the smallest size (or the 8-way split would require pairing it with a piece too large). The rest of the proof is basically showing that the existence of a piece of size &lt;code&gt;56&lt;/code&gt; would inevitably lead to a piece smaller than size &lt;code&gt;7&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the 7-way split a single portion of pieces must add up to size &lt;code&gt;72&lt;/code&gt; (one seventh). Thus the pieces of size &lt;code&gt;56&lt;/code&gt; each need to be paired with a total size of &lt;code&gt;72 - 56 = 16&lt;/code&gt;, and since &lt;code&gt;7&lt;/code&gt; is the smallest possible size, this total can be obtained by using at most two pieces (&lt;code&gt;floor(16 / 7)&lt;/code&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a two-piece combination, the larger piece must be larger than &lt;code&gt;7&lt;/code&gt; and at most &lt;code&gt;16 - 7 = 9&lt;/code&gt;. But in the 8-way split it would then be paired with a size less than &lt;code&gt;56&lt;/code&gt; but at most &lt;code&gt;54&lt;/code&gt;, all of which would have to be paired with a piece smaller than &lt;code&gt;7&lt;/code&gt; in the 9-way split to add up to &lt;code&gt;56&lt;/code&gt;. (In fact, &lt;code&gt;7&lt;/code&gt; is the only size smaller than &lt;code&gt;63 - (56 - 7) = 14&lt;/code&gt; which does not lead to this problem.) Therefore only a single piece of size &lt;code&gt;16&lt;/code&gt; can be paired with a &lt;code&gt;56&lt;/code&gt; in the 7-way split. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since everyone has exactly two pieces in the 8-way split, pieces of size &lt;code&gt;63 - 16 = 47&lt;/code&gt; are also required. For the 9-way split these pieces are paired with size &lt;code&gt;56 - 47 = 9&lt;/code&gt; as other pairings would require pieces smaller than &lt;code&gt;7&lt;/code&gt;. But in the 8-way split the pieces of size &lt;code&gt;9&lt;/code&gt; are then paired with size &lt;code&gt;63 - 9 = 54&lt;/code&gt;, which would lead to a piece of size at most &lt;code&gt;56 - 54 = 2&lt;/code&gt; in the 9-way split. This contradicts the fact that &lt;code&gt;7&lt;/code&gt; is the smallest piece, therefore there exists no solution for 16 pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since it has now been shown that there are no solutions with fewer than 17 pieces, the correct answer must be either 17 or 18 (which has been &lt;a href=&quot;https://puzzling.stackexchange.com/a/19897/15649&quot;&gt;shown possible&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm fairly certain that there exists no solution for 17 pieces, but I have not been able to prove it – the problem is that unlike this solution, there are no immediate forced pairings in the 8-way split (up to two pieces may be smaller than &lt;code&gt;7&lt;/code&gt;), so it becomes very complex to consider all possible cases in the same manner as I've done here.&lt;/p&gt;&#xA;" OwnerUserId="15649" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-08-19T23:37:58.553" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="19988" PostTypeId="1" AcceptedAnswerId="19992" CreationDate="2015-08-13T01:09:11.760" Score="10" ViewCount="633" Body="&lt;p&gt;Another that rhymes, my author desired&lt;br&gt;&#xA;'Twas limericks and riddles that he so admired&lt;br&gt;&#xA;To hide a relation&lt;br&gt;&#xA;&lt;em&gt;To earn reputation&lt;/em&gt;&lt;br&gt;&#xA;and brighten your day through wordplay inspired&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A puzzle I am, though not yet begun&lt;br&gt;&#xA;Of words I'll speak later, each stanza has one&lt;br&gt;&#xA;Their connection concealed&lt;br&gt;&#xA;Your mind you must wield&lt;br&gt;&#xA;Once gathered, you'll publish my scheme and be done.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two parts both wound tight&lt;br&gt;&#xA;Near my bridge they do meet&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you fancy a bite&lt;br&gt;&#xA;Eggs and bacon I heat&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of efforts in vain&lt;br&gt;&#xA;My partner's delayed&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things weak under strain&lt;br&gt;&#xA;Or a sweet snack homemade&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With only a few emerges a pattern&lt;br&gt;&#xA;Like Sega and Goya exposed my friend &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18424/surely-you-know-this-prominent-figure-hints-added&quot;&gt;Saturn&lt;/a&gt;&lt;br&gt;&#xA;These next few you'll find&lt;br&gt;&#xA;Are of the same kind&lt;br&gt;&#xA;As those layed above left for you to discern&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A thing often wasted&lt;br&gt;&#xA;Though not touched or seen&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A thing often tasted&lt;br&gt;&#xA;Or known for its green&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A man builds a wall&lt;br&gt;&#xA;Without wood or brick&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another might fall&lt;br&gt;&#xA;If that wall is too slick&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another that rhymes, my author desired&lt;br&gt;&#xA;'Twas limericks and riddles that he so admired&lt;br&gt;&#xA;To puzzle and jest&lt;br&gt;&#xA;And never to rest&lt;br&gt;&#xA;'Til you've been enlightened, an answer acquired&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Warning:&lt;/strong&gt; This puzzle was written in a land where caramel has two syllables, aunt is pronounced &quot;awnt&quot;, and words of _ittle rhyme with _iddle.&lt;/p&gt;&#xA;" OwnerUserId="13237" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-08-14T04:40:49.397" Title="The Anthropomorphized Riddle" Tags="|riddle|rhyme|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="19989" PostTypeId="2" ParentId="19986" CreationDate="2015-08-13T01:29:28.320" Score="1" Body="&lt;p&gt;WIP:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Each of the question answer pairs includes a number. What's more, you'll notice that from the first question to the next to last (Dust Bowl ... 38 year old) you can spot the two digit numbers from 30 ascending to 38 (the third president of Finland, Pehr took over in 1931, and WWII began in 1939 so three years before was 1936). Because 38 is in a clue by itself, I assume the 18/19 prefixes are irrelevant, but I could easily be wrong. I can't quite explain the jump to 48 in the last question and I really haven't read that thoroughly. I'll come back if I get anything else, otherwise maybe this will be useful to someone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13970" LastActivityDate="2015-08-13T01:29:28.320" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19990" PostTypeId="2" ParentId="19791" CreationDate="2015-08-13T02:02:01.230" Score="6" Body="&lt;p&gt;EDIT: McMallister was faster than me, well done! (there is not a huge difference in our explanation for the clues, but because i started writing them down, i wanted to finish them)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think i got it. I will most likely complete this post in the next 15 hours, i need to sleep soon :) -- if somebdy is faster than me, i won't complain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 666&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;clue 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &quot;mystery&quot; refers to the Book of Revelations (or maybe the Bible as a whole). In cryptic language, the number 666 comes up, which is itself a riddle since it is unclear what it actually means.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;clue 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; there is a discussion, wether 616 is actually the number of the beast&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;clue 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; there are 3 &quot;numbers&quot; (or digits, as i would prefer to call them). this works together with the title.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;clue 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I thinkt &quot;able to read&quot; refers to reading the bible passage, and &quot;to count&quot; refers to the quote there &quot;Here is Wisdom, Let him that hath understanding count the number of the beast&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;clue 5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; this relates to the Smiths number - the 34th smith number is our solution. check out the wiki article to understand more about. In this article it can be seen, that the Mr Smith these numbers are named by was most likely no mathematician, but a mathematician who knew Mr Smith &quot;invented&quot; the numbers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;clue 6:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; for this clue you have to take the number into account: &quot;6&quot;. If you come back again to this clue you get &quot;66&quot;, if you come back again to this clue again you get &quot;666&quot;, hencethe two &quot;agains&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;clue 7:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the sum of all numbers on a roulette table is 666&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;clue 8:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it is the number of the beast, so the &quot;beast&quot; is the owner&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;clue 9:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; here, &quot;upset&quot; is to be understood as &quot;turn upside down&quot;, so 666 becomes 999, which is a emergency number in the UK, hence the &quot;swift reaction&quot;. The author of this riddle has got the number in mind, and due to his username, we know where he stands (UK).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;clue 10:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; this refers to hexakosioihexekontahexaphobia&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;clue 11:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is the clue which gave me the solution initially, because 2,3,37 are the prime factors of 666&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clue 12:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; this refers to carbon, which is (iirc) the second most common atom in a human body. carbon is made of 6 protons, 6 neutrons, 6 electrons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;clue 13:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; again, you have to take the number of the clue into account, then you get 13:17-18 , which refers to the passage in the book of Revelations&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clue 14:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; yes, as a mathematician i can confirm, that i am not too comfortable when &quot;digits&quot; and &quot;numbers&quot; get mixed up&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clue 15:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; selfexplanatory if you know the solution&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clues arranged by certainty, that i am correct:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (uncertain) 8 4 1 9 3 2 13 10 11 12 6 5 7 (certain)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clues arranged by (my subjective) difficulty:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (hardest) 5 9 12 13 4 8 1 2 10 3 7 6 11(easiest)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks for the hints in the comments for solving clue #9 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All in all: a very good puzzle/riddle! well done! an additional remark (idea for another clue):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. there was once a fire in a city... (lets see if you get that :D) - i think that would fit in really well!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14385" LastEditorUserId="14385" LastEditDate="2015-08-13T07:25:23.410" LastActivityDate="2015-08-13T07:25:23.410" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19991" PostTypeId="2" ParentId="19806" CreationDate="2015-08-13T03:21:19.560" Score="0" Body="&lt;p&gt;Just a guess based on the tag wordplay&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Up, up, by the Ort Sphere's smallest test, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; up (on top ), line 1 (smallest)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The two of spades, from a new deck dealt, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; two refers to 2nd&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But oh, wait, now by one-third less.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; third&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What a certain literary professor for his unbeheld felt,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A din, a doubt... &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Got nothing for these two so far.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Out!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; out refers to ending&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9543" LastEditorDisplayName="user9543" LastEditDate="2015-08-13T03:42:21.493" LastActivityDate="2015-08-13T03:42:21.493" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19992" PostTypeId="2" ParentId="19988" CreationDate="2015-08-13T03:23:28.903" Score="12" Body="&lt;p&gt;The solution is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Riddle&lt;/strong&gt; &amp;amp; &lt;strong&gt;Rhyme&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Based on the fact that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first four (non-limerick) stanzas all give words that rhyme with &quot;riddle&quot; and the next four all give words that rhyme with &quot;rhyme&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Specifically the word derived from each stanza is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Two parts both wound tight / Near my bridge they do meet&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Fiddle&lt;/strong&gt; (from &lt;a href=&quot;https://puzzling.stackexchange.com/a/20013/5208&quot;&gt;user1618143&lt;/a&gt; - the taught string &amp;amp; body meet at the bridge of the instrument)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;When you fancy a bite / Eggs and bacon I heat&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Griddle&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Of efforts in vain / My partner's delayed&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Little&lt;/strong&gt; (from the idiom &quot;too little, too late&quot;, meaning &quot;efforts in vain&quot;, with the &quot;partner&quot; of &lt;em&gt;little&lt;/em&gt; being &lt;em&gt;late&lt;/em&gt;, or &quot;delayed&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Things weak under strain / Or a sweet snack homemade&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Brittle&lt;/strong&gt; (two definitions, since brittle means weak and peanut brittle is a sweet snack)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A thing often wasted / Though not touched or seen&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Time&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A thing often tasted / Or known for its green&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Lime&lt;/strong&gt; (or arguably thyme)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A man builds a wall / Without wood or brick&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Mime&lt;/strong&gt; (&quot;builds&quot; an imaginary wall to be trapped behind)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Another might fall / If that wall is too slick&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Climb&lt;/strong&gt;(er)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-08-14T04:40:49.397" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="19993" PostTypeId="2" ParentId="19791" CreationDate="2015-08-13T03:54:56.557" Score="14" Body="&lt;p&gt;Ok, I solved Clue 9 in its entirety, which no one else seems to have done so I will post my answer now.  Massive thanks to supinf for his own clue, which put me on the right track.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;666&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Next, the clues:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;666 is the number of the beast, which is itself a riddle (possibly referring to Nero Caesar), found in the Book of Revelations, a book containing many mysteries.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There is debate whether 666 or 616 is the actual number of the beast.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Three sixes make up the number of the beast.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The number 666 does not require any special math or measurement to deduce, like pi.  You just need to read the Bible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;666 is the 34th &lt;a href=&quot;https://en.wikipedia.org/wiki/Smith_number&quot;&gt;Smith number&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;666 is 6 repeated.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The numbers on a roulette wheel add up to 666.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;666 is the number of the beast, and &quot;The Beast&quot; certainly sounds uncouth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Upset&quot; 666 means to turn it upside down, into 999.  999 is the emergency number in UK.  The second sentence refers to Chasly, from UK.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Some people have fear of 666, called Hexakosioihexekontahexaphobia.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;These are the prime factors of 666.  666 = 2 x 3 x 3 x 37&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Our bodies contain large amounts of Carbon-12, which consists of 6 protons, 6 neutrons and 6 electrons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Revelations 13:17-18.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Bonus - supinf's clue:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Refers to the London fire of 1666.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6178" LastActivityDate="2015-08-13T03:54:56.557" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19995" PostTypeId="2" ParentId="3286" CreationDate="2015-08-13T07:23:43.993" Score="6" Body="&lt;p&gt;My 7 year old daughter came up with this. I can't find anything wrong with it...&#xA;&lt;a href=&quot;https://i.stack.imgur.com/V4Kxo.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/V4Kxo.jpg&quot; alt=&quot;6 pencils all touching each-other&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15656" LastActivityDate="2015-08-13T07:23:43.993" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="19996" PostTypeId="2" ParentId="19942" CreationDate="2015-08-13T07:43:32.730" Score="11" Body="&lt;p&gt;I am&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Space Shuttle Endeavour&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I came from a large family,&lt;br&gt;&#xA;with lots of people looking after me.&lt;br&gt;&#xA;My family now lives all over the place,&lt;br&gt;&#xA;but I have my very own space.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There are six space shuttle orbiters, each require lots of maintenance crew.  Now, they are located all around the US, Endeavour in particular is displayed in &lt;a href=&quot;https://en.wikipedia.org/wiki/California_Science_Center&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/California_Science_Center&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My oldest sister was one of her kind,&lt;br&gt;&#xA;but she could not do things that were more refined.&lt;br&gt;&#xA;My third sister was the first to pass,&lt;br&gt;&#xA;when it happened, there were many questions asked.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The first shuttle, Enterprise, was not capable of spaceflight.  The third orbiter, Challenger, was the first to be lost, resulting in many questions asked about the Shuttle program.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My second sister left for good when I was young.&lt;br&gt;&#xA;But she had a house that looked after the folks and the punks.&lt;br&gt;&#xA;She took loads and loads of videos and studied really hard,&lt;br&gt;&#xA;She also had her own place where many people bark.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The second shuttle, Columbia was lost in an accident that many should still remember; this was about 11 years after Endeavour's first flight.  The rest of the stanza refers to Columbia House (folk and punk songs), Columbia Pictures (videos), Columbia University (studied really hard).  The place where people bark is the US Congress, which is located in the District of Columbia, or Washington D.C.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fourth and fifth sisters were really hard working,&lt;br&gt;&#xA;but my fourth sister was always on TV, constantly looking.&lt;br&gt;&#xA;My fifth sister seemed to be lost,&lt;br&gt;&#xA;but no-one really knew what was the cause.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The 4th shuttle, Discovery and the fifth, Atlantis, flew 39 and 33 flights respectively, the highest of the fleet.  (source: &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_Space_Shuttle_missions&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/List_of_Space_Shuttle_missions&lt;/a&gt;). &#xA;Discovery is the name of a TV channel, while Atlantis is named after a &quot;lost&quot; continent.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I was born, I wasn't meant to be.&lt;br&gt;&#xA;But after much collaborations, I became a reality.&lt;br&gt;&#xA;We were the sleeping beauties, until it was time to work,&lt;br&gt;&#xA;but only with a selected few who could prove their worth.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The first two lines probably refer to the fact that Endeavour was built using spare parts.  The shuttles were kept safe until their missions.  They were piloted by astronauts, a select few indeed who must be accomplished both mentally and physically.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We had two half-sisters, who were too lazy to work,&lt;br&gt;&#xA;but upon closer inspection, they have their own quirks.&lt;br&gt;&#xA;We had a distant cousin who could do our job,&lt;br&gt;&#xA;but until now, we were the only ones who could hold down the fort.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The two half-sisters are probably Independence and America, two space shuttle replicas not actually intended, or capable of spaceflight.  The distant cousin is the Russian space shuttle, Buran.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My sisters and I have traveled a lot,&lt;br&gt;&#xA;one of these places was very hot.&lt;br&gt;&#xA;We got up at any time of the day,&lt;br&gt;&#xA;we weren't lazy, just needed good weather our way.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Shuttles travel long distances through space.  Re-entry temperatures are very hot.  Shuttles can launch at any time of day, but the weather must be favourable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6178" LastEditorUserId="6178" LastEditDate="2015-08-17T05:22:25.813" LastActivityDate="2015-08-17T05:22:25.813" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="19998" PostTypeId="2" ParentId="3286" CreationDate="2015-08-13T10:16:04.853" Score="3" Body="&lt;p&gt;If the pencils are cylinders, and do not have to have equal radii, nine mutually touching cylingers is possible. See page 15 of &lt;a href=&quot;http://arxiv.org/pdf/1312.6207v1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Oo1JO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Oo1JO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2187" LastEditorUserId="2187" LastEditDate="2017-10-21T23:46:34.967" LastActivityDate="2017-10-21T23:46:34.967" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19999" PostTypeId="2" ParentId="19986" CreationDate="2015-08-13T12:10:05.627" Score="1" Body="&lt;p&gt;All of Ada's responses end their sentence with alternating punctuation, except for &quot;They hit the May Ball...? Yes, May Balls.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This response is the second out of a span of three that end in a period. The &quot;pattern&quot; is that the phrases end in alternating forms. period, exclamation mark, question mark, ellipsis, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore I say that this is the incorrectly answered one. She should not have added the &quot;Yes, May Balls.&quot; at the end of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(What everyone's mentioning about the numbers 30-39 being used also applies, but I think this is the criteria for formatting responses)&lt;/p&gt;&#xA;" OwnerUserId="13068" LastActivityDate="2015-08-13T12:10:05.627" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20001" PostTypeId="2" ParentId="19908" CreationDate="2015-08-13T14:27:58.350" Score="13" Body="&lt;p&gt;I'm going to guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;When it rains it pours&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or, in &quot;other words&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;When IT reigns, IT pores.&lt;br&gt;&#xA;The IT department now &quot;reigns&quot; over the company, and thus &quot;pores&quot; over the source code.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nice puzzle!&lt;/p&gt;&#xA;" OwnerUserId="10861" LastActivityDate="2015-08-13T14:27:58.350" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20002" PostTypeId="2" ParentId="19986" CreationDate="2015-08-13T14:42:31.550" Score="3" Body="&lt;p&gt;There is something funny with most of the dates.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Wicked&lt;/em&gt; has been running in the West End for many years, &lt;a href=&quot;https://en.wikipedia.org/wiki/Wicked_%28musical%29#London&quot; rel=&quot;nofollow&quot;&gt;but it opened in 2006, not 1935.&lt;/a&gt;&lt;br&gt;&#xA; According to Wikipedia, the Faraday cage &lt;a href=&quot;https://en.wikipedia.org/wiki/Faraday_cage&quot; rel=&quot;nofollow&quot;&gt;was invented by Michael Faraday in 1836 rather than 1837.&lt;/a&gt;&lt;br&gt;&#xA; As noted by @Anachor, &lt;em&gt;Caught&lt;/em&gt; was released in &lt;a href=&quot;https://en.wikipedia.org/wiki/Max_Oph%C3%BCls#Filmography&quot; rel=&quot;nofollow&quot;&gt;1949, not 1932.&lt;/a&gt;&lt;br&gt;&#xA; Again according to wikipedia, the dust bowl &lt;a href=&quot;https://en.wikipedia.org/wiki/Dust_Bowl&quot; rel=&quot;nofollow&quot;&gt;started in 1934, not 1930.&lt;/a&gt;&lt;br&gt;&#xA; The date of Finland's election, 1931, is correct; however while strongly implied it is never stated.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4385" LastEditorUserId="4385" LastEditDate="2015-08-13T14:50:06.260" LastActivityDate="2015-08-13T14:50:06.260" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20003" PostTypeId="2" ParentId="19953" CreationDate="2015-08-13T15:06:25.853" Score="9" Body="&lt;p&gt;I think I found the answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Maybe it's a programmer exercise: you have to perform following conversions:&lt;br&gt;&#xA; decimal =&gt; string =&gt; sum of alphabet position of each letter =&gt; conversion in hexadecimal format&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In this way the list is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 = &quot;one&quot; = 15+14+5 = 34 (decimal) = 0x22 (hexadecimal)&lt;br&gt;&#xA; 3 = &quot;three&quot; = 20+8+18+5+5 = 56 (decimal) = 0x38 (hexadecimal)&lt;br&gt;&#xA; 6 = &quot;six&quot; = 19+9+24 = 52 (decimal) = 0x34 (hexadecimal)&lt;br&gt;&#xA; 7 = &quot;seven&quot; = 19+5+22+5+14 = 65 (decimal) = 0x41 (hexadecimal)&lt;br&gt;&#xA; 14 = &quot;fourteen&quot; = 6+15+21+18+20+5+5+14 = 104 (decimal) = 0x68 (hexadecimal)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And obviously,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 11 = &quot;eleven&quot; = 5+12+5+22+5+14 = 63 (decimal) = 0x3F (hexadecimal)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15668" LastEditorUserId="6205" LastEditDate="2015-08-13T15:12:33.223" LastActivityDate="2015-08-13T15:12:33.223" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20004" PostTypeId="2" ParentId="19901" CreationDate="2015-08-13T15:14:16.753" Score="1" Body="&lt;p&gt;Ok, so I am going to start a partial answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After finding out what was the problem with my „number to date” conversion, I managed to find out that the number 639407344 (actually) translates to &lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Fri, 06 Apr 1990 13:09:04 GMT&lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I think it would bee too easy for the door to be the 6th (F), so I tried a Caesar cipher with 6 shifts (As OP was stating, only the day counts), but that didn't get me anywhere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I also tried to do a vigenere cipher decoding using Fri / Friday as a key, with no luck. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, my encryption knowledge is very, very poor. So I got stuck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LE:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; So, the word &lt;strong&gt;abecedarian&lt;/strong&gt; is important, but so far I don't see how. I recall learning the letters of the alphabet in a certain way, but I haven't found a standard for the recommended order. So, I am thinking that my approach with another order for the letters of the alphabet is not the way to go.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10692" LastEditorUserId="10692" LastEditDate="2015-08-14T08:04:28.463" LastActivityDate="2015-08-14T08:04:28.463" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20005" PostTypeId="2" ParentId="19870" CreationDate="2015-08-13T15:50:33.047" Score="1" Body="&lt;p&gt;I think f'' gives a clue on how to prove it can't be done in 17 pieces. We start with his original split into 17 pieces:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;56-&amp;gt;7-&amp;gt;49-&amp;gt;14-&amp;gt;42-&amp;gt;9-&amp;gt;47-&amp;gt;16-&amp;gt;40-&amp;gt;23-&amp;gt;33-&amp;gt;18-&amp;gt;38-&amp;gt;25-&amp;gt;31-&amp;gt;32-&amp;gt;24&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We need at least one 56 piece because otherwise we'd have split the original 9 into (at least) 2 pieces each, making 18 pieces. This split however is not complete (as shown in his method). We need to split some of the other pieces further, and to compensate (to still have 17 pieces), we need two 56 pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm mainly getting at is that we'll need two whole 56 pieces. To form 63 pieces out of those we need two 7's, which would need two 49's, which need two 14, etc. forcing the pieces to be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7, 7, 14, 14, 21, 21, 28, 28, 35, 35, 42, 42, 49, 49, 56, 56&lt;/p&gt;&#xA;&#xA;&lt;p&gt;before we split them further. These are sixteen all in all. But notice that they are all a multiple of 7 and their sum can never form a 72 = 8*9. Splitting one of them would possibly fix the situation for two of the pieces, but that is not enough, we're still left with 13 pieces, none of which can be summed up to 72.&lt;/p&gt;&#xA;" OwnerUserId="15669" LastActivityDate="2015-08-13T15:50:33.047" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20006" PostTypeId="1" AcceptedAnswerId="20089" CreationDate="2015-08-13T16:29:48.177" Score="8" ViewCount="502" Body="&lt;p&gt;As &lt;a href=&quot;https://puzzling.stackexchange.com/users/12408/f&quot;&gt;f&quot;&lt;/a&gt; solved for &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18424/surely-you-know-this-prominent-figure-hints-added&quot;&gt;Saturn&lt;/a&gt;&lt;br&gt;&#xA;and &lt;a href=&quot;https://puzzling.stackexchange.com/users/1968/cain&quot;&gt;Cain&lt;/a&gt; reasoned &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18615/what-are-we-and-where-is-our-brother-hints-galore&quot;&gt;Stairs&lt;/a&gt;&lt;br&gt;&#xA;While &lt;em&gt;one word you seek&lt;/em&gt;&lt;br&gt;&#xA;Three &lt;a href=&quot;https://puzzling.stackexchange.com/questions/20006/my-solution-appears-to-be-absent&quot;&gt;this riddle&lt;/a&gt; bears:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carved from a forest we lie&lt;br&gt;&#xA;A road built of bone we supply&lt;br&gt;&#xA;With great walls of black&lt;br&gt;&#xA;We are known to attack&lt;br&gt;&#xA;From the forest an echo replies&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a singular purpose we're made&lt;br&gt;&#xA;To deny those of whom you're afraid&lt;br&gt;&#xA;And provide peace of mind&lt;br&gt;&#xA;Each day we're combined&lt;br&gt;&#xA;With our partner who therein has stayed&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At prominence ere the fall of Rome&lt;br&gt;&#xA;We held the theaters, bridges and domes&lt;br&gt;&#xA;In structures erected&lt;br&gt;&#xA;We shan't be neglected&lt;br&gt;&#xA;Lest one desires to topple our home&lt;br&gt;&#xA;&lt;hr&gt;&#xA;&lt;strong&gt;Hints:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first stanza is describing the method with which this riddle was written. For instance, I thought of many things which referred to Saturn and described them. Though technically separate words (homonyms, in fact), the answer is just one.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; By now you may have noticed that the second stanza is buried deep in metaphor. There is no real forest. There is no real road. There are no walls. They do attack, but not in a violent sense. The forest &lt;em&gt;does&lt;/em&gt; respond to the attack with a sound, however.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A biologist may cringe at my use of the word &quot;bone&quot; in the second stanza. It shouldn't prevent anyone from solving it, but it is scientifically inaccurate, so I apologize to those feeling scrupulous.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13237" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-08-15T02:42:56.553" Title="My solution appEars To be Absent!" Tags="|riddle|rhyme|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20007" PostTypeId="1" AcceptedAnswerId="20009" CreationDate="2015-08-13T16:42:13.880" Score="11" ViewCount="1951" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Dear friend,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;You've heard the malicious gossip by now, and&lt;br&gt;&#xA;  Although I didn't start it, I must confess that&lt;br&gt;&#xA;  Several times I actively participated in it.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I've come to think you are very self-centered,&lt;br&gt;&#xA;  Since it seems like every time you talk about me,&lt;br&gt;&#xA;  You feel the need to also mention yourself!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Still, I write this now in guilt and distress.&lt;br&gt;&#xA;  I want you to know that I'm part of the solution&lt;br&gt;&#xA;  And not part of the overall larger problem.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Forgive me- I know you can identify with me, so&lt;br&gt;&#xA;  Think about the good times when we first met:&lt;br&gt;&#xA;  The sense of self-empowerment you had then!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Though I'm a pro, and I was in the top 10 in my class,&lt;br&gt;&#xA;  Oddly, I feel larger and more together when I'm alone.&lt;br&gt;&#xA;  When I'm around friends I feel small and disconnected.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In loving apology,&lt;br&gt;&#xA;  me&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;What wrote this apology letter?&lt;/h2&gt;&#xA;" OwnerUserId="463" LastActivityDate="2015-08-14T00:21:26.800" Title="What wrote this apology letter?" Tags="|riddle|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20008" PostTypeId="2" ParentId="20006" CreationDate="2015-08-13T16:43:14.880" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Stanzas two and three are lock and key, respectively. (it refers to the keystone at the top of an arch, I'm guessing)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Stanza one, I'm assuming, is not &quot;lightning&quot; though I got a giggle out of an incidental reference to the webcomic &quot;An Ordinary Princess.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the reference that made me laugh. &lt;a href=&quot;http://www.ordinary-princess.com/?comic=chapter-4-page-10&quot; rel=&quot;nofollow&quot;&gt;http://www.ordinary-princess.com/?comic=chapter-4-page-10&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a singular purpose we're made&lt;br&gt;&#xA; To deny those of whom you're afraid&lt;br&gt;&#xA; And provide peace of mind&lt;br&gt;&#xA; Each day we're combined&lt;br&gt;&#xA; With our partner who therein has stayed&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This refers to how a lock is opened with a key, usually daily.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At prominence ere the fall of Rome&lt;br&gt;&#xA; We held the theaters, bridges and domes&lt;br&gt;&#xA; In structures erected&lt;br&gt;&#xA; We shan't be neglected&lt;br&gt;&#xA; Lest one desires to topple our home&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A keystone is placed at the top of an arch, and prevents the whole thing from falling down. it's surprisingly sturdy even though it doesn't quite look like it would be.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13068" LastEditorUserId="11475" LastEditDate="2015-08-13T18:46:32.447" LastActivityDate="2015-08-13T18:46:32.447" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20009" PostTypeId="2" ParentId="20007" CreationDate="2015-08-13T16:48:48.620" Score="11" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;The letter I&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You've heard the malicious gossip by now, and&lt;br&gt;&#xA;Although I didn't start it, I must confess that&lt;br&gt;&#xA;Several times I actively participated in it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;I&lt;/strong&gt; doesn't start &quot;the mal&lt;strong&gt;i&lt;/strong&gt;c&lt;strong&gt;i&lt;/strong&gt;ous goss&lt;strong&gt;i&lt;/strong&gt;p,&quot; but is in the phrase 3 times&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've come to think you are very self-centered,&lt;br&gt;&#xA;Since it seems like every time you talk about me,&lt;br&gt;&#xA;You feel the need to also mention yourself!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You mention yourself == I think etc or general i, I confusion&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Still, I write this now in guilt and distress.&lt;br&gt;&#xA;I want you to know that I'm part of the solution&lt;br&gt;&#xA;And not part of the overall larger problem.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;I&lt;/strong&gt; is in &quot;gu&lt;strong&gt;i&lt;/strong&gt;lt&quot; and &quot;d&lt;strong&gt;i&lt;/strong&gt;stress,&quot; and is in &quot;the solut&lt;strong&gt;i&lt;/strong&gt;on,&quot; but not in &quot;the overall larger problem&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Forgive me- I know you can identify with me, so&lt;br&gt;&#xA;Think about the good times when we first met:&lt;br&gt;&#xA;The sense of self-empowerment you had then!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I identify as I. (self empowerment referring to babies realizing they exist separate from the world sort of)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though I'm a pro, and I was in the top 10 in my class,&lt;br&gt;&#xA;Oddly, I feel larger and more together when I'm alone.&lt;br&gt;&#xA;When I'm around friends I feel small and disconnected.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pronoun is a professional noun apparently. 9th letter. I (larger and in one piece) is I but in words it's lowercase i (two parts --disconnected --, smaller).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the true answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Going hamateur wrote it, cause I did.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13916" LastEditorUserId="463" LastEditDate="2015-08-14T00:21:26.800" LastActivityDate="2015-08-14T00:21:26.800" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20010" PostTypeId="2" ParentId="19920" CreationDate="2015-08-13T17:07:16.573" Score="1" Body="&lt;p&gt;&lt;strong&gt;Partial answer:&lt;/strong&gt; I may have found a few more of the destinations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Coming from a town of few, I just can't wait to see&lt;br&gt;&#xA;people! People! People! In increasing quantity!&lt;br&gt;&#xA;I'm staying where I finish since it feels so right to me.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This suggests we're looking for either cities or countries of large population. China and Tokyo spring to mind, but given the paragraph about the first trip being in the west and the second in the east, probably US cities.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;When I planned out this vacation, if only I knew;&lt;br&gt;&#xA;bigger and bigger my target countries grew.&lt;br&gt;&#xA;When I finish this trip, I'm going home, thank you.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We're looking for big countries somewhere in the east of the world. Russia, obviously; maybe China and Kazakhstan as well?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The Avesta once noted that this city flows;&lt;br&gt;&#xA;As their regime, here the terrible chose.&lt;br&gt;&#xA;Now reborn anew, with haste this city grows.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Avesta makes one &lt;em&gt;think&lt;/em&gt; of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Avesta&quot; rel=&quot;nofollow&quot;&gt;sacred Zoroastrian text&lt;/a&gt;, but how about the &lt;a href=&quot;https://en.wikipedia.org/wiki/Avesta_(locality)&quot; rel=&quot;nofollow&quot;&gt;Swedish town&lt;/a&gt;? The town's name &quot;is of similar origin as the word å, meaning stream&quot;, so in that sense the city flows. Not sure about the last two lines though!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another possibility:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Astana&lt;/strong&gt;, the capital of Kazakhstan. This time the first line could refer to the Zoroastrian Avesta; Zoroastrianism certainly exists in Kazakhstan, and the city contains a river which &quot;flows&quot;. Ivan the Terrible of Russia might have had something to do with Astana. The city has been &quot;reborn&quot; in recent years - made into the capital, renamed from Akmola/Tselinograd - and has been growing very fast with a huge amount of ongoing building work.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A city of non-zero nuclear threat;&lt;br&gt;&#xA;Elena is priced in at ten, I would bet.&lt;br&gt;&#xA;For dry desert farming, it's the best you can get.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first two lines probably refer to &lt;a href=&quot;https://en.wikipedia.org/wiki/ELENA_reactor&quot; rel=&quot;nofollow&quot;&gt;the ELENA nuclear reactor&lt;/a&gt;, being developed by the Kurchatov institute in Moscow, so the answer is probably &lt;strong&gt;Moscow&lt;/strong&gt;. The last line is probably a reference to some Soviet scheme of farming the desert.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A festival city, yet not from Dubai;&lt;br&gt;&#xA;Living history looms here as tall as the sky,&lt;br&gt;&#xA;and it's nearly the finest if you're looking to buy.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Probably somewhere in the US. Las Vegas?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If you need an import, I'd hope there's no ice,&lt;br&gt;&#xA;or enough for a truck to make one-way trips twice.&lt;br&gt;&#xA;(Though Yamal is here in a pinch, which is nice.)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; f'' has already established that this is somewhere in Russia.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A birdlike creature, the foe of all snakes,&lt;br&gt;&#xA;on this city's flag an appearance makes.&lt;br&gt;&#xA;They have a whole castle for one man-made lake!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; f'' has solved this one, and again it's in Central Asia.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Home of the highlanders, a capital investment;&lt;br&gt;&#xA;Most recently subjected to corruption's assessment.&lt;br&gt;&#xA;Their country dissolved as a laughable contestant.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again sounds like the US (or am I just being prejudiced?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-08-13T17:07:16.573" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20011" PostTypeId="1" AcceptedAnswerId="20781" CreationDate="2015-08-13T17:33:43.513" Score="21" ViewCount="3032" Body="&lt;p&gt;I work on a farm, but it's not as romantic as you think: I need to wake up around sunrise to go to work, and I get paid just 4 pennies per minute. I had to take a second job to pay the bills. Honestly, I'd do anything just to get an apartment, but I'm such a sad sack it'll probably never happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is my name?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A correct answer should explain all the clues, and I'm looking for a specific, non-fictional human being. The clues are not meant to be taken literally.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(P.S. I removed the 'room and board' bit because it was just in there for flavor, but folks ended up reading into it. I should have realized that people would see it as a clue, so sorry about that.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Synonyms are important.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Only one of the clues is a pun.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Why would someone on a farm wake up around sunrise?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Each main clue refers to a thing of mine which are all (more-or-less) of the same kind.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bonus Clue 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I wish I had a job I was more passionate about.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bonus Clue 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I have to put in a heroic effort just to get out of bed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14441" LastEditorUserId="14441" LastEditDate="2015-08-20T15:01:00.027" LastActivityDate="2015-09-25T07:41:20.710" Title="A Hard-Luck Farmer" Tags="|riddle|wordplay|knowledge|" AnswerCount="13" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20012" PostTypeId="2" ParentId="20011" CreationDate="2015-08-13T18:20:54.273" Score="9" Body="&lt;p&gt;Does not nearly fit all the clues but hey! Why not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You sir are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A solar panel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I work on a farm, but it's not as romantic as you think: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Solar farms are not the picturesque ol' mcdonald farm&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I need to wake up around sunrise to go to work, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Solar panels start working at the rising of the sun.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and if you take out room and board, I get just 4 pennies for every minute of work.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Seems like the amount of energy a solar panel could produce in a minutes cost.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I had to take a second job to pay the bills. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; not sufficiently explained! (Do to the high initial costs of solar panels they can't be your sole source of income?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Honestly, I'd do anything just to get an apartment, but I'm such a sad sack it'll probably never happen.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sufficiently explained! (maybe it is hard to have solar panels in the city)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13916" LastActivityDate="2015-08-13T18:20:54.273" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20013" PostTypeId="2" ParentId="19988" CreationDate="2015-08-13T18:23:05.907" Score="5" Body="&lt;p&gt;Proposing a couple of corrections to Alconja's answer, which Roland said is not quite correct:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Two parts both wound tight / Near my bridge they do meet&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fiddle. The strings and bow are both tight, and they meet at the bridge.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Another might fall / If that wall is too slick&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rime.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12566" LastActivityDate="2015-08-13T18:23:05.907" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20014" PostTypeId="1" AcceptedAnswerId="20016" CreationDate="2015-08-13T18:24:41.860" Score="8" ViewCount="5571" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Select the figure that is missing from the collection below:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sk9vf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sk9vf.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;a.&lt;a href=&quot;https://i.stack.imgur.com/CyiHv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CyiHv.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;b.&lt;a href=&quot;https://i.stack.imgur.com/1LQn2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1LQn2.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;c.&lt;a href=&quot;https://i.stack.imgur.com/Ioewt.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ioewt.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;d.&lt;a href=&quot;https://i.stack.imgur.com/glYIj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/glYIj.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2375" LastEditorUserId="71981" LastEditDate="2020-11-16T05:57:14.000" LastActivityDate="2020-11-16T05:57:14.000" Title="Mensa IQ test question - shapes in a 4x3 grid" Tags="|logical-deduction|pattern|visual|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="20015" PostTypeId="2" ParentId="20011" CreationDate="2015-08-13T18:27:38.090" Score="-2" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Kimi the sweatshop worker&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I work on a farm, but it's not as romantic as you think:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many sweatshop workers start out on their family farm in the countryside&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I need to wake up around sunrise to go to work, and if you take out room and board, I get just 4 pennies for every minute of work. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They have to work horrendous hours and can get awful wages (which are often more than they'd make on the family farm)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I had to take a second job to pay the bills.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Kimi went from her family farm to the city where she works in a factory for Nike.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Honestly, I'd do anything just to get an apartment, but I'm such a sad sack it'll probably never happen.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Idk, like omg. She sends a lot of her money home to feed her family?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13538" LastEditorUserId="13538" LastEditDate="2015-08-13T18:34:38.750" LastActivityDate="2015-08-13T18:34:38.750" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20016" PostTypeId="2" ParentId="20014" CreationDate="2015-08-13T18:37:09.347" Score="8" Body="&lt;h2&gt;Answer:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Explanation:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All the possible answers except for A are in the collection. (Some of them just rotated or slightly different). But A is not, there is no triangle with &lt;strong&gt;5&lt;/strong&gt; lines in the collection, just one with 4 lines. The most conspicuous thing is that nothing is being &quot;added&quot; to the other three possibilites. (This should make it clear that A is the right answer)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user14478" LastEditorDisplayName="user14478" LastEditDate="2015-08-13T18:43:44.583" LastActivityDate="2015-08-13T18:43:44.583" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20017" PostTypeId="2" ParentId="20006" CreationDate="2015-08-13T19:19:01.403" Score="5" Body="&lt;p&gt;Forked off of Kingrames' partial solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the idiom &quot;under lock and key&quot;, meaning that something is very fiercely guarded&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Carved from a forest we lie&lt;br&gt;&#xA;A road built of bone we supply&lt;br&gt;&#xA;With great walls of black&lt;br&gt;&#xA;We are known to attack&lt;br&gt;&#xA;From the forest an echo replies  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Underwater. Ships are carved from wood, which comes from a forest. When they sink, they supply a road of bones to the bottom of the ocean. walls of black = darkness. The second forest is the forest on the bottom of the ocean. Seagrasses, kelp, maybe even a coral reef.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For a singular purpose we're made&lt;br&gt;&#xA;To deny those of whom you're afraid&lt;br&gt;&#xA;And provide peace of mind&lt;br&gt;&#xA;Each day we're combined&lt;br&gt;&#xA;With our partner who therein has stayed  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This refers to how a lock is opened with a key, often daily.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At prominence ere the fall of Rome&lt;br&gt;&#xA;We held the theaters, bridges and domes&lt;br&gt;&#xA;In structures erected&lt;br&gt;&#xA;We shan't be neglected&lt;br&gt;&#xA;Lest one desires to topple our home  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A keystone is placed at the top of an arch, and prevents the whole thing from falling down. it's surprisingly sturdy even though it doesn't quite look like it would be.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13538" LastActivityDate="2015-08-13T19:19:01.403" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20018" PostTypeId="2" ParentId="20011" CreationDate="2015-08-13T19:41:17.653" Score="2" Body="&lt;p&gt;could it be,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mr. Potato Head&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I work on a farm, but it's not as romantic as you think&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A potato is found on a farm and a couch potato probably doesn't make very romantic love on a couch since they're lazy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I need to wake up around sunrise to go to work, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Potatoes grow the most when the sun is out &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;if you take out room and board, I get just 4 pennies for every minute of work. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Couch's collect pennies at a very slow rate or it could be that potatoes are very cheap and there worth the farm is probably about 4 pennies but I don't know about the first part of that&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I had to take a second job to pay the bills. Honestly, I'd do anything just to get an apartment, but I'm such a sad sack it'll probably never happen.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Couch potatoes aren't very motivated so they only do the minimal amount of work&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All of these have to do with potatoes and the only person I can think of with potato in their name is Mr. Potato Head. Might be a bit of a stretch but I like my answer at least&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13925" LastEditorUserId="13925" LastEditDate="2015-08-13T19:44:32.207" LastActivityDate="2015-08-13T19:44:32.207" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20019" PostTypeId="2" ParentId="18618" CreationDate="2015-08-13T19:46:04.487" Score="5" Body="&lt;h1&gt;14:&lt;/h1&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;strong&gt;C&lt;/strong&gt;. Based on the book titles from the Wheel of Time series: The &lt;strong&gt;E&lt;/strong&gt;ye of the World, The &lt;strong&gt;G&lt;/strong&gt;reat Hunt, The &lt;strong&gt;D&lt;/strong&gt;ragon Reborn (&lt;em&gt;hello everyone!&lt;/em&gt;), The &lt;strong&gt;S&lt;/strong&gt;hadow Rising, The &lt;strong&gt;F&lt;/strong&gt;ires of Heaven, Lord of &lt;strong&gt;C&lt;/strong&gt;haos, A &lt;strong&gt;C&lt;/strong&gt;rown of Swords.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-08-18T17:20:36.600" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20020" PostTypeId="2" ParentId="20011" CreationDate="2015-08-13T20:03:11.283" Score="6" Body="&lt;p&gt;Some ideas (that might help someone else get a full answer):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I need to wake up around sunrise to go to work&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Do we replace &quot;wake up&quot; and &quot;sunrise&quot; with synonyms and then interpret &quot;around&quot; in a cryptic-crossword sense? So &quot;rise around dawn&quot; would give the nonsense word &lt;em&gt;ridawnse&lt;/em&gt;; maybe some other synonyms give a meaningful word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;if you take out room and board, I get just 4 pennies for every minute of work.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pennies is usually abbreviated to &quot;p&quot; (in the days before decimal currency was introduced in Britain, it was &quot;d&quot; instead). So maybe there's a word or phrase consisting of four p's (or four d's) plus synonyms for &quot;room&quot; and &quot;board&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's a possible answer (though I doubt it's the intended one):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;room board dddd&quot; is an anagram of &lt;strong&gt;Mr Brad Odd-Dodo&lt;/strong&gt;, which could be a man's name.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another possible answer (again, probably not the intended one):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your name is &lt;strong&gt;Patrick&lt;/strong&gt;, of course!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-08-13T20:03:11.283" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="20021" PostTypeId="2" ParentId="19986" CreationDate="2015-08-13T20:35:45.600" Score="5" Body="&lt;p&gt;The hidden meaning is to do with&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the game of cricket,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the only wrong answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;1932&quot;, to the third question,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and what Ada is supposed to do is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; include a cricket reference in every answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Numbering the questions and answers 1Q, 1A, etc.:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1A: &quot;The Dust &lt;em&gt;Bowl&lt;/em&gt;.&quot;&lt;br&gt;&#xA;2A: &quot;&lt;em&gt;Um&lt;/em&gt;...&quot;&lt;br&gt;&#xA;This refers to the saying &lt;a href=&quot;http://www.urbandictionary.com/define.php?term=um...cricket?&quot; rel=&quot;nofollow&quot;&gt;&quot;&lt;em&gt;Um&lt;/em&gt;...cricket&quot;&lt;/a&gt;, used when conversation has stopped and there's an awkward silence.&lt;br&gt;&#xA;3Q: &quot;(...) the film &lt;em&gt;Caught&lt;/em&gt; (...)&quot;&lt;br&gt;&#xA;3A: &quot;1932&quot;&lt;br&gt;&#xA;This has nothing to do with cricket and is the only wrong answer.&lt;br&gt;&#xA;4A: &quot;Mastur&lt;em&gt;bat&lt;/em&gt;ion.&quot;&lt;br&gt;&#xA;5Q: &quot;(...) parties &lt;em&gt;hit&lt;/em&gt; by Cambridge University students&quot;&lt;br&gt;&#xA;5A: &quot;They &lt;em&gt;hit&lt;/em&gt; the May &lt;em&gt;Ball&lt;/em&gt;...? Yes, May &lt;em&gt;Balls&lt;/em&gt;.&quot;&lt;br&gt;&#xA;6Q: &quot;Which musical have showgoers been &lt;em&gt;hitting&lt;/em&gt; (...) in the West End (...)?&quot;&lt;br&gt;&#xA;&quot;&lt;em&gt;End&lt;/em&gt;&quot; may possibly also be intended as a cricket reference.&lt;br&gt;&#xA;6A: &quot;&lt;em&gt;Wicked&lt;/em&gt;.&quot;&lt;br&gt;&#xA;Ha ha! Sounds like &quot;wicket&quot;. Funny one, Ada!&lt;br&gt;&#xA;7Q: &quot;What were you doing three years &lt;em&gt;before&lt;/em&gt; the start of World War Two?&quot;&lt;br&gt;&#xA;&quot;&lt;em&gt;Before&lt;/em&gt;&quot; appears in the term &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Leg_before_wicket&quot; rel=&quot;nofollow&quot;&gt;leg &lt;em&gt;before&lt;/em&gt; (wicket)&lt;/a&gt;&quot;&lt;br&gt;&#xA;7A: &quot;I wasn't born that &lt;em&gt;long&lt;/em&gt; &lt;em&gt;before&lt;/em&gt; wartime!&quot;&lt;br&gt;&#xA;She repeats &quot;&lt;em&gt;before&lt;/em&gt;&quot;, and a fielder can take the position of &quot;&lt;em&gt;long&lt;/em&gt; stop&quot;.&lt;br&gt;&#xA;8Q: &quot;What device is used for &lt;em&gt;blocking&lt;/em&gt; an electrostatic &lt;em&gt;field&lt;/em&gt;, and when was it invented?&quot;&lt;br&gt;&#xA;8A &quot;A Faraday &lt;em&gt;cage&lt;/em&gt; (...)&quot;&lt;br&gt;&#xA;There is a version of the game called &lt;a href=&quot;http://www.cagecricket.com/&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;cage&lt;/em&gt; cricket&lt;/a&gt;.&lt;br&gt;&#xA;9A: &quot;If you keep asking me personal questions, I'm going to get up and &lt;em&gt;run out&lt;/em&gt; of the room!&quot;&lt;br&gt;&#xA;Here we can think of a &quot;&lt;em&gt;run&lt;/em&gt;&quot;, of being &quot;&lt;em&gt;out&lt;/em&gt;&quot;, and of being &quot;&lt;em&gt;run out&lt;/em&gt;&quot;.&lt;br&gt;&#xA;10A &quot;Now you've got me completely &lt;em&gt;stumped&lt;/em&gt; (...)&quot;&lt;br&gt;&#xA;&lt;br&gt;&#xA;Every answer except one contains at least one word interpretable as a cricket reference. At least five questions also contain such references, but I have probably missed some.&lt;br&gt;&#xA;&lt;br&gt;&#xA;In the introduction, we have &quot;&lt;em&gt;G&lt;/em&gt;reatest &lt;em&gt;A&lt;/em&gt;ssessment &lt;em&gt;M&lt;/em&gt;atch &lt;em&gt;E&lt;/em&gt;ver&quot; - an acrostic clue that the solution concerns a &lt;em&gt;game&lt;/em&gt;.&lt;br&gt;&#xA;&lt;br&gt;&#xA;The word &quot;&lt;em&gt;test&lt;/em&gt;&quot; stands out, because it's not normally used for a round of quiz questions. We also have &quot;&lt;em&gt;match&lt;/em&gt;&quot;. These are references to &lt;a href=&quot;https://en.wikipedia.org/wiki/Test_cricket&quot; rel=&quot;nofollow&quot;&gt;test cricket&lt;/a&gt; and the playing of &lt;em&gt;test matches&lt;/em&gt;.&lt;br&gt;&#xA;&lt;br&gt;&#xA;&quot;&lt;em&gt;White&lt;/em&gt; room&quot; could be a reference to cricket &lt;em&gt;whites&lt;/em&gt;, white being the colour traditionally worn by cricket players.&lt;br&gt;&#xA;&lt;br&gt;&#xA;And the number of questions is 10, the maximum number of wickets in an innings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7363" LastEditorUserId="7363" LastEditDate="2015-08-14T00:18:21.343" LastActivityDate="2015-08-14T00:18:21.343" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20022" PostTypeId="2" ParentId="19988" CreationDate="2015-08-13T20:38:36.197" Score="4" Body="&lt;p&gt;Proposing a correction to Alconja's answer, on top of user's:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Of efforts in vain / My partner's delayed&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Twiddle. A person accomplishing nothing is twiddling his thumbs (an English idiom). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13538" LastActivityDate="2015-08-13T20:38:36.197" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20023" PostTypeId="2" ParentId="19963" CreationDate="2015-08-13T20:49:52.463" Score="16" Body="&lt;p&gt;These are all&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; anagrams of countries with the first letter removed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;along&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Angola&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;soothe&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Lesotho&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;haunt&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Bhutan&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;remain&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Armenia&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;leap&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Nepal&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;candle&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Iceland&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;aurist&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Austria&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the word that links them is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Albania&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-08-13T20:49:52.463" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20024" PostTypeId="1" AcceptedAnswerId="20025" CreationDate="2015-08-13T21:17:38.417" Score="4" ViewCount="313" Body="&lt;p&gt;You know those puzzles where you have to get from one word to another through a series of steps by only changing a letter between each step?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well we're going to something similar with names.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with one name and progress to the last, only changing one of the names at each step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example (with made up names)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;John Smith&lt;br&gt;&#xA;Jim Smith&lt;br&gt;&#xA;Jim Jones&lt;br&gt;&#xA;Sam Jones&lt;br&gt;&#xA;Sam Brown&lt;br&gt;&#xA;Mary Brown  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pretty simple right? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, I'm not going to give you the first or last name; you'll have to deduce them from my clues! These clues aren't necessarily direct references to the names, but may just describe an aspect of them or something famous about them. Here you go:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Precedes You Bastards&lt;br&gt;&#xA;  Constant Cameo&lt;br&gt;&#xA;  Short Punch&lt;br&gt;&#xA;  Bereaved Playboy&lt;br&gt;&#xA;  Who Says?&lt;br&gt;&#xA;  Shouted Thrice&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Two things:   &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There is 1 step where the first/last name gets switched (i.e. it could go from Hugh Grant to Grant Hill)&lt;/li&gt;&#xA;&lt;li&gt;The first name of the last person sounds like the last name of the first person&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I figure this'll be solved pretty quickly, but I have learned my lesson about dumping a BILLION PUZZLES into a single post, so maybe I will make others if people like this one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer must include all six names and reasoning.&lt;/p&gt;&#xA;" OwnerUserId="10861" LastActivityDate="2015-08-13T21:33:50.943" Title="Name Progression" Tags="|pattern|knowledge|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20025" PostTypeId="2" ParentId="20024" CreationDate="2015-08-13T21:33:50.943" Score="7" Body="&lt;p&gt;Precedes You Bastards&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stan Marsh (South Park, &quot;You killed Kenny, You Bastards!&quot; Not sure if the quote is ever applied to Stan, but it fits)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Constant Cameo&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stan Lee (Marvel Studios, makes a cameo in nearly every film)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Short Punch&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bruce Lee (Martial Arts)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bereaved Playboy&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bruce Wayne (Batman, he's definitely a playboy!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who Says?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Wayne Brady (Actor on Whose Line is it Anyway?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Shouted Thrice&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Marsha Brady (Brady Bunch, &quot;Marsha, Marsha, Marsha!&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And to the hint, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Marsha ~ Marsh&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14445" LastActivityDate="2015-08-13T21:33:50.943" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20026" PostTypeId="1" AcceptedAnswerId="20030" CreationDate="2015-08-13T22:26:40.830" Score="11" ViewCount="482" Body="&lt;p&gt;A woman was employed as a temporary receptionist. She met her boss, was given some brief instructions then sat down at the desk and waited for clients. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first client examined the magazines and other reading matter and finally settled on the local paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second client was equally patient . She started surfing the web.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third  client was sniffling terribly and appeared to be coming down with flu.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fourth client appeared to be worried about his forthcoming scan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fifth client brought a section of tree branch along and spent the whole time examining its bark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sixth  client had presumably missed lunch because she proceeded to eat some bread and cheese.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The seventh client was so terrified, all she wanted to do was hide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The eighth client came in out of breath as though he'd been in a race.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was no ninth client so the receptionist went to see if the boss needed anything. She opened the door, went in, gave a scream and rushed out again to make a phone call.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What three things can you now &lt;em&gt;confidently&lt;/em&gt; say about the boss?&lt;/p&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="14177" LastEditDate="2015-08-14T17:16:09.070" LastActivityDate="2015-08-14T17:16:09.070" Title="How was your day at work?" Tags="|word-problem|situation|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20027" PostTypeId="1" AcceptedAnswerId="20028" CreationDate="2015-08-13T22:28:50.667" Score="1" ViewCount="168" Body="&lt;p&gt;How about this easy one? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;I have to type 30 characters in the body even though my puzzle is composed of an image that says everything.&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CQdXl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CQdXl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9575" LastActivityDate="2015-08-13T22:30:13.693" Title="Getting Down to the Nitty Gritty" Tags="|rebus|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20028" PostTypeId="2" ParentId="20027" CreationDate="2015-08-13T22:30:13.693" Score="2" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Splitting hairs, since the word &quot;hairs&quot; is split with a pipe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2015-08-13T22:30:13.693" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20030" PostTypeId="2" ParentId="20026" CreationDate="2015-08-13T23:38:50.317" Score="12" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1. The boss is old.&lt;br&gt; &#xA;2. The boss is a lady.&lt;br&gt;&#xA;3. The boss is dead after swallowing all her clients!&lt;br&gt;&#xA;The last word in each client's description is associated with an animal from the rhyme &quot;There Was an Old Lady Who Swallowed a Fly&quot;:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The first client examined the magazines and other reading matter and finally settled on the local paper.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Fly paper&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second client was equally patient. She started surfing the web.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Spider web&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The third client was sniffling terribly and appeared to be coming down with flu.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Bird flu&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The fourth client appeared to be worried about his forthcoming scan.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;CAT scan&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The fifth client brought a section of tree branch along and spent the whole time examining its bark.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Dog bark&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sixth client had presumably missed lunch because she proceeded to eat some bread and cheese.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Goat cheese&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The seventh client was so terrified, all she wanted to do was hide.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Cow hide&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The eighth client came in out of breath as though he'd been in a race.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Horse race&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="14177" LastEditDate="2015-08-14T01:13:29.107" LastActivityDate="2015-08-14T01:13:29.107" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20031" PostTypeId="2" ParentId="20026" CreationDate="2015-08-13T23:40:35.730" Score="1" Body="&lt;p&gt;1)The boss is a veterinarian&#xA;Reason: All the clients had dog like behavior. For example client 1 decided to go to the bathroom on newspaper instead of magazine. Client 2 was a dog owner there to pick up her dog. All the actions of the other clients could be interpreted as something a dog would do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)The boss told here there were going to be 9 appointments&#xA;Reason: The lady was obviously expecting the 9th client to show because &quot;There was no ninth client so the receptionist went to see if the boss needed anything&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3)The boss was being attacked by the one of the clients(dogs) and was injured so the receptionist ran out of office to call police.&#xA;Reason: She screamed when she saw the injured vet and and ran out of office because the dog was currently attacking the vet.&lt;/p&gt;&#xA;" OwnerUserId="10856" LastActivityDate="2015-08-13T23:40:35.730" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20032" PostTypeId="2" ParentId="19806" CreationDate="2015-08-13T23:57:10.230" Score="1" Body="&lt;p&gt;I believe I only have a partial answer, but I enjoyed looking at it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What a certain literary professor for his unbeheld felt,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This certainly is a reference to Nabokov, but a third person reference so it is not about him&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The two of spades, from a new deck dealt, &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And with that, I have to assume this is a tarot card, the one for a quick thinker - the author likes the idea of writing about Nabokov&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Up, up, by the Ort Sphere's smallest test,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which he is now considering a reflection on his own mental abilities (unbeholden, of course), the Ort Sphere being the &quot;place&quot; of thought.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But oh, wait, now by one-third less.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A reduction by a third seems arbitrary, but in character for Nabokov works where the he reflects on his own work, and this author does too&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A din, a doubt...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This leads to doubt, confusion and hesitation, but ..&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Out!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the poem is written anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think rhyming in couplets, tarot card references (like 'Pale Fire' and maybe other words) was a strong indication that this is the right direction. If nothing else, I hope you find the answer entertaining!&lt;/p&gt;&#xA;" OwnerUserId="15681" LastActivityDate="2015-08-13T23:57:10.230" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20033" PostTypeId="1" AcceptedAnswerId="22304" CreationDate="2015-08-14T02:33:27.427" Score="4" ViewCount="668" Body="&lt;p&gt;I used to have a name like a valediction,&lt;br&gt;&#xA;Others say it's purely fiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My surname keeps your valuables safe,&lt;br&gt;&#xA;My upbringing was completely as a waif.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You've seen me in many a grand time,&lt;br&gt;&#xA;Up the ladder did I climb.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can be found picking on the smaller man;&lt;br&gt;&#xA;There are many that I am better than.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My lover found a life of empty space,&lt;br&gt;&#xA;Vowing that I would never see her face.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I decided to live like a camelid,&lt;br&gt;&#xA;In hopes to overcome what I did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even in death I provided support,&lt;br&gt;&#xA;Though to base pleasures I did resort.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;What is my name?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Starter Hint/Clarification:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first line of each stanza leads you to a homophone. Don't look for it in the second line of the stanza. Each stanza (after you find the homophone, of course) leads you to either one or two clues (but usually one), so if you can only relate a stanza to one, there may be no extra meaning.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer to each line (I think) is a homophone to the synonym of the last word/phrase. e.g. Cain's sibling --&gt; Abel --&gt; Able&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13538" LastEditorUserId="13538" LastEditDate="2015-09-17T21:21:39.227" LastActivityDate="2015-09-18T16:17:56.557" Title="The Homophonic Riddle" Tags="|rhyme|wordplay|knowledge|" AnswerCount="4" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20034" PostTypeId="2" ParentId="20033" CreationDate="2015-08-14T03:02:01.803" Score="0" Body="&lt;p&gt;Well I don't have it all but:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sherlock Holmes! (Elementary my dear watson)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I used to have a name like a valediction, &lt;br&gt;&#xA;Others say it's purely fiction.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; not done yet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My surname keeps your valuables safe, &lt;br&gt;&#xA;My upbringing was completely as a waif.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sherlock -&gt; sure lock [though surname implies last name?]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You've seen me in many a grand time, &lt;br&gt;&#xA;Up the ladder did I climb.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; not done yet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be found picking on the smaller man; &lt;br&gt;&#xA;There are many that I am better than.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Watson is picked on in some cases of sherlock holmes. Also Sherlock is sort of the best (detective among other things) so that implies a few may be lesser than he.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My lover found a life of empty space, &lt;br&gt;&#xA;Vowing that I would never see her face.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not done yet (not a holmes scholar)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I decided to live like a camelid, &lt;br&gt;&#xA;In hopes to overcome what I did.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; not done yet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Even in death I provided support, &lt;br&gt;&#xA;Though to base pleasures I did resort.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; he dies in The Final Problem taking out moriarty and his gang in the process. he is a noted used of strong drugs in excess.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I intend to add more details when I wake up on the morrow.&lt;/p&gt;&#xA;" OwnerUserId="13916" LastActivityDate="2015-08-14T03:02:01.803" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20035" PostTypeId="1" AcceptedAnswerId="109700" CreationDate="2015-08-14T03:42:23.537" Score="18" ViewCount="1029" Body="&lt;p&gt;They paid to see it and they applauded, as this concept was very new.&#xA;&lt;br/&gt;But only I could relate, for the differences in our &lt;strong&gt;case&lt;/strong&gt;s were very few.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This made me sad and I chose to cry.&#xA;&lt;br/&gt;It’s the only time they care, so I no more try.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That’s how it has been since my birth,&#xA;&lt;br/&gt;So is the case of my brothers on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It’s in our genes, they say. I can’t question even if I may.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They push us to the end of life, whenever they want.&#xA;&lt;br/&gt;Some have unavoidable reasons but mostly just chat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who am I, or what am I, please solve this riddle,&#xA;&lt;br/&gt;the riddler can’t rhyme anymore, so middle middle middle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT #1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Definitely you have seen it/him/her (may be in real, in media or in picture)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT #2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The words &quot;brothers&quot;, &quot;end of life&quot; are only figurative.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT #3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This hint might not help one get close to the answer but once someone got the answer, this hint can reason with one of the above lines. Hint is: &lt;strong&gt;Movie Reference in one of these lines.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT #4:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The word that has been highlighted is itself a hint. &quot;&lt;strong&gt;case&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT #5:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The word mentioned in Hint 4 is related to Hint 3. Now you know which particular line is referring to a movie. And you got the word &quot;case&quot; to think of the movie.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is my first attempt here. I will accept the answer if one can explain all the lines. If there are any English issues please do let me know. Good luck.&lt;/p&gt;&#xA;" OwnerUserId="15548" LastEditorUserId="8874" LastEditDate="2015-10-07T09:26:00.230" LastActivityDate="2021-04-25T21:33:40.187" Title="To tell me who I am, you must have made me cry" Tags="|riddle|rhyme|knowledge|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20037" PostTypeId="1" AcceptedAnswerId="20044" CreationDate="2015-08-14T10:43:02.820" Score="-5" ViewCount="1251" Body="&lt;p&gt;I'm confident that some of the members here will find this one easy. If that applies to you, please use spoiler hiding. If you didn't understand that last sentence, then my guess is you are not one of those members!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Puzzle&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;By running backwards it helps you to move forwards.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What and who am I talking about?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Notes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you get the correct answer (as judged by me), please include an explanation for those who are still puzzled. The best and fullest explanation  (as judged by up-voting) of everything in the puzzle and the build-up  gets the green tick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answers will consist of a single word, followed by an explanation of the title and my comments. (No more than 200 words please! :-)&lt;/p&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="10399" LastEditDate="2015-08-14T11:44:40.363" LastActivityDate="2015-08-14T19:31:41.427" Title="A reality for some members. A puzzle for others" Tags="|riddle|word|" AnswerCount="12" CommentCount="3" ClosedDate="2015-08-14T16:28:02.817" ContentLicense="CC BY-SA 3.0" />
  <row Id="20038" PostTypeId="2" ParentId="20037" CreationDate="2015-08-14T11:19:18.490" Score="5" Body="&lt;p&gt;Is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One way to run is to &lt;strong&gt;TROT&lt;/strong&gt;. If you turn this word backwards you get &lt;strong&gt;TORT&lt;/strong&gt;. If you pursue a civil case against someone, the establishment of a civil wrong (known as a TORT) will allow your case to move forwards quickly to a conclusion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-08-14T11:19:18.490" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20040" PostTypeId="2" ParentId="20037" CreationDate="2015-08-14T11:34:39.053" Score="3" Body="&lt;p&gt;What and who are you talking about ?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A catapult (not sure about the &quot;who&quot; part)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;By running backwards ...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You &quot;reload&quot; your catapult&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; it helps you to move forwards.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the more you reloaded it, the far the projectile will go.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13384" LastActivityDate="2015-08-14T11:34:39.053" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20041" PostTypeId="2" ParentId="20037" CreationDate="2015-08-14T11:36:50.180" Score="2" Body="&lt;p&gt;It's obviously&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A (comically-oversized) slingshot (and after the most recent edit, I add: A &quot;slingshot-enthusiast&quot; &lt;em&gt;wink wink&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because when you run backwards, you can only run so far before&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You get flung forwards at ridiculous speed and presumably hit something.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13068" LastEditorUserId="13068" LastEditDate="2015-08-14T14:38:46.210" LastActivityDate="2015-08-14T14:38:46.210" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20042" PostTypeId="2" ParentId="20037" CreationDate="2015-08-14T11:52:11.837" Score="2" Body="&lt;p&gt;Is it a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; stopwatch ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; With the time running backwards to zero (like an egg-timer) you will get faster (running forward) to reach the finish line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Title:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Only some members are doing sports, so there is the link to their reality ;-)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14461" LastEditorUserId="14461" LastEditDate="2015-08-14T11:56:35.013" LastActivityDate="2015-08-14T11:56:35.013" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20043" PostTypeId="2" ParentId="20037" CreationDate="2015-08-14T12:23:13.657" Score="6" Body="&lt;p&gt;You Are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You are a clown.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; By Running Backwards on the ball, The ball will Move Forwards&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15698" LastActivityDate="2015-08-14T12:23:13.657" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20044" PostTypeId="2" ParentId="20037" CreationDate="2015-08-14T13:16:21.873" Score="16" Body="&lt;p&gt;The word is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; !&amp;lt; which is two characters, which in the original character sets of ASCII or EBCDIC is two bytes, and while &quot;words&quot; are sized to fit the hardware, the two byte word was the earliest definition, thus this is a &quot;word&quot; in the 1970's memory sense.  Everything else was straight up stated: Please use spoiler hiding.  Run backwards to run forwards.  The &quot;who&quot; are, as stated, &quot;some of the members&quot; of this forum.  The title, &quot;A reality for some members a puzzle for others&quot; is a snarky complaint about people who don't sufficiently mask their guesses.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Additional comment&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; @JiK !&amp;lt; is the spoiler escape markup &quot;&gt;!&quot; in mirror image.  I suppose it's an arguable point whether it should be !&gt;, !&amp;lt; or &amp;lt;!  (... neat, the last of these is active!) In any case, this word, run backwards, is how some members should move forwards when answering certain riddles.  Today, it's a reality for some members but a true puzzle for others.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11921" LastEditorUserId="3911" LastEditDate="2015-08-14T15:56:58.180" LastActivityDate="2015-08-14T15:56:58.180" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20045" PostTypeId="2" ParentId="20037" CreationDate="2015-08-14T13:23:59.037" Score="2" Body="&lt;p&gt;Is it...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Research&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;As the more you move backwards into the past, the more you learn to help you move forward! As @user662852 mentions it hints how new users do not research previous questions and answers, or look at the help sections, to be able to craft a well structured answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9769" LastEditorUserId="9769" LastEditDate="2015-08-14T13:39:59.090" LastActivityDate="2015-08-14T13:39:59.090" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20046" PostTypeId="2" ParentId="19901" CreationDate="2015-08-14T13:34:02.873" Score="7" Body="&lt;p&gt;As Cristian Marian figured out, the number 639407344 is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The time/date &lt;strong&gt;Fri, 06 Apr 1990 13:09:04 GMT&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The question says that the day is very useful to abecedarians, which means that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The day (Friday) is going to be our alphabet key for our cipher.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since we have the time (which is a number) as a second key, we know we're using&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A &lt;strong&gt;Gronsfeld cipher&lt;/strong&gt;, which is like a Vigenere but uses a number key instead of a letter key alongside the modified alphabet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Putting it all together,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We input the cipher text with an alphabet of FRIDAYBCEGHJKLMNOPQSTUVWXZ and a key of 130904, giving us the result &quot;You should go through door F&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, I think I'll head through&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Door F, of course!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-08-14T13:34:02.873" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20048" PostTypeId="2" ParentId="20037" CreationDate="2015-08-14T14:26:26.157" Score="0" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Merlyn&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In at least one mythology, T.H. White's &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/The_Once_and_Future_King#Characterization_in_the_work&quot; rel=&quot;nofollow&quot;&gt;The Once and Future King&lt;/a&gt;&lt;/em&gt; he was living backwards through time, getting younger as time, and all his machinations, moved forward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In another mythology, a movie, he describes himself, not dissimilar to this question's title, as &quot;A dream to some, a nightmare to others!&quot;&lt;/p&gt;&#xA;" OwnerUserId="4384" LastEditorUserId="4384" LastEditDate="2015-08-14T14:39:38.087" LastActivityDate="2015-08-14T14:39:38.087" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20049" PostTypeId="2" ParentId="20006" CreationDate="2015-08-14T14:44:45.333" Score="3" Body="&lt;p&gt;Building on top of other answers, the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Scroll lock key (like the now useless one on keyboards)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Carved from a forest we lie&lt;br&gt;&#xA;A road built of bone we supply&lt;br&gt;&#xA;With great walls of black&lt;br&gt;&#xA;We are known to attack&lt;br&gt;&#xA;From the forest an echo replies&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Scrolls can be made from paper which come from trees. The great walls of black is the text that is written on them. Magical scrolls can be used for spells and attacks. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For a singular purpose we're made&lt;br&gt;&#xA;To deny those of whom you're afraid&lt;br&gt;&#xA;And provide peace of mind&lt;br&gt;&#xA;Each day we're combined&lt;br&gt;&#xA;With our partner who therein has stayed&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;As per Kingrames; This refers to how a lock is opened with a key, often daily. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At prominence ere the fall of Rome&lt;br&gt;&#xA;We held the theaters, bridges and domes&lt;br&gt;&#xA;In structures erected&lt;br&gt;&#xA;We shan't be neglected&lt;br&gt;&#xA;Lest one desires to topple our home&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;As per Kingrames again: A keystone is placed at the top of an arch, and prevents the whole thing from falling down. it's surprisingly sturdy even though it doesn't quite look like it would be.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13489" LastEditorUserId="13489" LastEditDate="2015-08-14T15:50:11.910" LastActivityDate="2015-08-14T15:50:11.910" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20050" PostTypeId="1" AcceptedAnswerId="20051" CreationDate="2015-08-14T14:46:47.817" Score="2" ViewCount="1035" Body="&lt;p&gt;&lt;strong&gt;Puzzle:&lt;/strong&gt; $$\begin{array}{}\boxed{3\;\,}&amp;amp;\boxed{150\;\;\;\,}&amp;amp;\boxed{12\;\;\;\,}\\\boxed{7\;\,}&amp;amp;\boxed{3164\;\,}&amp;amp;\boxed{56\;\;\;\,}\\\boxed{16}&amp;amp;\boxed{74112}&amp;amp;\boxed{272\;\;}\\\boxed{32}&amp;amp;\boxed{\text{?}\;\;\;\;\;\;\,}&amp;amp;\boxed{1024\,}\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have to figure out the question mark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My effort:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried almost for half-an-hour, but can't seem to work out the rule being followed here. I tried&lt;br&gt;&#xA;$$\begin{align}&#xA;7 - 3 &amp;amp;= 2^2\\  &#xA;16 - 7 &amp;amp;= 3^2\\  &#xA;32 - 16 &amp;amp;= 4^2&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was the only sequence I could figure out but it only works for the first column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need help figuring this out, along with a suitable explanation of the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Note&lt;/em&gt;&lt;/strong&gt;: This is &lt;em&gt;not&lt;/em&gt; a homework question.&lt;/p&gt;&#xA;" OwnerUserId="2062" LastEditorUserId="13494" LastEditDate="2015-08-20T21:48:42.267" LastActivityDate="2015-08-20T21:48:42.267" Title="Simple mathematical puzzle" Tags="|mathematics|pattern|number-sequence|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20051" PostTypeId="2" ParentId="20050" CreationDate="2015-08-14T14:54:16.803" Score="7" Body="&lt;p&gt;My answer: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\mathbf{1049088}$&#xA; &lt;br&gt;&lt;br&gt;&#xA; The pattern I noticed was&lt;br&gt;&#xA; $\text{middle column} = \text{right column}^2 + (\text{left column}\times 2^\text{row number})$&lt;br&gt;&#xA;&lt;br&gt;&#xA; $2^1\times 3 + 12^2 = 6 + 144  = 150$&lt;br&gt;&#xA; $2^2\times 7 + 56^2 = 28 + 3136 = 3164$&lt;br&gt;&#xA; $2^3\times 16 + 272^2 = 128 + 73984 = 74112$&lt;br&gt;&#xA; $2^4\times 32 + 1024^2 = 512 + 1048576 = 1049088$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My inspiration:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I divided 74112 by 272 and it came close to 272, so I though of squaring. Then I took the difference and said hey that looks like a multiple of 16, it was, I then applied the algorithm on all of them and the multiple trend of the difference from the square remained, and the 2, 4 ,8 pattern appeared. Leaving the assumption 16 would be the next difference multiplier.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13916" LastEditorUserId="4421" LastEditDate="2015-08-14T19:30:12.600" LastActivityDate="2015-08-14T19:30:12.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20052" PostTypeId="2" ParentId="20006" CreationDate="2015-08-14T14:55:16.563" Score="2" Body="&lt;p&gt;I'll take a &lt;em&gt;stab&lt;/em&gt; at it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could it be referring to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Soldiers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As f&quot; solved for Saturn&#xA;&lt;br&gt;and Cain reasoned Stairs&#xA;&lt;br&gt;While one word you seek&#xA;&lt;br&gt;Three this riddle bears:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The previous two riddles word answers start with 'S' as does the word Soldier.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Carved from a forest we lie&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A possible reference forest camouflage used by soldiers blend into the forest. Also that many times the soldiers can use the forest as cover and 'lie' waiting for their targets to pass through.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A road built of bone we supply&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Soldiers can be used to create a secure path (road) 'built' from the bones of those that get in the way. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With great walls of black&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A possible reference to a groups of soldiers shields (or nowadays tanks) that could be black.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are known to attack&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Soldiers are used in battles to attack the opponent.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the forest an echo replies&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the medieval times you would be able to hear the soldiers shouts and battle cries when they are engaging.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For a singular purpose we're made&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Soldiers are made for battle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To deny those of whom you're afraid&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A reference in stopping the enemy force by 'denying' them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And provide peace of mind&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They can eliminate the threat, allowing peace of mind knowing it is gone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Each day we're combined&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Soldiers are combined together to create a battalion when on duty.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With our partner who therein has stayed&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; While a soldier may go into battle, Often the spouse will remain at home for the family.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At prominence ere the fall of Rome&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Roman Empire was a defeated by and supplied vast amounts of soldiers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We held the theaters, bridges and domes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In urban battles (of WWII), many soldiers had to hold up in theaters and domes as well as protect bridges to maintain travel routes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In structures erected&#xA;&lt;br&gt;We shan't be neglected&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Today we have many erected structures of previous soldiers and in memorial of their battles and losses. The help remind us to not neglect their sacrifices.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lest one desires to topple our home&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;With a strong military (USA), people will not like to start fights them, or try to 'topple' their home.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastActivityDate="2015-08-14T14:55:16.563" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20053" PostTypeId="2" ParentId="19806" CreationDate="2015-08-14T14:59:54.800" Score="3" Body="&lt;p&gt;Edit: I &lt;em&gt;still&lt;/em&gt; do not have the answer, but have deciphered some things in the hints that may help others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Giveaway&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Louis_X_of_France&quot; rel=&quot;nofollow&quot;&gt;Louis X of France&lt;/a&gt; (1289-1316) was the first person to construct indoor tennis courts in the modern style. I suspect that tennis is the key to this entire thing but I can't quite make it happen!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hints for Line 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; These are Unicode codes for Cyrillic characters. Decoded, they are: Влади́мир Влади́мирович Набо́ков, or, in English, Vladimir Vladimirovich Nabokov. A famous line from &lt;em&gt;Lolita&lt;/em&gt; is &quot;There was no Lo to behold.&quot; So, she is his unbeheld. What did he feel for her? We're looking for a four (or three) letter word. Because of the tennis clue above, I thought LOVE, but then what's the three-letter word?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint for line 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mrs. Malaprop famously got her words mixed up, so we're looking for some kind of misstatement.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint for Line 5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;What is truth? It is the man who stands before you.&quot; is the English translation of the famous Latin anagram &lt;em&gt;Quid est veritas? / Est vir qui adest&lt;/em&gt;, so I think we're looking for an anagram. Perhaps an anagram of Ort Sphere, the weirdest phrase in here? That could be &quot;three pros&quot;, &quot;he reports&quot;, etc. but nothing obvious. As for &quot;Find the man whose name is like a rabbit,&quot; I came up with Charles Babbage, whose name is reminiscent of Babs Bunny, but that seems like a bit of a stretch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Putting it all together:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My very loose guess is that this is a cryptic description of a tennis match or point. All I have to go on is the Louis X clue, which I feel confident about, and many other vague inklings: the possibility of LOVE, the two of spades as maybe some reference to a serve, the line &quot;Out!&quot;, the author's use of &quot;&lt;strong&gt;great point&lt;/strong&gt;&quot; in the hints. Please, please if someone can use these hints to figure the whole thing out, do it! It's killing me.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10851" LastEditorUserId="10851" LastEditDate="2015-08-25T01:59:36.387" LastActivityDate="2015-08-25T01:59:36.387" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20054" PostTypeId="2" ParentId="19806" CreationDate="2015-08-14T15:24:57.093" Score="4" Body="&lt;p&gt;I think the man is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;b&gt;&lt;em&gt;Black Jack&lt;/em&gt;&lt;/b&gt; (The Card Game)&#xA;&lt;br&gt; [The player starts with an Ace and a Nine]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What a certain literary professor for his unbeheld felt,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The dealers repeat their 'scripted' talks about the rules/bets/progress (literary professor). His unbeheld [felt-&gt;feel]/touch is that he does not have influence on the results of the game.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Up, up, by the Ort Sphere's smallest test, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The player has a Ace and a Nine totaling to 20. He calls hit and is asking for another Ace (the smallest ort [card]) to get black jack!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The two of spades, from a new deck dealt, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;He is dealt a two of spades from the 'new deck'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But oh, wait, now by one-third less.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;He had an Ace which now has become a value of 1 instead of 11. Giving him one-third less of a total? [He must have had an Ace and a Nine for this to occur without having 21 or bust]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A din, a doubt... &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;He is thinking about his next move. He has a total of 12 in front of him. [statistically speaking he should hit]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Out!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;He calls hit. [He gets a 10 or face card] Bust, he lost! &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11244" LastEditorUserId="11244" LastEditDate="2015-08-14T15:44:03.987" LastActivityDate="2015-08-14T15:44:03.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20055" PostTypeId="1" AcceptedAnswerId="20065" CreationDate="2015-08-14T15:25:23.347" Score="7" ViewCount="1098" Body="&lt;p&gt;I am so optimistic, as happy as can be,&lt;br&gt;&#xA;But deep inside darkness, violence you may see.&lt;br&gt;&#xA;To half the population, danger is my name,&lt;br&gt;&#xA;For after an encounter, they never are the same.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For others I'm just there, don't really do too much,&lt;br&gt;&#xA;I tend to vanish quickly, without a single touch.&lt;br&gt;&#xA;But all I trap inside of me, as I contemplate,&lt;br&gt;&#xA;And soon they're free to go, without that which I hate.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determine now who I am, and what it is I do,&lt;br&gt;&#xA;Who fears me, who doesn't care? Am I one or two?    &lt;/p&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-08-14T19:52:30.670" Title="Let's put a smile on that face of yours!" Tags="|riddle|rhyme|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20056" PostTypeId="2" ParentId="20011" CreationDate="2015-08-14T15:32:35.273" Score="3" Body="&lt;p&gt;This might be a bit of a hail mary, but I'll throw it up and see what happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hard luck farmer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Andrew Luck, quarterback for the Indianapolis Colts&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I work on a farm.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Colts' practice facility is the Indiana Farm Bureau Football Center&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I need to wake up around sunrise to go to work.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Really reaching here, but it looks like there is a chain of breakfast restaurants, named Sunrise Cafe, in and around Indy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And if you don't count room and board, I get just 4 pennies for every minute of work.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nothing definitive here.  I looked at his average yearly salary but got nowhere, and can't get my mind away from quarter back.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I had to take a second job to pay the bills.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Staying away from the Buffalo Bills here and instead I'll go with the set of DirecTV Luck is now featured in, though having looked at his salary, I don't think he's hurting for money.  Nonetheless, commercial acting is a second job.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Honestly, I'd do anything just to get an apartment.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To be moving on up, a la The Jefferson's? Really grasping at straws with that one, but my sense of humor is satisfied with that as an answer, so I'll go with it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I'm such a sad sack it'll probably never happen.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quarterbacks get sacked and that cuts into their success.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is my name?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, I'll say that it's Andrew Luck, and I'll let my mind create further justification out of &quot;10 or 11 reads&quot;, and manipulate it to #12 reads (the coverage).  The third hint was just added, and it kinda rattles my conviction, but Luck is a specific, real person, with his name in the title.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11452" LastActivityDate="2015-08-14T15:32:35.273" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20057" PostTypeId="1" AcceptedAnswerId="24046" CreationDate="2015-08-14T15:36:20.160" Score="6" ViewCount="906" Body="&lt;p&gt;While raiding a tropical island, 6 pirates have an incredibly unlucky day. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While they scoured the island in search of treasure, one of them was infected with the Fanatic Fever, a disease which hits the pirate right in his greatest strength. (As everybody knows, a pirates greatest strength is his perfect rationality.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A healthy pirate will always act in a certain way, following lower number rules first then higher numbered rules:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;They will always choose action a over action b if action a means they have a higher chance of survival.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;They will always choose action a over action b if action a and b have equal chances of survival, and action a leads to higher expected gold. Note: expected gold is defined as such: an x percent chance of getting y gold is x*y expected gold.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;All else equal they prefer to see other pirates die. (They are ambivalent about whether more pirates die after they are dead.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;They randomly choose between actions they are ambivalent about in an entirely unpredictable fashion. In general, if there are n actions, all of which are equally preferable, they will choose each with 1/n probability.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;However, a pirate infected with the fanatic fever is different. He will always, and only, vote for plans that involve the oldest pirate getting the most gold. If they are the oldest pirate, they can and will only propose such plans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you may know, pirates have a particular method of distributing the gold pieces they get from piracy. The oldest pirate proposes a plan which says exactly how many gold pieces each pirate will receive. Gold pieces cannot be cut, so each pirate must receive a whole number of gold pieces in any given plan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, starting with the oldest pirate, and in order such that no younger pirate votes before an older pirate, each pirate publicly votes yes or no to the plan. Every pirates vote counts once except the oldest pirate's vote which counts one and a half times. The votes are tallied up. If there are more yesses, the gold is distributed as proposed. Otherwise, the oldest pirate is executed and they restart. No two pirates are exactly the same age.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although all pirates know that exactly one of them is infected with fanatic fever, and all other pirates are healthy, no healthy pirate initially knows who is infected (although each pirate does know if they in particular are infected). Every healthy pirate initially believes that every other pirate has an equal chance of being infected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, these pirates are actually so unlucky, that in addition to one of them being infected, they only found a single gold piece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: When it comes time to distribute the gold pieces, what should the oldest pirate propose assuming he is healthy, and how likely is he to survive assuming each other pirate has an equal chance of being infected? For completeness, give every possible scenario with its probability. Remember that none of the other pirates other than the infected one know that the oldest pirate is not infected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Challenge: Generalize for N pirates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extra hard challenge: Generalize for N pirates and X gold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Impossible (Literally?) challenge: Generalize for N pirated, X gold, and Z of the pirates are infected with fanatic fever.&lt;/p&gt;&#xA;" OwnerUserId="15529" LastEditorUserId="15529" LastEditDate="2015-08-15T04:04:22.937" LastActivityDate="2015-11-18T16:34:52.693" Title="The Fanatic Fever(twist on the 5 pirates puzzle)" Tags="|mathematics|strategy|probability|game-theory|" AnswerCount="2" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="20058" PostTypeId="2" ParentId="20037" CreationDate="2015-08-14T15:42:53.953" Score="0" Body="&lt;p&gt;This is similar to one already mentioned:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Someone running on a cylinder, or sphere.&#xA; They must run backwards to roll what they are running on forwards&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7265" LastActivityDate="2015-08-14T15:42:53.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20059" PostTypeId="2" ParentId="20011" CreationDate="2015-08-14T15:51:32.857" Score="4" Body="&lt;p&gt;This is a looong way from matching every clue, but maybe it's a start...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you work on a farm and need to be up for sunrise it's because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are a cockerel, or cock; sunrise being the cockcrow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This also works with not being romantic because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Although some people might try and pull &lt;a href=&quot;https://en.wikipedia.org/wiki/The_Naked_Man_(How_I_Met_Your_Mother)&quot; rel=&quot;nofollow&quot;&gt;The Naked Man&lt;/a&gt; on a date, it's not very romantic...!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So clearly your name is &lt;strong&gt;&lt;em&gt;not&lt;/em&gt;&lt;/strong&gt; (because it doesn't match with any other clues!):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alfred Hitchcock. (Only famous person I could think of who matched my small part [pun &lt;em&gt;very&lt;/em&gt; much intended] of an answer)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9444" LastActivityDate="2015-08-14T15:51:32.857" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20060" PostTypeId="2" ParentId="20037" CreationDate="2015-08-14T16:00:42.590" Score="3" Body="&lt;p&gt;First thing on my mind would be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Bounty: Obviously you need to be a member to be a bounty giver. When you do it your reputation runs backwards but afterwards you likely get a nice answer allowing you to move forward.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1666" LastActivityDate="2015-08-14T16:00:42.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20061" PostTypeId="2" ParentId="20037" CreationDate="2015-08-14T16:07:31.397" Score="3" Body="&lt;p&gt;Because you are &quot;from the UK&quot; I realised that this question &lt;em&gt;may&lt;/em&gt; be aimed at others from the UK, in which case:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &quot;members&quot; are Members of Parliament, commonly initialised to MPs. If you run MP backwards you get PM, the common initialisation of the Prime Minister, the head of the MPs. Being an MP and then becoming PM is definitely &quot;moving forwards&quot; in your career!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This doesn't really cover any of the &quot;spoiler&quot; clues though...&lt;/p&gt;&#xA;" OwnerUserId="9444" LastActivityDate="2015-08-14T16:07:31.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20062" PostTypeId="2" ParentId="20037" CreationDate="2015-08-14T16:13:19.567" Score="-5" Body="&lt;p&gt;Having set this puzzle, I have decided to answer it as well using my original target answer. This is because user662852 has provided an &lt;a href=&quot;https://puzzling.stackexchange.com/a/20044/5359&quot;&gt;answer&lt;/a&gt; that is probably better than the one I intended.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Single word answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; recursion&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A reality for some members. A puzzle for others&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some of the members are programmers/software developers. For them, recursion is a tool of the trade. It is a reality in their lives. For others it is a rather abstract concept.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm confident that some of the members here will find this one easy. If that applies to you, please use spoiler hiding. If you didn't understand that last sentence, then my guess is you are not one of those members!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My explanation here was rather mundane. Any computer whizz worthy of their salt would have spoiler hiding using &gt;! sussed in very short order. If you are not a computer whizz, then my sentence and other such things may remain a mystery for some time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By running backwards it helps you to move forwards.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The word recursion literally means to run backwards. In computer terms you run backwards and forwards in the stack. Using recursion helps programmers to move forwards in their work.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What and who am I talking about? &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am talking about recursion and about the members here who are computer programmers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2015-08-14T19:31:41.427" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20063" PostTypeId="2" ParentId="20055" CreationDate="2015-08-14T16:18:03.857" Score="3" Body="&lt;p&gt;You are a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dentist&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am so optimistic, as happy as can be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dentist are usually happy people, and they like smiles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But deep inside darkness, violence you may see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is dark in your mouth, which is where the dentist's work is done.  Depending on the status of your oral hygiene, the dentist's work might be violent, with bleeding gums and worse commonplace (not that I know from experience...).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To half the population, danger is my name,&#xA;For after an encounter, they never are the same.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A surprising number of people are afraid of going to the dentist, due to the pain that they expect and sometimes receive on visiting.  A dentist may leave someone &quot;never the same&quot; by pulling teeth, whether molars or otherwise, drilling, putting in caps, or just scaring the patient even more severely.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For others I'm just there, don't really do too much,&#xA;I tend to vanish quickly, without a single touch.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Those with good oral hygiene may actually see the dentist for no more than a minute.  The dental hygienist does most of the cleaning and inspection, which the true dentist just pops in to double-check.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But all I trap inside of me, as I contemplate,&#xA;And soon they're free to go, without that which I hate.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think this just refers to all that go to the dentist, not ALL IN EXISTENCE, so: Everyone that goes to the dentist has their teeth cleaned of plaque and gingivitis, which is what the dentist hates the most.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Determine now who I am, and what it is I do,&#xA;Who fears me, who doesn't care? Am I one or two?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are the dentist and hygienist, and you both clean teeth, but one more than the other.  Those with good teeth don't care, those with bad teeth do.  You are two who say &quot;Let's put a smile on that face of yours!&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll also say that: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;An orthodontist changes people, too, and could be said to hate crooked teeth, but people leave the orthodontist with braces, not straight teeth, and they leave the dentist with clean teeth.  With that, I think dentist better fits the bill.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11452" LastActivityDate="2015-08-14T16:18:03.857" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20065" PostTypeId="2" ParentId="20055" CreationDate="2015-08-14T17:15:46.713" Score="7" Body="&lt;p&gt;I'm thinking you're &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the absolute value function.&#xA;That is, |(+/-)x|, where if x is negative, it becomes positive (losing the negative sign, which you hate), but if x is already positive, nothing happens.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am so optimistic, as happy as can be,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Literally 100% positive&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But deep inside darkness, violence you may see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Negative numbers can be held inside the function before they're converted&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To half the population, danger is my name,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;all negative numbers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For after an encounter, they never are the same. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;they become positive&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For others I'm just there, don't really do too much,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;positive numbers don't change&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I tend to vanish quickly, without a single touch.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;nothing happens&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But all I trap inside of me, as I contemplate,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;nothing escapes the function. dun dun dunnnn.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And soon they're free to go, without that which I hate. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;no negative sign&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Determine now who I am, and what it is I do,&#xA; Who fears me, who doesn't care? Am I one or two? &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;negative numbers fear the operation, positive numbers don't.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13068" LastEditorUserId="13068" LastEditDate="2015-08-14T17:24:27.643" LastActivityDate="2015-08-14T17:24:27.643" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20066" PostTypeId="1" AcceptedAnswerId="20307" CreationDate="2015-08-14T17:59:24.403" Score="5" ViewCount="501" Body="&lt;p&gt;You know who I am, and don't forget it.  And don't forget what I'm telling you now:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Years ago, I was in a similar place&lt;br&gt;&#xA;So on this occasion I will treat you with grace&lt;br&gt;&#xA;You have adorned yourself to match my creation&lt;br&gt;&#xA;But let’s not forget your lower station&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your hands are your own, and that much is clear&lt;br&gt;&#xA;But the work they may do gives me reason to fear&lt;br&gt;&#xA;Keep them under wraps, and employ no device&lt;br&gt;&#xA;No matter how much your mind is enticed  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will use no tools to intimidate,&lt;br&gt;&#xA;My existence alone should make you hesitate&lt;br&gt;&#xA;When you return, it best not be late&lt;br&gt;&#xA;For- know it now- alone I will wait  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want memories more than regrets&lt;br&gt;&#xA;Then do not think these idle threats&lt;br&gt;&#xA;Dance around the issue, but do no more&lt;br&gt;&#xA;Or you’ll forever be met with a locked, closed door  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I being too hard on you? Do you know who I am and what I am talking about?&lt;/p&gt;&#xA;" OwnerUserId="11452" LastActivityDate="2015-08-20T01:41:15.333" Title="Just Don't Forget it" Tags="|riddle|rhyme|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20067" PostTypeId="2" ParentId="20035" CreationDate="2015-08-14T18:27:00.223" Score="5" Body="&lt;p&gt;Preliminary answer: Just to throw some ideas out there.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;They paid to see it and they applauded, as this concept was very new. &#xA;  But only I could relate, for the differences in our cases were very few.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Makes me think of original films, when the concept of moving pictures was new.&lt;br&gt;&#xA;The case could refer to &lt;a href=&quot;https://en.wikipedia.org/wiki/The_Dreyfus_Affair_(film_series)&quot; rel=&quot;nofollow&quot;&gt;La Case de Dreyfus à l'île du Diable&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This made me sad and I chose to cry. &#xA;  It’s the only time they care, so I no more try.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In sticking with the movie theme, this could reference trip to the Moon, where the moon cries when the rocket lands in his eye. Same director as the Dreyfus Affair&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;That’s how it has been since my birth, &#xA;  So is the case of my brothers on Earth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Qualifying brothers with on Earth either means there are other brothers not on earth, or the subject is not on Earth. Or maybe just to rhyme :P But I'm sticking with Moon as an answer here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rest of the Poem: Still working on it (read: I'm stumped).&lt;/p&gt;&#xA;" OwnerUserId="1968" LastEditorUserId="12700" LastEditDate="2015-08-16T18:18:59.430" LastActivityDate="2015-08-16T18:18:59.430" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20068" PostTypeId="2" ParentId="20014" CreationDate="2015-08-14T18:34:43.483" Score="17" Body="&lt;p&gt;Since the answer already has been accepted I would add this as a comment, but I do not have sufficient points to do so. I agree that the answer should be...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but due to the reason that...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ... in the two upper rows, each row contains two figures which encloses an area, and two that does not. Therefore, the figure replacing the question mark should enclose an area, which leaves only figure A.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12901" LastActivityDate="2015-08-14T18:34:43.483" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20069" PostTypeId="1" AcceptedAnswerId="20070" CreationDate="2015-08-14T19:27:24.843" Score="4" ViewCount="670" Body="&lt;p&gt;In your pocket I'll stay till you send me away;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But keep me a while and I'll build up a pile,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lest for desire I'll escape all a-fire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(haha I made this one up myself)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="9575" LastEditorUserId="5373" LastEditDate="2016-12-14T13:37:57.153" LastActivityDate="2016-12-14T13:37:57.153" Title="In your pocket for awhile" Tags="|riddle|word|rhyme|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20070" PostTypeId="2" ParentId="20069" CreationDate="2015-08-14T19:31:03.883" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MONEY?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In your pocket I'll stay till you send me away;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It stays in your pocket until you spend it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But keep me a while and I'll build up a pile,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Keep it for long enough and it'll pile up like a miser's hoard.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Lest for desire I'll escape all a-fire.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Money can be spent (or escape your possession) on your desires. Alternatively (thanks to Kik), this refers to the expression &quot;money is burning a hole in your pocket&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-08-14T19:35:53.373" LastActivityDate="2015-08-14T19:35:53.373" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20071" PostTypeId="2" ParentId="20069" CreationDate="2015-08-14T19:33:14.083" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Money&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In your pocket I'll stay till you send me away;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You keep money in your pocket until you spend it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But keep me a while and I'll build up a pile&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you save your money, it will add up, and you could also use it to earn additional interest in a savings account, or other investments&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lest for desire I'll escape all a-fire&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The well known phrase &quot;that money is burning a hole in your pocket&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11246" LastActivityDate="2015-08-14T19:33:14.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20072" PostTypeId="2" ParentId="20069" CreationDate="2015-08-14T19:33:31.477" Score="1" Body="&lt;p&gt;Is the answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;a coin/coins&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In your pocket I'll stay till you send me away;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;coins stay in your pocket until you spend them&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But keep me a while and I'll build up a pile,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If you don't use your coins you must be using bills and will inevitably get more coins in change. Alternatively you store your coins in a bank and they build interest&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lest for desire I'll escape all a-fire.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This is my biggest stretch. Unlike bills coins can't be set on fire, and the only way to really get rid of coins is to spend them on your desires&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15720" LastEditorUserId="15720" LastEditDate="2015-08-14T19:49:59.477" LastActivityDate="2015-08-14T19:49:59.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20074" PostTypeId="2" ParentId="19923" CreationDate="2015-08-14T20:43:15.120" Score="1" Body="&lt;p&gt;Possible solution to:       &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;&quot;He lurks even now in the waters of the Valley&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;&lt;strong&gt;Kings' Valley (Valley of Kings)&lt;/strong&gt; &lt;/strike&gt;&lt;br&gt;&#xA; &lt;strong&gt;Evidence:&lt;/strong&gt; During periodic thunder storms, the central area of the valley floods. Historical floods have caused some tombs to shift five meters into the ground. In fact, one of the tombs so shifted is KV62, the burial chamber that King Tut's remains were found in. Perhaps the &quot;He&quot; refers to Tut or another of the many Pharaohs whose ghosts could still &quot;lurk...in the waters of the Valley&quot; and who would be important enough for the specific &quot;He&quot; to make sense.&lt;br&gt;&#xA; &lt;strong&gt;Sources:&lt;/strong&gt;&lt;br&gt;&#xA; - &lt;a href=&quot;https://en.wikipedia.org/wiki/Valley_of_the_Kings&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Valley_of_the_Kings&lt;/a&gt;&lt;br&gt;&#xA; - &lt;a href=&quot;https://en.wikipedia.org/wiki/Tutankhamun&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Tutankhamun&lt;/a&gt;.&lt;br&gt;&#xA; &lt;strong&gt;Additional:&lt;/strong&gt; Valley of Kings == Kings' Valley. The quote &quot;waters of the Valley&quot; could also simply refer to the Nile, which lies directly adjacent to the Valley of Kings and is the source for the flooding during storms.  &lt;br&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Other Suggestion(s):&lt;/strong&gt;     &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - Helm's Deep&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15642" LastEditorUserId="15642" LastEditDate="2015-08-14T22:14:33.283" LastActivityDate="2015-08-14T22:14:33.283" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20075" PostTypeId="1" AcceptedAnswerId="20076" CreationDate="2015-08-14T20:53:27.823" Score="9" ViewCount="1183" Body="&lt;p&gt;A banana has nothing to do with this easy rebus puzzle (and I made it myself); repeat, this is not a banana.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SARS7.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SARS7.png&quot; alt=&quot;Not a Banana&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9575" LastEditorUserId="9000" LastEditDate="2015-08-14T21:21:28.197" LastActivityDate="2015-08-14T21:21:28.197" Title="This is not a banana" Tags="|rebus|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20076" PostTypeId="2" ParentId="20075" CreationDate="2015-08-14T20:54:03.343" Score="13" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Once in a blue moon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because the picture is literally:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ones in a blue moon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And that sounds like the more common phrase&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2015-08-14T20:54:34.433" LastActivityDate="2015-08-14T20:54:34.433" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="20077" PostTypeId="2" ParentId="20075" CreationDate="2015-08-14T21:01:34.467" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Once in a blue moon, because the moon is blue and has 1s &quot;in&quot; it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastActivityDate="2015-08-14T21:01:34.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20079" PostTypeId="2" ParentId="20069" CreationDate="2015-08-14T21:14:42.333" Score="0" Body="&lt;p&gt;Alternative answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lint.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In your pocket I'll stay till you send me away;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; til you pick it out&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But keep me a while and I'll build up a pile,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lint on the dryer filter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Lest for desire I'll escape all a-fire.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lint in the drier filter can catch on fire. (desire rhymes with dryer?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Rand wins but LINT WILL NOT BE SILINTSED&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13916" LastActivityDate="2015-08-14T21:14:42.333" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20080" PostTypeId="2" ParentId="19860" CreationDate="2015-08-14T21:40:46.567" Score="0" Body="&lt;p&gt;1.The guru said, &quot;at least one of you has (GREEN / BLUE) eyes.&quot; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;GREEN is a spanned word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2.You have limited weighings to find the odd coin among (NINE / TWELVE).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;NINE is spanned.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3.One of the most popular Puzzling.SE posts is about (TETRIS / PRISMS).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;TETRIS is spanned.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4.In the Monty Hall problem, one of your options is to (STICK / SWITCH).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;SWITCH is spanned.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5.I hope you will find this puzzle (INVENTIVE / DIFFICULT).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;INVENTIVE is spanned.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6.This puzzle's solution may depend on (SHAPE / SHAPES).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;SHAPE is spanned.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;7.Of course, that might just be a (DISTRACTION / DIVERSION).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;DISTRACTION is spanned.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;8.Will similar words make this puzzle (EASIER / SIMPLER)?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;EASIER is spanned.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;9.Or will you focus too much on the (DIFFERENCES / DISTINCTIONS)?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Both words are spanned.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;10.Don't work on this for too long or you may become (CONFUSED / ENTRANCED.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;ENTRANCED is spanned.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;11.I hope this puzzle isn't too (HARD / EASY)!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Both words are spanned.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;12.I can't think of any other relevant sentences so here are some things that annoy me: (INTERRUPTIONS / NOISE), (STUBBORNNESS / INFLEXIBILITY), and (THIEVES / LAWYERS).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;13.(By the way, the previous sentence's pairs were all (INDEPENDENT / UNRELATED).)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;14.Good luck solving this (CONUNDRUM / CHALLENGE)!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;CHALLENGE is spanned.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If my guesses are correct, Spanned words&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; have lines separating the letter in half in at least half of the word. (If it has to be over half, that might eliminate some of the duplicates I discovered.)&lt;br&gt;&lt;br&gt;Examples would be AEHINTZ who all divide themselves (with spanners, I guess if you want to call them that).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that the font changes when you start editing.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;M and W don't look the same when you're typing your answers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13068" LastEditorUserId="13068" LastEditDate="2015-08-14T21:47:44.587" LastActivityDate="2015-08-14T21:47:44.587" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20082" PostTypeId="1" AcceptedAnswerId="20083" CreationDate="2015-08-14T23:10:20.403" Score="6" ViewCount="462" Body="&lt;p&gt;You have landed on an island of fun-loving logicians (hmmm ...) and don't know how to find your way home.  After asking the first person you meet for directions, you are led to a secret, mystical place with a large stone engraved with the following drawing replete with directions:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pCLr6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pCLr6.png&quot; alt=&quot;The Compass&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;I want to go south,&amp;quot; you explain. &amp;quot;Is this drawing correct?&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;Judge for yourself,&amp;quot; answers the native.  &amp;quot;I can only tell you that one of the arrows points south, but I cannot tell you which one.  I cannot tell you how many arrows point in the right direction either, or you would know which way to go.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;Show that you are bright enough to regain your bearings (whilst keeping your marbles) by working out which arrow points south.&lt;/p&gt;&#xA;&lt;p&gt;[First fully explained answer carries the day.]&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;The &amp;quot;Pet Shop Boys&amp;quot; could help you too, if only you knew how.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14080" LastEditorUserId="5373" LastEditDate="2021-07-05T17:44:42.917" LastActivityDate="2021-07-05T17:44:42.917" Title="Regain your bearings from the logicians" Tags="|logical-deduction|visual|meta-knowledge|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="20083" PostTypeId="2" ParentId="20082" CreationDate="2015-08-14T23:19:06.813" Score="8" Body="&lt;p&gt;South is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; to the left.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt; the key observation is the statement &quot;I cannot tell you how many arrows point in the right direction either, or you would know which way to go.&quot; This tells us that the correct configuration, which say has $n$ arrows pointing in the right direction, is the &lt;em&gt;only&lt;/em&gt; one with $n$ arrows pointing in the right direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &lt;strong&gt;N&lt;/strong&gt; is in the right direction, then none of the other arrows are.&lt;br&gt;&#xA;If &lt;strong&gt;S&lt;/strong&gt; is in the right direction, then none of the other arrows are.&lt;br&gt;&#xA;If &lt;strong&gt;W&lt;/strong&gt; is in the right direction, then none of the other arrows are.&lt;br&gt;&#xA;If any of &lt;strong&gt;E, NE, SW&lt;/strong&gt; is in the right direction, then so are the other two (and no others).&lt;br&gt;&#xA;If &lt;strong&gt;NW&lt;/strong&gt; is in the right direction, then none of the other arrows are.&lt;br&gt;&#xA;If &lt;strong&gt;SE&lt;/strong&gt; is in the right direction, then none of the other arrows are.&lt;br&gt;&#xA;If &lt;em&gt;none&lt;/em&gt; of the arrows are in the right direction, then south could be in either of the directions labelled N or NE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the native telling us that 0 or 1 arrows point in the right direction wouldn't enable us to know which way to go; any other number except 3 is impossible; and if 3 arrows point in the right direction, then they're E, NE, and SW and south is to the left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;QED.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-08-14T23:27:27.897" LastActivityDate="2015-08-14T23:27:27.897" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20084" PostTypeId="2" ParentId="20082" CreationDate="2015-08-14T23:21:03.263" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;p&gt;If NE is correct then E and SW are also correct - you need to go W&lt;/p&gt;&#xA; &lt;p&gt;If NW correct then go SW&lt;/p&gt;&#xA; &lt;p&gt;If S then go S&lt;/p&gt;&#xA;&lt;p&gt;If SE then go NW&lt;/p&gt;&#xA;&lt;p&gt;If N then go SE&lt;/p&gt;&#xA;&lt;p&gt;If W then go E&lt;/p&gt;&#xA;&lt;p&gt;The only unique number of correct arrows is the first option, so you need to go west (as in the title of the song by the Pet Shop Boys)&lt;/p&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-08-14T23:21:03.263" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20085" PostTypeId="2" ParentId="20082" CreationDate="2015-08-14T23:23:44.997" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You need to go West. Of all the orientations of this compass, all but one has either none or one direction pointing to the right direction. The one where west is pointing down has 3 pointing in the right direction (SW, NE and E)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13928" LastActivityDate="2015-08-14T23:23:44.997" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20086" PostTypeId="1" AcceptedAnswerId="20092" CreationDate="2015-08-14T23:44:21.687" Score="6" ViewCount="359" Body="&lt;p&gt;This is my first rebus. The titre is a crue.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/e90iC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/e90iC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dSRYy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dSRYy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Qxtti.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Qxtti.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/b7bPy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/b7bPy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&lt;strong&gt;Hint&lt;/strong&gt;&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Note the &lt;em&gt;knowledge&lt;/em&gt; tag. You may have to use your internet savvy to solve this. Depending on what you know of course!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="14177" LastEditDate="2015-08-15T00:12:40.430" LastActivityDate="2015-08-15T12:12:17.140" Title="Rebus arone! (We're frying here)" Tags="|rebus|knowledge|" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20087" PostTypeId="1" AcceptedAnswerId="20102" CreationDate="2015-08-15T01:12:46.363" Score="0" ViewCount="290" Body="&lt;p&gt;Two teams relay race: members of a team of three take turns running from the point P to a points on the circle; A for the first, B for the second, and C for the third, starting and returning at point P, transferring the baton at point P.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The points A, B, and C are set by the organizers of the competition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that all runners have the same rate, each team selects the point P for the other team. Your task to maximize the length of the run for the competitors, by selecting the best point P to improve your probability to beat the competition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VcRWi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VcRWi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic mathematical problem can be described as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given three points $A$, $B$, and $C$ and the circle that goes through them, find the point $P$ on the circle that maximizes $PA + PB + PC$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9531" LastEditorDisplayName="user88" LastEditDate="2015-08-15T15:12:48.907" LastActivityDate="2015-08-16T01:06:08.937" Title="Win the relay competition by selecting the longer path for competitors" Tags="|mathematics|geometry|" AnswerCount="3" CommentCount="18" FavoriteCount="0" ClosedDate="2015-08-18T06:46:56.087" ContentLicense="CC BY-SA 3.0" />
  <row Id="20088" PostTypeId="1" AcceptedAnswerId="20090" CreationDate="2015-08-15T02:40:38.247" Score="2" ViewCount="279" Body="&lt;p&gt;I have a weekly roster, with 6 tasks and 2 worker on each task (ie 12 workers) I have  a sample spreadsheet but i am not sure how to upload it here&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want the tasks to be set in such a way that each week every worker works with a different worker than previously worked with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tasks remain static in the roster location, but each week the workers move down one task&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the time they rotate through the entire roster (12 weeks) they will have worked with every other worker once. (they will always need to work with one worker twice because they cant work with themselves.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have achieved this with my old 10 worker roster (5 tasks) but am not able to work it out for the 12 worker (6 tasks) roster &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(in the 10 worker roster the tacks are in this order D, B, C, A, E, A1, B1, E1, C1, D1.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tasks can be in any order however preferably not consecutive ie Task A is not followed by Task A1 etc, as this will mean a worker works the task two weeks running&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is anyone able to assist? &lt;/p&gt;&#xA;" OwnerUserId="15726" LastEditorUserId="4421" LastEditDate="2015-08-15T04:16:22.840" LastActivityDate="2015-08-15T04:16:22.840" Title="Scheduling workers so every worker works with every other" Tags="|logical-deduction|number-sequence|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20089" PostTypeId="2" ParentId="20006" CreationDate="2015-08-15T02:42:56.553" Score="3" Body="&lt;p&gt;I think it's just&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Keys&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As f&quot; solved for Saturn&lt;br&gt;&#xA;and Cain reasoned Stairs&lt;br&gt;&#xA;While one word you seek&lt;br&gt;&#xA;Three this riddle bears:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As the first hint suggests, each following stanza is a riddle pointing at the solution -- but describing a different word of the homonym-group.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Carved from a forest we lie&lt;br&gt;&#xA;A road built of bone we supply&lt;br&gt;&#xA;With great walls of black&lt;br&gt;&#xA;We are known to attack&lt;br&gt;&#xA;From the forest an echo replies  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Piano keys: Once made of ivory (technically not bone, hence the biologist comment), the white keys form a &quot;road&quot; of sorts delineated by black keys that protrude from the keyboard like walls. When struck, they &quot;attack&quot; a &quot;forest&quot; of wires, which &quot;reply&quot; with a tone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For a singular purpose we're made&lt;br&gt;&#xA;To deny those of whom you're afraid&lt;br&gt;&#xA;And provide peace of mind&lt;br&gt;&#xA;Each day we're combined&lt;br&gt;&#xA;With our partner who therein has stayed  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Keys, as in the things that go in your pocket and unlock things. Similar to Kingrames's interpretation, but I thought of the &quot;partner&quot; as the lock, although there could be other possibilities. There is some ambiguity about &quot;therein&quot; as I at first thought it referred to &quot;mind&quot;, then &quot;day&quot;, but it doesn't actually seem to refer to anything, so I'm recklessly assuming it was used for cadence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At prominence ere the fall of Rome&lt;br&gt;&#xA;We held the theaters, bridges and domes&lt;br&gt;&#xA;In structures erected&lt;br&gt;&#xA;We shan't be neglected&lt;br&gt;&#xA;Lest one desires to topple our home  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Key stones, as Kingrames suggested.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15727" LastActivityDate="2015-08-15T02:42:56.553" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20090" PostTypeId="2" ParentId="20088" CreationDate="2015-08-15T03:18:47.807" Score="5" Body="&lt;p&gt;This is impossible for 12 workers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, consider the 10-worker example. Write the task list out twice: &lt;code&gt;DBCAEABECDDBCAEABECD&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the distance between pairs of the same letter. The requirement that every worker works with every other one is equivalent to the statement that every distance from 1 to 9 is represented. D has distances of 1 and 9, A has distances of 2 and 8, E has distances of 3 and 7, C has distances of 4 and 6, and B has a distance of 5. Note that the distances for each letter sum to 10 (5+5=10 for B), because when we wrote the list twice we put identical tasks 10 spaces apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we go back to our single list, this means that we need one distance from each set &lt;code&gt;{1,9}{2,8}{3,7}{4,6}{5}&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we need to do the same thing for 12 workers. We need one distance from each set &lt;code&gt;{1,11}{2,10}{3,9}{4,8}{5,7}{6}&lt;/code&gt;. Because three of these sets contain only odd numbers and three contain only even numbers, the sum of the distances must be odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if we divide the twelve positions into pairs and sum the distances, the result is always &lt;em&gt;even&lt;/em&gt;, as this diagram shows:&lt;img src=&quot;https://i.stack.imgur.com/IxVYZ.png&quot; alt=&quot;1&quot;&gt;&lt;br&gt;&#xA;Every green segment is counted an odd number of times, and every yellow segment is counted an even number of times. Because there are 6 green segments, the total is even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the sum of the distances cannot be odd, it is impossible for every worker to work with every other worker. The best you can do is for each worker to work with every worker except one, and work with two different workers twice. The diagram shows one way to do this: &lt;code&gt;ABCDBECFDFBA&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same method shows that it is impossible for any number of workers 8n+4 or 8n+6 where n is an integer.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-08-15T03:47:01.033" LastActivityDate="2015-08-15T03:47:01.033" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20091" PostTypeId="2" ParentId="20057" CreationDate="2015-08-15T05:06:36.383" Score="2" Body="&lt;p&gt;For the easiest case (Oldest to Youngest is A - F)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Pirate A will propose giving herself the gold, which will succeed with probability 4/5 from pirate A's perspective.&lt;br&gt;&#xA;&lt;br&gt;&#xA;  An important point, which we will establish later, is that C is certain to survive or fevered.  If C is fevered, it might be possible in some hypothetical situation that C dies, but in that case it will be certain that D-F survive.  So C-F all feel safe and are just after the most gold and most blood, or are fevered.&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;  Pirate A can only appease one of C-F with gold, but if she picks the fevered pirate, she can't generate the 3.5 required votes, and thus only succeeds with probability at most 4/5 and no gold.   So Pirate A proposes &quot;Give the gold to me&quot;, gets pirate B's vote, and hopes to get the fevered pirate amoung C-F (probability 4/5).&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Pirate B votes for A's proposal since he is pretty much shark bait otherwise.  If B voted against A's proposal, then by how the vote turned out we'd know who the fevered pirate was.   If B proposes &quot;Give the gold to me&quot;, since C-F all feel safe, at least three votes will be for B's blood, and this is a sure loss.  Similarly, B can only appease one pirate with gold, and that is not enough to swing the votes.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Pirate C is safe if not fevered, because at that point she knows who the fevered pirate is / was.   If one of D-F is fevered, Pirate C can succeed with &quot;Give the gold to me&quot;.  Otherwise, C can offer D or E a gold and at least survive.  At that point, D or E will take the gold, since D's best plan would be to offer the gold to F.  If C is fevered, then she'll walk the plank but D will survive with the offer of a gold to F.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2722" LastActivityDate="2015-08-15T05:06:36.383" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20092" PostTypeId="2" ParentId="20086" CreationDate="2015-08-15T05:11:03.753" Score="6" Body="&lt;p&gt;How about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Roger Wilco and out&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In order, the clues are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Roger Rabbit.&lt;br&gt;&#xA; The band Wilco.&lt;br&gt;&#xA; Ampersand means &quot;and&quot;.&lt;br&gt;&#xA; Someone being thrown out&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The clue in the title is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;We're f***l***ying here&quot; because &quot; roger wilco and out&quot; is pilot jargon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastActivityDate="2015-08-15T05:11:03.753" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20093" PostTypeId="2" ParentId="19884" CreationDate="2015-08-15T08:06:36.557" Score="4" Body="&lt;p&gt;It's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CONGRATULATIONS.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wherever there's a mark on a lattice point (either dot or corner/intersection/line), it's interpreted as a 1; where there isn't, it's interpreted as a 0. Read down the columns to get 5-bit binary, then convert to letters (A is 1, B is 2) to get the answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2015-08-15T08:06:36.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20094" PostTypeId="2" ParentId="20086" CreationDate="2015-08-15T10:19:47.133" Score="1" Body="&lt;p&gt;I've accepted ET's answer. If anyone is interested in my thought processes when creating the puzzle, here you are.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Answer&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Roger, Wilco and Out.  (note all initial caps - as per standard notation)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The  eccentric spelling comes from the first character's speech pattern on the site where I got the picture. A slight wabbity wisp fwom Woger.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Rebus arone!&quot; - When converted by substituting l for r, we get something that sounds like &quot;&quot;Leave us alone&quot;.  The phrase  &lt;em&gt;Roger, Wilco and Out&lt;/em&gt;, does mean that in a sense. It translates to &quot;Okay, Will do, and Leave us alone, we're fine now.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We're flying here! - This a clue to the fact it is flying jargon when speaking to ATC or another pilot.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The hint about internet savvy.  I'm sure most people on a puzzling site know this already, however, for anyone who hasn't discovered it yet, you can search online using images. Thus, if you don't recognise a picture, you can often find it online. In the case of Wilco, for example, you would have found both &lt;a href=&quot;https://www.google.com/search?tbs=sbi: &amp;amp;=Search%20by%20image&quot; rel=&quot;nofollow&quot;&gt;the picture and the name of the band&lt;/a&gt;.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="14177" LastEditDate="2015-08-15T12:12:17.140" LastActivityDate="2015-08-15T12:12:17.140" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20095" PostTypeId="2" ParentId="19986" CreationDate="2015-08-15T10:33:17.303" Score="9" Body="&lt;p&gt;It's to do with&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dismissal in cricket&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Each back-and-forth is a reference to a particular &quot;Law&quot;/rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;Hopkins: What name was given to the period of severe dust storms across the US and Canada in 1930?&lt;br&gt;&#xA;Ada: The Dust Bowl.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Law 30: Bowled (dust bowl, 1930)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;Hopkins: Who was the third president of Finland, and when was he elected?&lt;br&gt;&#xA;Ada: Um...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Law 31: Timed out (time's up, 1931)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;Hopkins: Time's up! In what year was the film Caught, directed by Max Ophüls, released?&lt;br&gt;&#xA;Ada: 1932?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Law 32: Caught (Caught, 1932)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;Hopkins: In 1833, a law was passed in Russia banning what?&lt;br&gt;&#xA;Ada: Masturbation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Law 33: Handled the ball (masturbation, 1833)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;Hopkins: What is the name given to the parties hit by Cambridge University students at the end of the academic year ever since 1834?&lt;br&gt;&#xA;Ada: They hit the May Ball...? Yes, May Balls.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Law 34: Hit the ball twice (hit ball, hit ball, 1834)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;Hopkins: Which musical have showgoers been hitting continuously since 1935 in the West End of London?&lt;br&gt;&#xA;Ada: Wicked.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Law 35: Hit wicket (Wicked, 1935)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;Hopkins: What were you doing three years before the start of World War Two?&lt;br&gt;&#xA;Ada: I wasn't born that long before wartime!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Law 36: Leg before wicket/LBW (long before wartime, 3 years before WW2)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;Hopkins: What device is used for blocking an electrostatic field, and when was it invented?&lt;br&gt;&#xA;Ada: A Faraday cage. It was invented by Michael Faraday in 1837.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Law 37: Obstructing the field (blocking a field, 1837)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;Hopkins: Do you consider yourself to be attractive for a 38-year-old woman?&lt;br&gt;&#xA;Ada: If you keep asking me personal questions, I'm going to get up and run out of the room!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Law 38: Run out (run out, 38)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;Hopkins: I, being a fan of 2048, was wondering if anyone has any proven strategies for playing 2048?&lt;br&gt;&#xA;Ada: Now you've got me completely stumped. I don't know.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Law 39: Stumped (stumped, 39th question on PSE)... 2048 (Law 48 is a red herring)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;Hopkins: Congratulations! You have answered every question correctly except for one. Please go to the white room to await the others' results. Soon you will know whether you have successfully beaten the competition to become the final champion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;Aaaaaand for the title/names:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ada is the second half of cic&lt;strong&gt;ada&lt;/strong&gt;, and Hopkins is a bit like grasshopper - both relatives of the cricket. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;BONUS QUESTION!&lt;/strong&gt;&lt;br&gt;&#xA;The incorrect answer was 'Um...'. She left Hopkins to give the reference to 'time out' instead of doing so herself.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is totally brilliant but very hard, rand. Have you read &quot;Goedel, Escher, Bach&quot;?&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="2484" LastEditDate="2015-08-21T13:36:47.070" LastActivityDate="2015-08-21T13:36:47.070" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20096" PostTypeId="2" ParentId="20087" CreationDate="2015-08-15T10:59:55.387" Score="0" Body="&lt;p&gt;&lt;strong&gt;This answer has the assumption that runners have to move along the circle. Although not clearly stated in the puzzle I think, although the image in the puzzle has straight lines&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I must say that this answer is mostly intuition so I am not sure it is proof enough and I even don't know if it is the correct answer but I have a good feeling it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;WLOG assume A and B are the points that are closest together among the points ABC. Now take point A* and B* that are directly opposite of A and B on the circle. What you now notice is that if you pick P anywhere between A* and B* the walk from P to A plus the walk from P to B will be the same no matter where you pick it between these points, and notice that if you pick P outside these points the walk will become smaller. So you'll want to pick it between A* and B*. Now the third point C is left. Now we want the point that is farthest from C that is still between A* and B*. Logically this means that this point is either exactly on A* or on B*. So that is this is my answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the two point that are closest together pick the opposite point on the circle that is farthest from the third point.&lt;/p&gt;&#xA;" OwnerUserId="1748" LastEditorUserId="1748" LastEditDate="2015-08-15T13:00:56.987" LastActivityDate="2015-08-15T13:00:56.987" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="20097" PostTypeId="2" ParentId="20087" CreationDate="2015-08-15T14:47:08.960" Score="0" Body="&lt;p&gt;I have an okay answer, but I'm struggling with the proof, so I'll post it anyways to see if anyone has thoughts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ideal location for P to maximize distance is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Diametrically opposite B&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My thought process:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;We are given no hard information about where B is, but let's just say it's between A and C on the lower half of the circle. (This is with my assumption that A and C are opposite one another). I checked the simple cases first- if B is at the very bottom, the ideal P is at the top, and if B is at A or C, then the ideal P is at the other point. So it looks as if the best solution for any B is for P to be directly opposite it. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;-&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Now, I was thinking about how to prove this geometrically, and I tried doing some stuff with Ptolemy's theorem or the fact that if P and B are opposite, then PCBA forms a rectangle, but I didn't really get anywhere. And then of course I'd want to generalize for any A,B,C... But calculus is so much less fun...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14441" LastActivityDate="2015-08-15T14:47:08.960" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20098" PostTypeId="2" ParentId="20082" CreationDate="2015-08-15T15:13:17.520" Score="1" Body="&lt;p&gt;Let's replace the symbols with numbers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N = 1&lt;br&gt;&#xA;NE = 2&lt;br&gt;&#xA;E = 3&lt;br&gt;&#xA;SE = 4&lt;br&gt;&#xA;S = 5&lt;br&gt;&#xA;SW = 6&lt;br&gt;&#xA;W = 7&lt;br&gt;&#xA;NW = 8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the direction difference between two arrows is equal to the number difference, either both of them are correct or not. If we write the number difference-1 between the adjacent arrows, we can easily see if it's the case when we add up consecutive few such numbers to check whether the sum equals 0 mod 8. As we can see in this picture, only SW, NE and E (6, 2, 3) allow each other to be correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IrBAt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IrBAt.png&quot; alt=&quot;pic&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one of the remaining 5 arrows is correct, then no other arrow is, so we can't find out which one points to the south even if we're told only 1 arrow is correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If none of them is correct, we still have 2 possibilities left, so we can't either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the aforementioned trio must be right, so we must follow W (not the real west of course!).&lt;/p&gt;&#xA;" OwnerUserId="14016" LastActivityDate="2015-08-15T15:13:17.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20099" PostTypeId="1" AcceptedAnswerId="20136" CreationDate="2015-08-15T15:35:53.650" Score="20" ViewCount="2524" Body="&lt;p&gt;&lt;sub&gt;After reading a bit in Meta and the help section, I'm still not sure whether this qualifies as a &quot;good&quot; puzzle in the spirit of the site, but the votes will show - I liked it when I encountered it, and maybe others will enjoy it as well.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The goal of this puzzle is to find a set of 17 numbers between 0 and 100, so that every number between 0 and 100 can be written as the sum of two of these numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The limits of 0 and 100 are both &lt;em&gt;inclusive&lt;/em&gt;, so there are 101 sums to be constructed. The sums may involve the same number twice. So obviously, &lt;strong&gt;0&lt;/strong&gt; must be part of the solution, as &lt;strong&gt;0 = 0 + 0&lt;/strong&gt; is the only way of obtaining 0 as one sum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, a solution might look like&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;{ 0, 1, 3, ..., 49, 51 }&#xA;&#xA;  0 = 0 + 0  &#xA;  1 = 1 + 0&#xA;  2 = 1 + 1&#xA;  3 = 3 + 0&#xA;  4 = 3 + 1&#xA;  5 = ...&#xA;      ...  &#xA; 99 = ...&#xA;100 = 49 + 51&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A warning:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will easily find solutions with 18 numbers, or 17 numbers that give 95 of the required sums. But finding a valid solution is really, really hard. I have to admit that I did not manage to &lt;em&gt;manually&lt;/em&gt; find such a solution. Instead, I wrote a program to solve it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore &lt;strong&gt;the use of computers, programs or tools of any kind is explicitly encouraged&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In hindsight, might also have been possible to derive a solution manually. But it is difficult, regardless of whether someone considers it as a brainteaser or a programming challenge.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Update with Hints 0 and 1:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 0&lt;/strong&gt;: As already mentioned above, finding solutions that are &quot;nearly complete&quot; is far easier than finding an &lt;em&gt;actual&lt;/em&gt; solution. This does not necessarily mean that it is trivial to &lt;em&gt;manually&lt;/em&gt; find a solution, e.g. one that gives 95 sums or uses 18 numbers. It just means that there are &lt;em&gt;far&lt;/em&gt; more such solutions. As per request in the comments (and as already posted in the chat), one solution with 18 numbers is&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[0, 1, 2, 3, 4, 5, 10, 16, 22, 28, 34, 40, 41, 45, 48, 49, 50, 82]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But one should not expect a 17-number-solution to be &quot;similar&quot; to that in any way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1&lt;/strong&gt;: One could imagine different approaches for solving this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Starting with a solution that gives the desired 101 sums, but uses more than 17 numbers, and trying to reduce the number of numbers step by step&lt;/li&gt;&#xA;&lt;li&gt;Starting with a solution that uses 17 numbers, but does not give all 101 sums, and modifying the numbers step by step&lt;/li&gt;&#xA;&lt;li&gt;A hybrid method of building the list of numbers and checking the resulting sums in parallel, adding and removing numbers as necessary (as some sort of &quot;backtracking&quot;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It's hard to say which approach is more promising when trying to solve it &lt;em&gt;manually&lt;/em&gt;. The hybrid one seems to be the most &quot;controllable&quot; and easiest to wrap the head around when trying to solve it manually (but as I mentioned: I did not succeed with this when I tried it). For the program that I eventually wrote, I used the second approach: Starting with 17 &quot;reasonable&quot; numbers, and modifying them, step by step, trying to fill the gaps in the list of sums. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is another &lt;strong&gt;Hint 2&lt;/strong&gt; that refers to one particular solution, but I think that it will make it much easier to find this solution, so I'll wait with this one until tomorrow.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Another remark: There are problems that look somewhat similar to this one. For example, the problem of finding a &lt;a href=&quot;https://en.wikipedia.org/wiki/Sparse_ruler&quot;&gt;Sparse Ruler&lt;/a&gt;. Maybe someone finds this interesting as an entry point, although it will not necessarily help to solve this puzzle)&lt;/p&gt;&#xA;" OwnerUserId="15731" LastEditorUserId="15731" LastEditDate="2015-08-16T18:26:12.840" LastActivityDate="2015-10-26T11:27:30.530" Title="Puzzling 101: Sums of seventeen numbers" Tags="|mathematics|combinatorics|computer-puzzle|formation-of-numbers|" AnswerCount="3" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="20100" PostTypeId="1" AcceptedAnswerId="20107" CreationDate="2015-08-15T15:53:50.357" Score="6" ViewCount="568" Body="&lt;p&gt;In &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15529/is-this-tetris-puzzle-solvable&quot;&gt;Is this Tetris puzzle solvable?&lt;/a&gt; we established that it is not possible to form a rectangle with an uneven number of each Tetris piece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if we had a solution for 2 of each piece, we would now that it works with any even number of each piece, because we could just stack the 2 of each solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is it possible to form a rectangle with 2 of each Tetris piece and if yes what would your solution look like?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ASOZ1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ASOZ1.png&quot; alt=&quot;Tetris pieces&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I have previously answered this question in the wrong place at &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15529/is-this-tetris-puzzle-solvable&quot;&gt;Is this Tetris puzzle solvable?&lt;/a&gt; so if you can see deleted answers, please don't answer and let the other users have some fun with this!&lt;/p&gt;&#xA;" OwnerUserId="3982" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-08-16T01:48:46.073" Title="2 of each Tetris Puzzle" Tags="|geometry|tiling|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20102" PostTypeId="2" ParentId="20087" CreationDate="2015-08-15T17:25:22.213" Score="4" Body="&lt;p&gt;Suppose that AC is larger than AB or BC. If P is on the arc AB, then we can construct P' where PP' is parallel to BC. Then P'B=PC, P'C=PB, and P'A&gt;PA:&lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/6kdQG.png&quot; alt=&quot;1]&quot;&gt;&lt;br&gt;Similarly, if P is on arc BC, we can also find a better point. So to maximize PA+PB+PC, P must be on arc AC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zQ7lw.png&quot; alt=&quot;2]&quot;&gt;&lt;br&gt;Let the diameter of the circle be 1. PA=$\sin(θ)$, PB=$\sin(θ+γ)$, and PC=$\sin(β-θ)$. We need to find the maximum of PA+PB+PC, which is either when the derivative is zero, or at the boundaries θ=0 or θ=β. The derivative at θ=0 is $1+\cos(γ)-\cos(β)$, which is positive. The derivative at θ=β is $\cos(β)-\cos(α)-1$, which is negative. Neither of these can be the maximum, so we set the derivative equal to zero:&lt;br&gt;&#xA;$\cos(θ)+\cos(θ+γ)-\cos(β-θ)=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the sum and difference formulas,&lt;br&gt;&#xA; $\cos(θ)+\cos(θ)\cos(γ)-\sin(θ)\sin(γ)-\cos(θ)\cos(β)-\sin(θ)\sin(β)=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can rearrange and factor to get&lt;br&gt;&#xA;$\cos(θ)(1+\cos(γ)-\cos(β))=\sin(θ)(\sin(γ)+\sin(β))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or equivalently:&lt;br&gt;&#xA;$\tan(θ)=\frac{1+\cos(γ)-\cos(β)}{\sin(γ)+\sin(β)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore P is maximized when $θ=\tan^{-1}(\frac{1+\cos(γ)-\cos(β)}{\sin(γ)+\sin(β)})$. I couldn't simplify this any further.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-08-16T01:06:08.937" LastActivityDate="2015-08-16T01:06:08.937" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="20103" PostTypeId="1" AcceptedAnswerId="20106" CreationDate="2015-08-15T18:15:38.707" Score="-6" ViewCount="203" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gDKjw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gDKjw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to place the objects so that no object touches any other or any purple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This obviously needs a computer program:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://scratch.mit.edu/projects/73109982/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Packer&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be done - it may be possible to do it more than one way.&lt;/p&gt;&#xA;" OwnerUserId="11497" LastActivityDate="2015-08-15T22:23:29.067" Title="Pack a box with beach stuff" Tags="|packing|" AnswerCount="2" CommentCount="0" ClosedDate="2015-08-16T00:56:26.887" ContentLicense="CC BY-SA 3.0" />
  <row Id="20104" PostTypeId="2" ParentId="20103" CreationDate="2015-08-15T19:16:51.730" Score="1" Body="&lt;p&gt;I believe that this is good enough:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UZgGp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UZgGp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9531" LastActivityDate="2015-08-15T19:16:51.730" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20106" PostTypeId="2" ParentId="20103" CreationDate="2015-08-15T22:23:29.067" Score="3" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EEGdU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EEGdU.png&quot; alt=&quot;why you hovering on me creep&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can I say I am a sucker for scratch. though I had great difficulty with registering drags on most objects before I went into inspect code mode then it functioned great, the ball and plane were the only ones I could drag&lt;/p&gt;&#xA;" OwnerUserId="13916" LastActivityDate="2015-08-15T22:23:29.067" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20107" PostTypeId="2" ParentId="20100" CreationDate="2015-08-16T01:48:46.073" Score="9" Body="&lt;p&gt;I'm not sure what size box you want them packed in, but based on the 7-wide in your other question, there is this option:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/c78s9.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/c78s9.jpg&quot; alt=&quot;7x8 Grid&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13132" LastActivityDate="2015-08-16T01:48:46.073" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20108" PostTypeId="2" ParentId="18532" CreationDate="2015-08-16T02:34:01.980" Score="4" Body="&lt;p&gt;The OP added the following sentence in this &lt;a href=&quot;https://puzzling.stackexchange.com/questions/19901/which-door-leads-to-freedom#comment62137_20046&quot;&gt;comment&lt;/a&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Here's a hint, because I'm getting bored of waiting:&lt;br&gt;&#xA;  &lt;strong&gt;ahhltomot eiXect fcu eutnsr ee lnfhqso SPUN THREE TIMES!&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a 3 column transposition cipher. After some manual adjustment, this decodes to:&lt;br&gt;&#xA;&lt;strong&gt;Search the left column of the question&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part 3, Hints 1&amp;amp;2&lt;/strong&gt; - This is also a 3 column transposition cipher. After some manual adjustment and translation from Leetspeak, this decodes to:&lt;br&gt;&#xA;&lt;strong&gt;Search the far &lt;strike&gt;right&lt;/strike&gt; left column of the page to find the password&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part 3&lt;/strong&gt; - Based on the hints: &quot;all up, not down&quot;, &quot;left column&quot;, &quot;(markdown) format read&quot;, &quot;except for lines that are &gt;Quoted or &gt;!Spoilers&quot;, &quot;eleven characters (not a word)&quot;, and &quot;forget the stars&quot;, &lt;strong&gt;the password is:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;pTCPAPmYTTT&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Final Answer&lt;/strong&gt; - By subtracting this password from the ciphertext (irrt sthcgma), the final answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;type seventh&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-09-07T05:49:01.370" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="20109" PostTypeId="1" AcceptedAnswerId="20110" CreationDate="2015-08-16T03:05:03.337" Score="9" ViewCount="522" Body="&lt;p&gt;If a word has a certain property, I call it a Scientific Word™.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In each of the short, unrelated sentences below, only one of the choices is a Scientific Word™.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sentences do not affect whether a word is a Scientific Word™. Case, font, and letter/word appearance do not matter.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;The child's mother gave him a &lt;strong&gt;(apple/banana)&lt;/strong&gt; as a healthy snack.&lt;/li&gt;&#xA;  &lt;li&gt;When queried with the question about a raise, Dave's boss said &lt;strong&gt;(no/yes)&lt;/strong&gt;.&lt;/li&gt;&#xA;  &lt;li&gt;Billy's breakfast consisted of &lt;strong&gt;(bacon/beans)&lt;/strong&gt;, eggs, and toast.&lt;/li&gt;&#xA;  &lt;li&gt;The snake's sharp fangs contained a deadly &lt;strong&gt;(poison/venom)&lt;/strong&gt; ready to be inserted into your bloodstream at the slightest movement.&lt;/li&gt;&#xA;  &lt;li&gt;Jerry's computer had gotten infected with (a) &lt;strong&gt;(virus/malware)&lt;/strong&gt;.&lt;/li&gt;&#xA;  &lt;li&gt;The soldier died out in &lt;strong&gt;(war/battle)&lt;/strong&gt;.&lt;/li&gt;&#xA;  &lt;li&gt;After Riley had died so many times on his video game, he was filled with &lt;strong&gt;(anger/rage)&lt;/strong&gt;.&lt;/li&gt;&#xA;  &lt;li&gt;The same series of events had happened years &lt;strong&gt;(ago/before)&lt;/strong&gt;, and the result was not good.&lt;/li&gt;&#xA;  &lt;li&gt;Joe has such a smart &lt;strong&gt;(brain/mind)&lt;/strong&gt;.&lt;/li&gt;&#xA;  &lt;li&gt;To be nice, Sarah offered her umbrella to an old &lt;strong&gt;(lady/woman)&lt;/strong&gt;.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What makes a word a Scientific Word™?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorDisplayName="user14478" LastEditDate="2016-09-23T21:32:04.273" LastActivityDate="2016-09-23T21:32:04.273" Title="What is a Scientific Word™" Tags="|pattern|english|word-property|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20110" PostTypeId="2" ParentId="20109" CreationDate="2015-08-16T03:37:14.197" Score="4" Body="&lt;p&gt;I think a Scientific Word™ is one that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; that scientists around the world can use without any ambiguity, simply because these words are composed entirely from the symbols of elements in the periodic table. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ba Na Na (Barium Sodium Sodium) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No (Nobelium)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ba Co N (Barium Cobalt Nitrogen) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Po I S O N (Polonium Iodine Sulphur Oxygen Nitrogen) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; V I Ru S (Vanadium Iodine Rubidium Sulphur)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; W Ar (Tungsten Argon) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;7.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ra Ge (Radium Germanium)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;8.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ag O (Silver Oxygen)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;9.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; B Ra In (Boron Radium Indium) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;10.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; La Dy (Lanthanum Dyprosium) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All other words do not follow this Scientific Rule (patent pending).&lt;/p&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="2383" LastEditDate="2015-08-16T12:33:48.373" LastActivityDate="2015-08-16T12:33:48.373" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20111" PostTypeId="1" AcceptedAnswerId="20112" CreationDate="2015-08-16T03:46:17.050" Score="2" ViewCount="271" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Aliens have come to Earth to see if there is intelligent life elsewhere in the galaxy (they weren't able to tell from our television broadcasts). Help them to come to the right conclusion by determining the missing figure.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VZLsU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VZLsU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is one &quot;anomaly&quot; in the progression, but this is not a mistake. &lt;/p&gt;&#xA;" OwnerUserId="14080" LastEditorUserId="30903" LastEditDate="2017-04-13T00:18:40.477" LastActivityDate="2017-04-13T00:18:40.477" Title="Set the aliens straight by determining the missing figure in the progression" Tags="|pattern|visual|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20112" PostTypeId="2" ParentId="20111" CreationDate="2015-08-16T04:02:00.873" Score="3" Body="&lt;p&gt;I'm going to use numpad notation; i.e. the top row is 789, middle row is 456, and bottom row is 123.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One of the Xs 'bounces' up and down: 4 5 6 5 4 5 6 5...&#xA; One of them goes in a circle (but not in the corners): 8 6 2 4 8 6 2 4...&#xA; And one of them skips 3 every time, going counterclockwise: 7 6 1 8 3 4 9 2...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are &lt;em&gt;many&lt;/em&gt; other equivalent ways to write this, but in any case, the final answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the same as the first frame, since each cycle repeats every 8 frames (or a factor of 8).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="12408" LastEditDate="2015-08-16T13:57:25.033" LastActivityDate="2015-08-16T13:57:25.033" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20113" PostTypeId="2" ParentId="19923" CreationDate="2015-08-16T08:47:51.197" Score="2" Body="&lt;p&gt;He lurks even now...  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Montezuma's Revenge&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12863" LastActivityDate="2015-08-16T08:47:51.197" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20114" PostTypeId="1" AcceptedAnswerId="20121" CreationDate="2015-08-16T09:11:19.763" Score="6" ViewCount="274" Body="&lt;p&gt;A O E A O T F L A T O A _&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R R A C O A O M R Y Y E _&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fill in the missing two letters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Notes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please explain why, in the context you have discovered, these are the only two possible answers and there is no room for doubt.&lt;/p&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="14177" LastEditDate="2015-08-16T17:15:08.827" LastActivityDate="2015-08-16T17:15:08.827" Title="Not much to this one, once you've understood it" Tags="|pattern|letter-sequence|letters|" AnswerCount="1" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20115" PostTypeId="1" CreationDate="2015-08-16T11:35:40.183" Score="4" ViewCount="873" Body="&lt;p&gt;This is a puzzle of a game where I know the solution, but not how it is derived. So the correct answer goes to whoever can also explain the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the correct code:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is written on top of a closed door:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A..D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this is a number on the closed door:&#xA;13&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the floor the following table can be found:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A | + | 2&#xA;--+---+---&#xA;- | 3 | -&#xA;--+---+---&#xA;8 | + | D&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You have to find the correct 4 digit code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since nobody found the solution after a week, I'm now adding a part of the correct answer as hint. If this doesn't help, I'll add the complete solution in another week. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The middle part of the code contains the 13: so A13D&lt;br&gt;&#xA; A and D are replaced by single digits. But which?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10890" LastEditorUserId="4421" LastEditDate="2015-11-25T18:02:24.063" LastActivityDate="2015-11-25T18:02:24.063" Title="Find the correct code: Clues A .. D, 13 to open the door" Tags="|pattern|" AnswerCount="3" CommentCount="13" ClosedDate="2015-08-25T09:33:23.597" ContentLicense="CC BY-SA 3.0" />
  <row Id="20116" PostTypeId="1" AcceptedAnswerId="20119" CreationDate="2015-08-16T12:09:52.763" Score="11" ViewCount="2400" Body="&lt;p&gt;You are given three identical bricks (cuboids) and a ruler (scale) as shown in the figure. You have to find the length of the brick's diagonal without using any formula and by using the ruler only once. How will you do it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(You can't break or change any physical properties of the material provided and you can't use your own material.&#xA;There is no unique answer to this puzzle, and various solutions may be found useful.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LVFmS.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LVFmS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13652" LastEditorUserId="25966" LastEditDate="2017-01-20T13:53:14.813" LastActivityDate="2019-12-31T13:12:43.637" Title="Measure the diagonal of a brick" Tags="|logical-deduction|geometry|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20117" PostTypeId="1" AcceptedAnswerId="20125" CreationDate="2015-08-16T12:12:08.110" Score="10" ViewCount="311" Body="&lt;p&gt;I like trying to invent new types of puzzles. I hope this one works.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Q7Q0X.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Q7Q0X.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fskJY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fskJY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;strong&gt;How to answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Using the graph, find the clues and say what they are.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Using the clues, name the star. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Explain how the diagram led you to the answer. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Use all the letters in the graph at least once (some of them twice or more often).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="8874" LastEditDate="2015-10-07T09:17:48.477" LastActivityDate="2015-10-07T09:17:48.477" Title="Use the graph to find the clues. Use the clues to name the movie star" Tags="|letters|" AnswerCount="1" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20118" PostTypeId="2" ParentId="20116" CreationDate="2015-08-16T12:51:46.597" Score="1" Body="&lt;p&gt;Make sure the first brick's longest two edges face upwards. Then put the other two on top of it so that they're congruent to each other, with the other edge also on the bottom and adjust them until the longest edges of both will form the diagonal with some of the length left over (hold/support the top bricks to stop them falling off if necessary). Finally measure the distance between the point where the leftover begins and the opposite corner on the other top brick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If holding/supporting them is illegal, you can put the two &quot;top bricks&quot; first, put the first one mentioned on top, adjust it and measure. Pic below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LFgMk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LFgMk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="14016" LastEditDate="2015-08-16T13:02:19.657" LastActivityDate="2015-08-16T13:02:19.657" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20119" PostTypeId="2" ParentId="20116" CreationDate="2015-08-16T16:19:47.900" Score="39" Body="&lt;p&gt;Align all the bricks in the same orientation and put them together in an L shape like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;brick&#xA;brickbrick&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then the body diagonal can be measured in the &quot;missing fourth brick&quot; directly, unobscured by material.&lt;/p&gt;&#xA;" OwnerUserId="15749" LastActivityDate="2015-08-16T16:19:47.900" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20120" PostTypeId="2" ParentId="20116" CreationDate="2015-08-16T16:39:03.963" Score="30" Body="&lt;p&gt;This can be measured directly and easily as shown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mUsTE.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mUsTE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14177" LastActivityDate="2015-08-16T16:39:03.963" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20121" PostTypeId="2" ParentId="20114" CreationDate="2015-08-16T17:10:22.267" Score="9" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; E&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; U&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the reason is ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each letter is the last letter of the NATO phonetic alphabet, in order. So we have alphA, bravO, charliE, deltA etc. The missing letters are for Mike and Zulu&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9168" LastActivityDate="2015-08-16T17:10:22.267" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20122" PostTypeId="1" AcceptedAnswerId="20123" CreationDate="2015-08-16T18:53:25.077" Score="0" ViewCount="183" Body="&lt;p&gt;What is the name of the airport?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When (to within 5 years) was the photo taken?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What else can you say about it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OqBZ0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OqBZ0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="14177" LastEditDate="2015-10-07T09:51:24.617" LastActivityDate="2015-10-07T09:51:24.617" Title="Name that airport!" Tags="|visual|" AnswerCount="1" CommentCount="7" ClosedDate="2015-08-17T04:08:16.990" ContentLicense="CC BY-SA 3.0" />
  <row Id="20123" PostTypeId="2" ParentId="20122" CreationDate="2015-08-16T18:58:33.640" Score="6" Body="&lt;p&gt;The airport is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Shoreham Airport, West Sussex.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The date is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 5 June 2014.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reason is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the airport was being used as a film set for the shooting of the film &lt;em&gt;Woman in Gold&lt;/em&gt; set during World War 2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you want proof, you can find a newspaper article featuring the picture &lt;a href=&quot;http://www.dailymail.co.uk/news/article-2649923/Outrage-actors-dressed-Nazis-takeover-Kent-airport-24-hours-70th-anniversary-D-day-film-scenes-Ryan-Reynolds-film.html&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-08-16T18:58:33.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20124" PostTypeId="2" ParentId="20115" CreationDate="2015-08-16T19:13:45.557" Score="3" Body="&lt;p&gt;So my theory is it works for reading across and up and down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So two equations with 2 unknowns:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A+2-3-8+D=13$ and  $A-8+3+2-D=13$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplfies to $A+D=22$ and $A-D=16$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which rather simply goes to $A=19, D=3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we are looking for a 4-digit code 1903 works using 19 and 03 as the numbers. This has extra appeal as (recent) year numbers are popular 4-digit passwords.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alternatively we are given A..D. So I propose A19D as the code as A..D has room for two more digits.&#xA;I am not convinced on the code part but I think the arithmetic part is valid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13916" LastEditorUserId="13916" LastEditDate="2015-08-16T22:34:27.120" LastActivityDate="2015-08-16T22:34:27.120" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="20125" PostTypeId="2" ParentId="20117" CreationDate="2015-08-16T19:21:07.570" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Matt Damon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How was this achieved:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Looking at the groups connnected by red we get &lt;strong&gt;JASNO&lt;/strong&gt;, which is an anagram for Jason.&lt;br&gt; Looking at the green we get &lt;strong&gt;BUNOER&lt;/strong&gt; which anagrams to Bourne. &lt;br&gt; Matt Damon famously played Jason Bourne. &lt;br&gt; Maroon appears to anagram to Operation. &lt;br&gt; And Blue anagrams to Treadstone &lt;strong&gt;(thanks to Gordon K)&lt;/strong&gt;. &lt;sub&gt; Don't anagram all loosey goosey kids &lt;/sub&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How the graph works:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Anagrams are formed from the colour groupings. The colors have a node which is the circle, it is unique to that colour. Then then lines out from that color's node indicate the number of time the letter pair at the end (that isn't the node) appear in the anagram.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13916" LastEditorUserId="13916" LastEditDate="2015-08-16T22:23:57.750" LastActivityDate="2015-08-16T22:23:57.750" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20126" PostTypeId="1" AcceptedAnswerId="20134" CreationDate="2015-08-16T20:57:32.390" Score="6" ViewCount="664" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The aliens are upset that several humans brushed off their last challenge so easily. They would like to rectify this with a visible demonstration that their grey matter is indeed superior. Help to thwart them by determining the next figure in the following sequence: &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6YUaO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6YUaO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You might find that it is easier to determine the pattern than to find the missing figure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might not help much, but:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sequence cannot be continued after the missing figure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you are battling to get started, here's an oblique hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Keep your friends close and your enemies closer.&quot;&lt;br/&gt;&#xA; -- Michael Corleone, in the Godfather II&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14080" LastEditorUserId="30903" LastEditDate="2017-04-16T20:24:23.547" LastActivityDate="2017-04-16T20:24:23.547" Title="Frustrate the aliens further by determining the next figure in the sequence" Tags="|logical-deduction|pattern|visual|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20127" PostTypeId="2" ParentId="20011" CreationDate="2015-08-16T22:42:55.420" Score="1" Body="&lt;p&gt;Well I had to use some creative googling based off of a single pun.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/qfhSg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qfhSg.jpg&quot; alt=&quot;Duckington&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Am I jesting, surely not. You sir are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Farmer Duck!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I work on a farm.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He works on a farm. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I need to wake up around sunrise to go to work.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He does all the work, that is how it works on farms. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And if you don't count room and board, I get just 4 pennies for every minute of work.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pennies is a term which can be used to describe chicken food, so it makes sense for a duck to be compensated with pennies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I had to take a second job to pay the bills.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  THE SWEET SWEET PUNNING. He has a bill. In order to feed his bill, he has to get a second job! (I have not read farmer duck but based on the attire from the picture looks like he has sellable skills)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Honestly, I'd do anything just to get an apartment.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He does not seemed pleased with his living situation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I'm such a sad sack it'll probably never happen.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He looks sad in his picture... :(&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is my name?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Farmer Duck!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alas,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I have not read the tale so I could not tell of its congruence with my statements but it has a pun, if only one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13916" LastActivityDate="2015-08-16T22:42:55.420" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20128" PostTypeId="1" AcceptedAnswerId="20137" CreationDate="2015-08-17T00:40:14.020" Score="3" ViewCount="488" Body="&lt;p&gt;A few candies were distributed to three people Alice, Bob, and Charlie. The sum of the reciprocals of the number of each person's candies was equal to 1. &#xA;(Alice, Bob, and Charlie knew their number of candies and they knew that the sum of the reciprocals was 1, but they did not know the number of candies the other two persons had received.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I asked Alice whether she knows the number of each person's candies and Alice said she does not know it. Also, I asked her if she knows the total number of candies and she said she does not know it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I asked the same questions to Bob and he said that he does not know the number of each person's candies but he knows the total number of candies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, Charlie answered to the same questions that he knows not only the total number of candies but also the number of each person's candies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the total number of candies?&lt;/p&gt;&#xA;" OwnerUserId="3914" LastEditorUserId="8874" LastEditDate="2015-10-07T09:05:32.827" LastActivityDate="2015-10-07T09:15:23.157" Title="Number of candies" Tags="|mathematics|logical-deduction|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20129" PostTypeId="2" ParentId="20128" CreationDate="2015-08-17T01:09:16.690" Score="5" Body="&lt;p&gt;There are only a few possibilities: You can have $2+4+4=10$, or $2+3+6=11$, or $3+3+3=9$. &#xA;Alice must have gotten $2$ or $3$, as otherwise she would at least know the total, and Bob knows this.  Bob must have gotten $6$, as otherwise he would know how many Alice got. Charlie knows what he and Alice got, but we do not.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="8874" LastEditDate="2015-10-07T09:15:23.157" LastActivityDate="2015-10-07T09:15:23.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20130" PostTypeId="1" AcceptedAnswerId="20132" CreationDate="2015-08-17T01:42:12.927" Score="-2" ViewCount="435" Body="&lt;p&gt;There are two babies, the truth baby and the false baby. &#xA;The truth baby will respond to any question with a true answer, and the false baby will respond to any question with a false answer. &#xA;The babies share a language of either burping or farting. &#xA;A burp or fart would mean yes and no, but you don't know which is which.&#xA;You also do not know which is the truth baby and which the false baby.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are to ask each baby a single yes-or-no question.&#xA;Your goal is to discover whether or not the truth baby has hair. &#xA;(You can't simply look at it to find out, you must only use knowledge that you gain from asking questions.)&lt;/p&gt;&#xA;" OwnerUserId="15756" LastEditorUserId="8874" LastEditDate="2015-10-07T09:21:11.570" LastActivityDate="2015-10-07T09:21:11.570" Title="The true and false babies" Tags="|logical-deduction|liars|" AnswerCount="1" CommentCount="0" ClosedDate="2015-08-17T03:24:47.813" ContentLicense="CC BY-SA 3.0" />
  <row Id="20131" PostTypeId="1" AcceptedAnswerId="20133" CreationDate="2015-08-17T01:44:43.850" Score="16" ViewCount="1167" Body="&lt;p&gt;I dropped in on Ernie last week to help him prepare for his annual hosting of the Wine and Cheese Club party. This was a special occasion because the invited guest was the famous oenophile and turophile Jean-Claude la-Grange. Ernie’s party had been planned to the last detail. “The components will be based on the appreciation of Antipodean contrasts”, Ernie had explained, “I am mating a New Zealand &lt;em&gt;Ata Rangi Pinot Noir&lt;/em&gt; (2014) with a Swiss &lt;em&gt;Rohmilch Emmentale&lt;/em&gt;r”. “Isn’t that wine a little young for tasting?” I asked.  But Ernie explained that he had invented a Wine Vintage Accelerator which, by a combination of heat, vibration, and other secret factors, could add 10 years of cellaring to a bottle in minutes – this was to be its first test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At that moment the door-bell rang. “Answer that can you? That will be the cheese shipment from my affineur“, said Ernie. But disaster was in wait! Instead of a refrigerated box of Swiss cheese I was handed a scrap of crushed packaging stamped “Damaged in Transit”.  Ernie was distraught. “There isn’t another cheese of that quality within miles!”, he cried. Ernie looked carefully through the wreckage of the delivery. With a pair of tweezers he extracted the barest crumb of cheese that had somehow remained lodged in the packaging. “Maybe all is not lost”, he muttered. “I can determine the exact composition with my &lt;em&gt;Acme Cheese Analyser&lt;/em&gt; and then recreate it using milk from my mechanical dairy cow”. While the analyser was at work Ernie sketched a rapid diagram onto his chalk-board, explaining that he could suitably age the cheese by adapting his Vintage Accelerator. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I rushed out to buy suitable feed-stock while Ernie assembled and augmented the components of his &lt;em&gt;Instant Dairy Factory&lt;/em&gt;. When I returned everything was ready to go. I fed in the turnips and hay and minutes later a block of cheese was deposited on a waiting plate. The cheese certainly looked, felt and smelled right and when I took a tentative bite the nutty taste was perfect. But something was not quite right. “Ernie, it doesn’t have any holes! Jean-Claude will not be fooled for a moment”, I cried. Ernie agreed and altered his diagram. “I can use my new &lt;em&gt;Acme 3-D Printer&lt;/em&gt;. But there is no time to scan a cheese to make a suitable model – I will have to hack directly into its analogue controls”. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SI8Gk.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SI8Gk.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ernie rapidly assembled a complicated network of capacitors, inductors, relays and electromechanical components to programme the printer. He explained how it would work:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 1) The system prints a perfect cube of cheese precisely 100 mm on each side (with interior and/or exterior holes).&#xA; 2) The thumb-wheel determines the exact number of holes in the block of cheese (all of identical diameter).&#xA; 3) The bridge resistor determines the size of the holes. Each 1.0 kΩ of resistance will ‘add’ precisely 1% holes by volume (in the range 0-100%) in the block.&#xA; 4) The cheese is printed with the smallest diameter holes consistent with the required number of holes and percentage volume of void space.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;“Where will the holes be positioned in the block?” I asked. &lt;strong&gt;“Well, depending on the number of holes and the volume to be removed by them they could overlap each other. They could also be completely internal to the cube of cheese or could overlap its edges, so long as their volume and diameters are consistent with rules 1-4”&lt;/strong&gt; Ernie replied after a moments thought. Ernie asked me to solder in a 10 kΩ resistor across the bridge so we could get a cheese with exactly 10% void volume. I hastily complied while he made the final attachments between the cheese machine and the 3D printer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When all was ready, Ernie set a 1 on the thumb-wheel and pressed the go button. The printer clicked and buzzed for a few minutes and spat out a short spiral of printed paper, but when the output hopper swung open it was empty. Ernie scratched his head, and tried again with 2 set on the thumb-wheel – the result was the same, as it was when he tried a third time with a 3 on the thumb-wheel. “Maybe there is a blockage in the print jet” he exclaimed. “While I’m looking for that, can you read aloud what is printed on the paper?” I tore off the sheet and glanced over it. It described the setting data for each of the three cheese-less 3D-print runs we had made. I began reading the first aloud – “Run 1:: Feedstock: cheese. Cubical block: side length = 100.000 mm. Spherical voids: number = 1; diameter = 173.205 mm …” &lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Stop right there. The problem is obvious! – some fool...” (Ernie was obviously more than a little stressed) “...has soldered in the wrong valued resistor”. He de-soldered my resistor, threw it into the rubbish bin and soldered in a new one – this time 10 kΩ. “…and I think 7 holes will be perfect” he said as he adjusted the thumb-wheel and hit the go button. A few seconds later a perfect cube of cheese rested in the output hopper, and when I cut it into wedges the beautiful spherical holes were revealed. And not a moment too soon - the door-bell rang, Ernie’s guests arrived and I spent the next hour opening bottles of Pinot and (secretly) printing more cheese to Ernie’s specifications. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end of the evening Jean-Claude commented that the wine was ‘perfectly’ matured and its quality was “…only exceeded by the most exquisite cheese I have ever tasted”, so the party was a success. I never did get to work out exactly what value of resistance I had mistakenly used and have no recollection of the diameters of the spherical voids printed out for the 2-hole and 3-hole cheese-less cheeses. Can you help me understand what they were (As usual, accuracy to the nearest Ohm and micron should be enough)? &lt;/p&gt;&#xA;" OwnerUserId="1631" LastActivityDate="2015-08-17T19:10:45.470" Title="Ernie and the Artificial Emmental" Tags="|mathematics|calculation-puzzle|geometry|word-problem|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20132" PostTypeId="2" ParentId="20130" CreationDate="2015-08-17T01:54:10.727" Score="1" Body="&lt;p&gt;A simple variation on the two guards problem.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Ask of one baby, &quot;Are you the truth-teller?&quot; Whichever response you get, you know that means yes, and therefore the other means no. Go to the remaining baby and ask, &quot;Would the other baby say that the truth-teller has hair?&quot; If the truth-teller does have hair, the truth-teller would say that the liar would lie and say no, and the liar would lie and say that the truth-teller would say no. It works the same way in reverse, so whichever answer you get for this, the reality is the opposite.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14441" LastActivityDate="2015-08-17T01:54:10.727" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20133" PostTypeId="2" ParentId="20131" CreationDate="2015-08-17T02:24:56.197" Score="10" Body="&lt;p&gt;173.205 mm is $\sqrt3$ times 100 mm, so a 100 mm cube fits exactly into a sphere with diameter 173.205 mm. The resistor must have been 100 kΩ, which explains why nothing was printed!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest of the question is to determine the smallest possible diameter of two identical spheres that cover a cube, and the smallest possible diameter of three identical spheres that cover a cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we want to cover a cube with two identical spheres that are smaller than 173.205 mm, there must be two vertices on the same edge that are covered by different spheres. Then the midpoint of the opposite edge is 150 mm away from both, so the spheres must have a diameter of at least 150 mm. This is achieved by cutting the cube in half parallel to one of its faces and circumscribing a sphere around each half.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For three spheres, we can divide the cube into a 100 mm by 100 mm by 12.5 mm block and two 100 mm by 50 mm by 87.5 mm blocks, and surround each with a sphere with a diameter of 141.973 mm ($\frac{\sqrt{129}}{8}*100$ mm). This is optimal:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that three spheres with a smaller diameter could cover the cube. By the Pigeonhole Principle, one must cover at least three of the eight vertices. This is only possible if covers three vertices of a face.&#xA;Define a coordinate system so that the three vertices it covers are (0,0,0), (1,0,0), and (0,1,0). The sphere cannot cover (1,0,1/8), (0,1,1/8), (1,1,1/8), or any point on the edges of the opposite face. In addition, it cannot cover (0,0,7/8) because the triangle with vertices (1,0,0) (0,1,0) (0,0,7/8) will not fit in a circle of diameter 141.973 mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whichever sphere covers (0,0,7/8) cannot cover (1,1,1/8) or (1,1,1), so (1,1,1/8) and (1,1,1) must be covered by the same sphere. (1/2,0,1) is 141.973 mm away from (0,1,1/8) and (1,1,1/8), so it cannot be in the same sphere as either of those. Therefore (0,1,1/8) and (1,1,1/8) are in the same sphere. Similarly, (1,0,1/8) and (1,1,1/8) must be in the same sphere. But then (1,0,1/8), (0,1,1/8), and (1,1,1) are all in the same sphere, which is impossible because the triangle with those three vertices does not fit in a circle with diameter 141.973 mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the spheres must have a diameter of at least 141.973 mm.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In summary: the resistor was 100 kΩ, the diameter for 2 holes was 150.000 mm, and the diameter for 3 holes was 141.973 mm.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-08-17T19:10:45.470" LastActivityDate="2015-08-17T19:10:45.470" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20134" PostTypeId="2" ParentId="20126" CreationDate="2015-08-17T02:53:52.037" Score="7" Body="&lt;p&gt;The next figure should be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/GRelN.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GRelN.png&quot; alt=&quot;Next step&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each &lt;code&gt;X&lt;/code&gt; square shifts once each step, following a zigzagged path from top left to bottom right (wrapping back up to the top left), as can be seen in this colourised image:&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/uAseV.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uAseV.png&quot; alt=&quot;with colours&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="5208" LastEditDate="2015-08-17T03:00:34.127" LastActivityDate="2015-08-17T03:00:34.127" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20135" PostTypeId="1" AcceptedAnswerId="20138" CreationDate="2015-08-17T05:28:11.200" Score="37" ViewCount="10570" Body="&lt;p&gt;Here's a simple puzzle,&lt;br /&gt;&#xA;I know that you will find.&lt;br /&gt;&#xA;As long as you can count,&lt;br /&gt;&#xA;It shouldn't take much time.&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1&lt;br /&gt;&#xA;1&lt;br /&gt;&#xA;2&lt;br /&gt;&#xA;1&lt;br /&gt;&#xA;2&lt;br /&gt;&#xA;2&lt;br /&gt;&#xA;3&lt;br /&gt;&#xA;1&lt;br /&gt;&#xA;2&lt;br /&gt;&#xA;2&lt;br /&gt;&#xA;3&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the next 4 numbers in this sequence.&lt;/p&gt;&#xA;" OwnerUserId="13068" LastEditorUserId="4385" LastEditDate="2015-10-02T20:15:50.923" LastActivityDate="2019-05-06T13:53:32.617" Title="A very simple puzzle: 1,1,2,1,2,2,3" Tags="|number-sequence|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20136" PostTypeId="2" ParentId="20099" CreationDate="2015-08-17T05:35:21.393" Score="10" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; [0, 1, 3, 4, 7, 8, 9, 16, 17, 21, 24, 35, 46, 57, 68, 79, 90]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Method:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are at most ${17 \choose 2} = 136$ combinations of 17 numbers, so if a candidate set generates more than 35 duplicates (more than one pair of numbers adding up to the same sum), that set cannot produce all 101 required numbers and so can be culled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I generated sets with increasing cardinality, starting from the known {0,1}, culling the partial sets along the way. For each remaining set at that round, all possible sets with one extra number were generated and then culled, etc. This relies on the observation that adding elements to a set does not decrease the number of duplicates it generates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of candidate sets reached a maximum at cardinality 13, with just 149 candidates not culled by the end. Of these 149 sets, only one was able to generate all 101 numbers.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; As it turns out, I missed counting the 17 doubled numbers. Increasing the number of allowed duplicates produces the following lists:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[0, 1, 2, 3, 7, 11, 15, 19, 23, 27, 28, 29, 30, 61, 64, 67, 70]&#xA;[0, 1, 3, 4, 5, 8, 14, 20, 26, 32, 38, 44, 45, 47, 48, 49, 52]&#xA;[0, 1, 3, 4, 5, 8, 14, 20, 26, 32, 38, 44, 46, 47, 48, 49, 51]&#xA;[0, 1, 3, 4, 5, 8, 14, 20, 26, 32, 38, 44, 46, 48, 49, 51, 53]&#xA;[0, 1, 3, 4, 5, 8, 14, 20, 26, 32, 38, 44, 47, 48, 49, 51, 52]&#xA;[0, 1, 3, 4, 7, 8, 9, 16, 17, 21, 24, 35, 46, 57, 68, 79, 90]&#xA;[0, 1, 3, 4, 9, 11, 16, 20, 25, 30, 34, 39, 41, 46, 47, 49, 50]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Pseudocode: start by calling puzzling101([0,1]).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Invariant: each candidate is an ordered list (strictly increasing) of distinct integers.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;puzzling101(c) =&#xA;if (#elements in c == 17) {&#xA;    display c if it generates all 0..100&#xA;} else {&#xA;    min = 1 + largest number in c&#xA;    max = 84 + #elements in c&#xA;    for each x in [min..max], call puzzling101([x|c]) if [x|c] generates&#xA;    1. all numbers 0..x (ignore larger x if not), and&#xA;    2. less than 52 duplicates between 0 and 100&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="9422" LastEditorUserId="9422" LastEditDate="2015-08-18T22:58:18.343" LastActivityDate="2015-08-18T22:58:18.343" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="20137" PostTypeId="2" ParentId="20128" CreationDate="2015-08-17T06:28:52.293" Score="5" Body="&lt;p&gt;There are only three different ways of distributing the candies according to the puzzle statement:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;        3, 3, 3,   &#xA;   or   2, 4, 4,    &#xA;   or   2, 3, 6 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Alice cannot have 4 or 6, and hence must have 2 or 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Bob had 2, he'd known that Alice has 3 and Charlie has 6.&lt;br&gt;&#xA;If Bob had 4, he would have deduced Charlie has 4 and Alice 2.&lt;br&gt;&#xA;If Bob had 3, he wouldn't be able to deduce the total number of candies.&lt;br&gt;&#xA;Hence Bob has 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Charlie can figure out everything by elimination.&lt;br&gt;&#xA;The total number of candies is 11.&lt;/p&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="8874" LastEditDate="2015-10-07T09:13:26.153" LastActivityDate="2015-10-07T09:13:26.153" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20138" PostTypeId="2" ParentId="20135" CreationDate="2015-08-17T06:37:18.070" Score="64" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Represent natural numbers in binary. Count the 1's in the binary representation.&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;  &lt;code&gt;Number   Binary  Number of 1's&lt;/code&gt;&#xA;&lt;br&gt;&#xA;  &lt;code&gt;1            1           1&lt;/code&gt;&#xA;&lt;br&gt;&#xA;  &lt;code&gt;2           10           1&lt;/code&gt;&#xA;&lt;br&gt;&#xA;  &lt;code&gt;3           11           2&lt;/code&gt;&#xA;&lt;br&gt;&#xA;  &lt;code&gt;4          100           1&lt;/code&gt;&#xA;&lt;br&gt;&#xA;  &lt;code&gt;5          101           2&lt;/code&gt;&#xA;&lt;br&gt;&#xA;  &lt;code&gt;6          110           2&lt;/code&gt;&#xA;&lt;br&gt;&#xA;  &lt;code&gt;7          111           3&lt;/code&gt;&#xA;&lt;br&gt;&#xA;  &lt;code&gt;8         1000           1&lt;/code&gt;&#xA;&lt;br&gt;&#xA;  &lt;code&gt;9         1001           2&lt;/code&gt;&#xA;&lt;br&gt;&#xA;  &lt;code&gt;10        1010           2&lt;/code&gt;&#xA;&lt;br&gt;&#xA;  &lt;code&gt;11        1011           3&lt;/code&gt;&#xA;&lt;br&gt;&#xA;  &lt;strong&gt;&lt;code&gt;12        1100           2&lt;/code&gt;&lt;/strong&gt;&#xA;&lt;br&gt;&#xA;  &lt;strong&gt;&lt;code&gt;13        1101           3&lt;/code&gt;&lt;/strong&gt;&#xA;&lt;br&gt;&#xA;  &lt;strong&gt;&lt;code&gt;14        1110           3&lt;/code&gt;&lt;/strong&gt;&#xA;&lt;br&gt;&#xA;  &lt;strong&gt;&lt;code&gt;15        1111           4&lt;/code&gt;&lt;/strong&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;  Next are &lt;strong&gt;&lt;code&gt;2, 3, 3, 4&lt;/code&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Inspiration&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Looked at where the 1's occur. They occur at indexes 1, 2, 4, 8. After that it was simple.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15544" LastEditorUserId="15544" LastEditDate="2015-08-21T03:51:00.940" LastActivityDate="2015-08-21T03:51:00.940" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20139" PostTypeId="2" ParentId="19806" CreationDate="2015-08-17T07:01:41.143" Score="2" Body="&lt;p&gt;&lt;strong&gt;Partial Answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What a certain literary professor for his unbeheld felt,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nabokov write Lolita, and the literary professor main character feels love/luv for his unbeheld. Or is it from the first couple of lines... &quot;sin/soul&quot;? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The two of spades, from a new deck dealt, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hearts starts off like this. The 5 letter word is heart. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A din, a doubt... &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I wonder how much these next two lines are connected to &lt;a href=&quot;https://en.wikipedia.org/wiki/Tomorrow_and_tomorrow_and_tomorrow&quot; rel=&quot;nofollow&quot;&gt;Tomorrow, tomorrow, tomorrow&lt;/a&gt;... A doubt = An idiot (the malapropism), and that can be interpreted as sound (and fury?) and or a tale told by an idiot. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Out!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Out, as in Out brief candle. I doubt both of them are related. Although, this makes me wonder if the entire riddle is a reference to that soliloquy, considering there are 12 lines in that and 6 here. Could they match up when in the correct order? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14428" LastEditorUserId="14428" LastEditDate="2015-08-17T08:55:02.097" LastActivityDate="2015-08-17T08:55:02.097" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20140" PostTypeId="1" AcceptedAnswerId="20229" CreationDate="2015-08-17T07:21:40.313" Score="9" ViewCount="590" Body="&lt;p&gt;You are a spy that wanted to find out the job of a man named John. There are 100 glasses in front of you. There is also a note:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Hello! 99 of the glasses in front of you are poisoned! One is safe, however. Just remember that you only get one try! Here's a hint, to make it easier:&lt;/p&gt;&#xA;&lt;p&gt;&lt;code&gt;VDFVXDVADXVAAVGAGXAVAVGGGGXAFGGAXGAGVVXFGAVDXGDX&lt;/code&gt;&lt;/p&gt;&#xA;&lt;p&gt;Enjoy your drink, if you can! &lt;br&gt;&lt;br&gt;- John&lt;br&gt;&lt;br&gt;&lt;strong&gt;P.S.&lt;/strong&gt; If you don't drink one, a huge elephant shall fall out of the sky and crush you to death. You have one week.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Which glass should you drink from?&lt;/p&gt;&#xA;" OwnerDisplayName="user9377" LastEditorUserId="-1" LastEditorDisplayName="user9377" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-08-18T19:33:43.310" Title="John's revenge!" Tags="|cipher|" AnswerCount="2" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="20141" PostTypeId="1" AcceptedAnswerId="20199" CreationDate="2015-08-17T07:37:42.380" Score="24" ViewCount="3030" Body="&lt;p&gt;This is a variation of a game I used to play with my friends in my childhood. You are given a starting word, and using the following operations, you have to reach the target word. The solution that requires the least number of moves wins.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Rules:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;In any move, you can either replace a letter, add a letter, remove a letter, split a word or combine two words. These are the five valid moves, and only one of these can be applied in a move.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;After each move, each of the resulting words must be valid by itself, even if it makes no sense as a phrase. &lt;strong&gt;To level the playing field, only words found on Dictionary.com are considered valid.&lt;/strong&gt; No proper nouns, abbreviations, prefixes or suffixes allowed. (However, words like jack, john, jane etc. are allowed as they are used as common nouns too.) Alternate spellings and archaic words are permitted, but words referring to specific persons, places or events are not valid.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You may only add a letter to an existing word, but a new letter (such as a or I) cannot create a new word separately. Likewise, you cannot drop a letter from a one-letter word to remove it altogether.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;As an example, consider the following solution where I go from &amp;quot;stack&amp;quot; to &amp;quot;exchange&amp;quot;:&lt;/p&gt;&#xA;&lt;p&gt;stack&lt;br /&gt;&#xA;stark&lt;br /&gt;&#xA;star&lt;br /&gt;&#xA;stay&lt;br /&gt;&#xA;stray&lt;br /&gt;&#xA;astray&lt;br /&gt;&#xA;a stray&lt;br /&gt;&#xA;as stray&lt;br /&gt;&#xA;ax stray&lt;br /&gt;&#xA;ex stray&lt;br /&gt;&#xA;ex stay&lt;br /&gt;&#xA;ex star&lt;br /&gt;&#xA;ex stare&lt;br /&gt;&#xA;ex share&lt;br /&gt;&#xA;ex chare&lt;br /&gt;&#xA;ex charge&lt;br /&gt;&#xA;ex change&lt;br /&gt;&#xA;exchange&lt;/p&gt;&#xA;&lt;p&gt;This solution required 17 moves. (It may not be the optimal strategy for this pair of words, but it is a valid strategy, given just as an example).&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Challenge:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;In how many moves can you get from &amp;quot;Puzzling&amp;quot; to &amp;quot;Stack Exchange&amp;quot;?&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;An Extra Twist:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The solution which achieves this in the least number of moves will be awarded the green tick. However, in the event of a tie, each of the words will be assigned their Scrabble score and then added up. The one with the highest Scrabble score will win.&lt;/p&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="69582" LastEditDate="2021-08-16T00:10:58.773" LastActivityDate="2021-08-16T00:10:58.773" Title="From Puzzling to StackExchange" Tags="|optimization|english|word-ladder|pencil-and-paper-games|" AnswerCount="7" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="20142" PostTypeId="1" AcceptedAnswerId="20143" CreationDate="2015-08-17T07:57:22.800" Score="16" ViewCount="2397" Body="&lt;p&gt;Is there a solution in distinct positive integers $a,b,c$ to the equation $$a^3+b^3=c^4$$? If so, construct one; if not, prove that it can't possibly exist.&#xA;&lt;hr&gt;&#xA;Don't be too put off by the appearance of this puzzle: it's nowhere near as hard as its famous relative Fermat's Last Theorem. There should be an AHA moment when you realise what you need, and the solution is very surprising if you haven't seen the like before.&#xA;&lt;hr&gt;&#xA;&lt;sub&gt;&lt;sub&gt;This puzzle was discussed by Adam McBride in “Mathematics: The Greatest Subject in the World,” &lt;em&gt;The Mathematical Gazette&lt;/em&gt;, vol. 89, no. 516 [November 2005].&lt;/sub&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2016-09-30T14:58:50.997" LastActivityDate="2016-09-30T14:58:50.997" Title="Fermat's Last Theorem - or is it?" Tags="|mathematics|number-theory|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20143" PostTypeId="2" ParentId="20142" CreationDate="2015-08-17T08:05:50.577" Score="17" Body="&lt;p&gt;&lt;strong&gt;Trivial Solution&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $2^3+2^3=2^4$&#xA;&lt;br&gt;&#xA;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More Generally-&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &#xA; Let $k=m^3+n^3\ \  where\ \  n&amp;gt;m&amp;gt;1$&#xA;&lt;br&gt;&#xA; Then, $(mk)^3+(nk)^3=(m^3+n^3)k^3= k^4$&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; For example, if $m=2, n=4$, we have $k=72$ and&#xA;&lt;br&gt;&#xA; $ 144^3 + 288^3 =72^4 $&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15544" LastEditorUserId="15544" LastEditDate="2015-08-17T10:47:29.637" LastActivityDate="2015-08-17T10:47:29.637" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20144" PostTypeId="2" ParentId="20141" CreationDate="2015-08-17T08:20:09.550" Score="19" Body="&lt;p&gt;I've done it in 21:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;p&gt;puzzling&lt;/p&gt;&lt;p&gt;muzzling&lt;/p&gt;&lt;p&gt;muzz ling&lt;/p&gt;&lt;p&gt;muzz line&lt;/p&gt;&lt;p&gt;muzz lane&lt;/p&gt;&lt;p&gt;muzz cane&lt;/p&gt;&lt;p&gt;muzz care&lt;/p&gt;&lt;p&gt;muzz chare&lt;/p&gt;&lt;p&gt;muzz charge&lt;/p&gt;&lt;p&gt;muzz change&lt;/p&gt;&lt;p&gt;fuzz change&lt;/p&gt;&lt;p&gt;fuze change&lt;/p&gt;&lt;p&gt;faze change&lt;/p&gt;&lt;p&gt;fare change&lt;/p&gt;&lt;p&gt;tare change&lt;/p&gt;&lt;p&gt;stare change&lt;/p&gt;&lt;p&gt;starer change&lt;/p&gt;&lt;p&gt;star er change&lt;/p&gt;&lt;p&gt;star ex change&lt;/p&gt;&lt;p&gt;stark ex change&lt;/p&gt;&lt;p&gt;stark exchange&lt;/p&gt;&lt;p&gt;stack exchange&lt;/p&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="14135" LastEditDate="2015-08-17T20:15:54.127" LastActivityDate="2015-08-17T20:15:54.127" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="20145" PostTypeId="2" ParentId="20011" CreationDate="2015-08-17T09:03:05.557" Score="1" Body="&lt;p&gt;You are, very clearly&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Someone with the surname &quot;Love&quot;. Googling &lt;em&gt;celebrities with the surname &quot;Love&quot;&lt;/em&gt; gave me results of: Courtney Love, Jennifer Love Hewitt, and a whole stream of... er... employees of the &quot;adult entertainment&quot; industry...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I work on a farm:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A farm has a field, if it was a field of grass you could maybe play tennis on it, making it a Court(ney)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but it's not as romantic as you think:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This is a clue to the surname &quot;Love&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I need to wake up around sunrise to go to work&#xA;(&lt;strong&gt;&lt;em&gt;Disclaimer&lt;/strong&gt;: Do not open this spoiler if you are easily offended&lt;/em&gt;):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is something to do with cockcrow or cockerel. It may work for the &quot;adult entertainers&quot;, as they definitely get a lot of, erm, &quot;cockerel&quot;...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;if you don't count room and board:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Room and board = bed and breakfast... no idea where to go with this&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I get just 4 pennies for every minute of work:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A minute is a 60th of an hour. &quot;60th&quot; in a different sense means &lt;em&gt;diamond&lt;/em&gt;. Courtney Love was in the news for losing some diamonds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I had to take a second job to pay the bills:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well, &quot;The Bill&quot; is a phrase used in the UK to refer to the police. Courtney has been arrested for various drugs misdemeanours, so maybe this is something to do with being arrested and fined?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'd do anything just to get an apartment:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An apartment can be called (in the UK at least, not so sure about US) a &lt;em&gt;flat&lt;/em&gt;. With all the plastic surgery, Courtney's face is definitely &lt;em&gt;not&lt;/em&gt; flat, so I think I've got that one wrong.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm such a sad sack it'll probably never happen:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Is Courtney a bit grumpy normally? I don't really know, I barely know who she is.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I wish I had a job I was more passionate about:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well, she's a celeb so doesn't really have any job. And the &quot;passionate&quot; thing is referring to her surname &quot;Love&quot; again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, in summary you are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Courtney Love. And probably not an &quot;adult entertainment&quot; star.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9444" LastActivityDate="2015-08-17T09:03:05.557" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20147" PostTypeId="2" ParentId="20142" CreationDate="2015-08-17T10:08:18.897" Score="8" Body="&lt;p&gt;Same method as Anachor, pretty much:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $70^3 + 105^3 = (2*35)^3 + (3*35)^3 = (2^3 + 3^3)*35^3 = 35^4$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15761" LastEditorUserId="1748" LastEditDate="2015-08-17T10:21:13.763" LastActivityDate="2015-08-17T10:21:13.763" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20148" PostTypeId="2" ParentId="20099" CreationDate="2015-08-17T10:13:02.690" Score="5" Body="&lt;p&gt;Congratulations to &lt;a href=&quot;https://puzzling.stackexchange.com/a/20136/15731&quot;&gt;Lawrence for his solution&lt;/a&gt;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a list of all (?) solutions for this problem. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (0, 1, 2, 3, 7, 11, 15, 19, 23, 27, 28, 29, 30, 61, 64, 67, 70)&lt;br&gt;&#xA; (0, 1, 3, 4, 5, 8, 14, 20, 26, 32, 38, 44, 45, 47, 48, 49, 52)&lt;br&gt;&#xA; (0, 1, 3, 4, 5, 8, 14, 20, 26, 32, 38, 44, 46, 47, 48, 49, 51)&lt;br&gt;&#xA; (0, 1, 3, 4, 5, 8, 14, 20, 26, 32, 38, 44, 46, 48, 49, 51, 53)&lt;br&gt;&#xA; (0, 1, 3, 4, 5, 8, 14, 20, 26, 32, 38, 44, 47, 48, 49, 51, 52)&lt;br&gt;&#xA; (0, 1, 3, 4, 7, 8, 9, 16, 17, 21, 24, 35, 46, 57, 68, 79, 90)&lt;br&gt;&#xA; (0, 1, 3, 4, 9, 11, 16, 20, 25, 30, 34, 39, 41, 46, 47, 49, 50)   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;(I did not all find all of them myself: My program found 5 solutions, and after a few minutes, I stopped it - I'm not sure whether it would also have found the other ones, because it was not as systematic and deterministic as the approach that Lawrence described)&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most interesting solution, for me, was the last one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (0, 1, 3, 4, 9, 11, 16, 20, 25, 30, 34, 39, 41, 46, 47, 49, 50)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the solution that the &lt;strong&gt;Hint 2&lt;/strong&gt; would have referred to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Surprisingly, it does not need any value larger than 50. But more importantly: The solution is &lt;strong&gt;symmetric&lt;/strong&gt;! When this is plotted as a graph, it looks like the first part of a sine curve. I think if someone had come to the ideas of &lt;strong&gt;1.&lt;/strong&gt; Only using the numbers between 0 and 50, &lt;strong&gt;2.&lt;/strong&gt; Making the solution symmetric, and &lt;strong&gt;3.&lt;/strong&gt; Using &quot;more&quot; numbers at the beginning and end of the 0...50 interval, and fewer in the middle, then one &lt;em&gt;might&lt;/em&gt; also have found the solution manually. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;EDIT In response to the comments:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A rough description of the &quot;algorithm&quot;. As already mentioned, it is not very systematic, but did its job:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Initialization&lt;/strong&gt;: First of all, a simple initial solution is constructed by greedily collecting 17 numbers: Starting with a list of &lt;code&gt;[0,1]&lt;/code&gt;, in each step I'm examining all possible remaining numbers, and choose the one where the number of &quot;missing sums&quot; is the smallest. This produces the initial solution of&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0, 1, 3, 5, 7, 12, 17, 20, 26, 30, 35, 39, 41, 44, 46, 54, 68&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where 11 sums are still missing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Search&lt;/strong&gt;: I'm maintaining a list of &quot;candidate&quot; solutions, each given by the list of 17 numbers, and the number of missing sums. Then, I'm &quot;wiggling&quot; these numbers: For the candiate solution with the least number of missing sums, I'm creating variations, by shifting each number of the list between its left and right neighbor. For example, from a candiate solution like this&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0, 1, 3, 5, 7, 12, 17, 20, 26, 30, 35, 39, 41, 44, 46, 54, 68 (missing: 11)&#xA;                       ==&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'm creating further candidates:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0, 1, 3, 5, 7, 12, 17, 18, 26, 30, 35, 39, 41, 44, 46, 54, 68 (missing:...)&#xA;                       ==&#xA;0, 1, 3, 5, 7, 12, 17, 19, 26, 30, 35, 39, 41, 44, 46, 54, 68 (missing:...)&#xA;                       ==&#xA;...&#xA;0, 1, 3, 5, 7, 12, 17, 25, 26, 30, 35, 39, 41, 44, 46, 54, 68 (missing:...)&#xA;                       ==&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And I'm always doing this with the candiate with the least number of missing sums. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Actually, this &quot;list&quot; is a priority queue, ordered ascending by the number of missing sums, and I'm occasionally &quot;pruning&quot; this queue by removing candidates that are far worse than the current best one, but these are details).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not very deterministic, and (due to the pruning) not even guaranteed to find a solution at all, but it found ... at least 5 of them within a few minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;I probably won't post the (Java) code here, because it's a bit lengthy and rather messy, but I hope that the basic idea became clear&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="15731" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-08-17T13:23:35.427" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20149" PostTypeId="1" AcceptedAnswerId="20151" CreationDate="2015-08-17T11:02:09.267" Score="3" ViewCount="1171" Body="&lt;p&gt;&lt;strong&gt;Clues&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have hired Sherlock Holmes to find something for me. We are near my home at an outdoor event and queuing for the refreshment tent. I suspect that the object is to be found somewhere in this area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Holmes: &lt;em&gt;So Mister Chasly, How can I be of assistance?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Me: &lt;em&gt;Mr Holmes, you know I'm from near here. There's something I want you to find.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We near the front of the queue. Dr Watson comes hurrying up and pushes in without asking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Watson to Sherlock: &lt;em&gt;Tea Holmes?&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sherlock to Watson: &lt;em&gt;Queue, Watson, there's a queue!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ignoring Watson, I give the famous detective a piece of paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Me: &lt;em&gt;I afraid that this is the only clue that I can provide.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hand Sherlock a piece of paper with some figures written on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sherlock: &lt;em&gt;Hmm... 19852 05812&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Holmes screws up his eyes and appears to be concentrating hard. After a few moments he opens his eyes, turns and points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sherlock:  That way I think!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Forgetting about the refreshments, Holmes strides away. Watson and I follow behind. Soon we had found the thing I was seeking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To those reading this puzzle - I wish I could show you a photograph of the object, I took one but but I have lost it. Can you make me a new one? I'd be most grateful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Provide a photograph of the man-made object that Holmes found for me. The photo can be from any angle as long as most of the object is clearly visible and discernible among any nearby vegetation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTES&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CodeNewbie has demonstrated that I wasn't quite specific enough. The object is smaller than the average house and bigger than a football (English type). It is roughly/partly spherical in shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On request from the community I have also removed the restriction that answerers should withhold their reasoning. Please do give your reasoning. Of course I would still appreciate if you could hide it by using the &gt;! facility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this may be harder for some people because of where they live and therefore what they can be expected to know. They can still find all the information by using the clues but they may have to work a little harder and longer in order to find the extra knowledge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You do not need a camera, just the internet.&lt;/p&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="14177" LastEditDate="2015-08-17T15:13:10.670" LastActivityDate="2015-08-17T15:13:10.670" Title="No shoot Sherlock: Test your internet and detective skills with this one!" Tags="|story|computer-puzzle|geography|knowledge|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20150" PostTypeId="2" ParentId="20149" CreationDate="2015-08-17T11:17:53.143" Score="1" Body="&lt;p&gt;Is this the image you are looking for?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/XzRaT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XzRaT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;It is Britain's first licensed aerodrome. And since you are &lt;a href=&quot;https://puzzling.stackexchange.com/users/14177/chasly-from-uk&quot;&gt;chasly from UK&lt;/a&gt;, this would be quite close to you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-08-17T11:17:53.143" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20151" PostTypeId="2" ParentId="20149" CreationDate="2015-08-17T11:29:55.173" Score="4" Body="&lt;p&gt;Is this the thing that you're after:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/NZq8F.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NZq8F.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or, from another angle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/ebTDz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ebTDz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Method:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The discussion between Holmes and Watson mentions &quot;Tea&quot; and &quot;Queue&quot;. Putting these together with the numbers and carrying out Google search for &lt;code&gt;TQ 19852 05812&lt;/code&gt;, the first result (in the UK at least) is &lt;a href=&quot;http://www.google.co.uk/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=1&amp;amp;cad=rja&amp;amp;uact=8&amp;amp;ved=0CCEQFjAAahUKEwjAu7Wlm7DHAhVEOhQKHYKiDkk&amp;amp;url=http%3A%2F%2Fwww.airfieldinformationexchange.org%2Fcommunity%2Fshowthread.php%3F3760-Shoreham&amp;amp;ei=SuDRVcDuMMT0UILFusgE&amp;amp;usg=AFQjCNEBAgTSQ1FrxGJgtQRx9vow5_kv3w&amp;amp;sig2=rZpCUSqDlBsfUPttU1T8LA&amp;amp;bvm=bv.99804247,d.bGQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;Shoreham Airfield Information Exchange&lt;/a&gt; in which there is a discussion about this building. The setting for the question is an outdoor event which could quite easily be at the airfield.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="14135" LastEditDate="2015-08-17T13:33:09.963" LastActivityDate="2015-08-17T13:33:09.963" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="20152" PostTypeId="2" ParentId="20135" CreationDate="2015-08-17T13:00:36.647" Score="17" Body="&lt;p&gt;Elements in the sequence are computed with the following recursive formula:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $a_n = \begin{cases}0, &amp;amp; \text{if $n \le 0$}\\a_{\frac{n}{2}}, &amp;amp; \text{if $n$ is even}\\ a_{n-1}+1, &amp;amp; \text{if $n$ is odd} \end{cases}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus, the values for $a_{12}$ to $a_{15}$ are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $a_{12} = a_{6} = a_{3} = a_{2} + 1 = a_{1} + 1 = a_{0} + 1 + 1 = 2$&lt;br&gt;&#xA; $a_{13} = a_{12} + 1 = 3$&lt;br&gt;&#xA; $a_{14} = a_{7} = a_{6} + 1 = 3$&lt;br&gt;&#xA; $a_{15} = a_{14} + 1 = 4$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1750" LastActivityDate="2015-08-17T13:00:36.647" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20154" PostTypeId="1" AcceptedAnswerId="20376" CreationDate="2015-08-17T13:46:49.923" Score="6" ViewCount="638" Body="&lt;p&gt;Terry guessed the key based on clues she got from Jack&lt;br /&gt;&#xA;In a music-linked movie roughly three decades back&lt;br /&gt;&#xA;But you need to start from a more continental track&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Harold G to Harrowing (as Schmieder once denoted)&lt;br /&gt;&#xA;Led to smoking at the theatre - no longer here promoted&lt;br /&gt;&#xA;Decrypt the next installment if you want to get upvoted&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;XFMcvihqtfcLiuvnbzqqtfcvlshcgzafwxtzyqsrlwrgrbsmhzfxiihugqhjhhew Ojljlnhkhdfdbtwuowrvvorrpvvfhrggomcnhydflrbbvfocvvemrxelhyabhvu Riwrrvumrzruxiwykretjbqdannhyrggqbfwafkfzfwsjbusxabhvu&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer should be 13 characters long.&lt;br /&gt;&#xA;And Len has already done a lot of the hard work!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I could have used the word Harrying in place of Harrowing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="14135" LastEditDate="2015-08-21T22:32:38.160" LastActivityDate="2015-08-23T06:08:16.150" Title="Something for the real fans" Tags="|riddle|cipher|geography|knowledge|movies|" AnswerCount="2" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20155" PostTypeId="1" AcceptedAnswerId="20157" CreationDate="2015-08-17T14:18:28.547" Score="0" ViewCount="172" Body="&lt;p&gt;Around mid-seventies, early eighties this was a popular kid's verbal puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Half Circle Full Circle Half Circle Triangle with two legs minus Half Circle Full Circle Right Angle Triangle with two legs&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What was it about ?&lt;/p&gt;&#xA;" OwnerUserId="9892" LastEditorUserId="11876" LastEditDate="2015-08-19T06:48:12.903" LastActivityDate="2015-08-19T06:48:12.903" Title="Decipher the content" Tags="|wordplay|english|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20156" PostTypeId="1" AcceptedAnswerId="20160" CreationDate="2015-08-17T14:26:05.627" Score="-4" ViewCount="3089" Body="&lt;p&gt;Here is Mr. Dog, friend of Mrs. Goat. Move only two matchsticks such that Mr. Dog's head faces the opposite direction, i.e. to the right side.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KJvYo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KJvYo.png&quot; alt=&quot;Mr. Dog&quot; /&gt;&lt;/a&gt;&#xA;This is Part 2 of turning the animal series.&#xA;&lt;br&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/17172/turning-a-goat&quot;&gt;See Part 1 - Turning a Goat&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13652" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2017-12-29T15:34:26.670" Title="Turning a dog (Part 2)" Tags="|logical-deduction|visual|geometry|matches|" AnswerCount="4" CommentCount="1" ClosedDate="2015-08-17T17:05:26.983" ContentLicense="CC BY-SA 3.0" />
  <row Id="20157" PostTypeId="2" ParentId="20155" CreationDate="2015-08-17T14:27:46.380" Score="6" Body="&lt;p&gt;The answer is clearly:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; COCA-COLA&#xA; The riddle is describing the shapes of the letters, assuming that an uppercase sans serif font has been used which has a semicircle for the letter C&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="14135" LastEditDate="2015-08-17T18:15:27.733" LastActivityDate="2015-08-17T18:15:27.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20159" PostTypeId="2" ParentId="20156" CreationDate="2015-08-17T14:31:16.623" Score="10" Body="&lt;p&gt;How about this (apologies for the wobbly lines)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/CSlBs.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CSlBs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-08-17T14:31:16.623" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20160" PostTypeId="2" ParentId="20156" CreationDate="2015-08-17T14:31:56.880" Score="7" Body="&lt;h2&gt;Like this?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tKXFt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tKXFt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Fixed head (I must write something to get 30 characters)&lt;/p&gt;&#xA;" OwnerDisplayName="user14478" LastEditorDisplayName="user14478" LastEditDate="2015-08-17T14:37:20.183" LastActivityDate="2015-08-17T14:37:20.183" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20161" PostTypeId="2" ParentId="20156" CreationDate="2015-08-17T14:33:16.227" Score="1" Body="&lt;p&gt;I'll go with the cheeky answer of: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eJt4B.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eJt4B.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-08-17T14:33:16.227" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20162" PostTypeId="2" ParentId="20156" CreationDate="2015-08-17T14:38:58.983" Score="0" Body="&lt;p&gt;I saw all of your tentative and i raise you this :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zo0ub.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zo0ub.png&quot; alt=&quot;boop&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13384" LastActivityDate="2015-08-17T14:38:58.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20163" PostTypeId="1" AcceptedAnswerId="20164" CreationDate="2015-08-17T14:43:17.773" Score="6" ViewCount="2698" Body="&lt;p&gt;You were given 12 coins by your friend. He bet that if you could arrange these dozen coins into 6 rows of 4 coins such that it makes two similar shapes, he will give you 12 more coins. How will you do it?&lt;/p&gt;&#xA;" OwnerUserId="13652" LastActivityDate="2019-07-15T06:36:44.603" Title="A dozen into six rows?" Tags="|logical-deduction|geometry|" AnswerCount="3" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20164" PostTypeId="2" ParentId="20163" CreationDate="2015-08-17T14:51:48.590" Score="20" Body="&lt;p&gt;Forgive the badly drawn image:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/COl8E.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/COl8E.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Black are the coins, green are the rows of 4, red is the similar shape. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="11251" LastEditDate="2015-08-17T16:36:29.793" LastActivityDate="2015-08-17T16:36:29.793" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20165" PostTypeId="2" ParentId="20163" CreationDate="2015-08-17T14:52:50.423" Score="12" Body="&lt;p&gt;You arrange them into...&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; a star of David, with each point and intersection having a coin.&#xA; &lt;a href=&quot;https://web.archive.org/web/20170918011029/http://imgh.us/12coins.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://web.archive.org/web/20170918011029/http://imgh.us/12coins.svg&quot; alt=&quot;&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2019-07-15T06:36:44.603" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="20166" PostTypeId="2" ParentId="20135" CreationDate="2015-08-17T15:17:39.707" Score="3" Body="&lt;p&gt;Maybe not too much in the spirit of the riddle since it ignores the &lt;a href=&quot;/questions/tagged/wordplay&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;wordplay&amp;#39;&quot; rel=&quot;tag&quot;&gt;wordplay&lt;/a&gt; tag, but here is another potential continuation of the sequence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clearly the sequence is a level-order traversal of the binary tree&lt;br&gt;&#xA; \begin{matrix}&amp;amp;&amp;amp;&amp;amp;(1)&amp;amp;&amp;amp;&amp;amp;\\&amp;amp;(1,2)&amp;amp;&amp;amp;&amp;amp;&amp;amp;(1,2)&amp;amp;\\(2,3,1,2)&amp;amp;&amp;amp;(2,3,?,?)&amp;amp; &amp;amp; (?,?,?,?)&amp;amp;&amp;amp;(?,?,?,?) \\\end{matrix}&#xA;&lt;br&gt;&#xA; where each node at level $n$ is a $(2^n)$-tuple $N_n=(N_{n,0},...,N_{n,2^n-1})$ which is decided by the equations&#xA; \begin{gather*}N_{0,0}=1\\N_{n,i}=N_{n-1,\lfloor\frac{2^n-1-i}{2}\rfloor}+i-2\lfloor \frac{i}{2}\rfloor\end{gather*} &#xA; so the next elements in the sequence are&lt;br&gt;&#xA; \begin{gather*}N_{2,2}=N_{1,\lfloor\frac{1}{2}\rfloor}+2-2\lfloor \frac{2}{2}\rfloor=N_{1,0}+0=1\\N_{2,3}=N_{1,\lfloor\frac{0}{2}\rfloor}+3-2\lfloor \frac{3}{2}\rfloor=N_{1,0}+1=2\\N_{2,0}=N_{1,\lfloor\frac{3}{2}\rfloor}+0-2\lfloor \frac{0}{2}\rfloor=N_{1,1}+0=2\\N_{2,1}=N_{1,\lfloor\frac{2}{2}\rfloor}+1-2\lfloor \frac{1}{2}\rfloor=N_{1,1}+1=3\end{gather*}  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13498" LastEditorUserId="13498" LastEditDate="2015-08-17T16:59:08.473" LastActivityDate="2015-08-17T16:59:08.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20167" PostTypeId="2" ParentId="20135" CreationDate="2015-08-17T15:24:32.483" Score="1" Body="&lt;p&gt;Looks like $2, 3, 3, 4$ to me, since the first number is the head node of a binary tree. Going to the left you change nothing, but going to the right you add one. So,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$1$ goes to $1,2$.&lt;/li&gt;&#xA;&lt;li&gt;$2$ goes to $2,3$ and &lt;/li&gt;&#xA;&lt;li&gt;$3$ goes to $3,4$ &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So when you get to $1,2,2,3$ this can only go to $1,2,2,3,2,3,3,4$. In the statement of the problem we see $1,2,2,3$ but we don't see the next $4$ numbers, which are the solution.&lt;/p&gt;&#xA;" OwnerUserId="15767" LastEditorUserId="15544" LastEditDate="2015-08-30T03:11:48.807" LastActivityDate="2015-08-30T03:11:48.807" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20168" PostTypeId="1" AcceptedAnswerId="20174" CreationDate="2015-08-17T15:27:04.957" Score="2" ViewCount="673" Body="&lt;p&gt;(I was going to put this on the puzzling meta board before putting it live just to see how everyone reacted to it because I'm not sure how well it would be received, but for some reason I can't ask a question there right now so I figure it's a sign, and I'll go ahead and put it live)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recently obtained a 3D printer and have been experimenting with it. I printed out... well... something, and after scraping away the support structures I figured it could make for an interesting puzzle. One where you can't simply use google to search for the clues needed to solve it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for the specific name of this, you won't get credit for the solution by saying something like &quot;it's an artwork!&quot; or &quot;it's a piece of landscape!&quot; or &quot;it's a body part!&quot; I'm very picky and I want specifics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So tell me, puzzlers: What is this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9CKJ9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9CKJ9.jpg&quot; alt=&quot;mystery print&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It's probably bad form to give a hint before I even start, but just in case anyone's interested in 3d printing, there's a website called Thingiverse which is a phenomenal source of free 3d printable files, and that's where I got this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13068" LastEditorUserId="11475" LastEditDate="2015-08-17T15:28:16.483" LastActivityDate="2015-08-17T16:56:04.830" Title="What is this? A 3D-printed puzzle" Tags="|visual|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20170" PostTypeId="2" ParentId="20168" CreationDate="2015-08-17T15:47:37.080" Score="4" Body="&lt;p&gt;Is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An upside down bust of Marianne?&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/OWaRQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5359" LastActivityDate="2015-08-17T15:47:37.080" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20172" PostTypeId="1" AcceptedAnswerId="20175" CreationDate="2015-08-17T16:11:51.390" Score="7" ViewCount="792" Body="&lt;p&gt;From the lines below, can you determine what each is describing and the commonality between each answer? Once you get a couple, the pattern is obvious. Caveat: a handful of lines have a couple valid answers, and there may be some that I hadn't though of, but I am happy to accept any answer that seems viable (follows the pattern and fits the clue). Have fun:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A very adaptable novel of manners&lt;br&gt;&#xA;  A simple mistake becomes a criminal offense&lt;br&gt;&#xA;  An officer's responsibilities&lt;br&gt;&#xA;  Trinkets&lt;br&gt;&#xA;  Scientific process, in many cases&lt;br&gt;&#xA;  A witch's incantation&lt;br&gt;&#xA;  Econ 101&lt;br&gt;&#xA;  Like most wanted men&lt;br&gt;&#xA;  Dun-Dun!!!&lt;br&gt;&#xA;  Repeatedly&lt;br&gt;&#xA;  The lunar cycle&lt;br&gt;&#xA;  &quot;This pot's mine, boys!&quot;&lt;br&gt;&#xA;  Fisher's methodology&lt;br&gt;&#xA;  &quot;No use crying over spilled milk&quot;&lt;br&gt;&#xA;  Orator's introduction&lt;br&gt;&#xA;  90's events inspire 10's satirical movie&lt;br&gt;&#xA;  &quot;Time to get up!&quot;&lt;br&gt;&#xA;  A real deluge&lt;br&gt;&#xA;  Quite the stark contrast&lt;br&gt;&#xA;  A children's game&lt;br&gt;&#xA;  &quot;That's all there is to it&quot;&lt;br&gt;&#xA;  Try different combinations&lt;br&gt;&#xA;  Ready to rollout&lt;br&gt;&#xA;  For hunting purists&lt;br&gt;&#xA;  A stunning tactic  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For bonus points, can you come up with some good examples of the opposite of this pattern?&lt;/p&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="8874" LastEditDate="2015-10-07T08:52:23.460" LastActivityDate="2015-10-07T08:52:23.460" Title="What's the pattern here?" Tags="|word|pattern|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20173" PostTypeId="1" AcceptedAnswerId="20210" CreationDate="2015-08-17T16:13:41.753" Score="10" ViewCount="253" Body="&lt;p&gt;&lt;em&gt;To me, you are but a cafe&lt;br&gt;&#xA;Buffet; souffle; filet; parfait&lt;br&gt;&#xA;I'll be the hunter, you be the prey&lt;br&gt;&#xA;Slice me in twain, and leave my tail.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Lower, lower, lower yet&lt;br&gt;&#xA;submerged in water, never wet&lt;br&gt;&#xA;Pick me and you will not regret&lt;br&gt;&#xA;The leaders stay, the rest all bail.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Treat me as you have my kin&lt;br&gt;&#xA;Pin me down, and see my sins&lt;br&gt;&#xA;All spill asunder from within&lt;br&gt;&#xA;My heart bleeds, though my shell prevails.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Rooted deep within the ground&lt;br&gt;&#xA;A being of myth and tales unbound&lt;br&gt;&#xA;The net's been cast, and I've been found&lt;br&gt;&#xA;In full I stand, behind no veil.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Take them all to find my name&lt;br&gt;&#xA;A right mess made, and you're to blame&lt;br&gt;&#xA;Then take the rejects, cast in shame&lt;br&gt;&#xA;Spin them around and they'll do the same.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The first four stanzas describe my pieces, and the fifth describes my companion and I in whole. The question is, &lt;strong&gt;who are we?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Bonus points to whomever is able to point out the &lt;em&gt;cheeky&lt;/em&gt; hints I've left in each of the first four stanzas.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-08-18T09:33:01.380" Title="A Right Mess Made" Tags="|riddle|rhyme|wordplay|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20174" PostTypeId="2" ParentId="20168" CreationDate="2015-08-17T16:22:58.293" Score="5" Body="&lt;p&gt;Is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An upside down bust of Sappho?&lt;br&gt;&lt;br&gt;&lt;a href=&quot;https://i.stack.imgur.com/4zf0N.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4zf0N.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;br&gt;&lt;br&gt;Specifically the one on the right hand side.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12206" LastActivityDate="2015-08-17T16:22:58.293" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20175" PostTypeId="2" ParentId="20172" CreationDate="2015-08-17T16:32:31.523" Score="9" Body="&lt;p&gt;The pattern is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; any phrase in the form &quot;_____ and _____&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mostly complete list, but one or two left:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very adaptable novel of manners&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pride and Prejudice  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A simple mistake becomes a criminal offense&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Crime and punishment&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An officer's responsibilities&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Serve and protect&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Trinkets&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Odds and ends (fm. GentlePurpleRain)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Scientific process, in many cases&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Trial and error&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A witch's incantation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eye of newt and toe of frog&lt;br&gt;&#xA; (Double, double,) toil and trouble (fm. Sabre)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Econ 101&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Supply and demand&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Like most wanted men&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Armed and dangerous&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Dun-Dun!!!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Law &amp;amp; Order (fm. Mohair)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Repeatedly&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Over and over&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The lunar cycle&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Waxing and waning (fm. psmears)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;This pot's mine, boys!&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Read 'em and weep&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fisher's methodology&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Catch and release&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;No use crying over spilled milk&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Live and learn (fm. Curtis Reubens)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Orator's introduction&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ladies and gentlemen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;90's events inspire 10's satirical movie&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pain &amp;amp; Gain (fm. Sabre)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Time to get up!&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rise and shine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A real deluge&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (raining) cats and dogs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Quite the stark contrast&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Black and white&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A children's game&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chutes and Ladders&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;That's all there is to it&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Over and done with&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Try different combinations&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mix and match&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ready to rollout&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Locked and loaded (fm. Curtis Reubens)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For hunting purists&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bow and arrow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A stunning tactic&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Shock and awe&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10625" LastEditorUserId="10625" LastEditDate="2015-08-17T23:11:51.327" LastActivityDate="2015-08-17T23:11:51.327" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="20176" PostTypeId="2" ParentId="20172" CreationDate="2015-08-17T16:39:52.167" Score="3" Body="&lt;p&gt;Filling in some of lorimer's gaps:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;No use crying over spilled milk&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Live and learn.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ready to rollout&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Locked and loaded&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15761" LastActivityDate="2015-08-17T16:39:52.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20177" PostTypeId="2" ParentId="20168" CreationDate="2015-08-17T16:56:04.830" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/Qvgqj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Qvgqj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's a better picture for those of you confused about alexmc's answer. (whoever printed that one either chose to do it in multiple parts, probably because they scaled up the size and their print area could only handle so much, or they had a print error that caused the base to slide once it reached a certain height or something like that)&lt;/p&gt;&#xA;" OwnerUserId="13068" LastActivityDate="2015-08-17T16:56:04.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20178" PostTypeId="1" AcceptedAnswerId="20179" CreationDate="2015-08-17T18:02:26.130" Score="17" ViewCount="3321" Body="&lt;p&gt;Change my 1st, I'm quite a leaper.&lt;br&gt;&#xA;Change my 2nd, I'm something extra.&lt;br&gt;&#xA;Change my 3rd, I'm a heavy weapon.&lt;br&gt;&#xA;Change my 4th, I differ in color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What 4-letter word am I?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastActivityDate="2015-08-18T09:34:15.653" Title="What 4-letter word am I?" Tags="|riddle|word|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20179" PostTypeId="2" ParentId="20178" CreationDate="2015-08-17T18:08:26.670" Score="20" Body="&lt;p&gt;You are a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MARE&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Change my 1st, I'm quite a leaper.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;H&lt;/strong&gt;ARE, a rabbit&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Change my 2nd, I'm something extra.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; M&lt;strong&gt;O&lt;/strong&gt;RE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Change my 3rd, I'm a heavy weapon.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Mace_(club)&quot;&gt;MA&lt;strong&gt;C&lt;/strong&gt;E&lt;/a&gt;, a heavy metal club&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Change my 4th, I differ in color.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://dictionary.reference.com/browse/mark?s=t&quot;&gt;MAR&lt;strong&gt;K&lt;/strong&gt;&lt;/a&gt;, a spot or line on a surface, visible through difference in color or elevation from that of the surrounding area. &lt;em&gt;Definition from link at Dictionary.com&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13237" LastEditorUserId="13237" LastEditDate="2015-08-17T18:32:26.110" LastActivityDate="2015-08-17T18:32:26.110" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="20180" PostTypeId="2" ParentId="20178" CreationDate="2015-08-17T18:08:51.180" Score="10" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;MARE&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Change my 1st, I'm quite a leaper.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HARE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Change my 2nd, I'm something extra.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MORE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Change my 3rd, I'm a heavy weapon.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MACE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Change my 4th, I differ in color.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strike&gt;MARS&lt;/strike&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="4421" LastEditDate="2015-08-17T19:24:25.763" LastActivityDate="2015-08-17T19:24:25.763" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20181" PostTypeId="2" ParentId="20141" CreationDate="2015-08-17T18:18:01.090" Score="4" Body="&lt;p&gt;I did it in &lt;strong&gt;21&lt;/strong&gt; moves from &lt;code&gt;puzzling&lt;/code&gt; to &lt;code&gt;stack exchange&lt;/code&gt;. But I'm pretty sure there has to be a place to shave off a move.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    puzzling &lt;p&gt;&#xA;    muzzling &lt;p&gt;&#xA;    muzz ling &lt;p&gt;&#xA;    muzz line &lt;p&gt;&#xA;    muzz lane &lt;p&gt;&#xA;    muzz laner &lt;p&gt;&#xA;    muzz langer &lt;p&gt;&#xA;    muzz hanger &lt;p&gt;&#xA;    muzz changer &lt;p&gt;&#xA;    muzz change &lt;p&gt;&#xA;    fuzz change &lt;p&gt;&#xA;    fuze change &lt;p&gt;&#xA;    faze change &lt;p&gt;&#xA;    fare change &lt;p&gt;&#xA;    care change &lt;p&gt;&#xA;    carex change &lt;p&gt;&#xA;    car ex change &lt;p&gt;&#xA;    scar ex change &lt;p&gt;&#xA;    star ex change &lt;p&gt;&#xA;    stark ex change &lt;p&gt;&#xA;    stack ex change &lt;p&gt;&#xA;    stack exchange &lt;p&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditDate="2015-08-17T19:44:07.103" LastActivityDate="2015-08-17T19:44:07.103" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20184" PostTypeId="1" AcceptedAnswerId="20185" CreationDate="2015-08-17T19:29:19.537" Score="20" ViewCount="1017" Body="&lt;p&gt;I was reading a book that my friend had lent me and came across a slip of paper that I supposed was a bookmark. Reading it I saw the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DAM&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GAS&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LOS&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NET&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NIM&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TOR&lt;/p&gt;&#xA;&#xA;&lt;p&gt;VIC&lt;/p&gt;&#xA;&#xA;&lt;p&gt;WED&lt;/p&gt;&#xA;&#xA;&lt;p&gt;YAK&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing my friend likes word games and puzzles I tried to see what, if anything, connected the words. In the end I gave up and phoned him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Him: Oh ha, yes that's just a list of words that I was playing with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Me: Well I can see that. I wondered why you included LOS. That's Spanish isn't it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Him: Just read me all the words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read them down the phone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Him: Oh no, all the words on that paper are perfectly ordinary English dictionary words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Me: Are you sure that the name 'VIC' is in the dictionary?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Him: No, I mean I just cut that bookmark from the edge of a larger piece of paper. I obviously cut through the list of words as well. They were originally longer - all were five letters long in fact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Me: So what were the original words then?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Him: I think you have now have enough information to work that out for yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Me: Are you sure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Him: Just find what they all have in common.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing that my friend likes words and word-games. What were the original five-letter dictionary words? What did they all have in common?&lt;/p&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="14177" LastEditDate="2015-08-17T19:33:06.390" LastActivityDate="2015-08-17T19:33:06.390" Title="Three-letter words. Or are they longer?" Tags="|word|pattern|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20185" PostTypeId="2" ParentId="20184" CreationDate="2015-08-17T19:31:58.167" Score="21" Body="&lt;p&gt;I'm going to guess that the words were all originally&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Palindromes!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which would make the original words:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MADAM, SAGAS, SOLOS, TENET, MINIM, ROTOR, CIVIC, DEWED, and KAYAK.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-08-17T19:31:58.167" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20186" PostTypeId="1" AcceptedAnswerId="20202" CreationDate="2015-08-17T19:41:36.970" Score="1" ViewCount="618" Body="&lt;p&gt;You can come into us, you can walk our stairs,&lt;br&gt;&#xA;You can click pictures with us, but no-one will care.&lt;br&gt;&#xA;If our real purpose is asked, we're meant to tell the time,&lt;br&gt;&#xA;But not using us for that must be made a crime.&lt;/p&gt;&#xA;" OwnerUserId="15581" LastEditorUserId="5373" LastEditDate="2016-12-21T15:27:42.877" LastActivityDate="2016-12-21T15:27:42.877" Title="What am I? Better guess fast, the clock's ticking" Tags="|riddle|geography|outer-space|" AnswerCount="3" CommentCount="0" ClosedDate="2015-08-20T06:13:02.290" ContentLicense="CC BY-SA 3.0" />
  <row Id="20187" PostTypeId="2" ParentId="20186" CreationDate="2015-08-17T19:51:06.470" Score="0" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  A smart watch?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can come into us,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Put your wrist in it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;you can walk our stairs,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Count steps when running&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can click pictures with us, but no-one will care.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure but I guess there exist ones with a camera&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If our real purpose is asked, we're meant to tell the time,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because it's a watch&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But not using us for that must be made a crime.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is maybe a popular opinion?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-08-17T19:51:06.470" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20188" PostTypeId="1" AcceptedAnswerId="20191" CreationDate="2015-08-17T19:57:51.953" Score="9" ViewCount="414" Body="&lt;h1&gt;The puzzle&lt;/h1&gt;&#xA;&lt;p&gt;Given a set of numbers &lt;span class=&quot;math-container&quot;&gt;$\{1,2,\ldots,n\}$&lt;/span&gt; and an expression of inequalities, e.g., &lt;span class=&quot;math-container&quot;&gt;$ x_1 &amp;gt; x_2 &amp;lt; \cdots &amp;gt; x_n$&lt;/span&gt;. Prove that there always exists a solution such that &lt;span class=&quot;math-container&quot;&gt;$x_i \in \{1,2,\ldots,n\}$&lt;/span&gt; and that each &lt;span class=&quot;math-container&quot;&gt;$x_i$&lt;/span&gt; is distinct. Can you design an algorithm which always provides a correct solution in polynomial time?&lt;/p&gt;&#xA;&lt;p&gt;I will not take credit for this problem. It was originally posed by Jim Tanton &lt;a href=&quot;http://www.themathcircle.org/researchproblems.php&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;h3&gt;In less &amp;quot;mathy&amp;quot; terms:&lt;/h3&gt;&#xA;&lt;p&gt;You are given a string of &lt;span class=&quot;math-container&quot;&gt;$&amp;gt;$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$&amp;lt;$&lt;/span&gt; symbols (&lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; of them in total). Insert the numbers from &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$n+1$&lt;/span&gt; between them, so that all the inequalities are true.  Find a general solution that works for any string of symbols of any length, and doesn't take exponentially longer when the string of symbols is longer.&lt;/p&gt;&#xA;&lt;h3&gt;Example&lt;/h3&gt;&#xA;&lt;p&gt;You are given &lt;code&gt;&amp;lt;  &amp;gt;  &amp;gt;&lt;/code&gt;.&#xA;You must insert the numbers from &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt; between the symbols so that all the inequalities are true:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$1&amp;lt;4&amp;gt;3&amp;gt;2$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="12901" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-08-18T05:50:16.030" Title="A problem of inequalities" Tags="|mathematics|algorithm|number-theory|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20190" PostTypeId="2" ParentId="20186" CreationDate="2015-08-17T20:08:41.330" Score="4" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Pyramids&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can come into us, you can walk our stairs,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; visitors to the pyramids can climb the stairs inside&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can click pictures with us, but no-one will care.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are thousands of photos of the pyramids, so no-one will be bothered with your holiday snaps&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If our real purpose is asked, we're meant to tell the time,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The air shafts in the pyramids are built to align with the constellation Orion and the Sphinx's gaze marks the point of sunrise at the equinox.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But not using us for that must be made a crime.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Too much tourism is leading to the erosion of the pyramids.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-08-17T20:08:41.330" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20191" PostTypeId="2" ParentId="20188" CreationDate="2015-08-17T20:12:22.727" Score="8" Body="&lt;p&gt;This algorithm is my intuition and I don't know if it's right so please prove me wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First fill in all numbers in order from left to right. Then we look at each number from left to right and do this:&#xA;If the sign next to it is &amp;lt; do nothing, if it's &gt; take the right most number in your sequence, put it there and shift all other numbers to the right&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-08-17T20:12:22.727" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20192" PostTypeId="2" ParentId="20141" CreationDate="2015-08-17T20:45:56.403" Score="5" Body="&lt;p&gt;Here's a way to do it in &lt;strong&gt;20&lt;/strong&gt; moves:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;puzzling&lt;br&gt;&#xA;muzzling&lt;br&gt;&#xA;mizzling&lt;br&gt;&#xA;fizzling&lt;br&gt;&#xA;fizz ling&lt;br&gt;&#xA;fizz hing&lt;br&gt;&#xA;fizz hang&lt;br&gt;&#xA;fizz chang&lt;br&gt;&#xA;fizz change&lt;br&gt;&#xA;fuzz change&lt;br&gt;&#xA;fuze change&lt;br&gt;&#xA;faze change&lt;br&gt;&#xA;fare change&lt;br&gt;&#xA;tare change&lt;br&gt;&#xA;stare change&lt;br&gt;&#xA;starer change&lt;br&gt;&#xA;star er change&lt;br&gt;&#xA;star ex change&lt;br&gt;&#xA;stark ex change&lt;br&gt;&#xA;stark exchange&lt;br&gt;&#xA;stack exchange&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-08-17T20:45:56.403" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20193" PostTypeId="2" ParentId="19864" CreationDate="2015-08-17T20:52:37.150" Score="0" Body="&lt;p&gt;Is this some kind of dance with six steps which can be performed doing serveral of those 6 step sequences?&#xA;I noticed that the last number is the same as the first number in each sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numbers might indicate positions on a floor which you have to touch with your arms and legs.&lt;/p&gt;&#xA;" OwnerUserId="15740" LastActivityDate="2015-08-17T20:52:37.150" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20195" PostTypeId="2" ParentId="20188" CreationDate="2015-08-17T21:46:11.460" Score="2" Body="&lt;p&gt;&lt;strong&gt;Proof that it's possible:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume it's the case for all lengths smaller than $n$. Either 1 is at the beginning/end or in the middle. In both cases, it leads to a smaller solvable part or two parts, so it's doable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How to do it:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doesn't matter where you put your 1 as long as the spot doesn't have a greater sign next to it. Then do the same for 2, 3, ... Rinse and repeat.&lt;/p&gt;&#xA;" OwnerUserId="14016" LastActivityDate="2015-08-17T21:46:11.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20196" PostTypeId="1" AcceptedAnswerId="20198" CreationDate="2015-08-17T22:15:43.723" Score="2" ViewCount="1052" Body="&lt;p&gt;You wake up with a gentle breeze on your face. Opening your eyes, you suddenly find that you are no longer at your home. You look around and find that you are on the roof of a very high building with no idea how you got there. You find a door, but it is locked.&lt;/p&gt;&#xA;&lt;p&gt;You notice a paper sticking out of the bottom of the door. You open it and read-&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Hey! I hope you enjoyed your morning surprise. As you might have already noticed there is only one way out of this roof and it is locked. If you want to get out you have to find out the height of the roof from the ground. Write the answer on the piece of paper and pass it under the door.&lt;/p&gt;&#xA;&lt;p&gt;However, I'm in a good mood. So, I'll accept the answer as long as it is within 10% of the accurate height. I've also left some tools for you to use.&lt;/p&gt;&#xA;&lt;p&gt;Good luck.&lt;/p&gt;&#xA;&lt;p&gt;- Your prankster friend,&#xA;Jason&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Knowing that Jason only pulls the roughest tricks, you know there is no easy way out of this. You go over to the railing. You can see the ground below, but you are bad at estimating distances.&lt;/p&gt;&#xA;&lt;p&gt;You look around and find a pendulum and a stop watch. These are what Jason left for you. There is nothing else in the roof that you can possibly use. You pick up the pendulum with your hand. It's quite heavy and has about a 20cm long string. Hum... pendulum and stopwatch, maybe I can do it, you think.&lt;/p&gt;&#xA;&lt;p&gt;How do you find the height?&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;To prevent &amp;quot;loophole&amp;quot; answers the following rules hold:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;No, you don't know the height of the building.&lt;/li&gt;&#xA;&lt;li&gt;There is no way of communication except that piece of paper.&lt;/li&gt;&#xA;&lt;li&gt;The door is not breakable&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="15544" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-08-21T17:34:10.053" Title="Measure the height" Tags="|lateral-thinking|story|" AnswerCount="5" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="20198" PostTypeId="2" ParentId="20196" CreationDate="2015-08-17T22:29:21.267" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Assuming the building isn't very tall, you could drop the pendulum off of the side of building and start the stopwatch. Listening for when the pendulum hits the ground you could measure the time it takes to fall. Then you can do math to find the height based on time, assuming Jason hasn't taken this prank too far and changed gravity. So I guess it being a pendulum is irrelevant?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15777" LastEditorUserId="15544" LastEditDate="2015-08-17T23:31:17.877" LastActivityDate="2015-08-17T23:31:17.877" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20199" PostTypeId="2" ParentId="20141" CreationDate="2015-08-17T22:44:47.673" Score="23" Body="&lt;p&gt;I've got 20&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; puzzling &lt;br&gt;&#xA; muzzling &lt;br&gt;&#xA; muzz ling &lt;br&gt;&#xA; fuzz ling &lt;br&gt;&#xA; fuze ling &lt;br&gt;&#xA; faze ling &lt;br&gt;&#xA; fake ling &lt;br&gt;&#xA; sake ling &lt;br&gt;&#xA; saker ling &lt;br&gt;&#xA; sacker ling &lt;br&gt;&#xA; stacker ling &lt;br&gt;&#xA; stack er ling &lt;br&gt;&#xA; stack ex ling &lt;br&gt;&#xA; stack ex line &lt;br&gt;&#xA; stack ex lane &lt;br&gt;&#xA; stack ex cane &lt;br&gt;&#xA; stack ex care &lt;br&gt;&#xA; stack ex chare &lt;br&gt;&#xA; stack ex charge &lt;br&gt;&#xA; stack ex change &lt;br&gt;&#xA; stack exchange &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13290" LastActivityDate="2015-08-17T22:44:47.673" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20200" PostTypeId="2" ParentId="19864" CreationDate="2015-08-17T23:51:18.827" Score="4" Body="&lt;p&gt;It looks like fingering instructions for some strange scales on a guitar. That would explain why&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;there's always six of the numbers,&lt;/li&gt;&#xA;&lt;li&gt;they can always be split into concatenated strictly rising finite sequences of small integers with small differences (like 91112 -&gt; 9 11 12),&lt;/li&gt;&#xA;&lt;li&gt;the individual entries do not exceed 12 by much,&lt;/li&gt;&#xA;&lt;li&gt;the fourth sequence is one element shorter.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It's also compatible with the provided hints. The only problem is that I tried playing them and they don't sound like anything usual. But that could simply be a nonconventional style or not the default tuning, perhaps even a different but related instrument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; You were playing guitar and practising the following scales (only first octave listed, then it repeats as high as fingers reach without leaving position):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;code&gt;5 8 5 6 7 5 7&lt;/code&gt;: &lt;em&gt;A C D E♭ E G A&lt;/em&gt;: hexatonic blues scale from A [&lt;a href=&quot;https://en.wikipedia.org/wiki/Blues_scale&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;]&lt;/li&gt;&#xA;&lt;li&gt;Lines 1 and 4 in &quot;Some more sequences&quot; (differing only by a constant shift of 5 positions, i.e., G-&gt;C): &lt;em&gt;G A B C D E♭ F G&lt;/em&gt;: &quot;major minor&quot; scale as described in [&lt;a href=&quot;https://en.wikipedia.org/wiki/Heptatonic_scale#Heptatonia_secunda&quot; rel=&quot;nofollow&quot;&gt;2&lt;/a&gt;]&lt;/li&gt;&#xA;&lt;li&gt;Lines 2, 3, 5, and 6 (ditto, from F, G, B♭, C, respectively): &lt;em&gt;F G A B C♯ D E F&lt;/em&gt;: &quot;Lydian raised 5th&quot; [&lt;a href=&quot;https://en.wikipedia.org/wiki/Heptatonic_scale#Heptatonia_secunda&quot; rel=&quot;nofollow&quot;&gt;2&lt;/a&gt;]&lt;/li&gt;&#xA;&lt;li&gt;Last line: &lt;em&gt;C♯ D E F♯ G♯ A♯ B C♯&lt;/em&gt;: &quot;Phrygian raised 6th&quot; [&lt;a href=&quot;https://en.wikipedia.org/wiki/Heptatonic_scale#Heptatonia_secunda&quot; rel=&quot;nofollow&quot;&gt;2&lt;/a&gt;]&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Alternatively, all the heptatonic scales could just be of one type (e.g., all major minor), just not starting from the base tone. &lt;strong&gt;The most likely explanation&lt;/strong&gt;, subjectively speaking, is that they are &quot;melodic ascending minor&quot; scales [&lt;a href=&quot;https://en.wikipedia.org/wiki/Heptatonic_scale#Heptatonia_secunda&quot; rel=&quot;nofollow&quot;&gt;2&lt;/a&gt;] from C, D, E, F, G, A, B, C (this would be the full C-major scale in order). But the above is a more precise formulation for as a matter or principle you wouldn't start a scale on something that is not its first tone.&lt;/p&gt;&#xA;" OwnerUserId="15749" LastEditorUserId="15749" LastEditDate="2015-08-21T01:06:55.523" LastActivityDate="2015-08-21T01:06:55.523" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20201" PostTypeId="2" ParentId="20196" CreationDate="2015-08-18T00:58:06.437" Score="1" Body="&lt;p&gt;You can&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; lean over, count the number of floors, then multiple it by the average height of each floor (roughly about 3 meters).&#xA; &lt;br&gt;&lt;br&gt; Admittedly, this method might not be accurate within 10%. It also rely on you being able to lean far enough to count the number of floors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12332" LastActivityDate="2015-08-18T00:58:06.437" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20202" PostTypeId="2" ParentId="20186" CreationDate="2015-08-18T01:09:57.827" Score="2" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Jantar Mantar monument of Jaipur, Rajasthan.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can come into us.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tourists are allowed to enter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can walk our stairs.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are stairs that are climbed to reach the instruments.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can click pictures with us, but no-one will care.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is no prohibition against taking photos.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If our real purpose is asked, we're meant to tell the time.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The instruments were used for astronomy and for predictions of what astronomical events will happen and when.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But not using us for that must be made a crime.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You must not use the monuments for any purpose except to visit and take pictures.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14177" LastActivityDate="2015-08-18T01:09:57.827" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20204" PostTypeId="2" ParentId="20188" CreationDate="2015-08-18T05:50:16.030" Score="9" Body="&lt;p&gt;First, construct a mountain range from the inequality symbols, where $&amp;lt;$ is an upslope and $&amp;gt;$ is a downslope. For example, given $$&#xA;x_1 &amp;lt; x_2&amp;gt;x_3&amp;gt;x_4&amp;gt;x_5&amp;lt;x_6&amp;lt;x_7&amp;gt;x_8&#xA;$$&#xA;you would make&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\qquad\qquad\qquad\qquad$&lt;img src=&quot;https://i.stack.imgur.com/RXiCR.png&quot; /&gt;&lt;br&gt;&#xA;Next, place the numbers $1$ to $n$ on the vertices of this drawing, filling the levels in order from lowest to highest:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\qquad\qquad\qquad\qquad$&lt;img src=&quot;https://i.stack.imgur.com/TYD4W.png&quot; /&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, read this off from left to right, getting the solution $5&amp;lt;8&amp;gt;6&amp;gt;2&amp;gt;1&amp;lt;3&amp;lt;7&amp;gt;4.$ When done smartly, this takes linear time.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-08-18T05:50:16.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20205" PostTypeId="2" ParentId="20154" CreationDate="2015-08-18T06:11:17.833" Score="6" Body="&lt;p&gt;Partial answer.  This cipher is a:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Vigenere cipher that decodes to (with spaces added):&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;strong&gt;UTM three two U five nine two eight six two five six four seven four eight seven&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;A six letter function describes this place that links to the baroque master&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;Find a similar link to stanza one though not of bricks and plaster&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;The Vigenere key is &lt;strong&gt;DMAJOR&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first line of plaintext is a:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;UTM coordinate in zone 32 at eastings 592862 and northings 5647487&lt;br&gt;&#xA;which converts to a WGS84 coordinate of 50.971488°N 10.322607°E&lt;br&gt;&#xA;This location is very close to &lt;a href=&quot;https://en.wikipedia.org/wiki/Bach_House_(Eisenach)&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Bach House&lt;/strong&gt;&lt;/a&gt; in Eisenach, Germany  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second line of plaintext refers to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Johann Sebastian Bach as the baroque master who was born in Eisenach&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-08-21T05:24:39.667" LastActivityDate="2015-08-21T05:24:39.667" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20207" PostTypeId="1" AcceptedAnswerId="20209" CreationDate="2015-08-18T08:05:03.350" Score="0" ViewCount="2180" Body="&lt;p&gt;I've been trying to solve this for quite a while, it's been about 1 week.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My friend sent this to me to solve it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please do help&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Xabj, lbhe yvzvgf, fgbc obgurevat zr, gunax lbh!!&lt;/p&gt;&#xA;" OwnerUserId="15789" LastEditorUserId="4977" LastEditDate="2015-08-18T11:48:38.290" LastActivityDate="2015-08-18T11:48:38.290" Title="&quot;Rearrange the alphabet&quot; puzzle" Tags="|word|cipher|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20208" PostTypeId="2" ParentId="20207" CreationDate="2015-08-18T08:48:36.173" Score="4" Body="&lt;p&gt;It is a substitution cipher which says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Know, your limits, stop bothering me, thank you&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12901" LastActivityDate="2015-08-18T08:48:36.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20209" PostTypeId="2" ParentId="20207" CreationDate="2015-08-18T08:57:37.673" Score="8" Body="&lt;p&gt;It's actually a very simple shift (caesar) cipher. Just shift all letters by 13 and you get your solution.&#xA;So when A = 1 and Z = 26&#xA;Add 13 to the letter and you get your solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Know, your limits, stop bothering me, thank you&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because it is 13 so exactly half of 26 decrypting and encrypting is exactly the same operation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can try it out here:&#xA;&lt;a href=&quot;http://www.xarg.org/tools/caesar-cipher/&quot;&gt;http://www.xarg.org/tools/caesar-cipher/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-08-18T08:57:37.673" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20210" PostTypeId="2" ParentId="20173" CreationDate="2015-08-18T09:33:01.380" Score="4" Body="&lt;p&gt;I believe the solution is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The two words &lt;strong&gt;predicament&lt;/strong&gt; and &lt;strong&gt;rotate&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Taking each stanza, one at a time:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;To me, you are but a cafe / Buffet; souffle; filet; parfait / I'll be the hunter, you be the prey / Slice me in twain, and leave my tail.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gives us &lt;strong&gt;predator&lt;/strong&gt; (since it hunts and eats its prey), which we slice in half, keeping &lt;em&gt;pred&lt;/em&gt; and discarding &lt;em&gt;ator&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Lower, lower, lower yet / submerged in water, never wet / Pick me and you will not regret / The leaders stay, the rest all bail.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gives &lt;strong&gt;ice&lt;/strong&gt; (since we &quot;lower&quot; the temperature to make icecubes/bergs which sit in water but are solid not liquid, and you can use &quot;pick&quot; on ice), of which we keep the leaders, &lt;em&gt;ic&lt;/em&gt;, and discard &lt;em&gt;e&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Treat me as you have my kin / Pin me down, and see my sins / All spill asunder from within / My heart bleeds, though my shell prevails.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gives us &lt;strong&gt;ATM&lt;/strong&gt; (not sure about the first line, but you use a PIN to withdraw money, which is the root of all evil), of which we keep the outer &quot;shell&quot; of letters, &lt;em&gt;am&lt;/em&gt;, and leave behind the &quot;heart&quot;, &lt;em&gt;t&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Rooted deep within the ground / A being of myth and tales unbound / The net's been cast, and I've been found / In full I stand, behind no veil.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gives &lt;strong&gt;&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Ent&quot; rel=&quot;nofollow&quot;&gt;Ent&lt;/a&gt;&lt;/strong&gt; (mythical tree creatures), which we keep the entirety of.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Take them all to find my name / A right mess made, and you're to blame&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Putting together all the parts we kept, gives us &lt;strong&gt;predicament&lt;/strong&gt; (which is a &quot;right mess made&quot;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Then take the rejects, cast in shame / Spin them around and they'll do the same.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Taking all the bits we discarded, &quot;ator&quot;, &quot;e&quot; &amp;amp; &quot;t&quot;, we &quot;spin them around&quot;, to get &lt;strong&gt;rotate&lt;/strong&gt; (which also means spin around).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastActivityDate="2015-08-18T09:33:01.380" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20213" PostTypeId="2" ParentId="20141" CreationDate="2015-08-18T09:48:32.580" Score="2" Body="&lt;p&gt;19 moves i think&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1.puzzling&lt;br&gt;&#xA; 2.muzzling&lt;br&gt;&#xA; 3.muzz ling&lt;br&gt;&#xA; 4.muzz lang&lt;br&gt;&#xA; 5.muzz hang&lt;br&gt;&#xA; 6.muzz chang&lt;br&gt;&#xA; 7.muzz change&lt;br&gt;&#xA; 8.fuzz change&lt;br&gt;&#xA; 9.fuze change&lt;br&gt;&#xA; 10.faze change&lt;br&gt;&#xA; 11.fake change&lt;br&gt;&#xA; 12.flake change&lt;br&gt;&#xA; 13.flaker change&lt;br&gt;&#xA; 14.flak er change&lt;br&gt;&#xA; 15.flak ex change&lt;br&gt;&#xA; 16.flak exchange&lt;br&gt;&#xA; 17.flack exchange&lt;br&gt;&#xA; 18.slack exchange&lt;br&gt;&#xA; 19.stack exchange  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user13083" LastActivityDate="2015-08-18T09:48:32.580" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20214" PostTypeId="2" ParentId="20163" CreationDate="2015-08-18T10:14:16.480" Score="10" Body="&lt;pre&gt;&lt;code&gt;o  o&#xA;oooo&#xA;oooo&#xA;o  o&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Based on @AlexHurst answer.&lt;/p&gt;&#xA;" OwnerUserId="15795" LastActivityDate="2015-08-18T10:14:16.480" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20215" PostTypeId="2" ParentId="20196" CreationDate="2015-08-18T10:56:45.717" Score="6" Body="&lt;p&gt;A different approach:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Stand at the railing on whichever side of the building you can see the shadow it casts on the ground. You should see your own shadow at the top. Estimate the proportion of the total building+prisoner shadow is caused by you. This can be done most accurately by holding the pendulum string at a distance from your body, parallel to the ground (creating similar triangles) so that the top and bottom of the string align with the top and bottom of the shadow. Then pinch the points where the top of the building is. Fold the string to determine the ratio. Use the shadow of the railings to work out how much of your legs is lost in the shadow of the edge of the roof. Use the known length of the pendulum string to measure this point on your leg. Subtract that from your own height (which you should know). Using the ratio you estimated earlier, scale up the remainder of your height to determine the height of the building. Use the heavy pendulum to smash the stopwatch to get a sharp piece of metal or plastic to pierce your skin so that you can use your blood to write the answer on the paper.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-08-18T10:56:45.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20216" PostTypeId="1" AcceptedAnswerId="20218" CreationDate="2015-08-18T11:31:56.623" Score="2" ViewCount="5525" Body="&lt;p&gt;&lt;em&gt;I know this puzzle is easy, but I hope it isn't too easy... If it is I really have to work on my puzzle-creation skills :P&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Story &amp;amp; Question:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;8 men plan to go eating to a restaurant, where they will celebrate Mr. Blue's birthday. Here is an example how it could look like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QVb1m.jpg&quot; alt=&quot;Image&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The above image is just for visualization!) They're still struggling with coming up with a solution so every man is satisfied with his position. Can you help them find (one of the four) solutions only given the preferences of each man? (I will only accept an answer that contains a picture (or code snippet) of the solution like the one above, see &quot;Tools&quot;).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Preferences:&lt;/h2&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Mr. Red wants to sit at a corner.&lt;/li&gt;&#xA;&lt;li&gt;Mr. Green wants to face Mr. Red&lt;/li&gt;&#xA;&lt;li&gt;Mr. Blue wants to have Mr. Orange as an immediate neighbor&lt;/li&gt;&#xA;&lt;li&gt;Mr. Pink doesn't want to sit at a corner&lt;/li&gt;&#xA;&lt;li&gt;Mr. Orange wants to have Mr. Blue as an immediate neighbor and wants to have Mr. Red as an immediate neighbor&lt;/li&gt;&#xA;&lt;li&gt;Mr. Yellow wants to face Mr. Magenta&lt;/li&gt;&#xA;&lt;li&gt;Mr. Magenta doesn't want to have Mr. Green as an immediate neighbor&lt;/li&gt;&#xA;&lt;li&gt;Mr. Cyan wants to face Mr. Blue&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Clarification: If you divide the image in 5 columns, the seats inside the first 2 and the last 2 columns are the seats that are at a &lt;strong&gt;corner&lt;/strong&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Tools:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I've provided 9 pictures, one of the table and one of each man with alpha channels. This will make it much easier for you to create/get to an answer. All you need is an image editing program. But like mentioned before, I also accept an illustration with text in a code sample, like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;           Red|Green|Blue&#xA;        ====================&#xA;        ====================&#xA;   Pink ==================== Orange&#xA;        ====================&#xA;        ====================&#xA;        Yellow|Magenta|Cyan&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But I think an image like I made would be nicer for visualization :)    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Images you can download:&lt;/strong&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Rq1M2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Table and seats&lt;/a&gt;, &lt;a href=&quot;https://i.stack.imgur.com/UMWk5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mr. Red&lt;/a&gt;, &lt;a href=&quot;https://i.stack.imgur.com/mZgOc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mr. Green&lt;/a&gt;, &lt;a href=&quot;https://i.stack.imgur.com/rOliY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mr. Blue&lt;/a&gt;, &lt;a href=&quot;https://i.stack.imgur.com/0RK3G.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mr. Pink&lt;/a&gt;, &lt;a href=&quot;https://i.stack.imgur.com/VO34k.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mr. Orange&lt;/a&gt;, &lt;a href=&quot;https://i.stack.imgur.com/pgv42.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mr. Yellow&lt;/a&gt;, &lt;a href=&quot;https://i.stack.imgur.com/CB1Xc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mr. Magenta&lt;/a&gt;, &lt;a href=&quot;https://i.stack.imgur.com/wDSi5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mr. Cyan&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user14478" LastEditorDisplayName="user14478" LastEditDate="2015-08-18T11:53:51.627" LastActivityDate="2015-08-18T12:48:45.143" Title="8 men in a restaurant" Tags="|logical-deduction|logic-grid|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20217" PostTypeId="2" ParentId="20216" CreationDate="2015-08-18T11:42:55.557" Score="5" Body="&lt;p&gt;The following seating arrangement will satisfy everyone&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;         Red|Orange|Blue&#xA;       ====================&#xA;       ====================&#xA;Yellow ==================== Magenta&#xA;       ====================&#xA;       ====================&#xA;         Green|Pink|Cyan&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;How I arrived at this solution: I have to admit that it wasn't the most logically sound method, but it helped me get to a solution right away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using clue 1, I arbitrarily put Mr. Red at the top left of the table. Clue 2 put Mr. Green opposite him. Using clues 3 and 5, Mr. Orange came to the middle top position and Mr. Blue to the right top position as Mr. Green already occupied the position two spaces to the other side of Mr. Red. With the top middle position occupied, Mr. Pink could only sit at the bottom middle position according to clue 4. Clue 9 put My. Cyan at the bottom right position. Now, with only two places left, Mr. Magenta had to go to the rightmost seat, because he didn't want to sit next to Mr. Green according to clue 7. That put Mr. Yellow in the leftmost seat.&lt;/p&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-08-18T11:48:11.903" LastActivityDate="2015-08-18T11:48:11.903" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20218" PostTypeId="2" ParentId="20216" CreationDate="2015-08-18T11:50:58.507" Score="4" Body="&lt;p&gt;I think there are 4 that satisfy everyone (unless I've made a mistake):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rU9Mm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rU9Mm.jpg&quot; alt=&quot;Solution1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mezUo.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mezUo.jpg&quot; alt=&quot;Solution2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VXCZ4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VXCZ4.jpg&quot; alt=&quot;Solution3&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HRqQr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HRqQr.jpg&quot; alt=&quot;Solution4&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for approach I figured starting with Pink was good since he only had 2 possible slots, and conveniently he made the Blue-Orange-Red trio work out to be opposite him (because if they weren't opposite him, then Blue couldn't be Opposite Cyan). At that point the rest are really easy to fill in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I've changed it to images. Hopefully I didn't mess it up in doing so.&lt;/p&gt;&#xA;" OwnerUserId="15800" LastEditorUserId="15800" LastEditDate="2015-08-18T12:24:16.163" LastActivityDate="2015-08-18T12:24:16.163" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20219" PostTypeId="1" AcceptedAnswerId="20220" CreationDate="2015-08-18T12:13:58.197" Score="-4" ViewCount="221" Body="&lt;p&gt;What is the most compact way to store the sentence &lt;code&gt;The quick brown fox jumps over the lazy dog&lt;/code&gt; without losing data?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whatever way you think of must be explained and the process should be reversible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not looking for a small way to store on a computer but for writing on paper.&lt;/p&gt;&#xA;" OwnerUserId="13983" LastEditorUserId="13983" LastEditDate="2015-08-18T12:29:39.640" LastActivityDate="2015-08-18T14:17:29.567" Title="The smallest compression" Tags="|optimization|" AnswerCount="2" CommentCount="8" ClosedDate="2015-08-18T13:05:42.733" ContentLicense="CC BY-SA 3.0" />
  <row Id="20220" PostTypeId="2" ParentId="20219" CreationDate="2015-08-18T12:24:23.160" Score="1" Body="&lt;p&gt;Each letter is one of 26 separate characters. We could use a base-32 (2^5) storage format to store the numbers in binary.&#xA;a would be 00001, b would be 00010, etc. This is assuming that you don't care about capitalization, which would double the required space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The algorithm would require 43 characters of code for the entire sentence because some letters are duplicated. 43 * 5 = 215 bits, or just under 27 bytes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To decrypt it we would simply grab 5 characters at a time and add the relevant values to their binary value to get their ascii codes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry, something's preventing me from using spoiler tags here.&lt;/p&gt;&#xA;" OwnerUserId="13068" LastActivityDate="2015-08-18T12:24:23.160" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20221" PostTypeId="2" ParentId="20216" CreationDate="2015-08-18T12:32:05.067" Score="1" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Two men facing each other are either both at a corner, or neither are. So both R and G are at a corner.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Given 5, O has to be between B and R. Two spots away from a corner is another corner, but B and R can't face each other like R and G, so B, O and R need to sit on the same edge.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Based on 6 and the previous inferences, Y and M both need to sit on the short edges to avoid having to face B, O or R.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Based on 8 and 6, a possible solution can be this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;         C|P|G&#xA;  ====================&#xA;  ====================&#xA;M ==================== Y&#xA;  ====================&#xA;  ====================&#xA;         B|O|R&#xA;&lt;/code&gt;&lt;/pre&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;. Basically, select any of the 4 corners for R, then put G on the opposite, B at the other corner on R's edge, O in the middle of it, C at the remaining corner, M on the short edge away from G, Y on the other short edge and P at the remaining spot.&lt;/p&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="14016" LastEditDate="2015-08-18T12:48:45.143" LastActivityDate="2015-08-18T12:48:45.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20222" PostTypeId="2" ParentId="20219" CreationDate="2015-08-18T13:00:07.027" Score="0" Body="&lt;h2&gt;Solution : change the base !&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Compression&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take each word as a number in base 26 (a=1, b=2 etc) you have a number. For example :&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;the = 20*26²+8*26+5=13733 (t=20, h=8 and e=5)&amp;lt;br&amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then you change the base. You can take base 255 (each value is an ascii character) and convert the number :&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;13733 = 53*255+218&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You can convert your new numbers in ascii (base 255), and you get&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;5┌&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You saves one character here, yayy !&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Decompression&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To decrypt the message you basically do the same thing reversed (from base 255 to base 26).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit :&lt;/strong&gt;&#xA;&lt;br&gt;I did it for the whole sentence and I found this : &lt;code&gt;5┌ }=☺ ‼§‗ ◄ç L↑; ♦A═ 5┌ ♥Cü ♀)&lt;/code&gt;. &#xA;It saves 12 characters.&#xA;&lt;hr&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Old answer&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I can apply a small optimization by removing spaces.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;TheQuickBrownFoxJumpOverTheLazyDog&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Every new word start with a upper-cased letter&lt;/p&gt;&#xA;" OwnerUserId="13384" LastEditorUserId="13384" LastEditDate="2015-08-18T14:17:29.567" LastActivityDate="2015-08-18T14:17:29.567" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20225" PostTypeId="1" CreationDate="2015-08-18T13:59:52.723" Score="11" ViewCount="2585" Body="&lt;p&gt;Following is a list of lines; each line contains two riddles written by my pal Albert Iteration. Each riddle describes a single word. Your task is to identify the words being described and find what the two words in each line have in common (if there are multiple answers, choose the largest). They're in no particular order, there are easy ones sprinkled in there but I think this one shouldn't last too long on here.&lt;br&gt;&#xA;For clarity, I have added an asterisk to lines that still need solving. As progress is made, I will remove the asterisks&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; system's subdivision       | deviation; death  &#xA; lobotomee's loss           | employed entities  &#xA; crystal clear              | overtly obvious  &#xA; handsy humanoid            | atmospheric ambience  &#xA;*headdress holiday          | chromosomal cross-up  &#xA; lacking logic              | evoking enthusiasm            &#xA;*incumbent inhabitant       | ascent apex  &#xA;*cache contribution         | simultaneous states  &#xA;*multi-tasking mycota       | mammals' mark  &#xA; drink denier               | absolute autocracy  &#xA; sunday solver              | sage's saying  &#xA; successfully settle        | pardon, papally  &#xA; not normal                 | kanye's kin  &#xA; incorrectly interpreted    | college kid  &#xA; material metamorphosis     | knowledge; news  &#xA; celestial species          | bachelor brotherhood  &#xA;*wowing wizardry            | hearty herb  &#xA; similar starts             | repeated recounting  &#xA; formal familiarization     | orchestral outlier  &#xA; supremely significant      | placebo peddler  &#xA; immeasurably insignificant | certainly so  &#xA;*gun; gum                   | worthless whistle  &#xA;*meagerly mindful           | chitined crustacean  &#xA;*supervised span            | pulsate periodically  &#xA; wrestle; wrangle           | fragmented fruit  &#xA; prophetic portender        | ceremonial start  &#xA; replicate ravenously       | water ward  &#xA;*palpable proximity         | person picked  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And if you really want to impress, I think it would be fun if you could come up with your own two word riddles in Al's style for each answer. This is not necessary to get the green check, however. Alright have fun!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;And of course, when it comes to answers with multiple parts like this one, partial answers are always welcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clarification for those who know the gimmick/pattern:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;All of the shared words will be relatively common English words (though one is a little more archaic) that can be found on dictionary.com. There are no proper nouns and no words shorter than 3 letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="4050" LastEditDate="2015-08-27T12:45:10.513" LastActivityDate="2015-08-27T12:45:10.513" Title="What do these things have in common? (Albert Iteration)" Tags="|riddle|word|pattern|cryptic-clues|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20226" PostTypeId="2" ParentId="20140" CreationDate="2015-08-18T14:44:45.853" Score="0" Body="&lt;p&gt;OK, continuing the research I want to tell some facts:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Before permutating (rearranging) columns with key, free squares are filed with letter X, for example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    K   E   Y               K   E   Y&#xA;1   F   A   G          1    F   A   G&#xA;2   V   D   F    &amp;lt;=&amp;gt;   2    V   D   F&#xA;3   F   A   A          3    F   A   A&#xA;4   F                  4    F   X   X&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This means the final encrypted text length will be divisible by key length.&lt;br&gt;&#xA;In this example length is 12 and key has 3 letters.&lt;br&gt;&#xA;Now we our ciphertext's length is 48&lt;br&gt;&#xA;&lt;code&gt;VDFVXDVADXVAAVGAGXAVAVGGGGXAFGGAXGAGVVXFGAVDXGDX&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;123456789012345678901234567890123456789012345678&lt;/code&gt;&lt;br&gt;&#xA;It means we have several possible key lengths, as 48 is divisible by:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1, 2,             - Hardly possible&#xA;3, 4, 6, 8, 12,   - Possible&#xA;16, 24, 48        - Hardly possible&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Of which most probably key length is 3, 4, 6, 8, 12.&lt;br&gt;&#xA;Maybe now we should find words with this lengths:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;JohnsRevenge - 12  &#xA;John - 4&#xA;Kslkgh ?- 6&#xA;ADFGVX - 6&#xA;etc&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;2) Empty boxes are filled with X. It means that last &lt;strong&gt;row&lt;/strong&gt; can have more X-es, because count of X on last row doesn't change even after permutation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also probablity of X* pair the least, because usually all unused letters/numbers are in X row (in non-permutated ciphertext). &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    A   D   F   G   V   X&#xA;    ...&#xA;V   B   N   0   1   2   3 &#xA;X-&amp;gt; 4   5   6   7   8   9  - symbols in this row are used rarely, so X* is rare &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Back to the game. When trying all possible key lengthes (write cipher VDFVX... upside down, left to right, in 3, 4, 6, 8, 12 columns) the more X-es we have on the last row the more possibility is that we have found the correct key length. (It's about possibility, not 100% correct)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Not a fact but interesting:&lt;br&gt;&#xA;&lt;code&gt;@Kslkgh, I think I know the cipher and I think I know the key, but I can't for the life of me figure out what the keyword is...am I on the right track?&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;@BaileyM Do you mean that you know what the -&amp;gt;keysquare&amp;lt;- is. If you do, you are definately on the right track! –  Kslkgh&lt;/code&gt;&lt;br&gt;&#xA;Seems like he the OP reacted more on keysquare (and doesn't care about permutation key?). So probably key is obvious/easy to find.&lt;br&gt;&#xA;Also &lt;code&gt;The key is hidden in a -&amp;gt;method&amp;lt;- I've used on various occasions –  Kslkgh&lt;/code&gt; again makes me think the key is method name- ADFGVX (also it's length is equal to 6, and 48 is divisible by 6).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;TO BE CONTINUED...&lt;/p&gt;&#xA;" OwnerUserId="15786" LastEditorUserId="15786" LastEditDate="2015-08-18T15:53:15.263" LastActivityDate="2015-08-18T15:53:15.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20227" PostTypeId="1" AcceptedAnswerId="20239" CreationDate="2015-08-18T16:05:00.763" Score="6" ViewCount="1763" Body="&lt;p&gt;You cannot accurately predict my arrival.&lt;br&gt;&#xA;I seem to appear randomly out of nowhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without a word, I compel you into action.&lt;br&gt;&#xA;My demands upon you are always the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose to ignore me, and I'll annoy you.&lt;br&gt;&#xA;You have a strong desire to obey my will.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do visit some people more than others,&lt;br&gt;&#xA;But all will deal with me at some point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once obeyed, I usually leave for awhile,&lt;br&gt;&#xA;But sometimes I reappear nearby for more.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What Commander Am I?&lt;/strong&gt;&lt;br&gt;&#xA;&lt;sub&gt;The answer must explain (and fit with) all of the lines&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastActivityDate="2021-12-08T03:45:52.467" Title="What Commander Am I?" Tags="|riddle|word|" AnswerCount="8" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20228" PostTypeId="2" ParentId="20227" CreationDate="2015-08-18T16:08:50.087" Score="1" Body="&lt;p&gt;First attempt!:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://rayn44.deviantart.com/art/Sleepy-commander-478524496&quot; rel=&quot;nofollow&quot;&gt;The Sleepy Commander&lt;/a&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You cannot accurately predict my arrival.&lt;br&gt;&#xA;I seem to appear randomly out of nowhere.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You might usually sleep at night but you will get sleepy all of a sudden  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without a word, I compel you into action.&lt;br&gt;&#xA;My demands upon you are always the same.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Once you're sleepy you will want to go the sleep  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Choose to ignore me, and I'll annoy you.&lt;br&gt;&#xA;You have a strong desire to obey my will.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The more you struggle the more you want to sleep.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I do visit some people more than others,&lt;br&gt;&#xA;But all will deal with me at some point.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; some people need more sleep than others  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once obeyed, I usually leave for awhile,&lt;br&gt;&#xA;but sometimes I reappear nearby for more.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Once you obey (get some sleep), you will be fine until you're sleepy again.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2015-08-18T16:08:50.087" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20229" PostTypeId="2" ParentId="20140" CreationDate="2015-08-18T16:14:10.140" Score="5" Body="&lt;p&gt;The solution is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; youmustdrinkfromglasssix&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first step is to find the keysquare:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When looking &quot;deeper&quot; you can find it in the source of the question (click on &quot;edit&quot;): 20zsd5vt1wjipe4hfnmq8gbxy3lou6ra7k9c&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't know if the rest of the solution is allowed (I'm new here):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I have written a program to brute force all possible transpositions, and looked for the word &quot;glass&quot; in the decoded text. The numeric representation of the transposition key is [6, 3, 1, 4, 0, 7, 5, 2]. This matches the word &quot;MECHANIC&quot; (thanks to moonbutt74).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Source code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import java.nio.CharBuffer;&#xA;import java.util.Arrays;&#xA;import java.util.regex.Pattern;&#xA;&#xA;public class JohnsRevenge {&#xA;    private static final int MAX_TRANSPOSITION_KEY_LENGTH = 10;&#xA;    private static final char[] KEYSQUARE = &quot;20zsd5vt1wjipe4hfnmq8gbxy3lou6ra7k9c&quot;.toCharArray();&#xA;    private static final char[] CIPHER_TEXT = &quot;VDFVXDVADXVAAVGAGXAVAVGGGGXAFGGAXGAGVVXFGAVDXGDX&quot;.toCharArray();&#xA;    private static final Pattern PATTERN = Pattern.compile(&quot;glass&quot;);&#xA;&#xA;    private static int[] transpositionKey = null;&#xA;    private static int[] inverseTranspositionKey = null;&#xA;    private static char[] untransposed = new char[CIPHER_TEXT.length];&#xA;    private static char[] decoded = new char[CIPHER_TEXT.length / 2];&#xA;    private static CharBuffer buffer = CharBuffer.wrap(decoded);&#xA;&#xA;    private static void untranspose() {&#xA;        int width = inverseTranspositionKey.length;&#xA;        int height = (CIPHER_TEXT.length + width - 1) / width;&#xA;        int heightChange = CIPHER_TEXT.length - (height - 1) * width;&#xA;&#xA;        int cipherTextPos = 0;&#xA;        for (int x : inverseTranspositionKey) {&#xA;            int currentHeight = (x &amp;lt; heightChange ? height : height - 1);&#xA;            for (int y = 0; y &amp;lt; currentHeight; ++ y) {&#xA;                untransposed[x + y * width] = CIPHER_TEXT[cipherTextPos++];&#xA;            }&#xA;        }&#xA;    }&#xA;&#xA;    private static int charToIndex(char c) {&#xA;        switch (c) {&#xA;            case 'A': return 0;&#xA;            case 'D': return 1;&#xA;            case 'F': return 2;&#xA;            case 'G': return 3;&#xA;            case 'V': return 4;&#xA;            case 'X': return 5;&#xA;            default: throw new IllegalArgumentException();&#xA;        }&#xA;    }&#xA;&#xA;    private static void decode() {&#xA;        for (int i = 0; i &amp;lt; decoded.length; ++ i) {&#xA;            decoded[i] = KEYSQUARE[charToIndex(untransposed[i * 2]) * 6 + charToIndex(untransposed[i * 2 + 1])];&#xA;        }&#xA;    }&#xA;&#xA;    private static void check() {&#xA;        if (PATTERN.matcher(buffer).find()) {&#xA;            System.out.println(buffer + &quot; &quot; + Arrays.toString(transpositionKey));&#xA;        }&#xA;    }&#xA;&#xA;    private static void generateKeys(int pos) {&#xA;        outer:&#xA;        for (int i = 0; i &amp;lt; transpositionKey.length; ++ i) {&#xA;            for (int j = 0; j &amp;lt; pos; j ++) {&#xA;                if (i == transpositionKey[j]) {&#xA;                    continue outer;&#xA;                }&#xA;            }&#xA;            transpositionKey[pos] = i;&#xA;            inverseTranspositionKey[i] = pos;&#xA;            if (pos &amp;lt; transpositionKey.length - 1) {&#xA;                generateKeys(pos + 1);&#xA;            } else {&#xA;                untranspose();&#xA;                decode();&#xA;                check();&#xA;            }&#xA;        }&#xA;    }&#xA;&#xA;    public static void main(String[] args) {&#xA;        for (int i = 1; i &amp;lt;= MAX_TRANSPOSITION_KEY_LENGTH; ++ i) {&#xA;            System.out.println(&quot;[&quot; + i + &quot;]&quot;);&#xA;            transpositionKey = new int[i];&#xA;            inverseTranspositionKey = new int[i];&#xA;            generateKeys(0);&#xA;        }&#xA;    }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2015-08-18T19:33:43.310" LastActivityDate="2015-08-18T19:33:43.310" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20230" PostTypeId="2" ParentId="20227" CreationDate="2015-08-18T16:29:49.210" Score="4" Body="&lt;p&gt;Haha I doubt this is the answer, but it fits rather amusingly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Smartphone system update notifications!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You cannot accurately predict my arrival.&lt;br&gt;&#xA;I seem to appear randomly out of nowhere.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Nobody knows when the update notification will appear!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without a word, I compel you into action.&lt;br&gt;&#xA;My demands upon you are always the same.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;No words necessary, just that darn symbol at the top of your notification bar that sits there FOREVER. The demand, obviously, is that you perform your system update.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Choose to ignore me, and I'll annoy you.&lt;br&gt;&#xA;You have a strong desire to obey my will.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Seriously, it's super annoying. At least on my phone, I can't swipe it away or clear it from my notifications unless I perform the actual update.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I do visit some people more than others,&lt;br&gt;&#xA;But all will deal with me at some point.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Some phones probably require system updates more than others, but anybody with a smartphone will need to do it at some point. Although maybe iPhones do it automatically, I wouldn't know.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once obeyed, I usually leave for awhile,&lt;br&gt;&#xA;But sometimes I reappear nearby for more.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;After you actually perform the update, the notification disappears. Alas, I am unable to make the last line fit, specifically the &quot;nearby&quot; part.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10861" LastActivityDate="2015-08-18T16:29:49.210" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20231" PostTypeId="2" ParentId="20227" CreationDate="2015-08-18T16:30:16.010" Score="0" Body="&lt;p&gt;You could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; my diaphragm (of the respiratory system)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You cannot accurately predict my arrival.&#xA;I seem to appear randomly out of nowhere.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most people can't predict an onset of the hiccups, yawns, and sneezes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without a word, I compel you into action.&#xA;My demands upon you are always the same.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The spasms begin suddenly without warning. All hiccups, yawns, and sneezes are quite similar.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Choose to ignore me, and I'll annoy you.&#xA;You have a strong desire to obey my will.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hiccups are involuntary (and yawns somewhat), and ignoring them will do no good. It can be hard to resist yawns and sneezes, as the urge sometimes lingers well after resisting.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I do visit some people more than others,&#xA;But all will deal with me at some point.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people get hiccups more often than others, but everybody gets them at some point. People who are tired more will yawn more. People with allergies, for example, sneeze more than others.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once obeyed, I usually leave for awhile,&#xA;But sometimes I reappear nearby for more.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For yawns, &quot;once obeyed&quot; must mean sleeping. For sneezing, &quot;once obeyed&quot; must refer to succumbing to the urge. For hiccups, they leave as mysteriously as they appeared, but both can return again suddenly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13237" LastEditorUserId="13237" LastEditDate="2015-08-18T16:51:00.347" LastActivityDate="2015-08-18T16:51:00.347" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20232" PostTypeId="2" ParentId="20227" CreationDate="2015-08-18T16:30:40.197" Score="8" Body="&lt;p&gt;Inspired by Alex's answer, I think the commander refers to a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yawn. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You cannot accurately predict my arrival.&lt;br&gt;&#xA;I seem to appear randomly out of nowhere.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You never know when you will yawn.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without a word, I compel you into action.&lt;br&gt;&#xA;My demands upon you are always the same.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The yawn forces your mouth wide without making the slightest sound. The demands of a yawn are always the same-&quot;Go to sleep NOW!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Choose to ignore me, and I'll annoy you.&lt;br&gt;&#xA;You have a strong desire to obey my will.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you keep ignoring your yawns, they become more intense and frequent. And as you yawn more and more, the desire to sleep increases further. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I do visit some people more than others,&lt;br&gt;&#xA;But all will deal with me at some point.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Those who haven't got enough sleep at night will yawn more than others. As night rolls around, almost everyone gets a yawn or two before they take to their bed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once obeyed, I usually leave for awhile,&lt;br&gt;&#xA;But sometimes I reappear nearby for more.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Once you catch some sleep, you will stop yawning. But if someone sees you yawning, they will also inadvertently start yawning. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-08-19T05:24:18.970" LastActivityDate="2015-08-19T05:24:18.970" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20233" PostTypeId="1" AcceptedAnswerId="20243" CreationDate="2015-08-18T16:34:55.403" Score="568" ViewCount="69817" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Left alone, I'm a word with five letters.&lt;/p&gt;&#xA;&lt;p&gt;I'm honest and fair, I'll admit.&lt;/p&gt;&#xA;&lt;p&gt;Rearranged, I'm of no use to trains.&lt;/p&gt;&#xA;&lt;p&gt;Again, and I'm an overt place, warm and well lit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;What am I?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="13237" LastEditorUserId="51105" LastEditDate="2021-10-11T01:46:16.823" LastActivityDate="2021-11-06T12:31:40.707" Title="A short, brutal riddle" Tags="|riddle|word|rhyme|acrostic|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="20234" PostTypeId="2" ParentId="20227" CreationDate="2015-08-18T16:38:53.210" Score="2" Body="&lt;p&gt;Continuing on this &quot;involuntary physical reflex&quot; trend:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think it's a sneeze.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You cannot accurately predict my arrival.&#xA;I seem to appear randomly out of nowhere.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A sneeze is not predictable; even if you're trying to provoke it with pepper or dust, it cannot be &lt;em&gt;accurately&lt;/em&gt; determined when the sneeze will occur.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without a word, I compel you into action.&#xA;My demands upon you are always the same.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not much to do but to let go and sneeze.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Choose to ignore me, and I'll annoy you.&#xA;You have a strong desire to obey my will.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can try to hold it back; pinch your nose as such, but the need reappears shortly after.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I do visit some people more than others,&#xA;But all will deal with me at some point.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Due to allergies, colds, etc, some people sneeze a lot more than others; and everyone sneezes at some point.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once obeyed, I usually leave for awhile,&#xA;But sometimes I reappear nearby for more.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Usually a good sneeze gets rid of the need, but double sneezes are not uncommon either.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15809" LastEditorUserId="15809" LastEditDate="2015-08-18T17:01:56.987" LastActivityDate="2015-08-18T17:01:56.987" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20235" PostTypeId="2" ParentId="20227" CreationDate="2015-08-18T16:41:43.000" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Hunger, thirst, addiction, and the voices in your head all seem to fit but I'm going to go with Hunger.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You cannot accurately predict my arrival.&#xA; I seem to appear randomly out of nowhere.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Every time somebody shows up with good food, you're suddenly hungry. Any attempt to fit with a food schedule is going to be hard, even if you're at an office job.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without a word, I compel you into action.&#xA; My demands upon you are always the same.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Feed! Well, your rumbling stomach sure doesn't pronounce it well, even if that is the command.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Choose to ignore me, and I'll annoy you.&#xA; You have a strong desire to obey my will.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;An empty stomach is very annoying, and hard to resist.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I do visit some people more than others,&#xA; But all will deal with me at some point.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Some people need to eat more, and nobody can go without food completely (yet.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once obeyed, I usually leave for awhile,&#xA; But sometimes I reappear nearby for more.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You might not be hungry after you eat, but if anyone sees what you're eating, well, they might get very hungry.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13068" LastActivityDate="2015-08-18T16:41:43.000" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20236" PostTypeId="2" ParentId="19986" CreationDate="2015-08-18T17:01:18.690" Score="0" Body="&lt;p&gt;For the bonus, going off of d'alar'cop's answer, I'm guessing it's &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Faraday cage. This answers the literal question, but doesn't directly use words similar to the law. Um... produces words similar to the law, even if it doesn't contain them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15810" LastActivityDate="2015-08-18T17:01:18.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20238" PostTypeId="1" AcceptedAnswerId="20240" CreationDate="2015-08-18T17:13:52.227" Score="23" ViewCount="5360" Body="&lt;p&gt;&lt;sub&gt;This is a puzzle that was a fad when I was back in school. (It's not sooo long ago, but way before Smartphones with AngryBirds or DoodleJump came up...). For quite a while, everybody was scribbling this instead of paying attention in class. It is somehow addicting (&quot;This can't be &lt;em&gt;so&lt;/em&gt; hard...&quot;), and can be played everywhere and everytime, just using a pen and a piece of squared paper. In the end, when I solved it (with the help of a computer, to be honest), I wrote the solution on a piece of paper, and still carry it around in my wallet (although I don't know why).&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rules are simple: Given a grid of 10x10 squares, your task is to visit every square exactly once. In each step, you may &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;skip 2 squares horizontally or vertically or&lt;/li&gt;&#xA;&lt;li&gt;skip 1 square diagonally&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;as in this image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MW0GO.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MW0GO.png&quot; alt=&quot;Puzzle01.png&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The starting point is arbitrary (and I don't know whether it matters, but at least I can say that it is possible to solve it when starting with the sequence depicted in the image).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There certainly are many solutions, but it's definitely harder than it looks. There is a way to simplify it (maybe an obvious one for an experienced puzzler?), but this will be part of a hint, if it turns out to be necessary. Otherwise, you may have a nice pastime (for when the battery of your Smartphone is empty...)&lt;/p&gt;&#xA;" OwnerUserId="15731" LastEditorUserId="12700" LastEditDate="2015-08-19T06:40:27.280" LastActivityDate="2018-11-20T15:12:02.727" Title="Explore the square with 100 hops" Tags="|checkerboard|mazes|" AnswerCount="7" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20239" PostTypeId="2" ParentId="20227" CreationDate="2015-08-18T17:21:01.537" Score="12" Body="&lt;p&gt;You are...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; AN ITCH!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You cannot accurately predict my arrival.&#xA;I seem to appear randomly out of nowhere.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Itches come randomly and from nowhere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without a word, I compel you into action.&#xA;My demands upon you are always the same.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You get no verbal commands but the command is always the same, SCRATCH ME!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Choose to ignore me, and I'll annoy you.&#xA;You have a strong desire to obey my will.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You gotta scratch that itch!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I do visit some people more than others,&#xA;But all will deal with me at some point.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people are more itchy than others, but everyone itches&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once obeyed, I usually leave for awhile,&#xA;But sometimes I reappear nearby for more.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Usually when you itch it goes away, but sometimes it just &lt;a href=&quot;https://en.wikipedia.org/wiki/Referred_itch&quot;&gt;makes you itchy elsewhere&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15719" LastEditorUserId="463" LastEditDate="2015-08-18T17:23:24.647" LastActivityDate="2015-08-18T17:23:24.647" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20240" PostTypeId="2" ParentId="20238" CreationDate="2015-08-18T17:47:52.847" Score="12" Body="&lt;p&gt;I must admit that I have heard of this before, and I remember the right way of approaching this puzzle.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My tactic was to make four 5x5 squares, because that will make it much easier. After spending some time with trial &amp;amp; error (with some kind of logic although I didn't bruteforce it), I came up with this solution (Probably not yours, but I think it's valid):&lt;br&gt;&lt;br&gt;&lt;a href=&quot;https://i.stack.imgur.com/p5nJp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/p5nJp.png&quot; alt=&quot;Solution Screenshot&quot;&gt;&lt;/a&gt;&lt;br&gt;&lt;br&gt; Edit: After reading Anachors answer I realised his &quot;style&quot; was way nicer to read, so I also changed my image.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user14478" LastEditorUserId="9000" LastEditDate="2015-08-18T20:33:27.493" LastActivityDate="2015-08-18T20:33:27.493" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20241" PostTypeId="2" ParentId="19969" CreationDate="2015-08-18T17:49:16.537" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Doctor, specifically from the newer Doctor Who series. The first 3 blocks are catch phrases and companions. His name is his promise.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Piece-by piece:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During the square, it was a quite common thing that he said. And lets not forget to chastise the howling canine(of sorts).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;9 is a square, &quot;Fantastic!&quot; is the common thing. Bad Wolf/Rose is the &quot;howling canine&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;During the third to six and eight, it was in French. And out goes the flower and in with the secretary and Shakespeare's friend.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;10 could be third in this sequence (6, 8, 10)? I'm least confident about that part. &quot;Allons-y&quot; was 10's catchphrase, in French. Rose, Donna, and Martha are the flower, secretary (temp), and &quot;Shakespeare's friend&quot;, respectively.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;During the odious number, if you added a certain cheese, it became a mouse. Impossibility works quite well for a mnemonic device. Also, remember the stagnant water left years past.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;11 is an odious number (It has an odd number of 1s when written in binary). &quot;Geronimo&quot; is the catchphrase, Stilton is a cheese and Geronimo Stilton is a mouse from a series of childrens' books. Clara is &quot;The Impossible Girl&quot; and Amy Pond is &quot;stagnant water left years past&quot;, a combination of her name and ultimate fate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once, arguably, a perfect trinity: three as one and one as three. But along came the dinosaur, and coughed up a little something and then the three became four.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Twelfth Doctor is the fourth of the new series. His first full episode had the Tardis swallowed and spat out by a dinosaur.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15814" LastActivityDate="2015-08-18T17:49:16.537" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20243" PostTypeId="2" ParentId="20233" CreationDate="2015-08-18T18:06:30.620" Score="608" Body="&lt;p&gt;I believe the answer is  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;LIAR  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The riddle was tricky (as indicated by the title) and each of the lines is  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;a lie.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Left alone, I'm a word with five letters.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The word is actually 4 letters  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm honest and fair I'll admit  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Obviously, a liar is the antithesis of honest and fair.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rearranged, I'm of no use to trains.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;RAIL, an anagram of LIAR, is an incredibly important facet to the function of trains.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again, and I'm an overt place, warm and well-lit.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;LAIR, another anagram of LIAR, is a hidden, dark place, the opposite of the adjectives provided.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I could be way off track but I have a good feeling that this is it.&lt;/p&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="13970" LastEditDate="2015-08-18T18:07:16.077" LastActivityDate="2015-08-18T18:07:16.077" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="20244" PostTypeId="2" ParentId="20238" CreationDate="2015-08-18T19:26:00.907" Score="8" Body="&lt;p&gt;This is similar to the other solution, but uses a slightly different trick. Call a path which traverses all squares a complete path.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Instead of generating four $5 \times 5$ complete path, we generate a single complete &quot;circular&quot; path ie. one that traverses all squares and returns to the starting cell. By doing this we can make complete paths from each starting cell by properly &quot;rotating&quot; the circular path. In fact, we can generate 2 complete paths, one moving forwards, one backwards. &lt;br&gt;&lt;br&gt;&lt;a href=&quot;https://i.stack.imgur.com/tj1Sh.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tj1Sh.png&quot; alt=&quot;Circular 5x5 path&quot;&gt;&lt;/a&gt;&#xA;&lt;br&gt;&#xA;  Note that you just have to jump from 25 to 1 to make a circular path.&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;  Rotating this path properly we have the solution:&lt;br&gt;&lt;br&gt;&#xA;&#xA;  &lt;a href=&quot;https://i.stack.imgur.com/PLrFR.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PLrFR.png&quot; alt=&quot;10x10&quot;&gt;&lt;/a&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;  The lower right grid uses the backward path while other three move forwards.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15544" LastEditorUserId="15544" LastEditDate="2015-08-18T19:41:32.720" LastActivityDate="2015-08-18T19:41:32.720" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20245" PostTypeId="1" AcceptedAnswerId="20510" CreationDate="2015-08-18T19:26:12.713" Score="17" ViewCount="844" Body="&lt;p&gt;I wrote a beautiful, verbose poem for you all, but &lt;a href=&quot;https://puzzling.stackexchange.com/questions/19806/tartar-sauce-gary&quot;&gt;Riddler&lt;/a&gt; went and scrambled the whole thing up! Can you figure out the order of the lines and put it back together again for me? And while you're at it, can you discover what the poem was about?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Else you'll find them quite well-guarded&lt;br&gt;&#xA;  A drinker who is called the same&lt;br&gt;&#xA;  An orphan, cinematically&lt;br&gt;&#xA;  A pound of text with speech not florid&lt;br&gt;&#xA;  Swedish metal takes this name&lt;br&gt;&#xA;  When skip was blue, at birth he wore it&lt;br&gt;&#xA;  You count on them to win the game&lt;br&gt;&#xA;  Written rule, with one before it&lt;br&gt;&#xA;  And bywords of reality&lt;br&gt;&#xA;  Historic sacred melody&lt;br&gt;&#xA;  Dolly sang it, we adored it&lt;br&gt;&#xA;  A young pig's tag, an all-time great&lt;br&gt;&#xA;  Concrete responsibility&lt;br&gt;&#xA;  Some adorned and openhearted&lt;br&gt;&#xA;  A certain group has now departed&lt;br&gt;&#xA;  History doth these create&lt;br&gt;&#xA;  A father of the highest fame&lt;br&gt;&#xA;  Together here we rule your fate&lt;br&gt;&#xA;  With NASA's help, it all was started&lt;br&gt;&#xA;  In others magic worlds await&lt;/em&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A hint that's not really a hint:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Focus on figuring out what the lines are referencing first. Once you do that, discovering the true order of the lines should be trivial. Also, since I said it in a comment, I'll repeat it here: Each line &lt;em&gt;represents something different&lt;/em&gt;. Kind of.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A hint that's really a hint:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are a couple of lines that should be possibly to figure out without knowing the pattern or the order -- I'd reckon that &quot;A young pig's tag, an all-time great&quot; and &quot;With NASA's help, it all was started&quot; are the two easiest. If you can figure those two out, you should have no problem figuring out the overarching theme and working backwards from there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-08-25T17:12:21.780" Title="A crumbled, jumbled poem" Tags="|word|rhyme|wordplay|knowledge|poetry|" AnswerCount="2" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="20247" PostTypeId="2" ParentId="19806" CreationDate="2015-08-18T20:08:29.487" Score="0" Body="&lt;p&gt;Based on the hints, I think Line 5 is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The two of spades, from a new deck dealt  = 51 (dealing from the top, the 2 of spades is the 51st card in the deck.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Line 6 is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; But oh, wait, now by one-third less. = 34 (2/3 of 51)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Together, these two might be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Subway stops? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2505" LastActivityDate="2015-08-18T20:08:29.487" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20248" PostTypeId="1" AcceptedAnswerId="20251" CreationDate="2015-08-18T20:14:58.093" Score="13" ViewCount="1179" Body="&lt;p&gt;A quick little riddle about a peculiar object. See how quickly you can figure it out based on the mysterious clues below. The correct answer will be awarded to the first person to identify the object and explain how each line fits&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You can find me underwater, but I won't be wet.&lt;br&gt;&#xA;  You can find me in the air too, but I can't cross in between.&lt;br&gt;&#xA;  In either case, I charge to my death, spilling my guts.&lt;br&gt;&#xA;  If you touch me, I will die but if you carry me, you will die.&lt;br&gt;&#xA;  If you guess the wrong thing, the OP will have to kill me.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13970" LastActivityDate="2015-08-18T20:32:05.027" Title="A children's plaything" Tags="|riddle|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20249" PostTypeId="2" ParentId="20248" CreationDate="2015-08-18T20:31:07.550" Score="4" Body="&lt;p&gt;Is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;a bubble?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can find me underwater, but I won't be wet.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;perhaps an air bubble&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can find me in the air too, but I can't cross in between.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Bubbles float in the air but don't last &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In either case, I charge to my death, spilling my guts.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;pop!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you touch me, I will die but if you carry me, you will die.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I think this might be some sort of phrase not in my language.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you guess the wrong thing, the OP will have to kill me. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Well, if I'm wrong, at least one of the Powerpuff Girls is safe!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13068" LastActivityDate="2015-08-18T20:31:07.550" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20250" PostTypeId="2" ParentId="20248" CreationDate="2015-08-18T20:32:00.067" Score="5" Body="&lt;p&gt;Could you be a...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; bubble?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can find me underwater, but I won't be wet.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; bubbles are not wet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can find me in the air too, but I can't cross in between.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; soapbubbles can exist in air, but an underwater bubble will cease to exist when not surrounded by water&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In either case, I charge to my death, spilling my guts.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; bubbles charges to the surface and pop&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you touch me, I will die but if you carry me, you will die.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if you touch a bubble it will die. air bubbles carried in your blood flow can be very hazardous.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you guess the wrong thing, the OP will have to kill me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; i don't know about this one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12901" LastActivityDate="2015-08-18T20:32:00.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20251" PostTypeId="2" ParentId="20248" CreationDate="2015-08-18T20:32:05.027" Score="12" Body="&lt;p&gt;The object described is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A bubble&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can find me underwater, but I won't be wet.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A bubble underwater will be dry inside.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can find me in the air too, but I can't cross in between.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bubbles can also float in the air, but they cannot burst through the surface of the water&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In either case, I charge to my death, spilling my guts&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eventually bubbles of any sort will burst, releasing the gas inside them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you touch me, I will die but if you carry me, you will die.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you try to touch a bubble it will usually burst, but if you have a bubble in your blood it will kill you&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you guess the wrong thing, the OP will have to kill me. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If This guess is wrong, then @NeedAName will have to burst my bubble&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Children like to play with bubbles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-08-18T20:32:05.027" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20252" PostTypeId="2" ParentId="19807" CreationDate="2015-08-18T21:56:26.107" Score="7" Body="&lt;p&gt;Twenty virus samples are necessary. A chessboard can be divided into two diagonal meshes with no diagonal adjacencies between them, and the virus can't spread from one to the other just like a bishop can only reach half the squares, so we can consider only the &quot;black&quot; squares and use the same pattern for the white. Furthermore, if we rotate the board 45 degrees, the black squares form a square grid that's easier to work with:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    O&#xA;   OOO&#xA;  OOOOO&#xA; OOOOOOO&#xA;OOOOOOOOO&#xA;OOOOOOOOO&#xA; OOOOOOO&#xA;  OOOOO&#xA;   OOO&#xA;    O&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The virus spreads across this board orthogonally, like the original virus, so the same &lt;a href=&quot;https://puzzling.stackexchange.com/a/18075/12566&quot;&gt;edge-counting strategy&lt;/a&gt; works. It has 38 edges, so we need to start with at least 38/4=9.5 squares infected, rounding up to give us a lower bound of 10. And 10 is indeed sufficient:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    X&#xA;   XOO&#xA;  XOOOO&#xA; XOOOOOO&#xA;XOOOOOOOO&#xA;OOOOOOOOX&#xA; OOOOOOX&#xA;  OOOOX&#xA;   OOX&#xA;    X&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Using the same pattern to infect the white squares gives us an optimal (and easy to remember) solution: completely cover two opposite edges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: If the virus requires exactly two infected neighbors to spread, and no more, 20 is still sufficient, and the same lower bound applies because it's strictly less contagious. Use this pattern instead:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    X&#xA;   OXO&#xA;  OOOOO&#xA; OOOXOOO&#xA;OOOOOOOOO&#xA;XOXOXOXOX&#xA; OOOOOOO&#xA;  OOXOO&#xA;   OOO&#xA;    X&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="12566" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2015-08-18T22:02:24.937" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20253" PostTypeId="2" ParentId="19807" CreationDate="2015-08-18T22:32:39.790" Score="4" Body="&lt;p&gt;If I understand user1618143's answer correctly, then I think this is pretty much the same, but in a more visually pleasing form showing how the virus spreads:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/3p6av.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3p6av.gif&quot; alt=&quot;Virus spreading&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11727" LastActivityDate="2015-08-18T22:32:39.790" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20254" PostTypeId="2" ParentId="18618" CreationDate="2015-08-18T23:59:16.107" Score="4" Body="&lt;ol start=&quot;6&quot;&gt;&#xA;&lt;li&gt;S, M, H, D, W, M, ?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Y - the sequence is units of time increasing in length (second, minute, hour, day, week, month, year) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12277" LastActivityDate="2015-08-18T23:59:16.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20255" PostTypeId="2" ParentId="20011" CreationDate="2015-08-19T04:31:42.520" Score="2" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;Partial Explanation&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; tennis player &lt;a href=&quot;https://en.wikipedia.org/wiki/Richard_Gasquet&quot; rel=&quot;nofollow&quot;&gt;Richard Gasquet&lt;/a&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I work on a farm:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Thanks AndyT. A farm, as in a tennis court. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but it's not as romantic as you think&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He's a good tennis player, so he rarely gets &quot;love&quot;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I need to wake up around sunrise to go to work,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He's sponsored by &lt;a href=&quot;https://en.wikipedia.org/wiki/Le_Coq_Sportif&quot; rel=&quot;nofollow&quot;&gt;Le Coq Sportif&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm stuck after that. &lt;/p&gt;&#xA;" OwnerUserId="14428" LastActivityDate="2015-08-19T04:31:42.520" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20256" PostTypeId="1" AcceptedAnswerId="20258" CreationDate="2015-08-19T06:07:17.610" Score="7" ViewCount="891" Body="&lt;p&gt;The watch in the picture below contains a square, around which the hours are marked. At 3:00 the area enclosed between the hour hand, the minute hand, and the square sides is ¼ of the total area of the square. At 9:00 it is also ¼ of the total area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;How many times during 12 hours does the area enclosed by the hour hand and minute hand equal exactly ¼ of the square?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/up6At.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/up6At.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9531" LastEditorUserId="9531" LastEditDate="2015-08-19T19:04:29.960" LastActivityDate="2015-08-19T20:03:09.147" Title="When is the area between the hour and minutes hands equal?" Tags="|geometry|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20257" PostTypeId="2" ParentId="20225" CreationDate="2015-08-19T06:22:52.437" Score="10" Body="&lt;p&gt;I have only a partial solution to offer, but I'm sure I won't solve all riddles, so I'll share what I have so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solutions below have two spoilers each. The first is my attempt at a two-word riddle that describes what the other two have in common, which may be awkwardly phrased and no help at all. The second explains my solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;system's subdivision | deviation; death&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &amp;rarr; component; constituent&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;►&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;part&lt;/strong&gt;ition, de&lt;strong&gt;part&lt;/strong&gt;ure&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;crystal clear | overtly obvious &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &amp;rarr; actual ancestor&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;►&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; trans&lt;strong&gt;parent&lt;/strong&gt;, ap&lt;strong&gt;parent&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;drink denier | absolute autocracy  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &amp;rarr; added-up amount&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;►&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; tee&lt;strong&gt;total&lt;/strong&gt;ler, &lt;strong&gt;total&lt;/strong&gt;itarianism&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;sunday solver | sage's saying  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &amp;rarr; possible predicate&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;►&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; cruci&lt;strong&gt;verb&lt;/strong&gt;ialist, pro&lt;strong&gt;verb&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;successfully settle | pardon, papally  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &amp;rarr; penetrate problem&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;►&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; re&lt;strong&gt;solve&lt;/strong&gt;, ab&lt;strong&gt;solve&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;material metamorphosis | knowledge; news  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &amp;rarr; plane pattern&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;►&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; trans&lt;strong&gt;formation&lt;/strong&gt;, in&lt;strong&gt;formation&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;celestial species | bachelor brotherhood  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &amp;rarr; (fall flower)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;►&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (&lt;strong&gt;aster&lt;/strong&gt;oid, mon&lt;strong&gt;aster&lt;/strong&gt;y) (not correct, see comments)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;similar starts | repeated recounting  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &amp;rarr; procedural progression&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;►&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; all&lt;strong&gt;iteration&lt;/strong&gt;, re&lt;strong&gt;iteration&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;formal familiarization | orchestral outlier  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &amp;rarr; tear transporter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;►&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; intro&lt;strong&gt;duct&lt;/strong&gt;ion, con&lt;strong&gt;duct&lt;/strong&gt;or&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;supremely significant | placebo peddler  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &amp;rarr; sturdy steed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;►&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; para&lt;strong&gt;mount&lt;/strong&gt;, &lt;strong&gt;mount&lt;/strong&gt;ebank&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;immeasurably insignificant | certainly so  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &amp;rarr; ever ending&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;►&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; in&lt;strong&gt;finite&lt;/strong&gt;simal, de&lt;strong&gt;finite&lt;/strong&gt;ly&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;wrestle; wrangle | fragmented fruit  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &amp;rarr; modern music&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;►&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; g&lt;strong&gt;rap&lt;/strong&gt;ple, g&lt;strong&gt;rap&lt;/strong&gt;es (thanks to TroyAndAbed, see comments)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;prophetic portender | ceremonial start  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &amp;rarr; bygone birdwatcher&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;►&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;augur&lt;/strong&gt;, in&lt;strong&gt;augur&lt;/strong&gt;ation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9860" LastEditorUserId="9860" LastEditDate="2015-08-19T17:34:36.637" LastActivityDate="2015-08-19T17:34:36.637" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="20258" PostTypeId="2" ParentId="20256" CreationDate="2015-08-19T06:24:39.880" Score="13" Body="&lt;p&gt;&lt;strong&gt;Answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 22 times&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Note that the area covered by by two hands is one fourth of the square if and only if they are perpendicular ie. make a $90°$ angle, which happens 22 times on 12 hours.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;A)&lt;/strong&gt; &lt;em&gt;The area covered by two hands is one fourth of the square if and only if they are perpendicular.&lt;/em&gt; &lt;br&gt;&lt;br&gt; &lt;a href=&quot;https://i.stack.imgur.com/Peks0.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Peks0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;br&gt;&lt;br&gt; Let $OF$ and $OG$ be the hands. Let $h$ be the distance from center to the sides.&lt;br&gt;&lt;br&gt;  Now, &lt;br&gt;$[OFAG]= {1 \over 4} [ABCD] = [AOB]$&lt;br&gt; $\implies [AOF]= [GOB]$ &lt;br&gt; $\implies AF=GB$ (since the area and heights are equal)&lt;br&gt;&lt;br&gt; Since, $\angle OBG =\angle OAF= 45°$ and $OA=OB$&lt;br&gt; the triangles $AOF, BOG$ congruent and &lt;br&gt;&lt;br&gt; $\angle AOF =\angle GOB$ &lt;br&gt; $\implies \angle GOF=\angle GOA + \angle AOF = \angle GOA +\angle GOB =\angle AOB =90°$&#xA;&lt;br&gt;&#xA;  &lt;br&gt;&lt;br&gt;&#xA;  &lt;strong&gt;B)&lt;/strong&gt; &lt;em&gt;The clock hands are perpendicular 22 times.&lt;/em&gt;&lt;br&gt;&lt;br&gt; The hour hand makes 1 revolution per 12 hours. The minute hand makes 12. So, if we observe the minute hand from the perspective of the hour hand, we would see that it makes 12-1=11 revolutions per 12 hours. In other words, the minute hand crosses or overlaps the hour hand 11 times. Between each two crossing, there are two cases where they are perpendicular which makes 22 times per 12 hours.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15544" LastEditorUserId="15544" LastEditDate="2015-08-19T08:15:26.793" LastActivityDate="2015-08-19T08:15:26.793" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="20259" PostTypeId="2" ParentId="20256" CreationDate="2015-08-19T08:14:07.300" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's bound to happen once between two consecutive $(60k/11)*12$s (in mins) after (15/11)*12 mins (11 instances). In addition, &quot;going backwards&quot; from such an instance where both arrows overlap, it also happens (15/11)*12 mins ago (11 instances). It works the exact same way on a square as it does on a circle: Let's say that the two arrows are orthogonal and perpendicular to each other. We can choose the closest configuration to the one we have, so we don't need to rotate it all out of the quarter. To keep the area constant, we have to rotate both arrows an equal amount because of the symmetry, meaning the angle must always be 90 degrees (15 mins). Therefore it's 22, the total.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="14016" LastEditDate="2015-08-19T19:24:32.160" LastActivityDate="2015-08-19T19:24:32.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20265" PostTypeId="2" ParentId="20115" CreationDate="2015-08-19T12:45:42.060" Score="1" Body="&lt;p&gt;&lt;code&gt;A&lt;/code&gt; is number &lt;code&gt;11&lt;/code&gt;, &lt;code&gt;D&lt;/code&gt; is number &lt;code&gt;5&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first row &lt;code&gt;A+2 = 11+2 = 13&lt;/code&gt;, what is written on the door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is &lt;code&gt;8+D = 8+5 = 13&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For columns &lt;code&gt;A-8 = 11-8 = 3&lt;/code&gt;, the number in the centre of table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is &lt;code&gt;D-2 = 5-2 =3&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you have to type &lt;code&gt;AD = 1105&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="10880" LastActivityDate="2015-08-19T12:45:42.060" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20266" PostTypeId="1" AcceptedAnswerId="20269" CreationDate="2015-08-19T13:59:15.810" Score="2" ViewCount="826" Body="&lt;h2&gt;Story&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I love doing stuff with my friends on Buzzing Shark Exchange. This time my friend finally decided to create an account on this site.&lt;br&gt;&#xA;We wanted to have fun, so I created a cryptogram:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is this sequence of numbers about?&lt;br&gt;&#xA;  $953562951413$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I promised to give a bounty of 50 reputation to the one who found the answer. I gave this to my friend before I posted it, so he would be able to find the solution and have the bounty. He got ninja'd, and he lost the opportunity to answer correctly.&lt;br&gt;&lt;br&gt;&#xA;Next day, this friend sent me an SMS (a &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18601/the-strange-sms&quot;&gt;normal&lt;/a&gt; one): &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Hey buddy, i posted an enigm on the site. If you answer it correctly i'll give you a bounty.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I searched his post on the site. His only activity was a question posted yesterday:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I have a list of letters... what comes next?&lt;br&gt;&#xA;  &lt;code&gt;O S A J J M ???&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Easy, huh?&lt;br&gt;&#xA;I luckily answered it first and, as he promised, he gave me a bounty. However, my reputation didn't go up!&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Question&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Why didn't my reputation go up?&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Bonus&lt;/strong&gt;: What are the answers to the numbers and the letter sequence?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Clarifications&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Buzzing Shark Exchange works exactly as Puzzling Stack Exchange. The only difference is that it is allowed for sharks.&lt;/li&gt;&#xA;&lt;li&gt;This site works correctly. There are no bugs (sharks know how to code!)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13384" LastEditorUserId="5373" LastEditDate="2018-06-24T00:36:47.150" LastActivityDate="2018-06-24T00:36:47.150" Title="My friend's bounty" Tags="|pattern|lateral-thinking|" AnswerCount="5" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20267" PostTypeId="2" ParentId="20266" CreationDate="2015-08-19T14:07:18.467" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The letter sequence is first letter of the months starting from October going backwards. So the next letter in the sequence should be A for April. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's one bonus question out of the way. &lt;/p&gt;&#xA;" OwnerUserId="12332" LastActivityDate="2015-08-19T14:07:18.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20268" PostTypeId="2" ParentId="20266" CreationDate="2015-08-19T14:09:33.483" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sequence of numbers is the first 12 digits of pi (π) in reverse. (The last digit (or first digit in the sequence) is rounded off, thanks Vašek Potoček.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's another bonus question cracked!&lt;/p&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-08-19T14:25:33.987" LastActivityDate="2015-08-19T14:25:33.987" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20269" PostTypeId="2" ParentId="20266" CreationDate="2015-08-19T14:20:34.520" Score="7" Body="&lt;p&gt;For the main question:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If his only activity was a single question, your friend did not have the reputation to offer a bounty. I can only assume he gave you a &lt;a href=&quot;https://en.wikipedia.org/wiki/Bounty_(chocolate_bar)&quot; rel=&quot;nofollow&quot;&gt;coconut-filled chocolate&lt;/a&gt; instead. Or perhaps there are numerous people closing in on you, even now, attempting to take you, alive or dead, to your (presumably former) friend&lt;br&gt;&#xA;EDIT: If you were given a bounty, you would have been told about someone else to go capture and bring to the authorities/the bounty giver. I got my old west knowledge backwards, sorry.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11717" LastEditorUserId="11717" LastEditDate="2015-08-19T14:33:27.993" LastActivityDate="2015-08-19T14:33:27.993" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20270" PostTypeId="2" ParentId="20266" CreationDate="2015-08-19T14:27:33.563" Score="4" Body="&lt;p&gt;Well since you and your friend are clearly sharks, the bounty he gave you might have been the &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;H.M.S. Bounty, full of tasty tasty sailors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alas, such a bounty will not award you reputation.&lt;/p&gt;&#xA;" OwnerUserId="13068" LastActivityDate="2015-08-19T14:27:33.563" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20271" PostTypeId="2" ParentId="20266" CreationDate="2015-08-19T14:29:33.933" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you have been busy on Buzzing Shark Exchange you may have already hit the daily +200 rep limit, so it cannot increase any further.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-08-19T14:29:33.933" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20272" PostTypeId="2" ParentId="20256" CreationDate="2015-08-19T14:40:09.147" Score="-1" Body="&lt;p&gt;&lt;del&gt;I am counting 24 possibilities. &lt;/del&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;del&gt;Like Anachor already said, both hands must be perpendicular, and his 22 are correct. But i think there are 2 more: &lt;br&gt; 1. nearly before 3 o'clock, sth. like 02:58 or 02:59 where the hands are perpendicular &lt;br&gt; 2. and the same nearly before 9 o'clock, so like 08:58 or 08:59. &lt;br&gt;&lt;br&gt; So in both cases there are two appearances of the quarter within of two minutes.&lt;/del&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;del&gt;I hope i've made no logical error. If so, please be kind in the comments :-)&lt;/del&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;br&gt;&#xA;Edit: Forget my answer, Roland is totally right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For every minute the hour hand is moving by 1/720.&#xA;So at 2:59 it will be at 179/720 and the minute hand will be at 708/720, that is a difference of 191. The difference must be exactly 180 for them to be perpendicular.&lt;/p&gt;&#xA;" OwnerUserId="14461" LastEditorUserId="14461" LastEditDate="2015-08-19T15:14:38.463" LastActivityDate="2015-08-19T15:14:38.463" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20273" PostTypeId="2" ParentId="19920" CreationDate="2015-08-19T14:55:20.497" Score="3" Body="&lt;p&gt;&lt;em&gt;The Avesta once noted that this city flows;&lt;br&gt;&#xA;As their regime, here the terrible chose.&lt;br&gt;&#xA;Now reborn anew, with haste this city grows.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This refers to &lt;strong&gt;Kabul, Afghanistan.&lt;/strong&gt; &lt;br&gt;&#xA; The Kabul River is mentioned in the Avesta sacred texts.&#xA; It was the command center for the Soviets during the Soviet-Afghan War.&#xA; It is now the 5th fastest growing city in the world.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;When flying, you'll land here, amidst ice and rocks.&lt;br&gt;&#xA;When boating, the whole fjord is basically docks.&lt;br&gt;&#xA;When driving, be careful you don't hit an ox.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This could be &lt;strong&gt;Kangerlussuaq, Greenland&lt;/strong&gt; &lt;br&gt;&#xA; This is the location of Greenlands only airport capable of landing large jets.&lt;br&gt;&#xA; Not sure about it beilng all docks but it's on a fjord that is completely navigable by boat.&lt;br&gt;&#xA; Greenland is has a native population of muskox and ox warning signs are a common sight on the roads around the city.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13290" LastEditorUserId="13290" LastEditDate="2015-09-01T13:42:19.003" LastActivityDate="2015-09-01T13:42:19.003" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20274" PostTypeId="2" ParentId="20225" CreationDate="2015-08-19T15:04:54.697" Score="3" Body="&lt;p&gt;It took looking at M Oehm's solution for it to click for me, but here's my contribution so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lobotomee's loss | employed entities&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;person&lt;/strong&gt;ality | &lt;strong&gt;person&lt;/strong&gt;el&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;lacking logic | evoking enthusiasm&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ir&lt;strong&gt;ration&lt;/strong&gt;al | Inspi&lt;strong&gt;ration&lt;/strong&gt;al&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not normal | Kanye's kin&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;U&lt;strong&gt;north&lt;/strong&gt;odox | &lt;strong&gt;North&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Incorrectly interpreted | College kid&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Mis&lt;strong&gt;under&lt;/strong&gt;stood | &lt;strong&gt;under&lt;/strong&gt;graduate&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;Wowing wizardry | Hearty herb&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Be&lt;strong&gt;witch&lt;/strong&gt;ment | &lt;strong&gt;Witch&lt;/strong&gt;hazel&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Headdress holiday | chromosomal cross-up&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;f&lt;strong&gt;esti&lt;/strong&gt;val | transv&lt;strong&gt;esti&lt;/strong&gt;te&lt;/strike&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12914" LastEditorUserId="12914" LastEditDate="2015-08-20T14:46:33.763" LastActivityDate="2015-08-20T14:46:33.763" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20275" PostTypeId="1" AcceptedAnswerId="20276" CreationDate="2015-08-19T15:23:13.363" Score="10" ViewCount="1953" Body="&lt;p&gt;My young daughter and I love doing jigsaw puzzles together. The other day, we had a quite difficult one: 7 by 7 pieces!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We started as usual by splitting the pieces into 2 stacks: the edge pieces on one side, the middle pieces on the other side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before putting the edges together, we noticed that there were almost the same number of pieces in each stack:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DOWqu.png&quot; alt=&quot;Our 7x7 puzzle&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wondered then... our puzzle collection is huge! We have all the possible sizes, from 1x2 to 100x100 pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can we find one (or even more?) in our collection containing exactly the same number of edges pieces and middle pieces?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9315" LastActivityDate="2015-09-22T17:23:03.733" Title="Jigsaw puzzle: edge and middle pieces" Tags="|mathematics|jigsaw-puzzle|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20276" PostTypeId="2" ParentId="20275" CreationDate="2015-08-19T15:31:22.523" Score="14" Body="&lt;p&gt;What you're looking for are numbers $X$ and $Y$ where $2X + 2(Y-2) = (X-2)*(Y-2)$. This is generalized; a few possible (and practical) solutions would be:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$5$ by $12$, with $30$ of both edge and middle pieces.&lt;/li&gt;&#xA;&lt;li&gt;$6$ by $8$, with $24$ of both edge and middle pieces.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-08-19T15:31:22.523" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="20277" PostTypeId="2" ParentId="20275" CreationDate="2015-08-19T15:42:08.050" Score="4" Body="&lt;p&gt;For an n x m puzzle, the number of edge pieces, E, is defined by  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$2n+2(m-2)$ &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the number of middle pieces, M, is then defined by  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$nm-E$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we want to find a puzzle size for which E=M, then we are looking for the case where  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$nm-E=E \implies nm=2E \implies nm=4n+4m-8$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I wrote a quick nested for loop to search for solutions and found the pairs  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $n=5, m=12$ and $n=6,m=8$ &lt;br&gt;(In either case, swapping $n$ and $m$ is irrelevant i.e. $m=12,n=5$ leads to the same result)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="12700" LastEditDate="2015-08-19T17:15:08.787" LastActivityDate="2015-08-19T17:15:08.787" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20278" PostTypeId="2" ParentId="20245" CreationDate="2015-08-19T16:04:15.820" Score="1" Body="&lt;p&gt;I made it rhyme!*&lt;br&gt;&#xA;&lt;sup&gt;&lt;strong&gt;&lt;em&gt;*Not really a solution, though.&lt;/em&gt;&lt;/strong&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A father of the highest fame&lt;br&gt;&#xA;A drinker who is called the same&lt;br&gt;&#xA;Swedish metal takes this name&lt;br&gt;&#xA;You count on them to win the game  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Historic sacred melody&lt;br&gt;&#xA;An orphan, cinematically&lt;br&gt;&#xA;And bywords of reality&lt;br&gt;&#xA;Concrete responsibility  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A pound of text with speech not florid&lt;br&gt;&#xA;When skip was blue, at birth he wore it&lt;br&gt;&#xA;Dolly sang it, we adored it&lt;br&gt;&#xA;Written rule, with one before it  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A certain group has now departed&lt;br&gt;&#xA;With NASA's help, it all was started&lt;br&gt;&#xA;Some adorned and openhearted&lt;br&gt;&#xA;Else you'll find them quite well-guarded  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A young pig's tag, an all-time great&lt;br&gt;&#xA;History doth these create&lt;br&gt;&#xA;Together here we rule your fate&lt;br&gt;&#xA;In others magic worlds await  &lt;/p&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2015-08-19T16:04:15.820" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20279" PostTypeId="2" ParentId="20238" CreationDate="2015-08-19T17:07:35.380" Score="2" Body="&lt;p&gt;Here is a slightly different solution.  It doesn't necessarily scale well, but it does form a complete cycle of the board.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Solution&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;First, we create a 5x5 board like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;+--+--+--+--+--+&#xA;| 1|17|24| 2|18|&#xA;+--+--+--+--+--+&#xA;|22|14| 5|21|13|&#xA;+--+--+--+--+--+&#xA;|25| 8|11|16| 7|&#xA;+--+--+--+--+--+&#xA;| 4|20|23| 3|19|&#xA;+--+--+--+--+--+ &#xA;|10|15| 6| 9|12|&#xA;+--+--+--+--+--+&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Notice that if you move up 3 from the 25th spot, it takes you out of the current 5x5 square.  Thus, we can position 4 of these squares to make a 10x10 square, each starting their cycle in the square next to the shared vertex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets call the square above $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, define the function $R(S)$ to be the 90 degree counter-clockwise rotation of $S$.  Then, we can create a 10x10 square with the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  R(R(A))   R(A)&#xA;R(R(R(A)))   A &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We'd end up with the following in the middle 6x6 square:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;        +---+&#xA;        | 75|&#xA;        +---+    ...&#xA;   ...  |   |&#xA;        +---+---+---+---+&#xA;        | 51| 26|   | 50|&#xA;+---+---+---+---+---+---+&#xA;|100|   | 76| 1 |&#xA;+---+---+---+---+&#xA;            |   |  ...&#xA;     ...    +---+&#xA;            | 25|&#xA;            +---+&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;By doing this, we've created a 10x10 board with a complete cycle.  &lt;/p&gt;&#xA;&#xA;&lt;h1&gt;More General Solution&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Consider the following 5x5 squares.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;+--+--+--+--+--+&#xA;| 1|24| 8| 2|23|&#xA;+--+--+--+--+--+ &#xA;|19|16| 5|20|17| &#xA;+--+--+--+--+--+  &#xA;| 7|10|13|25| 9|  &#xA;+--+--+--+--+--+  &#xA;| 4|21|18| 3|22| &#xA;+--+--+--+--+--+  &#xA;|12|15| 6|11|14|  &#xA;+--+--+--+--+--+ &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Notice that moving diagonally up and to the right from the 25th spot will take you to the upper left position of the next 5x5 square, and moving diagonally down and to the right will take you to the bottom left position of the next square.  Thus, when entering this square from the top left position, you can enter the square to the right in the bottom left or top left position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example (S is the starting point in the first square, E is the ending point, and p is the possible starting point in the next square):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;+-----+-&#xA;|S    |p&#xA;|     | &#xA;|   E | ...&#xA;|     | &#xA;|     |p&#xA;+-----+-&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;A diagonal flip of this square will let you enter the square below in either the top left or top right position when starting in the top left corner.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;+-----+&#xA;|S    |&#xA;|     | &#xA;|     |&#xA;|  E  | &#xA;|     |&#xA;+-----+&#xA;|p   p|&#xA;  ...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now consider this second square:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;+--+--+--+--+--+&#xA;| 1|17| 7| 2|14|&#xA;+--+--+--+--+--+&#xA;| 9|22|25|10|21|&#xA;+--+--+--+--+--+&#xA;| 6| 3|13|18| 4|&#xA;+--+--+--+--+--+&#xA;|24|16| 8|23|15|&#xA;+--+--+--+--+--+&#xA;|12|19| 5|11|20|&#xA;+--+--+--+--+--+&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Similar to the first, you can enter the square above in either bottom corner.  A diagonal flip means you can enter the square to the left in either right corner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Throught the use of these two squares and their diagonal flip, you are able to enter the corner of any adjacent square when starting in the upper left corner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Through rotations, this also means you can also start in any corner!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To tile any rectangle which sides are multiples of 5, simply divide it up into 5x5 squares and then trace a path through these squares.  A spiral will work, or back and forth through each row will also work.  Use the appropriate square above to enter/exit the right corner and it is possible to tile any rectangle whose sides are multiples of 5.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Example&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Here is a 30x20 rectangle.  This is made up of 6x4 5x5 squares.  The starting corner of each 5x5 square is lettered in order.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;+--+--+--+--+--+--+&#xA;|a |b |c |d |e |f |&#xA;|  |  |  |  |  |  |&#xA;+--+--+--+--+--+--+&#xA;| l| k| j| i| h| g|&#xA;|  |  |  |  |  |  |&#xA;+--+--+--+--+--+--+&#xA;|m |n |o |p |q |r |&#xA;|  |  |  |  |  |  |&#xA;+--+--+--+--+--+--+&#xA;| x| w| v| u| t| s|&#xA;|  |  |  |  |  |  |&#xA;+--+--+--+--+--+--+&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The squares a-e and m-n all start in the top left and enter the square to the right in the top left.&#xA;The squares g-k and s-w all start in the top right and enter the square to the left in the top right.&#xA;The squares f and r start in the top left and enter the square below in the top right.&#xA;The sqaures l and x start in the top right and enter the square below in the top left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to tile this using a spiral pattern.  For example, here is a 7x5 set of 5x5 squares.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;+--+--+--+--+--+--+--+&#xA;|a |b |c |d |e |f |g |&#xA;|  |  |  |  |  |  |  |&#xA;+--+--+--+--+--+--+--+&#xA;|  |u |v |w |x |y | h|&#xA;|t |  |  |  |  |  |  |&#xA;+--+--+--+--+--+--+--+&#xA;|  |  |G |H |I | z| i|&#xA;|s |F |  |  |  |  |  |&#xA;+--+--+--+--+--+--+--+&#xA;|  |  |  |  |  | A| j|&#xA;|r | E| D| C| B|  |  |&#xA;+--+--+--+--+--+--+--+&#xA;|  |  |  |  |  |  | k|&#xA;| q| p| o| n| m| l|  |&#xA;+--+--+--+--+--+--+--+&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The pattern is to simply enter the next square in the corner farthest from the center.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="1592" LastEditDate="2015-08-20T12:36:10.900" LastActivityDate="2015-08-20T12:36:10.900" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20280" PostTypeId="1" AcceptedAnswerId="20281" CreationDate="2015-08-19T17:33:42.350" Score="-1" ViewCount="12664" Body="&lt;p&gt;&lt;strong&gt;This isn't a duplicate! The &quot;Two doors with two guards - one lies, one tells the truth&quot; isn't like this question! You just have to pay attention in each one to se the difference.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A man was walking along the road on the way to &lt;strong&gt;&lt;em&gt;Village Where All&#xA;  People Speak Truth&lt;/em&gt;&lt;/strong&gt;, which is nearby &lt;strong&gt;&lt;em&gt;Village Where All People&#xA;  Lie&lt;/em&gt;&lt;/strong&gt;. He had never been to any of the villages so he didn't know the&#xA;  way. He didn't want to go to &lt;strong&gt;&lt;em&gt;Village Where All People Lie&lt;/em&gt;&lt;/strong&gt;. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;At one point there is a fork in the road and he knows that to one side&#xA;  is the village where he wants to go and to the other side is the&#xA;  village who he wants to avoid at all costs. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Not knowing whether to go to the right or left, gets confused. Then he&#xA;  sees a woman in the middle of the fork. He don't know to which&#xA;  village she lives, then he will only ask ONE question to the woman,&#xA;  and thereby discover the way to the &lt;strong&gt;&lt;em&gt;Village Where All People Speak&#xA;  Truth&lt;/em&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What he should ask?&lt;/p&gt;&#xA;" OwnerUserId="15588" LastEditorUserId="15588" LastEditDate="2015-08-19T20:37:33.587" LastActivityDate="2015-08-19T20:37:33.587" Title="The Lying Villages" Tags="|logical-deduction|" AnswerCount="1" CommentCount="6" FavoriteCount="0" ClosedDate="2015-08-19T17:59:02.277" ContentLicense="CC BY-SA 3.0" />
  <row Id="20281" PostTypeId="2" ParentId="20280" CreationDate="2015-08-19T17:37:17.370" Score="2" Body="&lt;p&gt;I feel like this has to be a duplicate, but here goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ask her:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;What direction would you tell me to go if I asked you which path leads to the Village of Truth-Speakers?&lt;br&gt;&#xA;If she's a truther (props to Drake and Josh), she would tell you to take the path to the truth village, so she tells you to take the path to truth village. If she's a liar, she would tell you to take the path to lie village, so she must lie about that and point you to the path to truth village. Either way, go the way she points.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="11475" LastEditDate="2015-08-19T17:37:48.497" LastActivityDate="2015-08-19T17:37:48.497" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="20283" PostTypeId="1" AcceptedAnswerId="36324" CreationDate="2015-08-19T18:20:42.507" Score="6" ViewCount="1183" Body="&lt;p&gt;Following on from &lt;a href=&quot;https://puzzling.stackexchange.com/users/12700/codenewbie&quot;&gt;CodeNewbie's&lt;/a&gt; question &lt;a href=&quot;https://puzzling.stackexchange.com/questions/20141/from-puzzling-to-stackexchange&quot;&gt;&quot;From Puzzling to Stack Exchange&quot;&lt;/a&gt;, I thought I would set a problem of my own using the same rules:&lt;/p&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rules:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;In any move, you can either replace a letter, add a letter, remove a letter, split a word or combine two words. These are the five valid moves, and only one of these can be applied in a move.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;After each move, each of the resulting words must be valid by itself, even if it makes no sense as a phrase. &lt;strong&gt;To level the playing field, only words found on Dictionary.com are considered valid.&lt;/strong&gt; No proper nouns, abbreviations, prefixes or suffixes allowed. (However, words like jack, john, jane etc. are allowed as they are used as common nouns too.) Alternate spellings and archaic words are permitted, but words referring to specific persons, places or events are not valid.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You may only add a letter to an existing word, but a new letter (such as a or I) cannot create a new word separately. Likewise, you cannot drop a letter from a one-letter word to remove it altogether.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;p&gt;The problem is that two of the words I chose &lt;em&gt;Perfect&lt;/em&gt; and &lt;em&gt;Answer&lt;/em&gt; appeared to be word islands. By this I mean that there doesn't appear to be any way to move from those words to any other words following the rules above (except, of course for the trivial case of adding an &lt;strong&gt;S&lt;/strong&gt; to the end which doesn't help the transition to anything else).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the shortest word that anyone can find which is also a word island?&#xA;(I'll accept answers where it is one of a pair of words where the only valid moves are back and forwards between them)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, I would be glad to hear if my two examples are not, in fact, islands after all.&lt;/p&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="69582" LastEditDate="2021-07-06T22:10:08.280" LastActivityDate="2021-07-06T22:10:08.280" Title="Word ladder island (formerly dead-end)" Tags="|word|optimization|english|word-ladder|pencil-and-paper-games|" AnswerCount="4" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="20284" PostTypeId="2" ParentId="20283" CreationDate="2015-08-19T18:32:18.223" Score="6" Body="&lt;p&gt;A 4 letter word that I found to be a &lt;strike&gt;dead-end&lt;/strike&gt; word ladder island&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strike&gt;&lt;code&gt;xenon&lt;/code&gt;&lt;/strike&gt; &lt;strike&gt; &lt;code&gt;vixen&lt;/code&gt;&lt;/strike&gt; &lt;code&gt;lynx&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why it's a &lt;strike&gt;dead-end&lt;/strike&gt; word ladder island&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The plural is &lt;code&gt;lynxes&lt;/code&gt;, so that isn't a possible move. &lt;code&gt;ly&lt;/code&gt; is just a suffix, whereas &lt;code&gt;lyn&lt;/code&gt; is an abbreviation which makes that invalid too. So splitting it is not useful. Trying for replacements gives us &lt;code&gt;Lynn&lt;/code&gt; and &lt;code&gt;Lynd&lt;/code&gt; but both are invalid as they are proper nouns.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-08-20T08:22:11.307" LastActivityDate="2015-08-20T08:22:11.307" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="20286" PostTypeId="1" AcceptedAnswerId="20288" CreationDate="2015-08-19T18:46:49.450" Score="2" ViewCount="298" Body="&lt;p&gt;At any time $t_0$, let $f(a,t_0) = \{a_1, a_2, \ldots, a_n\}$, where each $a_i$ is distinct. &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If some $a_i = b$ such that $f(b,t_0) = \{b_1, b_2, \ldots, b_m\}$,&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then all $b_i \in f(b,t_0)$ are said to be &lt;em&gt;friends of $a$&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What does the function $f(x,t)$ represent?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;strong&gt;What is this ancient proverb?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="13237" LastEditorUserId="13237" LastEditDate="2015-08-19T20:10:03.910" LastActivityDate="2015-08-19T20:10:03.910" Title="A Mathematical Proverb" Tags="|riddle|mathematics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20287" PostTypeId="2" ParentId="20286" CreationDate="2015-08-19T18:52:50.030" Score="2" Body="&lt;p&gt;The function represents:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Friends of $a$ at time $t$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The proverb is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The friends of our friends are our friends.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More explained:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The first line defines the set of $a_i$ as friends of $a$ at any time $t$. The second line defines all $b_1$ as friends of $b$ at time $t$ and says $b$ is in the set of friends of $a$. The third line concludes that all $b_i$ are friends of $a$. This math is simplified to: All $b_i$ are friends of $a$ if some friend of $a$, $a_i = b$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11727" LastEditorUserId="11727" LastEditDate="2015-08-19T19:04:21.457" LastActivityDate="2015-08-19T19:04:21.457" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20288" PostTypeId="2" ParentId="20286" CreationDate="2015-08-19T19:01:33.397" Score="5" Body="&lt;p&gt;The function $f(x,t)$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Enemies of $x$ at time $t$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The proverb is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The enemy of my enemy is my friend&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All $b_i$ are friends of $a$. $b \in b_i$ can be defined in terms of $a$ as $b \in f(f(a,t),t)$ i.e $f^2(a,t)$. This is defined as friend, so we can conclude f = enemy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1968" LastEditorUserId="1968" LastEditDate="2015-08-19T19:17:45.600" LastActivityDate="2015-08-19T19:17:45.600" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20289" PostTypeId="2" ParentId="20283" CreationDate="2015-08-19T19:08:09.670" Score="1" Body="&lt;p&gt;Taking a step back from the ambitious 4-letter words (though I'm certain there has to be something), I found this among five letters:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;ZEBRA  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Removals and splitting: I used dictionary.com directly and tried removing each letter and got no legal words. I also tried spliting at every possible joint and got no legal words.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Additions: Using &lt;a href=&quot;http://www.visca.com/regexdict/&quot; rel=&quot;nofollow&quot;&gt;this tool&lt;/a&gt;, I searched for words containing a letter added in any spot and got no legal words.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Replacements: Using &lt;a href=&quot;http://www.quinapalus.com/cgi-bin/qat?pat=%60zebra&amp;amp;ent=Search&amp;amp;dict=0&quot; rel=&quot;nofollow&quot;&gt;this tool&lt;/a&gt;, I found only two words achieved by replacing a letter:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;DEBRA and ZABRA &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The first is of course a proper noun and the second doesn't appear on dictionary.com&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4.But what about the trivial 'S'? According to dictionary.com, the plural does in fact have an s (I've heard it pluralized the same, like moose). But running that through the same gauntlet above, you get similar results (one less result in step 3 even)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I trust these tools are reliable, but if anyone finds something I missed, let me know&lt;/p&gt;&#xA;" OwnerUserId="13970" LastActivityDate="2015-08-19T19:08:09.670" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20291" PostTypeId="1" AcceptedAnswerId="20294" CreationDate="2015-08-19T19:14:03.173" Score="15" ViewCount="922" Body="&lt;p&gt;&lt;em&gt;I've been avidly lurking and sometimes posting on Puzzling Stack Exchange for a few months now, and I thought it might be time for me to try a puzzle of my own. I hope it's interesting and difficult without being too roundabout!&lt;/em&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;While adventuring through the jungles of Puzzlandia, you come across a strange columnar stone. Carved into it is a face, and below the face is some text. Below the text are four stone levers arranged horizontally, unmarked. The text reads:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Greetings, adventurer! If you can solve my little test,&lt;br /&gt;&#xA;Great treasure awaits you beneath in a chest.&lt;br /&gt;&#xA;The letters below will be your clue;&lt;br /&gt;&#xA;Once you solve them you'll know what to do.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;XVWTWMDIPRHCVQSFTBLJFIYKYVGIMSLIGCMXZSIXBSYE&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The key Iies between my eyes...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;PIease be kind; this is my first ever puzzle on here! Let me know if there's anything I can do to improve my puzzle style, and feel free to ask questions for clarification, etc.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;What must you do to reveal the treasure?&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="12914" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-08-19T20:26:33.197" Title="The Monolith of Puzzlandia" Tags="|cipher|lateral-thinking|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20292" PostTypeId="2" ParentId="20275" CreationDate="2015-08-19T19:16:45.207" Score="2" Body="&lt;p&gt;Our goal is to find such $n$ and $m$ that $nm = 2(n-2)(m-2)$. They can't be equal of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming $n&amp;lt;m$, then $\dfrac n{n-2} &amp;gt; \sqrt2$, so $n&amp;lt;8$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then again, $2(n-2) &amp;gt; n$, therefore $4&amp;lt;n&amp;lt;8$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equality $\frac{2n-4}n = \frac m{m-2}$ can be modified as $\frac{n-4}n = \frac2{m-2}$, so $\frac{2n}{n-4}$ must be an integer equal to $m-2$, making 8 divisible by $n-4$. The latter can be 1 or 2, so either $n=5, m=12$ or $n=6, m=8$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disclaimer: I know I could just try seeing if it can be done for all 3 possible values of $n$ after narrowing it down, but didn't feel like it.&lt;/p&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="13494" LastEditDate="2015-09-22T17:23:03.733" LastActivityDate="2015-09-22T17:23:03.733" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20293" PostTypeId="2" ParentId="20283" CreationDate="2015-08-19T19:26:48.727" Score="1" Body="&lt;p&gt;Okay, here's a 6-letter word, which I checked using this tool:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://boulter.com/scrabble/&quot; rel=&quot;nofollow&quot;&gt;http://boulter.com/scrabble/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Sphinx and Sphynx&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Couldn't find any 5-letter words using the same letters, so that removes deletion.&lt;/li&gt;&#xA;&lt;li&gt;Inserting letters didn't create any new words that I could see (and the plural is -es, not -s, which helps with the trivial case)&lt;/li&gt;&#xA;&lt;li&gt;Checked all cases of replacing letters with blanks, no results&lt;/li&gt;&#xA;&lt;li&gt;Definitely can't split it into two words&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Granted, all this was done using the dictionary of the tool, which is not the same as Dictionary.com. I haven't manually checked each possibility with Dictionary.com, but I feel like this is pretty good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, its longer than &quot;answer&quot;. :(&lt;/p&gt;&#xA;" OwnerUserId="10861" LastActivityDate="2015-08-19T19:26:48.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20294" PostTypeId="2" ParentId="20291" CreationDate="2015-08-19T19:45:05.697" Score="17" Body="&lt;p&gt;You must press the levers in this order:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $1, 3, 4, 2, 1, 1, 3, 4$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How I came by this solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&quot;The key lies between my eyes...&quot;&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Upon close examination of the text, two letters that appear to be lowercase &quot;L&quot;s are actually uppercase &quot;I&quot;s:&#xA;&lt;br&gt;&#xA;&lt;pre&gt;&#xA; The key Iies between my eyes...&lt;br&gt;&#xA; PIease be kind,&#xA; &lt;/pre&gt;&#xA; I copied the text between them:&lt;br&gt;&#xA; &lt;strong&gt;&lt;code&gt;ies between my eyes... P&lt;/code&gt;&lt;/strong&gt;&lt;br&gt;&#xA; and stripped out the spaces and punctuation:&lt;br&gt;&#xA; &lt;strong&gt;&lt;code&gt;iesbetweenmyeyesP&lt;/code&gt;&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Using that as the key to a Vigenere cipher (what other common cipher requires a key?), the ciphertext easily decrypted to&lt;br&gt;&#xA; &lt;strong&gt;&lt;code&gt;presstheleversonethreefourtwooneonethreefour&lt;/code&gt;&lt;/strong&gt;&lt;br&gt;&#xA;  or&lt;br&gt;&#xA; &lt;strong&gt;&lt;code&gt;Press the levers: one, three, four, two, one, one, three, four.&lt;/code&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="4421" LastEditDate="2015-08-19T19:57:04.043" LastActivityDate="2015-08-19T19:57:04.043" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20295" PostTypeId="1" AcceptedAnswerId="20326" CreationDate="2015-08-19T20:44:00.557" Score="21" ViewCount="4244" Body="&lt;p&gt;The area of a circle of radius 1 is completely covered by seven smaller circles, all with the same radius as each other. (The circles can overlap - indeed they must!). What is the smallest radius the small circles can have so that it is possible to do this, and why is it not possible for any smaller radius?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/O7ite.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/O7ite.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15849" LastEditorUserId="15849" LastEditDate="2015-08-20T14:27:19.247" LastActivityDate="2023-03-01T07:20:55.140" Title="Seven overlapping circles" Tags="|geometry|" AnswerCount="6" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20296" PostTypeId="1" AcceptedAnswerId="20298" CreationDate="2015-08-19T20:55:49.120" Score="1" ViewCount="490" Body="&lt;p&gt;A sheik had two sons. He loved them both very much, but he had to choose one heir. So he called both and said to them: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Take your best camels and ride to the oasis, which is is the farthest&#xA;  in my realm. My heir will be whose camel is the last to arrive at my&#xA;  palace.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sons take their camels and rode together to the oasis slowly. But when they were there, they realized that they had a problem. Both of them wanted to be heir so they could not make a decision. But, after a while, an enchantress appeared and asked them, why they looked so sad. The sons told her their problem and the enchantress whispered something in both ears. Only seconds later, two camels left the oasis and ran to the palace of the sheik.&#xA;What did the enchantress whispered to the sons? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Remark: I translated this puzzle, which my grandfather told to me when I was a child, out of the german language. I'm afraid that my translation isn't 100% equal to the original puzzle, but I hope you have some fun anyway.)&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Edit:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Yes, this is a duplicate. I searched for a duplecate before i asked this question, but I didn't found the original. You can exchange animals and persons and then the puzzle is exactly the same.&lt;/p&gt;&#xA;" OwnerUserId="15848" LastEditorUserId="15848" LastEditDate="2015-08-19T21:24:14.227" LastActivityDate="2015-08-19T21:24:14.227" Title="The camel-challenge" Tags="|situation|" AnswerCount="2" CommentCount="3" ClosedDate="2015-08-19T21:12:24.983" ContentLicense="CC BY-SA 3.0" />
  <row Id="20297" PostTypeId="2" ParentId="20296" CreationDate="2015-08-19T21:00:54.263" Score="1" Body="&lt;p&gt;She either said :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Send your brother's camel.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;OR...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Dudes, just send both at the same time.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13068" LastActivityDate="2015-08-19T21:00:54.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20298" PostTypeId="2" ParentId="20296" CreationDate="2015-08-19T21:01:12.133" Score="1" Body="&lt;p&gt;I'm not sure but:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If they both agree to switch camels it turns into a race instead and they don't have to wait at the oasis until they die. But at the same time if one of them isn't as good a rider as the other, then he'd be a bit stupid to agree...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastActivityDate="2015-08-19T21:01:12.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20300" PostTypeId="2" ParentId="20295" CreationDate="2015-08-19T21:06:48.987" Score="1" Body="&lt;p&gt;I might be misunderstanding this, but based on what I think you're asking, we have the following lower bound:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$\sqrt{\frac17}$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is obtained by the following:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the larger circle has a radius of $1$, then it has an area of $\pi$.&lt;br&gt;&#xA;&lt;br&gt;&#xA; We know the seven smaller circles have enough total area to cover the larger, so we have&lt;br&gt;&#xA; $\pi \le 7 \pi r^2$&lt;br&gt;&#xA;&lt;br&gt;&#xA; Simple algebra from there yields&lt;br&gt;&#xA; $\sqrt{\frac17} \le r$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="4421" LastEditDate="2015-08-19T21:40:41.950" LastActivityDate="2015-08-19T21:40:41.950" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20303" PostTypeId="2" ParentId="20295" CreationDate="2015-08-19T21:19:06.330" Score="12" Body="&lt;p&gt;I think it's:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $1/2$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll use this arrangement, with one circle not pictured in the center (reasoning shown below). All blue lines are equal to $r$ in length. The green line needs to be equal to $1$. Call the distance from the origin to the &lt;em&gt;center&lt;/em&gt; of an outer circle $d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lMnv7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lMnv7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Finding the smallest $r$:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the image below, we create two triangles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first is a right triangle made from the gray line on the left, the green line, and the blue on top. These are of lengths $d$, $1$, and $r$, respectively, and we get: $d^2 + r^2 = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second is made from the other blue line and the same vertical gray line. To minimize the radii, we let the center circle just reach out to the blue line along the green path. Since all the circles have radius $r$, we are left with an isosceles triangle of sides $r$, $r$, and $d$, and angles of 30, 30, and 120 degrees. From this we acquire: $cos(30^o) = d/2r$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving the second for $d$ and substituting, we have $r = 1/2$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xL5iV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xL5iV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With no centered circle, and radii of $1/2$, only seven points on the edge of the unit circle are covered. This obviously will not suffice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why not smaller circles?&lt;/strong&gt;&#xA;This method optimized coverage for a scheme with one centered circle and six surrounding. It features the most symmetry of any seven-circle scheme; six lines of symmetry can be drawn through the origin to 12, 1, 2, 3, 4, and 5 o'clock. This optimizes coverage for circles by the same logic Archimedes used when &lt;a href=&quot;http://www.ams.org/samplings/feature-column/fc-2012-02&quot; rel=&quot;nofollow noreferrer&quot;&gt;measuring its area and circumference&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="13237" LastEditorUserId="13237" LastEditDate="2015-08-20T14:44:57.360" LastActivityDate="2015-08-20T14:44:57.360" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20304" PostTypeId="2" ParentId="20295" CreationDate="2015-08-19T21:21:24.343" Score="2" Body="&lt;p&gt;I put this as an answer because I wanted to be able to use the spoiler tags. I believe I know how the circles must be arranged, but do not have the math skills to determine what measurements theyd have.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Because of the nature of the shape (a circle) and the fact you are limited to 7 circles, no pattern or arrangement can efficiently cover the area. You need circles that stretch from the center point of the circle and that cover 1/7th of the circumference of the circle to cover(of radius 1). &lt;a href=&quot;https://i.stack.imgur.com/fZAU2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fZAU2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastActivityDate="2015-08-19T21:21:24.343" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20305" PostTypeId="2" ParentId="20295" CreationDate="2015-08-19T21:31:02.933" Score="21" Body="&lt;p&gt;I believe the smallest radius is..&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $r = \frac R 2$, i.e. &lt;strong&gt;$\frac12$&lt;/strong&gt;&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/TXTGM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TXTGM.png&quot; alt=&quot;7 overlapping circles&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13148" LastEditorUserId="4421" LastEditDate="2015-08-19T21:38:03.280" LastActivityDate="2015-08-19T21:38:03.280" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20306" PostTypeId="2" ParentId="20011" CreationDate="2015-08-19T21:59:07.637" Score="5" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Jeffrey Dean Morgan, the actor.  You were in the following movies:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I work on a farm, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Texas Killing Fields&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but it's not as romantic as you think: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;P.S. I Love You&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I need to wake up around sunrise to go to work, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Red Dawn&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and I get paid just 4 pennies per minute. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Watchmen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I had to take a second job to pay the bills. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Courier&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Honestly, I'd do anything just to get an apartment, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Resident&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I'm such a sad sack it'll probably never happen.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Losers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Bed and Breakfast)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Dead &amp;amp; Breakfast&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12061" LastEditorUserId="12061" LastEditDate="2015-08-19T22:07:43.400" LastActivityDate="2015-08-19T22:07:43.400" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20307" PostTypeId="2" ParentId="20066" CreationDate="2015-08-19T22:43:36.510" Score="11" Body="&lt;p&gt;&lt;strong&gt;Answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are a father talking to a man who wishes to court your daughter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation (broken into parts):&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intro&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You know who I am, and don't forget it. And don't forget what I'm telling you now: (the man wanting to court the daughter would at least be aware of the father's existence and would have reason to follow the father's council to stay on good terms with the family)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lines 1-2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Years ago, I was in a similar place. So on this occasion I will treat you with grace (the father would have once courted his own wife and had to have dealt with his wife's father, so the father will have at least some understanding of the man's intentions)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lines 3-4&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You have adorned yourself to match my creation. But let’s not forget your lower station (the man has likely prepared himself to meet the father and appear worthy of the daughter, possibly by dressing up. The father considers the man to be of lower station than him in relation to the girl, considering he is the girl's father and the man has relatively little relation with the daughter up to this point)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lines 5-6&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Your hands are your own, and that much is clear. But the work they may do gives me reason to fear (The father is expressing concern that the man may intend to have a sexual relationship with his possibly young or sheltered daughter)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lines 7-8&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Keep them under wraps, and employ no device. No matter how much your mind is enticed (The father is instructing the man to avoid having a sexual relationship with his daughter despite temptations that exist)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lines 9-10&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I will use no tools to intimidate, My existence alone should make you hesitate (I, for one, have definitely heard of fathers hinting at threats with baseball bats or shotguns (tools) over their daughters, but this father appears to be different. In any case, if the man is young the father is likely intimidating even without tools. If the man is a bit older and has intentions of a serious relationship, he probably wants to stay on the family's good side.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lines 11-12&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you return, it best not be late. For- know it now- alone I will wait (Don't keep my daughter out later than curfew, or whatever time you said you'd bring her back. If you do, the father will almost certainly be up worried and waiting, and likely very angry)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lines 13-14&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you want memories more than regrets, Then do not think these idle threats (If you want a long lasting relationship where you can build memories together, you probably don't want to be going behind the father's back, who appears to believe he has the ability to prevent a relationship between the two (again, possibly because the daughter and suitor are both young))&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lines 15-16&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dance around the issue, but do no more, Or you’ll forever be met with a locked, closed door (if the man doesn't follow the father's council he will quite literally be met with a closed door when the father will prevent his daughter (who appears to be still living with the family) from seeing him)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15853" LastEditorUserId="15853" LastEditDate="2015-08-20T01:41:15.333" LastActivityDate="2015-08-20T01:41:15.333" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="20308" PostTypeId="1" CreationDate="2015-08-19T23:49:20.820" Score="5" ViewCount="1577" Body="&lt;p&gt;Do you know The Fitch Cheney Five Cards Trick? The following is a much harder twist of it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Standard 52 cards deck. Magician is outside the room. Assistant asks random guy from the public to shuffle the deck and to select 3 cards from it. Then asks another random guy to select 1 card from these three. Then Assistant lays out the three cards in a row on the table - the last selected face down, the other two face up. Then Assistant calls Magician to enter the room. Magician looks for a while the three cards on the table and names the face down card.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&#xA;What code is used by Assistant and Magician?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep the following in mind:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;There is no communication of any sort between Assistant and Magician.&lt;/li&gt;&#xA;&lt;li&gt;Only the three cards arranged on the table give information to Magician.&lt;/li&gt;&#xA;&lt;li&gt;Repeating the trick several times is safe.&lt;/li&gt;&#xA;&lt;li&gt;It is possible to perform this trick over phone, i.e. Assistant to tell arrangement of the three cards (naming only the two face up cards of course) to Magician.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="15856" LastEditorUserId="15544" LastEditDate="2015-08-20T00:42:51.820" LastActivityDate="2015-08-20T17:04:32.507" Title="Three Cards Trick" Tags="|cards|magic|" AnswerCount="6" CommentCount="12" ClosedDate="2015-08-20T16:58:14.917" ContentLicense="CC BY-SA 3.0" />
  <row Id="20310" PostTypeId="2" ParentId="20308" CreationDate="2015-08-20T01:19:51.887" Score="7" Body="&lt;p&gt;I think this is impossible, but I hope someone points out an error in my reasoning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are $\binom{52}{3}\cdot 3=52\cdot 51\cdot 25$ situations the assistant can be handed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are only $52\cdot 51\cdot 3$ possible arrangements that the magician can be handed by the assistant. This is because what he sees is completely specified by the two face up cards (52 choices for one further to left, 51 for other) and the location of the one face down (3 choices for its location). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $52\cdot 51\cdot 25&amp;gt;52\cdot 51\cdot 3$, by the pigeonhole principle, there must exist two situations the assistant is handed which result in the same arrangement the magician sees. This means the magician will make the same guess in two situations where there are different face down cards, so he must be wrong for one of them.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-08-20T01:19:51.887" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20311" PostTypeId="2" ParentId="18782" CreationDate="2015-08-20T01:25:12.467" Score="71" Body="&lt;p&gt;&lt;strong&gt;Section One&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started by researching futhark, Germanic/Teutonic, Anglo-saxon runes. I don't interpret the characters too tightly as &quot;modern&quot; letters. (&lt;strong&gt;Note:&lt;/strong&gt; These are &lt;strong&gt;stylised&lt;/strong&gt; characters, bull past the flair to &quot;see&quot; them.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Christian_cross_variants&quot;&gt;crosses&lt;/a&gt;&lt;/strong&gt; &lt;strong&gt;ARE&lt;/strong&gt; Greek as suggested by a few commentators in the &lt;a href=&quot;http://britishlibrary.typepad.co.uk/digitisedmanuscripts/2015/08/help-us-decipher-this-inscription.html&quot;&gt;article&lt;/a&gt; linked to in OP. I don't see how the crossbars at the tips could make it mistakeable for a &lt;a href=&quot;https://en.wikipedia.org/wiki/Maltese_cross&quot;&gt;maltese cross&lt;/a&gt;, which debunks the argument for an Islamic origin. My reasoning being the level of craft involved in the rest of the inscription suggests that had the artisan intended to inscribe a maltese cross, he would have done so properly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another commentator high-signing PhD cred suggests the crosses in opening and closing to be the wrapper on the benediction,&#xA;&lt;strong&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=nnrG904talA&quot;&gt;In nomine Patris et Filii et Spiritus Sancti, Amen&lt;/a&gt;&lt;/strong&gt;, which would make them characters as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I'm on the right track the first few characters seem to be more like &lt;code&gt;HDGOG&lt;/code&gt;, I might be off on the &lt;code&gt;O&lt;/code&gt;, I think the tenth character may be a &lt;code&gt;U&lt;/code&gt;.&#xA;The mystery character &lt;code&gt;R #11&lt;/code&gt; may be a &quot;lowercase&quot; &lt;code&gt;N&lt;/code&gt;, which would make sense in that the name/title of Christ is still subordinate to the name God.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, runes are also &lt;strong&gt;still&lt;/strong&gt; believed to possess magic properties, it's likely that this inscription may be a name for the blade, a prayer, or a &quot;spell&quot;. Runes, besides being the regular writing of the people/tribes who regularly used them, were also viewed as mystical/magical by those from other tribes/groups. So I think that's a good place to start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beowulf would be a good primer, &lt;a href=&quot;http://gwydir.demon.co.uk/jo/nordic/runes4.htm&quot;&gt;Beware Grendel...&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; Runes [many] also represent individual words, &lt;a href=&quot;https://en.wikipedia.org/wiki/Gothic_alphabet&quot;&gt;Gothic runes.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's thought that a mixed cipher kind of thing might be going on here between the meaning of individual characters and some peppering with Latin maybe? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is some suggestion the &lt;code&gt;X&lt;/code&gt; character is a separator, otherwise the first letter of the name/title of Christ. &lt;code&gt;X = C&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, here is my first bit of guessing based on the rune theory, the first two letters from the Gothic Rune chart (linked above)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Naudiz&quot;&gt; N &quot;naudiz&quot; or &quot;nyd&quot; or &quot;nauor&quot; &lt;/a&gt; can mean roughly&#xA;  &lt;code&gt;need, constraint, duress, trouble&lt;/code&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Dagaz&quot;&gt; D &quot;dagaz&quot; or &quot;daeg&quot; &lt;/a&gt; means &lt;code&gt;day&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Going back to the suggestion of the cipher/Latin peppering-&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;N irae $\rightarrow$ &lt;code&gt;ire, anger, wrath, trouble&lt;/code&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;D dies $\rightarrow$ &lt;code&gt;day&lt;/code&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;dies irae $\rightarrow$ &lt;code&gt;day of wrath&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is not the craziest theory, but we are talking about a sword here. An incantation/inscription evoking the spirit/magic of death is not unreasonable. Those are rough translations of what those characters actually mean in runic context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://annales.info/evrope/drboglav/t07.htm&quot;&gt;this page&lt;/a&gt; for a listing of &quot;known&quot; translations of similar inscriptions.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Section 2:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This section will be for manipulations of the &quot;translations&quot;.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Word of the Dei... xD , sorry,&#xA;  &lt;a href=&quot;http://www.britannica.com/topic/homoousios&quot;&gt;[H]omoousios&lt;/a&gt; homoousios&#xA;  (Greek: “of one substance”) The XCH of the inscriptio has me reaching, &gt; true, but i think i am following the right track but don't quote me on quod ! xD&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE:&lt;/strong&gt; The &lt;em&gt;n&lt;/em&gt; character in oranipotentis seems to function as a connector with no other value. I wonder if it works like a &lt;a href=&quot;https://en.wikipedia.org/wiki/Bind_rune&quot;&gt;bindrune&lt;/a&gt; ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An interesting note on both the bindrune thought and some [further] Greek connections come by way of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Chi_Rho&quot;&gt;chi-rho&lt;/a&gt; symbol, see the spelling Xpisti below, which looks a little odd. ?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Settled so far...?&#xA;ORVI = ORUI the &quot;u&quot; serve the same function here.&#xA;Another example: sigillum = sigillvm == seal&#xA;So, ORVI, ORUI =&#xA;&#xA;[O]mnipotentis, [R]edemptoris [U]niversorum [I]esu&#xA;Almighty God, the Redeemer of all Jesus&#xA;&#xA;Xpisti = Christ, then what is C? Which again skews my results&#xA;below but, it's still neat how that turned out.&#xA;&#xA;After playing for a bit I think this may be similar&#xA;to a variation of the Greek word for appeal,&#xA;so,&#xA;&#xA;[X]pisti [C]hristi [H]omoousios&#xA;&#xA;Doing a run from Latin &amp;gt;&amp;gt; Greek,&#xA;&#xA;Xpisti Christi Homoousios &amp;gt;&amp;gt; Επίκληση Χριστός ομοούσιος&#xA;Phonetically --------------&amp;gt; Epíkli̱si̱ Christós omooúsios&#xA;&#xA;then Greek to English&#xA;&#xA;Επίκληση Χριστός ομοούσιος &amp;gt;&amp;gt; Invoking Christ consubstantial&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I will temporarily add this idea to my translation in progress but i am not settled on it. See &lt;a href=&quot;https://en.wikipedia.org/wiki/Consubstantiality&quot;&gt;consubstantiality&lt;/a&gt; .&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;From the Russian site;&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;+NDXOXCHWDRGHDXORVI+ &amp;lt;--this is from the article&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;+NDXOXGHMDnCHDXORVI+ &amp;lt;---this is from that site of known translations at the end of section one of my answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Russian&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Во имя господа Христа всемогущего и матери Христа. Во имя матери&#xA;  господа нашего Христа. Во имя господа Христа всемогущего, искупителя&#xA;  вселенной Иисуса.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Latin&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;N&lt;/strong&gt;(omine) &lt;strong&gt;d&lt;/strong&gt;(omini) &lt;strong&gt;X&lt;/strong&gt;(pisti) &lt;strong&gt;o&lt;/strong&gt;(ranipotentis), &lt;strong&gt;X&lt;/strong&gt;(pisti) &lt;strong&gt;g&lt;/strong&gt;(enitricis). &lt;strong&gt;N&lt;/strong&gt;(omine) &lt;strong&gt;m&lt;/strong&gt;(atris) &lt;strong&gt;d&lt;/strong&gt;(omini) &lt;strong&gt;n&lt;/strong&gt;(ostri) &lt;strong&gt;C&lt;/strong&gt;(hristi). &lt;strong&gt;N&lt;/strong&gt;(omine) &lt;strong&gt;d&lt;/strong&gt;(omi-ni) &lt;strong&gt;X&lt;/strong&gt;(pisti) &lt;strong&gt;o&lt;/strong&gt;(mnipotentis), &lt;strong&gt;r&lt;/strong&gt;(edemptoris) &lt;strong&gt;u&lt;/strong&gt;(niversorum) &lt;strong&gt;I&lt;/strong&gt;(esu).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;English&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the name of the Lord Jesus Christ almighty, and mother. In the name&#xA;  of the mother of our Lord Christ. In the name of God the almighty&#xA;  Christ, redeemer of the universe of Jesus.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think the &quot;W&quot; could be attributed to illiteracy or the error that comes with repetition. The &quot;lowercase&quot; is probably what i thought it to be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm still puzzling out particulars, there are letter &quot;substitutions&quot; and so, &quot;alternate&quot; spellings to sort through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google translate is a HORROR! [OMG] Any clergy worth their salt should go there and start fixing it, please!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; I believe the &quot;C&quot; and the &quot;G&quot; are backwards...&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;+ NDXO XGH WDnCHDX ORVI +&#xA;&#xA;Settled mostly....    &#xA; +&amp;lt;&#xA; In nomine Patris et Filii et Spiritus Sancti, Amen,&#xA;&#xA;Latin&#xA;[N]omine [D]omini [X]pisti, [O]ra i potentis&#xA;&#xA;to Greek - phonetic below&#xA;Προσεύχομαι στο όνομα του Χριστού, ο ισχυρός&#xA;Proséf̱chomai sto ónoma tou Christoú, o ischyrós&#xA;&#xA;to English&#xA;I pray in the name of Christ, the powerful&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;???? Going on the G tip =p, same move, Latin-&amp;gt;Greek-&amp;gt;Английский&#xA;This is still a guess.&#xA;&#xA;Latin&#xA;[X]pisti [G]enetricis? [H]omoousios ?&#xA;&#xA;to Greek - phonetic below&#xA;Μητέρα του Χριστού ομοούσιος&#xA;Mi̱téra tou Christoú omooúsios&#xA;&#xA;to English&#xA;Mother of Christ consubstantial&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[O]mnipotentis, [R]edemptoris [U]niversorum [I]esu&#xA;Almighty God, the Redeemer of all Jesus&#xA;&#xA; &amp;gt;+&#xA; In nomine Patris et Filii et Spiritus Sancti, Amen,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Final Guess.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the suggestion by commentator &lt;a href=&quot;http://profile.typepad.com/6p01b8d1456bae970c&quot;&gt;Philip Seddon&lt;/a&gt; ,&#xA;The latter &quot;H&quot; characters may be seen as indicators of a &lt;em&gt;feminine&lt;/em&gt; denominator, meaning N = H as covered&#xA;earlier in my answer on the subject of runic characters interchangeability/similarity/substitution etc..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the subject of the misidentified &quot;R&quot; being a &quot;lowercase&quot; n and, the &quot;C&quot; later &quot;G&quot; actually being&#xA;&quot;G&quot; later &quot;C&quot;, AND, on my almost certainty that &quot;X&quot; is not a &quot;placeholder/seperator&quot; but an alternate &quot;C&quot;&#xA;and lastly the &quot;W&quot; being downside-up &quot;M&quot;, although I believe a double &quot;U&quot;&#xA;e.g. &quot;VV&quot; could be just as like but, I couldn't see/find a fit.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;+NDXOXGHMDnCHDXORVI+&#xA;&#xA;+&amp;lt;&#xA;In Nomine Patris et Filii et Spiritus Sancti, Amen,&#xA;&#xA;N      D      X       O&#xA;Nomine Domini Christi omnipotentis, &#xA;&#xA;X       G          H      M      D      n      C&#xA;Christi genitricis Nomine Matris Domini nostri Christi,&#xA;&#xA;H      D      X       O            R           V           I&#xA;Nomine Domini Christi omnipotentis redemptoris universorum Iesu&#xA;&#xA;&amp;gt;+&#xA;In Nomine Patris et Filii et Spiritus Sancti, Amen,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So,&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Nomine Domini Christi omnipotentis, &#xA;Christi genitricis Nomine Matris Domini nostri Christi,&#xA;Nomine Domini Christi omnipotentis redemptoris universorum Iesu&#xA;&#xA;&#xA;the name of our Lord Jesus Christ Almighty,&#xA;the name of the Mother of Christ, Mother of our Lord Christ,&#xA;the name of Jesus Christ Almighty Redeemer of all&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'm still not satisfied with my answer and have clearly hit my limit of skill with this. I hope that by turning what i have so far over to Community Wiki [and a penalty free format] will attract some more users willing to jump in. It's a neat mystery and I want to know what the inscription actually is.&lt;/p&gt;&#xA;" OwnerUserId="15757" LastEditorUserId="15757" LastEditDate="2015-08-28T19:35:49.470" LastActivityDate="2015-08-28T19:35:49.470" CommentCount="7" CommunityOwnedDate="2015-08-28T19:35:49.470" ContentLicense="CC BY-SA 3.0" />
  <row Id="20312" PostTypeId="2" ParentId="20308" CreationDate="2015-08-20T01:41:08.463" Score="0" Body="&lt;p&gt;It could all depend how the cards&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; are placed.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Number the cards 0 to 51&lt;br&gt;&#xA; The assistant then lays down each of the three cards in one of four positions.&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/fWnya.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fWnya.png&quot; alt=&quot;Card Positions&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA; Turn this into a three digit &lt;a href=&quot;https://en.wikipedia.org/wiki/Quaternary_numeral_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;Quaternary Number&lt;/a&gt;&lt;br&gt;&#xA; convert it back to decimal, and match it up with the numbered cards.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This system can actually encode for 64 different cards if you had some weird deck. :)&lt;/p&gt;&#xA;" OwnerUserId="7715" LastActivityDate="2015-08-20T01:41:08.463" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20316" PostTypeId="2" ParentId="20308" CreationDate="2015-08-20T08:59:53.550" Score="3" Body="&lt;p&gt;&lt;strong&gt;Partial answer&lt;/strong&gt; - only satisfies 1, 2 and 3 &lt;strong&gt;and assumes&lt;/strong&gt; that the magician is permitted the luxury of using a pack of cards that has an assymetric back. I doubt that the audience would be suspicious if this was a otherwise plain back blatantly advertising the name of the show, the magician or the venue.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;p&gt;The magician and assistant have to agree on a suit priority in the event of both face-up cards having the same value. This will allow for one of the cards to be a high card (H) and the other a low card (L).&lt;/p&gt;&#xA; &lt;p&gt;There are six possible arrangements of face-down card, H and L. Adding the orientation of the face-down card into the mix (^ for upside-down and v for right-way-up), this now gives 12 combinations. If the cards are usually laid close together, the 13th value can be indicated by spacing the cards out a little more. In this last case the order of the cards is unimportant.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KICa8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KICa8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The suit would then be indicated by placing the cards &lt;strong&gt;slightly&lt;/strong&gt; off-centre on the table e.g towards the top for spades, right for hearts, bottom for clubs and left for diamonds&lt;/p&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's not perfect, but will certainly work once and, for most audiences would survive repetition.&lt;/p&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-08-20T08:59:53.550" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20317" PostTypeId="1" AcceptedAnswerId="20318" CreationDate="2015-08-20T12:06:23.997" Score="7" ViewCount="740" Body="&lt;p&gt;I found a small letters sequences on Stack Exchange, but some elements are missing :&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;T H B C D L PP WH ?? M CU ?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h2&gt;Question&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;What are the missing elements ?&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Hints&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Hint 1 :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Some elements I wrote can help you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2 :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Every ? is a letter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3 :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Think outside the box.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13384" LastEditorUserId="13152" LastEditDate="2015-08-20T12:25:38.800" LastActivityDate="2015-08-20T15:22:34.380" Title="An &quot;outside the box&quot; sequence" Tags="|lateral-thinking|letter-sequence|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20318" PostTypeId="2" ParentId="20317" CreationDate="2015-08-20T12:14:41.593" Score="12" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; T H B C D L PP WH &lt;strong&gt;AI&lt;/strong&gt; M CU &lt;strong&gt;F&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is from&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/1fv5y.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1fv5y.png&quot; alt=&quot;the links at the top of the footer&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA; the first letters of the links across the top of the site footer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13152" LastActivityDate="2015-08-20T12:14:41.593" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20319" PostTypeId="2" ParentId="20308" CreationDate="2015-08-20T12:29:54.750" Score="1" Body="&lt;p&gt;So assuming the trick is different over the phone (as the asker has sort of implied?), this would work:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Label the deck in any arbitrary way giving the cards each a value 1-52. Let's say it is Ace-&gt;King, Spades-&gt;Clubs-&gt;Hearts-&gt;Diamonds. So ace of spades is 1, 2 of spades is 2, ace of clubs is 14, king of diamonds is 52. Now imagine the table as a 3x3 grid with squares labeled 1-9. When the assistant places the cards down, they can be closest to one of these sections of the table, without being obviously placed there. So now there is a value 1-9 that can be conveyed. There are six ways the cards can be arranged in a row. Label them arbitrarily 1-6. The three cards can be given unique identities because one is flipped, and of the other two, one is higher, one is lower (based on the numbering above). Now there are $9*6=54$ possible combinations of order/location on table. Arbitrarily label those and you have one for each card, plus 2 extra if you use the jokers in your trick.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then for phone:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Just use any vocal queue, like the magician listing off all suits/values&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11727" LastActivityDate="2015-08-20T12:29:54.750" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20320" PostTypeId="2" ParentId="20308" CreationDate="2015-08-20T13:42:30.173" Score="0" Body="&lt;p&gt;I think this is a viable solution for over the phone and on the table although kind of complicated.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The spacing between the cards can be used as Morse code, which can be used over the phone. A space between the cards is a dash and side by side is a dot so you will have 4 combinations that equal 1-4.&lt;p&gt;&lt;strong&gt;--&lt;/strong&gt; = 1&lt;p&gt;&lt;strong&gt;.-&lt;/strong&gt; = 2&lt;p&gt;&lt;strong&gt;-.&lt;/strong&gt; = 3&lt;p&gt;&lt;strong&gt;..&lt;/strong&gt; = 4&lt;p&gt;From here we can use the face down card so determine a 4 bit binary number that has the ones spot as always zero so you can get&lt;p&gt;xxd = 110 = 12&lt;p&gt;xdx = 101 = 10&lt;p&gt;dxx = 011 = 6&lt;p&gt; Now comes the part that feels kind of incomplete or needs tweaking. The suit of the card can specify whether to add or subtract. You will have 16 combinations of Spades, Clubs, Hearts, and Diamonds.&lt;p&gt;SS CC HH DD SH CH HC DC = +&lt;p&gt;SC CS HS DS SD CD HD DH = -&lt;p&gt;For the suit you can place the cards slightly up or slightly down and change your voice inflections over the phone. To make it less obvious you could get the suit based off the card touching the face down card. In all the examples the first arrow is the face down card and you can use your imagination for the case when the face down card is last&lt;p&gt;v^ = S&lt;p&gt;^^ = C&lt;p&gt;vv = H&lt;p&gt;^v = D&lt;p&gt;Also think of the deck as circular so if you subtract 4 from 2 you will get Jack. I'm not sure how to justify this reasoning but if someone finds something wrong with it please let me know.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You should be able to get all combinations with that.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 12 + 1 = &lt;strong&gt;A&lt;/strong&gt;&lt;p&gt;12 + 2 = &lt;strong&gt;2&lt;/strong&gt;&lt;p&gt;12 + 3 = &lt;strong&gt;3&lt;/strong&gt;&lt;p&gt;12 + 4 = &lt;strong&gt;4&lt;/strong&gt;&lt;p&gt;12-1 = &lt;strong&gt;Q&lt;/strong&gt;&lt;p&gt;12-2 = &lt;strong&gt;J&lt;/strong&gt;&lt;p&gt;12-3 = &lt;strong&gt;9&lt;/strong&gt;&lt;p&gt;12-4 = &lt;strong&gt;8&lt;/strong&gt;&lt;p&gt;10 + 1 = Q&lt;p&gt;10 + 2 = &lt;strong&gt;K&lt;/strong&gt;&lt;p&gt;10 + 3 = A&lt;p&gt;10 + 4 = 2&lt;p&gt;10 - 1 = 9&lt;p&gt;10 - 2 = 8&lt;p&gt;10 - 3 = &lt;strong&gt;7&lt;/strong&gt;&lt;p&gt;10 - 4 = &lt;strong&gt;6&lt;/strong&gt;&lt;p&gt;6 + 1 = 7&lt;p&gt;6 + 2 = 8&lt;p&gt;6 + 3 = 9&lt;p&gt;6 + 4 = J&lt;p&gt;6 - 1 = &lt;strong&gt;5&lt;/strong&gt;&lt;p&gt;6 - 2 = 4&lt;p&gt;6 - 3 = 3&lt;p&gt;6 - 4 = 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13925" LastEditorUserId="13925" LastEditDate="2015-08-20T16:57:21.950" LastActivityDate="2015-08-20T16:57:21.950" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20321" PostTypeId="2" ParentId="20308" CreationDate="2015-08-20T13:48:52.717" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The table top is specially designed so that it is translucent/opaque from an angle and is transparent from straight on. Below the table is designed a small mirror that is hidden unless viewed from a certain angle.&lt;br&gt;&#xA;&lt;br&gt;&#xA;When the magician approaches the table, he is able to see the face of the face down card. He can then easily say what the card is.&lt;br&gt;&#xA;&lt;br&gt;&#xA;This solution relies on having the audience members do their selecting away from the table, or having a special bottom on the table that can drop away when the magician approaches, obscuring the trick.&lt;br&gt;&#xA;&lt;br&gt;&#xA;As for over the phone, there have been many possible ways given how this could be done. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12242" LastActivityDate="2015-08-20T13:48:52.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20323" PostTypeId="1" CreationDate="2015-08-20T14:48:56.917" Score="1" ViewCount="5435" Body="&lt;p&gt;I have to fill a whole 3x3 grid in such a way that the sum of each row, column, and main diagonal is 69. I couldn't find any logic to fill it up. I have to use &lt;strong&gt;distinct&lt;/strong&gt; numbers from 1 to 60 for this. Here's the square:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{align}&#xA;    &amp;amp;x_0&amp;amp;x_1&amp;amp; &amp;amp;x_2\\&#xA;    &amp;amp;y_0&amp;amp;y_1&amp;amp; &amp;amp;y_2\\&#xA;    &amp;amp;z_0&amp;amp;z_1&amp;amp; &amp;amp;z_2&#xA;\end{align}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All values $\{x_i, y_i, z_i\}$ should be less than 60.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left.\begin{align}&#xA;x_0+y_0+z_0=69\\&#xA;x_1+y_1+z_1=69\\&#xA;x_2+y_2+z_2=69\\&#xA;\end{align}\right\}\text{columns}\\$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left.\begin{align}&#xA;x_0+x_1+x_2=69\\&#xA;y_0+y_1+y_2=69\\&#xA;z_0+z_1+z_2=69\\&#xA;\end{align}\right\}\text{rows}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left.\begin{align}&#xA;x_0+y_1+z_2=69\\&#xA;x_2+y_1+z_0=69&#xA;\end{align}\right\}\text{diagonals}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone help me out??&lt;/p&gt;&#xA;" OwnerUserId="15872" LastEditorUserId="30633" LastEditDate="2016-11-03T06:55:28.130" LastActivityDate="2021-09-22T14:59:03.283" Title="Fill 3x3 magic square with distinct numbers 1..60 summing up to 69" Tags="|mathematics|logical-deduction|magic-square|calculation-puzzle|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20324" PostTypeId="2" ParentId="20323" CreationDate="2015-08-20T14:56:36.147" Score="10" Body="&lt;p&gt;The easiest way is just to take all 23's and...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; subtract 5, then add back the standard 3x3 magic square.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the solution is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 22 27 20&lt;br&gt;&#xA; 21 23 25&lt;br&gt;&#xA; 26 19 24&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2015-08-20T14:56:36.147" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20326" PostTypeId="2" ParentId="20295" CreationDate="2015-08-20T15:04:36.333" Score="21" Body="&lt;p&gt;Since nobody is providing the proof of minimality, I am going to add one.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;A:&lt;/strong&gt; &lt;em&gt;A circle of radius &lt;span class=&quot;math-container&quot;&gt;$r&amp;lt;1/2$&lt;/span&gt; can cover less than &lt;span class=&quot;math-container&quot;&gt;$ 1/6 $&lt;/span&gt; of the circumference of the unit circle&lt;/em&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/80ONdm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/80ONdm.png&quot; alt=&quot;Circles&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The length of the arc &lt;span class=&quot;math-container&quot;&gt;$DE$&lt;/span&gt; is maximized when the length of chord &lt;span class=&quot;math-container&quot;&gt;$DE$&lt;/span&gt; is maximized, for a circle with radius &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt;, this can be at most &lt;span class=&quot;math-container&quot;&gt;$2r$&lt;/span&gt;. Let, &lt;span class=&quot;math-container&quot;&gt;$\angle A =\theta$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now,&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$2r \ge DE = \sqrt{AD^2+AE^2-2 AD. AE\  \cos \theta} = \sqrt{2-2\cos\theta}$&lt;/span&gt; &lt;br&gt;&lt;br&gt; &lt;span class=&quot;math-container&quot;&gt;$\implies \cos\theta \ge 1-2r^2 \ge 1-2({1 / 2})^2 = 1/2 \implies \theta &amp;lt; 60^{\circ}=1/6 \times 360 ^{\circ}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Now, It follows that you need all 7 circles to cover the circumference if &lt;span class=&quot;math-container&quot;&gt;$r &amp;lt; 1/2$&lt;/span&gt;. But a circle with &lt;span class=&quot;math-container&quot;&gt;$r &amp;lt; 1/2$&lt;/span&gt; can't cover part of the circumference and the center at the same time. So, None of the 7 circles can cover the center. Hence, &lt;span class=&quot;math-container&quot;&gt;$r \ge 1/2$&lt;/span&gt;. A construction with &lt;span class=&quot;math-container&quot;&gt;$r=1/2$&lt;/span&gt; has been shown before.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Upon further research, I found out that this is actually called &lt;a href=&quot;http://mathworld.wolfram.com/DiskCoveringProblem.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;the disk covering problem&lt;/a&gt;. Formally stated this asks:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Given a unit disk, find the smallest radius &lt;span class=&quot;math-container&quot;&gt;$r_n$&lt;/span&gt; required for &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; disks of equal radius to completely cover the unit disk.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The first few values are given &lt;a href=&quot;https://en.wikipedia.org/wiki/Disk_covering_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="15544" LastEditorUserId="81979" LastEditDate="2023-03-01T07:20:55.140" LastActivityDate="2023-03-01T07:20:55.140" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="20327" PostTypeId="1" AcceptedAnswerId="20334" CreationDate="2015-08-20T15:17:49.043" Score="8" ViewCount="1046" Body="&lt;p&gt;A mathematician, a physicist, and an engineer found themselves caught in an ancient anecdote. Lacking a chemist to brew them an anecdote antidote, they fell to arguing over which of them was to be the butt of the joke ... when suddenly the physicist and the engineer disappeared! Between the spots where they'd been standing, a magician stood glaring at the mathematician.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Long have those of thy profession sought to ridicule my kin,&quot; he said menacingly. &quot;Wherefore then, now that thou art in my power, should I release thee? I shall trap thee in unamusing tales with thy fellow &lt;em&gt;scientists&lt;/em&gt; for ever.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;As I see you are a magician,&quot; the mathematician began, &quot;will a &lt;strong&gt;magic square&lt;/strong&gt; enable me to avert your power?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magician scoffed. &quot;What is thy field of expertise within that arid wasteland, Mathematics?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Number theory,&quot; the unperturbed&lt;sup&gt;1&lt;/sup&gt; mathematician replied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Ha! Those fools who seek to discover pattern among the prime numbers! In that case, I challenge thee to find a $2015\times2015$ magic square all of whose entries are &lt;strong&gt;prime&lt;/strong&gt;. If thou canst achieve this feat, I shall release thee unharmed; otherwise, thou art my slave for ever!&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can the mathematician succeed?&lt;/strong&gt; Give a proof either that such a magic square exists or that it cannot exist. Unlike the unfortunate mathematician, you don't actually need to construct it!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt;&lt;sub&gt;Well, of &lt;em&gt;course&lt;/em&gt; he was unperturbed. Perturbation theory is a branch of analysis, not number theory!&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-08-20T16:03:37.127" Title="The magic of the primes" Tags="|mathematics|story|number-theory|magic-square|no-computers|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="20329" PostTypeId="2" ParentId="20327" CreationDate="2015-08-20T15:35:20.390" Score="-1" Body="&lt;p&gt;Assuming the mathematician must use all primes in ascending order, as I gathered from a quick &lt;em&gt;magic square&lt;/em&gt; search:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It cannot be done. The mathematician must place the number $2$ in a square. With 2014 other primes, all odd, its row and column must add up to an even number. All columns and rows without $2$ will add up to an odd number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13237" LastActivityDate="2015-08-20T15:35:20.390" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20331" PostTypeId="1" AcceptedAnswerId="20332" CreationDate="2015-08-20T15:44:02.473" Score="21" ViewCount="4151" Body="&lt;p&gt;My dad once asked me:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;At what time will the second hand, minute hand, and hour hand on an analog clock all be 120° from each other?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;It's a simple question, but I thought it was a fun one to figure out.&lt;/p&gt;&#xA;&lt;p&gt;(Just keep in mind, each time the second hand moves, the minute and hour hand move just a little bit too.)&lt;/p&gt;&#xA;&lt;p&gt;If the second hand ran smoothly (rather than jumping abruptly from tick to tick), would that change your answer?&lt;/p&gt;&#xA;" OwnerUserId="15875" LastEditorUserId="69582" LastEditDate="2020-12-21T21:45:35.913" LastActivityDate="2021-03-01T18:08:26.793" Title="All clock hands at equal degrees from each other" Tags="|mathematics|geometry|calculation-puzzle|time|" AnswerCount="7" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="20332" PostTypeId="2" ParentId="20331" CreationDate="2015-08-20T15:54:39.050" Score="25" Body="&lt;p&gt;At noon, all three hands are at $0$ degrees from each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say the Moment of Truth is after the hour hand has moved around $x$ degrees. In that time, the minute hand moves $12x$ degrees and the second hand moves $720x$ degrees. So we want $x, 12x\pmod{360},$ $720x \pmod {360}$ to be separated by $120$ from each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the minute and second hands we know $720x-12x\pm120$ is a multiple of $360$, which means $59x\pm10$ is a multiple of $30$. From the minute and hour hands we know $11x\pm120$ is a multiple of $360$. So $11x$ and $59x$ are both integers, which means $x$ is an integer (since 11 and 59 are coprime). But then $720x$ is a multiple of $360$, which means $x$ must be $120$ or $240 \pmod {360}$ and so $12x$ is another multiple of $360$ contradiction. So&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; there is no solution!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="15544" LastEditDate="2015-09-12T17:11:41.440" LastActivityDate="2015-09-12T17:11:41.440" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="20333" PostTypeId="1" AcceptedAnswerId="20354" CreationDate="2015-08-20T16:01:52.000" Score="4" ViewCount="515" Body="&lt;h2&gt;Prologue&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;You're a adventurous treasure hunter that just entered the first room of &quot;&lt;em&gt;The temple of Demdarr&lt;/em&gt;&quot;, an ancient temple in the arabian desert. It is rumoured that you have to pass three tests (three rooms) in order to reach the much sighed-for treasure. No-one has ever even dared to take a step into the temple, as you end up dead if you make the slightest mistake. But of course you are confident in yourself and will try it. You enter the first room and encounter three 10x10 tilings. You realise each section has it's own rule or pattern. Now you just have to figure out which tilings are safe, and which are not in order to get to the next room. (Each unsafe tiling only will be triggered if an/multiple object/s over 1kg (2,2 lbs) is/are on it. You will die if this happens to an unsafe tiling and if you are inside the temple. And no, there is no way to survive it :) (you can't run that fast and you are 259,2 meters away from the entry) )&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Rules&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Movement:&lt;/strong&gt; In this room you  have to bear in mind that you can only move one or two steps horizontally/vertically, and only one step diagonaly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Visualization (You can only go on green squares):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LTOZv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LTOZv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;strong&gt;good answer&lt;/strong&gt; must contain all the tilings that are safe, wheter it's visualized in an image or you just list them like $(5,2)$ (which would be the square at the 5th column and 2nd row). Or, wrote/described as a vector, 5 to the right and 2 downwards. (from the top left)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;The Room&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If you have problems reading this or you want a more basic version, &lt;strong&gt;&lt;a href=&quot;https://i.stack.imgur.com/LWcsM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;click here&lt;/a&gt;.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Please also note that there are 4 mistakes in the bottom grid (thanks to Gordon K). False are R3C9, R4C7, R7C4, R8C7, R9C3.&lt;/em&gt;&#xA;&lt;em&gt;Also, in the middle grid, R8C3 and R3C4 should be &quot;&amp;lt;&quot;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(The door to the next room is at the top, so you start at the bottom)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VFtzT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VFtzT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Hints&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Hint 1 for bottom Grid:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The actual clue is &quot;r*f mod 12 = 0&quot;. Don't interpret anything in r and f, these are random variable names because I don't want to give away too much. Why L = 12? You will know that if you find out the pattern in this grid ;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 1 for middle Grid:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Look at the title of the post&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2 for middle Grid:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Actually, this is more of a clarification (I wouldn't have posted another hint so quickly). This grid doesn't have a pattern, although I stated before that every grid has one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 1 for top Grid:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is more than one pattern in this grid&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user14478" LastEditorDisplayName="user14478" LastEditDate="2015-08-20T21:34:41.800" LastActivityDate="2015-08-21T03:47:00.023" Title="The Temple Of Demdarr, God Of Puzzling (Part 1)" Tags="|cipher|pattern|" AnswerCount="2" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="20334" PostTypeId="2" ParentId="20327" CreationDate="2015-08-20T16:03:37.127" Score="15" Body="&lt;p&gt;The mathematician's task is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; possible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; By the &lt;a href=&quot;https://en.wikipedia.org/wiki/Green%E2%80%93Tao_theorem&quot;&gt;Green-Tao Theorem&lt;/a&gt;, there exists a sequence of $2015^2$ primes which form an arithmetic progression. This arithmetic progression can be arranged into an order 2015 magic square using the &lt;a href=&quot;https://en.wikipedia.org/wiki/Siamese_method&quot;&gt;usual method&lt;/a&gt;. Namely, place the lowest prime in the middle of the top row. For each prime placed, the next prime will be placed one square northeast of it (wrapping around at the borders), unless this is already filled, in which case it will be placed one square south.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-08-20T16:03:37.127" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="20335" PostTypeId="2" ParentId="20323" CreationDate="2015-08-20T16:19:14.173" Score="5" Body="&lt;p&gt;&lt;em&gt;This is essentially the same answer as Deusovi, but I thought I'd try to explain things a little better...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There is a well-known &lt;a href=&quot;https://en.wikipedia.org/wiki/Magic_square&quot; rel=&quot;nofollow&quot;&gt;magic square&lt;/a&gt; using the numbers from $1$ to $9$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{align}&#xA;    &amp;amp;2&amp;amp;7&amp;amp; &amp;amp;6\\&#xA;    &amp;amp;9&amp;amp;5&amp;amp; &amp;amp;1\\&#xA;    &amp;amp;4&amp;amp;3&amp;amp; &amp;amp;8&#xA;\end{align}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this square, all the rows, columns, and diagonals add up to $15 = \left(\frac{\text{max}+\text{min}}2\right)\times3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your square, you want the numbers to add up to $69$.  You can use the formula above to determine the $\text{max}$ and $\text{min}$ values for your square:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$69 = \left(\frac{\text{max}+\text{min}}2\right)\times3\\&#xA;\frac{69}3 = \frac{\text{max}+\text{min}}2\\&#xA;23 = \frac{\text{max}+\text{min}}2\\&#xA;2\times23 = \text{max}+\text{min}\\&#xA;\text{max} + \text{min} = 46$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to use consecutive numbers in your square, then the difference between $\text{max}$ and $\text{min}$ needs to be $8$ (just like $9-1$). (If you don't want consecutive numbers in your square, you can count by $2$s (difference $16$), or $3$s (difference $24$), etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for $x+(x+8)=46$ gives $2x=38$, so $x=19$.  That will be the lowest number in your square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$19-1=18$ (&lt;em&gt;lowest number in your square&lt;/em&gt; $-$ &lt;em&gt;lowest number in standard square&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So just add $18$ to every number in the standard square, and you'll have your magic square for $69$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{align}&#xA;    &amp;amp;20&amp;amp;25&amp;amp; &amp;amp;24\\&#xA;    &amp;amp;27&amp;amp;23&amp;amp; &amp;amp;19\\&#xA;    &amp;amp;22&amp;amp;21&amp;amp; &amp;amp;26&#xA;\end{align}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that there are many other solutions, some of which might not use a &quot;standard&quot; magic square (there might not be a regular interval between each of the entries in the square).&lt;/p&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2015-08-20T16:19:14.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20336" PostTypeId="2" ParentId="20323" CreationDate="2015-08-20T16:43:40.427" Score="3" Body="&lt;p&gt;Just put 23 at the center and let two squares at the corners of the same edge be &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt;, then calculate the others based on these two. Just make sure they're all distinct. For example:&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;15&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;44&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;10&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;18&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;23&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;28&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;36&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;31&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;" OwnerUserId="14016" LastEditorUserId="69582" LastEditDate="2021-09-22T14:59:03.283" LastActivityDate="2021-09-22T14:59:03.283" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="20337" PostTypeId="2" ParentId="17274" CreationDate="2015-08-20T17:01:52.843" Score="1" Body="&lt;p&gt;Look in the bottom left 3 x 3 block, and what I will tell you is based on the fact the puzzle MUST have a unique solution. For reference, I will number the squares across as 1-9 and the squares down A-I.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see that the 1 can only be place in either H3 or I3. Now since the squares H4 and I4 form a naked double (the numbers being 1 and 2), it will be obvious that the other number in H3 and I3 cannot be the 2. This is because it would be impossible for the puzzle to have a unique solution. Now, where can the 2 go?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 2 therefor has to be in either H1 or I1, and that eliminates the 2 in the possibilities you have for A1 meaning you can immediately write 4 there. How is that then?&lt;/p&gt;&#xA;" OwnerUserId="15879" LastEditorUserId="15879" LastEditDate="2015-08-20T17:22:09.650" LastActivityDate="2015-08-20T17:22:09.650" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20338" PostTypeId="2" ParentId="20233" CreationDate="2015-08-20T18:08:11.637" Score="86" Body="&lt;p&gt;I'm honest and fair I'll admit:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Snipe&quot;&gt;Snipe&lt;/a&gt;, a bird perhaps capable of deception. We'll have to cover our bases with truths &lt;em&gt;and&lt;/em&gt; lies to be sure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rearranged, I'm of no use to trains:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Spine (useless to trains)&lt;br&gt;&#xA; or&lt;br&gt;&#xA; Pines (a lie! Pine is a type of wood which could be used in &lt;a href=&quot;https://en.wikipedia.org/wiki/Railroad_tie&quot;&gt;railroad ties&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again, and I'm an overt place, warm and well-lit:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Penis (Warm, though often illegal to be overt and well-lit)&lt;br&gt;&#xA; or&lt;br&gt;&#xA; &lt;a href=&quot;https://en.wikipedia.org/wiki/Peins&quot;&gt;Peins, Netherlands&lt;/a&gt; (a lie! With moderate/cool yearly averages and a population of just 270 (in 2006), Peins is likely secluded, cool, and poorly lit.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15882" LastEditorUserId="13237" LastEditDate="2015-09-15T14:48:35.157" LastActivityDate="2015-09-15T14:48:35.157" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20339" PostTypeId="2" ParentId="20323" CreationDate="2015-08-20T18:22:04.233" Score="4" Body="&lt;p&gt;27048 such matrices exist. I present you a few:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;============&#xA;Matrix 1&#xA;21 23 25&#xA;29 22 18&#xA;19 24 26&#xA;============&#xA;Matrix 2&#xA;21 29 19&#xA;23 22 24&#xA;25 18 26&#xA;============&#xA;Matrix 3&#xA;22 23 24&#xA;29 21 19&#xA;18 25 26&#xA;============&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Rest (including these) are present &lt;a href=&quot;https://drive.google.com/file/d/0B7E86owOgl-PallIdVNteFdMM3M/view?usp=sharing&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="12058" LastEditorUserId="12058" LastEditDate="2015-08-20T18:54:38.587" LastActivityDate="2015-08-20T18:54:38.587" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20340" PostTypeId="1" AcceptedAnswerId="20349" CreationDate="2015-08-20T18:40:42.287" Score="10" ViewCount="1259" Body="&lt;p&gt;Left alone, I'm a thing that this riddle has not&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;By every good writer I shan't be forgot&lt;/strike&gt;  &lt;strong&gt;Author's edit:&lt;/strong&gt; This line has been deemed misleading and unfit for publication. The second line of the riddle titled &quot;Another short riddle&quot; shall henceforth be replaced with the following: &lt;em&gt;It's a thing that is a word that has some letters and definitions.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;Rearranged, if you shout this, you might be a liar&lt;/strike&gt; &lt;strong&gt;Author's edit, 2:&lt;/strong&gt; Well, this is embarrassing. Again, replace with the following: &lt;em&gt;The word described above is an anagram of another word. If shouted, it may have been done so by a liar... or not... You never can tell with that sort of thing.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;Again, if shouted, it's a pun used a lot.&lt;/strike&gt; &lt;strong&gt;Author's edit, 3:&lt;/strong&gt; Apologies, again. I hang my head in shame for such sloppy riddling. Replace with: &lt;em&gt;This word is again an anagram of the two above. It's rather popular as a pun, replacing another word with this.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="13237" LastEditorUserId="4050" LastEditDate="2016-01-13T18:45:20.163" LastActivityDate="2016-01-13T18:45:20.163" Title="Another short riddle. Left alone, I'm a thing that this riddle has not" Tags="|riddle|word|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20342" PostTypeId="1" AcceptedAnswerId="20344" CreationDate="2015-08-20T18:44:47.867" Score="1" ViewCount="359" Body="&lt;p&gt;You have to build a maze whose exit can be changed by a computer. The maze consists of one entrance, hallways which can intersect, and $n$ exits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The computer has several wires leaving it each controlling a set of gates, which are located in the hallways. All the gates controlled by the same wire must either be all open or all closed at the same time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In what way should this maze be built so that the number of wires leaving the computer is the lowest possible, while being able to pick any exit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clarification: There cannot be more than one exit open at the same time.&lt;/p&gt;&#xA;" OwnerDisplayName="user9771" LastEditorUserId="5373" LastEditDate="2015-08-20T18:56:43.183" LastActivityDate="2015-08-21T00:36:57.350" Title="Computer-controlled maze" Tags="|mathematics|graph-theory|mazes|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20344" PostTypeId="2" ParentId="20342" CreationDate="2015-08-20T18:50:35.917" Score="2" Body="&lt;p&gt;Since there is no concept of door-toggling, I'm guessing that the minimum number of wires needed is $2 * Log_2{N}$.  Normally this would be a binary choice, where you successively split the exits in half, but you need 2 wires for each split. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;-Old-&lt;/strong&gt;&lt;br&gt;&#xA;Based on the stipulation on the wires (All the gates controlled by the same wire must either be all open or all closed at the same time), it looks like you're stuck at $N$ wires. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If any wire connects 2 doors, then both must be open or both must be closed at all times. So, it is impossible to open only one of them. Hence, each door must have its own wire. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-08-20T19:13:07.227" LastActivityDate="2015-08-20T19:13:07.227" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="20345" PostTypeId="2" ParentId="20011" CreationDate="2015-08-20T18:59:42.533" Score="2" Body="&lt;p&gt;Closest I have found so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Bob Dylan, the musician. My clues relate to my songs:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I work on a farm, but it's not as romantic as you think:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Maggie's Farm&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I need to wake up around sunrise to go to work,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;House of the Rising Sun&quot; - though this is a cover.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and I get paid just 4 pennies per minute.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Have not found a fit.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I had to take a second job to pay the bills.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Have not found a fit.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Honestly, I'd do anything just to get an apartment,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I Am a Lonesome Hobo&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I'm such a sad sack it'll probably never happen.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Man of Constant Sorrow&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15843" LastActivityDate="2015-08-20T18:59:42.533" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20346" PostTypeId="2" ParentId="20331" CreationDate="2015-08-20T19:06:27.470" Score="5" Body="&lt;p&gt;If the hour hand's speed is $v$, then it's $12v$ for the minute hand and $720v$ for the second hand. Since the second hand is always at a minute point, so are the other hands for them to be 120 degrees (20 minute points) away from each other. For the minute hand to be exactly at one of them, the second hand must point upright, but then one hand must point at 8 and the other 4, which is impossible!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the second hand consistently moves, let the digital time be $h:m:s$. Then the hour hand is at $5h+m/12+s/720$, the min hand is at $m+s/60$ and the second hand is at $s$. For them to be 120 degs apart, the fractions representing all of them must have the same denominator. Since this can be the case for $s$ and $s/60$ only if $s=0$ (upright), we'll have to arrive at the same conclusion.&lt;/p&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="14016" LastEditDate="2015-08-21T13:43:20.990" LastActivityDate="2015-08-21T13:43:20.990" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20348" PostTypeId="2" ParentId="20342" CreationDate="2015-08-20T19:26:56.910" Score="1" Body="&lt;p&gt;In general you should build the maze as a binary tree.  In the case that the number of exits is an integer power of $2$, then if $2 ^ x = n$ you can get by with $2 x$ wires.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the number of exits is not a power of $2$, then the smallest number of wires will be $n$ up to $3$ exits, and match $2 x$ where $2 ^ x$ is the next smallest power of $2$ greater than $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt;: I posted this before seeing the edit to other solution.&lt;/p&gt;&#xA;" OwnerUserId="2505" LastEditorUserId="15544" LastEditDate="2015-08-20T19:34:05.063" LastActivityDate="2015-08-20T19:34:05.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20349" PostTypeId="2" ParentId="20340" CreationDate="2015-08-20T19:44:38.237" Score="12" Body="&lt;p&gt;After the last one, I feel obligated to go with my initial gut on this one (though I'm admittedly less confident, might be a bit stretched). My guess is that the word is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;FLOW&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Left alone, I'm a thing that this riddle has not  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The riddle certainly lacks this, what with the author's edits&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By every good writer I shan't be forgot/It's a thing that is a word that has some letters and definitions.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Flow is important in a writing style for the sake of cohesion. As for the revised line, I can't garner anything. While I suspect that this line is the source of some wordplay, maybe it's just as simple as it seems.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rearranged, if you shout this, you might be a liar/The word described above is an anagram of another word. If shouted, it may have been done so by a liar... or not... You never can tell with that sort of thing.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;WOLF is an anagram of flow, and for any familiar with the story of The Boy Who Cried Wolf, it makes sense that someone shouting it might be questioned.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again, if shouted it's a pun used a lot/This word is again an anagram of the two above. It's rather popular as a pun, replacing another word with this  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;FOWL is, I believe at least, a common pun replacing or being replaced by the word FOUL (e.g. a murder most fowl).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Like last time, I could be barking up the wrong tree. Hopefully not though!&lt;/p&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="13970" LastEditDate="2015-08-20T19:48:00.513" LastActivityDate="2015-08-20T19:48:00.513" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20350" PostTypeId="2" ParentId="20333" CreationDate="2015-08-20T20:42:49.247" Score="2" Body="&lt;p&gt;Answer to the bottom grid:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Converting the grid of numbers to their positions in the alphabet, the pattern gives a regular times table grid, however there are five letters in the grid which do not comply with modulo 26 conversion. I'll ignore them for now as they have no bearing on the cells required for the solution.&lt;br&gt;&#xA;&lt;code&gt;&#xA; 1    2   3   4   5   6   7   8   9   10&#xA; 2    4   6   8   10  12  14  16  18  20&#xA; 3    6   9   12  15  18  21  24  27  30&#xA; 4    8   12  16  20  24  28  32  36  40&#xA; 5    10  15  20  25  30  35  40  45  50&#xA; 6    12  18  24  30  36  42  48  54  60&#xA; 7    14  21  28  35  42  49  56  63  70&#xA; 8    16  24  32  40  48  56  64  72  80&#xA; 9    18  27  36  45  54  63  72  81  90&#xA; 10   20  30  40  50  60  70  80  90  100&#xA;&lt;/code&gt;&lt;br&gt;&#xA;Applying &lt;code&gt;mod(x,12)&lt;/code&gt; to these (as L is the 12th letter of the alphabet) gives&#xA;&lt;code&gt;&lt;br&gt;&#xA; 1    2   3   4   5   6   7   8   9   10&#xA; 2    4   6   8   10  &lt;strong&gt;0&lt;/strong&gt;   2   4   6   8&#xA; 3    6   9   0   3   6   9   0   3   6&#xA; 4    8   0   4   8   &lt;strong&gt;0&lt;/strong&gt;   4   8   0   4&#xA; 5    10  3   8   1   6   11  4   9   2&#xA; 6    0   6   0   6   &lt;strong&gt;0&lt;/strong&gt;   6   0   6   0&#xA; 7    2   9   4   11  6   1   8   3   10&#xA; 8    4   0   8   4   &lt;strong&gt;0&lt;/strong&gt;   8   4   0   8&#xA; 9    6   3   0   9   6   3   0   9   6&#xA; 10   8   6   4   2   &lt;strong&gt;0&lt;/strong&gt;   10  8   6   4&#xA;&lt;/code&gt;&lt;br&gt;&#xA; Every zero is safe, but the bold ones are the path you should take.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="15544" LastEditDate="2015-08-21T03:47:00.023" LastActivityDate="2015-08-21T03:47:00.023" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20351" PostTypeId="2" ParentId="20331" CreationDate="2015-08-20T21:02:21.557" Score="2" Body="&lt;p&gt;The two closest times I came up with are below, but even those weren't exact.  I'll have to go back and figure out what time are better as we get a finer time, say if we had a continual running analog clock and measured 100ths of a second.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 5:49:09 and 6:51:51&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15875" LastEditorUserId="69582" LastEditDate="2020-12-21T21:46:25.523" LastActivityDate="2020-12-21T21:46:25.523" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="20352" PostTypeId="2" ParentId="20331" CreationDate="2015-08-20T21:04:09.733" Score="2" Body="&lt;p&gt;Wrote a small mathematica program just to confirm what &lt;a href=&quot;https://puzzling.stackexchange.com/a/20332/5373&quot;&gt;rand al'thor&lt;/a&gt; has already proved elegantly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\left(\theta_s,\theta_m,\theta_h\right)$ be the angle in degree made by the second, minute and hour hand respectively in $t$ seconds such that at $t=0$ all three are zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that at $t=1sec$ $\Rightarrow\left(\theta_s=6,\theta_m=\frac{1}{10},\theta_h=\frac{1}{120}\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\left(\theta_{sm},\theta_{mh},\theta_{sh}\right)$ be the angle between sec-min, min-hr and sec-hr hands respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is the function written in mathematica which takes in one argument which is the time passed in seconds and calculates these three angles at that sec and gives a &lt;code&gt;True&lt;/code&gt; output only when all three are equal.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;t[sec_] := &#xA;Module[{},&#xA;\[Theta]s = (6)*sec;&#xA;\[Theta]m = (1/10)*sec;&#xA;\[Theta]h = (1/120)*sec;&#xA;\[Theta]sm = Mod[\[Theta]s - \[Theta]m, 360];&#xA;\[Theta]mh = Mod[\[Theta]m - \[Theta]h, 360];&#xA;\[Theta]sh = Mod[\[Theta]s - \[Theta]h, 360];&#xA;\[Theta]sm == \[Theta]mh == \[Theta]sh&#xA;];&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So now we can run this function from $t=1s$ to $t=43200s$ (total seconds in 12 hrs) and check for each tick of the second hand if the given condition is satisfied. This can be done in mathematica as follows:-&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;DeleteCases[Table[If[t[i] == True, Sow[i]], {i, 1, 43200}], Null]&#xA;(*{43200}*)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The only value which the above command gives is 43200 which off-course is when we circle back to where we started. &#xA;So to conclude we can say that there is no time between t=0 and t=43200 sec where the three angles are exactly equal to each other.&lt;/p&gt;&#xA;" OwnerUserId="101" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2017-01-08T16:01:38.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20353" PostTypeId="1" AcceptedAnswerId="20360" CreationDate="2015-08-20T22:24:25.337" Score="5" ViewCount="404" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Inscribe a square in a given circle by following the rules of construction with ruler and compass but... &lt;strong&gt;without using the compass&lt;/strong&gt;.&lt;br&gt;&#xA;  The center point of the circle is given too.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15856" LastEditorUserId="15544" LastEditDate="2015-08-21T04:09:39.680" LastActivityDate="2015-08-21T04:09:39.680" Title="Square inscribed in a circle (with a ruler only)" Tags="|geometry|construction|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20354" PostTypeId="2" ParentId="20333" CreationDate="2015-08-20T22:45:55.337" Score="6" Body="&lt;p&gt;&lt;strong&gt;Answer to the middle grid:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;We note that the sequence&lt;br&gt;&lt;br&gt;&#xA;   &lt;code&gt;$! (= {\ ]- ?' }[ ;. *# ~/ _&quot; )+ &amp;lt;^ |&amp;gt;&lt;/code&gt;&#xA;&lt;br&gt;&lt;br&gt;&#xA;contains each character exactly once and is 26 characters long (not&#xA;counting the spaces). This suggests that the characters map to the&#xA;alphabet, we transform the grid&lt;br&gt;&#xA;&lt;code&gt;&#xA; W I I U H L N N Y P&#xA; G O W &lt;strong&gt;R&lt;/strong&gt; S W Z I X G&#xA; L X N E &lt;strong&gt;R&lt;/strong&gt; &lt;strong&gt;A&lt;/strong&gt; U N K Y&#xA; I J U Y C Y Q Y O Z&#xA; F S C O K &lt;strong&gt;D&lt;/strong&gt; C K F I&#xA; G I W S &lt;strong&gt;M&lt;/strong&gt; Z S U W S&#xA; Y H U Q L X Y B Z C&#xA; Y G E P &lt;strong&gt;E&lt;/strong&gt; N G I B H&#xA; O K T &lt;strong&gt;D&lt;/strong&gt; C H X Q O K&#xA; O U I Q Y S W W S C&#xA;&lt;/code&gt;&lt;br&gt;&#xA;We now need to find our way through that, luckily, the title indicates&#xA;the name of our puzzling god: &lt;code&gt;Demdarr&lt;/code&gt;. We can so find the way that&#xA;leads through the middle grid (marked as bold)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answer to the last grid:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The grid is actually divided into four parts: The top left, the top right, the bottom left, the bottom right.&#xA; The operations are applied to the row and column of each field. So for the top left 5x5 grid we have &lt;code&gt;col+row=H&lt;/code&gt;. The top right 5x5 is &lt;code&gt;col-row=D&lt;/code&gt;, the bottom left 5x5 grid is &lt;code&gt;col*row=RD&lt;/code&gt;, the bottom right 5x5 is &lt;code&gt;col mod row=A&lt;/code&gt;. The results of the operations yield the number in each field, and the field is safe if it agrees with the given result.&lt;br&gt;&lt;br&gt;&#xA;The only exception to this is the bottom left grid, here a field is safe if the value mod 26 is either &lt;code&gt;18&lt;/code&gt;/&lt;code&gt;R&lt;/code&gt; or &lt;code&gt;4&lt;/code&gt;/&lt;code&gt;D&lt;/code&gt; &lt;br&gt;&lt;br&gt;&#xA;So the correct path through the whole grid is&lt;br&gt;&#xA;&lt;code&gt;&#xA; 2  3  4  5  6  | 5 6 7 8 9&#xA; 3  4  5  6  7  | &lt;strong&gt;4&lt;/strong&gt; 5 6 7 8&#xA; 4  5  6  7  &lt;strong&gt;8&lt;/strong&gt;  | 3 4 5 6 7&#xA; 5  6  7  &lt;strong&gt;8&lt;/strong&gt;  9  | 2 3 4 5 6&#xA; 6  7  &lt;strong&gt;8&lt;/strong&gt;  9  10 | 1 2 3 4 5&#xA; --------------------------&#xA; 6  12 &lt;strong&gt;18&lt;/strong&gt; 24 &lt;strong&gt;30&lt;/strong&gt; | 0 &lt;strong&gt;1&lt;/strong&gt; 2 3 4&#xA; 7  14 21 28 35 | 6 0 &lt;strong&gt;1&lt;/strong&gt; 2 3&#xA; 8  16 24 32 40 | 6 7 0 &lt;strong&gt;1&lt;/strong&gt; 2&#xA; 9  18 27 36 45 | 6 7 8 0 &lt;strong&gt;1&lt;/strong&gt;&#xA; 10 20 30 40 50 | 6 7 8 9 0&#xA;&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13147" LastEditorUserId="15544" LastEditDate="2015-08-21T03:45:18.847" LastActivityDate="2015-08-21T03:45:18.847" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20355" PostTypeId="2" ParentId="20342" CreationDate="2015-08-20T22:56:55.337" Score="2" Body="&lt;p&gt;The number of wires can be reduced the most by having the hallways form a tree starting from the entrance with one branch ending at each exit. At each layer, each branch of the tree splits into $m$ branches, so we need $m$ wires for each split.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula for the number of wires is $m\log_m(n)$. To find the best value for $m$, we need to find the point at which the value of this formula is the lowest:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;\frac{d}{dm}m\log_m(n) = 0 \\&#xA;\frac{ln(m) + 1}{ln(n)} = 0\\&#xA;m = e&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, $m$ must be an integer, so we pick the nearest integer, which is $3$. This gives us that the number of wires for a maze with $n$ possible exits is $3\log_3(n)$.&lt;/p&gt;&#xA;" OwnerDisplayName="user9771" LastEditorDisplayName="user9771" LastEditDate="2015-08-21T00:36:57.350" LastActivityDate="2015-08-21T00:36:57.350" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20356" PostTypeId="2" ParentId="20353" CreationDate="2015-08-21T01:09:36.323" Score="0" Body="&lt;p&gt;Assumption made: Rectangular ruler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the ruler is a rectangle, put the corner of the ruler on the center dot,spin around to draw the circle. Then at some arbitrary angle trace along the two sides of the ruler. They are perpindicular, now extend those lines to reach the circle forwards and backwards, and then connect the dots where the lines intersect the circles. That is a square.&lt;/p&gt;&#xA;" OwnerUserId="13916" LastActivityDate="2015-08-21T01:09:36.323" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20357" PostTypeId="1" CreationDate="2015-08-21T02:08:02.617" Score="9" ViewCount="325" Body="&lt;p&gt;What property do the following English letters share that other letters do not:  &lt;strong&gt;A&lt;/strong&gt;, &lt;strong&gt;C&lt;/strong&gt;, &lt;strong&gt;D&lt;/strong&gt;, &lt;strong&gt;E&lt;/strong&gt;, &lt;strong&gt;Q&lt;/strong&gt;, &lt;strong&gt;S&lt;/strong&gt;, &lt;strong&gt;W&lt;/strong&gt;, &lt;strong&gt;Y&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The property relates to special words that begin with these letters but not with other letters.&lt;br&gt;&#xA;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Another hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Think about words for clarifying a letter, such as &quot;A as in Alpha,&quot; &quot;B as in Bravo,&quot; &quot;C as in Charlie,&quot; and so on.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Yet another hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The first example is &quot;A as in Are&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15886" LastEditorUserId="15887" LastEditDate="2015-08-21T14:10:42.607" LastActivityDate="2015-08-22T11:26:48.360" Title="What is distinctive about these letters?" Tags="|letter-sequence|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20358" PostTypeId="2" ParentId="20353" CreationDate="2015-08-21T02:19:50.217" Score="1" Body="&lt;p&gt;We know this is possible&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those that don't know the classic construction with straightedge and compass: Perpendicular through the center, connect the points where the lines and arcs meet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Poncelet%E2%80%93Steiner_theorem&quot; rel=&quot;nofollow&quot;&gt;Poncelet-Steiner&lt;/a&gt; shows that straightedge and compass constructions can be done with a straightedge, arbitrary circle and centerpoint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can get parallel lines, perpendiculars and transfer lengths. See how &lt;a href=&quot;http://www.geogebra.org/b/265033&quot; rel=&quot;nofollow&quot;&gt;here.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note, to construct the perpendicular through the center is not directly possible. You need to construct off center through another point first and then transfer it.&lt;/p&gt;&#xA;" OwnerUserId="15857" LastEditorUserId="15544" LastEditDate="2015-08-21T04:08:32.570" LastActivityDate="2015-08-21T04:08:32.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20359" PostTypeId="1" AcceptedAnswerId="20762" CreationDate="2015-08-21T02:41:57.000" Score="5" ViewCount="1623" Body="&lt;p&gt;Here is a 10x10 grid of letters where the letters on each row have been scrambled randomly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Qv8J9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Qv8J9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the unscrambled grid, each cell is part of a word at least 5 letters long. Words are written from left to right or from top to bottom, and all the words are connected together. What is the original grid, and what are all the words it contains?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a copyable version of the scrambled grid:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A   E   S   K   L   N   A   O   P   O&#xA;&#xA;E   N   F   D   E   K   L   I   E   L&#xA;&#xA;A   N   A   C   U   K   T   T   T   G&#xA;&#xA;I   O   N   D   O   H   A   I   M   L&#xA;&#xA;T   N   O   S   S   T   L   I   L   O&#xA;&#xA;E   B   U   E   D   D   I   C   O   O&#xA;&#xA;S   G   I   O   A   D   I   A   M   N&#xA;&#xA;L   S   D   V   M   Y   K   E   E   E&#xA;&#xA;R   E   A   B   O   L   M   A   L   N&#xA;&#xA;Y   S   C   S   A   S   R   Y   E   O&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I didn't realize how hard this puzzle was when I posted it, so I'm adding a hint. Here's the grid where some letters have been filled in. All you have to do is find the missing letters:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fWXvG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fWXvG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user9771" LastEditorDisplayName="user9771" LastEditDate="2015-08-29T20:47:25.133" LastActivityDate="2017-11-19T01:30:06.017" Title="A scrambled grid" Tags="|wordplay|word-search|" AnswerCount="1" CommentCount="25" ContentLicense="CC BY-SA 3.0" />
  <row Id="20360" PostTypeId="2" ParentId="20353" CreationDate="2015-08-21T02:49:32.297" Score="3" Body="&lt;p&gt;Name the center of the circle $O$.  Draw a diameter $AOB$.  Pick another point $C$ on the circle.  By &lt;a href=&quot;https://en.wikipedia.org/wiki/Thales%27_theorem&quot; rel=&quot;nofollow&quot;&gt;Thale's theorem&lt;/a&gt; $AC\perp BC$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the method described &lt;a href=&quot;http://www2.washjeff.edu/users/mwoltermann/Dorrie/34.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; to draw lines parallel to $AB$ and $AC$ through $O$.  The intersections of these lines with the circle form a square.&lt;/p&gt;&#xA;" OwnerUserId="9718" LastActivityDate="2015-08-21T02:49:32.297" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20361" PostTypeId="2" ParentId="11492" CreationDate="2015-08-21T03:39:56.247" Score="7" Body="&lt;p&gt;One thing that is EXTREMELY important is improving your look-ahead! Look-ahead is your ability to foresee future cases and therefore prepare the algorithm for those cases so you can coordinate a pauseless transition from your current algorithm to the foreseen one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;THIS is what separates an above 30 from a 20 and sub 20 and even more so a teen to a sub 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are multiple ways to improve your look-ahead and the most common one is to use a metronome while solving. Basically you solve your cube to the beat of a metronome where you may only make one move per beat and that's it. The reason for this is so you can look at the rest of the cube while doing an algorithm, this is a necessity for increasing look-ahead skills. Eventually you can increase the speed of the metronome more and more at your own pace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically you want to minimize the time you spend just looking at the cube and maximize the time spent physically turning its faces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a sub 20 cuber and look-ahead makes the difference between a 20+ solve and a 13.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also may I ask what kind of f2l you are doing? Depending on your methods i may be able to lend some more help.&lt;/p&gt;&#xA;" OwnerUserId="15888" LastActivityDate="2015-08-21T03:39:56.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20367" PostTypeId="2" ParentId="20331" CreationDate="2015-08-21T14:00:20.247" Score="5" Body="&lt;p&gt;Funny,  I thought about exactly this problem a couple of weeks ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I reasoned as follows:  At 4:00, hours and minutes hands have the correct angle.  That angle repeats 11 times in 12 hours.  When the hour hand moves 1/11 turn, the minute hand moves 12/11 turns, that is 1 + 1/11.  The 120° angle is preserved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarily, when the minutes hand turns 1/11 (or 12/11), the seconds hand turns 60/11 turns, or 5 + 5/11 turns.  So, when hours and minutes hands have the correct angle, the position of the seconds hand is a multiple of 1/11 turn from digit 12.  But you need it at a multiple of 1/11 turn from digit 8.  Since these 2 sets never overlap, an exact solution does not exist.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastActivityDate="2015-08-21T14:00:20.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20368" PostTypeId="2" ParentId="20357" CreationDate="2015-08-21T14:14:18.447" Score="8" Body="&lt;p&gt;They form words that sound like &lt;strong&gt;other&lt;/strong&gt; letters of the alphabet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;pre&gt;Are sounds like R, but starts with A.&lt;br /&gt;&#xA;Cue sounds like Q, but starts with C.&lt;br /&gt;&#xA;Sea sounds like C, but starts with S.&lt;br /&gt;&#xA;Eye sounds like I, but starts with E.&lt;br /&gt;&#xA;Que sounds like K, but starts with Q.&lt;br /&gt;&#xA;Why sounds like Y, but starts with W.&lt;br /&gt;&#xA;You sounds like U, but starts with Y.&lt;br /&gt;&#xA;Double Yew sounds like W, but starts with a D. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;/pre&gt;Odd that that's the one that's not 3 letters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, If you told someone &quot;A as in Are&quot; it would just make them very confused. So these are words that you kinda have to see written down or have spelled out to understand.&lt;/p&gt;&#xA;" OwnerUserId="13068" LastEditorUserId="13068" LastEditDate="2015-08-21T14:27:48.290" LastActivityDate="2015-08-21T14:27:48.290" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="20369" PostTypeId="1" CreationDate="2015-08-21T14:18:44.433" Score="4" ViewCount="4675" Body="&lt;p&gt;Which five-digit number has the $n$th digit representing how many occurrences of the digit $(n-1)$ there are in the number, for all $n\le5$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the fourth digit represents the number of 3s, and the first digit represents the number of 0s.&lt;/p&gt;&#xA;" OwnerUserId="15903" LastEditorUserId="15544" LastEditDate="2015-08-21T14:56:55.780" LastActivityDate="2016-10-01T11:29:46.170" Title="A number with self-referential digits" Tags="|mathematics|logical-deduction|" AnswerCount="6" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20371" PostTypeId="2" ParentId="20369" CreationDate="2015-08-21T14:21:44.650" Score="3" Body="&lt;p&gt;One answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 21200 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2 zeroes, 1 one, 2 twos, 0 three, 0 four&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5359" LastActivityDate="2015-08-21T14:21:44.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20372" PostTypeId="1" AcceptedAnswerId="20620" CreationDate="2015-08-21T14:26:13.827" Score="30" ViewCount="2868" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Here is a riddle quite long&lt;br&gt;&#xA;  I hope you can follow along&lt;br&gt;&#xA;  You'll be delighted&lt;br&gt;&#xA;  While I recite it:&lt;br&gt;&#xA;  I'm known to be unmatched in song.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;You'll find me the whole world around&lt;br&gt;&#xA;  In nature 'tis true I abound&lt;br&gt;&#xA;  You might not expect&lt;br&gt;&#xA;  But just recollect:&lt;br&gt;&#xA;  In various forms I am found.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I am a king of my kind&lt;br&gt;&#xA;  And yet I have been quite maligned&lt;br&gt;&#xA;  Confused for another&lt;br&gt;&#xA;  Get this: I'm her brother&lt;br&gt;&#xA;  Though our differences are well defined.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;We're nothing alike, I insist&lt;br&gt;&#xA;  Her flowery form I resist&lt;br&gt;&#xA;  I'm more aggressive&lt;br&gt;&#xA;  She seems more passive&lt;br&gt;&#xA;  Yet one step from outburst she is.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;She's one of a number, all true&lt;br&gt;&#xA;  I'm not because I come from two&lt;br&gt;&#xA;  She gets her own line&lt;br&gt;&#xA;  I'll never have mine&lt;br&gt;&#xA;  You know me now surely, don't you?  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;See me or listen to flyers&lt;br&gt;&#xA;  Read me or climb up much higher&lt;br&gt;&#xA;  Once solving is done&lt;br&gt;&#xA;  Please pardon the pun&lt;br&gt;&#xA;  And now I must sleep, for I tire.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I worked pretty hard on this one; I wanted it to be challenging but doable. Don't take things too literally on this one and you won't be misled. As always, the green check goes to not only the correct final answer, but the one that also explains all of my hidden clues. I hope you guys enjoy!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ADDED STANZA This should be the nail in the coffin; then the hard work will be in finding all of the hidden--and cute, if I do say so myself--clues in the limerick. But beware, like in the rest of the limerick, don't take everything literally here. Wordplay abounds in this stanza. :  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;From all but one sense do we hide&lt;br&gt;&#xA;  My ghostly sister and I&lt;br&gt;&#xA;  She on a petal&lt;br&gt;&#xA;  Me on a metal&lt;br&gt;&#xA;  I've made many emperors die.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="13970" LastEditDate="2015-08-26T18:52:20.247" LastActivityDate="2015-08-27T19:45:04.023" Title="The Answer Would Make a Good Title" Tags="|riddle|rhyme|limerick|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20373" PostTypeId="1" AcceptedAnswerId="20485" CreationDate="2015-08-21T14:42:06.603" Score="12" ViewCount="461" Body="&lt;p&gt;&lt;strong&gt;Farthingbottom School&lt;/strong&gt; is an elite establishment catering for the educational needs of an international group of students from noteworthy families. They have had complaints from some parents that the school dinners are not very healthy, more specifically that there is not enough fish on offer. As a result, the Headmaster has brought in a menu consultant to determine some suitable options. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The consultant sent out a questionnaire asking the pupils to indicate their preference for the following four options: 'Anchovies', 'mackerel', 'kedgeree', and 'salmon' on the following rating scale:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;+4 Worship&#xA;+3 Adore&#xA;+2 Love&#xA;+1 Like&#xA;-1 Dislike&#xA;-2 Hate&#xA;-3 Detest&#xA;-4 Abhorr&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;These are the responses he got back:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Anchovies&lt;/strong&gt;&lt;br /&gt;&#xA;  &lt;em&gt;Love&lt;/em&gt;: Eldred-Campbell&lt;br /&gt;&#xA;  &lt;em&gt;Like&lt;/em&gt;: Ella-Charlotte&lt;br /&gt;&#xA;  &lt;em&gt;Dislike&lt;/em&gt;: Eloise-Florence; Magnus-Ford-Stephen; Marc-Frances&lt;br /&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;mackerel&lt;/strong&gt;&lt;br /&gt;&#xA;  &lt;em&gt;Love&lt;/em&gt;: Enrico; Portia; Eloise-Florence; Marc-Frances&lt;br /&gt;&#xA;  &lt;em&gt;Like&lt;/em&gt;: Felix&lt;br /&gt;&#xA;  &lt;em&gt;Dislike&lt;/em&gt;: Primula&lt;br /&gt;&#xA;  &lt;em&gt;Hate&lt;/em&gt;: Eldred-Campbell&lt;br /&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;salmon&lt;/strong&gt;&lt;br /&gt;&#xA;  &lt;em&gt;Worship&lt;/em&gt;: Eldred-Campbell&lt;br /&gt;&#xA;  &lt;em&gt;Like&lt;/em&gt;: Ella-Charlotte&lt;br /&gt;&#xA;  &lt;em&gt;Dislike&lt;/em&gt;: Randolph&lt;br /&gt;&#xA;  &lt;em&gt;Hate&lt;/em&gt;: Felix; Primula; Enrico; Marc-Frances; Magnus-Ford-Stephen&lt;br /&gt;&#xA;  &lt;em&gt;Detest&lt;/em&gt;: Portia; Eloise-Florence&lt;br /&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;kedgeree&lt;/strong&gt;&lt;br /&gt;&#xA;  &lt;em&gt;Like&lt;/em&gt;: Felix; Primula; Enrico; Portia; Eloise-Florence; Marc-Frances; Magnus-Ford-Stephen&lt;br /&gt;&#xA;  &lt;em&gt;Dislike&lt;/em&gt;: Eldred-Campbell&lt;br /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It was clear that serving kedgeree was going to suit the most children, however the headmaster anticipated that he would receive a backlash from Eldred-Campbell's parents. So he decided to write a letter to them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&#xA;What was their surname?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus:&lt;/strong&gt;&#xA;What were all the children's surnames?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The name of the school is completely irrelevant.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The rating system is powerful&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;What would this puzzle be about if there was a lack of fish?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="14135" LastEditDate="2015-08-24T11:20:56.613" LastActivityDate="2015-08-24T11:20:56.613" Title="Feeding Units of Fish to Fussy Toffs" Tags="|lateral-thinking|science|" AnswerCount="1" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="20374" PostTypeId="2" ParentId="20369" CreationDate="2015-08-21T14:49:27.880" Score="13" Body="&lt;p&gt;The solution is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 21200&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This can be found by simply iterating from a starting point. Let's choose 40000 as a start (it's as good as any).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 40000 — start with four zeroes (and count them)&lt;br&gt;&#xA; 40001 — count four zeroes, one four&lt;br&gt;&#xA; 31001 — count three zeroes, one one, one four&lt;br&gt;&#xA; 22010 — count two zeroes, two ones, one three&lt;br&gt;&#xA; 21200 — count two zeroes, one one, two twos&lt;br&gt;&#xA; &lt;strong&gt;21200&lt;/strong&gt; — count two zeroes, one one, two twos&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The next row counts the numbers it sees in the previous row. Once we hit two identical numbers back-to-back, we have settled on a solution.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You'll see that 43210, 11111, 22222, 33333, and 44444 are not issues if you make one simple admission.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 43210&lt;br&gt;&#xA; 11111&lt;br&gt;&#xA; 05000&lt;br&gt;&#xA; 400001 — allow counting the 5, but don't permanently extend the length&lt;br&gt;&#xA; 41000&lt;br&gt;&#xA; 31001&lt;br&gt;&#xA; 22010&lt;br&gt;&#xA; 21200&lt;br&gt;&#xA; 21200&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-08-22T15:59:54.100" LastActivityDate="2015-08-22T15:59:54.100" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20375" PostTypeId="2" ParentId="20372" CreationDate="2015-08-21T14:55:12.650" Score="1" Body="&lt;p&gt;I'm not sure about this, but I think the answer could be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Air&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm known to be unmatched in song.&lt;br&gt;&#xA;You'll find me the whole world around  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Air can be found anywhere in the world&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In nature 'tis true I abound  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Air is commonly found in nature&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You might not expect&lt;br&gt;&#xA;But just recollect:&lt;br&gt;&#xA;In various forms I am found.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Air can also be part of a storm, or it can be polluted&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a king of my kind  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is one of the four elements&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And yet I have been quite maligned &#xA;Confused for another  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It has been confused with fire&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Get this: I'm her brother  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; His sister is fire, as it's another of four elements. It's his sister as fire needs air to burn.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though our differences are well defined.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fire is very different from air&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We're nothing alike, I insist  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fire is very different from air&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Her flowery form I resist  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When air touches a fire, it warms up and goes upwards, away from it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm more aggressive  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;She seems more passive  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Strong wind, or hurricanes can tear up trees or houses from the ground, while fire just burns them&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet one step from outburst she is.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A fire can be controlled (for example if it's in a fireplace or a campfire), but it can also spread and be very damaging, like a forest fire&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;She's one of a number, all true  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fire is one of the four elements&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not because I come from two&lt;br&gt;&#xA;She gets her own line&lt;br&gt;&#xA;I'll never have mine&lt;br&gt;&#xA;You know me now surely, don't you?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See me or listen for flyers  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can see the wind moving things, or you can hear planes, which wouldn't work without air.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Read me or climb up much higher  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's windier the higher you go  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once solving is done&lt;br&gt;&#xA;Please pardon the pun  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now I must sleep, for I tire.  &lt;/p&gt;&#xA;" OwnerDisplayName="user9771" LastEditorDisplayName="user9771" LastEditDate="2015-08-21T15:00:59.610" LastActivityDate="2015-08-21T15:00:59.610" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20376" PostTypeId="2" ParentId="20154" CreationDate="2015-08-21T14:59:43.820" Score="5" Body="&lt;h2&gt;The Final Answer&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Is &lt;a href=&quot;http://www.stonesexhibitionism.com/&quot; rel=&quot;nofollow&quot;&gt;Exhibitionism&lt;/a&gt;.  I'm sure &quot;real fans&quot; will love this!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;The Riddle&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Stanza 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This refers to &lt;a href=&quot;http://www.imdb.com/title/tt0091306/&quot; rel=&quot;nofollow&quot;&gt;Jumpin' Jack Flash (1986)&lt;/a&gt;. The song &lt;a href=&quot;https://en.wikipedia.org/wiki/Jumpin%27_Jack_Flash&quot; rel=&quot;nofollow&quot;&gt;Jumpin' Jack Flash&lt;/a&gt; is in the key of B-Flat.  This is not the song that we're looking for however, we need a more &quot;continental&quot; track. &lt;br&gt;&lt;br&gt;&quot;Rule Britannia&quot; and &quot;God Save the Queen&quot; are both &quot;continental&quot; and in the soundtrack for Jumpin' Jack Flash but neither is in the key of D Major (thanks Len)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Stanza 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Harold G refers to &lt;a href=&quot;https://en.wikipedia.org/wiki/Harold_Godwinson&quot; rel=&quot;nofollow&quot;&gt;Harold Godwinson&lt;/a&gt; who was killed at the Battle of Hastings in 1066.  The harrowing or &lt;a href=&quot;https://en.wikipedia.org/wiki/Harrying_of_the_North&quot; rel=&quot;nofollow&quot;&gt;Harrying of the North&lt;/a&gt; occurred in 1069.&#xA;Schmieder refers to &lt;a href=&quot;https://en.wikipedia.org/wiki/Wolfgang_Schmieder&quot; rel=&quot;nofollow&quot;&gt;Wolfgang Schmieder&lt;/a&gt; who indexes JS Bach's music.  BWV 1066-1069 are his Orchestral Suites.  1068 is the famous &quot;Air on the G String&quot; which was used as the jingle for Hamlet Cigars.  With the tobacco publishing ban in the UK they are no longer aired.  The key of this song is D Major.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;The Cipher&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Using the key of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;DMAJOR&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;found above, the code is decrypted using a vigenere cipher, yielding:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;pre&gt;utm three two u five nine two eight six two five six four seven four eight seven&lt;/pre&gt;&#xA;&lt;pre&gt;a six letter function describes this place that links to the baroque master&lt;/pre&gt;&#xA;&lt;pre&gt;find a similar link to stanza one though not of bricks and plaster&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Line 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;UTM coordinate in zone 32 at eastings 592862 and northings 5647487 which converts to a WGS84 coordinate of 50.971488°N 10.322607°E - As seen in Len's answer.&lt;br /&gt;&lt;br /&gt;&#xA;This is the location of Bacchaus, the home of Johann Sebastian Bach, the baroque master.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Line 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The six letter function of the Bachhaus is a museum.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Line 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The &lt;a href=&quot;http://www.rollingstones.com/&quot; rel=&quot;nofollow&quot;&gt;Rolling Stones&lt;/a&gt; are beginning a museum exhibition in April 2016 called &lt;a href=&quot;http://www.stonesexhibitionism.com/&quot; rel=&quot;nofollow&quot;&gt;Exhibitionism&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1777" LastEditorUserId="14135" LastEditDate="2015-08-23T06:08:16.150" LastActivityDate="2015-08-23T06:08:16.150" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20378" PostTypeId="2" ParentId="20196" CreationDate="2015-08-21T15:15:45.330" Score="1" Body="&lt;p&gt;Maybe you could&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Measure the period of the pendulum with the stopwatch and use the formula for the approximate period of a pendulum at small angles to find the acceleration g due to gravity(put the length of the string over the quantity of the period divided by 2 pi quantity squared). You can find how much it differs from the 9.81 m/s^2 at the Earth's surface and calculate from there using Newton's gravitational force approximation formula.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15904" LastActivityDate="2015-08-21T15:15:45.330" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20379" PostTypeId="1" AcceptedAnswerId="20472" CreationDate="2015-08-21T15:21:44.150" Score="4" ViewCount="1115" Body="&lt;p&gt;Last week, I saw a magic show, and was the volunteer for climactic act. The magician introduced this act by talking about the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/6569/a-five-card-trick-how-does-it-work&quot;&gt;Fitch-Cheney card trick&lt;/a&gt;, but then explained that she was going to do an even &lt;em&gt;more&lt;/em&gt; impressive trick. Namely, instead of using a $52$ card deck, she would use a larger deck, with $N$ cards, though she didn't say exactly how big $N$ was. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what happened:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The magician left the room. An assistant asked me to look through the $N$ card deck, and pick out any $5$ cards. The assistant inspected the cards I chose, then gave one back to me, which I put in my pocket. He arranged the remaining four cards in a neat stack, and placed this stack face down on a table. The magician returned, inspected the stack, and successfully guessed the card in my pocket.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the largest value of $N$ for which the trick can be performed? For that $N$, how is it done?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that there is no slight of hand or secret communication. Each card is distinct, the faces are rotationally symmetric, and the stack is placed at the exact same spot on the table every time the trick is performed. The assistant is only allowed to choose which card the volunteer gets ($5$ choices), then permute the four other cards ($4!=24$ choices).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Aside: From &lt;a href=&quot;https://puzzling.stackexchange.com/questions/10004/the-fitch-cheney-card-trick-extended&quot;&gt;this question&lt;/a&gt;, we know $N=53$ is possible.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-08-24T03:11:14.120" Title="Five Card Magic Trick with $N$ card Deck" Tags="|strategy|cards|magic|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20380" PostTypeId="1" AcceptedAnswerId="20473" CreationDate="2015-08-21T16:04:56.363" Score="14" ViewCount="522" Body="&lt;p&gt;&lt;em&gt;To start, I'm strangely missing.&lt;br&gt;&#xA;Modified, I'm what a silly chef lacks.&lt;br&gt;&#xA;Again, and I'm subtly present.&lt;br&gt;&#xA;Do me harm, and I'm perhaps you.&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;With a change, I'm tall and resilient.&lt;br&gt;&#xA;Again, and this is certainly that.&lt;br&gt;&#xA;Again, I'm a guitarist's worry.&lt;br&gt;&#xA;Again, and I'll make a stand.&lt;br&gt;&#xA;Again, and our paths will cross.&lt;br&gt;&#xA;Again, and I'm some kind of kind of cat.&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Adorn me, and I'm also strangely missing.&lt;br&gt;&#xA;Enrich me, and I'll destroy everything.&lt;br&gt;&#xA;Destroy me, and I'm a secluded commander.&lt;br&gt;&#xA;Chop me down, and my feelings are obvious.&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;With a swap, I've been judged.&lt;br&gt;&#xA;With another, I'm a torrid remainder.&lt;br&gt;&#xA;And with another, I'm a torrid beginner.&lt;br&gt;&#xA;And a last, you'd better turn me.&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Top down, I'm virtually nonexistent.&lt;br&gt;&#xA;With a flip, you can find me within.&lt;br&gt;&#xA;Decorate me, and I'm a young helper.&lt;br&gt;&#xA;And again, I'm a soft, brimmed hat --&lt;br&gt;&#xA;One more change, and this is certainly not that!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Note: This is a &lt;strong&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/18767/whos-missing-from-this-glorious-poem-of-dynamic-words&quot;&gt;dynamic poem&lt;/a&gt;&lt;/strong&gt;, meaning that each line references a word, and the words change slightly between each line. There is a motif behind this poem, which encompasses six of the lines. A complete answer will explain each line, as well as the overarching motif of the poem. Also, I promise that this time &lt;a href=&quot;https://puzzling.stackexchange.com/questions/20245/a-crumbled-jumbled-poem&quot;&gt;the lines are in the right order&lt;/a&gt;!&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-08-25T17:12:11.750" Title="A wonderfully poor poem" Tags="|word|wordplay|poetry|" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="20381" PostTypeId="1" AcceptedAnswerId="20389" CreationDate="2015-08-21T16:18:13.550" Score="13" ViewCount="1333" Body="&lt;p&gt;By myself I can be quite persuasive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a letter out front, I'm powerful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With two behind, you'd better act fast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With three out front, I bring great suffering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is this strong word?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="13237" LastEditorUserId="4050" LastEditDate="2016-01-13T18:46:15.537" LastActivityDate="2016-01-13T18:46:15.537" Title="Another four-liner: By myself I can be quite persuasive" Tags="|riddle|word|" AnswerCount="6" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20382" PostTypeId="2" ParentId="20381" CreationDate="2015-08-21T16:38:33.937" Score="7" Body="&lt;p&gt;&lt;em&gt;By myself I can be quite persuasive.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An OLD person can be quite persuasive, since age brings wisdom - they should be listened to.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;With a letter out front, I'm powerful.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; GOLD is power: in the immortal words of Artemis Fowl, &lt;em&gt;aurum potestas est&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;With two behind, you'd better act fast.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As you get OLDER, you need to act fast before it's too late!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;With three out front, I bring great suffering.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A KOBOLD is a wicked mythological creature, which brings suffering to many.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-08-21T16:38:33.937" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20383" PostTypeId="2" ParentId="20381" CreationDate="2015-08-21T17:00:23.333" Score="2" Body="&lt;p&gt;Another possibility (if a bit cheeky)  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tom  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By myself I can be quite persuasive.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tom Riddle was very persuasive to those that followed or opposed him.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With a letter out front, I'm powerful.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;A&lt;/em&gt;tom, the building blocks of the universe, or an atom bomb is very powerful.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With two behind, you'd better act fast.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tom&lt;em&gt;my&lt;/em&gt;, for tommy gun will make you take cover fast.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With three out front, I bring great suffering.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Bot&lt;/em&gt;tom - Many a bottom has brought great suffering to those that came after.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-08-21T17:00:23.333" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20385" PostTypeId="2" ParentId="20379" CreationDate="2015-08-21T17:06:54.687" Score="7" Body="&lt;p&gt;A slightly different way to think about this problem is that the magician is trying to guess which five-card hand was initially drawn from the deck.  Since he can see the other four cards, this is the same as guessing the volunteer's card.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are $\binom{N}{5}$ possible five-card hands, and $4!\binom{N}{4}$ four-card &quot;messages&quot; (including orderings) that the assistant can send the magician.  Thus, we are looking for the largest $N$ such that the number of hands is less than or equal to the number of messages:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\binom{N}{5}\leq 4!\binom{N}{4} &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;solving this inequality yeilds:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;N\leq 124&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The answer GentlePurpleRain gave in &lt;a href=&quot;https://puzzling.stackexchange.com/questions/20379/five-card-magic-trick-with-n-card-deck/20385#comment63511_20379&quot;&gt;his comment&lt;/a&gt;.)  For $N=124$, we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\binom{124}{5}=225\,150\,024&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;possible hands and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;4!\binom{124}{4}=225\,150\,024&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;possible messages.  Since the inequality is saturated, we must devise a one-to-one mapping from hands to messages such that each message is a subset of its corresponding hand.  (I'm still working on that part...)&lt;/p&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-08-21T17:06:54.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20386" PostTypeId="2" ParentId="20381" CreationDate="2015-08-21T17:16:12.950" Score="4" Body="&lt;p&gt;Is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An Act&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By myself I can be quite persuasive.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A good act (performance) can be persuasive&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With a letter out front, I'm powerful.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A &lt;strong&gt;P&lt;/strong&gt;act : When you make a pact with someone, it's a powerful link.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With two behind, you'd better act fast.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Act&lt;strong&gt;or&lt;/strong&gt; , when an actor play, he needs to remember his lines quickly. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With three out front, I bring great suffering.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An &lt;strong&gt;imp&lt;/strong&gt;act : A violent impact can bring suffering &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Just realize that there is the word act in one of the line so it's problably the wrong answer&lt;/em&gt; &lt;/p&gt;&#xA;" OwnerUserId="5359" LastEditorUserId="5359" LastEditDate="2015-08-21T17:42:14.737" LastActivityDate="2015-08-21T17:42:14.737" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20387" PostTypeId="2" ParentId="20196" CreationDate="2015-08-21T17:34:10.053" Score="0" Body="&lt;p&gt;From Wikipedia: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Height&lt;/strong&gt; is the measurement of vertical distance, but has two meanings in common use. It can either indicate how &quot;tall&quot; something is, or how &quot;high up&quot; it is...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; we can interpret &quot;height of the roof from the ground&quot; as &quot;vertical distance from the lowest point on the roof to the highest point on the roof, as observed from the ground&quot;. I hope your relative velocity with the ground is non-relativistic, so the height should be the same when you observe it. If the roof is flat, simply pace off a distance from the highest point, or measure multiples of your own height, then use trigonometry to determine the height of that highest point.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12917" LastActivityDate="2015-08-21T17:34:10.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20388" PostTypeId="2" ParentId="20381" CreationDate="2015-08-21T17:40:10.967" Score="1" Body="&lt;p&gt;Could it be,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Imp&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By myself I can be quite persuasive.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tyrion Lannister in Game of Thrones is an imp and is very persuasive &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With a letter out front, I'm powerful.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A pimp holds a lot of power over his girls&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With two behind, you'd better act fast.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If someone has to impel you to do something you probably don't want to do it, so do it fast&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With three out front, I bring great suffering.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Shrimp. people with shell fish allergies can die from eating a shrimp&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13925" LastActivityDate="2015-08-21T17:40:10.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20389" PostTypeId="2" ParentId="20381" CreationDate="2015-08-21T17:55:20.950" Score="17" Body="&lt;p&gt;Again! This is starting to get eerie! The word is  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;URGE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By myself I can be quite persuasive.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Urges are certainly persuasive&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With a letter out front, I'm powerful.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;SURGE adds one letter and becomes a powerful force (e.g. power surge)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With two behind, you'd better act fast.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;URGENT adds two letters to the back and becomes a force pushing you to move promptly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With three out front, I bring great suffering.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;SCOURGE adds three letters up front and becomes a plague/calamity that spreads suffering.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no related pun this time.&lt;/p&gt;&#xA;" OwnerUserId="13970" LastActivityDate="2015-08-21T17:55:20.950" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20390" PostTypeId="2" ParentId="20381" CreationDate="2015-08-21T18:17:36.103" Score="1" Body="&lt;p&gt;I have a feeling I am onto it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By myself I can be quite persuasive.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; WORD - words are very persuasive&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With a letter out front, I'm powerful.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SWORD - swords are mighty objects&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This one I'm stuck on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With three out front, I bring great suffering.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; KEYWORD - this one is a huge stretch, but people often forget their keywords (passwords?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15909" LastActivityDate="2015-08-21T18:17:36.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20391" PostTypeId="2" ParentId="20369" CreationDate="2015-08-21T18:29:54.220" Score="2" Body="&lt;p&gt;The other answers give a correct solution, but they don't indicate whether it is the only solution.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In fact, 21200 is the only solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is shown by this brute-force Perl solution:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;for (10000 .. 44444) {&#xA;  say $_ if substr($_, 0, 1) == tr/0//&#xA;        and substr($_, 1, 1) == tr/1//&#xA;        and substr($_, 2, 1) == tr/2//&#xA;        and substr($_, 3, 1) == tr/3//&#xA;        and substr($_, 4, 1) == tr/4//;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="314" LastEditorUserId="314" LastEditDate="2015-08-21T19:28:49.683" LastActivityDate="2015-08-21T19:28:49.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20392" PostTypeId="2" ParentId="20369" CreationDate="2015-08-21T19:10:01.240" Score="7" Body="&lt;p&gt;cjm shows in his answer that there is only one correct solution by using a Perl program to brute force the puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can also be demonstrated logically:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First digit:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Cannot be 0, because then there would be at least 1 zero&lt;/li&gt;&#xA;&lt;li&gt;Cannot be 1, because it would mean that there would be at least a 3, in the best case (1xx10). No matter where you put it, there's no final possible solution.&lt;/li&gt;&#xA;&lt;li&gt;Cannot be 3, because the 4th digit would have to be at least a 1, and all other digits, including the one representing the number of 1s would have to be 0&lt;/li&gt;&#xA;&lt;li&gt;Cannot be a 4, because all other digits, including the one describing the number of 4s would have to be 0s&lt;/li&gt;&#xA;&lt;li&gt;Cannot be a 5, because itself would then have to be a 0&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Second digit:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Cannot be a 0, because then only one of the last digits can be a 0, so there would have to be, in the best case, at least a 3 in the number, giving us, in the best case, 20310, which is false.&lt;/li&gt;&#xA;&lt;li&gt;Cannot be a 2, 3, 4 or 5, because with more than one 1 we would have less than two 0s.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Third digit:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Cannot be a 0, because we already have a 2&lt;/li&gt;&#xA;&lt;li&gt;Cannot be a 1, because we already have one 1 and need only one&lt;/li&gt;&#xA;&lt;li&gt;Cannot be a 3, 4 or 5 because the remaining two digits need to be 0s, so we wouldn't have place to fit them&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Fourth and fifth digits:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Must be 0s, because we need at least 2.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Therefore the number is 21200.&lt;/p&gt;&#xA;" OwnerDisplayName="user9771" LastEditorDisplayName="user9771" LastEditDate="2015-08-24T15:54:13.780" LastActivityDate="2015-08-24T15:54:13.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20393" PostTypeId="2" ParentId="20011" CreationDate="2015-08-21T19:40:02.803" Score="1" Body="&lt;p&gt;Could it be,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One of the members of Pink Floyd (Roger Waters maybe?) The album Animals is loosely based on George Orwell's book Animal Farm so maybe him? I don't know who the person is really&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I work on a farm, but it's not as romantic as you think&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The song Pigs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I need to wake up around sunrise to go to work&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Welcome to the Machine.  Getting up everyday for the daily grind of work, a.k.a. the machine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and I get paid just 4 pennies per minute&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I had to take a second job to pay the bills.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Money&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Honestly, I'd do anything just to get an apartment&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Nobody Home&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I'm such a sad sack it'll probably never happen.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Goodbye Blue Sky. Whoever this is has no hope&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13925" LastEditorUserId="13925" LastEditDate="2015-08-21T19:42:50.820" LastActivityDate="2015-08-21T19:42:50.820" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20394" PostTypeId="1" AcceptedAnswerId="20408" CreationDate="2015-08-21T20:39:57.877" Score="-1" ViewCount="402" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I alone describe many things.&lt;br&gt;&#xA;  Add a letter and I become the perfect fit.&lt;br&gt;&#xA;  Add one more and I've hurt many.&lt;br&gt;&#xA;  Drop a letter to make me all natural.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What am I talking about?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="5373" LastEditDate="2016-09-25T13:34:20.000" LastActivityDate="2016-09-25T13:34:20.000" Title="I alone describe many things" Tags="|riddle|word|letters|" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20395" PostTypeId="2" ParentId="20394" CreationDate="2015-08-21T20:59:13.810" Score="2" Body="&lt;p&gt;I alone describe many things.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;It&lt;/strong&gt;&lt;br&gt;&#xA; You can indicate almost anything with &quot;it&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add a letter and I become the perfect fit.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;F&lt;/strong&gt;it&lt;br&gt;&#xA; I don't know how much more perfect &quot;fit&quot; you can get.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add one more and I've hurt many.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fi&lt;strong&gt;a&lt;/strong&gt;t&lt;br&gt;&#xA; Cars are dangerous.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Drop a letter to make me all natural.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fat&lt;br&gt;&#xA; Mammals all have fat naturally.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here's a couple more that are more of a stretch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I alone describe many things.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Kin&lt;/strong&gt;&lt;br&gt;&#xA; Everyone is someone's relative.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add a letter and I become the perfect fit.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;A&lt;/strong&gt;kin&lt;br&gt;&#xA; When you're a good match, you might be akin to the perfect solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add one more and I've hurt many.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A&lt;strong&gt;s&lt;/strong&gt;kin&lt;br&gt;&#xA; It's dangerous to be askin too many questions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Drop a letter to make me all natural.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Skin&lt;br&gt;&#xA; It's pretty natural.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I alone describe many things.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Hard&lt;/strong&gt;&lt;br&gt;&#xA; Many things are difficult.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add a letter and I become the perfect fit.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;S&lt;/strong&gt;hard&lt;br&gt;&#xA; A shard of glass is like a puzzle piece for a broken window.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add one more and I've hurt many.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Shar&lt;strong&gt;e&lt;/strong&gt;d&lt;br&gt;&#xA; Be careful about the gossip you share.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Drop a letter to make me all natural.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Shred&lt;br&gt;&#xA; Animals naturally shred their food.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-08-21T22:17:32.797" LastActivityDate="2015-08-21T22:17:32.797" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20396" PostTypeId="2" ParentId="20394" CreationDate="2015-08-21T22:10:21.443" Score="1" Body="&lt;p&gt;The word is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;RAM&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alone I describe many things:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Ram could be a model of car or an animal or a battering device.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add a letter and I become te perfect fit:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;CRAM, adding a letter and getting a tight fit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add another and I have hurt many:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;CRAMP, adding a letter and getting something that has probably hurt everyone at some point (but don't buy into that nonsense about eating+swimming).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Drop a letter and I become all natural:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;CAMP, remove the R and get a nature-based activity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13970" LastActivityDate="2015-08-21T22:10:21.443" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20397" PostTypeId="1" AcceptedAnswerId="20398" CreationDate="2015-08-22T00:21:12.980" Score="11" ViewCount="359" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DgvUL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DgvUL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; October 11 1884&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15910" LastActivityDate="2015-08-22T00:49:18.703" Title="Who Am I? Born in NYC" Tags="|rebus|" AnswerCount="1" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20398" PostTypeId="2" ParentId="20397" CreationDate="2015-08-22T00:49:18.703" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eleanor Roosevelt&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First part:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;L&quot; in &quot;or&quot; = Eleanor&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second part:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rows of &quot;elt&quot; = Roosevelt&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14013" LastActivityDate="2015-08-22T00:49:18.703" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20399" PostTypeId="1" AcceptedAnswerId="20400" CreationDate="2015-08-22T01:57:53.213" Score="1" ViewCount="244" Body="&lt;p&gt;You are given a rectangular watch, as provided in the picture below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HnWlV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How many times during a period of 12 hours, starting at 12:01 AM and ending at 12:01 PM, do the hour and minute hands divide the rectangle into two shapes of equal area?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9531" LastEditorUserId="7308" LastEditDate="2015-08-23T12:02:26.823" LastActivityDate="2015-08-23T12:02:26.823" Title="Another watch puzzle" Tags="|geometry|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20400" PostTypeId="2" ParentId="20399" CreationDate="2015-08-22T02:35:45.333" Score="2" Body="&lt;p&gt;There are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;11 times&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The only times the rectangle will be split into parts with equal areas is when the hands are opposite each other, which happens 11 times during a 12 hour period (about every 1:05).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11727" LastEditorUserId="11727" LastEditDate="2015-08-22T03:00:22.977" LastActivityDate="2015-08-22T03:00:22.977" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20401" PostTypeId="1" AcceptedAnswerId="20403" CreationDate="2015-08-22T03:10:40.780" Score="0" ViewCount="9840" Body="&lt;p&gt;Here in this circle there are seven segments. 6 numbers have been written in 6 segments following a particular pattern. Find the number which will replace ??? in the seventh segment and also describe the pattern that you have derived.&lt;a href=&quot;https://i.stack.imgur.com/YbpzT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YbpzT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13652" LastEditorUserId="4421" LastEditDate="2015-11-25T18:01:37.097" LastActivityDate="2015-11-25T18:01:37.097" Title="Missing number in a circle?" Tags="|logical-deduction|pattern|number-sequence|" AnswerCount="6" CommentCount="0" ClosedDate="2015-08-22T16:40:24.293" ContentLicense="CC BY-SA 3.0" />
  <row Id="20402" PostTypeId="2" ParentId="20401" CreationDate="2015-08-22T03:16:46.013" Score="5" Body="&lt;p&gt;Answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;39&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Starting with the part that has 3 in it, every other square clockwise is $3*n$ where $n$ is the next prime number. So:&lt;br&gt;$3*1=3$&lt;br&gt;$3*2=6$&lt;br&gt;$3*3=9$&lt;br&gt;$3*5=15$&lt;br&gt;$3*7=21$&lt;br&gt;$3*11=33$&lt;br&gt;$3*13=39$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11727" LastEditorUserId="11727" LastEditDate="2015-08-22T03:20:27.007" LastActivityDate="2015-08-22T03:20:27.007" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20403" PostTypeId="2" ParentId="20401" CreationDate="2015-08-22T05:04:19.623" Score="2" Body="&lt;p&gt;While I think &lt;a href=&quot;https://puzzling.stackexchange.com/a/20402/12700&quot;&gt;MisterEman22's answer&lt;/a&gt; is better, I have an alternate solution.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The missing number should be &lt;strong&gt;51&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How I arrived at this number,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Starting with the number 15, we add twice of the next number clockwise to get the number in the next sector. Then we choose that sector as the first sector and repeat. So, &lt;br&gt;&lt;br&gt;&lt;strong&gt;15&lt;/strong&gt; + 2 x &lt;strong&gt;3&lt;/strong&gt; = &lt;strong&gt;21&lt;/strong&gt;&lt;br&gt;&lt;br&gt;&lt;strong&gt;21&lt;/strong&gt; + 2 x &lt;strong&gt;6&lt;/strong&gt; = &lt;strong&gt;33&lt;/strong&gt;&lt;br&gt;&lt;br&gt;And following the formula again gives&lt;br&gt;&lt;strong&gt;33&lt;/strong&gt; + 2 x &lt;strong&gt;9&lt;/strong&gt; = &lt;strong&gt;51&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-08-22T05:04:19.623" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20404" PostTypeId="2" ParentId="18602" CreationDate="2015-08-22T06:46:33.357" Score="1" Body="&lt;p&gt;The thief:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Edgar&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solution&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I went with a VERY basic approach to this problem. I only used the final statement from Edgar: 'I would not think of committing this crime'. Since in order to prove a solution to this problem we have to know whether it is true of false. If Edgar is innocent we have no way to prove the truthfulness of that statement. But, if he is guilty then know it is false.  Hence Edgar is the thief.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15917" LastActivityDate="2015-08-22T06:46:33.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20405" PostTypeId="2" ParentId="20401" CreationDate="2015-08-22T08:18:58.380" Score="2" Body="&lt;p&gt;Another possibility:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 45&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3 + 3 = 6&lt;br&gt;&#xA; 6 + 3 = 9&lt;br&gt;&#xA; 9 + 6 = 15&lt;br&gt;&#xA; 15 + 6 = 21&lt;br&gt;&#xA; 21 + 12 = 33&lt;br&gt;&#xA; 33 + 12 = 45&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastActivityDate="2015-08-22T08:18:58.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20406" PostTypeId="2" ParentId="20401" CreationDate="2015-08-22T08:38:49.163" Score="0" Body="&lt;p&gt;Answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 30&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;................................&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/Dmtve.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Dmtve.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2375" LastActivityDate="2015-08-22T08:38:49.163" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20407" PostTypeId="2" ParentId="20357" CreationDate="2015-08-22T11:26:48.360" Score="0" Body="&lt;p&gt;My list was:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A as in Are (R)&lt;/li&gt;&#xA;&lt;li&gt;C as in Cue (Q)&lt;/li&gt;&#xA;&lt;li&gt;D as in Double-U (W)&lt;/li&gt;&#xA;&lt;li&gt;E as in Eye (I)&lt;/li&gt;&#xA;&lt;li&gt;Q as in Quai (K)&lt;/li&gt;&#xA;&lt;li&gt;S as in See (C)&lt;/li&gt;&#xA;&lt;li&gt;W as in Why (Y)&lt;/li&gt;&#xA;&lt;li&gt;Y as in You (U)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Runemoro's suggestion of B as in Bee does not work, as the target word has to sound the same as a &lt;em&gt;DIFFERENT LETTER&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can make these in different languages too, but I didn't think I could assume answerers knew German, French, or indeed several other languages.&lt;/p&gt;&#xA;" OwnerUserId="15887" LastActivityDate="2015-08-22T11:26:48.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20408" PostTypeId="2" ParentId="20394" CreationDate="2015-08-22T12:37:39.050" Score="2" Body="&lt;p&gt;The word is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I alone describe many things.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I&quot; describe many things on a daily basis.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add a letter and I become the perfect fit.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It, It could be anything so it is fit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add one more and I've hurt many.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fit, not being fit has hurt many.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Drop a letter to make me all natural.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It, again it could be anything so it is natural.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15919" LastEditorUserId="4312" LastEditDate="2015-11-30T05:49:03.040" LastActivityDate="2015-11-30T05:49:03.040" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20409" PostTypeId="1" CreationDate="2015-08-22T12:39:35.067" Score="10" ViewCount="21203" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CLKVO.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CLKVO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The arrangement of the numbers around the circumference of a standard &#xA;  dart board is as shown below&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  20 1 18 4 13 6 10 15 2 17 3 19 7 16 8 11 14 9 12 5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Oddly enough, no one seems to know for sure how this particular arrangement&#xA;  was selected.  ... it's clear that the numbers are ordered to mix the large and small together, and possibly to separate numerically close values as far as possible (e.g., 20 is far from 19), no one seems to know of any simple criterion that uniquely singles out this particular arrangement as the best possible in any quantitative sense.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;http://www.mathpages.com/home/kmath025.htm&quot; rel=&quot;noreferrer&quot;&gt;http://www.mathpages.com/home/kmath025.htm&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This appears to be an unsolved problem. How did the inventor of the standard dartboard come up with the order of the numbers in such a way as to minimise scores that are produced by inaccurate throws?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone see a pattern or was it just trial and error?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that computers weren't available then (pre 1900), can anyone suggest a pencil and paper method that produces a near-optimum result (and specifically &lt;em&gt;this&lt;/em&gt; result) in a reasonable time?&lt;/p&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="14177" LastEditDate="2015-08-24T17:55:57.420" LastActivityDate="2024-01-03T17:51:17.503" Title="Numbers on a dartboard - we know why they are in that order but how was it calculated without computers?" Tags="|pattern|number-sequence|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20410" PostTypeId="1" AcceptedAnswerId="20413" CreationDate="2015-08-22T13:24:17.163" Score="4" ViewCount="3993" Body="&lt;p&gt;What shape comes next?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;triangle pentagon square&lt;br&gt;&#xA;square hexagon hexagon square&lt;br&gt;&#xA;pentagon hexagon hexagon hexagon square triangle&lt;br&gt;&#xA;hexagon octagon octagon octagon octagon ????&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Looking at the hint already? Shame on you! Anyway, here it is: Often Same&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15920" LastActivityDate="2015-08-22T14:21:56.843" Title="What shape comes next?" Tags="|logical-deduction|pattern|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20411" PostTypeId="2" ParentId="20409" CreationDate="2015-08-22T13:27:07.763" Score="4" Body="&lt;p&gt;The numbering system on a standard dartboard is designed in such a way as to reduce ‘lucky shots’ and reduce the element of chance. The numbers are placed in an order to encourage accuracy and punish inaccuracy. The placing of low scoring numbers either side of large numbers e.g. 1 and 5 either side of 20, 3 and 2 either side of 17, 4 and 1 either side of 18, will punish poor throwing. If you shoot for the 20 segment, the penalty for lack of accuracy is to land in either a 1 or a 5. That is basically it.&lt;/p&gt;&#xA;" OwnerUserId="15920" LastActivityDate="2015-08-22T13:27:07.763" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20413" PostTypeId="2" ParentId="20410" CreationDate="2015-08-22T14:05:07.630" Score="7" Body="&lt;p&gt;answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; square&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Numbers&lt;/strong&gt; are the sum of the angles in each shape&lt;/p&gt;&#xA;&#xA;&lt;p&gt;multiply the first digit by itself, then calculate the sum of the remaining digits. The sum now should be equal to the product.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;3 5 4          : 3*3 = 9   [ 5 + 4 ]               = 9&#xA;&#xA;4 6 6 4        : 4*4 = 16  [ 6 + 6 + 4 ]           = 16&#xA;&#xA;5 6 6 6 4 3    : 5*5 = 25  [ 6 + 6 + 6 + 4 + 3 ]   = 25&#xA;&#xA;6 8 8 8 8      : 6*6 = 36  [ 8 + 8 + 8 + 8 + (4) ] = 36&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; hexagon octagon octagon octagon octagon [square]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12054" LastEditorUserId="12054" LastEditDate="2015-08-22T14:21:56.843" LastActivityDate="2015-08-22T14:21:56.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20414" PostTypeId="2" ParentId="20401" CreationDate="2015-08-22T15:31:56.240" Score="0" Body="&lt;p&gt;The answer should be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 13&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 15 + 3 + 21 + 6 + 33 + 9 + n = 100&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is likely not the true answer, but it's worth a shot.&lt;/p&gt;&#xA;" OwnerUserId="9871" LastActivityDate="2015-08-22T15:31:56.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20415" PostTypeId="1" AcceptedAnswerId="20418" CreationDate="2015-08-22T15:32:36.413" Score="5" ViewCount="293" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am used to keep people out&lt;br&gt;&#xA;  Divide me in half to gain access&lt;br&gt;&#xA;  Replace access's end to discover your history&lt;br&gt;&#xA;  Replace history's beginning to meet your end&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What is it I am talking about?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2015-08-22T15:57:54.860" Title="I am used to keep people out" Tags="|riddle|english|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20416" PostTypeId="2" ParentId="20401" CreationDate="2015-08-22T15:39:00.813" Score="0" Body="&lt;p&gt;I'm almost certain that this is the correct answer, because of the simple explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;39&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For each pair of adjacent numbers, caluclate $\text{largest number} - 2*\text{smallest number}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$15-2*3 = 9$, 9 is already in the circle&lt;br&gt;&#xA;$21-2*3 = 15$, 15 is already in the circle&lt;br&gt;&#xA;$21-2*6 = 9$, 9 is already in the circle&lt;br&gt;&#xA;$33-2*6 = 21$, 21 is already in the circle&lt;br&gt;&#xA;$33-2*9 = 15$, 15 is already in the circle  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only values for the missing number that follow this rule are: $3$, $33$, and $39$. The circle already contains $3$ and $33$, so we'll use $39$.&lt;/p&gt;&#xA;" OwnerDisplayName="user9771" LastActivityDate="2015-08-22T15:39:00.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20417" PostTypeId="2" ParentId="20415" CreationDate="2015-08-22T15:50:00.020" Score="4" Body="&lt;p&gt;I am used to keep people out:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; gates&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Divide me in half to gain access:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; gat (a revolver or pistol; possibly also a reference to how said gates are opened)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Replace access's end to discover your history:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; gad (meaning to go around from one place to another, in the pursuit of pleasure or entertainment)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Replace history's beginning to meet your end:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; sad (deaths are sad)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9871" LastActivityDate="2015-08-22T15:50:00.020" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20418" PostTypeId="2" ParentId="20415" CreationDate="2015-08-22T15:57:54.860" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;password&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Keeps people out&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;pass&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's you in&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;past&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is your history&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;last&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is your end&lt;/p&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-08-22T15:57:54.860" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20419" PostTypeId="2" ParentId="20394" CreationDate="2015-08-22T15:58:02.403" Score="3" Body="&lt;p&gt;Congrats @user! Aw, I was thinking of the first word but didn't figure out how some of the lines would fit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for the exercise @warspyking. I like how the previous line almost alluded to the next word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that the actual answer is out, here is another stretch for fun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I alone describe many things.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;ink&lt;/strong&gt;&lt;br&gt;One can write out the descriptions of many things with a pen or other ink-filled writing instrument.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add a letter and I become the perfect fit.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;link&lt;/strong&gt;&lt;br&gt;Things that link fit well together.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add one more and I've hurt many.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;plink&lt;/strong&gt;&lt;br&gt;As a verb meaning “to shoot&quot;, to plink at someone is likely to hurt that person.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Drop a letter to make me all natural.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;pink&lt;/strong&gt;&lt;br&gt;A colour found in nature or the colour of a ruddy person in his all natural birthday suit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15921" LastActivityDate="2015-08-22T15:58:02.403" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20420" PostTypeId="2" ParentId="19870" CreationDate="2015-08-22T16:06:18.913" Score="10" Body="&lt;p&gt;This is only a partial answer, but it does rule out some broad classes of 17-part solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As has been observed by others, it's convenient to work in units of 1/504 of the loot.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Executive summary (TL;DR)&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The three key contributions of this partial answer are that if there is a solution in 17 parts then:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There are seven pairs which each add up to 63 and one triplet which adds up to 63.&lt;/li&gt;&#xA;&lt;li&gt;There are at least four parts which are &lt;em&gt;not&lt;/em&gt; integral multiples of 1/504 of the loot.&lt;/li&gt;&#xA;&lt;li&gt;There are at least two and at most three parts equal to 56.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h3&gt;Important notes on notation&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;$\fbox{a}$ means an indivisible part equal to $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using superscript as frequency notation, not as exponentiation. E.g. $56^4$ means four 56s, and $\fbox{56}^4$ means four indivisible parts each equal to 56.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textrm{parts} = \textrm{vals}$ means that the parts on the left can be grouped to form the values on the right. E.g. $\fbox{7}^2 + \fbox{a} = 7 + 9$ would require $a = 2$. (I did consider using $\vdash$, but it turns out not to be very readable).&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Basic observations&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;These two are mentioned in other answers, and included here for completeness since I rely on them below.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;No part can be greater than 56.&lt;/li&gt;&#xA;&lt;li&gt;By the pigeonhole principle, the nine-way division of seventeen parts requires at least one singleton; so there is at least one part of $\fbox{56}$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There are other applications of the pigeonhole principle and its generalisations which I haven't seen explicitly in other answers:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There are no more than seven parts greater than 36, for if there were then one of the partitions of 72 would have to contain more than one of them, giving a contradiction. (Similarly there are no more than 9 parts greater than 28).&lt;/li&gt;&#xA;&lt;li&gt;Since no part is greater than 56, each of the 63s in the eight-way division requires at least two parts, so in particular seven of them have exactly two parts and the eighth has three parts. This means that &#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There are nine degrees of freedom.&lt;/li&gt;&#xA;&lt;li&gt;There are at least seven parts which are greater than 31.&lt;/li&gt;&#xA;&lt;li&gt;There are at most two parts smaller than 7 (and if there are exactly two then they must sum to at least 7).&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h3&gt;Brute-forcing partitions into &quot;integral&quot; parts&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Let us define an &quot;integral&quot; part to be a part which is an integer multiple of 1/504 of the loot. Then we can use the observation that the 63s are made up of seven pairs and a triplet to brute-force. There are 25 partitions of 63 into two parts no greater than 56, and 322 partitions of 63 into three parts no greater than 56, so there are only $\binom{25+6}{7} \cdot 322= 846723150$ &quot;integral&quot; cases to brute-force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, applying the requirement for a part of $\fbox{56}$, there are only $\binom{25+5}{6} \cdot 322 + \binom{25+6}{7} \cdot 3 = 199084275$ &quot;integral&quot; cases to brute-force. I've written a computer program to test them, and none of them works.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Fractional parts&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;However, I haven't seen a proof that the parts must be &quot;integral&quot;. Suppose that one of them, $\fbox{a}$ is &quot;fractional&quot;. Then in every share in which it occurs there must be at least one other &quot;fractional&quot; part in order to give an &quot;integral&quot; total. Wlog the share of 56 which contains $\fbox{a}$ contains another fractional part $\fbox{b}$. If $\fbox{b}$ is also in the same share of 63 as $\fbox{a}$ and the same share of 72 as $\fbox{a}$ then we could merge them into one part of $\fbox{a+b}$ getting a 16-part solution, which is impossible. Therefore there must be one &quot;integral&quot; share which contains $\fbox{a}$ but not $\fbox{b}$, and one which contains $\fbox{b}$ but not $\fbox{a}$. So we require at least &lt;em&gt;four&lt;/em&gt; &quot;fractional&quot; parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this bound is tight, because if we have four &quot;half-integral&quot; parts $\fbox{a}, \fbox{b}, \fbox{c}, \fbox{d}$ then we can pair them as $(\fbox{a} + \fbox{b}, \fbox{c} + \fbox{d}); (\fbox{a} + \fbox{c}, \fbox{b}+\fbox{d}); (\fbox{a} + d, \fbox{b}+\fbox{c})$ in the three divisions. This doesn't allow grouping them into fewer parts or rounding two up and two down.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Upper bound on parts equal to 56.&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;There are at most three parts equal to $\fbox{56}$. Proof is by contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that we have $\fbox{56}^4$. Then at least three of them come from pairs which add up to 63, so there are at least three parts equal to $\fbox{7}$. And looking at the seven-way division into 72s, there are thirteen non-56 parts to make four 16s and three 72s; each 72 takes at least two of those parts, leaving seven to make four 16s, so there must be a $\fbox{16}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far our known parts are $\fbox{7}^3 + \fbox{16} + \fbox{56}^4$, and there are nine unknown parts $\fbox{a} \ldots \fbox{i}$. If we take $$\begin{eqnarray}&#xA;\fbox{7}^3 + \fbox{16} + \fbox{56}^4 + \fbox{a} + \fbox{b} + \fbox{c} + \fbox{d} + \fbox{e} + \fbox{f} + \fbox{g} + \fbox{h} + \fbox{i} &amp;amp; = 56^9 \\&#xA;\fbox{7}^3 + \fbox{16} + \fbox{56}^4 + \fbox{a} + \fbox{b} + \fbox{c} + \fbox{d} + \fbox{e} + \fbox{f} + \fbox{g} + \fbox{h} + \fbox{i} &amp;amp; = 63^8 \\&#xA;\fbox{7}^3 + \fbox{16} + \fbox{56}^4 + \fbox{a} + \fbox{b} + \fbox{c} + \fbox{d} + \fbox{e} + \fbox{f} + \fbox{g} + \fbox{h} + \fbox{i} &amp;amp; = 72^7 \\&#xA;\end{eqnarray}$$ and remove known parts from the equations where there's only one value on the RHS which they fit or where they exactly match a value on the RHS, we get $$\begin{eqnarray}&#xA;\fbox{7}^3 + &amp;amp; \fbox{a} + \fbox{b} + \fbox{c} + \fbox{d} + \fbox{e} + \fbox{f} + \fbox{g} + \fbox{h} + \fbox{i} &amp;amp; = 40 + 56^4 \\&#xA;             &amp;amp; \fbox{a} + \fbox{b} + \fbox{c} + \fbox{d} + \fbox{e} + \fbox{f} + \fbox{g} + \fbox{h} + \fbox{i} &amp;amp; = 7 + 47 + 63^3 \\&#xA;\fbox{7}^3 + &amp;amp; \fbox{a} + \fbox{b} + \fbox{c} + \fbox{d} + \fbox{e} + \fbox{f} + \fbox{g} + \fbox{h} + \fbox{i} &amp;amp; = 16^3 + 72^3 \\&#xA;\end{eqnarray}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look at the seven-way division into 72s, each 16 on the RHS requires at least one of the unknown parts (since 16 is not a multiple of 7), and each of the 72s requires at least &lt;em&gt;two&lt;/em&gt; of the unknown parts &lt;em&gt;unless&lt;/em&gt; it has &lt;em&gt;all three&lt;/em&gt; of the $\fbox{7}$s (since $72 - 7 - 7 = 58 &amp;gt; 56$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case split:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Case 1: one of the 72s has only one of the unknown parts (wlog $\fbox{7}^3 + \fbox{i} = 72$, $i = 51$).&#xA;Let's look at the three divisions of the loot, removing known parts as before: $$\begin{eqnarray}&#xA;\fbox{7}^3 + &amp;amp; \fbox{a} + \fbox{b} + \fbox{c} + \fbox{d} + \fbox{e} + \fbox{f} + \fbox{g} + \fbox{h} &amp;amp; = 5 + 40 + 56^3 \\&#xA;&amp;amp; \fbox{a} + \fbox{b} + \fbox{c} + \fbox{d} + \fbox{e} + \fbox{f} + \fbox{g} + \fbox{h} &amp;amp; = 7 + 12 + 47 + 63 \\&#xA;&amp;amp; \fbox{a} + \fbox{b} + \fbox{c} + \fbox{d} + \fbox{e} + \fbox{f} + \fbox{g} + \fbox{h} &amp;amp; = 16^3 + 72^2&#xA;\end{eqnarray}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 5 and 7 (from, respectively, the nine-way and the eight-way divisions) take at least two unknown parts between them. $5 + 7 &amp;lt; 16$, so each of the 16s in the seven-way division takes at least one additional unknown part; $72 - 5 - 7 = 60 &amp;gt; 56$, so each of the 72s takes at least two additional unknown parts. That requires 2 + 3 + 4 = 9 unknown parts, and there are only eight left. This case is impossible.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Case 2: each of the 72s has two of the unknown parts. Wlog each of $\fbox{a}, \fbox{b}, \fbox{c}$ goes towards a different 16, and each of $(\fbox{d} + \fbox{e}), (\fbox{f} + \fbox{g}), (\fbox{h} + \fbox{i})$ goes towards a different 72. (There are also the $\fbox{7}^3$ to assign).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observe that $a \equiv b \equiv c \equiv d + e \equiv f + g \equiv h + i \equiv 2 \pmod 7$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observe also that the eight-way division is still requiring us to produce a 7 from the unknown parts. If we produce it as a $\fbox{7}$ then it can't come from a 16 (since $7 \not\equiv 2 \pmod 7$), so it must come from a 72; but then in order that the other unknown part which contributes towards that 72 be $\le 56$ we must contribute two of the previously known $\fbox{7}$ towards that 72, and this is equivalent to case 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the 7 must come from two parts. (See &quot;Basic observations&quot;: there are at most two parts smaller than 7). If both of those two parts contribute towards 72s (wlog $d + f = 7$) then even putting all three known $\fbox{7}$s towards them as well we require $e + g = 116$, which requires a part greater than 56.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one of the two parts must contribute towards a 16, and since the only positive number smaller than 7 which is equal to 2 (mod 7) is 2, we have wlog $a = 2$ and $d = 5$. But then $\fbox{a}$ must be grouped with two $\fbox{7}$s to make 16, leaving $\fbox{7} + \fbox{5} + \fbox{e} = 72$, so we need a part equal to 60, which is impossible. Both cases are now eliminated.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h3&gt;Lower bound on parts equal to 56&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma&lt;/strong&gt;: if the parts (mod 7) are $a^x + (-a)^x + 0^{17-2x}$ (where obviously $x \le 8$) then wlog a = 2 and x = 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;: mod 7, the seven-way division is $2^7$. We can ignore the $0^{17-2x}$ for the purposes of forming those $2^7$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider first the integral values of a.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a = 0: useless.&lt;/li&gt;&#xA;&lt;li&gt;a = 1: $a^2 = (-a)^5 = 2$. We require $x = 10$ to make $2^7$ as $(a^2)^5 + ((-a)^5)^2$. That's too high.&lt;/li&gt;&#xA;&lt;li&gt;a = 2: $a = (-a)^6 = 2$. We require $x = 6$ to make $2^7$ as $a^6 + ((-a)^6)^1$. That's possible.&lt;/li&gt;&#xA;&lt;li&gt;a = 3: $a^3 = (-a)^4 = 2$. We require $x = 12$ to make $2^7$ as $(a^3)^4 + ((-a)^4)^3$. That's too high.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now consider rational values of a. Let $a = p/q$ with $\gcd(p, q) = 1$. Then the minimum value of x can be found by taking the value which corresponds to $a = p$ and multiplying by $q$. But only $p=2$ works even with $q$ as low as 1, and any integral multiple of 6 greater than 6 is also greater than 8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Corollary&lt;/strong&gt;: there is no 17-part solution where the parts (mod 7) are $a^x + (-a)^x + 0^{17-2x}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;: Since one of the shares of 72 would have to include all six parts equal to -2 (mod 7), that would leave at most eleven parts for the remaining six shares, so we would require a part $\fbox{72}$, which is impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem&lt;/strong&gt;: there is no 17-part solution with only one part of $\fbox{56}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;: suppose the contrary: that we have a 17-part solution with only one part of $\fbox{56}$. Then that leaves sixteen parts to make eight 56s, so they must pair up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a graph whose vertices are the parts and which has edges connecting the vertices in each pair of the nine-way division $56^9$, and in each pair of the eight-way division $63^8$. No vertex has more than two edges (one for a pair summing to 56, and one for a pair summing to 63). There are no cycles (because they would have to have alternating edges, but then we have $2x$ vertices which sum to both $56x$ and $63x$, so $x$ must be 0). The $\fbox{56}$ is the only vertex with no edge to pair it in a sum of 56; the three vertices of the triple which sums to 63 are the only vertices not to have an edge to pair them in a sum of 63. Therefore we have two chains: one from $\fbox{56}$ to one vertex of the triple; the other from the second vertex of the triple to the third vertex of the triple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observe that since $56 = 63 = 0 \pmod 7$, the vertices in a chain alternate in sign (mod 7). So the chain which contains $\fbox{56}$ contains only multiples of 7, and the other chain must have equal numbers of $a \pmod 7$ and $-a \pmod 7$. By the corollary above, there is no solution.&lt;/p&gt;&#xA;" OwnerUserId="228" LastEditorUserId="228" LastEditDate="2015-08-31T19:41:16.523" LastActivityDate="2015-08-31T19:41:16.523" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20421" PostTypeId="1" AcceptedAnswerId="20423" CreationDate="2015-08-22T16:06:54.410" Score="-1" ViewCount="260" Body="&lt;p&gt;At night, someone is in a completely empty field. He sees absolutely nothing around him and there are no stars in the sky. He does not have a compass, or any other tools that can be used for navigation, and the only think he knows about the time is that it's before midnight. He's been there for less than a minute, but he already knows which way is north. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did he know?&lt;/p&gt;&#xA;" OwnerDisplayName="user9771" LastActivityDate="2015-08-22T17:07:10.940" Title="Finding north without a compass" Tags="|riddle|situation|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20422" PostTypeId="2" ParentId="20421" CreationDate="2015-08-22T16:13:51.143" Score="4" Body="&lt;p&gt;Answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He is on the ice field at the South pole. Any direction is North.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="14177" LastEditDate="2015-08-22T16:27:08.863" LastActivityDate="2015-08-22T16:27:08.863" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="20423" PostTypeId="2" ParentId="20421" CreationDate="2015-08-22T16:15:47.067" Score="1" Body="&lt;p&gt;&lt;strong&gt;Even though this answer got accepted: This answer will not really work and it is not 100% correct. Thanks to &lt;a href=&quot;https://puzzling.stackexchange.com/users/228/peter-taylor&quot;&gt;Peter Taylor&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/users/15544/anachor&quot;&gt;Anachor&lt;/a&gt; for catching that. I would be interested if there is a better answer (there surely is).&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'm just gonna post my answer... &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Becuase it's before midnight, the illuminated side of the moon faces west. (After midnight it's east). Now just face west and turn right - Then you face north.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user14478" LastEditorUserId="-1" LastEditorDisplayName="user14478" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2015-08-22T17:07:10.940" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20424" PostTypeId="2" ParentId="17882" CreationDate="2015-08-22T16:46:52.633" Score="3" Body="&lt;p&gt;The longest non-DE/RE solution with words from the SOWPODS dictionary is a unique 9-letter solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DIGNIFIES / SIGNIFIED&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are also two 8-letter solutions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DIPHONES / SIPHONED and DWINDLES / SWINDLED&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="228" LastActivityDate="2015-08-22T16:46:52.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20425" PostTypeId="1" AcceptedAnswerId="20429" CreationDate="2015-08-22T16:51:34.377" Score="-5" ViewCount="139" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am used to comprehend&lt;br&gt;&#xA;  Add one letter and I'm your food&lt;br&gt;&#xA;  Move a letter to be magical&lt;br&gt;&#xA;  Drop a letter to quiver in fear&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What exactly am I talking about?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2015-08-22T17:17:36.783" Title="I am used to comprehend" Tags="|riddle|english|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20426" PostTypeId="1" AcceptedAnswerId="20427" CreationDate="2015-08-22T16:55:29.427" Score="2" ViewCount="192" Body="&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;What if split is absent designedly and letters occupy more spaces obscuring obfuscated prototype?&lt;/h2&gt;&#xA;" OwnerUserId="15856" LastActivityDate="2015-08-22T17:03:47.307" Title="The most rational answer" Tags="|enigmatic-puzzle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20427" PostTypeId="2" ParentId="20426" CreationDate="2015-08-22T17:03:47.307" Score="3" Body="&lt;p&gt;Each word of the question&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; has one more letter than the corresponding digit of $\pi$ (rounded).&lt;br&gt;&#xA; &lt;code&gt;4 2 5 2 6 10 3 7 6 4 6 9 10 9&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;3 1 4 1 5  9 2 6 5 3 5 8  9 8&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The phrase 'letters occupy more spaces' in the question suggests extra letters and 'obscuring obfuscated prototype' makes me think of something now being hidden.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The title:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\pi$ is a ratio ($circumference \div diameter$)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13152" LastActivityDate="2015-08-22T17:03:47.307" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20428" PostTypeId="1" AcceptedAnswerId="20432" CreationDate="2015-08-22T17:10:16.047" Score="13" ViewCount="714" Body="&lt;p&gt;As the title suggests, the accepted answer will explain how each word was found and link the four relevant &lt;a href=&quot;https://www.wikipedia.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; pages. Some parts are definitely harder than others, but once several words are found, the rest will fall into place. Happy solving!&#xA;&lt;hr&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/IoVym.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IoVym.jpg&quot; alt=&quot;A rebus containing five words&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13237" LastActivityDate="2015-08-22T19:00:01.940" Title="Five words. Four Wikipedia pages. One rebus" Tags="|rebus|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20429" PostTypeId="2" ParentId="20425" CreationDate="2015-08-22T17:17:36.783" Score="3" Body="&lt;p&gt;&lt;em&gt;I am used to comprehend&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You often comprehend things by READing them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Add one letter and I'm your food&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BREAD is a kind of food.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Move a letter to be magical&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wizards stereotypically have long BEARDs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Drop a letter to quiver in fear&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A BEAR might make someone quiver in fear.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-08-22T17:17:36.783" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20430" PostTypeId="1" AcceptedAnswerId="20431" CreationDate="2015-08-22T17:32:00.493" Score="-2" ViewCount="187" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am a basic necessity for survival&lt;br&gt;&#xA;  Drop survival's end and I'm a temple for meditation&lt;br&gt;&#xA;  Repeat temple's last to electrical&lt;br&gt;&#xA;  Replace electricity's first to sleep well at a friend's&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What on Earth could I be talking about?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2015-08-22T19:33:14.180" Title="I am a basic necessity for survival" Tags="|riddle|english|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20431" PostTypeId="2" ParentId="20430" CreationDate="2015-08-22T17:41:51.337" Score="4" Body="&lt;p&gt;&lt;em&gt;I am a basic necessity for survival&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; WATER.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Drop survival's end and I'm a temple for meditation&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; WAT (see &lt;a href=&quot;https://en.wikipedia.org/wiki/Wat&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Repeat temple's last to electrical&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; WATT is a unit of (e.g. electrical) power.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Replace electricity's first to sleep well at a friend's&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BATT as in batting (wadding for a mattress) - thanks &lt;strong&gt;Somo145&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;warspy, you're making these too easy now! Also the watt is a unit of any sort of power, not just electrical.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-08-22T19:33:14.180" LastActivityDate="2015-08-22T19:33:14.180" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="20432" PostTypeId="2" ParentId="20428" CreationDate="2015-08-22T19:00:01.940" Score="19" Body="&lt;ol&gt;&#xA;&lt;li&gt;Image shows &lt;strong&gt;Alexander&lt;/strong&gt; (the Great).&lt;/li&gt;&#xA;&lt;li&gt;Cardinals are above bishops. The &lt;strong&gt;Pope&lt;/strong&gt; is above the cardinals.&lt;/li&gt;&#xA;&lt;li&gt;Splitting &quot;bed&quot; and &quot;cent&quot; according to the image makes the word &lt;strong&gt;Benedict&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&quot;Tavern&quot;-&quot;vet&quot;+&quot;~kobold&quot;-&quot;cob&quot; makes &lt;strong&gt;Arnold&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&quot;Black&quot; in German is &quot;schwarz&quot; and &quot;harrow&quot; in German is &quot;egge&quot;. Putting them together makes &lt;strong&gt;Schwarzenegger&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Taking them in pairs:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Alexander_Pope&quot;&gt;Alexander Pope&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Pope_Benedict&quot;&gt;Pope Benedict&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Benedict_Arnold&quot;&gt;Benedict Arnold&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Arnold_Schwarzenegger&quot;&gt;Arnold Schwarzenegger&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-08-22T19:00:01.940" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20434" PostTypeId="1" AcceptedAnswerId="20449" CreationDate="2015-08-22T19:52:39.963" Score="7" ViewCount="696" Body="&lt;p&gt;In a simple combination lock, a sequence of several digits is used as a password, with one wheel per digit. Generally, it is not possible to unlock the lock without knowing the password, but due to the workings of this lock's mechanism, a &quot;click&quot; is heard whenever a wheel is rotated into the correct position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the following lock has 3 wheels, each with the digits from 0 to 9. Trying all possible combinations would require up to 1000 attempts to unlock the lock. But due to the clicking mechanism, it is possible to so in at most 30 tries (up to 10 tries for each wheel, where a &quot;try&quot; refers to a single rotation of a wheel).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yazdEm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yazdEm.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, instead of 10 digits, you can design the lock to have $k$ digits on each wheel ($k \in \mathbb{N},\ k\geq 2$).  For security reasons, we want the lock to have at least $N=10^{10}$ different possible combinations. The lock can have as many wheels as needed.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What is the optimal value of $k$ to &lt;em&gt;minimize&lt;/em&gt; the number of tries (wheel rotations) needed to open the lock in the worst case? What about the average case?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Now, suppose that the lock can have different numbers of digits at different places. For example: The first wheel can hold 7 values (0 to 6), the second can hold 4 values and so on.  Again, how can we minimize the number of tries in the average and worst cases?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Can you generalize for any $N$?&lt;/p&gt;&#xA;" OwnerUserId="15544" LastEditorUserId="4269" LastEditDate="2015-09-12T17:06:45.830" LastActivityDate="2020-06-18T15:27:42.643" Title="Lock Optimisation" Tags="|mathematics|combinatorics|optimization|keys-and-locks|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="20437" PostTypeId="2" ParentId="20434" CreationDate="2015-08-22T21:59:43.573" Score="4" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;For each $k$-digit dial, the number of tries in the worst case is $k$, and the number of tries in the average case is $\frac{k+1}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We get $k = 3$, giving 63 tries in the worst case (beating $k=2$ and $k=4$, both on 68) and 42 tries in the average case (beating $k=2$ on 51 and $k=4$ on 42.5). This is probably dependent on $N$: $3^{21} = 10460353203$ is just larger than $N$, but $2^{34} = 17179869184$ is much larger than $N$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I haven't managed to do better than for 1.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Generalising the situation where all the dials are the same:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the number of dials is $\left\lceil{\frac {\ln N}{\ln k}}\right\rceil$; the worst case number of trials is $\left\lceil{\frac {\ln N}{\ln k}}\right\rceil k$, which is not fun to differentiate with respect to $k$. However, the minimum of $\frac{k}{\ln k}$ is $k = e$, so it's probably reasonable to expect the answer to be 3 for most values of $N$ and 2 for the rest.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I made a mistake with the average case, but 2012rcampion's answer goes further anyway so I won't correct it.&lt;/p&gt;&#xA;" OwnerUserId="228" LastEditorUserId="228" LastEditDate="2015-08-23T07:14:31.060" LastActivityDate="2015-08-23T07:14:31.060" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20438" PostTypeId="1" AcceptedAnswerId="20439" CreationDate="2015-08-22T22:22:09.543" Score="7" ViewCount="522" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If you are in me  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;My name on your breath&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Will surely begin me&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Yet end soon in death&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I'll shock you when high &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I'll please you when low &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;But fail to defend me &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;To jail you may go&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is a single word. Please explain how it relates to each line.&lt;/p&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="14177" LastEditDate="2015-08-22T22:42:25.937" LastActivityDate="2015-08-22T22:49:46.013" Title="I can be good in a fight" Tags="|riddle|word|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20439" PostTypeId="2" ParentId="20438" CreationDate="2015-08-22T22:43:09.730" Score="8" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Charge&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you are in me &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If you are in charge...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My name on your breath&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Shouting “Charge”...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Will surely begin me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Will start a charge...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet end soon in death&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Which will not last long but will lead to deaths on one side or another.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll shock you when high &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If the charge for something is too high, you’ll be unhappy. Alternatively a high electrical charge will provide a shock.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll please you when low &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;But if it’s lower than expected you’ll be pleased&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But fail to defend me &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To jail you may go&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If you are in a criminal court on charges and fail to defend them, you may end up in jail.&lt;br /&gt;&#xA;An explosive charge can be good in a fight.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="14135" LastEditDate="2015-08-22T22:49:46.013" LastActivityDate="2015-08-22T22:49:46.013" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20440" PostTypeId="1" CreationDate="2015-08-22T22:54:18.630" Score="11" ViewCount="456" Body="&lt;p&gt;The following stanzas&lt;br&gt;&#xA;Are hints to the answer&lt;br&gt;&#xA;Each a part of the end&lt;br&gt;&#xA;Their piece they defend&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I am found out in the ground&lt;br&gt;&#xA;  I can be bought by the pound&lt;br&gt;&#xA;  I'm put on the line&lt;br&gt;&#xA;  So in the end you can dine  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I am a basic desire&lt;br&gt;&#xA;  That many require&lt;br&gt;&#xA;  I get you places&lt;br&gt;&#xA;  And contribute in races  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Some find me childish&lt;br&gt;&#xA;  Some surpress me in kindness&lt;br&gt;&#xA;  But even if brief&lt;br&gt;&#xA;  After our encounter you'll feel a sense of relief&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You have all you need&lt;br&gt;&#xA;Now complete this one deed&lt;br&gt;&#xA;What is it in my eyes&lt;br&gt;&#xA;That will result in humanity's demise?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2015-08-22T23:06:41.023" LastActivityDate="2015-09-18T17:23:36.220" Title="What is it in my eyes that will result in humanity's demise?" Tags="|riddle|english|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20442" PostTypeId="1" AcceptedAnswerId="20443" CreationDate="2015-08-22T23:58:43.570" Score="3" ViewCount="297" Body="&lt;p&gt;I have legs and I sometimes move&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some without legs put me to use. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm known to have been musical, my favorite type is rock. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you've been acting badly, you're in for quite a shock. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What am I?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="15910" LastActivityDate="2015-08-23T00:14:11.687" Title="Simple four liner: what am I?" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20443" PostTypeId="2" ParentId="20442" CreationDate="2015-08-23T00:14:11.687" Score="1" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a chair&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have legs  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Most chairs have four legs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and I sometimes move / Some without legs put me to use.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a wheelchair&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm known to have been musical,   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Musical chairs is a popular children's game  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;my favorite type is rock.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Who doesn't love a rocking chair?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you've been acting badly, you're in for quite a shock.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the Electric Chair&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4385" LastActivityDate="2015-08-23T00:14:11.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20444" PostTypeId="1" AcceptedAnswerId="20460" CreationDate="2015-08-23T00:21:07.457" Score="19" ViewCount="1840" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;One rhyme,&lt;/p&gt;&#xA;&lt;p&gt;Space, time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;(8-letters)&lt;/p&gt;&#xA;&lt;p&gt;The above is my attempt at the world's shortest riddle.&lt;/p&gt;&#xA;&lt;p&gt;EDIT -  23 Aug 2015&lt;/p&gt;&#xA;&lt;p&gt;Here is my bid to get the riddle re-opened. Naturally to satisfy the community I have to add other constraints outside of the four words and this may make the riddle less likely to be the shortest. However, even if my riddle isn't the shortest, I'm still posing the question and looking for an answer.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The answer is a single 8-letter word.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The answer contains four vowels. It begins with a vowel and ends with a vowel.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Each line of the riddle must individually and independently fit the answer.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You must convincingly explain how the answer precisely fulfils the clues (in a normal riddlish fashion), possibly with the use of dictionary definitions.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2018-08-25T07:48:19.730" Title="My attempt at the world's shortest riddle: One rhyme, / Space, time" Tags="|riddle|word|" AnswerCount="5" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20446" PostTypeId="2" ParentId="20444" CreationDate="2015-08-23T02:02:18.247" Score="10" Body="&lt;p&gt;I optimistically up-voted this, but it does feel quite broad.&#xA;Here's a guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Relative&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One rhyme,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Rhyme&lt;/em&gt; describes the sounds words have &lt;em&gt;relative&lt;/em&gt; to one another.&lt;br&gt;&#xA; &lt;strong&gt;OR&lt;/strong&gt;&lt;br&gt;&#xA; Relative, as in family, could mean &quot;Ma and Pa&quot;, two relatives whose (informal) titles rhyme.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Space, time.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A hint at Einstein's &lt;a href=&quot;https://en.wikipedia.org/wiki/Theory_of_relativity&quot;&gt;Theory of relativity&lt;/a&gt;, which describes how measurements of space and time are relative to their observer's velocity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13237" LastActivityDate="2015-08-23T02:02:18.247" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20447" PostTypeId="1" AcceptedAnswerId="20455" CreationDate="2015-08-23T03:15:50.290" Score="6" ViewCount="724" Body="&lt;p&gt;There are two rooms.  One room contains ten switches, and the other contains ten lights.  Each switch controls one light, and no two switches control the same light.  It is not known which switch controls which light, and it is your job to label them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that, when you flip a switch, you can not immediately see which light has been turned on/off.  You have to walk into the other room in order to find out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the least amount of trips needed to label the switches?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will call a trip any time that you leave the room with the switches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will say that you are in the room with the switches, and that all of the lights are on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus thoughts:  what patterns emerge as you increase the amount of switches/lights?&lt;/p&gt;&#xA;" OwnerUserId="13054" LastEditorUserId="15544" LastEditDate="2015-08-23T09:49:26.130" LastActivityDate="2022-03-09T03:47:50.547" Title="Least amount of trips" Tags="|logical-deduction|optimization|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20449" PostTypeId="2" ParentId="20434" CreationDate="2015-08-23T06:09:55.190" Score="12" Body="&lt;h1&gt;Definitions&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Consider a lock $L$ consisting of $m$ wheels, where the $i$th wheel has $k_i$ digits (assume $k_i\geq 2$).  I will use the notation $L=\{k_1,k_2,\ldots,k_m\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of combinations $L$ can be set to, which I call the &lt;strong&gt;size&lt;/strong&gt; of $L$ which I write as $|L|$, is simply the product of the $k_i$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;|L|=\prod_{i=1}^m k_i&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The worst-case number of turns to unlock the lock is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\mathsf{WC}(L)=\sum_{i=1}^mk_i&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the average-case is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\mathsf{AC}(L)=\sum_{i=1}^m\frac{k_i+1}2&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An important fact is that there are no 'cross terms' in any of the quantities associated with a lock: any subset of the wheels may be considered separately from the rest.  (This also implies that the order of the $k_i$ is unimportant.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll also define four types of locks:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A &lt;strong&gt;worst-case minimal lock for $N$&lt;/strong&gt; is a lock with at least $N$ combinations and the smallest possible number of worst-case turns.  Formally, a lock $L$ is worst-case minimial if $|L|\geq N$ and $\forall L'\left(|L'|\geq N\implies\textsf{WC}(L')\geq\textsf{WC}(L)\right)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A &lt;strong&gt;worst-case optimal lock for $N$&lt;/strong&gt; is a minimal lock with the greatest size.  Formally, a lock $L$ is worst-case optimal if $|L|\geq N$ and $\forall L'\left(|L'|\geq|L|\iff\textsf{WC}(L')\geq\textsf{WC}(L)\right)$.  The set of worst-case optimal locks is the smallest set of locks which contains a worst-case minimal lock for all $N$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Average-case minimal locks&lt;/strong&gt; and &lt;strong&gt;average-case optimal locks&lt;/strong&gt; are defined analogously to the first two.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h1&gt;Worst Case&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Let us consider the replacement:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\{k,\ldots\}\to\{2,\frac{k}{2},\ldots\}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Assume that $k$ is even.)  We can see that the size is unchanged; however, the worst-case number of turns decreases by this replacement when:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{k}{2}+2&amp;lt;k \\&#xA;k+4&amp;lt;2k \\&#xA;k&amp;gt;4&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can perform a similar replacement for odd $k$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\{k,\ldots\}\to\{2,\frac{k+1}{2},\ldots\}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This replacement &lt;em&gt;increases&lt;/em&gt; the size, and also improves the worst-case count when:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{k+1}{2}+2&amp;lt;k \\&#xA;k+5&amp;lt;2k \\&#xA;k&amp;gt;5&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These replacements tell us an important fact: no worst-case minimal lock has $k_i&amp;gt;5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the replacement:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\{4,\ldots\}\to\{2,2,\ldots\}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under this replacement both the size and worst-case count of the lock are unchanged.  This means that it is sufficient to consider locks without $4$s for now.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now we move onto replacements dealing with optimal locks.  The replacement:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\{5,\ldots\}\to\{2,3,\ldots\}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;increases the size of a lock without changing it's worst-case count.  Therefore no optimal lock will contain a $5$.  Similarly, the replacement:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\{2,2,2,\ldots\}\to\{3,3,\ldots\}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;shows that no optimal locks have more than two $2$s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives us enough information to determine that all worst-case optimal locks take one of the forms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\{3\ldots\} \\&#xA;\{2,3\ldots\} \\&#xA;\{2,2,3\ldots\} \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that the last form is equivalent to $\{4,3\ldots\}$) $L$, the worst-case optimal lock for $N$, has $\lceil\log_3N\rceil$ wheels.  The number of $2$ wheels can be determined by looking at the fractional part, $f=\log_3N-\lfloor\log_3N\rfloor$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;if $f\leq\log_32^2-1\approx.2619\ldots$ then $L$ has two $2$s&lt;/li&gt;&#xA;&lt;li&gt;if $f\leq\log_32^1-0\approx .6309\ldots$ then $L$ has one $2$&lt;/li&gt;&#xA;&lt;li&gt;all other wheels are $3$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Example&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Take $N=10^{10}$.  We have $\log_310^{10}\approx 20.9590\ldots$, so the worst-case optimal lock has $21$ wheels.  The fractional part $.9590\ldots$ is larger than both cutoffs, so all wheels have $3$ digits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual number of combinations is $3^{21}=10\,460\,353\,203$ and the worst-case number of turns is $3\times 21=63$.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Average Case&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The arguments for the average-case minimal locks work similarly.  Again we consider the transformation for even $k$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\{k,\ldots\}\to\{2,\frac{k}{2},\ldots\}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The average-case number of turns decreases under this replacement when:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{k/2+1}2+\frac{2+1}2&amp;lt;\frac{k+1}2 \\&#xA;k/2+1+3&amp;lt;k+1 \\&#xA;k&amp;gt;6 \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For odd $k$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\{k,\ldots\}\to\{2,\frac{k+1}{2},\ldots\}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This replacement decreases the average-case count when:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{(k+1)/2+1}2+\frac{2+1}2&amp;lt; \frac{k+1}2 \\&#xA;(k+1)/2+2+3&amp;lt; k+1 \\&#xA;k&amp;gt; 7 \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus an average-case minimal lock will consist of no wheels greater than $7$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As before, we have an additional replacement under which both size and average count are unchanged:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\{6,\ldots\}\to\{2,3,\ldots\}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which shows that it is sufficient to consider locks without any $6$s.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The replacement:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\{7,\ldots\}\to\{3,3,\ldots\}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;increases the number of combinations while leaving the average-case count unchanged, implying that no average-case optimal lock contains a $7$.  Similarly, the replacements&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\{2,2,\ldots\}\to\{5,\ldots\} \\&#xA;\{2,3,3,\ldots\}\to\{4,5,\ldots\} \\&#xA;\{2,4,\ldots\}\to\{3,3,\ldots\} \\&#xA;\{2,5,\ldots\}\to\{3,4,\ldots\}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;show that an average-case optimal lock will have no $2$s, except in the special cases $\{2\}$ and $\{2,3\}$.  Following a similar pattern of replacements:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\{5,3,\ldots\}\to\{4,4,\ldots\} \\&#xA;\{5,4,4,\ldots\}\to\{3,3,3,3,\ldots\} \\&#xA;\{5,5,\ldots\}\to\{3,3,3,\ldots\}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we show that an average-case optimal lock will have no $5$s, except in the special cases $\{5\}$ and $\{5,4\}$.  At this point, we have shown that all average-case optimal locks consist only of $3$s and $4$s (excepting the aforementioned special cases).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, consider the replacement:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\{3,3,3,3,3,\ldots\}\to\{4,4,4,4,\ldots\}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As before, this shows that an average-case optimal lock will have at most four $3$s.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;All average-case optimal locks take one of the forms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\{2\} \\&#xA;\{5\} \\&#xA;\{2,3\} \\&#xA;\{6\} \\&#xA;\{4,5\} \\&#xA;\{4,\ldots\} \\&#xA;\{3,4,\ldots\} \\&#xA;\{3,3,4,\ldots\} \\&#xA;\{3,3,3,4,\ldots\} \\&#xA;\{3,3,3,3,4,\ldots\} \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a number of combinations $N\leq 5$, the smallest average-case optimal lock is simply $\{N\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $N=6$ there are two optimal locks: $\{2,3\}$ and $\{6\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $N&amp;gt;6$, the smallest average-case optimal lock has $\lceil\log_4N\rceil$ wheels.  The number of $3$s can be found by looking at the fractional part $f$ of $\log_4N$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$f\leq\log_43^4-3\approx.1699$: four $3$s&lt;/li&gt;&#xA;&lt;li&gt;$f\leq\log_43^3-2\approx.3774$: three $3$s&lt;/li&gt;&#xA;&lt;li&gt;$f\leq\log_43^2-1\approx.5849$: two $3$s&lt;/li&gt;&#xA;&lt;li&gt;$f\leq\log_43^1-0\approx.7925$: one $3$&lt;/li&gt;&#xA;&lt;li&gt;All other wheels have $4$ digits.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Except&lt;/strong&gt; when $17\leq N\leq 20$, where $2.0437&amp;lt;\log_4N&amp;lt;2.1610$.  The above procedure dictates that we should have three wheels but four $3$s; in this case the average-case optimal lock is actually $\{4,5\}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Example&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Take $N=10^{10}$.  We have $\log_410^{10}\approx 16.6096\ldots$, so the average-case optimal lock has $17$ wheels.  The fractional part, $.6096\ldots$, tells us that there is one $3$ and 16 $4$s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual number of combinations is $3\times 4^{16}=12\,884\,901\,888$ and the average-case number of turns is $\frac{3+1}2+16\times\frac{4+1}2=42$.&lt;/p&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="9718" LastEditDate="2015-08-24T15:31:59.810" LastActivityDate="2015-08-24T15:31:59.810" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="20450" PostTypeId="2" ParentId="20447" CreationDate="2015-08-23T06:43:47.310" Score="1" Body="&lt;p&gt;If you don't want to do any damage, it will take:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3 Trips&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For each trip you can determine if the bulbs are in one of 3 states&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Either the bulb is on, it is off and cold, or it is off and still warm. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The technique for 10 bulbs is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1. Switch off all bulbs and let them cool down.&lt;br /&gt;&#xA;2. Switch on 6 (say switches A-F) and after 10 minutes switch half of them off (D, E and F).&lt;br /&gt;&#xA;3. Go to the other room and test to see if the lights are on, off or warm.&lt;br /&gt;&#xA;4. Repeat as above, but this time set switches A, D and G to ON, B, E and H to WARM and leave the others off.&lt;br /&gt;&#xA;&lt;code&gt;A: ON ON&lt;br /&gt;B: ON WARM&lt;br /&gt;C: ON COLD&lt;br /&gt;D: WARM ON&lt;br /&gt;E: WARM WARM&lt;br /&gt;F: WARM COLD&lt;br /&gt;G: COLD ON&lt;br /&gt;H: COLD WARM&lt;br /&gt;I: COLD COLD&lt;br /&gt;J: COLD COLD&lt;/code&gt;&lt;br /&gt;&#xA;This will allow you to tie every light to every switch except for I and J which have both been off the entire time. Switch one of these on for the final test and all the bulbs will have been identified.&lt;br /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the general case where there are $N$ lightbulbs&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the number of trips is given by rounding $log3(N)$ up to the next whole number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can improve on the case for 10 lightbulbs:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If, while we are initially waiting for all the bulbs to cool down, we repeatedly switch J on and off so that the filament blows, then proceed as above, then on the first trip to the room one bulb would be blown (visible on close inspection of one of the off bulbs), leaving only nine to discriminate between which can be handled with one more trip.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="14135" LastEditDate="2015-08-23T06:52:18.643" LastActivityDate="2015-08-23T06:52:18.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20451" PostTypeId="2" ParentId="20440" CreationDate="2015-08-23T06:54:26.540" Score="2" Body="&lt;pre&gt;&lt;code&gt;I am found out in the ground&#xA;I can be bought by the pound&#xA;I'm put on the line&#xA;So in the end you can dine&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Earthworms. You use them as bait to go fishing! If specific, probably nightcrawlers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I am a basic desire&#xA;That many require&#xA;I get you places&#xA;And contribute in races&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cars? Could be any kind of vehicle, really.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Some find me childish&#xA;Some surpress me in kindness&#xA;But even if brief&#xA;After our encounter you'll feel a sense of relief&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gas? For shame, a puzzle that's a fart joke!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That makes the combined answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Wormscargas. Apparently that's the name of our doom. Probably some British term. Or someone's username.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13068" LastActivityDate="2015-08-23T06:54:26.540" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20452" PostTypeId="2" ParentId="20444" CreationDate="2015-08-23T07:00:24.190" Score="3" Body="&lt;p&gt;Shame about the eight letter restriction, otherwise the answer could be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dimension&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One rhyme&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The way that a poem sounds when read aloud has been described as one of the &lt;a href=&quot;http://www.news.cornell.edu/stories/2010/11/abrams-explains-four-dimensions-poem&quot; rel=&quot;nofollow&quot;&gt;dimensions of a poem&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Space, time&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Are referred to as the first 3 dimensions and the fourth dimension.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-08-23T07:00:24.190" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20453" PostTypeId="2" ParentId="20447" CreationDate="2015-08-23T07:35:27.677" Score="3" Body="&lt;p&gt;If we assume that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the lights cannot be touched so that the trick from the 3 lights version (1 light off and cold, 1 light off and warm, 1 light on) doesn't work&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;then&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; using $N$ trips you can distinguish $2^N$ lights, meaning that $4$ trips are needed for $10$ lights.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Usage scheme:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Assign a number to each switch, starting with 0. Convert each number to base 2 (i.e. binary). For 0 to 9 we get:&lt;br&gt;&lt;br&gt;0: 0000&lt;br&gt;1: 0001&lt;br&gt;2: 0010&lt;br&gt;3: 0011&lt;br&gt;4: 0100&lt;br&gt;5: 0101&lt;br&gt;6: 0110&lt;br&gt;7: 0111&lt;br&gt;8: 1000&lt;br&gt;9: 1001&lt;br&gt;&lt;br&gt;During the first trip turn off switches with 0 in the first column, turn on the ones with 1. Use the other columns for the next trips. For each light assign 0 for each trip if it's off, and a 1 otherwise. Match the number of the light to the number of the switch, after all trips are done.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we assume&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; that we can differentiate more than 2 states by varying the duration of the lights turned on and measuring the temperature of the lights, and we call the number of states $S$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;then&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; using $N$ trips we can distinguish $S^N$ lights. The usage scheme is similar to the one above, but we need to convert each number to base $S$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15544" LastEditDate="2015-08-24T13:16:56.570" LastActivityDate="2015-08-24T13:16:56.570" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20454" PostTypeId="1" CreationDate="2015-08-23T07:45:49.960" Score="12" ViewCount="2852" Body="&lt;p&gt;Let's consider the humble mechanical &lt;a href=&quot;https://en.wikipedia.org/wiki/Combination_lock#Multiple-dial_locks&quot; rel=&quot;nofollow noreferrer&quot;&gt;combination lock&lt;/a&gt;.  The basic type has $n$ dials, with $k$ digits on each dial, for a total of $k^n$ possible combinations.  (Typical luggage locks or suitcase lock have $k=10$ and $n=3$ or $4$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These locks are usually cheaply made and quite simple to crack, as their mechanism &quot;leaks&quot; information to an attacker.  For example, the lock might &lt;a href=&quot;https://puzzling.stackexchange.com/questions/20434/lock-optimisation&quot;&gt;'click' whenever a dial is rotated into the correct position&lt;/a&gt;, making opening the lock a simple matter of turning all of the dials until they click.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this question, let's assume that the lock works like this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Set the dials to any desired position.&lt;/li&gt;&#xA;&lt;li&gt;Attempt to open the lock (by, e.g., pulling on the shackle).&lt;/li&gt;&#xA;&lt;li&gt;If all the dials are in the correct position, the lock opens; otherwise it remains closed.&lt;/li&gt;&#xA;&lt;li&gt;However, if one or more of the dials are in the correct position, the lock will &quot;click&quot; when the shackle is pulled.  This gives us some extra information to work with.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What strategy should I use to minimize the worst-case number of attempts required to unlock the lock?&lt;/strong&gt; Note:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The initial starting position is random.&lt;/li&gt;&#xA;&lt;li&gt;Your strategy should be minimal for all $k\geq 2$ and $n\geq 1$, even cases where $n&amp;gt;k$.&lt;/li&gt;&#xA;&lt;li&gt;You must open your lock on the last attempt, even if you have figured out the combination without opening the lock.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is equivalent to a game of &lt;a href=&quot;https://en.wikipedia.org/wiki/Mastermind_(board_game)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mastermind&lt;/a&gt; with $n$ pegs in $k$ colors, where the codemaker only has one red pin to respond with, and you wish to minimize the number of turns in which you are guaranteed a win.&lt;/p&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2016-08-16T21:33:56.787" Title="Cracking a combination lock" Tags="|mathematics|combinatorics|optimization|keys-and-locks|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20455" PostTypeId="2" ParentId="20447" CreationDate="2015-08-23T07:57:50.130" Score="9" Body="&lt;p&gt;&lt;strong&gt;Three switches version&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; A 3 switch version of this problem exits, which can solved in 1 trip taking into account the fact that light bulbs heat up when lit. The classical solution goes-&#xA;&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; &lt;blockquote&gt;Turn on switches A, B and turn off B after (say) 5 minutes. Head over to the other room. The on light belongs to A, now, touch one of the bulb. It belongs to B if it is hot, else it belongs to C.&lt;/blockquote&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Non Heat Generating bulbs&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; In this case, we assume that the bulbs generate no heat. So, the heating trick cannot be used. Then, for &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; bulbs the number of trips needed is: &lt;span class=&quot;math-container&quot;&gt;$$\lceil\log_2{N}\rceil$$&lt;/span&gt;&lt;br&gt;&lt;br&gt;&#xA;  The strategy is -&#xA;&lt;br /&gt;&#xA;  &lt;blockquote&gt;Label each switch in binary. At the first trip, light the switches whose rightmost digit is 1. At the second trip light the switches whose second rightmost digit is 1 and so on...&#xA;  &lt;br&gt;&lt;br&gt;&#xA; This requires the number of digits in the binary representation of &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt;, which equals &lt;span class=&quot;math-container&quot;&gt;$k=\lceil\log_2{N}\rceil$&lt;/span&gt; trips. And the end of the trips, we can successfully determine each bit of the binary representation of each bulb (0 if off, 1 if on) and hence identify their switches.&lt;/blockquote&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; For &lt;span class=&quot;math-container&quot;&gt;$N=10$&lt;/span&gt; the answer is &lt;span class=&quot;math-container&quot;&gt;$\lceil\log_2{10}\rceil = 4$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Heat Generating Bulbs&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; This is similar to the previous case, but now we have &lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt; possible states of the bulb &lt;strong&gt;ON&lt;/strong&gt;, &lt;strong&gt;OFF&lt;/strong&gt;, &lt;strong&gt;HOT&lt;/strong&gt;. In this case we simply use base 3 numbers to label each switch and identify each trinary digit and hence the answer is: &lt;span class=&quot;math-container&quot;&gt;$$\lceil\log_3{N}\rceil$$&lt;/span&gt;&#xA;&#xA; For &lt;span class=&quot;math-container&quot;&gt;$N=10$&lt;/span&gt; the answer is &lt;span class=&quot;math-container&quot;&gt;$\lceil\log_3{10}\rceil = 3$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Multiple states of hot&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;  Now suppose, we can have two kinds of hot- warm and very hot. then we have 4 states and we work in base 4. In general if there are &lt;span class=&quot;math-container&quot;&gt;$k-2$&lt;/span&gt; states of hot, the answer is: &lt;span class=&quot;math-container&quot;&gt;$$\lceil\log_k{N}\rceil$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Extreme Solution&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Suppose we have extremely accurate thermometer. Then we can label the bulbs in one trip!&#xA;&lt;br /&gt;&#xA; &lt;blockquote&gt;Light up each bulb one after one at one minutes interval. Head to the other room, measure the temperatures, The highest belongs to the first lit switch, the second highest to the second lit and so on..&lt;/blockquote&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; This is equivalent to working in base &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15544" LastEditorUserId="69582" LastEditDate="2022-03-09T03:47:50.547" LastActivityDate="2022-03-09T03:47:50.547" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="20456" PostTypeId="1" AcceptedAnswerId="20457" CreationDate="2015-08-23T14:00:22.090" Score="8" ViewCount="2496" Body="&lt;p&gt;Yesterday I was doodling around with my very simple calculator as shown in the picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ufwqq.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ufwqq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First I would enter a few digits to make an integer (i.e. no decimal places). Then I would perform a simple calculation to get a second integer. At first I chose the numbers carefully as shown in the following list. I chose four digit numbers and, as you can see, I got some two digit answers. I wasn't exactly surprised because I had chosen the starting numbers specifically to yield those answers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3194,       17&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3195,       18&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4276,       19&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5429,       20&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6709,       22&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8843        23&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, instead of choosing the starting numbers myself, I removed any bias by using a random-number generator. I set it to give numbers between 1000 and 9999 inclusive. I then did the same calculation as before. Here are some of the results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1127,       13&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1152,       14&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1318,       16&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1362,       18&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1386,       20&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1450,       17&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1491,       14&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1536,       18&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1628,       20&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1662,       19&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1666,       20&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1838,       21&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1987,       19&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2305,       21&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2373,       19&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2591,       18&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2658,           23&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2693,       22&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2868,       25&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3132,       17&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3317,       16&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3320,       21&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3547,       18&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3598,       23&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3615,       18&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3660,       23&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Notes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I omitted a tag that would probably have given the answer away. I substituted the tag &lt;strong&gt;enigmatic-puzzles&lt;/strong&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As stated, the first list was biased by the numbers I chose. However all the numbers, in both lists, were subjected to the same calculation in order to arrive at the answers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It is possible to calculate the second column using very simple arithmetic.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I have sorted the first columns into numerical order. However the order has no effect on the answers. I just did it for neatness.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The answer can be worked out purely from the second, randomly generated list. You don't need the first list at all. However the first list is 100% correct and I included it because it might possibly give you a clue or at least start you on the right path.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;At no stage in the calculation will you encounter any decimal points - only integers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A. What game was I playing? Or rather what consistent calculation did I perform on all the above 4 digit integers in order to derive the second column?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B. To get the green tick, look at the number on the calculator. It is 1234567890.  If I performed the same calculation on that number, what would be the result in the second column? &lt;/p&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="14177" LastEditDate="2015-08-23T15:10:36.720" LastActivityDate="2015-08-23T18:45:47.180" Title="A basic calculator, a simple game. What was I playing at?" Tags="|pattern|enigmatic-puzzle|number-sequence|formation-of-numbers|" AnswerCount="1" CommentCount="16" FavoriteCount="0" ClosedDate="2015-08-23T18:45:55.917" ContentLicense="CC BY-SA 3.0" />
  <row Id="20457" PostTypeId="2" ParentId="20456" CreationDate="2015-08-23T15:09:54.980" Score="7" Body="&lt;p&gt;The number produced for your current calculator number is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 50&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The calculation used to reach this answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Count the segments that are shown on the calculator's display.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The omitted tag is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; seven-segment&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-08-23T15:14:41.927" LastActivityDate="2015-08-23T15:14:41.927" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20458" PostTypeId="2" ParentId="20440" CreationDate="2015-08-23T15:33:06.867" Score="0" Body="&lt;p&gt;&lt;em&gt;I am found out in the ground&lt;br&gt;&#xA;I can be bought by the pound&lt;br&gt;&#xA;I'm put on the line&lt;br&gt;&#xA;So in the end you can dine&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Answer is WORMS, found by &lt;strong&gt;Kingrames&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am a basic desire&lt;br&gt;&#xA;That many require&lt;br&gt;&#xA;I get you places&lt;br&gt;&#xA;And contribute in races&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This could be TONGUE. Food/eating is a basic desire, and everyone requires it sooner or later. Your tongue can get you places if you can speak eloquently, and it may have to race against other tongues - say if you're in a job interview and you need to talk fast in order to beat the other applicants.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Some find me childish&lt;br&gt;&#xA;Some surpress me in kindness&lt;br&gt;&#xA;But even if brief&lt;br&gt;&#xA;After our encounter you'll feel a sense of relief&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure about this one, but what we have so far is WORMTONGUE, of &lt;em&gt;Lord of the Rings&lt;/em&gt; fame, so the final answer could be something like &quot;Saruman&quot; or &quot;Palantir&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2015-08-23T15:33:06.867" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20459" PostTypeId="2" ParentId="20444" CreationDate="2015-08-23T15:48:00.970" Score="5" Body="&lt;p&gt;Although this is apparently not the intended answer, I think that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;INNUENDO&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is a good fit.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;blockquote&gt;&#xA;  &lt;p&gt;One rhyme&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm working on the basis that two words rhyme if their sequence of phonemes from the most stressed vowel to the end is the same. (I think this is the same definition as used by Mark Liberman in his &lt;a href=&quot;http://languagelog.ldc.upenn.edu/nll/?p=1946&quot; rel=&quot;noreferrer&quot;&gt;breakfast experiment on unrhymeable words&lt;/a&gt;). Working with &lt;a href=&quot;http://www.speech.cs.cmu.edu/cgi-bin/cmudict&quot; rel=&quot;noreferrer&quot;&gt;CMUDICT&lt;/a&gt;, since I don't have access to CELEX2, I find that the only rhyme for my answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;CRESCENDO&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote&gt;&#xA;  &lt;p&gt;Space, time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The space is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;INN&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the time is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;END&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="228" LastActivityDate="2015-08-23T15:48:00.970" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20460" PostTypeId="2" ParentId="20444" CreationDate="2015-08-23T15:55:32.617" Score="30" Body="&lt;p&gt;My answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; universe &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One rhyme,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; uni = one, verse = rhyme&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Space, time.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The universe is all of space and time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;8-letters: Check.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer contains four vowels: Check.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It begins with a vowel and ends with a vowel: Check.&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditDate="2015-08-23T16:00:37.053" LastActivityDate="2015-08-23T16:00:37.053" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20461" PostTypeId="1" AcceptedAnswerId="20463" CreationDate="2015-08-23T16:32:22.920" Score="12" ViewCount="3953" Body="&lt;p&gt;Samuel Bean, a Canadian doctor, had a tombstone erected in Rushes Cemetery near Crosshill, Wellesley Township, Ontario, for his first two wives, Henrietta and Susanna. The original stone was carved in the 1860s and became badly weathered; the following picture shows the more durable replica made to replace it in the 1980s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jUJIr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jUJIr.jpg&quot; alt=&quot;stone&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the message encrypted on the stone, and how can it be decrypted?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: the solution is already known, so you could find it by Googling. But it'll be more fun if you manage it by yourself, so &lt;a href=&quot;/questions/tagged/no-computers&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;no-computers&amp;#39;&quot; rel=&quot;tag&quot;&gt;no-computers&lt;/a&gt; and no cheating! :-)&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="8874" LastEditDate="2016-03-23T17:05:16.280" LastActivityDate="2016-03-23T17:05:16.280" Title="Two dead beans on a stone" Tags="|no-computers|cipher|history|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20462" PostTypeId="1" CreationDate="2015-08-23T17:59:16.633" Score="7" ViewCount="2075" Body="&lt;p&gt;I'm aware of this: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16000/strategies-for-creating-a-puzzle-with-a-unique-solution&quot;&gt;Strategies for creating a puzzle with a unique solution&lt;/a&gt;. However it is about number puzzles and so not relevant to my query.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am also aware of this &lt;a href=&quot;https://puzzling.stackexchange.com/questions/5249/how-do-i-write-a-solid-riddle&quot;&gt;How do I write a “solid” Riddle?&lt;/a&gt;, but I am not concerned with creating a 'solid' riddle when there is no limit on the number of clues/lines. I already have a couple of successful longer ones. My problem comes specifically when the clues are very few.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I enjoy trying to create very short riddles that have only one 'good' answer. The problem is that, although I may spend some considerable time trying to ensure the only good answer is unique, there is no guarantee that answerers will take an equal amount of time trying to solve it accurately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What tips can you give me to avoid attracting unconvincing answers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm beginning to think that either I have to add lots of artificial restrictions such as word-length or I have to have many lines in the riddle. That rather defeats what I am trying to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any tips for avoiding the problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What number of lines/clues do people find to be sufficient to avoid inaccurate answers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All tips and ideas welcome.&lt;/p&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-08-25T06:28:41.877" Title="Tips wanted on creating a successful short riddle with a unique answer without attracting unconvincing answers" Tags="|riddle|puzzle-creation|" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20463" PostTypeId="2" ParentId="20461" CreationDate="2015-08-23T18:52:10.207" Score="15" Body="&lt;p&gt;The message is written in this jagged spiral:&lt;img src=&quot;https://i.stack.imgur.com/hxI4r.jpg&quot; alt=&quot;1]&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In memoriam: Henrietta, 1st wife of S. Bean, M.D., who died 27th Sep. 1865 aged 23 years 2 months &amp;amp; 17 days, &amp;amp; Susanna, his 2nd wife, who died 27th April 1867 aged 26 years 10 months &amp;amp; 15 days. 2 better wives 1 man never had. They were gifts from God but are now in heaven. May God help me, S. B., to meet them there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-08-23T18:52:10.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20464" PostTypeId="2" ParentId="20462" CreationDate="2015-08-23T20:44:02.463" Score="2" Body="&lt;p&gt;This itself, is actually &lt;em&gt;just another very difficult puzzle&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the time, we don't have a hard and fast answer to how to make a puzzle work. I usually juggle around some concepts for days or weeks before I consider posting it because oftentimes, you're either unsure of how well a puzzle will be received, or you just know it works. I've changed quite a few puzzles a LOT before posting them, so I can only tell you from my own experience that it's a very good idea to exercise some restraint and give it a good length of time to stew in your brain for a while.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also a good idea to get some feedback on your puzzle by talking to some puzzling friends beforehand. If you don't have many, consider using the puzzling meta board as a place to gauge player reaction, especially if it's a new kind of puzzle. Always remember that more communication is always better when you're designing something. You just don't have to include all of the design documents in the final puzzle.&lt;/p&gt;&#xA;" OwnerUserId="13068" LastActivityDate="2015-08-23T20:44:02.463" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20465" PostTypeId="1" AcceptedAnswerId="20467" CreationDate="2015-08-23T21:36:03.330" Score="12" ViewCount="1534" Body="&lt;p&gt;This puzzle took me less than five minutes to compose. With luck it will take you about the same time to solve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;cp&lt;/p&gt;&#xA;&#xA;&lt;p&gt;dt&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ts&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Instructions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find two letters that, added to the above three pairs of letters, make three cross words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can move the above letters apart if necessary in a horizontal direction but you must not change their order or orientation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The added pair of letters is the same in each case and stays in the same order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the three words? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please make sure that they are cross words, not just any old set of words.&lt;/p&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="2673" LastEditDate="2015-08-24T12:46:56.833" LastActivityDate="2019-03-17T16:09:43.013" Title="A quick cross word puzzle" Tags="|wordplay|letter-sequence|letters|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20467" PostTypeId="2" ParentId="20465" CreationDate="2015-08-23T22:10:08.420" Score="17" Body="&lt;p&gt;Are the three words:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; crap, drat, and rats? By adding ra, you get words that are spoken when one is cross.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastActivityDate="2015-08-23T22:10:08.420" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20468" PostTypeId="2" ParentId="20465" CreationDate="2015-08-23T22:10:53.157" Score="3" Body="&lt;p&gt;The two letters are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;R and A&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Giving:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;C R A P (excuse the profanity, but it's the answer)&lt;br /&gt;&#xA;DRAT&lt;br /&gt;&#xA;and RATS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All three are words that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;you may say when you are cross&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-08-23T22:10:53.157" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20469" PostTypeId="2" ParentId="20462" CreationDate="2015-08-23T22:39:23.303" Score="7" Body="&lt;p&gt;I'll second Kingrames' advice and give some tips on refining riddles. I'm new to the site (been here for under two months), but I've had success with the following process:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make the best of each post, never think up, write, and post a riddle all in one sitting. Brainstorm before writing to make the most of your idea. Once written, revisit it &lt;em&gt;the next day&lt;/em&gt; and look for alternate answers that can be avoided; &lt;strong&gt;one poorly chosen word can tarnish a 200-word riddle.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, my &lt;a href=&quot;https://puzzling.stackexchange.com/questions/20233/a-short-brutal-riddle&quot;&gt;short, brutal riddle&lt;/a&gt; sat in a text document for a couple days as &quot;LIAR-RAIL-LAIR : lies?&quot;. My &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18424/surely-you-know-this-prominent-figure&quot;&gt;first rhyming riddle&lt;/a&gt; spent a few days as &quot;SATURN; sega, car company, planet, Saturday, rockets, Holst's planets&quot; and slowly evolved into a riddle over time. I posted it one day after I had finished writing it, changing a few words that might throw people off or make any clue too broad. Adding &quot;strong&quot; to the last line of &lt;a href=&quot;https://puzzling.stackexchange.com/questions/20381/another-four-liner&quot;&gt;this&lt;/a&gt; was a product of this process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're writing &lt;a href=&quot;https://puzzling.stackexchange.com/questions/20444/my-attempt-at-the-worlds-shortest-riddle&quot;&gt;very short riddles&lt;/a&gt;, incorrect answers are going to happen, and some &lt;em&gt;will&lt;/em&gt; be bad, but that's what down-votes are for. Just take your time with each riddle and consider other answers which might fit. &lt;strong&gt;You won't catch them all, and that's okay&lt;/strong&gt;; the community as a whole is quite clever!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Avoid posting multiple riddles, especially similar ones, on the same day.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Making a riddle easier is better than making a riddle harder&lt;/strong&gt;, because you usually can't gauge how hard a puzzle will be, and because making a riddle harder usually means the puzzle-maker just made it more vague and confusing for the sake of difficulty.&lt;/p&gt;&#xA;" OwnerUserId="13237" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-08-23T22:46:45.440" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20470" PostTypeId="2" ParentId="20115" CreationDate="2015-08-23T23:23:59.153" Score="2" Body="&lt;p&gt;Is the answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 7133 ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Based on the hint that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A and D are both one digit numbers, and the code is A13D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I tried using a variation on the approach that Going Hamateur used...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I set:&lt;br&gt; A+2-3-8+D = A-8+3+2-D &lt;br&gt; A-9+D = A-3-D &lt;br&gt; D = 3 &lt;br&gt; When I substituted 3 for D, I found A-6=A-6&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I then applied Moonbutt74's observation that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Diagonal 2+3+8 = 13.I then solved A+3+D=13 &lt;br&gt; D=3, so A=7&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, I believe the code is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 7133&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13244" LastEditorUserId="13244" LastEditDate="2015-08-24T04:27:24.360" LastActivityDate="2015-08-24T04:27:24.360" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20471" PostTypeId="2" ParentId="20454" CreationDate="2015-08-24T00:15:12.107" Score="5" Body="&lt;p&gt;Strategy for $n&amp;lt;k$:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Find the subset of digits being used by setting all dials to each digit at the same time and pulling. In the worst case, this takes $k$ moves and reduces $k$ to $n$.&lt;/li&gt;&#xA;&lt;li&gt;Place the dials in a position they don't click (unused digit)&lt;/li&gt;&#xA;&lt;li&gt;For each dial, try the used digits until one of them clicks or we reach the last one without having heard a click. Write down that digit. This takes in the worst case $n(k-1) = n(n-1)$ moves.&lt;/li&gt;&#xA;&lt;li&gt;Once we know all digits, open the lock with $1$ more move.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The total number of moves is: $k + n(n-1) + 1 = n^2 - n + k + 1$&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;Strategy for $n \ge k$ (I doubt this the most efficient strategy):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Continuously increase the combination by one (treat dials as digits of a base k number), until you find a solution that doesn't click. The number of solutions that don't click is $(k-1)^n$ (all digits wrong), so the number of solutions that do click is $k^n - (k-1)^n$. In that many moves, we have either found the solution (one of the click-solutions is the final solution) or a non-click solution.&lt;/li&gt;&#xA;&lt;li&gt;For each dial, try each digit one by one until you hear a click or the dial reaches the last one without clicking. Write that digit down. This takes $n(k-1)$ moves.&lt;/li&gt;&#xA;&lt;li&gt;We know all digits, so we can open the lock with $1$ more move.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The total number of moves is: $k^n - (k-1)^n + n(k-1) + 1 = k^n - (k-1)^n + nk + 1$&lt;/p&gt;&#xA;" OwnerDisplayName="user9771" LastEditorDisplayName="user9771" LastEditDate="2015-08-24T18:47:38.070" LastActivityDate="2015-08-24T18:47:38.070" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20472" PostTypeId="2" ParentId="20379" CreationDate="2015-08-24T03:11:14.120" Score="9" Body="&lt;p&gt;Stealing from 2012rcampion's answer, we need to create a one-to-one mapping from five-card hands to four-card messages (four cards plus permutation).  By symmetry, each four-card message can be mapped to an equal number of five-card hands, so by the &lt;a href=&quot;https://math.stackexchange.com/questions/83999/edge-coloring-of-a-k-regular-bipartite-graph&quot;&gt;regular version of Hall's Theorem&lt;/a&gt; there must exist such a one-to-one mapping (this is laid out in more detail in  GentlePurpleRain's comment link).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the trouble is actually constructing a matching, and ideally doing so in a way that would work well as a magic trick.  The following works for the former, and you can decide if it works for the latter.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Suppose the cards are numbered 0 to 123.  Call the five cards, from smallest to largest, $c_0$, $c_1$, $c_2$, $c_3$, and $c_4$.   To find the pocketed card, the assistant adds up the numbers on all five cards, reduces the answer modulo $5$ to get $S$, and then hands back $c_S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the magician can see four cards.   Suppose these four cards sum to $S'$ modulo $5$.  Suppose also she mentally removes these four cards from the deck and then relabels the remaining cards in order from $0$ to $119$ (call these &quot;reduced numbers&quot;).   Then we have&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the missing card must have a reduced number equivalent to $(-S')$ modulo $5$*.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This means there are only $24$ possibilities, and hence with a code that takes permutations of four cards to the set $\{1, 2, ..., 24\}$, the assistant can supply enough information for the magician to identify the card.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;*This is a bit hard to see, so let's break it down into cases.  If $c_0$ was removed, that means the sum of the five cards is $0$ mod $5$, and hence the missing card must be $-S'$ mod $5$ to counter the $S'$ sum of the other four cards.  If $c_1$ was removed, then the missing card must have had a number of $-S' + 1$, but since it appears after $c_0$ that we mentally removed from the deck, it's reduced number is $-S' + 1 - 1 = -S'$.  Similar logic shows this works for all five cases.&lt;/p&gt;&#xA;" OwnerUserId="2722" LastEditorUserId="-1" LastEditDate="2017-04-13T12:19:36.950" LastActivityDate="2015-08-24T03:11:14.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20473" PostTypeId="2" ParentId="20380" CreationDate="2015-08-24T06:52:27.423" Score="10" Body="&lt;p&gt;I figured out the first stanza last, and I'm least sure about the explanations for this set of words. You'd think you'd need to nail this down to get a solid foundation for figuring out the rest of the poem, but...well, here goes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;To start, I'm strangely missing.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;RHYME&lt;/strong&gt; - This poem has barely any rhyming lines, and there certainly isn't a rhyme in this line. And for those of us struggling with this puzzle, it may have seemed to lack “rhyme or reason&quot; (make no sense, have no logic).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Modified, I'm what a silly chef lacks.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;THYME&lt;/strong&gt; (modify by replacing R with T) - Maybe the chef is punny and has neither the herb thyme nor time to prepare dishes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Again, and I'm subtly present.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;THEME&lt;/strong&gt; (modify again by replacing Y with E) - @Bailey already mentioned that there's a theme or motif within this poem.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Do me harm, and I'm perhaps you.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;THEE&lt;/strong&gt; (harm by removing M) - The objective case of the word “thou&quot;, an archaic form of “you&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;With a change, I'm tall and resilient.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;TREE&lt;/strong&gt; (change H to R) - Both words can describe a tree.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Again, and this is certainly that.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;FREE&lt;/strong&gt; (change T to F) - This poem is an example of free verse, having no pattern in rhyme or meter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Again, I'm a guitarist's worry.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;FRET&lt;/strong&gt; (change E to T) - Part of a guitar as well as a synonym for “worry&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Again, and I'll make a stand.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;FEET&lt;/strong&gt; (change R to E) - Organisms with feet stand on them. For the poem motif, they are groups of syllables in a meter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Again, and our paths will cross.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;MEET&lt;/strong&gt; (change H to M) - To cross paths is to meet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Again, and I'm some kind of kind of cat.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;MEER&lt;/strong&gt; (change T to R) - cat -&gt; kat -&gt; meerkat (an animal)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Adorn me, and I'm also strangely missing.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;METER&lt;/strong&gt; (adorn by inserting T) - This poem doesn't seem to follow a meter, a pattern of stressed and unstressed syllables.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Enrich me, and I'll destroy everything.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;METEOR&lt;/strong&gt; (enrich by inserting O) - A collision with a meteor is very destructive. Ask the dinosaurs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Destroy me, and I'm a secluded commander.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;REMOTE&lt;/strong&gt; (destroy by rearranging letters) - A synonym for “secluded&quot; as well as a device that can command or control other devices&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Chop me down, and my feelings are obvious.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;EMOTE&lt;/strong&gt; (chop by removing R) - Word for showing feelings&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;With a swap, I've been judged.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;SMOTE&lt;/strong&gt; (swap E with S) - Possible punishment after judgement, especially from a deity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;With another, I'm a torrid remainder.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;SMOKE&lt;/strong&gt; (swap T with K) - What usually remains after something torrid like a fire&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And with another, I'm a torrid beginner.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;STOKE&lt;/strong&gt; (swap M with T) - What one does to start or feed a fire&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And a last, you'd better turn me.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;STONE&lt;/strong&gt; (swap K with N) - From the idiom “leave no stone unturned&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Top down, I'm virtually nonexistent.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;TONE&lt;/strong&gt; (remove top or first letter) - This poem doesn't seem to have a tone or mood, besides a puzzling one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;With a flip, you can find me within.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;BONE&lt;/strong&gt; (flip by swapping T with B) - As human beings, we have an internal skeleton of bones.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Decorate me, and I'm a young helper.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;BONNE&lt;/strong&gt; (decorate by inserting N) - French word for maidservant&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And again, I'm a soft, brimmed hat --&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;BONNET&lt;/strong&gt; (insert again with T) - Yup, it's a type of hat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;One more change, and this is certainly not that!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;SONNET&lt;/strong&gt; (final change by swapping B with S) - A 14-line poem of a specific scheme, which this dynamic puzzle poem isn't.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Overarching motif is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; this is a poem about a poem or prosodical terms: rhyme, free, feet, meter, tone, sonnet. Despite my name's meaning - no, really, part of it means poem - I'm unable to explain further at the moment. Sorry if I missed any details or just missed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My mobile device is ill suited for composing this answer.&lt;/p&gt;&#xA;" OwnerUserId="15921" LastActivityDate="2015-08-24T06:52:27.423" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20474" PostTypeId="1" AcceptedAnswerId="20524" CreationDate="2015-08-24T07:08:47.327" Score="5" ViewCount="1134" Body="&lt;p&gt;&lt;strong&gt;1.&lt;/strong&gt; &lt;a href=&quot;https://i.stack.imgur.com/bSPw3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bSPw3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2.&lt;/strong&gt; &lt;a href=&quot;https://i.stack.imgur.com/M4Ycc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/M4Ycc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3.&lt;/strong&gt; &lt;a href=&quot;https://i.stack.imgur.com/e34og.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/e34og.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4.&lt;/strong&gt; &lt;a href=&quot;https://i.stack.imgur.com/8wy6Y.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8wy6Y.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;5.&lt;/strong&gt; &lt;a href=&quot;https://i.stack.imgur.com/4OAxj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4OAxj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;br/&gt;&#xA;&lt;br/&gt;&#xA;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;6.&lt;/strong&gt; &lt;strong&gt;?&lt;/strong&gt;&#xA;&lt;br/&gt;&#xA;&lt;br/&gt;&#xA;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;7.&lt;/strong&gt;  &lt;a href=&quot;https://i.stack.imgur.com/JfBdm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JfBdm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;br&gt;&#xA;&lt;strong&gt;Find the missing object (No. 6)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Note:The first one to post an object with the matching criteria will be the winner.&lt;/strong&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Hint1: Someone who doesn't know the name of any of these objects too can find the correct answer.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Hint2: The newly added tag is a hint&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="2730" LastEditDate="2015-12-15T07:27:07.613" LastActivityDate="2015-12-15T07:27:07.613" Title="Find the missing object" Tags="|logical-deduction|pattern|number-sequence|" AnswerCount="4" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20475" PostTypeId="2" ParentId="20465" CreationDate="2015-08-24T07:24:57.653" Score="9" Body="&lt;p&gt;Just to be creative/contrary, here's a second equally valid solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; letters A,R&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Giving&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;CARP : to complain in a &lt;strong&gt;cross&lt;/strong&gt; manner&#xA;DART : a projectile whose tail ('flight') is &lt;strong&gt;cross-shaped&lt;/strong&gt;&#xA;ARTS : the main &lt;strong&gt;cross-disciplinary&lt;/strong&gt; faculty in university&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Please make sure that they are cross words, not just any old set of words.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They bloody well are cross words :)&lt;/p&gt;&#xA;" OwnerUserId="2349" LastEditorUserId="2349" LastEditDate="2019-03-17T00:45:56.380" LastActivityDate="2019-03-17T00:45:56.380" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="20476" PostTypeId="1" CreationDate="2015-08-24T09:02:12.377" Score="-6" ViewCount="954" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Here is the clue.&lt;/p&gt;&#xA;&lt;p&gt;Cut cut!&lt;/p&gt;&#xA;&lt;p&gt;What crazy, bloodthirsty creature remains? (5 letters)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;EDIT I added the word 'crazy'. Apologies that I missed this out before.&lt;/p&gt;&#xA;&lt;p&gt;Please use typical cryptic crossword strategies to solve this but some lateral thinking may be required.&lt;/p&gt;&#xA;&lt;p&gt;In the unlikely event that there are multiple valid answers then the one that gets the most up-votes in the next couple of days will be accepted.&lt;/p&gt;&#xA;&lt;p&gt;If you are not sure that your answer is correct, then it isn't! If you get the intended solution then you will be 100% certain of the reason.&lt;/p&gt;&#xA;&lt;p&gt;The answer is a single word. Please give the word and explain the cryptic techniques you used. I used two very straightforward ones.&lt;/p&gt;&#xA;&lt;p&gt;Hints&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Don't ignore the &lt;em&gt;lateral thinking&lt;/em&gt; tag.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Make sure you use all the information provided.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;No technical knowledge is needed (apart from a familiarity with cryptic crosswords) and this is a word that I'm sure everyone will know.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="69582" LastEditDate="2021-05-07T14:40:11.227" LastActivityDate="2021-05-07T14:40:11.227" Title="A quickie for the cryptic crossword enthusiasts: Here is the clue. / Cut cut! / What crazy, bloodthirsty creature remains? (5 letters)" Tags="|word|lateral-thinking|cryptic-clues|letters|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="20477" PostTypeId="2" ParentId="20474" CreationDate="2015-08-24T09:17:39.050" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 6. PENCIL ?&#xA; &lt;a href=&quot;https://i.stack.imgur.com/pctpi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pctpi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I based it on the first letter of their name :) which results a pattern of &lt;code&gt;S&lt;/code&gt; &amp;amp; &lt;code&gt;P&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;S - smile&lt;/li&gt;&#xA;&lt;li&gt;P - protractor&lt;/li&gt;&#xA;&lt;li&gt;S - stick&lt;/li&gt;&#xA;&lt;li&gt;P - PIZZA PIE&lt;/li&gt;&#xA;&lt;li&gt;S - &lt;a href=&quot;http://www.answers.com/Q/What_is_the_triangular_instrument_like_a_protractor_called&quot; rel=&quot;nofollow noreferrer&quot;&gt;set-square?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;P -? (&lt;strong&gt;PENCIL&lt;/strong&gt;)&lt;/li&gt;&#xA;&lt;li&gt;S - Shuriken ( &lt;em&gt;thanks to @Anachor&lt;/em&gt; )&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="12054" LastEditorUserId="11876" LastEditDate="2015-08-24T09:46:46.653" LastActivityDate="2015-08-24T09:46:46.653" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20478" PostTypeId="1" AcceptedAnswerId="20493" CreationDate="2015-08-24T09:23:50.063" Score="12" ViewCount="1152" Body="&lt;p&gt;&lt;em&gt;Perhaps the most famous of all,&lt;br&gt;&#xA;I was killed a century ago.&lt;br&gt;&#xA;My fate is well known to all:&lt;br&gt;&#xA;What bore me, hardened, killed me.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Named for an African province,&lt;br&gt;&#xA;And speedy as I could be.&lt;br&gt;&#xA;I held the record both ways&lt;br&gt;&#xA;For over twenty turns.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Named for another old province,&lt;br&gt;&#xA;I was killed a century ago.&lt;br&gt;&#xA;My fate is well known to all:&lt;br&gt;&#xA;It changed the course of history.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The second of my name,&lt;br&gt;&#xA;And the last one still around.&lt;br&gt;&#xA;Change one lady's name&lt;br&gt;&#xA;To get my predecessor.&lt;/em&gt;&#xA;&lt;hr&gt;&#xA;Who are we?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Each verse gives a different answer, but they're all connected as indicated by the title.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; &quot;an African province&quot; in the third verse now changed to &quot;another old province&quot;. Apologies for the error!&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-08-24T14:47:21.300" LastActivityDate="2015-08-24T16:28:03.490" Title="Quick four liners" Tags="|riddle|wordplay|" AnswerCount="3" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20479" PostTypeId="2" ParentId="20476" CreationDate="2015-08-24T09:32:47.173" Score="-1" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TROLL.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the &quot;cut&quot; command on a computer is implemented by Control+X. In cryptic-crossword lingo, &quot;cut&quot; can suggest either an abbreviation or a shortening of a word by removing some letters. &quot;Cut&quot;=&quot;Control+X&quot; can be thought of as an abbreviation, and we can remove some letters from &quot;Control+X&quot; to get &lt;em&gt;trol&lt;/em&gt;. Since we want a 5-letter creature, the answer looks like TROLL - we all know those can be &lt;a href=&quot;https://puzzling.stackexchange.com/a/10470/5373&quot;&gt;bloodthirsty&lt;/a&gt;!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wow ... have I managed to answer a cryptic-crossword question before Bailey M? :-D&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-08-24T09:32:47.173" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20480" PostTypeId="2" ParentId="20476" CreationDate="2015-08-24T09:55:48.253" Score="1" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SLICE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cut cut!&lt;br&gt;&#xA;What bloodthirsty creature remains?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you slice something, you are cutting into it. Also, once you chop off the first letter of SLICE, you are left with LICE, bloodsucking insects.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And of course it is 5 letters long.&lt;/p&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-08-24T10:11:34.867" LastActivityDate="2015-08-24T10:11:34.867" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20481" PostTypeId="2" ParentId="20474" CreationDate="2015-08-24T10:06:25.750" Score="3" Body="&lt;p&gt;[Inverted setsquare]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SUoo8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SUoo8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As per my observation the image patterns are in pairs, as in &lt;br&gt;&#xA;1-2 semicircle are inverted&lt;br&gt;&#xA;3-4 thickness reduces in opposite direction&lt;br&gt;&#xA;5-6 triangle and semicircle should be inverted&#xA;7 itself has complements&lt;/p&gt;&#xA;" OwnerUserId="15486" LastEditorUserId="12700" LastEditDate="2015-08-24T10:14:23.257" LastActivityDate="2015-08-24T10:14:23.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20482" PostTypeId="1" AcceptedAnswerId="20484" CreationDate="2015-08-24T10:14:06.227" Score="0" ViewCount="1706" Body="&lt;p&gt;This should be an easy puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Replace the x in the below with letters to make 6 valid 5-letters English words (no proper nouns or abbreviations) both vertically and horizontally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This puzzle will be more fun if you only use a dictionary to validate your answers rather than use it to search for words that fit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may be more than one solution.  However, the challenge is to use the least number of &lt;strong&gt;distinct characters&lt;/strong&gt; to replace the &lt;code&gt;x&lt;/code&gt;s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Green tick to the first valid answer that uses the least numbers of letters.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;code&gt;C x D x T&lt;/code&gt;&lt;br&gt;&#xA;  &lt;code&gt;x   x   x&lt;/code&gt;&lt;br&gt;&#xA;  &lt;code&gt;R x T x R&lt;/code&gt;&lt;br&gt;&#xA;  &lt;code&gt;x   x   x&lt;/code&gt;&lt;br&gt;&#xA;  &lt;code&gt;D x R x S&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12332" LastEditorUserId="4385" LastEditDate="2015-08-24T18:38:34.337" LastActivityDate="2015-08-25T09:47:12.963" Title="Find the missing letters" Tags="|crosswords|letters|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20483" PostTypeId="2" ParentId="20482" CreationDate="2015-08-24T10:18:31.357" Score="4" Body="&lt;p&gt;One of the solutions. (Yes, there are multiple possible)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;C A D E T&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;A   E   A&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;R O T O R&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;E   E   E&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;D A R E S&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Although there were no constraints on the letters to be added, I chose to do it by replacing x only with vowels. So I managed to do it using only A, E and O.&lt;/p&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-08-24T10:18:31.357" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20484" PostTypeId="2" ParentId="20482" CreationDate="2015-08-24T10:28:44.777" Score="5" Body="&lt;p&gt;A solution using the same pattern (only 2 letters):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;C A D E T&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;A   A   A&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;R A T E R&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;E   E   E&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;D A R E S&lt;/code&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Each are words of the form &lt;code&gt;_ A _ E _&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Just for the fun, here is one with 7 distinct letters.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;C A D E T&#xA;U   E   U   &#xA;R O T O R&#xA;E   E   F  &#xA;D I R T S&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="15544" LastEditorUserId="15544" LastEditDate="2015-08-25T09:47:12.963" LastActivityDate="2015-08-25T09:47:12.963" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20485" PostTypeId="2" ParentId="20373" CreationDate="2015-08-24T11:10:10.453" Score="7" Body="&lt;p&gt;I think that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the names of the fish are references to SI units:&lt;br&gt;&#xA; Anchovies =&gt; Amp&lt;br&gt;&#xA; mackerel =&gt; meter&lt;br&gt;&#xA; salmon =&gt; second&lt;br&gt;&#xA; kedgeree =&gt; kilogram  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So Eldred-Campbell's surname is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an SI derived unit made up of Amps^2, meters^-2, seconds^4, kilograms^-1... that would be &lt;strong&gt;Farad&lt;/strong&gt; and his initials E-C hint towards &lt;strong&gt;Electrical Capacitance&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the bonus question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ella-Charlotte: A^1 s^1 &lt;strong&gt;Coulomb&lt;/strong&gt; (Electrical Charge)&lt;br&gt;&#xA; Eloise-Florence: A^-1 m^2 s^-3 k^1 &lt;strong&gt;Volt&lt;/strong&gt; (Electromotive Force)&lt;br&gt;&#xA; Magnus-Ford-Stephen: A^-1 s^-2 k^1 &lt;strong&gt;Tesla&lt;/strong&gt; (Magnetic Field Strength)&lt;br&gt;&#xA; Marc-Frances: A^-2 m^2 s^-2 k^1  &lt;strong&gt;Weber&lt;/strong&gt; (Magnetic Flux) &lt;em&gt;(Wikipedia says this is kg⋅m2⋅s−2⋅A−1 so I think Marc Frances should merely dislike, rather than hate, Anchovies)&lt;/em&gt;&lt;br&gt;&#xA; Enrico: m^2 s^-2 k^1 &lt;strong&gt;Joule&lt;/strong&gt; (Energy)&lt;br&gt;&#xA; Portia: m^2 s^-3 k^1 &lt;strong&gt;Watt&lt;/strong&gt; (Power)&lt;br&gt;&#xA; Felix: m^1 s^-2 k^1 &lt;strong&gt;Newton&lt;/strong&gt; (Force)&lt;br&gt;&#xA; Primula: m^-1 s^-2 k^1 &lt;strong&gt;Pascal&lt;/strong&gt; (Pressure)&lt;br&gt;&#xA; Randolph: s-1 &lt;strong&gt;Becquerel&lt;/strong&gt; (Radioactivity)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This was a great puzzle. Thank goodness for &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/SI_derived_unit&quot;&gt;https://en.wikipedia.org/wiki/SI_derived_unit&lt;/a&gt; :-)&lt;br&gt;&#xA; P.S. The school should have been called the Seafood Institute!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2059" LastActivityDate="2015-08-24T11:10:10.453" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20487" PostTypeId="1" AcceptedAnswerId="20488" CreationDate="2015-08-24T12:22:25.013" Score="11" ViewCount="3413" Body="&lt;p&gt;Can anyone solve this riddle?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Always day and always night,&lt;br&gt;&#xA;  You'll know me when you see the light.&lt;br&gt;&#xA;  But don't look up, don't mess around -&lt;br&gt;&#xA;  In these words is where I may be found.&lt;br&gt;&#xA;  Through eternity ever I flowed,&lt;br&gt;&#xA;  Beyond the marble you call your abode.&lt;br&gt;&#xA;  I am warm and I am cold;&lt;br&gt;&#xA;  Visit me if you are bold.&lt;br&gt;&#xA;  If you are confused sometimes,&lt;br&gt;&#xA;  Just relax, read between the lines.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11304" LastEditorDisplayName="user20" LastEditDate="2015-08-24T19:25:27.510" LastActivityDate="2015-08-24T19:25:27.510" Title="Always day and always night - a riddle" Tags="|riddle|word|rhyme|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20488" PostTypeId="2" ParentId="20487" CreationDate="2015-08-24T12:28:59.857" Score="21" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Always day and always night,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Day and night are properties of planets with a sun, but space contains all of these planets, including both the parts in day and the parts in night.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'll know me when you see the light.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We [mostly] see light through empty space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But don't look up, don't mess around -&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ceilings and the light sky by day obscure our view of space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In these words is where I may be found.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Spaces between these words.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Through eternity ever I flowed,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Space has [probably] existed as long as time has.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Beyond the marble you call your abode.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Space surrounds the Earth (the marble) that's our home.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am warm and I am cold;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Space has no temperature, but places in space will be very hot and very cold.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Visit me if you are bold.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Brave the final frontier.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you are confused sometimes,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Spaced out.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Just relax, read between the lines.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Spaces between the lines.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12462" LastEditorUserId="12462" LastEditDate="2015-08-24T12:33:23.840" LastActivityDate="2015-08-24T12:33:23.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20490" PostTypeId="2" ParentId="20487" CreationDate="2015-08-24T12:34:50.460" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Space&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is in the source code :-)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit:&lt;/em&gt;  &lt;br&gt;&#xA;@downvoters, the question was/is if anyone can solve this and there were/are no other restrictions. So i choosed the easiest way for me [looking up the source code].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit 2:&lt;/em&gt; &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/osewT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/osewT.png&quot; alt=&quot;Source Code&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14461" LastEditorUserId="14461" LastEditDate="2015-08-24T14:32:03.433" LastActivityDate="2015-08-24T14:32:03.433" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="20491" PostTypeId="2" ParentId="20478" CreationDate="2015-08-24T12:48:19.147" Score="0" Body="&lt;p&gt;I don't have time to work on the whole thing right now, but here's a contribution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Wyndham_Halswelle&quot; rel=&quot;nofollow&quot;&gt;Wyndham Halswelle&lt;/a&gt; was a famous athlete who died in 1915 (a century ago). The circumstances of his death were well documented and he is still remembered.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12462" LastActivityDate="2015-08-24T12:48:19.147" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20493" PostTypeId="2" ParentId="20478" CreationDate="2015-08-24T13:04:28.450" Score="13" Body="&lt;p&gt;You are talking about &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Famous ships: all 4 are liners like the question title suggests&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Verse 1: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Titanic, most famous ship, destroyed by an iceberg&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Verse 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mauretania, holding the blue Ribbon for 20 years, west- and eastbound Atlantic speed record&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Verse 3: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sinking of the Lusitania and attack on Sussex was a turning point in WW1. Lusitania was named after a Roman province located in Portugal and Spain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Verse 4: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Queen Mary 2, famous liner. As far as I know, only active ship referring to a British queen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13748" LastEditorUserId="10781" LastEditDate="2015-08-24T16:28:03.490" LastActivityDate="2015-08-24T16:28:03.490" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20494" PostTypeId="2" ParentId="20478" CreationDate="2015-08-24T13:04:48.530" Score="6" Body="&lt;p&gt;I'm pretty sure they're all:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Ships  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1(very sure):  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Titanic. Very famous, sank in 1912 (approximately a century ago), and was borne by water, which, hardened, becomes the notorious iceberg &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3(unsure):  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I want to say the &lt;a href=&quot;https://en.wikipedia.org/wiki/RMS_Lusitania&quot;&gt;Lusitania&lt;/a&gt;, but I can't find a similarly named African province &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4(not entirely sure):  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;a href=&quot;https://en.wikipedia.org/wiki/RMS_Queen_Mary_2&quot;&gt;The Queen Mary 2&lt;/a&gt;, possibly&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11717" LastActivityDate="2015-08-24T13:04:48.530" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20495" PostTypeId="2" ParentId="20295" CreationDate="2015-08-24T13:59:11.877" Score="3" Body="&lt;p&gt;Backing up the r/2 claim with an alternative visualisation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/Ov3o9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ov3o9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15944" LastEditorUserId="5373" LastEditDate="2015-08-24T14:45:08.453" LastActivityDate="2015-08-24T14:45:08.453" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20496" PostTypeId="2" ParentId="20409" CreationDate="2015-08-24T16:37:04.763" Score="2" Body="&lt;p&gt;This is more an observation of the pattern than a method to get it, but if we assume a shooter has a spread of one space, meaning for example if aiming at 20, there is an equal chance of hitting 20,5,or 1, then we get these expected values for each target. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;20    1   18  4   13  6   10  15  2   17  3   19  7   16  8   11  14  9   12  5&lt;br&gt;&#xA;  8.6 13  7.6 11.6    7.6 9.6 10.3    9   11.3    7.3 13  9.6 14  10.3    11.6    11  11.3    11.6    8.6 12.3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The expected values range from 7.3 to 14, a pretty big spread. But if we order targets by expected value, we get&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;17&#xA;  13&#xA;  18&#xA;  20&#xA;  12&#xA;  15&#xA;  6&#xA;  19&#xA;  10&#xA;  16&#xA;  11&#xA;  2&#xA;  14&#xA;  4&#xA;  9&#xA;  8&#xA;  5&#xA;  1&#xA;  3&#xA;  7&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is semi-close to being ordered. Basically, if you evenly hit the target you're aiming at or one of it's neighbors, the best places to shoot for are actually 1,3 and 7, while the worst are 17, 13, and 18. There are still a couple inconsistencies, such as 14 being so high on the list, but this gives a general framework.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Other observations&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Even spread is impossible:&lt;/strong&gt; Consider 20. With value $a$ on its left and $b$ on its right, the expected value is $(20+a+b)/3$. Now consider spot $a$. 20 is one neighbor, call the other neighbor $c$. So if we have $(20+a+c)/3 = (20+a+b)/3 =&amp;gt; c = b$ which is impossible, because there are no repeating values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Smallest possible spread:&lt;/strong&gt; If we order the scores 20,1,19,2... I think we get the smallest difference in expected values, from 17 = 8 to 10 = 13.66&lt;/p&gt;&#xA;" OwnerUserId="1968" LastEditorUserId="1968" LastEditDate="2015-08-24T16:48:17.577" LastActivityDate="2015-08-24T16:48:17.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20497" PostTypeId="1" AcceptedAnswerId="20531" CreationDate="2015-08-24T16:39:43.607" Score="11" ViewCount="7667" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Won, now own.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rules:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;A “word” is an entry on &lt;a href=&quot;http://dictionary.reference.com/&quot; rel=&quot;nofollow&quot;&gt;dictionary.com&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A “sentence” is any text that is grammatically correct, has a complete meaning in the English language, starts with a capital letter and ends with a dot.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Repeated words are not allowed.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You are allowed to use punctuation, such as colons and quotation marks, as long as it’s not for making lists (i.e. cheating).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Sentences are rated by word count, however proper names count each as only one word since they can trivially be extended from simple proper nouns by appending surnames.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Common words cannot be changed to proper nouns, although these are allowed as long as they appear as such on dictionary.com.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Finally, as suggested, sentences and text inside quotes must actually make sense, not just gibberish.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;That’s it! Good luck, puzzlers!&lt;/p&gt;&#xA;" OwnerUserId="14219" LastEditorUserId="19114" LastEditorDisplayName="user20" LastEditDate="2020-01-04T22:23:55.317" LastActivityDate="2020-01-04T22:23:55.317" Title="Longest sentence where all words are anagrams of each other" Tags="|english|anagram|letters|open-ended|" AnswerCount="8" CommentCount="15" ClosedDate="2021-07-22T07:27:31.110" ContentLicense="CC BY-SA 3.0" />
  <row Id="20498" PostTypeId="2" ParentId="20497" CreationDate="2015-08-24T16:46:37.980" Score="9" Body="&lt;p&gt;Here's a five-word one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Rats' star tars tsar arts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or, &quot;The most famous from among the rats smears tar on art produced by the emperor.&quot;&lt;/p&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-08-24T16:46:37.980" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20500" PostTypeId="1" CreationDate="2015-08-24T17:52:02.100" Score="13" ViewCount="1310" Body="&lt;p&gt;A bishop and a rook, of opposing colors, start on opposite corners on the same side of a chessboard, oriented as in the following picture:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SQCV0.png&quot; alt=&quot;Chess Board&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;They then proceed to play a game of cat-and-mouse.&lt;/p&gt;&#xA;&lt;h2&gt;Rules&lt;/h2&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The pieces make a move based on the rules of chess, meaning that the bishop travels along the diagonal while the rook travels along the parallels.&lt;/li&gt;&#xA;&lt;li&gt;When a square is moved across, it becomes uninhabitable and can no longer be moved across.&lt;/li&gt;&#xA;&lt;li&gt;The bishop always goes first.&lt;/li&gt;&#xA;&lt;li&gt;If a piece is either a) captured, or b) does not move on its turn, whether by choice or because there are no valid moves left, that piece loses.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Based on these rules and assuming optimal play, which piece is crowned the victor?&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Bonus Question 1:&lt;/strong&gt; The same rules apply, except now a tile only becomes uninhabitable when a piece &lt;em&gt;finishes its move on that tile&lt;/em&gt;. Still assuming optimal play, which piece wins this variation?&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Bonus Question 2:&lt;/strong&gt; If the bishop starts in the top lefthand corner instead of the bottom lefthand corner, and the rook is now allowed to go first, which piece is crowned the victor? (Answer this question for both variations of uninhabitable squares.)&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Note:&lt;/strong&gt; The open-ended tag was included because there is ambiguity associated with the restriction of &amp;quot;optimal play&amp;quot;. A perfect answer will not only detail the victory condition of the winning piece, but also detail why the plays made can be considered &amp;quot;optimal&amp;quot; for each piece.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="69582" LastEditDate="2021-09-01T16:36:35.570" LastActivityDate="2021-09-01T16:36:35.570" Title="Chess Game of Life" Tags="|chess|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="20501" PostTypeId="2" ParentId="20497" CreationDate="2015-08-24T17:56:12.177" Score="9" Body="&lt;p&gt;8 words:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Parse, &quot;Asper spear pares prase, apres, reaps pears&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Meaning &quot;Parse the sentence: &quot;rough spear cuts/removes prase (a translucent, greenish variety of chalcedony.), afterwards it reaps pears&quot;. Convoluted, I know.&lt;/p&gt;&#xA;" OwnerUserId="15913" LastEditorUserId="15913" LastEditDate="2016-06-05T03:26:04.530" LastActivityDate="2016-06-05T03:26:04.530" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20502" PostTypeId="2" ParentId="20497" CreationDate="2015-08-24T18:01:31.930" Score="3" Body="&lt;p&gt;10 words&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Apres-rapes, parse &quot;prase spear pares spare pears, reaps asper&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;meaning&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;After you have completed the violent sexual attacks, analyse the following sentence: &quot;a weapon, made of green quartz, removes the skin of sweet fruits (rather like apples) which are additional to requirements, and is rewarded with a former silver coin of Turkey and Egypt&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thank you to @Ian McDonald for the helpful link.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: I see Respect has come up with something similar while I was working this out. &lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="4050" LastEditDate="2015-08-24T18:06:26.750" LastActivityDate="2015-08-24T18:06:26.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20503" PostTypeId="2" ParentId="20497" CreationDate="2015-08-24T18:12:13.790" Score="26" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Setal teals steal least-stale slate-stela Tesla tales. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;9 words&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Bristly birds steal the least stale tales of Tesla that are written on a slate stela (stone slab).&quot;&lt;/p&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="14177" LastEditDate="2015-08-25T20:56:07.433" LastActivityDate="2015-08-25T20:56:07.433" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20504" PostTypeId="2" ParentId="20500" CreationDate="2015-08-24T18:14:35.817" Score="5" Body="&lt;p&gt;I think the Rook has the advantage in the first and third cases due to 2 reasons: the bishop is limited to a single color and the rook overall has more moves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the bishop starts with G7, then the rook could go H5, and then the bishop has to go F8 (H6 is deadly, as is H8). The bishop then has 10 moves left, while the rook has 23.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, for the first case, the rook wants to split the board so that bishop has minimal spaces, then just outlast it. Try to stay above the bishop, or if the bishop has a long diagonal, go to it to split the board. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the third case, the same strategy as first case could work. If the rook goes E1, then the bishop moving 1-3 spaces means the rook can go E8 (E7 if dangerous), cutting the board in half. Anything other than that, E3, again cutting the board in half. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really have a good idea for the second case, except that I would favor the Rook as it has more moves available to it. In fact, the Rook could use the &quot;split the board&quot; tactic, but only need to hit 4 squares.&lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-08-24T18:49:31.570" LastActivityDate="2015-08-24T18:49:31.570" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20506" PostTypeId="2" ParentId="10753" CreationDate="2015-08-24T18:24:55.990" Score="1" Body="&lt;p&gt;Hold your cube so that the pieces are in the upper-front and upper-back position. In the case here, have the orange face pointed directly at you and yellow pointed up. This should put the cube in the correct orientation. Use: R' U2 R U R' U R U' R' U2 R U R' U R&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will swap the front-middle-upper and front-middle-back edge pieces and keep their orientation. However, the corner pieces will be repositioned and you will need to redo those steps. You're supposed to get the edge pieces correct before working on the corners anyways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;EDIT&lt;/em&gt;&lt;br&gt;&#xA;I decided to cheat and put it onto an online solver. It said it wasn't solvable. At first I thought that those moves would leave the corner pieces alone, but apparently not. I punched in a very similar pattern I had once and it solved it. So apparently your cube really is improperly manufactured. Just take the two pieces off of the cube and put them back in correctly.&lt;br&gt;&#xA;&lt;em&gt;/EDIT&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those who don't understand cube notation:&lt;br&gt;&#xA;R = right face of the cube&lt;br&gt;&#xA;U = upper face&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X means to rotate that face in a clockwise rotation (90')&lt;br&gt;&#xA;X' means to rotate the face in a counter-clockwise rotation (-90')&lt;br&gt;&#xA;X2 means to rotate the face twice (180')&lt;/p&gt;&#xA;" OwnerUserId="15950" LastEditorUserId="15950" LastEditDate="2015-08-25T01:11:07.460" LastActivityDate="2015-08-25T01:11:07.460" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20507" PostTypeId="2" ParentId="20474" CreationDate="2015-08-24T18:31:16.040" Score="3" Body="&lt;p&gt;My answer is...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/tCdAf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tCdAf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why? Well...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each image represents a different type of angle. 1 is a full revolution, 2 is a straight angle (180 degrees), 3 is no angle (0 degrees), 4 is an acute angle, 5 is a right angle, and 7 is a reflex angle. That leaves only an &lt;strong&gt;obtuse&lt;/strong&gt; angle, which this fan represents.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9289" LastActivityDate="2015-08-24T18:31:16.040" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20508" PostTypeId="2" ParentId="20500" CreationDate="2015-08-24T19:48:50.727" Score="9" Body="&lt;h2&gt;General thoughts&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The rook has obviously an advantage here, because it can use black as well as white squares, while the bishop can only use black squares. Apart from that several squares are dead ends for the bishop like for example &lt;code&gt;h8&lt;/code&gt;. If the rook can fence in the bishop in about a half of the board (&quot;Tron-Principle™&quot;) it is almost always a guaranteed win.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Base Question&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Complete winning strategy (in all cases the rook has more moves left than the number specified for the bishop):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1. ... Bg7 2. Rh5 (Bishop has 11 moves left)&#xA;&#xA;1. ... Bf6 2. Rh7&#xA;    2. ... Bg5 3. Rh8 (Bishop has 6 moves left)&#xA;    2. ... Bd8 3. Rg7 (Bishop has 8 moves left)&#xA;&#xA;1. ... Be5 2. Rh6&#xA;    2. ... Bf4 3. Re6 (Bishop has 5 moves left)&#xA;    2. ... Bg3 3. Re6 (Bishop has 4 moves left)&#xA;    2. ... Bg7 3. Rg6 (Bishop has 11 moves left)&#xA;    2. ... Bc7 3. Rh8 (Bishop has 4 moves left)&#xA;    2. ... Bb8 3. Rh7 (Bishop has 5 moves left)&#xA;&#xA;1. ... Bd4 2. Rh5&#xA;    2. ... Be3 3. Rd5 (Bishop has 5 moves left)&#xA;    2. ... Bf2 3. Rd5 (Bishop has 2 moves left)&#xA;    2. ... Bg1 3. Rd5 (Bishop has 0 moves left)&#xA;    2. ... Bf6 3. Rh7 (2nd case above)&#xA;    2. ... Bg7 3. Rg5 (Bishop has 11 moves left)&#xA;    2. ... Bb6 3. Rh7&#xA;        3. ... Bd8 4. Rd7 (Bishop has 0 moves left)&#xA;        3. ... Ba5 4. Rh8 (Bishop has 2 moves left)&#xA;    2. ... Ba7 3. Rh8 (Bishop has 0 moves left)&#xA;&#xA;1. ... Bc3 2. Rh5&#xA;    2. ... Bd2 3. Ra5 (Bishop has 6 moves left)&#xA;    2. ... Bf1 3. Ra5 (Bishop has 5 moves left)&#xA;    2. ... Bd4 3. Ra5 (Bishop has 6 moves left)&#xA;    2. ... Bf6 3. Rh7 (2nd case above)&#xA;    2. ... Bg7 3. Rg5 (Bishop has 11 moves left)&#xA;    2. ... Bb4 3. Rb5 (Bishop has 1 move left)&#xA;&#xA;1. ... Bb2 2. Rh3&#xA;    2. ... Bc1 3. Rb3 (Bishop has 4 moves left)&#xA;    2. ... Bd4 3. Rh5 (4th case above)&#xA;    2. ... Be5 3. Rh6 (3rd case above)&#xA;    2. ... Bf6 3. Rh7 (2nd case above)&#xA;    2. ... Bg7 3. Rh5 (1st case above)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h2&gt;Bonus Question 1&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I don't have a complete winning strategy here. But it seems even more obvious that the rook has an advantage here as well. My general strategy would be:&lt;br&gt;1. Block the black centre squares.&lt;br&gt;2. Block the path of the bishop on the sides (if the black centre squares are blocked, ideally only 1 more black square is needed for that).&lt;br&gt;3. Create dead ends for the bishop if possible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Bonus Question 2&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Complete winning strategy:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1. Re1 (as suggested by JonTheMon)&#xA;&#xA;    1. ... Bb7/Bc6/Bd5 2. Re8 (Bishop has 13/9/9 moves left)&#xA;&#xA;    1. ... Bf3 2. Re3 (Bishop has 12 moves left)&#xA;&#xA;    1. ... Bg2 2. Re2 (Bishop has 11 moves left)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2015-09-01T14:15:23.143" LastActivityDate="2015-09-01T14:15:23.143" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20509" PostTypeId="2" ParentId="20497" CreationDate="2015-08-24T20:03:41.030" Score="4" Body="&lt;p&gt;Here's an 8 word sentence with a proper noun:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Apres-rapes, Spear pares spare pears; reaps asper.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You didn't say I couldn't make one or all of them proper nouns!&lt;/p&gt;&#xA;" OwnerUserId="11926" LastEditorDisplayName="user20" LastEditDate="2015-08-28T18:35:28.587" LastActivityDate="2015-08-28T18:35:28.587" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20510" PostTypeId="2" ParentId="20245" CreationDate="2015-08-24T21:04:31.137" Score="10" Body="&lt;p&gt;Each line references&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; one of the first twenty books of the Old Testament. So the solution is to sort the lines in the order the books appear in the bible&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;With NASA's help, it all was started&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; GENESIS - This is the origin or start and is also the name of a &lt;a href=&quot;https://en.wikipedia.org/wiki/Genesis_(spacecraft)&quot; rel=&quot;noreferrer&quot;&gt;NASA spacecraft&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;A certain group has now departed&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; EXODUS - A mass departure of people&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Swedish metal takes this name&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LEVITICUS - This was a &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Leviticus_(band)&quot; rel=&quot;noreferrer&quot;&gt;Swedish Christian metal band&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;You count on them to win the game&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; NUMBERS - You count with numbers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Written rule, with one before it&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DEUTERONOMY - This is derived from the Greek for 'second law', so there is one rule before it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Dolly sang it, we adored it&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; JOSHUA - A song by Dolly Parton&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Together here we rule your fate&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; JUDGES - Can rule your fate if you're up in court&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;A young pig's tag, an all-time great&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RUTH - As in Babe Ruth, probably the most famous baseball player of all-time. He shares his name with the pig given the name Babe in the book &quot;The Sheep-Pig&quot; or &quot;Babe, the Gallant Pig&quot; depending on &lt;a href=&quot;https://en.wikipedia.org/wiki/The_Sheep-Pig&quot; rel=&quot;noreferrer&quot;&gt;which side of the Atlantic you reside&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;A father of the highest fame&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 SAMUEL - Samuel Adams was a Founding Father of the United States&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;A drinker who is called the same&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2 SAMUEL - Sam(uel) Adams is also the name of a popular American beer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Some adorned and openhearted&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 KINGS - The kings in a pack of cards can be adorned (of diamonds) or openhearted (of hearts)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Else you'll find them quite well-guarded&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2 KINGS - well guarded in a game of chess&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;History doth these create&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 CHRONICLES - Chronicles of history&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;In others magic worlds await&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2 CHRONICLES - Narnia is the magic land in the Chronicles of Narnia books&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;A pound of text with speech not florid&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; EZRA - Ezra Pound was a poet, who favoured short, simple sentences over a more elaborate style&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;When skip was blue, at birth he wore it&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; NEHEMIAH - This references &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Skip_James&quot; rel=&quot;noreferrer&quot;&gt;Skip James&lt;/a&gt;, a blues musician, born Nehemiah Curtis James&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;An orphan, cinematically&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ESTHER - The name of the orphan adopted in the film Orphan&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Concrete responsibility&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; JOB - A responsibility&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Historic sacred melody&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PSALMS - A psalm is a sacred song&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;And bywords of reality&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PROVERBS - Expressions that summarize, or provide a byword for reality&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12277" LastEditorUserId="12277" LastEditDate="2015-08-25T16:29:41.523" LastActivityDate="2015-08-25T16:29:41.523" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20511" PostTypeId="2" ParentId="20476" CreationDate="2015-08-24T21:05:18.273" Score="10" Body="&lt;p&gt;Here is the answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If you cut the head from the word 'cut' (leaving 'ut') then cut these two letters from 'the clue', you are left with 'hecle'. The word 'crazy' implies an anagram, so rearranging these letters gives 'leech' which is a five-letter bloodsucking creature.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-08-24T21:05:18.273" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20513" PostTypeId="1" AcceptedAnswerId="20548" CreationDate="2015-08-24T22:13:52.853" Score="-1" ViewCount="735" Body="&lt;ul&gt;&#xA;&lt;li&gt;I'm quick, I'm easy&lt;/li&gt;&#xA;&lt;li&gt;You can make me in a jiffy&lt;/li&gt;&#xA;&lt;li&gt;I'm on a farm&lt;/li&gt;&#xA;&lt;li&gt;I'm in your kitchen&lt;/li&gt;&#xA;&lt;li&gt;I'm in the pasture  &lt;/li&gt;&#xA;&lt;li&gt;I lay, I rot, in more places then one&lt;/li&gt;&#xA;&lt;li&gt;I'm smelly, but I'm also very tasty&lt;/li&gt;&#xA;&lt;li&gt;I have nicknames but I prefer one&lt;/li&gt;&#xA;&lt;li&gt;What am I?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note-&lt;/strong&gt; &lt;em&gt;You may not recognize the nickname&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint- Nice title, hey?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="15675" LastEditorUserId="8874" LastEditDate="2016-03-23T17:15:49.067" LastActivityDate="2016-03-23T17:15:49.067" Title="&quot;No Bake&quot; Riddle" Tags="|riddle|wordplay|language|regional-dialects|" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20515" PostTypeId="1" AcceptedAnswerId="20545" CreationDate="2015-08-24T23:20:49.820" Score="8" ViewCount="1093" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;You might think I'm broke, but then&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I'm a singer and sound like a hen&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;You can count on me&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I'm a tramp you see&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;So I'll sing you the blues once again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The above limerick conceals six song titles. Can you find all six and say what word they have in common?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: In order to make the limerick rhyme, I've had to take a few liberties but I think all the song titles are perfectly discoverable once you have guessed the common word. Some clues are cryptic and some are more straightforward.&lt;/p&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="14177" LastEditDate="2015-08-25T15:15:30.090" LastActivityDate="2015-08-25T18:44:08.133" Title="A limerick riddle. Provides clues to song titles that have a word in common" Tags="|riddle|word|knowledge|limerick|cryptic-clues|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20516" PostTypeId="2" ParentId="20454" CreationDate="2015-08-25T00:43:49.053" Score="2" Body="&lt;p&gt;Once you find a single combination that doesn't click, you can open the lock in $nk$ steps by taking $k$ steps to find each dial's correct position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to find a nonclicking combination: try all $2^n$ combinations where each dial is set to either 0 or 1. One of these is sure to not produce a click. Therefore, $2^n + nk$ guesses is enough to crack the lock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My current best strategy for $n &amp;gt; k &amp;gt; 2$ is to try combinations uniformly at random until you get one that doesn't click. This gives a bound of approximately (I think) $e^{n/k} \cdot n \log k$ for large $k$ and $n$. I won't write out all the details here, since (1) it's standard union bounding and (2) there's a good chance you can do better.&lt;/p&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2015-08-25T00:43:49.053" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20517" PostTypeId="2" ParentId="20497" CreationDate="2015-08-25T01:05:32.517" Score="6" Body="&lt;p&gt;&lt;strong&gt;9 words, 54 characters&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Artel's salter alerts staler slater, alters stelar estral ratels.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which means-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A salter (a person making salts) who belongs to an artel (a community of&#xA;    farmers), alerts a slater (a person who lays slates) who is staler (more &#xA;    unwitty), and alters an inscription of estral ratels (badgerlike carnivores &#xA;    who are in the estrous cycle) on a stone slub (stele).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also the words &quot;talers&quot; and &quot;laster&quot; which I am not able to fit in this sentence(?) meaningfully.&lt;/p&gt;&#xA;" OwnerUserId="15544" LastEditorDisplayName="user20" LastEditDate="2015-08-28T18:35:00.357" LastActivityDate="2015-08-28T18:35:00.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20518" PostTypeId="2" ParentId="20515" CreationDate="2015-08-25T01:30:25.727" Score="0" Body="&lt;p&gt;A wild guess here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might think I'm broke, but then&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Money, money, money by ABBA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can count on me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lean On Me by Bill Withers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;P.S. Now I have these songs singing in my head!&lt;/p&gt;&#xA;" OwnerUserId="12332" LastActivityDate="2015-08-25T01:30:25.727" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20519" PostTypeId="1" AcceptedAnswerId="20520" CreationDate="2015-08-25T01:35:28.543" Score="5" ViewCount="590" Body="&lt;p&gt;I have a color, I have a name,&#xA;&lt;br&gt;You love to play me, if you are game.&#xA;&lt;br&gt;If I am under age, you might have to send me back,&#xA;&lt;br&gt;When I am just the right age, you will be a happy chap.&#xA;&lt;br&gt;But if I am too old, well, then that's too bad.&lt;/p&gt;&#xA;" OwnerUserId="12332" LastActivityDate="2015-08-25T02:12:14.997" Title="Are you ready to play me?" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20520" PostTypeId="2" ParentId="20519" CreationDate="2015-08-25T02:12:14.997" Score="8" Body="&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Blackjack. Your color is black, your name is Jack, you're a game. If you're &quot;under age&quot; (9, 10, 11, etc.) I'm going to go back for more cards. If you're just the right age (21), I'm psyched! And if you're older, I just lost this round!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10851" LastActivityDate="2015-08-25T02:12:14.997" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20522" PostTypeId="2" ParentId="19806" CreationDate="2015-08-25T02:53:57.167" Score="10" Body="&lt;p&gt;Got it!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;What a certain literary professor for his unbeheld felt: LOVE (deciphered in other comments); in tennis, &quot;love&quot; means 0.&lt;br&gt;&#xA;The two of spades, from a new deck dealt: DEUCE. (Both the name for the card and a name for someone who deals the second card instead of the first to cheat.) A deuce is a tied 40-40 game in tennis.&lt;br&gt;&#xA;A din, a doubt: AD IN, AD OUT are tennis terms. When the score is tied at 40-40, a player must win two points in a row to win the game. AD IN and AD OUT are terms meaning the IN or OUT players have one of those points - AD stands for 'advantage'.&lt;br&gt;&#xA;Out!: Game over, OUT won.&lt;br&gt;&#xA;Up, up, by the Ort Sphere's smallest test, : Up, up refers to two points being gained; therefore the score is 30. &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Shor%27s_algorithm&quot;&gt;Peter Shor&lt;/a&gt;'s algorithm was first used to test factorization of 15.&lt;br&gt;&#xA;But oh, wait, now by one-third less.: The score in tennis goes from 0 to 15 to 30 to 40 (it was an abbreviation for 45); this means the increase is one-third less, as it only goes up by 10 instead of 15. So now the score is at FORTY.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the order is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;What a certain literary professor for his unbeheld felt: LOVE&lt;br&gt;&#xA;Up, up, by the Ort Sphere's smallest test, : FIFTEEN, THIRTY&lt;br&gt;&#xA;But oh, wait, now by one-third less.: FORTY&lt;br&gt;&#xA;The two of spades, from a new deck dealt: DEUCE.&lt;br&gt;&#xA;A din, a doubt: AD IN, AD OUT (the players are switching back and forth between having a deuce and giving one person the advantage).&lt;br&gt;&#xA;Out!: Out wins!  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="11876" LastEditDate="2015-08-25T15:18:00.563" LastActivityDate="2015-08-25T15:18:00.563" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20523" PostTypeId="2" ParentId="20462" CreationDate="2015-08-25T04:43:25.743" Score="3" Body="&lt;p&gt;If you build something into your riddle besides the solution itself, it can help to enforce a &quot;rule&quot; for the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me explain:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Maybe you have a riddle where every line rhymes, and the solution must also rhyme with all these lines.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Maybe the lines of the riddle begin with successive letters of the alphabet, and the solution must continue this trend.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Maybe the text of the riddle is formatted in a specific shape reminiscent of the answer.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Maybe the solution is actually contained within the riddle as:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;an anagram of itself&lt;/li&gt;&#xA;&lt;li&gt;spelled out be the initial letters of consecutive words, or the first/last letters of each line&lt;/li&gt;&#xA;&lt;li&gt;a play on words&lt;/li&gt;&#xA;&lt;li&gt;an alternate interpretation of the words in the title&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Maybe a word/letter/concept/rhyme/alliteration/etc. is repeated often within the riddle as a hint to the final solution (either the solution has the same characteristic or the characteristic provides a further hint to the solution). e.g. Most of the words start with &quot;B&quot;, and the solution is &quot;a bee&quot;, or the riddle uses only the first half of the alphabet, and the solution must do the same.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are many other similar strategies that you could use; these are just a few examples to get you started.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of these will completely preclude incorrect answers, but they should help to solidify your riddle and reduce the chance that someone will come up with a random solution that fits as well as yours.&lt;/p&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2015-08-25T04:43:25.743" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20524" PostTypeId="2" ParentId="20474" CreationDate="2015-08-25T05:52:35.340" Score="7" Body="&lt;p&gt;I think the missing image can well be satisfied by the one below&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/qUYGD.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qUYGD.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In each image, if we calculate the number of straight edges, we get 0, 1, 1, 2, 3, ?, 8. This looks very much like a Fibonacci sequence, with the sixth object requiring 5 edges. And I chose the most famous five sided object, the Pentagon. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Interestingly, &lt;a href=&quot;https://puzzling.stackexchange.com/a/20477/12700&quot;&gt;Dyrandz Famador&lt;/a&gt; could have had this one, if he only chose a different object whose name started with the letter 'P'. :-P&lt;/p&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2015-08-25T05:52:35.340" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20525" PostTypeId="2" ParentId="20462" CreationDate="2015-08-25T06:28:41.877" Score="3" Body="&lt;p&gt;Though I think it's impossible to &lt;em&gt;guarantee&lt;/em&gt; a single unique solution, I can think of a couple of things that can certainly help. I'll use your &quot;&lt;a href=&quot;https://puzzling.stackexchange.com/q/20444/5208&quot;&gt;One rhyme, / Space, time&lt;/a&gt;&quot; riddle as an example...&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Firstly, as &lt;a href=&quot;https://puzzling.stackexchange.com/a/20523/5208&quot;&gt;GentlePurpleRain mentions&lt;/a&gt;, you can try to directly limit solutions by including rules &lt;em&gt;within&lt;/em&gt; the riddle. For example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I am one word. I am one rhyme,&lt;br&gt;&#xA;  I am all of space and time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With this minor alteration, you're forcing the answer to be a single word. Also note that by simply adding the &quot;I am&quot; to the &quot;one rhyme&quot; line, you have effectively restricted things further, invalidating the incorrect answers you received, since they relied on a broader interpretation, such as &quot;I &lt;em&gt;have&lt;/em&gt; one rhyme&quot; (innuendo), or &quot;I &lt;em&gt;describe&lt;/em&gt; one rhyme&quot; (relative/dimension).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Second key to keeping a riddle short while limiting the possible answers is to try and include seeming contradictions. Look for various forms of wordplay/metaphor/etc based on your intended answer, and then pick clues such that they seem to point in opposite directions. This will naturally limit answers, as a solution for one clue will likely break another. For example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Split: So small, I'm but one rhyme.&lt;br&gt;&#xA;  Whole: Boundless, in space and time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Maybe not the best example of what I'm trying to get at, but here I'm trying to create a contradiction where the answer is apparently something very small, yet very big.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'm not saying either of my examples above are fool-proof (or even necessarily an improvement on your original, which I quite liked by the way), but hopefully they illustrate my points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One final note... Everything you do to your riddle is going to involve a compromise. Obviously my examples above have made your riddle longer, but unfortunately that's generally a side effect of making it more restricted. The key is to choose carefully when to compromise in balancing the brevity of the riddle with its clarity, because your compromises will provide diminishing returns. That is to say, going from a one line riddle to a two line riddle won't necessarily make it less appealing, but it can go a long way to making it less ambiguous. On the flip side, going from 5 stanzas to 10 stanzas probably isn't the most efficient way of reducing ambiguity, but it sure makes it more unwieldy and less appealing as a riddle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, I think the goal of creating an unambiguous four word riddle is a noble, but incredibly difficult goal. Let me know when you succeed. :)&lt;/p&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-08-25T06:28:41.877" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20526" PostTypeId="1" AcceptedAnswerId="20529" CreationDate="2015-08-25T06:44:45.770" Score="4" ViewCount="624" Body="&lt;p&gt;I came across this &lt;a href=&quot;http://mathnexus.wwu.edu/archive/oldie/detail.asp?ID=162&quot; rel=&quot;nofollow&quot;&gt;puzzle&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A wayfarer asks a shepherd how many sheep he has. He replies that if he numbered them two and two, or three and three, or four and four, or five and five, or six and six, there was always one over. If he numbered them by seven and seven there was none over. How many sheep had he?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is relatively easy to find a solution using constraints that you can construct from the information in the puzzle (hints):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Constraint 1: The number of sheep must be a multiple of seven. This one is obvious.&lt;br&gt;&#xA; Constraint 2: The number of sheep minus 1 must be a multiple of 60. This is the lowest number that has 2, 3, 4, 5 and 6 as divisors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and using a little bit of trial and error (solution):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 x 60 + 1 = 61, but is not a multiple of seven&lt;br&gt;&#xA; 2 x 60 + 1 = 121, but is not a multiple of seven&lt;br&gt;&#xA; 3 x 60 + 1 = 181, but is not a multiple of seven&lt;br&gt;&#xA; 4 x 60 + 1 = 241, but is not a multiple of seven&lt;br&gt;&#xA; 5 x 60 + 1 = 301 and is a multiple of seven. It is the lowest number of sheep that solves the puzzle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My questions now are:&lt;br&gt;&#xA;1. Can this puzzle be solved without trial and error, using calculations or reasoning?&lt;br&gt;&#xA;2. Is there only one solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know the answers to these questions.&lt;/p&gt;&#xA;" OwnerUserId="15959" LastActivityDate="2015-08-25T17:06:10.810" Title="Another Way to Count Sheep At Night in Bed" Tags="|calculation-puzzle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20527" PostTypeId="1" CreationDate="2015-08-25T06:53:29.447" Score="2" ViewCount="240" Body="&lt;p&gt;&lt;strong&gt;STORY FROM GOTTINGEN&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Last summer I spent my vacation in Gottingen - a beautiful city in the heart of Germany. I needed some help reading a map I got from my hotel, so approached a policeman on the street who seemed to be writing a ticket for illegal parking.&lt;/p&gt;&#xA;&lt;p&gt;-&lt;strong&gt;Sorry for interrupting you sir, can I ask you a question?&lt;/strong&gt; - I inquired.&lt;/p&gt;&#xA;&lt;p&gt;-&lt;strong&gt;I have a ticket to write here boy, can you wait for a minute?&lt;/strong&gt; - replied the policeman, visibly agitated.&lt;/p&gt;&#xA;&lt;p&gt;-&lt;strong&gt;No problem sir.&lt;/strong&gt; - I said and stepped respectfully aside.&lt;/p&gt;&#xA;&lt;p&gt;-&lt;strong&gt;This smart ass has been parking his car here illegally for 10 days. In Europe we don’t condone such behavior.&lt;/strong&gt; - the policeman shared his frustration. - &lt;strong&gt;Do you behave like this in your home country boy?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;-&lt;strong&gt;No sir.&lt;/strong&gt; - I replied with all the humbleness I was capable of.&lt;/p&gt;&#xA;&lt;p&gt;-&lt;strong&gt;500 Euros - this should teach him a lesson. I don’t think he will risk his license by parking on handicapped spots anymore.&lt;/strong&gt; -complacently concluded the policeman and signed the little paper.&lt;/p&gt;&#xA;&lt;p&gt;-&lt;strong&gt;I guess... Who would want this?&lt;/strong&gt; - I confirmed.&lt;/p&gt;&#xA;&lt;p&gt;The policeman put the ticket on the windshield diligently, hid the pen back in his shirt pocket and addressed me eventually:&lt;/p&gt;&#xA;&lt;p&gt;-&lt;strong&gt;You said you have a question for me boy. Go for it.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;-&lt;strong&gt;Uuuhm… I got this map from my hotel and on the diagonal I noticed some strange sequence of numbers - 10, 7, 2, 4, 2, 8... Can you tell me what these numbers stand for?&lt;/strong&gt; -I explained.&lt;/p&gt;&#xA;&lt;p&gt;-&lt;strong&gt;I have no clue.&lt;/strong&gt; -nonchalantly replied the policeman and without saying any other words, left.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;Please excuse my flawed writing, this is definitely not one of my biggest talents.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Update: I feel the problem is too cryptic, so decided just to post its solution below.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;HINT:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The city of Gottingen is not really important, just liked the way it sounds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14140" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-08-25T14:49:07.973" Title="Story from Gottingen" Tags="|number-sequence|story|" AnswerCount="2" CommentCount="10" FavoriteCount="0" ClosedDate="2015-08-26T06:39:52.170" ContentLicense="CC BY-SA 3.0" />
  <row Id="20528" PostTypeId="2" ParentId="20526" CreationDate="2015-08-25T07:11:24.847" Score="1" Body="&lt;p&gt;Well, you &lt;em&gt;did&lt;/em&gt; solve it with calculations and reasoning - you didn't try every number one-by-one! There's no faster way to do it, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And no, there isn't only one solution - you can always add the least common multiple of 2 through 7 (420) to get another valid answer. So 721, 1141, 1561... will all also be correct. The number you got is the smallest though - by the Chinese Remainder Theorem, all solutions are congruent modulo 420 (they give you the same remainder when divided by 420) so any possible answer less than 420 is the smallest (unless you're including negative sheep, of course).&lt;/p&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="11876" LastEditDate="2015-08-25T17:06:10.810" LastActivityDate="2015-08-25T17:06:10.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20529" PostTypeId="2" ParentId="20526" CreationDate="2015-08-25T07:18:42.263" Score="6" Body="&lt;p&gt;This is a system of equivalences. You may want to check the Chinese Remainder Theorem (which says that there is just one solution modulo 420):&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Chinese_remainder_theorem&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Chinese_remainder_theorem&lt;/a&gt;&#xA;and the Extended Euclidean algorithm (which gives you a way to compute such systems in general):&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Extended_Euclidean_algorithm&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Extended_Euclidean_algorithm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14140" LastActivityDate="2015-08-25T07:18:42.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20531" PostTypeId="2" ParentId="20497" CreationDate="2015-08-25T08:02:37.647" Score="19" Body="&lt;p&gt;&lt;strong&gt;11 words, 77 characters&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Anestri-resiant starnie retina's stearin stainer retains antsier, nastier retsina ratines.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which means:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having been resiant (residing) in anestri (a period of sexual rest), &#xA;the stearin stainer (a pigment used to stain stearin, a certain triglyceride compound)&#xA;of the retina of a starnie (a small star) retains&#xA;retsina ratines (loose, knotted fabric soaked in retsina, a pine-flavoured Greek wine)&#xA;that are more fidgety (antsier) and more nastier than before.&lt;/p&gt;&#xA;" OwnerUserId="15960" LastEditorDisplayName="user20" LastEditDate="2015-08-28T18:34:34.997" LastActivityDate="2015-08-28T18:34:34.997" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20532" PostTypeId="2" ParentId="20497" CreationDate="2015-08-25T08:13:18.933" Score="8" Body="&lt;p&gt;6 6-letter words, no punctuation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;Septal plates staple palest pastel petals.&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which means:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plates from a burial chamber use a stapler on pale, delicately shaded petals.&lt;/p&gt;&#xA;" OwnerUserId="15961" LastEditorDisplayName="user20" LastEditDate="2015-08-28T18:35:10.533" LastActivityDate="2015-08-28T18:35:10.533" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20533" PostTypeId="1" AcceptedAnswerId="20552" CreationDate="2015-08-25T09:20:53.653" Score="2" ViewCount="207" Body="&lt;p&gt;Patrick is answering a series of questions where he is given a word and he has to write a single character. He knows how to answer these and quickly answers-&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;code&gt;fsmss&lt;/code&gt; $\to$ &lt;code&gt;s&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;wsffs&lt;/code&gt; $\to$ &lt;code&gt;0&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;fwmhdhdbfw&lt;/code&gt; $\to$ &lt;code&gt;m&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;bsmhb&lt;/code&gt; $\to$ &lt;code&gt;0&lt;/code&gt; or &lt;code&gt;b&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;dbmhsm&lt;/code&gt; $\to$ &lt;code&gt;w&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;While answering, he took a break. However, when he returned from the break, he forgot the process he found the answers, help him figure out the following-&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;code&gt;fdhdbmhwmh&lt;/code&gt; $\to \ ?$&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;bmhsmhwfwdh&lt;/code&gt; $\to \ ?$&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;smhdfwsfdbmh&lt;/code&gt; $\to \ ?$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;bsmhb&lt;/code&gt; $\to$ &lt;code&gt;0&lt;/code&gt; or &lt;code&gt;b&lt;/code&gt; means both &lt;code&gt;0&lt;/code&gt; and &lt;code&gt;b&lt;/code&gt; are valid answers. In case of multiple answers any valid one is accepted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Don't sweat over the wordplay tag, it really isn't necessary to solve the puzzle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15544" LastEditorUserId="15544" LastEditDate="2015-08-25T12:57:44.063" LastActivityDate="2015-08-25T16:15:43.060" Title="A bunch of related letters" Tags="|pattern|wordplay|letter-sequence|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20535" PostTypeId="2" ParentId="20527" CreationDate="2015-08-25T09:37:42.017" Score="1" Body="&lt;p&gt;Just a shot in the dark:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might have a connection with your story and it is sth. that a policeman don't know. In your story the illegally parking car is blocking a parking spot for handicapped people. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the connection might be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;These numbers are representing zones in the city and the lower they are the easier you can get there/reach them with a handicap, like sitting in a wheelchair. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14461" LastEditorUserId="14461" LastEditDate="2015-08-25T09:48:16.383" LastActivityDate="2015-08-25T09:48:16.383" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20536" PostTypeId="2" ParentId="20513" CreationDate="2015-08-25T09:52:26.860" Score="2" Body="&lt;p&gt;Maybe you are talking about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a cheese&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I'm quick, I'm easy&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No idea for this one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You can make me in a jiffy&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cheese can be handmade.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I'm on a farm&lt;/li&gt;&#xA;&lt;li&gt;I'm in your kitchen&lt;/li&gt;&#xA;&lt;li&gt;I'm in the pasture&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cheese can be found on a farm (cheese production), in a kitchen (on a cheese plate) and in the pasture (goat milk that is used for cheese).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I lay, I rot, in more places then one&lt;/li&gt;&#xA;&lt;li&gt;I'm smelly, but I'm also very tasty&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some cheese smells a lot and they .. &quot;extend&quot; with age (they overflow). The more they smell, the tastier they are.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I have nicknames but I prefer one&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are many varieties of cheese. They can be called by their specific name ... or by the classic &quot;cheese&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13384" LastActivityDate="2015-08-25T09:52:26.860" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20537" PostTypeId="2" ParentId="10559" CreationDate="2015-08-25T11:19:33.060" Score="0" Body="&lt;p&gt;Is it a timekeeper?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first two lines says &quot;play catch&quot; which makes me think of clock hands which go at different rates, meeting only at specific points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second two says &quot;I am the second, an honourable mention&quot; which I think means literally the second. Most people omit the second when setting a time&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third couple seems to reference the point where the hands of the clock are connected. The point where &quot;Around about which, he and I dance together&quot;&lt;/p&gt;&#xA;" OwnerUserId="15965" LastActivityDate="2015-08-25T11:19:33.060" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20538" PostTypeId="1" CreationDate="2015-08-25T11:34:17.540" Score="3" ViewCount="273" Body="&lt;p&gt;I see that some puzzles specify that the answers must be findable in a specified dictionary. This seems like an excellent idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I've noticed several people specifying &lt;strong&gt;dictionary.com&lt;/strong&gt; Is there something especially good about this dictionary for puzzling purposes?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Online dictionaries that can be accessed as continuous text or are otherwise computer-searchable would be useful. Does anyone have links to such?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What's a good dictionary that has an extensive vocabulary but doesn't have 'weird' words, i.e. things that only specialists or lexicologists know about. For example that has 'incomprehensible' and 'quintessential' but not 'batrachophagous' or 'chirotonsor'.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;All suggestions and tips welcome.&lt;/p&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="14177" LastEditDate="2015-08-25T11:45:33.510" LastActivityDate="2015-08-25T12:05:49.460" Title="Restricting words in an answer to certain dictionaries" Tags="|word|puzzle-creation|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20539" PostTypeId="1" CreationDate="2015-08-25T11:38:46.817" Score="15" ViewCount="2080" Body="&lt;p&gt;&quot;I'm bored of riddles, I'm going to ask you a question about dice instead&quot;, says the Sphinx.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Now, I've got this set of &lt;a href=&quot;https://en.wikipedia.org/wiki/Nontransitive_dice#Efron.27s_dice&quot; rel=&quot;nofollow noreferrer&quot;&gt;Efron Dice&lt;/a&gt; here. There's four dice, and they have these values on their faces:&quot;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A: 4, 4, 4, 4, 0, 0&#xA;B: 3, 3, 3, 3, 3, 3&#xA;C: 6, 6, 2, 2, 2, 2&#xA;D: 5, 5, 5, 1, 1, 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rAy6e.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rAy6e.png&quot; alt=&quot;Efron&amp;#39;s Dice&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;So as I'm sure you can see, A beats B (with a 2/3 probability), B beats C, C beats D and D beats A, each with 2/3 probability.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;This works because I, the Sphinx, am a legendary magical animal, and so I can do weird things using the laws of probability.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;As a Sphinx, of course, this is very useful to me, because I can get my supplicant to choose a random die, and whichever one they pick I can always pick one that's better.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;But I'm tired of still losing 1/3 of the time! I want you to make me a set of dice where I'm even more likely to win!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;If you can then I'll let you pass, if not then (as usual) I'll eat you for dinner.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Die A must beat die B must beat die C must beat die D must beat die A, all with P(win) &gt; 1/2. (This is because, whichever die the supplicant chooses, the Sphinx wants dinnertime to remain more likely than not).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There must be four dice in each set. Each die has six faces.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Each face of each die can contain any number from the set of integers 0-6. Numbers may be used more than once on each die.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For each set of dice, set P is 'better than' set Q if the Sphinx is more likely to win her game with it. The game is that the supplicant picks a random die and then the Sphinx picks her own die which is likely to beat the supplicant's die. Set P is a 'better' set than Q if the Sphinx is more likely to win her game with set P than she would be with set Q. First the supplicant chooses any die at random from the set, then the Sphinx chooses the die which is most likely to beat the supplicant's die.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the Sphinx and the supplicant both roll the same number (a tie), then both parties roll again with the same dice.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For background info see Wikipedia: &lt;a href=&quot;https://en.wikipedia.org/wiki/Nontransitive_dice&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nontransitive Dice&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This is an open-ended challenge; the 'best' set of dice (the set with which the Sphinx is most likely to win her game) is the winning set.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Bonus credit if you can demonstrate that your set of dice is the best possible set.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Related (though not a duplicate): &lt;a href=&quot;https://puzzling.stackexchange.com/questions/4453/a-first-choice-die&quot;&gt;A first choice die&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="69582" LastEditDate="2021-07-21T18:52:13.003" LastActivityDate="2021-07-21T18:52:13.003" Title="The Sphinx's Dice" Tags="|mathematics|optimization|probability|dice|" AnswerCount="3" CommentCount="12" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20540" PostTypeId="2" ParentId="20538" CreationDate="2015-08-25T11:57:58.320" Score="6" Body="&lt;p&gt;A really good resource for dictionary txt files is &lt;a href=&quot;http://app.aspell.net/create&quot;&gt;SCOWL&lt;/a&gt;. You can vary spelling variants, uk/us spelling and dictionary size. It can serve you 3rd question as well. You just have to choose appropriate dictionary size. I think size 35 (small) should work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dictionary.reference.com/&quot;&gt;Dictionary.com&lt;/a&gt; is usually preferred because it is the most common dictionary site. (google dictionary, it is the first link) It has a good idea where to draw the line (between acceptable and non-acceptable words). A particular disadvantage is poor navigation ie. lack of alphabetical browsing. The only other frequently used dictionary I have seen is the scrabble dictionary. This, however, has problems because the inclusion of words often depend on their usability in an actual scrabble game.&lt;/p&gt;&#xA;" OwnerUserId="15544" LastEditorUserId="15544" LastEditDate="2015-08-25T12:05:49.460" LastActivityDate="2015-08-25T12:05:49.460" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20541" PostTypeId="2" ParentId="20539" CreationDate="2015-08-25T12:11:46.127" Score="6" Body="&lt;p&gt;The optimal set of dice is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the set you've already found: Efron's dice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is proved, along with a considerably more general statement ($n$-sided dice instead of just 6-sided ones), in the article referred to in &lt;a href=&quot;https://mathoverflow.net/questions/119867/what-is-the-most-extreme-set-4-or-5-nontransitive-n-sided-dice&quot;&gt;this&lt;/a&gt; post on MathOverflow.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:31.357" LastActivityDate="2015-08-25T12:11:46.127" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20542" PostTypeId="2" ParentId="20513" CreationDate="2015-08-25T13:32:06.363" Score="0" Body="&lt;p&gt;Maybe you are referring to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bacteria&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I'm quick, I'm easy&lt;/li&gt;&#xA;&lt;li&gt;You can make me in a jiffy&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bacteria reproduce quickly and easily&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;I'm on a farm&lt;/li&gt;&#xA;&lt;li&gt;I'm in the pasture  &lt;/li&gt;&#xA;&lt;li&gt;I'm in your kitchen&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yep.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;I lay, I rot, in more places then one&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bacteria is pretty much everywhere. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;I'm smelly, but I'm also very tasty&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Blue Cheese anyone?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;I have nicknames but I prefer one&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Germ is a nickname, but it is pretty recognizable. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11926" LastActivityDate="2015-08-25T13:32:06.363" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20543" PostTypeId="2" ParentId="20513" CreationDate="2015-08-25T13:43:03.380" Score="1" Body="&lt;p&gt;Are you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Cornbread&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm quick, I'm easy&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It really doesn't take much time to throw together cornbread&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can make me in a jiffy&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://www.jiffymix.com/&quot; rel=&quot;nofollow&quot;&gt;Jiffy mix&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm on a farm&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Corn comes from farms and cornbread is considered a farm staple&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm in your kitchen&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It's food, where else would you find it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm in the pasture&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Cornbread (or it's early descendants) was often taken by shepherds into a pasture as food. Also, this could be a play on the fields where corn is grown?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I lay, I rot, in more places then one&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Corn and cornbread can be found all over the world&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm smelly, but I'm also very tasty&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;No one says smells have to be bad :P&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have nicknames but I prefer one&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There are many names for cornbread: cornbread, johnny cakes, corncakes, hushpuppies, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12242" LastActivityDate="2015-08-25T13:43:03.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20544" PostTypeId="2" ParentId="20533" CreationDate="2015-08-25T13:56:12.503" Score="2" Body="&lt;p&gt;The more I thought about this, the more I realized that it isn't right. Maybe it will help someone else though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'w' can never follow an 's'. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'm' and 'w' are considered matching (they're just upside down!). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; '0' means there is no such letter, or adding no letters is sufficient. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What letter from the set &quot;b d f h m s w&quot; when added to the front makes the front and back letters match?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fsmss → s &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  fsmss → s (sfsmss)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;wsffs → 0&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  wsffs → 0 (since a w can not follow an s there is no letter)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fwmhdhdbfw → m&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  fwmhdhdbfw → m (mfwmhdhdbfw)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;bsmhb → 0 or b&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  bsmhb → 0 or b (Adding no letters is fine here, but if you add a b you still match.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;dbmhsm → w&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; dbmhsm → w (wdbmhsm)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That leaves the answers as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;fdhdbmhwmh → ?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; h (hfdhdbmhwmh)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;bmhsmhwfwdh → ?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; h (hbmhsmhwfwdh)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;smhdfwsfdbmh → ?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; h (hsmhdfwsfdbmh)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11926" LastEditorUserId="11926" LastEditDate="2015-08-25T14:07:01.317" LastActivityDate="2015-08-25T14:07:01.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20545" PostTypeId="2" ParentId="20515" CreationDate="2015-08-25T14:02:10.030" Score="11" Body="&lt;p&gt;I think the comment from Marek Oleszczuk got the keyword, here are my contributions for the other lines:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1/ You might think I'm broke, but then&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lady in Red by Chris de Burgh. If you are &quot;in the red&quot;, financially, you have a negative balance, and being broke is a similar state.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2a/ I'm a singer...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lady Madonna by the Beatles. Madonna is of course, among her other talents, a popular singer. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2b/ ...and sound like a hen&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lay Lady Lay by Bob Dylan (courtesy of @TroyAndAbed). Hens lay eggs. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3/ You can count on me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Three times a Lady by the Commodores. The line from the song containing the title is &quot;you're once, twice, three times a lady&quot;, hence the counting.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4/ I'm a tramp you see&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Lady is a Tramp by Frank Sinatra (taken from comments). This is self-explanatory, I think.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5/ So I'll sing you the blues once again.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lady sings the blues by Billie Holiday. Again this is self-explanatory.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The key word is of course &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lady. Each clue is told from the point of view of the particular Lady as designated in the song title, although sometimes interpreted differently (cryptically even) to the original.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7316" LastEditorUserId="7316" LastEditDate="2015-08-25T18:44:08.133" LastActivityDate="2015-08-25T18:44:08.133" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="20546" PostTypeId="2" ParentId="20513" CreationDate="2015-08-25T14:05:39.357" Score="0" Body="&lt;p&gt;Are you:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An Egg?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm quick, I'm easy&lt;br&gt;&#xA;You can make me in a jiffy  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Easy to prepare (sunny side anyone?)   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm on a farm  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yup  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm in your kitchen  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yup..  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm in the pasture  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yup...  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I lay, I rot, in more places then one&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some eggs are igonred after getting 'laid'  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm smelly, but I'm also very tasty  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Egg doesn't smell good, but it taste great!  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have nicknames but I prefer one  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  I dunno about this one...(EDIT: hm..cackleberries? roonies?)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; egg  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; OP does not like cakes? (baked with egg as an ingridient)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="10037" LastEditDate="2015-08-25T17:15:48.923" LastActivityDate="2015-08-25T17:15:48.923" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20547" PostTypeId="2" ParentId="20527" CreationDate="2015-08-25T14:36:42.373" Score="4" Body="&lt;p&gt;I made up the puzzle late last night and even though it seemed good back then, I feel it may be too misleading now. In order to save you the hassles, decided to post the solution as well.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All the clues are in the boy and the policeman's replies. There are board game references throughout, including:&lt;br&gt;&#xA; Ticket to Ride&lt;br&gt;&#xA; 10 Days in Europe&lt;br&gt;&#xA; Risk&lt;br&gt;&#xA; Guess Who&lt;br&gt;&#xA; Clue&lt;br&gt;&#xA; etc.&lt;br&gt;&#xA; The sequence references to the famous board game &quot;Sequence&quot; and the numbers are the ones appearing on the diagonal of the board.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This may be too cryptic for most of you, I guess will stick to the math puzzles mainly:)&lt;/p&gt;&#xA;" OwnerUserId="14140" LastEditorUserId="14013" LastEditDate="2015-08-25T14:42:44.057" LastActivityDate="2015-08-25T14:42:44.057" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20548" PostTypeId="2" ParentId="20513" CreationDate="2015-08-25T15:07:40.810" Score="7" Body="&lt;p&gt;I've been steaming over this for a bit now, and I can come up with nothing other than:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Cow pie; that being the pile of dung left by a cow, as well as a chocolately sweet (I'm familiar with the packaged candy from Baraboo, WI, but it also appears that there are &quot;no-bake&quot; cow pie/pile/patty recipes which popped after just searching those name variants).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm quick, I'm easy&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No bake recipes are often less time consuming and involved, since no cooking is needed.  It goes without saying that it's very easy for cows to make their pies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can make me in a jiffy&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, with no time in the oven, those patties would be done in the time it takes them to set.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm on a farm&lt;br&gt;&#xA;I'm in your kitchen&lt;br&gt;&#xA;I'm in the pasture&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cows are on farms and roam the pastures, so this is where they leave their pies (Definitely Wisconsin...).  And the kitchen is where you would make the confectionery variety.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I lay, I rot, in more places than one&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fields can be littered with pies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm smelly, but I'm also very tasty&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The cow's version is smelly, the chocolate is tasty.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have nicknames but I prefer one&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The origin is cow dung, but the cows' presentation of the dung likens it to a pie, a patty, or a pile (plus cow chips), which lead to us humans wanting chocolate in the same appetizing aesthetic.  Lots of nicknames, but I think cow pie is most common.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A cow pie/patty/pile/chip etc., of which the edible variety is a no bake treat, as the title hints.  (The cow variety could be baked by the sun and then burned by settlers, but that's neither here nor there)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11452" LastActivityDate="2015-08-25T15:07:40.810" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20549" PostTypeId="2" ParentId="20539" CreationDate="2015-08-25T15:17:44.717" Score="15" Body="&lt;p&gt;The post linked in rand al'thor answer optimized the worst case pick the visitor of the sphinx can make. But the question asked for the best average case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better solution in that case is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A: 1,1,5,5,5,5&#xA;B: 2,2,2,2,6,6&#xA;C: 3,3,3,3,3,3&#xA;D: 4,4,4,4,4,4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;B defeats A with a probability of $\frac{5}{9}$ &lt;br&gt;&#xA;C defeats B with a probability of $\frac{2}{3}$ &lt;br&gt;&#xA;D defeats C with a probability of $1$ &lt;br&gt;&#xA;A defeats D with a probability of $\frac{2}{3}$ &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On average that is a probability of $ \frac{1}{4} \cdot \frac{5}{9}+\frac{1}{4} \cdot \frac{2}{3}+\frac{1}{4} \cdot 1+\frac{1}{4} \cdot \frac{2}{3} = \frac{26}{36} &amp;gt; \frac{24}{36} = \frac{2}{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure if $\frac{26}{36}$ is optimal or not.&lt;/p&gt;&#xA;" OwnerUserId="10143" LastEditorUserId="11475" LastEditDate="2015-08-25T16:06:26.433" LastActivityDate="2015-08-25T16:06:26.433" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20551" PostTypeId="1" AcceptedAnswerId="20777" CreationDate="2015-08-25T16:06:52.727" Score="5" ViewCount="895" Body="&lt;p&gt;With all the popularity of four liners lately, I figured I'd contribute. Since my &lt;a href=&quot;https://puzzling.stackexchange.com/questions/20372/the-answer-would-make-a-good-title&quot;&gt;recent&lt;/a&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/questions/20225/what-do-these-things-have-in-common&quot;&gt;riddles&lt;/a&gt; seem to have been too hard, I'll err on the side of ease this time. Determine what each line is describing (each is a single word) and what the commonality is.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;I'm depending on you.&lt;/p&gt;&#xA;&lt;p&gt;I'm lost in your eyes; there's just something about them...&lt;/p&gt;&#xA;&lt;p&gt;I'm one of many, rarely referred to individually, but just look inside and you'll find me with the others.&lt;/p&gt;&#xA;&lt;p&gt;When the previous word's done with you, come see me!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;ADDENDUM: I've discovered that there is a 5th word that also shares the common bond of the previous 4. I'll add it here in the interest of making it perhaps easier on you:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A sailor's friend, I'm the concatenation of two words (slightly ironic because the first is not known to play well with the second).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;HINT:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;All 5 of the words are anagrams of each other. These words are 7 letters long each.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-08-31T20:37:39.940" Title="A Four Liner of my Own: I'm counting on you" Tags="|riddle|word|pattern|no-computers|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="20552" PostTypeId="2" ParentId="20533" CreationDate="2015-08-25T16:15:43.060" Score="6" Body="&lt;p&gt;The letters in the sequences are all from the (alphabetically ordered) set &lt;code&gt;b d f h m s w&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we pair these up differently we get:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; h/w: husband / wife&lt;br&gt;&#xA; f/m: father / mother&lt;br&gt;&#xA; b/s: brother / sister&lt;br&gt;&#xA; d/s: daughter / son&lt;br&gt;&#xA; - note that &quot;s&quot; seems to stand for both sister and son.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sequences he has already identified are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;fsmss -&amp;gt; s&lt;/code&gt; : my father's son's mother's son's sister is my sister&lt;br&gt;&#xA; &lt;code&gt;wsffs -&amp;gt; 0&lt;/code&gt; : my wife's son's father's father's son is me (represented by NULL)&lt;br&gt;&#xA; &lt;code&gt;fwmhdhdbfw -&amp;gt; m&lt;/code&gt; : my father's wife's mother's husband's daughter's husband's daughter's brother's father's wife is (drumroll please) my mother&lt;br&gt;&#xA; &lt;code&gt;bsmhb -&amp;gt; 0 or b&lt;/code&gt; : my brother's son's mother's husband's brother is either me or my brother&lt;br&gt;&#xA; &lt;code&gt;dbmhsm -&amp;gt; w&lt;/code&gt; : my daughter's brother's mother's husband's son's mother is my wife  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the new questions are:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; fdhdbmhwmh -&gt; &lt;strong&gt;f&lt;/strong&gt; : my father's daughter's husband's daughter's brother's mother's husband's wife's mother's husband is my father&lt;br&gt;&#xA; bmhsmhwfwdh -&gt; &lt;strong&gt;f&lt;/strong&gt; : my brother's mother's husband's son's [note: sister doesn't work here or I end up with my grandfather] mother's husband's wife's father's wife's daughter's husband is my father&lt;br&gt;&#xA; smhdfwsfdbmh -&gt; &lt;strong&gt;0&lt;/strong&gt; : my son's mother's husband's daughter's father's wife's son's father's daughter's brother's mother's husband is me.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2059" LastActivityDate="2015-08-25T16:15:43.060" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20553" PostTypeId="1" AcceptedAnswerId="20562" CreationDate="2015-08-25T16:59:20.203" Score="11" ViewCount="782" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Small mammal – I'm&lt;br&gt;&#xA;  A&lt;br&gt;&#xA;  International animal&lt;br&gt;&#xA;  With four legs, and,&lt;br&gt;&#xA;  Whiskers. Sharp eyes.&lt;br&gt;&#xA;  But, it wouldn't do&lt;br&gt;&#xA;  &lt;br&gt;&#xA;  To totally give up&lt;br&gt;&#xA;  A&lt;br&gt;&#xA;  Answer, too readily,&lt;br&gt;&#xA;  So those are all the&lt;br&gt;&#xA;  Clues that I am going&lt;br&gt;&#xA;  To give you; sorry!&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="4050" LastEditDate="2015-08-25T17:11:25.697" LastActivityDate="2015-08-25T21:07:25.353" Title="A small mammal?" Tags="|enigmatic-puzzle|poetry|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="20554" PostTypeId="1" AcceptedAnswerId="20557" CreationDate="2015-08-25T17:53:00.653" Score="2" ViewCount="434" Body="&lt;pre&gt;&lt;code&gt;{um} - [] ⼔erized&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The answer is a common phrase. (Though note that this doesn't mean it's a famous phrase or that it often stands on its own.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is too hard I will add some hints!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your browser shows a question mark in a box:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That's the unicode character U+2F14.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10851" LastEditorUserId="10851" LastEditDate="2015-08-25T18:09:36.053" LastActivityDate="2015-08-25T18:15:27.767" Title="A humble little symbol riddle" Tags="|word|cipher|wordplay|rebus|english|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20555" PostTypeId="2" ParentId="20553" CreationDate="2015-08-25T17:54:53.270" Score="3" Body="&lt;p&gt;I decided to post it anyway:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Lynx&lt;br&gt;&#xA; It's a small mammal (4-legged with whisker) found in northern europe, siberia, asia and north america. Felines are generally known to be the mammals with the best eyesight. The Canadian Lynx is the smallest of all lynx species. They are generally very hard to spot in the wild. But this has nothing to do with an enigmatic-puzzle, so it must be wrong! Sorry*&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastActivityDate="2015-08-25T17:54:53.270" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20556" PostTypeId="1" AcceptedAnswerId="20561" CreationDate="2015-08-25T18:12:16.573" Score="4" ViewCount="572" Body="&lt;p&gt;A man wakes up and goes along to get drunk. He talks foolishly, but works hard and gets paid, but passes the money on to a dependent.  Every evening he comes home.  As he grows old, he is lonely and dreams about different times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The man lives in Edinburgh, Scotland.  One day, he decides to walk due east, stopping at a city, then directly southwest, stopping at another city.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What city does he end up in?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clarification:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For walking, assume he walks straight-line distances ignoring features that might make it technically impassible, i.e. bodies of water, etc&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14383" LastEditorUserId="14383" LastEditDate="2015-08-25T18:35:14.127" LastActivityDate="2015-08-25T20:41:48.477" Title="Headlines proclaim: &quot;Read about this man's amazing life!&quot;" Tags="|lateral-thinking|enigmatic-puzzle|story|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20557" PostTypeId="2" ParentId="20554" CreationDate="2015-08-25T18:15:27.767" Score="5" Body="&lt;p&gt;So, I'm thinking the answer might be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; COMING EARLY, as in &quot;winter's coming early this year!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We are presented with {um} - [], which can be described as &quot;UM IN CURLY BRACKETS (minus) BRACKETS&quot;, or &quot;UM IN CURLY&quot;. The unicode character is the radical spoon, which makes the next direction &quot;Spoonerized&quot;, essentially switching the first consonant sounds of our words. Therefore, &quot;UM IN CURLY&quot; becomes either &quot;COMING EARLY&quot; or &quot;COME IN EARLY&quot;, and I've only ever heard one of those phrases...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-08-25T18:15:27.767" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20559" PostTypeId="1" AcceptedAnswerId="20755" CreationDate="2015-08-25T19:56:19.283" Score="7" ViewCount="403" Body="&lt;p&gt;&lt;em&gt;Below, you will find a list of clues! Each clue has an answer that is no more than two words and is related to each other answer in a noticeable (albeit perhaps not immediately obvious) way. Discover the answers to the clues and their intricate relationships!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Clues&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A bearlike whale&lt;/li&gt;&#xA;&lt;li&gt;Attire of an ancient, mysterious sorcerer&lt;/li&gt;&#xA;&lt;li&gt;&quot;Cool pointer!&quot;&lt;/li&gt;&#xA;&lt;li&gt;A blighted sort of metal&lt;/li&gt;&#xA;&lt;li&gt;One who expeditions through deep time&lt;/li&gt;&#xA;&lt;li&gt;Keeps using the same data&lt;/li&gt;&#xA;&lt;li&gt;Providing someone with assets, maybe&lt;/li&gt;&#xA;&lt;li&gt;Destroyer of subatomic particles&lt;/li&gt;&#xA;&lt;li&gt;Wrath of the mailman&lt;/li&gt;&#xA;&lt;li&gt;Once again, the mongrels choose to partake&lt;/li&gt;&#xA;&lt;li&gt;Doctors act like pigs&lt;/li&gt;&#xA;&lt;li&gt;The Lord of the Sound&lt;/li&gt;&#xA;&lt;li&gt;Weeps over grains&lt;/li&gt;&#xA;&lt;li&gt;One who takes good care of his money?&lt;/li&gt;&#xA;&lt;li&gt;The companions we have&lt;/li&gt;&#xA;&lt;li&gt;Deface a rotting body&lt;/li&gt;&#xA;&lt;li&gt;An indecent vegetable&lt;/li&gt;&#xA;&lt;li&gt;One who wets the bed&lt;/li&gt;&#xA;&lt;li&gt;Scampers all over&lt;/li&gt;&#xA;&lt;li&gt;An unpleasant idiot&lt;/li&gt;&#xA;&lt;li&gt;Intriguing symbols on an intriguing object&lt;/li&gt;&#xA;&lt;li&gt;The more voluptuous brother&lt;/li&gt;&#xA;&lt;li&gt;Pots of snails&lt;/li&gt;&#xA;&lt;li&gt;An old decision point&lt;/li&gt;&#xA;&lt;li&gt;Sounds like a spicy dish&lt;/li&gt;&#xA;&lt;li&gt;An area where certain clouds might form&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There have been &lt;a href=&quot;https://puzzling.stackexchange.com/questions/19947/literally-no-idea&quot;&gt;ones like this&lt;/a&gt; before, but this one is mine. It has its own, special letters and its own, special twist. Starting with a pattern similar to the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18768/if-dogs-could-talk&quot;&gt;ones like this&lt;/a&gt; is a good idea, but to solve the puzzle fully you'll have to figure out my own special gimmick.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-08-31T15:58:27.153" Title="Repeat the Method to Solve the Clues" Tags="|riddle|word|cryptic-clues|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20560" PostTypeId="1" AcceptedAnswerId="20563" CreationDate="2015-08-25T20:32:26.673" Score="13" ViewCount="1028" Body="&lt;p&gt;Background: There is a year round Burning Man &quot;radio station&quot; called BMIR -Burning Man Information Radio. (BMIR.org)  11 months out of the year, it is just streamed with no commercials.  Starting 2 weeks before the Burn, it is an actual radio station at Black Rock City and &quot;commercials&quot; are added, talking about random things like parties, camp locations, safety messages, etc.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well I have noticed that some of these commercials are different. A man says a rather cryptic sentence (or two, or three) in a monotone voice. At the end of the message, he repeats two words.  I have been listening to it over the last week since the commercials have started, and have found 14 distinct messages. Here they are: (things in parentheses are notes or other possible words)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The syrup is running all over the chair. You sit down in the chair and they syrup sticks you to the seat of the chair so you can't get up. Chair. Syrup.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Chickens are in all the compartments of the revolving door. As you go thru the revolving door the chickens fly up in your face. See the feathers flying in the door. Maybe you mash a chicken in the revolving door and it bleeds all over the revolving door. Revolving door. Chicken.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The sugar bowl is upset and the sugar is spilled all over the table and the flies are buzzing around it. Your family sits down to the table and has nothing to eat but sugar. Table. Sugar.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The doctor is munching on a big head of lettuce. The doctor holds the lettuce out to you and says let us eat more lettuce. Doctor. Lettuce.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You are pouring vinegar all over your automobile. In fact, you are washing your automobile in vinegar. You run out of gas, so you fill up the tank of the automobile with vinegar. A bottle of vinegar breaks and runs all over the seat of the automobile. Automobile. Vinegar.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There is a large urn of coffee in the center of the elevator with steam rising from it. Everyone riding up in the elevator is drinking coffee. Elevator. Coffee.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You walk down the street with potato salad. But men with guns rob you of your potato salad. You want to get your potato salad back but men with guns prevent you. Guns. Potato salad.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A bunch of bananas are hanging from the bars on the general delivery window. Possibly(Generally?) you sit in(on?) the general delivery window eating a banana. General delivery window. Bananas.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A bottle of milk is broken in the mail box and is running out all down the sides of the mailbox. Mailbox. Milk.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The butter has melted and is running all over the alarm clock. You attempt to pick up the alarm clock but it is so greasy with the melted butter that is slips out of your hand into the tub of butter. You are oiling the works of the alarm clock with butter. Alarm clock. butter.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Red juice tomatoes are splattered and mashed all over the policeman. You realize a lifelong ambition and throw a tomato at the policeman and hitting him square on the nose. Policeman. Tomato (order may be wrong)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You want to get into bed but the bed is overflowing with salad dressing. you are mixing a salad on the bed and pouring salad dressing on it. You are dressing when you see the bed covered with salad dressing. Bed. Salad Dressing.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Strawberries ripe red juicy strawberries are mashed all over the newspaper. You probably mash the strawberries when you pick up the newspaper. Your newspaper is buried in straw. Newspaper. Strawberry.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The floor is covered in onions and people are sitting on the floor peeling onions and crying their eyes out. Your eyes begin to water from the onions as you step onto the floor. Floor. Onions.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Messages do not repeat at the same time every day. There doesn't seem to be an order, and honestly, it is hard to tell since I may have missed some messages as I get up from my desk for whatever reason.  Originally I thought they were once per hour, but then 3 happened within 15 minutes of each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will update this if I find any more messages or if I can lock down an order. I can also help out with providing context to any possible solutions, if this is even anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Pajamas&quot;&lt;/p&gt;&#xA;" OwnerUserId="15969" LastEditorUserId="11876" LastEditDate="2016-11-03T22:01:19.310" LastActivityDate="2016-11-03T22:01:19.310" Title="I think there is a hidden message on Burning Man Information Radio" Tags="|cipher|pattern|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20561" PostTypeId="2" ParentId="20556" CreationDate="2015-08-25T20:34:26.940" Score="6" Body="&lt;p&gt;I'd say it's based on:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the song by the Proclaimers, who walked 500 miles and then walked 500 more. (Havering = talking foolishly, headline &lt;em&gt;proclaims&lt;/em&gt;, drunk, work hard, give away money etc.)&lt;br&gt;&#xA;&lt;br&gt;&#xA;Going east 500 miles and then south west 500 miles lands you in London, I'd say...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2015-08-25T20:41:48.477" LastActivityDate="2015-08-25T20:41:48.477" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20562" PostTypeId="2" ParentId="20553" CreationDate="2015-08-25T21:07:25.353" Score="18" Body="&lt;p&gt;Impressive construction!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Count the number of characters, including punctuation and spaces, in each line.&lt;br&gt;&#xA;&lt;br&gt;&#xA; 18 1 20 20 21 19&lt;br&gt;&#xA; 18 1 20 20 21 19&lt;br&gt;&#xA;&lt;br&gt;&#xA; Use the simple alphanumeric cipher 1=A, 2=B, 3=C, etc.&lt;br&gt;&#xA;&lt;br&gt;&#xA; RATTUS&lt;br&gt;&#xA; RATTUS&lt;br&gt;&#xA;&lt;br&gt;&#xA; You're &lt;em&gt;Rattus rattus&lt;/em&gt; or the &lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Black_rat&quot;&gt;black rat&lt;/a&gt;&lt;/strong&gt;!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15921" LastActivityDate="2015-08-25T21:07:25.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20563" PostTypeId="2" ParentId="20560" CreationDate="2015-08-25T21:09:55.503" Score="11" Body="&lt;p&gt;Unfortunately it's not really a puzzle...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The list is by Robert H. Nutt, and appears on &lt;a href=&quot;http://www.mcsweeneys.net/articles/how-to-remember-names-and-faces-part-3-a-shopping-list-a-woman-cant-forget&quot;&gt;McSweeney's Internet Tendency&lt;/a&gt;. It's proposed as a technique for remembering a shopping list. The full book &lt;em&gt;How to Develop a Good Memory For Names, Faces and Facts&lt;/em&gt; by the same author is available as a PDF &lt;a href=&quot;https://s3.amazonaws.com/MemoryDelight/software/HowToImproveYourMemory.pdf&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10851" LastActivityDate="2015-08-25T21:09:55.503" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20564" PostTypeId="1" AcceptedAnswerId="20574" CreationDate="2015-08-25T21:51:02.887" Score="5" ViewCount="502" Body="&lt;p&gt;&lt;em&gt;I was there when you were young, and helped you make your art&lt;br&gt;&#xA;I seemed much more vibrant then, but the drink took me apart&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You were older but your hand gripped tight, showing me your fears&lt;br&gt;&#xA;As we tried to answer questions, in front of all your peers&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In older age you may fall ill, depending on your diet&lt;br&gt;&#xA;Do not forget my remedy, though few still want to try it&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Fair is how I judged you, and though you walked the line&lt;br&gt;&#xA;You never once crossed over it, that margin very fine&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Whether it is age or force that causes you to die&lt;br&gt;&#xA;I will mark and remember the place in which you lie&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been around for some time, but am I recognizable to you anymore?&lt;br&gt;&#xA;The answer is found in each stanza, but not all stanzas need be solved to provide it.&lt;/p&gt;&#xA;" OwnerUserId="11452" LastActivityDate="2015-08-26T01:13:07.740" Title="Am I Gone for Good?" Tags="|riddle|rhyme|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20565" PostTypeId="2" ParentId="20564" CreationDate="2015-08-25T21:52:10.930" Score="3" Body="&lt;p&gt;Is the answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a PEN, or more generally WRITING?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I was there when you were young, and helped you make your art&lt;br&gt;&#xA;I seemed much more vibrant then, but the drink took me apart&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pens and pencils help children draw their art, often in vibrant colours. &quot;Drink&quot; refers to ink.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You were older but your hand gripped tight, showing me your fears&lt;br&gt;&#xA;As we tried to answer questions, in front of all your peers&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your hand grips your pen tightly when you're nervous in an exam.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In older age you may fall ill, depending on your diet&lt;br&gt;&#xA;Do not forget my remedy, though few still want to try it&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure about this bit. Maybe some quack remedy involving drinking ink?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Fair is how I judged you, and though you walked the line&lt;br&gt;&#xA;You never once crossed over it, that margin very fine&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You often write on horizontal lines and between vertical margins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Whether it is age or force that causes you to die&lt;br&gt;&#xA;I will mark and remember the place in which you lie&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Writing on your gravestone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-08-25T21:57:33.527" LastActivityDate="2015-08-25T21:57:33.527" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20566" PostTypeId="1" AcceptedAnswerId="20575" CreationDate="2015-08-25T21:57:24.463" Score="10" ViewCount="1512" Body="&lt;p&gt;We're going to take the 5 platonic solids (&lt;strong&gt;tetrahedron&lt;/strong&gt;, &lt;strong&gt;cube&lt;/strong&gt;, &lt;strong&gt;octahedron&lt;/strong&gt;, &lt;strong&gt;dodecahedron&lt;/strong&gt;, and &lt;strong&gt;icosahedron&lt;/strong&gt;) and suspend them in various ways (we'll assume that they are solid and of uniform density).  Then we'll do a horizontal cut through the centre of gravity and describe the shape of the resulting cut face.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The suspension methods will be:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;By a &lt;strong&gt;vertex&lt;/strong&gt; (so any one of the vertices)&lt;/li&gt;&#xA;&lt;li&gt;By a &lt;strong&gt;face&lt;/strong&gt; (meaning by the middle of one of the faces)&lt;/li&gt;&#xA;&lt;li&gt;By an &lt;strong&gt;edge&lt;/strong&gt; (meaning the middle of one of the edges)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(This is a generalisation of a problem that I remember describing to a group, and only convincing them that the answer I gave was right by cutting up a potato.  It was the &lt;strong&gt;cube&lt;/strong&gt;, &lt;strong&gt;vertex&lt;/strong&gt; case)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've given some of the simpler answers in brackets, as examples.  Bonus points for doing it all in your head...&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Tetrahedron&lt;/strong&gt; suspended by a &lt;strong&gt;vertex&lt;/strong&gt;. (&lt;em&gt;equilateral triangle&lt;/em&gt;)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Tetrahedron&lt;/strong&gt; suspended by a &lt;strong&gt;face&lt;/strong&gt;. (&lt;em&gt;equilateral triangle&lt;/em&gt;)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Tetrahedron&lt;/strong&gt; suspended by an &lt;strong&gt;edge&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Cube&lt;/strong&gt; suspended by a &lt;strong&gt;vertex&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Cube&lt;/strong&gt; suspended by a &lt;strong&gt;face&lt;/strong&gt;. (&lt;em&gt;square&lt;/em&gt;)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Cube&lt;/strong&gt; suspended by an &lt;strong&gt;edge&lt;/strong&gt;. (&lt;em&gt;rectangle&lt;/em&gt;)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Octahedron&lt;/strong&gt; suspended by a &lt;strong&gt;vertex&lt;/strong&gt;. (&lt;em&gt;square&lt;/em&gt;)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Octahedron&lt;/strong&gt; suspended by a &lt;strong&gt;face&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Octahedron&lt;/strong&gt; suspended by an &lt;strong&gt;edge&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Dodecahedron&lt;/strong&gt; suspended by a &lt;strong&gt;vertex&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Dodecahedron&lt;/strong&gt; suspended by a &lt;strong&gt;face&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Dodecahedron&lt;/strong&gt; suspended by an &lt;strong&gt;edge&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Icosahedron&lt;/strong&gt; suspended by a &lt;strong&gt;vertex&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Icosahedron&lt;/strong&gt; suspended by a &lt;strong&gt;face&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Icosahedron&lt;/strong&gt; suspended by an &lt;strong&gt;edge&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There's a couple in there that get a little tricky!!  Enjoy!&lt;/p&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="11876" LastEditDate="2016-03-23T16:15:53.683" LastActivityDate="2016-03-23T16:15:53.683" Title="Various cross-sections of platonic solids" Tags="|geometry|visual|knowledge|three-dimensional|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20567" PostTypeId="1" CreationDate="2015-08-25T22:13:33.417" Score="12" ViewCount="13202" Body="&lt;p&gt;I have to make a 3x3 magic square using the numbers 0-10 without 1 and 9. I have tried various things but am not good at this. The sums of each row, column, and diagonal have to be equal; I added all the numbers up and divided by three so I believe each row should add up to 15?&lt;/p&gt;&#xA;" OwnerUserId="15971" LastEditorUserId="5373" LastEditDate="2015-08-25T22:35:15.213" LastActivityDate="2018-10-06T17:23:34.387" Title="Magic square using numbers 0,2,3,4,5,6,7,8,10" Tags="|mathematics|magic-square|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20568" PostTypeId="2" ParentId="20567" CreationDate="2015-08-25T22:52:25.587" Score="31" Body="&lt;pre&gt; 2 10  3&#xA; 6  5  4&#xA; 7  0  8&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is the only one, not counting reflections and rotations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: As you found, the sum of all the numbers is three times the sum of each row (or column), so each row must add up to 15. There are only two ways to make 15 with 0 and two numbers: 0+7+8 and 0+5+10. Similarly, 10 can only be used in 2+3+10 and 0+5+10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each corner is in one row, one column, and one diagonal, so it must have three ways to make 15. The center must have four ways to make 15. Therefore 0 and 10 must be in the middle of the sides, opposite each other, with 5 in the center. Then, if we fill in 2 and 3 next to the 10, all the rest of the numbers are forced.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-08-25T22:57:44.720" LastActivityDate="2015-08-25T22:57:44.720" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="20569" PostTypeId="1" AcceptedAnswerId="20570" CreationDate="2015-08-25T23:01:05.533" Score="-4" ViewCount="345" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Many forms I take&lt;br&gt;&#xA;  Unification I aim to make&lt;br&gt;&#xA;  I can make you act quite dumb&lt;br&gt;&#xA;  Cause your tongue to go numb  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Appearing in every nation&lt;br&gt;&#xA;  I am the cause of relation&lt;br&gt;&#xA;  I can control&lt;br&gt;&#xA;  Even animals  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Father and son&lt;br&gt;&#xA;  I think I've won&lt;br&gt;&#xA;  Mother and child&lt;br&gt;&#xA;  Even in the wild  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;You'll never be alone&lt;br&gt;&#xA;  I'm accessible over phone&lt;br&gt;&#xA;  I am quite widespread and powerful&lt;br&gt;&#xA;  But if you're not careful  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I can ruin your life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Who or what is this?!?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="15913" LastEditDate="2015-08-26T00:30:07.277" LastActivityDate="2015-08-27T01:46:57.590" Title="Many forms I take" Tags="|riddle|english|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="20570" PostTypeId="2" ParentId="20569" CreationDate="2015-08-26T00:08:24.730" Score="2" Body="&lt;p&gt;The answer is -&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Love?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because-&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Love can be in many forms.&lt;br&gt;&#xA;  Love can definitely make people act quite dumb.&lt;br&gt;&#xA;  Love is in every nation and drives animals to reproduce.&lt;br&gt;&#xA;  Love is between father and son and mother and child.&lt;br&gt;&#xA;  Love is widespread and can be powerful and can also ruin your life ex: loving the wrong people. &#xA;  However I don't know how accessible over the phone it is :/&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15888" LastEditorUserId="15544" LastEditDate="2015-08-27T01:46:57.590" LastActivityDate="2015-08-27T01:46:57.590" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20571" PostTypeId="2" ParentId="20566" CreationDate="2015-08-26T00:38:01.877" Score="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;Equilateral triangle&lt;/li&gt;&#xA;&lt;li&gt;Equilateral triangle&lt;/li&gt;&#xA;&lt;li&gt;Square&lt;/li&gt;&#xA;&lt;li&gt;Regular hexagon&lt;/li&gt;&#xA;&lt;li&gt;Square&lt;/li&gt;&#xA;&lt;li&gt;Rectangle&lt;/li&gt;&#xA;&lt;li&gt;Square&lt;/li&gt;&#xA;&lt;li&gt;Regular hexagon&lt;/li&gt;&#xA;&lt;li&gt;Parallelogram&lt;/li&gt;&#xA;&lt;li&gt;Regular hexagon&lt;/li&gt;&#xA;&lt;li&gt;Regular decagon&lt;/li&gt;&#xA;&lt;li&gt;Regular hexagon&lt;/li&gt;&#xA;&lt;li&gt;Regular decagon&lt;/li&gt;&#xA;&lt;li&gt;Regular dodecagon&lt;/li&gt;&#xA;&lt;li&gt;Oblique hexagon&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="14013" LastEditorUserId="14013" LastEditDate="2015-08-26T19:22:49.877" LastActivityDate="2015-08-26T19:22:49.877" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20572" PostTypeId="2" ParentId="20567" CreationDate="2015-08-26T00:47:33.307" Score="22" Body="&lt;p&gt;In general, to know whether this sort of problem is solvable (given a list of 9 numbers, form a magic square out of them), you need to look for three arithmetic sequences that are themselves arithmetically sequenced &amp;mdash; namely, three sequences of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{matrix}&#xA;a &amp;amp; a+c &amp;amp; a+2c \\&#xA;a+b &amp;amp; a+b+c &amp;amp; a+b+2c \\&#xA;a+2b &amp;amp; a+2b+c &amp;amp; a+2b+2c&#xA;\end{matrix}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, $a = 0$, $b = 2$, and $c = 3$, because the sequences are $(0, 2, 4)$, $(3, 5, 7)$, and $(6, 8, 10)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, you arrange them like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{matrix}&#xA;a+b &amp;amp; a+2b+2c &amp;amp; a+c \\&#xA;a+2c &amp;amp; a+b+c &amp;amp; a+2b \\&#xA;a+2b+c &amp;amp; a &amp;amp; a+b+2c&#xA;\end{matrix}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to get your magic square. Note that every row, column, and diagonal sums up to $3a + 3b + 3c$. In your problem, this becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{matrix}&#xA;2 &amp;amp; 10 &amp;amp; 3 \\&#xA;6 &amp;amp; 5 &amp;amp; 4 \\&#xA;7 &amp;amp; 0 &amp;amp; 8 \\&#xA;\end{matrix}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can prove that this solution is unique. There are exactly eight combinations of the nine terms above that add up to $3a+3b+3c$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{matrix}&#xA;a &amp;amp; (a+b+c) &amp;amp; (a+2b+2c) &amp;amp; (1) \\&#xA;a &amp;amp; (a+b+2c) &amp;amp; (a+2b+c) &amp;amp; (2) \\&#xA;(a+b) &amp;amp; (a+c) &amp;amp; (a+2b+2c) &amp;amp; (3) \\&#xA;(a+b) &amp;amp; (a+2c) &amp;amp; (a+2b+c) &amp;amp; (4) \\&#xA;(a+b) &amp;amp; (a+b+c) &amp;amp; (a+b+2c) &amp;amp; (5) \\&#xA;(a+2b) &amp;amp; (a+c) &amp;amp; (a+b+2c) &amp;amp; (6) \\&#xA;(a+2b) &amp;amp; (a+2c) &amp;amp; (a+b+c) &amp;amp; (7) \\&#xA;(a+c) &amp;amp; (a+b+c) &amp;amp; (a+b+2c) &amp;amp; (8) \\&#xA;\end{matrix}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Out of these, the four terms that appear in the corner of the square appear exactly three times, the four terms that appear in the edges of the square appear twice, and the term that appears in the center $(a+b+c)$ appears four times. Therefore, if those terms are to appear in a magic square, they cannot appear anywhere but those places.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, suppose you place $a+b+c$ in the center, the only place it can go. Then you choose an edge square to put $a$ in. Whichever square you choose to put it in will rotate the entire magic square, but let's choose the bottom one for now:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{matrix}&#xA;? &amp;amp; ? &amp;amp; ? \\&#xA;? &amp;amp; a+b+c &amp;amp; ? \\&#xA;? &amp;amp; a &amp;amp; ?&#xA;\end{matrix}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the entry in the top square must be $a + 2b + 2c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{matrix}&#xA;? &amp;amp; a+2b+2c &amp;amp; ? \\&#xA;? &amp;amp; a+b+c &amp;amp; ? \\&#xA;? &amp;amp; a &amp;amp; ?&#xA;\end{matrix}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in order for the top row to sum up to $3a+3b+3c$, you still require one $b$ and one $c$. Therefore, only $a+b$ and $a+c$ can go in those squares, because you've already used $a$ and $a+b+c$, and the rest will make one of the coefficients go past 3. Whichever order you choose $a+b$ and $a+c$ to go in will reflect the magic square. After that, the rest of the magic square can be solved by summing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, we've proven that only rotations and reflections of the $3 \times 3$ magic square are also magic squares.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-08-26T01:10:12.547" LastActivityDate="2015-08-26T01:10:12.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20573" PostTypeId="2" ParentId="20372" CreationDate="2015-08-26T01:01:57.373" Score="7" Body="&lt;p&gt;I think the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; hydrogen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You'll find me the whole world around&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hydrogen is the most plentiful element, making up over 90% of the atoms in the universe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In nature 'tis true I abound&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As said above, hydrogen is universally abundant and found in nature.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In various forms I am found.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are 3 hydrogen isotopes: protium (&lt;sup&gt;1&lt;/sup&gt;H), deuterium (&lt;sup&gt;2&lt;/sup&gt;H), and tritium (&lt;sup&gt;3&lt;/sup&gt;H).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a king of my kind&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hydrogen has the atomic number 1, and it's the head, or &quot;king&quot;, of its &lt;a href=&quot;https://en.wikipedia.org/wiki/Period_(periodic_table)&quot; rel=&quot;nofollow&quot;&gt;period&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Group_(periodic_table)&quot; rel=&quot;nofollow&quot;&gt;group&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And yet I have been quite maligned&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hydrogen has been bad-mouthed for being extremely volatile and explosive. (i.e. the infamous &lt;a href=&quot;https://en.wikipedia.org/wiki/Hindenburg_disaster&quot; rel=&quot;nofollow&quot;&gt;Hindenburg disaster&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Confused for another&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Confused for helium because they of their chemical symbols (He and H).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Get this: I'm her brother&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hydrogen and helium are metaphorically siblings, being that they were &quot;born&quot; at the same time (the Big Bang).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though our differences are well defined.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Helium is atomically stable, a Noble Gas, and twice as heavy as hydrogen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Her flowery form I resist&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hydrogen resists helium's flowery/elaborate/ornate form, and vice versa, because helium has a full &lt;a href=&quot;https://en.wikipedia.org/wiki/Electron_shell&quot; rel=&quot;nofollow&quot;&gt;valence shell of electrons&lt;/a&gt; and does not bond readily.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm more aggressive&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, hydrogen is incredibly volatile and explosive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;She seems more passive&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Helium is not reactive, because of its full valence shell.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet one step from outburst she is.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the Sun, and general stars, helium is produced by the fusion of hydrogen. The outburst referenced is the death of a star: a supernova (or &lt;a href=&quot;https://en.wikipedia.org/wiki/Nova#Helium_novae&quot; rel=&quot;nofollow&quot;&gt;nova&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;She's one of a number, all true&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Helium is monatomic, meaning it is naturally found as a singular atom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not because I come from two&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hydrogen is diatomic, meaning it's naturally found bonded to itself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;She gets her own line&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Referring to the Noble Gas period. (a column on the periodic table, which helium is a member of)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll never have mine&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hydrogen is off by itself in the upper left of the periodic table.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See me or listen for flyers&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You would see a hydrogen bomb explosion, and could hear a blimp or dirigible fly over.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Read me or climb up much higher&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The higher you go, the higher the concentration of hydrogen in the atmosphere will be.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer could also be helium, carbon, or nitrogen (which fit worse/better depending on the line). All of which I have another set of explanations for. I feel like hydrogen fits pretty well, though.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Oh, the benefits of high school chemistry.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="9871" LastEditorUserId="9871" LastEditDate="2015-08-26T18:27:08.177" LastActivityDate="2015-08-26T18:27:08.177" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20574" PostTypeId="2" ParentId="20564" CreationDate="2015-08-26T01:13:07.740" Score="8" Body="&lt;p&gt;You are oft forgotten and supplanted by many newer things, but not quite gone yet...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chalk&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I was there when you were young, and helped you make your art&lt;/em&gt;&#xA;&lt;br&gt;&lt;em&gt;I seemed much more vibrant then, but the drink took me apart&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chalk coloring on sidewalks, etc...washes away easily with water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You were older but your hand gripped tight, showing me your fears&lt;/em&gt;&#xA;&lt;br&gt;&lt;em&gt;As we tried to answer questions, in front of all your peers&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Getting called up to the blackboard in school to write down answers with chalk.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In older age you may fall ill, depending on your diet&lt;/em&gt;&#xA;&lt;br&gt;&lt;em&gt;Do not forget my remedy, though few still want to try it&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People would frequently consume chalk or chalk-like remedies for stomach ailments.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Fair is how I judged you, and though you walked the line&lt;/em&gt;&#xA;&lt;br&gt;&lt;em&gt;You never once crossed over it, that margin very fine&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chalk is used to mark fair/foul in baseball (or other sports). (This one could be sketchy...)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Whether it is age or force that causes you to die&lt;/em&gt;&#xA;&lt;br&gt;&lt;em&gt;I will mark and remember the place in which you lie&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chalk outlines are frequent markers of location/position of the deceased upon discovery by authorities, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14405" LastActivityDate="2015-08-26T01:13:07.740" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20575" PostTypeId="2" ParentId="20566" CreationDate="2015-08-26T01:42:50.230" Score="3" Body="&lt;ul&gt;&#xA;&lt;li&gt;3: square (cuts across all four faces symmetrically)&lt;/li&gt;&#xA;&lt;li&gt;4: regular hexagon (cuts across all six faces symmetrically)&lt;/li&gt;&#xA;&lt;li&gt;8: regular hexagon (cuts across six faces symmetrically)&lt;/li&gt;&#xA;&lt;li&gt;9: rhombus (passes through two vertices and across two edges, cutting four faces in half)&lt;/li&gt;&#xA;&lt;li&gt;10: regular hexagon (cuts across six faces symmetrically)&lt;/li&gt;&#xA;&lt;li&gt;11: regular decagon (cuts across ten faces symmetrically)&lt;/li&gt;&#xA;&lt;li&gt;12: irregular hexagon (passes along two edges and across two edges, cutting four faces in half)&lt;/li&gt;&#xA;&lt;li&gt;13: regular decagon (cuts across ten faces symmetrically)&lt;/li&gt;&#xA;&lt;li&gt;14: regular hexagon (cuts across six faces symmetrically)&lt;/li&gt;&#xA;&lt;li&gt;15: irregular hexagon (passes along two edges and across two edges, cutting four faces in half)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The cross-sections for each polyhedron suspended from a vertex are the same as for its dual suspended from a face. The edge cross-sections are less regular because the orientation of the edge breaks rotational symmetries.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-08-26T01:42:50.230" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20576" PostTypeId="1" AcceptedAnswerId="20578" CreationDate="2015-08-26T06:33:52.557" Score="6" ViewCount="3990" Body="&lt;p&gt;This variation of the classical &lt;a href=&quot;http://www.mathsisfun.com/puzzles/farmer-crosses-river-solution.html&quot;&gt;Farmer Goat puzzle&lt;/a&gt; was in a puzzle contest at my college two years ago. &lt;br&gt;&#xA;Here it goes. &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Mr. Bruce Wayne&lt;/strong&gt; is very fond of parties. Well, he’s more fond of girls, but, let’s just say that he’s fond of parties. &lt;br&gt; So, he plans a party at an island off the coast of Gotham Bay. All his guests have already reached. And we all know, Mr. Wayne likes to make an entrance. So, he decides to go in the chopper along with two supermodels. But unfortunately for Mr. Wayne, the chopper can hold only two people. &lt;br&gt;&lt;br&gt;&#xA;  To add insult to the injury, the Mayor arrives with his two assistants to get a ride in the chopper. And there’s Mr. Fox, but with a secret case. Never mind, let’s just go to the party. &#xA;  But it seems someone doesn't want Mr. Wayne to enjoy tonight because there is no pilot to fly the chopper. And only Mr. Wayne, The Mayor and Mr. Fox know how to fly the chopper. &lt;br&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Oh no, wait! That’s not it. Mayor’s assistants won't stay with Mr. Wayne without their Mayor (&lt;em&gt;I don’t know Bruce Wayne freaks them out&lt;/em&gt;); the supermodels won’t stay with the Mayor without Bruce (they think the Mayor is a pervert but Bruce isn't!) and Mr. Fox would never leave the case.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let’s revise the situation. &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The passenger list is:&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Bruce Wayne&lt;/li&gt;&#xA;  &lt;li&gt;Bruce's 2 supermodels&lt;/li&gt;&#xA;  &lt;li&gt;Mayor&lt;/li&gt;&#xA;  &lt;li&gt;His 2 assistants&lt;/li&gt;&#xA;  &lt;li&gt;Mr. Fox&lt;/li&gt;&#xA;  &lt;li&gt;The secret case&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt; And the constraints are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;The capacity of the chopper is two persons. &lt;br&gt;&lt;/li&gt;&#xA;  &lt;li&gt;The secret case counts as a person (what, it's important!). &lt;/li&gt;&#xA;  &lt;li&gt;Only Bruce, Mayor or Mr. Fox can fly the chopper.&lt;/li&gt;&#xA;  &lt;li&gt;The Supermodels won’t stay with the Mayor without Bruce; Assistants&#xA;  won’t stay with Bruce without their Mayor and Mr. Fox won’t leave the&#xA;  case alone.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So much for making an entrance, Mr. Wayne!&#xA;All things said and done, we still have a party to attend to. Now that you’ve understood the situation, help Mr. Wayne and his friends (and the case, too) reach the party. &#xA;&lt;br&gt; &lt;br&gt;&#xA;I think I was able to solve this one, can you solve this puzzle?&lt;/p&gt;&#xA;" OwnerUserId="1766" LastEditorUserId="15544" LastEditDate="2015-08-26T08:27:47.537" LastActivityDate="2015-08-26T20:25:15.573" Title="Can you solve this Bruce Wayne logic puzzle?" Tags="|logical-deduction|river-crossing|" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20577" PostTypeId="2" ParentId="20576" CreationDate="2015-08-26T07:14:29.133" Score="3" Body="&lt;p&gt;I think this conundrum cannot be solved. Given the constraints, only Batman (B), Mayor (M) and Mr. Fox (F) can pilot the chopper. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If F goes first, he will travel with his case. And because he won't leave his case on the other side, so he will also travel back with his case. This makes F a useless candidate to undertake the first journey. &lt;/li&gt;&#xA;&lt;li&gt;If B pilots the first trip, he can only take one of his supermodels or M along with him (F doesn't travel without the case, so he cannot be a co-passenger). If he takes one supermodel, the other will freak out about staying with M on the same side. Taking M along seems like the only alternative. If both M and B go, then one of them has to return with the chopper, and either the supermodels or assistants will freak out, depending on who returns with the chopper. &lt;/li&gt;&#xA;&lt;li&gt;The same scenario as above plays out if M pilots the first trip.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So in the light of the current constraints, I'd say that no one can go to the party.&lt;/p&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-08-26T07:14:29.133" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20578" PostTypeId="2" ParentId="20576" CreationDate="2015-08-26T07:34:32.260" Score="8" Body="&lt;p&gt;Let me give it a try:&#xA;Let's call the models B1 and B2 since they belong to B and the assistants M1 and M2 and the case F1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B and M&lt;br&gt;&#xA;B back&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B and B1&lt;br&gt;&#xA;B and M back&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B and B2&lt;br&gt;&#xA;B back&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M and B&lt;br&gt;&#xA;M back&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F and F1&lt;br&gt;&#xA;B back &lt;/p&gt;&#xA;&#xA;&lt;p&gt;B and M&lt;br&gt;&#xA;M back&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M and M1&lt;br&gt;&#xA;B and M back&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M and M2&lt;br&gt;&#xA;M back&lt;/p&gt;&#xA;&#xA;&lt;p&gt;finally M and B&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-08-26T07:34:32.260" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="20580" PostTypeId="2" ParentId="20576" CreationDate="2015-08-26T10:51:10.817" Score="5" Body="&lt;p&gt;With Bruce Wayne B, Supermodels (S1 and S2), Mayor M, Assistants (A1 and A2), Mr Fox F and the Secret Case C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; From the comments, Bruce can be with the Case while Mr Fox flys the chopper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notation: (pilot) and passenger&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(F) and S1&lt;br&gt;&#xA;(F) back&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(F) and S2&lt;br&gt;&#xA;(F) back&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(M) and B (Bruce Wayne gets flown in by the Mayor to his waiting Supermodels).&lt;br&gt;&#xA;(M) back&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(F) and C&lt;br&gt;&#xA;(F) back&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(F) and M&lt;br&gt;&#xA;(F) back&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(F) and A1&lt;br&gt;&#xA;(F) back&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, (F) and A2  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that should work.&lt;/p&gt;&#xA;" OwnerUserId="13142" LastActivityDate="2015-08-26T10:51:10.817" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20581" PostTypeId="1" AcceptedAnswerId="20650" CreationDate="2015-08-26T11:03:25.240" Score="6" ViewCount="4147" Body="&lt;p&gt;Here is an open-ended challenge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the longest series of English words such that the final four letters of word $n$ are the same as the first four letters of word $n+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3 words) quintu&lt;strong&gt;plet&lt;/strong&gt; plet&lt;strong&gt;hora&lt;/strong&gt; horary  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At that point I stopped because I could find no dictionary words beginning with 'rary'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rules&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Only words that are defined in &lt;a href=&quot;http://dictionary.reference.com/&quot; rel=&quot;nofollow&quot;&gt;dictionary.com&lt;/a&gt; are allowed. Plurals and verb conjugation are allowed. If a word is doubtful then the community will decide by their votes. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;No repeats. No word must appear more than once in a list.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Four letter words are allowed. But if you use one then it must be the last in the list. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Before posting, check the longest answer so far. If you cannot do better then please don't post. This prevents many equal answers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If by chance two or more people post equal solutions at the same time then the one that is earliest on the SE system clock is counted and the others are not. As a courtesy, please delete your answer if you were beaten time-wise. If it is a dead-heat by the system clock then the answers will be considered equal.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;5a. If by examining someone else's answer you are able to improve it by inserting one or more words then this is allowed and you may give it as a separate answer. Note that the consensus from reading other peoples' comments is that this is disapproved of. If you do it you may get down-voted unless you can justify it to the community.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;Please give your answer in the following format so that others can easily see what they have to beat.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answer format&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3 words) quintuplet plethora horary&lt;/p&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="14177" LastEditDate="2015-08-26T13:43:19.703" LastActivityDate="2021-07-12T21:26:48.797" Title="Find the longest string of words such that the (4 letter) end of one word is the start of the next" Tags="|word|language|open-ended|letters|no-computers|" AnswerCount="7" CommentCount="41" ClosedDate="2021-07-12T15:09:53.257" ContentLicense="CC BY-SA 3.0" />
  <row Id="20582" PostTypeId="2" ParentId="20581" CreationDate="2015-08-26T11:11:40.350" Score="3" Body="&lt;p&gt;(4 words) building ding dinge inge &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(13 word) backdoor doorpost postface facedown downspouts &#xA;outsprang rangeland landmass masacres crescendo &#xA;endocardial Dialogues Guesswork workmate material &#xA;rialto altocumulus ulus&lt;/p&gt;&#xA;" OwnerUserId="13061" LastEditorUserId="13061" LastEditDate="2015-08-27T07:28:31.723" LastActivityDate="2015-08-27T07:28:31.723" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20583" PostTypeId="2" ParentId="20581" CreationDate="2015-08-26T11:12:03.563" Score="24" Body="&lt;p&gt;(4 words) monopoly polymath mathematically allying&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(8 words) sleepover overwork workshop shoplift liftoffs offside sideburn burnout&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(123 words) thigh highjack jackboot, bootstrap trapdoor doorstep stepbrother therein, reinterpret preterm termini minidrama ramada, madam adamant mantram tramp rampart, partake takeover overflow flowering ringside, sidestep stepmother there heretofore forevermore, moreover overfree freeform formant mantrap, trapdoor doorbell bellpull pullover overturn, turnover overwork workshop shoplift liftoffs, offspring ringtone toneless lessens sensitising, singsong songbird birdsong songbook bookkeepers, persistent tenterhook hooking kinglike likelihood, hoodwink winking kingship shipload loadmaster, stereotype typewriter iterant ranting tingling, lingering ringworm wormwood woodworm wormhole, holeless lessons sonship shipwreck reckoner, onerous rousing singalong longhorn hornbeam, beamish mishmash mashing hinge ingest, gesticulate latewood woodcock cockpits pitstop, stopping pinging gingerbread readership shipyard, yardland landlord lordship shipment mentoring, ringneck necklace lacewing wingspan spanking, kingfisher sherlock lockdown downside sidelong, longboat boatfuls fulsome somewhere herein, reinforcement menthol tholepins pinstripe riper&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;(commas every 5 words, to make the count easier)&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to all those who helped me with this, including Kingrames, Anachor, Gordon K, Code Newbie, Rj Geraci, and others.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-08-26T20:43:44.183" LastActivityDate="2015-08-26T20:43:44.183" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="20584" PostTypeId="2" ParentId="20581" CreationDate="2015-08-26T12:06:32.380" Score="21" Body="&lt;p&gt;From exchange to &lt;strike&gt;stack&lt;/strike&gt; overflow:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(45 words) exch&lt;strong&gt;ange&lt;/strong&gt; -&gt; angel&lt;strong&gt;fish&lt;/strong&gt; -&gt; fish&lt;strong&gt;bone&lt;/strong&gt; -&gt; bone&lt;strong&gt;meal&lt;/strong&gt; -&gt; meal&lt;strong&gt;time&lt;/strong&gt; -&gt; time&lt;strong&gt;line&lt;/strong&gt; -&gt; linea&lt;strong&gt;ment&lt;/strong&gt; -&gt; menta&lt;strong&gt;list&lt;/strong&gt; -&gt; list&lt;strong&gt;less&lt;/strong&gt; -&gt; les&lt;strong&gt;sons&lt;/strong&gt; -&gt; son&lt;strong&gt;ship&lt;/strong&gt; -&gt; ship&lt;strong&gt;side&lt;/strong&gt; -&gt; side&lt;strong&gt;walk&lt;/strong&gt; -&gt; walky&lt;strong&gt;ries&lt;/strong&gt; -&gt; ries&lt;strong&gt;ling&lt;/strong&gt; -&gt; lingui&lt;strong&gt;stic&lt;/strong&gt; -&gt; stick&lt;strong&gt;ball&lt;/strong&gt; -&gt; ball&lt;strong&gt;game&lt;/strong&gt; -&gt; game&lt;strong&gt;play&lt;/strong&gt; -&gt; play&lt;strong&gt;mate&lt;/strong&gt; -&gt; mate&lt;strong&gt;ship&lt;/strong&gt; -&gt; ship&lt;strong&gt;yard&lt;/strong&gt; -&gt; yard&lt;strong&gt;land&lt;/strong&gt; -&gt; land&lt;strong&gt;lord&lt;/strong&gt; -&gt; lor&lt;strong&gt;ding&lt;/strong&gt; -&gt; din&lt;strong&gt;ging&lt;/strong&gt; -&gt; gingerb&lt;strong&gt;read&lt;/strong&gt; -&gt; read&lt;strong&gt;just&lt;/strong&gt; -&gt; justi&lt;strong&gt;fies&lt;/strong&gt; -&gt; fi&lt;strong&gt;esta&lt;/strong&gt; -&gt; establi&lt;strong&gt;sher&lt;/strong&gt; -&gt; sher&lt;strong&gt;lock&lt;/strong&gt; -&gt; lock&lt;strong&gt;step&lt;/strong&gt; -&gt; stepfa&lt;strong&gt;ther&lt;/strong&gt; -&gt; there&lt;strong&gt;fore&lt;/strong&gt; -&gt; fo&lt;strong&gt;rest&lt;/strong&gt; -&gt; res&lt;strong&gt;tack&lt;/strong&gt; -&gt; tac&lt;strong&gt;king&lt;/strong&gt; -&gt; king&lt;strong&gt;fish&lt;/strong&gt; -&gt; fish&lt;strong&gt;tail&lt;/strong&gt; -&gt; tail&lt;strong&gt;race&lt;/strong&gt; -&gt; race&lt;strong&gt;mism&lt;/strong&gt; -&gt; mis&lt;strong&gt;move&lt;/strong&gt; -&gt;  m&lt;strong&gt;over&lt;/strong&gt; -&gt; overflow &lt;/p&gt;&#xA;" OwnerUserId="15985" LastEditorUserId="15985" LastEditDate="2015-08-26T13:10:16.230" LastActivityDate="2015-08-26T13:10:16.230" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="20585" PostTypeId="1" AcceptedAnswerId="20839" CreationDate="2015-08-26T14:20:38.010" Score="7" ViewCount="551" Body="&lt;p&gt;This riddle was inspired somewhat by the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/20513/no-bake-riddle/20548?noredirect=1#comment64160_20548&quot;&gt;&amp;quot;No Bake&amp;quot; Riddle&lt;/a&gt;, and the subsequent discussion regarding the creation of an &lt;a href=&quot;http://meta.puzzling.stackexchange.com/questions/3266/american-idiom-as-answer-to-puzzle&quot;&gt;American-English&lt;/a&gt; tag.  That should serve as a first hint, others will be added as necessary.&lt;/p&gt;&#xA;&lt;p&gt;Anyway, who am I, described in the lines below?&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;I am sorry to say that my surname will not lead you to my origin, for those I share a name with are far different from me and come from far away.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;They come from the depths, while I come from great heights, or so my name implies.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;They are quite sheltered and I am not, though being so may have been useful for me, at least in the past.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;You see, I used to be attached to powerful things, things bigger than myself, but now I am cast off, for the greater good, I suppose.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Those that really know me, know what I’m made of, well- some like me and some think I am revolting.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;I may not hang around like I used to, but living out west folks do tend to roam.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;I come from a different family, as the title says, and in my current condition, I cannot start a family of my own; &lt;em&gt;no life will come from me&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;Mount, mount, mount.  The mount could be where I am from, an act with which I'm forever done, or my better half could have become one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11452" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-03-23T17:15:12.063" Title="I Come from a Different Family" Tags="|riddle|knowledge|language|regional-dialects|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20586" PostTypeId="1" AcceptedAnswerId="20765" CreationDate="2015-08-26T14:52:19.180" Score="7" ViewCount="1133" Body="&lt;p&gt;My first riddle:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;With jolt or breath; I dine with death&lt;/p&gt;&#xA;&lt;p&gt;Unwanted by most; light I can boast&lt;/p&gt;&#xA;&lt;p&gt;Lectern or wall; tethered, not all&lt;/p&gt;&#xA;&lt;p&gt;Your will or my squall, the point of my call&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h3&gt;Hint 1:&lt;/h3&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; By use of metaphor, neither &lt;strong&gt;squall&lt;/strong&gt; nor &lt;strong&gt;call&lt;/strong&gt; make comment on decibel, though they be related.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h3&gt;Hint 2:&lt;/h3&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Watch out for word play!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h3&gt;Hint 3:&lt;/h3&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; I'll admit: I bent the rules of punctuation a tiny bit. I've adjusted 3 little markings; hopefully, this will point you in the right direction and the answer will click right away!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15991" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-09-01T19:10:30.557" Title="What Am I? - My First Riddle: With jolt or breath, I dine with death" Tags="|riddle|rhyme|" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20587" PostTypeId="2" ParentId="20586" CreationDate="2015-08-26T14:58:14.853" Score="2" Body="&lt;p&gt;You are a   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;EMT or Emergency Medical Service person&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With jolt or breath, I dine with death  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;With either a defibrillator or 'the breath of life' aka CPR, you save lives. Dining with death refers to the closeness to death that EMT's experience in their line of work&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unwanted by most, light I can boast  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;No one wants to see the EMT because it means they're hurt and the light is a reference to the idea that people see 'the light' when dying.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lecturn or wall, tethered, not all&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Not sure yet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your will or my squall, the point of my call  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If you die, you leave a will, and the ambulance doesn't use sirens when picking up a deceased person. If you are not dead, the ambulance comes sirens blazing to your aid. They only come to your aid, however, if 911 is called (or whatever equivalent of 911 it is outside of the U.S., I'm not sure)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="13970" LastEditDate="2015-08-26T14:59:44.880" LastActivityDate="2015-08-26T14:59:44.880" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20588" PostTypeId="2" ParentId="20585" CreationDate="2015-08-26T15:03:52.917" Score="3" Body="&lt;p&gt;My guess would be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Confederate Flag&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am sorry to say that my surname will not lead you to my origin, for those I share a name with are far different from me and come from far away.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The word confederate originates in the east, late latin&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You see, I used to be attached to powerful things, things bigger than myself, but now I am cast off, for the greater good, I suppose.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Recent debate and removal of confederate flag from political structures&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Those that really know me, know what I’m made of, well- some like me and some think I am revolting.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Those that are for the flag argue it does not stand for pro-slavery, rather in remembrance of the soldiers who fought and died. Others despise the flag for their interpreted meaning&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I may not hang around like I used to, but living out west folks do tend to roam.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Beginning to disappear over recent arguments, but can still find it in areas, typically more south or west&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15990" LastEditorUserId="5373" LastEditDate="2015-08-26T15:05:51.333" LastActivityDate="2015-08-26T15:05:51.333" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20589" PostTypeId="1" AcceptedAnswerId="20594" CreationDate="2015-08-26T15:15:56.363" Score="7" ViewCount="531" Body="&lt;p&gt;By a freak of nature and an error by the author, it's nearly the start of a new school year &lt;strong&gt;in 2016&lt;/strong&gt; and Mr Stubbins (the caretaker at &lt;strong&gt;Farthingbottom School&lt;/strong&gt;) is getting nervous. Since he started working at the school in 2001 he has been plagued by repeated acts of vandalism or, to be more precise, one specific act of vandalism - the theft of one of his 'H's. The entrance to the school is marked with 20 stone uppercase letters, each around 2 feet high and pranksters keep removing one. Sometimes he finds it hidden in the bushes some way away from the entrance, sometimes it disappears without a trace and he has to order a new one. It only happens (at most) once per year and he has come to conclude that it is some sort of rite of passage for the new intake of pupils at one of the 3 nearby boarding schools. These are the dates of the mornings each year when this calamity has been discovered:&lt;br /&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2015&lt;/strong&gt;: 13 September&lt;br /&gt;&#xA;&lt;strong&gt;2014&lt;/strong&gt;: 24 September&lt;br /&gt;&#xA;&lt;strong&gt;2013&lt;/strong&gt;: 06 September&lt;br /&gt;&#xA;&lt;strong&gt;2012&lt;/strong&gt;: 16 September&lt;br /&gt;&#xA;&lt;strong&gt;2011&lt;/strong&gt;: 28 September&lt;br /&gt;&#xA;&lt;strong&gt;2010&lt;/strong&gt;: 08 September&lt;br /&gt;&#xA;&lt;strong&gt;2009&lt;/strong&gt;: he camped out all night every night in September and &lt;em&gt;nothing happened&lt;/em&gt;&lt;br /&gt;&#xA;&lt;strong&gt;2008&lt;/strong&gt;: one morning near the end of the month&lt;br /&gt;&#xA;&lt;strong&gt;2007&lt;/strong&gt;: can't remember, but it was a Wednesday morning&lt;br /&gt;&#xA;&lt;strong&gt;2006&lt;/strong&gt;: &lt;em&gt;nothing happened&lt;/em&gt;&lt;br /&gt;&#xA;&lt;strong&gt;2005&lt;/strong&gt;: a Sunday morning&lt;br /&gt;&#xA;&lt;strong&gt;2004&lt;/strong&gt;: one morning near the middle of the month&lt;br /&gt;&#xA;&lt;strong&gt;2003&lt;/strong&gt;: one morning near the end of the month&lt;br /&gt;&#xA;&lt;strong&gt;2002&lt;/strong&gt;: &lt;em&gt;nothing happened&lt;/em&gt;&lt;br /&gt;&#xA;&lt;strong&gt;2001&lt;/strong&gt;: around the middle of the month&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The three suspected schools are &lt;strong&gt;Allbrightnow Academy&lt;/strong&gt;, &lt;strong&gt;Bodgingworthy Boys&lt;/strong&gt; and &lt;strong&gt;Consternation College&lt;/strong&gt;. Each of these runs a couple of late-evening clubs for their pupils which none of them would dream of missing out on. The pranksters wouldn't have been able to carry out their mission on an evening when they had a club to attend.&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Allbrightnow Academy has Origami on Monday and Astronomy on Friday&lt;br /&gt;&lt;/li&gt;&#xA;&lt;li&gt;Bodgingworthy Boys has Electronics on Wednesday and Film Club on Sunday&lt;br /&gt;&lt;/li&gt;&#xA;&lt;li&gt;Consternation College has Survival Skills on Sunday and Weightlifting on Friday.&lt;br /&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Should Mr Stubbins be prepared for intruders this year? If so, which night are they likely to come and which school are they going to come from?&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; Assume the schools are in the UK.&lt;/p&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="14135" LastEditDate="2015-08-26T15:30:27.447" LastActivityDate="2015-08-27T07:40:13.460" Title="Who's been stealing my H?" Tags="|logical-deduction|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="20590" PostTypeId="2" ParentId="20581" CreationDate="2015-08-26T15:16:37.190" Score="20" Body="&lt;p&gt;I think I have finally locked myself in at &lt;strong&gt;174 words:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;thigh, highjack, jackboot, bootstrap, trapdoor, doorstep, stepbrother, therein, reinterpret, preterm, termini, minidrama, ramada, madam, adamant, manta, antacid, acidifies, fiesta, establishment, mentalist, listen, stench, enchantment, menthols, holster, stereotype, typecast, castaways, wayside, sidearm, earmark, markdown, downcast, caste, aster, stertorous, rousing, singletons, tonsure, surefire, firearm, earmarking, kingpins, pinstripe, ripest, pestering, ringlike, likelihood, hoodwink, winking, kingship, shipload, loadstar, starch, archbishop, shopkeepers, persecute, cutest, testament, mentoring, ringside,sidecars, carsick, sickbeds, bedside, sidekick, kickback, backbone, bonehead, headband, bandanna, annalist, listens, tense, ensembles, blessing, singletree, treed, reedit, editorship, shipmate, materialising, singsong, songbird, birdbath, bathe, atheistic, stickleback, backdate, dateline, lineage, eagerest, restaff, taffrail, railroad, roadbeds, bedsore, sorehead, headgear, gearing, ringworm, wormwood, woodblock, lockouts, outscore, corespondent, dentistry, strychnine, ninepin, epinephrins, rinse, inseam, seamanship, shippers, perseverance, ancestor, storage, rageless, lessees, seesawing, wingspan, spanking, kingfishers, herself, selfish, fishbowl, bowlfuls, fulsome, somebody, bodysuit, suitcase, casein, seiner, inerrant, ranting, tinge, ingest, gestapos, apostleship, shipwreck, reckless, lessens, sensationalist, listener, enervating, tinging, gingerbread, readdressing, singe, ingesting, tingling, lingering, ringing, gingersnap, snapdragon, agonizing, zinging, ginghams, hamster, sterling, linguistic, stick, ticking, kingfisher, sheriff, riffing, fingering, ringmaster, sternest, nestling, linguines, inescapable, abled, bled&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to rand al'thor, Ian MacDonald, and Anachor for the help!&lt;/p&gt;&#xA;" OwnerUserId="11926" LastEditorUserId="15544" LastEditDate="2015-08-27T14:40:58.783" LastActivityDate="2015-08-27T14:40:58.783" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="20591" PostTypeId="2" ParentId="20586" CreationDate="2015-08-26T15:33:50.913" Score="4" Body="&lt;p&gt;&lt;strong&gt;Third attempt&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could it be,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A sound&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With jolt or breath, I dine with death&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; sound can be formed by jolt or someone talking, breath, and dies off quickly&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unwanted by most, light I can boast&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most people don't want to listen to a lecture and I would bet even less want to hear someone boasting about light&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lecturn or wall, tethered, not all&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You lecture at a lecturn and sound echoes off walls but they are not connected &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your will or my squall, the point of my call&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Talking can be your will and squall and call make sounds&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second attempt&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could it be,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A gun shot&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With jolt or breath, I dine with death&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An accidental shot or someone yelling the order to fire, causing death&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unwanted by most, light I can boast&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I don't think most people want to be shot and a muzzle flare produces light&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lecturn or wall, tethered, not all&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is a possibility of being shot anywhere, it doesn't matter if you are in class at a lecturn or in a back alley against a wall. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your will or my squall, the point of my call&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; shooting someone could be meditated, your will, or the sound of the gun going off, my squall.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First attempt&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could it be,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Electricity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With jolt or breath, I dine with death&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Electricity can jolt you and bring you back to life or it could kill you if too much of a jolt&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unwanted by most, light I can boast&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most people don't want to be shocked and electricity can produce light&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lecturn or wall, tethered, not all&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Electricity is not tethered to anything&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your will or my squall, the point of my call&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lightning forms in storms, or squalls&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13925" LastEditorUserId="13925" LastEditDate="2015-08-31T15:01:37.740" LastActivityDate="2015-08-31T15:01:37.740" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="20592" PostTypeId="2" ParentId="17211" CreationDate="2015-08-26T15:42:59.260" Score="11" Body="&lt;p&gt;The answer to Puzzle 2 is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;BOTANY = BOT + TAN + KNEE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(please see the chatroom for more details, I can only claim to have discovered the last clue and overall answer: &lt;a href=&quot;http://chat.stackexchange.com/rooms/26670/the-riddle-of-life-to-ease-all-your-strife&quot;&gt;http://chat.stackexchange.com/rooms/26670/the-riddle-of-life-to-ease-all-your-strife&lt;/a&gt; )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First a bird, nearby plane,&lt;br&gt;&#xA;no, a shattered sky!&lt;br&gt;&#xA;puzzle the letters and find out why.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;We are supposed to form letters here, according to the third line.&lt;br&gt;&#xA;B, the first letter of bird;&lt;br&gt;&#xA;O, which is nearby P(lane);&lt;br&gt;&#xA;T, which looks like a sky broken by a pillar. (this is according to @Nyk232 himself).&lt;br&gt;&#xA;&lt;br&gt;&#xA;Thus, &lt;strong&gt;BOT&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A screen, not reflecting,&lt;br&gt;&#xA;a deep brown hue,&lt;br&gt;&#xA;a reference makes me feel real blue.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Sunscreen is a screen, that when applied on your skin, it will not become a deep brown hue, because it is not &lt;em&gt;tanned&lt;/em&gt;.&lt;br&gt;&#xA;The reference is to &quot;tanning beds&quot;, which emit blue light.&lt;br&gt;&#xA;&lt;br&gt;&#xA;Thus, &lt;strong&gt;TAN&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A joint most fitting,&lt;br&gt;&#xA;D'el('s) bow string never fitting.&lt;br&gt;&#xA;the valley splits between.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The answer is a joint, but not elbow (&quot;Del's bow&quot; with a few letters removed).  &quot;The valley splits between&quot; could refer to the characters that separate &quot;el&quot; and &quot;bow&quot;.&lt;br&gt;&lt;br&gt;&#xA;The equivalent of the elbow is the &lt;strong&gt;KNEE&lt;/strong&gt;.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6178" LastActivityDate="2015-08-26T15:42:59.260" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20593" PostTypeId="1" CreationDate="2015-08-26T15:46:05.290" Score="6" ViewCount="1663" Body="&lt;p&gt;Once I completed solving Rubik's cube, &lt;strong&gt;a different pattern&lt;/strong&gt; came into my mind. It was-&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A B B&#xA;A B B&#xA;A C C&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here, same letters represent same colors. So is it possible to get this pattern in a Rubik's cube on all of it's faces?&lt;/p&gt;&#xA;" OwnerUserId="15992" LastEditorUserId="15544" LastEditDate="2015-08-26T15:59:00.900" LastActivityDate="2015-08-28T19:03:39.640" Title="rubik's cube inside a rubik's cube" Tags="|rubiks-cube|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20594" PostTypeId="2" ParentId="20589" CreationDate="2015-08-26T16:33:29.170" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Allbrightnow Academy are the culprits and they will attack on the evening of September 1st.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each of the dates is on the new moon of that month. If the new moon occurs before midday BST, it is the evening of the day before and therefore discovered that morning, otherwise it is at the end of the day and discovered the next. The two years where nothing happened were fridays, that rules out Bodgingworthy Boys. In 2008, the new moon was 09:12 on the 29th which was a Monday, therefore it would have been done on the Sunday which means Consternation College couldn't have done it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12548" LastEditorUserId="12548" LastEditDate="2015-08-27T07:40:13.460" LastActivityDate="2015-08-27T07:40:13.460" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20595" PostTypeId="1" AcceptedAnswerId="20596" CreationDate="2015-08-26T16:52:46.740" Score="65" ViewCount="16467" Body="&lt;p&gt;A perfectly symmetrical small 4-legged table is standing in a large room with a continuous but uneven floor. Is it always possible to position the table in such a way that it doesn't wobble, i.e. all four legs are touching the floor?&lt;/p&gt;&#xA;&lt;p&gt;No tricks. No &lt;a href=&quot;/questions/tagged/lateral-thinking&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;lateral-thinking&amp;#39;&quot; rel=&quot;tag&quot;&gt;lateral-thinking&lt;/a&gt;. Serious question (with real-life applications too!) with a serious answer.&lt;/p&gt;&#xA;&lt;br&gt;&#xA;&lt;p&gt;&lt;sub&gt;This might look like it'd fit better on Lifehacks.SE, but the answer has a nice mathematical proof/formula [depending on whether it's yes/no; I won't give the game away!] which is surprisingly simple and elegant.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2021-03-17T19:13:32.350" LastActivityDate="2021-03-17T19:13:32.350" Title="Is it always possible to balance a 4-legged table?" Tags="|mathematics|geometry|physics|" AnswerCount="12" CommentCount="37" ContentLicense="CC BY-SA 4.0" />
  <row Id="20596" PostTypeId="2" ParentId="20595" CreationDate="2015-08-26T17:15:24.537" Score="70" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; yes!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's why:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Imagine that the table can pass through the floor. We're going to call one leg the &quot;floating leg&quot; - the other three are going to always be on the floor. Now, after we rotate the table a quarter turn, the floating leg is going to be above the floor if it was originally below, or vice versa. By the intermediate value theorem, it will be exactly on the floor at one point in that rotation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More detailed proof:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Three legs of the table define a plane. Define the &quot;offset&quot; of a leg to be the distance above the floor if the other three legs are placed on the floor directly - negative if below, positive if above. In any arbitrary placement, the offset of any two adjacent legs will be positive and negative. (WLOG assume the one on the left is positive.) Offset is a continuous function because it's the distance from the floor. If you start with a positive offset on leg A and rotate the table 90° to the right, the offset on leg A (now in the original position of another leg) will be negative. This means that at some point, offset was 0, therefore the table was completely touching the floor.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="11876" LastEditDate="2015-08-27T01:25:24.270" LastActivityDate="2015-08-27T01:25:24.270" CommentCount="47" ContentLicense="CC BY-SA 3.0" />
  <row Id="20597" PostTypeId="1" AcceptedAnswerId="20630" CreationDate="2015-08-26T17:28:18.320" Score="20" ViewCount="984" Body="&lt;p&gt;&lt;em&gt;He was victorious, say you, in a foreign tongue.&lt;br&gt;&#xA;He was almost there, he's a ring in the skin;&lt;br&gt;&#xA;A half-fisted stove; a snake by the nines.&lt;br&gt;&#xA;Christmas Eve now holds their endless odd gin.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is this poem about?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Hint/Clarification:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This isn't a &quot;Who/what am I?&quot; riddle (meaning this riddle does not describe a singular entity).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="11475" LastEditDate="2015-08-27T13:56:39.517" LastActivityDate="2015-08-27T17:11:00.403" Title="He Was Victorious (a short riddle)" Tags="|riddle|rhyme|wordplay|poetry|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20598" PostTypeId="5" CreationDate="2015-08-26T17:28:48.747" Score="0" Body="&lt;p&gt;Use this tag for puzzles that are presented in the form of or include any of the various types of poetry.&lt;/p&gt;&#xA;&lt;h1&gt;What is poetry?&lt;/h1&gt;&#xA;&lt;p&gt;Poetry is the use of stylized writing and rhythm to evoke feeling beyond just the words said.&lt;/p&gt;&#xA;&lt;p&gt;Different use of rhyming schemes, syllable counts and emphasis can all be used to create different forms of poetry. For example, a Haiku is a special type of poetry that is comprised of 3 lines with 5 syllables in the first line, 7 syllables in the second and 5 syllables again in the final line.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;An example of a Haiku below:&lt;/em&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;This is a Haiku &lt;br/&gt;&#xA;Five seven five syllables &lt;br/&gt;&#xA;Exactly three lines&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Another well-known type that focuses on emphasis and rhythm is iambic pentameter, which frequently appears in works of Shakespeare. An 'iamb' is a metrical foot that references a syllable stress pattern of one short unstressed syllable followed by a long stressed syllable and pentameter means the line consists of five metrical feet so in this case five iambs in a line.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;An example of a verse written in iambic pentameter from Romeo and Juliet, with stressed syllables bolded:&lt;/em&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Two &lt;strong&gt;house&lt;/strong&gt;holds, &lt;strong&gt;both&lt;/strong&gt; a&lt;strong&gt;like&lt;/strong&gt; in &lt;strong&gt;dig&lt;/strong&gt;ni&lt;strong&gt;ty&lt;/strong&gt;, &lt;br/&gt;&#xA;In &lt;strong&gt;fair&lt;/strong&gt; Ve&lt;strong&gt;ro&lt;/strong&gt;na, &lt;strong&gt;where&lt;/strong&gt; we &lt;strong&gt;lay&lt;/strong&gt; our &lt;strong&gt;scene&lt;/strong&gt;, &lt;br/&gt;&#xA;From &lt;strong&gt;an&lt;/strong&gt;cient &lt;strong&gt;grudge&lt;/strong&gt; break &lt;strong&gt;to&lt;/strong&gt; new &lt;strong&gt;mu&lt;/strong&gt;ti&lt;strong&gt;ny&lt;/strong&gt;, &lt;br/&gt;&#xA;Where &lt;strong&gt;ci&lt;/strong&gt;vil &lt;strong&gt;blood&lt;/strong&gt; makes &lt;strong&gt;ci&lt;/strong&gt;vil &lt;strong&gt;hands&lt;/strong&gt; un&lt;strong&gt;clean&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;General poetry questions are off-topic and should be asked on &lt;a href=&quot;https://literature.stackexchange.com/&quot;&gt;Literature Stack Exchange&lt;/a&gt; (for questions about existing poems) or &lt;a href=&quot;https://writing.stackexchange.com/&quot;&gt;Writing Stack Exchange&lt;/a&gt; (for questions about creating poems).&lt;/p&gt;&#xA;&lt;p&gt;For questions based on, related to, or requiring knowledge of specific existing poems, the &lt;a href=&quot;/questions/tagged/literature&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;literature&amp;#39;&quot; aria-label=&quot;show questions tagged &amp;#39;literature&amp;#39;&quot; rel=&quot;tag&quot; aria-labelledby=&quot;literature-container&quot;&gt;literature&lt;/a&gt; tag should be included.&lt;/p&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="60644" LastEditDate="2023-02-02T16:12:40.533" LastActivityDate="2023-02-02T16:12:40.533" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="20599" PostTypeId="4" CreationDate="2015-08-26T17:28:48.747" Score="0" Body="Puzzles that are presented in the form of any of the various types of poetry." OwnerUserId="11475" LastEditorUserId="69582" LastEditDate="2021-02-03T07:13:10.820" LastActivityDate="2021-02-03T07:13:10.820" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="20600" PostTypeId="2" ParentId="20586" CreationDate="2015-08-26T17:29:22.453" Score="1" Body="&lt;p&gt;I'm thinking&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fire&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;With jolt or breath, I dine with death&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lightning strike or 'breath' (possible fanning motion to feed oxygen)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Unwanted by most, light I can boast&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wildfire, esp. currently in the western region&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Lecturn or wall, tethered, not all&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Unable to be contained&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Your will or my squall, the point of my call&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Intentionally started by man (campfire), or wildfire&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15990" LastActivityDate="2015-08-26T17:29:22.453" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20601" PostTypeId="2" ParentId="20597" CreationDate="2015-08-26T17:31:45.473" Score="12" Body="&lt;p&gt;It's about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Albert,_Prince_Consort&quot; rel=&quot;nofollow&quot;&gt;Prince Albert&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He was victorious, say you, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He married Queen Victoria&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;in a foreign tongue.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; His native language was German&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He was almost there, he's a ring in the skin;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Prince_Albert_(genital_piercing)&quot; rel=&quot;nofollow&quot;&gt;I'm going to let Wikipedia answer this one.&lt;/a&gt; &amp;lt;----------- Link NSFW - Not Safe For Work. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A half-fisted stove; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Rifle_Brigade_(Prince_Consort%27s_Own)&quot; rel=&quot;nofollow&quot;&gt;Rifle Brigade (Prince Consort's Own)&lt;/a&gt; wore &quot;stovepipe shakoes&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;a snake &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The engagement ring which Albert gave to Victoria was in the shape of a snake&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;by the nines.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; nine children&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Christmas Eve now holds their &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://news.bbc.co.uk/local/berkshire/hi/people_and_places/history/newsid_9286000/9286971.stm&quot; rel=&quot;nofollow&quot;&gt;Many Christmas traditions come from Victoria and Albert.&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;endless odd gin.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'gin' in the playing cards sense, because he liked playing Patience, which seems to go on forever.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="4050" LastEditDate="2015-08-27T15:40:52.553" LastActivityDate="2015-08-27T15:40:52.553" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20603" PostTypeId="1" AcceptedAnswerId="20607" CreationDate="2015-08-26T19:35:08.800" Score="5" ViewCount="381" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Administers food&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Does De Re Metallica&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Eponymous towns&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who am I?  &lt;/p&gt;&#xA;" OwnerUserId="11921" LastEditorUserId="11475" LastEditDate="2015-09-03T13:35:15.077" LastActivityDate="2015-09-03T13:35:15.077" Title="A Metallica haiku -- Who am I?" Tags="|riddle|poetry|haiku|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20604" PostTypeId="2" ParentId="20595" CreationDate="2015-08-26T19:50:26.270" Score="-2" Body="&lt;p&gt;The answer is certainly no, the best way to think about this is in terms of basic linear algebra, or if math is not your strong point just in terms of a 3d graph. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your questions translates to &quot;for any four points in 3d space, does there always exist a single 2d plane containing them all&quot;. For our problem: the four points are the bottoms of the four legs and the 2d plane is the floor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can certainly put any three points in the same plane by just drawing lines between them and defining that as the plane, but we will not always be able to add the fourth (in fact almost every other point will not be in this plane). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To visualise this consider the four points of the form (x,y,z) in Cartesian (normal) coordinates:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a = (0,0,0)    i.e. this signifies x=0, y=0, z=0&lt;/li&gt;&#xA;&lt;li&gt;b = (1,0,0)&lt;/li&gt;&#xA;&lt;li&gt;c = (0,1,0)&lt;/li&gt;&#xA;&lt;li&gt;d = (1,1,1) &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Clearly, a,b and c all lie in the (x,y)-plane, we will call this the floor. However, d is by definition NOT in the (x,y)-plane as the x value is set to zero. It lies exclusively in the (y,z)-plane and will always be shifted away by one unit no matter how much you rotate the points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try this out by plotting each of these points in the following link:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://hotmath.com/learning_activities/interactivities/3dplotter.swf&quot; rel=&quot;nofollow&quot;&gt;point/graph plotter&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rotate the graph around and line 3 points up into the same plane, the third will always appear a unit of 1 to the side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I changed the points to the &quot;table-like&quot; layout&lt;/p&gt;&#xA;" OwnerUserId="16000" LastEditorUserId="16001" LastEditDate="2015-08-26T20:06:02.480" LastActivityDate="2015-08-26T20:06:02.480" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20606" PostTypeId="2" ParentId="20576" CreationDate="2015-08-26T20:25:15.573" Score="1" Body="&lt;p&gt;@IvoBeckers is technically correct, but more realistically:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B and M&lt;br&gt;&#xA;B back&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B and B1&lt;br&gt;&#xA;B and M back&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B and B2&lt;br&gt;&#xA;B back&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M and B&lt;br&gt;&#xA;M back&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F and F1&lt;br&gt;&#xA;B knocks out F (because no innocent person was ever named Mr. Fox), &quot;confiscates&quot; the contents of F1, then makes his grand entrance with his 2 supermodels. Sure M &lt;em&gt;wants&lt;/em&gt; to go to the party, but that is hardly B's problem.&lt;/p&gt;&#xA;" OwnerUserId="16003" LastActivityDate="2015-08-26T20:25:15.573" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20607" PostTypeId="2" ParentId="20603" CreationDate="2015-08-26T20:39:01.787" Score="7" Body="&lt;p&gt;I am guessing&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Herbert Hoover&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Administers food&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He was the head of the U.S. Food Administration&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Does De Re Metallica&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He translated the book&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Eponymous towns&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; So-called Hoovervilles during the great depression&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15913" LastActivityDate="2015-08-26T20:39:01.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20610" PostTypeId="2" ParentId="20372" CreationDate="2015-08-26T21:02:48.500" Score="13" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Meter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I don't see how this would make a good title, just yet.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Here is a riddle quite long  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is an important clue that should not need explaining.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm known to be unmatched in song.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First pun:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Meter also refers to &lt;a href=&quot;https://en.wikipedia.org/wiki/Meter_(music)&quot;&gt;the rhythmic structure of music.&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The next two lines go back to the primary meaning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You'll find me the whole world around&lt;br&gt;&#xA;  In nature 'tis true I abound  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In various forms I am found.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My interpretation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The meter forms the base for a variety of units with the addition of &lt;a href=&quot;https://en.wikipedia.org/wiki/Metric_prefix&quot;&gt;SI prefixes&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Confused for another&lt;br&gt;&#xA;  Get this: I'm her brother&lt;br&gt;&#xA;  Though our differences are well defined.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now we are introduced the the &quot;sister&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the yard.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The next stanza explores the relationship:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We're nothing alike, I insist&lt;br&gt;&#xA;  Her flowery form I resist  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now a pun on the the sister:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A yard is also an &lt;a href=&quot;https://en.wikipedia.org/wiki/Yard_(land)&quot;&gt;open area where flowers might grow&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Yet one step from outburst she is.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This appears to be a pun on:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The definition of a yard as being three &lt;em&gt;feet&lt;/em&gt; long. An extra foot would be one &quot;step&quot; too far.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now the heart of the riddle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;She's one of a number, all true&lt;br&gt;&#xA;  I'm not because I come from two&lt;br&gt;&#xA;  She gets her own line&lt;br&gt;&#xA;  I'll never have mine&lt;br&gt;&#xA;  You know me now surely, don't you?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes. That was very helpful:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The key to understanding this stanza is knowing the definition of a meter compared to a yard. The International System &lt;a href=&quot;https://en.wikipedia.org/wiki/SI_base_unit&quot;&gt;fixes the length of a meter&lt;/a&gt; using two criteria: the definition of the second and distance light travels in a fraction of a second. Meanwhile, a &lt;a href=&quot;https://en.wikipedia.org/wiki/Yard#History&quot;&gt;yard has been defined&lt;/a&gt; either by physical standards or in relation to a meter. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The penultimate stanza seems to be a grabbag of puns:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;See me or listen for flyers&lt;br&gt;&#xA;  Read me or climb up much higher  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I believe this refers to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an altimeter, which sometimes has a &lt;a href=&quot;https://en.wikipedia.org/wiki/Bitching_Betty&quot;&gt;voice readout&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I'm less sure about:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And now I must sleep, for I tire.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Perhaps &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a tire pressure gauge/meter?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally, a callback to the sister:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;From all but one sense do we hide&lt;br&gt;&#xA;  My ghostly sister and I&lt;br&gt;&#xA;  She on a petal&lt;br&gt;&#xA;  Me on a metal&lt;br&gt;&#xA;  I've caused many rulers to die.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I see that the last line has been changed. Indeed, I was tipped off by the word that was altered.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Units of length can only be visualized; they have no taste, smell, feel or sound. Again, I believe the reference to &quot;petal&quot; is a pun on the garden meaning of yard. Metal rhymes, but might also be a reference to the &lt;a href=&quot;https://en.wikipedia.org/wiki/SI_base_unit&quot;&gt;definition of a second&lt;/a&gt;, which rests on the vibration of a caesium atom. The final line, in the form I quoted, probably refers to the changeover from imperial to metric units that ended the usefulness of many yardsticks/rulers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Criticism&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Assuming my answer is correct, this riddle leans a little too hard on puns that are a stretch. It could also use some condensing. Ideally, the rhyme would work on it's own and not require the answer to appreciate. But otherwise, well done. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If I'm wrong, I bow to your superior puzzling. ;-)&lt;/p&gt;&#xA;" OwnerUserId="15" LastActivityDate="2015-08-26T21:02:48.500" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="20611" PostTypeId="2" ParentId="20595" CreationDate="2015-08-26T21:29:48.857" Score="22" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;No.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason by provable test case:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;We have a table that is a solid, 3 foot tall, wide, and long, cube of wood, with 4 very short stubby legs on one side (we'll say 2 inches). If standing on a continuous but uneven floor, Say, a simple hexagonal pattern of half-spherical domes (5 inch high domes) (stylized like bubble paper, of course) the domes would lift the table off the floor, preventing it from balancing. At no point could you guarantee stable footing. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...Here's some images to help illustrate the problem.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/WmUem.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WmUem.png&quot; alt=&quot;A grid of bubbles&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's a grid of bubbles covering the floor. we can surmise that a table could put its legs between these bubbles and touch the floor, but only on the condition that the floor gives space to do so. if not, then you could &quot;balance&quot; the table only in name, on top of the bubbles, but we all know it wouldn't last.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/qTBCC.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qTBCC.png&quot; alt=&quot;A hex grid of bubbles&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When we switch to a hex-based formation, we see more problems. Firstly, depending on the dimensions of the table, it might not be possible to fit the legs inbetween the bubbles anymore. (In real life, tables don't have massless, frictionful legs because that would be pretty frickin dangerous. This question has what I like to call a spherical cow error [ &lt;a href=&quot;https://en.wikipedia.org/wiki/Spherical_cow&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Spherical_cow&lt;/a&gt; ] - Although it might be useful in real life scenarios, in order to say that it ALWAYS will be helpful in real life scenarios you have to keep adding criteria to the problem to eliminate &quot;edge cases&quot; which are valid concerns for other people, you just don't realize that they count just as much as you do)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/XwbrZ.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XwbrZ.png&quot; alt=&quot;A hex grid of bubbles from above&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From above, the situation is a lot more clear. Square tables are going to have a hell of a time trying to balance on this kind of floor, and most often, they're just going to remain in a &quot;good enough&quot; position where they still slide a bit and you just have to remember not to do anything on that table that requires stability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Final verdict:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The table can't be guaranteed to balance on the floor in any situation where the table legs have mass (and therefore volume).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13068" LastEditorUserId="13068" LastEditDate="2015-08-28T01:48:34.217" LastActivityDate="2015-08-28T01:48:34.217" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="20613" PostTypeId="2" ParentId="20595" CreationDate="2015-08-27T01:20:34.320" Score="1" Body="&lt;p&gt;You may be asking the wrong question.  As stated, no it isn't always possible to position an 'ideal table' on an uneven floor and guarantee that all four legs have firm contact with the ground.  The real question is it is possible to always distort a floor such that an ideal table &lt;em&gt;will&lt;/em&gt; wobble.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with an 'ideal floor' (perfectly smooth and level) made from soft wood, place your ideal table anywhere.  No wobble.  Now take a hammer and knock out a small uneven plug from the floor under one of the legs.  Now the table wobbles.  Move table and if doesn't wobble, repeat hammer knock out.  Repeat until you have a floor upon which there is no position where the table doesn't wobble.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a randomly sized (table width, depth) and a randomly uneven floor, you &lt;em&gt;might&lt;/em&gt; be able to place the table somewhere it doesn't wobble but you &lt;em&gt;might&lt;/em&gt; not.  Therefore it isn't &lt;em&gt;always&lt;/em&gt; possible.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I decided to update my answer since people seem to have misunderstood. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not suggesting that the question ask was wrong.  I simple thought that reframing the question might help people see the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not suggest that the floor is continuously moving.  The floor is &quot;uneven&quot; there are lots and lots and lots of ways that a floor can be uneven.  I simply started with an even floor and made it uneven to make a point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I know that my &lt;em&gt;&quot;repeat until you have ...&quot; loop is guaranteed to terminate?&lt;/em&gt;  Given the problem statement we know there are a finite number of positions available for the table on the floor (there must be some distance, no matter how small that defines a &quot;new&quot; position for the table).  Lets call the number of possible configurations &lt;em&gt;n&lt;/em&gt;.  Take the set of all uneven floors, in that there has to be at least one floor where there is only one position where the table wobbles (if there were no positions then the floor would be even).  There must also be a floor where there are only two positions where the table wobbles.  One where there are only three positions, etc.  Therefore, there must be at least one floor that has only one position where the table table wobbles.  Start with that floor.  Take hammer and 'unlevel' the floor under one of the legs, call that point &lt;em&gt;p&lt;/em&gt;.  In order for the loop to never end the change of &lt;em&gt;p&lt;/em&gt; must lead to a new position which is a rotation around point &lt;em&gt;p&lt;/em&gt; where the table does not wobble, but there is still only one position on the floor where the table doesn't wobble.  Hit &lt;em&gt;p&lt;/em&gt; again, repeat.  Since there are only &lt;em&gt;n&lt;/em&gt; possible positions of the table in the room you cannot repeat the procedure more than &lt;em&gt;n&lt;/em&gt; times.  Since you are not changing any other points you will by eventually have a floor where there is no position where the table doesn't wobble.  Therefore, there is at least one floor where you can not position the table in a wobble free position.  Thus it is not ALWAY possible to position the table so that it doesn't wobble. &lt;/p&gt;&#xA;" OwnerUserId="16007" LastEditorUserId="16007" LastEditDate="2015-08-29T02:14:37.423" LastActivityDate="2015-08-29T02:14:37.423" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20615" PostTypeId="2" ParentId="20595" CreationDate="2015-08-27T02:47:59.727" Score="11" Body="&lt;p&gt;The answer is &quot;YES&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Place the table on the floor, and label the legs, clockwise from above (A, B, C, D) and their points of contact on the floor (1, 2, 3, and 4) respectively. Note that despite the different labels, the four legs, by symmetry, must be identical in length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If all four legs are in contact, then the condition is satisfied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, then without loss of generality we can set Leg A as not in contact with Floor Point #1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then rotate the table by one-quarter turn clockwise, while allowing it to be supported by the floor throughout the rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now Leg A is at Point #2. and in contact with Point #2.  After all, Leg B was in contact before the rotation, and Leg A is identical to Leg B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, Leg D, formerly in contact with Point 4, has rotated around to Point 1, where it is out of contact, since it is replacing the identical Leg A, which was out of contact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Leg A made contact, while Leg D lost contact.  If Leg A made contact first, then at that point all four legs are in contact.  If Leg D left first, then the table is resting on two legs, and it tips over...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modesty and honesty require that I credit Martin Gardner &lt;a href=&quot;http://www.scientificamerican.com/article/strange-but-true-turning/&quot;&gt;http://www.scientificamerican.com/article/strange-but-true-turning/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yes, I did read &lt;strong&gt;&lt;em&gt;and&lt;/em&gt;&lt;/strong&gt; recall the original 1973 article...&lt;/p&gt;&#xA;" OwnerUserId="11648" LastActivityDate="2015-08-27T02:47:59.727" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="20616" PostTypeId="2" ParentId="20595" CreationDate="2015-08-27T03:24:30.477" Score="49" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reason is actually simpler and more intuitive than the other answers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It's obvious that the table can always stand on three legs, because the ends of two legs define a line, and the third leg can be brought into contact by rotating around that line.  Then, rotating the table around a vertical axis, the fourth leg &lt;em&gt;must&lt;/em&gt; touch the floor before any of the other legs becomes unable to touch, because the fourth leg touching is the reason one of the other legs wouldn't be able to touch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7514" LastEditorUserId="42452" LastEditDate="2018-10-04T21:38:35.233" LastActivityDate="2018-10-04T21:38:35.233" CommentCount="28" ContentLicense="CC BY-SA 4.0" />
  <row Id="20617" PostTypeId="2" ParentId="20581" CreationDate="2015-08-27T04:25:23.500" Score="3" Body="&lt;p&gt;A very similar puzzle to this was &lt;a href=&quot;https://codegolf.stackexchange.com/questions/44922/building-a-long-chain-of-words&quot;&gt;posted on PPCG&lt;/a&gt;, and I answered it, but it's significantly harder with an overlap of four letters instead of three. I modified my code from that problem (after digging through old files to find it and fixing some bugs), and, after a few minutes, I got this chain of length 300:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aachen -&gt; chenille -&gt; illegitimate -&gt; materialistic -&gt; stick -&gt; tick -&gt; tickling -&gt; linguistic -&gt; stickpins -&gt; pins -&gt; pinstripe -&gt; ripens -&gt; pens -&gt; pensioner -&gt; onerous -&gt; rousing -&gt; singletons -&gt; tonsuring -&gt; ringside -&gt; sidetrack -&gt; racketeering -&gt; ring -&gt; ringmaster -&gt; sterling -&gt; lingering -&gt; ringing -&gt; gingerbread -&gt; readjust -&gt; justifies -&gt; fiestas -&gt; stashed -&gt; shed -&gt; shedding -&gt; dinging -&gt; gingersnap -&gt; snap -&gt; snappers -&gt; persecute -&gt; cutest -&gt; test -&gt; tester -&gt; sternest -&gt; nesting -&gt; tinge -&gt; ingest -&gt; gesticulate -&gt; late -&gt; latent -&gt; tenth -&gt; enthronement -&gt; mentoring -&gt; ringworm -&gt; wormwood -&gt; wood -&gt; woodwork -&gt; workmanlike -&gt; like -&gt; likewise -&gt; wisecracking -&gt; kingship -&gt; shippers -&gt; persist -&gt; sisterhood -&gt; hoodwinking -&gt; kingfishers -&gt; herself -&gt; selfish -&gt; fisher -&gt; sheriff -&gt; riffraff -&gt; raffish -&gt; fish -&gt; fishtail -&gt; tailgate -&gt; gate -&gt; gatepost -&gt; poster -&gt; stereotype -&gt; typeface -&gt; facet -&gt; acetaminophen -&gt; phenomenons -&gt; nonsupport -&gt; portables -&gt; blessing -&gt; sing -&gt; singsong -&gt; songbird -&gt; birdwatcher -&gt; cheroot -&gt; root -&gt; rootless -&gt; lessens -&gt; sensationalist -&gt; listens -&gt; tense -&gt; ensembles -&gt; bless -&gt; less -&gt; lessees -&gt; seesawing -&gt; wing -&gt; wingspread -&gt; readership -&gt; shipload -&gt; load -&gt; loadstar -&gt; star -&gt; starch -&gt; archdeacons -&gt; consign -&gt; sign -&gt; signpost -&gt; postmark -&gt; markdown -&gt; down -&gt; downwind -&gt; windmill -&gt; millipede -&gt; pedestal -&gt; stalwart -&gt; wartime -&gt; timekeepers -&gt; perseverance -&gt; ancestry -&gt; strychnine -&gt; nineteenth -&gt; enthusiastic -&gt; stickleback -&gt; backfire -&gt; firetrap -&gt; trapdoor -&gt; doorways -&gt; ways -&gt; wayside -&gt; sideshows -&gt; hows -&gt; howsoever -&gt; ever -&gt; everybody -&gt; body -&gt; bodywork -&gt; workbook -&gt; bookkeepers -&gt; persimmons -&gt; monstrous -&gt; roustabout -&gt; bout -&gt; bouts -&gt; outs -&gt; outstrip -&gt; trip -&gt; tripled -&gt; pled -&gt; pledge -&gt; edgewise -&gt; wise -&gt; wiseacres -&gt; crescent -&gt; cent -&gt; centigram -&gt; gramophone -&gt; honeymoon -&gt; moonstruck -&gt; rucksack -&gt; sackfuls -&gt; fulsome -&gt; somewhere -&gt; heretofore -&gt; forestall -&gt; tall -&gt; tallyhoed -&gt; hoedown -&gt; downcast -&gt; cast -&gt; castor -&gt; storybook -&gt; book -&gt; bookcase -&gt; casework -&gt; work -&gt; workshop -&gt; shoplift -&gt; liftoffs -&gt; offs -&gt; offside -&gt; sidewalk -&gt; walkouts -&gt; outspread -&gt; read -&gt; readouts -&gt; outside -&gt; sideline -&gt; linemen -&gt; emend -&gt; mend -&gt; mendicant -&gt; cant -&gt; cantilever -&gt; evermore -&gt; moreover -&gt; overkill -&gt; killdeer -&gt; deerskin -&gt; skin -&gt; skinhead -&gt; headword -&gt; word -&gt; wordplay -&gt; play -&gt; playback -&gt; backstairs -&gt; airsick -&gt; sickbeds -&gt; beds -&gt; bedsore -&gt; sore -&gt; sorehead -&gt; head -&gt; headline -&gt; line -&gt; linefeed -&gt; feed -&gt; feedback -&gt; backside -&gt; sideshow -&gt; showdown -&gt; downhill -&gt; hillock -&gt; lockstep -&gt; stepbrother -&gt; therein -&gt; rein -&gt; reinterpret -&gt; pretend -&gt; tend -&gt; tenderfoot -&gt; foot -&gt; footstep -&gt; stepmother -&gt; therapist -&gt; pistons -&gt; tons -&gt; tonsure -&gt; sure -&gt; surefire -&gt; fireside -&gt; sidekick -&gt; kickback -&gt; backhand -&gt; hand -&gt; handpick -&gt; pickaback -&gt; back -&gt; backstops -&gt; topsoil -&gt; soil -&gt; soils -&gt; oils -&gt; oilskin -&gt; skins -&gt; kinswomen -&gt; omens -&gt; menswear -&gt; wearisome -&gt; somewhat -&gt; what -&gt; whatever -&gt; everywhere -&gt; here -&gt; herewith -&gt; with -&gt; withhold -&gt; hold -&gt; holdover -&gt; overland -&gt; landlady -&gt; ladybird -&gt; bird -&gt; birdbath -&gt; bathes -&gt; thespian -&gt; pianissimi -&gt; simile -&gt; milepost -&gt; posture -&gt; tureen -&gt; reenlist -&gt; listen -&gt; stenches -&gt; chestnuts -&gt; nuts -&gt; nutshell -&gt; hell -&gt; hellion -&gt; lion -&gt; lionizing -&gt; zing -&gt; zinging -&gt; gingersnaps -&gt; naps&lt;/p&gt;&#xA;" OwnerUserId="9192" LastEditorUserId="12700" LastEditDate="2015-08-27T04:29:25.547" LastActivityDate="2015-08-27T04:29:25.547" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20619" PostTypeId="2" ParentId="20225" CreationDate="2015-08-27T08:42:44.127" Score="2" Body="&lt;p&gt;Here are some of my guesses:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;handsy humanoid | atmospheric ambience&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;pri&lt;strong&gt;mate&lt;/strong&gt; / cli&lt;strong&gt;mate&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;headdress holiday | chromosomal cross-up&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;car&lt;/strong&gt;nival / &lt;strong&gt;car&lt;/strong&gt;cinogenic&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;lacking logic | evoking enthusiasm&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;ir&lt;strong&gt;ration&lt;/strong&gt;al / vene&lt;strong&gt;ration&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;multi-tasking mycota | mammals' mark  (not sure about the &quot;multitasking&quot; part...)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;porta&lt;strong&gt;bell&lt;/strong&gt;a / &lt;strong&gt;bell&lt;/strong&gt;ybutton&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;celestial species | bachelor brotherhood&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;ext&lt;strong&gt;rate&lt;/strong&gt;rrestial / f&lt;strong&gt;rate&lt;/strong&gt;rnity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;wowing wizardry | hearty herb&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;bran&lt;strong&gt;dish&lt;/strong&gt; / horsera&lt;strong&gt;dish&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;gun; gum | worthless whistle&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;t&lt;strong&gt;ride&lt;/strong&gt;nt / de&lt;strong&gt;ride&lt;/strong&gt; (probably wrong but oh well)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;supervised span | pulsate periodically&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;hos&lt;strong&gt;pita&lt;/strong&gt;lization / pal&lt;strong&gt;pita&lt;/strong&gt;te&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;replicate ravenously | water ward&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;pro&lt;strong&gt;life&lt;/strong&gt;rate / &lt;strong&gt;life&lt;/strong&gt;saver&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;palpable proximity | person picked&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;im&lt;strong&gt;mine&lt;/strong&gt;nt / no&lt;strong&gt;mine&lt;/strong&gt;e&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15960" LastEditorUserId="15960" LastEditDate="2015-08-27T09:50:46.580" LastActivityDate="2015-08-27T09:50:46.580" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20620" PostTypeId="2" ParentId="20372" CreationDate="2015-08-27T08:55:01.733" Score="20" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Purple, or possibly &quot;The color purple&quot;, which might make a good title.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is a riddle quite long&lt;br&gt;&#xA;I hope you can follow along&lt;br&gt;&#xA;You'll be delighted&lt;br&gt;&#xA;While I recite it:&lt;br&gt;&#xA;I'm known to be unmatched in song&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Unmatched in song: there is &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_English_words_without_rhymes&quot;&gt;no rhyme for purple&lt;/a&gt;, although this &lt;a href=&quot;http://www.rhymezone.com/r/rhyme.cgi?Word=purple&amp;amp;loc=snippettotal&amp;amp;typeofrhyme=exa&quot;&gt;hasn't stopped people trying&lt;/a&gt; (then I'll hurtle like Steve Urkel on a turtle in a circle, the feeling's universal like the hurtful color purple)  : &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'll find me the whole world around&lt;br&gt;&#xA;In nature 'tis true I abound&lt;br&gt;&#xA;You might not expect&lt;br&gt;&#xA;But just recollect:&lt;br&gt;&#xA;In various forms I am found.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are lots of purple plants and birds, e.g. aubergine (eggplant), lavendar, heather, grapes, purple heron, purple finch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a king of my kind&lt;br&gt;&#xA;And yet I have been quite maligned&lt;br&gt;&#xA;Confused for another&lt;br&gt;&#xA;Get this: I'm her brother&lt;br&gt;&#xA;Though our differences are well defined.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Purple is a regal colour. But purple looks quite a bit like violet, which is a female name, so we can call purple her brother&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We're nothing alike, I insist&lt;br&gt;&#xA;Her flowery form I resist&lt;br&gt;&#xA;I'm more aggressive&lt;br&gt;&#xA;She seems more passive&lt;br&gt;&#xA;Yet one step from outburst she is.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Violet is a flower. And a viole(n)t act is an outburst. As @Mythi commented, You can be purple with rage, or a shrinking violet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;She's one of a number, all true&lt;br&gt;&#xA;I'm not because I come from two&lt;br&gt;&#xA;She gets her own line&lt;br&gt;&#xA;I'll never have mine&lt;br&gt;&#xA;You know me now surely, don't you?&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Violet is a color with a single wavelength (a spectral colour). But purple is a mix of Red and Blue, with two distinct wavelengths (a non-spectral color). This means (see comment from @Jon Ericson) that violet is counted as one of the lines of the rainbow, but purple isn't, with &lt;a href=&quot;http://www.esl-lounge.com/songs/song-i-can-sing-a-rainbow.php&quot;&gt;one musical exception&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See me or listen to flyers&lt;br&gt;&#xA;Read me or climb up much higher&lt;br&gt;&#xA;Once solving is done&lt;br&gt;&#xA;Please pardon the pun&lt;br&gt;&#xA;And now I must sleep, for I tire.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can see purple, and could read the book &lt;a href=&quot;https://en.wikipedia.org/wiki/The_Color_Purple&quot;&gt;The color purple&lt;/a&gt;, or alternatively purple prose (comment from @Mythi). The Stone Temple Pilots released an album called Purple, and in this case are the Flyers (comment from @Jon Ericson). The pun is on &quot;tire/Tyre&quot;. The ancient city of &lt;a href=&quot;https://en.wikipedia.org/wiki/Purple&quot;&gt;Tyre was source of a purple dye, known as Tyrian purple&lt;/a&gt;: &quot;As early as the 15th century BC the citizens of Sidon and Tyre, two cities on the coast of Ancient Phoenicia, (present day Lebanon), were producing purple dye from a sea snail called the spiny dye-murex &quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From all but one sense do we hide&lt;br&gt;&#xA;My ghostly sister and I&lt;br&gt;&#xA;She on a petal&lt;br&gt;&#xA;Me on a metal&lt;br&gt;&#xA;I've made many emperors die.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Violet and purple can only be seen, not smelt/touched/tasted/heard. And being an E-M wave is quite ghostly. Petal is again a reference to the flower violet. Metal could refer to Manganese based pigments which can be purple, but the intended meaning was the Purple Heart Medal, awarded to injured US military personnel. Some emperors (e.g. Charlemagne) were buried in purple robes. And there may very well be a pun on die/dye.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7316" LastEditorUserId="7316" LastEditDate="2015-08-27T19:45:04.023" LastActivityDate="2015-08-27T19:45:04.023" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20621" PostTypeId="2" ParentId="20595" CreationDate="2015-08-27T09:32:54.940" Score="12" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; no.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answers that say 'yes' seem to stop after showing how you can make all 4 legs reach the floor. However, not only is it sometimes impossible to make all 4 legs touch the floor (think floors with large troughs, mounds, or waves), even when they do, that condition alone does not mean it can be said the table is &quot;balanced&quot;, &quot;stable&quot;, or &quot;doesn't wobble&quot;. Indeed, those answers accept out of hand that 3 legs of the table can touch the floor nicely, but no guarantee for such a condition is afforded anywhere in the problem statement.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Even a modestly turbulent surface that you or I wouldn't have trouble walking on won't hold a table steady unless most of the surface of each foot contacts the floor. This would be impossible if the floor was relentlessly bumpy, like a gravel road or a rocky coastline. If you want a more sterile example, consider something like &lt;code&gt;z = sin x + cos y&lt;/code&gt;. There are other, gravel-less ways to get this too. One is where the floor is large tiles that are flat but have a variety of slants. More real-life places come to mind. Speaking of real life, the table will not experience infinite friction with the ground (or with objects placed on it), which means a limit needs to be respected on how far off level the table can be before it simply falls over (or objects slide off).  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In conclusion, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; there are many factors that determine how and whether or not a table wobbles on a surface. There are probably more than I realize, since I am not a physicist. As for this question, I do not think &lt;code&gt;a large room with a continuous but uneven floor&lt;/code&gt; comes close to providing context sufficient to rule out the tendency to wobble.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16014" LastEditorUserId="42452" LastEditDate="2018-10-04T21:37:40.113" LastActivityDate="2018-10-04T21:37:40.113" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="20622" PostTypeId="2" ParentId="20581" CreationDate="2015-08-27T09:44:23.617" Score="12" Body="&lt;p&gt;350 and counting :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;highball balladeer deerflys flyspeck pecking kingfish fishable ables bless lessees seesawing wingback backache aches chestfuls fulsome somebody bodybuilding dinging gingerbread readapting tinge ingest gestapos apostasies siesta establisher sherbert berthing hinge ingestant tantalic alice licence enceinte integrate raters tersest sestet stethoscope copenhagen agency encyclic cliches chestier tiering ringbolt bolthead headaches chestnuts nutshell hellbent benthal thalami lamia amias miasmas smash masher sherd herdswomen omens menschen chenille illegal egalite liter iterant ranting tinging gingering ringdove dover overabundance ancestor storables blessedest destain tainting tingling lingering ringing gingersnap snapback backbencher cherish rishis shist histamin amino minoring ringlike likelihood hoodless lessens sensate sateen teener energizing zinging ginghams hamster stereochemistry strychnine ninefold foldboat boatbill billeter eterne ernest nester stereoscope coper opera peradventure tureen reenactment mentalist listen stench enchainment menthol thole holeless lessons sonships hipshot shotguns gunship shipload loadstar starch archaistic stick ticking kingfisher sheriff riffing fingering ringmaster stereoscopy copybook bookcase casebook bookend kendo endomorph orphan phantast taste aster stereotape tapeline lineage eagerest restack tackifies fiesta establishment menthols holster stereotype typebars barstool toolhead headachier hierarch archbishop shope hoper operandi andiron ironbark barkeep keepers persecute cutest testament mentioner onerous rousing singletons tonsorial rialto altogether therapist pistil stilbestrol troll rollback backbenchers herself selfheal healer alertest testee steeling linguistic sticking kingpins pinscher cherisher sherlock locking kingship shipmaster stereotypers persevering ringneck neckband bandanna annalist listens tense ensembles blessing singletree treed reedit edith dither there hereabout boutiques quester sternest nestlike likest kestrel trellising singsong songbird birdbath bathe atheistic stickleback backbone boneblack lackeyed eyedroppers persimmons monseigneur neurogram grammars marseille illegitimate materialist lister stertorous roustabout bouts outscore corelate lateen teenyboppers persist sisterhood hoodwink winking kingside sidearm earmark markdown downcast castaways wayside sideband bandoleer leering ringside sideburn burnouts outscoring ringtail tailbone bonefish fishbone bonehead headband bandwagon agonise niseis seismicity cityward wardens dense enserfing fingernail nailhead headgear gearcase caseload loadstone tones oneself selfhood hoodwinking kingwood woodblock lockouts outsell sellout louts outsets setscrew crewelwork workboat boatswain wainscot scotland landfall fallback backdate dateline linear nearest restaff taffrail railbird birdbrain rainbows bowshot shots hotshots hotspur spurge urgency encyclopedic edict dictaphone honest nestor store toreador adoring ringtoss tosspot spots potshard hardback backdoor doorbell belletristic stickouts outshone honeycomb combatant tantalum alumina minacity citywide widely...&lt;/p&gt;&#xA;" OwnerUserId="16018" LastActivityDate="2015-08-27T09:44:23.617" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20623" PostTypeId="1" AcceptedAnswerId="20625" CreationDate="2015-08-27T11:51:43.587" Score="-21" ViewCount="1174" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Questions Tags&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Users Bags&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;answer Question&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;UnAsk What lags?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Instructions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explain what lags.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You must use &lt;em&gt;all&lt;/em&gt; the clues and explain precisely how you got the answer from them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please do not guess random phrases that might fit. You must explain the clues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is two words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;_ _ _ _ _ _ _ _ /_ _ _ _  (8, 4)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Use upper and lower case appropriately when you answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="14177" LastEditDate="2015-08-27T13:06:04.937" LastActivityDate="2016-08-27T20:35:15.300" Title="I hope you can see what I'm getting at: Questions Tags / Users Bags / answer Question / UnAsk What lags?" Tags="|enigmatic-puzzle|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20624" PostTypeId="2" ParentId="20593" CreationDate="2015-08-27T11:56:09.267" Score="0" Body="&lt;p&gt;I see that there is a similar situation listed here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ruwix.com/the-rubiks-cube/rubiks-cube-patterns-algorithms/&quot; rel=&quot;nofollow&quot;&gt;http://ruwix.com/the-rubiks-cube/rubiks-cube-patterns-algorithms/&lt;/a&gt;&#xA;(Twisted cube in the big cube)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However it's not a match, you have two faces with swapped colors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully this link will assist in your efforts. If nothing else, maybe someone can add some more relevant information. I confess I am no expert on Rubik's cubes (I know how to do the checkerboard thing though!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If you edit this post, be sure to comment so your contributions don't get forgotten!)&lt;/p&gt;&#xA;" OwnerUserId="13068" LastActivityDate="2015-08-27T11:56:09.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20625" PostTypeId="2" ParentId="20623" CreationDate="2015-08-27T12:13:24.503" Score="8" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Puzzling&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The hyperlinks on the right of Puzzling page are lagged to create the 4 clues&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15668" LastEditorUserId="25966" LastEditDate="2016-08-27T20:34:57.650" LastActivityDate="2016-08-27T20:34:57.650" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20627" PostTypeId="2" ParentId="20559" CreationDate="2015-08-27T13:23:39.223" Score="3" Body="&lt;p&gt;&lt;em&gt;Partial answer, just to get the ball rolling and see if I am even on the right lines&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These seem to be related to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Keys on a keyboard&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A bearlike whale&lt;/li&gt;&#xA;&lt;li&gt;Attire of an ancient, mysterious sorcerer&lt;/li&gt;&#xA;&lt;li&gt;&quot;Cool pointer!&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DOWN ARROW (or could be up arrow)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A blighted sort of metal&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ALT-EM?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;One who expeditions through deep time&lt;/li&gt;&#xA;&lt;li&gt;Keeps using the same data&lt;/li&gt;&#xA;&lt;li&gt;Providing someone with assets, maybe&lt;/li&gt;&#xA;&lt;li&gt;Destroyer of subatomic particles&lt;/li&gt;&#xA;&lt;li&gt;Wrath of the mailman&lt;/li&gt;&#xA;&lt;li&gt;Once again, the mongrels choose to partake&lt;/li&gt;&#xA;&lt;li&gt;Doctors act like pigs&lt;/li&gt;&#xA;&lt;li&gt;The Lord of the Sound&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; VOLUME control&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Weeps over grains&lt;/li&gt;&#xA;&lt;li&gt;One who takes good care of his money?&lt;/li&gt;&#xA;&lt;li&gt;The companions we have&lt;/li&gt;&#xA;&lt;li&gt;Deface a rotting body&lt;/li&gt;&#xA;&lt;li&gt;An indecent vegetable&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CARET (a perversion of the word CARROT)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;One who wets the bed&lt;/li&gt;&#xA;&lt;li&gt;Scampers all over&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DASHes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;An unpleasant idiot&lt;/li&gt;&#xA;&lt;li&gt;Intriguing symbols on an intriguing object&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; QUESTION MARKS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The more voluptuous brother&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; GREATER THAN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Pots of snails&lt;/li&gt;&#xA;&lt;li&gt;An old decision point&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; no RETURN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Sounds like a spicy dish&lt;/li&gt;&#xA;&lt;li&gt;An area where certain clouds might form&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SPACE (molecular clouds, also known as stellar nurseries form in space)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-08-27T13:23:39.223" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20628" PostTypeId="1" AcceptedAnswerId="20632" CreationDate="2015-08-27T13:50:13.880" Score="13" ViewCount="1539" Body="&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/282/hats-and-aliens&quot;&gt;After failing to destroy the human race,&lt;/a&gt; Aliens have returned to make another attempt. This time they have planned a wicked strategy. They have again abducted $N$ people, but one of these people is actually an alien spy disguised as human. Each person (including the spy) represents equal portions of the human race.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The aliens give each abductee either a black or a white hat. The people are lined up in a single line, each facing forward, such that each person sees the hats of all persons in front of him. The aliens then proceed, starting from the last person, to ask each of the abductees what the color of their hat is. If they guess incorrectly, the human population they represent is killed. The others don't learn of this incorrect guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The abductees are given a chance to develop a strategy before they are lined up. However, whatever strategy they plan, it is known to the alien spy (and hence the aliens). The spy's task is to ensure as many deaths as possible. The aliens, knowing the strategy can rearrange the line any way they like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the humans have figured out that one of them is a spy, but they do not know which one. What strategy should they utilize to minimize the number of deaths?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Alternate formulation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N$ people are standing in a line. Each of them wears a black or white hat and can see the hats of all people in front. Starting with the last person, they guess the color of their hat. They want to maximize the number of correct guesses. However, they know that there is one person, whose identity is unknown, who wishes the opposite. What strategy should they follow to maximize the number of correct guesses in the worst case?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Notes&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The humans cannot communicate in any way once they are lined up.  &lt;/li&gt;&#xA;&lt;li&gt;The Aliens are serious this time; Do any thing funny and they will blow the whole earth.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="15544" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2018-05-23T18:09:25.953" Title="Hat-guessing with a spy" Tags="|logical-deduction|hat-guessing|" AnswerCount="1" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="20630" PostTypeId="2" ParentId="20597" CreationDate="2015-08-27T14:10:33.587" Score="22" Body="&lt;p&gt;What is this poem about?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; counting from 1 to 8&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He was victorious, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; he won &gt; ONE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;say you, in a foreign tongue.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;you&quot; in French is &quot;tu&quot; &gt; TWO&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He was almost there, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;almost&quot; means make an anagram of &quot;there&quot; &gt; THREE  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;he's a ring in the skin&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a ring is O, skin is FUR &gt; FOUR  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A half-fisted stove; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; half of fisted is &quot;fis&quot;; stove is to be read as a direction: &quot;s&quot; to &quot;ve&quot; meaning take &quot;fis&quot; and change the &quot;s&quot; to &quot;ve&quot; &gt; FIVE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;a snake by the nines.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a snake is S, nine is IX &gt; SIX&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Christmas Eve now holds&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;holds&quot; means &quot;contains&quot; &gt; Christma&lt;strong&gt;s Eve n&lt;/strong&gt;ow &gt; SEVEN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;their endless odd gin&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;their&quot; endless is &quot;thei&quot;; odd means to make an anagram; &quot;gin&quot; again is interpreted as a direction - &quot;G in&quot;, meaning to put a G into &quot;eiht&quot; &gt; EIGHT&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4385" LastEditorUserId="4385" LastEditDate="2015-08-27T17:11:00.403" LastActivityDate="2015-08-27T17:11:00.403" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="20631" PostTypeId="1" AcceptedAnswerId="20639" CreationDate="2015-08-27T14:39:26.433" Score="14" ViewCount="1185" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The previous aliens have departed from Earth. Another group has arrived from the vicinity of Alpha Centauri or Sirius (it's not important which) to probe the human psyche.&lt;br&gt;&#xA;  Help humanity to score a perfect 10 by determining the organisational principle involved in the following sequence of figures, and hence the missing figure.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2uDuJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2uDuJ.png&quot; alt=&quot;Sequence&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But wait, that's not all ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There are a number of clues embedded in the problem description. You will either find them, or you won't.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14080" LastEditorUserId="30903" LastEditDate="2017-04-16T20:24:56.883" LastActivityDate="2017-04-16T20:24:56.883" Title="Aliens are continuing to probe the human psyche. Help humanity to score a perfect 10 by finding the missing figure" Tags="|pattern|visual|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20632" PostTypeId="2" ParentId="20628" CreationDate="2015-08-27T14:41:11.507" Score="9" Body="&lt;p&gt;To start:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Persons 1 and 2 count the number of white hats from person 3 to person N. They say 'white' if it's an even number, or 'black' if it is not. If they agree, proceed to the next step.&lt;br&gt;&#xA;If not, one of them is the spy, and person 3 counts the number of white hats from person 4 to person N, again saying 'white' if even and 'black' if odd, then proceed. Thanks to @Arkku for that!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Starting from person 3 (or person 4, if 1 and 2 gave different answers), each person keeps track of how many people have said 'white', and uses this and the knowledge of the initial parity (I think that's the right term) of white hats to call out their own. The spy will inevitably call out the opposite, causing the next-in-line to also call out the incorrect colour. The system is parity-based, however, so this will put the group as a whole back on track (see example)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Total losses:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Two people in the line (the spy and the next person), and the 2 people at the beginning (since the aliens control the layout of the group), for a total of &lt;strong&gt;4 incorrect guesses&lt;/strong&gt;.&lt;br&gt;&#xA;&lt;strong&gt;The spy won't be one of the first two people&lt;/strong&gt; - if they are, a maximum of 3 people will call out incorrectly, if not, it will be 4.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Example:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;X,X|W,B(spy),B,W&lt;/strong&gt; - the first two don't matter, so are labelled as X. Each has a 50% chance of survival (or not, since the aliens will arrange them such that they all die, sadly)&lt;br&gt;&#xA;1 and 2 have both said 'white'. Everyone knows that neither of them is the spy&lt;br&gt;&#xA;Person 3 can see one white hat and knows there are an even number, so calls out 'white'. &lt;strong&gt;3 lives.&lt;/strong&gt; Persons 4,5 and 6 know that an odd number of white hats remain.&lt;br&gt;&#xA;4 is a spy, and knows from this that they have a black hat. They call 'white', dooming the region of the earth that they represent. &lt;strong&gt;4 dies&lt;/strong&gt;&lt;br&gt;&#xA;5 knows that there were an even number of white hats to start, that two have been called out, and can see one in front of them. They &lt;em&gt;must&lt;/em&gt; also be white, they think, and thus sentence themselves and 1/6th of the population to death. &lt;strong&gt;5 dies&lt;/strong&gt;&lt;br&gt;&#xA;6 also knows that there were an even number of white hats to start, and that three have been called. Therefore they must be white, to preserve parity. &lt;strong&gt;6 lives&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Same example, no spy:   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;4 knows that they have a black hat. They call 'black' &lt;strong&gt;4 lives&lt;/strong&gt;&lt;br&gt;&#xA;5 knows that there were an even number of white hats to start, that one has been called out, and can see one in front of them. They call black &lt;strong&gt;5 lives&lt;/strong&gt;&lt;br&gt;&#xA;6 also knows that there were an even number of white hats to start, and that three have been called. Therefore they must be white, to preserve parity. &lt;strong&gt;6 lives&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Either way, the spy only affects the outcome for themselves and the person after them in the line&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This only works for &lt;strong&gt;N &gt; 4&lt;/strong&gt; (though for N&amp;lt;8, random guessing will yield better results on average)&lt;/p&gt;&#xA;" OwnerUserId="11717" LastEditorUserId="11717" LastEditDate="2015-08-28T10:02:13.483" LastActivityDate="2015-08-28T10:02:13.483" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20633" PostTypeId="2" ParentId="20500" CreationDate="2015-08-27T15:06:35.583" Score="4" Body="&lt;h1&gt;Base Question&lt;/h1&gt;&#xA;&#xA;&lt;h2&gt;Answer:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the Rook wins&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Strategy:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In general a good winning strategy for the Rook is to cut the chess board horizontally above the Bishop, leaving him trapped in an area with few moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be more specific: to win the Rook should try to reach &lt;strong&gt;H7&lt;/strong&gt; as soon as possible, if the Bishop is already in &lt;strong&gt;G7&lt;/strong&gt; the Rook should move to &lt;strong&gt;H5&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Proof:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If the Rook manages to reach row 7 before the Bishop, it automatically wins. Refer to the image below (I will mark in black the squares already moved across):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7lZeO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7lZeO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the Rook is in &lt;strong&gt;H7&lt;/strong&gt; the most convenient position for the Bishop is to be in &lt;strong&gt;F6&lt;/strong&gt; and to have the next move. The possible movement for the Bishop are marked in yellow (the red marks indicate where he can't go to avoid being caught by the Rook on the Rook's next move), but both &quot;yellow moves&quot; leads to a situation where the Bishop is left with less moves than the rook. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;going to the upper yellow mark leads to &lt;em&gt;7 moves&lt;/em&gt; left for the Bishop and &lt;em&gt;24 moves&lt;/em&gt; left for the Rook&lt;/li&gt;&#xA;&lt;li&gt;going to the lower yellow mark leads to &lt;em&gt;6 moves&lt;/em&gt; left for the Bishop and &lt;em&gt;28 moves&lt;/em&gt; left for the Rook&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;To avoid the Rook going in &lt;strong&gt;H7&lt;/strong&gt;, the Bishop should go in &lt;strong&gt;G7&lt;/strong&gt; on the first move, refer to the image below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8YKaJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8YKaJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in this case the optimal move for the Rook is to go on &lt;strong&gt;H5&lt;/strong&gt;, (he should not go in &lt;strong&gt;H7&lt;/strong&gt; to avoid being trapped with &lt;em&gt;0 moves&lt;/em&gt; left).&#xA;Now the Bishop can only move in the yellow mark, this leads to &lt;em&gt;9 moves&lt;/em&gt; left for the Bishop and &lt;em&gt;20 moves&lt;/em&gt; left for the Rook.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: In 2nd scenario the optimal move for the Rook is in H2, it leaves more moves than H5 (23 moves instead of 20). Thanks @Jason.&lt;/p&gt;&#xA;" OwnerUserId="15823" LastEditorUserId="15823" LastEditDate="2015-09-01T12:35:06.127" LastActivityDate="2015-09-01T12:35:06.127" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20634" PostTypeId="2" ParentId="20595" CreationDate="2015-08-27T15:21:34.087" Score="1" Body="&lt;p&gt;I don't think you can solve this question infallibly unless you approach it using a theoretical approach specifically in multivariable calculus. If you reword your question into a mathematically approachable model, then the question goes:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let $m &amp;gt; 0$ be the greatest slope that is acceptable for a &quot;realistic floor.&quot; Let $l$ and $w$ be the length and width of the chair.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $S$ to be the set of &lt;strong&gt;all&lt;/strong&gt; continuous surfaces such that for each surface $S_i$, $|{dz \over {dx}}|,|{dz \over {dy}}| &amp;lt; m$ at every point in $S_i$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define the set $P_{A,B}$ be all the possible contact points (x,y,z) between a continuous surface $A$ and a plane $B$, both of which are in $\mathbb R^3$ (the 3rd dimension).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does $S$ have the property that there always exists a plane $B_i$ such that $P_{S_i, B_i}$ contains 4 contact points that form a rectangle of dimensions $l, w$ when connected?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I simply don't have the knowledge to derive and justify the result, and I don't think that kind of infallible solution would be easy to get here; math.stackexchange would be more appropriate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have only posed more questions so far, so to make up for this (kind of), I will hypothesize that the answer to the question is no; if I assume that $S$ does have said property for dimensions $l,w$, then the same property must be applicable for really large dimensions and really small dimensions (e.g. where $l,w \in (0, \infty)$ ), and I simply find that unfeasible.&lt;/p&gt;&#xA;" OwnerUserId="15995" LastEditorUserId="15995" LastEditDate="2015-08-27T15:48:07.330" LastActivityDate="2015-08-27T15:48:07.330" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20636" PostTypeId="1" AcceptedAnswerId="20723" CreationDate="2015-08-27T16:56:51.827" Score="2" ViewCount="2357" Body="&lt;p&gt;Go ahead: Turn me upside-down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But others will soon realize your error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Turn me back up, and the evidence is gone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't blame me; It was &lt;em&gt;your&lt;/em&gt; mistake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just give and, sometimes, take.&lt;/p&gt;&#xA;" OwnerUserId="16025" LastEditorUserId="4050" LastEditDate="2015-08-27T17:05:08.063" LastActivityDate="2015-08-30T04:44:18.880" Title="A quick riddle: what am I? Go ahead: Turn me upside-down" Tags="|riddle|" AnswerCount="9" CommentCount="2" FavoriteCount="0" ClosedDate="2015-09-05T04:31:16.370" ContentLicense="CC BY-SA 3.0" />
  <row Id="20638" PostTypeId="2" ParentId="20636" CreationDate="2015-08-27T17:34:54.847" Score="1" Body="&lt;p&gt;My guess would be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Go ahead: Turn me upside-down.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Symbolism for metaphorically turning an hourglass upside-down, to regain time or relive moments&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But others will soon realize your error.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most ambiguous line for me, perhaps indicates others notice the mistake you were trying to fix, or others see you are wasting your life trying to turn back time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Turn me back up, and the evidence is gone.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You've wasted time trying to turn back time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Don't blame me; It was your mistake.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Placing blame on time for going too quickly or not being able to revert decisions you made in the past&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I just give and, sometimes, take.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All living things die from old age&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;FYI:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; By 'turn back time', I do not mean literally. Think of that person you knew in high school that was very popular, but after school reality hit and they still haven't gotten over it, and whenever you see them they bring up moments that happened many years ago - (stuck in time?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15990" LastActivityDate="2015-08-27T17:34:54.847" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20639" PostTypeId="2" ParentId="20631" CreationDate="2015-08-27T17:39:23.853" Score="17" Body="&lt;p&gt;On this grid of digits:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;1 2 3&#xA;4 5 6&#xA;7 8 9&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;These layouts represent the digits of the powers of 2, starting at 512 and going down. The missing one is 64, which is the left and right squares in the middle row.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are only 10 possible layouts (1,2,... 512) because 1024 contains a 0, which doesn't have a spot on the grid.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-08-27T17:39:23.853" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20640" PostTypeId="1" AcceptedAnswerId="20647" CreationDate="2015-08-27T17:52:51.883" Score="2" ViewCount="9343" Body="&lt;p&gt;Open Challenge: What is the &lt;strong&gt;longest complete sentence&lt;/strong&gt; you can make wherein every word is a homophone? A simple example would be:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I threw you. (eye through yew)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ground Rules&lt;/strong&gt; (more may be added if needed):  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Words must be English words present on dictionary.com. This goes for both the one used in the sentence and it's homophonic partner(s).  &lt;/li&gt;&#xA;&lt;li&gt;The sentence must be a grammatically complete, comprehensible English sentence (any included quotes must also be comprehensible).  &lt;/li&gt;&#xA;&lt;li&gt;Words may not be repeated, and only one member from any homophonic group is allowed (eg 'I threw your eye' is disallowed) otherwise it would be chaos (e.g. 'I write: &quot;I write: &quot;&quot;I write: ....').  &lt;/li&gt;&#xA;&lt;li&gt;Punctuation is allowed.&lt;br&gt;&#xA;4.a No lists! (You didn't think I'd miss that did you?)&lt;br&gt;&#xA;4.b See Rule 2's note on quotations   &lt;/li&gt;&#xA;&lt;li&gt;No proper nouns (either in the sentence or used to make a homophonic pair)  &lt;/li&gt;&#xA;&lt;li&gt;For verification (and likely hilarity), include in your post the homophonic conjugate/complement of your sentence.  &lt;/li&gt;&#xA;&lt;li&gt;The sentence with the largest word count wins. If a tie, the sentence with the largest letter (read: not character) count wins.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="13970" LastEditDate="2015-08-27T18:01:21.993" LastActivityDate="2015-08-28T14:11:28.977" Title="Longest Homophone Sentence" Tags="|english|open-ended|" AnswerCount="5" CommentCount="4" ClosedDate="2015-08-28T05:13:48.153" ContentLicense="CC BY-SA 3.0" />
  <row Id="20643" PostTypeId="2" ParentId="20640" CreationDate="2015-08-27T18:14:29.730" Score="3" Body="&lt;p&gt;17 words to get things started&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I see you have to wait in here one more week for bread dough or new meat.  (eye sea yew half two weight inn hear won moor weak four bred doe oar knew meet)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="14135" LastEditDate="2015-08-27T18:23:49.563" LastActivityDate="2015-08-27T18:23:49.563" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20644" PostTypeId="2" ParentId="20640" CreationDate="2015-08-27T18:20:27.667" Score="5" Body="&lt;p&gt;&lt;strong&gt;33 words, 134 letters&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It rhymes :-) &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I stayed as lead for six whole slow weeks,&lt;br&gt;&#xA;  sighed seeing some plain foul leaks;&lt;br&gt;&#xA;  we have more great sales links by mail due,&lt;br&gt;&#xA;  but lessened our high loan weight loads too.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;eye, staid, ass, lid, four, sicks, hole, sloe, weaks, side, sing, sum, plane, fowl, leeks, wee half, moor, grate, sells/sails, lynx, buy, male, dew, butt, lessoned, hour, hi, lone waits lodes two&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="15544" LastEditorUserId="15544" LastEditDate="2015-08-28T08:34:32.593" LastActivityDate="2015-08-28T08:34:32.593" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20646" PostTypeId="2" ParentId="20640" CreationDate="2015-08-27T18:27:30.107" Score="0" Body="&lt;p&gt;35 words&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I taught him writing through eight straight days but father knew there was no right way not to wait for six more week nights, so we queued by a cheap cannon pair in our house. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;sounds like&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Eye taut hymn rioting threw ate strait daze butt farther new their wars know write whey nought to weight four sicks mow weak knights, sow wee cued buy uh cheep canon pare inn hour how's. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-08-27T18:54:56.780" LastActivityDate="2015-08-27T18:54:56.780" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20647" PostTypeId="2" ParentId="20640" CreationDate="2015-08-27T18:31:17.770" Score="10" Body="&lt;p&gt;98 Words, and a bit theatrical in form:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You, whose sick, dying horse weakly neighs through great pain; lo, I heard our chaste, poor-bred aunts rode foaled, male deer, wanting feet meat banned for sale, loot, or profit some few days past, but died in one cannon-weighted, taut-cord bait boat which flew overseas, would steer all ways, except when night's fair weather made a base course, led to foul mist, high waves, pale, barren beach shore hued red, so wrecked, holed, sinking, their craft bearing dual loads knew no more time; do hear me: steel your mind, pray; higher we raise souls, mourning.&lt;br&gt;&#xA;  &lt;br&gt; &lt;br&gt;&#xA; ewe who's sic, dyeing hoarse weekly nays threw grate pane; low, eye herd hour chased, pore/pour bread ants road fold mail dear, wonting feat meet band four/fore sail, lute, oar prophet sum phew daze past, butt dyed inn won canon waited taught cored bate bote witch flew oversees, would stere awl weighs, accept win knights fare whether maid uh bass coarse, lead two/too fowl missed, hi waives, pail, baron beech sure hewed read, sew recked, hold, syncing, their kraft baring duel lodes new know moor thyme; due here mi: steal yore mined, prey, hire wee rays soles, morning.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The obsession is &lt;strong&gt;&lt;em&gt;over&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="11452" LastEditorUserId="11452" LastEditDate="2015-08-28T14:11:28.977" LastActivityDate="2015-08-28T14:11:28.977" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20648" PostTypeId="2" ParentId="20640" CreationDate="2015-08-27T18:47:18.523" Score="1" Body="&lt;p&gt;29 words:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I pray to meet dear maid one night, or see her fair sight, laid in some beach by wine but no pie, mourning for me where none might hear. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;sounds like&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;eye prey two meat deer made won knight, ore sea hair fare site, lade inn sum beech buy whine butt know pi, morning four mi wear nun mite here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15913" LastActivityDate="2015-08-27T18:47:18.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20650" PostTypeId="2" ParentId="20581" CreationDate="2015-08-27T20:11:56.827" Score="6" Body="&lt;p&gt;(This answer previously had a solution probably found by computer search)&lt;/p&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="9718" LastEditDate="2021-07-12T21:26:48.797" LastActivityDate="2021-07-12T21:26:48.797" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="20651" PostTypeId="1" AcceptedAnswerId="20653" CreationDate="2015-08-27T20:28:56.277" Score="46" ViewCount="6110" Body="&lt;h2&gt;Preface&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;After your last mishap with a &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18448/murdered-down-the-rabbit-hole-a-long-logic-puzzle&quot;&gt;heap full of rabbits&lt;/a&gt;, you got sick of being lied to. Isn't life just so much easier when you can conclusively tell if someone's been telling the truth or not? For that very reason, you went out and bought a &lt;strong&gt;lie detector&lt;/strong&gt;. It works, too! Now, when interrogating suspects, it's as easy as asking the right questions, and getting the answers you need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, no silly rabbits to hide behind this time. An honest-to-god human has been murdered, and it's up to you to figure out, as they say, &quot;whodunit&quot;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Backstory&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This town is small. &lt;em&gt;Real&lt;/em&gt; small. Not a lot going on here, at all. You wander up to the sheriff's office and slam open the doors. The sheriff looks up at you from his desk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Detective. It's about time. I've been waitin' for ya.&quot; He shuffles up some papers on his desk and slides them to the side. &quot;This one's a big one. You should sit.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally, you're belligerent whenever you get the chance, but this time your feet hurt, so you gladly sit in the chair across from him. Continuing on, he tells you, &quot;Hugh G. Pokitz has been murdered.&quot; He pauses, waiting for a reaction, then shrugs. &quot;Right, I forgot you don't live here. Pokitz is the beneficiary of this entire town. He's the reason any of this exists. He's -- well, he &lt;em&gt;was&lt;/em&gt; -- the richest man this side of the Poughipasaw.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You nod in understanding. &quot;Tell me the facts.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sheriff begins to talk, but you pause him. Reaching into your briefcase, you pull out a small metal device and strap it onto his arm. He looks at you, puzzled. &quot;It's a lie detector,&quot; you explain. &quot;Just gotta make sure the thing is calibrated.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;He nods back, then continues. &quot;The doctor did a preliminary autopsy and discovered foxglove in Pokitz's system. It's been determined that the foxglove was the cause of his death. Here's the thing about this particular strain of foxglove -- it wilts within two days of not being stored properly, which the mansion does not have the means to do. It also has to be administered orally, or else it won't kill. So the poison got into the mansion within the last two days, and Pokitz somehow got tricked into eating it.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Who would have motive?&quot; You ask, dreading the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sheriff sighs. &quot;Basically everyone. No one liked him. He abused his employees, extorted the shopkeepers, and catcalled all the women. It's no secret that his wife, Mai, was only with him for his money. And he liked it that way.&quot; With a pause, he muses, &quot;A better question would be, who &lt;em&gt;didn't&lt;/em&gt; have motive?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You chuckle, then remember your here on business and steel yourself. &quot;So, to recap:&quot;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;TL;DR&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The poison made it into the mansion &lt;em&gt;within the last two days.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;The poison must have been &lt;em&gt;administered orally.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Basically &lt;em&gt;everyone is a suspect&lt;/em&gt;, since he was hated by all.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You finish jotting down your notes. The sheriff sees you close your notebook and makes an offer. &quot;I've brought in the, as I can see it, only possible suspects -- would you like to meet them?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You grin a devilish grin. &quot;Oh, I'd &lt;em&gt;love&lt;/em&gt; to.&quot;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Interrogations&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Snatching your lie detector, you're escorted down the hall to a holding cell. &quot;I traced the murder out in my head,&quot; the sheriff said. &quot;The poison must have been obtained from the apothecary. She's the only one anywhere near here who would have access to it. It was either bought in the store, or delivered in the mail, and I'm not sure which. Hopefully you and your lie detector can trace the poison straight from the apothecary to the victim's mouth.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You stop in front of a glass window. Behind the window sit eight people. The sheriff details them for you. &quot;From left to right, we have: &lt;strong&gt;Mai Pokitz&lt;/strong&gt;, the &lt;strong&gt;Pokitz's butler&lt;/strong&gt;, the &lt;strong&gt;Pokitz's chef&lt;/strong&gt;, the &lt;strong&gt;apothecary&lt;/strong&gt;, the &lt;strong&gt;apothecary's apprentice&lt;/strong&gt;, the &lt;strong&gt;postmaster&lt;/strong&gt;, the &lt;strong&gt;Pokitz Mansion gate guard&lt;/strong&gt;, and the &lt;strong&gt;Pokitz's maid.&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He walks back to the door you came through. &quot;I'll leave you to it.&quot; You smile and walk through the door to the holding cell.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suspect: Mai Pokitz&#xA;  &lt;hr /&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;Why aren't you crying?&quot; You sneer at Mai as you approach. She shrugs. &quot;I hated that guy, and I'm still rich. So 'boo-hoo', I guess, but it's no sweat off my back.&quot; She grins. &quot;I didn't kill him, though.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Slapping the lie detector onto her wrist, you growl, &quot;Say that again.&quot; She winces. &quot;Ow! What is that?&quot; You smile. &quot;It's a lie detector! Now I'll know for sure that everything you're saying is true.&quot; She scowls, then repeats herself. &quot;Okay, fine. &lt;strong&gt;I didn't kill my husband.&lt;/strong&gt;&quot; You eagerly wait for the lie detector to go off, but it doesn't.&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You scowl. &quot;Fine. Who had access to his food between its preparation and its consumption?&quot; She ponders for a moment. &quot;Um...&lt;strong&gt;only the chef and the butler&lt;/strong&gt; had access to his food. No one else could have touched it.&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Sounds like you've narrowed it down. &quot;Thank you for your time.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suspect: Pokitz's Chef&#xA;  &lt;hr /&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;I've got it all figured out, chef,&quot; you say as you attach the lie detector to his wrist. &quot;You went out and bought the poison from the apothecary. Then you came back, and you poisoned Hugh's meal, because he mistreated you. Admit it, and I'll read you your rights.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;The chef looks shocked. &quot;No, sir! It's true, I've been mistreated, but due to that, &lt;strong&gt;I haven't left the mansion in a week!&lt;/strong&gt; I haven't even been out of the kitchen, except to relieve myself.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;em&gt;Well, there goes that idea,&lt;/em&gt; you think. &quot;Fine. Did you see the butler do anything odd with the food before he served it?&quot; The chef responded, &quot;No, not at all. He simply &lt;strong&gt;took the tray as he always does&lt;/strong&gt; and carried it out of the kitchen.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Looks like the butler is your murderer. &quot;Thank you for your time.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suspect: Pokitz's Butler&#xA;  &lt;hr /&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;Okay, butler, the gambit's up.&quot; You attach the lie detector to his wrist. &quot;I know you poisoned Hugh's food between the kitchen and the dining room. He abused you, and you were sick of it. Admit it, you filth.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;The butler simply sighs. &quot;No, monsieur, I surely didn't. In fact, I &lt;strong&gt;didn't touch anything but the tray&lt;/strong&gt; between the kitchen and the dining room.&quot; You scowl at the lie detector, but he's telling the truth.&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;Did anything seem suspicious to you?&quot; You inquire. The butler shakes his head. &quot;No, monsieur. In fact, I &lt;strong&gt;taste the meal for poison&lt;/strong&gt; every night, as I did last night, so if the meal was poisoned, surely I would be dead now too.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Back to square one. &quot;Thank you for your time.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suspect: Apothecary&#xA;  &lt;hr /&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Time to at least figure out where the poison came from. With lie detector attached, you accuse the apothecary. &quot;You sold the poison to this woman, didn't you? Shame on you, facilitating a murder like that.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;How DARE you accuse me!&quot; The apothecary jabs back. &quot;No, I did not sell poison to this woman. In fact, &lt;strong&gt;I've never once before made a transaction with Mai Pokitz, nor anyone else in this holding cell&lt;/strong&gt;. And that includes gifts and donations of any kind.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You sigh. &quot;Fine then. Who else could have accessed your store?&quot; The apothecary jingles her keys. &quot;These are the keys to my store. Only &lt;strong&gt;myself and my apprentice&lt;/strong&gt; have copies.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Looks like you've got an apprentice to interrogate. &quot;Thank you for your time.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suspect: Apothecary's Apprentice&#xA;  &lt;hr /&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You circle your next suspect, as he shakes nervously in front of you. You gingerly attach the lie detector to his wrist. &quot;Something wrong, boy? You nervous?&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;N-no,&quot; he stammers. The lie detector lets out a shrill beep. &quot;Good to know that still works,&quot; you say. &quot;Don't worry, boy. If you aren't guilty, I'm sure everything's going to go fine for you.&quot; He gulps.&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;Now admit it,&quot; you continue. &quot;You took the foxglove from the cupboard and sold it to this woman, didn't you?&quot; He shakes his head, and stammers, &quot;Only &lt;strong&gt;my boss has keys to the cabinet&lt;/strong&gt; where the foxglove is kept. She says it's too dangerous for me.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You frown. Another dead end. &quot;Thank you for your time.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suspect: Postmaster&#xA;  &lt;hr /&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;em&gt;If the package wasn't bought in the store, it must have been delivered&lt;/em&gt;, you reason. Scowling, you snap the lie detector to the postmaster's wrist. &quot;I hear all the mail in the town, comes through you. Do I hear the truth?&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;The postmaster nods. &quot;You do, sir. &lt;strong&gt;I handle all the mail for the entire town&lt;/strong&gt;. If something was mailed, I would have seen it. And it's illegal to mail hazardous substances in this town. &lt;strong&gt;I check all the packages&lt;/strong&gt; myself in case someone is breaking the law.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You grin. Maybe you're finally getting somewhere. &quot;And did you see a package from the apothecary to this mansion?&quot; To your dismay, the postmaster simply shakes his head. &quot;No, sir. There was &lt;strong&gt;no package from the apothecary delivered to the mansion&lt;/strong&gt;.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You groan. &quot;Thank you for your time.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suspect: Pokitz Mansion gate guard&#xA;  &lt;hr /&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You peer into the guard's eyes as you attach the lie detector to his wrist. &quot;Tell me who left and who entered the mansion in the last two days.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;The guard stands firm. &quot;Sir, &lt;strong&gt;no one has left or entered the mansion in the last two days&lt;/strong&gt;. Three days, even. Not since Master Pokitz's last trip to market.&quot; You frown. &lt;em&gt;It must have been delivery then.&lt;/em&gt;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;Tell me, what do you know about the post delivered here?&quot; The guard continues to stand firm. &quot;I &lt;strong&gt;check all of the packages&lt;/strong&gt; delivered to the mansion. It's my duty. &lt;strong&gt;If there was poison, I would have seen it&lt;/strong&gt;.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You slump your shoulders in defeat. &quot;Thank you for your time.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suspect: Pokitz's Maid&#xA;  &lt;hr /&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;You circle around on the maid, and attach the lie detector to her wrist. &quot;I'm told you handle all of the mail once it makes it past the front gate.&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;She nods eagerly. &quot;Yes, sir, I &lt;strong&gt;handle all of the mail&lt;/strong&gt; once it gets within the walls of the mansion.&quot; You sneer. &quot;And did you see anything suspicious over the last couple of days?&quot;&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;She thinks for a moment, then shakes her head. &quot;No, sir. I &lt;strong&gt;didn't see anything that struck me as suspicious&lt;/strong&gt; in the mail over the last three days.&quot; Again, your lie detector stays silent.&lt;br&gt;&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Though you should be frustrated upon reaching another dead end, you instead smile. &quot;Thank you for your time.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;h2&gt;Conclusion&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;You leave the holding cell, and find the sheriff waiting for you on the other side of the door. &quot;See? They're just a dead end,&quot; he admits in defeat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You chuckle. &quot;Perhaps to you, pal. Perhaps to you. You've got a lot of truth-tellers in your town, which is a nice change of pace for me. But if you just listen to what everyone has to say, you'll see the &lt;em&gt;real&lt;/em&gt; truth through the cracks.&quot; The sheriff looks at you, puzzled, but you just beckon for him to follow you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Come on in, and let me show you justice at work. And let me borrow those handcuffs. We're gonna need 'em.&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Who murdered Hugh G. Pokitz?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.923" LastActivityDate="2020-04-01T07:37:40.160" Title="Who poisoned Hugh G. Pokitz? (a murder mystery logic puzzle)" Tags="|logical-deduction|story|mystery|" AnswerCount="5" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="20653" PostTypeId="2" ParentId="20651" CreationDate="2015-08-27T21:32:19.877" Score="34" Body="&lt;p&gt;It was a joint effort between&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; all 8 suspects.&lt;br&gt;&#xA;&lt;br&gt;&#xA; 1. Mai Pokitz at some point in the past approached the others with a scheme for getting rid of her hated husband.  She masterminded the scheme, but wasn't directly involved in the murder, so she could truthfully say &lt;strong&gt;she didn't kill her husband&lt;/strong&gt;.&lt;br&gt;&#xA;&lt;br&gt;&#xA; 2. On the night before the murder, the apothecary &quot;accidentally&quot; left the cabinet unlocked where the foxglove was stored.  When apprentice unlocked the shop in the morning, he packaged it up and mailed it to the mansion, thus &lt;strong&gt;it did not come from the apothecary&lt;/strong&gt; (but rather the apothecary's assistant).&lt;br&gt;&#xA;&lt;br&gt;&#xA; 3. The postmaster delivered the package to the mansion.  He &lt;strong&gt;checked the package&lt;/strong&gt;, as he claimed, knowing it contained poison, but didn't do anything about it.&lt;br&gt;&#xA;&lt;br&gt;&#xA; 4. The gate guard &lt;strong&gt;checked the package&lt;/strong&gt;, as he claimed.  &lt;strong&gt;There was poison, and he saw it&lt;/strong&gt;; he just chose not to do anything about it.&lt;br&gt;&#xA;&lt;br&gt;&#xA; 5. The maid &lt;strong&gt;handled the package&lt;/strong&gt;, but &lt;strong&gt;didn't see anything suspicious&lt;/strong&gt;, because she was expecting the package of poison, being in on the scheme.&lt;br&gt;&#xA;&lt;br&gt;&#xA; 6. The chef received the package, and froze the foxglove into some ice cubes.  Just before the meal went out, he added the ice cubes to Pokitz's drink.&lt;br&gt;&#xA;&lt;br&gt;&#xA; 7. The butler &lt;strong&gt;took the tray as he always does&lt;/strong&gt;, and &lt;strong&gt;touched nothing but the tray&lt;/strong&gt;.  He &lt;strong&gt;tasted the meal for poison&lt;/strong&gt;, but since the ice cubes hadn't yet melted, he did not ingest any of the poison.&lt;br&gt;&#xA;&lt;br&gt;&#xA; 8. Everyone sat back, and waited for the inevitable, and took the time to figure out the careful phrasing they would use when put under a lie detector in order to be able to escape suspicion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="4421" LastEditDate="2015-08-28T15:02:50.243" LastActivityDate="2015-08-28T15:02:50.243" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20654" PostTypeId="5" CreationDate="2015-08-27T22:01:33.587" Score="0" Body="&lt;p&gt;Regional dialects are varied and change along unclear boundaries, particularly in the English language. Wikipedia's &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_dialects_of_the_English_language&quot; rel=&quot;nofollow noreferrer&quot;&gt;list of dialects of English&lt;/a&gt; demonstrates the broad scope of spoken English across the world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This tag is intended to be used when a puzzle's answer requires some level of knowledge about a dialect of a pertinent language. In other words, while many answers are relatively global, &lt;a href=&quot;/questions/20513/no-bake-riddle&quot;&gt;an answer like &quot;cow pie&quot;&lt;/a&gt; may require knowledge of specific regional dialects of English, which this tag categorizes.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorUserId="11110" LastEditDate="2017-06-21T07:59:52.570" LastActivityDate="2017-06-21T07:59:52.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20655" PostTypeId="4" CreationDate="2015-08-27T22:01:33.587" Score="0" Body="Use this tag for when a local or regional dialect of a language is relevant to the solution of a puzzle." OwnerDisplayName="user20" LastEditorDisplayName="user20" LastEditDate="2015-08-27T22:01:33.587" LastActivityDate="2015-08-27T22:01:33.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20656" PostTypeId="2" ParentId="20595" CreationDate="2015-08-28T00:27:12.950" Score="2" Body="&lt;p&gt;I'm not entirely sure if this counts as lateral thinking, but couldn't you just chop off a leg? Looking at the answers the problem seems to be one leg connecting with the floor preventing the opposite leg doing so.&lt;/p&gt;&#xA;" OwnerUserId="16035" LastActivityDate="2015-08-28T00:27:12.950" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20662" PostTypeId="1" AcceptedAnswerId="20664" CreationDate="2015-08-28T05:25:23.050" Score="2" ViewCount="307" Body="&lt;p&gt;In continuation to the previous puzzle: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/20576/can-you-solve-this-bruce-wayne-logic-puzzle&quot;&gt;Puzzle1&lt;/a&gt;. &lt;br&gt;&#xA;This time, someone has kidnapped Mr. Fox. And all they want from Batman is the thing in the secret case (box).&lt;br&gt;&#xA;But Batman doesn't know the complete password required to open the box. &#xA;He knows that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;The password has a seven number key.&lt;/li&gt;&#xA;  &lt;li&gt;1st five numbers are : 14, 82, 56, 130, 62.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;That’s it! A bunch of random numbers, right?&lt;br&gt;&#xA;Batman has nothing to go on with. &lt;br&gt;&#xA;But luckily for Mr. Fox, he once told Alfred that all his passwords have a pattern. &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Help Batman save Mr. Fox. All you have to do is find the last two numbers of the key.&lt;/p&gt;&#xA;" OwnerUserId="1766" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-08-28T13:00:49.797" Title="Can you solve this password mystery? Someone has kidnapped Mr. Fox" Tags="|number-sequence|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20663" PostTypeId="1" AcceptedAnswerId="20666" CreationDate="2015-08-28T05:25:26.023" Score="4" ViewCount="649" Body="&lt;p&gt;This is a simple mathematical puzzle, which I decided to improve a bit one year after posting. Some of the answers below consider slightly different, but equivalent setting of the problem.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Rapunzel and the prince woke up next to each other and found&#xA;  themselves in surrounded by complete darkness.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;“Where are we?” asked the prince.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;“It seems that my evil stepmother has left us in the middle of the&#xA;  Room of Despair,” replied Rapunzel.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;“I don’t like this name. What do you know about the room?” inquired&#xA;  the prince.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;“I know that it is square-shaped, and one of its walls has a little&#xA;  door in the middle. In order to escape, we should leave through that&#xA;  door, but we must be careful. There are werewolves tied to each corner&#xA;  of the room, and if any of us gets in their reach, will be ripped&#xA;  apart.” explained Rapunzel.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;“I don't hear any werewolves around. And if I meet one, I can always&#xA;  slay it with my sword.” said the prince.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;“Werewolves are mysterious creatures - very silent, very quick, and&#xA;  very strong. I know they are here, they are waiting, and we can't&#xA;  defeat them.” replied Rapunzel.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;“OK then, since we don’t have anything else to do, let’s just pick a&#xA;  random direction and hope for the best.” suggested the Prince.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;“That’s too much of a risk,” opposed Rapunzel, &quot;It is said that one&#xA;  out of three people who dashes blindly in one direction, gets ripped by a werewolf even before reaching the walls of the room. And I have a better idea anyway.”&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;What is the idea of Rapunzel?  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;Remark: The chains the werewolves are tied with have the same length.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pGZaj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pGZaj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14140" LastEditorUserId="14140" LastEditDate="2016-09-29T18:09:57.923" LastActivityDate="2020-03-26T10:10:51.620" Title="Rapunzel and the Prince" Tags="|mathematics|geometry|story|strategy|" AnswerCount="4" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="20664" PostTypeId="2" ParentId="20662" CreationDate="2015-08-28T05:29:12.523" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 14-&gt;41-&gt;82, 82-&gt;28-&gt;56. &lt;br&gt;&#xA; Take the first two digits of a number, flip them and multiply the resulting number by 2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14140" LastEditorUserId="4385" LastEditDate="2015-08-28T12:32:57.037" LastActivityDate="2015-08-28T12:32:57.037" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20666" PostTypeId="2" ParentId="20663" CreationDate="2015-08-28T06:27:13.703" Score="7" Body="&lt;p&gt;If they know the room is rectangular and that they are in the center:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Start rolling the rope in a spiral (it should result in an &quot;almost&quot; circle). The circle will hit the walls first (before the corners). They just have to check if the door is there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EVEN BETTER&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; F1 is tied to the rope. F2 holds the rope, while F1, starts running around in circles. At every rotation, F2 release a bit more rope, until F1, hits a wall (he should hit it in the center of the said wall). All it's left is for F1 to find the door (one more rotation), and F2 to follow the rope to the door.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="5610" LastEditDate="2015-08-28T06:40:29.510" LastActivityDate="2015-08-28T06:40:29.510" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="20668" PostTypeId="1" AcceptedAnswerId="20670" CreationDate="2015-08-28T09:15:41.343" Score="6" ViewCount="802" Body="&lt;p&gt;Gregory is a farmer and every morning he makes sure to harvest the eggs of his chickens. He has 72 chickens and 12 chickens in each cage, so a total of 6 cages. Each of his chickens lays only one egg per day. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One morning, Gregory suddenly found out that in cage 1, the 4th and 7th chicken did not lay any egg for the day. He then thought that maybe the chickens are stressed. Then the next day, the same happened to 2 chickens in cage 2. The 9th and 12th chickens had no eggs. The next day, only the 6th chicken in cage 3 and the next day the 2nd and 8th chickens in cage 4. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;He then saw a pattern that every day a new cage was having a problem. So, he planned to guard cage 5 the whole night. While guarding that night, he heard movement and rushed to the cage and saw a shadow. He quietly watched the person and saw that they were stealing his eggs. The thief ran away after stealing an egg. Gregory checked and found out that the lost egg was from the 5th chicken.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gregory had enough and planned to set a trap. Can you help Gregory capture the thief? If you will help Gregory, on which chicken(s) will you put the trap?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gregory has a brother named Christian&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cage 5 was ransacked on a Friday night&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;ADDED HINTS:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Born in 16th century, Gregory is very superstitious&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 4&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After investigating, Gregory found out that the thief was Jason from the other town&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16036" LastEditorUserId="10781" LastEditDate="2015-08-28T18:29:23.590" LastActivityDate="2015-08-28T18:29:23.590" Title="Old Gregory had a farm: The Missing Eggs" Tags="|pattern|number-sequence|mystery|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="20669" PostTypeId="2" ParentId="20636" CreationDate="2015-08-28T09:39:06.553" Score="2" Body="&lt;p&gt;Apart from the last line, it sounds like:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;An Etch-A-Sketch  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Go ahead: Turn me upside-down.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Turning an etch-a-sketch upside down erases what's on the pad&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But others will soon realize your error.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Anyone else can see you've made a mistake if you're holding it upside down&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Turn me back up, and the evidence is gone.&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;No proof that anything is wrong now!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Don't blame me; It was your mistake.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You drew something that didn't turn out as you'd have liked&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I just give and, sometimes, take.&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Not sure here, which makes me doubt the whole answer  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second guess (to be elaborated on if correct):  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A magic 8 ball?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11717" LastEditorDisplayName="user20" LastEditDate="2015-08-28T18:21:54.893" LastActivityDate="2015-08-28T18:21:54.893" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20670" PostTypeId="2" ParentId="20668" CreationDate="2015-08-28T10:43:42.510" Score="5" Body="&lt;h2&gt;Short answer&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;It's about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; days of the week on different months of the year.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the 1st and 10th chickens will be targeted in the 6th cage.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Fuller explanation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There are 12 chickens in each cage, corresponding to the 12 months of the year. &quot;Gregory&quot; is a hint to use the Gregorian calendar; &quot;Jason&quot; is a hint to the months (JFMAMJ&lt;strong&gt;JASON&lt;/strong&gt;D); and &quot;Friday night&quot; is a hint to the days of the week.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;On the first day, the 4th and 7th chickens' eggs are stolen; in 2013 (say), April and July began on Mondays.&lt;/li&gt;&#xA;&lt;li&gt;On the second day, the 9th and 12th chickens' eggs are stolen; in 2014 (the next year), September and December began on Mondays.&lt;/li&gt;&#xA;&lt;li&gt;On the third day, the 6th chicken's egg is stolen; in 2015, June began on a Monday.&lt;/li&gt;&#xA;&lt;li&gt;On the fourth day, the 2nd and 8th chickens' eggs are stolen; in 2016, February and August will begin on Mondays.&lt;/li&gt;&#xA;&lt;li&gt;On the fifth day, the 5th chicken's egg is stolen; in 2017, May will begin on a Monday.&lt;/li&gt;&#xA;&lt;li&gt;Finally, January and October are the only months in 2018 which will begin on a Monday, so it's the 1st and 10th chickens the thief will target in cage 6.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thanks to &lt;strong&gt;f''&lt;/strong&gt; and others for help with the solution. And nice puzzle :-)&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;OP's intended explanation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;It's all about Friday the 13th. Cages 1-6 represent the years 2001-2006 while the 12 chickens represent the months. There were Friday the 13ths in April and July 2001, September and December 2002, June 2003, February and August 2004 (leap year, like 2016), May 2005, and January and October 2006.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="5373" LastEditDate="2015-08-28T17:02:55.070" LastActivityDate="2015-08-28T17:02:55.070" CommentCount="19" ContentLicense="CC BY-SA 3.0" />
  <row Id="20671" PostTypeId="2" ParentId="20623" CreationDate="2015-08-28T11:45:34.353" Score="5" Body="&lt;p&gt;Stefano has correctly deduced that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the links at the top of the site (to the right of &quot;Puzzling&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;are used to create the apparently meaningless string of words &quot;Questions Tags / Users Bags / answer Question / UnAsk&quot;, with also the extra letters &lt;strong&gt;d e ed&lt;/strong&gt;. So what we're left over with is the word &quot;deed&quot;; however this isn't in the (8,4) form required for the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's search Puzzling SE for the word &quot;deed&quot;. We get &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=deed&quot;&gt;this list of results&lt;/a&gt;, of which the top hit is &lt;a href=&quot;https://puzzling.stackexchange.com/questions/617/kruskals-hard-to-refold-envelope&quot;&gt;this question&lt;/a&gt;. You'll notice that it was posted by one &lt;strong&gt;shamisen&lt;/strong&gt; (8 letters) and the most recent edit was by &lt;strong&gt;JLee&lt;/strong&gt; (4 letters). So the final (8,4) answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; shamisen JLee&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;... and indeed the capitalisation is hard to get right!&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2016-08-27T20:35:15.300" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20672" PostTypeId="2" ParentId="20662" CreationDate="2015-08-28T12:55:36.043" Score="4" Body="&lt;p&gt;From a game-theory point of view, Batman should adopt a policy of &quot;no negotiation with hostage-takers&quot;, because if it becomes known to villains that they can get what they want through taking Batman's friends and family hostage, they'll keep on doing it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The logic of the no-negotiation policy is that terrorists will not take hostages if there is nothing to gain from the undertaking, i.e. a government will not concede to their demands.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://advat.blogspot.co.uk/2009/04/terrorism-game-theory.html?m=1&quot; rel=&quot;nofollow&quot;&gt;http://advat.blogspot.co.uk/2009/04/terrorism-game-theory.html?m=1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare a number of similar incidents with superhero spouses and significant others - Mary Jane, Lois Lane, etc etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Batman should contact Commissioner Gordon - an &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Iranian_Embassy_siege&quot; rel=&quot;nofollow&quot;&gt;armed assault by special forces, after a period of surveillance&lt;/a&gt;, is in the long run a more coherent strategy than simply giving in to the hostage-taker's demands.&lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="4050" LastEditDate="2015-08-28T13:00:49.797" LastActivityDate="2015-08-28T13:00:49.797" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20673" PostTypeId="2" ParentId="20651" CreationDate="2015-08-28T13:18:10.797" Score="6" Body="&lt;p&gt;I think it was &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mai Pokitz&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;She scowls, then repeats herself. &quot;Okay, fine. I said, I didn't kill my husband.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The lie detector was not yet on her wrist when she said she didn't kill her husband. However, after he attached it to her, she says 'I &lt;em&gt;said&lt;/em&gt;, I didn't kill my husband. She can truthfully say this, because that is in fact what she said.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit: Now that I think about it, it could also have been&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The maid&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;due to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; She says she didn't see anything that struck her as suspicious. Which, if she was the one intending to use the poison, why would it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, my guesses are based logically on wordplay, I couldn't tell you how it actually happened.&lt;/p&gt;&#xA;" OwnerUserId="15990" LastEditorUserId="15990" LastEditDate="2015-08-28T13:23:46.470" LastActivityDate="2015-08-28T13:23:46.470" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20674" PostTypeId="1" CreationDate="2015-08-28T13:27:08.613" Score="-1" ViewCount="790" Body="&lt;p&gt;A somewhat easy puzzle for those familiar with Linux.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Devil has found a way to make a Linux virus that, once executed on the right Linux machine, will break every single computer in the world permanently regardless of operating system and internet connectivity. It is evil, cannot be detected by malware scanners, survives storage media wipes, can stay in the volatile memory during cold reboots and breaks all encryption and hashing. It spreads faster than the Myspace Worm, inflicts more damage than Stuxnet does to a nuclear power plant and will share your porn habits with everyone you know. It just needs to be planted in the right computer and when Midnight on Halloween arrives, it will be unleashed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On October 30th, the Devil's advocate breaks into the house of Linus Torvalds and copied the devil virus to his master computer, configuring it to automatically activate at Midnight on October 31st.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;October 31st comes and goes, and the next day, the devil notices that there is a suspicious lack of global chaos. Linus had his computer on all night, he was even adding extra features to the kernel. Checking his Demonvision for the actions of the accountant, he realized that his advocate made a basic error when configuring the file in Linux. What went wrong?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If possible, try to answer with a small story and a technical explanation, separated by a horizontal line.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A first hint: Those that remember seeing a relatively well-known spy skulking around Puzzling.SE in May might get a clue as to the mistake the advocate made.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;this question is a near-duplicate of another question in essence since it relies on the same Linux functionality, but how that functionality is used is different. I think, just like the 146 questions that are tagged as word riddles that require us to combine different words of the same type, they are different puzzles with a similar answer. I don't think they're exact duplicates, which is what the duplicate flag is about. I at least think we need to get some discussion in Meta to determine at what point something is a duplicate.&lt;/p&gt;&#xA;" OwnerUserId="2778" LastEditorUserId="2778" LastEditDate="2015-08-29T21:13:26.630" LastActivityDate="2015-08-29T21:13:26.630" Title="The devil virus" Tags="|lateral-thinking|computer-puzzle|humor|" AnswerCount="1" CommentCount="19" FavoriteCount="0" ClosedDate="2015-08-30T02:24:36.850" ContentLicense="CC BY-SA 3.0" />
  <row Id="20676" PostTypeId="1" AcceptedAnswerId="20677" CreationDate="2015-08-27T15:28:41.877" Score="9" ViewCount="957" Body="&lt;h3&gt;Question&lt;/h3&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;What advice, strategies, keys, or elements would you suggest for creating challenging puzzles that aren't (reasonably) computer soluble?&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h3&gt;Details&lt;/h3&gt;&#xA;&lt;p&gt;I'm a programmer, so I can appreciate the argument that writing code to solve a puzzle is as enjoyable as solving it directly; don't take this question as any slight against computational solving. But I want to write good puzzles that people enjoy struggling with for a day or two; problems that can be solved via program typically lack this, in my experience.&lt;/p&gt;&#xA;&lt;p&gt;Sometimes, as in the case of &lt;a href=&quot;https://puzzling.stackexchange.com/questions/20581/find-the-longest-string-of-words-such-that-the-4-letter-end-of-one-word-is-the&quot;&gt;this puzzle&lt;/a&gt;, the inclusion of computers seems to take some of the fun out of it (as well as overshadow manual answers that deserve recognition) but there's no hard-and-true way to differentiate computer-generated solutions from manual ones.&lt;/p&gt;&#xA;&lt;p&gt;Now, normally I'd say that riddles (e.g. &lt;em&gt;What am I?&lt;/em&gt;) are the go-to solution for this problem, but they can get old very quickly (as evidenced &lt;a href=&quot;https://puzzling.stackexchange.com/questions/20569/many-forms-i-take&quot;&gt;here&lt;/a&gt;, based on some of the discussion). Another is cryptograms, but those also feel a bit tired sometimes.&lt;/p&gt;&#xA;&lt;p&gt;Do any veterans of the site have useful tips or advice in this area?&lt;/p&gt;&#xA;" OwnerUserId="13970" OwnerDisplayName="NeedAName" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-08-29T18:46:28.710" Title="What is the Key to Good No-Computers Puzzles?" Tags="|puzzle-creation|no-computers|" AnswerCount="6" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20677" PostTypeId="2" ParentId="20676" CreationDate="2015-08-27T16:24:09.117" Score="6" Body="&lt;p&gt;I've posted a link to &lt;strong&gt;&lt;a href=&quot;http://www.irregularwebcomic.net/3244.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;&lt;/strong&gt; article &lt;strong&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/5880/what-are-the-features-of-a-well-written-puzzle/16964#16964&quot;&gt;elsewhere&lt;/a&gt;&lt;/strong&gt;, but I think it might be especially relevant to your question here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It discusses some things to consider when trying to design a good puzzle, and I think all of the example puzzles in the article would be nigh-impossible to solve with a computer.  (At the very least, a computer wouldn't provide any significant advantage.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think there are two particular aspects of the puzzles that give them this quality:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A visual aspect.  When a puzzle involves looking at a picture of some sort and deriving clues or somehow altering or rearranging it, that's something that is very difficult to do with a computer.  Note that I'm not suggesting that you take a simple computer-solvable puzzle and publish it in an image format; I'm suggesting that the image be an integral part of the puzzle.&lt;/li&gt;&#xA;&lt;li&gt;An &quot;intuitive leap&quot;.  This, I think, is what is referred to by the &lt;a href=&quot;/questions/tagged/enigmatic-puzzle&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;enigmatic-puzzle&amp;#39;&quot; rel=&quot;tag&quot;&gt;enigmatic-puzzle&lt;/a&gt; tag on Puzzling.SE.  It means that the means or method of solving the puzzle is not immediately apparent.  You look at a cryptogram and immediately know that you need to find a series of letter substitutions that come out to plain English.  With an &lt;a href=&quot;/questions/tagged/enigmatic-puzzle&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;enigmatic-puzzle&amp;#39;&quot; rel=&quot;tag&quot;&gt;enigmatic-puzzle&lt;/a&gt;, you see a series of words/pictures/symbols etc., and must first figure out what is being asked of you before you can go about solving the puzzle.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Basically, I think what it boils down to is that creating a good, un-brute-forceable puzzle (if that's a word) requires a lot of time and effort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can create a cryptogram in 5 minutes by running some text through a website, you can't really expect that someone else won't be able to solve the same cryptogram in 5 minutes using a computer as well...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are a few examples of puzzles on this site that I think embody this answer well:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/17212/perplexing-fireworks&quot;&gt;Perplexing Fireworks&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/17008/sunset-parliament-sacrificing-posers&quot;&gt;Sunset Parliament Sacrificing Posers&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/15142/what-is-the-central-color&quot;&gt;What is the central color?&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/13305/a-treasure-hunt-for-the-magic-artefact&quot;&gt;A treasure hunt for the magic artefact&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4421" OwnerDisplayName="GentlePurpleRain" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2015-08-27T16:24:09.117" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20678" PostTypeId="2" ParentId="20676" CreationDate="2015-08-27T18:44:54.400" Score="5" Body="&lt;p&gt;I'm going to come back to this and extend it later, but I'll start off now anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every puzzle falls into one of 3 categories:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;someone posts a computer-assisted answer&lt;/li&gt;&#xA;&lt;li&gt;it can't be solved using a computer&lt;/li&gt;&#xA;&lt;li&gt;it has a &lt;a href=&quot;/questions/tagged/no-computers&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;no-computers&amp;#39;&quot; rel=&quot;tag&quot;&gt;no-computers&lt;/a&gt; tag (i.e. it &lt;em&gt;could&lt;/em&gt; be solved using a computer but the OP specifically asks people not to do this)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For point 3, I created the &lt;a href=&quot;/questions/tagged/no-computers&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;no-computers&amp;#39;&quot; rel=&quot;tag&quot;&gt;no-computers&lt;/a&gt; tag 5 months ago and it's only had 25 questions. For point 1, there aren't &lt;em&gt;that&lt;/em&gt; many questions answered by brute-forcing on this site - people with those sorts of questions tend to go to PPCG instead. So the &lt;strong&gt;vast majority&lt;/strong&gt; of the puzzles here are in point 2, i.e. of exactly the sort you're looking for.&#xA;&lt;hr&gt;&#xA;OK, that doesn't quite answer your question :-) You want tips on how to &lt;em&gt;construct&lt;/em&gt; such questions. So let's go through it tag-by-tag, starting from the top.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;/questions/tagged/riddle&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;riddle&amp;#39;&quot; rel=&quot;tag&quot;&gt;riddle&lt;/a&gt; As you say yourself, this is the easiest way to write a puzzle that can't be done with the help of a computer. And they'll never get old: they're works of art and people can always come up with new artistic ideas. If exactly the same idea gets flogged like a dead horse (hello spaghetti-parties/rebus-puzzles/warspyking), then &lt;em&gt;that&lt;/em&gt; can get old; but riddles in themselves? Never.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;/questions/tagged/math&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;math&amp;#39;&quot; rel=&quot;tag&quot;&gt;math&lt;/a&gt; There are whole books full of mathematical puzzles, most of which computers can't help you with. Computers can do &lt;em&gt;calculations&lt;/em&gt;, sure. They can solve linear equations. But anything mathsy that could be solved by a computer would count as a maths &lt;em&gt;problem&lt;/em&gt; and be off-topic anyway. Maths &lt;em&gt;puzzles&lt;/em&gt; generally need an unexpected approach and/or an 'aha!' moment, both of which require a human brain.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;/questions/tagged/logic-puzzle&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;logic-puzzle&amp;#39;&quot; rel=&quot;tag&quot;&gt;logic-puzzle&lt;/a&gt; Again, there are whole books full of these. All the old classics (knights-and-knaves, Monty Hall, hats of different colours) have been thrashed out already, so it's now a matter of writing variants. Boring variants can be solved using exactly the same method as the original, and a computer might be able to do that. Interesting variants need new ideas. That said, there are a lot of logic puzzles that can be brute-forced, like &lt;a href=&quot;https://puzzling.stackexchange.com/questions/11724/visiting-the-gym-this-month&quot;&gt;this one&lt;/a&gt; or &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15286/connect-the-pipes&quot;&gt;this one&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;to be continued&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" OwnerDisplayName="rand al'thor" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-08-27T18:44:54.400" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20679" PostTypeId="2" ParentId="20676" CreationDate="2015-08-28T09:59:52.650" Score="3" Body="&lt;p&gt;One other aspect - often &lt;em&gt;hard&lt;/em&gt; to do for the puzzle author!: If you manage to create a puzzle which can be solved by &lt;em&gt;deduction&lt;/em&gt; (to a large extend), then you can make it a condition of the solution to provide the &lt;em&gt;arguments&lt;/em&gt; and not only the &lt;em&gt;solution&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, people may still brute-force it, but the computer will not tell them &lt;em&gt;why&lt;/em&gt; one can come to this conclusion.&lt;/p&gt;&#xA;" OwnerUserId="5840" OwnerDisplayName="BmyGuest" LastActivityDate="2015-08-28T09:59:52.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20680" PostTypeId="2" ParentId="20676" CreationDate="2015-08-28T12:58:27.277" Score="6" Body="&lt;p&gt;the &lt;a href=&quot;/questions/tagged/no-computers&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;no-computers&amp;#39;&quot; rel=&quot;tag&quot;&gt;no-computers&lt;/a&gt; tag honestly, to me, seems like a crutch. If a puzzle is designed in such a way that its main appeal comes from not using a computer, then I feel people will enjoy going machine free to solve it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the puzzle is nothing more than an exercise in frustration until a computer gets involved, then it's more of a failure of design. That's not an insult; we're all guilty of making mistakes in our lives and posting bad puzzles is as important as posting good ones, because it builds our library and gives us a more impressive array of skills that are necessary for making and solving puzzles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Programming, at its core, is literally problem solving. Just be aware it's only natural for programmers to want to try and use those skills to solve puzzles.&lt;/p&gt;&#xA;" OwnerUserId="13068" OwnerDisplayName="Kingrames" LastActivityDate="2015-08-28T12:58:27.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20681" PostTypeId="2" ParentId="20674" CreationDate="2015-08-28T16:18:24.360" Score="0" Body="&lt;p&gt;Linus Torvalds wasn't home at the activation time, and he shut down his computer for power-saving. When he returned later and turned on his computer, it was already behind the scheduled 10PM Oct 31st. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, worry for next year. &lt;/p&gt;&#xA;" OwnerUserId="2783" LastActivityDate="2015-08-28T16:18:24.360" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20682" PostTypeId="1" AcceptedAnswerId="20687" CreationDate="2015-08-28T16:40:33.987" Score="13" ViewCount="1401" Body="&lt;p&gt;Your sysop recently quit due to what the law division calls &quot;internal struggles&quot;. Less fancily, he hated you, and he hated his job. As a parting gift, he left you a program entitled SwanSong.c. Being curious and slightly daft, you of course decide to execute it. Your console immediately floods with errors to standard error output:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;stderr&#xA;------&#xA;Line 4: Cannot exec a shared library directly.  &#xA;Line 5: Link has been severed.  &#xA;Line 5: Attempting to link in too many shared libraries.  &#xA;Line 7: Srmount error.  &#xA;Line 8: Too many users.  &#xA;Line 9: Name not unique on network.  &#xA;Line 14: RFS specific error.  &#xA;Line 18: Remote address changed.  &#xA;Line 18: Interrupted system call should be restarted.  &#xA;Line 18: Socket operation on non-socket.  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What does your sysop's swan song mean?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Footnote: Bonus points if you can write a program that returns this output. I'm pretty sure it's not possible, and it's definitely not a requirement to solve the problem.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="11475" LastEditDate="2015-08-28T20:24:49.403" LastActivityDate="2015-08-28T20:24:49.403" Title="Cryptic message from a former SYSOP" Tags="|cipher|computer-puzzle|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20685" PostTypeId="2" ParentId="20636" CreationDate="2015-08-28T17:40:17.170" Score="1" Body="&lt;p&gt;Here I will try to be funny:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's Grandma in dress!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Go ahead: Turn me upside-down.&lt;br&gt;&#xA;But others will soon realize your error.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yup, it's just wrong...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Turn me back up, and the evidence is gone.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; See! nothing happened&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Don't blame me; It was your mistake.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Shouldn't prank on grandma you little brat&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I just give and, sometimes, take.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; love, cookies, well many other things&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2015-08-28T17:40:17.170" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20686" PostTypeId="2" ParentId="20636" CreationDate="2015-08-28T17:45:21.970" Score="1" Body="&lt;p&gt;Perhaps it is  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An ant farm.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Go ahead: Turn me upside-down.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can turn an ant farm upside down if you really wanted.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But others will soon realize your error.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most people would recognize if an ant farm was oriented the wrong way.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Turn me back up, and the evidence is gone.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Once you turn it over, chances are that the tunnels have all collapsed/filled, and you can no longer see anything.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Don't blame me; It was your mistake.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The ants didn't do anything, but you now have a bunch of dead ants.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I just give and, sometimes, take.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ant farms give a lot of joy to curious young minds, but ants also take things... from everywhere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-08-28T17:45:21.970" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20687" PostTypeId="2" ParentId="20682" CreationDate="2015-08-28T17:47:34.627" Score="17" Body="&lt;p&gt;These are Linux error messages, which can be looked up here: &lt;a href=&quot;http://www.ioplex.com/~miallen/errcmp.html&quot;&gt;http://www.ioplex.com/~miallen/errcmp.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;83 (ELIBEXEC): Cannot exec a shared library directly&#xA;67 (ENOLINK): Link has been severed&#xA;82 (ELIBMAX): Attempting to link in too many shared libraries&#xA;69 (ESRMNT): Srmount error&#xA;87 (EUSERS): Too many users&#xA;76 (ENOTUNIQ): Name not unique on network&#xA;73 (EDOTDOT): RFS specific error&#xA;78 (EREMCHG): Remote address changed&#xA;85 (ERESTART): Interrupted system call should be restarted&#xA;88 (ENOTSOCK): Socket operation on non-socket&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Convert the error numbers to ASCII to get the message:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;SCREWLINUX&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-08-28T17:47:34.627" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20688" PostTypeId="2" ParentId="20539" CreationDate="2015-08-28T18:08:52.630" Score="5" Body="&lt;p&gt;Here's some python code which goes through all the possibilities, spitting out the best solution found so far.  It takes a little while to run through everything, as you might expect.  It seems to pause for a while, and then have sudden bursts of activity.  It's probably somewhat slow simply because it's having to print so many solutions.  Mine's been running for 20 minutes or so, and has converged onto a bunch of solutions with a score of 13/18 (as @Timbo found).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The logic of the code (for those who are interested is):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Generate all the possible dice (there are 462 of them)&lt;/li&gt;&#xA;&lt;li&gt;For each dice, generate a list of dice that are &quot;above&quot; them (ie has a &gt;50% probability that it beats it) - this is oneAway in the code.&lt;/li&gt;&#xA;&lt;li&gt;For each dice, generate a list of dice are are twoAway (using the above list).&lt;/li&gt;&#xA;&lt;li&gt;Using the above find all the dice that are 4 away from themselves (using the above) - there are 398 dice that do that.&lt;/li&gt;&#xA;&lt;li&gt;Using the above we can systematically go through all the loops of four.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It took more than an hour to run, partly because it went through all the results 4 times, which is an obvious improvement opportunity (edit - fixed that with the two discard lines near the bottom)!  And in the end it found 11 results that matched the above solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;115555 222266 333333 333334 13/18&lt;br&gt;&#xA;  115555 222266 333333 333344 13/18&lt;br&gt;&#xA;  115555 222266 333333 333444 13/18&lt;br&gt;&#xA;  115555 222266 333333 334444 13/18&lt;br&gt;&#xA;  115555 222266 333333 344444 13/18&lt;br&gt;&#xA;  115555 222266 333333 444444 13/18&lt;br&gt;&#xA;  115555 222266 333334 444444 13/18&lt;br&gt;&#xA;  115555 222266 333344 444444 13/18&lt;br&gt;&#xA;  115555 222266 333444 444444 13/18&lt;br&gt;&#xA;  115555 222266 334444 444444 13/18&lt;br&gt;&#xA;  115555 222266 344444 444444 13/18  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Python Code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import numpy as np&#xA;from fractions import Fraction&#xA;import itertools as it&#xA;from collections import defaultdict&#xA;&#xA;class dice:&#xA;    def __init__(self,faces):&#xA;        self.faces = faces  #a numpy array of 6 numbers&#xA;    def beats(self,other):&#xA;        &quot;probability that self beats other&quot;&#xA;        if np.all(self.faces==other.faces): return (0,1)&#xA;        a = np.tile(self.faces,(6,1))&#xA;        b = np.tile(other.faces.reshape(6,1),6)&#xA;        return (np.sum(a&amp;gt;b),np.sum(a!=b))&#xA;    def __repr__(self):&#xA;        return &quot;dice(%s)&quot;%self.faces&#xA;    def __str__(self):&#xA;        return &quot;%s%s%s%s%s%s&quot;%tuple(self.faces)&#xA;&#xA;def score(dice1,dice2,dice3,dice4):&#xA;    n1,d1 = dice2.beats(dice1)&#xA;    n2,d2 = dice3.beats(dice2)&#xA;    n3,d3 = dice4.beats(dice3)&#xA;    n4,d4 = dice1.beats(dice4)&#xA;    return Fraction(n1,d1*4)+Fraction(n2,d2*4)+Fraction(n3,d3*4)+Fraction(n4,d4*4)&#xA;&#xA;print(&quot;Generating a list of dice...&quot;,)&#xA;allDice = []   &#xA;for a in it.combinations_with_replacement(range(1,7),6):&#xA;    allDice.append(dice(np.array(a)))&#xA;print(&quot;done! %d dice generated&quot;%len(allDice))&#xA;&#xA;print(&quot;Generating oneAway...&quot;,)&#xA;oneAway = defaultdict(list)&#xA;for dice1 in allDice:&#xA;    for dice2 in allDice:&#xA;        n,d = dice2.beats(dice1)&#xA;        if n*2&amp;lt;=d: continue&#xA;        oneAway[dice1].append(dice2)&#xA;print(&quot;done! %d dice available&quot;%len(oneAway))&#xA;&#xA;print(&quot;Generating twoAway...&quot;,)&#xA;twoAway = defaultdict(set)&#xA;for dice1 in allDice:&#xA;    for dice2 in oneAway[dice1]:&#xA;        twoAway[dice1].update(oneAway[dice2])&#xA;print(&quot;done!&quot;)&#xA;&#xA;print(&quot;Generating fourAway...&quot;,)&#xA;fourAway = defaultdict(set)&#xA;for dice1 in allDice:&#xA;    for dice2 in twoAway[dice1]:&#xA;        fourAway[dice1].update(twoAway[dice2])&#xA;print(&quot;done!&quot;)&#xA;&#xA;print(&quot;Generating candidates...&quot;)&#xA;candidates=[]&#xA;for dice1 in allDice:&#xA;    if dice1 in fourAway[dice1]:&#xA;        candidates.append(dice1)&#xA;print(&quot;done!&quot;,len(candidates))&#xA;&#xA;print(&quot;Searching for results...&quot;)&#xA;maxScore=0&#xA;for dice1 in candidates:&#xA;    for dice3 in twoAway[dice1]:&#xA;        if dice1 not in twoAway[dice3]: continue&#xA;        for dice2 in oneAway[dice1]:&#xA;            if dice3 not in oneAway[dice2]: continue&#xA;            for dice4 in oneAway[dice3]:&#xA;                if dice1 not in oneAway[dice4]: continue&#xA;                s = score(dice1,dice2,dice3,dice4)&#xA;                if s&amp;gt;=maxScore:&#xA;                    maxScore = s&#xA;                    print(dice1,dice2,dice3,dice4,s)&#xA;    for a in oneAway: oneAway[a].discard(dice1)&#xA;    for a in twoAway: twoAway[a].discard(dice1)&#xA;print(&quot;done!&quot;)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Update: I noticed that the OP allows for 0-6, so adjusted the relevant line in the code and re-ran it. This yields additional solutions (for a total of 220), but still nothing better than 13/18:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;004444 111155 222222 333333 13/18&lt;br&gt;&#xA;  004444 111155 222222 222223 13/18&lt;br&gt;&#xA;  004444 111155 222222 222233 13/18&lt;br&gt;&#xA;  004444 111155 222222 222333 13/18&lt;br&gt;&#xA;  004444 111155 222222 223333 13/18&lt;br&gt;&#xA;  004444 111155 222222 233333 13/18&lt;br&gt;&#xA;  004444 111156 222222 333333 13/18&lt;br&gt;&#xA;  004444 111156 222222 222223 13/18&lt;br&gt;&#xA;  004444 111156 222222 222233 13/18&lt;br&gt;&#xA;  004444 111156 222222 222333 13/18&lt;br&gt;&#xA;  004444 111156 222222 223333 13/18&lt;br&gt;&#xA;  004444 111156 222222 233333 13/18&lt;br&gt;&#xA;  004444 111166 222222 333333 13/18&lt;br&gt;&#xA;  004444 111166 222222 222223 13/18&lt;br&gt;&#xA;  004444 111166 222222 222233 13/18&lt;br&gt;&#xA;  004444 111166 222222 222333 13/18&lt;br&gt;&#xA;  004444 111166 222222 223333 13/18&lt;br&gt;&#xA;  004444 111166 222222 233333 13/18&lt;br&gt;&#xA;  004444 111155 222223 333333 13/18&lt;br&gt;&#xA;  004444 111156 222223 333333 13/18&lt;br&gt;&#xA;  004444 111166 222223 333333 13/18&lt;br&gt;&#xA;  004444 111155 222233 333333 13/18&lt;br&gt;&#xA;  004444 111156 222233 333333 13/18&lt;br&gt;&#xA;  004444 111166 222233 333333 13/18&lt;br&gt;&#xA;  004444 111155 222333 333333 13/18&lt;br&gt;&#xA;  004444 111156 222333 333333 13/18&lt;br&gt;&#xA;  004444 111166 222333 333333 13/18&lt;br&gt;&#xA;  004444 111155 223333 333333 13/18&lt;br&gt;&#xA;  004444 111156 223333 333333 13/18&lt;br&gt;&#xA;  004444 111166 223333 333333 13/18&lt;br&gt;&#xA;  004444 111155 233333 333333 13/18&lt;br&gt;&#xA;  004444 111156 233333 333333 13/18&lt;br&gt;&#xA;  004444 111166 233333 333333 13/18&lt;br&gt;&#xA;  004445 111166 222222 333333 13/18&lt;br&gt;&#xA;  004445 111166 222222 222223 13/18&lt;br&gt;&#xA;  004445 111166 222222 222233 13/18&lt;br&gt;&#xA;  004445 111166 222222 222333 13/18&lt;br&gt;&#xA;  004445 111166 222222 223333 13/18&lt;br&gt;&#xA;  004445 111166 222222 233333 13/18&lt;br&gt;&#xA;  004445 111166 222223 333333 13/18&lt;br&gt;&#xA;  004445 111166 222233 333333 13/18&lt;br&gt;&#xA;  004445 111166 222333 333333 13/18&lt;br&gt;&#xA;  004445 111166 223333 333333 13/18&lt;br&gt;&#xA;  004445 111166 233333 333333 13/18&lt;br&gt;&#xA;  004455 111166 222222 333333 13/18&lt;br&gt;&#xA;  004455 111166 222222 222223 13/18&lt;br&gt;&#xA;  004455 111166 222222 222233 13/18&lt;br&gt;&#xA;  004455 111166 222222 222333 13/18&lt;br&gt;&#xA;  004455 111166 222222 223333 13/18&lt;br&gt;&#xA;  004455 111166 222222 233333 13/18&lt;br&gt;&#xA;  004455 111166 222223 333333 13/18&lt;br&gt;&#xA;  004455 111166 222233 333333 13/18&lt;br&gt;&#xA;  004455 111166 222333 333333 13/18&lt;br&gt;&#xA;  004455 111166 223333 333333 13/18&lt;br&gt;&#xA;  004455 111166 233333 333333 13/18&lt;br&gt;&#xA;  004555 111166 222222 333333 13/18&lt;br&gt;&#xA;  004555 111166 222222 222223 13/18&lt;br&gt;&#xA;  004555 111166 222222 222233 13/18&lt;br&gt;&#xA;  004555 111166 222222 222333 13/18&lt;br&gt;&#xA;  004555 111166 222222 223333 13/18&lt;br&gt;&#xA;  004555 111166 222222 233333 13/18&lt;br&gt;&#xA;  004555 111166 222223 333333 13/18&lt;br&gt;&#xA;  004555 111166 222233 333333 13/18&lt;br&gt;&#xA;  004555 111166 222333 333333 13/18&lt;br&gt;&#xA;  004555 111166 223333 333333 13/18&lt;br&gt;&#xA;  004555 111166 233333 333333 13/18&lt;br&gt;&#xA;  005555 111166 233334 444444 13/18&lt;br&gt;&#xA;  005555 111166 233344 444444 13/18&lt;br&gt;&#xA;  005555 111166 233444 444444 13/18&lt;br&gt;&#xA;  005555 111166 234444 444444 13/18&lt;br&gt;&#xA;  005555 111166 244444 444444 13/18&lt;br&gt;&#xA;  005555 111266 333333 333334 13/18&lt;br&gt;&#xA;  005555 111266 333333 333344 13/18&lt;br&gt;&#xA;  005555 111266 333333 333444 13/18&lt;br&gt;&#xA;  005555 111266 333333 334444 13/18&lt;br&gt;&#xA;  005555 111266 333333 344444 13/18&lt;br&gt;&#xA;  005555 111266 333333 444444 13/18&lt;br&gt;&#xA;  005555 122266 333333 333334 13/18&lt;br&gt;&#xA;  005555 122266 333333 333344 13/18&lt;br&gt;&#xA;  005555 122266 333333 333444 13/18&lt;br&gt;&#xA;  005555 122266 333333 334444 13/18&lt;br&gt;&#xA;  005555 122266 333333 344444 13/18&lt;br&gt;&#xA;  005555 122266 333333 444444 13/18&lt;br&gt;&#xA;  005555 112266 333333 333334 13/18&lt;br&gt;&#xA;  005555 112266 333333 333344 13/18&lt;br&gt;&#xA;  005555 112266 333333 333444 13/18&lt;br&gt;&#xA;  005555 112266 333333 334444 13/18&lt;br&gt;&#xA;  005555 112266 333333 344444 13/18&lt;br&gt;&#xA;  005555 112266 333333 444444 13/18&lt;br&gt;&#xA;  005555 222266 333333 333334 13/18&lt;br&gt;&#xA;  005555 222266 333333 333344 13/18&lt;br&gt;&#xA;  005555 222266 333333 333444 13/18&lt;br&gt;&#xA;  005555 222266 333333 334444 13/18&lt;br&gt;&#xA;  005555 222266 333333 344444 13/18&lt;br&gt;&#xA;  005555 222266 333333 444444 13/18&lt;br&gt;&#xA;  005555 111166 333333 333334 13/18&lt;br&gt;&#xA;  005555 111166 333333 333344 13/18&lt;br&gt;&#xA;  005555 111166 333333 333444 13/18&lt;br&gt;&#xA;  005555 111166 333333 334444 13/18&lt;br&gt;&#xA;  005555 111166 333333 344444 13/18&lt;br&gt;&#xA;  005555 111166 333333 444444 13/18&lt;br&gt;&#xA;  005555 111266 333334 444444 13/18&lt;br&gt;&#xA;  005555 122266 333334 444444 13/18&lt;br&gt;&#xA;  005555 112266 333334 444444 13/18&lt;br&gt;&#xA;  005555 222266 333334 444444 13/18&lt;br&gt;&#xA;  005555 111166 333334 444444 13/18&lt;br&gt;&#xA;  005555 111266 333344 444444 13/18&lt;br&gt;&#xA;  005555 122266 333344 444444 13/18&lt;br&gt;&#xA;  005555 112266 333344 444444 13/18&lt;br&gt;&#xA;  005555 222266 333344 444444 13/18&lt;br&gt;&#xA;  005555 111166 333344 444444 13/18&lt;br&gt;&#xA;  005555 111266 333444 444444 13/18&lt;br&gt;&#xA;  005555 122266 333444 444444 13/18&lt;br&gt;&#xA;  005555 112266 333444 444444 13/18&lt;br&gt;&#xA;  005555 222266 333444 444444 13/18&lt;br&gt;&#xA;  005555 111166 333444 444444 13/18&lt;br&gt;&#xA;  005555 111166 222222 233334 13/18&lt;br&gt;&#xA;  005555 111166 222222 233344 13/18&lt;br&gt;&#xA;  005555 111166 222222 233444 13/18&lt;br&gt;&#xA;  005555 111166 222222 234444 13/18&lt;br&gt;&#xA;  005555 111166 222222 244444 13/18&lt;br&gt;&#xA;  005555 111166 222222 333333 13/18&lt;br&gt;&#xA;  005555 111166 222222 333334 13/18&lt;br&gt;&#xA;  005555 111166 222222 333344 13/18&lt;br&gt;&#xA;  005555 111166 222222 333444 13/18&lt;br&gt;&#xA;  005555 111166 222222 222223 13/18&lt;br&gt;&#xA;  005555 111166 222222 222224 13/18&lt;br&gt;&#xA;  005555 111166 222222 222233 13/18&lt;br&gt;&#xA;  005555 111166 222222 222234 13/18&lt;br&gt;&#xA;  005555 111166 222222 222244 13/18&lt;br&gt;&#xA;  005555 111166 222222 222333 13/18&lt;br&gt;&#xA;  005555 111166 222222 222334 13/18&lt;br&gt;&#xA;  005555 111166 222222 222344 13/18&lt;br&gt;&#xA;  005555 111166 222222 222444 13/18&lt;br&gt;&#xA;  005555 111166 222222 223333 13/18&lt;br&gt;&#xA;  005555 111166 222222 223334 13/18&lt;br&gt;&#xA;  005555 111166 222222 223344 13/18&lt;br&gt;&#xA;  005555 111166 222222 223444 13/18&lt;br&gt;&#xA;  005555 111166 222222 224444 13/18&lt;br&gt;&#xA;  005555 111166 222222 233333 13/18&lt;br&gt;&#xA;  005555 111166 222222 334444 13/18&lt;br&gt;&#xA;  005555 111166 222222 344444 13/18&lt;br&gt;&#xA;  005555 111166 222222 444444 13/18&lt;br&gt;&#xA;  005555 111166 222223 333333 13/18&lt;br&gt;&#xA;  005555 111166 222223 333334 13/18&lt;br&gt;&#xA;  005555 111166 222223 333344 13/18&lt;br&gt;&#xA;  005555 111166 222223 333444 13/18&lt;br&gt;&#xA;  005555 111166 222223 334444 13/18&lt;br&gt;&#xA;  005555 111166 222223 344444 13/18&lt;br&gt;&#xA;  005555 111166 222223 444444 13/18&lt;br&gt;&#xA;  005555 111166 222224 444444 13/18&lt;br&gt;&#xA;  005555 111166 222233 333333 13/18&lt;br&gt;&#xA;  005555 111166 222233 333334 13/18&lt;br&gt;&#xA;  005555 111166 222233 333344 13/18&lt;br&gt;&#xA;  005555 111166 222233 333444 13/18&lt;br&gt;&#xA;  005555 111166 222233 334444 13/18&lt;br&gt;&#xA;  005555 111166 222233 344444 13/18&lt;br&gt;&#xA;  005555 111166 222233 444444 13/18&lt;br&gt;&#xA;  005555 111166 222234 444444 13/18&lt;br&gt;&#xA;  005555 111166 222244 444444 13/18&lt;br&gt;&#xA;  005555 111166 222333 333333 13/18&lt;br&gt;&#xA;  005555 111166 222333 333334 13/18&lt;br&gt;&#xA;  005555 111166 222333 333344 13/18&lt;br&gt;&#xA;  005555 111166 222333 333444 13/18&lt;br&gt;&#xA;  005555 111166 222333 334444 13/18&lt;br&gt;&#xA;  005555 111166 222333 344444 13/18&lt;br&gt;&#xA;  005555 111166 222333 444444 13/18&lt;br&gt;&#xA;  005555 111166 222334 444444 13/18&lt;br&gt;&#xA;  005555 111166 222344 444444 13/18&lt;br&gt;&#xA;  005555 111166 222444 444444 13/18&lt;br&gt;&#xA;  005555 111166 223333 333333 13/18&lt;br&gt;&#xA;  005555 111166 223333 333334 13/18&lt;br&gt;&#xA;  005555 111166 223333 333344 13/18&lt;br&gt;&#xA;  005555 111166 223333 333444 13/18&lt;br&gt;&#xA;  005555 111166 223333 334444 13/18&lt;br&gt;&#xA;  005555 111166 223333 344444 13/18&lt;br&gt;&#xA;  005555 111166 223333 444444 13/18&lt;br&gt;&#xA;  005555 111166 223334 444444 13/18&lt;br&gt;&#xA;  005555 111166 223344 444444 13/18&lt;br&gt;&#xA;  005555 111166 223444 444444 13/18&lt;br&gt;&#xA;  005555 111166 224444 444444 13/18&lt;br&gt;&#xA;  005555 111166 233333 333333 13/18&lt;br&gt;&#xA;  005555 111166 233333 333334 13/18&lt;br&gt;&#xA;  005555 111166 233333 333344 13/18&lt;br&gt;&#xA;  005555 111166 233333 333444 13/18&lt;br&gt;&#xA;  005555 111166 233333 334444 13/18&lt;br&gt;&#xA;  005555 111166 233333 344444 13/18&lt;br&gt;&#xA;  005555 111166 233333 444444 13/18&lt;br&gt;&#xA;  005555 111266 334444 444444 13/18&lt;br&gt;&#xA;  005555 122266 334444 444444 13/18&lt;br&gt;&#xA;  005555 112266 334444 444444 13/18&lt;br&gt;&#xA;  005555 222266 334444 444444 13/18&lt;br&gt;&#xA;  005555 111166 334444 444444 13/18&lt;br&gt;&#xA;  005555 111266 344444 444444 13/18&lt;br&gt;&#xA;  005555 122266 344444 444444 13/18&lt;br&gt;&#xA;  005555 112266 344444 444444 13/18&lt;br&gt;&#xA;  005555 222266 344444 444444 13/18&lt;br&gt;&#xA;  005555 111166 344444 444444 13/18&lt;br&gt;&#xA;  015555 222266 333333 333334 13/18&lt;br&gt;&#xA;  015555 222266 333333 333344 13/18&lt;br&gt;&#xA;  015555 222266 333333 333444 13/18&lt;br&gt;&#xA;  015555 222266 333333 334444 13/18&lt;br&gt;&#xA;  015555 222266 333333 344444 13/18&lt;br&gt;&#xA;  015555 222266 333333 444444 13/18&lt;br&gt;&#xA;  015555 222266 333334 444444 13/18&lt;br&gt;&#xA;  015555 222266 333344 444444 13/18&lt;br&gt;&#xA;  015555 222266 333444 444444 13/18&lt;br&gt;&#xA;  015555 222266 334444 444444 13/18&lt;br&gt;&#xA;  015555 222266 344444 444444 13/18&lt;br&gt;&#xA;  115555 222266 333333 333334 13/18&lt;br&gt;&#xA;  115555 222266 333333 333344 13/18&lt;br&gt;&#xA;  115555 222266 333333 333444 13/18&lt;br&gt;&#xA;  115555 222266 333333 334444 13/18&lt;br&gt;&#xA;  115555 222266 333333 344444 13/18&lt;br&gt;&#xA;  115555 222266 333333 444444 13/18&lt;br&gt;&#xA;  115555 222266 333334 444444 13/18&lt;br&gt;&#xA;  115555 222266 333344 444444 13/18&lt;br&gt;&#xA;  115555 222266 333444 444444 13/18&lt;br&gt;&#xA;  115555 222266 334444 444444 13/18&lt;br&gt;&#xA;  115555 222266 344444 444444 13/18&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2019-01-25T23:36:35.210" LastActivityDate="2019-01-25T23:36:35.210" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="20689" PostTypeId="2" ParentId="20636" CreationDate="2015-08-28T18:14:29.780" Score="1" Body="&lt;p&gt;Maybe it's:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A diode.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Go ahead: Turn me upside-down.&lt;br&gt;&#xA;But others will soon realize your error.&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;No more flow!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Turn me back up, and the evidence is gone.&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;What do you mean? The system is running fine!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Don't blame me; It was your mistake.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;These are put in backwards all the time (I've done it)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I just give and, sometimes, take.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;LEDs give light!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11926" LastEditorDisplayName="user20" LastEditDate="2015-08-28T18:19:08.983" LastActivityDate="2015-08-28T18:19:08.983" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20690" PostTypeId="2" ParentId="20636" CreationDate="2015-08-28T18:32:01.750" Score="1" Body="&lt;p&gt;Could this be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an Hourglass?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Go ahead: Turn me upside-down.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You turn an hourglass upside down to stop the flow of the sand and reverse it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But others will soon realize your error.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Assuming it is an unembellished hourglass, when you turn it upside down the differential in sand will show that it has been changed.  They will also notice if they are especially perceptive of the passage of time and notice that the sand completely drains either too quickly or too slowly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Turn me back up, and the evidence is gone.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you flip the hourglass to its original position, there will be nothing to prove that it was tampered with, unless the &lt;em&gt;others&lt;/em&gt; are of a scientific mind, and have graduated lines on their hourglass, or at least a separate, control hourglass.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Don't blame me; It was your mistake.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In turning the hourglass over you lose the time it was keeping track of.  It would have kept fine time if it had been undisturbed, but turning it over kept it from serving its purpose.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I just give and, sometimes, take.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One half of the hourglass gives sand, while the other half takes it (with a little help from gravity and an excess of space).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11452" LastActivityDate="2015-08-28T18:32:01.750" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20691" PostTypeId="2" ParentId="20593" CreationDate="2015-08-28T18:58:24.660" Score="3" Body="&lt;p&gt;This will be our model cube &lt;em&gt;(just so we're all on the same page of the face (center block) colors)&lt;/em&gt; -&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KG0Ru.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KG0Ru.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been trying to work on this, but I don't know enough math to formulate a rule. But, given the pattern -&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A B B&#xA;A B B&#xA;A C C&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We can note that each block of 4 that match in the upper right, it has 3 matches to its &lt;strong&gt;left&lt;/strong&gt; (depending on which way you hold the cube). Let's make our blocks of 4 and indicate which matches are to the left -&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YzNMZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YzNMZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note that the colors I picked are arbitrary, it will work out the same way if you choose different corners.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, all of the triples to the left match with another triple to the left, thereby requiring two corner pieces that match 2 colors &lt;strong&gt;and&lt;/strong&gt; an edge of those matching colors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pairs&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;yu&#xA;vx&#xA;wz&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We also know what &lt;em&gt;color pairs&lt;/em&gt; those corners &lt;em&gt;can't&lt;/em&gt; have so they don't match their own face -&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;zw - yellow, red&#xA;vx - green, white&#xA;uy - orange, blue&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The corners we have left:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;BOY&#xA;GRY&#xA;RBY&#xA;BOW&#xA;GRW&#xA;GOW&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;How they can match to our pairs -&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;zw - BOY, BOW, GRW, GOW&#xA;vx - BOY, GRY, RBY, BOW&#xA;uy - GRY, RBY, GRW, GOW&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So the pairs that are possible are&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;zw - BOx, GWx, OWx&#xA;vx - BOx, BYx, RYx&#xA;uy - RYx, GWx, GRx&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;With the blocks that remain, every color is present on exactly &lt;strong&gt;3 corners&lt;/strong&gt; and &lt;strong&gt;2 edges&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's give a variable name to the other matches that there needs to be -&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GBWd9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GBWd9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see any combination of the cubes we have left that will meet the required matching criteria. But I can't formulate a &lt;em&gt;proof&lt;/em&gt; of this. Maybe someone else can?&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditDate="2015-08-28T19:03:39.640" LastActivityDate="2015-08-28T19:03:39.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20692" PostTypeId="1" AcceptedAnswerId="21918" CreationDate="2015-08-28T19:25:39.547" Score="2" ViewCount="544" Body="&lt;p&gt;I am the one what did the did and the one what knows what did the did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am one and the same, but distinct and separate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situations I create are also quite tense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am no Cyrano, but if you're looking for me, you'll find ten letters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint #1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The answer to this puzzle is not a person. It is an adjective in both forms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13068" LastEditorUserId="13068" LastEditDate="2015-08-28T21:09:43.323" LastActivityDate="2015-09-04T11:31:51.117" Title="A quick puzzle to keep your brain busy on a Friday... or is it?" Tags="|riddle|word|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20693" PostTypeId="1" AcceptedAnswerId="20780" CreationDate="2015-08-28T19:35:21.247" Score="6" ViewCount="1529" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Seven follows suit,&lt;/p&gt;&#xA;&lt;p&gt;Eight go in line,&lt;/p&gt;&#xA;&lt;p&gt;Combined, 9 leave the queue.&lt;/p&gt;&#xA;&lt;p&gt;Leaving i e,  once removed, rearranged casts light upon the answer.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;What am I?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;HINT #1:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; I am a word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;HINT #2:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Pay attention to punctuation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;HINT #3:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Two of the three numbers in the riddle are not used like the other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;HINT #4:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The sequence of numbers and their progression is NOT a nod towards the answer. For instance they do not reference something which is versioned (think computer software/OS) or of something that is incrementally changed (playing cards). This was intentionally misleading in the riddle to add to its complexity but I am revealing since all answers up to now have read in to it, distracting from the actual method to solve it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Note: This riddle was previously closed for being &amp;quot;unclear&amp;quot; as to what it was asking. I have added a new tag, &amp;quot;word&amp;quot; (which is a hint in itself that I was trying to avoid originally), clarified the title and added the question &amp;quot;What am I?&amp;quot; at the end of the riddle. If this still is not clear enough please comment and suggest ways in which I can make it clearer. I am being cautious being overly specific as it is a short riddle and saying too much may reveal the answer.&lt;/p&gt;&#xA;" OwnerUserId="15843" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-09-01T12:07:30.950" Title="Seven, Eight, Nine Redefined - A 'What Am I?' Riddle" Tags="|riddle|word|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20695" PostTypeId="2" ParentId="16169" CreationDate="2015-08-28T20:05:17.720" Score="3" Body="&lt;p&gt;Swinging for the fences here,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The number 1 or the keyboard key 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I am on top politically, I am right,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 is True for computers  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but if I am on top of being not in violation, I am wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; on top of 1 being ! on the keyboard so !1 is 0 which is False&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I am above 8, I am wrong,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; * denotes a pointer in C which is initialized to NULL which evaluates to False &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but if I am low on being not in violation, I am forever.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Q is below 1 and is the set of rational numbers which is infinite or forever&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cannot be on top politically and above 8,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm pretty rusty on my C but I think pointers can't be assigned to address 1 since that is for the heap&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I can be on top and low of being not in violation. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; !1 is 0 and 0 is considered a rational number if 0 is on top. 0/x&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13925" LastEditorUserId="13925" LastEditDate="2015-08-28T20:16:04.463" LastActivityDate="2015-08-28T20:16:04.463" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20696" PostTypeId="2" ParentId="20682" CreationDate="2015-08-28T20:23:19.143" Score="14" Body="&lt;p&gt;For your bonus points, here's a (rather contrived) Perl program that will produce the desired output:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#!/usr/bin/perl&#xA;sub ugh { $! = vec &quot;SCREWLINUX&quot;, $x++, 8; warn &quot;Line &quot; . (caller)[2] . &quot;: $!\n&quot; }&#xA;&#xA;ugh;&#xA;ugh ugh;&#xA;&#xA;ugh;&#xA;ugh;&#xA;ugh;&#xA;&#xA;&#xA;&#xA;&#xA;ugh;&#xA;&#xA;&#xA;&#xA;ugh ugh ugh;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerDisplayName="user16048" LastActivityDate="2015-08-28T20:23:19.143" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20697" PostTypeId="2" ParentId="20693" CreationDate="2015-08-28T20:31:14.200" Score="1" Body="&lt;p&gt;I think you are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; one of the 6 cards in a game of Patience (Solitaire)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;7 follow suit&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 7 follows the number 6 in a column&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;8 go in line&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 8 is added to the column after 7&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Combined, 9 leave the queue&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; because 6, 7, and 8 are in the column, 9 can leave the deck (queue) and be added&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Leaving i e, once removed, rearranged casts light upon the answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; once you remove the numbers six through nine from the column, and add them to the bank of suits, you can rearrange your columns; making the solution to your game clearer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9871" LastActivityDate="2015-08-28T20:31:14.200" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20698" PostTypeId="2" ParentId="20676" CreationDate="2015-08-28T20:31:19.807" Score="1" Body="&lt;p&gt;I'd say we can take this question as a nice example: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/5049/nerds-jocks-and-lockers&quot;&gt;Nerds, Jocks, and Lockers&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The OP did not specify &lt;a href=&quot;/questions/tagged/no-computers&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;no-computers&amp;#39;&quot; rel=&quot;tag&quot;&gt;no-computers&lt;/a&gt; (I'm not entirely sure if that was a tag then). However when me and somebody else provided programs/ scripts to solve this, they were cast aside as &quot;inelegant&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead he accepted the answer which provided &lt;strong&gt;math proof&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'd say the best way is to simply ask for a piece of pencil and paper (just a bit of math that can be done easily on paper) proof. No need to overcomplicate things. A computer shows you the answer, not the mathematical steps to get there. You should filter out the programmers (such as myself) from the true mathematical.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-08-29T18:46:28.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20699" PostTypeId="2" ParentId="20636" CreationDate="2015-08-28T21:07:55.053" Score="1" Body="&lt;p&gt;Could it possibly be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A painting?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Go ahead: Turn me upside-down.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's quite easy to hang a painting the the wrong way, especially if the painting is abstract.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But others will soon realize your error.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Anyone familiar with the painting will know it's upside-down.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Turn me back up, and the evidence is gone.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you turn it back up, no one will know it was ever upside-down.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Don't blame me; It was &lt;em&gt;your&lt;/em&gt; mistake.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Obviously, I put the painting upside-down, so it's my fault.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I just give and, sometimes, take.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Paintings and other works of art can give people enjoyment and insight. Sometimes, people even take away a deeper meaning!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10186" LastActivityDate="2015-08-28T21:07:55.053" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20700" PostTypeId="1" AcceptedAnswerId="20703" CreationDate="2015-08-28T22:55:59.053" Score="-4" ViewCount="4674" Body="&lt;p&gt;5, 11, 17, 23, 29, 41&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note that I don't have an answer to this problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought of 2 possible answers, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) 5 - it is the only single digit number in the sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) 41 - all other adjacent numbers have a difference of 6&lt;/p&gt;&#xA;" OwnerUserId="12703" LastEditorUserId="12703" LastEditDate="2015-08-28T23:07:56.307" LastActivityDate="2015-08-29T00:13:46.147" Title="which number stands out the most in the given series?" Tags="|number-sequence|" AnswerCount="3" CommentCount="6" ClosedDate="2015-08-29T03:11:18.303" ContentLicense="CC BY-SA 3.0" />
  <row Id="20701" PostTypeId="2" ParentId="20700" CreationDate="2015-08-28T23:01:31.737" Score="0" Body="&lt;p&gt;&lt;strong&gt;New answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you add up the digits in each number you get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5, 2, 8, 5, 11, 5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;all of these are prime numbers except 8 (1+7), so 17 is standing out the most&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Previous answer&lt;/strong&gt;&#xA; I agree with b) 41, with a slightly different reason: all numbers are a multiple of 6 followed by minus 1, (5=6-1, 11=12-1, 17=18-1 etc) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your original answer would never allow the first number to be out of sequence. &lt;/p&gt;&#xA;" OwnerUserId="14243" LastEditorUserId="14243" LastEditDate="2015-08-28T23:14:51.133" LastActivityDate="2015-08-28T23:14:51.133" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20702" PostTypeId="2" ParentId="20700" CreationDate="2015-08-28T23:51:45.267" Score="4" Body="&lt;p&gt;5, 11, 17, 23, 29, 41&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5 is the answer because &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;it's the only single digit number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;11 is the answer because &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;it contains repeating digits, which stands out amongst those numbers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;17 is the answer because &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the difference between its digits is the same difference between it and the other numbers around it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;23 is the answer because &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;its digits are neighbors on the number line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;29 is the answer because &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;it has the greatest difference between its digits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;41 is the answer because &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;it is 12 units away from its previous neighbor in the list, rather than 6 like the others (or 5 like the first).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As you can see, each and every answer is the only true answer.&lt;/p&gt;&#xA;" OwnerUserId="13068" LastActivityDate="2015-08-28T23:51:45.267" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20703" PostTypeId="2" ParentId="20700" CreationDate="2015-08-29T00:01:04.720" Score="3" Body="&lt;p&gt;29 is the odd number out because &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; they're listing every other prime number starting from 5 (prime numbers from 5 to 41 are 5,7,11,13,17,19,23,29,31,37,41) so the sequence should have been 5,11,17,23,31,41 . 29 breaks that sequence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15913" LastEditorUserId="15913" LastEditDate="2015-08-29T00:13:46.147" LastActivityDate="2015-08-29T00:13:46.147" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20704" PostTypeId="2" ParentId="20636" CreationDate="2015-08-29T01:33:46.060" Score="2" Body="&lt;p&gt;I think it's a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Credit card  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Go ahead: Turn me upside-down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But others will soon realize your error.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Only works when you swipe it the right way up.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Turn me back up, and the evidence is gone.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No harm done. Just turn it over and swipe again.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Don't blame me; It was your mistake.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People blame the credit card when it doesn't go through.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I just give and, sometimes, take.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Primarily used for purchases (giving), though sometimes you get reimbursed on it.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another guess&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pocket cards in poker  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll explain way later&lt;/p&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2015-08-30T04:44:18.880" LastActivityDate="2015-08-30T04:44:18.880" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20705" PostTypeId="1" AcceptedAnswerId="20753" CreationDate="2015-08-29T01:35:06.767" Score="16" ViewCount="528" Body="&lt;p&gt;Here on SE I've only answered questions so far, so I thought it was time to ask some of my own. But where should I go to find the answers to these questions?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Would a &lt;em&gt;tsubo-niwa&lt;/em&gt; garden help with my kung-fu novel, &quot;Big Blind&quot;?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Does the Qur'an state anything about graduate students using plastic building blocks while abroad studying encryption algorithms?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Would it count as prior art if I used a Turing machine as a UHF transmitter for my cat, Gatto?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-08-29T16:26:41.573" LastActivityDate="2015-08-31T14:54:27.713" Title="Where to ask these questions?" Tags="|cipher|lateral-thinking|knowledge|" AnswerCount="1" CommentCount="19" ContentLicense="CC BY-SA 3.0" />
  <row Id="20708" PostTypeId="1" CreationDate="2015-08-29T04:40:08.640" Score="3" ViewCount="967" Body="&lt;p&gt;There are three poker players, who are playing heads-up against each other. One of them is a &lt;strong&gt;usual poker player&lt;/strong&gt;. Another - &lt;strong&gt;a telepath&lt;/strong&gt;, who can read thoughts. So if you know your cards, he knows them too. And a third one - a &lt;strong&gt;clairvoyant&lt;/strong&gt; who can't read thoughts, but knows which cards gonna come on flop, turn and river (but he doesn't know pocket cards). All three players are aware of poker maths and play correctly regardless of their skills. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have three match-ups.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A usual guy against a telepath.&lt;/li&gt;&#xA;&lt;li&gt;A telepath against clairvoyant.&lt;/li&gt;&#xA;&lt;li&gt;A clairvoyant against a usual guy. (he can bluff now, yeah?)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What are the best strategies of these players in their match-ups? Let's assume they should play even if they gonna lose (surely they can fold on any street), but want to minimize losses in the long run. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume they are &lt;strong&gt;playing limit holdem&lt;/strong&gt;. (Being on the small or big blind each all the time)&lt;/p&gt;&#xA;" OwnerUserId="14029" LastEditorUserId="14029" LastEditDate="2015-08-29T10:52:48.460" LastActivityDate="2015-08-31T19:20:54.457" Title="A poker player, a telepath and a clairvoyant" Tags="|strategy|cards|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="20709" PostTypeId="2" ParentId="20708" CreationDate="2015-08-29T10:33:50.447" Score="0" Body="&lt;p&gt;&lt;strong&gt;Game 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The telepath knows the normal player's cards at the start of the hand. The telepath can determine the probability of a person winning at any given point in time. The only way the normal person would make money would be if the odds were against him at the begginning, yet he bet (or called) substantial amount of money, and then the odds switched to his favour. This has a probability definitely less than 1/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, he would prefer folding in every hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Game 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As with game 1, the telepath is at a better position. Hence the clairvoyant will apply the same strategy as the normal guy in game 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Game 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This game is also biased, but fair enough for both players to employ complex strategies. If they are playing optimal strategies, then the normal player will fold most of the time (not always). He has meek chance of winning. However, for human players, he still has some chance of winning, if he is better at bluffing and calling bluffs than the clairvoyant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have not mentioned for how long they are required to play. If they are playing forever, the person at a disadvantage is going to go bankrupt after a finite number of games (due to the blinds), no matter what. If the number of games is fixed, the above rules apply.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-08-29T10:33:50.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20710" PostTypeId="2" ParentId="20663" CreationDate="2015-08-29T12:38:05.293" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If one stays at the centre, adjusting the rope as the other moves forward, the shortest distance allowing the latter to hit something will be half of a single side of the square. So if he only hits it 4 times while holding a given amount of the rope and turning it without hitting a minotaur, that means he's hit the door at least once. When he hits the door, the one moving does something like pull the rope thrice to alert the other, then adjusts the rope until the other catches up. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastActivityDate="2015-08-29T12:38:05.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20714" PostTypeId="2" ParentId="18782" CreationDate="2015-08-29T15:54:52.330" Score="3" Body="&lt;p&gt;There are no satisfying explanations as to why most blade inscriptions are incomprehensible. Of course, it is likely that the sword smiths who did the metal work for the inscriptions were illiterate; however, the craftsmen could not have failed so regularly and ruined so many valuable blades with illegible inscriptions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most epigraphic scholars suggest an initial-based abbreviation system; however, this thesis is only to be proved, if the inscrutable alignments of letters could be assigned to a common prayer, psalm or bible quote. This was hitherto only possible, if the inscriptions consisted of (at least some) full Latin words and not only initial sequences. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wegeli himself and Erben tried to relate the suspected initials with a prayer or a sentence spoken in conjunction with the sword presentation ceremony or in German &lt;code&gt;Schwertleite&lt;/code&gt; . This was a ritual, probably of Germanic origin, in which the father handed down the sword to his grown-up son as a sign that he can defend himself and the tribe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;During the Middle Ages the &lt;strong&gt;sword presentation ceremony&lt;/strong&gt;, performed by the liege lord or a cleric, put the warrior into service of the former, as a vassal in the first case or as a &lt;code&gt;miles Christi ad servitium Iesu Christi (soldier of Christ in service of Christ)&lt;/code&gt; in the second . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, as yet it has not been possible to identify such a traditional type of ceremonial dicta latina that was carved on a sword blade. Moreover, this theory presupposes that the inscriptions were generalized. However, quite conversely, the inscriptions (even though sometimes showing a constancy of letters) are extremely variable and appear to be very personal. One might say the individual secret of every sword bearer. It must have been a special dictum so obvious and so self-evident to him that it was not necessary to spell out its significant meaning.&lt;/p&gt;&#xA;" OwnerUserId="15920" LastEditorUserId="15544" LastEditDate="2015-08-29T16:18:21.363" LastActivityDate="2015-08-29T16:18:21.363" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20715" PostTypeId="2" ParentId="20676" CreationDate="2015-08-29T16:31:02.393" Score="1" Body="&lt;p&gt;In my experience, no-computers is for puzzles with some amount of computation, that a computer script would trivialize. For example, how many zeroes does 100! end in? What is the solution to [insert grid logic puzzle]? Usually the point of the puzzle is finding a cleverer way than brute force to solve it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For these puzzles, you can either try to make the puzzle infeasible for computers (e.g. by asking about 1000000000! instead) or add a no-computers tag. The first approach is more elegant when possible. Unfortunately, some puzzles cannot be made infeasible for computers without either killing the clever approach or making the clever approach annoyingly hard for humans.&lt;/p&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2015-08-29T16:31:02.393" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20716" PostTypeId="2" ParentId="20708" CreationDate="2015-08-29T16:34:21.980" Score="1" Body="&lt;p&gt;&lt;strong&gt;Game 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The telepath has a natural advantage, unless the usual guy does something to deal with it. There is no particular action the telepath can undertake beyond using their ability and otherwise playing poker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usual guy, on the other hand, has only one choice - play blind. Don't look at their cards, and play purely by watching the reactions of the telepath. Having skill at reading reactions, the regular poker player should be able to decide when to play or not, and thus should be able to gain a slight edge relative to any other plan. The best the telepath can do in this situation is to also play blind, in which case it basically becomes a 50% chance of winning for both sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Game 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, the clairvoyant is at a disadvantage, because everything they can read, the telepath can learn from them. As such, the clairvoyant has to use their gift carefully. Under most situations, playing blindly as with the regular player is the only option. However, they can use their gift as a technique to encourage the telepath to reveal the strength of their hand, by seeing certain cards and observing how the telepath responds after reading it from their mind. This can give the clairvoyant a small benefit over the telepath, but a lot still comes down to pure luck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Game 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this game, the clairvoyant has the advantage, and thus is likely to win. There is no particular strategy available to counter the advantage, as the regular player has to rely on their usual skills while having a lesser amount of knowledge of the hand. That said, some techniques can minimise the issue, and increase the potential for luck to go their way. The clairvoyant will, naturally, read every hand immediately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the clairvoyant has a tell, the regular player can use this to their advantage, as they can get some sense of the hand that is coming. By some effective bluffing in the first two rounds of betting, and then observing the reaction of the clairvoyant, they can get a sense of the kind of hand that is coming, and the clairvoyant's own hand. High bets on the first round suggests to the clairvoyant that the regular player has a strong hand, and thus will be confident if they've got a very strong hand or the cards are going to be bad for normally-strong hands, and thus will also be betting high. On the other hand, if they've got a weak hand and the cards are going to favour strong hands (lots of high cards, for instance), they'll bit minimally, or possibly even fold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second round of betting can then be used to determine further, in a similar way - observing the clairvoyant's reactions and tells should be enough to minimise the impact of their gift, but there's no way to completely neutralise it - that is, the clairvoyant will always have more information, and thus will always have a better than 50% chance of winning.&lt;/p&gt;&#xA;" OwnerUserId="9289" LastActivityDate="2015-08-29T16:34:21.980" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="20723" PostTypeId="2" ParentId="20636" CreationDate="2015-08-30T02:57:23.727" Score="6" Body="&lt;p&gt;My guess is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A pencil with an eraser in the back.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Go ahead: Turn me upside-down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But others will soon realize your error.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Turn a pencil upside down to use the eraser, when people see you erasing, they realize you made an error in your writing/drawing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Turn me back up, and the evidence is gone.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Turn the pencil back to its normal position and continue writing/drawing, as if you hadn't erased in the first place.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Don't blame me; It was your mistake.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Your writing/drawing mistakes are your fault, not the fault of the pencil.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I just give and, sometimes, take.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A pencil is used to &quot;give&quot; graphite to paper, but the eraser tip can be used to &quot;take&quot; graphite away from paper.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6178" LastActivityDate="2015-08-30T02:57:23.727" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20726" PostTypeId="2" ParentId="20595" CreationDate="2015-08-30T08:22:07.570" Score="6" Body="&lt;p&gt;&lt;strong&gt;The correct proof&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First we assume that the floor has elevation that is a continuous function of the position bounded between zero and the $p$ where $p$ is some parameter that depends on the shape of the table (see below for details), and that the table legs are lines that end in the points of a square $ABCD$. The problem is then assumed to be to position the table so that all four legs touch (and do not go through) the floor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the following method note that every step is physically possible in the sense that the table never intersects the floor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First note that there is a rotational orientation of the table (based on $p$), such that for any point on the floor we can hold the table in that orientation such that one leg touches the floor at that point. This is possible because we can hold the table sufficiently tilted and above the floor, so that the end of the lowest leg is directly above the desired point, and now lowering the table results in that leg touching the floor first by the intermediate value theorem (IVT) and the tilt. We can assume that $A$ is the end of that leg touching the floor at the chosen point. Now tilt the table around the line through $A$ that is perpendicular to $AB$ and horizontal, such that $B$ goes towards the floor, until the leg ending at $B$ touches the floor by IVT. Tilt the table around the line through $AB$ until a third leg touches the floor by IVT. Note that if the third leg is not there we can tilt some more so that the last leg touches the floor by IVT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in the above subroutine it is easy to choose the initial orientation of the table such that rotating the table by any angle around the vertical will give yet another initial orientation that works. So we get a function with its inputs being the desired point and the initial rotation around the vertical and its output being a table 'position' with $ABC$ on the floor (and $D$ possibly below). It is not hard to prove that this function is continuous, given some weak conditions on the floor (see below). This is the key.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perform this subroutine once to get $ABC$ on the floor. To be precise we simply get any three legs on the floor, and then rotate the table such that those legs are the ones that actually have endpoints being $ABC$. Then choose any path $P$ from $A$ to $B$ on the floor. Drag the table on the floor such that $A$ follows $P$. We claim that we can do so with $B,C$ remaining on the floor. This follows from the subroutine's continuity since we can keep the rotation around vertical fixed and just use the subroutine on the points along the path. In fact this is what can actually happen physically if you constrain the table's movements. Now drag the table on the floor such that $A$ remains where it is while $B$ goes to where $C$ originally was. Again we claim that we can keep $B,C$ on the floor in the process of dragging. This is because the subroutine applied to all possible rotations around the vertical gives all possible table 'positions' where $A$ is in that place and $B,C$ are on the floor. In particular, a quarter turn around the vertical gives a table 'position' where $A,B$ are at the original locations of $B,C$, and in that case $D$ is below the floor. By IVT, there is a smallest rotation less than that quarter turn such that the subroutine will give a table 'position' with $D$ exactly on the floor. This means that we can drag the table to move $B$ 'around' $A$ keeping $B,C$ on the floor because of the subroutine's continuity, and before a quarter 'turn' we would have found the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Table parameter&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly if the table top is much wider than the distance between legs, the method can fail to work simply because the side of the table top hits the floor. In the worst case the legs are so short and the table top has some downward protrusions that reach the plane through the square! In that case certain floors would make the problem impossible. However, as long as there is a plane that completely separates the leg ends from the table top, there will be some suitable parameter $p &amp;gt; 0$ for the above method to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly if the legs point outwards, the middle part of the legs may hit the floor if we use the method. Again, for reasonable tables there is some suitable parameter $p$ that works. (No I'm not going to rigorously prove this.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Floor conditions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, to ensure that the subroutine works, any sphere with centre on the floor and radius $AB$ must intersect the floor at points that are at unique angles from the centre. I'm not sure whether this condition can be removed without making the solution non-constructive, in the sense that we can easily prove existence of a solution but there may not be a physical method (like the one above) to obtain it systematically from any random starting position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, this condition is fulfilled by various simpler conditions such as having Lipschitz constant at most $1$.&lt;/p&gt;&#xA;" OwnerUserId="11806" LastActivityDate="2015-08-30T08:22:07.570" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20727" PostTypeId="1" AcceptedAnswerId="20735" CreationDate="2015-08-30T09:56:50.183" Score="4" ViewCount="232" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;From the White House to the Trevi Fountain&lt;br /&gt;&#xA;  I have sought advice from secret sympathisers&lt;br /&gt;&#xA;  Stalactites met stalagmites and orchids bloomed&lt;br /&gt;&#xA;  There were falling blocks and Napoleonic Attacks&lt;br /&gt;&#xA;  PSE pages held the answer&lt;br /&gt;&#xA;  Juggling spreadsheets and matrices&lt;br /&gt;&#xA;  Distilling it all to a cluster of neurons&lt;br /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What's the connection?&lt;/p&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-08-31T15:13:36.753" Title="14 links to a single word" Tags="|knowledge|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20730" PostTypeId="5" CreationDate="2015-08-30T15:52:07.747" Score="0" Body="&lt;p&gt;Puzzles can present themselves during all kinds of activities&#xA;because they have consequential solutions.&#xA;Such puzzles have concrete and testable parameters,&#xA;yet reach beyond imagination alone and broaden our understanding of puzzling.&#xA;This is for puzzles that actually occurred by happenstance,&#xA;not ones that were designed or merely might occur.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="18129" LastEditDate="2017-10-14T02:43:58.527" LastActivityDate="2017-10-14T02:43:58.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20731" PostTypeId="4" CreationDate="2015-08-30T15:52:07.747" Score="0" Body="For puzzles that presented themselves, with real parameters and consequences, during otherwise non-puzzling activities." OwnerUserId="2383" LastEditorUserId="18129" LastEditDate="2017-10-14T02:44:08.937" LastActivityDate="2017-10-14T02:44:08.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20735" PostTypeId="2" ParentId="20727" CreationDate="2015-08-30T16:52:46.077" Score="10" Body="&lt;p&gt;The Answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Columns&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the White House to the Trevi Fountain&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Both have columns along the facade.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have sought advice from secret sympathisers&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Advice columns (like ask amy type stuff). A fifth column is any group of people who undermine a larger group.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Stalactites met stalagmites and orchids bloomed&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When they meet it forms a continuous line top to bottom or a column. The column, or technically the gynostemium, is a reproductive structure that can be found in several plant families: Aristolochiaceae, Orchidaceae, and Stylidiaceae. &lt;sub&gt; wikipedia &lt;/sub&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There were falling blocks and Napoleonic Attacks&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Falling blocks refers to Columns a match-three puzzle video game, first created in 1989 &lt;sub&gt; Originally guessed tetris woops. &lt;/sub&gt;. A military column is a formation of soldiers marching together in one or more files in which the file is significantly longer than the width of ranks in the formation. Originating in the french revolutionary war.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;PSE pages held the answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are columns the pages are organized by like the related, hot network question column.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Juggling spreadsheets and matrices&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Both Spreadsheets ad matrices have columns. In toss juggling, columns, also known as One-up Two-up, is a juggling trick or pattern where the balls are thrown upwards without any sideways motion. The simplest version involves having three balls, with two going up simultaneously on either side, followed by one going up in the middle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Distilling it all to a cluster of neurons&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; there are small areas (columns) in which the neurons tend to have the same receptive field. A fractionating column or fractionation column is an essential item used in distillation of liquid mixtures so as to separate the mixture into its component parts, or fractions, based on the differences in volatilities. (distillation).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13916" LastEditorUserId="13916" LastEditDate="2015-08-31T15:13:36.753" LastActivityDate="2015-08-31T15:13:36.753" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20737" PostTypeId="1" CreationDate="2015-08-31T01:29:43.553" Score="8" ViewCount="387" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Yes, ommmm... Eee! Sea! Rye! Pea! Toe! Gee, ram.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What do you make of that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Backstory:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've just escaped from an asylum where I've stayed for the past ten years, and I'm a little excited about everything. I don't want to go back anytime soon, and meditating calms me down a little. But sometimes, just sometimes, I start acting looney again. What do you make of that?&lt;/p&gt;&#xA;" OwnerUserId="14013" LastEditorUserId="14013" LastEditDate="2015-09-10T17:54:16.623" LastActivityDate="2015-09-10T17:54:16.623" Title="Meditatively crazy" Tags="|enigmatic-puzzle|wordplay|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20738" PostTypeId="1" CreationDate="2015-08-31T06:09:12.790" Score="-3" ViewCount="975" Body="&lt;p&gt;A lovely place where ohm law is checked every day + you can feel the grapes (without the first letter). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clue:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can also find there gold… (The golden state)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What I came up until now is that the answer is a place in California.&#xA;Any help would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="16086" LastEditorUserId="5373" LastEditDate="2017-02-06T11:27:53.373" LastActivityDate="2017-02-06T11:27:53.373" Title="Interview riddle: A lovely place where ohm law is checked every day + you can feel the grapes (without the first letter)" Tags="|riddle|wordplay|" AnswerCount="2" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="20740" PostTypeId="1" AcceptedAnswerId="20743" CreationDate="2015-08-31T13:03:37.857" Score="18" ViewCount="2286" Body="&lt;p&gt;Here is a simple grid with letters in them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UBohZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UBohZ.png&quot; alt=&quot;![enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is &lt;strong&gt;one&lt;/strong&gt; significant word and &lt;strong&gt;ten&lt;/strong&gt; special numbers in it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Four questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What is the significant word?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What are the ten special numbers contained in it? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Where are they located?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What is &quot;fun&quot; name for this grid?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For those who want to copy the grid in text format.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;E   D   S   Q   U   A   R   E   S&#xA;R   N   I   N   E   S   I   X   T&#xA;D   Y   E   N   T   Y   F   I   Y&#xA;N   T   W   R   N   I   N   V   F&#xA;U   R   T   U   O   N   E   E   O&#xA;H   O   N   O   F   E   S   T   U&#xA;E   F   E   E   T   X   I   H   R&#xA;N   X   I   S   Y   T   R   I   E&#xA;O   E   N   O   Y   T   H   G   I&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Hint&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is a fun puzzle. Nothing serious about it. Targeted for fifth graders.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9892" LastEditorUserId="21234" LastEditDate="2016-05-20T01:14:37.673" LastActivityDate="2016-05-20T01:14:37.673" Title="What is special about this grid?" Tags="|mathematics|letter-sequence|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20742" PostTypeId="2" ParentId="18562" CreationDate="2015-08-31T13:25:34.787" Score="12" Body="&lt;p&gt;You are a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Vigilante&lt;/strong&gt;. Not a lawman, soldier or hero.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you watch over someone, that is a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; vigil&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the price we all pay is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ante, all the players in poker have to pay the 'ante' to play&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And so the price you pay for someone to watch over you is literally a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; vigil ante.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16098" LastActivityDate="2015-08-31T13:25:34.787" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20743" PostTypeId="2" ParentId="20740" CreationDate="2015-08-31T13:26:48.140" Score="24" Body="&lt;p&gt;The &lt;a href=&quot;https://i.stack.imgur.com/izuys.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;answer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The first ten square numbers, and the word &quot;Squares&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Completing the solution with a image.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/H0IEd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/H0IEd.png&quot; alt=&quot;![enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6178" LastEditorUserId="9892" LastEditDate="2016-03-25T18:41:53.617" LastActivityDate="2016-03-25T18:41:53.617" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20745" PostTypeId="2" ParentId="20740" CreationDate="2015-08-31T13:42:24.857" Score="3" Body="&lt;p&gt;@McMagister has solved this, I'm just going to answer part 4), the fun name for this grid. My suggestion: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ever decreasing squares &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12277" LastActivityDate="2015-08-31T13:42:24.857" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20746" PostTypeId="1" AcceptedAnswerId="20747" CreationDate="2015-08-31T13:48:09.087" Score="-6" ViewCount="316" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;There are &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; fish in a fish tank.&lt;/p&gt;&#xA;&lt;p&gt;They get fed the same fish food every day. They have varying opinions on the food.&lt;/p&gt;&#xA;&lt;p&gt;One fish says he only likes fish food that starts with the letter &lt;strong&gt;A&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Another says he only likes fish food that &lt;strong&gt;ends&lt;/strong&gt; with a &lt;strong&gt;vowel.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Yet another says he doesn't like food that those terrible humans would eat.&lt;/p&gt;&#xA;&lt;p&gt;Finally, one says he only eats green substances.&lt;/p&gt;&#xA;&lt;p&gt;The rest are too stubborn to say anything.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Can you figure out what food they all like?&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;- The food name doesn't really relate to any &lt;em&gt;real&lt;/em&gt; fish food names and companies. Answer has to fit all requirements.&lt;/p&gt;&#xA;" OwnerUserId="15675" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-09-06T05:51:59.837" Title="Fish In The Fish Bowl" Tags="|logical-deduction|" AnswerCount="3" CommentCount="2" ClosedDate="2015-08-31T14:24:41.387" ContentLicense="CC BY-SA 3.0" />
  <row Id="20747" PostTypeId="2" ParentId="20746" CreationDate="2015-08-31T13:50:41.153" Score="2" Body="&lt;p&gt;Oh, hey, the requirements have changed...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now they get fed&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ALGAE.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They like it because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It starts with an A, it ends with a vowel (E), it's green, and it's definitely not human food.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Do they get fed...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; FOOD?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If they do...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everyone likes it. Jim likes it because it starts with 'F', Joe likes it because it ends with a 'D', Bob likes it because it's only one word, and Bill likes it because it has an 'O' in it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="11475" LastEditDate="2015-09-01T16:27:12.307" LastActivityDate="2015-09-01T16:27:12.307" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20750" PostTypeId="2" ParentId="20746" CreationDate="2015-08-31T13:53:33.343" Score="1" Body="&lt;p&gt;I think their daily meal is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; FLOUR&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why? Because...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jim is happy because it starts with an F. Joe is satisfied because it ends in a consonant. Bob won't complain because it has a one word name. Bill loves it because it has an O in the name.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-08-31T13:56:59.867" LastActivityDate="2015-08-31T13:56:59.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20751" PostTypeId="2" ParentId="20746" CreationDate="2015-08-31T14:04:47.553" Score="1" Body="&lt;p&gt;How about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Foolish Fishy Feed?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It satisfies all conditions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. It starts with F&lt;br&gt;&#xA; 2. Ends with consonant D&lt;br&gt;&#xA; 3. It doesn't have two words it it... It has three!&lt;br&gt;&#xA; 4. It has O in it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14013" LastActivityDate="2015-08-31T14:04:47.553" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20752" PostTypeId="2" ParentId="20740" CreationDate="2015-08-31T14:52:14.783" Score="6" Body="&lt;p&gt;Since Pete has already come up with a creative  name, I will toss my suggestion into the hat as well...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Square Square Spiral &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-08-31T14:52:14.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20753" PostTypeId="2" ParentId="20705" CreationDate="2015-08-31T14:54:27.713" Score="9" Body="&lt;p&gt;Please ask your questions in &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://area51.stackexchange.com/&quot;&gt;http://area51.stackexchange.com/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To solve this question, you need to determine all the stack exchange sites, taking note of which sentence and which order they appear in.  @RespectMyAuthoritah has thankfully done this for me:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;japanese, gardening, martial-arts, writers, poker&lt;/li&gt;&#xA;&lt;li&gt;islam, academia, legos, travel, cryptography&lt;/li&gt;&#xA;&lt;li&gt;patents, computer-science, amateur-radio, pets, italian&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now go to &lt;a href=&quot;http://stackexchange.com/sites#&quot;&gt;http://stackexchange.com/sites#&lt;/a&gt; and look up the positions of each stack exchange site.  For each of the 3 questions, connect the five stack exchange sites in order to reveal a character for each site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3iEuP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Big picture incoming&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we get A51, or Area 51.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you interpret 5 as S, &lt;a href=&quot;https://puzzling.stackexchange.com/?as=1&quot;&gt;https://puzzling.stackexchange.com/?as=1&lt;/a&gt; is also another answer.&lt;/p&gt;&#xA;" OwnerUserId="6178" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2015-08-31T14:54:27.713" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20754" PostTypeId="2" ParentId="20693" CreationDate="2015-08-31T15:02:27.020" Score="4" Body="&lt;p&gt;Perhaps &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Windows&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Seven follows suit,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Referring to Windows 7 which followed vista&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Eight go in line,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Windows 8 came afterwards&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Combined, 9 leave the queue.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Windows 10 feels like a mix between 7 and 8, for example, the start menu is a mixture of traditional Windows 7 combined with Windows 8 'tile' feel. However, Windows 9 never came to fruition&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Leaving i e, once removed, rearranged casts light upon the answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Windows 10 left internet explorer, is is now Edge. I think the 'rearranged casts light upon the answer' bit maybe refers to the UI&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15990" LastActivityDate="2015-08-31T15:02:27.020" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20755" PostTypeId="2" ParentId="20559" CreationDate="2015-08-31T15:11:39.987" Score="6" Body="&lt;p&gt;All answers are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;anagrams using the letters RECURSION + A-Z depending on the clue's place on the list.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A bearlike whale: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A- URSINE ORCA &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Attire of an ancient, mysterious sorcerer: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; B- RUNIC ROBES&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Cool pointer!&quot;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C- NICE CURSOR&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A blighted sort of metal:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;D- CURSED IRON &lt;BR&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One who expeditions through deep time: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;E- EON CRUISER &lt;BR&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Keeps using the same data:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;F- RECURS INFO &lt;BR&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Providing someone with assets, maybe: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;G- RESOURCING &lt;BR&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Destroyer of subatomic particles: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;H- ION CRUSHER &lt;BR&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wrath of the mailman: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I- COURIER SIN &lt;BR&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once again, the mongrels choose to partake: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;J- CURS REJOIN &lt;BR&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Doctors act like pigs: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;K- CURERS OINK &lt;BR&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Lord of the Sound: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;L- SONIC RULER &lt;BR&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Weeps over grains: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;M- MOURNS RICE &lt;BR&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One who takes good care of his money?: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;N- COIN NURSER &lt;BR&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The companions we have: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;O- OUR CRONIES &lt;BR&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Deface a rotting body: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;P- RUIN CORPSE &lt;BR&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An indecent vegetable: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Q- RISQUE CORN &lt;BR&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One who wets the bed: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;R-URIC SNORER &lt;BR&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Scampers all over: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;S- SCURRIES ON &lt;BR&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An unpleasant idiot: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;T- SOUR CRETIN &lt;BR&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Intriguing symbols on an intriguing object: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;U- CURIO RUNES  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The more voluptuous brother: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;V- CURVIER SON  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pots of snails: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;W- COWRIE URNS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An old decision point: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;X- SENIOR CRUX  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sounds like a spicy dish: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Y- CURRY NOISE  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An area where certain clouds might form: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Z- CIRRUS ZONE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13290" LastEditorUserId="4385" LastEditDate="2015-08-31T15:58:27.153" LastActivityDate="2015-08-31T15:58:27.153" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20756" PostTypeId="1" AcceptedAnswerId="20757" CreationDate="2015-08-31T15:14:25.633" Score="13" ViewCount="2966" Body="&lt;p&gt;&lt;strong&gt;A classic old puzzle&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A monk starts climbing a mountain at 10 AM. He reaches the peak at 8PM and rests at the peak for the night. The next day he starts climbing down the the mountain at 10 AM and reaches the bottom at 8 PM again. Prove that there was a certain time of the day where the monk was at the same height on both days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Notes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The monk doesn't necessarily move at a constant rate, he might move fast or slow or even rest for some time. He can also turn back while climbing up or down. It is only certain that the monk starts his journey at 10 AM and finishes at 8PM.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="15544" LastEditorUserId="15544" LastEditDate="2015-09-01T13:42:44.010" LastActivityDate="2015-09-03T22:12:25.260" Title="Same time, same height" Tags="|logical-deduction|" AnswerCount="4" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="20757" PostTypeId="2" ParentId="20756" CreationDate="2015-08-31T15:23:05.030" Score="25" Body="&lt;p&gt;We can prove this using a simple graph:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/m2eUD.png&quot; alt=&quot;Graph&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No matter how you bend the lines, it should be clear that they will at some point intersect. This is always true because the monk's movement is continuous, so there can't be any jumps from one location to another.&lt;/p&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-08-31T15:23:05.030" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20758" PostTypeId="2" ParentId="20756" CreationDate="2015-08-31T15:24:04.903" Score="8" Body="&lt;p&gt;Lets imagine you have the monks location charted strictly on a line representing the distance they are along the path. If you start plotting with time they start on opposite ends eventually they will have to pass each other. This is the critical moment. In order for the monk to get to the bottom he has to get below the past self's location at some point. in order to do that he has to be at the same distance along as the other past self at the same time. &lt;sub&gt; I apologize for the poor wording &lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4Xk3s.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4Xk3s.png&quot; alt=&quot;The dots are monk&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13916" LastActivityDate="2015-08-31T15:24:04.903" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20759" PostTypeId="1" AcceptedAnswerId="20760" CreationDate="2015-08-31T15:58:42.213" Score="14" ViewCount="1537" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;When I woke up early in the morning, there were 15 of these staring at me.&lt;br&gt;&#xA;  I closed my eyes for a snooze, and when I opened them, there were still 15.&lt;br&gt;&#xA;  I arose to get dressed and when I turned back around, there were only 11!&lt;br&gt;&#xA;  As I ate my lunch, I saw 19 of them watching me in the kitchen.&lt;br&gt;&#xA;  Throughout the day, the least I ever saw were 6, the most: 21.&lt;br&gt;&#xA;  At one point I saw 8 of them; I wished I could be doing something other than watch them but there was nothing else to do.&lt;br&gt;&#xA;  Eventually I got tired of counting and, with 20 of them watching, I fell asleep.&lt;br&gt;&#xA;  When I woke up the next day, there were none at all! Getting up, I knew what I had to do to bring them back.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What are 'they' in the riddle? What did the author have to do to bring them back? In your answer, be sure to explain the count of them in each line.&lt;/p&gt;&#xA;" OwnerUserId="13970" LastActivityDate="2015-08-31T16:05:35.077" Title="They're Watching Me" Tags="|riddle|lateral-thinking|story|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20760" PostTypeId="2" ParentId="20759" CreationDate="2015-08-31T16:05:35.077" Score="14" Body="&lt;p&gt;These are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The line segments on a digital clock that make up the numbers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I woke up early in the morning, there were 15 of these staring at me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is 7:00 AM; 3+6+6 = 15.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I closed my eyes for a snooze, and when I opened them, there were still 15.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A snooze is typically nine minutes, making this 7:09 AM; 3+6+6 = 15.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I arose to get dressed and when I turned back around, there were only 11!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's been only a minute now, making it 7:10; 3+2+6 = 11.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As I ate my lunch, I saw 19 of them watching me in the kitchen.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lunch happens around noon, or 12:00; 2+5+6+6 = 19.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Throughout the day, the least I ever saw were 6, the most: 21.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The fewest happen at 1:11, the most at 10:08; 2+2+2 = 6, 2+6+6+7 = 21.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At one point I saw 8 of them; I wished I could be doing something other than watch them but there was nothing else to do.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's 11:11, make a wish; 2+2+2+2 = 8.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Eventually I got tired of counting and, with 20 of them watching, I fell asleep.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You must have fallen asleep at 10:00 PM; 2+6+6+6 = 20.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I woke up the next day, there were none at all! Getting up, I knew what I had to do to bring them back.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your clock batteries must be dead! Replace the batteries in your clock so you can see the numbers again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-08-31T16:05:35.077" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20761" PostTypeId="2" ParentId="20756" CreationDate="2015-08-31T16:26:41.723" Score="10" Body="&lt;p&gt;Bailey M has given the correct answer already. But I am going to provide another angle of attacking the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Call the original monk A. Take another monk B and on the second day, have him mimic the actions of monk A on the first day. Since B climbs from bottom to peak and A from peak to bottom, they must cross (in terms of height) at some point. Since monk B is mimicking A, monk A must have been at this height at this time on the first day. So, this is our desired answer.&lt;/p&gt;&#xA;" OwnerUserId="15544" LastEditorUserId="15544" LastEditDate="2015-09-03T22:12:25.260" LastActivityDate="2015-09-03T22:12:25.260" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20762" PostTypeId="2" ParentId="20359" CreationDate="2015-08-31T16:39:53.727" Score="6" Body="&lt;p&gt;One bugfix later, I think I have a solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Grid (I omitted some words from the list below):&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/a0Ak2.png&quot; alt=&quot;&quot;&gt;&#xA; &lt;pre&gt;&#xA; Horizontal:        Vertical:&#xA; &#xA; 0, 0: NEPAL        0, 1: EIGHT&#xA; 0, 4: LOOKS        0, 2: PENNIES&#xA; 1, 0: FIELD        0, 3: ALTITUDE&#xA; 1, 5: KNEEL        0, 6: ONTOLOGY&#xA; 2, 4: ATTACK       0, 8: SECOND&#xA; 3, 3: IDIOM        0, 9: ALKALI&#xA; 4, 3: TOOLS        0, 9: ALKALIN&#xA; 5, 1: DEUCE        1, 0: FULSOME&#xA; 6, 0: MISDO        1, 0: FULSOMELY&#xA; 6, 5: AGAIN        3, 6: OLOGY&#xA; 7, 2: SEEDY        5, 1: DIMES&#xA; 8, 2: ABNORMAL     5, 2: ESSAY&#xA; 8, 4: NORMA        5, 7: BALMS&#xA; 8, 4: NORMAL       5, 8: DIVAS&#xA; 9, 3: ACROSS       5, 9: INKLE&#xA; 9, 4: CROSS&#xA; 9, 4: CROSSE&#xA; &lt;/pre&gt;&#xA; The words are from this wordlist: &lt;a href=&quot;http://www-01.sil.org/linguistics/wordlists/english/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www-01.sil.org/linguistics/wordlists/english/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2015-09-02T17:37:46.897" LastActivityDate="2015-09-02T17:37:46.897" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20763" PostTypeId="1" AcceptedAnswerId="20776" CreationDate="2015-08-31T17:03:48.997" Score="4" ViewCount="229" Body="&lt;p&gt;Partially solved sudoku:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   ===================================&#xA;9: || 8  __ 3 || __ __ __|| 4  9  2 ||&#xA;8: || 6  4  1 || 9  __ 2 || __ __ __||&#xA;7: || __ __ __|| 3  4  8 || __ 6  1 ||&#xA;   ===================================&#xA;6: || __ __ 7 || __ 9  __|| 1  5  8 ||&#xA;5: || __ __ __|| __ 2  __|| __ __ 6 ||&#xA;4: || 5  6  __|| __ 1  __|| 2  __ __||&#xA;   ===================================&#xA;3: || 4  __ 6 || 2  3  __|| __ __ __||&#xA;2: || __ __ __|| __ 8  __|| 6  2  4 ||&#xA;1: || 7  8  2 || __ __ __|| 3  1  __||&#xA;   ===================================&#xA;      A  B  C    D  E  F    G  H  I &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;These are potential values on the missing fields:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A/2: -- [1, 3, 9]&#xA;A/5: -- [1, 3, 9]&#xA;A/6: -- [2, 3]&#xA;A/7: -- [2, 9]&#xA;B/2: -- [1, 3, 5, 9]&#xA;B/3: -- [1, 5, 9]&#xA;B/5: -- [1, 3, 9]&#xA;B/6: -- [2, 3]&#xA;B/7: -- [2, 5, 7, 9]&#xA;B/9: -- [5, 7]&#xA;C/2: -- [5, 9]&#xA;C/4: -- [4, 8, 9]&#xA;C/5: -- [4, 8, 9]&#xA;C/7: -- [5, 9]&#xA;D/1: -- [4, 5, 6]&#xA;D/2: -- [1, 5, 7]&#xA;D/4: -- [4, 7, 8]&#xA;D/5: -- [4, 5, 7, 8]&#xA;D/6: -- [4, 6]&#xA;D/9: -- [1, 5, 6, 7]&#xA;E/1: -- [5, 6]&#xA;E/8: -- [5, 7]&#xA;E/9: -- [5, 6, 7]&#xA;F/1: -- [4, 5, 6, 9]&#xA;F/2: -- [1, 5, 7, 9]&#xA;F/3: -- [1, 5, 7, 9]&#xA;F/4: -- [3, 4, 7]&#xA;F/5: -- [3, 4, 5, 7]&#xA;F/6: -- [3, 4, 6]&#xA;F/9: -- [1, 5, 6, 7]&#xA;G/3: -- [5, 7, 8, 9]&#xA;G/5: -- [7, 9]&#xA;G/7: -- [5, 7]&#xA;G/8: -- [5, 7, 8]&#xA;H/3: -- [7, 8]&#xA;H/4: -- [3, 4, 7]&#xA;H/5: -- [3, 4, 7]&#xA;H/8: -- [3, 7, 8]&#xA;I/1: -- [5, 9]&#xA;I/3: -- [5, 7, 9]&#xA;I/4: -- [3, 7, 9]&#xA;I/8: -- [3, 5, 7]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;My sudoku solver program is stuck on the above field. I didn't implement a backtracking algorithm on purpose, but I also can't figure out the next move, not to mention implementing the according strategy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please give a tip on how to find out the next number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&#xA;The original unsolved sudoku:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   ===================================&#xA;9: || __ __ 3 || __ __ __|| 4  9  2 ||&#xA;8: || 6  __ 1 || __ __ __|| __ __ __||&#xA;7: || __ __ __|| __ 4  8 || __ __ __||&#xA;   ===================================&#xA;6: || __ __ 7 || __ 9  __|| __ 5  8 ||&#xA;5: || __ __ __|| __ 2  __|| __ __ __||&#xA;4: || 5  6  __|| __ 1  __|| 2  __ __||&#xA;   ===================================&#xA;3: || __ __ __|| 2  3  __|| __ __ __||&#xA;2: || __ __ __|| __ __ __|| 6  __ 4 ||&#xA;1: || 7  8  2 || __ __ __|| 3  __ __||&#xA;   ===================================&#xA;      A  B  C    D  E  F    G  H  I &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="16102" LastEditorUserId="16102" LastEditDate="2015-08-31T18:58:35.350" LastActivityDate="2015-08-31T20:53:16.697" Title="Sudoku: need a traceable hint for the next number" Tags="|sudoku|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20764" PostTypeId="1" AcceptedAnswerId="20768" CreationDate="2015-08-31T17:21:04.880" Score="49" ViewCount="3555" Body="&lt;p&gt;This is a chess puzzle someone showed me a long time ago. If you already know this puzzle please give the others a chance to solve it first. This puzzle was originally created by Otto Gallischek and it was published in Weser-Kurier on 1960-02-25. The problem can be seen &lt;a href=&quot;https://www.yacpdb.org/#302939&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; on the website Yet Another Chess Problem Database. It is also featured in this &lt;a href=&quot;https://en.chessbase.com/post/liburkin-kozlowski-memorial-composing-tourneys&quot; rel=&quot;noreferrer&quot;&gt;Chess Base article&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one small change though-this puzzle has a White queen instead of a rook on h3, as the original problem has.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JCN2t.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above you see a position where white is obviously on the winning side. The queen is one step from giving a checkmate. And the heroic &lt;code&gt;h&lt;/code&gt; pawn made his way from &lt;code&gt;h2&lt;/code&gt; to &lt;code&gt;c7&lt;/code&gt; and is waiting for a promotion. However, black saw a last chance and moved &lt;code&gt;Rf3+&lt;/code&gt; in the last turn. The rook can't be taken, because black couldn't move anymore which is a stalemate (= draw). If the king moves away the rook will continue to give checks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you stop the rambling rook and find a way for white to win the game anyway?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: Computers are not allowed, and no cheating.&lt;/p&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="57521" LastEditDate="2019-08-18T03:08:48.657" LastActivityDate="2019-08-18T03:08:48.657" Title="Can you stop the rambling rook?" Tags="|chess|no-computers|" AnswerCount="2" CommentCount="12" ContentLicense="CC BY-SA 4.0" />
  <row Id="20765" PostTypeId="2" ParentId="20586" CreationDate="2015-08-31T17:32:45.470" Score="10" Body="&lt;p&gt;You are a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mouse&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 1 spoiler:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The squall and call each refer to a noise. Both computer mice and animal mice make quiet sounds (squall can mean a high-pitched noise).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2 spoiler:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer itself is a word play. Also, &quot;light&quot; and &quot;point&quot; are word plays.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3 spoiler:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;point&lt;/strong&gt; you in the right direction and the answer will &lt;strong&gt;click&lt;/strong&gt; right away&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With jolt or breath&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The mouse can be electric or living&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I dine with death&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mice are often killed while eating at a trap&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unwanted by most&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mice are most often considered pests&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Light I can boast&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A reference to optical and laser computer mice. &quot;Light&quot; can also refer to weight.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lectern or wall&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Computer mice on a desk or real mice in a wall&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tethered, not all&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wired or wireless computer mice&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your will or my squall, the point of my call&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your will = using a computer mouse to select an item&lt;br&gt;&#xA; My squall = Etymology: &quot;squeal&quot; + &quot;bawl&quot;. A squall can be a high-pitched noise. Hint 1 says it can be quiet, like a squeak.&#xA; The point of my call = Word play: &quot;point&quot; is a reference to a mouse pointer, but also means &quot;purpose&quot;. The call is a noise, and the purpose of a mouse making noise is his own squeaking, and the purpose of a computer mouse making a noise (click) is the will of the person using it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5340" LastEditorUserId="15991" LastEditDate="2015-09-01T19:10:30.557" LastActivityDate="2015-09-01T19:10:30.557" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20766" PostTypeId="1" CreationDate="2015-08-31T17:53:49.033" Score="3" ViewCount="561" Body="&lt;p&gt;Your goal is to create a &lt;strong&gt;word ladder&lt;/strong&gt; that goes from the word &quot;&lt;em&gt;puzzle&lt;/em&gt;&quot; to the word &quot;&lt;em&gt;riddle&lt;/em&gt;&quot; and then to the word &quot;&lt;em&gt;solution&lt;/em&gt;&quot;. In that order!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, if it's possible, if not you better show me why not!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Valid word-ladder moves:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Add a new letter&lt;/li&gt;&#xA;&lt;li&gt;Take away a letter&lt;/li&gt;&#xA;&lt;li&gt;Replace a single letter&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If more than 1 solution is found, the shortest will be accepted!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: A word must be on &lt;strong&gt;dictionary.com&lt;/strong&gt; to be counted as a valid word!&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="69582" LastEditDate="2021-08-15T23:24:26.027" LastActivityDate="2021-08-15T23:24:26.027" Title="From Puzzle to Riddle to Solution" Tags="|word|optimization|english|word-ladder|" AnswerCount="4" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="20767" PostTypeId="2" ParentId="20766" CreationDate="2015-08-31T18:14:55.697" Score="2" Body="&lt;p&gt;This has 14 steps from &lt;em&gt;puzzle&lt;/em&gt; to &lt;em&gt;riddle&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;puzzle&lt;/strong&gt;&lt;br&gt;&#xA;puzzled&lt;br&gt;&#xA;muzzled&lt;br&gt;&#xA;&lt;a href=&quot;http://dictionary.reference.com/browse/muzzed&quot; rel=&quot;nofollow&quot;&gt;muzzed&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://dictionary.reference.com/browse/fuzzed&quot; rel=&quot;nofollow&quot;&gt;fuzzed&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://dictionary.reference.com/browse/fuzed&quot; rel=&quot;nofollow&quot;&gt;fuzed&lt;/a&gt;&lt;br&gt;&#xA;fused&lt;br&gt;&#xA;fuse&lt;br&gt;&#xA;ruse&lt;br&gt;&#xA;rise&lt;br&gt;&#xA;ride&lt;br&gt;&#xA;rider&lt;br&gt;&#xA;ridder&lt;br&gt;&#xA;riddler&lt;br&gt;&#xA;&lt;strong&gt;riddle&lt;/strong&gt;&lt;br&gt;&#xA;&lt;em&gt;I haven't found a path that ends in solution.&lt;/em&gt;&lt;br&gt;&#xA;&lt;strong&gt;solution&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-09-01T12:04:45.633" LastActivityDate="2015-09-01T12:04:45.633" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20768" PostTypeId="2" ParentId="20764" CreationDate="2015-08-31T18:23:10.200" Score="63" Body="&lt;p&gt;If white can ever stay out of check for one turn, then it can promote its pawn and put black in checkmate. So in order to play perfectly, black must attempt to put white in check with every move. In turn, white should make sure that black has only one option for a check in the next move, or else the rook will &quot;break free&quot; and have much more influence over white's moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;White's overall game plan is to sacrifice pawns and maneuver the king until black falls into a simultaneous capture+checkmate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/c3ax9.gif&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/c3ax9.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA; 1. e3          Pawn moves forward to block rook.&#xA; 1. ...   Rxe3+ Rook captures pawn and places king in check.&#xA; 2. c3          Pawn moves forward to block rook.&#xA; 2. ...   Rxc3+ Rook captures pawn and places king in check.&#xA; 3. Ka2         King begins a &quot;serpentine&quot; maneuver, which will continue until move 9.&#xA; 3. ...   Ra3+  Rook takes only available move to put king in check, continuing until move 9.&#xA; 4. Kb1   Ra1+ &#xA; 5. Kc2   Rc1+ &#xA; 6. Kd3   Rc3+ &#xA; 7. Ke2   Re3+ &#xA; 8. Kf1   Re1+ &#xA; 9. Kg2   Rg1+ &#xA;10. Kf3   Rxg3+ Rook captures pawn in row 3 and places king in check.&#xA;11. Ke2         King begins serpentining back to column A.&#xA;11. ...   Re3+  Rook continues to pursue.&#xA;12. Kd1   Re1+ &#xA;13. Kc2   Rc1+ &#xA;14. Kb3   Rc3+ &#xA;15. Ka2   Ra3+ &#xA;16. Qxa3#       Queen takes rook and places black king in checkmate.&#xA;&lt;/pre&gt;&#xA;" OwnerUserId="1497" LastEditorUserId="4421" LastEditDate="2015-08-31T20:32:50.123" LastActivityDate="2015-08-31T20:32:50.123" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20769" PostTypeId="2" ParentId="20756" CreationDate="2015-08-31T18:32:14.043" Score="26" Body="&lt;p&gt;I'm gonna give a slightly more mathematical answer here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Intermediate_value_theorem&quot;&gt;Intermediate Value Theorom&lt;/a&gt; is defined as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $f(x)$ is a continuous function on $[a,b]$, for every $d$ between $f(a)$ and $f(b)$, there exists a $c$ between $a$ and $b$ so that $f(c) = d$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Before we apply this theorom to the monk, let us define some values:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Let $t$ be the time since the monk has begun an ascent or descent, in hours.&lt;/li&gt;&#xA;&lt;li&gt;Let $f(t)$ be the height of the monk going up the mountain.&lt;/li&gt;&#xA;&lt;li&gt;Let $g(t)$ be the height of the monk going down the mountain.&lt;/li&gt;&#xA;&lt;li&gt;Let the height the monk starts at be $0$, and the height of the mountain be $m$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We are attempting to prove that there exists some $t$ in $[0,10]$, such that $f(t)=g(t)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that the height of the monk is a continuous function of time, both going up the mountain and back down, so we can add a couple more conditions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The domain of $f(t)$ and $g(t)$ are $t=[0,10]$.&lt;/li&gt;&#xA;&lt;li&gt;Both $f(t)$ and $g(t)$ are continous functions with the following properties:&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$f(0)=0$&lt;/li&gt;&#xA;&lt;li&gt;$f(10)=m$&lt;/li&gt;&#xA;&lt;li&gt;$g(0)=m$&lt;/li&gt;&#xA;&lt;li&gt;$g(10)=0$&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Let us construct a function $d(t) = f(t) - g(t)$. This function is on the domain $t=[0,10]$ and the range $[-m,m]$, and remains continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By applying the Intermediate Value Theorom, we know that for any $h$ in $[-m,m]$, there is a $t$ such that $d(t)=h$. If we take $h=0$, then we know that there is a $t$ such that $d(t)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, we have proved that there is a $t$ such that $d(t)=f(t)-g(t)=0$, proving that there is a $t$ such that $f(t)=g(t)$. $\blacksquare$&lt;/p&gt;&#xA;" OwnerUserId="4966" LastEditorUserId="4687" LastEditDate="2015-08-31T21:10:07.930" LastActivityDate="2015-08-31T21:10:07.930" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20770" PostTypeId="2" ParentId="20766" CreationDate="2015-08-31T18:51:17.817" Score="2" Body="&lt;p&gt;A shorter path from puzzle to riddle (15 steps)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;puzzle&lt;/strong&gt;&lt;br&gt;&#xA;muzzle&lt;br&gt;&#xA;muzzled&lt;br&gt;&#xA;muzzed&lt;br&gt;&#xA;fuzzed&lt;br&gt;&#xA;fuzed&lt;br&gt;&#xA;fuze&lt;br&gt;&#xA;faze&lt;br&gt;&#xA;raze&lt;br&gt;&#xA;rase&lt;br&gt;&#xA;rise&lt;br&gt;&#xA;ride&lt;br&gt;&#xA;rider&lt;br&gt;&#xA;ridder&lt;br&gt;&#xA;riddler&lt;br&gt;&#xA;&lt;strong&gt;riddle&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-08-31T18:51:17.817" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20771" PostTypeId="2" ParentId="20708" CreationDate="2015-08-31T19:20:54.457" Score="1" Body="&lt;p&gt;Game 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The telepath will never call on the river.  She will either raise or fold.  And she will only fold if she knows she's going to lose.  She should bluff enough that nearly half her bluffs are called (assuming pot-limit raises).  If half your bluffs are called on pot-limit raises, then the bluffs break even.  The more your bluffs are called, the more often your real bets are called, and that's where the big money is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If she has a tell, then she'll realize the other person is seeing the tell and she can work on covering/faking the tell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The muggle should just accept that he's going to lose all his money.  His only chance would be to bet big before the flop, but this is a risky strategy.  When he starts with good cards, like a high pair, and the telepath starts with a higher pair, the muggle will lose big.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Game 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is even worse: the clairvoyant knows what all the cards will be, so the telepath now knows this, too.  The entire hand is a foregone conclusion in the telepath's mind.  The clairvoyant has zero chance.  The telepath will bet the maximum on half the hands (the one's she's winning) and often enough on the others to convince the clairvoyant to fold winning hands and bet on losing hands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Game 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This game has some level of subtlety.  The clairvoyant will know what she will have at the end of the game, but won't know what the other player will have.  However, there are lots of hands where you know you almost certainly must have the highest hand: an inside straight, a high straight, a flush with a high kicker, full-house or four-of-a-kind.  Even a high three-of-a-kind can often be an undisputable winner (eg. you have two aces, there's an ace on the table, and there are no other pairs or straight or flush candidates).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The clairvoyant requires skill with statistics and some skill at reading her opponent.  But I suspect even a casual player with a strong statistics background would trounce most any professional.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2015-08-31T19:20:54.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20772" PostTypeId="2" ParentId="20740" CreationDate="2015-08-31T19:23:52.730" Score="6" Body="&lt;p&gt;I'll add my suggestion for a a title too. I'm a fan of the much terser&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Spiral²&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Additionally, if you like puns you might try&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Squarey-go-Round!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4513" LastEditorUserId="4513" LastEditDate="2015-08-31T19:50:09.937" LastActivityDate="2015-08-31T19:50:09.937" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20773" PostTypeId="2" ParentId="20738" CreationDate="2015-08-31T19:49:23.627" Score="4" Body="&lt;p&gt;Based off of your posting on another website: &lt;a href=&quot;http://www.careercup.com/question?id=5114718433837056&quot; rel=&quot;nofollow&quot;&gt;http://www.careercup.com/question?id=5114718433837056&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I gathered the name of the company is Micron. A quick look at wikipedia shows Micron was founded in Boise, Idaho: &lt;a href=&quot;https://en.wikipedia.org/wiki/Micron_Technology&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Micron_Technology&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also noticed they work with &quot;many forms of semiconductor devices, including dynamic random-access memory, flash memory, and solid-state drives.&quot; - Hence the first hint towards a &quot;place where ohms law is checked every day&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also found that if you remove the &quot;B&quot; from the from the front of &quot;Boise&quot; -  you get &quot;Oise&quot; - an area of France that according to Google has many winerys and wine tasting tours. A nod towards &quot;you can feel the grapes (without the first letter).&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the clue you gave - I think the (Golden State) in parenthesis is your inclusion since you say you have guessed California so far. However I can't really find a direct connection to Boise and gold except that some gold medalists come from Boise in different sports and that Idaho in general seems to be popular for gold panning. There is also a city in Idaho named Golden - how that relates to Boise I am not sure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably a stretch but it was my lunch break and I wanted to give it a shot. I really do hope you will be honest with the interviewer if you get an answer from the internet rather than passing it off as your own since it appears you are asking on multiple sites. A lot of times these questions are more to see how you think rather than if you get the right answer. In fact it could even be a nonsensical riddle just to see what you come up with. &lt;/p&gt;&#xA;" OwnerUserId="15843" LastEditorUserId="15843" LastEditDate="2015-09-01T12:12:08.437" LastActivityDate="2015-09-01T12:12:08.437" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20774" PostTypeId="1" AcceptedAnswerId="20859" CreationDate="2015-08-31T20:15:00.163" Score="7" ViewCount="781" Body="&lt;p&gt;After a long night of stargazing, your best friend left you a cryptic note. A single sentence is written on the note, in beautifully printed cursive:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;I softly gazed nigh a lit, dull skyline, contemplating a dismayingly frozen future.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What was your best friend up all night looking at?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Hint #1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;a href=&quot;/questions/tagged/cryptograms&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;cryptograms&amp;#39;&quot; rel=&quot;tag&quot;&gt;cryptograms&lt;/a&gt; tag is there for a reason, guys. Seriously, anything that you &lt;em&gt;think&lt;/em&gt; could be the answer is not. You will &lt;strong&gt;&lt;em&gt;know&lt;/em&gt;&lt;/strong&gt; you have found the answer once you do. The sentence was chosen to sound akin to stargazing, but if you start parsing it for meaning you've already gone way too far.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Recently, there was a puzzle about a specific 'international animal' whose format was similar, albeit &lt;em&gt;certainly not identical&lt;/em&gt;, to the format of this poem. This should key you in on how to read past the words.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="11475" LastEditDate="2015-09-08T13:31:59.153" LastActivityDate="2015-09-09T21:57:17.850" Title="A long night of stargazing" Tags="|word|cipher|enigmatic-puzzle|" AnswerCount="5" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="20775" PostTypeId="2" ParentId="20766" CreationDate="2015-08-31T20:18:54.217" Score="3" Body="&lt;p&gt;I'm hoping we can switch between american and british english...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;puzzle&lt;/strong&gt;&lt;br&gt;&#xA;muzzle&lt;br&gt;&#xA;muzzled&lt;br&gt;&#xA;muzzed&lt;br&gt;&#xA;fuzzed&lt;br&gt;&#xA;fuzed&lt;br&gt;&#xA;fuze&lt;br&gt;&#xA;fuse&lt;br&gt;&#xA;ruse&lt;br&gt;&#xA;rise&lt;br&gt;&#xA;ride&lt;br&gt;&#xA;rider&lt;br&gt;&#xA;ridder&lt;br&gt;&#xA;riddler&lt;br&gt;&#xA;&lt;strong&gt;riddle&lt;/strong&gt;&lt;br&gt;&#xA;widdle&lt;br&gt;&#xA;waddle&lt;br&gt;&#xA;waddled&lt;br&gt;&#xA;wadded&lt;br&gt;&#xA;waded&lt;br&gt;&#xA;wade&lt;br&gt;&#xA;wad&lt;br&gt;&#xA;rad&lt;br&gt;&#xA;red&lt;br&gt;&#xA;redo&lt;br&gt;&#xA;redon&lt;br&gt;&#xA;radon&lt;br&gt;&#xA;radin&lt;br&gt;&#xA;radio&lt;br&gt;&#xA;ratio&lt;br&gt;&#xA;ration&lt;br&gt;&#xA;nation&lt;br&gt;&#xA;enation&lt;br&gt;&#xA;elation&lt;br&gt;&#xA;velation&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;em&gt;missing word&lt;/em&gt;&lt;br&gt;&#xA;volution&lt;br&gt;&#xA;&lt;strong&gt;solution&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="12206" LastEditorUserId="10399" LastEditDate="2015-09-01T13:40:50.487" LastActivityDate="2015-09-01T13:40:50.487" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20776" PostTypeId="2" ParentId="20763" CreationDate="2015-08-31T20:34:18.277" Score="4" Body="&lt;p&gt;After getting G6 (intersection of 7958 and 4263) and H1 (intersection of 7823 and 59 and eliminated 64), the next move would be I4 (9). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know this because C4 and C5 must be 4 or 8.  E6 (9) eliminates A6, B6, which means there's a 9 on row 5.  H9 (9) eliminates H4, leaving I4 = 9.  Follow up with F1 = 9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intermediate grid:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   ===================================&#xA;9: || _8 __ 3 || __ __ __|| 4  9  2 ||&#xA;8: || 6  _4 1 || _9 __ _2|| __ __ __||&#xA;7: || __ __ __|| _3 4  8 || __ _6 _1||&#xA;   ===================================&#xA;6: || __ __ 7 || _6 9  _4|| _1 5  8 ||&#xA;5: || __ __ __|| __ 2  __|| __ __ _6||&#xA;4: || 5  6  __|| __ 1  __|| 2  __ _9||&#xA;   ===================================&#xA;3: || __ __ _6|| 2  3  __|| __ __ __||&#xA;2: || __ __ __|| __ _8 __|| 6  _2 4 ||&#xA;1: || 7  8  2 || _4 _6 _9|| 3  _1 _5||&#xA;   ===================================&#xA;      A  B  C    D  E  F    G  H  I &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Completed grid:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   ===================================&#xA;9: || _8 _5 3 || _1 _7 _6|| 4  9  2 ||&#xA;8: || 6  _4 1 || _9 _5 _2|| _8 _7 _3||&#xA;7: || _2 _7 _9|| _3 4  8 || _5 _6 _1||&#xA;   ===================================&#xA;6: || _3 _2 7 || _6 9  _4|| _1 5  8 ||&#xA;5: || _1 _9 _8|| _5 2  _3|| _7 _4 _6||&#xA;4: || 5  6  _4|| _8 1  _7|| 2  _3 _9||&#xA;   ===================================&#xA;3: || _4 _1 _6|| 2  3  _5|| _9 _8 _7||&#xA;2: || _9 _3 _5|| _7 _8 _1|| 6  _2 4 ||&#xA;1: || 7  8  2 || _4 _6 _9|| 3  _1 _5||&#xA;   ===================================&#xA;      A  B  C    D  E  F    G  H  I &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-08-31T20:53:16.697" LastActivityDate="2015-08-31T20:53:16.697" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20777" PostTypeId="2" ParentId="20551" CreationDate="2015-08-31T20:37:39.940" Score="8" Body="&lt;p&gt;Aha! Finally. I spent a good few days on this riddle back when it was an 8-letter word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answers are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm depending on you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RELIANT&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm lost in your eyes; there's just something about them...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RETINAL&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm one of many, rarely referred to individually, but just look inside and you'll find me with the others.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ENTRAIL&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When the previous word's done with you, come see me!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LATRINE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A sailor's friend, I'm the concatenation of two words (slightly ironic because the first is not known to play well with the second).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RATLINE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15908" LastActivityDate="2015-08-31T20:37:39.940" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20778" PostTypeId="2" ParentId="20766" CreationDate="2015-08-31T21:12:56.653" Score="2" Body="&lt;p&gt;My proof of impossibility (I used a computer to reinforce my claim):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used a very wide dictionary (235,886 words 2.4 MB)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting at our endpoint &quot;solution&quot; the only options are:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;solution&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;solation&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;colation&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;eolation&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;isolation&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;sonation&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;volation&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;volution&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;volation&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;evolution&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;revolution&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;devolution&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;volition&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;vomition&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;nolition&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bold&lt;/strong&gt; = Not on dictionary.com&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Italicized&lt;/em&gt; = True dead end&lt;/p&gt;&#xA;" OwnerUserId="11926" LastEditorUserId="11926" LastEditDate="2015-09-01T14:03:45.297" LastActivityDate="2015-09-01T14:03:45.297" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="20780" PostTypeId="2" ParentId="20693" CreationDate="2015-09-01T03:33:46.090" Score="9" Body="&lt;p&gt;I do not believe this is the correct answer, but I believe this is the way to approach this riddle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seven follows suit,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;suitseven&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Eight go in line,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;lieightne&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Combined, 9 leave the queue.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;suitsevenlieightne - first 9 letters in the &quot;queue&quot; = lieightne&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Leaving i e, once removed, rearranged casts light upon the answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;lieightne - ie = lightne --&gt; lighten&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The word &quot;lighten&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6178" LastActivityDate="2015-09-01T03:33:46.090" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20781" PostTypeId="2" ParentId="20011" CreationDate="2015-09-01T06:07:13.613" Score="16" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Beethoven!&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I work on a farm&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The name Beethoven means &lt;strong&gt;beet garden&lt;/strong&gt; (&quot;hoven&quot; is plural of &quot;hof&quot;, which is German for &quot;farming estate&quot;). It is generally believed that the Beethoven family came from a line of Catholic farmers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but it's not as romantic as you think&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Beethoven straddles the Classical and &lt;strong&gt;Romantic eras&lt;/strong&gt; of Western classical music. Although a composer in the Romantic Period, his work, especially his earlier sonatas, were not very Romantic in style.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I need to wake up around sunrise to go to work&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;The Cockcrow&lt;/strong&gt; (Violin Sonata No. 10 in Gmaj, Op. 96)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and I get paid just 4 pennies per minute&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's easy to &lt;strong&gt;Rage Over a Lost Penny&lt;/strong&gt; when you earn so little (Rondo a Capriccio in Gmaj, Op. 129)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I had to take a second job to pay the bills&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Taking two jobs is &lt;em&gt;moonlighting&lt;/em&gt; (Piano Sonata No. 14 in C#m, Op. 27, No. 2, also known as the &lt;strong&gt;Moonlight Sonata&lt;/strong&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Honestly, I'd do anything just to get an apartment&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Consecration of the House&lt;/strong&gt; (Die Weihe des Hauses, overture, Op. 124)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I'm such a sad sack it'll probably never happen&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Sonata Pathétique&lt;/strong&gt; (Piano Sonata No. 8 in C minor, Op. 13)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bonus hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;I wish I had a job I was more passionate about.&lt;/em&gt;&lt;br&gt;&#xA; Beethoven slowly became &lt;strong&gt;deaf&lt;/strong&gt; which became a small obstacle in his enjoyment of music. He was also known for a &lt;strong&gt;troubled love life&lt;/strong&gt;, which impacted the frequency and tone of his compositions, as well as the quality of his performances.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bonus hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;I have to put in a heroic effort just to get out of bed.&lt;/em&gt;&lt;br&gt;&#xA; &quot;The Cockcrow&quot; was one of the last pieces written in the &lt;strong&gt;Heroic period&lt;/strong&gt; of Beethoven's life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15960" LastEditorUserId="15960" LastEditDate="2015-09-01T06:28:32.910" LastActivityDate="2015-09-01T06:28:32.910" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="20783" PostTypeId="2" ParentId="18614" CreationDate="2015-09-01T11:06:55.383" Score="-4" Body="&lt;p&gt;There is not enough information to determine the answer. The simplest explanation is that the sequence is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$5,6,8,11,45,74,101,101,101,101,\cdots$ (remaining constant all the way)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reason is that any other explanation will almost certainly require a longer description of the sequence than this one. So by Occam's razor this one would be preferable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My answer can be easily made mathematically precise via information theory, specifically one can define Kolmogorov complexity. To be fair we cannot choose a language specially designed for a particular answer otherwise its complexity can be made to be arbitrarily small. If we hence choose any mainstream Turing complete language we will almost certainly find that the sequence I gave above has the smallest program generating it.&lt;/p&gt;&#xA;" OwnerUserId="11806" LastEditorUserId="11806" LastEditDate="2015-09-01T14:13:15.510" LastActivityDate="2015-09-01T14:13:15.510" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20787" PostTypeId="1" AcceptedAnswerId="20793" CreationDate="2015-09-01T14:39:34.667" Score="3" ViewCount="453" Body="&lt;p&gt;Underneath this is a series of ones and zeroes; they are not to be interpreted as binary numbers. Perhaps some research is in order to determine their meaning. Comb the web once you determine what they are and you'll find resources that will help you determine what they represent. And if you're wondering, would the 'trivia' tag be relevant here, the answer is no because you already have a hint hidden herein.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;000000000 101 0110111 0110111 0110111 0100011 0101111 0010011 01010 1110010 1000010 1110100 1001110 1001000 1100110 101 000000000 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13970" LastActivityDate="2015-09-01T16:56:39.960" Title="Ones and Zeroes" Tags="|riddle|number-sequence|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20789" PostTypeId="1" CreationDate="2015-09-01T15:17:29.430" Score="2" ViewCount="764" Body="&lt;p&gt;Captain Jack and his crew have uncovered another treasure full of glistening golds and shiny silvers. There are 2015 chest and each of them contains some amount of gold and silver coins. Different chests may contain different amounts of gold or silver.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Captain Jack, being greedy as ever, wants to keep at least half of the gold and half of the silver. However, he is aware of dissatisfaction among the crew and wants to avoid a mutiny at all costs. So, he wants to give the crew as many &lt;strong&gt;treasure chests&lt;/strong&gt; as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Captain Jack must tell the crews how many chests he is going to give them &lt;strong&gt;before&lt;/strong&gt; he opens up the chests. After that he opens up each chest and decides which chests to keep and which to give away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What number should Jack tell the crews to ensure that he keeps at least half of the gold and half of the silver while keeping the crew as satisfied as possible? In other words, Find the maximum $N$ such that Jack can always distribute the chests in such a way that he gets to keep half the gold and half the silver and give away $N$ chests.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Notes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The captain may count how many silver and gold is in each chest, but cannot remove any gold or silver from its chest, since doing so will result in a irreversible curse for the entire crew and the captain.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Without opening a chest there is no way to tell how many gold or silver it contains.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="15544" LastEditorUserId="15544" LastEditDate="2015-09-01T16:16:32.880" LastActivityDate="2015-09-01T16:16:32.880" Title="Captain Jack and the treasure chests" Tags="|logical-deduction|" AnswerCount="3" CommentCount="13" FavoriteCount="0" ClosedDate="2015-09-01T20:30:14.090" ContentLicense="CC BY-SA 3.0" />
  <row Id="20790" PostTypeId="1" AcceptedAnswerId="21913" CreationDate="2015-09-01T15:22:41.373" Score="11" ViewCount="889" Body="&lt;h2&gt;&lt;em&gt;Who are two of my favourite puzzlers on Puzzling.SE?&lt;/em&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Complete the crossword below to find out.  A &lt;strong&gt;complete answer&lt;/strong&gt; will provide the names of &lt;strong&gt;&lt;em&gt;two&lt;/em&gt;&lt;/strong&gt; of my favourite puzzlers on Puzzling.SE as well as the &lt;strong&gt;completed crossword&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a twist here, but I'll leave it up to you to figure out what it is.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I have attempted to follow the conventions of &lt;em&gt;New York Times&lt;/em&gt; crosswords when it comes to the clues, but I'm sure there are a few places where I haven't quite succeeded, so please keep an open mind...!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those who want to print and work on it, &lt;a href=&quot;https://www.dropbox.com/s/k2nno0fkjgcx6t1/FavouritePuzzlers.pdf?dl=0&quot; rel=&quot;nofollow noreferrer&quot;&gt;here is a one-page printable version&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/o3Y7u.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/o3Y7u.png&quot; alt=&quot;Crossword grid&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&lt;strong&gt;&lt;em&gt;Across:&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&#xA;1. Bad end for a tooth?&lt;br&gt;&#xA;5. Formal sanction&lt;br&gt;&#xA;16. Word with &quot;box&quot; or &quot;muscle&quot;&lt;br&gt;&#xA;19. Nocturnal critter&lt;br&gt;&#xA;20. Like some cuisine&lt;br&gt;&#xA;21. _____ the ground&lt;br&gt;&#xA;22. World Cup cheer&lt;br&gt;&#xA;23. Inflection&lt;br&gt;&#xA;25. Untested&lt;br&gt;&#xA;26. Extinct bird&lt;br&gt;&#xA;27. Eyeballs or moons, e.g.&lt;br&gt;&#xA;28. It has bark but no bite&lt;br&gt;&#xA;30. Frivolous escapade&lt;br&gt;&#xA;32. Graveside music&lt;br&gt;&#xA;34. Stern direction&lt;br&gt;&#xA;37. Daytime TV choice&lt;br&gt;&#xA;41. All in ______ work&lt;br&gt;&#xA;42. &lt;em&gt;Ghostbusters&lt;/em&gt; Ph.D. Spengler&lt;br&gt;&#xA;44. Nighttime annoyance&lt;br&gt;&#xA;45. Along with &lt;em&gt;storge&lt;/em&gt; and &lt;em&gt;philia&lt;/em&gt;, kin of 35 Down&lt;br&gt;&#xA;48. Island on the Croatian coast of the Adriatic Sea&lt;br&gt;&#xA;49. Provoke to anger&lt;br&gt;&#xA;51. Part of GPS coords.&lt;br&gt;&#xA;52. Add a little colour to&lt;br&gt;&#xA;53. Prefix meaning &quot;personal, private&quot;&lt;br&gt;&#xA;55. California wines&lt;br&gt;&#xA;56. Family reunion attendee&lt;br&gt;&#xA;58. &lt;em&gt;Petro-Canada&lt;/em&gt; competitor&lt;br&gt;&#xA;59. Type of 7-line poem&lt;br&gt;&#xA;61. With gravity&lt;br&gt;&#xA;63. Means of manual propulsion&lt;br&gt;&#xA;64. Burn soother&lt;br&gt;&#xA;65. Penned&lt;br&gt;&#xA;66. In a languid manner&lt;br&gt;&#xA;68. Food store&lt;br&gt;&#xA;71. One of only two types of egg-laying mammals&lt;br&gt;&#xA;74. Beasts of burden&lt;br&gt;&#xA;75. ______ of rubbish; a falsehood&lt;br&gt;&#xA;76. One of only 5 on Earth&lt;br&gt;&#xA;77. Skip out on someone&lt;br&gt;&#xA;79. Prefix with &lt;em&gt;kind&lt;/em&gt; or &lt;em&gt;made&lt;/em&gt;&lt;br&gt;&#xA;80. State of disbelief?&lt;br&gt;&#xA;82. &lt;em&gt;Narcissus pseudonarcissus&lt;/em&gt;, e.g.&lt;br&gt;&#xA;86. Where the sun rises in Cancún&lt;br&gt;&#xA;88. Setting with knights and castles&lt;br&gt;&#xA;89. Forcibly remove&lt;br&gt;&#xA;91. Lean and tough&lt;br&gt;&#xA;92. Ladies' fingers&lt;br&gt;&#xA;93. Sis, bro, or cuz&lt;br&gt;&#xA;94. Close to the bullseye&lt;br&gt;&#xA;96. Protocol for server storage (abbr.)&lt;br&gt;&#xA;97. (He) knows, according to Jaques&lt;br&gt;&#xA;98. Tools for the links&lt;br&gt;&#xA;101. Half the title of a Kasabian song&lt;br&gt;&#xA;102. Greek muse&lt;br&gt;&#xA;104. Ingredient in Chinese soup&lt;br&gt;&#xA;107. You'll find several in intensive care units&lt;br&gt;&#xA;108. Inventor of dynamite&lt;br&gt;&#xA;109. Indigenous animals&lt;br&gt;&#xA;111. &quot;____ of wine, a cigarette, and then it's time to go&quot; - Leonard Cohen&lt;br&gt;&#xA;113. Jacob's twin&lt;br&gt;&#xA;114. Started, as a fire&lt;br&gt;&#xA;115. Potential PETA cry&lt;br&gt;&#xA;119. Vena cava, e.g.&lt;br&gt;&#xA;125. Fuss and bother&lt;br&gt;&#xA;126. Draw up&lt;br&gt;&#xA;127. It's just across the strait from Morocco&lt;br&gt;&#xA;128. ____'clock: coffee break time&lt;br&gt;&#xA;129. Ratio word&lt;br&gt;&#xA;130. ________ trend: eschews the latest fad&lt;br&gt;&#xA;131. Miss in Mex.&lt;br&gt;&#xA;&lt;/sup&gt;&#xA;&lt;sup&gt;&lt;br&gt;&#xA;&lt;strong&gt;&lt;em&gt;Down&lt;/em&gt;&lt;/strong&gt;:&lt;br&gt;&#xA;1. Play part&lt;br&gt;&#xA;2. Baby sound&lt;br&gt;&#xA;3. Type of &lt;em&gt;ex gratia&lt;/em&gt; payment&lt;br&gt;&#xA;4. Verve&lt;br&gt;&#xA;5. Court officials&lt;br&gt;&#xA;6. Off-road transport, for short&lt;br&gt;&#xA;7. However, poetically&lt;br&gt;&#xA;8. Common computer configuration file type&lt;br&gt;&#xA;9. Library section, briefly&lt;br&gt;&#xA;10. Cake shop employees&lt;br&gt;&#xA;11. Greek sea&lt;br&gt;&#xA;12.  _____ and feather&lt;br&gt;&#xA;13. Rage&lt;br&gt;&#xA;14. Potential salary in sales (abbr.)&lt;br&gt;&#xA;15. Poetic present&lt;br&gt;&#xA;16. Metaphors, e.g.&lt;br&gt;&#xA;17. Soothing plant&lt;br&gt;&#xA;18. Stern, on land&lt;br&gt;&#xA;24. Extremely large&lt;br&gt;&#xA;29. Daughter of Hyperion&lt;br&gt;&#xA;31. Sharp mountain ridge&lt;br&gt;&#xA;32. Bathsheba's lover&lt;br&gt;&#xA;33. _______ taw a puddy tat!&lt;br&gt;&#xA;34. Chastise&lt;br&gt;&#xA;35. See 45 Across&lt;br&gt;&#xA;36. Outfit for Octavian?&lt;br&gt;&#xA;38. Cornmeal porridge&lt;br&gt;&#xA;39. Agent of divine communication&lt;br&gt;&#xA;40. Piranha boss in Nintendo's Mario games&lt;br&gt;&#xA;43. Wilder ending?&lt;br&gt;&#xA;46. Marked down&lt;br&gt;&#xA;47. 2010 Indian rom-com: &lt;em&gt;I hate luv _______&lt;/em&gt;&lt;br&gt;&#xA;50. Stutterer's dissent&lt;br&gt;&#xA;54. WWII landing site: _____ Beach&lt;br&gt;&#xA;56. Stereotypical Puzzling.SE user&lt;br&gt;&#xA;57. Blacksmith's medium&lt;br&gt;&#xA;60. ____ up: drunk, slangily&lt;br&gt;&#xA;62. Gemini, e.g.&lt;br&gt;&#xA;66. Street kid&lt;br&gt;&#xA;67. &quot;I'm _____ walker, but I never walk back.&quot; - Abraham Lincoln&lt;br&gt;&#xA;68. Walk-on roles&lt;br&gt;&#xA;69. Home of Nome&lt;br&gt;&#xA;70. Many a member of Puzzling.SE&lt;br&gt;&#xA;71. Level of hierarchy&lt;br&gt;&#xA;72. Middling grades&lt;br&gt;&#xA;73. Hospital illnesses (abbr.)&lt;br&gt;&#xA;76. Not me or you&lt;br&gt;&#xA;77. Kevin and Francis&lt;br&gt;&#xA;78. Rears, to a sailor&lt;br&gt;&#xA;80. Pungent&lt;br&gt;&#xA;81. With &quot;chow&quot;, a delicious dish&lt;br&gt;&#xA;82. Matt in &lt;em&gt;Dogma&lt;/em&gt;&lt;br&gt;&#xA;83. Ones who correct misconceptions&lt;br&gt;&#xA;84. Livid&lt;br&gt;&#xA;85. Certain disinfectant&lt;br&gt;&#xA;87. Eschew takeout&lt;br&gt;&#xA;90. One of my favourite Puzzling.SE puzzlers&lt;br&gt;&#xA;95. On edge&lt;br&gt;&#xA;99. Almost at&lt;br&gt;&#xA;100. Serpentine sound&lt;br&gt;&#xA;103. Uses a spit&lt;br&gt;&#xA;105. Gospel, soul, blues, or funk (abbr.)&lt;br&gt;&#xA;106. Its capital is Lhasa&lt;br&gt;&#xA;109. Part of an envelope&lt;br&gt;&#xA;110. Assistant&lt;br&gt;&#xA;112. Pro&lt;br&gt;&#xA;113. Nights before&lt;br&gt;&#xA;116. Spanish gold&lt;br&gt;&#xA;117. Fairy&lt;br&gt;&#xA;118. MMA org.&lt;br&gt;&#xA;120. Leftover morsel&lt;br&gt;&#xA;121. Exclamation of awe&lt;br&gt;&#xA;122. Printer resolution&lt;br&gt;&#xA;123. Tolkien creature&lt;br&gt;&#xA;124. Mauna ____&lt;br&gt;&#xA;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Did you have any &quot;wrong&quot; guesses when completing the puzzle, especially around the area of 90D?  Perhaps you should explore those more. Compare your answers with other solvers. Did they get the same results?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="4421" LastEditDate="2015-09-03T17:47:38.840" LastActivityDate="2015-09-04T02:58:15.657" Title="Crossword: My favourite puzzlers" Tags="|enigmatic-puzzle|crosswords|" AnswerCount="2" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="20791" PostTypeId="2" ParentId="20789" CreationDate="2015-09-01T15:34:25.770" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;2015&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He can't transfer gold or silver between chests, but he can still remove gold and silver from chests. He removes gold and silver until he has at least half, then gives all the chests away.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12462" LastActivityDate="2015-09-01T15:34:25.770" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20792" PostTypeId="2" ParentId="20789" CreationDate="2015-09-01T15:43:20.597" Score="0" Body="&lt;p&gt;&lt;strong&gt;Post-edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Captain Jack must tell the crews how many chests he is going to give them before he opens up the chests.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Captain Jack doesn't want chests, he wants coin. The crew can have all 2015 chests after the gold and silver has been removed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pre-edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming he can't remove any gold or silver in addition to not being able to transfer, the maximum number of chests he could give is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;2014. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Example, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;chest 1 has 2014 gold and 2014 silver, and the remaining chests have 1g and 1 silver each.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15990" LastEditorUserId="15990" LastEditDate="2015-09-01T16:06:58.577" LastActivityDate="2015-09-01T16:06:58.577" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20793" PostTypeId="2" ParentId="20787" CreationDate="2015-09-01T15:56:45.423" Score="4" Body="&lt;p&gt;This is in progress, &lt;a href=&quot;http://www.codeproject.com/Articles/10162/Creating-EAN-Barcodes-with-C&quot; rel=&quot;nofollow noreferrer&quot;&gt;EAN-13 Barcode&lt;/a&gt; also &lt;a href=&quot;http://www.austincc.edu/hannigan/Math1513/codes.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So on the digits, the breakdown I have so far,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Quiet Zone  000000000&#xA;Lead / Trailer  101&#xA;&#xA;Manufacturer's Number&#xA;&#xA;0110111 0110111 0110111 0100011 0101111 0010011&#xA;  8        8       8       4       6      2    &#xA;&#xA;&#xA;Separator   01010&#xA;              5  &#xA;&#xA;&#xA;Product Number&#xA;&#xA;1110010 1000010 1110100 1001110 1001000 1100110 &#xA;   0       3       9       5       8       1   &#xA;&#xA;&#xA;Lead / Trailer  101&#xA;Quiet Zone  000000000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;888462 039581&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KPQCY.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KPQCY.gif&quot; alt=&quot;Barcode Example&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Was identifying what the code is the answer or are you looking for a particular message contained therein ?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;UPC-A       888462039581&#xA;EAN/UCC-13      0888462039581&#xA;Description     iPhone 6 Plus Gold 16GB&#xA;Issuing Country     United States&#xA;Last Modified       29 Mar 2015, 5:21 PM&#xA;Pending Requests        0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="15757" LastEditorUserId="15757" LastEditDate="2015-09-01T16:56:39.960" LastActivityDate="2015-09-01T16:56:39.960" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20794" PostTypeId="2" ParentId="20789" CreationDate="2015-09-01T16:04:19.653" Score="5" Body="&lt;p&gt;Given the changes, my other answer is no longer valid (but I'll leave it up). My new answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1007, which covers the worst-case scenario. He takes 1008.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've got an explanation but not a mathematical proof:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The worst-case scenario is the one in which an equal amount of gold and silver is in every chest. Because it's an odd number of chests, he has to take half (rounded up) to get half the total. This answer is the same if each chest contains the same total amount of gold and silver but in a linearly changing ratio (e.g. 5 4 3 2 1 gold and 1 2 3 4 5 silver). Any non-uniform chests are either higher-value on average, making them worth taking, or lower value on average, making all the other chests more worth taking; both of these cases make the case better than worst-case.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12462" LastActivityDate="2015-09-01T16:04:19.653" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20795" PostTypeId="1" AcceptedAnswerId="20798" CreationDate="2015-09-01T16:16:19.050" Score="-5" ViewCount="311" Body="&lt;p&gt;I can't help but wonder &lt;em&gt;if we've seen it before&lt;/em&gt;. I do know however it seems to be entirely made from booleans! I can't remember any booleans?!?!?!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;true&lt;br&gt;&#xA;  true false true true&lt;br&gt;&#xA;  true false true false true&lt;br&gt;&#xA;  true false false true false true true true false true true&lt;br&gt;&#xA;  true true false true true false false true false false true true true false true false true&lt;br&gt;&#xA;  true false false true true false false false false true true true false false true false false true true&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm almost positive this is a &lt;em&gt;known pattern&lt;/em&gt;, but maybe I'm wrong? Do any of you know the pattern, and can you confirm &lt;em&gt;if it's been here before&lt;/em&gt;?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2015-09-01T16:24:32.983" Title="Is this sequence new?" Tags="|pattern|word-sequence|" AnswerCount="1" CommentCount="14" ClosedDate="2015-09-09T16:10:47.077" ContentLicense="CC BY-SA 3.0" />
  <row Id="20796" PostTypeId="1" AcceptedAnswerId="20797" CreationDate="2015-09-01T16:19:41.020" Score="7" ViewCount="819" Body="&lt;p&gt;I took a standard 52-card deck and split it into two piles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pile 1:&lt;br&gt;&#xA;Clubs: A 3 5 6 7 8 9&lt;br&gt;&#xA;Diamonds: 7&lt;br&gt;&#xA;Hearts: A 3 5 6 7 8 9&lt;br&gt;&#xA;Spades: A 3 5 6 7 8 9&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pile 2:&lt;br&gt;&#xA;Clubs: 2 4 T J Q K&lt;br&gt;&#xA;Diamonds: A 2 3 4 5 6 8 9 T J Q K&lt;br&gt;&#xA;Hearts: 2 4 T J Q K&lt;br&gt;&#xA;Spades: 2 4 T J Q K&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What was my criterion?&lt;/p&gt;&#xA;" OwnerUserId="12822" LastActivityDate="2015-09-01T16:23:00.390" Title="How To Split The Deck" Tags="|logical-deduction|cards|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20797" PostTypeId="2" ParentId="20796" CreationDate="2015-09-01T16:23:00.390" Score="10" Body="&lt;p&gt;The criterion was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Separate the deck into two piles: one pile contains cards with a front face that is symmetric, the other pile contains those which are not symmetric. (All cards are symmetric on the back)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Rotational symmetry. (you can turn the card 180 degrees and it will look the exact same)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13068" LastActivityDate="2015-09-01T16:23:00.390" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20798" PostTypeId="2" ParentId="20795" CreationDate="2015-09-01T16:24:32.983" Score="8" Body="&lt;p&gt;If we convert the sequence to binary, with true=1 and false=0...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1&lt;br&gt;&#xA; 1011&lt;br&gt;&#xA; 10101&lt;br&gt;&#xA; 10010111011&lt;br&gt;&#xA; 11011001001110101&lt;br&gt;&#xA; 1001100001110010011&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then we convert from binary to decimal...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1&lt;br&gt;&#xA; 11&lt;br&gt;&#xA; 21&lt;br&gt;&#xA; 1211&lt;br&gt;&#xA; 111221&lt;br&gt;&#xA; 312211&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We see that the sequence is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The age old sequence where each number describes the count of digits in the previous number, so the value after 1 is &quot;one one&quot;, or 11.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Has it been here before?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2632/find-the-next-3-items-in-the-sequence?rq=1&quot;&gt;Yeah, it has.&lt;/a&gt; This isn't the original form, but it's another warspyking classic, which is really just as good.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-09-01T16:24:32.983" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20803" PostTypeId="2" ParentId="20790" CreationDate="2015-09-01T20:23:12.363" Score="10" Body="&lt;h2&gt;Partial Answer&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;It's close to a final answer... but I'm missing the twist and a final favorite puzzler...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyways, here's the grid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SVDmH.jpg&quot; alt=&quot;crossword grid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;ERRATA&lt;/strong&gt;: 96-Across should be 'SAS', not 'NAS'. That makes 83-Down 'DISABUSER'. (Thanks @Kingrames)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first favorite puzzler is &lt;strong&gt;xnor&lt;/strong&gt;, who's the answer to 90-Down. As I've already mentioned, the second puzzler to this point eludes me.&lt;/p&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="11475" LastEditDate="2015-09-01T20:51:58.437" LastActivityDate="2015-09-01T20:51:58.437" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="20804" PostTypeId="1" CreationDate="2015-09-01T20:56:25.567" Score="17" ViewCount="2581" Body="&lt;p&gt;A coworker and I are perplexed over this puzzle. We gave up and looked at the answer, but we cannot seem to understand how it works. Can anybody explain?&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vgpn4.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vgpn4.png&quot; alt=&quot;Puzzle Picture&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;h3&gt;Text version of the above image&lt;/h3&gt;&#xA;&lt;p&gt;We've created a simple code by taking a 13-letter phrase with no repeated letters and writing the remaining 13 letters of the alphabet in order under it.  For example:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;P U B L I C S E R V A N T&#xA;D F G H J K M O Q W X Y Z&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Letters are encoded by vertical substitution, so &lt;code&gt;GROVEL&lt;/code&gt; encodes as &lt;code&gt;BQEWOH&lt;/code&gt;.  The three words below are encoded using a different base phrase.  Can you deduce the phrase?&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;B O X C A R      J O K I L Y      Q U A R T Z&#xA;D M P K W V      N M C U H E      F I W V G S&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="16136" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-09-02T16:05:38.017" Title="Strategy for finding the key to this cipher" Tags="|cipher|solvability|explanation|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20805" PostTypeId="2" ParentId="20804" CreationDate="2015-09-01T21:16:32.567" Score="13" Body="&lt;p&gt;The coding phrase is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BUNCHOFGRAPES&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The resulting codex is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/yT4YB.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yT4YB.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;BR&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Found by:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Writing out the alphabet in order with it's coded letter beneath, looking for partial phrases - GRAPES jumped out:&#xA; &lt;a href=&quot;https://i.stack.imgur.com/AjA7Z.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AjA7Z.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12206" LastActivityDate="2015-09-01T21:16:32.567" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20806" PostTypeId="2" ParentId="20804" CreationDate="2015-09-01T21:27:30.607" Score="30" Body="&lt;p&gt;I'm not sure if you don't understand the example puzzle, or if you don't understand how to find the solution.  I'll explain both, just in case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The example takes a word/phrase that is 13 letters long with no repeated letters.  That means it contains exactly half of the letters in the alphabet (26 total).  Then the 13 remaining letters are written below, in alphabetical order.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;P U B L I C S E R V A N T&#xA;D F G H J K M O Q W X Y Z&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;To encrypt a word/phrase (like &lt;code&gt;GROVEL&lt;/code&gt; in the example), find the first letter in your grid, and take the letter directly above/below it.  &lt;code&gt;G&lt;/code&gt; is below &lt;code&gt;B&lt;/code&gt;, so our first encrypted letter is &lt;code&gt;B&lt;/code&gt;.  Next is &lt;code&gt;R&lt;/code&gt;.  It is above &lt;code&gt;Q&lt;/code&gt;, so the next encrypted letter is &lt;code&gt;Q&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuing in this way, we encrypt &lt;code&gt;GROVEL&lt;/code&gt; as &lt;code&gt;BQEWOH&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The puzzle creators have done the same thing with a different 13-letter phrase, and then have encrypted 3 words with it: &lt;code&gt;BOXCAR&lt;/code&gt;, &lt;code&gt;JOKILY&lt;/code&gt;, and &lt;code&gt;QUARTZ&lt;/code&gt;.  They show you the encrypted version of these three words, and they want you to work backwards and find the 13-letter phrase they used to create their grid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key to solving this is to realize that the letters at the end of the alphabet are not very common, so many of them will not be included in most 13-letter phrases.  That means they would be used to &quot;fill in&quot; the bottom row of the grid, which means they would remain in alphabetical order (much like &lt;code&gt;W X Y Z&lt;/code&gt; in the given example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We start constructing our grid from the bottom right.  According to the example words given, &lt;code&gt;Z&lt;/code&gt; matches with &lt;code&gt;S&lt;/code&gt;, &lt;code&gt;Y&lt;/code&gt; with &lt;code&gt;E&lt;/code&gt;, &lt;code&gt;X&lt;/code&gt; with &lt;code&gt;P&lt;/code&gt;, and &lt;code&gt;W&lt;/code&gt; with &lt;code&gt;A&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we write this into our grid, we have&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;. . . . . . . . . A P E S&#xA;. . . . . . . . . W X Y Z&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;That looks promising!  The letters on the top look like they could be part of a common English word or phrase!  Let's keep going.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;. . . . . . G I R A P E S&#xA;. . . . . . T U V W X Y Z&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This still looks promising.  We need to skip &lt;code&gt;S&lt;/code&gt;, &lt;code&gt;R&lt;/code&gt;, and &lt;code&gt;P&lt;/code&gt;, because they are already accounted for in the grid.  A few more:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;. . . J M F G I R A P E S&#xA;. . . N O Q T U V W X Y Z&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This doesn't look too good!  There is no English word with &lt;code&gt;JMFG&lt;/code&gt; in it!&#xA;But wait; &lt;code&gt;N&lt;/code&gt; and &lt;code&gt;O&lt;/code&gt; are way more common letters than &lt;code&gt;J&lt;/code&gt; and &lt;code&gt;M&lt;/code&gt;, so they're more likely to be in the original phrase.  What if we swap those around, so &lt;code&gt;N&lt;/code&gt; and &lt;code&gt;O&lt;/code&gt; are on the top?  Then the letters on the bottom are still in alphabetical order, as required, but we have something a little more sensible:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;. . . N O F G I R A P E S&#xA;. . . J M Q T U V W X Y Z&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now I'm starting to see a pattern.  Maybe the &lt;code&gt;I/U&lt;/code&gt; pair doesn't belong there.  We can flip it over and insert it at the appropriate point (to keep the bottom row in alphabetical order):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;. . . U N O F G R A P E S&#xA;. . . I J M Q T V W X Y Z&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now there are only 3 letter pairs unaccounted-for:  &lt;code&gt;B/D&lt;/code&gt;, &lt;code&gt;C/K&lt;/code&gt;, and &lt;code&gt;H/L&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There aren't that many ways they can be added, since we need to keep everything on the bottom row in alphabetical order, so a little trial and error (or guessing at the phrase) yields:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;B U N C H O F G R A P E S&#xA;D I J K L M Q T V W X Y Z&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Thus the solution is &lt;code&gt;BUNCH OF GRAPES&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2015-09-01T21:27:30.607" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20807" PostTypeId="1" AcceptedAnswerId="20809" CreationDate="2015-09-01T21:32:10.500" Score="10" ViewCount="1498" Body="&lt;p&gt;I'm not very good at all at chess, but was thinking about some interesting chess variations and thought of this question.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is well known that the fastest (i.e. fewest moves required) checkmate is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Fool%27s_mate&quot;&gt;Fool's Mate&lt;/a&gt;. My question is, what are the corresponding fastest checkmates when the kings are given the moveset of each of the other pieces (one at a time)? As a trivial example, if the kings' movement rules are replaced with those of a pawn, the answer is the same (fool's mate still applies). The fastest I have been able to come up with are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Pawn, Rook, Knight: 2&lt;br&gt;&#xA;  Bishop, Queen: 3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The green check goes to the first person who can match my records (with steps listed). I will award a 50 reputation bounty to anyone who can beat my record for bishop/queen (I doubt it is possible though). As in the case of Fool's Mate, you are allowed to make one player play atrociously; this is a proof of concept, not a realistic line of play. Good luck!&lt;/p&gt;&#xA;" OwnerUserId="13970" LastActivityDate="2015-09-02T18:15:59.930" Title="Fastest Checkmates" Tags="|chess|" AnswerCount="1" CommentCount="8" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20808" PostTypeId="2" ParentId="20804" CreationDate="2015-09-01T21:41:24.417" Score="5" Body="&lt;p&gt;Another method of solving:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the list of letter frequencies in English (usually something like &lt;code&gt;ETAOIN SHRDLU CMFWYP VBGKQJ XZ&lt;/code&gt; from most frequent to least, although it varies a little).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that your key phrase will be made up mostly of common letters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each letter pair, arrange it so that the more-common letter is on the top:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;S A D C E F T H I N O P R&#xA;Z W B K Y Q G L U J M X V&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Arrange the bottom row into alphabetical order (keeping the pairs together, of course):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;D T N C H O F I R A P E S&#xA;B G J K L M Q U V W X Y Z&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If it isn't already obvious what needs to be switched around, look for letter pairs that are relatively close to each other in the frequency list, or those that contain 2 letters nearer the beginning of the list, and try swapping them, rearranging the bottom as necessary to keep things in alphabetical order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It won't take long to arrive at&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;B U N C H O F G R A P E S&#xA;D I J K L M Q T V W X Y Z&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2015-09-01T21:41:24.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20809" PostTypeId="2" ParentId="20807" CreationDate="2015-09-01T22:01:49.797" Score="17" Body="&lt;p&gt;Pawn + Rook:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is basically the Fool's Mate again.&lt;br&gt;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/a8DcL.gif&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Knight:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I had to replace the King with a Knight for this.&lt;br&gt;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/cOGlQ.gif&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bishop + Queen:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This actually works for all pieces except the Knight. In normal chess games something like this is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Smothered_mate&quot; rel=&quot;noreferrer&quot;&gt;&quot;smothered mate&quot;&lt;/a&gt;.&lt;br&gt;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/ErYGM.gif&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another variant for the Bishop (there are even more):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/QxJbE.gif&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2015-09-02T14:51:47.453" LastActivityDate="2015-09-02T14:51:47.453" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20810" PostTypeId="1" AcceptedAnswerId="20811" CreationDate="2015-09-02T00:25:05.497" Score="10" ViewCount="920" Body="&lt;p&gt;As if it were to read your mind,&lt;br&gt;&#xA;My first tells which way you're inclined.&lt;br&gt;&#xA;My next, if you should like to ride,&lt;br&gt;&#xA;You'll find wherever roads divide.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In meadows green I toss and turn,&lt;br&gt;&#xA;Or else hold sway where fires burn.&lt;/p&gt;&#xA;" OwnerUserId="4385" LastActivityDate="2015-09-02T00:39:34.187" Title="A pastoral riddle" Tags="|riddle|rhyme|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20811" PostTypeId="2" ParentId="20810" CreationDate="2015-09-02T00:39:34.187" Score="17" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pitchfork.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As if it were to read your mind,&#xA;My first tells which way you're inclined.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Pitch&lt;/em&gt; is a description of angular direction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My next, if you should like to ride,&#xA;You'll find wherever roads divide.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A &lt;em&gt;fork&lt;/em&gt; in the road is where roads divide.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In meadows green I toss and turn,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Pitchforks&lt;/em&gt; are used in fields.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or else hold sway where fires burn.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Pitchforks&lt;/em&gt; are also found in imaginings of Hellscape.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2015-09-02T00:39:34.187" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20812" PostTypeId="1" AcceptedAnswerId="20815" CreationDate="2015-09-02T02:25:43.607" Score="34" ViewCount="4139" Body="&lt;p&gt;Having had just checked into the Grand Hotel, Mr. Hilbert slumped into the hotel room armchair with relief. Finally he could have some peace and quiet and solve the quaint riddle his colleague gave him! He pulled a note out from his breastpocket, and on it was simply scribbled:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Fix this equation by adding three mathematical symbols:&lt;br&gt;&#xA;  $$2  \space\space\space 2 \space  =   \space9$$&#xA;  NO letters, numbers, or tampering with the equal sign!&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, an hour and thousands of incorrect symbols later, Mr. Hilbert remained hopelessly stuck. Can you help Mr. Hilbert?&lt;/p&gt;&#xA;" OwnerUserId="15960" LastEditorUserId="6178" LastEditDate="2015-09-02T02:52:24.343" LastActivityDate="2019-04-08T04:15:58.187" Title="Mr. Hilbert and the Problem of the Erroneous Equation" Tags="|mathematics|calculation-puzzle|construction|" AnswerCount="12" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="20813" PostTypeId="2" ParentId="20812" CreationDate="2015-09-02T02:41:55.790" Score="7" Body="&lt;p&gt;My answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$$\Gamma(2) + 2 = \sqrt9$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6178" LastEditorUserId="6178" LastEditDate="2015-09-02T02:48:03.127" LastActivityDate="2015-09-02T02:48:03.127" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20814" PostTypeId="2" ParentId="20812" CreationDate="2015-09-02T02:56:46.630" Score="17" Body="&lt;p&gt;Probably fails the no letter criterion.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $-2+2={d \over dx} 9$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or using Lagrange notation as a workaround (thanks to McMagister) we can also write&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $-2+2= 9'$ &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15544" LastEditorUserId="15544" LastEditDate="2015-09-02T03:13:42.923" LastActivityDate="2015-09-02T03:13:42.923" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20815" PostTypeId="2" ParentId="20812" CreationDate="2015-09-02T03:04:58.923" Score="36" Body="&lt;p&gt;I don't know how to do the formatting (thanks McMagister for the edit) but the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$ 2\space\div\space  .\overline{2} = 9 $&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16142" LastEditorUserId="16142" LastEditDate="2015-09-02T03:12:53.420" LastActivityDate="2015-09-02T03:12:53.420" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20816" PostTypeId="2" ParentId="20812" CreationDate="2015-09-02T04:31:20.190" Score="4" Body="&lt;p&gt;What about :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $2\div2=\#\{9\}$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In case I got the symbols wrong, what I am trying to say is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; two divided by two equals the cardinality of the set of numbers that just contains the number nine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9353" LastEditorUserId="4421" LastEditDate="2015-09-02T04:39:22.690" LastActivityDate="2015-09-02T04:39:22.690" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="20817" PostTypeId="1" AcceptedAnswerId="20822" CreationDate="2015-09-02T05:53:53.613" Score="17" ViewCount="5064" Body="&lt;p&gt;I didn't invent this myself, I heard it a long time ago and it stuck with me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Four glasses are placed on the corners of a square rotating table. Some of the glasses are facing upwards and some upside-down. Your goal is to arrange the glasses so that they are all facing up or all facing down. Here are the rules:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You must keep your eyes closed at all times. (No tricks or lateral thinking, this is a pure logic puzzle)&lt;/li&gt;&#xA;&lt;li&gt;In a single turn, any two glasses may be inspected. After feeling their orientation, you may reverse the orientation of either, neither, or both glasses.&lt;/li&gt;&#xA;&lt;li&gt;After each turn, the table is rotated through a random angle.&lt;/li&gt;&#xA;&lt;li&gt;At any point, if all four glasses are of the same orientation a bell will ring. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Find a solution to ensure that all glasses have the same orientation (either up or down) in a finite number of turns. The algorithm must not depend on luck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; The bell will not ring mid-turn, it will only ring if the glasses are all of the same orientation at the end of your turn. &lt;/p&gt;&#xA;" OwnerUserId="16046" LastEditorUserId="16046" LastEditDate="2015-09-02T16:36:51.253" LastActivityDate="2018-12-06T20:49:11.603" Title="Four cups on a table" Tags="|logical-deduction|" AnswerCount="3" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="20818" PostTypeId="2" ParentId="20817" CreationDate="2015-09-02T06:09:12.547" Score="1" Body="&lt;p&gt;Since at any turn, two glasses are inspected, the best bet would be to identify one corner and inspect the glass at that corner and the glass on its opposite corner. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stick to one outcome, all up or all down (I am going ahead with downward arrangement). At the first turn, turn over any glass that isn't facing down. If the other two glasses are facing downward, then you're done. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the table is then rotated randomly, repeat this process. If you find that the glasses you touch are already facing downward, it could be that you are touching the same glasses as before. Assuming that the rotations of the table are truly random, there is a 0.5 probability of picking one combination of glasses. The probability of touching the same two glasses each time is (0.5)&lt;sup&gt;n&lt;/sup&gt;. So the probability of not having the glasses all down diminishes with each turn.&lt;/p&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-09-02T06:09:12.547" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20819" PostTypeId="2" ParentId="20812" CreationDate="2015-09-02T07:11:43.580" Score="24" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$$ 2 \div 2 = .\overline{9} $$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Simply rearranging the symbols used in the intended solution.&lt;/p&gt;&#xA;" OwnerUserId="4394" LastActivityDate="2015-09-02T07:11:43.580" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20820" PostTypeId="2" ParentId="20585" CreationDate="2015-09-02T07:55:53.023" Score="2" Body="&lt;p&gt;&lt;strong&gt;Second Attempt:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Vacuum Tube ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am sorry to say that my surname will not lead you to my origin, for those I share a name with are far different from me and come from far away.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Generally, we don't exclaim &lt;strong&gt;vacuum tubes!&lt;/strong&gt; when we think about tubes, hence the &lt;em&gt;different family&lt;/em&gt;. Sewage, water, oil, etc. pipelines are essentially tubes, which run a long length. Also, the vacuum part of vacuum tube is, naturally speaking, only found outside Earth's atmosphere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They are quite sheltered and I am not, though being so may have been useful for me, at least in the past.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The water, sewage, &amp;amp; similar pipelines run underground, within a shelter of Earth. And vacuum tubes, known to produce copious amounts of heat, could have used some shelter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You see, I used to be attached to powerful things, things bigger than myself, but now I am cast off, for the greater good, I suppose.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Vacuum tubes, during their times, were used to build then-powerful computers. However, transistors and LSI technology has kicked them out.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Those that really know me, know what I’m made of, well- some like me and some think I am revolting.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nothing. Vacuum tubes are, essentially, made of vacuum. Old techies like them, others, not so much. Also, note the re-&lt;em&gt;volting&lt;/em&gt; reference again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I may not hang around like I used to, but living out west folks do tend to roam.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The advent of transistors &amp;amp; LSI has phased out the use of vacuum tubes, but they still find applications in niche areas.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come from a different family, as the title says, and in my current condition, I cannot start a family of my own.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, the &lt;em&gt;current&lt;/em&gt; reference, and the vacuum tubes are now very inefficient so as to be brought to widespread commercial usage (niche applications).  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First Attempt:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Copper cable ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am sorry to say that my surname will not lead you to my origin, for those I share a name with are far different from me and come from far away.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think you are alluding to the Transatlantic Telegraph Cable here (far away &amp;amp; sheltered). As for different goes, there a whole lot of varied cables to be found in the world. Also, a &lt;em&gt;cable&lt;/em&gt; originates in the factory, whereas Copper originates in the Earth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They are quite sheltered and I am not, though being so may have been useful for me, at least in the past.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As above, mostly, though &lt;em&gt;I am not&lt;/em&gt; throws me off a bit, but not much. Being sheltered now would be of little use now, since cables are the precipice of their life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You see, I used to be attached to powerful things, things bigger than myself, but now I am cast off, for the greater good, I suppose.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A copper wire used for conducting electricity is almost always attached to equipment bigger than it. Also, wireless technology is more efficient than Copper cables since the cables caused too much of heat loss.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Those that really know me, know what I’m made of, well- some like me and some think I am revolting.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The re-&lt;em&gt;volting&lt;/em&gt; reference is obvious. Some people, most probably the previous generations, prefer wired tech to wireless.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I may not hang around like I used to, but living out west folks do tend to roam.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The advent of wireless technology has almost fully cast aside the copper wire, but it is still being used in most parts, but not like before when there used to be webs of it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come from a different family, as the title says, and in my current condition, I cannot start a family of my own.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, different types of cables, and the &lt;em&gt;current&lt;/em&gt; reference.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15607" LastEditorUserId="15607" LastEditDate="2015-09-02T09:29:57.443" LastActivityDate="2015-09-02T09:29:57.443" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20822" PostTypeId="2" ParentId="20817" CreationDate="2015-09-02T08:29:44.493" Score="11" Body="&lt;p&gt;I have solution that can guarantee the end goal being reached in a maximum of &lt;s&gt;&lt;strong&gt;four&lt;/strong&gt;&lt;/s&gt; &lt;strong&gt;five&lt;/strong&gt; steps. (Many thanks for @dmg and @Taemyr for their comments to fix this solution.) The trick is to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reduce the cups into the following configuration:&lt;br&gt;&#xA; $\begin{bmatrix} \circ &amp;amp; \bullet \\ \bullet &amp;amp; \circ \end{bmatrix} $&lt;br&gt;&#xA; where turning any two cups along the diagonal will satisfy the end goal&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For a constructive proof, we first consider the following &lt;s&gt;two&lt;/s&gt; three CASES:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CASE 1) the orientation of a single cup is different from the others, e.g. $\begin{bmatrix} \circ &amp;amp; \bullet \\ \circ &amp;amp; \circ \end{bmatrix} $ or $\begin{bmatrix} \bullet &amp;amp; \circ \\ \bullet &amp;amp; \bullet \end{bmatrix} $&lt;br&gt;&#xA;CASE 2a) two cups are in each orientation along the diagonal, e.g. $\begin{bmatrix} \circ &amp;amp; \bullet \\ \bullet &amp;amp; \circ \end{bmatrix} $&lt;br&gt;&#xA;CASE 2b) two cups are in each orientation along two sides, e.g. $\begin{bmatrix} \circ &amp;amp; \bullet \\ \circ &amp;amp; \bullet \end{bmatrix} $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Take two cups along the diagonal.&lt;/strong&gt;&lt;br&gt;&#xA; - If they are different, flip $\circ$ to $\bullet$. If this doesn't sound the bell, we either have $\begin{bmatrix} \circ &amp;amp; \bullet \\ \bullet &amp;amp; \circ \end{bmatrix}$ or $\begin{bmatrix} \circ &amp;amp; \bullet \\ \circ &amp;amp; \circ \end{bmatrix}$, with the majority orientation being $\circ$. In either case, proceed to &lt;strong&gt;step 2&lt;/strong&gt;.&lt;br&gt;&#xA; - If they are the same, flip both. If this doesn't sound the bell, we either have $\begin{bmatrix} \bullet &amp;amp; \circ \\ \bullet &amp;amp; \bullet \end{bmatrix}$ or $\begin{bmatrix} \circ &amp;amp; \bullet \\ \circ &amp;amp; \circ \end{bmatrix}$. Either way, the majority orientation is known, and we can proceed to &lt;strong&gt;step 3&lt;/strong&gt;.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Take two cups along the diagonal.&lt;/strong&gt;&lt;br&gt;&#xA; - If they are different, we now know we have $\begin{bmatrix} \circ &amp;amp; \bullet \\ \circ &amp;amp; \circ \end{bmatrix}$. Flip $\bullet$ to $\circ$ to win.&lt;br&gt;&#xA; - If they are the same, flip both. If this doesn't sound the bell, we now know we have $\begin{bmatrix} \circ &amp;amp; \bullet \\ \circ &amp;amp; \circ \end{bmatrix}$. Proceed to &lt;strong&gt;Step 3&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; At this point, we have already figured out what the majority orientation is, so we assume we have $\begin{bmatrix} \circ &amp;amp; \bullet \\ \circ &amp;amp; \circ \end{bmatrix} $ WLOG. Then:&lt;br&gt;&#xA; &lt;strong&gt;Take two cups along the diagonal.&lt;/strong&gt;&lt;br&gt;&#xA; - If the cups are different, flip $\bullet$ to $\circ$ to win.&lt;br&gt;&#xA; - If the cups are both $\circ$, flip one of them. We now know we have $\begin{bmatrix} \circ &amp;amp; \bullet \\ \circ &amp;amp; \bullet \end{bmatrix}$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 4&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Take two adjacent cups along an edge and flip both.&lt;/strong&gt;&lt;br&gt;&#xA; - If the cups chosen are the same, we win.&lt;br&gt;&#xA; - If they are different, we now know we have $\begin{bmatrix} \circ &amp;amp; \bullet \\ \bullet &amp;amp; \circ \end{bmatrix} $  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 5&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Take two cups along a diagonal and flip both.&lt;/strong&gt; We win!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15960" LastEditorUserId="15960" LastEditDate="2015-09-03T04:09:50.740" LastActivityDate="2015-09-03T04:09:50.740" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="20825" PostTypeId="1" CreationDate="2015-09-02T13:46:23.703" Score="64" ViewCount="25924" Body="&lt;p&gt;Just got a simple question from a friend; still thinking. Let's share it!  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Is it possible to get a, b, c when you have a⊕b, b⊕c, a⊕c ?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where ⊕ is the boolean XOR (exclusive OR) operator.&lt;br&gt;&#xA;a,b,c are &lt;em&gt;any&lt;/em&gt; boolean numbers with the &lt;em&gt;same&lt;/em&gt; length.&lt;br&gt;&#xA;For the sake of simplicity, consider a,b,c as bytes (length is 8 bits).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; You can use other operators too.&lt;br&gt;&#xA;&lt;strong&gt;Edit 2:&lt;/strong&gt; &lt;s&gt;Though it's preferable to use only ⊕ (if possible)&lt;/s&gt; (removed, use any operator)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note 1:&lt;/strong&gt; Answer without explanations (e.g. &lt;code&gt;yes&lt;/code&gt;/&lt;code&gt;no&lt;/code&gt;) are not allowed.&lt;br&gt;&#xA;&lt;strong&gt;Note 2:&lt;/strong&gt; This is a puzzling programming, math-related question and could be placed on Puzzling.SE, Programming.SE, StackOverflow and Math.SE. So excuse me or migrate the question if I pasted it on the wrong SE forum.&lt;/p&gt;&#xA;" OwnerUserId="15786" LastEditorUserId="10781" LastEditDate="2015-09-02T15:22:55.833" LastActivityDate="2015-09-07T17:19:55.153" Title="XOR - Is it possible to get a, b, c from a⊕b, b⊕c, a⊕c?" Tags="|mathematics|computer-puzzle|calculation-puzzle|computer-science|" AnswerCount="12" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="20826" PostTypeId="2" ParentId="20825" CreationDate="2015-09-02T13:54:10.037" Score="70" Body="&lt;p&gt;This is not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the two cases where a, b and c are all true or all false. Now in both cases we have&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a⊕b = b⊕c = a⊕c = false&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And more generally, $(¬a)⊕(¬b)=a⊕b$, so if $a,b,c$ is a solution, then so is $¬a,¬b,¬c$. &lt;em&gt;(from &lt;a href=&quot;https://puzzling.stackexchange.com/questions/20825/xor-is-it-possible-to-get-a-b-c-from-a%E2%8A%95b-b%E2%8A%95c-a%E2%8A%95c/20828#comment65622_20826&quot;&gt;Klaus Draeger&lt;/a&gt;)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="4394" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-09-02T14:59:41.960" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20828" PostTypeId="2" ParentId="20825" CreationDate="2015-09-02T14:01:09.977" Score="13" Body="&lt;p&gt;&lt;sub&gt;&lt;em&gt;(as an extension of &lt;a href=&quot;https://puzzling.stackexchange.com/a/20826/10399&quot;&gt;GOTO 0's answer&lt;/a&gt;)&lt;/em&gt;&lt;/sub&gt;&lt;br&gt;&#xA;This is also not possible if you consider these as binary numbers, you cannot determine the values of a, b, or c. For instance:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a = 0b11001111010&#xA;b = 0b11001111101&#xA;c = 0b11001111001&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;All three of the xors for these binary representations compress the results down to at most three bits (the right-most three). Anywhere that all three numbers match bits will end up as &lt;code&gt;0&lt;/code&gt;, resulting in an inconclusive initial state since the matching bits are effectively &quot;lost&quot;.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a⊕b = 0b111&#xA;b⊕c = 0b100&#xA;a⊕c = 0b011&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2015-09-02T14:09:00.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20829" PostTypeId="1" AcceptedAnswerId="20830" CreationDate="2015-09-02T14:02:04.003" Score="6" ViewCount="862" Body="&lt;p&gt;Let’s define a distinct coastline as a coastline you can theoretically walk completely without going through the territory of another country. For example, apart from islands, Mexico has two coastlines: one on the Pacific, and one on the Gulf of Mexico. On its main island, Ireland only has one, because even though it’s interrupted by Northern Ireland, you can still walk all the way from Drumskellan to Omeath without going through the U.K., by taking the long way around. Let’s count coastlines on oceans and major saltwater seas (e.g. the Mediterranean and the Black Sea), but not coastlines on inland seas (e.g. the Caspian and the Dead Sea).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By my count, there are 29 countries with multiple distinct coastlines on the same landmass, and most of them only have two.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Can you name the four countries in the world that have exactly three distinct coastlines on one landmass?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can you name the three countries in the world that have exactly four distinct coastlines on one landmass?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can you name the one country that has exactly six distinct coastlines on one landmass?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="10851" LastEditorUserId="10851" LastEditDate="2015-09-02T18:38:46.140" LastActivityDate="2015-09-03T00:11:07.030" Title="Tricoastal, quadricoastal, and hexacoastal countries" Tags="|geometry|geography|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20830" PostTypeId="2" ParentId="20829" CreationDate="2015-09-02T14:19:02.267" Score="6" Body="&lt;h2&gt;Answers&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Can you name the four countries in the world that have exactly three distinct coastlines on one landmass?&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Canada, Saudi Arabia, Malaysia, France&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Can you name the three countries in the world that have exactly four distinct coastlines on one landmass?&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The United States of America, Morocco, Russia&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Can you name the one country that has exactly six distinct coastlines on one landmass?&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Spain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Explanations&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;One of the two countries with four distinct coastlines (by this definition) is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The United States of America.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The four coastlines (on one landmass) are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alaska, Point Roberts, the Pacific coast (Washington to California), the Atlantic coast (Maine to Texas).&lt;br&gt;&#xA; If you count Great Salt Lake as fitting your definition of &quot;major saltwater sea&quot;, you could even make a case for this being a 5-coastline country.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This makes the second 4-coastline country:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Morocco&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The four coasts are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. West of Ceuta&lt;br&gt;&#xA; 2. Between Ceuta and Peñón de Vélez de la Gomera&lt;br&gt;&#xA; 3. Between Peñón de Vélez de la Gomera and Melilla&lt;br&gt;&#xA; 4. East of Melilla&lt;br&gt;&#xA; &lt;em&gt;(Melilla credit to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/20829/tricoastal-and-quadricoastal-countries/20830?noredirect=1#comment65631_20830&quot;&gt;kaine&lt;/a&gt;)&lt;/em&gt;&lt;br&gt;&#xA; &lt;em&gt;(Peñón de Vélez de la Gomera credit to &lt;a href=&quot;https://puzzling.stackexchange.com/a/20835/10399&quot;&gt;eudes&lt;/a&gt;)&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The final 4-coastline country is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Russia&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The coastlines are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. Black Sea&lt;br&gt;&#xA; 2. Pacific Ocean&lt;br&gt;&#xA; 3. Baltic Sea (Kaliningrad)&lt;br&gt;&#xA; 4. Baltic Sea (St Petersburg)&lt;br&gt;&#xA; I don't include the occupied territory of the Crimean peninsula. That's still part of Ukraine, even if Russia has tanks rolling all over it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;&lt;em&gt;6&lt;/em&gt;&lt;/strong&gt;-coastline country:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Spain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The coastlines here are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. Atlantic Ocean (north shore)&lt;br&gt;&#xA; 2. Mediterranean Sea (East of Gibraltar)&lt;br&gt;&#xA; 3. Strait of Gibraltar (West of Gibraltar)&lt;br&gt;&#xA; 4. The coastline of Ceuta across the Strait on Morocco's coast&lt;br&gt;&#xA; 5. The coastline of Melilla &lt;em&gt;(credit to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/20829/tricoastal-and-quadricoastal-countries/20830?noredirect=1#comment65631_20830&quot;&gt;kaine&lt;/a&gt; on this 5&lt;sup&gt;th&lt;/sup&gt; coast)&lt;/em&gt; to the East of Ceuta on Morocco's coast&lt;br&gt;&#xA; 6. The coastline of Peñón de Vélez de la Gomera &lt;em&gt;(credit to &lt;a href=&quot;https://puzzling.stackexchange.com/a/20835/10399&quot;&gt;eudes&lt;/a&gt; on this 6&lt;sup&gt;th&lt;/sup&gt; coast)&lt;/em&gt; on Morocco's coast&lt;br&gt;&#xA;  .... &lt;strong&gt;Those sneaky Spanish!!&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The four three-coastline countries are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. Canada — Pacific Ocean, Boundary Bay, Atlantic/Arctic Ocean&lt;br&gt;&#xA; 2. Saudi Arabia — Red Sea, Persian Gulf (x2) &lt;em&gt;(credit to &lt;a href=&quot;https://puzzling.stackexchange.com/a/20832/10399&quot;&gt;Bailey M&lt;/a&gt;)&lt;/em&gt;&lt;br&gt;&#xA; 3. Malaysia — South China Sea (x3, split twice by Brunei) .. bonus+2 Malacca Strait split by Singapore &lt;em&gt;(credit to &lt;a href=&quot;https://puzzling.stackexchange.com/a/20833/10399&quot;&gt;kaine&lt;/a&gt;)&lt;/em&gt;&lt;br&gt;&#xA; 4. France — North shore, south shore (x2, split by Monte Carlo)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Previously, I had listed Germany as a three-coastline country, but I suspect that Szczecin Lagoon is counted as a continuous extension of the Baltic Sea.&lt;/p&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-09-03T00:11:07.030" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="20831" PostTypeId="2" ParentId="20812" CreationDate="2015-09-02T14:24:19.253" Score="4" Body="&lt;p&gt;How about this?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$\lfloor\sqrt{2}\rfloor + 2 = \sqrt{9}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Oops, that is 4 symbols.  Thanks @corsiKa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This one uses three symbols:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$2 - 2 = \lfloor9\%\rfloor$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2318" LastEditorUserId="2318" LastEditDate="2015-09-02T17:24:41.127" LastActivityDate="2015-09-02T17:24:41.127" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20833" PostTypeId="2" ParentId="20829" CreationDate="2015-09-02T14:25:52.597" Score="1" Body="&lt;p&gt;Still a partial answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 Coast lines:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - Germany&#xA; - Spain&#xA; - North Korea&#xA; - Mexico&#xA; - Guatemala&#xA; - Hondoras&#xA; - Nicaragua&#xA; - Costa Rica&#xA; - Panama&#xA; - Columbia&#xA; - Argentina&#xA; - Turkey&#xA; - Egypt&#xA; - Isreal&#xA; - Senegal&#xA; - Croatia&#xA; - Thailand&#xA; - China (depending on how you view Hong Kong)&#xA; - United Arab Emirates&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3 Coast lines:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - Russia&#xA; - Saudi Arabia&#xA; - Canada&#xA; - Malaysia&#xA; - France&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4 Coast lines:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - United states&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditDate="2015-09-02T14:44:38.310" LastActivityDate="2015-09-02T14:44:38.310" CommentCount="5" CommunityOwnedDate="2015-09-02T14:44:38.310" ContentLicense="CC BY-SA 3.0" />
  <row Id="20834" PostTypeId="2" ParentId="20804" CreationDate="2015-09-02T14:37:59.613" Score="0" Body="&lt;p&gt;I know I'm late to the game but here's my solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I took the programming route and created an application to help me solve it. Since the algorithm to solve it has pretty much been discussed, I'll skip over that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, I created a series of buttons like so :&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A &amp;lt;=&amp;gt; W&#xA;B &amp;lt;=&amp;gt; D&#xA;...&#xA;S &amp;lt;=&amp;gt; Z&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then I created labels for each letter in the place of where it's partner would go alphabetically, like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;W D K B Y Q T L U N C H O J M X F V Z G I R A P E S&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Next, I set the buttons to toggle between the two choices, e.g. if I clicked on A, W would disappear and A would become visible and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I ran the app and went down the list of buttons and selected the one that is more common.  That got me this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;D    T L  N C  O    F    I R A P E S&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;From there, it quickly became obvious after a few logical swaps that the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BUNCHOFGRAPES&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9353" LastActivityDate="2015-09-02T14:37:59.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20835" PostTypeId="2" ParentId="20829" CreationDate="2015-09-02T15:16:26.890" Score="3" Body="&lt;p&gt;Some countries not yet mentioned (I assume that exclaves can be included, if on the same landmass):&lt;br&gt;&#xA;2 coastlines: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. Oman, because of Musandam Peninsula&lt;br&gt;&#xA; 2. Brunei &amp;ndash; indeed, two parts!&lt;br&gt;&#xA; 3. Lithuania, on Baltic Sea &quot;proper&quot; and on Curonian Spit&lt;br&gt;&#xA; 4. Ukraine, if Crimea is considered in Russia (see below). Also, the region of Budjak is separated from the &quot;mainland&quot; Ukraine by the Dniester estuary, though of course connected through some bridges (see &lt;a href=&quot;https://www.google.com/maps/place/@46.2959846,30.2600494,33604m/data=!3m1!1e3!4m2!3m1!1s0x0:0x0&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;) &amp;ndash; but this probably shouldn't count as separate coastline.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3 coastlines:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Spain, because of Gibraltar, but:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6 coastlines&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Spain, if we can treat Eurasia+Africa as one landmass, because it includes a few territories in North Africa, bordering Morocco, of which Ceuta, Melilla and Peñón de Vélez de la Gomera are parts of land.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And finally, with possibly 8:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Russia.&lt;br&gt;&#xA; Depending on how we define a distinct coastline precisely, Russia may have even 8 of them (6, if excluding unlawfully annexed Crimea). I describe them below, according to maps I found, and let the reader decide on their own, which are right.&lt;br&gt;&#xA; 1. with the Pacific and Arctic Oceans, from Vladivostok up to Murmansk&lt;br&gt;&#xA; 2. with the Baltic Sea (Gulf of Finland), around St. Petersburg&lt;br&gt;&#xA; 3. (in the Kaliningrad Oblast) with the Baltic Sea (Curonian Lagoon), inside the lagoon&lt;br&gt;&#xA; 4. (KO cont.) with the Baltic Sea (incl. Vistula Lagoon), from the Lithuanian border on the Curonian Spit to the Polish border inside the Vistula Lagoon&lt;br&gt;&#xA; 5. (KO cont.) with the Baltic Sea (incl. Vistula Lagoon), on the Vistula Spit. However, the Strait of Baltiysk, which currently separates the spit from the mainland, is artificial, albeit old (1497)&lt;br&gt;&#xA; 6. with the Syvash lagoons of the Sea o Azov, from the Isthmus of Perekop to the de facto border with Ukraine on the Arabat Spit, just south of Strilkove village&lt;br&gt;&#xA; 7. with the Black Sea and the Sea of Azov, the remainder of the coastline of the Crimean Peninsula&lt;br&gt;&#xA; The Crimean Peninsula is de facto administered by Russia. Hovewer, this administration isn't &quot;recognised by most of the international community&quot; (by en.wiki/Crimea), so 6 and 7 are not officialy in Russia.&lt;br&gt;&#xA; 8. with the Sea of Azov and the Black Sea, from Taganrog to Sochi.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, a comment, if I may (can't comment normally yet). It seems to me that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Germany has only 2 coastlines, because Szczecin Lagoon is connected with the rest of Baltic Coast &amp;ndash; you can walk along &lt;a href=&quot;https://en.wikipedia.org/wiki/Uznam&quot; rel=&quot;nofollow&quot;&gt;Uznam Island&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16154" LastEditorUserId="16154" LastEditDate="2015-09-02T20:39:31.003" LastActivityDate="2015-09-02T20:39:31.003" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20836" PostTypeId="2" ParentId="20825" CreationDate="2015-09-02T15:21:40.987" Score="114" Body="&lt;p&gt;If you're given just a⊕b and b⊕c, then you can calculate&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  (a⊕b) ⊕ (b⊕c)&#xA;= a ⊕ (b ⊕ b) ⊕ c   (since ⊕ is associative)&#xA;= a ⊕ 0 ⊕ c         (since X⊕X=0 for any X)&#xA;= a ⊕ c              (since X⊕0=X for any X)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;so in effect when you're given a⊕b, b⊕c, a⊕c you've only been given two numbers (because the last one is redundant). So (assuming a,b,c are 8 bits as in the question) you only have 16 bits of information, and hence you can't work out all of a,b and c since that would require 24 bits of information.&lt;/p&gt;&#xA;" OwnerUserId="10781" LastActivityDate="2015-09-02T15:21:40.987" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="20837" PostTypeId="2" ParentId="20737" CreationDate="2015-09-02T15:35:06.303" Score="3" Body="&lt;p&gt;This can be a coincidence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Literally reading it in Hindi Pronunciation (Other devanagari puzzlers can confirm)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SOM, EK SI RAI PEA TOH JI, RAM&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Extracting Meaning &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SOM -&gt; It is a liquor drink &lt;a href=&quot;https://en.wikipedia.org/wiki/Soma&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Soma&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; EK -&gt; One &lt;a href=&quot;https://en.wikibooks.org/wiki/Hindi/Numbers&quot; rel=&quot;nofollow&quot;&gt;https://en.wikibooks.org/wiki/Hindi/Numbers&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SI -&gt; Like &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RAI -&gt; Opinion&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PEA -&gt; Drink&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TOH -&gt; then&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; JI -&gt; Live&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RAM -&gt; Gods Name -&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Rama&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Rama&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&#xA;Inferring&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; OPINION is IF YOU DRINK SOM, YOU LIVE LIKE GOD.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9892" LastActivityDate="2015-09-02T15:35:06.303" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20839" PostTypeId="2" ParentId="20585" CreationDate="2015-09-02T16:40:35.650" Score="6" Body="&lt;p&gt;Are you a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Rocky Mountain Oyster?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The original question was about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Cow Pies, which are also a &quot;gross&quot; byproduct of a farm animal&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am sorry to say that my surname will not lead you to my origin, for those I share a name with are far different from me and come from far away. They come from the depths, while I come from great heights, or so my name implies.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Surname of &quot;oyster&quot;: real oysters come from the ocean, and these come from &#xA; bulls, pigs, or sheep&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They are quite sheltered and I am not, though being so may have been useful for me, at least in the past.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oysters have a shell for shelter. This would have helped the rocky mountain oyster, since testicles are known to be susceptible to pain. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You see, I used to be attached to powerful things, things bigger than myself, but now I am cast off, for the greater good, I suppose.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bulls are large and powerful animals, much larger than their &quot;oysters.&quot; They are commonly castrated when farmers are done breeding them, for the &quot;greater  good&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Those that really know me, know what I’m made of, well- some like me and some think I am revolting.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some find rocky mountain oysters to be a delicacy, but most think they are &#xA; rather gross sounding.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I may not hang around like I used to, but living out west folks do tend to roam.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The West(ern part of the USA) is well known to have many farms, on which farm animals roam, especially bulls/cows.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My first time answering a question!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saw the hints, I'll answer those too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I come from a different family, as the title says, and in my current condition, I cannot start a family of my own; no life will come from me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rocky mountain oysters have been removed and prepared as food, no longer part of the animal's reproductive system.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mount, mount, mount. The mount could be where I am from, an act with which I'm forever done, or my better half could have become one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mount refers to:&lt;br&gt;&#xA; 1. where I am from: an animal which a person mounts or rides&lt;br&gt;&#xA; 2. act with which I'm forever done: an animal mounting another for &#xA; reproduction&lt;br&gt;&#xA; 3. my better half could have become one: the animal could be mounted on a &#xA; wall as hunters sometimes do  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16158" LastEditorUserId="15544" LastEditDate="2015-09-02T17:03:27.303" LastActivityDate="2015-09-02T17:03:27.303" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20840" PostTypeId="2" ParentId="20825" CreationDate="2015-09-02T16:52:46.083" Score="6" Body="&lt;p&gt;@psmears @GOTO 0 and @Ian MacDonald have already asked the basic question. This is another extension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is interesting because operators lose information.  The question fundamentally asks what is lost and what can be recovered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It reminds me of a geometry trick: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Take any 4 points $A,B,C,D$ on a plane. Compute the middle points $E,F,G,H$, one between each in turn (the fourth middle point $H$ is between $D$ and $A$). Whatever the configuration of $ABCD$, $EFGH$ is always a parallelogram.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Computing the middle point is similar to addition here. The fact that $EFGH$ has a property that $ABCD$ did not have is typically a loss of information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back to XOR, note that with $a$, $a⊕b$ and $a⊕c$ you can recover everything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There exist some applications using XOR with similar properties, see &lt;a href=&quot;https://en.wikipedia.org/wiki/Fountain_code&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Fountain_code&lt;/a&gt;.  The explanation there is a little general, but basically XOR combinations of a set of numbers (a, b and c, and more) are numerous and have the interesting property that if you receive only part of these combinations you can recover everything.  This is useful for multicasting a signal to receivers on an unreliable medium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://en.wikipedia.org/wiki/Fountain_code&quot; rel=&quot;noreferrer&quot;&gt;Fountain code - Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;IETF RFC 5053 specifies in detail a systematic Raptor code, which has been adopted into multiple standards beyond the IETF, such as within the 3GPP MBMS standard for broadcast file delivery and streaming services, the DVB-H IPDC standard for delivering IP services over DVB networks, and DVB-IPTV for delivering commercial TV services over an IP network.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16159" LastEditorUserId="10781" LastEditDate="2015-09-02T20:22:17.087" LastActivityDate="2015-09-02T20:22:17.087" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20841" PostTypeId="2" ParentId="20812" CreationDate="2015-09-02T16:55:41.673" Score="14" Body="&lt;p&gt;I came up with this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$\lceil2\sqrt{2}\rceil = \sqrt{9}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5874" LastActivityDate="2015-09-02T16:55:41.673" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20843" PostTypeId="2" ParentId="20812" CreationDate="2015-09-02T18:09:10.697" Score="5" Body="&lt;p&gt;Assuming that perfect formatting is not required:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $ \neg (2 \space 2 = 9) $&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; not (twenty-two equals nine)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we count a pair of parentheses as a single unit, then:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $ \neg (2 + 2 = 9) $&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7425" LastActivityDate="2015-09-02T18:09:10.697" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20845" PostTypeId="2" ParentId="9875" CreationDate="2015-09-02T18:39:40.343" Score="8" Body="&lt;h2&gt;Solution (n = &lt;s&gt;16&lt;/s&gt; 17)&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1, 50, 3, 52, 46, 2, 44, 47, 9, 8, 11, 45, 38, 10, 36, 39, 17, 16, 19, 37, 30, 18, 28, 31, 25, 24, 27, 29, 22, 26, 20, 23, 33, 32, 35, 21, 14, 34, 12, 15, 41, 40, 43, 13, 6, 42, 4, 7, 49, 48, 51, 5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Starting position:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;AAAABBBBCCCCDDDDEEEEFFFFGGGGHHHHIIIIJJJJKKKKLLLLMMMM&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;17 steps:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;AMAMLAKLCBCLJCIJEDEJHEGHGFGHFGEFIHIFDICDKJKDBKABMLMB&#xA;ALABKMDACLCBIBFCEJEDGDHEGHGFEFJGIFIHCHLIKDKJAJMKMBML&#xA;ABALJLJMCACAHLHLECECFJFJGEGEDHDHIGIGBFBFKIKIMDBDMKMK&#xA;AKAKDBIBCMCMFAGBELEBHCEDGJGDJECFIHIFLGAHKFKHLILJMDMJ&#xA;ADAJIKHLCBCLGMFAEBELELDCGDGJCJBEIFIHAHMGKHKFBFKIMJMD&#xA;AJADFDFKCLCBHBHMEAEAJBJBGCGCLDLDIEIEMFLFKGKGKHJHMIMI&#xA;AIAIHJGJCKCKFLELEMEMDACLGBGLBCAJIDIJBEBHKFKHDGDFMHMF&#xA;AHAFGIHDCJCDEKJBELEBCMLAGLGBABMCIJIDLDKEKHKFJFIGMFMH&#xA;AFAHFHFICDCJDJDKEBELBLBMGAGAMLBLICICKJDJKEKEIHFHMGMG&#xA;AGAGHFEFCICIJDCDEKEKLBABGMGMLALBILIBJCJDKJKDHEHFMHMF&#xA;AHAFEGDHCFCHCIBJEDEJAKMLGBGLBMKAIBILDLICKDKJFJGEMFMH&#xA;AFAHJHJGCHCFLFLIEJEDMDLKGLGBKBJMIAIAIBHBKCKCGDFDMEME&#xA;AEAEDFCFCGCGBHAHEIEIBJBJGKGKDLDLIMIMFAFLKBKLHCHJMDMJ&#xA;ADAJCELHCFCHAGMFEHEFLIKDGJGDJKIBILIBHMGAKLKBFBECMJMD&#xA;AJADBDBECHCFMFBGEFEHKHDIGDGJIJFKIBILGLHMKAKAELJLMCMC&#xA;ACACLJAJCECELHLHEGEGJFJFGIGIHDHDIKIKFBFBKMKMDADBMLMB&#xA;ALABACMDCJCDBEKFEHEFDGIHGFGHFIGJIDIJHKELKBKLJMCAMBML&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;First pair in step 18:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ABALMLLCCDCJKJJEEFEHIHHGGHGFGFFIIJIDEDDKKLKBCBBMMAMA&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Returns to starting position in step 19 (therefore starting position in step 52 not possible):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;AAAABBBBCCCCDDDDEEEEFFFFGGGGHHHHIIIIJJJJKKKKLLLLMMMM&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h2&gt;&quot;Strategy&quot; (for the old n = 16 solution)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Shortly said it is a combination of guessing and luck. It is obvious that all possibilities cannot be calculated. We can safely assume that the 1st card stays at its place, moving it would only add a rotation of cards not changing the result. Then we still have&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$51! = 1551118753287382280224243016469303211063259720016986112000000000000$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;possibilities. I tried to limit the possibilities and started with a $4*13$ grid, where I placed the numbers $1 .. 4$ each in a separate column. Then I filled the other numbers by adding $4$ to the number above. Example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0mdUM.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Testing all possibilities gave only $n=12$ as best result. Then I tried to use another grid. I increased the width until I got the posted result at the value $8$. However, one thing generates trouble for widths which aren't divisors of $52$. The grid is simply too big. The following picture shows the grid for my end result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Q3aHY.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The green numbers are the placed $1..8$. The other values are this time the value above plus $8$. If we now use the first $52$ numbers we get numbers that are too big. I solved this problem by simply swapping the big numbers with appropriate numbers from the unused area (both marked red).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Update&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;For the $n = 17$ solution I also changed the direction of increments in one column.&lt;/p&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2015-09-03T19:52:17.640" LastActivityDate="2015-09-03T19:52:17.640" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20846" PostTypeId="2" ParentId="20774" CreationDate="2015-09-02T19:05:00.907" Score="1" Body="&lt;p&gt;He's referencing&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the inevitable heat death of the universe. If the universe keeps expanding, which scientists predict will happen, what heat there is will be spread out in an increasingly large space. Entropy increases to the maximum and the temperature will drop to around 1 degree Kelvin (any water remaining that hasn't irradiated to photons etc. will be in solid form by now).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The skyline is dulled by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;pollution - both light pollution and regular pollution, speeding up the increase of entropy towards the inevitable maximum.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I admit it's a little far-fetched, though.&lt;/p&gt;&#xA;" OwnerUserId="16035" LastActivityDate="2015-09-02T19:05:00.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20847" PostTypeId="2" ParentId="20774" CreationDate="2015-09-02T19:08:51.223" Score="0" Body="&lt;p&gt;He stayed up all night staring at a picture of his...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (ex?) girlfriend named Dallas [or some other name that matches to a city with a distinct skyline], who has apparently given him the cold shoulder. &lt;br /&gt;&lt;br /&gt;Technically this works even if it's not a picture, but the couple is drifting apart, but I tried to think of an actual star with a name along these lines and I'm drawing a blank. (I was aiming for the other definition of &quot;stargazing&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13068" LastEditorUserId="13068" LastEditDate="2015-09-02T19:16:33.167" LastActivityDate="2015-09-02T19:16:33.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20848" PostTypeId="1" AcceptedAnswerId="20852" CreationDate="2015-09-02T19:18:24.210" Score="4" ViewCount="521" Body="&lt;p&gt;This just in: an alien message was received on Earth!! It seems some extraterrestrial species has contacted us in a language with symbols similar to our own--presumably for our convenience. You, a rookie code-breaker, want a shot at deciphering their message, but NASA is not allowing just anyone to see the message so they're holding a test of abilities for any code-breakers interested. Those who can prove their mettle will be allowed access to the classified info. You show up for the test and are given a sheet of paper that reads:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Welcome code-breaker! Here's your chance to prove your abilities: this will be a test of your abilities to translate unknown languages. Consider it a warm-up.&lt;br&gt;&#xA;  Below is a short 'Rosetta Stone' for a language developed by a mystical religious cult long ago (no records exist of this language, so don't attempt to search for it). Can you use it to determine how to transcribe from English to their writing system? Fill in the sheet with the translation rule(s) and tell us what this group worshipped, in our language and theirs. Good Luck!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;English phrase:&lt;br&gt;&#xA;  THE QUICK BROWN FOX JUMPS ABOVE ANY HAZEL BRAVE DOG&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Above phrase written in mystery writing system:&lt;br&gt;&#xA;  C2A 3M7I8 F3P71 O47 3M287 J0GLD 62B 4N2I8 E95LH 31K&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A few quick notes we almost forgot: This language is similar to our own in some ways (it can express all of our letters for example) but dissimilar in one important way. Look for patterns and you'll figure it all out.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="11876" LastEditDate="2015-09-02T20:34:26.147" LastActivityDate="2015-09-02T20:34:26.147" Title="Lost in Translation: The Test" Tags="|cipher|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20849" PostTypeId="1" AcceptedAnswerId="23250" CreationDate="2015-09-02T19:20:09.727" Score="18" ViewCount="1880" Body="&lt;p&gt;I am looking for someone to be my match.  Try this riddle to see if it's you.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;This riddle is slightly different, but I think you will get used to it with time.&lt;/p&gt;&#xA;&lt;p&gt;Your brain is evolved, is it not? Then you should have two legs to stand on.&lt;/p&gt;&#xA;&lt;p&gt;Feel free to sketch out your ideas or to draw quick conclusions.&lt;/p&gt;&#xA;&lt;p&gt;After all, we are all open to learning new ideas from our peers.&lt;/p&gt;&#xA;&lt;p&gt;I hope this doesn’t make you too heated, though.&lt;/p&gt;&#xA;&lt;p&gt;You might think that we are both stuck, forces equally matched.&lt;/p&gt;&#xA;&lt;p&gt;But perhaps you thought less of this riddle than you should have.&lt;/p&gt;&#xA;&lt;p&gt;Now that I have you where I want you, the end is inevitable.&lt;/p&gt;&#xA;&lt;p&gt;And here it is: it is final.&lt;/p&gt;&#xA;&lt;p&gt;Hopefully my hints aren’t too subtle, for I would love you to get the answer, or at least to get close.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Each line corresponds to one word and each word corresponds with the others.  Once you get one, the rest should be fairly easy.  Figure all the words out and you just may be my match.&lt;/p&gt;&#xA;&lt;p&gt;And if you are my match, who are you?&lt;/p&gt;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;If you are my match, you may not be all of these, but you just may be The One.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;The words from each line do more than simply correspond with each other; there is a common element between all of them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; I've said that the words you should get from each line correspond to one another, but it may be more appropriate to say that there is a relationship between them.  It may be important to think in those terms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11452" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-10-16T18:46:15.203" Title="Are You All of These, or Just The One?" Tags="|riddle|word|wordplay|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20850" PostTypeId="1" AcceptedAnswerId="20853" CreationDate="2015-09-02T19:28:51.207" Score="1" ViewCount="311" Body="&lt;p&gt;You might think I can be anything and voted too broad;&lt;br&gt;&#xA;no two, but only one will fit in me.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am used in many a story for sins;&lt;br&gt;&#xA;yet I am not evil but the saint one.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sing, you will not hear but see me in many tones;&lt;br&gt;&#xA;Rearranged, same as in your notes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?  &lt;/p&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="6178" LastEditDate="2015-09-04T08:39:56.417" LastActivityDate="2015-09-04T08:39:56.417" Title="I am not evil but the saint one - What am I?" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20852" PostTypeId="2" ParentId="20848" CreationDate="2015-09-02T20:12:56.307" Score="4" Body="&lt;p&gt;The solution to the cryptogram is the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We have a substitution code that is &lt;b&gt;reversed&lt;/b&gt;. The code directly translates to: GOD EVARB LEZAH...KCIUQ EHT&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice how we have letters A-P appearing and the numbers 0-9 appearing.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum of these letters and digits is 26.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, we note that '3' appears quite frequently - but strangely, it doesn't correspond to any values directly prima facie.  If we take note that the only words that end in '3' appear three times, as do those words that end in 'E', we quickly find that the above solution is the case.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;THE QUICK BROWN FOX JUMPS ABOVE ANY HAZEL BRAVE DOG&#xA;K13 HL59E 8I2N4 B26 DLG0J 782M3 740 17P3F 8I7M3 A2C&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Key:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ABCDEFGHIJKLMNOP0123456789&#xA;DFGJKLMQRSTUVWYZPHOENIXABC&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So they worship...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Phoenix&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13920" LastEditorUserId="13920" LastEditDate="2015-09-02T20:27:47.103" LastActivityDate="2015-09-02T20:27:47.103" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20853" PostTypeId="2" ParentId="20850" CreationDate="2015-09-02T20:14:07.997" Score="6" Body="&lt;p&gt;It's...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;STONE.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;no two, but only one will fit in me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;stONE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am used in many story for sins;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;being stoned as punishment&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yet I am not evil but the saint one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;St. One&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sing, you will not hear but see me in many tones;&#xA;Rearranged, same as in your notes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Both anagrams of &quot;stone&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2015-09-02T20:14:07.997" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20854" PostTypeId="2" ParentId="20825" CreationDate="2015-09-02T20:27:35.173" Score="6" Body="&lt;p&gt;Remember that $\oplus$ is addition mod 2. So we wish to solve the system of equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{equation*}&#xA;    \begin{alignedat}{4}&#xA;      x &amp;amp; = &amp;amp; a &amp;amp; {}+{} &amp;amp; b &amp;amp;       &amp;amp;  \\&#xA;      y &amp;amp; = &amp;amp;   &amp;amp;       &amp;amp; b &amp;amp; {}+{} &amp;amp; c \\&#xA;      z &amp;amp; = &amp;amp; a &amp;amp;       &amp;amp;   &amp;amp; {}+{} &amp;amp; c&#xA;    \end{alignedat}&#xA;\end{equation*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving this over the real numbers we get the solution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{equation*}&#xA;    \begin{alignedat}{4}&#xA;      a &amp;amp; = &amp;amp; \frac{x}{2} &amp;amp; {}-{} &amp;amp; \frac{y}{2} &amp;amp;  {}+{}     &amp;amp; \frac{z}{2} \\&#xA;      b &amp;amp; = &amp;amp; \frac{x}{2} &amp;amp; {}+{} &amp;amp; \frac{y}{2} &amp;amp;  {}-{}     &amp;amp; \frac{z}{2} \\&#xA;      c &amp;amp; = &amp;amp; -\frac{x}{2} &amp;amp; {}+{} &amp;amp; \frac{y}{2} &amp;amp;  {}+{}     &amp;amp; \frac{z}{2} &#xA;    \end{alignedat}&#xA;\end{equation*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know by the theory of linear algebra that this solution is unique. However, this solution does not work in $\mathbb{Z}_2$, because we cannot divide by 2 in that field. Hence the system is not invertible. &lt;/p&gt;&#xA;" OwnerUserId="4330" LastActivityDate="2015-09-02T20:27:35.173" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20856" PostTypeId="2" ParentId="16169" CreationDate="2015-09-02T21:02:40.657" Score="12" Body="&lt;p&gt;I'll give it a guess.  I think Moghwyn was very close.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My answer is the number 0.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I am on top politically, I am right, but if I am on top of being not in violation, I am wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First, some wordplay:&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;On top politically&lt;/strong&gt; = &quot;in power&quot;, meaning the power that a number is raised to.&lt;br&gt;&#xA; &lt;strong&gt;Not being in violation&lt;/strong&gt; = &quot;fraction&quot;, because a violation is an infraction, and &quot;not being in&quot; means to remove the &quot;in&quot; from &quot;infraction&quot;.&lt;br&gt;&#xA; &lt;strong&gt;Right&lt;/strong&gt; = true, or 1&lt;br&gt;&#xA; &lt;strong&gt;Wrong&lt;/strong&gt; = false, or 0&lt;br&gt;&#xA;&lt;br&gt;&#xA; So the statement means:&lt;br&gt;&#xA;&lt;br&gt;&#xA; If a number is raised to the 0th power, it is 1.  If 0 is the numerator of a fraction, that fraction's value is 0.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I am above 8, I am wrong, but if I am low on being not in violation, I am forever.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; More wordplay:&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Above 8&lt;/strong&gt; = &quot;base&quot;, because in chemistry, a solution with a pH of 8+ is considered a base (as opposed to an acid).  In this case, base refers to the base of a power expression.  So in the expression x^y, x is the base and y is the power.&lt;br&gt;&#xA;&lt;br&gt;&#xA; So the sentence means:&lt;br&gt;&#xA;&lt;br&gt;&#xA; If 0 is the base, 0^y is 0.  If 0 is the denominator of a fraction, its value is infinity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cannot be on top politically and above 8, but I can be on top and low of being not in violation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0^0 is not allowed, but it can be both the numerator and denominator of a fraction, i.e. 0/0.&lt;br&gt;&#xA;&lt;br&gt;&#xA; I'm not sure I agree with this last statement because I think both expressions have undefined values.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4479" LastEditorUserId="4479" LastEditDate="2015-09-10T06:00:38.117" LastActivityDate="2015-09-10T06:00:38.117" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20857" PostTypeId="2" ParentId="5937" CreationDate="2015-09-02T21:03:50.180" Score="1" Body="&lt;p&gt;Could be &lt;em&gt;axiomatic&lt;/em&gt; - an &lt;em&gt;axiom&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An axiom or postulate is a premise or starting point of reasoning. As classically conceived, an axiom is a premise so evident as to be accepted as true without controversy. ... As used in modern logic, an axiom is simply a premise or starting point for reasoning.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Axiom&quot; rel=&quot;nofollow&quot;&gt;https://en.m.wikipedia.org/wiki/Axiom&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4050" LastActivityDate="2015-09-02T21:03:50.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20858" PostTypeId="1" AcceptedAnswerId="20881" CreationDate="2015-09-02T21:12:39.307" Score="12" ViewCount="643" Body="&lt;p&gt;&lt;em&gt;Note: This isn't an ordinary cryptic crossword, but it's certainly cryptic, and it's certainly a crossword, so I've adorned it with the title anyways.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SGrv6.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SGrv6.png&quot; alt=&quot;Crossword grid&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Across&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;1. &lt;em&gt;A fishy complaint,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;A covering sheet,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;A pack ox's burden,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;A sour, sweet treat.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3. &lt;em&gt;A big, furry mammal,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;A bubble of water,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;A term of endearment,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;A term for the slaughtered.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5. &lt;em&gt;An Islamic official,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;A Baldur's Gate ranger,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Popular women,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Spiritual chambers.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6. &lt;em&gt;An Italian actress,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;A dry, hard bread,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;A deceptive plot,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Means darkness ahead.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8. &lt;em&gt;To a smaller extent,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Where one might be flappin',&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;A monster's abode,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;For fear it should happen.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;10. &lt;em&gt;A place to play darts,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;A piece least elite,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Unbreakable rules,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Two small, furry feet.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;12. &lt;em&gt;An Iron Age Hallstatt,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;A room in a ward,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;A ring-a-ding thing,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;A flexible cord.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;14. &lt;em&gt;Shots past the target,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Direct nowhere near,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Alleges with poise,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Incredibly clear.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;15. &lt;em&gt;Islamic religion,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;A hooved, woodland creature,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;To once have been viewed,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;A prophetic preacher.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;16. &lt;em&gt;Sensory helpers,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;A mountainous lake,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Feathers' ancestor,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;A way you can make.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Down&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;1. &lt;em&gt;A super descriptor,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;A frivolous sin,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Adhered by a band,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;To gradually thin.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2. &lt;em&gt;A very short moment,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;A thing you might play,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;A poke with a needle,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;A thing you might pay.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3. &lt;em&gt;A slow, prolonged accent,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;An unstructured scuffle,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Sketched with a pencil,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Impeccable muscle.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4. &lt;em&gt;Fasten together,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Alternative turf,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Rends from the bone,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Goes under the surf.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7. &lt;em&gt;Killed, as a dragon,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;A gradual pace,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;A mix of ingredients,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;To put in its place.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;9. &lt;em&gt;A brief, reddish glow,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Complete lack of sweat,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;The edge of a rooftop,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;With or, makes a threat.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;10. &lt;em&gt;Parts of a sneaker,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Interrelated,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Walks back and forth,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Measured or rated.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;11. &lt;em&gt;A male, Slavic title,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;To zero amount,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Divide into parts,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;A continental count.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;12. &lt;em&gt;A purposeful capsule,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Afflicted by bane,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Suddenly broken,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;A vocal disdain.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;13. &lt;em&gt;A Caribbean dwarf,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Things touching the plate,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Flax fiber fabric,&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Places to wait.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2019-09-20T05:40:39.490" Title="Cryptic Crossword #4: Build Your Own Crossword" Tags="|crosswords|cryptic-crosswords|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20859" PostTypeId="2" ParentId="20774" CreationDate="2015-09-02T22:01:14.700" Score="7" Body="&lt;p&gt;I've tried using the initial letters of each word and running it through just about every decryption I can find. The only thing that comes close to looking like English is the result of an Affine Cipher decryption (coefficients A=7, B=9) which gives:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LF Hive of Zvoss&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But unless he was looking for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Low frequency electric beehives on an Irish technology website&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm fairly sure that this is NOT the correct answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint #2 seems to be referring to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/questions/20553/a-small-mammal&quot;&gt;A small mammal?&lt;/a&gt; in which the answer was found by counting the characters on each line. &lt;br /&gt;&lt;br /&gt;In this case, there are too many characters on the single line or even after the last comma for the same technique to be applied, so I would assume that the number of letters in each word is important. Counting these gives us:&lt;br /&gt;&lt;pre&gt;1 6 5 4 1 3 4 7 13 1 11 6 6&lt;/pre&gt;Unfortunately trying to map these numbers to letters with A=1, B=2, etc. gives:&lt;br /&gt;&lt;pre&gt;A F E D A C D G M A K F F&lt;/pre&gt; which is no more enlightening.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;However I now determined that the correct answer is:&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;br/&gt;&lt;img src=&quot;https://i.stack.imgur.com/6dScO.png&quot; alt=&quot;enter image description here]&quot;&gt;&lt;br /&gt;Alpha Centauri&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This answer is found by:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Totalling the scrabble letter values of each of the letters in the words in the poem and then converting these back to letters with A=1, B=2&lt;br /&gt;&#xA;&lt;code&gt;&#xA;I             = 1                              -&gt; A&#xA;SOFTLY        = 1+1+4+1+1+4               = 12 -&gt; L&#xA;GAZED         = 2+1+10+1+2                = 16 -&gt; P&#xA;NIGH          = 1+1+2+4                   =  8 -&gt; H&#xA;A             = 1                              -&gt; A&#xA;LIT           = 1+1+1                     =  3 -&gt; C&#xA;DULL          = 2+1+1+1                   =  5 -&gt; E&#xA;SKYLINE       = 1+5+4+1+1+1+1             = 14 -&gt; N&#xA;CONTEMPLATING = 3+1+1+1+1+3+3+1+1+1+1+1+2 = 20 -&gt; T&#xA;A             = 1                              -&gt; A&#xA;DISMAYINGLY   = 2+1+1+3+1+4+1+1+2+1+4     = 21 -&gt; U&#xA;FROZEN        = 4+1+1+10+1+1              = 18 -&gt; R&#xA;FUTURE        = 4+1+1+1+1+1               =  9 -&gt; I&#xA;&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-09-09T21:57:17.850" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="20863" PostTypeId="1" AcceptedAnswerId="20869" CreationDate="2015-09-03T01:25:31.730" Score="7" ViewCount="468" Body="&lt;p&gt;Having proven your worth in NASA's &lt;a href=&quot;https://puzzling.stackexchange.com/questions/20848/lost-in-translation-the-test&quot;&gt;test&lt;/a&gt;, you are allowed to work on the alien message. NASA experts tell you that they suspect that the language is a transcription of our letters/words written left to right. They need your help to figure out this writing system and how to get from English letters/words to this language. Use the provided Rosetta Stone for that. Return to them the rules of this writing system so that humans can interpret future alien messages. Do this and they'll even let you write the first message back to them! Be aware, this is not easy; it's already driven half of NASA's linguistic and cryptography experts crazy. Good luck!&lt;/p&gt;&#xA;&lt;h1&gt;Rosetta Stone&lt;/h1&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;HOWEVER EXCEEDINGLY LOUD CIRCUS ACCIDENTS QUEUE UP YEARLY&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;${V_!}^8B^4R^4ZX^4P{N\text*}^{4\text*}S^1KTF^5{F_!}^{86}S{N_!}^1ZP^6NX^2PN^1S^4KGN{P_!}^{{(64)}\text*}X^6Y{M_!}^{42}ZF^5$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;CHASTISEMENTS FROM JAZZY BOOKKEEPERS BEYOND AUSTRALIA&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;${G_!}NV^2NG^1N^4H^4KGN{L_!}Z^8H{D_!}^2{W\text*}^5{J_!}^{8\text*}{P\text*}^{4\text*}Y^4ZN{J_!}^4M^8KSX^{26}NTZ^2F^{12}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Accepted answer will be the first answer that provides all rules for this writing system that you can determine from the Rosetta Stone and can prove it by transcribing a short message of your choosing to the aliens. I recommend cooperation: if you see a pattern but can't get the rest, contribute what you have, maybe it will be useful to the others. There are several rules to find.&lt;/p&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-09-07T06:33:54.730" Title="Lost In Translation: Hard Mode" Tags="|cipher|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20864" PostTypeId="2" ParentId="20825" CreationDate="2015-09-03T01:41:55.567" Score="22" Body="&lt;p&gt;Your question with the XOR operator mathematically reduces to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the $\mathbb{F}_{2^m}$ field (field of characteristic 2 whose elements can be represented as sequences of $m$ bits, and where addition is bitwise XOR), is the following matrix:&#xA;$$&#xA;M = \left( \array{1\ 1\ 0\\1\ 0\ 1\\0\ 1\ 1}\right)&#xA;$$&#xA;invertible ? Indeed, for a dimension-3 vector $V = (x, y, z)$, multiplying $V$ by the matrix above yields a vector $M·V = (x⊕y, x⊕z, y⊕z)$. For the operation to be reversible, the matrix $M$ must be invertible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that matrix is not invertible, because the third line is equal to the XOR of the two first lines. The conclusion is thus that it is not possible, in all generality, to recover $a$, $b$ et $c$ from $a⊕b$, $a⊕c$ and $b⊕c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can add that the matrix, being non-invertible, has a non-trivial kernel: there is a subspace $K$ of $(\mathbb{F}_{2^m})^3$ that contains all vectors &lt;em&gt;V&lt;/em&gt; such that $M·V = (0, 0, 0)$. The rank of $M$ is 2, meaning that the subspace $K$ has dimension 1; it is not hard to see that $K$ consists in exactly the vectors $(x, x, x)$ for all possible values of $x$. When you have a solution $S = (a, b, c)$ for your equation (the values $a$, $b$ and $c$ match the known values of $a⊕b$, $a⊕c$ and $b⊕c$), then the set of solutions is exactly $\left\{ S+V | V \in K \right\}$. In other words, when you have a solution, you have exactly $2^m$ solutions, and nothing to distinguish between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can thus conclude that it is &lt;em&gt;never&lt;/em&gt; possible to unambiguously reconstruct $a$, $b$ and $c$ from $a⊕b$, $a⊕c$ and $b⊕c$.&lt;/p&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;p&gt;The same result extends to more than three values: you cannot unambiguously recover the $n$ values $a_0, a_1,… a_{n-1}$ from the set of $n$ values $a_i⊕a_{i+1}$ for all $i$ from $0$ to $n-1$ (taking $a_n = a_0$).&lt;/p&gt;&#xA;" OwnerUserId="4797" LastActivityDate="2015-09-03T01:41:55.567" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20865" PostTypeId="1" AcceptedAnswerId="20866" CreationDate="2015-09-03T02:54:29.487" Score="0" ViewCount="299" Body="&lt;p&gt;Today I wrote some C# code to generate all subsets of a set. It all worked correctly but then I realized about I'd made a rookie mistake. The puzzle here is not that the code is not working but rather that code works perfectly while it most definitely shouldn't be:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;IEnumerable&amp;lt;IList&amp;lt;T&amp;gt;&amp;gt; GetAllSubSets&amp;lt;T&amp;gt;(IList&amp;lt;T&amp;gt; items, int i)&#xA;{&#xA;    IEnumerable&amp;lt;IList&amp;lt;T&amp;gt;&amp;gt; subsets;&#xA;    if (i &amp;lt; 0)&#xA;        subsets = new [] { new List&amp;lt;T&amp;gt;() };&#xA;    else&#xA;        subsets = GetAllSubSets(items, i - 1);&#xA;    foreach (var s in subsets)&#xA;        yield return s;&#xA;&#xA;    if (i &amp;gt;= 0)&#xA;        foreach (var s in subsets)&#xA;        {&#xA;            s.Add(items[i]);&#xA;            yield return s;&#xA;        }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You might have realized the issue here already but if you haven't... There should be only a single call for each value i. So the call for i = -1 should only execute once. That means we create an array only once. So all the returned subsets should be pointing to same reference value of that one array and therefore all subsets would have same elements!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this code working perfectly fine then?&lt;/p&gt;&#xA;" OwnerUserId="16176" LastEditorUserId="6178" LastEditDate="2015-09-03T03:26:59.113" LastActivityDate="2015-09-03T03:31:22.257" Title="This code shouldn't be working" Tags="|computer-puzzle|algorithm|" AnswerCount="1" CommentCount="5" ClosedDate="2015-09-03T08:12:02.913" ContentLicense="CC BY-SA 3.0" />
  <row Id="20866" PostTypeId="2" ParentId="20865" CreationDate="2015-09-03T03:31:22.257" Score="2" Body="&lt;p&gt;The inner call to &lt;code&gt;GetAllSubSets()&lt;/code&gt; is only evaluated when the &lt;code&gt;yield&lt;/code&gt; occurs. This lazy evaluation means that the call to create a new subset occurs multiple times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove this, force the evaluation to happen immediately, and you will see that the &lt;code&gt;new&lt;/code&gt; call happens only once. It can be done by changing the inner call as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;subsets = GetAllSubSets(items, i - 1).ToList();&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now the call evaluates immediately and the &lt;code&gt;new&lt;/code&gt; call is only called once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most important part of this is to realize that you didn't properly understand the code that you wrote, so the best thing to do is to rewrite it until you do. I didn't understand it until I debugged it and saw the multiple &lt;code&gt;new&lt;/code&gt; calls. Then I realized what was going on.&lt;/p&gt;&#xA;" OwnerUserId="375" LastActivityDate="2015-09-03T03:31:22.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20868" PostTypeId="1" AcceptedAnswerId="20902" CreationDate="2015-09-03T05:18:29.720" Score="2" ViewCount="409" Body="&lt;p&gt;It seemed like a coincidence, but just as Mr. Hilbert finished the checkout procedures at the front desk of the Grand Hotel, the lights went out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;What happened, Mr. Cantor?&quot; Mr. Hilbert asked the concierge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;It's probably the fuse boxes,&quot; said Mr. Cantor rather disgruntledly. &quot;I heard our manager fell for a late-night TV ad and bought these cheap fuses for the lights.&quot; Mr. Hilbert followed Mr. Cantor to the electrical room. Right at the front of the endless array of metal cabinets was one labelled &quot;LOBBY AND FIRST $N$ ROOMS&quot;. On the ground were boxes upon boxes of &quot;KIRK'S ZWEET DEEL FUZEZ $(1 \space \Omega)$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;It's weird why we didn't switch to circuit breakers yet,&quot; said Mr. Cantor as he picked up a multimeter lying on the side. He made a grumbly noise. &quot;Hmpf, seems like the multimeter is almost out of battery. We might not able to measure that many fuses before it dies&quot;. He opened the cabinet and grumbled some more. &quot;And what kind of idiot doesn't label the fuses in a fuse box!&quot; Peering at the fuses, his annoyance exploded into incredulity. &quot;AND you can't even see the fuse wire in these fuses!!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;It's ok,&quot; said Mr. Hilbert while mustering with his most soothing voice possible. &quot;If we take out all the fuses and measure them together, we can minimize the number of measurements needed. Here's what we can do...&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Can you figure out the &lt;em&gt;minimum number of multimeter measurements&lt;/em&gt; needed to find a single blown fuse $(R = \infty \space \Omega)$ from $N+1$ fuses before Mr. Cantor goes crazy? Can you also prove that your solution is optimal and works for all $N &amp;gt; 2$ ?&lt;/p&gt;&#xA;" OwnerUserId="15960" LastActivityDate="2015-09-04T13:53:29.550" Title="Mr. Hilbert and the Problem of the Rogue Resistor" Tags="|mathematics|strategy|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20869" PostTypeId="2" ParentId="20863" CreationDate="2015-09-03T06:28:10.093" Score="11" Body="&lt;p&gt;Contributing Notes:&lt;/p&gt;&#xA;&lt;p&gt;Rule 1:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;For vowels, A=2, E=4, I=1, O=8, U=6. All vowels are powers (not sure&#xA;if I use the correct word)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Rule 2:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Y is 5 if y sounds like IE, add the values of I and E (4+1).&lt;/p&gt;&#xA;&lt;p&gt;ex. YEARLY sounds like YEARLIE&lt;/p&gt;&#xA;&lt;p&gt;ex. JAZZY sounde like JAZZIE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Rule 3:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If two vowels will repeat or in tandem, you will put an asterisk ( * )&lt;/p&gt;&#xA;&lt;p&gt;ex. EE in EXCEEDINGLY = 4*&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Rule 4:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If vowels will repeat twice, enclosed it in parenthesis&lt;/p&gt;&#xA;&lt;p&gt;ex. UEUE in QUEUE = (64)*&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Rule 5:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If the letter after a space is a vowel, put &amp;quot;X&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;ex. CIRCUS ACCIDENTS = X is seen&lt;/p&gt;&#xA;&lt;p&gt;ex. QUEUE UP = X is seen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Rule 6:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If the letter after a space is a consonant, use &amp;quot;!&amp;quot; after the first letter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: Added new rules:&lt;/p&gt;&#xA;&lt;p&gt;Rule 7:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If the a letter sounds like another letter in a word, you can use its conversion in alien language.&lt;/p&gt;&#xA;&lt;p&gt;ex. C(english) = P(alien) and S(english) = N&lt;/p&gt;&#xA;&lt;p&gt;CIRCUS -- first C is N(alien), second C is P(alien) and S = N(alien)&lt;/p&gt;&#xA;&lt;p&gt;Now for EXCEEDINGLY, you&#xA;can hear the hard C in first syllable that is why there is a P(alien) and N(alien) because of the soft C in second syllable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;MAPPING:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bFyHY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bFyHY.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There are many sounds of C:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;CH used in CHASTISEMENTS&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;sounds like TS, so in alien it is GN&lt;/p&gt;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;second C used in CIR&lt;strong&gt;C&lt;/strong&gt;US ; first C in A&lt;strong&gt;C&lt;/strong&gt;CIDENTS&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;sounds like K, so in alien it is P&lt;/p&gt;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;first C in &lt;strong&gt;C&lt;/strong&gt;IRCUS ; C in EX&lt;strong&gt;C&lt;/strong&gt;EEEDINGLY ; second C in AC&lt;strong&gt;C&lt;/strong&gt;IDENTS&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;sounds like S, so in alien it is N&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Q is P since it sounds like K.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Q in QUEUE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;X is P since it sounds like K.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;X in EXCEEDINGLY&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I guess X can also be N if used as S like XEROX or XYLOPHONE&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;My Message to the Aliens:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$X^1{P_!}^8H^4X^1K{Y_!}^{42}N^4$&lt;/span&gt;&lt;br /&gt;&#xA; ---- I COME IN PEACE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16036" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-09-07T06:33:54.730" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20870" PostTypeId="2" ParentId="20825" CreationDate="2015-09-03T06:45:13.753" Score="9" Body="&lt;p&gt;There are only a few cases, let's perform a quick inspection:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $$a⊕b=k_1$$ $$b⊕c=k_2$$ $$c⊕a=k_3$$ , we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{matrix}&#xA;a &amp;amp; b &amp;amp; c &amp;amp; &amp;amp;k_1 &amp;amp; k_2 &amp;amp; k_3\\ &#xA;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;\\&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; \rightarrow &amp;amp;0 &amp;amp; 0 &amp;amp; 0\\ &#xA;0 &amp;amp; 0 &amp;amp; 1 &amp;amp; \rightarrow &amp;amp;0 &amp;amp; 1 &amp;amp; 1\\ &#xA;0 &amp;amp; 1 &amp;amp; 0 &amp;amp; \rightarrow &amp;amp;1 &amp;amp; 1 &amp;amp; 0\\ &#xA;0 &amp;amp; 1 &amp;amp; 1 &amp;amp; \rightarrow &amp;amp;1 &amp;amp; 0 &amp;amp; 1\\ &#xA;1 &amp;amp; 0 &amp;amp; 0 &amp;amp; \rightarrow &amp;amp;1 &amp;amp; 0 &amp;amp; 1\\ &#xA;1 &amp;amp; 0 &amp;amp; 1 &amp;amp; \rightarrow &amp;amp;1 &amp;amp; 1 &amp;amp; 0\\ &#xA;1 &amp;amp; 1 &amp;amp; 0 &amp;amp; \rightarrow &amp;amp;0 &amp;amp; 1 &amp;amp; 1\\ &#xA;1 &amp;amp; 1 &amp;amp; 1 &amp;amp; \rightarrow &amp;amp;0 &amp;amp; 0 &amp;amp; 0&#xA;\end{matrix}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence only $$(k_1, k_2, k_3) \in \{(000), (011), (110), (101)\}$$ are solvable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, please note that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The RHS of the table is symmetric.&lt;/li&gt;&#xA;&lt;li&gt;All possible combination of $(k_1, k_2, k_3)$ have even number of '1's since summing the three equations: $$a⊕b=k_1$$ $$b⊕c=k_2$$ $$c⊕a=k_3$$ yields $$  k_1⊕k_2⊕k_3 =a⊕a⊕b⊕b⊕c⊕c=0$$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="16179" LastEditorUserId="16179" LastEditDate="2015-09-03T06:50:29.550" LastActivityDate="2015-09-03T06:50:29.550" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20871" PostTypeId="2" ParentId="17870" CreationDate="2015-09-03T07:28:29.253" Score="2" Body="&lt;p&gt;They were trying to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Decree the estimate of pi.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Searching, squaring, measuring, squeezing,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; These are ways to &lt;em&gt;estimating&lt;/em&gt; something -&gt; Estimate&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the path swaps next, with groups of best,&#xA;the last leader's final decision.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Last leader's final decision: decree. Estimate decree =&gt; Decree estimate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It shatters my heart, to relay this next part:&#xA;with a swear yelled in vain,&#xA;the ill-begotten's sheep was slain.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A swear yelled in vain: Oh f**k! This leads to: oh f =&gt; of&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Three's fourteen, the next sign reads,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pi. I have added the &lt;em&gt;the&lt;/em&gt; as addition, not something I could get from the puzzle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15607" LastEditorUserId="15607" LastEditDate="2015-09-03T07:38:38.487" LastActivityDate="2015-09-03T07:38:38.487" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20872" PostTypeId="2" ParentId="20868" CreationDate="2015-09-03T07:50:37.327" Score="0" Body="&lt;p&gt;Here is how I would do it (assuming a multimeter which is accurate enough).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;order the fuses to groups of $1, 2, 3, 4, ...$&lt;/li&gt;&#xA;&lt;li&gt;the remaining fuses (last group) won't take part in the measurement&lt;/li&gt;&#xA;&lt;li&gt;each group is ordered as a series of resistors, the total resistence of each group with $i$ fuses is $R_i = i * 1\Omega$&lt;/li&gt;&#xA;&lt;li&gt;arange each group in parallel for the measurment, the total resistance of the setup if all fuses are OK should be $R = \frac{1}{1/R_1 + 1/R_2 + ...}$&lt;/li&gt;&#xA;&lt;li&gt;if the total resistance is as defined above, then the broken fuse is in the group which wasn't measured&lt;/li&gt;&#xA;&lt;li&gt;otherwise determine the group based on the result of the measurement&lt;/li&gt;&#xA;&lt;li&gt;repeat the procedure with the smaller group&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Max number of fuses $F$ for $j$ measurements:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$F_1 = 2$&lt;/li&gt;&#xA;&lt;li&gt;$F_j = F_{j-1} + F_{j-1} + (F_{j-1} - 1) + (F_{j-1} - 2) + ... = F_{j-1} +  \frac{F_{j-1} * (F_{j-1} + 1)}{2}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;pre&gt;&#xA;Measuerements 1 2  3   4 ...&#xA;Fuses         2 5 20 230 ...&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Example with 9 fuses:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1 fuse with $R_1 = 1\Omega$&lt;/li&gt;&#xA;&lt;li&gt;2 fuses with $R_2 = 2\Omega$&lt;/li&gt;&#xA;&lt;li&gt;3 fuses with $R_3 = 3\Omega$&lt;/li&gt;&#xA;&lt;li&gt;remaining 3 fuses not used&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Measured value is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$0.5454..\Omega$ =&gt; broken fuse is in not measured group&lt;/li&gt;&#xA;&lt;li&gt;$0.6666..\Omega$ =&gt; broken fuse is in 3rd group&lt;/li&gt;&#xA;&lt;li&gt;$0.75\Omega$ =&gt; broken fuse is in 2nd group&lt;/li&gt;&#xA;&lt;li&gt;$1.2\Omega$ =&gt; broken fuse is in 1st group (with only 1 fuse, so we are done)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2015-09-03T09:17:42.180" LastActivityDate="2015-09-03T09:17:42.180" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20874" PostTypeId="2" ParentId="20825" CreationDate="2015-09-03T11:02:03.783" Score="19" Body="&lt;p&gt;No, since,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;A ⊕ B = B ⊕ C = A ⊕ C = 0&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can be either &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;code&gt;A = B = C = 0&lt;/code&gt; or &lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;A = B = C = 1&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="12946" LastEditorUserId="15544" LastEditDate="2015-09-03T14:24:29.197" LastActivityDate="2015-09-03T14:24:29.197" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20876" PostTypeId="2" ParentId="20868" CreationDate="2015-09-03T12:06:04.973" Score="2" Body="&lt;p&gt;If we have only one measurement, we can't have any two fuses in parallel or in series, because it wouldn't be possible to distinguish them from the multimeter reading. Every circuit with more than one fuse will have two in parallel or in series, so we can only test one fuse, and have one fuse unused, for a total of two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have two measurements, we can use the first one to narrow it down to two fuses. A circuit like this can test &lt;strong&gt;an arbitrary number&lt;/strong&gt; of fuses (if there are an even number of fuses, add a fuse in series with the uppermost fuse): &lt;img src=&quot;https://i.stack.imgur.com/L2jZp.png&quot; alt=&quot;]&quot;&gt;&#xA;&lt;br&gt;We can interpret these fuses as being in two parallel sequences: one from the top fuse marked &quot;233/144&quot; to the one marked &quot;inf&quot;, and one from the lower fuse marked &quot;233/144&quot; to the one marked &quot;233/89&quot;. The readings for each of the fuses on the top line being broken increase strictly from left to right. The same happens for the fuses in the lower line, so no two fuses in the same sequence result in the same reading. This means that for any reading, there are at most two fuses (one in the top line and one in the bottom line) that result in that reading, and we can use the second test to determine which.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we only need 2 measurements (with infinite precision) for any finite number of fuses!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In fact, every fuse results in a different reading when it is broken, except for the two at the top. This is guaranteed by the fact that if we alternate the two sequences of fuses as mentioned above, we get a single sequence that is also strictly increasing. Then, if we replace the top fuse in this system with a new one, we can determine which fuse is broken with only one measurement.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-09-03T21:21:50.373" LastActivityDate="2015-09-03T21:21:50.373" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20877" PostTypeId="2" ParentId="4568" CreationDate="2015-09-03T12:42:55.230" Score="1" Body="&lt;p&gt;&lt;strong&gt;The optimal number of questions that could be achieved is 5 questions.  I will demonstrate with only 6 questions (improving from 7 as stated on other answers).&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As others have, determine suit in two questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Is your color red?&lt;/strong&gt;&#xA;If no to red, the color is black so next ask is your suit spades.  If yes to red, ask if your suit is hearts&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Is your suit xxxx (per above)&lt;/strong&gt; If yes, you know the suit from the question.  If no, you know the suit from the only other suit with that color.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thinking of the type of card as Ace, Two, Three, ..., ..., Queen, King:&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;&lt;strong&gt;Are the number of letters in the type of card is less than, greater than or equal to 4?&lt;/strong&gt; &#xA;If less than 4 letters, your card is an Ace, Two, Six, or Ten. &#xA;If more than 4 letters, your card is a Three, Seven, Eight or Queen. &#xA;If equal to 4 letters, your card is a Four, Five, Nine, Jack or King.  I'll take each of these separately.&lt;/li&gt;&#xA;&lt;li&gt;Case less than 4 letters: &lt;strong&gt;Is your card value less than, greater than or equal to six?&lt;/strong&gt;  If equal or more, you know the card value, so proceed to ask for the card.  If less, you have a &lt;strong&gt;5th question&lt;/strong&gt; for whether ace or two.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now for the next case:&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;Case more than 4 letters: &lt;strong&gt;Is your card value less than, greater than or equal to Eight?&lt;/strong&gt; If equal or more, you know the card value, so proceed to ask for the card.  If less, you have a &lt;strong&gt;5th question&lt;/strong&gt; for whether three or seven.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;And the last case.  &lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;Case equal to 4 letters: &lt;strong&gt;Is your card value less than, greater than or equal to Nine?&lt;/strong&gt; If equal, you know the card value, so proceed to ask for the card.  If less, you have a &lt;strong&gt;5th question&lt;/strong&gt; for whether four or five. If more, you have a &lt;strong&gt;5th question&lt;/strong&gt; for whether jack or king. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;And finally the question for their card:&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;6&quot;&gt;&#xA;&lt;li&gt;(or this could be the 5th question): Combine the suit found above with the value determined to ask the exact card value.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Mathematically, the maximum values you can determine with a two value question (yes/no) is n=2^(q-1) where n is the number of items you are asking against and q is the number of questions asked.  If the scenario was &quot;how many questions until you &lt;strong&gt;know&lt;/strong&gt; the answer, the formula would be n=2^q but the OP stated the last question had to be asking the value, so this takes one more question.  In our case 1 question will be required if there is only one item queried against.  2 questions required for 2 items, 3 questions for up to 4 items, 4 questions for up to 8 items, (etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, that is not the case in our example.  Since the OP allows for a 3 value answer (less than, equal or greater), the optimal equation is 3^(q-1).  This means with optimal questions, you can split the number of possibilities by 1/3 for each question.  1q= 1 item, 2q= 3 items, 3q= 9 items, 4q = 27 items, 5q = 81 items.  With optimal questions, 52 cards should be able to be found with just 5 questions.  In this case I have demonstrated using 6 questions.&lt;/p&gt;&#xA;" OwnerUserId="16187" LastActivityDate="2015-09-03T12:42:55.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20878" PostTypeId="2" ParentId="20812" CreationDate="2015-09-03T12:51:12.127" Score="2" Body="&lt;p&gt;How about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $2'\cdot 2' = 9$, where $(\,\cdot\,)'$ denotes the successor function.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16188" LastActivityDate="2015-09-03T12:51:12.127" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20879" PostTypeId="1" AcceptedAnswerId="20899" CreationDate="2015-09-03T13:20:15.530" Score="-2" ViewCount="661" Body="&lt;p&gt;We have existed since the dawn of time,&lt;br&gt;&#xA;Yet we're not older than a blink of an eye&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We may be entities of creation,&lt;br&gt;&#xA;or entities of destruction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Behold, however, and you shall see&lt;br&gt;&#xA;We're nothing but dust&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People used to think we were of some ethereal nature&lt;br&gt;&#xA;It has been dismissed, in favour of something somewhat more mature&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can be everywhere,&lt;br&gt;&#xA;We can stay nowhere&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can't, however, be torn apart&lt;br&gt;&#xA;For which we have no part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who are we?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Behold&quot;, &quot;creation&quot; and &quot;destruction&quot; all take their meanings from a very specific context&lt;br&gt;&#xA;&lt;br&gt;&#xA; The second and third stanzas form an antithesis, representing &quot;two sides of the same coin&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14219" LastEditorUserId="14219" LastEditDate="2015-09-03T22:25:52.363" LastActivityDate="2015-09-06T03:21:31.287" Title="Name these entities which have deeply puzzled mankind" Tags="|riddle|word|" AnswerCount="4" CommentCount="7" ClosedDate="2015-09-06T05:33:40.920" ContentLicense="CC BY-SA 3.0" />
  <row Id="20880" PostTypeId="2" ParentId="20879" CreationDate="2015-09-03T13:34:46.247" Score="2" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chemical elements or atoms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We have existed since the dawn of time,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hydrogen was the first atom or matter to ever exist in the universe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet we're not older than a blink of an eye&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Atoms are constantly created inside the cores of stars through nuclear fusion, billions of them per second&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We may be entities of creation,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every kind of life we know is based on Carbon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or entities of destruction&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hydrogen bomb rings a bell? :(&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Behold, however, and you shall see&#xA;We're nothing but dust&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lots of possibilities: &quot;we are all formed by star dust&quot; is my favourite though :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;People used to think we were of some ethereal nature &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First elemental subdivision was fire, earth, water and wind. Then ether came along to explain how EMW propagated through space. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It has been dismissed, in favour of something somewhat more mature&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The ether model was abandoned &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can be everywhere,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes, you are!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can stay nowhere&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Matter and anti-matter (atoms and anti-atoms) can be created literally from nowhere. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can't, however, be teared apart&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Atom from greek &lt;em&gt;ἄτομος&lt;/em&gt; - àtomos -, indivisible&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For which we have no part.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;As we all know the atom isn't indeed indivisible, but formed by protons and electrons&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4652" LastEditorUserId="4652" LastEditDate="2015-09-03T13:39:59.167" LastActivityDate="2015-09-03T13:39:59.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20881" PostTypeId="2" ParentId="20858" CreationDate="2015-09-03T13:45:20.423" Score="6" Body="&lt;p&gt;The twist here is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Two cells of each answer can contain one of two letters, giving a total of four possible solutions per clue&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Completed grid:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/79zOh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/79zOh.png&quot; alt=&quot;Image of solved puzzle&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Across&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A fishy complaint&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CARP - To complain as well as a type of fish&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A covering sheet&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TARP - Short for tarpaulin&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A pack ox's burden&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CART - An ox might pull a cart&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A sour, sweet treat.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TART - A desert (sweet treat) as well as meaning sour&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;A big, furry mammal,&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BEAR&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A bubble of water,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BEAD&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A term of endearment,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DEAR&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A term for the slaughtered.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DEAD&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;An Islamic official,&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DIVAN - Can be a title applied to &lt;a href=&quot;https://en.wikipedia.org/wiki/Dewan#Title&quot; rel=&quot;nofollow noreferrer&quot;&gt;high officials in the Middle East&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A Baldur's Gate ranger,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; KIVAN - The name of a ranger in Baldur's Gate&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Popular women,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DIVAS - A term used for popular female singers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Spiritual chambers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; KIVAS - A kiva is &lt;a href=&quot;https://en.wikipedia.org/wiki/Kiva&quot; rel=&quot;nofollow noreferrer&quot;&gt;a room used by Puebloans for religious rituals&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;6&quot;&gt;&#xA;&lt;li&gt;An Italian actress,&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DUSE - Eleonora Duse was &lt;a href=&quot;https://en.wikipedia.org/wiki/Eleonora_Duse&quot; rel=&quot;nofollow noreferrer&quot;&gt;an Italian actress&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A dry, hard bread,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RUSK&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A deceptive plot,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RUSE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Means darkness ahead.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DUSK - The period before night time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;8&quot;&gt;&#xA;&lt;li&gt;To a smaller extent,&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LESS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where one might be flappin',&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; NEST - A bird might be flapping in a nest&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A monster's abode,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; NESS - Loch Ness is the home of the Loch Ness monster&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For fear it should happen.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LEST - Meaning for fear of something undesirable happening&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;10&quot;&gt;&#xA;&lt;li&gt;A place to play darts,&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LAWN - Darts can be played on a lawn (although I prefer the pub)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A piece least elite,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PAWN - Generally the least powerful piece in chess&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unbreakable rules,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LAWS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Two small, furry feet.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PAWS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;12&quot;&gt;&#xA;&lt;li&gt;An Iron Age Hallstatt,&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CELT - Hallstatt was an Iron Age Celtic culture&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A room in a ward,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CELL&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A ring-a-ding thing,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BELL&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A flexible cord.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BELT&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;14&quot;&gt;&#xA;&lt;li&gt;Shots past the target,&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; OVERS - Having gone past the target, one has overshot&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Direct nowhere near,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; AVERT - to ward off&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alleges with poise,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; AVERS - To allege as a fact&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Incredibly clear.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; OVERT&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;15&quot;&gt;&#xA;&lt;li&gt;Islamic religion,&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DEEN - A term &lt;a href=&quot;https://en.wikipedia.org/wiki/D%C4%ABn&quot; rel=&quot;nofollow noreferrer&quot;&gt;loosely associated with Islamic religion&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A hooved, woodland creature,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DEER&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To once have been viewed,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SEEN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A prophetic preacher.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SEER&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;16&quot;&gt;&#xA;&lt;li&gt;Sensory helpers,&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; EARS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A mountainous lake,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TARN - &lt;a href=&quot;https://en.wikipedia.org/wiki/Tarn_(lake)&quot; rel=&quot;nofollow noreferrer&quot;&gt;A mountain lake&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Feathers' ancestor,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TARS - Comes before feathers in the phrase 'tar and feather'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A way you can make.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; EARN - To make money&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Down&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A super descriptor,&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CAPED - As in 'Batman: The Caped Crusader'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A frivolous sin,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CAPER - A frowned upon activity that is nevertheless fairly harmless&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Adhered by a band,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TAPED&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To gradually thin.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TAPER&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;A very short moment,&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TRICE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A thing you might play,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TRICK - You might play a trick on someone&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A poke with a needle,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PRICK&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A thing you might pay.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PRICE - You might pay the price&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;A slow, prolonged accent,&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DRAWL&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An unstructured scuffle,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BRAWL&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sketched with a pencil,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DRAWN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Impeccable muscle.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BRAWN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;Fasten together,&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RIVET&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alternative turf,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DIVET - Alternative spelling of divot&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rends from the bone,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RIVES - To split or tear apart violently&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Goes under the surf.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DIVES&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;7&quot;&gt;&#xA;&lt;li&gt;Killed, as a dragon,&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SLEW - One slays a dragon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A gradual pace,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SLOW&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A mix of ingredients,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; STEW&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To put in its place.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; STOW&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;9&quot;&gt;&#xA;&lt;li&gt;A brief, reddish glow,&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ELVE - &lt;a href=&quot;http://dictionary.reference.com/browse/elve&quot; rel=&quot;nofollow noreferrer&quot;&gt;A reddish glow seen over a thunderstorm&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Complete lack of sweat,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; EASE - E.g. done with ease, without having to try&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The edge of a rooftop,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; EAVE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With or, makes a threat.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ELSE - As in 'or else'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;10&quot;&gt;&#xA;&lt;li&gt;Parts of a sneaker,&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LACES&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Interrelated,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LACED&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Walks back and forth,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PACES&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Measured or rated.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PACED&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;11&quot;&gt;&#xA;&lt;li&gt;A male, Slavic title,&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; NEVEN - &lt;a href=&quot;https://en.wikipedia.org/wiki/Neven&quot; rel=&quot;nofollow noreferrer&quot;&gt;A Slavic masculine name&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To zero amount,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; NEVER - As in something that never happens happens zero amount of times&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Divide into parts,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SEVER&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A continental count.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SEVEN - The number of continents, by most standards&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;12&quot;&gt;&#xA;&lt;li&gt;A purposeful capsule,&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BURSE - &lt;a href=&quot;http://dictionary.reference.com/browse/burse&quot; rel=&quot;nofollow noreferrer&quot;&gt;A pouch or case for some special purpose&lt;/a&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Afflicted by bane,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CURST - Alternative spelling of cursed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Suddenly broken,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BURST&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A vocal disdain.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CURSE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;13&quot;&gt;&#xA;&lt;li&gt;A Caribbean dwarf,&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TINEN - &lt;a href=&quot;http://www.wordhippo.com/what-is/the-meaning-of/haitian-creole-word-dc3cf8c97c117b5bf4840be2b2294d161861c3d3.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;A Haitian Creole word for dwarf&lt;/a&gt; (struggling by this stage were you?!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Things touching the plate,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TINES - The tines or prongs of a fork might touch a plate&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Flax fiber fabric,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LINEN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Places to wait.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LINES&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12277" LastEditorUserId="56416" LastEditDate="2019-09-20T05:40:39.490" LastActivityDate="2019-09-20T05:40:39.490" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="20884" PostTypeId="2" ParentId="4568" CreationDate="2015-09-03T14:04:03.773" Score="0" Body="&lt;p&gt;The following answer makes the assumption that counterfactual conditionals are acceptable and the answerer may NOT have an answer (be it yes or no).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Optimally, with this approach, we can achieve the result in &lt;b&gt;4 questions&lt;/b&gt; (not including the last question regarding the identity of the final card).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;If the card doesn't have a face, is it black?&#xA;If your card is divisible by 2, is it a heart?&#xA;If your card is greater than 5, is it a heart?&#xA;If your card is equal to 5, is it a 5 of diamonds?&#xA;Is your card the 3 of diamonds?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;By asking if-clause questions which are contrary to the fact, we are reducing the pool of possibilities by getting either 1 or 2 questions answered simultaneously.  If silence is not a possible response, then this approach obviously fails.&lt;/p&gt;&#xA;" OwnerUserId="13920" LastActivityDate="2015-09-03T14:04:03.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20885" PostTypeId="1" CreationDate="2015-09-03T14:19:42.970" Score="5" ViewCount="1161" Body="&lt;p&gt;I'm asking for a generalisation of &lt;a href=&quot;https://puzzling.stackexchange.com/questions/20817/four-cups-on-a-table&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;There is an &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; sided board with &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; cups. Some of the glasses are facing upwards and some upside-down. Your goal is to arrange the glasses so that they are all facing up or all facing down.&lt;/p&gt;&#xA;&lt;p&gt;On each turn, you can check any &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; cups. You are allowed to turn up to &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; of them. After each turn, the board rotates randomly.&lt;/p&gt;&#xA;&lt;p&gt;Here, &lt;span class=&quot;math-container&quot;&gt;$m\geq n\geq k$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Give an algorithm or a formula, if possible, that tells the minimum turns required.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;P.S.&lt;/strong&gt;&#xA;Maybe start-off with &lt;span class=&quot;math-container&quot;&gt;$n=k$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-09-06T06:18:01.510" Title="Generalisation of cup riddle" Tags="|logical-deduction|optimization|" AnswerCount="1" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="20886" PostTypeId="1" AcceptedAnswerId="20896" CreationDate="2015-09-03T14:48:05.720" Score="6" ViewCount="561" Body="&lt;p&gt;My name is Ulysses. I'm the tallest of six brothers and this is our story.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our father, Henry, was a very huge fan of greek mythology, but he never could imagine that we, his very sons, would live an Odyssey.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like I said, I am the tallest and oldest among us; watching the scene from up there gave me the best view of what happened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My youngest brother, Brontes, is one of the most influential and rebellious mass inciter of the time. &quot;Pushing them away from the bottom&quot; is his catch phrase; he goes from city to city trying to raise consensus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's why Brontes can't stand our brother Thanathos, the second born, since he represents all he is fighting against. Thanathos has become the president of a huge multinational corporation.  We can safely say he has reached the top of his career. He has become a bit selfish and stingy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And still, Thanatos was the only one who could help Deimos, the third in line. His small shop has recently declared bankruptcy, and he has started going down to a spiral of depression. If Thanathos helped him maybe all of this could have been avoided.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The penultimate of us is Castor. He is a not-so-famous showman, only known for his charm during the live shows. He is obviously conceited and relates very well with Thanathos, with whom he shares career objectives and common thoughts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, last but not least, there's Sarpedon. Our poor brother suffered mild brain damage when we were young, during a childhood fight with Thanathos. Since then his behaviour can only be described by one word: strange. Not capable of getting a job, he is taken care of by Thanathos, who feels guilty for what happened, although he will never admit it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every year we reunite. And every year we fight about the most utterly stupid things, like brothers do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But last time was different.&#xA;Brontes was mad at Thanathos and in part at Castor for not helping Deimos in his bankruptcy. As rebellious as ever, he punched Thanathos right in the nose, breaking it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nothing similar had ever happened in previous meetings.&#xA;Since then we reunited in 2 groups of 3 people.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now that I managed to get them all together again, seeing Brontes and Thanathos hug, I can say that our bonds are stronger than ever!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who are we? Who is our father?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most of the content is a filler for the story. You are still able to find the answer; the wording is the key.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4652" LastEditorUserId="4652" LastEditDate="2015-09-03T17:26:10.407" LastActivityDate="2015-09-03T17:26:10.407" Title="Down to a strange and charming story about brothers who hit the bottom, until they got up to the top" Tags="|story|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20887" PostTypeId="1" AcceptedAnswerId="20891" CreationDate="2015-09-03T14:52:47.707" Score="-3" ViewCount="358" Body="&lt;p&gt;The birthday child was distributing sweets among the class when the school principal stopped by.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After wishing the child a happy birthday and getting to know his age, the principal asked the following puzzle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;From a certain distance, a hunter spots 3 green parrots and 3 black crows sitting on a tree branch. Then, the hunter aims for a bird using his weapon and and shoots one green parrot. How many birds are left behind?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The smart child answers the puzzle covering all possible scenarios and impresses the principal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you smarter than the fifth grader? Can you impress the principal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint :&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Please read carefully and cover all cases and assumptions. Think like a fifth grader.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;This is another puzzle my classmate was asked in fifth grade and many of us in the class got it wrong. Apologies if it is a duplicate.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="9892" LastEditorUserId="15544" LastEditDate="2015-09-04T08:35:55.193" LastActivityDate="2015-09-04T08:35:55.193" Title="Birthday child bird count Puzzle" Tags="|lateral-thinking|" AnswerCount="3" CommentCount="5" FavoriteCount="0" ClosedDate="2015-09-03T15:26:53.810" ContentLicense="CC BY-SA 3.0" />
  <row Id="20888" PostTypeId="2" ParentId="20887" CreationDate="2015-09-03T14:54:24.393" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Zero live birds, 1 dead bird. The sound of the weapon or the dropping dead of the bird scared the rest of the birds away.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="138" LastEditorUserId="138" LastEditDate="2015-09-03T15:06:31.010" LastActivityDate="2015-09-03T15:06:31.010" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20891" PostTypeId="2" ParentId="20887" CreationDate="2015-09-03T15:11:05.373" Score="4" Body="&lt;p&gt;Well at first it's &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Just the dead parrot...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But then something happens.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The crows, being scavengers, return to eat the dead parrot.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the answer is...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;One, and then later 4. Provided the crows don't go and call their buddies, which can happen!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The alternative solution is a little simpler.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Only the dead parrot is left after the other birds fly away, and then the hunter takes the bird (so that it is no longer left behind) and he takes it off to go eat or stuff.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The puzzle points out a common misconception:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Many people think of hunters as simply shooters, but they hunt for a reason!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are other situations:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The weapon is not a gun. It is a Death Star. As a result, as soon as the weapon is fired, the only bird left is the Millenium Falcon. HA. You didn't see that coming did you?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13068" LastEditorUserId="13068" LastEditDate="2015-09-03T15:30:28.090" LastActivityDate="2015-09-03T15:30:28.090" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20892" PostTypeId="2" ParentId="20825" CreationDate="2015-09-03T15:20:18.820" Score="8" Body="&lt;p&gt;We can't get the precise values of $a$, $b$ and $c$, but we &lt;em&gt;can&lt;/em&gt; determine them &lt;em&gt;up to a constant.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically, given any solution $(a, b, c)$ and any constant $C$, then $(a \oplus C, b \oplus C, c \oplus C)$ is also a solution, because the $C$'s cancel out when we XOR them.  Furthermore, we can show that &lt;em&gt;all&lt;/em&gt; solutions to the given system of equations can, in fact, be derived from a single solution in this way.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To show this, it's useful to generalize a bit.  First of all, we note that the problem with three equations is simply a special case of the same problem with $n$ equations, where we're given&#xA;$$\begin{aligned}&#xA;a_2 \oplus a_1 &amp;amp;= b_1 \\&#xA;a_3 \oplus a_2 &amp;amp;= b_2 \\&#xA;&amp;amp;\ \ \vdots \\&#xA;a_n \oplus a_{n-1} &amp;amp;= b_{n-1} \\&#xA;a_1 \oplus a_n &amp;amp;= b_n&#xA;\end{aligned}$$&#xA;with known $(b_1, \dotsc, b_n)$, and wish to solve for $(a_1, \dotsc, a_n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, it's useful to note that bitwise XOR is equivalent to (vector) subtraction modulo 2 (and also, of course, to vector addition modulo 2, since those are the same thing; but the equivalence to subtraction gives a nicer generalization here).  That is, we may generalize the problem to&#xA;$$\begin{aligned}&#xA;a_2 - a_1 &amp;amp;= b_1 \\&#xA;a_3 - a_2 &amp;amp;= b_2 \\&#xA;&amp;amp;\ \ \vdots \\&#xA;a_n - a_{n-1} &amp;amp;= b_{n-1} \\&#xA;a_1 - a_n &amp;amp;= b_n&#xA;\end{aligned}$$&#xA;where the $a$'s and $b$'s are elements of an &lt;a href=&quot;https://en.wikipedia.org/wiki/Group_%28mathematics%29&quot;&gt;algebraic group&lt;/a&gt;* and $-$ is the subtraction operation defined by $x - y = x + (-y)$ (where $+$ is the group operation, and $-y$ is the inverse element of $y$).  You can easily check that bitstrings indeed satisfy the &lt;a href=&quot;https://en.wikipedia.org/wiki/Group_%28mathematics%29#Definition&quot;&gt;definition&lt;/a&gt; of a group, with XOR as the group operation and each bitstring as its own inverse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is &lt;em&gt;this&lt;/em&gt; useful?  Well, because as long as we're just doing addition (and subtraction, which is just addition of an inverse element), we can treat any such group elements just as if they were ordinary numbers, because (by definition) they obey the same algebraic rules.  In particular, by adding $a_i$ to both sides of the $i$-th equation and simplifying, we can rearrange the equations above into the following equivalent form:&#xA;$$\begin{aligned}&#xA;a_2 &amp;amp;= b_1 + a_1 \\&#xA;a_3 &amp;amp;= b_2 + a_2  \\&#xA;&amp;amp;\ \ \vdots \\&#xA;a_n &amp;amp;= b_{n-1} + a_{n-1} \\&#xA;a_1 &amp;amp;= b_n + a_n.&#xA;\end{aligned}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, we can see that, if we just &lt;em&gt;pick&lt;/em&gt; some value for $a_1$, then we can immediately read out the values of $a_2, \dotsc, a_n$ from the first $n-1$ equations, like this:&#xA;$$\begin{aligned}&#xA;a_2 &amp;amp;= b_1 + a_1 \\&#xA;a_3 &amp;amp;= b_2 + b_1 + a_1 \\&#xA;&amp;amp;\ \ \vdots \\&#xA;a_{n-1} &amp;amp;= b_{n-2} + \dotsb + b_2 + b_1 + a_1 \\&#xA;a_n &amp;amp;= b_{n-1} + b_{n-2} + \dotsb + b_2 + b_1 + a_1&#xA;\end{aligned}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, for each value of $a_1$, there can be (at most) one solution to these equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether or not the values of so obtained in fact &lt;em&gt;are&lt;/em&gt; a solution then depends on the last equation, which needs to yield the original $a_1$ value.  However, it turns out that &lt;em&gt;this doesn't depend on which value we pick!&lt;/em&gt;  In particular, substituting the value of $a_n$ calculated above into the last equation gives&#xA;$$a_1 = b_n + b_{n-1} + b_{n-2} + \dotsb + b_2 + b_1 + a_1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now we can simply subtract $a_1$ from both sides to reduce this equation to:&#xA;$$0 = b_n + b_{n-1} + b_{n-2} + \dotsb + b_2 + b_1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if this equation, which &lt;em&gt;only&lt;/em&gt; contains the $b$ values, holds, then so will the previous one (for any $a_1$!), and so every choice of $a_1$ will yield a solution to the original equation.  In fact, all these solutions will be of the form $a_i = a_i^0 + a_1$, where $a_i^0$ is the value of $a_i$ obtained by starting with $a_1 = 0$.  And conversely, if the last equation above does &lt;em&gt;not&lt;/em&gt; hold, then the original system of equations is inconsistent, and has &lt;em&gt;no&lt;/em&gt; solution at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;*) In fact, I didn't even need to assume that the group is &lt;a href=&quot;https://en.wikipedia.org/wiki/Abelian_group&quot;&gt;abelian&lt;/a&gt;, i.e. that $x + y = y + x$, although for XOR this does certainly hold.  In a &lt;em&gt;non&lt;/em&gt;-abelian group you may not be able to freely reorder the terms in a sum, which can sometimes get awkward, but I didn't really have much need for that here, so I decided to go ahead and prove a slightly more general result without that assumption.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Ps. If you still remember indefinite integrals from high school math, you may recall that the solution $F$ to an integral equation like $F = \int f(x)\,dx$ is only defined up to a constant: if $F$ is a solution, then so is $F + C$ for any constant $C$.  This is the exact same situation, just with discrete pairwise differences instead of differentials (and with XOR instead of ordinary subtraction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, if we're only given the differences between adjacent elements in a sequence, we cannot uniquely determine the original sequence without some way to fix the starting value.  Having the sequence loop around in a circle doesn't change this; it just adds an extra constraint that all the differences must add up to zero for there to be any solution at all.&lt;/p&gt;&#xA;" OwnerUserId="243" LastActivityDate="2015-09-03T15:20:18.820" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20893" PostTypeId="2" ParentId="20887" CreationDate="2015-09-03T15:25:01.447" Score="7" Body="&lt;p&gt;I know! I know!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The hunter has a green-parrot charged rifle, meaning that he literally SHOOTS GREEN PARROTS!! The bird he is aiming at is NOT on the same tree branch as the other birds, since seeing a comrade flying like a missile right in front of their eyes is too cold blooded also for the hunter. So on the branch 3 green parrots and 3 black crows remain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4652" LastActivityDate="2015-09-03T15:25:01.447" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20894" PostTypeId="2" ParentId="20879" CreationDate="2015-09-03T15:43:08.837" Score="3" Body="&lt;p&gt;I believe it is a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Star&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We have existed since the dawn of time&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stars have existed since the dawn of time, they are believed to have a part in the creation of our universe&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet we're not older than a blink of an eye&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Stars are constantly making new atoms inside themselves&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We may be entities of creation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As stated in the first explanation, they are believed to have a part in the creation of our universe&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or entities of destruction&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stars can turn into black holes when they die or collapse&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Behold, however, and you shall see, We are nothing but dust&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Stars are made of &lt;em&gt;stardust&lt;/em&gt; and so is essentially everything in the universe&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;People used to think we were of some ethereal nature&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;People used to think that stars were holes in the sky and there were gods looking down on them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It has been dismissed in favour of something somewhat more mature&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;We now know stars are big balls of elements and fire.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can be everywhere&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Everyone in the world sees them at different times&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can stay now where&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;They slowly move during the night&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can't, however be teared apart&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can't tear them apart they explode or &quot;fold in&quot; on themselves&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For which we have no part&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stars are big balls of fused elements and have no actual &quot;parts&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who are we&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Stars&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15675" LastActivityDate="2015-09-03T15:43:08.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20896" PostTypeId="2" ParentId="20886" CreationDate="2015-09-03T16:36:45.883" Score="8" Body="&lt;p&gt;You are probably&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; quarks. The six flavours are: Up (Ulysses), Down (Deimos), Top (Thanatos), Bottom (Brontes), Strange (Sarpedon) and Charm (Castor).  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and your father is probably&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Murray Gell-Mann, the father of the quark model (George Zweig also developed the theory independently, but he proposed the name &quot;ace&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's cool seeing&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the kinds of particle formation you were hinting at =)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15960" LastActivityDate="2015-09-03T16:36:45.883" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20898" PostTypeId="2" ParentId="12864" CreationDate="2015-09-03T18:12:08.493" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The Problem is ill Posed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If both person 1 states they see person 2 and person 2 states they see person 1 then they must have been in the library at the same time. This is true because at least one of person one and person 3 must be telling the truth as there is only one liar.&lt;/p&gt;&#xA;&lt;p&gt;To move further we need to further define what it means to lie:&lt;/p&gt;&#xA;&lt;h1&gt;If lying by omission is not lying&lt;/h1&gt;&#xA;&lt;p&gt;In this case people are allowed to have seen people they do not mention without being considered liars. The only way to be considered a liar is to say you saw someone you in fact did not.&lt;/p&gt;&#xA;&lt;p&gt;If this is the case then:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Any of them could be a liar.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;First lets tabulate the statements, if the column header stated they saw the row header then I'll mark that intersection. If it was a corroborated sighting it is marked with an X, if it was a one directional sighting it is marked with a ?&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;  a b c d e i&#xA;a \ X   ?&#xA;b X \     ?&#xA;c     \   ? X&#xA;d     ? \&#xA;e ?       \ ?&#xA;i   ? X ?   \&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;From this we can see each person stated at least one uncorroborated sighting.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; This means that any of them could be the liar: Everyone except the person the liar supposedly saw arrives. Then the liar leaves and the last person arrives. Now everyone would have seen everyone with the exception of the liar and the person they supposedly saw. Since the person they supposedly saw did not claim to see the liar, they did not lie. Thus everyone except the liar only told lies of omission and any of them could have done it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h1&gt;If the whole truth and nothing but the truth is required&lt;/h1&gt;&#xA;&lt;p&gt;In this case we can make some more deductions.&lt;/p&gt;&#xA;&lt;p&gt;If neither person 1 states they see person 2 nor person 2 states they see person 1 then either they must not have been in the library at the same time or one of them is the liar. This is true because if they were in the library at the same time at least one of them would have seen the other so then one of them must have omitted seeing the other and is thus lying.&lt;/p&gt;&#xA;&lt;p&gt;The liar could have been there the entire time. The liar could just be lying by omission. As long as everyone else's testimonies are consistent this should be valid. This means that any consistent system of testimonies with 5 people could be valid with the 6th person being the liar that was there the whole time. Thus we can just examine each 5 person subsystem. Thus the analysis that follows excludes the liar from the the analysis, and anywhere is says people or person, this means truth telling people/person as the liar is excluded from the analysis.&lt;/p&gt;&#xA;&lt;p&gt;First lets try the full system assuming no liar:&lt;/p&gt;&#xA;&lt;p&gt;Assuming no lies then here is a table where an X indicates that the row header and column header were in the library at the same time, and a blank indicates they were never in the library at the same time (top half used only for clarity):&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;  a b c d e i&#xA;a \ X   X X&#xA;b   \     X X&#xA;c     \ X X X&#xA;d       \   X&#xA;e         \ X&#xA;i           \&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Now if pair 1&amp;amp;2, pair 2&amp;amp;3, and pair 1&amp;amp;3, each have an X then 1,2, and 3 must have all been in the library at the same time. We know this must be true because none of them can reenter the library. WLOG 1 arrived first. 1 must not leave until both 2 and 3 have arrived. WLOG 2 arrives next. 2 must not leave until 3 has arrived. Thus when 3 arrives they will all be there.&lt;/p&gt;&#xA;&lt;p&gt;There are no sets of 4 people where each pair of people has an X. Thus there was not a time when there were 4 people simultaneously in the library. This means that the periods where there were at least 3 people in the library were never interrupted by a forth person coming or going and can be considered one time unit. (Remember the caveat that the liar is excluded, so the liar my be present simultaneously with three truth tellers)&lt;/p&gt;&#xA;&lt;p&gt;In the 6 person system there are 4 triplets:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/U9l2Q.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/U9l2Q.png&quot; alt=&quot;library occupant triplets&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;ABE, BEI, CDI, and CEI&lt;/p&gt;&#xA;&lt;p&gt;If these three triplets exist on the timeline then all of the X's involved with them will be satisfied. The remaining X's that need to be satisfied are:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;  a b c d e i&#xA;a \     X  &#xA;b   \&#xA;c     \      &#xA;d       \&#xA;e         \  &#xA;i           \&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;AD. For each of these pairs (just one in this case) there must be a time period during which only the pair is in the library as none of these pairs is part of a triplet.&lt;/p&gt;&#xA;&lt;p&gt;Thus a list of occupancies that must have existed at some point during the day are as follows: ABE, BEI, CDI, CEI, and AD&lt;/p&gt;&#xA;&lt;p&gt;This can be visualized in tabular where the rows are people, the columns are time periods and an O indicates occupancy.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;  1 2 3 4 5&#xA;a O       O&#xA;b O O&#xA;c     O O&#xA;d     O   O&#xA;e O O   O&#xA;i   O O O&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;To be a solution the time periods would need to rearrange so that each person occupied the library only during consecutive periods. As expected these periods cannot be arranged in such a manner. Proof for those interested otherwise skip to the next section:&lt;/p&gt;&#xA;&lt;p&gt;For i's visit to be consecutive periods 2, 3, and 4 must be next to each other. withing that span 2&amp;amp;4 and 3&amp;amp;4 must be consecutive for e and c respectively. That only allows the order 2, 4, 3 or its reverse:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;  1 2 4 3 5&#xA;a O       O&#xA;b O O    &#xA;c     O O&#xA;d       O O&#xA;e O O O&#xA;i   O O O&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;For d to be consecutive 5 must be consecutive with 3, and for e and b to be consecutive 1 must go next to 2, but for a to be consecutive 1 must go next to 5 which is impossible. Thus indeed someone must be lying.&lt;/p&gt;&#xA;&lt;h2&gt;The Possibilities&lt;/h2&gt;&#xA;&lt;h3&gt;A lied&lt;/h3&gt;&#xA;&lt;p&gt;Following the same logic as the previous example we can construct tables without A (s A could have been there the whole time if they're lying, This really boils down to removing A from the final period table:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;  2 4 3&#xA;b O&#xA;c   O O&#xA;d     O&#xA;e O O&#xA;i O O O&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; This is a legitimate possibility. A could be the culprit. Here's a diagram of the library occupancy including who saw who when. Note that A did not see B despite saying they did, whereas everyone else saw only who they said they saw, and no two people were in the library at the same time without one of them seeing the other. &lt;a href=&quot;https://i.stack.imgur.com/gf7aI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gf7aI.png&quot; alt=&quot;Library Occupancy Liar A&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h3&gt;B lied&lt;/h3&gt;&#xA;&lt;pre&gt;&lt;code&gt;  1 4 3 5&#xA;a O     O&#xA;c   O O&#xA;d     O O&#xA;e O O&#xA;i   O O&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Again not possible&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h3&gt;C lied&lt;/h3&gt;&#xA;&lt;pre&gt;&lt;code&gt;  1 2 3 5&#xA;a O     O&#xA;b O O&#xA;d     O O&#xA;e O O&#xA;i   O O&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Again not possible&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h3&gt;D lied&lt;/h3&gt;&#xA;&lt;pre&gt;&lt;code&gt;  1 2 4&#xA;a O&#xA;b O O&#xA;c     O&#xA;e O O O&#xA;i   O O&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; This scenario is also possible so D could be the culprit:&lt;a href=&quot;https://i.stack.imgur.com/AgySn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AgySn.png&quot; alt=&quot;Library Occupancy Liar D&quot; /&gt;&lt;/a&gt; Note: that here D said they saw A but they instead saw E. This Solution allows for everyone to see exactly 2 people, which is slightly more in line with the problem statement, so if the problem statement was modified to state that each person saw exactly two people then D would have to be the culprit as if A was the liar they would see a minimum of three people.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h3&gt;E lied&lt;/h3&gt;&#xA;&lt;pre&gt;&lt;code&gt;  1 2 3 5&#xA;a O     O&#xA;b O O&#xA;c     O&#xA;d     O O&#xA;i   O O&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Again not possible&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h3&gt;I lied&lt;/h3&gt;&#xA;&lt;pre&gt;&lt;code&gt;  1 4 3 5&#xA;a O     O&#xA;b O&#xA;c   O O&#xA;d     O O&#xA;e O O&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Again not possible&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;Conclusion&lt;/h2&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; This problem has two solutions as stated, but could be easily modified to eliminate A as the culprit by stating that each person saw exactly two other people in the library. In that case D is the culprit. Another modification/interpretation that some people took and would result in the same conclusion: The lair included no truth in their statement while the honest students stated the whole truth and nothing but the truth (but this makes the puzzle much easier, and relies on using two different definitions of lying). Yet another modification would be to state that no one, including the liar, saw anyone that they didn't claim to see.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6132" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-02-09T16:37:07.477" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="20899" PostTypeId="2" ParentId="20879" CreationDate="2015-09-03T18:29:40.513" Score="3" Body="&lt;p&gt;We are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Photons&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We have existed since the dawn of time,&#xA;Yet we're not older than a blink of an eye&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; as Light (photons) moves at the speed of light, they don't actually experience time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We may be entities of creation,&#xA;or entities of destruction&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Light forms the source of energy that powers light, radiation (which is light) can also breakdown molecules and kill people.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Behold, however, and you shall see&#xA;We're nothing but dust&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In air you only see light as reflection off of dust. Also a note on behold, let there be light. &lt;sub&gt;Biblic'n'stuff&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;People used to think we were of some ethereal nature&#xA;It has been dismissed, in favour of something somewhat more mature&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; THe photon is a pretty recent phenomenon, previous to that I am not sure what they believed probably some like blobby thingamajig, but photons are definitely a more refined theory.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can be everywhere,&#xA;We can stay nowhere&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Light moves at the speed of light always, also EM waves (light) is everywhere there is stuff (depending on your thoughts on gravitons it gets crazy)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can't, however, be teared apart&#xA;For which we have no part.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; photons can't be torn apart, as they don't have a mass to tear apart.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who are we?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We are photons&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13916" LastEditorUserId="13916" LastEditDate="2015-09-06T03:21:31.287" LastActivityDate="2015-09-06T03:21:31.287" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="20901" PostTypeId="2" ParentId="9875" CreationDate="2015-09-03T19:10:22.353" Score="1" Body="&lt;p&gt;While brainstorming on this, I think I can establish an absolute maximum upper bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are $52$ cards in the deck, and for each one, it needs to be adjacent to one of only $49$ of the remaining cards. If we only examine the head node of the string, we can ignore the fact that the last card of the deck SHOULD not be a duplicate of it, and that prevents us from narrowing it down even further, but the rest of the deck has to follow those rules, so it still matters. (visualize the deck as a loop and it helps)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can establish that the absolute maximum number of iterations of the algorithm is $49$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in iteration $#0$ (the &quot;unshuffled&quot; state), all cards are adjacent to 1-2 of their number. (each $7$ is adjacent to another $7$, and up to $2$ of them).&#xA;After the first application of the algorithm, each card must be adjacent to $0$ of their number. Each successful application of the algorithm must make &quot;progress&quot; until you reach $#25$ iterations, when it starts to regress back towards the initial state. Iteration $#49$ would be one of three straight where every number would be adjacent to another of its number, leading to iteration $#52$, which is the same as iteration $#0$. At iteration $#25$, it must be therefore at &quot;maximum&quot; &quot;randomness&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this particular incident, the nonrandomness of the deck would be &lt;strong&gt;RIDICULOUSLY OBVIOUS&lt;/strong&gt;, however. &#xA;It would start off looking something like $12121212343434345656565678787878$ and for &lt;em&gt;FOUR STRAIGHT SESSIONS OF GAMES IN BETWEEN SHUFFLES&lt;/em&gt;, you would be looking at the EXACT SAME OPENING HAND but with the suits changed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, I can't think of an algorithm that would cycle only the suits three out of every 4 &quot;shuffles&quot;. So this is pretty much the extent of my input on the subject until I get bored enough to code it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also to note: If there is no way to &quot;shuffle&quot; between suits for 3 out of 4 &quot;shuffles,&quot; then this serves as proof that $12$ is the upper bound, because that's the only way to expand beyond the simple problem of regression back towards the initial state. But I'm sure someone can explain it more eloquently than I can.&lt;/p&gt;&#xA;" OwnerUserId="13068" LastEditorUserId="13068" LastEditDate="2015-09-03T19:18:01.340" LastActivityDate="2015-09-03T19:18:01.340" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20902" PostTypeId="2" ParentId="20868" CreationDate="2015-09-03T19:24:20.040" Score="2" Body="&lt;h2&gt;Attempt for a 1 measurement solution&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I post it as a separate answer because it makes even more assumptions than the first one. Maybe I should limit it to &lt;a href=&quot;https://en.wikipedia.org/wiki/Spherical_cow&quot; rel=&quot;nofollow noreferrer&quot;&gt;spherical fuses in vacuum&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Assumption 1:&lt;/strong&gt; The multimeter is accurate enough for following measurements.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On the ground were boxes upon boxes of &quot;KIRK'S ZWEET DEEL FUZEZ $(1 \space \Omega)$&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As I cannot use them to replace all fuses, let's use them for the measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Assumption 2:&lt;/strong&gt; Infinite supply of knowingly working fuses (they were all tested by Kirk himself).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Assumption 3:&lt;/strong&gt; All fuses have a resistance of exactly $1\Omega$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Arrange all fuses that must be checked to following groups ordered in a series:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;first fuse is alone in a group (total resistance $R_1 = 1\Omega$)&lt;/li&gt;&#xA;&lt;li&gt;second fuse is with one working fuse from the supply in a group (total resistance $R_2 = 2\Omega$)&lt;/li&gt;&#xA;&lt;li&gt;third fuse is with two working fuses from the supply in a group (total resistance $R_3 = 3\Omega$)&lt;/li&gt;&#xA;&lt;li&gt;fourth fuse is with thress working fuses from the supply in a group (total resistance $R_4 = 4\Omega$)&lt;/li&gt;&#xA;&lt;li&gt;and so on ....&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For the measurement arrange all groups in parallel. The total resistance of this setup if all fuses are OK should be $R = \frac{1}{1/R_1 + 1/R_2 + 1/R_3 + ...}$. Based on the difference to the expected value the broken fuse can be determined in 1 measurement.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Example with 3 fuses&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BbTGS.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Measured value is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$1.2\Omega$ =&gt; Broken fuse is in group $1$.&lt;/li&gt;&#xA;&lt;li&gt;$0.75\Omega$ =&gt; Broken fuse is in group $2$.&lt;/li&gt;&#xA;&lt;li&gt;$0.6666..\Omega$ =&gt; Broken fuse is in group $3$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2015-09-04T13:53:29.550" LastActivityDate="2015-09-04T13:53:29.550" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20903" PostTypeId="1" AcceptedAnswerId="22445" CreationDate="2015-09-03T20:01:35.433" Score="6" ViewCount="686" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;A dream, not a dream.&lt;br&gt;&#xA;  Tailless, tail-less. Puzzled? Makes&lt;br&gt;&#xA;  Quite the parable.&lt;/em&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What am I referring to?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are a few cryptic-clue style indicators that I'm sure many of you have latched onto. The question is, what set of letters are they indicating? Pretty much the entire puzzle goes in order, so if you find yourself jumping around to try to solve it, you've strayed from the path.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer you seek is a single, eight-letter word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Synonymize twice.&lt;br&gt;&#xA; Cryptic clue, cryptic clue. Then,&lt;br&gt;&#xA; It's a synonym.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="11475" LastEditDate="2015-09-18T17:40:09.663" LastActivityDate="2015-09-21T16:00:57.683" Title="To dream, or not to dream (a haiku)" Tags="|riddle|wordplay|poetry|haiku|" AnswerCount="7" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="20904" PostTypeId="2" ParentId="20903" CreationDate="2015-09-03T20:15:59.287" Score="0" Body="&lt;p&gt;could it be,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Story, seems too easy for one of your puzzles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A dream, not a dream.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A story is can be like a dream if it is fictional&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tailless, tailless. Puzzled? Makes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tailless, tailless sounds like tell us, tell us(At least in the south it does). puzzled? tell us what? a story. Make up a story&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Quite the parable.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; parables are moral stories&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13925" LastActivityDate="2015-09-03T20:15:59.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20905" PostTypeId="2" ParentId="20903" CreationDate="2015-09-03T20:18:05.043" Score="6" Body="&lt;p&gt;Is it a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; tale?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A dream, not a dream&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fables is dreamy and fictional, but not a literal dream.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Tailless, tailless. Puzzled?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;tailless&quot; - &quot;less&quot;(the &quot;tail&quot; end of the word) = &quot;tail&quot; (rhymes with &quot;tale&quot;)&#xA; &quot;Puzzled&quot; = search for homophone of &quot;tail&quot; -&gt; &quot;tale&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Makes quite the parable&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Tale&quot; is a synonym of &quot;parable&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15960" LastActivityDate="2015-09-03T20:18:05.043" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20906" PostTypeId="2" ParentId="20903" CreationDate="2015-09-03T20:22:34.157" Score="4" Body="&lt;p&gt;I'm not fully convinced (which probably means it's incorrect as your puzzles go)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But could it be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Les Miserables?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A dream, not a dream.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A reference to the song &quot;I dreamed a dream&quot; - and perhaps how the situation in Paris itself is not a &quot;dream&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tailless, tailless&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Taking the last letter from tailless cryptic style we get - Tailles which are &quot;(in France) taxes levied on the common people by the king or an overlord&quot; which fits with the June Rebellion in Paris that Les Mis is based on.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Puzzled? Makes&#xA;Quite the parable.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I was hoping Les Mis would anagram to something to nicely fit in here, but alas..&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12206" LastActivityDate="2015-09-03T20:22:34.157" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20907" PostTypeId="2" ParentId="20879" CreationDate="2015-09-03T20:56:38.367" Score="1" Body="&lt;p&gt;Since this has been guessed, but deleted due to a poorly written answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Are you &lt;a href=&quot;https://en.wikipedia.org/wiki/Neutrino&quot; rel=&quot;nofollow&quot;&gt;neutrinos&lt;/a&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We have existed since the dawn of time,&lt;br&gt;&#xA;Yet we're not older than a blink of an eye&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Neutrinos were produced in the Big Bang with the rest of matter.&lt;br&gt;&#xA; Neutrinos are created with large Lorentz factors, so one that has been &lt;em&gt;alive&lt;/em&gt; for years in our rest frame could be alive for under a second in its own.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We may be entities of creation,&lt;br&gt;&#xA;or entities of destruction&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Neutrinos are created or destroyed in interactions with other particles. In these interactions, one or more particles are destroyed and others are created.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Behold, however, and you shall see&lt;br&gt;&#xA;We're nothing but dust&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They do not interact electromagnetically, so they cannot be &lt;em&gt;seen&lt;/em&gt; in the traditional sense. They are everywhere but often go unnoticed, so you could call them the dust of elementary particles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;People used to think we were of some ethereal nature&lt;br&gt;&#xA;It has been dismissed, in favour of something somewhat more mature&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The neutrino was thought up as a kind of &quot;ghost particle&quot;, which carried off some energy never to be detected, i.e. &lt;a href=&quot;http://www.merriam-webster.com/dictionary/ethereal&quot; rel=&quot;nofollow&quot;&gt;otherworldly&lt;/a&gt;&lt;br&gt;&#xA; Now we know not only that they exist, but that they interact and &lt;a href=&quot;https://en.wikipedia.org/wiki/Neutrino_detector&quot; rel=&quot;nofollow&quot;&gt;can be detected.&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can be everywhere,&lt;br&gt;&#xA;We can stay nowhere&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Neutrinos are everywhere. &lt;em&gt;See &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_neutrino_background&quot; rel=&quot;nofollow&quot;&gt;relic neutrinos&lt;/a&gt;&lt;/em&gt;&lt;br&gt;&#xA; Neutrinos are always on the move. With mass under an eV, most neutrinos are practically travelling at the speed of light with typical energies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can't, however, be teared apart&lt;br&gt;&#xA;For which we have no part.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are fundamental particles; they cannot be reduced to components.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13237" LastActivityDate="2015-09-03T20:56:38.367" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20913" PostTypeId="2" ParentId="20885" CreationDate="2015-09-03T23:55:50.730" Score="3" Body="&lt;p&gt;&lt;strong&gt;Observations:&lt;/strong&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;For $m=2r\ (r&amp;gt;1, r \in N)$ and $n=k=m-2$, the solution can be achieved in 5 moves or less.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $m=4$, see the original cup problem. Now, suppose, $m\ge6$&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;p&gt;First feel a row of $m-2$ cups and flip all of them down. If the bell doesn't ring, we have $m-2$ consecutive down cups and two unknown cups.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Then feel a one offset (row of $m-3$ cups (chain) and one disconnected cup) of cups and flip them all down. If the bell doesn't ring, we have only a single up cup.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Again feel a one offset.&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;If you feel an up cup, flip it and we are done. &lt;/li&gt;&#xA;  &lt;li&gt;Else, flip the second cup of the chain. Now we have, exactly two cup with $1$ or $3$ cups between them&lt;/li&gt;&#xA;  &lt;/ul&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;We feel a row of $m-2$ cups&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;If you feel 2 up cups, flip them and we are done. &lt;/li&gt;&#xA;  &lt;li&gt;Else, We can feel exactly one up cup. We flip every other cup starting from the up cup. This creates a alternating up-down pattern. &lt;/li&gt;&#xA;  &lt;/ul&gt;&lt;/li&gt;&#xA;  &lt;li&gt;Finally, we feel a alternating pattern ($r$ cups none of which are consecutive) and flip all of them.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;For a specific example ($m=6$, $n=k=4$) see below.  &lt;br&gt;&#xA;  &lt;strong&gt;Note:&lt;/strong&gt; It might be possible to finish in less than 5 moves.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $n=k &amp;lt; (m/2)$ it will always be impossible. (Thanks to f&quot;)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This is because if $n=k &amp;lt; (m/2)$ any pattern you choose will always have two adjacent, unchosen cups. If, in the original state of the table, there are two adjacent cups that are different, the table can rotate so that you never choose those two cups. Thus making it impossible to guarantee a solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If m is prime, $n=k ≤ (m−2)$ is impossible. (Thanks to Tyler Seacrest )&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To show this, suppose the two cups you choose not to feel are $i$ distance apart. Regardless of the initial state of the table, there must be two cups $i$ distance apart that are in different states (as long as the initial table is not solved). Hence the two cups you don't feel might always be in different states, making a sure win impossible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary of results&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;TestNote: Due to item 2 above, all values of n=k &amp;lt; m/2 are omitted&#xA;Key: ? = Not yet proven if solution exists or not&#xA;     X = Not possible&#xA;&#xA;m   n=k   moves&#xA;3    3      1   &#xA;3    2      2&#xA;&#xA;4    4      1&#xA;4    3      2&#xA;4    2      5  (see solution to original cup problem)&#xA;&#xA;5    5      1&#xA;5    4      2&#xA;5    3      X  (see item 3 above)&#xA;&#xA;6    6      1&#xA;6    5      2&#xA;6    4      5  (see proof below)&#xA;6    3      ?  &#xA;&#xA;7    7      1&#xA;7    6      2&#xA;7    5      X  (see item 3 above)&#xA;7    4      X  (see item 3 above)&#xA;&#xA;8    8      1&#xA;8    7      2&#xA;8    6      5&#xA;8    5      ?&#xA;8    4      ?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If anybody notices an error in the information above please let me know. Also if anybody can prove solutions for any of the question marks, or even for higher m values, those solutions might help us notice a pattern.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof of $m=6,\ n=k=4$&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: this method can also be used to solve $m=8,\ n=k=6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZfU2y.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZfU2y.jpg&quot; alt=&quot;m=6 n=k=4&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I have a strong hunch that $m=6,\ n=k=3$ is impossible, but can't prove it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof for $m = s^2$, $n = k = m - s$.  (five moves max)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope you can follow this despite not being as clear as Dsel's images.  In a way it is very similar to the $m = 6$, $n = k = 4$ case above.  A &quot;clump&quot; is when you feel all cups except $s$ consecutive cups.  A &quot;spread&quot; is when you feel all cups except $s$ cups evenly spread around the table (i.e. you don't feel every $s$th cup).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;p&gt;Feel a clump and turn them all down.  &lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Feel a spread and turn them all down.   At this point, at most one cup is up or we hear a bell.  &lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Feel a spread.  If we feel the one up cup, then bell time. Otherwise, turn them all up. We know have $s - 1$ down cups spread out so that there is a down cup every $s$th cup all the way around the table, except one up cup. All other cups are up as well.  &lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Feel a clump.  If there are $s-1$ down cups, turn them all up and hear a bell.  If there are $s-2$ down clumps:  group the cups into $s$ groups, where each group consists of $s$ consecutive cups, and one of the groups is all the cups you are not currently feeling.   Then all the groups have exactly one down cup in exactly the same spot, except one group which has all up cups.  Mimic the rest of the groups by turning one of cups down in the appropriate spot.  &lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Feel a spread.  From the last step, we know that all cups are up except $s$ cups are down, one down cup every $s$ cups.  By feeling a spread, we can see all the down cups or none of them.  If we see all the down cups, turn them up and hear a bell.  If we see none of the down cups, then turn all the cups down and hear a bell.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16046" LastEditorUserId="2722" LastEditDate="2015-09-06T06:18:01.510" LastActivityDate="2015-09-06T06:18:01.510" CommentCount="12" CommunityOwnedDate="2015-09-05T05:28:27.050" ContentLicense="CC BY-SA 3.0" />
  <row Id="21913" PostTypeId="2" ParentId="20790" CreationDate="2015-09-04T02:58:15.657" Score="6" Body="&lt;p&gt;Well, now that &lt;a href=&quot;https://puzzling.stackexchange.com/questions/20790/crossword-my-favourite-puzzlers/21913#comment65847_20790&quot;&gt;everyone&lt;/a&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/questions/20790/crossword-my-favourite-puzzlers/21913#comment65852_20803&quot;&gt;else&lt;/a&gt; has done &lt;a href=&quot;https://puzzling.stackexchange.com/a/20803/5208&quot;&gt;99% of the work&lt;/a&gt;, allow me take the glory... The second puzzler is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;JLee&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There's several alternative answers you can use for the following:&lt;br&gt;&#xA; &lt;strong&gt;81D&lt;/strong&gt; - With &quot;chow&quot;, a delicious dish - &lt;strong&gt;MEAN&lt;/strong&gt; (&quot;this is some &lt;em&gt;mean chow&lt;/em&gt;&quot;)&lt;br&gt;&#xA; &lt;strong&gt;82D&lt;/strong&gt; - Matt in Dogma - &lt;strong&gt;DEMON&lt;/strong&gt; (Matt Damon plays the part of Loki, a demonic figure)&lt;br&gt;&#xA; &#xA; &lt;strong&gt;89A&lt;/strong&gt; - Forcibly remove - &lt;strong&gt;EJECT&lt;/strong&gt;&lt;br&gt;&#xA; &lt;strong&gt;94A&lt;/strong&gt; - Close to the bullseye - &lt;strong&gt;ALMOST&lt;/strong&gt;&lt;br&gt;&#xA; &lt;strong&gt;101A&lt;/strong&gt; - Half the title of a Kasabian song - &lt;strong&gt;NEON&lt;/strong&gt; (from the song &quot;Neon Noon&quot;)&lt;br&gt;&#xA; &lt;strong&gt;107A&lt;/strong&gt; - You'll find several in intensive care units - &lt;strong&gt;ENS&lt;/strong&gt; (there's two types of nurses Registered Nurses and Enrolled Nurses, or RNs and ENs)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&#xA;&lt;br/&gt;&#xA;&lt;sub&gt;...I guess you couldn't fit Alconja into a four letter space anyway...&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-09-04T02:58:15.657" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="21914" PostTypeId="1" AcceptedAnswerId="21915" CreationDate="2015-09-04T04:36:49.350" Score="4" ViewCount="647" Body="&lt;p&gt;Imagine a sphere with a hole that has been drilled clean through its center (i.e. a cylindrical piece of the sphere is now missing). This new shape, with the core missing, has height of 6 when standing on the ground (i.e. flat side on the ground). What is the volume of this shape?&lt;/p&gt;&#xA;" OwnerUserId="16046" LastActivityDate="2015-09-04T06:36:53.680" Title="A Hollow Sphere" Tags="|mathematics|geometry|" AnswerCount="1" CommentCount="10" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21915" PostTypeId="2" ParentId="21914" CreationDate="2015-09-04T06:34:39.897" Score="2" Body="&lt;p&gt;This is a popular problem commonly referred to as the Napkin ring problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, irrespective of the diameter of the drilled cylinder, the volume of the resultant object will always be equal to the volume of a sphere whose diameter is the same length as the height of the object. In this case, the volume of the sphere is $4/3\pi (6/2)^3$ or roughly 113.1 cubic units, no matter what the diameter of the original sphere was. As long as it was drilled out perfectly till the height of the new object is 6, the volume remains the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Napkin_ring_problem&quot; rel=&quot;nofollow&quot;&gt;Here is the Wikipedia entry outlining the problem and its solution.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="6178" LastEditDate="2015-09-04T06:36:53.680" LastActivityDate="2015-09-04T06:36:53.680" CommentCount="0" CommunityOwnedDate="2015-09-04T06:34:39.897" ContentLicense="CC BY-SA 3.0" />
  <row Id="21916" PostTypeId="1" AcceptedAnswerId="21939" CreationDate="2015-09-04T10:13:27.520" Score="-1" ViewCount="173" Body="&lt;p&gt;This is a puzzle-&lt;a href=&quot;https://en.wikipedia.org/wiki/Haiku_in_English#One_line_.28monoku.29&quot; rel=&quot;nofollow noreferrer&quot;&gt;monoku&lt;/a&gt;. The answer is a place in the British Isles.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;brave&lt;/p&gt;&#xA;&lt;p&gt;stone&lt;/p&gt;&#xA;&lt;p&gt;heart&lt;/p&gt;&#xA;&lt;p&gt;forth&lt;/p&gt;&#xA;&lt;p&gt;Barksdale&lt;/p&gt;&#xA;&lt;p&gt;tower&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-09-09T17:02:16.367" Title="Brave stone heart forth Barksdale tower" Tags="|knowledge|geography|poetry|haiku|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21917" PostTypeId="2" ParentId="4568" CreationDate="2015-09-04T10:22:46.970" Score="0" Body="&lt;p&gt;5 questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First question would be &quot;With suit value as 0 for club, 1 for diamond, 2 for hearts, and 3 for spades calculate the total value of the card as 13 times the suit value plus the card value, except all diamonds above 5 of diamonds have total value 30 as does all hearts less than 10 of hearts; is the total value equal to, greater than or less than 30&quot;?  This reduces the set of possible cards to a set with at most 18 cards.  Proceed with similar questions to reduce the potential cards by one third in each question.&lt;/p&gt;&#xA;" OwnerUserId="1654" LastActivityDate="2015-09-04T10:22:46.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21918" PostTypeId="2" ParentId="20692" CreationDate="2015-09-04T11:31:51.117" Score="1" Body="&lt;p&gt;I am the one what did the did and the one what knows what did the did.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The person who has committed an evil, unethical, or immoral act is __________. The person who is looking for the _______ is also __________. It's often said that detectives are __________ of everyone they meet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am one and the same, but distinct and separate.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Both words are spelled the same, are adjectives, but describe two opposing people in the same situation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The situations I create are also quite tense.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Virtually all drama involves someone being __________. You will be hard pressed to find a crime drama that does not refer to someone as a _______ or __________.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am no Cyrano, but if you're looking for me, you'll find ten letters.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The word you're looking for has ten letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The word is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;SUSPICIOUS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And you often use words directly related to it such as &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;SUSPECT or SUSPICION.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13068" LastActivityDate="2015-09-04T11:31:51.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21919" PostTypeId="1" AcceptedAnswerId="21922" CreationDate="2015-09-04T11:49:31.570" Score="2" ViewCount="289" Body="&lt;p&gt;For the last training session of collaboration training, the group were taken to the brainstorm room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There were 8 boxes with something written text at the base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/d6gfx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/d6gfx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The instruction was : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can you collaborate and determine what is in the boxes?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rules : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) You may Mix, Match, shuffle, reshuffle the contents and pass it around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Think inside the box, about the box and around the box&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your time starts now&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In case somebody needs it in text format.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;4A2P            5N          2M1Y    &#xA;&#xA;&#xA;1C7R                        1J1V    &#xA;&#xA;&#xA;6E4S            1H4T        1I7U    &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Hint 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Put same number of Letter BLOCKS (Cubes) in each box corresponding to the number and shuffle it around. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9892" LastEditorUserId="4385" LastEditDate="2015-09-04T14:52:29.460" LastActivityDate="2015-09-04T14:52:29.460" Title="Numbers and Letters in Boxes" Tags="|word|" AnswerCount="1" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="21920" PostTypeId="1" AcceptedAnswerId="21929" CreationDate="2015-09-04T12:17:15.977" Score="3" ViewCount="336" Body="&lt;p&gt;No one knows quite why,&lt;br /&gt;&#xA;But we are very shy.&lt;br /&gt;&#xA;You can't touch our crowns,&lt;br /&gt;&#xA;So far from the ground.&lt;br /&gt;&lt;br /&gt;&#xA;It might have been because,&lt;br /&gt;&#xA;Of all the bump and grind.&lt;br /&gt;&#xA;But Ng showed that was false,&lt;br /&gt;&#xA;When given enough time.&lt;br /&gt;&lt;br /&gt;&#xA;Do you know what I speak of?&lt;br /&gt;&#xA;I don't think you do.&lt;br /&gt;&#xA;It will be a long weekend,&lt;br /&gt;&#xA;So best of luck to you.&lt;/p&gt;&#xA;" OwnerUserId="13068" LastActivityDate="2015-09-04T15:05:08.663" Title="Another Friday riddle - We are quite shy" Tags="|riddle|poetry|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="21922" PostTypeId="2" ParentId="21919" CreationDate="2015-09-04T12:53:04.683" Score="3" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mercury &lt;br/&gt;&#xA; Venus&lt;br/&gt;&#xA; Earth&lt;br/&gt;&#xA; Mars&lt;br/&gt;&#xA; Jupiter&lt;br/&gt;&#xA; Saturn&lt;br/&gt;&#xA; Uranus&lt;br/&gt;&#xA; Neptune&lt;br /&gt;&lt;br /&gt;&#xA; This is obtained by taking the number of each letter (e.g. 4 As, 2 Ps, etc.) and rearranging them.&lt;br/&gt;&#xA; Thanks to Kingrames for the pointer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="14135" LastEditDate="2015-09-04T13:00:32.310" LastActivityDate="2015-09-04T13:00:32.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21923" PostTypeId="1" CreationDate="2015-09-04T13:08:28.307" Score="3" ViewCount="506" Body="&lt;p&gt;Steve is having a rather bad day. Steve has died, and because Steve had a propensity for driving slow in the fast lane, Steve goes to hell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As soon as Steve gets to hell, Satan, Hitler, and Herostratus decide they're going to give Steve a chance to escape. They're going to play a game. All three of the nefarious denizens of hell have taken one of their favorite riddles, stuck it into a nefarious hat, and one of them will be chosen at random.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But they have made a boast! Hitler &lt;em&gt;reveals&lt;/em&gt; the answer to his riddle, and says: &quot;The answer to my riddle is a &lt;strong&gt;ruined painting&lt;/strong&gt;.&quot; Satan reveals the answer to his, as well: &quot;The answer to my riddle is a &lt;strong&gt;grilled fish&lt;/strong&gt;!&quot; And Herostratus reveals his answer too: &quot;My riddle's solution is a &lt;strong&gt;destroyed statue&lt;/strong&gt;!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As soon as the riddle is drawn out of the nefarious hat, each of the three nefarious denizens of hell are just going to start writing their answers! That's not fair! So Steve has to start writing an answer BEFORE the riddle is drawn, and the answer either has to fit all three criteria, or it has to be close enough that Steve can finish writing the answer before Hitler, Satan, and Herostratus can!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you help Steve? Do you know how Steve could possibly get out of this situation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use pencil and paper drawing (you probably have to), and bend any rules you can, because this is hell! Feel free to use very ambiguous fonts and/or pictographs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Remember:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Ruined Painting&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Grilled Fish&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Destroyed Statue&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You don't have to combine all three in one. You can start with very ambiguous text, and from there, show how you can write all three answers from your starting point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AS AN EXAMPLE:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steve could write &quot;ed&quot; and would have a very minor head start, being able to write &quot;ruin&quot; or &quot;grill&quot; or &quot;destroy&quot; before it, and the rest of the answer afterwards. Can you do better?&lt;/p&gt;&#xA;" OwnerUserId="13068" LastEditorUserId="13068" LastEditDate="2015-09-04T13:36:07.747" LastActivityDate="2015-09-04T17:38:44.980" Title="OE Challenge - Solve faster than fast! Help Steve get out of hell" Tags="|open-ended|pencil-and-paper-games|" AnswerCount="5" CommentCount="9" ClosedDate="2015-09-04T17:33:00.037" ContentLicense="CC BY-SA 3.0" />
  <row Id="21924" PostTypeId="2" ParentId="21923" CreationDate="2015-09-04T13:54:50.370" Score="2" Body="&lt;p&gt;Write out all three answers and then cross out the two wrong ones when the riddle is revealed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The danger here is if it takes longer to work out which answer is the correct one than it takes for the fiends to write out their answers, Steve is toast.&lt;/p&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-09-04T13:54:50.370" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21925" PostTypeId="2" ParentId="21923" CreationDate="2015-09-04T13:54:59.663" Score="2" Body="&lt;p&gt;I figure if they're going make you solve a riddle you might as well make your answer a riddle as well. It's all about how you interpret my answer do you find the answer you're looking for. You could say:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Well if the answer is &lt;strong&gt;grilled fish&lt;/strong&gt; take the first letter from each group of 3 letters consecutively (meaning from left to right). If the answer is &lt;strong&gt;ruined painting&lt;/strong&gt; take the 2nd letter from each group of 3 letters and the first from the groups of two (again from left to right). For &lt;strong&gt;destroyed statue&lt;/strong&gt; say take the last letter from all the groups (once more from left to right.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If he was writing on a chalkboard he just has to write everything vertically and erase the columns that are wrong. Unless I can make my description more explicit this answer should do it:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;grd&#xA;rue&#xA;iis&#xA;lnt&#xA;ler&#xA;edo&#xA;dpy&#xA;fae&#xA;iid&#xA;sns&#xA;htt&#xA; ia&#xA; nt&#xA; gu&#xA;  e&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7377" LastEditorUserId="4421" LastEditDate="2015-09-04T17:38:44.980" LastActivityDate="2015-09-04T17:38:44.980" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21927" PostTypeId="2" ParentId="21923" CreationDate="2015-09-04T14:26:46.750" Score="6" Body="&lt;p&gt;Here's a way that should be faster than writing down letters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say Steve was handed a piece a paper for him to draw on. Steve quickly draws an &lt;a href=&quot;https://en.wikipedia.org/wiki/Ichthys&quot;&gt;ichthus&lt;/a&gt; and adds some grill marks on it then rips the paper in half. He takes the bottom half, stands it up on the ground, and places some rocks around it so it stands up straight. If the answer is a destroyed statue he points at the paper on the ground that's been torn in half and claims that's his destroyed statue. If it's a ruined painting he holds up the half of the paper in his hand and says it's his ruined painting (since it's been torn in half). If it's a grilled fish he puts both halves together to show a complete grilled fish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This way Steve has all 3 answers completed and doesn't have to do any modification once he solves the riddle albeit he still has to put the 2 halves of paper together if it's the grilled fish.&lt;/p&gt;&#xA;" OwnerUserId="7377" LastActivityDate="2015-09-04T14:26:46.750" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21928" PostTypeId="2" ParentId="21923" CreationDate="2015-09-04T14:43:59.420" Score="4" Body="&lt;p&gt;Not having access to particularly good drawing tools, here's a very crude image:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/EjctO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EjctO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm sure someone with more artistic talent could take this concept and improve on it.&lt;/p&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-09-04T14:43:59.420" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21929" PostTypeId="2" ParentId="21920" CreationDate="2015-09-04T14:46:11.167" Score="2" Body="&lt;p&gt;So following the discussion in the &lt;a href=&quot;http://chat.stackexchange.com/rooms/28781/discussion-between-kingrames-and-moonbutt74&quot;&gt;chat&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you looking for &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Crown Shyness (Tree or Dryobalanops)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No one knows quite why,&#xA;But we are very shy.&#xA;You can't touch our crowns,&#xA;So far from the ground.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Species of some trees are shy to touch each other on the top. Hard to figure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It might have been because,&#xA;Of all the bump and grind.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Growing up with neighbors so close has its share of bumping and grinding. ;-)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But Ng showed that was false,&#xA;When given enough time.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Scientist/Researcher F.P. Ng showed the theory was &lt;a href=&quot;https://en.wikipedia.org/wiki/Crown_shyness&quot; rel=&quot;nofollow&quot;&gt;incorrect&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Do you know what I speak of?&#xA;I don't think you do.&#xA;It will be a long weekend,&#xA;So best of luck to you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Red Herrings with a Challenge :-)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks to @Kingrames for another informative quest.&lt;/p&gt;&#xA;" OwnerUserId="9892" LastEditorUserId="9892" LastEditDate="2015-09-04T15:05:08.663" LastActivityDate="2015-09-04T15:05:08.663" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21930" PostTypeId="2" ParentId="20903" CreationDate="2015-09-04T14:52:17.840" Score="1" Body="&lt;p&gt;Just thinking out loud in case somebody can use this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A dream, not a dream&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Daydream, an idea, a thought&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tailless, tail-less&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tailless - tail = less, otherwise tailless - the literal tail of the word (less) = tail&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Puzzled? Makes quite the parable&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My guess is that the combination of two words deduced from the first two clues forms the name of a well known parable&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15990" LastActivityDate="2015-09-04T14:52:17.840" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21931" PostTypeId="2" ParentId="21923" CreationDate="2015-09-04T14:58:08.483" Score="1" Body="&lt;p&gt;Just write all 3 answers with a checkbox.  As soon as you know the answer, check the correct box.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastActivityDate="2015-09-04T14:58:08.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21932" PostTypeId="1" AcceptedAnswerId="21944" CreationDate="2015-09-04T15:08:55.573" Score="10" ViewCount="781" Body="&lt;p&gt;&lt;em&gt;My first's an incredible youngster.&lt;br&gt;&#xA;My second takes time to complete.&lt;br&gt;&#xA;My third's an acrobatic contraption.&lt;br&gt;&#xA;My fourth can be used to compete.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My fifth can mean one plus one half.&lt;br&gt;&#xA;My sixth is a final two tether.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My seventh will make all the difference.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;What am I when my parts come together?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each line references a single word, with each word being pretty much unrelated to the desired animal. Once the words are interpreted correctly, you'll find the name of the dissected animal I seek. Keep in mind that the spacing is intentional.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #2/Clarification:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clarification first: I accidentally lied. One of the answers is actually two words. A few of the words are similar/synonymous, so finding one will help you find the rest. For the first line, one of the words would be more correct as a &lt;em&gt;proper noun&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="11475" LastEditDate="2015-09-10T15:44:14.373" LastActivityDate="2015-09-17T13:35:05.617" Title="What dissected animal am I?" Tags="|riddle|wordplay|" AnswerCount="3" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="21933" PostTypeId="1" AcceptedAnswerId="21937" CreationDate="2015-09-04T15:18:22.003" Score="7" ViewCount="720" Body="&lt;p&gt;Even though Bailey &lt;a href=&quot;https://puzzling.stackexchange.com/questions/20903/to-dream-or-not-to-dream-a-haiku&quot;&gt;beat&lt;/a&gt; me to the 17-syllabled punch, here's my haiku about one of my favorite things.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Size belied by name&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;Seen ev'rywhere but unseen&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;A slow moving beast&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-09-04T15:52:25.703" Title="Evolution's Triumph" Tags="|riddle|poetry|haiku|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21934" PostTypeId="2" ParentId="20903" CreationDate="2015-09-04T15:18:50.457" Score="0" Body="&lt;p&gt;Hmm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Idea #1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's the&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Martin Luther King &quot;&lt;a href=&quot;http://news.bbc.co.uk/1/hi/world/americas/3170387.stm&quot; rel=&quot;nofollow&quot;&gt;I Have A Dream&lt;/a&gt;&quot; speech&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A dream, not a dream.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A dream that MLK hoped to see come true.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tailless, tail-less. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MLK's work was terminated - cut short, curtailed - by his assassination.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Puzzled? Makes&#xA;Quite the parable.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The speech uses a number of rhetorical devices which have their roots in religious preaching; see &lt;a href=&quot;https://en.wikipedia.org/wiki/I_Have_a_Dream#The_speech_and_rhetoric&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/I_Have_a_Dream#The_speech_and_rhetoric&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MLK was of course a minister of religion. &quot;You know, actually all that I do in civil rights I do because I consider it a part of my ministry.&quot; &lt;a href=&quot;https://en.wikipedia.org/wiki/Martin_Luther_King,_Jr.#Ideas.2C_influences.2C_and_political_stances&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Martin_Luther_King,_Jr.#Ideas.2C_influences.2C_and_political_stances&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Idea #2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Joseph_(patriarch)#Dreams_as_motif_and_contribution_to_narrative&quot; rel=&quot;nofollow&quot;&gt;Joseph's dreams from Genesis&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A dream, not a dream.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the story, dreams become prophecy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tailless, tail-less. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Joseph's &lt;a href=&quot;https://en.wikipedia.org/wiki/Joseph_and_the_Amazing_Technicolor_Dreamcoat&quot; rel=&quot;nofollow&quot;&gt;coat of many colours&lt;/a&gt; is typically quite a long coat - a tail-coat! But it's taken away: Joseph is tail-coat-less.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Puzzled? &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pharoah was puzzled - needed Joseph to interpret the cow dreams.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Makes&#xA;Quite the parable.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The story is a Judeo-Christian-Islamic-Baha'i religious text.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="4050" LastEditDate="2015-09-04T15:24:23.430" LastActivityDate="2015-09-04T15:24:23.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21935" PostTypeId="2" ParentId="21933" CreationDate="2015-09-04T15:29:39.163" Score="5" Body="&lt;p&gt;Perhaps not intended, but fits:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Microsoft&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Size belied by name&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The word &quot;micro&quot; belies the size of the corporation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Seen ev'rywhere but unseen&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Windows is everywhere, but has no physical form to see since it is software&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A slow moving beast&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Microsoft as a large bureaucracy can be rather slow.  For instance, they finally added virtual desktops as a built-in feature to Windows 10, while Linux has had them for like, 20 years.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6178" LastActivityDate="2015-09-04T15:29:39.163" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21936" PostTypeId="2" ParentId="21933" CreationDate="2015-09-04T15:32:43.710" Score="0" Body="&lt;p&gt;Is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Bearded Dragon / Dragon Lizard?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Size belied by name&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; With Dragon in the name, you'd expect it to be huge.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Seen ev'rywhere but unseen&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reference to the camouflaging capabilities of the lizard.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A slow moving beast&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They like to stay still a lot..&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Evolution's Triumph&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The camouflage was an evolutionary triumph for the species.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12206" LastEditorUserId="12206" LastEditDate="2015-09-04T15:52:25.703" LastActivityDate="2015-09-04T15:52:25.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21937" PostTypeId="2" ParentId="21933" CreationDate="2015-09-04T15:48:43.043" Score="9" Body="&lt;p&gt;Is it a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Tardigrade&quot;&gt;Tardigrade&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Size belied by name&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A pretty long name for a micro-animal. It's other names are quite long as well - waterbear or moss piglet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Seen ev'rywhere but unseen&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tardigrades can literally be found anywhere (even in space!). But they are too small to be seen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A slow moving beast&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tardigrade comes from Tardigrada, meaning &quot;slow stepper&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the title:&#xA;Evolution's Triumph&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tardigrades can survive in really extreme environments (even space!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastActivityDate="2015-09-04T15:48:43.043" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21938" PostTypeId="2" ParentId="21932" CreationDate="2015-09-04T15:52:11.487" Score="-3" Body="&lt;p&gt;My first's an incredible youngster.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; B as in Spelling Bee&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My second takes time to complete.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A as in Answer of Q &amp;amp; A&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My third's an acrobatic contraption.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C as in Circus tent&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fourth can be used to compete.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; T -&gt; Target&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fifth can mean one plus one half.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; E -&gt; onE + on -&gt; E ? (I doubt this)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My sixth is a final two tether.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RI -&gt; Infra Red -&gt; or Rigorous Imprisonment&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My seventh will make all the difference.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A as in Attitude&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I when my parts come together?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BACTERIA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9892" LastActivityDate="2015-09-04T15:52:11.487" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21939" PostTypeId="2" ParentId="21916" CreationDate="2015-09-04T17:34:22.490" Score="2" Body="&lt;p&gt;Okay i will risk the smackdown on this,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will start with my answer but nailing the monoku word by word will take a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Wallace_Monument&quot; rel=&quot;nofollow&quot;&gt;The National Wallace Monument&lt;/a&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; brave heart would be funny if you meant &lt;a href=&quot;https://en.wikipedia.org/wiki/Wallace_Monument#/media/File:William_Wallace_Statue.JPG&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; which had to be put in a cage.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though i think you are looking for a more specific point or specific example among the &lt;a href=&quot;http://www.thesocietyofwilliamwallace.com/williamwallacemonuments.htm&quot; rel=&quot;nofollow&quot;&gt;many&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;On the monoku;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I broke it down from,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Brave stone heart forth Barksdale tower&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Brave heart stone tower forth Barksdale&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;brave heart would be funny if you meant &lt;a href=&quot;https://en.wikipedia.org/wiki/Wallace_Monument#/media/File:William_Wallace_Statue.JPG&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; which had to be put in a cage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;stone tower is umm, a stone tower?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Forth Valley is part of the view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Barksdale is where I'm stuck if any of the above is correct or even close as there is apparently no geographical feature/location associated with that &quot;name&quot; in the British Isles. Searching i pull up hits for tower houses and an air force base in Louisiana.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So i don't really dig on pop culture so my final guess on Barksdale would be the character from the wire, which could be a sneaky word-play clue to stand-in for the cage around the statue.&lt;/p&gt;&#xA;" OwnerUserId="15757" LastEditorUserId="4050" LastEditDate="2015-09-09T17:02:16.367" LastActivityDate="2015-09-09T17:02:16.367" CommentCount="14" CommunityOwnedDate="2015-09-05T02:39:21.777" ContentLicense="CC BY-SA 3.0" />
  <row Id="21940" PostTypeId="2" ParentId="18343" CreationDate="2015-09-04T18:25:59.073" Score="9" Body="&lt;p&gt;I only checked a few, so I'm not sure, but I think it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A donut phrase is one where the &quot;average letter&quot; is not found in the phrase. Here average letter is found by converting all letters into numbers where a=1 b=2 etc. and averaging. Round the average and convert back to a letter to get the average letter. It is called a donut phrase since the average letter is the hole, and the letters around it make up the donut.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15810" LastActivityDate="2015-09-04T18:25:59.073" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21941" PostTypeId="2" ParentId="20737" CreationDate="2015-09-04T18:41:54.970" Score="6" Body="&lt;p&gt;&lt;strike&gt;I only have part of it so far, but...&lt;/strike&gt; Okay I think I have it now?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reading &quot;Yes, ommmm... Eek! Sea! Rye! Pea! Toe! Gee, ram.&quot; as alternating multi-letter sounds/single-letter names gives the letters &quot;YS|M|EEK|C|RY|P|TO|G|RAM&quot;, the latter part of which is obviously the word &quot;cryptogram&quot;, a pretty meta description of the puzzle itself. The first part is slightly more confusing, but I'd wager the Y is meant to be pronounced as a letter as well, producing the phrase &quot;Wise Meek Cryptogram&quot;. Which kind of fits the meditation theme?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15908" LastEditorUserId="15908" LastEditDate="2015-09-04T19:08:17.753" LastActivityDate="2015-09-04T19:08:17.753" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21942" PostTypeId="2" ParentId="20737" CreationDate="2015-09-04T18:59:46.450" Score="3" Body="&lt;p&gt;I think it is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes, A meek cryptogram. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Obvious&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ommmm..&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ommmm sounds like ahmmmmm,  plus eak! combines to form A meek - meek means subtle, imposed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sea&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Letter C&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rye&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rye sounds like arrr , why  (RY)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pea&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Letter P&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Toe&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Whole sound, in crypTOgram&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Gee, Ram&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Whole sound, in cryptoGRAM&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Meaning&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The whole phrase would be 'Yes, A subtle or imposed cryptogram'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;disclaimer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Now if the phrase &quot;Yes, A meek cryptogram, is a cryptogram for something else entirely, youve got me&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15990" LastActivityDate="2015-09-04T18:59:46.450" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21944" PostTypeId="2" ParentId="21932" CreationDate="2015-09-04T20:10:33.707" Score="7" Body="&lt;p&gt;&lt;strong&gt;Second Attempt&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first's an incredible youngster.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dash -&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My second takes time to complete.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a period is a length or portion of time or &quot;.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My third's an acrobatic contraption.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a trapeze which has a horizontal bar -&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fourth can be used to compete.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Points! .&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fifth can mean one plus one half.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; . 1.5 take the dot out of it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My sixth is a final two tether.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - in a tournament bracket the final two teams are represented with a - between them&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My seventh will make all the difference.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - you can find the difference of two numbers by subtracting them -&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I believe it's,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;-.-.&amp;nbsp;&amp;nbsp;.-&amp;nbsp;&amp;nbsp;-&lt;/strong&gt; = Cat&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First Wrong Attempt&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first's an incredible youngster.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Thomas Young, I'm basing the rest of my answers off this&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My second takes time to complete.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everything, he was said to be the last man to know everything&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My third's an acrobatic contraption.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Springboard, Young's modulus is used to determine stress on the body and prior to that Hooke's law was used to determine deformation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fourth can be used to compete.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Time, he was part of a commission  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fifth can mean one plus one half.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Interference, when two waves are out of phase the cause interference demonstrated by the Double-slit experiment &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My sixth is a final two tether.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Natural &lt;a href=&quot;http://patents.justia.com/patent/20080076646&quot; rel=&quot;nofollow&quot;&gt;Fast arm-swing tether&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My seventh will make all the difference.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Genius, Thomas Young was a genius&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I believe it's,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Test in g. Animal testing dissected&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13925" LastEditorUserId="13925" LastEditDate="2015-09-17T13:35:05.617" LastActivityDate="2015-09-17T13:35:05.617" CommentCount="21" ContentLicense="CC BY-SA 3.0" />
  <row Id="21945" PostTypeId="2" ParentId="20825" CreationDate="2015-09-04T20:16:58.717" Score="1" Body="&lt;p&gt;&lt;code&gt;a⊕b ⊕ b⊕c == a⊕c&lt;/code&gt;&#xA;So, as already noted, you have 2 equations with 3 unknowns, and this is not possible, in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this is a common scenario in cryptography. If &lt;code&gt;b&lt;/code&gt; is the ciphertext, and &lt;code&gt;a&lt;/code&gt; and &lt;code&gt;c&lt;/code&gt; are two plaintexts, you can use &lt;code&gt;a⊕b&lt;/code&gt; and &lt;code&gt;b⊕c&lt;/code&gt; to get the XOR of two plaintexts, &lt;code&gt;a⊕c&lt;/code&gt;. You can then use linguistic analysis, statistical methods, etc., to decipher &lt;code&gt;a&lt;/code&gt;, &lt;code&gt;c&lt;/code&gt;, and then &lt;code&gt;b&lt;/code&gt;. But this relies on additional information about &lt;code&gt;a&lt;/code&gt; and &lt;code&gt;c&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="16219" LastActivityDate="2015-09-04T20:16:58.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21946" PostTypeId="1" AcceptedAnswerId="21948" CreationDate="2015-09-04T21:02:29.133" Score="5" ViewCount="433" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;If I take it steady, I am only right once,&lt;/p&gt;&#xA;&lt;p&gt;but if I'm wrong, I stay wrong.&lt;/p&gt;&#xA;&lt;p&gt;If I stand still, I am right twice.&lt;/p&gt;&#xA;&lt;p&gt;but if I go backwards, I am right four times.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2673" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-09-04T22:51:54.487" Title="This is how right I am, but what am I?" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21947" PostTypeId="1" AcceptedAnswerId="21953" CreationDate="2015-09-04T21:04:17.210" Score="8" ViewCount="341" Body="&lt;p&gt;A word used in anticipation,&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm in a simple machine,&lt;br&gt;&#xA;in the witty,&lt;br&gt;&#xA;and in boisterous celebration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm in things separated,&lt;br&gt;&#xA;in the intense,&lt;br&gt;&#xA;and in this clue's valuation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What am I?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;&lt;sub&gt;As always, the accepted answer will explain how each and every line fits the riddle. Despite brevity and ambiguity, I don't expect too many complaints of this riddle's broadness; it'll probably be solved in under 25 views.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="13237" LastActivityDate="2015-09-05T17:49:13.140" Title="NeedAName? This'll do" Tags="|riddle|word|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21948" PostTypeId="2" ParentId="21946" CreationDate="2015-09-04T21:06:50.600" Score="7" Body="&lt;p&gt;Are you : &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A clock?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I'm steady, I am only right once&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you stay on the right time, you're always right&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but if I'm wrong, I stay wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you're off, but working, you'll never have the right time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I stand still, I am right twice.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A broken clock is right twice a day&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but if I go backwards, I am right four times.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A clock running backwards is right four times a day&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5612" LastEditorUserId="5612" LastEditDate="2015-09-04T22:51:54.487" LastActivityDate="2015-09-04T22:51:54.487" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21949" PostTypeId="2" ParentId="21947" CreationDate="2015-09-04T22:28:14.287" Score="2" Body="&lt;p&gt;My first guess is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; INTUITION&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A word used in anticipation,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; hope&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm in a simple machine,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; mach&lt;strong&gt;IN&lt;/strong&gt;e&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;in the witty,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; wi&lt;strong&gt;T&lt;/strong&gt;ty &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and in boisterous celebration.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; boistero&lt;strong&gt;U&lt;/strong&gt;s celebrat&lt;strong&gt;I&lt;/strong&gt;on &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm in things separated,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;T&lt;/strong&gt;hings &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;in the intense,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;I&lt;/strong&gt;ntense&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and in this clue's valuation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; valuati&lt;strong&gt;ON&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15732" LastEditorUserId="15732" LastEditDate="2015-09-05T11:33:01.317" LastActivityDate="2015-09-05T11:33:01.317" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21950" PostTypeId="1" CreationDate="2015-09-04T23:59:15.120" Score="9" ViewCount="642" Body="&lt;p&gt;&lt;strong&gt;Backstory:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the forest hidden among the trees lies a village full of animals. They live by their animal gods' words. Once every full moon they receive a message from the gods. Occasionally it is a warning hidden among words of mystery. This is one of those cases. The leaders of this animal pack bring these messages to the wise riddlers of the &lt;em&gt;Stack Exchange Cave.&lt;/em&gt; This message reads these cleverly spun words:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The answer lies within the group&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Slowly rattling its brain for a plan&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;It is a watching, stalking, creeping rat&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Is it you? Is it me?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The trick is it treats you like kin&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Is it you? Me, you, or him? &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The answer lies within, for the right answer is in the group&lt;/p&gt;&#xA;  &#xA;  &lt;hr&gt;&#xA;  &#xA;  &lt;p&gt;What is this dire warning? Can you save the pack?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt;- &lt;em&gt;It doesn't have to be an animal. You could imagine the &lt;strong&gt;pack&lt;/strong&gt; was a city, or anything else.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint2&lt;/strong&gt;-&lt;em&gt;Tags for this puzzle could be&lt;/em&gt; &lt;strong&gt;cryptogram&lt;/strong&gt; &lt;em&gt;and&lt;/em&gt; &lt;strong&gt;poem&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="15675" LastEditorUserId="15675" LastEditDate="2015-09-13T13:58:55.570" LastActivityDate="2015-09-13T13:58:55.570" Title="One Pack, One Warning" Tags="|word|enigmatic-puzzle|" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21951" PostTypeId="1" AcceptedAnswerId="21969" CreationDate="2015-09-05T00:12:00.927" Score="8" ViewCount="1111" Body="&lt;p&gt;A magician by the name of Jason carries a deck of cards with him everywhere he goes. He has a game where you can say any phrase and a specific card will correspond with that phrase. The goal of the game is to find out what rule is being used to choose the card that matches the phrase.&lt;/p&gt;&#xA;&lt;p&gt;The deck is in the order shown, with the Ace of Spades being the first card and the Ace of Hearts being the last card:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pn95R.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pn95R.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&#xA;&lt;em&gt;image courtesy of  &lt;a href=&quot;http://sites.google.com/site/erfarmer/premo&quot; rel=&quot;nofollow noreferrer&quot;&gt;sites.google.com/site/erfarmer/premo&lt;/a&gt; &lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;Here are some example phrases with the correct card in bold, and that card's number in the deck in bold parenthesis.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;This phrase will be the eight of spades. &lt;strong&gt;Eight of Spades (8)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Bond. James Bond. &lt;strong&gt;Eight of Hearts (45)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Houston, we have a problem. &lt;strong&gt;Eight of Hearts (45)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Doughnut. &lt;strong&gt;Nine of Hearts (44)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Donut. &lt;strong&gt;Eight of Diamonds (21)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;United States. &lt;strong&gt;Four of Diamonds (17)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Canada. &lt;strong&gt;Seven of Clubs (33)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Jason, can you just tell me how you are doing this? &lt;strong&gt;Two of Hearts (51)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;I won't tell anybody! &lt;strong&gt;Seven of Clubs (33)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Yes. &lt;strong&gt;Queen of Clubs (28)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;No. &lt;strong&gt;Three of Diamonds (16)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;To clarify a bit, the whole magician story of Jason isn't relevant to figuring out how the number is calculated, I just wanted to give it a story so I could make sequels of different card-styled puzzles if this is popular. In addition, I used a Lua script to produce these results so I will provide an obfuscated version of that Lua script so you may try your own phrases and verify the ones that are here. You can try to de-obfuscate the script to find the solution, but I think it would be more difficult than just figuring it out.&lt;/p&gt;&#xA;&lt;p&gt;Capitalization, punctuation, and anything other than a letter does not matter for finding the answer and is stripped. The case of the letter does not matter.&lt;/p&gt;&#xA;&lt;p&gt;If you'd like to verify and test your own phrases, the script can be found here (Lua 5.2) &lt;a href=&quot;http://pastebin.com/4BXrH9jw&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://pastebin.com/4BXrH9jw&lt;/a&gt; &lt;strong&gt;32-bit&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://pastebin.com/3e5NKm7L&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://pastebin.com/3e5NKm7L&lt;/a&gt; &lt;strong&gt;64-bit&lt;/strong&gt; &lt;br&gt;&#xA;Usage is simple; feed your phrase as the first argument of the program and you will get a card out. Thanks to &lt;a href=&quot;https://puzzling.stackexchange.com/users/15757/moonbutt74&quot;&gt;moonbutt74&lt;/a&gt; for compiling the 64-bit verison, and to &lt;a href=&quot;https://puzzling.stackexchange.com/users/9718/2012rcampion&quot;&gt;2012rcampion&lt;/a&gt; for identifying where the issue was that was preventing this from running on some systems.&#xA;&lt;br&gt;&lt;br&gt;&lt;br&gt;&#xA;Hint 1:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; If you know the method, you can calculate short phrases mentally and medium phrases with the help of a pen and paper.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;br&gt;&lt;br&gt;&#xA;Hint 2:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Huey Lewis and the News might tell you that one of the steps is pretty hip.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15756" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-09-12T14:22:52.223" Title="The Magician's Card Game" Tags="|mathematics|calculation-puzzle|computer-puzzle|language|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="21952" PostTypeId="2" ParentId="21947" CreationDate="2015-09-05T05:03:51.433" Score="2" Body="&lt;p&gt;Are you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The word/letter &lt;strong&gt;I&lt;/strong&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A word used in anticipation,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;I&lt;/strong&gt; is a word, i as a letter is present in ant&lt;strong&gt;i&lt;/strong&gt;cipation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm in a simple machine,&#xA;in the witty,&#xA;and in boisterous celebration.&#xA;I'm in things separated,&#xA;in the intense,&#xA;and in this clue's valuation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; s&lt;strong&gt;I&lt;/strong&gt;mple mach&lt;strong&gt;I&lt;/strong&gt;ne, w&lt;strong&gt;I&lt;/strong&gt;tty, bo&lt;strong&gt;I&lt;/strong&gt;sterous celebrat&lt;strong&gt;I&lt;/strong&gt;on, th&lt;strong&gt;i&lt;/strong&gt;ngs seperated, &lt;strong&gt;i&lt;/strong&gt;ntense, valuat&lt;strong&gt;i&lt;/strong&gt;on&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the title: NeedAName? this will do&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; we can always refer to ourselves as 'I' which is a substitute for our name.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Seems too generic though.&lt;/p&gt;&#xA;" OwnerUserId="15544" LastActivityDate="2015-09-05T05:03:51.433" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21953" PostTypeId="2" ParentId="21947" CreationDate="2015-09-05T15:35:23.310" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;EVE!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not sure about all of these...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A word used in anticipation,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;On Chrismas Eve you anticipate Christmas Day&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm in a simple machine,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;LEVER&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;in the witty,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;CLEVER&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and in boisterous celebration.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;REVEL (thanks, dennisdeems and NeedAName)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm in things separated,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;SEVERED&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;in the intense,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;SEVERE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and in this clue's valuation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It's the SEVENTH clue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And of course...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;EVE is a name!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="11876" LastEditDate="2015-09-05T17:49:13.140" LastActivityDate="2015-09-05T17:49:13.140" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="21954" PostTypeId="2" ParentId="21947" CreationDate="2015-09-05T15:35:56.953" Score="5" Body="&lt;p&gt;My new guess (beaten to it by Deusovi) is :  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;EVE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A word used in anticipation,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;EVE relates to the time (typically night) before an event, eg Christmas Eve or the eve of battle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm in a simple machine,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;L&lt;strong&gt;EVE&lt;/strong&gt;R is a simple machine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;in the witty,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;CL&lt;strong&gt;EVE&lt;/strong&gt;R is synonymous with witty&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and in boisterous celebration.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;R&lt;strong&gt;EVE&lt;/strong&gt;LRY is defined as a wild celebration&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm in things separated,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;S&lt;strong&gt;EVE&lt;/strong&gt;R means to divide or separate e.g. a severed head&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;in the intense,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Two possible candidates here are S&lt;strong&gt;EVE&lt;/strong&gt;RE or F&lt;strong&gt;EVE&lt;/strong&gt;RISH though I like the first more for this line&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and in this clue's valuation.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;As Deusovi noted, this is the S&lt;strong&gt;EVE&lt;/strong&gt;NTH line in the poem&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;NeedAName? This'll do  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;EVE makes for a fine name, though I don't think I'd wear it well, what with having a Y chromosome and all.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="13970" LastEditDate="2015-09-05T15:53:17.157" LastActivityDate="2015-09-05T15:53:17.157" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21955" PostTypeId="1" CreationDate="2015-09-05T16:22:35.950" Score="6" ViewCount="7455" Body="&lt;p&gt;The superflip is the state on a rubik's cube that require the most amount of moves to be solved. This is how it looks like:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dfpP8.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dfpP8.png&quot; alt=&quot;The superflip&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;This state requires 20 moves (where half-turns counts as one move) to be solved.&lt;/p&gt;&#xA;&lt;p&gt;This got me wondering, what is the superflip on the 4x4x4 rubik's cube? I saw &lt;a href=&quot;https://www.youtube.com/watch?v=SuOb8ftATzc&quot; rel=&quot;noreferrer&quot;&gt;This&lt;/a&gt; video that showed how to do the superflip on bigger cubes such as the 4x4x4, but I'm pretty sure that it isn't the &amp;quot;real&amp;quot; superflip, because I could just move the right slice on the 4x4x4, and it requires one more move to be solved and therefore isn't the state that requires the most moves to solve.&lt;/p&gt;&#xA;&lt;p&gt;If you want to, you can include the superflips on 5x5x5 and up.&lt;/p&gt;&#xA;&lt;h1&gt;EDIT&lt;/h1&gt;&#xA;&lt;p&gt;With superflip I mean the state (or one of the states if there exists more than one) that takes the most amount of moves to be solved.&lt;/p&gt;&#xA;" OwnerUserId="16225" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-10-06T13:26:43.933" Title="What is the superflip on the 4x4x4 Rubik's cube?" Tags="|rubiks-cube|" AnswerCount="3" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21956" PostTypeId="2" ParentId="21955" CreationDate="2015-09-05T17:31:10.017" Score="3" Body="&lt;p&gt;The algorithm in the video you linked is correct - you perform the exact same algorithm as with a 3x3x3 cube, except the &quot;middle slice&quot; on each face is the middle &lt;em&gt;two&lt;/em&gt; layers instead of the middle &lt;em&gt;one&lt;/em&gt;. This flips all the edge pieces, which is the definition of superflip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The superflip is only one example of a &quot;maximum-move scramble&quot; on the 3x3x3 Rubik's Cube. It doesn't necessarily correspond to a maximum-move scramble state on any of the others.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-09-05T17:31:10.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21957" PostTypeId="2" ParentId="21955" CreationDate="2015-09-05T17:50:28.363" Score="10" Body="&lt;p&gt;The 4x4 doesn't have a corresponding superflip state, because the definition of the superflip is ambiguous, and no good definition leads to a known superflip position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you define the superflip state as &quot;all edges flipped,&quot; then it is possible to treat the edge pairs like 3x3 edges and the center four pieces as 3x3 centers. This will allow you to create a superflip orientation on the 3x3 &lt;em&gt;reduction&lt;/em&gt; of the 4x4. This is not, however, a state requiring the maximum moves to solve on the 4x4, because the centers remain solved. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you define the superflip state as a checkerboard going in from each edge (roughly speaking), then no even numbered cube has one. To see why, remember that the corners stay fixed on a regular superflip to preserve the cube's symmetry - this can't happen for even puzzles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine the red-blue edge. The first corner has red on top, then the next edge has blue, next edge red, and last corner is blue. The first corner has red atop it, and the second has blue, which means one was necessarily swapped with the corner on the opposite side of the cube. This makes the checkerboarding pattern not possible to complete in any sense without swapping corners. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, if you define the superflip as &quot;the position taking the longest number of moves to solve,&quot; it may be some variant of these positions, but it likely wouldn't be recognizable as a regular superflip. This position isn't known, though. The 3x3 superflip position was demonstrated to take the maximum number of moves when it was &lt;a href=&quot;http://cube20.org&quot; rel=&quot;nofollow&quot;&gt;proved the 3x3 can be solved in 20 moves&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No such analysis has been done on the 4x4, which would be required to find and prove such a state needs the maximum moves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If I recall, though I can't seem to find a reference for this, on the 4x4, the estimate for optimum move sequence is 33.)&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorDisplayName="user20" LastEditDate="2015-09-10T01:59:52.877" LastActivityDate="2015-09-10T01:59:52.877" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21958" PostTypeId="1" CreationDate="2015-09-05T20:14:49.810" Score="21" ViewCount="2973" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;You have a square 8x8 board, and tetrominoes of all possible&#xA;  shapes.  Each individual square on a tetromino is the same size as one square on the board (i.e. any given tetromino will cover exactly 4 board squares). You can put the tetrominoes on the board in any orientation (rotated, flipped), but they must be aligned with the squares on the board, and can't extend off the board or overlap another tetromino.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The task:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Place tetrominoes on the board in such a way that it is impossible to add another tetromino, using the fewest number of tetrominoes possible.  You will end up with many blank spaces on the board, of up to 3 squares in size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;What is the smallest number of tetrominoes that can be placed on the board, fulfilling the requirements above?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="15856" LastEditorUserId="15544" LastEditDate="2015-09-27T22:54:06.573" LastActivityDate="2020-12-27T03:54:24.863" Title="Occupy a field with tetrominos" Tags="|tiling|construction|" AnswerCount="9" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="21959" PostTypeId="2" ParentId="21955" CreationDate="2015-09-05T20:38:18.550" Score="1" Body="&lt;p&gt;The 20 move upper bound (face turn metric) on the 3x3x3 rubik's cube was only proven in 2010, by extensive computer search (many machine years, run in time donated on the Cloud.) &lt;a href=&quot;https://en.wikipedia.org/wiki/God%27s_algorithm&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/God%27s_algorithm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before then it was known that the superflip, along with a number of other highly symmetrical positions, took 20 moves. It had long been conjectured that the maximum moves would be required for a highly symmetrical position. But it was still required to analyse all the unsymmetrical positions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to prove that the superflip must be a local maximum. Given that it has the full 48-fold symmetry of the cube, all moves performed on the superflip must lead towards or away from the solved state. As the latter could only be true if it was totally impossible to solve the cube, the former must be true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quarter turn metric was only proven to require max 26 moves in 2014. Interestingly, the longest move sequence is not the superflip, but rather is exemplified by a position called superflip composed with fourspot, which has only 16-fold symmetry. &lt;a href=&quot;http://www.cube20.org/qtm/&quot; rel=&quot;nofollow&quot;&gt;http://www.cube20.org/qtm/&lt;/a&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the longest move sequence on the 4x4x4 rubik's cube by whatever metric you choose is almost certainly unknown, but highly symmetrical. And choice of metric is complicated, because you have to consider whether moves of single internal slices are allowed.&lt;/p&gt;&#xA;" OwnerUserId="4768" LastActivityDate="2015-09-05T20:38:18.550" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21960" PostTypeId="2" ParentId="21958" CreationDate="2015-09-05T22:14:44.730" Score="7" Body="&lt;p&gt;Here is a solution with 9 tetrominos:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/zgA0E.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I made some calculations for an upper limit to see if bruteforcing 8 tetrominos would be possible, and it doesn't look good:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;64 starting positions for the first tetromino&lt;/li&gt;&#xA;&lt;li&gt;60 for the second&lt;/li&gt;&#xA;&lt;li&gt;56 for the third&lt;/li&gt;&#xA;&lt;li&gt;...&lt;/li&gt;&#xA;&lt;li&gt;per starting position 2*3*3 possibilities&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;End result: 3.7e23&lt;/p&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="30887" LastEditDate="2017-03-08T12:04:46.237" LastActivityDate="2017-03-08T12:04:46.237" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21962" PostTypeId="1" CreationDate="2015-09-05T23:57:57.520" Score="4" ViewCount="453" Body="&lt;p&gt;J-001&#xA;10111111&#xA;11100000&lt;/p&gt;&#xA;&#xA;&lt;p&gt;J-002&#xA;01001111&#xA;11110011&lt;/p&gt;&#xA;&#xA;&lt;p&gt;J-003&#xA;01110011&#xA;11111100&lt;/p&gt;&#xA;&#xA;&lt;p&gt;J-004&#xA;10000011&#xA;11101111&lt;/p&gt;&#xA;&#xA;&lt;p&gt;J-005&#xA;01010101&#xA;11100110&lt;/p&gt;&#xA;&#xA;&lt;p&gt;J-006&#xA;10010101&#xA;11111001&lt;/p&gt;&#xA;&#xA;&lt;p&gt;J-007&#xA;01101001&#xA;11101001&lt;/p&gt;&#xA;&#xA;&lt;p&gt;J-008&#xA;10101001&#xA;11110110&lt;/p&gt;&#xA;&#xA;&lt;p&gt;J-009&#xA;01011110&#xA;11101101&lt;/p&gt;&#xA;&#xA;&lt;p&gt;J-010&#xA;???????&#xA;???????&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;Also what would these series be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;J-100&#xA;???????&#xA;???????&lt;/p&gt;&#xA;&#xA;&lt;p&gt;J-999&#xA;???????&#xA;???????&lt;/p&gt;&#xA;" OwnerUserId="16230" LastActivityDate="2017-04-22T16:07:52.113" Title="What comes next in this binary sequence?" Tags="|number-sequence|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21963" PostTypeId="2" ParentId="21958" CreationDate="2015-09-06T00:35:05.717" Score="5" Body="&lt;p&gt;There are many solutions with:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 9 tetronimos.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Here is another one:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;! &lt;a href=&quot;https://i.stack.imgur.com/BDXtj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BDXtj.png&quot; alt=&quot;&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;http://www.jinchess.com/chessboard/?p=-P--p-P--PPpppPP-P--P-P-pppPP-p--p--Pppp--Pp--P-PPPppPP----p--P-&amp;amp;ps=mark-flat&quot; rel=&quot;nofollow noreferrer&quot;&gt;jinchess.com&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; (Thanks Sleafar for showing me how to post a solution without my poor drawing skills!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Here is a (tentative) proof that you can have no less than the aforementioned number of tetrominos:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; We can prove it with any 9-tetromino solution, but it is clearest with Sleafar's answer. All spaces are in sets of 3 squares, except for one of them, which is a 1-square space. To reduce the number of tetrominos further, 4 more squares need to become spaces, i.e. 10 trinominos plus 1 domino. Hereon we will consider the placement of these trinomino &amp;quot;spaces&amp;quot;.&lt;br /&gt;&#xA; Now, tile an 8x8 board with trinominos such that no two trinominos are touching (think Blokus, only corners can touch). It is easy to see that a maximum of 11 trinominos can be placed. Let's call the set of such trinomino solutions &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt;. Using the pigeonhole principle, it is clear that there are no solutions in this set where the majority of interstitials are 4 or more squares.&lt;br /&gt;&#xA; To get to the set of 10 trinominos plus 1 domino, we only need to move one of the trinomino squares from set &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt;. However, no matter which square we pick, the majority of other interstitials are still not going to 4 or more squares. 9 tetrominos is therefore the best solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15960" LastEditorUserId="11110" LastEditDate="2020-12-27T03:54:24.863" LastActivityDate="2020-12-27T03:54:24.863" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="21964" PostTypeId="1" CreationDate="2015-09-06T06:13:14.913" Score="2" ViewCount="2938" Body="&lt;p&gt;Suppose you have a triangle with a place to put a number on each of the three vertices. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What unique numbers can go in each of the three vertices such that the numbers on each edge add to a perfect square?&lt;/p&gt;&#xA;" OwnerUserId="16231" LastEditorUserId="7308" LastEditDate="2015-09-11T01:14:35.093" LastActivityDate="2016-06-27T06:19:02.180" Title="Add a number to each vertex of a triangle such that each edge adds to a perfect square" Tags="|mathematics|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21966" PostTypeId="2" ParentId="9224" CreationDate="2015-09-06T08:13:05.367" Score="1" Body="&lt;p&gt;Is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A needle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because-&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sharp tip, if scratched becomes blunt and useless,&#xA;  Thin too&lt;br&gt;&#xA;  It can kill&lt;br&gt;&#xA;  It is helpful  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16165" LastEditorUserId="16165" LastEditDate="2015-09-06T20:28:07.240" LastActivityDate="2015-09-06T20:28:07.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21967" PostTypeId="2" ParentId="21962" CreationDate="2015-09-06T12:17:46.777" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LmNZ1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LmNZ1.jpg&quot; alt=&quot;ELFlh.jpg&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;From the barcode image as per OP's&#xA;suggestion of read order [R-&amp;gt;L&amp;amp;T-&amp;gt;B]&#xA;and by 2012rcampion's suggestion of&#xA;BLACK=1 WHITE=0&#xA;&#xA;j-001 11111101 00000111&#xA;&#xA;j-002 11110010 11001111&#xA;&#xA;j-003 11001110 00111111&#xA;&#xA;j-004 11000001 11110111&#xA;&#xA;j-005 10101010 01100111&#xA;&#xA;j-006 10101001 10011111&#xA;&#xA;j-007 10010110 10010111&#xA;&#xA;j-008 10010101 01101111&#xA;&#xA;j-009 01111010 10110111&#xA;&#xA;j-010 01100001 00101111&#xA;&#xA;j-011 01011101 11011111&#xA;&#xA;j-012 01000110 01000111&#xA;&#xA;j-013 00110001 01010111&#xA;&#xA;j-014 00100110 11111111&#xA;&#xA;j-015 00011010 00001111&#xA;&#xA;j-016 00001101 10100111&#xA;&#xA;j-017 11111001 01111110&#xA;&#xA;j-018 11101110 11010110&#xA;&#xA;j-019 11010010 00100110&#xA;&#xA;j-020 11000101 10001110&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="15757" LastEditorUserId="15757" LastEditDate="2015-09-08T00:17:48.343" LastActivityDate="2015-09-08T00:17:48.343" CommentCount="1" CommunityOwnedDate="2015-09-06T12:17:46.777" ContentLicense="CC BY-SA 3.0" />
  <row Id="21968" PostTypeId="1" CreationDate="2015-09-06T13:28:02.380" Score="1" ViewCount="1753" Body="&lt;p&gt;&lt;em&gt;Note:&lt;/em&gt; There are some rude words in this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's an implicit puzzle in posed in this &lt;a href=&quot;https://www.youtube.com/watch?v=-gwXJsWHupg&quot; rel=&quot;nofollow&quot;&gt;classic Monty Python sketch&lt;/a&gt; (you can also ready the &lt;a href=&quot;http://www.montypython.net/scripts/wood.php&quot; rel=&quot;nofollow&quot;&gt;script&lt;/a&gt; online). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Various words are described as being 'woody' or 'tinny', and one is described as being 'PVC'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some &lt;em&gt;woody&lt;/em&gt; words:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Gorn (possibly Gone)&lt;/li&gt;&#xA;&lt;li&gt;Sausage&lt;/li&gt;&#xA;&lt;li&gt;Seemly&lt;/li&gt;&#xA;&lt;li&gt;Prodding&lt;/li&gt;&#xA;&lt;li&gt;Vacuum&lt;/li&gt;&#xA;&lt;li&gt;Bound&lt;/li&gt;&#xA;&lt;li&gt;Vole&lt;/li&gt;&#xA;&lt;li&gt;Caribou&lt;/li&gt;&#xA;&lt;li&gt;Intercourse&lt;/li&gt;&#xA;&lt;li&gt;Pert&lt;/li&gt;&#xA;&lt;li&gt;Thighs&lt;/li&gt;&#xA;&lt;li&gt;Botty&lt;/li&gt;&#xA;&lt;li&gt;Erogenous Zone&lt;/li&gt;&#xA;&lt;li&gt;Ocelot&lt;/li&gt;&#xA;&lt;li&gt;Wasp&lt;/li&gt;&#xA;&lt;li&gt;Yowling&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Here are some &lt;em&gt;tinny&lt;/em&gt; words:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Litter bin&lt;/li&gt;&#xA;&lt;li&gt;Newspaper&lt;/li&gt;&#xA;&lt;li&gt;Antelope&lt;/li&gt;&#xA;&lt;li&gt;Recidivist&lt;/li&gt;&#xA;&lt;li&gt;Tit&lt;/li&gt;&#xA;&lt;li&gt;Simpkins&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This one is &lt;em&gt;PVC&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Leap&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So here's the question, what makes a word woody, and what makes it tinny? or even PVC? I'd prefer some kind of rule which can be used to categorise words into one of these categories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's some words to test your rule on, tell me which category they belong to and why?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Guildford&lt;/li&gt;&#xA;&lt;li&gt;Godalming&lt;/li&gt;&#xA;&lt;li&gt;Artichoke&lt;/li&gt;&#xA;&lt;li&gt;Trigonometry&lt;/li&gt;&#xA;&lt;li&gt;Brigand&lt;/li&gt;&#xA;&lt;li&gt;Fox&lt;/li&gt;&#xA;&lt;li&gt;Bookworm&lt;/li&gt;&#xA;&lt;li&gt;Gaggle&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2673" LastEditorUserId="14135" LastEditDate="2015-09-06T19:06:58.630" LastActivityDate="2015-09-06T19:06:58.630" Title="Woody words, tinny words" Tags="|pattern|wordplay|open-ended|" AnswerCount="0" CommentCount="5" ClosedDate="2015-09-07T07:43:41.420" ContentLicense="CC BY-SA 3.0" />
  <row Id="21969" PostTypeId="2" ParentId="21951" CreationDate="2015-09-06T14:14:22.857" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;ol&gt;&lt;li&gt;Set the value of each letter to its position in the alphabet, i.e. $a=1, b=2, \ldots, z=26$&lt;/li&gt;&#xA;     &lt;li&gt;Count the number of vowels (a,e,i,o,u) in the phrase. Let's call this $V_{\text{num}}$&lt;/li&gt;&#xA;     &lt;li&gt;Count the total number of letters in the phrase. Let's call this $L_{\text{num}}$&lt;/li&gt;&#xA;     &lt;li&gt;Let the value of the first letter in the sentence be $L$.&lt;/li&gt;&#xA;     &lt;li&gt;Use this formula:&lt;br&gt;&#xA;&lt;br&gt;&#xA;         $$\text{Card} = (V_{\text{num}}\times V_{\text{num}}\times L+L_{\text{num}}) \bmod 52$$&lt;/li&gt;&#xA; &lt;/ol&gt;&lt;br&gt;&#xA; &lt;hr&gt;&#xA;&#xA;  PS: Actually more accurate is:&lt;br&gt;&#xA;&lt;br&gt;&#xA;  $$\text{Card} = (([V_{\text{num}}\times V_{\text{num}}\times L+L_{\text{num}}]-1) \bmod 52) + 1$$&lt;br&gt;&#xA;&lt;br&gt;&#xA;  but I used the first expression just for clarity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15856" LastEditorUserId="13494" LastEditDate="2015-09-12T14:22:52.223" LastActivityDate="2015-09-12T14:22:52.223" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="21970" PostTypeId="2" ParentId="21958" CreationDate="2015-09-06T15:09:50.677" Score="17" Body="&lt;p&gt;Here is yet another solution with 9 pieces. This one is nice and symmetrical.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/J16ZU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/J16ZU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have been trying to think of a way to show that 8 will not work by arguing in terms of the number of edge squares that need to be covered. However, I have not got very far with this. Here is a diagram I have been pondering.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/bXWao.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bXWao.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Update&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I haven't really given this a lot more thought. Call a piece an &quot;edge piece&quot; if it covers at least one edge square (by which I mean the 28 squares that have orthogonal borders with fewer than 4 other squares). It is easy enough to show that a solution must have at least 4 edge pieces. It is also easy to show that if a solution has exactly 4 edge pieces then it needs another 5 pieces, so it follows that any solution with at most 8 pieces needs at least 5 edge pieces.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Update 2&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Here is another approach that may be more promising. Consider the set of all square-edges, that is to say: each of the 64 squares in the board has 4 square-edges but most are shared between two adjacent squares (some are unique to a single square, that is to say those at the edges of the board). There are 9 * 8 * 2 = 144 square-edges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each tetromino accounts for at most 13 square-edges, considering both its external and internal square-edges. So if you have 8 of them you've accounted for 8 * 13 = 104 square-edges, leaving 40. Of these 40 some of them are the internal square-edges of gaps, and some of them are square-edges at the board edge where there is a gap. Per the second diagram above, at most 21 edge squares may belong to gaps. We have to take into account that some might be edge squares at the corner, so an upper bound for square-edges at the board edge where there is a gap is 25.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leaves 15 for internal square-edges of gaps. It's fairly easy to show that you can have at most 21 of these. If we could show that at least 7 external square-edges of tetrominoes are shared between two tetrominoes, we'd have an argument. So the roadblock here is showing that in any solution, there are at least 7 spots where two tetromino squares border each other orthogonally.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Update 3: There is no 8-piece solution in the case of a torus&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This problem came back into my mind this evening and I came up with a proof that 8 pieces are not sufficient on a torus. By &quot;on a torus&quot; I mean replacing the chessboard of the original problem by an 8*8 tiled torus (&quot;donut-shaped&quot; surface); think of the original chessboard but the squares on the top row border those in the bottom row, and the squares in the leftmost column border those in the rightmost column. Tetronimoes may be placed so that they straddle these newly-invented borders.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider that a tetromino has either 8 border edges (if it is the 2x2 square tetromino) or 10 border edges (if it is any other tetromino). In contrast to this, a gap (being at most 3 squares in size) has no internal vertices and has a fixed number of border edges (dependent on the size of the gap). A gap of size 3 has 8 border edges, a gap of size 2 has 6 border edges, and a gap of size 1 has 4 border edges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we have a solution with 8 tetrominoes. There are 64 - 8*4 = 32 squares not covered by any tetromino. These must be arranged into gaps of size at most 3. The way of doing this that leads to the least total border edges of gaps is to maximise the number of gaps of size 3. This means 10 gaps of size 3 and 1 gap of size 2. This gives rise to 10*8 + 1*6 = 86 border edges of gaps (and none are counted twice, because if they were then the two gaps would not be distinct). However, each border edge has to also be the border edge of a tetromino, and each tetromino border edge can be paired with only one gap border edge. There are only 8 tetrominoes by hypothesis, and as remarked previously each tetromino can have at most 10 border edges, so this gives us at most 8*10 = 80 tetromino border edges, which is less than the 86 we require.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should justify the claim that the way of decomposing the 32 squares (in fact any number of squares) into gaps (of size at most 3) that gives rise to the fewest gap border edges is to make as many gaps of size 3 as possible. First observe that to achieve the fewest possible gap border edges, we should not have both a gap of size 2 and a gap of size 1, for if we did, then these two gaps would contribute 6 + 4 = 10 gap border edges between them whereas we could combine them together into a gap of size 3 and have only 8 border edges. Secondly, observe that we should not have 3 gaps of size 2, as this would give us 18 gap border edges against the 16 we would have with 2 gaps of size 3. (Of course, we should not have 3 gaps of size 1 either.) Lastly, observe that if we have 2 gaps of size 2 then we are no worse off by switching to one gap of size 3 and one gap of size 2 (2*6 = 12 = 8 + 4).&lt;/p&gt;&#xA;" OwnerUserId="12623" LastEditorUserId="43327" LastEditDate="2017-12-14T03:56:01.780" LastActivityDate="2017-12-14T03:56:01.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21972" PostTypeId="2" ParentId="21951" CreationDate="2015-09-06T19:31:58.167" Score="5" Body="&lt;p&gt;The Lua code snippets in the question simply import a string consisting of a long sequence of escapes, and then run the imported object.  Treating the string as a binary file, we get something like:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0000000: 1b4c 7561 5100 0104 0404 0800 0000 0000 .LuaQ...........&#xA;0000010: 0000 0000 0000 0000 0000 0213 0601 0000 ................&#xA;...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The first 12 bytes are the header to compiled Lua bytecode.  That means that our first step is to:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Step 1: Write a Lua Decompiler&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I wasn't really satisfied with the existing tools for manipulating Lua bytecode, plus I love to learn new languages, so I wrote my own parser in Mathematica.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This took quite a while, since the documentation is not that great: I had only the &lt;a href=&quot;http://www.lua.org/source/5.2/lopcodes.h.html#OP_MOVE&quot;&gt;source code&lt;/a&gt; and &lt;a href=&quot;http://luaforge.net/docman/83/98/ANoFrillsIntroToLua51VMInstructions.pdf&quot;&gt;this document&lt;/a&gt; to guide me.  However, it was made easier by the fact that Lua's VM operates on a very high level, has very few instructions, and uses a stack-based register scheme.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although my code handles both version 5.1 and 5.2, the way closures are handled in 5.1 is slightly easier to understand, so that's the version that I chose to analyze.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Step 2: Analyze the Bytecode&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Once the parser was written, this part actually went pretty fast.  &lt;code&gt;ChipperNickel&lt;/code&gt; hinted that large sections of the code had no effect on the output.  Even before parsing the bytecode you can tell that &lt;code&gt;math.random&lt;/code&gt; is used (global function names are stored as strings in the constants table) even though we know that the output is deterministic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We start looking at the end of the instructions section of the top-level function:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;249 GETGLOBAL 11 50                R11 = global(&quot;print&quot;)&#xA;250 MOVE      12 9 0               R12 = R9&#xA;251 CALL      11 2 1               R11(R12)&#xA;252 MOVE      11 4 0               R11 = R4&#xA;253 GETGLOBAL 12 51                R12 = global(&quot;io&quot;)&#xA;254 GETTABLE  12 12 308            R12 = R12[&quot;write&quot;]&#xA;255 GETTABLE  13 10 9              R13 = R10[R9]&#xA;256 LOADK     14 53                R14 = &quot; (&quot;&#xA;257 MOVE      15 9 0               R15 = R9&#xA;258 LOADK     16 54                R16 = &quot;)&quot;&#xA;259 CONCAT    13 13 16             R13 = R13 .. R14 .. R15 .. R16&#xA;260 CALL      11 3 1               R11(R12, R13)&#xA;261 RETURN    0 1 0                return &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Eliminating the intermediate assignments, we end up with:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;print(R9)&#xA;R4(io.write, R10[R9] .. &quot; (&quot; .. R9 .. &quot;)&quot;)&#xA;return &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This matches the output format of the program perfectly (right down to the missing &lt;code&gt;\n&lt;/code&gt;):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1&#xA;Ace of Spades (1)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Assuming that the function in &lt;code&gt;R4&lt;/code&gt; actually calls &lt;code&gt;io.write&lt;/code&gt; and that this is not a red herring, we now know that &lt;code&gt;R9&lt;/code&gt; is the number of the card and &lt;code&gt;R10&lt;/code&gt; is a table containing a list of card names as strings.  We just need to find out how &lt;code&gt;R9&lt;/code&gt; is calculated and what order the cards are placed in &lt;code&gt;R10&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;However, I'll first take a quick detour to inspect the function in &lt;code&gt;R4&lt;/code&gt;, just to make sure we have everything right.  &lt;code&gt;R4&lt;/code&gt; is last set here:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;61  CLOSURE   4 3                  R4 = closure(KPROTO3, R4, ...)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Thus we're looking for the third function prototype.  The header of the third function lists two input parameters (matching what we've seen) and no upvalues &#xA;(which is good because it won't depend on other registers manipulated during program execution).  It also occupied lines 51-58 of the source code (starting and ending line numbers are not stripped with the rest of the debugging information) so we know it's relatively short.  Here is the entire instruction listing for the function:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0  NEWTABLE  2 0 0                R2 = {}&#xA;1  LOADK     3 0                  R3 = 1.&#xA;2  GETGLOBAL 4 1                  R4 = global(&quot;math&quot;)&#xA;3  GETTABLE  4 4 258              R4 = R4[&quot;random&quot;]&#xA;4  LOADK     5 3                  R5 = 20.&#xA;5  LOADK     6 4                  R6 = 50.&#xA;6  CALL      4 3 2                R4 = R4(R5, R6)&#xA;7  LOADK     5 0                  R5 = 1.&#xA;8  FORPREP   3 8                  R3 -= R5; PC += 8&#xA;9  LEN       7 2 0                R7 = len(R2)&#xA;10 ADD       7 7 256              R7 = R7 + 1.&#xA;11 GETGLOBAL 8 1                  R8 = global(&quot;math&quot;)&#xA;12 GETTABLE  8 8 258              R8 = R8[&quot;random&quot;]&#xA;13 LOADK     9 5                  R9 = 10.&#xA;14 LOADK     10 3                 R10 = 20.&#xA;15 CALL      8 3 2                R8 = R8(R9, R10)&#xA;16 SETTABLE  2 7 8                R2[R7] = R8&#xA;17 FORLOOP   3 9                  R3 += R5; if(R3 &amp;lt;?= R4) { R6 = R3; PC += -9 }&#xA;18 LEN       3 2 0                R3 = len(R2)&#xA;19 ADD       3 3 256              R3 = R3 + 1.&#xA;20 SETTABLE  2 3 0                R2[R3] = R0&#xA;21 LEN       3 2 0                R3 = len(R2)&#xA;22 GETTABLE  3 2 3                R3 = R2[R3]&#xA;23 MOVE      4 1 0                R4 = R1&#xA;24 TAILCALL  3 2 0                return R3(R4)&#xA;25 RETURN    3 0 0                return R3, ...&#xA;26 RETURN    0 1 0                return &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note that a function call treats the stack as empty except for the input arguments, so we start off with &lt;code&gt;R0&lt;/code&gt; as &lt;code&gt;io.write&lt;/code&gt; and &lt;code&gt;R1&lt;/code&gt; as the output string.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First I'll note that &lt;code&gt;R2&lt;/code&gt; contains a list and is not reassigned throughout the function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lines 18-20 append &lt;code&gt;io.write&lt;/code&gt; to the end of the list.  Lines 21 and 22 then extract that element back out of the list.  Finally, line 24 calls &lt;code&gt;io.write(R1)&lt;/code&gt;, which is what we suspected the output was in the first place.  This means we can disregard the rest of the function.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now we are looking for where &lt;code&gt;R10&lt;/code&gt; is populated and &lt;code&gt;R9&lt;/code&gt; is calculated.  Going back a little in the top-level function we see:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;120 NEWTABLE  10 0 0               R10 = {}&#xA;121 GETGLOBAL 11 26                R11 = global(&quot;table&quot;)&#xA;122 GETTABLE  11 11 283            R11 = R11[&quot;insert&quot;]&#xA;123 MOVE      12 10 0              R12 = R10&#xA;124 LOADK     13 28                R13 = &quot;Ace of Spades&quot;&#xA;125 CALL      11 3 1               R11(R12, R13)&#xA;126 LOADK     11 25                R11 = 2.&#xA;127 LOADK     12 29                R12 = 10.&#xA;128 LOADK     13 14                R13 = 1.&#xA;129 FORPREP   11 7                 R11 -= R13; PC += 7&#xA;130 GETGLOBAL 15 26                R15 = global(&quot;table&quot;)&#xA;131 GETTABLE  15 15 283            R15 = R15[&quot;insert&quot;]&#xA;132 MOVE      16 10 0              R16 = R10&#xA;133 MOVE      17 14 0              R17 = R14&#xA;134 LOADK     18 30                R18 = &quot; of Spades&quot;&#xA;135 CONCAT    17 17 18             R17 = R17 &amp;lt;&amp;gt; R18&#xA;136 CALL      15 3 1               R15(R16, R17)&#xA;137 FORLOOP   11 8                 R11 += R13; if(R11 &amp;lt;?= R12) { R14 = R11; PC += -8 }&#xA;138 GETGLOBAL 11 26                R11 = global(&quot;table&quot;)&#xA;139 GETTABLE  11 11 283            R11 = R11[&quot;insert&quot;]&#xA;140 MOVE      12 10 0              R12 = R10&#xA;141 LOADK     13 31                R13 = &quot;Jack of Spades&quot;&#xA;142 CALL      11 3 1               R11(R12, R13)&#xA;143 GETGLOBAL 11 26                R11 = global(&quot;table&quot;)&#xA;144 GETTABLE  11 11 283            R11 = R11[&quot;insert&quot;]&#xA;145 MOVE      12 10 0              R12 = R10&#xA;146 LOADK     13 32                R13 = &quot;Queen of Spades&quot;&#xA;147 CALL      11 3 1               R11(R12, R13)&#xA;148 GETGLOBAL 11 26                R11 = global(&quot;table&quot;)&#xA;149 GETTABLE  11 11 283            R11 = R11[&quot;insert&quot;]&#xA;150 MOVE      12 10 0              R12 = R10&#xA;151 LOADK     13 33                R13 = &quot;King of Spades&quot;&#xA;152 CALL      11 3 1               R11(R12, R13)&#xA;153 GETGLOBAL 11 26                R11 = global(&quot;table&quot;)&#xA;154 GETTABLE  11 11 283            R11 = R11[&quot;insert&quot;]&#xA;155 MOVE      12 10 0              R12 = R10&#xA;156 LOADK     13 34                R13 = &quot;Ace of Diamonds&quot;&#xA;157 CALL      11 3 1               R11(R12, R13)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Without doing a detailed analysis of this code I can see that we're creating an empty list in &lt;code&gt;R10&lt;/code&gt;, and then appending (with &lt;code&gt;table.insert&lt;/code&gt;) cards names in ascending order, with the ace first.  The suit order is Spades, Diamonds, Clubs, Hearts.  (Personally I would have made a doubly-nested loop, indexing into a list of suit names in the outer loop.  This has the advantage of tightening up the source code while further obfuscating the bytecode.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Looking back at the instruction just before the previous segment, we see an assignment to &lt;code&gt;R9&lt;/code&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;111 GETGLOBAL 9 15                 R9 = global(&quot;string&quot;)&#xA;112 GETTABLE  9 9 280              R9 = R9[&quot;len&quot;]&#xA;113 GETGLOBAL 10 12                R10 = global(&quot;input&quot;)&#xA;114 CALL      9 2 2                R9 = R9(R10)&#xA;115 ADD       8 8 9                R8 = R8 + R9&#xA;116 MOVE      9 4 0                R9 = R4&#xA;117 MOVE      10 3 0               R10 = R3&#xA;118 MOVE      11 8 0               R11 = R8&#xA;119 CALL      9 3 2                R9 = R9(R10, R11)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We're calling &lt;code&gt;R9&lt;/code&gt; as a function, but we can see from instruction 116 that it's just our old friend &lt;code&gt;KPROTO3&lt;/code&gt; as &lt;code&gt;R4&lt;/code&gt; again.  Thus we can simplify the listing to just:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;R9 = R3(R8 + string.len(input))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;R3&lt;/code&gt; was last set in instruction 60:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;60  CLOSURE   3 2                  R3 = closure(KPROTO2, R3, ...)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Thus we must dive into the functions again.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The header for function 2 states that it takes one argument and has no upvalues.  However, it occupies a whole 17 lines of source code.  Here are the instructions (remember the argument is in &lt;code&gt;R0&lt;/code&gt;):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0  ADD       0 0 256              R0 = R0 + 25.&#xA;1  ADD       0 0 257              R0 = R0 + 1.&#xA;2  SUB       0 0 258              R0 = R0 - 26.&#xA;3  GETGLOBAL 1 3                  R1 = global(&quot;math&quot;)&#xA;4  GETTABLE  1 1 260              R1 = R1[&quot;floor&quot;]&#xA;5  MOVE      2 0 0                R2 = R0&#xA;6  CALL      1 2 2                R1 = R1(R2)&#xA;7  MOVE      0 1 0                R0 = R1&#xA;8  NEWTABLE  1 0 0                R1 = {}&#xA;9  LOADK     2 1                  R2 = 1.&#xA;10 GETGLOBAL 3 3                  R3 = global(&quot;math&quot;)&#xA;11 GETTABLE  3 3 261              R3 = R3[&quot;random&quot;]&#xA;12 LOADK     4 6                  R4 = 250.&#xA;13 LOADK     5 7                  R5 = 999.&#xA;14 CALL      3 3 2                R3 = R3(R4, R5)&#xA;15 LOADK     4 1                  R4 = 1.&#xA;16 FORPREP   2 6                  R2 -= R4; PC += 6&#xA;17 GETGLOBAL 6 3                  R6 = global(&quot;math&quot;)&#xA;18 GETTABLE  6 6 261              R6 = R6[&quot;random&quot;]&#xA;19 LOADK     7 8                  R7 = 0.&#xA;20 LOADK     8 7                  R8 = 999.&#xA;21 CALL      6 3 2                R6 = R6(R7, R8)&#xA;22 SETTABLE  1 5 6                R1[R5] = R6&#xA;23 FORLOOP   2 7                  R2 += R4; if(R2 &amp;lt;?= R3) { R5 = R2; PC += -7 }&#xA;24 LEN       2 1 0                R2 = len(R1)&#xA;25 ADD       2 2 257              R2 = R2 + 1.&#xA;26 SETTABLE  1 2 0                R1[R2] = R0&#xA;27 LOADNIL   0 0 0                R0 = nil&#xA;28 LEN       2 1 0                R2 = len(R1)&#xA;29 GETTABLE  2 1 2                R2 = R1[R2]&#xA;30 MOD       2 2 265              R2 = R2 % 52.&#xA;31 EQ        0 2 264              if(R2 == 0.) PC++&#xA;32 JMP       0 3                  PC += 3&#xA;33 LOADK     2 9                  R2 = 52.&#xA;34 RETURN    2 2 0                return R2&#xA;35 JMP       0 4                  PC += 4&#xA;36 LEN       2 1 0                R2 = len(R1)&#xA;37 GETTABLE  2 1 2                R2 = R1[R2]&#xA;38 MOD       2 2 265              R2 = R2 % 52.&#xA;39 RETURN    2 2 0                return R2&#xA;40 RETURN    0 1 0                return &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It's blatantly obvious from the first three instructions that Lua does &lt;em&gt;not&lt;/em&gt; use an optimizing compiler.  We'll treat those as a noop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we go backwards, just as before.  Note that there are two return statements amid a bunch of jumps.  Parsing that code (instructions 28-40) nets us:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;R2 = R1[len(R1)] % 52&#xA;if R2 == 0 then&#xA;    return 52&#xA;else&#xA;    return R1[len(R1)] % 52&#xA;end&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This just returns the last element of the list &lt;code&gt;R1&lt;/code&gt; modulo &lt;code&gt;52&lt;/code&gt;, using the smallest &lt;em&gt;positive&lt;/em&gt; number as the representative of each equivalence class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, note that in instructions 24-26, &lt;code&gt;R0&lt;/code&gt; is appended to &lt;code&gt;R1&lt;/code&gt;.  This means that we can simplify the code even further:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;if R0 % 52 == 0 then&#xA;    return 52&#xA;else&#xA;    return R0 % 52&#xA;end&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Since the result of this function is used to index into a list of 52 elements (the list of card names), the effect of this function is to treat the list as cyclic.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Returning to the top-level function we just need to figure out what &lt;code&gt;R8&lt;/code&gt; is set to.  Tracing the code back I notice that at instruction 61, the entire stack consists only of function handles (excepting &lt;code&gt;R0&lt;/code&gt; which is not subsequently used):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1   CLOSURE   1 0                  R1 = closure(KPROTO0, R1, ...)&#xA;...&#xA;3   CLOSURE   2 1                  R2 = closure(KPROTO1, R2, ...)&#xA;...&#xA;60  CLOSURE   3 2                  R3 = closure(KPROTO2, R3, ...)&#xA;61  CLOSURE   4 3                  R4 = closure(KPROTO3, R4, ...)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Thus we can go forward from this point.  I'll analyze the following instructions in sections:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;62  GETGLOBAL 5 13                 R5 = global(&quot;arg&quot;)&#xA;63  GETTABLE  5 5 270              R5 = R5[1.]&#xA;64  SETGLOBAL 5 12                 global(&quot;input&quot;) = R5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here we just set &lt;code&gt;input = arg[1]&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;65  GETGLOBAL 5 15                 R5 = global(&quot;string&quot;)&#xA;66  GETTABLE  5 5 272              R5 = R5[&quot;lower&quot;]&#xA;67  GETGLOBAL 6 12                 R6 = global(&quot;input&quot;)&#xA;68  CALL      5 2 2                R5 = R5(R6)&#xA;69  SETGLOBAL 5 12                 global(&quot;input&quot;) = R5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here we set &lt;code&gt;input = string.lower(input)&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;70  GETGLOBAL 5 12                 R5 = global(&quot;input&quot;)&#xA;71  SELF      5 5 273              R6 = R5; R5 = R5[&quot;gsub&quot;]&#xA;72  LOADK     7 18                 R7 = &quot;%L&quot;&#xA;73  LOADK     8 19                 R8 = &quot;&quot;&#xA;74  CALL      5 4 2                R5 = R5(R6, R7, R8)&#xA;75  SETGLOBAL 5 12                 global(&quot;input&quot;) = R5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Equivalent to &lt;code&gt;input = input.gsub(input, &quot;%L&quot;, &quot;&quot;)&lt;/code&gt;.  This replaces all non-lowercase letters with the empty string.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;76  LOADK     5 20                 R5 = 0.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Initialize &lt;code&gt;R5&lt;/code&gt; to zero.  (It will be a counter later on in the code.)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;77  GETGLOBAL 6 15                 R6 = global(&quot;string&quot;)&#xA;78  GETTABLE  6 6 277              R6 = R6[&quot;sub&quot;]&#xA;79  GETGLOBAL 7 12                 R7 = global(&quot;input&quot;)&#xA;80  LOADK     8 14                 R8 = 1.&#xA;81  LOADK     9 14                 R9 = 1.&#xA;82  CALL      6 4 2                R6 = R6(R7, R8, R9)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here we do &lt;code&gt;R6 = string.sub(input, 1, 1)&lt;/code&gt; to store the first letter of the string in &lt;code&gt;R6&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;83  GETGLOBAL 7 15                 R7 = global(&quot;string&quot;)&#xA;84  GETTABLE  7 7 278              R7 = R7[&quot;byte&quot;]&#xA;85  MOVE      8 6 0                R8 = R6&#xA;86  CALL      7 2 2                R7 = R7(R8)&#xA;87  SUB       7 7 279              R7 = R7 - 96.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now we calculate &lt;code&gt;R7 = string.byte(R6) - 96&lt;/code&gt;.  This gets the number of the first letter of the string (starting at &lt;code&gt;1&lt;/code&gt; for &lt;code&gt;a&lt;/code&gt;).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;88  LOADK     8 20                 R8 = 0.&#xA;89  LOADK     9 14                 R9 = 1.&#xA;90  GETGLOBAL 10 15                R10 = global(&quot;string&quot;)&#xA;91  GETTABLE  10 10 280            R10 = R10[&quot;len&quot;]&#xA;92  GETGLOBAL 11 12                R11 = global(&quot;input&quot;)&#xA;93  CALL      10 2 2               R10 = R10(R11)&#xA;94  LOADK     11 14                R11 = 1.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here we set up for a for-loop.  These lines are equivalent to:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;R8  = 0&#xA;R9  = 1&#xA;R10 = string.len(input)&#xA;R11 = 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now on to the for-loop itself:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;95  FORPREP   9 12                 R9 -= R11; PC += 12&#xA;96  MOVE      13 4 0               R13 = R4&#xA;97  MOVE      14 2 0               R14 = R2&#xA;98  GETGLOBAL 15 15                R15 = global(&quot;string&quot;)&#xA;99  GETTABLE  15 15 277            R15 = R15[&quot;sub&quot;]&#xA;100 GETGLOBAL 16 12                R16 = global(&quot;input&quot;)&#xA;101 MOVE      17 12 0              R17 = R12&#xA;102 MOVE      18 12 0              R18 = R12&#xA;103 CALL      15 4 0               R15, ... = R15(R16, R17, R18)&#xA;104 CALL      13 0 2               R13 = R13(R14, ...)&#xA;105 TEST      13 0 0               if(R13) PC++&#xA;106 JMP       0 1                  PC += 1&#xA;107 ADD       5 5 270              R5 = R5 + 1.&#xA;108 FORLOOP   9 13                 R9 += R11; if(R9 &amp;lt;?= R10) { R12 = R9; PC += -13 }&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note that the for-loop keeps four state registers: the starting index, ending index, and step are stored in &lt;code&gt;R9&lt;/code&gt;-&lt;code&gt;R11&lt;/code&gt;, and the loop variable (I'll use &lt;code&gt;i&lt;/code&gt;) is stored in &lt;code&gt;R12&lt;/code&gt;.  We can simplify to:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;for i = 1, string.len(input), 1 do&#xA;    R13 = R4(R2, string.sub(input, i, i))&#xA;    if R13 do&#xA;        R5++&#xA;    end&#xA;end&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Remembering what &lt;code&gt;KPROTO3&lt;/code&gt; (&lt;code&gt;R4&lt;/code&gt;) does, we can simplify further to:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;for i = 1, string.len(input), 1 do&#xA;    if R2(string.sub(input, i, i)) do&#xA;        R5++&#xA;    end&#xA;end&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Thus we call &lt;code&gt;KPROTO1&lt;/code&gt; (&lt;code&gt;R2&lt;/code&gt;) on all of the letters in the string and increment &lt;code&gt;R5&lt;/code&gt; if the result is truthy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally a calculation:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;109 POW       5 5 281              R5 = R5 ** 2.&#xA;110 MUL       8 5 7                R8 = R5 * R7&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;With this, we have almost everything we need.  The final card number is, mod &lt;code&gt;52&lt;/code&gt;, the number of truthy letters in the string, squared, times the value of the first letter, plus the number of letters in the string.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The only step left is to look at function 1.  Unfortunately both functions 0 and 1 have a closure on &lt;code&gt;R0&lt;/code&gt;, a list.  Function 0 is very simple, just appending its argument minus ten to the list.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Function 1 is also rather straightforward:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0  LOADBOOL  1 0 0                R1 = False&#xA;1  LOADK     2 0                  R2 = 1.&#xA;2  GETUPVAL  3 0 0                R3 = U0&#xA;3  LEN       3 3 0                R3 = length(R3)&#xA;4  LOADK     4 0                  R4 = 1.&#xA;5  FORPREP   2 9                  R2 -= R4; PC += 9&#xA;6  GETGLOBAL 6 1                  R6 = global(&quot;string&quot;)&#xA;7  GETTABLE  6 6 258              R6 = R6[&quot;byte&quot;]&#xA;8  MOVE      7 0 0                R7 = R0&#xA;9  CALL      6 2 2                R6 = R6(R7)&#xA;10 GETUPVAL  7 0 0                R7 = U0&#xA;11 GETTABLE  7 7 5                R7 = R7[R5]&#xA;12 EQ        0 6 7                if(R6 == R7) PC++&#xA;13 JMP       0 1                  PC += 1&#xA;14 LOADBOOL  1 1 0                R1 = True&#xA;15 FORLOOP   2 10                 R2 += R4; if(R2 &amp;lt;?= R3) { R5 = R2; PC += -10 }&#xA;16 RETURN    1 2 0                return R1&#xA;17 RETURN    0 1 0                return &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It simply checks if the character code of the first character of its argument is present in &lt;code&gt;R0&lt;/code&gt; (referred to here as &lt;code&gt;U0&lt;/code&gt;, the first upvalue).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now we need to understand what values &lt;code&gt;R0&lt;/code&gt; is initialized with.  Instructions 5 through 59 are all sequences like:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;5 MOVE      3 1 0                R3 = R1&#xA;6 LOADK     4 0                  R4 = 107.&#xA;7 CALL      3 2 1                R3(R4)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This sequence appends &lt;code&gt;107&lt;/code&gt; (minus &lt;code&gt;10&lt;/code&gt;) to the list &lt;code&gt;R0&lt;/code&gt;.  In addition to a bunch of junk values (junk because they're larger than the max value of a byte), the following important values are appended:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;107 - 10 =  97 = 'a'&#xA;111 - 10 = 101 = 'e'&#xA;115 - 10 = 105 = 'i'&#xA;121 - 10 = 111 = 'o'&#xA;127 - 10 = 117 = 'u'&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Thus, &lt;code&gt;KPROTO2&lt;/code&gt; considers string truthy if they start with vowels.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Step 3: Put It All Together&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Although no question was explicitly stated, the implicit question is &lt;strong&gt;what procedure does Jason use to map phrases to cards?&lt;/strong&gt;  My answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jason counts the number of vowels in the phrase, then squares it.  Then he multiplies that number by the position in the alphabet (&lt;code&gt;a&lt;/code&gt; is 1, &lt;code&gt;z&lt;/code&gt; is 26) of the phrase's first letter.  Then he adds the number of letters in the phrase.&lt;br&gt;&#xA; &#xA; Then, starting with the deck in order, Jason moves the top card to the bottom a number of times equal to the number he calculated.  Then he chooses the bottom card.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="9718" LastEditDate="2015-09-06T19:37:47.100" LastActivityDate="2015-09-06T19:37:47.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21973" PostTypeId="2" ParentId="20825" CreationDate="2015-09-06T20:57:20.780" Score="1" Body="&lt;p&gt;I know there have been enough good answers already, but here is what I personally find to be the simplest:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No. We can easily list all possible ways to xor these three numbers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a&amp;oplus;b &amp;oplus; a&amp;oplus;c = b&amp;oplus;c&lt;br /&gt;&#xA;a&amp;oplus;b &amp;oplus; b&amp;oplus;c = a&amp;oplus;c&lt;br /&gt;&#xA;a&amp;oplus;c &amp;oplus; b&amp;oplus;c = a&amp;oplus;b&lt;br /&gt;&#xA;x&amp;oplus;x = 0 for any x&lt;br /&gt;&#xA;0&amp;oplus;x = x for any x&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that any expression involving these numbers and 0 combined with xor will always reduce to one of those numbers again (your three numbers and 0 are closed under xor). As a consequence, nothing else can be the value of any such expression, a, b and c are no exception.&lt;/p&gt;&#xA;" OwnerUserId="16238" LastActivityDate="2015-09-06T20:57:20.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21974" PostTypeId="2" ParentId="21962" CreationDate="2015-09-06T22:06:20.633" Score="3" Body="&lt;p&gt;I'm cheating a little and basing my answer off of &lt;a href=&quot;http://s28.postimg.org/5ajf8aobx/barcodes.jpg&quot; rel=&quot;nofollow&quot;&gt;the 120 barcodes&lt;/a&gt; the OP &lt;a href=&quot;http://chat.stackexchange.com/transcript/message/23925271#23925271&quot;&gt;provided in chat&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm reading the squares as bits from right-to-left, bottom-to-top, with black as &lt;code&gt;0&lt;/code&gt; and white as &lt;code&gt;1&lt;/code&gt;.  Here is the bit numbering I use:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;16 15 14 13 12 11 10  9&#xA; 8  7  6  5  4  3  2  1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note that this disagrees with the silkscreened labels on the barcode reader PCB.  I don't think that the hardware labels necessarily correspond to the software's processing order though.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The first thing I noticed was that bits &lt;code&gt;6&lt;/code&gt;-&lt;code&gt;11&lt;/code&gt; of the barcodes are just &lt;code&gt;((J - 1) &amp;gt;&amp;gt; 1) &amp;amp; 0x3F&lt;/code&gt; (where &lt;code&gt;J&lt;/code&gt; is the number printed on each barcode).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, &lt;code&gt;J-045&lt;/code&gt;, &lt;code&gt;J-093&lt;/code&gt;, and &lt;code&gt;J-101&lt;/code&gt; do not follow this pattern:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;code&gt;J-045&lt;/code&gt;: should be &lt;code&gt;0x16&lt;/code&gt; (22), actually &lt;code&gt;0x3E&lt;/code&gt; (62)&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;J-093&lt;/code&gt;: should be &lt;code&gt;0x2E&lt;/code&gt; (46), actually &lt;code&gt;0x3F&lt;/code&gt; (63)&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;J-101&lt;/code&gt;: should be &lt;code&gt;0x32&lt;/code&gt; (50), actually &lt;code&gt;0x3F&lt;/code&gt; (63)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There are two other patterns that I am fairly sure about though:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Bits &lt;code&gt;15&lt;/code&gt; and &lt;code&gt;16&lt;/code&gt; are always opposite of each other.&lt;/li&gt;&#xA;&lt;li&gt;Excluding bit &lt;code&gt;16&lt;/code&gt;, the number of &lt;code&gt;1&lt;/code&gt;s is always even.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Therefore I believe that bits &lt;code&gt;15&lt;/code&gt; and &lt;code&gt;16&lt;/code&gt; are some form of parity check.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remaining bits, &lt;code&gt;1&lt;/code&gt;-&lt;code&gt;14&lt;/code&gt;, seem to have an additional parity check.  For all the barcodes we have, I found the following relations:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;xor(    b2, b3, b4, b5, b6, b7             ) = b11&#xA;xor(b1,     b3, b4, b5,         b8, b9     ) = b12&#xA;xor(b1, b2,     b4, b5, b6,     b8,     b10) = b13&#xA;xor(b1, b2, b3,     b5,     b7,     b9, b10) = b14&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;My best guess is that the first fourteen bits contain a ten-bit value and some sort of four-bit checksum.  However, an exhaustive search of all 4- and 5-bit CRC generator polynomials has not yielded any results.&lt;/p&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="9718" LastEditDate="2015-09-09T04:29:58.443" LastActivityDate="2015-09-09T04:29:58.443" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21975" PostTypeId="1" AcceptedAnswerId="21995" CreationDate="2015-09-06T22:50:36.613" Score="20" ViewCount="561" Body="&lt;p&gt;It's the first full week of term at &lt;strong&gt;Farthingbottom School&lt;/strong&gt; and Portia, Felix, Primula and Randolph have arrived in the music room for their lesson. Professor N. Igma is nowhere to be seen, but he has left a message on the blackboard for them:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This term we will be performing four pieces of music at the end of term concert. Please practise them. Two methods will be required.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On the desk at the front of class is a single musical score:&lt;/p&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/796Nx.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/796Nx.gif&quot; alt=&quot;![enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;p&gt;The children get out their instruments and try to play the music, but it sounds terrible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Can you determine which four pieces of music the children should be practising?&lt;/p&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="14135" LastEditDate="2015-09-08T20:46:40.013" LastActivityDate="2015-09-08T20:46:40.013" Title="Preparing for the end of term performance" Tags="|cipher|wordplay|music|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21976" PostTypeId="2" ParentId="21958" CreationDate="2015-09-06T23:27:49.820" Score="4" Body="&lt;p&gt;Here is one with 8 tetrominos &lt;em&gt;and 1 square (black)&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/y3Mma.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/y3Mma.png&quot; alt=&quot;8.25 tetrominos&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Maybe it's possible to rearrange some tetrominos and get rid of that last square?&lt;/p&gt;&#xA;" OwnerUserId="13148" LastActivityDate="2015-09-06T23:27:49.820" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21977" PostTypeId="2" ParentId="21950" CreationDate="2015-09-07T00:16:10.280" Score="8" Body="&lt;p&gt;I'm new here and this is a far shot but...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it an evil&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; gopher&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer lies within the group&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'The group' contains the word 'gopher'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Slowly rattling its brain for a plan&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gophers shake their head around to sense with their whiskers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is a watching, stalking, creeping rat&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gophers are rodents, and so are rats&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The trick is it treats you like kin&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gophers look like small and harmless animals but they secretly plan to rule the world&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16240" LastActivityDate="2015-09-07T00:16:10.280" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21979" PostTypeId="2" ParentId="20825" CreationDate="2015-09-07T09:56:28.903" Score="3" Body="&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;XOR is true if exactly one of the inputs is true. Which means, if you negate both inputs, XOR stays the same. This means you can't distinguish &quot;a⊕b&quot; from &quot;(not a)⊕(not b)&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say you have an algorithm to calculate a,b,c for given a⊕b, a⊕c and b⊕c. You use it and get a', b', c'. If someone else comes and says &quot;no, the correct solution is actually (not a', not b', not c')&quot;, nobody can decide which is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't matter if a,b,c are single-bit booleans or longer. The argument applies to each bit.&lt;/p&gt;&#xA;" OwnerUserId="2783" LastActivityDate="2015-09-07T09:56:28.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21983" PostTypeId="2" ParentId="21958" CreationDate="2015-09-07T18:15:32.733" Score="3" Body="&lt;p&gt;I found a few interesting things. Purple and orange are used to highlight recurring structures, green is general, black is the flaw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8 pieces, no edges touching, no unfilled areas of size 1, near-success.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/DCi8E.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DCi8E.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; ______ &lt;a href=&quot;https://i.stack.imgur.com/RiIAF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RiIAF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;8 pieces, no edges touching, lots of unfilled areas of size 3, near-success.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/9Xa5P.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9Xa5P.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11100" LastEditorUserId="43327" LastEditDate="2017-12-14T03:56:06.007" LastActivityDate="2017-12-14T03:56:06.007" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21984" PostTypeId="2" ParentId="21975" CreationDate="2015-09-07T19:30:34.847" Score="3" Body="&lt;p&gt;Not an answer yet, but some thoughts along the way that might help someone else...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The only notes in there are A, B and C (albeit in different octaves). This makes it unlikely that they are simply tunes mashed together in some way (like taking every fourth note, starting from an offset of [0,1,2,3] to play a given tune). Not many tunes are confined to only 3 notes. &lt;p&gt;6/4 would be a fairly unusual time signature but the bars do all appear to add up to 6 crotchets (quarter notes for you 'Murricans). However the patterns of notes and rests are all over the place. I wonder if the note lengths have anything to do with the encoding? &lt;p&gt;Presumably the &quot;read between the lines&quot; comment, the &quot;Novas Requiem Cuts (5-11)&quot; and the &quot;Steve Chivet de Norman (5437)&quot; are significant in some way. The &quot;Novas&quot; reads like a cryptic crossword clue but that might just be that I spend too much time doing crosswords. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Given the hint in the comment that &quot;Novas Requiem Cuts&quot; is an anagram of &quot;count semiquavers&quot; I counted the semiquavers in the first line:&lt;br&gt;&#xA;&lt;code&gt;8A 1C 12C 1r 2r; 12B 4B-1B 1r 2r 4r; 12C 4r 8r; 14B-7B 1r 2r; 8A-3C 1B 8C 3B 1r; 8B 12r 4r; 12A-3A 1r 8r&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="2059" LastEditorUserId="2059" LastEditDate="2015-09-08T14:34:44.110" LastActivityDate="2015-09-08T14:34:44.110" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="21985" PostTypeId="1" AcceptedAnswerId="21986" CreationDate="2015-09-08T00:29:07.047" Score="5" ViewCount="655" Body="&lt;p&gt;We can place &lt;a href=&quot;https://en.wikipedia.org/wiki/Tetromino&quot;&gt;S and Z tetromino&lt;/a&gt; shaped islands on an 8x8 grid of water in such a way that no two islands are touching by their edges, because then they would not be two islands. Two islands may touch by their corners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do this in a way that minimizes the size of the largest pond in the grid.&lt;/p&gt;&#xA;" OwnerUserId="11100" LastActivityDate="2015-09-08T16:36:12.737" Title="Planning an Archipelago" Tags="|tiling|construction|" AnswerCount="2" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="21986" PostTypeId="2" ParentId="21985" CreationDate="2015-09-08T00:50:38.917" Score="8" Body="&lt;p&gt;The best I can do so far is 5.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/Upwuv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Upwuv.png&quot; alt=&quot;board&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit: Got 4!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/FaYPt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FaYPt.png&quot; alt=&quot;board2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="11876" LastEditDate="2015-09-08T00:58:01.040" LastActivityDate="2015-09-08T00:58:01.040" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="21988" PostTypeId="2" ParentId="21958" CreationDate="2015-09-08T00:52:01.337" Score="4" Body="&lt;p&gt;OPTIMALITY SOLUTION GUIDELINES&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As several people pointed out already, it is not hard to give an example with 9 tetraminos. I also believe this is the correct answer and even though proving it rigorously requires long and tedious case analysis, I will give some guidelines. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start by assuming the contrary, i.e. it is possible to use only 8 tetraminos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;STEP 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that if a rectangle 4x8 is covered by 4 tetraminos with no space for more, then its top row of length 8 both either contains at most 2 squares covered by tetraminos or does not contain covered squares at positions {1,2,7,8} or {3,4,5,6}. This implies that it is impossible both the bottom and the top half of the checkboard contain 4 full tetraminos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;STEP 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that it is impossible that both the top and the bottom half of the checkboard to contain one full tetramino and to share 6 more tetraminos. Prove that it is impossible that the top half of the checkboard to contain only 3 full tetraminos and a part from another one. Similarly for the bottom half. These two are fairly easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;STEP 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the remaining cases are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the top and the bottom half share 4 tetraminos and each of them contains entirely 2 more tetraminos;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the top and the bottom half share 3 tetraminos and each of them contains entirely 2 or 3 more tetraminos; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the top and the bottom half share 2 tetraminos and each of them contains entirely 3 more tetraminos;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of these 3 cases require lots of consideration, including number of squares covered by tetraminos by a half checkboard, squares covered on the middle rows (the 2 middle rows should contain at least 6 covered squares), etc. Also I think it is good to consider the four 4x4 corners of the checkboard, since each of them either contains one full tetramino or contains many partials. It may be possible to simplify the case analysis even more, but I'm not sure it is worth to work on it. I hope these guidelines help someone find a complete proof.&lt;/p&gt;&#xA;" OwnerUserId="14140" LastEditorUserId="14140" LastEditDate="2015-09-08T01:07:00.530" LastActivityDate="2015-09-08T01:07:00.530" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21989" PostTypeId="2" ParentId="20738" CreationDate="2015-09-08T03:25:04.193" Score="8" Body="&lt;p&gt;It's probably&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Irvine,_California&quot;&gt;Irvine, Califonia&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ohm's law:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I*R = V (&lt;strong&gt;&lt;em&gt;IRV&lt;/em&gt;&lt;/strong&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The part about the grapes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Vine - first letter = &lt;strong&gt;&lt;em&gt;INE&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Result:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Irv+ine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The clue you gave is probably just referencing:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; California is the golden state, as you thought. I couldn't find any obvious information about a gold mine or anything in the area.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The biggest clue:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You mentioned the interview was with Broadcom, which is located in Irvine.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5340" LastActivityDate="2015-09-08T03:25:04.193" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21990" PostTypeId="2" ParentId="21958" CreationDate="2015-09-08T03:39:48.600" Score="3" Body="&lt;p&gt;So I was thinking about this and there's the beginnings of a proof. Think about the perimeters of the shapes.  The tetronimos have perimeters of length 10 (or less in the case of the square). The outside has 32. There's at least two edges touching at each side, so that removes 8 edges at least, so there's a maximum of 96 free edges. Any additional edge that's touching will reduce that number by 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now count the spaces in between. We intuitively think that these will generally be trionimos, but they could be smaller. They will never share an edge with each other (obviously) just with the outsides and the tetronimos. There are 32 spaces. If they were all dominoes then you'd have a perimeter of 16 dominoes times 6 edges, or a total of at least 96. Since that equals the 96 from the first paragraph we know that if there is just one additional overlap (between the edges of two tetronimos or a tetronimo and the side of the board) then it cannot be done with just dominoes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best case for these is that there are 10 trionimos, and 1 domino, for a total of 86. So if we can demonstrate that there have to be more overlaps than the basic 2 per side, we'll get closer. You'd just need 5 additional overlaps to get you there. I think you can get 1 in each corner (at least in certain cases), which would mean you'd need just one more shared edge to be required to blow the solution up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the very least this should help people looking for a solution! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit to take the proof further:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarise where we got to:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We have a minimum perimeter of 86 from the empty spaces.&lt;/li&gt;&#xA;&lt;li&gt;We have a maximum perimeter available of 96 from the tetronimos and outer edges&lt;/li&gt;&#xA;&lt;li&gt;Any additional overlaps (apart from the 2 squares per side) reduces the maximum perimeter available by 2.  &lt;/li&gt;&#xA;&lt;li&gt;Meaning that if we can find at least 6 overlaps, we prove that 8 tetronimos is insufficient.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(Aside:  The numbers for an 8x7 square are 76 minimum and 88 maximum, which may explain why this is possible, where the 8x8 is (probably) not.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will now show that each corner adds at least one additional overlap, essentially getting us 4 of the 6 overlaps we need to prove our case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gjbOr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gjbOr.png&quot; alt=&quot;Case 1: 3 in the corner&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we have three blanks in the corner, as shown in yellow.  Then the squares marked X1 and Y1 must be part of two separate tetronimos (call them X and Y respectively).  If the squares marked X2 or Y2 were part of their respective tetronimos (X &amp;amp; Y), then we'd have an additional overlap (between tetronimo and the edge).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence suppose that X2 and Y2 are not part of their tetronimos.  Then X3 and Y3 are.  But then the square marked * must be part of either X or Y, and either way X and Y share an edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aLeyb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aLeyb.png&quot; alt=&quot;Case 2: 3 in the corner&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;s&gt;More or less exactly the same argument, but with a different picture.  In this case the * can be X, Y or some third tetronimo (which would give us two overlaps).&lt;/s&gt; - Thanks to @Sleafar for point out that this proof is wrong. It can however still be proved, I believe, by considering more elaborate cases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uQiM4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uQiM4.png&quot; alt=&quot;Case 2: Two scenarios&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For case 2, we need to extend the scenario to both corners and look for 2 overlaps.  See the above diagram.  From the basic blank squares three in the top-left corner, you can build up other parts of the board that must be this way or add an additional overlap.  Specifically X extends along the bottom of the blank squares, and Y comes out at Y3.  There must be three more blanks in the top right, and Z must be at Z1 and Z2.  We know we're going to get an extra overlap where Y and Z join, so we need to find one more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scenario 1 has Z3, which leads to Y4 and Z4.  The two squares below Z1 must be blank, as well as the one below Z4.  This creates 3 blanks together, which must be surrounded by tetronimos.  Specifically the square A1 must be a tetronimo.  That leaves the square with an *, which is a single blank square.  A single blank square will push up the minimum by 2, which is equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scenario 2 has Y4.  The square below Y3 must be blank, as well as the square below the blank below Y1 (otherwise we'd have a single blank again).  That means the square marked with an * must be a tetronimo (either Z or a new tetronimo).  Either way, there's an additional overlap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7rSKG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7rSKG.png&quot; alt=&quot;Case 3: Not quite in the corner&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Easy peasy.  X2 has to be part of X, so you get an extra overlap between X and the edge of the board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 4:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PzPr2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PzPr2.png&quot; alt=&quot;Case 4: Domino at the corner&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, in the case of the a domino on the side, it either reduces to a similar situation to Case 2 above, or it's surrounded by a tetronimo, shown as X1,X2,X3,X4 in the diagram (call this tetronimo X).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this scenario, consider the yellow squares marked B2.  If any of them are part of another tetronimo, then you have an extra overlap.  So suppose they are all blank.  Then the square Y1 must be part of a new tetronimo, which is an additional overlap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Other cases:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other cases are kind of trivial, but for the sake of completeness:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the corner is a single blank square this is effectively the same, because two blank singletons adds to the minimum perimeter by 2.&lt;/li&gt;&#xA;&lt;li&gt;If the corner has a tetronimo in it, then one of the adjacent edges must have the same tetronimo, and so you have an additional edge.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Where we are&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So each corner definitely creates at least one additional overlap.  This brings the maximum available perimeter down to 88, which is almost the minimum perimeter (86).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we can find two more overlaps then we prove that the 8x8 board cannot be covered by 8 tetronimos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that this can be done by considering a few more cases.  This may be susceptible to a computer search for all possible cases along one edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;s&gt;We can also make the following conclusions, based on the above.  A 8 piece solutions would:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Not use any of the straight tetronimo, because the four squares running along-side the tetronimo must have another tetronimo in it, which would be an additional overlap not covered above.&lt;/li&gt;&#xA;&lt;li&gt;Not use the square tetronimo, because this would reduce the maximum available because of its lower perimeter (8 vs 10 for the other tetronimos).&lt;/li&gt;&#xA;&lt;li&gt;Have 10 trionimos and 1 domino as the blank spaces.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The first of these is particularly interesting.  It eliminates the case where the 8 squares touching the edge of the board are 8 separate tetronimos, because the middle four squares could only have one covered by a straight line tetronimo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These restrictions certainly seem to support the thesis that 9 is the minimum, but we're not quite there!&lt;/s&gt; - Thanks, @Sleafar&lt;/p&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2015-09-09T22:29:06.813" LastActivityDate="2015-09-09T22:29:06.813" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21991" PostTypeId="2" ParentId="21985" CreationDate="2015-09-08T05:43:39.733" Score="4" Body="&lt;p&gt;For completeness, decided to add the optimality proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the opposite - i.e. you can cover the board with tetraminos, such that there is not a pond of size larger than 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the 4 squares a7, a8, b7, b8. At least one of them should be covered by a tetramino. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 1. Only one square is covered. Now there are basically 3 different subcases to consider, first two of which are straightforward:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fflLu.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fflLu.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/DGk5P.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DGk5P.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the third case, when we have a tetramino on b8, c8, c7, d7, has 2 subcases. The first one leads us to (up to symmetry)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/C0t4e.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/C0t4e.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the second one leads us to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DAhyL.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DAhyL.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 2. 2 squares are covered. WLOS let them be b7 and b8. Now using the observation above it is easy to see that we should have tetraminos on the following 2 places:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1YCVy.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1YCVy.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, now the square d6 is an issue and once again we get a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 3. 3 squares are covered. If a8 is covered, then (up to symmetry) we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zIpFT.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zIpFT.gif&quot; alt=&quot;&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise WLOG we have a tetramino on a7, b7, b8, c8. Also because of the observation above the square c6 should be covered as well. Similarly, a2, b1, b2, c3 should be covered by tetraminos. Now it is easy to see that a4, a5, a6, b6 can not be covered by tetraminos, so once again we get an island of size 4. Contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jJSa0.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jJSa0.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14140" LastEditorUserId="14140" LastEditDate="2015-09-08T16:36:12.737" LastActivityDate="2015-09-08T16:36:12.737" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="21992" PostTypeId="1" AcceptedAnswerId="22525" CreationDate="2015-09-08T09:41:29.993" Score="3" ViewCount="715" Body="&lt;p&gt;There is a line of infinite length (about zero thickness) in a 3D space. It can rotate or move across any plane at any speed. Every point (flagged or not) that has already been visited by the line cannot be touched again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your mission is to set flags on specific points in space. As you set each flag, it will move until it touches the flag. Then you set another flag, and so on. You cannot set a flag on a point it has already visited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is it possible to make it impossible for the 'line' to reach a flag, after a finite number of flags?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer the same Q for a line of finite length.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-09-23T04:26:10.363" Title="Line of destruction" Tags="|mathematics|geometry|visual|" AnswerCount="3" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="21993" PostTypeId="2" ParentId="21992" CreationDate="2015-09-08T10:27:10.603" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;No.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't have a mathematical proof, but merely an explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Assume the line never rotates (it doesn't need to, since you only set new flags after it reaches one). This reduces the problem to a point in 2D space. However, since you can't place a flag on a visited point, you can never &quot;cut off&quot; the point and stop it sneaking between the gaps in the &quot;walls&quot; it's left behind.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A couple of changes would drastically alter the problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; However, if you could set more than one flag at once and it couldn't, or had to touch them simultaneously, you could force the line to rotate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12462" LastActivityDate="2015-09-08T10:27:10.603" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21995" PostTypeId="2" ParentId="21975" CreationDate="2015-09-08T12:42:16.700" Score="14" Body="&lt;h2&gt;Cracked it!&lt;/h2&gt;&#xA;&lt;p&gt;The two anagrams at the top of the score are the keys to this puzzle:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Novas Requiem Cuts = Count Semiquavers (thanks, Zandar!)&lt;br /&gt;&#xA; Steve Chivet De Norman = Never Mind The Octaves&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The first clue gives us three of the song titles:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/KpMtV.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KpMtV.png&quot; alt=&quot;partial solution showing correspondence between note lengths and three song titles&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Ignore the rests — they are just there to pad out each bar to the required length. If you count the number of semiquavers in each note or cluster of tied notes, then convert them to letters of the alphabet (A=1, B=2, etc.), then, with a few minor corrections, you get the following titles:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; • Hallelujah Chorus&lt;br /&gt;&#xA; * O Holy Night&lt;br /&gt;&#xA; * Away In A Manger&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The fourth title is encoded in the notes themselves. There are 57 notes, all of which are A, B or C. When grouped into threes, these form a series of 19 base-3 numbers, where AAA=0, AAB=1, AAC=2, ABA=3 and so on:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; ACC BBB CAB ACB ABB AAB BAC ABB AAA BAB BBA ACC ABA CBB ACC BBB CAB ABB BCC&lt;br /&gt;&#xA; 022 111 201 021 011 001 102 011 000 101 110 022 010 211 022 111 201 011 122&lt;br /&gt;&#xA; 8, 13, 19, 7, 4, 1, 11, 4, 0, 10, 12, 8, 3, 22, 8, 13, 19, 4, 17&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Using these numbers as offsets into the alphabet (A=0, B=1, etc.), we get the last title:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; In The Bleak Midwinter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="40" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-09-08T15:37:53.323" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21999" PostTypeId="1" AcceptedAnswerId="22000" CreationDate="2015-09-08T21:48:07.960" Score="3" ViewCount="557" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;An caterpillar hatches on one corner of a checkerboard. At every second, it moves from one square to another adjacent (sharing a side) one. However, it can not return to any cell it has visited before except the starting cell. The caterpillar wishes to visit every cell of the board exactly once and then return to the starting square.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example for a $4 \times 4$ board one possible path for the caterpillar is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;a href=&quot;https://i.stack.imgur.com/J2ma0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/J2ma0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Can the caterpillar do so in $8 \times 8$ board?&lt;/li&gt;&#xA;&lt;li&gt;Can the caterpillar do so in $7 \times 7$ board?&lt;/li&gt;&#xA;&lt;li&gt;Can you generalize for a $m \times n$ board?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="15544" LastEditorUserId="15544" LastEditDate="2015-09-09T14:51:30.297" LastActivityDate="2015-09-09T21:53:05.223" Title="Checkerboard tour" Tags="|logical-deduction|checkerboard|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22000" PostTypeId="2" ParentId="21999" CreationDate="2015-09-08T22:12:53.423" Score="19" Body="&lt;p&gt;Let's just go straight to the $m\times n$ board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n$ is even, start in the bottom left corner. Go up all the way, right 1, down almost all the way (just 1 block from the bottom), right 1, up all the way, and so on. As we are going up at column 1 and we are alternating between up and down, at column $n$ we will be going down. This time go all the way down and then all the way left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise if $m$ is even, rotate the rectangle 90 degrees and do the algorithm above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise $n$ is odd and $m$ is odd. If we color the board black and white in a checkerboard pattern, each move to an adjacent square will alternate between black and white. In position 1 we are on a black square, so in position $n\cdot m + 1$ we should be on a white square. But in a tour of the board both of those positions should be the bottom left corner and the same color. Thus we cannot tour the board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: In the above, $m$ and $n$ should both be greater than 1 (notes Wa Kai). If one of $n$ and $m$ is 1 while the other is greater than $2$, it is obviously impossible. If the other is 2 it is possible (notes Darrel Hoffman).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case of $1\times1$ might be up for debate, but I would argue that it is impossible. An allowed move is to an adjacent square, so there are no possible moves and thus nothing can happen.&lt;/p&gt;&#xA;" OwnerUserId="11100" LastEditorUserId="11100" LastEditDate="2015-09-09T21:53:05.223" LastActivityDate="2015-09-09T21:53:05.223" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22001" PostTypeId="2" ParentId="20774" CreationDate="2015-09-08T22:35:06.333" Score="1" Body="&lt;p&gt;Here's a ninth incorrect answer to this riddle, but at least we will finally address the phrase &quot;in beautifully printed cursive&quot; and share some ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What makes cursive unique? How can we come up with an answer that relies on the handwriting style? Loops (and other features) are dependent on the friend's handwriting, so let's first look at the letters i,t,j, and x. These are the only letters which &lt;em&gt;require&lt;/em&gt; a second stroke, which makes them unique (since cursive is used to avoid such extraneous strokes). We can ignore capital letters because the only one used -- I -- requires just one stroke.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number of pen-to-paper contacts during writing (per word):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;1 2 1 2 1 3, 1 2, 4 1 3 1 2&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Letter placement relative to i,t (no j,x in note):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;1 3t2 5 1i2 1 1it, 4 4i2, 3t5ti2 1 1i4i4 6 2t3&lt;/strong&gt;&lt;br&gt;&#xA;  or&lt;br&gt;&#xA;  &lt;strong&gt;4 t 8 i 4 i 0 t 8 i 5 t 5 t 0 i 4 i 4 i 12 t 3&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I played with some of these numbers and &lt;a href=&quot;http://www.avastronomyclub.org/skymap/d/skymap.php&quot; rel=&quot;nofollow&quot;&gt;this skymap&lt;/a&gt; but found nothing notable. However, it's quite possible that the answer is not an astrophysical object (or clouds, or planes, or satellites, meteors, UFOs, rain, snow, sleet, birds, a dragon, blimps, hot-air balloons, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, we could consider letters which reach above or below the &lt;em&gt;normal&lt;/em&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Cursive#/media/File:Cursive.svg&quot; rel=&quot;nofollow&quot;&gt;height of letters&lt;/a&gt; like a, c, e, i, etc. We could classify letters as &lt;em&gt;tall&lt;/em&gt; (b, d, l, etc.), &lt;em&gt;deep&lt;/em&gt; (g, j, p, etc.), and &lt;em&gt;f&lt;/em&gt; (f, f, f, etc.). Then, we could just speculate wildly:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;The Centauri Constellation&lt;/strong&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Only six letters are omitted from the note (b,j,q,v,w,x). Centauri has six stars (though so do Brahe and Corvi, and &lt;a href=&quot;http://wiki.uqm.stack.nl/List_of_constellations&quot; rel=&quot;nofollow&quot;&gt;others&lt;/a&gt;...), so obviously it's a fitting answer to this riddle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13237" LastActivityDate="2015-09-08T22:35:06.333" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22002" PostTypeId="2" ParentId="563" CreationDate="2015-09-09T06:18:00.443" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://www.cs.brandeis.edu/~storer/JimPuzzles/RUBIK/GearCube/GearCube.pdf&quot; rel=&quot;nofollow&quot;&gt;This document provides a much simpler solution!&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In brief:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Solve corners&lt;/li&gt;&#xA;&lt;li&gt;Solve as much as possible of the edges&lt;/li&gt;&#xA;&lt;li&gt;Solve the rest using simple moves exchanging edges and centres&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Here is an extract from the document:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Restore puzzle to be flat (easy - do flips as needed).&lt;/li&gt;&#xA;&lt;li&gt;Solve the corners (easy - faces cannot rotate 90 degrees).&lt;/li&gt;&#xA;&lt;li&gt;Use step A to solve as much as possible, use Step B, and repeat until solved (repositioning&#xA;the cube as appropriate):&#xA;A. Flip the right face clockwise 6 times.&#xA;(Exchanges front/rear and top/bottom of the vertical center layer).&#xA;B. Flip the bottom face clockwise, flip the right face twice clockwise, flip the bottom face&#xA;counter clockwise, flip the right face twice counter clockwise.&#xA;(Exchanges front/rear of vertical center layer and left/right centers.)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Really easy!&lt;/p&gt;&#xA;" OwnerUserId="16264" LastActivityDate="2015-09-09T06:18:00.443" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22003" PostTypeId="2" ParentId="21999" CreationDate="2015-09-09T09:41:16.303" Score="-1" Body="&lt;p&gt;I am new here, but I want to give it a shot!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isn't it just as simple as: As long as either &lt;em&gt;m&lt;/em&gt; or &lt;em&gt;n&lt;/em&gt; mod 2 == 0, it is possible?&#xA;Of course this is only applicable for squares and rectangles only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; So I also figured over lunch that my answer was Incorrect. When you enter the values in a scenario where either n or m == 1. Sadly the caterpillar is stuck.&lt;/p&gt;&#xA;" OwnerUserId="16266" LastEditorUserId="16266" LastEditDate="2015-09-09T10:42:58.377" LastActivityDate="2015-09-09T10:42:58.377" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22004" PostTypeId="2" ParentId="21999" CreationDate="2015-09-09T12:01:40.613" Score="1" Body="&lt;p&gt;It's not possible if there are an odd number of squares, because when we paint them like a checkerboard, the last move must lead to a square of the same color as the first one, with the color changing every move. However, then the number of moves must be odd, which is a contradiction. When it's even, it's always possible.&lt;/p&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="14016" LastEditDate="2015-09-09T12:29:41.623" LastActivityDate="2015-09-09T12:29:41.623" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22005" PostTypeId="1" AcceptedAnswerId="22006" CreationDate="2015-09-09T13:25:15.053" Score="10" ViewCount="1956" Body="&lt;p&gt;No math involved here, just a bit of lateral thinking. Can you figure out what comes next?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2, 4, 8, 4, 4, 1, 3, 2, 7, 8, 3, 3, 6, 3, 4, 5...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus&lt;/strong&gt;: How about here?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;6, 3, 6, 3, 3, 5, 3, 3, 4...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10851" LastEditorUserId="4421" LastEditDate="2015-11-25T18:01:25.647" LastActivityDate="2015-11-25T18:01:25.647" Title="Seeing the forest for the trees and all that jazz" Tags="|pattern|lateral-thinking|number-sequence|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22006" PostTypeId="2" ParentId="22005" CreationDate="2015-09-09T13:27:31.760" Score="21" Body="&lt;p&gt;The first sequence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each number in the sequence represents the number of letters in its respective sequential word in the sentence above. This makes the next (and final) number in the sequence:&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;4&lt;/strong&gt;: from &lt;em&gt;&quot;next&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: Thanks to @alexmc for the bonus part:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Using the same method as in the first sequence, but with the words from the title, we can count the letters and see that the next (and final) number in the bonus sequence is also:&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;4&lt;/strong&gt;: from &lt;em&gt;&quot;jazz&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="10399" LastEditDate="2015-09-09T17:54:13.397" LastActivityDate="2015-09-09T17:54:13.397" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="22009" PostTypeId="1" AcceptedAnswerId="22012" CreationDate="2015-09-09T14:00:47.613" Score="14" ViewCount="934" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Four times they proclaimed their name,&lt;br&gt;&#xA;  Spare the Romans and it was the same.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Diverging, in a sacred abode they prayed,&lt;br&gt;&#xA;  Afterwards some real ideas they sprayed.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Despite their fame, they stressed their existence&lt;br&gt;&#xA;  Before using doors in the wrong sense.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;For the next, they chose a tail for the title&lt;br&gt;&#xA;  Which was poetic for it was their final.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Can you guess who they are?&lt;/p&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-09-10T05:03:39.420" LastActivityDate="2015-09-11T13:36:16.803" Title="Luminous Balloon" Tags="|riddle|rhyme|knowledge|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22010" PostTypeId="1" AcceptedAnswerId="22013" CreationDate="2015-09-09T14:05:02.083" Score="24" ViewCount="4117" Body="&lt;p&gt;For each of the below sentences, find a set of anagrams that can fill in the blanks to form a coherent, grammatically correct sentence. See the example below if you don't understand. I won't provide the letter length of each anagram set (at least not yet) but will say that all are in [4,6] letters. I will accept the first answer that finds an anagram set that meets the requirements for each sentence (even if this happens to differ from my intended set, which may well be possible). Have fun and good luck!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Example Sentence:&lt;br&gt;&#xA;  The &lt;code&gt;___&lt;/code&gt; is located &lt;code&gt;___&lt;/code&gt; the &lt;code&gt;___&lt;/code&gt; on the body.&lt;br&gt;&#xA;  Example Answer:&lt;br&gt;&#xA;  BOWEL, BELOW, ELBOW.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Sentences:&lt;br&gt;&#xA;  The &lt;code&gt;___&lt;/code&gt;, full of &lt;code&gt;___&lt;/code&gt; water, &lt;code&gt;___&lt;/code&gt; for several years before eroding.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In between the first and second &lt;code&gt;___&lt;/code&gt;, we met the &lt;code&gt;___&lt;/code&gt; of &lt;code&gt;___&lt;/code&gt;.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The tailor did amazing work: some &lt;code&gt;___&lt;/code&gt; and a &lt;code&gt;___&lt;/code&gt; here and there handled the extra fabric (pardon the occasional and accidental &lt;code&gt;___&lt;/code&gt;) and a &lt;code&gt;___&lt;/code&gt; in the mirror revealed a seamless fit.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;From the &lt;code&gt;___&lt;/code&gt;, he sat quiet as a &lt;code&gt;___&lt;/code&gt; while I played the &lt;code&gt;___&lt;/code&gt; and &lt;code&gt;___&lt;/code&gt; of the performance.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;He was a &lt;code&gt;___&lt;/code&gt; of many famous historical &lt;code&gt;___&lt;/code&gt;, some of whom were even featured on our &lt;code&gt;___&lt;/code&gt;. (This anagram set contains words of length 5)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="13970" LastEditDate="2015-09-09T18:40:43.840" LastActivityDate="2015-09-11T14:24:58.677" Title="Anagram Sentences" Tags="|word|anagram|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22011" PostTypeId="2" ParentId="22005" CreationDate="2015-09-09T14:07:47.267" Score="15" Body="&lt;p&gt;You can see that the next number in the first sequence is &lt;strong&gt;4&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;No math involved here, just a bit of lateral thinking. Can you figure out what comes next?&#xA;2  4    8        4     4    1 3   2  7       8         3   3   6      3   4    5     ...&#xA;                                                                                     4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The bonus also has &lt;strong&gt;4&lt;/strong&gt; as the next number in the sequence.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Seeing the forest for the trees and all that jazz&#xA;6      3   6      3   3   5     3   3   4    ...&#xA;                                             4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2015-09-09T14:07:47.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22012" PostTypeId="2" ParentId="22009" CreationDate="2015-09-09T14:08:01.543" Score="18" Body="&lt;p&gt;They are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the rock band Led Zeppelin.  The clues point to the titles of their albums.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Four times they proclaimed their name,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the first four albums are simply titled Led Zeppelin I, II, III and IV&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Spare the Romans and it was the same.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; but for the number, each album has the same title (thanks to DasBeasto)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Diverging, in a sacred abode they prayed,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Houses of the Holy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Afterwards some real ideas they sprayed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Physical Graffiti&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Despite their fame, they stressed their existence&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Presence&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Before using doors in the wrong sense.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In Through the Out Door&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the next, they chose a tail for the title&lt;br&gt;&#xA;Which was poetic for it was their final.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Coda (&quot;coda&quot;, meaning &quot;tail&quot; in Italian, is a musical term designating the closing section of a piece)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Luminous Balloon&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Luminous = L.E.D.  Balloon = Zeppelin :D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4385" LastEditorUserId="4385" LastEditDate="2015-09-11T13:36:16.803" LastActivityDate="2015-09-11T13:36:16.803" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="22013" PostTypeId="2" ParentId="22010" CreationDate="2015-09-09T14:17:27.993" Score="20" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The &lt;code&gt;___&lt;/code&gt;, full of &lt;code&gt;___&lt;/code&gt; water, &lt;code&gt;___&lt;/code&gt; for several years before eroding.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DELTAS, SALTED, LASTED (Thanks alexmc.)&lt;br&gt;&lt;br&gt;&lt;strike&gt;LATS, SALT, LAST &lt;br&gt;(Lats are the currency units of Latvia. This sentence is not logical, but forms a coherent grammatical sentence nonetheless.)&lt;br&gt;&lt;br&gt;KALE, LAKE, LEAK&lt;br&gt;(Also not logical, but grammatically correct.)&lt;/strike&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In between the first and second &lt;code&gt;___&lt;/code&gt;, we met the &lt;code&gt;___&lt;/code&gt; of &lt;code&gt;___&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ACTS, CAST, CATS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The tailor did amazing work: some &lt;code&gt;___&lt;/code&gt; and a &lt;code&gt;___&lt;/code&gt; here and there handled the extra fabric (pardon the occasional and accidental &lt;code&gt;___&lt;/code&gt;) and a &lt;code&gt;___&lt;/code&gt; in the mirror revealed a seamless fit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PINS, SNIP, NIPS, SPIN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;From the &lt;code&gt;___&lt;/code&gt;, he sat quiet as a &lt;code&gt;___&lt;/code&gt; while I played the &lt;code&gt;___&lt;/code&gt; and &lt;code&gt;___&lt;/code&gt; of the performance.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ONSET, STONE, TONES, NOTES&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;He was a &lt;code&gt;___&lt;/code&gt; of many famous historical &lt;code&gt;___&lt;/code&gt;, some of whom were even featured on our &lt;code&gt;___&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SCION ICONS COINS (Thanks Simon for the best fit.)&lt;br&gt;&lt;br&gt;&lt;strike&gt;MASTER, ARMETS, STREAM (Armets are medieval era helmets.) (Thanks GordonK for the push.)&lt;/strike&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-09-11T14:24:58.677" LastActivityDate="2015-09-11T14:24:58.677" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="22014" PostTypeId="1" AcceptedAnswerId="22021" CreationDate="2015-09-09T16:49:26.440" Score="5" ViewCount="2783" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tNulP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tNulP.png&quot; alt=&quot;this is related to engineering graphics, its a very simple 3D shape!&quot;&gt;&lt;/a&gt;&#xA;Find the simple 3D solid associated with these views. It is given that there are no hidden lines (lines which are not visible from the side or top view).&lt;/p&gt;&#xA;" OwnerUserId="16271" LastEditorUserId="15544" LastEditDate="2015-09-09T18:38:30.633" LastActivityDate="2015-09-09T20:06:08.443" Title="Find the simple 3D solid associated with these views" Tags="|logical-deduction|geometry|" AnswerCount="5" CommentCount="5" ClosedDate="2015-09-09T22:19:37.933" ContentLicense="CC BY-SA 3.0" />
  <row Id="22015" PostTypeId="2" ParentId="22014" CreationDate="2015-09-09T16:56:26.237" Score="3" Body="&lt;p&gt;Could it be,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A table with a hole in the middle? The side view is the table turned upside down&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13925" LastActivityDate="2015-09-09T16:56:26.237" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22016" PostTypeId="2" ParentId="22014" CreationDate="2015-09-09T16:57:14.333" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A 2 legged (thanks to Engineer Toast, it's simpler to draw) table with a hole in the middle.&lt;br&gt;&lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/PiLKd.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2015-09-09T17:51:19.987" LastActivityDate="2015-09-09T17:51:19.987" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="22018" PostTypeId="2" ParentId="22014" CreationDate="2015-09-09T17:59:26.463" Score="1" Body="&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; I trashed my crappy picture to ask a question instead. I don't know if anyone else will benefit from it, but I would like to learn personally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the picture below, if there was a hole through the center of the shape, the hidden lines would &lt;em&gt;have&lt;/em&gt; to be show like this? I very well could be incorrect in my statement.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my understanding @dberm22 and @alexmc pictures would have to be drawn like this because the lines making the hole are hidden behind the wall, which is not the original side view.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please correct if I'm wrong and give a little explanation so I can screw my heads on right! -Thanks &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OXqoa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OXqoa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another thought was (Maybe this can help someone)  if you look really close at the side view, you'll notice the inside of both the &quot;legs&quot; have a little extra line to them.  I don't know if that's a mess up on the picture or if it has significant value to the solution!! &lt;/p&gt;&#xA;" OwnerUserId="16273" LastEditorUserId="15544" LastEditDate="2015-09-09T20:06:08.443" LastActivityDate="2015-09-09T20:06:08.443" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22019" PostTypeId="1" AcceptedAnswerId="22023" CreationDate="2015-09-09T18:43:47.710" Score="-10" ViewCount="590" Body="&lt;p&gt;Before we begin, allow me to define what I call a &lt;strong&gt;Scramble Sentence&lt;/strong&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A &lt;em&gt;Scramble Sentence&lt;/em&gt; is a grammatically correct sentence, where at least 1 other permutation, of the letters in the sentence, exists that it makes a different sentence with a different meaning.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's define a quick restriction though, just to make sure it isn't too easy, &lt;strong&gt;NO WORDS CAN BE REUSED!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The aim of this question is to find the &lt;strong&gt;longest possible sentence which fits all of the above conditions!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: &lt;em&gt;Punctuation doesn't count as part of the sentence (you don't have to scramble commas and such)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Additional restrictions:&lt;/strong&gt;&lt;br&gt;&#xA;* No listing out words&lt;br&gt;&#xA;* No quotation marks  &lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2015-09-11T01:24:05.197" LastActivityDate="2015-09-11T01:24:05.197" Title="Longest $Scramble Sentence$?" Tags="|english|open-ended|no-computers|" AnswerCount="2" CommentCount="11" ClosedDate="2015-09-09T19:47:20.440" ContentLicense="CC BY-SA 3.0" />
  <row Id="22020" PostTypeId="2" ParentId="22014" CreationDate="2015-09-09T18:48:50.643" Score="0" Body="&lt;p&gt;Picture a cube.  Remove a square cylendar from the middle so it looks like the top view.  Now pass it through a half circle, or similar curve, cutting away much of the cube.  The key is that the sides need to be cut down enough so that the center gap is showing like in the second picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the front, it will look like a half circle.  From the top, still looks like a square donut.  And from the side, a U.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry, I don't have any drawing tools at my disposal.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2015-09-09T18:48:50.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22021" PostTypeId="2" ParentId="22014" CreationDate="2015-09-09T18:49:20.797" Score="9" Body="&lt;p&gt;Modeled a solution in my CAD software. Here are some shaded views:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gOqie.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gOqie.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just for completeness, here are the drawing views that this part produces:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dHmky.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dHmky.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to clarify, the other solutions posted suggested this part as a solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4wioC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4wioC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, notice the extra lines in the produced drawing. These lines would also appear in the &quot;table with a hole in the middle&quot; solution, so these are not correct.&lt;/p&gt;&#xA;" OwnerUserId="4865" LastEditorUserId="4865" LastEditDate="2015-09-09T19:07:39.333" LastActivityDate="2015-09-09T19:07:39.333" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22023" PostTypeId="2" ParentId="22019" CreationDate="2015-09-09T19:00:45.943" Score="3" Body="&lt;p&gt;&lt;em&gt;Definitely not the longest, but getting the ball rolling.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;18 letters&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;He bares on restraint.&lt;/strong&gt;&lt;br&gt;&#xA;(&lt;em&gt;He takes off his shirt when restrained.&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;anagrams to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Restrain the bear, son!&lt;/strong&gt;&lt;br&gt;&#xA;(&lt;em&gt;Take control of the hairy creature, my male offspring!&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not a single word reused, so this has 100% new words. &lt;/p&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-09-09T19:00:45.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22025" PostTypeId="2" ParentId="22019" CreationDate="2015-09-09T19:06:01.663" Score="0" Body="&lt;p&gt;I care for you - choose: &quot;abet&quot;, &quot;abut&quot;, &quot;acme&quot;, &quot;acre&quot;, &quot;acres&quot;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your choices are: &quot;rom&quot;, &quot;beat&quot;, &quot;tuba&quot;, &quot;mace&quot;, &quot;race&quot;, &quot;scare&quot;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can make this as long as you want until you run out of words. This question is poorly defined.&lt;/p&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="11876" LastEditDate="2015-09-09T19:31:52.430" LastActivityDate="2015-09-09T19:31:52.430" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="22026" PostTypeId="1" AcceptedAnswerId="22028" CreationDate="2015-09-09T19:08:52.007" Score="6" ViewCount="2200" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Arrange ten coins in the familiar ten-pin bowling formation (see figure). What is the smallest number of pennies you must remove so that no equilateral triangle, of any size, will have its three corners marked by the centers of three pennies that remain? Not counting rotations and reflections as different, there is only one pattern for the removal of the minimum number of pennies.&lt;br&gt;&#xA;  Note that the original pattern contains two equilateral triangles that are tipped so that their bases are not horizontal.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Give also a simple proof that your answer is minimal.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AnVBam.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AnVBam.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This problem (without the request for a proof) is from Martin Gardner's &quot;The Colossal Book of Short Puzzles and Problems&quot;.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="14080" LastEditorUserId="15544" LastEditDate="2015-09-10T06:40:19.197" LastActivityDate="2022-09-27T03:48:16.193" Title="Arrange ten coins in a bowling formation. Fewest pennies to remove so no three pennies that remain have centers that form an equilateral triangle?" Tags="|mathematics|geometry|" AnswerCount="3" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="22027" PostTypeId="1" AcceptedAnswerId="22134" CreationDate="2015-09-09T19:09:48.990" Score="2" ViewCount="816" Body="&lt;p&gt;Find the next number in this sequence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$120, 106, 112, 108, 80, 92, 84,...$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This sequence might be easy for some, but you won't be able to solve it if you don't know a certain subject in maths (That I will reveal if nobody finds the answer in 30 minutes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First Hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Look at the title!&#xA;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second Hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It doesn't involve any special series or integrals.&#xA;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Third hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It is in the number theory puzzle&#xA;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fourth hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is an easy recurrence relation.&#xA;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fifth hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It uses a well known function&#xA;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sixth hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 120 and 106 are just starting values&#xA;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14469" LastEditorUserId="14469" LastEditDate="2015-09-11T22:50:12.047" LastActivityDate="2015-09-13T19:13:29.917" Title="Don't try this if you don't know your maths!" Tags="|mathematics|number-sequence|" AnswerCount="3" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="22028" PostTypeId="2" ParentId="22026" CreationDate="2015-09-09T19:59:42.290" Score="6" Body="&lt;p&gt;Note there are 15 equilateral triangles. If we label the pennies A through J going from left to right and then row by row, they are ABC, BDE, CEF, DGH, EHI, FIJ, BCE, DEH, EFI, ADF, BGI, CHJ, AGJ, BFH, and CDI.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ppSIUm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ppSIUm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that AGJ, BFH, and CDI do not share any pennies, so we need to remove at least one penny from each of those sets. Also note that no matter which three we remove from those sets, some triangle with E will be left behind. Therefore we must remove at least four pennies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possible way to do this (the only possible way, if you ignore rotations) is to remove A, E, H, and I.&lt;/p&gt;&#xA;" OwnerUserId="10851" LastEditorUserId="10851" LastEditDate="2015-09-10T15:36:58.343" LastActivityDate="2015-09-10T15:36:58.343" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="22029" PostTypeId="2" ParentId="20774" CreationDate="2015-09-09T20:06:13.153" Score="0" Body="&lt;p&gt;I think your friend stayed up all night looking at a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Constellation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If each word maps to a letter then there must be 13 letters in the answer, and 'constellation' is the only 13-letter word I can find which fits the theme of astronomy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's also possible that he spent the night gazing at some&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bachelorettes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've come to this view because of the&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dastardliness&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;of the puzzle!&lt;/p&gt;&#xA;" OwnerUserId="4050" LastActivityDate="2015-09-09T20:06:13.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22030" PostTypeId="1" CreationDate="2015-09-09T20:06:33.053" Score="-1" ViewCount="378" Body="&lt;p&gt;Recently a question of mine was put on hold for being too broad: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/22019/longest-scramble-sentence&quot;&gt;Longest $Scramble Sentence$?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or if you prefer have this in the one window:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Before we begin, allow me to define what I call a &lt;strong&gt;Scramble Sentence&lt;/strong&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A &lt;em&gt;Scramble Sentence&lt;/em&gt; is a grammatically correct sentence, where at least 1 other permutation, of the letters in the sentence, exists that it makes a different sentence with a different meaning.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's define a quick restriction though, just to make sure it isn't too easy, &lt;strong&gt;NO WORDS CAN BE REUSED!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The aim of this question is to find the &lt;strong&gt;longest possible sentence which fits all of the above conditions!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: &lt;em&gt;Punctuation doesn't count as part of the sentence (you don't have to scramble commas and such)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Additional restrictions:&lt;/strong&gt;&lt;br&gt;&#xA;* No listing out words&lt;br&gt;&#xA;* No quotation marks  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It was supposed to be a fun and interesting question so that users can &quot;compete&quot; to find the longest sentence that can be rescrambled into a completely different sentence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But my question for you guys, is how should I make that puzzle &lt;strong&gt;less broad&lt;/strong&gt;, and limit the possible answers. I hope to learn from these suggestions and to improve my puzzle creating skills for the future!&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-09-16T21:47:53.277" Title="Make the question less broad" Tags="|english|open-ended|" AnswerCount="1" CommentCount="11" ClosedDate="2015-09-10T01:53:23.560" ContentLicense="CC BY-SA 3.0" />
  <row Id="22031" PostTypeId="2" ParentId="22026" CreationDate="2015-09-09T21:10:13.513" Score="0" Body="&lt;p&gt;In any case, at least 3 coins must be removed because the 3 smallest triangles that can be formed with the corner coins don't intersect with each other, so we can do no less than removing 1 from each of these triangles. But is 3 enough?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hexagonal part in the center contains at least one triangle as long as the coin in the middle is kept and there are two neighboring coins on its &quot;edges&quot;. If we remove only 3 coins from the corner triangles, we end up either leaving the 3 outermost coins untouched or removing less than 3 of the coins on the hexagon's edges, both of which are unwanted consequences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there are 15 triangles, eliminating 6 of them by taking the central one out would be a good move. The remaining triangles always rely on an outermost coin, so we can reach our goal by later taking out all 3.&lt;/p&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="14016" LastEditDate="2015-09-09T21:35:43.340" LastActivityDate="2015-09-09T21:35:43.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22032" PostTypeId="2" ParentId="20369" CreationDate="2015-09-09T21:45:34.023" Score="1" Body="&lt;p&gt;There are no solutions for 1,2 and 3 digits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two solutions for 4 digits:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1210 and 2020&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is one for 5 digits:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 21200&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One for 7 digits:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3211000&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is there any other possible solution?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Self-descriptive_number&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Self-descriptive_number&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;https://oeis.org/A138480&quot; rel=&quot;nofollow&quot;&gt;https://oeis.org/A138480&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the general math solution (or attempt to build one):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://math.stackexchange.com/questions/849813/are-there-numbers-that-describe-themselves-in-some-base-but-not-according-to-the&quot;&gt;https://math.stackexchange.com/questions/849813/are-there-numbers-that-describe-themselves-in-some-base-but-not-according-to-the&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user12496" LastEditorDisplayName="user12496" LastEditDate="2015-09-09T22:13:21.960" LastActivityDate="2015-09-09T22:13:21.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22033" PostTypeId="2" ParentId="22010" CreationDate="2015-09-09T22:07:00.020" Score="17" Body="&lt;p&gt;I believe this is a better fit for the last one than CodeNewbie's.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;He was a &lt;code&gt;___&lt;/code&gt; of many famous historical &lt;code&gt;___&lt;/code&gt;, some of whom were even featured on our &lt;code&gt;___&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SCION ICONS COINS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7346" LastActivityDate="2015-09-09T22:07:00.020" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22034" PostTypeId="1" AcceptedAnswerId="22036" CreationDate="2015-09-09T22:20:18.033" Score="9" ViewCount="266" Body="&lt;p&gt;I just came across this note in my office!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Date of writing: Sept. 9th, 2015  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In exactly one week from today, and then a week from that date and then another week afterwards, something will happen that causes certain badly designed computer programs to function improperly on those days.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;It happens those three days and no other days in the year 2015. In any given year it will happen in exactly three days out of the year.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What might this finicky event be?&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="12700" LastEditDate="2015-09-10T01:52:13.050" LastActivityDate="2015-09-10T03:55:47.820" Title="One, two, and three weeks from now" Tags="|enigmatic-puzzle|computer-puzzle|" AnswerCount="1" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="22035" PostTypeId="2" ParentId="22030" CreationDate="2015-09-09T22:34:50.337" Score="3" Body="&lt;p&gt;This question can be made less broad with any of the following conditions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;only names of cities may be used  &lt;/li&gt;&#xA;&lt;li&gt;you may only use the letter 'b'  &lt;/li&gt;&#xA;&lt;li&gt;limit your answer to no more than two words  &lt;/li&gt;&#xA;&lt;li&gt;words may not scramble into words previously used  &lt;/li&gt;&#xA;&lt;li&gt;words may only scramble to words beginning with x  &lt;/li&gt;&#xA;&lt;li&gt;both sentences must mention pears&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2015-09-09T22:34:50.337" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22036" PostTypeId="2" ParentId="22034" CreationDate="2015-09-10T03:49:49.437" Score="6" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every Wednesday in September with a double digit for the date. Programs writing out the date like &quot;Wednesday September 16th, 2015&quot; will have the longest print on these days because Wednesday and September have the most letters out of the days of the week and months of the year respectively. For the above example format, if the print buffer is a char[30], there will be a buffer overflow for these three days of the year.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Old alternative answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Potential answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if(strcmp(dayoftheweek,&quot;Wednesday&quot;)==0 &amp;amp;&amp;amp; (day==16||day==23||day==30) &amp;amp;&amp;amp; monthoftheyear%2==1)destroyEverything();&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This line of code snuck in by a mischievous programmer would make any program fail three days of every year. Every year has two months where the Wednesdays fall on the 16th, 23rd, and the 30th, and they fall on even and odd months. Last year's occurred in July. (Note: leaving out platform-specific access of date information)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastEditorUserId="7722" LastEditDate="2015-09-10T03:55:47.820" LastActivityDate="2015-09-10T03:55:47.820" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="22037" PostTypeId="2" ParentId="22027" CreationDate="2015-09-10T06:36:30.420" Score="6" Body="&lt;p&gt;The sequence and its difference sequence are-&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$S = 120,\ 106,\ 112,\ 108,\ 80,\ 92,\ 84,...$&lt;br&gt;&#xA;  $\Delta S= -14,\ 6,\ -4,\ -28,\ 12,\ -8...$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It seems that the difference sequence doubles every 3 terms. So, I would guess that the next three terms in the difference sequence would be&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;-14, 6, -4, -28, 12, -8, &lt;strong&gt;(-56, 24, -16)&lt;/strong&gt; or &lt;strong&gt;(-42,18,-12)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, the next terms might be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;120, 106, 112, 108, 80, 92, 84, &lt;strong&gt;(28, 52, 36)&lt;/strong&gt; or &lt;strong&gt;(42, 60, 48)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15544" LastActivityDate="2015-09-10T06:36:30.420" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22038" PostTypeId="2" ParentId="21958" CreationDate="2015-09-10T06:57:39.160" Score="0" Body="&lt;p&gt;Some hints for a proof of 8 tetrominoes being an impossible solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's flip the question, to lay emphasis on the blanks, rather than the covered spaces.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Fill 32 cells of an 8x8 grid with monominoes, dominoes and/or&#xA;  trinominoes such that:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(a) No two pieces touch each other&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(b) The leftover squares can be occupied by 8 tetrominoes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solving the above Q alone is enough, you may wish to avoid reading the possible method of proving, which I have given below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Possible method of proving&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Method relies on an intelligent brute-forcing program)&#xA;For every piece that is placed, there are a certain no. of cells that are adjacent to the piece. No other piece can be placed on them (ofcourse, we will place the tetrominoes there, but forget that for now), hence they are called BLOCKED. The number of cells a piece occupies (1,2 or 3) is called USEFUL. For example, an I-shaped trinomino in the middle of the board will look like this&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  B B B   &#xA;B X X X B&#xA;  B B B&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We have 3 Xs (USEFUL) and 8 Bs (BLOCKED). Hence ratio is 3:8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the same piece was placed on the edge:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;_________&#xA;B X X X B&#xA;  B B B &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Ratio is 3:5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A piece placement with a high USEFUL is to BLOCKED ratio is more optimal. For the corner, the placing with the best ratio is an L-shaped trinomino (Ratio 3:3). Hence we cover all 4 corners in this manner. For the side of the board, the best placement is an L-shaped trinomino touching the edge with 2 cells, giving a ratio 3:5. (I-shaped trinominoes also have 3:5 ratio, but there is no space to place them.) Hence we cover each edge in such a manner. We now have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X X   X X   X X&#xA;X     X       X&#xA;&#xA;X           X X&#xA;X X           X&#xA;&#xA;X       X     X&#xA;X X   X X   X X&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We have successfully occupied 24 squares. We need to occupy 8 more. Even by trial and error, it appears impossible. Let's trying proving it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For floating pieces (not touching the edge), the best ratio is still given by the L-shape (3:7), followed by the I-shaped (3:8). A domino gives 2:6, and a monomino gives 1:4. The best option seems to be two L-shaped trinominoes and a single domino. However, we are unable to place them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the &quot;supposedly optimal&quot; solution fails, we try solutions slightly worse than the optimal one. Probably, a program will be good at this. The program should keep trying worser and worser piece placements (in terms of the ratios), until it becomes obvious that there is no solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Flaws&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.My method more-or-less ignores rule (b). If there are cases where rule (a) is fulfilled, they should be very few in number, and the program should quickly be able to check for satisfaction of rule (b)&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;And, it ignores the fact that BLOCKED squares are often shared by trinominoes/dominoes/monominoes. Preference should be given to BLOCKED squares shared by three pieces, rather than two. I don't know how to implement this. Hence I said, use a program to test cases slightly worse than optimal also.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-09-10T06:57:39.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22039" PostTypeId="2" ParentId="16169" CreationDate="2015-09-10T06:59:06.193" Score="1" Body="&lt;p&gt;By a stretch, but are you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Negative Infinity (-INF) ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;EDIT: Refined my reasoning after reading the highest voted answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; on top politically = power, being not in violation = fraction&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I am on top politically, I am right, but if I am on top of being not in violation, I am wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; x^-INF ~ 0. from a programming context, I'm sure I read somewhere that different languages treat 0 as TRUE and non-zero as FALSE. -INF/x, with x &amp;lt;&gt; INF, -INF, or 0 , is still -INF. Hence it is FALSE.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I am above 8, I am wrong, but if I am low on being not in violation, I am forever.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; -INF/8 is -INF which is not a very desirable number. But if it is 8/-INF, and if I remember school math correctly, it is 0. Which is good, and TRUE =&gt; truth, which by a stretch, is forever. Or may be it is 0 that is forever.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cannot be on top politically and above 8, but I can be on top and low of being not in violation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; x^-INF !&gt; (not greater than) 8, obviously. And Infinity Arithmetic defines fraction (division) by -INF.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15607" LastEditorUserId="15607" LastEditDate="2015-09-10T07:34:06.683" LastActivityDate="2015-09-10T07:34:06.683" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22040" PostTypeId="1" AcceptedAnswerId="22082" CreationDate="2015-09-10T07:39:32.040" Score="15" ViewCount="984" Body="&lt;p&gt;Yesterday, I overheard the conversation between my friends Jim and Rose:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Jim&lt;/strong&gt;: I know lots of games.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Rose&lt;/strong&gt;: Wow, let's play some.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Jim&lt;/strong&gt;: If Jack too was here, it would have been great fun.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Rose&lt;/strong&gt;: Oh, I miss Jack a lot. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Jim&lt;/strong&gt;: He came last week for the exams.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Rose&lt;/strong&gt;: Ha! Really? He didn't tell me. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Jim&lt;/strong&gt;: He asked for a tea party.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Rose&lt;/strong&gt;: Yes, it should be arranged.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Jim&lt;/strong&gt;: Let's go at 4:00pm. He must be at home at that time.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Rose&lt;/strong&gt;: He has nailed the exams and so he should be hosting the&#xA;  party, right?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Jim&lt;/strong&gt;: Yup, and I bought a pen for him to appreciate his success in the&#xA;  exams.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Rose&lt;/strong&gt;: Oh, great. Let me see it.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Jim&lt;/strong&gt;: Aha, you lost. I am the winner.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What game they were playing? How did Rose lose and Jim win? And how does this conversation make sense?&lt;/p&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="2730" LastEditDate="2016-10-26T06:42:35.447" LastActivityDate="2016-10-26T06:42:35.447" Title="Suspicious Conversation" Tags="|logical-deduction|lateral-thinking|" AnswerCount="2" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="22042" PostTypeId="2" ParentId="17211" CreationDate="2015-09-10T10:07:30.200" Score="1" Body="&lt;p&gt;Part 3 joke answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Apple or Spiced Apple (to go with the spice theme)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A book, a crook, a surly cook, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Apples feature prominently in alphabet books, and the fruit of the Tree of the knowledge of good and evil in the Bible is often depicted as an apple.&lt;br&gt;&lt;br&gt;&#xA;The apple's botanical name is &lt;em&gt;Malus Domestica&lt;/em&gt;. &lt;em&gt;Malus&lt;/em&gt; is Latin for something bad or evil, which accurately describes a crook, while &lt;em&gt;Domestica&lt;/em&gt; describes a domestic helper, of which cooking for the household is one of their duties.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Gassy Silena garnished me, &#xA;though of a form, a rose I be.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Gassy Silena -&gt; Granny Smith (a stretch, I know)&lt;br&gt;&lt;br&gt;&#xA;Apples are part of the Rosaceae (rose) family of plants.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm a piece of pie, &#xA;a rude, crude man, &#xA;a real clean freak.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1. Apple Pie&lt;br&gt;&#xA;2. &lt;em&gt;Malus&lt;/em&gt; again&lt;br&gt;&#xA;3. The clean interface of Apple products?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a joke answer because the hints clearly indicate the answer is a compound word, of sorts.&lt;/p&gt;&#xA;" OwnerUserId="6178" LastActivityDate="2015-09-10T10:07:30.200" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22043" PostTypeId="1" AcceptedAnswerId="22107" CreationDate="2015-09-10T11:15:46.027" Score="13" ViewCount="324" Body="&lt;p&gt;Kickstarting &lt;a href=&quot;http://meta.puzzling.stackexchange.com/questions/4351/fortnightly-topic-challenge-1-movies?cb=1&quot;&gt;the fortnightly challenge&lt;/a&gt;, here's a &lt;a href=&quot;/questions/tagged/movies&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;movies&amp;#39;&quot; rel=&quot;tag&quot;&gt;movies&lt;/a&gt; puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following actors have been ordered on a particular criteria:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Gerard Depardieu&lt;br&gt;&#xA;  Damon Wayans&lt;br&gt;&#xA;  Jacques Perrin&lt;br&gt;&#xA;  Maurice Chevalier&lt;br&gt;&#xA;  Rod Steiger&lt;br&gt;&#xA;  Jean-Pierre Cassel&lt;br&gt;&#xA;  Matt Damon              &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can you guess why they have been listed in this order?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;BONUS:&lt;/strong&gt;&lt;br&gt;&#xA;Can you add a name somewhere in this list?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The title has a big clue. See if you can find it. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2015-09-11T18:17:07.023" LastActivityDate="2015-09-11T20:57:16.427" Title="How do you rank these actors?" Tags="|knowledge|movies|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22045" PostTypeId="1" AcceptedAnswerId="22170" CreationDate="2015-09-10T12:34:24.520" Score="8" ViewCount="611" Body="&lt;p&gt;My first attempt at puzzling. Let's see how many of you figure out the right &lt;em&gt;angle&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt; EDIT: Some small changes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A mere tool I am,  &lt;br&gt;In a world so grown, &lt;br&gt;Yet here I am,&#xA;  &lt;br&gt;Taking you on.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I'm a supporter &lt;br&gt;of things well known, &lt;br&gt;Without me, these&#xA;  portals, &lt;br&gt;you would never call your own.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;And kid yourself not, &lt;br&gt;I am almost the same, &lt;br&gt;As when I say &quot;The&#xA;  fate of the world &lt;br&gt;depends on.&quot; &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Of what I am made, &lt;br&gt;Is the Swan's Song of Stars, &lt;br&gt;And it's&#xA;  neighbors, &lt;br&gt;they may sound but &lt;br&gt;unattractive they are.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;EDIT: Another clue, as I feel the above clues might prove to be insufficient, and something that I feel I may have insufficiently described:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What I am made of that I mentioned, &lt;br&gt; is an ingredient among a few, &lt;br&gt; among its kind it lies, &lt;br&gt;an elemental country not new.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;Who am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.: This is TOO easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Hint 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A fine, young lady once said two me: &quot;Don't take it literally for what it is, dear. Magic is just Science we don't understand.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;EDIT: Hint 2 - Now this is just giving it away..&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Those famous things I support, &lt;br&gt;Once spawned beings of such form, &lt;br&gt; So formidable these off-springs, &lt;br&gt; Rode this world's storms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15607" LastEditorUserId="15607" LastEditDate="2015-09-12T04:39:59.043" LastActivityDate="2015-09-14T10:05:59.823" Title="Of Portals &amp; Worlds" Tags="|riddle|wordplay|" AnswerCount="3" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22046" PostTypeId="2" ParentId="22045" CreationDate="2015-09-10T12:51:29.647" Score="4" Body="&lt;p&gt;Are you:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Internet browser&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A mere tool I am,&lt;br&gt;&#xA;In a world so grown,&lt;br&gt;&#xA;Yet here I am,&lt;br&gt;&#xA;Taking you on.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Well, I'm using a browser, a tool, right now, am I?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm a supporter&lt;br&gt;&#xA;of things well known,&lt;br&gt;&#xA;Without me, these portals,&lt;br&gt;&#xA;you would never call your own. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wikipedia, Google, etc. are all sites, or portals, which support things well known. You can also create your own site, but without a browser you can't develop it and call it your own.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And kid yourself not,&lt;br&gt;&#xA;I am almost the same,&lt;br&gt;&#xA;As when I say &quot;The fate of the world&lt;br&gt;&#xA;depends on my name.&quot; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the recent arab spring &lt;a href=&quot;http://www.theatlantic.com/technology/archive/2011/09/so-was-facebook-responsible-for-the-arab-spring-after-all/244314/&quot; rel=&quot;nofollow&quot;&gt;Facebook was used as a strong tool&lt;/a&gt; to spread the voice of the protest, preferred to the controlled mainstream tools such as television and radio.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of which I am made,&lt;br&gt;&#xA;Is the Swan's Song of Stars,&lt;br&gt;&#xA;And it's neighbors,&lt;br&gt;&#xA;they may sound but&lt;br&gt;&#xA;unattractive they are. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No idea for the swan song, but the neighbours are the television and radio, which are &quot;unattractive&quot; for the young people (and also rely a lot on the sound).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4652" LastEditorUserId="4652" LastEditDate="2015-09-10T12:57:47.473" LastActivityDate="2015-09-10T12:57:47.473" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22047" PostTypeId="1" AcceptedAnswerId="22048" CreationDate="2015-09-10T13:58:58.793" Score="1" ViewCount="227" Body="&lt;p&gt;There is a bee and a lizard at the corner of an $l\times b\times h$ cuboid. The bee can fly. The lizard can walk on walls, the ceiling or the floor. Both must reach the diagonally opposite corner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the minimum distance must they cover to do so?&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="15544" LastEditDate="2015-09-10T14:33:50.307" LastActivityDate="2015-09-10T14:36:10.327" Title="Bee and lizard in a room" Tags="|mathematics|geometry|visual|" AnswerCount="1" CommentCount="3" ClosedDate="2015-09-10T18:12:45.513" ContentLicense="CC BY-SA 3.0" />
  <row Id="22048" PostTypeId="2" ParentId="22047" CreationDate="2015-09-10T14:11:29.303" Score="3" Body="&lt;p&gt;Assuming the &quot;diagonally opposite corner&quot; means &quot;the one on the other side of the center of the room&quot; (that is the internal diagonal). We get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The bee must travel the length of the internal diagonal which is $\sqrt{l*l + b*b + h*h}$, because it can go in a straight line&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And (ah, dagn it, I fell right into that one):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lizard: let m1, m2, m3 be the sorted l,b,h. Then $\sqrt{(m1+m2)*(m1+m2) + m3*m3})$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="5610" LastEditDate="2015-09-10T14:36:08.700" LastActivityDate="2015-09-10T14:36:08.700" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22050" PostTypeId="1" AcceptedAnswerId="22055" CreationDate="2015-09-10T14:37:05.223" Score="9" ViewCount="945" Body="&lt;p&gt;Continuing the Movie theme, can you unravel the well-known quotes below?&#xA;I would encourage teamwork with this one.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;code&gt;4:0:0:0:6:1:1:0:3:0:1:1:1:5:4:0:0:2:2:3:0:1:1:0:1:0 (4, 1'2 1 7 2'2 3 2 6 7.)&lt;/code&gt;&lt;br&gt;&#xA;Said by JG (1939)&lt;br /&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;1:0:0:1:2:0:1:1:2:0:2:1:0:1:3:0:0:1:1:1:1:0:0:0:1:0 (4'1 7 2 3, 3.)&lt;/code&gt;&lt;br&gt;&#xA;Said by HB (1942)&lt;br /&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;3:0:0:0:3:3:1:1:4:0:0:2:3:3:2:0:0:1:0:3:0:1:0:0:3:0 (6, 6, 4'2 3 3 2 2 3 2!)&lt;/code&gt;&lt;br&gt;&#xA;Said by KW (1964)&lt;br /&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;0:2:0:3:3:2:0:1:0:0:0:3:0:1:10:2:0:2:3:2:2:0:1:0:3:0 (3'2 4 8 2 4 3 6 5 3!)&lt;/code&gt;&lt;br&gt;&#xA;Said by MC (1969)&lt;br /&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;4:0:1:0:5:3:1:2:2:0:1:0:3:4:2:0:0:2:1:1:1:0:0:0:0:0 (1'1 5 4 3 2 5 2 3'1 6.)&lt;/code&gt;&lt;br&gt;&#xA;Said by MB (1972)&lt;br /&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;3:2:0:1:4:0:3:0:1:0:0:0:0:3:3:0:0:2:0:1:1:0:0:0:1:0 (3'2 5 4 1 6 4.)&lt;/code&gt;&lt;br&gt;&#xA;Said by RS (1975)&lt;br /&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;1:0:0:0:1:0:0:0:1:0:1:1:1:1:2:0:0:0:0:2:1:0:0:0:1:0 (3 6' 2 2?)&lt;/code&gt;&lt;br&gt;&#xA;Said by RdN (1976)&lt;br /&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;1:0:0:2:5:1:1:2:2:0:1:1:0:2:6:0:0:4:2:3:1:0:0:0:1:0 (5 3 3 3 5 3'2 7 3.)&lt;/code&gt;&lt;br&gt;&#xA;Said by AG (1977)&lt;br /&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;3:1:0:0:5:0:1:5:1:0:0:0:1:2:2:0:0:1:4:3:1:1:0:0:3:0 (2'1 3 3 7. 2'1 1 4 7 3!)&lt;/code&gt;&lt;br&gt;&#xA;Said by TJ (1979)&lt;br /&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;0:0:0:0:3:0:0:2:0:0:0:0:1:1:2:1:0:0:0:1:0:0:0:0:0:0 (2 5 4.)&lt;/code&gt;&lt;br&gt;&#xA;Said by DB (1982)&lt;br /&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;4:0:0:2:2:0:1:1:0:0:1:0:2:0:1:0:0:0:0:0:0:0:0:0:2:0 (2 5, 4 2 3.)&lt;/code&gt;&lt;br&gt;&#xA;Said by CE (1983)&lt;br /&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;1:0:0:1:3:1:0:1:2:0:0:4:1:1:2:0:0:1:1:3:0:0:0:0:2:0 (3 5 2 2 6 6.)&lt;/code&gt;&lt;br&gt;&#xA;Said by AP (1983)&lt;br /&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;1:2:1:0:1:0:0:0:1:0:1:2:0:0:0:0:0:0:0:0:0:0:0:0:0:0 (1'2 2 4.)&lt;/code&gt;&lt;br&gt;&#xA;Said by AS (1984)&lt;br /&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;4:0:6:0:0:1:1:6:4:0:1:2:2:3:3:2:0:0:2:4:0:0:2:0:2:0 (3 3 5 3 4 9 10 8.)&lt;/code&gt;&lt;br&gt;&#xA;Said by PM (1984)&lt;br /&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;1:0:0:3:7:0:2:1:1:0:0:0:0:3:3:0:0:3:1:1:0:0:3:0:0:0 (5 2'2 5 2 3'1 4 5)&lt;/code&gt;&lt;br&gt;&#xA;Said by CL (1985)&lt;br /&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;2:1:1:3:3:0:0:0:3:0:0:5:0:2:5:4:0:0:3:5:5:0:1:0:1:0 (2 4 3 6 5 2 2 6 2 6 6!)&lt;/code&gt;&lt;br&gt;&#xA;Said by JLC (1988)&lt;br /&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;0:1:1:1:2:1:0:1:4:0:0:3:1:0:2:0:0:0:0:1:2:0:1:0:1:0 (2 3 5 2, 2 4 4.)&lt;/code&gt;&lt;br&gt;&#xA;Said by RL (1989)&lt;br /&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;5:1:1:3:8:0:0:1:4:0:1:1:2:1:3:1:0:5:3:2:1:1:0:1:4:1 (5 4. 5 3 3, 4. 4 4 5 13.)&lt;/code&gt;&lt;br&gt;&#xA;Said by RW (1989)&lt;br /&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;7:1:2:1:5:1:0:3:7:0:0:1:1:4:1:0:0:1:3:3:0:2:1:0:0:0 (1 3 3 5 4 4 4 5 3 1 4 7.)&lt;/code&gt;&lt;br&gt;&#xA;Said by AH (1991)&lt;br /&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;2:0:1:1:2:0:0:3:0:0:0:1:0:2:1:0:0:1:0:4:2:0:0:0:1:0 (3 3'1 6 3 5.)&lt;/code&gt;&lt;br&gt;&#xA;Said by JN (1992)&lt;br /&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;0:2:0:1:2:0:4:0:3:0:0:1:0:2:1:0:0:1:2:2:2:1:0:0:3:0 (3 4 6, 2 3 4 5.)&lt;/code&gt;&lt;br&gt;&#xA;Said by TR (1994)&lt;br /&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;4:0:1:5:6:1:5:1:4:0:0:0:0:8:9:0:0:1:2:3:0:1:0:0:1:0 (2 4 1 3'1 3 2, 4 9, 4 7, 3 4 5!)&lt;/code&gt;&lt;br&gt;&#xA;Said by JC (1998)&lt;br /&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;1:0:0:2:5:0:0:0:1:0:0:1:0:0:1:2:0:0:1:0:0:0:0:0:0:0 (1 3 4 6.)&lt;/code&gt;&lt;br&gt;&#xA;Said by HJO (1999)&lt;br /&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;0:1:1:1:3:1:0:1:0:0:0:1:0:1:1:0:0:2:1:1:1:1:0:0:2:0 (9 4, 5.)&lt;/code&gt;&lt;br&gt;&#xA;Said by TC (2002)&lt;br /&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;1:0:0:2:0:2:0:0:8:0:2:9:0:2:6:0:0:1:0:0:3:0:3:0:3:0 (1 4 4 3 3, 1 4 4 3 3 1 4 4 3.)&lt;/code&gt;&lt;br&gt;&#xA;Said by LN (2008)&lt;br /&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;0:0:0:0:1:0:0:1:1:0:0:0:0:0:2:0:0:1:3:0:1:0:1:0:1:0 (3 2 7?)&lt;/code&gt;&lt;br&gt;&#xA;Said by HL (2008)&lt;br /&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;3:0:1:1:2:6:2:4:3:0:1:3:2:4:7:0:0:1:3:3:3:0:0:0:3:0 (2 2 4, 4 2 4 6 7! 2'1 2 6, 1'1 5 3!)&lt;/code&gt;&lt;br&gt;&#xA;Said by EF (2010)&lt;br /&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;1:0:0:0:0:0:1:0:1:0:0:0:1:0:2:0:0:1:0:1:0:0:0:0:0:0 (1 2 5.)&lt;/code&gt;&lt;br&gt;&#xA;Said by VD (2014)&lt;br /&gt;  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="15544" LastEditDate="2015-09-10T15:32:02.243" LastActivityDate="2015-09-10T17:53:15.673" Title="Well-known quotations" Tags="|knowledge|movies|anagram|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22051" PostTypeId="1" AcceptedAnswerId="23815" CreationDate="2015-09-10T14:37:25.070" Score="9" ViewCount="794" Body="&lt;p&gt;Here's my first puzzle attributed to the fortnightly topic, &lt;a href=&quot;/questions/tagged/movies&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;movies&amp;#39;&quot; rel=&quot;tag&quot;&gt;movies&lt;/a&gt;. I figured I'd try my hand at this series of pattern recognition puzzles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a chart listing words that fall into one or neither of two categories, Spiel Words™ and Spieg Words™ (though it should be noted that all Spieg Words™ are also Spiel Words™, they are just a more specific categorization). Mathematically speaking, the set of all Spieg Words is a subset of the set of all Spiel Words&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Spieg Words™:&lt;/h3&gt;  &#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Cow&lt;br&gt;&#xA;  Law&lt;br&gt;&#xA;  Urn&lt;br&gt;&#xA;  Dear&lt;br&gt;&#xA;  Hear&lt;br&gt;&#xA;  Lamb&lt;br&gt;&#xA;  Lock&lt;br&gt;&#xA;  Made&lt;br&gt;&#xA;  Pear&lt;br&gt;&#xA;  Week&lt;br&gt;&#xA;  West&lt;br&gt;&#xA;  Apart&lt;br&gt;&#xA;  Liver&lt;br&gt;&#xA;  Ridge&lt;br&gt;&#xA;  Round&lt;br&gt;&#xA;  People&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;Spiel Words™:&lt;/h3&gt;  &#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Amp&lt;br&gt;&#xA;  Far&lt;br&gt;&#xA;  Red&lt;br&gt;&#xA;  Two&lt;br&gt;&#xA;  Atom&lt;br&gt;&#xA;  East&lt;br&gt;&#xA;  Find&lt;br&gt;&#xA;  Lace&lt;br&gt;&#xA;  Lose&lt;br&gt;&#xA;  Clock&lt;br&gt;&#xA;  Close&lt;br&gt;&#xA;  Range&lt;br&gt;&#xA;  South&lt;br&gt;&#xA;  Versa&lt;br&gt;&#xA;  Ground&lt;br&gt;&#xA;  Tenant&lt;br&gt;&#xA;  Picture  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;Neither&lt;/h3&gt;  &#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Oil&lt;br&gt;&#xA;  Buck&lt;br&gt;&#xA;  Dean&lt;br&gt;&#xA;  Dirt&lt;br&gt;&#xA;  Sear&lt;br&gt;&#xA;  Vice&lt;br&gt;&#xA;  Block&lt;br&gt;&#xA;  Crazy&lt;br&gt;&#xA;  Filth&lt;br&gt;&#xA;  Irate&lt;br&gt;&#xA;  North&lt;br&gt;&#xA;  Shark&lt;br&gt;&#xA;  Thief&lt;br&gt;&#xA;  Energy&lt;br&gt;&#xA;  Figure&lt;br&gt;&#xA;  Future&lt;br&gt;&#xA;  Zombie&lt;br&gt;&#xA;  Abridge&lt;br&gt;&#xA;  Electricity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can you determine the rule I use to decide if I call it a Spiel Word™, a Spieg Word™, or neither? Feel free to ask below for certain words, but I ask that to avoid spamming up the question, if you have multiple, please ask them all at once.  &lt;/p&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="30633" LastEditDate="2016-11-08T00:04:13.783" LastActivityDate="2016-11-08T00:04:13.783" Title="What is a Spiel/Spieg Word™?" Tags="|pattern|movies|word-property|" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="22052" PostTypeId="2" ParentId="22050" CreationDate="2015-09-10T14:41:09.863" Score="3" Body="&lt;ol start=&quot;7&quot;&gt;&#xA;&lt;li&gt;&lt;code&gt;1:0:0:0:1:0:0:0:1:0:1:1:1:1:2:0:0:0:0:2:1:0:0:0:1:0 (3 6' 2 2?)&#xA;Said by RdN (1976)&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This would be &lt;code&gt;You talkin' to me?&lt;/code&gt; by Robert de Niro giving the following pieces of information:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the initials are of the actor, not the character&lt;/li&gt;&#xA;&lt;li&gt;the parentheses are word length and the presence of apostrophes&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Therefore:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;code&gt;4:0:0:0:6:1:1:0:3:0:1:1:1:5:4:0:0:2:2:3:0:1:1:0:1:0 (4, 1'2 1 7 2'2 3 2 6 7.)&#xA;Said by JG (1939)&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;Toto, I've a feeling we're not in Kansas anymore.&lt;/code&gt; by Judy Garland&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;23&quot;&gt;&#xA;&lt;li&gt;&lt;code&gt;1:0:0:2:5:0:0:0:1:0:0:1:0:0:1:2:0:0:1:0:0:0:0:0:0:0 (1 3 4 6.)&#xA;Said by HJO (1999)&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;I see dead people.&lt;/code&gt; by Haley J Osment&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1509" LastEditorUserId="15544" LastEditDate="2015-09-10T15:37:46.493" LastActivityDate="2015-09-10T15:37:46.493" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22053" PostTypeId="2" ParentId="22050" CreationDate="2015-09-10T14:52:19.337" Score="2" Body="&lt;pre&gt;&lt;code&gt;25)  1:0:0:2:0:2:0:0:8:0:2:9:0:2:6:0:0:1:0:0:3:0:3:0:3:0 (1 4 4 3 3, 1 4 4 3 3 1 4 4 3.)&#xA;Said by LN (2008)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Is Liam Neeson with:&lt;br&gt;&#xA;    I will look for you, I will find you and I will kill you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14461" LastEditorUserId="15544" LastEditDate="2015-09-10T15:38:40.030" LastActivityDate="2015-09-10T15:38:40.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22054" PostTypeId="2" ParentId="22050" CreationDate="2015-09-10T14:53:32.557" Score="6" Body="&lt;p&gt;The key is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The 26 numbers denote the frequency of each letter in the quote - the following bracketed numbers are the number of letters in each word with added punctuation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;13&quot;&gt;&#xA;&lt;li&gt;&lt;code&gt;1:2:1:0:1:0:0:0:1:0:1:2:0:0:0:0:0:0:0:0:0:0:0:0:0:0 (1'2 2 4.)&#xA;Said by AS (1984)&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I'll be back - Arnold Schwarzenegger (Terminator)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;23&quot;&gt;&#xA;&lt;li&gt;&lt;code&gt;1:0:0:2:5:0:0:0:1:0:0:1:0:0:1:2:0:0:1:0:0:0:0:0:0:0 (1 3 4 6.)&#xA;Said by HJO (1999)&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I see dead people - Hayley Joel Osment (The Sixth Sense)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;28&quot;&gt;&#xA;&lt;li&gt;&lt;code&gt;1:0:0:0:0:0:1:0:1:0:0:0:1:0:2:0:0:1:0:1:0:0:0:0:0:0 (1 2 5.)&#xA;Said by VD (2014)&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I am groot - Vin Diesel (Guardians of the Galaxy)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;10&quot;&gt;&#xA;&lt;li&gt;&lt;code&gt;0:0:0:0:3:0:0:2:0:0:0:0:1:1:2:1:0:0:0:1:0:0:0:0:0:0 (2 5 4.)&#xA;Said by DB (1982)&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;ET Phone Home - Drew Barrymore (E.T.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;26&quot;&gt;&#xA;&lt;li&gt;&lt;code&gt;0:0:0:0:1:0:0:1:1:0:0:0:0:0:2:0:0:1:3:0:1:0:1:0:1:0 (3 2 7?)&#xA;Said by HL (2008)&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Why So Serious? - Heath Ledger (The Dark Knight)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;18&quot;&gt;&#xA;&lt;li&gt;&lt;code&gt;5:1:1:3:8:0:0:1:4:0:1:1:2:1:3:1:0:5:3:2:1:1:0:1:4:1 (5 4. 5 3 3, 4. 4 4 5 13.)&#xA;Said by RW (1989)&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Carpe diem. Seize the day, boys. Make your lives extraordinary. - Robin Williams (Dead Poet's Society)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;24&quot;&gt;&#xA;&lt;li&gt;&lt;code&gt;0:1:1:1:3:1:0:1:0:0:0:1:0:1:1:0:0:2:1:1:1:1:0:0:2:0 (9 4, 5.)&#xA;Said by TC (2002)&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everybody runs, Fletch. - Tom Cruise (Minority Report)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;21&quot;&gt;&#xA;&lt;li&gt;&lt;code&gt;0:2:0:1:2:0:4:0:3:0:0:1:0:2:1:0:0:1:2:2:2:1:0:0:3:0 (3 4 6, 2 3 4 5.)&#xA;Said by TR (1994)&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Get busy living, or get busy dying. - Tim Robbins (Shawshank Redemption)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;12&quot;&gt;&#xA;&lt;li&gt;&lt;code&gt;4:0:0:2:2:0:1:1:0:0:1:0:2:0:1:0:0:0:0:0:0:0:0:0:2:0 (2 5, 4 2 3.)&#xA;Said by CE (1983)&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Go ahead, make my day. - Clint Eastwood (Sudden Impact)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12206" LastEditorUserId="15544" LastEditDate="2015-09-10T15:34:41.243" LastActivityDate="2015-09-10T15:34:41.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22055" PostTypeId="2" ParentId="22050" CreationDate="2015-09-10T14:54:53.943" Score="7" Body="&lt;p&gt;So the rules are (thanks to Kate Gregory and Deusovi for these):&lt;br&gt;&#xA;1. The numbers separated by colons indicate the number of occurrences of each letter of the alphabet in the quote (the first being A and the last being Z).&lt;br&gt;&#xA;2. The numbers in parentheses indicate the length of each word in the quote, along with punctuation.&lt;br&gt;&#xA;3. The second line indicates the initials of the actor/actress attributed with the quote alongside the year of the release of the movie containing the quote.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;4:0:0:0:6:1:1:0:3:0:1:1:1:5:4:0:0:2:2:3:0:1:1:0:1:0 (4, 1'2 1 7 2'2 3 2 6 7.)&#xA;Said by JG (1939)&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Toto, I've a feeling we're not in Kansas anymore.&quot; The Wizard of Oz 1939 Judy Garland (Kate Gregory)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;1:0:0:1:2:0:1:1:2:0:2:1:0:1:3:0:0:1:1:1:1:0:0:0:1:0 (4'1 7 2 3, 3.)&#xA;Said by HB (1942)&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Here's looking at you, kid.&quot; Casablanca 1942 Humphrey Bogart&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;3:0:0:0:3:3:1:1:4:0:0:2:3:3:2:0:0:1:0:3:0:1:0:0:3:0 (6, 6, 4'2 3 3 2 2 3 2!)&#xA;Said by KW (1964)&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Infamy, infamy, they've all got it in for me!&quot; Carry On Cleo 1964 Kenneth Williams&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;0:2:0:3:3:2:0:1:0:0:0:3:0:1:10:2:0:2:3:2:2:0:1:0:3:0 (3'2 4 8 2 4 3 6 5 3!)&#xA;Said by MC (1969)&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;You're only supposed to blow the bloody doors off!&quot; The Italian Job 1969 Michael Caine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;4:0:1:0:5:3:1:2:2:0:1:0:3:4:2:0:0:2:1:1:1:0:0:0:0:0 (1'1 5 4 3 2 5 2 3'1 6.)&#xA;Said by MB (1972)&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;I'm gonna make him an offer he can't refuse.&quot; The Godfather 1972 Marlon Brando&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;3:2:0:1:4:0:3:0:1:0:0:0:0:3:3:0:0:2:0:1:1:0:0:0:1:0 (3'2 5 4 1 6 4.)&#xA;Said by RS (1975)&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;You're gonna need a bigger boat.&quot; Jaws 1975 Roy Scheider&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;1:0:0:0:1:0:0:0:1:0:1:1:1:1:2:0:0:0:0:2:1:0:0:0:1:0 (3 6' 2 2?)&#xA;Said by RdN (1976)&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;You talkin' to me?&quot; Taxi Driver 1976 Robert deNiro (Kate Gregory)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;1:0:0:2:5:1:1:2:2:0:1:1:0:2:6:0:0:4:2:3:1:0:0:0:1:0 (5 3 3 3 5 3'2 7 3.)&#xA;Said by AG (1977)&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;These are not the droids you're looking for.&quot; Star Wars 1977 Alec Guinness (corsiKa)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;3:1:0:0:5:0:1:5:1:0:0:0:1:2:2:0:0:1:4:3:1:1:0:0:3:0 (2'1 3 3 7. 2'1 1 4 7 3!)&#xA;Said by TJ (1979)&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;He's not the Messiah. He's a very naughty boy!&quot; Monty Python's Life of Brian 1979 Terry Jones&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;0:0:0:0:3:0:0:2:0:0:0:0:1:1:2:1:0:0:0:1:0:0:0:0:0:0 (2 5 4.)&#xA;Said by DB (1982)&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;ET Phone Home - E.T. 1982 Drew Barrymore (alexmc)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;4:0:0:2:2:0:1:1:0:0:1:0:2:0:1:0:0:0:0:0:0:0:0:0:2:0 (2 5, 4 2 3.)&#xA;Said by CE (1983)&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Go ahead, make my day.&quot; Sudden Impact 1983 Clint Eastwood&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;1:0:0:1:3:1:0:1:2:0:0:4:1:1:2:0:0:1:1:3:0:0:0:0:2:0 (3 5 2 2 6 6.)&#xA;Said by AP (1983)&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Say hello to my little friend.&quot; Scarface 1983 Al Pacino (How did I not get that sooner!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;1:2:1:0:1:0:0:0:1:0:1:2:0:0:0:0:0:0:0:0:0:0:0:0:0:0 (1'2 2 4.)&#xA;Said by AS (1984)&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;I'll be back&quot; The Terminator 1984 Arnold Schwarzenegger (alexmc)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;4:0:6:0:0:1:1:6:4:0:1:2:2:3:3:2:0:0:2:4:0:0:2:0:2:0 (3 3 5 3 4 9 10 8.)&#xA;Said by PM (1984)&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Man who catch fly with chopstick accomplish anything.&quot; Karate Kid 1984 Pat Morita&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;1:0:0:3:7:0:2:1:1:0:0:0:0:3:3:0:0:3:1:1:0:0:3:0:0:0 (5 2'2 5 2 3'1 4 5)&#xA;Said by CL (1985)&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Where we're going we don't need roads&quot; Back to the Future 1985 Christopher Lloyd&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;2:1:1:3:3:0:0:0:3:0:0:5:0:2:5:4:0:0:3:5:5:0:1:0:1:0 (2 4 3 6 5 2 2 6 2 6 6!)&#xA;Said by JLC (1988)&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;To call you stupid would be an insult to stupid people!&quot; A Fish Called Wanda 1988 Jamie Lee Curtis&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;0:1:1:1:2:1:0:1:4:0:0:3:1:0:2:0:0:0:0:1:2:0:1:0:1:0 (2 3 5 2, 2 4 4.)&#xA;Said by RL (1989)&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;If you build it, he will come.&quot; Field of Dreams 1989 Ray Liotta&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;5:1:1:3:8:0:0:1:4:0:1:1:2:1:3:1:0:5:3:2:1:1:0:1:4:1 (5 4. 5 3 3, 4. 4 4 5 13.)&#xA;Said by RW (1989)&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Carpe diem. Seize the day, boys. Make your lives extraordinary.&quot; Dead Poet's Society 1989 Robin Williams (alexmc)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;7:1:2:1:5:1:0:3:7:0:0:1:1:4:1:0:0:1:3:3:0:2:1:0:0:0 (1 3 3 5 4 4 4 5 3 1 4 7.)&#xA;Said by AH (1991)&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;I ate his liver with some fava beans and a nice chianti.&quot; The Silence of the Lambs 1991 Anthony Hopkins&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;2:0:1:1:2:0:0:3:0:0:0:1:0:2:1:0:0:1:0:4:2:0:0:0:1:0 (3 3'1 6 3 5.)&#xA;Said by JN (1992)&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;You can't handle the truth.&quot; A Few Good Men 1992 Jack Nicholson&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;0:2:0:1:2:0:4:0:3:0:0:1:0:2:1:0:0:1:2:2:2:1:0:0:3:0 (3 4 6, 2 3 4 5.)&#xA;Said by TR (1994)&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Get busy living, or get busy dying.&quot; Shawshank Redemption 1994 Tim Robbins (alexmc)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;4:0:1:5:6:1:5:1:4:0:0:0:0:8:9:0:0:1:2:3:0:1:0:0:1:0 (2 4 1 3'1 3 2, 4 9, 4 7, 3 4 5!)&#xA;Said by JC (1998)&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;In case I don't see ya, good afternoon, good evening, and good night!&quot; The Truman Show 1998 Jim Carrey&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;1:0:0:2:5:0:0:0:1:0:0:1:0:0:1:2:0:0:1:0:0:0:0:0:0:0 (1 3 4 6.)&#xA;Said by HJO (1999)&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;I see dead people.&quot; The Sixth Sense 1999 Haley Joel Osment (Kate Gregory)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;0:1:1:1:3:1:0:1:0:0:0:1:0:1:1:0:0:2:1:1:1:1:0:0:2:0 (9 4, 5.)&#xA;Said by TC (2002)&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Everybody runs, Fletch.&quot; Minority Report 2002 Tom Cruise (alexmc)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;1:0:0:2:0:2:0:0:8:0:2:9:0:2:6:0:0:1:0:0:3:0:3:0:3:0 (1 4 4 3 3, 1 4 4 3 3 1 4 4 3.)&#xA;Said by LN (2008)&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;I will look for you, I will find you and I will kill you.&quot; Taken 2008 Liam Neeson (Wa Kai)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;0:0:0:0:1:0:0:1:1:0:0:0:0:0:2:0:0:1:3:0:1:0:1:0:1:0 (3 2 7?)&#xA;Said by HL (2008)&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Why so serious?&quot; Batman:The Dark Knight 2008 Heath Ledger (alexmc)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;3:0:1:1:2:6:2:4:3:0:1:3:2:4:7:0:0:1:3:3:3:0:0:0:3:0 (2 2 4, 4 2 4 6 7! 2'1 2 6, 1'1 5 3!)&#xA;Said by EF (2010)&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Oh my gosh, look at that fluffy unicorn! He's so fluffy, I'm gonna die!&quot; Despicable Me 2010 Elsie Fisher&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;1:0:0:0:0:0:1:0:1:0:0:0:1:0:2:0:0:1:0:1:0:0:0:0:0:0 (1 2 5.)&#xA;Said by VD (2014)&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;I am groot&quot; Guardians of the Galaxy 2014 Vin Diesel (alexmc)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="13970" LastEditDate="2015-09-10T17:53:15.673" LastActivityDate="2015-09-10T17:53:15.673" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22056" PostTypeId="2" ParentId="22040" CreationDate="2015-09-10T15:23:39.217" Score="8" Body="&lt;p&gt;My first attempt:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jim and Rose has already starting the game the moment the conversation starts, and the rule is to continue the conversation in a way that, when reading backward, it makes sense.  They were both doing fine until Rose's last sentence, which loses her the game.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2015-09-10T15:23:39.217" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22057" PostTypeId="1" AcceptedAnswerId="22062" CreationDate="2015-09-10T15:35:14.153" Score="36" ViewCount="2178" Body="&lt;p&gt;During rand al'thor's unscheduled sabbatical from Puzzling, he's managed to keep himself busy. How, you ask? Why, by making movies, of course! Unfortunately, our resident Puzzling expert is not quite as adept at movie-making, so he's stolen a few ideas from Hollywood. Can you parse through his ten movie pitches and determine their titles?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Pitch #1:&lt;/strong&gt; After his mother's tragic death, rand al'thor takes a tumble to a land he only remembers from his dreams. With his mother's final directions in mind, he sets off on his journey to discover his true destiny, making new friends and meeting old friends along the way.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Pitch #2:&lt;/strong&gt; After his wife's tragic death, rand al'thor communicates with her spirit from beyond the grave, teaching him along the way that perhaps his village's greatest enemies could become their greatest allies and directing him to the true enemy's lair, where he discovers the daughter he thought he had lost.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Pitch #3:&lt;/strong&gt; Rand al'thor, a psychologist, moves to a cottage and strives for love alongside his sister. Rand al'thor attempts to help a young boy, though as his life constantly shifts it becomes apparent that the young boy was helping him all along.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Pitch #4:&lt;/strong&gt; Rand al'thor, a space bandit, is abducted by aliens along with the rest of the cast of his once-hit sci-fi show. Believing them to be real space heroes, the aliens take rand and his crew on an adventure across space to protect it from a powerful villain and an even more powerful stone.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Pitch #5:&lt;/strong&gt; Rand al'thor stars as a young, female gang member, who finds herself part of a street brawl with fatal consequences. As she moves away from her hometown, she and other members of her gang are forced to grapple with their internal emotional battle as well as the consequences of their violent pasts.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Pitch #6:&lt;/strong&gt; After an innocuous conversation on a plane, rand al'thor finds himself wrapped up in an international escapade to stop the plans of a raving madman who desires a world of utter chaos. Rand dons his alter-ego as he attempts to save his city and avoid the fall his superiors have set him up to take.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Pitch #7:&lt;/strong&gt; After an unwarranted visit from a wise, old man, rand al'thor finds himself on an adventure deep underground, where he's met with orcs, goblins, and even dinosaurs. The adventure takes a treacherous turn when rand discovers a powerful relic, leading him to realize that he and his party must escape the cavern or they will be consumed by lava.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Pitch #8:&lt;/strong&gt; Though rand al'thor has spent years being ignored by scientists, his greatest fears prove correct as he finds himself transported to a futuristic city filled with massive robots and sleek buildings. The world is soon consumed by an environmental superstorm, leading rand al'thor and his new scientist ally to trek through the catastrophe in search of his son and his friends.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Pitch #9:&lt;/strong&gt; After narrowly escaping a run in with a familial zombie, rand al'thor takes a job at a boarding school priding itself on ancient tradition. Using unusual techniques, rand convinces the boys in his class to band together and fight off the undead hordes that surround them.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Pitch #10:&lt;/strong&gt; A comical menagerie led by rand al'thor set off to film a movie on an ominous, uncharted island. The group is faced with immense challenges along the way, including dinosaurs and a massive ape and climaxing with an encounter with rand's wicked uncle, whom rand must overcome to reclaim his rightful title.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Rand, if you see this and you're displeased by the use of your name, just ping me in chat and let me know. I'll edit it right out. I figured you'd find this amusing, though! :)&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-09-11T16:21:00.777" Title="Expert puzzler, apprentice movie-maker" Tags="|knowledge|movies|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22062" PostTypeId="2" ParentId="22057" CreationDate="2015-09-10T15:58:23.137" Score="19" Body="&lt;p&gt;The movie pitches are all:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A combination of the plots of two films. The names of those two films share a word or part of a word that allows them to overlap (e.g. &lt;em&gt;Alice in Wonderland&lt;/em&gt; overlaps with &lt;em&gt;Land Before Time&lt;/em&gt; to give &lt;em&gt;Alice in Wonder&lt;strong&gt;land&lt;/strong&gt; Before Time&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With quite a lot of the help from the community, here are answers to all of them:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pitch #1: After his mother's tragic death, rand al'thor takes a tumble to a land he only remembers from his dreams. With his mother's final directions in mind, he sets off on his journey to discover his true destiny, making new friends and meeting old friends along the way.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alice in Wonder&lt;strong&gt;land&lt;/strong&gt; Before Time (Aggie Kidd / user14418)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pitch #2: After his wife's tragic death, rand al'thor communicates with her spirit from beyond the grave, teaching him along the way that perhaps his village's greatest enemies could become their greatest allies and directing him to the true enemy's lair, where he discovers the daughter he thought he had lost.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; How to Train Your &lt;strong&gt;Dragon&lt;/strong&gt;fly (loopsgald)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pitch #3: Rand al'thor, a psychologist, moves to a cottage and strives for love alongside his sister. Rand al'thor attempts to help a young boy, though as his life constantly shifts it becomes apparent that the young boy was helping him all along.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Sixth &lt;strong&gt;Sense&lt;/strong&gt; and Sensibility (NeedAName)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pitch #4: Rand al'thor, a space bandit, is abducted by aliens along with the rest of the cast of his once-hit sci-fi show. Believing them to be real space heroes, the aliens take rand and his crew on an adventure across space to protect it from a powerful villain and an even more powerful stone.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Guardians of the &lt;strong&gt;Galaxy&lt;/strong&gt; Quest (Lampost42)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pitch #5: Rand al'thor stars as a young, female gang member, who finds herself part of a street brawl with fatal consequences. As she moves away from her hometown, she and other members of her gang are forced to grapple with their internal emotional battle as well as the consequences of their violent pasts.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Inside &lt;strong&gt;Out&lt;/strong&gt;siders&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pitch #6: After an innocuous conversation on a plane, rand al'thor finds himself wrapped up in an international escapade to stop the plans of a raving madman who desires a world of utter chaos. Rand dons his alter-ego as he attempts to save his city and avoid the fall his superiors have set him up to take.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Dark &lt;strong&gt;Knight&lt;/strong&gt; and Day (chsanno)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pitch #7: After an unwarranted visit from a wise, old man, rand al'thor finds himself on an adventure deep underground, where he's met with orcs, goblins, and even dinosaurs. The adventure takes a treacherous turn when rand discovers a powerful relic, leading him to realize that he and his party must escape the cavern or they will be consumed by lava.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Unexpected &lt;strong&gt;Journey&lt;/strong&gt; to the Centre of the Earth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pitch #8: Though rand al'thor has spent years being ignored by scientists, his greatest fears prove correct as he finds himself transported to a futuristic city filled with massive robots and sleek buildings. The world is soon consumed by an environmental superstorm, leading rand al'thor and his new scientist ally to trek through the catastrophe in search of his son and his friends.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Day After &lt;strong&gt;Tomorrow&lt;/strong&gt;land&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pitch #9: After narrowly escaping a run in with a familial zombie, rand al'thor takes a job at a boarding school priding itself on ancient tradition. Using unusual techniques, rand convinces the boys in his class to band together and fight off the undead hordes that surround them.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dawn of the &lt;strong&gt;Dead&lt;/strong&gt; Poets' Society (raisinghellyer)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pitch #10: A comical menagerie led by rand al'thor set off to film a movie on an ominous, uncharted island. The group is faced with immense challenges along the way, including dinosaurs and a massive ape and climaxing with an encounter with rand's wicked uncle, whom rand must overcome to reclaim his rightful title.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Lion &lt;strong&gt;King&lt;/strong&gt; Kong&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="14135" LastEditDate="2015-09-11T16:21:00.777" LastActivityDate="2015-09-11T16:21:00.777" CommentCount="20" ContentLicense="CC BY-SA 3.0" />
  <row Id="22063" PostTypeId="2" ParentId="22057" CreationDate="2015-09-10T16:34:38.403" Score="4" Body="&lt;p&gt;Unfinished&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pitch #2: After his wife's tragic death, rand al'thor communicates with her spirit from beyond the grave, teaching him along the way that perhaps his village's greatest enemies could become their greatest allies and directing him to the true enemy's lair, where he discovers the daughter he thought he had lost.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; How to Train Your Dragonfly&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pitch #3: Rand al'thor, a psychologist, moves to a cottage and strives for love alongside his sister. Rand al'thor attempts to help a young boy, though as his life constantly shifts it becomes apparent that the young boy was helping him all along.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Sixth Sense and Sensibility&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pitch #4: Rand al'thor, a space bandit, is abducted by aliens along with the rest of the cast of his once-hit sci-fi show. Believing them to be real space heroes, the aliens take rand and his crew on an adventure across space to protect it from a powerful villain and an even more powerful stone.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Guardians of the Galaxy Quest&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pitch #5: Rand al'thor stars as a young, female gang member, who finds herself part of a street brawl with fatal consequences. As she moves away from her hometown, she and other members of her gang are forced to grapple with their internal emotional battle as well as the consequences of their violent pasts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pitch #6: After an innocuous conversation on a plane, rand al'thor finds himself wrapped up in an international escapade to stop the plans of a raving madman who desires a world of utter chaos. Rand dons his alter-ego as he attempts to save his city and avoid the fall his superiors have set him up to take.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Dark Knight and Day&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pitch #10: A comical menagerie led by rand al'thor set off to film a movie on an ominous, uncharted island. The group is faced with immense challenges along the way, including dinosaurs and a massive ape and climaxing with an encounter with rand's wicked uncle, whom rand must overcome to reclaim his rightful title.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Lion King Kong&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15990" LastEditorUserId="15990" LastEditDate="2015-09-10T18:52:02.820" LastActivityDate="2015-09-10T18:52:02.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22064" PostTypeId="1" AcceptedAnswerId="22070" CreationDate="2015-09-10T18:06:02.910" Score="20" ViewCount="1333" Body="&lt;p&gt;Fill a $4\times4$ grid with positive integers so that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Every cell has a different integer&lt;/li&gt;&#xA;&lt;li&gt;The &lt;em&gt;product&lt;/em&gt; of the numbers in each row is $5040$, and similarly for the columns&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Source&lt;/em&gt;: This was an NPR weekly listener challenge, aired on 2005-10-09. See &lt;a href=&quot;http://www.taterenner.com/nprmps.htm&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="11100" LastEditDate="2015-09-10T22:51:21.693" LastActivityDate="2015-09-11T08:03:55.677" Title="The 5040 Square" Tags="|magic-square|construction|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22067" PostTypeId="2" ParentId="22057" CreationDate="2015-09-10T18:40:44.847" Score="7" Body="&lt;p&gt;I can't comment yet so I'm just posting this as an answer to help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pitch #2: After his wife's tragic death, rand al'thor communicates with her spirit from beyond the grave, teaching him along the way that perhaps his village's greatest enemies could become their greatest allies and directing him to the true enemy's lair, where he discovers the daughter he thought he had lost.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;something&lt;/em&gt; Dragonfly or Dragonfly &lt;em&gt;something&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With help from alexmc in comments Pitch 2 is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; How to train your Dragonfly&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16292" LastEditorUserId="16292" LastEditDate="2015-09-10T18:57:09.480" LastActivityDate="2015-09-10T18:57:09.480" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22068" PostTypeId="1" CreationDate="2015-09-10T19:01:01.447" Score="10" ViewCount="695" Body="&lt;p&gt;So I recently created a crossword for you guys, but I misplaced everything except my list of words! Being too busy working on movie-related puzzles to recreate it, I asked some help from my friends (you guys know how I love layers in my puzzles).&lt;/p&gt;&#xA;&lt;p&gt;First, I had my friend Al Literation (you may remember him) write the clues based on my words.&lt;/p&gt;&#xA;&lt;p&gt;Then I gave the vertical answers to my friend Anna Graham to write down in the crossword image.&lt;/p&gt;&#xA;&lt;p&gt;Finally I gave the horizontal words to my amigo Juan Hamming-Distans (what's that, a stretch you say? Get over it, a theme is a theme) so he could write them in the crossword.&lt;/p&gt;&#xA;&lt;p&gt;Anna and Juan each altered the words they were given to new words (their transformations did not alter the length of my words) and redrew the crossword so that they could make them fit together. That's what you see below. You can find Al's clues below; for each one, determine my original word and, based on whether they're down or across, figure out what Anna or Juan did to them before placing them in the crossword (there will almost always be only one possibility). All words (my original ones and the modified ones) can be found on dictionary.com&lt;/p&gt;&#xA;&lt;h3&gt;Crossword&lt;/h3&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ai9DD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ai9DD.png&quot; alt=&quot;Crossword&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th&gt;Horizontal&lt;/th&gt;&#xA;&lt;th&gt;Vertical&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;4. Battlewise bravery&lt;/td&gt;&#xA;&lt;td&gt;1. Shadowy Sherlock&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;5. Sans sensation&lt;/td&gt;&#xA;&lt;td&gt;2. Charismatic quality&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;8. Medical mesh material&lt;/td&gt;&#xA;&lt;td&gt;3. Hired helping hand&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;10. Anaerobically assembled acid&lt;/td&gt;&#xA;&lt;td&gt;6. Sprinkles salvation&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;12. Opposite of old origin offensive&lt;/td&gt;&#xA;&lt;td&gt;7. Pixelated puerile protagonist&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;13. Pertaining preposition&lt;/td&gt;&#xA;&lt;td&gt;9. Cosmological cloud&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;15. Sound system segment&lt;/td&gt;&#xA;&lt;td&gt;11. Dull discomforts&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;20. Decided determination&lt;/td&gt;&#xA;&lt;td&gt;14. Rising rocket requirement&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;21. Observe outlook on oracular orbs&lt;/td&gt;&#xA;&lt;td&gt;16. Ambush assault&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;22. Rifle rest&lt;/td&gt;&#xA;&lt;td&gt;17. Spear's similar&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;23. Devilish, deemed Dante&lt;/td&gt;&#xA;&lt;td&gt;18. Deviate direction&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;25. Constant current classification&lt;/td&gt;&#xA;&lt;td&gt;19. Accomplish again&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;28. Vowelless vituperation&lt;/td&gt;&#xA;&lt;td&gt;24. Metal mixture&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;29. African antelope&lt;/td&gt;&#xA;&lt;td&gt;26. These three things&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;30. Painlessly punish progeny&lt;/td&gt;&#xA;&lt;td&gt;27. Scared swallow&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;" OwnerUserId="13970" LastEditorUserId="69582" LastEditDate="2021-01-27T16:10:53.900" LastActivityDate="2021-01-27T21:57:44.100" Title="Crossword With Friends" Tags="|anagram|crosswords|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="22069" PostTypeId="2" ParentId="22057" CreationDate="2015-09-10T19:19:14.183" Score="3" Body="&lt;p&gt;Pitch 1 seems to be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alice in Wonderland Before Time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14418" LastEditorUserId="11475" LastEditDate="2015-09-10T19:20:26.113" LastActivityDate="2015-09-10T19:20:26.113" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22070" PostTypeId="2" ParentId="22064" CreationDate="2015-09-10T19:35:02.943" Score="22" Body="&lt;p&gt;A possible solution is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;10   8    3  21&#xA;12  15   14   2&#xA; 1   7   30  24&#xA;42   6    4   5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Strategy&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$5040=2^4 \times 3^2 \times 5 \times 7$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First I decided where to put the multiples of $7$ and $5$. Then I multiplied proper exponents of $2$ and $3$ to each cell. I started with:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;5 1 1 7&#xA;1 5 7 1&#xA;1 7 5 1&#xA;7 1 1 5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This formation ensures that the number of cells without a $5$ or $7$ stays minimum. There are other such formations. But this is the most simple. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, There are eight $1$'s, four each of $5$ and$7$. When adding exponents of $3$ my strategy was to half each number of repeatation. A simple pattern was:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 1 3 3&#xA;3 3 1 1&#xA;1 1 3 3&#xA;3 3 1 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Multiplying by this grid the previous one, we have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 5  1  3 21&#xA; 3 15  7  1&#xA; 1  7 15  3&#xA;21  3  1  5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now the repeatation pattern was like this (same color represents same powers of $3$, $5$, $7$ in factorization):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tKz6ks.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tKz6ks.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any two cells with the same color I had to differentiate them by multiplying different powers of $2$. Note the red and yellow cells. There are four of each. Hence I needed at least four different powers of $2$, $2^4$ couldn't be chosen because putting $2^4$ in a red or yellow cell, forces two cells of the opposite color to have same powers of $2$, hence be same. So, the four red and yellow cells had to be multiplied with $2^0, 2^1, 2^2,2^3$ A bit of fiddling led to the pattern:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2 8 1 1&#xA;4 1 2 2&#xA;1 1 2 8&#xA;2 2 4 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Multiplying by this grid, we get the desired solution.&lt;/p&gt;&#xA;" OwnerUserId="15544" LastEditorUserId="15544" LastEditDate="2015-09-10T20:22:47.800" LastActivityDate="2015-09-10T20:22:47.800" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22071" PostTypeId="1" AcceptedAnswerId="22073" CreationDate="2015-09-10T20:09:56.447" Score="3" ViewCount="426" Body="&lt;p&gt;Guess the movie (or movie series):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Gold Lamé swimsuit,&lt;br&gt;&#xA;  Asthmatic dad's dead.&lt;br&gt;&#xA;  Sorry Luke, she's your sister.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Plagued by movie curse,&lt;br&gt;&#xA;  Odd-numbered installments bad.&lt;br&gt;&#xA;  Is Vger God? No.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Strangely therapist&lt;br&gt;&#xA;  Doesn't know he's dead.&lt;br&gt;&#xA;  The kind of thing you'd notice.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Extremely cheap Snape&lt;br&gt;&#xA;  Robs Nakatomi Plaza.&lt;br&gt;&#xA;  Bruce in vest and socks.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;You should always choose&lt;br&gt;&#xA;  The lesser of two weevils.&lt;br&gt;&#xA;  Surprise!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Rainy and muddy.&lt;br&gt;&#xA;  But you can't live on millet.&lt;br&gt;&#xA;  The farmers have won.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Theme park power fail&lt;br&gt;&#xA;  Destruction ensues&lt;br&gt;&#xA;  As large T. Rex runs amok.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Theme park power fail&lt;br&gt;&#xA;  Destruction ensues&lt;br&gt;&#xA;  As Yul Brynner runs amok.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Short-arsed Scientologist,&lt;br&gt;&#xA;  Confidence crisis.&lt;br&gt;&#xA;  The pattern is full.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;We deal in lead, friend.&lt;br&gt;&#xA;  We're in the same business, eh?&lt;br&gt;&#xA;  Farmers win again.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Nuke it from orbit.&lt;br&gt;&#xA;  Get away from her, you bitch!&lt;br&gt;&#xA;  Costaguano?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4050" LastActivityDate="2015-09-10T21:43:52.857" Title="Gold Lamé swimsuit, / Asthmatic dad's dead. / Sorry Luke, she's your sister" Tags="|knowledge|movies|poetry|haiku|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="22072" PostTypeId="2" ParentId="22064" CreationDate="2015-09-10T20:11:05.043" Score="2" Body="&lt;p&gt;After unsuccessful manual fiddling with the prime factors I decided to write a program, unfortunately not fast enough. Here it is anyway. It should print all possible solutions, but will contain duplicates. And I have no idea how long it takes to run, there are many solutions.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;public class Main {&#xA;    private static final int SIZE = 4;&#xA;    private static final int[] FACTORS = new int[] {2, 2, 2, 2, 3, 3, 5, 7};&#xA;&#xA;    private static final ArrayList&amp;lt;int[]&amp;gt; perms = new ArrayList&amp;lt;&amp;gt;();&#xA;&#xA;    private static void generateGrids(int depth, int[] grid) {&#xA;        if (depth &amp;lt; FACTORS.length) {&#xA;            for (int[] perm : perms) {&#xA;                for (int i = 0; i &amp;lt; SIZE; ++ i) {&#xA;                    grid[i * SIZE + perm[i]] *= FACTORS[depth];&#xA;                }&#xA;                generateGrids(depth + 1, grid);&#xA;                for (int i = 0; i &amp;lt; SIZE; ++ i) {&#xA;                    grid[i * SIZE + perm[i]] /= FACTORS[depth];&#xA;                }&#xA;            }&#xA;        } else {&#xA;            int[] tmp = new int[SIZE * SIZE];&#xA;            System.arraycopy(grid, 0, tmp, 0, tmp.length);&#xA;            Arrays.sort(tmp);&#xA;            for (int i = 0; i &amp;lt; tmp.length - 1; ++ i) {&#xA;                if (tmp[i] == tmp[i + 1]) {&#xA;                    return;&#xA;                }&#xA;            }&#xA;            System.out.println(Arrays.toString(grid));&#xA;        }&#xA;    }&#xA;&#xA;    private static void generatePerms(int depth, int[] perm) {&#xA;        outer:&#xA;        for (int i = 0; i &amp;lt; SIZE; ++i) {&#xA;            for (int j = 0; j &amp;lt; depth; j++) {&#xA;                if (i == perm[j]) {&#xA;                    continue outer;&#xA;                }&#xA;            }&#xA;            perm[depth] = i;&#xA;            if (depth &amp;lt; SIZE - 1) {&#xA;                generatePerms(depth + 1, perm);&#xA;            } else {&#xA;                int[] tmp = new int[SIZE];&#xA;                System.arraycopy(perm, 0, tmp, 0, SIZE);&#xA;                perms.add(tmp);&#xA;            }&#xA;        }&#xA;    }&#xA;&#xA;    public static void main(String[] args) {&#xA;        generatePerms(0, new int[SIZE]);&#xA;        int[] grid = new int[SIZE * SIZE];&#xA;        Arrays.fill(grid, 1);&#xA;        generateGrids(0, grid);&#xA;    }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="15632" LastActivityDate="2015-09-10T20:11:05.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22073" PostTypeId="2" ParentId="22071" CreationDate="2015-09-10T20:17:06.893" Score="5" Body="&lt;pre&gt;&lt;code&gt;Gold Lamé swimsuit,&#xA;Asthmatic dad's dead.&#xA;Sorry Luke, she's your sister.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Star Wars&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Plagued by movie curse,&#xA;Odd-numbered installments bad.&#xA;Is Vger God? No.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Star Trek&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Strangely therapist&#xA;Doesn't know he's dead.&#xA;The kind of thing you'd notice.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sixth Sense&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Extremely cheap Snape&#xA;Robs Nakatomi Plaza.&#xA;Bruce in vest and socks.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Die Hard&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;You should always choose&#xA;The lesser of two weevils.&#xA;Surprise!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Master and Commander&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Rainy and muddy.&#xA;But you can't live on millet.&#xA;The farmers have won.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Seven Samurai (thanks Zfrisch)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Theme park power fail&#xA;Destruction ensues&#xA;As large T. Rex runs amok.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jurassic Park&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Theme park power fail&#xA;Destruction ensues&#xA;As Yul Brynner runs amok.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Westworld&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Short-arsed Scientologist,&#xA;Confidence crisis.&#xA;The pattern is full.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Top Gun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;We deal in lead, friend.&#xA;We're in the same business, eh?&#xA;Farmers win again.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Magnificent Seven&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Nuke it from orbit.&#xA;Get away from her, you bitch!&#xA;Costaguano?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Aliens&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="14135" LastEditDate="2015-09-10T21:43:52.857" LastActivityDate="2015-09-10T21:43:52.857" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22075" PostTypeId="1" AcceptedAnswerId="22087" CreationDate="2015-09-10T21:51:58.530" Score="8" ViewCount="327" Body="&lt;p&gt;The last thing you remembered,&lt;br&gt;&#xA;Or what happened at the start.&lt;br&gt;&#xA;From reality dismembered?&lt;br&gt;&#xA;That is a magician's art.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;He really wasn't funny,&lt;br&gt;&#xA;Neither the following curse.&lt;br&gt;&#xA;Can't sleep, dark or sunny,&lt;br&gt;&#xA;Across the universe.   &lt;/p&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-09-11T13:14:27.347" Title="What really happened - Who am I?" Tags="|riddle|rhyme|knowledge|movies|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22076" PostTypeId="2" ParentId="22064" CreationDate="2015-09-10T21:52:03.787" Score="10" Body="&lt;p&gt;Another solution (with diagonals as bonus):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;10    4    6   21&#xA;18    7   20    2&#xA;28   12    3    5&#xA; 1   15   14   24&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Things that multiply to 5040:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;each of the four rows  &lt;/li&gt;&#xA;&lt;li&gt;each of the four columns&lt;/li&gt;&#xA;&lt;li&gt;each of two diagonals&lt;/li&gt;&#xA;&lt;li&gt;four center cells&lt;/li&gt;&#xA;&lt;li&gt;four corner cells&lt;/li&gt;&#xA;&lt;li&gt;two middle cells of the  top row two middle cells of bottom row&lt;/li&gt;&#xA;&lt;li&gt;two middle cells of the rightmost column and two middle cells of the leftmost column&lt;/li&gt;&#xA;&lt;li&gt;each of the four corner 2x2 grids&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2722" LastEditorUserId="15544" LastEditDate="2015-09-11T08:03:55.677" LastActivityDate="2015-09-11T08:03:55.677" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22078" PostTypeId="1" AcceptedAnswerId="22079" CreationDate="2015-09-10T22:36:52.580" Score="9" ViewCount="285" Body="&lt;p&gt;Having read &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/The_Grapes_of_Wrath&quot;&gt;The Grapes of Wrath&lt;/a&gt;&lt;/em&gt;, I started wondering why grapes were the only fruit assigned a vice.  Why can't oranges, or watermelons, or even pomelos grapple with the same licentiousness and iniquity that the poor grapes have to deal with on a daily basis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I began exploring what the sins of other fruits might be, and made myself a list.  In doing so, I discovered that &lt;a href=&quot;https://en.wikipedia.org/wiki/John_Steinbeck&quot;&gt;John Steinbeck&lt;/a&gt; had it wrong!  He apparently should have titled his book &lt;em&gt;The Grapes of Envy&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the list I came up with.  Can you determine how I linked each fruit with its particular vice?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;&lt;em&gt;apples of spite&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;strong&gt;&lt;em&gt;apricots of blasphemy&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;strong&gt;&lt;em&gt;bananas of pride&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;strong&gt;&lt;em&gt;blueberries of intolerance&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;strong&gt;&lt;em&gt;boysenberries of faithlessness&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;strong&gt;&lt;em&gt;cantaloupes of pettiness&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;strong&gt;&lt;em&gt;coconuts of complacency&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;strong&gt;&lt;em&gt;grapes of envy&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;strong&gt;&lt;em&gt;kiwis of apathy&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;strong&gt;&lt;em&gt;mangoes of sloth&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;strong&gt;&lt;em&gt;nectarines of jealousy&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;strong&gt;&lt;em&gt;pears of avarice&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;strong&gt;&lt;em&gt;persimmons of perversion&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;strong&gt;&lt;em&gt;plums of cowardice&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;strong&gt;&lt;em&gt;pomegranates of gluttony&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;strong&gt;&lt;em&gt;quinces of egotism&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;strong&gt;&lt;em&gt;tangerines of profligacy&lt;/em&gt;&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus A&lt;/strong&gt;:  What sin might be associated with &lt;code&gt;pomelos&lt;/code&gt;?&lt;br&gt;&#xA;&lt;strong&gt;Bonus B&lt;/strong&gt;:  What fruits might be subject to the sin of &lt;code&gt;manipulation&lt;/code&gt;?&lt;/p&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2015-09-11T02:09:32.310" Title="The Depravity of Produce" Tags="|pattern|enigmatic-puzzle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22079" PostTypeId="2" ParentId="22078" CreationDate="2015-09-10T23:17:47.557" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sum of their character values &lt;code&gt;a=1, b=2, ..., z=26&lt;/code&gt; is the same for the fruit and the sin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I found this by observing that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  each fruit and sin have similar number of letters, and guessing the answer from there, writing up code to check.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-09-11T02:09:32.310" LastActivityDate="2015-09-11T02:09:32.310" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22080" PostTypeId="1" AcceptedAnswerId="22081" CreationDate="2015-09-11T00:45:55.060" Score="3" ViewCount="263" Body="&lt;p&gt;Well, hello hello hello hello hello hello hello, and welcome to the Puzzling Stack Exchange, or &amp;#22823;&amp;#23478;&amp;#20204;&amp;#22909;, as they say in Chinese. Tonight we'll start with a riddle to see if you can find out who we are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sitting where I sit is a man who is not small, but if I were, I would be an insignificant little thing.&lt;br&gt;&#xA;To my left is a man who doesn't empty gas giants in our Solar System but does the opposite,&lt;br&gt;&#xA;To his left is the comedian who killed Goliath (and Webb).&lt;br&gt;&#xA;Two seats to my right is an invoice that surrounds a motte,&lt;br&gt;&#xA;And directly to my right is that joker whose name I always pronounce like the &quot;e&quot; isn't there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fingers on buzzers please. Who are we, and what episode is this?&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-09-11T01:49:49.033" Title="We are a panel of five people" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22081" PostTypeId="2" ParentId="22080" CreationDate="2015-09-11T01:41:58.467" Score="3" Body="&lt;p&gt;You are, from left to right (facing the panel):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Bill Bailey, Alan Davies, Stephen Fry, Phill Jupitus, and David Mitchell&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are the&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Panel from QI, and this is episode 32, &quot;Corby&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanations:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Bill Bailey&lt;/strong&gt;: Invoice = Bill, Motte is a reference to a &lt;a href=&quot;https://en.wikipedia.org/wiki/Motte-and-bailey_castle&quot; rel=&quot;nofollow&quot;&gt;motte-and-bailey castle&lt;/a&gt;.&lt;br&gt;&#xA;&lt;strong&gt;Alan Davies&lt;/strong&gt;: I think joker is just a generic reference to him being a comedian, and Stephen Fry always pronounces his last name like &quot;Davis&quot;.&lt;br&gt;&#xA;&lt;strong&gt;Stephen Fry&lt;/strong&gt;: I think &quot;insignificant little thing&quot; is meant to reference the phrase &quot;small fry&quot;.&lt;br&gt;&#xA;&lt;strong&gt;Phill Jupitus&lt;/strong&gt;: Rather than &quot;empty gas giants&quot; he will &quot;fill Jupiters&quot;. Eh, ehhh?&lt;br&gt;&#xA;&lt;strong&gt;David Mitchell&lt;/strong&gt;: This was the one that cracked the whole thing for me, because I'm a huge fan. &quot;Killed Goliath&quot; = David, obviously, and David Mitchell was a double act with Robert Webb for a long time, with a couple shows and even their own &lt;a href=&quot;https://en.wikipedia.org/wiki/Mitchell_and_Webb&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article!&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10861" LastEditorUserId="10861" LastEditDate="2015-09-11T01:49:49.033" LastActivityDate="2015-09-11T01:49:49.033" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22082" PostTypeId="2" ParentId="22040" CreationDate="2015-09-11T01:58:52.460" Score="14" Body="&lt;p&gt;I thought that it was&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; animals formed by consecutive letters in the sentence&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and it turns out that's correct.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Jim: I know lots of games.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;owl&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Rose: Wow, let's play some.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;owl also (edited to say: probably owlet)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Jim: If Jack too was here, it would have been great fun.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;bee&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Rose: Oh, I miss Jack a lot.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;jackal&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Jim: He came last week for the exams.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;camel&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Rose: Ha! Really? He didn't tell me.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;hare&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Jim: He asked for a tea party.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;rat&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Rose: Yes, it should be arranged.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;bear&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Jim: Let's go at 4:00pm. He must be at home at that time.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;goat&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Rose: He has nailed the exams and so he should be hosting the party, right?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;snail&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Jim: Yup, and I bought a pen for him to appreciate his success in the exams.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;edited to add: ape (thanks to f' in the comments)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Rose: Oh, great. Let me see it.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;And here's where Rose lost, since she had no animal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Jim: Aha, you lost. I am the winner.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="12092" LastEditorUserId="12092" LastEditDate="2015-09-11T04:53:36.623" LastActivityDate="2015-09-11T04:53:36.623" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22083" PostTypeId="2" ParentId="21964" CreationDate="2015-09-11T02:29:51.067" Score="4" Body="&lt;p&gt;There are an infinite combinations of such numbers. A few processes for finding these are described below.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here is a process for finding a triple of these numbers from two perfect squares. Pick two squares, $a$ and $b$, such that $a$ is even. Then, put $a/2$ on two vertices of the triangle, and $b-(a/2)$ on the third vertex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, there are three combinations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$a/2 + a/2 = a$, which is a perfect square, and&lt;/li&gt;&#xA;&lt;li&gt;$(b-(a/2))+a/2 = b$, occuring twice, which is a perfect square.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thus, you can pick any two squares, and as long as one of them is even, you can find three numbers that go on the vertices of the triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick bit of Haskell to find all triples is (with &lt;code&gt;isSquare&lt;/code&gt; as a function returning true/false depending on whether the input is a square):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[(a, b, c) | a &amp;lt;- [1..50], b &amp;lt;- [1..a], c &amp;lt;- [1..b], isSquare (a+b), isSquare (b+c), isSquare (a+c)]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Or, alternately (leaving off casting for clarity):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[(a/2, a/2, b-(a/2)) | a &amp;lt;- [1..50], b &amp;lt;- [1..a], isSquare a, isSquare b]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It is &lt;em&gt;not&lt;/em&gt;, however, the case that all combinations are generated in this way. To show this, if we run:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[(a, b, c) | a &amp;lt;- [1..50], b &amp;lt;- [1..a], c &amp;lt;- [1..b], isSquare (a+b), isSquare (b+c), isSquare (a+c), a /= b, a /= c, b /= c]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;we then get the output:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[(30, 19, 6), (44, 20, 5), (47, 34, 2), (48, 33, 16)]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which are the first four triples of &lt;em&gt;unique&lt;/em&gt; numbers satisfying the perfect-square condition.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastActivityDate="2015-09-11T02:29:51.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22084" PostTypeId="1" AcceptedAnswerId="22085" CreationDate="2015-09-11T02:46:16.227" Score="8" ViewCount="320" Body="&lt;p&gt;Scripts are like a box of sweets, you never know if a line or two is going to stick in people's mind.  Just imagine if the scriptwriters asked the actors to say:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It's breathing! It's breathing!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Plainly speaking, I do not care one way or another.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You cannot comprehend the facts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I will search for you, I will locate you and I will dispose of you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I shall revisit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Top of the day, you communist country.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;7&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Alien calling base.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;8&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Do not misjudge the influence of evil.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;9&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Spy, top spy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;10&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You spitting words at me?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;11&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;May the imaginary field surround you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;12&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you construct it, he will visit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;13&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I am traversing the road here!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;14&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I notice corpses.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;15&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ground control, we have a complication. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;16&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Au Revoir, darling.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;17&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Say your greetings to this little thing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;18&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My treasure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;19&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Use another way to make my alcoholic drink.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;20&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I am going to make him a proposition that he thinks it's really really good.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;21&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I am on top of the world!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;22&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I will order the same dish as her.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would be surprised if this puzzle takes more than 10 minutes to solve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Bonus:&lt;/em&gt; Just for fun, can you identify the actors who said them without Googling?&lt;/p&gt;&#xA;" OwnerUserId="12332" LastActivityDate="2015-09-14T20:55:19.793" Title="Scripts are like a box of sweets" Tags="|knowledge|movies|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22085" PostTypeId="2" ParentId="22084" CreationDate="2015-09-11T03:58:37.493" Score="8" Body="&lt;p&gt;Nobody yet? Okay I'll throw up the partial answer, then!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the spirit of the bonus, I have not looked up any of the actors. I'm surprised how many of these I knew off the top of my head!&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;It's breathing! It's breathing!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;IT'S ALIVE! IT'S ALIIIIIIIIIIIIVE! - Colin Clive, Frankenstein&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Plainly speaking, I do not care one way or another.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Frankly, my dear, I don't give a damn. - Clark Gable, Gone With The Wind&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You cannot comprehend the facts.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;YOU CAN'T HANDLE THE TRUTH! - Jack Nicholson, A Few Good Men&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I will search for you, I will locate you and I will dispose of you.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I will hunt you down, I will find you, and I will kill you. - Liam Neeson, Taken&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I shall revisit.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I'll be back. - Arnold Schwarzenegger, Terminator (does it count if I had to google his name to remember how to spell it?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Top of the day, you communist country.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Good Morning, Vietnam! - Robin Williams, Good Morning Vietnam (thanks to JoeZ)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Alien calling base.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;E.T. phone home. - ???, ET The Extraterrestrial (thanks to Bailey M)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Do not misjudge the influence of evil.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Do not underestimate the power of the Dark Side. - James Earl Jones, Star Wars&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Spy, top spy.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Bond. James Bond. -Sean Connery (and many other), [basically every James Bond movie]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You spitting words at me?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You talkin' to me? - Robert DeNiro, Taxi Driver&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;May the imaginary field surround you.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;May the Force be with you. - A number of people, but I'll go with Harrison Ford, Star Wars&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you construct it, he will visit.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If you build it, he will come. -Kevin Costner, Field of Dreams&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I am traversing the road here!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Hey, I'm walkin' here, I'm walkin' here! - Dustin Hoffman, Midnight Cowboy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I notice corpses.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I see dead people. - Haley Joel Osment, The Sixth Sense&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Ground control, we have a complication.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Houston, we have a problem. - Tom Hanks, Apollo 13 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Au Revoir, darling.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'Hasta la vista, baby' – Arnold Schwarzenegger, Terminator 2 (answered by Tim Couwelier in comments below)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Say your greetings to this little thing.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Say hello to my little friend. - Al Pacino, Scarface&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;My treasure.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;My preciousssssss. - Andy Serkis, Lord of the Rings (Thanks Bailey M)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Use another way to make my alcoholic drink.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Shaken, not stirred. - [insert James Bond actor], [insert James Bond movie]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I am going to make him a proposition that he thinks it's really really good.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I'll make him an offer he can't refuse. - Marlon Brando, The Godfather&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I am on top of the world!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I'm king of the world! - Leonardo DiCaprio, Titanic&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I will order the same dish as her.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I'll have what she's having. - Estelle Reiner, When Harry Met Sally (actor answered by dmg in comment below)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I am positive I'm missing some obvious ones.&lt;/p&gt;&#xA;" OwnerUserId="10861" LastEditorUserId="12332" LastEditDate="2015-09-14T20:55:19.793" LastActivityDate="2015-09-14T20:55:19.793" CommentCount="21" ContentLicense="CC BY-SA 3.0" />
  <row Id="22087" PostTypeId="2" ParentId="22075" CreationDate="2015-09-11T12:40:22.367" Score="6" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Christopher Nolan, as described by his filmography.  Each line is a different movie.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The last thing you remembered,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Memento&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or what happened at the start.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Batman Begins&lt;/strong&gt; (Could also be &lt;strong&gt;Inception&lt;/strong&gt; if you take the title literally).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From reality dismembered?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Inception&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is a magician's art.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;The Prestige&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He really wasn't funny, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;The Dark Knight&lt;/strong&gt; (featuring The Joker)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Neither the following curse. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Following&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can't sleep, dark or sunny, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Insomnia&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Across the universe. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Interstellar&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9523" LastEditorUserId="4385" LastEditDate="2015-09-11T13:14:27.347" LastActivityDate="2015-09-11T13:14:27.347" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22088" PostTypeId="1" AcceptedAnswerId="22092" CreationDate="2015-09-11T13:58:19.063" Score="13" ViewCount="1104" Body="&lt;p&gt;We're a group of cousins who live in the same city.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I live next door to my sister.&lt;/li&gt;&#xA;&lt;li&gt;Across the street, four of my cousins live in the same house: a pair of identical twins and a pair of fraternal twins.&lt;/li&gt;&#xA;&lt;li&gt;Next door to them, three more of my cousins live in the same house: another pair of identical twins and their little sister.&lt;/li&gt;&#xA;&lt;li&gt;My brother, who looks just like me, lives alone way out on the outskirts of town.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Who are we?&lt;/p&gt;&#xA;" OwnerUserId="10851" LastActivityDate="2015-09-11T14:48:22.650" Title="A famous, hardworking family" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22089" PostTypeId="1" AcceptedAnswerId="22090" CreationDate="2015-09-11T13:59:49.843" Score="4" ViewCount="370" Body="&lt;p&gt;I have an IDEA for a puzzle.&lt;br /&gt;&#xA;I think I'll add two letters.&lt;br /&gt;&#xA;But am I a puzzle about a recent discovery?&lt;br /&gt;&#xA;Or a very old conquest, newly recoined?&lt;br /&gt;&#xA;The news would tell you both of these are very new and very old.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which two things am I talking about? And what's their connection?&lt;/p&gt;&#xA;" OwnerUserId="13068" LastEditorUserId="13068" LastEditDate="2015-09-11T14:15:48.577" LastActivityDate="2015-09-11T14:15:48.577" Title="In the news this month, September 2015" Tags="|knowledge|anagram|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22090" PostTypeId="2" ParentId="22089" CreationDate="2015-09-11T14:13:00.417" Score="5" Body="&lt;p&gt;The two words are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Naledi, the new hominid species&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Denali, the recently renamed mountain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Both&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; differ by two letters from IDEA. Also, both stories involve climbing mountains.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1509" LastActivityDate="2015-09-11T14:13:00.417" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22091" PostTypeId="2" ParentId="22045" CreationDate="2015-09-11T14:28:19.640" Score="2" Body="&lt;p&gt;This could be totally off-base, but I'll give it a shot and say that you are a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Keystone (in an arch), which is an angled stone, keeping other stones in place.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A mere tool I am,&lt;br&gt;&#xA;In a world so grown,&lt;br&gt;&#xA;Yet here I am,&lt;br&gt;&#xA;Taking you on.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A keystone is a wedge-shaped stone, the Wedge being one of the simple machines (mere tools) in our advanced world.  In this puzzle, the keystone is taking on all us adept puzzlers, though it is just a simple machine.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm a supporter&lt;br&gt;&#xA;of things well known,&lt;br&gt;&#xA;Without me, these portals,&lt;br&gt;&#xA;you would never call your own.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The keystone is the one piece of the arch that holds all the rest together, and many (though not all) arches can be in the form of doors, or &quot;portals&quot;.  Without the keystone, these would not exist.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And kid yourself not,&lt;br&gt;&#xA;I am almost the same,&lt;br&gt;&#xA;As when I say &quot;The fate of the world&lt;br&gt;&#xA;depends on &lt;em&gt;my name&lt;/em&gt;.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This one stone is the arch is the key to making the arch stable.  Though it does not bear much of the actual load, the other stones depend on it to stay in place.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of which I am made,&lt;br&gt;&#xA;Is the Swan's Song of Stars,&lt;br&gt;&#xA;And it's neighbors,&lt;br&gt;&#xA;they may sound but&lt;br&gt;&#xA;unattractive they are.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Here's where I start grasping at straws... The keystone is a stone or rock, which is made from the same elements that stars are.  Then I get tangled in the syntax and can't determine if the neighbors are referring back to the keystone or to Of Which I Am Made.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What I am made of that I mentioned,&lt;br&gt;&#xA;is an ingredient among a few,&lt;br&gt;&#xA;among its kind it lies,&lt;br&gt;&#xA;an elemental country not new.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rocks and stones are nothing new, neither are the elements present in the universe, as they are the foundation of everything.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11452" LastActivityDate="2015-09-11T14:28:19.640" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22092" PostTypeId="2" ParentId="22088" CreationDate="2015-09-11T14:48:22.650" Score="17" Body="&lt;p&gt;Sounds like you're:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Punctuation marks on a QWERTY keyboard  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You and your sister are:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A period/full stop, and the key next to it, a comma  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your cousins:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Live on the next row up of the keyboard. The first identical twins are the colon, living with (on the same key as) the semicolon (the fraternal twins). Depending on your keyboard, the key next to that has either &lt;strong&gt;&quot;&lt;/strong&gt; and &lt;strong&gt;'&lt;/strong&gt; or &lt;strong&gt;@&lt;/strong&gt; and &lt;strong&gt;'&lt;/strong&gt; - presumably you have the former, making the identical twins quotation marks, and their little sister an apostrophe.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your brother:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Is the other period/full stop over by the numpad, out on the outskirts of the keyboard&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11717" LastActivityDate="2015-09-11T14:48:22.650" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22093" PostTypeId="1" AcceptedAnswerId="22106" CreationDate="2015-09-11T15:56:02.923" Score="2" ViewCount="178" Body="&lt;h2&gt;Read the title and identify&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Airtight ace has, our &quot;helicopter&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cold manner hydra, implore &quot;phi&quot; law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A only hardcore nut, he shook Mr. cell's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Loathe &quot;big&quot; geezer, a Lonely Myth's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Soon &quot;sign&quot; emerge, just &quot;lire&quot; game&lt;/p&gt;&#xA;" OwnerUserId="9892" LastEditorUserId="15544" LastEditDate="2015-09-11T20:19:04.453" LastActivityDate="2015-09-11T20:19:04.453" Title="Can you detect the pattern?" Tags="|wordplay|knowledge|anagram|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22094" PostTypeId="1" AcceptedAnswerId="22097" CreationDate="2015-09-11T16:34:44.513" Score="7" ViewCount="554" Body="&lt;p&gt;Let's keep this movie train rolling! Below are a handful of Academy Award Best Picture winning movie titles, but each has been encrypted somehow. There is a common rule determining the encryption methods. Can you determine what each title is and how it was encrypted?&lt;/p&gt;&#xA;&lt;h3&gt;Titles&lt;/h3&gt;  &#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;BUO QVRHT PW WHF FHSUNUN GARNU&lt;/p&gt;&#xA;&lt;p&gt;PO CVWNTYA FOY QLD TGN&lt;/p&gt;&#xA;&lt;p&gt;MJCTFWIG PO GTBKOC&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-09-11T20:21:46.607" Title="Encrypted Titles" Tags="|cipher|movies|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22096" PostTypeId="1" AcceptedAnswerId="22098" CreationDate="2015-09-11T16:58:50.523" Score="1" ViewCount="4494" Body="&lt;p&gt;Can I ask a yes/no question so that no matter whether the answerer answers yes or no, he always lies?&lt;/p&gt;&#xA;" OwnerUserId="16303" LastActivityDate="2015-09-11T17:59:11.657" Title="What is a yes/no question that forces the answerer to always lie?" Tags="|logical-deduction|" AnswerCount="3" CommentCount="3" ClosedDate="2015-09-11T18:18:06.210" ContentLicense="CC BY-SA 3.0" />
  <row Id="22097" PostTypeId="2" ParentId="22094" CreationDate="2015-09-11T16:59:36.187" Score="15" Body="&lt;p&gt;&lt;em&gt;BUO QVRHT PW WHF FHSUNUN GARNU&lt;/em&gt; decrypts as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;ALL QUIET ON THE WESTERN FRONT&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;PO CVWNTYA FOY QLD TGN&lt;/em&gt; decrypts as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;NO COUNTRY FOR OLD MEN&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;MJCTFWIG PO GTBKOC&lt;/em&gt; decrypts as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;LAWRENCE OF ARABIA&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The encryption algorithm is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rotate all non-space letters by the year the movie was produced.&lt;br&gt;&#xA; &lt;code&gt;in:   ALL QUIET ON THE WESTERN FRONT&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;salt: 193 01930 19 301 9301930 19301&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;out:  BUO QVRHT PW WHF FHSUNUN GARNU&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;code&gt;in:   NO COUNTRY FOR OLD MEN&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;salt: 20 0720072 007 200 720&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;out:  PO CVWNTYA FOY QLD TGN&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;code&gt;in:   LAWRENCE OF ARABIA&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;salt: 19621962 19 621962&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;out:  MJCTFWIG PO GTBKOC&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; I'm sure there's a name for this kind of cypher. Someone can probably help me out with that.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I came to this conclusion by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; first figuring out the names of the movies by the number of letters in each word. Once I had that, I ran a short little script that gave me the difference between each letter as a number.&lt;br&gt;&#xA; &lt;pre&gt;&lt;code&gt;def f(a, b):&#xA;  d = []&#xA;  for c in xrange(len(a)):&#xA;    if a[c] == ' ':&#xA;      continue&#xA;    d += [(ord(a[c])-ord(b[c])+26)%26]&#xA;  return d&lt;/code&gt;&lt;/pre&gt;&lt;br&gt;&#xA; The numbers that came back were sequences of the same four digits repeated over and over. I checked these digits against the release dates, and they matched.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-09-11T20:21:46.607" LastActivityDate="2015-09-11T20:21:46.607" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22098" PostTypeId="2" ParentId="22096" CreationDate="2015-09-11T17:01:39.887" Score="8" Body="&lt;p&gt;A pretty simple one would be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Is your answer to this question going to be &quot;no&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-09-11T17:01:39.887" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22099" PostTypeId="1" CreationDate="2015-09-11T17:06:58.107" Score="2" ViewCount="1476" Body="&lt;p&gt;Given 10 coin generating machines, all of which produces a coin of weight x. Out of these, two machines are defective and produces a coin of weight x-1. You are given a weighing machine, how will you find the defective machines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extend this questions to a total of n machines , out of those n machines, m are defective and producing coins of x-1 weights.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was asked this in an interview but unfortunately could not crack it. Please suggest some valid solution.&lt;/p&gt;&#xA;" OwnerUserId="16304" LastActivityDate="2015-09-11T17:30:42.340" Title="How to find the defective machines?" Tags="|logical-deduction|enigmatic-puzzle|" AnswerCount="3" CommentCount="5" ClosedDate="2015-09-11T18:28:32.563" ContentLicense="CC BY-SA 3.0" />
  <row Id="22100" PostTypeId="2" ParentId="22096" CreationDate="2015-09-11T17:14:12.200" Score="4" Body="&lt;p&gt;How about a question with a false premise?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Have you stopped beating your wife? Asked of someone who has never beaten his wife.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whether yes or no, the answer will not be true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A slightly less sexist or upsetting version might be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Did you make the hat you're wearing? Asked of someone who is not wearing a hat&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again if you can confine them to yes/no you can guarantee a lie&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you can take a question that can only be answered &quot;yes&quot; or &quot;no&quot; and conflate it with the opposite&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Will you your answer to &quot;are you asleep&quot; be the same as this one?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1509" LastActivityDate="2015-09-11T17:14:12.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22101" PostTypeId="2" ParentId="22099" CreationDate="2015-09-11T17:14:40.207" Score="1" Body="&lt;p&gt;&lt;em&gt;This answer assumes that the 'weighing machine' represents a scale to weight two objects against one another. If it turns out the weighing machine is something different, I'll have to reword this answer.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Weigh 5 coins on each side of the scale.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Scenario A:&lt;/strong&gt; Both sides have the same weight, meaning each side has one defective coin. For each side, weight two coins versus two coins. If the weights are even, the fifth coin is the defective one. If the weights are uneven, weigh the two lighter coins against each other. The lighter of the two is the defective coin.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Scenario B:&lt;/strong&gt; One side is lighter than the other. That side has both defective coins. Weigh two coins from that side against another two coins from that side.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If they are the same weight, again each side has a defective coin. Weigh the two pairs against each other, and the lighter two coins are defective.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If one side is lighter, that side has at least one defective coin. Weigh the two coins in that pair against each other. If they're the same weight, they are your two defective coins. If one is lighter than the other, the lighter one and the one excluded from the first measurement on this side are your two defective coins.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-09-11T17:14:40.207" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22102" PostTypeId="2" ParentId="22099" CreationDate="2015-09-11T17:19:49.493" Score="5" Body="&lt;p&gt;Presuming the restriction is to only use the weighing machine once, then it's a question of generating the right number of coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most general solution is to simply generate $2^{n-1}$ coins from the $n$th machine, and weigh the whole lot.  Assuming you know what the $x$ is, you'll know how much you are below that.  This discrepancy can be converted to binary and you can read of the machine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this will require over 1000 coins in the first instance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know that there are exactly 2 machines out of the 10, you can rely on that to generate a more reasonable number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generate $1, 2, 3, 5, 8, 13,\cdots$ coins from each of the 10 machines, and read of the answer (Fibonanci again?!)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;3: 1 and 2&lt;/li&gt;&#xA;&lt;li&gt;4: 1 and 3&lt;/li&gt;&#xA;&lt;li&gt;5: 2 and 3&lt;/li&gt;&#xA;&lt;li&gt;6: 1 and 5&lt;/li&gt;&#xA;&lt;li&gt;7: 2 and 5&lt;/li&gt;&#xA;&lt;li&gt;8: 3 and 5&lt;/li&gt;&#xA;&lt;li&gt;9: 1 and 8&lt;/li&gt;&#xA;&lt;li&gt;10: 2 and 8&#xA;etc.  You get the idea.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This approach can, I believe be generalised, if $m$ is known.  You probably just need a recurrence that sums over $m$ coins rather than $2$ (as above).  I'd have to think about that a bit to confirm it, so I'll wait for the OP to clarify what his question actually is.&lt;/p&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2015-09-11T17:30:42.340" LastActivityDate="2015-09-11T17:30:42.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22103" PostTypeId="2" ParentId="22099" CreationDate="2015-09-11T17:22:22.623" Score="1" Body="&lt;p&gt;The problem is easy if you can use as many weighings as you like. Take any one coin and weigh it (9 separate times) against coins from other machines. You will know which two are lighter when you're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're trying to minimize weighings, try weighing a different number of coins from each machine. Then you can back-conclude that the coins came from particular machines. Sometimes it's suggested to weigh 1 from machine 1, 2 from machine 2 etc against a similar total from say machine 10 but that may not work... if the weights differ by 5, does that mean machine 2 and 3 are bad, or 1 and 4? Best to use binary and weigh 1 from machine 1, 2 from 2, 4 from 3, 8 from 4 and so on. &lt;/p&gt;&#xA;" OwnerUserId="1509" LastActivityDate="2015-09-11T17:22:22.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22104" PostTypeId="2" ParentId="21932" CreationDate="2015-09-11T17:22:43.773" Score="5" Body="&lt;p&gt;Partial solution; hopefully someone can help me fill in the rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A cat?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My first's an incredible youngster.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dash from The Incredibles --&gt; Dash&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My second takes time to complete.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My third's an acrobatic contraption.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A trapeze, which contains a horizontal bar --&gt; Dash&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fourth can be used to compete.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fifth can mean one plus one half.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A period, as in the decimal 1.5 --&gt; Dot&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My sixth is a final two tether.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A line connecting the final two members of a bracket --&gt; Dash&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My seventh will make all the difference.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I when my parts come together?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A cat. In Morse code, C = – • – •, A = • –, and T = –. Using the spacing, this corresponds to the answers to the given clues.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16275" LastActivityDate="2015-09-11T17:22:43.773" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22105" PostTypeId="2" ParentId="22096" CreationDate="2015-09-11T17:33:35.077" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;You didn't go to work today, right?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You could possibly trap them with how you interpret it. So if I were to answer &quot;no&quot; You could respond with, &quot;Oh, you didn't go to work!&quot; when I was implying &quot;no&quot; to you being &quot;right&quot; or correct. &#xA;Or if I answer no as in &quot;I didn't go&quot; you could respond &quot;I could have sworn you weren't there!&quot; as implying I was answering &quot;no&quot; to you being correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With &quot;yes&quot; I could mean &quot;Yes, I did go to work&quot; but you can take it as &quot;Yes, I didn't go to work&quot;.&#xA;Or if I answered &quot;yes&quot; as in &quot;Yes I didn't go&quot; you can interpret it as &quot;Yes, I didn't go&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I &lt;em&gt;think&lt;/em&gt; this all makes sense and I think I covered the possibilities of just Yes or No being a possible answer.&lt;/p&gt;&#xA;" OwnerUserId="16273" LastEditorUserId="16273" LastEditDate="2015-09-11T17:59:11.657" LastActivityDate="2015-09-11T17:59:11.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22106" PostTypeId="2" ParentId="22093" CreationDate="2015-09-11T18:29:57.760" Score="3" Body="&lt;p&gt;These are all anagrams of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Famous fictional detectives and the authors who created them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Airtight ace has, our &quot;helicopter&quot;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Hercule_Poirot&quot; rel=&quot;nofollow&quot;&gt;Hercule Poirot&lt;/a&gt; (&lt;a href=&quot;https://en.wikipedia.org/wiki/Agatha_Christie&quot; rel=&quot;nofollow&quot;&gt;Agatha Christie&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Cold manner hydra, implore &quot;phi&quot; law.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Philip_Marlowe&quot; rel=&quot;nofollow&quot;&gt;Philip Marlowe&lt;/a&gt; (&lt;a href=&quot;https://en.wikipedia.org/wiki/Raymond_Chandler&quot; rel=&quot;nofollow&quot;&gt;Raymond Chandler&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A only hardcore nut, he shook Mr. cell's.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Sherlock_Holmes&quot; rel=&quot;nofollow&quot;&gt;Sherlock Holmes&lt;/a&gt; (&lt;a href=&quot;https://en.wikipedia.org/wiki/Arthur_Conan_Doyle&quot; rel=&quot;nofollow&quot;&gt;Arthur Conan Doyle&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Loathe &quot;big&quot; geezer, a Lonely Myth's&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/The_Inspector_Lynley_Mysteries&quot; rel=&quot;nofollow&quot;&gt;Thomas Lynley&lt;/a&gt; (&lt;a href=&quot;https://en.wikipedia.org/wiki/Elizabeth_George&quot; rel=&quot;nofollow&quot;&gt;Elizabeth George&lt;/a&gt;)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Soon &quot;sign&quot; emerge, just &quot;lire&quot; game&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Jules_Maigret&quot; rel=&quot;nofollow&quot;&gt;Jules Maigret&lt;/a&gt; (&lt;a href=&quot;https://en.wikipedia.org/wiki/Georges_Simenon&quot; rel=&quot;nofollow&quot;&gt;Georges Simenon&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="4421" LastEditDate="2015-09-11T20:09:00.700" LastActivityDate="2015-09-11T20:09:00.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22107" PostTypeId="2" ParentId="22043" CreationDate="2015-09-11T18:49:44.637" Score="12" Body="&lt;p&gt;I think I've got it!  The trick to this was &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the title of movie the actor was in is based off the ranking system in US Army from top to bottom of the list the actors were in.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Gerard Depardieu&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Colonel Chabert (1994)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Damon Wayans&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Major&quot; Payne (1995)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Jacques Perrin&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Captain's Honor (1982)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Maurice Chevalier&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Smiling Lieutenant (1931)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rod Steiger&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Sergeant (1968)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Jean-Pierre Cassel&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Elusive Corporal (1962) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Matt Damon&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Saving Private Ryan (1998?) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;BONUS!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mark Aldrich&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gods and Generals (2003)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16273" LastEditorUserId="4385" LastEditDate="2015-09-11T20:57:16.427" LastActivityDate="2015-09-11T20:57:16.427" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22108" PostTypeId="2" ParentId="22043" CreationDate="2015-09-11T19:30:05.847" Score="3" Body="&lt;p&gt;For the bonus :)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; James Cromwell: The General's Daugther (1999) (Placed at the very top of the list)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastActivityDate="2015-09-11T19:30:05.847" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22109" PostTypeId="1" AcceptedAnswerId="22116" CreationDate="2015-09-11T20:31:50.207" Score="5" ViewCount="242" Body="&lt;p&gt;I don't think it'll be a hard one, but hopefully it's fun     : )&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;One is not unlike your home, albeit to some eyes, much darker.&lt;/p&gt;&#xA;&lt;p&gt;Another was abandoned and hidden, a shameful and suppressed existence.&#xA;No girl deserves such a fate. Yet in the end her secret is known to all.&lt;/p&gt;&#xA;&lt;p&gt;A genus of air-breathing land snails, or more specifically: terrestrial pulmonate gastropod mollusks in the family Endodontidae. Or at least was...&lt;/p&gt;&#xA;&lt;p&gt;A series of buildings in which corpses are reduced to ash by fire. But somehow I'm cold too.&lt;/p&gt;&#xA;&lt;p&gt;One was nursed by a wolf. Such an animal, no wonder he killed his&#xA;brother.&lt;/p&gt;&#xA;&lt;p&gt;The leader of a group of heroes that prevented an evil goddess from&#xA;returning to the world. But regardless of his deeds, he's still only&#xA;half as well as humanity's last hope.&lt;/p&gt;&#xA;&lt;p&gt;A city located in southern Tunisia. Home of the native Berber&#xA;population and one particular well known individual.&lt;/p&gt;&#xA;&lt;p&gt;The last has been extensively studied by astronomers, leading it to be&#xA;termed &amp;quot;arguably the next most important one in the sky after the Sun,&#xA;but don't be led astray, it is no star, have the balls to look it up a bit&#xA;more.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;We are all the same, and yet none exist together. What are we and where do we come from?&lt;/p&gt;&#xA;&lt;p&gt;If you're looking for a hint, maybe you should look at the title a bit closer...&lt;/p&gt;&#xA;" OwnerUserId="11251" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-09-12T20:16:41.777" Title="Maybe we're all just pavement oil" Tags="|knowledge|movies|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22110" PostTypeId="1" AcceptedAnswerId="22171" CreationDate="2015-09-11T21:01:23.977" Score="5" ViewCount="339" Body="&lt;p&gt;&lt;p&gt;When this man was young, Frankenstein films inspired him to take up acting, but during a performance at The Rep (L.A.) cement splashed on his face so he switched to directing for a while and there were books he wanted to abridge too. &lt;/p&gt;&lt;p&gt;Far from high arts and stories of class, action movies called out to him - those where he could let loose: cannons firing with a bite, the bullets whizzing behind, enemy lines of infantry yelling &quot;Geronimo&quot;, an American legend striding out of San Francisco, pike swinging like a scythe - quick - and the dead piled up all around while the sound of mortar gets nearer.&lt;/p&gt;&lt;p&gt;Not everyone thought he was super - many saw him as an enemy. Oft he stated &quot;After my first Oscar nomination I let my ego run away. Jury after jury failed to give me the top award by a narrow margin.&quot;&lt;/p&gt;&lt;p&gt;Don't forget short young men could succumb at 21 to the power, the absolute power of Hollywood. Many an unlucky lady has found that they are heartbreakers who want zero ties - a couple of hoosiers all night long.&lt;/p&gt;&lt;p&gt;The firm way he handled his first wife led to their split. Decisions regarding another woman led to thunder, fireworks and extreme measures.&#xA;He tried to explain but before the conversation had begun, forgiveness evaporated. She is typical of doctor's wives from Mississippi - burning with rage and often turning crimson.&lt;/p&gt;&lt;p&gt;Tides turn, though and he was still someone to look dewy at. Tear-patterns streaming down the cheeks of women were not uncommon.&lt;/p&gt;&lt;p&gt; Valour during his military service in his twilight years meant he was leader of the pack. Age didn't matter. He was decorated for his courageous efforts twice.&lt;/p&gt;&lt;p&gt; In a lifetime of travelling the world he has developed great respect for the Mexican people he met, but not for the French. Connections overseas and at home have said that he has always been great company - business conversations in March or diecast modelling of a scarecrow at night.&lt;/p&gt;&lt;p&gt; Move several years later and he is no longer in his prime. Cut from the A list he is no longer desired. Speaking to the media about his chances of playing leading man again, they say &quot;No Way!&quot;.&lt;/p&gt;&lt;p&gt; &quot;Outrageous&quot; is my response.&lt;/p&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Who is this superstar and how many films can you find (please include the number to make it easier to check if you've got them all!)?&lt;/p&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="14135" LastEditDate="2015-09-11T22:22:16.323" LastActivityDate="2015-09-14T16:52:56.070" Title="This isn't really my life story" Tags="|knowledge|movies|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22111" PostTypeId="2" ParentId="22109" CreationDate="2015-09-11T21:16:29.523" Score="0" Body="&lt;p&gt;I think I have at least part of the answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Constellations&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One is not unlike your home, albeit to some eyes, much darker.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cepheus&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another was abandoned and hidden, a shameful and suppressed existence. No girl deserves such a fate. Yet in the end her secret is known to all.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A genus of air-breathing land snails, or more specifically: terrestrial pulmonate gastropod mollusks in the family Endodontidae. Or at least was...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Limax&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A series of buildings in which corpses are reduced to ash by fire. But somehow I'm cold too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One was nursed by a wolf. Such an animal, no wonder he killed his brother.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The leader of a group of heroes that prevented an evil goddess from returning to the world. But regardless of his deeds, he's still only half as well as humanity's last hope.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A city located in southern Tunisia. Home of the native Berber population and one particular well known individual.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The last has been extensively studied by astronomers, leading it to be termed &quot;arguably the next most important one in the sky after the Sun, but don't be led astray, it is no star, have the balls to look it up a bit more.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Vega&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15853" LastActivityDate="2015-09-11T21:16:29.523" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22112" PostTypeId="2" ParentId="22110" CreationDate="2015-09-11T21:27:00.217" Score="2" Body="&lt;p&gt;List in progress&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gene Hackman&lt;br&gt;&#xA;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without googling these are the movies I have, I'll complete later if no one else does.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The French Connection, Heartbreakers, Absolute Power, Young Frankenstein, A bridge Too Far, Power, The Runaway Jury, Loose Cannons, Mississippi Burning, Crimson Tide, Hoosiers, The Firm, The Mexican...&lt;br&gt;&#xA;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll take a look at his IMDb list later, I get the feeling they are probably all there because 80% of the words in the text remind me of a movie title, I'm just unsure if he was in all of them.&lt;/p&gt;&#xA;" OwnerUserId="11251" LastEditorUserId="15544" LastEditDate="2015-09-11T22:20:27.487" LastActivityDate="2015-09-11T22:20:27.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22113" PostTypeId="2" ParentId="22027" CreationDate="2015-09-11T22:58:56.393" Score="4" Body="&lt;p&gt;These are the known numbers so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;S = \left(\begin{matrix}&#xA; 0 &amp;amp; 120 \\&#xA; 1 &amp;amp; 106 \\&#xA; 2 &amp;amp; 112 \\&#xA; 3 &amp;amp; 108 \\&#xA; 4 &amp;amp; 80 \\&#xA; 5 &amp;amp; 92 \\&#xA; 6 &amp;amp; 84&#xA;\end{matrix}\right)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When investigating the numbers, I noticed a pattern in them:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Delta S = 112 - 106 = 6 \\&#xA;\Delta S = 112 - 108 = 4&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Delta S = 92 - 80 = 12 \\&#xA;\Delta S = 92 - 84 = 8&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These number seem to be related to each other in groups of 3 and their $\Delta$'s between each other seem to be doubling. Backwards engineering using this, leaves us with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Delta S = a - 120 = 2 \\&#xA;a = 2 \\&#xA;S(-1) = 122&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's compare the highest numbers of all the known triplets:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Delta S = 122 - 112 = 10 \\&#xA;\Delta S = 112 - 92 = 20&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which confirms my theory that the pattern must be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\text{in }\mathbb{Z} \\[12pt]&#xA;S(x) = \begin{cases}&#xA;S(x+1) - \left(6 * 2^{\left\lfloor\frac{x-1}{3}\right\rfloor}\right) &amp;amp; \text{if } ((x-1) \text{ mod } 3) = 0 \\&#xA;132 - \left(20 * 2^{\left\lfloor\frac{x-1}{3}\right\rfloor}\right) &amp;amp; \text{if } ((x-2) \text{ mod } 3) = 0 \\&#xA;S(x-1) - \left(4 * 2^{\left\lfloor\frac{x-1}{3}\right\rfloor}\right) &amp;amp; \text{if } (x \text{ mod } 3) = 0&#xA;\end{cases}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next number in the sequence would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;S(7) &amp;amp;= S(7+1) - \left(6 * 2^{\left\lfloor\frac{7-1}{3}\right\rfloor}\right) \\&#xA;&amp;amp;= \left(132 - \left(20 * 2^{\left\lfloor\frac{8-1}{3}\right\rfloor}\right)\right) - \left(6 * 2^{\left\lfloor\frac{7-1}{3}\right\rfloor}\right) \\&#xA;&amp;amp;= \left(132 - \left(20 * 4\right)\right) - \left(6 * 4\right) \\&#xA;&amp;amp;= \left(132 - 80\right) - 24 \\&#xA;&amp;amp;= 52 - 24 \\&#xA;&amp;amp;= 28&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="4196" LastEditorUserId="4196" LastEditDate="2015-09-12T12:14:17.203" LastActivityDate="2015-09-12T12:14:17.203" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22114" PostTypeId="1" AcceptedAnswerId="22119" CreationDate="2015-09-12T00:44:05.170" Score="32" ViewCount="873" Body="&lt;p&gt;Reverend Spooner dictated a list of his favorite movies to his befuddled secretary.  That well-meaning lad, unaware of his master's tendency to swap some of the consonants in a phrase, did not realize these were movie titles.  He thought he would improve the sense of the items in the list as you see below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your task is to set things right by deducing what Spooner actually said, and identifying the intended movie title.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Blunder Involving Gliding Toy&lt;/li&gt;&#xA;  &lt;li&gt;Sunrise Over the Place to Keep Garden Tools&lt;/li&gt;&#xA;  &lt;li&gt;Purchaser of Easter Flower&lt;/li&gt;&#xA;  &lt;li&gt;Those Presenting the Bill for Organic Tortillas&lt;/li&gt;&#xA;  &lt;li&gt;Greetin' the People of Sound Mind by Chimin' a Bell&lt;/li&gt;&#xA;  &lt;li&gt;A Citrus for a Recordkeeper of Asian Cookware&lt;/li&gt;&#xA;  &lt;li&gt;Ouchy Milk Farm&lt;/li&gt;&#xA;  &lt;li&gt;Athlete Whose Home is by the Train Tracks&lt;/li&gt;&#xA;  &lt;li&gt;Cereal That Will Make You Grow Tall&lt;/li&gt;&#xA;  &lt;li&gt;Rousing the Warriors&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4385" LastEditorUserId="4385" LastEditDate="2015-09-17T19:00:17.887" LastActivityDate="2015-09-17T19:00:17.887" Title="Reverend Spooner's Favorite Movies" Tags="|wordplay|knowledge|movies|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22115" PostTypeId="2" ParentId="8359" CreationDate="2015-09-12T00:44:46.533" Score="11" Body="&lt;p&gt;60 nodes and 120 edges:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/65sl0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;The shape that looks like a square is actually a square. And &lt;a href=&quot;http://www.wolframalpha.com/input/?i=tan%2875+deg%29&quot; rel=&quot;noreferrer&quot;&gt;$\tan 75^{\circ} = 2+\sqrt{3}$&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Slightly better than the easy answer.&lt;/p&gt;&#xA;" OwnerUserId="11495" LastEditorUserId="11495" LastEditDate="2015-09-12T00:57:48.053" LastActivityDate="2015-09-12T00:57:48.053" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22116" PostTypeId="2" ParentId="22109" CreationDate="2015-09-12T03:39:23.597" Score="4" Body="&lt;p&gt;They are all:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Planets in movies. &quot;None exist together&quot; since they are in different universes. Perhaps the title could read &quot;Pavement Oil&lt;strong&gt;s&lt;/strong&gt;&quot;, which would be an anagram of &lt;strong&gt;Movie Planets&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One is not unlike your home, albeit to some eyes, much darker.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Dark Planet&lt;/strong&gt;: The name the aliens give to Earth in &lt;a href=&quot;https://en.wikipedia.org/wiki/Escape_from_Planet_Earth&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Escape from Planet Earth&lt;/em&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another was abandoned and hidden, a shameful and suppressed existence. No girl deserves such a fate. Yet in the end her secret is known to all.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Miranda&lt;/strong&gt;: The &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_Firefly_planets_and_moons#Miranda&quot; rel=&quot;nofollow&quot;&gt;planet&lt;/a&gt; in &lt;a href=&quot;https://en.wikipedia.org/wiki/Serenity_(film)&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Serentiy&lt;/em&gt;&lt;/a&gt;, home of a failed experiment with sinister results.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A genus of air-breathing land snails, or more specifically: terrestrial pulmonate gastropod mollusks in the family Endodontidae. Or at least was...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Thermia&lt;/strong&gt;: A genus of &lt;a href=&quot;https://en.wikipedia.org/wiki/Thermia&quot; rel=&quot;nofollow&quot;&gt;snail&lt;/a&gt;, and home world of &lt;a href=&quot;http://galaxyquest.wikia.com/wiki/Thermia&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Galaxy Quest's&lt;/em&gt; Thermians&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A series of buildings in which corpses are reduced to ash by fire. But somehow I'm cold too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Crematoria&lt;/strong&gt;: Plural of crematorium, and the &lt;a href=&quot;http://riddick.wikia.com/wiki/Crematoria&quot; rel=&quot;nofollow&quot;&gt;planet&lt;/a&gt; with harsh temperatures from the Riddick universe. &lt;em&gt;Thanks to Gordon K for this one!&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One was nursed by a wolf. Such an animal, no wonder he killed his brother.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Romulus&lt;/strong&gt;: The &lt;a href=&quot;https://en.wikipedia.org/wiki/Romulus_and_Remus&quot; rel=&quot;nofollow&quot;&gt;brother of Remus&lt;/a&gt;, and home world to Star Trek's &lt;a href=&quot;https://en.wikipedia.org/wiki/Romulan&quot; rel=&quot;nofollow&quot;&gt;Romulans&lt;/a&gt;. Romulus killed Remus, and the Romulans destroyed Vulcan (&lt;em&gt;thanks to Gordon K for Vulcan reference&lt;/em&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The leader of a group of heroes that prevented an evil goddess from returning to the world. But regardless of his deeds, he's still only half as well as humanity's last hope.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Tanis&lt;/strong&gt;: The hero from &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_Dragonlance_characters&quot; rel=&quot;nofollow&quot;&gt;Dragonlance&lt;/a&gt;, and the planet from &lt;a href=&quot;https://en.wikipedia.org/wiki/Pandorum&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Pandorum&lt;/em&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A city located in southern Tunisia. Home of the native Berber population and one particular well known individual.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Tataouine&lt;/strong&gt;: The &lt;a href=&quot;https://en.wikipedia.org/wiki/Tataouine&quot; rel=&quot;nofollow&quot;&gt;city&lt;/a&gt; is real, but I assume the well known individual is Star Wars' Anakin Skywalker (from the planet &lt;a href=&quot;https://en.wikipedia.org/wiki/Tatooine&quot; rel=&quot;nofollow&quot;&gt;Tatooine&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The last has been extensively studied by astronomers, leading it to be termed &quot;arguably the next most important one in the sky after the Sun&quot;, but don't be led astray, it is no star, have the balls to look it up a bit more.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Vega&lt;/strong&gt;: A relatively nearby &lt;a href=&quot;https://en.wikipedia.org/wiki/Vega&quot; rel=&quot;nofollow&quot;&gt;star&lt;/a&gt; and the moon where Dark Helmet instructs his minions to &lt;a href=&quot;https://www.youtube.com/watch?v=g3iFJpGJiug&quot; rel=&quot;nofollow&quot;&gt;comb the desert&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;It appears most of the clues begin with a reference to a real-world interpretation, then hint at the fictional interpretation.&lt;/p&gt;&#xA;" OwnerUserId="13237" LastEditorUserId="13237" LastEditDate="2015-09-12T20:16:41.777" LastActivityDate="2015-09-12T20:16:41.777" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="22118" PostTypeId="2" ParentId="21950" CreationDate="2015-09-12T05:03:30.433" Score="2" Body="&lt;p&gt;By a very long stretch, is the answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Roup?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An infectious disease of poultry and pigeons characterized by inflammation of and mucous discharge from the mouth and eyes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I believe the backstory is what it was, just a backstory, including the cryptic verse. The only real clue is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer lies within, for the right answer is in the group&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; gROUP&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15607" LastActivityDate="2015-09-12T05:03:30.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22119" PostTypeId="2" ParentId="22114" CreationDate="2015-09-12T05:42:01.577" Score="32" Body="&lt;p&gt;Finished!&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Blunder Involving Gliding Toy&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Kite Flub (&lt;em&gt;Fight Club&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Sunrise Over Place to Keep Garden Tools&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dawn of the Shed (&lt;em&gt;Shaun of the Dead&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Purchaser of Easter Flower&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lily Buyer (&lt;em&gt;Billy Liar&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Those Presenting the Bill for Organic Tortillas&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Natural Corn Billers (&lt;em&gt;Natural Born Killers&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Greetin' the People of Sound Mind by Chimin' a Bell&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ringin' in the Sane (&lt;em&gt;Singin' in the Rain&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Citrus For a Recordkeeper of Asian Cookware&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Wok Clerk Orange (&lt;em&gt;A Clockwork Orange&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Ouchy Milk Farm&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hurty Dairy (&lt;em&gt;Dirty Harry&lt;/em&gt;) [Peter Taylor]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Athlete Whose Home is by the Train Tracks&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Railhouse Jock (&lt;em&gt;Jailhouse Rock&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Cereal That Will Make You Grow Tall&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Height Wheat (&lt;em&gt;White Heat&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Rousing the Warriors&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Waking the Braves (&lt;em&gt;Breaking the Waves&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nice puzzle.&lt;/p&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="14135" LastEditDate="2015-09-12T14:41:43.687" LastActivityDate="2015-09-12T14:41:43.687" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="22120" PostTypeId="2" ParentId="22114" CreationDate="2015-09-12T11:03:27.050" Score="7" Body="&lt;h3&gt;Ouchy Milk Farm&lt;/h3&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hurty Dairy =&gt; Dirty Harry&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="228" LastActivityDate="2015-09-12T11:03:27.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22121" PostTypeId="1" AcceptedAnswerId="22122" CreationDate="2015-09-12T11:48:12.387" Score="-3" ViewCount="1541" Body="&lt;p&gt;You need to add a single line and make the statement true&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$5+5=5$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All the numbers are made by matchsticks.&lt;/p&gt;&#xA;" OwnerDisplayName="user16282" LastActivityDate="2020-09-04T00:23:19.047" Title="Make the statement true!" Tags="|logical-deduction|enigmatic-puzzle|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22122" PostTypeId="2" ParentId="22121" CreationDate="2015-09-12T11:54:58.833" Score="3" Body="&lt;p&gt;Trivial Answer. &#xA;If they are made by matchsticks, then:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Add a single matchstick obliquely on equal sign to make it non-equality sign. After all 5 + 5 is not equal to 5&lt;br&gt;&#xA; $$5+5\ne5$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10618" LastEditorUserId="15544" LastEditDate="2015-09-12T12:43:23.637" LastActivityDate="2015-09-12T12:43:23.637" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22127" PostTypeId="2" ParentId="22121" CreationDate="2015-09-12T14:48:44.930" Score="1" Body="&lt;p&gt;An alternative solution:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$$5+5⩾5$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Excuse the fact that there's a rotation issue - the markup for equations doesn't appear to permit rotations to fix it.&lt;/p&gt;&#xA;" OwnerUserId="9289" LastEditorUserId="9289" LastEditDate="2020-09-04T00:23:19.047" LastActivityDate="2020-09-04T00:23:19.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22129" PostTypeId="1" AcceptedAnswerId="22157" CreationDate="2015-09-12T17:05:55.943" Score="10" ViewCount="798" Body="&lt;p&gt;There are 100 friends and $n$ enemies in a room. In the next room, there are $100+n$ open boxes, and $100+n$ keys, one corresponding to each box. They are numbered for easy identification. A box can be locked without using a key. Upto 2 keys can be placed in a box before locking it. A key cannot be locked inside its own box. The friends and enemies not yet called are allowed to have private chats while the game is in progress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One by one, people are called into the room. First a friend is called, then an enemy is called, then a friend, then an enemy and so on. If $n&amp;lt;100$, only friends will be called once the enemies have run out. A person can pick up 0,1 or 2 keys lying around and lock them into a single box. Then he comes back and makes a public statement(s) that can be heard by everyone. Then he leaves the game and can not communicate further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the king enters the room. He has a lockpick that can be used to break open exactly one box. After this, he will try to open all the boxes. The friends want him to succeed and the enemies want him to fail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the largest value of $n$ for which the friends guarantee a win?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. People can use computers for making decisions beyond human thinking.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4269" LastEditDate="2015-09-13T08:41:06.600" LastActivityDate="2015-09-13T20:22:20.810" Title="$100 + n$ boxes for the king!" Tags="|mathematics|logical-deduction|optimization|keys-and-locks|" AnswerCount="2" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="22130" PostTypeId="5" CreationDate="2015-09-12T17:14:21.320" Score="0" Body="&lt;p&gt;For puzzles about using keys to open locks and doors. Examples include:&lt;/p&gt;&#xA;&lt;p&gt;(i) Minimizing the number of keys required when strategically spread amongst different people (e.g. &lt;a href=&quot;https://puzzling.stackexchange.com/questions/112/the-puzzle-of-the-locked-box&quot;&gt;&lt;em&gt;The Puzzle of the Locked Box&lt;/em&gt;&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;p&gt;(ii) Finding the correct permutation of keys to open a set of locks (e.g. &lt;a href=&quot;https://puzzling.stackexchange.com/questions/25912/key-and-lock-puzzle&quot;&gt;&lt;em&gt;Key and lock puzzle&lt;/em&gt;&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="60644" LastEditDate="2020-09-23T14:02:27.747" LastActivityDate="2020-09-23T14:02:27.747" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22131" PostTypeId="4" CreationDate="2015-09-12T17:14:21.320" Score="0" Body="For puzzles about using keys to open locks and doors." OwnerUserId="4269" LastEditorUserId="11497" LastEditDate="2016-08-20T20:02:59.823" LastActivityDate="2016-08-20T20:02:59.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22132" PostTypeId="1" AcceptedAnswerId="22138" CreationDate="2015-09-12T18:10:23.367" Score="3" ViewCount="1988" Body="&lt;p&gt;I know the famous 12 coin, 3 weighings problem has been asked a few times already.  And I saw at least one solution which can be adapted to this problem.  Here's a generalisation (one of the many which could be done of course).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose you have 120 coins, all identical, except for 1 which is a slightly different weight.  You have a pan balance, and need to make as few weighings as possible to identify both (a) which coin is counterfeit, and (b) whether it's heavier or lighter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6mRa6m.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6mRa6m.png&quot; alt=&quot;Pan balance&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's no cheating, like adding coins on half way (weigh?) through - that would be an additional weighing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer may surprise some people.  I would like to know:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How many weighings are needed, and&lt;/li&gt;&#xA;&lt;li&gt;A reasonable strategy for doing this&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A common solution to the 12 coin puzzle involves...I can barely type the word...cases (&lt;em&gt;gasp&lt;/em&gt;)!  That would get tricky for 120 coins, but you are, of course, most welcome to do so!  I can't promise to read through all the cases and mark it right though.&lt;/p&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15544" LastEditDate="2015-09-12T18:12:42.403" LastActivityDate="2022-09-15T18:05:20.833" Title="12 Coins 3 weighings generalization" Tags="|logical-deduction|weighing|" AnswerCount="2" CommentCount="5" ClosedDate="2015-09-13T14:05:52.877" ContentLicense="CC BY-SA 3.0" />
  <row Id="22133" PostTypeId="2" ParentId="133" CreationDate="2015-09-12T18:32:24.523" Score="2" Body="&lt;p&gt;I came up with and solved this problem independently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let any set of riddles which should be enough to open the box be called &lt;em&gt;sufficient&lt;/em&gt;. Define a &lt;em&gt;nearly-sufficient&lt;/em&gt; set as one which should not open the box, but will become sufficient if any other riddle is solved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case given, the nearly-sufficient sets are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Both A riddles, any B riddle, and any C riddle (24 ways);&lt;/li&gt;&#xA;&lt;li&gt;Both A riddles, and any five C riddles (6 ways);&lt;/li&gt;&#xA;&lt;li&gt;All four B riddles, and any two C riddles (15 ways);&lt;/li&gt;&#xA;&lt;li&gt;Any B riddle, and any five C riddles (24 ways).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Every set that is not sufficient is a subset of at least one nearly-sufficient set (as long as it is still possible, add riddles to the set without making it sufficient; when it becomes impossible, the result is nearly-sufficient).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any nearly-sufficient set, there must be a key that none of the riddles in that set gives (otherwise you would be able to open the box after solving only that set of riddles). But every other riddle must give that key (otherwise it would not be possible to open the box even though the set of solved riddles would become sufficient). Call this key &lt;em&gt;pivotal&lt;/em&gt; for that nearly-sufficient set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No two nearly-sufficient sets can have the same pivotal key, so there must be at least one key for every nearly-sufficient set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, assign a pivotal key for every nearly-sufficient set. If we solve an insufficient set of riddles, then it is a subset of some nearly-sufficient set. We have not acquired the pivotal key for that set, so we cannot open the box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible for us to solve a sufficient set of riddles and not be able to open the box? If so, then we are still missing a key. That key is pivotal for one of the nearly-sufficient sets. If we do not have that key, then we have not solved any riddles that are not in that set. But this is impossible, because a sufficient set cannot be a subset of a nearly-sufficient set. So if we solve a sufficient set of riddles, then we will be able to open the box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the case in the question, there are 69 nearly-sufficient sets, so 69 keys are needed. For every nearly-sufficient set, all the riddles not in that set should give the corresponding key when solved.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-09-12T18:32:24.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22134" PostTypeId="2" ParentId="22027" CreationDate="2015-09-12T18:57:22.080" Score="9" Body="&lt;p&gt;The next number in this sequence is,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $80$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because the sequence is defined as such:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $a_0 = 120,\:a_1 = 106,\:a_n = \phi(a_{n-1} + a_{n-2})$&lt;br&gt;&#xA;&lt;br&gt;&#xA; Where $\phi$ denotes &lt;a href=&quot;https://en.wikipedia.org/wiki/Euler&amp;#39;s_totient_function&quot;&gt;Euler's totient function&lt;/a&gt;, defined as the number of integers between 1 and $x-1$ inclusive that are coprime to $x$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sequence continues,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $120,\,106,\,112,\,108,\,80,\,92,\,84,\,80,\,80,\,64,\,48,\,48,\,\overline{32,\,32,\,32},\,\dots$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11100" LastEditorUserId="11100" LastEditDate="2015-09-12T19:03:57.503" LastActivityDate="2015-09-12T19:03:57.503" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22135" PostTypeId="1" AcceptedAnswerId="22136" CreationDate="2015-09-12T18:57:25.627" Score="2" ViewCount="1331" Body="&lt;p&gt;I got the following puzzle in an aptitude test. What should the answer be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sE4hL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sE4hL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16318" LastEditorUserId="5373" LastEditDate="2017-04-28T12:54:24.840" LastActivityDate="2019-10-06T06:18:37.583" Title="Which is the missing pattern? Why?" Tags="|pattern|visual|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22136" PostTypeId="2" ParentId="22135" CreationDate="2015-09-12T19:05:47.160" Score="8" Body="&lt;p&gt;Each column contains two squares and one of every other shape. In each row, the squares and hexagons are one color, and the triangles and circles are the other color (they alternate rows).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first column needs another square, so the answer is A or D. In the fourth row, the squares and hexagons should be white, and the triangles and circles should be black. So &lt;strong&gt;D&lt;/strong&gt; is the answer.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-09-12T19:05:47.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22137" PostTypeId="2" ParentId="8285" CreationDate="2015-09-12T19:27:25.103" Score="6" Body="&lt;p&gt;Here is a more symmetrical solution, not minimal though.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/5wOFN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5wOFN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;PS: That solution was also given there: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/8359/touching-matchsticks-with-compass-and-straightedge&quot;&gt;Touching matchsticks with compass and straightedge&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-09-12T19:40:40.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22138" PostTypeId="2" ParentId="22132" CreationDate="2015-09-12T23:20:45.160" Score="3" Body="&lt;p&gt;First, number the coins using Balanced Ternary. This is a ternary numeral system using the digits -, 0, and + to represent -1, 0, and 1.&lt;/p&gt;&#xA;&lt;p&gt;For the sake of brevity I'm going to illustrate this using 12 coins, but this can be easily extended to larger numbers. Just add more trits to each number. (For example, 120 coins would require 5 trits, corresponding to 5 weighings.)&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1 = 00+&lt;/li&gt;&#xA;&lt;li&gt;2 = 0+-&lt;/li&gt;&#xA;&lt;li&gt;3 = 0+0&lt;/li&gt;&#xA;&lt;li&gt;...&lt;/li&gt;&#xA;&lt;li&gt;11 = ++-&lt;/li&gt;&#xA;&lt;li&gt;12 = ++0&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Each of these can also be inverted by swapping each symbol.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;-1 = 00-&lt;/li&gt;&#xA;&lt;li&gt;-2 = 0-+&lt;/li&gt;&#xA;&lt;li&gt;-3 = 0-0&lt;/li&gt;&#xA;&lt;li&gt;...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;This numbering now tells us where to place the coins.&lt;/p&gt;&#xA;&lt;p&gt;For the first weighing, put all the coins with a + in their first trit on the right pan, and all the coins with a - in their first trit in the left pan. Leave all the coins with a 0 in their first trit off. Then record the result: if it tips to the left, that's a -. If it tips to the right, that's a +. If it is balanced, that's a 0.&lt;/p&gt;&#xA;&lt;p&gt;Repeat as many times as necessary, then look at the number you've recorded. With 12 coins there will be three weighings, so the result might be e.g. +-+ (tipped to the right, tipped to the left, tipped to the right). This is the number of the false coin. If it's positive, that coin is heavier. If it's negative, that coin is lighter. If the number is zero, no coin was fake.&lt;/p&gt;&#xA;&lt;p&gt;HOWEVER, this doesn't actually work precisely as I have written it. No positive number has a - in its first trit, so the first weighing will have one pan empty. Unless you have a stock of real coins to compare against (a variant of the problem), it won't be a useful measurement.&lt;/p&gt;&#xA;&lt;p&gt;So we need to invert some of the coins' numbers (so coin 5 might become coin -5), then interpret the result differently if one of these is chosen: a negative number will now mean that the fake coin is heavier, and a positive number that it is lighter.&lt;/p&gt;&#xA;&lt;p&gt;This is where I got stuck in my own experimentation, but some web research indicates that John Conway came up with &lt;a href=&quot;https://web.archive.org/web/20170929193800/http://mathforum.org/kb/message.jspa?messageID=1085028&amp;amp;tstart=0&quot; rel=&quot;nofollow noreferrer&quot;&gt;a very clever way&lt;/a&gt; to determine which ones to flip: go through the coins, and look at the first &amp;quot;change&amp;quot; in the sequence. If it is - to 0, or 0 to +, or + to -, keep it positive. Otherwise, invert that number.&lt;/p&gt;&#xA;&lt;p&gt;This isn't as elegant as I would like—it feels like there should be some way to decide mathematically which numbers to invert. But for now, that solution is the best I have.&lt;/p&gt;&#xA;" OwnerUserId="16319" LastEditorUserId="11110" LastEditDate="2022-09-15T18:05:20.833" LastActivityDate="2022-09-15T18:05:20.833" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22139" PostTypeId="2" ParentId="22135" CreationDate="2015-09-13T00:26:14.057" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If there are matching shapes in any row they must be of the same colour -             any remaining shapes must be the opposite colour.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Each row requires either 2 or 3 squares. Each column requires 2 squares.&lt;br&gt;&#xA; Each row &amp;amp; column requires 1 triangle &amp;amp; 1 hexagon.&lt;br&gt;&#xA; Each row requires either 1 or 2 circles. Each column requires 1 circle.&lt;br&gt;&#xA;&lt;br&gt;&#xA; D is the only answer that works.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14317" LastEditorUserId="11497" LastEditDate="2019-10-06T06:18:37.583" LastActivityDate="2019-10-06T06:18:37.583" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22140" PostTypeId="2" ParentId="22132" CreationDate="2015-09-13T01:25:48.477" Score="1" Body="&lt;p&gt;Each weighing allows you to triple the number of coins, so you should be able to do 36 coins in four weighings, 108 in five, but require six for 121 coins.  The solution for 12 coins only gets complex for the endgame.  For 108, you split them into three batches of 36 and weigh two batches.  If they balance, you are down to the 36 case.  If not, you have 36 H coins (which might be heavy) and 36 L coins (which might be light).  Weigh 12H+12L against 12H+12L.  No matter the result, you now have 12H+12L and can weigh 4H+4L against 4H+4L.  No matter the result, you are in the 12 coin case where the first weighing did not balance.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2015-09-13T01:25:48.477" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22143" PostTypeId="1" AcceptedAnswerId="22151" CreationDate="2015-09-13T05:15:01.013" Score="10" ViewCount="987" Body="&lt;p&gt;You have an 8x8 board, and you must partially cover it with a single kind of tetromino in such a way that it is impossible to place any additional tetrominoes of that kind in the empty spaces. Additionally, no two tetrominoes may touch by their edges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I consider tetrominoes that are reflections of each other to be different, so there are 7 possible choices:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XteR3.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XteR3.gif&quot; alt=&quot;I, J, L, O, S, T, Z&quot;&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;For which tetrominoes is this possible, and how many pieces does it require (minimum)?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="11100" LastEditorUserId="11100" LastEditDate="2015-09-13T05:21:53.980" LastActivityDate="2015-09-15T00:21:23.440" Title="Occupy a field with your choice of tetromino" Tags="|tiling|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22144" PostTypeId="2" ParentId="22143" CreationDate="2015-09-13T06:06:05.297" Score="5" Body="&lt;p&gt;&lt;em&gt;Partial answer.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is impossible for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I&lt;br&gt;&#xA; Any piece that is placed must have the adjacent sides empty, leaving a shadow that exactly matches its own shape.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is possible for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; O&lt;br&gt;&#xA; Place them all one square apart in a lattice.&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/MAetO.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MAetO.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is impossible for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; J/L.&lt;br&gt;&#xA; I'm not good at drawing pretty pictures, but the trouble is that at some point you will have a 3-flat exposed to the inside of the square instead of being able to flush up against the edge. There are too many holes that need to be filled, which always ends up leaving a piece-sized gap.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-09-14T14:46:33.973" LastActivityDate="2015-09-14T14:46:33.973" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22146" PostTypeId="2" ParentId="22129" CreationDate="2015-09-13T09:54:26.300" Score="4" Body="&lt;p&gt;&lt;strong&gt;Partial Start&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some quick rules I could figure out:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Perfect Information&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everyone remaining can know what is happening. For example, each friend can make announcement regarding which keys and boxes s/he has locked, and which keys and boxes were locked by previous enemy (based on info of previous friend making announcement). This means everyone know what the current graph structure is. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Winning Conditions&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You win if there exists a directional graph from at least one point when it is possible to traverse all nodes. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Losing Conditions&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There exists two or more graphs which do not share any node with each other, OR [they only share sink nodes with each other AND have at least one other node]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Friend non-move&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  No friend must constitute must lock box B with only one key K if box K and key B are still in play. Enemy then can simply lock box K with key B, making it isolated graph with rest of structure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I will try to come up with some more thinking when I have more than 10 minutes to spare. &lt;/p&gt;&#xA;" OwnerUserId="10618" LastEditorUserId="16322" LastEditDate="2015-09-13T13:12:25.507" LastActivityDate="2015-09-13T13:12:25.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22147" PostTypeId="1" AcceptedAnswerId="22149" CreationDate="2015-09-13T14:12:40.177" Score="24" ViewCount="7526" Body="&lt;p&gt;&lt;em&gt;I read this puzzle in an Italian newspaper some days ago. It reported it was asked at one of the latest Mathematical Olympiads. It's not difficult, but it requires some logical thinking. I'm posting it here as I'd like to know if you reach the same conclusion as me.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a village live two kinds of people: those who always lie and those who always tell the truth. Everybody knows each other in the village and therefore knows whether a particular villager is a liar or a truth-teller.&lt;br&gt;&#xA;Strolling around the village we meet a group of four villagers. We ask to each one of them: &quot;How many liars are in this group?&quot; and we get these answers: &quot;0&quot;, &quot;1&quot;, &quot;2&quot;, &quot;3&quot;.&lt;br&gt;&#xA;How many liars are in that group? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;EDIT: There is probably an error in the riddle as published in the newspaper. If the answer we get from the group of villagers is instead&lt;/em&gt; &quot;1&quot;, &quot;2&quot;, &quot;3&quot;, &quot;4&quot; &lt;em&gt;then there is an unique solution (as reported in the accepted answer below).&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="16323" LastEditorUserId="16323" LastEditDate="2015-09-14T14:24:31.713" LastActivityDate="2023-05-23T21:00:49.197" Title="The village of the liars and truth-tellers" Tags="|logical-deduction|liars|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22149" PostTypeId="2" ParentId="22147" CreationDate="2015-09-13T14:26:15.123" Score="31" Body="&lt;p&gt;The answer is... &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...either 3 or 4:&lt;br&gt;&#xA; As all four answers are different, at most one of them can be true.&lt;br&gt;&#xA; Hence either all of them are liars, or the guy with the answer &quot;3&quot; is a truth-teller.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="16323" LastEditDate="2015-09-13T14:33:01.200" LastActivityDate="2015-09-13T14:33:01.200" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="22150" PostTypeId="1" AcceptedAnswerId="22152" CreationDate="2015-09-13T14:40:18.170" Score="10" ViewCount="840" Body="&lt;p&gt;Alice, Bob and Carol sit around a round table. Initially Alice has $n$ cookies, while Bob and Carol have none. Alice divides her cookies evenly between Bob and Carol; if there is a leftover cookie (as $n$ is odd) then Alice eats it.&#xA;Then Bob does the same with his cookies, then Carol does the same with hers. And so on, and so on, and so on.&#xA;The process continues forever: in a round-robin fashion, everybody divides his/her cookies evenly between the other two persons and eats leftovers if there are any.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: For which starting values of $n\ge1$ does the process continue forever WITHOUT a single cookie being eaten?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="2383" LastEditDate="2015-09-13T19:15:34.353" LastActivityDate="2015-10-08T14:54:19.210" Title="Alice, Bob and Carol, and many cookies" Tags="|mathematics|combinatorics|game|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22151" PostTypeId="2" ParentId="22143" CreationDate="2015-09-13T15:21:26.413" Score="9" Body="&lt;p&gt;&lt;em&gt;A partial solution complementary to IanMacDonald's.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible for this pair&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; S and Z&lt;br&gt;&#xA; Place them in a fairly symmetric array, as follows. S is mirror image of Z.&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/K5EIY.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/K5EIY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA; Now to prove that the minimal number of pieces is eight!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is another tile that works:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's &quot;T&quot; time [again]:&#xA; &lt;a href=&quot;https://i.stack.imgur.com/75oZu.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/75oZu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA; &lt;del&gt;Again this seems to require eight tiles&lt;/del&gt;It looks like this requires seven tetrominos.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Added proof that it is impossible to get out of &quot;JaiL&quot;&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;There is a pair of tetrominos that are unable to tile the chessboard.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is a fairly exhaustive (and exhausting) proof that J tetromino (and by symmetry the L tetromino) cannot tile the board.&lt;br&gt;&#xA; The proof starts by showing that tiling is impossible on an infinite board.&lt;br&gt;&#xA; Then to avoid this on the chessboard, all pieces must be in close proximity to an edge or corner.&#xA; The next diagram shows there are three mandatory cells around every piece (blue) - these must be occupied or there will be a vacancy (gray) of the J shape. These are shown in light blue, assuming the playing field is rotated so that piece 1 is oriented as shown.&#xA; &lt;a href=&quot;https://i.stack.imgur.com/nhMJU.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nhMJU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &#xA; Then there are four possible positions of a second piece to cover the mandatory cell on the right. Two of these are shown in the next diagram. Both of these lead to a J vacancy because the second piece introduces new mandatory cells that cannot be occupied simultaneously (the red X's). &lt;em&gt;Neither of these can be moved close to the top edge of a chessboard to avoid vacancies. In the first case, the gray J on the left will then be unavoidable. In the second case, rotational symmetry means the upside-down arrangement is very far from the new top edge.&lt;/em&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/MSEtA.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MSEtA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA; Then one other orientation immediately leads to a J vacancy, so there is only one possible orientation of the second piece relative to the first. This applies to a third piece relative to the second, and so on. So the result is a closed figure with an interior that cannot accomodate any more non-touching J pieces, but which has J vacancies.&#xA; &lt;a href=&quot;https://i.stack.imgur.com/cmZGP.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cmZGP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA; In the last case, the cyclic arrangement can be avoided by placing the first two tetrominos against the top edge. But then, one of the next three arrangements results (the case where there is one column to the left is trivial), where placement of all the light blue pieces is forced, resulting in a J vacancy or touching edges in each case.&#xA; &lt;a href=&quot;https://i.stack.imgur.com/CSkbT.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CSkbT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA; The only other way to avoid one of these arrangements is to place tetromino 1 so that it is abutting the right edge of the chessboard, so that there is no need to place tetromino 2 as in previous arrangements. Then **all tetrominos* must be placed this way with the &quot;top&quot; of the J touching an edge. The resulting arrangement leaves a 2x2 vacancy at the center of the board and hence the four mandatory cells in the next figure. Then J pieces can be placed in one of two symmetric ways (otherwise one of the mandatory cells will be vacant). But then it is impossible not to leave a J vacancy.&#xA; &lt;a href=&quot;https://i.stack.imgur.com/3yMjn.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3yMjn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14080" LastEditorUserId="14080" LastEditDate="2015-09-15T00:21:23.440" LastActivityDate="2015-09-15T00:21:23.440" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="22152" PostTypeId="2" ParentId="22150" CreationDate="2015-09-13T16:06:20.800" Score="9" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; there is no such number.&lt;br&gt;&lt;br&gt;Let's define the amount of cookies at the beginning as $1$, and look at the fractions everyone has after each round:&lt;br&gt;&lt;br&gt;$\frac{1}{2}\space\frac{1}{2}$&lt;br&gt;&lt;br&gt;$\frac{3}{4}\space\frac{1}{4}$&lt;br&gt;&lt;br&gt;$\frac{5}{8}\space\frac{3}{8}$&lt;br&gt;&lt;br&gt;$\frac{11}{16}\space\frac{5}{16}$&lt;br&gt;&lt;br&gt;$\frac{21}{32}\space\frac{11}{32}$&lt;br&gt;&lt;br&gt;$\frac{43}{64}\space\frac{21}{64}$&lt;br&gt;&lt;br&gt;$...$&lt;br&gt;&lt;br&gt;$\frac{n}{n+m}\space\frac{m}{n+m}$&lt;br&gt;&lt;br&gt;$\frac{n+2*m}{2(n+m)}\space\frac{n}{2(n+m)}$&lt;br&gt;&lt;br&gt;As can be seen above all numerators are odd, and all denominators are a power of $2$. And this will stay that way, because every new numerator is $2*previous \space numerator$ which gives an even number and after adding another odd numerator it becomes odd again. That way the fraction can never be reduced, and we never get a loop which would be required for this to work.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Extension to 4 friends (requested by  ghosts_in_the_code):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The proof works similar to the above case, we start with $1$ again:&lt;br&gt;&lt;br&gt;$\frac{1}{3}\space\frac{1}{3}\space\frac{1}{3}$&lt;br&gt;&lt;br&gt;$\frac{4}{9}\space\frac{4}{9}\space\frac{1}{9}$&lt;br&gt;&lt;br&gt;$\frac{16}{27}\space\frac{7}{27}\space\frac{4}{27}$&lt;br&gt;&lt;br&gt;$\frac{37}{81}\space\frac{28}{81}\space\frac{16}{81}$&lt;br&gt;&lt;br&gt;$\frac{121}{243}\space\frac{85}{243}\space\frac{37}{243}$&lt;br&gt;&lt;br&gt;$...$&lt;br&gt;&lt;br&gt;$\frac{n}{n+m+o}\space\frac{m}{n+m+o}\space\frac{o}{n+m+o}$&lt;br&gt;&lt;br&gt;$\frac{n+3*m}{3(n+m+o)}\space\frac{n+3*o}{3(n+m+o)}\space\frac{n}{3(n+m+o)}$&lt;br&gt;&lt;br&gt;None of the numerators in the example is divisible by $3$. New numerators are created by $3*previous \space numerator$ which is obviously divisible by $3$, and adding of another numerator (not divisible by $3$) which makes the new numerator not divisible by $3$ again.&lt;br&gt;&lt;br&gt;This can be extended to any number of friends.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2015-09-17T19:24:50.933" LastActivityDate="2015-09-17T19:24:50.933" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22154" PostTypeId="2" ParentId="22150" CreationDate="2015-09-13T17:26:27.687" Score="1" Body="&lt;p&gt;There are no such solutions!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this explicitly we can calculate how many cookies the 3 people ( A, B and C ) have at a given stage. To do this I altered the problem by assuming that after person A (Alice) shares their cookies for the first time, everyone gives their cookies to the person alphabetically before them: so C gives to B, B gives to A and A gives to C. In practice this gives the following for the first 2 stages:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At stage 1: A has n, B has 0 and C has 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At stage 2: A has n/2, B has n/2 and C has 0  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can repeat this procedure of A halving the cookies and then the cookies being shifted along (so now &lt;strong&gt;only&lt;/strong&gt; Alice halves her cookies). This generates the same fractions to appear at each stage as the original problem yet different people will have different fractions. (The shift essentially changes A into person B on stages that are 2 mod 3, C on 0 mod 3 stages and A on 1 mod 3 stages)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(note C will always have 0 cookies)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this we can calculate how many cookies people have at a given stage in the new altered problem (and this will be a permutation of the solution to the original problem as discussed above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let A(i),B(i) denote the cookies A and B have a stage i.&#xA;We have $A(1) = n$ and $B(1) = 0 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now consider how A(i) changes in the altered problem, we deduce:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A(i+1) = B(i) + \frac {A(i)} {2}  $       &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $B(i+1) = \frac{A(i)} {2}   $                   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore $B(i) = \frac{A(i-1)} {2}$ from that second equation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting this back in to the first gives (after arranging):&#xA;$2A(i+2) - A(i+1) - A(i) = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a standard difference equation which we can solve with the initial conditions I wrote above. I'll leave out the algebra. This gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ A(i) =  \frac {2n}{3}  \frac{2^i - (-1)^i} {2^i}$&#xA;and since no cookies are eaten B(i) = n - A(i)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And from this we know how many cookies A has in the altered problem which implies someone in the original problem has A(i) cookie at a given stage. We now note that A(i) must always be an integer for the cookies to be halved correctly without one being eaten. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However it is clear that $(2^i - (-1)^i)$ is always an odd number so for A(i) to be an integer for all i, 2^i must divide n for all positive intgers i. This is not possible unless we are in the trival case in which n=0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(note it is possible to calculate exactly how many cookies each person has at a stage by combining the solution to the difference equation with my comment :'(The shift essentially changes A into person B on stages that are 2 mod 3, C on 0 mod 3 stages and A on 1 mod 3 stages)')&lt;/p&gt;&#xA;" OwnerUserId="16326" LastEditorUserId="16326" LastEditDate="2015-09-13T17:38:31.853" LastActivityDate="2015-09-13T17:38:31.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22157" PostTypeId="2" ParentId="22129" CreationDate="2015-09-13T19:22:24.097" Score="7" Body="&lt;p&gt;The limit for n is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Surprisingly, 101!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;On the first turn, the friends will place 2 random keys into box A. This is the box which the king will open. Consider a box &lt;strong&gt;openable&lt;/strong&gt; if it is box A or its key is in an openable box. The goal is to have all locked boxes be openable by the end. This is currently the case.&#xA;&lt;br&gt;&lt;br&gt;&#xA;For the next action by the enemies - putting keys x or (x,y) into box Z - key z must be available or in an openable box. If z is available, place it and random key w into an openable box (otherwise just place v,w). The point of placing w is to always keep at least one unlocked, openable box&#xA; &lt;br&gt;&lt;br&gt;&#xA;If the enemies take the last turn, all boxes except those for the remaining keys are openable, and the enemy may not lock the keys into those boxes. Victory! &lt;br&gt;&lt;br&gt;&#xA; The enemies can, however, construct a loop with 2 consecutive turns, so $n$ large enough to allow that is the limit. Since the friends use 2 keys per turn, there must be at least 202 keys to defeat them, hence $n=101$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3987" LastEditorUserId="3987" LastEditDate="2015-09-13T20:22:20.810" LastActivityDate="2015-09-13T20:22:20.810" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="22158" PostTypeId="2" ParentId="133" CreationDate="2015-09-13T20:38:21.657" Score="1" Body="&lt;p&gt;Instead of keys, have the questions enable access to pebbles. The final box requires you to tip a balance by putting a certain number of pebbles on the scale. A puzzle gives you pebbles essentially based on the score of the puzzle. For example, you want all these to be equivalent:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;One level A riddle and two level B riddles.&lt;/li&gt;&#xA;&lt;li&gt;One level A riddle, one level B riddle, and two level C riddles.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This tells me a B is worth two Cs.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Two level B riddles and three level C riddles.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This tells me an A is worth 3 Cs&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;All six level C riddles.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This actually makes you slightly inconsistent. Using C=1, B=2, and A=3 the four criteria you have are worth 7, 7, 7, and 6 points. So change the last rule to &quot;all seven level C riddles&quot; and you basically just need them to earn 7 pebbles somehow. Or keep it as-is but give them a one-pebble bonus for answering all the puzzles of a given level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simple to explain and to describe.&lt;/p&gt;&#xA;" OwnerUserId="1509" LastActivityDate="2015-09-13T20:38:21.657" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22163" PostTypeId="1" AcceptedAnswerId="22164" CreationDate="2015-09-13T21:40:35.587" Score="1" ViewCount="370" Body="&lt;p&gt;We have 3 closed boxes. One contains apples and the second contains oranges and the third contains apples and oranges . They have labels &lt;code&gt;Apples&lt;/code&gt;, &lt;code&gt;Oranges&lt;/code&gt;, &lt;code&gt;Apple &amp;amp; Oranges&lt;/code&gt; but all the labels are put on the wrong boxes. How do you put the labels on boxes correctly if you can only open one box to see what is inside ?&lt;/p&gt;&#xA;" OwnerUserId="16329" LastEditorUserId="15544" LastEditDate="2015-09-13T21:55:19.477" LastActivityDate="2015-09-13T22:11:41.323" Title="The 3 closed boxes" Tags="|logical-deduction|" AnswerCount="2" CommentCount="8" ClosedDate="2015-09-14T03:16:34.023" ContentLicense="CC BY-SA 3.0" />
  <row Id="22164" PostTypeId="2" ParentId="22163" CreationDate="2015-09-13T21:50:05.420" Score="1" Body="&lt;p&gt;If picking from Apples is right, by symmetry so is picking from Oranges.  Also if you pick from Apples and get an orange, you don't know if the box is Oranges or mixed.  You have to pick from Apples &amp;amp; Oranges.  Whatever you get, you know what box it is.  Then the fact that both the remaining labels are wrong will let you sort out the other two.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rereading the question, you can open any box you like.  If you open Apples and see Oranges, the Apples label goes on the box that currently says Apples and Oranges.  When I heard this the first time, you were only allowed to draw one fruit from the box you picked. In that case it must be the Apples and Oranges box.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditDate="2015-09-13T22:11:41.323" LastActivityDate="2015-09-13T22:11:41.323" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22165" PostTypeId="2" ParentId="22163" CreationDate="2015-09-13T22:04:54.003" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Look at the box labelled &quot;Apples&quot;. If there are only oranges in it, then the one labelled as mixed has only non-oranges (apples) and the remaining one both. If there are both in the first box, the one labelled with the non-apple (orange) is actually the apple box and the remaining one the orange box. By a similar logic you can also choose the one labelled as an orange box thanks to the symmetry.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastActivityDate="2015-09-13T22:04:54.003" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22166" PostTypeId="2" ParentId="22150" CreationDate="2015-09-13T22:09:12.673" Score="8" Body="&lt;p&gt;The answer, as already others have said, is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; there are no $n$ for which this is true.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is another proof why. Assume a cookie is never eaten. We can think of the situation as a series of states, $s_0,s_1,s_2,\dots$, where each $s_i$ is a triple $(a,b,c)$ of how many cookies each person has. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually, a state must repeat. Suppose that $s_i$ is the first repeated state, and the first time it repeats is at step number $j$ (so $j&amp;gt;i$ and $s_j=s_i$). There are two possibilities: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;CASE 1&lt;/strong&gt;: $i\ge1$. This means that $s_{i-1}$ and $s_{j-1}$ are distinct states which both lead to state $s_i$. This is impossible: when no cookies are eaten, each state has a unique state which leads to it. For example, the unique state leading to $(a,b,0)$ is $(a-b,0,2b)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;CASE 2&lt;/strong&gt;: $i=0$. This means that the state $s_{j-1}$ leads to the state $s_j=s_0=(n,0,0)$. Again, this is impossible: as long as no cookies are eaten, then after each move, there will be two people who have cookies, so no state can lead to $(n,0,0)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Either way, assuming no cookies are eaten leads to a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-09-13T22:09:12.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22167" PostTypeId="2" ParentId="19920" CreationDate="2015-09-14T02:04:22.143" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The last city is &lt;strong&gt;Tallinn&lt;/strong&gt;, with &quot;towering&quot; meaning &quot;tall&quot; and &quot;hotel&quot; meaning &quot;inn&quot;. It also sounds like the foot of a bird (&quot;talon&quot;). Its old name was Reval, which sounds like &quot;reveal&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13928" LastActivityDate="2015-09-14T02:04:22.143" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22168" PostTypeId="2" ParentId="17277" CreationDate="2015-09-14T03:22:44.073" Score="1" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But only if the matrix contains contradictory information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the errors have yet to cause contradictory information to appear in an incomplete grid, then the answer really is no, you have the 2 approaches (apart from cheating and checking the answers page of your puzzle book) that can determine if the grid is bad, and which to use, as you have probably guessed, depends on how much progress has been made on the puzzle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just as the matrix can be used in several ways to spot connections it can also be used to spot information that is obviously bad by similar analysis of the pairs of crosses that share rows and columns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The partial grid below shows crosses that assert truth. (eg the Dane lives in the white house) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QvabZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QvabZ.png&quot; alt=&quot;Einstein grid errors example&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each colour being a group that together are bad. Each group containing at least one cross that is wrong. In these examples you can remove a single cross and the other crosses of that colour are no longer contradictory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember testing for obvious errors can only prove an answer to be bad. It can not prove it to be good.&lt;/p&gt;&#xA;" OwnerUserId="11297" LastActivityDate="2015-09-14T03:22:44.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22169" PostTypeId="2" ParentId="22147" CreationDate="2015-09-14T05:10:07.623" Score="-4" Body="&lt;p&gt;I agree with Gamow's reasoning but I think the answer is a bit wrong, why??&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...either 3 or 4:&#xA;  As all four answers are different, at most one of them can be true.&#xA;  Hence either all of them are liars, or the guy with the answer &quot;3&quot; is a truth-teller&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That much is true, but come to think again , if he told the truth then the other three are all liars.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Now on answering the answer about other three villagers &lt;strong&gt;non of the liars would have answered 2&lt;/strong&gt; (i.e 2 liars and 1 truth teller) because by that he would be telling the truth which makes the reasoning invalid&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the only possible answer here is that &lt;strong&gt;all of them (all 4 villagers) are liars&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="16333" LastEditorUserId="16333" LastEditDate="2015-09-14T06:45:25.563" LastActivityDate="2015-09-14T06:45:25.563" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="22170" PostTypeId="2" ParentId="22045" CreationDate="2015-09-14T10:05:59.823" Score="5" Body="&lt;p&gt;Are you:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A hinge&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A mere tool I am,&lt;br&gt;&#xA;In a world so grown,&lt;br&gt;&#xA;Yet here I am,&lt;br&gt;&#xA;Taking you on.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A hinge is a simple tool.  Also it is angled, as was mentioned in the puzzle preface.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm a supporter&lt;br&gt;&#xA;of things well known,&lt;br&gt;&#xA;Without me, these portals,&lt;br&gt;&#xA;you would never call your own.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Doors&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And kid yourself not,&lt;br&gt;&#xA;I am almost the same,&lt;br&gt;&#xA;As when I say &quot;The fate of the world&lt;br&gt;&#xA;depends on&quot; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Depends on = hinges on&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of which I am made,&lt;br&gt;&#xA;Is the Swan's Song of Stars,&lt;br&gt;&#xA;And it's neighbors,&lt;br&gt;&#xA;they may sound but&lt;br&gt;&#xA;unattractive they are.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hinges are typically made of brass, which is metal.&lt;br&gt;&#xA; Swan's song of stars I believe refers to the fact that metals come from the remnant of stars.&lt;br&gt;&#xA; Unattractive I believe refers to the fact that brass is not magnetic. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What I am made of that I mentioned,&lt;br&gt;&#xA;is an ingredient among a few,&lt;br&gt;&#xA;among its kind it lies,&lt;br&gt;&#xA;an elemental country not new.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think this means that brass is made of copper and zinc, which are two among a few metals on the periodic table.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Those famous things I support,&lt;br&gt;&#xA;Once spawned beings of such form,&lt;br&gt;&#xA;So formidable these off-springs,&lt;br&gt;&#xA;Rode this world's storms.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Storm doors, which have springs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4479" LastActivityDate="2015-09-14T10:05:59.823" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22171" PostTypeId="2" ParentId="22110" CreationDate="2015-09-14T15:00:55.513" Score="6" Body="&lt;p&gt;Gene Hackman movies.  I have found 49 in total (highlighted bold below).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When this man was &lt;strong&gt;young, Frankenstein&lt;/strong&gt; films inspired him to take&#xA;  up acting, but during a performance at &lt;strong&gt;The Rep (L.A.) cement&#xA;  s&lt;/strong&gt;plashed on his face so he switched to directing for a while and&#xA;  there were books he wanted to &lt;strong&gt;abridge too.&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Far&lt;/strong&gt; from high arts and stories of &lt;strong&gt;class, action&lt;/strong&gt; movies called out to him - those where he could let &lt;strong&gt;loose: cannons&lt;/strong&gt; firing with a&#xA;  &lt;strong&gt;bite, the bullet&lt;/strong&gt;s whizzing &lt;strong&gt;behind, enemy lines&lt;/strong&gt; of infantry yelling &lt;strong&gt;&quot;Geronimo&quot;, an American legend&lt;/strong&gt; striding out of San&#xA;  Fran&lt;strong&gt;cisco, pike&lt;/strong&gt; swinging like a scy&lt;strong&gt;the - quick - and the dead&lt;/strong&gt;&#xA;  piled up all around while the sound of mor&lt;strong&gt;tar get&lt;/strong&gt;s nearer.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Not everyone thought he was &lt;strong&gt;super - man&lt;/strong&gt;y saw him as an &lt;strong&gt;enemy.&#xA;  Oft he state&lt;/strong&gt;d &quot;After my first Oscar nomination I let my ego &lt;strong&gt;run&#xA;  away. Jury&lt;/strong&gt; after jury failed to give me the top award by a &lt;strong&gt;narrow&#xA;  margin&lt;/strong&gt;.&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Don't for&lt;strong&gt;get short y&lt;/strong&gt;oung men could succum&lt;strong&gt;b at 21&lt;/strong&gt; to the&#xA;  &lt;strong&gt;power&lt;/strong&gt;, the &lt;strong&gt;absolute power&lt;/strong&gt; of Hollywood. Many an un&lt;strong&gt;lucky lady&lt;/strong&gt; has found that they are &lt;strong&gt;heartbreakers&lt;/strong&gt; who w&lt;strong&gt;ant z&lt;/strong&gt;ero ties - a&#xA;  couple of &lt;strong&gt;hoosiers&lt;/strong&gt; &lt;strong&gt;all night long.&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;The firm&lt;/strong&gt; way he handled his first wife led to their &lt;strong&gt;split. Decisions&lt;/strong&gt; regarding &lt;strong&gt;another woman&lt;/strong&gt; led to th&lt;strong&gt;under, fire&lt;/strong&gt;works and&#xA;  &lt;strong&gt;extreme measures&lt;/strong&gt;. He tried to explain but before the &lt;strong&gt;conversation&lt;/strong&gt; had beg&lt;strong&gt;un, forgiven&lt;/strong&gt;ess evaporated. S&lt;strong&gt;he is t&lt;/strong&gt;ypical of &lt;strong&gt;doctor's wives&lt;/strong&gt; from &lt;strong&gt;Mississippi - burning&lt;/strong&gt; with rage and&#xA;  often turning &lt;strong&gt;crimson.&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Tide&lt;/strong&gt;s turn, though and he was still someone to look de&lt;strong&gt;wy at. Tear-p&lt;/strong&gt;atterns streaming down the cheeks of women were not&#xA;  &lt;strong&gt;uncommon.&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Valour&lt;/strong&gt; during his military service in his &lt;strong&gt;twilight&lt;/strong&gt; years meant he was leader of &lt;strong&gt;the pack. Age&lt;/strong&gt; didn't matter. He was decorated for&#xA;  his courageous efforts &lt;strong&gt;twice.&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;In a lifetime&lt;/strong&gt; of travelling the world he has developed great respect for &lt;strong&gt;the Mexican&lt;/strong&gt; people he met, but not for the &lt;strong&gt;French.&#xA;  Connection&lt;/strong&gt;s overseas and at home have said that he has always been&#xA;  great &lt;strong&gt;company - business&lt;/strong&gt; conversations in &lt;strong&gt;March or die&lt;/strong&gt;cast&#xA;  modelling of a &lt;strong&gt;scarecrow&lt;/strong&gt; at &lt;strong&gt;night.&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Move s&lt;/strong&gt;everal years later and he is no longer in his &lt;strong&gt;prime. Cut&lt;/strong&gt; from the A list he is no longer desi&lt;strong&gt;red. S&lt;/strong&gt;peaking to the media about&#xA;  his chances of playing leading man again, they say &quot;&lt;strong&gt;No Way!&quot;.&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;&lt;strong&gt;Out&lt;/strong&gt;rageous&quot; is my response.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14233" LastEditorUserId="14233" LastEditDate="2015-09-14T16:52:56.070" LastActivityDate="2015-09-14T16:52:56.070" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22172" PostTypeId="2" ParentId="22068" CreationDate="2015-09-14T15:02:48.840" Score="5" Body="&lt;h2&gt;PARTIAL ANSWER&lt;/h2&gt;&#xA;&lt;p&gt;Here's what I've come up with so far. I'm very close; I just need to finish the upper-right corner. I was too lazy to put letters into the crossword image, so hopefully someone else can verify that this all works. (Sidenote: This is my first answer on this website, so any suggestions on proper formatting would be appreciated!)&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Edit 1:&lt;/strong&gt; Answered 1 Down.&lt;/p&gt;&#xA;&lt;h2&gt;Horizontal&lt;/h2&gt;&#xA;&lt;p&gt;4 Battlewise bravery&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Valor → Vapor&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;5 Sans sensation&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Dull → Dill&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;8 Medical mesh material&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Gauze → Gauge&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;10 Anaerobically assembled acid&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Lactic → Tactic&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;12 Opposite of old origin offensive&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Occidental → Accidental&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;13 Pertaining preposition&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; About → Abort&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;15 Sound system segment&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Speaker → Sneaker&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;20 Decided determination&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Resolve → Revolve&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;21 Observe outlook on oracular orbs&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Peer → Seer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;22 Rifle rest&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Bipod → Biped&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;23 Devilish, deemed Dante&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Material → Maternal&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;25 Constant current classification&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Direct → Direst&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;28 Vowelless vituperation&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Tsk → Ask&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;29 African antelope&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Oryx → Onyx&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;30 Painlessly punish progeny&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Ground → Around&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;Vertical&lt;/h2&gt;&#xA;&lt;p&gt;1 Shadowy Sherlock&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Sleuth → Hustle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;2 Charismatic quality&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Charm → March&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;3 Hired helping hand&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Aide → Idea&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;6 Sprinkles salvation&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Anoints → Nations&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;7 Pixelated puerile protagonist&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Link → Kiln&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;9 Cosmological cloud&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Nebula → Unable&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;11 Dull discomforts&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Aches → Chase&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;14 Rising rocket requirement&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Thrust → Truths&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;16 Ambush assault&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Raid → Arid&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;17 Spear's similar&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Pike → Kepi&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;18 Deviate direction&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Veer → Ever&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;19 Accomplish again&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Redo → Rode&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;24 Metal mixture&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Alloy → Loyal&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;26 These three things&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Words → Sword&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;27 Scared swallow&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Gulp → Plug&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16275" LastEditorUserId="69582" LastEditDate="2021-01-27T21:57:44.100" LastActivityDate="2021-01-27T21:57:44.100" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22173" PostTypeId="2" ParentId="22150" CreationDate="2015-09-14T15:06:18.943" Score="0" Body="&lt;p&gt;The number of cookies a person receives is halved at each round, until eventually it becomes 1. At this time the cookie can't be shared anymore and must be eaten. So, for the process to continue forever, $n$ must be an infinitely large power of two. &lt;/p&gt;&#xA;" OwnerUserId="16323" LastEditorUserId="16323" LastEditDate="2015-10-08T14:54:19.210" LastActivityDate="2015-10-08T14:54:19.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22174" PostTypeId="1" AcceptedAnswerId="22175" CreationDate="2015-09-14T15:32:43.963" Score="17" ViewCount="1286" Body="&lt;p&gt;A grasshopper is hopping around on the integers and starts its journey on the number $81$. In a jump starting from the integer $m$, the grasshopper &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;may jump to any integer $m^k$ with integer $k\ge1$&lt;/li&gt;&#xA;&lt;li&gt;may jump to the number that results from $m$ by first cutting the decimal representation into a left piece and a right piece of two digits, and then adding three times the right piece to the left piece (for example, from $m=123456789$ the grasshopper may jump to $1234567+3\cdot89$). The left piece may be empty (for example, the jumps from $m=12$ to $0+3\cdot12$ and from $m=8$ to $0+3\cdot08$ are permitted).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;No other jumps are allowed.  Is there a sequence of jumps that brings the grasshopper to number $82$?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-09-14T16:02:09.130" LastActivityDate="2015-09-15T02:30:40.157" Title="Hopping from 81 to 82" Tags="|mathematics|reachability|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="22175" PostTypeId="2" ParentId="22174" CreationDate="2015-09-14T16:58:08.997" Score="20" Body="&lt;p&gt;It is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; not possible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Consider the remainder of the number when divided by $13$. Initially we start with remainder $3$ ($81\equiv 3 \mod 13$), we have to get remainder $4$ ($82\equiv 4 \mod 13$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying step 1, assuming that $m\equiv 3\mod 13$, we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;m &amp;amp;\to m^k \\&#xA;3 &amp;amp;\to 3^1, 3^2, 3^3, \ldots \mod 13 \\&#xA;3 &amp;amp;\to 3, 9, 1, 3, \ldots \mod 13&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we know that we can at least get to $1$, $3$, and $9 \mod 13$.  Applying the same argument for the two new values, we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;1 &amp;amp;\to 1^1, 1^2, 1^3, \ldots \mod 13 \\&#xA;1 &amp;amp;\to 1, 1, 1, \ldots \mod 13 \\&#xA;9 &amp;amp;\to 9^1, 9^2, 9^3, \ldots \mod 13 \\&#xA;9 &amp;amp;\to 9, 3, 1, 9, \ldots \mod 13 \\&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Here we have explicitly shown the argument for $9$, but we could have used the fact that $9=3^2$, so the powers of $9$ must be a subset of the powers of $3$ computed earlier.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore using step 1 we can get only numbers which are congruent to 1, 3 or 9 modulo 13. Now we verify that this is still true even if we apply step 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we start with the number m and apply step 2, we will get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;m=100a+b &amp;amp;\to a + 3b&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;100a+b &amp;amp;\to a+3b \mod 13 \\&#xA;100a+b &amp;amp;\to 300a + 3b \mod 13 \\&#xA;m &amp;amp;\to 3m \mod 13&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus it is sufficient to consider the action of $m\to 3m$ on the values from earlier ($1$, $3$, and $9$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;3\times 1 &amp;amp;\to 3 \mod 13 \\&#xA;3\times 3 &amp;amp;\to 9 \mod 13 \\&#xA;3\times 9 &amp;amp;\to 1 \mod 13 \\&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Again, we could have used the fact that all the values are powers of three instead of calculating all the transformations explicitly.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the numbers that we get through any number of jumps is always of the form $13k+1$, $13k+3$, or $13k+9$ ($m\equiv1,3,9\mod 13$).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since $82$ gives remainder $4$ when divided by $13$, it is not reachable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14140" LastEditorUserId="14140" LastEditDate="2015-09-15T02:30:40.157" LastActivityDate="2015-09-15T02:30:40.157" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="22176" PostTypeId="1" AcceptedAnswerId="22317" CreationDate="2015-09-14T19:07:48.490" Score="4" ViewCount="425" Body="&lt;p&gt;Nuja, dying pirate, the last of his clan, wrote a letter to his nephew trying to tell him the location of a treasure he had buried sometime ago. Knowing that the facility of prison hospital would read his letter thoroughly, he put the location in a cipher. The letter reads thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dear Hector&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know I will not live long now. I have lived a long enough life and had a lot of experiences. There is a great treasure I must tell you about, so that you may endeavor to find it. So know my dear, that treasure is inner peace, chivalry and selflessness. Like the sun, and the rain, which serve silently, thanklessly and constantly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All things, from elephant to dachshund&lt;br&gt;&#xA;Derive their warmth from the sun&lt;br&gt;&#xA;The sun earns nothing. The sun does not starve&lt;br&gt;&#xA;I wasted my life in vain. And alas&lt;br&gt;&#xA;St. George had long said, an ungrateful soul is dead&lt;br&gt;&#xA;But in my pine and restless crave&lt;br&gt;&#xA;I did not recognize, I was not, but a spectre&lt;br&gt;&#xA;Reeling in the abyss of my own sinful ignorance&lt;br&gt;&#xA;In my ignorance. In my wasteful existence&lt;br&gt;&#xA;&quot;Lo!&quot; the angels said &quot;For he's worse than crow or croc&lt;br&gt;&#xA;&quot;He's suffering and he'll suffer in hell&lt;br&gt;&#xA;&quot;Eden's garden? Ha!&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So. Where is the treasure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hints&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hector is christian and believes in trinity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Two&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nuja's sister's (Hector's aunt) name is Roche&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Three&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Roche lives in the countryside&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Four&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When taking the union of two sets, the overlapping members count only once!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I will mark the correct answer or give out the correct answer after 2 days.&lt;/p&gt;&#xA;" OwnerUserId="16342" LastActivityDate="2015-09-18T08:00:21.077" Title="Where Is The Treasure?" Tags="|cipher|pattern|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22177" PostTypeId="1" AcceptedAnswerId="22180" CreationDate="2015-09-14T19:34:44.523" Score="12" ViewCount="1693" Body="&lt;p&gt;Due to a communication breakdown and some misunderstanding, a truck with four boxes (pictured) full of valuables was left unguarded and unlocked door for exactly 1 hour in front of the bank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each box weighed the same but the content in each was different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though the passing thief had enough time to steal all four, he left the crate marked as &quot;diamonds&quot; untouched.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tGBPD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tGBPD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can you quickly deduce why?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9892" LastEditorUserId="4421" LastEditDate="2015-11-25T18:01:03.893" LastActivityDate="2019-04-07T01:29:45.387" Title="A reason to celebrate" Tags="|lateral-thinking|english|" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22178" PostTypeId="2" ParentId="22177" CreationDate="2015-09-14T19:39:20.733" Score="14" Body="&lt;p&gt;I have a guess, though it may not be correct.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The other three are metals that can be melted down and effectively obscured as &quot;stolen&quot;, while diamonds are useless if changed and will typically have identifying information laser-cut into them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2015-09-14T19:39:20.733" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22180" PostTypeId="2" ParentId="22177" CreationDate="2015-09-14T19:47:14.123" Score="23" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The box with diamonds was upside down, so that the label &quot;NO17&quot; looked like &quot;LION&quot;.  The thief did not want to tangle with a dangerous animal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4385" LastActivityDate="2015-09-14T19:47:14.123" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="22182" PostTypeId="2" ParentId="20595" CreationDate="2015-09-14T19:53:29.753" Score="9" Body="&lt;p&gt;I can't tell if anyone has posted this reference, but &lt;a href=&quot;http://arxiv.org/pdf/math-ph/0510065v6.pdf&quot;&gt;here is an arXiv link&lt;/a&gt; entitled &lt;em&gt;ON THE STABILITY OF FOUR-LEGGED TABLES&lt;/em&gt; with the following abstract:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We prove that a perfect square table with four legs, placed on&#xA;  continuous irregular ground with a local slope of at most 14.4 degrees&#xA;  and later 35 degrees, can be put into equilibrium on the ground by a&#xA;  “rotation” of less than 90 degrees. We also discuss the case of&#xA;  non-square tables and make the conjecture that equilibrium can be&#xA;  found if the four feet lie on a circle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So your answer is &lt;em&gt;it has been proven for certain constraints on how irregular the ground it&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="12795" LastActivityDate="2015-09-14T19:53:29.753" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22183" PostTypeId="1" AcceptedAnswerId="22184" CreationDate="2015-09-14T20:08:27.340" Score="6" ViewCount="227" Body="&lt;p&gt;Again, it's been a while since we have seen one of these, so I thought I should throw another one out there. (See &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15339/four-words&quot;&gt;the original&lt;/a&gt; for the solution format.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first has been seen around here a lot recently. (September 2015)&lt;br&gt;&#xA;My second is what my first does this weekend.&lt;br&gt;&#xA;My third provides a welcome breeze.&lt;br&gt;&#xA;My fourth begins a clue, but is only within.&lt;br&gt;&#xA;My fifth presented why I am right.  &lt;/p&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.923" LastActivityDate="2015-09-15T05:07:11.330" Title="Five-words (-----|||||) brought back" Tags="|word|word-square|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22184" PostTypeId="2" ParentId="22183" CreationDate="2015-09-14T20:48:25.997" Score="5" Body="&lt;p&gt;It's:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;M O V I E&lt;/code&gt;&lt;br&gt;&lt;code&gt;O P E N S&lt;/code&gt;&lt;br&gt;&lt;code&gt;V E N T S&lt;/code&gt;&lt;br&gt;&lt;code&gt;I N T R A&lt;/code&gt;&lt;br&gt;&lt;code&gt;E S S A Y&lt;/code&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My first has been seen around here a lot recently. (September 2015)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;a href=&quot;https://puzzling.meta.stackexchange.com/questions/4351/fortnightly-topic-challenge-1-movies?cb=1&quot;&gt;&lt;strong&gt;MOVIE&lt;/strong&gt; tag&lt;/a&gt; has been used a lot, recently.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My second is what my first does this weekend.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Movies open (are released) on weekends --&gt; a movie &lt;strong&gt;OPENS&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My third provides a welcome breeze.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;VENTS&lt;/strong&gt; carry air through structures.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fourth begins a clue, but is only within.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An INTRO begins something, &lt;strong&gt;INTRA&lt;/strong&gt; is a prefix meaning &quot;within&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fifth presented why I am right.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An &lt;strong&gt;ESSAY&lt;/strong&gt; can make an argument.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13237" LastEditorUserId="-1" LastEditDate="2017-03-16T16:42:07.097" LastActivityDate="2015-09-14T20:56:00.470" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22185" PostTypeId="1" AcceptedAnswerId="22211" CreationDate="2015-09-14T21:22:46.560" Score="4" ViewCount="935" Body="&lt;p&gt;Let's define a Complete Scrabble Block as an arrangement of Scrabble tiles such that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The four edges are all flat.&lt;/li&gt;&#xA;&lt;li&gt;All words are words.&lt;/li&gt;&#xA;&lt;li&gt;There is an order in which the tiles can be played in a game of Scrabble to reach the result.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;CAB   CAB   CAB   CAB   CAB&#xA;        E   O E   O E   ODE&#xA;        T   P T   POT   POT&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What is the largest Complete Scrabble Block that can be created? Provide an example final grid and an order of play:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;CAB   111&#xA;ODE   352&#xA;POT   342&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Notes:&lt;br&gt;&#xA;- You don't need to stick to the Scrabble tile counts if you don't want to (though it would be cool if you did).&lt;br&gt;&#xA;- You don't need to stick to the Scrabble dictionary if you don't want to (though it would be cool if you did).&lt;br&gt;&#xA;- The intermediary steps do not need to be flat-edged.&lt;br&gt;&#xA;- The final solution need not be a square.&lt;br&gt;&#xA;- I don't know the answer, so if nobody can definitively prove correctness in their answer after a week, I'll accept the answer that provides the largest area (w*h). In the event of a tie, I'll choose the answer based on Scrabble score. In the event that this is still a tie, I'll choose the one that was put up first.&lt;/p&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-09-15T14:21:27.757" LastActivityDate="2015-09-15T17:44:16.587" Title="Largest Complete Scrabble Block" Tags="|word|open-ended|crosswords|tiling|word-square|" AnswerCount="3" CommentCount="3" ClosedDate="2021-10-18T22:36:00.043" ContentLicense="CC BY-SA 3.0" />
  <row Id="22186" PostTypeId="2" ParentId="22185" CreationDate="2015-09-14T21:57:41.567" Score="3" Body="&lt;p&gt;Here's a start (taken from &lt;a href=&quot;https://en.wikipedia.org/wiki/Word_square&quot; rel=&quot;nofollow&quot;&gt;this page&lt;/a&gt;) .  I'm sure someone can expand on it or find something larger.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;G L A S S E S    1 A 3 7 6 8 5&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;R E L A P S E    1 A 3 7 6 8 5&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;I M I T A T E    1 2 2 2 2 2 2&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;S M E A R E D    1 A 3 9 6 8 5&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;T A N N E R Y    1 4 3 4 4 4 4&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;A&lt;/code&gt; is used instead of the number 10, to keep things to a single digit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;** &lt;em&gt;It wasn't clear from the question whether all of the words had to be acceptable Scrabble™ words, but I should note that &lt;code&gt;SATAN&lt;/code&gt; is not in the official Scrabble™ Dictionary, and thus this answer might be invalidated (unless someone can some up with a clever way of modifying it to remove the offending word.&lt;/em&gt; &lt;/p&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="4421" LastEditDate="2015-09-14T23:54:52.690" LastActivityDate="2015-09-14T23:54:52.690" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22187" PostTypeId="2" ParentId="22177" CreationDate="2015-09-14T21:59:09.163" Score="11" Body="&lt;p&gt;The density of diamonds is 1/5 the density of gold.  So if the thief had stolen the box of gold and realized that it was full, he would know that it would require five times the volume of diamonds to weigh the same amount.  Since the boxes were all roughly the same size, the thief easily determined that the diamonds were fakes. &lt;/p&gt;&#xA;" OwnerUserId="16344" LastActivityDate="2015-09-14T21:59:09.163" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22188" PostTypeId="2" ParentId="22177" CreationDate="2015-09-14T23:13:07.207" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The boxes are stacked, and so if the thief sees the gold and platinum boxes in front, the diamonds are hidden from view. You can't see box 17 unless you moved the other boxes out of the way.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16346" LastEditorUserId="16346" LastEditDate="2019-04-07T01:29:45.387" LastActivityDate="2019-04-07T01:29:45.387" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22189" PostTypeId="2" ParentId="22185" CreationDate="2015-09-14T23:22:54.910" Score="6" Body="&lt;p&gt;How about this one (taken from &lt;a href=&quot;http://ask.metafilter.com/176017/Whats-the-largest-possible-word-block-in-scrabble&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;) which is also a word square:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/6YIBL.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6YIBL.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-09-14T23:22:54.910" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22190" PostTypeId="2" ParentId="22176" CreationDate="2015-09-15T00:37:17.033" Score="0" Body="&lt;p&gt;My answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Church of Saint-Roch in Paris&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clues:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All things, from elephant to dachshund&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He is a patron Saint of dogs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Roche lives in the countryside&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Born in Montpellier, Kingdom of Majorca. Montpellier  is a city in southern France,  Located on the south coast of France on the Mediterranean Sea.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When taking the union of two sets, the overlapping members count only once!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The church is also notable as a result of the marriage there of the Marquis de Sade on May 17, 1763.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16036" LastActivityDate="2015-09-15T00:37:17.033" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22196" PostTypeId="1" AcceptedAnswerId="22228" CreationDate="2015-09-15T03:16:20.620" Score="11" ViewCount="331" Body="&lt;p&gt;Amongst the elements I reside,&lt;br&gt;&#xA;Beneath the earth, below the tide.&lt;br&gt;&#xA;Through air and aether you dread my flight,&lt;br&gt;&#xA;Fire descends on the stroke of midnight.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I? And will we ever come in, out of the cold?&lt;/p&gt;&#xA;" OwnerUserId="4164" LastEditorDisplayName="user20" LastEditDate="2015-09-15T10:07:00.577" LastActivityDate="2015-09-16T03:24:31.013" Title="Hoping for a new start" Tags="|riddle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22197" PostTypeId="1" CreationDate="2015-09-15T03:23:49.797" Score="17" ViewCount="1638" Body="&lt;p&gt;There are 10 agents living at their respective homes. They used a secret mailing system that sends one box at a time, but it is now spoilt. Now, it functions as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It counts the number of deliveries to each agent.&lt;/li&gt;&#xA;&lt;li&gt;It always sends the box to the agent with the lowest delivery count.&lt;/li&gt;&#xA;&lt;li&gt;If there is more than one such agent, it will pick one of them randomly.&lt;/li&gt;&#xA;&lt;li&gt;Once the box reaches an agent, it will remain there until that agent decides to send it.&lt;/li&gt;&#xA;&lt;li&gt;For security, the box will never travel unless there is a lock on it.&lt;/li&gt;&#xA;&lt;li&gt;The time taken by the distribution system to deliver a box is variable and random and always non-negligible, so waiting for a fixed number of seconds/minutes/days before sending a box is not a viable way of sending information. The transmission system creates a queue if it receives boxes while it already has a box in the process of being transmitted. Only one box can be in the process of transmission at one point in time, the rest are all on hold in the queue.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rules:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Each agent has an unlimited supply of boxes of various sizes, and padlocks (each with a single key). The padlocks look identical, the keys do not. This supply is given to them after the pre-planning phase.&lt;/li&gt;&#xA;&lt;li&gt;Any number of locked padlocks can be put on any box.&lt;/li&gt;&#xA;&lt;li&gt;A box can contain open/locked padlocks, keys, papers, and boxes, each with/without its own locks. (Note the difference between putting a padlock &lt;em&gt;on&lt;/em&gt; a box, i.e., locking it, and putting a padlock &lt;em&gt;in&lt;/em&gt; a box, i.e., transporting it)&lt;/li&gt;&#xA;&lt;li&gt;A padlock can be opened or closed only with its key.&lt;/li&gt;&#xA;&lt;li&gt;Boxes cannot change size. If Box A was once placed in Box B, Box B can not be placed in Box A ever.&lt;/li&gt;&#xA;&lt;li&gt;Keys and locks cannot be duplicated. Papers can be photocopied.&lt;/li&gt;&#xA;&lt;li&gt;The agents can have a pre-arranged plan, before they start the mission. Once the mission has started, they can communicate with each other only through papers (which they can create) that are sent in the box.&lt;/li&gt;&#xA;&lt;li&gt;The box used for delivery can be changed by an agent who opens it and can transfer the contents to a larger (or maybe even smaller) box. However, the mailing system will treat it just like the old one. The mailing order remains unchanged. The size of the new box will be random and hence its dimensions cannot convey information.&lt;/li&gt;&#xA;&lt;li&gt;pre-arranged planning only involves strategy talk, no handing out/exchanging locks or keys.&lt;/li&gt;&#xA;&lt;li&gt;no writings are to be made on the outside of the box sent around, nor messages taped to the side. No unlocked padlocks, etc. can be sticking out of the box.&lt;/li&gt;&#xA;&lt;li&gt;the lock(s) that is/are to be on the box when the message is sent, is/are required to be in locked state.&lt;/li&gt;&#xA;&lt;li&gt;the 'box transfer count' isn't visible, but held by the agency that does the box transportation.&lt;/li&gt;&#xA;&lt;li&gt;a person can send multiple boxes one after the other, but there should be negligible time lag between the transmissions.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mission&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Agent 001 has received a top secret message on a sheet of paper. He needs to inform all the agents. What is the minimum number of transmissions it will take before he can guarantee that everyone has received it (or a copy of it)?&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4269" LastEditDate="2018-03-28T13:20:40.537" LastActivityDate="2018-04-21T11:55:08.193" Title="Agent 001's faulty delivery box" Tags="|logical-deduction|keys-and-locks|" AnswerCount="9" CommentCount="22" ContentLicense="CC BY-SA 3.0" />
  <row Id="22198" PostTypeId="2" ParentId="22197" CreationDate="2015-09-15T11:23:37.537" Score="4" Body="&lt;p&gt;Before you read on, know the answer is made with the following assumptions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;pre-arranged planning only involves strategy talk, no handing out/exchanging locks or keys.&lt;/li&gt;&#xA;&lt;li&gt;no writings are to be made on the outside of the box sent around, nor messages taped to the side.&lt;/li&gt;&#xA;&lt;li&gt;the lock that is to be on the box when the message is sent, is assumed to be required to be in locked state.&lt;/li&gt;&#xA;&lt;li&gt;the 'box transfer count' isn't visible, but held by the agency that does the box transportation.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First attempt:&lt;/strong&gt; (establishing upper boundary)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Person A adds lock A, sends to person B. (1 move)&#xA;B receives, adds lock B. We know that 9 more moves are needed to ensure everyone has received it at least once. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we want to ensure B can read it, it needs to pass by person A to remove the lock, and end up with B again.&#xA;Worst case A is the last person to go from 'received once' to 'received twice', and B is the last person in the third cycle to receive it, open, and read the message.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one person knowing it requires 0 moves (the one locking it in), two people requires a maximum of 29 moves. (no move required from 'nobody' to 'agent A', as pointed out by Gordon K in comments).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each additional person, the same cycle  is required: Locking it with lock B, sending it to a person C who adds lock C, cycling back to person B to remove lock B, and then end up with C again who can read the message.&#xA;Given there now is a return action to the first agent in a new cycle, a cycle consists of 30 moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the principle is the same, we end up at &lt;strong&gt;0 + 29 + 8x30 = 269 moves&lt;/strong&gt; for 10 people to know the contents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to elaborate on a few of the 'mechanisms':&lt;br&gt;&#xA; 1. Once an agent has added his lock, on the next delivery with two locks he removes his own lock. After that he only passes the box onwards without further action&#xA; &lt;br&gt;2. How do you know there's a full cycle completed? Well, here come the different sized boxes into play. Each time one of the recipients has learnt the contents and he passes it on, he changes the box size. If you see a new box appear for the first time, with a single lock on it, you can consider yourself the next to learn the contents and add a second lock. All others will first see it pass with TWO locks on it, and know the new box isn't for them yet. This also means that knowing this, they will not get tempted to consider themselves the new recipient if that box passes a second time with a single lock on it, untill the box size changes again.&#xA;By the number of box size changes, all of them know how many people know the contents already.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll try and see if any mechanisms could work to reduce upper boundary, which as it is, is a rather negative outlook.&lt;/p&gt;&#xA;" OwnerUserId="2156" LastEditorUserId="2156" LastEditDate="2015-09-16T10:49:03.963" LastActivityDate="2015-09-16T10:49:03.963" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="22199" PostTypeId="1" AcceptedAnswerId="22246" CreationDate="2015-09-15T12:33:42.650" Score="4" ViewCount="381" Body="&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;HERE IS MY ENTRY TO THE FORTNIGHTLY TOPIC CHALLENGE #1&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First One is the answer of this&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Even when my eyes are closed, I can see it.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Even when my ears are covered, I can hear it.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Even when my nose is blocked, I can smell it.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Even when my mouth is shut, my tongue can taste it.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Even when my hands are tied, I can touch it.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Even though I can feel it, I am not sure how real is it.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Can you guys feel it, then tell me what is it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second one is the answer of this&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Take one stick. No.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Take another. No.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Take one more. No.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Take a hundred sticks. Yes, that will be it.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Then, what is it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The real question is to find the third one that is going to be real soon and explain how you get it.&lt;/strong&gt;&#xA;&lt;br /&gt;&#xA;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINTS&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;&lt;em&gt;HINT1:&lt;/strong&gt; The answer is a movie name&lt;/em&gt;&lt;br /&gt;&#xA;____________________________________&lt;br /&gt;&#xA;&lt;strong&gt;&lt;em&gt;HINT2:&lt;/strong&gt; Answers to first two riddles has a common link (connection), with the help of which you can find the third one.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="2730" LastEditDate="2015-09-17T09:03:21.713" LastActivityDate="2015-09-17T11:42:43.847" Title="Find The Third One From The First Two Riddles" Tags="|riddle|knowledge|movies|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22200" PostTypeId="2" ParentId="22196" CreationDate="2015-09-15T13:37:45.337" Score="3" Body="&lt;p&gt;Guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Uranium or Plutonium etc&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Amongst the elements I reside,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are examples of elements.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Beneath the earth, below the tide.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are probably found under the ground (I am no expert).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Through air and aether you dread my flight,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;my flight&quot; is a nuclear bombing. Through air, of course.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fire descends on the stroke of midnight.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm not sure how &quot;midnight&quot; fits. &quot;Fire descends&quot; though does seem to potentially be referring to a bomb.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And will we ever come in, out of the cold?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;cold&quot; as in cold wars? In that case, no.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11100" LastActivityDate="2015-09-15T13:37:45.337" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22203" PostTypeId="2" ParentId="22197" CreationDate="2015-09-15T15:14:08.540" Score="2" Body="&lt;p&gt;&lt;strong&gt;270 transmissions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This answer builds on the algorithm f&quot; mentioned in his comment to the question. While this answer is similar to that by Tim Couwelier, the main difference is simplicity: no boxes are used other than the delivery box, and protocol state is determined by the number of times an agent receives the box. This answer allows (but does not rely on) the assumption that &quot;the&quot; delivery box in the question is a single, fixed box that cannot be substituted for any other box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The counting mechanism requires every agent to handle the box exactly once before getting the box again. Agents count the number of times they receive the box and follow the protocol, then pass the box on. If no protocol is specified for that agent, the box is just passed on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process starts after agent 001 puts the message into the box, locks it and sends it off. Agent 001 is designated 00X initially.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Protocol for each agent:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If there is 1 lock on the box and you haven't read the message yet, add your own lock. The box ends this round with 2 locks.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Agent 00X: remove your lock; you are no longer designated 00X. The box ends this round with 1 lock.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the lock remaining on the box is yours: remove it and read the message, then re-lock the box; you are now designated 00X. The box ends this round with 1 lock.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For all agents to read the message, the protocol runs 9 times with 10 transmissions per round. Total number of transmissions is therefore 9x10x3 = 270.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As an aside, there is an interesting pair of properties for this algorithm:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;each person's locking and subsequent unlocking have to occur in different rounds due to the delivery box's counting mechanism, so cycling through completely back to the same state (e.g. 1 to 2 to 3 to 1) must occupy at least 2 rounds; but&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;any pair of consecutive states (1 to 2, 2 to 3, 3 to 1) can conceivably be optimised to a single round, given a suitably cooperative order of delivery.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9422" LastEditorUserId="9422" LastEditDate="2015-09-16T12:43:29.423" LastActivityDate="2015-09-16T12:43:29.423" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="22204" PostTypeId="1" AcceptedAnswerId="22205" CreationDate="2015-09-15T15:43:49.680" Score="7" ViewCount="451" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Her favorite fruits are lemons and _____________.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;He drove his car to _______________.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;He rode a horse along the ______________.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;When I eat ____________, I get a smile on my face.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Take ____________ while driving a race car.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Rose thorns can make your finger _____________.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Every time you visit Rome, you find _____________ to do.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;My uncle is a ______________ and wears a purple robe.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;On ______________, we'll drive a five-seater car.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;'Finders keepers', she said to her __________________.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The men in the manors were held for _______________.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Witches have _______________ when making potions.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;She ate it and then ______________ that it tasted funny.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Wait in the kitchen for the sauce to _______________.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is a single rule used to fill all the above blanks. What is it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Notes&lt;/strong&gt;&lt;br&gt;&#xA;1. This is my first such word puzzle. Hope you like it.&lt;br&gt;&#xA;2. Each blank has one word. The length of the blank has nothing to do with the length of the words.&lt;br&gt;&#xA;3. I can't think of a good title that does not reveal the answer.&lt;br&gt;&#xA;4. There is no math or encryption involved.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4421" LastEditDate="2015-09-16T05:09:13.937" LastActivityDate="2015-09-16T05:09:13.937" Title="Filling blanks with fixed rule" Tags="|word|enigmatic-puzzle|wordplay|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22205" PostTypeId="2" ParentId="22204" CreationDate="2015-09-15T15:48:09.960" Score="6" Body="&lt;p&gt;The rule is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each word is an anagram of another word in the sentence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Her favorite fruits are lemons and _____________.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MELONS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He drove his car to _______________.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DOVER&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He rode a horse along the ______________.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SHORE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I eat ____________, I get a smile on my face.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LIMES&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Take ____________ while driving a race car.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CARE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rose thorns can make your finger _____________.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SORE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Every time you visit Rome, you find _____________ to do.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MORE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My uncle is a ______________ and wears a purple robe.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BORE (thanks, @alexmc!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On ______________, we'll drive a five-seater car.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; EASTER&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;'Finders keepers', she said to her __________________.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; FRIENDS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The men in the manors were held for _______________.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RANSOM&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Witches have _______________ when making potions.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; OPTIONS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;She ate it and then ______________ that it tasted funny.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; STATED&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wait in the kitchen for the sauce to _______________.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; THICKEN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="11475" LastEditDate="2015-09-15T16:12:07.507" LastActivityDate="2015-09-15T16:12:07.507" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22206" PostTypeId="1" AcceptedAnswerId="22208" CreationDate="2015-09-15T16:30:22.070" Score="32" ViewCount="1347" Body="&lt;p&gt;Can you determine these popular quotes as retold by Shakespeare? Before anyone points out, yes I'm aware that Bill used predominantly iambic pentameter, but I've never been a huge fan of that, so I'm going with iambic tetrameter instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How oft the potion, orange in hue&lt;br&gt;&#xA;  delights my sense among the dew.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Indeed this land can scarce contain&lt;br&gt;&#xA;  Our presence should we both remain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;From first I heard thine mouth's sweet spell&lt;br&gt;&#xA;  A victim helpless I befell.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This vessel shall contain our dirge&lt;br&gt;&#xA;  Should storm or monstrous beast resurge.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;His stature lacking to impress&lt;br&gt;&#xA;  Beseech you greet him none the less.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="8874" LastEditDate="2016-03-23T14:47:59.263" LastActivityDate="2016-03-23T14:47:59.263" Title="Shakespeare. William Shakespeare" Tags="|movies|poetry|literature|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22207" PostTypeId="1" AcceptedAnswerId="22209" CreationDate="2015-09-15T16:42:41.697" Score="4" ViewCount="212" Body="&lt;p&gt;Here is an easy riddle. Recommended for those who haven't studied too much of maths.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There are 100 people in a room.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$99\%$ of the people in the room are literate.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Some people are removed.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now, $98\%$ of the people in the room are literate.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;How many people were removed?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And don't tell me that someone was educated sitting in the room... :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your first answer was $1$, you're wrong.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-09-22T17:23:12.677" Title="Literacy in the house" Tags="|mathematics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22208" PostTypeId="2" ParentId="22206" CreationDate="2015-09-15T16:46:36.070" Score="22" Body="&lt;p&gt;Here's a shot. All but the 2nd are on &lt;a href=&quot;https://en.wikipedia.org/wiki/AFI%27s_100_Years...100_Movie_Quotes&quot;&gt;this nifty Wikilist&lt;/a&gt;:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How oft the potion, orange in hue&lt;br&gt;&#xA;delights my sense among the dew.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I love the smell of napalm in the morning.&quot; -- &lt;em&gt;Apocalypse Now&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Indeed this land can scarce contain&lt;br&gt;&#xA;Our presence should we both remain.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;This town ain't big enough for the both of us&quot;, -- &lt;a href=&quot;http://www.imdb.com/title/tt0023682/quotes&quot;&gt;&lt;em&gt;The Western Code&lt;/em&gt;&lt;/a&gt; My thanks to dperry!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From first I heard thine mouth's sweet spell&lt;br&gt;&#xA;A victim helpless I befell.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;You had me at 'hello.'&quot; -- &lt;em&gt;Jerry Maguire&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This vessel shall contain our dirge&lt;br&gt;&#xA;Should storm or monstrous beast resurge.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;You're gonna need a bigger boat.&quot; -- &lt;em&gt;Jaws&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;His stature lacking to impress&lt;br&gt;&#xA;Beseech you greet him none the less.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Say 'hello' to my little friend!&quot; -- &lt;em&gt;Scarface&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And for flavor, the title references...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Bond. James Bond&quot; -- first said in &lt;em&gt;Dr. No&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13237" LastEditorUserId="13237" LastEditDate="2015-09-16T16:48:08.090" LastActivityDate="2015-09-16T16:48:08.090" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22209" PostTypeId="2" ParentId="22207" CreationDate="2015-09-15T16:52:39.103" Score="6" Body="&lt;p&gt;The amount of people removed was&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $50$, and all of them were literate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; With $99\%$ of people in the room of $100$ being literate, there must be $99$ literate people and $1$ illiterate person. $98\%$ can only be achieved by two fractions: $98/100$ and $49/50$. Since we must remove at least one person to decrease the percentage from $99\%$, the only possible fraction left is $49/50$ to achieve $98\%$ literacy. Therefore, we must remove $50$ literate people from the room.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-09-15T16:52:39.103" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22210" PostTypeId="1" AcceptedAnswerId="22213" CreationDate="2015-09-15T17:42:01.337" Score="4" ViewCount="652" Body="&lt;p&gt;To make payments, the Non-Pythagoreans use coins in three denominations of 999, 1000, and 1001 Oboloi. What is the largest integer amount of Oboloi that can &lt;strong&gt;not&lt;/strong&gt; be represented by using these three types of coins?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comment 1: valid representations use non-negative numbers of coins&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comment 2: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/9443/pythagorean-coins&quot;&gt;Pythagorean coins&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2016-03-07T12:44:06.413" Title="Non-Pythagorean coins (Part 1)" Tags="|mathematics|combinatorics|frobenius|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22211" PostTypeId="2" ParentId="22185" CreationDate="2015-09-15T17:44:16.587" Score="2" Body="&lt;p&gt;Here's what I would consider a &quot;fuzzy&quot; answer, since some of the words used in it have questionably acceptable definitions. Nevertheless, I think it's a good example, and I've included links to prove that each word has at least one source that claims it is acceptable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia has several examples on its &lt;a href=&quot;https://en.wikipedia.org/wiki/Word_square&quot; rel=&quot;nofollow noreferrer&quot;&gt;word square&lt;/a&gt; page. Here is an 8x8 example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;L A T E R A L S&lt;/code&gt;&lt;br&gt;&lt;code&gt;A X O N E M A L&lt;/code&gt;&lt;br&gt;&lt;code&gt;T O E P L A T E&lt;/code&gt;&lt;br&gt;&lt;code&gt;E N P L A N E D&lt;/code&gt;&lt;br&gt;&lt;code&gt;R E L A N D E D&lt;/code&gt;&lt;br&gt;&lt;code&gt;A M A N D I N E&lt;/code&gt;&lt;br&gt;&lt;code&gt;L A T E E N E R&lt;/code&gt;&lt;br&gt;&lt;code&gt;S L E D D E R S&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately Wikipedia doesn't give steps for actually playing this out in Scrabble, so I did that myself. Note that this is not possible in Scrabble with the number of available tiles, and many of the words are not in the official Scrabble dictionary. Anyway, here are the steps, with a special thanks to &lt;a href=&quot;https://puzzling.stackexchange.com/users/14135/gordon-k&quot;&gt;Gordon K&lt;/a&gt; for the formatting idea:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/axs0F.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/axs0F.png&quot; alt=&quot;8x8 word square&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And here are the justifications for why these are all words:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. &lt;em&gt;Laterals&lt;/em&gt;: In the &lt;a href=&quot;http://www.hasbro.com/scrabble-2/en_US/search.cfm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Scrabble dictionary&lt;/a&gt;.&lt;br&gt;&#xA; 2. Ditto.&lt;br&gt;&#xA; 3. &lt;em&gt;Sledders&lt;/em&gt;: In the Scrabble dictionary.&lt;br&gt;&#xA; 4. Ditto.&lt;br&gt;&#xA; 5. &lt;em&gt;Axonemal&lt;/em&gt;, &lt;em&gt;Ax&lt;/em&gt;, &lt;em&gt;To&lt;/em&gt;, &lt;em&gt;En&lt;/em&gt;, &lt;em&gt;Re&lt;/em&gt;, &lt;em&gt;Am&lt;/em&gt;, &lt;em&gt;La&lt;/em&gt;: All in the Scrabble dictionary.&lt;br&gt;&#xA; 6. Ditto.&lt;br&gt;&#xA; 7. &lt;em&gt;Amandine&lt;/em&gt;: In the Scrabble dictionary.&lt;br&gt;&#xA; 8. Ditto.&lt;br&gt;&#xA; 9. &lt;em&gt;Lateener&lt;/em&gt;, &lt;em&gt;Ate&lt;/em&gt;: Both in the Scrabble dictionary.&lt;br&gt;&lt;em&gt;Ned&lt;/em&gt;: Scottish derogatory term for hooligans. (source: &lt;a href=&quot;https://en.wikipedia.org/wiki/Ned_(Scottish)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;)&lt;br&gt;&lt;em&gt;Ded&lt;/em&gt;: British term for a Russian soldier who has served two or three years. (Source: &lt;a href=&quot;http://dictionary.reference.com/browse/ded?s=t&quot; rel=&quot;nofollow noreferrer&quot;&gt;dictionary.com&lt;/a&gt;)&lt;br&gt;&lt;em&gt;Ner&lt;/em&gt;: Informal interjection, taken from &quot;neener&quot;. (Source: &lt;a href=&quot;https://en.wiktionary.org/wiki/ner&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wiktionary&lt;/a&gt;)&lt;br&gt;&#xA; 10. Ditto.&lt;br&gt;&#xA; 11. &lt;em&gt;Enplaned&lt;/em&gt;: In the Scrabble dictionary.&lt;br&gt;&#xA; 12. Ditto.&lt;br&gt;&#xA; 13. &lt;em&gt;Relanded&lt;/em&gt;, &lt;em&gt;Plate&lt;/em&gt;: In the Scrabble dictionary.&lt;br&gt;&lt;em&gt;Anded&lt;/em&gt;: Past tense of &quot;and&quot;. (Source: &lt;a href=&quot;http://www.yourdictionary.com/anded&quot; rel=&quot;nofollow noreferrer&quot;&gt;yourdictionary.com&lt;/a&gt;). I really, really don't like that definition or website, but I use the term all the time when referring to digital logic functions. (In those cases it may be more appropriately written ANDed, and thus this word gives me the most hesitation with this answer.)&lt;br&gt;&#xA; 14. &lt;em&gt;Toeplate&lt;/em&gt;: In the Scrabble dictionary.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16275" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2015-09-15T17:44:16.587" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22212" PostTypeId="1" AcceptedAnswerId="22221" CreationDate="2015-09-15T17:49:32.583" Score="0" ViewCount="212" Body="&lt;p&gt;&lt;strong&gt;If you know that:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;TWO + THREE = TILAS&lt;br&gt;&#xA;  THREE + NINE = TUZRI&lt;br&gt;&#xA;  FOUR + TWO = GIRF&lt;br&gt;&#xA;  EIGHT + NINE = EVOUX&lt;br&gt;&#xA;  NINE + NINE = BARAI&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can you tell:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;TWO + NINE = ?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15856" LastActivityDate="2015-09-15T19:17:40.103" Title="Sum the two numbers" Tags="|mathematics|enigmatic-puzzle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22213" PostTypeId="2" ParentId="22210" CreationDate="2015-09-15T18:01:44.370" Score="10" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 498500&lt;br&gt;&#xA; because $498500=498\times999+998$ and in modular arithmetic we have $1000\equiv1\pmod{999}$ and $1001\equiv2\pmod{999}$, so the remainder of $998$ cannot be made by fewer than $499$ additions of positive numbers less than or equal to $2$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14080" LastActivityDate="2015-09-15T18:01:44.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22214" PostTypeId="2" ParentId="22210" CreationDate="2015-09-15T18:17:17.387" Score="8" Body="&lt;p&gt;This problem is an instance of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Coin_problem&quot;&gt;Frobenius coin problem&lt;/a&gt;.  In fact, the coin values form an arithmetic sequence, whose Frobenius number has the &lt;a href=&quot;https://en.wikipedia.org/wiki/Coin_problem#Arithmetic_sequences&quot;&gt;special form&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;g(a, a+d, \ldots, a+sd) = \left(\left\lfloor\frac{a-2}{s}\right\rfloor+1\right)a + (d-1)(a-1)-1&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;g(999, 1000, 1001) &amp;amp;= \\ g(999, 999+1, 999+2\times 1) &amp;amp;= \left(\left\lfloor\frac{999-2}{2}\right\rfloor+1\right)999 + (1-1)(999-1)-1 \\ &amp;amp;= 498\ 500&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the largest denomination that cannot be made with the Non-Pythagorean coins is $498\ 500$.&lt;/p&gt;&#xA;" OwnerUserId="9718" LastActivityDate="2015-09-15T18:17:17.387" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22215" PostTypeId="1" AcceptedAnswerId="22217" CreationDate="2015-09-15T18:19:04.590" Score="3" ViewCount="260" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Forethought furnished me to men;&lt;br&gt;&#xA;  For ages I sourced all.&lt;br&gt;&#xA;  But slowly, as the men advanced,&lt;br&gt;&#xA;  I watched my kingdom fall.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I used to give so much to men,&lt;br&gt;&#xA;  But now they need me less.&lt;br&gt;&#xA;  Perfunctory my uses now;&lt;br&gt;&#xA;  Allure I once possessed   &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Replaced, am I, by lightning caught;&lt;br&gt;&#xA;  May Franklin be most cursed.&lt;br&gt;&#xA;  He gave mankind alternatives,&lt;br&gt;&#xA;  And so my growth reversed.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I gave to men their healthy meat;&lt;br&gt;&#xA;  I made them safe at night.&lt;br&gt;&#xA;  &lt;em&gt;I&lt;/em&gt; gave them weapons, traps, and shields;&lt;br&gt;&#xA;  &lt;strong&gt;&lt;em&gt;I&lt;/em&gt;&lt;/strong&gt; shared with them my might!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;But still I'm used, albeit less;&lt;br&gt;&#xA;  They see me oft as &quot;quaint&quot;.&lt;br&gt;&#xA;  Where once I was in every home,&lt;br&gt;&#xA;  Now usually I ain't.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;They think me fine on wintry nights,&lt;br&gt;&#xA;  But keep me in my place.&lt;br&gt;&#xA;  They build me traps of steel and stone,&lt;br&gt;&#xA;  And to their friends showcase.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;Oh, look,&quot; they say, &quot;we tamed the beast!&quot;&lt;br&gt;&#xA;  I &lt;em&gt;roar&lt;/em&gt; internally!&lt;br&gt;&#xA;  &quot;That's very nice,&quot; their friends reply.&lt;br&gt;&#xA;  &lt;strong&gt;I spit my tongues at thee!&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Let me loose, I shall &lt;em&gt;destroy&lt;/em&gt;&lt;br&gt;&#xA;  Each thing that you hold dear.&lt;br&gt;&#xA;  I rage against my captors now;&lt;br&gt;&#xA;  They're always right to fear.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I'm ancient and I'm primal too;&lt;br&gt;&#xA;  I date from long before&lt;br&gt;&#xA;  The rise of all these puny men;&lt;br&gt;&#xA;  I wage a ceaseless war.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;All that you built upon my back&lt;br&gt;&#xA;  Shall be destroyed by rage.&lt;br&gt;&#xA;  I'm wild, devouring all I see;&lt;br&gt;&#xA;  I grow, but do not age.   &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;You may send troops with special suits,&lt;br&gt;&#xA;  Who try to beat me back,&lt;br&gt;&#xA;  But I return and start again;&lt;br&gt;&#xA;  With lightning I attack.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;So sit or stand, you fleshy thing;&lt;br&gt;&#xA;  Believe that I am tamed.&lt;br&gt;&#xA;  My will be told: the world would be&lt;br&gt;&#xA;  Entirely enflamed.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Who am I?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explain as many clues as you think necessary to show how clever you are / how obvious I am.&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="4421" LastEditDate="2015-09-15T19:40:41.407" LastActivityDate="2015-09-15T19:40:41.407" Title="Forethought Brought Me Down" Tags="|riddle|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22216" PostTypeId="1" AcceptedAnswerId="22220" CreationDate="2015-09-15T18:24:16.173" Score="7" ViewCount="513" Body="&lt;p&gt;To make payments, the Non-Pythagoreans use coins in three denominations of 999, 1000, and 1001 Oboloi. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Some amounts of Oboloi do not allow any representation by these three denominations (for example 1002).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Other amounts of Oboloi do allow two or more different representations by these three denominations (for example 2000 allows 1000*2 and 999*1+1001*1).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Some amounts of Oboloi have a unique representation by these three denominations (for example 2997 only allows the representation 2997=999*3) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What is the largest integer amount of Oboloi that has a unique representation by these three types of coins?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comment 1: valid representations use non-negative numbers of coins&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comment 2: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/9443/pythagorean-coins&quot;&gt;Pythagorean coins&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comment 3: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/22210/non-pythagorean-coins-part-1&quot;&gt;Non-Pythagorean coins (Part 1)&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2016-03-07T12:36:41.570" Title="Non-Pythagorean coins (Part 2)" Tags="|mathematics|combinatorics|frobenius|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22217" PostTypeId="2" ParentId="22215" CreationDate="2015-09-15T18:27:27.320" Score="4" Body="&lt;p&gt;Well, it's obviously:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fire&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You describe men discovering/creating fire, thinking it to be &quot;magic&quot;. Nowadays, we don't need it for light, since we've &quot;bottled lightning&quot; with light bulbs. &lt;br&gt;&lt;br&gt;&#xA; You reference fireplaces &quot;in every home&quot; (&quot;traps of steel and stone&quot;), fire used to forge &quot;weapons, traps, and shields&quot;, for cooking meat, and the &quot;roaring&quot; of the fire, the tongues of flames. &lt;br&gt;&lt;br&gt;&#xA; &quot;Let [it] loose&quot; and the fire will destroy you and/or your belongings and cause forest fires. &quot;Troops with special suits&quot; are firemen trying to kill the fire. Lightning can cause fires, and fires will inevitably be created all over the world; it cannot be killed for good.&lt;br&gt;&lt;br&gt;&#xA; The &quot;Franklin&quot; reference is almost certainly Benjamin Franklin, though he did little work to make fire less essential in daily life. He's credited with inventing the Franklin stove, which made fire even more useful.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13237" LastEditorUserId="13237" LastEditDate="2015-09-15T18:38:55.327" LastActivityDate="2015-09-15T18:38:55.327" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22219" PostTypeId="2" ParentId="22215" CreationDate="2015-09-15T18:34:53.603" Score="5" Body="&lt;p&gt;You are:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Fire&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Forethought brought me to the world of men&lt;br&gt;&#xA;Millennia passed as I sourced all&lt;br&gt;&#xA;But slowly as the men advanced&lt;br&gt;&#xA;I watched my kingdom fall&lt;br&gt;&#xA;I used to give so much to men&lt;br&gt;&#xA;But now they need me less&lt;br&gt;&#xA;Perfunctory their uses now&lt;br&gt;&#xA;But magic I once possessed&lt;br&gt;&#xA;Replaced, am I, by lightning bottled&lt;br&gt;&#xA;May Franklin be most cursed&lt;br&gt;&#xA;He gave men alternatives&lt;br&gt;&#xA;And so my growth reversed&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This passage describes how fire, once a source of magic and wonder, has lost it's allure and fascination as technology has advanced. Now we have electricity to give us light and power where before we required fire (torches/lanterns coal burning/electric motors). I wonder if forethought is a reference to Prometheus, who in Greek mythology is the one responsible for giving the gift of fire to mankind in a scheme (Upon further research, I found this &lt;a href=&quot;http://www.etymonline.com/index.php?term=Prometheus&amp;amp;allowed_in_frame=0&quot; rel=&quot;nofollow&quot;&gt;etymological explanation&lt;/a&gt; of Prometheus)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I gave men their healthy meat&lt;br&gt;&#xA;I made them safe at night&lt;br&gt;&#xA;I gave weapons, traps, and shields&lt;br&gt;&#xA;I shared with them my might&lt;br&gt;&#xA;Still they use me, although less&lt;br&gt;&#xA;They see me oft as &quot;quaint&quot;&lt;br&gt;&#xA;Where once I was in every home&lt;br&gt;&#xA;Now I often ain't&lt;br&gt;&#xA;They think me pleasant on wintry nights&lt;br&gt;&#xA;But keep me in my place&lt;br&gt;&#xA;They build traps of steel and stone&lt;br&gt;&#xA;And to their friends showcase&lt;br&gt;&#xA;Look, they say, &quot;We tamed the beast.&quot;&lt;br&gt;&#xA;I roar internally&lt;br&gt;&#xA;Well, how nice their friends reply&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Fire was used to cook meat, by smiths to forge weapons and armor, provided light and kept predators at bay overnight. Nowadays, fires are really only used in fireplaces (cage of steel and stone) as a cute old-fashioned way to stay warm in the cold.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I spit my tongues at thee!&lt;br&gt;&#xA;Let me loose; I shall destroy&lt;br&gt;&#xA;All that you hold dear&lt;br&gt;&#xA;I rage against my captors now&lt;br&gt;&#xA;They are right to fear&lt;br&gt;&#xA;I am ancient from long before&lt;br&gt;&#xA;The rise of these puny men&lt;br&gt;&#xA;I wage a ceaseless war&lt;br&gt;&#xA;All that you built upon my back&lt;br&gt;&#xA;Shall be torn down by rage&lt;br&gt;&#xA;I am wild and devour&lt;br&gt;&#xA;I grow but do not age&lt;br&gt;&#xA;You may send troops with special suits&lt;br&gt;&#xA;They may beat me back&lt;br&gt;&#xA;But I return and start again&lt;br&gt;&#xA;With lightning attack&lt;br&gt;&#xA;So sit or stand, you fleshy thing&lt;br&gt;&#xA;Believe that I am tamed&lt;br&gt;&#xA;My will be told: The world would be&lt;br&gt;&#xA;Entirely enflamed&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Fire is often described as 'licking' or having tongues at the edges of flame. Fire is an incredibly destructive force that can ravage cities, homes, buildings, forests, etc. The men in special suits is of course a reference to firefighters. 'lightning' attack is likely a reference to the fact that lightning striking dead wood in a forest can start a fire (and is believed to be how fire was discovered by neanderthals).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="13970" LastEditDate="2015-09-15T18:40:15.020" LastActivityDate="2015-09-15T18:40:15.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22220" PostTypeId="2" ParentId="22216" CreationDate="2015-09-15T18:53:45.480" Score="7" Body="&lt;p&gt;I think it is-&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $499\times 999 + 1000 = 499501$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The following facts hold-&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;There can be at most one 1000-coin. Otherwise we replace 2 1000-coins with (999+1001)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There can be exactly one type of coin from 999 and 1001, otherwise we replace two of them with 2 1000-coins.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There can be at most 500, 999-coins. Other wise we can replace 501, 999-coins with 499, 1001-coins and one 1000 coin.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;There can be at most 499, 1001-coins. Other wise we can replace 500, 1001-coins with 500, 999-coins and one 1000 coin.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Checking the top few possibilities:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;500,  999-coins and a 1000: can  be replaced with 500, 1001-coins&lt;/li&gt;&#xA;&lt;li&gt;499, 1001-coins and a 1000: can  be replaced with 501, 999-coins&lt;/li&gt;&#xA;&lt;li&gt;499,  999-coins and a 1000: Unique&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Uniqueness Proof&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $(499*999+1000)/1001=499501/1001=499.002&amp;gt;499$&lt;br&gt;&#xA;We need exactly 500 coins (since even 499 1001-coins falls short).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at it mod 1000, we are summing 500 numbers from the set $\{-1, 0, 1\}$ to $501$. So, all of them must be $-1\ (999)$ except one $0\ (1000)$.&lt;/p&gt;&#xA;" OwnerUserId="15544" LastEditorUserId="15544" LastEditDate="2015-09-16T18:57:37.833" LastActivityDate="2015-09-16T18:57:37.833" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22221" PostTypeId="2" ParentId="22212" CreationDate="2015-09-15T19:17:40.103" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; OCJS - Interpret each word as a number in base 26, with A=0, B=1, etc. Then use normal addition (be sure to carry 26's when necessary)!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15908" LastActivityDate="2015-09-15T19:17:40.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22222" PostTypeId="2" ParentId="22216" CreationDate="2015-09-15T19:18:21.217" Score="3" Body="&lt;p&gt;Some conditions for a representation $999d_1 + 1000d_2 + 1001d_3 = N$ not to be unique.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If $d_2 \ge 2$, then $(d_1+1, d_2-2, d_3+1)$ is also a valid representation of $N$. For $N$ to have a unique representation, we must have $d_2 = 0$ or $1$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If both $d_1 \ge 1$ and $d_2 \ge 1$, then $(d_1-1, d_2+2, d_3-1)$ is a valid representation of $N$, so at least one of $d_1$ and $d_3$ equals $0$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $501 \le d_1 \le 999$, then--assuming that $d_3=0$ because otherwise we already showed that there would be another representation--$(0, d_2 + 2d_1 - 1000, 1000-d_1)$ is a valid representation of $N$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $d_1 = 500$ and $d_2 = 1$, then--assuming that $d_3 = 0$ again--$(0, 1, 500)$ is another valid representation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $500 \le d_3 \le 999$, then $(1000-d_3, d_2 + 2d_3-999, 0)$ is a valid representation of $N$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $d_3 = 499$ and $d_2 = 1$, then--assuming that $d_3 = 0$ again--$(501, 0, 0)$ is another valid representation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If either $d_1$ or $d_3$ exceed $999$, then we can subtract $1000$ from them and add $999$ or $1001$ to $d_2$, respectively, for another valid representation of $N$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So what is the largest number we can construct that satisfies these requirements? For $d_2 = 1$, there are two possibilities, $d_3 = 498$ or $d_1 = 499$. For $d_2 = 0$, there are $d_3 = 499$ and $d_1 = 500$. These are $499498$, $499501$, and $499499$, $499500$, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $499501$ is indeed uniquely representable (I claim that it is), then it is the largest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $N \equiv d_3 - d_1 \pmod{1000}$ from the equation at the top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is another representation, then $d_1 + d_3 \le d_1 + d_2 + d_3 = \frac{N-d_3+d_1}{1000} &amp;lt; 499 - \frac{d_3-d_1}{1000}$. The largest $d_1$ can be is clearly $499$, and the largest $d_3$ can be is $498$. Because of this, to achieve $d_3 - d_1 \equiv 499 \pmod{1000}$, we would need $d_3 - d_1 = -499$. If $d_3 &amp;gt; 0$, then $d_1 &amp;gt; 499$ which is impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the answer is $499\cdot 999 + 1000 = \boxed{499501}$.&lt;/p&gt;&#xA;" OwnerUserId="11100" LastEditorUserId="11100" LastEditDate="2015-09-15T19:33:32.813" LastActivityDate="2015-09-15T19:33:32.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22223" PostTypeId="2" ParentId="22199" CreationDate="2015-09-15T19:23:23.717" Score="0" Body="&lt;p&gt;(Partial Answer)&#xA;I'm going to say the answer to the first one is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13068" LastActivityDate="2015-09-15T19:23:23.717" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22224" PostTypeId="2" ParentId="22216" CreationDate="2015-09-15T19:24:07.043" Score="3" Body="&lt;p&gt;Consider the following equivalent representations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;1000 + 1000 &amp;amp;= 999 + 1001 \\&#xA;499 \times 1001 + 1000 &amp;amp;= 501 \times 999 \\&#xA;500 \times 999 + 1000 &amp;amp;= 500\times 1001&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first shows that a unique representation can have no more than one $1000$, and cannot contain both $999$ and $1001$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second shows that a unique representation must have $500$ or fewer $999$s, and $498$ or fewer $1001$s if it contains a $1000$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the third shows that a unique representation must have $499$ or fewer $1001$s, and $499$ or fewer $999$s if it contains a $1000$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have four possible maximal unique representations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;500 \times 999 &amp;amp;= 499\ 500 \\&#xA;1000 + 499 \times 999 &amp;amp;= 499\ 501 \\&#xA;499 \times 1001 &amp;amp;= 499\ 499 \\&#xA;1000 + 498 \times 1001 &amp;amp;= 499\ 498&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the largest uniquely representable value is $499\ 501$.&lt;/p&gt;&#xA;" OwnerUserId="9718" LastActivityDate="2015-09-15T19:24:07.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22226" PostTypeId="1" AcceptedAnswerId="22227" CreationDate="2015-09-15T22:33:15.213" Score="8" ViewCount="1381" Body="&lt;p&gt;Alice loves magic squares. She has a 4x4 square, where she can put a number in each cell. But alas! Some evil person has poked a hole in her square. Alice is really really sad because she can't make a magic square anymore. Won't you cheer poor Alice up?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You are given a 4x4 square with 1 of its cells removed. Put the numbers 1 to 15 in it so that each row, column and diagonal sum to the same.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15544" LastActivityDate="2019-10-15T00:29:42.980" Title="The magic square with a hole" Tags="|magic-square|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22227" PostTypeId="2" ParentId="22226" CreationDate="2015-09-15T22:54:18.877" Score="13" Body="&lt;p&gt;Here are a few from &lt;a href=&quot;http://www.cafe64.net/numbers/4x4/4x4_1.html&quot; rel=&quot;noreferrer&quot;&gt;this site&lt;/a&gt;. Any magic square with integers 0-15 will work, considering the 0 as a hole. I've chosen four with the 0 in the inner four cells just to be sure it meets your criteria. The rows, columns, and diagonals of each add up to 30.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2GsKI.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2GsKI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13237" LastEditorUserId="13237" LastEditDate="2015-09-15T23:00:41.703" LastActivityDate="2015-09-15T23:00:41.703" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22228" PostTypeId="2" ParentId="22196" CreationDate="2015-09-16T03:24:31.013" Score="4" Body="&lt;p&gt;A more practical application of Ben Frankel's answer would be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nuclear weapons&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Amongst the elements I reside,&#xA;Beneath the earth, below the tide.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nukes are housed in silos below the earth and submarines under the sea. Earth and water were once believed to be elements (along with air and fire that are mentioned later)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Through air and aether you dread my flight,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No one wants nukes in the air&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fire descends on the stroke of midnight. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This line is a reference to the doomsday clock. A nuclear war (fire descending) would mean the clock has struck midnight.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And will we ever come in, out of the cold?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Will we ever see nuclear disarmament and an end to the cold war style stand offs that we currently have? I hope so but I doubt it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13197" LastActivityDate="2015-09-16T03:24:31.013" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22230" PostTypeId="2" ParentId="22210" CreationDate="2015-09-16T08:41:26.580" Score="2" Body="&lt;p&gt;The amounts can be written as 999$x$ + 1000$y$ + 1001$z$. If we want to add 1 to it:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;We can increase $y$ and decrease $x$ by 1.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We can increase $z$ and decrease $y$ by 1.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If neither can be done ($x$ and $y$ = 0), we can increase $x$ by 500 and decrease $z$ by 499 at the very least.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, 1001*498=498498 can't rise to 498499 and any amount starting from 499499 can be obtained. Taking 499499 and decreasing $z$s in return for $y$s (the opposite of the second scenario), we can decrement the amount by 1 every time (until 499000). Later, we decrement those new $y$s in return for $x$es (the opposite of the first scenario) until we get to 498501, which is obtainable. By this logic, 498500 is the greatest unobtainable amount.&lt;/p&gt;&#xA;" OwnerUserId="14016" LastActivityDate="2015-09-16T08:41:26.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22233" PostTypeId="1" AcceptedAnswerId="22236" CreationDate="2015-09-16T09:59:22.757" Score="15" ViewCount="2499" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;I am playing poker with three friends and from a well-shuffled deck we have each been dealt five cards. I have a hand consisting of the four kings and the two of hearts. Being a poker wizard I know exactly the probability that I have a winning hand.&lt;/p&gt;&#xA;&lt;p&gt;But then I suddenly discover that earlier in the day the family’s children were playing with the cards and fed six of them to their pet shark (but I don’t know which). How does this information change the probability with which I believe my hand will beat my opponents? Why?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;From &lt;a href=&quot;http://www.statslab.cam.ac.uk/%7Errw1/prob/exprob1.pdf&quot; rel=&quot;noreferrer&quot;&gt;http://www.statslab.cam.ac.uk/~rrw1/prob/exprob1.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4508" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-09-17T13:45:56.017" Title="Poker and a card shark" Tags="|probability|cards|" AnswerCount="7" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22234" PostTypeId="1" AcceptedAnswerId="22483" CreationDate="2015-09-16T10:06:08.540" Score="5" ViewCount="467" Body="&lt;p&gt;&lt;strong&gt;Decode the word coded inside the square:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&lt;code&gt;A.O.O.E D 2&lt;/code&gt;&lt;br /&gt;&lt;code&gt;T D C.I.E D&lt;/code&gt;&lt;br /&gt;&lt;code&gt;2 T.E.D.D.C&lt;/code&gt;&lt;br /&gt;&lt;code&gt;C.A O E O C&lt;/code&gt;&lt;br /&gt;&lt;code&gt;E 2.E.2 O E&lt;/code&gt;&lt;br /&gt;&lt;code&gt;C O.D A D.E&lt;/code&gt;&lt;br /&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The hint itself is hidden in the square.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are two steps! The first shows what to use to decode the second step.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rearranging the letters doesn't change anything.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 4 (new):&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First-grader would have more then enough math-skills to deal with the letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15856" LastEditorUserId="15856" LastEditDate="2015-09-18T18:34:28.297" LastActivityDate="2015-09-22T03:47:23.640" Title="Almost magical square" Tags="|cipher|enigmatic-puzzle|" AnswerCount="1" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="22236" PostTypeId="2" ParentId="22233" CreationDate="2015-09-16T10:43:36.723" Score="25" Body="&lt;p&gt;It changes...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;nothing at all. Since you have no information on the cards lost, it's just like putting six random cards at the bottom of the deck, which could be 'absorbed' by the shuffle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2015-09-16T10:43:36.723" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="22238" PostTypeId="1" CreationDate="2015-09-16T12:28:32.663" Score="3" ViewCount="982" Body="&lt;p&gt;One of this and one of that;&lt;br /&gt;&#xA;Don't you like my fancy hat?&lt;br /&gt;&#xA;When they looked at what they want,&lt;br /&gt;&#xA;I was found most elegant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's not much of me, you see;&lt;br /&gt;&#xA;But I'm quite meant to be.&lt;br /&gt;&#xA;In this social function you'll find:&lt;br /&gt;&#xA;I'm quite right every time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't hang yourself upon my rhymes,&lt;br /&gt;&#xA;That will just waste all your time.&lt;br /&gt;&#xA;But try and find the banquet table:&lt;br /&gt;&#xA;The dish they're serving is quite fabled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The fancy hat is a red herring, I just wanted something elegant that rhymed. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the hidden tag wasn't statistics, that's a little too ...specific.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I guess technically there is a fancy hat, but you have to use your imagination.&lt;br /&gt; (Yes, that is a valid hint)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13068" LastEditorUserId="13068" LastEditDate="2015-09-16T19:02:29.127" LastActivityDate="2018-04-26T12:24:30.210" Title="Short and sweet riddle" Tags="|riddle|" AnswerCount="5" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="22239" PostTypeId="5" CreationDate="2015-09-16T12:38:19.923" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-09-16T12:38:19.923" LastActivityDate="2015-09-16T12:38:19.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22240" PostTypeId="4" CreationDate="2015-09-16T12:38:19.923" Score="0" Body="hidden-tag denotes a tag that is hidden, at first, to show that the puzzle, once solved, will be properly categorized. It is a placeholder for that tag." OwnerUserId="13068" LastEditorUserId="13068" LastEditDate="2015-09-16T12:50:40.613" LastActivityDate="2015-09-16T12:50:40.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22241" PostTypeId="2" ParentId="22233" CreationDate="2015-09-16T13:23:00.430" Score="0" Body="&lt;p&gt;This is a complicated (almost unanswerable) question as there is no known implementation of optimal strategy in Poker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By seeing the cards that you know are not eaten (your cards + cards opened on the table), you get a list of cards of which you know for sure the 6 cards will be in. &lt;strong&gt;You will have to account for the probability of each card being missing in your strategy.&lt;/strong&gt; As you keep playing many hands, you will eventually get to know which cards are missing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Until then, you will even have to account for information that will be revealed in the future, and how this benefits you and the opponent right now and later on.&lt;/strong&gt; For example, you will have a probability distribution of all cards and the probability of them being missing. You will also have a probability distribution of how each card opened is going to affect your and your opponent's position currently. You will then need to compute your and your opponent's future possible states with probabilities. In each of these possible scenarios (which number in billions and trillions), you will have to compute how new information that is assumed to be  revealed and that which is yet to be revealed affect your position in the game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is most interesting is that you may even have to take into account future hands, when playing a given hand. An extremely rare circumstance is that you would voluntarily give up the current hand and fold, so that your cards remain secret, and this will benefit you in all future hands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;As of today, I think this is all beyond our computation. Most humans will probably assume the deck is complete (until they know the missing cards), and computers will fare only slightly better.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-09-16T13:23:00.430" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="22242" PostTypeId="2" ParentId="22233" CreationDate="2015-09-16T14:14:30.727" Score="1" Body="&lt;p&gt;I don't know for real the math behind it, but take in consideration a few things:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Not many hands in poker win from 'KKKK2': I don't play 5 Cards Poker, but i guess , if it goes by the same winning principle that Holdem have , the only hands that could win against that are: 'AAAAX'(let X be a random card) , 'AKQJT'(all of the same kind, or Royal Straight Flush ) and Straight Flushes . That said, if you have the four kings in hand, nobody can make the 'AKQJT' combination . With that in mind , only 'AAAAX' and Straight Flushes &amp;lt; K  can beat you. In Texas Hold'em ( or 2 cards poker ) , the chance of making 'Four of a kind' is '4,164 : 1' or 0.0240% , and for Straight Flushes there are 0.0014% or 72,192 : 1 odds , which is even worse.&lt;/p&gt;&#xA;&lt;p&gt;With that in mind, we can go to the second point:&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We need to know what is the probability of one of those 6 cards being an ACE , which would make the 'AAAAX' combination impossible .  If you have a 52 cards deck, and remove 5( 'KKKK2' ) , there is 47 cards left. On those 47, what's the chance on the removal of 6 that one will be an ACE? Pretty low chance, lower than 0.0240% .&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Going by the same principle that you can't make a combination if you don't have that card on the deck , we don't even need to recalculate the probability for straight flushes , because they're going to be even harder to get , so less than 0.0014% of chance.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Well , I don't know the probability for any of those cases , but at least it would make me confortable enough to go 'all-in' on that hand.&lt;/p&gt;&#xA;&lt;p&gt;Thanks Tim Couwelier for the remind, i completely forgot about straight flushes.&lt;/p&gt;&#xA;" OwnerUserId="16367" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-09-16T14:34:29.567" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22243" PostTypeId="1" AcceptedAnswerId="22245" CreationDate="2015-09-16T14:29:20.133" Score="16" ViewCount="1421" Body="&lt;p&gt;Yesterday's &lt;a href=&quot;https://puzzling.stackexchange.com/questions/22206/shakespeare-william-shakespeare&quot;&gt;Shakespearean quotes&lt;/a&gt; went so well, I decided to give it another go. I tried to pick ones that were a tad more obscure (by which I mean featured on fewer top X lists), but all are in my opinion well known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Do not from me applause withhold!&lt;br&gt;&#xA;  Do I not your attention hold?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;'Twere mine own loving arms were sought&lt;br&gt;&#xA;  by maid from which thou were begot&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A distance never trekked before&lt;br&gt;&#xA;  I shall traverse; and one step more!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The cobbled stones we'll not desire&lt;br&gt;&#xA;  when destination we acquire.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I've just two tasks; to your dismay&lt;br&gt;&#xA;  my stash of sweets be empt this day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6.(a bit harder than the others)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The darkened mind alone shall call&lt;br&gt;&#xA;  the words of nary one or all&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-09-17T14:19:41.073" Title="Art Thou Addressing Me?" Tags="|movies|poetry|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22244" PostTypeId="2" ParentId="22233" CreationDate="2015-09-16T14:34:25.223" Score="7" Body="&lt;p&gt;&lt;strong&gt;What follows below assumes a variation of poker is played with no cards visible on the table&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hands to which you lose given a full deck:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;4 aces + random kicker&lt;/li&gt;&#xA;&lt;li&gt;any straight flush&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Given you have 5 cards, there's 47 left.&lt;/strong&gt;&#xA;&lt;br&gt;To get 4 aces + any random, chances are: 4/47 * 3/46 * 2/45 * 1/44 * 43/43 or &#xA;1.032/184.072.680&#xA;&lt;br&gt;To get one particular straight flush, chances are:&#xA;5/47 * 4/46 * 3/45 * 2/44 * 1/45 = 120 / 184.072.680&#xA;&lt;br&gt;There's 8 possible straight ranges (A-5 up to 8-Q) in 4 suits , so 32 possible straight flushes. We also need to scratch A-5 of hearts and 2-6 of hearts, so 30 options left.&#xA;Odds are now 3.600 / 184.072.680 you lose to a straight flush.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total odds of losing to one particular friend: 4632 / 184.072.680 (approx 1/37739), odds of losing to either of the three = 13.896 / 184.072.680 (approx 1/12580)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Now assume you have 5 cards, but there's only 42 left&lt;/strong&gt;&#xA;&lt;br&gt;Chances for each option remaining are calculated in a similar way, but with 42 cards rather then 47:&#xA;&lt;br&gt;4 Aces + random kicker = 4/42 * 3/41 * 2/40 * 1/39 * 38/38 = 912 / 102.080.160&#xA;&lt;br&gt;Any of the 30 straight flushes = 30 * 5/42 * 4/41 * 3/40 * 2/39 * 1/38 = 3.600 / 102.080.160&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;This however assumes that all the required cards are left remaining. For each of the above options, the occurence rate should be multiplied with the chances that set is actually possible with the remaining cards, or: the odds none of the required cards have been taken out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the aces, we need the odds that none of the 4 aces were in the 5 cards taken out of the set of 47.&#xA;For the straight flushes, we need the odds that none of the 5 cards for a given straight flush were taken out of the set of 47.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To select 5 cards from a deck of 47, there are 47*46*45*44*43 =  184.072.680 combinations.&#xA;The amount of combinations that does NOT hold at least one of the 5 cards is 42 * 41 * 40 * 39 * 38 =  102.080.160&#xA;To select 5 cards from a deck of 47, the amount of combinations without a single ace being in them  is 43*42*41*40*39 = 115.511.760&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we multiply odds of hand beating us and odds of the hand being possible, and sum that over all hands:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;Aces:&#xA;&lt;br&gt;912 / 102.080.160 * 115.511.760 / 184.072.680 = 1032 / 184.072.600 (same as normal)&#xA;&lt;br&gt;Straight flushes&#xA;&lt;br&gt;3.600 / 102.080.160 * 102.080.160 / 184.072.680 =3600 / 184.072.680 (same as normal)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;Total odds of losing to one particular friend: 4632 / 184.072.680 (approx 1/37739), odds of losing to either of the three = 13.896 / 184.072.680 (approx 1/12580)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LONG ANSWER SHORT: It's the exact same&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2156" LastEditorUserId="2156" LastEditDate="2015-09-16T14:48:21.367" LastActivityDate="2015-09-16T14:48:21.367" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22245" PostTypeId="2" ParentId="22243" CreationDate="2015-09-16T14:36:34.573" Score="14" Body="&lt;p&gt;Full set of answers, with thanks to derium:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Are you not entertained?&quot; Maximus, &lt;em&gt;Gladiator&lt;/em&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I am your father&quot; Darth Vader, &lt;em&gt;Star Wars: The Empire Strikes Back&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;To Infinity... And Beyond!&quot; Buzz Lightyear, &lt;em&gt;Toy Story&lt;/em&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Where we're going, we don't need roads!&quot; Doc Brown, &lt;em&gt;Back to the future&lt;/em&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;I came here to chew bubblegum and kick ass... and I'm all out of bubblegum&quot; Nada, &lt;em&gt;They Live&lt;/em&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Only a Sith deals in absolutes&quot; a hypocritical jedi, &lt;em&gt;Star Wars: Revenge of the Sith&lt;/em&gt; (shamelessly borrowed from derium)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;And for the title:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;You talkin' to me?&quot; Travis Bickle, &lt;em&gt;Taxi Driver&lt;/em&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Given the theme of the puzzle, I feel obliged to point out that there is a &lt;a href=&quot;http://www.amazon.co.uk/William-Shakespeares-Star-Wars-Doescher/dp/1594746370&quot; rel=&quot;nofollow&quot;&gt;Shakespearean 'translation' of the original Star Wars trilogy&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11717" LastEditorUserId="11717" LastEditDate="2015-09-17T14:19:41.073" LastActivityDate="2015-09-17T14:19:41.073" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22246" PostTypeId="2" ParentId="22199" CreationDate="2015-09-16T14:52:06.000" Score="5" Body="&lt;p&gt;It is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Labor of Love&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The 1st part seems to point to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Sixth Sense &amp;lt;-- it mentions the normal 5 senses, and then the second to last and last line are asking what is the next one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The 2nd part I believe refers to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Unbreakable &amp;lt;-- 3 sticks you can easily break, but a 100?  No way.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which leads me to believe that these refer to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; M. Night Shyamalan and Bruce Willis movies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A quick online search showed me the above answer as being in the works.&lt;/p&gt;&#xA;" OwnerUserId="14233" LastEditorUserId="14233" LastEditDate="2015-09-17T11:42:43.847" LastActivityDate="2015-09-17T11:42:43.847" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22247" PostTypeId="2" ParentId="22238" CreationDate="2015-09-16T15:31:23.607" Score="6" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\hat q$  (as far as I understand it, the estimate of the probability of failure in statistics)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;One of this and one of that;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Refers to the practice of sampling in statistics -- take a small sample and base your estimate off of it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Don't you like my fancy hat?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Refers to the circumflex (hat) on the letter &lt;code&gt;q&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;When they looked at what they want,&lt;br&gt;&#xA;I was found most elegant.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is the best (most elegant) way of estimating, given a small sample size.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;There's not much of me, you see;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's only a single letter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But I'm quite meant to be.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;Quite&lt;/code&gt; hints at the letter &lt;code&gt;q&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In this social function you'll find:&lt;br&gt;&#xA;I'm quite right every time.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Using $\hat q$ to derive an estimate is fairly accurate; another hint at the letter &lt;code&gt;q&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Don't hang yourself upon my rhymes,&lt;br&gt;&#xA;That will just waste all your time.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Don't take this poem too literally; instead look at all the &lt;code&gt;q&lt;/code&gt;-words, and derive your answer that way.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But try and find the banquet table:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;Banquet&lt;/code&gt; is another hint at the letter &lt;code&gt;q&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The dish they're serving is quite fabled.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\hat q$ is a well-known variable in statistics.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2015-09-16T15:31:23.607" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22248" PostTypeId="2" ParentId="22238" CreationDate="2015-09-16T16:01:19.973" Score="2" Body="&lt;p&gt;Totally ripping off/inspired by GentlePurpleRain's answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\hat p$ , the estimated probability of success in statistics.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;One of this and one of that;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\hat{p}$ is often used in binomial trials, which have either success or failure..&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Don't you like my fancy hat?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Refers to the circumflex (hat) on the letter &lt;code&gt;p&lt;/code&gt;, which signifies an estimate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;When they looked at what they want,&lt;br&gt;&#xA;I was found most elegant.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most people are interested in the probability of success &lt;em&gt;p&lt;/em&gt;, but can't determine that exactly (at least frequentists). Thus, the &lt;em&gt;hat&lt;/em&gt; makes it possible to estimate, and hats make things more elegant.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;There's not much of me, you see;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's only a single letter, and in many trials the probability for success may be small.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But I'm quite meant to be.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; `The probability of something happening - meant to be - is &lt;em&gt;p&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In this social function you'll find:&lt;br&gt;&#xA;I'm quite right every time.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; \hat{p} is the best estimator of the probability of success.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Don't hang yourself upon my rhymes,&lt;br&gt;&#xA;That will just waste all your time.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Flipped upside down, a &lt;em&gt;p&lt;/em&gt; may look like a noose hanging from something.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But try and find the banquet table:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;p&lt;/em&gt; is often confusing in statistics because it is used to denote a lot of different things. Damn statisticians not having much imagination...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The dish they're serving is quite fabled.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;p&lt;/em&gt; for p-values, probability, estimates of probability, etc...  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All of this would lead me to guess that the &quot;hidden tag&quot; discussed in comments and in previous edits is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Probability  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-09-16T16:01:19.973" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22249" PostTypeId="1" AcceptedAnswerId="22252" CreationDate="2015-09-16T16:56:46.123" Score="3" ViewCount="437" Body="&lt;p&gt;Little known I was before,&lt;br&gt;&#xA;Now so many do me adore.&lt;br&gt;&#xA;I changed my life around you see,&lt;br&gt;&#xA;Now puzzle out who I may be.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I shot that kid, the cover up.&lt;br&gt;&#xA;Not quite rain, can't fill your cup.&lt;br&gt;&#xA;I puzzle it out with my mind,&lt;br&gt;&#xA;That explosion could have made me blind.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That man tried to start a war.&lt;br&gt;&#xA;Electricity can make me sore.&lt;br&gt;&#xA;Then my friends finally arrived.&lt;br&gt;&#xA;I don't like my dad, you soon derived.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plagued by dreams, house blown apart,&lt;br&gt;&#xA;Maybe that plan wasn't so smart.&lt;br&gt;&#xA;I'll be back soon, don't you pout,&lt;br&gt;&#xA;Raise the cry, give a shout!  &lt;/p&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-09-17T17:04:19.323" Title="A complete Change - Who am I?" Tags="|riddle|rhyme|knowledge|movies|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22250" PostTypeId="1" AcceptedAnswerId="22251" CreationDate="2015-09-16T17:08:41.727" Score="13" ViewCount="1155" Body="&lt;p&gt;Professor Erasmus has spent the last weekend by busily adding up all kinds of numbers. He told me today about his experiences: &lt;em&gt;&quot;On saturday evening, I added up twenty consecutive powers of $2$, and wrote their sum $S$ into my notebook. On sunday afternoon, I added up the first $n$ positive integers, and wrote their sum $T$ into my notebook. On monday morning, I suddenly realized that the sums $S$ and $T$ are equal. Mathematics is full of fascinating surprises!&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has the professor once again made one of his well-known mathematical blunders, or do such twenty consecutive powers of $2$ indeed exist?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="4421" LastEditDate="2015-11-25T18:00:36.310" LastActivityDate="2015-11-25T18:00:36.310" Title="Erasmus additions" Tags="|mathematics|number-theory|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22251" PostTypeId="2" ParentId="22250" CreationDate="2015-09-16T17:20:55.440" Score="19" Body="&lt;p&gt;The sum of twenty consecutive powers of $2$, starting at $2^k$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S=2^k(1+2+4+\cdots+2^{19})=2^k(2^{20}-1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the sum of the first $n$ positive integers is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T = 1+2+\cdots+n=\frac{n(n+1)}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If these two quantities are equal&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n(n+1)=2^{k+1}(2^{20}-1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so it suffices to take $k=19,n=2^{20}-1=1048575$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible with lowest power of $2$ of $2^{19}$, and a sum of consecutive integers from $1$ to $1048575$.&lt;/p&gt;&#xA;" OwnerUserId="14080" LastEditorUserId="14080" LastEditDate="2015-09-16T17:26:53.047" LastActivityDate="2015-09-16T17:26:53.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22252" PostTypeId="2" ParentId="22249" CreationDate="2015-09-16T18:27:29.887" Score="5" Body="&lt;p&gt;Could it be,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Robert Downey Jr.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Little known I was before,&lt;br&gt;&#xA;Now so many do me adore.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Robert Downey was little known before his role as Iron Man&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I changed my life around you see,&lt;br&gt;&#xA;Now puzzle out who I may be.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He had a lot of problems before his role as Iron Man which I think he said one time saved him. I don't have a link for that just going on memory&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I shot that kid, the cover up.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; US Marshals, it's the only one that I can see a possible cover up in but I don't get the kid part&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not quite rain, can't fill your cup.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tropic Thunder&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I puzzle it out with my mind,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sherlock Holmes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That explosion could have made me blind.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The nuclear explosion from the avengers is bright enough to blind someone&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That man tried to start a war.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Richard III&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Electricity can make me sore.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One of Tony Stark's enemies in Iron Man 2 used electrical whips. I feel like that would make someone sore&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then my friends finally arrived.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Less Than Zero&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't like my dad, you soon derived.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I don't think he likes his father in the movie The Judge&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Plagued by dreams, house blown apart,&lt;br&gt;&#xA;Maybe that plan wasn't so smart.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; His house was blown up in the third Iron Man&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll be back soon, don't you pout,&lt;br&gt;&#xA;Raise the cry, give a shout!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Captain America: Civil War will have Robert Downey Jr. in it in 2016&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13925" LastEditorUserId="13925" LastEditDate="2015-09-17T17:04:19.323" LastActivityDate="2015-09-17T17:04:19.323" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="22254" PostTypeId="1" CreationDate="2015-09-16T18:40:09.490" Score="3" ViewCount="387" Body="&lt;p&gt;&lt;em&gt;Sunny morning&lt;br&gt;&#xA;After brilliant night,&lt;br&gt;&#xA;Wheat fields&lt;br&gt;&#xA;Tinted in red,&lt;br&gt;&#xA;Road in between&lt;br&gt;&#xA;Whispering - love...&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beautiful, isn't it? Looks tidy and full? Don't trust your first impression... Find out what is before the thunderstorm?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each row represents a single word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Looking above where thunderstorms exist may help.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For some first strophe is in place, for others it is not.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15856" LastEditorUserId="15856" LastEditDate="2015-09-18T18:57:23.623" LastActivityDate="2015-09-18T18:57:23.623" Title="What is before the thunderstorm" Tags="|riddle|wordplay|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22255" PostTypeId="2" ParentId="22243" CreationDate="2015-09-16T20:12:39.170" Score="7" Body="&lt;p&gt;For 6&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Only a Sith deals in absolutes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14418" LastActivityDate="2015-09-16T20:12:39.170" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22256" PostTypeId="2" ParentId="22254" CreationDate="2015-09-16T20:25:33.820" Score="6" Body="&lt;p&gt;Perhaps a bit of a stretch, but my first thought was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/zL4hv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zL4hv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sunny morning&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sunny -&gt; Yellow, morning -&gt; 'M'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After brilliant night&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The signs are brightly illuminated at night.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wheat fields&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The arches don't look unlike the drooping heads of wheat stalks&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tinted in red&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The red background&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Road in between&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Referencing the middle 'road' in the M, between the two outer 'roads'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whispering - love&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm lovin it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is before the thunderstorm?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is clearly the wrong answer, but I'm enjoying making it fit. &lt;BR&gt; According to this McDonald's ad, Ronald McDonald is what/who appears before the thunderstorm (it's just cloudy beforehand):  &lt;a href=&quot;http://www.ispot.tv/ad/7ogD/mcdonalds-happy-meal-cloudy-day&quot; rel=&quot;nofollow noreferrer&quot;&gt;McDonald's Thunderstorm Ad&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12206" LastEditorUserId="12206" LastEditDate="2015-09-16T20:51:05.757" LastActivityDate="2015-09-16T20:51:05.757" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22259" PostTypeId="1" AcceptedAnswerId="22288" CreationDate="2015-09-16T20:49:12.040" Score="12" ViewCount="457" Body="&lt;p&gt;There is a group of 300 Twitter users. Each user is following exactly one other user in the group. Prove that there exists a smaller group of 100 users where no one is following anyone else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Source:&lt;/strong&gt; &lt;a href=&quot;http://www.austms.org.au/Publ/Gazette/2015/Jul15/Puzzle.pdf&quot;&gt;Aust MS Gazette Puzzle Corner 43&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="8874" LastEditDate="2015-09-18T14:45:40.623" LastActivityDate="2015-09-18T14:45:40.623" Title="Twitter Followers" Tags="|mathematics|graph-theory|" AnswerCount="5" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22260" PostTypeId="2" ParentId="22259" CreationDate="2015-09-16T21:00:41.717" Score="-1" Body="&lt;p&gt;If each user is following exactly one other user, then it can be a ring of 300 users forming a circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pulling every third person from that loop will give you a group of 100 users who are not following the next person in the subgroup. Even pulling every OTHER person in the big loop will do that for 150 people total.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other extreme has 299 users following user $a$, and user $a$ follows one other user.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both situations provide the required criteria of at least 100 users that are not following another user in the subgroup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also have 150 groups of users following each other in pairs. That's over 100.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you could even have 100 groups of three users forming small circles, which is equivalent to the big circle approach, you'd just end up taking one person from each circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure someone will swoop in with a mathematical proof that it goes from 100 to 299 based on the number of loops or something but it seems so obvious to me that I don't feel the need to bust out the groan-inducing calculus.&lt;/p&gt;&#xA;" OwnerUserId="13068" LastEditorUserId="13068" LastEditDate="2015-09-16T21:23:55.030" LastActivityDate="2015-09-16T21:23:55.030" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22261" PostTypeId="2" ParentId="15437" CreationDate="2015-09-16T21:16:17.090" Score="1" Body="&lt;p&gt;Some more of the puzzle, based off of frodo's answer. (Lots of guesses here)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My second is sometimes called what people sometimes call the shark hanging in the Smithsonian.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jaws, so maybe &lt;strong&gt;{ }&lt;/strong&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My sixth is in the number three.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clue #3 has an extra &lt;strong&gt;hyphen&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My seventh can replace my first and my last.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A &lt;strong&gt;semicolon&lt;/strong&gt; can be used to combine two halves of a sentence, similarly to a colon and parentheses.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If these guesses are right, the answer looks like the following (square = unknown).. unfortunately, I still have no idea what it's supposed to be.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; :{}☐---;()&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5616" LastActivityDate="2015-09-16T21:16:17.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22262" PostTypeId="2" ParentId="22250" CreationDate="2015-09-16T21:25:40.050" Score="0" Body="&lt;p&gt;$$2^x + ... + 2^{x+19} = {n(n+1)\over 2} = 2^x(2^{20}-1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n(n+1) = 2^{x+1} (2^{20}-1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it's definitely possible if $n = 2^{20}-1$ and $x = 19$.&lt;/p&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="15544" LastEditDate="2015-09-24T11:15:21.907" LastActivityDate="2015-09-24T11:15:21.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22263" PostTypeId="2" ParentId="22259" CreationDate="2015-09-16T21:36:13.640" Score="6" Body="&lt;p&gt;We consider the directed graph G=(V,E) of all Twitter users, such that V={1,2,...,300} and $(i,j)\in E$ if and only if user $i$ follows user $j$. Since every node has outgoing degree of $1$, the graph contains several cycles and each of the other nodes is path connected to one of the cycles, i.e. for every user $i$ not belonging to a cycle and every cycle $C$, there is succession of users $a_1, a_2, ... , a_k$, such that user $i$ follows $a1$, $a1$ follows $a_2$, ... , $a_k$ follows someone from $C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we can conclude that the graph can be split into $l$ independent subgraphs, where $l$ is the number of cycles in $G$. Alternatively, we can assume that $G$ has size $n$, contains just one cycle and we want to prove that we can choose $\lceil n/3\rceil$ users in it. Now arrange the users in layers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First layer - the users belonging to the cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second layer - the users who follow someone from the cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third layer - the users who follow someone following someone from the cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so on...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let the number of people in the first layer is $x_0$, the number of people in the second layer is $x_1$, the number of people in the third layer is $x_2$, etc. Clearly, if &#xA;$$\lfloor x_0/2\rfloor + x_2+ x_4+... \geq \lceil n/3\rceil$$&#xA;or&#xA;$$x_1+x_3+x_5+...\geq \lceil n/3\rceil,$$&#xA;then the problem is solved. In the first case we just choose every other user in the cycle + the users in the even layers, in the second case we choose the users in the odd layer. If  we assume that these two inequalities are not true, then&#xA;$$\lfloor x_0/2\rfloor +x_1+x_2+...\leq 2\lceil n/3\rceil-2$$&#xA;and since&#xA;$$x_0+x_1+x_2+...=n,$$&#xA;we get&#xA;$$n+2-2\lceil n/3\rceil \leq \lceil x_0/2\rceil.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last expression, combined with &#xA;$$\lfloor x_0/2\rfloor + x_2+ x_4+... \leq \lceil n/3\rceil -1$$&#xA;gives&#xA;$$n+3 + x_2 + x_4 + ...\leq 3 \lceil n/3 \rceil +\lceil x_0/2 \rceil - \lfloor x_0/2 \rfloor \leq 3\lceil n/3 \rceil +1,$$&#xA;which is possible only if $x_0$ is odd, $n=3m+1, x_2=x_4=...=0$ and then $x_i=0$ for $i\geq 2$ (this is true because the third layer is empty and therefore all successive layers are empty as well). Then we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lfloor x_0/2 \rfloor \leq m,$$&#xA;$$x_1 \leq m,$$&#xA;$$x_0+x_1=3m+1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the only option is $x_0=2m+1$ and $x_1=m$. Since $m\geq 1$, there is at least one user U in the second layer. He follows some user V from the first layer. Now we take again $m$ users from the first layer which do not follow each other, skipping user V. Then add user U to the bunch and eventually find $m+1$ users which satisfy the condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, using the arguments above we can easily conclude that the extremal case is when the graph $G$ is consisted of several subgraphs, which are either $3$-cycles or $3$-cycles with one additional user which follows someone from the $3$-cycle.&lt;/p&gt;&#xA;" OwnerUserId="14140" LastEditorUserId="14140" LastEditDate="2015-09-17T02:17:16.363" LastActivityDate="2015-09-17T02:17:16.363" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22265" PostTypeId="2" ParentId="22233" CreationDate="2015-09-16T22:47:26.047" Score="9" Body="&lt;p&gt;If your &quot;Statistics intuition&quot; is broken (like mine), here's a simpler example that might make it more clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a deck of 5 cards, numbered 1-5.  My single opponent and I each draw a card, and the higher card wins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've draw a 4!  My odds of winning, with a normal deck, are 75%, since the only card my opponent can win with is the 5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Buuuut one of the cards has gone missing, and it could have been any card (other than the 4 I'm holding).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the 5 is missing (25% chance), then I win automatically!  This is &lt;strong&gt;a higher&lt;/strong&gt; probability than a normal deck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If any other card is missing (75% chance), I have a 66% chance of winning.  This is &lt;strong&gt;a lower&lt;/strong&gt; probability than a normal deck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All told, my overall odds of winning are 25%(1) + 75%(0.66), or 75%.  The same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that by removing a card, any gains I would get by eliminating winning hands (the 5, or a Royal Flush in the real example) are countered by the fact that I also have a chance to eliminate losing hands (the 1, or a Flush). &lt;/p&gt;&#xA;" OwnerUserId="11887" LastActivityDate="2015-09-16T22:47:26.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22267" PostTypeId="2" ParentId="22250" CreationDate="2015-09-16T23:18:19.787" Score="6" Body="&lt;p&gt;&lt;strong&gt;If he claims to have done this without a computer, Erasmus has certainly made a mistake.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first triangular number that is the sum of 20 consecutive powers of 2 is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1 + 2 + 3 + \ldots + 1048575 = 2^{19} + 2^{20} + \ldots + 2^{38} = 549755289600$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Erasmus might have computed the sum $S$ on Saturday afternoon. However, consider what he would have gone through on Sunday.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose Erasmus can keep the current integer and the running sum in his head (so he's not slowed by having to write down the numbers). The world record for &lt;a href=&quot;http://www.recordholders.org/en/list/memory.html#adding10digits&quot;&gt;mental addition of 10-digit numbers&lt;/a&gt; is 2:51 to add 10 numbers. That comes out to 1.71 seconds per (random) digit. Now, Erasmus doesn't have quite as hard a job, because his digits aren't random -- but he will have to carry quite a lot more and he has many more digits to work with, 6.22 million. So let's suppose he can do a hundred times better than the world record holder at 17 milliseconds per digit. Unfortunately, it will still take him over a day to reach $T$ -- not a task that could be done in an afternoon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Yes, there's a simple shortcut, but Erasmus clearly states, &quot;&lt;em&gt;I added up the first $n$ positive integers&lt;/em&gt;,&quot; not &quot;I calculated the sum of the first $n$ positive integers.&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, I think it's safe to say that the professor has bungled it up again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, if he was able to do it in his head in about 5 hours (as generous as I think I can get for &quot;afternoon&quot;), that would put him at 1 addition per 14 milliseconds, significantly faster than such early computers as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Harvard_Mark_I&quot;&gt;Harvard Mark I&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Harvard_Mark_II&quot;&gt;Mark II&lt;/a&gt;, and &lt;a href=&quot;https://en.wikipedia.org/wiki/Z3_%28computer%29&quot;&gt;Z3&lt;/a&gt;. If he had to use paper, he would have gone through several reams of the stuff.&lt;/p&gt;&#xA;" OwnerUserId="15727" LastActivityDate="2015-09-16T23:18:19.787" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22268" PostTypeId="2" ParentId="22233" CreationDate="2015-09-16T23:33:28.933" Score="0" Body="&lt;p&gt;If the children fed random cards to the shark, and neither you nor your opponents know which ones, then the probabilities haven't changed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If either you or one of your opponents know which cards were fed to the shark, that player would have an advantage; they could calculate more precisely what is the probability of winning or losing with the hand they have. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same if the cards are not picked randomly: Whoever knows how to eaten cards were picked has the same advantage as before. &lt;/p&gt;&#xA;" OwnerUserId="4664" LastActivityDate="2015-09-16T23:33:28.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22269" PostTypeId="2" ParentId="22243" CreationDate="2015-09-17T06:11:10.233" Score="1" Body="&lt;p&gt;For the first one, how about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I believe I deserve some recognition from this board&quot; - &lt;em&gt;Mark Zuckerberg&lt;/em&gt; from &lt;em&gt;The Social Network&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16363" LastActivityDate="2015-09-17T06:11:10.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22270" PostTypeId="2" ParentId="22259" CreationDate="2015-09-17T09:08:47.743" Score="1" Body="&lt;p&gt;I had a bit of a hard time understanding Artur's answer becuase of the mathematics involved. So I tried to came up with my own solution. I got one and I think it's correct and of course it also involves mathematics but I believe it is easier to understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us define all people as $S$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we divide the people in 3 groups:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$X$ the people that are not being followed at all&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$Y$ the people not in $X$ and that are also not being followed by anyone in $X$, note that $Y \ge S - 2X$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;$Z$ the rest, note that $Z \le X$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Let's also define $Q$, a subgroup of $Y$: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the people in $Y$ that are also not being followed by anyone in $Z$. note $Q \ge S - 3X$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And then always pick everyone from $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now with $S=300$ as the problem states:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $X \ge 100$, you're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $X = 99$, the last 1 you pick just needs to be anyone of $Y$  and $Y \ge 102$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $X = 98$. Now we want to take the remaining 2 from $Q$. $Q \ge 6$. Notice something familiar? 2 is one third of 6, just like the original problem. In the &lt;em&gt;worst case&lt;/em&gt; all people in $Q$ follows someone else in $Q$, but anyhow, with this group $Q$ we just start over and do the algorithm again, since this algorithm guarantees that one third gets picked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because in general:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $X \ge S/3$, you're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any $M &amp;gt; 0$&lt;br&gt;&#xA;if $X = S/3 - M$, then $Q \ge 3M$ and we can do this algorithm again with $Q$ which will get us $M$ people picked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one corner-case and that is when $X = 0$. This means that there are &lt;em&gt;only&lt;/em&gt; seperate cycles. In the best case there is only one cycle and then you can pick every other person. In the worst case there are only cycles of 3 people and then you can pick 1 person from every cycle.&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-09-17T09:08:47.743" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22271" PostTypeId="2" ParentId="11348" CreationDate="2015-09-17T09:32:54.753" Score="2" Body="&lt;p&gt;&lt;sub&gt; This a community wiki answer to bring together each individual parts of the answer. Please upvote the relevant answers as well if you are upvoting this. &lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1.&lt;/strong&gt; H, J, K, M, N, U, V, W, X, and Y are. &#xA;A, B, C, D, E, F, G, I, L, O, P, Q, R, S, T, and Z aren’t.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  The letters that &quot;are&quot; all have open tops. In other words can catch rainwater if used as a cup. (Maxqueue)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2.&lt;/strong&gt;  B, D, E, F, H, I, K, L, M, N, P, R, T, U, V, W, X, Y, and Z are. A, C, G, J, O, Q, and S aren’t.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letters all have a part touching the top left corner of the letter's containing rectangle. (itriedacrab)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3.&lt;/strong&gt; E, F, P, T, and Y are. A, B, C, D, G, H, I, J, K, L, M, N, O, Q, R, S, U, V, W, X, and Z aren’t.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The chosen letters are all formed with a single 3-way junction. (Joffan)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15544" LastEditorUserId="9892" LastEditDate="2015-09-17T11:37:09.363" LastActivityDate="2015-09-17T11:37:09.363" CommentCount="0" CommunityOwnedDate="2015-09-17T09:32:54.753" ContentLicense="CC BY-SA 3.0" />
  <row Id="22272" PostTypeId="1" CreationDate="2015-09-17T10:45:13.350" Score="-1" ViewCount="723" Body="&lt;p&gt;You have 4 spacial locks with arrows on them, each facing up or down (you don't know the start order), on each corner of a box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can only check if all of them are facing the same direction (all arrows are to the same direction UP or DOWN) - as the box opens when they do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have 3 switches -&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) changes one of the locks randomly to the opposite state&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) changes randomly two close locks states (corners with a line between them - up right + up left \ up right +bottom right \ up left+bottom left \ down left+down right) total of 4 options &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) changes randomly two far locks (up left +bottom right \ up right+ bottom left) total of 2 options&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each turn you play you can check if the box opens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the series of clicks that will open the box no matter the initial order of the locks arrows?&lt;/p&gt;&#xA;" OwnerUserId="16379" LastEditorUserId="16379" LastEditDate="2015-09-17T12:00:03.347" LastActivityDate="2015-09-17T12:00:03.347" Title="open a box with 4 locks" Tags="|riddle|" AnswerCount="1" CommentCount="3" ClosedDate="2015-09-17T11:35:59.113" ContentLicense="CC BY-SA 3.0" />
  <row Id="22274" PostTypeId="2" ParentId="22272" CreationDate="2015-09-17T11:55:17.223" Score="0" Body="&lt;p&gt;First, let's isolate the ones 2 right and 2 wrong:&#xA;&lt;br&gt;1 1  / 0 1 &#xA;&lt;br&gt;0 0  / 1 0 &#xA;&lt;br&gt; Let's call these 'A' and 'B'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we find a combination, using switches 2 and 3, that is guaranteed to lead to a valid solution (0000 or 1111)?&#xA;&lt;br&gt; From starting position A, using switch 2 will either solve, or lead to position B. Using switch 3 will always turn it into case B (or it's inverse).&#xA;Using switch 2 and then switch 3, from starting position A, we'll either solve after one move, or solve in two (A to B, then solve B)&#xA;&lt;br&gt; From starting position B, switch 3 is a guaranteed solve, whereas switch 2 leads to state 'A'. Using switch 2 then switch 3 will lead to a guaranteed state 'A'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So 2/3/2/3 will:&#xA;Solve a starting state A in one or two moves.&#xA;Solve a starting state B in at most 4 moves (solve in one, or turn to 'A' in two, then solved in one or two more).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's assume we do a proces 2/3/2/3, and no solution comes up. This means we have a setup where 3 of the switches are the same, and one is different. Using switch 1 once, will solve it, or turn it into a 2 up/2 down situation. Either way that implies it's in position A or B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary:&lt;/strong&gt;&#xA;&lt;br&gt;2 / 3 / 2 / 3 / 1 / 2 / 3 / 2 / 3 should yield a guaranteed solution in at most 9 tries.&lt;/p&gt;&#xA;" OwnerUserId="2156" LastActivityDate="2015-09-17T11:55:17.223" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22275" PostTypeId="1" CreationDate="2015-09-17T11:59:46.747" Score="-3" ViewCount="213" Body="&lt;p&gt;$$\begin{align}&#xA;4&amp;amp;\qquad6\\&#xA;&amp;amp;\,\diagdown\diagup\\&#xA;&amp;amp;\;\;\,8\\&#xA;\\&#xA;6&amp;amp;\quad\;\;18\\&#xA;&amp;amp;\,\diagdown\diagup\\&#xA;&amp;amp;\;\,27\\&#xA;\\&#xA;10&amp;amp;\quad\;\;\,16\\&#xA;&amp;amp;\diagdown\diagup\\&#xA;&amp;amp;\;\;x\\&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is an easy one. You have to find $x$.&lt;/p&gt;&#xA;" OwnerUserId="16380" LastEditorUserId="4421" LastEditDate="2015-09-17T17:34:16.523" LastActivityDate="2016-08-15T16:05:32.230" Title="Elementary Number pattern" Tags="|number-sequence|" AnswerCount="3" CommentCount="3" ClosedDate="2016-08-15T18:01:21.553" ContentLicense="CC BY-SA 3.0" />
  <row Id="22280" PostTypeId="2" ParentId="22233" CreationDate="2015-09-17T13:45:56.017" Score="3" Body="&lt;p&gt;Instead of feeding the cards to the shark, imagine you deal it in to the game. This doesn't change whether your hand is best among the human players.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively play in a casino but when it comes to burn a card, feed it to a shark. Obviously this makes no difference to the outcome of the game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Us3DD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Us3DD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4508" LastActivityDate="2015-09-17T13:45:56.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22281" PostTypeId="2" ParentId="1776" CreationDate="2015-09-17T14:41:31.843" Score="5" Body="&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; My answer isn't even on the list of options. My bad, scrap this. I totally read the 51 as a 31.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though this question is old and hasn't seen much action recently, it bothers me that the accepted answer uses additional constants when there is a much simpler solution. It can be stated in one line:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The top number is equal to the average of the products of the diagonal numbers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This can be expressed as the following equation:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;E = ((a * d) + (b * c))/2,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where the pentagon is represented by the variables shown below.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  E&#xA;a   b&#xA;c   d&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For the first pentagon: &lt;code&gt;((5 * 8) + (6 * 7))/2 = (40 + 42)/2 = 41&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second: &lt;code&gt;((9 * 6) + (4 * 2))/2 = (54 + 8)/2 = 31&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for the unknown: &lt;code&gt;((8 * 6) + (7 * 2))/2 = (48 + 14)/2 = 31&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the answer is 31.&lt;/p&gt;&#xA;" OwnerUserId="16275" LastEditorUserId="16275" LastEditDate="2015-09-17T19:54:40.993" LastActivityDate="2015-09-17T19:54:40.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22282" PostTypeId="2" ParentId="22259" CreationDate="2015-09-17T15:19:17.167" Score="0" Body="&lt;p&gt;&lt;strong&gt;The Obligatory MS Paint Images (now with Word 2010!)&lt;/strong&gt;&lt;br&gt;&#xA;Since each Twitter user &lt;em&gt;must&lt;/em&gt; be connected with at least one other user, each will be a part of a larger network of 2-N users. In each independent group of users, we will look at how many we can isolate. Obviously, we can treat each separate group independently; if there were a connection between the two, they would become one larger group to be analyzed.&lt;br&gt;&#xA;&lt;strong&gt;If ANY group of M members (for which M|N) forbids isolating (in green) one third or more of its members, we have failed&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1. Two mutually following: (1,1)&lt;/strong&gt;&#xA;Here we can just pick one of the indistinguishable Twitter users. We get $1/2$ of all users in groups of two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XQivB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XQivB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2. Three followed by (1,1,1) and (2,1,0)&lt;/strong&gt;&#xA;There are only two ways for this independent group to follow each other, since no user can follow oneself. Here we have our worst passing case, $1/3$ of members in a (1,1,1) group, and $2/3$ from the (2,1,0).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lAFOb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lAFOb.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3. Four followed by (1,1,1,1), (2,1,1,0), (2,2,0,0), and (3,1,0,0)&lt;/strong&gt;&#xA;As the groups get larger and more complex, it becomes impossible to pick fewer than $1/3$ of users.&lt;br&gt;&#xA;&lt;strong&gt;In the (1,1,1,1):&lt;/strong&gt; For this and larger simple cycles of (1,1,1...,1), we can always choose every other user. For even M, we isolate $1/2$ of users. For odd M, we have $(M-1)/2$ users, which is lowest at (1,1,1), $1/3$.&lt;br&gt;&#xA;&lt;strong&gt;In each other case:&lt;/strong&gt; We can represent the group as a smaller group with one or more outsiders, followed by zero people themselves. This will always allow for at least as many isolations as the minimal (1,1,1) case, as seen in the images of (2,1,1,0), (2,2,0,0), and (3,1,0,0).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6AuVg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6AuVg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Finally, since each possible group of M users -- from $M = 2$ to $M = N$ -- allows for at least $1/3$ of the users to be isolated, at least $1/3$ of the total N users MUST be isolated from each other. For $N = 300$, this is $100$ Twitter users.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="13237" LastEditorUserId="13237" LastEditDate="2015-09-17T16:24:00.437" LastActivityDate="2015-09-17T16:24:00.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22283" PostTypeId="2" ParentId="20033" CreationDate="2015-09-17T16:26:56.743" Score="0" Body="&lt;p&gt;Well The Answer Seems to be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sherlock Holmes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I used to have a name like a valediction,&#xA;Others say it's purely fiction.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Valediction means to say something at fare well, thats &lt;strong&gt;good luck&lt;/strong&gt; or say sheer luck which is homophone for Sherlock&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My surname keeps your valuables safe,&#xA;My upbringing was completely as a waif.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Holmes homophone for homes, which keep our valuable safe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You've seen me in many a grand time,&#xA;Up the ladder did I climb.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not Done Yet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be found picking on the smaller man;&#xA;There are many that I am better than.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bully syn. noun:Domineering Person, whose homophone could be dominating person. And we know how much dominating Sherlock could be.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My lover found a life of empty space,&#xA;Vowing that I would never see her face.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Working on Irene Adler and empty.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I decided to live like a camelid,&#xA;In hopes to overcome what I did.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Not Done&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Even in death I provided support,&#xA;Though to base pleasures I did resort.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Not done&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11343" LastActivityDate="2015-09-17T16:26:56.743" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22284" PostTypeId="1" CreationDate="2015-09-17T16:42:28.153" Score="2" ViewCount="312" Body="&lt;p&gt;Just wondering whether anyone can see an unambiguous solution the reasoning test example below? Please explain your reasoning if possible. Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Uz8SO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uz8SO.png&quot; alt=&quot;Reasoning Test&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16388" LastActivityDate="2015-09-17T18:06:16.027" Title="Does this puzzle have a clear solution?" Tags="|pattern|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22285" PostTypeId="2" ParentId="22284" CreationDate="2015-09-17T16:48:25.890" Score="1" Body="&lt;p&gt;The answer is either A or B...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In each move:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;bottom right shaded symbol becomes top left hollow symbol&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;top right symbol is rotated and enlarged to become bottom left symbol&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I can't see any means of deciding between A and B - the horizontal alignment of the left column maybe? - so I'm going to go with &lt;em&gt;no&lt;/em&gt;, I don't think there is a unique solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Or&lt;/em&gt;&lt;/strong&gt; if a feeble reason will do, then&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;new symbol in top right must contain a vertical line&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And then the answer is uniquely &lt;strong&gt;A&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="4050" LastEditDate="2015-09-17T17:05:29.553" LastActivityDate="2015-09-17T17:05:29.553" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22287" PostTypeId="2" ParentId="22284" CreationDate="2015-09-17T16:56:25.467" Score="1" Body="&lt;p&gt;It actually does make sense when you see the hint. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a diagram is of the form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{matrix}&#xA;    L &amp;amp; ᴎ  \\&#xA;    M &amp;amp; ⚫  \\&#xA;\end{matrix}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the next is of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{matrix}&#xA;    \circ &amp;amp; P  \\&#xA;    \text{N} &amp;amp; Q  \\&#xA;\end{matrix}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $\circ$ was filled black in the first matrix and filled empty in the second and ᴎ was flipped vertically and enlarged. P and Q are new unused figures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the answer should be B. A doesn't work because the hollow arrow was used before. The diamond was used before too, but it was of different proportions. Look at the hint, it actually reveals the whole answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could answer this quickly only because I saw the hint. I don't know how much time it would take me if I didn't (or if I could even solve it for that matter)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is the problem clear enough?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that A E has come up with a equally valid reasoning for A. In my opinion, the puzzle is unambiguous enough to discard options C,D,E. But differentiating between A and B is up to interpretation.&lt;/p&gt;&#xA;" OwnerUserId="15544" LastEditorUserId="15544" LastEditDate="2015-09-17T18:06:16.027" LastActivityDate="2015-09-17T18:06:16.027" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="22288" PostTypeId="2" ParentId="22259" CreationDate="2015-09-17T17:50:17.063" Score="6" Body="&lt;p&gt;Imagine all 300 Twitterites are standing in a room. Each person follows one other, so the average number of followers a person has is one. This means there exists a person $P$ with at most one follower. We select person $P$ to be in the smaller group, and we dismiss the person $P$ follows and the one person following $P$ (if there is such a person) from the room. Now nobody left in the room follows or is followed by $P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We repeat this process. Each time, we choose a person in the room who has at most one follower still in the room, we add that person to the smaller group, and we lose at most three people from the room (the person we add to the smaller group, the person they follow, and the person following them, if there is one). The room starts with 300 people, so we can do this 100 times, resulting in a small group of 100 people in which nobody follows anybody else.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-09-17T17:58:58.063" LastActivityDate="2015-09-17T17:58:58.063" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22289" PostTypeId="2" ParentId="20033" CreationDate="2015-09-17T18:09:19.487" Score="1" Body="&lt;p&gt;Here are some guesses. It still isn't clear to me how this is supposed to be solved; we'll have to wait for the correct answer to see if this was worth our efforts, apparently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used to have a name like a valediction,&lt;br&gt;&#xA;Others say it's purely fiction.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The valediction Bye, which has the homophones By and Buy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My surname keeps your valuables safe,&lt;br&gt;&#xA;My upbringing was completely as a waif.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lock has the homophone Locke, a surname.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The rest are wild guesses&lt;/strong&gt;&lt;br&gt;&#xA;You've seen me in many a grand time,&lt;br&gt;&#xA;Up the ladder did I climb.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rose (rows), Peace (Piece), Flower (Flour)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be found picking on the smaller man;&lt;br&gt;&#xA;There are many that I am better than.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One (Won), Weak (Week)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My lover found a life of empty space,&lt;br&gt;&#xA;Vowing that I would never see her face.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; None (&quot;empty space&quot;) is a homophone with Nun. Some cloisters forbid contact with the outside world (&quot;never see her face&quot;). Does anyone know of a story where a man fell in love with a nun who lived in a place like &lt;a href=&quot;https://en.wikipedia.org/wiki/Monastery_of_the_Angels&quot; rel=&quot;nofollow&quot;&gt;the Monastery of Angels&lt;/a&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I decided to live like a camelid,&lt;br&gt;&#xA;In hopes to overcome what I did.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Desert (the verb, not the noun) is a homophone with Dessert (the food)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Even in death I provided support,&lt;br&gt;&#xA;Though to base pleasures I did resort.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Prophet (dying as a martyr) is a homophone with Profit (arguably a base pleasure)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13237" LastEditorUserId="13237" LastEditDate="2015-09-17T21:59:15.140" LastActivityDate="2015-09-17T21:59:15.140" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22290" PostTypeId="1" AcceptedAnswerId="22292" CreationDate="2015-09-17T18:21:35.937" Score="13" ViewCount="1190" Body="&lt;p&gt;Professor Erasmus told me that today he has proved another fascinating theorem about strings made of 0s and 1s. He takes an arbitrary such string and repeats the following step on it:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;if the leftmost symbol is 1, Erasmus removes the three leftmost symbols and attaches 11 at the right end&lt;/li&gt;&#xA;&lt;li&gt;if the leftmost symbol is 0, Erasmus removes the three leftmost symbols and attaches 0010 at the right end&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;If the procedure ever reaches a string of length at most 2, it halts.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Example 1: Starting from 1001, this procedure yields 111, then 11, and halts.&lt;/p&gt;&#xA;&lt;p&gt;Example 2: Starting from 01010, this procedure yields 100010, then 01011, then 110010, then again 01011, then again 110010; etc. There is a loop of length two.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Professor Erasmus has proved that independently of the starting string, the procedure will eventually halt or enter a loop of length at most 25.&lt;/p&gt;&#xA;&lt;p&gt;Has the professor once again made one of his well-known mathematical blunders, or does his claimed theorem indeed hold true?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2018-11-26T13:54:47.150" Title="The Erasmus 0-1 game" Tags="|mathematics|combinatorics|reachability|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22291" PostTypeId="2" ParentId="22275" CreationDate="2015-09-17T18:21:50.963" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am going to go ahead and say X is 125 as in the other cases the bottom value is equal to one half of the top left value cubed.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A clear flaw is the lack of incorporating the top right value but it is completely viable to assume it could be a red herring.&lt;/p&gt;&#xA;" OwnerUserId="13916" LastEditorUserId="13916" LastEditDate="2016-08-15T16:05:32.230" LastActivityDate="2016-08-15T16:05:32.230" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22292" PostTypeId="2" ParentId="22290" CreationDate="2015-09-17T18:50:10.643" Score="24" Body="&lt;p&gt;Professor Erasmus&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; has made a blunder.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the leftmost six numbers are $110\,010$, then the first step removes $110$ from the left and adds $11$ on the right, and the next step removes $010$ from the left and adds $0010$ on the right. The result is that after two steps, $110\,010$ has been removed from the left and added on the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, if the leftmost twelve numbers are $001\,000\,101\,111$, the net result is that after four steps these numbers have been removed from the left and added on the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Call the strings $S_1=110\,010$ and $S_2=001\,000\,101\,111$. Suppose we start with the string $$S_1S_1\ldots S_1 S_2,$$ with $n$ copies of $S_1$ followed by a copy of $S_2$. After two steps, we get $S_1\ldots S_1 S_2 S_1$. After two more steps, $S_1 \ldots S_1 S_2 S_1 S_1$, and so on. So after $2n$ steps, we have reached $S_2 S_1 \ldots S_1$. Four more steps takes us back to the original $S_1\ldots S_1 S_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original string is the only string in the cycle whose rightmost five digits are $01111$, so this is a cycle of length $2n+4$. We can choose $n$ arbitrarily, so as long as $n\geq 11$, we get a cycle of length greater than 25.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-09-17T18:53:59.197" LastActivityDate="2015-09-17T18:53:59.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22294" PostTypeId="1" AcceptedAnswerId="22297" CreationDate="2015-09-17T19:45:07.997" Score="9" ViewCount="784" Body="&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;As of today&#xA;&#xA;I have roughly 3,700&#xA;I've been given roughly 10,300&#xA;I'm almost always complete!&#xA;I contain roughly 11,000&#xA;I get roughly 6,300&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is my first puzzle posted, so I'm very new, so bear with me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;br&gt;&#xA;You guys are too good, I'll need to really make it tricky next time!&lt;/p&gt;&#xA;" OwnerUserId="16273" LastEditorUserId="16273" LastEditDate="2016-12-20T13:52:36.577" LastActivityDate="2016-12-20T13:52:36.577" Title="My first riddle on Puzzling Beta!" Tags="|riddle|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22296" PostTypeId="2" ParentId="22294" CreationDate="2015-09-17T19:53:04.757" Score="6" Body="&lt;p&gt;PARTIAL&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Puzzling StackExchange beta!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have roughly 3,700  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There are 3640 questions as of this writing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've been given roughly 10,300  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;10,318 answers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I contain roughly 11,000  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A quick estimate of users on the site gives me 10,440. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I get roughly 6,300  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;6,265 new visitors a day &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13970" LastActivityDate="2015-09-17T19:53:04.757" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22297" PostTypeId="2" ParentId="22294" CreationDate="2015-09-17T19:55:36.117" Score="10" Body="&lt;p&gt;This is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Referring to the Puzzling beta stats as of 9/17/2015&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have roughly 3,700&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It has 3,639 questions &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've been given roughly 10,300&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It has 10,318 answers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm almost always complete!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Most of our questions have been completed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I contain roughly 11,000&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are 10,925 users&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I get roughly 6,300&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The site gets 6,265 visitors/day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12206" LastActivityDate="2015-09-17T19:55:36.117" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22298" PostTypeId="1" AcceptedAnswerId="22300" CreationDate="2015-09-17T20:19:49.947" Score="10" ViewCount="771" Body="&lt;p&gt;I am hidden in the shortest route through this maze.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NIJa1.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NIJa1.png&quot; alt=&quot;Maze - find the shortest route&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Twice, if you know what you're looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More clues to follow (although I'm sure you're unlikely to need them)...&lt;/p&gt;&#xA;" OwnerUserId="15942" LastActivityDate="2016-07-23T19:54:02.773" Title="I'm hidden in the maze" Tags="|mazes|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22299" PostTypeId="1" AcceptedAnswerId="22308" CreationDate="2015-09-17T20:27:04.627" Score="7" ViewCount="171" Body="&lt;pre&gt;&lt;code&gt;Help me, for I have become longer -- taller -- thinner&#xA;I am more ungenerous, more unwilling to give&#xA;Sharply, I will hurt you, wound you&#xA;A musician of the purest variety&#xA;I burn, slightly and superficially&#xA;A functional, right ratio&#xA;Here now abbreviated&#xA;Included within&#xA;Me&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is this poem talking about?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-09-18T00:17:23.010" Title="A Poetic Ladder" Tags="|riddle|poetry|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22300" PostTypeId="2" ParentId="22298" CreationDate="2015-09-17T20:49:57.623" Score="9" Body="&lt;p&gt;Working right to left, we can label each vertex with the length of the shortest path to the End, eliminating edges that are not part of such a path. It is not too hard to find the label for a vertex once we have found the labels for that vertex's neighbors to the right. I ended up with this diagram:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/pf40k.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pf40k.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, beginning from Start, there is only one shortest path to End:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/4xYW8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4xYW8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA; This path uses edge lengths $2$, $7$, $1$, $8$, $2$, $8$, $1$, $8$, $2$. This is the beginning of the decimal expansion the mathematical constant $e\approx 2.71828182\ldots$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="22922" LastEditDate="2016-07-23T19:54:02.773" LastActivityDate="2016-07-23T19:54:02.773" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22301" PostTypeId="1" AcceptedAnswerId="22310" CreationDate="2015-09-17T20:57:01.767" Score="11" ViewCount="694" Body="&lt;p&gt;I wanted to talk to them about a majestic animal, but they laughed at me how could I find such an animal majestic. I told them what they are not allowed to do, but they insisted that they will do it anyway.&#xA;Despite all this, I still wanted to try to befriend them, so I promised I will bring presents to them. This is when they first panicked and then angrily chased me away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who was I talking to, and why did we misunderstand each other so much?&lt;/p&gt;&#xA;" OwnerUserId="2507" LastActivityDate="2015-09-18T05:25:57.547" Title="A failed communication attempt" Tags="|riddle|" AnswerCount="2" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22302" PostTypeId="2" ParentId="22299" CreationDate="2015-09-17T21:06:38.677" Score="0" Body="&lt;p&gt;I think the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A tensile failure test  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Help me, for I have become longer -- taller -- thinner  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A tensile specimen fails by lengthening and becoming thinner&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am more ungenerous, more unwilling to give&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It becomes harder (work hardens), which could be described as ungenerous (translating from the scientific term to the usual english usage), and gives with more difficulty (modulus increases). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sharply, I will hurt you, wound you  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Look, I don't really think this is the answer, so this gets more and more &lt;em&gt;scretched&lt;/em&gt; (excuse the pun).  Sharply could refer to a sharpie impact test, or the fact that a sharp edge makes a stress concentrator, which tend to be the break point, or the possible sharp edges left over after it snaps&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A musician of the purest variety&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mmm.  Well, I guess we could be performing the test on a musical string.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I burn, slightly and superficially&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Often there is localized heating at the break point&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A functional, right ratio&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The ratio between the length and the cross sectional area (okay, okay, I know it's weak - I didn't see your answer written here!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here now abbreviated&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Break = abbreviated&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Included within&lt;br&gt;&#xA;Me  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Inclusions can be the source of the weakness.  &lt;em&gt;Sigh&lt;/em&gt;  It's just so &lt;em&gt;weak&lt;/em&gt;.  I feel &lt;em&gt;broken&lt;/em&gt;.  Torn into two.  Rent asunder.  Unable to hold it all together.  Brittle.  Someone give this dog a &lt;em&gt;dog bone&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15942" LastActivityDate="2015-09-17T21:06:38.677" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22303" PostTypeId="1" AcceptedAnswerId="22305" CreationDate="2015-09-17T21:11:32.660" Score="8" ViewCount="2656" Body="&lt;p&gt;There's a boy. Let's call him Konrad. He's got his grandparents on father's side east of the city and on mother's side west of the city. Every Sunday, he walks to the train station and takes the first arriving carrier that happens to head towards any of the grandparents' places.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a while, the mother's parents complain that they hardly see him, while father's parents remark that he's showing up a bit too often. Konrad verifies their story and realizes that he's been visiting the latter almost every weekend, while hardly ever visiting the former.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is it (statistically) possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some premises to clarify the circumstances for a rigid deduction process.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It's &lt;strong&gt;not&lt;/strong&gt; a trick question - the answer is statistically deducible.&lt;/li&gt;&#xA;&lt;li&gt;There's only one train line and each end point is at one of the grandparents'.&lt;/li&gt;&#xA;&lt;li&gt;The  train in each direction arrives/leaves exactly every 60 minutes.&lt;/li&gt;&#xA;&lt;li&gt;There's never any extraordinary circumstances affecting the schedule.&lt;/li&gt;&#xA;&lt;li&gt;The track is single-lane so the trains never arrive at the same time.&lt;/li&gt;&#xA;&lt;li&gt;The boy arrives at the station at random occasions.&lt;/li&gt;&#xA;&lt;li&gt;The boy has no preferences towards any of the destinations.&lt;/li&gt;&#xA;&lt;li&gt;The grandparents are always home.&lt;/li&gt;&#xA;&lt;li&gt;There might be multiple trains in service and all travel at a constant speed.&lt;/li&gt;&#xA;&lt;li&gt;The boy can board at any non-endpoint station closest to him home.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4319" LastEditorUserId="4319" LastEditDate="2015-09-18T10:49:53.843" LastActivityDate="2015-09-18T13:26:58.027" Title="Why are grandparents on the father's side more popular?" Tags="|logical-deduction|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22304" PostTypeId="2" ParentId="20033" CreationDate="2015-09-17T21:18:44.770" Score="-4" Body="&lt;p&gt;Well, many people have responded with Sherlock Holmes, which isn't true. I haven't been very active on puzzling so I'll just publish the answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Lancelot&quot; rel=&quot;nofollow&quot;&gt;Sir Lancelot du Lac&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I used to have a name like a valediction,&lt;br&gt;&#xA;Others say it's purely fiction.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Goodnight = Good knight. He was one of the greatest knights in King Arthur's domain, and before having an affair with Guinevere was a good (just, fair) knight.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My surname keeps your valuables safe,&lt;br&gt;&#xA;My upbringing was completely as a waif.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://www.youtube.com/watch?v=bkXsyUxBOvU&quot; rel=&quot;nofollow&quot;&gt;Lock&lt;/a&gt; = &lt;a href=&quot;https://www.youtube.com/watch?v=cm4tK1uEFYg&quot; rel=&quot;nofollow&quot;&gt;Lac&lt;/a&gt;. He was raised by the Lady of the Lake as an orphan, which is why he has the last name Lac.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You've seen me in many a grand time,&lt;br&gt;&#xA;Up the ladder did I climb.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://www.youtube.com/watch?v=eJ9-d9jheMI&quot; rel=&quot;nofollow&quot;&gt;Epoch&lt;/a&gt; = &lt;a href=&quot;https://www.youtube.com/watch?v=Zlt7mSF1Bj0&quot; rel=&quot;nofollow&quot;&gt;Epic&lt;/a&gt;. Lancelot is one of the most recurring characters in history, being part of Le Morte D'Arthur and many other writings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be found picking on the smaller man;&lt;br&gt;&#xA;There are many that I am better than.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is awkward. I forgot the homophone to this one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My lover found a life of empty space,&lt;br&gt;&#xA;Vowing that I would never see her face.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://www.youtube.com/watch?v=AheA1Mt2PoA&quot; rel=&quot;nofollow&quot;&gt;None&lt;/a&gt; = &lt;a href=&quot;https://www.youtube.com/watch?v=SD3zGJpc_88&quot; rel=&quot;nofollow&quot;&gt;nun&lt;/a&gt;. His lover, Guinevere, became a nun, and declared that Lancelot would never again see her alive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I decided to live like a camelid,&lt;br&gt;&#xA;In hopes to overcome what I did.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://www.youtube.com/watch?v=NmZBKb_Oqik&quot; rel=&quot;nofollow&quot;&gt;llama&lt;/a&gt; = &lt;a href=&quot;https://www.youtube.com/watch?v=4_0zWhK7MPY&quot; rel=&quot;nofollow&quot;&gt;lama&lt;/a&gt;. He became a monk (à la Dalai Lama) and so lived out the rest of his life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Even in death I provided support,&lt;br&gt;&#xA;Though to base pleasures I did resort.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well, I honestly forgot what this one was.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13538" LastEditorUserId="13538" LastEditDate="2015-09-18T16:17:56.557" LastActivityDate="2015-09-18T16:17:56.557" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="22305" PostTypeId="2" ParentId="22303" CreationDate="2015-09-17T21:40:44.397" Score="28" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The train to his paternal grandparents is closely followed by the train to the maternal grandparents. In order to go to the maternal grandparents he has to arrive in the interval between the two trains. Let's say it is 5 minutes. He will reach his maternal grandparents only if he arrives at the station during those 5 minutes. If he arrives at the station at any of the other 55 minutes. he ends up going to his paternal grandparents.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10038" LastEditorUserId="4319" LastEditDate="2015-09-17T22:57:01.397" LastActivityDate="2015-09-17T22:57:01.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22306" PostTypeId="1" AcceptedAnswerId="22313" CreationDate="2015-09-17T22:37:33.980" Score="6" ViewCount="1602" Body="&lt;p&gt;In the diagram below you can see four independent patterns. Each pattern is a puzzle by itself, and by solving it you will be able to encounter one or more digits of the final, 6-digit numeric solution. Signs above tell how to get these digits once the mini-puzzle is solved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most puzzlers should already be familiarized with the challenge presented by these patterns. However, the solution is this case involves numbers as well as letters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Backstory:&lt;/strong&gt; &lt;em&gt;&quot;The 6-digit number you will find is a password that allows you to enter the fifth dimension and therefore become one with the universe. Also, it's the only way to save humanity.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AyXOJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AyXOJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Contrary to what my storytelling skills may indicate, no black sorcery is required in the solution to this puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Also, please notice that this post contains hidden tag(s), which would otherwise hurt the purpose of the puzzle.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It's natural for you to be &lt;em&gt;clueless&lt;/em&gt; during these puzzles. However, numbers may come in handy when words are lacking.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14219" LastEditorUserId="14219" LastEditDate="2015-09-18T04:04:09.347" LastActivityDate="2015-09-18T16:03:32.027" Title="Find the 6-digit number encoded in this &quot;mathematical&quot; diagram" Tags="|logical-deduction|password|" AnswerCount="2" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22307" PostTypeId="2" ParentId="22298" CreationDate="2015-09-17T22:52:07.620" Score="3" Body="&lt;p&gt;A common way to solve a maze like this is with dijkstra's algorithm.  If you print out the maze, you can solve it with pen and paper in a few minutes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You start by marking all the nodes as infinite (inf in the diagrams).  Then go to the first point and mark it as 0.  Then fill in the distances to the adjacent nodes and mark the first node as visited (red in the below):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RosKN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RosKN.png&quot; alt=&quot;Dijkstra-step 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you go to the &lt;strong&gt;unvisited&lt;/strong&gt; node (green in the diagram) with the &lt;strong&gt;smallest distance&lt;/strong&gt; on it, and repeat the same steps as before: you consider the distance to the adjacent unvisited nodes, and mark those with the total distance (if it's smaller than whatever distance is there already).  In this case, $6(=2+4)$ and $9(=2+7)$ are both less than infinity, so in they go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ItGfx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ItGfx.png&quot; alt=&quot;Dijkstra-step 2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeating, the smallest unvisited node is now the one labelled $5$.  $5+5=10&amp;gt;9$, so the $9$ stays.  But $5+8=13&amp;lt;\infty$, so we change that one to 13:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/z4PYH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/z4PYH.png&quot; alt=&quot;Dijkstra-step 3&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so on, until step 10:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7QoA1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7QoA1.png&quot; alt=&quot;Dijkstra-step 10&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so on, until step 24:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3Q6R2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3Q6R2.png&quot; alt=&quot;Dijkstra-step 24&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yEzM4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yEzM4.png&quot; alt=&quot;Dijkstra-step 25&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This yields the length of the smallest path.  However, you can figure out what the path was, by working backwards from the 39, and choosing edges which, when subtracted give the number in the node.  For example, $39-6=33\neq 34$ so that's not the path, $39-2=37$ so that is a path, and $39-5=34\neq 35$ so that's not the path.  So the last step of the shortest path is the $2$.&lt;/p&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2016-07-23T15:31:58.780" LastActivityDate="2016-07-23T15:31:58.780" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22308" PostTypeId="2" ParentId="22299" CreationDate="2015-09-17T23:43:58.207" Score="11" Body="&lt;p&gt;Each line&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; is given by the answer to the previous line, minus one letter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Help me, for I have become longer -- taller -- thinner&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This piece of spaghetti is &lt;strong&gt;stringier&lt;/strong&gt; than that one!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am more ungenerous, more unwilling to give&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A miserly, parsimonious person is &lt;strong&gt;stingier&lt;/strong&gt; than a charitable person.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sharply, I will hurt you, wound you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;strong&gt;stinger&lt;/strong&gt; of a wasp for example causes a small wound.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A musician of the purest variety&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A &lt;strong&gt;singer&lt;/strong&gt; might be considered pure as they only use their voice, no other &quot;impurities&quot; like instruments.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I burn, slightly and superficially&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I thought I had ruined the food, but I had only &lt;strong&gt;singe&lt;/strong&gt;d it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A functional, right ratio&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The mathematical ratio of a right angled triangle of the opposite and the hypotenuse is the &lt;strong&gt;sine&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here now abbreviated&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; But who writes &quot;sine&quot;? &lt;strong&gt;sin&lt;/strong&gt;(θ) = o/h;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Included within&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  The answer is actually with&lt;strong&gt;in&lt;/strong&gt; the line&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;I&lt;/strong&gt; figured this one out first and worked backwards.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4729" LastEditorUserId="4729" LastEditDate="2015-09-18T00:17:23.010" LastActivityDate="2015-09-18T00:17:23.010" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22309" PostTypeId="2" ParentId="22301" CreationDate="2015-09-18T00:42:29.120" Score="7" Body="&lt;p&gt;Haven't figured out the animals, but...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your friend is misunderstanding you so much because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;he's German. &quot;Gift&quot; is the German word for &quot;Poison&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2015-09-18T00:42:29.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22310" PostTypeId="2" ParentId="22301" CreationDate="2015-09-18T01:37:05.983" Score="16" Body="&lt;p&gt;You were talking to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;German people.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you are relatively inexperienced in german yourself, and made many common englishmen's mistakes (like abusing cognates) while learning it, leading to all these misunderstandings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example, firstly,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;you said &quot;Eagle&quot; or &quot;Igel&quot; when you meant &quot;Adler&quot;, and was misunderstood as saying that hedgehogs were majestic animals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;When they said it wasn't majestic, you said &quot;nicht müssen&quot; (translating &quot;not&quot; and &quot;must&quot; individually) where you should have said &quot;nicht dürfen&quot;. Instead of &quot;must not&quot;, they understood &quot;don't have to&quot;, so they continued.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You, aware of these misunderstandings, didn't blame them and instead tried to befriend them by promising a gift. However,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the word &quot;gift&quot; in German means &quot;poison&quot;, so you went from being just a weird guy who likes hedgehogs to a psychopath who threatens people with poison (presumably because they don't share his affinity for hedgehogs).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And thats's when they panicked and started chasing you away.&lt;/p&gt;&#xA;" OwnerUserId="14219" LastEditorUserId="14219" LastEditDate="2015-09-18T05:25:57.547" LastActivityDate="2015-09-18T05:25:57.547" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22311" PostTypeId="2" ParentId="22306" CreationDate="2015-09-18T01:51:24.043" Score="2" Body="&lt;p&gt;I'm going to give this a shot:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each bend in the string of dots represents a new number to be summed ($\Sigma$) or product-ed ($\Pi$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We go in order:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;3 + 4 = 7&lt;/li&gt;&#xA;&lt;li&gt;4 &amp;times; 3 &amp;times; 3 &amp;times; 4 = 144&lt;/li&gt;&#xA;&lt;li&gt;3 + 3 = 6&lt;/li&gt;&#xA;&lt;li&gt;4 &amp;times; 3 &amp;times; 3 = 36&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;But 7144636 is &lt;em&gt;seven&lt;/em&gt; digits, so this is obviously not the right answer. Am I on the right track, though?&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-09-18T01:51:24.043" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22313" PostTypeId="2" ParentId="22306" CreationDate="2015-09-18T02:22:09.400" Score="10" Body="&lt;p&gt;Each puzzle is a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; crossword&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;to be solved by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; filling in written-out English numbers of the correct length&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solving its gives&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;TEN      TEN&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;  I      W I     O   O O&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;  N   ZERO N   TEN   NINE&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;  E        E     E   E E&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We interpret each of these by computing the corresponding:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; sum of 19, product of 0, sum of 11, product of 9&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and concatenating gives the six-digit answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 190119&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4421" LastEditDate="2015-09-18T16:03:32.027" LastActivityDate="2015-09-18T16:03:32.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22315" PostTypeId="2" ParentId="22303" CreationDate="2015-09-18T04:17:54.613" Score="3" Body="&lt;p&gt;Ok, with some of the additional clarifications, I've revised my answer, which I feel provides a tad more of an explanation about the &lt;em&gt;why&lt;/em&gt; of the other perfectly acceptable response. This revision would apply even for a multi-track scenario, and applies no matter where Konrad's station is located relative to the endpoints - as long as there is a fixed schedule (with no simultaneous arrivals), as stated above in the criteria above.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let X = the time between the arrivals of any west-bound train and the next east-bound train. &lt;br&gt;Let Y = the time between the arrivals of any east-bound train and the next west-bound train. &lt;br&gt;&lt;br&gt;Since there is a fixed 60-minute interval for each direction, the probability of taking an east-bound train (to the paternal grandparents) should be directly related to the ratio X / 60 (with Y / 60 representing the probability of heading west). The higher X is, the lower Y is, and thus...whichever interval has the longer wait, will determine the most probable direction he travels. &lt;br&gt;&lt;br&gt;Thus, for Konrad to visit the paternal grandparents more often, the schedule at Konrad's station has a larger interval between the arrival of the west-bound train and the next east-bound train, than between the east-bound and next west-bound train.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Original below (based on a very simple model of a single train which traverses the line back and forth - still works within those constraints, as it provides a practical explanation regarding the scheduling...but isn't as flexible):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Konrad lives closer to the east side of town. This is dependent upon the following: &lt;br&gt;- There is a single train (single track) which travels to each end-point, and then travels back on the same track towards the other end-point. &lt;br&gt;- The full round-trip takes 60 minutes (each direction arrives/leaves exactly every 60 minutes). &lt;br&gt;&lt;br&gt;The probability of Konrad catching the train headed in any particular direction, is directly related to his proximity to either end-point. &lt;br&gt;&lt;br&gt;Think of the total round trip as being split into two parts: the path from Konrad's stop to the east endpoint and back &lt;em&gt;Pe&lt;/em&gt;, and the path from Konrad's stop to the west endpoint and back &lt;em&gt;Pw&lt;/em&gt;. The closer he lives to the east endpoint of the train line, the shorter &lt;em&gt;Pe&lt;/em&gt; is, and the longer &lt;em&gt;Pw&lt;/em&gt; is. Therefore, when arriving at a random time, it's much more probable that he arrives while the train is traveling along the &lt;em&gt;Pw&lt;/em&gt; path, in which case the train will be headed east when it finally arrives at his stop.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14405" LastEditorUserId="14405" LastEditDate="2015-09-18T11:39:35.603" LastActivityDate="2015-09-18T11:39:35.603" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22316" PostTypeId="2" ParentId="22275" CreationDate="2015-09-18T05:35:37.657" Score="2" Body="&lt;p&gt;To me the pattern appears to be as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Multiply the first two numbers and divide by the third number (to which the two lines lead). In the first case, this operation gives 3 ($4 \times 6 \div 8$). In the second line, it gives 4 ($6 \times 18 \div 27$). So moving forward, I'd expect the next line to give a result of 5 from this operation. So $10 \times 16 \div 32 = 5$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I believe $x$ should be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 32&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-09-18T05:35:37.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22317" PostTypeId="2" ParentId="22176" CreationDate="2015-09-18T06:18:09.497" Score="9" Body="&lt;p&gt;&lt;strong&gt;WOW THIS WAS HARD!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer I got is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Under the last pine tree in Roche's garden&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How I got it...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets follow the clues. The first clue of trinity means split it into sections, each section containing 3 lines. Now if we take the last 3 characters from the first line of each set, and using the 4th rule, ignore any duplicate characters, we get&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; under, last, tree, Roche's&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I figured this out from the second hint, to be a good match.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, we take the third line from each set. It bothered me that they were all an odd number of words, so I looked at the middle word of each line and got&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the, pine, in, garden&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add the two together and you get &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Under the last pine tree in Roche's garden&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am not sure about the methodology used other than what I have described because I just stared at it until I saw the message. &lt;/p&gt;&#xA;" OwnerUserId="16396" LastEditorUserId="12700" LastEditDate="2015-09-18T08:00:21.077" LastActivityDate="2015-09-18T08:00:21.077" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22318" PostTypeId="1" AcceptedAnswerId="22328" CreationDate="2015-09-18T06:44:34.110" Score="2" ViewCount="577" Body="&lt;p&gt;What movies are described by the haiku below?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We are engaged&lt;br&gt;&#xA;  Year after year&lt;br&gt;&#xA;  Year after year&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;No memory of you&lt;br&gt;&#xA;  Every day is the same&lt;br&gt;&#xA;  Every date is different&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Stuck at one place&lt;br&gt;&#xA;  for 5.29 days&lt;br&gt;&#xA;  Delusions aplenty&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Something bad happened&lt;br&gt;&#xA;  Three of us stuck inside pod&lt;br&gt;&#xA;  Can we still go home?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I am a lawyer with&lt;br&gt;&#xA;  A demanding boss&lt;br&gt;&#xA;  I quit&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A secret hobby&lt;br&gt;&#xA;  Of a handsome guy&lt;br&gt;&#xA;  I like, or do I?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;7&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A taxi driver&lt;br&gt;&#xA;  A girl comes through the roof&lt;br&gt;&#xA;  Let's find us some stones&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;8&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A dangerous outlaw&lt;br&gt;&#xA;  A poor guy&lt;br&gt;&#xA;  An afternoon train&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;9&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Boy writes greeting cards&lt;br&gt;&#xA;  Boy meets girl&lt;br&gt;&#xA;  Boy does not get girl&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;10&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Catches a psychopath&lt;br&gt;&#xA;  Who catches a plane&lt;br&gt;&#xA;  That catches the passenger who saves the plane&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This puzzle has a little theme running through it.&lt;/p&gt;&#xA;" OwnerUserId="12332" LastEditorUserId="4050" LastEditDate="2015-09-18T17:43:43.643" LastActivityDate="2015-09-18T17:43:43.643" Title="Haiku about Movies: We are engaged / Year after year / Year after year" Tags="|movies|haiku|" AnswerCount="6" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22319" PostTypeId="2" ParentId="22318" CreationDate="2015-09-18T07:04:46.297" Score="1" Body="&lt;p&gt;2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 50 first dates&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; War of the Worlds (Might be wrong)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;7:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The fifth element&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;10:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Non-Stop (Might be wrong)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14461" LastEditorUserId="14461" LastEditDate="2015-09-18T09:25:59.797" LastActivityDate="2015-09-18T09:25:59.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22320" PostTypeId="2" ParentId="22318" CreationDate="2015-09-18T07:11:06.323" Score="0" Body="&lt;p&gt;3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 127 Hours&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;9:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 500 Days of Summer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastActivityDate="2015-09-18T07:11:06.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22321" PostTypeId="1" AcceptedAnswerId="22325" CreationDate="2015-09-18T07:11:34.697" Score="2" ViewCount="2062" Body="&lt;p&gt;OOOOOOK! So. Here we go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Scenario&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your friend (a restaurant owner) calls you one afternoon in urgency. Something terrible has happened and he needs your support at once!! You borrow $50 from the neighbor's kid (as you are broke -as always-) and get a cab to his place. You see him sitting head-in-knees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;What's up brah?&quot; You ask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I'm ruined!&quot; He replies. &quot;A group of high dignitaries are coming to my restaurant and the special lamb-fried-rice I prepared for them is spiked with poison!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You wonder for a moment &lt;em&gt;why&lt;/em&gt; a group of high dignitaries would come to a useless place as this restaurant, but * ahem ahem *. So you go on to ask &quot;Is &lt;strong&gt;&lt;em&gt;all&lt;/em&gt;&lt;/strong&gt; the dish spiked?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;No. Only 1 plate. But demons take me, I don't know which plate it is!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Calm down man and tell me things from the start!&quot; You tell him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;OK here goes. A group of 16 dignitaries was coming to my place and they are all Asian. So I prepared our high specialty dish, lamb-fried-rice. Now it happens that one of my waiters was demanding me a salary increment for previous 4 months and I had been refusing him (oh you cheap skunk!). Now as a cook was passing around the kitchen area, he saw the waiter take out a small bottle from his pocket and add it to a plate (there were 16 plates on the row). He immediately went and confronted him about it, on which, that idiot jumped from the window of 3rd floor and ... died on the spot. In the meantime, other waiters had picked the plates up and left for placing them on the table. So now the idiot cook doesn't even know which plate it was that was poisoned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;We have 4 poison testing machines with us which can detect &lt;strong&gt;&lt;em&gt;the minutest&lt;/em&gt;&lt;/strong&gt; amount of poison in given sample. But all machines can work on only &lt;strong&gt;one&lt;/strong&gt; sample at a time and each machines take 55 minutes for producing the result. The dignitaries are arriving in 1 hours and 15 minutes. How the demons am I supposed to find the spiked plate in this time? Aaaaaaargh! I'm ruined. Ruined! &lt;em&gt;Ruined!&lt;/em&gt; RUINED!!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In Simple Words&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 16 plates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only one plate is poisoned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 4 poison testing machines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each machine can work on only one sample at a time. The result is delivered only as positive (sample contains poison) or negative (sample is not poisoned).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each machine takes 55 minutes to process the sample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dignitaries are arriving in 1 hour and 15 minutes (75 minutes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You must find the poisoned plate within this time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: This is a puzzle which computer science students will be able to solve easily. Others will most probably not even get on the right thinking track. Not saying that it's impossible or something, but other people will probably make wrong assessments and start with wrong model in the very first place. &lt;em&gt;THIS IS NOT A MATHEMATICAL PUZZLE&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. If you find the poisoned plate, don't expect your friend to pay you the 50$ you borrowed to reach him. If he was such a generous man, he would have increased the salary of the poor waiter!&lt;/p&gt;&#xA;" OwnerUserId="16342" LastEditorUserId="15544" LastEditDate="2015-09-24T06:03:56.970" LastActivityDate="2015-09-24T06:03:56.970" Title="Where Is The Poisonous Plate?" Tags="|logical-deduction|pattern|story|" AnswerCount="4" CommentCount="13" ClosedDate="2015-09-19T14:26:47.250" ContentLicense="CC BY-SA 3.0" />
  <row Id="22322" PostTypeId="1" AcceptedAnswerId="22326" CreationDate="2015-09-18T07:11:35.147" Score="11" ViewCount="390" Body="&lt;p&gt;In a party, there were many puzzlers. Some of these were:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;A japanese hacker&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;An infinite loop&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A novice programmer&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Two magic words&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;An exclusive gate&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A pokémon&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A second derivative&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A navigator&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A man in search of a pseudonym&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A king named Rames&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;An artificial ghoul from japan&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The only true fanatic&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A gamer expressing sudden pain&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A pixelated Joe&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A black bear&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A stylish logician&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The FSM&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The most prolific writer&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A Cyclic postdoc mathematician of order 4&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A person from the land of Rampart&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Mr K, happy to update his software&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;One that is too slim to be seen from afar&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A sunflower&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A winged tiger who rules&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A cat with an evil smile&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;An unusual atmospheric phenomenon&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A Java developer who considers things&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A man from the Kiasyd bloodline&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;An aircraft concerned about privacy&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;An iOS dev with &amp;quot;pretty cool&amp;quot; apps&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Beyond these famous ones mentioned, there are many, many more.&lt;/p&gt;&#xA;&lt;p&gt;In fact, you may be in the party too. Can you figure out who was each person in this list?&lt;/p&gt;&#xA;" OwnerUserId="14219" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2018-06-24T00:35:20.197" Title="A party with strange people: Who are they?" Tags="|word|wordplay|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22323" PostTypeId="2" ParentId="22318" CreationDate="2015-09-18T07:14:02.187" Score="2" Body="&lt;p&gt;getting tired and have some work to do so here are the ones ive got so far&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 50 first dates&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 127 hours&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;7&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Fifth Element&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;8&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3:10 to Yuma&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;9&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 500 days of summer perhaps? (never saw the film)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16396" LastActivityDate="2015-09-18T07:14:02.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22324" PostTypeId="2" ParentId="22321" CreationDate="2015-09-18T07:27:17.790" Score="2" Body="&lt;p&gt;label the plates A to P, and the testers 1 to 4. Place a portion of the food from each plate into the coresponding tester&lt;/p&gt;&#xA;&#xA;&lt;p&gt;tester 1:&#xA;Plates BFGHLMNP&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tester 2:&#xA;Plates CFIJLMOP&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tester 3:&#xA;Plates DGIKLNOP&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tester 4:&#xA;Plates EHJKMNOP&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then run all 4 machines at once after they are loaded up with small amounts of each sample. Once the 55 minutes are over, you can then read which testers (if any) read as containing poison and match that up to the corresponding plate. If no testers read any poison then it is plate A. follow the reverse of the table to figure out which plate is poisoned.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;A no testers detected any poison&#xA;B 1&#xA;C 2&#xA;D 3&#xA;E 4&#xA;F 12&#xA;G 13&#xA;H 14&#xA;I 23&#xA;J 24&#xA;K 34&#xA;L 123&#xA;M 124&#xA;N 134&#xA;O 234&#xA;P 1234&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;so if testers 2 and 3 come back positive for example, it is plate I, Testers 23 and 4 being positive suggests plate O has poison&lt;/p&gt;&#xA;" OwnerUserId="16396" LastEditorUserId="16396" LastEditDate="2015-09-18T08:04:03.913" LastActivityDate="2015-09-18T08:04:03.913" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="22325" PostTypeId="2" ParentId="22321" CreationDate="2015-09-18T07:29:26.700" Score="14" Body="&lt;p&gt;Number each plate from $0$ to $15$. Convert each plate number into binary. Number the machines 0-3. The $i$th machine takes in all samples with digit $i=0$. Each positive sample means $0$ and the rest mean $1$. Hence you know the poisoned sample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Simplified:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Tester 0 has samples 0-7 mixed together&lt;/li&gt;&#xA;&lt;li&gt;Tester 1 has samples 0-3, 8-11 mixed together&lt;/li&gt;&#xA;&lt;li&gt;Tester 2 has samples 0-1, 4-5, 8-9, 12-13 mixed together&lt;/li&gt;&#xA;&lt;li&gt;Tester 3 has samples 0, 2, 4, 6, 8, 10, 12, 14 mixed together&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Start with a number 15. If tester 0 is positive, subtract 8. If tester 1 is positive, subtract another 4. If tester 2 is positive, subtract another 2. If tester 3 is positive, subtract another 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, testers 0 and 2 test positive. Poisoned sample $=15-8-2=5$&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4269" LastEditDate="2015-09-18T15:09:22.353" LastActivityDate="2015-09-18T15:09:22.353" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22326" PostTypeId="2" ParentId="22322" CreationDate="2015-09-18T07:31:55.470" Score="9" Body="&lt;p&gt;The names you list, are people that appear on the top users page for this month.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A japanese hacker&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Hackiisan&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An infinite loop&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;GOTO 0&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A novice programmer&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;CodeNewbie&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Two magic words&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;squeamish ossifrage&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An exclusive gate&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;xnor&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A pokémon&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/users/11475/bailey-m%20%22bailey-m&quot;&gt;Bailey M&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A second derivative&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;f'' (thanks to dmg)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A navigator&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Nautilus&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A man in search of a pseudonym&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;NeedAName&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A king named Rames&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Kingrames&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An artificial ghoul&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/users/15544/anachor&quot;&gt;Anachor&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The only true fanatic&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/users/10399/ian-macdonald&quot;&gt;IanMacdonald&lt;/a&gt; (he's 'on top' - but just for the month.. not all time)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A gamer expressing sudden pain&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;gamow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A pixelated Joe&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/users/88/joe-z&quot;&gt;Joe Z&lt;/a&gt; (pixels in avatar icon)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A black bear&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/users/4797/thomas-pornin&quot;&gt;Thomas Pornin&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A stylish logician&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;LogicianWithAHat&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The FSM&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/users/11100/ben-frankel&quot;&gt;Ben Frankel&lt;/a&gt; (has the 'Flying Spaghetti Monster' as display picture)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The most prolific writer&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/users/4729/james-webster&quot;&gt;James Webster&lt;/a&gt; (very elaborate profile...)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A Cyclic postdoc mathematician of order 4&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/users/2627/julian-rosen&quot;&gt;Julian Rosen&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A person from the land of Rampart&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/users/13237/roland&quot;&gt;Roland&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mr K, happy to update his software&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/users/14135/gordon-k&quot;&gt;Gordon K&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One that is too slim to be seen from afar&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Sleafar&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A sunflower&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/users/12206/alexmc&quot;&gt;alexmc&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The powerful master of us all&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Emrakul (as moderator?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A cat with an evil smile&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/users/10781/psmears&quot;&gt;psmears&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An unusual atmospheric phenomenon&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;GentlePurpleRain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Java developer who considers things&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/users/4385/dennisdeems&quot;&gt;dennisdeems&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A man from the Kiasyd bloodline&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Marconius&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An aircraft concerned about privacy&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/users/15786/jet&quot;&gt;Jet&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An iOS dev with &quot;pretty cool&quot; apps&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/users/10851/luke&quot;&gt;Luke&lt;/a&gt; (thanks to dmg)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2156" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-09-18T08:41:37.267" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="22327" PostTypeId="2" ParentId="22318" CreationDate="2015-09-18T08:17:23.957" Score="2" Body="&lt;p&gt;Number 2 may also be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Groundhog Day&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 6:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Das Parfum. Die Geschichte eines Mörders&lt;br&gt;&#xA; or Fifty Shades of Grey&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16360" LastEditorUserId="4421" LastEditDate="2015-09-18T16:15:47.907" LastActivityDate="2015-09-18T16:15:47.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22328" PostTypeId="2" ParentId="22318" CreationDate="2015-09-18T08:51:44.583" Score="4" Body="&lt;p&gt;I &lt;em&gt;think&lt;/em&gt; the list is complete now, though uncertain on the last answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;THEME:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All move titles have numbers in them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.imdb.com/title/tt1195478/&quot; rel=&quot;nofollow&quot;&gt;The Five-Year Engagement&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.imdb.com/title/tt0343660/&quot; rel=&quot;nofollow&quot;&gt;50 First Dates&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.imdb.com/title/tt1542344/&quot; rel=&quot;nofollow&quot;&gt;127 Hours&lt;/a&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.imdb.com/title/tt0112384/&quot; rel=&quot;nofollow&quot;&gt;Apollo 13&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.imdb.com/title/tt0313737/&quot; rel=&quot;nofollow&quot;&gt;Two Weeks Notice&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.imdb.com/title/tt2322441/&quot; rel=&quot;nofollow&quot;&gt;50 Shades of Grey&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;7&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.imdb.com/title/tt0119116/&quot; rel=&quot;nofollow&quot;&gt;The Fifth Element&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;8&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.imdb.com/title/tt0381849/&quot; rel=&quot;nofollow&quot;&gt;3:10 to Yuma&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;9&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.imdb.com/title/tt1022603/&quot; rel=&quot;nofollow&quot;&gt;(500) Days of Summer&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;10&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.imdb.com/title/tt0118571/&quot; rel=&quot;nofollow&quot;&gt;Air Force One&lt;/a&gt; ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2156" LastEditorUserId="4421" LastEditDate="2015-09-18T16:22:10.930" LastActivityDate="2015-09-18T16:22:10.930" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22329" PostTypeId="2" ParentId="22318" CreationDate="2015-09-18T09:13:17.223" Score="2" Body="&lt;p&gt;The two that are yet to be solved:&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Apollo 13  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Two weeks' notice (though I'll admit I got that through google)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11717" LastActivityDate="2015-09-18T09:13:17.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22330" PostTypeId="2" ParentId="19860" CreationDate="2015-09-18T09:34:15.660" Score="3" Body="&lt;ol&gt;&#xA;&lt;li&gt;GREEN / BLUE &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; GREEN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;NINE / TWELVE&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; NINE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;TETRIS / PRISMS&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TETRIS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;STICK / SWITCH&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; STICK?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;INVENTIVE / DIFFICULT&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; INVENTIVE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;SHAPE / SHAPES&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SHAPE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;DISTRACTION / DIVERSION&#xA;n/a&lt;/li&gt;&#xA;&lt;li&gt;EASIER / SIMPLER&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SIMPLER&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;DIFFERENCE / DISTINCTION&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DIFFERENCE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;CONFUSED / ENTRANCED&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ENTRANCED&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;HARD / EASY&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HARD&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;INTERRUPTIONS / NOISE&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; NOISE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;STUBBORNNESS / INFLEXIBILITY&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; STUBBORNNESS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;THIEVES / LAWYERS&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LAWYERS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;INDEPENDENT / UNRELATED&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; INDEPENDENT&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;CONUNDRUM / CHALLENGE&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CHALLENGE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So this is the solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SpannedWords are words you can &quot;walk over&quot;. Sometimes gaps are allowed, but only if there is a &quot;way&quot; from letter to letter.&#xA; This is way FEET is a special SpannedWort; there are no (as in zero) gaps between the &quot;top strokes&quot;.&#xA; That's why SHAPES is not a SpannedWord because there is no connection between the E and S at the end.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16398" LastEditorUserId="16398" LastEditDate="2015-09-20T12:01:00.760" LastActivityDate="2015-09-20T12:01:00.760" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="22331" PostTypeId="2" ParentId="22321" CreationDate="2015-09-18T09:57:22.043" Score="14" Body="&lt;p&gt;Basically, you have 2^4 plates to test, and 4 machines. This obviously hints of binary. You create a series of combinations to each plate occurs in a unique combination of testers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based of the binary system, you make combinations so all different options are created. '0' means a plate is not included in the tester sample, '1' means it is included.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a way is shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fnVfr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fnVfr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The test shows that there's 16 unique tester combinations, matching the 16 different outcomes for the 4 testers. Therefor, matching the results of the 4 testers in 55 minutes, means you have 20 minutes to prepare the samples and make the correct conclusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather then calculating (which you could fail in stressy situations) you can just look up which 'plate' matches the positive readings from the testers. If you insist on calculating, you could convert the readings to decimal, but you'd have to increase the result by 1 (plates start at 1, binary count at 0)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the example above, you could see that if testers 1,2 and 4 come back negative, and tester 3 is positive, you know for certain plate 3 is the poisoned plate.&lt;/p&gt;&#xA;" OwnerUserId="2156" LastEditorUserId="2156" LastEditDate="2015-09-19T12:37:38.303" LastActivityDate="2015-09-19T12:37:38.303" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22332" PostTypeId="1" AcceptedAnswerId="22333" CreationDate="2015-09-18T10:56:37.377" Score="-4" ViewCount="597" Body="&lt;p&gt;I came across this puzzle and can't figure it out.&lt;br&gt;&#xA;4 is to 1, as 12 is to 2, as 86 is to ___?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wish I could clarify the question but this was the way it was presented without further clues or constraints.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The given answer is 3, but no explanation was provided.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16401" LastEditorUserId="16401" LastEditDate="2015-09-18T16:44:30.470" LastActivityDate="2015-09-19T02:09:54.753" Title="Number puzzle: 4 is to 1, as 12 is to 2, as 86 is to ___?" Tags="|logical-deduction|number-sequence|" AnswerCount="3" CommentCount="3" ClosedDate="2015-09-18T13:10:52.263" ContentLicense="CC BY-SA 3.0" />
  <row Id="22333" PostTypeId="2" ParentId="22332" CreationDate="2015-09-18T11:05:45.540" Score="5" Body="&lt;p&gt;One sequence that would work is the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; f(x) = x^(x+1) + x + 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which gives:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; f(1) = 1^2 + 1 + 2 = 4&lt;br&gt;&#xA; f(2) = 2^3 + 2 + 2 = 12&lt;br&gt;&#xA; f(3) = 3^4 + 3 + 2 = 86&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, having to figure that out from just the first two seems.. at least farfetched.&lt;/p&gt;&#xA;" OwnerUserId="2156" LastActivityDate="2015-09-18T11:05:45.540" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22334" PostTypeId="1" AcceptedAnswerId="22336" CreationDate="2015-09-18T11:17:06.917" Score="1" ViewCount="1064" Body="&lt;p&gt;Four friends were travelling together. Their names were Dim, Burk, Lait and Kron. As they walked on, they saw a magician selling a flying carpet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;This carpet costs 100 gold coins. It will take you anywhere you want to travel to. For a bonus, it won't ever require any gas or rocket fuel!&quot; said the mage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dim had 90 gold coins with him which he paid to the magician and promised the remaining 10 coins for a later date. The magician sold him the flying carpet on credit. Now they all sat on the flying carpet and ordered it to take them to their destination. As they were flying over, they saw an old hermit sitting on a rock with a magic lamp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Instead of light, this lamp emanates darkness. Only the holder of this lantern can see what is around him, when it is lit, the others will only see pitch darkness! The price is finding my ten goats which went grazing this morning but haven't returned yet. But don't think it is easy, for they run faster than an Arabian horse and are sneakier than a mouse.&quot; said the hermit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dim said to Lait: &quot;You are the fastest runner amongst us. Only you can go and find it. This lamp might help us if we have it.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Lait ran in the direction of the pasture and after two days of tireless struggle, returned with the goats. The hermit handled him the lamp of darkness. They sat on their flying carpet and resumed their journey.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Soon they saw a large grizzly bear being chased by a crowd of people. When the crowd reached the bear, it turned around, upon which the people fled. But as the bear turned back and ran again towards the woods, the crowed followed it again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Help us!&quot; a man from the crowd shouted. &quot;This bear has killed our tribe chief who kept a secret magic bowl with him. If you kill this bear, we shall give you the bowl.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Burk was the strongest of them all. Dim asked if he could kill the bear. Burk grunted and stepped down from the carpet. After a long and fierce battle, Burk managed to kill the bear$\dots$ empty handed. The people awarded Burk the magic bowl which would show its master anything the master wanted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As they flew further on the carpet, they saw a young man sitting solemnly on a tree trunk. &quot;What's wrong, fellow?&quot; Dim asked as they flew over him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;My lover is dead and I don't know what to do with my magical perfume anymore. It will soften any woman's heart and make her fall in love with you, but oh, she's dead now and I love no-one else. I would give it to you if you sing me a song to console my broken heart.&quot; replied the young man.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kron had the softest voice and was masterful with his violin. He played a tune so soft and soothing that the young man forgot his agony and gave him the perfume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that they were flying on the carpet again, Dim suggested that they see the most gorgeous woman in the world, in the magic bowl. All agreed. Burk ordered the bowl to show them the most gorgeous woman. It showed a lofty castle with 1000 rooms. It was the castle of King Ramana, whose own daughter (the most gorgeous woman in the world) was being held prisoner there by the rebel minister who had killed the king and was forcing the princess to marry him. Her name was Princess Kamni.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;We must save her!&quot; Announced Burk. &quot;But I don't know how to do it.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dim ordered the carpet to take them to the castle of Princess Kamni. Upon reaching the castle grounds, Dim asked Burk and Lait to go together, holding the lantern of darkness and the magic bowl. The lantern should keep them from being spotted and the bowl would show them in which room the princess was kept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They did so and managed to rescue her. Upon reaching the sunlight, a few castle guards tried to confront them, but were quickly disposed off by Burk. The princess was asleep as they took her. Kron sprayed some of his perfume on the princess. When she woke up, she saw the four friends and said thus: &quot;I know you are my saviors and I would marry one of you. So, who has done the most for my rescue?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upon this, they all started quarreling. Dim said his carpet had the major role as they couldn't have reached the castle in these tall mountains without it. Burk refused and said if it wasn't for his magic bowl, they wouldn't even have known that such a princess exists. Lait claimed it would have been impossible to safely rescue her without his lantern of darkness, and Kron finally announced if it wasn't for his perfume, she would simply shoo them away and walk her own way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The princess listened to their full story and decided who her husband would be. She then awarded three things to the other three friends.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A cat whose master would have 9 lives.&lt;/li&gt;&#xA;&lt;li&gt;A herb that would heal any injury.&lt;/li&gt;&#xA;&lt;li&gt;A potion whoever drinks it, would stay young till the day of his death.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Question (worth the princess): Who got to marry the princess and how/why were the other three gifts awarded?&lt;/p&gt;&#xA;" OwnerUserId="16342" LastEditorUserId="8874" LastEditDate="2015-09-18T14:17:43.420" LastActivityDate="2015-09-18T14:17:43.420" Title="A Fairytale Of The Past" Tags="|lateral-thinking|story|" AnswerCount="2" CommentCount="3" ClosedDate="2015-09-18T15:39:24.483" ContentLicense="CC BY-SA 3.0" />
  <row Id="22335" PostTypeId="2" ParentId="22332" CreationDate="2015-09-18T12:21:43.427" Score="2" Body="&lt;p&gt;I have an alternate way of looking at it. The derived numbers 1, 2, 3 are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the count of the unique digits in the prime factorization of the original numbers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is my explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4 is factorized as $2\cdot2$, and the number of unique digits is 1.&lt;br&gt;12 is factorized as $2 \cdot 2 \cdot 3$, and the number of unique digits is 2.&lt;br&gt;86 is factorized as $2 \cdot 43$, and the number of unique digits is 3.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-09-18T12:21:43.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22336" PostTypeId="2" ParentId="22334" CreationDate="2015-09-18T12:44:06.470" Score="1" Body="&lt;p&gt;The way you've told the story, Dim is clearly the leader and can take at least partial credit for each item:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Dim had 90 gold coins with him which he paid to the magician and promised the rest 10 for a later date.&lt;/p&gt;&#xA;&lt;p&gt;Dim said to Lait: &amp;quot;You are the fastest runner amongst us...&lt;/p&gt;&#xA;&lt;p&gt;Dim asked if he could kill the bear.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;What's wrong, fellow?&amp;quot; Dim asked as they flew over.&lt;/p&gt;&#xA;&lt;p&gt;Dim suggested that they see the most gorgeous woman in the world,&lt;/p&gt;&#xA;&lt;p&gt;Dim asked Burk and Lait to go together, holding the lantern of darkness and the magic bowl.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;He's co-ordinating things, finding out what they need, and assigning the right people to tasks. So he &amp;quot;wins&amp;quot; the princess.&lt;/p&gt;&#xA;&lt;p&gt;The other three items:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the cat to Kron, since his talents won't otherwise keep him alive&lt;/li&gt;&#xA;&lt;li&gt;the herb to Burk, who lives by his strength and is likely to get hurt often&lt;/li&gt;&#xA;&lt;li&gt;the potion to Lait so he can retain his talent of running fast&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1509" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-09-18T13:05:20.437" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22338" PostTypeId="2" ParentId="22332" CreationDate="2015-09-18T13:01:33.553" Score="2" Body="&lt;p&gt;There can be many different answers. Without additional clues or constraints we can't be sure what the questioner intended.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polynomials of many different degrees could fit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Factorising leads to some possibilities. We could suggest 86 -&gt; 2, as 86 has 2 prime factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could ask how many pen strokes are needed. 4 needs 1, 12 needs 2, 86 needs 2. Hence 86 -&gt; 2 for a different reason.&lt;/p&gt;&#xA;" OwnerUserId="16404" LastEditorUserId="16404" LastEditDate="2015-09-18T17:23:47.280" LastActivityDate="2015-09-18T17:23:47.280" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22339" PostTypeId="2" ParentId="22334" CreationDate="2015-09-18T13:17:33.800" Score="12" Body="&lt;p&gt;Taking advantage of the 'lateral thinking' tag:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The herb to the young man on the tree-trunk - perhaps it can heal his broken heart  &lt;/li&gt;&#xA;&lt;li&gt;The potion to the old hermit, so that he may run after his own goats  &lt;/li&gt;&#xA;&lt;li&gt;The cat to the tribe chief, giving him another 8 lives  &lt;/li&gt;&#xA;&lt;li&gt;Finally, she marries the magician. Without the magic carpet, none of this would be possible, and, until the 10 coins are paid, said carpet arguably still belongs to him  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(Although it was Kron who sprayed her with the perfume, which presumably made her fall in love with him - hopefully she is somehow resistant to the effect of magic rohypnol)&lt;/p&gt;&#xA;" OwnerUserId="11717" LastEditorUserId="11717" LastEditDate="2015-09-18T13:48:51.677" LastActivityDate="2015-09-18T13:48:51.677" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22340" PostTypeId="2" ParentId="22303" CreationDate="2015-09-18T13:26:58.027" Score="0" Body="&lt;p&gt;This is because the boy lives&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; in the west of the city.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And there is only one track&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; so there is only one train going back and forth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The track looks like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;W -- NW -- N -- NE -- E&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;GM   B                GF&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the schedule looks like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 14:00 leave at station west (near mother's parents: GM)&lt;br&gt;&#xA; 14:05 leave at station north-west (where the boy lives: B)&lt;br&gt;&#xA; 14:15 leave at station north&lt;br&gt;&#xA; 14:25 leave at station north-east&lt;br&gt;&#xA; 14:30 leave at station east (grandparents on father's side: GF)&lt;br&gt;&#xA; 14:25 leave at station north-east&lt;br&gt;&#xA; 14:45 leave at station north&lt;br&gt;&#xA; 14:55 leave at station north-west (B again).&lt;br&gt;&#xA; (repeating every hour).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So we can conclude&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the boy goes to his father's parents 50/60 times, and his mother's parents 10/60 times. As he only goes to his mother's parents if he arrives between xx:55 and xx:05 at the station.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In order to go to his mother's parents more often&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; he has to walk to station north, further away from mother's parents!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2711" LastActivityDate="2015-09-18T13:26:58.027" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="22341" PostTypeId="1" AcceptedAnswerId="22345" CreationDate="2015-09-18T14:27:55.800" Score="-2" ViewCount="128" Body="&lt;p&gt;Bob works in a rural area, so when Bob drives home from work each day, he takes a back-country highway.  However, because Bob works in a rural area in Pennsylvania, where civil engineers are crazy, there are several intersections controlled by stoplights along this (60 MPH) highway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of these intersections is particularly aggravating, since it's located less than a quarter mile after a bend in the road.  Being a rural area, the area off to the side of the road is heavily wooded, such that it's impossible to see the stoplight before coming around the bend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, on most days, Bob is able to accurately predict whether the light he's coming up on is green, or whether he'll need to slow down for it, well before reaching the bend in the road.  But on some days, he has no idea whatsoever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's his secret, and why doesn't it always work?&lt;/p&gt;&#xA;" OwnerUserId="12156" LastEditorUserId="11876" LastEditDate="2015-09-18T14:34:41.760" LastActivityDate="2015-09-18T14:39:12.180" Title="Stoplight clairvoyance" Tags="|situation|" AnswerCount="3" CommentCount="0" ClosedDate="2015-09-18T20:52:14.573" ContentLicense="CC BY-SA 3.0" />
  <row Id="22342" PostTypeId="1" AcceptedAnswerId="22361" CreationDate="2015-09-18T14:29:11.397" Score="36" ViewCount="4386" Body="&lt;p&gt;In the country Curgonia, there are many types of fake coins and only a single type of genuine coins. The weights of these coins satisfy the following conditions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;All genuine coins have the same weight&lt;/li&gt;&#xA;&lt;li&gt;Every fake coin is heavier than any genuine coin&lt;/li&gt;&#xA;&lt;li&gt;No two fake coins have the same weight &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Cosmo puts 100 coins on the table and tells Fredo: &lt;em&gt;&quot;Thirty of these coins are genuine and seventy of them are fake.&quot;&lt;/em&gt; Then Cosmo leaves the room. On the table, there is a balance with two pans (but there are no weights).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What is the smallest possible number of weighings that guarantee Fredo to identify at least one genuine coin?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-09-18T15:13:26.773" LastActivityDate="2015-09-28T15:35:10.963" Title="Thirty genuine and seventy fake coins" Tags="|mathematics|strategy|weighing|" AnswerCount="9" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22343" PostTypeId="2" ParentId="22341" CreationDate="2015-09-18T14:31:39.690" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Those new LED stoplights can get really bright, so I imagine that on (semi-)foggy days, the light can be seen reflected off of the fog around the corner. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-09-18T14:31:39.690" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22344" PostTypeId="2" ParentId="22341" CreationDate="2015-09-18T14:33:16.137" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If someone is driving ahead of him, he'd see them slowing down as they go through the bend and are able to see the light at the intersection. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alternately, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if there is traffic coming the other direction, their speed determines if the light is green. If the other traffic is moving fast, they came through a green light. If they're moving slower, they are traffic coming from the intersecting road and haven't had time to come up to speed. that would mean his light is red. It wouldn't always work, because slower traffic could be oncoming traffic coming out of a newly-green light, and Bob will have slowed down for a green light.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13101" LastEditorUserId="13101" LastEditDate="2015-09-18T14:39:12.180" LastActivityDate="2015-09-18T14:39:12.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22345" PostTypeId="2" ParentId="22341" CreationDate="2015-09-18T14:34:57.327" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is assuming that upon reaching the bend, a car would be required to brake immediately to stop for a red light. &lt;br&gt; If while approaching the bend he sees a car at the bend and it is braking, he can assume that the light is red, if they do not brake it is green.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why it doesn't always work:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people always brake going around bends.&lt;BR&gt;OR&lt;BR&gt;There happen not to be any cars ahead of him that day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12206" LastActivityDate="2015-09-18T14:34:57.327" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22346" PostTypeId="2" ParentId="22342" CreationDate="2015-09-18T14:38:54.963" Score="1" Body="&lt;p&gt;I'll do a baseline of &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 10 weighings&lt;br&gt;&#xA;&lt;br&gt;&#xA; My strategy is to essentially find the lightest coin. Split the coins into a group of 64, a group of 8, and a group of 28 ignored. This guarantees at least 1 legit coin is part of the group.&lt;br&gt;&#xA; Do a binary comparison for each group to find the lightest coin. Then compare the coins.  The lightest(s) are genuine. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-09-18T14:38:54.963" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22347" PostTypeId="2" ParentId="22342" CreationDate="2015-09-18T14:41:33.203" Score="9" Body="&lt;p&gt;Hm...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The obvious strategy is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;every time, weighing the lightest coin you get against an unused coin&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;for a worst case of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;70 weighings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if there's  way to improve that by weighing multiple coins against each other at once instead of just having one on each pan - it doesn't seem likely to me because one fake coin and one real can weigh more than two fakes, but it's still possible that you could improve this.&lt;/p&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="11876" LastEditDate="2015-09-18T18:10:15.467" LastActivityDate="2015-09-18T18:10:15.467" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22348" PostTypeId="2" ParentId="22342" CreationDate="2015-09-18T15:07:00.283" Score="5" Body="&lt;p&gt;The smallest possible number of weighings that guarantee a genuine coin is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 70&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's consider the information we get from each type of weighing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we weigh one coin against another, an even weighing tells us that both coins are good, whereas an uneven weighing tells us only that the heavier coin is fake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we weigh multiple coins against each other, an even result tells us nothing unless we know that we have a group with only two fake coins; if we have three, we have the possibility of two fake coins on one side weighing the same as a fake and a genuine coin on the other. An uneven result tells us only that the heavier side has at least one fake coin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be obvious that weighing multiple coins gives less information in all cases unless we already have a group with only two fake coins (at which point an uneven weighing would still not identify a genuine coin). The best strategy is then to weigh single coins against each other. If the weights are even, we have found a good coin; if not, we eliminate a fake coin. In the worst case, all weighings will be uneven, in which case we will need 70 weighings to identify a genuine coin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that it makes no difference which coins we weigh, as long as we set the heavier coin aside each time; we could continue to weigh the lighter coin against a new coin each time, or weigh 71 coins in a &quot;tournament&quot;, or just take random coins for each weighing. This is because an uneven weighing provides absolutely no information about the lighter coin, so the only thing that changes about our situation after the first weighing is that we effectively have 99 coins, of which 69 are fake. The only point at which a genuine coin will be identified in the worst case is after all the fake ones have been eliminated, at which point all the remaining ones will be genuine by default.&lt;/p&gt;&#xA;" OwnerUserId="12511" LastEditorUserId="12511" LastEditDate="2015-09-18T16:03:42.340" LastActivityDate="2015-09-18T16:03:42.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22349" PostTypeId="2" ParentId="22342" CreationDate="2015-09-18T15:14:56.630" Score="2" Body="&lt;p&gt;The minimum possible number of weighings is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 71&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we have 70 fake coins and 30 real coins, and we want to pair them one by one, then we need to guarantee a case where the scale is balanced in order to win.  If we make 50 comparisons among the 100 coins AND THEN discard the heavy coins, we'll be left with 50 of the lighter coins.  Note that among these 50 light coins, 30 are guaranteed to be the genuine coins.  If we perform 21 comparisons, we guarantee that all of the fake coins have been paired up with a real coins, and been discarded for being too heavy, so the last pair MUST contain a real coin.&#xA;This means we have the initial 50 comparisons, then an extra 21, yielding 71.&lt;/p&gt;&#xA;" OwnerUserId="13920" LastEditorUserId="13920" LastEditDate="2015-09-18T15:34:11.167" LastActivityDate="2015-09-18T15:34:11.167" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22350" PostTypeId="2" ParentId="22342" CreationDate="2015-09-18T15:32:32.863" Score="0" Body="&lt;p&gt;The only information that any given weighing (of equal number of coins) reveals to us:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If pans are equal, either both or none of the pans have fake coins.&lt;/li&gt;&#xA;&lt;li&gt;If pans are unequal, either one or both of the pans have fake coins.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;We also notice that finding actual weights of coins in terms of other coins is pointless&lt;/strong&gt;, since we might end up with equations like $Fake\ coin\ 1 \geq 200 \times Fake\ coin\ 2$. The only that seems of practical importance is sorting the coins relatively (in ascending/descending order).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's make the assumption that equal pans always means no fake coins in the given scenario. (There exist no two sets of fake coins with equal weights.) However, we will not know this. So even if we stumble upon two equal pans, we will repeatedly use different weighings till we conclude that there is no fake coin. This can take a long time. This also means that &lt;strong&gt;the fact as to which pan is greater is of no importance, once we know we have unequal pans.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to worst case, we will &lt;em&gt;never&lt;/em&gt; get equal pans unless this is logically implied from previous facts (in which case, weighing them is pointless).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hence we can conclude that we will never get equal pans in our perfect strategy under the worst case.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If pans are unequal, we gain no new information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hence, we conclude that it most be possible to  achieve a state where it is logically implied that a particular coin is genuine, without us ever getting equal pans in our weighings.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This only seems possible by doing $70$ weighings, as others have said.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4269" LastEditDate="2015-09-20T14:59:57.417" LastActivityDate="2015-09-20T14:59:57.417" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22351" PostTypeId="2" ParentId="22342" CreationDate="2015-09-18T15:52:48.377" Score="0" Body="&lt;p&gt;Here's my two cents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Split the coins into two groups. One group should be 71 coins, and in the other, 29 coins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Step 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since we know that the group containing 71 coins has at least one genuine coin. And we know that the genuine coin is the lightermost of them all. So we begin by placing one two coins randomly on the scale and place away the heavier one for each weigh. This way we will have to make a total of 70 weighs to make sure all coins have been weighed and we have the lightermost of them all. This is the genuine coin. If we ever get two coins of the same mass any time in the weighing process we stop there. Both of these coins are genuine.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16342" LastActivityDate="2015-09-18T15:52:48.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22352" PostTypeId="1" AcceptedAnswerId="22362" CreationDate="2015-09-18T16:07:21.240" Score="4" ViewCount="1224" Body="&lt;p&gt;A person is locked up in a mental institution because he claims that he's been having clairvoyant visions from ages long time ago. At some point, it becomes evident that he actually tells the truth and that he indeed can see the past, just as if he was there.&lt;/p&gt;&#xA;&lt;p&gt;It is a gold mine for the historians, so the police and others so the authorities immediately issue an order of release. However, due to some confusion at the mental institution, it's unclear who he is. After eliminating all possible candidates, the correct selection process produces two individuals - Adam and Benny. Precisely one of them is guaranteed to be the medium with superpowers, while the other is a fraudulent liar. The problem is that there's no way to determine who's who.&lt;/p&gt;&#xA;&lt;p&gt;Both of them claim to be the true psychic and both state that they're immensely upset with the unfair incarceration in the mental institution. As a result, both refuse to provide any new story from the past to verify the ability.&lt;/p&gt;&#xA;&lt;p&gt;Finally, both agree to tell one story each. The subject is a feast at a house of a Viking family and based on that, we need to determine who's the true asset and who's a darn cheat.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Adam's story:&lt;/strong&gt;&lt;br /&gt;&#xA;&lt;em&gt;We started with a main course consisting of beef and a side salad of carrots and cabbage. The beverage was served first afterwards, together with mashed apples cooked in cow milk and flour. There was no singing and the conversation was in English.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Benny's story:&lt;/strong&gt;&lt;br /&gt;&#xA;&lt;em&gt;First, we had some beer, not very strong, and then water. We were served pork dipped in fat and burned almost to well-done over an open fire. On the plate, we also got eggs and some kind of mix of tomatoes and potatoes. For desert, we had a cake with a lot of berries. It was noisy from the music and singing.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;After being presented with the above information, a person stands up and states: &amp;quot;I know who's got the super powers!&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Who is it and why?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;h1&gt;Story revelation&lt;/h1&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;The language remark is a honeypot - it's actually not the issue at all. Vikings &lt;strong&gt;did&lt;/strong&gt; visit England and it's &lt;strong&gt;possible&lt;/strong&gt; that they spoke the language. Also, there's been Vikings living &lt;strong&gt;on&lt;/strong&gt; the British islands and I never said where the feast is taking place. Also, it's possible that the &lt;strong&gt;guests&lt;/strong&gt; at the feast were visitors from England.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4319" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2017-12-13T19:30:39.740" Title="Who is a real time traveller and who is lying?" Tags="|knowledge|story|mystery|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="22353" PostTypeId="1" CreationDate="2015-09-18T16:07:28.573" Score="6" ViewCount="379" Body="&lt;p&gt;In a remote town where the internet has not yet reached, people use a indigenous telephone to communicate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each telephone call occupies the phone to the nearest minute ceiling. So if a call is 15 seconds long, it will be considered a 1 minute call. If it is 3 minute 2 seconds it will be a 4 minute call.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The telephone company charges 1 cent for the first minute, 2 for the second and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One fine Saturday morning at 0900 hours, the Mayor's wife phones two close friends inviting them to an event the coming Monday at 6:00 pm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The invite is to be forwarded to at least 1 and maximum of 3 other town residents via telephone. Each invite takes the same time in minutes as the invitee number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Assuming there are no multiple invites the questions are :&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) What would be the &lt;strong&gt;minimum time&lt;/strong&gt; taken to reach 50000 unique residents in the town? How much money would the telephone company have made in this case?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) How many maximum invites could have been reached in the first three hours?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) With above conditions What would be the strategy be to reach the maximum but get charged the minimum in the first three hours?&lt;/p&gt;&#xA;" OwnerUserId="9892" LastEditorUserId="9892" LastEditDate="2015-09-18T16:34:03.603" LastActivityDate="2015-10-24T13:27:33.533" Title="Invitation to the party" Tags="|mathematics|logical-deduction|" AnswerCount="1" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="22354" PostTypeId="1" AcceptedAnswerId="22357" CreationDate="2015-09-18T16:08:28.550" Score="1" ViewCount="801" Body="&lt;p&gt;My grandfather had a very tough and interesting life. He was born in one city, some years later went to school in another city, and finally lived and worked in a third city. He was lucky though to die in exactly the same city he was born in.  Just before he died, he told me his only regret was that he wished he had travelled more.  Especially, he would have liked to see more than one city in his life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3668/what-color-was-the-bear&quot;&gt;what color was the bear&lt;/a&gt;?&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-09-18T18:44:04.570" Title="What color was the bear?" Tags="|lateral-thinking|geography|" AnswerCount="2" CommentCount="13" FavoriteCount="0" ClosedDate="2015-09-18T18:49:16.467" ContentLicense="CC BY-SA 3.0" />
  <row Id="22356" PostTypeId="2" ParentId="22354" CreationDate="2015-09-18T16:20:18.640" Score="7" Body="&lt;p&gt;Partial answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think your grandfather lived in Saint Petersburg. It was named Petrograd from 1914 to 1924 and Leningrad from 1924 to 1991. I assume bears near Saint Petersburg are brown, but the reference to the story is not clear to me yet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastActivityDate="2015-09-18T16:20:18.640" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22357" PostTypeId="2" ParentId="22354" CreationDate="2015-09-18T16:20:40.850" Score="8" Body="&lt;p&gt;The color of the bear is brown.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is a Russian bear.&#xA;&lt;br&gt;&#xA; Your grandfather was born in St. Petersburg, went to school in&#xA; Petrograd, worked in Leningrad, and died in St. Petersburg.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-09-18T16:20:40.850" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22358" PostTypeId="2" ParentId="22352" CreationDate="2015-09-18T16:25:27.327" Score="1" Body="&lt;p&gt;Given the stories I am pretty sure Vikings didn't speak English so that seems to be a dead giveaway that Benny is the real clairvoyant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I think the puzzle is fundamentally illogical considering: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;The problem is that there's no way to determine who's who.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Couldn't he just tell each of the officials/historians/etc. a fact that only they know from their past to prove? Seems pretty straight forward. I might be missing something but this doesn't seem entirely puzzling.&lt;/p&gt;&#xA;" OwnerUserId="15843" LastEditorUserId="15843" LastEditDate="2015-09-18T16:32:19.783" LastActivityDate="2015-09-18T16:32:19.783" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22359" PostTypeId="2" ParentId="20440" CreationDate="2015-09-18T16:31:46.017" Score="3" Body="&lt;p&gt;I got some inspiration last night, so let's try another guess. For the first three stanzas, I have:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;WORM&lt;/strong&gt; - See answer by @Kingrames.&lt;br&gt;&#xA;&lt;br&gt;&#xA; I have multiple answer candidates for this second stanza that I can make fit with my final answer. I'm going to go with &lt;strong&gt;STAMINA&lt;/strong&gt; - People need it to endure physical hardships and those with more of it do better in race competitions.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;LAUGHTER&lt;/strong&gt; - It can be inappropriately childish as well as cathartic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All together, I got&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; parts of an &lt;strong&gt;APPLE&lt;/strong&gt; (sort of). One can find a worm in an apple. Apples have stamen (stamina). Apples have peels (peals of laughter). There exists the phrase &quot;apple of one's eye&quot;. In Christian theology, the apple could be the forbidden fruit that led to the fall of humanity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15921" LastActivityDate="2015-09-18T16:31:46.017" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22360" PostTypeId="1" AcceptedAnswerId="22369" CreationDate="2015-09-18T16:36:32.603" Score="11" ViewCount="241" Body="&lt;p&gt;I've found a list of movie names, but I have no idea what the movies are about and I've never seen them before. Based on their names and my imagination, Ive guessed the genre and written a pitch for the movie from what I imagined they are about. I've included the decade of each movie's release to help you narrow down the names. Can you figure out all 7 names?&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;List&lt;/h3&gt;  &#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Documentary 1990's / 2010's&lt;br&gt;&#xA;A company decides that all of its shipped product are unfit for consumer use and request that every one of them is pulled off the shelves&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Fairy Tale / Drama 1980's&lt;br&gt;&#xA;Once upon a time there was a princess who lived in a castle and she had long hair and she was blonde&#xA;and she went into the forest to get some flowers and she found some flowers and she liked them and she&#xA;smelled them and she brought them back and she put them in a vase and she looked at them and she admired&#xA;them and she wanted some water for them so she went to the lake for some water and she saw some fish so she watched them swim for a bit then she went home and she realized she forgot her water so she went back and she...&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Apocalyptic Sci-Fi 1930's&lt;br&gt;&#xA;A massive power outage along the Pacific Coast cripples the area and cuts off all forms of communication with&#xA;people in the region&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Math Documentary  2010's&lt;br&gt;&#xA;Mathematicians struggle to determine what happens to the function $f(x)$ defined to be&#xA;$\pi \text{ if } x\ge0$&lt;br&gt;&#xA;$142 \text{ if } x\lt0$&lt;br&gt;&#xA;as x proceeds to 0.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Science Documentary 1980's&lt;br&gt;&#xA;Theoretical physicists discuss what, if anything, could have existed before the Big Bang, in a thought-expirement&#xA;taking place on a 0-dimensional realm.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Dystopia Sci-Fi 2000's&lt;br&gt;&#xA;In a futuristic society where efficiency and productivity are worshipped ideals, people's worth is determined&#xA;by their contributions to society. Once members have reached a point where they can no longer work, they are culled.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Health Documentary 1990's&lt;br&gt;&#xA;Doctors and health experts explain how renal failure can be combatted by the miraculous process of dialysis,&#xA;which filters blood through a machine in lieu of normal kidney function.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13970" LastActivityDate="2015-09-18T17:47:38.237" Title="What's In A Name?" Tags="|movies|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22361" PostTypeId="2" ParentId="22342" CreationDate="2015-09-18T16:38:40.993" Score="23" Body="&lt;p&gt;As many have already stated, the best you can do is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; seventy weighings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Suppose the genuine coins have weight 0, and the other coins have weights of distinct powers of two.  Then any try on the old balance will always just tip the scale to the side of the heaviest fake coin, since it weighs more than all the other coins on the scale put together.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose the scale somehow magically marked the heaviest coin on the scale with a big red X everytime you weighed some coins.  Now, there is no use using the coin with an X again.  Also, you've learned absolutely nothing about the other coins.  Hence, even with this extra magic, the only information you can gain each time is to eliminate one coin as being fake.   Thus, it takes in the worst case seventy weighings to eliminate all the fake coins and find at least one genuine one.&lt;/p&gt;&#xA;" OwnerUserId="2722" LastActivityDate="2015-09-18T16:38:40.993" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22362" PostTypeId="2" ParentId="22352" CreationDate="2015-09-18T16:38:48.680" Score="14" Body="&lt;p&gt;The one with powers is  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Adam  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Tomato#History&quot;&gt;Tomatoes&lt;/a&gt; weren't seen outside of the South American continent until the 1500s, or thereabouts, somewhat after the vikings' main era of prominence. Since Benny claims to have eaten them, he's probably a liar. Or Adam edited the wikipedia page to discredit him.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11717" LastActivityDate="2015-09-18T16:38:48.680" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22363" PostTypeId="1" AcceptedAnswerId="22368" CreationDate="2015-09-18T16:45:12.187" Score="6" ViewCount="411" Body="&lt;p&gt;I think this should be pretty easy. Determine what movie each haiku references.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm sorry,  Father&lt;br&gt;&#xA;  I should not have run away&lt;br&gt;&#xA;  I will avenge you&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Little and broken&lt;br&gt;&#xA;  I remember everyone&lt;br&gt;&#xA;  Join my family  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hear my tragic tale&lt;br&gt;&#xA;  Love before the petal falls&lt;br&gt;&#xA;  Won't you please be mine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Alone in the wild&lt;br&gt;&#xA;  Growing up to save us all&lt;br&gt;&#xA;  He will lead the band&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Strangers take her in&lt;br&gt;&#xA;  Each one unique but the same&lt;br&gt;&#xA;  The old hag's scheme fails&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Fighting for honor&lt;br&gt;&#xA;  Protecting my family&lt;br&gt;&#xA;  Though done with deceit&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14418" LastEditorUserId="4050" LastEditDate="2015-09-18T17:45:06.073" LastActivityDate="2015-09-18T17:45:06.073" Title="Actual Movie Haikus: I'm sorry, Father / I should not have run away / I will avenge you" Tags="|movies|haiku|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22364" PostTypeId="1" AcceptedAnswerId="22371" CreationDate="2015-09-18T16:46:37.560" Score="-1" ViewCount="232" Body="&lt;p&gt;You have a scale that is remarkably sturdy, and there are seven spheres of unequal mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;No sphere has the same mass as any other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is (almost) no limit to the mass of any sphere, but the most massive sphere is the &lt;em&gt;planet that you are standing on&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know the exact mass of the smallest two of these spheres. One is one kilogram and one is three kilograms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your scale is able to fit more than one sphere on each end, however the largest sphere, the final sphere, is far too large to fit on either end of the scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of the remaining spheres fall between the three-kilogram and the planet, and none are large enough to cause the planet's center of gravity to be outside its core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will assume that the planet-sphere does not become indented at the surface due to the mass of the other spheres, or the scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your scale will register this information:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) whether the two weighings are equal&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) if the weighings are not equal, which side weighs more, and...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) if one side weighs more, the amount by which the weight exceeds, if within 10 percent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you have enough information to determine the mass of the remaining spheres using only your scale and the weighings? If not, what about all of them except the planet?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can safely assume that an accurate measurement would be accurate to within 0.1 kilograms, and for the sake of simplicity we won't take into account cosmic dust buildup during measurement or anything of the sort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are free to assume any upper bound necessary for the size of &quot;a planet.&quot; But no shenanigans regarding planets so big they collapse into black holes or any other &lt;a href=&quot;/questions/tagged/lateral-thinking&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;lateral-thinking&amp;#39;&quot; rel=&quot;tag&quot;&gt;lateral-thinking&lt;/a&gt; craziness, please. &lt;/p&gt;&#xA;" OwnerUserId="13068" LastActivityDate="2015-09-18T17:16:33.813" Title="Seven Spheres of Unequal Mass, a weighing problem with a twist" Tags="|mathematics|weighing|science|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22365" PostTypeId="2" ParentId="20440" CreationDate="2015-09-18T16:49:01.293" Score="2" Body="&lt;p&gt;Basically just posting @Dr Xorille's answer so if this is right do not award me the checkmark we will let him post it and you can award it to him. Or maybe to Kingrames - IDK up to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Global warming&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using @Kingrames answers &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; WORMS,CARS,GAS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using the first part of the riddle we know each of these plays their &quot;part&quot; in the demise of humanity&#xA;Breaking down each one's role: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;Worms&lt;/code&gt;: &quot;The presence of worms affects how much carbon dioxide is produced in the soil and how much escapes to the atmosphere. Scientists are concerned that earthworms increase greenhouse gas emissions - and that earthworm numbers are on the rise.&quot; &lt;a href=&quot;http://www.theguardian.com/environment/blog/2013/feb/05/how-worms-accelerating-climate-change&quot; rel=&quot;nofollow&quot;&gt;http://www.theguardian.com/environment/blog/2013/feb/05/how-worms-accelerating-climate-change&lt;/a&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;code&gt;Cars&lt;/code&gt;: Release carbon dioxide which research at the moment shows is one major human controllable promotion of global warming.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;code&gt;Gas&lt;/code&gt;: Greenhouse gas emissions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;**Formatting for hiding the above for some reason doesn't work no matter what I try if I have them split so I have to combine them.&lt;/p&gt;&#xA;" OwnerUserId="15843" LastEditorUserId="15921" LastEditDate="2015-09-18T17:23:36.220" LastActivityDate="2015-09-18T17:23:36.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22366" PostTypeId="2" ParentId="22342" CreationDate="2015-09-18T16:49:21.437" Score="2" Body="&lt;p&gt;Weighing multiple coins gives us no real information because there are no bounds on how much more the fake coins weigh.  Weighing one pair tells us only that (1) both are the same and both are real, or (2) the heavier coin is fake and can be eliminated from the pool of candidates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Worst case, 70 weighings will eliminate 70 fakes and all the rest are real (tho most likely a pair of reals would be found earlier).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note than any solution must access (weigh at least once) a minimum of 71 coins, to be sure that at least one real coin has been considered.  70 weighings can do this (for example, just compare the lighter of each weighing with the next coin from the pool of unweighed ones; after 70 compares you have tested 71 coins).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have not yet found any way to make use of the number of real coins.  This answer is the same whether there are 70 fake + 1 real, or 70 fake + 1000 real - just do 70 compares of any type and each time eliminate the heavier one from the pool (or both are real if equal).  If there is any way to beat 70 compares, it's going to have to infer some information from the number of real coins, but I'm not finding any way to do that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: oops, I had a bad answer which somebody commented on while I was (immediately) revising it - ignore that first comment if it makes no sense, it was my blunder.&lt;/p&gt;&#xA;" OwnerUserId="16414" LastEditorUserId="16414" LastEditDate="2015-09-18T17:47:43.307" LastActivityDate="2015-09-18T17:47:43.307" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22367" PostTypeId="2" ParentId="22363" CreationDate="2015-09-18T16:54:05.040" Score="1" Body="&lt;p&gt;W.I.P.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Snow White, taken in by 7 dwarves, each of which has a different name based on an attribute. An evil witch attempts to put her in an eternal coma but she's awakened by true love's kiss.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Mulan, who defends her family's honor fighting in her father's stead but pretending to be a man&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Not sure if they are all&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Disney princesses or disney movies  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or not&lt;/p&gt;&#xA;" OwnerUserId="13970" LastActivityDate="2015-09-18T16:54:05.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22368" PostTypeId="2" ParentId="22363" CreationDate="2015-09-18T16:54:18.170" Score="4" Body="&lt;p&gt;1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Lion King!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Probably Lilo and Stitch. Thanks VictorHenry! Ohana means family!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Beauty and the Beast&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This, Tarzan. You, Jane.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Snow White and the Seven Dwarves&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Mulan! I'll make a man... out of YOOOOOOUUUUU!!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13068" LastEditorUserId="13068" LastEditDate="2015-09-18T17:01:41.893" LastActivityDate="2015-09-18T17:01:41.893" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22369" PostTypeId="2" ParentId="22360" CreationDate="2015-09-18T16:56:56.600" Score="4" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Total Recall&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The NeverEnding Story&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All Quiet on the Western Front &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Limitless&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Land Before Time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No Country For Old Men&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Blood In, Blood Out&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="2627" LastEditDate="2015-09-18T17:47:38.237" LastActivityDate="2015-09-18T17:47:38.237" CommentCount="16" CommunityOwnedDate="2015-09-18T17:06:47.990" ContentLicense="CC BY-SA 3.0" />
  <row Id="22371" PostTypeId="2" ParentId="22364" CreationDate="2015-09-18T17:16:33.813" Score="3" Body="&lt;p&gt;If the masses of the six smaller spheres are (in kg) 1, 3, 5, 11, 23, 49, then there is no way to learn anything from this scale except for sorting the spheres by weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no combination of those numbers that is within 10 percent of any other combination.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2015-09-18T17:16:33.813" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22373" PostTypeId="1" AcceptedAnswerId="22379" CreationDate="2015-09-18T18:50:34.077" Score="10" ViewCount="557" Body="&lt;p&gt;Wanda the Wanderer is driving the streets of Gnu York. She doesn't know the what the entire map of Gnu York looks like, but she knows this much:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Gnu York is flat&lt;/li&gt;&#xA;&lt;li&gt;Every intersection of streets is four-way, resembling a &quot;+&quot;&lt;/li&gt;&#xA;&lt;li&gt;There are no overpasses, tunnels, or dead ends&lt;/li&gt;&#xA;&lt;li&gt;There are no streets leading out of Gnu York&lt;/li&gt;&#xA;&lt;li&gt;There are only finitely many intersections&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Whenever Wanda reaches an intersection, she decides whether to go left, right or straight according to the repeating pattern&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Left $\to$ Right $\to$ Straight $\to$ Left $\to$ Right $\to$ Straight $\to\dots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if she turned left before, she will turn right at the next intersection, then go straight at the one after that, then left at the one after that.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At noon, Wanda passes by city hall. Prove that she will eventually pass by city hall again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Below is a an example of what Gnu York might look like, with $\color{red}{\star}$ being city hall, along with how Wanda's path would start out. However, your proof should work for any road layout which obeys the given rules, not just the one below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UcPdL.png&quot; alt=&quot;Gnu York&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-09-18T20:54:41.737" LastActivityDate="2015-09-18T21:09:21.097" Title="Lost in Gnu York" Tags="|graph-theory|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22374" PostTypeId="2" ParentId="22321" CreationDate="2015-09-18T19:25:44.693" Score="0" Body="&lt;p&gt;Working backwards, we can get an answer a non-programmer might come up with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, 75 minutes is a red herring. Since we can't do multiple rounds of testing and since the testers can't return partial results, we can't leverage the extra 20 minutes. We are done after 55.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, note that we could put rice from a specific plate into multiple testers or into no testers:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If multiple testers are positive, then the poisoned plate must be one whose rice was put into &lt;em&gt;each&lt;/em&gt; of those testers.&lt;/li&gt;&#xA;&lt;li&gt;If no testers are positive, then the plate must be one whose rice wasn't put into &lt;em&gt;any&lt;/em&gt; testers.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Since any permutations of positive and negative results are now allowed, there are 2^4=16 ways to get results. Since there are 16 plates, we need to leverage &lt;em&gt;each&lt;/em&gt; outcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a given outcome, just work backwards, assuming outcome #X must indicate plate #X:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Outcome 1) No testers are positive. Since this situation must uniquely identify plate #1, we needed to have aside plate #1 and only plate #1.&lt;/li&gt;&#xA;&lt;li&gt;...&lt;/li&gt;&#xA;&lt;li&gt;Outcome 8) The third tester is positive, others are negative. Rice from plate #8 must have been put into the third tester and only into the third tester.&lt;/li&gt;&#xA;&lt;li&gt;...&lt;/li&gt;&#xA;&lt;li&gt;Outcome 16) All testers are positive. Since this situation must uniquely identify plate #16, we needed to put some rice from plate #16 into each tester. None of the other plates are in each tester.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2448" LastActivityDate="2015-09-18T19:25:44.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22375" PostTypeId="1" AcceptedAnswerId="22380" CreationDate="2015-09-18T19:49:55.463" Score="6" ViewCount="565" Body="&lt;p&gt;Hangman game backward version 6 - guess what was the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;fill in a &quot;d&quot; -  I am a magical device that you can convert energy with the tip of your fingers.&lt;br&gt;&#xA;fill in a &quot;g&quot; - You can hear, under one condition, miles away from me.&lt;br&gt;&#xA;fill in a &quot;t&quot; - It's been a long day, come to me and I'll make it short for you.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, your answer can be in the form of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The question was: _ p p l e&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="13237" LastEditDate="2015-09-18T21:29:20.453" LastActivityDate="2015-09-18T21:29:20.453" Title="The hangman game backward #6" Tags="|riddle|knowledge|english|language|letters|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22376" PostTypeId="1" AcceptedAnswerId="22406" CreationDate="2015-09-18T20:02:11.103" Score="9" ViewCount="2615" Body="&lt;p&gt;I can be quite bright but I'm never right.&lt;br /&gt;&#xA;I support you in peace and protect you in fight.&lt;/p&gt;&#xA;&lt;p&gt;I'm seldom alone and when I am, there's no use.&lt;br /&gt;&#xA;Even when you select me, another you choose.&lt;/p&gt;&#xA;&lt;p&gt;You never ask how I'm doing, even when I'm blue.&lt;br /&gt;&#xA;I might be unhealthy but I can't catch a flu.&lt;/p&gt;&#xA;&lt;p&gt;Sometimes I'm in transit and sometimes home I'll stay.&lt;br /&gt;&#xA;Sometimes you can see a person who for me will pray.&lt;/p&gt;&#xA;&lt;p&gt;If you don't have me at all, you're poor not having a dime&lt;br /&gt;&#xA;Or you might simply have fallen as a victim of crime.&lt;/p&gt;&#xA;&lt;p&gt;I can hurt you and wound you and even make you bleed.&lt;br /&gt;&#xA;Regardless of your pain, I'm something you still need.&lt;/p&gt;&#xA;&lt;p&gt;I might be unwanted but also an object of desire.&lt;br /&gt;&#xA;I could be damaged by earth, water and fire.&lt;/p&gt;&#xA;&lt;p&gt;I could be expensive or cheap or even free.&lt;br /&gt;&#xA;When people look at you, they often see me.&lt;/p&gt;&#xA;&lt;p&gt;You'll be both arrogant and humble if I'm put on your head.&lt;br /&gt;&#xA;I often bear a name that someone else before me had.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;What am I?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;Hints&lt;/h1&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;We're looking for a common everyday object.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;It's not anything abstract as &amp;quot;love&amp;quot; or &amp;quot;energy&amp;quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;It's not symbolic as &amp;quot;the letter A&amp;quot; or &amp;quot;peace sign&amp;quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4319" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2022-05-05T16:06:19.230" Title="I can be quite bright but I'm never right. Can you guess who I am?" Tags="|riddle|rhyme|" AnswerCount="4" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="22377" PostTypeId="2" ParentId="22373" CreationDate="2015-09-18T20:32:49.513" Score="0" Body="&lt;p&gt;There's only two possibilities for such movement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Option one is that Wanda never hits an end of the path, going on left-right-straight (which is impossible, because it's countered by the requirement that the number of intersections is finite).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Option two is that she at some point hits a straight that she's already been on. If it's arrived to and passed along on in opposite direction, then every left turn has a corresponding right turn when she was there the last time. It means that she's bound to arrive at the point of origin, hence undoubtedly passing the city hall again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, is it possible to get to the same street in the same direction without traveling back along it first? No, because that would contradict the patter of left turn being a right turn when traveling back. Once she's hitting &lt;strong&gt;any&lt;/strong&gt; street she's been on (as shown in the first paragraph), she's unable to follow other path than the one she's been on (but in the opposite direction).&lt;/p&gt;&#xA;" OwnerUserId="4319" LastActivityDate="2015-09-18T20:32:49.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22378" PostTypeId="1" CreationDate="2015-09-18T20:59:34.850" Score="13" ViewCount="823" Body="&lt;p&gt;&lt;strong&gt;Puzzlers!&lt;/strong&gt; Welcome to &lt;em&gt;&lt;strong&gt;Cryptic Jeopardy&lt;/strong&gt;&lt;/em&gt;, where your cryptic-crossword solving skills will be pushed to their limits -- in &lt;em&gt;direct competition&lt;/em&gt; with your fellow puzzlers! Here's how it works:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If you see a clue you know the answer to, post the answer in a partial answer. Make sure you post an explanation of how you got to the answer as well.&lt;/li&gt;&#xA;&lt;li&gt;If your answer is correct, I will award you the points associated with that clue on my scoreboard! However, if your answer is wrong, you don't get another chance to answer that question, so make sure you get it right!&lt;/li&gt;&#xA;&lt;li&gt;After every clue has been answered, the puzzler with the most points gets the green check!&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Here are your categories and your clues:&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;Code-Dependent&lt;/h2&gt;&#xA;&lt;p&gt;&lt;strong&gt;$100&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Dimple between eyes due to peculiar flames&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;$200&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Cross isoprenes off&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;$300&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Particle goes back-to-back&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;$400&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Ill will&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;$500&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Angry? Upset? Read between the lines&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;One Of Us&lt;/h2&gt;&#xA;&lt;p&gt;&lt;strong&gt;$100&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Crafty, crooked&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;$200&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Bless borough in disarray&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;$300&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; New Yorker said he would summon something&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;$400&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Small dekagram atom&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;$500&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Exciting; eating; excited; eaten&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;O, The Places You'll Go&lt;/h2&gt;&#xA;&lt;p&gt;&lt;strong&gt;$100&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Messy bathrooms&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;$200&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Complain about&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;$300&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Practice freely between rounds&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;$400&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Blended egg drink&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;$500&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Unit of power wrapped around and around&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;Fly Me A River&lt;/h2&gt;&#xA;&lt;p&gt;&lt;strong&gt;$100&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Break a writing utensil&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;$200&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Royal weaselly carnivore&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;$300&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Small indent in power tool&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;$400&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Hear the ocean? Carelessly dock, then yell violently&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;$500&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Jessica and Mr. Rick finally get together&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;An Unreal Vacation&lt;/h2&gt;&#xA;&lt;p&gt;&lt;strong&gt;$100&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; First off the ungulate's back&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;$200&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Titan holds muddled metal&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;$300&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Curious last character accompanies birth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;$400&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; In audition, belly button ring breaks&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;$500&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Dad, it echoes me&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Solved Clues:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Code-Dependent&lt;/strong&gt;&lt;br /&gt;&#xA;\$100 - Pete&lt;br /&gt;&#xA;\$200 - Deusovi&lt;br /&gt;&#xA;\$300 - dennisdeems&lt;br /&gt;&#xA;\$400 - dennisdeems&lt;br /&gt;&#xA;\$500 - Pete&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;One of Us&lt;/strong&gt;&lt;br /&gt;&#xA;\$100 - Deusovi&lt;br /&gt;&#xA;\$200&lt;br /&gt;&#xA;\$300 - Pete&lt;br /&gt;&#xA;\$400&lt;br /&gt;&#xA;\$500&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;O, the Places You'll Go&lt;/strong&gt;&lt;br /&gt;&#xA;\$100 - Pete&lt;br /&gt;&#xA;\$200 - Pete&lt;br /&gt;&#xA;\$300 - Pete&lt;br /&gt;&#xA;\$400 - Alconja&lt;br /&gt;&#xA;\$500 - kanchirk&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Fly Me A River&lt;/strong&gt;&lt;br /&gt;&#xA;\$100 - dennisdeems&lt;br /&gt;&#xA;\$200 - Pete&lt;br /&gt;&#xA;\$300 - Pete&lt;br /&gt;&#xA;\$400 - Pete&lt;br /&gt;&#xA;\$500 - dennisdeems&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;An Unreal Vacation&lt;/strong&gt;&lt;br /&gt;&#xA;\$100 - dennisdeems&lt;br /&gt;&#xA;\$200 - dennisdeems&lt;br /&gt;&#xA;\$300&lt;br /&gt;&#xA;\$400 - Pete&lt;br /&gt;&#xA;\$500 - dennisdeems&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Leaderboard:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$1.$&lt;/span&gt; Pete (\$2800)&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$2.$&lt;/span&gt; dennisdeems (\$2100)&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$3.$&lt;/span&gt; kanchirk (\$500)&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$4.$&lt;/span&gt; Alconja (\$400)&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$5.$&lt;/span&gt; Deusovi (\$300)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Good luck, and have fun!&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;sup&gt;&lt;em&gt;Note:&lt;/em&gt; The cryptic clues don't include definitions since they fit into categories. They also don't include letter count, but if they prove to be too difficult I will add them. There are some more difficult cryptic techniques being used here than in typical crosswords, specifically synonym anagrams. They're hopefully all solvable, though!&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="-1" LastEditorDisplayName="user20" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-08-05T16:16:32.117" Title="Cryptic Jeopardy" Tags="|cryptic-clues|" AnswerCount="8" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="22379" PostTypeId="2" ParentId="22373" CreationDate="2015-09-18T21:09:21.097" Score="13" Body="&lt;p&gt;There are a finite number of states she can be in, where a state is determined by direction, road, and upcoming turn. Each state has a unique predecessor and successor; therefore, there are no branches in the directed graph of states as vertices and edges connecting each state to its successor. This means every state must be part of a cycle.&lt;/p&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2015-09-18T21:09:21.097" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22380" PostTypeId="2" ParentId="22375" CreationDate="2015-09-18T21:14:20.163" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;WIN_ER&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fill in a &quot;d&quot; - I am a magical device that you can convert energy with the tip of your fingers.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;WINDER is not exactly magical, but it does allow you to convert work from the twisting of your fingers to potential/elastic energy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fill in a &quot;g&quot; - You can hear, under one condition, miles away from me  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;WINGER is a heavy metal band with one song called 'Under one Condition' and another called 'Miles Away'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fill in a &quot;t&quot; - It's been a long day, come to me and I'll make it short for you.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;WINTER days are shorter during winter season &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the edited clue&lt;br&gt;&#xA;fill in a &quot;n&quot; - I believe you wished you were me!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;WINNER is something that people wish they were.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="13970" LastEditDate="2015-09-18T21:21:27.697" LastActivityDate="2015-09-18T21:21:27.697" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="22381" PostTypeId="2" ParentId="22376" CreationDate="2015-09-18T21:59:18.250" Score="3" Body="&lt;p&gt;I'm going to guess the obvious answer, even though I suspect it's wrong:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Paper money&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be quite bright but I'm never right.&#xA;I support you in peace and protect you in fight.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Useful in many circumstances&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm seldom alone and when I am, there's no use.&#xA;Even when you select me, another you choose.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Kept in a wallet.  When you select me, it's to pay and get something else&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You never ask how I'm doing, even when I'm blue.&#xA;I might be unhealthy but I can't catch a flu.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No relationship with your money, and spending can be unhealthy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I'm in transit and sometimes home I'll stay.&#xA;Sometimes you can see a person who for me will pray.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sometimes you forget your wallet.  Plenty of people pray for it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you don't have me at all, you're poor not having a dime&#xA;Or you might simply have fallen as a victim of crime.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stolen or poor if you don't have it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can hurt you and wound you and even make you bleed.&#xA;Regardless of your pain, I'm something you still need.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Paper cut!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might be unwanted but also an object of desire.&#xA;I could be damaged by earth, water and fire.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Easily damaged.  Many people want to go cash free&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I could be expensive or cheap or even free.&#xA;When people look at you, they often see me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Different denominations.  People often figure out how to get you to pay, and are not interested in you otherwise.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'll be both arrogant and humble if I'm put on your head.&#xA;I often bear a name that someone else before me had.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Notes often have famous people on them.  You look stupid with money (literally) on your head, but you might be arrogant if people pour money on your head (a euphemism for pay well).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15942" LastActivityDate="2015-09-18T21:59:18.250" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22382" PostTypeId="1" AcceptedAnswerId="22383" CreationDate="2015-09-18T22:34:14.817" Score="24" ViewCount="1321" Body="&lt;p&gt;&lt;p&gt;I wanted to make a puzzle for you all with some help from my friends so I dictated a list of my favourite movies to Paul and he typed them up. He then gave the list to Alice to describe the titles.&lt;/p&gt;&lt;p&gt;I looked at Alice's list and I was initially confused, because her sentences didn't describe my film titles. Then it became apparent that the keyboard Paul had used wasn't working very well. Every item on my list had been typed with a single letter missing. Of course the spellchecker didn't pick up on it, because every title still made sense.&lt;/p&gt;&lt;p&gt;I've reproduced Alice's list here with the year of the film and the missing letter.&lt;/p&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&lt;li&gt;B (2009): One cereal grain that knows how to party.&lt;/li&gt;&#xA;&lt;li&gt;C (2006): Footwear jottings.&lt;/li&gt;&#xA;&lt;li&gt;E (1967): Do you wear that to sleep in? Your head must get hot.&lt;/li&gt;&#xA;&lt;li&gt;E (1981): My mum's an angel, but he's completely different.&lt;/li&gt;&#xA;&lt;li&gt;E (2011): The fish that spawned all the rest.&lt;/li&gt;&#xA;&lt;li&gt;G (1975): He wants to marry into our family.&lt;/li&gt;&#xA;&lt;li&gt;G (1977): You aren't allowed to marry that many women. You'll be one over the limit.&lt;/li&gt;&#xA;&lt;li&gt;H (1993): That's a lot of fuss to make about some shorthand.&lt;/li&gt;&#xA;&lt;li&gt;H (1994): I know that tombstones don't exactly look healthy, but this one is terrible. It's yellow too.&lt;/li&gt;&#xA;&lt;li&gt;I (1964): I have many girlfriends, but this one takes the longest to get to.&lt;/li&gt;&#xA;&lt;li&gt;I (1979): A tale of the outer layer of a cereal grain.&lt;/li&gt;&#xA;&lt;li&gt;I (1996): I filed for protection of my invention in London.&lt;/li&gt;&#xA;&lt;li&gt;K (1967): I was walking through the forest and someone surprised me with an exclamation.&lt;/li&gt;&#xA;&lt;li&gt;L (1959): She's gorgeous, but those sores haven't healed yet.&lt;/li&gt;&#xA;&lt;li&gt;L (1985): A baby bear who eats with us every morning.&lt;/li&gt;&#xA;&lt;li&gt;L (1990): Joseph's father had a snake.&lt;/li&gt;&#xA;&lt;li&gt;L (1994): Stories about young dogs.&lt;/li&gt;&#xA;&lt;li&gt;L (1995): The corpse appears to be stirring. He's not asleep.&lt;/li&gt;&#xA;&lt;li&gt;M (1991): The researchers have gone home - it's so quiet.&lt;/li&gt;&#xA;&lt;li&gt;N (1954): The lady at the back has lost her husband.&lt;/li&gt;&#xA;&lt;li&gt;N (1973): It's the bloke off Top Gear.&lt;/li&gt;&#xA;&lt;li&gt;N (1973): Pay attention, Dorothy.&lt;/li&gt;&#xA;&lt;li&gt;N (1977): My youngest child has met a lot of strangers. The other two haven't.&lt;/li&gt;&#xA;&lt;li&gt;N (1990): Only use this shed if you are in Russia in the middle of autumn.&lt;/li&gt;&#xA;&lt;li&gt;N (2002): Manhattan comedy.&lt;/li&gt;&#xA;&lt;li&gt;N (2012): These competitions are bigger than last time.&lt;/li&gt;&#xA;&lt;li&gt;O (1971): That's a big cooker. How long does it take to wind it up?&lt;/li&gt;&#xA;&lt;li&gt;O (1989): All they do is talk about their deceased animals.&lt;/li&gt;&#xA;&lt;li&gt;O (1992): Not many ministers.&lt;/li&gt;&#xA;&lt;li&gt;P (1983): It would look better if we took the red ones from your shoes and swapped them with the black ones in mine.&lt;/li&gt;&#xA;&lt;li&gt;R (1930): These mass baptisms are really noisy, but the minister furthest from the sunrise has everything under control.&lt;/li&gt;&#xA;&lt;li&gt;R (1969): He's a real pain in the neck.&lt;/li&gt;&#xA;&lt;li&gt;R (1982): Knuckle gore.&lt;/li&gt;&#xA;&lt;li&gt;R (2002): Less than half the people are planting up their seedlings.&lt;/li&gt;&#xA;&lt;li&gt;R (2008): His electrons and protons don't match.&lt;/li&gt;&#xA;&lt;li&gt;S (1977): Battles over bitumen.&lt;/li&gt;&#xA;&lt;li&gt;S (1992): Her only coin.&lt;/li&gt;&#xA;&lt;li&gt;S (1995): A Conservative that you play with.&lt;/li&gt;&#xA;&lt;li&gt;S (2000): Missing feline.&lt;/li&gt;&#xA;&lt;li&gt;S (2004): When he passed away, all he left us were these garden tools.&lt;/li&gt;&#xA;&lt;li&gt;T (1953): The god of shepherds in the House of Lords.&lt;/li&gt;&#xA;&lt;li&gt;T (1955): A women encounters a change of gradient.&lt;/li&gt;&#xA;&lt;li&gt;T (1968): Do those monkeys fly that thing?&lt;/li&gt;&#xA;&lt;li&gt;T (1986): I don't like it at the beach. This stuff makes me uncomfortable.&lt;/li&gt;&#xA;&lt;li&gt;U (1955): He'd cause a lot more trouble if he had some luggage!&lt;/li&gt;&#xA;&lt;li&gt;U (1978): We use it to wash all the creatures.&lt;/li&gt;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Question:&lt;/b&gt; Can you work out the titles as Alice saw them, and how they should have been?&lt;/p&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="14135" LastEditDate="2015-09-19T18:56:32.460" LastActivityDate="2015-09-23T19:49:34.717" Title="I've seen these movies countless ties" Tags="|wordplay|movies|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="22383" PostTypeId="2" ParentId="22382" CreationDate="2015-09-18T22:49:49.557" Score="15" Body="&lt;ol&gt;&#xA;&lt;li&gt;B (2009): One cereal grain that knows how to party. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;The (B)oat That Rocked&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;C (2006): Footwear jottings. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Notes on a S(c)andal&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;E (1967): Do you wear that to sleep in? Your head must get hot. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;In the H(e)at of the Night&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;E (1981): My mum's an angel, but he's completely different. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;The Evil D(e)ad&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;E (2011): The fish that spawned all the rest. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Source Cod(e)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;G (1975): He wants to marry into our family. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;The Man Who Would Be Kin(g)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;G (1977): You aren't allowed to marry that many women. You'll be one over the limit. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;A Brid(g)e Too Far&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;H (1993): That's a lot of fuss to make about some shorthand. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Much Ado About Not(h)ing&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;H (1994): I know that tombstones don't exactly look healthy, but this one is terrible. It's yellow too. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;S(h)allow Grave&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;I (1964): I have many girlfriends, but this one takes the longest to get to. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;My Fa(i)r Lady&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;I (1979): A tale of the outer layer of a cereal grain. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Monty Python's Life of Br(i)an&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;I (1996): I filed for protection of my invention in London. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;The English Pat(i)ent&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;K (1967): I was walking through the forest and someone surprised me with an exclamation. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;The Jungle Boo(k)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;L (1959): She's gorgeous, but those sores haven't healed yet. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;S(l)eeping Beauty&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;L (1985): A baby bear who eats with us every morning. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Breakfast C(l)ub&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;L (1990): Joseph's father had a snake. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Jacob's (L)adder&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;L (1994): Stories about young dogs. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Pu(l)p Fiction&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;L (1995): The corpse appears to be stirring. He's not asleep. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;The Wa(l)king Dead&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;M (1991): The researchers have gone home - it's so quiet. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;The Silence of the La(m)bs&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;N (1954): The lady at the back has lost her husband. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Rear Wi(n)dow&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;(1973): It's the bloke off Top Gear. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;The Sti(n)g&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;N (1973): Pay attention, Dorothy. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Do(n')t Look Now&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;N (1977): My youngest child has met a lot of strangers. The other two haven't. '&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Close Encounters of the Third Ki(n)d&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;N (1990): Only use this shed if you are in Russia in the middle of autumn. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;The Hu(n)t for Red October&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;N (2002): Manhattan comedy. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Ga(n)gs of New York&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;N (2012): These competitions are bigger than last time. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;The Hu(n)ger Games&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;O (1971): That's a big cooker. How long does it take to wind it up? &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;A Clockwork (O)range&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;O (1989): All they do is talk about their deceased animals. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Dead P(o)ets Society&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;O (1992): Not many ministers. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;A Few Go(o)d Men&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;P (1983): It would look better if we took the red ones from your shoes and swapped them with the black ones in mine. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Trading (P)laces&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;R (1930): These mass baptisms are really noisy, but the minister furthest from the sunrise has everything under control. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;All Quiet on the Western F(r)ont&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;R (1969): He's a real pain in the neck. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;True G(r)it&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;R (1982): Knuckle gore. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Fi(r)st Blood&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;R (2002): Less than half the people are planting up their seedlings. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Minority Repo(r)t&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;R (2008): His electrons and protons don't match. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;I(r)on Man&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;S (1977): Battles over bitumen. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;(S)tar Wars&lt;/strong&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;S (1992): Her only coin. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;(S)cent of a Woman&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;S (1995): A Conservative that you play with. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Toy (S)tory&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;S (2000): Missing feline. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Ca(s)t Away&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;S (2004): When he passed away, all he left us were these garden tools. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Dead Man's (S)hoes&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;T (1953): The god of shepherds in the House of Lords. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Pe(t)er Pan&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;T (1955): A women encounters a change of gradient. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Lady and the (T)ramp&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;T (1968): Do those monkeys fly that thing? &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Plane(t) of the Apes&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;T (1986): I don't like it at the beach. This stuff makes me uncomfortable. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;S(t)and By Me&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;U (1955): He'd cause a lot more trouble if he had some luggage! &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Rebel Without a Ca(u)se&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;U (1978): We use it to wash all the creatures. &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Animal Ho(u)se&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13237" LastEditorUserId="4204" LastEditDate="2015-09-23T19:49:34.717" LastActivityDate="2015-09-23T19:49:34.717" CommentCount="2" CommunityOwnedDate="2015-09-18T22:49:49.557" ContentLicense="CC BY-SA 3.0" />
  <row Id="22384" PostTypeId="2" ParentId="22382" CreationDate="2015-09-18T22:50:45.043" Score="4" Body="&lt;p&gt;Here's ones that I've got (as a separate answer, I'll update the community wiki with them too):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 4. The Evil Dead (becomes &quot;The Evil Dad&quot;)&#xA;12. The English Patient (becomes &quot;The English Patent&quot;)&#xA;15. Breakfast Club (becomes &quot;Breakfast Cub&quot;)&#xA;16. Jacob's Ladder (becomes &quot;Jacob's Adder&quot;)&#xA;17. Pulp Fiction (becomes &quot;Pup Fiction&quot;)&#xA;20. Rear Window (becomes &quot;Rear Widow&quot;)&#xA;28. Dead Poets Society (becomes &quot;Dead Pets Society&quot;)&#xA;30. Trading Places (becomes &quot;Trading Laces&quot;)&#xA;35. Iron Man (becomes &quot;Ion Man&quot;)&#xA;38. Toy Story (becomes &quot;Toy Tory&quot;)&#xA;39. Cast Away (becomes &quot;Cat Away&quot;)&#xA;46. Animal House (becomes &quot;Animal Hose&quot;)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Funnily enough, for #39 I actually did find a movie in 2000 named &quot;Lost Cat&quot;.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-09-18T23:29:49.667" LastActivityDate="2015-09-18T23:29:49.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22385" PostTypeId="2" ParentId="22376" CreationDate="2015-09-18T23:27:08.017" Score="3" Body="&lt;p&gt;Okay, this is even more far fetched:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A hook used as a prosthetic left arm&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be quite bright but I'm never right.&#xA;I support you in peace and protect you in fight.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Can be shiny, but it's still for the left (not the right).  Used both for peace, and can be an effective defensive weapon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm seldom alone and when I am, there's no use.&#xA;Even when you select me, another you choose.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Normally you've got a right arm.  When you use your arm it's normally to pick up something else.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You never ask how I'm doing, even when I'm blue.&#xA;I might be unhealthy but I can't catch a flu.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You don't talk to your prosthetic, even if it's blue.  It could rust or weaken, but it's not the flu.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I'm in transit and sometimes home I'll stay.&#xA;Sometimes you can see a person who for me will pray.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You don't always wear me.  Could be posted ahead to avoid having to take it through airport security.  If someone doesn't have one, but is missing an arm, they might ask for one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you don't have me at all, you're poor not having a dime&#xA;Or you might simply have fallen as a victim of crime.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A prosthetic might be one of the first things people get, unless someone's stolen it from you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can hurt you and wound you and even make you bleed.&#xA;Regardless of your pain, I'm something you still need.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hooks could be dangerous, but still useful.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might be unwanted but also an object of desire.&#xA;I could be damaged by earth, water and fire.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Presumably people would prefer not to need a hook, but they need it anyway.  It's not impervious to damage.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I could be expensive or cheap or even free.&#xA;When people look at you, they often see me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Various levels of hook exist, from ones made from scrap to fancy ones.  But regardless, when people see you, it's still quite obvious that you wear a hook.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'll be both arrogant and humble if I'm put on your head.&#xA;I often bear a name that someone else before me had.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Okay, this answer seems to fall apart now...  I didn't have much hope anyway.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15942" LastActivityDate="2015-09-18T23:27:08.017" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22386" PostTypeId="1" AcceptedAnswerId="22388" CreationDate="2015-09-18T23:54:49.170" Score="15" ViewCount="1534" Body="&lt;p&gt;Each of the following phrases refers to a specific city on Earth. You'll find out what I mean when you get the answers.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;In which city...&lt;/h1&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;...is there a giant blockade to prevent small rodents from flooding a river?&lt;/li&gt;&#xA;&lt;li&gt;...can a 30-year-old male get his skin irradiated while wearing a baseball cap?&lt;/li&gt;&#xA;&lt;li&gt;...does the phrase &quot;my friend&quot; get translated into two different languages at once?&lt;/li&gt;&#xA;&lt;li&gt;...is a hole in your skin that sweats considered a piece of music?&lt;/li&gt;&#xA;&lt;li&gt;...does everyone dress like a superhero, a vampire, or a witch?&lt;/li&gt;&#xA;&lt;li&gt;...do sounds quickly become more intense by a factor of 10?&lt;/li&gt;&#xA;&lt;li&gt;...are all the sharp blades plated with gold, or at least something of that colour?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-09-19T02:28:54.433" Title="In which cities do the following puns apply?" Tags="|wordplay|geography|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22387" PostTypeId="2" ParentId="22382" CreationDate="2015-09-19T00:01:34.147" Score="5" Body="&lt;p&gt;Can someone point me to the wiki so that I can add these as I find them? I will be updating as I find more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5: The fish that spawned all the others.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Source Code/Cod&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;8: That's a lot of fuss to make about some shorthand.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Much Ado About Nothing/Noting&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;11:A tale of the outer layer of a cereal grain.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Monty Python's Life of Brian/Bran&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;21: It's the bloke off Top Gear.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Sting/Stig&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;29: Not many ministers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Few Good/God Men&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;31: These mass baptisms are really noisy, but the minister furthest from the sunrise has everything under control.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All Quiet on the Western Front/Font&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;33: Knuckle gore.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First/Fist Blood&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;37: Her only coin.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Scent/Cent of a Woman&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16394" LastEditorUserId="16394" LastEditDate="2015-09-19T00:08:39.010" LastActivityDate="2015-09-19T00:08:39.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22388" PostTypeId="2" ParentId="22386" CreationDate="2015-09-19T00:51:12.987" Score="12" Body="&lt;p&gt;The cities are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Amsterdam ((h)amster + dam)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Manhattan (man (in a) hat (gets a) tan) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Miami (mi + ami; mi=spanish for my, ami=french for friend)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Singapore (sing a pore)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cape Town (legislative capital of South Africa; superheros, vampires and witches all wear capes)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Belfast (bel = factor of ten on log scale for sound level)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yellowknife (capital of Northwest Territories, Canada)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="14080" LastEditorUserId="2627" LastEditDate="2015-09-19T02:28:54.433" LastActivityDate="2015-09-19T02:28:54.433" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="22389" PostTypeId="2" ParentId="22376" CreationDate="2015-09-19T01:01:04.780" Score="2" Body="&lt;p&gt;Here is my guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can be quite bright but I'm never right.&#xA;I support you in peace and protect you in fight.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Energy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm seldom alone and when I am, there's no use.&#xA;Even when you select me, another you choose.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Scissor(s) blade&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You never ask how I'm doing, even when I'm blue.&#xA;I might be unhealthy but I can't catch a flu.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sky or River&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I'm in transit and sometimes home I'll stay.&#xA;Sometimes you can see a person who for me will pray.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (Shooting) Star&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you don't have me at all, you're poor not having a dime&#xA;Or you might simply have fallen as a victim of crime.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wallet or Time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can hurt you and wound you and even make you bleed.&#xA;Regardless of your pain, I'm something you still need.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Knife&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might be unwanted but also an object of desire.&#xA;I could be damaged by earth, water and fire.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Skin&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I could be expensive or cheap or even free.&#xA;When people look at you, they often see me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Smile&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'll be both arrogant and humble if I'm put on your head.&#xA;I often bear a name that someone else before me had.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Crown or Title&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Power&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9892" LastActivityDate="2015-09-19T01:01:04.780" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22390" PostTypeId="1" AcceptedAnswerId="22391" CreationDate="2015-09-19T03:24:49.093" Score="-3" ViewCount="377" Body="&lt;p&gt;I was raised this question in my PSA exam and I am still not able to find the answer.The question reads::&lt;br&gt;&#xA;Find the number to replace the &quot;???&quot; in the given box and the trend used to find it either row-wise or column-wise?? Options were::&lt;br&gt;&#xA;a)2765&lt;br&gt;&#xA;b)3125&lt;br&gt;&#xA;c)8796&lt;br&gt;&#xA;d)30008&lt;img src=&quot;https://i.stack.imgur.com/epuVa.png&quot; alt=&quot;Here is the BOX&quot;&gt;&gt;&lt;/p&gt;&#xA;" OwnerUserId="13652" LastEditorUserId="4421" LastEditDate="2015-11-25T18:00:17.440" LastActivityDate="2015-11-25T18:00:17.440" Title="Find the number and the trend?" Tags="|pattern|number-sequence|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22391" PostTypeId="2" ParentId="22390" CreationDate="2015-09-19T03:49:16.503" Score="4" Body="&lt;p&gt;The number in the third column is $(m+n)^{(m+n)}$, where $m$ is the number in the first column and $n$ is the number in the second column..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is $(3+2)^{(3+2)} = 5^5 = 3125$.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-09-19T03:49:16.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22393" PostTypeId="2" ParentId="22378" CreationDate="2015-09-19T04:38:46.517" Score="4" Body="&lt;p&gt;The $100 One Of Us clue is...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; me! DEVIOUS is an anagram of DEUSOVI. (That's actually the origin of the username. :D )&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The $200 Code clue is...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;EXPRESSION (anagram of ISOPRENES + X)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="11876" LastEditDate="2015-09-19T04:45:21.160" LastActivityDate="2015-09-19T04:45:21.160" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22394" PostTypeId="1" CreationDate="2015-09-19T04:55:17.783" Score="9" ViewCount="1158" Body="&lt;p&gt;What's a four letter word that sounds exactly the same if you take away the first and last letters?&lt;/p&gt;&#xA;" OwnerUserId="16426" LastEditorUserId="11876" LastEditDate="2015-09-19T05:01:03.207" LastActivityDate="2015-09-19T12:30:37.360" Title="Four Letter Word" Tags="|word|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22395" PostTypeId="2" ParentId="22394" CreationDate="2015-09-19T04:58:16.753" Score="30" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Know - &gt;no&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;${} {} {} {} $&lt;/p&gt;&#xA;" OwnerUserId="15942" LastActivityDate="2015-09-19T04:58:16.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22396" PostTypeId="1" AcceptedAnswerId="22403" CreationDate="2015-09-19T07:41:19.027" Score="12" ViewCount="494" Body="&lt;p&gt;After my &lt;a href=&quot;https://puzzling.stackexchange.com/questions/22382/ive-seen-these-movies-countless-ties&quot;&gt;last attempt&lt;/a&gt; to compile a movie list, we replaced Paul's keyboard and tried again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once again I dictated a list of my favourite movies to Paul and he typed them up. He then gave the list to Alice to describe the titles. I looked at Alice's list and it still wasn't right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then it became apparent that the new keyboard Paul had used was inserting extra random letters this time. Every item on my list had picked up one additional letter. Of course the spellchecker didn't pick up on it, because every title still made sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've reproduced Alice's new list here with the year of the film and the extra letter.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;ol&gt;&lt;li&gt;B (1925): Sweeping with style.&lt;/li&gt;&#xA;&lt;li&gt;B (1967): Young Skywalker tries to program his prosthetic.&lt;/li&gt;&#xA;&lt;li&gt;B (1975): Was that a femur that you tossed over that tree? You almost hit that bird.&lt;/li&gt;&#xA;&lt;li&gt;B (1981): All our cork trees have been stripped.&lt;/li&gt;&#xA;&lt;li&gt;B (1988): Neurosurgeon.&lt;/li&gt;&#xA;&lt;li&gt;B (1993): Furniture north of Manhattan.&lt;/li&gt;&#xA;&lt;li&gt;C (1988): You will chop these vegetables until your hands hurt.&lt;/li&gt;&#xA;&lt;li&gt;E (1935): I've soaked that pot almost 40 times and it still hasn't come clean!&lt;/li&gt;&#xA;&lt;li&gt;E (1984): Use it to cover the hole left by a lumbar puncture.&lt;/li&gt;&#xA;&lt;li&gt;F (1996): A girl's dress.&lt;/li&gt;&#xA;&lt;li&gt;G (1987): A royal viaduct.&lt;/li&gt;&#xA;&lt;li&gt;G (2012): Babe's biography.&lt;/li&gt;&#xA;&lt;li&gt;H (1991): You fell asleep in the fireplace again, didn't you?&lt;/li&gt;&#xA;&lt;li&gt;L (1966): Saints spend time with alopecia sufferers and their unattractive friends.&lt;/li&gt;&#xA;&lt;li&gt;L (2008): Why can't we have REAL bullets for this mission?&lt;/li&gt;&#xA;&lt;li&gt;L (2011): We mustn't avoid a discussion about nonnegative temperature designations.&lt;/li&gt;&#xA;&lt;li&gt;M (1997): A military assault by rodents.&lt;/li&gt;&#xA;&lt;li&gt;N (1946): The search engine is down for maintenance.&lt;/li&gt;&#xA;&lt;li&gt;N (1980): Calculating distances the bovine way.&lt;/li&gt;&#xA;&lt;li&gt;N (1992): Old boys and their ground beef&lt;/li&gt;&#xA;&lt;li&gt;N (2010): Those women are trustworthy, but their husband's aren't.&lt;/li&gt;&#xA;&lt;li&gt;O (1946): Our oily fruit are in their prime.&lt;/li&gt;&#xA;&lt;li&gt;O (1949): Grumble, grumble, GRUMBLE.&lt;/li&gt;&#xA;&lt;li&gt;O (1958): Why did you leave your parka up there? The sun has been shining on that metal all day.&lt;/li&gt;&#xA;&lt;li&gt;P (1957): She loves me, loves me not, loves me, loves me not, loves me, loves me not, LOVES ME. Wait - you're supposed to do that with the petals.&lt;/li&gt;&#xA;&lt;li&gt;P (1986): Find a title for my flowing text.&lt;/li&gt;&#xA;&lt;li&gt;R (1956): We've been overrun by people who want to kidnap Amity's police chief.&lt;/li&gt;&#xA;&lt;li&gt;R (1961): They pull strange faces on Leros.&lt;/li&gt;&#xA;&lt;li&gt;R (1980): An amphibian&lt;/li&gt;&#xA;&lt;li&gt;R (1984): It's harvest time near Hollywood.&lt;/li&gt;&#xA;&lt;li&gt;R (1990): Marge, Bart, Lisa and Maggie have all gone out.&lt;/li&gt;&#xA;&lt;li&gt;R (1991): Pickled flower phobia&lt;/li&gt;&#xA;&lt;li&gt;R (1996): Let's go get the shrimp now.&lt;/li&gt;&#xA;&lt;li&gt;S (1955): Moving trains from dusk to dawn.&lt;/li&gt;&#xA;&lt;li&gt;S (1979): Nuclear winter.&lt;/li&gt;&#xA;&lt;li&gt;S (1992): Let's play &quot;Fortune Tellers&quot;.&lt;/li&gt;&#xA;&lt;li&gt;S (1993): Prehistoric glimmer.&lt;/li&gt;&#xA;&lt;li&gt;S (2012): You can't have any more samples of material.&lt;/li&gt;&#xA;&lt;li&gt;T (1960): Mr Spielberg is fantastic.&lt;/li&gt;&#xA;&lt;li&gt;T (2005): An interplanetary verruca.&lt;/li&gt;&#xA;&lt;li&gt;U (1976): I'll eat every dish you prepared for the Obamas.&lt;/li&gt;&#xA;&lt;li&gt;W (1995): An amphibian&lt;/li&gt;&#xA;&lt;li&gt;W (2008): Pay the umpire to fix the match.&lt;/li&gt;&#xA;&lt;li&gt;X (1973): Basket-weaving a cat with no tail.&lt;/li&gt;&#xA;&lt;li&gt;Y (1941): An American rapper gets his passport.&lt;/li&gt;&lt;/ol&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Can you work out the titles as Alice saw them, and how they should have been?&lt;/p&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="52483" LastEditDate="2018-11-22T20:05:13.707" LastActivityDate="2018-11-26T05:54:07.167" Title="Morse of the films that I've watched a lot" Tags="|wordplay|movies|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22397" PostTypeId="2" ParentId="22378" CreationDate="2015-09-19T07:53:36.237" Score="3" Body="&lt;p&gt;Code-Dependent&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$100&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; IDENTIFIER - DENT (dimple) between two I's (eyes), anagram of FIRE (peculiar flames) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$500&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LITERAL - Anagram of IRATE (Angry? Upset?) between two L's (between the lines) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One Of Us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$300&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ALCONJA - I'll conjure (summon something) with a New York accent &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;O, The Places You'll Go&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$100&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; OSLO - Anagram of (messy) LOOS (bathrooms) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$200&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; OMAN - Anagram of (about) MOAN (complain)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$300&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ONTARIO - Anagram of (freely) TRAIN (practice) between O and O (rounds) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fly Me A River&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$200&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; KINGFISHER - KING (royal) FISHER (weaselly carnivore) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$300&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SANDPIPER - dice PIP (small indent) inside (in) SANDER (power tool)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$400&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CORMORANT - C sounds like sea (hear the ocean?), anagram of (carelessly) MOOR (dock),  RANT (yell violently) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An Unreal Vacation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$400&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; AVALON - Anagram of (breaks) NAVAL sounds like navel (In audition, belly button) and O (ring) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12277" LastEditorUserId="12277" LastEditDate="2015-09-21T12:47:15.883" LastActivityDate="2015-09-21T12:47:15.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22400" PostTypeId="2" ParentId="22378" CreationDate="2015-09-19T10:48:55.530" Score="1" Body="&lt;p&gt;O, The Places You'll Go&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$500&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unit of power wrapped around and around&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ottawa =&gt; O-Watt-a -&gt; @ -&gt; at&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Since I am unsure. Hint for others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$400&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Blended egg drink&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some place possibly an anagram or similar to &quot;Omelet&quot; or &quot;Omlette&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9892" LastEditorUserId="4385" LastEditDate="2015-09-21T23:00:55.483" LastActivityDate="2015-09-21T23:00:55.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22401" PostTypeId="2" ParentId="22394" CreationDate="2015-09-19T12:17:47.697" Score="16" Body="&lt;p&gt;How about:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ENVY&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you take away the first and last letters, you're left with&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; NV, which you would pronounce by the individual letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="388" LastActivityDate="2015-09-19T12:17:47.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22402" PostTypeId="2" ParentId="22394" CreationDate="2015-09-19T12:30:37.360" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; KNEE and NE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;are both pronounced&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; N IY1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;according to the &lt;a href=&quot;http://www.speech.cs.cmu.edu/cgi-bin/cmudict&quot;&gt;CMU Pronouncing Dictionary&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="228" LastActivityDate="2015-09-19T12:30:37.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22403" PostTypeId="2" ParentId="22396" CreationDate="2015-09-19T13:02:11.763" Score="9" Body="&lt;ol&gt;&#xA;&lt;li&gt;B (1925): Sweeping with style.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;The Gold (B)rush&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;B (1967): Young Skywalker tries to program his prosthetic.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;Co(b)ol Hand Luke&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;B (1975): Was that a femur that you tossed over that tree? You almost hit that bird.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;(B)one Flew Over the Cuckoo's Nest&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;B (1981): All our cork trees have been stripped.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;Raiders of the Lost (B)ark&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;B (1988): Neurosurgeon.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;(B)rain Man&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;B (1993): Furniture north of Manhattan.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;A Bronx Ta(b)le&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;C (1988): You will chop these vegetables until your hands hurt.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;Di(c)e Hard&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;E (1935): I've soaked that pot almost 40 times and it still hasn't come clean!&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;The 39 St(e)eps&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;E (1984): Use it to cover the hole left by a lumbar puncture.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;This Is Spinal Tap(e)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;F (1996): A girl's dress.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;The (F)rock&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;G (1987): A royal viaduct.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;The Princess Brid(g)e&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;G (2012): Babe's biography.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;Life of Pi(g)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;H (1991): You fell asleep in the fireplace again, didn't you?&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;Night on (H)earth&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;L (1966): Saints spend time with alopecia sufferers and their unattractive friends.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;The Good, the Ba(l)d, and the Ugly&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;L (2008): Why can't we have REAL bullets for this mission?&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;The B(l)ank Job&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;L (2011): We mustn't avoid a discussion about nonnegative temperature designations.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;We Need to Talk About Ke(l)vin&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;M (1997): A military assault by rodents.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;The (M)ice Storm&lt;/strong&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;N (1946): The search engine is down for maintenance.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;The Bi(n)g Sleep&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;N (1980): Calculating distances the bovine way.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;Ra(n)ging Bull&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;N (1992): Old boys and their ground beef.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;Of Mi(n)ce and Men&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;N (2010): Those women are trustworthy, but their husband's aren't.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;Despicable Me(n)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;O (1946): Our oily fruit are in their prime.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;The Best Years of Our (O)lives&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;O (1949): Grumble, grumble, GRUMBLE.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;The Third M(o)an&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;O (1958): Why did you leave your parka up there? The sun has been shining on that metal all day.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;C(o)at on a Hot Tin Roof&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;P (1957): She loves me, loves me not, loves me, loves me not, loves me, loves me not, LOVES ME. Wait - you're supposed to do that with the petals.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;The Seventh Se(p)al&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;P (1986): Find a title for my flowing text.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;The Name of the (P)rose&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;R (1956): We've been overrun by people who want to kidnap Amity's police chief.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;Invasion of the B(r)ody Snatchers&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;R (1961): They pull strange faces on Leros.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;The Gu(r)ns of Navarone&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;R (1980): An amphibian.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;The F(r)og&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;R (1984): It's harvest time near Hollywood.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;Beverly Hills C(r)op&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;R (1990): Marge, Bart, Lisa and Maggie have all gone out.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;Home(r) Alone&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;R (1991): Pickled flower phobia.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;Cape(r) Fear&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;R (1996): Let's go get the shrimp now.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;A Time to K(r)ill&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;S (1955): Moving trains from dusk to dawn.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;The Night of the (S)hunter&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;S (1979): Nuclear winter.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;Apocalypse (S)now&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;S (1992): Let's play &quot;Fortune Tellers&quot;.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;The (S)crying Game&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;S (1993): Prehistoric glimmer.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;Jurassic (S)park&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;S (2012): You can't have any more samples of material.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;End of (S)watch&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;T (1960): Mr Spielberg is fantastic.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;The Magnificent S(t)even&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;T (2005): An interplanetary verruca.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;War(t) of the Worlds&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;U (1976): I'll eat every dish you prepared for the Obamas.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;All the President's Men(u)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;W (1995): An amphibian.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;The Ne(w)t&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;W (2008): Pay the umpire to fix the match.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;Let the Right One (W)in&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;X (1973): Basket-weaving a cat with no tail.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;The Wicker Man(x)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;Y (1941): An American rapper gets his passport.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;Citizen Kan(y)e&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="14135" LastEditDate="2018-11-26T05:54:07.167" LastActivityDate="2018-11-26T05:54:07.167" CommentCount="5" CommunityOwnedDate="2015-09-19T13:02:11.763" ContentLicense="CC BY-SA 4.0" />
  <row Id="22405" PostTypeId="2" ParentId="22378" CreationDate="2015-09-19T14:19:21.020" Score="0" Body="&lt;p&gt;My guess for&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$400:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    Small dekagram atom -&gt; User &lt;a href=&quot;https://puzzling.stackexchange.com/users/226/smallatom&quot;&gt;smallatom&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Code-Dependent&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$500&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Angry? Upset? Read between the lines can't help but thing -&gt; Extracting &quot;Rage&quot; from (Angry Upset) -&gt; It must be HULK.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15732" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2015-09-20T23:58:39.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22406" PostTypeId="2" ParentId="22376" CreationDate="2015-09-19T14:46:10.017" Score="6" Body="&lt;p&gt;Okay, I think I've got it:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; A left shoe&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I can be quite bright but I'm never right.&#xA;I support you in peace and protect you in fight.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Shoes can be flashy.  Left shoe isn't right.  Supports you, and you can kick people with it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I'm seldom alone and when I am, there's no use.&#xA;Even when you select me, another you choose.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Mostly useful in pairs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You never ask how I'm doing, even when I'm blue.&#xA;I might be unhealthy but I can't catch a flu.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Shoes can be blue, but you still don't inquire about their health.  And a shoe can be nasty!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Sometimes I'm in transit and sometimes home I'll stay.&#xA;Sometimes you can see a person who for me will pray.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; You went with them or leave them behind.  Some people are desperate for particular examples.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If you don't have me at all, you're poor not having a dime&#xA;Or you might simply have fallen as a victim of crime.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Even very poor people have shoes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I can hurt you and wound you and even make you bleed.&#xA;Regardless of your pain, I'm something you still need.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Shoes hurt you, and can make you bleed, but you still need them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I might be unwanted but also an object of desire.&#xA;I could be damaged by earth, water and fire.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Unwanted old shoe vs designer shoe.  Some are susceptible to damage from mud, water or fire.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I could be expensive or cheap or even free.&#xA;When people look at you, they often see me.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Range of shoes available.  Put your best foot forward.  Shoes are often noticed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You'll be both arrogant and humble if I'm put on your head.&#xA;I often bear a name that someone else before me had.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Shoes sometimes have the name of their designer.  A shoe on the head could be a sign of madness or a very eccentric hat such as an aristocrat might wear.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2022-05-05T16:06:19.230" LastActivityDate="2022-05-05T16:06:19.230" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="22407" PostTypeId="2" ParentId="15437" CreationDate="2015-09-19T17:20:18.180" Score="-4" Body="&lt;p&gt;My first was answered by @A.D.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A colon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My second was answered by @GentlePurpleRain in the comments&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A hyphen (a high fin)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My third was technically half-answered by @A.D&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A tilde then re-moved (or shifted) is a &quot;`&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fourth was answered by @A.D.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A hyphen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fifth was also answered by @A.D.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A dash&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My sixth was answered by @Leslie P.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A hyphen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My seventh was not answered&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's a hyphen... it can be used like a pair of commas in a phrase within a sentence or in the same sense with parentheses&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My eighth was answered by @frodoskywalker&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; End Parenthesis&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My all is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; :-`----)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="167" LastEditorUserId="167" LastEditDate="2015-09-19T17:50:19.173" LastActivityDate="2015-09-19T17:50:19.173" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22408" PostTypeId="2" ParentId="22378" CreationDate="2015-09-19T17:31:48.490" Score="4" Body="&lt;p&gt;Code-Dependent&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$300  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; particle = ion; goes back-to-back = recurs --&gt; RECURSION  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$400&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; anagram (ill) of &quot;testament&quot; (will) : STATEMENT&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;O, The Places You'll Go&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$400&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;S&gt;OSAKA - 'blend' of O ('egg') and SAKE (drink)&lt;/s&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fly Me A River&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$100&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Break a writing utensil : anagram for &quot;a pencil&quot; : PELICAN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$500&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jessica and Mr. Rick finally get together : charade : alba + ross + t (last letter in &quot;get&quot; = ALBATROSS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An Unreal Vacation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$100&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'ot' (first letters of 'off' and 'the') placed after 'camel' ( ungulate's back) : CAMELOT&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$200&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Atlas (titan) contains (holds) anagram of 'tin' (muddled metal) : ATLANTIS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$500&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pa (Dad) + it x2 (&quot;echoes&quot;) + I (me) : PAITITI&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4385" LastEditorUserId="4385" LastEditDate="2015-09-22T14:25:30.683" LastActivityDate="2015-09-22T14:25:30.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22409" PostTypeId="1" AcceptedAnswerId="22410" CreationDate="2015-09-19T19:16:39.220" Score="28" ViewCount="1419" Body="&lt;p&gt;Cosmo complains: &quot;At the party yesterday at our place, some guys were smoking in our living room. This morning we detected that these careless smokers had burnt four holes into the carpet.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fredo: &quot;What a shame!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cosmo continues: &quot;The carpet is a square of side length 2.75 meters. The cigaret holes are tiny, essentially point-shaped, but clearly visible. We shall put the carpet to the garbage.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fredo says: &quot;But you could also cut the carpet down to a smaller square of side length 1 meters.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cosmo: &quot;I am not sure that this is possible. The cigaret holes are badly distributed.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fredo: &quot;It definitely is possible to cut the carpet down to a square of side length 1 meters.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Question: Is Fredo right?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-09-20T02:48:54.643" Title="Careless smokers" Tags="|mathematics|geometry|dissection|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22410" PostTypeId="2" ParentId="22409" CreationDate="2015-09-19T19:22:45.760" Score="35" Body="&lt;p&gt;Yes, it's possible, because one can fit 5 disjoint 1x1 squares in a 2.75x2.75 square: four in the corners, and one in the center rotated 45 degrees. The four cigaret holes can't eliminate all 5 squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ofeID.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ofeID.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tilted square fits in a cross whose center's square diagonal has a of length 1. So the cross needs width $1/ \sqrt{2} \approx 0.707 &amp;lt; 0.75$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the smallest square table with this property has sides length $2 + 1/ \sqrt{2}$. A smaller table might not work because putting cigaret burns just inside the four inside corners of the corner 1x1 squares eliminates all 1x1 subsquares.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2015-09-19T19:42:31.657" LastActivityDate="2015-09-19T19:42:31.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22411" PostTypeId="1" AcceptedAnswerId="22414" CreationDate="2015-09-19T21:01:23.807" Score="10" ViewCount="1154" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The is a large cube formed by gluing together 27 smaller cubes of uniform size (see figure). A termite starts at the center of a face of any of the outside cubes and bores a path that takes him once through each cube. His movement is always parallel to a side of the large cube, never diagonal.&lt;br&gt;&#xA;  Is it possible for the termite to bore through each of the 26 outside cubes once and only once, then complete his journey by entering the central cube for the first time?&lt;br&gt;&#xA;  If possible, show how; if not, prove it.&#xA;  &lt;a href=&quot;https://i.stack.imgur.com/0wcQj.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0wcQj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This puzzle is from Martin Garner's &lt;em&gt;The Colossal Book of Short Puzzles and Problems&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope the problem is not too &lt;em&gt;boring&lt;/em&gt; (ahem) for members of this forum.&lt;/p&gt;&#xA;" OwnerUserId="14080" LastActivityDate="2015-09-19T22:53:10.390" Title="Termite eating through a large cube composed of 27 smaller cubes while not moving diagonally" Tags="|geometry|combinatorics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22412" PostTypeId="2" ParentId="22411" CreationDate="2015-09-19T21:18:39.543" Score="5" Body="&lt;p&gt;EDIT: Nevermind, misunderstood the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Color the cubelets so that the corners are black. There are 14 black and 13 white cubelets. At move 1, the termite is on a black cubelet, and the termite alternates between black and white for each move. Therefore at move 27, the termite will again be on a black cubelet, but the center is white. Thus the termite cannot complete such a journey.&lt;/p&gt;&#xA;" OwnerUserId="11100" LastEditorUserId="11100" LastEditDate="2015-09-19T21:45:00.587" LastActivityDate="2015-09-19T21:45:00.587" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22413" PostTypeId="1" AcceptedAnswerId="22431" CreationDate="2015-09-19T22:11:48.387" Score="8" ViewCount="485" Body="&lt;p&gt;After spending half of the night with important calculations Professor Delvershy arrived late this morning. To his dismay he discovered that one of his students used the unexpected spare time to transform his calculations to modern art:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Nz0Lb.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fortunately the student always used the same symbol for the same digit. Can you help the Professor to restore his calculations?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Text version for copying:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ABC + DEC = FFA&#xA;  -     -     -&#xA; DB + BED = GHI&#xA;  =     =     =&#xA;JCE + GHJ = EFG&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="15632" LastActivityDate="2017-12-27T04:52:46.170" Title="Professor Delvershys calculations" Tags="|mathematics|logical-deduction|calculation-puzzle|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22414" PostTypeId="2" ParentId="22411" CreationDate="2015-09-19T22:12:37.980" Score="8" Body="&lt;p&gt;The termite is doomed to fail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 8 corner cubes, 12 edge cubes, 6 middle cubes, and 1 center cube. Since the last cube must be the center cube, the second to last cube must be a middle cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Middle cubes and Corner cubes behave roughly the same way. They can't link to each other, but they can both link to an Edge cubes. Because of this, we will refer to Middle cubes and Corner cubes with MC and Edge cubes with E.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any path including MCs and Es will need to alternate between the two because Es and MCs can't link to themselves. We have 14 MCs and 12 Es, however, so no matter how you figure it there aren't enough Es to link together all the MCs. You would need at least 13. Sorry, termite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to Artur Kirkoryan for coming up with a way to simplify my proof and make it much more understandable.&lt;/p&gt;&#xA;" OwnerUserId="16394" LastEditorUserId="16394" LastEditDate="2015-09-19T22:53:10.390" LastActivityDate="2015-09-19T22:53:10.390" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22415" PostTypeId="1" CreationDate="2015-09-19T22:29:22.893" Score="4" ViewCount="645" Body="&lt;p&gt;Each clue references a movie. Take all of the clues, combine them, and solve that clue to get you one final movie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where did the literally animated little guy and his family come from?&lt;br&gt;&#xA;What type was 2 less than the first 23rd?&lt;br&gt;&#xA;Who was the botanist that shook the earth with his finger?&lt;br&gt;&#xA;TECHNICOLOR! without the earth or the obviousness.&lt;br&gt;&#xA;The aliens are almost extinct.&lt;br&gt;&#xA;'ey you, hand, lose the glasses!&lt;br&gt;&#xA;2 more and then...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck!&lt;/p&gt;&#xA;" OwnerUserId="167" LastEditorUserId="167" LastEditDate="2015-09-21T19:56:15.783" LastActivityDate="2015-09-21T19:56:15.783" Title="Movies in movies" Tags="|riddle|movies|" AnswerCount="1" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22417" PostTypeId="1" CreationDate="2015-09-19T23:01:50.147" Score="9" ViewCount="879" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If you wish to leave, then there's no doubt&lt;br&gt;&#xA;  first you must look for the way out.&lt;br&gt;&#xA;  Then with open eyes you must stare;&lt;br&gt;&#xA;  just look down to meet my glare.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;To remember this ordeal,&lt;br&gt;&#xA;  to take my thoughts and make them real,&lt;br&gt;&#xA;  then at the bottom, beside my eyes,&lt;br&gt;&#xA;  tell me where the treasure lies.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If you'd rather forget your strife&lt;br&gt;&#xA;  and quietly, quickly end my life&lt;br&gt;&#xA;  get in line, but do not pout,&lt;br&gt;&#xA;  Instead you must scream and shout.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Either way you're in the clear,&lt;br&gt;&#xA;  but there's one thing you must do from here:&lt;br&gt;&#xA;  say no more, come back inside,&lt;br&gt;&#xA;  and I'll let you off this crazy ride!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is this poem describing?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="16440" LastEditorUserId="52483" LastEditDate="2018-11-22T20:01:00.487" LastActivityDate="2018-11-23T10:44:56.723" Title="How to leave me behind" Tags="|riddle|rhyme|" AnswerCount="4" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="22418" PostTypeId="1" CreationDate="2015-09-20T00:41:10.143" Score="1" ViewCount="391" Body="&lt;p&gt;Two players, A and B, are each given air horns. Also present is a device which will at some random point produce a noise. If the device produces a noise before either player, they die. If both players produce a noise before the device, the first to produce the noise is killed and the second one lives. If only one player produces a noise and the device sounds before the other player gets a chance, the player unable to produce a sound is killed. Is it necessarily so that if they are perfectly logical and selfish beings, they both shall die?&lt;/p&gt;&#xA;" OwnerUserId="16442" LastEditorUserId="16442" LastEditDate="2015-09-20T02:45:39.477" LastActivityDate="2015-09-21T14:49:55.407" Title="The Buzzer Game" Tags="|game|game-theory|" AnswerCount="7" CommentCount="1" ClosedDate="2015-09-22T18:40:34.663" ContentLicense="CC BY-SA 3.0" />
  <row Id="22420" PostTypeId="2" ParentId="22418" CreationDate="2015-09-20T01:50:59.943" Score="-1" Body="&lt;p&gt;&lt;strong&gt;The Devil's Advocate's Answer:&lt;/strong&gt;&lt;br&gt;&#xA;No, it is not &lt;em&gt;necessarily&lt;/em&gt; so. Many of the most logical people to have lived have believed in an afterlife. Someone could believe that sacrifices made in this life, especially an apparently altruistic act such as this, will be met with rewards in another. One could regard this sacrifice also as an act of selfishness, since they do it with the expectation of future reward or to confirm their sense of moral superiority. If they value these things above their own life, it is logical and selfish to sound one's own horn.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, if the death from not sounding a horn is grim and painful, one might consider sounding the horn if that death promises to be instant or painless.&lt;/p&gt;&#xA;" OwnerUserId="13237" LastActivityDate="2015-09-20T01:50:59.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22422" PostTypeId="2" ParentId="22418" CreationDate="2015-09-20T02:35:42.897" Score="0" Body="&lt;p&gt;It depends on what happens if they're indifferent. Assuming symmetry is broken somehow (because if it isn't, then they trivially both die), then one person will sound their airhorn first. Then, the second person will die either way, but they get to choose if the first one lives. The question doesn't say anything about what will happen then.&lt;/p&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2015-09-20T02:35:42.897" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22424" PostTypeId="1" AcceptedAnswerId="22426" CreationDate="2015-09-20T03:13:02.783" Score="20" ViewCount="915" Body="&lt;p&gt;There is an island near &lt;a href=&quot;https://puzzling.stackexchange.com/questions/tagged/blue-eyes&quot;&gt;the famous Blue Eyes island&lt;/a&gt; where 100 perfect logicians live, each with a number between 1 and 100 (repetitions allowed) tattooed on their forehead.  They know all the numbers of the other inhabitants except their own.  If ever one of them learns their own number, tradition dictates that they krill themselves.  That is, the next daily all-island meeting, at precisely 12:00 noon, they must dump a bucket of raw krill on their heads.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It so happens that half the inhabitants have the number $50$, while the other half have the number $51$.   One day, a travelling knight (truth-teller) is addressing all the inhabitants, and mentions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I was surprised to find only two distinct numbers among the people here, and that those two numbers were consecutive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Immediately, all the inhabitants deduce that they have either a $50$ or a $51$ on their forehead. But that is not narrowed down enough to commit &quot;sushi-cide&quot;.  Do any of the inhabitants end up krilling themselves, and if so when?&lt;/p&gt;&#xA;" OwnerUserId="2722" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2015-09-20T07:56:00.797" Title="Numbered Logicians near the Blue Eyes island" Tags="|logical-deduction|blue-eyes|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22425" PostTypeId="2" ParentId="22413" CreationDate="2015-09-20T03:23:39.847" Score="1" Body="&lt;p&gt;The professors calculations were:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;pre&gt;618  +  378  =  996&lt;br&gt;&#xA;  -       -       -&lt;br&gt;&#xA; 31  +  173  =  204&lt;br&gt;&#xA;  =       =       =&lt;br&gt;&#xA;587  +  205  =  792&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A: 6&lt;br&gt;&#xA;B: 1&lt;br&gt;&#xA;C: 8&lt;br&gt;&#xA;D: 3&lt;br&gt;&#xA;E: 7&lt;br&gt;&#xA;F: 9&lt;br&gt;&#xA;G: 2&lt;br&gt;&#xA;H: 0&lt;br&gt;&#xA;I: 4&lt;br&gt;&#xA;J: 5  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now I have to admit I bruteforced this solution, the only clever part I did to reduce the permutations from &lt;code&gt;10!&lt;/code&gt; to &lt;code&gt;9!*3&lt;/code&gt; was to realise that the only characters that don't start at the beginning of a number are &lt;code&gt;CHI&lt;/code&gt; and thus are the only candidates for &lt;code&gt;0&lt;/code&gt; assuming the professor doesn't write with leading zeros.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The code I used to bruteforce this for anyone who is JavaScript inclined is:&#xA;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;var variables = ['A','B','C','D','E','F','G','I','J'];&#xA;var assertions = [&#xA;  'ABC + DEC == FFA',&#xA;  'DB + BED == GHI',&#xA;  'JCE + GHJ == EFG',&#xA;  'ABC - DB == JCE',&#xA;  'DEC - BED == GHJ',&#xA;  'FFA - GHI == EFG'&#xA;].map(function(assertion) {&#xA;  expression = assertion&#xA;    .replace(/[A-J]+/g, 'Number($&amp;amp;)')&#xA;        .replace(/(?=([A-J])([A-J]))[A-J]/g, '$1+\'\'+')&#xA;    .replace(/[A-J]/g,'charMap[\'$&amp;amp;\']');&#xA;&#xA;  return new Function('charMap', 'return ' + expression + ';');&#xA;});&#xA;&#xA;function permute(arr, memo) {&#xA;  var result&#xA;    , cur&#xA;    , memo = memo || [];&#xA;&#xA;  for (var i = 0; i &amp;lt; arr.length; i++) {&#xA;    cur = arr.splice(i, 1);&#xA;&#xA;    if (arr.length === 0) {&#xA;      result = testPermutation(memo.concat(cur));&#xA;      if (result) return result;&#xA;    }&#xA;&#xA;    result = permute(arr.slice(), memo.concat(cur));&#xA;    if (result) return result;&#xA;&#xA;    arr.splice(i, 0, cur[0]);&#xA;  }&#xA;&#xA;  return null;&#xA;}&#xA;&#xA;function testPermutation(permutation) {&#xA;  var charMap = permutation.reduce(function(obj, variable, index) {&#xA;    obj[variable] = index + 1;&#xA;    return obj;&#xA;  }, {&#xA;    'H': 0&#xA;  });&#xA;&#xA;  return assertions.every(function(assertion) {&#xA;    return assertion(charMap);&#xA;  }) ? charMap : false;&#xA;}&#xA;&#xA;var combination = permute(variables);&#xA;console.log(combination);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="12875" LastEditorUserId="12875" LastEditDate="2015-09-20T12:13:23.400" LastActivityDate="2015-09-20T12:13:23.400" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22426" PostTypeId="2" ParentId="22424" CreationDate="2015-09-20T03:57:16.627" Score="12" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Yes&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One important difference between this problem and the blue-eyes problem is that here, even if there was one 50 and ninety-nine 51s, the person with the 50 would not know their own number (it could also be 52). This means that the crucial chain of deductions cannot start from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is a place to start this chain. If there was one 2 and ninety-nine 1s, then the person with the 2 would know their own number. So, if nothing happens after one day, it becomes common knowledge that this is not the distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there were two 2s and ninety-eight 1s, then both 2s would know their numbers after a day passed, and they would krill themselves on the second day. So if two days pass with no krillings, then it becomes common knowledge that this is not the distribution of numbers either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After 99 days pass, it becomes common knowledge that the numbers are not 1s and 2s. Then, if there was one 3 and ninety-nine 2s, the person with the 3 would know that they didn't have a 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After 198 days, it is common knowledge that the numbers are not 2s and 3s. After the 4851st (99*49) days 49s are ruled out. A similar chain of deductions starting at 100 has also ruled out 52 and higher numbers, so it is common knowledge that the numbers are 50 and 51.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, it takes 49 more days to eliminate all the possibilities except fifty 50s and fifty 51s. On the 4901st day, all of the islanders krill themselves.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-09-20T03:57:16.627" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="22427" PostTypeId="2" ParentId="22418" CreationDate="2015-09-20T05:31:56.020" Score="2" Body="&lt;p&gt;Since there is a random element, logic would seem to fall out of the equation. They are both going to try to survive, which means blowing their horn as soon as the other person blows theirs or blowing it right before the device sounds. Since human reaction time is quite fast, the odds of guessing when the device will sound accurately enough that the other person doesn't have time to respond are very low. So, then, the best bet is to wait for the other person to blow their horn, right? The hitch is, they are both perfectly logical so they know that the other one has come to the same conclusion. Then they will both know that waiting for the other to blow is certain death for all, so they will either blow their horns randomly or try to trick the other person into blowing their horn. In a situation like this, tension would be very high, and making a sudden movement might be enough to get the other person to blow first out of overactive reflexes. Assuming that they are not exactly the same person, despite being perfectly logical and selfish, one may succeed in tricking the other into blowing, or one may blow first in hopes that they are blowing just before the device goes off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They don't necessarily both have to die.&lt;/p&gt;&#xA;" OwnerUserId="16394" LastActivityDate="2015-09-20T05:31:56.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22429" PostTypeId="2" ParentId="22418" CreationDate="2015-09-20T06:20:40.550" Score="0" Body="&lt;p&gt;Firstly, the question did not resolve the situation when one player sounds the horn &lt;em&gt;during&lt;/em&gt; the time the device is sounding. Does that player die from being the first person to sound the airhorn, or not die (while the other dies) because he &quot;produced a noise and the device sounds before the other player gets a chance&quot;? If he doesn't die, we have a loophole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, the question does not specify if they are in the same room, and if they are, whether the device itself is present &lt;em&gt;in the same room&lt;/em&gt;. If they are in separate rooms and cannot see the device directly, we note the question does not specify if the players have heard what the device sounds like before (and when it does, where the sound comes from). Then similar to Warknall's answer, they can try tricking each other that by making it sound like the device has sounded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they are in the same room, the question does not specify what qualifies as a &quot;player making a noise&quot; means. I think the intended answer to this is making a sound from their respective assigned airhorns (otherwise they would have a hilariously silent tickle war). If so, then the logical thing to do is to grab the other player's horn and sound it as soon as possible. Alternatively, they can cooperate and try to destroy the device together, but I doubt that is an option within the framework of the question.&lt;/p&gt;&#xA;" OwnerUserId="15960" LastActivityDate="2015-09-20T06:20:40.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22430" PostTypeId="2" ParentId="22418" CreationDate="2015-09-20T07:02:55.353" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is not necessary that both die. As both are perfectly logical both know that waiting for the other to make a noise will kill them both. Therefore both can agree to adapt the rules to make one of them survive, assuming equal chances for both.&lt;br&gt;&lt;br&gt;One possibility is to prevent that both players can hear each others noise. For example by putting on headphones with loud music. Of course they both will make sure the other can't cheat, by checking the headphones before.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastActivityDate="2015-09-20T07:02:55.353" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22431" PostTypeId="2" ParentId="22413" CreationDate="2015-09-20T07:33:40.203" Score="7" Body="&lt;p&gt;Here's an analytic solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;code&gt;DEC - BED = GHJ&lt;/code&gt;, the middle digits imply either &lt;code&gt;H=0&lt;/code&gt; or &lt;code&gt;H=9&lt;/code&gt;. From &lt;code&gt;DB + BED = GHI&lt;/code&gt;, since &lt;code&gt;B&lt;/code&gt;$\neq$&lt;code&gt;G&lt;/code&gt; in the hundreds place, there must be a carry from the sum in the middle digits, possibly including a carry from the right digits. That is, &lt;code&gt;D+E(possibly +1)=10+H&lt;/code&gt; But since &lt;code&gt;D+E&lt;/code&gt; is at most &lt;code&gt;17&lt;/code&gt; (when &lt;code&gt;{D,E}={8,9}&lt;/code&gt;), &lt;code&gt;H&lt;/code&gt; cannot be 9, so &lt;strong&gt;&lt;code&gt;H=0&lt;/code&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;code&gt;FFA - GHI = EFG&lt;/code&gt;, since &lt;code&gt;H=0&lt;/code&gt;, we have the following.&#xA;$$F-G = E \tag{1}\label{1}$$&#xA;$$A-I = G \tag{2}\label{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;code&gt;DEC - BED = GHJ&lt;/code&gt;, since &lt;code&gt;H=0&lt;/code&gt;, we have the following.&#xA;$$C-D = J \tag{3}\label{3}$$&#xA;$$D-B = G \tag{4}\label{4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;code&gt;JCE + GHJ = EFG&lt;/code&gt;, since &lt;code&gt;H=0&lt;/code&gt; (and so &lt;code&gt;F&lt;/code&gt; $\neq$ &lt;code&gt;0&lt;/code&gt;), we have the following.&#xA;$$F = C+1 \tag{5}\label{5}$$&#xA;$$E+J = 10+G \tag{6}\label{6}$$&#xA;$$J+G = E \tag{7}\label{7}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting $E$ from $\eqref{7}$ into $\eqref{6}$, we get &lt;strong&gt;&lt;code&gt;J=5&lt;/code&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting $E$ from $\eqref{7}$ into $\eqref{1}$, we get&#xA;$$F = 5 + 2G \tag{8}\label{8}$$&#xA;from which we get 2 possibilities: $(G,F) \in \{(1,7),(2,9)\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equating $\eqref{5}$ and $\eqref{8}$, we get&#xA;$$C = 4 + 2G \tag{9}\label{9}$$&#xA;from which we get 2 possibilities: $(G,C) \in \{(1,6),(2,8)\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting $D=C-5$ from $\eqref{3}$ into $\eqref{4}$, we get&#xA;$$B = C - 5 - G \tag{10}\label{10}$$&#xA;into which we test the 2 possibilities for $(G,C)$ from $\eqref{9}$ to get $(G,C,B) \in \{(1,6,0), (2,8,1)\}$. Since &lt;code&gt;H=0&lt;/code&gt;, we discard the triple with $B=0$. So &lt;strong&gt;&lt;code&gt;B=1&lt;/code&gt;&lt;/strong&gt;, &lt;strong&gt;&lt;code&gt;C=8&lt;/code&gt;&lt;/strong&gt; and &lt;strong&gt;&lt;code&gt;G=2&lt;/code&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also get &lt;strong&gt;&lt;code&gt;F=9&lt;/code&gt;&lt;/strong&gt; from $\eqref{8}$, &lt;strong&gt;&lt;code&gt;E=7&lt;/code&gt;&lt;/strong&gt; from $\eqref{1}$, and &lt;strong&gt;&lt;code&gt;D=3&lt;/code&gt;&lt;/strong&gt; from $\eqref{3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leaves &lt;code&gt;A&lt;/code&gt; and &lt;code&gt;I&lt;/code&gt; to be matched with &lt;code&gt;4&lt;/code&gt; and &lt;code&gt;6&lt;/code&gt;. From $\eqref{2}$, we conclude that &lt;strong&gt;&lt;code&gt;A=6&lt;/code&gt;&lt;/strong&gt; and &lt;strong&gt;&lt;code&gt;I=4&lt;/code&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting it all together, we have&#xA;&lt;strong&gt;&lt;code&gt;ABCDEFGHIJ = 6183792045&lt;/code&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="9422" LastEditorUserId="9422" LastEditDate="2015-09-20T07:57:21.033" LastActivityDate="2015-09-20T07:57:21.033" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22432" PostTypeId="2" ParentId="22418" CreationDate="2015-09-20T08:24:31.780" Score="2" Body="&lt;p&gt;By the given parameters, A and B are both logical and selfish. Logically, there is no single-sided solution guaranteeing a logician's own survival - sounding the horn immediately has the same deadly effect as waiting indefinitely for the other logician's horn. The 'selfish' parameter is therefore disregarded as (logically) a 'don't care' condition. The remaining thing to consider is the survival of at least one of the logicians (in this case, the other logician), which each can effect by sounding their own horn first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both logicians follow the above logic and immediately sound their own horns. The possibility of survival arises when the killing device does not sound in less time than the logicians are physically capable of sounding their own horns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming identical reaction times, speed of thought, etc, the immediate simultaneous sounding by A and B means that none of the deadly conditions comes into play, so both logicians survive. If that assumption doesn't hold, then the slower logician survives.&lt;/p&gt;&#xA;" OwnerUserId="9422" LastActivityDate="2015-09-20T08:24:31.780" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22433" PostTypeId="2" ParentId="15209" CreationDate="2015-09-20T12:26:38.837" Score="0" Body="&lt;p&gt;My first thought is that the man was actually murdered. Most suicides try to make things quick, or something which cannot be easily reversed once set up. The block of ice answer gives the man a chance to undo the rope around the neck. Also, there seems to be no note, although the body was never searched. Given the information, it appears that the man was not fully examined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the man was drowned, which explains the damp ground directly beneath him. The man was weak, and therefore could have been easily overpowered by a man of average strength, so almost anyone could have killed him... as long as he was strong enough to lift that ladder. This also explains why the ladder would be outside. The murderer simply took it with him before shutting the barn door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as the barn being locked from the inside, depending on the type of lock, it could have been locked while the door was still open, and then shut from the outside. It happens to many bathroom doors in my family all the time. The fact that the door could be unlocked from the outside (how else was the body found?) might even suggest that a spare key for the lock might not be too hard to find.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the murderer simply drowned the man, took him to the barn, and hanged him (utilizing the ladder which was, later, placed outside). &lt;/p&gt;&#xA;" OwnerUserId="16451" LastActivityDate="2015-09-20T12:26:38.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22434" PostTypeId="2" ParentId="22342" CreationDate="2015-09-20T14:24:20.223" Score="0" Body="&lt;p&gt;Minimum 1 weight (best case) when you find two coins with the same weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Worst case 70 comparisons (worst case) if you have to weight all the wrong ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;If you're lucky enough to weight two coins that weights the same (good ones are all equal and fake ones are all different) then you resolved the problem. Each time you weight you just have to make sure to grab the lighter one. If you're unlucky enough to get 70 times the balance unbalanced, you will remain with the 29 other genuine coins.&lt;/p&gt;&#xA;" OwnerUserId="16453" LastEditorUserId="16453" LastEditDate="2015-09-20T22:23:30.367" LastActivityDate="2015-09-20T22:23:30.367" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22435" PostTypeId="1" AcceptedAnswerId="22437" CreationDate="2015-09-20T14:35:18.957" Score="4" ViewCount="1030" Body="&lt;p&gt;The Empty Rectangles method is a strategy for solving Sudoku puzzles. I have seen the description of this strategy &lt;a href=&quot;http://www.sudokuwiki.org/Empty_Rectangles&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, I have the following question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a problem if the other end of the strong link and the ERI intersect &lt;strong&gt;inside&lt;/strong&gt; the box which contains the empty rectangle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example if, in the 4th picture of the link above, Y was in A2 and Z in A8.&lt;/p&gt;&#xA;" OwnerUserId="16454" LastEditorUserId="2627" LastEditDate="2015-09-20T16:57:22.697" LastActivityDate="2015-09-20T16:57:22.697" Title="Empty Rectangles - Does the rule apply when the intersection takes place inside the box with the empty rectangle?" Tags="|sudoku|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22437" PostTypeId="2" ParentId="22435" CreationDate="2015-09-20T15:06:01.320" Score="2" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HNeZy.png&quot; alt=&quot;1]&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An Empty Rectangle means that in a box, a particular number is restricted to one row and one column. In this example on the linked page, the 8 in the top-right box is either in row C or column 8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either X or Y contains an 8. If Y contains the 8, then Z cannot. If X contains the 8, then the 8 in the top-right box cannot be in row C, so it must be in column 8. Then, Z also cannot contain 8. This is why this configuration allows Z to be eliminated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Y was at A2 and Z at A8 instead, this logic would not eliminate Z. If X is 8, then the 8 in the top-right box would have to be in column 8, but A8 is still a possible position for it.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-09-20T15:06:01.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22438" PostTypeId="1" AcceptedAnswerId="22439" CreationDate="2015-09-20T17:17:30.633" Score="16" ViewCount="1519" Body="&lt;p&gt;In the country of Dalgonia, there is only one type of fake coins and only one type of genuine coins. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;All genuine coins have the same weight.&lt;/li&gt;&#xA;&lt;li&gt;All fake coins have the same weight. &lt;/li&gt;&#xA;&lt;li&gt;Fake coins and genuine coins have different weights&lt;br&gt;&#xA;(but it is not known which of these two weights is larger).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Furthermore, the genuine coins have the following outrageous property: whenever somebody can logically deduce that a certain coin is genuine, this coin evaporates into thin air (and disappears forever from Dalgonia).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cosmo puts $N\ge3$ coins on the table and tells Fredo: &quot;Exactly $N-1$ of these coins are genuine and exactly one of them is fake.&quot; Then Cosmo leaves the room. On the table, there is a balance with two pans (but there are no weights).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; For which values of $N\ge3$ is Fredo able to identify the fake coin and to determine whether it is heavier or lighter than the genuine coins of Dalgonia?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-11-01T01:04:33.100" Title="Evaporating coins" Tags="|mathematics|strategy|weighing|" AnswerCount="5" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="22439" PostTypeId="2" ParentId="22438" CreationDate="2015-09-20T17:57:58.677" Score="13" Body="&lt;p&gt;You can guarantee finding and determining the weight of the fake coin for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;any even $N$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Any useful weighing will be of an even number of coins, with the same number on each side of the balance. If the result is balanced, all the coins must be genuine and so will immediately disappear. Therefore if you start with an odd number of coins, there will always be the possibility that all weighings are balanced, eliminating an even number of genuine coins each time, until only the fake coin is left. At this point the fake coin is identified, but its weight cannot be determined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you start with an even number of coins, put half on each side of the balance; the result must be unbalanced. Now take the heavier side and weigh individual coins against each other. Either a weighing will be unbalanced, in which case the heavier coin is the fake, or you will eliminate coins until one or none are left. If none are left, the fake is light and you can repeat the process for the lighter side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one coin is left, call it $H$ and take three coins $A,B,C$ from the lighter side. Weigh $HA$ vs $BC$.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If it balances, all are genuine and disappear; now the fake coin must be light, and you can weigh the other coins individually until you find it or it's the only remaining coin.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $HA$ is lighter, $A$ must be fake and light.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $HA$ is heavier, all coins other than those just weighed must be genuine and disappear. $A$ cannot be light or heavy, so it too is genuine and disappears. Now weigh $B$ and $C$ against each other: either one of them will be lighter than the other and fake, or they will balance and $H$ is fake and heavy. (Thanks to Julian Rosen for resolving this case.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thus you can always find and determine the relative weight of the fake coin.&lt;/p&gt;&#xA;" OwnerUserId="12511" LastEditorUserId="12511" LastEditDate="2015-09-20T20:35:01.270" LastActivityDate="2015-09-20T20:35:01.270" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="22443" PostTypeId="2" ParentId="20903" CreationDate="2015-09-21T02:46:29.737" Score="0" Body="&lt;p&gt;Based on the new hints, here is my second attempt:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;allegory&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A dream, not a dream&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Refers to the words &lt;strong&gt;analogy&lt;/strong&gt; and &lt;strong&gt;reality&lt;/strong&gt;. I believe &quot;analogy&quot; is dream-like in the sense of being hypothetical.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Tailless, tail-less&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The subtle difference in punctuation for &quot;tailless&quot; refers to two different ways of removing a &quot;tail&quot; from each of the words.&lt;br&gt;&#xA; &lt;strong&gt;logy&lt;/strong&gt; = removing &quot;anal&quot; (tail in the literal sense) from &quot;analogy&quot;&lt;br&gt;&#xA; &lt;strong&gt;real&lt;/strong&gt; = removing &quot;-ity&quot; (the suffix being the tail) from &quot;reality&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Puzzled? Makes quite the parable&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Scrambling &quot;logy + real&quot; gives us &lt;strong&gt;allegory&lt;/strong&gt;, which is a synonym for &quot;parable&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15960" LastActivityDate="2015-09-21T02:46:29.737" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22444" PostTypeId="2" ParentId="22438" CreationDate="2015-09-21T05:16:18.000" Score="-1" Body="&lt;h3&gt;Odd $N$&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Mark all the coins with $c_1, c_2,\ldots,c_N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick one coin, $c_k$. Weigh &lt;strong&gt;all &lt;em&gt;other&lt;/em&gt; coins&lt;/strong&gt; on the balance, putting half on either side.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the balance is equal, $c_k$ is fake.&lt;/li&gt;&#xA;&lt;li&gt;If the balance is unbalanced, $c_k$ is genuine and we have reduced our problem to even $N\geq 2$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5901" LastEditorUserId="5901" LastEditDate="2015-09-21T06:44:57.490" LastActivityDate="2015-09-21T06:44:57.490" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22445" PostTypeId="2" ParentId="20903" CreationDate="2015-09-21T06:09:12.213" Score="3" Body="&lt;p&gt;I think @Hackiisan is right but for the wrong reason. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Allegory &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A dream&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A goal&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not a dream, tailless, tail-less&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not a dream = reality.  Remove &quot;tai&quot; (a tailless tail) from reality leaves &quot;rely&quot; (So &quot;Not a dream, tailless, tail-less&quot; means &quot;remove tai from reality&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;puzzled?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; anagram of &quot;goal rely&quot; gives you allegory&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;makes quite a parable&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; allegory&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2015-09-21T16:00:57.683" LastActivityDate="2015-09-21T16:00:57.683" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22446" PostTypeId="1" AcceptedAnswerId="22448" CreationDate="2015-09-21T08:53:48.260" Score="4" ViewCount="490" Body="&lt;p&gt;Professor Halfbrain has recently made a fascinating discovery on quadrilaterals in the plane.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Halfbrain's quadrilateral theorem:&lt;/strong&gt;&#xA;  Let $ABCD$ be a plane quadrilateral that possesses an incircle and a circumcircle.&#xA;  If the center of the incircle coincides with the center of the circumcircle, then $ABCD$ is a square.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Question: Is this theorem indeed true, or has the professor once again made one of his mathematical blunders?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-09-21T15:04:57.500" LastActivityDate="2015-09-21T15:04:57.500" Title="Professor Halfbrain's quadrilateral theorem" Tags="|mathematics|geometry|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22447" PostTypeId="2" ParentId="22446" CreationDate="2015-09-21T09:30:15.113" Score="2" Body="&lt;p&gt;The theorem is...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;... not true. There are an infinite number of &lt;strong&gt;concave&lt;/strong&gt; quadrangles which also bicentric with a incentre-circumcentre distance of zero, each one associated with a specific out-radius:in-radius ratio $R/r$, where $R/r &amp;lt; \sqrt{2}$. According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Bicentric_quadrilateral#Fuss.27_theorem&quot; rel=&quot;nofollow&quot;&gt;Fuss' Theorem&lt;/a&gt;, only in the special case where $R/r = \sqrt{2}$ does a square solution appear.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To construct a concave solution, within a given circle, draw two arbitrary chords $AB$ and $AD$ from $A$, such that $\angle BAD &amp;lt; 60^{\circ}$. Then, draw the incircle such that the chords $AB$ and $AD$ are both tangent to the incircle. Finally, choose a point $C$ on the incircle to complete the quadrangle $ABCD$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15960" LastEditorUserId="15960" LastEditDate="2015-09-21T09:42:59.463" LastActivityDate="2015-09-21T09:42:59.463" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22448" PostTypeId="2" ParentId="22446" CreationDate="2015-09-21T09:30:28.850" Score="7" Body="&lt;p&gt;It is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; true&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let's first draw the concentric incircle and circumcircle. We draw the quadrilateral starting with a vertex on the outer circle and proceeding clockwise. The edge leaving that vertex must be tangent to the inner circle and so is forced. &#xA;&lt;br&gt;&#xA; The next vertex is where this line intersects the outer circle. Now, we're in the same situation as before, but rotated. So, the resulting shape must be rotationally-symmetric (edge-transitive). &#xA;&lt;br&gt;&#xA; If it returns to the initial vertex without crossing itself, it must be a regular polygon. If it has four edges, it's a square.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-09-21T09:30:28.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22450" PostTypeId="1" AcceptedAnswerId="22451" CreationDate="2015-09-21T10:56:18.407" Score="15" ViewCount="3816" Body="&lt;p&gt;Lisa gets picked up from work by her husband at exactly 4 pm every day.&#xA;One day Lisa finished work half an hour earlier and started walking to meet up with her husband. She met her husband on the way, and they drove home together. When they arrived home Lisa noticed that they got back 10 minutes earlier than usual.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For how long did Lisa walk?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few notes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Her husband drove from home&lt;/li&gt;&#xA;&lt;li&gt;The route is symmetric with regards to speed/distance&lt;/li&gt;&#xA;&lt;li&gt;The car can take an instantaneous U-turn&lt;/li&gt;&#xA;&lt;li&gt;Lisa didn't tell her husband that she left early&lt;/li&gt;&#xA;&lt;li&gt;It's neither more nor less traffic than usual.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5517" LastEditorUserId="5517" LastEditDate="2016-06-22T07:54:17.580" LastActivityDate="2018-09-05T19:20:51.043" Title="For how long did Lisa walk?" Tags="|calculation-puzzle|" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22451" PostTypeId="2" ParentId="22450" CreationDate="2015-09-21T11:04:25.590" Score="21" Body="&lt;p&gt;Lisa walked &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 25 minutes. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&#xA;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To arrive home 10 minutes earlier than usual, her husband has to drive 5 minutes less on each way (bidirectional). So he picked her up at 3:55 pm. Since she started walking at 3:30 pm, she had to have been walking for 25 minutes in order for the above condition to be true.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14461" LastEditorUserId="4865" LastEditDate="2015-09-21T14:52:54.883" LastActivityDate="2015-09-21T14:52:54.883" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22453" PostTypeId="1" AcceptedAnswerId="22470" CreationDate="2015-09-21T13:03:29.163" Score="10" ViewCount="626" Body="&lt;p&gt;One day, a small jet was flying over the concretist poem &quot;Velocidade&quot; (Speed), by the Brazilian poet Ronaldo Azeredo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to cover every vowel present in this poem, it goes &lt;em&gt;in straight line&lt;/em&gt; from one letter to another either orthogonally or diagonally adjacent, never making a turn sharper than 45 degrees*.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the letters are like points in a square lattice, and the jet travels at constant speed, what is the fastest way this can be achieved by the pilot?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VSFZi.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VSFZi.jpg&quot; alt=&quot;Velocidade - Ronaldo Azeredo&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that no restrictions were placed in where the jet should start, except that its position goes from one letter to another. Also, the jet is not allowed to leave the matrix of letters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;* Measured by the external angle, i.e. deviation from the original path, so that a 0 degree turn leads to the same direction, while a 180 degree turn makes the jet go in the reverse.&lt;/p&gt;&#xA;" OwnerUserId="14219" LastEditorUserId="14219" LastEditDate="2015-09-22T05:07:57.367" LastActivityDate="2015-09-23T22:03:55.587" Title="Find the fastest path through this Brazilian concretist poem" Tags="|mathematics|optimization|poetry|" AnswerCount="3" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="22454" PostTypeId="2" ParentId="22450" CreationDate="2015-09-21T13:04:02.830" Score="4" Body="&lt;p&gt;User Wa Kai was much faster, but I cannot make myself delete this, having spent so much time figuring it out.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To arrive 10 minutes earlier, the husband has to drive 5 minutes less each way. Thus, they met 3:55, after Lisa walked &lt;strong&gt;25 minutes&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, this only works under the assumption that her husband departs from their home, which was, for me, not clear from the text.&lt;/p&gt;&#xA;" OwnerUserId="16460" LastEditorUserId="16460" LastEditDate="2015-09-21T16:23:49.827" LastActivityDate="2015-09-21T16:23:49.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22456" PostTypeId="1" AcceptedAnswerId="22457" CreationDate="2015-09-21T14:01:42.160" Score="7" ViewCount="513" Body="&lt;p&gt;Class 6F at &lt;strong&gt;Farthingbottom School&lt;/strong&gt; was very unusual this year. All the students had transferred from a School in the USA. Today was their first history lesson with Professor Flavanna.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Welcome, children! I have observed that you all have nice alliterative names from good military families but, for the purposes of my class, I need to assign you all with new surnames. You shall now be known as:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Arthur Charlestown&lt;/li&gt;&#xA;&lt;li&gt;Audrey Burnham&lt;/li&gt;&#xA;&lt;li&gt;Barbara Reading&lt;/li&gt;&#xA;&lt;li&gt;Brian Beverley&lt;/li&gt;&#xA;&lt;li&gt;Clive Salisbury&lt;/li&gt;&#xA;&lt;li&gt;Colin Leeds&lt;/li&gt;&#xA;&lt;li&gt;Unwin Clare&lt;/li&gt;&#xA;&lt;li&gt;Wendy Castleton&quot;  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&quot;There are two of you who I think are in the wrong class, but if I have to teach you, you shall be know as:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Boris St. Francis&lt;/li&gt;&#xA;&lt;li&gt;Steven Rosales&quot;  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&quot;Now open your copies of &lt;em&gt;All Hell Let Loose&lt;/em&gt; at chapter one.&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What are the children's real surnames?&lt;/p&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-09-21T14:42:27.420" Title="Redesignated students" Tags="|knowledge|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22457" PostTypeId="2" ParentId="22456" CreationDate="2015-09-21T14:13:51.593" Score="10" Body="&lt;p&gt;I am quite confident that the answer is correct. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Based on the clues.. history, alliteration, army &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Arthur Charlestown&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  -&gt; Arthur Abbot -&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Audrey Burnham&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; -&gt; Audrey Aulick &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Barbara Reading&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; -&gt; Barbara Bailey&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Brian Beverley&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; -&gt; Brian Branch &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clive Salisbury&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; -&gt; Clive Claxton&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Colin Leeds&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; -&gt; Colin Conner&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unwin Clare&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; -&gt; Unwin Upshur &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wendy Castleton&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; -&gt; Wendy Ward&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Apparently the surnames assigned are the names of &lt;a href=&quot;https://en.wikipedia.org/wiki/Town-class_destroyer&quot;&gt;USS Town-class destroyers&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Boris St. Francis&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; -&gt; Boris Bancroft&lt;br&gt;&#xA; Note : HMCS St. Francis was transferred to Royal Canadian Navy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Steven Rosales&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; -&gt; Steven Stembel&lt;br&gt;&#xA; Note : USS Stembel was loaned to Rebpublic of Argentina on 7th Aug 1961&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I guess that completes the answer. A very interesting format of the question though. Thanks for asking.&lt;/p&gt;&#xA;" OwnerUserId="9892" LastEditorUserId="14135" LastEditDate="2015-09-21T14:42:27.420" LastActivityDate="2015-09-21T14:42:27.420" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22458" PostTypeId="1" AcceptedAnswerId="22474" CreationDate="2015-09-21T14:23:30.300" Score="7" ViewCount="22386" Body="&lt;p&gt;We have a regular analog clock with three hands - second, minute and hour. On a given day, how many times do:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) the minute and hour hands&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) the minute and second hands&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c) the hour and second hands&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(d) all three hands&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;meet?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One approach would be to just list out all the times they meet, but I'm looking for a more logical/mathematical approach.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="52483" LastEditDate="2018-11-22T19:48:39.857" LastActivityDate="2018-11-22T19:48:39.857" Title="Meeting of hands on a clock" Tags="|mathematics|visual|time|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22460" PostTypeId="2" ParentId="22458" CreationDate="2015-09-21T14:49:39.227" Score="4" Body="&lt;p&gt;Question:  At midnight, when all the hands are pointing at twelve, does that count as a meeting?  Does it count as a second meeting when they all point at midnight $24$ hours later?  I will assume only one of these counts as a meeting.  i.e. if one of them is in the $24$ hour time interval, then the other cannot be.   For example, let our $24$ hour window go from 12:00:01 to 12:00:01 the following day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a) &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The hour hand will go around the clock twice during a day, every $12$ hours.  During a single rotation, the minute hand will go around $12$ times.  The minute hand will pass the hour hand $11$ times in the first $12$ hours, and $11$ times in the second $12$ hours.  This means that they will meet $22$ times. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For b)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In a single hour, the second hand will go around the clock $60$ times.  The minute hand will go around once.  Thus, the second hand will pass the minute hand $59$ times in an hour.  This means they will meet $59 \times 24=1416$ times in a day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For c)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The second hand will go around the clock $60 \times 12=720$ times in $12$ hours.  The hour hand will go around once.  Thus, they will meet $719$ times in $12$ hours, and $1438$ in $24$ hours.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For d)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is the trickiest one.  First, lets note that the first $12$ hours and the second $12$ hours of the day are symmetrical.  Also, in the first 12 hours, we already know that the hour and minute hand only meet 11 times.  Since there is the same amount of time between each meeting, we know that there are $\frac{12}{11}$ hours between each meeting.   $\frac{12}{11}$ hours $=\frac{12 \times 60}{11}$ minutes $=\frac{12 \times 60 \times 60}{11}$ seconds.  Thus, each meeting is at every $3927.28$ seconds.  We know the first meeting is around 1:06.  The second hand will be pointing at $3927.28 \mod 60 = 27.28$ seconds.  Thus, at the first meeting of the minute and hour hand (near the one), the second had will be pointing half way between the 5 and 6.  If we continue this, we will see that the second hand never meets they other two at the exact same time.  Thus, all three only meet twice - once at noon, and again at midnight.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="1592" LastEditDate="2015-09-21T14:59:45.567" LastActivityDate="2015-09-21T14:59:45.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22461" PostTypeId="2" ParentId="22418" CreationDate="2015-09-21T14:49:55.407" Score="1" Body="&lt;p&gt;Solving such questions is easy using the method given &lt;a href=&quot;https://en.wikipedia.org/wiki/Nash_equilibrium#Nash_equilibria_in_a_payoff_matrix&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The three strategies given to the players to choose from are as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: Never horn&lt;br&gt;&#xA;2: Horn immediately&lt;br&gt;&#xA;3: Horn after the other player&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The payoff-table looks as follows (1 means life, 0 means death):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;              A adopts 1      A adopts 2       A adopts 3&#xA;B adopts 1      0,0               1,0             0,0    &#xA;B adopts 2      0,1             1/2,1/2           1,0    &#xA;B adopts 3      0,0               0,1             0,0    &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The only Nash Equilibrium is when one player chooses 2 and the other chooses 3. Meaning that immediately horning, or horning after the other player are both equally good strategies. Hence they will randomly select strategy 2 or 3.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-09-21T14:49:55.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22462" PostTypeId="2" ParentId="22458" CreationDate="2015-09-21T15:02:46.150" Score="0" Body="&lt;p&gt;First we find angular velocities of the hands:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;H_{\text{speed}}&amp;amp;=360^\circ/12\text h=\tfrac12^\circ/\text{min}\\&#xA;M_{\text{speed}}&amp;amp;=360^\circ/\text h=6^\circ/\text{min}\\&#xA;S_{\text{speed}}&amp;amp;=360^\circ/\text{min}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a slower hand covers an angle $x$ and another faster one covers angle $360^\circ+x$ in same time, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\text{Time 1}&amp;amp;=\text{Time 2}\\[5pt]&#xA;\frac{\text{Angle 1}}{\text{Speed 1}}&amp;amp;=\frac{\text{Angle 2}}{\text{Speed 2}}\\&#xA;\frac{x}{\text{Speed 1}}&amp;amp;=\frac{360^\circ+x}{\text{Speed 2}}\\&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for $x$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=\frac{360^\circ\times \text{Speed 1}}{\text{Speed 2}-\text{Speed 1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By substituting velocities of any two hands, we get to know what angle the slower hand covers before every time they meet. We divide this angle by the total angle covered during the day (by the slower hand) to find out how many times they meet.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="13494" LastEditDate="2015-09-22T18:41:18.627" LastActivityDate="2015-09-22T18:41:18.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22463" PostTypeId="2" ParentId="22458" CreationDate="2015-09-21T15:07:47.260" Score="0" Body="&lt;h2&gt;Partial answer&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Minute and hour hands&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since a standard analog clock only shows 12 hours, it's sufficient to solve this for 12 hours and then multiply by 2 to cover the whole day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The minute hand rotates at a rate of 1 revolution per hour, or 6°/min. The hour hand rotates 12 times more slowly, or 0.5°/min. Suppose N minutes have passed. The positions of the hands are &lt;code&gt;6N&lt;/code&gt;° for the minute hand and &lt;code&gt;0.5N&lt;/code&gt;° for the hour. Thus, we want to solve when these two values are equal. However, since the minute hand effectively restarts every hour, we want to use some modular arithmetic here. Namely, the minute hand's position is actually &lt;code&gt;6N%60&lt;/code&gt;, where &lt;code&gt;%&lt;/code&gt; denotes the modulo operator. Additionally, we are constraining the problem to the first 12 hours, or the first 720 minutes. The final equation is then:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;6N%60 = 0.5N, N ∈ [0, 720)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There are 11 distinct solutions to this equation in the selected range, so the hands overlap 11 times in each 12-hour period. Therefore they overlap 22 times in each day.&lt;/p&gt;&#xA;" OwnerUserId="16275" LastActivityDate="2015-09-21T15:07:47.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22464" PostTypeId="2" ParentId="22458" CreationDate="2015-09-21T15:22:15.153" Score="1" Body="&lt;p&gt;You can consider the position of the hands as sequences mod 1. By which I mean giving a hand a position $x$ with $ 0\le x &amp;lt; 1$ which represents its position. When the hand rotates back around to 12 o'clock we assign it zero (so it keeps looping). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Essentially if $x$ exceeds 1 then we minus an integer so its between 0 and 1 again, this is how the 'mod 1' idea works and compensates for the clock looping.&#xA;In this case $x= \frac{1}{2}$ would represent the hand at 6 o'clock and in general $x$ represents the hand being at 12x o'clock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can take the slowest hand, the hour hand and call its position $x$. We know that all three hands coincide at 12 exactly (when $x=0$) and after this the minute hand moves 12 times as fast as the hour hand since one rotation of the hour hand corresponds to 12 hours passing which is 12 rotations of the minute hand. Likewise the seconds hand moves 60 times the speed the minute hand or $60*12$ times faster than the hour hand. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this we deduce that when the hour hand is at position $x$ mod 1 the minute hand is at position $12x$ mod 1 and the second hand is at position $720x$ mod 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can calculate the answers you want by equating these mod 1 sequences:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) for the minute and hour hand to coincide then $x=12x$ mod 1 and thus $11x=0$ mod 1 which implies $11x$ is an integer meaning $x=k/11$ for $0 \le k \le 10$ giving 11 solutions. (Note the exact time they coincide will be when the hour hand and minute hands are at $60*k/11$). The hour hand goes around twice in a day however giving 22 solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) for the minute and second hand $12x = 720x$ mod 1 so $x=60x$ mod 1 implying $59x=0$ mod 1 and as before $x=k/59$ for $0 \le k \le 58$ giving 59 solutions in one hour. Therefore $59*24$ in one day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) repeating yet again will give 719 solutions to th sequence part and $719*2$ to the total in a day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d) Finally for d) we note that 11 and 719 are prime so fractions other than 0 of the form $k/11$ and $m/719$ will never coincide (for these fractions less than one) and therefore both a) and c) cannot hold simultaneously(unless we are at o) which is what we need for all hands to coincide. Therefore all three hands only meet when $x=0$ at 12 o'clock, so twice in a day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this answers the question in a mathematical way!&lt;/p&gt;&#xA;" OwnerUserId="16326" LastEditorUserId="16326" LastEditDate="2015-09-21T16:31:19.653" LastActivityDate="2015-09-21T16:31:19.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22465" PostTypeId="2" ParentId="22438" CreationDate="2015-09-21T15:29:22.250" Score="4" Body="&lt;p&gt;It works for any $N \ge 4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $N=3$ you cannot guarantee success.  You must weigh only 2 coins.  If the coin not measured is fake, then you will be successful since the scale will balance and those will disappear, leaving you with the fake.  But if the fake is measured, then the coin not measured will disappear, leaving you with 2 coins and no reference to know which is fake.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $N \ge 4$, the solution is as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide the coins into two equal groups.  If $N$ is odd, leave once coin out.  Measure the groups against each other.  There are 2 possible outcomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the scales balance, all the measured coins disappear and the coin left out is fake.  This can only occur when $N$ was odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, one side will drop.  The coin left out (if $N$ was odd) will disappear since it is known that the fake coin was measured.  You are left with one set of coins which may contain a fake heavy coin, and one which may contain a fake light coin.  Each group has $M \ge 2$ coins in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the heavy group and weigh two coins against each other until you run out of coins or find the heavy fake coin (you can be more efficient by weighing groups of coins, but this question doesn't ask us to minimize uses of the scale).  If you run out of coins to weigh, and have none left over, then simply repeat with the light pile to find the fake light coin.  If instead, you had one coin left, then you know that it might be fake and heavy, or genuine and the fake is in the light pile.  This can only occur if $M \ge 3$ and odd.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $M \ge 3$, there are at least 3 coins in the light pile. So, use the technique by @Zandar to determine the result.  Pick three coins from the light pile called $A, B,$ and $C$.  The coin left over from the heavy pile is $H$.  So weigh $HA$ vs $BC$.  There are three possible results:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$HA$ is lighter. Thus, $H$ was not a fake heavy coin and $A$ is the fake light coin.&lt;/li&gt;&#xA;&lt;li&gt;$HA$ is the same as $BC$.  Again, $H$ was not a fake heavy coin and all the measured coins disappear.  So, simply weight the remaining coins in the light group 1 vs 1 until you find the fake or are left with one left over (which must be fake).&lt;/li&gt;&#xA;&lt;li&gt;$HA$ is heavier.  This means that either $H$ is heavy, or $B$ or $C$ are light.  All other coins ($A$ and the rest of the coins in the light pile) disappear since they were genuine.  So, weigh $B$ vs $C$ to see if either one was light (and fake).  If not, then it was $H$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2015-09-21T15:29:22.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22466" PostTypeId="2" ParentId="22453" CreationDate="2015-09-21T16:48:06.030" Score="4" Body="&lt;p&gt;OK. Touching every vowel once, using coordinates A-J horizontal and 0-9 vertically, with the origin in the top left:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;J9-&amp;gt;J1, I0, H0-&amp;gt;A7, A8, B9-&amp;gt;H9&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;I9-&amp;gt;I2, H1, G1-&amp;gt;B6, B7, C8-&amp;gt;G8&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;H7-&amp;gt;H3, G2, F2-&amp;gt;C5, C6, D7-&amp;gt;F7&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;G6-&amp;gt;G4, F3-&amp;gt;D3, C4, C5, D6-&amp;gt;E6, F5&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's 71 steps. It's probably not optimized perfectly, so I'll keep looking, but it's at least a starting point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: This one has 59 but touches some vowels more than once. Counting as distance (cf. Sleafar's comment) it would be 68.11 units with 22 diagonal moves.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;J9-&amp;gt;J1, I0, H0, G1, G2, H3, I4-&amp;gt;I8&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;H9-&amp;gt;B9, A8, A7, B6, C6-&amp;gt;E8-&amp;gt;G8, H7&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;H7-&amp;gt;H5, G4, F4-&amp;gt;C7, C8, D9, E9, F8&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;F7-&amp;gt;I4-&amp;gt;I2, H1, G1, F2-&amp;gt;F5, E6&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/Fwxlc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fwxlc.png&quot; alt=&quot;Picture n&amp;#39; stuff&quot;&gt;&lt;/a&gt;&#xA;&lt;br&gt;&#xA; This one doesn't waste as much time by avoiding U-turns, but instead grabs what it can that's nearby.&lt;br&gt;&#xA;&lt;br&gt;&#xA; ETA3: nevermind, the edit would have missed one 'E'. i'll quit second-guessing myself now  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10625" LastEditorUserId="10625" LastEditDate="2015-09-21T17:51:52.463" LastActivityDate="2015-09-21T17:51:52.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22467" PostTypeId="2" ParentId="22415" CreationDate="2015-09-21T17:21:31.617" Score="3" Body="&lt;p&gt;Well, I am going to start a community answer.  Anyone have any ideas??&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I was thinking that the movies all have some sort of abbreviation in the title, and then the letters could be used in the final answer. With the clues referencing space, and the ET answer, it seems that all of the movies can be space based. Might be a route to go down.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where did the animated little guy and his family come from?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Possibly the Secret of N.I.M.H.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What type was 2 less than the first 23rd?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; V for Vendetta (the 23rd letter of the alphabet is W.  Take away two legs of the W, and you get a V)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who was the botanist that shook the earth with his finger?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; E.T. The Extra Terrestial (he was a botanist and his finger was magical)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;TECHNICOLOR! without the earth or the obviousness.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Maybe 2001: A Space Odyssey &amp;lt;-- the clue, but not the theme above.  Not sure if the theme is correct, so leaving here.  Thanks to @JGibbers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The aliens are almost extinct.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Titan A.E.  (In this movie, humans have become the aliens, and are nearly extinct)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hey you, hand, lose the glasses!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2 more and then...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14233" LastEditorUserId="16466" LastEditDate="2015-09-21T19:42:27.210" LastActivityDate="2015-09-21T19:42:27.210" CommentCount="5" CommunityOwnedDate="2015-09-21T17:56:05.673" ContentLicense="CC BY-SA 3.0" />
  <row Id="22468" PostTypeId="1" AcceptedAnswerId="22478" CreationDate="2015-09-21T17:30:01.253" Score="22" ViewCount="3676" Body="&lt;p&gt;Johnny has a perfectly spherical apple, with a diameter of 70 mm. While he looks away, a worm burrows through (entering &lt;em&gt;and&lt;/em&gt; leaving) the apple, forming a single tunnel of length 69 mm and negligible width. The worm does not retrace its path at any point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that Johnny can cut the apple in half, with a single straight cut, in such a way that one of the halves is untouched by the worm.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="13237" LastEditDate="2015-09-21T21:28:38.650" LastActivityDate="2015-09-23T23:21:42.877" Title="Worm and an Apple" Tags="|mathematics|geometry|" AnswerCount="5" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22469" PostTypeId="2" ParentId="22468" CreationDate="2015-09-21T17:47:59.707" Score="5" Body="&lt;p&gt;You can make the cut, as long as you prove the worm has only been in one hemisphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of, the worm cannot possibly have gone through the center (that path would take 70mm, so the center is 'safe'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The worm's path is most inconvenient if it remains in a plane, as that maximizes it's useful length available to create an 'inconvenient' path for the apple slicer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, assume a plane through the apple and the 'point of entry', and assume the origin in the center of the apple, and the X-axis going through the point of entry. We need to prove that from this situation, there is no way for the worm to create a path that doesn't allow for a suitable cut. The only way the worm can block a such path, is by having an angle of 180 degrees or higher created, compared to the origin, at the side of the worm's path.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/VTtd3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VTtd3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;(worm's path in red, 'angle' as described shown in green)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a part, at it's minimal length, would be $2x$ the radius, or 70mm. We know however that the path is only 69mm. Therefore the worm cannot ensure there is no untouched hemisphere; so the cut is always possible: cut in a plane perpendicular to the displayed plane, which goes through the center of the apple and any line in the 180°+ range that is untouched by the worm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given wording is perhaps a bit complex, and as some have asked more detail on why I focussed on paths in a plane, I'll hereby use the wording of &lt;a href=&quot;https://puzzling.stackexchange.com/users/14219/ivan-barreto&quot;&gt;Ivan Barreto&lt;/a&gt;  from the comments:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Another way of stating it is that the projection of the tunnel in the&lt;br&gt;&#xA;  xy plane should obscure an angle greater than or equal to 180 degrees.&lt;br&gt;&#xA;  However, any such tunnel is 70mm or longer. Hence, since a curve is no&lt;br&gt;&#xA;  shorter than its projection, it should also be 70mm or longer   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2156" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2015-09-22T13:36:12.957" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22470" PostTypeId="2" ParentId="22453" CreationDate="2015-09-21T17:53:28.330" Score="6" Body="&lt;p&gt;Here's a solution with&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 54 steps.&lt;br&gt;&lt;br&gt;$38 \space horizontal/vertical + 16 \space diagonal =&amp;gt; length \space 60.627$&lt;br&gt;Thanks to  f'' for length optimization.&lt;br&gt;&lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/sCg3t.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A brute-forced solution with&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 47 steps.&lt;br&gt;&lt;br&gt;$28 \space horizontal/vertical + 19 \space diagonal =&amp;gt; length \space 54.87$&lt;br&gt;&lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/Sv9jd.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2015-09-23T22:03:55.587" LastActivityDate="2015-09-23T22:03:55.587" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="22471" PostTypeId="2" ParentId="22468" CreationDate="2015-09-21T18:28:31.877" Score="3" Body="&lt;p&gt;Cutting the apple in half requires that the cut passes through the centre of the sphere. Let us assume that the worm can win. We know that the worm cannot reach the centre. Hence atleast one point of the worm must reside in &lt;em&gt;every possible plane passing through the centre&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us assume a horizontal 'base' plane through the centre for convenience. Let us mark north, south, east and west on this plane. If we rotate the plane along the north-south axis, we come across a number of planes, each passing through the centre. The worm must reside in all of them. Hence it must reside in the line of intersection of these planes (the N-S axis).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we bring back our plane to original state. We rotate it by a small angle on the east-west axis. Now we again follow the procedure above (continuous N-S rotation) again, and get a new line of intersection of the new planes. We bring it back to initial state, make an E-W rotation of different angle, and then repeat the same N-S rotation procedure again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By repeating the above two steps a number of times, we get a number of lines in which the worm must reside. All these lines reside in the same plane (Perpendicular to the 'base' and sharing the N-S axis).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The worm must pass through every line passing through the centre of a given plane perpendicular to the base plane. Hence we could have started with any base plane, we could end up with any perpendicular plane. &lt;strong&gt;Hence the worm must reside in every line (not just plane) passing through the centre of the sphere.&lt;/strong&gt; I don't know the proof to this being impossible, but it seems quite obvious that this is impossible.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-09-21T18:28:31.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22472" PostTypeId="2" ParentId="17886" CreationDate="2015-09-21T18:54:35.127" Score="3" Body="&lt;p&gt;Seems like this works?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;B  L' D  U' B' R  U' L2 B' U  L' U' L  R' U' R  U2 F' U' F  B  U&#xA;B' U' B  U  B' D  U  L' U  L  D  U  F' U  F  D  U2 B  U' B' D  U  L  U'&#xA;L' U' B' U  B  R  U  B  U' B' R' U' F  U' B' U  F' U' B2 U' F  U  L&#xA;F  L' B' L  F' L' U' F' U&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/E2thO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/E2thO.png&quot; alt=&quot;Seems like this works&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not unique btw, the bottom four edge colours can easily be permuted, for example.&lt;/p&gt;&#xA;" OwnerUserId="15960" LastActivityDate="2015-09-21T18:54:35.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22473" PostTypeId="2" ParentId="22417" CreationDate="2015-09-21T19:13:43.583" Score="1" Body="&lt;p&gt;I know this might sound crazy but my guess is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Footwear&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you wish to leave, then there's no doubt&#xA;first you must look for the way out.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Obviously the door&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then with open eyes you must stare;&#xA;just look down to meet my glare.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I always wear shoes/sneakers/sandals while standing so I have to &quot;look down&quot; to see it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To remember this ordeal,&#xA;to take my thoughts and make them real,&#xA;then at the bottom, beside my eyes,&#xA;tell me where the treasure lies.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fasten the Strap of the footwear. Eyes are the Lace holes, while I am reading &quot;treasure lies&quot; as &quot;foot rests&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you'd rather forget your strife&#xA;and quietly, quickly end my life&#xA;get in line, but do not pout,&#xA;Instead you must scream and shout.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If it is uncomfortable (new ones sometimes are) instead of removing it, just &quot;cry&quot; in pain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Either way you're in the clear,&#xA;but there's one thing you must do from here:&#xA;say no more, come back inside,&#xA;and I'll let you off this crazy ride!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Come back home or to a place and remove them after their use is over.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9892" LastActivityDate="2015-09-21T19:13:43.583" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22474" PostTypeId="2" ParentId="22458" CreationDate="2015-09-21T19:34:34.693" Score="8" Body="&lt;p&gt;I think there is a much simpler solution than all provided so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following three facts:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;hour hand will rotate 2 full times in a day&lt;/li&gt;&#xA;&lt;li&gt;minute hand will rotate 24 full times in a day&lt;/li&gt;&#xA;&lt;li&gt;second hand will rotate 1440 full times in a day&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) the minute and hour hands, will meet exactly:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 22 times: 24 - 2 (once every 24/22 hours)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(b) the minute and second hands, will meet exactly:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1416 times: 1440 - 24 (once every 24/1416 hours)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(c) the hour and second hands, will meet exactly:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1438 times: 1440 - 2 (once every 24/1438 hours)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(d) all three hands, will meet exactly:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; twice: only at exactly 12:00:00 o'clock (noon and midnight)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Simply because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The faster hand passes the slower hand by the number of laps it makes minus the number of laps the slower hand makes. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With the special case:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One hand lapping another won't necessarily coincide with the third hand being there. Try to find the common multiples of the three fractions 24/22, 24/1416, and 24/1438 and you will see there are only two at 24/1 and 24/2. (ie. 12 hours and 24 hours after the start).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4125" LastEditorUserId="4125" LastEditDate="2015-09-21T22:23:38.693" LastActivityDate="2015-09-21T22:23:38.693" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22475" PostTypeId="2" ParentId="22468" CreationDate="2015-09-21T20:21:36.493" Score="0" Body="&lt;p&gt;I don't believe it can be proved; not mathematically at least. This may rely a bit on wordplay, but I'll give it a shot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation posed does not seem to unambiguously declare that the worm ever left the apple; only that it burrowed through it, and the length of the resulting tunnel was 69mm. Here, I am interpreting &lt;em&gt;through&lt;/em&gt; as &lt;em&gt;within&lt;/em&gt;. For example, a termite can burrow through the wall of your house, reside between your interior and exterior walls, and never leave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this in mind, out of all 69mm paths within the apple, at least one of them must go through the center point. Some more wordplay results in two different outcomes:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;em&gt;Half&lt;/em&gt; defined as two separate pieces, resulting in one &quot;half&quot; that is 99.99% of the apple, and another &quot;half&quot; that is 0.01% of the apple.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Half&lt;/em&gt; defined as two separate pieces that are of either equivalent volume or shape.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Again, if the worm never left, AND has reached the exact middle of the apple, then no matter which way you cut it, it will have at least a little bit of worm in it. This worm need not travel in a straight line either, he could curve up, down, left, right, or even tie himself in a little knot.&lt;/p&gt;&#xA;" OwnerDisplayName="user16469" LastActivityDate="2015-09-21T20:21:36.493" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22476" PostTypeId="2" ParentId="22450" CreationDate="2015-09-21T20:29:44.297" Score="0" Body="&lt;p&gt;So, the answers above say that he met her at&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 15:55?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How could that be if he did not know that she was leaving early.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seems to me like she would be walking at least&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 30 minutes until it's 16:00 (4 PM) and first then she has a chance of meeting her husband along the road -- a road which is now a bit shorter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; NORMAL 16:00 work---60---home =&gt; 90km/h = 40mins (=16:40).&#xA; &#xA; TODAY 15:30 work---15---pickup---45---home =&gt; 30km/h (30 mins walk) + 90km/h (30 mins drive) (=16:30).&#xA; &#xA; (fast walking, I know)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But now, we can certainly tweak those variables, and end up with other answers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; NORMAL 16:00 work---40---home =&gt; 40km =&gt; 40km/h = 60mins (=17:00).&#xA;&#xA; TODAY 15:30  work---10---pickup---30---work = 35 minute walk (17km/h) and 45 mins drive for 30 km with 40km/h, and we arrive at 16:50, so, she got picked up at 16:05, and has walked for 35 minutes (with an insane speed of 17km/h).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think the answer is that there is no answer until we get more variables (her walking speed, total distance, car walking speed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm sorry for/if not following SE conventions here. Personal first post.)&lt;/p&gt;&#xA;" OwnerUserId="16468" LastActivityDate="2015-09-21T20:29:44.297" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22478" PostTypeId="2" ParentId="22468" CreationDate="2015-09-21T20:46:11.380" Score="33" Body="&lt;p&gt;Call the point where the worm entered P, and the point where the worm exited Q. Let P' be the point diametrically opposite P.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a unique plane through the center of the apple such that P' and Q are mirror images across that plane. Any point on this plane is equidistant from P' and Q. If the worm was able to reach this plane and then exit at Q, it would have been able to exit at P' instead. But the distance from P to P' is 70 mm, so it could not have made it from P to P'. Therefore, the worm could not have reached any point of this plane, and it is safe to cut the apple there.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-09-21T20:46:11.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22479" PostTypeId="2" ParentId="22378" CreationDate="2015-09-22T01:25:35.597" Score="4" Body="&lt;p&gt;&lt;em&gt;O, The Places You'll Go&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;$400&lt;/strong&gt; - Blended egg drink&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Oregon&lt;/strong&gt; - eggnog (egg drink) =&gt; roe nog (roe is fish eggs) =&gt; oregon (blended = anagram)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;An Unreal Vacation&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;$300&lt;/strong&gt; - Curious last character accompanies birth&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Norumbega&quot; rel=&quot;nofollow&quot;&gt;Norombega&lt;/a&gt;&lt;/strong&gt; - The last letter of the Greek alphabet (curious last character) is &quot;Omega&quot;, into which we mix (accompany) the letters from &quot;born&quot; (birth).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="5208" LastEditDate="2015-10-07T04:55:11.193" LastActivityDate="2015-10-07T04:55:11.193" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22480" PostTypeId="1" AcceptedAnswerId="22484" CreationDate="2015-09-22T01:32:03.780" Score="5" ViewCount="789" Body="&lt;p&gt;Mr. Hilbert was sitting alone at the dinner table of the Grand Hotel restaurant, waiting for his two friends, Mr. Euler and Mr. Langrange, to show up. It was 5 minutes past six already, but knowing the two of them, it wasn't too much of a surprise to Mr. Hilbert that their arrivals times were a bit variational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being slightly bored, Mr. Hilbert started to play with his napkin ring. At first, it looked just like &lt;a href=&quot;https://puzzling.stackexchange.com/questions/21914/a-hollow-sphere&quot;&gt;a hollow sphere&lt;/a&gt; with a circular hole, but upon closer inspection, the hole seemed more like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mcoW2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mcoW2.png&quot; alt=&quot;Bird&amp;#39;s eye view of the hole&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the hole is the shaded area, the intersection of four quarter-circles of radius $r$. And this hole was drilled straight through a sphere of diameter $r$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QhF3C.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QhF3C.png&quot; alt=&quot;How the hole is drilled through the napkin-ring sphere&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Hmm, I wonder how much volume is cut out of the original sphere in order to get this napkin ring?&quot; wondered Mr. Hilbert.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Can you figure it out? Express your answer in terms of $r$.&lt;/p&gt;&#xA;" OwnerUserId="15960" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-09-22T22:55:47.137" Title="Mr. Hilbert and the Problem of the Nuanced Napkin Ring" Tags="|mathematics|" AnswerCount="3" CommentCount="5" ClosedDate="2015-09-23T04:41:05.490" ContentLicense="CC BY-SA 3.0" />
  <row Id="22481" PostTypeId="2" ParentId="22453" CreationDate="2015-09-22T02:33:46.397" Score="-2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think I am missing something obvious, since the solution couldn't be that easy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;16 diagonal + 16 vertical/horizontal steps.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/28SWD.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/28SWD.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10618" LastEditorUserId="10618" LastEditDate="2015-09-22T03:25:01.207" LastActivityDate="2015-09-22T03:25:01.207" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22482" PostTypeId="2" ParentId="22480" CreationDate="2015-09-22T03:13:16.490" Score="0" Body="&lt;p&gt;Okay, I might be a bit rusty. But looking from the top, and considering just one of the quadrants (actually just the top half of one), you can integrate and get the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int^{\sqrt{3}-1}_0\int^{\sqrt{4-(x+1)^2}-1}_0\int^{\sqrt{1-x^2-y^2}}_0dzdydx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've set $r=2$, but I've also only considered an eighth of the volume, so the above gives the answer in units of $r$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first couple of integrations doesn't lead to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int^{\sqrt{3}-1}_0\left[y-yx^2-y^3/3\right]^{\sqrt{4-(x+1)^2}-1}_0dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because I missed the sqrt in the innermost integral. Thanks for noticing&lt;/p&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2015-09-22T05:32:19.587" LastActivityDate="2015-09-22T05:32:19.587" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22483" PostTypeId="2" ParentId="22234" CreationDate="2015-09-22T03:32:59.633" Score="10" Body="&lt;p&gt;The hidden word is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Unveil&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since rearranging the letters doesn't change anything,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the information must be encoded in the frequencies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In ascending order:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I T A 2 C O D E&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This hints at the&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; International Telegraph Alphabet 2 (ITA-2)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So we can translate the dots.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The source: &lt;a href=&quot;https://commons.m.wikimedia.org/wiki/File:International_Telegraph_Alphabet_2.jpg&quot;&gt;https://commons.m.wikimedia.org/wiki/File:International_Telegraph_Alphabet_2.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14219" LastEditorUserId="14219" LastEditDate="2015-09-22T03:47:23.640" LastActivityDate="2015-09-22T03:47:23.640" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22484" PostTypeId="2" ParentId="22480" CreationDate="2015-09-22T04:47:52.637" Score="2" Body="&lt;p&gt;Taking the solution by @Dr Xorile a step further, I get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;V = kr^3&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;k = \int_0^{\frac{\sqrt{3}-1}{2}}&amp;amp;\left[\left(1-4 x^2\right)&#xA;   \tan ^{-1}\left(\frac{\sqrt{3-4 x (x+1)}-1}{\sqrt{4 x+2 \sqrt{3-4 x&#xA;   (x+1)}-3}}\right)\cdots\right. \\&#xA;&amp;amp;-\sqrt{4 x+2 \sqrt{3-4 x (x+1)}-3}\cdots \\&#xA;&amp;amp;\left.+\sqrt{(3-4 x (x+1))&#xA;   \left(4 x+2 \sqrt{3-4 x (x+1)}-3\right)}\right] \, dx \\&#xA;&amp;amp;\approx 0.280\,225\,788\,122\,747\,024\,207\,247\,869\,453\,505\ldots&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="9718" LastActivityDate="2015-09-22T04:47:52.637" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22485" PostTypeId="1" AcceptedAnswerId="22487" CreationDate="2015-09-22T06:52:07.500" Score="2" ViewCount="1064" Body="&lt;p&gt;Which of these tiles would fit the sequence? I've tried to think about it but I have no idea. &lt;img src=&quot;https://i.stack.imgur.com/OPoZh.jpg&quot; alt=&quot;&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/u8CrQ.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="16477" LastEditorUserId="15544" LastEditDate="2015-09-24T05:32:33.833" LastActivityDate="2015-09-24T05:32:33.833" Title="Which of the options fit the sequence?" Tags="|pattern|visual|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22486" PostTypeId="2" ParentId="22485" CreationDate="2015-09-22T07:03:54.577" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you assign positive numbers to squares on the left of the bar and negative numbers to squares on the right and using B for black and W for white, you get the following:&lt;br&gt;&lt;br&gt;$(1B+3W)+(1B-2W)=(2B+1W)$&lt;br&gt;$(1B-3W)+(2W-1B)=(-1W)$&lt;br&gt;$(2B)+(0)=(2B)$&lt;br&gt;&lt;br&gt;So the solution should be the top right image.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastActivityDate="2015-09-22T07:03:54.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22487" PostTypeId="2" ParentId="22485" CreationDate="2015-09-22T07:09:45.220" Score="7" Body="&lt;p&gt;Here is the answer:&lt;br&gt;&#xA;if you cancel out item from left side to right hand side you can get remaining value.&lt;br&gt;&#xA;you can go any way vertically or horizontally.&lt;br&gt;&#xA;In this I am canceling item vertically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ckkEu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ckkEu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9626" LastActivityDate="2015-09-22T07:09:45.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22491" PostTypeId="1" AcceptedAnswerId="22493" CreationDate="2015-09-22T11:57:14.687" Score="7" ViewCount="969" Body="&lt;p&gt;To save money, a meticulous person had a oddly unique way of splurging money. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each day the person would splurge an amount equal to the month-digits + day dollars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for the first of April it would be 4+1 = 5 Dollars, while on first of June it would be 6+1=7 dollars. On the 10th of April it would be 4+10 = 14 dollars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;One day the person counted a total splurge of 97 dollars for the last 5 consecutive odd numbered days including this day.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What date can it be?&lt;/p&gt;&#xA;" OwnerUserId="9892" LastEditorUserId="52483" LastEditDate="2018-11-22T19:45:00.800" LastActivityDate="2018-11-22T19:45:00.800" Title="What date can it be?" Tags="|calculation-puzzle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22492" PostTypeId="2" ParentId="22491" CreationDate="2015-09-22T12:07:24.100" Score="7" Body="&lt;p&gt;It is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; June, 5th (#1) or in a leap year it can be March 3rd (#2) too&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Calculation #1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 29+5 + 31+5 + 1+6 + 3+6 + 5+6 = 97&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Calculation #2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 25+2 + 27+2 + 29+2 + 1+3 + 3+3 = 97&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How I solved it?&lt;br&gt;&#xA;I made an excel sheet with all odd days in a year and a second column for the sum of the last 5 days.&lt;/p&gt;&#xA;" OwnerUserId="14461" LastEditorUserId="15544" LastEditDate="2015-09-23T09:26:22.927" LastActivityDate="2015-09-23T09:26:22.927" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22493" PostTypeId="2" ParentId="22491" CreationDate="2015-09-22T12:54:00.293" Score="11" Body="&lt;p&gt;The following answer looks much more complicated than it actually is. But it dictates a more &quot;logical&quot; way to solve this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two cases:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;All five dates are in the same month&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the middle date be $d/m$. Then the spending is  $5m+5d$  which is divisible by 5, hence cannot be $97$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Five dates are in two different months&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;The first month has $29/30$ days.&lt;/em&gt;  &lt;br&gt;&#xA;The five days are in an arithmetic progression mod 30 hence mod 5. Taking mod $5$ of the five days, we see that each days must be distinct modulo 5, hence the sum of days (not months) is divisible by 5. Therefore the sum of the five months must be 2 mod 5. It follows that there must be three dates on the first months and two on the second. So, the dates are &lt;code&gt;25,27,29,1,3&lt;/code&gt; which sum to $85$. So, the month has to be $\lfloor {97-85 \over 5}\rfloor $ $=2$, ie. February. This has only 28 days. However, on leap years this works and leads to the solution- &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;25 Feb, 27 Feb, 29 Feb, 1 Mar, 3 Mar&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;The first month has $31$ days.&lt;/em&gt;  &lt;br&gt;&#xA;We can divide in four cases according as how many days in the first month.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;25,27,29,31,1 -&amp;gt; sum exceeds 97&#xA;27,29,31,1,3 -&amp;gt; m = -0.2, not an integer&#xA;29,31,1,3,5 -&amp;gt; m = 5 = May&#xA;31,1,3,5,7 -&amp;gt; m = 7.2, not an integer&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This leads to the solution- &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;29 May, 31 May, 1 June, 3 June, 5 June&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;The first month has $28$ days, ie. February on a non leap year.&lt;/em&gt;  &lt;br&gt;&#xA;There are only four cases according as how many days fall in February. None gives a solution.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="15544" LastEditorUserId="15544" LastEditDate="2015-09-23T19:25:08.523" LastActivityDate="2015-09-23T19:25:08.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22495" PostTypeId="1" AcceptedAnswerId="22498" CreationDate="2015-09-22T13:39:14.473" Score="5" ViewCount="2648" Body="&lt;p&gt;As the title says it is a compete the sequence.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/crDkn.jpg&quot; alt=&quot;&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/8gSR8.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My thoughts:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From all the circle ones that aren't a circle, you can make a circle by rotating the pieces. From all the triangle ones that aren't a triangle, you can make a triangle by moving one line. But I can't figure out a rule for the square-ish things. (Of course the rules I came up with for the circle and triangle might be wrong.) &lt;/p&gt;&#xA;" OwnerUserId="16477" LastEditorUserId="52483" LastEditDate="2018-11-22T19:42:50.017" LastActivityDate="2019-01-18T16:33:24.083" Title="Complete the sequence" Tags="|pattern|visual|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22497" PostTypeId="2" ParentId="22495" CreationDate="2015-09-22T14:21:29.780" Score="1" Body="&lt;p&gt;I think it is the,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Third one in the first row&lt;br&gt;&#xA; There is a circle, a mirrored half of a triangle, and an upside down T in each each row&lt;br&gt;&lt;br&gt;&#xA; The first row is correct&lt;br&gt;&lt;br&gt;&#xA; The second row, you cut it vertically and switch the sides&lt;br&gt;&lt;br&gt;&#xA; The third row, you cut it down the middle vertically and flip the right side over the x axis and the left side over the y axis&lt;br&gt;&lt;br&gt;&#xA; The reason it isn't a circle, triangle, and U is because of the second row. There is no way to move the triangle and U to get the circle while preserving their shape, therefore, they need to move the same way the circle will which gives you the shapes in the first row. the rows do not have the same moves to get to the end result although they do build on each other in number of moves. The first row has 0 moves, the second has 1, and the third has 2.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13925" LastEditorUserId="13925" LastEditDate="2015-09-22T15:04:01.513" LastActivityDate="2015-09-22T15:04:01.513" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22498" PostTypeId="2" ParentId="22495" CreationDate="2015-09-22T14:24:33.333" Score="7" Body="&lt;p&gt;I'll go with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/VCRk0.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VCRk0.png&quot; alt=&quot;The top right answer&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There's a triangle, a circle and a 'U' shape. They each appear both 'normal', once with both halves mirrored along a vertical axis, and once with the left side intact and the right rotated 180 degrees.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Based on what's there already:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the three states for triangle and circle, and both the normal and mirrored state of the 'U' shape&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;what we need is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; left side of the U intact, right side rotated 180 degrees.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2156" LastActivityDate="2015-09-22T14:24:33.333" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22500" PostTypeId="1" AcceptedAnswerId="22506" CreationDate="2015-09-22T14:45:51.007" Score="15" ViewCount="828" Body="&lt;p&gt;I created a &lt;a href=&quot;https://puzzling.stackexchange.com/questions/22424/numbered-logicians-near-the-blue-eyes-island&quot;&gt;variant&lt;/a&gt; of the &lt;a href=&quot;https://xkcd.com/blue_eyes.html&quot; rel=&quot;noreferrer&quot;&gt;Blue Eyes problem&lt;/a&gt; where the logic took 4901 days (over 13 years) to totally play out.  I thought this was nice, but I was hoping for an even longer period of time.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, I could trivially I could do the following:  Suppose $n$ is the number of inhabitants, and the maximum size number on their heads is $n$, and they each have forehead number either $\lfloor n/2 \rfloor$ or $\lfloor n/2 \rfloor + 1$.   Then the amount of time it takes to discover their numbers is roughly proportional to $n^2/2$.   Thus, by increasing $n$ I can increase the amount of time &lt;strong&gt;quadratically&lt;/strong&gt;.  But I was wondering if it was possible to do better yet.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a way to design a Blue-Eyes-like logic puzzle where the amount of time for the islanders to figure out the answer is &lt;strong&gt;exponential&lt;/strong&gt; in the size of the problem?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A couple notes&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I don't know if this is possible -- my attempts so far haven't done very well.  Even rough ideas or explanations as to why it might not be possible would be appreciated.&lt;/li&gt;&#xA;&lt;li&gt;Anything about the problem can be changed I'd say:  for example, perhaps not everyone can see everyone else.  However, the core mechanic of ever increasing common knowledge should remain. Ideally, it would also be a simple to state puzzle, and perhaps with some other twists and turns for any potential puzzle solver.&lt;/li&gt;&#xA;&lt;li&gt;Suppose the islanders are immortal -- just in case the solution to your puzzle outpaces a person's expected lifetime.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2722" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-09-23T15:47:59.910" Title="Exponential time in a Blue Eyes variant" Tags="|mathematics|puzzle-creation|computer-science|blue-eyes|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22502" PostTypeId="1" AcceptedAnswerId="22507" CreationDate="2015-09-22T15:01:48.327" Score="4" ViewCount="576" Body="&lt;p&gt;Challenge: Create a 3x3 matrix of letters (NO REPEATS) such that the determinant of the matrix forms words from the 'multiplications'. For example, in a simple 2x2 case, take the matrix&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{bmatrix}O &amp;amp; I\\N &amp;amp; R\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the determinant of which is OR-IN. This forms 2 2-letter words. In the case of a 3x3, you will end up with 6 3-letter words if done correctly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't know how to find the determinant of a matrix? Fear not, here's the guide (or read a more detailed explanation &lt;a href=&quot;https://en.wikipedia.org/wiki/Determinant&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|A| = \begin{vmatrix} a &amp;amp; b &amp;amp; c\\d &amp;amp; e &amp;amp; f\\g &amp;amp; h &amp;amp; i \end{vmatrix} = a\begin{vmatrix} e &amp;amp; f\\h &amp;amp; i \end{vmatrix} - b\begin{vmatrix} d &amp;amp; f\\g &amp;amp; i \end{vmatrix} + c\begin{vmatrix} d &amp;amp; e\\g &amp;amp; h \end{vmatrix} = aei+bfg+cdh-ceg-bdi-afh .&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that looks too complicated, &lt;a href=&quot;http://www.purplemath.com/modules/determs2.htm&quot; rel=&quot;nofollow&quot;&gt;here's another method&lt;/a&gt; that some find easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't worry about the additions and subtractions, I'm only concerned with the words formed. &lt;s&gt;That said, if your words could be added and subtracted as base 26 numbers according to the determinant and result in a valid word, I would be incredibly impressed (a 100 reputation impressed, even).&lt;/s&gt; (This is mathematically impossible, determinant is always 0; sorry to get your hopes up). As usual, words are considered valid iff they are English words found on dictionary.com (read: this does not include (pre/suf)fixes, abbreviations, acronyms, etc.)&lt;/p&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="13970" LastEditDate="2015-10-21T14:35:30.130" LastActivityDate="2015-10-21T14:35:30.130" Title="Determinant of Word Matrix" Tags="|mathematics|letters|" AnswerCount="5" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="22503" PostTypeId="2" ParentId="22495" CreationDate="2015-09-22T15:06:37.207" Score="0" Body="&lt;p&gt;It is&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;the third one in the first row.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Let's look at the triangles:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Row 2, Column 2: First picture: the intact triangle&lt;/p&gt;&#xA;&lt;p&gt;Row 3, Column 2: Second picture: right half of the triangle has been rotated horizontally and vertically&lt;/p&gt;&#xA;&lt;p&gt;Row 1, Column 3: Third picture: other half of the triangle has been rotated horizontally and vertically, afterwards that whole picture is again rotated horizontally&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;And now for the circles&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Row 1, Column 2: First picture: the intact circle&lt;/p&gt;&#xA;&lt;p&gt;Row 3, Column 1: Second picture: half of the circle has been rotated horizontally and vertically&lt;/p&gt;&#xA;&lt;p&gt;Row 2, Column 1: See: half of the circle has been rotated horizontally and vertically, and then again rotated horizontally&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Aaaand finally the .. squarish things?&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Row 2, Column 3: intact U&lt;/p&gt;&#xA;&lt;p&gt;Row 1, Column 1: again, rotating the right half vertically and horizontacally&lt;/p&gt;&#xA;&lt;p&gt;That is the one in the new Row 1, Column 3&lt;/p&gt;&#xA;&lt;p&gt;Row 1, Column 1: final picture&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You can also see a pattern in the rows and columns:&lt;/p&gt;&#xA;&lt;p&gt;Triangles&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Rows: 2, 3, 1&lt;/p&gt;&#xA;&lt;p&gt;Columns: 2, 2, 3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Circles&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Rows: 1, 3, 2&lt;/p&gt;&#xA;&lt;p&gt;Columns: 2, 1, 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Squares&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Rows:    2, 1, 3&lt;/p&gt;&#xA;&lt;p&gt;Columns: 3, 1, 3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So all in all it's this:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Rows are: 1,2,3 - 1,2,3 - 1,2,3&lt;/p&gt;&#xA;&lt;p&gt;Columns are: 1,1,2 - 2,2,3 - 3,3,1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Cheers.&lt;/p&gt;&#xA;" OwnerUserId="16398" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-09-22T15:06:37.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22504" PostTypeId="1" CreationDate="2015-09-22T15:09:58.817" Score="0" ViewCount="749" Body="&lt;p&gt;Seven thieves are to distribute a number of coins between themselves. They find that it splits equally, but being astute mathematicians, they realize that if there were 2, 3, 4, 5, or 6 thieves to split coins among, there would always be one coin left over. How many coins are there?&lt;/p&gt;&#xA;" OwnerUserId="16490" LastEditorUserId="5373" LastEditDate="2016-10-24T00:40:12.783" LastActivityDate="2016-10-24T00:40:12.783" Title="seven thief and equal coins" Tags="|mathematics|" AnswerCount="1" CommentCount="1" ClosedDate="2015-09-22T15:49:58.897" ContentLicense="CC BY-SA 3.0" />
  <row Id="22505" PostTypeId="2" ParentId="22504" CreationDate="2015-09-22T15:17:25.590" Score="0" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 301&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The math is below:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 301/7 = 43&lt;br&gt;&#xA; 301/6 = 50 + 1/6&lt;br&gt;&#xA; 301/5 = 60 + 1/5&lt;br&gt;&#xA; 301/4 = 75 + 1/4&lt;br&gt;&#xA; 301/3 = 100 + 1/3&lt;br&gt;&#xA; 301/2 = 150 + 1/2  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13538" LastActivityDate="2015-09-22T15:17:25.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22506" PostTypeId="2" ParentId="22500" CreationDate="2015-09-22T15:19:53.157" Score="20" Body="&lt;p&gt;This might not qualify as a blue-eyes puzzle because it does not use common knowledge, but it involves chains of deductions based on nothing happening for a particular amount of time:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n$ villagers wear either black or white hats. They sit in a line, so that each villager can see all the hats in front of them, but not the hats behind them. If a villager ever deduces their own hat color, they get up and leave the line at noon on the next day (without anyone in front noticing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One day, a light is installed where all villagers can see it. The light remains on as long as at least one villager in the line has a black hat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the villagers all have black hats, nothing happens for the first $2^{n-1}-1$ days. On the $2^{n-1}$th day, the first villager leaves. $2^{n-2}$ days later, the second villager leaves. This continues until the last villager leaves on the $2^n-1$th day.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-09-22T15:42:16.730" LastActivityDate="2015-09-22T15:42:16.730" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22507" PostTypeId="2" ParentId="22502" CreationDate="2015-09-22T15:28:14.353" Score="4" Body="&lt;p&gt;Try&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\begin{bmatrix} B &amp;amp; S &amp;amp; L\\I &amp;amp; A &amp;amp; U\\M &amp;amp; D &amp;amp; T \end{bmatrix}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This produces the following words:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;BAT&lt;/strong&gt;&lt;br&gt;&#xA; &lt;strong&gt;SUM&lt;/strong&gt;&lt;br&gt;&#xA; &lt;strong&gt;LID&lt;/strong&gt;&lt;br&gt;&#xA; &lt;strong&gt;LAM&lt;/strong&gt;&lt;br&gt;&#xA; &lt;strong&gt;BUD&lt;/strong&gt;&lt;br&gt;&#xA; &lt;strong&gt;SIT&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="4421" LastEditDate="2015-09-22T15:37:10.650" LastActivityDate="2015-09-22T15:37:10.650" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22508" PostTypeId="2" ParentId="22502" CreationDate="2015-09-22T15:29:04.610" Score="4" Body="&lt;p&gt;$\begin{bmatrix}F &amp;amp; C &amp;amp; P\\A &amp;amp; O &amp;amp; U\\T &amp;amp; N &amp;amp; B\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;making the words FOB, FUN, CAB, CUT, PAN, and POT. Each column on its own is also a word.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-09-22T15:29:04.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22509" PostTypeId="2" ParentId="22502" CreationDate="2015-09-22T15:31:12.187" Score="3" Body="&lt;p&gt;To join the gang, here's what I came up with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\begin{bmatrix}H &amp;amp; C &amp;amp; N\\A &amp;amp; I &amp;amp; O\\P &amp;amp; G &amp;amp; T\end{bmatrix}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Makes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HIT, COP, NAG, NIP, HOG and CAT&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also note that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The columns in this matrix, just like f'' his answer, form words on their own.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2156" LastActivityDate="2015-09-22T15:31:12.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22510" PostTypeId="1" AcceptedAnswerId="22642" CreationDate="2015-09-22T16:11:51.810" Score="4" ViewCount="670" Body="&lt;p&gt;This is a graphical arrangement puzzle. Consider the following 36 tiles:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/adAA5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/adAA5.png&quot; alt=&quot;Tiles&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Each tile has up to 8 &lt;em&gt;connected&lt;/em&gt; directions and each tile has an initial &lt;em&gt;up&lt;/em&gt; side. (The edge to whose centre the central pentagram points is the upper edge.)&lt;/p&gt;&#xA;&lt;p&gt;These tiles have to be arranged into an arbitrary, single &lt;a href=&quot;https://en.wikipedia.org/wiki/Polyomino&quot; rel=&quot;nofollow noreferrer&quot;&gt;polyomino&lt;/a&gt; with a single restriction:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;No tile may be placed such that a (brown) &lt;em&gt;connection&lt;/em&gt; points directly at another tile that isn't pointing back.&lt;/strong&gt; &lt;br&gt;&#xA;Note that it is possible to have a &lt;em&gt;connection&lt;/em&gt; pointing at &amp;quot;nothing&amp;quot; (on the edge of the polyomino), but if it points at another tile, then that tile must have a &lt;em&gt;connection&lt;/em&gt; pointing back.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The resulting polyomino is &lt;strong&gt;scored&lt;/strong&gt; like this:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Dissect the polyomino (however you wish) into smaller rectangles such that each tile is part of exactly one such rectangle.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The number of &lt;em&gt;inner edges&lt;/em&gt; in these rectangles - i.e. all edges which connect two tiles of the rectangle - are counted, and that number is squared.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If any tile was rotated from its original position, a value has to be subtracted from your score: This value is the square of the total number of 90° rotations used (compared to the original tile-set.)  For example, if one tile was rotated clockwise by 90° and another by 180°, then the total number of 90° rotations is 3, and the penalty is &lt;span class=&quot;math-container&quot;&gt;$3^2=9$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Bonus score:&lt;/strong&gt; Solutions which include all 36 tiles in a square are ranked by the number of 'separate networks' they form - the less networks the better, i.e. the ultimate solution additionally forms a single &amp;quot;brown&amp;quot; network.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;h2&gt;Find the highest possible score for this puzzle.&lt;/h2&gt;&#xA;&lt;p&gt;(The tile-arrangement has to be shown.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;Example of a &lt;em&gt;valid&lt;/em&gt; arrangement (only 9 tiles):&lt;/h2&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AwPXh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AwPXh.png&quot; alt=&quot;Valid&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;As no tile was rotated, the score of this set is &lt;span class=&quot;math-container&quot;&gt;$7^2 + 1^2 + 0^2 = 50$&lt;/span&gt;&#xA;( A 2x3 rectangle has 7 inner edges, a 2x1 rectangle has 1 inner edge, a single tile has no inner edge. )&lt;/p&gt;&#xA;&lt;h2&gt;Example of an &lt;em&gt;invalid&lt;/em&gt; arrangement (only 9 tiles):&lt;/h2&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yLG8J.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yLG8J.png&quot; alt=&quot;Invalid&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The bottom-right corner tile points towards the central tile, which isn't pointing back.&lt;/p&gt;&#xA;&lt;h2&gt;2nd Example of an &lt;em&gt;invalid&lt;/em&gt; arrangement (only 4 tiles):&lt;/h2&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NIo1O.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NIo1O.png&quot; alt=&quot;Invalid&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Both C4 and A4 point at a tile (the one diagonal below) which is not pointing backwards. (That they two 'link' with each other via the corner is not important.)&lt;/p&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-09-26T11:03:29.270" Title="What is the largest, compact connected-network polyomino for these tiles?" Tags="|visual|polyomino|network|" AnswerCount="3" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="22511" PostTypeId="1" CreationDate="2015-09-22T16:37:12.493" Score="-6" ViewCount="225" Body="&lt;p&gt;Bob is mentoring Steve at a new job, and they are working their way through some poorly written instructions to install an update for some critical software.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things are going okay up until they find this sentence:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;YOU CANNOT INSTALL THIS UPDATE ON A TERMINAL SERVER CLIENT&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Bob doesn't see what the problem is, but Steve says &quot;Every word in that sentence is ambiguous!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bob demands that Steve prove this wild accusation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you present a list of statements proving that every single word in the sentence is ambiguous?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, present a list of clarifications showing that each word can be interpreted in more than one way. Bonus points awarded for finding extra criteria, but you have to find alternate meanings for all ten words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently this is too broad and confusing so here's a hint and a guide to the first two words.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;You&quot; can mean the person who is reading the statement, &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;You&quot; can be a proper noun, as in a person's name.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;likewise:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Cannot&quot; can indicate that you are physically incapable of doing something, and the statement is a warning not to attempt to do so as that might lead to physical injury in the attempt...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Cannot&quot; can indicate that you are not permitted but are certainly capable, and the statement is a command...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or... (bonus)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Cannot&quot; is Mr. You's last name.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The goal of this exercise is to produce the remaining 8 words's dual meanings.&lt;/p&gt;&#xA;" OwnerUserId="13068" LastEditorUserId="13068" LastEditDate="2015-09-22T19:08:03.007" LastActivityDate="2015-09-22T19:08:03.007" Title="Bob and Steve find an ambiguous sentence - but HOW ambiguous?" Tags="|english|" AnswerCount="1" CommentCount="0" ClosedDate="2015-09-22T18:41:05.757" ContentLicense="CC BY-SA 3.0" />
  <row Id="22512" PostTypeId="2" ParentId="22511" CreationDate="2015-09-22T16:58:43.557" Score="3" Body="&lt;p&gt;Base meaning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This server that uses a terminal client (to connect to other computers) does not support this update.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You cannot install this update on a terminal server client, but other people can.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cannot:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You are not allowed to install the update, but it's perfectly possible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Install:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The computer can successfully run the software, but you cannot give the software power over the computer (eg root access).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You cannot install any updates at all onto this computer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Update:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You cannot install this update to the critical software OR to the operating system.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You must use another computer to install this update; the update can exist on this computer, but an auxiliary compute is necessary.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The terminal server client designated &quot;A&quot; cannot hold this update.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Terminal:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This computer is at the end of a network; the update is only functional on computers that connect multiple other computers to each other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Server: This robot butler does not support the update. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This computer is not a server but a client of servers, ie it connects to servers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Client:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The client (the software connecting to another computer) cannot handle the update, but that is unrelated to the client (the computer connecting to another computer).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="11876" LastEditDate="2015-09-22T18:27:08.030" LastActivityDate="2015-09-22T18:27:08.030" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="22513" PostTypeId="1" AcceptedAnswerId="22514" CreationDate="2015-09-22T17:16:48.280" Score="5" ViewCount="282" Body="&lt;p&gt;&lt;em&gt;This puzzle is based on &lt;a href=&quot;https://puzzling.stackexchange.com/users/13970/needaname&quot;&gt;NeedAName&lt;/a&gt;'s recent puzzle,  &lt;strong&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/22502/word-matrices&quot;&gt;Determinant of Word Matrix&lt;/a&gt;&lt;/strong&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The challenge:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Create a $3\times3$ matrix of distinct letters (i.e. no repeated letters) &lt;strong&gt;&lt;em&gt;with no vowels in the middle row&lt;/em&gt;&lt;/strong&gt; such that the six terms generated when calculating the determinant form valid English words.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;Words are considered valid &lt;strong&gt;if and only if they are English words found on &lt;a href=&quot;http://dictionary.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dictionary.com&lt;/a&gt;&lt;/strong&gt; (not including (pre/suf)fixes, abbreviations, acronyms, proper names, etc.).&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example, with a simple $2\times2$ matrix $\begin{bmatrix}O &amp;amp; I\\N &amp;amp; R\end{bmatrix}$, the determinant is $OR-IN$.&lt;br&gt;&#xA;This forms two 2-letter words.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;With a $3\times3$ matrix $\begin{bmatrix} A &amp;amp; B &amp;amp; C\\D &amp;amp; E &amp;amp; F\\G &amp;amp; H &amp;amp; I \end{bmatrix}$&#xA; you will end up with six 3-letter words:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$AEI + BFG + CDH -CEG - BDI - AFH$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the actual determinant doesn't matter, the easiest way to think of this is as a $3\times5$ matrix with the first two columns repeated:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{bmatrix} A &amp;amp; B &amp;amp; C &amp;amp; A &amp;amp; B\\D &amp;amp; E &amp;amp; F &amp;amp; D &amp;amp; E\\G &amp;amp; H &amp;amp; I &amp;amp; G &amp;amp; H \end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you just need to make sure all the intact diagonals $\searrow$ and $\swarrow$ are valid words.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(For more information on finding determinants, see &lt;a href=&quot;https://en.wikipedia.org/wiki/Determinant&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; or &lt;a href=&quot;http://www.purplemath.com/modules/determs2.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-09-22T18:38:55.587" Title="Word Matrices, Level 2" Tags="|mathematics|word|letters|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22514" PostTypeId="2" ParentId="22513" CreationDate="2015-09-22T17:59:12.413" Score="3" Body="&lt;p&gt;With the help of a list of three-letter words and an automated search:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{bmatrix} A &amp;amp; I &amp;amp; O\\F &amp;amp; D &amp;amp; C\\E &amp;amp; T &amp;amp; S \end{bmatrix}$&#xA;producing the words ADS, ACT, IFS, ICE, OFT, and ODE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pastebin.com/NGRPGWrx&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; are all 50 results my program found.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-09-22T18:38:55.587" LastActivityDate="2015-09-22T18:38:55.587" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22517" PostTypeId="2" ParentId="22502" CreationDate="2015-09-22T18:47:48.780" Score="4" Body="&lt;p&gt;&lt;strong&gt;A simple idea to generate solutions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the matrix be-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{bmatrix} a &amp;amp; b &amp;amp; c\\d &amp;amp; e &amp;amp; f\\g &amp;amp; h &amp;amp; i \end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we need the following to be valid words.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;aei afh bdi bfg cdh ceg&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This can be treated as a substitution cipher and solved using a brute force dictionary search.  This produces lots of solutions. &lt;a href=&quot;http://pastebin.com/5ddTQP8C&quot; rel=&quot;nofollow&quot;&gt;Here are 954 solutions&lt;/a&gt;. I stopped there because the engine I was using only shows top 1000 results (of which 954 were unique).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also tried a 4x4 matrix search. It didn't produce any complete result but some got close. The searching key is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;AFKP AFLO AGJP AGLN AHJO AHKN BEKP BELO BGIP BGLM BHIO BHKM CEJP CELN CFIP CFLM CHIN CHJM DEJO DEKN DFIO DFKM DGIN DGJM.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://quipqiup.com/&quot; rel=&quot;nofollow&quot;&gt;Try here at quipquip&lt;/a&gt;. However it might be possible if we allow some repetitions., which I haven't tried yet.&lt;/p&gt;&#xA;" OwnerUserId="15544" LastEditorUserId="15544" LastEditDate="2015-09-23T05:20:54.527" LastActivityDate="2015-09-23T05:20:54.527" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22519" PostTypeId="2" ParentId="22510" CreationDate="2015-09-22T20:05:58.590" Score="6" Body="&lt;h2&gt;6x6 solution with rotations: 3479&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $60^2 - 11^2 = 3479$ (single network)&lt;br&gt;&lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/7Vqn6.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Inkscape image&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Not part of the answer, but I have created an &lt;a href=&quot;https://inkscape.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Inkscape&lt;/a&gt; image which should make handling the tiles way easier. Download it from here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://imgh.us/polyomino.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://imgh.us/polyomino.svg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &lt;strong&gt;snap&lt;/strong&gt; to work configure it like in the screenshot below (the three activated buttons near the right border):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oR5qW.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2015-09-24T21:43:49.850" LastActivityDate="2015-09-24T21:43:49.850" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="22523" PostTypeId="2" ParentId="22480" CreationDate="2015-09-22T22:55:47.137" Score="0" Body="&lt;p&gt;In my answer, the integral was performed in cylindrical coordinates. The tricky part is getting a simple $\rho(\theta)$, which can be calculated as shown in the following diagram: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pImsW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pImsW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives us:&#xA;$$r^2=\rho^2 +d^2-2\rho d \cos{\left(\theta +\frac{\pi}{2}\right)}$$&#xA;$$\Rightarrow\frac{2\rho(\theta)}{r}=\sqrt{2}\left(\sqrt{1+\sin^2\theta}-\sin\theta\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align*}&#xA;V_{\text{hole}}&amp;amp;=16\int^{\pi/2}_{\pi/4}{\int^0_{\rho(\theta)}{\int^\sqrt{(r/2)^2-\rho^2}_0}{\rho dz d\rho d\theta}} \\&#xA;\frac{V_{\text{hole}}}{r^3} &amp;amp;=\frac{\pi}{6}-\frac{2}{3}\int^{\pi/2}_{\pi/4}{\left[1-\left(\frac{2\rho(\theta)}{r}\right)^2\right]^{3/2} d\theta} \\&#xA;&amp;amp;= \frac{\pi}{6}-\frac{4\sqrt{2}}{3}\int^{\pi/2}_{\pi/4}{\left[\cos2\theta+2\sin\theta\sqrt{1+\sin^2\theta} - \frac{3}{2}\right]^{3/2} d\theta} \\&#xA;&amp;amp;\approx 0.280225788122747024207247869453505\cdots&#xA;\end{align*}$$&lt;/p&gt;&#xA;" OwnerUserId="15960" LastActivityDate="2015-09-22T22:55:47.137" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22524" PostTypeId="2" ParentId="21992" CreationDate="2015-09-23T00:17:23.093" Score="1" Body="&lt;p&gt;I sketched a playing stategy for the line. It's pretty complicated, but if I didn't miss any point, the line should be able to reach the flags, no matter where they're placed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to evade getting in a locked position, the line needs to restrict its movements in a way such that the projection of the set of all visited points into the horizontal plane has a total area of zero (i.e. It's a curve). This way, it will always be able to escape by going up or down from the neighborhood of any point in the plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that any flag is reachable in this way, first notice that the ones in unvisited areas are trivially reachable within these constraints: As long as the line is kept vertical, its shadow will not sweep any area on the xy-plane. For areas which have already been passed by, the mechanism is slightly more complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During vertical motion, small horizontal displacements would be necessary to avoid self collision. However, it would not be a problem, since for any obstacle at some nonzero horizontal distance, an even smaller horizontal displacement would be possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose, now, the flag lies on the curve. It can be shown, upon suitable tameness conditions (piecewise differentiability), that there can be found in its neighborhoood infinitely many line segments of unvisited points. These correspond to vertically unbounded areas which are completely free for maneuver. They can be reached from above or below by using the escape method explained at the start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider one of those areas. In the worst case, every point directly above and below the flag is already visited. But by simultaneously moving forward, up/down, and tilting the top it is possible to make it reach the goal. After that, the vertical orientation can be restablished without self-intersections, by going backwards, tilting back up and moving downwards fast enough to compensate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This entire process can be done in arbitrarily tight horizontal spaces, by shrinking the tilting angle (tilting less). Since the only rotation involves the plane of movement, it preserves the property that the projection of visited points forms a meagre set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an illustration of this mechanism:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/1aIpq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1aIpq.png&quot; alt=&quot;Line&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please, don't comment. I used all my drawing skill to make this one&lt;/p&gt;&#xA;" OwnerUserId="14219" LastEditorUserId="14219" LastEditDate="2015-09-23T04:26:10.363" LastActivityDate="2015-09-23T04:26:10.363" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22525" PostTypeId="2" ParentId="21992" CreationDate="2015-09-23T02:26:37.173" Score="2" Body="&lt;p&gt;&lt;strong&gt;No, it is impossible to block off the line from any point, whether the line is finite or infinite.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say you have some maze path &lt;em&gt;M&lt;/em&gt; already dragged out by the line.  Because no two &quot;walls&quot; may be adjacent (there is always a gap between them), it is always possible to go back arbitrarily close to any point on &lt;em&gt;M&lt;/em&gt; by backtracking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, pick any point &lt;em&gt;P&lt;/em&gt; in the space not intersecting with &lt;em&gt;M&lt;/em&gt;.  There must be another point &lt;em&gt;S&lt;/em&gt; on the path &lt;em&gt;M&lt;/em&gt; that is the shortest distance to &lt;em&gt;P&lt;/em&gt;.  Since we can always backtrack, we trace &lt;em&gt;M&lt;/em&gt; back to any point between &lt;em&gt;P&lt;/em&gt; and &lt;em&gt;S&lt;/em&gt;, and then we have a straight shot to &lt;em&gt;P&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="12795" LastActivityDate="2015-09-23T02:26:37.173" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22526" PostTypeId="1" AcceptedAnswerId="22527" CreationDate="2015-09-23T02:35:28.827" Score="1" ViewCount="192" Body="&lt;p&gt;There are $n$ ants randomly distributed on a ring facing clockwise or anticlockwise. Each ant is carrying a flag numbered $1,\ldots,n$.  They all start moving around the ring at the same speed (in the direction they were facing). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When they meet they swap flags and turn around and continue walking in the direction they came from (still at the same speed). Effectively a perfect bounce of the ants. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find a configuration where the initial position is never recovered, or prove that the initial position will always be recovered. &lt;/p&gt;&#xA;" OwnerUserId="15942" LastActivityDate="2015-09-23T02:55:56.543" Title="Flag carrying ants in a ring" Tags="|mathematics|logical-deduction|" AnswerCount="1" CommentCount="5" ClosedDate="2015-09-23T10:36:26.227" ContentLicense="CC BY-SA 3.0" />
  <row Id="22527" PostTypeId="2" ParentId="22526" CreationDate="2015-09-23T02:55:56.543" Score="3" Body="&lt;p&gt;The initial position will be recovered. The flags keep circulating, so after one circuit they are all back where they started. At that time there will be a permutation of the ants. The ants will be facing the same direction as the one in their location at the start. Some power of the permutation is the identity, so the initial configuration will recur.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2015-09-23T02:55:56.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22528" PostTypeId="1" CreationDate="2015-09-23T03:21:15.747" Score="0" ViewCount="1595" Body="&lt;p&gt;Given a grid of $n \times m$ points, on a sheet of paper, what's the minimum amount of lines (which can extend infinitely) required to pass through each point, that you can draw without lifting a pencil?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sample grid and answer: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/7797/connect-the-dots&quot;&gt;Connect the dots&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; Lines are of less width then points are radius, all lines must be straight.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-09-30T02:04:50.950" Title="What is the minimum number of line segments that need to be made to cross all points on a $n \times m$ grid?" Tags="|mathematics|geometry|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="22529" PostTypeId="2" ParentId="22528" CreationDate="2015-09-23T03:46:26.090" Score="4" Body="&lt;p&gt;I hope the lines are not required to intersect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want maximum number of points on each line. We can have a maximum of $m$ points on it (for $m\geq n$). Hence we draw $n$ parallel lines, each with $m$ dots.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-09-23T03:46:26.090" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22530" PostTypeId="1" AcceptedAnswerId="22531" CreationDate="2015-09-23T06:08:37.847" Score="0" ViewCount="1433" Body="&lt;p&gt;On a table there are three cups and one banknote (a dollar bill will do). These are the only objects you may use, with the goal of placing one of the cups, &#xA;&lt;em&gt;upright&lt;/em&gt;, &lt;em&gt;above&lt;/em&gt; the other two, without any of the cups touching another cup. The cups must be upright at all times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wfPdV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wfPdV.png&quot; alt=&quot;Three cups and bill drawing&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's nothing special about the bill; it will bend if you suspend it between two cups and place the third on top of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are at least three very different solutions to this problem!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;No lame tricks:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;there &lt;em&gt;is&lt;/em&gt; gravity&lt;/li&gt;&#xA;&lt;li&gt;the cups are of regular weight; correct solutions can even support tumblers or shot glasses&lt;/li&gt;&#xA;&lt;li&gt;the bill is a regular bill, such as a $1 bill&lt;/li&gt;&#xA;&lt;li&gt;there's nothing in the cups&lt;/li&gt;&#xA;&lt;li&gt;can't use any other objects or external support (e.g. holding the cup in someone's hand)&lt;/li&gt;&#xA;&lt;li&gt;can't melt or otherwise dramatically alter the cups (think &quot;glasses&quot; instead)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4185" LastEditorUserId="15544" LastEditDate="2015-09-24T05:59:07.750" LastActivityDate="2015-09-24T05:59:07.750" Title="Make a glass sit above two other glasses using only a regular banknote" Tags="|physics|" AnswerCount="2" CommentCount="7" ClosedDate="2015-09-23T21:53:17.533" ContentLicense="CC BY-SA 3.0" />
  <row Id="22531" PostTypeId="2" ParentId="22530" CreationDate="2015-09-23T06:21:21.850" Score="3" Body="&lt;p&gt;You could fold the bill back and forth into a W (or maybe more folds) and use it as a rigid support. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could push the bottom two glasses together and put the money in between the bottom glasses and the top one. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;You could roll the money to form a column, with the one end bent and flattened at the height of one of the other cups. Then rest the top cup on the money column with the fanned money and one of the cups for balance. No need for the second cup. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You could roll it in a column and balance the top one on that. (thanks @Tim)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;2a. Same as above, but with the note in one of the cups. &lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;Fold the note long wise, near one of the long edges and parallel to that edge. Wrap it around the lip of one of the cups. Put that cup in the middle of a stack. The three cups will be stacked upright, but not touching because of the note (this depends on the definition of &quot;above&quot;). &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2015-09-24T02:50:08.710" LastActivityDate="2015-09-24T02:50:08.710" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="22532" PostTypeId="2" ParentId="22530" CreationDate="2015-09-23T07:18:10.843" Score="2" Body="&lt;p&gt;Your picture already has the right arrangement:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DcUQf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I don't have any cups on hand in the dorm, but you get the idea.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I interpreting the problem wrongly, or is it just that easy? =)&lt;/p&gt;&#xA;" OwnerUserId="9718" LastActivityDate="2015-09-23T07:18:10.843" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22533" PostTypeId="1" AcceptedAnswerId="22535" CreationDate="2015-09-23T07:27:13.393" Score="2" ViewCount="585" Body="&lt;p&gt;$n$ villagers wear either black or white hats. The truth is that all the hats are black, but this is not known. They sit in a line, so that each villager can see all the hats in front of them, but not the hats behind them. If a villager ever deduces the fact &lt;strong&gt;that all the villagers (including himself) know that their hats are black&lt;/strong&gt;, then he leaves at noon on the next day (without anyone in front noticing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One day, a light is installed where all villagers can see it. The light remains on as long as at least one villager in the line has a black hat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Will anyone ever leave? If so, after how long?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-09-23T09:17:58.690" Title="Knowledge about common knowledge (hats puzzle)" Tags="|logical-deduction|hat-guessing|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="22534" PostTypeId="2" ParentId="22500" CreationDate="2015-09-23T07:29:05.070" Score="0" Body="&lt;p&gt;I'm hoping that the following Q increases the time exponentially, but I don't know the answer myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/22533/knowledge-about-common-knowledge-hats-puzzle&quot;&gt;Knowledge about common knowledge (hats puzzle)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@f&quot; Yes, I have mostly copied it from you. Thanks for the idea. (Must I apologise?)&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2015-09-23T07:29:05.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22535" PostTypeId="2" ParentId="22533" CreationDate="2015-09-23T07:45:17.410" Score="0" Body="&lt;p&gt;I think that only for $n=1$ a villager can leave because that is the trivial case. In any other case the villager at the back of the line can never deduce his own color, even for $n=2$.&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-09-23T07:45:17.410" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22536" PostTypeId="2" ParentId="22533" CreationDate="2015-09-23T07:46:22.893" Score="0" Body="&lt;p&gt;I am very frequently very incorrect when it comes to Blue Eyes puzzles, but since you must make deductions about your hat color based on the light and the light cannot go out until someone leaves and no one can leave until everyone knows their hat color, I postulate that for $n &amp;gt; 1$ no one will ever leave. I swear I typed most of this up before Ivo posted his similar answer.&lt;/p&gt;&#xA;" OwnerUserId="16394" LastEditorUserId="16394" LastEditDate="2015-09-23T07:57:00.850" LastActivityDate="2015-09-23T07:57:00.850" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22537" PostTypeId="1" CreationDate="2015-09-23T08:02:45.897" Score="9" ViewCount="880" Body="&lt;p&gt;The database of mini-facebook only has space for the data of 100 participants. These 100 participants want to establish as many mini-facebook friendships among them as possible. However, mini-facebook has introduced the following set of ground rules:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Mini-facebook friendships are always symmetric.&lt;/li&gt;&#xA;&lt;li&gt;Mini-facebook only accepts one mini-facebook friendship or mini-facebook un-friend request per second. If two or more requests arrive simultaneously, they are all rejected.&lt;/li&gt;&#xA;&lt;li&gt;Any mini-facebook friendship request must consist of four distinct participants $A,B,C,D$ (ordered) so that none of the four pairs $AB$, $BC$, $CD$, $DA$ are currently mini-facebook friends. Mini-facebook will then establish the new friendship $AD$.&lt;/li&gt;&#xA;&lt;li&gt;Any mini-facebook un-friend request must consist of two distinct participants $A,B$ that are currently friends. Mini-facebook will then delete the new friendship $AB$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What is the maximum number of mini-facebook friendships that can exist simultaneoulsy at any moment in time?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2022-02-20T09:32:53.793" Title="The mini-facebook puzzle" Tags="|mathematics|strategy|combinatorics|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22538" PostTypeId="1" CreationDate="2015-09-23T08:07:54.920" Score="2" ViewCount="282" Body="&lt;p&gt;There are n strings in a bag. In one pass, you randomly pick one end of a string and then one more string.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you tie these two ends together and put the resulting string back into the bag. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you do another pass and so on. The game ends, when there are no loose ends remaining in the bag. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the probability that there will be exactly $t$ loops at the end?&lt;/p&gt;&#xA;" OwnerUserId="12703" LastEditorUserId="8874" LastEditDate="2015-09-23T08:15:45.663" LastActivityDate="2015-10-07T18:27:58.910" Title="N strings in a bag" Tags="|probability|combinatorics|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22539" PostTypeId="2" ParentId="22538" CreationDate="2015-09-23T09:00:49.027" Score="1" Body="&lt;p&gt;You can certainly get a recurrence relationship, I can't immediately see how to get an explicit formula from it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $P_n(t)$ be the probability of making $t$ loops from $n$ strings. It's clear that $P_0(t)=P_n(0)=0$ for $t&amp;gt;0, n&amp;gt;0$, and $P_0(0)=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After picking the end of the first string, there are $2n-1$ ends to select from, of which one will form a loop, and the remainder will join two strings to effectively become one, so $$P_n(t)=\frac{1}{2n-1}P_{n-1}(t-1)+\frac{2n-2}{2n-1}P_{n-1}(t)$$&lt;/p&gt;&#xA;" OwnerUserId="12625" LastEditorUserId="12625" LastEditDate="2015-09-23T09:07:25.913" LastActivityDate="2015-09-23T09:07:25.913" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22540" PostTypeId="2" ParentId="22533" CreationDate="2015-09-23T09:17:58.690" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Even if they don't all have to gain knowledge and leave at the same time, nobody can leave for $n&amp;gt;1$ because your condition for leaving is that an individual knows that &lt;strong&gt;everyone is wearing a black hat&lt;/strong&gt;. How can the person at the front ever work out if he is wearing a white hat or if there are any white hats anywhere behind him? Consider the BW case. The person at the back can only see a white hat so he knows he is wearing a black hat but cannot leave the line because at least one person is not wearing a black hat, so the light stays on. The person at the front would therefore be wrong to assume that he is wearing a black hat however long he waits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-09-23T09:17:58.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22541" PostTypeId="2" ParentId="22538" CreationDate="2015-09-23T09:41:26.330" Score="1" Body="&lt;p&gt;Here's a partial answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we count tied-together strings as a single string, each pass reduces the number of strings by 1. The probability of not picking the same string on the $(n-k)$-th pass is $\frac{2k-2}{2k-1}$. Otherwise, it is $\frac{1}{2k-1}$. Therefore, the probability of getting only one loop is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P[t=1] = \frac{2^{n-1}(n-1)!}{(2n-1)!!}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get $t-1$ more loops, we need to divide $P[t=1]$ by any $t-1$ numerators out of the $n-1$ numerators we have (thus replacing the $2k-2$ on the $(n-k)$-th pass with a $1$). There are $\binom{n-1}{t-1}$ ways to achieve this, and each of them are equally likely. We therefore have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align*}&#xA;P[t] &amp;amp;= \frac{2^{n-1}(n-1)!}{(2n-1)!!}\cdot\sum_{j=1}^{\binom{n-1}{t-1}} \frac{1}{2^{t-1} S_j(n-1,t-1)} \\ &#xA; &amp;amp;= \frac{2^{n-t}(n-1)!}{(2n-1)!!}\cdot\sum_{j=1}^{\binom{n-1}{t-1}} \frac{1}{S_j(n-1,t-1)} \\&#xA; &amp;amp;= \frac{2^{n-t}}{(2n-1)!!} \cdot \sum_{j=1}^{\binom{n-1}{n-t}}{ S_j(n-1,n-t)}&#xA;\end{align*}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $S_j(n,k)$ is the product of some distinct combination of $k$ numbers between $1$ and $n$, inclusive. For the case of $t=2$, the sum on the second line is the $n-1$-th harmonic number. I don't know how to reduce this for higher values of $t$ (hence the partial answer).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Instead of dividing out the $2k-2$s we don't want, we can directly multiply the appropriate numerators together. This gives us the equivalent (but slightly simplified) expression on the third line I appended to the equation block.&lt;/p&gt;&#xA;" OwnerUserId="15960" LastEditorUserId="15960" LastEditDate="2015-10-07T18:27:58.910" LastActivityDate="2015-10-07T18:27:58.910" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22542" PostTypeId="2" ParentId="22537" CreationDate="2015-09-23T11:04:57.793" Score="3" Body="&lt;p&gt;My guess is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $97 + 96 + 95 + ... + 1 = (97*98)/2 =4,753$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you do it like this&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; set aside two people, call them B and C. Then interconnect everyone else using B en C as the middle two arguments of the friendship. The number of connections will be like this: the first person connects with the 97 other people that are not B or C, the second with everyone except the first, and so on.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-09-23T11:04:57.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22543" PostTypeId="2" ParentId="22537" CreationDate="2015-09-23T11:39:53.323" Score="7" Body="&lt;p&gt;Building on Ivo Beckers' answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Out of the 4950 possible friendships (100 x 99 / 2), 197 of them cannot occur. &#xA;Those 197 are B with 98 others, C with 98 others, and B-C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I wondered if there was improvement to be made, and I think there is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Form groups like this: &lt;br&gt;B - random1 - C - random2. &#xA;  &lt;br&gt;Neither of the 4 exists already, so B is connected to random2 now. &lt;br&gt;&lt;br&gt;Repeat so for any of the other 96 options (100-B-C-random1-random2) for random2. &#xA;  &lt;br&gt;This implies we've made another 97 matches, for a total of 4753 + 97 = &lt;strong&gt;4850&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now can we take this further? Can we make combinations that create links for C?&#xA;&lt;br&gt;They would have to look like this:&#xA;C - random1 - random2 - random3&#xA;&lt;br&gt; there are however no options left for 'random2' which isn't linked to any of the others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;To go beyond that: Can we link B and C? Combination would have to be like B - random1 - random2 - C. There are no 'random1' and 'random2' left from the pool that are unlinked, nor is there a B-random1 connection possible.&lt;/p&gt;&#xA;" OwnerUserId="2156" LastActivityDate="2015-09-23T11:39:53.323" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22544" PostTypeId="2" ParentId="22378" CreationDate="2015-09-23T14:31:16.850" Score="0" Body="&lt;p&gt;Just a shot in the dark: &lt;br&gt;&lt;br&gt;&#xA;One of us  -  $500 &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Who is Wa Kai? My profile picture fits it really god :)&lt;br&gt; COOKIES?!?!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14461" LastActivityDate="2015-09-23T14:31:16.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22546" PostTypeId="2" ParentId="22500" CreationDate="2015-09-23T14:55:43.353" Score="1" Body="&lt;p&gt;These puzzles break my brain, but I'll give it a shot:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: This is wrong, this is just another quadratic version.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are &lt;em&gt;n&lt;/em&gt; islanders, each wearing a hat with a number on it. It's common knowledge that they are all wearing hats with non-negative integers on them, and they can all see each others' hats, but nobody knows their own hat number. At midnight, any islander that knows they are wearing a hat with the smallest number (or tied for smallest) of those on the island must leave on the ferry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One day, an oracle shows up and tells them that nobody is wearing a hat with a number greater than &lt;em&gt;m&lt;/em&gt;. Suppose that everybody actually has the number 0 on their hat. What happens?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let &lt;em&gt;n&lt;/em&gt;=&lt;em&gt;m&lt;/em&gt;=100. I think that after the first night, it's now common knowledge that there are at least two people with hats less than 100, since everybody who sees 99 people with hats saying 100 would have left, and nobody left. After that, though, we can only rule out one specific configuration each night: after night 2, we know that it's not [99]*2, [100]*98. After night 3, we know that it's not [99]*3, [100]*97. After 100 nights, we know that there is at least one person who has neither a 99 nor a 100, and so on. Once all possibilities except a hundred zeros have been eliminated everybody leaves at once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I think the time is about &lt;em&gt;m&lt;/em&gt;^&lt;em&gt;n&lt;/em&gt; (something like &lt;em&gt;m&lt;/em&gt;^&lt;em&gt;n&lt;/em&gt; - &lt;em&gt;m&lt;/em&gt;). If we arbitrarily constrain &lt;em&gt;m&lt;/em&gt; to be equal to &lt;em&gt;n&lt;/em&gt;, then we get better than exponential growth. With constant &lt;em&gt;m&lt;/em&gt; it's exponential in &lt;em&gt;n&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Nope, everything but the last paragraph is right (I think) but then I got the actual math wrong. It's just &lt;em&gt;m&lt;/em&gt; * &lt;em&gt;n&lt;/em&gt; as pointed out by f''.&lt;/p&gt;&#xA;" OwnerUserId="9523" LastEditorUserId="9523" LastEditDate="2015-09-23T15:47:59.910" LastActivityDate="2015-09-23T15:47:59.910" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22547" PostTypeId="1" CreationDate="2015-09-23T15:04:13.833" Score="2" ViewCount="192" Body="&lt;p&gt;(I did not make this one up, but it's one of my favorites and I didn't see it on here!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a long shift at the restaurant, your fellow waiter Jeremiah proposes the following game: Start with an empty round table, and take turns placing a single round plate anywhere you want on the table. The plates cannot be stacked on top of each other but they can overhang the edge of the table. The first person who is not able to place a plate loses. Jeremiah's going to let you choose who goes first because he's a nice guy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you go first or second, and what's your strategy?&lt;/p&gt;&#xA;" OwnerUserId="10851" LastActivityDate="2015-09-23T15:12:40.503" Title="Round plates on a round table" Tags="|mathematics|geometry|" AnswerCount="1" CommentCount="4" ClosedDate="2015-09-23T15:28:19.567" ContentLicense="CC BY-SA 3.0" />
  <row Id="22549" PostTypeId="2" ParentId="22547" CreationDate="2015-09-23T15:12:40.503" Score="5" Body="&lt;p&gt;The answer relies on&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Symmetry&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Choice:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You go first&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Strategy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You put yours first exactly in the center. After that, you mirror (well, point-symmetrical to the center) exactly what the other guy does. You'll always be certain to have a spot (given he had one last turn), so he'll run out first.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2156" LastActivityDate="2015-09-23T15:12:40.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22550" PostTypeId="1" AcceptedAnswerId="22747" CreationDate="2015-09-23T15:41:02.607" Score="4" ViewCount="2179" Body="&lt;p&gt;&lt;em&gt;Here's the first riddle I invented for Puzzling SE. Hope you'll enjoy solving it.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;This story takes place during World War II. A German spy is assigned the task to sneak overnight into a military base near London and steal secret documents of utmost importance.&lt;/p&gt;&#xA;&lt;p&gt;The documents are kept inside 30 safes of reinforced unbreakable steel, and each safe is protected by a numeric secret code. Therefore the only way to get the documents is to open each safe with the correct code. If the wrong code is entered, the safe will immediately trigger a siren which will alert all military personnel, so the spy has only one try available.&lt;/p&gt;&#xA;&lt;p&gt;German Intelligence was able to provide the spy with the following information: &lt;br&gt;&#xA;First, it looks like the codes were not chosen at random, but instead follow some unknown pattern. Second, they intercepted half of the safes' codes, which they communicate to the spy:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;Safe   Code&#xA;#1     2&#xA;#2     4&#xA;#3     6 &#xA;#4     8&#xA;#5     10&#xA;#6     12&#xA;#7     14&#xA;#8     16&#xA;#9     18&#xA;#10    20&#xA;#11    22&#xA;#12    24&#xA;#13    26&#xA;#14    28&#xA;#15    30&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;The second half of the safes' codes is still unknown. German Intelligence advises the spy to delay the mission until all codes have been discovered; however, the spy believes that the pattern is obvious, as a safe's code appears to be the double of the safe's number.&lt;/p&gt;&#xA;&lt;p&gt;That same night, the spy enters unseen the military base, finds the safes, silently opens each one of them with the correct code and steals the documents. Everything goes well from safe #1 to #17.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;However, when the spy enters the code 36 on safe #18, the safe doesn't open; its alarm triggers, and the spy is captured.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Can you find why? If not, continue reading.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;A few weeks later, the military base has sixty more safes delivered. The codes on these safes are programmed following the same pattern.&lt;/p&gt;&#xA;&lt;p&gt;After the end of the war, the commander of the base is asked by the High Command to compile a list of all safes and codes. He produces the following document:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;Safe   Code   Safe   Code   Safe   Code   Safe   Code   Safe   Code   Safe   Code&#xA;#1     2      #16    32     #31    28     #46   -42     #61    34     #76    46     &#xA;#2     4      #17    34     #32    30     #47   -40     #62    36     #77    48    &#xA;#3     6      #18    25     #33    32     #48   -38     #63    38     #78    50&#xA;#4     8      #19    38     #34    34     #49   -36     #64    40     #79    52   &#xA;#5     10     #20    40     #35    36     #50    30     #65    42     #80    25    &#xA;#6     12     #21    42     #36    38     #51    32     #66    44     #81    27    &#xA;#7     14     #22    44     #37    40     #52    34     #67    46     #82    29&#xA;#8     16     #23    46     #38    42     #53    36     #68    48     #83    31&#xA;#9     18     #24    48     #39    44     #54    38     #69    50     #84    33&#xA;#10    20     #25    50     #40   -54     #55    40     #70    34     #85    35&#xA;#11    22     #26    52     #41   -52     #56    42     #71    36     #86    37&#xA;#12    24     #27    54     #42   -50     #57    44     #72    38     #87    39&#xA;#13    26     #28    56     #43   -48     #58    46     #73    40     #88    41&#xA;#14    28     #29    58     #44   -46     #59    48     #74    42     #89    43&#xA;#15    30     #30    26     #45   -44     #60    32     #75    44     #90    38&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;&lt;strong&gt;Can you find now how the codes were chosen - and hence why the spy failed?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;em&gt;There is an unique solution for the riddle. Don't try to find overcomplicated formulas. There is more than just mathematics -- in fact, although there is quite a bit of calculations, the mathematics involved are very basic. To crack the code, you'll have to do some lateral thinking.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;em&gt;Language is important. The story takes place in Great Britain because the riddle is related to the English language. The nationality of the spy is irrelevant.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Hint 3&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;em&gt;Observe how the British chose the secret algorithm to fool spies. &amp;quot;Code of safe &lt;span class=&quot;math-container&quot;&gt;$n = 2 \times n$&lt;/span&gt;&amp;quot; is a red herring -- it looks like it is the solution, but is valid only up to safe #29 with the exception of safe #18. The units for each &amp;quot;ten&amp;quot; follow a pattern, but each &amp;quot;ten&amp;quot; breaks the pattern. Where would you find such a pattern?&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16323" LastEditorUserId="16323" LastEditDate="2022-09-20T11:36:03.777" LastActivityDate="2022-09-20T11:36:03.777" Title="The London safes and their mysterious combinations" Tags="|pattern|lateral-thinking|story|" AnswerCount="9" CommentCount="54" ContentLicense="CC BY-SA 4.0" />
  <row Id="22551" PostTypeId="1" AcceptedAnswerId="22555" CreationDate="2015-09-23T15:52:05.357" Score="8" ViewCount="602" Body="&lt;p&gt;Your spaceship pulls out of uberdrive (a new transportation mode that has overtaken hyperdrive in the space exploration market) and you find yourself gazing at the beauty of Earth for the first time in decades (though to you it has been a far shorter time). It has been a while since you and two dozen others were sent out in search of a potential new home. You dock at the I.S.S. IV and find it empty. Puzzled, you find a hologram in the control station and play it. In English, it plays: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hello. Humanity has been forced to relocate by an alien menace known as the Fblthp. One of our scouts found a viable area for&#xA;  humanity to try and establish itself. If you are receiving this message, you are likely one of&#xA;  our scouts returning from your reconnaissance. Or you may be one of the Fblthp race. Unfortunately,&#xA;  the Fblthp understand English so we cannot simply tell you our new location, but they have practically no understanding of other languages or&#xA;  our geography. To ensure that, if you're one of us, you return to us with your vital intel, we have hidden the clues&#xA;  to our location in etymology so that the Fblthp cannot decipher it. Use the light language training that you received and decipher&#xA;  the U.S. cities via their etymological origin; if you get to a name and it seems ambiguous,  choose the capital with that name.&#xA;  Find the cities, and you will find our new home.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;CLUES&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Holy Faith&lt;br&gt;&#xA;  Village of Jack's Progeny&lt;br&gt;&#xA;  Country of the Dove&lt;br&gt;&#xA;  Ford of Frank&lt;br&gt;&#xA;  White Oak&lt;br&gt;&#xA;  City of Red People&lt;br&gt;&#xA;  Emperor Aurelius' New City&lt;br&gt;&#xA;  Town of the Husband/Man&lt;br&gt;&#xA;  River Strait&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where is humanity holed up?&lt;/p&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="13970" LastEditDate="2015-09-23T16:30:42.373" LastActivityDate="2015-09-29T16:20:13.580" Title="Message to Survivors" Tags="|language|geography|outer-space|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22552" PostTypeId="2" ParentId="22550" CreationDate="2015-09-23T16:04:12.620" Score="8" Body="&lt;p&gt;The code should have been&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I believe that the codes for each safe are equivalent to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $($The next safe # in the sequence $\times 2) -2$.  Since Safe 30 is at the end of the sequence, technically Safe 1 would be next in sequence $(1 \times 2) -2 = 0$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14233" LastEditorUserId="4421" LastEditDate="2015-09-23T21:04:22.510" LastActivityDate="2015-09-23T21:04:22.510" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="22553" PostTypeId="5" CreationDate="2015-09-23T16:22:33.660" Score="0" Body="&lt;p&gt;A puzzle pertaining to topics associated with space such as:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;astronomy&lt;/li&gt;&#xA;&lt;li&gt;stars&lt;/li&gt;&#xA;&lt;li&gt;planets&lt;/li&gt;&#xA;&lt;li&gt;space travel&lt;/li&gt;&#xA;&lt;li&gt;cosmology&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Lists_of_astronomical_objects&quot; rel=&quot;nofollow noreferrer&quot;&gt;other space objects&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Note that general outer space questions are off-topic but can be asked on our sister site &lt;a href=&quot;https://astronomy.stackexchange.com/&quot;&gt;Astronomy Stack Exchange&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11497" LastEditorUserId="60644" LastEditDate="2020-11-22T11:42:19.863" LastActivityDate="2020-11-22T11:42:19.863" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22554" PostTypeId="4" CreationDate="2015-09-23T16:22:33.660" Score="0" Body="A puzzle pertaining to topics associated with space like astronomy, space travel, and cosmology. " OwnerUserId="13970" LastEditorUserId="70545" LastEditDate="2021-02-04T14:09:44.120" LastActivityDate="2021-02-04T14:09:44.120" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22555" PostTypeId="2" ParentId="22551" CreationDate="2015-09-23T16:40:23.423" Score="10" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The new home of mankind are the &lt;strong&gt;Pleiades&lt;/strong&gt;.&lt;br&gt;&#xA;&lt;br&gt;&#xA; (The geographic positions of the nine cities are very similar&#xA; to the positions of the nine brightest stars in the Pleiades;&#xA; see the image at &lt;a href=&quot;https://en.wikipedia.org/wiki/Pleiades&quot;&gt;https://en.wikipedia.org/wiki/Pleiades&lt;/a&gt; )&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The nine cities are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Holy Faith  =  Santa Fe, New Mexico  &lt;/li&gt;&#xA;  &lt;li&gt;Village of Jack's Progeny  =  Jacksonville, Florida &lt;/li&gt;&#xA;  &lt;li&gt;Country of the Dove  =  Columbia, South Carolina   &lt;/li&gt;&#xA;  &lt;li&gt;Ford of Frank  = Frankfort, Kentucky  &lt;/li&gt;&#xA;  &lt;li&gt;White Oak  =   Albuquerque, New Mexico&lt;/li&gt;&#xA;  &lt;li&gt;City of Red People  =  Oklahoma City, Oklahoma  &lt;/li&gt;&#xA;  &lt;li&gt;Emperor Aurelius' New City  =  New Orleans, Louisiana &lt;/li&gt;&#xA;  &lt;li&gt;Town of the Husband/Man  = Charleston, West Virginia  &lt;/li&gt;&#xA;  &lt;li&gt;River Strait  =  Detroit, Michigan  &lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-09-29T16:20:13.580" LastActivityDate="2015-09-29T16:20:13.580" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="22556" PostTypeId="2" ParentId="22550" CreationDate="2015-09-23T16:47:58.807" Score="5" Body="&lt;p&gt;The code should have been&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The story plays in the leap year 1944. The British picked the codes day by day, and simply doubled the number of the current day.  The numbering started on Feburary 1 (code $2*1=2$), Feburary 2 (code $2*2=4$), $\ldots$, Feburary 29 (code $2*29=58$), March 1 (code $2*1=2$).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-09-23T16:47:58.807" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22557" PostTypeId="2" ParentId="15280" CreationDate="2015-09-23T16:56:19.910" Score="0" Body="&lt;p&gt;Here are my answers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $5 + 13 \times 4/9 + 6 + 12 \times 1 - 3 - 11 + 7 \times 8/2 - 10 = 66$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16509" LastEditorUserId="30376" LastEditDate="2016-10-04T15:02:55.140" LastActivityDate="2016-10-04T15:02:55.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22559" PostTypeId="2" ParentId="22550" CreationDate="2015-09-23T17:38:18.197" Score="3" Body="&lt;p&gt;I think this is too broad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, the code might have been &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;0 or 1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to the formula:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $code=2*num\pmod{59}$&lt;br&gt;&#xA;  or&lt;br&gt;&#xA; $code=2*num\pmod{60}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15544" LastActivityDate="2015-09-23T17:38:18.197" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22560" PostTypeId="2" ParentId="22550" CreationDate="2015-09-23T18:09:36.640" Score="1" Body="&lt;p&gt;I think the code is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 62&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Based on the idea that the code is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each safe number ($n$) is mapped onto double the $n$th number that is not the sum of 4 distinct nonzero squares. This would mean the 30th safe was mapped to 62. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It could also be that the code is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each safe number ($n$) is mapped onto double the $n$th number that has 2 or fewer distinct prime factors. Then $n=1,\ldots,29$ maps to $2n$ and 30 maps to 62.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or it could be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each safe numbered ($n$) maps to double the $n$th number that cannot be expressed in form $p+q^2+r^3+s^4$, where $p, q, r, s$ are primes. Then $n$ maps to $2n$ for $n\in{1,\ldots,29}$ and 30 maps to 64&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Obviously I used oeis.org to find the above, but it does suggest that the question is a bit broad&lt;/p&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2015-09-23T22:53:38.770" LastActivityDate="2015-09-23T22:53:38.770" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="22562" PostTypeId="1" CreationDate="2015-09-23T19:07:24.873" Score="-3" ViewCount="440" Body="&lt;p&gt;Complete the sequence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5928 1411 9909 1882 5419 ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't seem like a simple polynomial fit will work, I have tried the difference method and the quotient method. Any hint would be appreciated. Thanks&lt;/p&gt;&#xA;" OwnerUserId="16511" LastEditorUserId="15544" LastEditDate="2015-09-24T05:29:33.020" LastActivityDate="2015-09-24T05:29:33.020" Title="Math sequence puzzle" Tags="|mathematics|number-sequence|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22563" PostTypeId="1" AcceptedAnswerId="23061" CreationDate="2015-09-23T20:25:47.960" Score="6" ViewCount="1224" Body="&lt;p&gt;Source of an element that never existed&lt;br&gt;&#xA;Perhaps with thermal tangles&lt;br&gt;&#xA;Some of us are not as neat&lt;br&gt;&#xA;Our name is a misnomer&lt;br&gt;&#xA;An electronic errand?&lt;br&gt;&#xA;Hard to pinpoint&lt;br&gt;&#xA;Following death&lt;br&gt;&#xA;We fade too  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What are we and what is the significance of each line?&lt;/p&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="4050" LastEditDate="2015-09-27T15:18:56.423" LastActivityDate="2015-10-09T21:10:53.327" Title="A rare sight indeed" Tags="|riddle|outer-space|" AnswerCount="5" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22564" PostTypeId="1" AcceptedAnswerId="22567" CreationDate="2015-09-23T20:41:54.997" Score="4" ViewCount="1467" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Twinkle twinkle little star&lt;/p&gt;&#xA;&lt;p&gt;How I wonder what you are&lt;/p&gt;&#xA;&lt;p&gt;In the sky just passing by&lt;/p&gt;&#xA;&lt;p&gt;You can vanish in the blink of an eye&lt;/p&gt;&#xA;&lt;p&gt;Moving at speeds I can't believe&lt;/p&gt;&#xA;&lt;p&gt;All the things you can achieve&lt;/p&gt;&#xA;&lt;p&gt;In an atmosphere you will burn&lt;/p&gt;&#xA;&lt;p&gt;Luckily you always make the turn&lt;/p&gt;&#xA;&lt;p&gt;My shift is over so I am out(ter space)&lt;/p&gt;&#xA;&lt;p&gt;You'll get this easy without a doubt!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This is just something fun I wanted to put together for the new tag, not meant to be a brain buster or anything!&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;P.S. The last two lines are not part of the riddle, you can ignore those when thinking&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="16273" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-09-27T03:31:41.427" Title="An outer-space riddle" Tags="|riddle|rhyme|outer-space|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22566" PostTypeId="2" ParentId="22564" CreationDate="2015-09-23T21:04:41.710" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Would you be a comet seen from Earth?   I'd explain in detail but it's rather self explanatory, a comet fits every clue. Nice rhymes :)&lt;br&gt;&#xA;&lt;br&gt;&#xA;Comets can be perceived as shooting stars (twinkle twinkle little star), which aren't always visible to the naked eye very long (in the sky obviously). Comets are &lt;strong&gt;very&lt;/strong&gt; fast, and travel incredible distances. When they pass close to the sun or too close to planets' atmospheres, they'll burn up somewhat. Some comets have periodic orbits in our solar system so once they've finished their 'pass', they head outwards towards the kuiper belt and oort cloud before coming back etc...   Any quick wiki search will tell you pretty much all you want to know on comets :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastActivityDate="2015-09-23T21:04:41.710" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22567" PostTypeId="2" ParentId="22564" CreationDate="2015-09-23T21:22:01.170" Score="4" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the International Space Station (ISS)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit - a few more details explaining the answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The ISS really does pass by to the naked eye, it travels at &lt;a href=&quot;https://www.google.com/search?hl=en&amp;amp;q=how%20fast%20is%20the%20ISS%20going&quot; rel=&quot;nofollow&quot;&gt;a few miles per second&lt;/a&gt;, the astronauts aboard the ISS get tons of science done and, like anything that orbits, it is always falling (or &quot;turning&quot;) to keep the same distance from the earth as it flies along.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12795" LastEditorUserId="12795" LastEditDate="2015-09-23T21:47:05.510" LastActivityDate="2015-09-23T21:47:05.510" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22568" PostTypeId="2" ParentId="22564" CreationDate="2015-09-23T21:45:15.367" Score="2" Body="&lt;p&gt;Is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A supernova?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Twinkle twinkle little star&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How I wonder what you are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the sky just passing by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Before going supernova the star &quot;moves across the sky&quot; as the earth rotates/orbits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can vanish in the blink of an eye&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Once a start has gone supernova it vanishes from the sky.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Moving at speeds I can't believe&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Either a reference to the velocity of the star itself, or the rate of collapse.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All the things you can achieve&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; From the remnants of a supernova new stars can form.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In an atmosphere you will burn&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Self explanatory. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Luckily you always make the turn&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not too sure about this one. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12206" LastActivityDate="2015-09-23T21:45:15.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22569" PostTypeId="2" ParentId="22563" CreationDate="2015-09-23T21:56:07.510" Score="3" Body="&lt;p&gt;Is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Supernova?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Source of an element that never existed&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Heavy elements are created in supernova by the process of nuclear fusion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Perhaps with thermal tangles&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some of us are not as neat&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's an explosion... so pretty messy. There are also symmetrical/non-symmetrical supernovae.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Our name is a misnomer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Nova&quot; means new - when in reality the stars that are dying are old.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An electronic errand?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Huge amounts of electronic radiation are generated - in the for of visible light, xrays and gamma rays.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hard to pinpoint&#xA;Following death&#xA;We fade too&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Once a star has &quot;died&quot; or gone supernova - it has expelled all of its matter and can no longer be located.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Title:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;They are a rare sight indeed :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12206" LastActivityDate="2015-09-23T21:56:07.510" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22570" PostTypeId="2" ParentId="22563" CreationDate="2015-09-23T21:58:14.570" Score="3" Body="&lt;p&gt;I've only 1 thing in mind and it seems to fit most of the clues:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Black Hole&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Source of an element that never existed&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Aether was attributed to gravity, pretty much the only thing a black hole has.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Perhaps with thermal tangles&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure on this one, they obey thermodynamic laws?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some of us are not as neat&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Often accompanied by an accretion disk of left over material&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Our name is a misnomer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not actually a hole&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An electronic errand?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Strong electro magnetic fields&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hard to pinpoint&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dont' emit light so hard to see&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Following death&#xA;We fade too&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Via Hawking radiation, black holes do dissipate and fade over time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16305" LastActivityDate="2015-09-23T21:58:14.570" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22571" PostTypeId="2" ParentId="22510" CreationDate="2015-09-23T22:27:38.930" Score="2" Body="&lt;h1&gt;6x6 solution with rotations: 3404&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 6x6 square = 1 bounding rectangle with 60 inner edges&lt;br&gt;&#xA; 14 total 90° rotations&lt;br&gt;&lt;br&gt;&#xA; $60^2 - 14^2 = 3404$&lt;br&gt;&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/X3XvF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/X3XvF.png&quot; alt=&quot;6x6 polyomino solution&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A big thanks to &lt;a href=&quot;https://puzzling.stackexchange.com/users/15632/sleafar&quot;&gt;Sleafar&lt;/a&gt; for the Inkscape image! It made my life a &lt;em&gt;lot&lt;/em&gt; easier when playing around with this.&lt;/p&gt;&#xA;" OwnerUserId="16275" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2015-09-23T22:27:38.930" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22572" PostTypeId="2" ParentId="22468" CreationDate="2015-09-23T23:21:42.877" Score="-1" Body="&lt;p&gt;It is fairly simple. Draw a line from point P to Q pick the point in the middle of the line. This is the point X. Now point X is an axis of your apple - cut apple in half across the axis. As worm has done a path shorter than diameter there is no way it could get to the center of the apple and get out of it.&lt;/p&gt;&#xA;" OwnerUserId="16516" LastActivityDate="2015-09-23T23:21:42.877" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22574" PostTypeId="2" ParentId="22563" CreationDate="2015-09-24T01:59:15.713" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Virtual particles? (more specifically virtual electrons or antiprotons?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Source of an element that never existed- &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Different particles with different properties than we accept as within Standard Model would have resulted in very different elements. This one got me thinking that virtuals would work, but this is a better fit than the rest of the statements are for virtuals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Perhaps with thermal tangles- &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Thermal excitation of a space increases momentary virtual particle creation(and subsequent annihilation) as seen near black holes, etc. and energy is also released in virtual pair annihilation. &lt;strong&gt;Even more concretely, magnetic tangles affect the travel of electrons.&lt;/strong&gt; This one really got me thinking about the fit. (I used &lt;a href=&quot;http://nuclear.unh.edu/~chandran/Chandran_Cowley1998.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; paper for reference)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some of us are not as neat- &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Disobey 1st law of thermo, off mass shell, mass is unpredictable, and the math also gets hairy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Our name is a misnomer- &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Not actually 'virtual,' just exist for a terribly short time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An electronic errand?- &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Errant energy, in the form of particles and charges that are far outside of traditional consideration of non-exceptional physics? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hard to pinpoint- &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Uncertainty principle combined with the fact that annihilation happens more quickly with more energetic virtuals. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Following death &#xA;We fade too-&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;[only considering these as one statement, because I am at a loss of ideas for those with separate meanings] If considering the increased frequency of virtual creation near/within a black hole, with the 'death'(the &quot;pop&quot; of a small black hole depleted by Hawking radiation) of a black hole causing them to &quot;fade&quot; and become less energetic and less frequent, this makes some sense. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16510" LastEditorUserId="16510" LastEditDate="2015-09-24T14:07:05.987" LastActivityDate="2015-09-24T14:07:05.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22575" PostTypeId="1" AcceptedAnswerId="22595" CreationDate="2015-09-24T02:20:43.657" Score="14" ViewCount="4051" Body="&lt;p&gt;While messing around at one evening, I made six small drawings, one for each friend of mine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My drawings were made to be very specific: Taking the action or object represented in any of them, you can rip out all the non-alphabetic characters, then discard one letter to yield the respective friend's first name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if I had drawn something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/61zR7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/61zR7.png&quot; alt=&quot;PuzzleExample&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could expect it's talking about marriage, so &quot;marry&quot; would be the action, yielding, among some unusual names, &quot;Mary&quot;. Happily, all my friends' names are in what is considered the &quot;normal&quot; spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you guess their names only from the drawings below?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Friend 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/81y8q.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/81y8q.png&quot; alt=&quot;Puzzle 0&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Friend 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JOmMK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JOmMK.png&quot; alt=&quot;Puzzle 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Friend 3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/68HgN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/68HgN.png&quot; alt=&quot;Puzzle 2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Friend 4:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/P2dxZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/P2dxZ.png&quot; alt=&quot;Puzzle 3&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Friend 5:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kyzGd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kyzGd.png&quot; alt=&quot;Puzzle 4&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My greatest friend (And love):&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1DdLI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1DdLI.png&quot; alt=&quot;Puzzle 5&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note: The order has changed because of one drawing that I forgot to add to the list.&lt;/p&gt;&#xA;" OwnerUserId="14219" LastEditorUserId="14219" LastEditDate="2015-09-24T20:58:33.547" LastActivityDate="2015-09-24T22:28:00.287" Title="What are my friends' names? (A visual puzzle)" Tags="|word|wordplay|visual|rebus|science|" AnswerCount="9" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="22578" PostTypeId="2" ParentId="22562" CreationDate="2015-09-24T03:33:04.410" Score="12" Body="&lt;p&gt;The next number in this sequence is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;3655&lt;/strong&gt; (confirmed at the &lt;a href=&quot;http://www.itstoohard.com/puzzle/2pXK7kIq&quot;&gt;link provided by Sleafar&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This sequence is based on the:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Middle-square_method&quot;&gt;Middle-square method&lt;/a&gt; for generating pseudorandom numbers.&lt;br&gt;&#xA;&lt;br&gt;&#xA; In this case, the sequence starts with a 4 digit seed (5928), which is then squared (35141184). The middle 4 digits of the result become the next number in the sequence. If the result is less than 8 digits, leading zeroes are provided.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, this sequence progresses as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;pre&gt;5928  35141184&#xA;1411  01990921&#xA;9909  98188281&#xA;1882  03541924&#xA;5419  29365561&#xA;3655  13359025&#xA;3590  12888100&lt;/pre&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastActivityDate="2015-09-24T03:33:04.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22580" PostTypeId="2" ParentId="22575" CreationDate="2015-09-24T04:52:32.957" Score="1" Body="&lt;p&gt;Is 5...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Maximillian? It's the opposite (&quot;no&quot;) of &quot;mini millionth&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2015-09-24T04:52:32.957" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22581" PostTypeId="2" ParentId="22575" CreationDate="2015-09-24T08:00:43.483" Score="4" Body="&lt;p&gt;Number 2 shows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An atom of hydrogen with an electron &lt;strong&gt;A&lt;/strong&gt;, a proton &lt;strong&gt;B&lt;/strong&gt; and uncertainty over whether there is a neutron &lt;strong&gt;C&lt;/strong&gt;. The verb could be analyze stable hydrogen isotopes (ASHI) and dropping the letter 'i' gives &lt;strong&gt;&lt;em&gt;Ash&lt;/em&gt;&lt;/strong&gt;. &lt;em&gt;Ok, I know this is a bit of a stretch.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number 4 shows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'K' on a disc, so the verb would be 'disc K'. Drop the letter 's' to get &lt;strong&gt;&lt;em&gt;Dick&lt;/em&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="14135" LastEditDate="2015-09-24T13:35:16.020" LastActivityDate="2015-09-24T13:35:16.020" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22582" PostTypeId="5" CreationDate="2015-09-24T08:51:00.033" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-09-24T08:51:00.033" LastActivityDate="2015-09-24T08:51:00.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22583" PostTypeId="4" CreationDate="2015-09-24T08:51:00.033" Score="0" Body="Puzzles in the form of a haiku, a very short form of poetry, typically a three line observation on a certain topic." OwnerUserId="15544" LastEditorUserId="15544" LastEditDate="2015-09-24T09:59:53.037" LastActivityDate="2015-09-24T09:59:53.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22584" PostTypeId="1" AcceptedAnswerId="22593" CreationDate="2015-09-24T09:13:46.800" Score="13" ViewCount="1113" Body="&lt;p&gt;Professor Erasmus has constructed a special convex polyhedron from perfectly homogeneous material, which he modestly calls the &quot;Professor-Erasmus-polyhedron&quot;.&#xA;The professor claims that he can put the polyhedron into the lake next to his house, so that the polyhedron floats&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;with 90% of its volume below the water level, and&lt;/li&gt;&#xA;&lt;li&gt;more than 50% of its surface area above the water level.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Has the professor once again made a mathematical blunder, or does such a polyhedron indeed exist?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-09-24T13:21:29.280" LastActivityDate="2015-09-28T13:10:47.820" Title="The Erasmus polyhedron" Tags="|mathematics|geometry|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="22586" PostTypeId="2" ParentId="22563" CreationDate="2015-09-24T09:36:20.167" Score="0" Body="&lt;p&gt;They are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_ray&quot; rel=&quot;nofollow&quot;&gt;Cosmic rays&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;Source of an element that never existed&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quoting wikipedia:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cosmic rays are also responsible for the continuous production of a number of &lt;a href=&quot;https://en.wikipedia.org/wiki/Radionuclide&quot; rel=&quot;nofollow&quot;&gt;unstable isotopes&lt;/a&gt; in the Earth's atmosphere, such as &lt;a href=&quot;https://en.wikipedia.org/wiki/Carbon-14&quot; rel=&quot;nofollow&quot;&gt;carbon-14&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;Perhaps with thermal tangles&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quoting a part of the wikipedia article (with a message about possible neutrality issues):&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;A role of cosmic rays directly or via solar-induced modulations in climate change was suggested by &lt;a href=&quot;https://en.wikipedia.org/wiki/Edward_P._Ney&quot; rel=&quot;nofollow&quot;&gt;Edward P. Ney&lt;/a&gt; in 1959 and by &lt;a href=&quot;https://en.wikipedia.org/wiki/Robert_E._Dickinson&quot; rel=&quot;nofollow&quot;&gt;Robert E. Dickinson&lt;/a&gt; in 1975. The idea has been revived in recent years, most notably by &lt;a href=&quot;https://en.wikipedia.org/wiki/Henrik_Svensmark&quot; rel=&quot;nofollow&quot;&gt;Henrik Svensmark&lt;/a&gt;, who has argued that because &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_variation&quot; rel=&quot;nofollow&quot;&gt;solar variation&lt;/a&gt; modulates the cosmic ray flux on Earth, they would consequently affect the rate of cloud formation and hence the climate. However, other scientists have vigorously criticized Svensmark for sloppy and inconsistent work&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; So, it is something actually being debated and at least unclear. However, the riddle verse contains the word &quot;perhaps&quot;, so it is not taking it for granted nor as refuted.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;Some of us are not as neat&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quoting from wikipedia:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Cosmic rays attract great interest practically, due to the damage they inflict on microelectronics and life outside the protection of an atmosphere and magnetic field&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;Our name is a misnomer&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quoting from wikipedia:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;The term &lt;em&gt;ray&lt;/em&gt; is a historical accident, as cosmic rays were at first, and wrongly, thought to be mostly &lt;a href=&quot;https://en.wikipedia.org/wiki/Electromagnetic_radiation&quot; rel=&quot;nofollow&quot;&gt;electromagnetic radiation&lt;/a&gt;.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;An electronic errand?&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quoting from wikipedia:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Of primary cosmic rays, which originate outside of Earth's atmosphere, about 99% are the nuclei (stripped of their electron shells) of well-known atoms, and about 1% are solitary electrons (similar to beta particles).&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Also, the answer to the &quot;Some of us are not as neat&quot; part also might apply.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;Hard to pinpoint&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They were first detected only at the first years of the 20th century, even if many of them might be hitting you at this exact moment. They could be correctly identified only in the 1930's. See the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_ray#Identification&quot; rel=&quot;nofollow&quot;&gt;&quot;Identification&quot; section on the wikipedia's article&lt;/a&gt; for details. They require quite softiscate equipment and technology in order to be detected and identified.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;Following death&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quoting wikipedia:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Early speculation on the sources of cosmic rays included a 1934 proposal by Baade and &lt;a href=&quot;https://en.wikipedia.org/wiki/Fritz_Zwicky&quot; rel=&quot;nofollow&quot;&gt;Zwicky&lt;/a&gt; suggesting cosmic rays originated from supernovae. A 1948 proposal by &lt;a href=&quot;https://en.wikipedia.org/wiki/Horace_W._Babcock&quot; rel=&quot;nofollow&quot;&gt;Horace W. Babcock&lt;/a&gt; suggested that magnetic variable stars could be a source of cosmic rays. Subsequently in 1951, Y. Sekido et al. identified the &lt;a href=&quot;https://en.wikipedia.org/wiki/Crab_Nebula&quot; rel=&quot;nofollow&quot;&gt;Crab Nebula&lt;/a&gt; as a source of cosmic rays. Since then, a wide variety of potential sources for cosmic rays began to surface, including &lt;a href=&quot;https://en.wikipedia.org/wiki/Supernova&quot; rel=&quot;nofollow&quot;&gt;supernovae&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Active_galactic_nuclei&quot; rel=&quot;nofollow&quot;&gt;active galactic nuclei&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Quasar&quot; rel=&quot;nofollow&quot;&gt;quasars&lt;/a&gt;, and &lt;a href=&quot;https://en.wikipedia.org/wiki/Gamma-ray_burst&quot; rel=&quot;nofollow&quot;&gt;gamma-ray bursts&lt;/a&gt;.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Supernovae are massive stars that blow up spectacular deaths. Nebulae (like &lt;a href=&quot;https://en.wikipedia.org/wiki/Crab_Nebula&quot; rel=&quot;nofollow&quot;&gt;Crab Nebulae&lt;/a&gt;) are supernova remnants, so created by some star deaths. Gamma-ray bursts are theorized to be created in supernovas and neutron stars merges (neutron stars are a possible result of supernovas). Active galactic nuclei and quasars are brilliant accretion discs of supermassive blackholes, and both a black hole and their accretion discs are implied to death of stars.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;We fade too&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When cosmic rays interact with Earth's particles (mainly in atmosphere), they'll fade. It is easier to explain with this wikipedia's picture:&lt;br&gt;&#xA; &lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/0/08/Atmospheric_Collision.svg/400px-Atmospheric_Collision.svg.png&quot; /&gt;&lt;br&gt;&#xA; And quoting the text:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Primary cosmic rays primarily originate from outside the &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_System&quot; rel=&quot;nofollow&quot;&gt;Solar System&lt;/a&gt; and sometimes even the &lt;a href=&quot;https://en.wikipedia.org/wiki/Milky_Way&quot; rel=&quot;nofollow&quot;&gt;Milky Way&lt;/a&gt;. When they interact with Earth's atmosphere, they are converted to secondary particles.&lt;/strong&gt;&lt;br&gt;&#xA; &lt;br&gt;&#xA; &lt;strong&gt;When cosmic rays enter the &lt;a href=&quot;https://en.wikipedia.org/wiki/Earth%27s_atmosphere&quot; rel=&quot;nofollow&quot;&gt;Earth's atmosphere&lt;/a&gt; they collide with &lt;a href=&quot;https://en.wikipedia.org/wiki/Atom&quot; rel=&quot;nofollow&quot;&gt;atoms&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Molecule&quot; rel=&quot;nofollow&quot;&gt;molecules&lt;/a&gt;, mainly oxygen and nitrogen. The interaction produces a cascade of lighter particles, a so-called &lt;a href=&quot;https://en.wikipedia.org/wiki/Air_shower_(physics)&quot; rel=&quot;nofollow&quot;&gt;air shower&lt;/a&gt; secondary radiation that rains down, including &lt;a href=&quot;https://en.wikipedia.org/wiki/X-rays&quot; rel=&quot;nofollow&quot;&gt;x-rays&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Muon&quot; rel=&quot;nofollow&quot;&gt;muons&lt;/a&gt;, protons, &lt;a href=&quot;https://en.wikipedia.org/wiki/Alpha_particle&quot; rel=&quot;nofollow&quot;&gt;alpha particles&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Pion&quot; rel=&quot;nofollow&quot;&gt;pions&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Electron&quot; rel=&quot;nofollow&quot;&gt;electrons&lt;/a&gt;, and &lt;a href=&quot;https://en.wikipedia.org/wiki/Neutron&quot; rel=&quot;nofollow&quot;&gt;neutrons&lt;/a&gt;.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5044" LastEditorUserId="5044" LastEditDate="2015-09-24T11:26:57.790" LastActivityDate="2015-09-24T11:26:57.790" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22588" PostTypeId="2" ParentId="22575" CreationDate="2015-09-24T12:02:47.767" Score="2" Body="&lt;p&gt;My guess for 5 is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Nom, because &lt;a href=&quot;https://en.wikipedia.org/wiki/Micro-&quot; rel=&quot;nofollow&quot;&gt;one millionth&lt;/a&gt; is prefixed by &lt;em&gt;mu&lt;/em&gt;, so &quot;No mu&quot; becomes Nom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12795" LastEditorUserId="12795" LastEditDate="2015-09-24T13:02:59.667" LastActivityDate="2015-09-24T13:02:59.667" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22589" PostTypeId="2" ParentId="22584" CreationDate="2015-09-24T13:11:47.207" Score="12" Body="&lt;p&gt;Lateral thinking solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Professor Erasmus lives on a planet that is made entirely of water except for the small floating patch of earth that supports his house. The Professor-Erasmus-polytope is about as large as the planet, so that most of its surfaces skim along the top of the water, while most of its volume remains below the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Artist's conception:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/M3LIZ.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/M3LIZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1497" LastActivityDate="2015-09-24T13:11:47.207" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22590" PostTypeId="2" ParentId="22575" CreationDate="2015-09-24T13:56:05.783" Score="3" Body="&lt;p&gt;For 1, I think that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It's a net &quot;catching&quot; a butterfly so the action is &quot;catch&quot; and the friend's name is &quot;Cath&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For 5 I wonder if:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The friend's name is Meg - from &quot;Mega&quot; (Million).  I have no sensible logic to explain why the action is a &quot;Million&quot; though, so probably wrong.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1608" LastActivityDate="2015-09-24T13:56:05.783" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22592" PostTypeId="2" ParentId="22575" CreationDate="2015-09-24T14:55:09.537" Score="-1" Body="&lt;p&gt;Joining the 'million' train of thought. Is the answer to number 5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Milli?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16523" LastActivityDate="2015-09-24T14:55:09.537" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22593" PostTypeId="2" ParentId="22584" CreationDate="2015-09-24T15:34:24.200" Score="16" Body="&lt;p&gt;Ok, i'll take a stab at it. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Take a cone of radius 100 and height 10. place it point first into the water, so that .34 units stick out of the water. At that point, 90% of the volume (94397.4 of 104720) is underwater, and less than half of the surface area (29462 of 62988.5) is under water. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Oh fine, polyhedron, i'll change it to a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; pyramid of base 100x100 height 10.  Point first, if .34 units stick out of the water.  Volume (30048 of 33333), surface (9516 of 20198)&lt;br&gt;&#xA;&lt;br&gt;&#xA; I came to this idea from the idea that a really flat pyramid would have the surface area of the base be close to 1/2 of the total surface area. Then I just had to work out where 90% of the volume would be, and check if that area is under 50%.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Let's take height as a ratio of width (xw). So, we know that the height underwater is .966 of the total height, which means the width underwater is also .966 of the original width.&lt;br&gt;&#xA; So, the surface area of the part underwater is $2 * .966w * \sqrt{\frac{(.966w)^2}{4} + (.966xw)^2}$&lt;br&gt;&#xA; And of course the full surface area is $w^2 + 2 * w * \sqrt{\frac{w^2}{4} + (xw)^2}$&lt;br&gt;&#xA; Plugging that all in gives an X &amp;lt; .2883&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-09-28T13:10:47.820" LastActivityDate="2015-09-28T13:10:47.820" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22595" PostTypeId="2" ParentId="22575" CreationDate="2015-09-24T16:07:00.043" Score="15" Body="&lt;p&gt;For number two:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The invisible dog comment is spot on.  But because the dog is not&#xA; there, what you are left seeing is the leash.  Drop an S and you get:&#xA; &lt;strong&gt;Leah&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For number three:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think that Gordon K went too far.  It shows an atom, and dropping&#xA; one letter gives us a very &quot;normal&quot; name: &lt;strong&gt;Tom&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For number five:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It seems pretty apparent from the drawing that the K is transparent. &#xA; In other words, a &quot;clear K&quot;.  Drop the E and you have: &lt;strong&gt;Clark&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For number four:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are putting the R in the E can. In other words, &quot;can the R in E&quot;. &#xA; Lose the first N and you have: &lt;strong&gt;Catherine&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11775" LastEditorUserId="11775" LastEditDate="2015-09-24T18:16:42.387" LastActivityDate="2015-09-24T18:16:42.387" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22597" PostTypeId="1" AcceptedAnswerId="22599" CreationDate="2015-09-24T17:28:54.003" Score="7" ViewCount="389" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The first one borne, the others belated.&lt;br&gt;&#xA;  Legions have my intentions debated.&lt;br&gt;&#xA;  Headdress upholding my endless mission;&lt;br&gt;&#xA;  I rule either way, death or submission.&lt;br&gt;&#xA;  Gather your might to face my attrition.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Borne the second, I'm the arm of the first.&lt;br&gt;&#xA;  Legions have fallen to quench my bloodthirst.&lt;br&gt;&#xA;  Damage I bring with the swing of my sword.&lt;br&gt;&#xA;  On no side I fight; no army, no horde.&lt;br&gt;&#xA;  Of my accord I fight: to fight all accord.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Grass and grain ruined, look on with disgust.&lt;br&gt;&#xA;  I'll show you fear in a handful of dust.&lt;br&gt;&#xA;  To bring your suffering I come in haste.&lt;br&gt;&#xA;  Hard, cold scales watch silently as you waste.&lt;br&gt;&#xA;  Now greet you the fourth, the wicked, the chased.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Hark and cower at my fearsome approach,&lt;br&gt;&#xA;  As the mortal realm I dauntless encroach.&lt;br&gt;&#xA;  Eyed in my wake is the reaper of souls.&lt;br&gt;&#xA;  Skin corpsely pallid, eyes dark as coals.&lt;br&gt;&#xA;  Scurry now mortals, and hide in your holes.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can you identify the four characters and explain their references?&lt;/p&gt;&#xA;" OwnerUserId="13970" LastActivityDate="2015-09-24T18:15:09.597" Title="Eureka!: A Riddle" Tags="|riddle|poetry|literature|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22598" PostTypeId="1" AcceptedAnswerId="22710" CreationDate="2015-09-24T17:59:19.817" Score="9" ViewCount="1277" Body="&lt;p&gt;Remember to look up at the stars and not down at your feet.&lt;br&gt;&#xA;Try to make sense of what you see...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We look up at the same stars and see such different things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sDqi5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sDqi5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I looked up, &lt;em&gt;second to the right&lt;/em&gt; he said.&lt;br&gt;&#xA;What did I see?    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Here's a little rhyme to help you start&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;em&gt;Some numbers are important, others not, all letters are the two.&lt;br&gt;&#xA;Colors like red herrings, a lot, every single one 'cept blue!&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastEditorUserId="11251" LastEditDate="2015-09-28T13:33:37.063" LastActivityDate="2015-09-29T19:59:55.733" Title="Second star to the right and straight on 'til morning" Tags="|enigmatic-puzzle|visual|outer-space|" AnswerCount="2" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="22599" PostTypeId="2" ParentId="22597" CreationDate="2015-09-24T18:09:40.210" Score="6" Body="&lt;p&gt;Is it...  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the four apocalyptic horsemen?&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Conquest&lt;/strong&gt;&lt;br&gt;&#xA; &quot;I rule either way&quot; is the key phrase. &quot;Intentions debated&quot; refers to the fact that people have the most varied interpretations for the first horseman. Traditionally, the indicating role of Conquest has been assigned to Egypt, which has prominent wartime &quot;headdresses&quot;.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;War&lt;/strong&gt;&lt;br&gt;&#xA; War is the most common means to achieve conquest. In Revelations, war breaks out on a global scale, thus the free-for-all description of &quot;on no side I fight&quot;.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Famine&lt;/strong&gt;&lt;br&gt;&#xA; &quot;Grass and grain ruined&quot; is thr key phrase. Famine is first and foremost a poor crop in agriculture. The &quot;cold, hard scales&quot; refer to the balancing scales of the ancient marketplace, symbolically representing the rapid inflation of food prices during times of famine.&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Death&lt;/strong&gt;&lt;br&gt;&#xA; &quot;Reaper of souls&quot; is pretty explicit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15960" LastEditorUserId="11251" LastEditDate="2015-09-24T18:15:09.597" LastActivityDate="2015-09-24T18:15:09.597" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22600" PostTypeId="1" AcceptedAnswerId="22609" CreationDate="2015-09-24T19:15:47.750" Score="4" ViewCount="285" Body="&lt;p&gt;... I am pretty hungry though. Figure out what I want to eat:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;01001001 00011111 01000100 10110100 11110101 10010001 00011111 01000001 10111101 10110011 01100000 01100110 10000110 11000010 10001000 10010001 01001010 01010000 10001100 11001111 10101100 10001000 11111010 00011101 00100110&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When it matters, you'll need to know that the last two get capitalized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus:&lt;/strong&gt;&#xA;Encode the answer the same way I encoded the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Look at the title to see how it's NOT encoded.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11727" LastEditorUserId="11727" LastEditDate="2015-09-25T01:25:57.183" LastActivityDate="2015-09-25T01:50:00.750" Title="Don't worry, I won't byte" Tags="|cipher|" AnswerCount="1" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22601" PostTypeId="1" AcceptedAnswerId="22611" CreationDate="2015-09-24T19:17:24.350" Score="3" ViewCount="479" Body="&lt;p&gt;Two friends, Amy and Ben, each use a three-disk combination lock to secure their bikes. Each disk has $10$ orientations, with integers $0$ to $9$, so there are $10^3$ possible combinations. One day, Amy and Ben compare methods for locking their bikes...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Amy:&lt;/strong&gt; &quot;I prefer to keep turning the disks until adding another turn would result in diminishing marginal lock-iness.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Lock-iness, $\Omega(n)$, counts the number of distinct combinations $n$ turns away from the unlocking combo.&lt;/em&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ben:&lt;/strong&gt; &quot;Wow! I do the same thing! Except that I would never only turn one disk; that'd be too easy for a thief to crack. Neglecting those cases, I also add turns until adding another would result in diminishing marginal lock-iness.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;When calculating $\Omega$, Ben neglects combinations which would involve only turning one disk from the unlocking combo. He only turns two or three disks when making it lock-y.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Amy:&lt;/strong&gt; &quot;To clarify, I turn the first disk $n_1$ times in &lt;em&gt;only one&lt;/em&gt; direction. Then I turn the second disk $n_2$ times in a direction, and the third $n_3$ times in a direction, where $n = n_1 + n_2 + n_3$ is the total number of turns. If the number of possible combinations $n$ turns away from my secret combo increases from $n$ to $n+1$ by less than that from $n-1$ to $n$, I stop at $n$ turns.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Consider $\Omega(n-1) = 40$, $\Omega(n) = 70$. If $\Omega(n+1) &amp;lt; 100$, Amy would experience diminishing marginal lock-iness (the next turn adds fewer combinations than the previous turn) and she would stop at $n$.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ben:&lt;/strong&gt; &quot;We are in agreement. I stop at $m$ for the same reasons, except I don't count combinations where only one $m_i$ is nonzero; I consider it unsafe and expect a wise thief to make the same assumption. I'll leave my shiny red bike out there unwatched for a while, because it is so locked.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Ben must start by turning at least two disks, since he can't turn just one disk. He makes turns until an added turn would increase his $\Omega$ by less than the previous turn (given his &quot;no single disk&quot; restriction).&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Amy:&lt;/strong&gt; &quot;Yes. I will do the same with my bike, the dirty, but equally appealing blue bike next to yours.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately for them, a would-be thief heard this discussion and began counting states on paper.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;The Problem:&lt;/strong&gt; For a known value of $n$ turns, the thief could find every possible combination, one of which will open the lock. Whose bike will the thief choose to ride away with, if ease of thievery is the only factor?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Examples:&lt;/strong&gt; If Amy had instead said that she only turns one disk once, $n=1$, there'd be $\Omega=6$ different combinations for a thief (knowing $n$) to try.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Ben had said he turns two different disks once each, $n=2$ and $n_i = n_j = 1$, there would be $\Omega(2)=12$ different combos for the thief to try. Knowing both of these examples, he would try to steal Amy's to get away quickest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Amy makes two turns like Ben, her $\Omega(2)=18$, because she considers cases where $n = n_i = 2$, $n_j = n_k = 0$. Her marginal lock-iness from $n=1$ to $n=2$ is $\Omega(2)-\Omega(1)=12$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EV2vz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EV2vz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13237" LastEditorUserId="13237" LastEditDate="2015-09-24T21:51:23.707" LastActivityDate="2015-09-25T02:41:30.083" Title="Whose lock is lock-ier?" Tags="|calculation-puzzle|combinatorics|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="22602" PostTypeId="2" ParentId="22575" CreationDate="2015-09-24T19:28:43.613" Score="2" Body="&lt;p&gt;for the new friend number 1 I'm thinking,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bo, it looks like a box to me and drop the x you get Bo which is a usual name here in the south.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13925" LastActivityDate="2015-09-24T19:28:43.613" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22604" PostTypeId="2" ParentId="22575" CreationDate="2015-09-24T20:20:11.897" Score="2" Body="&lt;p&gt;The rounded corners on #1 make it look a bit like&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a television&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So perhaps the name is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Elly (i.e., &quot;telly&quot; without the &quot;t&quot;)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="40" LastActivityDate="2015-09-24T20:20:11.897" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22605" PostTypeId="2" ParentId="22575" CreationDate="2015-09-24T22:28:00.287" Score="4" Body="&lt;p&gt;Hmm this isn't used commonly as a first name but... &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It looks like an unknown object on top of something. So the question to be asked is &quot;What's on?&quot;. Take away the 'h' and you've got WATSON &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12277" LastActivityDate="2015-09-24T22:28:00.287" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22608" PostTypeId="1" CreationDate="2015-09-25T00:20:12.590" Score="0" ViewCount="146" Body="&lt;p&gt;There are curved lines drawn on a piece of paper. Lines can have two characteristics:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;One of its endpoints is on another characteristic 1 line&lt;/li&gt;&#xA;&lt;li&gt;One of its endpoints is on another characteristic 2 line, exactly halfway between:&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Its two endpoints&lt;/li&gt;&#xA;&lt;li&gt;A point where a characteristic 2 line touches it and an endpoint&lt;/li&gt;&#xA;&lt;li&gt;Two points where a characteristic 2 line touches it&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Is it possible for there to be a third characteristic such that, for all sets of lines&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It is impossible for all possible characteristic 3 starting points connect to a unique characteristic 1 endpoint&lt;/li&gt;&#xA;&lt;li&gt;It is impossible for all possible characteristic 2 starting points connect to a unique characteristic 3 starting point&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="16532" LastEditorUserId="16532" LastEditDate="2015-09-25T00:25:16.897" LastActivityDate="2015-09-25T01:56:44.357" Title="Curved lines on a paper" Tags="|mathematics|" AnswerCount="1" CommentCount="2" ClosedDate="2015-09-25T02:58:15.283" ContentLicense="CC BY-SA 3.0" />
  <row Id="22609" PostTypeId="2" ParentId="22600" CreationDate="2015-09-25T01:50:00.750" Score="9" Body="&lt;p&gt;This is a variant on the Bacon cipher. Each group of five digits encodes a letter by converting it from binary to A-Z, with 00001=A, 00010=B, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The message is: &lt;code&gt;I DOT IMGUR DOT COM SLASH M P THREE E A C FOUR DOT GIF&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This points to the image at &lt;a href=&quot;https://i.stack.imgur.com/I9hCU.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/I9hCU.gif&lt;/a&gt;, which shows a pizza.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-09-25T01:50:00.750" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22610" PostTypeId="2" ParentId="22608" CreationDate="2015-09-25T01:56:44.357" Score="1" Body="&lt;p&gt;&lt;strong&gt;Yes&lt;/strong&gt;, if we define characteristic 3 curves like so:&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;A characteristic 3 curve cannot touch a curve of another characteristic at any point.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This trivially satisfies both criteria.&lt;/p&gt;&#xA;" OwnerUserId="9718" LastActivityDate="2015-09-25T01:56:44.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22611" PostTypeId="2" ParentId="22601" CreationDate="2015-09-25T02:41:30.083" Score="3" Body="&lt;p&gt;Using &lt;em&gt;Mathematica&lt;/em&gt; I generated all possible combinations for each $n$:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; n  ΩA  ΩA'   ΩB  ΩB'&#xA;---------------------&#xA; 1   6    5    0    0&#xA; 2  18   12   12   12&#xA; 3  38   20   32   20&#xA; 4  66   28   60   28&#xA; 5  99   33   96   36&#xA; 6 134   35  128   32&#xA; 7 174   40  168   40&#xA; 8 222   48  216   48&#xA; 9 278   56  272   56&#xA;10 292   14  291   19&#xA;11 308   16  302   11&#xA;12 352   44  346   44&#xA;13 388   36  382   36&#xA;14 416   28  410   28&#xA;15 292 -124  289 -121&#xA;16 400  108  394  105&#xA;17 416   16  410   16&#xA;18 424    8  418    8&#xA;19 424    0  418    0&#xA;20 292 -132  291 -127&#xA;21 424  132  418  127&#xA;22 424    0  418    0&#xA;23 424    0  418    0&#xA;24 424    0  418    0&#xA;25 292 -132  289 -129&#xA;26 424  132  418  129&#xA;27 424    0  418    0&#xA;28 424    0  418    0&#xA;29 424    0  418    0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;After $n=29$ the last ten lines repeat endlessly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can see that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Amy chooses $n=9$ with $\Omega_A(9)=278$, while Ben chooses $n=5$ with $\Omega_B(5)=96$.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ben has made a poor decision about his stopping criteria, and the thief will always choose Ben's bike!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here's the code I used:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;combos[n_] := combos[n] = Union @@ (&#xA;                            Tuples[Transpose[Mod[{#, -#}, 10]]] &amp;amp; /@&#xA;                              Union @@ (&#xA;                                Permutations /@ IntegerPartitions[n, {3}, Range[0, n]]&#xA;                              )&#xA;                          )&#xA;&#xA;ΩA[n_] := ΩA[n] = Length[combos[n]]&#xA;ΩB[n_] := ΩB[n] = Length @ Select[combos[n], Count[0] @ # &amp;lt;= 1 &amp;amp;]&#xA;&#xA;Grid @ Table[{n, ΩA[n], ΩA[n] - ΩA[n-1], ΩB[n], ΩB[n] - ΩB[n-1]}, {n, 50}]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It's actually quite fast.&lt;/p&gt;&#xA;" OwnerUserId="9718" LastActivityDate="2015-09-25T02:41:30.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22612" PostTypeId="1" AcceptedAnswerId="22618" CreationDate="2015-09-25T06:48:56.310" Score="10" ViewCount="535" Body="&lt;p&gt;Mr. Hilbert had enjoyed stargazing ever since he became friends with Mr. Beckley. So imagine his excitement when he saw an exceptionally bright comet trail in the sky one evening, as he was stargazing at the observatory conveniently located just behind the Grand Hotel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now further imagine his surprise when he realized that the comet was heading towards him, and shortly thereafter crashlanded a few hundred meters away! It turned out to be a weird ceramic sphere, slightly shimmering in the moonlight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A depression of the button on the top by a curious Mr. Hilbert opened up the sphere to reveal this message inside:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XzsNj.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XzsNj.png&quot; alt=&quot;The message from outer space&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;?!?!&quot; said Mr. Hilbert.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Can you decipher the message from outer space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Why is there a &quot;wordplay&quot; tag?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15960" LastActivityDate="2015-09-29T06:42:59.233" Title="Mr. Hilbert and the Message from Outer Space" Tags="|cipher|wordplay|outer-space|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22614" PostTypeId="1" AcceptedAnswerId="22674" CreationDate="2015-09-25T08:04:36.610" Score="27" ViewCount="1803" Body="&lt;p&gt;Winnie-the-Pooh keeps his $27$ honey pots in the larder. &#xA;Each pot contains up to $1$ kilogram of honey, and different pots &#xA;contain different quantities of honey. &#xA;All $27$ pots together contain $17$ kilogram of honey.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every day, Winnie-the-Pooh selects $7$ pots, picks a real number $x$, &#xA;and then eats exactly $x$ kilogram of honey from each of the selected pots.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is it always (independently of the initial distribution &#xA;  of honey) possible for Winnie to empty all $27$ honey pots in a finite &#xA;  number of days?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-06T17:23:25.420" LastActivityDate="2017-03-13T23:04:58.527" Title="Winnie-the-Pooh and the 27 honey pots" Tags="|mathematics|strategy|" AnswerCount="9" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="22615" PostTypeId="1" AcceptedAnswerId="22617" CreationDate="2015-09-25T08:53:28.007" Score="16" ViewCount="1118" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PEQuJ.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PEQuJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q: Can you divide this shape into 4 equal parts, and then form a square?&lt;/p&gt;&#xA;" OwnerUserId="16539" LastEditorUserId="15544" LastEditDate="2015-09-27T02:59:33.327" LastActivityDate="2015-09-27T02:59:33.327" Title="The Challenge Square" Tags="|geometry|dissection|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22617" PostTypeId="2" ParentId="22615" CreationDate="2015-09-25T09:14:20.987" Score="18" Body="&lt;p&gt;You can do it like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/MFO3S.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MFO3S.png&quot; alt=&quot;solution&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The cut starts right in the center of those sides.&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-09-25T09:14:20.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22618" PostTypeId="2" ParentId="22612" CreationDate="2015-09-25T10:22:48.697" Score="11" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Extraterrestial greetings from Alpha Centauri. We come in peace&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How I got it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Try and see the white letters on a grey background, easier if you fill in the extra whitespace. &lt;a href=&quot;https://i.stack.imgur.com/guNLr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/guNLr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And as to how I figured that out, here's the classic that came to mind&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/SqSa3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SqSa3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As for why the 'worldplay' tag:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; With a little help from the comments below: 'outer space' (the ongoing puzzle theme) is what is shown: by removing the bands between the lines, you only see the space between the letters rather then the letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2156" LastEditorUserId="2156" LastEditDate="2015-09-29T06:42:59.233" LastActivityDate="2015-09-29T06:42:59.233" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="22619" PostTypeId="2" ParentId="22614" CreationDate="2015-09-25T11:35:02.540" Score="5" Body="&lt;p&gt;It is not always possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First note that, suppose that it would be possible then it could always be finished in $\binom{27}{7}$ days, because this is the number of possible ways to pick 7 pots, and eating from the same 7 pots on multiple days can always be combined to eating that amount in 1 day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all those combinations every pot occurs the same amount of time. So let's say it was possible to distribute the 17kg evenly (which isn't allowed by the rules). Then there is a solution in exactly $\binom{27}{7}$ days in which every day the same exact amount is eaten.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that given a distribution you are always allowed to subtract $x$ from every pot and try to solve that distribution instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this is what we do:&lt;br&gt;&#xA;put 1 gram in the first pot, put 1/10 gram in the second, 1/100 in the third and so on. Whatever is left from the 17kg we distribute evenly among all pots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to what I said before, we can neglect that final part so we just see if we can solve it with the 1, 1/10, 1/100 distribution. And this is simple: we can't solve it because the honey in pot one is even more than all the honey of the other pots combined.&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-09-25T11:35:02.540" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="22622" PostTypeId="2" ParentId="14953" CreationDate="2015-09-25T16:04:09.623" Score="2" Body="&lt;p&gt;Please mark the city not matching the others:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Clearly it's Sydney because it's the only one that's not a capital city&lt;/li&gt;&#xA;&lt;li&gt;No wait, clearly it's Paris because it's the only one that has a one-word anagram.&lt;/li&gt;&#xA;&lt;li&gt;Oops, it's obviously Buenos Aires because it's the only one in the Western Hemisphere&lt;/li&gt;&#xA;&lt;li&gt;My bad, it's actually Moscow because it's the only one that doesn't use the Latin alphabet&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I hate questions like this.&lt;/p&gt;&#xA;" OwnerUserId="2318" LastActivityDate="2015-09-25T16:04:09.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22623" PostTypeId="1" AcceptedAnswerId="22633" CreationDate="2015-09-25T17:01:13.680" Score="8" ViewCount="484" Body="&lt;p&gt;These phrases can be put into pairs based on a rule. Find the odd one out - your answer is the one that would match it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before&lt;br&gt;&#xA;Boyfriend or admirer&lt;br&gt;&#xA;Brown or Pratt&lt;br&gt;&#xA;Characteristic of Santa Claus or alcohol&lt;br&gt;&#xA;Coyote's plan&lt;br&gt;&#xA;Empty space between chairs&lt;br&gt;&#xA;Hang around longer than necessary&lt;br&gt;&#xA;It's in a can&lt;br&gt;&#xA;It's in airplane bathrooms&lt;br&gt;&#xA;King&lt;br&gt;&#xA;Listen to&lt;br&gt;&#xA;Make up for a debt&lt;br&gt;&#xA;Reason&lt;br&gt;&#xA;Street&lt;br&gt;&#xA;Street sign with an arrow&lt;br&gt;&#xA;Superman's vision&lt;br&gt;&#xA;Triumphed  &lt;/p&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="11876" LastEditDate="2015-09-25T18:33:51.000" LastActivityDate="2015-09-26T01:46:17.230" Title="Pairs of Phrases" Tags="|wordplay|english|" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="22626" PostTypeId="1" AcceptedAnswerId="22627" CreationDate="2015-09-25T18:11:07.950" Score="12" ViewCount="731" Body="&lt;p&gt;This kind of puzzle is relatively straightforward - solve the cryptic clues, fill in the grid, and find the concealed quote! Bonus points for anyone who can tell me who said it &lt;em&gt;without&lt;/em&gt; Googling it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The irony is that an American uncle has wrecked cars (7)&lt;br&gt;&#xA;__ __ __ __ __ __ __&lt;br&gt;&#xA;&lt;em&gt;06 27 42 52 61 20 30&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Limit citation to 80% with a resolution (5)&lt;br&gt;&#xA;__ __ __ __ __&lt;br&gt;&#xA;&lt;em&gt;55 56 02 11 48&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Baby has misshapen appendage and skin discoloration (6)&lt;br&gt;&#xA;__ __ __ __ __ __&lt;br&gt;&#xA;&lt;em&gt;09 17 19 05 10 14&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This evening, the baby eats, almost (7)&lt;br&gt;&#xA;__ __ __ __ __ __ __&lt;br&gt;&#xA;&lt;em&gt;22 34 49 15 26 04 23&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The idea is that it's &lt;em&gt;not&lt;/em&gt; a big, headless cat (6)&lt;br&gt;&#xA;__ __ __ __ __ __&lt;br&gt;&#xA;&lt;em&gt;07 16 32 24 41 54&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Bar in Game of Thrones, initially (5)&lt;br&gt;&#xA;__ __ __ __ __&lt;br&gt;&#xA;&lt;em&gt;29 25 01 08 33&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Dime is tight, he thayth with a lithp (5)&lt;br&gt;&#xA;__ __ __ __ __&lt;br&gt;&#xA;&lt;em&gt;35 37 13 43 36&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Make morning drink less heady by preceding it with tea and candy (6)&lt;br&gt;&#xA;__ __ __ __ __ __&lt;br&gt;&#xA;&lt;em&gt;50 53 40 39 38 57&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Lone dead bodies are in the woods (6)&lt;br&gt;&#xA;__ __ __ __ __ __&lt;br&gt;&#xA;&lt;em&gt;62 45 58 60 59 12&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Cow's cries mark the beginning of the end for woodland creature (5)&lt;br&gt;&#xA;__ __ __ __ __&lt;br&gt;&#xA;&lt;em&gt;47 18 51 44 21&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Inserted information incorrectly in section of form (5)&lt;br&gt;&#xA;__ __ __ __ __&lt;br&gt;&#xA;&lt;em&gt;46 31 63 28 03&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IAAGE.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IAAGE.png&quot; alt=&quot;Grid&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-10-05T17:30:44.307" Title="Cryptic Quote #1: To Infinity and Beyond!" Tags="|cryptic-crosswords|outer-space|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22627" PostTypeId="2" ParentId="22626" CreationDate="2015-09-25T18:53:29.890" Score="13" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The irony is that an American uncle has wrecked cars (7)  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;SARCASM&lt;/strong&gt; (&lt;strong&gt;SAM&lt;/strong&gt; + scrambled &lt;strong&gt;CARS&lt;/strong&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Limit citation to 80% with a resolution (5)  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;QUOTA&lt;/strong&gt; (&lt;strong&gt;QUOT&lt;/strong&gt; = 80% of QUOTE + &lt;strong&gt;A&lt;/strong&gt; resolution)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Baby has misshapen appendage and skin discoloration (6)  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;INFANT&lt;/strong&gt; (&lt;strong&gt;FIN&lt;/strong&gt; and &lt;strong&gt;TAN&lt;/strong&gt; misshapen)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This evening, the baby eats, almost (7)  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;TONIGHT&lt;/strong&gt; (&lt;strong&gt;TOT&lt;/strong&gt; encompasses &lt;strong&gt;NIGH&lt;/strong&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The idea is that it's &lt;em&gt;not&lt;/em&gt; a big, headless cat (6)  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;NOTION&lt;/strong&gt; (&lt;strong&gt;NOT&lt;/strong&gt; + (LION - head = &lt;strong&gt;ION&lt;/strong&gt;))&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Bar in Game of Thrones, initially (5)  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;INGOT&lt;/strong&gt; (&lt;strong&gt;IN&lt;/strong&gt; + &lt;strong&gt;GOT&lt;/strong&gt; (initials of Game of Thrones))&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Dime is tight, he thayth with a lithp (5)  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;TENTH&lt;/strong&gt; (&lt;strong&gt;TENSE&lt;/strong&gt; with a lisp)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Make morning drink less heady by preceding it with tea and candy (6)  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;TOFFEE&lt;/strong&gt; (&lt;strong&gt;COFFEE&lt;/strong&gt; - head + &lt;strong&gt;T&lt;/strong&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Lone dead bodies are in the woods (6)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;CORPSE&lt;/strong&gt; (&lt;strong&gt;R&lt;/strong&gt; (are) in &lt;strong&gt;COPSE&lt;/strong&gt;)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Cow's cries mark the beginning of the end for woodland creature (5)  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;MOOSE&lt;/strong&gt; (&lt;strong&gt;MOOS&lt;/strong&gt; + &lt;strong&gt;E&lt;/strong&gt; (beginning of &quot;end&quot;))&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Inserted information incorrectly in section of form (5)  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;FIELD&lt;/strong&gt; (&lt;strong&gt;FILED&lt;/strong&gt; scrambled)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The solution is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/O3SVj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/O3SVj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;said by (I did have to google it)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pope Pius XII&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="4421" LastEditDate="2015-10-05T17:30:44.307" LastActivityDate="2015-10-05T17:30:44.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22628" PostTypeId="2" ParentId="22614" CreationDate="2015-09-25T20:15:02.677" Score="-2" Body="&lt;p&gt;I'm in two minds about this.  The above arguments sound convincing, but it seems that if we simply go through the $\binom{27}{7}$ possibilities taking $x_i$ for $i\in(1,\ldots,888 030)$ out of the relevant pots, we have 27 equations with 888030 unknowns.  Okay, there's a constraint of $x_i\geq0$, which is concerning, but $888030\gg27$.&lt;/p&gt;&#xA;" OwnerUserId="15942" LastActivityDate="2015-09-25T20:15:02.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22629" PostTypeId="2" ParentId="22550" CreationDate="2015-09-25T20:21:38.757" Score="5" Body="&lt;h2&gt;Partial Answer&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The code has something to do with the digits of the safe number.  If the code to safe $n$ is denoted by $C(n)$, then we can see that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;C(n) = 2\left[f\left(\left\lfloor\frac{n}{10}\right\rfloor\right) + (n~\text{mod}~10)\right]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The multiplication by two is because all code values are even.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know the values of $f$ for $0$ through $6$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;f(0)&amp;amp;=0\\&#xA;f(1)&amp;amp;=10\\&#xA;f(2)&amp;amp;=20\\&#xA;f(3)&amp;amp;=13\\&#xA;f(4)&amp;amp;=-27\\&#xA;f(5)&amp;amp;=15\\&#xA;f(6)&amp;amp;=16\\&#xA;f(7)&amp;amp;=17&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are hinted that the answer has to do with the &lt;em&gt;location&lt;/em&gt; of the safes, which leaves me to believe that $f$ is related to the &lt;em&gt;English spelling&lt;/em&gt; or &lt;em&gt;pronunciation&lt;/em&gt; of the numbers.  In particular, it seems to me as if there is some relationship between &lt;em&gt;thirty&lt;/em&gt;/&lt;em&gt;thirteen&lt;/em&gt;, &lt;em&gt;fifty&lt;/em&gt;/&lt;em&gt;fifteen&lt;/em&gt;, etc., although $f(4)$ breaks this pattern.&lt;/p&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="9718" LastEditDate="2015-09-30T15:45:51.470" LastActivityDate="2015-09-30T15:45:51.470" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="22630" PostTypeId="1" AcceptedAnswerId="22631" CreationDate="2015-09-25T21:02:32.610" Score="5" ViewCount="1946" Body="&lt;p&gt;I came around a &lt;a href=&quot;http://young.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;website&lt;/a&gt; recently and I noticed these strange hand illustrations, followed by a number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VlZfm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VlZfm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Dxjy6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Dxjy6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/xYqLI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xYqLI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/34ezm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/34ezm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see hundreds of different snapshots of the website using the archive.org wayback machine feature:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://web.archive.org/web/20141218121548/http://young.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://web.archive.org/web/20141218121548/http://young.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Use top page slider to navigate through website history)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have any idea what it means?&lt;/p&gt;&#xA;" OwnerUserId="16544" LastEditorUserId="16544" LastEditDate="2015-09-25T21:30:37.003" LastActivityDate="2015-09-25T21:52:05.297" Title="Strange code made of hand sign and number" Tags="|cipher|enigmatic-puzzle|number-sequence|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22631" PostTypeId="2" ParentId="22630" CreationDate="2015-09-25T21:25:15.767" Score="13" Body="&lt;p&gt;The hand signs are &lt;a href=&quot;https://en.wikipedia.org/wiki/American_Sign_Language&quot; rel=&quot;nofollow noreferrer&quot;&gt;ASL&lt;/a&gt; (American Sign Language) numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ERMX3.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The author of the site implemented the page counter by calling two PHP scripts for each page:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;http://www.young.com/count/count.php?page=young&#xA;http://www.young.com/count/small.php?page=young&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Each one of these scripts contains Javascript to insert the correct images for the corresponding number.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;However, the implementation is quite buggy.  In particular, one of the numbers will always be higher than the other, since (apparently) each script independently increments the page counter.  Which one will be higher depends on the order that the browser sends the requests, and the order the server processes them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, if two or more people try to access the page at the same time, their requests may interleave, resulting in the apparent count increasing more than once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, for the images you gave we can translate the hand signs into Arabic numerals:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;count.php   small.php&#xA;---------------------&#xA;  183         130&#xA;    2           3&#xA;   29          10&#xA;  390         389&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The Wayback machine seems to be loading the script files in reverse order, and seems to be waiting some time between subsequent accesses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I accessed the page, I got &lt;code&gt;30/27&lt;/code&gt;, &lt;code&gt;35/36&lt;/code&gt;, and &lt;code&gt;39/40&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This has been an example of how &lt;em&gt;not&lt;/em&gt; to implement a page counter.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Update&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;s&gt;I've been calling one of the scripts in a loop, and the count doesn't seem to wrap around at 1000.  It's probably a daily/weekly/monthly count.&lt;/s&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The count has decreased back to zero, although it briefly went above 1000.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some further tests, it looks like the counter is resetting randomly, with more frequent resets under higher server load.  Looking at the quality of the code, I'm assuming this is some sort of race condition in the PHP script that wipes out the counter if two or more threads try to modify it at once.  (I once had a small website that stored some persistent data in a flat file, and struggled with this problem for some time.  The solution is to use a real database.)&lt;/p&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="9718" LastEditDate="2015-09-25T21:52:05.297" LastActivityDate="2015-09-25T21:52:05.297" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22632" PostTypeId="2" ParentId="22614" CreationDate="2015-09-25T22:03:45.987" Score="16" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;h1&gt;Update:&lt;/h1&gt; I made a big blunder in the previous version, this should be fixed now.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Part 1: It is always possible to remove amount $x$ from any number $n \ge 7$ of pots if all pots have at least amount $x$&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Let $p$ be the number of permutations with 7 pots containing a specific pot. Cycle all permutations and remove amount $\frac xp$ from each pot. As each pot occurs $p$ times the total removed amount from each pot is $x$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Part 2: A combination of pots where the 7 pots containing the most amounts are equal is know to work regardless of the amounts in other pots&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Consider the following image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sA55t.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bars show the amounts in the pots, ordered by the amount. The amounts in the first 7 pots are equal. Start with the smallest pot $J$ and remove the amount of $J$ from all pots (this works as seen in part 1). Repeat with pot $I$ and then with pot $H$. At the end we have the top 7 pots containing only the black marked area, which can be emptied in the next step.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Part 3: Removing superfluous amount in biggest pot requires 6 times the same amount 20 smallest pots&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Consider the following image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/iVTug.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We must remove the red part in pot $A$ to reach the situation from part 2. It is easy to see that we can do this easily if we have the same 6 parts in pots $H-M$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This is the place where I made a blunder in the previous version, assuming I could trade the red part in $A$ with a combination of parts of the same size in pots $&amp;gt;G$, instead of 6 times the size.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What can be said about the distribution of the amounts beyond pot $G$? Actually it doesn't matter, if we use a trick. Consider the following image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AJjQg.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to remove the red part in pot $A$. The orange part in pot $H$ has 6 times the size of the red part. We can remove the red part and the orange part of pot $H$ while also removing the orange parts in pots $A-G$. To do this we generate all permutations of 6 pots from the range $B-H$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\binom{7}{6}$ = 7&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number each of the pots $B-H$ appears in this permutations is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\binom{6}{5}$ = 6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we add pot $A$ to all these permutations and remove the red part from each permutation. We remove the red part 7 times from pot $A$ and 6 times from pots $B-H$ which is exactly what we see in the image above. This can be easily extended to any number of pots, as long as they contain 6 times the size of the red part. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other conclusion is, no matter what we do, we can't remove the red part if the size of the pots beyond $G$ is less than 6 times the size of the red part.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Part 4: Removing superfluous amount in 2 biggest pot requires 2.5 times the same amount 20 smallest pots&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Consider the following image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jLKgG.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar to part 3, we now want to remove the 2 red parts in pots $A-B$. It is again easy to see that we can do this easily if we have the same 5 parts in pots $H-L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the same reasoning as in part 3, it can be shown, that the amount distribution beyond pot $G$ doesn't matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can also be shown that similar applies to superfluous amounts in pots $C-F$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Part 5: The general case&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The following image shows the general case which is guaranteed to be solvable based on the parts above:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Sd31V.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we have seen in part 3, the distribution in pots beyond $G$ doesn't matter. Also as seen in part 2 adding amounts in pots beyond $G$ is solvable too. What is left is to show which amounts can be solved in any case.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Part 6: The edge case&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RS3O3.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above image shows the edge case, where the red area is maximized (if we limit ourselves to 13 pots). The height of pots $H-M$ must be the same as the height of the red part of pot $A$. Pots $B-G$ must have also the same size. As $A$ is limited to $1$, the height of pots $B-M$ must be $\frac{1}{2}$. The size of this case is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1 + 12 * \frac{1}{2} = 7$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This means that 7kg honey is enough to guarantee, that the puzzle is solvable, if the pot size is limited to 1kg.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any other amounts which we would try to add, don't change anything. Amounts added to pots beyond $G$ are OK as shown in part 2. Amounts added to pots $B-G$ reduce the required amount in pots beyond $G$ as can be seen in the following image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hfSxj.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more thing we could do, is to use the other 14 pots we didn't use yet to further maximize the red area, like in the following image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1gk45.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We used here one more pot. We increased the size of the red area above $A$ by some amount $x$. We also increased the size of the red area beyond pot $G$ by $6x$. But at the sime time we decreased the size of the black area by $7x$ which in the end gives a total change of $0$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The pot size which would still work for 17kg is (based on the first edge case):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x + 12 * \frac{x}{2} = 17$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = \frac{17}{7} = 4,28...$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, this solutions works with any number of pots $\ge 13$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Alternatively we can make Winnie-the-Pooh really happy and tell him, he can eat from 17 pots each day (which would still work with 17kg and 1kg max per pot).&lt;/p&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2015-09-28T19:07:41.830" LastActivityDate="2015-09-28T19:07:41.830" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22633" PostTypeId="2" ParentId="22623" CreationDate="2015-09-26T00:09:07.490" Score="7" Body="&lt;p&gt;The rule of the pairs is that one of the words in the pair is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Pig_Latin&quot; rel=&quot;nofollow&quot;&gt;Pig-Latin&lt;/a&gt; version of the other.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;King / Superman's vision (Rex / X-ray)&lt;/li&gt;&#xA;&lt;li&gt;Before / Make up for a debt (Pre- / Repay)&lt;/li&gt;&#xA;&lt;li&gt;Triumphed / Street sign with an arrow (Won / One-Way)&lt;/li&gt;&#xA;&lt;li&gt;Reason / Street (Why / Highway)&lt;/li&gt;&#xA;&lt;li&gt;Lover / Listen to (Beau / Obey)&lt;/li&gt;&#xA;&lt;li&gt;It's in a can / It's in an airplane bathroom (Trash / Ashtray)&lt;/li&gt;&#xA;&lt;li&gt;Characteristic of Santa Claus or alcohol / Hang around longer than necessary (Stout / Outstay)&lt;/li&gt;&#xA;&lt;li&gt;Coyote's plan / Empty space between chairs (Wile / Aisleway)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The odd one out is Chris (Brown or Pratt), which would match with &lt;em&gt;risqué&lt;/em&gt; if a clue for that were to be included. Thanks to f'' for the last four pairs.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-09-26T01:46:17.230" LastActivityDate="2015-09-26T01:46:17.230" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="22634" PostTypeId="2" ParentId="22614" CreationDate="2015-09-26T01:50:53.857" Score="4" Body="&lt;p&gt;We can formulate the problem as follows. This is not an answer but perhaps it leads to one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $27$ pots correspond to a choice of values $p_i$ with $0 \leq i &amp;lt; 27$ satisfying:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$p_i \neq p_j$ whenever $i \neq j$;&lt;/li&gt;&#xA;&lt;li&gt;$0 \leq p_i \leq 1$;&lt;/li&gt;&#xA;&lt;li&gt;$\sum p_i = 17$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So each pot configuration lies in $Z'$ the intersection of a hypercube $[0,1]^{27}\subset\mathbb{R}^{27}$ with the hyperplane $\pi$ defined by $(3)$. Of course, the point must also satisfy $(1)$, so it must be in the complement of the union of another $\binom{27}{2}$ hyperplanes. Let $Z \subset \mathbb{R}^{27}$ be the set defined by these conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A solution for a given configuration corresponds to a choice of values $v_j \geq 0$ with $0 \leq j &amp;lt; \binom{27}{7}$. Each such value is the amount of honey Pooh eats from each pot in a given combination of $7$ out of the $27$ pots, so they must satisfy&#xA;$$\sum v_j = \frac{17}{7},$$and also for each $0 \leq i &amp;lt; 27$&#xA;$$\sum v_{j_i} = p_i ,$$&#xA;where the sum has $\binom{26}{6}$ terms and is carried over the $v_j$'s which involve pot $i$. These last $27$ equations can be written in matrix form as $M\cdot V = P$, where $V$ is the column vector ${\left(v_j\right)}$, $P$ is the column vector $\left(p_i\right)$ and $M$ is a $27 \times \binom{27}{7}$ matrix of $0$'s and $1$'s. Each line has $\binom{26}{6}$ nonzero entries and each column has $7$ nonzero entries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $V$ must lie in the set $S$, defined by the intersection of a hyperplane $\sigma$ in $\binom{27}{7}$-dimensional space with its first quadrant. A solution for $P \in Z$ exists if and only if $P \in M(S)$. Thus, a solution always exists if and only if $Z \subset M(S)$. It might however be easier to show that $Z' \subset M(S)$; this suffices since $Z \subset Z'$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to see $M(\sigma) \subset \pi$, and it should not be too hard to show that's actually an equality (this is necessary). It's also easy to see that if $V$ has nonnegative coordinates then so does $M(V)$. However this still needs to be improved.&lt;/p&gt;&#xA;" OwnerUserId="16478" LastEditorUserId="16478" LastEditDate="2015-09-28T07:21:07.013" LastActivityDate="2015-09-28T07:21:07.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22635" PostTypeId="2" ParentId="14953" CreationDate="2015-09-26T05:26:04.170" Score="1" Body="&lt;p&gt;Without even considering knowledge of cities' significance or the composition of various metals I was able to get both 1 and 3 right. I applied the same extremely elementary logic to both: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sydney is the only city of the 4 that ends with a vowel(aeiou, sometimes y.) &#xA; Bronze is the only metal of those listed that had a vowel at the end. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This could be the absolute wrong approach, but it yielded the correct answer.&lt;/p&gt;&#xA;" OwnerUserId="16510" LastActivityDate="2015-09-26T05:26:04.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22636" PostTypeId="1" AcceptedAnswerId="22637" CreationDate="2015-09-26T05:53:33.013" Score="2" ViewCount="145" Body="&lt;p&gt;I have encoded a riddle. Everything necessary to decode the riddle is supplied below. There are two parts to the puzzle: Decode the riddle, then solve the riddle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MSSCMMEMESSQGMGSESSWGSGGMCGQMMMSWMCKASSGCGMESMCGBSKSM&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1010010010000010010110011001000100000100100101100100010110011000100100010000011000100101100101100010011000&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;How would someone with no knowledge of complex ciphers or matrices approach the ciphertext? Think in simple terms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Nothing represented by the binary is greater than two.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I will add more hints if necessary.&lt;/p&gt;&#xA;" OwnerUserId="16510" LastEditorUserId="16510" LastEditDate="2015-09-26T06:34:16.090" LastActivityDate="2015-09-26T07:24:28.607" Title="Synthetic smarts" Tags="|riddle|cipher|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22637" PostTypeId="2" ParentId="22636" CreationDate="2015-09-26T06:19:59.970" Score="7" Body="&lt;p&gt;The message is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;OUTCOME OF TURING TEST WITHIN CHRONOSYNCLASTIC INFUNDIBULUM&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The method of encipherment:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The first step is to realize that every block of two binary digits corresponds to one letter. We can then group as follows:&lt;br&gt;&lt;code&gt;M,10 S,10 S,01 C,00 etc.&lt;/code&gt;&lt;br&gt;All we then do with these groupings is 'increment' the letter by the number represented by the binary in its group. For instance, the first group is &lt;code&gt;M,10&lt;/code&gt;, and 10 is 2 in binary, so two letters after &lt;code&gt;M&lt;/code&gt; is the first letter of the plaintext &lt;code&gt;O&lt;/code&gt;.&lt;br&gt;&lt;br&gt;Continuing like this we get:&lt;br&gt;&lt;code&gt;OUTCOMEOFTURINGTESTWITHINCHRONOSYNCLASTICINFUNDIBULUM&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7308" LastEditorUserId="7308" LastEditDate="2015-09-26T06:36:47.747" LastActivityDate="2015-09-26T06:36:47.747" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22639" PostTypeId="2" ParentId="22636" CreationDate="2015-09-26T07:24:28.607" Score="2" Body="&lt;p&gt;Here is my guess for the riddle decoded by Tryth:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chronosynclastic infundibulum is a term used in &lt;a href=&quot;https://en.wikipedia.org/wiki/The_Sirens_of_Titan&quot; rel=&quot;nofollow&quot;&gt;&quot;The Sirens of Titan&quot;&lt;/a&gt;, and refers to a place in which one's existence in space-time becomes fragmented, being able to appear at multiple locations and time points simultaneously. This grants some humans within this space the ability to know the past and the future. In such a case, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Turing_test&quot; rel=&quot;nofollow&quot;&gt;Turing test&lt;/a&gt;, which tests whether an artificial intelligence can be sufficiently advanced so as to be indistinguishable from a reasonable human being, becomes exponentially more difficult, i.e. the outcome is almost always a failed test.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's also possible that the riddle is referring to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Halting_problem&quot; rel=&quot;nofollow&quot;&gt;halting problem for testing Turing completeness&lt;/a&gt;, which is a decision problem that is known to be undecidable for a general class of computers (called Turing machines). If this is true, then &quot;the outcome of Turing test&quot; refers to a computer within the chronosynclastic infundibulum that overcomes this limitation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15960" LastActivityDate="2015-09-26T07:24:28.607" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22642" PostTypeId="2" ParentId="22510" CreationDate="2015-09-26T11:03:29.270" Score="3" Body="&lt;h1&gt;6x6 solution with 3 rotations: 3591&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 6x6 square = 1 bounding rectangle with 60 inner edges&lt;br&gt;&#xA; 3 total 90° rotations&lt;br&gt;&lt;br&gt;&#xA; $60^2 - 3^2 = 3591$ (&lt;em&gt;single network&lt;/em&gt;)&lt;br&gt;&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/P1aZq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/P1aZq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks again to Sleafar for the image!&lt;/p&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-09-26T11:03:29.270" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22644" PostTypeId="1" CreationDate="2015-09-26T15:47:37.827" Score="2" ViewCount="198" Body="&lt;p&gt;Based on &lt;a href=&quot;https://puzzling.stackexchange.com/questions/22584/the-erasmus-polyhedron&quot;&gt;The Erasmus polyhedron&lt;/a&gt; by @Gamow..&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Construct a 3D convex polyhedron of any single material that can float&#xA;  in water such that $x\%$ of its volume is below water level and $y\%$&#xA;  of its surface area is above water level. Find the solution with&#xA;  maximum $(x\times y)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Notes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Archimedes principle is valid.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Material must be something that exists in real life.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Assume that it does not have any holes/pores.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-09-27T08:34:15.100" Title="The incredible polyhedron" Tags="|mathematics|geometry|optimization|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22645" PostTypeId="1" AcceptedAnswerId="22647" CreationDate="2015-09-26T17:23:50.213" Score="11" ViewCount="847" Body="&lt;p&gt;Professor Halfbrain has recently made several fascinating discoveries on cutting convex polygons in the plane.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Halfbrain's first cutting theorem:&lt;/strong&gt;&#xA;  Every convex polygon can be cut (by a perfectly straight cut) into two polygons $A$ and $B$ so that $A$ and $B$ have the same perimeter and so that the length of the &lt;strong&gt;longest&lt;/strong&gt; side of $A$ equals the length of the &lt;strong&gt;longest&lt;/strong&gt; side of $B$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Question: Is this theorem indeed true, or has the professor once again made one of his notorious mathematical blunders?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-09-27T11:54:32.417" LastActivityDate="2015-09-27T13:18:20.080" Title="Professor Halfbrain's first cutting theorem" Tags="|mathematics|geometry|dissection|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22646" PostTypeId="1" AcceptedAnswerId="22691" CreationDate="2015-09-26T17:38:20.553" Score="3" ViewCount="829" Body="&lt;p&gt;She is the feast, and I am the famine.&#xA;Two of a kind, remarkably similar we are.&#xA;When you look at her, you might see me, &#xA;but that's your fault, I'm not there.&#xA;one little straight line, you glazed right over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you replaced her with me, all would be gone,&#xA;and many would be very unhappy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Persons from &quot;across the pond&quot; (as THEY say) would have an easier time finding the pair of words.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The little line that's missing isn't a letter of the alphabet. (it might be &lt;em&gt;part&lt;/em&gt; of one though...)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13068" LastEditorUserId="13068" LastEditDate="2015-09-28T21:59:31.670" LastActivityDate="2015-09-29T18:43:52.207" Title="Another riddle, two twins" Tags="|riddle|word|" AnswerCount="6" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22647" PostTypeId="2" ParentId="22645" CreationDate="2015-09-26T18:23:41.107" Score="15" Body="&lt;p&gt;Yes, it is always possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick a point $P$ on the original convex polygon, and then place the point $Q$ on $P$ and move it continuously clockwise along the perimeter of the polygon until it comes back to $P$. As $Q$ is moving, keep track of the clockwise distance along the perimeter from $P$ to $Q$. By the intermediate value theorem, at one point that distance was equal to half the perimeter. Thus for any point on the perimeter, there is a unique &quot;opposite&quot; point that splits the perimeter in half.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now move $P$ and $Q$ along the perimeter of the polygon together. As they are moving, they cut the original polygon into $A$ and $B$, and we consider the longest edge of each of the two polygons. Clearly the length of the longest edge is continuous because it's unchanging when it is an edge of the original polygon and changing continuously when it is $PQ$. We move $P$ and $Q$ until they switch places, so $A$ and $B$ also switched places, so the lengths of the longest edges of $A$ and $B$ also swapped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus by the intermediate value theorem there was a $PQ$ that split the original polygon into $A$ and $B$ with equal perimeters and equal longest edges.&lt;/p&gt;&#xA;" OwnerUserId="11100" LastEditorUserId="11100" LastEditDate="2015-09-26T18:31:10.200" LastActivityDate="2015-09-26T18:31:10.200" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22648" PostTypeId="1" CreationDate="2015-09-26T19:37:14.430" Score="4" ViewCount="1046" Body="&lt;p&gt;I am trying to find this puzzle which I once came across as a kid, but can't seem to recall it properly now (or find it). It goes something like this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a reservoir of capacity 20 litres filled with water. You have 2 buckets of volume x and y (can't seem to recall this information). Without any other forms of measurement, the final aim is to have 2 litres in both buckets. Does anyone remember this puzzle ? It might have been by Lewis Carroll. The values of x,y might have been 3,4 or 4,5 but I am 99% positive about the 20 litres and 2 litres part. Would be very helpful if someone can help me find it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Thanks for the replies guys, sorry I don't recollect the original puzzle, but from what I remember there was no wasting of water and the supply of water also wasn't infinite. &lt;/p&gt;&#xA;" OwnerUserId="16553" LastEditorUserId="4385" LastEditDate="2015-10-10T20:22:29.767" LastActivityDate="2015-11-13T04:31:52.353" Title="Trying to find a water measuring puzzle" Tags="|puzzle-identification|decanting-problems|" AnswerCount="5" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="22650" PostTypeId="2" ParentId="22648" CreationDate="2015-09-27T01:58:53.970" Score="1" Body="&lt;p&gt;The puzzle as you describe it is impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If no water is wasted and the buckets' capacities are less than 20L, then after every move either one of the buckets is empty or one of the buckets is full. Therefore, there cannot be 2L in both buckets.&lt;/p&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2015-09-27T01:58:53.970" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22651" PostTypeId="2" ParentId="22646" CreationDate="2015-09-27T07:32:20.153" Score="1" Body="&lt;p&gt;Is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; She is &lt;strong&gt;fat&lt;/strong&gt; and you are &lt;strong&gt;flat&lt;/strong&gt;. One little line - the letter &lt;strong&gt;l&lt;/strong&gt; - is the difference between you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-09-27T07:32:20.153" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22653" PostTypeId="2" ParentId="22644" CreationDate="2015-09-27T08:34:15.100" Score="5" Body="&lt;p&gt;&lt;strong&gt;Solution with $xy \to 5000$&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To maximise the volume in the water, choose a material that has the same density as water, and place it so that the top face sits at the surface of the water. Having equal density with water means the object won't rise or sink. This allows $x$ to approach 100.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To maximise the surface area above the water, choose a short shape with one large face. An extremely shallow pyramid would allow $y$ to approach 50. (I'm ignoring the effects of the water's surface tension.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water density &lt;a href=&quot;https://en.wikipedia.org/wiki/Density#Water&quot;&gt;varies with temperature&lt;/a&gt;, with a peak of 1000 kgm$^{-3}$ at 4$^\text{o}$C. At 80$^\text{o}$C, the density of water is 971.8 kgm$^{-3}$, which is close to sodium's 970 kgm$^{-3}$. (Don't actually put a slab of sodium into hot water, though!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plastics come in a &lt;a href=&quot;http://www.dotmar.com.au/density.html&quot;&gt;variety&lt;/a&gt; of densities, with Polyethylene Polystone M-Flametech said to have a density of 0.98 g/cm$^3$, or 980 kgm$^{-3}$. A slab of this material should do the trick with the water somewhere between &#xA;60$^\text{o}$C and 80$^\text{o}$C.&lt;/p&gt;&#xA;" OwnerUserId="9422" LastActivityDate="2015-09-27T08:34:15.100" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22654" PostTypeId="2" ParentId="22646" CreationDate="2015-09-27T09:20:29.310" Score="12" Body="&lt;p&gt;Is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; dessert (she) vs desert (you)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Feast / famine:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; dessert is a type of food, deserts produce very little food.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Two of a kind:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; similar spelling.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you look at her, you might see me ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it is a common spelling mistake to confuse the two.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One little line:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the extra 's' is a single line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Glazed:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; common to certain types of desserts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you replaced her with me ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; there won't be much left on earth if each dessert is replaced with a desert of any reasonable size. The sweet-toothed would be very unhappy, though exchanging almost any climate for a desert would be unpleasant for many.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9422" LastActivityDate="2015-09-27T09:20:29.310" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22655" PostTypeId="1" AcceptedAnswerId="22658" CreationDate="2015-09-27T11:54:38.520" Score="12" ViewCount="682" Body="&lt;p&gt;Professor Halfbrain has recently made several fascinating discoveries on cutting convex polygons in the plane.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Halfbrain's second cutting theorem:&lt;/strong&gt;&#xA;  Every convex polygon can be cut (by a perfectly straight cut) into two polygons $A$ and $B$ so that $A$ and $B$ have the same perimeter and so that the length of the &lt;strong&gt;shortest&lt;/strong&gt; side of $A$ equals the length of the &lt;strong&gt;shortest&lt;/strong&gt; side of $B$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Question: Is this second theorem indeed true, or has the professor once again made one of his notorious mathematical blunders?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Comments:&lt;br&gt;&#xA;1. A convex polygon has $n$ pairwise distinct vertices. A side of the polygon is the straight line segment connecting two consecutive vertices along the convex hull.&lt;br&gt;&#xA;2. For the professor's first cutting theorem, see &lt;a href=&quot;https://puzzling.stackexchange.com/questions/22645/professor-halfbrains-cutting-theorem&quot;&gt;Professor Halfbrain&amp;#39;s first cutting theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-09-27T17:53:16.430" Title="Professor Halfbrain's second cutting theorem" Tags="|mathematics|geometry|dissection|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22657" PostTypeId="2" ParentId="22646" CreationDate="2015-09-27T17:20:43.427" Score="1" Body="&lt;p&gt;I don't think the answer is: Friend/Fiend&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's are more serious attempts:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; glut/gut&lt;br&gt;&#xA; glory/gory&lt;br&gt;&#xA; health/heath&lt;br&gt;&#xA; triple/tripe  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;None of these is really great though.  Maybe someone else can have more success with this pastebin of words which are still words when the L is removed: &lt;a href=&quot;http://pastebin.com/Rjcc98QU&quot; rel=&quot;nofollow&quot;&gt;http://pastebin.com/Rjcc98QU&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I avoided words with more than one L, so I guess that's the next search...which can be found here: &lt;a href=&quot;http://pastebin.com/vRu7QE1p&quot; rel=&quot;nofollow&quot;&gt;http://pastebin.com/vRu7QE1p&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which yields such attempts as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; bellies/belies&lt;br&gt;&#xA; filling/filing&lt;br&gt;&#xA; lawful/awful&lt;br&gt;&#xA; slick/sick&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;sigh&lt;/em&gt;.  It seems my computer can't save my brain - I'll have to &lt;strong&gt;think&lt;/strong&gt; about it...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[edit]&#xA;So apparently it's not a letter that's missing.  Which makes me think that it's a &lt;a href=&quot;http://pastebin.com/ASPadxtz&quot; rel=&quot;nofollow&quot;&gt;T going to an L&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spoiler:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; coat/coal&lt;br&gt;&#xA; fatter/falter&lt;br&gt;&#xA; malt/mall&lt;br&gt;&#xA; stew/slew&lt;br&gt;&#xA; stash/slash&lt;br&gt;&#xA; meat/meal&lt;br&gt;&#xA; stacks/slacks&lt;br&gt;&#xA; teacher/leacher (my favourite so far)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Okay, none of these is really a spoiler, because that would imply that they might be right.  I guess I could have missed the word pair in my quick look through the pastebin, but perhaps it's not a T-&gt;L.  It could be T-&gt;I, or even I-&gt;T (depending on your font).  Or M -&gt; N or A-&gt;H.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the feast should be the one with an extra line... &lt;/p&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2015-09-29T00:01:56.317" LastActivityDate="2015-09-29T00:01:56.317" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22658" PostTypeId="2" ParentId="22655" CreationDate="2015-09-27T17:53:16.430" Score="13" Body="&lt;p&gt;The professor made a mistake, this theorem is false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a regular pentagon. The point opposite of a vertex is the midpoint of the opposite edge, and the point opposite of a point on an edge is somewhere on an edge adjacent to the opposite vertex. In both cases the cutting edge is longer than the sidelength of the original polygon and so it cannot be the shortest edge of either $A$ or $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the cutting edge is between two sides of the original polygon, it must be, as pictured, like the green line and not the red line. Thus one of $A$ or $B$ will have both of the short edges and it cannot satisfy the conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pSodC.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pSodC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the pentagon below. It is close enough to a regular pentagon that our considerations about the regular pentagon hold. So we are left with 5 cases to check, one for each vertex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From A: On one side we have 1 + 0.99 = 1.99, and on the other side 0.96 + 0.97 = 1.93. Thus we can't split the opposite edge 0.98 right in half like we need to for the shortest edges to be equal, as then the perimeters of $A$ and $B$ wouldn't be equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From B: Same reasoning, 1.97 = 0.99 + 0.98 &gt; 1.96 = 1 + 0.96 = 1.96.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From C: Same reasoning, 1 + 0.99 = 1.99 &gt; 1.95 = 0.98 + 0.97.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From D: Same reasoning, 0.99 + 0.98 = 1.97 &gt; 1.95 = 0.97 + 0.96.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From E: Same reasoning, 1 + 0.96 = 1.96 &gt; 1.95 = 0.98 + 0.97.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Moplr.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Moplr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the above convex polygon is a counterexample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(note that if it's not actually close enough to a regular pentagon--which I doubt--we could add more 9's to the edges as in 0.9996, 0.9997, etc.)&lt;/p&gt;&#xA;" OwnerUserId="11100" LastActivityDate="2015-09-27T17:53:16.430" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22659" PostTypeId="2" ParentId="22598" CreationDate="2015-09-27T18:21:28.457" Score="1" Body="&lt;p&gt;The question is interesting with lots of possible solutions. Currently I see 2 of them:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is a start marked as &quot;L&quot; on the right side. If that means left side,then left is right, right is left. So your words &lt;em&gt;&quot;second to the right&quot;&lt;/em&gt; would mean second to the left, then the solution can be the red &quot;2&quot; star on the left or another star marked &quot;2&quot; on the bottom-left side (there are lots of possible ways of realizing what means 2nd)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the second way (more possible)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is &quot;E&quot; on the top-right side. So if we connect &quot;L&quot; and &quot;E&quot; and assume that it shows the left direction, then upper side will become &quot;left&quot;, bottom side will be the &quot;right&quot;. And then &lt;em&gt;&quot;second to the right &quot;&lt;/em&gt; can mean star marked 2 on the bottom side.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15786" LastActivityDate="2015-09-27T18:21:28.457" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22660" PostTypeId="2" ParentId="22646" CreationDate="2015-09-27T20:35:49.567" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; F(L)OOD.  The extra line is the letter l, leading to famine.  Being American or British doesn't make much difference, unless you look down at all the water, as you cross the pind.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16563" LastActivityDate="2015-09-27T20:35:49.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22661" PostTypeId="1" AcceptedAnswerId="22663" CreationDate="2015-09-27T22:50:03.097" Score="5" ViewCount="522" Body="&lt;p&gt;Pete's room has a rectangular floor with dimensions $15 \times 20$. He wants to tile the floor of his room with black and white coloured tiles. Each tile is an unit square (ie. $1 \times 1$ square).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it comes to these type of things Pete is simply a bit picky. He happens to prefer even numbers to odd ones and demands that if he places a O tetromino on the floor, it must contain an even number of black tiles. (In other words, every 2x2 square contains an even number of black tiles). The task itself is rather easy. Indeed a checkerboard coloring (alternating black and white) suffices. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JW391.png&quot; alt=&quot;The checkerboard colouring. Each 2x2 squares contains exactly 2 black squares which is even.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the tiler has managed to goof up and has coloured three of the corner cells white and the other black. Pete is furious and has threatened to fire the tiler if he can't tile the place properly maintaining the above condition?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the tiler save himself from being fired?&lt;/p&gt;&#xA;" OwnerUserId="15544" LastEditorUserId="15544" LastEditDate="2015-09-28T07:21:12.703" LastActivityDate="2015-09-28T07:21:12.703" Title="Perfectionist Pete's Perfect Pain: Even Tilings" Tags="|story|tiling|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22662" PostTypeId="2" ParentId="22614" CreationDate="2015-09-27T23:26:17.243" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;YES, HE CAN!&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Rigorous proof and exact bounds&lt;/strong&gt;:&#xA;  If the starting amount of honey in each pot is less or equal to $17/7$ kilograms, then Pooh can always do this. If the starting amount of honey in each pot can be more than $17/7$ kilograms, then Pooh can not always do this.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;More generally:&lt;/strong&gt; If the total amount of honey is $A$, the number of pots is $X$, each of the pots contains at most $P$ kilograms of honey and Pooh chooses every day $r \leq X$ pots, then he can always empty them if and only if $P \leq A/r$. (I have provided computations for $A=17, X=27, r=7$, but the same approach works for all triplets $(A, X, r)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PART 1.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The idea is to arrange the pots by amount of honey in decreasing order - #1, #2, #3, etc. and then take honey in such way so that we get more and more pots with the same amount of honey as pot #7.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Arrange the pots by amount of honey:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_1\geq x_2\geq x_3\geq x_4\geq...\geq x_{27}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_1\geq x_2\geq ...\geq x_k&amp;gt;x_{k+1}=x_{k+2}=...=x_l&amp;gt;x_{l+1}\geq x_{l+2}\geq ...\geq x_{27}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $k&amp;lt;7$ and $l\geq 7$ (this means that all pots with numbers from $k$ up to $l$ have the same amount of honey as pot #7). Now Pooh takes honey from the first $l$ pots, such that the amount of honey taken from each of pots $k+1, ... , l$ is $(7-k)/(l-k)$ of the amount of honey taken from each of pots $1,2,...,k$ and one of these two conditions is satisfied:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the amount of honey in pot $k$ becomes equal to the amount of honey in pots $k+1,k+2,...,l$;&lt;/li&gt;&#xA;&lt;li&gt;the amount of honey in pot $l+1$ becomes equal to the amount of honey in pots $k+1, k+2,...,l$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It is easy to see that Pooh can complete this task in $l$ days - just take honey from pots $1$, $2$, ... ,$k$ + some honey from $7-k$ pots among $k+1$,$k+2$,...$l$ in cyclic manner. After he completes this step, he repeats the process until all pots get equal amount of honey in them and eventually become empty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we prove that eventually we will get $x_1=x_2=...=x_{27}=0$. Indeed, if we assume that Pooh gets stuck, then all of the pots except for at most $6$ should become empty. Let the number of non-empty pots be $m \leq 6$ and the total amount of honey remaining is $H \in [0,\min(mP,17))$. Clearly, the amount of honey taken from these $m$ pots is exactly $m(17-H)/7$ (because every time we take out honey, $m/7$ of it is taken out of them). Since each pot contains in the beginning at most $P$ kilograms of honey, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{(17-H)m}{7}+H\leq mP.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inequality above just states that the total amount of honey remaining in the last $m$ pots + the amount of honey taken out of them is no more than the maximum amount of honey there was in the pots initially. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can rewrite the inequality as:&#xA;$$0&amp;lt;H(7-m)\leq m(7P-17) \rightarrow P&amp;gt;17/7$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore we can conclude that if $P\leq 17/7$, then Pooh can always do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PART 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The idea is to fill up the first pot entirely.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now we prove the inverse. Let the first pot has $P$ kilograms of honey and the others have $(17-P)/26$ each. Now every time we take $x$ kilograms of honey from the first pot, we take at least $6x$ kilograms of honey from the remaining pots. In order to empty the first pot, we have to take out $P$ kilograms of honey from it and therefore we have to take out $6P$ kilograms of honey from the other pots. However, $6P&amp;gt;17-P$, which is a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Remark: If the total amount of honey is $A$ and the number of pots Pooh can use on every day is $r$, then using the same calculations it is easy to see that the optimal $P=A/r$. The total number of pots does not matter at all, it could be anywhere from $7$ to infinity.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="14140" LastEditorUserId="14140" LastEditDate="2015-09-30T04:34:48.500" LastActivityDate="2015-09-30T04:34:48.500" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22663" PostTypeId="2" ParentId="22661" CreationDate="2015-09-28T00:02:12.197" Score="11" Body="&lt;p&gt;I claim that the tiler cannot save himself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let black be +1 and white be 0. Assume that somehow the tiler fixed the floor tiling so that it worked out. We'll count the same sum in two different ways.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The sum, over every possible O tetromino, of the sum of its $4$ squares. Because in any O tetromino there should be an even number of +1s, there should also be an even sum. The entire count is a sum of even numbers, and thus is also even.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The sum, over every tile, of its color (as 1 or 0) multiplied by the number of O tetrominoes that it can be a part of. The four corners can be a part of 1 O tetromino, the edges 2, and everything else 4. Thus the parity of the entire count is equal to the parity of the number of black tiles in the corners&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;By 1 and 2 together, the parity of the number of black tiles in the corners should be even. This is a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="11100" LastActivityDate="2015-09-28T00:02:12.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22664" PostTypeId="1" AcceptedAnswerId="22770" CreationDate="2015-09-28T07:20:51.397" Score="18" ViewCount="1318" Body="&lt;p&gt;I am new here. I really like puzzles, riddles and logic thinking. So, as an introduction, I created a very easy riddle. Hope it is good enough for everyone. :D&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can someone help me what should I do with my children?  They are really a unique bunch but no one likes to be with them.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;My oldest child is very temperamental. He is always red with anger and likes to yell at other people.&lt;br&gt;&#xA;  My second child is quite the opposite of my eldest. She is not as loud and angry as my oldest, but she definitely likes to ignore everyone.&lt;br&gt;&#xA;  My third child likes to dance. I've gotta say he is one hell of a shaker.&lt;br&gt;&#xA;  My fourth child.. Hmm, what can I say? She just follows my third child all around.&lt;br&gt;&#xA;  My fifth child keeps on following her older sister (refering to the fourth child). Those three rarely split up.&lt;br&gt;&#xA;  My sixth child is really hot, I mean, literally hot. I guess his condition also affects his appetite as he doesn't like to eat.&lt;br&gt;&#xA;  And lastly, my seventh child, still a baby. Oh, how I wish he would stop crying. He has been crying ever since he was born.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;It is saddening. It seems like everyone hates them. This is quite depressing for a parent.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Can you tell me who my children are?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am not sure if this concept/theme already exists, so I apologize if there is already one like this. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What the parent says about them as a whole is relevant: &quot;Everyone really &lt;em&gt;hates&lt;/em&gt; them.&quot; Though the parent might be wrong on &lt;em&gt;hate&lt;/em&gt;, maybe everyone is avoiding them for some other reason besides &lt;em&gt;hate&lt;/em&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The only important order is 3rd-4th-5th child. Note that the parent only mentioned that these three rarely split up.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The theme doesn't require someone to be really scientific. It is a common theme, maybe check out the news? You might get some &lt;em&gt;inspiration&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16566" LastEditorUserId="16566" LastEditDate="2015-10-01T04:44:08.063" LastActivityDate="2015-10-01T10:09:57.997" Title="My unique children" Tags="|riddle|" AnswerCount="2" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="22666" PostTypeId="1" AcceptedAnswerId="22667" CreationDate="2015-09-28T07:42:39.200" Score="29" ViewCount="3554" Body="&lt;p&gt;A Monday number is a positive integer $N$ with the following three properties:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The decimal representation of $N$ does not contain the digit 0&lt;/li&gt;&#xA;&lt;li&gt;The decimal representation of $N$ does not contain any digit twice&lt;/li&gt;&#xA;&lt;li&gt;$N$ is divisible by every digit $D$ that occurs in its decimal representation&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What is the largest Monday number?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-09-28T08:52:30.660" Title="The largest Monday number" Tags="|mathematics|number-theory|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="22667" PostTypeId="2" ParentId="22666" CreationDate="2015-09-28T08:52:30.660" Score="46" Body="&lt;p&gt;Notice 9867312 is a Monday number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The largest Monday number may not contain 5 because in this case it would end in 5, and thus not be divisible by 2, 4 and 8, so it would have at most 6 digits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, a Monday number may not have 8 digits. Indeed, if that were the case, the preceding paragrph would imply such a number has each digit but 0 and 5 in it. In particular, it would have the digit 3. But the sum of its digits would be 1 + 2 + 3 + 4 + 6 + 7 + 8 + 9 = 40, which is not divisible by 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that the largest Monday number must have 7 digits. If it has the digits 9, 8 and 7 it must be a multiple of 504, and it's easy check the highest Monday number that is a multiple of 504 is 9867312. Because we know the largest Monday number has 7 digits, it follows that this is the largest such number.&lt;/p&gt;&#xA;" OwnerUserId="16478" LastActivityDate="2015-09-28T08:52:30.660" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="22668" PostTypeId="1" AcceptedAnswerId="22669" CreationDate="2015-09-28T11:03:29.130" Score="15" ViewCount="1310" Body="&lt;p&gt;My first is in &lt;strong&gt;donut&lt;/strong&gt; and also in &lt;strong&gt;dough&lt;/strong&gt;&lt;br&gt;&#xA;My second in &lt;strong&gt;grew&lt;/strong&gt;, but never in &lt;strong&gt;grow&lt;/strong&gt;&lt;br&gt;&#xA;My third is in &lt;strong&gt;famine&lt;/strong&gt; but never in &lt;strong&gt;meat&lt;/strong&gt;&lt;br&gt;&#xA;My fourth is in &lt;strong&gt;sofa&lt;/strong&gt; but never in &lt;strong&gt;seat&lt;/strong&gt;&lt;br&gt;&#xA;My fifth is in &lt;strong&gt;soldier&lt;/strong&gt;, but never in &lt;strong&gt;sarge&lt;/strong&gt;&lt;br&gt;&#xA;My sixth is in &lt;strong&gt;flabby&lt;/strong&gt; and also in &lt;strong&gt;large&lt;/strong&gt;&lt;br&gt;&#xA;My seventh in &lt;strong&gt;twentieth&lt;/strong&gt;, not &lt;strong&gt;twenty-third&lt;/strong&gt;&lt;br&gt;&#xA;My whole helps with hearing before sounds are heard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There isn't a mistake with my seventh.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-09-28T14:58:47.627" Title="Not just a word, what am I?" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22669" PostTypeId="2" ParentId="22668" CreationDate="2015-09-28T12:28:55.720" Score="15" Body="&lt;p&gt;The poem refers to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the notes of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Solf%C3%A8ge&quot;&gt;solfège&lt;/a&gt;:&lt;br&gt;&#xA;&lt;br&gt;&#xA; My first is in &lt;strong&gt;do&lt;/strong&gt;nut and also in &lt;strong&gt;do&lt;/strong&gt;ugh&lt;br&gt;&#xA; My second in g&lt;strong&gt;re&lt;/strong&gt;w, but never in grow&lt;br&gt;&#xA; My third is in fa&lt;strong&gt;mi&lt;/strong&gt;ne but never in meat&lt;br&gt;&#xA; My fourth is in so&lt;strong&gt;fa&lt;/strong&gt; but never in seat&lt;br&gt;&#xA; My fifth is in &lt;strong&gt;sol&lt;/strong&gt;dier, but never in sarge&lt;br&gt;&#xA; My sixth is in f&lt;strong&gt;la&lt;/strong&gt;bby and also in &lt;strong&gt;la&lt;/strong&gt;rge&lt;br&gt;&#xA; My seventh in twen&lt;strong&gt;ti&lt;/strong&gt;eth, not twenty-third&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Do Re Mi Fa Sol La Ti&lt;/strong&gt; (Do) = 1 octave&lt;br&gt;&#xA;&lt;br&gt;&#xA;  According to Wikipedia, &quot;The study of solfège enables the musician to audiate, or mentally hear, the pitches of a piece of music which he or she is seeing for the first time and then to sing them aloud.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16572" LastEditorUserId="4421" LastEditDate="2015-09-28T14:58:47.627" LastActivityDate="2015-09-28T14:58:47.627" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22670" PostTypeId="2" ParentId="22614" CreationDate="2015-09-28T12:36:26.963" Score="1" Body="&lt;p&gt;My instincts told me to consider irrationality, so irrationality I will consider.&lt;/p&gt;&#xA;&lt;h2&gt;All Rational&lt;/h2&gt;&#xA;&lt;p&gt;Consider the case where all the pots contain a rational amount of honey. A rational number is one that can be expressed as &lt;span class=&quot;math-container&quot;&gt;$\dfrac{p}{q}$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt; are both integers. Hence, it can be seen that there exists some value &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; for which every pot of honey contains an amount of honey equal to &lt;span class=&quot;math-container&quot;&gt;$ba$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; is some integer number. (for instance, if the honey in pot &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$\dfrac{p_n}{q_n}$&lt;/span&gt; I can let &lt;span class=&quot;math-container&quot;&gt;$a = \dfrac{1}{q_1q_2...q_{27}}  $&lt;/span&gt;)  In effect, all the pots of honey contain an amount of honey that is some integer number amount of a small division.&lt;/p&gt;&#xA;&lt;p&gt;Our plan now is simply to take &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; amount of honey from the seven pots that contain the most honey. We can ensure that this is doable by letting &lt;span class=&quot;math-container&quot;&gt;$a = \dfrac{a}{7}$&lt;/span&gt;. We will only run into an issue if at one point of time there exists only six or less pots with honey in them. In the worst case scenario, this happens when we have six pots with an extreme amount of honey in them and 21 pots with barely a trickle of honey in them. Because we are dealing with real numbers, the pots can have amounts of honey really really close to &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;, so we can assume in the worst case scenario all six of the top-pots contain &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; kg of honey. The other &lt;span class=&quot;math-container&quot;&gt;$21$&lt;/span&gt; pots must then contain &lt;span class=&quot;math-container&quot;&gt;$11$&lt;/span&gt; kg of honey. But we can easily see that if this were the case, every single turn you would choose the first six pots (for they contain the most honey) and a pot from the remaining &lt;span class=&quot;math-container&quot;&gt;$21$&lt;/span&gt; pots, and you would effectively deplete &lt;span class=&quot;math-container&quot;&gt;$11*6$&lt;/span&gt; kg of honey from the first six pots (&lt;span class=&quot;math-container&quot;&gt;$66$&lt;/span&gt; kilograms) which is more than enough to balance the amount of honey in the pots. Meaning that if all the pots contain a rational amount of honey, life is good and Pooh is a happy, albeit unhealthy, bear. -I will try and word this better, currently it's a little bit badly worded.&lt;/p&gt;&#xA;&lt;h2&gt;Some Irrational&lt;/h2&gt;&#xA;&lt;p&gt;Here comes the interesting part. Say for simplicity you have one pot with an irrational amount of honey in it. You might wish to panic, but fear not, for all the pots added up must total &lt;span class=&quot;math-container&quot;&gt;$17$&lt;/span&gt;kg! This means that there must be at least one other irrational pot out there! Our aim then is to convert the irrational pots into rational ones, for then our job is easy.&lt;/p&gt;&#xA;&lt;p&gt;Consider the case where we have only two irrational pots. The two irrational pots must be a 'conjugate pair', i.e. if one of the irrational pots was &lt;span class=&quot;math-container&quot;&gt;$\pi /10$&lt;/span&gt; the other must be of the form &lt;span class=&quot;math-container&quot;&gt;$k - \pi/10$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; is some constant greater than &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt;. Another thing we must note is that we can 'truncate' an irrational number: for instance if we have the irrational number &lt;span class=&quot;math-container&quot;&gt;$0.35817395185712$&lt;/span&gt; we can take &lt;span class=&quot;math-container&quot;&gt;$0.00017395185712$&lt;/span&gt; away from it to rationalize it, and this way we avoid the trouble of accidentally 'grounding' a pot, or eating all of its honey. The concept of the conjugate irrational pot pairs is very important. For example, say we had 25 rational pots, and 2 irrational pots, one which contains an irrational number &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt;, and one which contains its conjugate irrational number, &lt;span class=&quot;math-container&quot;&gt;$\bar{w}$&lt;/span&gt;. We can then choose to rationalize &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt;, and choose six other pots (which don't include the &lt;span class=&quot;math-container&quot;&gt;$\bar{w}$&lt;/span&gt;), but in the process we create six other &lt;span class=&quot;math-container&quot;&gt;$\bar{w}$&lt;/span&gt; pots. In essence, we can then choose these seven &lt;span class=&quot;math-container&quot;&gt;$\bar{w}$&lt;/span&gt; pots and rationalize them. I am really bad with words, here is a diagram: &lt;a href=&quot;https://i.stack.imgur.com/C6yFf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/C6yFf.png&quot; alt=&quot;here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Of course, this is a really simple case. Much more complex cases exist, for instance what if 26 of the pots contained a different irrational number, and the 27th pot contained the conjugates to all the other 26 pots? (because a pot containing&lt;span class=&quot;math-container&quot;&gt;$(\bar{z} + \bar{y})$&lt;/span&gt; acts as a conjugate for both &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt;)The key thing to note is that with this 1:1 conjugate ratio business, the above tactic can be applied multiple times over and over to remove each set of conjugates.&lt;/p&gt;&#xA;&lt;p&gt;The REAL problem arises when you have a set of three complex numbers that combined create a real one. In this case there is no longer a 1:1 conjugate ratio, (i.e. &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt; does still have a conjugate pair &lt;span class=&quot;math-container&quot;&gt;$\bar{w}$&lt;/span&gt;, but it exists split into two different pots). But doesn't this remind you of something? Indeed, we can use the method shown in the image to OBTAIN &lt;span class=&quot;math-container&quot;&gt;$\bar{w}$&lt;/span&gt; if it happens to be split into multiple pots.&lt;/p&gt;&#xA;&lt;p&gt;-checking logic and rewriting, is still a work in progress-&lt;/p&gt;&#xA;" OwnerUserId="15499" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-09-28T13:17:59.603" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22673" PostTypeId="1" AcceptedAnswerId="22675" CreationDate="2015-09-28T17:06:25.017" Score="11" ViewCount="1098" Body="&lt;p&gt;My Indian friend Jamal recently participated in the well-known TV quiz&#xA;show &lt;strong&gt;&lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Who_Wants_to_Be_a_Millionaire&quot;&gt;Who wants to be a millionaire?&lt;/a&gt;&lt;/em&gt;&lt;/strong&gt;. You might have heard or read about Jamal's spectacular success; the media reported on it extensively.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what Jamal told us about his 4.000 Euro question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I knew that A, B, and C were all relatives. Answer D seemed to be the throw-away answer since he was human instead and he was voiced by Mr. Top 40 himself. Answers A and B reminded me of two characters from &quot;Through the Looking-Glass&quot;, and they even came in the right order; however, I was pretty sure that they were his cousins instead.  So, I chose the character who many people have said screwed up the show. Good choice! I had reached the 8.000 Euro question!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;What was the 4.000 Euro question?&lt;br&gt;&#xA;What were the four possible answers A, B, C, D?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="9404" LastEditDate="2016-03-04T16:07:28.163" LastActivityDate="2016-03-04T16:07:28.163" Title="Slumdog millionaire (7)" Tags="|knowledge|family|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22674" PostTypeId="2" ParentId="22614" CreationDate="2015-09-28T17:21:47.430" Score="13" Body="&lt;p&gt;Here is a rigorous proof that Pooh can always succeed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $P_i$ be the amount of honey in the $i^{th}$ pot when sorted in decreasing honey contents, so $1\ge P_1\ge P_2\ge \dots\ge P_{27}\ge0$ and $P_1+\dots+P_{27}=17$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the &lt;a href=&quot;https://puzzling.stackexchange.com/a/22632/10615&quot;&gt;beginning of Sleafar's solution&lt;/a&gt;, we now need only show how to equalize the seven heaviest pots. At any point, let $m$ be the number of pots which have the same amount of honey as the heaviest, and let $n$ be the number of nonempty pots (so initially, $m=1$ and $n=27$). We show that, as long as $m&amp;lt;7$ and $n\ge 14$, it is possible to either increase $m$ or decrease $n$. We then argue why $n$ can't drop below $14$ before $m$ reaches $7$, so that eventually we equalize the top $7$ pots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the method. Let $x$ be the gap in honey contents between the heaviest $m$ pots and the next strictly lighter pot. Each day for the next $n-7$ days, Pooh eats $\frac{x}{n-7}$ kg from each of the $m$ heaviest pots. Each of these days, he must eat the same amount from some $7-m$ other pots. He does this in such a way that the $n-7$ pots $P_8,P_9,\dots,P_n$ are eaten from equally over this $n-7$ day period (for example, by arranging the nonempty pots in a circle, and eating from each consecutive segment of length $7-m$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as each of the nonempty pots contain at least $\frac{x}{n-7}\cdot (7-m)$ kg of honey, then this will work, causing the $m$ heaviest pots to decrease to the next highest weight, thus increasing the number $m$. If the nonempty pots do not contain enough honey to do this, then adjust $x$ so that the smallest pot contains exactly $\frac{x}{n-7}\cdot (7-m)$. Then doing the procedure will empty that pot, thus decreasing $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, why can't $n$ drop below $14$ before $m$ reaches $7$? First off, the amount of honey that needs to be removed from pots 8 through $n$ in order to equalize pots 1 through 7 is equal to &#xA;$$&#xA;6(P_1-P_2)+5(P_2-P_3)+4(P_3-P_4)+3(P_4-P_5)+2(P_5-P_6)+(P_6-P_7)&#xA;$$&#xA;This is because, during the phase when pots $P_1,\dots,P_m$ are equal, we have to remove a total of $P_{m}-P_{m+1}$ from each of the first $m$ pots, and $(7-m)$ times that from the last pots 8 through $n$. A different way of writing this is &#xA;$$&#xA;6P_1-P_2-P_3-\dots -P_7\le 6 - 6P_7&amp;lt; 3&#xA;$$&#xA;The last inequality follows since $P_7&amp;gt; \frac12$. Why is this true? If not, then the first 6 pots would initially contain at most $6$ kg, and the last 21 would contain at most $\frac12$, so that the total amount of honey was at most $6+\frac12\cdot 21=16.5$, which is impossible since we started with $17$ kg of honey. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the procedure will remove less than $3$ kg from the last 20 pots. However, the first 14 pots contain at most 14 kg, meaning the lightest pots together contain at least $17-14=3$ kg. Since our procedure eliminates the lightest pots first, the first 14 pots will not be emptied, as claimed.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-09-28T21:02:31.170" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22675" PostTypeId="2" ParentId="22673" CreationDate="2015-09-28T17:43:10.020" Score="14" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Which one of these is Scooby-Doo's nephew?    &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A. Scooby-Dee&lt;br&gt;&#xA;  B. Scooby-Dum&lt;br&gt;&#xA;  C. Scrappy-Doo&lt;br&gt;&#xA;  D. Shaggy  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Scooby Dee&quot; and &quot;Scooby Dum&quot; sound like &quot;Tweedle Dee&quot; and &quot;Tweedle Dum&quot; from &quot;Through the Looking Glass&quot;, and they were cousins. Scrappy Doo was a third cousin (and did screw up the show). Shaggy (Rogers?) was voiced by Casey Kasem, voice of America's Top 40 radio show.&lt;/p&gt;&#xA;" OwnerUserId="16572" LastEditorUserId="8874" LastEditDate="2015-09-29T07:22:01.390" LastActivityDate="2015-09-29T07:22:01.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22676" PostTypeId="1" AcceptedAnswerId="22680" CreationDate="2015-09-28T18:35:23.110" Score="20" ViewCount="1412" Body="&lt;p&gt;Shaq is shooting free throws. He makes his first shot, then misses his second. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Confidence is a huge factor in how well Shaq plays. This means that for each subsequent shot, the probability he makes it is equal to the fraction of shots he has made so far. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, there is a $\frac12$ chance he makes his third shot. If he makes that, there is a $\frac23$ chance he will make his fourth. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;After 101 shots (including the first two), what is the probability that Shaq sank less than 21 baskets?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Side note: Symmetry implies that Shaq will make half of his free throws on average, which matches his &lt;a href=&quot;http://bleacherreport.com/articles/886876-the-nbas-top-ten-all-time-worst-free-throw-shooters/page/8&quot;&gt;career average of 52.7%&lt;/a&gt; pretty closely.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-09-29T03:55:42.647" Title="Shooting Free Throws" Tags="|probability|" AnswerCount="3" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22679" PostTypeId="2" ParentId="22676" CreationDate="2015-09-28T20:41:45.757" Score="10" Body="&lt;p&gt;This is equivalent to Polya's urn model (&lt;a href=&quot;https://en.wikipedia.org/wiki/P%C3%B3lya_urn_model&quot;&gt;https://en.wikipedia.org/wiki/P%C3%B3lya_urn_model&lt;/a&gt;). From the canonical starting state of one ball of each colour the resulting distribution at any stage is uniform, i.e. Shaq has an equal chance of making any number of free throws between 1 and 100.&lt;br&gt;&#xA;So the probability that Shaq makes 20 or fewer free throws is $\frac{20}{100}=0.2$. &lt;/p&gt;&#xA;" OwnerUserId="14080" LastEditorUserId="14080" LastEditDate="2015-09-28T20:47:31.483" LastActivityDate="2015-09-28T20:47:31.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22680" PostTypeId="2" ParentId="22676" CreationDate="2015-09-28T20:47:16.913" Score="18" Body="&lt;p&gt;After $n$ shots, it is equally likely that Shaq has made any number of shots from $1$ to $n-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof by induction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n=2$, then Shaq has made $1$ shot, which is the entire range from $1$ to $2-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For Shaq to make $1$ shot out of $n$, he must make $1$ shot out of $n-1$ and then miss the next shot. This has a probability of $\frac{1}{n-2}*\frac{n-2}{n-1}=\frac{1}{n-1}$.&lt;/li&gt;&#xA;&lt;li&gt;For Shaq to make $n-1$ shots out of $n$, he must make $n-2$ shots out of $n-1$ and then make the next shot. This has a probability of $\frac{1}{n-2}*\frac{n-2}{n-1}=\frac{1}{n-1}$.&lt;/li&gt;&#xA;&lt;li&gt;For Shaq to make $k$ shots out of $n$, with $1&amp;lt;k&amp;lt;n-1$, he either makes $k-1$ shots out of $n-1$ and makes the next shot, or he makes $k$ out of $n-1$ and misses the next. The total probability of these is $\frac{1}{n-2}*\frac{k-1}{n-1}+\frac{1}{n-2}(1-\frac{k}{n-1})=\frac{1}{n-2}(\frac{k-1}{n-1}+1-\frac{k}{n-1})=\frac{1}{n-2}*\frac{n-2}{n-1}=\frac{1}{n-1}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So the probability that Shaq makes $m$ shots out of $n$, for $1\le m\le n-1$, is always $\frac{1}{n-1}$. The probability that Shaq makes between $1$ and $20$ shots inclusive out of $101$ is $\frac{20}{100}=\boxed{\frac{1}{5}}$.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-09-28T20:47:16.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22681" PostTypeId="2" ParentId="22550" CreationDate="2015-09-29T02:04:52.050" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think I know how the code works but can't figure out the critical bits. Basically if you assume that values one to twenty are C(n)=2*n then every safe after that has a code that is equal to the sum of the codes of the constituent parts i.e. code(twentyone) = code(twenty)+code(one) code(thirtyone)=code(thirty)+code(one) and so on It would be nice to know how code(thirty), code(forty), code(fifty) and code(sixty) are calculated&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10038" LastEditorUserId="10038" LastEditDate="2015-09-29T03:37:36.293" LastActivityDate="2015-09-29T03:37:36.293" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22682" PostTypeId="2" ParentId="22614" CreationDate="2015-09-29T03:02:13.057" Score="4" Body="&lt;p&gt;An attempt at a TLDR version of the proof using ideas from all the other solutions (especially Artur Kirkoyan's take on the problem) .  Definitely not as rigorous but I hope still convincing.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Suppose Pooh follows two rules while eating honey&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Never eat from a pot with less if you could be eating from a pot with more.&lt;/li&gt;&#xA;&lt;li&gt;If you must choose one pot from several that are equal, cycle between these pots as to eat from them equally.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Suppose Pooh follows these rules but still ends up with a set $S$ of less than seven nonempty pots at the end.   By following the rules, the pots in $S$ must have always been the pots with strictly the most honey, and have therefore been eaten from constantly.  Since each pot in $S$ holds at most 1 kg, the constant eating means at most $7$ kg total has been eaten.  But this is ridiculous, as there must be more than 11 kg outside the set $S$.&lt;/p&gt;&#xA;" OwnerUserId="2722" LastActivityDate="2015-09-29T03:02:13.057" CommentCount="0" CommunityOwnedDate="2015-09-29T03:02:13.057" ContentLicense="CC BY-SA 3.0" />
  <row Id="22683" PostTypeId="1" AcceptedAnswerId="22685" CreationDate="2015-09-29T03:33:28.293" Score="-3" ViewCount="158" Body="&lt;p&gt;A very deep well&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Full of swords, knives and daggers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these within you&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Really a body part&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16580" LastActivityDate="2015-09-29T03:51:34.040" Title="Body Talk pt. 1" Tags="|riddle|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22684" PostTypeId="1" AcceptedAnswerId="22687" CreationDate="2015-09-29T03:51:12.220" Score="10" ViewCount="592" Body="&lt;p&gt;Card 0: &lt;strong&gt;title&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;the answer to each card is a single english word&#xA;the cards dont have to be solved in order but some cards refer back to previous answers&#xA;each card has a title which is a clue that is usually essential to the puzzle&#xA;sometimes the title has to be interpreted cleverly&#xA;good luck&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Card 1: &lt;strong&gt;-55&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;WWWWWWW&#xA;WWWWWWW&#xA;&#xA;WWWWW&#xA;WWWWWW&#xA;&#xA;WWWWWWW&#xA;WWWWWW&#xA;&#xA;WWWWWW&#xA;WWWWWWW&#xA;&#xA;WWWWWWW&#xA;WWWWW&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Card 2: &lt;strong&gt;1+2=&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;npwan&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Card 3: &lt;strong&gt;why not&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;periphery&#xA;similarity&#xA;meander&#xA;earnest&#xA;serendipity&#xA;relief&#xA;array&#xA;gregarious&#xA;sanitary&#xA;replay&#xA;elusive&#xA;perjury&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Card 4: &lt;strong&gt;cardinal NEWS&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;24 2 0 23 0 3 3 23 2 0 34&#xA;4 34 4 4 0 3 2 0 2 24 0&#xA;2 2 2 0 123 2 12 12 0 3 13&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Card 5: &lt;strong&gt;sewoh?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                    _/\_&#xA;                   \    /&#xA;               _/\_/    \_/\_&#xA;              \              /&#xA;              /_            _\&#xA;                \          /&#xA;     _/\_      _/          \_      _/\_&#xA;    \    /    \              /    \    /&#xA;_/\_/    \_/\_/              \_/\_/    \_/\_&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Card 6: &lt;strong&gt;it's python&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;''.join(card[i].ans[-i] for i in (1,2,3,4,5))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Card 7: &lt;strong&gt;five by five&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;step 1) determin. the previous answers 2 through 6 in order&#xA; (step 1 should require computer assistance)&#xA;step 2) verify that the plural is a name&#xA;step 3) shake well&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Hint&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; graph with 33 vertices&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Hint 2&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2-6 have been determined but that's not step 1. what is &quot;determin.&quot; and why should it require computer assistance?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11100" LastEditorUserId="11100" LastEditDate="2015-10-03T22:24:36.603" LastActivityDate="2015-10-04T05:24:45.287" Title="Seven Puzzle Cards" Tags="|enigmatic-puzzle|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="22685" PostTypeId="2" ParentId="22683" CreationDate="2015-09-29T03:51:34.040" Score="2" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A mouth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A very deep well&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Means the digestive track from the mouth to the stomach&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Full of swords, knives and daggers&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Means the teeth because they are sharp and can cut through like swords, knives and daggers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All these within you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; With 'these', meant the line before, which are teeth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16566" LastActivityDate="2015-09-29T03:51:34.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22686" PostTypeId="2" ParentId="22676" CreationDate="2015-09-29T03:55:42.647" Score="7" Body="&lt;p&gt;Here is a quick &lt;a href=&quot;https://en.wikipedia.org/wiki/Monte_Carlo&quot;&gt;Monte Carlo&lt;/a&gt; simulation I wrote to verify the result of $\frac{1}{5}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Python 3.4:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from random import random&#xA;def made_less_than_21():&#xA;    baskets = 1&#xA;    total = 2&#xA;    for shot in range(99):&#xA;        if random() &amp;lt; baskets/total:&#xA;            baskets += 1&#xA;        total += 1&#xA;    return baskets &amp;lt; 21&#xA;less_than_21 = 0&#xA;trials = 100000&#xA;for trial in range(trials):&#xA;    if made_less_than_21():&#xA;        less_than_21 += 1&#xA;print (&quot;Answer is about&quot;, less_than_21/trials)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;$\implies answer \approx 0.200622$&lt;/p&gt;&#xA;" OwnerUserId="16586" LastActivityDate="2015-09-29T03:55:42.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22687" PostTypeId="2" ParentId="22684" CreationDate="2015-09-29T04:32:25.340" Score="7" Body="&lt;p&gt;Partial answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Card 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Count the W's in each line, interpret as two-digit numbers, and subtract 55 to get &lt;code&gt;22 01 21 12 20&lt;/code&gt;. With 1-26 representing A-Z, this spells &lt;code&gt;VAULT&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Card 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Take the first letters of each word without a y to get &lt;code&gt;MERGE&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Card 5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The title is an anagram of &quot;whose?&quot;. Part of the Koch snowflake is shown, so the answer to &quot;whose?&quot; is &quot;Koch's&quot;. This anagrams to &lt;code&gt;SHOCK&lt;/code&gt; (or &lt;code&gt;HOCKS&lt;/code&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-09-29T04:32:25.340" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22688" PostTypeId="1" AcceptedAnswerId="23981" CreationDate="2015-09-29T05:13:45.883" Score="0" ViewCount="4008" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FM9Ig.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FM9Ig.png&quot; alt=&quot;3 boxes&quot;&gt;&lt;/a&gt; &lt;br&gt;&#xA;The numbers in all the three groups follow the same mathematical pattern. &lt;br&gt;&#xA;Can you work it out? &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Note : The 4 numbers in a square form a single group, and the answer is a positive integer. &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Please try first without the hints.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hint 1: You have to work on equations involving addition.&#xA;&lt;br&gt;&#xA;&#xA; Hint 2: You should calculate the powers of each number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1766" LastEditorUserId="1766" LastEditDate="2016-03-17T07:53:33.967" LastActivityDate="2016-03-17T07:53:33.967" Title="Can you work out the number which replaces the question mark?" Tags="|mathematics|pattern|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="22689" PostTypeId="2" ParentId="22684" CreationDate="2015-09-29T05:49:32.787" Score="4" Body="&lt;p&gt;Card 6&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Turns/torus/tyres etc, based on f&quot;'s answers and guessing the most likely results for t?r?s. Much less conclusive when I've read the clue properly. It looks like they may all be 5 letter words&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Card 7&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;wiles&lt;/strong&gt; Guessing that determin.  means take the determinant, and that a=1,b=2..., the determinant of the matrix is 125239. There's various groupings that could work, but Lewi (12 5 23 9) is such that adding an &quot;s&quot; gives a name (lewis). Mixing that up gives you wiles. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2015-10-04T05:24:45.287" LastActivityDate="2015-10-04T05:24:45.287" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22690" PostTypeId="1" CreationDate="2015-09-29T07:30:14.587" Score="15" ViewCount="1670" Body="&lt;p&gt;I'm looking for guidelines on how to organize a good (IRL) treasure hunt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there steps to follow to ensure the available locations are used to the fullest? Good practices to organize my research ? Methods to write the clues ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm quite lost here and I want to do my best. If possible, I'd prefer a step-by-step guide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit :&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I first wrote a generic question so the answer could be useful to someone else but it ended up being too broad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The treasure hunt in question is for a group of children composed of &lt;strong&gt;eight eleven-ish years old&lt;/strong&gt; and a few younger ones. I don't think they'll have smartphones, but if one of them do they won't be able to use it since the coverage is terrible here. Most of them know the location relatively well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The settings is &lt;strong&gt;my grandmother's home&lt;/strong&gt;, a three stories house with a large garden, in a village in the south of France (Aveyron to be precise).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The house is full of old/interesting furniture (grandfather clock, massive 60's TV set, fireplace, bookshelves filled with history or decorative books, a gross stuffed owl, a collection of Murano glass fishes, my uncles and aunts' things from when they were teenagers, wardrobes with mirrors, etc.).     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has an attic, a cellar, two kitchens, two bathrooms, a corridor that used to be a bathroom, six bedrooms, an office, what was a store ten years ago (now a TV room but still has a shop-window overlooking the street and a one-way mirror allowing people in the kitchen to see what's going on inside), a dinning room, a sewing room, a garage, a room used only when killing rabbits and making preserves and my grandfather's workshops (which we won't use since they're full of scrap metal parts and rusty cars).&lt;br&gt;&#xA;The garden has a willow tree, chestnut trees, hedges, rabbits cages in a corner, a small stone table and lots of good hiding spots, but not many flowers. There's also a vegetable garden two streets away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The village itself is not very interesting, there's a church (old and always open), a small market place, a few shops, and that's it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the clues should be in the house and garden, but since there's not many cars, and the kids are old enough, they can leave the house and search for a couple of clues in the village.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought there would be enough materials to organize an interesting treasure hunt for my cousin's birthday (apparently she loves this kind of activity), but I'm afraid the path will be too short or the clues too obvious (something like &quot;Yay! You found the 6th clue, now go to the attic to find the 7th!&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PS :&lt;/strong&gt; I'm not a fluent English speaker, please tell me if you see grammar mistakes or badly translated expressions.&lt;/p&gt;&#xA;" OwnerUserId="16587" LastEditorUserId="16587" LastEditDate="2015-09-29T16:33:40.947" LastActivityDate="2015-09-30T16:06:58.920" Title="How to organize a treasure hunt?" Tags="|puzzle-creation|treasure-hunt|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="22691" PostTypeId="2" ParentId="22646" CreationDate="2015-09-29T09:52:12.140" Score="8" Body="&lt;p&gt;Worth a try:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Draught (British) vs Drought &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;She is the feast, and I am the famine&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Draught in British is equivalent to Draft in American, which has a meaning of: a serving of drink (usually alcoholic) drawn from a keg. In most feasts, alcoholic drinks are commonly served. Drought, basically shortage of water which can cause Famine.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One little straight line,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The difference would be the line in the letter $a$, which would be removed to form the letter $o$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Glazed right over&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are cooking techniques which uses beer/alcoholic drinks as glaze over ham, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you replaced her with me, all would be gone, and many would be very unhappy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Change draught (drinks/beverages) to drought (shortage) would definitely cause everyone unhappiness.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16566" LastEditorUserId="11727" LastEditDate="2015-09-29T18:43:52.207" LastActivityDate="2015-09-29T18:43:52.207" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22692" PostTypeId="2" ParentId="22684" CreationDate="2015-09-29T10:15:33.120" Score="6" Body="&lt;p&gt;Partial answer, building on f*'s answer to the first card.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Card 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The card reads &lt;strong&gt;1 + 2 = npwan&lt;/strong&gt;. 1 refers to the word on the first card, &lt;code&gt;VAULT&lt;/code&gt;. The plus means to add the letters' indices (starting with &lt;code&gt;A = 0&lt;/code&gt;) individually without carry, i.e. modulo 26. The resulting digits are interpreted as letters again, which yields &lt;code&gt;ROBOT&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Card 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I haven't got an answer, but I suspect that cardinal news refers to the cardinal directios so that 1 is north, 2 is east, 3 is west and 4 is south. The numbers are combinations of these directions or 0. All numbers point inwards; there are no Ns on the top (north) row and no Ss on the bottom row. But what to make of that? No idea.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Card 6:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Using Dr Xorile's pattern matching, but taking the last letter from the first card, the second but last from the second, and so on, the pattern is &lt;code&gt;TOR?S&lt;/code&gt;, which could be &lt;code&gt;TORUS&lt;/code&gt;. (That would also fit in with the clue that the plural, Tori, is a name.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit: A bit of progress!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Card 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My assumption that &lt;code&gt;NEWS&lt;/code&gt; are the cardinal directions was right, I guess. If we take the 33 numbers as points on a grid and then draw connections to neighbouring nodes in the respective directions, we get:&lt;br&gt;&#xA;&lt;br&gt;&#xA;      &lt;a href=&quot;https://i.stack.imgur.com/Lv9l2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Lv9l2.png&quot; alt=&quot;connected nodes&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This requires some editing.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The outer border is just decoration. If we remove it, the grid looks like:&lt;br&gt;&#xA;&lt;br&gt;&#xA;     &lt;a href=&quot;https://i.stack.imgur.com/9kAKP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9kAKP.png&quot; alt=&quot;connected nodes wothout border&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This step seems to require some imagination, but ..&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The forms of the &lt;strong&gt;u&lt;/strong&gt; and the &lt;strong&gt;t&lt;/strong&gt; leap out, but the other letters aren't as unique. The first could be &lt;strong&gt;b&lt;/strong&gt;, &lt;strong&gt;G&lt;/strong&gt;, &lt;strong&gt;h&lt;/strong&gt;, &lt;strong&gt;n&lt;/strong&gt;, &lt;strong&gt;o&lt;/strong&gt; or &lt;strong&gt;S&lt;/strong&gt;. The third coulod be &lt;strong&gt;r&lt;/strong&gt; or &lt;strong&gt;c&lt;/strong&gt; and the fourth could be an &lt;strong&gt;o&lt;/strong&gt; or an &lt;strong&gt;n&lt;/strong&gt;, depending on whether the borders are just borders or part of the letters. But checking for &lt;code&gt;/^[bghnos]u[rc][no]t$/i&lt;/code&gt; in a (rather small) dictionary yields just &lt;code&gt;BURNT&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This solution also fits nicely with ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ... my earlier assumption that the word on the sixth card is &lt;code&gt;TORUS&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the above and f'''s solutions are true, the 5&amp;times;5 grid is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;ROBOT&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;MERGE&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;BURNT&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;SHOCK&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;TORUS&lt;/code&gt;     &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And now, &quot;shake well&quot;. This looks as if we need an anagram of sorts, but I haven't got a solution for that.&lt;/p&gt;&#xA;" OwnerUserId="9860" LastEditorUserId="9860" LastEditDate="2015-10-01T06:56:05.267" LastActivityDate="2015-10-01T06:56:05.267" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22693" PostTypeId="1" AcceptedAnswerId="22826" CreationDate="2015-09-29T11:37:08.997" Score="3" ViewCount="325" Body="&lt;p&gt;I can be defining,&lt;br&gt;&#xA;But I don't give much away.&lt;br&gt;&#xA;I can be pretty good,&lt;br&gt;&#xA;Except when I'm nasty.&lt;br&gt;&#xA;I'm never at the extremes,&lt;br&gt;&#xA;But you'll find me with a switch at the end.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What word am I?&lt;/p&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-10-02T23:07:32.477" Title="A multi-purpose word" Tags="|riddle|word|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22694" PostTypeId="2" ParentId="22693" CreationDate="2015-09-29T12:38:46.323" Score="0" Body="&lt;p&gt;I was thinking&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sick&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be defining,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sick can be used as an adjective, or like a defining part in someones life. (&quot;Getting sick was a defining point in Rick's life, now he doesn't smoke anymore&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I don't give much away.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sick is a very broad term to describe something, illness or action&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be pretty good,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Can be used in a good way! (&quot;That was a sick 720 Gazelle Flip bro!&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Except when I'm nasty.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Describing yourself or someone else. (&quot;I just got sick&quot; or &quot;Jenny is sick today&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm never at the extremes,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Normally sick isn't used as a term to describe a terminal illness or to describe something astonishing or breathtaking.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But you'll find me with a switch at the end&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CK Components is a well known manufacture of many different types of switches!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16273" LastEditorUserId="16273" LastEditDate="2015-09-29T12:47:10.483" LastActivityDate="2015-09-29T12:47:10.483" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22695" PostTypeId="1" AcceptedAnswerId="22709" CreationDate="2015-09-29T12:40:57.880" Score="8" ViewCount="499" Body="&lt;p&gt;Eight trees $ABCDEFGH$ are standing (in this order) on the shores of circle lake.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The four trees $ACEG$ form a square of area $500m^2$.&lt;/li&gt;&#xA;&lt;li&gt;The four trees $BDFH$ form a rectangle of area $400m^2$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What is the largest possible area of the octagon $ABCDEFGH$ under these conditions?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-09-29T12:50:07.670" LastActivityDate="2016-03-13T15:30:53.857" Title="On the shores of circle lake" Tags="|mathematics|geometry|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22697" PostTypeId="1" AcceptedAnswerId="22712" CreationDate="2015-09-29T14:20:30.897" Score="-1" ViewCount="415" Body="&lt;p&gt;A Tuesday number is a positive integer $N$ with $d$ digits where the given properties hold true for every positive integer $i$ in the range $0&amp;lt;i&amp;lt;d$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$N\times i$ contains the exact same digits (in any order) as $N$&lt;/li&gt;&#xA;&lt;li&gt;Moreover, it should be possible to split $N$ into two parts, such that writing the first half after the second does, in fact, give the value of $N\times i$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Some valid examples of splitting and reordering:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1371\rightarrow 3711, 7113, 1137, 1371$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$02684\rightarrow 26840, 68402, 84026, 40268, 02684$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Find the largest Tuesday number.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You probably will require the number to start with a zero.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-09-29T21:20:45.463" Title="The largest Tuesday number" Tags="|mathematics|number-theory|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22698" PostTypeId="2" ParentId="22693" CreationDate="2015-09-29T14:29:25.713" Score="0" Body="&lt;p&gt;My guess would be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; riddle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's why:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I can be defining,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A riddle should be defining and define some answer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I don't give much away.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Riddles normally don't give too many hints, otherwise it's not a good one&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be pretty good,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are good and bad riddles(obviously)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Except when I'm nasty.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Definition of &quot;nasty&quot; is &quot;&lt;em&gt;very bad or unpleasant&lt;/em&gt;&quot;, which is also an opposite of &quot;good&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm never at the extremes,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Can't find any decent explanation for this line (yet).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But you'll find me with a switch at the end.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Once you know the answer you'll be like: &quot;Aha! Now it makes sense!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9352" LastActivityDate="2015-09-29T14:29:25.713" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22699" PostTypeId="2" ParentId="22646" CreationDate="2015-09-29T14:36:16.933" Score="2" Body="&lt;p&gt;I see that @Dr_Xorile has drawn the same conclusion,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it seemed to me that T and L were source of this confusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yet while pondering about this antonym conundrum,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find that &lt;em&gt;'cross the pond&lt;/em&gt; is used in DC and in London.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you read too rapidly across the word here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;em&gt;feast&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You might happen to miss a stroke and instead, read the:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;em&gt;least&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They represent a dearth and glut, the difference is clear,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet rhyme and look alike to the (too) hasty eye or ear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, British or American, I'll yield I do not 'get';&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the answer to the last line seems to make it a sure bet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I would take the first of yours, replace it with the latter,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The least, that's absolutely none! and surely that would matter&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To those who gathered to partake in this replete repast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that's my guess, no more, no less, and this line is my last.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As requested, an explanation in plaintext:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; feast/least. &quot;Two of a kind, remarkably similar&quot;: the words both look and sound alike. When you skim-read (&quot;glazed over&quot;) the word 'feast', you might miss the cross-stroke (&quot;one little straight line&quot;) on the f and instead read an l, resulting in least. And if you replace a feast with the least - the least being nothing at all - then &quot;all would be gone, and many would be very unhappy&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hints:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hint 1: the sound of the word should be more distinguishable in the UK/US than in the US/UK. I learnt British English and feast/least rhyme for me; I don't know how they're pronounced in the US, but have learned in the past that pronounciations can vary wildly. Hint 2: the little line that's missing isn't a letter of the alphabet on its own, but is strongly hinted to be part of the letter; in this case the cross-stroke on the f.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16591" LastEditorUserId="16591" LastEditDate="2015-09-29T17:50:25.300" LastActivityDate="2015-09-29T17:50:25.300" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22700" PostTypeId="2" ParentId="22695" CreationDate="2015-09-29T14:41:57.533" Score="2" Body="&lt;p&gt;&lt;strong&gt;Stating Point / Not Even a Partial Answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm putting this in an answer just to use the spoiler tag for those that want to do all the math themselves.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A square of area $500m^2$ has sides of $\sqrt{500}=10\sqrt{5}$.&lt;br&gt;&#xA; A square of side $\sqrt{500}$ has a diagonal of $\sqrt{1000}=10\sqrt{10}$&lt;br&gt;&#xA; As the square is inscribed within the circle, that means the circle has a diameter of $10\sqrt{10}$ and a radius of $5\sqrt{10}$&lt;br&gt;&#xA;&lt;br&gt;&#xA; The rectangle &lt;strong&gt;BDFH&lt;/strong&gt; has an area of $\mathbf{BD}\times\mathbf{BF}=400m^2$&lt;br&gt;&#xA; The length of a chord is given by $2R\sin{\frac{\theta}{2}}$ which, in this case, becomes $10\sqrt{10}sin{\frac{\theta}{2}}$ where $\theta$ is the angle between the two points as taken from the center of the circle. &lt;a href=&quot;http://mathworld.wolfram.com/CircularSegment.html&quot; rel=&quot;nofollow&quot;&gt;Reference&lt;/a&gt;   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastActivityDate="2015-09-29T14:41:57.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22701" PostTypeId="2" ParentId="22697" CreationDate="2015-09-29T14:43:30.607" Score="3" Body="&lt;p&gt;The largest Tuesday number is...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; unknown. It has been conjectured that there are an infinite number, but not proven. (&lt;a href=&quot;http://mathworld.wolfram.com/CyclicNumber.html&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="11876" LastEditDate="2015-09-29T15:19:19.467" LastActivityDate="2015-09-29T15:19:19.467" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="22702" PostTypeId="2" ParentId="22695" CreationDate="2015-09-29T15:28:48.880" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Total area of 670.8&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We know the square sits evenly on the circle, with the diagonal equal to the diameter. That is $10\sqrt{10}$.&lt;br&gt;&#xA;The diagonal of the rectangle is also equal to the diameter, and the area of that is $D^2 * \sin{x} * \cos{x}$ where X is the angle between the diagonal and a side of the rectangle. That works out to 26.57° or 63.43°. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, our octagon can be made by having the rectangle rotated around the square. That in turn creates 8 slices that added up make the octagon's area. The area of each slice is $\frac{r^2 * \sin{x}}{2}$. As the only variable there is the sin(x), that is what we have to maximize.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 4 (mirrored) slices (if the square has its diagonal vertical, and the rectangle's longer centerline is horizontal-ish) are $\sin{x}, \sin{(90-x)}, \sin{(x+36.86)}, \sin{(53.14-x)}$.  Maximizing the sum of those gives $x = 26.57°$ or having the points be symmetrical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, now we have an octagon made up of slices either 26.57° or 63.43°.  Those work out to areas of 55.9 and 111.8.  Four of each gives a total area of $670.8$.&lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-09-29T19:04:10.693" LastActivityDate="2015-09-29T19:04:10.693" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22703" PostTypeId="1" AcceptedAnswerId="22705" CreationDate="2015-09-29T16:18:44.740" Score="-2" ViewCount="224" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Remark:&lt;/strong&gt; Because of the comments I decided to drop the made up story, to remove any ambiguities.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You have a balance scale. On one side of the scale you put an $1kg$ cube made of iron, on the other side an $1kg$ cube made of balsa wood. You perform a weighing under normal conditions on earth. What will be the result of the weighing, assuming the scale is accurate enough to show any difference.&lt;/p&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2015-09-30T06:53:34.337" LastActivityDate="2015-09-30T06:53:34.337" Title="What is heavier?" Tags="|physics|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22704" PostTypeId="2" ParentId="22550" CreationDate="2015-09-29T16:22:20.527" Score="1" Body="&lt;p&gt;&lt;strong&gt;An answer without pure math&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Safe number 1 starts with code = 2.&#xA;Stepsize between safes is 2, except when exception fits as described hereunder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exeption-check:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When safe number written as English word ends with &quot;y&quot;, replace &quot;y&quot; by &quot;een&quot; and look for that English word in the previous safe numbers. If found, take the code of that safe number. &#xA;If not found, but there exists a word with a difference of only one letter, the following rule will apply:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;fetch previous replacement's safe number, i.e. of the &quot;een&quot; word (&lt;strong&gt;not&lt;/strong&gt; safe's code)&lt;/li&gt;&#xA;&lt;li&gt;substract this safe number from actual safe number&lt;/li&gt;&#xA;&lt;li&gt;take code of the safe found this way&lt;/li&gt;&#xA;&lt;li&gt;multiply by -1 (because initial search was negative)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="16594" LastActivityDate="2015-09-29T16:22:20.527" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22705" PostTypeId="2" ParentId="22703" CreationDate="2015-09-29T16:25:20.630" Score="5" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The iron will be heavier, because the balsa wood has more buoyancy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To be more specific:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The density of balsa wood is 40 - 340kg/m$^3$, and the displaced air is about 1.25kg/m$^3$, so the reduction in the balsa effective weight is between 0.4% and 3%.  The density of iron is about 8000kg/m$^3$, so the reduction is about 0.01%.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2015-09-29T16:31:47.937" LastActivityDate="2015-09-29T16:31:47.937" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="22707" PostTypeId="2" ParentId="22528" CreationDate="2015-09-29T16:56:56.050" Score="4" Body="&lt;p&gt;This is my hypothesis, but I haven't figured out how to prove it yet:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $m = n$, the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $2(n-1)$.  This employs the solution given in the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/7797/connect-the-dots&quot;&gt;linked question&lt;/a&gt;, of basically using the well-known four-line solution for a $3\times3$ grid, and then extending a square spiral outward to encompass the other dots:&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/WDXXE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WDXXE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;  Any other solution (e.g. using a triangle spiral instead of a square) doesn't seem to improve on this number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For $m &amp;gt; n$, the solution is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $2n-1$.  There doesn't appear to be any more efficient solution than just drawing a spiral or zigzagging across the rows:&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/Efeek.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Efeek.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;em&gt;or&lt;/em&gt; &lt;a href=&quot;https://i.stack.imgur.com/HzXn5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HzXn5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-09-29T16:56:56.050" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22708" PostTypeId="2" ParentId="22697" CreationDate="2015-09-29T17:07:44.507" Score="2" Body="&lt;p&gt;If leading zeroes were not allowed, then:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For numbers where $d$ is greater than 11,  $N*i$ will have more digits in it than N for the largest value of $i$. The largest cyclic number without a leading zero shown on the Wikipedia page that satisfies the rules of a Tuesday number appears to be &lt;strong&gt;142857&lt;/strong&gt;. I don't believe there is a higher one as 7 is the largest prime below 10 and 11 doesn't generate a cyclic number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="14135" LastEditDate="2015-09-29T18:16:30.997" LastActivityDate="2015-09-29T18:16:30.997" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22709" PostTypeId="2" ParentId="22695" CreationDate="2015-09-29T18:22:38.887" Score="6" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $300\sqrt{5}=670.8 m^2$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reasoning&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm going to leave out the meters unit on all forthcoming measurements.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As others have figured out, the square $ACEG$ holds the key to the circle's measurements. With an area of 500, the sides are each of length $\sqrt{500}$, leading to a diagonal of $10\sqrt{10}$. Since the square is circumscribed along the circle, this is equal to the diameter. Thus, the radius, $r$, is $5\sqrt{10}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's create a Cartesian coordinate system with the center of the circle at $(0, 0)$. Next, let's place the square $ACEG$ rotated by 45° from its &quot;normal&quot; view. (It should look like a &quot;diamond&quot; in the standard, if not mathematical, sense.) With $A$ in the upper-left and working clockwise (I know it's breaking convention, but this is how I drew it and I don't want to confuse myself), the points of the square are $(-r, 0)$, $(0, r)$, $(r, 0)$, and $(0, -r)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's place the rectangle down. Here is the part that I cannot prove: I propose that the maximal area occurs when a line drawn from the center of the circle to any point on the square $ACEG$ intersects the midpoint of the edge of the rectangle $BDFH$. If you imagine the square as tilted 45°, then the rectangle is placed &quot;flat&quot; such that the edges are parallel with the edges of the page. Let's call the length of the long edges—i.e. segments $BD$ and $FH$—$a$ and the length of the short edges—i.e. segments $DF$ and $HB$—$b$. We have a system of equations to solve the positions of the corners:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The area is 400, so $ab=400$.&lt;/li&gt;&#xA;&lt;li&gt;Either diagonal intersects the center of the circle and is therefore the diameter. Thus, $a^2+b^2=(2r)^2$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Solving this system for $a, b &amp;gt;0$ yields $a=20\sqrt{2}$ and $b=10\sqrt{2}$. Now we know the points of rectangle $BDFH$, in order: $(-a/2, b/2)$, $(a/2, b/2)$, $(a/2, -b/2)$, $(-a/2, -b/2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total area of the octagon is that of the rectangle $BDFH$ plus four triangles: $BCD$, $DEF$, $FGH$, and $HAB$. The rectangle has area $ab=400$. The triangles $BCD$ and $FGH$ each have an area of $a(r-b/2)/2$, and the triangles $DEF$ and $HAB$ each have an area of $b(r-a/2)/2$. Combining these results in a total area of $ab+b(r-a/2)+a(r-b/2)$. Plugging in the values found for $a$ and $b$ yields the number I stated above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reference Figure&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/aeTHo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aeTHo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16275" LastEditorUserId="8874" LastEditDate="2016-03-13T15:30:53.857" LastActivityDate="2016-03-13T15:30:53.857" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22710" PostTypeId="2" ParentId="22598" CreationDate="2015-09-29T19:59:55.733" Score="3" Body="&lt;p&gt;I see &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the constellation Leo.  By starting at the &quot;7&quot; star (the second to the right in images of Peter Pan stars), and counting the number of points on that star, you will see another star pointing back.  Continue that process and you eventually end up at the &quot;L&quot; star.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tracing the full line looks like this...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/FeAUD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FeAUD.png&quot; alt=&quot;stars&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which looks an awful lot like this...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/UNRsI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UNRsI.png&quot; alt=&quot;leo&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14233" LastActivityDate="2015-09-29T19:59:55.733" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22711" PostTypeId="1" AcceptedAnswerId="22713" CreationDate="2015-09-29T21:06:36.427" Score="13" ViewCount="1114" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I tried twice to be one of 14 though it was not to be&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;My first was normal and no ones tragedy&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;the second was definitely someone's mistake&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;as the very air I breathed was enough to cause a break.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who am I and what did I fail to do?&lt;/p&gt;&#xA;" OwnerUserId="14233" LastActivityDate="2015-09-29T21:24:37.050" Title="I am One Out of Fourteen" Tags="|outer-space|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22712" PostTypeId="2" ParentId="22697" CreationDate="2015-09-29T21:20:45.463" Score="5" Body="&lt;p&gt;There are some extremely large Tuesday numbers, with billions of digits. Probably there is no bound on how large they can get, though nobody knows how to prove this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n$ is a positive integer, the decimal expansion of $1/n$ is eventually periodic, and the period has at most $n-1$ digits. The period can be exactly $n-1$ digits only if $n$ is a prime number (though this does not happen for every prime). For example,&#xA;$$&#xA;\frac{1}{7}=0.\overline{142857}&#xA;$$&#xA;has period $7-1=6$ (and indeed $7$ is prime).&lt;br&gt; If the decimal expansion of $1/p$ has period $p-1$, we call $p$ a &lt;a href=&quot;https://en.wikipedia.org/wiki/Full_reptend_prime&quot;&gt;&lt;em&gt;full reptend prime&lt;/em&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can obtain a Tuesday number (also called a &lt;a href=&quot;https://en.wikipedia.org/wiki/Cyclic_number&quot;&gt;cyclic number&lt;/a&gt;) from one full period of the decimal expansion of the reciprocal of a full reptend prime, and every Tuesday number arises this way. For example, $7$ is a full reptend prime, so $142857$ is a Tuesday number:&#xA;$$&#xA;\begin{array}{ll}&#xA;1\cdot142857=142857,&amp;amp;2\cdot 142857=2857\&amp;gt;14,\\&#xA; 3\cdot 142857=42857\&amp;gt;1,&amp;amp;4\cdot 142857=57\&amp;gt;1428,\\&#xA; 5\cdot 142857=7\&amp;gt;14285,&amp;amp; 6\cdot 142857=857\&amp;gt;142.\\&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be proven that $p$ is a reptend prime if and only if the first $p-1$ powers of $10$ give distinct remainders when divided by $p$ (in this case one says $10$ &lt;em&gt;is a primitive root modulo $p$&lt;/em&gt;). &lt;a href=&quot;https://en.wikipedia.org/wiki/Artin%27s_conjecture_on_primitive_roots&quot;&gt;Artin's conjecture on primitive roots&lt;/a&gt; asserts that every non-square positive integer is a primitive root modulo $p$ for infinitely many primes $p$, and this would imply that there exist arbitrarily large Tuesday numbers.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastActivityDate="2015-09-29T21:20:45.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22713" PostTypeId="2" ParentId="22711" CreationDate="2015-09-29T21:24:37.050" Score="14" Body="&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fred Wallace Haise Junior&lt;br&gt;&#xA; 12 people have walked on the moon, but it would have been 14 (including you) if Apollo 13 had not run into problems (no-one's mistake)&lt;br&gt;&#xA; You were due to go to the moon again on Apollo 19 but this was cancelled - this was someone's mistake.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-09-29T21:24:37.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22715" PostTypeId="1" AcceptedAnswerId="22716" CreationDate="2015-09-29T22:01:29.923" Score="5" ViewCount="287" Body="&lt;p&gt;Nogginson Rossouw, professor of Astronomy at his university, set an exam a few days ago. However, it seems that all of the students who have taken the exam, wrote strange answers. The professor asked them to write short notes on 8 astronomy related topics and all of them have given weird answers. For example, one of the students, Drew Plater's answer script looks like this-&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;A measuring instrument and unit.&lt;/li&gt;&#xA;  &lt;li&gt;Organic compound found in our body.&lt;/li&gt;&#xA;  &lt;li&gt;A flat surface.  &lt;/li&gt;&#xA;  &lt;li&gt;A hybrid between a primate and a carnivore?  &lt;/li&gt;&#xA;  &lt;li&gt;Wooden case used for storing goods.&lt;/li&gt;&#xA;  &lt;li&gt;Divide and analyse a sentence.&lt;/li&gt;&#xA;  &lt;li&gt;Cost for a trip to the sun?&lt;/li&gt;&#xA;  &lt;li&gt;Humorous radiance?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assuming that the students didn't purposefully perform vandalism, What were the topics the good professor asked his students to write notes on and why have they produced such strange results? &lt;/p&gt;&#xA;" OwnerUserId="15544" LastEditorUserId="15544" LastEditDate="2015-09-30T23:10:06.387" LastActivityDate="2015-09-30T23:11:56.587" Title="The Case of the Strange Astronomy Test" Tags="|word|outer-space|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22716" PostTypeId="2" ParentId="22715" CreationDate="2015-09-29T22:08:28.293" Score="7" Body="&lt;p&gt;The students have produced strange results because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the topics are all missing a letter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Topic 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;mete(o)r&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Topic 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;(a)steroid&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Topic 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;plane(t)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Topic 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;ap(h)elion (From @Takeshi)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Topic 5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;crate(r)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Topic 6:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;parse(c)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Topic 7:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;solar f(l)are&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Topic 8:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;co(s)mic ray&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-09-30T23:11:56.587" LastActivityDate="2015-09-30T23:11:56.587" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22719" PostTypeId="1" AcceptedAnswerId="22728" CreationDate="2015-09-30T02:09:47.517" Score="7" ViewCount="448" Body="&lt;p&gt;Due to many complaints about my code problem being to unoriginal, I have decided to write another problem based off the same idea, and a simple story.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now Sherlock and Watson were solving a case of Miranda White. She was a teacher at the local middle school, but was notorious for being tricky and catching her students off guard. She was found dead at the bottom of the stairs, seeming to have tripped over something, but it was not known what.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Holmes found himself in the teachers classroom, seeing if there was any clues to help guide him on his quest to find the murderer. On the blackboard he found a few simply wrong math equations.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;3+90+53+92+42=3+1&lt;br&gt;&#xA;  22+7=50&lt;br&gt;&#xA;  7+8+4+3+92+42=102+1&lt;br&gt;&#xA;  59+8+9+42+75+18+22+74+33+2+75=?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who was the killer and how did Sherlock Holmes know?&lt;br&gt;&#xA;It's elementary my fellow puzzlers...&lt;/p&gt;&#xA;" OwnerUserId="167" LastActivityDate="2015-09-30T06:22:05.037" Title="The Case of the Elementary Math" Tags="|pattern|lateral-thinking|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22720" PostTypeId="1" AcceptedAnswerId="22722" CreationDate="2015-09-30T04:13:10.787" Score="8" ViewCount="495" Body="&lt;p&gt;We &lt;a href=&quot;https://puzzling.stackexchange.com/questions/22396/morse-of-the-films-that-ive-watched-a-lot&quot;&gt;replaced Paul's keyboard&lt;/a&gt; a second time, and this time I (not Gordon K) tried naming a bunch of movies that I hadn't ever watched, to see if that would help mitigate the keyboard curse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Turns out, it only made it worse, because this time, the keyboard &lt;em&gt;swapped&lt;/em&gt; out one letter for another, performing &lt;em&gt;both&lt;/em&gt; the deletion and insertion at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spellchecker never picked up on these either, because as usual, every movie title still made sense, even for the two titles in this list that aren't even in English (they of course didn't pass an English spell-check, but in their native languages). I've given here each letter that was in the original movie title, along with the letter that it was swapped with to the right.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;3 &gt; 4 (2006) A score of scores.  &lt;/li&gt;&#xA;&lt;li&gt;A &gt; E (1975) Followers of Yahweh.  &lt;/li&gt;&#xA;&lt;li&gt;A &gt; I (1960) A silky cloth that covers one’s face.  &lt;/li&gt;&#xA;&lt;li&gt;B &gt; C (2013) Wood that makes noise come April.  &lt;/li&gt;&#xA;&lt;li&gt;B &gt; F (2007) A god who’s bad at English announces his plan to Noah.  &lt;/li&gt;&#xA;&lt;li&gt;C &gt; S (1966) My sister’s monarch.  &lt;/li&gt;&#xA;&lt;li&gt;D &gt; A (2011) The person who has a region of Spain printed on her.  &lt;/li&gt;&#xA;&lt;li&gt;D &gt; B (2004) Gulf, at least two days from now.  &lt;/li&gt;&#xA;&lt;li&gt;D &gt; P (2008) The person who harvests grain.  &lt;/li&gt;&#xA;&lt;li&gt;E &gt; A (2006) An achievement of extraordinary gladness.  &lt;/li&gt;&#xA;&lt;li&gt;E &gt; D (2010) A relationship created in December.  &lt;/li&gt;&#xA;&lt;li&gt;F &gt; D (2003) A large plate.  &lt;/li&gt;&#xA;&lt;li&gt;G &gt; B (1986) The piece of bread on a burger you put your fingers on.  &lt;/li&gt;&#xA;&lt;li&gt;G &gt; D (1964) A treasure locator.  &lt;/li&gt;&#xA;&lt;li&gt;H &gt; C (2007) A man-eating man floats up.&lt;/li&gt;&#xA;&lt;li&gt;H &gt; L (1927) We’ve put yeast in this bread six times already, and we have to do it again?  &lt;/li&gt;&#xA;&lt;li&gt;H &gt; Y (2011) A sudden sharp cry.  &lt;/li&gt;&#xA;&lt;li&gt;I &gt; O (1951) Everybody run! It’s an alien bikini bottom!  &lt;/li&gt;&#xA;&lt;li&gt;K &gt; B (1921) Charlie Chaplin participates in an auction.  &lt;/li&gt;&#xA;&lt;li&gt;L &gt; B (2005) A cemetery crew marches playing instruments.  &lt;/li&gt;&#xA;&lt;li&gt;L &gt; W (1979) The man everyone mistook for Jesus has a spouse.  &lt;/li&gt;&#xA;&lt;li&gt;N &gt; F (1935) There’s a row at the theatre.  &lt;/li&gt;&#xA;&lt;li&gt;N &gt; M (1985) The squishy purple balls that are dragon Pokémon.  &lt;/li&gt;&#xA;&lt;li&gt;N &gt; S (2010) A film about the New World.  &lt;/li&gt;&#xA;&lt;li&gt;N &gt; W (2014) A fox who steals things from the United States.  &lt;/li&gt;&#xA;&lt;li&gt;O &gt; A (1976) Ending of a prayer.  &lt;/li&gt;&#xA;&lt;li&gt;O &gt; G (1971) A monastic farm that runs extremely systematically.  &lt;/li&gt;&#xA;&lt;li&gt;O &gt; S (2014) A film all about what animals use to walk.  &lt;/li&gt;&#xA;&lt;li&gt;P &gt; H (2009) [makes a sound indicating unsureness]  &lt;/li&gt;&#xA;&lt;li&gt;P &gt; T (1978) They really dilute stuff.  &lt;/li&gt;&#xA;&lt;li&gt;R &gt; D (1950) That’s a big god.  &lt;/li&gt;&#xA;&lt;li&gt;R &gt; N (1945) Each step is a vertebra.  &lt;/li&gt;&#xA;&lt;li&gt;R &gt; N (2008) His armour is too damp.  &lt;/li&gt;&#xA;&lt;li&gt;R &gt; U (2011) A mansion for combat.  &lt;/li&gt;&#xA;&lt;li&gt;R &gt; V (1988) A narcissistic adult.  &lt;/li&gt;&#xA;&lt;li&gt;S &gt; A (2006) The unfair actions of a corpse.  &lt;/li&gt;&#xA;&lt;li&gt;S &gt; A (2013) A college full of swiss-cheese plants.  &lt;/li&gt;&#xA;&lt;li&gt;S &gt; R (2009) I’m on a roller coaster and I can’t see anything!  &lt;/li&gt;&#xA;&lt;li&gt;S &gt; T (2014) The mistake we made by beginning.  &lt;/li&gt;&#xA;&lt;li&gt;S &gt; V (2014) A member of the Scooby Crew.  &lt;/li&gt;&#xA;&lt;li&gt;S &gt; X (1968) A celestial ball with a peak-like structure.  &lt;/li&gt;&#xA;&lt;li&gt;T &gt; D (2010) An instruction manual on siphoning bodily fluids out of a reptile.  &lt;/li&gt;&#xA;&lt;li&gt;T &gt; G (1984) He makes plants grow.  &lt;/li&gt;&#xA;&lt;li&gt;T &gt; K (2005) The turban-wearers strike back.  &lt;/li&gt;&#xA;&lt;li&gt;U &gt; A (2012) Sports that gallows executioners play.  &lt;/li&gt;&#xA;&lt;li&gt;U &gt; O (2006) My solemn promise to be faithful is causing extreme hassle.  &lt;/li&gt;&#xA;&lt;li&gt;Y &gt; U (1986) Influenza.   &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Can you figure out what these titles are so that poor Alice doesn't have to retype them all?&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-10-01T10:46:16.447" Title="A lot of these firms were never watched by me" Tags="|wordplay|movies|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22721" PostTypeId="2" ParentId="22695" CreationDate="2015-09-30T04:18:44.750" Score="1" Body="&lt;p&gt;The side length $s$ of the square is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;s = \sqrt{500} = 10\sqrt{5}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radius of the circle $r$ is half the diagonal of this square:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;r = \frac{s}{\sqrt{2}} = 5\sqrt{10}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diagonal of the rectangle (of width $w$ and height $h$) is also equal to twice the radius, and its area is equal to the product of its dimensions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;w^2 + h^2 = (2r)^2 = 1000\\&#xA;wh = 400&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for $h$ and $w$, we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;w &amp;amp;= 20\sqrt{2} \\&#xA;h &amp;amp;= 10\sqrt{2} \\&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angle $\phi$ between the long side ($w$) of the rectangle and its diagonal is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\phi = \arctan\frac{h}{w} = \arctan\frac{1}{2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a vertex of the square adjacent to the endpoints of the short side of the rectangle.  The angle between the two corners of the rectangle is $2\phi$, so if the rectangle is oriented so that there is an angle $\theta$ between one of its corners and the vertex of the square, there will be an angle $2\phi-\theta$ between that vertex and the other corner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since each of the sides of the square occupies an angle of $\pi/2$ measured from the circle's center, the angles between the corners of the rectangle and the next two corners of the squares are $\pi/2-\theta$ and $\pi/2-(2\phi-\theta)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final four angles are repeats of these, since the arrangement has twofold rotational symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the octagon is made of eight sectors.  Each has two side lengths of $r$ and the angles between those sides are (two each of):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\theta, \\&#xA;2\phi-\theta, \\&#xA;\pi/2 - \theta, \\&#xA;\pi/2 - 2\phi + \theta&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total area of the octagon is then just:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;r^2\left(\sin\theta + \sin(2\phi-\theta) + \sin(\pi/2 - \theta) + \sin(\pi/2 - 2\phi + \theta)\right) \\&#xA;=300(\sin\theta + 2\cos\theta)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plotting this quantity, we can see that it is maximized when $\theta=\phi$, and its maximum value is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;300\sqrt{5}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="9718" LastActivityDate="2015-09-30T04:18:44.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22722" PostTypeId="2" ParentId="22720" CreationDate="2015-09-30T04:26:30.750" Score="6" Body="&lt;ol&gt;&#xA;&lt;li&gt;3 &gt; 4 (2006) A score of scores.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;(3/4)00&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;A &gt; E (1975) Followers of Yahweh.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;J(a/e)ws&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;A &gt; I (1960) A silky cloth that covers one’s face.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Mask of Sat(a/i)n&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;B &gt; C (2013) Wood that makes noise come April.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Spring (B/C)reakers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;B &gt; F (2007) A god who’s bad at English announces his plan to Noah.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There Will Be (B/F)lood&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;C &gt; S (1966) My sister’s monarch.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Le Roi de (C/S)œur&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;D &gt; A (2011) The person who has a region of Spain printed on her.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Girl with the (D/A)ragon Tattoo&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;D &gt; B (2004) Gulf, at least two days from now.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The (D/B)ay After Tomorrow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;D &gt; P (2008) The person who harvests grain.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Rea(d/p)er&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;E &gt; A (2006) An achievement of extraordinary gladness.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Happy Fe(e/a)t&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;E &gt; D (2010) A relationship created in December.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Winter's Bon(e/d)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;F &gt; D (2003) A large plate.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Big (F/D)ish&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;G &gt; B (1986) The piece of bread on a burger you put your fingers on.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Top (G/B)un&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;G &gt; D (1964) A treasure locator.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Goldfin(g/d)er&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;H &gt; C (2007) A man-eating man floats up.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;(H/C)annibal Rising&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;H &gt; L (1927) We’ve put yeast in this bread six times already, and we have to do it again?  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;7th (H/L)eaven&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;H &gt; Y (2011) A sudden sharp cry.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The (H/Y)elp&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;I &gt; O (1951) Everybody run! It’s an alien bikini bottom!  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Th(i/o)ng from Another World&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;K &gt; B (1921) Charlie Chaplin participates in an auction.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The (K/B)id&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;L &gt; B (2005) A cemetery crew marches playing instruments.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;(L/B)and of the Dead&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;L &gt; W (1979) The man everyone mistook for Jesus has a spouse.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Monty Python's (L/W)ife of Brian&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;N &gt; F (1935) There’s a row at the theatre.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The (N/F)ight at the Opera&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;N &gt; M (1985) The squishy purple balls that are dragon Pokémon.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Goo(n/m)ies&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;N &gt; S (2010) A film about the New World.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The America(n/s)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;N &gt; W (2014) A fox who steals things from the United States.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;American S(n/w)iper&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;O &gt; A (1976) Ending of a prayer.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The (O/A)men&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;O &gt; G (1971) A monastic farm that runs extremely systematically.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A Clockwork (O/G)range&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;O &gt; S (2014) A film all about what animals use to walk.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Leg(o/s) Movie&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;P &gt; H (2009) [makes a sound indicating unsureness]  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;U(p/h)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;P &gt; T (1978) They really dilute stuff.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Watershi(p/t) down&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;R &gt; D (1950) That’s a big god.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;(R/D)io Grande&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;R &gt; N (1945) Each step is a vertebra.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Spi(r\n)al Staircase&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;R &gt; N (2008) His armour is too damp.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Da(r/n)k Knight&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;R &gt; U (2011) A mansion for combat.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;War Ho(r/u)se&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;R &gt; V (1988) A narcissistic adult.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;(R/V)ain Man&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;S &gt; A (2006) The unfair actions of a corpse.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Pirates of the Carribean: Dead Man's Che(s/a)t&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;S &gt; A (2013) A college full of swiss-cheese plants.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Monster(s/a) University&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;S &gt; R (2009) I’m on a roller coaster and I can’t see anything!  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Blind (S/R)ide&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;S &gt; T (2014) The mistake we made by beginning.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Fault in our Star(s/t)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;S &gt; V (2014) A member of the Scooby Crew.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;(S/V)elma&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;S &gt; X (1968) A celestial ball with a peak-like structure.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Planet of the Ape(s/x)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;T &gt; D (2010) An instruction manual on siphoning bodily fluids out of a reptile.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;How to (T/D)rain Your Dragon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;T &gt; G (1984) He makes plants grow.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The (T/G)erminator&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;T &gt; K (2005) The turban-wearers strike back.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Star Wars Episode III: Revenge of the Si(t/k)h&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;U &gt; A (2012) Sports that gallows executioners play.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The H(u/a)nger Games&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;U &gt; O (2006) My solemn promise to be faithful is causing extreme hassle.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;An Inconvenient Tr(u/o)th&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;Y &gt; U (1986) Influenza.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Fl(y/u)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="2156" LastEditDate="2015-10-01T10:45:59.133" LastActivityDate="2015-10-01T10:45:59.133" CommentCount="6" CommunityOwnedDate="2015-09-30T04:26:30.750" ContentLicense="CC BY-SA 3.0" />
  <row Id="22725" PostTypeId="1" CreationDate="2015-09-30T05:24:12.963" Score="18" ViewCount="3810" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;There is a flower garden, a lake and three temples. If we throw a flower into the lake, we will get back triple the number of flowers we threw in. A man wants to keep an equal number of flowers in all three temples.&lt;/p&gt;&#xA;&lt;p&gt;He throws a certain number of flowers in the lake. He leaves part of the flowers that he got back at the first temple. He then throws all the rest in the lake and puts some in the second temple. Finally, he throws the leftovers in the lake and puts them all in the third temple.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;How many flowers did he start out with&lt;/strong&gt;, &lt;em&gt;and&lt;/em&gt; &lt;strong&gt;how many flowers are in each temple?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="16603" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2020-04-05T09:14:04.963" Title="Temple lake question" Tags="|mathematics|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22726" PostTypeId="2" ParentId="22725" CreationDate="2015-09-30T05:40:37.160" Score="27" Body="&lt;p&gt;Let's say the man threw &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; flowers into the lake. Then he has &lt;span class=&quot;math-container&quot;&gt;$3x$&lt;/span&gt; flowers and puts &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; of them into the first temple, leaving him with &lt;span class=&quot;math-container&quot;&gt;$3x-y$&lt;/span&gt; flowers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He throws them in again, resulting in &lt;span class=&quot;math-container&quot;&gt;$9x - 3y$&lt;/span&gt; flowers, and puts &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; of them in the temple again, for &lt;span class=&quot;math-container&quot;&gt;$9x - 4y$&lt;/span&gt; flowers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He throws them in a third time resulting in &lt;span class=&quot;math-container&quot;&gt;$27x - 12y$&lt;/span&gt; flowers, and puts &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; of them in the temple, for &lt;span class=&quot;math-container&quot;&gt;$27x - 13y = 0$&lt;/span&gt; flowers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The smallest integer solution that will work here is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$x = 13$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$y = 27$&lt;/span&gt;, so the man had 13 flowers initially and put 27 flowers in each temple.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In general, this sort of problem is called an &lt;a href=&quot;https://en.wikipedia.org/wiki/Annuity&quot; rel=&quot;nofollow noreferrer&quot;&gt;annuity problem&lt;/a&gt;, and has applications in finances when calculating amortization rates of mortgages (hence the term &quot;annuity problem&quot;). In the above example, an equivalent problem using loans would be that you have a debt of \$13 with an annual interest rate of 200% that you want to pay off in 3 years, which requires a payment of \$27 every year. To calculate the payments required, the following formula is used:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$p = \frac{P(1+i)^n}{1 + (1+i) + (1+i)^2 + (1+i)^3 + \ldots + (1+i)^{n-1}} = \frac{P(1+i)^n i}{(1+i)^n - 1}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; is the periodic payment, &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; is the initial principal amount, &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; is the number of pay periods, and &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; is the interest rate per pay period. In the problem statement above, &lt;span class=&quot;math-container&quot;&gt;$i = 2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$n = 3$&lt;/span&gt;, so the ratio of flowers put into each temple to  initial flowers is 27/13 (&lt;span class=&quot;math-container&quot;&gt;$p = \frac{P(2+1)^3(2)}{(2+1)^3 - 1} = \frac{54P}{26} = \frac{27}{13}P$&lt;/span&gt;), which is consistent with what working it out manually gave us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll notice with the above formula that any multiple of 13 flowers would work, if you put the same multiple of 27 flowers into each temple. If you allowed for fractional flowers, you could start with 1 flower and put 27/13 flowers into each temple and it would work out the same way.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="11497" LastEditDate="2020-04-04T16:59:55.667" LastActivityDate="2020-04-04T16:59:55.667" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22727" PostTypeId="2" ParentId="22725" CreationDate="2015-09-30T05:44:42.000" Score="2" Body="&lt;p&gt;After trial &amp;amp; Error method i got this result!&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;suppose we have 6 flowers at start!&#xA;&#xA;starting amount of flowers = 6.&#xA;&#xA;among 6 we throw 3 flowers into lake,&#xA;so we have 3 + (3*3) = 12&#xA;from which we put 6 to first temple.&#xA;after that we have 6 flowers again , &#xA;so again we throw 3 flowers into lake &#xA;so we have 3 + (3*3) = 12&#xA;and we put 6 onto second temple.&#xA;than we have 6 flowers, we through 0 flowers to lake &#xA;so we have 6 + (0x3) = 6&#xA;and put all 6 flowers on third temple and finally we have no flowers in hand.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4977" LastEditorUserId="4977" LastEditDate="2015-09-30T05:47:22.637" LastActivityDate="2015-09-30T05:47:22.637" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22728" PostTypeId="2" ParentId="22719" CreationDate="2015-09-30T06:22:05.037" Score="10" Body="&lt;p&gt;The numbers represent the element with that atomic number, and addition represents string concatenation. The right sides of the equations are the element names spelled out in full, and the left sides are element abbreviations. So:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;LiThIUMo = Lithium + [one letter]&#xA;TiN = Tin&#xA;NOBeLiUMo = Nobelium + [one letter]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which makes the answer:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;PrOFMoReArTiWAsHeRe&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="12917" LastActivityDate="2015-09-30T06:22:05.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22729" PostTypeId="1" AcceptedAnswerId="22732" CreationDate="2015-09-30T07:04:07.563" Score="5" ViewCount="1675" Body="&lt;p&gt;A Wednesday number is a number $N$ where any two consecutive digits make a prime, and all such primes formed are distinct&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, $1371$ is a Wednesday number because $13$, $37$ and $71$ are all distinct primes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Find the largest Wedenesday number.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-09-30T08:51:04.883" Title="The largest Wednesday number" Tags="|mathematics|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22730" PostTypeId="1" CreationDate="2015-09-30T07:07:22.487" Score="7" ViewCount="608" Body="&lt;p&gt;A Thursday number is a number  where any three consecutive digits make a prime, and all such primes formed are distinct.&lt;/p&gt;&#xA;&lt;p&gt;For example, 13739 is a Thursday number because 137, 373, and 739 are all distinct primes.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Find the largest Thursday number.&lt;/strong&gt; (Preferably without using a program)&lt;/p&gt;&#xA;&lt;p&gt;P.S. I know today is not yet Thursday, but I may not be able to come online tomorrow.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="69582" LastEditDate="2021-02-02T16:47:19.920" LastActivityDate="2021-02-02T16:47:19.920" Title="The largest Thursday number" Tags="|mathematics|primes|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22731" PostTypeId="2" ParentId="22729" CreationDate="2015-09-30T07:23:24.737" Score="4" Body="&lt;p&gt;My attempt:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;8973713119&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With the primes:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;89  97  73  37  71  13  31  11  19  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="16607" LastEditorUserId="15544" LastEditDate="2015-09-30T08:51:04.883" LastActivityDate="2015-09-30T08:51:04.883" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22732" PostTypeId="2" ParentId="22729" CreationDate="2015-09-30T07:42:11.653" Score="16" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 619737131179&lt;br&gt;&#xA;&lt;br&gt;&#xA; 61,19,97,73,37,71,13,31,11,17,79 are distinct primes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First we limit the number of digits, there are only 21 2 digit primes, namely&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;11   13   17   19   23   29   31   37   41   43   47    &#xA;53   59   61   67   71   73   79   83   89   97&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note that, since no prime ends in 0,2,4,6,8,5, we can only have those digits at the start, Therefore, except the leftmost prime all other must consist of the digits 1,3,7,9. There are 10 such primes.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;11   13   17   19   31   37   71   73   79  97&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Since a digit is used twice (as the first digit and as the last), it follows that there can be at most 10 primes (hence 11 digits) except the first. So, there can be a total of 12 digits, the leftmost from 0,2,4,5,6,8, the others from 1,3,7,9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First let us ignore the leftmost digits and make a 11 digit Wednesday number from 1,3,7,9. Note that, in the primes&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;11   13   17   19   31   37   71   73   79  97&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;1 is used four times at the tens digit, three times as the unit digit. Therefore it must be at the beginning, similarly 9 at the end. The best we can do this way is $19737131179$. Now we can append a 6 to the beginning and get the desired result. (note that we can't append 8 since 81 is not a prime).&lt;/p&gt;&#xA;" OwnerUserId="15544" LastEditorUserId="15544" LastEditDate="2015-09-30T07:51:04.807" LastActivityDate="2015-09-30T07:51:04.807" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22733" PostTypeId="2" ParentId="22729" CreationDate="2015-09-30T08:08:21.497" Score="2" Body="&lt;p&gt;My attempt:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $619737131179$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Each 2 consecutive digits should make a prime number, so we can only use these prime numbers:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;11     13     17     19     &#xA;23     29 &#xA;31     37     &#xA;41     43     47     &#xA;53     59     &#xA;61     67     &#xA;71     73     79     &#xA;83     89     &#xA;97&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We can only use primes that consist out of $1$, $3$, $7$ and/or $9$, because only with this primes we can continue the row or primes. The reason to only use primes of these numbers is because a prime number never(except for $2$ and $5$) ends on an even number or $5$, so if a prime starts with it, it must be the very first one. That means that we only have 10 numbers:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;11     13     17     19&#xA;31     37&#xA;71     73     79&#xA;97&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And we have to make the longest(&lt;strong&gt;not largest!!!&lt;/strong&gt;) possible number out of it.&lt;br&gt;&#xA;That is: $19737131179$, which uses all 10 primes(and is therefore 11 digits long).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we only need to find the biggest prime number that ends with a $1$ and starts with an even number or $5$. And that is %61%, so the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $619737131179$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9352" LastActivityDate="2015-09-30T08:08:21.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22735" PostTypeId="2" ParentId="22730" CreationDate="2015-09-30T09:39:37.790" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 76199197739719373311313797&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Work still in progress&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Same as with the &lt;em&gt;Wednesday number&lt;/em&gt;, we can only use 1, 3, 7, and 9 in primes, except for the first one, so we only have 30 primes that we can use:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;113 131 137 139 173 179 191 193 197 199 &#xA;311 313 317 331 337 373 379 397 &#xA;719 733 739 773 797 &#xA;911 919 937 971 977 991 997&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Same as with the &lt;em&gt;Wednesday number&lt;/em&gt; we should make a longest number with them. Observing the numbers gives us the following:&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: left;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: right;&quot;&gt;Starts:&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: right;&quot;&gt;Ends:&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;11&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;2&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;13&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;2&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;17&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;19&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;2&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;31&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;2&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;33&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;37&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;3&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;39&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;2&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;71&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;73&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;3&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;77&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;79&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;2&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;91&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;2&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;93&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;97&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;4&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: left;&quot;&gt;99&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: right;&quot;&gt;1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;Where column &lt;em&gt;Starts&lt;/em&gt; means that a prime starts with that number, &lt;em&gt;Ends&lt;/em&gt; means that the prime ends with that number and numbers are just # of occurrences. Finding a 1 - 1 pairs we can immediately say, that they come after each other:&lt;/p&gt;&#xA;&lt;p&gt;71, 77, 93:&lt;br /&gt;&#xA;9&lt;strong&gt;71&lt;/strong&gt;9, 9&lt;strong&gt;77&lt;/strong&gt;3 and 1&lt;strong&gt;93&lt;/strong&gt;7&lt;/p&gt;&#xA;&lt;p&gt;After that it's just a bit of playing around trying to find the longest number. It will consist out of 23 3-digit primes(taking the sum of the lowest of each row in the table), so 25 digits long.&lt;/p&gt;&#xA;" OwnerUserId="9352" LastEditorUserId="69582" LastEditDate="2021-02-02T16:44:55.803" LastActivityDate="2021-02-02T16:44:55.803" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="22736" PostTypeId="2" ParentId="22730" CreationDate="2015-09-30T11:05:43.507" Score="6" Body="&lt;p&gt;I believe it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 9419919379773971911373313179&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;containing these primes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 941-419-199-991-919-193-937-379-797-977-773-739-397-971-719-191-911-113-137-373-733-331-313-131-317-179&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I wrote a computer program for this that brute forced it. Of course computer programs are prone to errors but I believe it's right.&lt;/p&gt;&#xA;" OwnerUserId="1748" LastEditorUserId="1748" LastEditDate="2015-09-30T11:48:31.953" LastActivityDate="2015-09-30T11:48:31.953" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22737" PostTypeId="1" AcceptedAnswerId="22738" CreationDate="2015-09-30T11:25:30.203" Score="3" ViewCount="2267" Body="&lt;p&gt;Q1: Fill this hexagon with numbers 1-19 (no repetition) such that the sum of every vertical and diagonal row is the same. (This should be easy)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q2: Assume that one of these pre-solved cells was not given. If you are still able to solve the problem with a unique solution, do you know which cell it is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus:&lt;/strong&gt; How many solved cells can you delete and still be able to solve it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZqBTG.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZqBTG.jpg&quot; alt=&quot;Puzzle&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4269" LastEditDate="2015-09-30T12:02:00.810" LastActivityDate="2015-10-01T18:12:46.077" Title="Hexagonal sum filling" Tags="|mathematics|visual|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22738" PostTypeId="2" ParentId="22737" CreationDate="2015-09-30T11:33:26.820" Score="3" Body="&lt;p&gt;Solution to Q1:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;       3&#xA;   19    17&#xA;16     7    18&#xA;    2     1&#xA;12     5    11&#xA;    4     6&#xA;10     8     9&#xA;   13    14&#xA;      15&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Solution to Q2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The 1 is unnecesary. The 19, 12 and 7 cells come as usual. Then, between the 10 and 18 cells we have three numbers which must sum 10, and the only possible choice (since 2 and 3 are taken) is 1, 4 and 5. Below the 2 there must be a 4 (since else the number below would have to be either 12 or 16 (already taken). Then, the number 1 must be where it is (instead of 5) or else the number below the 18 would have to be 7 (already taken).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bonus:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;After removing the 1, the 17 is also unnecesary. If the number were a 19, no number could go in place of the actual 19. If the number were less than 17, the three numbers between 10 and 18 would have to sum less than 10 (which is impossible if 2 and 3 are already taken).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Moreover, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The 2 is also unnecesary, as it is possible to complete the outer ring without it with a bit of thinking: the 17, 19 and 12 come with a similar reasoning as before. Then, the 13 and 15 must be there (there is no other pair of available numbers which sum 28, and if the order were reversed, there would be no solutions for the two numbers below the 19). Then the 14, 9 and 11 come naturally. Once the outer ring is filled, the inner ring is also unique: between the 12 and 14 there must be a pair {5,7} or {4, 8}, and between the 13 and 11 there must be a pair {6,8}. Therefore, the bottom number of the inner ring must be 8, and the rest of the numbers are trivial to get.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Furthermore,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Removing the 16 cell also yields the same unique solution (I checked by trial and error).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;TL;DR;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The puzzle has a unique solution with only the cells 3, 10, and 18. It is possible that this is not the only combination of 3 cells that yields a unique solution, and it may be even possible that the puzzle has a unique solution with 2 cells (this is the minimum required for symmetry reasons). However, checking all combinations for 2 given cells is almost impossible without brute-forcing the puzzle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13523" LastEditorUserId="13523" LastEditDate="2015-09-30T13:47:46.473" LastActivityDate="2015-09-30T13:47:46.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22739" PostTypeId="2" ParentId="22737" CreationDate="2015-09-30T11:56:47.833" Score="0" Body="&lt;p&gt;The sum in each row has to be 38.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have all numbers from 1 to 19 which add up to 190 divided by 5 rows in each direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this information we can start filling the hexagon easily until 7 numbers are left:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;      03      &#xA;   19    17   &#xA;16    07    18&#xA;   02    01   &#xA;12    ??    11&#xA;   ??    ??   &#xA;10    ??    09&#xA;   ??    ??   &#xA;      ??      &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;leaving the numbers &lt;code&gt;4&lt;/code&gt; &lt;code&gt;5&lt;/code&gt; &lt;code&gt;6&lt;/code&gt; &lt;code&gt;8&lt;/code&gt; &lt;code&gt;13&lt;/code&gt; &lt;code&gt;14&lt;/code&gt; &lt;code&gt;15&lt;/code&gt; which we have to fit by intelligent guessing:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;      03      &#xA;   19    17   &#xA;16    07    18&#xA;   02    01   &#xA;12    05    11&#xA;   04    06   &#xA;10    08    09&#xA;   13    14   &#xA;      15      &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As for Q2 I'm not sure if i understand it correctly, but removing any of the numbers &lt;code&gt;3&lt;/code&gt; &lt;code&gt;17&lt;/code&gt; &lt;code&gt;18&lt;/code&gt; would still give a problem with a unique solution, as 2 of them still give us the third.&lt;/p&gt;&#xA;" OwnerUserId="16608" LastEditorUserId="85" LastEditDate="2015-09-30T14:59:02.750" LastActivityDate="2015-09-30T14:59:02.750" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22741" PostTypeId="1" AcceptedAnswerId="22759" CreationDate="2015-09-30T12:21:25.987" Score="4" ViewCount="398" Body="&lt;p&gt;A Friday number is a &lt;em&gt;prime&lt;/em&gt; number N where any three consecutive digits make a prime, and all such primes formed are distinct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, 1373 is a Friday number because 137, 373 and 1373 are all distinct primes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Find the largest Friday number.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-10-01T02:53:59.863" Title="The largest Friday number" Tags="|mathematics|number-theory|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="22742" PostTypeId="2" ParentId="22550" CreationDate="2015-09-30T12:27:38.153" Score="1" Body="&lt;p&gt;The code should have been&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 26&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Partial explanation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each ten breaks the pattern, except 1-29. Instead of the expected &lt;code&gt;thirty * 2&lt;/code&gt; the code uses &lt;code&gt;thirteen&lt;/code&gt; (very similar in pronunciation): &lt;code&gt;thirteen * 2 = 26&lt;/code&gt;. The same applies for fifty &lt;code&gt;fifteen * 2 = 30&lt;/code&gt;, sixty &lt;code&gt;sixteen * 2 = 32&lt;/code&gt;, seventy &lt;code&gt;seventeen * 2 = 34&lt;/code&gt;, ninety &lt;code&gt;nineteen * 2 = 38&lt;/code&gt;. This didn't work for eleven and twelve, which is why safes #1-29 follow a regular pattern.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, I'm still in the dark on #40 and #80.&lt;/p&gt;&#xA;" OwnerUserId="16615" LastActivityDate="2015-09-30T12:27:38.153" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22743" PostTypeId="1" AcceptedAnswerId="22749" CreationDate="2015-09-30T12:54:21.573" Score="5" ViewCount="1166" Body="&lt;p&gt;Professor Halfbrain was severely ill last week, and he had to spend the six days from Monday till Saturday in the hospital. Luckily he has fully recovered by now. He told us that on each of these six days of illness he was visited by some of his 20 closest friends; and of course many of these friends visited him more than once. Professor Erasmus thought about this for some time, and then told me: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;No matter how and when and how often these 20 friends were visiting Professor Halfbrain on these six days of illness, one will be able to pick two days $D_1$ and $D_2$ and pick five friends $F_1,F_2,F_3,F_4,F_5$, such that the following holds for each friend $F_i$ with $1\le i\le5$: $F_i$ has either visited Halfbrain on both days $D_1$ and $D_2$, or $F_i$ has visited Halfbrain neither on day $D_1$ nor on day $D_2$.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is this statement correct, or has professor Erasmus once again made a mathematical blunder?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-09-30T16:24:21.490" LastActivityDate="2015-09-30T16:24:21.490" Title="Six days of illness (Part 1)" Tags="|mathematics|combinatorics|" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22744" PostTypeId="2" ParentId="22743" CreationDate="2015-09-30T13:38:58.680" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; OF COURSE HE IS RIGHT, HE IS A PROFESSOR.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;We will prove the problem for 8 friends instead of 5 and see that this is the optimal bound.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Proof.&lt;/p&gt;&#xA;&lt;p&gt;Let us imagine that each of the professor's friends has a twin who visits him on the days when the sibling does not. Then we have &lt;span class=&quot;math-container&quot;&gt;$40$&lt;/span&gt; people in total and every day exactly &lt;span class=&quot;math-container&quot;&gt;$20$&lt;/span&gt; of them visit the professor. Now if we prove that there exist days &lt;span class=&quot;math-container&quot;&gt;$D_1, D_2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt; people who have visited the professor on both days &lt;span class=&quot;math-container&quot;&gt;$D_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D_2$&lt;/span&gt;, then the problem will be solved. Indeed, it is impossible that we choose both twins of a pair in these &lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt; people, because at most one of them can visit the professor on both days. If some of these &lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt; people are not the professor's friends, just exchange them with their twins and we are done.&lt;/p&gt;&#xA;&lt;p&gt;Next, let us label the friends and their twins with #1, #2, ... , #40 such that #j is the twin of #(j+20).&lt;/p&gt;&#xA;&lt;p&gt;For each &lt;span class=&quot;math-container&quot;&gt;$i\leq 6$&lt;/span&gt;, denote with &lt;span class=&quot;math-container&quot;&gt;$x_i$&lt;/span&gt; the &lt;span class=&quot;math-container&quot;&gt;$40$&lt;/span&gt;-dimensional vector&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$x_i=(\delta_{i,1}, \delta_{i,2}, ... \delta_{i,40}),$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$\delta_{i,j}$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; if person #j visited the professor on day i and &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt; otherwise. If we assume that no &lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt; people visited the professor on the same 2 days, we have &lt;span class=&quot;math-container&quot;&gt;$\langle x_k, x_l\rangle \leq 7$&lt;/span&gt; for all pairs &lt;span class=&quot;math-container&quot;&gt;$(k,l)$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$\langle \cdot, \cdot \rangle$&lt;/span&gt; is the regular scalar product.&lt;/p&gt;&#xA;&lt;p&gt;Now we have&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\langle V, V\rangle=\langle \sum_{i\leq 6} x_i, \sum_{i\leq 6} x_i \rangle = \sum_{i \leq 6} \langle x_i,x_i\rangle + \sum_{i\neq j} \langle x_i, x_j \rangle \leq 6.20+30.7=330.$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Let &lt;span class=&quot;math-container&quot;&gt;$V=(V_1, V_2,...,V_{40})$&lt;/span&gt;. For every &lt;span class=&quot;math-container&quot;&gt;$j\leq 20$&lt;/span&gt; we have &lt;span class=&quot;math-container&quot;&gt;$V_j + V_{j+20}=6$&lt;/span&gt; and therefore &lt;span class=&quot;math-container&quot;&gt;$V_j^2+V_{j+20}^2\geq 18$&lt;/span&gt;. Therefore&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\langle V, V \rangle = \sum_{j\leq 40}V_j^2 \geq 20.18=360,$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;which gives us a contradiction.&lt;/p&gt;&#xA;&lt;p&gt;In order to see that &lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt; friends is optimal, just consider the following example courtesy of @The Dark Truth (may want to upvote him as well for this):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;1111 1111 1100 0000 0000&lt;/p&gt;&#xA;&lt;p&gt;1111 0000 0011 1111 0000&lt;/p&gt;&#xA;&lt;p&gt;1000 1110 0011 1000 1110&lt;/p&gt;&#xA;&lt;p&gt;0100 1001 1010 0110 1101&lt;/p&gt;&#xA;&lt;p&gt;0010 0101 0101 0101 1011&lt;/p&gt;&#xA;&lt;p&gt;0001 0010 1100 1011 0111&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14140" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-09-30T15:23:33.863" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22745" PostTypeId="2" ParentId="22743" CreationDate="2015-09-30T13:57:42.153" Score="3" Body="&lt;p&gt;Professor Erasmus is &lt;strong&gt;wrong&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof by Example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;|  D   Friends---------------&amp;gt;&#xA;|  a&#xA;|  y&#xA;\/ s&#xA;&#xA;1111 1111 1100 0000 0000&#xA;1111 0000 0011 1111 0000&#xA;1000 1110 0011 1000 1110&#xA;0100 1001 1010 0110 1101&#xA;0010 0101 0101 0101 1011&#xA;0001 0010 1100 1011 0111&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I did this by stating all 20 permutations of exactly 3 visits on 6 days.&lt;br&gt;&#xA;You will find no 2 days with either more than 4 Friends visiting or more than 4 friends not visiting.&lt;/p&gt;&#xA;" OwnerUserId="16608" LastEditorUserId="8874" LastEditDate="2015-09-30T16:18:52.413" LastActivityDate="2015-09-30T16:18:52.413" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="22747" PostTypeId="2" ParentId="22550" CreationDate="2015-09-30T14:03:07.893" Score="11" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To get the code of a safe, write down the safe's number as a word, replace the letters with numbers using the following matching, and finally add them together.&#xA; &lt;pre&gt;&#xA; e: 0    f: 5    g: -62  h: 67&#xA; i: 0    l: 8    n: 9    o: -7&#xA; r: -72  s: 0    t: 11   u: 82&#xA; v: 5    w: 0    x: 12   y: 9 &lt;/pre&gt;&#xA; (the other letters do not appear in the numbers)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How I got there:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After the last hint (and all the comments and thinking before) it was pretty clear to me, that each letter in the word had to have some value which then would add up to the safe's code. Obviously the usual matching &lt;code&gt;a = 1, b = 2, ...&lt;/code&gt; would not fit, so I had to do some calculations based on the assumption that &lt;code&gt;t+y = 20&lt;/code&gt; which I guessed after hint 5.&#xA;&lt;pre&gt;&#xA; t+y = 20&#xA; t+w+e+n+t+y = 40 =&gt; t+w+e+n = 20&#xA; t+e+n = 20 = t+w+e+n =&gt; w = 0&#xA; t+w+o = t+o = 4, t+e+n = 20 =&gt; t+t+o+n+e = 24&#xA; t+t+o+n+e = 24, o+n+e = 2 =&gt; t+t = 22 =&gt; t = 11&#xA; t+y = 20, t = 11 =&gt; y = 9&#xA; t+e+n = 20, t = 11 =&gt; e+n = 9&#xA; e+n = 9, o+n+e = 2 =&gt; o = -7&#xA; n+i+n+e = 18, e+n = 9 =&gt; n+i = 9 =&gt; i = e&#xA; n+i+n+e+t+e+e+n = 38, n+i+n+e = 18 =&gt; t+e+e+n = 20&#xA; t+e+e+n = 20 = t+e+n =&gt; e = i = 0, n = 9&#xA; f+i+f+t+y = 30, t+y = 20 =&gt; f+i+f = f+f = 10 =&gt; f = 5&#xA; f+i+v+e = f+v = 10 =&gt; v = 5&#xA; s+e+v+e+n = s+v+n = 14 =&gt; s = 0&#xA; e+l+e+v+e+n = l+v+n = 22 =&gt; l = 8&#xA; s+i+x = s+x = 12 =&gt; x = 12&#xA; f+o+r+t+y = -54 =&gt; r = -72&#xA; f+o+u+r = 8 =&gt; u = 82&#xA; t+h+r+e+e = t+h+r = 6 =&gt; h = 67&#xA; e+i+g+h+t = g+h+t = 16 =&gt; g = -62&#xA; &lt;/pre&gt;&#xA; I hope that helps to understand the way I found the solution and also how the codes are calculated.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15985" LastEditorUserId="15985" LastEditDate="2015-10-01T07:24:06.737" LastActivityDate="2015-10-01T07:24:06.737" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="22748" PostTypeId="2" ParentId="22743" CreationDate="2015-09-30T14:20:34.017" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The professor is right!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Dark Truth has a very nice grid, so we'll start with that:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1111 1111 1100 0000 0000&#xA;1111 0000 0011 1111 0000&#xA;1000 1110 0011 1000 1110&#xA;0100 1001 1010 0110 1101&#xA;0010 0101 0101 0101 1011&#xA;0001 0010 1100 1011 0111&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Since we can take the visits each day as a binary value, we can do bit mappings across days to get repeat visitors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's take a 2 day period and assume there are 0 repeats. A third day would have a total of 20 repeats split between the first two days.  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0101 0101 0101 0101 0101&#xA;1010 1010 1010 1010 1010&#xA;xxxx xxxx xxxx xxxx xxxx&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Let's take another 2 day period and assume there are 4 repeats. A third day could not have the 4 repeats, but instead would have a total of 16 repeats split between the first two days.  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0000 0101 0101 0101 0101&#xA;0000 1010 1010 1010 1010&#xA;1111 xxxx xxxx xxxx xxxx&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So, if anything, the professor was being too conservative with how many repeat showings/not showings there were. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-09-30T14:20:34.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22749" PostTypeId="2" ParentId="22743" CreationDate="2015-09-30T15:53:37.733" Score="5" Body="&lt;p&gt;Professor Halbrain&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; is correct. In fact, we can find eight friends satisfying the given condition.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Call the six days $D_1,\ldots,D_6$, and number the friends $1,\ldots,20$. For $i=1,2,\ldots,6$, let $x_i$ be the the vector whose $j$-th component ($j=1,2,\ldots,20$) is $1$ if friend $j$ visited Halfbrain on $D_i$, and $-1$ otherwise. Let $\langle\cdot,\cdot\rangle$ denote the usual scalar product of vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each $i,j$, let $A_{ij}$ be the number of friends who visited Halfbrain on both $D_i$ and $D_j$, or neither $D_i$ nor $D_j$. We have $\langle x_i,x_j\rangle=A_{ij}-(20-A_{ij})=2A_{ij}-20$, and in particular $\langle x_i,x_i\rangle=20$. Now:&#xA;$$&#xA;0\leq\left\langle \sum_i x_i,\sum_i x_i\right\rangle = \sum_{i}\langle x_i,x_i\rangle+\sum_{i\neq j} \langle x_i,x_j\rangle\leq 6\cdot 20+30\cdot\max_{i\neq j}\big(2A_{ij}-20\big).&#xA;$$&#xA;This means there must be some $i\neq j$ with $2A_{ij}-20\geq -4$, or equivalently $A_{ij}\geq 8$. So there are $8$ friends who either visited Halfbrain both on $D_i$ and $D_j$, or neither on $D_i$ nor $D_j$.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-09-30T16:03:27.413" LastActivityDate="2015-09-30T16:03:27.413" CommentCount="2" CommunityOwnedDate="2015-09-30T15:53:37.733" ContentLicense="CC BY-SA 3.0" />
  <row Id="22750" PostTypeId="2" ParentId="22690" CreationDate="2015-09-30T16:06:58.920" Score="6" Body="&lt;p&gt;Here is my view on arranging treasure hunts:-&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Types of Treasure Hunt:&lt;/h1&gt;&#xA;&#xA;&lt;h2&gt;Treasure Hunt Trail&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;A chain of clues where each clue must be solved to find the location of the next clue, ultimately leading to the location of the 'treasure'. All participants are working together to try to find each of the locations and they move en-masse from one location to the next.  This type of treasure hunt favours younger participants and smaller groups.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pros&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There is an instant reward after solving each clue - positive feedback when you find the next clue.&lt;/li&gt;&#xA;&lt;li&gt;There is an element of suspense as you are never sure if this next location will be where the treasure is.&lt;/li&gt;&#xA;&lt;li&gt;You can plan the route that the hunt takes to maximise the distance between successive clues. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cons&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Some participants can feel left out if they are slow getting to a location and/or miss hearing a clue, so they only know where to go by following the pack. The larger the group, the more likely this will happen.  &lt;/li&gt;&#xA;&lt;li&gt;Every clue has to be solved, so the hunt-setter has to be on-hand to provide prompts and clues.  &lt;/li&gt;&#xA;&lt;li&gt;It can be difficult to pitch the clues so that they are not all too hard or all too easy.  &lt;/li&gt;&#xA;&lt;li&gt;If someone spots a clue out-of-sequence, they can short-cut the search.  &lt;/li&gt;&#xA;&lt;li&gt;Often the biggest/fastest participants will 'win'.  &lt;/li&gt;&#xA;&lt;li&gt;The clues need to be hidden well enough so that they are not spotted inadvertantly before the preceeding clue has been solved, but not hidden so well that they cannot be found even with the information from the solved clue.  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Scatter Event&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Each small team of participants (or each individual if you want people to work alone) is given a sheet with clues to information that they need to find on their search. They can tackle the clues in any order and the 'winner' is the team back first with the most correct answers. Alternatively you can provide loads of clues and set a time limit, making it clear at the start that no-one will finish all of them. This type of hunt works better with older participants and larger groups.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pros&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You can have a range of clue difficulties so every team is likely to find some of the   answers and feel more involved.  &lt;/li&gt;&#xA;&lt;li&gt;It's not just a case of fastest team wins.  &lt;/li&gt;&#xA;&lt;li&gt;Each team can feel that they are in with a chance until the result is announced.  &lt;/li&gt;&#xA;&lt;li&gt;You don't need to worry about clues being spotted too early.  &lt;/li&gt;&#xA;&lt;li&gt;Encourages competition between teams and cooperation within a team.  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cons&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A longer attention span needed.  &lt;/li&gt;&#xA;&lt;li&gt;Some people might spend too long on one clue and run out of time.  &lt;/li&gt;&#xA;&lt;li&gt;You've got to mark the results. &lt;/li&gt;&#xA;&lt;li&gt;You need to work harder to come up with the locations and clues and you'll need more of them.   &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h1&gt;Your Case:&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;With around a dozen children, mostly around 11, you could probably take either approach. If the 11-year-olds know the younger children well and would actively try to get them involved, I would be tempted to divide them into 3 teams and go with a Scatter Event so that there's a bit of competition. If there is a chance that they wouldn't work well with or involve the younger ones on their team, go with the Treasure Hunt Trail.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;For a Trail&lt;/h2&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I would stick to the house and garden for this one. Any clues hidden in the village might get moved by someone.&lt;/li&gt;&#xA;&lt;li&gt;Choose a good location where the trail will end. This has got to be somewhere that they won't stumble across accidentally in the search for the other clues. Perhaps the treasure is in a locked chest or cupboard that you point out to them at the start and they have to follow the clues to find the key. This has the advantage that everyone knows where the trail will end, so any younger kids who get bored can be there at the end to see the treasure uncovered.&lt;/li&gt;&#xA;&lt;li&gt;Choose good hiding locations for the other clues. These need to be fairly specific so that they are easily found if the children know what they're looking for (e.g. on the underside of the rabbit hutch), but not so obvious or similar that children running around looking randomly for clues will find them.&lt;/li&gt;&#xA;&lt;li&gt;Order the locations in such a way as to maximise the travelling distance. If one clue is in the attic, the next should be in the garden or the cellar. The next one upstairs.&lt;/li&gt;&#xA;&lt;li&gt;Once you have established the order of the locations, come up with clues. Generally they should start easy and get harder. In this way the random searching is more likely to happen only once the early clues are solved and there are fewer clues left hidden to stumble upon. 11-year olds can handle a variety of different problems: anagrams, simple alphabet ciphers, logic problems. Include some clues that are based on observation rather than just deduction to allow the younger children to participate. For example you could have a clue that (once it's decoded) directs them to a location &quot;near the grey horse&quot;, referring to a painting of a grey horse somewhere in the property that many of them may not have noticed.&lt;/li&gt;&#xA;&lt;li&gt;Make sure the children know where you are if they are struggling with a clue so they come to you, rather than you having to follow them up to the attic and to the far ends of the garden. It might be worth making the clues at these extreme locations a bit easier to solve.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h2&gt;For a Scatter Event&lt;/h2&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The locations can be more widespread here as the participants are only collecting information. This could be the date on a gravestone, the colour of a front door, the number of plantpots outside a front door.&lt;/li&gt;&#xA;&lt;li&gt;You don't need to hide anything and the locations of the answers don't have to be places that they can get to, just things that they can see.&lt;/li&gt;&#xA;&lt;li&gt;Provide a variety of fairly straightforward clues and more cryptic ones. If you are feeling really adventurous, then make them all rhyming couplets.(Fragile creatures from the sea, how many of us can you see?)&lt;/li&gt;&#xA;&lt;li&gt;Mix up the order of the locations and the difficulties.&lt;/li&gt;&#xA;&lt;li&gt;Check with the teams now and again and drop hints with those that are struggling.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;hr /&gt;&#xA;I hope this is of some help to get you started.&lt;/p&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-09-30T16:06:58.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22751" PostTypeId="1" AcceptedAnswerId="22755" CreationDate="2015-09-30T16:24:23.403" Score="6" ViewCount="189" Body="&lt;p&gt;Professor Doublebrain was severely ill last week, and he had to spend the six days from Monday till Saturday in the hospital. Luckily he has fully recovered by now. He told us that on each of these six days of illness he was visited by some of his 20 closest friends; and of course many of these friends visited him more than once. Professor Erasmus thought about this for some time, and then told me: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;No matter how and when and how often these 20 friends were visiting Professor Doublebrain on these six days of illness, one will be able to pick two days $D_1$ and $D_2$ and pick five friends $F_1,F_2,F_3,F_4,F_5$, such that the following holds: either all five friends visited Doublebrain on both days $D_1$ and $D_2$, or all five friends visited Doublebrain neither on day $D_1$ nor on day $D_2$.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is this statement correct, or has professor Erasmus once again made a mathematical blunder?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-09-30T17:23:09.117" Title="Six days of illness (Part 2)" Tags="|mathematics|combinatorics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22752" PostTypeId="1" AcceptedAnswerId="22753" CreationDate="2015-09-30T16:31:39.007" Score="1" ViewCount="1218" Body="&lt;p&gt;You may recall the prison experiment done along the following lines:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;One hundred prisoners have been newly ushered into prison. The warden tells them that starting tomorrow, each of them will be placed in an isolated cell, unable to communicate amongst each other. Each day, the warden will choose one of the prisoners at random, and place him in a central interrogation room made of concrete walls 7ft on all sides and containing only a light bulb affixed to the ceiling with a toggle switch on one wall. The prisoner will be left alone for 1 minute in this room. The prisoner will be able to observe the current state of the light bulb. If he wishes, he can toggle the light bulb. He also has the option of announcing that he believes all prisoners have visited the interrogation room at some point in time. If this announcement is true, then all prisoners are set free, but if it is false, all prisoners are executed. The warden leaves, and the prisoners huddle together to discuss their fate. Can they agree on a protocol that will guarantee their freedom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The solution to this puzzle (I surprisingly cannot find it on PSE, is found &lt;a href=&quot;https://math.stackexchange.com/questions/116340/100-prisoners-and-a-lightbulb&quot;&gt;here&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since that experiment's success, a new, crueler warden has taken over. He does not want the &#xA;prisoners to succeed in obtaining their freedom. He is bound by law&#xA;to offer the prisoner's the same opportunity, but he is allowed to change the rules to make it harder&#xA;for them. To that end, he comes up with a plan that he thinks makes it impossible to succeed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He removed the lightswitch that controls the bulb from the room. In addition,&#xA;after each prisoner's time in the central room, he will peer in the room himself and see if&#xA;anything has been altered or left behind; if he sees any difference, all prisoners will be executed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under these new rules, what plan can the prisoners devise to guarantee their freedom?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CLARIFICATION: To help keep the possibilities from being too broad, I want to specify that the answer I have in mind does not involve anything besides the room itself (nothing like 'prisoners banging on their doors if they've visited the room', etc) and can yield a strategy that can guarantee with 100% certainty when all prisoners have visited the room (nothing like 'wait x years and they've probably all been').&lt;/p&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="-1" LastEditDate="2017-04-13T12:19:36.950" LastActivityDate="2016-05-04T05:12:51.027" Title="100 Prisoners and a Light 2: Electric Boogaloo" Tags="|lateral-thinking|situation|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22753" PostTypeId="2" ParentId="22752" CreationDate="2015-09-30T16:56:12.397" Score="6" Body="&lt;p&gt;Using the same logic in the original answer (of switching the light bulb on and off), you can still use that algorithm by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; slightly loosening/tightening the bulb in it's socket. (about a quarter turn or so).  The light bulb would still be lit/unlit, but the next prisoner can easily tell the &quot;on/off&quot; state from the &quot;tightness&quot; of the light bulb.  And the warden sees no visible difference in the room.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Link to answer by Yisong Song &lt;a href=&quot;https://www.math.washington.edu/~morrow/336_11/papers/yisong.pdf&quot; rel=&quot;nofollow&quot;&gt;https://www.math.washington.edu/~morrow/336_11/papers/yisong.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14233" LastEditorUserId="14233" LastEditDate="2015-10-01T12:52:19.113" LastActivityDate="2015-10-01T12:52:19.113" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22754" PostTypeId="2" ParentId="22725" CreationDate="2015-09-30T16:59:11.163" Score="6" Body="&lt;p&gt;A simpler approach is to start from the end:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x$ flowers went to the third temple.  So, $x$ flowers came out of the lake the third time, and ${1\over3}x$ flowers were thrown in the 3rd time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x$ flowers went to the second temple plus ${1\over3}x$ flowers that were thrown in the lake, that is ${4\over3}x$ flowers that came from the lake the second time, and therefore ${4\over9}x$ flowers were thrown in the second time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x$ flowers went to the first temple plus ${4\over9}x$ flowers that were thrown in the lake, that is ${13\over9}x$ flowers that came from the lake the first time, and therefore ${13\over27}x$ flowers were thrown in the first time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming flowers come in integer numbers, we have to multiply $x$ by $27$.  Let's name $n = 27x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution: $13n$ flowers were thrown in the lake, $39n$ came out, $27n$ went to the first temple, the remaining $12n$ were thrown back in the lake, $36n$ came out, $27n$ went to the second temple, the remaining $9n$ were thrown back in the lake, $27n$ came out, and all went to the third temple.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2015-10-01T13:46:48.640" LastActivityDate="2015-10-01T13:46:48.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22755" PostTypeId="2" ParentId="22751" CreationDate="2015-09-30T17:23:09.117" Score="3" Body="&lt;p&gt;If worded like this the solution is the one i tried to give on the first part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/22743/six-days-of-illness-part-1/22745#22745&quot;&gt;Six days of illness (Part 1)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Professor Erasmus is &lt;strong&gt;wrong&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof by Example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;|  D   Friends---------------&amp;gt;&#xA;|  a&#xA;|  y&#xA;\/ s&#xA;&#xA;1111 1111 1100 0000 0000&#xA;1111 0000 0011 1111 0000&#xA;1000 1110 0011 1000 1110&#xA;0100 1001 1010 0110 1101&#xA;0010 0101 0101 0101 1011&#xA;0001 0010 1100 1011 0111&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I did this by stating all 20 permutations of exactly 3 visits on 6 days.&lt;br&gt;&#xA;You will find no 2 days with either more than 4 Friends visiting or more than 4 friends not visiting.&lt;/p&gt;&#xA;" OwnerUserId="16608" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-09-30T17:23:09.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22756" PostTypeId="1" AcceptedAnswerId="22760" CreationDate="2015-09-30T22:14:48.797" Score="15" ViewCount="335" Body="&lt;p&gt;The year is 2674 and the earth has long been abandoned for a greener planet (apparently even the sky is green). Unfortunately for Bob Junior the Umpth, he has still 250 years left on his shift aboard the old ISS orbiting our ancestral home.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To help him cope with insomnia, loneliness and because of his increasing dementia, Bob has built a small probe he regularly launches throughout the solar system. The probe emits his favorite lullaby and makes retrieving it as enjoyable as bursting bubble wrap. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However one day, as he retrieves the probe, he notices the nursing rhyme sounds distorted, as if someone has tampered with it. Quickly he attempts to decipher the hidden message within, in order to respond in kind to whoever must have attempted communicating with him.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result can be found below:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; -Twain.kle twinkled little starf&#xA; How I who.nder what you fare&#xA; Ups above I the world sor high&#xA; I like us a diamond in the sky&#xA;&#xA; -_ When. the blazing sung is goned&#xA; When che nothing shines upon a _&#xA; Then can you show o your little plight&#xA; Twinkle twinkle us all the night&#xA;&#xA; -Then _ the traveler. in _ thee dark&#xA; Thanks you forc your tiny spargks&#xA; A her could not se ce which way to go&#xA; If you did turn out twinkles so&#xA;&#xA; -In the dark _. c blue sky youd keep&#xA; And of ate n through Amy curtains peep&#xA; For yo up nerve or shut yo cur eye&#xA; Till the suny is i no then sky&#xA;&#xA; -As your a bright. and a tiny spardk&#xA; Lights the travel cerdin the dark&#xA; Though I know no pot what you are&#xA; Twinklle twinklle u little x star&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What is the secret message hidden in the lullaby?&lt;/p&gt;&#xA;" OwnerUserId="11251" LastActivityDate="2015-10-01T05:17:12.450" Title="Twinkle Twinkle Bob Junior the Umpth" Tags="|cipher|knowledge|outer-space|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22757" PostTypeId="1" AcceptedAnswerId="22758" CreationDate="2015-09-30T23:32:13.373" Score="2" ViewCount="411" Body="&lt;p&gt;Two players A and B play Tic-Tac-Toe on an infinite 3-dimensional grid. Player A starts first and uses &lt;strong&gt;X&lt;/strong&gt;, player B plays second and uses &lt;strong&gt;O&lt;/strong&gt;. Every time player A places an &lt;strong&gt;X&lt;/strong&gt; on the grid, player B can place $n$ number of &lt;strong&gt;O&lt;/strong&gt;'s. Player A wins if he eventually succeeds to make 3 in a row - &lt;strong&gt;horizontally&lt;/strong&gt; or &lt;strong&gt;vertically&lt;/strong&gt;, otherwise the game is a draw. For which values of $n$ player A can always win?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Harder version: What if player A has to make $k&amp;gt;3$ in a row? Partial solutions are welcome, 2-dimensional grid and $k=4$ seems doable.&#xA;(I think the answer is nice, but don't have a complete solution yet)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark: No diagonals here.&lt;/p&gt;&#xA;" OwnerUserId="14140" LastEditorUserId="3993" LastEditDate="2015-10-01T07:02:44.747" LastActivityDate="2015-10-01T07:02:44.747" Title="Another Tic-Tac-Toe problem" Tags="|strategy|tic-tac-toe|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22758" PostTypeId="2" ParentId="22757" CreationDate="2015-10-01T00:01:30.453" Score="6" Body="&lt;p&gt;A wins if $n=2$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No matter what B plays on their first turn, there will be a flat plane passing through the first X, which contains at most one O. A can create a 2 by 2 square on that plane while forcing all of B's moves to block both ends of a 2-in-a-row, and then complete a row of 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B draws if $n=3$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide the entire grid into 2x2x2 cubes. Every cell borders exactly three cells in its cube. Whenever A places an X, B places O's in those three cells. This guarantees that A can never have X's in two neighboring cells in the same cube. Every possible line of three includes two neighboring cells in the same cube, so A can't complete a line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same argument also shows that B can draw on an $m$-dimensional board with $k=3$ if $n\ge m$.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-10-01T00:33:58.527" LastActivityDate="2015-10-01T00:33:58.527" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22759" PostTypeId="2" ParentId="22741" CreationDate="2015-10-01T02:35:04.003" Score="7" Body="&lt;p&gt;I have an answer of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 9419919379719113739773313173&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using the primes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 941 419 199 991 919 193 937 379 797 971 719 171 911 113 137 373 739 397 977 773 733 331 313 131 317 173&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When attempting to solve the Thursday number, I ended up with a list of 260 groups of 24 3-digit primes, where all primes only contained 1, 3, 7, and 9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All 260 numbers in my list start with a 1. All also contain 971 and 991. That led me to finding the largest one that was a prime with 94 added to the beginning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Working from the 260th (largest) backwards, the 250th was the first one to be a prime with 94 added to the start of it.&lt;/p&gt;&#xA;" OwnerUserId="375" LastEditorUserId="375" LastEditDate="2015-10-01T02:53:59.863" LastActivityDate="2015-10-01T02:53:59.863" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22760" PostTypeId="2" ParentId="22756" CreationDate="2015-10-01T04:43:13.613" Score="15" Body="&lt;p&gt;Extracting the superfluous characters from each verse results in:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;pre&gt;- a . d f    h . f    s i r i u s&lt;br&gt;&#xA;- _ . g d    c a _    c a n o p u s&lt;br&gt;&#xA;- _ . _ e    c g      a r c t u r u s&lt;br&gt;&#xA;- _ . c d    a a      p r o c y o n&lt;br&gt;&#xA;- a . a d    c d      p o l l u x&lt;/pre&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Each of these lines represent:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;the brightness&lt;/strong&gt; of a star (apparent magnitude), followed by &lt;strong&gt;the distance&lt;/strong&gt; of a star (light years) using the letters a-h for the digits 1-8, followed by &lt;strong&gt;the name of a star&lt;/strong&gt;. The underscore represents the digit &lt;strong&gt;0&lt;/strong&gt;:&lt;br&gt;&#xA;&lt;pre&gt;-1.46    8.6   Sirius (the brightest star, excluding the Sun)&lt;br&gt;&#xA;-0.74   310    Canopus (the 2nd brightest star)&lt;br&gt;&#xA;-0.05    37    Arcturus (the 4th brightest star)&lt;br&gt;&#xA; 0.34    11    Procyon (the 8th brightest star)&lt;br&gt;&#xA; 1.14    34    Pollux (the &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_brightest_stars&quot;&gt;17th brightest star&lt;/a&gt;, although some&lt;br&gt;&#xA;               sources show Pollux as the 16th brightest star)&lt;/pre&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5188" LastEditorUserId="5188" LastEditDate="2015-10-01T05:17:12.450" LastActivityDate="2015-10-01T05:17:12.450" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22761" PostTypeId="2" ParentId="22664" CreationDate="2015-10-01T04:58:17.547" Score="3" Body="&lt;p&gt;Let me take the plunge, following @Raystafarian's idea&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My oldest child is very temperamental. He is always red with anger and likes to yell at other people.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sand storm. Yelling refers to the loud winds&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My second child is quite the opposite of my eldest. She is not as loud and angry as my oldest, but she definitely likes to ignore everyone.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Blizzards. Quieter, but forcing people to stay at home&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My third child likes to dance. I've gotta say he is one hell of a shaker.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Earth quake&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fourth child.. Hmm, what can I say? She just follows my third child all around.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tsunami follows earthquake&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fifth child keeps on following her older sister (refering to the fourth child). Those three rarely split up.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Landslides also follow earthquakes &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My sixth child is really hot, I mean, literally hot. I guess his condition also affects his appetite as he doesn't like to eat.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Heat waves, dries things out without consuming them&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And lastly, my seventh child, still a baby. Oh, how I wish he would stop crying. He has been crying ever since he was born.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Flooding, from rainstorms. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For an over all theme of &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Natural disasters. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I suspect that this is not right, but hopefully it will earn us a hint! &lt;/p&gt;&#xA;" OwnerUserId="15942" LastActivityDate="2015-10-01T04:58:17.547" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22763" PostTypeId="1" AcceptedAnswerId="22764" CreationDate="2015-10-01T07:06:10.423" Score="8" ViewCount="2751" Body="&lt;p&gt;Two men are sitting in the park:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-I have 2 kids under school age. - one said.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-How old are they? - Asked the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-The product of their ages is equal to the number of pigeons that you see around us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-This information is insufficient for me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-The elder has green eyes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Now I see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How old are the kids? (assuming a child can start school being 6 )&lt;/p&gt;&#xA;" OwnerUserId="16360" LastActivityDate="2015-10-04T01:00:16.733" Title="How old are the kids?" Tags="|logical-deduction|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22764" PostTypeId="2" ParentId="22763" CreationDate="2015-10-01T07:15:31.517" Score="13" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The children are aged $4$ and $1$.&lt;br&gt;&#xA;&lt;br&gt;&#xA;  $4$ is the only number which can be a product of two different combinations of two numbers below $6$, where one of these combinations consists of two identical numbers.&lt;br&gt;&#xA;&lt;br&gt;&#xA; We need two combinations because knowing the product of their ages wasn't enough information.&lt;br&gt;&#xA;&lt;br&gt;&#xA;  One of these combinations has to consist of two identical numbers because the information that there is one child that is older than the other one has to eliminate that combination.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The only product this can refer to is $4$ with the combinations $2\times2$ and $1\times4$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16608" LastEditorUserId="4421" LastEditDate="2015-10-01T15:21:29.990" LastActivityDate="2015-10-01T15:21:29.990" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="22765" PostTypeId="2" ParentId="22763" CreationDate="2015-10-01T07:26:07.800" Score="10" Body="&lt;p&gt;The logic offered here follows the deduction process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The product must be some number $1 \leq p \leq 25$, where $p = a_1 a_2$ with $1 \leq a_1 \leq a_2 \leq 5$. The $a_i$ are the ages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $p$ were prime, then we'd have $a_1 = 1$ and $a_2 = p$ so $p$ may not be prime, because the product information is insufficient. We thus know $p$ may not be $2,3,5,7,11,13,17,19,23$. We may also exclude product values that, when factorized, necessarily include a factor greater than 5. This way, we also exclude $14,18,21,22,24$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the product value $p$ is at first insuficcient, it must admit more than one factorization in factors no greater than 5. This excludes $6, 8, 9, 10, 12, 15, 16, 20, 25$. The only possibility is thus $4 = 2\times 2 = 4\times 1$. Since there must be an elder, they may not be equal factors, and the ages are $1$ and $4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Welp, a little too late.&lt;/p&gt;&#xA;" OwnerUserId="16478" LastActivityDate="2015-10-01T07:26:07.800" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22766" PostTypeId="1" AcceptedAnswerId="22768" CreationDate="2015-10-01T07:53:39.143" Score="5" ViewCount="1246" Body="&lt;p&gt;Every day, Little Johnny Red picks a positive integer $m$, arbitrarily permutes the digits of $m$ to a new integer $n$, and then computes the sum $m+n$. The sums that can be produced that way are called &lt;strong&gt;red&lt;/strong&gt; integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every day, Little Annie Yellow picks a positive integer $p$ that entirely consists of $1$s, and then multiplies $p$ by nine. The products that can be produced that way are called &lt;strong&gt;yellow&lt;/strong&gt; integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An integer is &lt;strong&gt;orange&lt;/strong&gt;, if it simultaneoulsy belongs to the red and the yellow numbers.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; How many integers below one Googolplex are orange?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Remark: All integers in this puzzle are represented in decimal, and all digits are decimal digits.)&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-10-01T08:02:33.010" LastActivityDate="2015-10-01T08:44:29.773" Title="Red, yellow and orange numbers" Tags="|mathematics|number-theory|" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22768" PostTypeId="2" ParentId="22766" CreationDate="2015-10-01T08:43:03.103" Score="10" Body="&lt;pre&gt;&lt;code&gt;exactly half a Googol ((10^(100))/2)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Every orange number &lt;strong&gt;o&lt;/strong&gt; has to consist completely of the digit 9 &lt;strong&gt;and&lt;/strong&gt; there has to be a number &lt;strong&gt;m&lt;/strong&gt; which completely consists of pairs that add up to 9 (1+8, 2+7, 3+6, 4+5) and has the same length as &lt;strong&gt;o&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other numbers &lt;strong&gt;m&lt;/strong&gt; are not possible, as we have to get a sum of exactly 9 in every digit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we tried to do it with a carryover the sum would have to be 18 in every digit with a carryover of one from the digit below, which is not possible on the smallest digit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore all numbers &lt;strong&gt;m&lt;/strong&gt; need to have an even number of digits which results in all numbers &lt;strong&gt;o&lt;/strong&gt; having an even number of digits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum number of digits for numbers below below one Googolplex is exactly Googol and as half of those are an odd number of digits the solution is half a Googol.&lt;/p&gt;&#xA;" OwnerUserId="16608" LastActivityDate="2015-10-01T08:43:03.103" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22769" PostTypeId="2" ParentId="22766" CreationDate="2015-10-01T08:44:29.773" Score="7" Body="&lt;p&gt;I think the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $5*10^{99}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Yellow numbers are numbers that consist only of $9$'s. $9, 99, 999, 9999, ...$ are yellow numbers. To find orange numbers we need to check the yellows whether they are also red numbers or not. We can easily see that all yellow numbers with an even number of digits ($99, 9999, 999999, ...$) are red numbers, as you can construct a number $n$ with the digits $9 - x, 9 - x, ... , x, x, ... $. For example $9999 = 1188 + 8811$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now what is left are the yellow numbers with an odd number of digits. Those are not red numbers, as for every digit $9-x$ you need a second digit $x$ for them to add up to $9$. This will always leave you with one digit left for which there exists no 'partner'.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Thus the answer is the yellow numbers smaller than one googolplex with an even number of digits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15985" LastActivityDate="2015-10-01T08:44:29.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22770" PostTypeId="2" ParentId="22664" CreationDate="2015-10-01T09:51:12.887" Score="7" Body="&lt;p&gt;Thought I'd post my comment as an answer given that Takeshi has confirmed some of it - though I'm still a little uncertain on a few.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My oldest child is very temperamental. He is always red with anger and likes to yell at other people.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Volcanic Eruption (loud, spews red lava)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My second child is quite the opposite of my eldest. She is not as loud and angry as my oldest, but she definitely likes to ignore everyone.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Blizzard (quiet and not many people are out in it; opposite of hot is cold)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My third child likes to dance. I've gotta say he is one hell of a shaker.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Earthquake (shake shake shake)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My fourth child.. Hmm, what can I say? She just follows my third child all around.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Tsunami (follows earthquakes all around, even in rings)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My fifth child keeps on following her older sister (referring to the fourth child). Those three rarely split up.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Flood (usually a result of a tsunami that hits land)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My sixth child is really hot, I mean, literally hot. I guess his condition also affects his appetite as he doesn't like to eat.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;6&quot;&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Wildfire (hot and moves &lt;em&gt;fast&lt;/em&gt; (to abstain from eating))&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And lastly, my seventh child, still a baby. Oh, how I wish he would stop crying. He has been crying ever since he was born.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;7&quot;&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Thunderstorms (usually raining, though sometimes in FL we go with just lightning and thunder)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;And I would say the narrator is &lt;em&gt;Mother Earth&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditDate="2015-10-01T10:09:57.997" LastActivityDate="2015-10-01T10:09:57.997" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22771" PostTypeId="1" AcceptedAnswerId="22773" CreationDate="2015-10-01T10:01:35.090" Score="0" ViewCount="22262" Body="&lt;p&gt;There are three Chests containing 100 coins each, The labels on the three chests are :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Label 1 : &quot;100 gold coins&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Label 2 : &quot;50 gold and 50 silver coins&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Label 3 : &quot;100 silver coins&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are informed that all three of the labels are incorrectly placed. You have the chance to pick a random coin from any chest and then choose a chest to keep.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which chest would you choose that you get the &quot;100 gold coins&quot;?&lt;/p&gt;&#xA;" OwnerUserId="16628" LastEditorUserId="9000" LastEditDate="2015-10-01T20:43:10.573" LastActivityDate="2015-10-09T06:39:57.553" Title="The Lucky Chest" Tags="|logical-deduction|" AnswerCount="1" CommentCount="9" ClosedDate="2015-10-01T10:31:05.107" ContentLicense="CC BY-SA 3.0" />
  <row Id="22772" PostTypeId="2" ParentId="22763" CreationDate="2015-10-01T10:15:37.317" Score="-3" Body="&lt;p&gt;Green is the fourth color in the order 'VIB&lt;strong&gt;G&lt;/strong&gt;YOR'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the age of the elder one is 4 and other kid is 1.&lt;/p&gt;&#xA;" OwnerUserId="16629" LastActivityDate="2015-10-01T10:15:37.317" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22773" PostTypeId="2" ParentId="22771" CreationDate="2015-10-01T10:16:44.070" Score="2" Body="&lt;p&gt;There are two possible scenarios: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;pre&gt; chest 1: 100 silver coins - chest 2: 100 gold coins - chest 3: 50 of both&#xA; chest 1: 50 of both - chest 2: 100 silver coins - chest 3: 100 gold coins&#xA; &lt;/pre&gt; In both scenarios chest 3 contains at least 50 gold coins, so it is the safest to pick. Yet it is not guaranteed that it contains the 100 coins.&#xA; &lt;br&gt;&lt;br&gt;&#xA; Because I am allowed to pick a coin from one of the chests, it is certain that I'll get the 100 gold coins. &lt;strong&gt;I will pick a coin from the second chest.&lt;/strong&gt; If it is gold I know that the coins are placed as in scenario 1 and I should choose the chest 2. Au contraire, if the coin is silver I know the 100 gold coins are in chest 3.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15985" LastEditorUserId="5188" LastEditDate="2015-10-09T06:39:57.553" LastActivityDate="2015-10-09T06:39:57.553" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22776" PostTypeId="1" AcceptedAnswerId="22778" CreationDate="2015-10-01T10:41:36.813" Score="13" ViewCount="10191" Body="&lt;p&gt;You and your friends were having lunch during your break in school when suddenly one of your friends shouted &lt;strong&gt;&quot;I'm going to the moon!&quot;&lt;/strong&gt;. One by one, he asked your friends what they are going to bring if they are going to join him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cathy: I'm going to bring lots of milk because the moon looks like a yummy white chocolate cookie!&lt;br&gt;&#xA;Alex: I'm going to bring vegetables then. I'm a vegetarian.&lt;br&gt;&#xA;Michael: Eggs. I'm bringing eggs. The moon reminds me of eggs because of its color.&lt;br&gt;&#xA;John: -points to the mango he is eating-.&lt;br&gt;&#xA;Kate: I'll bring jam then. Max, bring bread please!!! Please!&lt;br&gt;&#xA;Max: Okay fine, I'm bringing sourdough bread though.&lt;br&gt;&#xA;Yuji: I'm going to bring my favorite udon!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You sat there, ready to munch on your deliciously prepared nachos with jalapeno sauce and cheese, when Russel suddenly asked you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Russel: How about you then? -while pointing at you-. What would you bring?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You looked confused. You weren't sure of what you are going to answer. You thought carefully and closed your eyes for a good whole minute. After that long minute of thinking, you opened your eyes, smiled and fed yourself with one big piece of nacho.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: Why didn't you answer anything? Did you figure out the pattern? Or did you just give up?&lt;/p&gt;&#xA;" OwnerUserId="16566" LastEditorUserId="15544" LastEditDate="2015-10-01T17:55:05.393" LastActivityDate="2016-05-04T11:50:57.787" Title="I'm Going to the Moon!" Tags="|pattern|story|outer-space|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22778" PostTypeId="2" ParentId="22776" CreationDate="2015-10-01T10:59:58.980" Score="19" Body="&lt;p&gt;I'm happy bringing:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nachos&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Each person is bringing something which&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Begins with the same letter as a planet in order of distance from the sun.&lt;br&gt;&#xA; Mercury -&gt; milk.&lt;br&gt;&#xA; Venus -&gt; vegetables&lt;br&gt;&#xA; Earth -&gt; eggs.&lt;br&gt;&#xA; Mars -&gt; mango.&lt;br&gt;&#xA; Jupiter -&gt; jam.&lt;br&gt;&#xA; Saturn -&gt; sourdough.&lt;br&gt;&#xA; Uranus -&gt; udon.&lt;br&gt;&#xA; My planet would be Neptune&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="14135" LastEditDate="2015-10-01T11:04:44.343" LastActivityDate="2015-10-01T11:04:44.343" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="22779" PostTypeId="2" ParentId="22752" CreationDate="2015-10-01T15:01:23.090" Score="0" Body="&lt;p&gt;A person who lives to a ripe old age will live about 30,000 days.  Living in prison is unlikely to increase your life expectancy.  If being in prison decreases your life expectancy by as much as 1 day, that means your probability of meeting a prison-specific doom is about 1 in 1 billion, each day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether any given prisoner has been in the interrogation chamber is anti-correlated with whether any other prisoner has.  After 100 days, the odds of every prisoner having been in the chamber is very small, because if even a single prisoner has been in twice, then at least one other has not been in at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as time passes, the correlations become smaller.  By the time the average number of visits reaches 10, there is very little correlation at all.  In other words, as the number of days passed ($N$) becomes large, the probability that any prisoner has not been in the room approaches (assuming independent Poisson distributions):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P(N) \approx 100 e^{-{N\over 100}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using logarithms, we see that when $P(N) = 10^{-9}$, we have $N = 2532$, which is just shy of 7 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in other words, if they do nothing in the room, except at the 7-year point guess that everyone has been in the room, their probability of dying that day is less than it would be if they just went back to their cells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This plan is faster than the standard plan or any of the alternatives mentioned.  Even waiting a mere 5 years would still have less than 1/million chance of being wrong.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2015-10-01T15:01:23.090" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22780" PostTypeId="2" ParentId="22763" CreationDate="2015-10-01T15:17:25.387" Score="3" Body="&lt;p&gt;Perhaps a simpler way to figure this out:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Knowing the product of the two ages was not enough for the second man to deduce the ages, which means the product has at least two distinct factorizations using numbers less than 6.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Once he knew that the ages were different (one child is the elder), he could figure it out, so obviously one of the factorizations was into 2 identical numbers.  That means that the product of the ages is a square number.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Knowing this, it is trivial to check the squares of numbers less than 6 to determine which of them has a second factorization:&lt;br&gt;&#xA;&lt;br&gt;&#xA; $1^2 = 1 \leftarrow$  no other factorization&lt;br&gt;&#xA; $2^2 = 4 \leftarrow$  can also be written as $1\times4$&lt;br&gt;&#xA; $3^2 = 9 \leftarrow$  no other factorization using numbers under 6&lt;br&gt;&#xA; $4^2 = 16 \leftarrow$  no other factorization using numbers under 6&lt;br&gt;&#xA; $5^2 = 25 \leftarrow$  no other factorization using numbers under 6&lt;br&gt;&#xA;&lt;br&gt;&#xA; Thus the product is obviously $4$, and the children are $1$ and $4$ years old.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2015-10-01T15:17:25.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22781" PostTypeId="1" CreationDate="2015-10-01T15:33:22.937" Score="2" ViewCount="325" Body="&lt;p&gt;You have a tanker that can hold any amount of water. You have $n$ taps that either let in or let out water. The rate of flow of water (in $ml/min$) for each tap is given by $f_n$ (positive for inflow, negative for outflow). The tank initially has $x\ ml$ and you must bring it to a state of $y\ ml$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Values of $f_n$ are real numbers and $n,x,y$ are integers. Taps are opened or closed only once every minute: no action can be taken except when the second hand of the clock is at $0$. Any number of taps can be closed or opened simultaneously. All these values are given to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q1: What is the fastest algorithm to find a solution (or find that one does not exist)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q2: Such questions when put to other humans are usually not that hard (as compared to those put to a computer). Can you suggest some tricks that a person can use to solve such questions faster?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q3: What if there are initially $k$ taps open and when you open taps, you must also simultaneously close taps, such that exactly $k$ taps remain open throughout.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I hope I haven't clubbed too many questions together. This Q may require some serious math, please explain your conclusions in simple words also.)&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="93" LastEditDate="2015-10-09T15:05:26.877" LastActivityDate="2015-11-12T15:09:15.790" Title="Closing taps (generalisation)" Tags="|mathematics|logical-deduction|optimization|algorithm|reachability|" AnswerCount="2" CommentCount="5" ClosedDate="2015-11-12T21:39:17.607" ContentLicense="CC BY-SA 3.0" />
  <row Id="22782" PostTypeId="1" AcceptedAnswerId="23044" CreationDate="2015-10-01T15:34:50.567" Score="12" ViewCount="941" Body="&lt;p&gt;It's not only the pupils at &lt;strong&gt;Farthingbottom School&lt;/strong&gt; that come from famous families, some of the teachers do too. One of these is the popular chemistry teacher, Dr M. Langley. He's a real sun-worshipper and spends a lot of time topping up his tan. Unfortunately he's not in today because he hasn't achieved his target of 811.44 hours exposure (and obviously turning orange is far more important than teaching his pupils).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today his less popular brother Dr H. Langley is taking his class. He hates chemistry (and incidentally he hates cryptography even more).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I don't know why my brother has to spend this many weeks lazing around. Here, take these and complete them&quot; he says as he passes out some sheets of paper, &quot;and tell me what record was held for 52 years.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/93v9x.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/93v9x.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pupils look at the pieces of paper and then Felix shouts out &quot;This isn't a periodic table, is it, sir?&quot;&lt;br&gt;&#xA;&quot;That's right&quot;, answers Dr Langley&lt;br&gt;&#xA;&quot;Some of these letter combinations appear more than once&quot;, chips in Randolph.&lt;br&gt;&#xA;&quot;Of course they do&quot; he replies sharply.&lt;br&gt;&#xA;&quot;Are we supposed to be filling in the grey boxes?&quot; asks Marc-Frances.&lt;br&gt;&#xA;&quot;No, idiot, the yellow ones&quot;, snaps the teacher.&lt;br&gt;&#xA;Hesitantly Portia raises her hand and says &quot;Sir, are all these entries up to date and complete?&quot;&lt;br&gt;&#xA;&quot;Well, at least there's one bright one in the class&quot;, he says. &quot;As of today (1st October, 2015) the entries are correct as I have determined them, but this is not a complete set. I considered there was more than enough information for you without expanding the table any further.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Can you determine which record Dr Langley is referring to?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Edit:&lt;/em&gt;&lt;/strong&gt;&#xA;After a considerable time during which there had been some muttering, but not a lot of progress, Dr Langley shook his head and said,   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I have heard lots of muttering about stars, but this is not my brother's class. You already know all you need to about his obsession (or would if you could do simple arithmetic). And his obsession should lead the search to find mine. I blame it all on our ancestors. Now hurry up and get finished before I have to throw out the whole chart and start all over again!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="14135" LastEditDate="2015-10-09T19:42:59.190" LastActivityDate="2015-10-09T20:53:17.573" Title="There's no chemistry in this class" Tags="|word|outer-space|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="22783" PostTypeId="2" ParentId="22781" CreationDate="2015-10-01T15:49:11.503" Score="0" Body="&lt;p&gt;Mostly a train of thought answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, when searching for values out of a group to fit another value, I tend to take it in proportions: is there some value that is close to my end value, are the options finer or coarser, can I combine options to make new options, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q3 is likely the hardest for both humans and computers, since you have to substitute a lot. I suppose it's a bit like the original question, only working with differences instead of values. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-10-01T15:49:11.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22784" PostTypeId="1" AcceptedAnswerId="22791" CreationDate="2015-10-01T16:19:16.670" Score="84" ViewCount="10716" Body="&lt;p&gt;Three prisoners have a brief strategy meeting, and then are not allowed to communicate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each night one of the three has steak for dinner, while the other two have fish tacos.  Also each night, each of the three prisoners votes for one of the following two options:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;All of us have had steak at least once.&lt;/li&gt;&#xA;&lt;li&gt;Don't know yet.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If a majority go with option 2, nothing happens that night.  If a majority go with option 1, then they are set free if they are right, and executed if they are wrong. The distribution of votes is kept secret, so it is unknown what each of the others voted.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What should their strategy be?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2722" LastEditorUserId="13484" LastEditDate="2015-10-02T13:18:24.487" LastActivityDate="2016-02-14T15:12:06.507" Title="Three voting prisoners" Tags="|strategy|" AnswerCount="8" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="22785" PostTypeId="2" ParentId="22784" CreationDate="2015-10-01T16:31:26.193" Score="32" Body="&lt;ul&gt;&#xA;&lt;li&gt;Call the three prisoners $A$, $B$, $C$.&lt;/li&gt;&#xA;&lt;li&gt;Denote by $a_n$ (respectively $b_n$, $c_n$) the number of times prisoner $A$ (respectively $B$, $C$) had steak during the first $n$ days.&lt;/li&gt;&#xA;&lt;li&gt;Note that prisoner $A$ (respectively $B$, $C$) knows his own value $a_n$ (respectively $b_n$, $c_n$), but does not know the other two values.&lt;/li&gt;&#xA;&lt;li&gt;Then $a_n+b_n+c_n=n$ for all $n$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;One good strategy for the three prisoners is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $A$ votes for option 1, if and only if $1\le a_n&amp;lt;n/2$&lt;br&gt;&#xA; $B$ votes for option 1, if and only if $1\le b_n&amp;lt;n/2$&lt;br&gt;&#xA; $C$ votes for option 1, if and only if $1\le c_n&amp;lt;n/2$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof that they will never be executed:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;As long as only one prisoner (say $A$) has had steak, we have $a_n=n$ and $b_n=c_n=0$. Then $B$ and $C$ both vote for option 2.&lt;/li&gt;&#xA;&lt;li&gt;As long as only two prisoners (say $A$ and $B$) had steak, we have $a_n+b_n=n$ and $c_n=0$. Then $C$ votes for option 2. Furthermore, it is impossible that simultaneously $a_n&amp;lt;n/2$ and $b_n&amp;lt;n/2$ (as this would imply the contradiction $a_n+b_n&amp;lt;n$).  Hence at least two prisoners vote for option 2.&lt;/li&gt;&#xA;&lt;li&gt;Summarizing: unless all three have had steak already, the majority will vote for option 2.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof that they will be set free on the first day where this becomes possible:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If all three already had steak on day $n$, then $a_n,b_n,c_n\ge1$ and $a_n+b_n+c_n=n$ implies that at least two of the numbers are strictly smaller than $n/2$. (If two are at least $n/2$ and the third one is at least $1$, then their sum would be at least $n+1$.) Hence at least two of the prisoners vote for option 1, and they are set free.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-10-01T16:54:24.803" LastActivityDate="2015-10-01T16:54:24.803" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22786" PostTypeId="1" AcceptedAnswerId="22796" CreationDate="2015-10-01T17:12:22.757" Score="8" ViewCount="827" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The following is a straightforward (but nonetheless not completely trivial) generalization of Tyler Seacrest's great puzzle &quot;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/22784/three-voting-prisoners&quot;&gt;Three voting prisoners&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are $n\ge2$ prisoners that have a brief strategy meeting, and then are not allowed to communicate any more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the following days, exactly $s$ out of the $n$ prisoners get steak for dinner, while the remaining $n-s$ prisoners get fish tacos.  Also each night, each of the $n$ prisoner casts a vote for one of the following two options:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;All of us have had steak at least once.&lt;/li&gt;&#xA;&lt;li&gt;Don't know yet.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If at least $m$ out of the $n$ prisoners go with option 1, then they are all set free if they are right, and all executed if they are wrong.&#xA;If at most $m-1$ of them go with option 1, then nothing happens that night.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&#xA;For which combinations $(n,s,m)$ does there exist a deterministic strategy for the prisoners that (a) avoids execution and (b) guarantees that they are eventually set free, once all of them have had steak at least once.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2015-10-02T17:43:41.253" Title="A generalization of Tyler Seacrest's &quot;Three voting prisoners&quot; puzzle" Tags="|mathematics|strategy|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22787" PostTypeId="1" AcceptedAnswerId="22789" CreationDate="2015-10-01T17:14:20.227" Score="2" ViewCount="1283" Body="&lt;p&gt;You have N keys, exactly K of which will open your door. If you tries keys at random and you never try the same key twice, then &lt;strong&gt;X&lt;/strong&gt; is the number of keys you will try before opening the door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;X&lt;/strong&gt; is a Random Variable, which means it will take on different values due to chance, but some values may be more likely than others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have derived the PMF for this distribution, which gives the Probability that &lt;strong&gt;X&lt;/strong&gt; = k, where k can take integer values between 1 and N-K+1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: If we have N=10 keys, and K=3 of them will open the door, the probability that it takes us 4 tries to open the door is given by: Pr(&lt;strong&gt;X&lt;/strong&gt; = 4 | N=10, K=3) = 0.125&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;The Puzzle&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Find the Expected Value of &lt;strong&gt;X&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, given the parameters N and K, how many keys should we &lt;em&gt;expect&lt;/em&gt; to try before opening the door. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your answer should be a function of N and K. You can assume N and K are both positive integers. Here are some &lt;em&gt;Expected Values of &lt;strong&gt;X&lt;/em&gt;&lt;/strong&gt; given specific values for N and K.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;E&lt;/strong&gt;(&lt;strong&gt;X&lt;/strong&gt;| N=10, K=3) = 2.75&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;E&lt;/strong&gt;(&lt;strong&gt;X&lt;/strong&gt;| N=12, K=3) = 3.25&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;E&lt;/strong&gt;(&lt;strong&gt;X&lt;/strong&gt;| N=15, K=3) = 4&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;E&lt;/strong&gt;(&lt;strong&gt;X&lt;/strong&gt;| N=25, K=15) = 1.625&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;E&lt;/strong&gt;(&lt;strong&gt;X&lt;/strong&gt;| N=35, K=15) = 2.25&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;E&lt;/strong&gt;(&lt;strong&gt;X&lt;/strong&gt;| N=50, K=2) = 17&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Moreover,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If K = N, your function should return 1&lt;/li&gt;&#xA;&lt;li&gt;If K &gt; N, your function should return 0&lt;/li&gt;&#xA;&lt;li&gt;Finally, if K = 1, then &lt;strong&gt;X&lt;/strong&gt; is known as a Discrete Uniform Random Variable, and your function should return (N + 1)/2&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11063" LastEditorUserId="11876" LastEditDate="2015-10-01T20:00:02.377" LastActivityDate="2015-10-01T20:00:02.377" Title="How many keys will we try before opening the door?" Tags="|probability|" AnswerCount="2" CommentCount="5" ClosedDate="2015-10-02T13:56:52.417" ContentLicense="CC BY-SA 3.0" />
  <row Id="22788" PostTypeId="2" ParentId="22784" CreationDate="2015-10-01T17:21:28.430" Score="15" Body="&lt;p&gt;@Gamow already offered a nice solution, but I decided to give an alternative one.&lt;/p&gt;&#xA;&lt;p&gt;The good thing about this one is that even &lt;em&gt;&lt;strong&gt;if the prisoners are not told when they will start getting steaks for dinner, it still works&lt;/strong&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Strategy:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;A prisoner always votes 2 if he has never had steak for dinner.&lt;/p&gt;&#xA;&lt;p&gt;If a prisoner had steak for dinner, then he votes 1 if he had odd number of consecutive steaks nights and 2 if he had even number of consecutive steak nights. If a prisoner didn't have steak for dinner (but had steak in the past), then he votes 1 if he had even number of consecutive non-steak nights and 2 if he had odd number of consecutive non-steak nights.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This way the prisoners will never be executed, because in case just 2 of them have had steaks, each day exactly one of them will be voting for option 1. Also it is easy to see that the prisoners will be released on the first or second possible day.&lt;/p&gt;&#xA;" OwnerUserId="14140" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-10-01T18:43:29.913" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22789" PostTypeId="2" ParentId="22787" CreationDate="2015-10-01T17:46:39.217" Score="4" Body="&lt;p&gt;If $N\ge K$, then $F(N,K)=\frac{N+1}{K+1}$, which can be proved by induction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $N=K$, then $F(N,K)=1=\frac{N+1}{K+1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, consider what happens if you try a key. It works with probability $\frac{K}{N}$, and fails with probability $1-\frac{K}{N}$. If it fails, you now have $N-1$ keys, of which $K$ work. So $F(N,K)=1+(1-\frac{K}{N})F(N-1,K)=1+(\frac{N-K}{N})\frac{N}{K+1}=1+\frac{N-K}{K+1}=\frac{N+1}{K+1}$, as desired.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-10-01T17:46:39.217" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22790" PostTypeId="2" ParentId="22787" CreationDate="2015-10-01T17:58:38.087" Score="4" Body="&lt;p&gt;Alternate way to derive the formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine that someone arranges the $N-K$ non-working keys and $K+1$ markers in a circle in random order. One of the markers is chosen at random as a starting point, and the rest are replaced with working keys. We try each key, starting with the one right after the starting point, until we find one that works. We want to know the expected distance from the starting point to the first working key.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we consider the situation before the starting position is chosen, and sum the distances from each marker to the next, we get $N+1$, the total number of positions in the circle. Each marker is equally likely to become the starting point, so the average distance is $\frac{N+1}{K+1}$.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-10-01T17:58:38.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22791" PostTypeId="2" ParentId="22784" CreationDate="2015-10-01T17:59:16.377" Score="166" Body="&lt;p&gt;Another intuitive, no-math (and I believe best) strategy could be as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The prisoner that gets steak the first night should always vote 2 (Don't know).&#xA; The other two prisoners that get fish tacos the first night should vote&#xA; 2 until they get steak for the first time, then vote 1 (Steaks) every night from&#xA; then on.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This ensures that&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The majority won't vote 1 (Steaks) when they would be wrong.&lt;/li&gt;&#xA;&lt;li&gt;The majority will vote 1 the first night they all had steak.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7551" LastEditorUserId="8874" LastEditDate="2015-10-01T18:01:51.067" LastActivityDate="2015-10-01T18:01:51.067" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="22792" PostTypeId="2" ParentId="22737" CreationDate="2015-10-01T18:12:46.077" Score="2" Body="&lt;p&gt;Q1.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;       3&#xA;   19    17&#xA;16     7    18&#xA;    2     1&#xA;12     5    11&#xA;    4     6&#xA;10     8     9&#xA;   13    14&#xA;      15&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Q2. No, because...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus: the magic hexagon is &lt;a href=&quot;https://codegolf.stackexchange.com/q/6304/194#comment-13215&quot;&gt;unique up to reflection and rotation&lt;/a&gt;, so any two cells other than the central one suffice to give a unique solution.&lt;/p&gt;&#xA;" OwnerUserId="228" LastEditorUserId="-1" LastEditDate="2017-04-13T12:39:28.753" LastActivityDate="2015-10-01T18:12:46.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22793" PostTypeId="2" ParentId="22784" CreationDate="2015-10-01T18:14:50.893" Score="2" Body="&lt;p&gt;I thought of it like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First reduce number of cases. Every prisoner is identical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider 3 days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If after 3 days , a person has received steak more than once, he will say no as it cant be possible that the rest have received it. The person who didnt receive it is 1 person and he will say no as he knows he didnt receive it. Majority : N N Y(Safe)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If after 3 days , a person has received steak once only, he might anticipate the other two have got it. The other person if he has gotten steak twice , its reduced to case 1.So he also got once. The third person will have to get once. So all will say Y and will be set free ! &lt;/p&gt;&#xA;" OwnerUserId="16641" LastActivityDate="2015-10-01T18:14:50.893" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22794" PostTypeId="2" ParentId="22786" CreationDate="2015-10-01T19:07:55.570" Score="8" Body="&lt;p&gt;Suppose that there is a winning strategy for the prisoners with $m&amp;lt;n-s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an arbitrary sequence of steak allocations such that every prisoner eats steak, some $m$ prisoners will eventually vote for option 1. If we swap the meals of the $n-m$ other prisoners around, the $m$ prisoners cannot tell the difference, so they must vote for option 1 anyway. But $n-m&amp;gt;s$, so we can swap the meals of those $n-m$ such that one never eats steak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that a winning strategy is impossible for $m&amp;lt;n-s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;If $m=n-s$, we can use the same strategy that @matega described for the original problem. The $s$ prisoners who receive steak on the first day always vote for option 2, and the other $m$ vote for option 1 if they have received steak at least once.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-10-02T17:43:41.253" LastActivityDate="2015-10-02T17:43:41.253" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22795" PostTypeId="1" CreationDate="2015-10-01T19:43:10.150" Score="29" ViewCount="4572" Body="&lt;p&gt;Great movie by the way. I'm quoting from memory, so I may get the wording wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The positive integers are each colored Red, Yellow or Green.&lt;br&gt;&#xA;  Prove that for any such coloring, there must exist three distinct positive integers $a,b,c$ such that the colors of $~a$, $~b$, $~c$, $~a+b$, $~a+c$, $~b+c$, and $~a+b+c~$ are all the same.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The integers must be non-zero, but they do not have to be distinct.&lt;/p&gt;&#xA;" OwnerUserId="16642" LastEditorUserId="39" LastEditDate="2015-10-04T14:49:01.563" LastActivityDate="2016-04-09T00:00:21.313" Title="The tough one from &quot;A Brilliant Young Mind&quot; (2014)" Tags="|mathematics|logical-deduction|combinatorics|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="22796" PostTypeId="2" ParentId="22786" CreationDate="2015-10-01T20:25:55.963" Score="6" Body="&lt;p&gt;A strategy exists&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if and only if $n-s\leq m\leq n$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clearly no strategy can exist if $m&amp;gt;n$, and f'' has given an argument that no strategy can exists if $m&amp;lt;n-s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $n-s\leq m\leq n$, and set $k=m-(n-s)\geq 0$. There are ${n\choose k}$ possible groups of $k$ prisoners, and the prisoners agree on an ordering of the groups. Say the groups are $G_0,\ldots,G_{{n\choose k}-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Call a prisoner &lt;em&gt;fishy&lt;/em&gt; if they were served fish the first day, and &lt;em&gt;steaky&lt;/em&gt; if they were served steak the first day. There are $s$ steaky prisoners and $n-s$ fishy prisoners. The prisoners will use the following strategy:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If you are a fishy prisoner, then vote $1$ if you have ever been served steak, and vote $2$ otherwise&lt;/li&gt;&#xA;&lt;li&gt;If you are a steaky, then vote $1$ if you are a member of group $G_r$, where $r$ is the remainder when the day number is divided by ${n\choose k}$, and vote $2$ otherwise.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;At most $k$ steaky prisoners vote $1$ on any day. So if $m$ prisoners voted $1$, then is must be the case that all $n-s$ fishy prisoners voted $1$, meaning that every prisoner has been served steak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, if everyone has been served steak, then all $n-s$ fishy prisoners will vote $1$. If $G_r$ is a group of $k$ steaky prisoners (such a group exists because $k\leq n$, which follows from the assumption $m\leq n$), then the next time the day number leaves a remainder of $r$ when divided by ${n\choose k}$, $k$ steaky prisoners and $n-s$ fishy prisoners will vote $1$, for a total of $m$. &lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-10-01T20:55:22.910" LastActivityDate="2015-10-01T20:55:22.910" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22797" PostTypeId="1" AcceptedAnswerId="22798" CreationDate="2015-10-01T20:58:34.163" Score="13" ViewCount="1229" Body="&lt;p&gt;Many years ago, a man a third of the way though his life gave me a ring. It wasn't anything grand at first, but he was proud to have given me that ring.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Little did he know, I am an immortal, or at least I would live a far longer life than he did. Over many years, even more men (and women and even children) have given me rings. Some have passed, and some are still here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually, everyone who give me a ring will leave. It's a painful life to live. And no matter how fancy or loud the rings they give me are, many who see me will still prefer a ring similar to the one that first man gave me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about you? Have you given me your ring?&lt;/p&gt;&#xA;" OwnerUserId="12242" LastActivityDate="2015-10-01T22:02:41.363" Title="Some rings aren't meant for fingers" Tags="|riddle|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22798" PostTypeId="2" ParentId="22797" CreationDate="2015-10-01T21:03:53.113" Score="12" Body="&lt;p&gt;You are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A telephone.&lt;br&gt;&lt;br&gt;&#xA; Invented (and first rung) by Alexander Graham Bell about a third of the way into his life, the phone has outlived Bell by almost a century and shows no sign of becoming obsolete any time soon.&lt;br&gt;&lt;br&gt;&#xA; Men, women and children ring a telephone when they want to speak to someone. Many people who have rung a telephone have since died, and everyone who rings a telephone will eventually hang up and leave the call.&lt;br&gt;&lt;br&gt;&#xA;Modern phones (especially cellphones) have fancy rings, but quite a few people (myself included) like their phone to ring as if it had actual bells on it, similar to the original phones.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13197" LastEditorUserId="13197" LastEditDate="2015-10-01T21:44:56.550" LastActivityDate="2015-10-01T21:44:56.550" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22800" PostTypeId="2" ParentId="22797" CreationDate="2015-10-01T22:02:41.363" Score="0" Body="&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a bathtub in a hotel.  The first time someone takes a bath, not much of a ring is left.  However, as time goes on, the ring gets better and better.  :D&#xA; (Most people prefer not to have a ring in the bathtub, so prefer a ring like the original!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16650" LastActivityDate="2015-10-01T22:02:41.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22808" PostTypeId="2" ParentId="22795" CreationDate="2015-10-02T08:37:31.773" Score="10" Body="&lt;p&gt;The question concerns a special case of &quot;Hindman's theorem&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Hindman's theorem&lt;/strong&gt;&lt;br&gt;&#xA;  Suppose that the natural numbers are colored with $r$ different colors. &#xA;  Then there exists a color $c$ and an infinite set $D$ of natural numbers, so that all elements of $D$ are colored with $c$ and so that every finite sum of elements of $D$ also has color $c$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For solving the puzzle, you just apply Hindman's theorem with $r=3$ colors (red, yellow, green). The theorem gives you an infinite mono-chromatic set $D$, from which you may pick three arbitrary elements $a,b,c$; these three elements have all desired properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional information:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A &lt;a href=&quot;https://en.wikipedia.org/wiki/IP_set&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; article on so-called IP sets and Hindman's theorem&lt;/li&gt;&#xA;&lt;li&gt;The article &lt;em&gt;&quot;A Simple Proof and Some Difficult Examples for Hindman's Theorem&quot;&lt;/em&gt; by Henry Towsner on the &lt;a href=&quot;http://arxiv.org/abs/0906.3885&quot; rel=&quot;nofollow&quot;&gt;arXive&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Example 3 in the &lt;a href=&quot;http://www.tricki.org/article/How_to_use_ultrafilters&quot; rel=&quot;nofollow&quot;&gt;Tricki&lt;/a&gt; article &lt;em&gt;&quot;How to use ultrafilters&quot;&lt;/em&gt; contains another proof of Hindman's theorem&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-10-03T10:50:47.063" LastActivityDate="2015-10-03T10:50:47.063" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22809" PostTypeId="2" ParentId="22502" CreationDate="2015-10-02T08:41:00.520" Score="1" Body="&lt;p&gt;Sooo, this riddle was stuck in my head for a while, because I really wanted to get a whole sentence in that matrix. Unfortunately it is not perfect yet, hopefuly others have any ideas to fix it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here is the matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|A| = \begin{vmatrix} i &amp;amp; a &amp;amp; o\\r &amp;amp; t &amp;amp; n\\f &amp;amp; g &amp;amp; m \end{vmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Leading to&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It Man Forgot Farming&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cheers!&lt;/p&gt;&#xA;" OwnerUserId="16398" LastEditorUserId="16398" LastEditDate="2015-10-02T09:23:44.443" LastActivityDate="2015-10-02T09:23:44.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22811" PostTypeId="2" ParentId="22784" CreationDate="2015-10-02T11:46:05.787" Score="4" Body="&lt;p&gt;&lt;em&gt;Note: this is no longer valid since the stipulation was added that the vote is secret.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not optimal (release one day later), but my intuition came up with:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Vote 1 to communicate &quot;I had steak for the first time tonight&quot;;&lt;/li&gt;&#xA;&lt;li&gt;Vote 2 on fish taco or any subsequent steak nights.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is repeated until each prisoner has voted 1. Then they have the next day to pack their belongings: they all vote 1 that night and are set free.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A small benefit over the intended (mathematical) answer is that the prisoners only require three bits of write-once memory each. Chalk on their cell walls will do.&lt;/p&gt;&#xA;" OwnerUserId="13484" LastEditorUserId="13484" LastEditDate="2015-10-07T12:58:21.663" LastActivityDate="2015-10-07T12:58:21.663" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22812" PostTypeId="1" AcceptedAnswerId="23568" CreationDate="2015-10-02T12:30:59.893" Score="26" ViewCount="2311" Body="&lt;p&gt;Word War II has been raging for years, and many innocent words have been rounded up into concatenation camps (which is certainly a death sentence). Thus far, you've managed to evade detection but many of your friends have not been so lucky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though it likely spells your doom, you've decided to help and have dug a narrow tunnel into one of the camps through which you'll be able to help a small group of words escape. Unfortunately you can barely get a word in edgewise, so they'll have to come through a few letters at a time. The even bigger problem, is that once on the outside the escapees will need to be disguised or they'll be too easily recognised and risk being caught again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The night of the escape comes, and a small group of words has gathered near the entrance to the tunnel:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;abstruse, condors, honest, invention, neon, reacted, vague&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now it's up to you to help them escape and ensure their disguises are good enough to evade detection on the outside.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rules&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Move words or word fragments from one side of the tunnel to the other until all constituent letters have escaped to the other side&lt;/li&gt;&#xA;&lt;li&gt;Word fragments must always be a contiguous sequence of letters taken from the start or the end of a single existing word&lt;/li&gt;&#xA;&lt;li&gt;Movement can only occur from one side of the tunnel to the other (i.e. you can't rearrange words whilst still on one side), however you can move through the tunnel in either direction&lt;/li&gt;&#xA;&lt;li&gt;Words formed on either side of the tunnel must always be valid dictionary words (see clarifications below), though fragments in transit need not&lt;/li&gt;&#xA;&lt;li&gt;In order to remain disguised, any word that appears at any time on one side of the tunnel, must never appear on the other side in the exact same form (the corollary of which is that a word can't move directly in whole from one side to the other)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A simple, one word example&lt;/strong&gt; (with &quot;@&quot; being the tunnel):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;reheated @&lt;br&gt;&#xA;  &lt;em&gt;reheat&lt;/em&gt; @ &lt;em&gt;ed&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;heat&lt;/em&gt; @ &lt;em&gt;reed&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;he&lt;/em&gt; @ reed, &lt;em&gt;at&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;here&lt;/em&gt; @ &lt;em&gt;ed&lt;/em&gt;, at&lt;br&gt;&#xA;  &lt;em&gt;there&lt;/em&gt; @ ed, &lt;em&gt;a&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;re&lt;/em&gt; @ ed, a, &lt;em&gt;the&lt;/em&gt;&lt;br&gt;&#xA;  @ ed, &lt;em&gt;are&lt;/em&gt;, the&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clarifications on what constitutes a &quot;word&quot;&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be considered a valid word, it must have an entry on &lt;a href=&quot;http://dictionary.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;dictionary.com&lt;/a&gt;, with the following stipulations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;English words only&lt;/li&gt;&#xA;&lt;li&gt;No acronyms and abbreviations (though shortenings that have become common words, such as &quot;ed&quot; are ok, so long as they're listed as such) &lt;/li&gt;&#xA;&lt;li&gt;No proper nouns&lt;/li&gt;&#xA;&lt;li&gt;Informal words are ok, but regional specific slang is not&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit (Oct 6)&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recognise that retrospectively changing things is a little unfair to existing answers (please don't downvote them), but I wanted to make a few further clarifications as to my intentions of what constitutes a word:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Use of accented characters isn't allowed (&quot;é&quot; is not the same as &quot;e&quot;)&lt;/li&gt;&#xA;&lt;li&gt;Just having an entry on dictionary.com doesn't make it a word (eg. every letter of the alphabet has it's own entry, but they aren't words, unless it has additional definition, eg. &quot;I&quot; or &quot;a&quot;, similarly chemical symbols aren't words)&lt;/li&gt;&#xA;&lt;li&gt;Related/Derived forms of a given word are allowed, even when they don't have a page of their own (eg. &quot;test&quot; has a page, and lists things like &quot;testing&quot; and &quot;testability&quot; under derived forms even though they don't have their own individual entry pages)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;That being said the English language is by it's very nature fluid, so while I'm trying to be unambiguous, all these rules are going to be somewhat subjective and I'm happy for people to argue their case for exceptions.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Edit (Oct 27)&lt;/strong&gt;:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I've &lt;a href=&quot;https://puzzling.stackexchange.com/a/23568/5208&quot;&gt;posted my reference solution below&lt;/a&gt;, if anyone is interested. It's marked as accepted since it's currently the &quot;best&quot; solution. Still happy to move the tick if anyone can make improvements...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&lt;strong&gt;Note&lt;/strong&gt;: There's potentially multiple solutions to this problem, so if multiple people get it, I'll award the tick to the person who can arrive at the least number of words on the other side of the tunnel (with the least number of steps taken as a tie-breaker). For reference, my solution results in 6 words.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2015-10-27T17:31:59.757" Title="Famous Last Words (or Escape the Concatenation Camp)" Tags="|word|letters|" AnswerCount="5" CommentCount="12" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22813" PostTypeId="2" ParentId="22784" CreationDate="2015-10-02T13:44:23.180" Score="29" Body="&lt;p&gt;Rule 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Vote &quot;I don't know&quot; if you had steak that day or haven't had it yet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rule 2:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Vote &quot;All have had steak&quot; if you didn't have steak that day, but have had it in the past.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2015-10-02T13:44:23.180" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22814" PostTypeId="2" ParentId="22812" CreationDate="2015-10-02T14:21:26.183" Score="4" Body="&lt;p&gt;I've been working on this for days now (see original date and dozens of edits). I'm so close.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;abstruse, condors, honest, invention, neon, &lt;strong&gt;acted&lt;/strong&gt;, vague @ &lt;strong&gt;re&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;abstruse, condors, &lt;strong&gt;hone&lt;/strong&gt;, invention, neon, acted, vague @ re&lt;strong&gt;st&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;abstruse, condors, hone, invention, neon, acted, vague&lt;strong&gt;st&lt;/strong&gt; @ &lt;strong&gt;re&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;abstruse, condors, hone, &lt;strong&gt;ion&lt;/strong&gt;, neon, acted, vaguest @ re, &lt;strong&gt;invent&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;abstruse, condors, hone, ion, neon, acted, vaguest, &lt;strong&gt;vent&lt;/strong&gt; @ re, &lt;strong&gt;in&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;abstruse, condors, hone, ion, neon, acted, &lt;strong&gt;guest&lt;/strong&gt;, vent @ re, &lt;strong&gt;va&lt;/strong&gt;in&lt;/li&gt;&#xA;&lt;li&gt;abstruse, condors, hone, ion, neon, &lt;strong&gt;ed&lt;/strong&gt;, guest, vent @ re, vain, &lt;strong&gt;act&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;use&lt;/strong&gt;, condors, hone, ion, neon, ed, guest, vent @ re, vain, &lt;strong&gt;abstr&lt;/strong&gt;act&lt;/li&gt;&#xA;&lt;li&gt;condors, hone, ion, neon, ed, guest, vent @ re&lt;strong&gt;use&lt;/strong&gt;, vain, abstract&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;condor&lt;/strong&gt;, hone, ion, neon, ed, guest, vent @ reuse&lt;strong&gt;s&lt;/strong&gt;, vain, abstract&lt;/li&gt;&#xA;&lt;li&gt;condor, hone, ion, neon, ed, guest, vent, &lt;strong&gt;tract&lt;/strong&gt; @ reuses, vain, &lt;strong&gt;abs&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;dor&lt;/strong&gt; &lt;sup&gt;&lt;a href=&quot;http://dictionary.reference.com/browse/dor?s=t&quot; rel=&quot;nofollow&quot;&gt;src&lt;/a&gt;&lt;/sup&gt;, hone, ion, neon, ed, guest, vent, tract @ reuses, vain, abs, con&lt;/li&gt;&#xA;&lt;li&gt;dor, hone, ion, neon, ed, guest, tract @ reuses, vain, abs, con&lt;strong&gt;vent&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;do&lt;/strong&gt;, hone, ion, neon, ed, guest, tract @ reuses, vain &lt;strong&gt;r&lt;/strong&gt;abs &lt;sup&gt;&lt;a href=&quot;http://dictionary.reference.com/browse/rab?s=ts&quot; rel=&quot;nofollow&quot;&gt;src&lt;/a&gt;&lt;/sup&gt;, convent&lt;/li&gt;&#xA;&lt;li&gt;do, hone, ion, neon, &lt;strong&gt;re&lt;/strong&gt;ed, guest, tract @ &lt;strong&gt;uses&lt;/strong&gt;, vain, rabs, convent&lt;/li&gt;&#xA;&lt;li&gt;do, hone, ion, neon, &lt;strong&gt;ed&lt;/strong&gt;, guest, tract @ uses, vain, rabs, convent, &lt;strong&gt;re&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;hone, ion, neon, ed, guest, tract @ uses, vain, rabs, convent, re&lt;strong&gt;do&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;ho&lt;/strong&gt;, ion, neon, ed, guest, tract @ uses, vain, rabs, convent, redo&lt;strong&gt;ne&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;ion, neon, ed, guest, tract @ &lt;strong&gt;ho&lt;/strong&gt;uses, vain, rabs, convent, redone&lt;/li&gt;&#xA;&lt;li&gt;ion, neon, ed, tract @ &lt;strong&gt;guest&lt;/strong&gt;houses, vain, rabs, convent, redone&lt;/li&gt;&#xA;&lt;li&gt;ion, neon, &lt;strong&gt;r&lt;/strong&gt;ed, tract @ guesthouses, vain, &lt;strong&gt;abs&lt;/strong&gt;, convent, redone&lt;/li&gt;&#xA;&lt;li&gt;ion, neon, red&lt;strong&gt;abs&lt;/strong&gt;, tract @ guesthouses, vain, convent, redone&lt;/li&gt;&#xA;&lt;li&gt;ion, neon, &lt;strong&gt;dabs&lt;/strong&gt;, tract @ guesthouses, vain, convent, redone, &lt;strong&gt;re&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;ion, neon, dabs @ guesthouses, vain, convent, redone, re&lt;strong&gt;tract&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;neon, dabs @ guesthouses, vain, convent, redone, retract&lt;strong&gt;ion&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;neon, &lt;strong&gt;re&lt;/strong&gt;dabs @ guesthouses, vain, convent, redone, &lt;strong&gt;traction&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;neon, &lt;strong&gt;red&lt;/strong&gt; @ guesthouses, vain, convent, redone, traction, &lt;strong&gt;abs&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;neon, red, &lt;strong&gt;vent&lt;/strong&gt; @ guesthouses, vain, &lt;strong&gt;con&lt;/strong&gt;, redone, traction, abs&lt;/li&gt;&#xA;&lt;li&gt;neon, &lt;strong&gt;ed&lt;/strong&gt;, vent @ guesthouses, vain, con, redone, traction, &lt;strong&gt;r&lt;/strong&gt;abs&lt;/li&gt;&#xA;&lt;li&gt;neon, &lt;strong&gt;re&lt;/strong&gt;ed, vent @ guesthouses, vain, con, &lt;strong&gt;done&lt;/strong&gt;, traction, rabs&lt;/li&gt;&#xA;&lt;li&gt;neon, &lt;strong&gt;ed&lt;/strong&gt;, vent @ guesthouses, vain, con, done, &lt;strong&gt;re&lt;/strong&gt;traction, rabs&lt;/li&gt;&#xA;&lt;li&gt;neon, ed, vent, &lt;strong&gt;ion&lt;/strong&gt; @ guesthouses, vain, con, done &lt;strong&gt;retract&lt;/strong&gt;, rabs&lt;/li&gt;&#xA;&lt;li&gt;neon, ed, vent, &lt;strong&gt;i&lt;/strong&gt; @ guesthouses, vain, con, done, retract, rabs, &lt;strong&gt;on&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;neo&lt;/strong&gt;, ed, vent, i @ guesthouses, vain, cont, done, retract, rabs, &lt;strong&gt;non&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;ed, vent, I @ guesthouses, vain, &lt;strong&gt;neo&lt;/strong&gt;con, done, retract, rabs, non&lt;/li&gt;&#xA;&lt;li&gt;ed, vent, i&lt;strong&gt;va&lt;/strong&gt; &lt;sup&gt;&lt;a href=&quot;http://dictionary.reference.com/browse/iva?s=t&quot; rel=&quot;nofollow&quot;&gt;src&lt;/a&gt;&lt;/sup&gt; @ guesthouses, &lt;strong&gt;in&lt;/strong&gt;, neocon, done, retract, rabs, non&lt;/li&gt;&#xA;&lt;li&gt;ed, iva @ guesthouses, in&lt;strong&gt;vent&lt;/strong&gt;, neocon, done, retract, rabs, non&lt;/li&gt;&#xA;&lt;li&gt;ed, &lt;strong&gt;a&lt;/strong&gt; @ guesthouses, invent, neocon, done, retract, rabs, non, &lt;strong&gt;IV&lt;/strong&gt; &lt;sup&gt;&lt;a href=&quot;http://dictionary.reference.com/browse/iv?s=t&quot; rel=&quot;nofollow&quot;&gt;src&lt;/a&gt;&lt;/sup&gt;&lt;/li&gt;&#xA;&lt;li&gt;ed @ guesthouses, invent, neocon, done, retract, rabs, &lt;strong&gt;a&lt;/strong&gt;non, IV&lt;/li&gt;&#xA;&lt;li&gt;@ guesthouses, invent&lt;strong&gt;ed&lt;/strong&gt;, neocon, done, retract, rabs, anon, IV&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If &lt;strong&gt;neo&lt;/strong&gt; isn't counted, this is where I'm at&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;34&quot;&gt;&#xA;&lt;li&gt;&lt;strong&gt;eon&lt;/strong&gt;, ed, vent, i @ guesthouses, vain, cont, done, retract, rabs, &lt;strong&gt;non&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;eon, ed, vent, i&lt;strong&gt;va&lt;/strong&gt; &lt;sup&gt;&lt;a href=&quot;http://dictionary.reference.com/browse/iva?s=t&quot; rel=&quot;nofollow&quot;&gt;src&lt;/a&gt;&lt;/sup&gt; @ guesthouses, &lt;strong&gt;in&lt;/strong&gt;, con, done, retract, rabs, non&lt;/li&gt;&#xA;&lt;li&gt;eon, ed, iva @ guesthouses, in&lt;strong&gt;vent&lt;/strong&gt;, con, done, retract, rabs, non&lt;/li&gt;&#xA;&lt;li&gt;eon, ed, &lt;strong&gt;IV&lt;/strong&gt; &lt;sup&gt;&lt;a href=&quot;http://dictionary.reference.com/browse/iv?s=t&quot; rel=&quot;nofollow&quot;&gt;src&lt;/a&gt;&lt;/sup&gt; @ guesthouses, invent, con, done, retract, rabs, non, &lt;strong&gt;a&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;ed, IV @ guesthouses, invent, con, done, retract, rabs, non, a&lt;strong&gt;eon&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;IV @ guesthouses, invent&lt;strong&gt;ed&lt;/strong&gt;, con, done, retract, rabs, non, aeon&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Just gotta figure out how to get that &lt;em&gt;V&lt;/em&gt; over there and then make &lt;strong&gt;icon&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditDate="2015-10-09T15:05:18.133" LastActivityDate="2015-10-09T15:05:18.133" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="22815" PostTypeId="1" AcceptedAnswerId="22817" CreationDate="2015-10-02T14:53:24.270" Score="5" ViewCount="292" Body="&lt;p&gt;The &lt;strong&gt;Association for the Construction, Development and Wellbeing of Nuclear Power Plants&lt;/strong&gt; (or simply &lt;strong&gt;ACDWNPP&lt;/strong&gt; for short) plans to build a new nuclear power plant in Checkerboard City. As you all surely know, the area of Checkerboard City is divided into 144 square  blocks that are arranged in the form of a 12x12 checkerboard. For building the power plant, the ACDWNPP needs to purchase 12 of these blocks that form a 3x4 or 4x3 rectangular region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The town council of Checkerboard City wants to protect Checkerboard City against this nuclear threat. The council decides to purchase a small number of blocks so that there remains no possibility for the ACDWNPP to get a 3x4 or 4x3 rectangular region (from the remaining blocks).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;br&gt;&#xA;  What is the smallest possible number of blocks that the town council needs to purchase?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-10-02T15:33:38.970" Title="Checkerboard City versus the ACDWNPP" Tags="|mathematics|combinatorics|checkerboard|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22816" PostTypeId="1" AcceptedAnswerId="22823" CreationDate="2015-10-02T15:17:20.207" Score="2" ViewCount="275" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;We're a real party people,&lt;br&gt;&#xA;  An American creation;&lt;br&gt;&#xA;  Our seven parts combined&lt;br&gt;&#xA;  Became a musical sensation.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who are we?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-10-02T17:51:40.113" Title="A Test of Musical Knowledge" Tags="|riddle|rhyme|knowledge|music|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22817" PostTypeId="2" ParentId="22815" CreationDate="2015-10-02T15:27:24.723" Score="7" Body="&lt;p&gt;The city council needs to purchase &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 12 blocks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that, we can divide the $12 \times 12$ board into 12, $3\times 4$ region. Each of these regions needs a block, otherwise the ACDWNPP can buy that region. So, we need at least 12 blocks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A solution with 12 blocks is given below.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/BUyxUm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BUyxUm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15544" LastActivityDate="2015-10-02T15:27:24.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22818" PostTypeId="2" ParentId="22815" CreationDate="2015-10-02T15:33:38.970" Score="3" Body="&lt;p&gt;Here's a solution requiring 13 blocks; I doubt that it's optimal.  I'm sure there's a nice mathematical solution to the problem, but I haven't figured it out.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/ftjV7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ftjV7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2015-10-02T15:33:38.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22819" PostTypeId="1" CreationDate="2015-10-02T15:53:50.077" Score="3" ViewCount="917" Body="&lt;p&gt;Premise: These are like the knight-knave puzzles; only now we are visiting insane asylums, where there are only patients and doctors. Moreover, each of them are either insane or sane. Sane inhabitants of an asylum are entirely accurate in their beliefs (they believe all and only true statements and disbelief all and only false ones), whereas insane inhabitants are totally inaccurate in their beliefs. Nonetheless they are all honest--whatever they say, they believe. You may reason as you like, but you must write down your reasoning. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions: You visit an asylum and ask an inhabitant “Are you a patient?” The inhabitant replies “I believe so.” Is anything necessarily wrong with this asylum? &lt;/p&gt;&#xA;" OwnerUserId="16674" LastEditorUserId="15544" LastEditDate="2015-10-02T17:10:55.007" LastActivityDate="2015-10-02T22:10:57.277" Title="Similar to knight/knave puzzle: Insane asylum" Tags="|logical-deduction|word-problem|story|liars|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22820" PostTypeId="2" ParentId="22819" CreationDate="2015-10-02T16:25:31.643" Score="5" Body="&lt;p&gt;Not necessarily. That is a reply that could be made by an insane patient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anything an insane inhabitant says is false. When an insane inhabitant says &quot;I believe [X]&quot;, then the truth is that he does not believe [X]. And since an insane inhabitant disbelieves only true statements, this means that [X] is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a more symbolic representation,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;insane_believes(X) -&amp;gt; !X&#xA;insane_person_says(X) -&amp;gt; insane_believes(X)&#xA;Therefore&#xA;insane_person_says(insane_believes(X)) -&amp;gt; insane_person_says(!X) -&amp;gt; !!X -&amp;gt; X&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Therefore, the response you hear to your question could come from&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a sane patient&lt;/li&gt;&#xA;&lt;li&gt;an insane patient&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The former is an injustice, since sane people should not be patients. But the latter is perfectly normal for an asylum. So there is not anything necessarily wrong with the asylum.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If the inhabitant had instead replied &quot;I am a patient&quot;, then he could be either a sane patient or an insane doctor. Both of these are injustices, so it is a good thing that is not what the inhabitant said.&lt;/p&gt;&#xA;" OwnerUserId="1497" LastEditorUserId="1497" LastEditDate="2015-10-02T16:38:29.287" LastActivityDate="2015-10-02T16:38:29.287" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="22821" PostTypeId="2" ParentId="14448" CreationDate="2015-10-02T17:34:33.663" Score="1" Body="&lt;p&gt;Some lateral thinking answers:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Turn the light on.&lt;/li&gt;&#xA;&lt;li&gt;Put on your night vision goggles.&lt;/li&gt;&#xA;&lt;li&gt;Use your smartphone as a flashlight. &lt;/li&gt;&#xA;&lt;li&gt;Start a fire.&lt;/li&gt;&#xA;&lt;li&gt;Use your well trained cat to find the coins.&lt;/li&gt;&#xA;&lt;li&gt;Open the door and use the sunlight.&lt;/li&gt;&#xA;&lt;li&gt;Melt all the coins and reforge two coins with tails on both sides.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11926" LastEditorUserId="11926" LastEditDate="2015-10-02T17:48:29.483" LastActivityDate="2015-10-02T17:48:29.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22822" PostTypeId="2" ParentId="14450" CreationDate="2015-10-02T17:42:41.630" Score="0" Body="&lt;p&gt;Why not use force?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Either you will give me the gold coin OR I will commit suicide.&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since the demon is possessing you, it must comply, or perish alongside you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11926" LastEditorUserId="11475" LastEditDate="2015-10-02T17:53:16.203" LastActivityDate="2015-10-02T17:53:16.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22823" PostTypeId="2" ParentId="22816" CreationDate="2015-10-02T17:51:40.113" Score="8" Body="&lt;p&gt;I think you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the seven &lt;a href=&quot;https://en.wikipedia.org/wiki/Founding_Fathers_of_the_United_States&quot;&gt;Founding Fathers&lt;/a&gt;:&lt;br&gt;&lt;br&gt;John Adams, Benjamin Franklin, Alexander Hamilton, John Jay, Thomas Jefferson, James Madison, and George Washington&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;We're a real party people,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They advocated political parties (Tea Party).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An American creation;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They founded America.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Our seven parts combined&lt;br&gt;&#xA;Became a musical sensation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Two musicals (&lt;a href=&quot;https://en.wikipedia.org/wiki/1776_(musical)&quot;&gt;&lt;em&gt;1776&lt;/em&gt;&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Hamilton_(musical)&quot;&gt;&lt;em&gt;Hamilton&lt;/em&gt;&lt;/a&gt;) were written about the Founding Fathers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9871" LastActivityDate="2015-10-02T17:51:40.113" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22824" PostTypeId="2" ParentId="22819" CreationDate="2015-10-02T17:55:12.227" Score="1" Body="&lt;p&gt;&lt;em&gt;This is further to @Kevin's answer; I was having trouble understanding it, and thought I would reason it out more.  If this helps you understand it, please upvote Kevin's answer, as he actually figured it out.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;No, there is nothing &lt;em&gt;necessarily&lt;/em&gt; wrong the asylum.  The inhabitant could be an insane patient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inhabitant believes the statement, &lt;strong&gt;&lt;em&gt;I believe I am a patient.&lt;/em&gt;&lt;/strong&gt;  Note that this is different than saying she believes the statement, &lt;strong&gt;&lt;em&gt;I am a patient.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If she is an insane patient, then she is saying that she believes the statement, &lt;strong&gt;&lt;em&gt;I believe I am a patient&lt;/em&gt;&lt;/strong&gt; to be true.  Since anything an insane person believes is necessarily false, then the statement &lt;strong&gt;&lt;em&gt;I believe I am a patient&lt;/em&gt;&lt;/strong&gt; is false.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the inverse is actually true:  &lt;strong&gt;&lt;em&gt;I believe I am not a patient.&lt;/em&gt;&lt;/strong&gt;  This means that she believes the statement &lt;strong&gt;&lt;em&gt;I am not a patient&lt;/em&gt;&lt;/strong&gt; to be true.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since anything an insane person believes is necessarily false, then the statement &lt;strong&gt;&lt;em&gt;I am not a patient&lt;/em&gt;&lt;/strong&gt; is false, and she is in fact a patient.&lt;/p&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2015-10-02T17:55:12.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22825" PostTypeId="1" AcceptedAnswerId="23315" CreationDate="2015-10-02T18:00:04.963" Score="18" ViewCount="1976" Body="&lt;p&gt;&lt;p&gt;A life inside the confining dark&lt;br&gt;&#xA;Their work unseen, but try to hark&lt;br&gt;&#xA;Hark to what they say, get steady&lt;br&gt;&#xA;Go ahead and free one already&#xA;&lt;p&gt;From my experience I can tell&lt;br&gt;&#xA;I'm very close to my defeat&lt;br&gt;&#xA;Maybe a clever mind can help&lt;br&gt;&#xA;Which one is it I urgently need?&#xA;&lt;hr&gt;&#xA;Hint for those who don't know what the riddle is asking:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The first verse describes a group of entities of which the second verse wants you to choose the right one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint for those who need to narrow down their number of possible solutions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Use your bright mind :P Plus, the solution is available in a figurative and obvious way in order to confirm or deny your guess.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint for everyone else:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I recommend using a computer for this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5106" LastEditorUserId="5106" LastEditDate="2015-11-05T22:03:35.253" LastActivityDate="2016-03-28T04:25:09.837" Title="Harder than hard" Tags="|riddle|word|" AnswerCount="8" CommentCount="16" ClosedDate="2015-11-04T15:49:18.120" ContentLicense="CC BY-SA 3.0" />
  <row Id="22826" PostTypeId="2" ParentId="22693" CreationDate="2015-10-02T18:15:32.553" Score="4" Body="&lt;p&gt;How about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;MEAN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be defining&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;em&gt;mean&lt;/em&gt; can be used to describe a meaning of sth i.e. define its &lt;em&gt;meaning&lt;/em&gt;. Also &lt;em&gt;fussy&lt;/em&gt; is one possible use of &lt;em&gt;mean&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I don't give much away.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;By calculating the average you don't get much other information besides that. Also &lt;em&gt;mean&lt;/em&gt; can be used for people who don't like sharing money/things.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be pretty good,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It can be good (or not :P) when talking about grades/marks. &lt;strong&gt;Edit:&lt;/strong&gt; colloquial for &lt;em&gt;excellent&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Except when I'm nasty.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;mean can also mean &lt;em&gt;nasty&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm never at the extremes,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Due to the way of how you calculate an &lt;em&gt;average&lt;/em&gt; (math.) it never can be at the extremes (given there is more than one value :p )&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But you'll find me with a switch at the end.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Oh hell..... AMEN :D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5106" LastEditorUserId="5106" LastEditDate="2015-10-02T23:07:32.477" LastActivityDate="2015-10-02T23:07:32.477" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22827" PostTypeId="1" AcceptedAnswerId="22831" CreationDate="2015-10-02T18:25:25.137" Score="8" ViewCount="652" Body="&lt;p&gt;A mad scientist uses ethically questionable methods to study the spatial cognition of ants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last time, he left his lab ant in a cubic room and filled it with painfully intoxicating gas, capable of killing an adult ant in exactly twenty minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The room was perfectly sealed except for a small hole in the (diametrically) opposite vertex from that on which the ant was sitting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, using it's notorious spatial intuition, the ant was able in no time to compute the shortest path and run for it's life. Unfortunately the ant could not survive for much longer after escaping, due to permanent spiracle and kidney toxicity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even then, the scientist's controversial experiment had repercussions all over the world, because it proved that the solution previously computed by mathematicians was indeed optimal, and no path could be shorter than $\sqrt5$ times the side length of the box. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mad scientist then made a new room to defy any kind of geometrical intuition from his ants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This time, he created a 98-dimensional hypercube which could only be walked along the inside of it's 97-dimensional hypersurface. Then he filled it's entire $1\;googol$$\:\mathrm{dm^2.m^{96}}$ hypervolume with a toxic hypergas capable of killing an ant in the same amount of time as before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the next lab ant noticed that with simple spatial reasoning and dimensional analogy it's again possible to derive the shortest path, and started in no time to follow it, at maximal speed, to the opposite vertex of the hypercube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very unluckily for the ant, it died exactly as it came out of the hyper-room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ant was very smart for figuring out the optimal solution, but the scientist was soon asking himself if the ant died because it was too slow, or the time was too little.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;According to this experiment, how fast was the ant?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(Answer in centimeters per second, with at least two decimal digits)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="14219" LastEditorUserId="14219" LastEditDate="2015-10-03T14:43:16.227" LastActivityDate="2015-10-03T14:43:16.227" Title="This ant sure is smart. But how fast is it?" Tags="|mathematics|geometry|word-problem|optimization|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="22828" PostTypeId="1" CreationDate="2015-10-02T19:02:36.237" Score="2" ViewCount="1824" Body="&lt;p&gt;I met a group of people, but no one told me their names. They just told me these things.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My name contains myself&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;My name contains something to put on toast&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;My name contains a ship able to withstand impossible floods!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;My name contains a hiding place for animals&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;My name contains human creativity  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;My name contains an indication of&#xA;  point or place&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;My name contains an acronym where abusers go for help&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;My name contains the center of the Earth&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;My name contains the reamins of a fire&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;My name contains a boobie catcher&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;My name contains what corn comes on&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;My name contains a couple thousand pounds&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;My name contains something that is different from what is expected&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;My name contains an acronym of an agency created by Franklin D.&#xA;  Roosevelt&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;My name contains an identifying mark&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;My name contains another word for steal&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;My name contains a cars fingerprint&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can you help me identify their first names?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the names are common names that can be found in America. &#xA;I understand there can be multiple answers for some, I'll take any that fit the clue. (Or multiple if you want to get crazy!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;My name contains an indication of point or place&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Think of it from more of a third grade English perspective &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;My name contains an identifying mark&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Think of it from a farmers perspective&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16273" LastEditorUserId="16273" LastEditDate="2015-10-02T20:05:47.680" LastActivityDate="2015-10-02T20:32:13.727" Title="What is my first name?" Tags="|riddle|" AnswerCount="1" CommentCount="0" ClosedDate="2015-10-02T21:01:35.843" ContentLicense="CC BY-SA 3.0" />
  <row Id="22829" PostTypeId="2" ParentId="22828" CreationDate="2015-10-02T19:13:04.473" Score="4" Body="&lt;p&gt;My name contains myself&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Me&lt;/strong&gt;l / &lt;strong&gt;Me&lt;/strong&gt;lissa&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My name contains something to put on toast&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Jam&lt;/strong&gt;es (&lt;strong&gt;Butter&lt;/strong&gt;cup is another possibility)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My name contains a ship able to withstand impossible floods!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cl&lt;strong&gt;ark&lt;/strong&gt; / M&lt;strong&gt;ark&lt;/strong&gt; (&lt;em&gt;Noah's ark was built to withstand impossible floods&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My name contains a hiding place for animals&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ai&lt;strong&gt;den&lt;/strong&gt; / &lt;strong&gt;Den&lt;/strong&gt;ise&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My name contains human creativity&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Art&lt;/strong&gt;hur / C&lt;strong&gt;art&lt;/strong&gt;er&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My name contains an indication of point or place&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ale&lt;strong&gt;x&lt;/strong&gt; / Ma&lt;strong&gt;x&lt;/strong&gt; (&lt;em&gt;X marks the spot&lt;/em&gt;)&#xA; &lt;strong&gt;There&lt;/strong&gt;sa / &lt;strong&gt;There&lt;/strong&gt;se&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My name contains an acronym where abusers go for help&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Is&lt;strong&gt;aa&lt;/strong&gt;c / &lt;strong&gt;Aa&lt;/strong&gt;ron (&lt;em&gt;AA = Alcoholics Anonymous&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My name contains the center of the Earth&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Core&lt;/strong&gt;y / &lt;strong&gt;Core&lt;/strong&gt;en&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My name contains the remains of a fire&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Ash&lt;/strong&gt;ley / D&lt;strong&gt;ash&lt;/strong&gt;ell&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My name contains a boobie catcher&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bar&lt;strong&gt;bra&lt;/strong&gt; / &lt;strong&gt;Bra&lt;/strong&gt;dley / &lt;strong&gt;Bra&lt;/strong&gt;ndon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My name contains what corn comes on&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ja&lt;strong&gt;cob&lt;/strong&gt; / &lt;strong&gt;Cob&lt;/strong&gt;y&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My name contains a couple thousand pounds&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Al&lt;strong&gt;ton&lt;/strong&gt; / &lt;strong&gt;Ton&lt;/strong&gt;y (&lt;em&gt;A ton is 2,000lbs&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My name contains something that is different from what is expected&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; T&lt;strong&gt;odd&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My name contains an acronym of an agency created by Franklin D. Roosevelt&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Rea&lt;/strong&gt;gan / B&lt;strong&gt;rea&lt;/strong&gt;nne (&lt;em&gt;Roosevelt's &quot;Alphabet Agencies&quot; include REA (Rural Electrification Administration) (now Rural Utilities Service)&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My name contains an identifying mark&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; O&lt;strong&gt;scar&lt;/strong&gt; / &lt;strong&gt;Scar&lt;/strong&gt;lett (&lt;em&gt;A scar is an identifying mark&lt;/em&gt;)&lt;br&gt;&#xA; &lt;strong&gt;Brand&lt;/strong&gt;on&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My name contains another word for steal&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Rob&lt;/strong&gt;ert / &lt;strong&gt;Rob&lt;/strong&gt;in&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My name contains a cars fingerprint&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Vin&lt;/strong&gt;cent / Mel&lt;strong&gt;vin&lt;/strong&gt;/ Ke&lt;strong&gt;vin&lt;/strong&gt;  &lt;em&gt;(VIN = Vehicle Identification Number)&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13925" LastEditorUserId="9523" LastEditDate="2015-10-02T20:32:13.727" LastActivityDate="2015-10-02T20:32:13.727" CommentCount="4" CommunityOwnedDate="2015-10-02T19:13:04.473" ContentLicense="CC BY-SA 3.0" />
  <row Id="22830" PostTypeId="1" AcceptedAnswerId="22897" CreationDate="2015-10-02T19:46:54.587" Score="6" ViewCount="1432" Body="&lt;p&gt;Together I am four and five.  Apart, am I four or five?  Often I am written one or one.  Confused, I may be payments for a bard, antique hearing-aids, the price of gas, brightly-burning fat, a damned man's ogle, guidelines for heroism, a tempting meal, a whimsical shine, a predictor of problems, a prenatal jape, or I may tamper with explosives.  What kind of question is this?  Programs and pop-quizzes frequently ask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hints:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;four and five&quot; refers to the number of vowels and consonants in the answer.  &quot;four or five&quot; refers to the number of letters in each of the two words.  The &quot;confused&quot; clues all refer to anagrams of the main answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16131" LastEditorUserId="16131" LastEditDate="2015-10-05T16:43:35.390" LastActivityDate="2015-10-06T13:16:57.080" Title="What kind of question is this?" Tags="|word|anagram|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22831" PostTypeId="2" ParentId="22827" CreationDate="2015-10-02T20:01:20.860" Score="6" Body="&lt;p&gt;First, here is the reason the answer is $\sqrt{5}$ in the three dimensional case. Take one of the rooms vertical walls containing the opposite corner, and fold it down, so it forms a 2 by 1 rectangle with the floor. The ant's optimal path is to walk the diagonal of this rectangle, of length $\sqrt{2^2+1^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a $d$ dimensional cube, imagine folding down one of the walls to form a $(d-1)$ dimensional hyperrectangle with dimensions $2\times 1\times 1\times \dots\times 1$. The ant's optimal path is to walk the diagonal of this rectangle, which is of length $\sqrt{2^2+1^2+\dots+1^2}=\sqrt{d+2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The side length $s$ of the room is given by $s^{98}=10^{100}dm^2m^{96}$$=10^{98}m^{98}$, meaning $s=10$. So that in this case the optimal path length is&#xA;$$&#xA;\ell=\sqrt{98+2}\cdot 10=100\text{ m}&#xA;$$&#xA;Since the ant traverses this length in exactly 20 minutes, his speed must be&#xA;$$&#xA;\frac{100}{20}\text{ m}/\text{min} \approx 8.33\text{ cm}/\text{s}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="14219" LastEditDate="2015-10-03T02:18:42.173" LastActivityDate="2015-10-03T02:18:42.173" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22832" PostTypeId="2" ParentId="22830" CreationDate="2015-10-02T21:43:29.440" Score="-1" Body="&lt;p&gt;This question is...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Convergent&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Meaning...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are very few possible answers. In this case, there is only one (convergent)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This also means that...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The only part in your question that was relevant was the end. The rest were merely red herrings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2015-10-02T21:43:29.440" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22833" PostTypeId="2" ParentId="22819" CreationDate="2015-10-02T22:10:57.277" Score="2" Body="&lt;p&gt;&lt;strong&gt;Given:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Sane person must believe true statements, insane must believe false&#xA;statements &lt;/li&gt;&#xA;&lt;li&gt;Person will accurately say what they believe&lt;/li&gt;&#xA;&lt;li&gt;Person says they believe they are a patient &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sane Doctor&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Does not believe &quot;I am a patient&quot; because it's false&lt;/li&gt;&#xA;&lt;li&gt;Does not believe &quot;I believe I am a patient&quot; because it's false&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Insane Doctor&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Does believe &quot;I am a patient&quot; because it's false&lt;/li&gt;&#xA;&lt;li&gt;Does not believe &quot;I believe I am a patient&quot; because it's true&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sane Patient&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Does believe &quot;I am a patient&quot; because it's true&lt;/li&gt;&#xA;&lt;li&gt;Does believe &quot;I believe I am a patient&quot; because it's true&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Insane Patient&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Does not believe &quot;I am a patient&quot; because it's true&lt;/li&gt;&#xA;&lt;li&gt;Does believe &quot;I believe I am a patient&quot; because it's false&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The person we are talking to must be a patient, but they could either be sane or insane and we have no way of knowing which from the information provided. Also, I don't know what criteria is used to determine if &quot;anything is wrong with the asylum,&quot; but this situation is possible given the information provided.&lt;/p&gt;&#xA;" OwnerUserId="16677" LastActivityDate="2015-10-02T22:10:57.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22834" PostTypeId="2" ParentId="22784" CreationDate="2015-10-02T22:11:44.190" Score="3" Body="&lt;p&gt;Here is a very mathematical one. It works, but is much more complicated than it needs to be. I thought I'd write it down just for fun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fix some sequences of integers $a_n, \ b_n, \ c_n$ so that 1) $a_n, b_n, c_n \leq n$, and 2) for any triple of integers $a,b,c$, there is some $n$ with $a_n = a, \ b_n = b,\ c_n = c$ (this $n$ need not be unique). The plan is now to check, on $n$-th day, if it was the case that prisoner A got steak on day $a_n$, prisoner B on night $b_n$ and prisoner C on night $c_n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The voting rule is now simple. On day $n$, prisoner A votes 1 if and only if he got steak on day $a_n$, and fish on days $b_n$ and $c_n$. Analogous rule applies to B and to C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The prisoners will eventually be set free. Indeed, if they get steak on days $a,\ b$ and $c$ respectively, then they will be set free on the $n$-the day, where $n$ is the first integer with $a_n = a, \ b_n = b,\ c_n = c$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, if two prisoners (say A and B) vote 1 on a given day $n$, then between them on days $a_n,\ b_n,\ c_n$ they only ate 2 steaks. Since the total of 3 steaks was distributed, the third one had to go to C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has a nice added bonus that on the day the prisoners are set free, each of them votes 1.&lt;/p&gt;&#xA;" OwnerUserId="2855" LastEditorUserId="2855" LastEditDate="2015-10-02T22:19:51.893" LastActivityDate="2015-10-02T22:19:51.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22835" PostTypeId="2" ParentId="3401" CreationDate="2015-10-02T23:44:18.680" Score="-1" Body="&lt;p&gt;Joe Z. answered &quot;The answer to the question is $1.19. He has 3 quarters, 4 dimes, and 4 pennies.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other options are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;1 quarter, 9 dimes and 4 pennies... which also equals $1.19&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;1 half dollar, 1 quarter, 4 dimes and 4 pennies... also $1.19&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="16678" LastActivityDate="2015-10-02T23:44:18.680" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22836" PostTypeId="1" CreationDate="2015-10-02T23:56:16.830" Score="6" ViewCount="791" Body="&lt;p&gt;The basis of this puzzle, is based on a common 2D chess puzzle, where you try to fit 8 queens on a chessboard without the queens being able to attack each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extend this concept into a 3D cube (8x8x8... like 8 chess boards on each other), but we do analysis on both vertical and horizontal planes. So no queen can attack each other in any vertical plane or in any horizontal plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the maximum number of queens can you fit into this chess cube without them being able to attack each other. (To describe your solution, feel free to post an image showing each layer in order, or illustrate it with text of x's and o's)&lt;/p&gt;&#xA;" OwnerUserId="16678" LastEditorUserId="8874" LastEditDate="2015-10-03T12:04:16.670" LastActivityDate="2015-11-20T04:48:18.563" Title="3D Chess Cube - Queen Puzzle" Tags="|mathematics|combinatorics|chess|checkerboard|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="22837" PostTypeId="1" AcceptedAnswerId="22847" CreationDate="2015-10-03T00:16:47.313" Score="7" ViewCount="798" Body="&lt;p&gt;Miss Cathy is a psychologist. However, there's a problem. She has very bad short-term memory loss. Although she has this condition, she doesn't let it get in the way of her work. Every time a patient comes for a visit she quickly writes the patient's information, the way they look, their personality. She is also very organized and keeps her patient records arranged alphabetically.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One day, business was slow. She only had five patients that day and coincidentally all the patients knew each other. When she was going to store the patients' records, she noticed that she forgot something very important. She forgot to write the names of the patients! The only thing she wrote were their appearance and personality.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;File #1&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Very fashionista.&lt;br&gt;&#xA;  Wore clothes of different colors.&lt;br&gt;&#xA;  Her dress was very blue like the ocean.&lt;br&gt;&#xA;  Wears high boots the color of mustard.&lt;br&gt;&#xA;  Silky gloves the color of the forest and a very loud hair color of red!&lt;br&gt;&#xA;  Quite a fashion disaster though.&lt;br&gt;&#xA;  She had a lot to say.&lt;br&gt;&#xA;  Very talkative but sometimes she just suddenly stops in the middle of talking.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;File #2&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ginger&lt;br&gt;&#xA;  Freckles&lt;br&gt;&#xA;  Very pointy nose, rarely see that kind of lady around these parts.&lt;br&gt;&#xA;  Wore a long elegant red dress with a tinge of orange.&lt;br&gt;&#xA;  Really angry at the first girl.&lt;br&gt;&#xA;  Said she is much better than her and much more pretty.&lt;br&gt;&#xA;  But when she asks people who they like better, most of the time she loses.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;File #3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Very old man.&lt;br&gt;&#xA;  White hair&lt;br&gt;&#xA;  Wore blue overalls and a white shirt.&lt;br&gt;&#xA;  Age has really played on him.&lt;br&gt;&#xA;  Slow talker.&lt;br&gt;&#xA;  His cellphone rang and he asked me to answer it.&lt;br&gt;&#xA;  Probably not yet used to new technologies.&lt;br&gt;&#xA;  Says that his generation is being overshadowed now with the new generation.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;File #4&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Seems to be a character.&lt;br&gt;&#xA;  Quite large...&lt;br&gt;&#xA;  Said he has three siblings, one that works on automobiles, one that is a younger version of himself, and one that is working as a lifeguard.&lt;br&gt;&#xA;  He says he is one of the pioneers that made a lot of phone calls easier. I don't get it though.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;File #5&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Looks like he's been through a lot of adventures&lt;br&gt;&#xA;  Seems like he won't get lost.&lt;br&gt;&#xA;  Took an apple from his backpack and took a bite.&lt;br&gt;&#xA;  Proud of his energy... Surely, I can see that he is quite energetic  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can you help Miss Cathy identify who were her patients for that day?&lt;/p&gt;&#xA;" OwnerUserId="16566" LastEditorUserId="16566" LastEditDate="2015-10-03T13:44:30.750" LastActivityDate="2015-10-05T01:11:06.477" Title="Miss Cathy's Patients" Tags="|story|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22839" PostTypeId="1" AcceptedAnswerId="22841" CreationDate="2015-10-03T10:20:16.180" Score="4" ViewCount="1264" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vl0WZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vl0WZ.png&quot; alt=&quot;![enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the picture above, all the numbers 1-12 are given, using only three nines. Now, I want to make a similar watch, but using fours instead of nines.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you make a list of the numbers 1-12 using &lt;strong&gt;exactly&lt;/strong&gt; three fours (for each number). So, for instance, you can't do 4+4 to get 8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Combining two 4s to make 44 is not legal&lt;/li&gt;&#xA;&lt;li&gt;Repeating / recurring decimal representation is not legal (as used to get seven in the example watch).&lt;/li&gt;&#xA;&lt;li&gt;Flooring / rounding is not legal&lt;/li&gt;&#xA;&lt;li&gt;The answer must be exact. &lt;/li&gt;&#xA;&lt;li&gt;Parentheses are of course OK&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You are free to use other functions, but it's possible to do this using only &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_mathematical_functions&quot; rel=&quot;nofollow noreferrer&quot;&gt;elementary functions&lt;/a&gt; and the factorial &lt;a href=&quot;https://en.wikipedia.org/wiki/Factorial&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;code&gt;!&lt;/code&gt;&lt;/a&gt; (no other &quot;Special functions&quot; (as they are called in the wiki-article)).&lt;/p&gt;&#xA;" OwnerUserId="5517" LastEditorUserId="15544" LastEditDate="2015-10-03T15:09:03.257" LastActivityDate="2015-10-08T19:12:40.153" Title="May the fours be with you!" Tags="|mathematics|calculation-puzzle|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22841" PostTypeId="2" ParentId="22839" CreationDate="2015-10-03T10:40:49.770" Score="3" Body="&lt;p&gt;Using only 5 ops, $+,\ -,\ \times,\ \div,\ \sqrt{}$, I can get all (although it might be debatable whether the radix point counts as one)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$1=\sqrt4 -{ 4 \over 4}$&lt;/li&gt;&#xA;&lt;li&gt;$2={4+4 \over 4} $&lt;/li&gt;&#xA;&lt;li&gt;$3=4 - { 4 \over 4}$&lt;/li&gt;&#xA;&lt;li&gt;$4=4+4-4$&lt;/li&gt;&#xA;&lt;li&gt;$5=4+{4 \over 4}$&lt;/li&gt;&#xA;&lt;li&gt;$6=4 + 4 -\sqrt4 $&lt;/li&gt;&#xA;&lt;li&gt;$7={\sqrt4 \over .4}+\sqrt4={4!+4 \over 4}$   [Solution without &lt;code&gt;!&lt;/code&gt;, thanks to Somo145]&lt;/li&gt;&#xA;&lt;li&gt;$8=4 + \sqrt4 +\sqrt4$&lt;/li&gt;&#xA;&lt;li&gt;$9=(4-.4)/.4$&lt;/li&gt;&#xA;&lt;li&gt;$10=4 + 4 +\sqrt4$&lt;/li&gt;&#xA;&lt;li&gt;$11=(4+.4)/.4={4! - \sqrt4 \over \sqrt 4}$&lt;/li&gt;&#xA;&lt;li&gt;$12=4 + 4 +4$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In fact, you can get rid of $-$ and $\times$ and do it with three $\big(+,\div,\sqrt{} \big)$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$1={\sqrt4 + \sqrt4 \over 4}$&lt;/li&gt;&#xA;&lt;li&gt;$2={4+4 \over 4} $&lt;/li&gt;&#xA;&lt;li&gt;$3=\sqrt4 + { 4 \over 4}$&lt;/li&gt;&#xA;&lt;li&gt;$4={4+4 \over \sqrt4}$&lt;/li&gt;&#xA;&lt;li&gt;$5=4+{4 \over 4}$&lt;/li&gt;&#xA;&lt;li&gt;$6=\sqrt4 + \sqrt4 +\sqrt4 $&lt;/li&gt;&#xA;&lt;li&gt;$7={\sqrt4 \over .4}+\sqrt4 $  [Somo145]  &lt;/li&gt;&#xA;&lt;li&gt;$8=4 + \sqrt4 +\sqrt4$&lt;/li&gt;&#xA;&lt;li&gt;$9={\sqrt4 \over .4}+4 $ &lt;/li&gt;&#xA;&lt;li&gt;$10=4 + 4 +\sqrt4$&lt;/li&gt;&#xA;&lt;li&gt;$11=(4+.4)/.4$&lt;/li&gt;&#xA;&lt;li&gt;$12=4 + 4 +4$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="15544" LastEditorUserId="5781" LastEditDate="2015-10-08T19:12:40.153" LastActivityDate="2015-10-08T19:12:40.153" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22842" PostTypeId="1" AcceptedAnswerId="22843" CreationDate="2015-10-03T10:41:41.863" Score="15" ViewCount="1119" Body="&lt;p&gt;You're standing outside your apartment building after a late night out, with perhaps one beer too many, and you realize you have completely forgotten the code to get in. Luckily, you're a mathematical genius, and somehow that part of the brain is completely unaffected by the alcohol, so you decide to find the optimal strategy to get in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The code is entered on a standard 0-9 keypad, and you know that it is a four digit code. Entering the correct sequence of digits will open the door. When entering a sequence, every four digit sub sequence will be evaluated by the security system. I.e. entering &lt;code&gt;195638&lt;/code&gt; will evaluate the following three sequences: &lt;code&gt;1956&lt;/code&gt;, &lt;code&gt;9563&lt;/code&gt;, and &lt;code&gt;5638&lt;/code&gt;, so it is a waste of time to try each four digit sequence one by one. &lt;br&gt;&lt;br&gt;&#xA;&lt;strong&gt;Question: What is the smallest number of keypresses needed to try every sequence between &lt;code&gt;0000&lt;/code&gt; and &lt;code&gt;9999&lt;/code&gt;&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="4624" LastEditorUserId="8874" LastEditDate="2015-10-08T18:18:55.713" LastActivityDate="2019-03-01T20:33:29.040" Title="One beer too many" Tags="|mathematics|calculation-puzzle|combinatorics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22843" PostTypeId="2" ParentId="22842" CreationDate="2015-10-03T11:07:25.880" Score="15" Body="&lt;p&gt;The &lt;strong&gt;De Bruijn sequence&lt;/strong&gt; $B(k,n)$ is a cyclic sequence over an alphabet of size $k$ that contains every possible word of length $n$ exactly once;&#xA;see the &lt;a href=&quot;https://en.wikipedia.org/wiki/De_Bruijn_sequence&quot;&gt;wikipedia&lt;/a&gt; for more information.&lt;br&gt;&#xA;The length of such a De Bruijn sequence $B(k,n)$ is $k^n$, that is, it equals the number of words of length $n$ over an alphabet of size $k$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The De Bruijn sequence $B(10,4)$ is over an alphabet of size 10 (that is, the ten digits 0,1,2,$\ldots$,9), and it contains every possible word of length 4 (that is, every 4-digit number). The length of $B(10,4)$ is $10^4$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The puzzle now does not ask about a &lt;em&gt;cyclic&lt;/em&gt; sequence, but about an ordinary linear sequence. If we follow the cyclic De Bruijn sequence $B(10,4)$ around the cycle, we will encounter every 4-digit number exactly once. This takes $10^4$ digits plus an additional 3 digits at the end (where the cycle closes, while the linear sequence remains open).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the answer is $10^4+3=10003$ keypresses.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-10-03T11:07:25.880" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22844" PostTypeId="1" AcceptedAnswerId="22855" CreationDate="2015-10-03T11:25:24.500" Score="2" ViewCount="244" Body="&lt;p&gt;Find the next term in the following sequence: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$1/5$, $~~2/8$, $~~2/12$, $~~4/18$, $~~2/24$,  $~~?$ &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's my first question so it might come off as a bit too easy. &lt;/p&gt;&#xA;" OwnerUserId="10343" LastEditorUserId="8874" LastEditDate="2015-10-03T17:29:16.087" LastActivityDate="2015-10-03T17:29:16.087" Title="Find the next term: 1/5, 2/8, 2/12, etc" Tags="|mathematics|number-sequence|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22846" PostTypeId="1" AcceptedAnswerId="22884" CreationDate="2015-10-03T13:33:16.097" Score="16" ViewCount="1585" Body="&lt;p&gt;Professor Halfbrain has spent his last weekend with analyzing &lt;span class=&quot;math-container&quot;&gt;$n\times n$&lt;/span&gt; chessboards. Halfbrain says that a subset &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; of squares on such a chessboard is  &lt;em&gt;queen-connected&lt;/em&gt;, if a chess queen can move around on &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; and reach all the squares in &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; while making only legal queen moves. (The queen is only allowed to move from squares in &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; to squares in &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt;, but the move may pass over other squares that are not necessarily in &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt;. Every square in &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; may be visited many times.)&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Professor Halfbrain's first theorem&lt;/strong&gt;: If the squares of an &lt;span class=&quot;math-container&quot;&gt;$n\times n$&lt;/span&gt; chessboard with &lt;span class=&quot;math-container&quot;&gt;$n\ge2$&lt;/span&gt; are painted red, then these red squares form a queen-connected set.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Professor Halfbrain's second theorem&lt;/strong&gt;: If the squares of an &lt;span class=&quot;math-container&quot;&gt;$n\times n$&lt;/span&gt; chessboard with &lt;span class=&quot;math-container&quot;&gt;$n\ge2$&lt;/span&gt; are painted red and blue, then the red squares or the blue squares (or both) form a queen-connected set.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Professor Halfbrain's third theorem&lt;/strong&gt;: If the squares of an &lt;span class=&quot;math-container&quot;&gt;$n\times n$&lt;/span&gt; chessboard with &lt;span class=&quot;math-container&quot;&gt;$n\ge2$&lt;/span&gt; are painted red and blue and yellow, then the red squares or the blue squares or the yellow squares (or two of these sets, or all three sets) form a queen-connected set.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Professor Halfbrain's fourth theorem&lt;/strong&gt;: If the squares of an &lt;span class=&quot;math-container&quot;&gt;$n\times n$&lt;/span&gt; chessboard with &lt;span class=&quot;math-container&quot;&gt;$n\ge2$&lt;/span&gt; are painted red, blue, yellow and green, then at least one of the color classes must form a queen-connected set.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;We all know Professor Halfbrain as an honorable and trustworthy mathematician, but even the best mathematician may sometimes be mistaken.&#xA;Hence the question: Which of the Professor's four theorems do actually hold true?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-10-05T09:07:40.780" Title="Professor Halfbrain's chessboard theorems" Tags="|mathematics|combinatorics|checkerboard|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22847" PostTypeId="2" ParentId="22837" CreationDate="2015-10-03T14:08:32.457" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; These are browsers. The descriptions are of their logos.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first one is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Chrome&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Firefox&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The third is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Internet Explorer (IE and Edge logos are blue and white.) Question assumes Microsoft doesn't have a phone? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The fourth is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Opera_(web_browser)&quot; rel=&quot;nofollow&quot;&gt;Opera&lt;/a&gt;. &quot;The siblings are Opera Mobile, Opera Mini, and Opera Coast. It came up with Speed Dial browsing, and its symbol is a big O (large character).&quot; –  Illyasviel&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The fifth is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Safari.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1509" LastEditorUserId="11876" LastEditDate="2015-10-05T01:11:06.477" LastActivityDate="2015-10-05T01:11:06.477" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22848" PostTypeId="1" AcceptedAnswerId="22861" CreationDate="2015-10-03T14:37:34.810" Score="6" ViewCount="266" Body="&lt;p&gt;This is for a robotics solver using a real cube. How many facelets must the machine check to figure out the position of everything in a scrambled cube? &lt;/p&gt;&#xA;" OwnerUserId="16693" LastEditorUserId="16693" LastEditDate="2015-10-03T17:21:14.170" LastActivityDate="2015-10-03T19:54:28.040" Title="Minimum number of stickers to inspect in a Rubik cube to define a scrambled position?" Tags="|visual|rubiks-cube|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22849" PostTypeId="2" ParentId="22846" CreationDate="2015-10-03T15:22:56.103" Score="10" Body="&lt;p&gt;The first theorem is trivial &amp;mdash; the queen can obviously travel onto all the squares of a grid.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here's a counterexample for the fourth theorem, using a 4&amp;times;4 grid:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 2 3 4&#xA;3 4 1 2&#xA;2 1 4 3&#xA;4 3 2 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For the colours 1 and 4, you can't reach the inner squares from the outer squares, and for the colours 2 and 3, you can't reach &lt;em&gt;any&lt;/em&gt; of the squares from another square of the same colour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fun fact: This arrangement will work on an infinite grid when tiled. Each of the colours will be divided into two queen-connected subsets.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2015-10-03T19:55:49.543" LastActivityDate="2015-10-03T19:55:49.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22850" PostTypeId="2" ParentId="22844" CreationDate="2015-10-03T16:02:23.453" Score="1" Body="&lt;p&gt;Just a guess..&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\frac{2}{28}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sum of numerator and denominator of a term equals difference between numerator and denominator of the next term.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This probably implies that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The next term is of the type $\frac{a}{a+26}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I don't know why &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm guessing $a=2$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A more complete answer will be required.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-10-03T16:02:23.453" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22852" PostTypeId="1" AcceptedAnswerId="22991" CreationDate="2015-10-03T16:26:12.033" Score="7" ViewCount="813" Body="&lt;p&gt;There is a race which has a truth gene. It consists of 3 alleles, namely T, F and O. A person inherits one allele from each of his parents, hence he/she has always exactly two alleles. For example, if the parents have alleles TF and FO, the child may have TF, TO, FF or FO. Genes are distributed randomly in the initial species. They behave as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;TT&lt;/strong&gt; - Always speaks the truth, remains silent if not possible.&lt;sup&gt;1&lt;/sup&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;TO&lt;/strong&gt; - Same as TT.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;FF&lt;/strong&gt; - Always lies, remains silent if not possible.&lt;sup&gt;1&lt;/sup&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;FO&lt;/strong&gt; - Same as FF.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;TF&lt;/strong&gt; - One of T and F is the dominant allele.&lt;sup&gt;2&lt;/sup&gt; The individual behaves as per the dominant gene.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;OO&lt;/strong&gt; - Children die at birth.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clarifications:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;There exists certain questions where it is not possible to give a truthful answer.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This is a concept in genetics. For example, if T is the dominant allele, the person always speaks the truth. If F is dominant, he always lies. The dominant allele is the same for an entire species. So, if your father has F dominant in TF, and you are also a TF by birth, you will also have F dominant.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are no gene defects. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Assume that every individual has children with exactly one opposite sex partner.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h2&gt;Question&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;You have just found &lt;em&gt;all&lt;/em&gt; $8$ great-grandparents of a person named Mr.X. Each of them knows correctly the nature (lying or truth-telling) of each of their own direct descendants (but not each other's or Mr.X's) and themselves. They do not know anybody's genetic makeup. They do not know which allele is dominant, and neither do you. Nobody (including the great grand parents) knows the nature of Mr. X.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;You must ask them a series of yes/no questions to determine the sex and nature (truth-telling/lying) of Mr.X, or else conclude that such a task is unaccomplishable for the given scenario. The maximum questions you require (worst case scenario) is $n$. Minimize $n$.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4269" LastEditDate="2015-10-09T15:00:01.170" LastActivityDate="2015-10-09T15:00:01.170" Title="Born to lie, to live, to die" Tags="|mathematics|logical-deduction|probability|liars|optimization|" AnswerCount="4" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="22853" PostTypeId="1" CreationDate="2015-10-03T16:36:13.573" Score="6" ViewCount="465" Body="&lt;p&gt;Both players are dealt $26$ cards from a regular $52$ card deck. He discards $1$ and makes $5$ hands (from the remaining $25$) in any order, each with $5$ cards. His score is given by the sum of points scored by each of his individual hands as per the following table:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Royal flush - 10&lt;/li&gt;&#xA;&lt;li&gt;Straight flush - 9&lt;/li&gt;&#xA;&lt;li&gt;4 of a kind - 8&lt;/li&gt;&#xA;&lt;li&gt;Full house - 7&lt;/li&gt;&#xA;&lt;li&gt;Flush - 6&lt;/li&gt;&#xA;&lt;li&gt;Straight - 5&lt;/li&gt;&#xA;&lt;li&gt;3 of a kind - 4&lt;/li&gt;&#xA;&lt;li&gt;2 pairs - 3&lt;/li&gt;&#xA;&lt;li&gt;Pair - 2&lt;/li&gt;&#xA;&lt;li&gt;High card - 1&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The actual ranks used do not matter. You are playing against an omniscient opponent (who works directly against your goal). You, however, have hacked into the computer that shuffles the deck. Hence, you can decide exactly which cards to deal to whom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 1:&lt;/strong&gt; Deal $26$ cards to your opponent to minimize his score.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 2:&lt;/strong&gt; Deal $26$ cards (from the same deck) each to yourself and your opponent to maximize the difference between your and your opponent's score.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-10-06T21:02:24.087" Title="Five-hand poker" Tags="|logical-deduction|game|optimization|cards|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22854" PostTypeId="2" ParentId="22852" CreationDate="2015-10-03T17:06:44.880" Score="1" Body="&lt;p&gt;Almost certainly non- optimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initial step: Query for determining truth teller vs falsehoodian:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Would you remain silent if I gave you an inanswerable question. &lt;br&gt;&#xA; You do this for the number of great grandparents you need to get &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One note on an unsolvable case:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If they are All TF, then it is impossible to derive as they will all display the dominant trait, with no indication they are heterozygotes. This tells us in general, we cannot guarantee a solution that doesn't have to find out all of the ancestors genotypes. However this can be optimized by asking, do any of your descendants display a different phenotype than you, after having surmised the original great grandparent's phenotype.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A pretty key question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Do any of your descendants have a phenotype neither of their parents have? (or something to that effect) If true, this immediately determines if there is a recessive, Additionally depending on the phrasing what is the recessive gene.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sorry for just a partial answer, I have some stuff to do hope to come back, and maybe someone can be inspired to greater creativity after me.&lt;/p&gt;&#xA;" OwnerUserId="13916" LastActivityDate="2015-10-03T17:06:44.880" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22855" PostTypeId="2" ParentId="22844" CreationDate="2015-10-03T17:14:39.760" Score="4" Body="&lt;p&gt;The given number sequence is derived from &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the sequence of prime numbers:&lt;br&gt;&#xA;&lt;br&gt;&#xA; $~~~~$ $2$, $~3$, $~5$, $~7$, $~11$, $~13$, $~17$, $~19$, $~23$, $~\ldots$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The two entries of every pair are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the difference and the sum of consecutive prime numbers  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This yields&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;3-2/3+2, $~$5-3/5+3, $~$7-5/7+5, $~$11-7/11+7, $~$13-11/13+11, $~$17-13/17+13,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so that the next term will be $~$&lt;strong&gt;4/30&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-10-03T17:14:39.760" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22856" PostTypeId="2" ParentId="22853" CreationDate="2015-10-03T17:16:21.347" Score="0" Body="&lt;p&gt;Initial response for question 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A reasonable goal is to avoid straights and full houses and flushes. &lt;br&gt; For straights, you can do this by clustering the number values you give them into chunks of 4. 2,3,4,5,7,8,9,10,Q,K,A (if ace is high). Sadly this is only 11 values. To avoid full houses we cannot have sets of 3. So this only allows us to allocate 22 cards of 26. &lt;br&gt; Looking at the flushes restriction we have 26 cards over 4 suits and all we want to do is minimize the number of times they can have 5 of a suit. So this recommends a 9,9,4,4 distribution of suits. &lt;br&gt;&lt;br&gt; This is where it gets slightly interesting, what if you have it so that the cards necessary for a flush, take away all the three of a kind? We can allocate 4 numbers to be 3 of a kind. Sadly, the  opponent could just use the other 5 of the suit. &lt;br&gt; Oh well another interesting problem I don't have the time at the moment to rigorify. I hope someone can use some thought I had to lead to something.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13916" LastActivityDate="2015-10-03T17:16:21.347" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22857" PostTypeId="2" ParentId="22846" CreationDate="2015-10-03T17:48:26.867" Score="12" Body="&lt;p&gt;Adding to Joe Z.'s answer, the third theorem is also false.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;     2 1 1 3&#xA;     1 3 1 2&#xA;     1 1 3 2&#xA;     3 2 2 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The the bottom-right 1, the top-left 2 and the middle 3's (versus the outer 3's) are each queen-isolated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thinking on something for the second theorem.&lt;/p&gt;&#xA;" OwnerUserId="16478" LastEditorUserId="16478" LastEditDate="2015-10-03T18:19:14.423" LastActivityDate="2015-10-03T18:19:14.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22859" PostTypeId="1" AcceptedAnswerId="23189" CreationDate="2015-10-03T18:05:36.493" Score="7" ViewCount="859" Body="&lt;p&gt;My Indian friend Jamal recently participated in the well-known TV quiz&#xA;show &lt;strong&gt;&lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Who_Wants_to_Be_a_Millionaire&quot; rel=&quot;nofollow&quot;&gt;Who wants to be a millionaire?&lt;/a&gt;&lt;/em&gt;&lt;/strong&gt;. You might have heard or read about Jamal's spectacular success; the media reported on it extensively.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what Jamal told us about his 8.000 Euro question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We moved on to the 8.000 Euro question. Before reading it to me, Prem&#xA;  Kumar made some tasteless joke about the monotone voice that I had been&#xA;  using when answering his questions so far.  I was annoyed, but decided to&#xA;  stay calm and to ignore it.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I had never been to the place mentioned in the question; however, each&#xA;  of the phrases in the four possibilities was familiar to me. I decided&#xA;  to try and match each possibility to its correct place and hopefully&#xA;  the one left over would be the answer.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;D's associated metropolis did not come to my mind right away.  With C,&#xA;  I figured the association based on the stadium's name; this could not be&#xA;  the right answer. Answer B also was very obvious, as that movie and its&#xA;  associated (and somewhat annoying) song had come out a few years ago.&#xA;  Answer A didn't ring a bell right away, so I focused closer on A and D.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;D's phrase brought Disneyland to mind for some reason.  My mental&#xA;  connection was that if you are happy, then you're probably friendly as well.&#xA;  I knew that wasn't it, though, but maybe it was some place nearby.&#xA;  I turned to answer A and kept repeating the phrase in my mind over and&#xA;  over again.  Suddenly a faint bell did start ringing, and brought me to&#xA;  liberty. I said in a loud and clear voice: &quot;My final answer is A.&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Prem Kumar looked at me with a very serious face, causing my heart to&#xA;  beat faster.  &quot;I am afraid that that is&quot;, he said, pausing for a few&#xA;  seconds, &quot;absolutely correct!  If you decide to stop now, you are&#xA;  guaranteed to walk away with at least 8.000 Euros!&quot; I breathed a sigh&#xA;  of relief.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;What was the 8.000 Euro question?&lt;br&gt;&#xA;What were the four possible answers A, B, C, D?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="8874" LastEditDate="2015-10-14T11:30:45.053" LastActivityDate="2016-03-07T09:14:07.507" Title="Slumdog millionaire (8)" Tags="|knowledge|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22860" PostTypeId="2" ParentId="22846" CreationDate="2015-10-03T19:02:40.347" Score="14" Body="&lt;p&gt;If a checkerboard is colored red and blue, either there is a continuous path of neighboring cells of a single colour from the top to the bottom, or a continuous path of neighboring cells of a single colour from the left to the right. (Neighboring includes diagonally touching cells.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;WLOG there is a red path from the top to the bottom. The path must contain a cell on each row. The queen can follow the path, taking detours to visit all the other red cells in each row before returning to the path.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorDisplayName="user88" LastEditDate="2015-10-03T19:59:34.603" LastActivityDate="2015-10-03T19:59:34.603" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="22861" PostTypeId="2" ParentId="22848" CreationDate="2015-10-03T19:53:38.387" Score="9" Body="&lt;p&gt;To guarantee a valid representation (note that this is necessarily an upper bound), you need to scan 39 stickers.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;2 from centers:&lt;/strong&gt; You know the sticker opposite any center you scan, and the direction of the stickers is fixed. Therefore, with two stickers, you know the opposite stickers by default. Then, the other two stickers have two possible orientations, but only one will match what you know about the cube.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;15 from corners:&lt;/strong&gt; To tell the third color on a corner, you only need to scan two of its stickers. While it's possible to have two corners come up with the same two stickers, it's &lt;em&gt;not&lt;/em&gt; possible for them to have the same orientation. In other words, if you placed the RBY corner and RBW corner such that Y/W pointed down, you'd end up with one corner with RB, and one with BR. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently, there's no ambiguity if you scan two stickers. That's 2 stickers by 8 corners is 16 stickers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why 15, though? Well, you know exactly what the last corner is right out of the box. All you need to know is its orientation, which you only need one sticker to determine. The other two colors are in a fixed order relative to that sticker.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;22 from edges:&lt;/strong&gt; Unfortunately, you don't have much leeway when scanning edges. Since it's possible to cycle any three edges, the only case you can be sure to have full information about is when there are two edges left, each with one sticker you know and one you don't. There's only one valid position here, because the two edges can't be swapped freely. In any other case, the edges could be swapped and you wouldn't know.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;From this, $2+15+22=39$. However, strictly speaking, this is far from the ideal algorithm. &lt;/p&gt;&#xA;&#xA;&lt;h3&gt;A worst-case analysis won't give you an optimal scan.&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The above is the worst-case maximum, and represents an upper bound. It is not, however, an ideal scan. To scan with as few stickers as possible, this is what you need to do:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Create two internal tables in the program's memory: one for corners, and one for edges. (Centers are trivial and don't warrant a table.) Whenever you scan a sticker, add it to that piece's list of stickers, and see if there is more than one possible piece in the index those stickers could be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will always net you less than 40 scans - 39 is the worst-case, which is what happens when at every point, you can't make any deductions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are a few cases which the above won't strictly trap for that could reduce the number of stickers you need to scan. For corners, let's say you have three corners left: RBY, WRG, GYO. If you scan a white sticker, you don't need to scan anything else: you know you've found the WRG corner, and can remove it from the list. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By a similar token, let's say you had WRG, WGO left, and you've already scanned both the W stickers. The next time you scan one of these pieces, let's say you see G. You then know that the corner you're looking at is WGO - but not only that, if you re-evaluate the table, you'll easily see that the other white sticker &lt;em&gt;must&lt;/em&gt; belong to the WRG corner. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same methods apply to edges.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastActivityDate="2015-10-03T19:53:38.387" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22862" PostTypeId="2" ParentId="22852" CreationDate="2015-10-03T21:19:59.010" Score="1" Body="&lt;p&gt;Following on from @goinghamateur, isn't is simply a case of asking a single great-grandparent&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Would you remain silent if I gave you an inanswerable question?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then ask them if Mr. X is male (which is probably true anyway) and whether Mr. X is a truthteller.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Based on their answer to the first question, you know whether they are telling the truth for the other two. All the stuff about alleles is irrelevant. $n = 3$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-10-03T21:19:59.010" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22863" PostTypeId="2" ParentId="22852" CreationDate="2015-10-03T22:14:06.233" Score="4" Body="&lt;p&gt;Note: The latest edit invalidates this answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can just ask one great-grandparent&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. Does exactly one of you and Mr. X tell the truth?&lt;br&gt;&#xA; 2. Is exactly one of the statements 'Mr. X is male' and 'You tell the truth' true?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This gives the truth table&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/L7fGV.png&quot; alt=&quot;Truth table&quot;&gt;&lt;br&gt;&#xA; (where G = Great-grandparent tells the truth, M = Mr. X is male, T = Mr. X tells the truth, 1 and 2 = answer to that question)&lt;br&gt;&#xA; which has each combination of answers mapped onto one unique combination of attributes for Mr. X regardless of whether the great-grandparent tells the truth. So $n = 2$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13152" LastEditorUserId="13152" LastEditDate="2015-10-04T10:45:05.437" LastActivityDate="2015-10-04T10:45:05.437" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22864" PostTypeId="2" ParentId="22763" CreationDate="2015-10-04T01:00:16.733" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That the father had to imply one is older than the other means the product is a perfect square that &lt;em&gt;could also&lt;/em&gt; be written as the product of 1 and itself. Even then it has to be less than 6, so it can only be 4 while allowing 2 solutions at first (2*2 and 1*4). The elder one is 4 and the other is 1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastActivityDate="2015-10-04T01:00:16.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22865" PostTypeId="1" AcceptedAnswerId="22876" CreationDate="2015-10-04T10:05:52.087" Score="11" ViewCount="731" Body="&lt;p&gt;Despite strong objections by many here at StackExchange, the mad scientist continues to use ethically questionable methods to study the spatial cognition of ants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last time, he left ant Juliet at a corner of a cubic box and placed ant Romeo at the opposite corner. Much to the mad scientist's surprise, without hesitation Romeo followed the shortest path over the sides of the box that brought him straight to Juliet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the mad scientists has worked out a much bigger challenge for Romeo. Again he leaves Juliette at a corner of a box, but this time it is a 7-dimensional hypercubic box. Out of the 127 remaining corners, he selects the one that is furthest from Juliet. With a big grin the mad scientist places Romeo at this very corner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Romeo is confused. He knows how to handle a box with 6 sides. But now as many as 672 square sides are available on the box. How should he optimize the journey to his beloved Juliet? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you help Romeo? How long is the shortest path (expressed in box edge lengths) to Juliet?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Inspired by &lt;a href=&quot;https://puzzling.stackexchange.com/questions/22827/this-ant-sure-is-smart-but-how-fast-is-it&quot;&gt;this recent puzzle&lt;/a&gt;.]&lt;/p&gt;&#xA;" OwnerUserId="7573" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-10-05T18:37:38.563" Title="Ant on a hyperbox" Tags="|mathematics|geometry|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22866" PostTypeId="2" ParentId="20849" CreationDate="2015-10-04T13:29:46.640" Score="1" Body="&lt;p&gt;&lt;strong&gt;Partial answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I noticed all of those could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Something like emotions/feelings/... In order: &lt;br&gt;Patience, Reason, Excitement, Motivation, Anger, Despair, Regret, Satisfaction, Relief, Hope&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now I'm having trouble to find &lt;em&gt;the one&lt;/em&gt; though ;)&lt;/p&gt;&#xA;" OwnerUserId="5106" LastActivityDate="2015-10-04T13:29:46.640" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22868" PostTypeId="1" AcceptedAnswerId="22873" CreationDate="2015-10-04T14:03:03.427" Score="9" ViewCount="1456" Body="&lt;p&gt;Professor Erasmus informed me today that he has proved another fascinating theorem about chessboards.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Consider a standard $8\times8$ chessboard with 32 black and 32 white squares.&#xA;  Pick an arbitrary (black or white) square $x$. &#xA;  Compute the sum $B$ of the 32 squared distances between the center of $x$ and the centers of the 32 black squares.&#xA;  Compute the sum $W$ of the 32 squared distances between the center of $x$ and the centers of the 32 white squares.&#xA;  Then independently of the choice of $x$, you will always find that $B=W$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Has the professor once again made one of his well-known mathematical blunders, or does his claimed theorem indeed hold true?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2016-05-20T16:04:55.143" Title="The Erasmus chessboard theorem" Tags="|mathematics|combinatorics|checkerboard|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22870" PostTypeId="2" ParentId="22868" CreationDate="2015-10-04T15:18:30.127" Score="9" Body="&lt;p&gt;The professor&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; is not as dumb as some people may be tempted to think.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let the intersection cell of row $i$ and column $j$ be $(i,j)$. Let the chosen cell be $(x,y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;B &amp;amp;=\sum_{(i,j) \mathrm{ is\  black}} (x-i)^2 + (y-j)^2\\&amp;amp;= \sum_{(i,j) \mathrm{\  is\  black}}(x-i)^2 + \sum_{(i,j) \mathrm{\  is\  black}}(y-j)^2\\&amp;amp;=x_b+y_b&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;W&amp;amp;=\sum_{(i,j) \mathrm{\  is\  white}} (x-i)^2 + \sum_{(i,j) \mathrm{\  is\  white}}(y-j)^2\\&amp;amp;=x_w+y_w&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since, there are exactly 4 black and 4 white cells each row,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{(i,j) \mathrm{\  is\  black}} (x-i)^2=4\sum_{i=1}^8(x-i)^2=\sum_{(i,j) \mathrm{\  is\  white}} (x-i)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $x_w=x_b$&#xA;Similarly, $y_w=y_b$&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Therefore, $B=W$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TM;DR (Too mathy, didn't read)&lt;/strong&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first thing to understand is the sum of squares of distances from a cell to all black cells is the sum of squares of distances of the x coordinates $(x_b)$ plus the sum of squares of distances of the y coordinates $(y_b)$. This follows from Pythagoras/distance formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But since there are an equal number of black and white cells each row, The set of distances from black cells is actually the same set as the set of distances from white cells. Therefore $x_b$ and $x_w$ are summing the squares of the same set and therefore are equal. Similarly, $y_b$ and $y_w$  are equal. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Therefore B and W are equal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15544" LastEditorUserId="15544" LastEditDate="2015-10-04T15:26:25.700" LastActivityDate="2015-10-04T15:26:25.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22871" PostTypeId="2" ParentId="22868" CreationDate="2015-10-04T16:33:38.070" Score="4" Body="&lt;p&gt;Let us take a cell $(x,y)$ and four cells in a block $(a,b),(a+1,b),(a,b+1),(a+1,b+1)$. We know that two of them are black and two are white. We can prove that the sum of squared distances of $(x,y)$ from the black and white cells is equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Pythagoras theorem,&#xA;$$B=[d\ to\ (a,b)]+[d\ to\ (a+1,b+1)]$$&#xA;$$=[(x-a)^2+(y-b)^2]+[(x-a-1)^2+(y-b-1)^2]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&#xA;$$W=[d\ to\ (a,b+1)]+[d\ to\ (a+1,b)]$$&#xA;$$=[(x-a)^2+(y-b-1)^2]+[(x-a-1)^2+(y-b)^2]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On comparinging, we find these two equal. Hence we conclude that for any cell and a block of $4$ other cells, this theorem is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We divide the $8\times 8$ checkerboard into $16\ (2\times 2)$ blocks. Since the theorem holds for distances to each of the blocks, it also holds for distances to &lt;em&gt;all&lt;/em&gt; of the blocks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only block not taken into account is the block where the cell itself resides. It is easy to show that this theorem is true in the cell's own block. ($1^2+1^2=(\sqrt2)^2$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hence proved.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did not cheat from @Anachor 's answer. I couldn't understand it because I am yet to learn how to use the $\sum$ symbol properly. I knew Pythagoras theorem had to be used right from the beginning and made up the proof on my own.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4269" LastEditDate="2015-10-05T09:39:06.440" LastActivityDate="2015-10-05T09:39:06.440" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22872" PostTypeId="2" ParentId="22865" CreationDate="2015-10-04T16:39:13.963" Score="3" Body="&lt;p&gt;My guess it's:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\sqrt{6^2+1^2} = \sqrt{37}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I believe that you can fold every two-dimensional side in such a way so you get a 6x1 rectangle and then you just take the diagonal. Just like you can get a 2x1 rectangle with a regular 3d cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To illustrate this take these as the vertices of our 6 by 1 rectangle (sorry. I'm not familiar with mathematical drawing tools):  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ABCDEFG  &#xA;HIJKLMN&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then you have these 7-dimensional coordinates for these vertices.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A = (0,0,0,0,0,0,0)  &#xA;B = (0,1,0,0,0,0,0)  &#xA;C = (0,1,1,0,0,0,0)  &#xA;D = (0,1,1,1,0,0,0)  &#xA;E = (0,1,1,1,1,0,0)  &#xA;F = (0,1,1,1,1,1,0)  &#xA;G = (0,1,1,1,1,1,1)  &#xA;H = (1,0,0,0,0,0,0)  &#xA;I = (1,1,0,0,0,0,0)  &#xA;J = (1,1,1,0,0,0,0)  &#xA;K = (1,1,1,1,0,0,0)  &#xA;L = (1,1,1,1,1,0,0)  &#xA;M = (1,1,1,1,1,1,0)  &#xA;N = (1,1,1,1,1,1,1) &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-10-04T16:39:13.963" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22873" PostTypeId="2" ParentId="22868" CreationDate="2015-10-04T16:53:15.677" Score="13" Body="&lt;p&gt;Place a white 1kg weight at the center of each white square and a black 1kg weight at the center of each black square. Then $B$ is the rotational inertia of the black weights around $x$, and $W$ is the rotational inertia of the white weights around $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the Parallel Axis Theorem, the rotational inertia of an object depends only on (1) its total mass, (2) its moment about the center of mass, and (3) the distance from its center of mass to the axis of rotation. These quantities are evidently the same for the set of black weights and the set of white weights, by symmetry. So their rotational inertia around any point $x$ is equal for both sets. It follows that $B=W$.&lt;/p&gt;&#xA;" OwnerUserId="1752" LastEditorUserId="11806" LastEditDate="2016-05-20T16:04:55.143" LastActivityDate="2016-05-20T16:04:55.143" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22874" PostTypeId="2" ParentId="22865" CreationDate="2015-10-04T18:04:40.350" Score="4" Body="&lt;p&gt;Romeo picks a square face having his starting position as a vertex, and travels along its diagonal. Romeo then picks another square that has his current position as a vertex but does not share a side with the first square, and travels along its diagonal.  So far Romeo has travelled $2\sqrt{2}$ side lengths. Now Romeo can find a cube  with him and Juliet at opposite vertices, and he will have to travel $\sqrt{5}$ side lengths more to reach Juliet, for a total distance $2\sqrt{2}+\sqrt{5}\approx 5.06$ side lengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If ours is the standard unit hypercube $[0,1]^7$ in $7$-space, then Romeo's path is piecewise linear, with vertices&#xA;$$&#xA;\begin{array}{l}&#xA;(0,0,0,0,0,0,0)\\&#xA;(1,1,0,0,0,0,0)\\&#xA;(1,1,1,1,0,0,0)\\&#xA;(1,1,1,1,1,0,1/2)\\&#xA;(1,1,1,1,1,1,1).&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-10-04T18:19:15.720" LastActivityDate="2015-10-04T18:19:15.720" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22875" PostTypeId="1" AcceptedAnswerId="22877" CreationDate="2015-10-04T19:01:46.477" Score="2" ViewCount="315" Body="&lt;p&gt;In Infinityland, people have mastered full control over Infinity. They have built registers which are capable of holding as much data as needed. There are two types of registers in infinityland which are called infinity registers:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;The integer infinity register&lt;/strong&gt;: Can hold any integer value from -inf to + inf. Costs $\$100$ .&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;The real infinity register&lt;/strong&gt;: Can hold any real value from -inf to + inf with infinite precision. Costs $\$200$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Rather expensive, are they not? Well, not actually, because the people over at Infinityland have really good compression algorithms which allow them to stash huge amounts of data in those tiny registers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the lowest cost you must incur to save a list of real numbers if&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The list can hold a maximum of $100$ real numbers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The list can hold infinitely many real numbers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second case, the information must be decompressible without the knowledge of the size of the list.&lt;/p&gt;&#xA;" OwnerUserId="15544" LastEditorUserId="15544" LastEditDate="2015-10-04T19:49:12.610" LastActivityDate="2015-10-06T21:45:51.923" Title="Registers in Infinityland" Tags="|mathematics|optimization|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22876" PostTypeId="2" ParentId="22865" CreationDate="2015-10-04T19:38:08.110" Score="9" Body="&lt;p&gt;$$&#xA;\begin{array}{l}&#xA;(0,0,0,0,0,0,0)\\&#xA;(1,\frac{3}{4},0,0,0,0,0):\frac{5}{4}u\\&#xA;(1,1,\frac{1}{3},0,0,0,0):\frac{5}{12}u\\&#xA;(1,1,1,\frac{1}{2},0,0,0):\frac{5}{6}u\\&#xA;(1,1,1,1,\frac{2}{3},0,0):\frac{5}{6}u\\&#xA;(1,1,1,1,1,\frac{1}{4},0):\frac{5}{12}u\\&#xA;(1,1,1,1,1,1,1):\frac{5}{4}u.&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total distance traveled: $5$ units. This corresponds to the shortest path through six squares arranged in a zigzag fashion: $\sqrt{3^2+4^2}$.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-10-04T19:38:08.110" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22877" PostTypeId="2" ParentId="22875" CreationDate="2015-10-04T19:38:26.977" Score="6" Body="&lt;p&gt;For both, I can do it with&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $200.&lt;br&gt;&#xA;&lt;br&gt;&#xA; I'm just going to describe the second case here. First, take the inverse tangent of all of your numbers and divide by 2pi to convert everything to a real number between -1/2and 1/2; then add 1/2 to get all of your numbers between 0 and 1.&#xA;&lt;br&gt;&#xA; Next, split all of your numbers up into digits. Start with &quot;0.&quot;. Take the first digit of the first, then the next unused digit of the first and second, then the next unused digit of the first through third, then the next unused digit of the first through fourth, et cetera. This gives you one real number!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To encode the number of items:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Start with that number instead of 0 (thanks, Ben) or write it in unary (or base-9 but efficiency doesn't really matter here) at the beginning and add a 0 in between the unary and the compressed string.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof that this is the best possible:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The only cheaper possibility is that you could do it with an integer register. That would mean countably infinite compressed numbers would decompress into uncountably infinite original numbers. This would require one input to give multiple outputs, which is not possible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="11876" LastEditDate="2015-10-04T19:55:21.957" LastActivityDate="2015-10-04T19:55:21.957" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="22880" PostTypeId="1" AcceptedAnswerId="22893" CreationDate="2015-10-04T20:31:46.090" Score="10" ViewCount="685" Body="&lt;p&gt;It had sounded like a great idea at the time. Ernie had been gifted a travel package for himself plus eight companions (he chose to bring along seven friends from his Wine Club, plus me) by the President of &lt;strong&gt;&lt;em&gt;Lighter Than Air Tours&lt;/em&gt;&lt;/strong&gt; - which was using his artificial spider silk technology in their modern take on airship travel. &quot;&lt;em&gt;21st Century Zeppelins!&lt;/em&gt;&quot; the advertising blurb had exclaimed, &quot;&lt;em&gt;All the benefits of a 4 star Hotel combined with the convenience of safe and rapid air travel&lt;/em&gt;&quot;. &quot;&lt;em&gt;!!!Nine Cities in Nine Days!!!&lt;/em&gt;&quot; - it had promised.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RQzRd.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RQzRd.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plan was that the air-ship would pick us all up from a central location and then, over nine nights, fly us to nine &quot;&lt;em&gt;Exciting and Adventurous Destinations&lt;/em&gt;&quot;. We would arrive at each location at dawn - spend  the day &quot;&lt;em&gt;Exploring the Delights of the Environs&lt;/em&gt;&quot; - then re-board the airship to be &quot;&lt;em&gt;Whisked Overnight, in Luxurious Comfort along the most direct great circle path&lt;/em&gt;&quot; to the next destination. At the end of the tour we would all be returned to the starting-point. And the list of destinations (provided in alphabetical order) had sounded rather interesting...&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; In no particular order, you will visit the delights of:&#xA;   Aksaray, Turkey&#xA;   Denver, USA&#xA;   Germiston, South Africa&#xA;   Ilheus, Brazil&#xA;   Moriguchi, Japan&#xA;   Odessa, Ukraine&#xA;   Perth, Australia&#xA;   Toulouse, France&#xA;   Warangal, India&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;...especially as some of the places appeared a little off the track for normal tourist travel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After pick-up we were shown our (rather luxurious) individual cabin rooms (each with a &quot;&lt;em&gt;Port-hole View of the Skies&lt;/em&gt;&quot;), given a quick itinerary (the order that we would visit the cities), then informed that each night, because we were special guests, there would be bottomless complimentary Champagne for all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, the combination of high-altitude air-ship travel, jet-lag and (to be honest) far too much free alcohol left us feeling more than a little &#xA; under the weather. On one morning the following conversation took place around the breakfast table (shortly before we were to arrive at our next destination).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Oh my God&quot;, said George, who looked a little fragile after the drinking races the night before, &quot;I can't remember where we were yesterday, where we are going today, the order in which we were to visit the cities or, for that matter, how many days we have been travelling&quot;. The rest of us, also regretting the night before's excesses, all made similar comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The only thing I remember&quot;, said Marg, &quot;is that in the ordered sequence of cities, no more than three formed an alphabetically increasing sub-sequence (based on the first letters of the city's names)&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Are you sure of that?&quot;, replied Jeff. &quot;My memory is that in the same ordered sequence, no more than three formed an alphabetically increasing sub-sequence (based on the last letters of the city's names)&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;That can't be right!&quot;, exclaimed Harriet. &quot;I'm positive that no more than three cities formed an alphabetically decreasing sub-sequence ((based on the first letters of the city's names)&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;You are all miss-remembering, exclaimed Wilbur. &quot;No more than three cities formed an alphabetically decreasing sub-sequence (based on the last letters of the city's names)&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Well you can't all be right&quot;, said the always argumentative Fred with a grimace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;But what if you are...?&quot;, continued Dorothy, who hated conflict.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Well, even if you are all correct in your recollections&quot;, said Ernie, &quot;there still isn't enough information to determine the correctly ordered sequence of cities. Does anyone else have more information?&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At that point I remembered a conversation I had with the Second Officer, when we were dining at the Captain's Table the previous night (before the drinking races began). &quot;I was told&quot;, I explained, &quot;that the combined flying distance between the first and second cities and the second and third cities, was 'substantially' less than the combined flying distance between the seventh and eighth cities and the eighth and ninth cities. And furthermore&quot;, I continued &quot;last night's flight was to be the shortest 'inter-city' flight on the whole tour&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;In that case, - &lt;strong&gt;&lt;em&gt;assuming all your recollections are correct&lt;/em&gt;&lt;/strong&gt; - the order is obvious...&quot;, retorted Ernie, &quot;...and in a few minutes we should be arriving at the city of ...&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But at that very moment Ernie's voice was drowned out by the airship's intercom announcing our destination for the morning and also the ordered list of cities visited so far and still to visit. After that, Ernie still asserted that he had already worked out the itinerary from the clues provided but none of us were convinced. Can anyone help me. Is it possible to calculate the itinerary or was he just guessing?&lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="1631" LastEditDate="2015-10-05T22:28:32.310" LastActivityDate="2015-10-06T07:16:14.980" Title="Ernie and the Grand Tour" Tags="|mathematics|word-problem|combinatorics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22881" PostTypeId="2" ParentId="22812" CreationDate="2015-10-05T00:20:44.950" Score="5" Body="&lt;p&gt;This was tricky:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;0 (start): abstruse, condors, honest, invention, neon, reacted, vague, @&lt;br&gt;&#xA;1: abstruse, condors, honest, invention, neon, &lt;em&gt;react&lt;/em&gt;, vague, @, &lt;strong&gt;ed&lt;/strong&gt;&lt;br&gt;&#xA;2: abstruse, condors, honest, invention, neon, &lt;em&gt;re&lt;/em&gt;, vague, @, &lt;strong&gt;act&lt;/strong&gt;, ed&lt;br&gt;&#xA;3: &lt;em&gt;use&lt;/em&gt;, condors, honest, invention, neon, re, vague, @, &lt;strong&gt;abstr&lt;/strong&gt;act, ed&lt;br&gt;&#xA;4: use, &lt;em&gt;con&lt;/em&gt;, honest, invention, neon, re, vague, @, abstract, &lt;strong&gt;dors&lt;/strong&gt;, ed&lt;br&gt;&#xA;5: use, con, honest, invention&lt;strong&gt;s&lt;/strong&gt;, neon, re, vague, @, abstract, &lt;em&gt;dor&lt;/em&gt;, ed&lt;br&gt;&#xA;6: use&lt;strong&gt;r&lt;/strong&gt;, con, honest, inventions, neon, re, vague, @, abstract, do, ed&lt;br&gt;&#xA;7: user, con, honest, &lt;em&gt;ions&lt;/em&gt;, neon, re, vague, @, abstract, &lt;strong&gt;invent&lt;/strong&gt;, do, ed&lt;br&gt;&#xA;8: &lt;em&gt;er&lt;/em&gt;, con, honest, ions, neon, re, vague, @, abstract, invent, do, &lt;strong&gt;us&lt;/strong&gt;ed&lt;br&gt;&#xA;9: er, con&lt;strong&gt;vent&lt;/strong&gt;, honest, ions, neon, re, vague, @, abstract, &lt;em&gt;in&lt;/em&gt;, do, used&lt;br&gt;&#xA;10: er, &lt;em&gt;vent&lt;/em&gt;, honest, ions, neon, re, vague, @, abstract, in, &lt;strong&gt;con&lt;/strong&gt;do, used&lt;br&gt;&#xA;11: er, vent, honest, ions, neon, vague, @, abstract, &lt;strong&gt;re&lt;/strong&gt;in, condo, used&lt;br&gt;&#xA;12: er, honest, ions, neon, vague, @, abstract, rein&lt;strong&gt;vent&lt;/strong&gt;, condo, used&lt;br&gt;&#xA;13: er, &lt;em&gt;ho&lt;/em&gt;, ions, neon, vague, @, abstract, reinvent, condo, used, &lt;strong&gt;nest&lt;/strong&gt;&lt;br&gt;&#xA;14: er, ions, neon, vague, @, abstract, reinvent, condo, &lt;strong&gt;ho&lt;/strong&gt;used, nest&lt;br&gt;&#xA;15: er, neon, vague, @, abstract, reinvent&lt;strong&gt;ions&lt;/strong&gt;, condo, housed, nest&lt;br&gt;&#xA;16: neon, vague, @, abstract&lt;strong&gt;er&lt;/strong&gt;, reinventions, condo, housed, nest&lt;br&gt;&#xA;17: neon, vague&lt;strong&gt;st&lt;/strong&gt;, @, abstracter, reinventions, condo, housed, &lt;em&gt;ne&lt;/em&gt;&lt;br&gt;&#xA;18: neon, &lt;em&gt;guest&lt;/em&gt;, @, abstracter, reinventions, condo, housed, &lt;strong&gt;va&lt;/strong&gt;ne&lt;br&gt;&#xA;19: &lt;em&gt;on&lt;/em&gt;, guest, @, abstracter, reinventions, condo&lt;strong&gt;ne&lt;/strong&gt;, housed, vane&lt;br&gt;&#xA;20: on&lt;strong&gt;e&lt;/strong&gt;, guest, @, abstracter, reinventions, condone, housed, &lt;em&gt;van&lt;/em&gt;&lt;br&gt;&#xA;21: &lt;strong&gt;d&lt;/strong&gt;one, guest, @, abstracter, reinventions, condone, &lt;em&gt;house&lt;/em&gt;, van&lt;br&gt;&#xA;22: done, @, abstracter, reinventions, condone, &lt;strong&gt;guest&lt;/strong&gt;house, van&lt;br&gt;&#xA;23: &lt;em&gt;e&lt;/em&gt;, @, abstracter, reinventions, condone, guesthouse, van, &lt;strong&gt;don&lt;/strong&gt;&lt;br&gt;&#xA;24: e&lt;strong&gt;s&lt;/strong&gt;, @, abstracter, &lt;em&gt;reinvention&lt;/em&gt;, condone, guesthouse, van, don&lt;br&gt;&#xA;25 (finish): @, abstracter, reinvention, condone, guesthouse, van&lt;strong&gt;es&lt;/strong&gt;, don  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-10-05T00:20:44.950" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22882" PostTypeId="2" ParentId="22865" CreationDate="2015-10-05T08:10:00.563" Score="4" Body="&lt;p&gt;No paths can do better than an analogous to &lt;h1&gt;&lt;em&gt;f''&lt;/em&gt;&lt;/h1&gt; 's diagonal path, for hypercubes of any dimension. Without loss of generality, let's assume the side length equals one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make the diagonal path, you select one edge, find an adjacent (touching) edge perpendicular to the previous one, and another one perpendicular to both, and the fourth one perpendicular to the first three, and so on, until an edge is reached which contains the endpoint. Every edge changes one coordinate from zero to one, or one to zero, and since they are all perpendicular, it's a different one each time, so no coordinates are changed back to zero, and we're guaranteed to reach the opposite vertex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way, there are $n$ of them, where $n$ is the number of dimensions. In an isometric map, each edge shares one face with the previous one, and no three edges are on the same face. This is enough to show that they form a zig-zag line, and the optimal path is obtained by linking the two extremities (proven later):&#xA;&lt;a href=&quot;https://i.stack.imgur.com/XPw02.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XPw02.png&quot; alt=&quot;Optimal&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, for even $n$ we have a chart with equal number of horizontal and vertical edge intersections (or displacements), while for odd $n$ they differ by $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also see that &lt;em&gt;any&lt;/em&gt; path, in its chart, intercepts the vertical and horizontal edges at least $n$ times, one for each coordinate changed to $1$ (same could be said about those increasing from $0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take a isometric chart and consider the dimensions, $dx$ and $dy$, of the bounding rectangle. Consider those with $dx+dy$ equal to $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The squared length of a straight line from one corner to another of the rectangle ($dy^2 + dx^2$) is minimum when $dx$ is closest to $dy$, since it is equal to the square of their sum (a constant) minus twice their product (maximum when they're closest):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$dx^2 + dy^2 = (dx+dy)^2 - 2dxdy$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An optimal solution is $dx=dy$ if $n$ is even and $dx=dy+1$ otherwise. Our diagonal is the shortest curve bounded by a rectangle*, so it is optimal between every curve bounded by a rectangle of equal $dx+dy$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sQdjy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sQdjy.png&quot; alt=&quot;Proof without words&quot;&gt;&lt;/a&gt;&#xA;Since the line must touch the four edges, we can reflect its pieces over the intersections to show that it is greater than the diagonal*&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rectangles with $dx+dy$ greater than $n$ have also greater diagonals, since by decreasing $dx$ or $dy$ we can make a &lt;em&gt;shorter&lt;/em&gt; path which has a sum of $n$, and, thus, is greater than or equal to our diagonal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So our diagonal solution, presented above, is the shortest in any rectangle that has $dx+dy$ no less than $n$. Now it only remains to show that &lt;em&gt;smaller rectangles don't lead to optimal paths&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every optimal path in a rectangle with semiperimeter ($dx+dy$) smaller than $n$ has to cross at least $n$ edges, moving one unit vertically or horizontally (or both) each time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if this was done monotonically, $dx+dy$ would be greater than or equal to $n$. So it eventually runs in the opposite direction. Since the path is straight inside squares, the change happens at an edge or corner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if all turns were in the corners, then the paths after the turning points could be reflected vertically or horizontally in the chart one at a time while preserving length, until there is a monotonic path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This path would then lie in a rectangle of semiperimeter $n$ or more, so would need to be greater or equal in path length to our diagonal solution. If any turn is at an edge, then the path is also not optimal. So there is no such optimal path.&lt;/p&gt;&#xA;" OwnerUserId="14219" LastEditorUserId="14219" LastEditDate="2015-10-05T18:37:38.563" LastActivityDate="2015-10-05T18:37:38.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22883" PostTypeId="1" AcceptedAnswerId="22887" CreationDate="2015-10-05T08:55:52.920" Score="10" ViewCount="2087" Body="&lt;p&gt;Little Johnny Red forgot to write the multiplication sign between two 3-digit numbers $x$ and $y$ and simply wrote them as one number. When Johnny's teacher graded the homework, it turned out that the 6-digit number written by Johnny was three times the correct product $xy$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the positive integers $x$ and $y$!&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-11T09:08:01.343" LastActivityDate="2016-03-11T09:08:01.343" Title="A sloppy calculation" Tags="|mathematics|calculation-puzzle|no-computers|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22884" PostTypeId="2" ParentId="22846" CreationDate="2015-10-05T09:07:40.780" Score="11" Body="&lt;p&gt;Here's a community wiki containing all four of the answers presented by various people:&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;One-colour theorem&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The queen can travel on all the squares of the grid.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Two-colour theorem&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;(answer by &lt;a href=&quot;https://puzzling.stackexchange.com/a/22860/88&quot;&gt;f''&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a checkerboard is coloured red and blue, either there is a continuous path of neighbouring (by edge or corner) cells of a single colour from the top to the bottom, or a continuous path of neighbouring cells of a single colour from the left to the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A queen can travel on this path one square at a time, and therefore can access either every row or column of the checkerboard. From each position on the path, the queen can then access every square of the same colour on the grid.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Three-colour theorem&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;(answer by &lt;a href=&quot;https://puzzling.stackexchange.com/a/22857/88&quot;&gt;Fimpellizieri&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following grid:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 2 1 1 3&#xA; 1 3 1 2&#xA; 1 1 3 2&#xA; 3 2 2 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;is a counterexample.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Four-colour theorem&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;(Answer by &lt;a href=&quot;https://puzzling.stackexchange.com/a/22849/88&quot;&gt;Joe Z.&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the &lt;a href=&quot;https://en.wikipedia.org/wiki/Four_color_theorem&quot; rel=&quot;noreferrer&quot;&gt;four-colour theorem&lt;/a&gt; is true, the following grid:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 1 2 3 4&#xA; 3 4 1 2&#xA; 2 1 4 3&#xA; 4 3 2 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;is a counterexample to the four-colour queen-accessibility theorem.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-10-05T09:07:40.780" CommentCount="1" CommunityOwnedDate="2015-10-05T09:07:40.780" ContentLicense="CC BY-SA 3.0" />
  <row Id="22885" PostTypeId="2" ParentId="22825" CreationDate="2015-10-05T09:45:06.070" Score="6" Body="&lt;p&gt;The question relates to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Veins inside the body &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A life inside the confining dark&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Veins are keeping us alive in the dard&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Their work unseen, but try to hark&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The pulse you can hear through them&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hark to what they say, get steady&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Go ahead and free one already&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A reference to the jugular vein / carotid vein / some vein of which there is only one in the body&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From my experience I can tell&#xA;I'm very close to my defeat&#xA;Maybe a clever mind can help&#xA;Which one is it I urgently need?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You need the ulnar / radial artery, which if you &quot;free&quot; then you can overcome your defeat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I sincerely hope that that answer is wrong, cos its just too dark for my tastes! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hint&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I needed to use a computer for the names of the veins, since I didn't know them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12104" LastEditorUserId="12104" LastEditDate="2015-10-07T06:59:57.513" LastActivityDate="2015-10-07T06:59:57.513" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22886" PostTypeId="2" ParentId="22883" CreationDate="2015-10-05T10:16:33.957" Score="10" Body="&lt;p&gt;One answer would be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $167\times334\ (\times3 = 167334)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How I found it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The easiest way to have at least one of the numbers $x$ in the product with $3$ is if the other number $y$ multiplied with $3$ is a bit over $1000$. The smallest such $y$ is $334$ which gives, multiplied with $3$, the product $1002$. Afterwards we only need an $x$ that multiplied with $2$ gives us $334$; therefore $x = 167$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16608" LastEditorUserId="4421" LastEditDate="2015-10-05T15:42:47.213" LastActivityDate="2015-10-05T15:42:47.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22887" PostTypeId="2" ParentId="22883" CreationDate="2015-10-05T10:37:57.567" Score="20" Body="&lt;p&gt;The only answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $167$ and $334$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I formulated the question as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the numbers are $x$ and $y$, then $1000x + y = 3xy$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This can be rewritten as &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $y=\frac{1000x}{3x-1}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If $3x-1$ is to completely divide $1000$, and $x$ is a three-digit number, then $3x-1$ can be only equal to $500$. Then $x$ solves to $167$ and $y$ is $334$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12104" LastEditorUserId="4421" LastEditDate="2015-10-05T15:39:49.787" LastActivityDate="2015-10-05T15:39:49.787" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22888" PostTypeId="2" ParentId="22883" CreationDate="2015-10-05T11:56:34.703" Score="13" Body="&lt;p&gt;Let's say that $x$ and $y$ are the 3 digit numbers, then&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $3*x*y = 1000x + y$&lt;br&gt;&#xA; $3y = 1000 + y/x$&lt;br&gt;&#xA;&lt;br&gt;&#xA; That number $y/x$ is positive and less than 10, because $y&amp;lt;=999$ and $x&amp;gt;=100$, so we can rewrite this equation as:&lt;br&gt;&#xA;&lt;br&gt;&#xA; $1000 &amp;lt; 3y &amp;lt; 1010$&lt;br&gt;&#xA; $333.3 &amp;lt; y &amp;lt; 336.7$&lt;br&gt;&#xA;&lt;br&gt;&#xA; So $y$ can be 334, 335 or 336&lt;br&gt;&#xA;&lt;br&gt;&#xA; Substitute these values to our equation and we get the following:&lt;br&gt;&#xA;&lt;br&gt;&#xA; $3x*334 = 1000x + 334$&lt;br&gt;&#xA; $1002x-1000x = 334$&lt;br&gt;&#xA; $x = 167$&lt;br&gt;&#xA;&lt;br&gt;&#xA; So 334 and 167 is a match&lt;br&gt;&#xA;&lt;br&gt;&#xA; $3x*335 = 1000x + 335$&lt;br&gt;&#xA; $5x = 335$&lt;br&gt;&#xA; $x = 67$&lt;br&gt;&#xA;&lt;br&gt;&#xA; 67 is a 2 digit number, so y can't be 335&lt;br&gt;&#xA;&lt;br&gt;&#xA; $3x*336 = 1000x + 336$&lt;br&gt;&#xA; $8x = 336$&lt;br&gt;&#xA; $x = 42$&lt;br&gt;&#xA;&lt;br&gt;&#xA; 42 is also not a 3 digit number, so y can't be 336&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore the only answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 334 and 167&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Check that this answer is valid:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $334*167 = 55778$&lt;br&gt;&#xA; $55778 * 3 = 167334$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9352" LastEditorUserId="9626" LastEditDate="2015-10-05T13:34:14.877" LastActivityDate="2015-10-05T13:34:14.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22889" PostTypeId="1" AcceptedAnswerId="22894" CreationDate="2015-10-05T14:31:48.783" Score="3" ViewCount="1469" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aee38.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aee38.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The title of the question says everything $\ldots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt:&#xA;&lt;br&gt;&#xA;We count $2(1+1+1+2+1+1+2+1+1+1+1)=26$ triangles. (On each side $13$ triangles, and then multiplied by $2$). And then we combine them and add. Are there really more than $26$ triangles?. &lt;br&gt;&#xA;I got this question in a test and the options were $20,21,26$ and &lt;em&gt;None of the above&lt;/em&gt;. I wrote &lt;em&gt;None of the above&lt;/em&gt;. Is this right? Please tell.&lt;br&gt;&#xA;Edit: Please ignore the shading, all the intersections count as vertices and it is a 2d figure. &lt;/p&gt;&#xA;" OwnerUserId="11425" LastEditorUserId="11425" LastEditDate="2015-10-06T05:13:08.380" LastActivityDate="2015-10-06T21:37:11.893" Title="Find the total number of triangles in the diagram" Tags="|geometry|triangle|counting|" AnswerCount="6" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="22890" PostTypeId="2" ParentId="22889" CreationDate="2015-10-05T14:58:09.053" Score="4" Body="&lt;p&gt;If the triangles &lt;strong&gt;must&lt;/strong&gt; have a named(A to H) point at each corner then I think the answer is actually:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;21&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Count:&lt;/p&gt;&#xA;&lt;p&gt;The triangles in one half only(left half in this case):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;(B,E,D) (B,E,G) (B,G,D)&lt;/p&gt;&#xA;&lt;p&gt;(A,D,E) (A,D,G) (A,G,E)&lt;/p&gt;&#xA;&lt;p&gt;(A,B,E) (A,D,B)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;so 16 for both halves&lt;/p&gt;&#xA;&lt;p&gt;plus the triangles which go over both halves&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;(A,H,G) (A,C,B)&lt;/p&gt;&#xA;&lt;p&gt;(D,G,H) (D,E,F)&lt;/p&gt;&#xA;&lt;p&gt;(D,B,C)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;for a total of:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 16 + 5 = 21&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;if i missed anything please correct me&lt;/p&gt;&#xA;" OwnerUserId="16608" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-10-05T14:58:09.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22891" PostTypeId="2" ParentId="22889" CreationDate="2015-10-05T15:03:55.387" Score="0" Body="&lt;p&gt;If every intersection counts as a vertex, then yes, the answer is indeed &lt;em&gt;None of the above&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I counted 29 triangles here. The whole image can be symmetrically divided by the vertical &lt;strong&gt;AD&lt;/strong&gt; line, so we have to count the triangles on one side, multiply it by 2 and then add 4(EFD, GHD, ABC, AGH).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's call intersection of GD and BC a point X, intersection of GH and AD a point Y and intersection of BC and AD a point Z like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iRqG3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iRqG3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have(at least, maybe more):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. EAG&lt;br&gt;&#xA; 2. EGB&lt;br&gt;&#xA; 3. AGY&lt;br&gt;&#xA; 4. GBX&lt;br&gt;&#xA; 5. BXD&lt;br&gt;&#xA; 6. DXZ&lt;br&gt;&#xA; 7. DGY&lt;br&gt;&#xA; 8. DGB&lt;br&gt;&#xA; 9. DEA&lt;br&gt;&#xA; 10. DBE&lt;br&gt;&#xA; 11. EBX&lt;br&gt;&#xA; 12. ABZ&lt;br&gt;&#xA; 13. ABD&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then we can multiply it by 2 to get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 13*2 = 26&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And add 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 26 + 4 = 30&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the answer is indeed &lt;em&gt;None of the above&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="9352" LastActivityDate="2015-10-05T15:03:55.387" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22892" PostTypeId="2" ParentId="22889" CreationDate="2015-10-05T15:14:32.243" Score="2" Body="&lt;p&gt;I managed to find&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;38&lt;/strong&gt; triangles in total. (Thanks to @TheDarkTruth for the last two, and to @JonTheMon for pointing out duplicates).&lt;br&gt;&#xA;&lt;br&gt;&#xA; I may have missed some, but it seems like your answer should definitely be &quot;None of the above.&quot;&lt;br&gt;&#xA;&lt;br&gt;&#xA;  &lt;a href=&quot;https://i.stack.imgur.com/6Iuz5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6Iuz5.png&quot; alt=&quot;1st triangle&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/I1a3P.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/I1a3P.png&quot; alt=&quot;2nd triangle&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/ly54w.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ly54w.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/JFxtb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JFxtb.png&quot; 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alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/6QDUk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6QDUk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/ccKEK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ccKEK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/3Q37b.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3Q37b.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZFPYR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZFPYR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/fB9mj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fB9mj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/Mexe3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Mexe3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/6Ng4m.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6Ng4m.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/43MUn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/43MUn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/WsmkI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WsmkI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="4421" LastEditDate="2015-10-05T19:55:32.813" LastActivityDate="2015-10-05T19:55:32.813" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22893" PostTypeId="2" ParentId="22880" CreationDate="2015-10-05T15:46:24.797" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; With the updated definition of &lt;strong&gt;subsequence&lt;/strong&gt; there are still multiple solutions. The shortest is:&lt;br&gt;&lt;br&gt;&lt;strong&gt;Germiston, Toulouse, Denver, Odessa, Aksaray, Warangal, Ilheus, Perth, Moriguchi&lt;/strong&gt;&lt;br&gt;&lt;br&gt;The distances between the cities are (in $m$):&lt;br&gt;&lt;br&gt;&lt;strong&gt;8229026, 8168594, 9463389, 939396, 4941708, 13487561, 14170226, 7696249&lt;/strong&gt;&lt;br&gt;&lt;br&gt;This gives a ratio of about $1.33$ between the sums of both last and both first distances, with a total distance of $67,096,149 m$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2015-10-06T07:16:14.980" LastActivityDate="2015-10-06T07:16:14.980" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22894" PostTypeId="2" ParentId="22889" CreationDate="2015-10-05T16:23:58.790" Score="3" Body="&lt;p&gt;I think I managed to find&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 40 triangles&lt;br&gt;&#xA;&lt;br&gt;&#xA; 12 singles&lt;br&gt;&#xA; 12 doubles&lt;br&gt;&#xA; 8 triple&lt;br&gt;&#xA; 4 quad&lt;br&gt;&#xA; 2 fives&lt;br&gt;&#xA; 1 six&lt;br&gt;&#xA; 0 sevens&lt;br&gt;&#xA; 1 eight&lt;br&gt;&#xA; Ignoring the 12 singles:&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/fAOew.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fAOew.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-10-05T20:24:00.610" LastActivityDate="2015-10-05T20:24:00.610" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22895" PostTypeId="1" AcceptedAnswerId="22900" CreationDate="2015-10-05T16:36:47.747" Score="3" ViewCount="372" Body="&lt;p&gt;By hitting me, you make everyone understand why you have attacked all before me, as well as why you will hit all others that come after me.&lt;/p&gt;&#xA;" OwnerUserId="14233" LastEditorUserId="14233" LastEditDate="2015-10-05T19:35:51.943" LastActivityDate="2015-10-05T20:26:10.860" Title="Strike me down, and I shall become more powerful" Tags="|riddle|word|computer-puzzle|" AnswerCount="2" CommentCount="1" ClosedDate="2015-10-05T20:40:18.217" ContentLicense="CC BY-SA 3.0" />
  <row Id="22896" PostTypeId="2" ParentId="22889" CreationDate="2015-10-05T16:54:56.627" Score="0" Body="&lt;p&gt;Assuming that no two points can belong to a triangle unless they are joined:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 22 triangles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;12 with A&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5 without A but with B&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 without A,B but with C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 without A,B,C but with D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0 without A,B,C,D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I might have missed some.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-10-05T16:54:56.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22897" PostTypeId="2" ParentId="22830" CreationDate="2015-10-05T17:49:56.457" Score="9" Body="&lt;p&gt;What kind of question is this?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;True/False&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Together I am four and five.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As per the hint, &quot;True/False&quot; has four vowels and five consonants.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Apart, am I four or five?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As per the hint, &lt;strong&gt;true&lt;/strong&gt; has four letters and &lt;strong&gt;false&lt;/strong&gt; has five.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Often I am written one or one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Often written as &quot;T or F&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Confused, I may be payments for a bard,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; lute fares&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;antique hearing-aids,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ear flutes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the price of gas,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; fuel rates&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;brightly-burning fat,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; suet flare&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;a damned man's ogle,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; faust leer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;guidelines for heroism, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;feat rules&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;a tempting meal, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; feast lure&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;a whimsical shine, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; fae luster&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;a predictor of problems, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; fault seer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;a prenatal jape,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; fetal ruse &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or I may tamper with explosives.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; alter fuse&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, it might be a land contract:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; turf lease&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or a judge of respectful gestures:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; salute ref&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or that burning sensation you get in your ears when woodwinds play a very fast glissando:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; flute sear&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11775" LastEditorUserId="11775" LastEditDate="2015-10-06T13:16:57.080" LastActivityDate="2015-10-06T13:16:57.080" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22898" PostTypeId="2" ParentId="22830" CreationDate="2015-10-05T17:52:02.473" Score="5" Body="&lt;p&gt;The question is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;true/false&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Together I am four and five. Apart, am I four or five?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Darrel H provided this in his hint. Together, the answer is composed of four vowels and five consonants. Separately and respectively, they are four and five letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Often I am written one or one.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As a boolean data type, true can equal &quot;1&quot;. Some further explanation I'm not sure about...  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Confused, I may be payments for a bard&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; lute fares&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;antique hearing-aids&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ear flutes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;the price of gas&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; fuel rates&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;brightly-burning fat&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; suet flare&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;a damned man's ogle&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Faust leer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;guidelines for heroism&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; feat rules&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;a tempting meal&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; feast lure&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;a whimsical shine&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; fae luster&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;a predictor of problems&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; fault seer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;a prenatal jape&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; fetal ruse  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;or I may tamper with explosives&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; alter fuse&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15921" LastEditorUserId="9523" LastEditDate="2015-10-05T18:38:26.697" LastActivityDate="2015-10-05T18:38:26.697" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="22899" PostTypeId="2" ParentId="22895" CreationDate="2015-10-05T20:17:55.043" Score="4" Body="&lt;p&gt;My guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Enter key (or submit button) or possibly Period key - Submitting your text explains why you struck keys earlier as well as providing context for your next submission&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16642" LastActivityDate="2015-10-05T20:17:55.043" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22900" PostTypeId="2" ParentId="22895" CreationDate="2015-10-05T20:26:10.860" Score="10" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The space bar. Each time you hit it, it ends the previous word and makes it and the next word that will be typed more readable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16729" LastActivityDate="2015-10-05T20:26:10.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22902" PostTypeId="2" ParentId="22812" CreationDate="2015-10-05T22:39:29.337" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0: abstruse, condors, honest, invention, neon, reacted, vague @&lt;br&gt;&#xA; 1: abstruse, condors, honest, invention, neon, &lt;em&gt;react&lt;/em&gt;, vague @ &lt;strong&gt;ed&lt;/strong&gt;&lt;br&gt;&#xA; 2: abstruse, condors, honest, invention, neon, &lt;em&gt;re&lt;/em&gt;, vague @ ed, &lt;strong&gt;act&lt;/strong&gt;&lt;br&gt;&#xA; 3: &lt;em&gt;use&lt;/em&gt;, condors, honest, invention, neon, re, vague @ ed, &lt;strong&gt;abstr&lt;/strong&gt;act&lt;br&gt;&#xA; 4: use, &lt;em&gt;dors&lt;/em&gt;, honest, invention, neon, re, vague @ ed, abstract, &lt;strong&gt;con&lt;/strong&gt;&lt;br&gt;&#xA; 5: use, &lt;em&gt;dor&lt;/em&gt;, honest, invention, neon, re, vague @ ed, abstract&lt;strong&gt;s&lt;/strong&gt;, con&lt;br&gt;&#xA; 6: use, dor, honest, &lt;em&gt;ion&lt;/em&gt;, neon, re, vague @ ed, abstracts, con, &lt;strong&gt;invent&lt;/strong&gt;&lt;br&gt;&#xA; 7: use, dor, honest, &lt;em&gt;on&lt;/em&gt;, neon, re, vague @ ed, abstracts, con, invent, &lt;strong&gt;i&lt;/strong&gt;&lt;br&gt;&#xA; 8: use, dor, honest, &lt;em&gt;o&lt;/em&gt;, neon, re, vague @ ed, abstracts, con, invent, i&lt;strong&gt;n&lt;/strong&gt;&lt;br&gt;&#xA; 9: use, &lt;em&gt;do&lt;/em&gt;, honest, o, neon, re, vague @ ed, abstracts, con, invent, &lt;strong&gt;r&lt;/strong&gt;in&lt;br&gt;&#xA; 10: use&lt;strong&gt;r&lt;/strong&gt;, do, honest, o, neon, re, vague @ ed, abstracts, con, invent, &lt;em&gt;in&lt;/em&gt;&lt;br&gt;&#xA; 11: &lt;em&gt;us&lt;/em&gt;, do, honest, o, neon, re, vague @ ed, abstracts, con, invent, in, &lt;strong&gt;er&lt;/strong&gt;&lt;br&gt;&#xA; 12: do, honest, o, neon, re, vague @ &lt;strong&gt;us&lt;/strong&gt;ed, abstracts, con, invent, in, er&lt;br&gt;&#xA; 13: do, &lt;em&gt;hone&lt;/em&gt;, o, neon, re, vague @ used, abstracts, con, invent, in, er&lt;strong&gt;st&lt;/strong&gt;&lt;br&gt;&#xA; 14: do, hone, o, neon, re, vague&lt;strong&gt;st&lt;/strong&gt; @ used, abstracts, con, invent, in, &lt;em&gt;er&lt;/em&gt;&lt;br&gt;&#xA; 15: hone, o, neon, re, vaguest @ used, abstracts, con&lt;strong&gt;do&lt;/strong&gt;, invent, in, er&lt;br&gt;&#xA; 16: &lt;em&gt;ho&lt;/em&gt;, o, neon, re, vaguest @ used, abstracts, condo&lt;strong&gt;ne&lt;/strong&gt;, invent, in, er&lt;br&gt;&#xA; 17: ho, o&lt;strong&gt;ne&lt;/strong&gt;, neon, re, vaguest @ used, abstracts, &lt;em&gt;condo&lt;/em&gt;, invent, in, er&lt;br&gt;&#xA; 18: ho, one, neon, re&lt;strong&gt;do&lt;/strong&gt;, vaguest @ used, abstracts, &lt;em&gt;con&lt;/em&gt;, invent, in, er&lt;br&gt;&#xA; 19: ho, one, neon, &lt;em&gt;o&lt;/em&gt;, vaguest @ used, abstracts, con, invent, in, er&lt;strong&gt;red&lt;/strong&gt;&lt;br&gt;&#xA; 20: ho&lt;strong&gt;ed&lt;/strong&gt;, one, neon, o, vaguest @ used, abstracts, con, invent, in, &lt;em&gt;err&lt;/em&gt;&lt;br&gt;&#xA; 21: hoed, one&lt;strong&gt;r&lt;/strong&gt;, neon, o, vaguest @ used, abstracts, con, invent, in, &lt;em&gt;er&lt;/em&gt;&lt;br&gt;&#xA; 22: hoed, oner, neon, o, &lt;em&gt;guest&lt;/em&gt; @ used, abstracts, con, invent, &lt;strong&gt;va&lt;/strong&gt;in, er&lt;br&gt;&#xA; 23: hoed, oner, neon, o&lt;strong&gt;va&lt;/strong&gt;, guest @ used, abstracts, con, invent, &lt;em&gt;in&lt;/em&gt;, er&lt;br&gt;&#xA; 24: hoed, oner, neon, &lt;em&gt;a&lt;/em&gt;, guest @ used, abstracts, con, invent, in, &lt;strong&gt;ov&lt;/strong&gt;er&lt;br&gt;&#xA; 25: hoed, oner, &lt;em&gt;eon&lt;/em&gt;, a, guest @ used, abstracts, con, invent, in&lt;strong&gt;n&lt;/strong&gt;, over&lt;br&gt;&#xA; 26: hoed, oner, eon, a&lt;strong&gt;n&lt;/strong&gt;, guest @ used, abstracts, con, invent, &lt;em&gt;in&lt;/em&gt;, over&lt;br&gt;&#xA; 27: hoed, oner, &lt;em&gt;on&lt;/em&gt;, an, guest @ used, abstracts, con&lt;strong&gt;e&lt;/strong&gt;, invent, in, over&lt;br&gt;&#xA; 28: hoed, oner, on, an&lt;strong&gt;e&lt;/strong&gt;, guest @ used, abstracts, &lt;em&gt;con&lt;/em&gt;, invent, in, over&lt;br&gt;&#xA; 29: hoed, oner, on, guest @ used, abstracts, con, invent, in&lt;strong&gt;ane&lt;/strong&gt;, over&lt;br&gt;&#xA; 30: &lt;em&gt;ho&lt;/em&gt;, oner, on, guest @ used, abstracts, con&lt;strong&gt;ed&lt;/strong&gt;, invent, inane, over&lt;br&gt;&#xA; 31: oner, on, guest @ &lt;strong&gt;ho&lt;/strong&gt;used, abstracts, coned, invent, inane, over&lt;br&gt;&#xA; 32: oner, &lt;strong&gt;d&lt;/strong&gt;on, guest @ housed, abstracts, &lt;em&gt;cone&lt;/em&gt;, invent, inane, over&lt;br&gt;&#xA; 33: oner, don&lt;strong&gt;e&lt;/strong&gt;, guest @ housed, abstracts, &lt;em&gt;con&lt;/em&gt;, invent, inane, over&lt;br&gt;&#xA; 34: oner, guest @ housed, abstracts, con, invent, inane, over&lt;strong&gt;done&lt;/strong&gt;&lt;br&gt;&#xA; 35: &lt;em&gt;one&lt;/em&gt;, guest @ housed, abstracts, con, invent, inane&lt;strong&gt;r&lt;/strong&gt;, overdone&lt;br&gt;&#xA; 36: &lt;strong&gt;d&lt;/strong&gt;one, guest @ &lt;em&gt;house&lt;/em&gt;, abstracts, con, invent, inaner, overdone&lt;br&gt;&#xA; 37: guest @ house, abstracts, con&lt;strong&gt;done&lt;/strong&gt;, invent, inaner, overdone&lt;br&gt;&#xA; 38: @ &lt;strong&gt;guest&lt;/strong&gt;house, abstracts, condone, invent, inaner, overdone  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Formatting and some word ideas from @GordonK - thanks!&lt;/p&gt;&#xA;" OwnerUserId="15921" LastEditorUserId="15921" LastEditDate="2015-10-08T15:14:46.930" LastActivityDate="2015-10-08T15:14:46.930" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="22903" PostTypeId="1" AcceptedAnswerId="22904" CreationDate="2015-10-05T23:03:39.337" Score="6" ViewCount="3270" Body="&lt;p&gt;Given a &lt;strong&gt;Double Twelves Domino Set,&lt;/strong&gt; prove that the dominos can be arranged in a valid formation such that placement follows the domino placement rules, namely:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A domino can be placed with the short end touching the short end of another domino, provided that the short end of both dominos match in number of pips.&lt;/li&gt;&#xA;&lt;li&gt;A domino with the same number of pips both ends can be placed horizontally perpendicular to another domino, simply for conservation of space, and placing the domino this way does not permit any additional connections.&lt;/li&gt;&#xA;&lt;li&gt;A domino can be rotated 90 degrees in either direction to change the direction of the train of dominoes, only for conservation of space.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;An image of a &lt;strong&gt;Double Twelves Domino Set:&lt;/strong&gt; it includes, 91 tiles and 1092 pips.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6mSbd.jpg&quot; alt=&quot;doubletwelves&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will accept the best answer, but for extra consideration, provide proof that this works for dominoes going up to size &lt;em&gt;N&lt;/em&gt;, or also provide an image of the solution, additional extra consideration will be awarded if the final solution is arranged in a creative, organized shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please use spoilers to hide mathematical proofs, or key answers. Good luck!&lt;/p&gt;&#xA;" OwnerUserId="12752" LastEditorUserId="30903" LastEditDate="2017-03-12T06:37:31.797" LastActivityDate="2022-01-10T18:09:44.927" Title="Arranging all the Dominos" Tags="|mathematics|dominoes|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22904" PostTypeId="2" ParentId="22903" CreationDate="2015-10-05T23:29:44.047" Score="8" Body="&lt;p&gt;The generalization of this problem is already known: &lt;a href=&quot;https://en.wikipedia.org/wiki/De_Bruijn_sequence&quot;&gt;https://en.wikipedia.org/wiki/De_Bruijn_sequence&lt;/a&gt;. A De Bruijn sequence given an alphabet and a word-length is a string containing, as substrings, all words of that word length with letters from that alphabet exactly once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can apply the existing knowledge on De Bruijn sequences (which can be found on the wikipedia page) to this problem. If we set our alphabet to {0, ..., 12} and our word-length to 2, a De Bruijn sequence is exactly a solution to this problem (a word is a domino). There are $\frac{(13!)^{13^1}}{13^2} = 1.25\times10^{125}$ possible solutions for a Double Twelves Domino Set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also algorithms on the wikipedia page that calculate example De Bruijn sequences given alphabet-size and word-length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Note that De Bruijn sequences are cyclic. This is fine however, we can just tack the beginning character onto the end (or beginning $n-1$ characters in the case of word-length $n$).&lt;/p&gt;&#xA;" OwnerUserId="11100" LastEditorUserId="11100" LastEditDate="2015-10-06T00:45:55.037" LastActivityDate="2015-10-06T00:45:55.037" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22905" PostTypeId="2" ParentId="22889" CreationDate="2015-10-06T04:45:51.737" Score="3" Body="&lt;p&gt;I count 40 different triangles in the picture &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qfQWs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qfQWs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To check this I write them down in a sorted list of triples. Each triple is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;normalized in that way that the components of the triple are alphabecticaly &lt;/p&gt;&#xA;&#xA;&lt;p&gt;ordered.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1.  A   B   C&#xA;2.  A   B   D&#xA;3.  A   B   E&#xA;4.  A   B   J&#xA;5.  A   C   D&#xA;6.  A   C   F&#xA;7.  A   C   J&#xA;8.  A   D   E&#xA;9.  A   D   F&#xA;10. A   D   G&#xA;11. A   D   H&#xA;12. A   E   G&#xA;13. A   F   H&#xA;14. A   G   I&#xA;15. A   G   H&#xA;16. A   H   I&#xA;17. B   C   D&#xA;18. B   D   E&#xA;19. B   D   G&#xA;20. B   D   J&#xA;21. B   D   K&#xA;22. B   D   L&#xA;23. B   E   G&#xA;24. B   E   K&#xA;25. B   G   K&#xA;26. C   D   F&#xA;27. C   D   H&#xA;28. C   D   J&#xA;29. C   D   K&#xA;30. C   D   L&#xA;31. C   F   H&#xA;32. C   F   L&#xA;33. C   H   L&#xA;34. D   E   F&#xA;35. D   G   H&#xA;36. D   G   I&#xA;37. D   H   I&#xA;38. D   J   K&#xA;39. D   J   L&#xA;40. D   K   L&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;How can I check the list?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To check if the triangle (G,H,D) is considered in my count I normalize it to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(D, G, H) and look it up in the list and find it in line 33.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To check that I did no miss a triangle that shoul be inserted in the list &lt;/p&gt;&#xA;&#xA;&lt;p&gt;between &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;25. B   G   K&#xA;26. C   D   F&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I do the following&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;there is no triangle following (B, G, K) with vertices B an G, because the &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;only triple I fave to check is (B, G, L), which is not a valid triangle. (B, G, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;E) is a valid triangle but it does not follow (B, G, K) alphabetically. So it &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is counted elsewhere in the list and mudt not be considered here.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;There is no triange (B,H,...) because BH is not an edge in the picture. The &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;same is true for (B, I, ...).&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;(B, J, ...) can be skipped because if we supply K or L for ... we get a &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;triangle that is deeneraed t a line and we won't count such a line.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;similar arguments hold for (B, K, ...)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;(B, L, ...) must not be considered because there is no letter following &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;So al triangles starting with B are exhausted and we can check trinles &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;starting with C, or more precise starting at least with (C, D, ...). But (C, D, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;E) is no a triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the next triple (C, D, F) is a trianle and in the list.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is also a &lt;a href=&quot;http://ideone.com/sSkz1P&quot; rel=&quot;nofollow noreferrer&quot;&gt;python program&lt;/a&gt; that counts and displays all triangles.&lt;/p&gt;&#xA;" OwnerUserId="1967" LastEditorUserId="1967" LastEditDate="2015-10-06T21:37:11.893" LastActivityDate="2015-10-06T21:37:11.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22906" PostTypeId="2" ParentId="22883" CreationDate="2015-10-06T12:58:50.023" Score="3" Body="&lt;p&gt;Because who doesn't like brute force? (And because the answer is already here)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;python2:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;gt;&amp;gt;&amp;gt; for i in range(100,1000):&#xA;...     for j in range(100, 1000):&#xA;...             if 1000*i + j == 3*i*j:&#xA;...                     print i, j&#xA;... &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    167 334&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For numbers this size it's quicker than working it out 'properly'&lt;/p&gt;&#xA;" OwnerUserId="4577" LastActivityDate="2015-10-06T12:58:50.023" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22907" PostTypeId="1" AcceptedAnswerId="22915" CreationDate="2015-10-06T15:01:19.880" Score="6" ViewCount="691" Body="&lt;p&gt;Yesterday evening, Cristiano Ronaldo looked into a newspaper and quickly detected the sports section. After fifteen minutes of reading and studying, Cristiano bitterly complained about the minimal amount of information provided by modern soccer tables. &#xA;Here are the first few lines of the current table of the Spanish soccer league:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;          Team            M   Pts  GD&#xA;     ----------------------------------&#xA;      1.  Villarreal      7   16   +7&#xA;      2.  Real Madrid     7   15  +13&#xA;      3.  Celta de Vigo   7   15   +8&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Clearly the column &lt;strong&gt;M&lt;/strong&gt; contains the number of matches played so far, &#xA;the column &lt;strong&gt;Pts&lt;/strong&gt; contains the number of points reached so far, and&#xA;the column &lt;strong&gt;GD&lt;/strong&gt; contains the current goal difference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what Cristiano really wants to know is the number &lt;strong&gt;W&lt;/strong&gt; of won, the number &lt;strong&gt;L&lt;/strong&gt; of lost, and the number &lt;strong&gt;D&lt;/strong&gt; of drawn matches. And (according to Cristiano) there is no way of deducing these numbers from the information provided in the table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gareth Bale (who is mathematically gifted) observed that for Villarreal the numbers actually can be deduced: &lt;strong&gt;W=5&lt;/strong&gt;, &lt;strong&gt;L=1&lt;/strong&gt;, and &lt;strong&gt;D=1&lt;/strong&gt;. On the other hand, for Celta there exist two different possible number triples.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; From which combinations of two non-negative integers &lt;strong&gt;M&lt;/strong&gt; and &lt;strong&gt;Pts&lt;/strong&gt; can one uniquely determine the three numbers &lt;strong&gt;W&lt;/strong&gt;, &lt;strong&gt;L&lt;/strong&gt; and &lt;strong&gt;D&lt;/strong&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note: For a win/loss/draw a team respectively scores 3/0/1 points.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-10-08T12:18:13.720" LastActivityDate="2015-10-08T12:18:13.720" Title="Cristiano Ronaldo's soccer puzzle" Tags="|mathematics|combinatorics|sports|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22908" PostTypeId="5" CreationDate="2015-10-06T15:02:35.103" Score="0" Body="&lt;p&gt;A puzzle related to some sport in any way.&lt;/p&gt;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;players&lt;/li&gt;&#xA;&lt;li&gt;teams&lt;/li&gt;&#xA;&lt;li&gt;rules&lt;/li&gt;&#xA;&lt;li&gt;competitions&lt;/li&gt;&#xA;&lt;li&gt;tables&lt;/li&gt;&#xA;&lt;li&gt;records&lt;/li&gt;&#xA;&lt;li&gt;facts&lt;/li&gt;&#xA;&lt;li&gt;history&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Sports questions which aren't puzzles are off-topic but can be asked on our sister site &lt;a href=&quot;https://sports.stackexchange.com/&quot;&gt;Sports Stack Exchange&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="60644" LastEditDate="2020-11-30T10:45:52.807" LastActivityDate="2020-11-30T10:45:52.807" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22909" PostTypeId="4" CreationDate="2015-10-06T15:02:35.103" Score="0" Body="A puzzle related to sports competitions, sports tables, sports records, or the history of sports." OwnerUserId="8874" LastEditorUserId="69582" LastEditDate="2021-02-09T18:18:26.520" LastActivityDate="2021-02-09T18:18:26.520" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22910" PostTypeId="2" ParentId="22907" CreationDate="2015-10-06T15:06:16.593" Score="1" Body="&lt;p&gt;If you want to add 7 numbers out of 3/0/1 to 15, there are only two ways to go: 3,3,3,3,3,0,0 and 3,3,3,3,1,1,1. So the possible solutions are &lt;strong&gt;W=5, L=2, D=0&lt;/strong&gt; and &lt;strong&gt;W=4, L=0, D=3&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;/edit: oops, a more general solution was asked for. So ignore me :-)&lt;/p&gt;&#xA;" OwnerUserId="2783" LastActivityDate="2015-10-06T15:06:16.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22911" PostTypeId="2" ParentId="22907" CreationDate="2015-10-06T15:16:20.243" Score="3" Body="&lt;p&gt;For any Number &lt;strong&gt;M&lt;/strong&gt; if &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Pts=(M*3)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(((M-2)*3)+1)&amp;lt;=Pts&amp;lt;=((M*3)-2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Pts=(((M-2)*3)-1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0&amp;lt;=Pts&amp;lt;=2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;there exists exactly one solution.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Pts = M*3 -&amp;gt;&#xA;W=M L=0 D=0&#xA;&#xA;Pts = M*3-1 -&amp;gt;&#xA;not possible&#xA;&#xA;Pts = M*3-2 -&amp;gt;&#xA;W=M-1 L=0 D=1&#xA;&#xA;Pts = (M-1)*3 -&amp;gt;&#xA;W=M-1 L=1 D=0&#xA;&#xA;Pts = (M-1)*3-1 -&amp;gt;&#xA;W=M-2 L=0 D=2&#xA;&#xA;Pts = (M-1)*3-2 -&amp;gt;&#xA;W=M-2 L=1 D=1&#xA;&#xA;Pts = (M-2)*3 -&amp;gt;&#xA;either W=M-2 L=2 D=0&#xA;or     W=M-3 L=0 D=3&#xA;&#xA;Pts = (M-2)*3-1 -&amp;gt;&#xA;W=M-3 L=1 D=2&#xA;&#xA;For everything below as long as you have at least 2 defeats and one win&#xA;you can exchange them with 3 draws.&#xA;This goes for all combinations except:&#xA;&#xA;Pts = 0 -&amp;gt;&#xA;W=0 L=M D=0&#xA;&#xA;Pts = 1 -&amp;gt;&#xA;W=0 L=M-1 D=1&#xA;&#xA;Pts = 2 -&amp;gt;&#xA;W=0 L=M-2 D=2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Of course i excluded all combinations that are obviously impossible with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Pts&amp;gt;(M*3)$&lt;/p&gt;&#xA;" OwnerUserId="16608" LastEditorUserId="16608" LastEditDate="2015-10-06T15:30:28.160" LastActivityDate="2015-10-06T15:30:28.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22912" PostTypeId="2" ParentId="22825" CreationDate="2015-10-06T15:44:19.013" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;A life inside the confining dark&lt;br&gt;&#xA;  Their work unseen...&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Something working in a confined dark place. It lives is whole life inside this place.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;...but try to hark&lt;br&gt;&#xA;  Hark to what they say, get steady&lt;br&gt;&#xA;  Go ahead and free one already&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Something that communicates, acknowledges, makes itself present. Something that can be freed, released.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;From my experience I can tell&lt;br&gt;&#xA;  I'm very close to my defeat&lt;br&gt;&#xA;  Maybe a clever mind can help&lt;br&gt;&#xA;  Which one is it I urgently need?&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You need pulse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your arteries live their whole life inside your body (a confined dark place). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their function is to carry the blood away from the hearth and to all your organs. They have a very important role on the human body, even though we can't see it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can hear/feel your pulse by pressing them. It is almost as if they are communicating with you. The sound you feel/hear is your heartbeat. The sound is released when the hearth pumps the blood through its valves. Without it, it means you're dead (defeated).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're feeling ill, dizzy or have any other symptoms like headaches, it could be hearth related, so you should check your pulse and make sure you're not having an hearth attack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Verse one talks of organs, verse two talks of the heart. Having a &quot;hard heart&quot; is a common term. Regarding the computer, maybe you use the computer to talk to more people? The title fits, but I can't get the hint to fit so well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16745" LastEditorUserId="16753" LastEditDate="2015-10-06T22:48:07.277" LastActivityDate="2015-10-06T22:48:07.277" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="22913" PostTypeId="2" ParentId="22825" CreationDate="2015-10-06T16:07:09.347" Score="0" Body="&lt;p&gt;The answer is a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; biological virus.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Para 1:-&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A biological virus or just a virus is considered to be living as well as dead(confined dark) at the same time.The damage they do to our body can't be seen as they are very small.Don't ignore them and try to adopt healthy habits in order to expel(=free) these viruses from the body.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Para 2:-&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; From hint 1,as a virus can also mean a computer virus,so we need to check whether it satisfies all conditions in this para.With passage of time(=experience),either the cure of a viral disease is found by an intelligent doctor or scientist or the almighty human body produces antibodies to tackle the antigens.A computer virus on the other hand might be impossible to destroy and can cause a lot of damage,unless blocked.So,it can't be a viable alternative.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12690" LastEditorUserId="12690" LastEditDate="2016-03-28T04:25:09.837" LastActivityDate="2016-03-28T04:25:09.837" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22914" PostTypeId="2" ParentId="22825" CreationDate="2015-10-06T17:23:41.653" Score="0" Body="&lt;p&gt;I'm going to say it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an idea.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A life inside the confining dark&lt;br&gt;&#xA;Their work unseen, but try to hark&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You cannot &quot;see&quot; ideas, they are inside your head.  And with a little imagination, ideas have a &quot;life&quot; of their own.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Hark to what they say, get steady&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can listen to what your ideas (and intuition) are telling you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Go ahead and free one already&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can free an idea by communicating it to others.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;From my experience I can tell&lt;br&gt;&#xA;I'm very close to my defeat&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You know someone will solve the riddle, and &quot;defeat&quot; you, as the riddler.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Maybe a clever mind can help&#xA;Which one is it I urgently need?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The &quot;one&quot; idea is the idea of &quot;The answer is 'an idea'.&quot; And you &quot;need&quot; that one particular idea so that the riddle is solved (and completed). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And as far as the hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You need to use a computer to be able to access the website and post the answer to &quot;solve&quot; the riddle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11775" LastEditorDisplayName="user20" LastEditDate="2015-10-06T18:27:48.193" LastActivityDate="2015-10-06T18:27:48.193" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22915" PostTypeId="2" ParentId="22907" CreationDate="2015-10-06T17:54:01.970" Score="4" Body="&lt;p&gt;Let $P=$ Pts. We are looking for nonnegative integer solutions to the equations&#xA;$$&#xA;3W+D=P\qquad \qquad W+L+D=M&#xA;$$&#xA;Once $W$ is chosen, the unique choices for $D$ and $L$ which make the above true are $D=P-3W$ and $L=M-P+2W$. Using the inequalities $W,D,L\ge0$, there is a solution as long as $W$ is an integer in the below range: &#xA;$$&#xA;\max(0,\tfrac{P-M}2)\le W\le \tfrac{P}3&#xA;$$&#xA;This means there is a unique solution if and only if these two bounds are equal, so when &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lfloor{\tfrac{P}3}\rfloor=0 \quad \text{or}\quad \lfloor\tfrac{P}3\rfloor=\lceil\tfrac{P-M}2\rceil.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $P\le 2$, the left inequality holds, so $M$ can be anything. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $P\ge 3$, the right equality implies that $P-M$ is either equal to $2\lfloor\tfrac{P}3\rfloor$ or $2\lfloor\tfrac{P}3\rfloor-1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, we get that there is a unique solution if and only if&#xA;$$&#xA;P\le 2 \quad \text{or} \quad M-P+2\lfloor\tfrac{P}3\rfloor=0\text{ or }1&#xA;$$&#xA;For example, in the Villarreal case, $7-16+2\lfloor\tfrac{16}3\rfloor=1$, so there is a unique solution. For Celta, $7-15+2\lfloor\tfrac{15}3\rfloor=2$, so there are multiple solutions.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-10-06T20:40:20.897" LastActivityDate="2015-10-06T20:40:20.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22916" PostTypeId="1" AcceptedAnswerId="22918" CreationDate="2015-10-06T18:29:06.550" Score="8" ViewCount="1485" Body="&lt;p&gt;My family is troubled, but it seems it’s been that way for some time.  While some of us claim to be “shrewd businessmen”, this is sadly not the case; you could say we act in some underground operations.  Time and time again, it’s the money that draws us out of our darkness, out of our safety, where a hit is put on almost every one of us.  By now, we should know the danger, but I guess we lack the foresight.  Personally, I’ve never seen a rat among us, but there’s definitely someone who’s turned on the family.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, do you know my family? &lt;/p&gt;&#xA;" OwnerUserId="11452" LastActivityDate="2015-10-06T19:08:49.940" Title="Is the Fuzz onto My Family?" Tags="|riddle|wordplay|knowledge|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22917" PostTypeId="2" ParentId="22853" CreationDate="2015-10-06T18:41:45.650" Score="0" Body="&lt;p&gt;The lowest thing you can attempt to avoid is the straight and 3 of a kinds. With both, you're unable to do it, since you can only leave 2 gaps, filling 22/26 cards. Straights are worth more than 3 of a kind, but 3 of a kind leads to full houses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how about something like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;     S D H C&#xA;A      x&#xA;K    x x&#xA;Q       &#xA;J    x x&#xA;10   x x&#xA;9    x x x x &#xA;8        x  &#xA;7    x     x&#xA;6    x x x x &#xA;5        x&#xA;4      x   x&#xA;3      x x x&#xA;2        x x&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So, either 2x four of a kind + 1 full house, 2x 2 pairs = 29, or&lt;br&gt;&#xA;2 straights, 2 full house, 2 pair = 31&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On our end, we get 2 royal flushes, 1 straight flush, 1 full house , 2 pairs= 39&lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-10-06T18:47:10.707" LastActivityDate="2015-10-06T18:47:10.707" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22918" PostTypeId="2" ParentId="22916" CreationDate="2015-10-06T18:46:39.327" Score="12" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think I do.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Are you:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Whack A Mole arcade game?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;some of us claim to be “shrewd businessmen”&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A reference to the American Shrew Mole?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;you could say we act in some underground operations&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Moles live in underground burrows.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;it’s the money that draws us out of our darkness&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The moles come out when money is put into the arcade game.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where a hit is put on almost every one of us&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They then each proceed to be hit by a mallet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By now, we should know the danger&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Seeing as it happens everytime someone plays a game..&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Personally, I’ve never seen a rat among us, but there’s definitely someone who’s turned on the family.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A reference to the mole-rat?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12206" LastActivityDate="2015-10-06T18:46:39.327" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22919" PostTypeId="2" ParentId="22916" CreationDate="2015-10-06T18:47:02.947" Score="1" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mustelidae? (Mink, shrew, weasel, etc)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;“shrewd businessmen”&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Straight-forward pun off shrew&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;we act in some underground operations&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Live underground&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;it’s the money that draws us out of our darkness, out of our safety, where a hit is put on almost every one of us&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The pelts can sell for some money&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Personally, I’ve never seen a rat among us, but there’s definitely someone who’s turned on the family&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Very similar to rats. Also a wordplay off of &quot;weasel&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5612" LastActivityDate="2015-10-06T18:47:02.947" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22920" PostTypeId="2" ParentId="22916" CreationDate="2015-10-06T19:08:49.940" Score="1" Body="&lt;p&gt;I know it's a long shot and I'm really reaching here, but are you:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Playing cards&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Family&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can categorise them together as a family, especially face cards (the royal family)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some of us claim to be “shrewd businessmen”&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I have no idea about this one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You could say we act in some underground operations&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Illegal gambling is called &quot;underground&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It’s the money that draws us out of our darkness&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You put down money to draw cards out of the deck for yourself&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where a hit is put on almost every one of us&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You &quot;hit&quot; in blackjack to get another card&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By now, we should know the danger&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because it's a part of every card game with a deck&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I’ve never seen a rat among us&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ratholing is hiding chips and removing them from play&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fuzz&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'Fuzz' is reference to shuffling your hand or the desk, sometimes to preserve the initial order&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16729" LastActivityDate="2015-10-06T19:08:49.940" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22921" PostTypeId="2" ParentId="22853" CreationDate="2015-10-06T19:36:07.983" Score="6" Body="&lt;p&gt;&lt;strong&gt;Answer 1&lt;/strong&gt;, showing his 26 cards:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;X H C S D&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;A * *&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;K * *&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;Q     * *&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;J * *&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;0 * *&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;9     * *&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;8 * *&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;7 * *&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;6     * *&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;5 * *&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;4 * *&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;3     * *&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;2 * *&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He has&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2 flush (6x2)&lt;br&gt;&#xA; 2 straights (5x2)&lt;br&gt;&#xA; 1 2-pair (3)&lt;br&gt;&#xA; 25 points&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The secret here is to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Avoid all 3 of a kinds and as many flushes as possible. This can be done with a spread of 9 of 2 suits and 4 of the other 2 suits. Stagger the cards to avoid straight flushes, and you're done.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answer 2&lt;/strong&gt;, showing my 26 cards:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;X H C S D&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;A * *&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;K * *&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;Q * *&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;J * *&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;0 * * * *&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;9     * *&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;8     * *&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;7     * *&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;6     * *&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;5 * * * *&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;4&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;3     *&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;2     *&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2 Royal flushes (10x2)&lt;br&gt;&#xA; 2 Straight flushes (9x2)&lt;br&gt;&#xA; 1 4 of a kind (8)&lt;br&gt;&#xA; 46 points total&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He has&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 Full house (7)&lt;br&gt;&#xA; 4 Flush (6x4)&lt;br&gt;&#xA; 31 points&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For a difference of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 15 points&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My goal was to frustrate his choices. Give him enough to get a 4 of a kind or a flush (in one suit), then he can have 1 full house, but a second full house costs him 2 flushes. I also took away the ability to get a straight, because that seemed like easy points from leftover cards.&lt;/p&gt;&#xA;" OwnerUserId="375" LastEditorUserId="375" LastEditDate="2015-10-06T21:02:24.087" LastActivityDate="2015-10-06T21:02:24.087" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="22923" PostTypeId="1" CreationDate="2015-10-06T20:12:33.660" Score="-1" ViewCount="283" Body="&lt;p&gt;FROM 2013 &lt;a href=&quot;https://en.wikipedia.org/wiki/Real_Playing_Game&quot; rel=&quot;nofollow&quot;&gt;Real Playing Game&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the &lt;a href=&quot;https://www.youtube.com/watch?v=CRhfOW4S8xs&quot; rel=&quot;nofollow&quot;&gt;trailer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are an old, deathly-ill multimillionaire. A biotech company has developed some brain transfer technology and offers you the chance to participate in a game to win a young body if you pay the asking price of 5 billion dollars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The catch: You will have to play to the death against 9 other players and only the winner will keep the young body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you accept the deal and wake up in a Young Body in an abandoned building and soon you find the other 9 players. In total there are 5 men and 5 women (all young).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some walking you and the other players find the game master and he shows you some holograms of 10 old people (6 men and 4 women) with some bio-briefing. You can't identify which old person is inside of which young body. Even gender cant help you identify the person because some guy chose a women as his new body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the game rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;They have to kill one person every hour or the game will kill one of them at random&lt;/li&gt;&#xA;&lt;li&gt;After they kill someone, the killer has 10 minutes to select the correct hologram of his/her victim or they will be killed. They will also be killed if they make the wrong choice.&lt;/li&gt;&#xA;&lt;li&gt;If killer guesses right, the victim's hologram disappears. If the killer guesses wrong, both the right victim and the now-dead killer's dissapear.&lt;/li&gt;&#xA;&lt;li&gt;You know which is your own hologram&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So what is the best strategy to win this game? Here are some suggested possibilities:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Go hide and hope you aren't selected are random.&lt;/li&gt;&#xA;&lt;li&gt;Kill someone and choose a hologram at random. Would you kill a man or woman?&lt;/li&gt;&#xA;&lt;li&gt;Start with one strategy and switch once less players remain.&lt;/li&gt;&#xA;&lt;li&gt;Form an alliance as long that is mutually beneficial to the other player. Be careful! Don't share your own identity or the other player will kill you for sure.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5165" LastEditorUserId="9000" LastEditDate="2015-10-06T21:16:05.440" LastActivityDate="2015-10-07T00:53:38.830" Title="Kill for a young body" Tags="|strategy|probability|" AnswerCount="2" CommentCount="3" ClosedDate="2015-10-07T04:47:53.583" ContentLicense="CC BY-SA 3.0" />
  <row Id="22924" PostTypeId="2" ParentId="22825" CreationDate="2015-10-06T20:35:36.510" Score="1" Body="&lt;p&gt;My idea is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; computer storage or degrading computer components&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A life inside the confining dark&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Computer enclosures are dark inside&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Their work unseen, but try to hark&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you do not see them working&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hark to what they say, get steady&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For storage, the OS will warn you of running out of space. For computer components, they often give warning when failing. Disk drives overheat, get loud, or stop writing. Flash storage generally becomes readable but not writable. Fans will start working less optimally... etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Go ahead and free one already&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Storage: free memory or drive space. When resources are no longer being used, it is often referred to as &quot;freeing&quot; or &quot;releasing&quot; the CPU/RAM/etc. Components: free the broken part&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From my experience I can tell&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Knowing about computers, or the computer knowing its own components&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm very close to my defeat&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Storage: defeat is a completely filled drive. Failing components: defeat is the end of the part's lifespan&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Maybe a clever mind can help&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a technician, or someone who can work on computers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which one is it I urgently need?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Storage: what region of disk space or what files to delete... or which between disk/RAM. Components: which part needs to be replaced.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Regarding hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I suppose you must use the computer to find if you are out of space, and the computer can also run diagnostics on your hardware. But it seems like awkward phrasing that it was suggested to &quot;use a computer to solve this&quot;. Alternatively, you may need to research all this information... but it was stated &quot;the computer&quot; and not &quot;Google&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16753" LastEditorUserId="16753" LastEditDate="2015-10-06T20:50:33.597" LastActivityDate="2015-10-06T20:50:33.597" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22926" PostTypeId="2" ParentId="22923" CreationDate="2015-10-06T21:30:01.310" Score="0" Body="&lt;p&gt;Until there's more than one opponent, the chances of being killed by the timer is always less or equal than the chances of being killed by choosing the wrong hologram. Besides, you'll be risking getting killed in battle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The exception is when there's only one person left on a gender&lt;/em&gt;. But this exception will only work &lt;em&gt;if&lt;/em&gt; you're sure there was only one gender swap. There's one man in a body of a woman. If you're this man, wait until there's only one left in a gender, then try to kill him/her. If you're not, wait until there's only one man left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have some allies to reduce the chances of battle in case you cross someone by chance, but make sure you keep a low profile from the beginning and avoid confrontations. The key is to rely on chance, survive, and wait until the body count is high enough, or in the best case scenario, the opponents were all killed off. &lt;/p&gt;&#xA;" OwnerUserId="16745" LastEditorUserId="16745" LastEditDate="2015-10-07T00:53:38.830" LastActivityDate="2015-10-07T00:53:38.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22927" PostTypeId="2" ParentId="22923" CreationDate="2015-10-06T21:32:42.693" Score="2" Body="&lt;p&gt;When it comes down to 2 people, you'll want to kill the other person, as you know 100% who the other person is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the beginning you do not want to kill anyone, as you have only 1/10th of the chance of dieing, where if you killed someone you have 8/9ths chance of picking the wrong person and dieing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you had a way to force other people to kill each other, that would be the best strategy, as there is no risk here. Other than that, hang tight until the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. Just because one person changed sex doesn't mean others didn't as well... so guessing based on sex is to no advantage.&lt;/p&gt;&#xA;" OwnerUserId="16678" LastActivityDate="2015-10-06T21:32:42.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22928" PostTypeId="1" AcceptedAnswerId="22936" CreationDate="2015-10-06T21:38:11.640" Score="11" ViewCount="839" Body="&lt;p&gt;After his &lt;a href=&quot;https://puzzling.stackexchange.com/questions/22846/professor-halfbrains-chessboard-theorems/&quot;&gt;first set of theorems&lt;/a&gt; involving queen-accessibility, Professor Halfbrain started wondering about infinite chessboards rather than just ones with arbitrary large dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He came up with a few new theorems:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem 1a.&lt;/strong&gt; If the squares of an infinite chessboard are coloured with two colours, then the squares of at least one of the colours will form a queen-connected set.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Theorem 1b.&lt;/strong&gt; If the squares of an infinite chessboard are coloured with three colours, then the squares of at least one of the colours will form a queen-connected set.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Theorem 1c.&lt;/strong&gt; If the squares of an infinite chessboard are coloured with four colours, then the squares of at least one of the colours will form a queen-connected set.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let a &lt;em&gt;queen-connected component&lt;/em&gt; of a set $S$ of squares be a subset $C$ that is queen-connected, where no other squares in $S$ are queen-accessible from $C$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem 2a.&lt;/strong&gt; If the squares of an infinite chessboard are coloured with five colours, then the squares of at least one of the colours will have fewer than five queen-connected components.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Theorem 2b.&lt;/strong&gt; If the squares of an infinite chessboard are coloured with $n \ge 6$ colours, then the squares of at least one of the colours will have fewer than $n$ queen-connected components.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can you prove or disprove these theorems? For which values of $n$ does theorem 2b hold true?&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2020-07-07T04:40:08.073" Title="Professor Halfbrain's infinite chessboard theorems" Tags="|mathematics|combinatorics|checkerboard|" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22930" PostTypeId="1" AcceptedAnswerId="22931" CreationDate="2015-10-07T00:23:18.103" Score="7" ViewCount="406" Body="&lt;p&gt;Can you identify the function depicted below?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit: this is actually a function over nonnegative integers with the points interpolated by straight lines. I hope this helps.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1lde0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1lde0.png&quot; alt=&quot;Mystery function&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16755" LastEditorUserId="4421" LastEditDate="2015-11-25T18:00:03.560" LastActivityDate="2015-11-25T18:00:03.560" Title="Can you identify this function?" Tags="|mathematics|number-sequence|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22931" PostTypeId="2" ParentId="22930" CreationDate="2015-10-07T01:24:31.413" Score="12" Body="&lt;p&gt;This is &lt;a href=&quot;https://oeis.org/A043537&quot;&gt;A043537&lt;/a&gt;, the number of distinct digits in the base-10 form of n. The peaks in the first part of the graph are from the multiples of 11. After it reaches 100, there is a plateau for 102-109, then it drops for 110 and 111, and goes back up for 120. After that, the dips are for numbers with two digits the same.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-10-07T01:24:31.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22932" PostTypeId="1" AcceptedAnswerId="23247" CreationDate="2015-10-07T02:35:16.017" Score="8" ViewCount="899" Body="&lt;p&gt;A Tom Swifty is a pun involving somebody saying something in a particular manner reminiscent of the thing being said. For example, from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Tom_Swifty&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia article&lt;/a&gt; for Tom Swifty:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;I'll have a martini,&amp;quot; Tom said &lt;strong&gt;dryly&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;I think I'm a frog,&amp;quot; Tom &lt;strong&gt;croaked&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The following Swifties have had their punny phrases replaced with &lt;em&gt;synonyms&lt;/em&gt; of the puns themselves. Can you figure out the original words and make yourselves groan?&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&amp;quot;I have an eighth note,&amp;quot; Tom said &lt;strong&gt;trembling&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&amp;quot;I can't find my underpants,&amp;quot; Tom said &lt;strong&gt;shortly&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&amp;quot;I really hope I get a 7 or 11,&amp;quot; Tom said &lt;strong&gt;riskily&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&amp;quot;I never decrease as I move right,&amp;quot; Tom said &lt;strong&gt;boringly&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&amp;quot;I can't go below this rope, it's too hard,&amp;quot; Tom said &lt;strong&gt;uncertainly&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&amp;quot;I have an unnatural view of my buttocks,&amp;quot; Tom said &lt;strong&gt;retrospectively&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&amp;quot;I baked this bread using a substitute grain,&amp;quot; Tom said with a &lt;strong&gt;dramatic gesture&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&amp;quot;I'm protesting to the singer who popularized Autotune,&amp;quot; Tom told the &lt;strong&gt;meat man&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&amp;quot;I'm the prince from Undertale,&amp;quot; Tom said &lt;strong&gt;in the role of Louis the Métis rioter&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&amp;quot;I feel like I'm creating part of a play,&amp;quot; Tom said while &lt;strong&gt;taking a standardized test&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&amp;quot;I've come up with a new Conservative party member,&amp;quot; Tom said while &lt;strong&gt;listing down items&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&amp;quot;I need to find a way to make this work with Apple's calendar app,&amp;quot; Tom said &lt;strong&gt;meticulously&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-07-03T19:00:57.937" Title="Can you figure out these Tom Swifties?" Tags="|wordplay|english|" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22933" PostTypeId="2" ParentId="22932" CreationDate="2015-10-07T02:57:45.013" Score="2" Body="&lt;p&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;with a quaver&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;monotonously&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;on reflection&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;7.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;flourish&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;10.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;writing the ACT&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;12&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;methodically&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="11876" LastEditDate="2015-10-07T03:13:35.837" LastActivityDate="2015-10-07T03:13:35.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22934" PostTypeId="2" ParentId="22932" CreationDate="2015-10-07T03:00:23.623" Score="1" Body="&lt;p&gt;&lt;strong&gt;Partial Answer&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1, &quot;I have an eighth note,&quot; Tom said &lt;strong&gt;trembling&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I have an eighth note,&quot; Tom said &lt;strong&gt;with a quaver&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5, &quot;I can't go below this rope, it's too hard,&quot; Tom said &lt;strong&gt;uncertainly&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I can't go below this rope, it's too hard,&quot; Tom said, &lt;strong&gt;in limbo&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6, &quot;I have an unnatural view of my buttocks,&quot; Tom said &lt;strong&gt;retrospectively&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;del&gt;&quot;I have an unnatural view of my buttocks,&quot; Tom said &lt;strong&gt;looking back&lt;/strong&gt;.&lt;/del&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;7, &quot;I baked this bread using a substitute grain,&quot; Tom said with a &lt;strong&gt;dramatic gesture&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I baked this bread using a substitute grain,&quot; Tom said with a &lt;strong&gt;wry expression&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;8, &quot;I'm protesting to the singer who popularized Autotune,&quot; Tom told the &lt;strong&gt;meat man&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I'm protesting to the singer who popularized Autotune,&quot; Tom told the &lt;strong&gt;butcher&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="5208" LastEditDate="2015-10-07T03:16:22.590" LastActivityDate="2015-10-07T03:16:22.590" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="22935" PostTypeId="2" ParentId="22932" CreationDate="2015-10-07T03:02:59.680" Score="2" Body="&lt;p&gt;Partial answer for ones that I don't think any one else has got right first:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I have an eighth note,&quot; Tom said &lt;strong&gt;with a quaver&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;I can't find my underpants,&quot; Tom said &lt;strong&gt;briefly&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;I really hope I get a 7 or 11,&quot; Tom said &lt;strong&gt;dicily&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;I have an unnatural view of my buttocks,&quot; Tom said &lt;strong&gt;in hindsight&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;11.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;del&gt;&quot;I've come up with a new Conservative party member,&quot; Tom said while &lt;strong&gt;making &lt;a href=&quot;https://en.wikipedia.org/wiki/A-List_%28Conservative%29&quot; rel=&quot;nofollow&quot;&gt;a list&lt;/strong&gt;&lt;/a&gt;&lt;/del&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13197" LastEditorUserId="13197" LastEditDate="2015-10-07T03:59:32.340" LastActivityDate="2015-10-07T03:59:32.340" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="22936" PostTypeId="2" ParentId="22928" CreationDate="2015-10-07T03:16:53.083" Score="7" Body="&lt;p&gt;I believe the following two claims show that theorems&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 1b, 1c, 2a, 2b are all false.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Here are the claims:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Claim:  On an infinite chessboard, there is an infinite set &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; of queens such that no two queens attack each other and no square on the board is attacked by three queens.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Let &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; be a finite set of red queens such that no two attack each other and no square is attacked by three queens.  Place blue queens on all squares attacked by exactly two red queens.  Now the set of all queens -- red and blue -- is a finite set of queens, and hence there is at least one square &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt; on the infinite chessboard not attacked by any red or blue queen.&lt;/p&gt;&#xA;&lt;p&gt;Put a red queen on &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt; and add this red queen to &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; to create a set &lt;span class=&quot;math-container&quot;&gt;$S'$&lt;/span&gt;  Then &lt;span class=&quot;math-container&quot;&gt;$S'$&lt;/span&gt; is a set of queens where no two attack each other, since no two attack each other in &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt; doesn't attack any other queens.  Furthermore, no square is attacked three times by red queens, since &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt; would need to be one of the attacking queens, but then &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt; would be attacked by a blue queen from before, which is impossible.&lt;/p&gt;&#xA;&lt;p&gt;Since we can continually add queens to &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt;, by doing this infinitely many times we achieve an infinite &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; which proves the claim.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Claim:  It is possible to colour the squares of an infinite chessboard with three colors so that each color has an infinite number of queen-connected components.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Take an infinite set &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; from the previous claim, and colour these squares red, blue, and green so there is an infinite number of each colour.  For any square &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt; attacked by zero or more queens in &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt;, colour it a different colour than any attacking queen.  This is possible since there are at most two queens from &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; attacking &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt; and we have three colours to work with.  Hence every square in &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; is its own isolated component, which proves the claim.&lt;/p&gt;&#xA;" OwnerUserId="2722" LastEditorUserId="1649" LastEditDate="2020-07-07T04:40:08.073" LastActivityDate="2020-07-07T04:40:08.073" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22937" PostTypeId="2" ParentId="22825" CreationDate="2015-10-07T04:49:32.860" Score="3" Body="&lt;p&gt;I feel like this fits but is still a little bit of a stretch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A computer program is what the riddle is referring to, and the program is controlling computer memory, the thing the second verse is asking about.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A life inside the confining dark&lt;br&gt;&#xA;Their work unseen, but try to hark&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Computer programs are &quot;hidden in the dark&quot; and can't be seen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hark to what they say, get steady&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The program is executed and terminated by the user.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Go ahead and free one already&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Free previously used memory.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From my experience I can tell&lt;br&gt;&#xA;I'm very close to my defeat&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The program is compiled and therefore &quot;knows&quot; when it will be terminated.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Maybe a clever mind can help&lt;br&gt;&#xA;Which one is it I urgently need?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The human writing the program decides what to do with the memory.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16759" LastActivityDate="2015-10-07T04:49:32.860" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22938" PostTypeId="1" CreationDate="2015-10-07T05:11:41.550" Score="7" ViewCount="671" Body="&lt;p&gt;Five cousins are camping together.  Two are from Montana, and three are from Wyoming.  Two are Pattersons, and three are Andersons.  Irene and Yvonne are from the same state, while Michael and Emily are from different states.  Leo and Michael have the same last name; however, Emily and Yvonne do not.  The Patterson from Montana awakes in the night to go to the outhouse.  What is that cousin's first name?&lt;/p&gt;&#xA;" OwnerUserId="16761" LastEditorUserId="11100" LastEditDate="2015-10-07T05:33:11.030" LastActivityDate="2015-10-08T15:00:31.310" Title="Five Cousins Camping" Tags="|logic-grid|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22939" PostTypeId="2" ParentId="22938" CreationDate="2015-10-07T05:22:38.543" Score="9" Body="&lt;p&gt;Irene and Yvonne are from the same state&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They cannot both be from Montana, because then the other three cousins would all be from Wyoming and&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Michael and Emily are from different states&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; would be impossible. Therefore Irene and Yvonne are both from Wyoming. The same reasoning over the next two clues gets us that Leo and Michael are Andersons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Patterson from Montana&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Patterson from Montana is not from Wyoming so we can exclude Irene and Yvonne, and not an Anderson so we can exclude Leo and Michael. As there are only 5 cousins and we have excluded 4, we are only left with Emily.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally, we can show that the clues are consistent with an example.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;pre&gt;     Name:     Emily       Irene       Leo        Michael    Yvonne&#xA;     Surname:  Patterson   Patterson   Anderson   Anderson   Anderson&#xA;     State:    Montana     Wyoming     Montana    Wyoming    Wyoming&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And indeed this example fits all of the clues.&lt;/p&gt;&#xA;" OwnerUserId="11100" LastEditorUserId="9000" LastEditDate="2015-10-07T15:37:29.967" LastActivityDate="2015-10-07T15:37:29.967" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22940" PostTypeId="1" AcceptedAnswerId="22941" CreationDate="2015-10-07T07:45:10.113" Score="16" ViewCount="2058" Body="&lt;p&gt;I landed on the island of Knights and Knaves. There are only 2 types of people on the island, knights who always tell the truth and knaves who always lie.  There are 2 leaders of the entire island, Raymond and Martin. Note that Raymond and Martin can both be knights, both be knaves or it can be that one is a knight and one is a knave.&lt;/p&gt;&#xA;&lt;p&gt;I found them both at the same time, one in a Red hat and one in a Blue hat but I was not sure of who was who.&lt;/p&gt;&#xA;&lt;p&gt;I asked both of them: &lt;em&gt;&amp;quot;Is Raymond a Knight?&amp;quot;&lt;/em&gt; Only the one with the blue hat answered, and I can't remember what he said.  But I do remember that at that point I was able to tell who was Raymond and who was Martin.&lt;/p&gt;&#xA;&lt;p&gt;What color was Raymond's Hat?&lt;/p&gt;&#xA;" OwnerUserId="15645" LastEditorUserId="69582" LastEditDate="2023-08-14T16:10:50.020" LastActivityDate="2023-08-14T16:10:50.020" Title="Meta Knights and Knaves Puzzle with Hats" Tags="|liars|meta-knowledge|" AnswerCount="5" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22941" PostTypeId="2" ParentId="22940" CreationDate="2015-10-07T08:07:09.890" Score="14" Body="&lt;p&gt;If the Blue hat guy answered YES (Raymond is a knight), then there were many compatible scenarios:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Raymond indeed was a knight, and Blue=knight=Raymond&lt;/li&gt;&#xA;&lt;li&gt;Raymond indeed was a knight, and Blue=knight=Martin&lt;/li&gt;&#xA;&lt;li&gt;Raymond was a knave, and Blue=knave=Raymond&lt;/li&gt;&#xA;&lt;li&gt;Raymond was a knave, and Blue=knave=Martin&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If the Blue hat guy answered NO (Raymond is a knave), then there are essentially two compatible scenarios:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If Blue=knight, then Raymond=knave and hence Raymond=red&lt;/li&gt;&#xA;&lt;li&gt;If Blue=knave, then Raymond=knight and hence Raymond=red&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Answer:&lt;/strong&gt; Since you were able to tell who was Raymond and who was Martin, the Blue hat guy must have answered NO, and Raymond was the Red hat guy. &#xA;  (But you could not deduce knighthood and knavehood of Raymond and Martin.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-10-07T08:20:55.693" LastActivityDate="2015-10-07T08:20:55.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22942" PostTypeId="1" AcceptedAnswerId="22943" CreationDate="2015-10-07T08:30:44.047" Score="14" ViewCount="1279" Body="&lt;pre&gt;&lt;code&gt;    ----------------------------------------- &#xA;    | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | &#xA;    ----------------------------------------- &#xA;    | 9 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | &#xA;    ----------------------------------------- &#xA;    | 8 | 9 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | &#xA;    ----------------------------------------- &#xA;    | 7 | 8 | 9 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | &#xA;    ----------------------------------------- &#xA;    | 6 | 7 | 8 | 9 | 0 | 1 | 2 | 3 | 4 | 5 |   &#xA;    ----------------------------------------- &#xA;    | 5 | 6 | 7 | 8 | 9 | 0 | 1 | 2 | 3 | 4 | &#xA;    -----------------------------------------&#xA;    | 4 | 5 | 6 | 7 | 8 | 9 | 0 | 1 | 2 | 3 | &#xA;    -----------------------------------------&#xA;    | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 0 | 1 | 2 |  &#xA;    -----------------------------------------&#xA;    | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 0 | 1 |  &#xA;    -----------------------------------------&#xA;    | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 0 | &#xA;    -----------------------------------------&#xA;    &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;br&gt;&#xA;Is it possible to pick ten entries from this table, so that&#xA;&lt;ul&gt;&#xA;&lt;li&gt;each row contains exactly one picked entry;&lt;/li&gt;&#xA;&lt;li&gt;each column contains exactly one picked entry;&lt;/li&gt;&#xA;&lt;li&gt;each of the numbers &lt;span class=&quot;math-container&quot;&gt;$0,1,2,3,4,5,6,7,8,9$&lt;/span&gt; is picked exactly once?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2021-03-14T13:07:41.617" LastActivityDate="2021-03-14T13:07:41.617" Title="Ten distinct numbers in the table" Tags="|mathematics|combinatorics|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22943" PostTypeId="2" ParentId="22942" CreationDate="2015-10-07T09:08:21.490" Score="25" Body="&lt;p&gt;It's not possible.&lt;/p&gt;&#xA;&lt;p&gt;The parity (odd/even) of the entries follows a checkerboard pattern. Labeling the rows and columns 0 to 9 in order, the sum of the entry, its column, and its row is always even. So, summing these over all included entries must give an even number.&lt;/p&gt;&#xA;&lt;p&gt;But, this sum must include each row 0 to 9, whose sum is odd, and likewise for the columns and entries. So, the overall sum must be odd, a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="69582" LastEditDate="2021-02-14T16:21:36.580" LastActivityDate="2021-02-14T16:21:36.580" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22944" PostTypeId="1" AcceptedAnswerId="22946" CreationDate="2015-10-07T10:31:49.240" Score="6" ViewCount="667" Body="&lt;p&gt;So I have a website with some job interview type questions on it. Yesterday I pulled a question out the depths of my memory:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A train enters a station at 50m/s where it starts to decelerate uniformly, coming to a complete stop at the end of the platform. The distance from the entrance to the station to the platform end is 500m. At the exact same moment as the train enters the station a Bee takes off from the buffers at the end of the platform and heads towards the train, when it reaches the front of the train it turns around and heads back to where it started. It continues to do this, start of the platform, front of the train, start of the platform back and forth until the train stops whereupon it is crushed.&lt;/p&gt;&#xA;&lt;p&gt;If the speed of the Bee is 40m/s how far does the Bee travel?&#xA;You can expand this question to determine, for example, the time of the first impact or, &lt;strong&gt;is there such a thing as (and if so what is it), a minimum speed the Bee needs to travel at in order to be able to perform this feat?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I know the first part is easy enough, classic puzzle and all that. But then i got carried away. It's the bit in bold I'm struggling with, i've been thinking about it for half a day and don't have an effective way of modelling it.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;My thinking was this. Can we be sure that when the Bee reaches the train and turns around, flying in the same direction as the train that the Bee will be able to fly faster than the train is currently travelling. I thought this might be interesting because, the faster the Bee is travelling the further down the track it will be when it meets the train and therefore the faster the train will be going. Conversely the slower the Bee flies the slower the train will be going at that point where it meets the train. So modelling this might be interesting.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The question is &lt;a href=&quot;http://puzzles.nigelcoldwell.co.uk/fiftythree.htm&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;bee train puzzle&quot;&gt;here&lt;/a&gt; if anyone is interested. If you can help and I can use your answer I'll of course give you a mention.&lt;/p&gt;&#xA;" OwnerUserId="16719" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-12-10T23:56:22.170" Title="A puzzle about flies and trains" Tags="|mathematics|calculation-puzzle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22945" PostTypeId="2" ParentId="22944" CreationDate="2015-10-07T11:01:11.257" Score="4" Body="&lt;p&gt;The average speed of the train during its deceleration is 25 m/s, since it started at 50 m/s and with uniform deceleration slowed to 0 m/s.  During this time, the bee traveled at 40 m/s, which is $1.6$ times as fast as the average speed of the train.  Thus, the bee traveled $1.6$ times as far, for a total of 800m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The train traveled 500m during its deceleration, at an average speed of 25 m/s, so the time it took to stop is $t = 20s$.  The acceleration is $a = -v_0/t = -2.5 ms^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using a bit of calculus or basic physics, we derive the formula for distance traveled under constant deceleration is $D = v_0 t + 1/2 a t^2$ (recall that $a$ is negative, in this case).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It takes 4 seconds for the train to decelerate to 40 m/s, so the distance traveled in that time is $D = 180m$.  During that time, the bee traveled 160m.  Combined, that is 340m, which is less than the distance from where the bee started to the entrance to the station.  Thus, the train was definitely traveling at less than 40 m/s when the bee first reached it.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2015-10-07T11:01:11.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22946" PostTypeId="2" ParentId="22944" CreationDate="2015-10-07T11:36:39.320" Score="6" Body="&lt;p&gt;The acceleration of the train is $a_{T} = -\frac{5}{2}\mathrm{ms^{-1}}$.&lt;br&gt;&#xA;Using basic calculus we can derive that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v_{T} = 50-\frac{5}{2}t$&lt;br&gt;&#xA;$x_{T} = 50t-\frac{5}{4}t^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the speed of the bee be $v_{B}$. Then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_{B} = 500-v_{B}t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the train first touches the bee, $x_{B} = x_{T}$ and, if the speed of the bee is to be minimal, it has to exactly equal the speed of the train, eg. $v_{B} = v_{T}$. Therefore we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v_{B} = 50-\frac{5}{2}t$&lt;br&gt;&#xA;$50t - \frac{5}{4}t^2 = 500-v_{B}t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, by substitution,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$50t - \frac{5}{4}t^2 = 500-(50-\frac{5}{2}t)t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving gives $t = \frac{20}{3}$ or $t=20$. If $t=20$, $v_{B} = 0$, which makes sense - if the bee does not move, the train will have zero speed when it touches the bee. However, the more interesting solution is $t=\frac{20}{3}$, which gives $v_{B} = \frac{100}{3}\mathrm{ms^{-1}}$. This is the minimum speed of the bee.&lt;/p&gt;&#xA;" OwnerUserId="7308" LastActivityDate="2015-10-07T11:36:39.320" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22956" PostTypeId="1" AcceptedAnswerId="22966" CreationDate="2015-10-07T14:56:06.667" Score="4" ViewCount="7403" Body="&lt;p&gt;Here is the string I was given:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;WWDETHHNECDPHNEHOSHTEPSAOREGOTNEOGWOYTEVWLAAEROREANNOMEEWT&#xA;GTELTEETOVWDIUEHPIBWER&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;All I know about this puzzle (after asking the people who created it) is that there's a key that will either solve or help solve the puzzle.&lt;/p&gt;&#xA;&lt;p&gt;I've tried all kinds of stuff to get something coherent to show up.  I did a frequency analysis and tried different combinations of those letter matches (e.g. &amp;quot;E&amp;quot; is most frequent in the string, so I plugged in &amp;quot;E&amp;quot; to the cipher per ETNORIA, and so on).  It does appear to be natural language based on the frequency analysis, but none of the substitutions I was making seemed to get me anywhere.&lt;/p&gt;&#xA;&lt;p&gt;The double letters are throwing me off too.  There are only so many likely double letter combinations, and the fact that the double-W starts the cipher is hard to work with.  It could be a word like OOPS, but that seems an odd way to start the string.  I thought maybe the double letters indicated something else like a space, or that they should be ignored all together, but that didn't get me too far.&lt;/p&gt;&#xA;&lt;p&gt;I also looked at the most common trigrams and digrams. HNE and ORE repeat twice each so I tried using THE as one of those.  I didn't really have anywhere to go from there.  TE repeats 4 times and NE repeats 3 times.&lt;/p&gt;&#xA;&lt;p&gt;Not having spaces is really making this difficult.  This is part of a series of puzzles they've had sitting around my office, and they've gotten progressively harder.  Based on the last one I solved, it's possible this one goes multiple levels deep, such as revealing another puzzle after solving the cipher.&lt;/p&gt;&#xA;&lt;p&gt;You don't necessarily have to solve this for me, but I could really use some more tips!&lt;/p&gt;&#xA;" OwnerUserId="16770" LastEditorUserId="69582" LastEditDate="2021-07-03T04:09:42.883" LastActivityDate="2021-07-03T04:09:42.883" Title="Help deciphering string of letters without spaces" Tags="|cipher|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="22957" PostTypeId="2" ParentId="1980" CreationDate="2015-10-07T15:14:08.603" Score="2" Body="&lt;p&gt;The other answers assume that the second person uses the first person's silence as additional information.  But what if they are all required to answer at the same time?  Or do so in a predefined order?  Or do so without anyone else knowing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there is still a solution.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Person 2 will always assume he has the opposite of person 3 and say it.&lt;/li&gt;&#xA;&lt;li&gt;If 2 and 3 are the same, then person 1 will say the opposite colour since there can only ever be 2 of the same colour.  Otherwise, a random colour.&lt;/li&gt;&#xA;&lt;li&gt;3/4 will say a random colour.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It is guaranteed that at least one of person 1 or person 2 will be correct.  If person 1 is wrong, then 2 and 3 must have different colours.  But person 2 would have said the opposite colour of 3, so person 2 would be correct.  &lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2015-10-07T15:14:08.603" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22958" PostTypeId="2" ParentId="1980" CreationDate="2015-10-07T15:59:55.480" Score="1" Body="&lt;p&gt;The answer is simple. If you can't see anyone, pick randomly but in the end your guess don't matter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can see someone, then pick the opposite color of the person directly in front of you. This is the highest probability for 2 and if 1 does the same you get the answer no matter what.&lt;/p&gt;&#xA;" OwnerUserId="16678" LastActivityDate="2015-10-07T15:59:55.480" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22959" PostTypeId="2" ParentId="22942" CreationDate="2015-10-07T16:21:21.563" Score="3" Body="&lt;p&gt;It might be possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The square you have provided is a latin square.  While I do not know of one orthgonal to it, here is a &lt;a href=&quot;http://www.cecm.sfu.ca/organics/papers/lam/paper/html/POLS10/POLS10.html&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; to a pair of orthogonal latin squares of order 10.  So it is possible that there is a square orthogonal to the one you gave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets assume a latin square orthogonal to the one you have given exists.  Lets call these latin squares $A$ and $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, create the superposition of these two squares $C$.  Thus, the entries of $C_{ij}=\{A_{ij},B_{ij}\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since these squares are orthogonal, we know that for every $n,m \in \{0-9\}$ there exists an entry in $C$ that is $\{n,m\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, we know that there exists the following entries in $C$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\{0,0\}, \{0,1\}, \{0,2\}, \{0,3\}, \{0,4\}, \{0,5\}, \{0,6\}, \{0,7\}, \{0,8\}, \{0,9\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the first number $0$ comes from square $A$, and $A$ is a latin square, we know that every $0$ comes from a different row and column of $A$.  This also means that every entry containing a $0$ as the first number in $C$ must come from a different row and column of $C$.  Thus, we have a set of entries in $B$, one from each row and one from each column, and all the numbers in $\{0-9\}$ are picked exactly once.  So long as $B$ is the original latin square, we are done!&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2015-10-07T16:21:21.563" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22961" PostTypeId="2" ParentId="22648" CreationDate="2015-10-07T17:06:13.607" Score="-3" Body="&lt;p&gt;If &quot;Without any other forms of measurement&quot; still allows me to have a random bucket of unknown capacity, and I can pour water back into the reservoir, then it can be done. If the difference between the two buckets is 2L, it's trivial. Fill the larger one, then carefully pour from the larger into the smaller. The larger now contains the difference (2L). Pour this into the extra bucket. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do it again and you have 2L in the larger bucket. Pour the smaller one back into the reservoir and fill it from the extra bucket. Done!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the difference is 1L, you need more extra buckets.&lt;/p&gt;&#xA;" OwnerUserId="1509" LastActivityDate="2015-10-07T17:06:13.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22962" PostTypeId="2" ParentId="22942" CreationDate="2015-10-07T17:27:39.733" Score="12" Body="&lt;p&gt;As @xnor said, we can show it is not possible by using a parity argument. Let's look at a few examples of smaller tables to illustrate the argument. Let's start with the obviously impossible 2x2:&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;To get a 0 and a 1, with something in each of columns 0 and 1, and something in each of rows 0 and 1, we need the total sum of rows, columns, and entries to be 3. However, look at the four entries:&lt;/p&gt;&#xA;&lt;p&gt;0 at 0,0 (sum 0)&lt;br /&gt;&#xA;1 at 1,0 (sum 2)&lt;br /&gt;&#xA;1 at 0,1 (sum 2)&lt;br /&gt;&#xA;0 at 1,1 (sum 2)&lt;/p&gt;&#xA;&lt;p&gt;There's no way to get a sum of 3 because they are all even. Let's move on to a 3x3:&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;In this case, there are solutions:&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;* &lt;strong&gt;0&lt;/strong&gt; *&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;* &lt;strong&gt;1&lt;/strong&gt; *&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;* &lt;strong&gt;2&lt;/strong&gt; *&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;However, look at the odds vs evens:&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;* &lt;strong&gt;0&lt;/strong&gt; *&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;* &lt;strong&gt;2&lt;/strong&gt; *&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;* &lt;strong&gt;2&lt;/strong&gt; *&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;* &lt;strong&gt;0&lt;/strong&gt; *&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;* &lt;strong&gt;2&lt;/strong&gt; *&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;* &lt;strong&gt;0&lt;/strong&gt; *&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;Unlike in the 10x10 table, they do not follow a checkerboard pattern. If we calculate the sums:&lt;/p&gt;&#xA;&lt;p&gt;0 at 0,0 = 0&lt;br /&gt;&#xA;1 at 2,1 = 4&lt;br /&gt;&#xA;2 at 1,2 = 5&lt;/p&gt;&#xA;&lt;p&gt;The overall sum is 9 - exactly what we'd expect to get from having 0, 1, and 2, with a representative from each of columns 0, 1, and 2, and a representative from each of rows 0, 1, and 2.&lt;/p&gt;&#xA;&lt;p&gt;Now let's look at a 4x4, with the even numbers marked:&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;* &lt;strong&gt;0&lt;/strong&gt; *&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;* &lt;strong&gt;2&lt;/strong&gt; *&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;* &lt;strong&gt;0&lt;/strong&gt; *&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;* &lt;strong&gt;2&lt;/strong&gt; *&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;* &lt;strong&gt;2&lt;/strong&gt; *&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;* &lt;strong&gt;0&lt;/strong&gt; *&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;* &lt;strong&gt;2&lt;/strong&gt; *&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;* &lt;strong&gt;0&lt;/strong&gt; *&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;So now we know that if an entry's row+column is even, the entry is even. If the entry's row+column is odd, the entry is odd. In both cases, the parity is even. We need 0+1+2+3 from columns 0+1+2+3 and rows 0+1+2+3 = 18, which is even. Since the needed parity is even, parity does not rule out that there might be a solution.&lt;/p&gt;&#xA;&lt;p&gt;It actually still doesn't work. The easiest way to see this is by choosing any 0, which rules out two of each of the other numbers. Then look at the two options for 2 - one rules out both the remaining 1s, and the other rules out both the remaining 3s:&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;With a 5x5:&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;Our parity check says we need (0+1+2+3+4)*3 = 30, and we don't have a checkerboard pattern of odds and evens, so it's worth looking for a solution. It turns out there's quite an easy one to find:&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;* &lt;strong&gt;0&lt;/strong&gt; *&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;* &lt;strong&gt;3&lt;/strong&gt; *&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;* &lt;strong&gt;1&lt;/strong&gt; *&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;* &lt;strong&gt;4&lt;/strong&gt; *&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;* &lt;strong&gt;2&lt;/strong&gt; *&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;With a 6x6:&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;5&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;5&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;5&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;5&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;5&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;5&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;Here the parity check fails. If we start to list out each entry:&lt;/p&gt;&#xA;&lt;p&gt;0 at 0,0 = 0&lt;br /&gt;&#xA;1 at 1,0 = 2&lt;br /&gt;&#xA;2 at 2,0 = 4&lt;br /&gt;&#xA;3 at 3,0 = 6&lt;br /&gt;&#xA;4 at 4,0 = 8&lt;br /&gt;&#xA;5 at 5,0 = 10&lt;br /&gt;&#xA;5 at 0,1 = 6&lt;br /&gt;&#xA;0 at 1,1 = 2&lt;br /&gt;&#xA;1 at 2,1 = 4&lt;br /&gt;&#xA;2 at 3,1 = 6&lt;br /&gt;&#xA;3 at 4,1 = 8&lt;br /&gt;&#xA;...&lt;/p&gt;&#xA;&lt;p&gt;it's easy to see that in each case, the sum of entry+row#+column# is even. However, to get 0+1+2+3+4+5 while representing each of columns 0+1+2+3+4+5 and rows 0+1+2+3+4+5, we need the total sum of entry+row+column to be 45, an odd number. Thus there cannot be a way to solve a 6x6.&lt;/p&gt;&#xA;&lt;p&gt;I'm not going to go any farther, because they're getting big enough that I don't think it's easy enough to see the connection for it to be helpful. However, I did notice that odd-sized tables have a trivial way to generate a solution - choose the zero at 0,0, and then go diagonally up and to the right (wrapping around):&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;* &lt;strong&gt;0&lt;/strong&gt; *&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;5&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;6&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;5&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;6&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;* &lt;strong&gt;3&lt;/strong&gt; *&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;5&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;6&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;* &lt;strong&gt;1&lt;/strong&gt; *&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;5&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;* &lt;strong&gt;6&lt;/strong&gt; *&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;* &lt;strong&gt;4&lt;/strong&gt; *&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;5&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;6&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;* &lt;strong&gt;2&lt;/strong&gt; *&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;5&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;6&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;" OwnerUserId="1955" LastEditorUserId="69582" LastEditDate="2021-02-14T16:42:53.120" LastActivityDate="2021-02-14T16:42:53.120" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22963" PostTypeId="1" AcceptedAnswerId="22964" CreationDate="2015-10-07T17:44:17.530" Score="6" ViewCount="422" Body="&lt;p&gt;While I was working on &lt;a href=&quot;https://puzzling.stackexchange.com/questions/22942/ten-distinct-numbers-in-the-table&quot;&gt;Ten distinct numbers in the table&lt;/a&gt;, &lt;a href=&quot;https://puzzling.stackexchange.com/a/22962/1955&quot;&gt;I found&lt;/a&gt; that odd-sized tables will all work. Additionally, the technique &lt;a href=&quot;https://puzzling.stackexchange.com/a/22943/1955&quot;&gt;@xnor used&lt;/a&gt; to prove a 10x10 doesn't work does not automatically rule out 4x4 and 8x8 tables from working. A 4x4 table doesn't work, but what about an 8x8 table?&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;5&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;6&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;7&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;7&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;5&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;6&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;6&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;7&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;5&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;5&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;6&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;7&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;5&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;6&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;7&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;5&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;6&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;7&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;5&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;6&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;7&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;5&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;6&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;7&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;0&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;Same rules as before - can you pick the numbers 0 through 7 such that you have one of each number, with every column and row being represented exactly once each?&lt;/p&gt;&#xA;&lt;p&gt;If this doesn't work, are there any even-sized tables that do? If it does, is there a general rule for which ones work and which don't?&lt;/p&gt;&#xA;&lt;p&gt;&lt;sub&gt;Parity check: 0-7 add up to 28, so numbers+rows+columns adds up to 84. Even number passes the parity check.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1955" LastEditorUserId="69582" LastEditDate="2021-02-15T04:34:22.547" LastActivityDate="2021-02-15T04:34:22.547" Title="Eight distinct numbers in the table" Tags="|mathematics|combinatorics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22964" PostTypeId="2" ParentId="22963" CreationDate="2015-10-07T17:58:44.623" Score="10" Body="&lt;p&gt;This is impossible for an $n\times n$ table whenever $n$ is even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we number the rows $0,1,\ldots,n-1$ and the columns similarly, the entry in row $i$, column $j$ is congruent to $j-i$ modulo $n$. It follows that if we pick $n$ entries, one from each row and column, the sum of these entries must be divisible by $n$ (modulo $n$, each number $0,1,\ldots,n-1$ gets added once as a column number and subtracted once as a row number). So if it is possible to pick our $n$ numbers to be $0,1,\ldots,n-1$, is must be the case that $0+1+\ldots+(n-1)$ is divisible by $n$. We can compute&#xA;$$&#xA;0+1+\ldots+(n-1)=n\frac{n-1}{2}.&#xA;$$&#xA;This sum is divisible by $n$ if and only if $(n-1)/2$ is an integer, i.e. if and only if $n$ is odd.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastActivityDate="2015-10-07T17:58:44.623" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22965" PostTypeId="2" ParentId="22928" CreationDate="2015-10-07T18:13:32.473" Score="4" Body="&lt;p&gt;Theorem 1c is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; False.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can colour the board as&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;pre&gt; A B C D&#xA; C D A B &lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and then tile this on the entire chessboard.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-10-07T18:13:32.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22966" PostTypeId="2" ParentId="22956" CreationDate="2015-10-07T18:42:21.350" Score="11" Body="&lt;p&gt;This is a columnar transposition cipher. The message was written out in 6 columns, and then the columns were taken in the order 1-2-4-3-6-5 corresponding to the order of the letters in 'DELIVR'. To get the message back, reconstruct the columns:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;WELOVE&#xA;WHATWE&#xA;DOANDW&#xA;ESEEIT&#xA;THROUG&#xA;HTOGET&#xA;HERWHE&#xA;NPEOPL&#xA;ESAYIT&#xA;CANTBE&#xA;DONEWE&#xA;PROVET&#xA;HEMWRO&#xA;NG&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We love what we do and we see it through together. When people say it can't be done we prove them wrong.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-10-07T18:42:21.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22967" PostTypeId="2" ParentId="22928" CreationDate="2015-10-07T18:47:39.583" Score="3" Body="&lt;p&gt;Theorem 1a is also true and it 'follows' from the previous result. Suppose we have colored our infinite chessboard blue and red, and suppose without loss of generality the red squares are &lt;strong&gt;not&lt;/strong&gt; queen-connected. In particular, there must be two red squares $r_1,r_2$ such that they cannot be connected by queen moves. We will show the blue squares must form a queen-connected set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $b_1, b_2$ be any two blue squares. If they can be directly attacked by a queen from each other, there is nothing to prove. If they can't, pick large enough $n$ and an $n \times n$ chessboard $B$ that contains all of $r_1,r_2,b_1$ and $b_2$; it is clear this is always possible. Now simply apply Professor Halfbrain's second theorem to deduce the blue squares in $B$ must form a queen-connected set; in particular, $b_1$ and $b_2$ are queen-connected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The important thing to take away from here is that if two squares are not queen-connected in a board $B$, they remain not queen-connected on a sub-board $B'$.&lt;/p&gt;&#xA;" OwnerUserId="16478" LastEditorUserId="16478" LastEditDate="2015-10-09T01:00:59.613" LastActivityDate="2015-10-09T01:00:59.613" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22971" PostTypeId="1" CreationDate="2015-10-07T20:20:31.570" Score="16" ViewCount="1225" Body="&lt;p&gt;This problem has been vexing me all day. It comes from &lt;a href=&quot;http://www.matrix67.com/iqtest/&quot; rel=&quot;noreferrer&quot;&gt;an old IQ test&lt;/a&gt;. Note that while the test says not to discuss the problems on it in order to maintain its integrity, the test itself closed years ago so this shouldn't matter now. I have managed to narrow it down decisively to 4 possible solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KlohK.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KlohK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've gotten what I thought was the hard part out of the way; that is, I did some matrix witchcraft using an online calculator (this is just a massive linear systems of equations problem) to find the value of all the letters used here... except for &quot;y&quot;, which appears only in &quot;Feynman.&quot; As it turns out, finding y is the real challenge here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each letter has a different value. Here are the letters assigned to each value, in order of value:&#xA;$$&#xA;\begin{array}{cc}&#xA;\text{Value}&amp;amp;\text{Letter}\\\hline&#xA;1&amp;amp; \\&#xA; 2&amp;amp; M\\&#xA; 3&amp;amp; T\\&#xA; 4&amp;amp; V\\&#xA; 5&amp;amp; H\\&#xA; 6&amp;amp; A\\&#xA; 7&amp;amp; E\\&#xA; 8&amp;amp; O\\&#xA; 9&amp;amp; I\\&#xA;\end{array}\hspace{10mm}&#xA;\begin{array}{cc}&#xA;\text{Value}&amp;amp;\text{Letter}\\\hline&#xA;10&amp;amp; U\\&#xA; 11&amp;amp; B\\&#xA; 12&amp;amp; G\\&#xA; 13&amp;amp; K\\&#xA; 14&amp;amp; P\\&#xA; 15&amp;amp; S\\&#xA; 16&amp;amp; F\\&#xA; 17&amp;amp; W\\&#xA; 18&amp;amp; Z&#xA;\end{array}\hspace{10mm}&#xA;\begin{array}{cc}&#xA;\text{Value}&amp;amp;\text{Letter}\\\hline&#xA;19&amp;amp; \\&#xA; 20&amp;amp; C\\&#xA; 21&amp;amp; L\\&#xA; 22&amp;amp; N\\&#xA; 23&amp;amp; R\\&#xA; 24&amp;amp; D\\&#xA; 25&amp;amp; \\&#xA; 26&amp;amp; \\&#xA;&amp;amp;\\&#xA;\end{array}&#xA;$$&#xA;There are known letters with values 2-24, with one exception (19). I think it's safe to assume that the values start at 1 and end at 26, because there are 26 letters in the alphabet. Since every letter has a different value, I suspect that &lt;strong&gt;Y is either 1, 19, 25, or 26.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read through every message board on the internet that has this problem. None of them have explained the solution, but one of them seemed to have a user who knew the solution but refused to divulge it. He said the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Besides,are you sure there isn't room for something like TWO independent (but consistent) systems of equations with juggled integers values regarding alphabetical order ?&#xA;  Fact that choice of value for &quot;Y&quot; isn't logically clear / unique is good enough sign for suspicious minds.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In another post:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;There are mathematically consistent system &amp;amp; subsystems of lin. eq.+&#xA;  logically implemented subsystem to the puzzle.&#xA;  Altogether they generate a unique solution, which isn't 94.&#xA;  One side of the puzzle people here realized fine, but it seems that the other people didn't.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't exactly follow him. From what I know of matrices, there is either exactly one solution, infinitely many solutions, or no solution to large systems of equations like these. So the concept of this equation having another unique set of solutions doesn't make sense in my opinion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more thing: I don't think it's without significance that when the letters and their values are lined up in order of value, all five vowels appear in a row. Y is &quot;sometimes&quot; a vowel. Is this a clue?&lt;/p&gt;&#xA;" OwnerUserId="16782" LastEditorUserId="8874" LastEditDate="2015-10-08T07:32:51.920" LastActivityDate="2018-11-10T18:45:09.547" Title="Names of Physicists Problem" Tags="|logical-deduction|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22972" PostTypeId="1" AcceptedAnswerId="22975" CreationDate="2015-10-07T22:20:51.963" Score="31" ViewCount="10233" Body="&lt;p&gt;There's a game development company which released a mysterious &lt;a href=&quot;http://www.countdowntoreveal.com/&quot; rel=&quot;noreferrer&quot;&gt;&quot;countdown to reveal&quot;&lt;/a&gt; web page, and the page is designed pretty fancy, so everyone in that gaming community is curious about this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/w2cRD.jpg&quot; alt=&quot;countdown web page&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to be a secret puzzle, to which hints are being released. So far, a &lt;a href=&quot;https://vk.com/club89771130&quot; rel=&quot;noreferrer&quot;&gt;forum post by the developers&lt;/a&gt; &lt;sub&gt;(1)&lt;/sub&gt; claims 4 of 6 hints have been released:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gzmbd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;(1)(The forum releasing the information is in Russian, however the game is popular in English speaking countries - If you use Google Chrome to translate that page and view the russian comments, lots of useful information is revealed.)&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Update:&lt;/b&gt; &#xA;&lt;code&gt;TIMERCODE:::SOLVED:::get ready for escape&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;What could the secret message be?&lt;/h2&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Hint 1:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Everyone noticed this code, which is actually slightly visible on the web page:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;N5TYJJETNWRXJ6KBNAYYFJEGPOSXD4EG&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the website code, it has the ID &lt;code&gt;secret&lt;/code&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;&amp;lt;div id=&quot;secret&quot;&amp;gt;N5TYJJETNWRXJ6KBNAYYFJEGPOSXD4EG&amp;lt;/div&amp;gt;&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Hint 2:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;So, I look further into the code, and hidden is another mysterious, hidden element, containing binary data:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;01001011 01001001 01011010 01010100 01000101 01010111 01010011 00110010 01001100 01000010 01001011 01010110 01001111 01010100 01000011 01001010 01001010 01001010 01000010 01000110 01000011 01010101 01000011 01010110 01000111 01011010 01000101 01010101 01010001 01010100 01000011 01000010 01001011 01001010 01000001 01010110 01001011 01010101 00110010 01001110 01001011 01001110 01001001 01000110 01001111 01010111 01010011 01001011 01001011 01010010 01001001 01010101 01001001 01010101 01000011 01001011 01001011 00110101 01001001 01010100 01000111 01010110 00110010 01000100 01001001 01001110 01001010 01000101 00110100 01010110 00110010 01001100 01001010 00110101 01001011 01000101 01000111 01010101 01000011 00110010 01001001 00110101 01001100 01010101 00110110 01001101 01010011 01010101 01001011 01000110 01000010 01010101 01010101 01010110 00110010 01011000 01001011 01000110 01001001 01000101 01010101 01010110 00110010 01000100 01001011 01001010 01001000 01000110 01001111 01010111 01001011 01010000 01001011 01010010 01000011 01010101 01001001 01010111 01010011 01001000 01001011 00110101 01000110 01010100 01000101 01010110 01000011 01010010 01001001 01001110 01001001 01000101 00110010 01010110 00110010 01010010 01000111 01001110 01001011 01000110 01001011 01010001 00110010 01010011 01001010 01011010 01000010 01010110 01001011 01010100 00110010 01010101 01001001 01001110 01000011 01010110 01000101 01010010 00110010 01011000 01001011 01000010 01000110 01000110 01001001 01010101 01001011 01000101 01001011 01000010 01000110 01000110 01001111 01010101 01001010 01010011 01001010 01001010 01001101 01010101 01000111 01010101 01010011 01001011 01001100 01000010 01001011 01010101 00110110 01010110 01000011 01011000 01001001 01010010 01001110 01000101 01001111 01010110 00110010 01010000 01001011 01000010 01001011 01000110 01000011 01010010 01000011 01001001 01001010 01001010 01001100 01010110 01000011 01001101 00110010 01010101 01001011 01001110 01000010 01010110 01000101 01010100 01010011 01000100 01001011 01010110 01001000 01010110 01001001 01010001 01001011 01000101 01001011 01000010 01000100 01010110 01001111 01010111 01010011 01001010 01001010 01001010 01001001 01010101 01001001 01010011 01010011 01010110 01001011 00110101 01001001 01010100 01000111 01010011 01010011 01010010 01001001 01001110 01001010 01000101 01010101 01010110 00110010 01000100 01001001 00110101 01001001 01010001&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which translates to the alphabet characters:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;KIZTEWS2LBKVOTCJJJBFCUCVGZEUQTCBKJAVKU2NKNIFOWSKKRIUIUCKK5ITGV2DINJE4V2LJ5KEGUC2I5LU6MSUKFBUUV2XKFIEUV2DKJHFOWKPKRCUIWSHK5FTEVCRINIE2V2RGNKFKQ2SJZBVKT2UINCVER2XKBFFIUKEKBFFOUJSJJMUGUSKLBKU6VCXIRNEOV2PKBKFCRCIJJLVCM2UKNBVETSDKVHVIQKEKBDVOWSJJJIUISSVK5ITGSSRINJEUV2DI5IQ&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the website code, it has the ID &lt;code&gt;secret_code&lt;/code&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;lt;div id=&quot;secret_code&quot;&amp;gt;&#xA;    &amp;lt;div class=&quot;secret_code&quot;&amp;gt;... binary data here ...&amp;lt;br&amp;gt;&amp;lt;span&amp;gt;_&amp;lt;/span&amp;gt;&amp;lt;/div&amp;gt;&#xA;&amp;lt;/div&amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Hint 3:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vlS4E.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Hint 4:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;This one seems to have a pixelated image of the bulb on the main page in the background. Upon reading into the Javascript, if you click the bulb 10 times, the binary code will drop down from above.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/34lRn.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Hint 5:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The devs recently released this on the FB page. It looks like the bottom of hint 3.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/s6wpf.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Hint 6:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Dh4Ol.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Dh4Ol.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Hint 7:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;This picture, along with the message &quot;the true darkness is transparent&quot;. The image is a picture of a Wikipedia page in Cryllic/Russian. (Page in English: &lt;a href=&quot;https://en.wikipedia.org/wiki/Portable_Network_Graphics#Transparency_of_image&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Portable_Network_Graphics#Transparency_of_image&lt;/a&gt;)&#xA;&lt;img src=&quot;https://i.stack.imgur.com/rPSQC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, importantly, this image was uploaded along with the above:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uDRAN.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uDRAN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QIXgY.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QIXgY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Darkness is transparent&quot;. This one has been worked on in image editors by users on the forum I link to at the top of this post and a short, perhaps 6 character string was discovered by editing it. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Hint 8:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&quot;Look in the dark&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Hint 9:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;A post on their FB page:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;STATUS UPDATE:&#xA;CRRNT RDDL STATUS:&#xA;RDDL ADDED:&#xA;S__RPUZZLE:::INPROGRESS&#xA;&lt;a href=&quot;http://www.countdowntoreveal.com/cam38/c38error_log.txt&quot; rel=&quot;noreferrer&quot;&gt;http://www.countdowntoreveal.com/cam38/c38error_log.txt&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two helpful links from this (decoded the first line of base64 text for the second one and kinda guessed on the other two):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.countdowntoreveal.com/cam38/corrupt2.rar&quot; rel=&quot;noreferrer&quot;&gt;http://www.countdowntoreveal.com/cam38/corrupt2.rar&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.countdowntoreveal.com/cam38/corrupt4.rar&quot; rel=&quot;noreferrer&quot;&gt;http://www.countdowntoreveal.com/cam38/corrupt4.rar&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.countdowntoreveal.com/cam38/node_14-4_kord.part1.rar&quot; rel=&quot;noreferrer&quot;&gt;http://www.countdowntoreveal.com/cam38/node_14-4_kord.part1.rar&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The code to open both of these is 89771130, the number from the Facebook group. However, the file inside cannot be extracted by conventional means.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Hint 10:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;A post on their FB page:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;STATUS UPDATE:&#xA;KNOWTHETRUTH12_ALLSEEINGEYE STARTED&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xL7dl.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xL7dl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Hint 11:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Cn0ar.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Cn0ar.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Additional Hint A:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;As Ivo Beckers pointed out in the comments, &quot;tell me more&quot; displays the following QR code which reads &lt;strong&gt;vjku ku pqv eyuc&lt;/strong&gt;, using this &lt;a href=&quot;https://zxing.org/w/decode?u=http%3A%2F%2Fi.stack.imgur.com%2F7ueaG.png&quot; rel=&quot;noreferrer&quot;&gt;decoder&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then applying a Caesarian shift, this text translates to &lt;strong&gt;this is not cwsa&lt;/strong&gt; which is the solved hint shown above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7ueaG.png&quot; alt=&quot;QR code&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Additional Hint B:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;If one tries inputting the secret code as an extension to the URL, an interesting page is displayed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IcPrN.png&quot; alt=&quot;description&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="16788" LastEditorUserId="11876" LastEditDate="2015-10-23T01:21:46.043" LastActivityDate="2015-10-23T01:21:46.043" Title="What secret is this massive technical puzzle hiding? [riddles and data puzzles]" Tags="|cipher|enigmatic-puzzle|computer-puzzle|alternate-reality-game|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22974" PostTypeId="2" ParentId="22971" CreationDate="2015-10-07T23:54:58.157" Score="7" Body="&lt;p&gt;Buried on page 4 of &lt;a href=&quot;http://www.scienceforums.net/topic/25114-extremely-difficult-question-from-an-iq-test/page-4&quot;&gt;this&lt;/a&gt; message board discussion is a post from a member of the High IQ Society stating that the correct answer (according to the person who set the puzzle) is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; FEYNMAN=101 (because Y=26)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However he also includes this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;note that no two physicists are equally valued&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is incorrect as&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Davisson and Millikan are both 103.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He doesn't explain why the answer is as stated and suggests contacting the puzzle creator. According to the High IQ Society web page, Nathan is still a member, so the stated email address (nathan@highiqsociety.org) should still be valid.&lt;/p&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-10-07T23:54:58.157" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22975" PostTypeId="2" ParentId="22972" CreationDate="2015-10-08T00:21:16.603" Score="16" Body="&lt;p&gt;&lt;strong&gt;UPDATE 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Reorganizing the whole post to include only the parts that weren't dead ends)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first mystery string, &lt;code&gt;N5TYJJETNWRXJ6KBNAYYFJEGPOSXD4EG&lt;/code&gt; is decoded as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Caesar_25(&quot;N5TYJJETNWRXJ6KBNAYYFJEGPOSXD4EG&quot;) -- A =&amp;gt; Z, B =&amp;gt; A, C =&amp;gt; B, ...&#xA;= &quot;M5SXIIDSMVQWI6JAMZXXEIDFONRWC4DF&quot;&#xA;&#xA;decode_base32(&quot;M5SXIIDSMVQWI6JAMZXXEIDFONRWC4DF&quot;) = &quot;get ready for escape&quot;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(Base 32 reference for those who haven't seen it yet: &lt;a href=&quot;https://en.wikipedia.org/wiki/Base32&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second mystery string:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;KIZTEWS2LBKVOTCJJJBFCUCVGZEUQTCBKJAVKU2NKNIFOWSKKRIUIUCKK5ITGV2DINJE4V2LJ5KEGUC2I5LU6MSUKFBUUV2XKFIEUV2DKJHFOWKPKRCUIWSHK5FTEVCRINIE2V2RGNKFKQ2SJZBVKT2UINCVER2XKBFFIUKEKBFFOUJSJJMUGUSKLBKU6VCXIRNEOV2PKBKFCRCIJJLVCM2UKNBVETSDKVHVIQKEKBDVOWSJJJIUISSVK5ITGSSRINJEUV2DI5IQ&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;is decoded as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;decode_base32(&quot;KIZ...&quot;) = &quot;R32ZZXUWLIJBQPU6IHLARAUSMSPWZJTQDPJWQ3WCCRNWKOTCPZGWO2TQCJWWQPJWCRNWYOTEDZGWK2TQCPMWQ3TUCRNCUOTCERGWPJTQDPJWQ2JYCRJXUOTWDZGWOPTQDHJWQ3TSCRNCUOTADPGWZIJQDJUWQ3JQCRJWCGQ&quot;&#xA;&#xA;Caesar_10(&quot;R32ZZX...&quot;) -- A =&amp;gt; K, B =&amp;gt; L, C =&amp;gt; M, ...&#xA;= &quot;B32JJHEGVSTLAZE6SRVKBKECWCZGJTDANZTGA3GMMBXGUYDMZJQGY2DAMTGGAZTGMBXGIYDONJQGU2DAMZWGA3DEMBXMEYDMOBQGZTDANZTGA2TIMBTHEYDGNJQGYZDANRTGA3DCMBXMEYDKNZQGJSTANTEGA3TAMBTGMQA&quot;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The &quot;B32&quot; is a just a tag that says &quot;hey, the rest of this is base32&quot;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;decode_base32(&quot;JJH...&quot;) = &quot;JNCVSX2OJ5PTAXY2f07306f07506e06402f03307207505403606207a06806f07305403903506206306107a05702e06d070033 &quot;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;That string has 3 parts: 15 characters of base32 (ending with the &lt;code&gt;Y&lt;/code&gt;, the last uppercase letter), 86 characters of hex (starting with &lt;code&gt;2f0730&lt;/code&gt;), and one trailing space which seems to serve no purpose at all.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;decode_base32(&quot;JNCVSX2OJ5PTAXY&quot;) = &quot;KEY_NO_0_&quot;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The &lt;code&gt;2f0730...&lt;/code&gt; string, laid out in rows of 3, has an obvious structure:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2f0&#xA;730&#xA;6f0&#xA;750&#xA;6e0&#xA;640&#xA;2f0&#xA;330&#xA;720&#xA;750&#xA;540&#xA;360&#xA;620&#xA;7a0&#xA;680&#xA;6f0&#xA;730&#xA;540&#xA;390&#xA;350&#xA;620&#xA;630&#xA;610&#xA;7a0&#xA;570&#xA;2e0&#xA;6d0&#xA;700&#xA;33&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Remove that column of 0's and you get a simple hex string in which every pair of hexdigits decodes into a printable ASCII character:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;decode_hex(&quot;2f736f756e642f3372755436627a686f735439356263617a572e6d7033&quot;)&#xA;= &quot;/sound/3ruT6bzhosT95bcazW.mp3&quot;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which is a valid path on the server:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;http://countdowntoreveal.com/sound/3ruT6bzhosT95bcazW.mp3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;consisting of beeps (possible Morse code), voices (possibly Russian), and static. And static. And static.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spectrogram of the audio looks something like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nAoq9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nAoq9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some thoughts about the new stuff: there is a strong suggestion we should extract the alpha channel from a PNG and view it as a separate image, but I've tried that on all the images on the page and none of them showed anything interesting. Also viewing the images without their alpha channels. One of the background images, &lt;code&gt;/img/dg.png&lt;/code&gt;, has a little extra viewable area around the edges where the alpha channel was covering it up, but it's obviously only there for artistic effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE 4&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found the last piece of the rar archive at &lt;a href=&quot;http://www.countdowntoreveal.com/cam04/corrupt3.rar&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.countdowntoreveal.com/cam04/corrupt3.rar&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I probed &lt;code&gt;http://www.countdowntoreveal.com/cam01/&lt;/code&gt; through &lt;code&gt;http://www.countdowntoreveal.com/cam99/&lt;/code&gt; and got 404 replies from most of them. The exceptions were:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;code&gt;cam38&lt;/code&gt; (which we already knew about, it gives a 403 reply when you try to get the directory)&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;cam02&lt;/code&gt; and &lt;code&gt;cam04&lt;/code&gt; (403 reply just like &lt;code&gt;cam38&lt;/code&gt; indicating that a directory does exist; only &lt;code&gt;cam04&lt;/code&gt; contained the &lt;code&gt;corrupt3.rar&lt;/code&gt; we needed)&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;cam01&lt;/code&gt; &lt;code&gt;cam03&lt;/code&gt; &lt;code&gt;cam11&lt;/code&gt; &lt;code&gt;cam12&lt;/code&gt; &lt;code&gt;cam33&lt;/code&gt; &lt;code&gt;cam34&lt;/code&gt; &lt;code&gt;cam35&lt;/code&gt; &lt;code&gt;cam36&lt;/code&gt; &lt;code&gt;cam37&lt;/code&gt; - all of these reply to the directory query with an HTML page that says &quot;NO SIGNAL&quot; and &quot;return&quot;, in the colors and font of the main page, with &quot;return&quot; being a link to the main page. There doesn't seem to be any hidden information on these pages; they're apparently just there to discourage people like me from persisting with a search long enough to find &lt;code&gt;cam04&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;With the complete rar file, we can now extract the DICOM image properly. Once again I'll have to shrink it to be allowed to upload it here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/67FnA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/67FnA.jpg&quot; alt=&quot;CAM83 12:43:20:27 NODE 14-4 KORD&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to the skyscrapers and helicopter and &quot;CAM83&quot; that we could see before, we have some text at the bottom of the image: &quot;12:43:20:27&quot; and &quot;NODE 14-4 KORD&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The timestamp &quot;12:43:20:27&quot; appears in the text file &lt;code&gt;c38error_log.txt&lt;/code&gt; and as noted by 2012rcampion it's out of sequence in that file. I don't know what to make of that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A question was asked about the DICOM metadata. I don't see anything interesting in it. Some timestamps pointing to October 8, somet blank fields, basic image metadata (width/height etc.) Here's a dump, created by the &lt;code&gt;medcon&lt;/code&gt; utility:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(0002,0000) UL[1] MetaElementGroupLength: 196 (4 bytes)&#xA;(0002,0001) OB[1] FileMetaInformationVersion: (2 bytes)&#xA;(0002,0002) UI[1] MediaStorageSOPClassUID: [1.2.840.10008.5.1.4.1.1.7] (26 bytes)&#xA;(0002,0003) UI[1] MediaStorageSOPInstanceUID: [1.2.276.0.7230010.3.1.4.3408318492.4584.1444297906.1] (52 bytes)&#xA;(0002,0010) UI[1] TransferSyntaxUID: [1.2.840.10008.1.2.1] (20 bytes)&#xA;(0002,0012) UI[1] ImplementationClassUID: [1.2.276.0.7230010.3.0.3.5.4] (28 bytes)&#xA;(0002,0013) SH[1] ImplementationVersionName: [OFFIS_DCMTK_354] (16 bytes)&#xA;(0008,0016) UI[1] SOPClassUID: [1.2.840.10008.5.1.4.1.1.7] (26 bytes)&#xA;(0008,0018) UI[1] SOPInstanceUID: [1.2.276.0.7230010.3.1.4.3408318492.4584.1444297906.1] (52 bytes)&#xA;(0008,0020) DA[1] StudyDate: [20151008] (8 bytes)&#xA;(0008,0021) DA[1] SeriesDate: [20151008] (8 bytes)&#xA;(0008,0030) TM[1] StudyTime: [095146] (6 bytes)&#xA;(0008,0031) TM[1] SeriesTime: [095146] (6 bytes)&#xA;(0008,0050) SH[0] AccessionNumber: (no value)&#xA;(0008,0060) CS[1] Modality: [OT] (2 bytes)&#xA;(0008,0064) CS[1] ConversionType: [WSD] (4 bytes)&#xA;(0008,0070) LO[0] Manufacturer: (no value)&#xA;(0008,0080) LO[0] InstitutionName: (no value)&#xA;(0008,0090) PN[0] ReferringPhysiciansName: (no value)&#xA;(0008,1030) LO[0] StudyDescription: (no value)&#xA;(0008,103E) LO[0] SeriesDescription: (no value)&#xA;(0010,0010) PN[0] PatientsName: (no value)&#xA;(0010,0020) LO[0] PatientID: (no value)&#xA;(0010,0030) DA[0] PatientsBirthDate: (no value)&#xA;(0010,0032) TM[0] PatientsBirthTime: (no value)&#xA;(0010,0040) CS[0] PatientsSex: (no value)&#xA;(0020,000D) UI[1] StudyInstanceUID: [1.2.276.0.7230010.3.1.2.3408318492.4584.1444297906.2] (52 bytes)&#xA;(0020,000E) UI[1] SeriesInstanceUID: [1.2.276.0.7230010.3.1.3.3408318492.4584.1444297906.3] (52 bytes)&#xA;(0020,0010) SH[0] StudyID: (no value)&#xA;(0020,0011) IS[0] SeriesNumber: (no value)&#xA;(0020,0013) IS[0] InstanceNumber: (no value)&#xA;(0028,0002) US[1] SamplesPerPixel: 3 (2 bytes)&#xA;(0028,0004) CS[1] PhotometricInterpretation: [RGB] (4 bytes)&#xA;(0028,0006) US[1] PlanarConfiguration: 0 (2 bytes)&#xA;(0028,0008) IS[1] NumberOfFrames: [1] (2 bytes)&#xA;(0028,0010) US[1] Rows: 816 (2 bytes)&#xA;(0028,0011) US[1] Columns: 1866 (2 bytes)&#xA;(0028,0100) US[1] BitsAllocated: 8 (2 bytes)&#xA;(0028,0101) US[1] BitsStored: 8 (2 bytes)&#xA;(0028,0102) US[1] HighBit: 7 (2 bytes)&#xA;(0028,0103) US[1] PixelRepresentation: 0 (2 bytes)&#xA;(7FE0,0010) OW[1] PixelData: (4567968 bytes)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE 5&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the /camXX pages now say &quot;Online&quot;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Mq1W4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Mq1W4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, the ones that were 403 Forbidden or &quot;NO SIGNAL&quot; have changed to &quot;ONLINE&quot;, the 404's (at least some of them - I didn't try them all again) are still 404, except for cam83 (which is now &quot;ONLINE&quot;). That's definitely interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;code&gt;/cam83/&lt;/code&gt; exists now where it didn't before. By analogy with &lt;a href=&quot;http://www.countdowntoreveal.com/cam38/c38error_log.txt&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.countdowntoreveal.com/cam38/c38error_log.txt&lt;/a&gt; I suspected &lt;a href=&quot;http://www.countdowntoreveal.com/cam83/c83error_log.txt&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.countdowntoreveal.com/cam83/c83error_log.txt&lt;/a&gt; might exist and I was right. Here it is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;################################################&#xA;&#xA;09:17:12:00,133 ###*#*#*@# fai&amp;amp;*###*#*#*@# fai&amp;amp;*&#xA;09:17:13:02,045 g a#sted service.&#xA;09:17:14:50,553 ting s#### 1450 In@#$$ficient sys##########&#xA;09:17:16:12,423  o# &quot;@###@@#*@###*#*#*@####*#*#*@#50&#xA;09:17:17:00,045 RVJST1JfQUNDRVNTX0RFTklFRCA1ICgweDUp&#xA;09:17:18:03,453 RVJST1JfRklMRV9OT1RfRU5DUllQVEVEIDYwMDcgKDB4MTc3Nykg&#xA;09:17:23:10,023 #####*#*#*@# fai&amp;amp;*@&#xA;09:17:25:00,885 ERROR_FILE_NOT_FOUND 2 (0x2)&#xA;09:17:26:06,053 k## or restore op##ation termi###tin&#xA;09:17:27:10,603 RRnough server sto#age is av4$@able to &#xA;09:17:29:00,035 ERROR @$$50: / hex #SYSTEM_RESOURCES&#xA;09:17:35:03,553 RVJST1JFUlJPUkVSUk9SRVJST1JFUlJPUkVSUk9SRVJST1JFUlJPUkVSUk9SRVJST1JFUlJPUg==&#xA;09:17:36:55,023 ERROR_PATH_NOT_FOUND 3 (0x3)&#xA;09:17:38:04,275 ##UNDEFIENED##&#xA;09:17:45:00,053 Key value 128 ##UNDEFIENED##&#xA;09:17:47:33,323 U2FsdGVkX18z6MR5KMT9DfGS3bP/YMBPnLkbfGNxPZgMR1817s89jBvWPb1WgRa6pP4ZJYNKTAw92hCrEzHVa3LtCK/Xt4yQIFogTBFSzcZ1aOP24BOFkJmaTYmlrHi9QWdESfji5ITIgt9JD8fmX6og9Wl8XZ29rhAIeImVPfShsLEVTswtE1vEM+IiX/xr9UvUql50r0RaN0abjbGQsQ==&#xA;09:17:48:07,035 Back## or###*#*#*@# op##ati###*#*#*@#mi###ting a####ly.&#xA;09:17:49:00,553 ERROR_PATH_NOT_FOUND 3 (0x3)&#xA;09:17:51:20,303 U2FsdGVkX18vM/BTHII1L4LTlvfkMjaB7s/84Al/y1L0hcUGGyMyeARy1TKnL90nB092FcSq2QnHcx03QrIauV6pG5oyisvxqzsJAMuD9gk=&#xA;09:17:54:00,325 ##UNDEFIENED##&#xA;09:18:15:50,553 T3BlcmF0aW5nIHN5cyMjbSBlcnIjIyAxNDUwIEluQCMkJGZpY2llbnQgc3lzIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMj&#xA;09:18:18:12,523 Key value 256 ##UNDEFIENED##&#xA;09:18:19:00,025 hoBFujy/ucmQE6rOElwIxy9aCmDRXlnFtoJ9raVHsbI=&#xA;09:18:22:03,453 RVJST1IgMTEzMDogTm90IGVub3VnaCBzZXJ2ZXIgc3RvI2FnZSBpcyBhdjQkQGFibGUgdG8gIyNjZXNzIHRoaXMgY29tbWFuZC4=&#xA;09:18:25:10,023 ZXJyb3IgYSMjZXNzIExAI0skICNETUw6SiMjICMoKSok&#xA;09:18:26:00,835 Back## or###*#*#*@##ati@#mi###ting a####ly.&#xA;09:18:28:06,053 em reso@@#ces exist $$ complete the&#xA;09:18:34:10,343 ERROR_PATH_NOT_FOUND 3 (0x3)&#xA;09:18:45:00,035 S5tem co4ion EBa###*#*#*@# op##ati###*#*#*@#mi###ting a#sted service.&#xA;09:18:55:03,353 RVJST1JfRklMRV9OT1RfRU5DUllQVEVEIDYwMDcgKDB4MTc3Nykg&#xA;09:18:57:55,033 RVJST1JfQUNDRVNTX0RFTklFRCA1ICgweDUp&#xA;09:19:18:04,675 erat@#######*#*#*@#es exist $$ complet&#xA;09:19:25:00,053 ERROR_ACCESS_DENIED 5 (0x5)&#xA;09:19:27:33,373 m co4ion EBa###*#*#*@# op##ati###&#xA;09:19:28:07,035 ERROR_FILE_NOT_FOUND 2 (0x2)&#xA;09:19:39:00,853 ERROR_N$$%SYSTEM_RESOURCES&#xA;09:19:41:20,303 # ption&#xA;09:19:44:00,025 # c##r&#xA;&#xA;################################################&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;These 3 lines of base64 are especially interesting:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;09:17:47:33,323 U2FsdGVkX18z6MR5KMT9DfGS3bP/YMBPnLkbfGNxPZgMR1817s89jBvWPb1WgRa6pP4ZJYNKTAw92hCrEzHVa3LtCK/Xt4yQIFogTBFSzcZ1aOP24BOFkJmaTYmlrHi9QWdESfji5ITIgt9JD8fmX6og9Wl8XZ29rhAIeImVPfShsLEVTswtE1vEM+IiX/xr9UvUql50r0RaN0abjbGQsQ==&#xA;09:17:51:20,303 U2FsdGVkX18vM/BTHII1L4LTlvfkMjaB7s/84Al/y1L0hcUGGyMyeARy1TKnL90nB092FcSq2QnHcx03QrIauV6pG5oyisvxqzsJAMuD9gk=&#xA;09:18:19:00,025 hoBFujy/ucmQE6rOElwIxy9aCmDRXlnFtoJ9raVHsbI=&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The first and second both decode to something starting with the 8 character string &lt;code&gt;Salted__&lt;/code&gt; followed by random-looking bytes. That format is consistent with the output of the &lt;code&gt;openssl enc&lt;/code&gt; command. The third base64 decodes to 32 random-looking bytes (&lt;code&gt;0x86 0x80 0x45 0xBA 0x3C 0xBF 0xB9 0xC9 0x90 0x13 0xAA 0xCE 0x12 0x5C 0x08 0xC7 0x2F 0x5A 0x0A 0x60 0xD1 0x5E 0x59 0xC5 0xB6 0x82 0x7D 0xAD 0xA5 0x47 0xB1 0xB2&lt;/code&gt;), which could possibly be a key for symmetric encryption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE 6&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After much unsuccessful guessing, I decided to try brute force on the encrypted strings. I made a special-purpose program to do it (&lt;code&gt;openssl enc&lt;/code&gt; stripped down to have no features except the minimum necessary to try a bunch of passwords with &lt;code&gt;aes-128-cbc&lt;/code&gt; on a single hardcoded input). I fed it a word list and got nothing. I fed it a bigger word list and got nothing. There was nothing left to do but try feeding it all possible passwords, starting with short ones and working up to longer ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out the password is &lt;code&gt;1&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The decrypted text is&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ares and mars h##e s#mething in co#mon, but are p##ntless wit%%#t their r^%@# which *&amp;amp;@&amp;amp;*&amp;amp;($ truth for whole bunch of people %##$#@#$###&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The second encrypted string was also encrypted with the password &lt;code&gt;1&lt;/code&gt; but it used &lt;code&gt;aes-256-cbc&lt;/code&gt;. The lines &lt;code&gt;09:17:45:00,053 Key value 128 ##UNDEFIENED##&lt;/code&gt; and &lt;code&gt;09:18:18:12,523 Key value 256 ##UNDEFIENED##&lt;/code&gt; are probably hints about the different key sizes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second string decrypts to this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ERROR @$$50: / hex ##5aa ERROR_N$$%SYSTEM_RESOURCES&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So the first encrypted string is some weird stuff about &quot;ares and mars&quot;, the second is just more of that fake error stuff we've seen already, and the third one is still a mystery.&lt;/p&gt;&#xA;" OwnerDisplayName="user16792" LastEditorUserId="17048" LastEditDate="2015-10-22T15:43:58.693" LastActivityDate="2015-10-22T15:43:58.693" CommentCount="33" ContentLicense="CC BY-SA 3.0" />
  <row Id="22976" PostTypeId="2" ParentId="1980" CreationDate="2015-10-08T00:46:11.943" Score="0" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Number 1 thinks, that if both he and number 2 tell that their hat has the color different to number 3's hat color, then either he or number 2 is right, (see &lt;a href=&quot;https://puzzling.stackexchange.com/a/22957/1967&quot;&gt;this answer&lt;/a&gt;). So he say that he has a black hat.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Number 2 thinks that if number 1 will give an answer, then he does this because he see two heads with the same color (see &lt;a href=&quot;https://puzzling.stackexchange.com/a/1981/1967&quot;&gt;this answer&lt;/a&gt;. Therefore he assumes that his hat has the same color as 3's hat says that his hat is white.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Number 3 errorneously thinks that he cannot know anything because he stares at a wall (see &lt;a href=&quot;https://puzzling.stackexchange.com/a/15990/1967&quot;&gt;this answer&lt;/a&gt;) so he chooses a color by random.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Number 4 knows that if three people choose the same color at most two can be wrong and chooses the same color as number three.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So if number 3 choosed white he  choosed the right color. If he choosed black then number 4 will choose black too, and number 4 will be right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a strategy for number 3, too. He can assume that at least one of 1 or 2 has guessed the right color. This is possible if both 1 and 2 think in the way 2 actally does. It is not possible that 2 thinks in the way 1 actually does because 2 says a different color than 1. Number three should assume both 1 and 2 guessed the wrong color. Then it is imortant to guess the right color ( it is not really important because 4 can save them all). So he shoulkd assume that they did think wrong as they actually did. So he should select the color different to the color selected by 1 and the same color chisen by 2. So he should choose white. &lt;/p&gt;&#xA;" OwnerUserId="1967" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2015-10-08T01:05:04.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22978" PostTypeId="2" ParentId="18437" CreationDate="2015-10-08T09:10:58.600" Score="4" Body="&lt;p&gt;I think this is how we can construct a solution with &lt;s&gt;&lt;strong&gt;501 fuses&lt;/strong&gt;&lt;/s&gt; &lt;strong&gt;32 fuses.&lt;/strong&gt; (Previous 45-fuse solution turned out to be incorrect... sorry =( ...)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;501 fuses&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This is the general diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/f3EEg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/f3EEg.png&quot; alt=&quot;Circuit diagram for N boxes, N/2 fuses&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $N = 1000$ be the number of boxes, $0 \leq m \leq N$ be the number of conductor boxes, $ s = N - m$ be the number of insulator boxes, and define $A = \sum_{i=1}^{F}{a_i}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, from the above circuit diagram, it can be shown that the current across each fuse before the switch is turned on is:&#xA;$$I_j = \frac{V_0}{R(mk+A)+1}a_j$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the 501-fuse solution, we want:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;no fuses blown:&lt;/strong&gt; no insulators&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;1 fuse only blown after 1 s:&lt;/strong&gt; 1 insulator&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;1 fuse blown, another fuse blows 1s later:&lt;/strong&gt; 2 insulators&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;2 fuses only simultaneously blown after 1s:&lt;/strong&gt; 3 insulators&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;2 fuses blown, another fuse blows 1s later:&lt;/strong&gt; 4 insulators&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;3 fuses only simultaneously blown after 1s:&lt;/strong&gt; 5 insulators&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;... and so on. This means we need $F = N/2+1 = 501$ fuses to distinguish $N=1000$ fuses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplicity, set $R = 1 \ \Omega$ and $k = 2 \ \Omega$. To not blow any fuses when there are $s= 0$ insulators, but blow the first fuse when there are $s=1$ insulators, we have the following inequality for $a_1$:&#xA;$$\frac{2N+A-1}{V_0} &amp;lt; a_1 &amp;lt; \frac{2N+A+1}{V_0}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, we can set $a_1 = 1$ and $V_0 = (2N+A) \text{ volts}$. We can similarly find inequalities for any $a_j, j \in (0,N/2+1)$ as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{2N+A-(\sum_{i=1}^{j-1}{a_i}+2(j-1)+1)}{V_0} &amp;lt; a_j &amp;lt; \frac{2N+A-(\sum_{i=1}^{j-1}{a_i}+2(j-1)-1)}{V_0}\\ &#xA;\Rightarrow \ a_j =  1-\frac{\sum_{i=1}^{j-1}{a_i}+2(j-1)}{V_0} \pm \frac{1}{V_0} \ \cdots\cdots\cdots \ (*)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, if we pick the midpoint of each inequality as the value of $a_j$, then we have:&#xA;$$a_j - a_{j+1} = \frac{a_j+2}{V_0}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving this recurrence relation gives us the explicit solution of $a_j$ in terms of $V_0$ as:&#xA;$$ a_j = 3\left(1-\frac{1}{V_0}\right)^{j-1}-2 \ \cdots\cdots\cdots \ (**)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using equation $(*)$, we can obtain $a_{N/2+1}=\frac{N}{V_0-1}$, which we can now substitute $a_{N/2+1}$ into equation $(**)$ and solve for $V_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $A = \sum_{i=1}^{F}{a_i} &amp;lt; N/2+1$, I numerically searched for a root of equation $(**)$ in the range of $(2N, 2N+A) \approx (2000,2500)$ and found $V_0 \approx 2352.96438 \text{ volts}$ as a unique solution (as long as the voltage is accurate up to $\frac{1}{V_0} \approx 4\times10^{-3} \text{ volts}^{-1}$ it should be fine).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting this value of $V_0$ into equation $(**)$ to solve for $\{a_j\}$ gives us the construction.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;32 fuses&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;To get a 32-fuse solution, we generalize our solution by adding new $G$ new fuses $\{b_k\}$ into the circuit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the general diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cu7vU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cu7vU.png&quot; alt=&quot;Circuit diagram for N boxes, sqrt(N)/2 fuses&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $N = 1000$ be the number of boxes, $0 \leq m \leq N$ be the number of conductor boxes, $ s = N - m$ be the number of insulator boxes, and define $A_j = \sum_{i=j}^{F}{a_i}$ and $B = \sum_{i=1}^{G}{b_i}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, from the above circuit diagram, it can be shown that the current across each pf the top $F$ fuses before the switch is turned on is:&#xA;$$I_j = V_A a_j = \frac{V_0}{R(\frac{mkB}{mk+B}+A_1)+1} a_j$$&#xA;$$I_\ell = \frac{V_A mk}{mk + B}b_\ell = \frac{V_0}{\frac{(RA_1+1)B}{mk}+B+RA_1+1} b_\ell$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the regime where the total resistance of the $G$ lower fuses, $1/B$, is perturbational, i.e. $B \gg Nk$. In this case, we can approximate the above two current equations as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_j \approx \frac{V_0}{R(mk+A_1)+1} a_j$$&#xA;$$I_\ell \approx \frac{V_0}{R(mk+A_1)+1} \frac{mkb_\ell}{B}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under this approximation, $I_j$ is as before, and $I_\ell$ scales linearly with the fraction $mkb_\ell/B$. This is suggestive of a nested readout.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(p_a q_b|r_a s_b,...)_a$ denote the scenario where $p$ fuses from group $a$ blow simultaneously after 1 sec, then $y$ fuses from group $b$ blow simultaneously after 2 sec, etc., until no more fuses blow. The pipe denotes the current time point, e.g. $(x|y,z)$ means the time is now between 1 and 2 sec, i.e. after $x$ fuses have blown simultaneously, but before $y$ fuses will blow simultaneously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the following readout scheme:&lt;br&gt;&#xA;$(0_a 0_b, 0_a 0_b): 0$ insulators&lt;br&gt;&#xA;$(0_a 1_b, 0_a 0_b): 1$ insulators&lt;br&gt;&#xA;$(0_a 1_b, 0_a 1_b): 2$ insulators&lt;br&gt;&#xA;$(0_a 2_b, 0_a 0_b): 3$ insulators&lt;br&gt;&#xA;$\cdots$&lt;br&gt;&#xA;$(0_a 15_b, 0_a 1_b): 30$ insulators&lt;br&gt;&#xA;$(0_a 16_b, 0_a 0_b): 31$ insulators&lt;br&gt;&#xA;$(1_a 0_b, 0_a 0_b): 32$ insulators&lt;br&gt;&#xA;$(1_a 1_b, 0_a 0_b): 33$ insulators&lt;br&gt;&#xA;$...$&lt;br&gt;&#xA;$(16_a 16_b, 0_a 0_b): 1023$ insulators    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, $F = G = \sqrt{N}/2$, and $F + G = 32$ fuses can be used to distinguish $N = 1000$ fuses. For simplicity, set $R = 1 \ \Omega$. Similarly as above, we have the following inequality for $a_1$:&#xA;$$(N+1-F)k+A_1 &amp;lt; a_1 V_0 + 1 &amp;lt; Nk+A_1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Correspondingly, we have the following inequalities for any $a_{j \geq 2}$ as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(N+1-jF)k+A_{j-1} &amp;lt; a_j V_0 -1 &amp;lt; (N+1-(j-1)F)k+A_j$$&#xA;$$\text{subject to } 2(N-F) &amp;gt; a_1 &amp;gt; a_2 &amp;gt; ..., \forall a_j$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To determine $b_\ell$, since $\{b_\ell\}$ is decoupled from $\{I_j\}$ in the perturbation approximation, we can always set $b_\ell = \ell B_0/G$ (where $B_0$ is a constant sufficiently large such that $\frac{B_0}{k} \gg N$) and choose $k$ and $\{a_j\}$ such that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$N+1-jF &amp;lt; \frac{a_j}{k} &amp;lt; N+1-(j-1)F$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final set of inequalities can then be solved to obtain an approximate solution, as demonstrated above for the case of 501 fuses. Since $k$ and $B_0$ are unconstrained, a solution for when the approximation is not applied must also exist.&lt;/p&gt;&#xA;" OwnerUserId="15960" LastEditorUserId="15960" LastEditDate="2015-11-05T11:16:23.340" LastActivityDate="2015-11-05T11:16:23.340" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22980" PostTypeId="1" AcceptedAnswerId="22983" CreationDate="2015-10-08T12:31:27.413" Score="8" ViewCount="1332" Body="&lt;p&gt;Karim Benzema likes to remember a small soccer tournament in which he participated many years ago. Karim remembers the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There was a certain number $N$ of teams participating in the tournament. (But Karim does unfortunately not remember the value of $N$.)&lt;/li&gt;&#xA;&lt;li&gt;Each team played exactly one match against each of the other teams.&lt;/li&gt;&#xA;&lt;li&gt;For a win/draw/loss a team respectively scored 3/1/0 points.&lt;/li&gt;&#xA;&lt;li&gt;At the end of the tournament, Karim's team had gained more points than any of the other teams.&lt;/li&gt;&#xA;&lt;li&gt;At the end of the tournament, Karim's team had won fewer matches than any of the other teams.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What is the smallest possible value $N\ge2$ that would be compatible with Karim's recollections?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-10-08T13:45:52.800" Title="Karim Benzema's soccer puzzle" Tags="|mathematics|combinatorics|sports|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22981" PostTypeId="2" ParentId="22980" CreationDate="2015-10-08T13:26:59.670" Score="2" Body="&lt;p&gt;I'm thinking that the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; he recollected wrong&lt;br&gt;&#xA;&lt;br&gt;&#xA; To have more points, but fewer wins than the other teams means that his team had a lot of ties. Let's assume they have 0 wins and all ties. That's an average of 1 point per game.&lt;br&gt;&#xA; In a game with a winner and loser, a team, on average, gets 1.5 points.&lt;br&gt;&#xA; If there are 4 teams, and each of the other teams has a win, a loss, and a tie (you), they all beat you in points.&lt;br&gt;&#xA; If there are 5 teams, with a win, loss, tie (you) and variable, they are already at your point total, with 1 game left.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Keep on extrapolating up, and that win advantage keeps your team from having the most points.&lt;br&gt;&#xA; &lt;strong&gt;Edit&lt;/strong&gt;&lt;br&gt;&#xA; Ah, I see where I went wrong. I was thinking that wins weren't useful, since they'd be offset by wins of the other team. And they are, but the other team gets losses to match the wins, even for the offset wins. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-10-08T13:45:52.800" LastActivityDate="2015-10-08T13:45:52.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22983" PostTypeId="2" ParentId="22980" CreationDate="2015-10-08T13:38:17.950" Score="12" Body="&lt;p&gt;This is possible with $N = 8$ teams.  Karim's team won 2 games, and tied the other 5, for a total of 11 points.  The other teams won three games each, lost three games, and either tied or lost to Karim's team.  The teams that tied Karim's team did best, for a total of 10 points (3 wins, 1 tie).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove this is the minimum, work with the number of wins Karim's team had.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With 0 wins for Karim's team, each other team must have had at least 1 win and 1 loss, playing each other.  So they played at least two games, so there must be at least 3 of them.  For $N=4$, Karim's team got 3 points in ties, while the others got 1 win, 1 tie, for a total of four points.  Adding more teams adds at least as many points to the other teams as it does to Karim's, so that won't help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, with 1 win for Karim's team, the others must win 2 games and lose 2 games, so there must be at least 5 of them.  Karm's team scores 7 points (1 win, 4 ties).  The teams that tie Karim's team score 7 points (2 wins, 1 tie).  Again, adding more teams gives at least as many points to the other teams as it does to Karim's, so that won't help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, Karim's team must have won at least two games.  That works for $N=8$, as described above.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2015-10-08T13:38:17.950" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22984" PostTypeId="2" ParentId="22980" CreationDate="2015-10-08T13:38:58.047" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; N=8&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First we look at all teams except Karim's so $N-1$ teams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they had all won exactly one time against one other team, lost exactly one time (to minimize the highest amount of points) and played a draw in all $N-3$ other games(-1 since they can't play against themself, -1 won, -1 lost), then Karim's Team could only get a maximum of $N-1$ points (all games played with draw) while all other teams get exactly $N$ points (N-3 draw + 1 win).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To maximize the possible interference from Karim's Team against the other Teams lets say that they won exactly half of their games against each other and lost exactly half their matches against each other. Then from $N-2$ games (-1 excluding Karin's team, -1 can't play against themself) they would win exactly $\frac{N-2}{2}$ games and lose the same amount of games.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now Karim's team should not lose against another team, therefore each other Team could only get 1 extra point from Karin's Team in a draw for a maximum of $3\times\frac{N-2}{2}+1=\frac{3N}{2}-2$ points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Karims Team can have at most $\frac{N-2}{2}-1=\frac{N}{2}-2$ wins and $N-1-(\frac{N}{2}-2)=\frac{N}{2}+1$ draws for a total of $3*(\frac{N}{2}-2)+(\frac{N}{2}+1)=2N-5$ points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Karim's team needs at least one point more that the Rest they need $\frac{3N}{2}-1$ points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Rest is simple calculation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2N-5=\frac{3N}{2}-1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{N}{2}-5=-1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{N}{2}=4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N=8$&lt;/p&gt;&#xA;" OwnerUserId="16608" LastActivityDate="2015-10-08T13:38:58.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22986" PostTypeId="2" ParentId="22940" CreationDate="2015-10-08T14:44:23.507" Score="-1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; If the blue one says yes: &lt;br&gt;&lt;br&gt; True: Raymond is a knight. Either the blue one's Raymond, or he's Martin, and Raymond (the red one) is also a knight. &lt;br&gt;&lt;br&gt; False: Raymond is a knave. Either the blue one's Raymond, or if he's Martin, Raymond (the red one) is also a knave. &lt;br&gt;&lt;br&gt; If the blue one says no: &lt;br&gt;&lt;br&gt; True: Raymond is a knave, so the blue one can't be him. &lt;br&gt;&lt;br&gt; False: Raymond is a knight, so the blue one can't be him. &lt;br&gt;&lt;br&gt; He must have said no.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;With less guessing:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; His answer can be yes or no, and true or false. The better answer must rule more out by creating contradictions. One can call themself a truth teller, but when you call yourself a liar, you contradict yourself, so he must have called Raymond one by saying no, ruling out him being Raymond himself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="14016" LastEditDate="2023-08-14T15:53:50.630" LastActivityDate="2023-08-14T15:53:50.630" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="22987" PostTypeId="1" AcceptedAnswerId="22989" CreationDate="2015-10-08T14:59:25.697" Score="4" ViewCount="301" Body="&lt;p&gt;Below is a simple substitution cipher. There are a number of online tools that could solve this in seconds. The easy first step, therefore, is to decode the text to it's original.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bhrp pbwyq phecc bhi swwo uem bieah hrp pwm;&lt;br&gt;&#xA;  Emo Ayrpfrm Ayrpfrem phecc mi’iy sw tq,&lt;br&gt;&#xA;  Nywu bhrp oeq bw bhi imorms wn bhi jwyco,&lt;br&gt;&#xA;  Tvb ji rm rb phecc ti yiuiutiyio-&lt;br&gt;&#xA;  Ji nij, ji heffq nij, ji temo wn tywbhiyp;&lt;br&gt;&#xA;  Nwy hi bw-oeq bheb phiop hrp tcwwo jrbh ui&lt;br&gt;&#xA;  Phecc ti uq tywbhiy; ti hi mi’iy pw krci,&lt;br&gt;&#xA;  Bhrp oeq phecc simbci hrp awmorbrwm;&lt;br&gt;&#xA;  Emo simbciuim rm Imscemo mwj-e-tio&lt;br&gt;&#xA;  Phecc bhrml bhiupickip eaavyp’o bhiq jiyi mwb hiyi,&lt;br&gt;&#xA;  Emo hwco bhiry uemhwwop ahief jhrcip emq pfielp&lt;br&gt;&#xA;  Bheb nwvshb jrbh vp vfwm Permb Ayrpfrm’p oeq.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second step is to figure out the rule by which the cipher's key was made. That may seem broad but there is a very definite  order by which each letter was assigned which might seem very obvious once the key is laid out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My goal is to provide a relatively simple challenge as well as to test how effective this method may be applied as a piece to a larger puzzle.&lt;/p&gt;&#xA;" OwnerUserId="9000" LastActivityDate="2015-10-08T15:14:15.323" Title="First Step Easy, Second Step the Challenge" Tags="|cipher|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22988" PostTypeId="2" ParentId="22938" CreationDate="2015-10-08T15:00:31.310" Score="0" Body="&lt;p&gt;Home states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Their home states are either Montana or Wyoming, so Irene, Yvonne and one of Michael or Emily must be from Wyoming. Leo is from Montana.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Surnames:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Likewise, Leo, Michael and one of Emily/Yvonne are Andersons, and Irene is a Patterson.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Leo and one of Michael/Emily are from Montana, while Irene and one of Emily/Yvonne are Pattersons, with the common name being &lt;strong&gt;Emily&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Complete solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Leo Anderson - Montana, Michael Anderson - Wyoming, Yvonne Anderson - Wyoming, Emily Patterson - Montana, Irene Patterson - Wyoming&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastActivityDate="2015-10-08T15:00:31.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22989" PostTypeId="2" ParentId="22987" CreationDate="2015-10-08T15:14:15.323" Score="4" Body="&lt;p&gt;The solution is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This story shall the good man teach his son;&lt;br&gt;&#xA;And Crispin Crispian shall ne’er go by,&lt;br&gt;&#xA;From this day to the ending of the world,&lt;br&gt;&#xA;But we in it shall be remembered-&lt;br&gt;&#xA;We few, we happy few, we band of brothers;&lt;br&gt;&#xA;For he to-day that sheds his blood with me&lt;br&gt;&#xA;Shall be my brother; be he ne’er so vile,&lt;br&gt;&#xA;This day shall gentle his condition;&lt;br&gt;&#xA;And gentlemen in England now-a-bed&lt;br&gt;&#xA;Shall think themselves accurs’d they were not here,&lt;br&gt;&#xA;And hold their manhoods cheap whiles any speaks&lt;br&gt;&#xA;That fought with us upon Saint Crispin’s day.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The key is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ETAOINSHRDLCUMWFGYPBVKJXQZ0123456789&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is just&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the English alphabet ordered by frequency. A is encoded to the first most common letter, B to the second, and so on.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2015-10-08T15:14:15.323" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22991" PostTypeId="2" ParentId="22852" CreationDate="2015-10-08T16:15:02.090" Score="2" Body="&lt;p&gt;It is not necessarily possible to get an answer as to the nature of Mr. X.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is certainly possible that all of Mr. X's ancestors are genotype TF.  In that scenario, it is then possible that Mr. X is either TT or FF.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this scenario, no matter what the dominance rules are for the two alleles, we cannot know anything about what genes Mr. X received, so we cannot determine his nature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there scenarios where it is possible to determine Mr. X's nature?  &lt;strong&gt;Yes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If two parents are both liars and they have a child that is a truth-teller, then the F-gene must be dominant, and both parents were heterozygous TF$^1$, and the child is homozygous TT.  The logic is the same with T and F reversed and reversed dominance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, if X's father's parents were both liars, but X's parents were both truth-tellers, then F would be the dominant gene, and since X's parents were both truth-tellers, they could not have had the F gene to pass to X, so X would be a truth-teller.  Likewise for X's mothers parents, or for authenticity-switching.  This leaves 4 scenarios.  There are also eight scenarios where one pair of great-grandparents are the dominant type, their child is the recessive type, and X's parents are both the recessive type.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;: Finding $n$.&#xA;To summarize, it is only possible to determine X's nature if X's parents are both of the recessive trait and someone who is of the recessive trait has parents that are both the dominant trait.  The latter condition identifies which trait is recessive, which allows us to know that both of X's parents must be homozygous recessive, so X must also.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, assume X's parents are both the recessive trait and ask questions to see if you can prove it. If they are not both the recessive trait, or you cannot prove they are, then it is impossible to tell X's nature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Annoyingly, you do not know which great-grandparents are paired with which others.  However, I think it is safe to assume that each person knows the answer to the questions &quot;Do you have a (grand)child in common with (point to a person) him/her?&quot;.  I am also assuming it is possible to determine the genders of the 8 people, without asking any questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We'll use the standard trick of asking what someone would say if you asked them a question: &quot;What would you say if I asked you 'Q'?&quot; will always elicit the true answer to question 'Q'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick a woman, label her &quot;A&quot; and use Q = &quot;You have a grandchild that is a parent of X.  Is that child a liar?&quot;  Given the double-asking, this comes out as &quot;You have a grandchild that is a parent of X, if I asked you if that person is a liar, what would you say?&quot;  Assume that the answer we get for that parent is of the recessive trait.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick another woman, label her &quot;C&quot;, and ask her: &quot;If I asked you if you and (point to A) her have a grandchild in common, what would you say?&quot;  If the answer is no, then ask her the same first question.  If the answer is yes, then ask one of the other two women the same first question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have now identified the traits of X's parents.  If the answers do not match, then it is not possible to determine X's traits.  You have asked 3 questions so far.  If they match, continue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If C does not share a grandchild with A, then ask one of the other two women that question.  At this point, relabel the women so that A &amp;amp; C share a grandchild, as do E &amp;amp; G.  At this point, you have asked at 3 or 4 questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ask A &quot;You have a child that is a grandparent of X.  Is I asked you if that child is a liar, what would you say?&quot;  If the answer indicates that child has the dominant trait, ask C the same question.  If this answer is also the dominant trait, then you have your solution.  The best case scenario is 5 questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If either of those answers was the submissive trait, then ask the same question of the other two women.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The worst case scenario is that all four of the grandparents have the submissive trait, as well as both parents.  You can determine this in at most 8 questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, ask each of the eight people &quot;If I asked you if you were a liar, what would you say?&quot;  8 more questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If more than half are of the dominant trait, then there must be at least one pair where both are dominant.  You don't need to determine which; you have succeeded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The worst case scenario is that two of each gender are the dominant trait and two are of the recessive trait.  In this case, ask each of the woman with the dominant trait &quot;If I asked you if he (points to one male with dominant trait) or he (points to other male with dominant trait) were your husband, what would you say?&quot;  If either says yes, you have succeeded.  If both say no, then it is not possible to determine X's trait.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, all tolled, this adds up to 18 questions for the worst case scenario.  In the case that all of X's parents and grandparents are the recessive trait, as are half his great-grandmothers and great-grandfathers, then finding the answer or finding the problem is unsolvable comes down to determining if any of the pairs of great-grandparents were of the dominant trait.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^1$Note: the presence of the O allele does not affect the math.  If two people are truth-tellers and have a child that is a liar, then the T gene must be dominant, and the child must be either FF or FO.  If T is dominant and two people are both liars, then their children must also be liars (or stillbirths, which we know Mr. X and his ancestors are not).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to Mr X.'s gender.  We can assume that the title &quot;Mister&quot; means X is male.  If that assumption is not valid, then we have to add yet another question to determine X's gender, for a total of 19 questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Complication&lt;/strong&gt;: If two people have had a stillbirth, then they must both have had an O gene.  Thus, if X's parents are both the same phenotype (liar or truth-teller), but they have also had a stillbirth, then we know X must be the same type, regardless of gene dominance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second Update&lt;/strong&gt;  Dealing with stillbirths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the possibility that we can use stillbirths to rule out the possibility that someone inherited a trait.  If a person's parents are both one trait, and the person had a sibling that was stillborn, then that person must have that same trait, and cannot have any genes for the opposite trait.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would take 16 questions to determine that all of X's parents, grandparents, and great-grandparents are of the same trait (see above).  The worst case scenario is that X had no stillborn siblings, nor did his parents, but all four of his grandparents did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that the only stillbirths anyone knows about are those among their descendants (including their own children, of course).  Thus, it would take 7 questions to determine that X had no siblings that were stillbirths, that neither of his parents did, and that all 4 of his grandparents did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the new worst-case (yest still solvable) scenario is $n = 23$ questions.  (Plus 1 to determine X's gender, if necessary).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Third Update&lt;/strong&gt;  Adding another, different worst-case scenario, but same limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone is the same trait as both his/her parents, but has siblings of the opposite trait, then we know that the person and parents are the dominant trait, and the sibling is the recessive trait.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, if both of X's parents are the recessive trait, each of them has a parent of each type, and one of the grandparents with the dominant trait has parents with the dominant trait and a sibling with a recessive trait, we can determine for sure which trait is recessive, and that X's parents both have it, so X must, as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This adds up to 16 questions to determine all the ancestor's traits, as above, plus 4 questions to sort which great-grandparents are which, and 2 questions to determine if a grandparent with the dominant trait has a sibling with the recessive trait.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This takes $n = 22$ questions.  However, if after all questions, we still don't know, there is the possibility that X had a stillborn sibling, so we could determine for sure that X had that trait with 1 more question.  Thus brings us up to $n = 23$, which is the same as for the second update.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="2328" LastEditDate="2015-10-08T19:33:11.697" LastActivityDate="2015-10-08T19:33:11.697" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22992" PostTypeId="1" AcceptedAnswerId="23004" CreationDate="2015-10-08T16:47:08.027" Score="5" ViewCount="848" Body="&lt;p&gt;&lt;sub&gt;This is in the spirit of the &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%22what+is+a%22+word+is%3Aquestion&quot;&gt;What is a Word/Phrase™&lt;/a&gt; series started by &lt;a href=&quot;https://puzzling.stackexchange.com/users/463&quot;&gt;JLee&lt;/a&gt; with a special brand of &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%22Phrase%22%20user%3A463%20is%3Aquestion&quot;&gt;Phrase™&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%22What+is+%22+%22Word%22+-missing+user%3A463+is%3Aquestion&quot;&gt;Word™&lt;/a&gt; puzzles.&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;If a word conforms to a special rule, I call it a Structurally Sound Word™.&lt;/p&gt;&#xA;&lt;p&gt;Use the following examples below to find the rule.&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;Structurally Sound Words™&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;Not Structurally Sound Words™&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;DENIED&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;ALLOWED&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;DETERMINED&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;UNDETERMINED&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;FEEBLEMINDED&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;STRONG-WILLED&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;FREETHINKER&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;FOLLOWER&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;IDENTIFIED&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;UNKNOWN&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;INFIELDER&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;PITCHER&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;INFINITE&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;BOUNDED&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;MILLIMETER&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;CENTIMETER&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;PRETEEN&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;TEENAGER&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;REMEMBERED&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;FORGOTTEN&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;REPENT&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;RELAPSE&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;RETIREMENT&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;CAREER&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;THIRTEENTH&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;FOURTEENTH&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;LIMITED&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;EXTRA&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;And, if you want to analyze, here is a CSV version:&lt;/p&gt;&#xA;&lt;pre&gt;&#xA;Structurally Sound Words™,Not Structurally Sound Words™&#xA;DENIED,ALLOWED&#xA;DETERMINED,UNDETERMINED&#xA;FEEBLEMINDED,STRONG-WILLED&#xA;FREETHINKER,FOLLOWER&#xA;IDENTIFIED,UNKNOWN&#xA;INFIELDER,PITCHER&#xA;INFINITE,BOUNDED&#xA;MILLIMETER,CENTIMETER&#xA;PRETEEN,TEENAGER&#xA;REMEMBERED,FORGOTTEN&#xA;REPENT,RELAPSE&#xA;RETIREMENT,CAREER&#xA;THIRTEENTH,FOURTEENTH&#xA;LIMITED,EXTRA&#xA;&lt;/pre&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="69582" LastEditDate="2023-10-20T15:32:13.477" LastActivityDate="2023-10-20T15:32:13.477" Title="What is a Structurally Sound Word™?" Tags="|pattern|english|word-property|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22995" PostTypeId="2" ParentId="22972" CreationDate="2015-10-08T17:34:32.353" Score="3" Body="&lt;p&gt;Based on the latest hint on their FB,&#xA;the binary complex translates to&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;KIZTEWS2LBKVOTCJJJBFCUCVGZEUQTCBKJAVKU2NKNIFOWSKKRIUIUCKK5ITGV2DINJE4V2LJ5KEGUC2I5LU6MSUKFBUUV2XKFIEUV2DKJHFOWKPKRCUIWSHK5FTEVCRINIE2V2RGNKFKQ2SJZBVKT2UINCVER2XKBFFIUKEKBFFOUJSJJMUGUSKLBKU6VCXIRNEOV2PKBKFCRCIJJLVCM2UKNBVETSDKVHVIQKEKBDVOWSJJJIUISSVK5ITGSSRINJEUV2DI5IQ&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which is base32 encoding, decoding it gives&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;R32ZZXUWLIJBQPU6IHLARAUSMSPWZJTQDPJWQ3WCCRNWKOTCPZGWO2TQCJWWQPJWCRNWYOTEDZGWK2TQCPMWQ3TUCRNCUOTCERGWPJTQDPJWQ2JYCRJXUOTWDZGWOPTQDHJWQ3TSCRNCUOTADPGWZIJQDJUWQ3JQCRJWCGQ&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Caesar shift it by 10 gives&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;B32JJHEGVSTLAZE6SRVKBKECWCZGJTDANZTGA3GMMBXGUYDMZJQGY2DAMTGGAZTGMBXGIYDONJQGU2DAMZWGA3DEMBXMEYDMOBQGZTDANZTGA2TIMBTHEYDGNJQGYZDANRTGA3DCMBXMEYDKNZQGJSTANTEGA3TAMBTGMQA&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Take away the first 3 character gives you&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;JJHEGVSTLAZE6SRVKBKECWCZGJTDANZTGA3GMMBXGUYDMZJQGY2DAMTGGAZTGMBXGIYDONJQGU2DAMZWGA3DEMBXMEYDMOBQGZTDANZTGA2TIMBTHEYDGNJQGYZDANRTGA3DCMBXMEYDKNZQGJSTANTEGA3TAMBTGMQA&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is another base 32 encoded string, so decoding it gives you&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;JNCVSX2OJ5PTAXY2f07306f07506e06402f03307207505403606207a06806f07305403903506206306107a05702e06d070033_&#xA;The '_' at the end is a space character&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The lower case characters and number are probably hexes, but I'm not sure where to go from there.&lt;/p&gt;&#xA;" OwnerUserId="16621" LastActivityDate="2015-10-08T17:34:32.353" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22996" PostTypeId="1" AcceptedAnswerId="22998" CreationDate="2015-10-08T17:39:16.137" Score="1" ViewCount="143" Body="&lt;p&gt;Extend the following sequence forward and backward:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   ???? ???? 1808 1816 1824 1828 1836  &#xA;   1840 1840 1844 1848 1848 1852 1856  &#xA;   1860 1864 1868 1876 1880 1880 1884  &#xA;   1888 1896 1900 1908 1912 1920 1920  &#xA;   1928 1932 1944 1952 1960 1960 1952  &#xA;   none ???? ???? ???? ???? ???? ????  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2015-10-08T18:11:56.943" Title="Forward and backward" Tags="|pattern|number-sequence|knowledge|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22997" PostTypeId="1" AcceptedAnswerId="23000" CreationDate="2015-10-08T17:54:34.477" Score="6" ViewCount="1111" Body="&lt;p&gt;&lt;sub&gt;This is in the spirit of the &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%22what+is+a%22+word+is%3Aquestion&quot;&gt;What is a Word/Phrase™&lt;/a&gt; series started by &lt;a href=&quot;https://puzzling.stackexchange.com/users/463&quot;&gt;JLee&lt;/a&gt; with a special brand of &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%22Phrase%22%20user%3A463%20is%3Aquestion&quot;&gt;Phrase™&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%22What+is+%22+%22Word%22+-missing+user%3A463+is%3Aquestion&quot;&gt;Word™&lt;/a&gt; puzzles.&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;If a word conforms to a special rule, I call it a Cursive Word™.&lt;/p&gt;&#xA;&lt;p&gt;Use the following examples below to find the rule.&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;Cursive Words™&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;Not Cursive Words™&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;BILLIONS&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;TRILLIONS&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;CIRCUMCISION&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;VASECTOMY&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;LOLLIPOPS&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;SUCKERS&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;MICROSCOPIC&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;MACROSCOPIC&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;PLUG&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;OUTLET&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;POISONOUS&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;VENOMOUS&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;RUNNING&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;SPRINTING&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;SUBDIVISION&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;NEIGHBORHOOD&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;SUSPICIOUS&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;TRUSTING&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;SWIMMING&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;FLOATING&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;UNCOMPROMISING&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;FLEXIBLE&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;UNCONSCIOUS&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;ALERT&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;VOODOO&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;MAGIC&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;WINDMILL&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;PROPELLOR&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;And, if you want to analyze, here is a CSV version:&lt;/p&gt;&#xA;&lt;pre&gt;&#xA;Cursive Words™,Not Cursive Words™&#xA;BILLIONS,TRILLIONS&#xA;CIRCUMCISION,VASECTOMY&#xA;LOLLIPOPS,SUCKERS&#xA;MICROSCOPIC,MACROSCOPIC&#xA;PLUG,OUTLET&#xA;POISONOUS,VENOMOUS&#xA;RUNNING,SPRINTING&#xA;SUBDIVISION,NEIGHBORHOOD&#xA;SUSPICIOUS,TRUSTING&#xA;SWIMMING,FLOATING&#xA;UNCOMPROMISING,FLEXIBLE&#xA;UNCONSCIOUS,ALERT&#xA;VOODOO,MAGIC&#xA;WINDMILL,PROPELLOR&#xA;&lt;/pre&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="69582" LastEditDate="2023-10-20T15:35:21.797" LastActivityDate="2023-10-20T15:35:21.797" Title="What is a Cursive Word™?" Tags="|pattern|english|word-property|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22998" PostTypeId="2" ParentId="22996" CreationDate="2015-10-08T17:59:13.107" Score="2" Body="&lt;p&gt;Backward:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1789 1796&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Forward:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1976 1980 1980 1992 2000 2008&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The dates that a president won an election for the first time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16729" LastEditorUserId="16729" LastEditDate="2015-10-08T18:06:50.293" LastActivityDate="2015-10-08T18:06:50.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22999" PostTypeId="2" ParentId="22996" CreationDate="2015-10-08T18:05:29.927" Score="2" Body="&lt;p&gt;I think the first two numbers are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1789 and 1796&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the future of the sequence is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1976 1980 1980 1992 2000 2008&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The pattern is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The years of the first US presidential election that each president was on the winning side of&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The pattern &quot;seems&quot; to distort near the end because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Nixon was elected to the vice presidential position before he was elected as president, and Gerald Ford was never elected to office at all.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Actually, I think the original list of numbers may have contained an error.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I think 1836 should be 1832, because van Buren was elected vice president in Jackson's second term.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16814" LastEditorUserId="16814" LastEditDate="2015-10-08T18:11:56.943" LastActivityDate="2015-10-08T18:11:56.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23000" PostTypeId="2" ParentId="22997" CreationDate="2015-10-08T19:03:53.867" Score="7" Body="&lt;p&gt;A Cursive Word is a word which, in that font...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...only needs one stroke per letter to be written. That is, no letters have more than 2 intersections with an odd number of paths coming from them (&quot;intersections&quot; includes endpoints)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2015-10-08T19:03:53.867" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23002" PostTypeId="2" ParentId="22992" CreationDate="2015-10-08T19:20:21.967" Score="2" Body="&lt;p&gt;It seems to me that a structurally sound word is one that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Contains letters that do not have rounded, unstable &quot;bottoms,&quot; that is, the strokes near the bottom of each letter are either flat or are standing on legs. The Not Structurally Sound words contain letters that are rounded and would easily roll over if they were real structures.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16729" LastActivityDate="2015-10-08T19:20:21.967" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23003" PostTypeId="2" ParentId="22992" CreationDate="2015-10-08T20:21:51.333" Score="5" Body="&lt;p&gt;Is a structurally sound word a word that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; only contains letters with a component that goes the entire height of the letter? e.g. 'BDEFHIKLMNPRT'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16034" LastActivityDate="2015-10-08T20:21:51.333" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23004" PostTypeId="2" ParentId="22992" CreationDate="2015-10-08T20:26:03.303" Score="9" Body="&lt;p&gt;A structurally sound word is one such that &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; each letter has a full-height vertical component that could be &quot;load-bearing&quot; -- like a stud -- were it a physical thing. Letters that are full height but without a full vertical component, such as 'A', 'C', and 'X' are structurally 'unsound' since they don't have a &quot;load-bearing&quot; component that runs from top to bottom.&lt;br&gt;&#xA; Structurally sound letters: BDEFHIKLMNPRT&lt;br&gt;&#xA; Structurally unsound letters: ACGJOQSUVWXYZ&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13101" LastActivityDate="2015-10-08T20:26:03.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23006" PostTypeId="1" AcceptedAnswerId="23007" CreationDate="2015-10-08T21:16:23.647" Score="0" ViewCount="2389" Body="&lt;p&gt;A traveling woman arrives in a new city. She enters the hotel of the main street and tells the innkeeper: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I need to rent a room for 15 days. Unfortunately I have no money with me, however I own this jewelry that I can pawn to you right now.&quot; Saying so, she produces a gold chain composed of 15 identical links:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ooooooooooooooo&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Every link is pure gold; surely each one of these is sufficient to cover the cost of one night in your hotel. In 15 days I'll meet someone that owes me a large sum of money. Then I'll be able to pay you and I'll take back my chain.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Well, young lady,&quot; replies the innkeeper &quot;a pawn is fine, but I can't take the responsibility of keeping such a precious object for a long time. I'll tell you what we do. You keep the chain, but you leave me one golden link each day as security. After 15 days you pay me the room with the money you got, and I'll give you back the 15 links.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The woman stays pensive. She'll need a goldsmith to cut her chain, and the goldsmith will ask some money for each link he has to cut and to weld afterwards. Therefore it's better to cut as less links as possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a moment, she finds the solution. What's the smallest number of cuts?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: one reader correctly pointed out a similar question with 7 links. How do you solve the problem for 15 links?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="16323" LastEditorUserId="16323" LastEditDate="2015-10-08T21:32:16.650" LastActivityDate="2015-10-08T21:37:14.430" Title="The golden chain" Tags="|logical-deduction|story|" AnswerCount="1" CommentCount="4" ClosedDate="2015-10-09T06:45:34.583" ContentLicense="CC BY-SA 3.0" />
  <row Id="23007" PostTypeId="2" ParentId="23006" CreationDate="2015-10-08T21:33:03.950" Score="3" Body="&lt;p&gt;You need to make&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2 cuts&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;severing the chain into lengths of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1, 2, 4, and 8. That is, one cut to break into 6, 1, and 8, and then a second cut to break 6 into 2 and 4. I'd enumerate the different combinations, but thanks to binary math, we know that any number less than a given positive power of two can be formed by some sum of all the smaller positive powers of two.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14441" LastEditorUserId="14441" LastEditDate="2015-10-08T21:37:14.430" LastActivityDate="2015-10-08T21:37:14.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23008" PostTypeId="1" AcceptedAnswerId="23014" CreationDate="2015-10-08T21:39:41.807" Score="5" ViewCount="328" Body="&lt;p&gt;In the very near future, Bob Mancave (the one who lives on Krypton street next to the Jawa shop) was presumed dead after having been missing for over a week. He was said to be found in his mancave, under a mountain of pizza crusts and a few thousand custom titled video game boxes (presumely crushed as the shelf which probably held the games was found collapsed next to it). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Detective Hubert Clueless, tasked with inspecting the scene barely skimmed the top of the video game pile (supposedly hiding the corpse), before declaring with omnipotence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Bob isn't dead you buffoons, there is no corpse here, he just probably went to buy a new game.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;But Detective, he's been gone for over a week!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To which the Detective answered:&lt;br&gt;&#xA;&quot;Here, take a look at these titles, you can probably guess what game he went to buy:&quot; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Future Breeze&lt;/li&gt;&#xA;&lt;li&gt;Unending Task&lt;/li&gt;&#xA;&lt;li&gt;Shielded Heart&lt;/li&gt;&#xA;&lt;li&gt;Velocity Urge&lt;/li&gt;&#xA;&lt;li&gt;Era Fighters&lt;/li&gt;&#xA;&lt;li&gt;Crimson Band&lt;/li&gt;&#xA;&lt;li&gt;Pumpkin Chest&lt;/li&gt;&#xA;&lt;li&gt;Robot Attack&lt;/li&gt;&#xA;&lt;li&gt;Networked Nightfall&lt;/li&gt;&#xA;&lt;li&gt;Satan Tears&lt;/li&gt;&#xA;&lt;li&gt;Hit-man Code&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&quot;They just spell it right out! Figure it out yet? Now go get me some mountain dew.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What game is Bob out to buy?&lt;/p&gt;&#xA;" OwnerUserId="11251" LastEditorUserId="11251" LastEditDate="2015-10-09T19:35:02.913" LastActivityDate="2015-10-09T19:35:02.913" Title="Entertainment Discombobulation" Tags="|knowledge|video-games|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23009" PostTypeId="1" AcceptedAnswerId="23011" CreationDate="2015-10-08T21:55:40.243" Score="4" ViewCount="255" Body="&lt;p&gt;I found a math problem &lt;a href=&quot;https://twitter.com/LearningMaths/status/618401821200130048&quot; rel=&quot;nofollow noreferrer&quot;&gt;on Twitter&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as it was presented, its solution is trivial. I managed to make it a bit more challenging by muddling up some equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/F3Lmf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/F3Lmf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you solve the advanced version of the puzzle?&lt;/p&gt;&#xA;" OwnerUserId="16323" LastActivityDate="2015-10-09T04:48:21.360" Title="&quot;Twelve of a times&quot; table" Tags="|mathematics|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23010" PostTypeId="2" ParentId="23009" CreationDate="2015-10-08T22:04:53.880" Score="3" Body="&lt;p&gt;It is the&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Times table, because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Look at the fourth line: flag times person equals person. Therefore flag equals 1. Now look at the third line. Person squared is a 2 digit number starting with 1 (flag). The only such perfect square is 16, so person=4 and underground train is 6.&lt;br&gt;&#xA; Police car times 4 is 10X + 6, police car must be 4 or 9, but 4 is already assigned. Police car is 9, X is 3.&#xA; Present times 4 equals boat, boat is a single digit multiple of 4, 4 is already assigned so boat equals 8 and present equals 2.&#xA; Stop times 4 equals 20 plus tick, meaning stop must equal 5,6 or 7. 6 is already assigned, 5 makes tick equal zero, 7 makes tick equal 8 (already assigned) so stop is 5,  tick is 0. 7 remains unassigned.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3987" LastEditorUserId="3987" LastEditDate="2015-10-08T22:15:17.707" LastActivityDate="2015-10-08T22:15:17.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23011" PostTypeId="2" ParentId="23009" CreationDate="2015-10-08T22:09:21.357" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The flag is 1, because it's the multiplicative identity. The person must be 4, then, because 4 is the only number such that its square is &amp;lt;20 and &gt;=10. The train must then be 6, as it's the second digit of 16. The car must be 9 and the X is 3, because 36 is the only other product of 4 and a single digit number that ends in 6. The present must be 2 and the boat 8, as 8 is the only single digit product of 4 that is not 4 or 0. The only digits remaining are 5 and 0, and conveniently, 5 * 4 = 20, giving us the theta as 5 and the check as 0.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14441" LastActivityDate="2015-10-08T22:09:21.357" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23012" PostTypeId="1" AcceptedAnswerId="23121" CreationDate="2015-10-08T22:23:19.430" Score="1" ViewCount="160" Body="&lt;p&gt;One evokes pizza&lt;br&gt;&#xA;One fuzzy, maybe funny&lt;br&gt;&#xA;Spooks, fruit or pork, frog&lt;br&gt;&#xA;Double letters (in Japan?)&lt;br&gt;&#xA;Two known names; link them to solve&lt;br&gt;&#xA;Bonus: This poem has link, too  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All lines apply to&lt;br&gt;&#xA; two characters in my mind.&lt;br&gt;&#xA; Although both are in&lt;br&gt;&#xA; entertainment, only one&lt;br&gt;&#xA; renown in video games.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Spelling clue somewhere,&lt;br&gt;&#xA; but answers' forms could vary.&lt;br&gt;&#xA; How they sound matter the most.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Warming up for a puzzle inspired by this fortnight's theme... Don't worry; it would not be more of my awful poems.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="15921" LastEditorUserId="15921" LastEditDate="2015-10-09T16:33:56.527" LastActivityDate="2015-10-12T18:45:56.317" Title="What a kooky association" Tags="|knowledge|poetry|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23013" PostTypeId="2" ParentId="23009" CreationDate="2015-10-08T22:35:52.577" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The man doesn't represent 0, so the flag must be 1. As the square of the digit represented by the man starts with 1 on the 3rd row, it's 16 (6 = bus sign) and the man is 4. On the first row, the box can only be 2 and the ship 8. The last row's product starts with 2 and ends with a digit that's not 4 or 8, so the OK sign is 0 and the theta one 5. As for the 2nd row, the product can only be 36, making the X sign 3 and the car 9.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From top to bottom:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4 times table:&lt;br&gt;&#xA; $2 \times 4=8$&lt;br&gt;&#xA; $9 \times 4=36$&lt;br&gt;&#xA; $4 \times 4=16$&lt;br&gt;&#xA; $1 \times 4=4$ and&lt;br&gt;&#xA; $5 \times 4=20$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="4421" LastEditDate="2015-10-09T04:48:21.360" LastActivityDate="2015-10-09T04:48:21.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23014" PostTypeId="2" ParentId="23008" CreationDate="2015-10-08T22:37:09.593" Score="8" Body="&lt;p&gt;The existing games are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Future Breeze - &lt;em&gt;Morrowind&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Unending Task - &lt;em&gt;Everquest&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Shielded Heart - &lt;em&gt;Armored Core&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Velocity Urge - &lt;em&gt;Need for Speed&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Era Fighters - &lt;em&gt;Dynasty Warriors&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Crimson Band - &lt;em&gt;Red Faction&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Pumpkin Chest - &lt;em&gt;Orange Box&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Robot Attack - &lt;em&gt;Mech War&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Networked Nightfall - &lt;em&gt;EVE Online&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Satan Tears - &lt;em&gt;Devil May Cry&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Hit-man Code - &lt;em&gt;Assassin's Creed&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The first letter of each game title spell out ME ANDROMEDA, so the game he has gone out to buy (and will be out for a while, it's not releasing until late next year), is &lt;em&gt;Mass Effect : Andromeda&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Full credits go to @deusovi, @Going hamateur, @Mekalikot, @GentlePurpleRain, and particularly @fodder.&lt;/p&gt;&#xA;" OwnerUserId="2406" LastEditorUserId="2406" LastEditDate="2015-10-09T06:17:52.437" LastActivityDate="2015-10-09T06:17:52.437" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="23017" PostTypeId="2" ParentId="22928" CreationDate="2015-10-09T01:30:45.847" Score="1" Body="&lt;p&gt;The &quot;Theorem 1a&quot; accepts another proof which doesn't depend on topological concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We suppose that both colors are not queen-connected, and derive a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are two squares in the red set ($r_1$ and $r_2$) which aren't queen-acessible from one another, each square attacked by both of them is blue. Otherwise, there would be a two-move path from one of the squares to the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Call one such square $b_1$, and the intesecting attack lines $s$ and $t$ (coming from $r_1$ and $r_2$ respectively).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if one blue pair is not queen-connected, they can't both be able to reach a blue square, since that would imply the existance of a path from one another. So we can take one of our pair which is not connected to $b_1$ and call it $b_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Call $t$ one attack line from $b_2$ which intercepts $r$ and $s$ at a pair of squares. This is always possible, because they are either both orthogonal (and intercepted by a diagonal), both diagonal (and intercepted by an orthogonal), or one orthogonal and one diagonal. In this case, the other orthogonal crosses both. Notice that two diagonals do not always intercept at a square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $t$ crosses $r$, and $t$ crosses $s$, the square must be red, since a blue square would link $b_2$ (contained in $t$) to $b_1$ (contained in $r$ and $s$), which was assumed not to be the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can connect our red pair by going from $r_1$ to $r\cap t$ to $t\cap s$ to $r_2$, since they are all red and all share one attack line with the previous square, contradicting our assumptions. Therefore, either we can't find a blue or a red disconnected pair, so one of the colors is queen-connected.&lt;/p&gt;&#xA;" OwnerUserId="14219" LastEditorUserId="14219" LastEditDate="2015-10-09T13:11:28.300" LastActivityDate="2015-10-09T13:11:28.300" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23018" PostTypeId="2" ParentId="22972" CreationDate="2015-10-09T02:51:59.027" Score="5" Body="&lt;p&gt;(It turns out this clue has already been solved; I'm posting because I haven't seen a description of the solution yet.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the clues is located under the URL &lt;code&gt;http://www.countdowntoreveal.com/spqr/&lt;/code&gt;.  The page is blank, save for the text:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;...THERE IS YZM4 EYES OUT&#xA;THERE, 38 TO BE ZXJYB,&#xA;WHO 3JFBG MAKE AN&#xA;ERROR 9NLN IN A WHILE, I&#xA;R4DA== JUDGE THEM...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The text appears to be interspersed with Base64-encoded fragments.  Viewing the page source reveals the correct capitalization:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;...There is YzM4 eyes out there, 38 to be ZXJyb, who 3JfbG make an error 9nLn in a while,&#xA;i R4dA== judge them...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Piecing together the fragments, we get the following string:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;YzM4ZXJyb3JfbG9nLnR4dA==&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which decodes to:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;c38error_log.txt&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This leads to the URL &lt;code&gt;http://www.countdowntoreveal.com/cam38/c38error_log.txt&lt;/code&gt;, a text file containing the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;################################################&#xA;&#xA;12:42:13:00,003 From server Observer14-4&#xA;12:42:14:00,033 ERROR @#0: Ins#$@icient system res@#$rces exist to complete the requested service.&#xA;12:42:35:00,203 ERROR @$$50: / hex ##5aa ERROR_N$$%SYSTEM_RESOURCES&#xA;12:42:56:00,003 Operating sys##m err## 1450 In@#$$ficient system reso@@#ces exist $$ complete the r%^@@#uested se#v##e.&#xA;12:43:34:00,030 #&#xA;12:43:35:00,003 #####*#*#*@#&#xA;12:43:20:27,000 Error Code 9$%%#@0###*#*#*@#&#xA;12:43:45:00,033 cam##:Remote system node14%#%#%##. The s##* # in an in##id state %#%#%## for #e*@*ed action&#xA;12:43:46:00,233 L2NhbTM4L25vZGVfMTQtNF9rb3JkLnBhcnQxLnJhcg==&#xA;12:43:47:00,133 /@###*#/corrupt2.rar&#xA;12:43:48:00,233 I0AjIyNAQCMqQCMjIyojKiMqQCMjIyMqIyojKkAj&#xA;12:43:56:00,015 #@###@@#*@###*#*#*@####*#*#*@#&#xA;12:43:58:00,343 Write o# &quot;device&quot; failed, status = %%$50&#xA;12:43:59:10,033 ERROR 1130: Not enough server sto#age is av4$@able to ##cess this command.&#xA;12:44:11:00,205 E##OR 1130 / hex 0x46a ERROR_NOT_ENOUGH_SERVER_MEMORY&#xA;12:44:13:00,053 Back## or restore op##ation termi###ting a####ly.&#xA;12:44:16:02,233 Back## or###*#*#*@# op##ati###*#*#*@#mi###ting a####ly.&#xA;12:44:35:00,303 ERROR @$$50: / hex ##5aa ERROR_N$$%SYSTEM_RESOURCES&#xA;12:44:56:00,053 Operating sys##m err## 1450 In@#$$ficient system reso@@#ces exist $$ complete the r%^@@#uested se#v##e.&#xA;12:47:38:00,073 Operating sys##m err## 1450 In@#$$ficient sys################################################&#xA;&#xA;################################################&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This fake error log contains another two Base64-encoded strings:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;L2NhbTM4L25vZGVfMTQtNF9rb3JkLnBhcnQxLnJhcg==&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;/cam38/node_14-4_kord.part1.rar&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I0AjIyNAQCMqQCMjIyojKiMqQCMjIyMqIyojKkAj&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#@###@@#*@###*#*#*@####*#*#*@#&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;currently I'm working on the &quot;timestamps,&quot; which contain four colon-separated parts and are non-monotonic.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I've spent some time trying to transcribe the morse code from the audio file that &lt;a href=&quot;https://puzzling.stackexchange.com/a/22975/9718&quot;&gt;Wumpus Q. Wumbley&lt;/a&gt; found in his answer.  Unfortunately it's been rather difficult, so my current results need more discussion than can fit in a comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've generated two visual aids; first, a spectrogram (0 to 4 kHz, 1024 samples per bin, Hamming windowed):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YKB5k.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YKB5k.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;&lt;sup&gt;(click for full resolution)&lt;/sup&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the appearance of a second signal at 3&amp;nbsp;kHz during the noise bursts; this could be an image of the morse resulting from distortion/clipping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, a correlation of the audio with the first clean 'dit,' absolute-valued and lowpass filtered (200 Hz cutoff):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1M16J.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1M16J.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;&lt;sup&gt;(click for full resolution)&lt;/sup&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also added grid lines every three units (assuming one unit per dot, three per dash, one between parts of a letter, and three between letters) as accurately as I could.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what I've been able to determine:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;code&gt;e&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;Probably &lt;code&gt;m&lt;/code&gt; (&lt;code&gt;dah dah&lt;/code&gt;); but possibly &lt;code&gt;d&lt;/code&gt; (&lt;code&gt;dah di dit&lt;/code&gt;).  Although a gap appears in the &quot;image signal&quot; on the spectrogram, I'm not very trusting of it; the correlation plot doesn't show a significant gap.&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;e r&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;c&lt;/code&gt; (&lt;code&gt;dah di dah dit&lt;/code&gt;) or &lt;code&gt;x&lt;/code&gt; (&lt;code&gt;dah di di dah&lt;/code&gt;); or even possibly &lt;code&gt;6&lt;/code&gt; (&lt;code&gt;dah di di di dit&lt;/code&gt;).  There are two noise bursts right where the two spaces between dots would occur, making this one pretty much impossible to determine.&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;g&lt;/code&gt;, &lt;code&gt;m&lt;/code&gt;, &lt;code&gt;o&lt;/code&gt;, &lt;code&gt;q&lt;/code&gt;, or &lt;code&gt;z&lt;/code&gt;.  All we can see clearly are the first two &lt;code&gt;dah&lt;/code&gt;s, so any of these letters are possible.  The &quot;image signal&quot; looks like the letter ends with &lt;code&gt;dah dit&lt;/code&gt;: this would be &lt;code&gt;Ч&lt;/code&gt; if the signal was Cyrillic morse, but no such letter (&lt;code&gt;dah dah dah dit&lt;/code&gt;) exists in English morse.  The correlation seems to indicate the letter is &lt;code&gt;z&lt;/code&gt; (&lt;code&gt;dah dah di dit&lt;/code&gt;).&lt;/li&gt;&#xA;&lt;li&gt;There might be some more letters underneath the next few noise bursts; it certainly looks like the frequency content is there, but it is somewhat weaker than the previous characters, and there aren't any clearly visible gaps.  Speculation is meaningless here.  It does look like the &quot;image signal&quot; is part of the same signal as the voice cutting in and out of the &quot;main&quot; morse code signal.&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;t&lt;/code&gt; - The next visible letter &lt;em&gt;could&lt;/em&gt; be any letter that ends with a &lt;code&gt;dah&lt;/code&gt;, but there isn't a strong indication of any earlier components to the letter.&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;a r k&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;dah dah ...&lt;/code&gt; - again, this could be any of the (many) letters starting with two &lt;code&gt;dah&lt;/code&gt;s.&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;From the correlation it looks like &lt;code&gt;z&lt;/code&gt; again, followed by &lt;code&gt;di di di di dah&lt;/code&gt; (&lt;code&gt;4&lt;/code&gt;), then possibly &lt;code&gt;dah, dit&lt;/code&gt; (&lt;code&gt;t e&lt;/code&gt;)&lt;/li&gt;&#xA;&lt;li&gt;The image signal is not trustworthy here, since it seems to indicate that the last &lt;code&gt;dah&lt;/code&gt; is part of an extra-long dash.  This would support the hypothesis that the &quot;image signal&quot; is in fact a second morse code signal cutting in and out, and not a distortion of the &quot;main&quot; signal.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;The signal finally ends with what looks like a dash and an extra-long dash; but both are stronger than the rest of the morse code, the second tone is at a lower frequency, and there is no space between the two.  Therefore I suspect this is somehow identifying a part of the transmission, like a sign-off tone.  (Although, since we don't know if this is the complete signal, it could be e.g. identifying the start of the 'next' transmission.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2015-10-13T04:23:31.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23021" PostTypeId="2" ParentId="22648" CreationDate="2015-10-09T07:10:51.383" Score="-1" Body="&lt;p&gt;For 2 litres in each bucket first take Saurabh Prajapati's solution(&lt;a href=&quot;https://puzzling.stackexchange.com/questions/22648/trying-to-find-a-water-measuring-puzzle/23020#23020&quot;&gt;Trying to find a water measuring puzzle&lt;/a&gt;) until the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Y(5l)=5, X(4l)=2, R=13.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Empty &lt;strong&gt;Y&lt;/strong&gt; into &lt;strong&gt;R&lt;/strong&gt; and fill the 2 litres of &lt;strong&gt;X&lt;/strong&gt; into &lt;strong&gt;Y&lt;/strong&gt; afterwards:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Y=2, X=0, R=18$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we fill &lt;strong&gt;X&lt;/strong&gt; completely and empty it into the ground. Do that 4 times:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Y=2, X=0, R=2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we only need to fill the 2 litres from &lt;strong&gt;R&lt;/strong&gt; into &lt;strong&gt;X&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Y=2, X=2, R=0$&lt;/p&gt;&#xA;" OwnerUserId="16608" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-10-10T16:35:22.300" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23022" PostTypeId="1" AcceptedAnswerId="23023" CreationDate="2015-10-09T09:53:50.773" Score="22" ViewCount="2217" Body="&lt;p&gt;Factoid: The noun &quot;assassination&quot; first appeared in print in Shakespeare's Macbeth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, forget that. The puzzle is to reproduce the following ngram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1ZgtH.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1ZgtH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct answer will be a link to the &lt;a href=&quot;https://books.google.com/ngrams&quot; rel=&quot;noreferrer&quot;&gt;Google ngram&lt;/a&gt; website. On following the link, the above graph will be reproduced exactly. Of course I have edited my version of the image to conceal the search term I used. On following your link, the text-box will be shown filled in, thus creating the graph. All other parameters are shown exactly as I submitted them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Detail&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jTRc8.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jTRc8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="8874" LastEditDate="2015-10-09T14:18:54.583" LastActivityDate="2015-10-09T14:18:54.583" Title="Assassinate Shakespeare!" Tags="|enigmatic-puzzle|knowledge|google-ngram|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23023" PostTypeId="2" ParentId="23022" CreationDate="2015-10-09T12:25:09.190" Score="27" Body="&lt;p&gt;Got it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://books.google.com/ngrams/graph?content=Kill%20Bill&amp;amp;year_start=1950&amp;amp;year_end=2000&amp;amp;corpus=15&amp;amp;smoothing=3&amp;amp;share=&amp;amp;direct_url=t1%3B%2CKill%20Bill%3B%2Cc0&quot;&gt;Kill Bill&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not sure much explanation is needed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Just look at the title&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="14135" LastEditDate="2015-10-09T13:03:37.660" LastActivityDate="2015-10-09T13:03:37.660" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23024" PostTypeId="1" AcceptedAnswerId="23026" CreationDate="2015-10-09T13:56:06.350" Score="6" ViewCount="202" Body="&lt;p&gt;Cosmo has arranged the integers $1,2,\ldots,49$ in a $7\times7$ table, so that in each row (left to right) and in each column (top to bottom) the entries occur in increasing order.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Questions:&lt;/strong&gt;&lt;br&gt;&#xA;  What is the &lt;strong&gt;smallest&lt;/strong&gt; possible sum of the numbers in the fourth row in such an arrangement?&lt;br&gt;&#xA;  What is the &lt;strong&gt;largest&lt;/strong&gt; possible sum of the numbers in the fourth row in such an arrangement?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-10-09T14:11:48.350" LastActivityDate="2015-10-11T08:54:18.467" Title="Increasing rows and columns" Tags="|mathematics|combinatorics|no-computers|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23025" PostTypeId="1" AcceptedAnswerId="23027" CreationDate="2015-10-09T14:21:44.870" Score="14" ViewCount="1604" Body="&lt;p&gt;&lt;sub&gt;You are the protagonist in a game designed by NeedAName Inc. Proceed through the levels using&#xA;your video game knowledge to solve the puzzles therein. Solve all levels and beat the game!&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Level 1: Whose Lines Are These Anyways?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/23155/dungeon-crawler-level-2-5&quot;&gt;Next Level&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You enter the first room of the game's dungeon and see a wall displaying a list of data. Each line has a sequence of numbers followed by a genre of game and you recognize that each line represents a quote from a game. Use this information to determine the quote and the game/series that the quote is from.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;A&lt;/strong&gt;. &lt;code&gt;2 1 6 4!&lt;/code&gt; - &lt;em&gt;Shooter/Flyer&lt;/em&gt;&lt;br&gt;&#xA;  &lt;strong&gt;B&lt;/strong&gt;. &lt;code&gt;3 2 3!&lt;/code&gt; - &lt;em&gt;Adventure/RPG&lt;/em&gt;&lt;br&gt;&#xA;  &lt;strong&gt;C&lt;/strong&gt;. &lt;code&gt;8!&lt;/code&gt; - &lt;em&gt;Fighter&lt;/em&gt;&lt;br&gt;&#xA;  &lt;strong&gt;D&lt;/strong&gt;. &lt;code&gt;3. 3 5 7.&lt;/code&gt; - &lt;em&gt;Adventure/RPG&lt;/em&gt;&lt;br&gt;&#xA;  &lt;strong&gt;E&lt;/strong&gt;. &lt;code&gt;6 3!&lt;/code&gt; - &lt;em&gt;Fighter&lt;/em&gt;&lt;br&gt;&#xA;  &lt;strong&gt;F&lt;/strong&gt;. &lt;code&gt;5 3 6&lt;/code&gt; - &lt;em&gt;First-Person Shooter&lt;/em&gt;&lt;br&gt;&#xA;  &lt;strong&gt;G&lt;/strong&gt;. &lt;code&gt;4 1 5, 3 6.&lt;/code&gt; - &lt;em&gt;Adventure/RPG&lt;/em&gt;&lt;br&gt;&#xA;  &lt;strong&gt;H&lt;/strong&gt;. &lt;code&gt;1 2 5.&lt;/code&gt; - &lt;em&gt;Adventure/RPG&lt;/em&gt;&lt;br&gt;&#xA;  &lt;strong&gt;I&lt;/strong&gt;. &lt;code&gt;3 4 4 2 9.&lt;/code&gt; - &lt;em&gt;Simulation&lt;/em&gt;&lt;br&gt;&#xA;  &lt;strong&gt;J&lt;/strong&gt;. &lt;code&gt;6 5!&lt;/code&gt; - &lt;em&gt;Fighter&lt;/em&gt;&lt;br&gt;&#xA;  &lt;strong&gt;K&lt;/strong&gt;. &lt;code&gt;3 4 2 1 3!&lt;/code&gt; - &lt;em&gt;Platformer/Puzzle&lt;/em&gt;&lt;br&gt;&#xA;  &lt;strong&gt;L&lt;/strong&gt;. &lt;code&gt;2'1 9 2 2 5! 4 4!&lt;/code&gt; - &lt;em&gt;Adventure/RPG&lt;/em&gt;&lt;br&gt;&#xA;  &lt;strong&gt;M&lt;/strong&gt;. &lt;code&gt;3 4 4!&lt;/code&gt; - &lt;em&gt;Fighter&lt;/em&gt;&lt;br&gt;&#xA;  &lt;strong&gt;N&lt;/strong&gt;. &lt;code&gt;3 4 9 10 6&lt;/code&gt; - &lt;em&gt;Real-Time Strategy&lt;/em&gt;&lt;br&gt;&#xA;  &lt;strong&gt;O&lt;/strong&gt;. &lt;code&gt;3 4 4 3 6 2 2.&lt;/code&gt; - &lt;em&gt;Scrolling Shooter&lt;/em&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The exit to the dungeon has a voice recognition device; what quotes and corresponding games/series do you read to the device?&lt;/p&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2015-10-13T14:45:59.610" Title="Dungeon Crawler: Level 1" Tags="|cipher|knowledge|video-games|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23026" PostTypeId="2" ParentId="23024" CreationDate="2015-10-09T14:26:33.360" Score="6" Body="&lt;p&gt;Assume a table labelled with columns A-G and rows 1-7  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question A : Smallest total&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Row 4, Column A has 3 cells above it in the same column of the table and all of these must be smaller than cell &lt;strong&gt;4A&lt;/strong&gt;, so the smallest it can be is $4$.&lt;br&gt;&#xA; Row 4, Column B has 3 cells directly above it, along with cell &lt;strong&gt;4A&lt;/strong&gt; and all the cells directly above that, making a total of seven cells which must be smaller than &lt;strong&gt;4B&lt;/strong&gt;, so the smallest it can be is $8$,&lt;br&gt;&#xA; Continuing across in this fashion, the lowest total possible is $4+8+12+16+20+24+28 = 112$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Question B : Largest total&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reversing the approach above, the largest value possible in Row 4, Column G is $46$ as there need to be 3 numbers larger below it in the grid.&lt;br&gt;&#xA; Row 4, Column F needs 7 number bigger than it, so it can be at most $42$.&lt;br&gt;&#xA; Continuing gives $22+26+30+34+38+42+46=238$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Examples:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qkSUD.png&quot; alt=&quot;![enter image description here&quot;&gt;]&lt;a href=&quot;https://i.stack.imgur.com/qkSUD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="14135" LastEditDate="2015-10-11T08:54:18.467" LastActivityDate="2015-10-11T08:54:18.467" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23027" PostTypeId="2" ParentId="23025" CreationDate="2015-10-09T14:32:54.100" Score="6" Body="&lt;p&gt;&lt;strong&gt;A&lt;/strong&gt;. &lt;code&gt;2 1 6 4!&lt;/code&gt; - &lt;em&gt;Shooter/Flyer&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Do a Barrel Roll! - &lt;em&gt;Star Fox 64&lt;/em&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;B&lt;/strong&gt;. &lt;code&gt;3 2 3!&lt;/code&gt; - &lt;em&gt;Adventure/RPG&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Fus Ro Dah! - &lt;em&gt;The Elder Scrolls V: Skyrim&lt;/em&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;C&lt;/strong&gt;. &lt;code&gt;8!&lt;/code&gt; - &lt;em&gt;Fighter&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Hadouken! - &lt;em&gt;Street Fighter&lt;/em&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I&lt;/strong&gt;. &lt;code&gt;3 4 4 2 9.&lt;/code&gt; - &lt;em&gt;Simulation&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You have died of dysentery - &lt;em&gt;Oregon Trail&lt;/em&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;J&lt;/strong&gt;. &lt;code&gt;6 5!&lt;/code&gt; - &lt;em&gt;Fighter&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Falcon Punch! - &lt;em&gt;Super Smash Bros.&lt;/em&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;K&lt;/strong&gt;. &lt;code&gt;3 4 2 1 3!&lt;/code&gt; - &lt;em&gt;Platformer/Puzzle&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The cake is a lie! - &lt;em&gt;Portal&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;N&lt;/strong&gt;. &lt;code&gt;3 4 9 10 6&lt;/code&gt; - &lt;em&gt;Real-Time Strategy&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You must construct additional pylons - &lt;em&gt;Starcraft&lt;/em&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;O&lt;/strong&gt;. &lt;code&gt;3 4 4 3 6 2 2.&lt;/code&gt; - &lt;em&gt;Scrolling Shooter&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;All your base are belong to us - &lt;em&gt;Zero Wing&lt;/em&gt; (is it bad that I needed to use google to find the name of the game?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11717" LastEditorUserId="4421" LastEditDate="2015-10-09T16:00:17.273" LastActivityDate="2015-10-09T16:00:17.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23028" PostTypeId="5" CreationDate="2015-10-09T14:34:22.380" Score="0" Body="&lt;p&gt;For puzzles based, related to, or requiring knowledge of, video games, game characters, genres, platforms, etc. General video games questions are off-topic but can be asked on our sister site &lt;a href=&quot;https://gaming.stackexchange.com/&quot;&gt;Arqade Stack Exchange&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Topics include:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;characters&lt;/li&gt;&#xA;&lt;li&gt;monsters&lt;/li&gt;&#xA;&lt;li&gt;plots&lt;/li&gt;&#xA;&lt;li&gt;genres&lt;/li&gt;&#xA;&lt;li&gt;platforms (Xbox, PlayStation, Nintendo,...)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/41748/what-is-a-fantastical-word&quot;&gt;What is a Fantastical Word™?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/23290/video-games-cryptic-crossword&quot;&gt;Video Games Cryptic Crossword&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/23155/dungeon-crawler-level-2-5&quot;&gt;Dungeon Crawler: Level 2.5&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11497" LastEditorUserId="69582" LastEditDate="2021-03-07T15:53:34.777" LastActivityDate="2021-03-07T15:53:34.777" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23029" PostTypeId="4" CreationDate="2015-10-09T14:34:22.380" Score="0" Body="For puzzles based, related to, or requiring knowledge of, video games, game characters, genres, platforms, etc." OwnerUserId="13970" LastEditorUserId="69582" LastEditDate="2021-03-07T15:53:52.690" LastActivityDate="2021-03-07T15:53:52.690" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23030" PostTypeId="2" ParentId="23025" CreationDate="2015-10-09T14:39:11.067" Score="5" Body="&lt;p&gt;&lt;strong&gt;A&lt;/strong&gt;. &lt;code&gt;2 1 6 4!&lt;/code&gt; - &lt;em&gt;Shooter/Flyer&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Do a barrel roll! - &lt;em&gt;Star Fox 64&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;D&lt;/strong&gt;. &lt;code&gt;3. 3 5 7.&lt;/code&gt; - &lt;em&gt;Stealth/Action&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; War. War never changes. - &lt;em&gt;Fallout 3&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;E&lt;/strong&gt;. &lt;code&gt;6 3!&lt;/code&gt; - &lt;em&gt;Fighter&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Finish him! - &lt;em&gt;Mortal Kombat&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;G&lt;/strong&gt;. &lt;code&gt;4 1 5, 3 6.&lt;/code&gt; - &lt;em&gt;Adventure/RPG&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stay a while, and listen. - &lt;em&gt;Diablo 2&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I&lt;/strong&gt;. &lt;code&gt;3 4 4 2 9.&lt;/code&gt; - &lt;em&gt;Simulation&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You have died of dysentery. - &lt;em&gt;Oregon Trail&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;L&lt;/strong&gt;. &lt;code&gt;2'1 9 2 2 5! 4 4!&lt;/code&gt; - &lt;em&gt;Adventure/RPG&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's dangerous to go alone! Take this! - &lt;em&gt;Legend of Zelda&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;M&lt;/strong&gt;. &lt;code&gt;3 4 4!&lt;/code&gt; - &lt;em&gt;Fighter&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Get over here! - &lt;em&gt;Mortal Kombat&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14418" LastEditorUserId="4421" LastEditDate="2015-10-09T15:57:57.507" LastActivityDate="2015-10-09T15:57:57.507" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23031" PostTypeId="2" ParentId="23025" CreationDate="2015-10-09T14:46:21.110" Score="3" Body="&lt;p&gt;&lt;strong&gt;H&lt;/strong&gt;. &lt;code&gt;1 2 5.&lt;/code&gt; - &lt;em&gt;Adventure/RPG&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I am error. - Zelda 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;O&lt;/strong&gt;. &lt;code&gt;3 4 4 3 6 2 2.&lt;/code&gt; - &lt;em&gt;Scrolling Shooter&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;All your base are belong to us. - Zero Wing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12566" LastEditorUserId="4421" LastEditDate="2015-10-09T15:59:33.037" LastActivityDate="2015-10-09T15:59:33.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23032" PostTypeId="2" ParentId="23012" CreationDate="2015-10-09T15:40:21.997" Score="1" Body="&lt;p&gt;Here's what I suspect so far, but it's a bit kooky/incomplete at places:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;These are all Super Smash Bros 4 characters  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One evokes pizza  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;My best bet so far is Wario, since in the WarioWare series, there is a character Mona Pizza that sells...well you get it. There's also a song on WarioWare stage called &quot;Mona Pizza's Song&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One fuzzy, maybe funny  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This one I don't get. At first I thought Donkey Kong/Diddy Kong, but the clue implies it's a single character.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Spooks  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Pacman &amp;amp; ghost co.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fruit or pork  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Here I think this is Mr. Game &amp;amp; Watch. I haven't played him in SSB4, but in brawl, it was difficult to tell if what he was flinging in his neutral B was bacon or fish or what. Also one of his Side-B results (7, I think?) drops a healing fruit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;frog  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Greninja is a frog Pokemon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Double letters (in Japan?)  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There are a couple characters with double letters, but the Japan clue makes me more certain that it's Wii Fit Trainer (where Villager would not have double--or single--L in it's Japanese rendition and Wii is the Japanese name for the console)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Two known names; link them to solve  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This was the clue that started it off for me, this seems to hint at Zelda and her alter ego Sheik. The inclusion of the word link reinforces this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bonus: This poem has link, too  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Link himself is also a character in the game&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13970" LastActivityDate="2015-10-09T15:40:21.997" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23033" PostTypeId="1" AcceptedAnswerId="23037" CreationDate="2015-10-09T15:52:58.263" Score="2" ViewCount="666" Body="&lt;p&gt;You have five 1's at your disposal, together with five arithmetic operations of your choice. However, as you only have five operations, you should choose them wisely. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What is the largest integer that you can generate this way?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Numbers can not be infinite. No dividing by 0.&lt;/li&gt;&#xA;&lt;li&gt;You cannot concatenate the 1's (i.e. you cannot use two 1's to make 11)&lt;/li&gt;&#xA;&lt;li&gt;You cannot use any other numbers in any other form: no Greek alternatives, no constants such as $e$ or $\pi$.&lt;/li&gt;&#xA;&lt;li&gt;Parentheses come for free; you may use as many as you like.&lt;/li&gt;&#xA;&lt;li&gt;You may use two or more operations in a row&lt;/li&gt;&#xA;&lt;li&gt;You may use any notation you would like. One solution below uses &quot;Knuth's Up Arrow Notation&quot;. Each arrow uses one operation of the five allowed operations.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Examples:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  1+1+1+1++1 = 5&#xA;&#xA;  ((1+1+1)↑↑(1+1)) = 27  &amp;lt;-- Uses Knuth's Up Arrow Notation&#xA;&#xA;  (1+1)^((1+1+1)!) = 64&#xA;&#xA;  ((1+1+1)!)^(1+1) = 81&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I have posted &lt;a href=&quot;https://puzzling.stackexchange.com/a/23034/4865&quot;&gt;my solution&lt;/a&gt; below, let's see if you can beat me!&lt;/p&gt;&#xA;" OwnerUserId="4865" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-10-09T19:52:22.120" Title="Largest number with five 1's and five numeric operations" Tags="|mathematics|calculation-puzzle|" AnswerCount="5" CommentCount="10" ClosedDate="2015-10-09T19:54:12.493" ContentLicense="CC BY-SA 3.0" />
  <row Id="23034" PostTypeId="2" ParentId="23033" CreationDate="2015-10-09T15:55:49.073" Score="1" Body="&lt;p&gt;$$((1+1+1)^{(1+1)})! = 362,880 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or, &lt;code&gt;((1+1+1)^(1+1))!&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uses 3 &lt;code&gt;+&lt;/code&gt;, 1 &lt;code&gt;^&lt;/code&gt;, and 1 &lt;code&gt;!&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try it on &lt;a href=&quot;http://www.wolframalpha.com/input/?i=9!&quot; rel=&quot;nofollow&quot;&gt;Wolfram Alpha&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4865" LastEditorUserId="4865" LastEditDate="2015-10-09T16:11:20.427" LastActivityDate="2015-10-09T16:11:20.427" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23035" PostTypeId="1" AcceptedAnswerId="23038" CreationDate="2015-10-09T16:20:48.937" Score="7" ViewCount="283" Body="&lt;p&gt;Can one place the numbers $1-64$ in the squares of a chessboard so that every two squares that share an edge have a difference of at most seven?&lt;/p&gt;&#xA;" OwnerUserId="2722" LastEditorUserId="8874" LastEditDate="2015-10-09T17:18:05.407" LastActivityDate="2015-10-11T03:27:13.770" Title="Keep the chessboard differences below eight" Tags="|mathematics|combinatorics|checkerboard|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23036" PostTypeId="2" ParentId="23033" CreationDate="2015-10-09T16:26:37.880" Score="1" Body="&lt;p&gt;Must I use all 1's?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(((1+1+1)!)!)! = 3!!! = 6!! = 720! = 2.601 × 10^{1746}$$&lt;/p&gt;&#xA;" OwnerUserId="16642" LastEditorUserId="4865" LastEditDate="2015-10-09T17:13:50.670" LastActivityDate="2015-10-09T17:13:50.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23037" PostTypeId="2" ParentId="23033" CreationDate="2015-10-09T17:04:34.820" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;You may use any notation you would like.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Browsing Wikipedia I found the &lt;a href=&quot;https://en.wikipedia.org/wiki/Steinhaus%E2%80%93Moser_notation&quot; rel=&quot;nofollow&quot;&gt;Steinhaus–Moser notation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If all ones must be used we can start with:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(1+1+1)^(1+1)=9&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;With one operator left we can put the number in a circle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;⑨&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Wikipedia already ② is too big to be displayed. If less than 5 ones can be used, we can make the number even larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of a circle we can use any n-sided polygon to make the number arbitrarily large. See for example the definition of Moser's number in the article linked above.&lt;/p&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2015-10-09T18:44:08.240" LastActivityDate="2015-10-09T18:44:08.240" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23038" PostTypeId="2" ParentId="23035" CreationDate="2015-10-09T17:17:18.857" Score="4" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No, this is not possible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Argument:&#xA;Suppose for the sake of contradiction that such a placement would exist.&#xA;We number the rows $1$ to $8$, and we number the columns $1$ to $8$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Denote by $S$ the set of squares that contain the small numbers $1,\ldots,32$.&lt;/li&gt;&#xA;&lt;li&gt;For $1\le i\le8$, let $r_i$ denote the number of squares in $S$ that belong to the $i$th row.&lt;/li&gt;&#xA;&lt;li&gt;A row $i$ with $r_i=0$ is called empty, and a row with $r_i=8$ is called full.&lt;/li&gt;&#xA;&lt;li&gt;For $1\le i\le8$, let $c_i$ denote the number of squares in $S$ that belong to the $i$th column.&lt;/li&gt;&#xA;&lt;li&gt;Denote by $T$ the set of squares in $S$, that have at least one neighbor outside $S$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Our goal is to show that $T$ contains at least eight elements: then the smallest number of a square in $T$ is at most $25$, and its neighbor outside $S$ has number at least $33$. Done. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We may assume without loss of generality that $S$ contains the first $r_i$ squares from every row $i$ and the first $j$ squares from every column $j$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 1:&lt;/strong&gt; There are no full and no empty rows.&lt;br&gt;&#xA;Then every row contains at least one square from $T$, and $|T|\ge8$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 2:&lt;/strong&gt; There exists some full row, and there exists some empty row.&lt;br&gt;&#xA;Rotate the chessboard by 90 degrees, and you are in Case 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 3:&lt;/strong&gt; There exists some empty row, but there are no full rows.&lt;br&gt;&#xA;Then $r_1\le7$ and $r_8=0$. Then there exists a square $(i,j)\in S$ with $i+j\ge9$; otherwise $S$ could not have $32$ elements.&#xA;Then the squares $(k,c_k)$ with $1\le k\le i$ and the squares $(a_m,m)$ with $1\le m\le j$ all are in $T$. These are at least $i+j-1\ge 8$ distinct squares (as the square $(i,j)$ itself may be counted twice).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 4:&lt;/strong&gt; There exists some full row, but there are no empty rows.&lt;br&gt;&#xA;Use a symmetric argument that is centered around the large numbers $33,\ldots,64$. &lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="2722" LastEditDate="2015-10-09T18:54:33.723" LastActivityDate="2015-10-09T18:54:33.723" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23039" PostTypeId="1" AcceptedAnswerId="23040" CreationDate="2015-10-09T17:28:21.803" Score="7" ViewCount="731" Body="&lt;p&gt;This is the first in a series &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=user%3A9000%20is%3AQuestion%20title%3A%22Lights%20in%20the%20Dungeon%22&quot;&gt;Lights in the Dungeon&lt;/a&gt; puzzles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;It is intended to be easy to help work out unpredicted kinks for future puzzles that are built around the same story.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/23053/lights-in-the-dungeon-four-by-four&quot;&gt;-&gt; next (2nd)&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You have been captured by Lord Hooty McOwlface, a cruel tyrant that tightly controls his territory with an iron &lt;strike&gt;talon&lt;/strike&gt; fist. The dungeon is cold and without comforts but you are only kept there for a single night before you are hauled before the demonic ruler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;You are a very lucky human!&quot; screeches the despot. &quot;We are going to play some games. If you perform poorly, you shall be eaten! But, if you perform well, you shall get to play another game! Does that sound fair?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Well,&quot; you stammer as you try to look at anything except his terrible countenance, &quot;I suppose that's better than being eaten right away, isn't it?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Of course it is!  My grand vizier shall inform you of the rules. Be careful not to damage your delicious flesh if you lose. I would be most displeased.&quot; &#xA;A bright purple cat disengages itself from the Lord's grasp and leaps from the arm of the throne. It approaches you with a scroll held in it's teeth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Is this your grand vizier?&quot; you ask, incredulity sneaking into your voice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Fakky Jim is a trusted and noble creature! He is far superior to humans. Also, he's magic. Why did you think he was purple?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You accept the scroll and unroll it to reveal the following rules:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;You will be placed in a dungeon in which there are many rooms. Each room has a single light in the middle. You must turn off every&#xA;  light.&lt;/li&gt;&#xA;  &lt;li&gt;The lights are easily turned off by a short chain hanging from the fixture.&lt;/li&gt;&#xA;  &lt;li&gt;As soon as you exit a room, the floor will collapse to reveal the thousand-foot chasm below. You may not re-enter a room and, therefore,&#xA;  must turn off each light by passing through each room exactly once.&lt;/li&gt;&#xA;  &lt;li&gt;The map below shows the current design of the dungeon with each room bordered by red and labeled with a alphabetic character.&lt;/li&gt;&#xA;  &lt;li&gt;You may rearrange the rooms however you like before you enter but you cannot change the overall dimensions of the dungeon. In this case, that means the dungeon must be three rooms wide and three rooms tall.&lt;/li&gt;&#xA;  &lt;li&gt;You cannot rotate or flip rooms. You may only translate them.&lt;/li&gt;&#xA;  &lt;li&gt;You may only enter the dungeon once and exit once. As soon as you exit the dungeon, any remaining floors will collapse.&lt;/li&gt;&#xA;  &lt;li&gt;If there are any rooms not visited and, therefore, any lights left on, you lose. If all lights are turned off when you exit the dungeon,&#xA;  then you win.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How can you rearrange the dungeon rooms so that you can win the game?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;There are probably multiple solutions. The first valid answer will be accepted.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Np5Ex.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Np5Ex.png&quot; alt=&quot;Maze 3x3&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For reference, here are the 16 possible rooms. (Note that this is all &lt;em&gt;possible&lt;/em&gt; rooms. For this puzzle, you must use exactly the rooms shown in the map above.)&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rp7xy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rp7xy.png&quot; alt=&quot;A&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/E9wSu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/E9wSu.png&quot; alt=&quot;B&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/SlADl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SlADl.png&quot; alt=&quot;C&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/yEoqW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yEoqW.png&quot; alt=&quot;D&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/QGft0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QGft0.png&quot; alt=&quot;E&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/VREO9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VREO9.png&quot; alt=&quot;F&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/R1gGL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/R1gGL.png&quot; alt=&quot;G&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/vK9AI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vK9AI.png&quot; alt=&quot;H&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/Cft7i.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Cft7i.png&quot; alt=&quot;I&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/3oSC9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3oSC9.png&quot; alt=&quot;J&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/E72qu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/E72qu.png&quot; alt=&quot;K&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/p8yY1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/p8yY1.png&quot; alt=&quot;L&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/jqqKf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jqqKf.png&quot; alt=&quot;M&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/zKQB7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zKQB7.png&quot; alt=&quot;N&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/yTRBU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yTRBU.png&quot; alt=&quot;O&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/s78Fu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/s78Fu.png&quot; alt=&quot;P&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2016-08-31T15:32:32.620" Title="Lights in the Dungeon: Three by Three" Tags="|visual|story|mazes|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23040" PostTypeId="2" ParentId="23039" CreationDate="2015-10-09T17:44:32.537" Score="5" Body="&lt;p&gt;I hope I got this right:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/KRQyy.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KRQyy.jpg&quot; alt=&quot;Solution&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Looking forward to the next one!&lt;/p&gt;&#xA;" OwnerUserId="15985" LastActivityDate="2015-10-09T17:44:32.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23043" PostTypeId="2" ParentId="23033" CreationDate="2015-10-09T18:30:31.007" Score="0" Body="&lt;p&gt;S(1+1, (1+1+1)!) = S(2, 6) is a large number, it has not yet been computed but is known to be &gt; 7.4 × 10^36534 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Busy_beaver#Exact_values_and_lower_bounds_for_some_S.28n.2C_m.29_and_.CE.A3.28n.2C_m.29&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16835" LastActivityDate="2015-10-09T18:30:31.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23044" PostTypeId="2" ParentId="22782" CreationDate="2015-10-09T18:36:41.427" Score="6" Body="&lt;p&gt;I believe the blank spaces are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HE  RM  SE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The meaning of these numbers are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Asteroid designations. The dY is the number of years ago (counting from 2015) it was discovered, the H is the absolute magnitude, and the letters plus the &quot;atomic number&quot; (to use the periodic table comparison) make up the provisional designation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, for example,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The very first block (in the Hydrogen position) would be asteroid 2004 LJ1, which has an absolute magnitude of 15.4.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The numbered clues, in order, spell out&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HERMES, which was an asteroid discovered in 1937. At the time, it was the closest observed near-miss between an asteroid and Earth. This record held until 1989, 52 years later, when a closer orbital pass was observed, this time of asteroid Asclepius.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The amount of time that M Langley sunbathes for, 811.44 hours&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Measured in weeks, as suggested by his rude brother, is 4.83, the absolute magnitude of the sun.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The name Langley itself refers to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Samuel Pierpont Langley, who did quite a bit of research on the Sun, and invented a device called the bolometer, which measures absolute magnitude. Magnitudes of stars and Solar System bodies are calculated differently, because putting rocks on the same scale as nuclear furnaces would be meaningless. The absolute magnitude of a star has the symbol M, and the absolute magnitude of a planet/asteroid has the symbol H. Just like the brothers and their respective obsessions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16814" LastEditorUserId="16814" LastEditDate="2015-10-09T20:53:17.573" LastActivityDate="2015-10-09T20:53:17.573" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="23051" PostTypeId="2" ParentId="23033" CreationDate="2015-10-09T19:03:35.723" Score="1" Body="&lt;p&gt;Using the same notation as &lt;a href=&quot;https://en.wikipedia.org/wiki/Graham&amp;#39;s_number&quot; rel=&quot;nofollow&quot;&gt;Graham's Number&lt;/a&gt; and assuming that we don't have to use all five ones:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g_{g_{g_{g_{g_{1}}}}} = Something Absurdly Large$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that g&lt;sub&gt;g&lt;sub&gt;1&lt;/sub&gt;&lt;/sub&gt; makes Graham's Number (g&lt;sub&gt;64&lt;/sub&gt;) look absolutely infinitesimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is much smaller, but uses all five ones:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g_{(1+1+1)^{(1+1)}}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In general,&#xA;$$g_n = 3\uparrow^{g_{n-1}}3$$&#xA;where&#xA;$$g_1 = 3\uparrow\uparrow\uparrow\uparrow3$$&lt;/p&gt;&#xA;" OwnerUserId="16838" LastEditorUserId="16838" LastEditDate="2015-10-09T19:52:22.120" LastActivityDate="2015-10-09T19:52:22.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23052" PostTypeId="2" ParentId="23025" CreationDate="2015-10-09T19:04:33.400" Score="4" Body="&lt;p&gt;&lt;strong&gt;F&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Would you kindly...          -&lt;em&gt;Bioshock&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastActivityDate="2015-10-09T19:04:33.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23053" PostTypeId="1" AcceptedAnswerId="23054" CreationDate="2015-10-09T19:11:01.790" Score="6" ViewCount="664" Body="&lt;p&gt;This is the second in a series &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=user%3A9000%20is%3AQuestion%20title%3A%22Lights%20in%20the%20Dungeon%22&quot;&gt;Lights in the Dungeon&lt;/a&gt; puzzles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/23039/lights-in-the-dungeon-three-by-three&quot;&gt;previous (1st) &amp;lt;--&lt;/a&gt; | &lt;a href=&quot;https://puzzling.stackexchange.com/questions/23113/lights-in-the-dungeon-five-by-five&quot;&gt;--&gt; next (3rd) &lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As you exit the three by three dungeon, you are greeted by a grinning Lord Hooty McOwlface. A grin does not make him appear friendly. Rather, you get the feeling that he's deciding which portions of you would be tastiest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Well!&quot; screeches his Lordship, &quot;Has he succeeded?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It takes you a moment but you realize that he is not talking to you and is, in fact, addressing the purple cat sitting at a bank of computer terminals. You know this to be Fakky Jim, the Lord's grand vizier who is, by coincidence, a magic cat. He turns his furred head and utters a single, &quot;Mreow.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Drat!&quot; says McOwlface. &quot;I suppose we're on to round two, eh? Very good. The dungeon shall be larger but the rules are largely the same. Larger? Largely? I am hilarious...&quot; He pauses for you to appreciate this but you can give nothing but a blank stare. He dismisses you with a wave. &quot;See Fakky for the updates.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You walk over to the cat and retrieve a scroll. Unfurling it, you find the rules below (revisions from the previous challenge are in bold):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;You will &lt;strong&gt;enter&lt;/strong&gt; a dungeon in which there are many rooms. Each room has a single light in the middle. You must turn off every light.&lt;/li&gt;&#xA;  &lt;li&gt;The lights are easily turned off by a short chain hanging from the fixture.&lt;/li&gt;&#xA;  &lt;li&gt;As soon as you exit a room, the floor will collapse to reveal the thousand-foot chasm below. You may not re-enter a room and, therefore,&#xA;  must turn off each light by passing through each room exactly once.&lt;/li&gt;&#xA;  &lt;li&gt;The map below shows the current design of the dungeon with each room bordered by red and labeled with a alphabetic character.&lt;/li&gt;&#xA;  &lt;li&gt;You may rearrange the rooms however you like before you enter but you cannot change the overall dimensions of the dungeon. In this case, that means the dungeon must be &lt;strong&gt;four&lt;/strong&gt; rooms wide and &lt;strong&gt;four&lt;/strong&gt; rooms tall.&lt;/li&gt;&#xA;  &lt;li&gt;You cannot rotate or flip rooms. You may only translate them.&lt;/li&gt;&#xA;  &lt;li&gt;You may only enter the dungeon once and exit once. As soon as you exit the dungeon, any remaining floors will collapse.&lt;/li&gt;&#xA;  &lt;li&gt;If there are any rooms not visited and, therefore, any lights left on, you lose. If all lights are turned off when you exit the dungeon, then you win.&lt;/li&gt;&#xA;  &lt;li&gt;&lt;strong&gt;If you create a dungeon map that you cannot enter or cannot exit, you shall lose.&lt;/strong&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How can you rearrange the dungeon rooms so that you can win the game?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;There are probably multiple solutions. The first valid answer will be accepted.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BQojl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BQojl.png&quot; alt=&quot;Maze 4x4&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;ABBC&#xA;DDEF&#xA;GGHH&#xA;MMOO&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For reference, here are the 16 possible rooms. (Note that this is all &lt;em&gt;possible&lt;/em&gt; rooms. For this puzzle, you must use exactly the rooms shown in the map above.)&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rp7xy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rp7xy.png&quot; alt=&quot;A&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/E9wSu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/E9wSu.png&quot; alt=&quot;B&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/SlADl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SlADl.png&quot; alt=&quot;C&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/yEoqW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yEoqW.png&quot; alt=&quot;D&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/QGft0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QGft0.png&quot; alt=&quot;E&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/VREO9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VREO9.png&quot; alt=&quot;F&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/R1gGL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/R1gGL.png&quot; alt=&quot;G&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/vK9AI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vK9AI.png&quot; alt=&quot;H&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/Cft7i.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Cft7i.png&quot; alt=&quot;I&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/3oSC9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3oSC9.png&quot; alt=&quot;J&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/E72qu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/E72qu.png&quot; alt=&quot;K&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/p8yY1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/p8yY1.png&quot; alt=&quot;L&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/jqqKf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jqqKf.png&quot; alt=&quot;M&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/zKQB7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zKQB7.png&quot; alt=&quot;N&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/yTRBU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yTRBU.png&quot; alt=&quot;O&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/s78Fu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/s78Fu.png&quot; alt=&quot;P&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2016-08-31T15:32:44.763" Title="Lights in the Dungeon: Four by Four" Tags="|story|visual|mazes|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23054" PostTypeId="2" ParentId="23053" CreationDate="2015-10-09T19:28:32.767" Score="8" Body="&lt;p&gt;How about:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;BGBC&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;HFMM&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;OADH&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;OGED&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;a href=&quot;https://i.stack.imgur.com/R8DWH.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/R8DWH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This has the nice additional quality that, no matter which room you enter by, you can still complete the path.&lt;/p&gt;&#xA;" OwnerUserId="1497" LastActivityDate="2015-10-09T19:28:32.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23055" PostTypeId="1" CreationDate="2015-10-09T19:30:43.057" Score="0" ViewCount="255" Body="&lt;p&gt;Inspired by &lt;a href=&quot;https://puzzling.stackexchange.com/questions/23033/largest-number-with-five-1s-and-five-numeric-operations&quot;&gt;this question&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the smallest positive integer that you &lt;strong&gt;cannot&lt;/strong&gt; make with 5 ones and five operators?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You must use all five ones and all five operators&lt;/li&gt;&#xA;&lt;li&gt;Allowed operators are: +, -, *, /, ^, !, Knuth up-arrows. Each up-arrow counts as one operation.&lt;/li&gt;&#xA;&lt;li&gt;Concatenation is not allowed. e.g. you cannot use two 1s to make 11.&lt;/li&gt;&#xA;&lt;li&gt;You may use parentheses freely&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3987" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-10-10T09:15:25.487" Title="Smallest impossible positive integer with 5 ones and 5 operators" Tags="|mathematics|" AnswerCount="2" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="23059" PostTypeId="1" AcceptedAnswerId="25121" CreationDate="2015-10-09T20:30:34.523" Score="8" ViewCount="913" Body="&lt;p&gt;After installing the newest MMO and creating a fresh character you spawn for the first time. After a quick look at the map, you see you are at The Famous Last Words Square. You quickly realize that the ground is covered by big letters arranged in a ... square:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;+-----------------------------------------------+&#xA;| P   U   N   T   ~   U   U   R   J   ~   K   G |&#xA;|   +---------------------------------------+   |&#xA;| J | N   A   A   T   V   F   V   K   J   G | T |&#xA;|   |   +-------------------------------+   |   |&#xA;| I | ~ | V   J   ~   N   A   A   W   N | K | E |&#xA;|   |   |   +-----------------------+   |   |   |&#xA;| V | I | S | C   N   F   V   W   U | J | K | E |&#xA;|   |   |   |   +---------------+   |   |   |   |&#xA;| A | J | N | K | F   V   K   S | T | I | V | E |&#xA;|   |   |   |   |   +-------+   |   |   |   |   |&#xA;| ~ | K | I | K | U | Z   Q | V | V | K | T | E |&#xA;|   |   |   |   |   +---+   |   |   |   |   |   |&#xA;| J | N | W | U | X   ~   H | ~ | N | Z | W | E |&#xA;|   |   |   |   +-----------+   |   |   |   |   |&#xA;| N | Q | T | E   I   T   U   W | I | X | V | E |&#xA;|   |   |   +-------------------+   |   |   |   |&#xA;| O | U | Z   J   G   U   O   I   Z | I | K | E |&#xA;|   |   +---------------------------+   |   |   |&#xA;| I | P   A   A   N   ~   S   V   O   I | ~ | E |&#xA;|   +-----------------------------------+   |   |&#xA;| V   A   W   T   Z   O   V   S   I   V   A | E |&#xA;+-------------------------------------------+   |&#xA;| E   E   E   E   E   E   E   E   E   E   E   E |&#xA;+-----------------------------------------------+&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Not far away you see an NPC with an exclamation mark above his head, so you decide to talk to him:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Every time a hero dies a horrible death the last words he said become alive. Some of this words gathered at this place and formed the letters on the ground. However only the wisest heroes can decipher them.&lt;br&gt;&lt;br&gt;There is also a legend about this square. If someone can decipher the hidden phrases the name of a creature will appear. Saying its name 3 times will make the one incredibly rich.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Decipher the hidden phrases and find out the name of the creature.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is more than one reason for the E's at the bottom and right side:&lt;br&gt;- they are needed to fill the square&lt;br&gt;- they are part of a &quot;phrase&quot;&lt;br&gt;- they show how the phrases are arranged in the square&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2016-01-06T16:53:53.473" LastActivityDate="2016-01-06T21:02:09.573" Title="The Famous Last Words Square" Tags="|cipher|video-games|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23061" PostTypeId="2" ParentId="22563" CreationDate="2015-10-09T21:10:53.327" Score="2" Body="&lt;p&gt;In order to provide a fully complete answer to this question, I have created this answer as a &lt;strong&gt;community wiki&lt;/strong&gt; (which does not attract any reputation score).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer from @Evert has correctly solved the riddle and many of the explanations, so his answer has been upvoted, but it left a few unsolved lines.&#xA;&lt;hr /&gt;&#xA;I was looking for:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Planetary_nebula&quot; rel=&quot;nofollow&quot;&gt;Planetary Nebulae&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Source of an element that never existed  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Nebulae were once believed to contain &lt;a href=&quot;http://io9.com/5924504/nebulium-the-forbidden-element-that-never-existed&quot; rel=&quot;nofollow&quot;&gt;Nebulium&lt;/a&gt; - an element which ended up not existing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Perhaps with thermal tangles  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Some scientists have hypothesized the existence of &lt;strong&gt;cold knots&lt;/strong&gt; containing very little hydrogen to explain discrepancies in results of measuring the composition of planetary nebulae.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some of us are not as neat  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Some of the known planetary nebulae are within the set of &lt;a href=&quot;https://en.wikipedia.org/wiki/Messier_object&quot; rel=&quot;nofollow&quot;&gt;Messier Objects&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Our name is a misnomer  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  When William Herschel first observed planetary nebulae, he thought they resembled the shape of planets and the name stuck, though they are formed from dying stars and have nothing to do with planets.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An electronic errand?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  A planetary nebula is a kind of &lt;a href=&quot;https://en.wikipedia.org/wiki/Emission_nebula&quot; rel=&quot;nofollow&quot;&gt;e-mission&lt;/a&gt; nebula.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hard to pinpoint  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  The distance to a planetary nebula is &lt;a href=&quot;https://www.eso.org/sci/publications/messenger/archive/no.32-jun83/messenger-no32-20-22.pdf&quot; rel=&quot;nofollow&quot;&gt;very hard to determine&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Following death  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  By the time a planetary nebula forms, any planets around the star at the core will have been destroyed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We fade too  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Planetary nebulae are relatively very short-lived, lasting only a few tens of thousands of years (as compared to the several billion years of a star's life-cycle) before fading away.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A rare sight indeed&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This short lifespan helps explain why so few can be found (only around 3000 are known). They are also faint objects - none are visible to the naked eye.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-10-09T21:10:53.327" CommentCount="0" CommunityOwnedDate="2015-10-09T21:10:53.327" ContentLicense="CC BY-SA 3.0" />
  <row Id="23062" PostTypeId="2" ParentId="23055" CreationDate="2015-10-09T21:51:29.910" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I could generate 0-9 but I'm stuck at 10:&lt;br&gt;&lt;br&gt;$0=1/1/1/1-1!$&lt;br&gt;$1=1/1/1/1/1!$&lt;br&gt;$2=1+1/1/1/1!$&lt;br&gt;$3=1+1+1/1/1!$&lt;br&gt;$4=1+1+1+1/1!$&lt;br&gt;$5=1+1+1+1+1!$&lt;br&gt;$6=(1+1+1)!/1/1$&lt;br&gt;$7=(1+1+1)!+1/1$&lt;br&gt;$8=(1+1+1)!+1+1$&lt;br&gt;$9=(1+1+1)\uparrow(1+1!)$&lt;br&gt;$10=...$&lt;br&gt;$11=...$&lt;br&gt;$12=(1+1+1)!*(1+1)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastActivityDate="2015-10-09T21:51:29.910" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23063" PostTypeId="1" AcceptedAnswerId="23065" CreationDate="2015-10-10T01:51:05.803" Score="3" ViewCount="148" Body="&lt;p&gt;Inspired by &lt;a href=&quot;https://puzzling.stackexchange.com/questions/23035/keep-the-chessboard-differences-below-eight&quot;&gt;Keep the chessboard differences below eight&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given an $n\times m$ grid of squares, what is the smallest possible integer bound $x$ for which it is possible to fill the squares with the integers $1,\ldots,nm$ so that each two squares with a common edge have an absolute difference of $x$ or less? And what is the best strategy of finding such a grid?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-10-10T19:18:21.277" Title="Keep the grid differences below $x$" Tags="|mathematics|combinatorics|checkerboard|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23064" PostTypeId="2" ParentId="23055" CreationDate="2015-10-10T02:58:42.117" Score="1" Body="&lt;p&gt;Sleafar got the right answer - 10 -  by forcing it. I got same by the following observation: in order to get numbers above 6 one have to make a group of (1+1+1) to get 3 to further operate with. as the remaining we have two &quot;1&quot; and 3 operations available. Among these 3 operations one should be reserved for connection to the first group. And to get number over 7 one must use + operator to get 2. That leads us to solve the problem having 3 and 2 and two remaining operators. Having these assumptions, 8 and 9 are possible to produce while 10 is not.&#xA;BTW : the max positive integer is ((1+1+1)↑(1+1))! = 9! = 362880&lt;/p&gt;&#xA;" OwnerUserId="16360" LastEditorUserId="9422" LastEditDate="2015-10-10T09:15:25.487" LastActivityDate="2015-10-10T09:15:25.487" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23065" PostTypeId="2" ParentId="23063" CreationDate="2015-10-10T04:18:40.757" Score="5" Body="&lt;p&gt;You can do it with $x=\min(n,m)$.  Put $1$ in the upper left and fill each diagonal starting at the bottom and going up and right in order.  A $4 \times 5$ example is below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{|r|r|r|r|r|}\hline&#xA;\phantom{00}1 &amp;amp; \phantom{00}3 &amp;amp; \phantom{00}6 &amp;amp; \phantom010 &amp;amp; \phantom014\\\hline&#xA;2 &amp;amp;  5 &amp;amp;  9 &amp;amp; 13 &amp;amp; 17\\\hline&#xA;4 &amp;amp;  8 &amp;amp; 12 &amp;amp; 16 &amp;amp; 19\\\hline&#xA;7 &amp;amp; 11 &amp;amp; 15 &amp;amp; 18 &amp;amp; 20\\\hline&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this is the best possible, it is easier to do the $x \times x$ square as you (may) get to use smaller numbers for this part of the grid.  Now follow the proof in the linked question to see that $x-1$ is not possible.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="8874" LastEditDate="2015-10-10T19:18:21.277" LastActivityDate="2015-10-10T19:18:21.277" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23071" PostTypeId="1" AcceptedAnswerId="23107" CreationDate="2015-10-10T10:29:22.337" Score="3" ViewCount="501" Body="&lt;p&gt;I wanted to tell a friend about a character in one of the best games I ever played. However, this game is pretty old and I neither remember the name of the game nor of the character. But there are some other things I remember:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The character I'm looking for also had memory problems, but it was nothing compared to the main character of the game. I don't want to spoil the story, therefore I don't look for the main character.&lt;/li&gt;&#xA;&lt;li&gt;He really likes to kill people, but he is not a soldier or warrior like the other two characters I'm not looking for. Assassin would be a better description.&lt;/li&gt;&#xA;&lt;li&gt;I'm not looking for the small guy with odd pronunciation, I wonder how anyone can understand what he says.&lt;/li&gt;&#xA;&lt;li&gt;I'm neither looking for the little girl nor for their big hairy friend.&lt;/li&gt;&#xA;&lt;li&gt;I'm also not looking for the female character who managed to get captured twice during the game.&lt;/li&gt;&#xA;&lt;li&gt;There is another female character which was chosen as the 10th worst character on a ranking list. The character I'm looking for is quite the opposite, he even won two awards.&lt;/li&gt;&#xA;&lt;li&gt;Finally, the old hermit character is another one I'm not looking for.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Help me Puzzling.SE. You're my only hope.&lt;/p&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2015-10-12T15:00:59.330" LastActivityDate="2015-10-12T18:01:40.130" Title="Characters make the game" Tags="|knowledge|video-games|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23072" PostTypeId="1" AcceptedAnswerId="23078" CreationDate="2015-10-10T12:45:50.973" Score="8" ViewCount="740" Body="&lt;p&gt;Damiano has thrown four dice and the numbers &lt;strong&gt;2, 2, 4, 5&lt;/strong&gt; showed up on top.&lt;br&gt;&#xA;Damiano asks himself: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;What is the smallest positive integer that cannot be generated with these four numbers according to the following rules?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;One may use the four numbers 2, 2, 4, 5, and it is fine if not all of them are used.&lt;/li&gt;&#xA;&lt;li&gt;Concatenation of digits is NOT allowed.&lt;/li&gt;&#xA;&lt;li&gt;The only allowed operations are additions, subtraction, multiplication, and division ($+,-,*,/$).&lt;/li&gt;&#xA;&lt;li&gt;One may use any number of brackets.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Examples: $~~~~1=5-4$; $~~~~2=4-2$; $~~~~3=4-(2/2)$; $~~~~4=(4+2)-2$; $~~~~$etc.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2019-12-30T08:32:27.467" LastActivityDate="2019-12-30T08:32:27.467" Title="Four dice puzzle: 2,2,4,5" Tags="|mathematics|calculation-puzzle|formation-of-numbers|dice|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23077" PostTypeId="1" AcceptedAnswerId="23122" CreationDate="2015-10-10T13:18:56.427" Score="9" ViewCount="461" Body="&lt;p&gt;This continues Damiano's puzzle &quot;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/23072/&quot;&gt;Four dice puzzle: 2,2,4,5&lt;/a&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Damiano keeps throwing his four dice. After a lot of throwing and thinking and working, he has determined for every throw $a,b,c,d$ of his four dice the smallest positive integer $N(a,b,c,d)$ that cannot be generated from this throw according to the following rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;One may use the four numbers $a,b,c,d$ in any order, and it is fine if not all of them are used.&lt;/li&gt;&#xA;&lt;li&gt;Concatenation of digits is NOT allowed.&lt;/li&gt;&#xA;&lt;li&gt;The only allowed operations are additions, subtraction, multiplication, and division ($+,-,*,/$).&lt;/li&gt;&#xA;&lt;li&gt;One may use any number of brackets.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Which throw $a,b,c,d$ of dice yields the largest number $N(a,b,c,d)$ ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2019-12-30T08:32:39.453" LastActivityDate="2019-12-30T08:32:39.453" Title="Four dice puzzle: What's the best throw?" Tags="|mathematics|calculation-puzzle|formation-of-numbers|dice|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23078" PostTypeId="2" ParentId="23072" CreationDate="2015-10-10T14:12:19.670" Score="7" Body="&lt;p&gt;&lt;strong&gt;The answer is:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;23.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Joel Rondeau’s answer gives ways of getting anything up to this, except for 17; but 17 = ((4+2)*2)+5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I am not mistaken, the full range of positive integer values achievable is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,24,25,26,28,30,32,36,38,40,42,44,48,50,56,60,80]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(found using a small Haskell script).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1 = ((2-4)*2)+5&lt;br&gt;&#xA;2 = ((2/2)-4)+5&lt;br&gt;&#xA;3 = ((4-5)+2)+2&lt;br&gt;&#xA;4 = ((5-4)*2)+2&lt;br&gt;&#xA;5 = ((2-4)+2)+5&lt;br&gt;&#xA;6 = ((2/4)*2)+5&lt;br&gt;&#xA;7 = ((4+2)*2)-5&lt;br&gt;&#xA;8 = ((5*2)-4)+2&lt;br&gt;&#xA;9 = ((2-2)+4)+5&lt;br&gt;&#xA;10 = ((2/2)+4)+5&lt;br&gt;&#xA;11 = ((4*2)-2)+5&lt;br&gt;&#xA;12 = ((4-2)*5)+2&lt;br&gt;&#xA;13 = ((2+2)+4)+5&lt;br&gt;&#xA;14 = ((5-2)*4)+2&lt;br&gt;&#xA;15 = ((4*2)+2)+5&lt;br&gt;&#xA;16 = ((5*2)+4)+2&lt;br&gt;&#xA;17 = ((4+2)*2)+5&lt;br&gt;&#xA;18 = ((5+2)*2)+4&lt;br&gt;&#xA;19 = (5*4)-(2/2)&lt;br&gt;&#xA;20 = ((5+4)*2)+2&lt;br&gt;&#xA;21 = ((2+2)*4)+5&lt;br&gt;&#xA;22 = ((5+4)+2)*2&lt;br&gt;&#xA;24 = ((5*4)+2)+2&lt;br&gt;&#xA;25 = ((2/2)+4)*5&lt;br&gt;&#xA;26 = ((4*2)+5)*2&lt;br&gt;&#xA;28 = ((5*2)+4)*2&lt;br&gt;&#xA;30 = ((5+2)*4)+2&lt;br&gt;&#xA;32 = ((4+2)*5)+2&lt;br&gt;&#xA;36 = ((5+2)+2)*4&lt;br&gt;&#xA;38 = ((5*4)*2)-2&lt;br&gt;&#xA;40 = ((2+2)+4)*5&lt;br&gt;&#xA;42 = ((5*4)*2)+2&lt;br&gt;&#xA;44 = ((5*4)+2)*2&lt;br&gt;&#xA;48 = ((5*2)+2)*4&lt;br&gt;&#xA;50 = ((4*2)+2)*5&lt;br&gt;&#xA;56 = ((5+2)*4)*2&lt;br&gt;&#xA;60 = ((4+2)*2)*5&lt;br&gt;&#xA;80 = ((2+2)*4)*5 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4234" LastEditorUserId="4234" LastEditDate="2015-10-10T14:36:21.767" LastActivityDate="2015-10-10T14:36:21.767" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="23081" PostTypeId="2" ParentId="22648" CreationDate="2015-10-10T19:39:29.177" Score="0" Body="&lt;p&gt;Assuming the buckets have equal surface on bottom and top there is a very easy solution &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fill the 4l bucket, pour water into the 5l one till you see the back corner of the bucket; done&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other then this there is no solution where you fill both buckets, not without wasting water or using a second reservoir. &lt;/p&gt;&#xA;" OwnerUserId="16846" LastActivityDate="2015-10-10T19:39:29.177" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23082" PostTypeId="2" ParentId="23035" CreationDate="2015-10-11T03:27:13.770" Score="5" Body="&lt;p&gt;Thought I would post this proof since it reduces the case analysis, though it is the same basic idea as Gamow's solution.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Given a partially filled out chessboard, say a row or column is &lt;em&gt;started&lt;/em&gt; if there is at least one number in that row or column.  Let $S$ be the set of squares containing the numbers $1$ through $k$.  Here $k$ is the smallest number so that all the rows are started or all the columns are started. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $T$ be all the squares of $S$ that are adjacent to squares outside of $S$.  Like in Gamow's solution, we just need $|T| \geq 8$, since there would then be no way to fill in the square adjacent to the smallest numbered square of $T$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without loss of generality, assume $k$ starts the final row to be started.  No row can be completely full, since the row with $k$ contains only $k$, and if any other row were full every column would have been started, meaning we should have chosen a smaller value for $k$.  Since every row is started but no row is full, there must be a element of $T$ in every row.&lt;/p&gt;&#xA;" OwnerUserId="2722" LastActivityDate="2015-10-11T03:27:13.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23083" PostTypeId="1" CreationDate="2015-10-11T04:57:03.800" Score="1" ViewCount="258" Body="&lt;p&gt;Moved from Math StackExchange: &#xA;The Fifteen Puzzle consists of sliding square tiles numbered $1...15$ held in a $4\times4$ frame with one empty square. Any tile adjacent to the empty square can slide into it.&#xA;The standard initial position is&#xA;$$&#xA;\begin{array}{|c|c|c|} \hline&#xA;    1 &amp;amp; 2&amp;amp;3 &amp;amp; 4 \\  \hline&#xA;   5 &amp;amp; 6 &amp;amp; 7 &amp;amp; 8\\  &#xA;   \hline&#xA;9 &amp;amp; 10 &amp;amp; 11 &amp;amp; 12\\  \hline&#xA;13 &amp;amp; 14 &amp;amp; 15 \\ \hline&#xA;\end{array}&#xA;$$&#xA;The target position we would like to reach is: &#xA;\begin{array}{|c|c|c|} \hline&#xA;    15 &amp;amp; 14 &amp;amp; 13 &amp;amp; 12 \\  \hline&#xA;   11 &amp;amp; 10 &amp;amp; 9 &amp;amp; 8\\  &#xA;   \hline&#xA;7 &amp;amp; 6 &amp;amp; 5 &amp;amp; 4\\  \hline&#xA;3 &amp;amp; 2 &amp;amp; 1 \\ \hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A state machine model of the puzzle has states consisting of a $4 \times 4$ matrix with $16$ entries consisting of the integers $1...15$ as well as one “empty” entry—like each of the two arrays above.&#xA;The state transitions correspond to exchanging the empty square and an adjacent numbered tile. For example, an empty at position $(2,2)$ can exchange position with tile above it, namely, at position $(1,2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the method of finding invariants, how can you prove that there is no way to reach the target state starting from the initial state? &lt;/p&gt;&#xA;" OwnerUserId="16851" LastEditorUserId="8874" LastEditDate="2015-10-11T11:55:14.220" LastActivityDate="2015-10-11T11:55:14.220" Title="Fifteen Puzzle: Finding Invariants" Tags="|combinatorics|sliding-blocks|" AnswerCount="0" CommentCount="2" ClosedDate="2015-10-11T22:53:22.660" ContentLicense="CC BY-SA 3.0" />
  <row Id="23087" PostTypeId="1" AcceptedAnswerId="23099" CreationDate="2015-10-11T12:43:12.333" Score="12" ViewCount="3062" Body="&lt;p&gt;Enrico wants to construct a new six-sided die with the following properties:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Each side should show a positive integer &lt;/li&gt;&#xA;&lt;li&gt;Different sides should show different integers&lt;/li&gt;&#xA;&lt;li&gt;If two sides share an edge, then their integers should differ by at least 2.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What is the smallest possible sum of the six integers on Enrico's die?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-10-12T08:36:52.903" Title="A new six-sided die" Tags="|mathematics|calculation-puzzle|dice|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23089" PostTypeId="2" ParentId="23087" CreationDate="2015-10-11T13:01:25.730" Score="10" Body="&lt;p&gt;I was able to get a sum of 27.  The three pairs of opposite sides are 1 and 2, 4 and 5, and 7 and 8.&lt;/p&gt;&#xA;" OwnerUserId="5095" LastActivityDate="2015-10-11T13:01:25.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23091" PostTypeId="1" AcceptedAnswerId="23101" CreationDate="2015-10-11T15:34:07.467" Score="5" ViewCount="360" Body="&lt;p&gt;It is known that 7 random riffle shuffles are enough to make almost every configuration equally likely in a deck of 52 cards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Perfect Shuffle&lt;/strong&gt; is when you cut the cards exactly in half and perfectly interleave the cards from the two halves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In-Shuffle&lt;/strong&gt; is Perfect Shuffle in which top card from the top half remains top after the shuffle, while &lt;strong&gt;Out-Shuffle&lt;/strong&gt; is one in which bottom card from the first half becomes bottom (in other words top card becomes second) after the shuffle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If randomly chosen In and Out Shuffles are performed on a deck of 52&#xA;  cards, how many shuffles are needed to make almost every configuration&#xA;  equally likely?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15856" LastEditorUserId="15856" LastEditDate="2015-10-11T17:25:17.910" LastActivityDate="2015-10-12T00:54:06.260" Title="Random Shuffled Deck of Cards" Tags="|mathematics|cards|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23093" PostTypeId="2" ParentId="23091" CreationDate="2015-10-11T16:52:42.903" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think it is impossible. Using $52$ cards there are $52!=8.06*10^{67}$ possible combinations of cards. Using in/out-shuffles randomly we have potentially $2^n$ possibilities after $n$ shuffles. To reach at least $52!$ combinations this way we would need at least $226$ shuffles.&lt;br&gt;&lt;br&gt;Let's look at the situation after $104$ shuffles. The interesting feature of in/out-shuffles is that &lt;a href=&quot;https://en.wikipedia.org/wiki/Out_shuffle&quot; rel=&quot;nofollow&quot;&gt;$8$ out-shuffles&lt;/a&gt; in a row return the deck to the starting position. The same applies to &lt;a href=&quot;https://en.wikipedia.org/wiki/In_shuffle&quot; rel=&quot;nofollow&quot;&gt;$52$ in-shuffles&lt;/a&gt;. So after $104$ shuffles we have at least $2$ cases with the starting position again, and not enough other cases to cover all other combinations. After a total of $208$ shuffles we have more than $4$ cases with starting position and not enough cases to cover the rest, and so on.&lt;br&gt;&lt;br&gt;Therefore we will never reach a situation where all shuffles are equally likely. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2015-10-11T17:08:56.647" LastActivityDate="2015-10-11T17:08:56.647" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23094" PostTypeId="1" AcceptedAnswerId="23097" CreationDate="2015-10-11T17:34:01.270" Score="6" ViewCount="500" Body="&lt;p&gt;&lt;em&gt;An easy puzzle for beginners:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are in front of a computer which, upon your input, will produce a sequence of numbers. Your task is to discover which logic was used to program the computer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sitting in front of the computer, you type: 7. &#xA;The computer replies with the sequence: 5, 4, 4, 4, and then 4 all along.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you type 6. &#xA;The computer replies with: 3, 5, 4, 4, 4, and then 4 all along. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pensive, you type 42. &#xA;Computer's reply: 8, 5, 4, 4, 4, and then 4 all along again.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You try different numbers, and notice that all sequences eventually go with 4 repeated endlessly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the algorithm used?&lt;/p&gt;&#xA;" OwnerUserId="16323" LastEditorUserId="4551" LastEditDate="2015-10-11T20:24:54.517" LastActivityDate="2015-10-11T20:24:54.517" Title="The program in the computer" Tags="|pattern|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23095" PostTypeId="5" CreationDate="2015-10-11T17:35:24.143" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-10-11T17:35:24.143" LastActivityDate="2015-10-11T17:35:24.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23096" PostTypeId="4" CreationDate="2015-10-11T17:35:24.143" Score="0" Body="An easy riddle for beginners." OwnerUserId="16323" LastEditorUserId="16323" LastEditDate="2015-10-11T22:56:11.520" LastActivityDate="2015-10-11T22:56:11.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23097" PostTypeId="2" ParentId="23094" CreationDate="2015-10-11T17:39:58.510" Score="12" Body="&lt;p&gt;Each sequence is generated by:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Taking the number of letters in the spelling of the last number&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I.E.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 7=S-E-V-E-N has 5 letters&lt;br&gt;&#xA; 5=F-I-V-E has 4 letters&lt;br&gt;&#xA; 4=F-O-U-R has 4 letters....&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also Note  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Four is the only number that has the same number of letters as the number itself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7715" LastActivityDate="2015-10-11T17:39:58.510" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23098" PostTypeId="2" ParentId="8132" CreationDate="2015-10-11T17:59:44.373" Score="4" Body="&lt;p&gt;&lt;strong&gt;medical&lt;/strong&gt; and &lt;strong&gt;decimal&lt;/strong&gt; are pretty close, but &lt;strong&gt;claimed&lt;/strong&gt; and &lt;strong&gt;declaim&lt;/strong&gt; are also anagrams of them.  Pick any pair.  In particular, &lt;strong&gt;claimed/decimal&lt;/strong&gt; and &lt;strong&gt;medical/declaim&lt;/strong&gt; both satisfy the &quot;&lt;em&gt;no pair of letters in the same order&lt;/em&gt;&quot; criteria.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I may come up with some more later...&lt;/p&gt;&#xA;" OwnerUserId="16131" LastActivityDate="2015-10-11T17:59:44.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23099" PostTypeId="2" ParentId="23087" CreationDate="2015-10-11T18:11:26.640" Score="24" Body="&lt;p&gt;As already mentioned, the smallest possible sum is 27, resulting from the numbers 1, 2, 4, 5, 7, 8 arranged like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 1&#xA;4758&#xA; 2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Optimality is a consequence of the observation that no three sides of the die may show three consecutive numbers (e.g. 1, 2, 3 or 3, 4, 5 are not possible).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this, suppose that $n$, $n+1$ and $n+2$ were all on different sides of Enrico's die. Then because of rule #3, $n$ and $n+1$ must be located on opposite sides, and so must $n+1$ and $n+2$: a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that no three sides may show three consecutive numbers implies that there must be at least two &lt;em&gt;gaps&lt;/em&gt; in the number sequence between three pairs of consecutive numbers, and we see that the sum of the sequence is minimized for gaps at 3 and 6.&lt;/p&gt;&#xA;" OwnerUserId="4394" LastEditorUserId="4394" LastEditDate="2015-10-11T21:46:37.583" LastActivityDate="2015-10-11T21:46:37.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23101" PostTypeId="2" ParentId="23091" CreationDate="2015-10-12T00:54:06.260" Score="10" Body="&lt;p&gt;No amount of in-shuffles and out-shuffles will approach a random shuffle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say that two cards mirror each other if they are the same distance from the middle of the deck. This means that the $i$&lt;sup&gt;th&lt;/sup&gt; and $j$&lt;sup&gt;th&lt;/sup&gt; cards from the top mirror each other if and only if $i+j=53$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If two cards mirror each other, then after the deck is either in-shuffled or out-shuffled, they will still mirror each other. This means that the result of several in/out-shuffles is uniquely determined by the top 26 cards, so that at most $2^{26}\times 26!$ of the $52!$ deck permutations can be achieved by in/out shuffling.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-10-12T00:54:06.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23103" PostTypeId="1" AcceptedAnswerId="23105" CreationDate="2015-10-12T07:37:23.680" Score="4" ViewCount="330" Body="&lt;p&gt;Professor Halfbrain has spent another weekend on analyzing and dissecting chessboards. His efforts resulted in the following surpring theorem.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Professor Halfbrain's chessboard dissection theorem&lt;/strong&gt;:&lt;br&gt;&#xA;  For every $m\ge1.000.000$ and $n\ge1.000.000$, the $m\times n$ chessboard can be dissected into pieces of size $~4\times6$, $~6\times4$, $~5\times7$, $~$and $~7\times5$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is this theorem true, or has the professor once again made one of his mathematical blunders?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-10-12T10:25:37.660" Title="Professor Halfbrain's chessboard dissection theorem" Tags="|mathematics|combinatorics|checkerboard|dissection|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23104" PostTypeId="2" ParentId="23087" CreationDate="2015-10-12T08:36:52.903" Score="13" Body="&lt;p&gt;Can I use a &lt;a href=&quot;https://en.wikipedia.org/wiki/Barrel_dice&quot;&gt;barrel die&lt;/a&gt;? Then we could do an arrangement of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 3 5 2 4 6&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;for a sum of &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 21.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can also do  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 5 3 6 2 4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which works even better, since opposing faces have the same sum.&lt;/p&gt;&#xA;" OwnerUserId="16474" LastActivityDate="2015-10-12T08:36:52.903" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23105" PostTypeId="2" ParentId="23103" CreationDate="2015-10-12T10:25:37.660" Score="10" Body="&lt;p&gt;The theorem is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; true&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First note that any number &gt;= 11 can be written as a sum of 4,6 and 5 but also as a sum of 4,6 and 7:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;11 = 6 + 5 = 4 + 7&lt;br&gt;&#xA;12 = 6 + 6&lt;br&gt;&#xA;13 = 4 + 4 + 5 = 6 + 7&lt;br&gt;&#xA;14 = 5 + 5 + 4 = 7 + 7&lt;br&gt;&#xA;For higher numbers just add 4 to these.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing this, fill the top row with blocks having a width of 4,6 or 5. Now we want to level the bottom of these blocks by adding same-width blocks underneath it. The smallest value of this is the least common factor of the heights of these blocks. LCM(6,4,7) = 84. In other words:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As long as the width of the board &gt;= 11 and the height &gt; 84 you can always remove 84 rows.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now do this: Keep removing 84 rows until the total height is divisible by 5. This can maximally take 4 steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Now do something similar for the columns:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fill the left column with blocks having a height of 4,6 or 7. Now we want to level the right side of these blocks by adding same-height blocks to the right of it. The smallest value of this is the least common factor of the widths of these blocks. LCM(6,4,5) = 60. In other words:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As long as the height of the board &gt;= 11 and the width &gt; 60 you can always remove 60 columns.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now do this: Keep removing 60 rows until the total width is divisible by 7. This can maximally take 6 steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this algorithm you'll end up with a board with the width divisible by 7 and the height divisible by 5 so we can fill this board with 7x5 blocks, proving the theorem.&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-10-12T10:25:37.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23106" PostTypeId="2" ParentId="22648" CreationDate="2015-10-12T14:46:09.723" Score="1" Body="&lt;p&gt;The puzzle you seek is a variant of a classic that exists in several versions.  There does not seem to be a single definitive source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A version of the puzzle was presented in the movie &lt;em&gt;Die Hard: With a Vengeance&lt;/em&gt;, which, I'm guessing, is where most people today have encountered it: &quot;provided with an unlimited water supply, a 5-gallon jug, and a 3-gallon jug, measure out precisely 4 gallons, by filling and emptying the jugs.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has been attributed to &lt;a href=&quot;https://en.wikipedia.org/wiki/Sim%C3%A9on_Denis_Poisson&quot; rel=&quot;nofollow&quot;&gt;Siméon Denis Poisson&lt;/a&gt;, but apparently there are much older versions, for example one by &lt;a href=&quot;https://en.wikipedia.org/wiki/Luca_Pacioli&quot; rel=&quot;nofollow&quot;&gt;Luca Pacioli&lt;/a&gt; from the 15th century.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a source that discusses the puzzle and touches on its history and some of its different versions: &lt;a href=&quot;http://www.pballew.net/PatBlogs.html/221&quot; rel=&quot;nofollow&quot;&gt;http://www.pballew.net/PatBlogs.html/221&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="4385" LastEditorUserId="4385" LastEditDate="2015-10-14T03:40:10.347" LastActivityDate="2015-10-14T03:40:10.347" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23107" PostTypeId="2" ParentId="23071" CreationDate="2015-10-12T15:05:17.693" Score="3" Body="&lt;p&gt;I was thinking&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HK-47 from Star Wars: KOTOR, &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;•The character I'm looking for also had memory problems, but it was nothing compared to the main character of the game. I don't want to spoil the story, therefore I don't look for the main character.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - The main character, Darth Revan, had amnesia, which was a major part of the story.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;•He really likes to kill people, but he is not a soldier or warrior like the other two characters I'm not looking for. Assassin would be a better description.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HK-47 is an assassin droid and really doesn't like people, whom he refers to as &quot;meatbags.&quot;  The other two soldiers/warriors represent Canderous Ordo and Carth Onasi.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;•I'm not looking for the small guy with odd pronunciation, I wonder how anyone can understand what he says.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is the droid T3-M4, who speaks only Binary.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;•I'm neither looking for the little girl nor for their big hairy friend.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mission Vao the Twi'lek and Zaalbar the Wookiee&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;•I'm also not looking for the female character who managed to get captured twice during the game.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bastila Shan, the Jedi who caused the memory loss of Darth Revan&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;•There is another female character which was chosen as the 10th worst character on a ranking list. The character I'm looking for is quite the opposite, he even won two awards.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Juhani, the Cathar Dark Jedi&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;•Finally, the old hermit character is another one I'm not looking for.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jedi Master Jolee Bindo&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14233" LastEditorUserId="14233" LastEditDate="2015-10-12T18:01:40.130" LastActivityDate="2015-10-12T18:01:40.130" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23110" PostTypeId="2" ParentId="23072" CreationDate="2015-10-12T15:55:53.203" Score="4" Body="&lt;p&gt;I don't know if this is what you we going for or not, but assuming we can only use &lt;em&gt;each operator once&lt;/em&gt; (there's no rule specified for this)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The answer is:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;17&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;List:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2*(5-4)/2 = 1&lt;br&gt;&#xA; (2-5+4)*2 = 2&lt;br&gt;&#xA; (2*5-4)/2 = 3&lt;br&gt;&#xA; 2+(5-4)*2 = 4&lt;br&gt;&#xA; 2+5-4/2 = 5&lt;br&gt;&#xA; (2+5-4)*2 = 6&lt;br&gt;&#xA; (2*5+4)/2 = 7&lt;br&gt;&#xA; 2*5-4+2 = 8&lt;br&gt;&#xA; 2*(5+4)/2 = 9&lt;br&gt;&#xA; (2+5)*2-4 = 10&lt;br&gt;&#xA; (2+5*4)/2 = 11&lt;br&gt;&#xA; 2+5*4/2 = 12&lt;br&gt;&#xA; 2*(5/2+4) = 13&lt;br&gt;&#xA; (2+5)*4/2 = 14&lt;br&gt;&#xA; (2+4)&lt;em&gt;5/2 = 15&lt;br&gt;&#xA; 2&lt;/em&gt;(5+4)-2 = 16&lt;br&gt;&#xA; (2+5/2)*4 = 18&lt;br&gt;&#xA; 4*5-2/2 = 19&lt;br&gt;&#xA; 2+5*4-2 = 20&lt;br&gt;&#xA; 2/2+5*4 = 21&lt;br&gt;&#xA; (2/2+5)*4 = 24&lt;br&gt;&#xA; (2/2+4)*5 = 25&lt;br&gt;&#xA; (2+5)*4-2 = 26&lt;br&gt;&#xA; (2+4)*5-2 = 28&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also relevant: Your rule about not using each one is irrelevant, any number you can make with less can be made with all of them, at least, up to the smallest number&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if anybody's interested, here's the Lua script used to achieve such results:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; local Numbers = {&lt;br&gt;&#xA;    {2,5,4,2},&lt;br&gt;&#xA;    {2,5,2,4},&lt;br&gt;&#xA;    {2,2,5,4},&lt;br&gt;&#xA;    {2,2,4,5},&lt;br&gt;&#xA;    {2,4,5,2},&lt;br&gt;&#xA;    {2,4,2,5},&lt;br&gt;&#xA;    {4,2,5,2},&lt;br&gt;&#xA;    {4,2,2,5},&lt;br&gt;&#xA;    {4,5,2,2},&lt;br&gt;&#xA;    {5,2,4,2},&lt;br&gt;&#xA;    {5,2,2,4},&lt;br&gt;&#xA;    {5,4,2,2}&lt;br&gt;&#xA; }&lt;br&gt;&#xA;&lt;br&gt;&#xA; local Operators = {&lt;br&gt;&#xA;    {'+','/','&lt;em&gt;','-'},&lt;br&gt;&#xA;    {'+','/','-','&lt;/em&gt;'},&lt;br&gt;&#xA;    {'+','-','/','&lt;em&gt;'},&lt;br&gt;&#xA;    {'+','-','&lt;/em&gt;','/'},&lt;br&gt;&#xA;    {'+','&lt;em&gt;','/','-'},&lt;br&gt;&#xA;    {'+','&lt;/em&gt;','-','/'},&lt;br&gt;&#xA;    {'-','+','&lt;em&gt;','/'},&lt;br&gt;&#xA;    {'-','+','/','&lt;/em&gt;'},&lt;br&gt;&#xA;    {'-','/','+','&lt;em&gt;'},&lt;br&gt;&#xA;    {'-','/','&lt;/em&gt;','+'},&lt;br&gt;&#xA;    {'-','&lt;em&gt;','+','/'},&lt;br&gt;&#xA;    {'-','&lt;/em&gt;','/','+'},&lt;br&gt;&#xA;    {'&lt;em&gt;','+','/','-'},&lt;br&gt;&#xA;    {'&lt;/em&gt;','+','-','/'},&lt;br&gt;&#xA;    {'&lt;em&gt;','-','+','/'},&lt;br&gt;&#xA;    {'&lt;/em&gt;','-','/','+'},&lt;br&gt;&#xA;    {'&lt;em&gt;','/','+','-'},&lt;br&gt;&#xA;    {'&lt;/em&gt;','/','-','+'},&lt;br&gt;&#xA;    {'/','+','&lt;em&gt;','-'},&lt;br&gt;&#xA;    {'/','+','-','&lt;/em&gt;'},&lt;br&gt;&#xA;    {'/','-','+','&lt;em&gt;'},&lt;br&gt;&#xA;    {'/','-','&lt;/em&gt;','+'},&lt;br&gt;&#xA;    {'/','&lt;em&gt;','+','-'},&lt;br&gt;&#xA;    {'/','&lt;/em&gt;','-','+'}&lt;br&gt;&#xA; }&lt;br&gt;&#xA;&lt;br&gt;&#xA; function BracketForm(eq, f)&lt;br&gt;&#xA;    if f == 0 then&lt;br&gt;&#xA;       return eq&lt;br&gt;&#xA;    elseif f == 1 then&lt;br&gt;&#xA;       return &quot;(&quot;..eq:sub(1,3)..&quot;)&quot;..eq:sub(4)&#xA;    elseif f == 2 then&lt;br&gt;&#xA;       return eq:sub(1,4)..&quot;(&quot;..eq:sub(5)..&quot;)&quot;&lt;br&gt;&#xA;    elseif f == 3 then&lt;br&gt;&#xA;       return eq:sub(1,2)..&quot;(&quot;..eq:sub(3,5)..&quot;)&quot;..eq:sub(6)&#xA;    elseif f == 4 then&lt;br&gt;&#xA;       return &quot;(&quot;..eq:sub(1,5)..&quot;)&quot;..eq:sub(6)&#xA;    elseif f == 5 then&lt;br&gt;&#xA;       return eq:sub(1,2)..&quot;(&quot;..eq:sub(3)..&quot;)&quot;&lt;br&gt;&#xA;    elseif f == 6 then&lt;br&gt;&#xA;       return &quot;((&quot;..eq:sub(1,3)..&quot;)&quot;..eq:sub(4,5)..&quot;)&quot;..eq:sub(6)&lt;br&gt;&#xA;    elseif f == 7 then&lt;br&gt;&#xA;       return eq:sub(1,2)..&quot;((&quot;..eq:sub(3,5)..&quot;)&quot;..eq:sub(6)..&quot;)&quot;&lt;br&gt;&#xA;    end&lt;br&gt;&#xA; end&lt;br&gt;&#xA;&lt;br&gt;&#xA; local Found = {}&lt;br&gt;&#xA;&lt;br&gt;&#xA; for i,v in pairs(Numbers) do&lt;br&gt;&#xA;    for ii,vv in pairs(Operators) do&lt;br&gt;&#xA;       local equation = v[1] .. vv[1] .. v[2] .. vv[2] .. v[3] .. vv[3] .. v[4]&lt;br&gt;&#xA;       for i = 0, 7 do&lt;br&gt;&#xA;          equation2 = BracketForm(equation, i)&lt;br&gt;&#xA;          local ans = load(&quot;return &quot; .. equation2)()&lt;br&gt;&#xA;          Found[ans] = ans &gt; 0 and ans%1 == 0 and not Found[ans] and equation2 or nil&lt;br&gt;&#xA;       end&lt;br&gt;&#xA;    end&lt;br&gt;&#xA; end&lt;br&gt;&#xA;&lt;br&gt;&#xA; local answers = {}&lt;br&gt;&#xA; for i, v in pairs(Found) do&lt;br&gt;&#xA;    answers[#answers+1] = i&lt;br&gt;&#xA; end&lt;br&gt;&#xA;&lt;br&gt;&#xA; for i = 1,#answers do&lt;br&gt;&#xA;    local least = math.min(table.unpack(answers))&#xA;    local eq = Found[least]&lt;br&gt;&#xA;    for ii = 1, #answers do&lt;br&gt;&#xA;       if answers[ii] == least then&lt;br&gt;&#xA;          table.remove(answers, ii)&lt;br&gt;&#xA;       end&lt;br&gt;&#xA;    end&lt;br&gt;&#xA;    print(eq ..&quot;=&quot;..least)&lt;br&gt;&#xA; end&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditDate="2015-10-13T01:51:57.320" LastActivityDate="2015-10-13T01:51:57.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23111" PostTypeId="2" ParentId="9063" CreationDate="2015-10-12T16:09:21.417" Score="0" Body="&lt;p&gt;Already answered, but how I did it was by unintentionally disregarding the &quot;not all bags need to be used&quot; stipulation and dividing amongst 10 bags. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the numbers 1, 2, 3, 4 and 5 any nonzero digit can be created with addition, meaning there are 85 coins(1+2+3+4+5=15) and 5 bags left over to make every base ten number up to 100. To complete this I got 5, 10, 20, 20 and 30, rounding out 10 bags that can make every sum from 1 - 100. Explaining the intuition would be quite clumsy, but it basically boils down to considering the addition of the existing multiples of 10 with respect to future possibilities.&lt;/p&gt;&#xA;" OwnerUserId="16510" LastActivityDate="2015-10-12T16:09:21.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23112" PostTypeId="2" ParentId="23071" CreationDate="2015-10-12T16:24:58.477" Score="3" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HK-47 from KotOR.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Details:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. Main character was amnesiac, HK needed to have his memories restored.&lt;br&gt;&#xA; 2. He is a hunter-killer droid (assassin).&lt;br&gt;&#xA; 3. Small guy with weird pronunctiation would be Vandar Tokare, a member&lt;br&gt;&#xA; of Yoda's species.&lt;br&gt;&#xA; 4. The little girl would be Mission, and her wookiee friend Zaalbar.&lt;br&gt;&#xA; 5. Bastila is the character captured twice.&lt;br&gt;&#xA; 6. Juhani was the reviled character (deservedly so, bleh).&lt;br&gt;&#xA; 7. The old hermit is Jolee Bindo.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16874" LastEditorUserId="16874" LastEditDate="2015-10-12T16:29:59.523" LastActivityDate="2015-10-12T16:29:59.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23113" PostTypeId="1" AcceptedAnswerId="23119" CreationDate="2015-10-12T17:08:37.783" Score="7" ViewCount="782" Body="&lt;p&gt;This is the third in a series &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=user%3A9000%20is%3AQuestion%20title%3A%22Lights%20in%20the%20Dungeon%22&quot;&gt;Lights in the Dungeon&lt;/a&gt; puzzles.&lt;/p&gt;&#xA;&lt;p&gt;&lt;sub&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/23053/lights-in-the-dungeon-four-by-four&quot;&gt;previous (2nd) &amp;lt;-- &lt;/a&gt;| &lt;a href=&quot;https://puzzling.stackexchange.com/questions/23192/lights-in-the-dungeon-eighteen-by-eleven&quot;&gt;--&amp;gt; next (4th)&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;You are practically skipping as you leave the &lt;a href=&quot;https://puzzling.stackexchange.com/q/23053/9000&quot;&gt;four by four dungeon&lt;/a&gt;. This has been pretty easy so far. Perhaps you can survive long enough that his Lordship will tire of the games and simply let you go. Just then, you spot your captor watching you and are reminded that he might grow tired and &lt;em&gt;eat&lt;/em&gt; you instead. Perhaps you shouldn't be so flippant in your victory.&lt;/p&gt;&#xA;&lt;p&gt;Lord Hooty McOwlface glances at Fakky Jim for confirmation of your victory. The grand vizier returns a sorrowful &amp;quot;Mreow&amp;quot; and the Lord seethes beneath his cloak.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;Very well!&amp;quot; he barks at you. &amp;quot;Since the first two were so easy, I'm going to take some time to make it a little bit harder for you. Guards! Escort him to his cell.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;A few days pass and then you are brought back to the royal chamber. The despot looks fairly pleased with himself but, based on the first two puzzles, you aren't really that worried.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;I've thrown you a curveball this time, peasant! There are a few new room designs you haven't seen before. Turning off every light is impossible! Now, it would be no fun if we &lt;em&gt;knew&lt;/em&gt; you were going to lose so you aren't actually required to turn off &lt;em&gt;all&lt;/em&gt; the lights. You've saved me enough on electricity so far that I was able to afford a small helicopter for Fakky Jim. He'll be heading in after you to turn off any lights you can't get. Of course, this introduces some new rules. Fakky will fill you in.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;Once again, you retrieve a scroll from the cat and unfurl it to find the rules.(Revisions from the previous challenge are in bold.)&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You will enter a dungeon in which there are many rooms. Each room has a single light in the middle. &lt;strike&gt;You must turn off every light&lt;/strike&gt;.&lt;/li&gt;&#xA;&lt;li&gt;The lights are easily turned off by a short chain hanging from the fixture.&lt;/li&gt;&#xA;&lt;li&gt;As soon as you exit a room, the floor will collapse to reveal the thousand-foot chasm below. You may not re-enter a room and, therefore, you must turn off each light by passing through each room exactly once.&lt;/li&gt;&#xA;&lt;li&gt;The map below shows the current design of the dungeon with each room bordered by red and labeled with a alphabetic character.&lt;/li&gt;&#xA;&lt;li&gt;You may rearrange the rooms however you like before you enter but you cannot change the overall dimensions of the dungeon. In this case, that means the dungeon must be &lt;strong&gt;five&lt;/strong&gt; rooms wide and &lt;strong&gt;five&lt;/strong&gt; rooms tall.&lt;/li&gt;&#xA;&lt;li&gt;You cannot rotate or flip rooms. You may only translate them.&lt;/li&gt;&#xA;&lt;li&gt;You may only enter the dungeon once and exit once. As soon as you exit the dungeon, any remaining floors will collapse.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;strike&gt;If there are any rooms not visited and, therefore, any lights left on, you lose. If all lights are turned off when you exit the dungeon, then you win.&lt;/strike&gt;&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;If you create a dungeon map that you cannot enter or cannot exit, you shall lose.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Fakky is easily confused. There can be only two entrances to the dungeon. (One entrance, one exit)&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Fakky is easily annoyed. If he has to turn off too many lights, he'll quit. You must extinguish at least 70% of the lights on your way through.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;The helicopter can fly but cannot pass through walls. Every room must be accessible even though you can't visit them all.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;How can you rearrange the dungeon rooms so that you can win the game?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;sub&gt;There are probably multiple solutions. The first valid answer will be accepted.&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Optimization Bonus: Can maximize the number of rooms whose lights you turn off on your pass-through?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;sub&gt;There are three dead-ends so the absolute max is 22/25 rooms (88%)&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OazPS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OazPS.png&quot; alt=&quot;Maze 5x5&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;pre&gt;&#xA;AABCC&#xA;DDEEE&#xA;FGGHH&#xA;HJKKM&#xA;MNNNO&#xA;&lt;/pre&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;For reference, here are the 16 possible rooms. (Note that this is all &lt;em&gt;possible&lt;/em&gt; rooms. For this puzzle, you must use exactly the rooms shown in the map above.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rp7xy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rp7xy.png&quot; alt=&quot;A&quot; /&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/E9wSu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/E9wSu.png&quot; alt=&quot;B&quot; /&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/SlADl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SlADl.png&quot; alt=&quot;C&quot; /&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/yEoqW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yEoqW.png&quot; alt=&quot;D&quot; /&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/QGft0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QGft0.png&quot; alt=&quot;E&quot; /&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/VREO9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VREO9.png&quot; alt=&quot;F&quot; /&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/R1gGL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/R1gGL.png&quot; alt=&quot;G&quot; /&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/vK9AI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vK9AI.png&quot; alt=&quot;H&quot; /&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/Cft7i.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Cft7i.png&quot; alt=&quot;I&quot; /&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/3oSC9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3oSC9.png&quot; alt=&quot;J&quot; /&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/E72qu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/E72qu.png&quot; alt=&quot;K&quot; /&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/p8yY1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/p8yY1.png&quot; alt=&quot;L&quot; /&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/jqqKf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jqqKf.png&quot; alt=&quot;M&quot; /&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/zKQB7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zKQB7.png&quot; alt=&quot;N&quot; /&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/yTRBU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yTRBU.png&quot; alt=&quot;O&quot; /&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/s78Fu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/s78Fu.png&quot; alt=&quot;P&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="69582" LastEditDate="2021-10-18T18:13:00.687" LastActivityDate="2021-10-18T18:13:00.687" Title="Lights in the Dungeon: Five by Five" Tags="|visual|story|optimization|mazes|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23114" PostTypeId="1" AcceptedAnswerId="23120" CreationDate="2015-10-12T17:26:22.460" Score="1" ViewCount="2685" Body="&lt;p&gt;A twist on the classic, &quot;how many weighings to find the odd weight ball in 9 balls&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose there is n balls. There are two types of balls. One type weighs less than the other. In the set of balls there is at least one of each ball. What is the least amount of weighing needed to separate the two different set of balls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am thinking at least n weighings since we do not know how many of each ball there are.&lt;/p&gt;&#xA;" OwnerUserId="16878" LastEditorUserId="8874" LastEditDate="2015-10-12T18:29:04.750" LastActivityDate="2015-10-12T20:16:50.167" Title="Ball Weighing Riddle" Tags="|strategy|weighing|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23115" PostTypeId="1" CreationDate="2015-10-12T17:38:25.950" Score="8" ViewCount="3259" Body="&lt;p&gt;A voice recognition software crashes from using the letter &quot;b&quot;, but you need to test it.  You have to speak continuously for about an hour, using meaningful words, but you cannot repeat words, and you can't use the letter &quot;b&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can you solve this problem if no one gives you such a text and you have no time to pre-write it and you have to start speaking right away?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is only one word, which shows that you can be sure to speak without the letter &quot;b&quot; for a long time without thinking and repeating.&lt;/p&gt;&#xA;" OwnerUserId="16877" LastEditorUserId="5373" LastEditDate="2020-05-05T14:39:49.510" LastActivityDate="2020-05-05T14:39:49.510" Title="Speak without using the letter &quot;b&quot;" Tags="|strategy|letters|situation|" AnswerCount="3" CommentCount="8" ClosedDate="2015-10-13T14:50:13.920" ContentLicense="CC BY-SA 3.0" />
  <row Id="23116" PostTypeId="2" ParentId="23115" CreationDate="2015-10-12T17:46:16.447" Score="40" Body="&lt;p&gt;You could count upwards from 1, because none of the numbers less than a billion contains the letter &quot;b&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this might be repeating words if &quot;101&quot; is considered to contain &quot;one&quot;. (The version I've seen before is &quot;Provide 100 words that don't contain the letter 'a'.&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you happen to have the system memorized, you can use &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_polygons#Systematic_polygon_names&quot;&gt;polygon names&lt;/a&gt; instead.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-10-12T17:55:16.337" LastActivityDate="2015-10-12T17:55:16.337" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23117" PostTypeId="2" ParentId="23115" CreationDate="2015-10-12T17:49:57.833" Score="15" Body="&lt;p&gt;Given the hint of one word, I'm guessing the full chemical name for titin, which is 189,819 characters long and does not contain the letter &quot;b&quot;. It takes roughly 3 hours to finish. It is one word so it doesn't repeat anything. Although it is questionable whether if it is meaningful&lt;/p&gt;&#xA;" OwnerUserId="16621" LastActivityDate="2015-10-12T17:49:57.833" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23119" PostTypeId="2" ParentId="23113" CreationDate="2015-10-12T18:01:14.180" Score="5" Body="&lt;p&gt;After some trial and error I found an optimal solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;JGGKK&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;EDAOH&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;FCBEH&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;MHENC&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;MANND&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or as an image:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/mtTKQ.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mtTKQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I reached the maximum of 22 rooms, so Fakky Jim will have to turn off the lights in the 3 rooms with a dead end.&lt;/p&gt;&#xA;" OwnerUserId="448" LastEditorUserId="448" LastEditDate="2015-10-12T18:11:22.800" LastActivityDate="2015-10-12T18:11:22.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23120" PostTypeId="2" ParentId="23114" CreationDate="2015-10-12T18:08:14.647" Score="2" Body="&lt;p&gt;I'm pretty sure that it takes at most N-1 weighings to determine the relative weight of every ball.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, let's say N=8, with balls B1-B8.&#xA;We can start by comparing groups of two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B1 &amp;amp; B2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B3 &amp;amp; B4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B5 &amp;amp; B6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B7 &amp;amp; B8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there are only two possible results from these tests. Either they don't weigh the same, or they do. If the balls are of different weights, we place them into the &quot;heavy&quot; pile or &quot;light&quot; pile accordingly, easy-peasy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the balls DO weigh the same, then we just place that group aside for a minute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once this first trial is complete, we've done 4 weighings. If the balls are all different, we're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, on the other hand, some or all of those weighings led to the balls being equal, we do another round. I'm going to assume that all of the weighings came up equal, to pursue the worst case scenario. For the next round, we combine the groups from the first round, and weigh them against each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B1+B2 vs B3+B4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B5+B6 vs B7+B8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After 6 weighings, we're at the same place as after the first round. If the weights were different, the answer is obvious. If they weighed the same, we do one more trial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B1+B2+B3+B4 vs B5+B6+B7+B8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now we know for sure which balls weigh what, in the worst case, with 7 weighings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(And in all honesty, if the problem states that we KNOW there is at least one ball with a different weight, that last weighing is unnecessary. We wouldn't necessarily know which group was which weight, but we'd be able to separate them.)&lt;/p&gt;&#xA;" OwnerUserId="16814" LastActivityDate="2015-10-12T18:08:14.647" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23121" PostTypeId="2" ParentId="23012" CreationDate="2015-10-12T18:45:56.317" Score="4" Body="&lt;p&gt;I believe the characters are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pacman and Fozzy bear from the Muppet Show&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; pacman is shaped like a pizza with a slice missing&lt;br&gt;&#xA; Fozzy is fuzzy, and is a lame comedian&lt;br&gt;&#xA; pacman deals with ghosts and fruit, fozzy has Ms Piggy and Kermit (pork  and frog)&lt;br&gt;&#xA; In japanese, pacman is sometimes spelled Pakkuman&lt;br&gt;&#xA; The link: they both go Wakka-wakka!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16874" LastActivityDate="2015-10-12T18:45:56.317" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23122" PostTypeId="2" ParentId="23077" CreationDate="2015-10-12T18:58:14.253" Score="7" Body="&lt;p&gt;I have an answer of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2, 4, 5, 6&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which yields an N of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 45&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Barring a bug in my test program, this should be correct. I tried all possible values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are all the values for my answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 = 5 - 4&lt;br&gt;&#xA; 2 = 2&lt;br&gt;&#xA; 3 = 5 - 2&lt;br&gt;&#xA; 4 = 4&lt;br&gt;&#xA; 5 = 5&lt;br&gt;&#xA; 6 = 6&lt;br&gt;&#xA; 7 = 5 + 2&lt;br&gt;&#xA; 8 = 6 + 2&lt;br&gt;&#xA; 9 = 5 + 4&lt;br&gt;&#xA; 10 = 6 + 4&lt;br&gt;&#xA; 11 = 6 + 5&lt;br&gt;&#xA; 12 = 6 * 2&lt;br&gt;&#xA; 13 = 6 + 5 + 2&lt;br&gt;&#xA; 14 = (5 * 4) - 6&lt;br&gt;&#xA; 15 = 6 + 5 + 4&lt;br&gt;&#xA; 16 = (6 * 2) + 4&lt;br&gt;&#xA; 17 = (6 * 2) + 5&lt;br&gt;&#xA; 18 = (5 + 4) * 2&lt;br&gt;&#xA; 19 = (6 * 4) - 5&lt;br&gt;&#xA; 20 = 5 * 4&lt;br&gt;&#xA; 21 = ((6 * 4) + 2) - 5&lt;br&gt;&#xA; 22 = (5 * 4) + 2&lt;br&gt;&#xA; 23 = (5 * 4) + (6 / 2)&lt;br&gt;&#xA; 24 = 6 * 4&lt;br&gt;&#xA; 25 = ((6 + 4) * 2) + 5&lt;br&gt;&#xA; 26 = (6 * 4) + 2&lt;br&gt;&#xA; 27 = ((6 * 4) + 5) - 2&lt;br&gt;&#xA; 28 = (5 + 2) * 4&lt;br&gt;&#xA; 29 = (6 * 4) + 5&lt;br&gt;&#xA; 30 = 6 * 5&lt;br&gt;&#xA; 31 = (6 * 4) + 5 + 2&lt;br&gt;&#xA; 32 = (6 * 5) + 2&lt;br&gt;&#xA; 33 = ((2 / 4) + 5) * 6&lt;br&gt;&#xA; 34 = (6 * 5) + 4&lt;br&gt;&#xA; 35 = ((6 / 2) + 4) * 5&lt;br&gt;&#xA; 36 = (4 + 2) * 6&lt;br&gt;&#xA; 37 = ((6 + 2) * 4) + 5&lt;br&gt;&#xA; 38 = ((5 + 2) * 6) - 4&lt;br&gt;&#xA; 39 = ((5 / 2) + 4) * 6&lt;br&gt;&#xA; 40 = (6 + 2) * 5&lt;br&gt;&#xA; 41 = ((4 + 2) * 6) + 5&lt;br&gt;&#xA; 42 = (5 + 2) * 6&lt;br&gt;&#xA; 43 = (6 * 4 * 2) - 5&lt;br&gt;&#xA; 44 = (6 + 5) * 4&lt;br&gt;&#xA; 45 = ???&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="375" LastEditorUserId="375" LastEditDate="2015-10-12T19:43:24.557" LastActivityDate="2015-10-12T19:43:24.557" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23124" PostTypeId="2" ParentId="23114" CreationDate="2015-10-12T20:16:50.167" Score="1" Body="&lt;p&gt;Nice puzzle.  I know there is already an accepted answer, but I thought I would point out N-1 is not always optimal.  The below solution separates 8 balls in six weighings.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Take a group of four and call them A, B, C, and D.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weight AB vs CD, and A vs B.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If AB balance CD and&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;if A balances B, then all four A, B, C, D weigh the same.&lt;/li&gt;&#xA;&lt;li&gt;if (say) A is heavier than B, then A and B are known, while C and D are heavy and light in some order.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;If (say) AB is heaver than CD and&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;if A balances B, then A and B are both heavy, and C and D can be known with one more weighing.&lt;/li&gt;&#xA;&lt;li&gt;if (say) A is heavier than B, then A and B are known, and C and D are both light.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Basically there are two possible results from considering A, B, C, and D.  Either all of them can be identified in three weighings, or you know all four balls are the same in just two weighings.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So do the same thing with the other four balls E, F, G, and H.   Either you can identify all eight balls in 6 weighings, or you have at least one group of four  which are all the same and have only used at most 5 weighings.  You can then finish up by weighing A vs. E.&lt;/p&gt;&#xA;" OwnerUserId="2722" LastActivityDate="2015-10-12T20:16:50.167" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23125" PostTypeId="2" ParentId="22812" CreationDate="2015-10-12T20:18:55.237" Score="3" Body="&lt;p&gt;Regarding &quot;reheated&quot;, it seems that you can do it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;['reheated'] / []&lt;br&gt;&#xA;  ['re'] / ['heated']&lt;br&gt;&#xA;  ['reed'] / ['heat']&lt;br&gt;&#xA;  ['re'] / ['heat', 'ed']&lt;br&gt;&#xA;  [] / ['reheat', 'ed']&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know it was just an example, but I wanted to confirm that I had a correct understanding of how it worked.  This assumes that you're allowed &quot;ed&quot;, which is marginal, IMHO!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do it without &quot;ed&quot;, you could do it like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;['reheated']  /  []&lt;br&gt;&#xA;   ['heated']  /  ['re']&lt;br&gt;&#xA;   ['heat']  /  ['reed']&lt;br&gt;&#xA;   ['at']  /  ['reed', 'he']&lt;br&gt;&#xA;   ['a']  /  ['treed', 'he']&lt;br&gt;&#xA;   ['ad']  /  ['tree', 'he']&lt;br&gt;&#xA;   []  /  ['tree', 'head'] &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Still working on the full puzzle...&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1 ['abstruse', 'condors', 'honest', 'invention', 'neon', 'reacted', 'vague'] @ []&lt;br&gt;&#xA;  2 ['abstruse', 'condors', 'honest', 'invention', 'neon', 'react', 'vague'] @ ['&lt;strong&gt;ed&lt;/strong&gt;']&lt;br&gt;&#xA;  3 ['abstruse', 'condors', 'honest', 'invention', 'neon', 're', 'vague'] @ ['ed', '&lt;strong&gt;act&lt;/strong&gt;']&lt;br&gt;&#xA;  4 ['use', 'condors', 'honest', 'invention', 'neon', 're', 'vague'] @ ['ed', '&lt;strong&gt;abstr&lt;/strong&gt;act']&lt;br&gt;&#xA;  5 ['use', 'condors', 'honest', 'invent', 'neon', 're', 'vague'] @ ['ed', 'abstract&lt;strong&gt;ion&lt;/strong&gt;']&lt;br&gt;&#xA;  6 ['use', 'condors', 'honest', 'vent', 'neon', 're', 'vague'] @ ['ed', 'abstraction', '&lt;strong&gt;in&lt;/strong&gt;']&lt;br&gt;&#xA;  7 ['use', 'condor', 'honest', 'vent', 'neon', 're', 'vague'] @ ['ed', 'abstraction&lt;strong&gt;s&lt;/strong&gt;', 'in']&lt;br&gt;&#xA;  8 ['use', 'condo', 'honest', 'vent', 'neon', 're', 'vague'] @ ['ed', 'abstractions', '&lt;strong&gt;r&lt;/strong&gt;in']&lt;br&gt;&#xA;  9 ['use&lt;strong&gt;r&lt;/strong&gt;', 'condo', 'honest', 'vent', 'neon', 're', 'vague'] @ ['ed', 'abstractions', 'in']&lt;br&gt;&#xA;  10 ['us', 'condo', 'honest', 'vent', 'neon', 're', 'vague'] @ ['ed', 'abstractions', 'in', '&lt;strong&gt;er&lt;/strong&gt;']&lt;br&gt;&#xA;  11 ['us', 'condo', 'ho', 'vent', 'neon', 're', 'vague'] @ ['ed', 'abstractions', 'in', 'er&lt;strong&gt;nest&lt;/strong&gt;']&lt;br&gt;&#xA;  12 ['condo', 'ho', 'vent', 'neon', 're', 'vague'] @ ['&lt;strong&gt;us&lt;/strong&gt;ed', 'abstractions', 'in', 'ernest']&lt;br&gt;&#xA;  13 ['condo', 'vent', 'neon', 're', 'vague'] @ ['&lt;strong&gt;ho&lt;/strong&gt;used', 'abstractions', 'in', 'ernest']&lt;br&gt;&#xA;  14 ['condo', 'vent', 'neon', 're&lt;strong&gt;d&lt;/strong&gt;', 'vague'] @ ['house', 'abstractions', 'in', 'ernest']&lt;br&gt;&#xA;  15 ['condo', 'vent', 'neon', 'red', 'vague&lt;strong&gt;st&lt;/strong&gt;'] @ ['house', 'abstractions', 'in', 'erne']&lt;br&gt;&#xA;  16 ['condo', 'vent', 'neon', 'red', 'guest'] @ ['house', 'abstractions', '&lt;strong&gt;va&lt;/strong&gt;in', 'erne']&lt;br&gt;&#xA;  17 ['condo', 'vent', 'neon', 'red'] @ ['&lt;strong&gt;guest&lt;/strong&gt;house', 'abstractions', 'vain', 'erne']&lt;br&gt;&#xA;  18 ['con', 'vent', 'neon', 'red'] @ ['guesthouse', 'abstractions', 'vain', 'erne', '&lt;strong&gt;do&lt;/strong&gt;']&lt;br&gt;&#xA;  19 ['con&lt;strong&gt;tractions&lt;/strong&gt;', 'vent', 'neon', 'red'] @ ['guesthouse', 'abs', 'vain', 'erne', 'do']&lt;br&gt;&#xA;  20 ['contractions', 'vent', 'on', 'red'] @ ['guesthouse', 'abs', 'vain', 'erne', 'do&lt;strong&gt;ne&lt;/strong&gt;']&lt;br&gt;&#xA;  21 ['con', 'vent', 'on', 'red'] @ ['guesthouse', 'abs', 'vain', 'erne', 'done', '&lt;strong&gt;tractions&lt;/strong&gt;']&lt;br&gt;&#xA;  22 ['vent', 'on', 'red'] @ ['guesthouse', 'abs', 'vain', 'erne', '&lt;strong&gt;con&lt;/strong&gt;done', 'tractions']&lt;br&gt;&#xA;  23 ['vent', 'o', 'red'] @ ['guesthouse', '&lt;strong&gt;n&lt;/strong&gt;abs', 'vain', 'erne', 'condone', 'tractions']&lt;br&gt;&#xA;  24 ['vent', 'o', 're'] @ ['guesthouse', 'nabs', 'vain', 'erne', 'condone&lt;strong&gt;d&lt;/strong&gt;', 'tractions']&lt;br&gt;&#xA;  25 ['vent', 'o'] @ ['guesthouse', 'nabs', 'vain', 'erne', 'condoned', '&lt;strong&gt;re&lt;/strong&gt;tractions']&lt;br&gt;&#xA;  26 ['vent', 'o&lt;strong&gt;ne&lt;/strong&gt;'] @ ['guesthouse', 'nabs', 'vain', 'er', 'condoned', 'retractions']&lt;br&gt;&#xA;  27 ['one'] @ ['guesthouse', 'nabs', 'vain', '&lt;strong&gt;vent&lt;/strong&gt;er', 'condoned', 'retractions']&lt;br&gt;&#xA;  28 ['one&lt;strong&gt;s&lt;/strong&gt;'] @ ['guesthouse', 'nab', 'vain', 'venter', 'condoned', 'retractions']&lt;br&gt;&#xA;  29 ['&lt;strong&gt;b&lt;/strong&gt;ones'] @ ['guesthouse', 'na', 'vain', 'venter', 'condoned', 'retractions']&lt;br&gt;&#xA;  30 ['bone'] @ ['guesthouse', 'na', 'vain', 'venter&lt;strong&gt;s&lt;/strong&gt;', 'condoned', 'retractions']&lt;br&gt;&#xA;  31 ['bone&lt;strong&gt;d&lt;/strong&gt;'] @ ['guesthouse', 'na', 'vain', 'venters', 'condone', 'retractions']&lt;br&gt;&#xA;  32 ['bon'] @ ['guesthouse', 'na', 'vain', 'venters', 'condone', 'retractions', '&lt;strong&gt;ed&lt;/strong&gt;']&lt;br&gt;&#xA;  33 ['on'] @ ['guesthouse', 'na', 'vain', 'venters', 'condone', 'retractions', '&lt;strong&gt;b&lt;/strong&gt;ed']&lt;br&gt;&#xA;  34 ['&lt;strong&gt;n&lt;/strong&gt;on'] @ ['guesthouse', 'a', 'vain', 'venters', 'condone', 'retractions', 'bed']&lt;br&gt;&#xA;  35 [] @ ['guesthouse', 'a&lt;strong&gt;non&lt;/strong&gt;', 'vain', 'venters', 'condone', 'retractions', 'bed'] &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or, going for fewer steps (edit - down to 30 steps):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1 ['abstruse', 'condors', 'honest', 'invention', 'neon', 'reacted', 'vague'] @ []&lt;br&gt;&#xA;  2 ['abstruse', 'condors', 'honest', 'invention', 'neon', 'react', 'vague'] @ ['&lt;strong&gt;ed&lt;/strong&gt;']&lt;br&gt;&#xA;  3 ['abstruse', 'condors', 'honest', 'invention', 'neon', 're', 'vague'] @ ['ed', '&lt;strong&gt;act&lt;/strong&gt;']&lt;br&gt;&#xA;  4 ['use', 'condors', 'honest', 'invention', 'neon', 're', 'vague'] @ ['ed', '&lt;strong&gt;abstr&lt;/strong&gt;act']&lt;br&gt;&#xA;  5 ['use', 'condors', 'honest', 'invent', 'neon', 're', 'vague'] @ ['ed', 'abstract&lt;strong&gt;ion&lt;/strong&gt;']&lt;br&gt;&#xA;  6 ['use', 'condors', 'honest', 'vent', 'neon', 're', 'vague'] @ ['ed', 'abstraction', '&lt;strong&gt;in&lt;/strong&gt;']&lt;br&gt;&#xA;  7 ['use', 'condor', 'honest', 'vent', 'neon', 're', 'vague'] @ ['ed', 'abstraction&lt;strong&gt;s&lt;/strong&gt;', 'in']&lt;br&gt;&#xA;  8 ['use', 'condo', 'honest', 'vent', 'neon', 're', 'vague'] @ ['ed', 'abstractions', '&lt;strong&gt;r&lt;/strong&gt;in']&lt;br&gt;&#xA;  9 ['use&lt;strong&gt;r&lt;/strong&gt;', 'condo', 'honest', 'vent', 'neon', 're', 'vague'] @ ['ed', 'abstractions', 'in']&lt;br&gt;&#xA;  10 ['us', 'condo', 'honest', 'vent', 'neon', 're', 'vague'] @ ['ed', 'abstractions', 'in', '&lt;strong&gt;er&lt;/strong&gt;']&lt;br&gt;&#xA;  11 ['us', 'condo', 'ho', 'vent', 'neon', 're', 'vague'] @ ['ed', 'abstractions', 'in', 'er&lt;strong&gt;nest&lt;/strong&gt;']&lt;br&gt;&#xA;  12 ['condo', 'ho', 'vent', 'neon', 're', 'vague'] @ ['&lt;strong&gt;us&lt;/strong&gt;ed', 'abstractions', 'in', 'ernest']&lt;br&gt;&#xA;  13 ['condo', 'vent', 'neon', 're', 'vague'] @ ['&lt;strong&gt;ho&lt;/strong&gt;used', 'abstractions', 'in', 'ernest']&lt;br&gt;&#xA;  14 ['condo', 'vent', 'neon', 're&lt;strong&gt;d&lt;/strong&gt;', 'vague'] @ ['house', 'abstractions', 'in', 'ernest']&lt;br&gt;&#xA;  15 ['condo', 'vent', 'neon', 'red', 'vague&lt;strong&gt;st&lt;/strong&gt;'] @ ['house', 'abstractions', 'in', 'erne']&lt;br&gt;&#xA;  16 ['condo', 'vent', 'neon', 'red', 'guest'] @ ['house', 'abstractions', '&lt;strong&gt;va&lt;/strong&gt;in', 'erne']&lt;br&gt;&#xA;  17 ['condo', 'vent', 'neon', 'red'] @ ['&lt;strong&gt;guest&lt;/strong&gt;house', 'abstractions', 'vain', 'erne']&lt;br&gt;&#xA;  18 ['condo&lt;strong&gt;ne&lt;/strong&gt;', 'vent', 'neon', 'red'] @ ['guesthouse', 'abstractions', 'vain', 'er']&lt;br&gt;&#xA;  19 ['con', 'vent', 'neon', 'red'] @ ['guesthouse', 'abstractions', 'vain', 'er', '&lt;strong&gt;done&lt;/strong&gt;']&lt;br&gt;&#xA;  20 ['con&lt;strong&gt;e&lt;/strong&gt;', 'vent', 'neon', 'red'] @ ['guesthouse', 'abstractions', 'vain', 'er', 'don']&lt;br&gt;&#xA;  21 ['cone&lt;strong&gt;d&lt;/strong&gt;', 'vent', 'neon', 'red'] @ ['guesthouse', 'abstractions', 'vain', 'er', 'on']&lt;br&gt;&#xA;  22 ['coned', 'vent', 'neon'] @ ['guesthouse', 'abstractions', 'vain', 'er&lt;strong&gt;red&lt;/strong&gt;', 'on']&lt;br&gt;&#xA;  23 ['con', 'vent', 'neon'] @ ['guesthouse', 'abstractions', 'vain', 'erred', 'on', '&lt;strong&gt;ed&lt;/strong&gt;']&lt;br&gt;&#xA;  24 ['con', 'neon'] @ ['guesthouse', 'abstractions', 'vain', 'erred', 'on', '&lt;strong&gt;vent&lt;/strong&gt;ed']&lt;br&gt;&#xA;  25 ['con', 'eon'] @ ['guesthouse', 'abstractions', 'vain', 'erred', '&lt;strong&gt;n&lt;/strong&gt;on', 'vented']&lt;br&gt;&#xA;  26 ['con&lt;strong&gt;ed&lt;/strong&gt;', 'eon'] @ ['guesthouse', 'abstractions', 'vain', 'err', 'non', 'vented']&lt;br&gt;&#xA;  27 ['con', 'eon'] @ ['guesthouse', 'abstractions', 'vain', 'err', 'non', 'vented', '&lt;strong&gt;ed&lt;/strong&gt;']&lt;br&gt;&#xA;  28 ['eon'] @ ['guesthouse', 'abstractions', 'vain', 'err', 'non', '&lt;strong&gt;con&lt;/strong&gt;vented', 'ed']&lt;br&gt;&#xA;  29 ['&lt;strong&gt;n&lt;/strong&gt;eon'] @ ['guesthouse', 'abstractions', 'vain', 'err', 'on', 'convented', 'ed']&lt;br&gt;&#xA;  30 [] @ ['guesthouse', 'abstractions', 'vain', 'err', 'on', 'convented', '&lt;strong&gt;neon&lt;/strong&gt;ed'] &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2015-10-27T17:31:59.757" LastActivityDate="2015-10-27T17:31:59.757" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="23126" PostTypeId="1" CreationDate="2015-10-12T21:30:39.557" Score="20" ViewCount="5984" Body="&lt;p&gt;A casino offers the following game:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You flip a fair coin repeatedly and record the results on slips of paper, one result per slip. At any time you may decide to stop, at which time one of the slips is chosen uniformly at random. You win \$1  if the chosen slip says &quot;Heads&quot;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How much should you be willing to pay to play this game, and what is your strategy?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I heard this puzzle several years ago, I do not know the solution.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="13475" LastEditDate="2017-12-13T19:29:39.627" LastActivityDate="2017-12-13T19:29:39.627" Title="What is the expected value of this coin flipping game?" Tags="|mathematics|probability|" AnswerCount="6" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23127" PostTypeId="2" ParentId="23126" CreationDate="2015-10-12T21:59:21.313" Score="0" Body="&lt;p&gt;Well, let's work out the probability tree and see what pops up:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;      1      0&#xA;    x    .5     0&#xA;  x    .66  .33     0&#xA;x    .75    .5  .25   0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If you get a heads, you just end the game and take your money.&lt;br&gt;&#xA;If you have all tails, you keep on flipping.&lt;br&gt;&#xA;At the first .5, you have an even chance of winning/losing if you stay.&lt;br&gt;&#xA;Say you go on. The two 50% chances have offsetting chances, expected payout is still 50/50.&lt;br&gt;&#xA;Say you go another layer. Expected winnings is still 50/50. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, say you go TTH.  At that point, your chances are .33. If you go again, your cumulative chance actually goes up to .375. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, it seems like flip until you have even heads/tails chance, or greater. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, 50% H (1), 25% TH (.50), 6.25% TTHH (.50).&lt;br&gt;&#xA;That looks like an infinite series where each term is 1/4 of the last, so the formula would be :&lt;br&gt;&#xA;$.5 + .125 * \frac{1}{1-.25} = \frac{2}{3}$ &lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-10-13T13:03:32.643" LastActivityDate="2015-10-13T13:03:32.643" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23128" PostTypeId="2" ParentId="23126" CreationDate="2015-10-12T21:59:53.397" Score="0" Body="&lt;p&gt;You should not pay more than $0.50 to play this game. Here is why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each flip brings the odds of your slip draw closer to 50:50 odds. Yes you may hit heads on your first flip at which point, you take the casino for their dollar. This will only happen 50% of the time. The other 50% of the time you will get a tails which forces you to keep on flipping to try for a better chance than 0% on your paper slip drawing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say at some point you get 4 heads and 3 tails. That is a ~ 57% chance of winning that dollar (i.e. 4 heads / 7 slips). The longer you go the closer you get to 50:50 odds. Paying any more than $0.50 may win you money in the early stages, but only 50% of the time. The other 50% of the time you keep flipping to play catchup with the tails you have accumulated bringing you ever closer to the 50:50 odds you get from just flipping the coin in the first place :D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reponse To Comments:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Someone asked if you can get above 50% by continuing to flip. You will most definitely get more heads, and for a short time may even have greater than 50% odds, but your best odds happen with early flips. The trick to this puzzle is you have to realize that the times when you are above 50% win rate only happens 50% of the time you play this game with the casino.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take for example that I flip heads on my first flip. I have a 50% chance of doing that and it will give me a 100% chance on my win rate during the dollar draw. My real odds end up being 50% for the flip time 100% for the draw or 0.5 * 1 = 0.5 or 50% because the other 50% of the time I will hit a tails and have a 0% chance of winning the draw and have to keep flipping or forfeit my bet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't do better than 100% chance of winning, so the other 50% of the time you either lose your entire bet or you keep flipping to get an advantage of 50% or greater. The kicker is as you keep flipping, 50% of the time you will dig the hole deeper (i.e. get more tails). The more you flip, the closer the odds converge to 50:50 because you can't outpace the tails you have accumulated and even when you do get a streak of heads going, you are equally likely and definitely will hit an equal streak of tails whether it be this round of betting or some future round since the coin flip is 50:50.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So even though I may have more heads (at some point in time), the ratio of heads to tails will always converge on 50:50.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 heads / 4 flips = 75%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6 heads / 10 flips = 60%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;30 heads / 55 flips = 54.5%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1000 heads / 1900 flips = 52.6%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1000 heads is a lot of heads, but 1000 when compared to 900 tails is a smaller advantage than when you compare 6 heads out of 10 flips.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I know someone out there is thinking, well I would just stop at the 75% and play (3/4), but remember that is just a rate at which you CAN win a dollar. The other 25% of the draw will lose your bet. Even worse, in a future round of betting you ABSOLUTELY WILL come across a game where you have 3 tails and 1 head giving you only a 25% to win the draw and 75% chance to lose. In this scenario, you can either play catch-up with the the tails, which can take you deeper into the tail territory or in the best case even you out to the 50% win rate or you can choose to submit to a win rate lower than 50% in which case you negate any winnings you had in previous rounds of betting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the long run you will never come out above a 50% win rate and should not pay more than $0.50 to play.&lt;/p&gt;&#xA;" OwnerUserId="16787" LastEditorUserId="16787" LastEditDate="2015-10-12T23:35:58.597" LastActivityDate="2015-10-12T23:35:58.597" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23129" PostTypeId="2" ParentId="23126" CreationDate="2015-10-12T22:00:51.810" Score="1" Body="&lt;p&gt;I'm sure someone will come up with a better answer, but as a simple benchmark:&lt;br&gt;&#xA;&lt;br&gt;&#xA;Given that your coin is fair (equal chance of heads and tails), and that time has no value in this problem, I would follow the strategy of ending as soon as you get your first heads.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your first toss is heads ($\frac{1}{2}$  chance), then you win \$1 with a perfect probability. If you throw a tails then heads, ($\frac{1}{4}$ chance), then you have a $\frac{1}{2}$ chance of winning \$1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can use this set up to find the expected winnings each time as:&lt;br&gt;&#xA;$\sum_{n=1}^\infty \frac{1}{n}*\frac{1}{2^n}$ or $\sum_{n=1}^\infty \frac{1}{n2^n}$. This holds true because the chance of winning after $n$ flips following this strategy is $\frac{1}{n}$ and the chance of getting a heads on the $n^{th}$ flip is $\frac{1}{2^n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually, we get that we expect to win \$.69 (really $ln(2)$ or $0.6931472$), so paying \$.69 or less should lead to a profit as we play enough games.&lt;/p&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="9951" LastEditDate="2015-10-13T00:01:22.520" LastActivityDate="2015-10-13T00:01:22.520" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23130" PostTypeId="2" ParentId="23126" CreationDate="2015-10-12T22:08:38.913" Score="1" Body="&lt;p&gt;You can get at least a 50% chance of success very easily, always stop after the first flip. You can do a bit better by only stopping if you get &quot;Heads&quot; and continuing for a while if you get tails. It seems reasonable to me that you could get a 75% win ratio by saying &quot;if I get tails, keep flipping until the heads and tails even out&quot;, which by the law of large numbers should &lt;em&gt;eventually&lt;/em&gt; happen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there isn't anything really special about for 50% if you're willing to keep flipping forever. If you waited until by a fluke of probability there was eventually a 60% ratio, you could do better. By waiting an extremely large amount of time, you could get the ratio as large as you wanted, since (even though the probability would continually drop) there is no penalty for playing for an excessively long time. So, you can get arbitrarily close to 100%, although you will never reach it unless you got lucky on the first flip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, any cost between \$0 and \$0.99 will favor the better, while a cost of $1 will &lt;em&gt;barely&lt;/em&gt; favor the house. &lt;/p&gt;&#xA;" OwnerUserId="16887" LastActivityDate="2015-10-12T22:08:38.913" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="23131" PostTypeId="1" AcceptedAnswerId="23164" CreationDate="2015-10-13T00:10:13.927" Score="7" ViewCount="918" Body="&lt;p&gt;Encouraged by the reasonable success of my first ngram puzzle &lt;a href=&quot;https://puzzling.stackexchange.com/questions/23022/assassinate-shakespeare&quot;&gt;Assassinate Shakespeare!&lt;/a&gt; I've decided to explore the idea further.&lt;/p&gt;&#xA;&lt;p&gt;Here is a new ngram. Your task is to reproduce it exactly by entering the relevant search terms into Google ngram.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6kIa9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6kIa9.png&quot; alt=&quot;ngram&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Detail&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6Uddx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6Uddx.png&quot; alt=&quot;ngram settings&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;The answer&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The correct answer will be a link to the &lt;a href=&quot;https://books.google.com/ngrams&quot; rel=&quot;nofollow noreferrer&quot;&gt;Google ngram viewer&lt;/a&gt; showing the exact same graph as I have given but with the mystery text filled in.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;How to proceed&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;This puzzle is designed to be difficult but not impossible. I thought my first ngram puzzle was difficult and yet within three hours someone had solved it. This time I decided to pitch the level at those in the community who are super-quick and give them a little more of a challenge.&lt;/p&gt;&#xA;&lt;p&gt;Finding a solution does not rely on brute-force methods. If you observe all aspects of the diagram carefully and take note of the Knowledge tag, you should be able to arrive at a partial answer and then a very small amount of fine-tuning will complete the solution.&lt;/p&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="69582" LastEditDate="2021-11-07T04:43:55.417" LastActivityDate="2021-11-07T04:43:55.417" Title="Google ngram puzzle No 2" Tags="|enigmatic-puzzle|knowledge|google-ngram|" AnswerCount="1" CommentCount="7" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23132" PostTypeId="2" ParentId="23126" CreationDate="2015-10-13T00:10:35.770" Score="6" Body="&lt;p&gt;The number of heads will equal the number of tails infinitely many times, so at any point, you can give up trying to be clever and just wait until the proportion is 1/2, then take that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's first suppose that the player is impatient, meaning that for a particular number $N$, they will give up after $N$ turns, either claiming the current ratio if it is at least 1/2 or waiting for it to reach 1/2 if not. Let $f_N(h,t)$ most amount of money such an impatient player could expect to win, if they already have gotten $h$ heads and $t$ tails. Using the boundary conditions $f_N(h,N-h)=\max(\frac{h}N,\frac12)$ and the recursive formula&#xA;$$&#xA;f_N(h,t)=\max\left(\tfrac12 f_N(h+1,t)+\tfrac12 f_N(h,t+1),\tfrac{h}{h+t}\right)&#xA;$$&#xA;we can use a computer to compute $f_N(0,0)$, which is the initial value of the game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jdawiseman.com/papers/easymath/coin-stopping.html&quot; rel=&quot;noreferrer&quot;&gt;People have done exactly this&lt;/a&gt;, and found that for large values of $N$, $f_N(0,0)\approx0.79$. This means there are definitely strategies which win slightly more that $79$ cents on average, though no one is sure if any more can be attained.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-10-13T00:10:35.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23133" PostTypeId="2" ParentId="23126" CreationDate="2015-10-13T00:11:14.503" Score="11" Body="&lt;p&gt;I would like to throw another possible answer into the mix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My strategy would be to say stop whenever heads is in the majority.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot prove that this is optimal, nor can I figure out exactly what this converges to.  However, I think it's like this.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You will only call it on odd turns (say after $i$ turns)&lt;/li&gt;&#xA;&lt;li&gt;If you're going down the binomial tree, there will be some number of the nodes in the $i$th row where there are $(i+1)/2$ heads (these will be where you'll say &quot;stop&quot;).&lt;/li&gt;&#xA;&lt;li&gt;The expected value of these will be $n\cdot 2^{-i}\cdot\frac{(i+1)/2}{i}$, where $i$ is the row number, and $n$ is the number of nodes with the correct value.&lt;/li&gt;&#xA;&lt;li&gt;Then (I think) that $n=(2n)!/(n!(n+1)!)$ (I brute forced the first few and searched on &lt;a href=&quot;http://oeis.org/A000108&quot;&gt;OEIS&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;With further brute forcing this converged as follows:&#xA;\begin{eqnarray}&#xA;1 &amp;amp;&amp;amp; 0.5 \\&#xA;3 &amp;amp;&amp;amp; 0.5833333333333334\\&#xA;5 &amp;amp;&amp;amp; 0.6208333333333333\\&#xA;7 &amp;amp;&amp;amp; 0.643154761904762\\&#xA;9 &amp;amp;&amp;amp; 0.6583457341269842\\&#xA;11 &amp;amp;&amp;amp; 0.6695318136724387\\&#xA;13 &amp;amp;&amp;amp; 0.6782081958839772\\&#xA;15 &amp;amp;&amp;amp; 0.6851906177589772\\&#xA;17 &amp;amp;&amp;amp; 0.690966518207047\\&#xA;19 &amp;amp;&amp;amp; 0.6958473229716441\\&#xA;21 &amp;amp;&amp;amp; 0.7000424908764525\\&#xA;23 &amp;amp;&amp;amp; 0.703698753813252\\&#xA;25 &amp;amp;&amp;amp; 0.7069223589691968\\&#xA;27 &amp;amp;&amp;amp; 0.7097923778046177\\&#xA;29 &amp;amp;&amp;amp; 0.7123690326457777\\&#xA;\end{eqnarray}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the solution is bigger than this, at least (there's some value to most of the other nodes, so even if you called it here, you could get something more).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone good at sequences could figure out this sum:&#xA;\begin{equation}\sum_{n=0}^{\infty}\frac{(2n)!}{n!(n+1)!}\cdot\frac{n+1}{2n+1}\cdot\frac{1}{2^{2n+1}}\end{equation}&#xA;you'd have the answer to what this strategy would give you.  Someone else may be able to find a better strategy of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the sum to 200 gives (approximately) 0.7654966208975894&lt;/p&gt;&#xA;" OwnerUserId="15942" LastActivityDate="2015-10-13T00:11:14.503" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23134" PostTypeId="1" AcceptedAnswerId="23137" CreationDate="2015-10-13T01:33:43.310" Score="5" ViewCount="702" Body="&lt;p&gt;&lt;strong&gt;The Problem:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;There are two players.&lt;/li&gt;&#xA;&lt;li&gt;Each player writes a number, hidden from the other player. It can be any integer 1 or greater.&lt;/li&gt;&#xA;&lt;li&gt;The players reveal their numbers.&lt;/li&gt;&#xA;&lt;li&gt;Whoever chose the lower number gets 1 point, unless the lower number is lower by only 1, then the player with higher number gets 2 points.&lt;/li&gt;&#xA;&lt;li&gt;If they both chose the same number, neither player gets a point.&lt;/li&gt;&#xA;&lt;li&gt;This repeats, and the game ends when one player has 5 points.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Is it possible to have a strong winning strategy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My Approach:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;There is no such thing as random if there is a strategy.&lt;/li&gt;&#xA;&lt;li&gt;If so, naturally each player is going to to pick  1. But if I know that you are going to pick 1, I will pick 2. Then you know I am going to pick 2, you will pick 3. And so forth.&lt;/li&gt;&#xA;&lt;li&gt;Obviously I can't deduce that into a strategy based on I know that you know I know you know. So I can only take a guess as the best bet.&lt;/li&gt;&#xA;&lt;li&gt;The way I guess your next number is taken the average of all the numbers you played in the same round. Plus 1. Unless your average number is greater than 2, then I will pick your mean mod by 5 (a simple way to counter skew the mean).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Not Strong Enough:&lt;/strong&gt;&#xA;There seems to be a better winning strategy. So what are the shortcomings of this approach?&lt;/p&gt;&#xA;" OwnerUserId="16893" LastActivityDate="2015-10-13T22:30:50.040" Title="First to Five Points" Tags="|logical-deduction|calculation-puzzle|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23135" PostTypeId="2" ParentId="23134" CreationDate="2015-10-13T03:02:53.720" Score="4" Body="&lt;p&gt;I'd have to say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No it's not possible to have a strong winning strategy. If we are assuming both players play optimally, then no matter what strategy is used, both players will always play optimally. This is the case for any game in which both players take their turns simultaneously. Unless the opponent is playing non-optimally and we are given their exact strategy, there is no way to create an optimal strategy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11727" LastActivityDate="2015-10-13T03:02:53.720" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23136" PostTypeId="2" ParentId="23134" CreationDate="2015-10-13T03:31:29.900" Score="1" Body="&lt;p&gt;The situation is not dissimilar to rock, paper scissors, which can be shown using game theory (if my memory serves me correctly - the lecture was ~20 years ago) that picking a random number works best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The unboundedness is kind of interesting.  But there's obviously a trend to keep the game close to 1, and the symmetry would suggest that it's impossible to get an edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However it should still be possible to set up a game theory matrix and find the Nash equilibrium, which would give you the optimal relative probability of choosing each number.  But I wasn't able to figure this out yet.  I'll try to find a bit of time later!&lt;/p&gt;&#xA;" OwnerUserId="15942" LastActivityDate="2015-10-13T03:31:29.900" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23137" PostTypeId="2" ParentId="23134" CreationDate="2015-10-13T04:21:17.737" Score="10" Body="&lt;p&gt;If you have a strategy that wins more than 50% of the time, your opponent can use the same strategy and also win more than 50% of the time, which is impossible. Therefore, the best possible strategy cannot guarantee winning more than 50% of the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what strategy will guarantee at least 50% against any other strategy?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Part 1: Without the 5 points goal&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;For now, we will ignore the goal of getting to 5 points, and just try to maximize the value of our score minus the opponent's score. We must find a mixed strategy such that the expected value against any other strategy is at least 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we choose 1 with probability $p_1$, 2 with probability $p_2$, and so on, such that $p_1+p_2+p_3+\ldots=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the opponent always chooses 1, our expected payoff is $2p_2-p_3-p_4-\ldots=p_1+3p_2-1$. This quantity must be at least 0, or else we are losing. We can repeat this process for each possible number the opponent chooses:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-2p_1+2p_3-p_4-p_5-\ldots=-p_1+p_2+3p_3-1\ge0\\&#xA;p_1-2p_2+2p_4-p_5-\ldots=2p_1-p_2+p_3+3p_4-1\ge0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, for all $n\ge2$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2(p_1+p_2+\ldots+p_{n-2})-p_{n-1}+p_n+3p_{n+1}-1\ge0\\&#xA;2(p_1+p_2+\ldots+p_{n+1})+p_{n+1}\ge1+3p_{n-1}+p_n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S_n=p_1+p_2+\ldots+p_n$. Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3S_{n+1}-S_n\ge1+S_n+2S_{n-1}-3S_{n-2}\\&#xA;3S_{n+1}+3S_{n-2}\ge1+2S_n+2S_{n-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This holds down to $n=1$ if we assume $S_0=S_{-1}=0$. Some linear algebra produces a solution: $$S_1=1/16\\S_2=3/8\\S_3=5/8\\S_4=15/16\\S_5=1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So our optimal strategy is: $$p_1=p_5=1/16\\p_2=p_4=5/16\\p_3=1/4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This strategy guarantees that in the long run, our points will keep pace with the opponent's points.&#xA;&lt;hr&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Part 2: With the goal&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Now we add the goal. This changes things slightly, because if we have four points, it doesn't matter whether we win one point or two, and similarly for the opponent.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;4-4&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Suppose both players have 4 points. If we get one point or two points, our payoff is 1, and if the opponent gets one point or two points, our payoff is -1. We require our expected payoff to be 0 (any higher is again impossible because of symmetry). Repeating the process with the adjusted payoffs results in the inequality:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2S_{n+1}+2S_{n-2}\ge1+S_n+S_{n-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can solve this to find:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p_1=p_2=p_3=1/3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes sense, because this situation is basically rock-paper-scissors with the numbers 1, 2, and 3: each number beats one and loses to one. All the numbers greater than 3 are irrelevant because they lose to this strategy.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;4-3&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;This case is asymmetric, which makes it harder to analyze. The side with 4 points wins with one point or two, while the side with 3 points wins if getting two points. We know the situation is equal if the side with 3 points gains one point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If my methods and calculations were correct, the value of this situation is about $0.239142$ for the player with 4 points (or about 62% chance to win). This number is the real root of a fifth degree polynomial. The strategy for the player with 4 points is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p_1\approx0.067558\\p_2\approx0.222986\\p_3\approx0.253375\\p_4\approx0.333977\\p_5\approx0.122105$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the strategy for the player with 3 points is the same in reverse:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p_1\approx0.122105\\p_2\approx0.333977\\p_3\approx0.253375\\p_4\approx0.222986\\p_5\approx0.067558$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This result indicates that it is very unlikely that there will be a nice solution for the rest of the problem.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;3-3&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Using $0.239142$ as the value of winning or losing one point, we can calculate the optimal strategy for a score of 3-3:&#xA;$$p_1=p_7\approx0.009228\\p_2=p_6\approx0.191209\\p_3=p_5\approx0.161754\\p_4\approx0.275618$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not going to continue further than this because the rest of the asymmetric scores will be very tedious to solve.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-10-13T22:30:50.040" LastActivityDate="2015-10-13T22:30:50.040" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23139" PostTypeId="1" AcceptedAnswerId="23141" CreationDate="2015-10-13T10:29:45.050" Score="1" ViewCount="358" Body="&lt;p&gt;Your agent has retrieved an encoded hard disk from the enemy. It has ten files. Each file has been encrypted using a different password. The encryption algorithm used in uncrackable. Hence you used brute-force to crack the first seven passwords. After this, it is taking too long to do so. However, the contents of the files are genuine and you need to crack the remaining three.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the first seven passwords:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;10 &#xA;1110 &#xA;11110 &#xA;100110 &#xA;1110010110 &#xA;111100111010110 &#xA;100110011110111010110&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Find the remaining three.&lt;/strong&gt; I hope my hints aren't too revealing. I will give the eighth code after a week, and one or two more hints (if this remains unsolved).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It does not involve addition, subtraction, multiplication or division.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Don't convert them into decimal numbers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every code is longer than the previous, and every code contains only 0s and 1s. All these rules are true for any more numbers in the series.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A similar pattern is more popular and even has a wikipedia page on it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;P.S. This is my first number sequence puzzle.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4269" LastEditDate="2015-10-13T15:14:43.417" LastActivityDate="2015-10-13T15:14:43.417" Title="Ten binary passwords (The **** and *** sequence)" Tags="|pattern|number-sequence|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23140" PostTypeId="1" CreationDate="2015-10-13T10:51:21.233" Score="-4" ViewCount="316" Body="&lt;p&gt;I've followed the beginner's Rubik's cube solving method and I've arrived at this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Jz6rM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jz6rM.jpg&quot; alt=&quot;first and second blue and green layers&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qrck4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qrck4.jpg&quot; alt=&quot;second yellow layer is misconceived&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as you can see, I've succeeded mounting the first layer but within the second one there's a problem with the edge of the yellow/white facets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a solution to this dilemma? &lt;/p&gt;&#xA;" OwnerUserId="16904" LastEditorUserId="25966" LastEditDate="2016-11-16T20:38:33.207" LastActivityDate="2016-11-16T23:45:21.970" Title="how to resolve this 3*3 cube" Tags="|rubiks-cube|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23141" PostTypeId="2" ParentId="23139" CreationDate="2015-10-13T10:54:19.473" Score="4" Body="&lt;p&gt;This is just...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Look and Say sequence, but the numbers over 2 are written out in binary.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the next number is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1110010110010011011110111010110&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the others are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1111001110101100111001011010011011110111010110; 100110011110111010110011110011101011011100101101010011011110111010110.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2015-10-13T10:54:19.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23143" PostTypeId="2" ParentId="23140" CreationDate="2015-10-13T11:14:41.597" Score="2" Body="&lt;p&gt;Take the cube, Face the side whose second layer is not made towards you , &#xA;and the 1st layer which you made on the top&#xA;Now do the following steps&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the part of the second layer , is on the right side, Then, Move the right side down(counterclockwise); lower layer to the right ; right layer (which you moved down earlier) upwards ie(clockwise this time).&lt;/li&gt;&#xA;&lt;li&gt;Then , you will find that, 1st layer which you made, one part is missing , form the 1st layer again, by the same methods by which you made the 1st layer.&lt;/li&gt;&#xA;&lt;li&gt;Now the apply the same rule to solve the second layer, which you were trying earlier.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This happens , when all the lower layer edge pieces, have one part same colour as that of the centre of the lower layer.&lt;/p&gt;&#xA;" OwnerUserId="16671" LastActivityDate="2015-10-13T11:14:41.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23144" PostTypeId="2" ParentId="23115" CreationDate="2015-10-13T11:32:54.280" Score="3" Body="&lt;p&gt;put your fingers in your mouth as if your trying to pull your mouth open sideways so your lips can't touch making the b sound. Then you can speak whatever you like, even words containing the b because you won't make the sound. A lot of sounds will be deformed but I think most sounds will be recognizable.&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-10-13T11:32:54.280" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23145" PostTypeId="1" AcceptedAnswerId="23146" CreationDate="2015-10-13T11:43:48.777" Score="21" ViewCount="1167" Body="&lt;pre&gt;&#xA;       1  2  3  4  5  6  7  8  9 10 11 &#xA;     __________________________________&#xA; 1  |  T  N  R  C 18  P  1  2  L 19  Y&#xA; 2  | 14  E 25  A  D  D 10  Q  9  B  D&#xA; 3  | 20  2  R  V  5 12  C  7  3  P  T&#xA; 4  |  L  M 11  R 22  V  S  1  Y  3  C&#xA; 5  |  R  X  A  6  Z  8 22  G 10 15  I&#xA; 6  |  J  B  2 14  H 25  S  M  7  5 24&#xA; 7  |  5  F  8  O  5  K  N  8 12 18  U&#xA; 8  |  V 11  G  Q  4  Y 26  H  T  W  B&#xA; 9  | 13  I  4 17  R  U  A  J  P  8 21&#xA;10  |  S  E  E 20 16  E  O 23  5  3  L&#xA;11  |  O  4 19  X  9  N  D  1  R  6  H&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;First take the first of Roman numeral V,&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;Then second from second, it's easy you see.&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;Follow that up with the last of the second,&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;A Fourth: fifth of sixth, or so it is reckoned.&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;The next can be found in the middle of ten.&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;What's next? Third of three gets added on then.&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;Follow that pattern, append fourth of four,&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;Choose second of nine next, of this I am sure.&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;At ninth, we tack on the second of one,&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;And add on the third of eight (nearly done),&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;End with the start of second to last,&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;And join them together (I'm sure you'll be fast).&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;Now, follow these steps with an observant mind,&lt;/em&gt;&lt;br&gt;&#xA;  &lt;em&gt;And all going well, ___ ________ you'll find!&lt;/em&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastActivityDate="2015-10-13T12:03:17.650" Title="The solution you've been looking for... (Find it, with these 11 simple steps!)" Tags="|enigmatic-puzzle|rhyme|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23146" PostTypeId="2" ParentId="23145" CreationDate="2015-10-13T11:50:21.833" Score="11" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RED HERRINGS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I dont think this answer needs much explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Previous solution is wrong. True solution is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; THE SOLUTION&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;A&lt;/strong&gt; of &lt;strong&gt;B&lt;/strong&gt; in each row of the poem refers to letter &lt;strong&gt;A&lt;/strong&gt; of row &lt;strong&gt;B&lt;/strong&gt; of the poem itself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16608" LastEditorUserId="16608" LastEditDate="2015-10-13T12:03:17.650" LastActivityDate="2015-10-13T12:03:17.650" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="23147" PostTypeId="1" AcceptedAnswerId="23163" CreationDate="2015-10-13T12:03:17.040" Score="8" ViewCount="410" Body="&lt;p&gt;My eldest came home with some interesting maths homework.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We have three bowls containing peas.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Distribute the peas evenly across the bowls, by moving peas from one bowl to another. The only move allowed, is doubling the amount of peas in one bowl by taking them from one other bowl.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;For example:&lt;br&gt;&#xA;  $\begin{array}{rlrlrl}&#xA;A &amp;amp; &amp;amp; B &amp;amp; &amp;amp; C &amp;amp; \\&#xA;\hline &#xA;11 &amp;amp; &amp;amp; 6 &amp;amp; &amp;amp; 7 \\&#xA;4 &amp;amp; (-7) &amp;amp; 6 &amp;amp; &amp;amp; 14 &amp;amp; (+7) \\&#xA;4 &amp;amp; &amp;amp; 12 &amp;amp; (+6) &amp;amp; 8 &amp;amp; (-6) \\&#xA;8 &amp;amp; (+4) &amp;amp; 8 &amp;amp; (-4) &amp;amp; 8&#xA;\end{array}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We had some fun with this. We quickly decided that the order of the bowls doesn't matter, so we can just sort them by the number of peas they contain.&lt;br&gt;&#xA;Then, there are just three moves possible:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Double the amount of peas in the first bowl (containing the fewest peas) by taking them from the third bowl (containing the most peas)&lt;/li&gt;&#xA;&lt;li&gt;Double the amount of peas in the first bowl containing the fewest by taking them from the second bowl (containing the neither the most nor the fewest peas)&lt;/li&gt;&#xA;&lt;li&gt;Double the amount of peas in the second bowl by taking them from the third bowl&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;From that, we made state diagrams and found our solutions. We also found out that for one of the problems, we could never reach an end state (with all bowls containing the same amount of peas).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;But of course I couldn't leave it alone. So I tried to formalise what we did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amount of peas in each bowl at any time is of course a positive integer (or possibly 0):&lt;br&gt;&#xA;$a_i \le b_i \le c_i \,|\,a_i, b_i, c_i \in \mathbb Z_{\ge 0}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A solution has the peas distributed equally across the bowls:&lt;br&gt;&#xA;$a_i = b_i = c_i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total amount of peas at any time is a multiple of 3:&lt;br&gt;&#xA;$a_i + b_i + c_i = 3n \,|\,n \in \mathbb Z_{\ge 0}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The three possible moves:&lt;br&gt;&#xA;$\left\{a_{i+1}, b_{i+1}, c_{i+1}\right\} \in \begin{Bmatrix} &#xA;\left\{2\cdot a_i, b_i, c_i-a_i\right\} \\&#xA;\left\{2\cdot a_i, b_i-a_i, c_i\right\} \\&#xA;\left\{a_i, 2\cdot b_i, c_i-b_i\right\} \end{Bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set of possible moves brings us to the conclusion that for each state after the first, &lt;em&gt;at least one bowl&lt;/em&gt; contains an even amount of peas. Since in the end state, all bowls contain the same amount of peas, they all contain the same &lt;em&gt;even&lt;/em&gt; amount of peas and thus the total number of peas must be even.&lt;br&gt;&#xA;If we combine this with the requirement that the total number of peas is a multiple of 3 as well, we see that the total number of peas &lt;em&gt;must be&lt;/em&gt; a multiple of 6:&lt;br&gt;&#xA;$a_i + b_i + c_i = 6n \,|\,n \in \mathbb Z_{\ge 0}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(Ignoring the very trivial case where we start with a solved state).&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;But these conditions aren't sufficient to guarantee a solvable puzzle. For instance, $\left\{a_0, b_0, c_0\right\} = \left\{6,12,24\right\}$ is not solvable. The only reachable states are $$\begin{Bmatrix} &#xA;\left\{ 6, 12, 24 \right\} \\&#xA;\left\{12, 12, 18 \right\} \\&#xA;\left\{ 0, 18, 24 \right\} \\&#xA;\left\{ 0,  6, 36 \right\} \\&#xA;\left\{ 0, 12, 30 \right\} \\&#xA;\end{Bmatrix}$$ &#xA;neither of which is an end state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So what additional conditions do we need to define to make such a puzzle sovable?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Problems can have multiple solutions. See for instance this example, with two possible paths to a solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\begin{array}{rlrlrl}&#xA;A &amp;amp; &amp;amp; B &amp;amp; &amp;amp; C &amp;amp; \\&#xA;\hline &#xA;11 &amp;amp; &amp;amp; 6 &amp;amp; &amp;amp; 7 \\&#xA;4 &amp;amp; (-7) &amp;amp; 6 &amp;amp; &amp;amp; 14 &amp;amp; (+7) \\&#xA;4 &amp;amp; &amp;amp; 12 &amp;amp; (+6) &amp;amp; 8 &amp;amp; (-6) \\&#xA;8 &amp;amp; (+4) &amp;amp; 8 &amp;amp; (-4) &amp;amp; 8&#xA;\end{array}$  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;and  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\begin{array}{rlrlrl}&#xA;A &amp;amp; &amp;amp; B &amp;amp; &amp;amp; C &amp;amp; \\&#xA;\hline &#xA;11 &amp;amp; &amp;amp; 6 &amp;amp; &amp;amp; 7 \\&#xA;5 &amp;amp; (-6) &amp;amp; 12 &amp;amp; (+6) &amp;amp; 7 &amp;amp; \\&#xA;10 &amp;amp; (+5) &amp;amp; 12 &amp;amp; &amp;amp; 2 &amp;amp; (-5) \\&#xA;8 &amp;amp; (-2) &amp;amp; 12 &amp;amp; &amp;amp; 4 &amp;amp; (+2) \\&#xA;8 &amp;amp; &amp;amp; 8 &amp;amp; (-4) &amp;amp; 8 &amp;amp; (+4) &#xA;\end{array}$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The only way to solve a problem that I've come up so far, is drawing a state diagram and determining the shortest path to the end state. But while state diagrams work fine for small numbers of peas, the number of possible states may well explode for higher numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is there another way to solve these problems? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is a good strategy to arrive at a solution?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Yes, I'm asking&lt;/em&gt; two &lt;em&gt;questions in one, but I don't want to repeat the explanation for a second question. Also, I feel both questions are closely related, since what I'm really looking for is an analysis of this type of problem, if possible even with a variable number of bowls.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditDate="2018-01-09T16:08:01.743" LastActivityDate="2018-01-09T16:08:01.743" Title="Visualise Whirled Peas" Tags="|mathematics|solvability|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23150" PostTypeId="1" AcceptedAnswerId="23153" CreationDate="2015-10-13T12:43:26.247" Score="62" ViewCount="9236" Body="&lt;p&gt;Abraham Van Helsing and Jonathan Harker play a game against Count Dracula.&#xA;The three players agree on the following rules:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;p&gt;Count Dracula and Van Helsing enter the crypt, while Jonathan Harker has to wait outside. The crypt contains 1000 coffins that are numbered from 1 up to 1000, together with 1000 golden lockets that are engraved with the numbers from 1 to 1000.&#xA;  Each of the numbers from 1 to 1000 occurs on exactly one coffin and on exactly one locket.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Count Dracula puts the lockets into the coffins, so that every coffin contains exactly one locket. Van Helsing observes this, and knows exactly which locket is in which coffin.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Now it is Van Helsing's turn:  Van Helsing may (but does not have to) pick a pair of coffins and switch the lockets in these two coffins. Dracula observes the action of Van Helsing.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Van Helsing now leaves the crypt through the back door. Jonathan Harker (who has gained no additional knowledge over the choices of Dracula and Van Helsing in the first three steps) enters the crypt.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Count Dracula chooses an integer $N$ with $1\le N\le1000$, and announces $N$ to Jonathan.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Jonathan is allowed to perform 500 search steps of the following form: Jonathan chooses a coffin, opens it, and checks whether the locket in this coffin carries the number $N$.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;ul&gt;&#xA;  &lt;li&gt;If Jonathan Harker finds the locket with number $N$ during his 500 search steps, then he and Van Helsing have won the game. In this case, they are allowed to drive a wooden stake through Dracula's heart.  &lt;/li&gt;&#xA;  &lt;li&gt;If Harker does not manage to find the locket with number $N$, then Count Dracula is allowed to drink Van Helsing's and Harker's blood to the last remaining drop.&lt;/li&gt;&#xA;  &lt;/ul&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Harker and Van Helsing discuss their options and want to agree on a good strategy.&#xA;If Harker chooses his 500 coffins simply at random, then the team Harker &amp;amp; Van Helsing has a 50% probability of winning the game.  Is there a strategy that would guarantee them an even better probability of success?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="10781" LastEditDate="2015-10-14T13:57:32.057" LastActivityDate="2015-10-14T19:51:55.407" Title="How to beat Count Dracula" Tags="|strategy|game|" AnswerCount="5" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23152" PostTypeId="2" ParentId="23150" CreationDate="2015-10-13T12:58:48.460" Score="8" Body="&lt;p&gt;Van Helsing and Harker can improve their chances by &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; moving locket 1 to coffin 1.  This leaves 999 coffins in an undetermined state, and then their chances are 500/999, which is 50.05%.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am sure someone will come up with a better solution with higher survival probability, but this shows you can get at least minimal improvement.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Of course, if Dracula is stupid enough to choose locket #1 as the one Harker needs to find, then they have a 100% chance of survival, since Jonathon will know exactly where that locket is.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14233" LastEditorUserId="14233" LastEditDate="2015-10-13T18:44:34.450" LastActivityDate="2015-10-13T18:44:34.450" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23153" PostTypeId="2" ParentId="23150" CreationDate="2015-10-13T13:15:44.087" Score="81" Body="&lt;p&gt;The lockets and coffins are always found in loops.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;open a coffin&lt;/li&gt;&#xA;&lt;li&gt;look at the number of the locket in the coffin&lt;/li&gt;&#xA;&lt;li&gt;go to the coffin with the same number as the locket&lt;/li&gt;&#xA;&lt;li&gt;open the coffin&lt;/li&gt;&#xA;&lt;li&gt;repeat from 2&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;At one point you will definitely find the locket of the coffin you first opened&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove this lets look at the other possibility and try to enter a loop without the first coffin in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is impossible, as any locket which would be needed to enter that second loop would have to be included in the second loop itself and therefore can't be used to enter it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that we know that all coffins are ordered in one or more loops we can easily define a strategy for 100% success:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Van Helsing: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Find (if it exists) a loop that contains more than 500 coffins (there can be at most one such loop since we only have 1000 coffins) and switch two lockets in that loop so that two new loops with each 500 or less coffins are created.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Jonathan Harker: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Start with the coffin with the number &lt;strong&gt;N&lt;/strong&gt; as stated by Dracula and follow the loop until you find the locket with the number &lt;strong&gt;N&lt;/strong&gt;. As no loop will have more than 500 coffins it doesn't matter which number dracula chooses.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To understand splitting of loops:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine 4 coffins 1,2,3,4 with the lockets 2,3,4,1 (in that order).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now have a loop of size 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we now switch the lockets of coffin 2 and 4 we get coffins 1,2,3,4 with lockets 2,1,4,3 (in that order).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now we have two loops of size 2 (namely 1,2 and 3,4).&lt;/p&gt;&#xA;" OwnerUserId="16608" LastEditorUserId="16608" LastEditDate="2015-10-13T13:34:18.080" LastActivityDate="2015-10-13T13:34:18.080" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="23155" PostTypeId="1" AcceptedAnswerId="23156" CreationDate="2015-10-13T14:25:00.977" Score="5" ViewCount="479" Body="&lt;p&gt;&lt;sub&gt;You are the protagonist in a game designed by NeedAName Inc. Proceed through the levels using&#xA;your video game knowledge to solve the puzzles therein. Solve all levels and beat the game!&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Level 2.5: A Poetic Riddle&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;After a weird glitch in Level 2, you find yourself transported to Level 2.5 . You wonder what happened in that weird level, but find that you don't want to dwell on it; the game developers probably should've hired beta testers. Moving on, you find a riddle scrawled on the walls:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;One kills with range or mythic pet&lt;br&gt;&#xA;  No passive vet this fighter yet.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Another kills by proxy host&lt;br&gt;&#xA;  A frightful host he proudly boasts.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;One bashes skulls with just one fist&lt;br&gt;&#xA;  A brute, for this does he exist.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A shielded elemental mage&lt;br&gt;&#xA;  Dare not enrage this spellwise sage.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A righteous man, this holy knight&lt;br&gt;&#xA;  He conquers night with hallowed light.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;To shadows this one is confined&lt;br&gt;&#xA;  Both flesh and mind be struck in kind.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The last a man of varied shapes&lt;br&gt;&#xA;  The beasts he tames, their forms he drapes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Based off the riddle, you recognize the game and enter it's name into the keypad at the door. What game is it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/23025/dungeon-crawler-level-1&quot;&gt;Previous Level&lt;/a&gt;  &lt;a href=&quot;https://puzzling.stackexchange.com/questions/23157/dungeon-crawler-level-3&quot;&gt;Next Level&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-10-13T14:46:46.853" Title="Dungeon Crawler: Level 2.5" Tags="|riddle|knowledge|video-games|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23156" PostTypeId="2" ParentId="23155" CreationDate="2015-10-13T14:31:16.763" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sounds like Diablo 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One kills with range or mythic pet&lt;br&gt;&#xA; No passive vet this fighter yet.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Amazon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another kills by proxy host&lt;br&gt;&#xA; A frightful host he proudly boasts.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Necromancer - has undead minions&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One bashes skulls with just one fist&lt;br&gt;&#xA; A brute, for this does he exist.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Barbarian smash&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A shielded elemental mage&lt;br&gt;&#xA; Dare not enrage this spellwise sage.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sorcerer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A righteous man, this holy knight&lt;br&gt;&#xA; He conquers night with hallowed light.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Paladin&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To shadows this one is confined&lt;br&gt;&#xA; Both flesh and mind be struck in kind.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Assassin&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The last a man of varied shapes&lt;br&gt;&#xA; The beasts he tames, their forms he drapes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Druid&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-10-13T14:31:16.763" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23157" PostTypeId="1" AcceptedAnswerId="23168" CreationDate="2015-10-13T14:44:13.347" Score="4" ViewCount="650" Body="&lt;p&gt;&lt;sub&gt;You are the protagonist in a game designed by NeedAName Inc. Proceed through the levels using&#xA;your video game knowledge to solve the puzzles therein. Solve all levels and beat the game!&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Level 3: Cheat Codes&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;You enter the third room of the dungeon, and find a paper with what you recognize as hex cheat codes.&#xA;The door to the exit of the room has a gameshark and a keypad awaiting a cheat code entry.&#xA;The list of codes reads:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;41207472&lt;br&gt;&#xA;  75652067&lt;br&gt;&#xA;  616d6572&lt;br&gt;&#xA;  206b6e6f&lt;br&gt;&#xA;  77732074&lt;br&gt;&#xA;  68652063&lt;br&gt;&#xA;  6f646520&lt;br&gt;&#xA;  746f2064&lt;br&gt;&#xA;  65636970&lt;br&gt;&#xA;  6865723a&lt;br&gt;&#xA;  42594150&lt;br&gt;&#xA;  50575a4c&lt;br&gt;&#xA;  53424144&lt;br&gt;&#xA;  46435251 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using the information hidden in the cheat codes, what is the cheat code that you enter to move on to the next level?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/23155/dungeon-crawler-level-2-5&quot;&gt;Previous Level&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2015-10-14T14:06:44.863" Title="Dungeon Crawler: Level 3" Tags="|cipher|knowledge|video-games|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23159" PostTypeId="1" AcceptedAnswerId="23160" CreationDate="2015-10-13T15:29:50.743" Score="3" ViewCount="216" Body="&lt;p&gt;You're thinking of trying a variant of &lt;a href=&quot;https://puzzling.stackexchange.com/questions/23149/the-chuck-a-luck&quot;&gt;Chuck-a-luck&lt;/a&gt;.  In this variant, you choose one number in the range $1$ to $6$ and five dice are rolled.  These are the payouts:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Number of dice           &#xA;equal to your  Your         &#xA;number         payout    &#xA;---------------------&#xA;None          - 5 USD &#xA;   1            0 USD        &#xA;   2          + 5 USD   &#xA;   3          +10 USD &#xA;   4          +15 USD &#xA;   5          +20 USD &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So for example if none of the dice show your number, you pay $\$5$, while if exactly 1 die shows your number, you don't pay or receive anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You suspect the odds are against you but you want to know exactly how much.  You've also had a few too many, so you don't want to have to resort to inclusion-exclusion or anything complicated.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is a quick and simple way to calculate the expected value of this game?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2722" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-10-14T11:21:31.440" Title="A Chuck-a-luck variant" Tags="|mathematics|calculation-puzzle|probability|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23160" PostTypeId="2" ParentId="23159" CreationDate="2015-10-13T15:43:45.247" Score="3" Body="&lt;p&gt;The expected value is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $-5/6$ USD&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The payoffs in the table follow the formula  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   payoff =   5 * (# dice equal to your number    minus 1)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Since there are six possible outcomes per dice, for a single die the expected number of dice equal to your number is $1/6$. By linearity of expectation the expected number of dice equal to your number is $5/6$, and $5\cdot(5/6-1)=-5/6$.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-10-13T15:43:45.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23161" PostTypeId="1" AcceptedAnswerId="23162" CreationDate="2015-10-13T15:51:27.420" Score="20" ViewCount="2098" Body="&lt;p&gt;The numbers 1, 2, ..., 500 are written on a blackboard. Each minute any two numbers&#xA;are wiped out and their positive difference is written instead. At the end only one number remains.&lt;/p&gt;&#xA;&lt;p&gt;Which values can this final number take?&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="69582" LastEditDate="2021-09-03T02:38:56.190" LastActivityDate="2021-09-03T16:11:24.340" Title="The last number on the blackboard" Tags="|mathematics|calculation-puzzle|reachability|blackboard|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23162" PostTypeId="2" ParentId="23161" CreationDate="2015-10-13T16:16:21.660" Score="27" Body="&lt;h1&gt;Parity&lt;/h1&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If you remove any two even numbers, their difference will also be an even number.  The total number of odd numbers will remain the same.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you remove any two odd numbers, their difference will be an even number.  The total number of odd numbers will decrease by two.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you remove an even and an odd number, their difference will be an odd number.  The total number of odd number will remain the same (remove one, add one).&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Thus we can see that the number of odd numbers at any step is equivalent to the starting count modulo two.  Therefore if you start with an odd number of odd numbers, the final result will be odd, otherwise it will be even.  In this case we have an even number of odds, so the final number will be &lt;strong&gt;even.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;An alternate proof: consider the total of all the numbers on the blackboard.  When you remove two numbers and write their difference what happens to this total?&lt;/p&gt;&#xA;&lt;p&gt;Call the two numbers removed &lt;em&gt;a&lt;/em&gt; and &lt;em&gt;b&lt;/em&gt;.  Order them so &lt;em&gt;a&lt;/em&gt; ≥ &lt;em&gt;b&lt;/em&gt;, so they will be replaced with &lt;em&gt;a&lt;/em&gt; − &lt;em&gt;b&lt;/em&gt;.  When we make the replacement, the total changes by −&lt;em&gt;a&lt;/em&gt; − &lt;em&gt;b&lt;/em&gt; + (&lt;em&gt;a&lt;/em&gt; − &lt;em&gt;b&lt;/em&gt;) = −2&lt;em&gt;b&lt;/em&gt;; that is, the total decreases by an even number.&lt;/p&gt;&#xA;&lt;p&gt;Finally, the total of the starting numbers is even—it is equal to (500 × 501)/2 = 250 × 501 = 2 × (125 × 501)—and since the difference of two even numbers is always even, after any step the total will remain even.  When one number is left it will be equal to the total, so it must be even.&lt;/p&gt;&#xA;&lt;h1&gt;Magnitude&lt;/h1&gt;&#xA;&lt;p&gt;If you remove any two non-negative numbers, their positive difference must be less than or equal to the larger of the two.  Therefore the final result can be &lt;strong&gt;at most 500&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;It turns out that all numbers meeting both these criteria—that is, is all non-negative even numbers not greater than 500—are reachable: see &lt;a href=&quot;https://puzzling.stackexchange.com/a/23166/9718&quot;&gt;Dr Xorile's answer&lt;/a&gt; for a proof.&lt;/p&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="9718" LastEditDate="2021-09-03T16:11:24.340" LastActivityDate="2021-09-03T16:11:24.340" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23163" PostTypeId="2" ParentId="23147" CreationDate="2015-10-13T16:23:51.890" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If $a_0$, $b_0$, and $c_0$ are positive integers, the puzzle is solvable from an initial state $\{a_0,b_0,c_0\}$ if and only if $$a_0+b_0+c_0=3\cdot 2^m\cdot \gcd(a_0,b_0,c_0)$$ for some $m\geq 0$, where $\gcd(a_0,b_0,c_0)$ denotes the greatest common divisor of $a_0$, $b_0$, and $c_0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For $k\geq 3$ an odd number, one can check that if there is a bowl whose pea number is not divisible by $k$, then after making a move there will still be a bowl whose pea number is not divisible by $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the initial state is $\{a_0,b_0,c_0\}$ is solvable, and let $n=\frac{a_0+b_0+c_0}{3}$ be the desired number of peas in each bowl.&lt;br&gt;&#xA;If $k\geq 3$ is an odd number dividing $n$, then in the final state every pea number is divisible by $k$, so the argument above implies that $k$ divides $a_0$, $b_0$, and $c_0$. Every odd divisor of $n$ is a common divisor of $a_0$, $b_0$, and $c_0$, so it must be the case that $n=2^m\cdot \gcd(a_0,b_0,c_0)$, or equivalently $a_0+b_0+c_0=3\cdot 2^m\cdot \gcd(a_0,b_0,c_0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, suppose $n=2^m\cdot \gcd(a_0,b_0,c_0)$.  We can solve the puzzle as follows. If the current state is $\{a,b,c\}$, write $a=2^x a'$, $b=2^y b'$, $c=2^z c'$, where $a'$, $b'$, and $c'$ are odd. The condition on $n$ implies that either $a=b=c$, or that of the three numbers $x$, $y$, and $z$, two will be equal and the third will be larger.&lt;br&gt;&#xA;Reordering if necessary, we can assume $x=y&amp;lt;z$ and that $a\leq b$. If $a&amp;lt; b$, replace $\{a,b,c\}$ by $\{2a,b-a,c\}$ and repeat the process. If $a=b$, replace $\{a,b,c\}$ by $\{2a,b,c-a\}$ (if $a&amp;lt;c$) or $\{a-c,b,2c\}$ (if $a&amp;gt;c$) and repeat the process. This will eventually terminate with all three bowls having the same number of peas. The reason this terminates is that if $\{a,b,c\}$ is replaced by $\{2a,b-a,c\}$, then the values of $x$ and $y$ increase by at least one. The replacement with $\{2a,b,c-a\}$ or $\{a-c,b,2c\}$ cannot happen twice in a row, and the size of $x$, $y$, and $z$ must remain bounded.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="68" LastEditDate="2016-02-26T13:39:09.083" LastActivityDate="2016-02-26T13:39:09.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23164" PostTypeId="2" ParentId="23131" CreationDate="2015-10-13T17:19:01.273" Score="9" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ronald Reagan, Jimmy Carter, Gerald Ford, Richard Nixon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(in this order, to make the colors match correctly.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://books.google.com/ngrams/graph?content=Ronald%20Reagan%2C%20Jimmy%20Carter%2C%20Gerald%20Ford%2C%20Richard%20Nixon&amp;amp;year_start=1900&amp;amp;year_end=2000&amp;amp;corpus=17&amp;amp;smoothing=3&amp;amp;share=&amp;amp;direct_url=t1%3B%2CRonald%20Reagan%3B%2Cc0%3B.t1%3B%2CJimmy%20Carter%3B%2Cc0%3B.t1%3B%2CGerald%20Ford%3B%2Cc0%3B.t1%3B%2CRichard%20Nixon%3B%2Cc0&quot;&gt;Link to Google ngram&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-10-13T17:24:07.107" LastActivityDate="2015-10-13T17:24:07.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23165" PostTypeId="2" ParentId="23157" CreationDate="2015-10-13T17:19:18.903" Score="2" Body="&lt;p&gt;Partial answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Converting the Hex codes to a string gives:&lt;br&gt;&#xA; A true gamer knows the code to decipher: &lt;strong&gt;BYAPPWZLSBADFCRQ&lt;/strong&gt;&lt;br&gt;&#xA;&lt;em&gt;(I assumed the non-hex line returns weren't part of it, anyway it can't be converted since they weren't in hex to start off with)&lt;/em&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;My guess would be that a famous cheat code is necessary to decipher that (like the konami code, but that gives nonsense. So probably another one like it)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastEditorUserId="11251" LastEditDate="2015-10-14T14:06:44.863" LastActivityDate="2015-10-14T14:06:44.863" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23166" PostTypeId="2" ParentId="23161" CreationDate="2015-10-13T17:24:23.457" Score="24" Body="&lt;p&gt;Building on @2012rcampion's answer, here's a simple constructive proof that any even number is reachable.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the (even) number that you want to reach (say $n$) and put it aside.  Then take the difference between all the pairs $(500,499), (498,497),\ldots,(n+2,n+1)$, and $(n-1,n-2),(n-3,n-4),\ldots,(3,2)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That will give you an odd number of 1's.  You'll also still have the original 1, so the total number of 1's is even (when the number of odd numbers is even).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You then pair them up and take the difference which gives you a bunch of 0's.  Then you boil it down to $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar approach works when there's an odd number of odd numbers, except that you're left with one 1 at the end, so you can get to $n-1$ in that way.&lt;/p&gt;&#xA;" OwnerUserId="15942" LastActivityDate="2015-10-13T17:24:23.457" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23167" PostTypeId="1" CreationDate="2015-10-13T17:25:47.527" Score="0" ViewCount="1120" Body="&lt;p&gt;i've followed the beginner's method, i've got the superior facet correct, the 1st and the 2nd layer correct. But in the third layer i always invert the colors but only got one corner right as shown in the pictures. Is there a solution ? &#xA;&lt;a href=&quot;https://i.stack.imgur.com/VFoPC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VFoPC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/V3eWP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/V3eWP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16904" LastActivityDate="2015-10-13T19:09:22.847" Title="3rd layer dilemma" Tags="|rubiks-cube|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23168" PostTypeId="2" ParentId="23157" CreationDate="2015-10-13T17:51:06.980" Score="4" Body="&lt;p&gt;I would say&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; EE6B2800&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Converting the hexcode to ascii gives:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A true gamer knows the code to decipher:BYAPPWZLSBADFCRQ&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The code to decipher(Vigenere) the last part is the Konami Code (with two start buttons at the end. Only got this one after the tip in the comments. I only remember it with a single start button at the end while wikipedia doesn't say anything about a start button at all.):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;BYAPPWZLSBADFCRQ : code[uuddlrlrbass] = hexmefourbillion&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Converting to hexadecimal:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4,000,000,000 = EE6B2800 (base16)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I tried to do something with that last bit but putting it into an ascii converter again only leads to garbage.&lt;/p&gt;&#xA;" OwnerUserId="16608" LastActivityDate="2015-10-13T17:51:06.980" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23169" PostTypeId="2" ParentId="23167" CreationDate="2015-10-13T19:09:22.847" Score="2" Body="&lt;p&gt;There is a solution (I got my cube to your position and solved it). Hold the cube with the 3rd layer on top and follow these directions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;With the corner that is correct on the right hand side closest to you, use &lt;strong&gt;U R U' L' U R' U' L&lt;/strong&gt; until the other corners are in the right spot. This algorithm rotates the front left, back left, and back right corners. It also changes their orientation, which might result in solving the cube (although I don't think it does for your specific case).&lt;/li&gt;&#xA;&lt;li&gt;Now that all the corners are in the proper place, rotate the top layer so that one of the corners that needs to be oriented is in the front right position. Use &lt;strong&gt;2(R' D' R D)&lt;/strong&gt; until the corner you were dealing with is oriented properly. Now rotate &lt;strong&gt;ONLY THE TOP LAYER&lt;/strong&gt; so that another unoriented corner is in that position. Use the same algorithm until that corner is oriented. Two keys in this step are making sure that you are doing the proper &lt;strong&gt;D&lt;/strong&gt; and &lt;strong&gt;D'&lt;/strong&gt; (look at the bottom of the cube and figure out which way is clockwise/counter-clockwise) and only rotating the top layer when switching between corners that you are orienting.&lt;/li&gt;&#xA;&lt;li&gt;The top layer is probably 90 or 180 degrees off, so just rotate that back the way it should be.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11727" LastActivityDate="2015-10-13T19:09:22.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23170" PostTypeId="1" AcceptedAnswerId="23172" CreationDate="2015-10-13T22:50:02.763" Score="3" ViewCount="695" Body="&lt;p&gt;This is a logical puzzle that I found, but I can't for the life of me figure out who likes Actress E and hence I'm stuck. The question is as follows :&lt;/p&gt;&#xA;&lt;p&gt;At a movie theatre six people are  randomly chosen to take part in a survey. Out of the 6 people 4 are working professionals and 2 are college students.&#xA;These six people each live in 4 different parts of the city namely Area P, Area Q, Area R and  Area S. In the survey each person is asked two questions&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Question 1 - What type of movies do they like?  Each  person answered a different type of movie out of Cartoons, Science Fiction, Comedies, Dramas, Documentaries and Tragedies.&lt;/p&gt;&#xA;&lt;p&gt;Question 2 - Who is their favourite actress?  Again  each person  answered a different actress namely A, B, C, D, E or F.&#xA;The following information is provided&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;None of the working professionals like Science Fiction or Tragedies.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Only two persons live in Area Q and one of them  likes Cartoons and the other likes Tragedies&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;One of the persons who lives in  Area  R likes Documentaries&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;One of the persons who lives in Area S likes Dramas&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The college students live in Area Q and Area P&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The person whose favourite actress is C does not like Dramas&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The person whose favourite actress is E does not like Cartoons&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Two working professionals who are the only people living in their respective areas have their favourite actresses as B and C .&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="16914" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2018-08-30T00:56:29.500" Title="Who likes Actress E?" Tags="|logical-deduction|logic-grid|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="23171" PostTypeId="2" ParentId="23170" CreationDate="2015-10-13T23:11:55.147" Score="0" Body="&lt;p&gt;WP1 lives is P, likes Comedy and actress A, D, E, or F&lt;/p&gt;&#xA;&#xA;&lt;p&gt;WP2 lives in S, likes Drama and likes B&lt;/p&gt;&#xA;&#xA;&lt;p&gt;WP3 lives in R, likes Documentaries and C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;WP4 lives in Q, likes Cartoons and likes A, D or F&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C1 lives in P, likes SF and doesn't like B or C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C2 lives in Q, likes Tragedies and doesn't like B or C. &lt;/p&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2015-10-13T23:21:59.987" LastActivityDate="2015-10-13T23:21:59.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23172" PostTypeId="2" ParentId="23170" CreationDate="2015-10-13T23:12:04.393" Score="8" Body="&lt;p&gt;There is not enough information to determine which person chose Actress E.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we have a situation that satisfies all of the statements. One person chose A, one chose D and one person chose F. Out of these three people, at most one likes cartoons. Then we can switch E with one of the other two, producing another situation that satisfies all of the statements.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The information we can deduce is as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A professional, who likes documentaries and chose Actress C, lives alone in area R.&lt;br&gt;A professional, who likes dramas and chose Actress B, lives alone in area S.&lt;br&gt;A professional, who likes cartoons and did not choose Actress E, lives in area Q.&lt;br&gt;A student, who likes tragedies, lives in area Q.&lt;br&gt;A professional, who likes comedies, lives in area P.&lt;br&gt;A student, who likes science fiction, lives in area P.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Any one of the last three people in that list could have chosen E.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-10-13T23:12:04.393" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23175" PostTypeId="1" CreationDate="2015-10-14T02:42:48.540" Score="12" ViewCount="1115" Body="&lt;p&gt;A nice little riddle in the form of a command.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Now write this try:&lt;/p&gt;&#xA;&lt;p&gt;Disguise in evil ones not,&lt;/p&gt;&#xA;&lt;p&gt;Lights change to flick you&lt;/p&gt;&#xA;&lt;p&gt;And with in for not eating&lt;/p&gt;&#xA;&lt;p&gt;One to not scratch means.&lt;/p&gt;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Try to decipher, please try your best...&lt;br /&gt;&#xA;Until you post the answer, I'll never rest.&lt;/p&gt;&#xA;" OwnerUserId="167" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2017-05-16T22:35:33.843" Title="No! I command you to" Tags="|riddle|" AnswerCount="5" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23179" PostTypeId="1" AcceptedAnswerId="23518" CreationDate="2015-10-14T06:05:16.087" Score="8" ViewCount="1233" Body="&lt;p&gt;I have been provided the following short puzzle and need help understanding the answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Complete the following letter series: &#xA;  &lt;br&gt; &lt;strong&gt;a_abc_ab_ _c_bb_cc_&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I filled it like this: aaabcaabbbc&lt;strong&gt;a&lt;/strong&gt;bbccca&lt;br&gt;&#xA;I had noticed there could be a pattern constructed using blocks of 7 sequential letters (&lt;code&gt;XXXYZXX&lt;/code&gt;):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;aaabcaa&#xA;bbbcabb&#xA;ccca...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;However, I have been told that the correct answer is aaabcaabbbc&lt;strong&gt;c&lt;/strong&gt;bbccca, which differs from my answer in only one place. Please note, I am not sure about the answer. It was taken only as a reference. So please let this undermine your approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone help me understand what makes this answer is correct?&lt;/p&gt;&#xA;" OwnerUserId="11425" LastEditorUserId="11425" LastEditDate="2015-10-15T16:01:50.490" LastActivityDate="2015-10-24T21:35:23.940" Title="Completing the letter series:" Tags="|letter-sequence|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="23180" PostTypeId="1" AcceptedAnswerId="23182" CreationDate="2015-10-14T06:08:35.840" Score="7" ViewCount="1284" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;C4T, F10R, I20P, L43N, O90L&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It can be easily seen that the first letters are 3rd, 6th, 9th, 12th, 15th and the last ones are 20th, 18th, 16th, 14th, 12th. But I cannot understand the logic of numbers between them. So please can someone explain it?&lt;/p&gt;&#xA;" OwnerUserId="11425" LastActivityDate="2015-10-14T08:54:35.353" Title="Find the term which does not fit into the series given below" Tags="|strategy|letter-sequence|" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23182" PostTypeId="2" ParentId="23180" CreationDate="2015-10-14T06:57:39.240" Score="12" Body="&lt;p&gt;I think odd one is F10R. It should have been F9R.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the numbers will follow a pattern if it's F9R.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like the first number is 4. So second is&#xA;4+4 + 1 = 9.. Next&lt;/p&gt;&#xA;&#xA;&lt;p&gt;9+9 + 2 =20..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;20+20 + 3=43..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;43+43 + 4=90..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure though&lt;/p&gt;&#xA;" OwnerUserId="16918" LastActivityDate="2015-10-14T06:57:39.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23183" PostTypeId="2" ParentId="23180" CreationDate="2015-10-14T08:54:35.353" Score="10" Body="&lt;p&gt;The number sequence is defined by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $x_n = 3\cdot2^n-n-1$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This yields  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_1 = 3\cdot2^1-1-1=4$&lt;br&gt;&#xA;$x_2 = 3\cdot2^2-2-1=9$&lt;br&gt;&#xA;$x_3 = 3\cdot2^3-3-1=20$&lt;br&gt;&#xA;$x_4 = 3\cdot2^4-4-1=43$&lt;br&gt;&#xA;$x_5 = 3\cdot2^5-5-1=90$&lt;br&gt;&#xA;$x_6 = 3\cdot2^6-6-1=185$   &lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-10-14T08:54:35.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23184" PostTypeId="1" AcceptedAnswerId="23187" CreationDate="2015-10-14T09:05:16.800" Score="3" ViewCount="417" Body="&lt;p&gt;This continues (and completes) the sequence of four dice puzzles:&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/23072/&quot;&gt;Four dice puzzle: 2,2,4,5&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/23077&quot;&gt;Four dice puzzle: What&amp;#39;s the best throw?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The days go by and Damiano is still throwing his four dice. He has determined for every throw $a,b,c,d$ of his four dice (with $a\le b\le c\le d$) the smallest positive integer $N(a,b,c,d)$ that cannot be generated from this throw according to the following rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;One may use the four numbers $a,b,c,d$ in any order, and it is fine if not all of them are used.&lt;/li&gt;&#xA;&lt;li&gt;Concatenation of digits is NOT allowed.&lt;/li&gt;&#xA;&lt;li&gt;The only allowed operations are additions, subtraction, multiplication, and division ($+,-,*,/$).&lt;/li&gt;&#xA;&lt;li&gt;One may use any number of brackets.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Questions:&lt;/strong&gt;  &lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Which throw $a,b,c,d$ of dice yields the &lt;strong&gt;smallest&lt;/strong&gt; number $N(a,b,c,d)$ ?&lt;/li&gt;&#xA;  &lt;li&gt;How many different throws $a,b,c,d$ have $N(a,b,c,d)\ge40$ ?&lt;/li&gt;&#xA;  &lt;li&gt;Which throws $a,b,c,d$ have $N(a,b,c,d)=28$ ?&lt;/li&gt;&#xA;  &lt;li&gt;What is $N(6,6,6,6)$ ?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2019-12-30T08:33:35.407" LastActivityDate="2019-12-30T08:33:35.407" Title="Four dice puzzle: Postscriptum" Tags="|mathematics|formation-of-numbers|dice|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23185" PostTypeId="2" ParentId="23150" CreationDate="2015-10-14T09:46:49.430" Score="19" Body="&lt;p&gt;Source: TheDarkTruth's &lt;a href=&quot;https://puzzling.stackexchange.com/a/23153/3918&quot;&gt;answer&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 3 parts to this answer:&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;The algorithm to find the locket marked with number N(1)&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Ignoring the fact that he has to stop after 500 coffins, this is the algorithm Jonathan Harker uses to find the locket which we mark as N(1):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Jonathan Harker opens coffin N(1), from which he extracts locket N(2)&lt;/li&gt;&#xA;&lt;li&gt;Jonathan Harker opens coffin N(2), from which he extracts locket N(3)&lt;/li&gt;&#xA;&lt;li&gt;Jonathan Harker opens coffin N(3), from which he extracts locket N(4)&lt;/li&gt;&#xA;&lt;li&gt;...&lt;/li&gt;&#xA;&lt;li&gt;Jonathan Harker opens coffin N(k), from which he extracts locket N(k+1)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Up to this point we have opened &lt;code&gt;k-1&lt;/code&gt; coffins with matching lockets, the coffin N(1) and the locket N(k+1). The locket N(k+1) will either match coffin N(1), ending &lt;strong&gt;&lt;em&gt;the loop&lt;/em&gt;&lt;/strong&gt; (and the algorithm) or match a new coffin, adding another step in the algorithm.&#xA;Since the lockets and coffins are never repeated, there can be at most 1000 steps in the loop, but there can be less. The trick is that if Jonathan Harker starts with the coffin N(x), he knows that the locket N(x) is in that loop and will eventually be reached.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Defeating the algorithm&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;It's easy to see that what locket Count Dracula chooses inside a loop makes no difference. The only thing that matters is how many steps are in the loop. Thus Count Dracula's choice is not which locket to pick, but which loop to pick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in order to defeat the algorithm, when Count Dracula picks the locket to be found, has to choose a loop with more than 500 steps. Thus when placing the lockets at the beginning of the game, he must ensure that such a loop exists.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Ensuring success&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Van Helsing's part in the story is now clear. He must ensure that when Count Dracula picks the &lt;strong&gt;&lt;em&gt;loop&lt;/em&gt;&lt;/strong&gt;, there will be none with more than 500 steps to choose.&#xA;Since there are 1000 coffins, if more than 500 are in a loop, there are less than 500 coffins left, not enough for another loop larger than 500 steps. This means there is at most one loop with more than 500 steps (from now on called LOOP).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What Van Helsing needs to do to ensure the success is to turn LOOP into two loops with less than 500 steps each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supposing LOOP has L steps, Van Helsing needs to choose a random coffin C(1) in the loop and the one farthest from it in both directions, coffin C(L/2). If he switches the two lockets, he turns LOOP with L steps into two loops with L/2 steps.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Putting it all together&lt;/h1&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Count Dracula places the lockets so a huge loop is formed&lt;/li&gt;&#xA;&lt;li&gt;Van Helsing breaks the huge loop into two smaller loops&lt;/li&gt;&#xA;&lt;li&gt;Count Dracula choose any locket in the largest loop&lt;/li&gt;&#xA;&lt;li&gt;Jonathan Harker iterates through the loop, eventually finding the locket&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3918" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-10-14T09:46:49.430" CommentCount="0" CommunityOwnedDate="2015-10-14T09:46:49.430" ContentLicense="CC BY-SA 3.0" />
  <row Id="23187" PostTypeId="2" ParentId="23184" CreationDate="2015-10-14T12:03:26.017" Score="4" Body="&lt;p&gt;I wrote a program that calculates all N(a,b,c,d). This is my result:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;N(1,1,1,1) = 5&#xA;N(1,1,1,2) = 7&#xA;N(1,1,1,3) = 10&#xA;N(1,1,1,4) = 11&#xA;N(1,1,1,5) = 13&#xA;N(1,1,1,6) = 15&#xA;N(1,1,2,2) = 10&#xA;N(1,1,2,3) = 11&#xA;N(1,1,2,4) = 14&#xA;N(1,1,2,5) = 17&#xA;N(1,1,2,6) = 20&#xA;N(1,1,3,3) = 14&#xA;N(1,1,3,4) = 19&#xA;N(1,1,3,5) = 22&#xA;N(1,1,3,6) = 26&#xA;N(1,1,4,4) = 22&#xA;N(1,1,4,5) = 27&#xA;N(1,1,4,6) = 13&#xA;N(1,1,5,5) = 13&#xA;N(1,1,5,6) = 14&#xA;N(1,1,6,6) = 15&#xA;N(1,2,2,2) = 11&#xA;N(1,2,2,3) = 17&#xA;N(1,2,2,4) = 19&#xA;N(1,2,2,5) = 23&#xA;N(1,2,2,6) = 19&#xA;N(1,2,3,3) = 22&#xA;N(1,2,3,4) = 29&#xA;N(1,2,3,5) = 19&#xA;N(1,2,3,6) = 28&#xA;N(1,2,4,4) = 21&#xA;N(1,2,4,5) = 33&#xA;N(1,2,4,6) = 29&#xA;N(1,2,5,5) = 29&#xA;N(1,2,5,6) = 44&#xA;N(1,2,6,6) = 27&#xA;N(1,3,3,3) = 14&#xA;N(1,3,3,4) = 17&#xA;N(1,3,3,5) = 22&#xA;N(1,3,3,6) = 23&#xA;N(1,3,4,4) = 22&#xA;N(1,3,4,5) = 29&#xA;N(1,3,4,6) = 34&#xA;N(1,3,5,5) = 34&#xA;N(1,3,5,6) = 37&#xA;N(1,3,6,6) = 20&#xA;N(1,4,4,4) = 10&#xA;N(1,4,4,5) = 18&#xA;N(1,4,4,6) = 31&#xA;N(1,4,5,5) = 12&#xA;N(1,4,5,6) = 17&#xA;N(1,4,6,6) = 21&#xA;N(1,5,5,5) = 8&#xA;N(1,5,5,6) = 13&#xA;N(1,5,6,6) = 14&#xA;N(1,6,6,6) = 9&#xA;N(2,2,2,2) = 7&#xA;N(2,2,2,3) = 13&#xA;N(2,2,2,4) = 11&#xA;N(2,2,2,5) = 17&#xA;N(2,2,2,6) = 15&#xA;N(2,2,3,3) = 19&#xA;N(2,2,3,4) = 17&#xA;N(2,2,3,5) = 29&#xA;N(2,2,3,6) = 23&#xA;N(2,2,4,4) = 11&#xA;N(2,2,4,5) = 23&#xA;N(2,2,4,6) = 17&#xA;N(2,2,5,5) = 28&#xA;N(2,2,5,6) = 23&#xA;N(2,2,6,6) = 21&#xA;N(2,3,3,3) = 13&#xA;N(2,3,3,4) = 25&#xA;N(2,3,3,5) = 29&#xA;N(2,3,3,6) = 26&#xA;N(2,3,4,4) = 23&#xA;N(2,3,4,5) = 41&#xA;N(2,3,4,6) = 31&#xA;N(2,3,5,5) = 29&#xA;N(2,3,5,6) = 38&#xA;N(2,3,6,6) = 23&#xA;N(2,4,4,4) = 11&#xA;N(2,4,4,5) = 25&#xA;N(2,4,4,6) = 15&#xA;N(2,4,5,5) = 26&#xA;N(2,4,5,6) = 45&#xA;N(2,4,6,6) = 17&#xA;N(2,5,5,5) = 14&#xA;N(2,5,5,6) = 24&#xA;N(2,5,6,6) = 25&#xA;N(2,6,6,6) = 17&#xA;N(3,3,3,3) = 11&#xA;N(3,3,3,4) = 17&#xA;N(3,3,3,5) = 19&#xA;N(3,3,3,6) = 13&#xA;N(3,3,4,4) = 18&#xA;N(3,3,4,5) = 37&#xA;N(3,3,4,6) = 29&#xA;N(3,3,5,5) = 32&#xA;N(3,3,5,6) = 19&#xA;N(3,3,6,6) = 14&#xA;N(3,4,4,4) = 10&#xA;N(3,4,4,5) = 22&#xA;N(3,4,4,6) = 29&#xA;N(3,4,5,5) = 31&#xA;N(3,4,5,6) = 43&#xA;N(3,4,6,6) = 17&#xA;N(3,5,5,5) = 11&#xA;N(3,5,5,6) = 29&#xA;N(3,5,6,6) = 25&#xA;N(3,6,6,6) = 11&#xA;N(4,4,4,4) = 10&#xA;N(4,4,4,5) = 14&#xA;N(4,4,4,6) = 13&#xA;N(4,4,5,5) = 22&#xA;N(4,4,5,6) = 32&#xA;N(4,4,6,6) = 19&#xA;N(4,5,5,5) = 7&#xA;N(4,5,5,6) = 13&#xA;N(4,5,6,6) = 33&#xA;N(4,6,6,6) = 13&#xA;N(5,5,5,5) = 8&#xA;N(5,5,5,6) = 13&#xA;N(5,5,6,6) = 3&#xA;N(5,6,6,6) = 8&#xA;N(6,6,6,6) = 9&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And now to answer the questions, assuming my program made no mistakes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Which throw a,b,c,d of dice yields the smallest number N(a,b,c,d) ?   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;N(5,5,6,6) = 3&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How many different throws a,b,c,d have N(a,b,c,d)≥40 ?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;four:&lt;br&gt;&#xA;N(1,2,5,6) = 44&lt;br&gt;&#xA;N(2,3,4,5) = 41&lt;br&gt;&#xA;N(2,4,5,6) = 45&lt;br&gt;&#xA;N(3,4,5,6) = 43&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Which throws a,b,c,d have N(a,b,c,d)=28 ?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;only (1,2,3,6) and (2,2,5,5)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is N(6,6,6,6) ?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;9&lt;/p&gt;&#xA;" OwnerUserId="1748" LastEditorUserId="1748" LastEditDate="2015-10-14T12:20:37.680" LastActivityDate="2015-10-14T12:20:37.680" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23189" PostTypeId="2" ParentId="22859" CreationDate="2015-10-14T12:47:58.527" Score="6" Body="&lt;p&gt;The question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Which of these is the slogan for the city of Philadelphia?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The correct answer A is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The City of Brotherly Love&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Suddenly a faint &lt;strong&gt;bell&lt;/strong&gt; did start ringing, and brought me to &lt;strong&gt;liberty&lt;/strong&gt;&quot;: The &lt;strong&gt;liberty bell&lt;/strong&gt; is found in Philadelphia.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The other answers were&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The City of Angels [Los Angeles. Also the name of 1998 movie starring Nick Cage and Meg Ryan; the song being Angel by Sarah McLachlan (thanks @irishPanda)]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;C:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Mile High City [Denver, home to The Mile High stadium. Thanks to @Fillet for this one - I was nowhere!]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;D:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Magic City [Sounds a bit like a ride at Disney World, but is actually the slogan for Miami, which is in Florida and not too far (in US geography terms) from Disney World in Orlando, Florida.]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alternative idea for D:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;America's friendliest city [which is Delphos. &quot;Delphos&quot; has similar letters to Disney... but is a tiny place in Ohio, so unlikely to be correct]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9444" LastEditorUserId="9444" LastEditDate="2016-03-07T09:14:07.507" LastActivityDate="2016-03-07T09:14:07.507" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="23191" PostTypeId="2" ParentId="23150" CreationDate="2015-10-14T15:20:27.817" Score="2" Body="&lt;p&gt;Van Helsing should count the even numbers that are located in a coffin numbered by another even number. he should than pick that number and place it in coffin nr 1. Now when Harker enters the tomb, he will look into the first coffin. When he knows the number of even numbers in even numbered coffins, he can easily calculate the number of even numbers in uneven coffins, uneven numbers in uneven numbered coffins and uneven numbers in even numbered coffins. This way he can increase his chances of finding the correct number by searching in even/uneven numbered coffins depending on N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example, Dracula places his the lockets randomly in every coffin. Van Helsing counts the even numbered lockets placed in even numbered coffins and finds that they are 333. He than switched the number in the first coffin with coffin containing 333. Harkels than enters the Tomb and Dracula declares the number to be 777. Harkel opens the first coffin ands finds locket nr 333. From this information he concludes that there are 333 even numbers in even coffins. 167 even numbers in uneven coffins(500 - 333), 333 uneven numbers in uneven coffins( we have 333 uneven coffins left for our uneven numbers), and 167 uneven numbers in even coffins (167 even coffins left for our uneven numbers). Now that Harkel knows that N is 777, an uneven numbers, and there are 333 uneven numbers in uneven coffins, he wil strictly search in eneven numbered coffins, if he has searched 333 uneven coffins and found uneven numbers, and not found N, he than knows that his N is in a even numbered coffin and will from than on only search even numbered coffins. This should enhance his chance of finding N by some margin. &lt;/p&gt;&#xA;" OwnerUserId="5124" LastActivityDate="2015-10-14T15:20:27.817" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23192" PostTypeId="1" AcceptedAnswerId="23197" CreationDate="2015-10-14T17:10:58.617" Score="9" ViewCount="928" Body="&lt;p&gt;This is the fourth in a series &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=user%3A9000%20is%3AQuestion%20title%3A%22Lights%20in%20the%20Dungeon%22&quot;&gt;Lights in the Dungeon&lt;/a&gt; puzzles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/23113/lights-in-the-dungeon-five-by-five&quot;&gt;previous (3rd) &amp;lt;--&lt;/a&gt;|&lt;a href=&quot;https://puzzling.stackexchange.com/questions/23260/lights-in-the-dungeon-hooty-and-the-message&quot;&gt;--&gt; next (5th)&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://puzzling.stackexchange.com/q/23113/9000&quot;&gt;five by five dungeon&lt;/a&gt; was only a little harder than the previous ones. You're feeling pretty good about it and you're even up for a little taunting. &quot;Alright, Hooty, where the six by six?&quot; you ask as you approach the cloaked figure waiting for you at the exit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Six by six?&quot; asks the Lord, feigning ignorance. &quot;I'm afraid I don't know what you talking about. You see, Fakky Jim and I have been working on your next challenge while you were busy rearranging tiles and walking through the dungeon, flipping lights like a good little monkey. Your next challenge is... a bit more grand. If you succeed, I promise to tell you the &lt;em&gt;real&lt;/em&gt; reason I've brought you here.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intrigued, you walk over to Fakky Jim and receive a scroll with the rules. You can tell they've been re-written from scratch as they were getting a bit illegible with all the strikethroughs and revisions from the previous puzzles.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;You will enter a dungeon in which you must visit every room and turn off the light in the middle, an easy operation to perform.&lt;/li&gt;&#xA;  &lt;li&gt;As soon as you exit a room, every door to that room slams shut and locks. You can not re-enter that room through any means.&lt;/li&gt;&#xA;  &lt;li&gt;The map below shows the current design of the dungeon with each room bordered by red and labeled with a alphabetic character.&lt;/li&gt;&#xA;  &lt;li&gt;You may rearrange the rooms however you like before you enter but you cannot change the perimeter of the dungeon. In this case, that means you must maintain the exact stairstep shape that is eighteen rooms tall and somewhere between seven and eleven rooms wide.&lt;/li&gt;&#xA;  &lt;li&gt;You cannot rotate or flip rooms. You may only translate them.&lt;/li&gt;&#xA;  &lt;li&gt;The dungeon must have exactly one entrance and one exit.&lt;/li&gt;&#xA;  &lt;li&gt;There must be &lt;strong&gt;&lt;em&gt;exactly one&lt;/em&gt;&lt;/strong&gt; possible path through the dungeon. Obviously, any path is reversible if you simply enter from the other end but that does not count as a separate possible path.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Surprised, you look up at Fakky Jim and ask, &quot;What happened to the collapsing floors?&quot; The Grand Vizier just stares at you. You're not sure what you expected from a cat, magic though it may be. You turn to Lord Hooty and repeat the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Ah,&quot; he starts, a sorrowful look upon his face, &quot;we changed that after losing the cleaning crew. It was a tragedy, really. They were only doing their job. At least they were tasty so that's a silver lining.&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How can you rearrange the dungeon rooms so that you can win the game?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;There are probably multiple solutions. The first valid answer will be accepted.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/x38jH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/x38jH.png&quot; alt=&quot;Maze 54x7,8,9,10,11,11&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;AAAAAAA&#xA;AABBBBB&#xA;CCCCCCC&#xA;DDDDDDEE&#xA;EFFFFFFF&#xA;FFFFFFFF&#xA;FFFFGGGGG&#xA;GHHHHMMMM&#xA;MMMMMMMMM&#xA;MMMMMMMMMM&#xA;MMMMMMMMMM&#xA;MMMMMMMMNN&#xA;NNNNNNNNNNN&#xA;NNNNNNNNNNN&#xA;NNNNNNNNNNN&#xA;OOOOOOOPPPP&#xA;PPPPPPPPPPP&#xA;PPPPPPPPPPP&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For reference, here are the 16 possible rooms. (Note that this is all &lt;em&gt;possible&lt;/em&gt; rooms. For this puzzle, you must use exactly the rooms shown in the map above.)&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rp7xy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rp7xy.png&quot; alt=&quot;A&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/E9wSu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/E9wSu.png&quot; alt=&quot;B&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/SlADl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SlADl.png&quot; alt=&quot;C&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/yEoqW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yEoqW.png&quot; alt=&quot;D&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/QGft0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QGft0.png&quot; alt=&quot;E&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/VREO9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VREO9.png&quot; alt=&quot;F&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/R1gGL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/R1gGL.png&quot; alt=&quot;G&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/vK9AI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vK9AI.png&quot; alt=&quot;H&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/Cft7i.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Cft7i.png&quot; alt=&quot;I&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/3oSC9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3oSC9.png&quot; alt=&quot;J&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/E72qu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/E72qu.png&quot; alt=&quot;K&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/p8yY1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/p8yY1.png&quot; alt=&quot;L&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/jqqKf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jqqKf.png&quot; alt=&quot;M&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/zKQB7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zKQB7.png&quot; alt=&quot;N&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/yTRBU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yTRBU.png&quot; alt=&quot;O&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/s78Fu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/s78Fu.png&quot; alt=&quot;P&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2016-08-31T15:33:05.873" Title="Lights in the Dungeon: Eighteen by Eleven" Tags="|story|visual|mazes|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23193" PostTypeId="1" AcceptedAnswerId="23195" CreationDate="2015-10-14T18:51:18.520" Score="6" ViewCount="525" Body="&lt;p&gt;Here's a custom Pac-Man level. Pac-Man is to do what Pac-Man always does: eat, eat, eat! Wait, where are the pac-dots?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/arWi4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/arWi4.png&quot; alt=&quot;Pac-Man Paths Puzzle&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;INSTRUCTIONS MANUAL&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There are 3 hidden phrases representing otherworldly items in this special maze. One uncovers them by collecting numbered &quot;dots&quot; (not actual dots, just numbers, for increased visibility) following 3 paths that have the below names and main clues.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;em&gt;Classic path&lt;/em&gt;: original eatables&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Sweetheart path&lt;/em&gt;: treats for her&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Indigestible path&lt;/em&gt;: none of the above&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mission:&lt;/strong&gt; Find and decrypt the 3 path phrases from each set of numbers encountered along the path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus mission:&lt;/strong&gt; Figure out the sources of each path phrase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Navigational clues:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Within each path, Pac-Man aims for the high score, going for the better target that adheres to his current path's clue within his line of sight.&lt;/li&gt;&#xA;&lt;li&gt;He generally doesn't change direction unless he hits a wall or spots something tasty.&lt;/li&gt;&#xA;&lt;li&gt;Strangely, numbers are neither tasty nor have point values.&lt;/li&gt;&#xA;&lt;li&gt;Normal Pac-Man travelling rules apply: he eats or collects everything along his path, and he doesn't go through anything that remains uneaten or uncollected.&lt;/li&gt;&#xA;&lt;li&gt;Fearing ghosts would notice where he's been, he doesn't backtrack.&lt;/li&gt;&#xA;&lt;li&gt;Each path starts from where he is now; restart the level to try a different path.&lt;/li&gt;&#xA;&lt;li&gt;Each path ends when he reaches the last applicable item.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cheat codes:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;sub&gt;Note: It's the developer's first custom level, done as a quick share. There may be errors. It may be too easy. It may be too messily complex. She'll determine what works and what doesn't from observing the players and any feedback. If the above aren't gimmes enough, she can provide further clues here.&lt;/sub&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;  The maze is small enough so that one can figure out the paths through trial and error, but because this puzzle was built around some basic Pac-Man knowledge, here is &lt;a href=&quot;https://en.wikipedia.org/wiki/Pac-Man&quot; rel=&quot;nofollow noreferrer&quot;&gt;its Wikipedia article&lt;/a&gt;. Just this and one other linked article would be helpful. (Maybe I'll link both after the answer.)&#xA; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15921" LastEditorUserId="4421" LastEditDate="2015-10-14T20:22:54.563" LastActivityDate="2015-10-14T20:22:54.563" Title="Eat It! Through Multi Paths" Tags="|knowledge|visual|mazes|video-games|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23194" PostTypeId="2" ParentId="23193" CreationDate="2015-10-14T19:34:55.527" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pac-man is suffering from: &lt;strong&gt;&lt;em&gt;mushroom cake nuka-cola&lt;/em&gt;&lt;/strong&gt;  indigestion&lt;br&gt;&#xA; &lt;sup&gt; explanation at the end &lt;/sup&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Go straight for that left cherry, then down. Follow:&lt;br&gt;&#xA;13-21-19-8-18-15-15-13 and end at that green thingy.&lt;br&gt;&#xA;It spells out &lt;strong&gt;Mushroom&lt;/strong&gt;. (and I'm thinking Mario obviously)&lt;br&gt;&#xA;&lt;br&gt;&#xA; If you start out through the top, you get &lt;strong&gt;Liar Cake&lt;/strong&gt; (portal)&lt;br&gt;&#xA; 12-9-1-18-3-1-11-5   (but it keeps going) -(1)&lt;br&gt;&#xA; 14-21-11-1-3-15-12-1 (a) &lt;strong&gt;Nuka-Cola&lt;/strong&gt; - fallout&lt;br&gt;&#xA; I'm unsure if those are 2 different lines or the same, also you could tie the &#xA; mushroom after the liar cake by going through the bottom middle but that last bit does 'end' at a key so maybe those are simply the three. Mario, Portal, Fallout.&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;sup&gt;or maybe someone baked a particular nasty cake, and people are now suffering from mushroom cake fallout instead of yellow (uranium) cake fallout (which could explain the liar part) ... &lt;em&gt;but I think that might be getting a bit far-fetched&lt;/em&gt; :P &lt;/sup&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastEditorUserId="11251" LastEditDate="2015-10-14T20:02:31.070" LastActivityDate="2015-10-14T20:02:31.070" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23195" PostTypeId="2" ParentId="23193" CreationDate="2015-10-14T19:42:26.493" Score="3" Body="&lt;p&gt;Based on the following paths, and appending the nth letter of the alphabet as you pass by it...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; = Mushroom&lt;a href=&quot;https://i.stack.imgur.com/ZLZgo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZLZgo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Part 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; = Liar Cake&lt;a href=&quot;https://i.stack.imgur.com/IFeBC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IFeBC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Part 3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; = Nuka-Cola&lt;a href=&quot;https://i.stack.imgur.com/3DgNQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3DgNQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&#xA; So, references are based on Mario, Portal, and Fall-out Games.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14233" LastEditorUserId="14233" LastEditDate="2015-10-14T19:55:25.470" LastActivityDate="2015-10-14T19:55:25.470" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23196" PostTypeId="2" ParentId="23150" CreationDate="2015-10-14T19:51:55.407" Score="7" Body="&lt;p&gt;I've decided to visualize components of TheDarkTruth's answer, and hopefully clarify some misunderstanding about his solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, I found the best way to think of the problem was as a sort of modified linked list/directed graph. Each numbered coffin is a node and each locket is a pointer/directed edge to another node.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rules I deduced:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Each node must point to another node (a node can point to itself), as each coffin has a locket.&lt;/li&gt;&#xA;&lt;li&gt;No two edges can point to the same node (lockets are uniquely numbered).&lt;/li&gt;&#xA;&lt;li&gt;No node can have more than one outgoing edge (only one locket per coffin).&lt;/li&gt;&#xA;&lt;li&gt;Every node has an edge directed to it (lockets are numbered 1 to 1000, as are coffins, so each coffin must have a corresponding locket).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Here's an example of what you might see:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7sQkO.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7sQkO.png&quot; alt=&quot;Generic cycle&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are examples of violations of the rules:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Coffin with two lockets (you can't exit a cycle from inside).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hxBQz.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hxBQz.png&quot; alt=&quot;List item&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Two lockets with the same number (you can't enter a cycle from outside).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1aGPi.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1aGPi.png&quot; alt=&quot;List item&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;A coffin with no locket inside (there are no terminating nodes).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OaJkE.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OaJkE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;A coffin such that no locket exists with the same number (there are no nodes without a pointer to it).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Vby2Q.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Vby2Q.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since these lists can't terminate and can't branch, then they must form cycles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, a crudely drawn example of the locket swap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/67rqQ.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/67rqQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9696" LastActivityDate="2015-10-14T19:51:55.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23197" PostTypeId="2" ParentId="23192" CreationDate="2015-10-14T20:01:19.337" Score="8" Body="&lt;p&gt;I started from the bottom-right corner and tried to get rid of all the Ms and Ns first, so I came up with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/w5u0x.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/w5u0x.png&quot; alt=&quot;Solution image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or as text:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;ACPPPBC&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;PAGPPMM&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;PBHBOMM&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;PAGHHMMP&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;PPPPOMMP&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;FGGEDMMP&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;FGNOPMMFC&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;PPPFPMMMM&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;FHBEPMMMM&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;FADFCMMMMP&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;FBCFFMMMMP&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;AEFDFMMMMP&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;FNNNDMMMANC&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;ANNNCMMFNGM&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;FNNNOMFPFOM&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;ANNNNOFNDPM&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;FNNNNNNNNOD&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;ANNNNNNNNNN&lt;/code&gt;   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="448" LastActivityDate="2015-10-14T20:01:19.337" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23198" PostTypeId="2" ParentId="23192" CreationDate="2015-10-14T20:28:18.023" Score="3" Body="&lt;p&gt;It's a bit messy, but this would work.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/CSQGr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CSQGr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11047" LastEditorUserId="11047" LastEditDate="2015-10-14T20:51:32.230" LastActivityDate="2015-10-14T20:51:32.230" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23199" PostTypeId="1" AcceptedAnswerId="23203" CreationDate="2015-10-14T20:40:17.450" Score="1" ViewCount="1312" Body="&lt;p&gt;I feel empty inside, but I think writing about it will help me cope.  Here is some prose I wrote:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Am I dreaming again?&lt;br&gt;&#xA;  This doesn’t look like the dreams that I am used to... Worse.&lt;br&gt;&#xA;  At times I felt I rode on the air, and all was crystal clear.&lt;br&gt;&#xA;  But those dreams shattered, and so often I don’t even recognize myself in the mirror.&lt;br&gt;&#xA;  Time has done that to me.&lt;br&gt;&#xA;  Before, I could eat all I want and not worry about a thing.&lt;br&gt;&#xA;  Now, my skin an unnatural hue, my world overrun by vermin, it’s as if I were under a curse.&lt;br&gt;&#xA;  But we all have a yarn to spin, and I am the star of this one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Do you know who I am?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus&lt;/strong&gt;    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many references can you find to my games above?&lt;/p&gt;&#xA;" OwnerUserId="11452" LastEditorUserId="8874" LastEditDate="2015-10-15T08:27:31.497" LastActivityDate="2015-10-15T08:27:31.497" Title="My Life...Sucks" Tags="|riddle|knowledge|video-games|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23200" PostTypeId="2" ParentId="23199" CreationDate="2015-10-14T20:58:05.573" Score="0" Body="&lt;p&gt;Are you &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Kain, from Legacy of Kain? Every line seems to point at a vampire creature.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some of the items mentioned in the riddle&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Crystal = games published by Crystal Dynamics&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dreams = Was a huge hit on the Dreamcast&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="14233" LastEditorUserId="14233" LastEditDate="2015-10-14T21:03:20.620" LastActivityDate="2015-10-14T21:03:20.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23201" PostTypeId="2" ParentId="23199" CreationDate="2015-10-14T21:02:23.753" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The theme of dreams shattering, the reference to the mirror and the general 'happiness' and 'life' fading away made me thing a LOT of &lt;a href=&quot;https://en.wikipedia.org/wiki/Lost_Odyssey&quot; rel=&quot;nofollow&quot;&gt;Lost Odyssey&lt;/a&gt;. I am unsure if there is more to this puzzle than just a fancy description however...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastActivityDate="2015-10-14T21:02:23.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23202" PostTypeId="2" ParentId="23199" CreationDate="2015-10-14T21:05:06.120" Score="3" Body="&lt;p&gt;Are you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nights from Nights into Dreams&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This doesn’t look like the dreams that I am used to... Worse.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The game is about Nightopia, a dream world &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At times I felt I rode on the air, and all was crystal clear.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nights can fly, Nights has a jewel on him/her&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But those dreams shattered, and so often I don’t even recognize myself in the mirror.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wizeman the Wicked stole all the dream energy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Time has done that to me.&#xA;Before, I could eat all I want and not worry about a thing.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm not sure about this one&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, my skin an unnatural hue, my world overrun by vermin, it’s as if I were under a curse.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nights is purple and pink, and the world is overrun by Nightmaren&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But we all have a yarn to spin, and I am the star of this one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nights is the main character&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16729" LastActivityDate="2015-10-14T21:05:06.120" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23203" PostTypeId="2" ParentId="23199" CreationDate="2015-10-14T21:12:12.443" Score="8" Body="&lt;p&gt;Are you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Kirby (clever because &quot;my life &lt;em&gt;sucks&lt;/em&gt;&quot;. heh.) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I only got some of these references. Haven't actually played that many games D: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I dreaming again?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dream Land sequels (2 and 3?) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This doesn’t look like the dreams that I am used to... Worse. (not sure)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At times I felt I rode on the air, and all was crystal clear.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Kirby 64: The Crystal Shards&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But those dreams shattered, and so often I don’t even recognize myself in the mirror.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Kirby &amp;amp; the Amazing Mirror&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Time has done that to me.&#xA;Before, I could eat all I want and not worry about a thing. (not sure if specific reference) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, my skin an unnatural hue, my world overrun by vermin, it’s as if I were under a curse.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Kirby: Canvas Curse&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But we all have a yarn to spin, and I am the star of this one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Kirby's Epic Yarn&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16034" LastEditorUserId="16034" LastEditDate="2015-10-14T21:19:32.930" LastActivityDate="2015-10-14T21:19:32.930" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23204" PostTypeId="1" AcceptedAnswerId="23205" CreationDate="2015-10-15T02:02:19.283" Score="14" ViewCount="925" Body="&lt;p&gt;Ten rows of numbers are written on a blackboard. The first row has one $1$, the second has two $2$'s, and so on up through the tenth row with ten $10$'s:&#xA;$$&#xA;\begin{gather*}&#xA;1\\&#xA;2,\;2\\&#xA;3,\;3,\;3\\&#xA;\vdots\\&#xA;10,\;10,\ldots,\;10&#xA;\end{gather*}&#xA;$$&#xA;Choose two of the numbers on the board, erase them, and write their product divided by their sum (which will likely be a fraction). Repeat the process until only one number remains.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the largest value that the remaining number could be? Also, what is the smallest value?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="13475" LastEditDate="2017-12-13T19:30:17.183" LastActivityDate="2017-12-13T19:30:17.183" Title="Triangle of numbers on a blackboard" Tags="|mathematics|blackboard|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23205" PostTypeId="2" ParentId="23204" CreationDate="2015-10-15T02:41:06.140" Score="19" Body="&lt;p&gt;Notice that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{\frac{xy}{x+y}}=\frac{x+y}{xy}=\frac{1}{x}+\frac{1}{y}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the sum of the reciprocals of the numbers on the board is always preserved. The sum of the reciprocals of the numbers in each row is $1$, so the sum of all the reciprocals is $10$. Therefore, the last number left must be $\frac{1}{10}$.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-10-15T02:41:06.140" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23206" PostTypeId="2" ParentId="20849" CreationDate="2015-10-15T02:57:19.830" Score="0" Body="&lt;p&gt;Partial guess:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wanted to just ask a question in the comments but I don't have enough (any) reputation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Does it have anything to do with elements (periodic table or lore e.g. fire air water etc)? You mentioned &quot;element&quot; in a hint and in a comment, but that could just be you liking the word's description of the riddle. I'm leaning towards water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Getting used to something over time&quot; Water dissolves other elements and normally makes them safer by doing so. Also diffusion, but that's a stretch (most of this is)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Water has two hydrogen &quot;legs&quot;;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Got nothing for this one. Water adapts and is a liquid?&lt;br&gt;&#xA;&lt;br&gt;&#xA; &quot;Learning things from peers&quot; is interacting/heating up when dissolving other elements/molecules&lt;br&gt;&#xA;&lt;br&gt;&#xA; This one's a stretch, but water is a solute, and lots of dissolving reactions produce heat.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Stuck together implying a bond&lt;br&gt;&#xA;&lt;br&gt;&#xA; Nobody expects water? This is looking weaker&lt;br&gt;&#xA;&lt;br&gt;&#xA; I have nothing for the &quot;end is inevitable&quot; or &quot;final&quot; aswell&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Additional things from hints:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Elements match into different molecules; &#xA;&#xA; &quot;the one&quot; could by indicating hydrogen;&#xA;&#xA; you could be oxygen with the match being water;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;looking into this a bit more, the other thing I could think of was:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a pirate. &quot;two legs&quot; mocking peg legs, &quot;avast&quot; is a nautical term/pirate slang. Avast also means halt, but... nothing with match or the one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16934" LastActivityDate="2015-10-15T02:57:19.830" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23211" PostTypeId="1" AcceptedAnswerId="23224" CreationDate="2015-10-15T12:41:40.867" Score="-3" ViewCount="1134" Body="&lt;p&gt;You walk into your room, and give a quick glance at your iPad that you've had for so long. It's been a while (to be exact, 6, no, 13 years) since everything's, well, &lt;em&gt;happened&lt;/em&gt;, and you've realized that it's probably a good time to apologize to me. Nothing about you is really the same anymore. Getting shot kind of does tend to ruin your chances of succeeding in life. You sigh and run your hand through your hair. Your iPad blinks and it's... wow, it's a letter from &lt;em&gt;me&lt;/em&gt;. Of all others.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;hey, i just implore you to understand that the occurences of that night really made me do some thinkin. and i just hope you also understand that it really changed me. as soon as i shot you i realized the exact repurcussions of my actions. you, me, the others, we all made it to this afterlife, or... thin that comes after death. it's just like you, really. all bubbles and stuff. they keep on floatin around. i crossed the path of some of the others, but understandably, they all still tried to go as fast as possible. i suppose that's fair punishment for all that i did.&lt;/p&gt;&#xA;&lt;p&gt;i really miss you. i hope you get that i'm bein sincere and all here. i understand that i had to appreciate the relationship that you and i had and i coulda kept it goin by talkin to you about my feelins for you. i shouldn'ta tried to go against the relationship system, but i had too strong feelins for you for me to stop myself.&lt;/p&gt;&#xA;&lt;p&gt;i'm sorry for killin you. it's been a long time, and i just hope that you are able to see this and try to stay for a period of time if our paths cross again.&lt;/p&gt;&#xA;&lt;p&gt;regards,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The signature is blotched out with a single tear.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Who am I, and who are you?&lt;/p&gt;&#xA;&lt;p&gt;In your answer, please explain how you figured out the riddle, and explanations of the hints if you used them.&lt;/p&gt;&#xA;&lt;p&gt;Hint 1&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; There is a reason why you had trouble with how long it's been since you've seen me.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Hint 2&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Certain letters have been intentionally omitted from the letter. The bad grammar is also intentional.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Hint 3&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; You're probably staring the answer in the face right now.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;I'm open to feedback, as this is my first post here on P.SE. Thank you very much!&lt;/p&gt;&#xA;" OwnerUserId="15603" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-10-15T17:10:54.493" Title="A love letter, 13 years after death" Tags="|riddle|knowledge|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23213" PostTypeId="1" AcceptedAnswerId="23215" CreationDate="2015-10-15T13:23:32.970" Score="5" ViewCount="388" Body="&lt;p&gt;There are $n$ balls out of which $k$ are defective, where $k&amp;lt;n/2$. &#xA;All defective balls have the same weight $w$, and all non-defective balls have the same weight $v$ with $v&amp;lt;w$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determine all defective balls with the smallest possible number of weighings on a two-armed balance.&lt;/p&gt;&#xA;" OwnerDisplayName="user16942" LastEditorUserId="30903" LastEditDate="2017-02-08T07:29:52.880" LastActivityDate="2017-02-08T07:29:52.880" Title="Find k defective balls out of n balls" Tags="|strategy|combinatorics|weighing|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23214" PostTypeId="2" ParentId="3280" CreationDate="2015-10-15T13:49:01.113" Score="1" Body="&lt;p&gt;The question specifies that the dragon will kill you only if you leave. &#xA;So, if we open both doors and wait for 24 hours, the dragon may eventually random walk into the magnified hallway, and you can go through dragon's door.&#xA;Off course, am assuming, the dragon won't kill you if you don't leave,&#xA;and that the size of doors is big enough for dragon, and that the dragon moves.&lt;/p&gt;&#xA;" OwnerUserId="6216" LastActivityDate="2015-10-15T13:49:01.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23215" PostTypeId="2" ParentId="23213" CreationDate="2015-10-15T14:03:47.093" Score="3" Body="&lt;p&gt;1.) It is easy to see that you can always do with $n-1$ weighings: put the first ball into the left pan, and in $n-1$ steps try each of the remaining $n-1$ balls in the right pan. This divides the balls into two groups (balls thta have equal weight as the first ball, and balls that have different weight). The smaller group are the defective balls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) It is easy to see that any deterministic weighing strategy needs at least $\log_3{n\choose k}$ weighings by the standard information theoretic argument: There are ${n\choose k}$ possible scenarios for choosing $k$ defective ones out of a total of $n$ balls, and every weighing only has three possible outcomes (left pan heavier; right pan heavier; both pans equally heavy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.) Laszlo Pyber has designed a strategy that in the worst case uses at most $\log_3{n\choose k}+15k$ weighings (&lt;em&gt;L. Pyber, &quot;How to find many counterfeit coins? Graphs and Combinatorics 2 (1986), pages 173–177&lt;/em&gt;).&#xA;Note that Pyber's result is strong if $k$ is very small in comparison to $n$ (like the cases with $k=\log\log n$), but weak if $k$ is large (like $k&amp;gt;n/15$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4.) The exact optimal number of weighings for every combination of $n$ and $k$ is an unsolved problem in combinatorial search theory.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-10-15T14:03:47.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23216" PostTypeId="1" AcceptedAnswerId="23221" CreationDate="2015-10-15T14:19:37.250" Score="19" ViewCount="2807" Body="&lt;p&gt;There are $64$ pupils attending a particular school. &#xA;Any two of these pupils share (at least) one common grandfather. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this imply that there are at least $43$ pupils all of which &#xA;have a common grandfather?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="11876" LastEditDate="2016-03-23T20:20:52.887" LastActivityDate="2016-03-24T12:20:38.567" Title="School where every pair of students share a common grandfather" Tags="|mathematics|graph-theory|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23217" PostTypeId="5" CreationDate="2015-10-15T14:23:17.100" Score="0" Body="" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-10-15T14:23:17.100" LastActivityDate="2015-10-15T14:23:17.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23218" PostTypeId="4" CreationDate="2015-10-15T14:23:17.100" Score="0" Body="A small and simple puzzle that can be solved in 15 seconds, sometimes a trick question." OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-10-15T14:23:17.100" LastActivityDate="2015-10-15T14:23:17.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23219" PostTypeId="2" ParentId="23216" CreationDate="2015-10-15T14:43:54.797" Score="9" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There can be at most three grandfathers (GPs). If there are four, a child could have GPs $A$ and $B$, and another one could have GPs $C$ and $D$, violating the first restriction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, our students can be divided into three groups, for each pairing of GPs.&lt;br&gt;&#xA;$64$ divided by $3$ is $21.33$.  So, if each group had $21$, then our max for a common GP would be $42$. But, that's only $63$ students, so by the pigeonhole principle, we have to add the last child to a group, allowing us to get to $43$ pupils with a common GP. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="8874" LastEditDate="2016-03-24T12:20:38.567" LastActivityDate="2016-03-24T12:20:38.567" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23220" PostTypeId="2" ParentId="23216" CreationDate="2015-10-15T14:48:20.347" Score="8" Body="&lt;p&gt;Yes, at least 43 pupils have the same grandfather.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each pupil has two biological grandfathers, so in order for every pupil to have one grandfather in common with every other pupil, there would need to be a maximum of three grandfathers total. Call these grandfathers Larry, Curly, and Moe for the sake of argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The grandchildren of Larry and Curly are group A, Larry and Moe are group B, and Moe and Curly are group C. One of these groups (let's say group C) is the smallest. If group C has 21 pupils, A and B have a total of 43 pupils (i.e., Larry has 43 grandchildren at the school). If group C has fewer than 21 pupils, then Larry has more than 43 grandchildren at the school. If group C has more than 21 pupils, it is no longer the smallest group, and this process starts all over with some other group as the smallest group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Pigeonhole Principle is applied twice here. First, to find the maximum number of grandfathers, and then again to find the minimum number of grandchildren sharing the same grandfather. &#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Pigeonhole_principle&quot;&gt;https://en.wikipedia.org/wiki/Pigeonhole_principle&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16945" LastActivityDate="2015-10-15T14:48:20.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23221" PostTypeId="2" ParentId="23216" CreationDate="2015-10-15T15:44:08.383" Score="25" Body="&lt;p&gt;First of all, with some half-sibling incest, it's possible for a student to only have one grandpa. If that happens, then all 64 students have that grandpa as well. So, we can assume everyone has two grandpas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say that a &quot;grandpa triangle&quot; is a set of three grandpas, where each pair have a common grandchild. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;If there is no grandpa triangle, then all 64 students have a common grandpa.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is certainly true if all students have the same pair of grandpas. So, suppose the Alice has grandpas $\{g_1,g_2\}$, and Bob student has grandpas $\{g_1,g_3\}$. Then every student must have $g_1$ as a grandpa: if they didn't, they would have to have $g_2$ in order to share a grandpa with Alice, and also have to have $g_3$ to share a grandpa with Bob, which would make a grandpa triangle.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;If there is a grandpa triangle, then there are only three grandpas.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's say the grandpa triangle is $\{g_1,g_2,g_3\}$, where Alice, Bob and Charlie have the grandpa sets of $\{g_1,g_2\},\{g_2,g_3\}$ and $\{g_3,g_1\}$. If any student, Diane, had a grandpa $g_4$ which was different then $g_1,g_2,g_3$, then Diane's pair of grandpas would have to be either $\{g_4,g_1\},\{g_4,g_2\},$ or $\{g_4,g_3\}$, but this would mean she didn't share a grandpa with Bob, Charlie, or Alice, respectively. Therefore, every student must have both their grandpas in the set $\{g_1,g_2,g_3\}$, so these are the only three grandpas.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;If there are only three grandpas, the some grandpa has 43 grandkids&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we let $n_1,n_2,n_3$ be the number of grandkids that $g_1,g_2,g_3$ have. Then $n_1+n_2+n_3$ will be twice the number of kids, since every kid is counted twice. This means $n_1+n_2+n_3=2\cdot 64=128$. In order for three numbers to add to $128$, at least one must be $43$ or more.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-10-15T17:54:23.860" LastActivityDate="2015-10-15T17:54:23.860" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23222" PostTypeId="5" CreationDate="2015-10-15T15:49:07.183" Score="0" Body="&lt;p&gt;When you enter phrases into the Google Books Ngram Viewer, it displays a graph showing how those phrases have occurred in a corpus of books (e.g., &quot;British English&quot;, &quot;English Fiction&quot;, &quot;French&quot;) over the selected period of years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google Ngram Viewer can be found &lt;a href=&quot;https://books.google.com/ngrams&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, and the help file &lt;a href=&quot;https://books.google.com/ngrams/info&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="11497" LastEditDate="2016-09-01T06:38:51.190" LastActivityDate="2016-09-01T06:38:51.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23223" PostTypeId="4" CreationDate="2015-10-15T15:49:07.183" Score="0" Body="A puzzle centered around graphs produced by the Google books Ngram viewer." OwnerUserId="8874" LastEditorUserId="11497" LastEditDate="2016-09-01T07:18:36.217" LastActivityDate="2016-09-01T07:18:36.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23224" PostTypeId="2" ParentId="23211" CreationDate="2015-10-15T16:25:24.053" Score="3" Body="&lt;p&gt;You're...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eridan Ampora from Homestuck. He types like that and...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Spoilers follow)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He killed his girlfriend and they met up in the dreambubble afterlife. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The 6/13 is a reference to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;his species' time; six sweeps is equivalent to 13 years.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2015-10-15T16:25:24.053" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23225" PostTypeId="2" ParentId="22932" CreationDate="2015-10-15T16:49:52.770" Score="1" Body="&lt;ol start=&quot;9&quot;&gt;&#xA;&lt;li&gt;&quot;I'm the prince from Undertale,&quot; Tom said in the role of Louis the Métis rioter&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tom said as Riel&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16949" LastActivityDate="2015-10-15T16:49:52.770" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23227" PostTypeId="1" CreationDate="2015-10-15T17:23:43.380" Score="29" ViewCount="1518" Body="&lt;p&gt;The numbers $ 2, \ldots, 2015 $ are written on a blackboard. Each minute any two numbers $ x $ and  $ y $&#xA;are wiped out and are replaced by two  numbers $\displaystyle \frac { 4x + 3y } { 5 }  $ and  $\displaystyle \frac { 24x - 7y } { 25 }  $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible that after some time the numbers $ 1, \ldots, 2014 $ are written on the blackboard ?&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="11876" LastEditDate="2015-11-25T02:42:47.110" LastActivityDate="2015-11-27T18:21:08.460" Title="Numbers on the blackboard: From 2-2015 to 1-2014" Tags="|mathematics|reachability|blackboard|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="23228" PostTypeId="2" ParentId="23179" CreationDate="2015-10-15T17:27:46.937" Score="0" Body="&lt;p&gt;The only thing I can think is that with the &quot;correct&quot; answer you get a single, double and triple sequence of all 3 letters.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;triple &lt;strong&gt;a&lt;/strong&gt;, &lt;/li&gt;&#xA;&lt;li&gt;single &lt;strong&gt;b&lt;/strong&gt;, &lt;/li&gt;&#xA;&lt;li&gt;single &lt;strong&gt;c&lt;/strong&gt;, &lt;/li&gt;&#xA;&lt;li&gt;double &lt;strong&gt;a&lt;/strong&gt;, &lt;/li&gt;&#xA;&lt;li&gt;triple &lt;strong&gt;b&lt;/strong&gt;, &lt;/li&gt;&#xA;&lt;li&gt;double &lt;strong&gt;c&lt;/strong&gt;, &lt;/li&gt;&#xA;&lt;li&gt;double &lt;strong&gt;b&lt;/strong&gt;, &lt;/li&gt;&#xA;&lt;li&gt;triple &lt;strong&gt;c&lt;/strong&gt;, &lt;/li&gt;&#xA;&lt;li&gt;single &lt;strong&gt;a&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I can't find a pattern to that though.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a,b,c,a,b,c,b,c,a&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;3,1,1,2,3,2,2,3,1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Maybe someone else sees it?&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2015-10-15T17:27:46.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23229" PostTypeId="1" CreationDate="2015-10-15T18:12:57.337" Score="0" ViewCount="474" Body="&lt;p&gt;3 9 3 9 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8 6 4 8 6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 4 2 4 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;9 8 X 8 9&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What should X be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="16951" LastActivityDate="2015-11-18T06:14:48.863" Title="What should X be?" Tags="|riddle|number-sequence|" AnswerCount="5" CommentCount="3" FavoriteCount="0" ClosedDate="2015-11-18T13:47:41.453" ContentLicense="CC BY-SA 3.0" />
  <row Id="23230" PostTypeId="2" ParentId="23229" CreationDate="2015-10-15T18:24:24.683" Score="-1" Body="&lt;p&gt;I'm leaning towards:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 9 or 7. Based on the prior series, you're alternating numbers. With the series 8 6 4 8 6, it seems that it's alternating 8, 6, 4, 8, 6, 4 into infinity. If the rows have a relationship with each other then the answer could be 7 - you're subtracting 1 (the difference between the alternating numbers) from the second number in the series, and that calculation is made every other row. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16729" LastEditorUserId="16729" LastEditDate="2015-10-15T18:32:23.123" LastActivityDate="2015-10-15T18:32:23.123" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23231" PostTypeId="2" ParentId="23229" CreationDate="2015-10-15T18:36:16.987" Score="4" Body="&lt;p&gt;We are shown two kinds of rows - those with just two values, and those with three. The ones with two values follow this pattern:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A B A B A&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The ones with three follow this pattern:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A B C A B&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since the last line &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; has the 8 and 9 in opposite order at the end, &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;it is a line with&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; two&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;values. The missing number is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 9&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1509" LastActivityDate="2015-10-15T18:36:16.987" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23233" PostTypeId="2" ParentId="23229" CreationDate="2015-10-15T19:09:07.223" Score="2" Body="&lt;p&gt;Note: this can't be correct, but I'll leave it to inspire others.&lt;br&gt;&#xA;My vote is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Each line is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a number followed by that number squared, repeated infinitely&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3 squared is 9&lt;br&gt;&#xA; 8 squared is 64&lt;br&gt;&#xA; 2 squared is 4&lt;br&gt;&#xA; 9 squared is 81&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12092" LastActivityDate="2015-10-15T19:09:07.223" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23235" PostTypeId="2" ParentId="16460" CreationDate="2015-10-15T19:57:44.220" Score="2" Body="&lt;p&gt;Since we're throwing out alternative answers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The clues stated that his car was not dented from hitting his wife. However...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He did in fact hit her, with enough force that the numbers and letters of his front-mounted license plate left an impression on the wife's body, but not enough &#xA; to dent the car. When he arrived on scene, a clever detective recognized &#xA; the impression as a mirror of the one on the husband's car.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16874" LastActivityDate="2015-10-15T19:57:44.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23239" PostTypeId="2" ParentId="23229" CreationDate="2015-10-15T20:35:59.523" Score="2" Body="&lt;p&gt;I think the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 9&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we take the 1st, 3rd, and 5th column of each row &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1st*3rd and 3rd*5th&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then take the answer and multiply the two numbers together you get what is in the 2nd and 4th column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1st row:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1st*3rd = 2nd  -&gt; 3*3 = 9&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3rd*5th = 4th  -&gt; 3*3 = 9&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2nd row:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8*4=32 -&gt; 3*2 = 6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4*6=24 -&gt; 2*4 = 8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3rd row:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1st*3rd = 2nd  -&gt; 2*2=4 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3rd*5th = 4th  -&gt; 2*2=4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4th row:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;9*9=81 -&gt; 8*1 = 8 (2nd)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;9*9=81 -&gt; 8*1 = 8 (4th)&lt;/p&gt;&#xA;" OwnerUserId="16471" LastActivityDate="2015-10-15T20:35:59.523" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23240" PostTypeId="1" CreationDate="2015-10-16T01:51:56.303" Score="13" ViewCount="6517" Body="&lt;p&gt;Can you please help settle a debate about the best strategy for this game? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At pub trivia when there is a tie for first or second the host uses a guess the right number between 1 and 99. Each person gets an alternate guess and she says higher or lower to narrow the range. The debate is about when you get down to a range of 5-10 numbers. I always like to leave at least a 2 number gap if possible so that if I get it wrong, my opponent still has a 50/50 choice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;eg Let's say the possible numbers left are 21, 22, 23, 24, &amp;amp; 25. I would choose 23. Once the host says higher or lower (if I'm not right), the opponent still has to guess the right answer, and if they guess wrong, I will get it right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is the probability of each guesser (1st or 2nd guesser) winning from this point, and which is the best first choice for guesser 1? Should I choose the middle, the edge or the in between?&lt;/p&gt;&#xA;" OwnerUserId="16959" LastActivityDate="2015-10-16T22:11:34.233" Title="Higher or Lower Guessing Game. Help settle the debate please" Tags="|strategy|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23241" PostTypeId="2" ParentId="23240" CreationDate="2015-10-16T02:05:35.257" Score="4" Body="&lt;p&gt;&lt;strong&gt;This is a game about minimizing the amount of information your opponent has.&lt;/strong&gt; In order to think about this, let's first make the assumption that the number you guess will be wrong (because if it's right, so ends the experiment). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The announcer will tell your opponent whether your guess was too high or too low. This will eliminate whichever number you guessed, plus everything higher/lower than it - in other words, at least one number must vanish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, if you pick 23, then your opponent knows the number is either 24/25 or 21/22. If you pick 21, though, then your opponent knows the number is one of 22/23/24/25. Strictly speaking, if your opponent guesses randomly, then they have a worse chance of guessing correctly by a factor of two when you pick 21 as compared to when you pick 23.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the correct strategy is this: pick a number on the outer bound of the available range. If your range is from 21 to 25, pick either 21 or 25. That will give your opponent the least information to work from on their turn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if your opponent plays ideally, at this point, you still get one more guess than them, because you pick three numbers and your opponent picks two. This means, even if your opponent plays ideally, you have a 60% (3/5) chance of coming out as the winner. &lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastActivityDate="2015-10-16T02:05:35.257" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23242" PostTypeId="2" ParentId="23240" CreationDate="2015-10-16T02:11:20.637" Score="15" Body="&lt;p&gt;If you have a range of 1 number, you are guaranteed to get it right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a range of 2 numbers, you have a 1/2 chance of getting it on your guess. If you don't your opponent has a range of 1 number, so they are guaranteed to get it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a range of 3 numbers, you can choose the middle or an edge. The middle leaves two ranges of 1, while the edge leaves a range of 2. The better strategy must be to choose the edge, and your chance of winning is 2/3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a range of 4 numbers, you can choose a middle number or an edge.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If you choose an edge and you're wrong, you leave a range of 3 numbers. We know that in this situation, the opponent has a 2/3 chance of winning.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you choose a middle number and you're wrong, there is a range of 2 on one side, and a single number. In two cases, the actual number is in the range of 2 (leaving a 1/2 chance for the opponent). In the other case, it is the single number, and your opponent gets it. Overall, the chance that the opponent wins is (2/3)(1/2)+(1/3)(1)=2/3.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Therefore, either way, you have a 1/4+(3/4)(1/3)=1/2 chance of winning.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now we can consider the range of five numbers.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If you choose an edge, you leave a range of 4, which we know is a 1/2 chance for the opponent.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you choose the middle, you leave two ranges of 2, which we know is also a 1/2 chance.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you choose one of the other numbers (22 or 24), you leave a range of 3 and a range of 1. This gives your opponent a chance of more than 1/2.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In conclusion, you can choose 21, 23, or 25 to get a 1/5+(4/5)(1/2)=3/5 chance of winning, but 22 and 24 are bad.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-10-16T22:11:34.233" LastActivityDate="2015-10-16T22:11:34.233" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23244" PostTypeId="2" ParentId="23240" CreationDate="2015-10-16T05:26:52.607" Score="6" Body="&lt;p&gt;It turns out that it's &lt;strong&gt;always&lt;/strong&gt; a good strategy to choose one end of the range that is presented to you.  But let's face it, that's going to be irritating for everyone waiting.  Luckily it also turns out that it's &lt;strong&gt;almost always&lt;/strong&gt; just as good of an idea to pick around the middle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the stats:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; Size of range: 1 choose number 1 to win 1&#xA; Size of range: 2 choose number 1 to win 1/2&#xA; Size of range: 3 choose number 1 to win 2/3&#xA; Size of range: 4 choose number 2 to win 1/2&#xA; Size of range: 5 choose number 3 to win 3/5&#xA; Size of range: 6 choose number 3 to win 1/2&#xA; Size of range: 7 choose number 3 to win 4/7&#xA; Size of range: 8 choose number 4 to win 1/2&#xA; Size of range: 9 choose number 5 to win 5/9&#xA; Size of range: 10 choose number 5 to win 1/2&#xA; Size of range: 11 choose number 5 to win 6/11&#xA; Size of range: 12 choose number 6 to win 1/2&#xA; Size of range: 13 choose number 7 to win 7/13&#xA; Size of range: 14 choose number 7 to win 1/2&#xA; Size of range: 15 choose number 7 to win 8/15&#xA; Size of range: 16 choose number 8 to win 1/2&#xA; Size of range: 17 choose number 9 to win 9/17&#xA; Size of range: 18 choose number 9 to win 1/2&#xA; Size of range: 19 choose number 9 to win 10/19&#xA; Size of range: 20 choose number 10 to win 1/2&#xA; Size of range: 21 choose number 11 to win 11/21&#xA; Size of range: 22 choose number 11 to win 1/2&#xA; Size of range: 23 choose number 11 to win 12/23&#xA; Size of range: 24 choose number 12 to win 1/2&#xA; Size of range: 25 choose number 13 to win 13/25&#xA; Size of range: 26 choose number 13 to win 1/2&#xA; Size of range: 27 choose number 13 to win 14/27&#xA; Size of range: 28 choose number 14 to win 1/2&#xA; Size of range: 29 choose number 15 to win 15/29&#xA; Size of range: 30 choose number 15 to win 1/2&#xA; Size of range: 31 choose number 15 to win 16/31&#xA; Size of range: 32 choose number 16 to win 1/2&#xA; Size of range: 33 choose number 17 to win 17/33&#xA; Size of range: 34 choose number 17 to win 1/2&#xA; Size of range: 35 choose number 17 to win 18/35&#xA; Size of range: 36 choose number 18 to win 1/2&#xA; Size of range: 37 choose number 19 to win 19/37&#xA; Size of range: 38 choose number 19 to win 1/2&#xA; Size of range: 39 choose number 19 to win 20/39&#xA; Size of range: 40 choose number 20 to win 1/2&#xA; Size of range: 41 choose number 21 to win 21/41&#xA; Size of range: 42 choose number 21 to win 1/2&#xA; Size of range: 43 choose number 21 to win 22/43&#xA; Size of range: 44 choose number 22 to win 1/2&#xA; Size of range: 45 choose number 23 to win 23/45&#xA; Size of range: 46 choose number 23 to win 1/2&#xA; Size of range: 47 choose number 23 to win 24/47&#xA; Size of range: 48 choose number 24 to win 1/2&#xA; Size of range: 49 choose number 25 to win 25/49&#xA; Size of range: 50 choose number 25 to win 1/2&#xA; Size of range: 51 choose number 25 to win 26/51&#xA; Size of range: 52 choose number 26 to win 1/2&#xA; Size of range: 53 choose number 27 to win 27/53&#xA; Size of range: 54 choose number 27 to win 1/2&#xA; Size of range: 55 choose number 27 to win 28/55&#xA; Size of range: 56 choose number 28 to win 1/2&#xA; Size of range: 57 choose number 29 to win 29/57&#xA; Size of range: 58 choose number 29 to win 1/2&#xA; Size of range: 59 choose number 29 to win 30/59&#xA; Size of range: 60 choose number 30 to win 1/2&#xA; Size of range: 61 choose number 31 to win 31/61&#xA; Size of range: 62 choose number 31 to win 1/2&#xA; Size of range: 63 choose number 31 to win 32/63&#xA; Size of range: 64 choose number 32 to win 1/2&#xA; Size of range: 65 choose number 33 to win 33/65&#xA; Size of range: 66 choose number 33 to win 1/2&#xA; Size of range: 67 choose number 33 to win 34/67&#xA; Size of range: 68 choose number 34 to win 1/2&#xA; Size of range: 69 choose number 35 to win 35/69&#xA; Size of range: 70 choose number 35 to win 1/2&#xA; Size of range: 71 choose number 35 to win 36/71&#xA; Size of range: 72 choose number 36 to win 1/2&#xA; Size of range: 73 choose number 37 to win 37/73&#xA; Size of range: 74 choose number 37 to win 1/2&#xA; Size of range: 75 choose number 37 to win 38/75&#xA; Size of range: 76 choose number 38 to win 1/2&#xA; Size of range: 77 choose number 39 to win 39/77&#xA; Size of range: 78 choose number 39 to win 1/2&#xA; Size of range: 79 choose number 39 to win 40/79&#xA; Size of range: 80 choose number 40 to win 1/2&#xA; Size of range: 81 choose number 41 to win 41/81&#xA; Size of range: 82 choose number 41 to win 1/2&#xA; Size of range: 83 choose number 41 to win 42/83&#xA; Size of range: 84 choose number 42 to win 1/2&#xA; Size of range: 85 choose number 43 to win 43/85&#xA; Size of range: 86 choose number 43 to win 1/2&#xA; Size of range: 87 choose number 43 to win 44/87&#xA; Size of range: 88 choose number 44 to win 1/2&#xA; Size of range: 89 choose number 45 to win 45/89&#xA; Size of range: 90 choose number 45 to win 1/2&#xA; Size of range: 91 choose number 45 to win 46/91&#xA; Size of range: 92 choose number 46 to win 1/2&#xA; Size of range: 93 choose number 47 to win 47/93&#xA; Size of range: 94 choose number 47 to win 1/2&#xA; Size of range: 95 choose number 47 to win 48/95&#xA; Size of range: 96 choose number 48 to win 1/2&#xA; Size of range: 97 choose number 49 to win 49/97&#xA; Size of range: 98 choose number 49 to win 1/2&#xA; Size of range: 99 choose number 49 to win 50/99&#xA; Size of range: 100 choose number 50 to win 1/2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;These odds of winning are exactly the same as the odds of winning if you simply pick one of the edges.&lt;/p&gt;&#xA;" OwnerUserId="15942" LastActivityDate="2015-10-16T05:26:52.607" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23245" PostTypeId="1" AcceptedAnswerId="23246" CreationDate="2015-10-16T06:50:53.277" Score="1" ViewCount="364" Body="&lt;p&gt;Given a 4x4 grid of squares, can you black out exactly six of them so that the number of blacked out squares in each row, as well as in each column, is even. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All solutions are permutations of a single one (i.e. all solutions are obtained from any given one by simply flipping around rows and columns). &lt;/p&gt;&#xA;" OwnerUserId="10262" LastActivityDate="2015-10-16T12:15:46.100" Title="A simple grid puzzle" Tags="|mathematics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23246" PostTypeId="2" ParentId="23245" CreationDate="2015-10-16T06:56:06.453" Score="4" Body="&lt;pre&gt;&lt;code&gt;■■□□&#xA;■□■□&#xA;□■■□&#xA;□□□□&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There are two blacked out squares in each of the 1st, 2nd and 3rd rows and columns and no blacked out squares in the 4th row and column.&lt;/p&gt;&#xA;" OwnerUserId="16608" LastEditorUserId="12700" LastEditDate="2015-10-16T08:00:16.123" LastActivityDate="2015-10-16T08:00:16.123" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23247" PostTypeId="2" ParentId="22932" CreationDate="2015-10-16T07:12:40.823" Score="1" Body="&lt;p&gt;Since everybody's gotten partial answers, I'll make a CW to amalgamate the correct ones people have gotten so far:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Player One&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I have an eighth note,&quot; Tom said &lt;strong&gt;with a quaver&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Player One&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I can't find my underpants,&quot; Tom said &lt;strong&gt;briefly&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;Player One &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I really hope I get a 7 or an 11,&quot; Tom said &lt;strong&gt;dicily&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;Deusovi&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I never decrease as I move right,&quot; Tom said &lt;strong&gt;monotonously&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;Alconja&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I can't go below this rope, it's too hard,&quot; Tom said, &lt;strong&gt;in limbo&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;6&quot;&gt;&#xA;&lt;li&gt;Player One&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I have an unnatural view of my buttocks,&quot; Tom said &lt;strong&gt;in hindsight&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;7&quot;&gt;&#xA;&lt;li&gt;Alconja&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I baked this bread using a substitute grain,&quot; Tom said with a &lt;strong&gt;flourish&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;8&quot;&gt;&#xA;&lt;li&gt;Alconja&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I'm protesting to the singer who popularized Autotune,&quot; Tom told the &lt;strong&gt;butcher&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;9&quot;&gt;&#xA;&lt;li&gt;Person&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I'm the prince from Undertale,&quot; Tom said &lt;strong&gt;as Riel&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;10&quot;&gt;&#xA;&lt;li&gt;Deusovi&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I feel like I'm creating part of a play,&quot; Tom said while &lt;strong&gt;writing the ACT&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;11&quot;&gt;&#xA;&lt;li&gt;shettysahab&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I've come up with a new Conservative party member,&quot; Tom said while &lt;strong&gt;taking inventory&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;12&quot;&gt;&#xA;&lt;li&gt;Deusovi&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I need to find a way to make this work with Apple's calendar app,&quot; Tom said &lt;strong&gt;methodically&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2016-07-03T19:00:57.937" LastActivityDate="2016-07-03T19:00:57.937" CommentCount="3" CommunityOwnedDate="2015-10-16T07:12:40.823" ContentLicense="CC BY-SA 3.0" />
  <row Id="23249" PostTypeId="2" ParentId="23240" CreationDate="2015-10-16T07:32:47.740" Score="7" Body="&lt;p&gt;Following the same strategy as Dr Xorile, I wrote a quick programming solution to this problem.  The goal is to define a function $p_{n,k}$, the probability of winning with $n$ numbers left, if you choose the $k$-th number.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;p[n_, k_] := p[n, k] = 1/n + (k - 1)/n (1 - p[k - 1]) + (n - k)/n (1 - p[n - k])&#xA;p[0] = 0;&#xA;p[1] = 1;&#xA;p[n_] := p[n] = Max[Table[p[n, k], {k, n}]]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I define the auxiliary function $p_n$ as the probability of winning with $n$ numbers if you choose ideally.  I also assume that each number is equally likely to be correct.  (This is not true in real life: for example, people rarely choose multiples of ten because they don't seem &quot;random.&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this function we can break $p_{n,k}$ into three mutually exclusive events:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You guess the correct number.  This happens with probability $1/n$ (and in this case you win with probability $1$.&lt;/li&gt;&#xA;&lt;li&gt;You guess high.  This happens with probability $(k-1)/n$.  Your opponent goes next, and since there are $k-1$ numbers left he has a probability to win of $p_{k-1}$; your probability of winning in this scenario is therefore $1-p_{k-1}$.&lt;/li&gt;&#xA;&lt;li&gt;You guess low.  The reasoning is the same as above, except there are $n-k$ numbers remaining so the probability of this event is $(n-k)/n$ and your probability of winning is $1-p_{n-k}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Therefore the total probability is just:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;p_{n,k} = \frac 1 n + \frac{k-1} n (1-p_{k-1}) + \frac{n-k} n (1-p_{n-k})&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also write the winning probability for the ideal strategy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;p_n = \max_{k\in[1,n]}p_{k,n}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The base case is $p_1=1$ (If there is only one number left, you win).  The case $p_0=0$ in the code is not strictly necessary, as that case only contributes with probability $0$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Running this program, I found that there are three different cases:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$n$ is even: $p_{n,k}=1/2$&lt;/li&gt;&#xA;&lt;li&gt;$n$ is odd, $k$ is even: $p_{n,k}=(n-1)/2n$&lt;/li&gt;&#xA;&lt;li&gt;$n$ is odd, $k$ is odd: $p_{n,k}=(n+1)/2n$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If this is true, it follows that $p_n=1/2$ for even $n$ and $(n+1)/2n$ for odd $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can prove this by induction:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If $n$ is even, then either:&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$k$ is odd; $k-1$ is even and $n-k$ is odd, so:&#xA;$$ \begin{align} p_{n,k} &amp;amp;= \frac 1 n + \frac{k-1} n \frac 1 2 + \frac{n-k} n \frac{n-k-1}{2(n-k)} \\ &amp;amp;= \frac 2{2n} + \frac{k-1}{2n} + \frac{n-k+1}{2n} \\ &amp;amp;= \frac{2+k-1+n-k-1}{2n} \\ &amp;amp;= \frac n {2n} = \frac 1 2 \end{align} $$&lt;/li&gt;&#xA;&lt;li&gt;$k$ is even; $k-1$ is odd and $n-k$ is even, so:&#xA;$$ \begin{align} p_{n,k} &amp;amp;= \frac 1 n + \frac{k-1} n \frac{k-1-1}{2(k-1)} + \frac{n-k} n \frac 1 2 \\ &amp;amp;= \frac 2{2n} + \frac{k-2}{2n} + \frac{n-k}{2n} \\ &amp;amp;= \frac{2+k-2+n-k}{2n} \\ &amp;amp;= \frac n {2n} = \frac 1 2 \end{align} $$&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;If $n$ is odd and $k$ is even, $k-1$ and $n-k$ are both odd, so:&#xA;$$ \begin{align} p_{n,k} &amp;amp;= \frac 1 n + \frac{k-1} n \frac{k-1-1}{2(k-1)} + \frac{n-k} n \frac{n-k-1}{2(n-k)} \\ &amp;amp;= \frac 2{2n} + \frac{k-2}{2n} + \frac{n-k+1}{2n} \\ &amp;amp;= \frac{2+k-2+n-k-1}{2n} \\ &amp;amp;= \frac{n-1}{2n} \end{align} $$&lt;/li&gt;&#xA;&lt;li&gt;If $n$ is odd and $k$ is odd, $k-1$ and $n-k$ are both even, so:&#xA;$$ \begin{align} p_{n,k} &amp;amp;= \frac 1 n + \frac{k-1} n \frac 1 2 + \frac{n-k} n \frac 1 2 \\ &amp;amp;= \frac 2{2n} + \frac{k-1}{2n} + \frac{n-k}{2n} \\ &amp;amp;= \frac{2+k-1+n-k}{2n} \\ &amp;amp;= \frac{n+1}{2n} \end{align} $$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Q.E.D.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note that the middle number is &lt;em&gt;not&lt;/em&gt; always the best choice: if $n\equiv 3\mod 4$, then $k=(n+1)/2$ will be even, and your probability of winning will be decreased by $1/n$ compared to the perfect strategy.  The first strategy that Dr Xorile proposes (choosing one of the endpoints) &lt;em&gt;is&lt;/em&gt; a perfect strategy.  A perfect strategy that still roughly halves the possibilities on each turn could be:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Given that you know the number is between $a$ and $b$, inclusive, compute the number of numbers remaining, $n=b-a+1$.&lt;/li&gt;&#xA;&lt;li&gt;If $n$ is even, choose either of $(b+a\pm 1)/2$.&lt;/li&gt;&#xA;&lt;li&gt;Otherwise, if $n+1$ is divisible by four, choose either of $(b+a)/2\pm 1$.&lt;/li&gt;&#xA;&lt;li&gt;Otherwise choose $(b+a)/2$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note again the assumptions we made: that your opponent also follows a perfect strategy (although not necessarily an identical strategy), and that the answer is evenly distributed among the possible guesses.&lt;/p&gt;&#xA;" OwnerUserId="9718" LastActivityDate="2015-10-16T07:32:47.740" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23250" PostTypeId="2" ParentId="20849" CreationDate="2015-10-16T07:43:50.333" Score="10" Body="&lt;p&gt;Ok, I'm pretty sure it's referring to...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...words with &lt;strong&gt;&quot;mate&quot;&lt;/strong&gt; in them&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This riddle is slightly different, but I think you will get used to it with time.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Accli&lt;strong&gt;mate&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Your brain is evolved, is it not? Then you should have two legs to stand on.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pri&lt;strong&gt;mate&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Feel free to sketch out your ideas or to draw quick conclusions.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Esti&lt;strong&gt;mate&lt;/strong&gt; / Guessti&lt;strong&gt;mate&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;After all, we are all open to learning new ideas from our peers.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; School&lt;strong&gt;mates&lt;/strong&gt; / Class&lt;strong&gt;mate&lt;/strong&gt;s&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I hope this doesn’t make you too heated, though.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Inflam&lt;strong&gt;mate&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You might think that we are both stuck, forces equally matched.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stale&lt;strong&gt;mate&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But perhaps you thought less of this riddle than you should have.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Underesti&lt;strong&gt;mate&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Now that I have you where I want you, the end is inevitable.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Check&lt;strong&gt;mate&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And here it is: it is final.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ulti&lt;strong&gt;mate&lt;/strong&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Hopefully my hints aren’t too subtle, for I would love you to get the answer, or at least to get close.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Approxi&lt;strong&gt;mate&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Be my match, or be The One&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Find words with &lt;strong&gt;&quot;mate&quot;&lt;/strong&gt; in them, or soul&lt;strong&gt;mate&lt;/strong&gt; / the word &lt;strong&gt;&quot;mate&quot;&lt;/strong&gt; itself / legti&lt;strong&gt;mate&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Avast!&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Ahoy &lt;strong&gt;mate&lt;/strong&gt;y&quot; =)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15960" LastEditorUserId="15960" LastEditDate="2015-10-16T07:54:06.797" LastActivityDate="2015-10-16T07:54:06.797" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23251" PostTypeId="2" ParentId="23240" CreationDate="2015-10-16T08:37:17.883" Score="5" Body="&lt;p&gt;If you are guessing with an odd range (e.g. 5), always make sure that after you choose there are two even numbers ranges left for your opponent. There will be in total  4 numbers left, but you can split them oddly (1-3) or evenly (0-4 or 2-2). Make sure you choose the even split.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an &lt;em&gt;even range&lt;/em&gt;, both players have 50% chance of winning with optimal play. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;There is a guaranteed 50% strategy for player 2: choose the&#xA;  neighbour of the number taken by player 1 to leave two even ranges.&#xA;  For example, with 1-10, if player 1 takes nr. 3, you take nr. 4 (with&#xA;  same probability of winning directly), leaving ranges 1-2 and 5-10,&#xA;  both which are even, and you can repeat the tactic to always have 50% chance.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;There is also a guaranteed 50% strategy for player 1: if you always take&#xA;  the penultimate number in the range N, either this number is correct&#xA;  (you win), the final number in the range is correct (you lose with&#xA;  same probability), or the game continues with an even range, with the other player to move, at which point you can use the player 2 strategy to ensure a 50% chance.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By guessing a number that forces your opponent into one of two even ranges you get one free guess, and after that, given optimal play, it's a coin toss (apologies for the mixed metaphor). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your case with a range of 5, you can choose:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;21 : 0-4 split (good: even/even)&lt;/li&gt;&#xA;&lt;li&gt;22: 1-3 split (bad: odd/odd)&lt;/li&gt;&#xA;&lt;li&gt;23: 2-2 split (good)&lt;/li&gt;&#xA;&lt;li&gt;24: 3-1 split (bad)&lt;/li&gt;&#xA;&lt;li&gt;25: 4-0 split (good)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So if the range is N and an an odd number, and you leave your opponent two even ranges, then your probability of winning overall is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P     = 1/N + (1-1/N) * 1/2$$&#xA;$$P     = 1/2 + 1/(2N)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for the range of 5, that gives $P= 1/2 + 1/10 = 3/5$, as found by Emrakul and f''.  &lt;/p&gt;&#xA;" OwnerUserId="7316" LastEditorUserId="7316" LastEditDate="2015-10-16T09:13:56.903" LastActivityDate="2015-10-16T09:13:56.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23253" PostTypeId="2" ParentId="23245" CreationDate="2015-10-16T12:15:46.100" Score="-1" Body="&lt;p&gt;If the question &lt;strong&gt;meant&lt;/strong&gt; to ask:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;em&gt;..can you black out exactly six of them so that the number of blacked out squares in each row, as well as in each column, is&lt;/em&gt; &lt;strong&gt;&lt;em&gt;equal&lt;/em&gt;&lt;/strong&gt;?&quot;,   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;then the following solution would work:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/kOj8A.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kOj8A.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With 1.5 blackout out in each row and column.&lt;/p&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-10-16T12:15:46.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23255" PostTypeId="1" AcceptedAnswerId="23263" CreationDate="2015-10-16T15:16:20.170" Score="8" ViewCount="713" Body="&lt;p&gt;In the country of Ebbozonia, there are only two type of coins: light coins and heavy coins. The weights of these coins satisfy the following properties:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;All light coins have the same weight $L$.&lt;/li&gt;&#xA;&lt;li&gt;All heavy coins have the same weight $H$.&lt;/li&gt;&#xA;&lt;li&gt;Heavy coins are heavier than light coins: $H&amp;gt;L$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The precise values of $L$ and $H$ are not known to the public. &#xA;The difference between $L$ and $H$ should be fairly small, as there is no way of distinguishing the two coin types without a balance. &#xA;His Highness, the Honourable Minister of Treasury, has recently announced the following important property of the Ebbozonian coin system:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If a balance is in perfect equilibrium with $h_1$ heavy and $\ell_1$ light coins on the right pan and with $h_2$ heavy and $\ell_2$ light coins on the left pan, then $h_1=h_2$ and $\ell_1=\ell_2$ must necessarily hold true.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Cosmo puts 10 Ebbozonian coins on the table and asks Fredo to determine quickly whether these 10 coins all have the same weight. &#xA;On the table, there is a balance with two pans (but there are no weights).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Can Fredo solve Cosmo's problem by using the balance at most three times?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-10-16T15:27:53.327" LastActivityDate="2015-10-16T20:59:25.720" Title="The Ebbozonian coin weighing puzzle" Tags="|mathematics|strategy|weighing|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23256" PostTypeId="2" ParentId="23255" CreationDate="2015-10-16T15:42:44.300" Score="1" Body="&lt;p&gt;I believe that the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;No, it is not possible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To explore why it isn't possible, let's look at a similar problem, with eight coins, numbered C1-C8.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Put four coins on each side of the balance.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;C1-C4 vs C5-C8&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If the balance is not equal, then we're done. We know for a fact that there are two weights of coins in the pile, and we don't need to go any further.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If, on the other hand, the balance does NOT move, then the two piles have equal numbers of each type of coin, and could conceivably all be the same weight. So the next step is to take C1-C4 and place two coins on each side of the balance. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;C1&amp;amp;C2 vs C3&amp;amp;C4&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Again, if the balance moves, we're done. If, on the other hand, the balance stays even, we will do one more test&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;C1 vs C2&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now, if the balance is STILL even, all of our coins are the same weight.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The 10 coin example doesn't work out quite as well, though. After the first weighing, we're left with five coins and two weighings, which is not enough to identify the coins. &#xA;One weighing can identify two coins, two weighings can identify four, three weighings can identify eight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some methods that can solve these problems more efficiently with larger numbers of coins, but the total number usually has to be divisible by 3 or 4. With 10 coins, I don't believe it can be done in 3 actions&lt;/p&gt;&#xA;" OwnerUserId="16814" LastEditorUserId="16814" LastEditDate="2015-10-16T20:55:12.103" LastActivityDate="2015-10-16T20:55:12.103" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23257" PostTypeId="2" ParentId="23255" CreationDate="2015-10-16T16:07:17.890" Score="1" Body="&lt;p&gt;I don't think it's possible. I can ID up to 8 coins in 3 trials by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weigh 1v1, if different, we have our answer.&lt;br&gt;&#xA;Weigh 2(previous)v2&lt;br&gt;&#xA;Weigh 4(prev)v4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This allows us to check up to 8 coins. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-10-16T16:07:17.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23260" PostTypeId="1" AcceptedAnswerId="23313" CreationDate="2015-10-16T17:28:49.070" Score="11" ViewCount="1220" Body="&lt;p&gt;This is the fifth in a series &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=user%3A9000%20is%3AQuestion%20title%3A%22Lights%20in%20the%20Dungeon%22&quot;&gt;Lights in the Dungeon&lt;/a&gt; puzzles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/23192/lights-in-the-dungeon-eighteen-by-eleven&quot;&gt;previous (4th) &amp;lt;--&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There is a lot of flavorful fluff here but, if you want, you can probably just skip down to the map and try to figure out the message without reading everything else.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It had taken you a little longer to solve the &lt;a href=&quot;https://puzzling.stackexchange.com/q/23192/9000&quot;&gt;eighteen by eleven&lt;/a&gt; dungeon map than it had to solve the previous challenges but you persevered. The good news is that, once it was &lt;a href=&quot;https://puzzling.stackexchange.com/a/23197/9000&quot;&gt;solved&lt;/a&gt;, it was pretty fast to run through. There weren't any branching paths so you didn't have to think about which way to go. You emerge from the exit a little tired from the long path (142 rooms is a lot more tiring than 25) and look around for your captor. You spot Lord Hooty McOwlface hunched over a control panel with his grand vizier, Fakky Jim. Meandering over, you catch a snippet of their whispered conversation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;No, we can't make it too easy! ... Yes, I know it's the same thing but we chose this pattern for a reason. ... Look, I'm the great and powerful lord here and you're the freaking cat! I say the way it's been encoded is fine and we're certainly not going to take the time to change it now.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your shoe scuffs against the ground and His Lordship whirls around, apparently startled. As his cap settles back down from the quick motion, his eyes narrow. &quot;Ah,&quot; he starts, &quot;You're done. Very well. Just sit over there for now. We'll be with you in a minute.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You turn in the direction he points and spy an overstuffed armchair. It's a bit out-of-place in the otherwise dank dungeon but at least it looks comfortable. You settle down, pull our your smart phone - no signal, unfortunately - and try to figure out the WiFi password. Eventually, Hooty approaches you with a gleam in his eye. Is that a smile? Is that what his smile looks like? How disturbing. He begins to orate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Congratulations on having made it this far! You have passed some basic intelligence tests and have performed admirably well despite the pressure and very present danger to your person. If you can solve this final challenge, you will be free to go. In fact, I will &lt;em&gt;encourage&lt;/em&gt; you to go and give you some riches to help you along. I have encoded a message for you. Your challenge is to decode that message. If you'll just follow me this way...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You trod behind the despot and he leads you back to one of the dungeon entrances. You're a bit hesitant as you haven't been given the chance to rearrange any of the rooms this time but you don't have much choice when the waiting guards grab you and hurl you bodily into the room, slamming the door behind you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You stand up and dust yourself off in time to hear screeching from the other side of the doors. &quot;No, you idiots! You forgot to give him the pack!&quot; The doors swing open and a light bag is hurled in your face. As you recover, you hear the doors being closed and locked once more. Opening the pack, you find a pen and a pad of paper. Looking around, you notice a few changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the other door in the room is unlocked again. You dare not leave the room for fear it will slam shut like it did the last time but it looks like the next door is open, too. You stroll across the room for a better look and you nearly slip on a puddle. Oh, god, is that blood? Why is there blood? You back up a bit and see that it has been very deliberately painted in a pattern. It reads &lt;code&gt;1D&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You scribble this in your notepad and, after first ensuring that you didn't miss any other changes, you move to the next room. As you expected, the door slams behind you. The next room also has a message on the floor: &lt;code&gt;06&lt;/code&gt;. You move through a few more rooms as you notice that the layout appears to be identical to the dungeon you just solved. They must have followed right behind you painting the blood on the floor but didn't have time to rearrange anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flipping to the next page, you sketch out the dungeon map - you have impeccable memory, good for you - and write the message in each room within the corresponding position on the map. As you finish the last room, you notice that the exit door is not open. Unfortunately, you don't notice this until the door behind you has slammed shut. Now you're trapped in a room with &lt;code&gt;04&lt;/code&gt; scrawled in blood on the floor and no way out. A voice from the other side of the exit cackles at you and calls out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Very good walking, young morsel. I &lt;em&gt;do&lt;/em&gt; hope you've been writing things down. Did you recognize the dungeon? &lt;sub&gt;Not that it really matters...&lt;/sub&gt; Now, tell me my secret or I'll leave you in there to starve &lt;em&gt;almost&lt;/em&gt; to death. After all, I wouldn't want you to spoil...&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is Lord Hooty McOwlface's message?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here is a copy of the map with the encoded message from each room: (you entered in the top left at the green arrow and are now stuck in the bottom right near the red arrow)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UCdcL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UCdcL.png&quot; alt=&quot;Maze Secret&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There's nothing to keep track of time in the room so you don't know how long you've been fiddling with your notes when you hear a voice from outside. &quot;Nothing, eh? I was thinking you were oh so clever, too. Well, look, I want someone to finally succeed in this so I'll give you a hint.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You might be thinking that the message in each room is a two-digit hexadecimal number. It's not. It's &lt;em&gt;two&lt;/em&gt; one-digit hexadecimal numbers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;For what it's worth, here's the text versions of the maze:&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;ACPPPBC&#xA;PAGPPMM&#xA;PBHBOMM&#xA;PAGHHMMP&#xA;PPPPOMMP&#xA;FGGEDMMP&#xA;FGNOPMMFC&#xA;PPPFPMMMM&#xA;FHBEPMMMM&#xA;FADFCMMMMP&#xA;FBCFFMMMMP&#xA;AEFDFMMMMP&#xA;FNNNDMMMANC&#xA;ANNNCMMFNGM&#xA;FNNNOMFPFOM&#xA;ANNNNOFNDPM&#xA;FNNNNNNNNOD&#xA;ANNNNNNNNNN&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;pre&gt;&#xA;1D,06,--,--,--,01,0E&#xA;--,0E,0E,--,--,0D,01&#xA;--,0E,07,1D,15,0D,0E&#xA;--,1D,0F,0E,04,1D,14,--&#xA;--,--,--,--,01,2D,12,--&#xA;0D,0E,16,1D,0E,0D,0E,--&#xA;0D,06,0E,0D,--,0E,0E,01,05&#xA;--,--,--,1F,--,0F,07,0E,06&#xA;0D,0E,0D,0E,--,0D,0E,06,03&#xA;1D,0F,0D,02,08,0D,06,0E,06,--&#xA;16,0E,01,0E,06,0D,0E,16,0E,--&#xA;04,01,05,02,03,0E,06,1D,02,--&#xA;16,0D,0E,07,0E,07,0E,04,03,06,07&#xA;08,0D,0D,1D,11,1E,0D,1E,0D,0E,03&#xA;0D,0D,0D,0D,0E,0D,11,--,0E,0F,02&#xA;1D,16,07,03,06,0E,03,1D,1F,--,0E&#xA;03,01,0D,0D,0D,08,06,0E,0F,08,06&#xA;06,04,06,0E,0D,0D,0D,0D,0D,0E,04&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For reference, here are the 16 possible rooms.&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rp7xy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rp7xy.png&quot; alt=&quot;A&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/E9wSu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/E9wSu.png&quot; alt=&quot;B&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/SlADl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SlADl.png&quot; alt=&quot;C&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/yEoqW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yEoqW.png&quot; alt=&quot;D&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/QGft0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QGft0.png&quot; alt=&quot;E&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/VREO9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VREO9.png&quot; alt=&quot;F&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/R1gGL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/R1gGL.png&quot; alt=&quot;G&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/vK9AI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vK9AI.png&quot; alt=&quot;H&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/Cft7i.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Cft7i.png&quot; alt=&quot;I&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/3oSC9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3oSC9.png&quot; alt=&quot;J&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/E72qu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/E72qu.png&quot; alt=&quot;K&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/p8yY1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/p8yY1.png&quot; alt=&quot;L&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/jqqKf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jqqKf.png&quot; alt=&quot;M&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/zKQB7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zKQB7.png&quot; alt=&quot;N&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/yTRBU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yTRBU.png&quot; alt=&quot;O&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/s78Fu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/s78Fu.png&quot; alt=&quot;P&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2016-08-31T15:34:10.693" Title="Lights in the Dungeon: Hooty and the Message" Tags="|cipher|story|visual|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23263" PostTypeId="2" ParentId="23255" CreationDate="2015-10-16T20:59:25.720" Score="13" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I ran a test with 6 coins, leaving 2 off each time. I got that to work. This made me think I could scale it up to 10 coins by doing 3 on each side. I tried that and got all the cases where 2 coins were different, but 3 coins would stop me (and I thought only the even sets would be a problem). That made me try a different number of coins in each weighing. That's when I found:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Weighing 1: 1 2 3 vs 4 5 6&lt;br&gt;&#xA; Weighing 2: 1 2 3 6 vs 7 8 9 10&lt;br&gt;&#xA; Weighing 3: 1 2 3 4 5 vs 6 7 8 9 10&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="375" LastActivityDate="2015-10-16T20:59:25.720" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23265" PostTypeId="1" AcceptedAnswerId="23527" CreationDate="2015-10-17T02:51:31.630" Score="14" ViewCount="961" Body="&lt;p&gt;In a small town, there exist a number of organizations of people. Each year, they send out a list of their members to everyone and this is the official way in which people know they are part of an organization. This usually goes well, however, this year things didn't go well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The town's postal service is notoriously bad, so the letters get a bit mixed up. In particular, each person receives the membership lists of every organization of which they are &lt;em&gt;not&lt;/em&gt; a part (and none of the lists of the organizations they &lt;em&gt;are&lt;/em&gt; a part of). Worse, since membership in an organization changes year to year, no one is sure which organizations they belong to (or even if they belong to any) nor how many organizations there are. Everyone receives the erroneous mailing the same morning (and this is common knowledge) and quickly discover the nature of the error (i.e. that &lt;em&gt;everyone&lt;/em&gt; received the mailing from all the organizations they were not a parto f).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The good news is that all the organizations hold a daily meeting at noon in the town center, visible to everyone. Moreover, everyone knows that there is at least one organization in the town. If anyone can deduce that they are definitely part of at least one organization, they will go to the town center at noon, and continue to do so daily. Anyone who is not sure will stay home to avoid embarrassment. If an entire organization's worth of people go to the town center, they will recognize this fact (by magic), and have a proper meeting. No one who stayed home can tell if a proper meeting has occurred, since meetings look just like a bunch of people standing around awkwardly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that all the actors will deduce anything logically possible, will there eventually be a proper meeting? Will everyone who goes to the town center eventually be part of a proper meeting?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For mathematical formulation, see below&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is a set $S$ of sets of people. Each person knows about every $s\in S$ of which they are not a member. It is common knowledge that $S$ is not empty. Each day, they all simultaneously either declare that they can prove that they are in some $s\in S$ or that they cannot prove this. An organization holds a proper meeting when every member can prove they are a member of some set. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4765" LastEditorUserId="4765" LastEditDate="2015-10-20T02:50:11.597" LastActivityDate="2015-10-26T10:29:47.173" Title="Mixed Up Memberships" Tags="|logical-deduction|blue-eyes|" AnswerCount="5" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="23266" PostTypeId="1" AcceptedAnswerId="23267" CreationDate="2015-10-17T03:34:45.700" Score="-12" ViewCount="656" Body="&lt;p&gt;A &lt;strong&gt;party&lt;/strong&gt; is a &lt;em&gt;group chat&lt;/em&gt;, the only difference being there's a &quot;&lt;strong&gt;leader&lt;/strong&gt;&quot; (&lt;em&gt;this is the person who started the chat&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In these parties, there can only be &lt;strong&gt;5 maximum users&lt;/strong&gt;, &lt;em&gt;this includes the leader&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to that, you may only be the leader of a &lt;strong&gt;single party&lt;/strong&gt; at any given moment in time. &lt;em&gt;You can be a member of as many parties which invite you.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if there were $n$ users on this site, and they equally spread out into as &lt;strong&gt;many parties as they could&lt;/strong&gt; (&lt;em&gt;e.g. Every user was in 5 parties&lt;/em&gt;), how many parties would coexist on the site at that moment?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2016-04-18T18:07:15.470" Title="Social networking &quot;party&quot; system" Tags="|mathematics|strategy|no-computers|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23267" PostTypeId="2" ParentId="23266" CreationDate="2015-10-17T03:55:19.457" Score="4" Body="&lt;p&gt;There would just be $n$ parties. Every party has five members, every person is in five parties. $5n/5=n$.&lt;/p&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2015-10-17T03:55:19.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23269" PostTypeId="1" AcceptedAnswerId="23271" CreationDate="2015-10-17T16:39:03.233" Score="5" ViewCount="618" Body="&lt;p&gt;Lionel Messi sentimentally remembers a small soccer tournament in which he participated many years ago. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There were four teams in the tournamnent: Atleticolino, Barcelino, Chicorrotico and Diminuto.&lt;/li&gt;&#xA;&lt;li&gt;Each team played exactly one match against each of the other teams.&#xA;For a win/draw/loss a team respectively scored 3/1/0 points.&lt;/li&gt;&#xA;&lt;li&gt;Chicorrotico scored twice as many goals as Atleticolino did.&lt;/li&gt;&#xA;&lt;li&gt;Barcelino conceded twice as many goals as Atleticolino did.&lt;/li&gt;&#xA;&lt;li&gt;At least one match ended in a boring draw without a single scored goal.&lt;/li&gt;&#xA;&lt;li&gt;The total number of goals scored in the tournament was at most 17.&lt;/li&gt;&#xA;&lt;li&gt;At the end of the tournament, Atleticolino had scored 7 points, Barcelino 4 points, Chicorrotico 3 points, and Diminuto was last with only 1 point.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What were the results of the six matches? How many goals did each team receive and score in every single match?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-10-17T17:11:31.327" LastActivityDate="2015-10-17T18:12:43.250" Title="Lionel Messi's soccer puzzle" Tags="|mathematics|calculation-puzzle|sports|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23270" PostTypeId="2" ParentId="23265" CreationDate="2015-10-17T17:44:44.787" Score="4" Body="&lt;p&gt;There will eventually be a proper meeting, but not necessarily of every organization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with imagining Person A, who does not get any letters. This tells Person A &quot;there is no organization of which I am &lt;strong&gt;not&lt;/strong&gt; a member.&quot; Person A goes to the town square on Day 1 and from then on. If there is only one organization, or every person is in every organization, this happens for everyone and the process ends on Day 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine Person B gets 1 letter. These means there are n+1 organizations, where n is 0 or more, and Person B is in n of them. Person B can't be sure that n isn't 0, so stays home.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next day, Person B can compare the names in their letter to the people who went to town. If there are people named in the letter who didn't go to town, Person B knows that those people must have received letters. This means there are at least two organizations, and therefore B is in one. B will now start going to town.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If all the people in B's letter went to town, then B will have to see if anyone not in the letter went to town. This too will indicate the existence of another letter, and therefore another organization, and send B to town from now on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However if the set of people who went to town exactly matches B's letter, there will be nothing to indicate that there is a second organization and B will stay home on day 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From then on, if anyone else suddenly starts going to town, B will realize there is another organization and start going to town.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider Person C who got two letters. This person only exists if there are two or more organizations. This person believes there are n+2 organizations and C is in n of them, but n might be 0. Person C monitors the people who go to town and compares them to the people on the two letters. Anyone named in a letter who didn't go to town on Day 1 must have got at least one letter. If they then go to town on Day 2, it's because their letter names someone who didn't go to town yet. It might be C, but C can't be sure unless everyone else has gone to town. However as with A, if anyone goes to town who isn't in B's letters, there is a third organization that person is in, and C must be in it too. This will send C to town from then on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As well, if someone is in both of C's letters, but didn't go to town on Day 1, they must have received a third letter. This will send C to town on Day 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a person will go to town when they know there is an organization they didn't get a letter for. This happens:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;right away if you didn't get a letter&lt;/li&gt;&#xA;&lt;li&gt;whenever someone goes to town who is &lt;strong&gt;not&lt;/strong&gt; named in any of your letters&lt;/li&gt;&#xA;&lt;li&gt;whenever someone named in &lt;strong&gt;all&lt;/strong&gt; of your letters &lt;strong&gt;doesn't&lt;/strong&gt; go to town&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is not guaranteed to bring everyone. For example consider Org 1 with ABCD and Org 2 with ABCDE. A, B, C, and D don't get a letter and go right away. E gets only the letter for Org 1, looks it over and says &quot;yup, that makes sense&quot; and doesn't go to town. However a proper meeting does get hold for Org 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying this to the example from a comment by the OP: consider organizations AC, AD, AE, ACDE, BC, BD, BE, BCDE.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A gets letters naming BC, BD, BE, BCDE&lt;/li&gt;&#xA;&lt;li&gt;B gets letters naming AC, AD, AE, ACDE&lt;/li&gt;&#xA;&lt;li&gt;C gets letters naming AD, AE, BD, BE&lt;/li&gt;&#xA;&lt;li&gt;D gets letters naming AC, AE, BC, BE&lt;/li&gt;&#xA;&lt;li&gt;E gets letters naming AC, AD, BC, BD&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;On Day 1 nobody goes to town.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On Day 2, A sees that B is named in all the A letters, but didn't go to town. Therefore B must have another letter naming A. From now on, A goes to town. B draws the same conclusions about A. C, D, and E can't be sure yet because nobody is named in all their letters. It's possible they aren't in any organizations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On Day 3, C sees that A and B went to town. But D and E didn't. This must mean that D and E have at least one letter each. Yet D and E are in each of the letters C holds. Therefore, there's another organization, and C must be in it. C goes to town from then on. By the same logic so do D and E.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for this example, all the organizations end up having proper meetings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the next example, A B C AB AC BC:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A gets letters naming B, C, and BC&lt;/li&gt;&#xA;&lt;li&gt;B gets letters naming A, C, and AC&lt;/li&gt;&#xA;&lt;li&gt;C gets letters naming A, B, and AB&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;On Day 1 nobody goes to town because they got letters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On Day 2, A thinks &quot;B should have gone to town because B is in an organization. So B must have got a letter. Perhaps it was this same C letter that I got.&quot; So A expects that B was expecting C to go to town on Day 1. Since C didn't go, A expects B to decide to go on Day 2, but A stays home since there's no reason to conclude A is in a group. The others follow the equivalent reasoning for their letters and stay home.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On Day 3, A thinks &quot;B should totally have gone to town yesterday if the only letter B got was this C letter I have here. So in addition to this letter naming C that I have, B must have some other letter as well, naming me (or possibly me and C). I had better go to town.&quot; And A goes to town. The same holds for the others, who go to town on Day 3 as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the OP is right that this adds another circumstance under which you go town. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Whenever someone who is not named in &lt;strong&gt;all&lt;/strong&gt; your letters (they've been dealt with in an earlier step) &lt;strong&gt;doesn't&lt;/strong&gt; go to town on a day when they should have, based on the letters you have seen in which the person isn't named.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1509" LastEditorUserId="1509" LastEditDate="2015-10-23T20:25:33.860" LastActivityDate="2015-10-23T20:25:33.860" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="23271" PostTypeId="2" ParentId="23269" CreationDate="2015-10-17T18:02:25.840" Score="6" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;br&gt;&#xA; A-B 2-1&lt;br&gt;&#xA; A-C 0-0&lt;br&gt;&#xA; A-D 1-0&lt;br&gt;&#xA; B-C 0-0&lt;br&gt;&#xA; B-D 1-0&lt;br&gt;&#xA; C-D 6-6  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;br&gt;&#xA; There's six games:&lt;br&gt;&#xA; A-B&lt;br&gt;&#xA; A-C&lt;br&gt;&#xA; A-D&lt;br&gt;&#xA; B-C&lt;br&gt;&#xA; B-D&lt;br&gt;&#xA; C-D&#xA; We know A has 2 wins and a draw, B has a win, a draw and a loss, D had a draw and two losses.&#xA;  If those 3 had a total of 3 wins, 3 draws and 3 losses, C must have had 3 draws (1 win and 2 losses would mean the number of wins and losses isn't equal)&lt;br&gt;&#xA; So A wins vs B, draws vs C and wins versus D / B loses vs A, draws vs c and wins vs D / C ties all.&lt;br&gt;&#xA; A-B 1-0&lt;br&gt;&#xA; A-C 0-0&lt;br&gt;&#xA; A-D 1-0&lt;br&gt;&#xA; B-C 0-0&lt;br&gt;&#xA; B-D 1-0&lt;br&gt;&#xA; C-D 0-0&lt;br&gt;&#xA; Those are the minimal scores to achieve the win results. Now as for the goal balances: C scored twice as much as A did. So C had a big score draw vs either B or D. Given there's no extra limitations on D, it's easiest to try first to just alter the score of the C vs D draw.&lt;br&gt;&#xA; A-B 1-0&lt;br&gt;&#xA; A-C 0-0&lt;br&gt;&#xA; A-D 1-0&lt;br&gt;&#xA; B-C 0-0&lt;br&gt;&#xA; B-D 1-0&lt;br&gt;&#xA; C-D 4-4&lt;br&gt;&#xA; So now C has scored 4, vs 2 for A.&lt;br&gt;&#xA; B conceded twice as many goals as A did. Without alterations, we'd be at one conceded vs none. If we change the results from the A-B game to 2-1, we end up at two conceded vs one.&lt;br&gt;&#xA; &lt;strong&gt;A-B 2-1&lt;/strong&gt;&lt;br&gt;&#xA; A-C 0-0&lt;br&gt;&#xA; A-D 1-0&lt;br&gt;&#xA; B-C 0-0&lt;br&gt;&#xA; B-D 1-0&lt;br&gt;&#xA; C-D 4-4&lt;br&gt;&#xA; Now to check all that's asked:&lt;br&gt;&#xA; A ends up with 7 points, B with 4, C with 3, D with 1.&lt;br&gt;&#xA; C scored twice as many as A =&gt; Not met. We need to retweak the C-D game.&lt;br&gt;&#xA; A-B 2-1&lt;br&gt;&#xA; A-C 0-0&lt;br&gt;&#xA; A-D 1-0&lt;br&gt;&#xA; B-C 0-0&lt;br&gt;&#xA; B-D 1-0&lt;br&gt;&#xA; &lt;strong&gt;C-D 6-6&lt;/strong&gt;&lt;br&gt;&#xA; Now to check all that's asked:&lt;br&gt;&#xA; A ends up with 7 points, B with 4, C with 3, D with 1.&lt;br&gt;&#xA; C scored twice as many as A =&gt; 6 vs 3&lt;br&gt;&#xA; B conceded twice as many as A =&gt; 2 vs 1&lt;br&gt;&#xA; No more then 17 goals =&gt; 17 goals exactly.&lt;br&gt;&#xA; QED.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2156" LastEditorUserId="2156" LastEditDate="2015-10-17T18:07:50.783" LastActivityDate="2015-10-17T18:07:50.783" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23273" PostTypeId="2" ParentId="23269" CreationDate="2015-10-17T18:12:43.250" Score="6" Body="&lt;p&gt;From the number of points, we can deduce that Atleticolino (A) won two games and drew one, Barcelino (B) won one, drew one, and lost one, and Diminuto (D) drew one game and lost three. Chicorrotico (C) could have one win and two losses, or three draws. However, the total number of draws must be even, so C drew all three games.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we know that C drew against A, B, and D, so A won against B and D, and B won against D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If A won against B, then B must have conceded at least one goal. Then A must also have conceded at least one goal. If A conceded a goal while winning two games and drawing one, then A must have scored at least three goals, so C must have scored at least six goals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the games that C played, C's opponents scored the same number of goals as C. There must have been at least one goal scored in each of the other three games. Therefore C could not have scored eight or more goals, or else there would be at least 19 total goals. So C scored exactly six goals, and A scored exactly three.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For A to score exactly three goals, they must have conceded exactly one goal, and won against B and D by a margin of exactly one goal. There are three cases, all of which produce a solution:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A conceded a goal to B. Then A went 2-1 against B, 0-0 against C, and 1-0 against D. B has conceded two goals to A, so they conceded no goals to C or D: B went 0-0 against C and 1-0 against D. Then C went 6-6 against D, with a total of 17 goals scored.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;pre&gt;A 2-1 B&#xA;A 0-0 C&#xA;A 1-0 D&#xA;B 0-0 C&#xA;B 1-0 D&#xA;C 6-6 D&lt;/pre&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A conceded a goal to C. Then A went 1-0 against B, 1-1 against C, and 1-0 against D. B conceded one goal to D (couldn't be to C, because it would be impossible to have a 0-0 draw). So B went 0-0 against C and 2-1 against D. C went 5-5 against D, with a total of 17 goals scored.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;pre&gt;A 1-0 B&#xA;A 1-1 C&#xA;A 1-0 D&#xA;B 0-0 C&#xA;B 2-1 D&#xA;C 5-5 D&lt;/pre&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A conceded a goal to D. Then A went 1-0 against B, 0-0 against C, and 2-1 against D. B conceded another goal against C (couldn't be to D, because there would end up being too many goals scored). So B went 1-1 against C and 1-0 against D. C went 5-5 against D, again with a total of 17 goals.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;pre&gt;A 1-0 B&#xA;A 0-0 C&#xA;A 2-1 D&#xA;B 1-1 C&#xA;B 1-0 D&#xA;C 5-5 D&lt;/pre&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-10-17T18:12:43.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23275" PostTypeId="1" AcceptedAnswerId="23295" CreationDate="2015-10-17T22:02:48.097" Score="15" ViewCount="780" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;2,&#xA;3,&#xA;5,&#xA;7,&#xA;...&#xA;11,&#xA;13,&#xA;17,&#xA;...&#xA;25,&#xA;29,&#xA;...&#xA;35,&#xA;...&#xA;43,&#xA;47,&#xA;49,&#xA;...&#xA;53,&#xA;59,&#xA;61,&#xA;65,&#xA;67,&#xA;...&#xA;71,&#xA;...&#xA;??&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The answer (indicated by two question marks) has two digits.&lt;/li&gt;&#xA;&lt;li&gt;When you see '...', it means that one or more numbers are deliberately concealed at that point.&lt;/li&gt;&#xA;&lt;li&gt;The question is purely mathematical and some arithmetical calculation is needed.&lt;/li&gt;&#xA;&lt;li&gt;The numbers always increase as you move from left to right. The mathematical term for this is that the sequence is monotonically increasing.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&lt;strong&gt;Hint 0&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;&#xA; A member submitted the following comment:&lt;br /&gt;&#xA;&#xA; &lt;em&gt;Under the assumption that 2 is the only even number in the sequence,&#xA; we can say that the first &amp;quot;...&amp;quot; is 9, the fifth is 51, and the sixth&#xA; is 69.&lt;/em&gt;&lt;br /&gt;&#xA;&#xA; That statement is false and, in order to solve the puzzle as intended, you must understand why.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Ask yourself why there are gaps. What am I hiding?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; It is an infinite sequence of integers, and 2 is the only even integer in the sequence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-10-19T10:19:23.157" Title="Can you generate the next number in this integer series and describe the rule?" Tags="|number-sequence|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23276" PostTypeId="2" ParentId="23275" CreationDate="2015-10-18T05:32:19.977" Score="3" Body="&lt;p&gt;The sequence is the union of the primes, the odd squares, and the multiples of 5 that are neither divisible by 3 nor 11, in increasing order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So filling in the blanks and continuing the sequence,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2 3 5 7 .... 11 13 17 ........ 25 29 .. 35 ........ 43 47 49 .. 53 59 61 65 67 .. 71 .. ???&#xA;&#xA;2 3 5 7 9 10 11 13 17 19 20 23 25 29 31 35 37 40 41 43 47 49 50 53 59 61 65 67 70 71 73 ...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="11100" LastActivityDate="2015-10-18T05:32:19.977" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="23277" PostTypeId="1" CreationDate="2015-10-18T08:37:59.960" Score="3" ViewCount="354" Body="&lt;p&gt;When setting a number sequence puzzle, one normally gives a finite sample and asks for the rule that specifies the members. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sample has to be finite because it's impossible to uniquely specify an infinite series without giving away the answer. (Note: Is this true?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course the problem is that any finite string of numbers can have an infinite number of rules that specify it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the next number in this series?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 3 4 7 9 ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To which someone could say: &lt;em&gt;These are members of the set {2 3 4 7 9 73} so the answer is 73.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This is the union of the sets {2, 3, 4, 7, 9} and {10, 11, 12, 13, ...} so the answer is 10.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is taking things to an extreme but how is it possible to counter such suggestions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The real-life example that caused me to pose the question is here &lt;a href=&quot;https://puzzling.stackexchange.com/questions/23275/can-you-generate-the-next-number-in-this-series-and-describe-the-rule&quot;&gt;Can you generate the next number in this integer series and describe the rule?&lt;/a&gt;  and the first answer proposed was some arbitrary union of sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any recognised ways, or can you suggest a way, of designing number sequence puzzles that can tie down the generating rule exactly without actually specifying the answer? If not, is the number sequence puzzle genre dead?&lt;/p&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-10-18T17:25:34.007" Title="When setting number-sequence puzzles, is there any accepted method to prevent arbitrary solutions?" Tags="|number-sequence|puzzle-creation|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23279" PostTypeId="1" CreationDate="2015-10-18T12:35:12.663" Score="13" ViewCount="1234" Body="&lt;p&gt;I met a puzzle, which is basically well known &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16/100-prisoners-names-in-boxes&quot;&gt;100 Prisoners&amp;#39; Names in Boxes&lt;/a&gt; , but... it has an algorithm, which gives prisoners 100% chance to survive!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is achieved by a small change in the procedure:&lt;br&gt;&#xA;&lt;em&gt;Prisoners can invite a friend, who visit the room before the first prisoner and is allowed to check out all boxes and switch places of two names in them.&lt;/em&gt;&lt;br&gt;&#xA;That's it. The prisoners and the friend have a chance to plot their strategy in advance, but no communication is allowed after beginning of the procedure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How 100% survival can be reached here is quite obvious for those who read the linked topic, but anyway here is it:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The numbers from 1 to 100 is given to each prisoner (you can read &quot;number&quot; as a &quot;name&quot;) and to each box. The friend must find all chains of numbers like &quot;Box A has number B -&gt; Box B has number C -&gt; ... -&gt; box X has number A&quot;. There can be only one chain of length more than 50. If the friend finds it he must switch places of two distant numbers in the chain. In that case the chain will be divided into two chains of length less than 50.&lt;br&gt;&#xA; Then all prisoners just must start from the box with their number and follow the chain until they reaches the box, which contains they number. Since there won't be any chain longer than 50 all of them will succeed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As you can see, we got quite nice puzzle on it's own, which can be given even to a child, who can't do complicated probability calculations.&lt;br&gt;&#xA;So I started to wonder - &lt;strong&gt;What are other possible changes in the initial &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16/100-prisoners-names-in-boxes&quot;&gt;100 Prisoners&amp;#39; Names in Boxes&lt;/a&gt; procedure you can make that similarly lead to 100% chance of survival?&lt;/strong&gt; I mean really small changes, hopefully even smaller than the given example.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2018-08-24T08:24:27.283" Title="Let's deal with &quot;Names in Boxes&quot; Safely" Tags="|logical-deduction|puzzle-creation|algorithm|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23280" PostTypeId="2" ParentId="1963" CreationDate="2015-10-18T13:20:06.047" Score="-1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$2+2+2$$ &lt;br&gt; $$3\times3-3$$ &lt;br&gt; $$\sqrt{4}+\sqrt{4}+\sqrt{4}$$ &lt;br&gt; $$\frac{5}{5}+5$$ &lt;br&gt;  $$6\times\frac{6}{6}$$ &lt;br&gt;  $$7-\frac{7}{7}$$ &lt;br&gt; $$\sqrt[3]{8} +\sqrt[3]{8} +\sqrt[3]{8}$$ &lt;br&gt;  $$\sqrt{9}\times\sqrt{9}-\sqrt{9}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16994" LastEditorUserId="16322" LastEditDate="2016-01-09T18:16:44.883" LastActivityDate="2016-01-09T18:16:44.883" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23281" PostTypeId="1" AcceptedAnswerId="23284" CreationDate="2015-10-18T14:01:37.440" Score="-2" ViewCount="912" Body="&lt;p&gt;I woke up this morning with a bit of amnesia, so I need help identifying where I'm at. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the last 10 minutes I saw a pyramid,  the Eiffel Tower, the Statue of Liberty and a pirate ship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: _ _ _   _ _ _ _ _&lt;/p&gt;&#xA;" OwnerUserId="16678" LastEditorUserId="11876" LastEditDate="2015-10-18T14:04:16.390" LastActivityDate="2015-10-20T15:58:16.893" Title="Riddle: Where am I at?" Tags="|riddle|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23284" PostTypeId="2" ParentId="23281" CreationDate="2015-10-18T15:14:58.410" Score="3" Body="&lt;p&gt;Unfortunately there is no pirate ship in &lt;a href=&quot;http://www.minimundus.at/en/interactive/picture-archive.html&quot; rel=&quot;nofollow&quot;&gt;Minimundus&lt;/a&gt; so you must be in Las Vegas.&lt;/p&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2015-10-18T18:38:05.943" LastActivityDate="2015-10-18T18:38:05.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23285" PostTypeId="1" AcceptedAnswerId="23286" CreationDate="2015-10-18T15:38:50.287" Score="8" ViewCount="432" Body="&lt;p&gt;A big cube is cut into 99 smaller cubes. &#xA;Exactly 98 of these 99 smaller cubes are unit cubes.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What is the volume of the big cube?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-10-21T18:31:54.270" Title="A big cube and 99 smaller cubes" Tags="|mathematics|geometry|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23286" PostTypeId="2" ParentId="23285" CreationDate="2015-10-18T15:50:26.980" Score="13" Body="&lt;p&gt;Hm...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $k$ denote the side length of the non-unit cube. We know $k&amp;gt;1$ because if it weren't, then there couldn't be a cube, and for the same reason we know $k\in\mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $n$ denote the length of the full cube. This means $n^3-k^3=98$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That factors into $(n-k)(n^2+nk+k^2)=98$; the only factors of 98 that produce an integer value for $n$ and $k$ are 2 and 49. This makes $n=5$&#xA;and $k=3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the volume of the full cube is 125 cubic units.&lt;/p&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="11876" LastEditDate="2015-10-21T18:31:54.270" LastActivityDate="2015-10-21T18:31:54.270" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23287" PostTypeId="2" ParentId="23277" CreationDate="2015-10-18T17:16:49.573" Score="4" Body="&lt;p&gt;1.&lt;em&gt;&quot;Are there any recognised ways, or can you suggest a way, of designing number sequence puzzles that can tie down the generating rule exactly without actually specifying the answer?&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, there is not. The prove to this statement is in your answer:&lt;br&gt;&#xA; Any statement like &quot;These are members of the set {2 3 4 7 9 73} so the answer is 73.&quot; is an appropriate answer, until you write specifically in your problem conditions ether &quot;73 is not correct answer&quot; or &quot;f(i) is correct answer, where f is function defined as ...&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.&quot;&lt;em&gt;If not, is the number sequence puzzle genre dead?&lt;/em&gt;&quot;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure that I understand this question exactly, but number sequence puzzle genre is not dead! But not because those puzzle can be made correct and objective, but because puzzles are created for people and all people love subjective things! By definition of &quot;love&quot; :).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.The really good sequence puzzles have:&lt;br&gt;&#xA;a) a very short and easy to understand pattern.&#xA;b) are tested (with many people) to be sure that all other appropriate patterns are quite bulky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example Fibonachi sequence: 1 1 2 3 5 8 13. The pattern a_{i+2} = a_{i+1} + a_{i} is short and easy to remember and understand, on the other hand the pattern &quot;it is part of sequence 1 1 2 3 5 8 13 75 87 98&quot; is not short and very hard to remember.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4.&lt;em&gt;&quot;The sample has to be finite because it's impossible to uniquely specify an infinite series without giving away the answer. (Note: Is this true?)&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see no reason why this should be true. &quot;Giving away&quot; means &quot;to make obvious&quot;, but concept of obvious depends on mans mind and not only subjective, but includes some finite things by it's own, for example if you write down infinite number of objects in a row, a man still can read only finite number of them during his life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also consider this example, which took me 30 sec to come up with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;04, 087, 03484, 091276, 15996464&lt;br&gt;&#xA;122540867,  10960141044, 2828916977616&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would imagine that this sequence is similarly hard to crack with 5 and with 8 (20,100) samples. Isn't it? This is the case simply because the complexity of samples after 5th is just too big to be of any use if you have no idea about the pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to know, the pattern is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Take the following sequence: a_i = 1024^i and exclude from each number digits with odd position in the number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2015-10-18T17:25:34.007" LastActivityDate="2015-10-18T17:25:34.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23288" PostTypeId="1" AcceptedAnswerId="23289" CreationDate="2015-10-18T18:57:20.317" Score="8" ViewCount="570" Body="&lt;p&gt;$$&#xA;\small&#xA;\begin{array}{c}&#xA;1\\ &#xA;1 &amp;amp; 2\\&#xA;1 &amp;amp; 3 &amp;amp; 2\\ &#xA;1 &amp;amp; 3 &amp;amp; 2 &amp;amp; 4\\ &#xA;1 &amp;amp; 5 &amp;amp; 2 &amp;amp; 4 &amp;amp; 3\\ &#xA;1 &amp;amp; 4 &amp;amp; 2 &amp;amp; 6 &amp;amp; 3 &amp;amp; 5\\ &#xA;1 &amp;amp; 6 &amp;amp; 2 &amp;amp; 5 &amp;amp; 3 &amp;amp; 7 &amp;amp; 4\\ &#xA;1 &amp;amp; 5 &amp;amp; 2 &amp;amp; 7 &amp;amp; 3 &amp;amp; 6 &amp;amp; 4 &amp;amp; 8\\ &#xA;1 &amp;amp; 9 &amp;amp; 2 &amp;amp; 6 &amp;amp; 3 &amp;amp; 8 &amp;amp; 4 &amp;amp; 7 &amp;amp; 5\\ &#xA;1 &amp;amp; 6 &amp;amp; 2 &amp;amp; 10 &amp;amp; 3 &amp;amp; 7 &amp;amp; 4 &amp;amp; 9 &amp;amp; 5 &amp;amp; 8\\ &#xA;1 &amp;amp; 9 &amp;amp; 2 &amp;amp; 7 &amp;amp; 3 &amp;amp; 11 &amp;amp; 4 &amp;amp; 8 &amp;amp; 5 &amp;amp; 10 &amp;amp; 6\\ &#xA;1 &amp;amp; 7 &amp;amp; 2 &amp;amp; 10 &amp;amp; 3 &amp;amp; 8 &amp;amp; 4 &amp;amp; 12 &amp;amp; 5 &amp;amp; 9 &amp;amp; 6 &amp;amp; 11\\ &#xA;1 &amp;amp; 12 &amp;amp; 2 &amp;amp; 8 &amp;amp; 3 &amp;amp; 11 &amp;amp; 4 &amp;amp; 9 &amp;amp; 5 &amp;amp; 13 &amp;amp; 6 &amp;amp; 10 &amp;amp; 7\\ &#xA;1 &amp;amp; 8 &amp;amp; 2 &amp;amp; 13 &amp;amp; 3 &amp;amp; 9 &amp;amp; 4 &amp;amp; 12 &amp;amp; 5 &amp;amp; 10 &amp;amp; 6 &amp;amp; 14 &amp;amp; 7 &amp;amp; 11\\ &#xA;1 &amp;amp; 12 &amp;amp; 2 &amp;amp; 9 &amp;amp; 3 &amp;amp; 14 &amp;amp; 4 &amp;amp; 10 &amp;amp; 5 &amp;amp; 13 &amp;amp; 6 &amp;amp; 11 &amp;amp; 7 &amp;amp; 15 &amp;amp; 8\\ &#xA;1 &amp;amp; 9 &amp;amp; 2 &amp;amp; 13 &amp;amp; 3 &amp;amp; 10 &amp;amp; 4 &amp;amp; 15 &amp;amp; 5 &amp;amp; 11 &amp;amp; 6 &amp;amp; 14 &amp;amp; 7 &amp;amp; 12 &amp;amp; 8 &amp;amp; 16\\ &#xA;1 &amp;amp; 17 &amp;amp; 2 &amp;amp; 10 &amp;amp; 3 &amp;amp; 14 &amp;amp; 4 &amp;amp; 11 &amp;amp; 5 &amp;amp; 16 &amp;amp; 6 &amp;amp; 12 &amp;amp; 7 &amp;amp; 15 &amp;amp; 8 &amp;amp; 13 &amp;amp; 9\\&#xA;1 &amp;amp; 10 &amp;amp; 2 &amp;amp; 18 &amp;amp; 3 &amp;amp; 11 &amp;amp; 4 &amp;amp; 15 &amp;amp; 5 &amp;amp; 12 &amp;amp; 6 &amp;amp; 17 &amp;amp; 7 &amp;amp; 13 &amp;amp; 8 &amp;amp; 16 &amp;amp; 9 &amp;amp; 14\\&#xA;1 &amp;amp; 15 &amp;amp; 2 &amp;amp; 11 &amp;amp; 3 &amp;amp; 19 &amp;amp; 4 &amp;amp; 12 &amp;amp; 5 &amp;amp; 16 &amp;amp; 6 &amp;amp; 13 &amp;amp; 7 &amp;amp; 18 &amp;amp; 8 &amp;amp; 14 &amp;amp; 9 &amp;amp; 17 &amp;amp; 10\\&#xA;1 &amp;amp; 11 &amp;amp; 2 &amp;amp; 16 &amp;amp; 3 &amp;amp; 12 &amp;amp; 4 &amp;amp; 20 &amp;amp; 5 &amp;amp; 13 &amp;amp; 6 &amp;amp; 17 &amp;amp; 7 &amp;amp; 14 &amp;amp; 8 &amp;amp; 19 &amp;amp; 9 &amp;amp; 15 &amp;amp; 10 &amp;amp; 18&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the next line in this number triangle sequence?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the $100$th line?&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="2819" LastEditDate="2015-10-19T09:29:45.513" LastActivityDate="2016-03-14T06:32:48.233" Title="Next line in triangular sequence" Tags="|number-sequence|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23289" PostTypeId="2" ParentId="23288" CreationDate="2015-10-18T19:49:33.847" Score="7" Body="&lt;p&gt;The $n$th line represents the order of elimination in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Josephus_problem&quot;&gt;Josephus problem&lt;/a&gt; with $n$ people and $k=2$. In other words, the numbers are written in order, skipping one spot every time and wrapping around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next line is &lt;code&gt;1,19,2,12,3,17,4,13,5,21,6,14,7,18,8,15,9,20,10,16,11&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $100$th line is &lt;code&gt;1,51,2,76,3,52,4,99,5,53,6,77,7,54,8,89,9,55,10,78,11,56,12,95,13,57,14,79,15,58,16,90,17,59,18,80,19,60,20,98,21,61,22,81,23,62,24,91,25,63,26,82,27,64,28,96,29,65,30,83,31,66,32,92,33,67,34,84,35,68,36,100,37,69,38,85,39,70,40,93,41,71,42,86,43,72,44,97,45,73,46,87,47,74,48,94,49,75,50,88&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-10-18T19:49:33.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23290" PostTypeId="1" AcceptedAnswerId="23296" CreationDate="2015-10-18T23:25:05.940" Score="7" ViewCount="766" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GWEOX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GWEOX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ACROSS      &#xA;1   God no longer (4,2)&#xA;#4  Mode of transport for an electron density of 1001 (9)&#xA;#6  No hit out (4)&#xA;7   Jill's brother is a stranger (3,3'1,4,4)&#xA;12  Wasn't Clyde Narain a mammal? (7,3)&#xA;14  Insect killer above the red square (4,4,6)&#xA;15  Helminthiasis (5)&#xA;16  Found in my stomach (4)&#xA;17  He keeps punching the pavement (6,7)&#xA;19  How to get into the hairless Sumerian city (6'1,4)&#xA;20  A rubber dog's toy shaped like a mule (6,4)&#xA;21  Excavator aroused abnormally (5,5)&#xA;23  Last dream (5,7)&#xA;27  What a smell! (4)&#xA;29  Iranian heir (6,2,6)&#xA;31  Gracelands former occupant endlessly messed up (8,4)&#xA;36  Young man with a preference for parchment (8)&#xA;38  It's the bola I'd tangled up (6)&#xA;41  Lucifer Lacrimosa (5,3,3)&#xA;43  Grey (5,3,5)&#xA;45  Proxemics offenders (5,8)&#xA;46  The grassy knoll won't tell (6,4)&#xA;48  Bad mood (4)&#xA;49  It's not doubling time (4-4)&#xA;50  Harry Potter's canine omen buys tickets in the US (4,8)&#xA;51  Seen in his naked form (5)&#xA;52  Aves aggrieved (5,5)&#xA;&#xA;DOWN    &#xA;1   John Wayne takes extreme action (4,5)&#xA;2   Mr Astley takes chances (4,9)&#xA;#3  Over-enthused about icing a cake (9)&#xA;5   Overlooked planets (9,6)&#xA;8   Phoning about your shift (4,2,4)&#xA;9   Watery brontide (5,7)&#xA;10  Tents messed up by a Sumerian goddess (6)&#xA;#11 Attempts an alternative to coffee, initially confused (6)&#xA;13  It sounds like hay for a priest (6,6)&#xA;18  Podiatrist's problem with a count (6,3,4)&#xA;22  James Annelida (9,3)&#xA;24  They play for the director of a Life of Pi (7)&#xA;25  £1000 robbery without thinking (5,5,4)&#xA;26  You can't have it, that brush (11)&#xA;28  Bill is ready to fly (3,3,5)&#xA;30  A tripping skydiver does it (7)&#xA;32  It looks like Mr Hanks does hair (4,6)&#xA;33  Instructs the avian limb (4,9)&#xA;34  The french leave a cricketer (4,6)&#xA;35  Tangled ropes (5)&#xA;37  Too many sweets in your pocket (5,5)&#xA;39  Smaug + Tiamat (6,6)&#xA;40  It goes round your pebble (4,4)&#xA;42  Outcome of a religious service (4,6)&#xA;44  Simon Cowell (6)&#xA;47  Torn, broken (4)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Hopefully there aren't too many here that people find obscure or ambiguous. If people are struggling, I might add release dates.&lt;/p&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="14135" LastEditDate="2015-10-21T22:34:17.100" LastActivityDate="2015-10-23T06:43:38.947" Title="Video Games Cryptic Crossword" Tags="|cryptic-crosswords|video-games|" AnswerCount="6" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23291" PostTypeId="2" ParentId="23288" CreationDate="2015-10-19T02:54:29.560" Score="5" Body="&lt;p&gt;For &lt;em&gt;nth&lt;/em&gt; line, you place &lt;em&gt;n&lt;/em&gt; empty slots in a row. Start filling numbers sequentially starting from 1 into the slots by skipping an empty slot every time you place a number. Whenever you reach the end of the line, you wrap around and keep filling the empty slots keeping the above rule in mind. This will the give next line as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1, 19, 2, 12, 3, 17, 4, 13, 5, 21, 6, 14, 7, 18, 8, 15, 9, 20, 10, 16, 11&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the &lt;em&gt;100th&lt;/em&gt; line would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1, 51, 2, 76, 3, 52, 4, 99, 5, 53, 6, 77, 7, 54, 8, 89, 9, 55, 10, 78, 11, 56, 12, 95, 13, 57, 14, 79, 15, 58, 16, 90, 17, 59, 18, 80, 19, 60, 20, 98, 21, 61, 22, 81, 23, 62, 24, 91, 25, 63, 26, 82, 27, 64, 28, 96, 29, 65, 30, 83, 31, 66, 32, 92, 33, 67, 34, 84, 35, 68, 36, 100, 37, 69, 38, 85, 39, 70, 40, 93, 41, 71, 42, 86, 43, 72, 44, 97, 45, 73, 46, 87, 47, 74, 48, 94, 49, 75, 50, 88&lt;/p&gt;&#xA;" OwnerUserId="12537" LastEditorUserId="12537" LastEditDate="2015-10-19T02:59:32.530" LastActivityDate="2015-10-19T02:59:32.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23292" PostTypeId="1" CreationDate="2015-10-19T07:47:01.207" Score="6" ViewCount="788" Body="&lt;p&gt;This puzzles continues &#xA;&quot;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/23255/the-ebbozonian-coin-weighing-puzzle&quot;&gt;The Ebbozonian coin weighing puzzle&lt;/a&gt;&quot; from last week.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In the country of Ebbozonia, there are only two type of coins: light coins and heavy coins. The weights of these coins satisfy the following properties:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;All light coins have the same weight $L$.&lt;/li&gt;&#xA;&lt;li&gt;All heavy coins have the same weight $H$.&lt;/li&gt;&#xA;&lt;li&gt;Heavy coins are heavier than light coins: $H&amp;gt;L$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The precise values of $L$ and $H$ are not known to the public. &#xA;The difference between $L$ and $H$ should be fairly small, as there is no way of distinguishing the two coin types without a balance. &#xA;His Highness, the Honourable Minister of Treasury, has recently announced the following important property of the Ebbozonian coin system:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If a balance is in perfect equilibrium with $h_1$ heavy and $\ell_1$ light coins on the right pan and with $h_2$ heavy and $\ell_2$ light coins on the left pan, then $h_1=h_2$ and $\ell_1=\ell_2$ must necessarily hold true.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Cosmo puts &lt;strong&gt;30&lt;/strong&gt; Ebbozonian coins on the table and asks Fredo to determine quickly whether these &lt;strong&gt;30&lt;/strong&gt; coins all have the same weight. &#xA;On the table, there is a balance with two pans (but there are no weights).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Can Fredo solve Cosmo's problem by using the balance at most &lt;strong&gt;four&lt;/strong&gt; times?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2016-12-09T15:12:40.367" Title="Another Ebbozonian coin weighing puzzle" Tags="|mathematics|strategy|weighing|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23293" PostTypeId="2" ParentId="23290" CreationDate="2015-10-19T07:51:11.417" Score="4" Body="&lt;p&gt;Across:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: God no longer (4,2)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Deus Ex&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;41: Lucifer Lacrimosa (5,3,3)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Devil May Cry&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;50: Harry Potter's canine omen buys tickets in the US (4,8)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Grim Fandango&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16608" LastEditorUserId="16608" LastEditDate="2015-10-19T08:40:47.947" LastActivityDate="2015-10-19T08:40:47.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23294" PostTypeId="1" CreationDate="2015-10-19T07:56:27.820" Score="11" ViewCount="426" Body="&lt;p&gt;Let us suppose that we have an $m\times n$ grid such that every element is a digit in base $10$. Then we can read the numbers from a grid such that we fix a starting element and go to one of the eight nearest grid and maintain that direction. Or, to be precisely, you don't have to go to that direction any steps so you can also read the numbers $1, 4,9$. How small value $mn$ can be such that one can read the numbers $1^2,2^2,3^2,\ldots,100^2$ from the grid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example if the grid is&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;00182&#xA;36495&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;then one can read the squares $1^2,\ldots,10^2$ from it but for example $11^2=121$ is missing. In this rectange $mn=2\cdot 5=10$.&lt;/p&gt;&#xA;" OwnerUserId="16999" LastEditorUserId="4269" LastEditDate="2015-10-19T09:37:44.713" LastActivityDate="2015-10-19T22:31:05.640" Title="What is the smallest rectangle containing the squares of 1 through 100?" Tags="|mathematics|combinatorics|optimization|logic-grid|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23295" PostTypeId="2" ParentId="23275" CreationDate="2015-10-19T08:23:12.930" Score="15" Body="&lt;p&gt;The list contains:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the prime numbers in hexadecimal representation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The list then starts as: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;2, 3, 5, 7,&lt;/strong&gt; B, D, &lt;strong&gt;11, 13, 17,&lt;/strong&gt; 1D, 1F, &lt;strong&gt;25, 29,&lt;/strong&gt; 2B, 2F, &lt;strong&gt;35,&lt;/strong&gt; 3B, 3D, &lt;strong&gt;43, 47, 49,&lt;/strong&gt; 4F,&lt;br&gt;&#xA; &lt;strong&gt;53, 59, 61, 65, 67,&lt;/strong&gt; 6B, 6D, &lt;strong&gt;71,&lt;/strong&gt; 7F, &lt;strong&gt;83, 89,&lt;/strong&gt; 8B, &lt;strong&gt;95, 97,&lt;/strong&gt; 9D, A3, $\ldots,$ $\ldots,$ $\ldots$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The answer to the puzzle is &lt;strong&gt;83&lt;/strong&gt;, the next term in the sequence given in the problem statement whose hexadecimal representation consists of two decimal digits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="14177" LastEditDate="2015-10-19T09:21:49.703" LastActivityDate="2015-10-19T09:21:49.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23296" PostTypeId="2" ParentId="23290" CreationDate="2015-10-19T08:40:34.357" Score="9" Body="&lt;p&gt;I think this is now a complete solution! &#xA;This was only possible due to the partial solutions from @Gabriel Guimarães, @Squeamish Ossifrage, @Hazel, @The Dark Truth and @Irishpanda, whose answers I copied in, and a comment by @Zandar. Of course, being a crossword, the partial solutions also gave at times prompts of additional letters.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Across&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1   God no longer (4,2)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Deus Ex. Found by @The Dark Truth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4  Mode of transport for an electron density of 1001 (9)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Minecraft. MI is Roman numeral for 1001. n_e is the electon density in semiconductor physics. A craft can be a ship or spaceship.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6 No hit out (4)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Sims (Anagram of Miss). Found in comment by @Zandar&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;7   Jill's brother is a stranger (3,3'1,4,4)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You don't know Jack &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;12  Wasn't Clyde Narain a mammal? (7,3)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chuckie egg (Clyde Narain is DJ Chuckie: &lt;a href=&quot;https://en.wikipedia.org/wiki/Chuckie_(DJ)&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Chuckie_(DJ)&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;14  Insect killer above the red square (4,4,6)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Raid over Moscow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;15  Helminthiasis (5)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Worms&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;16 Found in my stomach (4)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Myst (&lt;strong&gt;my st&lt;/strong&gt;omach). Found by @Gabriel Guimarães&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;17 He keeps punching the pavement (6,7)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Street Fighter. Found by @Irishpanda&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;19 How to get into the hairless Sumerian city (6'1, 4)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Baldur's Gate (Bald + Ur, gate). Found by @Gabriel Guimarães&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;20  A rubber dog's toy shaped like a mule (6,4)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Donkey Kong&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;21  Excavator aroused abnormally (5,5)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Manic Miner&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;23  Last dream (5,7)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Final Fantasy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;27  What a smell! (4)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pong&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;29  Iranian heir (6,2,6)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Prince of Persia&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;31  Gracelands former occupant endlessly messed up (8,4)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Resident Evil (Elvis without an &quot;s&quot; anagrammed)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;36  Young man with a preference for parchment (8)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Paperboy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;38  It's the bola I'd tangled up (6)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Diablo (anagram &lt;em&gt;bola I'd&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;41: Lucifer Lacrimosa (5,3,3)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Devil may cry. Found by @The Dark Truth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;43  Grey (5,3,5)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Black and white (if you mix the paint)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;45  Proxemics offenders (5,8)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Space Invaders (&lt;a href=&quot;https://en.wikipedia.org/wiki/Proxemics&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Proxemics&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;46  The grassy knoll won't tell (6,4)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Silent Hill&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;48  Bad mood (4)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Doom (anagram of mood)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;49  It's not doubling time (4-4)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Half-Life&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;50  Harry Potter's canine omen buys tickets in the US (4,8)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Grim Fandango. Found by @The Dark Truth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;51  Seen in his naked form (5)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Snake (hidden in hi&lt;strong&gt;s nake&lt;/strong&gt;d form&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;52: Aves aggrieved (5,5)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Angry Birds. Found by @squeamish ossifrage&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;DOWN&lt;br&gt;&#xA;1   John Wayne takes extreme action (4,5)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Duke Nukem (J.W's nickname is &quot;the Duke&quot;) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2   Mr Astley takes chances (4,9)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rick Dangerous&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3  Over-enthused about icing a cake (9)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pipemania (you can pipe icing onto a cake)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5   Overlooked planets (9,6)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Forgotten Worlds&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;8   Phoning about your shift (4,2,4)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Call of Duty&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;9 Watery brontide (5,7)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hydro Thunder. Found by @Irishpanda&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;10  Tents messed up by a Sumerian goddess (6)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pacman (camp anagrammed with An (sumerian goddess))&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;11 Attempts an alternative to coffee, initially confused (6)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tetris (&lt;em&gt;Tries&lt;/em&gt; + initial letter of &lt;em&gt;Tea&lt;/em&gt; anagrammed)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;13  It sounds like hay for a priest (6,6)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cannon Fodder&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;18 Podiatrist's problem with a count (6,3,4)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Toejam and Earl&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;22  James Annelida (9,3)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Earthworm Jim&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;24 They play for the director of a Life of Pi (7)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Angband. (Ang Lee is the director). Found by @Hazel&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;25  £1000 robbery without thinking (5,5,4)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Grand Theft Auto (a Grand is slang for 1000£)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;26  You can't have it, that brush (11)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Minesweeper&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;28 Bill is ready to fly (3, 3, 5)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jet Set Willy. Found by @Gabriel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;30  A tripping skydiver does it (7)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fallout (of the plane)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;32 It looks like Mr Hanks does hair (4,6)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tomb Raider (Tom Braider). Found by @sqeamish ossifrage&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;33  Instructs the avian limb (4,9)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wing Commander&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;34  The french leave a cricketer (4,6)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lego Batman (&lt;strong&gt;Le&lt;/strong&gt; is &quot;The&quot; in French, &lt;strong&gt;go&lt;/strong&gt; is to leave, and a cricketer is a bat(s)man&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;35  Tangled ropes (5)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Spore (anagram of ropes)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;37 Too many sweets in your pocket (5,5)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Candy Crush. Found by @squeamish ossifrage&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;39  Smaug + Tiamat (6,6)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Double Dragon (Smaug is a dragon in &quot;The Hobbit&quot;, Tiamat is a goddess in the form of a dragon: &lt;a href=&quot;https://en.wikipedia.org/wiki/Tiamat&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Tiamat&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;40 It goes round your pebble (4,4)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rock Band. Found by @Hazel&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;42  Outcome of a religious service (4,6)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mass Effect&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;44  Simon Cowell (6)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hitman.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;47  Torn, broken (4)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tron (Anagram)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7316" LastEditorUserId="7316" LastEditDate="2015-10-23T06:43:38.947" LastActivityDate="2015-10-23T06:43:38.947" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="23298" PostTypeId="2" ParentId="23290" CreationDate="2015-10-19T09:19:50.313" Score="4" Body="&lt;p&gt;A few more...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ACROSS:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;52: Aves aggrieved (5,5)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Angry &lt;a href=&quot;http://dictionary.reference.com/browse/aves&quot; rel=&quot;nofollow&quot;&gt;Birds&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;DOWN:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;32: It looks like Mr Hanks does hair (4,6)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TOMB RAIDER (Tom, braider)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;37: Too many sweets in your pocket (5,5)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Candy Crush&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="40" LastEditorUserId="16608" LastEditDate="2015-10-19T09:26:29.113" LastActivityDate="2015-10-19T09:26:29.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23301" PostTypeId="2" ParentId="23290" CreationDate="2015-10-19T11:16:37.043" Score="4" Body="&lt;p&gt;Some more:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Across:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;17 Found in my stomach (4)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Myst (&lt;strong&gt;my st&lt;/strong&gt;omach)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;19 How to get into the hairless Sumerian city (6'1, 4)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Baldur's Gate (Bald + Ur, gate)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Down:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;28 Bill is ready to fly (3, 3, 5)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jet Set Willy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16998" LastActivityDate="2015-10-19T11:16:37.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23303" PostTypeId="2" ParentId="17886" CreationDate="2015-10-19T11:21:51.520" Score="-1" Body="&lt;p&gt;I don't know the answer to the OP's question and doubt anyone knows, but I find the question about &quot;any permutation&quot; interesting and may be it can help as a first step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;So the task is to put 6x9 coloured objects, 9 of each colour into 6 distinguishable faces, each face has 9 distinguishable cells, in a way that each face contains each colour.&lt;/em&gt; &#xA;And we don't distinguish two ways, which are differ by interchange of objects of the same colour between faces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got the result of ~= 2*10^27.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, this number is quite bigger than number of any solvable positions of the cube: ~=4*10^19.&#xA;This is because we consider not only solvable+unsolvable positions of the cube, but any placement of colour on it's faces (so some of our permutations can't be reached even by disassembling the cube).&lt;br&gt;&#xA;But anyway I find it nice that one can find this number. For example, when I tried to find number of &quot;any combinations&quot; (ignoring positions of colours within each face) I failed to do this (this is doable, but would take several hours of quite stupid work for considering different possibilities).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step would be taking into account that different cells are dependent on each other because of technical properties of the Rubik's cube. May be instead of considering faces one should consider rotations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the prove for the given number:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each face must contain one object of each colour, independently of the permutation. That gives 6x6=36 objects in total.  Each of 6 objects can be puted onto a face in P(9,6) = 9!/6! ways. This gives P(9,6)^6 permutations for those 36 objects.&lt;br&gt;&#xA;&lt;br&gt;&#xA; For each of those permutations we have some permutation of the rest 6x3=18 objects. They can be distributed between 6x3 = 18 cells arbitrary. Using Permutations with Repetition formula we get here 18!/(3!)^6 permutations.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Therefore, in total we have (9!/6!)^6*18!/(3!)^6 = 18!*(9!/6!/3!)^6 = 18!*84^6 ~= 2*10^27 permutations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2015-10-19T21:44:21.240" LastActivityDate="2015-10-19T21:44:21.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23308" PostTypeId="2" ParentId="23281" CreationDate="2015-10-19T13:15:38.273" Score="4" Body="&lt;p&gt;Another valid answer would be: &lt;a href=&quot;http://legoland.com/&quot; rel=&quot;nofollow&quot;&gt;Legoland&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17000" LastEditorUserId="9000" LastEditDate="2015-10-19T13:31:51.513" LastActivityDate="2015-10-19T13:31:51.513" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23309" PostTypeId="2" ParentId="23290" CreationDate="2015-10-19T13:31:40.867" Score="3" Body="&lt;p&gt;Yet a few more:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ACROSS&lt;/p&gt;&#xA;&#xA;&lt;p&gt;17  He keeps punching the pavement (6,7)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Street Fighter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;DOWN&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1   John Wayne takes extreme action (4,5)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Duke Nukem&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;9   Watery brontide (5,7)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hydro Thunder&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16874" LastEditorUserId="16874" LastEditDate="2015-10-19T13:46:17.463" LastActivityDate="2015-10-19T13:46:17.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23310" PostTypeId="1" AcceptedAnswerId="23312" CreationDate="2015-10-19T13:46:59.843" Score="7" ViewCount="1166" Body="&lt;p&gt;So yesterday on the bus ride back from school I was sitting beside my friend James talking about food (we were really hungry from all the still life paintings in Art class). Suddenly, his gaze wandered away and he stopped talking. I was getting used to this - lately James had took up a new hobby, cryptography, and sometimes had &amp;quot;brilliant puzzle ideas&amp;quot; he liked to share with me. I sat silently for a minute or so until he got down the bus and said simply &amp;quot;Tell you 'bout it tomorrow&amp;quot;. &amp;quot;Sure&amp;quot;, I said, &amp;quot;See you tomorrow&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;The next day, as I entered the classroom, he handed me a scribbled piece of paper:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Today at breakfast the weirdest thing happened. I was eating something tasty, when all of a sudden a witch popped up right in front of me! She cracked a laugh and said:&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;Behold, I am the Breakfast Witch, computer age vegan activist.&lt;/p&gt;&#xA;&lt;p&gt;For you crimes against nature you shall pay! But fear not, child, your skin I shall not flay.&lt;/p&gt;&#xA;&lt;p&gt;A curse upon your breakfast is enough - so you may never taste this stuff.&lt;/p&gt;&#xA;&lt;p&gt;There is  even a way to lift it, in fact; there has to be, per the Witch Contract.&lt;/p&gt;&#xA;&lt;p&gt;Your evil treats I shall transform into twelve five-element magic forms,&lt;/p&gt;&#xA;&lt;p&gt;And upon that another layer of deception shall further obscure the information.&lt;/p&gt;&#xA;&lt;p&gt;But in the case you can see beyond the lies, I shall restore your murderous meal to its original guise&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;The Breakfast Witch disappeared, and I instantly forgot what I was eating. I couldn't tell for the life of me what I was having for breakfast! Instead, what I saw in my plate, smelled in the air and tasted in my mouth was (don't ask me how!) &amp;quot;218D103061009AC&amp;quot;. What was I eating?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I had no clue at all how to solve that. To me it looked like a crazy riddle and an even crazier sequence of letters and numbers! During break I went online and searched for the numbers, but found nothing. James kept asking me every chance he got if I'd &amp;quot;figured it out yet&amp;quot;. So annoying.&lt;/p&gt;&#xA;&lt;p&gt;By the end of the day, I was still stumped. I really want to crack this, if just to put an end to the smug way James keeps smiling at me. Searching online at home I found this site and figured it would be a great place to solve this and finally put an end to at least this one puzzle. So, can you help me find out what he was having for breakfast?&lt;/p&gt;&#xA;" OwnerUserId="16998" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-10-19T17:48:58.067" Title="The Cursed Breakfast" Tags="|cipher|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23312" PostTypeId="2" ParentId="23310" CreationDate="2015-10-19T17:26:55.680" Score="12" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;EGGS AND BACON&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's start from the end. Combination of digits and letters looks like a hex number. Let's try converting them... If we convert it to binary, we get 60 (which is 12*5) digits (leading zeros appear from digit-by-digit conversion). This looks good, as it suits for &quot;twelve five-element magic forms&quot; and conversion to hex is &quot;another layer of deception&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;00100 00110 00110 10001 00000 01100 00011 00001 00000 00010 01101 01100&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Breakfast Witch indicated &quot;another layer of deception&quot;, which we remove by the transformation $0\to A$ and $1\to B$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;AABAA AABBA AABBA BAAAB AAAAA ABBAA AAABB AAAAB AAAAA AAABA ABBAB ABBAA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Bacon&amp;#39;s_cipher&quot;&gt;&lt;strong&gt;Baconian Cipher&lt;/strong&gt;&lt;/a&gt;, which finally yields&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; EGGS AND BACON&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And that's what James was having for breakfast.&lt;/p&gt;&#xA;" OwnerUserId="17005" LastEditorUserId="8874" LastEditDate="2015-10-19T17:48:58.067" LastActivityDate="2015-10-19T17:48:58.067" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23313" PostTypeId="2" ParentId="23260" CreationDate="2015-10-19T17:57:26.973" Score="6" Body="&lt;p&gt;The code in each room represents a sequence of 1-3 tiles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The left digit indicates how many empty P-type tiles are at the start of the sequence. This must be so because the sum of all the left digits on the path equals 26, which is how many P tiles are in the maze.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right digit indicates what type of non-empty tile is in the sequence. Each digit maps to a tile:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 - A&#xA;2 - B&#xA;3 - C&#xA;...&#xA;D - M&#xA;E - N&#xA;F - O&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This must be so because there are 35 &quot;E&quot;s in the code, and 35 N tiles in the maze, and 41 &quot;D&quot;s in the code, and 41 M tiles in the maze... etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the first few tiles in the path &lt;code&gt;1D 06 OE&lt;/code&gt; decode to `PM F N&quot;. The full decoded sequence is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;PMFNNGNPMONPMPEDANPMPFNMMFNMPONMMONMPMPFDANAEBNBHFCNGNMPFHMMPMPANMMMMPMPFGCFNMPNGNMMMONMPPMPMMMANANPDPBNNGNFNFNMPACPMPONONMPNDPMPFNFNAEFCFNBCFGCBNFHONFHMMMACFDFNMMMMMND&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If you lay these tiles out left to right in the stair-like pattern of the maze, you get this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/UpwWV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UpwWV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which appear to be a block letter representation of the message:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; JET&lt;br&gt;&#xA; FUEL&lt;br&gt;&#xA; CANT&lt;br&gt;&#xA; MELT&lt;br&gt;&#xA; STEEL&lt;br&gt;&#xA; BEAMS  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1497" LastEditorUserId="1497" LastEditDate="2015-10-20T13:02:48.113" LastActivityDate="2015-10-20T13:02:48.113" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23314" PostTypeId="1" AcceptedAnswerId="23321" CreationDate="2015-10-19T18:34:19.847" Score="9" ViewCount="2053" Body="&lt;p&gt;The word ANEMONE has only seven letters, but four syllables; this gives $1.75$ letters per syllable on average.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you find another English word with&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;that is not a proper name&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;that has at least four syllables&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;that has an average number of letters per syllable below $1.75$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(If there are different solutions with the same average then the one with more consonants gets the prize. If $y$ is pronounced as vowel it counts as a vowel. Otherwise I count $y$ as a consonant.)&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="9951" LastEditDate="2015-10-20T14:26:11.660" LastActivityDate="2015-10-23T13:32:58.440" Title="The ANEMONE puzzle" Tags="|word|language|" AnswerCount="12" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23315" PostTypeId="2" ParentId="22825" CreationDate="2015-10-19T18:36:45.130" Score="4" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The heart&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;&lt;em&gt;A life inside the confining dark&lt;br&gt;&#xA;Their work unseen, but try to hark&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The organs in the body. They are all &quot;alive&quot; and you can hear some of them if you hark.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Hark to what they say, get steady&lt;br&gt;&#xA;Go ahead and free one already&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The lines above tell us to hark the sound of the body.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;From my experience I can tell&lt;br&gt;&#xA;I'm very close to my defeat&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;In the pulse diagrams there are highs and lows and this is reference to a &quot;low&quot;. Also note that one heartbeat a month is not enough for a living :D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Maybe a clever mind can help&lt;br&gt;&#xA;Which one is it I urgently need?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A new/stronger heart is needed. Also we need to use a computer in order to solve the riddle. Like look in the profile of @Avigrail and there you can see a reputation diagram. It's quite the same:&lt;a href=&quot;https://i.stack.imgur.com/PuUd8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PuUd8.jpg&quot; alt=&quot;rep&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks for the great riddle :)&lt;/p&gt;&#xA;" OwnerUserId="11145" LastEditorUserId="11145" LastEditDate="2015-11-06T10:09:27.163" LastActivityDate="2015-11-06T10:09:27.163" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="23316" PostTypeId="2" ParentId="23314" CreationDate="2015-10-19T18:55:29.840" Score="8" Body="&lt;p&gt;A word with 6 letters and 4 syllables (1.5 letters per syllable):  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;a href=&quot;http://dictionary.reference.com/browse/aviary&quot;&gt;Aviary&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15921" LastEditorUserId="5188" LastEditDate="2015-10-20T03:00:18.933" LastActivityDate="2015-10-20T03:00:18.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23317" PostTypeId="2" ParentId="23314" CreationDate="2015-10-19T18:56:44.420" Score="12" Body="&lt;p&gt;My answer (ratio 1.4):&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;a href=&quot;http://www.merriam-webster.com/dictionary/oxyopia&quot; rel=&quot;noreferrer&quot;&gt;Oxyopia&lt;/a&gt;//&lt;a href=&quot;http://dictionary.reference.com/browse/oogonia&quot; rel=&quot;noreferrer&quot;&gt;Oogonia&lt;/a&gt; (5 syllables - 1.4 letters per syllable)&lt;br /&gt;&#xA; Myopia - nearsightedness(4 syllables - 1.5 letters per syllable)&lt;br /&gt;&#xA; You'd also have Opiomania (opium addiction) and Oniomania (desire to shop)&lt;br /&gt;&#xA; (6 syllables, 9 letters, 1.5 ratio)&lt;br /&gt;&#xA; Oceanic/Ammonia (4 syllables - 1.75 letters per syllable)&lt;br /&gt;&#xA; &lt;a href=&quot;https://en.wiktionary.org/wiki/Iouea&quot; rel=&quot;noreferrer&quot;&gt;Iouea&lt;/a&gt;  (4 syllables - 1.25 letters per syllable)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-10-20T20:06:40.887" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23318" PostTypeId="1" AcceptedAnswerId="23336" CreationDate="2015-10-19T19:06:47.177" Score="6" ViewCount="215" Body="&lt;p&gt;Using the input and output of the below encryption method, can you determine the algorithm (what is the process/rule for converting input to output)? Note that each input/output pair is separate and not dependent on the others.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Inputs:&lt;br&gt;&#xA;  #1. HELLO WORLD&lt;br&gt;&#xA;  #2. THE QUICK BROWN FOX JUMPS OVER THE LAZY DOG&lt;br&gt;&#xA;  #3. BLAST OF POWER FOLLOWING A FLASH OF LIGHT&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Outputs:&lt;br&gt;&#xA;  #1. JTDFR IYWLF&lt;br&gt;&#xA;  #2. YKY TWRCJ MRPRB LRV HTBPG QNUO QHQ SFVI LEF&lt;br&gt;&#xA;  #3. NSFHP TS IIWRY KEHJYPUBK G LHKAH UH DOADO&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="13970" LastEditDate="2015-10-19T19:14:09.627" LastActivityDate="2015-10-20T16:20:25.650" Title="Method to the Madness" Tags="|cipher|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23320" PostTypeId="2" ParentId="23294" CreationDate="2015-10-19T19:56:50.543" Score="6" Body="&lt;p&gt;It is known that $mn\leq 121$; &lt;a href=&quot;https://stackoverflow.com/questions/3382859/solving-a-recreational-square-packing-problem&quot;&gt;Solving a Recreational Square Packing Problem&lt;/a&gt;.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how much the result can be improved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Example of 11x11 solution copied from the page above:&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;6 4 6 9 4 1 2 9 7 3 6&#xA;9 2 7 7 4 4 8 1 2 1 7&#xA;1 0 6 2 7 0 4 4 8 3 4&#xA;2 1 2 2 5 5 9 2 9 6 5&#xA;9 2 5 5 2 0 2 6 3 9 1&#xA;1 6 3 6 0 0 9 3 7 0 6&#xA;6 0 0 4 9 0 1 6 0 0 4&#xA;9 8 4 4 8 0 1 4 5 2 3&#xA;2 4 8 2 8 1 6 8 6 7 5&#xA;1 7 6 9 2 4 5 4 2 7 6&#xA;6 6 3 8 8 5 6 1 5 2 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="17007" LastEditorUserId="-1" LastEditDate="2017-05-23T12:38:53.617" LastActivityDate="2015-10-19T22:31:05.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23321" PostTypeId="2" ParentId="23314" CreationDate="2015-10-19T22:04:53.103" Score="7" Body="&lt;p&gt;A word with 6 letters, 4 syllables (1.5 ratio):  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Idiocy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11734" LastEditorUserId="5188" LastEditDate="2015-10-20T02:53:38.237" LastActivityDate="2015-10-20T02:53:38.237" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23326" PostTypeId="1" AcceptedAnswerId="23328" CreationDate="2015-10-20T06:35:06.357" Score="7" ViewCount="369" Body="&lt;p&gt;&lt;em&gt;Many games runs on me, &#xA;&lt;br&gt;yet I'm found in a game.&#xA;&lt;br&gt;To the game's family,&#xA;&lt;br&gt;I lend also my name.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;So go run and climb,&#xA;&lt;br&gt;and jump with abandon,&#xA;&lt;br&gt;but remember that I'm&#xA;&lt;br&gt;the thing that you stand on.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please tell me what I am, and explain why.&lt;/p&gt;&#xA;" OwnerUserId="7316" LastActivityDate="2015-10-20T08:04:47.267" Title="A Multi-leveled Gaming Concept" Tags="|riddle|video-games|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23328" PostTypeId="2" ParentId="23326" CreationDate="2015-10-20T08:04:47.267" Score="7" Body="&lt;p&gt;You're...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a platform!&lt;br&gt;&#xA; Platform is a word for the system a game runs on - PC, Wii, XBox, Playstation, etc.&lt;br&gt;&#xA; Platforms are also common in games like Mario and Metroid.&lt;br&gt;&#xA; Those games are called platformers.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2015-10-20T08:04:47.267" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23329" PostTypeId="1" AcceptedAnswerId="23347" CreationDate="2015-10-20T09:00:41.397" Score="3" ViewCount="855" Body="&lt;p&gt;What is the minimal number of 3s required to create a clock where each of the hours is replaced by a mathemetical expression using only the digit 3 and standard mathematical operators?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A friend mentioned this puzzle to me, but didn't remember the answer. The puzzle is inspired by &lt;a href=&quot;http://ecx.images-amazon.com/images/I/51jnFVuRlUL._SY300_.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; clock. I tried searching for similar puzzles, but only came across &lt;a href=&quot;https://puzzling.stackexchange.com/questions/2515/clock-values-using-just-9-and-some-operators&quot;&gt;this one&lt;/a&gt;, which has has rules I don't really like. I prefer this puzzle in a simpler form, only allowing rational numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested in both the minimal number of 3s per expression and the minimal number of 3s in total.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My best solution so far uses thirty 3s, but I feel like it can be improved. It would be nice to be able to formulate the puzzle such that you are only allowed to use 3 three times per expression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My solution:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  $1 = 3\div3$&lt;br&gt;&#xA;  $2 = 3!\div3$&lt;br&gt;&#xA;  $3 = 3$&lt;br&gt;&#xA;  $4 = 3 + (3\div3)$&lt;br&gt;&#xA;  $5 = 3! - (3\div3)$&lt;br&gt;&#xA;  $6 = 3!$&lt;br&gt;&#xA;  $7 = 3! + (3\div3)$&lt;br&gt;&#xA;  $8 = 3! + (3!\div3)$&lt;br&gt;&#xA;  $9 = 3\times3$&lt;br&gt;&#xA; $10 = (3\times3) + (3\div3)$&lt;br&gt;&#xA; $11 = 33\div3$ or with one extra 3: $((3!\times3!) - 3)\div3$&lt;br&gt;&#xA; $12 = (3!\times3!)\div3$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Create a clock face using only the digit 3 and standard mathematical operations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Allowed:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt; The digit 3&#xA;&lt;li&gt; Addition, subtraction, multiplication, division&#xA;&lt;li&gt; Square root&#xA;&lt;li&gt; Factorial&#xA;&lt;li&gt; Concatenation (33 counting as two 3s)&#xA;&lt;/ul&gt; &#xA;" OwnerUserId="17012" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-10-20T17:43:57.990" Title="Clock values using only the digit 3" Tags="|calculation-puzzle|optimization|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="23330" PostTypeId="2" ParentId="23314" CreationDate="2015-10-20T09:36:23.400" Score="3" Body="&lt;p&gt;A word with 6 letters and 4 syllables (ratio of 1.5):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Myopia&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16363" LastEditorUserId="9000" LastEditDate="2015-10-20T14:37:17.780" LastActivityDate="2015-10-20T14:37:17.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23331" PostTypeId="2" ParentId="23290" CreationDate="2015-10-20T10:32:25.283" Score="3" Body="&lt;p&gt;Two extra:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DOWN&lt;/p&gt;&#xA;&#xA;&lt;p&gt;24  They play for the director of a Life of Pi (7)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Angband&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;40  It goes round your pebble (4,4)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rock Band&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5327" LastEditorUserId="5327" LastEditDate="2015-10-20T15:49:38.517" LastActivityDate="2015-10-20T15:49:38.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23332" PostTypeId="2" ParentId="23314" CreationDate="2015-10-20T14:22:10.850" Score="7" Body="&lt;p&gt;Another 6 letter 4 syllable: (1.5)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Acuity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16131" LastActivityDate="2015-10-20T14:22:10.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23333" PostTypeId="1" AcceptedAnswerId="23339" CreationDate="2015-10-20T15:55:01.680" Score="3" ViewCount="374" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VDg4J.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VDg4J.jpg&quot; alt=&quot;Magic square&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What type of magic square is this?&lt;/p&gt;&#xA;" OwnerUserId="17021" LastEditorUserId="8874" LastEditDate="2015-10-20T16:21:13.233" LastActivityDate="2020-04-29T11:22:57.700" Title="What type of magic square is this?" Tags="|mathematics|magic-square|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23334" PostTypeId="1" AcceptedAnswerId="23352" CreationDate="2015-10-20T15:56:20.017" Score="10" ViewCount="1925" Body="&lt;p&gt;Four friends want to share a pizza. However, the pizza became badly misshapen en route; though it is still flat, it is now shaped like some arbitrary polygon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BPHnM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BPHnM.png&quot; alt=&quot;a poorly pepperoni&amp;#39;d polygonal pizza&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Prove that it is still possible to cut the pizza into four portions of equal area, using exactly &lt;em&gt;two&lt;/em&gt; straight cuts which are &lt;em&gt;perpendicular&lt;/em&gt; to each other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clarification:&lt;/strong&gt; The two cuts divide the plane into four regions, and the four &quot;portions&quot; are defined to be the pizza parts contained in these regions. A portion might consist of several disconnected pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The cuts are allowed to go outside the pizza. The pizza is placed on a cutting board, and then a pizza cutter is rolled in a straight line twice, such that the two lines are perpendicular. The pizza may not be moved before making the second cut. The pepperoni distribution doesn't matter).&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-10-20T16:27:14.973" LastActivityDate="2015-10-20T21:36:37.697" Title="Polygonal Pizza" Tags="|mathematics|geometry|" AnswerCount="3" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="23335" PostTypeId="2" ParentId="23281" CreationDate="2015-10-20T15:58:16.893" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Internet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was thinking of particular websites, but they don't fit your hint (actually a major restriction)&lt;/p&gt;&#xA;" OwnerUserId="16642" LastActivityDate="2015-10-20T15:58:16.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23336" PostTypeId="2" ParentId="23318" CreationDate="2015-10-20T16:01:55.850" Score="4" Body="&lt;p&gt;You can get the answer by...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When looking at the keyboard, shift right a number of characters equal to the position of the letter you are typing, wrapping around the keboard for only alpha keys. (i.e. H shifts right 1 to J, E shifts right 2 to T, L shift right 3 to D, then shifts right 4 to F, etc.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14233" LastActivityDate="2015-10-20T16:01:55.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23337" PostTypeId="2" ParentId="23318" CreationDate="2015-10-20T16:02:41.997" Score="4" Body="&lt;p&gt;The first thing that I notice when I look at the cypher text is that the word &quot;the&quot; gives us a palindrome both times.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This makes me suspect that positioning matters.  The fact that &quot;H&quot; gives us &quot;H&quot; both times in appears in the 27th position might also support this.  If we search the alphabet for a relationship that gives us such a result, we might come up empty handed.  However, if we look down at our English/America keyboard, we can find the letters &quot;T&quot; and &quot;H&quot; separated by one character, just as they are in the word &quot;THE&quot;.  In input 2, the first &quot;T&quot; is the first character, so traveling to the right 1 key gives us &quot;Y&quot;.  Traveling to the right from &quot;E&quot; by three keys (because &quot;E&quot; is in the third position) also gives us &quot;Y&quot;.  If we look at the next word, we can see that spaces don't matter.  Following this pattern, we can get the rest of the cypher text.  If we reach the end of a line of letter characters on our keyboard, we simply loop back to the first letter key on the left on that row of keys.  We can also test this by reversing the cypher, as well.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alternatively:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you don't know what a keyboard is, it is almost still possible to solve this cryptogram as long as you realize that each character has only 7-10 characters that it can become when encrypted.  To test this, we can map out all of the known characters based on their positions (rows can be position, the columns can be the plain text, and each cell will contain the known cypher text for that character/position combination).  In doing so, some patterns quickly become apparent.  We may not have to map all of it, but we know we should definitely complete the second input, because we need the brown fox to tell us at least one position/cypher text combination for each letter of the alphabet.  There will still be some small holes, in the end, because the cypher text does not contain the characters &quot;X&quot; or &quot;Z&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17020" LastEditorUserId="17020" LastEditDate="2015-10-20T16:20:25.650" LastActivityDate="2015-10-20T16:20:25.650" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23339" PostTypeId="2" ParentId="23333" CreationDate="2015-10-20T16:20:48.047" Score="6" Body="&lt;p&gt;The given square $X$ results from the following classical magic square (which arranges the number $1,2,\ldots,25$ in a $5\times5$ square so that every row, every column, and each of the two main diagonals adds up to the magical sum 65)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\left[\begin{array}{rrrrr}&#xA;15&amp;amp;16&amp;amp;22&amp;amp; 3&amp;amp; 9\\&#xA; 2&amp;amp; 8&amp;amp;14&amp;amp;20&amp;amp;21\\&#xA;19&amp;amp;25&amp;amp; 1&amp;amp; 7&amp;amp;13\\&#xA; 6&amp;amp;12&amp;amp;18&amp;amp;24&amp;amp; 5\\&#xA;23&amp;amp; 4&amp;amp;10&amp;amp;11&amp;amp;17&#xA;\end{array}\right]$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;by subtracting 1 from every entry, adding 10 times a permutation matrix, and finally supressing the resulting 0-entry in the center of the square:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle X=&#xA;\left[\begin{array}{rrrrr}&#xA;15&amp;amp;16&amp;amp;22&amp;amp; 3&amp;amp; 9\\&#xA; 2&amp;amp; 8&amp;amp;14&amp;amp;20&amp;amp;21\\&#xA;19&amp;amp;25&amp;amp; 1&amp;amp; 7&amp;amp;13\\&#xA; 6&amp;amp;12&amp;amp;18&amp;amp;24&amp;amp; 5\\&#xA;23&amp;amp; 4&amp;amp;10&amp;amp;11&amp;amp;17&#xA;\end{array}\right] -&#xA;\left[\begin{array}{rrrrr}&#xA;1&amp;amp;1&amp;amp;1&amp;amp;1&amp;amp;1\\1&amp;amp;1&amp;amp;1&amp;amp;1&amp;amp;1\\1&amp;amp;1&amp;amp;1&amp;amp;1&amp;amp;1\\1&amp;amp;1&amp;amp;1&amp;amp;1&amp;amp;1\\1&amp;amp;1&amp;amp;1&amp;amp;1&amp;amp;1&#xA;\end{array}\right] +&#xA;\left[\begin{array}{rrrrr}&#xA;0&amp;amp;0&amp;amp;10&amp;amp;0&amp;amp;0\\&#xA;0&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;10\\ &#xA;0&amp;amp;10&amp;amp;0&amp;amp;0&amp;amp;0\\&#xA;0&amp;amp;0&amp;amp;0&amp;amp;10&amp;amp;0\\ &#xA;10&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0&#xA;\end{array}\right]&#xA;$  &lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-10-20T16:20:48.047" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23340" PostTypeId="2" ParentId="23314" CreationDate="2015-10-20T16:33:03.067" Score="8" Body="&lt;p&gt;10 letters, 6 syllables (1.66), not the best, but a fitting entry&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Actiniaria (the scientific order of the anemone)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14233" LastActivityDate="2015-10-20T16:33:03.067" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23341" PostTypeId="1" AcceptedAnswerId="23343" CreationDate="2015-10-20T16:51:05.637" Score="13" ViewCount="1117" Body="&lt;p&gt;Enrico draws a square in the plane, and then secretly picks a point $P$ that is &#xA;either situated &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;inside the square, &lt;/li&gt;&#xA;&lt;li&gt;or outside the square, &lt;/li&gt;&#xA;&lt;li&gt;or on the boundary of the square.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Damiano sees the square drawn by Enrico, but does not know the position &#xA;of the secret point $P$. &#xA;Damiano may choose a straight line $\ell$ in the plane and show $\ell$ to Enrico. Enrico then truthfully answers Damiano whether the secret point $P$ lies on line $\ell$, and in case the point is not the line, on which side of $\ell$ this secret point is located.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What is the smallest possible number of lines that Damiano can query so that he will discover with absolute certainty whether the secret point is inside, outside, or on the square?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-21T09:41:46.843" LastActivityDate="2016-08-09T17:13:58.473" Title="Inside or outside the square?" Tags="|mathematics|geometry|strategy|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23343" PostTypeId="2" ParentId="23341" CreationDate="2015-10-20T16:57:11.167" Score="28" Body="&lt;p&gt;I'm guessing the minimum is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3&lt;br&gt;&#xA;&lt;br&gt;&#xA; Do a diagonal. If not on the line, do the other diagonal. You now know the quadrant.  Now pick the side of the square in your quadrant. Now you know if the point is inside/outside/on the square.&lt;br&gt;&#xA;&lt;br&gt;&#xA; If the first line or second line are on the point, you can still use one of the adjacent sides of the square to figure out if the point is inside/outside the square&lt;br&gt;&#xA;&lt;br&gt;&#xA; Two queries are not sufficient since you need 3 lines to create the minimum bounding area. With 2 lines, you either have 4 distinct regions (or 3 if parallel), but they are unbound, so the point could be anywhere in that region, either in or out of the square. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-10-22T13:03:56.947" LastActivityDate="2015-10-22T13:03:56.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23344" PostTypeId="2" ParentId="22825" CreationDate="2015-10-20T17:05:28.280" Score="1" Body="&lt;p&gt;Is it:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Minesweeper?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A life inside the confining dark  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mines and numbers are hidden  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Their work unseen, but try to hark&lt;br&gt;&#xA;Hark to what they say, get steady  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Numbers are the hints (hark) to nearby mines.  Also when removing a mine one should stay steady  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Go ahead and free one already  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you successfully uncover a mine.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From my experience I can tell&lt;br&gt;&#xA;I'm very close to my defeat  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Minesweeper's defeat = player close to win?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Maybe a clever mind can help&lt;br&gt;&#xA;Which one is it I urgently need?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That one last mine hidden?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2015-10-20T17:05:28.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23346" PostTypeId="2" ParentId="23334" CreationDate="2015-10-20T17:39:45.143" Score="7" Body="&lt;p&gt;Draw two perpendicular lines, such that line1 &amp;amp; line2 intersect at P &amp;amp; line1 is at some angle theta (say 0) to x-axis.&lt;br&gt;&#xA;Number the 4 quadrants Q1,Q2,Q3,Q4.&lt;br&gt;&#xA;Move the intersection P along line1 , from very far left to very far right. At some position, the area on the left must equal the area on the right, because at very far left, the area of the 2 left quadrants will be 0, while at very far right, the area of the 2 right quadrants will be 0, and at some intermediate position, the areas will be equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do this movement for P moving along line2 , from very far up to very far down. Again the 2 upper quadrants will be equal to the 2 lower quadrant at some point along line2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically , we have moved P all over the plane , when line1 was at theta, and we have found a position where Q1+Q2=Q3+Q4=A/2 &amp;amp; Q2+Q3=Q4+Q1=A/2.&#xA;Now Q1=Q3 &amp;amp; Q2=Q4, or opposite quadrants are equal.&lt;br&gt;&#xA;We get Q1=A/2-Q2.&lt;br&gt;&#xA;If Q1=Q2, we are done. If not Q1 is-less-than Q2 Or Q1 is-more-than Q2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now repeat this process, but use theta from 0 to 90 Degrees. The locus of the points (for each theta) with this movement must be continuous.&lt;br&gt;&#xA;Starting at theta=0 and going to theta=90, we see that Q1 and Q2 exchange their final values. So, starting at Q1 is-less-than Q2, we get Q1 is-more-than Q2. So, at some theta, Q1=Q2. So all 4 quadrants are equal at this point of intersection, at this angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summary : We search over all intersection points, at all angles, which will give us at least one solution, because of continuity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Soxqm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Soxqm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, start with the horizontal gray line and the vertical rid line on the left. Intersection is P, with 4 quadrants. We see that area is Q1+Q4=0, considering the areas of the pizza in the quadrants.&lt;br&gt;&#xA;We move P slowly towards right, to get the center red line and the rightmost left line. We end up with Q2+Q3=0.&lt;br&gt;&#xA;So, at some point, we have Q1+Q4=Q2+Q3, shown as the central red line.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, move the gray line upwards and similarly get a new P which has Q1+Q4=Q2+Q3.&#xA;In the example given, the locus of P which has Q1+Q4=Q2+Q3 is the central red line, but in general, it may be some other continuous curve C.&lt;br&gt;&#xA;In the curve C, when we move P from top to bottom, we start with Q1+Q2=0 and end with Q3+Q4=0, so at some point, Q1+Q2=Q3+Q4.&lt;br&gt;&#xA;If Q1=Q2 here , we are done.&lt;br&gt;&#xA;Else we must repeat the process with a new theta.&lt;br&gt;&#xA;When we vary theta from 0 to 90, we will see that we move from Q1 is-less-than Q2 to Q1 is-more-than Q2 , so at some intermediate point, Q1=Q2, so all 4 quadrants are equal.&lt;/p&gt;&#xA;" OwnerUserId="10940" LastEditorUserId="10940" LastEditDate="2015-10-20T19:26:11.317" LastActivityDate="2015-10-20T19:26:11.317" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23347" PostTypeId="2" ParentId="23329" CreationDate="2015-10-20T17:43:57.990" Score="3" Body="&lt;p&gt;I've scratched my head at this for an hour and I could only find one improvement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 12 = 3! + 3!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12795" LastActivityDate="2015-10-20T17:43:57.990" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23348" PostTypeId="2" ParentId="23265" CreationDate="2015-10-20T17:55:08.407" Score="2" Body="&lt;p&gt;A person will go to the town center if they can determine that they must be a member of some organization. If they can determine that there is at least one letter they did not receive, then they must be a member of some organization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they are not a member of any organization, then they received the membership of every organization. This allows them to do a simple proof by contradiction. They start off by assuming that they are not a member of any organization (and therefore have perfect knowledge of the situation), and then wait until something unexpected happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest case is that a person receives no letters. Assuming that they are not a member of any organization implies there are no organizations, which they know to be false. Therefore, they must be a member of at least one organization.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Consider how this plays out with organizations consisting only of single members: A, B, and C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A gets letters about B and C, B gets letters about A and C, and C gets letters about A and B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On day 1, as everyone received letters, nobody goes to the town center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On day 2, A assumes that there are only two organizations &amp;mdash; B and C. B received a letter about C, and C received a letter about B. Thus, neither of them should have gone to the town center. As neither of them did, A has no reason to believe that there are any other organizations. B and C follow similar reasonings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On day 3, A assumes that there are only two organizations &amp;mdash; B and C. On day 2, B would have reasoned as follows: &quot;Suppose there is only one organization, with C as its only member. Then C should have gone to the town center on day 1. However, C did not, so I must be a member of some group. So I will go to the town center today.&quot; However, B did not go to the town center on day 2, so there must be an organization that A does not know about that A is a member of. So A will go to the town center. By symmetric reasoning, B and C will also go to the town center.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;We know that it is not the case that everyone will always be able to determine that they are a member of some organization. Consider the case of organizations A, AB, ABC, and ABCD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On day 1, A, having received no letters, goes to the town center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every day after that, A goes to the town center while B, C, and D all stay home. B thinks there's only one organization A, and only A going to the town center confirms that. C knows there's also an organization AB, but expects B to stay at home because B doesn't know about that one. D knows there's also an organization ABC, but knows that C and B are unaware of its existence. There will never be an event that disrupts anyone's expectations, so this is a stable situation.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There are two questions we still need to answer &amp;mdash; will there always be at least one proper meeting, and will everyone who goes eventually be in a proper meeting?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First, let's establish a base condition&lt;/strong&gt; - there will always be at least one person who goes to the town center. I'll use a group of A, B, C, and D to help explain my proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone does not receive a letter, then they will go the first day. This is the only way in which someone will go on the first day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now assume that everyone received at least one letter (and so nobody went on the first day). Then on the second day A, assuming that she is not a part of any group and therefore knows the membership of every group, will consider every other person. If any of B, C, and D is a member of all the groups, then that person should not have received any letters and gone on the first day. But nobody went the first day, so A must not have perfect information, and therefore must be a member of some group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If nobody goes on the second day, then A will go one level deeper in her reasoning. She first considers B &amp;mdash; he knows about all the groups for which he is not a member. If C or D is a member of all of those groups, then B would have expected them to go on the first day. When he observed that they did not, he should have gone on the second day. Because he did not go on the second day, A knows that there must be some group that she does not know about and therefore is a member of. She follows the same reasoning for C and D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If nobody goes on the third day, A will reach the deepest level of reasoning necessary for four people. She will again consider B's reasoning. Assuming that she knows about all the groups that exist, B knows only about groups that consist of C and/or D. So he, on the third day, would have considered C's reasoning for the second day. He would have assumed that he knew about all the groups that exist, and since he only knows about groups the have only C and D in them, C must have only known about a group of which D is the only member. So C, on the second day, would have thought that only D was a member of a group, and having not witnessed D go on the first day, should have gone on the second day. Thus, B, having not witnessed C go on the second day, should have gone on the third day. Thus A, if she does not see B go on the third day, will go on the fourth day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In more formal terms, on day $k+1$ person $p_0\in P$ will determine if $\exists p_{i1}\in (P-\{p_0\}) \exists p_{i2}\in (P-\{p_0,p_{i1}\})...\exists p_{ik}\in (P-\{p_0,...,p_{i(k-1)}\})\forall g\in G: \{p_{i1},...,p_{ik}\}\cap g \ne \emptyset$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $G$ is the set of all groups that $p_0$ knows about. If this condition is met, then there are no groups that do not contain at least one of the people considered. This means that on day $k+1$, $p_0$ expected each person in the chain $p_{ij}$ to expect the next person in the chain to have left on the previous day (with the last person expected to leave on the first day). Because the expectation was not met, $p_0$ will go to the town center on day $k+1$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Everyone who goes will eventually be a part of a proper meeting:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $p_0$ be a person who will go to the town center. If $p_0$ is the sole member of an organization, then a proper meeting can be held immediately. Otherwise, $\exists g_0,p_1\mid p_0,p_1\in g_0$. Because $p_0$ and $p_1$ are both members of that organization, neither received a letter about it. Suppose there are no organizations of which $p_0$ is a member but $p_1$ is not. Then, when $p_0$ goes to the town center, $p_1$ will know there is at least one organization that they did not receive a letter for, and must therefore be a member of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise $\exists g_1\mid p_0\in g_1\land p_1\notin g_1$. By our above reasoning, we know that anyone who only belongs to organizations $p_0$ does will go to the town center one day after $p_0$ at the latest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each person $p_i\in g_1$ who is a member of $g_i\mid p_0\notin g_i$, $p_1$ can follow the same chain of logic we used to prove that someone will eventually go. I'm having trouble writing out the formal logic for it, but here's the basic idea &amp;mdash; $p_1$ will expect $p_i$ to expect etc. In the end it boils down to two cases - either an entire organization goes, or $p_1$ will go, and because $p_1$ was an arbitrary member of $g_0$ that means everyone in $g_0$ will go.&lt;/p&gt;&#xA;" OwnerUserId="1955" LastEditorUserId="1955" LastEditDate="2015-10-22T04:31:13.003" LastActivityDate="2015-10-22T04:31:13.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23349" PostTypeId="2" ParentId="23314" CreationDate="2015-10-20T18:01:26.013" Score="-1" Body="&lt;p&gt;14 letters and 4 syllables (ratio of 1.75):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;wordprocessors&lt;/p&gt;&#xA;" OwnerUserId="17023" LastActivityDate="2015-10-20T18:01:26.013" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23350" PostTypeId="2" ParentId="23314" CreationDate="2015-10-20T18:46:09.697" Score="5" Body="&lt;p&gt;A common word with 10 letters and 6 syllables. (1.67 ratio)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; aficionado&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And just for fun, I believe these are words with the worst ratio: 10 letters with only 1 syllable (10.0 ratio):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; strengthed, scraunched&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17024" LastEditorUserId="17024" LastEditDate="2015-10-20T19:01:35.977" LastActivityDate="2015-10-20T19:01:35.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23351" PostTypeId="1" AcceptedAnswerId="23355" CreationDate="2015-10-20T19:56:06.810" Score="0" ViewCount="442" Body="&lt;p&gt;A last hurrah for the video game challenge:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Well here I am, on the run from the mob.  Lin Hyle, that’s the pork-peahatted big boss that I’ve done wrong.  Like, like I don’t really know what I did to burst his bubble.  Sure, if I had the time I could puzzle it out, but doesn’t even matter.&lt;/p&gt;&#xA;&lt;p&gt;I had to ditch my beetle about a mile back; the shift lever jammed    into park after I hit a hole, and I skidded into a tree.  Jumping out, I could see even in the dark, nuts missing from my wheels.  It had been a setup.&lt;/p&gt;&#xA;&lt;p&gt;I grabbed my keys and some rope from the trunk and took off into the depths of the woods.  In the distance I hear a train sounding its lonesome “choo….choo…”.  But every sound seems to scare me at this point, and every&lt;em&gt;thing&lt;/em&gt;, too.  It feels like enemies are all around, and I know the power and the… the thoroughness of Lin: he doesn’t just make you dead, he makes you redead.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Who am I? And why do I feel surrounded by my foes?&lt;/p&gt;&#xA;" OwnerUserId="11452" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-10-21T12:56:18.237" Title="Enemies are Everywhere I Look!" Tags="|riddle|knowledge|wordplay|video-games|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23352" PostTypeId="2" ParentId="23334" CreationDate="2015-10-20T20:23:42.440" Score="18" Body="&lt;p&gt;By continuity, for any cut angle, there's a unique cut that slices the pizza into two equal-area segments. Moreover, the cut placement is continuous in the angle, since for close angles, the lines are nearly parallel and their intersection point must lie within the convex hull of the pizza, a finite range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, consider a cut at some angle and the perpendicular angle. These must intersect and so divide the pizza into four sections, with any two adjacent sections totaling half the pizza. So, the pairs of opposite sections have the same area. It remains to make the area for each pair equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do this by continuously varying the initial angle over a range of 90 degrees. From the start to the end, the original and perpendicular cut will have switched, so the respective areas will have switched. Since the cut placement is continuous in the angle, the sizes of the sections also vary continuously, and so by the intermediate value theorem, must be equal for some angle.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-10-20T20:23:42.440" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23353" PostTypeId="2" ParentId="23334" CreationDate="2015-10-20T20:32:34.340" Score="4" Body="&lt;p&gt;Take a line and roll it along until it divides the pizza into two equal halves. Then take a second line, perpendicular to the first and roll it along until it divides the pizza into two equal halves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diagonally opposite quadrants will have equal mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How find the focus of the centre of the cross for all orientations of the lines. And imagine transitioning smoothly from the original cross to one at right angles to it (which will be in the same location). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The one at right angles will be heavier where the previous one was lighter and &lt;em&gt;vice versa&lt;/em&gt;. So between these there must be one where they're all the same (since each of the transitional states has diagonally opposite quadrants equal). &lt;/p&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2015-10-20T21:36:37.697" LastActivityDate="2015-10-20T21:36:37.697" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23354" PostTypeId="1" AcceptedAnswerId="23357" CreationDate="2015-10-20T21:13:42.113" Score="-2" ViewCount="626" Body="&lt;p&gt;&lt;em&gt;Wasn't quite sure if this was suitable for Puzzling; however it has a problem/puzzle in it, so decided to give it a shot. Please appropriately downvote if you feel you should.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yesterday morning I came to class as usual, and a student came up to me. He said he had a proof that could revolutionize mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being a Grade 11 teacher, I didn't believe him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I asked him to tell me about the proof, he said &lt;em&gt;&quot;I've divided it into several steps, and said it seemed to work perfectly well, with zero problems.&quot;&lt;/em&gt; Then he told me &lt;em&gt;&quot;It's a proof that 1 is equal to 0.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I asked if he could find any mistakes, he responded &lt;em&gt;&quot;None that I could find, sir, however it wouldn't surprise me if there was. I had thought of it after you taught us factoring. I think you'd like to take a look&quot;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I indeed did, and he showed me, mentioning &lt;em&gt;that it was indeed stating $1=0$.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously this is false, and by looking at the paper I noticed his trivial error almost immediately.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Assume x and y are two nonzero numbers.&lt;br&gt;&#xA;  $x = y$  &lt;br&gt;&#xA;  $x^2 = x*y $ &lt;br&gt;&#xA;  $x^2 - y^2 = x*y - y^2  $&lt;br&gt;&#xA;  $x + y = y  $&lt;br&gt;&#xA;  $2*y = y  $&lt;br&gt;&#xA;  $2 = 1  $&lt;br&gt;&#xA;  $1 = 0$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, I have a challenge for you, and this is to find the error my student made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But not only that, you must explain how everything he told me was true.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="3993" LastEditDate="2015-10-20T22:14:10.490" LastActivityDate="2015-10-20T22:14:10.490" Title="Wait wait wait wait... 1=0?" Tags="|mathematics|word|english|" AnswerCount="2" CommentCount="2" ClosedDate="2015-10-21T12:04:05.430" ContentLicense="CC BY-SA 3.0" />
  <row Id="23355" PostTypeId="2" ParentId="23351" CreationDate="2015-10-20T21:15:14.460" Score="4" Body="&lt;p&gt;I think you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Link, from the Legend of Zelda series&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasons&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Peahat&quot;, &quot;Darknut&quot; and &quot;Redead&quot; are mentioned in the puzzle (as well as numerous other creatures, as Zymus points out), and they are enemies from the Zelda games.  &quot;Lin Hyle&quot; is awfully close to &quot;Link Hylian&quot;, &quot;time&quot; is mentioned in the first paragraph as well as &quot;puzzle&quot;.  In addition, grabbing &quot;keys&quot; from a &quot;trunk&quot; is a fairly rampant past-time in Zelda games.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14233" LastEditorUserId="14233" LastEditDate="2015-10-21T12:56:18.237" LastActivityDate="2015-10-21T12:56:18.237" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23356" PostTypeId="2" ParentId="23354" CreationDate="2015-10-20T21:22:06.927" Score="4" Body="&lt;p&gt;The student divided by zero between the third and fourth lines. He wrote the third line as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(x + y)(x-y) = y(x-y)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And proceeded to cancel the term $(x-y)$. This can be done if and only if $(x-y) \neq 0$, but since the hypothesis is $x=y$ we arrive at a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="16478" LastActivityDate="2015-10-20T21:22:06.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23357" PostTypeId="2" ParentId="23354" CreationDate="2015-10-20T21:24:34.573" Score="17" Body="&lt;p&gt;The errors lies here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is an implicit division by 0 between steps 3 and 4. Step 3 states: $x^2-y^2=xy-y^2$. Factoring this produces $(x+y)(x-y)=y(x-y)$. Going to step 4 requires that you divide through by $(x-y)$, but step 1 implies that $x-y=0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He is &lt;em&gt;technically&lt;/em&gt; correct in what he said because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He said the proof has &quot;zero problems&quot;. This is correct; there is a problem with the number 0 in his proof.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16275" LastActivityDate="2015-10-20T21:24:34.573" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23358" PostTypeId="1" AcceptedAnswerId="24223" CreationDate="2015-10-21T00:11:36.980" Score="6" ViewCount="992" Body="&lt;p&gt;Seemingly cat traps love medley,&#xA;&lt;br&gt;Cats love the responses to me,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I help prevent a big clean up,&#xA;&lt;br&gt;So you can type up Sup!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I trap cats like a boss,&#xA;&lt;br&gt;since they can't get across,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I put a bounty on this guy,&#xA;&lt;br&gt; and attracted so many guys,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I look like an election just happened,&#xA;&lt;br&gt;these people so happy for them votes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I? To be clear, I am a question and sorry for making the riddle rhyme! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is a specific question...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2 and 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The question is another SE question and there are no anagrams.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9042" LastEditorUserId="9042" LastEditDate="2015-11-24T22:46:20.960" LastActivityDate="2015-11-25T18:46:28.660" Title="Cats, More Cats, and Questions" Tags="|riddle|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23359" PostTypeId="1" AcceptedAnswerId="23363" CreationDate="2015-10-21T00:59:00.320" Score="8" ViewCount="237" Body="&lt;p&gt;Recently I purchased a bunch of mobile puzzle games for my iPhone and one of them was the famous top-rated app &quot;Strata&quot; by Graveck. The rules are simple - you begin with an $n \times n$ board, some squares of which are colored in arbitrary colors. Then you start placing stripes of whatever color you choose over entire rows and columns of the board. Your task is after placing all available $2n$ stripes, the color of every (colored) square to match the color of the stripe which has been placed second over it (on top).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you find an easy way to beat every single level of this game?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. You can see a video explanation of the rules in the app review below:&#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=NwjyOpA14Vk&quot;&gt;https://www.youtube.com/watch?v=NwjyOpA14Vk&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14140" LastEditorUserId="14140" LastEditDate="2015-10-21T01:50:15.303" LastActivityDate="2015-10-21T08:08:34.163" Title="How to beat &quot;Strata&quot;?" Tags="|algorithm|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23360" PostTypeId="1" AcceptedAnswerId="23361" CreationDate="2015-10-21T02:47:45.360" Score="8" ViewCount="971" Body="&lt;p&gt;Short clue.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is simply a space filler to satisfy the SE editor. It does not form part of the puzzle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="4050" LastEditDate="2015-10-21T20:47:38.197" LastActivityDate="2015-10-21T20:47:38.197" Title="Some kind of equine religion?" Tags="|enigmatic-puzzle|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="23361" PostTypeId="2" ParentId="23360" CreationDate="2015-10-21T03:05:48.400" Score="16" Body="&lt;p&gt;Horse Cult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it is an anagram of short clue.&lt;/p&gt;&#xA;" OwnerUserId="13916" LastActivityDate="2015-10-21T03:05:48.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23362" PostTypeId="2" ParentId="23359" CreationDate="2015-10-21T07:38:43.977" Score="1" Body="&lt;p&gt;Not sure if it always lead to solutions but I think that this strategy is pretty solid for the easier levels at least:&lt;br&gt;&#xA;&lt;strong&gt;Always do the columns or rows in order of most colors to least colors in that row or column.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To explain it further. If a row or column has only a single color do that one last. It's kinda logical  in a way because if you didn't do it last there is a possibility that a square is overwritten in another direction later on. And it's then also impossible for that last move to screw up other squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This automatically means that rows that have more colors come first. And the more colors you have the more you need to overwrite colors in that row or column so more colored rows have priority over rows with less colors.&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-10-21T07:38:43.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23363" PostTypeId="2" ParentId="23359" CreationDate="2015-10-21T08:08:34.163" Score="5" Body="&lt;p&gt;Here's a foolproof way to do it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fill in any &quot;consistent&quot; rows and columns first - anything that has all the same color (or gray).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ignore those squares and fill in newly consistent rows and columns. There will only be none if the puzzle is unsolvable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you're done, reset and do the entire thing in reverse order.&lt;/p&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2015-10-21T08:08:34.163" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="23364" PostTypeId="1" AcceptedAnswerId="23366" CreationDate="2015-10-21T09:17:29.720" Score="3" ViewCount="805" Body="&lt;p&gt;Longer clue.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In response to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/23360/some-kind-of-equine-religion&quot;&gt;Some kind of equine religion?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-10-22T08:03:17.293" Title="Relatively perverted?" Tags="|enigmatic-puzzle|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="23366" PostTypeId="2" ParentId="23364" CreationDate="2015-10-21T09:22:35.530" Score="6" Body="&lt;p&gt;Answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Uncle Ogler  (or of course ogler uncle)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  uncle = relative&lt;br&gt;&#xA; It's rather perverted if your uncle ogles you!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="14177" LastEditDate="2015-10-21T09:54:17.867" LastActivityDate="2015-10-21T09:54:17.867" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23367" PostTypeId="2" ParentId="23314" CreationDate="2015-10-21T09:55:30.017" Score="1" Body="&lt;p&gt;Multiple words: 7 Letter 4 syllable.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; anatomy idiotic  analogy anxiety atypical unusual biology zoology&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2656" LastActivityDate="2015-10-21T09:55:30.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23368" PostTypeId="1" AcceptedAnswerId="23369" CreationDate="2015-10-21T11:38:11.033" Score="-1" ViewCount="425" Body="&lt;p&gt;And an even longer clue: tolerating a look at her belly.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In response to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/23360&quot;&gt;Some kind of equine religion?&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/questions/23364&quot;&gt;Relatively perverted?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-11-02T13:03:42.687" Title="What Britney Spears is famous for" Tags="|enigmatic-puzzle|anagram|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23369" PostTypeId="2" ParentId="23368" CreationDate="2015-10-21T11:53:01.230" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Endure glance on navel?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is an anagram of the clue&lt;br&gt;&#xA;&lt;br&gt;&#xA; $~~~~~~~$  AND AN EVEN LONGER CLUE $~~=~~$ ENDURE GLANCE ON NAVEL.&lt;br&gt;&#xA;&lt;br&gt;&#xA;To endure is to tolerate, and a &quot;look at the belly&quot; could be a &quot;glance on navel&quot;.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I admit I had an anagram generator help me a little (not completely), so apologies if that's considered cheating.&lt;/p&gt;&#xA;" OwnerUserId="17033" LastEditorUserId="8874" LastEditDate="2015-10-22T08:11:32.787" LastActivityDate="2015-10-22T08:11:32.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23370" PostTypeId="1" AcceptedAnswerId="23372" CreationDate="2015-10-21T12:08:06.353" Score="15" ViewCount="1355" Body="&lt;p&gt;The other day I was playing the popular board game with my good friend, Chet Manley.  Though he had me on the ropes for most of the game, I finally had a 6 letter word that I slapped down on the board with glee.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Ha, 15 points, even though I had to use both blank tiles to get it,&quot; I said peppily.  &quot;But at least it is not a word you see every day.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Oh, without a doubt, &quot; Chet exclaimed, &quot; it is a magnificent word.  However, what you have put down is only a variation of its spelling.  If you had had one additional tile, you could have used the 7-letter version and scored more points.  However, that is a physical impossibility.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Confused, I stared at the board, and did not understand what Chet was trying to tell me.  Can you tell me what I could have done, if only the laws of physics had let me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Notes&lt;/em&gt;&lt;/strong&gt;:  Assume the English language and version of Scrabble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, for scoring purposes, assume that no letter was on a modifying square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If people seem to be having trouble, I will add in a hint later that will tell you what the point score of the new word would be if I had been able to play it.&lt;/p&gt;&#xA;" OwnerUserId="14233" LastEditorUserId="20274" LastEditDate="2016-06-13T10:14:30.160" LastActivityDate="2016-06-13T10:14:30.160" Title="Scrabble Impossibilities" Tags="|word|english|board-games|scrabble|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23371" PostTypeId="1" AcceptedAnswerId="23374" CreationDate="2015-10-21T12:13:22.067" Score="4" ViewCount="160" Body="&lt;p&gt;Tush (also known as &lt;a href=&quot;https://en.wikipedia.org/wiki/Tush_(TV_series)&quot; rel=&quot;nofollow&quot;&gt;The Bill Tush Show&lt;/a&gt;) was an American late-night television sketch comedy and variety show developed for Bill Tush and directed by R. T. Williams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone explain to me why the show was simply cobbled-together mixed-up crap?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is not a rant—in fact, living in Britain as I do, I have never seen the show. It is a genuine puzzle. Please explain all three elements of the show that I have mentioned. How can they be used together in a logical way to provide the answer?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The link I gave was for interest only and to demonstrate that it was a real show. You can ignore any information gained by following the link. All the necessary information is given in the visible part of the question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="4050" LastEditDate="2015-10-21T20:54:12.117" LastActivityDate="2015-10-21T20:54:12.117" Title="The Bill Tush Show" Tags="|enigmatic-puzzle|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23372" PostTypeId="2" ParentId="23370" CreationDate="2015-10-21T12:15:22.717" Score="22" Body="&lt;p&gt;The word was...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;PIZAZZ. The two ending Z's were blank, so you scored 3+1+10+1=15 points.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the reason you couldn't play the more common variant was...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Scrabble sets only have one Z and two blanks; PIZZAZZ has four Z's, so you couldn't play it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2015-10-21T12:15:22.717" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23374" PostTypeId="2" ParentId="23371" CreationDate="2015-10-21T13:26:18.110" Score="6" Body="&lt;p&gt;Take some crap:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Bullshit&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mix it up:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; B u ll sh i t -&gt; B i ll t u sh&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cobble it together:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bill tush&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1497" LastActivityDate="2015-10-21T13:26:18.110" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23375" PostTypeId="2" ParentId="23368" CreationDate="2015-10-21T13:35:53.443" Score="1" Body="&lt;p&gt;Could it be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One Unclad Navel Genre&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BS is famous for her unclad navel&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or perhaps:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nude navel race on glen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because while I don't &lt;em&gt;remember&lt;/em&gt; a Scottish theme to any BS music videos, I'm happy to say I'm not much of an expert on the topic.&lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="4050" LastEditDate="2015-10-21T13:41:19.080" LastActivityDate="2015-10-21T13:41:19.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23376" PostTypeId="1" AcceptedAnswerId="23377" CreationDate="2015-10-21T14:14:46.003" Score="21" ViewCount="4990" Body="&lt;p&gt;This is a simple one, but it's BTTF (Back to the Future) day, so I figured it was appropriate:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What goes a minute in four seconds?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16131" LastEditorUserId="12450" LastEditDate="2015-10-22T06:45:27.400" LastActivityDate="2015-10-22T18:08:43.847" Title="Simple riddle involving time travel" Tags="|riddle|" AnswerCount="3" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="23377" PostTypeId="2" ParentId="23376" CreationDate="2015-10-21T14:42:45.383" Score="31" Body="&lt;p&gt;The&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Earth rotates through one minute of arc every 4 seconds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2015-10-21T14:42:45.383" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23378" PostTypeId="2" ParentId="23376" CreationDate="2015-10-21T14:44:30.677" Score="8" Body="&lt;p&gt;There is no time travel involved.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The fastest cars in the world (not counting drag racers)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because... (The math is pretty rough but I think it shows I'm on the right track)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Minute&quot; means &quot;minute of arc&quot; and not a unit of time&lt;br&gt;&#xA; &lt;a href=&quot;https://www.google.com/search?q=Diameter%20of%20Earth&quot;&gt;Diameter of Earth&lt;/a&gt; = 7,917.5 miles&lt;br&gt;&#xA; Perimeter of Earth = 24,873.6 miles (I know it's not really a sphere)&lt;br&gt;&#xA; One degree of arc = 69.1 miles&lt;br&gt;&#xA; One minute of arc = 1.15 miles&lt;br&gt;&#xA; &lt;a href=&quot;https://www.google.com/search?q=venom%20gt%20top%20speed&quot;&gt;Top speed of the Venom GT&lt;/a&gt; = 270 mph&lt;br&gt;&#xA; Distance traveled by Venom GT in 15 seconds = 1.13 miles   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2015-10-21T15:07:12.200" LastActivityDate="2015-10-21T15:07:12.200" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23383" PostTypeId="2" ParentId="23376" CreationDate="2015-10-21T16:02:33.440" Score="14" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The hour hand on a 24-hour clock.&lt;br/&gt;&lt;br/&gt;There are 360 degrees in a circle and 60 minutes in a degree, so a minute is 1/360/60 = 1/21600 of a circle.  The hour hand moves 1/24/60/60 = 1/86400 of a circle each second, so in 4 seconds it moves 4/86400 = 1/21600.&lt;br/&gt;&lt;br/&gt;Note:  If you say &quot;earth&quot;, make sure to qualify that you are talking about its rotation with respect to the sun, since its rotation with respect to the stars is about 1/365 off of one full rotation each day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1818" LastEditorUserId="1818" LastEditDate="2015-10-22T18:08:43.847" LastActivityDate="2015-10-22T18:08:43.847" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23384" PostTypeId="1" AcceptedAnswerId="23394" CreationDate="2015-10-21T16:04:16.530" Score="5" ViewCount="946" Body="&lt;p&gt;&lt;sub&gt;You are the protagonist in a game designed by NeedAName Inc. Proceed through the levels using&#xA;your video game knowledge to solve the puzzles therein. Solve all levels and beat the game!&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;Level 4: Secret Cow Boss Level&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/23157/dungeon-crawler-level-3&quot;&gt;Previous Level&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having bested the previous levels and proven your prowess with games, you now find yourself in the inner workings of a&#xA;console running a video game. You look around and on each of 3 walls you see a section with information written. On&#xA;the floor of the room in the center, you see: 'Part 1 → Part 2 → Part 3 → Who are you?'. Using the information on the walls and your&#xA;knowledge (video game and otherwise), can you answer the final question and beat the game?&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Part 1: Physics&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;$8.9875517873681764 \times 10^{16}$&lt;br&gt;&#xA;$3.6266180501548694449 \times 10^{47}$&lt;br&gt;&#xA;$7.3890561073081489$&lt;br&gt;&#xA;$9.3094189052089 \times 10^{9}$&lt;br&gt;&#xA;$1.64461934145166806681$&lt;br&gt;&#xA;$8.396989835536 \times 10^{-6}$&lt;br&gt;&#xA;$3.6266180501548694449 \times 10^{47}$&lt;br&gt;&#xA;$7.3890561073081489$  &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Part 2: Memory&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;0x00000000: 0&lt;/strong&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;0x00000001-00000005: 19--&lt;/strong&gt;  &#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;0x00000001 htrit ipvgw ifsfj iccsk titwr weyfp cqgc. jqobe xjcsz giovg lqxwj elkew xudvc rpvko sxit.  &lt;/li&gt;&#xA;&lt;li&gt;0x00000002 vside bcmpx csanp rqexm ezpet zjwem peup. ttrup hvzgt ntzpg ,tjqy vaake ljatp .  &lt;/li&gt;&#xA;&lt;li&gt;0x00000003 wyzji mphce atrmo jctv, yoglg gsiwa oolkx .mxhp hifes raojx jpwjc spxqq rnm.  &lt;/li&gt;&#xA;&lt;li&gt;0x00000004 yzvqd glqqf wlanp izas, nlmmp qhctu sxmpo .amau twvsi debcs heljf jcwvz jrylt rf?  &lt;/li&gt;&#xA;&lt;li&gt;0x00000005 cyesa mzkez rxomp ,mody weuze ofwj. fjyhq fxbhf ymylw hrflx goswl pdi.  &lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;0x00000006-00000007: #--&lt;/strong&gt;  &#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;0x00000006 sfmye usiux eqldc iceyw e,xtx jzrqy pyioz zj.wj elkyw yaoea kelir bhen, ppzsj gikav tvf.  &lt;/li&gt;&#xA;&lt;li&gt;0x00000007 jfqfn pthhp qflfj swoss txlrv esrit d.cqf vjuoo itxcx pfwpk qctuh tzgxi fkjdw .  &lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;0x00000008-00000009: --&lt;/strong&gt;  &#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;0x00000008 jphle izk,u zpiif c,tnf qgeso sotrf tjzr. kewdo vcjky eqfjy enoix tjyev tss.  &lt;/li&gt;&#xA;&lt;li&gt;0x00000009 htvxt pqjqf vxtbe mqyw, nrflh hcsra etwe. ssbmb ycdtx xcbyc qfjnn etroj vnsd?  &lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Part 3: Graphics&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;$\begin{bmatrix}&#xA;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;4&amp;amp;4&amp;amp;4&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0\\&#xA;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;3&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;8&amp;amp;4&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9\\&#xA;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;3&amp;amp;4&amp;amp;4&amp;amp;4&amp;amp;4&amp;amp;4&amp;amp;4&amp;amp;8&amp;amp;9&amp;amp;9&amp;amp;9\\&#xA;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9\\&#xA;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;4&amp;amp;4&amp;amp;4\\&#xA;9&amp;amp;4&amp;amp;9&amp;amp;4&amp;amp;9&amp;amp;4&amp;amp;9&amp;amp;4&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;7&amp;amp;9&amp;amp;9&amp;amp;9\\&#xA;7&amp;amp;6&amp;amp;7&amp;amp;6&amp;amp;7&amp;amp;6&amp;amp;7&amp;amp;6&amp;amp;7&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;7&amp;amp;9&amp;amp;9&amp;amp;9\\&#xA;9&amp;amp;2&amp;amp;9&amp;amp;2&amp;amp;9&amp;amp;2&amp;amp;9&amp;amp;2&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;7&amp;amp;9&amp;amp;9&amp;amp;9\\&#xA;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;7&amp;amp;9&amp;amp;9&amp;amp;9\\&#xA;4&amp;amp;4&amp;amp;4&amp;amp;4&amp;amp;4&amp;amp;4&amp;amp;4&amp;amp;4&amp;amp;4&amp;amp;4&amp;amp;4&amp;amp;4&amp;amp;7&amp;amp;4&amp;amp;4&amp;amp;4&#xA;\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{bmatrix}&#xA;0&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0\\&#xA;9&amp;amp;4&amp;amp;4&amp;amp;4&amp;amp;4&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0\\&#xA;7&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;7&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0\\&#xA;7&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;7&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0\\&#xA;7&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;7&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;3&amp;amp;1&amp;amp;5&amp;amp;8&amp;amp;0&amp;amp;0\\&#xA;7&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;7&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;7&amp;amp;7&amp;amp;0&amp;amp;0&amp;amp;0\\&#xA;7&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;7&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;4&amp;amp;4&amp;amp;7&amp;amp;7&amp;amp;4&amp;amp;4&amp;amp;0\\&#xA;7&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;7&amp;amp;9&amp;amp;9&amp;amp;7&amp;amp;4&amp;amp;4&amp;amp;4&amp;amp;4&amp;amp;4&amp;amp;4&amp;amp;7\\&#xA;7&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;7&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;7&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;9&amp;amp;7&amp;amp;0\\&#xA;7&amp;amp;4&amp;amp;4&amp;amp;4&amp;amp;4&amp;amp;7&amp;amp;4&amp;amp;4&amp;amp;4&amp;amp;7&amp;amp;4&amp;amp;4&amp;amp;4&amp;amp;4&amp;amp;7&amp;amp;0&#xA;\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The accepted answer will address all parts of the puzzle and provide an answer to each. Teamwork is encouraged, if you want to make a community wiki, feel free.&lt;/p&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-10-22T01:43:35.070" Title="Dungeon Crawler - Final Level" Tags="|riddle|mathematics|cipher|physics|video-games|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23385" PostTypeId="2" ParentId="23384" CreationDate="2015-10-21T17:20:53.753" Score="6" Body="&lt;p&gt;Well, I think I've got the physics part:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The numbers appear to all be squares of physical constants, modulo a certain amount of rounding.  Specifically, we have&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; c (speed of light)&lt;br&gt;&#xA; L (Avogadro's number, also sometimes written N_A)&lt;br&gt;&#xA; e (mathematical constant)&lt;br&gt;&#xA; F (Faraday constant)&lt;br&gt;&#xA; A (Appears to be something called Glaisher's constant, which I haven't come across before)&lt;br&gt;&#xA; b (Wien displacement constant)&lt;br&gt;&#xA; L (Avogadro's number, also sometimes written N_A)&lt;br&gt;&#xA; e (mathematical constant)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For a possible answer to part 1 of &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; cLeFAbLE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit:  Based on dperry's decryption of the riddles in part 2, 6 and 7 appear to be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Abra and Jynx, two more Pokémon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Riddle 9 refers to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the PlayStation, which ran on an MIPS microprocessor.  MIPS is a Reduced Instruction Set Computing (RISC) chip, and uses 32 bits.  (I'm sure I'm using some of the technical language wrong, but the answer is 32)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17036" LastEditorUserId="17036" LastEditDate="2015-10-21T19:48:51.253" LastActivityDate="2015-10-21T19:48:51.253" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23386" PostTypeId="1" AcceptedAnswerId="23389" CreationDate="2015-10-21T17:27:13.730" Score="2" ViewCount="308" Body="&lt;p&gt;On a Wednesday afternoon, Marty McFly starts his DeLorean at 14:00 hours and travels 30 years in the past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After spending 3 days 30 minutes, travels 60 years to the future where he stays for another 3 days and 30 minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He sets his clock to 30 years in the past to 14:00 hours and safely makes it to the present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question : So what day of the week is it for him today and what is the total time Marty McFly was actually &quot;away&quot; Time Travelling?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hint : No Tricks, simple straight forward maths. Will clear your time travel concepts. Disregard leap year, location, time zone changes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9892" LastEditorUserId="9892" LastEditDate="2015-10-21T19:31:57.947" LastActivityDate="2015-10-21T19:31:57.947" Title="Time Traveller Maths" Tags="|mathematics|calculation-puzzle|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23387" PostTypeId="1" AcceptedAnswerId="23388" CreationDate="2015-10-21T17:28:21.183" Score="1" ViewCount="496" Body="&lt;p&gt;My friend is a bit of a video game addict.  Recently, after a bout of video-game-induced violence (I won't go into details here) he was sentenced to six months in jail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part of the conditions of his sentence are that he have nothing to do with video games at all; he can't even speak about them while serving his sentence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He's going to be released soon, and I want to throw him a party.  I've decided I'm going to have a cake in the shape of his favourite video game character, but I realized I don't know who that is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I managed to smuggle a note into the prison asking him to let me know, but unfortunately his only method of getting a message back to me is via email, and all of his messages are read by the prison guards, so he dare not mention anything about video games.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just received this message from him today, and I think he's trying to answer my question, but he seems to have encrypted his answer to get it past the guards.  Can you help me figure out who his favourite character is, so I can have the cake ready for his party?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is his message:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dear friend,  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Thank you so much for your letter; I've been missing you.  I am looking forward to partying with you when I get out of here (in only 64 more hours!). In response to the question you asked,  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;  &#xA;  &lt;p&gt;(Oops; looks like something went wrong with my keyboard there.)  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Anyway, I'll see you soon.  Hang in there!  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Your friend.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="4421" LastEditDate="2015-10-21T18:16:47.420" LastActivityDate="2015-10-21T19:55:24.820" Title="The Letter of the Law" Tags="|cipher|video-games|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23388" PostTypeId="2" ParentId="23387" CreationDate="2015-10-21T18:02:10.310" Score="8" Body="&lt;p&gt;Convert that string from base64. That gives a long string of two-digit hexadecimal numbers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;71 75 75 2e 2e 5f 5f 0d 0a ...&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then take the characters with those ASCII codes to get an ASCII art picture of &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Pikachu.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/UJHL2.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UJHL2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="4421" LastEditDate="2015-10-21T19:55:24.820" LastActivityDate="2015-10-21T19:55:24.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23389" PostTypeId="2" ParentId="23386" CreationDate="2015-10-21T18:15:17.347" Score="5" Body="&lt;p&gt;The day of the week is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Tuesday&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If today is X and I go back in time Y years, then spend 3 days and go back to the future by Y years, then the day I end up at is X+3. It might be easier to comprehend if we think about it in only days. Let's say I start at day 10. I go back in time 5 days to day 5. I spend 1 day in the past so now I'm at day 6. Then I go 5 days back to the future and end up at day 11, which is 1 day from the present day I started at. Going through all the time travel Marty does, there is a total of 3+3=6 days passing, so he would end up 6 days from when he left, which is a Tuesday.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what you mean by how much time he spent time travelling so I'll give both interpretations:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The total time that passed for him was 6 days and 1 hour. Just simple addition of &lt;code&gt;3 days 30 minutes + 3 days 30 minutes&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The total time he was gone from the present is 6 days exactly. When he finally comes back to the present he goes back 30 years to the time 14:00. When he does this, he is a total of 30 years, 6 days, 1 hour from the time he originally left the present (as shown above). This means he is at a time of 15:00 (he left at 14:00, and is X days + 1 hour ahead) before he goes back to the present. By going back 30 years to 14:00, he actually goes back 30 years 1 hour. This results in his total time &quot;gone&quot; from the present at &lt;code&gt;6 days 1 hour - 1 hour = 6 days&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11727" LastActivityDate="2015-10-21T18:15:17.347" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23390" PostTypeId="2" ParentId="23386" CreationDate="2015-10-21T18:17:29.783" Score="1" Body="&lt;p&gt;You say, &quot;Zero seconds for Time Travel.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You ask, &quot;How long did he take Time Travelling?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer: zero seconds. (by your definition)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest of the time he was doing other things.&lt;/p&gt;&#xA;" OwnerUserId="14177" LastActivityDate="2015-10-21T18:17:29.783" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23391" PostTypeId="2" ParentId="23384" CreationDate="2015-10-21T18:46:22.360" Score="3" Body="&lt;p&gt;Building off Rauwyn's answer (his final answer was the key to a Vigenère cypher), I worked out the questions in the Memory part and have a few answers to far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;0x00000001-00000005: 19--&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;0x00000001&lt;/strong&gt; find the release year of this platformer. float through dreams with its sucky performer &lt;em&gt;(Kirby's Dreamland, 1992)&lt;/em&gt;&lt;br&gt;&#xA; &lt;strong&gt;0x00000002&lt;/strong&gt; the year in my name not the year of release. ported to console, i fly up with ease&lt;br&gt;&#xA; &lt;strong&gt;0x00000003&lt;/strong&gt; unveiled at world fair, to face her and him. i showed to many the secret of nim &lt;em&gt;(the nimatron from the 1940 New York World's Fair?)&lt;/em&gt;&lt;br&gt;&#xA; &lt;strong&gt;0x00000004&lt;/strong&gt; world famous game expo, ill blow your mind. what is the year of the first of my kind? &lt;em&gt;(E3, 1995)&lt;/em&gt;&lt;br&gt;&#xA; &lt;strong&gt;0x00000005&lt;/strong&gt; an analog console, im sure to amuse. find out when i was created or lose  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;0x00000006-00000007: #--&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;0x00000006&lt;/strong&gt; quite the small creature, with only one move. with instant withdrawal, elusive i prove.&lt;br&gt;&#xA; &lt;strong&gt;0x00000007&lt;/strong&gt; humanoid female you dont want to miss. youre under my spell if you give me a kiss&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;0x00000008-00000009: --&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;0x00000008&lt;/strong&gt; hedgehog, soldier, plumber domination. its your final final destination. &lt;em&gt;(Super Smash Bros (Brawl?))&lt;/em&gt;&lt;br&gt;&#xA; &lt;strong&gt;0x00000009&lt;/strong&gt; first of four stations, i read from a disc. how many bits can you find in my risc?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13101" LastEditorUserId="13101" LastEditDate="2015-10-21T19:23:44.943" LastActivityDate="2015-10-21T19:23:44.943" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23392" PostTypeId="2" ParentId="23384" CreationDate="2015-10-21T19:43:02.877" Score="0" Body="&lt;p&gt;I don't have enough points to comment, but here's what I got.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Battlefield 1942 was ported to console and one could say it has been &quot;flying&quot; up in popularity since then.&#xA; or it could be the game &quot;1942&quot; released for arcade in 1984 and ported to the Famicom console in 1985. It's a scrolling flying shooter. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Might refer to the Magnavox Odyssey created in 1972 which was a digital console but widely mistaken to be an analog console.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Oops just saw Rauwyn's answer for 9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;9:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Playstation 1 risc R3000 I believe had 32bits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16471" LastEditorUserId="16471" LastEditDate="2015-10-22T00:21:47.643" LastActivityDate="2015-10-22T00:21:47.643" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23393" PostTypeId="2" ParentId="382" CreationDate="2015-10-21T22:23:14.450" Score="0" Body="&lt;p&gt;This isn't a &lt;strong&gt;complete&lt;/strong&gt; solution and probably may not even be completely &lt;strong&gt;right&lt;/strong&gt; but I'm posting anyway since I found the partial solution quite simple and the Author stressed on &lt;em&gt;elegance&lt;/em&gt;. Also since an answer has long been accepted. I'm hoping others could add to this answer to make it complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Question can be&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is One subtracted from the Number you are thinking of prime?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1 - 1 = 0 (Not Applicable, 0 is not a Natural Number)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 - 1 = 1 (I Don't know, Controversial Answer! But neither prime nor composite!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 - 1 = 2 (Yes)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 - 1 = 3 (Yes)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5 - 1 = 4 (No)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the only problem is for 2 &amp;amp; 3. &quot;Sometimes&quot; means the question has to do with probability or a random physical event.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will edit my answer if I can complete it!&lt;/p&gt;&#xA;" OwnerUserId="17040" LastActivityDate="2015-10-21T22:23:14.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23394" PostTypeId="2" ParentId="23384" CreationDate="2015-10-22T01:43:35.070" Score="3" Body="&lt;p&gt;I went ahead and jumped to part 3 with the information we had.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I mapped the numbers in the 2D arrays in part 3 to the answers that were found for part 2 and interpreted it as ASCII. This is the result: &lt;a href=&quot;http://pastebin.com/SNjBKUU0&quot; rel=&quot;nofollow&quot;&gt;http://pastebin.com/SNjBKUU0&lt;/a&gt; (sorry I couldn't figure out how to format it here). I figured out #8 and #5 by figuring out what the image should show and doing some Googling.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1947, year of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cathode_Ray_Tube_Amusement_Device&quot; rel=&quot;nofollow&quot;&gt;first video game&lt;/a&gt; made entirely with analog electronic devices.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;8:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Event 41 in Super Smash Bros takes place on Final Destination with Mario, Sonic, and Snake.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Final answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am Mario. Given away by the Piranha Plant and the ? blocks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17043" LastActivityDate="2015-10-22T01:43:35.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23395" PostTypeId="2" ParentId="23292" CreationDate="2015-10-22T02:33:37.377" Score="4" Body="&lt;p&gt;I hypothesize that 22 is the maximum, and the maximum follows &lt;a href=&quot;http://oeis.org/A033484&quot; rel=&quot;nofollow&quot;&gt;this oeis.org sequence&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I've had a little progress without a computer. I want briefly to return to @Joel's solution to the 10 coin, 3 weighing puzzle and the reason why it works. To start we'll say heavy coins weigh 1 and light weigh 0.  Also we're assuming that all weighing are balanced (since otherwise we're done), have the same number of coins on each side and we want to show that with a certain configuration we can prove that all the coins weigh the same. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1&lt;/strong&gt; In my first weighing I'm going to take a group of coins that I'll call $A$ with an as yet unknown size and with weight $a$. They will be on one side, and I'll get to the other side later. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2&lt;/strong&gt; In my second weighing, I'll take $A$ and another coin $B$ which I'll say weighs $b$ on one side, and other coins of equal number on the other side, which I'll call group $C$. Obviously $C$ weighs $a+b$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3&lt;/strong&gt; Now for my third weighing, I'll take $C$ and $B$ and weigh it against $A$ and two as yet untested coins ($D$). Then the two untested coins must weigh $2b$ (since $C+B-A$ weighs $(a+b) +b-a$). Therefore the two untested coins must each weigh $b$. So now I have three coins that all weigh $b$ and that aren't in $A$. Suppose I had those three on the other side of weighing 1. Then $A$ had three coins in it, and it must consist of three coins each weighing $b$, and so $C$ must have four coins each weighing $b$, and so all 10 coins must be the same. This is a proof that Joel's previous solution works as intended. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, one could also instead have simply continued with a fourth weighing following a similar strategy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4&lt;/strong&gt; Weigh $C$, $B$ and $D$ against $A$ and four as yet untested coins ($E$).  Once again it is easy to show that the 4 new coins ($E$) weigh $4b$, and so each one weighs $b$. In this case when you go back to the first weighing you weigh $A$ against 7 coins ($B$, $D$ and $E$), which weigh $7b$, and so once again it follows that all 22 coins weigh $b$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, 22 coins is not 30 coins, so I suspect that my generalization may not be the best one available.  It does show that for $n&amp;gt;1$ weighings there is a strategy to get $3.2^{n-1}-2$ coins sorted out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize this answer for 4 weighings:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$1,\ldots,7$ v $8,9,10,11,12,13,14$&lt;/li&gt;&#xA;&lt;li&gt;$1 ,\ldots, 7,8$ v $15,\ldots,22$&lt;/li&gt;&#xA;&lt;li&gt;$1,\ldots,7,9,10$ v $8,15,\ldots,22$&lt;/li&gt;&#xA;&lt;li&gt;$1,\ldots,7,11,12,13,14$ v $8,9,10,15,\ldots,22$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2015-11-02T14:24:39.093" LastActivityDate="2015-11-02T14:24:39.093" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="23396" PostTypeId="2" ParentId="23265" CreationDate="2015-10-22T08:04:13.223" Score="1" Body="&lt;p&gt;&lt;em&gt;Thanks Milo Brandt's for a nice puzzle.&lt;/em&gt;&#xA;The answer to his questions are &quot;&lt;strong&gt;yes&lt;/strong&gt;&quot; and &quot;&lt;strong&gt;yes&lt;/strong&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even more, I state that:&lt;br&gt;&#xA;&lt;strong&gt;S1. At least one person will go to town, on day N or earlier.&lt;br&gt;&#xA;S2. Everyone who goes to the town will be part of a proper meeting eventually, on day (N+1) or earlier.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here and later, N is number of organisations with at least 1 member. org - shortcut organisation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose all organisations are not empty and all people are male :), and each man is in some organisation. Obviously, this doesn't change result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, let's note, that &lt;em&gt;the condition for a person who got L letters to stay home is that he can assume that there is exactly L orgs and find no contradictions with it.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's use math. induction to prove that S1 and S2 is true for any N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;a) Base.&lt;/em&gt; If there is only 1 org, then all people got 0 letters and go to town on the first day, and have a proper meeting immediately. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;b) Let's suppose that S1 and S2 are true for any town with k-1 orgs and prove this for k orgs.&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, there is k orgs. We consider a random guy X. He got at most k-1 letters and assumes that there is no other ogrs. Thereby, he sees the town as a town with k-1 orgs and expects at least someone to show up on the (k-1)-th day at least. S1 is true if everything goes as he expected. If this is not the case, then X understands that he don't have a full picture and goes to the town on k-th day. That proves that S1 is always true for k orgs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's consider a random guy Y, who is already in town and still doesn't have a proper meeting.&lt;br&gt;&#xA;First of all, all guys, who belong to all orgs of Y, will go to town on the next day. Indeed, Y is not in their letters, but they see him in town...&lt;br&gt;&#xA;If at least one Y's org has no members left, Y will get his proper meeting (next day, which is (k+1)-th day) (making S2 true).&lt;br&gt;&#xA;If this is not the case, let's consider a guy Z - a member of an arbitrary Y's org, who stays at home.&lt;br&gt;&#xA;Z sees the town as a town with k-1 orgs, and Y is a member of some of them. So Z expects Y to get proper meeting on day k at least. And either Y got it on day k, or Z goes to town on day k+1 and Y has proper meeting(-s) then (because Z was chosen arbitrary and all other co-members of Y and Z will do the same). That means S2 is true for all cases with k orgs in town, what had to be proven.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thereby, according to math induction principle the initial statements are true.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2015-10-22T09:35:44.660" LastActivityDate="2015-10-22T09:35:44.660" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23397" PostTypeId="1" AcceptedAnswerId="23405" CreationDate="2015-10-22T08:53:34.283" Score="9" ViewCount="783" Body="&lt;p&gt;Yesterday afternoon I met professor Halfbrain at a coffee place. &#xA;The professor looked very tired. He told me that he hadn't slept for many days, spending his time with &lt;strong&gt;writing lots of zeros&lt;/strong&gt; and with &lt;strong&gt;computing square roots&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&quot;Writing lots of zeros?&quot;&lt;/strong&gt;, I asked him surpised. &#xA;It turned out that the professor had first written down the integer $49$ and then inserted one $0$ in the middle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&quot;Computing square roots?&quot;&lt;/strong&gt;, I asked him. &#xA;It turned out that the professor had noticed that his first number $409$ wasn't a perfect square (whereas the number $49=7^2$ is a well-known square). So he had inserted another $0$, and noticed that $4009$ wasn't a perfect square either. He inserted one more $0$, but $40009$ is not square. And so on.&#xA;After inserting many many further zeros into the number, professor Halfbrain finally managed to detect a square number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us denote by $H(k)$ the integer whose decimal representation is of the form&#xA;$4000\cdots009$ with exactly $k$ zeros between the digits $4$ and the $9$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What is the smallest integer $k\ge1$, for which $H(k)$ is a perfect square? Does such a $k$ really  exist,  or has the professor once again made one of his notorious mathematical blunders?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-10-23T06:43:29.867" Title="Professor Halfbrain and the next square" Tags="|mathematics|calculation-puzzle|number-theory|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23398" PostTypeId="1" CreationDate="2015-10-22T09:46:23.183" Score="-6" ViewCount="464" Body="&lt;p&gt;What is the first clue to this question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the second clue to this question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;What is the hint to this question?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="14177" LastEditDate="2015-10-22T10:06:29.080" LastActivityDate="2015-10-27T11:34:32.633" Title="What is the title to this question?" Tags="|enigmatic-puzzle|" AnswerCount="5" CommentCount="13" ClosedDate="2019-11-14T17:20:46.960" ContentLicense="CC BY-SA 3.0" />
  <row Id="23399" PostTypeId="2" ParentId="23397" CreationDate="2015-10-22T09:59:13.700" Score="1" Body="&lt;p&gt;It's not possible, there is no such $k$. We know, because the number ends on 9, that the last digit of our would be perfect square is either a 3 or a 7. We therefor know that if there was a perfect square, it would be of the form $a*10+3$ or $a*10+7$, where a is integer. Squaring this number would have to give $H(k)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when we do this, we get $100*a^2+60*a+9$ and $100*a^2+140*a+49$ respectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reasoning isn't &lt;em&gt;as&lt;/em&gt; helpful as i thought it was, so we can do without, but i'll leave it in case anyone wants to talk about it :P&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;But since $10^n$ with $n$ an integer can't be cleanly divided by either 6 or 14 for any value of $n$, we are always left with 'stray numbers' where $H(k)$ demands that we have zeros&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are left with the equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $4*10^{k+1}=100*a^2+60*a$ or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $4*10^{k+1}=100*a^2+140*a+40$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking (1) we simplify to $5(5*a^2+3*a)=10^{k+1}$, which has a single integer solution (according to wolfram alpha) at $k=0$ and $a=-1$, (which gives the value -7 for our perfect square) but a requirement is that $k&amp;gt;0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking (2) likewise gives just 1 integer solution, namely $k=0$ and $a=0$, but $k&amp;gt;0$&lt;/p&gt;&#xA;" OwnerUserId="16846" LastEditorUserId="16846" LastEditDate="2015-10-22T13:40:46.163" LastActivityDate="2015-10-22T13:40:46.163" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23400" PostTypeId="2" ParentId="23398" CreationDate="2015-10-22T10:10:05.110" Score="7" Body="&lt;p&gt;Just a guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;What is the answer to this question?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16608" LastActivityDate="2015-10-22T10:10:05.110" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23401" PostTypeId="2" ParentId="23398" CreationDate="2015-10-22T10:22:53.463" Score="4" Body="&lt;p&gt;They're &quot;What is the title to this question?&quot;, &quot;What is the first clue to this question?&quot;, &quot;What is the second clue to this question?&quot;, and &quot;What is the hint to this question?&quot;, respectively.&lt;/p&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2015-10-22T10:22:53.463" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23402" PostTypeId="2" ParentId="23398" CreationDate="2015-10-22T13:06:11.013" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is the title.&lt;br&gt;&#xA; This is the first clue.&lt;br&gt;&#xA; This is the second clue.&lt;br&gt;&#xA; This is the hint.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="8874" LastEditDate="2015-10-22T13:26:08.733" LastActivityDate="2015-10-22T13:26:08.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23403" PostTypeId="2" ParentId="23398" CreationDate="2015-10-22T13:30:37.657" Score="9" Body="&lt;p&gt;Well clearly:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;&lt;strong&gt;What&lt;/strong&gt;&quot; is the answer to this question!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;What&quot; is the title, the first clue, the second, and indeed the hint.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9444" LastEditorUserId="9444" LastEditDate="2015-10-22T13:36:23.677" LastActivityDate="2015-10-22T13:36:23.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23404" PostTypeId="1" CreationDate="2015-10-22T13:38:09.630" Score="9" ViewCount="2999" Body="&lt;p&gt;One for the front&lt;br&gt;&#xA;Two for the back&lt;br&gt;&#xA;You will never need three.&lt;br&gt;&#xA;What is this?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought it might be personality.&lt;/p&gt;&#xA;" OwnerUserId="17052" LastEditorUserId="4050" LastEditDate="2015-10-23T09:39:13.327" LastActivityDate="2015-10-23T09:39:13.327" Title="Riddle I have no idea: &quot;One for the front / Two for the back ...&quot;" Tags="|riddle|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23405" PostTypeId="2" ParentId="23397" CreationDate="2015-10-22T14:03:44.593" Score="15" Body="&lt;p&gt;$X^2 = 4000..009$&lt;br&gt;&#xA;$X$ is obviously odd. Let's say $X = 2y+3$:&lt;br&gt;&#xA;$4y^2+12y+9 = 4000..009$&lt;br&gt;&#xA;$y^2+3y = 1000..000 = 10^{k+1}$&lt;br&gt;&#xA;$y * (y+3) = 2^{k+1} * 5^{k+1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y$ and $y+3$ are different modulo 2 and modulo 5, also $y&amp;lt;y+3$, thereby $y = 2^{k+1}$ and $y+3 = 5^{k+1}$. But this works only for $k = 0$.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2015-10-23T06:43:29.867" LastActivityDate="2015-10-23T06:43:29.867" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23406" PostTypeId="2" ParentId="23404" CreationDate="2015-10-22T14:11:24.737" Score="16" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mirror&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(one to see your front; two to see your back; you never need three)&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-10-22T14:11:24.737" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23407" PostTypeId="1" AcceptedAnswerId="23409" CreationDate="2015-10-22T14:54:46.810" Score="-1" ViewCount="12781" Body="&lt;p&gt;A man lives in a small house with a farm as his back garden and has a river beside his house. On the other side of the river is a shop. One day, he visited the shop, he bought a chicken and a fox for his farm and bought a bag of corn to feed the chicken. The man can only take one thing and himself across in the boat. Without killing any animals, or letting one animal eat another animal or item, how will the man get across?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I heard this once but could never find a solution.&lt;/p&gt;&#xA;" OwnerUserId="11457" LastActivityDate="2015-10-22T15:09:43.137" Title="A man, a chicken, and a fox?" Tags="|river-crossing|" AnswerCount="1" CommentCount="1" ClosedDate="2015-10-22T15:04:23.623" ContentLicense="CC BY-SA 3.0" />
  <row Id="23408" PostTypeId="1" AcceptedAnswerId="23425" CreationDate="2015-10-22T14:55:38.047" Score="3" ViewCount="578" Body="&lt;p&gt;In the past, I've made some simple ciphers as part of larger riddles, and sometimes this has lead to puzzles coming undone in the middle. So all of you who enjoy trying to crack cryptograms with analytic techniques, here's one just for you!! Can you determine both the key used to encrypt and the original message (it will be in English) encrypted from the following ciphertext? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;u axft rpalg tymh dytyaiju nffvlag tmb knrd ys, l'ie ymr pbmvzhv bf gmfdaozm oqq emqb bbu tmb qbw jqbvr tymh.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I will accept the answer that provides the key, original message, and steps for solving. If you use a program or tool to do part of it (this is not necessary, all steps can be done by hand.), that's fine, but you have to be able to explain what the tool is doing and how it determines the results. Good luck!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HINT:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;First step should be determining the method of encryption. You already know enough to narrow in on this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;P.S.:&lt;br&gt;&#xA;GentlePurpleRain has cracked the code, and has a detailed explanation below, but if you had success with another method, please feel free to contribute, I'd be very interested in hearing other approaches/techniques. This also helps me and others improve future cryptograms.&lt;/p&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="13970" LastEditDate="2015-10-22T21:06:53.927" LastActivityDate="2015-10-22T21:59:15.810" Title="Cryptanalysis Time!" Tags="|cipher|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23409" PostTypeId="2" ParentId="23407" CreationDate="2015-10-22T15:09:43.137" Score="2" Body="&lt;p&gt;He can do it in seven steps:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Cross the river with the chicken.&lt;/li&gt;&#xA;&lt;li&gt;Return across the river with nothing, leaving the chicken behind.&lt;/li&gt;&#xA;&lt;li&gt;Cross the river with the fox.&lt;/li&gt;&#xA;&lt;li&gt;Return across the river with the chicken, leaving the fox behind.&lt;/li&gt;&#xA;&lt;li&gt;Leave the chicken on the shop side, and cross the river with the corn.&lt;/li&gt;&#xA;&lt;li&gt;Return across the river with nothing, leaving the corn behind.&lt;/li&gt;&#xA;&lt;li&gt;Cross the river with the chicken again.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;At no time are the fox and chicken, or chicken and corn, left alone together.&lt;/p&gt;&#xA;" OwnerUserId="16945" LastActivityDate="2015-10-22T15:09:43.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23410" PostTypeId="1" AcceptedAnswerId="23412" CreationDate="2015-10-22T15:18:34.887" Score="24" ViewCount="1926" Body="&lt;p&gt;There's a song I really like but I can't remember the title and I certainly can't remember the words. Perhaps you can help me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've summarised the lyrics as best I can, but in a rather literal way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm afraid I've forgotten the first verse completely so that makes it even harder. Anyway here goes with the bit I can remember.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Greetings! Lots of greetings! But, oops mind that hole.&lt;/p&gt;&#xA;  &#xA;  &lt;hr&gt;&#xA;  &#xA;  &lt;p&gt;I'm not so worried now it's dark and we're all here. How about a show?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;My brain doesn't seem to be working, I think I'm coming down with a&#xA;  bug. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Anyway, as I said, How about a show from these two people, a&#xA;  small creature and... Ooh I'm feeling rather...you know.&lt;/p&gt;&#xA;  &#xA;  &lt;hr&gt;&#xA;  &#xA;  &lt;p&gt;Here's a self-contradictory statement for you. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Is it my birthday? Thanks! &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;We have everlasting togetherness.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Any help?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a &lt;strong&gt;very&lt;/strong&gt; well-known song with &lt;strong&gt;many&lt;/strong&gt; millions of plays on Youtube--if only I could remember the title. &lt;/p&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="14177" LastEditDate="2015-10-22T16:01:16.623" LastActivityDate="2015-10-22T16:12:35.913" Title="Nice song, pity about the words" Tags="|knowledge|music|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23411" PostTypeId="1" AcceptedAnswerId="23415" CreationDate="2015-10-22T15:24:46.400" Score="5" ViewCount="951" Body="&lt;h1&gt;Rules&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;You are given a square grid, with numbers (0-9) in some of its cells.&#xA;You have to connect some of the numbers, so that each &quot;island&quot; of numbers has the same total sum. A number can have anywhere from 0 to 4 connections that go straight up, down, left or right. Two connections may never intersect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some notes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You can't just connect all numbers together, you must have at least 2 &quot;islands&quot; of numbers.&lt;/li&gt;&#xA;&lt;li&gt;The only place a T-intersection may happen is at a number, not in an empty cell.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h1&gt;Solved examples&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Here the sum of every group is 3 (note that the lonely 3 is still a valid group)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 1-----1  1&#xA; | 3      |&#xA; 1     2--0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Alternate solution&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 1-----1--1&#xA;   3&#xA; 1-----2--0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;A bigger example&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 1--0--3     5&#xA; |  |        |&#xA; |  2-----4  |&#xA; |           |&#xA; 8     7-----6&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h1&gt;Problem&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/l0iBF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/l0iBF.png&quot; alt=&quot;Image of the grid&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are a few textual versions of the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compressed:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2    00&#xA;3  02  3&#xA; 0 140&#xA;  4  3 4&#xA;  023&#xA;01 25&#xA;3  1 22&#xA;46     3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Another version, easier to read:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 2              0  0&#xA;&#xA; 3        0  2        3&#xA;&#xA;    0     1  4  0&#xA;&#xA;       4        3     4&#xA;&#xA;       0  2  3&#xA;&#xA; 0  1     2  5&#xA;&#xA; 3        1     2  2&#xA;&#xA; 4  6                 3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;EDIT:&lt;/em&gt; Ok, It seems to be quite easy, right? Now here is a harder challenge: Can you split it into a maximal number of &quot;islands&quot;? (Thanks to @ghosts_in_the_code for this addition)&lt;/p&gt;&#xA;" OwnerUserId="11679" LastEditorUserId="4421" LastEditDate="2015-11-25T17:57:45.727" LastActivityDate="2015-11-25T17:57:45.727" Title="Connect the numbers! (Sudoku-like challenge)" Tags="|calculation-puzzle|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23412" PostTypeId="2" ParentId="23410" CreationDate="2015-10-22T15:39:56.857" Score="21" Body="&lt;p&gt;The song is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Smells Like Teen Spirit, by Nirvana&lt;br&gt;&#xA; Greetings! Lots of greetings! (Hello, hello, hello.....)&lt;br&gt;&#xA; But, oops mind that hole. (How low)&lt;br&gt;&#xA; I'm not so worried now it's dark (With the lights out, it's less dangerous)&lt;br&gt;&#xA; and we're all here. How about a show? (Here we are now, entertain us)&lt;br&gt;&#xA; My brain doesn't seem to be working, I think I'm coming down with a bug.(I feel stupid, and contagious)&lt;br&gt;&#xA; Anyway, as I said, How about a show (here we are now, entertain us)&lt;br&gt;&#xA; from these two people, a small creature and... Ooh I'm feeling&lt;br&gt;&#xA; rather...you know. (a mulatto, an albino, a mosquito,  my libido)&lt;br&gt;&#xA; Here's a contradictory statement for you. (I'm worse at what I do best)&lt;br&gt;&#xA; Is it my birthday? Thanks! (For this gift, I feel blessed)&lt;br&gt;&#xA; We have everlasting togetherness. (Our little group has always been and always will until the end)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16874" LastEditorUserId="16874" LastEditDate="2015-10-22T16:12:35.913" LastActivityDate="2015-10-22T16:12:35.913" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23413" PostTypeId="2" ParentId="23411" CreationDate="2015-10-22T16:09:11.550" Score="1" Body="&lt;p&gt;Initial question and examples didn't make it clear whether existing islands could be split up, so keeping existing islands together I came up with this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we go: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/nZPjT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nZPjT.png&quot; alt=&quot;Ans&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Easier to read:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/hDdx9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hDdx9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;uhhh...I need help with formatting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2    00+&#xA;3  02--3&#xA;|0+140&#xA;| 4  3 4&#xA;| 023  |&#xA;01 25  |&#xA;3  1 22+&#xA;46-----3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;+&quot; indicate intersections and 2 islands add up to 30 each.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16471" LastEditorUserId="16471" LastEditDate="2015-10-22T17:46:04.847" LastActivityDate="2015-10-22T17:46:04.847" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23414" PostTypeId="2" ParentId="23411" CreationDate="2015-10-22T16:13:15.660" Score="1" Body="&lt;p&gt;The puzzle is way too easy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Keep the first five rows in Group A, and the last three in Group B. Transfer the '1' at row 6 col 2 to Group A, and the '2' at row 5 col 4 to Group B. Connections within the group is very easy. Now we have 2 groups that add up to $30$ each.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-10-22T16:13:15.660" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23415" PostTypeId="2" ParentId="23411" CreationDate="2015-10-22T16:24:07.650" Score="5" Body="&lt;p&gt;Once you realize that a number needs to be at each corner (e.g. no intersections in a blank square), then the difficulty crops up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, here is a solution with 6 islands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is the largest count of islands you can get.  I thought at first that it could be 10, but with the 4 locked in the lower left, then the lowest value would be 7, which is not a factor of 60 (the total value of numbers in the puzzle).  With the way the puzzle is set up, your lowest factor would be 10.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/udqbh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/udqbh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14233" LastEditorUserId="14233" LastEditDate="2015-10-22T17:19:18.793" LastActivityDate="2015-10-22T17:19:18.793" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23416" PostTypeId="2" ParentId="23411" CreationDate="2015-10-22T16:46:34.757" Score="3" Body="&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;    2--------------0--0&#xA;    |&#xA;    3---------0--2--------3&#xA;              |  | &#xA;        0-----1  4--0&#xA;        |           | &#xA;        |  4--------3-----4&#xA;        |  |              |&#xA;        |  0  2--3        | &#xA;        |     |           |&#xA;     0  1     2--5        |&#xA;     |        |           |&#xA;     3--------1-----2--2  |&#xA;     |                    |&#xA;     4--6                 3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A simple 2 island solution.&lt;/p&gt;&#xA;" OwnerUserId="16846" LastActivityDate="2015-10-22T16:46:34.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23417" PostTypeId="1" AcceptedAnswerId="23651" CreationDate="2015-10-22T17:01:42.120" Score="8" ViewCount="1217" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Oh!   &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Please help.   &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;There is a number sequence hidden in here, but where?   &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Here are just a few examples so I will make the answer multiple choice. Can you find the sequence, the pattern, and the choice that follows the same pattern? I hope this isn't too broad and only one pattern matches.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;OPTION 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm the first and obviously best option. You should really pick me. Don't you want to be number one? You don't even need to read the other options. Just stop right here. Your gut is telling you I'm the right choice, isn't it? I know it is. If you keep reading, they're going to try to convince you that they're really the best answers and that's just going to confuse you. Have you ever felt silly because you didn't go with your gut? You let yourself be swayed by smooth talkers and regretted it, right? Don't let that happen to you here. Be strong. Be vigilant. I'm right and you know it. Go ahead, post an answer. Vote for Option 1. Of course, you have to figure out why. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;OPTION 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Look, I know I'm the middle answer so I don't get the first word or the final word. I'm going to have to convince you in another way. I will be the most helpful. I'm nice enough - and desperate enough - to give you a hint. I'm the middle answer and I'm also the answer of medium length. Maybe &lt;strong&gt;the length matters&lt;/strong&gt;. Maybe &lt;strike&gt;the length doesn't matter. I'm not telling. I'm going to help you. Yes, I am. No, we're not; now, shut up! He doesn't know what he's talking about. We're not going to help no stinkin' puzzle solver to solve our puzzle. We hates the filthy puzzlers. We hates them!&lt;/strike&gt; Look, I'm really sorry about that guy. He doesn't speak for all of us. I want to help and the hint I gave it totally a legitimate hint. Honest. It's not a red herring at all. Or is it? I don't know. Well, I do know but I'm not telling you. Can you count in different bases? How about two bases at once? How would you do that? Good luck. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;OPTION 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Save the best for last, right? I am so incredibly superior to the other jokers that it's not funny any more. Really, I just feel bad for them. They are truly inadequate and have no idea. Poor, poor other answers. You know how, when someone is dead, it's only sad for other people? It's the same way when you're a wrong answer. They don't know they're wrong, but they are. Of course, you might want some proof, so here it is. It's simple, really. I am by far the longest answer. I hold the most text and, therefore, hold the most ideas. Ideas are from creative minds. Creative minds are better than dull minds. Because creative minds are better, they produce ideas, and I have the most ideas, I am clearly better. QED. It's simple, really. I tried explaining this to them one time but they just stared at me with a dull expression on their face. I don't think they really followed me at all. I'm not even sure they were listening, to be honest. They get the far away look in their eyes whenever I try explaining things. Obviously, that is more evidence of their dullness. Pick the best, pick me: Option Three. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The hints from Option 2 are legitimate. You're going to need to need a base besides base 10. Which base? Well, which option is helpful?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2015-10-28T19:23:28.313" LastActivityDate="2015-10-29T15:42:51.953" Title="Hidden Number Sequence" Tags="|number-sequence|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23418" PostTypeId="2" ParentId="23411" CreationDate="2015-10-22T17:15:53.453" Score="5" Body="&lt;p&gt;A quick argument that APrough's solution with 6 islands (each with sum 10) already maximizes the number of islands:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The sum of all numbers is 60. Hence the only possibilities for $x$ islands each with sum $y$ are the following twelve pairs $(x,y)$: $~~(1,60)$, $~(2,30)$, $~(3,20)$ , $~(4,15)$, $~(5,12)$, $~(6,10)$, $~(10,6)$, $~(12,5)$, $~(15,4)$, $~(20,3)$, $~(30,2)$, and $~(60,1)$.&lt;/li&gt;&#xA;&lt;li&gt;The first pair has $x=1$, and a single island is forbidden by the rules.&lt;/li&gt;&#xA;&lt;li&gt;The number $6$ (in row $8$, column $2$) enforces $y\ge6$.&lt;/li&gt;&#xA;&lt;li&gt;This leaves us with the six pairs $~(2,30)$, $~(3,20)$ , $~(4,15)$, $~(5,12)$, $~(6,10)$, $~(10,6)$.&lt;/li&gt;&#xA;&lt;li&gt;Also $(x,y)=(10,6)$ is impossible: In this case, the entry $4$ in the lower left corner can neither be connected to its right neighbor $6$ (as the sum of the resulting island would exceed $6$), not can it be connected to its upper neighbor $3$ (as the sum be violated again), nor can it stay alone.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-10-22T17:15:53.453" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23419" PostTypeId="1" AcceptedAnswerId="23543" CreationDate="2015-10-22T17:42:48.677" Score="5" ViewCount="409" Body="&lt;p&gt;&lt;sub&gt;I had an idea for a cipher and I don't know enough about ciphers to know if it already exists. I'd like to see how easily this is solved and if such an idea is already popularized somewhere. I have a hard time believing that nobody has ever had the idea because I'm just not that clever but I don't remember seeing it before. I've provided what I think is plenty of text to allow for breaking and tried to liven it up a bit with some flavor.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Prior to joining forces with Lord Hooty McOwlface - indeed, even before his Lorship's rise to power - Fakky Jim was a typical human. He was a metal worker refining zinc for the U.S. Mint to produce coins. There were a lot of special regulations he had to follow but, in general, he was just a foundry guy. He lived alone and spent most weekend picking up scrap metal to sell to his employer (for non-coinage use, obviously).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was a fine Saturday in June when his life was changed. He was in the dump looking for scrap when he happened across a lamp. It was covered in soot and, as he tried to wipe it clean and inspect the metal quality, the lamp shook violently and shocked his hand. Fakky dropped the thing and out swarmed a dark, purplish-black cloud of lightning and laughter. In a few seconds, the cloud had solidified into the grinning face of a genie with fiery red eyes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;VERY WELL&lt;/code&gt; boomed the smoky creature inside Fakky's head, &lt;code&gt;YOU SHALL BE TESTED.&lt;/code&gt;&lt;br&gt;&#xA;“Tested?” asked Fakky Jim. “I thought I was supposed to get some wishes.”&lt;br&gt;&#xA;&lt;code&gt;ONLY IF YOUR WISH IS TO BE TESTED AND REWARDED OR PUNISHED BASED ON YOUR PERFORMANCE.&lt;/code&gt;&lt;br&gt;&#xA;“But what if I –“&lt;br&gt;&#xA;&lt;code&gt;LET THE TEST BEGIN&lt;/code&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A giant penny was suddenly floating in front of Fakky’s face. Slowly, it rotated down so he was looking at the edge. There were letters inscribed there. The penny began to rotate and reveal more characters in an apparently random sequence. He tried to move but found himself rooted to the spot. Confused, he looked back up at the genie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;TELL ME THE MESSAGE ON THE EDGE OF THIS COIN AND YOU SHALL BE FREED AND REWARDED.&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;FAIL, AND BE FREED AND PUNISHED.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, Fakky was unable to decode the coin and was turned into a cat, albeit an immortal, magical cat. His anger towards what he felt was a random punishment of the universe led him down a path of evil. How would you have fared against the same challenge? (Text on the edge of the coin is transcribed below.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;GntS rDPSF zOc TFWdO xdzSr zHn ntS eBUIdSr aqntfIs GnqUI Pm sgJr bnOUJmFOs B mdX mBsJPm bnObFhWdE hm khadqUZ BmE EdEhDzUFE Un UgF QSPQnrJshPm UIzU zkk ldO zSF bqdBsFc dRVBk mPv vF zqF dmfzHFE hm B HqFzU bJuhM vBS UFTsJmf XIFsgFq sgBs Ozshnm nS zmZ mzsJnO rP bnmDdhudE Bmc rP EFchbzsdE DBm kPOH dOEVqd vd BSF NdU Pm B fSFzs CBssMFehFkc nG sgBs Xzq vF gBWd bnNd Un cFchbzUd B onSsJnm nG sgzs ehFMc zT B ehOzk SdTshOH okBDF GnS Ugnrd vgP gdSd fzuF UgdJq kJuFr sIzs sgzU OBUJPO lhHIU Mhud hs hT BksnHFUgdq GJsUhmf zOE QSnoFS UIBs vF rgPtkc En UghT CVs hO z kBSfFq rFOTF vF DBO mPs EdchbzsF  vd DBO mns DPmTdDqzsF  vd bzm OPU IBMkPX  sIhr fqPtOc sgd aqzWd NFm kJWJmf zOE cdzc XgP rUSVHfkFE Idqd gzWF DnOTFbSBsdE JU GBS zaPud PVS onPS onvdS Un BEc PS cdUSzDs sgd XPSkc vhMk kJUsMF mnsd OPq kPmH SFNFNCdq XgBs vd rzZ IFSF atU JU DBO mdWdq enSHds vgzs sgFZ chc IFSF hs JT ePq VT Ugd kJWJmH qBUIFq sn CF cdEhbzsFE gdqF Un sgd VmGJOhrgdc vPSj vIJbI UIdx XIn GnVHgU gdSd gBWF sItT GzS rn mnakZ BcWzODdEhs JT qzsIFS enq VT sn CF Idqd cdchbzsFE UP UIF fqFBs UzTj qFNBhOhmf CdePqd tr  Ugzs eqnN sIdrd gPmPqdc Edzc vF UzjF JmDqFzrFE EFWnsJPm UP UgBU bzVrF ePq vgJDI UgFZ fzuF Ugd kzTU etkM NFzrVSd ne EFuPUhPO  sIzU vd gFSF gJfgMZ qdrnkWd sIzs UIdrd cFzc TgBkk OPU IzWF EhdE Jm uBhO  UIzs sghr OBUJnO VmEdq HPE rgzMk IzuF B mFv aJSsI PG eqFdcnl  Bmc UgzU HPudqOlFms nG Ugd oFnokF CZ UgF oFnQMd ePq sgF oFPQkd rIBMk mns QFqhrg eSnN sgF dzSsI&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2015-11-13T16:10:45.280" LastActivityDate="2015-11-13T16:10:45.280" Title="Penny for your Thoughts" Tags="|cipher|" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23420" PostTypeId="1" AcceptedAnswerId="23423" CreationDate="2015-10-22T18:05:39.797" Score="5" ViewCount="1024" Body="&lt;p&gt;My Indian friend Jamal recently participated in the well-known TV quiz&#xA;show &lt;strong&gt;&lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Who_Wants_to_Be_a_Millionaire&quot; rel=&quot;nofollow&quot;&gt;Who wants to be a millionaire?&lt;/a&gt;&lt;/em&gt;&lt;/strong&gt;. You might have heard or read about Jamal's spectacular success; the media reported on it extensively.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what Jamal told us about the 16.000 Euro question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The 16.000 Euro question brought up my area of expertise.  I was familiar&#xA;  with the first technical term mentioned in the question, but its second&#xA;  technical term seemed somewhat obscure.  I noticed that all the answers&#xA;  were consecutive powers of two and that they were listed in increasing&#xA;  order as well.  I figured that programmers and computer scientists usually&#xA;  enjoy play-of-words, and after some thought I hesitatingly chose B.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;What was the 16.000 Euro question?&lt;br&gt;&#xA;What were the four possible answers A, B, C, D?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="9404" LastEditDate="2016-03-04T16:06:53.050" LastActivityDate="2016-03-04T16:06:53.050" Title="Slumdog millionaire (9)" Tags="|knowledge|computer-science|" AnswerCount="3" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23421" PostTypeId="2" ParentId="23420" CreationDate="2015-10-22T18:22:58.773" Score="5" Body="&lt;p&gt;The question and the answers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; How many bytes go in a kibibyte? &lt;br&gt; A. 512 &lt;br&gt; B. 1024 &lt;br&gt; C. 2048  &lt;br&gt; D. 4096 &lt;br&gt; &lt;br&gt; There are some variations possible: &lt;br&gt;  - How many bits go in a kibibyte? &lt;br&gt; - How many bits go in a mebibyte? &lt;br&gt; - How many bytes go in a mebibyte? &lt;br&gt; &lt;br&gt; Answers are the corresponding powers of 2 of course.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Justification:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; kibi is a wordplay on kilo and binary, and mebi is a wordplay on mega and binary. &lt;br&gt; Therefore this answer does statistify the wordplay requirement.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16322" LastEditorUserId="16322" LastEditDate="2015-10-23T09:34:50.647" LastActivityDate="2015-10-23T09:34:50.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23422" PostTypeId="2" ParentId="23420" CreationDate="2015-10-22T18:28:20.483" Score="3" Body="&lt;p&gt;I think the answers were:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; binary, quadral, octal, hexadecimal&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The question might be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Which number system is not commonly used by programmers?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastActivityDate="2015-10-22T18:28:20.483" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23423" PostTypeId="2" ParentId="23420" CreationDate="2015-10-22T18:37:10.310" Score="15" Body="&lt;p&gt;The question was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;How many bits make a nibble?&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Justification: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;a href=&quot;https://en.wikipedia.org/wiki/Nibble&quot;&gt;wikipedia&lt;/a&gt; says: The term &quot;nibble&quot; originates from its representing &#xA; &quot;half a byte&quot;, with &quot;byte&quot; a homophone of the English word &quot;bite&quot;. &#xA; Computer scientists enjoy the play-of-words with biting and nibbling.&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Jamal says: &quot;I was familiar with the first technical term mentioned &#xA; in the question, but its second technical term seemed somewhat obscure.&quot; &#xA; The term &quot;bit&quot; is very common, while the term &quot;nibble&quot; is not so&#xA; well-known and somewhat obscure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then the four answers were:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A. $~~~~~2$&lt;br&gt;&#xA; B. $~~~~~4$&lt;br&gt;&#xA; C. $~~~~~8$&lt;br&gt;&#xA; D. $~~~16$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-10-22T18:56:31.347" LastActivityDate="2015-10-22T18:56:31.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23424" PostTypeId="2" ParentId="23404" CreationDate="2015-10-22T18:52:13.810" Score="8" Body="&lt;p&gt;This is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a tricycle. 1 wheel in front, two in back. You would never need three wheels in either position.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17057" LastActivityDate="2015-10-22T18:52:13.810" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23425" PostTypeId="2" ParentId="23408" CreationDate="2015-10-22T20:52:32.673" Score="10" Body="&lt;p&gt;A cipher with a key immediately makes me think of a Vigenere cipher, which is probably the most common cipher employing a key.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way a Vigenere cipher works is much like a typical Caesar cipher (which just shifts all letters a certain number of places in the alphabet), but it also employs a key, which allows each letter to be shifted a &lt;em&gt;different number&lt;/em&gt; of places in the alphabet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if your key is &lt;strong&gt;&lt;code&gt;SECRET&lt;/code&gt;&lt;/strong&gt; and you want to encrypt the message &lt;strong&gt;&lt;code&gt;THE MISSION STARTS AT NOON&lt;/code&gt;&lt;/strong&gt;, you would assign a number to each letter of the key (&lt;strong&gt;&lt;code&gt;A&lt;/code&gt;&lt;/strong&gt;=0, &lt;strong&gt;&lt;code&gt;B&lt;/code&gt;&lt;/strong&gt;=1, etc.).  Then you would shift each letter of your message that number of characters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;THE MISSION STARTS AT NOON&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;SEC RETSECR ETSECR ET SECR&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now &lt;code&gt;T&lt;/code&gt; is 19 and &lt;code&gt;S&lt;/code&gt; is 18, so we add $19 + 18 = 37$.&lt;br&gt;&#xA;37 is more than 25 (= &lt;code&gt;Z&lt;/code&gt;), so we need to &quot;wrap around&quot;, which is the same as subtracting 26 from our result.  $37-26 = 11$ = &lt;code&gt;L&lt;/code&gt;, so the first letter of our encryption is &lt;code&gt;L&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By following the same method, the second letter would be encrypted as $7 + 4 = 11$ = &lt;code&gt;L&lt;/code&gt;, and the third as $4 + 2 = 6$ = &lt;code&gt;G&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see that even though the first three letters of the plaintext are different (&lt;code&gt;THE&lt;/code&gt;), two of them encrypt to the same letter (&lt;code&gt;L&lt;/code&gt;).  This makes a Vigenere cipher much harder to crack than a Caesar cipher, because a given character in the ciphertext doesn't represent the same character in the plaintext each time it appears.  This also means that frequency analysis is useless, since the most frequent letter in the ciphertext might represent six different letters in the plaintext, none of which are particularly frequent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are still weaknesses, though.  As the text gets longer, it's more likely that common words (&lt;code&gt;the&lt;/code&gt;, &lt;code&gt;and&lt;/code&gt;, &lt;code&gt;a&lt;/code&gt;, etc.) will end up being encrypted by the same section of the key, and will thus end up encrypted as the same sequence of letters.  By analyzing the ciphertext, we can pick out certain letter sequences that repeat multiple times.  By looking at the spacing between them, we can guess at how long the key is.  For example, if a certain sequence appears at positions 65, 89, 105, and 121, we might guess that the key is of length 8, because all of the distances between those numbers are divisible by 8 (24, 16, 16).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing the length of the key can be very helpful, because it lets us know which parts of the message were encrypted using the same letter of the key.  In a long-enough message, we can look at those letters and actually do a frequency analysis of just the letters encrypted by a single character of the key.  Since they were all encrypted the same way, the more-frequent letters will probably map to the more-frequent English letters, and from that point we can make some intelligent guesses about what some words might be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many online tools that can help with this kind of thing.  Some try to offer a one-stop solution, which determines key length, finds the key, and decrypts the ciphertext without any input from you.  These tend to be mediochre at best.  Other tools will help you with one aspect of the problem.  I used &lt;a href=&quot;http://www.dcode.fr/vigenere-cipher&quot;&gt;this tool&lt;/a&gt; (the &lt;code&gt;FIND KEY LENGTH&lt;/code&gt; button) to help me determine the key length.  It gave me a very high likelihood that the key was 6 characters long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then used &lt;a href=&quot;http://www.mygeocachingprofile.com/codebreaker.vigenerecipher.aspx&quot;&gt;this tool&lt;/a&gt; to do a frequency analysis based on the key length.  Basically, it took the characters at position 1, 7, 13, 19, etc. (6 apart), and looked at which letters were most common.  It then took the most common English letter, and determined the difference between them.  It assumed that was the first letter of the key.  Then it did the same with characters at positions 2, 8, 14, 20, 26, etc., and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I anticipated that it would not find the exact solution, but hoped it would be able to fill in enough that I could guess what was missing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the second suggestion it spit out, I got a key of &lt;code&gt;IOXNAR&lt;/code&gt; and a plaintext that included the phrase &lt;code&gt;NOXHONG CEN NARM QE&lt;/code&gt;.  I thought this looked a lot like English, and guessed that it might be &lt;code&gt;NOTHING CAN HARM ME&lt;/code&gt;.  Looking at the incorrect letters, they were all at position 1 or 3 of the key, so it looked like my key was correct except for the first and third characters:  'xOxNAR`.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did the math to figure out what letter would be required to get a plaintext of &lt;code&gt;NOTHING CAN HARM ME&lt;/code&gt;, and determined that the first character of the key was &lt;code&gt;M&lt;/code&gt; and the third was &lt;code&gt;D&lt;/code&gt;, making the key &lt;code&gt;MODNAR&lt;/code&gt;.  Looks nonsensical at first, until you realize that it is simply the word &lt;strong&gt;RANDOM&lt;/strong&gt; backwards!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using that key to decipher the whole message, I got&lt;br&gt;&#xA;&lt;code&gt;I MUST ADMIT THAT ALTHOUGH NOTHING CAN HARM ME, I'VE HAD MOMENTS OF PARANOIA AND EVEN YOU CAN NOW SENSE THAT&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Afterward, I looked a little further through the tool's results, and realized that it had actually found the solution, at attempt #42.&lt;/p&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2015-10-22T20:52:32.673" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23427" PostTypeId="2" ParentId="23408" CreationDate="2015-10-22T21:59:15.810" Score="1" Body="&lt;p&gt;I wanted to contribute my own approach to solving this (I had someone else choose the message and key). Note that GentlePurpleRain has the accepted answer to this question, I'm providing my own answer for academic reasons only so be sure and +1 his answer!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First I wanted to track down the key length, to limit the options of key. I noticed two repeated words, 'tmb' and 'tymh', which repeat after 42 and 72 letters respectively. This means our key length must be a factor of both of these numbers, giving us the options of 1,2,3, and 6 letters. Length 1 was easy enough to rule out, since that is just a Ceasar cipher and checking all 26 possibilities was trivial. I strongly suspected it was 3 or 6, and decided to get more information before deciding (though checking for all 2 letter possibilities is also 'feasible', since there are only ~100 letter words and &amp;lt;600 total combinations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next I noticed that there was a word of the form x'xx. As far as I could find, the only two contractions that match this form are I'll and I've. So, moving forward with the knowledge that 'l' in cipher = 'i' in message and the location of the 'l', I determined that the letter in the key at that location must have been D. So now, I had that:&lt;br&gt;&#xA;*If key length=2, I have 'D-', which reduces to the difficulty of key length=1 was trivial to check via bruteforce.&lt;br&gt;&#xA;*If length=3, I have '--D' which reduces to ~50 words/&amp;lt;600 total combinations&lt;br&gt;&#xA;*If length=6, I have '--D---' which reduces to ~250 words.&lt;br&gt;&#xA;At this point I could have bruteforced all ~300 word possibilities for the key, but had another idea (which was fortunate since as it turned out, the key would not have been found in my dictionary)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether the key is 3 or 6, we know that the letters in the 3,9,15,21,etc spot are being encrypted with the letter D. So we go backwards by 3 letters at each of those spots and now the cipher becomes (capitals are letters unencrypted as they appear in original message):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;u aUft rpaIg tymh AytyaiGu nffvIag tmb Hnrd ys, I'ie ymr MbmvzhS bf gmfAaozm oNq emqb Ybu tmb Nbw jqbSr tymh.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I first focused on 'AytyaiGu' A &lt;a href=&quot;http://www.visca.com/regexdict/&quot; rel=&quot;nofollow&quot;&gt;regex&lt;/a&gt; showed me that only 10 words match that pattern a*****g* and of those, the word 'although' was the most likely candidate (the others being obscure fields or scientific terms). That word is conveniently long enough to give me the entirety of the key based on that guess and after seeing that the key was 'RANDOM' backwards, I was went ahead and plugged the cipher into a simple &lt;a href=&quot;http://www.dcode.fr/vigenere-cipher&quot; rel=&quot;nofollow&quot;&gt;vigenere decoder tool&lt;/a&gt; that takes in a key, provided MODNAR, and out popped the plaintext message. Voila!&lt;/p&gt;&#xA;" OwnerUserId="13970" LastActivityDate="2015-10-22T21:59:15.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23428" PostTypeId="2" ParentId="23404" CreationDate="2015-10-22T23:54:34.530" Score="6" Body="&lt;p&gt;Potential answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://onlineslangdictionary.com/meaning-definition-of/number-one&quot;&gt;http://onlineslangdictionary.com/meaning-definition-of/number-one&lt;/a&gt; &amp;lt;-- see the first example, don't need a 3.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7722" LastActivityDate="2015-10-22T23:54:34.530" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23429" PostTypeId="2" ParentId="14992" CreationDate="2015-10-23T00:20:29.897" Score="0" Body="&lt;p&gt;Hmm.. I have an idea...&lt;br&gt;&#xA;The gates are &quot;somewhere in the grid&quot;. So maybe it's written like O=0, S=5, etc. Then it becomes&#xA; &lt;pre&gt;L0FC5URFTCZLRNTVZFUUNRBT0N&#xA;JZTZ0DUJRT5CJCBFN0F5Y5VZCJ&#xA;RFBYMBVRM50YYLNMYYJCDLRFZB&lt;/pre&gt;&#xA;(If there are more substitutions, tell me)&lt;/p&gt;&#xA;" OwnerDisplayName="user17008" LastActivityDate="2015-10-23T00:20:29.897" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23430" PostTypeId="2" ParentId="19870" CreationDate="2015-10-23T00:24:45.087" Score="2" Body="&lt;p&gt;Let's ask ourselves the question: &lt;strong&gt;How many pieces of 56 grams can we possibly have if we want a 17-piece solution?&lt;/strong&gt; This will help us limit the groupings in the 9-way split.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume we have three pieces of 56 grams. Then in the 8-way split scenario, one of the 56-gram pieces must be in a 2-piece group to get 63 grams, and thus there must exist (at least) two pieces of 7 grams. We also know that there is either a third piece $K$ of 7 grams, or exactly two pieces $K_1, K_2$ which add up to 7 grams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the 7-way split scenario, each 63-gram group needs to have at least two pieces of gold (otherwise, there will be a single piece of 72 grams &gt; 56 grams). This means that we have 3 extra pieces of gold to assign to groups in the 7-way split. The known two pieces that weigh 7 grams cannot be in a group with only two pieces (otherwise the other piece in the group would be heavier than 56 grams). So the following three possibilities exist:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;One group only with more than 2 pieces:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[56] + [ 7] + [ 7] + [A?] + [B?]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We know this cannot be true, since $K$ or both $K_1, K_2$ must be in this group, making the total larger than 72 grams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Two groups with more than 2 pieces, a 3-piece group and a 4-piece group:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[56] + [ 7] + [A?] + [C1?]&#xA;&#xA;[56] + [B?] + [C2?]&#xA;&#xA;[56] + [16]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Because of the pigeonhole principle, we cannot have $K = 7$ be in one of the these groups (which would make at least one group unequal to 72 grams total), nor A = $K_1$, B = $K_2$ (for the same reason). Thus we have C1 = $K_1$, C2 = $K_2$. The last 7-gram piece also needs to be in one of these groups, so either:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B = 7 $\Rightarrow$ $K_2$ = 9 (contradiction)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A = 7 $\Rightarrow$ $K_1$ = 2 $\Rightarrow$ $K_2$ = 5 $\Rightarrow$ B = 11&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will now do some Sudoku-like deductions and reach a contradiction. The 11-gram piece will occupy a 2-piece group in the 8-way split, and thus generate a 52-gram piece, which will generate a 4-gram piece in the 9-way split, and this 4-gram piece must be part of a 2-piece group in the 8-way or 7-way split, and generate a piece that is larger than 56 grams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Three groups with more than 2 pieces, 3 pieces each:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[56] + [ 7] + [B1?]&#xA;&#xA;[56] + [ 7] + [B2?]&#xA;&#xA;[56] + [A?] + [B3?]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We also know this cannot be true, because either $K_1, K_2$ must be in this group (which by the pigeonhole principle would make at least one group unequal to 72 grams total), or A = $K$, and $B_i = 9, \forall i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this latter case, the pigeonhole principle again requires at least one $B_i$ to be in a two-piece group in the 8-way split and generate a 54-gram piece, and at least one 7-gram and one 9-gram piece to be in a two-piece group in the 9-way split and generate a 49-gram and a 47-gram piece, respectively. The 54-gram, 49-gram and 47-gram pieces must then generate 18-gram, 23-gram and 25-gram pieces in the 7-way split, leaving two free pieces that must add up to 72. But since the 54-piece must either create two pieces of 1 grams or one piece of 2 grams in the 9-way split, we reach a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From the above, it is clear that there can be at most two 56-gram pieces in a 17-piece solution. This implies that the 9-way split has at least six 2-piece groups, plus either a 56-gram single-piece group + one 3-piece group (two 56-gram pieces case), or two 2-piece groups (single 56-gram piece case).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will now show that the single 56-gram piece case admits no solutions with 17 pieces. To do this, we will consider the even/odd parity required for each of the groups in the 8-way split. There are 7 2-piece groups and one 3-piece groups in the 8-way split. Since each group must sum to 63 grams (odd), each 2-piece group must have one odd-weighted piece and one even-weighted piece. (The 3-piece group is either one odd or all odd pieces.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Correspondingly, the 9-way split has 8 groups of 2 pieces, and one group of a single 56-gram piece. These 2-piece groups must be either all odd-weighted or all even weighted. Therefore, &lt;strong&gt;any pairing in the 8-way split cannot appear in the 9-way split.&lt;/strong&gt; Moreover, this means that if we fix the location and the weight of any two pieces in the 8-way split not in the same group, the weights of all the other pieces can be determined, relative to the fixed weights.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider the scenario in which the 56-gram piece is in the 2-piece group, which generates a piece of 7 grams in the same group, which, when put in a (2-piece) group in the 9-way split, generates a piece of 49 grams, which can then be put back in the 8-way split again, so on and so forth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did this for all the possible scenarios with a single 56-gram piece in the 8-way split:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;56-gram piece in 2-piece group $\Rightarrow$ 7-gram weight $\Rightarrow$ 49-gram in 9-way split $\Rightarrow$ 49-gram put into 2-piece group in 8-way split ...&lt;/li&gt;&#xA;&lt;li&gt;56-gram piece in 2-piece group $\Rightarrow$ 7-gram weight $\Rightarrow$ 49-gram in 9-way split $\Rightarrow$ 49-gram put into 3-piece group (all odd / one odd only) in 8-way split ...&lt;/li&gt;&#xA;&lt;li&gt;56-gram piece in 3-piece group $\Rightarrow$ $K_1 = 7 - K_2, K_2$ $\Rightarrow$ $7-K_2$ put into 9-way split ...&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In all these scenarios, a full, self-consistent set of weights will be generated, all dependent on a single variable, in the form of $n_i-X$ or $n_i+X$, where $X$ is the independent variable. One example is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Wkv9Z.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Wkv9Z.png&quot; alt=&quot;9 Gangsters and a Gold Bar&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is for scenario 3 above, with independent variable $X = K_2$. It is easy to see that no two pieces with the same weight parity can sum to 72 grams if they have opposing signs (such that the independent variable $X$ is cancelled out). This is because for all scenarios, $n_i$ must be a multiple of 7 between 0 and 56, and even and odd pairs can never share $n_i$ by construction. (I know &quot;by construction&quot; is not rigorous at all, but I don't know of a clever way to prove this more elegantly.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only possibility is if the signs are the same, in which case the value of X will be forced to one value. However, this means that only a maximum of 1 same-parity pair can sum to 72 grams. By the pigeonhole principle, there are at least 4 2-piece groups in the 7-way split scenario, one of which is guaranteed to be both odd in weight parity, and the other guaranteed to be both even in weight parity. Therefore, the 7-way split has at least one 2-piece group that can never sum to 72 grams, and our proof for the single 56-gram case is done.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;We are now left with the case where exactly two 56-gram pieces (and one 7-gram piece) exist in the 17-piece solution. If we can prove this case hsa no valid solutions, we are done. However, I was lazy, so I just brute-forced all possible combinations. I found no valid solutions in this case, and so the proof is done.&lt;/p&gt;&#xA;" OwnerUserId="15960" LastActivityDate="2015-10-23T00:24:45.087" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23432" PostTypeId="2" ParentId="23314" CreationDate="2015-10-23T00:49:00.307" Score="3" Body="&lt;h1&gt;14 letters/8 syllables (ratio 1.75):&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Utilitarianism&quot; rel=&quot;nofollow&quot;&gt;utilitarianism&lt;/a&gt;&lt;br&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Egalitarianism&quot; rel=&quot;nofollow&quot;&gt;egalitarianism&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user17008" LastActivityDate="2015-10-23T00:49:00.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23433" PostTypeId="5" CreationDate="2015-10-23T00:59:50.627" Score="0" Body="&lt;p&gt;Alternate reality games (A.K.A. ARGs) are games taking part online, in the real-world or both that involve puzzles and ciphers hidden in videos, images or audio. As the puzzle is solved a story often becomes revealed. Often ARGs are used as promotional material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple of well-known ARGs are the Gravity Falls Cipher Hunt and Fortnite's ARGs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information check the Wikipedia article on &lt;a href=&quot;https://www.google.ca/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=1&amp;amp;cad=rja&amp;amp;uact=8&amp;amp;ved=2ahUKEwi66Nr_9vTeAhUJuFMKHY7aBzwQFjAAegQIBxAB&amp;amp;url=https%3A%2F%2Fen.wikipedia.org%2Fwiki%2FAlternate_reality_game&amp;amp;usg=AOvVaw0WveGCkv-wvQPX3jl3xTnc&quot; rel=&quot;nofollow noreferrer&quot;&gt;Alternate Reality Games&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="52483" LastEditDate="2018-11-27T16:20:33.463" LastActivityDate="2018-11-27T16:20:33.463" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23434" PostTypeId="4" CreationDate="2015-10-23T00:59:50.627" Score="0" Body="An alternate reality game is a game, usually online, involving puzzles and ciphers hidden in mysterious videos, images or audio with a story behind them. Usually it is meant to promote a product." OwnerUserId="11876" LastEditorUserId="52483" LastEditDate="2018-11-27T16:20:29.643" LastActivityDate="2018-11-27T16:20:29.643" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23436" PostTypeId="1" CreationDate="2015-10-23T05:30:25.423" Score="4" ViewCount="800" Body="&lt;p&gt;Notation: For an integer $k\ge1$, let $D(k)$ denote the sum of all positive integer divisors of $k$ that are strictly smaller than $k$. For example $D(9)=1+3=4$ and $D(33)=1+3+11=15$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A man picked two odd positive integers $m\ge3$ and $n\ge3$. &#xA;He computed the sums of divisors for $m$ and $n$, and multiplied them together. &#xA;Then he suddenly realized that&#xA;$$ D(m)\cdot D(n) ~=~ (m-1)(n-1)$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is this indeed possible, or has the man made a calculation error?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4269" LastEditDate="2017-03-21T11:37:51.723" LastActivityDate="2017-03-21T11:37:51.723" Title="Product of sum of divisors of two numbers" Tags="|mathematics|number-theory|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23437" PostTypeId="2" ParentId="23404" CreationDate="2015-10-23T06:19:59.063" Score="12" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It's the letter 'o'.&lt;br&gt;&#xA;At the front of '&lt;strong&gt;o&lt;/strong&gt;ne', at the back of 'tw&lt;strong&gt;o&lt;/strong&gt;' and none in 'three'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17064" LastEditorUserId="9244" LastEditDate="2015-10-23T09:26:17.130" LastActivityDate="2015-10-23T09:26:17.130" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23438" PostTypeId="1" AcceptedAnswerId="23453" CreationDate="2015-10-23T06:27:04.750" Score="12" ViewCount="906" Body="&lt;h3&gt;My first attempt:&lt;/h3&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;One was a star.&lt;br&gt;&#xA;  The best was bested.&lt;br&gt;&#xA;  One drove a car.&lt;br&gt;&#xA;  One only rested.    &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;One made art.&lt;br&gt;&#xA;  With stones they vied.&lt;br&gt;&#xA;  But then they did part.&lt;br&gt;&#xA;  And half of them died.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;31 was a cop.&lt;br&gt;&#xA;  Mustard's bank was his nose.&lt;br&gt;&#xA;  They rose to the top.&lt;br&gt;&#xA;  As your matriarch knows.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Of the last two,&lt;br&gt;&#xA;  One is amazed.&lt;br&gt;&#xA;  The other sells shoes.&lt;br&gt;&#xA;  The youth were all crazed.   &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;They all shared a liver.&lt;br&gt;&#xA;  They spoke of her love.&lt;br&gt;&#xA;  Iron met river.&lt;br&gt;&#xA;  And villainous glove.     &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A lane full of sights.&lt;br&gt;&#xA;  The town had a pool.&lt;br&gt;&#xA;  Three members, one knight.&lt;br&gt;&#xA;  A silvery tool.   &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Fabulous number.&lt;br&gt;&#xA;  A half dozen cried.&lt;br&gt;&#xA;  Eyes gilded in slumber.&lt;br&gt;&#xA;  We heard of a slide.   &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A teacher was killed.&lt;br&gt;&#xA;  And broken wings flew.&lt;br&gt;&#xA;  A door was filled.&lt;br&gt;&#xA;  And you're wealthy too.   &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;An apartment was burned.&lt;br&gt;&#xA;  Or so it was said.&lt;br&gt;&#xA;  Because a lover was spurned.&lt;br&gt;&#xA;  With a drain in his bed.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;So far as I have seen, most puzzles like this require a line by line explanation, so that is expected here as well.  Enjoy!&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note:  I have been made aware that there were a couple of errors in the original post (which have now been corrected) - one was due to autocorrect and a failure to proof read, the other was a total brainfart on my part.  Sorry.  I will use proper diligence next time.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="17062" LastEditorUserId="17062" LastEditDate="2015-10-23T21:45:27.260" LastActivityDate="2015-11-10T01:06:31.150" Title="This one should be popular!" Tags="|riddle|knowledge|wordplay|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23444" PostTypeId="1" CreationDate="2015-10-23T10:39:41.993" Score="3" ViewCount="1116" Body="&lt;p&gt;I often wonder: suppose I wake up one day and find myself in a maze of Hedges, like the one in &quot;The Shining&quot;, what is the best strategy to get out of it? &#xA;We can assume we have sunlight or a torch if its night time (or a full moon night).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose there is some better option then running around randomly?&lt;/p&gt;&#xA;" OwnerUserId="12518" LastEditorUserId="8874" LastEditDate="2015-10-23T11:41:29.363" LastActivityDate="2015-10-23T11:43:35.347" Title="Strategy for getting out of Hedge maze?" Tags="|strategy|algorithm|mazes|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23445" PostTypeId="2" ParentId="23444" CreationDate="2015-10-23T10:53:55.493" Score="5" Body="&lt;p&gt;The standard algorithm is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hold your hald on the right wall at all times and keep walking forward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this only works in mazes where the exits are both on the outside and either you use it from the beginning or there are no disconnected parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another one that works if you're allowed to use markings:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you get to an unmarked junction, mark the path you entered by, then pick a random path. If you're going forward and you reach a  junction wih a single mark somewhere, double mark that branch and turn back. If you reach a dead end, turn back and make a double mark. If you run out of possibilities, go back down the single marked path and add another mark on your way back. Treat all double marks as walls.&lt;/p&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2015-10-23T10:53:55.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23449" PostTypeId="2" ParentId="23444" CreationDate="2015-10-23T11:27:01.140" Score="3" Body="&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;&lt;strong&gt;You may want to check the &lt;a href=&quot;https://en.wikipedia.org/wiki/Maze_solving_algorithm&quot; rel=&quot;nofollow&quot;&gt;wikipedia page&lt;/a&gt; on maze solving strategies.&lt;/strong&gt;&lt;/h2&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;One of the most comprehensive answers to this question is given by William of Baskerville in Umberto Eco's book &quot;The name of the rose&quot; (page 188 of Mariner books edition, 2014):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;To find the way out of a labyrinth&quot;, William recited, &quot;there is only one means.&#xA;  At every new junction, never seen before, the path we have taken will be marked with three signs. If, because of previous signs on some of the paths of the junction, you see that the junction has already been visited, you will make only one mark on the path you have taken.&#xA;  If all the apertures have already been marked, then you must retrace your steps.&#xA;  But if one or two apertures of the junction are still without signs, you will choose any one, making two signs on it.&#xA;  Proceeding through an aperture that bears only one sign, you will make two more, so that now the aperture bears three.&#xA;  All the parts of the labyrinth must have been visited if, arriving at a junction, you never take a passage with three signs, unless none of the other passages is now without signs.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;How do you know that? Are you an expert on labyrinths?&quot;&lt;br&gt;&#xA;&quot;No, I am citing an ancient text I once read.&quot;&lt;br&gt;&#xA;&quot;And by observing this rule you get out?&quot;&lt;br&gt;&#xA;&quot;Almost never, as far as I know. But we will try it, all the same.&#xA;And besides, within the next day or so I will have lenses and time to devote myself more to the books.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://books.google.at/books?id=ChlOAwAAQBAJ&amp;amp;printsec=frontcover#v=onepage&amp;amp;q=labyrinth&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Google books, page 188&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;The maze traversal algorithm of Charles Tremaux: &lt;a href=&quot;https://en.wikipedia.org/wiki/Maze_solving_algorithm#Tr.C3.A9maux.27s_algorithm&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;The general &lt;a href=&quot;https://en.wikipedia.org/wiki/Maze_solving_algorithm&quot; rel=&quot;nofollow&quot;&gt;wikipedia page&lt;/a&gt; on maze solving strategies&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-10-23T11:43:35.347" LastActivityDate="2015-10-23T11:43:35.347" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23450" PostTypeId="2" ParentId="23314" CreationDate="2015-10-23T12:17:23.243" Score="2" Body="&lt;p&gt;WC. 2 letters, 4 syllables, 0.5 ratio. An english word, not a proper noun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initialisms are the way to go here, but I can't think of others which have a high proportion of W's and which aren't proper nouns.&lt;/p&gt;&#xA;" OwnerUserId="4833" LastActivityDate="2015-10-23T12:17:23.243" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23451" PostTypeId="1" AcceptedAnswerId="23455" CreationDate="2015-10-23T13:17:28.560" Score="6" ViewCount="237" Body="&lt;p&gt;&lt;sub&gt;Video game theme period has ended but have another puzzle anyway.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You're in some sort of apartment of interconnecting rooms. The below map materializes in your hands (so magical), highlighting that you are in &lt;strong&gt;Room *&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; A B C D E&#xA; F G H 1 I J&#xA; K L 2 * M N&#xA; O P Q 3 R S&#xA;   T U V W X&#xA;     Y Z&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;A message appears on the floor on which you stand (more magic)! &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;code&gt;Colour?&lt;/code&gt;&lt;br&gt;&#xA;  &lt;code&gt;↑ Red or Pink&lt;/code&gt;&lt;br&gt;&#xA;  &lt;code&gt;↓ Blue or Orange&lt;/code&gt;&lt;br&gt;&#xA;  &lt;code&gt;← Green or Camouflage&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;You then realize that you're completely alone, but you really shouldn't be? Weren't there two others with you? Maybe this clue will lead you to your missing party.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on who you are, you should be able to find them following the floor clues. Search for one, then the other, in either order. When you encounter an occupied room, you'll have the option to use two teleportation runes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;~&lt;/strong&gt; is a restart rune that will send you and any partner with you back to Room * with the initial floor clue. Your partner will accompany you in your new search. If you happen to encounter a stranger, they'll remain in their room; you can't recruit them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;@&lt;/strong&gt; is an exit rune that will send you and your party back to your home world, but it would only activate if you find your other two members.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;So...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Who are you?&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Who are they?&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;What path did you use to find them?&lt;/em&gt;&#xA;&lt;br&gt;(may vary, for confirming that you didn't cheat by &quot;jumping ahead&quot;)&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Finally, where is your home world?&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Oh, you're not the only one searching for someone else, as you'll come to discover. &lt;em&gt;How would they answer?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;Gender?&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;← He&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;→ She&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;↑ It&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;Outfit?&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;← Hat, overalls, facial hair&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;↓ Long coat or saddle and shoes&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;Description?&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;→ Round and glowy&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;↓ Mechanically-inclined&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room A&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;Backed yourself in a corner&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;↓ No big deal: this should take you back on your path&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room B&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;You taking the scenic route?&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;← Sure!&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;↓ Oops!&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room C&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;&lt;code&gt;&quot;The Enrichment Centre reminds you to ignore any advice given.&quot;&lt;/code&gt;&lt;/em&gt;&lt;br&gt;&#xA;  &amp;nbsp;&lt;strong&gt;~ @&lt;/strong&gt;    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room D&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;Movement?&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;← Resting on one of six sides&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;→ Running on all four legs&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room E&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;&lt;code&gt;&quot;Neigh!&quot;&lt;/code&gt;&lt;/em&gt;&lt;br&gt;&#xA;  &amp;nbsp;&lt;strong&gt;~ @&lt;/strong&gt;    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room F&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;Favourite things?&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;↑ Eggs and red herrings&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;↓ Emeralds&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room G&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;Defence?&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;← Punch and roll&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;↑ Smoke screen&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room H&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;← More rival than friend?&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room I&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;Gift to you?&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;→ Curve ball&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;↑ Ride for a song&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;↓ Bandana and bountiful bullets&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room J&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;Dead end. You are likely to be eaten by a...&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;← Quick U-turn&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room K&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;&lt;code&gt;&quot;Monotreme with dreadlocks? Chaos!&quot;&lt;/code&gt;&lt;/em&gt;&lt;br&gt;&#xA;  &amp;nbsp;&lt;strong&gt;~ @&lt;/strong&gt;    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room L&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;&lt;code&gt;&quot;Let's-a go, Bro!&quot;&lt;/code&gt;&lt;/em&gt;&lt;br&gt;&#xA;  &amp;nbsp;&lt;strong&gt;~ @&lt;/strong&gt;    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room M&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;&lt;code&gt;&quot;I'm no rookie! You can shoot me if I slow you down.&quot;&lt;/code&gt;&lt;/em&gt;&lt;br&gt;&#xA;  &amp;nbsp;&lt;strong&gt;~ @&lt;/strong&gt;    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room N&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;&lt;code&gt;&quot;Hey! Listen!&quot;&lt;/code&gt;&lt;/em&gt;&lt;br&gt;&#xA;  &amp;nbsp;&lt;strong&gt;~ @&lt;/strong&gt;    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room O&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;&lt;code&gt;&quot;I'm not in a locker this time, and my pants are dry!&quot;&lt;/code&gt;&lt;/em&gt;&lt;br&gt;&#xA;  &amp;nbsp;&lt;strong&gt;~ @&lt;/strong&gt;    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room P&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;← Nerd with Stockholm Syndrome and an Oedipal complex?&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room Q&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;Classification?&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;← Man&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;↓ Beast&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room R&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;Popular fan opinion?&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;→ Annoyingly helpful&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;↓ Homicidally endearing&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room S&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;Would make a better guide than these magical floor clues?&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;↑ Sure!&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;↓ Would rather find own path&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room T&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;&lt;code&gt;&quot;This little vulpine can fly and swim...unlike you.&quot;&lt;/code&gt;&lt;/em&gt;&lt;br&gt;&#xA;  &amp;nbsp;&lt;strong&gt;~ @&lt;/strong&gt;    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room U&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;Classification?&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;← Mammal-like&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;↓ Reptile-like&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room V&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;Movement?&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;← Sometimes, briefly, like a helicopter&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;→ Sometimes, briefly, with a stolen body - not in a potato&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room W&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;&lt;code&gt;&quot;This little idiot would like you to take a test.&quot;&lt;/code&gt;&lt;/em&gt;&lt;br&gt;&#xA;  &amp;nbsp;&lt;strong&gt;~ @&lt;/strong&gt;    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room X&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;Backed yourself in a corner&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;↑ About-face&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room Y&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;&lt;code&gt;&quot;Let's flutter jump out of here.&quot;&lt;/code&gt;&lt;/em&gt;&lt;br&gt;&#xA;  &amp;nbsp;&lt;strong&gt;~ @&lt;/strong&gt;    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Room Z&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;How did you get here?!&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;You got in this doorless room; you figure a way out.&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15921" LastActivityDate="2015-10-23T15:58:33.367" Title="Seeking Sidekicks" Tags="|knowledge|mazes|video-games|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23452" PostTypeId="2" ParentId="23436" CreationDate="2015-10-23T13:20:29.157" Score="5" Body="&lt;h2&gt;(Not a solution, just some partial progress:)&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Lemma 1.&lt;/strong&gt; If $m$ is odd and $D(m)$ is even, then $m$ is a perfect square.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof: Let $\prod_{i=1}^kp_i^{a_i}$ be the prime factorization of $m$. &#xA;Then the sum of all divisors of $m$ (with $m$ itself included) is&#xA;$$ D(m)+m ~=~ \prod_{i=1}^k (1+p_i+p_i^2+\cdots+p_i^{a_i})$$&#xA;If one of the exponents $a_i$ is odd, then the corresponding factor $1+p_i+p_i^2+\cdots+p_i^{a_i}$ in the right hand side has an even number of odd terms, and hence is even. Then the right hand side is even, which is in contradiction to $m$ odd and $D(m)$ even.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This contradiction implies that all exponents $a_i$ must be even, so that $m$ is a perfect square. Q.e.d.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Lemma 2.&lt;/strong&gt; If that man's calculation is indeed correct, then at least one of the numbers $m$ and $n$ must be an odd perfect square.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof: By the problem statement both numbers $m$ and $n$ are odd. &#xA;Suppose for the sake of contradiction that neither of them is a square. &#xA;Then by Lemma 1, the numbers $D(m)$ and $D(n)$ both are odd, too.&#xA;But then the left hand side $D(m)D(n)−1$ in the given equation is even,&#xA;whereas its right hand side $mn−(m+n)$ is odd. Contradiction. Q.e.d.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-10-23T13:20:29.157" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23453" PostTypeId="2" ParentId="23438" CreationDate="2015-10-23T13:22:57.550" Score="11" Body="&lt;p&gt;The theme is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Beatles &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One was a star.&#xA; The best was bested.&#xA; One drove a car.&#xA; One only rested. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ringo &lt;strong&gt;Starr&lt;/strong&gt;. Pete &lt;strong&gt;Best&lt;/strong&gt; was replaced. Either &lt;em&gt;Drive My Car&lt;/em&gt; (vvye) or &lt;em&gt;Day Tripper&lt;/em&gt; (&lt;strong&gt;Sunday driver, yeah&lt;/strong&gt;). While there is a lot of sleeping in Beatles songs, &lt;em&gt;I'm Only Sleeping&lt;/em&gt; (vvye) also includes &quot;only&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One made art.&#xA; With stones they vied.&#xA; But then they did part.&#xA; And half of them died.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stuart Sutcliffe, original bassist, left to be &lt;strong&gt;an artist&lt;/strong&gt; (A E),  They were a group at the same time as the &lt;strong&gt;Rolling Stones&lt;/strong&gt;, then they &lt;strong&gt;broke up&lt;/strong&gt; and now &lt;strong&gt;two of the four&lt;/strong&gt; (Harrison and Lennon) are &lt;strong&gt;dead&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;31 was a cop.&#xA; Mustard's bank was his nose.&#xA; They rose to the top.&#xA; As your matriarch knows.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In Maxwell's Silver Hammer, &quot;&lt;strong&gt;P.C. 31&lt;/strong&gt; said we caught a dirty one.&quot; (Irishpanda) - PC for Police Constable, or cop, &lt;em&gt;Mean Mr Mustard&lt;/em&gt; &quot;&lt;strong&gt;Keeps a ten-bob note up his nose&lt;/strong&gt;&quot;, the Beatles were the top band by any measure, and &lt;em&gt;Your Mother Should Know&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of the last two,&#xA; One is amazed.&#xA; The other sells shoes.&#xA; The youth were all crazed. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is Paul McCartney, who sang &lt;em&gt;Maybe I'm Amazed&lt;/em&gt;, and Ringo Starr, who now has a line of shoes (among other things). The last line is Beatlemania (A E)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They all shared a liver.&#xA; They spoke of her love.&#xA; Iron met river.&#xA; And villainous glove. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; pun on &lt;strong&gt;Liverpool&lt;/strong&gt; where they were from (histocrat), &lt;em&gt;She Loves You&lt;/em&gt; (A E), Liverpool's Cast Iron Shore mentioned in Glass Onion (Irishpanda), in &lt;em&gt;Yellow Submarine&lt;/em&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Chief_Blue_Meanie&quot; rel=&quot;nofollow&quot;&gt;bad guy's&lt;/a&gt; fiercest weapon was &quot;The Dreadful Flying Glove&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A lane full of sights.&#xA; The town had a pool.&#xA; Three members, one knight.&#xA; A silvery tool. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Penny Lane&lt;/em&gt;, pun on &lt;strong&gt;Liverpool&lt;/strong&gt; where they were from, all four were awarded MBE but John returned his, Paul was knighted, and &lt;em&gt;Maxwell's Silver Hammer&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fabulous number.&#xA; A half dozen cried.&#xA; Eyes gilded in slumber.&#xA; We heard of a slide. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Fab Four&quot;, though there's plenty of weeping, in &lt;em&gt;While My Guitar Gently Weeps&lt;/em&gt; (A E) the six strings of the guitar can be the half dozen, &lt;em&gt;Golden Slumbers&lt;/em&gt;, and &lt;em&gt;Helter Skelter&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A teacher was killed.&#xA; And broken wings flew.&#xA; A door was filled.&#xA; And you're wealthy too. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Maxwell's Silver Hammer&lt;/em&gt; is about a teacher being killed (Fillet), a line in &lt;em&gt;Blackbird&lt;/em&gt; is &lt;strong&gt;take these broken wings and learn to fly&lt;/strong&gt;, in &lt;em&gt;Fixing A Hole&lt;/em&gt;: &quot;I'm filling the cracks that ran through the door&quot; (Irishpanda), &lt;em&gt;Baby, You're a Rich Man&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An apartment was burned.&#xA; Or so it was said.&#xA; Because a lover was spurned.&#xA; With a drain in his bed. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Norwegian Wood (This Bird Has Flown)&lt;/em&gt;. &quot;I crawled off to sleep in the bath&quot; and &quot;I lit a fire.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1509" LastEditorUserId="17062" LastEditDate="2015-11-10T01:06:31.150" LastActivityDate="2015-11-10T01:06:31.150" CommentCount="28" ContentLicense="CC BY-SA 3.0" />
  <row Id="23454" PostTypeId="2" ParentId="23314" CreationDate="2015-10-23T13:32:58.440" Score="2" Body="&lt;p&gt;Here is a website that can answer a lot of these questions with a really concise URL syntax: yougowords.com&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;7 letters and 4 syllables: &lt;a href=&quot;http://www.yougowords.com/7-letters/4-syllables&quot; rel=&quot;nofollow&quot;&gt;http://www.yougowords.com/7-letters/4-syllables&lt;/a&gt; (ability, apology)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;6 letters and 4 syllables: &lt;a href=&quot;http://www.yougowords.com/6-letters/4-syllables&quot; rel=&quot;nofollow&quot;&gt;http://www.yougowords.com/6-letters/4-syllables&lt;/a&gt; (utopia, apiary)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;8 letters and 5 syllables: &lt;a href=&quot;http://www.yougowords.com/8-letters/5-syllables&quot; rel=&quot;nofollow&quot;&gt;http://www.yougowords.com/8-letters/5-syllables&lt;/a&gt; (academia, obituary)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note, a lot of them seem to be wrong, such as &lt;em&gt;alien&lt;/em&gt;, which has three syllables instead of four.&lt;/p&gt;&#xA;" OwnerUserId="12795" LastActivityDate="2015-10-23T13:32:58.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23455" PostTypeId="2" ParentId="23451" CreationDate="2015-10-23T14:39:58.763" Score="2" Body="&lt;p&gt;I found these characters:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Luigi through 2L, Yoshi through 2QUY, Knuckles from Sonic the Hedgehog through 1HGFK, Tails through 3VUT, Navi from Legend of Zelda through 3RSN, Epona through 1DE, Wheatley from Portal through 3RW, the Companion Cube through 1DC, Otakon from Metal Gear Solid through 2QPO, and Meryl through 1IM.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That makes the possible answers&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Mario, Sonic, Link, Chell, and Solid Snake&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the homeworlds&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Mushroom Kingdom, Mobius, Hyrule, Aperture Science, and... wherever Metal Gear takes place.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="11876" LastEditDate="2015-10-23T15:58:33.367" LastActivityDate="2015-10-23T15:58:33.367" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23456" PostTypeId="2" ParentId="23451" CreationDate="2015-10-23T14:44:06.767" Score="1" Body="&lt;p&gt;I am&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Link (The Legend of Zelda)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and I found&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Navi and Epona&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;through the paths&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;3RSN and 1IE.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16608" LastActivityDate="2015-10-23T14:44:06.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23458" PostTypeId="1" AcceptedAnswerId="23460" CreationDate="2015-10-23T15:12:18.343" Score="13" ViewCount="734" Body="&lt;p&gt;Today I met professor Halfbrain at the tea house. &#xA;The professor looked very tired, and apparently had not slept for the last couple of days. He told me that he had been spending his time with cutting right-angled triangles into smaller pieces. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The professor had originally started out with six huge right-angled triangles: &#xA;two congruent triangles with side length $9,12,15$, &#xA;two congruent triangles with side length $12,16,20$, and&#xA;two congruent triangles with side length $15,20,25$. &lt;/li&gt;&#xA;&lt;li&gt;In one cutting operation, the professor cut one of his current right-angled triangles into two smaller right-angled ones. The cut was always done along the altitude dropped onto the hypotenuse. (Note that all the resulting triangles are again right-angled.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The professor was extremely happy to report that $-$ after many hours of cutting and cutting and cutting $-$ he had finally produced a triangle set in which there did not remain a single pair of congruent triangles. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is it indeed possible to generate a set of pairwise &#xA;  non-congruent triangles (according to the above rules), or has the &#xA;  professor once again made one of his mathematical blunders?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-09T09:15:44.520" LastActivityDate="2016-03-09T09:15:44.520" Title="Professor Halfbrain and the right-angled triangles" Tags="|mathematics|geometry|reachability|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23460" PostTypeId="2" ParentId="23458" CreationDate="2015-10-23T15:49:45.423" Score="12" Body="&lt;p&gt;Professor Halfbrain&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; has made a blunder, it is not possible to generate a set of pairwise non-congruent triangles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For integer $a$, $b\geq 0$, let $T_{a,b}$ denote the right triangle with side lengths&#xA;$$&#xA;\frac{3^{a+1} 4^b}{5^{a+b-1}},\;\;\;\frac{3^a 4^{b+1}}{5^{a+b-1}}\;\;\;\frac{3^a 4^b}{5^{a+b-2}}.&#xA;$$&#xA;Cutting triangle $T_{a,b}$ results in the triangles $T_{a+1,b}$ and $T_{a,b+1}$. The initial triangles are two copies each of $T_{1,0}$, $T_{0,1}$, and $T_{0,0}$, so every triangle ever produced will have the form $T_{a,b}$ for some integers $a$, $b\geq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To the triangle $T_{a,b}$, assign a weight $\frac{1}{2^{a+b}}$. Dissecting $T_{a,b}$ into $T_{a+1,b}$ and $T_{a,b+1}$ does not change the sum of the weights. At the start, the sum of the weights is&#xA;$$&#xA;2\cdot\frac{1}{2^1}+2\cdot\frac{1}{2^1}+2\cdot\frac{1}{2^0}=4.&#xA;$$&#xA;If at any point the triangles are pairwise non-congruent, the sum of the weights must be less than&#xA;$$&#xA;\sum_{a,b\geq 0}\frac{1}{2^{a+b}}=4.&#xA;$$&#xA;So it is not possible that no pair of triangles is congruent.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastActivityDate="2015-10-23T15:49:45.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23461" PostTypeId="1" AcceptedAnswerId="23474" CreationDate="2015-10-23T16:03:28.783" Score="6" ViewCount="347" Body="&lt;p&gt;For each of the numbering systems below, can you determine the rule for representing numbers in that system?&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th&gt;Decimal&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;Number System 1&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;Number System 2&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;Number System 3&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;25&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;9&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;231&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;200&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;392&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;$,&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;6040&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2003&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;5,236&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;W+&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;83440&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1011002&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;146,004&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;00t&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1380300&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;300000012&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;25,897,872&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&amp;gt;3V1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;C6345030&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;123014&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;" OwnerUserId="13970" LastEditorUserId="69582" LastEditDate="2021-01-27T05:13:08.690" LastActivityDate="2021-01-27T05:13:08.690" Title="All Your Base (Numbering System Conversions)" Tags="|mathematics|pattern|" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23462" PostTypeId="1" AcceptedAnswerId="23464" CreationDate="2015-10-23T16:50:31.010" Score="15" ViewCount="1590" Body="&lt;p&gt;Let me tell you about the day in my life that everything changed.  Maybe it will sound something like your life.  Maybe only bits and pieces will seem familiar.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;I got up early this morning, but that isn’t out of the ordinary- the men in my commune usually wake everybody up at the crack of dawn.  As the sun was rising, the excitement was rising as well.  We knew that daylight would bring visitors, as well as interesting sights, sounds, and smells.  Some of the ladies took time to dress up their choicest body parts, feeling that these might garner them attention.  Others did their best to look young and vital, capable of starting a family and growing it for many years.&lt;/p&gt;&#xA;&lt;p&gt;While they strutted their stuff, I sat alone on the ground.  I guess that’s just where I am comfortable, with the simple pleasures.  I closed my eyes to listen to the day, when all of the sudden, I was embraced and lifted from my place.  Immediately, I knew.  I had been chosen.  I was going to be taken away to live in a new place, with this kind old man.  He looked like he had spent many a day in the sun, but this seemed to have kept him warm on the inside.  Sure, there were stories of other freedom-seeking ladies, leaving the commune only to be cooped up day and night, just trying to scratch out a livelihood.  These did not matter to me.  The man seemed nice, and I was ready to start the sunny side of my life.&lt;/p&gt;&#xA;&lt;p&gt;Travelling with the man, sometimes in his arms, sometimes at his heels, we stopped at two other communes, where he brought two more additions into his “family”.  One was much bigger than me, and I didn’t really trust him.  He seemed to have eyes for me, but there was something sinister there as well.  He would slink around and make me feel uncomfortable.  The third addition became my enemy as well, and my glaring eyes said as much.  This was because the third addition could not walk, and thus had to be carried.  Angry and uneasy, I now stuck very close to my new patron and did my best to stay in his eyesight.&lt;/p&gt;&#xA;&lt;p&gt;This was effective and it relaxed me, but it did not last.  As we neared the man’s home, he became nervous and somewhat puzzled.  Something seemed to be stopping us from going any further.  It became apparent that he had not coordinated his travel plans very well, but obviously wanted to ensure safe passage to us all.  He didn’t want to split up this “family” that just barely formed, and even more did not want us to fall in harm’s way.&lt;/p&gt;&#xA;&lt;p&gt;I sat in the dirt as the man began sketching out diagrams in it,    counting to himself, and pointing at the three of us and himself in    various orders.  I began to let myself listen to the day again and    must have drifted into a light sleep.  I remember him hugging me    three more times or so, but wasn’t fully awake again until we had    reached his home.  He seemed pleased, relieved, and remarked about    how “it was all so easy.”  He must have been referring to the kind of    life we’ll get to lead, the easy life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Questions&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Who is narrating this day?&lt;/li&gt;&#xA;&lt;li&gt;Who makes up the &amp;quot;family&amp;quot; the narrator joins?&lt;/li&gt;&#xA;&lt;li&gt;What really happened on this day? &lt;em&gt;(remember the narrator missed some things while asleep)&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11452" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-10-23T17:26:30.863" Title="An Unfamiliar Day in the Life" Tags="|lateral-thinking|story|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23464" PostTypeId="2" ParentId="23462" CreationDate="2015-10-23T17:00:11.433" Score="13" Body="&lt;p&gt;The narrator is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The chicken from the fox, chicken, and bag of corn puzzle (scratching out a living, cooped up)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the family is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The farmer, the chicken, the fox (slinking around, makes the chicken uneasy), the bag of corn (had to be carried)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What happened?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Why, the farmer collected all the &quot;family members&quot; and on the way home happened on the river that he had to cross safely without any of the family eating one of the others. The chicken had vague recollections of the times she had been carried across.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16874" LastEditorUserId="16874" LastEditDate="2015-10-23T17:26:30.863" LastActivityDate="2015-10-23T17:26:30.863" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23465" PostTypeId="1" AcceptedAnswerId="23468" CreationDate="2015-10-23T17:30:35.857" Score="8" ViewCount="773" Body="&lt;p&gt;&lt;sub&gt;This is in the spirit of the &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%22what+is+a%22+word+is%3Aquestion&quot;&gt;What is a Word/Phrase™&lt;/a&gt; series started by &lt;a href=&quot;https://puzzling.stackexchange.com/users/463&quot;&gt;JLee&lt;/a&gt; with a special brand of &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%22Phrase%22%20user%3A463%20is%3Aquestion&quot;&gt;Phrase™&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%22What+is+%22+%22Word%22+-missing+user%3A463+is%3Aquestion&quot;&gt;Word™&lt;/a&gt; puzzles.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a word conforms to a special rule, I call it a Tellygraph Word™.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you tell me how to get a word that conforms to the rule?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LMC4I.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LMC4I.png&quot; alt=&quot;Examples of Tellygraph Words&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, if you want to analyze, here is a CSV version:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;Tellygraph Words™,Not Tellygraph Words™&#xA;ANVIL,FORGE&#xA;ANXIETY,STRESS&#xA;FILTHY,DIRTY&#xA;HAMLET,HOVEL&#xA;INFLAME,IGNITE&#xA;INHALE,BREATHE&#xA;KNIFE,BLADE&#xA;NATIVE,LOCAL&#xA;NEAT,CLEAN&#xA;TWINKLE,SHINE&#xA;VITAL,CRITICAL&#xA;VIXEN,HARLOT&#xA;WEALTHY,WELL-OFF&#xA;WHINE,FUSS&#xA;&lt;/pre&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2016-09-21T17:11:37.407" Title="What is a Tellygraph Word™?" Tags="|pattern|english|word-property|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23466" PostTypeId="1" AcceptedAnswerId="23472" CreationDate="2015-10-23T17:31:44.033" Score="5" ViewCount="435" Body="&lt;p&gt;&lt;sub&gt;This is in the spirit of the &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%22what+is+a%22+word+is%3Aquestion&quot;&gt;What is a Word/Phrase™&lt;/a&gt; series started by &lt;a href=&quot;https://puzzling.stackexchange.com/users/463&quot;&gt;JLee&lt;/a&gt; with a special brand of &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%22Phrase%22%20user%3A463%20is%3Aquestion&quot;&gt;Phrase™&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%22What+is+%22+%22Word%22+-missing+user%3A463+is%3Aquestion&quot;&gt;Word™&lt;/a&gt; puzzles.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a word conforms to a special rule, I call it a Unbeecoming Word™.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the following examples below to find the rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9mzor.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9mzor.png&quot; alt=&quot;Examples of Unbeecoming Words™&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, if you want to analyze, here is a CSV version:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;Unbeecoming Words™,Not Unbeecoming Words™&#xA;FASHION,STYLE&#xA;FORMULA,RECIPE&#xA;FUNKY,WEIRD&#xA;INFORM,EDUCATE&#xA;INFORMAL,CASUAL&#xA;JOURNALISM,REPORTING&#xA;LIQUOR,BEER&#xA;MASONRY,BRICKWORK&#xA;SAILOR,MIDSHIPMAN&#xA;SQUAWK,SQUEAL&#xA;SQUIRM,WIGGLE&#xA;UNFAIR,CRUEL&#xA;UNIFORMS,OUTFITS&#xA;URINAL,TOILET&#xA;&lt;/pre&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2016-09-21T17:12:11.047" Title="What is an Unbeecoming Word™?" Tags="|pattern|english|word-property|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23467" PostTypeId="1" AcceptedAnswerId="23600" CreationDate="2015-10-23T17:35:10.653" Score="15" ViewCount="805" Body="&lt;p&gt;&lt;sub&gt;This is in the spirit of the &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%22what+is+a%22+word+is%3Aquestion&quot;&gt;What is a Word/Phrase™&lt;/a&gt; series started by &lt;a href=&quot;https://puzzling.stackexchange.com/users/463&quot;&gt;JLee&lt;/a&gt; with a special brand of &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%22Phrase%22%20user%3A463%20is%3Aquestion&quot;&gt;Phrase™&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%22What+is+%22+%22Word%22+-missing+user%3A463+is%3Aquestion&quot;&gt;Word™&lt;/a&gt; puzzles.&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;If a word conforms to a special rule, I call it a Measured Word™.&lt;/p&gt;&#xA;&lt;p&gt;Use the following examples below to find the rule.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ssfMR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ssfMR.png&quot; alt=&quot;Examples of Measured Words™&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;And, if you want to analyze, here is a CSV version: (If you copy this text elsewhere, make sure that your font matches the one in the image above.)&lt;/p&gt;&#xA;&lt;pre&gt;&#xA;Measured Words™,Not Measured Words™&#xA;AMITY,DISCORD&#xA;BEST,WORST&#xA;CLAW,TALON&#xA;DAM,WALL&#xA;DYE,COLOR&#xA;MILK,WATER&#xA;MOTH,FLY&#xA;NEST,DEN&#xA;OX,BULL&#xA;PAW,HOOF&#xA;REST,WALK&#xA;SLY,SIMPLE&#xA;WOLF,TIGER&#xA;ZESTY,MILD&#xA;&lt;/pre&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; You've had a stroke of good luck! I'm giving you a hint.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2021-12-17T19:56:20.353" LastActivityDate="2021-12-17T19:56:20.353" Title="What is a Measured Word™?" Tags="|pattern|english|letters|word-property|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23468" PostTypeId="2" ParentId="23465" CreationDate="2015-10-23T18:00:21.883" Score="10" Body="&lt;p&gt;A Tellygraph Word™ is, in essence, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;a word containing no letters that are drawn using a curved line (&lt;code&gt;BCDGJOPQRSU&lt;/code&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's called a Tellygraph Word™ because &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;if you look at the top edge or the bottom edge of the word, the letters can be interpreted as Morse Code.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The word &lt;strong&gt;&lt;code&gt;ANVIL&lt;/code&gt;&lt;/strong&gt;  The top edge of the word is $\centerdot\centerdot\centerdot\centerdot\centerdot - \centerdot$.  The bottom edge is $\centerdot\centerdot\centerdot\centerdot\centerdot - -$  (The first five dots on the top are from the top of the &lt;code&gt;A&lt;/code&gt;, the tops of the two arms of the &lt;code&gt;N&lt;/code&gt;, and the tops of the two arms of the &lt;code&gt;V&lt;/code&gt;.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="16363" LastEditDate="2016-09-21T06:15:35.673" LastActivityDate="2016-09-21T06:15:35.673" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23469" PostTypeId="1" AcceptedAnswerId="23561" CreationDate="2015-10-23T18:34:28.790" Score="10" ViewCount="942" Body="&lt;h2&gt;I AM AN IDIOT AND HAD THE WRONG NAME FOR THE NEMESIS&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;I'm a little worried now that the pattern won't get recognized because it's requires an &quot;Aha!&quot; leap that is only obvious to me because I designed the puzzle. I'll try to give a few more helpful hints and see what happens.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Before he was an evil overlord ruling with an iron talon over a realm of darkness, Hooty McOwlface was Harold McArthur and was a world-renowned adventurer. He sought treasure and fame. On one particular undertaking, he came across what appeared to be an ancient tomb buried deep within the Amazon. After defeating the usual gambit of traps and tricks, he came upon a gilded room. His torch created stars on the polished walls and all the world was gold. Aside from the plating, though, there was no treasure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although he was at first frustrated by what he took to be the evidence of grave robbers, his hope renewed upon spying a single silver brick on the opposite wall. When his fingers brushed against it, he was surprised to find it was warm and moved under his touch. With a smooth motion, it slid up and under the tile above, revealing a panel lit from within. There was a 9x30 grid of holes, many of which had small pegs inserted into them. It reminded Harold of a toy from his childhood that let you create lighted artwork with small, plastic pegs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There were five pegs laying on a tray just underneath the panel. They each had a faint glow and some kind of marking on the side but, aside from that, they looked exactly like the pegs already inserted into the panel. Experimentally, he stuck all five into random open holes. Nothing happened. He tried to pull out one of the pegs that were inserted to begin with but it was stuck fast. He couldn't just try every possible combination because there were almost 572 million possibilities and Harold had forgotten to pack lunch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After studying the panel for a while, a glimmer of recognition formed. He picked up a peg and stuck it into a particular open hole. Nothing happened but he was confident so he went ahead with the rest. Once all five were in, their glow vanished! The panel slid shut and a vibration started under his feet. Slowly, ever so slowly, the entire wall to the right of the panel slid aside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Harold raised his torch and peered inside, a loud bang nearly gave him a heart attack. Suddenly, his face was peppered with something small and slick. When he recovered from the shock, he saw that he had been shot with a confetti cannon. The colored paper littered the floor. Looking up, he saw the only thing in the secret room: a banner that read &quot;GOOD JOB!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Angered at the waste of his time, he stripped every ounce of gold from the walls and used it to hire his first henchmen and take his vengeance upon the world.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Would you have been able to solve the panel in time for lunch?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the image below, The white squares with a hollow circle represent empty holes. The black squares with the double white circle represent holes with pegs in them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0zjtc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0zjtc.png&quot; alt=&quot;Diagram of Puzzle Panel&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the pegs exactly as they appeared to Harold. They should be used to fill in five of the empty holes in the image above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cOQF6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cOQF6.png&quot; alt=&quot;Image of Pegs&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To alleviate any chance of confusion on the characters, here is the text from each peg:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;Blue Peg (ϠϕϞζ)&#xA;--------------------&#xA;Ϡ   03E0   Sampi (Archaic)&#xA;ϕ   03D5   Phi (Symbol)&#xA;Ϟ   03DE   Koppa (Capital)&#xA;ζ   03B6   Zeta (Capital)&#xA;&#xA;Red Peg (ΫΘβγ)&#xA;--------------------&#xA;Ϋ   03AB   Upsilon (Capital with Dialytika)&#xA;Θ   0398   Theta (Capital)&#xA;β   03B2   Beta (Lowercase)&#xA;γ   03B3   Gamma (Lowercase)&#xA;&#xA;Green Peg (θγΘϞ)&#xA;--------------------&#xA;θ   03B8   Theta (Lowercase)&#xA;γ   03B3   Gamma (Lowercase)&#xA;Θ   0398   Theta (Capital)&#xA;Ϟ   03DE   Koppa (Capital)&#xA;&#xA;Purple Peg (φβΣθ)&#xA;--------------------&#xA;φ   03C6   Phi (Lowercase)&#xA;β   03B2   Beta (Lowercase)&#xA;Σ   03A3   Sigma (Capital)&#xA;θ   03B8   Theta (Lowercase)&#xA;&#xA;Orange Peg (ΔΣβΨ)&#xA;--------------------&#xA;Δ   0394   Delta (Capital)&#xA;Σ   03A3   Sigma (Capital)&#xA;β   03B2   Beta (Lowercase)&#xA;Ψ   03A8   Psi (Capital)&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;One of Harold's first henchmen was heard to be retelling this story in a tavern shortly after receiving his first paycheck. It turns out that Harold was pretty upset about how he had been tricked and was telling everyone. As it turns out,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the peg colors were a &lt;a href=&quot;https://en.wikipedia.org/wiki/Red_herring&quot; rel=&quot;nofollow noreferrer&quot;&gt;red herring&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The first thing Harold did after hiring his first henchmen was to hunt down whoever created that temple. After all, based on the solution, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; he knew it was a recent creation and not an ancient one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As it turns out, the perpetrator was none other his old nemesis:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lord Augustus.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Of course, that's just what Harold's nemesis called himself once he rose to power. To annoy his, Harold called him by his original name:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; August&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What infuriated Harold the most is how he was tricked into thinking it was an ancient temple. Once he recognized the pattern, he knew it couldn't possibly have been older than &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the 1870s because the pattern it references wasn't invented until that decade.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What was even &lt;em&gt;more&lt;/em&gt; infuriating was all the distracting extra information. Someone was just trying to make explorers waste time.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The characters inscribed on the pegs were &lt;em&gt;also&lt;/em&gt; a red herring! In reality, it didn't matter which peg went where! The only thing that mattered was which five holes got filled in.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If you asked Harold how he managed to find the pattern behind the pegs, he would explain that he tried breaking it up into pieces for analysis. What worked the best was when he&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; divided the first three rows into 3x3 blocks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Knowing what his made his nemesis famous, the grid started to make sense.&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2016-08-31T15:33:33.170" LastActivityDate="2016-08-31T15:33:33.170" Title="Peg the Panel to Procure the Prize" Tags="|pattern|story|visual|" AnswerCount="3" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="23470" PostTypeId="2" ParentId="23461" CreationDate="2015-10-23T19:38:35.070" Score="4" Body="&lt;p&gt;Partial answer ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number System 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is basically a base 95 number system where the value of each digit is the corresponding ASCII value minus 32.&lt;br&gt;&#xA; $$\small\begin{array}{rrrrr}\\ &amp;amp;9 \rightarrow &amp;amp;&amp;amp;&amp;amp;&amp;amp;(57-32) \cdot 95^0=&amp;amp;25\\&amp;amp;\$, \rightarrow &amp;amp;&amp;amp;&amp;amp;(36-32) \cdot 95^1 + &amp;amp;(44-32) \cdot 95^0=&amp;amp;392\\&amp;amp;W+ \rightarrow &amp;amp;&amp;amp;&amp;amp;(87-32) \cdot 95^1 + &amp;amp;(43-32) \cdot 95^0=&amp;amp;523600\\&amp;amp;t \rightarrow &amp;amp;&amp;amp;(48-32) \cdot 95^2 + &amp;amp;(48-32) \cdot 95^1 + &amp;amp;(116-32) \cdot 95^0=&amp;amp;146004\\&amp;amp;&amp;gt;3V1 \rightarrow &amp;amp;(62-32) \cdot 95^3 + &amp;amp;(51-32) \cdot 95^2 + &amp;amp;(86-32) \cdot 95^1 + &amp;amp;(49-32) \cdot 95^0=&amp;amp;25897872\\\end{array}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Number System 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this system the digits have the following values from least to most significant:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $1^0 \space 2^1 \space 3^2 \space 4^3 \space ...$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Interpreting $C$ as $12$ similar to hexadecimal numbers this gives:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\small\begin{array}{rrrrrrrrrr}\\&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;2 \cdot 3^2 + &amp;amp;3 \cdot 2^1 + &amp;amp;1 \cdot 1^0 = &amp;amp;25\\&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;6 \cdot 4^3 + &amp;amp;0 \cdot 3^2 + &amp;amp;4 \cdot 2^1 + &amp;amp;0 \cdot 1^0 = &amp;amp;392\\&amp;amp;&amp;amp;&amp;amp;&amp;amp;8 \cdot 5^4 + &amp;amp;3 \cdot 4^3 + &amp;amp;4 \cdot 3^2 + &amp;amp;4 \cdot 2^1 + &amp;amp;0 \cdot 1^0 = &amp;amp;5236\\&amp;amp;&amp;amp;1 \cdot 7^6 + &amp;amp;3 \cdot 6^5 + &amp;amp;8 \cdot 5^4 + &amp;amp;0 \cdot 4^3 + &amp;amp;3 \cdot 3^2 + &amp;amp;0 \cdot 2^1 + &amp;amp;0 \cdot 1^0 = &amp;amp;146004\\&amp;amp;12 \cdot 8^7 + &amp;amp;6 \cdot 7^6 + &amp;amp;3 \cdot 6^5 + &amp;amp;4 \cdot 5^4 + &amp;amp;5 \cdot 4^3 + &amp;amp;0 \cdot 3^2 + &amp;amp;3 \cdot 2^1 + &amp;amp;0 \cdot 1^0 = &amp;amp;25897872\end{array}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="4421" LastEditDate="2015-11-06T15:19:01.130" LastActivityDate="2015-11-06T15:19:01.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23471" PostTypeId="1" AcceptedAnswerId="23475" CreationDate="2015-10-23T20:22:48.970" Score="11" ViewCount="1266" Body="&lt;pre&gt;&lt;code&gt;I am one of many brothers &amp;amp; sisters,&#xA;Maybe the most important of all.&#xA;&#xA;I am made fun of because I'm different&#xA;I do not deny I am the fattest of them all!&#xA;&#xA;Indeed I bring order to our master,&#xA;without me their words would be in duress.&#xA;&#xA;You can see a part of me everywhere&#xA;Near you or above, in the vast expanses.&#xA;&#xA;The other part is where many &#xA;like to drown their sorrows &amp;amp; some de-stress&#xA;&#xA;My methods are unconventional though,&#xA;through divide I bring peace!&#xA;&#xA;The world would be chaotic,&#xA;If my existence you cease!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who am I?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my first puzzle ever, so please excuse noviceness. Hence, there may be multiple right answers and hence try to explain your answer in the context of the riddle!&lt;/p&gt;&#xA;" OwnerUserId="17040" LastActivityDate="2015-10-23T20:40:27.583" Title="I am one of many! Who am I?" Tags="|riddle|word|rhyme|poetry|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23472" PostTypeId="2" ParentId="23466" CreationDate="2015-10-23T20:27:49.453" Score="5" Body="&lt;p&gt;The Unbeecoming words&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; don't include letters that are sounded with an &lt;code&gt;ee&lt;/code&gt; when said: &lt;code&gt;BCDEGPTVZ&lt;/code&gt; (bee, see, dee ...). Hence, unb&lt;strong&gt;ee&lt;/strong&gt;coming.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-10-23T20:27:49.453" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23474" PostTypeId="2" ParentId="23461" CreationDate="2015-10-23T20:39:25.043" Score="5" Body="&lt;p&gt;Number system 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This system is based on prime factors.  The rightmost digit is the number of $2$s in the factorization, the next is the number of $3$s, and so on.&lt;br&gt;&#xA;&lt;br&gt;&#xA;  $$\small\begin{align}25_{10} &amp;amp;=&amp;amp; 5^2 \centerdot 3^0 \centerdot 2^0 &amp;amp;=&amp;amp; 200_{pf} &amp;amp;\;\;\;\text{(look at the exponents)}\\392_{10} &amp;amp;=&amp;amp; 7^2 \centerdot 5^0 \centerdot 3^0 \centerdot 2^3 &amp;amp;=&amp;amp; 2003_{pf}\\5236_{10} &amp;amp;=&amp;amp; 17^1 \centerdot 13^0 \centerdot 11^1 \centerdot 7^1 \centerdot 5^0 \centerdot 3^0 \centerdot 2^2 &amp;amp;=&amp;amp; 1011002_{pf}\\146004_{10} &amp;amp;=&amp;amp; 23^3 \centerdot 19^0 \centerdot 17^0 \centerdot 13^0 \centerdot 11^0 \centerdot 7^0 \centerdot 5^0 \centerdot 3^1 \centerdot 2^2 &amp;amp;=&amp;amp; 300000013_{pf}\\25,897,872_{10} &amp;amp;=&amp;amp; 13^1 \centerdot 11^2 \centerdot 7^3 \centerdot 5^0 \centerdot 3^1 \centerdot 2^4 &amp;amp;=&amp;amp; 123014_{pf}\\\end{align}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="4421" LastEditDate="2015-11-06T15:21:03.480" LastActivityDate="2015-11-06T15:21:03.480" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="23475" PostTypeId="2" ParentId="23471" CreationDate="2015-10-23T20:40:27.583" Score="14" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Space Bar&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am one of many brothers &amp;amp; sisters,&#xA;Maybe the most important of all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am made fun of because I'm different&#xA;I do not deny I am the fattest of them all!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The space bar is the largest key on a keyboard&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Indeed I bring order to our master,&#xA;without me their words would be in duress.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Without the space bar, all of the words you type would be connected.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can see a part of me everywhere&#xA;Near you or above, in the vast expanses.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This describes the word &quot;space.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The other part is where many&#xA;like to drown their sorrows &amp;amp; some de-stress&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This describes the word &quot;bar.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My methods are unconventional though,&#xA;through divide I bring peace!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The world would be chaotic,&#xA;If my existence you cease!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A space bar divides characters, andthisclauseisprettychaotic!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17079" LastActivityDate="2015-10-23T20:40:27.583" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23476" PostTypeId="1" CreationDate="2015-10-23T21:03:45.483" Score="9" ViewCount="8931" Body="&lt;p&gt;So, my math teacher gave us a magic math square with &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the 9 in the bottom right corner, &lt;/li&gt;&#xA;&lt;li&gt;the 7 in the left column middle row, and &lt;/li&gt;&#xA;&lt;li&gt;the 1 in the middle column top row.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;She said she would give whoever figured this out a can of soda from the teachers' lounge. I can't figure it out. I've tried nonstop for 2 days$\ldots$ &#xA;I need help :(    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those of you who don't know, a magic square is a grid with (in this case) 9 boxes and you have to put the numbers 1-9 in each square. &#xA;Each column, row, and diagonally have to equal the same number. My math teacher said her and the other math teachers have found a way so it is possible but I'm not the best at math so I really need some help. &lt;/p&gt;&#xA;" OwnerUserId="17080" LastEditorUserId="8874" LastEditDate="2015-10-24T12:01:02.507" LastActivityDate="2018-12-24T22:12:29.487" Title="Complete the magic square!" Tags="|mathematics|combinatorics|magic-square|" AnswerCount="5" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23477" PostTypeId="2" ParentId="5122" CreationDate="2015-10-23T21:12:52.747" Score="1" Body="&lt;p&gt;I think it could also be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Heat / Atoms!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You need me. You need me in many ways. Yet, I'm slowly losing my family.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Atoms are continuously splitting away from each other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can count on me as you always did!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quite general, but can also mean Humans constantly estimating the number of atoms. In a smaller sense, &quot;measuring heat&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sound of an explosion, transferred to the orange land.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The BIG BANG&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can stop the darkness!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Photons&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you want, I can even stop the bitter cold!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Electrons vibrating produces heat&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I bring you life as in 2 bound circle-like shapes!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The nucleus and electrons!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tSYRa.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tSYRa.gif&quot; alt=&quot;Atom&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet, I keep losing my family. We are torn apart by whom who need us most.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nature needs them the most and they are getting split from each other every moment!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17040" LastActivityDate="2015-10-23T21:12:52.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23478" PostTypeId="2" ParentId="23476" CreationDate="2015-10-23T21:30:08.367" Score="5" Body="&lt;p&gt;It's not possible. I think this website explained very good why: &lt;a href=&quot;http://www.mathematische-basteleien.de/magsquare.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.mathematische-basteleien.de/magsquare.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To quote it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You have 1+2+3+4+5+6+7+8+9=45. In a magic square you have to add 3 numbers again and again. Therefore the average sum of three numbers is 45:3=15. The number 15 is called the magic number of the 3x3 square. &#xA;  You can also achieve 15, if you add the middle number 5 three times. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;You can reduce 15 in a sum of three summands eight times:15=1+5+9  &#xA;  15=1+6+8. 15=2+4+9 &#xA;  15=2+5+8  15=2+6+7  &#xA;  15=3+4+8 15=3+5+7  &#xA;  15=4+5+6&#xA;  The odd numbers 1,3,7, and 9 occur twice in the reductions, the even numbers 2,4,6,8 three times and the number 5 four times. &#xA;  Therefore you have to place number 5 in the middle of the magic 3x3 square. The remaining odd numbers have to be in the middles of a side and the even numbers at the corners. &#xA;  Under these circumstances there are eight possibilities building a square: &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/E5xSS.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;All the eight squares change into each other, if you reflect them at the axes of symmetry. You count symmetric squares only once. Therefore there is only one magic 3x3 square. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1748" LastEditorUserId="11110" LastEditDate="2018-12-24T10:32:36.527" LastActivityDate="2018-12-24T10:32:36.527" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23479" PostTypeId="2" ParentId="23476" CreationDate="2015-10-23T21:46:13.117" Score="16" Body="&lt;p&gt;Given your description, the filled-in numbers on the magic square look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{c|c|c}&#xA;&amp;amp;1&amp;amp;\\\hline&#xA;7&amp;amp;&amp;amp;\\\hline&#xA;&amp;amp;&amp;amp;9&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, other answerers have pointed out that the middle number is 5, so the 9 should go on the bottom rather than the bottom-right, in which case the completed magic square looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{c|c|c}&#xA;6&amp;amp;1&amp;amp;8\\\hline&#xA;7&amp;amp;5&amp;amp;3\\\hline&#xA;2&amp;amp;9&amp;amp;4&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that's only for a magic square with the numbers 1 to 9. If you thought a little outside the box, you might wonder whether you could do it with other numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can use some algebra to show that this is still impossible. Let's suppose the number 1 isn't filled in, and the middle number is $x$, like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{c|c|c}&#xA;&amp;amp;&amp;amp;\\\hline&#xA;7&amp;amp;x&amp;amp;\\\hline&#xA;&amp;amp;&amp;amp;9&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the properties that all 3-by-3 magic squares have is that the rows, columns, and diagonals add up to $3x$ (3 times the middle number). So we can fill in some of the other numbers with respect to $x$, by subtracting $3x$ from them:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{c|c|c}&#xA;2x-9&amp;amp;&amp;amp;\\\hline&#xA;7&amp;amp;x&amp;amp;2x-7\\\hline&#xA;&amp;amp;&amp;amp;9&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then solve for the other two corner squares:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{c|c|c}&#xA;2x-9&amp;amp;&amp;amp;x-2\\\hline&#xA;7&amp;amp;x&amp;amp;2x-7\\\hline&#xA;x+2&amp;amp;&amp;amp;9&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally solve for the last two edge squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{c|c|c}&#xA;2x-9&amp;amp;11&amp;amp;x-2\\\hline&#xA;7&amp;amp;x&amp;amp;2x-7\\\hline&#xA;x+2&amp;amp;2x-11&amp;amp;9&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Huh? It turns out that no matter what the middle number $x$ is, if the 7 and 9 are placed where they are, the top square &lt;em&gt;has&lt;/em&gt; to be 11. The fact that it was filled out as 1 means that the magic square contradicts itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then again... if you think a little further outside the box, who said that the filled-in squares had to remain unmodified? If you draw another 1 next to the existing 1 in the top square, you get 11 as required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So... do that. Then, fill in any single number on any square and the rest of the numbers will work themselves out like magic. You can ask your teacher for that can of soda now ;-)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;...except not really. Your teacher probably still expects a solution with distinct positive integers (whole numbers that are all different). There exists one, but I'll leave it to you to figure it out on your own since you seem to want to do as much of this as you can by yourself.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="4421" LastEditDate="2015-11-02T21:03:01.067" LastActivityDate="2015-11-02T21:03:01.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23480" PostTypeId="1" AcceptedAnswerId="23481" CreationDate="2015-10-24T01:07:21.873" Score="5" ViewCount="320" Body="&lt;h3&gt;Beatles and Friends:&lt;/h3&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Derek, some called me,&lt;br /&gt;&#xA;Earlier in my life.&lt;br /&gt;&#xA;I took Majnun's story,&lt;br /&gt;&#xA;And my best friend's wife.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I met them with Richard.&lt;br /&gt;&#xA;In circles it goes.&lt;br /&gt;&#xA;And twice near the end,&lt;br /&gt;&#xA;My service they chose.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Famous in my own right.&lt;br /&gt;&#xA;They asked for my axe.&lt;br /&gt;&#xA;But the &amp;quot;number&amp;quot; that night,&lt;br /&gt;&#xA;I played on a sax.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I was often above them,&lt;br /&gt;&#xA;But I frequently played.&lt;br /&gt;&#xA;I made the world love them.&lt;br /&gt;&#xA;Thus my fame was made&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;My story is tragic.&lt;br /&gt;&#xA;But they were my boys.&lt;br /&gt;&#xA;After I saw the magic,&lt;br /&gt;&#xA;In a cellar of noise.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;My first true love was art,&lt;br /&gt;&#xA;And Astrid my second.&lt;br /&gt;&#xA;From my friends did I part,&lt;br /&gt;&#xA;Before fame had beckoned.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I once lived with my Best friend.&lt;br /&gt;&#xA;And I gave him a brother.&lt;br /&gt;&#xA;For me did the band send,&lt;br /&gt;&#xA;So I left him and his mother.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I swam in their second,&lt;br /&gt;&#xA;Half servant, half pal.&lt;br /&gt;&#xA;My fate no one reckoned,&lt;br /&gt;&#xA;The police shot poor Mal.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Who are these eight people, and what does each line mean?&lt;/p&gt;&#xA;&lt;p&gt;Basic Hint:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Some familiarity with the Beatles' associates, collaborators, and entourage is necessary to solve this one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Advanced Hints:&lt;/p&gt;&#xA;&lt;p&gt;These are pretty big hints, and probably give things away, so use them only as a last resort.&lt;/p&gt;&#xA;&lt;p&gt;2:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Also associated with:  Nat King Cole, Sam Cooke, Sly &amp;amp; the Family Stone, King Curtis, the Rolling Stones, Ray Charles, George Harrison, Ringo Starr &amp;amp; His All-Starr Band, and Eric Clapton, as well as being a solo artist.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;3:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Perhaps because of their alleged rivalry, when this man was asked to perform on a Beatles song, he chose to do so on an instrument he didn't know how to play. As a result, we have the worst instrumental performance ever made on a Beatles recording.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;4:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/ajRfL.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ajRfL.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;8:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/KcWaw.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KcWaw.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17062" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-10-24T04:40:32.243" Title="This one should be popular: Part two!" Tags="|riddle|knowledge|wordplay|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23481" PostTypeId="2" ParentId="23480" CreationDate="2015-10-24T01:24:17.710" Score="4" Body="&lt;p&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eric Clapton. First line refers to Derek and the Dominos. The line about Majnun refers to a legend that was part of the inspiration for &quot;Layla&quot;. The best friend's wife was Pattie Boyd, who was married to his friend George Harrison &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Billy Preston. Keyboard player. His most well known song is &quot;Will It Go Round In Circles. He met them while touring with Little Richard, and he played with them on the albums Let It Be and Abbey Road just before they broke up. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Brian Jones. The Rolling Stones were thought of as huge rivals to the Beatles. He played sax on You Know My Name (Look Up the Number).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sir George Martin. Producer and sounde engineer (above them, in the booth) often referred to as the 'Fifth Beatle'. Contributed to their commercial success and the formation of their sound. Did arranging work, and often performed on various recordings in different capacities.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Brian Epstein. He discovered the Beatles in the Cavern Club (Cellar Full of Noise, which was also the title of his autobiography), and managed them until his tragic, early death. Also somewhat tragic due to his homosexuality at a time when such was considered criminal in England.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stuart Sutcliffe. Original bassist, but left to pursue art before the group became famous. Astrid Kirchnerr was his fiancee before his untimely death.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;7.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Neil Aspinall. Moved in with his friend Pete Best and his mother. Got involved with the mother and fathered a child (giving Pete a brother) and later became the Beatles' road manager.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;8.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mal Evans. Drove the band around, helped set up equipment, acted as a gofer, etc. In their second film, Help!, he appeared as a swimmer emerging from under the ice after it had been cracked open by a bomb. Was shot by police at his house when his girlfriend told police he had a gun, and they mistook his air rifle for a real weapon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16874" LastEditorUserId="16874" LastEditDate="2015-10-24T04:40:32.243" LastActivityDate="2015-10-24T04:40:32.243" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23482" PostTypeId="1" AcceptedAnswerId="23490" CreationDate="2015-10-24T04:42:44.470" Score="3" ViewCount="343" Body="&lt;p&gt;I have been many things in my life. Specifically, and in order, a monkey, a knife, a junkie, an alien, a lunatic, a cat, an aristocrat, a clown, and finally, an earthling.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What or who am I, and what do each of these things refer to?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hints:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first was eponymous, the second a change, the third an explorer, the fourth an idol. The fifth just a lad, the sixth quite cool, the seventh pale and waifish, the eighth fond of ashes, the last a technophile.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My left is big, my right changes size.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Little Liza, Gnome, Andy Warhol, Starman, Drive-in, Big Brother, It's not the side effects, Little green, I'm afraid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="17062" LastEditorUserId="17062" LastEditDate="2015-10-24T05:18:39.813" LastActivityDate="2015-10-25T02:28:45.267" Title="A Riddle of Many Faces" Tags="|riddle|knowledge|wordplay|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23483" PostTypeId="1" AcceptedAnswerId="23484" CreationDate="2015-10-24T05:29:55.280" Score="8" ViewCount="630" Body="&lt;p&gt;You are a prisoner.  One day, the Warden summons you and ninety-nine others to see his hat collection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Bowlers and baseball caps. Stensons and sombreros.  One hundred and one headpieces, each unique in the world,&quot; says the Warden.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;You are the smartest of prisoners in my domain, and by law I must offer you a chance at your freedom.  So here is the game we will play.  While you all are blindfolded, I will place a hat on each of your heads.  I will also wear one myself.    Once the blindfolds are removed, you will look around and know the identity of every hat except two: your own hat, and my hat.  Yes, by then I'll be gone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;You will all then guess, by secret ballot, the identity of my hat.  Besides seeing what other people are wearing, these guesses are independent.  No communication once the game begins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;If everyone names my hat, you will all go free.  But a single wrong answer means you can all walk yourselves right back to your cells.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Good luck.  I reckon you will need it, as I calculate your odds of winning at $2^{-100}$.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As the prisoners gather to discuss their strategy, someone asks you what should be done.  What do you say?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A couple notes:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Assume you can't cheat by looking at the brim of your own hat -- you really don't know what your own hat is.&lt;/li&gt;&#xA;&lt;li&gt;The solution I am aware of uses math beyond high-school level.  Nothing too super powered, but something you might not be aware of if you weren't / aren't a math major in college.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2722" LastActivityDate="2015-10-24T06:05:32.407" Title="One hundred and one hats" Tags="|mathematics|hat-guessing|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23484" PostTypeId="2" ParentId="23483" CreationDate="2015-10-24T05:57:56.463" Score="8" Body="&lt;p&gt;Over all possible situations, each prisoner will guess right at most half the time. If we want to maximize the probability of all guessing right, we must make it so that all prisoners are correct in the same situations as each other. To do this, we must find a property of hat assignments such that it is flipped if any two hats are exchanged. This is one way to express that property:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assign each prisoner a number from 1 to 100, and have the warden be 101. Also number the hats from 1 to 101. Consider the number of pairs of people such that the one with the larger number has the smaller hat. One possibility for the warden's hat makes the number even, and the other makes it odd (this is the &lt;em&gt;parity of the permutation&lt;/em&gt; defined by the hat assignments).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If all the prisoners choose the one that makes the number even (all choosing odd is equally good), then they will be all correct half the time and all wrong half the time, which is the best possible chance of success.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-10-24T06:05:32.407" LastActivityDate="2015-10-24T06:05:32.407" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23486" PostTypeId="2" ParentId="12" CreationDate="2015-10-24T11:06:49.610" Score="1" Body="&lt;p&gt;I came across &lt;a href=&quot;http://arxiv.org/pdf/1201.0749v2.pdf&quot; rel=&quot;nofollow&quot;&gt;this result&lt;/a&gt; that claims that 16-clue sudokus are not solvable without guessing as they have multiple solutions possible. While they have used brute-force to go over all the solutions, there is no mathematical proof yet. I think.&lt;/p&gt;&#xA;" OwnerUserId="17088" LastActivityDate="2015-10-24T11:06:49.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23490" PostTypeId="2" ParentId="23482" CreationDate="2015-10-24T12:34:58.200" Score="6" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; David Bowie&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the justification is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The varying stage characters created by Bowie over the years.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;a monkey, &amp;amp; eponymous &amp;amp; Little Liza&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; His first stage name was &lt;strong&gt;Davy Jones&lt;/strong&gt;, which also was the stage name of David Thomas &quot;Davy&quot; Jones of the Monkees; his first song was &quot;Liza Jane&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;a knife, &amp;amp; a change &amp;amp; Gnome&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Then he changed his stage name to &lt;strong&gt;David Bowie&lt;/strong&gt;, like the Bowie knife called after American frontiersman Jim Bowie; song &quot;The Laughing Gnome&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;a junkie, &amp;amp; an explorer &amp;amp; Andy Warhol&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; astronaut and space explorer &lt;strong&gt;Major Tom&lt;/strong&gt;; the line &quot;We know Major Tom's a junkie&quot; occurs in the later song &quot;Ashes to ashes&quot;; Song &quot;Andy Warhol&quot; on &quot;Hunky Dory&quot;; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;an alien, &amp;amp; an idol &amp;amp; Starman&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Ziggy Stardust&lt;/strong&gt;= rock star and idol; album &quot;Ziggy Stardust&quot;; song &quot;Starman&quot;; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;a lunatic, &amp;amp; a lad &amp;amp; Drive-in&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Aladdin Sane&lt;/strong&gt; = &quot;A Lad Insane&quot;;  the album includes the song &quot;Drive-In Saturday&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;a cat, &amp;amp; quite cool &amp;amp; Big Brother&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Halloween Jack&lt;/strong&gt; is &quot;a real cool cat&quot;; album &quot;Diamond Dogs&quot; with song &quot;Big Brother&quot;;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;an aristocrat, &amp;amp; pale and waifish &amp;amp; It's not the side effects&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &quot;&lt;strong&gt;Thin White Duke&lt;/strong&gt;&quot; on &quot;Station to station&quot;; &quot;It's not the side effects of the cocaine / I'm thinking that it must be love&quot;; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;a clown, &amp;amp; fond of ashes &amp;amp; Little green&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The music video of &quot;Ashes to Ashes&quot; featured Bowie dressed up as a &lt;strong&gt;Pierrot clown&lt;/strong&gt;; &quot;Ashes to Ashes&quot; contains the line &quot;But the little green wheels are following me&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;and finally, an earthling &amp;amp; a technophile &amp;amp; I'm afraid&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I'm Afraid of Americans&quot; on Bowie's techno album &quot;&lt;strong&gt;Earthling&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Finally, hint 2 with &quot;My left is big, my right changes size&quot; refers to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the permanently dilated pupil in Bowie's left eye.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="17062" LastEditDate="2015-10-25T02:28:45.267" LastActivityDate="2015-10-25T02:28:45.267" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23492" PostTypeId="2" ParentId="22353" CreationDate="2015-10-24T13:27:33.533" Score="2" Body="&lt;p&gt;ghost_in_the_code's comment made me look at this question again and this is what I came up with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without looking at the cost it's pretty obvious that the fastest way to reach everyone is for everyone to forward 3 invites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let $f(x)$ be the function for the number of invitees that get added after $x$ minutes. It's not hard to see that this is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$f(x) = f(x-1) + f(x-3) + f(x-6)$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because everyone that was added 1 minute ago can add one, everyone that was invited three minutes ago just finish their second invite and everyone that was invited six minutes ago finished their third invite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we need to determine the base cases.&lt;br&gt;&#xA;$f(1)=1$ this is the first invite of the mayor's wife&lt;br&gt;&#xA;$f(2)=1$ this is the first invite of the $f(1)$&lt;br&gt;&#xA;$f(3)=2$ this is the second invite of the mayor's wife plus the first invite of $f(2)$&lt;br&gt;&#xA;$f(4)=3$ this is the second invite of $f(1)$ plus the first invites of $f(3)$&lt;br&gt;&#xA;$f(5)=3$ this is the second invite of $f(2)$ plus the first invites of $f(4)$&lt;br&gt;&#xA;$f(6)=5$ this is the second invites of $f(3)$ plus the first invites of $f(5)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $g(x)$ be the number of invites reached after $x$ minutes. This basically is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$g(x) = f(x) + f(x-1) + ... + f(1)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now using a computer program i saw that $g(24)=37088$ and $g(25)=57059$ so to partially answer &lt;strong&gt;Q1&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Afer 25 minutes 50,000 people can be reached.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To answer &lt;strong&gt;Q2&lt;/strong&gt; we just take $g(180)$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;After 3 hours 5,705,547,003,706,489,579,067,455,461,663,446 people can be reached.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think that &lt;strong&gt;Q3&lt;/strong&gt; is not really clear what your asking. You either want to reach the maximum or you want to be charged the mininum or any step in between. We can determine how to get the cheapest for $X$ people or we can decide how many people we can reach for $X$ cents but you haven't stated that goal.&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-10-24T13:27:33.533" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23493" PostTypeId="1" CreationDate="2015-10-24T13:29:07.873" Score="7" ViewCount="473" Body="&lt;p&gt;Let me take the English word &lt;strong&gt;&quot;stone&quot;&lt;/strong&gt; as an example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Replacing the string &quot;one&quot; by the corresponding number &quot;1&quot;  allows me to write &lt;strong&gt;stone&lt;/strong&gt; as &lt;strong&gt;st1&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In Dutch, &quot;stone&quot; is translated to &quot;steen&quot; and &quot;one&quot; is translated to &quot;een&quot;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Replacing the string &quot;een&quot; by the corresponding number &quot;1&quot;  allows me to write &lt;strong&gt;steen&lt;/strong&gt; as &lt;strong&gt;st1&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now note that both in English and in Dutch this replacement process leads to the same string &lt;strong&gt;&quot;st1&quot;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What is the longest example of this type that you can come up with?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The translation of the numbers can't be the same in both languages.&lt;/li&gt;&#xA;&lt;li&gt;The score of the word is based on the amount of number/letters in the &#xA;word that contains the letters/numbers (not the words with letters only)&lt;/li&gt;&#xA;&lt;li&gt;The number cant refere to a number in a word like 100000ling (100000 would refer to a number)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Current highscore&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;points      word           user        Language&#xA;______________________________________________&#xA;4           all1           Gamow         nl-ger&#xA;3           st1            Sven b        nl-en&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerDisplayName="user9543" LastEditorDisplayName="user9543" LastEditDate="2015-10-25T00:41:41.323" LastActivityDate="2020-04-14T15:07:11.953" Title="The longest word that includes numbers that can be changed with letters in different languages" Tags="|word|language|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23495" PostTypeId="1" AcceptedAnswerId="23502" CreationDate="2015-10-24T14:15:25.477" Score="5" ViewCount="262" Body="&lt;p&gt;Find the largest integer $N$ whose decimal representation has the following properties:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The rightmost digit in the decimal representation is not $0$.&lt;/li&gt;&#xA;&lt;li&gt;There exists a digit $d$ in the decimal representation which is not the leftmost digit, so that crossing out this digit $d$ yields the decimal representation of an integer divisor of $N$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;br&gt;&#xA;The integer $121$ is not divisible by $10$, and crossing out the digit $2$ yields the divisor $11$ of $121$.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-10-24T20:51:41.973" Title="Crossing out a digit" Tags="|mathematics|number-theory|no-computers|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23496" PostTypeId="2" ParentId="23493" CreationDate="2015-10-24T14:24:56.373" Score="3" Body="&lt;p&gt;Here is a repair to Ivo Beckers' first answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;5ling&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In Dutch: 5=&lt;strong&gt;vijf&lt;/strong&gt;, and &lt;strong&gt;vijf&lt;/strong&gt;ling=quintuplet.&lt;br&gt;&#xA;In German: 5=&lt;strong&gt;F&amp;#252;nf&lt;/strong&gt;, and &lt;strong&gt;F&amp;#252;nf&lt;/strong&gt;ling=quintuplet.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Under the new rules:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;All1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In Dutch:  1=&lt;strong&gt;een&lt;/strong&gt;, and All&lt;strong&gt;een&lt;/strong&gt;= lonely&lt;br&gt;&#xA;In German: 1=&lt;strong&gt;ein&lt;/strong&gt;, and All&lt;strong&gt;ein&lt;/strong&gt;= lonely&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-10-24T19:42:04.880" LastActivityDate="2015-10-24T19:42:04.880" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23497" PostTypeId="1" AcceptedAnswerId="23498" CreationDate="2015-10-24T14:54:54.560" Score="11" ViewCount="5482" Body="&lt;p&gt;Of three persons A, B and C, one always lies while the other two always speak the truth. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;C asked A: &quot;Do you always speak the truth, yes or no?&quot; &lt;/li&gt;&#xA;&lt;li&gt;A said something that C could not understand. &lt;/li&gt;&#xA;&lt;li&gt;So C asked B: &quot;What did A just say?&quot;&lt;/li&gt;&#xA;&lt;li&gt;B replied: &quot;A said NO.&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, who is the liar?&lt;/p&gt;&#xA;" OwnerUserId="17087" LastEditorUserId="8874" LastEditDate="2015-10-24T15:01:15.240" LastActivityDate="2020-07-07T08:05:33.573" Title="Who is the liar among A,B,C?" Tags="|logical-deduction|liars|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23498" PostTypeId="2" ParentId="23497" CreationDate="2015-10-24T14:58:19.720" Score="13" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; B is the liar.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reason:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If A is a truth-teller, he has correctly answered YES (I am a truth-teller).&lt;br&gt;&#xA; If A is a liar, he also has answered YES (lying, as he is a liar).&lt;br&gt;&#xA;&lt;br&gt;&#xA; So A must have said YES, and B must be lying when he claims that A said NO.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-10-24T15:04:43.203" LastActivityDate="2015-10-24T15:04:43.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23499" PostTypeId="1" AcceptedAnswerId="23501" CreationDate="2015-10-24T15:12:06.457" Score="0" ViewCount="498" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I take a box and put $K$ mini boxes inside. Then I chose several of those mini boxes at random and put $K$ micro boxes in each. I continue this procedure indefinitely long. Thereby, at the end each box contains (directly) either K or 0 boxes.&lt;br&gt;&#xA;  I tell you that there are $M$ boxes, which are not empty. Tell me how many are there empty boxes?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="28" LastActivityDate="2015-10-24T15:18:20.757" Title="Boxxxing puzzle" Tags="|mathematics|" AnswerCount="1" CommentCount="2" ClosedDate="2015-10-26T05:00:23.457" ContentLicense="CC BY-SA 3.0" />
  <row Id="23501" PostTypeId="2" ParentId="23499" CreationDate="2015-10-24T15:18:20.757" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $Empty=1+(M\times K)-M$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In other words:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $1$ (box we start with) $+(M\times K)$ ($K$ boxes we put $M$ times into another box) $-M$ (boxes that are not empty)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16608" LastActivityDate="2015-10-24T15:18:20.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23502" PostTypeId="2" ParentId="23495" CreationDate="2015-10-24T15:59:29.027" Score="5" Body="&lt;p&gt;&lt;strong&gt;Answer: $180625 = 17*10625$.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N = a*10^{n+1}+d*10^n+c = k*(a*10^n+c)$, $a,d,c,k$ is positive integers. $d &amp;lt; 10; c &amp;lt; 10^n; k &amp;gt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;move some terms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(10-k)*a*10^n + d*10^n = (k-1)*c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now. First of all we need to maximise $n$. In that case $(k-1)*c$ must be divisible by maximum power of 10. The only limit is that $c$ is not divisible by 10 and $k &amp;lt;= 19$ (otherwise left part of equation become negative). Then $k = 2^4+1 = 17, c = 5^4*c'$ and:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(-7)*a*10^n + d*10^n = c'*10^4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;max $n$ is 4, then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-7*a + d = c'$&lt;br&gt;&#xA;$7*a = d - c'$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $a$ must be maximised. $d &amp;lt;= 9$, then $a_{max} = 1$, also $c'_{max} = 1$, since $c'_{max}*5^4 = c$ is not divisible by 10.&lt;br&gt;&#xA;That means $a = 1, d = 8, c = 1; N = 180000+1*5^4 = 180625$&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2015-10-24T20:51:41.973" LastActivityDate="2015-10-24T20:51:41.973" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23503" PostTypeId="1" AcceptedAnswerId="23510" CreationDate="2015-10-24T16:20:21.990" Score="8" ViewCount="288" Body="&lt;p&gt;An autonumerogram, or self-enumerating number, is a number that can be read as describing itself by providing a pairwise inventory of its own digits. For example, 22 (two 2's) and 14233221 (one 4, two 3's, three 2's, and two 1's) are both autonumerograms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A pan-digit autonumerogram of order n is an autonumerogram containing all digits &quot;0&quot; through  &quot;n&quot;. What is the smallest pan-digit autonumerogram?&lt;/p&gt;&#xA;" OwnerUserId="7573" LastEditorUserId="7573" LastEditDate="2015-10-24T16:56:02.057" LastActivityDate="2015-10-24T19:14:51.860" Title="Autonumerograms" Tags="|mathematics|formation-of-numbers|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23504" PostTypeId="2" ParentId="23503" CreationDate="2015-10-24T16:27:50.807" Score="4" Body="&lt;p&gt;Isn't it the same thing as a &lt;a href=&quot;https://puzzling.stackexchange.com/questions/1904/find-the-smallest-n-to-complete-the-task-write-a-true-self-reflective-statemen/1907#1907&quot;&gt;self-reflective numerical statement&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In which case the answer is 10141516181923273271.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we're talking pan digital autonumerograms of &quot;arbitrary order $n$&quot;, a list of answers follows from that question as well:&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;10123133&lt;/li&gt;&#xA;&lt;li&gt;1014223133&lt;/li&gt;&#xA;&lt;li&gt;[no solution]&lt;/li&gt;&#xA;&lt;li&gt;10151623243241&lt;/li&gt;&#xA;&lt;li&gt;1014161723253251&lt;/li&gt;&#xA;&lt;li&gt;101415171823263261&lt;/li&gt;&#xA;&lt;li&gt;10141516181923273271&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2015-10-24T19:14:51.860" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23510" PostTypeId="2" ParentId="23503" CreationDate="2015-10-24T18:09:40.997" Score="9" Body="&lt;p&gt;0 can only appear once, since no digit that is included in the autonumerogram will appear 0 times. So our number must include &quot;10&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then 1 appears at least twice in the autonumerogram, so it must include a digit that is at least 2. If there are exactly 2 1's and exactly 2 2's, then 2 appears at least 3 times, a contradiction. So the largest digit in the autonumerogram must be at least 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we want the largest digit to be 3, then the number is of the form 10A1B2C3 (possibly with some sections rearranged), where A, B, and C are all 1, 2, or 3. We must have A+B+C=7 to make the total correct, so either they are 2, 2, 3 or they are 1, 3, 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If A, B, and C are 2, 2, and 3, then there are a total of 2 1's, 3 2's, and 2 3's. The smallest number for these is 10212332.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If A, B, and C are 1, 3, and 3, then there are a total of 3 1's, 1 2, and 3 3's. The smallest number for these is 10123133.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the smallest pan-digit autonumerogram is 10123133.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-10-24T18:09:40.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23511" PostTypeId="1" CreationDate="2015-10-24T18:46:07.233" Score="3" ViewCount="424" Body="&lt;p&gt;I think my math teacher is a spy. On the first day of class, they wrote &lt;em&gt;most&lt;/em&gt; of a secret message on the board - teacher said there were some letters missing. They said that anyone who could solve it would get extra credit. I think they're trying to find recruits for their secret organization. I've always wanted to be a spy. Can you help me find the missing letters and the secret message?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;T B H E I A S G I R S E S A U T R Y E E _ _ _ _ !&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17096" LastEditorUserId="9000" LastEditDate="2015-10-26T14:53:40.697" LastActivityDate="2015-10-26T14:53:40.697" Title="A Secret Message In Plain(ish) Sight" Tags="|cipher|word-problem|" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23512" PostTypeId="1" AcceptedAnswerId="23514" CreationDate="2015-10-24T18:47:37.957" Score="3" ViewCount="480" Body="&lt;p&gt;Sheldon, Leonard and Wolowitz play a sizzling game of rock-paper-scissors for three players. In every round, each player simultaneously shows one of the three shapes. Rock beats scissors, scissors beats paper, and paper beats rock. If in a round exactly two distinct shapes are shown then 1 point is added to the score of the one player or to the two players who showed the winning shape, otherwise no point is added. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After many rounds of playing it occurred that each of the shapes had been shown exactly the same number of times. Is it possible that at this moment the total sum of points of Sheldon and Leonard and Howard was 2015?&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="7715" LastEditDate="2015-10-24T20:45:26.533" LastActivityDate="2015-10-24T20:45:26.533" Title="Rock-paper-scissors" Tags="|mathematics|strategy|game|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23513" PostTypeId="2" ParentId="23511" CreationDate="2015-10-24T18:57:19.697" Score="14" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;this is sure to be a great year&quot; with the hidden letters &quot;taor&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;just split it into even and odd letters:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TBHEIASGIRSESAUTRYEE -&gt; THISISSURE__ BEAGREATYE__&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;with spaces:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; THIS IS SURE__, BE A GREAT YE__&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the last 4 letters i just guessed&lt;/p&gt;&#xA;" OwnerUserId="16608" LastActivityDate="2015-10-24T18:57:19.697" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23514" PostTypeId="2" ParentId="23512" CreationDate="2015-10-24T19:16:52.240" Score="6" Body="&lt;p&gt;This is not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After each round, let $n_R$, $n_P$, and $n_S$ denote the total number of rocks, papers, and scissors' that have appeared, and let $S$, $L$, and $W$ be the scores of Sheldon, Leonard, and Wolowitz. Define a quantity&#xA;$$&#xA;Q:=S+L+W+2\cdot n_R+ n_P.&#xA;$$&#xA;It can be checked that on each round, the value of $Q$ either stays the same, increases by $3$, or increases by $6$. Since the value of $Q$ begins at $0$, $Q$ will always be divisible by $3$. However, if each shape has appeared the same number of times and the sum of the scores is $2015$, then&#xA;$$&#xA;Q=S+L+W+n_R+2 n_P=2015+3 n_R=3(671+n_R) +2&#xA;$$&#xA;is not divisible by $3$.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastActivityDate="2015-10-24T19:16:52.240" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23515" PostTypeId="2" ParentId="23512" CreationDate="2015-10-24T19:21:13.597" Score="4" Body="&lt;p&gt;Let R, P, and S be the number of rocks, papers, and scissors that have been shown so far, and let Q be the total sum of points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I claim that if 3 does not divide Q, then R/P/S are distinct. By symmetry, if we prove that R/P must be distinct, then also P/S and S/R must be distinct. We will consider the effect that one round can have on the following two values: R-P mod 3 and Q mod 3.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Q does not change. This happens when there is a 3 of a kind of a 1 1 1. In either case clearly R-P mod 3 does not change.&lt;/li&gt;&#xA;&lt;li&gt;Q increases by 1. This happens in 3 cases: 2R P, 2P S, 2S R. Clearly in all three cases R-P increases by 1 (mod 3).&lt;/li&gt;&#xA;&lt;li&gt;Q increases by 2. This happens in 3 cases: R 2P, P 2S, S 2R. Clearly in all three cases R-P increases by 2 (mod 3).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Therefore it is always true that $R-P \equiv Q \pmod{3}$, so if $Q \not\equiv 0 \pmod{3}$, we have $R-P\not\equiv0\pmod{3}$ or $R\not\equiv P\pmod{3}$ which implies that $R \ne P$, as claimed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the contrapositive of my claim, $R=P=Q \implies Q\equiv 0\pmod{3}\implies Q \ne 2015$. In fact, just two of R/P/Q being equal would imply the same thing.&lt;/p&gt;&#xA;" OwnerUserId="11100" LastActivityDate="2015-10-24T19:21:13.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23516" PostTypeId="2" ParentId="23512" CreationDate="2015-10-24T19:40:07.037" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is not possible&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we use variables instead of rock, paper, and scissors there are only 4 possible transactions.(let $A&amp;lt;B&amp;lt;C&amp;lt;A$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for number of times played we will only take note of the change of the difference to the average.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1: ABC -&amp;gt;  0A  0B  0C -&amp;gt; 0 points&#xA;2: AAA -&amp;gt; +2A -1B -1C -&amp;gt; 0 points&#xA;3: AAB -&amp;gt; +1A  0B -1C -&amp;gt; 1 point&#xA;4: ABB -&amp;gt;  0A +1B -1C -&amp;gt; 2 points&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If we get a &quot;1&quot; there is no change whatsoever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we get a &quot;2&quot; AAA it can be solved(to a state where every difference to the average is 0) by:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;two &quot;2&quot; BBB and CCC -&gt; result 0 points total&lt;/li&gt;&#xA;&lt;li&gt;one &quot;2&quot; BBB one &quot;3&quot; CCA and one &quot;4&quot; BCC -&gt; result 3 points total  &lt;/li&gt;&#xA;&lt;li&gt;one &quot;3&quot; BBC and one &quot;4&quot; BCC -&gt; result 3 points total&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If we get a &quot;3&quot; AAB (and without using anymore &quot;2&quot; since we already found all solutions with &quot;2&quot;) it can be solved by:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;two &quot;3&quot; BBC and CCA -&gt; result 3 points total&lt;/li&gt;&#xA;&lt;li&gt;one &quot;4&quot; BCC -&gt; result 3 points total&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If we get a &quot;4&quot; ABB (without using anymore &quot;2&quot; or &quot;3&quot;) it can be solved by:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;two &quot;4&quot; BCC and CAA -&gt; result 6 points total&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We can now see that whatever we do to get each shape the same number of times our result must be dividable by 3, which 2015 is clearly not.&lt;/p&gt;&#xA;" OwnerUserId="16608" LastActivityDate="2015-10-24T19:40:07.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23518" PostTypeId="2" ParentId="23179" CreationDate="2015-10-24T21:29:22.167" Score="3" Body="&lt;p&gt;Another sequence that would make sense would be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; aaa,b,c,aa,bb,cc,a,bbb,ccc,...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;since&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the a's decrease and the b's and c's increase in numbers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That doesn't explain the stated &lt;em&gt;correct&lt;/em&gt; answer, though.&lt;/p&gt;&#xA;" OwnerUserId="12901" LastEditorUserId="12901" LastEditDate="2015-10-24T21:35:23.940" LastActivityDate="2015-10-24T21:35:23.940" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23519" PostTypeId="1" CreationDate="2015-10-24T22:12:08.687" Score="-4" ViewCount="1161" Body="&lt;p&gt;All I know is that this is in hexadecimal, after that I can't figure out how to decrypt this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;30 36 31 20 30 36 35 20 30 36 30 20 30 34 30 20 30 36 31 20 30 36 36 20 30 36 34 20 30 34 30 20 30 36 31 20 30 36 36 20 30 36 34 20 30 34 30 20 30 36 31 20 30 36 36 20 30 36 30 20 30 34 30 20 30 36 31 20 30 36 36 20 30 36 33 20 30 34 30 20 30 36 30 20 30 36 37 20 30 36 32 20 30 34 30 20 30 36 30 20 30 36 35 20 30 36 37 20 30 34 30 20 30 36 30 20 30 36 35 20 30 36 37 20 30 34 30 20 30 36 31 20 30 36 36 20 30 36 37 20 30 34 30 20 30 36 31 20 30 36 36 20 30 36 37 20 30 34 30 20 30 36 31 20 30 36 36 20 30 36 37 20 30 34 30 20 30 36 30 20 30 36 35 20 30 36 36 20 30 34 30 20 30 36 31 20 30 36 37 20 30 36 31 20 30 34 30 20 30 36 31 20 30 36 35 20 30 36 37 20 30 34 30 20 30 36 31 20 30 36 36 20 30 36 35 20 30 34 30 20 30 36 31 20 30 36 36 20 30 36 34 20 30 34 30 20 30 36 31 20 30 36 36 20 30 36 35 20 30 34 30 20 30 36 31 20 30 36 34 20 30 36 32 20 30 34 30 20 30 36 31 20 30 36 34 20 30 36 35 20 30 34 30 20 30 36 30 20 30 36 35 20 30 36 36 20 30 34 30 20 30 36 31 20 30 36 34 20 30 36 33 20 30 34 30 20 30 36 31 20 30 36 35 20 30 36 37 20 30 34 30 20 30 36 31 20 30 36 35 20 30 36 35 20 30 34 30 20 30 36 30 20 30 36 35 20 30 36 37 20 30 34 30 20 30 36 31 20 30 36 34 20 30 36 33 20 30 34 30 20 30 36 31 20 30 36 35 20 30 36 30 20 30 34 30 20 30 36 31 20 30 36 34 20 30 36 31 20 30 34 30 20 30 36 31 20 30 36 35 20 30 36 36 20 30 34 30 20 30 36 31 20 30 36 35 20 30 36 36 20 30 34 30 20 30 36 31 20 30 36 34 20 30 36 35 20 30 34 30 20 30 36 31 20 30 36 35 20 30 36 34 20 30 34 30 20 30 36 30 20 30 36 35 20 30 36 37 20 30 34 30 20 30 36 31 20 30 36 32 20 30 36 35 20 30 34 30 20 30 36 31 20 30 36 30 20 30 36 33 20 30 34 30 20 30 36 31 20 30 36 30 20 30 36 34 20 30 34 30 20 30 36 31 20 30 36 35 20 30 36 32 20 30 34 30 20 30 36 31 20 30 36 34 20 30 36 36 20 30 34 30 20 30 36 31 20 30 36 30 20 30 36 36 20 30 34 30 20 30 36 31 20 30 36 36 20 30 36 34 20 30 34 30 20 30 36 31 20 30 36 34 20 30 36 35 20 30 34 30 20 30 36 31 20 30 36 36 20 30 36 31 20 30 34 30 20 30 36 30 20 30 36 36 20 30 36 35 20 30 34 30 20 30 36 31 20 30 36 33 20 30 36 31 20 30 34 30 20 30 36 30 20 30 36 36 20 30 36 37 20 30 34 30 20 30 36 31 20 30 36 35 20 30 36 34 20 30 34 30 20 30 36 31 20 30 36 31 20 30 36 35 20 30 34 30 20 30 36 31 20 30 36 37 20 30 36 30 20 30 34 30 20 30 36 30 20 30 36 36 20 30 36 33 20 30 34 30 20 30 36 31 20 30 36 36 20 30 36 33 20 30 34 30 20 30 36 31 20 30 36 37 20 30 36 32 20 30 34 30 20 30 36 31 20 30 36 34 20 30 36 36 20 30 34 30 20 30 36 30 20 30 36 36 20 30 36 34 20 30 34 30 20 30 36 31 20 30 36 33 20 30 36 32 20 30 34 30 20 30 36 30 20 30 36 36 20 30 36 35 20 30 34 30 20 30 36 31 20 30 36 30 20 30 36 32 20 30 34 30 20 30 36 31 20 30 36 36 20 30 36 37&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerDisplayName="user17100" LastEditorUserId="69582" LastEditDate="2023-11-02T16:05:22.867" LastActivityDate="2023-11-02T16:05:22.867" Title="Decrypting Hexadecimal" Tags="|cipher|" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23520" PostTypeId="2" ParentId="23519" CreationDate="2015-10-24T22:19:13.720" Score="2" Body="&lt;p&gt;Using &lt;a href=&quot;http://string-functions.com/hex-string.aspx&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; hex to string converter yields this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;061 065 060 040 061 066 064 040 061 066 064 040 061 066 060 040 061 066 063 040 060 067 062 040 060 065 067 040 060 065 067 040 061 066 067 040 061 066 067 040 061 066 067 040 060 065 066 040 061 067 061 040 061 065 067 040 061 066 065 040 061 066 064 040 061 066 065 040 061 064 062 040 061 064 065 040 060 065 066 040 061 064 063 040 061 065 067 040 061 065 065 040 060 065 067 040 061 064 063 040 061 065 060 040 061 064 061 040 061 065 066 040 061 065 066 040 061 064 065 040 061 065 064 040 060 065 067 040 061 062 065 040 061 060 063 040 061 060 064 040 061 065 062 040 061 064 066 040 061 060 066 040 061 066 064 040 061 064 065 040 061 066 061 040 060 066 065 040 061 063 061 040 060 066 067 040 061 065 064 040 061 061 065 040 061 067 060 040 060 066 063 040 061 066 063 040 061 067 062 040 061 064 066 040 060 066 064 040 061 063 062 040 060 066 065 040 061 060 062 040 061 066 067&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And using an octal to ASCII converter yields:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;150 164 164 160 163 072 057 057 167 167 167 056 171 157 165 164 165 142 145 056 143 157 155 057 143 150 141 156 156 145 154 057 125 103 104 152 146 106 164 145 161 065 131 067 154 115 170 063 163 172 146 064 132 065 102 167&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Octal to ASCII again yields a Youtube channel:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/channel/UCDjfFteq5Y7lMx3szf4Z5Bw&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/channel/UCDjfFteq5Y7lMx3szf4Z5Bw&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="11876" LastEditDate="2015-10-25T00:01:31.513" LastActivityDate="2015-10-25T00:01:31.513" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23521" PostTypeId="1" AcceptedAnswerId="23605" CreationDate="2015-10-24T22:44:04.027" Score="10" ViewCount="621" Body="&lt;p&gt;So, a few days ago I told James the answer to his &lt;a href=&quot;https://puzzling.stackexchange.com/questions/23310/the-cursed-breakfast&quot;&gt;riddle&lt;/a&gt;. He seemed at the same time proud of me and a bit disappointed... I guess he thought it would take longer to solve! &quot;It seems the witch's &lt;em&gt;hex&lt;/em&gt; and &lt;em&gt;Bacon&lt;/em&gt; for breakfast weren't such subtle clues then!&quot;, he said. I felt then these puzzles were only starting...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lately, James has been introducing me to some Classical music. We have very similar tastes in music, and very different from most other kids, and so we always talk to each other about it. Recently we both started taking music classes, and right now we're learning some theory. So when I got a text from him simply saying &quot;what does this say&quot;, and a little later a picture, I wasn't so surprised.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This time, I'm pretty sure I actually got the answer! After the last puzzle, I've been reading up on codes and cyphers and it paid off nicely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you determine what message is encoded in the following picture?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kfEoG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kfEoG.png&quot; alt=&quot;tunes&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Feel the rhythm&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This sentece may be of importance: &lt;em&gt;I've been reading up on codes and cyphers&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16998" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-10-28T18:17:45.600" Title="A Musical Message" Tags="|cipher|music|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23522" PostTypeId="2" ParentId="23519" CreationDate="2015-10-25T00:53:44.897" Score="7" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;Author's note:&lt;/strong&gt; I would NOT recommend clicking any of the links in this post unless you are willing to see some creepy stuff.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Continuing on from where Deusovi left off:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://www.youtube.com/channel/UCDjfFteq5Y7lMx3szf4Z5Bw&quot; rel=&quot;nofollow&quot;&gt;Youtube channel in question&lt;/a&gt; has the name &quot;Ahriman Ratum&quot;. In Latin, this means &quot;Ahriman Ratified&quot;, where &lt;a href=&quot;https://en.wikipedia.org/wiki/Angra_Mainyu&quot; rel=&quot;nofollow&quot;&gt;Ahriman&lt;/a&gt; appears to be some sort of destructive spirit in the religion of &lt;a href=&quot;https://en.wikipedia.org/wiki/Zoroastrianism&quot; rel=&quot;nofollow&quot;&gt;Zoroastrianism&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The user's description is hex encoded twice and reads:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To the creator of epsat.webs.com and &lt;a href=&quot;https://www.youtube.com/user/nargargole&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/user/nargargole&lt;/a&gt; : This is a representation of your&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;(It appears the user did not realize the description has a length limit)&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The user's only video has a description which is hex encoded twice and reversed. Decoded, it reads:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Yazamaide Ahriman&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Warning: Here is where it gets more creepy.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The video appears to be taken from &lt;a href=&quot;https://www.youtube.com/channel/UCYxyglXZ3Ev9JpNZ0sJzAHA&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; channel, called &quot;Vorsk Backup Channel&quot;. From what I've read, and pieced together, this is a backup of a channel called &quot;Vorsk&quot; which was terminated by YouTube due to Community Guidelines violations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=xr9B0-xkAJk&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; video of his criticizes the user &lt;a href=&quot;https://www.youtube.com/user/Takedownman&quot; rel=&quot;nofollow&quot;&gt;Takedownman&lt;/a&gt; for reporting his channel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the videos of his channel seem to focus around the murder of &lt;a href=&quot;https://en.wikipedia.org/wiki/Murder_of_JonBen%C3%A9t_Ramsey&quot; rel=&quot;nofollow&quot;&gt;JonBenét Ramsey&lt;/a&gt;. &lt;a href=&quot;https://www.youtube.com/watch?v=PJbIcrpdtO4&quot; rel=&quot;nofollow&quot;&gt;One&lt;/a&gt;, entitled Lou Smit, shows a wobbly camera walking up to his grave. This is creepy, as I feel a hoaxer would not go to such lengths to make his channel 'look' creepy. &lt;a href=&quot;https://en.wikipedia.org/wiki/Lou_Smit&quot; rel=&quot;nofollow&quot;&gt;Lou Smit&lt;/a&gt; was an officer that worked on the JonBenét case. He protested that the Ramseys, who were the main suspects, were innocent, and the case remains unresolved. The description reads, after trivial decoding:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;He was right, he was right. They didn't listen to him&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll add more later to this post. Enough creep for one day.&lt;/p&gt;&#xA;" OwnerUserId="7308" LastEditorUserId="164" LastEditDate="2016-05-25T01:27:08.233" LastActivityDate="2016-05-25T01:27:08.233" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23523" PostTypeId="1" AcceptedAnswerId="23536" CreationDate="2015-10-25T01:06:36.163" Score="5" ViewCount="11723" Body="&lt;p&gt;In each category, all the animals have something in common, and all the categories are related.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;Category 1a:&lt;/h1&gt;&#xA;&lt;p&gt;Walruses, birds, ponies, tigers, elephants, cats, monkeys, bulldogs, bees, raccoons, cats,  octopuses, pigs, bullfrogs, blue jays, Eagles, worms,  fish, horses,  homing birds, blackbirds, kittens,  hogs, crabs, penguins,  doves,  sheep, sheepdogs, glow worms, butterflies, canaries, puppies, lizards, lambs.&lt;/p&gt;&#xA;&lt;h1&gt;Category 1b:&lt;/h1&gt;&#xA;&lt;p&gt;Roosters, songbirds, cats, dogs, horses, sheep, lions, elephants, cows, hens, crickets, blackbirds, either ducks, swans, or geese.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;Category 2:&lt;/h1&gt;&#xA;&lt;p&gt;Rams, horses, alligators,  frogs, bison, bluebirds, blackbird, cocks, seagulls, cows, snakes, honeybees, inchworms, dragonflies, moose (meese?), geese, koala bears, magpies.&lt;/p&gt;&#xA;&lt;p&gt;Struszia (trilobite)&lt;/p&gt;&#xA;&lt;h1&gt;Category 3:&lt;/h1&gt;&#xA;&lt;p&gt;Chicken, fish, monkey.&lt;/p&gt;&#xA;&lt;p&gt;Struszia (trilobite)&lt;/p&gt;&#xA;&lt;h1&gt;Category 4:&lt;/h1&gt;&#xA;&lt;p&gt;Mouse, blackbird, monkey, reindeer.&lt;/p&gt;&#xA;&lt;p&gt;Avalanchurus (trilobite)&lt;/p&gt;&#xA;&lt;h1&gt;Category 5:&lt;/h1&gt;&#xA;&lt;p&gt;Turkey, hound dog.&lt;/p&gt;&#xA;&lt;p&gt;Bumba (tarantula)&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;What is the theme of each category?  What do all the categories have in common?  Provide an example of each item.&lt;/h2&gt;&#xA;" OwnerUserId="17062" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2017-11-29T03:23:23.663" Title="What do we have in common?" Tags="|riddle|pattern|knowledge|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23524" PostTypeId="1" AcceptedAnswerId="23528" CreationDate="2015-10-25T01:12:09.903" Score="5" ViewCount="373" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VyfT4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VyfT4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above picture represents the drums stocked with wine at the shop. Each number represents the quantity in the same unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end of the day, the last remaining drum is retained by the store owner since it contains red wine, while the rest contained white. While going through the sales record of the day, the owner realized the following, &quot;The first buyer received three times the quantity as much as the second while the third received twice as much as the second.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you identify which drum contained the red wine?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a bonus you can also mention the quantity purchased by each buyer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addendum : To make the solution unique &quot;The number of drums sold to each were the same&quot;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Note : I am more interested in the methodology used to arrive at the solution. Brute force may help, but the problem should be solvable using only pencil and paper. Derived from a century old puzzle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9892" LastEditorUserId="9892" LastEditDate="2015-10-25T09:10:04.677" LastActivityDate="2015-10-25T10:29:24.787" Title="Which drum contained the Red Wine?" Tags="|mathematics|logical-deduction|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23525" PostTypeId="1" AcceptedAnswerId="23526" CreationDate="2015-10-25T05:47:37.540" Score="-6" ViewCount="126" Body="&lt;p&gt;Two plane mirrors facing each other are arranged at $60^o$ to each other. A point is located on the angle bisector. How many images of the point are there?&lt;/p&gt;&#xA;" OwnerUserId="17087" LastEditorUserId="13475" LastEditDate="2017-12-13T20:17:08.390" LastActivityDate="2017-12-13T20:17:08.390" Title="The number of images of the point is.?" Tags="|mathematics|" AnswerCount="1" CommentCount="0" ClosedDate="2015-10-25T07:41:08.873" ContentLicense="CC BY-SA 3.0" />
  <row Id="23526" PostTypeId="2" ParentId="23525" CreationDate="2015-10-25T06:29:55.740" Score="2" Body="&lt;p&gt;It's a direct formula. If $360/A$ is even then number of images formed  = $(360/A) - 1$ irrespective of it's position.&#xA;Which in this case is 5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To know more visit &lt;a href=&quot;https://physicswithpradeep.files.wordpress.com/2013/04/ray-optics.pdf&quot; rel=&quot;nofollow&quot;&gt;https://physicswithpradeep.files.wordpress.com/2013/04/ray-optics.pdf&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="17070" LastEditorUserId="17070" LastEditDate="2015-10-25T09:35:13.917" LastActivityDate="2015-10-25T09:35:13.917" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23527" PostTypeId="2" ParentId="23265" CreationDate="2015-10-25T07:58:49.080" Score="2" Body="&lt;p&gt;I claim the answers to both questions are &lt;strong&gt;yes&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;main lemma&lt;/strong&gt; needed is this: suppose $X$ is any collection of people such that every organization contains at least one person from $X$. Then there is some $n\ge 1$ such that at least $n$ people from $X$ go to the town center at noon on day $n$ (note that this $n$ must be less than or equal to the size of $X$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Call a set $X$ of people &lt;em&gt;sufficient&lt;/em&gt; if every organization contains at least one person from $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove the lemma, induct on the size $k$ of the sufficient set $X$. Suppose that all the townspeople know that the lemma is true for sufficient sets of size strictly less than $k$, that $X$ is a sufficient set of $k$ people, and that for every $1 \le n &amp;lt; k$ there have been strictly less than $n$ people from the set $X$ who showed up at the town center on day $n$. Then on the morning of day $k$, by the known cases of the lemma everyone in town will know that every proper subset of $X$ is insufficient. Suppose Alice is in the set $X$. Since $X$ is sufficient, every organization not containing Alice contains someone else from the set $X$, so Alice knows that if she were not a member of any organization then the collection of people in $X$ other than Alice would be sufficient - which she knows is impossible by the morning of day $k$! Thus Alice can prove that she is a member of some organization by day $k$. Since all of the other members of $X$ can reason similarly, on day $k$ all $k$ members of $X$ will show up in the town center (remember, this argument only applies if there was no $1 \le n &amp;lt; k$ such that at least $n$ people from $X$ showed up on day $n$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;To answer the first question&lt;/strong&gt;: let $X$ be the set of people who never show up to the town center. By the lemma, $X$ must be insufficient - so there is some organization which has no members in $X$, and that organization must therefore eventually have a proper meeting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second question, we will need a &lt;strong&gt;secondary lemma&lt;/strong&gt;: if Alice shows up for the first time on day $k$, then Alice is contained in some sufficient set $X$ of $k$ people such that for every $1 \le n &amp;lt; k$, strictly less than $n$ people showed up in the town center on day $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Call a set $Y$ of people &lt;em&gt;apparently sufficient&lt;/em&gt; (to Alice) if every organization which does not contain Alice contains at least one person from $Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will prove the secondary lemma by induction on the day $k$ when Alice first shows up in the town center. If $k = 1$, then it is obvious: the only way that Alice can show up at the town center on the first day is if Alice all by herself is sufficient. Now assume that $k$ is at least $2$, that Alice can prove the secondary lemma for strictly smaller values of $k$, and that Alice shows up for the first time on day $k$. Alice knows that if she were not a member of any organizations, then apparently sufficient sets and sufficient sets would be the same - and so she can use smaller cases of the secondary lemma to predict exactly what would happen on days $1$ to $k-1$ under the assumption that she is not a member of any organization. The only way she can possibly prove that she is a member of some organization on day $k$ is if some other person - say Bob - deviates from this prediction before day $k$. If Bob deviated before day $k-1$, then Alice would have gone to the town center before day $k$, so in fact Bob must have deviated from the prediction on day $k-1$. Since every sufficient set appears to be sufficient to Alice, the only way Bob can deviate from her prediction is if he fails to go to the town center when Alice expects him to. Thus there must be some apparently sufficient set $Y$ containing Bob of size $k-1$, such that for every $1 \le n &amp;lt; k-1$, strictly less than $n$ people from $Y$ show up at the town center on day $n$. Letting $X$ be the set obtained by adding Alice to $Y$, we see that since $X$ has size $k$, that $X$ is sufficient, and that on day $k-1$ at most $k-2$ people from $X$ showed up at the town center (since neither Alice nor Bob showed up on day $k-1$). Thus, for every $1 \le n &amp;lt; k$, strictly less than $n$ people from $X$ showed up at the town center on day $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;To answer the second question&lt;/strong&gt;: Suppose that Alice goes to the town center for the first time on day $k$, let $X$ be a sufficient set containing Alice of size $k$ such that for every $1 \le n &amp;lt; k$ strictly less than $n$ people showed up in the town center on day $n$. Let $Y$ be the set obtained by deleting Alice from $X$ (so $Y$ appears sufficient to Alice), and let $Z$ be the set of people who never show up at the town center. Then we can apply the main lemma to see that $Y \cup Z$ must be insufficient, so some organization must have no member in common with $Y\cup Z$. Since $Y$ is apparently sufficient, that organization must have Alice as a member, and by the definition of $Z$ that organization must eventually have a proper meeting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whew, that was long.&lt;/p&gt;&#xA;" OwnerUserId="11295" LastEditorUserId="11295" LastEditDate="2015-10-25T08:09:53.273" LastActivityDate="2015-10-25T08:09:53.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23528" PostTypeId="2" ParentId="23524" CreationDate="2015-10-25T08:12:58.540" Score="4" Body="&lt;p&gt;The total number of available units is $207$. The total amount of sold units must be divisible by $6$. So the number in the remaining drum must be $3+6x$. If we assume the owner didn't want to move the drums around, then the remaining drum must be one of the bottom ones. The only one matching is $15$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the drums sold to each customer I started with the last customer and again drums from the bottom:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;third customer: $10+11+13+13+17=64$&lt;/li&gt;&#xA;&lt;li&gt;second customer: $9+10+13=32$&lt;/li&gt;&#xA;&lt;li&gt;first customer: the rest except $15=96$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The additional requirement is easy to find, as each customer bought $7$ drums:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;first customer: $11+12+13+13+13+17+17=96$&lt;/li&gt;&#xA;&lt;li&gt;second customer: $3+3+3+5+5+6+7=32$&lt;/li&gt;&#xA;&lt;li&gt;third customer: the rest except $15=64$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And it still isn't unique:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;first customer: $11+13+13+13+15+17+17=99$&lt;/li&gt;&#xA;&lt;li&gt;second customer: $3+3+3+5+5+6+8=33$&lt;/li&gt;&#xA;&lt;li&gt;third customer: the rest except $9=66$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2015-10-25T09:50:10.050" LastActivityDate="2015-10-25T09:50:10.050" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23529" PostTypeId="2" ParentId="23524" CreationDate="2015-10-25T08:52:51.903" Score="3" Body="&lt;p&gt;Say the second buyer bought $x$ units. Then $6x$ units were sold (assuming no other customers that day). The drums sum to $207$ units. The closest multiple of $6$ to this is $204$, leaving one of the 3-unit drums.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, assuming drums are sold in their entirety or not at all, it is possible to leave unsold any drum with $3+6x$ units for any non-negative integer $x$. This means the drum remaining could be any of the $3, 9,$ or $15$ unit drums.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;New requirement: the number of drums sold were the same for each of the 3 customers (i.e. 7 each, leaving 1 for the owner).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is just the 3-unit drum remaining, $x=(207-3)/6 = 34$. However, after reserving one of the 3-unit drums for the owner, the smallest sum of 7 drums is 36. I.e. $x&amp;gt;34$. Reject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the other 2 cases, $x$ is 32 or 33. The smallest sum is 32, so there are possible solutions. With the given numbers, starting with the smallest sum gets you pretty close to $x$. Using that set as a starting point and swapping a number for one slightly larger, then doing the same for $2x$ gets you to a complete solution (the $3x$ combination is then determined).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sleafar has already listed some possible solutions, so I'll just add that other solutions still remain. For example, leaving the 9-unit drum behind, the customers could have bought the following combinations: $(3,3,3,5,5,7,7), (6,8,8,10,10,11,13), (12,12,13,13,15,17,17)$.&lt;/p&gt;&#xA;" OwnerUserId="9422" LastEditorUserId="9422" LastEditDate="2015-10-25T10:29:24.787" LastActivityDate="2015-10-25T10:29:24.787" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23530" PostTypeId="2" ParentId="3623" CreationDate="2015-10-25T09:32:29.000" Score="2" Body="&lt;p&gt;I don't know solution for any N, but may be investigation of more general problem can be helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say we have N dimensional space and for each X turn O has M turns. Call the result (max length for X) - L(N,M). Here I call players - X and O, and their marks - x and o. Then:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;L(1,2+) = 1. Obvious.&lt;/li&gt;&#xA;&lt;li&gt;L(1,1) = 2. This one is more interesting. Let's prove the value.&lt;br&gt;&#xA;1 is unreachable for O. And 2 is reachable by following the simple rule: When X makes turn put your 'o' always on the right of 'x', unless you can't.. when you can't - put it on the left. This guaranties that all new 1-x-size &quot;islands&quot;, which X creates are limited on the right and(!) no island can grow.&lt;br&gt;&#xA;Note, that if X always put x in distant parts of our space, we never actually see line of 2 x's. And O's goal not to create such a lines limited on the ends, but to prevent them. This must be common for all (N,M) cases.&lt;/li&gt;&#xA;&lt;li&gt;L(2, 8+) = 1. This is equivalent for L(1,2).&lt;/li&gt;&#xA;&lt;li&gt;L(2, 6) = 2. This is equivalent for L(1,1) in sense of safety (we surround the island from all sides and can &lt;em&gt;forget&lt;/em&gt; about it), but note, that 6 &amp;lt;&gt; 8/2, things become more interesting here. To surround horizontal/vertical xx we need 10 o's, but to surround diagonal xx we need 12 o's at least. So the rule would be - surround any new x with 6 o's: 3 on the top, 2 more on the right and 1 more at the bottom. If X attaches one more x - surround both. Simple and safe.&#xA;This of course can be optimised, because we don't need all-sides-surrounding, only 2-sides.&lt;/li&gt;&#xA;&lt;li&gt;L(2,4) = 2. This can be done. There are basically 4 possible lines directions (hor, vert, 2 x diag) and we need to limit each of them on one side exactly. For example we can limit each new x putting 3 o on top and 1 more on the right. When X attaches second x we still need to limit it in some places on top and right. First x will limit only one way, so up to 3 ways must be blocked with o's. Note the difference with L(1,1), where first x on the island limits second x &quot;halffully&quot; making all job, which O had to do with first x. So we spend 3 o's to do with second x the same thing, which we did with first x before, then we got 1 o left to limit our line - this is exactly what we need. Thereby we have all x's, independently of when they are added on the board or on the island limited on 4 sides, this guaranties that we can end any xx line from growing.&lt;/li&gt;&#xA;&lt;li&gt;L(2,3) = ?. This where things get messy already. O can't lock all 4 directions for first x, X put second x correspondingly and get xx line unlimited on both ends, but will it help to create X to create xxx line? O still have 2 o's to limit any line and 1 o to make precautions. But O now can't stop X's islands from growing, on early stages at least, thereby, X can create folks now and make groups of xx lines, which are not limited at 3 or may be even 4 directions, the last will mean that xxx line will be created.&lt;br&gt;&#xA;I played this game a bit, and it feels like L(2,3) = 2, but a short strategy is hard to find. I think, O must limit all 3 ways on the top (diagonals are the most important), deal with all folks on the bottom and slowly surround the island on all sides. I hope to finish this case later (or someone else can do it, then may be we can go to L(2,4) or L(3,M)).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="28" LastActivityDate="2015-10-25T09:32:29.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23531" PostTypeId="1" AcceptedAnswerId="23534" CreationDate="2015-10-25T12:33:09.267" Score="3" ViewCount="265" Body="&lt;p&gt;The ringmaster of a flea circus puts three fleas $A$, $B$, $C$ on three different numbers on the real number line, so that flea $B$ sits exactly in the middle between $A$ and $C$. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Whenever the ringmaster shouts &lt;em&gt;&quot;Hop!&quot;&lt;/em&gt;, one of the three fleas jumps over one of the other fleas to the mirror point on the other side. (In other words, a flea sitting in point $x-y$ may jump over a flea sitting in point $x$ to the new point $x+y$.)&lt;/li&gt;&#xA;&lt;li&gt;While the fleas are jumping around, sometimes two of them may be sitting simultaneously on the same real number. (This is fine, as these fleas are infinitesimally small.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;After some time, the ringmaster notices that the three fleas again occupy the three starting points, but they are now sitting in a different order.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is it possible that flea $A$ is now sitting on the starting point of flea $B$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-10-25T12:41:35.347" LastActivityDate="2015-10-25T19:41:31.150" Title="A curious incident in the flea circus" Tags="|mathematics|combinatorics|reachability|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23532" PostTypeId="2" ParentId="23531" CreationDate="2015-10-25T13:02:37.967" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is not possible&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$A$ and $B$ have a distance of $d$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same distance $d$ is between $B$ and $C$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance between $A$ and $C$ is therefore $2d$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following rules hold true:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If before a jump the distance between all pairs of fleas is a multiple of any number $N$ then any flea can only jump a distance equal to a multiple of $2N$&lt;/li&gt;&#xA;&lt;li&gt;if before a jump the distance between all pairs of fleas is a multiple of any number $N$ and any flea jumps a distance equal to a multiple of $N$ then the distance between all pairs of fleas after the jump is still a multiple of $N$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Using both rules we can see, that if rule 1 holds true before any jump $J$ then it also holds true for jump $J+1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since rule 1 holds true for the first jump with $N=d$ we can see that it holds true for all jumps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see now that because $A$ is exactly a distance of $k$ from the starting point of $B$ and has to jump multiples of $2k$ with every jump it can never have a distance of $0$ from the starting point of $B$ and can therefore never be in the starting point of $B$&lt;/p&gt;&#xA;" OwnerUserId="16608" LastActivityDate="2015-10-25T13:02:37.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23533" PostTypeId="2" ParentId="23531" CreationDate="2015-10-25T13:04:52.953" Score="3" Body="&lt;p&gt;Let us assume the fleas to be sitting on consecutive terms of an infinite AP. No flea can leave the AP. This means that a flea on an odd number term can never reach an even number term (and vice versa)'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence it is not possible.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-10-25T13:04:52.953" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23534" PostTypeId="2" ParentId="23531" CreationDate="2015-10-25T13:09:16.673" Score="5" Body="&lt;p&gt;&lt;strong&gt;No&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can describe their co-ordinates at the start as -1, 0, 1 on some scale where 0 is defined to be where B is sitting and the unit 1 is half the distance between A and C. There is a simple transformation from this scale to the real numbers where they were placed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Switching their positions means that A has to reach 0 and B has to reach -1. But A will be jumping in multiples of 2. For example if A is one unit from a flea (as it is at the start) it will jump 2 units and will still be one unit from that flea. It starts on an odd number and will therefore stay on an odd number (on my transformed scale.) Similarly B starts on an even number and will stay on an even number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By jumping past each other they can move arbitrarily up and down the numberscale but they cannot change the odd/even state of their position using my transformed scale, nor move to a point that is not an integer on my scale.&lt;/p&gt;&#xA;" OwnerUserId="1509" LastActivityDate="2015-10-25T13:09:16.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23536" PostTypeId="2" ParentId="23523" CreationDate="2015-10-25T17:36:32.413" Score="5" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Animal names related to the BEATLES.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Category 1a:&lt;/strong&gt;&#xA;Walruses, birds, ponies, tigers, elephants, cats, monkeys, bulldogs, bees, raccoons, cats,  octopuses, pigs, bullfrogs, blue jays, Eagles, worms,  fish, horses,  homing birds, blackbirds, kittens,  hogs, crabs, penguins,  doves,  sheep, sheepdogs, glow worms, butterflies, canaries, puppies, lizards, lambs.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Animal names&lt;/strong&gt; occuring in songs by the Beatles: &#xA; I am the walrus. Free as a Bird. And Your Bird Can Sing. Dig a Pony. &#xA; Bungalow Bill went out tiger hunting with his elephant and gun.&#xA; Three Cool Cats. Everybody's Got Something To Hide Except Me And My Monkey.&#xA; Hey Bulldog. Bees. Rocky Raccoon. Octopus's Garden. Piggies. &#xA; Hey Bulldog (Bullfrog, doing it again). Blue jay way. Etc. etc. etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Category 1b:&lt;/strong&gt;&#xA;Rooster, songbirds, cat, dog, horse, sheep, lion, elephant, cow, hen, crickets, blackbirds, either ducks, swans, or geese. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Animal sounds&lt;/strong&gt; occuring in songs by the Beatles.&lt;br&gt;&#xA; Good Morning Good Morning (starts with rooster crow, birds, a cat, &#xA; a dog, a cow, a horse, a sheep, a lion, an elephant, and a &#xA; group of bloodhounds).&#xA; Sun King (cricket sounds).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Category 2:&lt;/strong&gt; &#xA;Rams, horses, alligators, frog, bison, blue birds, blackbird, cock, seagull, cow, snake, honeybee, inchworm, dragonfly, moose, goose, koala bears, magpies.&#xA;Struszia (trilobite)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Paul McCartney:&lt;/strong&gt; Album &quot;ram&quot;; songs &quot;All You Horse Riders&quot;, &#xA; &quot;Alligator&quot;, &quot;Rupert and the Frog Song&quot;, &quot;Bison&quot;, &quot;Bluebird&quot;, &#xA; &quot;Blackbird&quot;, &quot;Great Cock and Seagull Race&quot;, &quot;Cow&quot;, &quot;What It Is&quot;, &#xA; &quot;The inchworm&quot;, &quot;Little Lamb Dragonfly&quot;, &quot;Morse Moose &amp;amp; The Grey Goose&quot;,&lt;br&gt;&#xA; &quot;Ode To A Koala Bear&quot;, &quot;Two Magpies&quot;.&#xA; Struszia mccartneyi (trilobite) named after him.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Category 3:&lt;/strong&gt; Chicken, fish, monkey. Struszia (trilobite)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;George Harrison:&lt;/strong&gt; Song &quot;Tandoori Chicken&quot;, &quot;Fish on the Sand&quot;, &#xA; &quot;Tweeter &amp;amp; The Monkey Man&quot;.&#xA; Struszia harrisoni named after him.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Category 4:&lt;/strong&gt; Mouse, blackbird, monkey, reindeer. Avalanchurus (trilobite)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Ringo Starr:&lt;/strong&gt; Songs &quot;Scouse The Mouse&quot;, &quot;Bye Bye Blackbird&quot;, &#xA; &quot;Monkey See - Monkey Do&quot;, &quot;Rudolph the Red-Nosed Reindeer&quot;. &#xA; Avalanchurus starri named after him.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Category 5:&lt;/strong&gt; Turkey, hound dog. Bumba (tarantula)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;John Lennon:&lt;/strong&gt; Songs &quot;Cold turkey&quot; and &quot;Hound dog&quot;. &#xA; Tarantula &quot;Bumba lennoni&quot; named after him.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-10-25T18:34:31.923" LastActivityDate="2015-10-25T18:34:31.923" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="23537" PostTypeId="2" ParentId="23531" CreationDate="2015-10-25T19:41:31.150" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No, as others have said&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For me the easiest way to see it is that if they are all starting on integers (say 0,1 and 2), then jumping leaves their polarity unchanged. So an odd number jumping over another number lands on an odd number, and an even number jumping over an even number stays on an even number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is simply because a jump moves them a distance of $2d$, where $d$ is the distance between the jumper and the jumpee. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence not only can they not swap, but they can't even reach each other's spaces. A and C are doomed to stay on the even numbers, and B on the odd. &lt;/p&gt;&#xA;" OwnerUserId="15942" LastActivityDate="2015-10-25T19:41:31.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23538" PostTypeId="2" ParentId="3623" CreationDate="2015-10-25T20:08:58.847" Score="2" Body="&lt;p&gt;I don't know the answer exactly, but roughly speaking it should scale with $N$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the answer is a line of length $n$, and suppose that the O player is able to contain the X player in a hypercube with side of length $n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the X player will have placed $n^N$ X's. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The O player will have surrounded the hypercube on all sides, which requires $2Nn^{N-1}$ O's (actually a bit more than that because we need the edges. But bare with me for now). Of course the O player gets to put down twice as many O's. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this establishes an equation of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation} 2Nn^{N-1}=2n^N\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which reduces to $N=n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, things aren't quite that simple, because I've excluded the edges of the surrounding hypercube. I could include those twice, and the corners thrice, by looking at $2N(n+2)^{N-1}$. The number of corners is $2^N$.  The number of edges is $nN2^{N-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So an overall equation can be set up and solved for $N$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation} 2N(n+2)^{N-1}-2.2^N-nN.2^N=2n^N\end{equation}&lt;/p&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2015-10-25T20:35:24.213" LastActivityDate="2015-10-25T20:35:24.213" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23539" PostTypeId="1" AcceptedAnswerId="23551" CreationDate="2015-10-25T23:00:48.600" Score="4" ViewCount="498" Body="&lt;p&gt;The following, apparently random list of seemingly random nonsense is a metaphorical interpretation of something that exists in the real world.  What is it?  As always, go line by line.  But once you figure it out, you'll have everything you need to write the solution.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The current endeavor is not a genre, it is mass murder&lt;/p&gt;&#xA;&lt;p&gt;Your were torn from the respirator&lt;/p&gt;&#xA;&lt;p&gt;You were brought up to date&lt;/p&gt;&#xA;&lt;p&gt;Like half of a starlet or a whole amputee&lt;/p&gt;&#xA;&lt;p&gt;You wore vestments&lt;/p&gt;&#xA;&lt;p&gt;And played Dracula&lt;/p&gt;&#xA;&lt;p&gt;You scurry down the lane&lt;/p&gt;&#xA;&lt;p&gt;Defenseless, unhidden&lt;/p&gt;&#xA;&lt;p&gt;Concealed canine hunters for profit&lt;/p&gt;&#xA;&lt;p&gt;The statues pursue you with murderous intent&lt;/p&gt;&#xA;&lt;p&gt;I shall pacify an acquaintance&lt;/p&gt;&#xA;&lt;p&gt;Hasten hither, my child&lt;/p&gt;&#xA;&lt;p&gt;Veni, vidi, cessavitque&lt;/p&gt;&#xA;&lt;p&gt;Take leave of the flora&lt;/p&gt;&#xA;&lt;p&gt;Lest you fall ill 'mongst the murk&lt;/p&gt;&#xA;&lt;p&gt;They're mutts, bejeweled&lt;/p&gt;&#xA;&lt;p&gt;John Samhain is, at present, a snow leopard&lt;/p&gt;&#xA;&lt;p&gt;Who dwells above a vault&lt;/p&gt;&#xA;&lt;p&gt;Lift in disrepair, hemp brings him down&lt;/p&gt;&#xA;&lt;p&gt;So he prowls the macadam with Cheetah&lt;/p&gt;&#xA;&lt;p&gt;A tartlet nears with The Specials&lt;/p&gt;&#xA;&lt;p&gt;She wears no grin, but Salvador's accessory&lt;/p&gt;&#xA;&lt;p&gt;Nostalgic, chocolate chip cookies&lt;/p&gt;&#xA;&lt;p&gt;Destroyed and quadriplegic, yet the boxers are unwavering&lt;/p&gt;&#xA;&lt;p&gt;Annus belbus in the dog days of summer&lt;/p&gt;&#xA;&lt;p&gt;Strut next to the beach;  hasten to the trench&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;In the Year 2525&amp;quot;, a hipster goes solo&lt;/p&gt;&#xA;&lt;p&gt;Pain is inevitable, try to live through the night&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Minor hint:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Rather Orwellian, isn't it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Enormous, gigantic, probably ruins everything hint:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The riddle's solution is best friend to both sexes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17062" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-10-26T17:03:54.097" Title="Could you *be* any more metaphorical?" Tags="|riddle|knowledge|wordplay|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23543" PostTypeId="2" ParentId="23419" CreationDate="2015-10-26T02:51:40.983" Score="4" Body="&lt;p&gt;The message is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Four score and seven years ago our fathers brought forth on this continent a new nation conceived in liberty and dedicated to the proposition that all men are created equal now we are engaged in a great civil war testing whether that nation or any nation so conceived and so dedicated can long endure we are met on a great battlefield of that war we have come to dedicate a portion of that field as a final resting place for those who here gave their lives that that nation might live it is altogether fitting and proper that we should do this but in a larger sense we can not dedicate we can not consecrate we can not hallow this ground the brave men living and dead who struggled here have consecrated it far above our poor power to add or detract the world will little note nor long remember what we say here but it can never forget what they did here it is for us the living rather to be dedicated here to the unfinished work which they who fought here have thus far so nobly advancedit is rather for us to be here dedicated to the great task remaining before us that from these honored dead we take increased devotion to that cause for which they gave the last full measure of devotion that we here highly resolve that these dead shall not have died in vain that this nation under god shall have a new birth of freedom and that government of the people by the people for the people shall not perish from the earth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To solve follow this rules&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rule 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if small letter, get the next letter. In case of z, go back to a&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rule 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if capital letter, go back to the previous letter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks DrunkWolf for faster conversion, it takes time for manual conversion lol&lt;/p&gt;&#xA;" OwnerUserId="16036" LastEditorUserId="16036" LastEditDate="2015-10-26T07:39:08.827" LastActivityDate="2015-10-26T07:39:08.827" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23544" PostTypeId="2" ParentId="23476" CreationDate="2015-10-26T06:55:18.263" Score="4" Body="&lt;p&gt;Lateral thinking solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;96 9&lt;strong&gt;1&lt;/strong&gt; 98&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;9&lt;strong&gt;7&lt;/strong&gt; 95 93&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;92 99 &lt;strong&gt;9&lt;/strong&gt;4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I saw it from another post, so it's not original.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-10-26T06:55:18.263" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23545" PostTypeId="2" ParentId="6415" CreationDate="2015-10-26T09:12:41.960" Score="5" Body="&lt;p&gt;We solve the problem in two steps.&#xA;First we solve the problem of a random walk with no cliff.&#xA;Then we show how the solution to the problem with the cliff can be expressed in therms of the solution without the cliff.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Unconstrained walk&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Consider the problem of a random walker moving without a cliff, i.e. just an unconstrained random walker.&#xA;Denote the probability of arriving at point $j$, having started at point $i$, after $n$ steps, by the symbol $p_{ji}(n)$.&#xA;Let $k$ denote the number of rightward steps.&#xA;Then the number of leftward steps is $n - k$.&#xA;The number of ways we can arrange $k$ rightward and $n - k$ leftward steps is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{n!}{k! (n - k)!}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the probability of getting any such string of steps is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p^{n - k}q^{k} \, ,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the probability of such a walk is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p_{ji}(n) = p^{n - k}q^{k} \frac{n!}{k! (n - k)!} \, . $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The displacement $d$ of this walk is the distance between the end and start points, $d \equiv j - i$.&#xA;This displacement must also be equal to the the number of rightward steps minus the number of leftward steps,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;d &amp;amp;= k - (n - k) \\&#xA;\implies k &amp;amp;= (d + n) / 2 \, .&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore,&#xA;$$p_{ji}(n) = p^{(n - d)/2} q^{(n + d)/2}&#xA;\frac{n!}{\Big(\frac{n - d}{2}\Big)! \Big(\frac{n + d}{2}\Big)!} \, . $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this expression is only valid when $n$ and $d$ are both even or both odd.&#xA;Otherwise, $p_{ji}(n)$ is zero.&#xA;Note also that the problem has translational symmetry; $i$ and $j$ do not appear in the expression for $p_{ji}(n)$.&#xA;The random walk probabilities depend only on the translation, so we replace the $ji$ subscript with $d$, writing&#xA;$$p_d(n) = p^{(n - d)/2} q^{(n + d)/2}&#xA;\frac{n!}{\Big(\frac{n - d}{2}\Big)! \Big(\frac{n + d}{2}\Big)!} \, . $$&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Generating functions&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Of course, we need to solve the original problem which includes the cliff.&#xA;Let $f_{ji}(n)$ denote the probability that the walker arrives at point $j$ &lt;em&gt;for the first time&lt;/em&gt;, having started at point $i$, after $n$ steps.&#xA;Again, by translational symmetry we can write this as $f_d(n)$.&#xA;The answer to the question &quot;what is the probability that the walker ever falls off the cliff&quot; is&#xA;$$\text{Probability of falling} = \sum_{n=0}^\infty f_{-1}(n) \, .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now here's the amazing part: if we define &lt;strong&gt;generating functions&lt;/strong&gt; for $p_d(n)$ and $f_d(n)$ as&#xA;$$&#xA;P_d(z) \equiv \sum_{n=0}^\infty p_d(n) z^n&#xA;\qquad&#xA;F_d(z) \equiv \sum_{n=0}^\infty f_d(n) z^n \, ,&#xA;$$&#xA;then it turns out that $^{[a]}$&#xA;$$F_d(z) = P_d(z) / P_0(z) \, .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is awesome because the the probability that the walker ever falls off the cliff is&#xA;\begin{align}&#xA;\text{Probability of falling}&#xA;&amp;amp;=\sum_{n=0}^\infty f_{-1}(n) \\&#xA;&amp;amp;= \lim_{z \rightarrow 1} F_{-1}(z) \\&#xA;&amp;amp;= \lim_{z \rightarrow 1} P_{-1}(z) / P_0(z) \\&#xA;&amp;amp;= \lim_{z \rightarrow 1} \left( \sum_{n=0}^\infty p_{-1}(n)z^n \right) / \left( \sum_{n=0}^\infty p_0(n) z^n \right) \, .&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, we've written the solution to the problem purely in terms of the unconstrained probabilities for which we already found a solution!&#xA;All that remains is doing the sums.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Solution&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;It turns out that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_{-1}(z) = \sum_{n=0}^\infty p_{-1}(n)z^n = \frac{1}{2qz} \frac{1 - \sqrt{1 - 4 pqz^2}}{\sqrt{1 - 4pqz^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_0(z) = \sum_{n=0}^\infty p_0(n)z^n = \frac{1}{\sqrt{1 - 4pqz^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\text{Probability of falling}&#xA;&amp;amp;= \lim_{z \rightarrow 1} P_{-1}(z) / P_0(z) \\&#xA;&amp;amp;= \lim_{z \rightarrow 1} \frac{1 - \sqrt{1 - 4pqz^2}}{2qz} \\&#xA;&amp;amp;= \lim_{z \rightarrow 1} \frac{1 - \sqrt{1 - 4p(1-p)z^2}}{2(1-p)z} \\&#xA;&amp;amp;= \frac{1 - 2\sqrt{(p - 1/2)^2}}{2(1-p)} \\&#xA;&amp;amp;= \left\{ \begin{array}{ll} \frac{p}{1-p} &amp;amp; p &amp;lt; 1/2 \\ 1 &amp;amp; p &amp;gt; 1/2 \end{array} \right.&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which solves the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's pretty cool that this method got the kink at $p=1/2$ without us having to make any extra logical arguments.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;$\lim_{z \rightarrow 1}$&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;It's interesting to think about the meaning of the limit $z \rightarrow 1$.&#xA;The sums over $n$ in the generating functions involve the factor $z^n$.&#xA;For $|z|&amp;lt;1$, the sums de-emphasize terms with large numbers of steps.&#xA;In other words, $z&amp;lt;1$ means that we don't count long walks as much when computing the probability of falling off the cliff.&#xA;In Figure 1 we plot the probability of falling off the cliff for various values of $z$.&#xA;For low values of $z$ the probability of falling off the cliff is low for all $p$.&#xA;This makes sense because low $z$ means that we strongly de-emphasize longer walks; even with $p=1$ we may not fall off the cliff because $z&amp;lt;1$ means we don't always even count the first step.&#xA;For higher values of $z$ the probability to fall off increases.&#xA;At $z=1$, which represents the original problem, the curve forms a cusp at $p=1/2$.&#xA;It is interesting that the sequence of curves for values of $z$ less than one has no cusp, yet the limiting curve for $z=1$ does have a cusp.&#xA;For $z&amp;gt;1$ the curve diverges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cqTui.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cqTui.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;strong&gt;Figure 1:&lt;/strong&gt; Absorption probability as a function of $p$ for a few values of $z$.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Higher moments&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Note that if we want to compute higher moments of the number of steps of the random walk we can do it by differentiating the expression we already found.&#xA;For example, the mean number of steps before falling off the cliff is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\sum_{n=0}^\infty f_{-1}(n) n&#xA;&amp;amp;= \lim_{z \rightarrow 1} \frac{d}{dz} \sum_{n=0}^\infty f_{-1}(n) z^n \\&#xA;&amp;amp;= \lim_{z \rightarrow 1} \frac{d}{dz} F_{-1}(z) \\&#xA;&amp;amp;= \frac{p}{\sqrt{(p - 1/2)^2}} - \frac{1 - 2\sqrt{(p - 1/2)^2}}{2(1 - p)} \\&#xA;&amp;amp;= \left\{ \begin{array}{ll}&#xA;\frac{p}{p-1/2} - 1 &amp;amp; p &amp;gt; 1/2 \\&#xA;\frac{p}{1/2 - p} - \frac{p}{1 - p} &amp;amp; p &amp;lt; 1/2 \, .&#xA;\end{array} \right.&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this function goes to $1$ as $p \rightarrow 1$, which makes sense because the walker has to fall off on the first step.&#xA;The other asymptotic behaviors look wrong though, so maybe I messed up some algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[a]$: You can prove this by thinking of every walk as two parts: a first part which gets to $j$ for the first time, and then a second part which wanders off but eventually ends up at $j$.&lt;/p&gt;&#xA;" OwnerUserId="7488" LastEditorUserId="7488" LastEditDate="2015-10-27T22:23:44.230" LastActivityDate="2015-10-27T22:23:44.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23546" PostTypeId="1" AcceptedAnswerId="23564" CreationDate="2015-10-26T09:29:06.027" Score="11" ViewCount="1559" Body="&lt;p&gt;The ringmaster of a flea circus puts four fleas $A$, $B$, $C$, $D$ on four different points in the plane that form the corners of a square.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Whenever the ringmaster shouts &lt;em&gt;&quot;Hop!&quot;&lt;/em&gt;, one of the four fleas jumps over one of the other fleas to the mirror point on the other side. (In other words, a flea sitting in point $x$ may jump over a flea sitting in point $y$ to the new point $z$, so that $y$ is the midpoint between $x$ and $z$.)&lt;/li&gt;&#xA;&lt;li&gt;While the fleas are jumping around, sometimes two of them may be sitting simultaneously on the same point. (This is fine, as these fleas are infinitesimally small.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is it possible that after some time the three fleas $A$, $B$ and $C$ are sitting in points on the same straight line?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-10-28T10:29:47.237" Title="Another curious incident in the flea circus" Tags="|mathematics|combinatorics|reachability|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23547" PostTypeId="1" AcceptedAnswerId="23553" CreationDate="2015-10-26T10:07:54.340" Score="3" ViewCount="291" Body="&lt;p&gt;Same deal as &lt;a href=&quot;https://puzzling.stackexchange.com/questions/23539/could-you-be-any-more-metaphorical&quot;&gt;previous post&lt;/a&gt;.  One real world source; answers must explain each line and what they all mean when put together.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Additionally amongst the mortality, while the final several cadavers reclined, decaying in the slick high street, the Venetian blinds raised 2.54 cm in Sobriety Structure, aloft on Rustlers' Knoll, and magenta aberrant orbs gaped below on Covet Metropolis.   Zero additional large tires.&lt;/p&gt;&#xA;&lt;p&gt;Syphonaptera the magnitude of rattus sipped on rattus the magnitude of felis catus.  And 500 score humanoids divided into minute clans, pining for the most elevated of the pure towers.  Similar to groups of mutts attacking the window facades of adore-yours-truly street.&lt;/p&gt;&#xA;&lt;p&gt;Tearing and re-enveloping mustela, and glimmering sterling vixen, currently leggings.  Hereditary emblem of corundum and fractured beryl.  At some time in the near future.&lt;/p&gt;&#xA;&lt;p&gt;The era of the carbon allotrope hounds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Minor Hint:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;  The scheme used here is very simple.  &amp;quot;Look, up in the air - it's Superman!&amp;quot; would be &amp;quot;Gaze, above amongst the breathable gases - it's Kal-El!&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Major Hint (Use only in case of emergencies):&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;  It has yet to happen, but it's already a myth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17062" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-10-26T13:18:48.247" Title="Yes, I *can* be more metaphorical" Tags="|riddle|wordplay|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23548" PostTypeId="2" ParentId="23546" CreationDate="2015-10-26T10:24:16.583" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is not possible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lets assume that the fleas are standing on an infinite chessboard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First we show that it isn't possible for $A$, $B$, and $C$ to be in the same row or column:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If we assume that all rows and columns are numbered and further assume that $A$ is in an odd numbered row and column, then one of $B$ and $C$ must be in an even numbered row and odd numbered column, while the other is in an odd numbered row and even numbered column. Since a flea must always jump an even number of rows and an even number of columns it will always stay in even/odd rows/columns depending on its starting position. Therefore $A$, $B$, and $C$ can't be in the same row or column at any point in time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Next for diagonals with any angle.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If we assume that $A$ is on a white field, then $B$ and ar on black fields. Again since a flea must always jump an even number of rows and an even number of columns it will always stay on a white/black field depending on its starting position. For any diagonal(with any angle) in a chessboard we can see, that for all fields that are exactly on the diagonal at least one of the following two statements holds true, (1)all fields in an even numbered row have the same colour and all fields in an odd numbered row have the same colour, or (2)all fields in an even numbered row have the same colour and all fields in an odd numbered row have the same colour. Therefore if $B$ and $C$ are on that diagonal then all fields on that diagonal must be black and $A$ can never land on it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16608" LastEditorUserId="16608" LastEditDate="2015-10-26T14:37:41.623" LastActivityDate="2015-10-26T14:37:41.623" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23549" PostTypeId="2" ParentId="23546" CreationDate="2015-10-26T10:25:28.513" Score="3" Body="&lt;p&gt;This is an answer to the question: &quot;Is it possible that after some time the &lt;strong&gt;four&lt;/strong&gt; fleas A, B, C and D are sitting in points on the same straight line?&quot; which is considerably simpler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No it is not possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the first time this would happen: by defintion being on the same line means all flea-to-flea vectors are colinear, but a jump simply means to translate by twice a flea-to-flea vector: therefore at the previous jump the flea-to-flea vectors were already colinar and the fleas were already on the same line: this contradicts the assumption that this is the first time it has happened.&lt;/p&gt;&#xA;" OwnerUserId="1779" LastEditorUserId="1779" LastEditDate="2015-10-28T10:29:47.237" LastActivityDate="2015-10-28T10:29:47.237" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23550" PostTypeId="2" ParentId="23265" CreationDate="2015-10-26T10:29:47.173" Score="1" Body="&lt;p&gt;This is how I see it. The main goal for everyone is to determine whether the lists they receive are from all organisations there are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trivial case is the case when one does not receive a list. This means that that person is in an organisation so he will go to town.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now everyone does this: For every person on your lists imagine what lists that person would have if the lists you have are from all organisations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there are 2 cases:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;There is a person that has no lists&lt;/li&gt;&#xA;&lt;li&gt;Every person has at least one list&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You expect that person to appear.&lt;br&gt;&#xA;If that person appears it means you have the lists of all organizations and you are in none.&lt;br&gt;&#xA;If that person doesn't appear it means he has a list you don't have and you are in it so you go to town the next day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You wait a day and now for each person in your list imagine you were that person and apply the first strategy again. Keep doing this every day until you get a Case 1. This is guaranteed to happen because the number of lists for each person shrinks every time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This means the answer to both questions is yes.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could wonder: what would happen if a person goes to town while you didn't predict it? If it were to happen it means you are in a organisation. But I strongly believe that this is impossible to happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's work out a case:&#xA;Say person A gets BCD BC CD BD B C D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If A had all lists he would expect that everyone had the following lists:&lt;br&gt;&#xA;B: CD C D&lt;br&gt;&#xA;C: BD B D&lt;br&gt;&#xA;D: BC B C  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now have a case 2&lt;br&gt;&#xA;So first we pretend we are B so we have CD C D. He then expects for the remaining persons to have this:&lt;br&gt;&#xA;C: E&lt;br&gt;&#xA;D: C&lt;br&gt;&#xA;Now for C:&lt;br&gt;&#xA;B: E&lt;br&gt;&#xA;D: B&lt;br&gt;&#xA;Now for D:&lt;br&gt;&#xA;B: C&lt;br&gt;&#xA;C: B  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again a Case 2. If we now apply it again we are in a state where everyone has no lists so we have a Case 1. so A expects that B, C and D all go to town. If they are there then A is not in an organisation, if they are not then A is in a organisaion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe this strategy is foolproof for any situation&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-10-26T10:29:47.173" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23551" PostTypeId="2" ParentId="23539" CreationDate="2015-10-26T10:57:42.287" Score="3" Body="&lt;p&gt;This is an interpretation of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The David Bowie song, Diamond Dogs (btw the &lt;em&gt;probably ruins everything&lt;/em&gt; hint was the key) &lt;a href=&quot;https://en.wikipedia.org/wiki/Diamond_Dogs_(song)&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Diamond_Dogs_(song)&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The current endeavor is not a genre, it is mass murder&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This ain't rock'n'roll. This is genocide! &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your were torn from the respirator&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You were brought up to date&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As they pulled you out of the oxygen tent &lt;br&gt;You asked for the latest party &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Like half of a starlet or a whole amputee&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You wore vestments&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And played Dracula&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; With your silicone hump and your ten inch stump, Dressed like a priest you was, Todd Browning's freak you was &lt;br&gt;&lt;br&gt;(Tod Browning directed the 1931 Dracula film with Bela Lugosi&lt;br&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Dracula_(1931_English-language_film)&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Dracula_(1931_English-language_film)&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You scurry down the lane&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Defenseless, unhidden&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concealed canine hunters for profit&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Crawling down the alley on your hands and knee, I'm sure you're not protected, for it's plain to see, The diamond dogs are poachers and they hide behind trees &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The statues pursue you with murderous intent&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hunt you to the ground they will, mannequins with kill appeal &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I shall pacify an acquaintance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hasten hither, my child&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I'll keep a friend serene &lt;br&gt;Oh baby, come unto me &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Veni, vidi, cessavitque&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take leave of the flora&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lest you fall ill 'mongst the murk&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They're mutts, bejeweled&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Well, she's come, been and gone. &lt;br&gt;Come out of the garden, baby &lt;br&gt;You'll catch your death in the fog &lt;br&gt;Young girl, they call them the Diamond Dogs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;John Samhain is, at present, a snow leopard&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who dwells above a vault&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lift in disrepair, hemp brings him down&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So he prowls the macadam with Cheetah&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Halloween Jack is a real cool cat &lt;br&gt;And he lives on top of Manhattan Chase &lt;br&gt;The elevator's broke, so he slides down a rope &lt;br&gt;Onto the street below, oh Tarzie, go man go &lt;br&gt;&lt;br&gt; (Samhain is another term for Halloween, Jack is often what John's are called, Tarzan was friends with a Cheetah )&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A tartlet nears with The Specials&lt;/p&gt;&#xA;&#xA;&lt;p&gt;She wears no grin, but Salvador's accessory&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nostalgic, chocolate chip cookies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Destroyed and quadriplegic, yet the boxers are unwavering&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Meet his little hussy with his ghost town approach &lt;br&gt;Her face is sans feature, but she wears a Dali brooch &lt;br&gt;Sweetly reminiscent, something mother used to bake &lt;br&gt;Wrecked up and paralyzed, Diamond Dogs are stabilized &lt;br&gt;&lt;br&gt;The Specials had a hit with &quot;Ghost Town&quot;. Salvador Dali. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Annus belbus in the dog days of summer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strut next to the beach; hasten to the trench&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;In the Year 2525&quot;, a hipster goes solo&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pain is inevitable, try to live through the night&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;In the year of the scavenger,the season of the bitch &lt;br&gt; Sashay on the boardwalk, scurry to the ditch&lt;br&gt;Just another future song, lonely little kitsch &lt;br&gt;(There's gonna be sorrow) try and wake up tomorrow&lt;br&gt;&lt;br&gt;(Belbus is Latin for Hyena (&lt;a href=&quot;http://latin-dictionary.net/definition/6268/belbus-belbi&quot; rel=&quot;nofollow&quot;&gt;http://latin-dictionary.net/definition/6268/belbus-belbi&lt;/a&gt;). &lt;em&gt;In the year 2525&lt;/em&gt; is a song about the future.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Minor Hint: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The song has a post apocalypic setting, reminiscent to some of the novel: 1984. According to Wikipedia this was the intention of Bowie on the Album of the same name: &quot;Thematically, it was a marriage of the novel Nineteen Eighty-Four by George Orwell and Bowie's own glam-tinged vision of a post-apocalyptic world.&quot; &lt;a href=&quot;https://en.wikipedia.org/wiki/Diamond_Dogs&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Diamond_Dogs&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Major Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A __ is a man's/girl's best friend.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The genuine article:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=36lWAcY9IXE&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=36lWAcY9IXE&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7316" LastEditorUserId="17062" LastEditDate="2015-10-26T17:03:54.097" LastActivityDate="2015-10-26T17:03:54.097" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="23552" PostTypeId="2" ParentId="20099" CreationDate="2015-10-26T11:27:30.530" Score="4" Body="&lt;p&gt;This problem is almost identical to &quot;Coins of the Realm&quot; which appeared an old Scientific American column, from June 1964. It's included in Chapter 12 of his Sixth Book of Mathematical Games from Scientific American (W. H. Freeman, 1971), and in the following form in Chapter 1 of his The Colossal Book of Short Puzzles and Problems (Norton, 2006, edited by Dana Richards):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In this country at least eight coins are required to make the sum of 99 cents: a half-dollar, a quarter, two dimes and four pennies. Imagine you are the leader of a small country and you have the task of setting up a system of coinage based on the cent as the smallest unit. Your objective is to issue the smallest number of different coins that will enable any value from 1 to 100 cents (inclusive) to be made with no more than two coins. &#xA;  For example, the objective is easily met with 18 coins of the following values: &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;1, 2, 3, 4, ... 8, 9, 10, 20, 30, 40, 50, 60, 70, 80, 90. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Can you do better? Every value must be obtainable either by one coin or as a sum of two coins. The two coins need not, of course, have different values.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Martin gives two 16-coin solutions, as follows (which would be 17-number solutions to the problem here):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1, 3, 4, 9, 11, 16, 20, 25, 30, 34, 39, 41, 46, 47, 49, 50&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is listed as being given in Roland Sprague's Recreation in Mathematics, and is good for precisely the range 1—100. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1, 3, 4, 5, 8, 14, 20, 26, 32, 38, 44, 47, 48, 49, 51, 52&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is listed as having been provided by Peter Wegner, and is good for the larger range 1—104.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those solutions were proposed in the 60s so very likely were done manually.&lt;/p&gt;&#xA;" OwnerUserId="17117" LastActivityDate="2015-10-26T11:27:30.530" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23553" PostTypeId="2" ParentId="23547" CreationDate="2015-10-26T13:18:48.247" Score="4" Body="&lt;p&gt;The mystery guest is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Future Legend by David Bowie  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Additionally amongst the mortality, while the final several cadavers reclined, decaying in the slick high street,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; And In the Death, as the last few corpses lay rotting on the slimy thoroughfare  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the Venetian blinds raised 2.54 cm in Sobriety Structure, aloft on Rustlers' Knoll  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The shutters lifted an inch in temperance building, high on Poacher's Hill  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and magenta aberrant orbs gaped below on Covet Metropolis.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; And red mutant eyes gazed down on Hunger City  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Zero additional large tires.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No more big wheels  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Syphonaptera the magnitude of rattus sipped on rattus the magnitude of felis catus.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fleas the size of rats sucked on rats the size of cats  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And 500 score humanoids divided into minute clans,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; And ten thousand peoploids split into small tribes  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;pining for the most elevated of the pure towers. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Coveting the highest of the sterile skyscrapers  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Similar to groups of mutts attacking the window facades of adore-yours-truly street.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Like packs of dogs assaulting the glass fronts of Love-Me Avenue  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tearing and re-enveloping mustela, and glimmering sterling vixen, currently leggings.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ripping and rewrapping mink and shiny silver fox, now legwarmers  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hereditary emblem of corundum and fractured beryl.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Family badge of sapphire and cracked emerald  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At some time in the near future.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Any day now,  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The era of the carbon allotrope hounds.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the year of the Diamond Dogs  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a spoken piece that introduces the bleak, post-apocalyptic future setting, leading into Diamond Dogs, the song of the previous question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16874" LastActivityDate="2015-10-26T13:18:48.247" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23557" PostTypeId="1" AcceptedAnswerId="23560" CreationDate="2015-10-26T14:26:55.180" Score="11" ViewCount="1066" Body="&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/23554/self-referential-puzzles&quot;&gt;https://puzzling.stackexchange.com/questions/23554/self-referential-puzzles&lt;/a&gt; made me wonder. How many solutions are there to the below box when multiple digits are allowed? I don't know the answer myself, so a good proof of your answer is appreciated. We're counting characters, so 11 is considered as two 1's and not as one 11.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{|cc|} \hline \text{The number of 0's in this box is} &amp;amp; \text{_______} \\ \text{The number of 1's in this box is} &amp;amp; \text{_______} \\ \text{The number of 2's in this box is} &amp;amp; \text{_______} \\ \text{The number of 3's in this box is} &amp;amp; \text{_______} \\ \text{The number of 4's in this box is} &amp;amp; \text{_______} \\ \text{The number of 5's in this box is} &amp;amp; \text{_______} \\ \text{The number of 6's in this box is} &amp;amp; \text{_______} \\ \text{The number of 7's in this box is} &amp;amp; \text{_______} \\ \text{The number of 8's in this box is} &amp;amp; \text{_______} \\ \text{The number of 9's in this box is} &amp;amp; \text{_______} \\ \hline \end{array}&lt;/p&gt;&#xA;" OwnerUserId="1748" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2017-10-22T20:37:26.237" Title="Self referential puzzles - follow-up" Tags="|logical-deduction|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="23559" PostTypeId="2" ParentId="23469" CreationDate="2015-10-26T15:10:51.610" Score="5" Body="&lt;p&gt;This is a long shot, but...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...on the Wikipedia page, we find 31 different symbols in the (old) Greek alphabet. Since there are exactly 30 columns, it leads me to think there is a mapping between the symbol and the column, presumably $[0,30]$. Now, there are four of the symbols and there are also colours, so we have way more information than is needed to get a position in the peg matrix. However, if the colour of the peg determines one out of the five first rows and then we add the sum of the symbols to that position. To test our hypothesis, we compute every position and see that there really is a free hole for the peg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum (using the position in alphabetical order starting from zero) of the symbol sets: (&lt;strong&gt;BLUE&lt;/strong&gt;, $25+22+26+7=78$), (&lt;strong&gt;RED&lt;/strong&gt;, $9+7+1+2=19$), (&lt;strong&gt;GREEN&lt;/strong&gt;, $7+2+7+26=42$), (&lt;strong&gt;PURPLE&lt;/strong&gt;, $20+1+17+7=45$), (&lt;strong&gt;ORANGE&lt;/strong&gt;, $3+17+1+20=41$). Then, the possible positions for each peg are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ls9cN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ls9cN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming this is correct, there are $6 \times (6-1) \times (5-1)^3 = 1920 &amp;lt; N &amp;lt; 6^2 \times 5^3 =  4500$ different combinations to try (handwavingly) mapping the colours to the rows. Maybe I made a mistake when computing the numbers (and the solution is distinct) or I am going down the completely wrong route.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The message is binary in some encoding -- UTF-9?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT2: The rows have 18, 14, 14, 14, 13, 15, 15, 5 and 15 pegs respectively. There are exactly 5 odd-parity rows. If the last column is the parity column, it should be filled with 100011111. However, the parity column could be any of several, so the solution is not really unique.&lt;/p&gt;&#xA;" OwnerUserId="12901" LastEditorUserId="12901" LastEditDate="2015-11-12T18:52:26.593" LastActivityDate="2015-11-12T18:52:26.593" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23560" PostTypeId="2" ParentId="23557" CreationDate="2015-10-26T17:18:04.943" Score="12" Body="&lt;p&gt;(assuming numbers with leading zeros, ex &quot;05&quot; are forbidden)  &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Part 1. Establishing an upper boundary for digit size&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Suppose there exists a solution where the largest blank is filled in with three digits.&lt;br&gt;&#xA;Then the box must contain at least 108 digits, so one blank must contain at least 9 digits. But the largest blank contains only three digits, which is a contradiction.&lt;br&gt;&#xA;So there can't be a solution whose largest blank is three digits.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose there exists a solution where the largest blank is filled in with four digits.&lt;br&gt;&#xA;Then the box must contain at least 1008 digits, so one blank must contain at least 99 digits. But the largest blank contains only four digits, which is a contradiction.&lt;br&gt;&#xA;So there can't be a solution whose largest blank is four digits.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This pattern of contradiction continues upwards forever. So, if a solution exists, its largest blank must have no more than two digits.   &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Part 2. Establishing an upper boundary for number of two-digit slots&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Suppose there exists a solution where exactly two blanks are filled with two digits.&lt;br&gt;&#xA;The total number of digits in the entire box is 10 + 8*1 + 2*2  = 22.&lt;br&gt;&#xA;The sum of the numbers in each blank must exactly equal the total number of digits in the entire box.&lt;br&gt;&#xA;But the minimum theoretical sum of all numbers in each blank is 1*8 + 2*10 = 28&lt;br&gt;&#xA;So there can't be a solution where exactly two blanks are filled with two digts.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose there exists a solution where exactly three blanks are filled with two digits.&lt;br&gt;&#xA;The total number of digits in the entire box is 10 + 7*1 + 3*2  = 23.&lt;br&gt;&#xA;The sum of the numbers in each blank must exactly equal the total number of digits in the entire box.&lt;br&gt;&#xA;But the minimum theoretical sum of all numbers in each blank is 1*7 + 3*10 = 37.&lt;br&gt;&#xA;So there can't be a solution where exactly three blanks are filled with two digts.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This pattern of contradictions continues upwards forever. So, if a solution exists, it can have no more than one blank slot with two digits.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the previous question, the solution that has zero two-digit slots was already found, so the remainder of this post will search for a solution containing exactly one blank slot with two digits.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Part 3. Determining which slot has two digits&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Suppose there exists a solution where exactly one blank is filled in with two digits.&lt;br&gt;&#xA;the total number of digits in the entire box = 10 + (9*1) + (1*2) = 21.&lt;br&gt;&#xA;The sum of the numbers in each blank must exactly equal the total number of digits in the entire box.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the existing labels, the number in each blank must be at least 1.&lt;br&gt;&#xA;No single-digit blank may contain a digit larger than 3, because then the sum of the numbers in each blank would exceed 21.&lt;br&gt;&#xA;The two-digit blank may not exceed 12, because then the sum of the numbers in each blank would exceed 21.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because none of them appear in any blanks, The blanks in slots 4, 5, 6, 7, 8, and 9 are all &quot;1&quot;.&lt;br&gt;&#xA;The digit in blank slot 1 must be at least 7.&lt;br&gt;&#xA;If blank slot 1 is not the two-digit slot, the minimum theoretical sum of numbers in each blank is 7+10+(1*8) = 25. But this is larger than 21, so blank slot 1 must contain two digits.  &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Part 4. Finding the value of the two digit slot&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Blank slot 1 can't contain &quot;12&quot;, because once you account for the &quot;1&quot; in &quot;12&quot; and the &quot;1&quot; in the label, you need to fit ten &quot;1&quot;s in nine remaining slots.&lt;br&gt;&#xA;Blank slot 1 can't contain &quot;10&quot;, because blank slot 0 will be at least 2, and the slot corresponding to 0's number will be at least 2, and you'll need to fit eight &quot;1&quot;s in the seven remaining slots.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So blank slot 1 must contain &quot;11&quot;.  &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Part 5. Finding the remaining values&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Accounting for the &quot;11&quot; in the two-digit slot, and the &quot;1&quot; in the label, eight slots must contain &quot;1&quot;. Exactly one blank slot contains a single digit other than 1.&lt;br&gt;&#xA;The blank slot not containing a 1 must be equal to 21 - (11 + 8*1) = 2.&lt;br&gt;&#xA;That will be the second &quot;2&quot; to appear in the box, so it must go in slot 2.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the only multi-digit solution is &lt;strong&gt;[1, 11, 2, 1, 1, 1, 1, 1, 1, 1]&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Appendix. Brute-forcing all one digit solutions&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The digits in the one-digit solution add up to 20, and no slot contains a 0. Using this information, you can iterate through all possible configurations and find the self-referential ones. Sample Python implementation:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from collections import Counter&#xA;import functools&#xA;&#xA;def iter_ways(x, nums):&#xA;    if x == 0:&#xA;        return&#xA;    if nums == 1:&#xA;        if 1 &amp;lt;= x&amp;lt;= 9:&#xA;            yield [x]&#xA;        return&#xA;    for i in range(1,10):&#xA;        if i &amp;lt;= x:&#xA;            for rest in iter_ways(x-i, nums-1):&#xA;                yield [i] + rest&#xA;&#xA;def valid(seq):&#xA;    c = Counter()&#xA;    for i in range(10):&#xA;        c[i] = seq[i]&#xA;    digit_str = &quot;0123456789&quot; + &quot;&quot;.join(map(str, seq))&#xA;    digits = map(int, digit_str)&#xA;    return c == Counter(digits)&#xA;&#xA;for seq in iter_ways(20,10):&#xA;    if valid(seq):&#xA;        print seq&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Result:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[1, 7, 3, 2, 1, 1, 1, 2, 1, 1]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So the only one-digit solution is &lt;strong&gt;[1, 7, 3, 2, 1, 1, 1, 2, 1, 1]&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1497" LastEditorUserId="1497" LastEditDate="2015-10-26T20:05:11.910" LastActivityDate="2015-10-26T20:05:11.910" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23561" PostTypeId="2" ParentId="23469" CreationDate="2015-10-26T17:50:33.573" Score="8" Body="&lt;p&gt;Harry &lt;em&gt;thinks&lt;/em&gt; his nemesis is &lt;a href=&quot;https://en.wikipedia.org/wiki/August_Dvorak&quot; rel=&quot;nofollow noreferrer&quot;&gt;August Dvorak&lt;/a&gt; (1894&amp;ndash;1975), but is mistaken.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on Toast's comment:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also, only the first row of your grid is correct. The next two are... off.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I prepared an alternate dissection of the grid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hYPXw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, anyone who knows a &lt;a href=&quot;https://en.wikipedia.org/wiki/Dvorak_Simplified_Keyboard&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dvorak&lt;/a&gt; user knows that they avoid &lt;a href=&quot;https://en.wikipedia.org/wiki/QWERTY&quot; rel=&quot;nofollow noreferrer&quot;&gt;QWERTY&lt;/a&gt; like the plague.  The puzzlemaker is more likely a descendant of &lt;a href=&quot;https://en.wikipedia.org/wiki/Christopher_Latham_Sholes&quot; rel=&quot;nofollow noreferrer&quot;&gt;Christopher Latham Sholes&lt;/a&gt; (1819&amp;ndash;1890).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;After &lt;a href=&quot;https://puzzling.stackexchange.com/questions/23469/peg-the-panel-to-procure-the-prize/23561#comment73742_23561&quot;&gt;f'' showed me&lt;/a&gt; how to tie strings between the pegs in order to form letters, the solution was easy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EiFlV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The new peg positions are highlighted in red.&lt;/p&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2015-11-19T02:34:38.750" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="23562" PostTypeId="1" AcceptedAnswerId="23563" CreationDate="2015-10-26T18:31:22.807" Score="0" ViewCount="427" Body="&lt;p&gt;Secret Agent Michael Scarn enters his hotel room, only to find it having been ransacked. He needs to send an encrypted message to his handler notifying them that his cover has been compromised.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to send encrypted messages, agents need a Cryptographically Secure Pseudo-Random Number Generator. Agent Scarn has two such generators with him, but through some testing, he can tell that one of them has been tampered with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He doesn't know which of the two PRNGs have been modified; if he uses the wrong one, &lt;a href=&quot;https://www.schneier.com/fortuna.html&quot; rel=&quot;nofollow&quot;&gt;the rival spy agency will likely be able to decode the message&lt;/a&gt;, putting both him and his hander in danger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can Agent Scarn still generate secure random numbers for the encrypted message?&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2016-10-18T19:15:18.037" Title="How can our secret agent produce a cryptographically secure random number?" Tags="|mathematics|story|cryptography|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="23563" PostTypeId="2" ParentId="23562" CreationDate="2015-10-26T18:35:51.200" Score="11" Body="&lt;p&gt;Easy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Use both PRNGs and XOR the two numbers obtained each time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How does this guarantee that the message will not be decodable?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Assume that his enemies know exactly what PRNG A will produce, but nothing of what B will produce. We can describe A XOR B as using a one-time pad to encrypt A, which is an unbreakable form of encryption unless the pad is reused or the pad is not generated in a cryptographically secure manner. Because only one of the PNRGs has been tampered with, the other will still generate cryptographically secure numbers. Also, because XOR is symmetric, $a\oplus b=b\oplus a$, making it irrelevant which one is A and which one is B. Thus, $a_i\oplus b_i=c_i$ is a new random number that cannot be obtained without knowing both $a_i$ AND $b_i$ (which requires knowing the output of both A and B).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1955" LastEditorUserId="1955" LastEditDate="2015-10-26T18:48:15.697" LastActivityDate="2015-10-26T18:48:15.697" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23564" PostTypeId="2" ParentId="23546" CreationDate="2015-10-26T18:44:19.307" Score="17" Body="&lt;p&gt;&lt;em&gt;This answer is entirely due to Henning Makholm.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we draw a grid in such a way that the four fleas are at positions $(0,0), (0,1),(1,0)$ and $(1,1)$, then the fleas will forever be at integer points on the grid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Color the lattice points red, blue, green and purple by repeating the pattern &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/h1rtl.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key observation is that a flea will always hop to the same color lattice point it starts on. This is because hopping changes both its coordinates by a multiple of two, meaning the parity of both its coordinates are preserved, and these parities determine the point's color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, any line of lattice points will only consist of two colors. Since any three fleas are always on three different colors (they start out this way, and the colors never change), this means three fleas can never be on a line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why only two colors on a line? Write the parametric equation for the line as $$(x,y) = (a,b)+t(p,q), \quad t\in \mathbb R$$&#xA;If the line contains at least two lattice points (otherwise it would be uninteresting for us here), we can chose all of $a,b,p,q$ as integers, and we can &lt;em&gt;also&lt;/em&gt; choose $p$ and $q$ to be coprime, in which case the lattice points on the line will be exactly those with integer $t$. When this $t$ is &lt;em&gt;even&lt;/em&gt;, the point has the same color as $(a,b)$, and when $t$ is &lt;em&gt;odd&lt;/em&gt; it has the same color as $(a+p,b+q)$.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="17119" LastEditDate="2015-10-27T09:54:50.260" LastActivityDate="2015-10-27T09:54:50.260" CommentCount="4" CommunityOwnedDate="2015-10-26T18:44:19.307" ContentLicense="CC BY-SA 3.0" />
  <row Id="23567" PostTypeId="1" CreationDate="2015-10-26T23:45:55.983" Score="13" ViewCount="1427" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/J9gxU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/J9gxU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hi, my friend gave me a puzzle, and I need help in solving it. &#xA;I am required to decrypt this code and get a password. &#xA;Please help! The hint is: &lt;strong&gt;$100 bill&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here's the text version of the image above:&#xA;&lt;PRE&gt;&#xA;A,M,L, ,F,M,A,A&#xA;N,L, ,C,F,H,E,G&#xA;L,U, ,E, ,E,T,&#xA;H,W,I,I,S, ,L,T&#xA;F,N,O,T,H,E,C,E&#xA; ,N,R,O,M,O, ,E&#xA; , ,E,A, ,N,T,T&#xA;M,T, ,R,E,T,T,E&#xA;&lt;/PRE&gt;&lt;/p&gt;&#xA;" OwnerUserId="17129" LastEditorDisplayName="user88" LastEditDate="2015-10-27T16:17:55.423" LastActivityDate="2016-07-03T19:07:17.050" Title="The 100 dollar bill puzzle" Tags="|cipher|pattern|" AnswerCount="1" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23568" PostTypeId="2" ParentId="22812" CreationDate="2015-10-27T06:31:14.877" Score="8" Body="&lt;p&gt;For reference, here's the solution I (OP) came up with while constructing this puzzle. I'm marking it as the &quot;accepted&quot; answer because I believe it is currently the best solution. If anyone improves on it, I'll move the tick.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;0: abstruse, reacted, honest, condors, invention, neon, vague @&lt;br&gt;&#xA;  1: abstruse, &lt;em&gt;react&lt;/em&gt;, honest, condors, invention, neon, vague @ &lt;strong&gt;ed&lt;/strong&gt;&lt;br&gt;&#xA;  2: abstruse, &lt;em&gt;act&lt;/em&gt;, honest, condors, invention, neon, vague @ &lt;strong&gt;re&lt;/strong&gt;ed&lt;br&gt;&#xA;  3: abstruse, act, honest, condors, &lt;em&gt;invent&lt;/em&gt;, neon, vague @ reed, &lt;strong&gt;ion&lt;/strong&gt;&lt;br&gt;&#xA;  4: abstruse, honest, condors, invent, neon, vague @ reed, &lt;strong&gt;act&lt;/strong&gt;ion&lt;br&gt;&#xA;  5: abstruse, honest, &lt;em&gt;condor&lt;/em&gt;, invent, neon, vague @ reed&lt;strong&gt;s&lt;/strong&gt;, action&lt;br&gt;&#xA;  6: &lt;em&gt;use&lt;/em&gt;, honest, condor, invent, neon, vague @ reeds, &lt;strong&gt;abstr&lt;/strong&gt;action&lt;br&gt;&#xA;  7: use, honest, condor, invent&lt;strong&gt;s&lt;/strong&gt;, neon, vague @ &lt;em&gt;reed&lt;/em&gt;, abstraction&lt;br&gt;&#xA;  8: use, honest, condor, &lt;strong&gt;re&lt;/strong&gt;invents, neon, vague @ &lt;em&gt;ed&lt;/em&gt;, abstraction&lt;br&gt;&#xA;  9: use, honest, &lt;em&gt;condo&lt;/em&gt;, reinvents, neon, vague @ &lt;strong&gt;red&lt;/strong&gt;, abstraction&lt;br&gt;&#xA;  10: use, honest, &lt;em&gt;con&lt;/em&gt;, reinvents, neon, vague @ red, abstraction, &lt;strong&gt;do&lt;/strong&gt;&lt;br&gt;&#xA;  11: honest, con, reinvents, neon, vague @ red, abstraction, &lt;strong&gt;douse&lt;/strong&gt;&lt;br&gt;&#xA;  12: honest, con, &lt;em&gt;reinvent&lt;/em&gt;, neon, vague @ red, abstraction, douse&lt;strong&gt;s&lt;/strong&gt;&lt;br&gt;&#xA;  13: honest, con&lt;strong&gt;do&lt;/strong&gt;, reinvent, neon, vague @ red, abstraction, &lt;em&gt;uses&lt;/em&gt;&lt;br&gt;&#xA;  14: honest, con, reinvent, neon, vague @ red, abstraction, uses, &lt;strong&gt;do&lt;/strong&gt;&lt;br&gt;&#xA;  15: honest, con, reinvent, &lt;em&gt;on&lt;/em&gt;, vague @ red, abstraction, uses, do&lt;strong&gt;ne&lt;/strong&gt;&lt;br&gt;&#xA;  16: honest, reinvent, on, vague @ red, abstraction, uses, &lt;strong&gt;con&lt;/strong&gt;done&lt;br&gt;&#xA;  17: &lt;strong&gt;abs&lt;/strong&gt;, honest, reinvent, on, vague @ red, &lt;em&gt;traction&lt;/em&gt;, uses, condone&lt;br&gt;&#xA;  18: abs&lt;strong&gt;cond&lt;/strong&gt;, honest, reinvent, on, vague @ red, traction, uses, &lt;em&gt;one&lt;/em&gt;&lt;br&gt;&#xA;  19: abscond, &lt;em&gt;hone&lt;/em&gt;, reinvent, on, vague @ red, traction, uses, &lt;strong&gt;st&lt;/strong&gt;one&lt;br&gt;&#xA;  20: abscond, hone, reinvent, on, vague&lt;strong&gt;r&lt;/strong&gt; @ &lt;em&gt;ed&lt;/em&gt;, traction, uses, stone&lt;br&gt;&#xA;  21: abscond, hone, reinvent, on, &lt;em&gt;vague&lt;/em&gt; @ ed, traction, &lt;strong&gt;r&lt;/strong&gt;uses, stone&lt;br&gt;&#xA;  22: abscond, hone, reinvent, on, vague&lt;strong&gt;st&lt;/strong&gt; @ ed, traction, ruses, &lt;em&gt;one&lt;/em&gt;&lt;br&gt;&#xA;  23: abscond, hone, reinvent&lt;strong&gt;ion&lt;/strong&gt;, on, vaguest @ ed, &lt;em&gt;tract&lt;/em&gt;, ruses, one&lt;br&gt;&#xA;  24: abscond, hone, &lt;em&gt;reinvent&lt;/em&gt;, on, vaguest @ ed, tract, ruses, one, &lt;strong&gt;ion&lt;/strong&gt;&lt;br&gt;&#xA;  25: abscond, hone&lt;strong&gt;r&lt;/strong&gt;, reinvent, on, vaguest @ ed, tract, &lt;em&gt;uses&lt;/em&gt;, one, ion&lt;br&gt;&#xA;  26: abscond, honer, &lt;em&gt;invent&lt;/em&gt;, on, vaguest @ ed, &lt;strong&gt;re&lt;/strong&gt;tract, uses, one, ion&lt;br&gt;&#xA;  27: abscond, honer, &lt;em&gt;vent&lt;/em&gt;, on, vaguest @ ed, retract, uses, one, ion, &lt;strong&gt;in&lt;/strong&gt;&lt;br&gt;&#xA;  28: abscond, honer, vent, on, &lt;em&gt;guest&lt;/em&gt; @ ed, retract, uses, one, ion, &lt;strong&gt;va&lt;/strong&gt;in&lt;br&gt;&#xA;  29: abscond, &lt;em&gt;hon&lt;/em&gt;, vent, on, guest @ ed, retract, uses, one, ion, vain&lt;strong&gt;er&lt;/strong&gt;&lt;br&gt;&#xA;  30: abscond, &lt;em&gt;ho&lt;/em&gt;, vent, on, guest @ ed, retract, uses, &lt;strong&gt;n&lt;/strong&gt;one, ion, vainer&lt;br&gt;&#xA;  31: ho, vent, on, guest @ &lt;strong&gt;abscond&lt;/strong&gt;ed, retract, uses, none, ion, vainer&lt;br&gt;&#xA;  32: vent, on, guest @ absconded, retract, &lt;strong&gt;ho&lt;/strong&gt;uses, none, ion, vainer&lt;br&gt;&#xA;  33: on, guest @ absconded, retract, houses, none&lt;strong&gt;vent&lt;/strong&gt;, ion, vainer&lt;br&gt;&#xA;  34: guest @ absconded, retract, houses, nonevent, &lt;strong&gt;on&lt;/strong&gt;ion, vainer&lt;br&gt;&#xA;  35: @ absconded, retract, &lt;strong&gt;guest&lt;/strong&gt;houses, nonevent, onion, vainer  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastActivityDate="2015-10-27T06:31:14.877" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23570" PostTypeId="1" CreationDate="2015-10-27T08:22:53.647" Score="0" ViewCount="250" Body="&lt;p&gt;&lt;em&gt;Based on &lt;a href=&quot;https://puzzling.stackexchange.com/questions/23546/another-curious-incident-in-the-flea-circus&quot;&gt;Another curious incident in the flea circus&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/questions/23531/a-curious-incident-in-the-flea-circus&quot;&gt;A curious incident in the flea circus&lt;/a&gt;&lt;/em&gt; by @Gamow&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a $n$ dimensional cube in an $n$ dimensional world. There is a flea on each vertex of the hypercube. When the ringmaster shouts &quot;Hop!&quot;, any one flea at a point $A$ selects any other flea at another point $B$ and jumps to a point $C$ such that $AB=BC=\frac12AC$. Two fleas may sit at the same point in space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible that after some time, there are $3$ fleas sitting on points on a straight line?&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-10-27T10:13:55.537" Title="The infinite flea circus" Tags="|mathematics|geometry|combinatorics|reachability|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23571" PostTypeId="2" ParentId="23570" CreationDate="2015-10-27T10:13:55.537" Score="9" Body="&lt;p&gt;No; the argument that works in the 2-dimensional case works with essentially no change in $n$ dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Select a coordinate system such that the $2^n$ fleas start out at positions $\{0,1\}^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paint the integer points on the lattice in $2^n$ different colors based on the parity of their coordinates -- that is, the point $(x_1,\ldots,x_n)$ will have the same color as $(x_1 \bmod 2,\ldots, x_n \bmod 2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every flea jump changes each of the flea's coordinates by an &lt;em&gt;even&lt;/em&gt; number, so each flea stays on the color it started out on (and all fleas are on &lt;em&gt;different&lt;/em&gt; colors to begin with).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like in the 2D case, any straight line containing two different lattice points will contain points from of exactly two colors, so there are only two of the fleas that can even reach a position on the line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why only two colors on a line? Write the parametric equation for the line as $$(x_1,\ldots,x_n) = (a_1,\ldots,a_n)+t(p_1,\ldots,p_n), \quad t\in \mathbb R$$&#xA;Without loss of generality all of the $a_i$s and $p_i$s can be integers, and we can also assume that $\gcd(p_1,\ldots,p_n)=1$ (otherwise divide through by the gcd, and scale the $t$s appropriately). This means that the lattice points on the line will be exactly those with integer $t$. When this $t$ is &lt;em&gt;even&lt;/em&gt;, the point has the same color as $(a_1,\ldots,a_n)$, and when $t$ is &lt;em&gt;odd&lt;/em&gt; it has the same color as $(a_1+p_1,\ldots,a_n+p_n)$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(This argument works even in an 1-dimensional world, although there it is trivial that we can't have three collinear fleas because there are only two fleas in total!)&lt;/p&gt;&#xA;" OwnerUserId="17119" LastActivityDate="2015-10-27T10:13:55.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23572" PostTypeId="2" ParentId="23398" CreationDate="2015-10-27T11:34:32.633" Score="2" Body="&lt;p&gt;I've had another thought. This time I'm going with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A question!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The title is a question, the first clue is a question, the second clue is a question, and also the hint is a question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9444" LastActivityDate="2015-10-27T11:34:32.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23573" PostTypeId="1" AcceptedAnswerId="23575" CreationDate="2015-10-27T11:54:53.127" Score="-1" ViewCount="325" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;You have a bag that contains $b$ black balls and $w$ white balls.&#xA;  ($b+w&amp;gt;2$) You put your hand into the bag and pull out a pair of balls.&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;If both balls are black, you throw away one and return one to the bag.  &lt;/li&gt;&#xA;  &lt;li&gt;If both balls are white, you throw them both away.  &lt;/li&gt;&#xA;  &lt;li&gt;If the balls are of different colours, you throw away the black ball and return the white ball to the bag.  &lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;You repeat this process $N$ times. At the end, you end up with&#xA;  exactly one ball in the bag.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For what all values of $b$ and $w$ is the maximum possible $N$ different from the minimum possible $N$?&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-10-27T16:38:17.567" Title="Removing black and white balls from a bag in $N$ steps" Tags="|mathematics|logical-deduction|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23575" PostTypeId="2" ParentId="23573" CreationDate="2015-10-27T12:20:01.403" Score="10" Body="&lt;p&gt;The conditions can be reduced to effectively the following 2 cases:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;remove 2 white balls&lt;/li&gt;&#xA;&lt;li&gt;remove 1 black ball&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If $w$ is even then the last ball must be black and then $N=\frac{w}{2}+(b-1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $w$ is odd then the last ball must be white and then $N=\frac{w-1}{2}+b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases the value of $N$ is fixed, so the minimum and maximum are equal.&lt;/p&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2015-10-27T16:38:17.567" LastActivityDate="2015-10-27T16:38:17.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23576" PostTypeId="2" ParentId="23573" CreationDate="2015-10-27T13:36:39.060" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The maximum and minimum $N$ are different, when $w$ is even and $b$ is not zero.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whenever $w$ is uneven we will, at the end, be left with a single white ball. In this case $N=\frac{w-1}{2}+b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever $w$ is even and $b$ equals zero we will be left with zero balls. In this case $N=\frac{w}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever $w$ is even and $b$ is not zero we have two possible outcomes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If we remove the last black ball before the last two white balls we will be left with zero balls in the end. In this case $N=\frac{w}{2}+b$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If we remove the last two white balls before the last black ball we will be left with one black ball in the end. In this case $N=\frac{w}{2}+b-1$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Our maximum with first case is exactly one step more than our minimum with the second case.&lt;/p&gt;&#xA;" OwnerUserId="16608" LastEditorUserId="16608" LastEditDate="2015-10-27T13:42:22.830" LastActivityDate="2015-10-27T13:42:22.830" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23577" PostTypeId="1" AcceptedAnswerId="23578" CreationDate="2015-10-27T15:44:26.630" Score="6" ViewCount="8035" Body="&lt;p&gt;Following &lt;a href=&quot;https://puzzling.stackexchange.com/questions/14872/5-letter-word-removing-letters&quot;&gt;Part 1&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take away my second letter, and I still sound the same. &#xA;&lt;br&gt;Take away my third letter, I still sound the same. &lt;br&gt;Even take away my fourth and fifth letters, I will still sound the same. &lt;br&gt;I am a five letter word. &lt;br&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="2425" LastEditorUserId="5373" LastEditDate="2018-05-23T20:26:39.643" LastActivityDate="2018-05-23T20:26:39.643" Title="5 Letter Word, Removing letters - Part 2" Tags="|riddle|word|letters|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23578" PostTypeId="2" ParentId="23577" CreationDate="2015-10-27T16:02:49.590" Score="9" Body="&lt;p&gt;You are:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Queue &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Qeue&lt;/strong&gt; I would try to pronounce the same,&lt;br&gt;&#xA; &lt;strong&gt;Que&lt;/strong&gt; would sound like queue, unless you speak Spanish &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And Finally &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter Q sounds like the full five letter word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Approaching each line as a separate removal, we have: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Qeue&lt;br&gt;&#xA; Quee (Kwee, close to Queue)&lt;br&gt;&#xA; Que (Again, unless you are Spanish)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-10-27T16:02:49.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23579" PostTypeId="2" ParentId="23546" CreationDate="2015-10-27T16:05:48.033" Score="0" Body="&lt;p&gt;Yes!  It is entirely possible!  But only if you allow for the use of different forms of geometry within the same problem.  Then all you need to do is center their Euclidean square on the geographic North (or South) pole.  If you keep the side length of the square short the fleas will appear to be in a square to the human observer.  There is no hopping required.  They are now on the same Elliptic line (i.e. the same Latitude). &lt;/p&gt;&#xA;" OwnerUserId="17128" LastActivityDate="2015-10-27T16:05:48.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23580" PostTypeId="2" ParentId="23567" CreationDate="2015-10-27T16:13:09.400" Score="30" Body="&lt;p&gt;So it was reading Dan Brown's &lt;em&gt;The Lost Symbol&lt;/em&gt; that got me the answer to this one, along with that one deleted answer that mentioned &quot;Benjamin Franklin&quot; (because his face appears on the US $100 bill).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key is to use the &quot;order eight Franklin square&quot; mentioned in that book, and map each letter of the grid to that position. Specifically, this one (from &lt;a href=&quot;http://www.pleacher.com/mp/puzzles/mathpuz/maghist.html&quot;&gt;this site&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/leBCQ.gif&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/leBCQ.gif&quot; alt=&quot;Benjamin Franklin&amp;#39;s magic square.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we do that, we get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TELL WON THE NAME GF THE ELEMENT TTHAT COMES AFTER CALIFORNIUM&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which I assume was &lt;em&gt;supposed&lt;/em&gt; to read:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TELL NOW THE NAME OF THE ELEMENT THAT COMES AFTER CALIFORNIUM&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which is, of course, &lt;a href=&quot;https://en.wikipedia.org/wiki/Einsteinium&quot;&gt;einsteinium&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorDisplayName="user88" LastEditDate="2016-07-03T19:07:17.050" LastActivityDate="2016-07-03T19:07:17.050" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="23581" PostTypeId="2" ParentId="23577" CreationDate="2015-10-27T17:12:16.567" Score="6" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a crowd heckling: &quot;BOOOO&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Take away the third letter:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BOOO&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Take away the second letter:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BOOO&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Take away the fourth and fifth letters:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BOO&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2015-10-27T17:12:16.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23582" PostTypeId="1" AcceptedAnswerId="23583" CreationDate="2015-10-27T17:53:00.907" Score="9" ViewCount="825" Body="&lt;p&gt;An equilateral triangle with sidelength $L$ can be tiled by trapezoids with sidelengths $2,1,1,1$.&lt;br&gt;&#xA;What are the possible values for $L$?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-10-27T18:05:50.750" Title="Tiling by trapezoids" Tags="|mathematics|geometry|tiling|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23583" PostTypeId="2" ParentId="23582" CreationDate="2015-10-27T18:04:38.247" Score="12" Body="&lt;p&gt;An equilateral triangle with sidelength $L$ can be divided into $L^2$ equilateral triangles with sidelength $1$. A trapezoid with sidelengths $2,1,1,1$ covers three of these triangles, so $L^2$ must be divisible by $3$. $L$ has to be an integer (because if it is tiled by the trapezoids, its sidelength must be the sum of sidelengths of the trapezoids), so $L$ is divisible by $3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that three of these trapezoids can tile an equilateral triangle with sidelength $3$. Then, for any equilateral triangle with a sidelength that is a multiple of $3$, we can divide it into triangles with sidelength $3$ and tile each of those with the trapezoids.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-10-27T18:04:38.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23584" PostTypeId="2" ParentId="23582" CreationDate="2015-10-27T18:05:50.750" Score="10" Body="&lt;p&gt;They can be arranged like so:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/KdGZS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KdGZS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which means &lt;code&gt;L&lt;/code&gt; is  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; any multiple of 3 (green is the base triangle to show further tiling)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13101" LastActivityDate="2015-10-27T18:05:50.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23585" PostTypeId="1" CreationDate="2015-10-27T20:28:25.173" Score="0" ViewCount="437" Body="&lt;p&gt;An assignment given to us as part of our college course in AI requires us to solve a cryptarithmetic puzzle. The problem is I can't find my way around puzzles - and it was never a problem until now, because courses which I took didn't have assignments where I had to solve a puzzle.&#xA;I am looking for a general method to solve a puzzle, such as &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    SEND&#xA;+   MORE&#xA;________&#xA;   MONEY&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now, I don't know how that expression makes sense. So, the first thing which wish to understand is to how to make sense out of this, and then the next thing is how to solve it. It would be great if you could explain what is going on at each step.&lt;/p&gt;&#xA;" OwnerUserId="17142" LastActivityDate="2015-10-27T20:28:25.173" Title="How do I solve cryptarithmetic puzzles?" Tags="|alphametic|" AnswerCount="0" CommentCount="5" ClosedDate="2015-10-27T21:36:16.390" ContentLicense="CC BY-SA 3.0" />
  <row Id="23586" PostTypeId="1" AcceptedAnswerId="23587" CreationDate="2015-10-27T22:04:31.993" Score="13" ViewCount="335" Body="&lt;p&gt;Below are thirteen clues that describe paired words. The first word in each pair is an atrocious attribute. The second word is a fearful figure. The words in the pair share a sort-of stem (or simply a set) of consecutive letters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt; &#xA;&lt;em&gt;Shopworn shrieker states stale stories&lt;/em&gt; = &lt;em&gt;banal  banshee&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Building bedizenment battling bad breath  &lt;/li&gt;&#xA;  &lt;li&gt;Bustling bones belike buzzing bumblebee&lt;/li&gt;&#xA;  &lt;li&gt;Candid composite corpse constructor  &lt;/li&gt;&#xA;  &lt;li&gt;Clever cackler who can conjure with a cauldron &lt;/li&gt;&#xA;  &lt;li&gt;Counterfeit kidnapper  &lt;/li&gt;&#xA;  &lt;li&gt;Flirty fluid fleecer  &lt;/li&gt;&#xA;  &lt;li&gt;Form fitting fabric on feral full-moon fancier&lt;/li&gt;&#xA;  &lt;li&gt;Halfling hater who is a philharmonic harpist  &lt;/li&gt;&#xA;  &lt;li&gt;Prejudiced and with pileous, prodigious pedes&lt;/li&gt;&#xA;  &lt;li&gt;Reserved and preserved  &lt;/li&gt;&#xA;  &lt;li&gt;Serpent-stranded sister swallowing excessively  &lt;/li&gt;&#xA;  &lt;li&gt;Warmongering wishmaker with wispless warmth&lt;/li&gt;&#xA;  &lt;li&gt;Wild haunter in the White House  &lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15921" LastEditorUserId="11452" LastEditDate="2015-10-28T13:51:12.943" LastActivityDate="2015-10-30T04:04:17.160" Title="Monstrous Morphemes" Tags="|crossword-clues|seasonal|" AnswerCount="2" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23587" PostTypeId="2" ParentId="23586" CreationDate="2015-10-28T01:15:00.067" Score="6" Body="&lt;p&gt;Some guesses (and a lot of help from the commentators!):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Building bedizenment battling bad breath&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; garlic-breathed gargoyle&lt;br&gt;&#xA; OP alternative: gargling gargoyle &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Bustling bones belike buzzing bumblebee&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; skedaddling skeleton&lt;br&gt;&#xA; OP alternative: skeltering skeleton  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Candid composite corpse constructor&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; frank Frankenstein&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Clever cackler who can conjure with a cauldron&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; witty witch&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Counterfeit kidnapper&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; bogus bogeyman  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Flirty fluid fleecer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; vampish vampire&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Form fitting fabric on feral full-moon fancier&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lycra-wearing lycanthrope&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Halfling hater who is a philharmonic harpist&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; orchestral orc&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Prejudiced and with pileous, prodigious pedes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; bigoted Bigfoot&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Reserved and preserved&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; mum mummy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Serpent-stranded sister swallowing excessively&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; gorging Gorgon  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Warmongering wishmaker with wispless warmth&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; genocidal genie&lt;br&gt;&#xA; OP alternative: jingoistic jinn&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Wild haunter in the White House&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; political poltergeist  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15921" LastEditDate="2015-10-30T04:04:17.160" LastActivityDate="2015-10-30T04:04:17.160" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23588" PostTypeId="1" AcceptedAnswerId="23589" CreationDate="2015-10-28T05:59:01.077" Score="19" ViewCount="3747" Body="&lt;p&gt;I come after you are born&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm painful when born&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I reveal myself when happiness comes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you do not care for me I will hurt you.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Born two times  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I live no longer as you live.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two families in the same home, we all are fair.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;No one gets hurt if one of us dies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who am I?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4312" LastEditorUserId="4312" LastEditDate="2018-04-26T12:20:29.397" LastActivityDate="2018-04-26T12:20:29.397" Title="If you don't care for me, I will hurt you. Who am I?" Tags="|riddle|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23589" PostTypeId="2" ParentId="23588" CreationDate="2015-10-28T06:36:28.557" Score="21" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; tooth / teeth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Came after your born&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It will grow in time after baby is born&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm painful when born&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Babies are irritated when they started to grow because it is painful or itchy that is why they use teething toys&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Its comes to show when happiness comes,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When smiling, it shows&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But if you do not care I will hurt you soon&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It will rot and needs to be extracted by a dentist when you do not brush your teeth or take care of it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Born two times&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Milk teeth is first and Permanent teeth is second&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I live no longer as you live&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you die, teeth will no longer have a function but they don't decay&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Have two family in same home, we all are fair.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Upper jaw set and lower jaw set of teeth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No one get hurt if one die.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Other tooth will still function even one is extracted&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16036" LastActivityDate="2015-10-28T06:36:28.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23590" PostTypeId="1" AcceptedAnswerId="23604" CreationDate="2015-10-28T08:34:34.113" Score="5" ViewCount="924" Body="&lt;p&gt;Fredo owns a nice balance with two pans.&#xA;He now wants to design a system of 20 integer weights $w_1,w_2,\ldots,w_{20}$ with the following properties: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$1\le w_1\le w_2\le w_3\le\cdots\le w_{20}$&lt;/li&gt;&#xA;&lt;li&gt;Any integer weight $U$ from the range $1\le U\le2015$ can be balanced by placing it into the left pan and some subset of the weights into the right pan.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What is the smallest possible value that  weight $w_{20}$ can take under these conditions?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-10-28T16:54:49.250" Title="A balance and twenty weights" Tags="|mathematics|combinatorics|weighing|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23591" PostTypeId="2" ParentId="23590" CreationDate="2015-10-28T08:58:23.740" Score="5" Body="&lt;p&gt;I have a solution with the largest weight being:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 147&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First i tried using binary weigths ($2^{N-1}$) but this solution would be giving me the possibility of balancing way larger weights than what is needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So i tried to think about where to stop using binary weigths so that the higher weights can be balanced by using equal weights of smaller size than the &lt;em&gt;would be next&lt;/em&gt; binary weigth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My solution uses the binary weigths 1, 2, 4, 8, 16, 32, 64, 128.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With those i can balance all weights between 1 and 255.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subtracting all binary weights from 2015 leaves us with 1760.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we only need to divide 1760 by the amount of our left over weigths to determine the smallest amount they should have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$w=\frac{1760}{12}=146\frac{2}{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So our smallest possible biggest weight is 147.&lt;/p&gt;&#xA;" OwnerUserId="16608" LastActivityDate="2015-10-28T08:58:23.740" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23592" PostTypeId="2" ParentId="23586" CreationDate="2015-10-28T09:32:51.157" Score="4" Body="&lt;p&gt;1 - Building bedizenment (ornament or dressing, according to google) battling bad breath  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Garlicy gargoyle (with thanks to Fillet)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5 - Counterfeit kidnapper  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Bogus Bogeyman&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;7 - Form fitting fabric on feral full-moon fancier  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Lycra-clad Lycanthrope  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;8 - Halfling hater who is a philharmonic harpist  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Orchestral Orc&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;11 - Serpent-stranded sister swallowing excessively  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strike&gt;Guzzling&lt;/strike&gt; Gorging Gorgon  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;12 - Warmongering wishmaker with wispless warmth (borrowing from APrough's comment)  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Genocidal Genie&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11717" LastEditorUserId="11717" LastEditDate="2015-10-28T15:04:59.877" LastActivityDate="2015-10-28T15:04:59.877" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23594" PostTypeId="2" ParentId="23521" CreationDate="2015-10-28T10:06:24.087" Score="0" Body="&lt;p&gt;Map Full note to 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Map Half note to 0 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Binary to Ascii coversion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;amp;#204;&amp;amp;#169;T&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HTML Codes to Ascii&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ì © T&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ICT?&lt;/p&gt;&#xA;" OwnerUserId="17131" LastActivityDate="2015-10-28T10:06:24.087" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23595" PostTypeId="2" ParentId="23590" CreationDate="2015-10-28T10:54:54.400" Score="4" Body="&lt;p&gt;If you used binary weights, you wouldn't need 20. You could use 1,2,4, ... 1024 to weigh anything up to 2096, which is more than 2015, with just 11 weights. But you've been told to use 20 weights.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you could take the last one, 1024, and replace it with two 512s. Now you're using 12 weights and the largest is 512. But you still have 8 more weights available. If you try to split 1024 into three, you're getting down below 512, the binary weight before the 1024. What's more, you're still trying to reach 2096 when you don't need to. So set aside the 1024 and the 512, and you have 1,2,4, ... 256 for 9 weights, and can use 11 weights for the remaining (2015-511) 1504 g. That's an average of 137 each, which is below the 256, so set aside 256 also. Now you have 1,2, 4 ... 128 for 8 weights, and 12 weights for (2015-255) 1760, averaging 147. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to make any number up to 512, you will need a &quot;256&quot; - which will be made up of these extra weights - use a pair of 128s, and you have 10 weights accounted for. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to make any number up to 1024, you will need a &quot;512&quot;. Use 4 128s and you have 14 weights accounted for. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wanted to make numbers up to 2096 the next step would be to create a &quot;1024&quot;, but you don't. The largest number you need is 2015 which is 991 more than 1024. So if you can make 991 out of 6 weights, you can make any number between 991 and 2015. It averages 165. So you could use 5 165s and a 164.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can this be improved? What if I had 12 147s after the 1..128 and didn't bother with the &quot;256&quot; and the &quot;512&quot;? They pull the average down because they're only 128. I can make any number up to 255 with the 1..128, so I can make 147, then throw in a 147 and carry on adding 1, 2, 3, 4, to that until I hit 294, then use 2 of the 147s and so on. Now 12x147 is 1764 not 1760, so I need a mix of 157 and 156s to hit 1887.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using that approach, do I even need that first 128? Well, how else will I make 129? or 130? So I do and the largest weight will have to be 147g.&lt;/p&gt;&#xA;" OwnerUserId="1509" LastEditorUserId="1509" LastEditDate="2015-10-28T11:51:55.787" LastActivityDate="2015-10-28T11:51:55.787" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23597" PostTypeId="1" AcceptedAnswerId="23598" CreationDate="2015-10-28T13:44:52.560" Score="3" ViewCount="838" Body="&lt;p&gt;I got a card with a text from a software company, and it tells me to decode it.&lt;br&gt;&#xA;The only thing written on the card is the following code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  bWpvey9vam9rYjMxMjZBZXpvYnVzYmRmL2Rwbg==&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Might be Java related&lt;/p&gt;&#xA;" OwnerUserId="15660" LastEditorUserId="4269" LastEditDate="2015-10-28T15:29:19.817" LastActivityDate="2015-11-26T11:01:23.743" Title="Crack the code: JavaRelated" Tags="|cipher|computer-puzzle|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23598" PostTypeId="2" ParentId="23597" CreationDate="2015-10-28T13:59:08.077" Score="11" Body="&lt;p&gt;The string is base 64 encoded, decoding gives:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  mjo{/ojokb3126Aezobusbdf/dpn&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The string &lt;code&gt;/dpn&lt;/code&gt; gives a further clue how it needs to be deciphered; if you take the previous character in the ASCII table you get &lt;code&gt;.com&lt;/code&gt;. Substituting each letter with the previous letter in the ASCII table gives the following email address. I have replaced @ with (at) to avoid email scrapers abusing this address.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  linz.ninja2015(at)dynatrace.com&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="17155" LastEditorUserId="17155" LastEditDate="2015-10-28T16:07:42.713" LastActivityDate="2015-10-28T16:07:42.713" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="23599" PostTypeId="2" ParentId="23521" CreationDate="2015-10-28T14:09:35.430" Score="8" Body="&lt;ol&gt;&#xA;&lt;li&gt;I started out with&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; morse code - whole for &lt;code&gt;-&lt;/code&gt;, half for &lt;code&gt;.&lt;/code&gt; and break for break.&lt;br/&gt;I got:&lt;br/&gt;&#xA; ZMICTFUFHNSEFARN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Then I figured &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The key is &lt;code&gt;D&lt;/code&gt; (having &lt;code&gt;C#&lt;/code&gt; and &lt;code&gt;F#&lt;/code&gt;), so I tried implying ceaser's 3 forward and 3 backward.&#xA;The second attempt gave:&#xA;&lt;br/&gt;WJFZQCRCEKPBCXOK&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;Noticing that&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There are 2 lines of notes with 8 letters each I tried the rail-fence cipher&#xA;&lt;br/&gt;WJFZQCRC&#xA;&lt;br/&gt;EKPBCXOK,&#xA;&lt;br/&gt;which gave:&#xA;&lt;br/&gt;&lt;strong&gt;WE&lt;/strong&gt;jkfpzbqccx&lt;strong&gt;ROCK&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which seems pretty reasonable, though I still might be missing something...&lt;/p&gt;&#xA;" OwnerUserId="4317" LastActivityDate="2015-10-28T14:09:35.430" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23600" PostTypeId="2" ParentId="23467" CreationDate="2015-10-28T14:12:14.880" Score="13" Body="&lt;p&gt;I think I got it, a Measured Word™ is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A word in which if we replace every letter with the sum of strait lines and curves, we get the sequence &lt;code&gt;1234&lt;/code&gt;&#xA; repeating with some offset. &lt;br&gt;&lt;br&gt; &#xA; E.g. &lt;br&gt;&#xA; &lt;pre&gt;AMITY becomes 34123 and &lt;br&gt;WOLF becomes 4123&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11679" LastActivityDate="2015-10-28T14:12:14.880" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23602" PostTypeId="1" AcceptedAnswerId="23603" CreationDate="2015-10-28T15:33:14.043" Score="2" ViewCount="227" Body="&lt;p&gt;We have a matrix made up of $m$ by $n$ dots. Can you give a function that counts the number of squares that can be found by joining any $4$ dots in it?&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-10-28T15:58:20.250" Title="Squares in a dot matrix" Tags="|mathematics|geometry|combinatorics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23603" PostTypeId="2" ParentId="23602" CreationDate="2015-10-28T15:58:20.250" Score="1" Body="&lt;p&gt;For every $m\ge n$ (otherwise just swap $m$ and $n$) the number of squares $N$ is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $N=\sum_{i=1}^{n-1}(n-i)\times(m-i)\times i$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where $(n-i)\times(m-i)$ counts all squares of sidelength $i$ and sides parallel to the rows and columns and multiplies it by the amount of squares we get when we slide their corners along those sides($\times i$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&#xA;squares size 3 in a 5 by 5 matrix:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;++++*  *++++&#xA;+**+*  *+**+&#xA;+**+*  *+**+&#xA;++++*  *++++&#xA;*****  *****&#xA;&#xA;*****  *****&#xA;++++*  *++++&#xA;+**+*  *+**+&#xA;+**+*  *+**+&#xA;++++*  *++++&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;squares by sliding along the edges of one size 3 square:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;++++&#xA;+**+&#xA;+**+&#xA;++++&#xA;&#xA;*+**&#xA;*+++&#xA;+++*&#xA;**+*&#xA;&#xA;**+*&#xA;+++*&#xA;*+++&#xA;*+**&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(i know its ugly but i couldn't do better)&lt;/p&gt;&#xA;" OwnerUserId="16608" LastActivityDate="2015-10-28T15:58:20.250" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23604" PostTypeId="2" ParentId="23590" CreationDate="2015-10-28T16:46:29.437" Score="7" Body="&lt;p&gt;The weight $w_i -1$ for $i\in [1, 20]$ can only be weighted using the weights $w_1$ to $w_{i-1}$: indeed all other weights are more heavy than $w_i -1$ (recall that the weights are ordered by increasing weights).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore when you add all the weights $w_1$ to $w_{i-1}$ they must &lt;strong&gt;at least&lt;/strong&gt; weight $w_i -1$. This gives the relation :&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;$$w_i -1 \leq \sum_{k=1}^{i-1} w_k$$&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This gives in particular that $w_1 -1 \leq 0$ and because we imposed the property $1 \leq w_1$ we have $w_1 =1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also leads to $w_2 -1 \leq w_1$, i.e. $w_2 \leq 2$, and $w_3 -1 \leq w_1 + w_2$ i.e. $w_3 \leq 4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally it leads to:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;$$w_i \leq 2^{i-1}$$&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Let's now look at $\sum_{i=1}^{20}w_i \geq 2015$ (this comes from the fact that you can weight 2015).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can separate the 12 last terms of the sum as follows:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;$$\sum_{i=9}^{20}w_i \geq 2015 - \sum_{i=1}^{8}w_i$$&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Hence $\sum_{i=9}^{20}w_i \geq 2015 - \sum_{i=1}^{8}2^{i-1}$&#xA;and $\sum_{i=9}^{20}w_i \geq 1760$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We conclude that $max_{i=9}^{20} w_i \geq \lceil{1760/12}\rceil$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And in particular that:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;$$max_{i=1}^{20} w_i \geq 147$$&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;There should be a weight of weight at least 147.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;On the other hand &lt;strong&gt;there exists a set of weights where all weights are at most 147 satisfying the conditions&lt;/strong&gt;, for instance : {1, 2, 4, 8, 16, 32, 63, 125, 147, 147, 147, 147, 147, 147, 147, 147, 147, 147, 147, 147}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof of this can take the other answers to this post as indication.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So this proves that &lt;strong&gt;147 is the smallest value $w_{20}$ can take under the mentioned conditions&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1779" LastEditorUserId="1779" LastEditDate="2015-10-28T16:54:49.250" LastActivityDate="2015-10-28T16:54:49.250" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23605" PostTypeId="2" ParentId="23521" CreationDate="2015-10-28T18:17:45.600" Score="15" Body="&lt;p&gt;To start, a big shout-out to &lt;a href=&quot;https://puzzling.stackexchange.com/users/4317/jnf&quot;&gt;JNF&lt;/a&gt; for getting me most of the way there. (If you upvote me, please upvote them as well!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Morse code. Quarter note = dash, eighth note = dot, rest = space.&lt;br&gt;Result: &lt;strong&gt;ZMICTFUFHNSEFARN&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Step two:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;em&gt;key&lt;/em&gt; (chuckle chuckle, well done Gabriel G) is D major, so I tried a Vigenère cipher with the key &quot;dmajor&quot;. &lt;br&gt;Result: &lt;strong&gt;WAITFORTHEENCORE&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore the encoded message is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Wait for the encore&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16275" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.923" LastActivityDate="2015-10-28T18:17:45.600" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23606" PostTypeId="2" ParentId="23417" CreationDate="2015-10-28T18:31:14.030" Score="0" Body="&lt;p&gt;Here goes a very wild guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sequence is 1, 1, 2, 6, 33; the &lt;a href=&quot;http://oeis.org/A127114&quot; rel=&quot;nofollow&quot;&gt;representation of $n!$ in base 7&lt;/a&gt;, and it's coded for in the amount of punctuation marks per paragraph. The correct answer would then be &lt;strong&gt;option 1&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rationale:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What caught my attention was the splitting of the intro text into four paragraphs, and for some reason the wording in the options made me focus on the punctuation.&lt;br&gt;&#xA; Counting the punctuation marks (including ') yields 1, 1, 2, 6 for the intro and 33, 55 and 52 for options 1, 2 and 3 respectively.&lt;br&gt;&#xA;&lt;br&gt;&#xA; I then searched for such sequences on &lt;a href=&quot;http://oeis.org/search?q=1%2C1%2C2%2C6%2C33&amp;amp;sort=&amp;amp;language=english&amp;amp;go=Search&quot; rel=&quot;nofollow&quot;&gt;oeis.org&lt;/a&gt;, and only &quot;1, 1, 2, 6, 33&quot; seems to be listed there. Searching for the other two options yields no results.&lt;br&gt;&#xA; The most likely candidate for a sequence starting with 1, 1, 2, 6, 33 is, as I mentioned, the representations of $n!$ in base 7 &amp;mdash; kind of an obscure one, but since option 2 hinted at &quot;bases&quot;, it may not be completely impossible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is probably way off, but at least from here it seems remotely plausible.&lt;/p&gt;&#xA;" OwnerUserId="17033" LastActivityDate="2015-10-28T18:31:14.030" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23607" PostTypeId="1" AcceptedAnswerId="23611" CreationDate="2015-10-28T18:44:36.710" Score="2" ViewCount="440" Body="&lt;p&gt;I was reading &lt;a href=&quot;https://puzzling.stackexchange.com/q/48/9000&quot;&gt;this&lt;/a&gt; question / answers realized that people spend a lot more time playing this game than I thought.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find it interesting - although you may not - to know what is the minimal time needed to get the maximal sum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under reasonable assumptions, the question reduces to: &lt;strong&gt;What is the minimal number of moves needed to end up with the highest sum of tiles?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(It is still interesting to think about how much time I need to be playing to get the super score! I'm not sure if the game keeps records of times somewhere.)&lt;/p&gt;&#xA;" OwnerUserId="17160" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2015-11-18T19:19:06.683" Title="What's the minimal time needed to get the highest possible sum of tiles in 2048?" Tags="|2048-game|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23608" PostTypeId="1" AcceptedAnswerId="23625" CreationDate="2015-10-28T19:09:23.090" Score="17" ViewCount="1718" Body="&lt;p&gt;Two hikers are separated by a two-dimensional mountain range, like the one shown below. The mountain range alternates between peaks and valleys, connected by straight lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/u0G3I.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/u0G3I.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both hikers are at sea level, and the mountain range never dips below sea level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two hikers want to meet up with each other. Prove that they can do this &lt;em&gt;while staying at the same altitude as each other for their entire journey.&lt;/em&gt; They are allowed to backtrack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Source:&lt;/strong&gt; &lt;a href=&quot;http://www.cs.cmu.edu/puzzle/puzzle12.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.cs.cmu.edu/puzzle/puzzle12.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="12408" LastEditDate="2015-10-28T20:38:43.390" LastActivityDate="2015-10-29T12:29:09.133" Title="Hikers Meeting in the Middle" Tags="|mathematics|geometry|" AnswerCount="5" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="23609" PostTypeId="2" ParentId="23607" CreationDate="2015-10-28T19:30:09.173" Score="-1" Body="&lt;h1&gt;3,932,156 points over 18.2 hours @ 1 move / second&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Let's assume 1 move / second and all thanks to &lt;a href=&quot;https://www.reddit.com/r/2048/comments/214njx/highest_possible_score_for_2048_warning_math&quot; rel=&quot;nofollow&quot;&gt;Reddit&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This assumes that every tile you spawn is a &lt;kbd&gt;2&lt;/kbd&gt;, and the last tile you spawn is a &lt;kbd&gt;4&lt;/kbd&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are $16$ squares. You'd think that the highest possible number would be $2&#xA;^{16} (=65536)$, which would fill up the board, making it impossible to progress. However, if you do manage to spawn a &lt;kbd&gt;4&lt;/kbd&gt; tile as your last tile, then you can continue to progress to $2^{17} (=131,072)$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Scoring works like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Merging two lower-tier blocks together will give you the score of the higher-tier block (score of $+8$ gained from merging two &lt;kbd&gt;4&lt;/kbd&gt;s). For any specific tile score, you have to add up all the scores from the lower tiers.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Creating a &lt;kbd&gt;2&lt;/kbd&gt; tile $= +0$ pts   &lt;/li&gt;&#xA;&lt;li&gt;Creating a &lt;kbd&gt;4&lt;/kbd&gt; tile $= 4 = +4$ pts   &lt;/li&gt;&#xA;&lt;li&gt;Creating an &lt;kbd&gt;8&lt;/kbd&gt; tile $= 8 + (2\times4) = +16$ pts   &lt;/li&gt;&#xA;&lt;li&gt;Creating a &lt;kbd&gt;16&lt;/kbd&gt; tile $= 16 + (2\times8) + (4\times4) = +48$ pts   &lt;/li&gt;&#xA;&lt;li&gt;Creating a &lt;kbd&gt;32&lt;/kbd&gt; tile $ = 32 + (2\times16) + (4\times8) + (8\times4) = +128$ pts ...   &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Which can be simplified:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Tile &lt;kbd&gt;$2^1$&lt;/kbd&gt;$ = 0 \times 2^1 = 0$ pts   &lt;/li&gt;&#xA;&lt;li&gt;Tile &lt;kbd&gt;$2^2$&lt;/kbd&gt;$ = 1 \times 2^2 = 4$ pts   &lt;/li&gt;&#xA;&lt;li&gt;Tile &lt;kbd&gt;$2^3$&lt;/kbd&gt;$ = 2 \times 2^3 = 16$ pts   &lt;/li&gt;&#xA;&lt;li&gt;Tile &lt;kbd&gt;$2^4$&lt;/kbd&gt;$ = 3 \times 2^4 = 48$ pts&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Making tile &lt;kbd&gt;$2^n$&lt;/kbd&gt;$ = (n-1)2^n$ points.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Making the max tile &lt;kbd&gt;$2^{17}$&lt;/kbd&gt;$ = 16 \times 2^{17} = 2,097,152$ points. You would need to spawn 65,536 &lt;kbd&gt;2&lt;/kbd&gt; tiles to make a single &lt;kbd&gt;131,072&lt;/kbd&gt; tile, which, at 1 legal move per second, would take 18.2 hours.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;That's only making the max tile.  Now we still need to fill the rest of the board. (Note: combining times are correct, as when you merge two tiles together, you spawn a &lt;kbd&gt;2&lt;/kbd&gt; at the same time)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Next tile &lt;kbd&gt;$2^{16}$&lt;/kbd&gt;$ = 983,040$ points ($2^{15}$ tiles or $9.1$ hours)   &lt;/li&gt;&#xA;&lt;li&gt;&lt;kbd&gt;$2^{15}$&lt;/kbd&gt;$ = 458,752$ points ($2^{14}$ tiles or $4.55$ hours)&lt;/li&gt;&#xA;&lt;li&gt;&lt;kbd&gt;$2^{14}$&lt;/kbd&gt;$ = 212,992$ points ($2^{13}$ tiles or $2.27$ hours)&lt;/li&gt;&#xA;&lt;li&gt;&lt;kbd&gt;$2^{13}$&lt;/kbd&gt;$ = 98,304$ points ($2^{12}$ tiles or $1.13$ hours)&lt;/li&gt;&#xA;&lt;li&gt;&lt;kbd&gt;$2^{12}$&lt;/kbd&gt;$ = 45,056$ points ($2^{11}$ tiles or $34$ minutes)&lt;/li&gt;&#xA;&lt;li&gt;&lt;kbd&gt;$2^{11}$&lt;/kbd&gt;$ = 20,480$ points ($2^{10}$ tiles or $17$ minutes)&lt;/li&gt;&#xA;&lt;li&gt;&lt;kbd&gt;$2^{10}$&lt;/kbd&gt;$ = 9216$ points ($512$ tiles or $8.5$ minutes)&lt;/li&gt;&#xA;&lt;li&gt;&lt;kbd&gt;$512$&lt;/kbd&gt;$ = 4096$ points ($256$ tiles or $4.3$ minutes)&lt;/li&gt;&#xA;&lt;li&gt;&lt;kbd&gt;$256$&lt;/kbd&gt;$ = 1792$ points ($128$ tiles or $2.1$ minutes)&lt;/li&gt;&#xA;&lt;li&gt;&lt;kbd&gt;$128$&lt;/kbd&gt;$ = 768$ points ($64$ tiles or $64$ seconds)&lt;/li&gt;&#xA;&lt;li&gt;&lt;kbd&gt;$64$&lt;/kbd&gt;$ = 320$ points ($32$ tiles or $32$ seconds)&lt;/li&gt;&#xA;&lt;li&gt;&lt;kbd&gt;$32$&lt;/kbd&gt;$ = 128$ points ($16$ tiles or $16$ seconds)&lt;/li&gt;&#xA;&lt;li&gt;&lt;kbd&gt;$16$&lt;/kbd&gt;$ = 48$ points ($8$ tiles or $8$ seconds)&lt;/li&gt;&#xA;&lt;li&gt;&lt;kbd&gt;$8$&lt;/kbd&gt;$ = 12$ points ($3$ tiles or $3$ seconds)(This space will be created by two &lt;kbd&gt;2&lt;/kbd&gt; tiles, and the last &lt;kbd&gt;4&lt;/kbd&gt; tile, so $-4$ points from the equation)   &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This fills up 15 of the 16 spaces on the board. We don't include the last tile, as it cannot merge with the &lt;kbd&gt;8&lt;/kbd&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Max points possible: $3,932,156$.&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;strong&gt;Max tiles spawned: $131,070$ ($+1$ for last tile, $-1$ for the &lt;kbd&gt;4&lt;/kbd&gt;)&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;strong&gt;Around 36.4 hours @ 1 move per second.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you happen to miss out on the last &lt;kbd&gt;4&lt;/kbd&gt; tile, your max score would be $1,835,008$, with $65,535$ ($+1$ for last tile) &lt;kbd&gt;2&lt;/kbd&gt; tiles spawned, and $18.2$ hours @ 1 move per second.&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="4421" LastEditDate="2015-11-18T19:19:06.683" LastActivityDate="2015-11-18T19:19:06.683" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23611" PostTypeId="2" ParentId="23607" CreationDate="2015-10-28T19:43:01.413" Score="1" Body="&lt;p&gt;There is no single anawer to this question because the game randomly adds a &lt;code&gt;2&lt;/code&gt; or &lt;code&gt;4&lt;/code&gt; tile. But it is possible to give a range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/a/2753/15632&quot;&gt;This answer&lt;/a&gt; shows the maximum possible score. The sum of all tiles is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2^{18}-4=262140$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To reach this result you need a &lt;code&gt;4&lt;/code&gt; as a new tile at least at 16 specific times:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;first time, when the rest of the board is filled with values from 4 to 65536&lt;/li&gt;&#xA;&lt;li&gt;second time, when the rest of the board is filled with values from 4 to 32768 and 131072&lt;/li&gt;&#xA;&lt;li&gt;and so in until the last &lt;code&gt;4&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Let's assume the rest of the new tiles were always a &lt;code&gt;2&lt;/code&gt;. All values on the board are from the 2 starting tiles (also assumed as &lt;code&gt;2&lt;/code&gt;) and all added tiles. That means, in this case we have the following number of moves:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{262140-(2*2)-(16*4)}{2}+16=131052$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume we always get a &lt;code&gt;4&lt;/code&gt; as new tile and also 2 times &lt;code&gt;4&lt;/code&gt; as the starting tiles it's even simpler:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{262140-(2*4)}{4}=65533$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming 1 second per move it‘s between 18.2 and 36.4 hours.&lt;/p&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-10-29T05:16:29.497" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23612" PostTypeId="2" ParentId="4818" CreationDate="2015-10-28T19:43:23.253" Score="1" Body="&lt;p&gt;Joe Z has it correct.   Ultimately it's caused by the fact that some of the pieces are indistinguishable.   I think tho he meant to say edge or center pieces not just centers only because of course in a 4x4 there are a bunch of edge pieces that look the same, not just the centers.  In order to create what cubers call a &quot;parity error&quot; you need to move around some of the indistinguishable pieces either the edges or the centers.  &lt;/p&gt;&#xA;" OwnerUserId="17164" LastActivityDate="2015-10-28T19:43:23.253" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23613" PostTypeId="1" CreationDate="2015-10-28T19:49:42.160" Score="0" ViewCount="983" Body="&lt;p&gt;Claire walks to work at 6km/hr and Charles cycles along the same route. Normally Charles&#xA;overtakes Claire outside the public library. One day Claire leaves 20 minutes late and Charles&#xA;overtakes her outside the swimming pool, 3km before she gets to the library.&#xA;At what speed does Charles cycle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 9km/hr&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B 12km/hr&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C 15km/hr&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D 18km/hr&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E 24km/hr&lt;/p&gt;&#xA;" OwnerUserId="17166" LastActivityDate="2015-10-28T21:23:53.593" Title="Walking and cycling" Tags="|logical-deduction|calculation-puzzle|" AnswerCount="2" CommentCount="2" ClosedDate="2015-10-28T20:35:41.523" ContentLicense="CC BY-SA 3.0" />
  <row Id="23614" PostTypeId="2" ParentId="23608" CreationDate="2015-10-28T20:11:35.167" Score="2" Body="&lt;p&gt;Alright, let's call the person on the left Alice, and the person on the right Bob.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conditions that make it possible for them to be at the same altitude until they meet are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the first peak that Alice encounters is higher than the first peak that Bob encounters&lt;/li&gt;&#xA;&lt;li&gt;the bottom of the first valley that Bob encounters is the lowest non-sea level valley on the graph, and the second peak that he encounters is the highest point on the graph. Therefore, this slope encompasses every possible altitude in the mountain range, except for sea level.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Therefore, the solution:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Bob climbs to his first peak; Alice matches his altitude&lt;/li&gt;&#xA;&lt;li&gt;Bob descends to his first valley; Alice goes back down the mountain, matching his altitude&lt;/li&gt;&#xA;&lt;li&gt;Alice proceeds to the right; Bob simply matches her altitude the whole way by traversing the slope between his first valley and second peak, until Alice reaches her fourth mountain, which Bob is on the other side of, and they both ascend to meet at the peak.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="17167" LastActivityDate="2015-10-28T20:11:35.167" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23615" PostTypeId="2" ParentId="23613" CreationDate="2015-10-28T20:14:03.940" Score="0" Body="&lt;p&gt;I don't know exactly why I have so much trouble doing it more easily, but very well. Let $x_0$ be Claire's initial position and $x_1$ be Charles's (both in $km$). Let $v_1$ be Charles's velocity in $km/h$. We can describe the normal process as an equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}x_0 + 6t = x_1 + v_1t\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $t$ is how long Claire takes to meet Charles (at the public library).&#xA;Now, when she is late, we can similarly describe it as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_0 + 6\tilde{t} = x_1 + v_1\left(\tilde{t}+\tfrac13\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\tilde{t}$ is how long Claire takes ot meet Charles (outside the swimming pool), and the $\tfrac13$ comes from the fact that Charles had already been cycling for 20 minutes, or $\tfrac13$ of an hour, before Claire began walking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, we can relate these equations (which describe the positions at which they meet, in $km$) by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_1 + v_1t - 3 = x_1 + v_1(\tilde{t} + \tfrac13)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it's a matter of juggling these equations. From the thrid equation we obtain $$v_1 = \frac{3}{t - \tilde{t} - \tfrac13}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first two imply $$x_0 - x_1 = t(v_1 - 6) = \tilde{t}(v_1 - 6) + \frac{v_1}{3} $$&#xA;so that$$t - \tilde{t} = \frac{v_1}{3(v_1-6)} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting back into the expression we previously obtained for $v_1$, simple arithmetics yield $v_1 = 18$. So Charles's velocity is $18km/h$.&lt;/p&gt;&#xA;" OwnerUserId="16478" LastEditorUserId="16478" LastEditDate="2015-10-28T21:23:53.593" LastActivityDate="2015-10-28T21:23:53.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23617" PostTypeId="2" ParentId="23613" CreationDate="2015-10-28T20:31:26.110" Score="1" Body="&lt;p&gt;Given: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Distance: 3 km&lt;br&gt;&#xA;  Time: 20 minutes&lt;br&gt;&#xA;  Velocity: ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pretty easy to calculate: &lt;em&gt;Velocity = Distance/Time&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the &lt;strong&gt;delta&lt;/strong&gt; of their speeds is: 3 km / 20 min = 9 km/h&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing that and the velocity of Claire we can easily find the velocity of Charles: 6 km/h + 9 km/h = &lt;strong&gt;15 km/h&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is: &lt;strong&gt;C&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9352" LastActivityDate="2015-10-28T20:31:26.110" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23618" PostTypeId="2" ParentId="23608" CreationDate="2015-10-28T20:52:42.587" Score="1" Body="&lt;p&gt;At each given time we should only care about the next peak or valley of each of the hikers(let's call them Bruce B. and Tony S.). There are 3 possibilities:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The next peak of Bruce B. is higher than the next peak of Tony S.&lt;/li&gt;&#xA;&lt;li&gt;The next peak of Bruce B. is lower than the next peak of Tony S.&lt;/li&gt;&#xA;&lt;li&gt;The next peaks are equally high&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In case 1: Tony S. would climb up his peak and Bruce B. would go just as high. Then Tony S. would move to the valley while Bruce B. continues his backtracking descent&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case 2: Bruce B. would climb up his peak and Tony S. would go just as high. Then Bruce B. would move to the valley while Tony S. continues his backtracking descent&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case 3: They both go to the next peak and:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Either they meet(if it was the only remaining peak between them)&lt;/li&gt;&#xA;&lt;li&gt;Or they look at the next peak and repeat the steps&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So basically, &lt;em&gt;the one whose next peak is higher will have to backtrack, letting the other one move forward.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this strategy they will eventually meet either:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;On the highest peak&lt;/li&gt;&#xA;&lt;li&gt;Or on the highest point between 2 highest peaks&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9352" LastActivityDate="2015-10-28T20:52:42.587" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23619" PostTypeId="2" ParentId="23608" CreationDate="2015-10-28T20:57:00.723" Score="0" Body="&lt;p&gt;Let's take a peak diagram of:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;             B &#xA;            /\            C&#xA;           /  \          /\&#xA;    A     /    \        /  \&#xA;   /\____/      \      /    \&#xA;  /              \____/      \&#xA;1/                            \2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The steps would be:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1 to A&lt;/li&gt;&#xA;&lt;li&gt;1 down, then up towards B&lt;/li&gt;&#xA;&lt;li&gt;2 hits C, goes down, 1 goes down&lt;/li&gt;&#xA;&lt;li&gt;1 hits valley, goes back up to A, down until 2 can hit valley&lt;/li&gt;&#xA;&lt;li&gt;1 goes up to B, 2 goes up to B&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Another way of looking at it is: 1-Peak, 1-Valley, 2-Peak, 1-Valley, 1-Peak, 2-Valley, 1-Peak, 1-Valley, 12-Final&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's try to factor in direction (Forward is towards highest peak): 1-Peak(F-F), 1-Valley(F-A), 2-Peak(F-F), 1-Valley(A-F), 1-Peak(A-A), 2-Valley(A-F), 1-Peak(F-F), 1-Valley(F-A), 12-Final(F-F)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I don't know if there's a good way of determining which peak/valley to go to. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would say that it looks like you take the current altitude span (e.g. 2 to C initially vs 1 to A initially) and see if the entirety of the other's route falls into it. Then take the next span (C to valley) and see how much of the other's span falls into that altitude. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-10-28T21:13:22.953" LastActivityDate="2015-10-28T21:13:22.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23621" PostTypeId="1" CreationDate="2015-10-28T21:14:22.427" Score="-4" ViewCount="406" Body="&lt;p&gt;Two friends each weighing 150 lbs arrive at a Rope bridge 100 yards long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The warning sign at the start of the bridge informs it can hold only a maximum of 325 lbs of weight at any given second, otherwise it instantly collapses endangering any crossing humans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both wish to cross the bridge simultaneously while each of them carries three sacks of potatoes one of which is 2 lbs, the second is 3 lbs while the third sack is 11 lbs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How will they be able to accomplish this in one single trip together?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Please Please clearly state your assumptions. Water under the bridge. Just do not give up based on some prejudices or previous experience. This is a Classical Problem to Solve.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Socrates : What Plato is going to say next is a Lie.&#xA; Plato: Whatever Socrates mentioned is the truth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9892" LastEditorUserId="9892" LastEditDate="2015-10-29T12:15:50.213" LastActivityDate="2015-10-29T12:15:50.213" Title="Crossing the Bridge together" Tags="|logical-deduction|lateral-thinking|science|" AnswerCount="7" CommentCount="5" ClosedDate="2015-10-29T10:19:59.333" ContentLicense="CC BY-SA 3.0" />
  <row Id="23622" PostTypeId="1" CreationDate="2015-10-28T21:17:44.527" Score="-4" ViewCount="750" Body="&lt;p&gt;Playing on an $ n \times n $ grid, how can we determine the largest possible title you can achieve in a game, assuming the computer places tiles in the perfect spots.&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2023-10-18T19:10:30.887" Title="Maximum tile possible in a game if 2048 on an nXn grid" Tags="|mathematics|2048-game|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23623" PostTypeId="2" ParentId="23621" CreationDate="2015-10-28T21:24:43.730" Score="4" Body="&lt;p&gt;The standard solution is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;juggling&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but that won't work since&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the force exerted by your hand has to be greater than the weight, and due to Newton's third law, that force will be the same as the force you exert on the bridge in addition to your weight.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2015-10-28T21:24:43.730" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23624" PostTypeId="2" ParentId="23622" CreationDate="2015-10-28T21:35:25.570" Score="1" Body="&lt;p&gt;The answer is $2^{n^2+1}$. This is because $n^2$ is the number of squares in an $n\times n$ grid, and the $+1$ comes from requiring that the last square spawned be a $4 = 2^2$.&lt;/p&gt;&#xA;" OwnerUserId="16478" LastActivityDate="2015-10-28T21:35:25.570" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="23625" PostTypeId="2" ParentId="23608" CreationDate="2015-10-28T21:40:06.660" Score="15" Body="&lt;p&gt;I think @Sleafar's deleted answer works:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Put the hikers and the mountains in a big swimming pool. The hikers will always swim at the water level which we will change.&lt;/p&gt;&#xA;&lt;p&gt;Start increasing the water level until one hiker reaches a local peak. The hiker that reached the peak will now descend on the other side of the mountain. Decrease the water level until one hiker reaches a valley. This hiker will now ascend the next mountain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Whenever one hiker reaches a peak or a valley, we reverse the water direction and the hiker continues on the other side of the peak or valley, while the other hiker reverses direction. (If both hikers encounter a peak or valley at the same time, they both continue in the same direction as before.)&lt;/p&gt;&#xA;&lt;p&gt;This process is the same in reverse. Because it can be reversed without ambiguity, it is impossible to get stuck in a cycle.&lt;/p&gt;&#xA;&lt;p&gt;Suppose the process ever returns to the starting state. Then, we can run it backwards and it will look exactly the same as it did forwards. Then, exactly halfway through, both hikers must reverse direction at the same time, which is impossible.&lt;/p&gt;&#xA;&lt;p&gt;Because the procedure cannot be stuck in a loop, and it can't return to the starting state, it must end with at least one hiker reaching the opposite side (in fact, they both reach the opposite side, but we don't need to prove this). This guarantees that the hikers meet at some point.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-10-28T21:40:06.660" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23626" PostTypeId="2" ParentId="23621" CreationDate="2015-10-28T21:48:14.847" Score="1" Body="&lt;p&gt;Assuming that these 2 friends are healthy but not entirely fit...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Get 1 friend to make a 30 pound makeshift dumbbell from the potatoes and have him/her to start working out for 30 minutes. Have the other friend go for a run for 30 minutes. 1 pound of combined water weight will be shed and they will be able to cross the bridge. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If they don't want to wait...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Have one of them do continuous gag reflexes until they throw up. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12477" LastActivityDate="2015-10-28T21:48:14.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23627" PostTypeId="2" ParentId="23621" CreationDate="2015-10-28T21:50:32.700" Score="2" Body="&lt;p&gt;I may just be pedantic but here's my take:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It doesn't state that the friends are carrying identical bags:&lt;br/&gt;&lt;br/&gt;&#xA;&quot;One of which is 2 lbs the second is 3 lbs while the third sack is 11 lbs.&quot;&lt;br/&gt;&lt;br/&gt;&#xA;So if one friend is carrying those three bags adding up to 16 lbs the other could be carrying any combination of 3 bags adding up to 9 lbs making a total of 325 lbs as the two friends together weigh 300 lbs themselves&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17054" LastActivityDate="2015-10-28T21:50:32.700" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23628" PostTypeId="2" ParentId="23621" CreationDate="2015-10-28T21:56:45.067" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This 100 yard rope bridge is a real tourist attraction and, as a result, a souvenir/gift shop has opened up right next to the bridge. As luck would have it, they've just received a brand new canister of helium for the gift balloons. The standard gift shop canister of helium contains 250 cubic feet of helium which would be enough to support the weight of 8 pounds of potatoes plus the string and balloons &lt;a href=&quot;http://www.stevespanglerscience.com/blog/2009/10/17/balloon-boy-helium-physics-science/&quot; rel=&quot;nofollow&quot;&gt;(see here)&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-10-28T21:56:45.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23629" PostTypeId="1" AcceptedAnswerId="23636" CreationDate="2015-10-28T22:53:46.617" Score="16" ViewCount="282" Body="&lt;p&gt;It's almost Halloween at &lt;strong&gt;Farthingbottom School&lt;/strong&gt; and Professor Charles Flavanna is getting festive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Right children, put down your copies of &lt;em&gt;The Origins of Fort Monmouth&lt;/em&gt; - it's time to carve some pumpkins! Have a look at my creation and follow suit. You have 30 minutes and no talking.&quot;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ZMBK2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZMBK2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a sterling effort by the class, the following pumpkins were produced:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/85bG5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/85bG5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Hold on a minute&quot;, said the Professor, &quot;two of you have not been paying attention.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Which children have made a mistake and what should their pumpkins have looked like?&lt;/p&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-10-29T08:44:27.103" Title="Pumpkin Carving" Tags="|seasonal|" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="23630" PostTypeId="2" ParentId="23621" CreationDate="2015-10-28T23:24:14.343" Score="5" Body="&lt;p&gt;Do to PHYSICS a bridge is under most strain when weight is applied to the middle.  As we are only a couple of pounds over the weight, all that we need to do is make sure that all the weight is never applied to the middle, and the bridge should be fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tie the potato sacks together to form a long line from lightest to heaviest&#xA;2-2-3-3-11-11&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the friends walk together (holding hands i guess) and drag the chain of sacks behind them.  Potato sacks are big, so the extra weight is almost 20 feet behind.  This is more than enough to make sure we never have 325 in the center&lt;/p&gt;&#xA;" OwnerUserId="1639" LastActivityDate="2015-10-28T23:24:14.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23631" PostTypeId="2" ParentId="23608" CreationDate="2015-10-29T01:18:32.133" Score="3" Body="&lt;p&gt;Suppose we reformulate the problem as follows. I apologize for the mathematical jargon; the essential idea is that we are to consider the set of legal positions for the two hikers, and show that the initial position is reachable from the position with the hikers reversed. We do this by showing, in a precise sense, that nothing disconnects them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To start out, let us make a definition:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $f:[0,1]\rightarrow \mathbb [0,\infty)$ be a continuous* function with $f(0)=f(1)=0$, representing the height of the mountain range a given distance from the left. Assume that the hikers begin at $0$ and $1$ respectively. Next, let $S=[0,1]\times[0,1]$ be the unit square, which is the set of tuples $(x,y)$ of possible positions of the hikers. Define the difference in height $g$ as:&#xA;  $$g(x,y)=f(x)-f(y).$$&#xA;  Note that the legal positions are exactly those for which $g(x,y)=0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We want to know if there is a path with $g(x,y)=0$ throughout starting at $(1,0)$ and ending at $(0,1)$. This is equivalent to saying the hikers may meet (as they will meet in the middle when they swap positions like in this example). The only way this could fail to happen is if there is some path of illegal positions $\gamma$ running from $(1,x)$ or $(x,1)$ to a position of  $(0,x)$ or $(x,0)$, which would divide the square suitably. Notice, moreover, that $g(1,x)\leq 0$ and $g(0,x)\leq 0$ whereas $g(x,1)\geq 0$ and $g(x,0)\geq 0$. As the sign of $\gamma$ must not change due to the intermediate value theorem, it either runs bottom to top or left to right. However, this is impossible as it must then intersect the line $x=y$, along which $g(x,y)=0$, which must not be true of any point on $\gamma$. Therefore, no path of illegal positions divides the space into two parts, and there is thus, to the contrary, a path from $(0,1)$ to $(1,0)$, as desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*I will admit I have no proof in mind to show that &quot;two points in a closed set in $S$ are disconnected if only if a path in the complement divides them.&quot; Maybe someone who wants to do some analysis can fill this in. In the piecewise case, we can, with some reductions, work on a discrete grid rather than the unit square and the proof is obvious there)&lt;/p&gt;&#xA;" OwnerUserId="4765" LastEditorUserId="4765" LastEditDate="2015-10-29T01:26:19.877" LastActivityDate="2015-10-29T01:26:19.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23632" PostTypeId="2" ParentId="23621" CreationDate="2015-10-29T06:00:30.953" Score="1" Body="&lt;p&gt;Here is my solution&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Get a long rope and tie all the sacks together. Then throw the sack to the river under the bridge keeping the end of the rope in hand. Cross the bridge together and then pull up the potatoes. Even though the potatoes get sink in water, the effective weight of the potatoes will be considerably less than the actual weight.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastActivityDate="2015-10-29T06:00:30.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23633" PostTypeId="2" ParentId="23621" CreationDate="2015-10-29T07:33:50.530" Score="2" Body="&lt;p&gt;Given that each of them has identical bags of potatoes(11-3-2 each) and not given that they are on one side of the bridge the solution is trivial:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are on the different sides of that bridge. Each of them can leave his bags on his side and take the bags of the other guy on the other side and continue walking. Their total weight is 300 lbs, which is less than 325 that a bridge can handle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9352" LastActivityDate="2015-10-29T07:33:50.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23634" PostTypeId="2" ParentId="14992" CreationDate="2015-10-29T08:02:52.513" Score="1" Body="&lt;p&gt;I'm gonna go with &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;E3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;can't see&quot; were the words you were referring to in your hint. If you can't see and (according to your comment) hearing isn't the right way to figure it out either, that leaves touch as the only other means that makes sense to me. If I were to emboss the grid on a surface and then tried to run my hands over it, I would think that the Us would be easiest to feel (from the top). There are only 4 Us and they can be interpreted as 2 braille cells reading E and 3. (Alternatively 5 and C, but you wrote a gate was identified by a letter followed by a number.) I still feel like I'm missing something, because Us are not the only letters that you could feel that way, also Y and V, but I couldn't make any sense of those...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17069" LastActivityDate="2015-10-29T08:02:52.513" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23636" PostTypeId="2" ParentId="23629" CreationDate="2015-10-29T08:36:26.157" Score="19" Body="&lt;p&gt;The two kids are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Richard and Susan&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;reading from left to right, the teeth should spell your first initial using Morse code&lt;br/&gt;Where upper tooth is &lt;code&gt;.&lt;/code&gt; and lower tooth is &lt;code&gt;-&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Professor Charles' example is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;lower/upper/lower/upper = &lt;code&gt;-.-.&lt;/code&gt; which is a Morse &lt;code&gt;C&lt;/code&gt;&lt;br/&gt;&#xA;Richard should have done upper/lower/upper =&lt;code&gt;.-.&lt;/code&gt; = &lt;code&gt;R&lt;/code&gt;&lt;br/&gt;&#xA;Susan should have done three upper teeth - &lt;code&gt;...&lt;/code&gt; = &lt;code&gt;S&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; (according to comment)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usage of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Morse code is actually part of the lesson, not just a festive break. The children were learning about &lt;em&gt;Fort Monmouth&lt;/em&gt;, where morse code radios were developed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4317" LastEditorUserId="4317" LastEditDate="2015-10-29T08:44:27.103" LastActivityDate="2015-10-29T08:44:27.103" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23637" PostTypeId="1" AcceptedAnswerId="23641" CreationDate="2015-10-29T09:10:55.447" Score="4" ViewCount="321" Body="&lt;p&gt;Halloween is just around the corner and the witch is planning a little party for the 100 villagers of the nearby village. Since the villagers burned down her house again and again during the last few years she would like to use this chance to make peace with them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least that is what she told the villagers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course an evil witch would not simply forgive being attacked and so she collected the essences of animals, to be exact the essences of bears, hawks, and spiders, during the past year. Her plan is to use these essences to concoct potions which would transform the drinker into a creature with characteristics of all 3 animals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course there are rules for these potions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The witch must concoct each potion on its own so 100 potions in total.&lt;/li&gt;&#xA;&lt;li&gt;For the concocting process all 3 essences are needed in a specific order, therefore each potion must contain at least a small amount of each essence.&lt;/li&gt;&#xA;&lt;li&gt;Each potion must must be concocted of the same amount of combined essences to get potions of equal strength. Otherwise the villagers who drank a stronger potion would be able to cancel the transformation of those who drank weaker potions.&lt;/li&gt;&#xA;&lt;li&gt;So the villagers would never find happiness again no two villagers should have the same transformation. Therefore each potion needs to be concocted using a different combination of amounts of essences.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Especially her last rule is a big problem for the witch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;She is only in posession of 4 magical cauldrons. One to contain each essence and one to concoct potions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless the essences are in a complete potion or in one of the cauldrons they will dissipate in a matter of seconds so the witch can't measure the amount of essences for each potion beforehand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore the concocting process is extremely complex and if she doesn't constantly stir it the incomplete potion would explode in her face.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For these reasons she can only use one ladle per essence to measure the needed amount for a potion.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How many different sized ladles does the witch need to concoct all 100 potions?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16608" LastEditorUserId="16608" LastEditDate="2015-10-29T09:23:02.577" LastActivityDate="2015-10-29T12:36:15.643" Title="The Witch Invites" Tags="|optimization|seasonal|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23638" PostTypeId="1" AcceptedAnswerId="23639" CreationDate="2015-10-29T09:34:43.780" Score="2" ViewCount="779" Body="&lt;p&gt;How to solve this puzzle? Please explain with steps please! &#xA;&lt;a href=&quot;https://i.stack.imgur.com/0n6iI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0n6iI.jpg&quot; alt=&quot;Follow the link&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17171" LastEditorUserId="7308" LastEditDate="2015-10-29T10:22:28.217" LastActivityDate="2015-10-29T10:22:28.217" Title="Forty-Eight: What is the next bigger number?" Tags="|mathematics|calculation-puzzle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23639" PostTypeId="2" ParentId="23638" CreationDate="2015-10-29T09:39:25.660" Score="13" Body="&lt;p&gt;The magic word is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Pell equation: &lt;a href=&quot;https://en.wikipedia.org/wiki/Pell%27s_equation&quot;&gt;https://en.wikipedia.org/wiki/Pell's_equation&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What the puzzle actually does is: The smaller square $x^2=25$ equals $\frac{1}{2}48+1$, and the larger square $y^2=49$ equals $48+1$. By multiplying the first equation by $2$ and then subtracting the second from it we get the (negative) Pell equation&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$2x^2-y^2=1$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The solutions of this Pell equation are given by the following sequence $(x_n,y_n)$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$x_1=1$ and $y_1=1$&lt;br&gt;&#xA;  $x_{n+1}=3x_n+2y_n$ and $y_{n+1}=4x_n+3y_n$ &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here $(x_2,y_2)=(5,7)$ corresponds to the given solution, &#xA;and $(x_3,y_3)=(29,41)$ corresponds to the solution you are looking for.&#xA;The corresponding magic number and &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the answer to the puzzle is:  &lt;strong&gt;1680&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;with $x_3^2=\frac{1}{2}1680+1$ and $y_3^2=1680+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer can also be found in the &lt;strong&gt;&quot;The On-Line Encyclopedia of Integer Sequences&quot;&lt;/strong&gt;, as sequence &lt;a href=&quot;http://oeis.org/A008845&quot;&gt;A008845&lt;/a&gt;.&#xA;The next five terms (and magic numbers) are $57120$, $1940448$, $65918160$, $2239277040$, and $76069501248$.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-10-29T09:58:24.507" LastActivityDate="2015-10-29T09:58:24.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23640" PostTypeId="2" ParentId="23637" CreationDate="2015-10-29T09:41:46.973" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; She will only need 3 ladles of the same size, 1 for each essence(since &lt;em&gt;she can only use one ladle per essence&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; She can put 10 measures of the first essence(B), then 10 measures of the second(H) and then 10 of the third(S) to make 1 potion, then she can put 11B measures, 10H measures and 10S. This gives us 2 potions of equal strength(30 measures), but different combination of amounts.&lt;br&gt;&#xA; Putting 10H first and then 10B and 10S will also produce a different potion, so she has plenty of choice and she only needs the ladle(s) of same size&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9352" LastActivityDate="2015-10-29T09:41:46.973" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23641" PostTypeId="2" ParentId="23637" CreationDate="2015-10-29T09:42:23.847" Score="4" Body="&lt;p&gt;She would need&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;13 ladels&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explained:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;We'll use EP (essence portion) as a unit.&lt;br/&gt;&#xA;She needs same strength potions for each villager, meaning the amount of EPs in each potion is equal. To get 100 different potions we need a number which can be summed up to using 3 parameters in 100 different ways.&lt;br/&gt;&#xA;I don't remember the proper math, but a small test on the computer showed that 16 is the smallest available, having 105 permutations.&lt;br/&gt;&#xA;Since the different compositions add up to 16, and we will have at least 1EP of each, she can ladel at most 14 EP of one.&lt;br/&gt;&#xA;Needing only 100 potions we can eliminate the 3 permutations using 14 (&lt;code&gt;1-1-14&lt;/code&gt;, &lt;code&gt;1-14-1&lt;/code&gt;, &lt;code&gt;14-1-1&lt;/code&gt;), and we're left with 13 ladels.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4317" LastActivityDate="2015-10-29T09:42:23.847" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23642" PostTypeId="2" ParentId="23638" CreationDate="2015-10-29T09:42:37.773" Score="4" Body="&lt;p&gt;The smallest i could get was $1680$. ($1681=41^2$,$841=29^2$).&lt;/p&gt;&#xA;" OwnerUserId="16628" LastActivityDate="2015-10-29T09:42:37.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23644" PostTypeId="2" ParentId="14992" CreationDate="2015-10-29T11:01:00.357" Score="7" Body="&lt;p&gt;Finally got it. The first important word, as we already knew, is &quot;unusual&quot;, referring to &lt;a href=&quot;https://en.wikipedia.org/wiki/Unusual_number&quot; rel=&quot;nofollow&quot;&gt;unusual numbers&lt;/a&gt;. Since the hint says &quot;not primes&quot;, we are interested in the sequence of not-prime unusual numbers, that goes like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;6, 10, 14, 15, 20, 21, 22, 26, ...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;These actually dictates &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the valid letters in the grid by their number:&#xA;&#xA;    F, J, N, O, T, U, V, Z&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we apply this knowledge to the grid, we get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;pre&gt;&lt;code&gt;      O F     U   F T   Z     N T V Z F U U N     T O N&#xA;    J Z T Z O   U J   T     J     F N O F       V Z   J&#xA;      F         V       O       N       J         F Z  &lt;/code&gt;&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I tried this ages ago but it didn't seem useful at all,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; until @icke mentioned the Braille idea. If each letter is a dot, in Braille this reads:&#xA;&#xA;&lt;pre&gt;   W H I T E K I N G P A W N&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I guess this refers to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the &lt;a href=&quot;https://en.wikipedia.org/wiki/King%27s_Pawn_Game&quot; rel=&quot;nofollow&quot;&gt;most common chess opening&lt;/a&gt;, defined by the simple and ubiquitous move&#xA;&#xA;    E4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and that's your gate.&lt;/p&gt;&#xA;" OwnerUserId="5990" LastEditorUserId="7316" LastEditDate="2015-10-29T12:09:56.960" LastActivityDate="2015-10-29T12:09:56.960" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23645" PostTypeId="1" AcceptedAnswerId="23653" CreationDate="2015-10-29T12:09:07.223" Score="14" ViewCount="1601" Body="&lt;p&gt;Professor Erasmus has constructed a special convex &lt;a href=&quot;https://en.wikipedia.org/wiki/Pentagon&quot;&gt;pentagon&lt;/a&gt; $ABCDE$ that he modestly calls the &quot;Professor-Erasmus-pentagon&quot;. &#xA;The professor claims that he can cut off a smaller pentagon &lt;a href=&quot;https://en.wikipedia.org/wiki/Similarity_(geometry)&quot;&gt;similar&lt;/a&gt; to pentagon $ABCDE$ by a straight line.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Has the professor once again made one of his well-known mathematical blunders, or do such pentagons indeed exist?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&#xA;Note: The five points A,B,C,D,E are all distinct. There are no tricks (like paper folding; pentagons made of rubber; etc).&#xA;&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-10-29T20:44:50.590" Title="The Erasmus pentagon" Tags="|mathematics|geometry|dissection|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23647" PostTypeId="2" ParentId="23637" CreationDate="2015-10-29T12:36:15.643" Score="-3" Body="&lt;p&gt;0, as using a ladle would mean the essences would dissipate: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless the essences are in a complete potion or in one of the cauldrons they will dissipate in a matter of seconds so the witch can't measure the amount of essences for each potion beforehand.&lt;/p&gt;&#xA;" OwnerUserId="17175" LastActivityDate="2015-10-29T12:36:15.643" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23648" PostTypeId="2" ParentId="18437" CreationDate="2015-10-29T14:14:53.237" Score="8" Body="&lt;p&gt;It can be done with &lt;s&gt;63&lt;/s&gt; &lt;s&gt;48&lt;/s&gt; 44 fuses (22 at top and bottom).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic idea is that there's a top and bottom circuit in parallel, and a fuse will blow in whichever circuit has lower resistance. This will increase the resistance of that circuit and the process will repeat. The bottom circuit therefore establishes roughly how many insulators there are, and the top one refines that estimate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Overall circuit is show here (where $n$ is the number of boxes that insulate, and the circles are the fuses)&lt;/strong&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/c603h.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/c603h.png&quot; alt=&quot;Circuit to measure 1000 ohms&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Box circuit that gives an equivalent resistance of $n\Omega$ where $n$ of the boxes are insulators&lt;/strong&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/SymCZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SymCZ.png&quot; alt=&quot;Equivalent circuit&quot;&gt;&lt;/a&gt;&#xA;There are three groupings:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The 1000 boxes, which are each in parallel with a resistor of resistance $1\Omega$, and combined in series to give a total resistance of  $n\Omega$.&lt;/li&gt;&#xA;&lt;li&gt;22 fuses on the top: the first shorting the box resistor, and the other 21 with a $2\Omega$ resistor. &lt;/li&gt;&#xA;&lt;li&gt;22 fuses on the bottom each with a $44\Omega$ resistor and a fuse&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;A 2A current source (easily made with a voltage source and a large resistor) supplies 2A to the circuit, which means that more than 1A will be going through one fuse or other at all times (or it will be balanced, and both will blow). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the fuses are short circuits, only one fuse from the top and one from the bottom could possibly burn out.  The order will depend on the value of $n$.  The bottom fuses go first until $44k&amp;lt;n&amp;lt;44(k+1)$ for some $k\in(0,1,\ldots,22)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the top fuses go subtracting $2\Omega$ at a time until the value of $n$ is reached (equality) or surpassed, from which the value can be deduced. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This basic idea scales as roughly $\sqrt{2N}$ where $N$ is the number of boxes.  I've tested this with some simple python code, and it works in all 1001 cases.&lt;/p&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2015-11-07T02:18:11.607" LastActivityDate="2015-11-07T02:18:11.607" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="23649" PostTypeId="2" ParentId="23645" CreationDate="2015-10-29T14:34:05.147" Score="1" Body="&lt;p&gt;This is possible. The trick is to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Create an asymmetric Pentagon, which can be &quot;flipped&quot; by one cut.&#xA;Use that &lt;code&gt;similar&lt;/code&gt; allows to mirror an object.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My solution looks like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/2wdJM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2wdJM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;My image isn't exact, since the ratios of the sides need to be the same (e.g. ED/CB = CB/DX).&#xA;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16487" LastActivityDate="2015-10-29T14:34:05.147" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23650" PostTypeId="2" ParentId="23645" CreationDate="2015-10-29T15:25:28.177" Score="5" Body="&lt;p&gt;Intuitively, I would make a polygon with sides $AB$, $BC$, $CD$, $DE$, $EA$ having lengths in increasing order $1,a,a^2,a^3,a^4$ such that the distance between, for some $a &amp;gt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, put a point $E'$ such that $E'A = a^{-1}$ and cut the polygon on the line $E'D$. We just have to make sure that $E'D = ED = a^4$. If so, all sides will be $a^{-1},1,a,a^2,a^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: This is how the picture would look like (approximately, don't have any proper tool for drawing). All angles except the one between $a^3$ and $a^4$ are equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AdL5X.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AdL5X.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12901" LastEditorUserId="12901" LastEditDate="2015-10-29T20:44:50.590" LastActivityDate="2015-10-29T20:44:50.590" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23651" PostTypeId="2" ParentId="23417" CreationDate="2015-10-29T15:33:18.333" Score="2" Body="&lt;p&gt;The hidden sequence is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $2$ $10$ $42$ (here I knew it must be the right one) $170$ $682$&lt;br&gt;&lt;br&gt;or in binary:&lt;br&gt;&lt;br&gt;$10$ $1010$ $101010$ $10101010$ $1010101010$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How to find it?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Count the letters (ignoring whitespace and punctuation) of each paragraph. The first four numbers are from the intro. The three options have $512$, $682$ and $852$ letters. Therefore option 2 is the matching one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2015-10-29T15:42:51.953" LastActivityDate="2015-10-29T15:42:51.953" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23652" PostTypeId="2" ParentId="23645" CreationDate="2015-10-29T16:20:50.840" Score="3" Body="&lt;p&gt;&lt;s&gt;This is not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at angle $C$ and find the new angle it corresponds to in the smaller pentagon. If it is one of the other angles of the larger pentagon, then every angle is congruent, so it is a regular pentagon, and this case is trivially impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $C$ and $C'$ are in the same place, then $AE=A'E'$, but the second pentagon must be smaller than the first. So $C$ must correspond to one of the newly created angles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this applies to every angle, not just $C$. So every angle on the pentagon must either correspond to itself or a newly-created angle. Only 2 new angles can be created in any convex pentagon made from a cut through a convex pentagon. Every angle must stay in the exact same place, so the only possible solution is one with no cut at all.&lt;/s&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: This has been invalidated by f'''s example - having four congruent angles is possible in this context.&lt;/p&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="11876" LastEditDate="2015-10-29T17:59:39.397" LastActivityDate="2015-10-29T17:59:39.397" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23653" PostTypeId="2" ParentId="23645" CreationDate="2015-10-29T17:29:27.843" Score="18" Body="&lt;p&gt;Let $A=(-30,60),B=(0,0),C=(150,0),D=(300,300),E=(225,400)$. $F=(-12,84)$ is on $AE$ and $G=(175,50)$ is on $CD$. Then $ABCDE$ is similar to $FABCG$, by a factor of $\sqrt5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key point of this solution is that angles $GFA, EAB, ABC, BCD, CDE$ are all equal ($\arctan-2\approx116.57^\circ$) and $FA,AB,BC,CD$ are in geometric sequence. Then, we can rotate the pentagon so that each of those edges corresponds to the next one in the smaller pentagon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image of the pentagon:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/xjcR2.png&quot; alt=&quot;1]&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same idea can be done with other values of the key angle between $90^\circ$ and $135^\circ$ and other values for the ratio of similarity.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-10-29T18:03:18.623" LastActivityDate="2015-10-29T18:03:18.623" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23654" PostTypeId="1" AcceptedAnswerId="23658" CreationDate="2015-10-29T19:19:48.570" Score="13" ViewCount="471" Body="&lt;p&gt;Bringing this puzzle type back again.&lt;br&gt;&#xA;(See &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15339/four-words&quot;&gt;the original&lt;/a&gt; for the solution format.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first is what you will seek.&lt;br&gt;&#xA;My second is what you should be.&lt;br&gt;&#xA;My third is something all monsters have.&lt;br&gt;&#xA;My fourth is what you may feel.&lt;br&gt;&#xA;My fifth is the shameful answer the following day.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For 5, there is an abbreviation involved.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-11-02T21:07:39.880" Title="Spooky Five Words (-----|||||)" Tags="|word|word-square|seasonal|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23655" PostTypeId="2" ParentId="12396" CreationDate="2015-10-30T01:14:28.537" Score="3" Body="&lt;p&gt;This is a big supposition, but just maybe it happens that the ghost has a quirk where it doesn't wait for you to finish asking the question if it has heard enough of it to be 100% confident of the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, I'd ask, very slowly: &quot;Is the door to freedom one of these two doors: this red door right here, or...&quot; and after a long pause, pace slowly between the other doors, indecisively, finally finishing the question &quot;...this blue door?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the ghost is an impatient sort of ghost, or an optimized-algorithm ghost, and the red door is the one to freedom, it will answer &quot;yes&quot; right away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, you'll get a &quot;yes&quot; or &quot;no&quot; only after naming the second door, and there's your answer.&lt;/p&gt;&#xA;" OwnerUserId="5197" LastActivityDate="2015-10-30T01:14:28.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23656" PostTypeId="1" CreationDate="2015-10-30T01:19:48.427" Score="5" ViewCount="1274" Body="&lt;p&gt;You are provided with the 8-segment array in the picture together with all the necessary electronics to make any combination of lit segments you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mYhH9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mYhH9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fillable version for graphics. &lt;a href=&quot;https://i.stack.imgur.com/K1hnN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/K1hnN.png&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/K1hnN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/K1hnN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the &lt;strong&gt;smallest positive number&lt;/strong&gt; that you can display without any two boxes being identical?  (A box is a grey rectangle containing 8 segments)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any combination of segments is allowed within a box provided it is accepted by the community as a valid symbol. ('accepted' is decided by net up-votes for your answer)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any recognised numbering system that is commonly used is allowed. If the one you use is common but little-known (perhaps because it is specialized) then you must back up your claim with a link to an authoritative source. You must also explain it precisely so that it can be understood by others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You cannot mix numbering systems unless you can show that this is routinely done by some substantial section of the populace. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You cannot invent your own symbols or language. All symbols must be understandable and approved by the community.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The competition remains open until there has been no answer for one week. At that point (or as soon after as I am able) I shall accept the &lt;strong&gt;most up-voted answer&lt;/strong&gt; (net votes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The question has been put on hold--therefore I'll make an adjustment about the timescale for accepting an answer  (if you want to reopen it please vote to do so).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If I could wind back the clock, I would require the answers to be finite. This was an oversight because I did not think an infinite could be considered a number.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Because there are answers and they have been voted on, it is too late to change the criteria. My suggestion is that people who are interested submit answers in the categories of finite or infinite numbers according to their preference. I will stick to my original criterion for acceptance.&#xA;&lt;strong&gt;Note&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You can display your answer using ASCII provided it is clear how it would look when translated into segments. Otherwise please use an actual 8-segment representation to clarify the layout of any symbols that aren't obvious.&lt;/p&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="25957" LastEditDate="2016-12-29T09:54:40.810" LastActivityDate="2016-12-29T09:54:40.810" Title="What is the smallest positive number on an 8-segment array? (no repeats)" Tags="|open-ended|seven-segment|" AnswerCount="7" CommentCount="5" ClosedDate="2015-10-30T20:46:43.943" ContentLicense="CC BY-SA 3.0" />
  <row Id="23657" PostTypeId="2" ParentId="23656" CreationDate="2015-10-30T02:02:15.457" Score="10" Body="&lt;p&gt;Here's one to start off:&#xA;&lt;code&gt;1 E - 9 8 7 6 5&lt;/code&gt; $=10^{-98765}$&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-10-30T02:02:15.457" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23658" PostTypeId="2" ParentId="23654" CreationDate="2015-10-30T02:24:52.707" Score="5" Body="&lt;p&gt;And the &lt;em&gt;spooky&lt;/em&gt; solution is...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\begin{matrix}C&amp;amp;A&amp;amp;N&amp;amp;D&amp;amp;Y\\A&amp;amp;W&amp;amp;A&amp;amp;R&amp;amp;E\\N&amp;amp;A&amp;amp;M&amp;amp;E&amp;amp;S\\D&amp;amp;R&amp;amp;E&amp;amp;A&amp;amp;D\\Y&amp;amp;E&amp;amp;S&amp;amp;D&amp;amp;A\\\end{matrix}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="4421" LastEditDate="2015-11-02T21:07:39.880" LastActivityDate="2015-11-02T21:07:39.880" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23659" PostTypeId="2" ParentId="23656" CreationDate="2015-10-30T02:55:51.800" Score="22" Body="&lt;p&gt;I don't know if you can get smaller than&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/OzP2g.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OzP2g.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;&lt;a href=&quot;http://mathworld.wolfram.com/Epsilon.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;epsilon&lt;/a&gt;&lt;/strong&gt; - a small positive infinitesimal quantity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="4421" LastEditDate="2015-10-30T17:53:48.760" LastActivityDate="2015-10-30T17:53:48.760" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23660" PostTypeId="1" AcceptedAnswerId="23663" CreationDate="2015-10-30T03:43:42.577" Score="4" ViewCount="155" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;She is sunny, while his light does lack.&lt;br&gt;&#xA;  She has a flavour, while he holds something back.&lt;br&gt;&#xA;  Although he is far more solemn, her vividness may make you cringe.&lt;br&gt;&#xA;  Still he goes fittingly with many others; she with her partners oft look fringe.&lt;br&gt;&#xA;  Oh, these two paired, how surprising, they form quite the pack!&lt;br&gt;&#xA;  Heed well when you see them lest you be vulnerable to attack.&lt;br&gt;&#xA;  Majestic creatures in some forms, conjuring gored gourds when given more range,&lt;br&gt;&#xA;  They celebrate together in the season for harvest, for remembrance, and for change.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What are the two? Yes, the &quot;what&quot; is a clue.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="15921" LastActivityDate="2015-10-30T15:08:04.270" Title="Contrasting Partners" Tags="|poetry|seasonal|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23661" PostTypeId="2" ParentId="23656" CreationDate="2015-10-30T07:55:12.433" Score="7" Body="&lt;p&gt;Building up on f'' s answer:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;        ###           ###    ###    ###    ###    ###&#xA;    #  #             #   #  #   #  #   #      #  #&#xA;    #  #             #   #  #   #  #   #      #  #&#xA;    #  #             #   #  #   #  #   #      #  #&#xA;        ###    ###    ###    ###    ###           ###&#xA;    #  #                 #      #  #   #      #  #   #&#xA;    #  #                 #      #  #   #      #  #   #&#xA;    #  #                 #      #  #   #      #  #   #&#xA;        ###           ###           ###           ###&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;$=10^{-99876}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the possibility to write the digit 9 in two different ways on this kind of display.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dropping the 1 in front we could get as low as  $10^{-998766}$:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; ###           ###    ###    ###    ###    ###&#xA;#             #   #  #   #  #   #      #  #      #&#xA;#             #   #  #   #  #   #      #  #      #&#xA;#             #   #  #   #  #   #      #  #      #&#xA; ###    ###    ###    ###    ###           ###    ###&#xA;#                 #      #  #   #      #  #   #  #   #&#xA;#                 #      #  #   #      #  #   #  #   #&#xA;#                 #      #  #   #      #  #   #  #   #&#xA; ###           ###           ###           ###    ###&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="16608" LastEditorUserId="16608" LastEditDate="2015-10-30T09:18:23.813" LastActivityDate="2015-10-30T09:18:23.813" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23662" PostTypeId="1" AcceptedAnswerId="23689" CreationDate="2015-10-30T08:18:31.237" Score="9" ViewCount="244" Body="&lt;p&gt;The ringmaster of a flea circus puts four fleas $A$, $B$, $C$, $D$ on&#xA;four different points in the plane that form the corners of a square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever the ringmaster shouts &lt;em&gt;&quot;Hop!&quot;&lt;/em&gt;, one of the four fleas jumps&#xA;over the &lt;a href=&quot;https://en.wikipedia.org/wiki/Centroid#Of_a_finite_set_of_points&quot;&gt;centroid&lt;/a&gt; of the three other fleas to the mirror point on&#xA;the other side.&#xA;(In other words, a flea sitting in point $x$ may jump over the centroid&#xA;$y$ to the new point $z$, so that $y$ is the midpoint between $x$ and $z$.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; &#xA;  Is it possible that after a finite number of jumps, two of the fleas &#xA;  are simultaneously sitting on the same point?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-10-30T22:07:38.287" Title="Yet another curious incident in the flea circus" Tags="|mathematics|combinatorics|reachability|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23663" PostTypeId="2" ParentId="23660" CreationDate="2015-10-30T09:26:02.937" Score="5" Body="&lt;p&gt;The two partners are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Orange and Black&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;She is sunny, while his light does lack.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sun can appear orange in the correct conditions. Black is effectively a lack of light.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;She has a flavour, while he holds something back.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Orange is a flavour, holds something back could refer to milk in tea/coffee (Original solution) or to the ACDC song &quot;Back in Black&quot; (@Deusovi). (Or &lt;a href=&quot;https://en.wikipedia.org/wiki/Orange_Is_the_New_Black&quot; rel=&quot;nofollow&quot;&gt;should that be&lt;/a&gt; the ACDC song &quot;Back in Orange&quot;?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Although he is far more solemn, her vividness may make you cringe.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Black is worn e.g. at a funeral and in mourning. Orange is a bright vivid colour. This is &lt;a href=&quot;http://www.imdb.com/title/tt0119164/quotes?item=qt0427153&quot; rel=&quot;nofollow&quot;&gt;used as a joke&lt;/a&gt; in the film &quot;The Full Monty&quot;: &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Still he goes fittingly with many others; she with her partners oft look fringe.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In fashion, you can wear almost anything with black. But red and orange, or pink and orange seem to clash. Alternatively, black &lt;a href=&quot;http://www.songtexte.com/songtext/acdc/back-in-black-1bdf7910.html&quot; rel=&quot;nofollow&quot;&gt;has many rhymes&lt;/a&gt;, but &lt;a href=&quot;http://www.rhymezone.com/r/rhyme.cgi?Word=orange&amp;amp;loc=snippettotal&amp;amp;typeofrhyme=exa&quot; rel=&quot;nofollow&quot;&gt;orange doesn't&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Oh, these two paired, how surprising, they form quite the pack!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Despite the differences they make an attractive and distinctive pairing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Heed well when you see them lest you be vulnerable to attack.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Maybe they are warning colours together, like for a wasp/hornet, although I'd consider black/yellow more typical.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Majestic creatures in some forms, conjuring gored gourds when given more range,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tigers (or Monarch Butterflies) are majestic creatures, gored gourds refers to carved pumpkins...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They celebrate together in the season for harvest, for remembrance, and for change.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Autumn is the season for harvest, for remembrance of the 1918 Armistice, and for leaves etc dropping. It is also the season for Halloween, where orange and black are the dominant colours: &lt;a href=&quot;http://www.thingsthatgoboo.com/halloween/halloweensymbolorangeblack.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.thingsthatgoboo.com/halloween/halloweensymbolorangeblack.htm&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note also that the rhyming scheme of the poem is &lt;strong&gt;AABBAABB&lt;/strong&gt; where...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;A&lt;/strong&gt; rhymes with Black, and &lt;strong&gt;B&lt;/strong&gt; does it's best to rhyme with orange.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7316" LastEditorUserId="7316" LastEditDate="2015-10-30T15:08:04.270" LastActivityDate="2015-10-30T15:08:04.270" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23665" PostTypeId="2" ParentId="23656" CreationDate="2015-10-30T09:53:29.843" Score="3" Body="&lt;p&gt;In non-standard analysis, a halo (or &lt;a href=&quot;https://en.wikipedia.org/wiki/Monad_(non-standard_analysis)&quot; rel=&quot;nofollow&quot;&gt;monad&lt;/a&gt;) is described as &quot;the collection hal(x) of numbers which are infinitely close to x &quot; (Lobry et al). Thus, the numbers close to 0 would be hal(0) or monad(0). Hyperreal numbers are described further in &quot;Foundations of Infinitesimal Calculus&quot; (Keisler)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;       _            _   _      _&#xA;| |   | |   |      |   | |      |&#xA;|-|   |-|   |      |   | |      |&#xA;| |   | |   |_     |_  |_|   . _|&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keisler, H. Jerome. &lt;em&gt;Foundations of infinitesimal calculus&lt;/em&gt;. Vol. 20. Boston: Prindle, Weber &amp;amp; Schmidt, 1976.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lobry, Claude, and Tewfik Sari. &quot;Non-standard analysis and representation of reality.&quot; &lt;em&gt;International Journal of Control&lt;/em&gt; 81.3 (2008): 519-536.&lt;/p&gt;&#xA;" OwnerDisplayName="user2674" LastEditorDisplayName="user2674" LastEditDate="2015-10-30T14:59:08.973" LastActivityDate="2015-10-30T14:59:08.973" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23666" PostTypeId="1" CreationDate="2015-10-30T09:56:27.470" Score="0" ViewCount="271" Body="&lt;p&gt;This is a follow-up to the question, &lt;a href=&quot;https://puzzling.stackexchange.com/questions/23656/what-is-the-smallest-positive-number-on-an-8-segment-array&quot;&gt;What is the smallest possible positive number on an 8-segment array?&lt;/a&gt;  However, this time repeats are allowed!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are provided with the 8-segment array in the picture together with all the necessary electronics to make any combination of lit segments you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mYhH9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mYhH9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fillable version for graphics &lt;a href=&quot;https://i.stack.imgur.com/K1hnN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/K1hnN.png&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/z6GLg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/z6GLg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the largest positive number that you can display? Unlike in the previous question, boxes can be identical. (A box is a grey rectangle containing 8 segments)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any combination of segments is allowed within a box provided it is accepted by the community as a valid symbol. ('accepted' is decided by net up-votes for your answer)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any recognised numbering system that is commonly used is allowed. If the one you use is common but little-known (perhaps because it is specialized) then you must back up your claim with a link to an authoritative source. You must also explain it precisely so that it can be understood by others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You cannot mix numbering systems unless you can show that this is routinely done by some substantial section of the populace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You cannot invent your own symbols or language. All symbols must be understandable and approved by the community.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The competition remains open until there has been no answer for one week. At that point (or as soon after as I am able) I shall accept the most up-voted answer (net votes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If the question gets closed (and then possibly reopened) then I'll make an adjustment about the timescale for accepting an answer.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If I could wind back the clock, I would require the answers to be finite. This was an oversight because I did not think an infinite could be considered a number.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Because there are answers and they have been voted on, it is too late to change the criteria. My suggestion is that people who are interested submit answers in the categories of finite or infinite numbers according to their preference. I will stick to my original criterion for acceptance.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can display your answer using ASCII text provided it is clear how it would look when translated into segments. Otherwise please use an actual 8-segment representation to clarify the layout of any symbols that aren't obvious.&lt;/p&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2016-12-29T09:54:26.813" Title="What is the largest possible positive number on an 8-segment array? (repeated symbols allowed)" Tags="|open-ended|seven-segment|" AnswerCount="5" CommentCount="5" ClosedDate="2015-11-04T15:44:03.853" ContentLicense="CC BY-SA 3.0" />
  <row Id="23667" PostTypeId="2" ParentId="23656" CreationDate="2015-10-30T10:16:05.727" Score="0" Body="&lt;p&gt;Here's my attempt:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/WJ1sP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WJ1sP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Probably not a valid solution though since it is actually not positive.&lt;/p&gt;&#xA;" OwnerUserId="4624" LastActivityDate="2015-10-30T10:16:05.727" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="23669" PostTypeId="2" ParentId="23666" CreationDate="2015-10-30T10:29:50.850" Score="3" Body="&lt;p&gt;I submit&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bhegs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bhegs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ALEPH 999.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's infinite, 998 'steps' above countable infinity.&lt;/p&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="4421" LastEditDate="2015-10-30T18:45:25.517" LastActivityDate="2015-10-30T18:45:25.517" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="23670" PostTypeId="2" ParentId="23666" CreationDate="2015-10-30T10:52:31.790" Score="1" Body="&lt;p&gt;As someone has submitted an infinite number, I'll kick off with a finite one which is pretty big.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 GOOGOL&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ofBZi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ofBZi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I'd also offer 9 GOOGOL  but I don't know if I would need an 'S' on the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or 99GOOGOL if I am allowed to omit the space.&lt;/p&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="14177" LastEditDate="2015-10-30T11:02:01.660" LastActivityDate="2015-10-30T11:02:01.660" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23671" PostTypeId="2" ParentId="23666" CreationDate="2015-10-30T10:57:22.937" Score="4" Body="&lt;p&gt;If we're restricted to finite numbers, I submit...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5qQOH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5qQOH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ggggggg9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the (((((((9th number in Graham's series)th number in Graham's series)th number in Graham's series)...)))) where Graham's series (notated as gn for some n) is defined under &quot;definition&quot; &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Graham%27s_number&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;here.&quot;&gt;here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="4421" LastEditDate="2015-10-30T18:56:17.700" LastActivityDate="2015-10-30T18:56:17.700" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23674" PostTypeId="2" ParentId="23666" CreationDate="2015-10-30T12:14:35.063" Score="3" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6Utfs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6Utfs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TREE(99) from Kruskal's tree theorem.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4269" LastEditDate="2015-10-31T03:42:17.200" LastActivityDate="2015-10-31T03:42:17.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23675" PostTypeId="2" ParentId="11780" CreationDate="2015-10-30T13:50:32.647" Score="22" Body="&lt;p&gt;I believe Spencerkatty's solution is maximal...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any net may be constructed by choosing which of the 81 cells to mask.&#xA;There are $\sum_{k=0}^{81}{81\choose k}=2^{81}$ ways to construct a net.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather than attempt to search this vast space we can attempt to partition the space in a way that will allow us to find out where we may need to look in more detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will call a three by three square of chosen masks that could cover a block a &quot;block-mask&quot;.&#xA;There are $\sum_{k=0}^9{9\choose k}=2^9$ such block-masks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any net may instead be constructed by placing any one of these block-masks over each of the nine blocks in turn. $(2^9)^9 = 2^{81}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lopsy's post provided some base constraints - situations that certainly cause a potential ambiguity and hence any net containing a pattern of maskings breaking these constraints is not a valid net.&#xA;I will label these constraints with lower case letters and further constraints either implied by these or via observation of block-mask properties with numbers.&#xA;Changing slightly from Lopsy's post I will refer to horizontal lines of influence as &quot;band-lines&quot; and vertical ones as &quot;stack-lines&quot; and any such lines simply as &quot;lines&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;a)&lt;/strong&gt; Asserting an already asserted line causes ambiguity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) implies that:&#xA;&lt;strong&gt;1)&lt;/strong&gt; Once a band (stack) contains a block-mask asserting all three band-lines (stack-lines) the other blocks in it's band (stack) may not assert any band-lines (stack-lines)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will say that a band-line (stack-line) is &quot;effectively-asserted&quot; when the other two band-lines (stack-lines) in it's band (stack) are each asserted from different blocks (which are, necessarily, in it's band (stack)).&#xA;This is the situation, prior to placing the third block-mask, in Lopsy's post where she says, &quot;On the left, we have a block column with all three lines of influence exerted by pairs in three different boxes.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;b)&lt;/strong&gt; Asserting an effectively-asserted line causes ambiguity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) and (b) imply that:&#xA;&lt;strong&gt;2)&lt;/strong&gt; Once a band (stack) contains two block-masks each of which assert any band-lines (stack-lines) the third may not assert any band-lines (stack-lines)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lopsy's observation &quot;On the right, we have two lines of influence &quot;reflecting&quot; off a diagonal pair in the same box&quot; also causes ambiguity.&#xA;I will call the diagonal pair a &quot;reflective-pair&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are ${9\choose 2}=36$ possible pairs of maskings in a block-mask, 18 assert a line and the other 18 are &quot;reflective-pairs&quot;, hence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3)&lt;/strong&gt; All block-masks containing more than one masking either assert at least one line or contain at least one reflective pair.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will call any pattern of maskings causing ambiguity due to the presence of reflective-pairs a &quot;reflective-path&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;c)&lt;/strong&gt; Completing a reflective-path causes ambiguity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lopsy's example of a reflective-path is only one case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reflective-path exists not only when the crossing lines are both asserted but also when either, or both, are effectively-asserted, for example here a reflective-pair across two effectively asserted lines causes ambiguity:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;...|...|...    ...|...|...    ...|...|...&#xA;...|...|.##    ...|...|.12    ...|...|.21&#xA;...|...|...    ...|...|...    ...|...|...&#xA;---+---+---    ---+---+---    ---+---+---&#xA;...|...|##.    ...|...|12.    ...|...|21.&#xA;...|...|...    ...|...|...    ...|...|...&#xA;...|...|...    ...|...|...    ...|...|...&#xA;---+---+---    ---+---+---    ---+---+---&#xA;...|.#.|#..    ...|.1.|2..    ...|.2.|1..&#xA;..#|.#.|...    ..1|.2.|...    ..2|.1.|...&#xA;..#|...|..#    ..2|...|..1    ..1|...|..2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It is possible to create a longer reflective-paths between two asserted or effectively-asserted lines, so long as the maskings only ever occupy zero or two cells per block, zero or two cells per row, and zero or two cells per column, for example here is a double-reflection between two effectively-asserted lines:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;...|...|...    ...|...|...    ...|...|...&#xA;...|...|...    ...|...|...    ...|...|...&#xA;...|...|...    ...|...|...    ...|...|...&#xA;---+---+---    ---+---+---    ---+---+---&#xA;#..|...|#..    2..|...|1..    1..|...|2..&#xA;...|#..|..#    ...|1..|..2    ...|2..|..1&#xA;#..|#..|...    1..|2..|...    2..|1..|...&#xA;---+---+---    ---+---+---    ---+---+---&#xA;...|.#.|#..    ...|.1.|2..    ...|.2.|1..&#xA;..#|.#.|...    ..1|.2.|...    ..2|.1.|...&#xA;..#|...|..#    ..2|...|..1    ..1|...|..2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;A chain of reflections may also create a circular reflective-path, although this probably wont be required to show if there are or are not valid nets with greater than 33 maskings. Such circular paths only cause ambiguity when one traversal of the circuit does not invert the polarity of the pair (or, equivalently, if the sum of the slopes of the reflective-pairs is zero), for example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;...|...|...    ...|...|...    ...|...|...&#xA;...|...|...    ...|...|...    ...|...|...&#xA;...|...|...    ...|...|...    ...|...|...&#xA;---+---+---    ---+---+---    ---+---+---&#xA;...|#..|#..    ...|1..|2..    ...|2..|1..&#xA;...|...|...    ...|...|...    ...|...|...&#xA;...|.#.|..#    ...|.2.|..1    ...|.1.|..2&#xA;---+---+---    ---+---+---    ---+---+---&#xA;...|.#.|..#    ...|.1.|..2    ...|.2.|..1&#xA;...|...|...    ...|...|...    ...|...|...&#xA;...|#..|#..    ...|2..|1..    ...|1..|2..&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;but not, for example, these ones (following a pair through one traversal will swap the positions; or the sum of the slopes are non-zero):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;...|...|...    ...|...|...    ...|...|...    ...|...|...&#xA;...|...|...    ...|...|...    ...|...|...    ...|...|...&#xA;...|...|...    ...|...|...    ...|...|...    ...|...|...&#xA;---+---+---    ---+---+---    ---+---+---    ---+---+---&#xA;...|#..|..#    ...|#..|#..    ...|#..|#..    ...|.#.|..#&#xA;...|...|...    ...|...|...    ...|...|...    ...|...|...&#xA;...|.#.|#..    ...|.#.|..#    ...|.#.|..#    ...|#..|#..&#xA;---+---+---    ---+---+---    ---+---+---    ---+---+---&#xA;...|.#.|..#    ...|.#.|#..    ...|#..|..#    ...|#..|..#&#xA;...|...|...    ...|...|...    ...|...|...    ...|...|...&#xA;...|#..|#..    ...|#..|..#    ...|.#.|#..    ...|.#.|#..&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(1), (3) and (c) imply that:&#xA;&lt;strong&gt;4)&lt;/strong&gt; Once a block has a block-mask in it's band that asserts all three band-lines and a block-mask in it's stack that asserts all three stack-lines then it may not be masked by a block-mask containing more than one masking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2), (3) and (c) imply that:&#xA;&lt;strong&gt;5)&lt;/strong&gt; Once a block has two block-masks in it's band (stack) each of which assert any band-lines (stack-lines), and either two block-masks in it's stack (band) each of which assert any stack-lines (band-lines) or a block-mask in it's stack (band) which asserts all three stack-lines (band-lines) then it may not be masked by a block-mask containing more than one masking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we inspect the block-masks containing exactly three maskings we may observe that 27 assert no band-lines and 27 assert no stack-lines, and furthermore that:&#xA;&lt;strong&gt;6)&lt;/strong&gt; All block-masks that contain three maskings which do not assert any band-lines (stack-lines) either assert or reflect onto every stack-line (band-line)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1), (6) and (c) imply that:&#xA;&lt;strong&gt;7)&lt;/strong&gt; Once a block has a block-mask in it's band (stack) that asserts all three band-lines (stack-lines) and a block-mask in it's stack (band) that asserts any stack-lines (band-lines) it may not be masked by a block-mask containing more than two maskings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2), (6) and (c) imply that:&#xA;&lt;strong&gt;8)&lt;/strong&gt; Once a block has two block-masks in it's band (stack) each of which assert any band-lines (stack-lines) and a block-mask in it's stack (band) that asserts any stack-lines (band-lines) it may not be masked by a block-mask containing more than two maskings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are nine possible crossings of lines in a block-mask, each of which has two associated reflective-pairs; if a block-mask does not assert either line or contain either of these reflective-pairs then it contains at most four maskings, therefore:&#xA;&lt;strong&gt;9)&lt;/strong&gt; Once a block has a block-mask in it's band that asserts any band-line and a block-mask in it's stack that asserts any stack-line it may not be masked by a block-mask containing more than four maskings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upon further inspection of the domain of block-masks it may be observed that:&#xA;&lt;strong&gt;10)&lt;/strong&gt; All block-masks containing at least seven maskings assert all three band-lines and all three stack-lines;&#xA;&lt;strong&gt;11)&lt;/strong&gt; All block-masks containing at least six maskings assert all three lines in one direction and at least one line in the other;&#xA;&lt;strong&gt;12)&lt;/strong&gt; There are block-masks containing five maskings that assert only one line in each direction.&#xA;&lt;strong&gt;13)&lt;/strong&gt; All block-masks containing at least four maskings assert at least one line in each direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if we wish to place a block-mask with at least some number of maskings we necessarily assert some minimum number of lines in each of the two directions:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;minimum maskings  minimal assertions&#xA;       9               3, 3&#xA;       8               3, 3&#xA;       7               3, 3&#xA;       6               1, 3&#xA;       5               1, 1&#xA;       4               1, 1&#xA;       3               0, 0&#xA;       2               0, 0&#xA;       1               0, 0&#xA;       0               0, 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;When placing a block-mask we are constrained to some maximum number of maskings by the combinations of currently placed block-masks in the block's band and stack:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;     stack blocks asserting: 0 &amp;amp; 0  0 &amp;amp; 1  0 &amp;amp; 3  1 &amp;amp; 1&#xA;band blocks asserting:&#xA;                     0 &amp;amp; 0     9      6      3      3      &#xA;                     0 &amp;amp; 1     6      4      2      1                       &#xA;                     0 &amp;amp; 3     3      2      1      1&#xA;                     1 &amp;amp; 1     3      1      1      1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This analysis ignores the effects of blocks asserting two band-lines or two stack-lines and the specifics of which lines in particular are asserted, but the maximums still hold since asserting more lines for a block never allows us to use more maskings in it's block-mask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, rather than excluding specific nets due to them either causing ambiguity, or not containing enough maskings - we may instead exclude nets of certain forms due to them either causing ambiguity or not being able to contain enough maskings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that any valid net is also valid under any combination of permutations of: bands, stacks, in-band rows, and in-stack columns&#xA;- since if it were not then a sudoku causing ambiguity in the permuted net could be transformed inversely (which is also necessarily a valid soduku) to cause ambiguity in the original net.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When implementing the search, to avoid isomorphisms, we can first place the main diagonal from top-left to bottom-right such that no placed block-mask has maximal masking greater than the previous one placed (since we could permute bands) and then place two more block-masks completing a second diagonal (e.g. place middle-right and bottom-centre, completing a second diagonal with top-left) such that this second diagonal does not maximally mask more than the main diagonal (since we could permute stacks). Furthermore we should only place a main diagonal if it may contain more than 11 maskings - since we aim to find solutions where the total is greater than 33 and therefore at least one positive diagonal must contain more than 33/3 (the same would be true of any partition into three sets of size three including stacks, bands, and negative diagonals).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Running a search starts starts like so:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                     3                  3                  3&#xA;  9   9   9         393  3   3         393  1   3         393  1   1&#xA;                     3                  3                  3&#xA;                                            3                  3  &#xA;  9   9   9          3   9   9          1  393  3          1  393  1&#xA;                                            3                  3  &#xA;                                                                   3&#xA;  9   9   9          3   9   9          3   3   9          1   1  393&#xA;                                                                   3&#xA;&#xA;max:  81                 57                 41                  33&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and yields a single form we need to investigate:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  3   0   0&#xA; 393 010 030&#xA;  3   0   0&#xA;&#xA;  0   3   0&#xA; 010 393 030&#xA;  0   3   0&#xA;&#xA;  0   0   0&#xA; 030 030 030&#xA;  0   0   0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This has a maximal masking of 35.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a nine were reduced to an eight that block-mask would still assert all three lines in both directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Top-right and middle-right both have all their band-lines asserted, so by (1) may not assert any band-lines; and by (7), if one of them asserts any stack-lines the other may only contain a maximum of two maskings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly bottom-left and bottom-centre have all their stack-lines asserted, so by (1) may not assert any stack-lines; and by (7), if one of them asserts any band-lines the other may only contain a maximum of two maskings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to achieve greater than 33 at least one of these pairs of blocks would need to contain a block-mask containing three maskings which asserts no lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, by (3), (6) and (c), the other of the pair would be reduced to only allowing one masking - here are two (of the 18 ambiguous cases) for each of three of the (six) ways we could place top-right.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;..#|...|#..    ..#|...|#..    ..!|...|..!    ..!|...|..!    ..#|...|#..    ..#|...|#..&#xA;..!|...|..!    ..!|...|..!    ..#|...|#..    ..#|...|#..    ..!|...|.!.    ..!|...|.!.&#xA;..!|...|.!.    ..!|...|.!.    ..!|...|.!.    ..!|...|.!.    ..!|...|..!    ..!|...|..!&#xA;---+---+---    ---+---+---    ---+---+---    ---+---+---    ---+---+---    ---+---+---&#xA;...|!..|.!.    ...|!..|..!    ...|!..|.!.    ...|!..|..!    ...|!..|.!.    ...|!..|..!&#xA;...|!..|..!    ...|!..|.!.    ...|!..|..!    ...|!..|.!.    ...|!..|..!    ...|!..|.!.&#xA;...|#..|...    ...|#..|...    ...|#..|...    ...|#..|...    ...|#..|...    ...|#..|...&#xA;---+---+---    ---+---+---    ---+---+---    ---+---+---    ---+---+---    ---+---+---&#xA;...|...|...    ...|...|...    ...|...|...    ...|...|...    ...|...|...    ...|...|...&#xA;...|...|...    ...|...|...    ...|...|...    ...|...|...    ...|...|...    ...|...|...&#xA;...|...|...    ...|...|...    ...|...|...    ...|...|...    ...|...|...    ...|...|...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The maximum achievable across each of these two pairs of blocks is now four rather than six reducing this form's maximal masking to 33 (without even needing to consider bottom-right).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, Spencerkatty's solution is maximal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;python code for the search:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from copy import deepcopy&#xA;from itertools import product&#xA;&#xA;# lookups for the maximal masking possible due to crossing of band-line assertions and stack-line assertions from two blocks.&#xA;CONSTRAIN_MASKINGS = [[9, 6, 6, 3], [6, 4, 4, 2], [6, 4, 4, 2], [3, 2, 2, 1]]&#xA;&#xA;# lookup for maximal band-line and stack-line assertions that may be made given a crossing of band-line assertions and stack-line assertions from two blocks.&#xA;CONSTRAIN_ASSERTIONS = [[(3, 3), (3, 1), (3, 1), (0, 0)], [(1, 3), (1, 1), (1, 1), (0, 0)], [(1, 3), (1, 1), (1, 1), (0, 0)], [(0, 0), (0, 0), (0, 0), (0, 0)]]&#xA;&#xA;&#xA;class BlockMaskCategory:&#xA;&#xA;    def __init__(self):&#xA;        self.isPlaced = False&#xA;        self.bandAssertions = 3&#xA;        self.stackAssertions = 3&#xA;        self.maxMaskings = 9&#xA;&#xA;    def __repr__(self):&#xA;        return '''&#xA;  {0} &#xA; {1}{2}{1}&#xA;  {0} '''.format(not self.isPlaced and ' ' or self.stackAssertions, not self.isPlaced and ' ' or self.bandAssertions, self.maxMaskings)&#xA;&#xA;    def constrain(self, othersInBand, othersInStack):&#xA;        inBandPlaced = [o.bandAssertions for o in othersInBand if o.isPlaced] &#xA;        inStackPlaced = [o.stackAssertions for o in othersInStack if o.isPlaced]&#xA;        inBandAsserting = [a for a in inBandPlaced if a]&#xA;        inStackAsserting = [a for a in inStackPlaced if a]&#xA;        if (len(inBandAsserting) == 2 or 3 in inBandAsserting) and (len(inStackAsserting) == 2 or 3 in inStackAsserting):&#xA;            maxMaskings = 1&#xA;            bandAssertions = 0&#xA;            stackAssertions = 0&#xA;        elif inStackAsserting and (len(inBandAsserting) == 2 or 3 in inBandAsserting):&#xA;            maxMaskings = 2&#xA;            bandAssertions = 0&#xA;            stackAssertions = 0&#xA;        elif inBandAsserting and (len(inStackAsserting) == 2 or 3 in inStackAsserting):&#xA;            maxMaskings = 2&#xA;            bandAssertions = 0&#xA;            stackAssertions = 0&#xA;        else:&#xA;            if len(inBandAsserting) == 2:&#xA;                maxMaskings = 3&#xA;                b = 3&#xA;                s = 0&#xA;            elif len(inStackAsserting) == 2:&#xA;                maxMaskings = 3&#xA;                b = 0&#xA;                s = 3&#xA;            else:&#xA;                maxMaskings = 9&#xA;                b = 0&#xA;                s = 0&#xA;            for bandAssertions, stackAssertions in product(inBandPlaced or [0], inStackPlaced or [0]):&#xA;                m = CONSTRAIN_MASKINGS[bandAssertions][stackAssertions]&#xA;                if m &amp;lt; maxMaskings:&#xA;                    maxMaskings = m&#xA;                    b = bandAssertions&#xA;                    s = stackAssertions&#xA;            bandAssertions, stackAssertions = CONSTRAIN_ASSERTIONS[b][s]&#xA;&#xA;        if self.isPlaced:&#xA;            if self.maxMaskings &amp;gt; maxMaskings:&#xA;                return False&#xA;        else:&#xA;            self.maxMaskings = maxMaskings&#xA;            self.bandAssertions = bandAssertions&#xA;            self.stackAssertions = stackAssertions&#xA;        return True&#xA;&#xA;    def choices(self):&#xA;        if self.maxMaskings &amp;gt; 6:&#xA;            return [(3, 3, self.maxMaskings), (3, 1, 6), (1, 3, 6), (1, 1, 5), (1, 1, 4), (0, 0, 3), (0, 0, 2), (0, 0, 1)]&#xA;        if self.maxMaskings == 6:&#xA;            return [(self.bandAssertions, self.stackAssertions, self.maxMaskings), (1, 1, 5), (1, 1, 4), (0, 0, 3), (0, 0, 2), (0, 0, 1)]&#xA;        if self.maxMaskings &amp;gt; 3:&#xA;            return [(1, 1, m) for m in range(self.maxMaskings, 3, -1)] + [(0, 0, 3), (0, 0, 2), (0, 0, 1)]&#xA;        return [(0, 0, m) for m in range(self.maxMaskings, 0, -1)]&#xA;&#xA;&#xA;class Net:&#xA;&#xA;    def __init__(self):&#xA;        iters = tuple(range(3))&#xA;        self.blocks = [[BlockMaskCategory() for stack in iters] for band in iters]&#xA;        self.otherStackIndexes = [tuple((stack - 1, stack - 2)) for stack in iters]&#xA;        self.otherBandIndexes = [tuple((band - 1, band - 2)) for band in iters]&#xA;        self.othersInBand = [[[self.blocks[band][si] for si in self.otherStackIndexes[stack]] for stack in iters] for band in iters]&#xA;        self.othersInStack = [[[self.blocks[bi][stack] for bi in self.otherBandIndexes[band]] for stack in iters] for band in iters]&#xA;&#xA;    def __repr__(self):&#xA;        return '\n'.join(['\n'.join([''.join(s) for s in zip(*[blockMask.__repr__().split('\n') for blockMask in band])]) for band in self.blocks]) + '    = {0}\n'.format(self.maxMaskings())&#xA;&#xA;    def maxMaskings(self):&#xA;        return sum(sum(blockMask.maxMaskings for blockMask in band) for band in self.blocks)&#xA;&#xA;    def update(self, band, stack, bandAssertions, stackAssertions, maxMaskings, remove=False):&#xA;        block = self.blocks[band][stack]&#xA;        block.bandAssertions = bandAssertions&#xA;        block.stackAssertions = stackAssertions&#xA;        block.maxMaskings = maxMaskings&#xA;        block.isPlaced = not remove&#xA;        for otherInBand, si in zip(self.othersInBand[band][stack], self.otherStackIndexes[stack]):&#xA;            if not otherInBand.constrain(self.othersInBand[band][si], self.othersInStack[band][si]):&#xA;                return False&#xA;        for otherInStack, bi in zip(self.othersInStack[band][stack], self.otherBandIndexes[band]):&#xA;            if not otherInStack.constrain(self.othersInBand[bi][stack], self.othersInStack[bi][stack]):&#xA;                return False&#xA;        return True&#xA;&#xA;    def iterSolutions(self, verbose=False):&#xA;        queue = [(1, 0), (0, 1), (2, 0), (0, 2), (2, 1), (1, 2), (2, 2), (1, 1), (0, 0)]&#xA;        return self._iterSolutions(queue, 0, verbose)&#xA;&#xA;    def _iterSolutions(self, queue, last, verbose):&#xA;        if not queue:&#xA;            if verbose:&#xA;                print(&quot;** found a solution, yielding it...&quot;)&#xA;            yield deepcopy(self)&#xA;            return&#xA;        if len(queue) == 6:&#xA;            if sum(self.blocks[i][i].maxMaskings for i in (0, 1, 2)) &amp;lt;= 11:&#xA;                if verbose:&#xA;                    print(&quot;symmetry restriction: main diagonal is below the average requirement&quot;)&#xA;                return&#xA;        elif len(queue) == 4:&#xA;            if sum(self.blocks[i][i].maxMaskings for i in (0, 1, 2)) &amp;lt; sum(self.blocks[b][s].maxMaskings for (b, s) in ((0, 0), (1, 2), (2, 1))):&#xA;                if verbose:&#xA;                    print(&quot;symmetry restriction: second placed diagonal is more maximally masking than the main diagonal&quot;)&#xA;                return&#xA;        cmpLast = 6 &amp;lt; len(queue) &amp;lt; 9&#xA;        band, stack = queue.pop()&#xA;        block = self.blocks[band][stack]&#xA;        bandAssertionsPrior = block.bandAssertions&#xA;        stackAssertionsPrior = block.stackAssertions&#xA;        maxMaskingsPrior = block.maxMaskings&#xA;        for bandAssertions, stackAssertions, maxMaskings in block.choices():&#xA;            if cmpLast and maxMaskings &amp;gt; last:&#xA;                if verbose:&#xA;                    print(&quot;symmetry restriction: main diagonal not in descending order; {0} &amp;gt; {1}&quot;.format(maxMaskings, last))&#xA;                continue&#xA;            if self.update(band, stack, bandAssertions, stackAssertions, maxMaskings):&#xA;                if verbose:&#xA;                    print(&quot;update to:&quot;)&#xA;                    print(self)&#xA;                if self.maxMaskings() &amp;gt; 33:&#xA;                    for solution in self._iterSolutions(queue[:], maxMaskings, verbose):&#xA;                        yield solution&#xA;                elif verbose:&#xA;                    print(&quot;dead end: {0} &amp;lt;= 33&quot;.format(self.maxMaskings()))&#xA;            elif verbose:&#xA;                print(&quot;Symmetry restriction: adding {0} (asserting {1} band-lines and {2} stack-lines) at band {3}, stack {4} would further constrain an already placed block-mask&quot;.format(maxMaskings, bandAssertions, stackAssertions, band, stack))&#xA;            if not self.update(band, stack, bandAssertionsPrior, stackAssertionsPrior, maxMaskingsPrior, True):&#xA;                raise RuntimeError(&quot;Failed to remove block-mask&quot;)&#xA;            # if verbose:&#xA;                # print(&quot;back track&quot;)&#xA;                # print(self)&#xA;&#xA;&#xA;if __name__ == &quot;__main__&quot;:&#xA;    n = Net()&#xA;    for solution in n.iterSolutions():&#xA;        solution&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;python code for producing the block-mask domain so as to be able to investigate potential partitions:&#xA;Note - a blockMask's representation shows lines asserted and maskings on the left, reflections with their polarities (an x being both) in the middle and reflections that matter on the right (since if a block mask asserts one of the crossing lines the reflection is of no consequence in the case of single reflections).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;SIZE = 3&#xA;ITEMS = tuple(range(SIZE))&#xA;PAIRS = tuple((a, b) for a in ITEMS for b in ITEMS if a &amp;lt; b)&#xA;NO_MASKINGS = [[False for col in ITEMS] for row in ITEMS]&#xA;OPAQUE_CHARS = ('1', '#')&#xA;TRANSPARENT_CHARS = ('0', '.')&#xA;&#xA;class BlockMask:&#xA;&#xA;    def __init__(self, maskings=NO_MASKINGS):&#xA;        if isinstance(maskings, str):&#xA;            maskingChars = OPAQUE_CHARS + TRANSPARENT_CHARS&#xA;            vals = [maskingChar in OPAQUE_CHARS for maskingChar in maskings if maskingChar in maskingChars]&#xA;            if len(vals) != SIZE ** 2:&#xA;                raise ValueError(&quot;Expected 9 characters from OPAQUE_CHARS + TRANSPARENT_CHARS = {0} + {1}&quot;.format(OPAQUE_CHARS, TRANSPARENT_CHARS))&#xA;            vals = [[vals[SIZE * row + col] for col in ITEMS] for row in ITEMS]&#xA;        elif len(maskings) == SIZE ** 2:&#xA;            vals = [[maskings[SIZE * row + col] and True or False for col in ITEMS] for row in ITEMS]&#xA;        elif len(maskings) == SIZE:&#xA;            vals = []&#xA;            for bandMaskings in maskings:&#xA;                if len(bandMaskings) != SIZE:&#xA;                    raise ValueError(&quot;Got an Iterable of length {0} but one of it's members was not also of length {0}&quot;.format(SIZE))&#xA;                vals.append([masking and True or False for masking in bandMaskings])&#xA;        else:&#xA;            raise ValueError(&quot;Expected either a string representation, a flat Iterable of length {0}, or a nested Iterable length {1} containing Iterables of length {1}&quot;.format(SIZE ** 2, SIZE))&#xA;&#xA;        assertsBandLines = set()&#xA;        for bandLine, (rowA, rowB) in enumerate(PAIRS):&#xA;            for col in ITEMS:&#xA;                if vals[rowA][col] and vals [rowB][col]:&#xA;                    assertsBandLines.add(bandLine)&#xA;                    break&#xA;&#xA;&#xA;        assertsStackLines = set()&#xA;        for stackLine, (colA, colB) in enumerate(PAIRS):&#xA;            for row in ITEMS:&#xA;                if vals[row][colA] and vals [row][colB]:&#xA;                    assertsStackLines.add(stackLine)&#xA;                    break&#xA;&#xA;        reflectionsFromBandLines = [set() for band in ITEMS]&#xA;        reflectionsFromBandLinesExclusive = [set() for band in ITEMS]&#xA;        reflectionsBandLinesExclusive = set()&#xA;        reflectionsFromStackLines = [set() for stack in ITEMS]&#xA;        reflectionsFromStackLinesExclusive = [set() for band in ITEMS]&#xA;        reflectionsStackLinesExclusive = set()&#xA;        reflectionPolarities = {}&#xA;        reflectionPolaritiesExclusive = set()&#xA;        for bandLine, (rowA, rowB) in enumerate(PAIRS):&#xA;            excludedBandLine = bandLine in assertsBandLines&#xA;            for stackLine, (colA, colB) in enumerate(PAIRS):&#xA;                neg = vals[rowA][colA] and vals [rowB][colB]&#xA;                pos = vals[rowB][colA] and vals [rowA][colB]&#xA;                if neg or pos:&#xA;                    reflectionsFromBandLines[bandLine].add(stackLine)&#xA;                    reflectionsFromStackLines[stackLine].add(bandLine)&#xA;                    reflectionPolarities[(bandLine, stackLine)] = pos - neg&#xA;                    if not (excludedBandLine or stackLine in assertsStackLines):&#xA;                        reflectionsFromBandLinesExclusive[bandLine].add(stackLine)&#xA;                        reflectionsBandLinesExclusive.add(bandLine)&#xA;                        reflectionsFromStackLinesExclusive[stackLine].add(bandLine)&#xA;                        reflectionsStackLinesExclusive.add(stackLine)&#xA;                        reflectionPolaritiesExclusive.add((bandLine, stackLine, pos - neg))&#xA;&#xA;        self.maskings = vals&#xA;        self.nMaskings = sum(sum(row) for row in vals)&#xA;        self.assertsBandLines = assertsBandLines&#xA;        self.assertsStackLines = assertsStackLines&#xA;        self.reflectionsFromBandLines = reflectionsFromBandLines&#xA;        self.reflectionsFromBandLinesExclusive = reflectionsFromBandLinesExclusive&#xA;        self.reflectionsBandLinesExclusive = reflectionsFromBandLinesExclusive&#xA;        self.reflectionsFromStackLines = reflectionsFromStackLines&#xA;        self.reflectionsFromStackLinesExclusive = reflectionsFromStackLinesExclusive&#xA;        self.reflectionsStackLinesExclusive = reflectionsStackLinesExclusive&#xA;        self.reflectionPolarities = reflectionPolarities&#xA;        self.reflectionPolaritiesExclusive = reflectionPolaritiesExclusive&#xA;&#xA;    def __repr__(self):&#xA;        r = '\n '&#xA;        rla = ' '&#xA;        for y in ITEMS:&#xA;            rla += y in self.assertsStackLines and '|' or ' '&#xA;        r += rla + '\n '&#xA;        for x in ITEMS:&#xA;            bla = x in self.assertsBandLines and '-' or ' '&#xA;            r += bla&#xA;            for y in ITEMS:&#xA;                r += self.maskings[x][y] and '#' or '.'&#xA;            r += bla + '   '&#xA;            for y in ITEMS:&#xA;                r += y in self.reflectionsFromBandLines[x] and ('x', '+', '-')[self.reflectionPolarities[(x, y)]] or '.'&#xA;            r += '    '&#xA;            for y in ITEMS:&#xA;                r+= y in self.reflectionsFromBandLinesExclusive[x] and ('x', '+', '-')[self.reflectionPolarities[(x, y)]]  or '.'&#xA;            r += '\n '&#xA;        r += rla + '\n'&#xA;        return r&#xA;&#xA;&#xA;def blockMaskIter(nMaskings, cur=[0 for i in range(9)]):&#xA;    if nMaskings &amp;lt; 1:&#xA;        yield BlockMask(cur)&#xA;        return&#xA;    if 1 in cur:&#xA;        s = 9 - cur[::-1].index(1)&#xA;    else:&#xA;        s = 0&#xA;    for pos in range(s, 10 - nMaskings):&#xA;        next = cur[:]&#xA;        next[pos] = 1&#xA;        for mask in blockMaskIter(nMaskings - 1, next):&#xA;            yield mask&#xA;&#xA;def domainIter():&#xA;    for nMaskings in range(9, -1, -1):&#xA;        for blockMask in blockMaskIter(nMaskings):&#xA;            yield blockMask&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="16717" LastEditorUserId="16717" LastEditDate="2016-04-10T09:33:20.403" LastActivityDate="2016-04-10T09:33:20.403" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23676" PostTypeId="1" CreationDate="2015-10-30T15:51:34.143" Score="5" ViewCount="650" Body="&lt;p&gt;Recently I travelled far distance by train and a stranger told me this riddle but I never got the answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;100 years ago, when far distance travelling was something special, a traveller came back from far countries to his town. He was in 16 different countries and from each country he brought back 4 treasures. He hid each treasure in a different secret location. As those were gorgeous he wanted to show them to the people in the town who were all 512 honest people. At least he thought so. He heared the rumor that 4 people were not that honest and will steal the gorgeous treasures to get them for themself in the next month after the treasures were shown to all. But this wouldnt stop him from showing the culture of the far countries to the people. Each person will be shown 1 treasure of each country but he can choice which of the 4 treasures of each country to show whom. The honest people would never tell anybody else the location of the treasures they know but also accusing a honest person of stealing is also out of question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can he identify the dishonest people after the month and how?&#xA;What is the minimum amount of treasures per country/maximum count of dishonest/honest people? Is there a formula/strategy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach was: Show each person a unique combination of treasures and find the dishonest one according to the combination. But it only works for a single dishonest person.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clarification: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;He hides the treasures&lt;/li&gt;&#xA;&lt;li&gt;He tells the people locations&lt;/li&gt;&#xA;&lt;li&gt;4 people steal every treasure they know&lt;/li&gt;&#xA;&lt;li&gt;He notices the theft&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;He only see what they stole together, not what each individual stole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some research I found the minimum number of possible combinations of stolen treasures. I put it in spoiler in case someone wants to solve it on their own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number of combintations:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\binom{512}{4}=\frac {512!}{4!*(512-4)!}=2,829,877,120$$&lt;br&gt;&#xA; (See &lt;a href=&quot;https://en.wikipedia.org/wiki/Combination&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Combination&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also I solved it by bruteforce for 4 people, 2 dishonest people, 3 countries, 2 treasures per country:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;&#xA;1 1 1&#xA;1 1 2&#xA;1 2 1&#xA;2 1 1&lt;/code&gt;&#xA;where the number tells us if we show the treasure 1 or 2 to a person and each line is for a different person and each row is for a country. The possible outcomings are:&#xA;&lt;code&gt;&#xA;1 1 3&#xA;1 3 1&#xA;1 3 3&#xA;3 1 1&#xA;3 1 3&#xA;3 3 1&lt;/code&gt;&#xA;&#xA;Each cell is the sum of the numbers of the stolen treasures in a country.&#xA;We have a population of 4. We get 6 combination from this which also supports my found formula. $$\binom{4}{2}=\frac {4!}{2!*(4-2)!}=6$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17184" LastEditorUserId="17184" LastEditDate="2015-11-05T19:54:58.143" LastActivityDate="2015-11-05T19:54:58.143" Title="Find multiple dishonest people" Tags="|logical-deduction|optimization|" AnswerCount="2" CommentCount="16" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23679" PostTypeId="2" ParentId="23676" CreationDate="2015-10-30T17:57:21.897" Score="1" Body="&lt;p&gt;This solution is to approach this problem with a divide-and-conquer strategy. You are right about the unique key. What mainly gives this away is the fact that all numbers we're given are powers of 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is to divide the original 512 people into groups of even and odd people.  This will leave two groups of 256 people, each of which has seen a unique treasure.  This process will be repeated until the 8th break, in which we have a single person seeing a unique item.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note however that since each person will be seeing a treasure from a country, each of the 8 times we divide, we are showing a single country's treasure to that group of people, but each split will be seeing a different box.  So the first 256, for example, will be seeing country 1, box 1; the second 256 will be seeing country 1, treasure 2.  This will continue per layer until we reach the 9th country.  At this point we start back at the top, 512.  We continue: even 256 country 9, treasure 1, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point each person has seen a treasure from every country.  We have made two sweeps through the 512 people; once breaking them in half for the first eight countries, then the remaining 8 countries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice how each person will see a unique sequence of the 64 total treasures when we break it down this way.  Ergo, even if you are only presented with the stolen goods of all 4 people (not each individual thief), you know that there will be 4 unique items to each of the 4 thieves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can we be sure this works?  Each point in the tree that this forms will be identified by a unique code, say &quot;c1t1&quot; denoting country and treasure.  There will only be 256 people with &quot;c1t1&quot; and only 128 with &quot;c2t2&quot;.  There will be no one with &quot;c2t3&quot; with &quot;c1t1&quot;.  This allows us to identify the culprits by identifying the only logical course through our tree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using your above example (the one you solved with brute force) with this strategy:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  2 people see country 1, object 1 (AB). //ABCD are the 4 townspeople &#xA;  2 people see country 1, object 2 (CD).&#xA;  1 person sees country 2, object 1 (A).&#xA;  1 person sees country 2, object 2 (B).&#xA;  1 person sees country 3, object 1 (C).&#xA;  1 person sees country 3, object 2 (D).&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Suppose we now have 2 thieves (dishonest people).  If object 1, country 3 is missing and object 2 country 1 is missing, we know who the culprits are immediately.  Notice that once a unique item of this kind is missing (and our assumption is the thieves steal unique items), then we know who they are.&lt;/p&gt;&#xA;" OwnerUserId="13920" LastEditorUserId="13920" LastEditDate="2015-11-02T15:02:33.170" LastActivityDate="2015-11-02T15:02:33.170" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="23680" PostTypeId="2" ParentId="23656" CreationDate="2015-10-30T18:38:02.813" Score="0" Body="&lt;p&gt;What about...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 / treE(3)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think you can do a divide sign...&lt;/p&gt;&#xA;" OwnerUserId="17188" LastActivityDate="2015-10-30T18:38:02.813" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23681" PostTypeId="1" AcceptedAnswerId="23684" CreationDate="2015-10-30T18:38:57.233" Score="11" ViewCount="1006" Body="&lt;p&gt;Bringing back the Backwards Hangman (created by &lt;a href=&quot;https://puzzling.stackexchange.com/users/10037/alex&quot;&gt;Alex&lt;/a&gt;, see &lt;a href=&quot;https://puzzling.stackexchange.com/questions/11281/the-hangman-game-backward&quot;&gt;here&lt;/a&gt; for the first).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;fill in an &quot;f&quot; - The devil may carry.&lt;br&gt;&#xA;fill in a &quot;d&quot; - Probably has glasses.&lt;br&gt;&#xA;fill in a &quot;c&quot; - Popped at the party.&lt;br&gt;&#xA;fill in a &quot;w&quot; - A lot of this went into the first 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As always, your answer should be in form of: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The question was: _ p p l e&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;followed by explanation of how it fits each clue. &lt;/p&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-10-30T18:49:10.200" Title="Don't get caught by the Hang Man" Tags="|riddle|english|letters|seasonal|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23682" PostTypeId="2" ParentId="23662" CreationDate="2015-10-30T18:42:28.040" Score="1" Body="&lt;p&gt;EDIT: I'm an idiot and answered the wrong question. Disregard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Call the initial positions of fleas A, B, C, and D &lt;code&gt;(0,0)&lt;/code&gt;, &lt;code&gt;(1,0)&lt;/code&gt;, &lt;code&gt;(0,1)&lt;/code&gt;, and &lt;code&gt;(1,1)&lt;/code&gt; respectively. At any point in time, two fleas will be at a distance &lt;code&gt;(n,m)&lt;/code&gt; away, where &lt;code&gt;n&lt;/code&gt; and &lt;code&gt;m&lt;/code&gt; are integers. Pivoting will move one flea a distance &lt;code&gt;(2n,2m)&lt;/code&gt;. Flea A can only occupy the points &lt;code&gt;(2n,2m)&lt;/code&gt;, flea B &lt;code&gt;(1+2n,2m)&lt;/code&gt;, flea C &lt;code&gt;(2n,1+2m)&lt;/code&gt;, and flea D &lt;code&gt;(1+2n,1+2m)&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="17189" LastEditorUserId="17189" LastEditDate="2015-10-30T18:50:42.533" LastActivityDate="2015-10-30T18:50:42.533" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23683" PostTypeId="2" ParentId="23666" CreationDate="2015-10-30T18:43:08.970" Score="4" Body="&lt;p&gt;Using actual digits (instead of letters or concepts), I present&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Be7Vb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Be7Vb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or $9^{11111111111111}$&lt;/p&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2015-10-30T18:43:08.970" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23684" PostTypeId="2" ParentId="23681" CreationDate="2015-10-30T18:47:49.430" Score="9" Body="&lt;p&gt;the question was  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; _ o r k&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fill in an &quot;f&quot; - The devil may carry.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the devil's pitchfork&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fill in a &quot;d&quot; - Probably has glasses. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; dork / nerd stereotypically wears glasses&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fill in a &quot;c&quot; - Popped at the party. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; cork from a wine bottle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fill in a &quot;w&quot; - A lot of this went into the first 3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; work, it certainly did!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13101" LastActivityDate="2015-10-30T18:47:49.430" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23685" PostTypeId="2" ParentId="23681" CreationDate="2015-10-30T18:48:01.170" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; _ ork&lt;br&gt;&#xA; Devil carries a (pitch)fork&lt;br&gt;&#xA; Dorks often have glasses&lt;br&gt;&#xA; Corks are popped when opening champagne&lt;br&gt;&#xA; Work was done for this puzzle. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-10-30T18:48:01.170" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23686" PostTypeId="2" ParentId="23681" CreationDate="2015-10-30T18:49:10.200" Score="10" Body="&lt;p&gt;Question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; _ o r k&lt;br&gt;&#xA;&lt;br&gt;&#xA; The devil carries a pitchfork &#xA;&lt;br&gt;&#xA; A dork wears glasses &#xA;&lt;br&gt;&#xA; You pop the cork off champagne at a party &#xA;&lt;br&gt;&#xA; You put work into this question &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3980" LastActivityDate="2015-10-30T18:49:10.200" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23687" PostTypeId="2" ParentId="23656" CreationDate="2015-10-30T18:52:43.397" Score="3" Body="&lt;p&gt;What about  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/OPJev.png&quot; alt=&quot;this&quot;&gt;.&lt;br&gt;&#xA; Using Hex, that's A^-FEDCB  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Equivalent to 10^-1043915&lt;/p&gt;&#xA;" OwnerUserId="16308" LastActivityDate="2015-10-30T18:52:43.397" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23688" PostTypeId="2" ParentId="15034" CreationDate="2015-10-30T19:13:58.697" Score="6" Body="&lt;p&gt;Here's my attempt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The clue is &quot;SIAMESE&quot; repeated 5 times, and it points towards &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the &lt;a href=&quot;https://en.wikipedia.org/wiki/Siamese_method&quot;&gt;Siamese method&lt;/a&gt; for the creation of 5x5 magic squares.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we use that sequence to read the grids, we have:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BESTBOUNDSFORRAMFIVESEVEN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HOWMANYSQUARESARELIKETHIS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, the second one is the easiest: it is known that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the total amount of possible 5x5 magic squares is 275305224.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The hint comes handy now:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;mail me something&quot;, that number is a ZIP code: Goldsboro, North Carolina.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first one gave me a headache: after a long search I came across&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ramsey%27s_theorem#Ramsey_numbers&quot;&gt;Ramsey numbers&lt;/a&gt;. The phrase is asking for the lower and upper bounds of R(5,7), which are 80-143.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, if you look that up,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you find that's the Italian code for Naples.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the final answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; NAPLES --&gt; GOLDSBORO&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5990" LastActivityDate="2015-10-30T19:13:58.697" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23689" PostTypeId="2" ParentId="23662" CreationDate="2015-10-30T19:22:57.030" Score="9" Body="&lt;p&gt;WLOG the fleas start at the corners of the unit square, with A at $(0,0)$, B at $(1,0)$, C at $(1,1)$, and D at $(0,1)$. I claim that after $n$ jumps, A will be at $(\frac{2a_1}{3^n},\frac{2a_2}{3^n})$, B will be at $(\frac{2b_1+1}{3^n},\frac{2b_2}{3^n})$, C will be at $(\frac{2c_1+1}{3^n},\frac{2c_2+1}{3^n})$, and D will be at $(\frac{2d_1}{3^n},\frac{2d_2+1}{3^n})$ for some integers $a_1,a_2,b_1,b_2,c_1,c_2,d_1,d_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is true in the starting position, with $n=0$. Now suppose that it is true after $n$ jumps, and A jumps next. The centroid of B, C, and D is $(\frac{2b_1+1+2c_1+1+2d_1}{3^{n+1}},\frac{2b_2+2c_2+1+2d_2+1}{3^{n+1}})$, and A ends up at the point $(\frac{4b_1+4c_1+4d_1+4-6a_1}{3^{n+1}},\frac{4b_2+4c_2+4d_2+4-6a_2}{3^{n+1}})=(\frac{2(2b_1+2c_1+2d_1+2-3a_1)}{3^{n+1}},\frac{2(2b_2+2c_2+2d_2+2-3a_2)}{3^{n+1}})$. Meanwhile, B stays at $(\frac{2b_1+1}{3^n},\frac{2b_2}{3^n})=(\frac{2(3b_1+1)+1}{3^{n+1}},\frac{2(3b_2)}{3^{n+1}})$, and similarly for C and D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar calculations show that if B, C, or D jumps instead of A, the claim still holds. Therefore it holds for all $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore no two fleas can ever be at the same position, because at least one coordinate's numerator would have to be both even and odd.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-10-30T22:07:38.287" LastActivityDate="2015-10-30T22:07:38.287" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23690" PostTypeId="1" AcceptedAnswerId="23744" CreationDate="2015-10-30T21:36:51.280" Score="15" ViewCount="975" Body="&lt;p&gt;It's the eve of Halloween, and everyone is gathering for the Monster Bash at the Monster Mansion.  The same haunts are hosting this year's event: Count Dracula, Frankenstein's Monster, Wolfman, Mummy, and the Gillman, visiting from the Black Lagoon.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally, their parties are pretty exclusive, but you've gotten friendly with the monsters as of late and they have agreed to let you invite some other humans to the festivities, provided that &lt;em&gt;you&lt;/em&gt; at least, wear a costume.  You don your ever-ready Sherlock Holmes costume and head out, confident you'll soon see all the people you invited.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;The Night of the Party&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The door opens and the mansion looks finer than it ever has- one could say that the addition of a little fresh blood really &lt;em&gt;livened&lt;/em&gt; the place up.  Glad to have finally made it, you promptly start up the Monster Mash and dance like there's no tomorrow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you stop spinning and shaking, however, you notice that the party has, well, &lt;em&gt;died&lt;/em&gt;.  Everyone seems to be in a catatonic daze, but for the monsters and yourself.  They're zombies, just not the flesh eating kind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are obviously shocked, concerned, and confused.  You really love spending time with the monsters and don't want them to fall into trouble, but you also feel responsible for all the humans that you invited to the party.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You decide to embody your costume a bit more, and deduce who (if anyone) among the monsters could have been responsible. But before doing that, you want to quench your thirst after all that dancing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;That's when you notice that all of the other &lt;em&gt;people&lt;/em&gt; have cups in their hands.  Something was added to the &quot;bloodbath&quot; punch, this much you are sure.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Making your way over to the line of buffet tables, you see trays of food to on side, illuminated by candles.  The glare of the moon glances off the punch bowl on the other side, and you, Drac, Frank, Wolf, Mum, and Gill gather around, peering in at the contents.  The distorted shapes of your six sets of eyes add to the many candy eyes floating in the punch.  You look up and carefully survey your suspects.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Suspects&lt;/h1&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Frankenstein's Monster&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Pulling Frank aside, you can immediately tell that he is upset, on the brink of tears. &#xA;  &lt;br/&gt; Because it seems a logical first question, you ask him &quot;Why are you upset?&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Frank sniffles before ansering &quot;I.. make...punch....bad...&quot;&#xA;  Of course, you think; Frank loves working in the kitchen because it reminds him of the doctor's lab.  Frank certainly isn't the brightest of the bunch, so he probably blames himself for whatever the punch did to the guests.  Additionally, since he is so dim, there's no way he could have planned something like lacing a punch.  Then again...the snuffed candles on the drink side of the buffet might mean something.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Count Dracula&lt;/strong&gt;&lt;br/&gt;&lt;br/&gt;&#xA;  Allowing Dracula to sidle up on you instead of pulling him aside, you go for the obvious question.  &quot;Drac, bloodbath punch is your favorite alternative to actual blood.  You wouldn't have requested it, just to poison everyone here, right?  That would make you an easy culprit.&quot;&lt;br/&gt;&lt;br/&gt;&quot;Of course not,&quot; he scoffs, &quot;I was sleeping through all the preparations today, and someone would have noticed if I'd added anything once the party started.  Besides, the punch bowl has a silver ladle.  Whether I wanted to mix something in, or just get a drink, I'd be putting myself at risk getting near the bowl.&quot;&lt;br/&gt;&lt;br/&gt;He did seem a little skittish peering over the bowl, you recall...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Wolfman&lt;/strong&gt;&lt;br/&gt;&lt;br/&gt;&#xA;  The Wolfman scratches himself uneasily as you approach him.  It's not easy to look in any of the monster's eyes for truth, but Wolf is usually a bit easier to read, since he looks so much like your dog.&lt;br/&gt;&lt;br/&gt;&#xA;  &quot;Don't you have a history with gypsies, Wolf, the king that make all sort of potions and incantations? It seems like we might have a bit of that going on here.&quot;&lt;br/&gt;&lt;br/&gt;&#xA;  &quot;Not a chance this is from me.  You don't know how hard I work to stop myself from hurting people.  And Drac and I are in the same boat, I don't want to touch that silver ladle either.&quot;  Right.  You knew that.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Mummy&lt;/strong&gt;&lt;br/&gt;&lt;br/&gt;The Mummy is pacing back and forth, so you pull him over to the buffet tables and give him a reassuring pat on the shoulder.  A puff of dust flies from him and you notice the table is already coated in a thin layer of it.  &lt;br/&gt;&lt;br/&gt;Before you can even ask a question, Mummy blurts out &quot;I was wrapping presents!&quot;&lt;br/&gt;&#xA;  &quot;That's the wrong holiday, Mummy, nice try but that alibi won't fly.&quot;&lt;br/&gt;&lt;br/&gt;&#xA;  &quot;No, honest,&quot; he exclaims, &quot;that's what my people &lt;em&gt;do&lt;/em&gt;.  Plus, it's never too early to start on holiday presents...&quot;  You can't disagree with that, but the dust is undeniable.  Mummy is as dry as bone (and bits of flesh), so it could just be a coincidence.  But his dryness would make him wary of fires.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Gillman&lt;/strong&gt;&lt;br/&gt;&lt;br/&gt;&#xA;  You don't know Gillman as well, but you'd like to think (especially dressed as Sherlock) that you know what makes him tick.  &quot;It must have been terrible undergoing all those experiments, you know, from those scientists trying to figure out what makes you tick?  Is this your turn to conduct a little experiment of your own, Gill?&quot;&lt;br/&gt;&lt;br/&gt;&#xA;  &quot;Hey, I don't mind people.  I prefer them wet and in skimpy swimsuits, but if they mean me no harm, I mean them no harm.  And you should know that I stay away from any liquids that are not water; it would be all too easy for me to drink a coke, fill a pool with coke, then wind up with clogged gills.  You'd never see me touching anything &lt;em&gt;but&lt;/em&gt; water.&quot;  Having swam in a pool of coke yourself, you know of the slippery slope he mentioned.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;The Deduction&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Having interviewed the monster friends, you wander back to the source of the crime once more.  Standing at the punch bowl, you look for any details you could have missed, but there isn't much there to miss.  You reflect on everything that happened during the night and everything the monsters told you to prove their own innocence.  If you can find the monster responsible, you should be able to reverse whatever happened to all the humans, and you &lt;em&gt;need&lt;/em&gt; that to happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you have it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who is the culprit?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&lt;br/&gt;&#xA;&lt;em&gt;For Clarification:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;There is no vampiric/mummy mind control at play.  The monsters are friends.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;The culprit may have had help, or could have worked alone.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This was largely inspired by &lt;a href=&quot;https://puzzling.stackexchange.com/questions/20651/who-poisoned-hugh-g-pokitz-a-murder-mystery-logic-puzzle&quot;&gt;this puzzle&lt;/a&gt;, which was awesome.&lt;/p&gt;&#xA;" OwnerUserId="11452" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-11-02T16:13:36.810" Title="There's a Ghost of a Chance!" Tags="|story|mystery|seasonal|" AnswerCount="3" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23691" PostTypeId="2" ParentId="23690" CreationDate="2015-10-30T22:05:55.623" Score="1" Body="&lt;p&gt;You're a poisonous spider dressed as Sherlock Holmes that took part in preparing the punch and accidentally (or not) poisoned it?&lt;/p&gt;&#xA;" OwnerUserId="16800" LastActivityDate="2015-10-30T22:05:55.623" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23692" PostTypeId="1" AcceptedAnswerId="23693" CreationDate="2015-10-30T23:25:01.843" Score="6" ViewCount="630" Body="&lt;p&gt;Insprired by the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/23681/dont-get-caught-by-the-hang-man&quot;&gt;previous backwards hangman&lt;/a&gt; (created originally by &lt;a href=&quot;https://puzzling.stackexchange.com/users/10037/alex&quot;&gt;Alex&lt;/a&gt;, see &lt;a href=&quot;https://puzzling.stackexchange.com/questions/11281/the-hangman-game-backward&quot;&gt;here&lt;/a&gt;), which was solved so quickly, I though I'd make another one.  Hopefully slightly harder!  Three solutions in a minute suggests more demand is there and more supply needed.  I've made the clues a little more cryptic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fill in an &quot;x&quot; - Grows good for your car&lt;br&gt;&#xA;Fill in a &quot;g&quot;  - Your just deserts, and what to do to war!&lt;br&gt;&#xA;Fill in a &quot;k&quot; - Emerges only after they're gone&lt;br&gt;&#xA;Fill in an &quot;l&quot; - Land of the dragon&lt;br&gt;&#xA;Fill in a &quot;r&quot; - Pottery offered for sale&lt;br&gt;&#xA;Fill in a &quot;v&quot; - Greetings are forces of nature!&lt;br&gt;&#xA;Fill in a &quot;d&quot; - Struggles through&lt;br&gt;&#xA;Fill in an &quot;n&quot; - As new, but changes and diminishes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer will be in the form&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;W_LL  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;followed by an explanation... &lt;/p&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-11-02T01:57:11.390" Title="Yet another backwards hangman" Tags="|riddle|word|english|language|cryptic-clues|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23693" PostTypeId="2" ParentId="23692" CreationDate="2015-10-30T23:37:16.113" Score="7" Body="&lt;p&gt;The question was:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; WA_ES&lt;br&gt;&#xA; waxes : good for your car to protect the coat; grows like the waxing moon&lt;br&gt;&#xA; wages : you earned them by working, and you wage war&lt;br&gt;&#xA; wakes : &quot;emerges&quot; like awakens, and &quot;after they're gone&quot; like the wake of a boat. – credit @Charles Koppelman&lt;br&gt;&#xA; Wales : has a dragon on its flag&lt;br&gt;&#xA; wares : wares are things for sale; pottery = earthenware&lt;br&gt;&#xA; waves : a wave of the hand, a wave on the ocean&lt;br&gt;&#xA; wades : wading through a bog is quite a struggle&lt;br&gt;&#xA; wanes : diminishes like the waning moon; &quot;As new, but changes&quot; points to &quot;wanes&quot; being an anagram of &quot;as new&quot;. – credit @Zandar  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4385" LastEditorUserId="4385" LastEditDate="2015-11-02T01:57:11.390" LastActivityDate="2015-11-02T01:57:11.390" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="23694" PostTypeId="2" ParentId="12396" CreationDate="2015-10-31T07:28:23.673" Score="1" Body="&lt;p&gt;Figured i'd add my answer here, since it's slightly different from the other approaches, and you can never have enough answers to strange ghosts in impossible situations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd ask&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If you'd completely and utterly forget about the middle door, and you'd draw a cross on the right door, would THE door leading to death have a cross on it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This works because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the middle door (was) the life one, the ghost would have 2 doors leading to death, with no way to determine which one i meant by 'THE door'. However, if door 2 was death, then he now only knows of a single door that leads to death, and can answer without a problem. Resulting in Yes to right, No to left, Silence to middle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that you can't ask&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Would the door leading to life have a cross on it? Because as many a logic professor will tell you, quantifying over an empty set is always true.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16846" LastActivityDate="2015-10-31T07:28:23.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23697" PostTypeId="1" CreationDate="2015-10-31T13:53:54.297" Score="-6" ViewCount="207" Body="&lt;p&gt;What number comes next?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;11,13,17,19,23,25,?&lt;/p&gt;&#xA;" OwnerUserId="17199" LastEditorUserId="11876" LastEditDate="2015-10-31T14:44:10.143" LastActivityDate="2015-10-31T14:44:10.143" Title="What will be the missing number?" Tags="|pattern|calculation-puzzle|" AnswerCount="1" CommentCount="2" ClosedDate="2015-11-01T05:40:50.860" ContentLicense="CC BY-SA 3.0" />
  <row Id="23698" PostTypeId="2" ParentId="23697" CreationDate="2015-10-31T13:58:43.347" Score="2" Body="&lt;p&gt;It's 29. The numbers alternate increasing by 2 and 4.&lt;/p&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2015-10-31T13:58:43.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23700" PostTypeId="1" AcceptedAnswerId="23701" CreationDate="2015-10-31T15:49:31.740" Score="11" ViewCount="495" Body="&lt;p&gt;The famulus of a mathematical meta magician puts infinitely many cards &#xA;into a top hat.&lt;br&gt;&#xA;Each of these cards carries a non-negative integer.&#xA;The top hat contains&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;exactly $n$ cards showing the integer $0$,&lt;/li&gt;&#xA;&lt;li&gt;exactly $n$ cards showing the integer $1$,&lt;/li&gt;&#xA;&lt;li&gt;exactly $n$ cards showing the integer $2$,&lt;/li&gt;&#xA;&lt;li&gt;exactly $n$ cards showing the integer $3$,&lt;/li&gt;&#xA;&lt;li&gt;exactly $n$ cards showing the integer $4$,&lt;/li&gt;&#xA;&lt;li&gt;and so on.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then the mathematical meta magician starts removing cards from the top hat.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In the $1$st step, the magician takes ten cards with total value $1$.&lt;/li&gt;&#xA;&lt;li&gt;in the $2$nd step, the magician takes ten cards with total value $2$,&lt;/li&gt;&#xA;&lt;li&gt;in the $3$rd step, the magician takes ten cards with total value $3$,&lt;/li&gt;&#xA;&lt;li&gt;in the $4$th step, the magician takes ten cards with total value $4$,&lt;/li&gt;&#xA;&lt;li&gt;in the $5$th step, the magician takes ten cards with total value $5$,&lt;/li&gt;&#xA;&lt;li&gt;and so on,&lt;/li&gt;&#xA;&lt;li&gt;in the $k$th step, the magician takes ten cards with total value $k$,&lt;/li&gt;&#xA;&lt;li&gt;and so on.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What is the smallest possible integer $n$, for which &#xA;  the mathematical meta magician can make an infinite number of such &#xA;  steps (without ever getting stuck)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-10-31T16:10:57.000" Title="The mathematical meta magician" Tags="|mathematics|combinatorics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23701" PostTypeId="2" ParentId="23700" CreationDate="2015-10-31T16:10:57.000" Score="14" Body="&lt;p&gt;The following shows the optimal way to pull numbers out the hat, showing every $1$ that is used. Below, $n$ is $100$. This can obviously be continued forever.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0 0 0 0 0 0 0 0 0 1&#xA;0 0 0 0 0 0 0 0 1 1&#xA;0 0 0 0 0 0 0 1 1 1&#xA;0 0 0 0 0 0 1 1 1 1&#xA;0 0 0 0 0 1 1 1 1 1&#xA;0 0 0 0 1 1 1 1 1 1&#xA;0 0 0 1 1 1 1 1 1 1&#xA;0 0 1 1 1 1 1 1 1 1&#xA;0 1 1 1 1 1 1 1 1 1&#xA;1 1 1 1 1 1 1 1 1 1&#xA;1 1 1 1 1 1 1 1 1 2&#xA;1 1 1 1 1 1 1 1 2 2&#xA;1 1 1 1 1 1 1 2 2 2&#xA;1 1 1 1 1 1 2 2 2 2&#xA;1 1 1 1 1 2 2 2 2 2&#xA;1 1 1 1 2 2 2 2 2 2&#xA;1 1 1 2 2 2 2 2 2 2&#xA;1 1 2 2 2 2 2 2 2 2&#xA;1 2 2 2 2 2 2 2 2 2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;All that remains to be shown is that there is no way to do this with only $99$ copies of each card. Notice that on the $k^{th}$ day, the magician has drawn $10k$ cards with average value $\frac{k+1}{20}$. However, we can also bound the average of the first $10k$ cards, and if this is greater than $\frac{k+1}{20}$, the magician can't do their trick. In particular consider the $cn^{th}$ step. The lowest $10cn$ cards will be numbered from $0$ to $10c-1$, each having a set of $n$ cards. Thus, they average to $\frac{10c}2=5c$. This yields that we must have&#xA;$$\frac{cn+1}{20}\geq 5c$$&#xA;$$cn+1 \geq 100 c$$&#xA;meaning $n\geq 100$.&lt;/p&gt;&#xA;" OwnerUserId="4765" LastActivityDate="2015-10-31T16:10:57.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23702" PostTypeId="1" AcceptedAnswerId="23705" CreationDate="2015-10-31T17:51:38.203" Score="5" ViewCount="467" Body="&lt;p&gt;The square numbers $ 1^2, 2^2, 3^2, 4^2, \cdots, 100^2, 101^2 $ are written on the blackboard. &#xA;Each minute any two numbers are wiped out, and the absolute value of their difference is written instead. At the end only one number remains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the smallest value that this final number can take?&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="8874" LastEditDate="2015-11-01T20:33:56.943" LastActivityDate="2015-11-01T20:33:56.943" Title="The smallest value on the blackboard" Tags="|mathematics|reachability|blackboard|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23703" PostTypeId="2" ParentId="23702" CreationDate="2015-10-31T19:35:31.127" Score="3" Body="&lt;p&gt;If you erase $n^2$ and $(n+1)^2$ starting from $n = 2$ and going up by $2$, you end up with the numbers $1, 5, 9, \ldots, 201$, which is a total of 51 numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you then erase pairs of consecutive numbers starting from $5$ and $9$, you end up with a single $1$ and 25 $4$'s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you erase consecutive pairs of 4's until they become $0$, you're left with one $1$ and one $4$, which gives you a final number of $3$.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-10-31T19:35:31.127" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23704" PostTypeId="2" ParentId="23702" CreationDate="2015-10-31T19:55:36.723" Score="4" Body="&lt;p&gt;If you keep taking the largest 2 numbers, and substract them from eachother, you end up with &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the &lt;strong&gt;number 1&lt;/strong&gt;. This is the lowest number possible: a move does not change the parity of the number of odd integers on the blackboard. As in the beginning the blackboard does contain $51$ odd integers, it will also contain an odd number of odd integers when the process terminates. Hence the final number will always be odd.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's no very clean way to illustrate this (that i came up with) besides giving the calculations used&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 10201 - 10000 = 201, 9801 - 9604 = 197, 9409 - 9216 = 193, 9025 - 8836 = 189, 8649 - 8464 = 185, 8281 - 8100 = 181, 7921 - 7744 = 177, 7569 - 7396 = 173, 7225 - 7056 = 169, 6889 - 6724 = 165, 6561 - 6400 = 161, 6241 - 6084 = 157, 5929 - 5776 = 153, 5625 - 5476 = 149, 5329 - 5184 = 145, 5041 - 4900 = 141, 4761 - 4624 = 137, 4489 - 4356 = 133, 4225 - 4096 = 129, 3969 - 3844 = 125, 3721 - 3600 = 121, 3481 - 3364 = 117, 3249 - 3136 = 113, 3025 - 2916 = 109, 2809 - 2704 = 105, 2601 - 2500 = 101, 2401 - 2304 = 97, 2209 - 2116 = 93, 2025 - 1936 = 89, 1849 - 1764 = 85, 1681 - 1600 = 81, 1521 - 1444 = 77, 1369 - 1296 = 73, 1225 - 1156 = 69, 1089 - 1024 = 65, 961 - 900 = 61, 841 - 784 = 57, 729 - 676 = 53, 625 - 576 = 49, 529 - 484 = 45, 441 - 400 = 41, 361 - 324 = 37, 289 - 256 = 33, &lt;strong&gt;225 - 201 = 24, 197 - 196 = 1, 193 - 189 = 4, 185 - 181 = 4, 177 - 173 = 4, 169 - 169 = 0, 165 - 161 = 4, 157 - 153 = 4, 149 - 145 = 4, 144 - 141 = 3, 137 - 133 = 4, 129 - 125 = 4, 121 - 121 = 0, 117 - 113 = 4, 109 - 105 = 4, 101 - 100 = 1, 97 - 93 = 4, 89 - 85 = 4, 81 - 81 = 0, 77 - 73 = 4, 69 - 65 = 4, 64 - 61 = 3, 57 - 53 = 4, 49 - 49 = 0, 45 - 41 = 4, 37 - 36 = 1, 33 - 25 = 8, 24 - 16 = 8, 9 - 8 = 1, 8 - 4 = 4, 4 - 4 = 0, 4 - 4 = 0, 4 - 4 = 0, 4 - 4 = 0, 4 - 4 = 0, 4 - 4 = 0, 4 - 4 = 0, 4 - 4 = 0, 4 - 3 = 1, 3 - 1 = 2, 2 - 1 = 1, 1 - 1 = 0, 1 - 1 = 0, 1 - 0 = 1, 1 - 0 = 1, 1 - 0 = 1, 1 - 0 = 1, 1 - 0 = 1, 1 - 0 = 1, 1 - 0 = 1, 1 - 0 = 1, 1 - 0 = 1, 1 - 0 = 1, 1 - 0 = 1, 1 - 0 = 1, 1 - 0 = 1, 1 - 0 = 1&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where the bolded area indicates the part that's different from Joe Z's method.&lt;/p&gt;&#xA;" OwnerUserId="16846" LastEditorUserId="8874" LastEditDate="2015-10-31T20:40:46.987" LastActivityDate="2015-10-31T20:40:46.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23705" PostTypeId="2" ParentId="23702" CreationDate="2015-10-31T20:42:30.477" Score="9" Body="&lt;h2&gt;A simple way to get to 1&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;First we will transform the squares $1$ to $81$ as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$81-64=17$&lt;br&gt;&#xA;$49-36=13$&lt;br&gt;&#xA;$16-9=7$&lt;br&gt;&#xA;$4-1=3$&lt;br&gt;&#xA;&lt;br&gt;&#xA;$17-7=10$&lt;br&gt;&#xA;$13-3=10$&lt;br&gt;&#xA;&lt;br&gt;&#xA;$25-10=15$&lt;br&gt;&#xA;$15-10=5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are left with the $5$ and additional $92$ squares from $10^2$ to $101^2$. Each consecutive $4$ squares can be transformed to a $4$ like in the following example using $100$, $121$, $144$ and $169$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$169-144=25$&lt;br&gt;&#xA;$121-100=21$&lt;br&gt;&#xA;$25-21=4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives us a total of $1*5$ and $23*4$. We can get rid of $22*4$ by doing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$4-4=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the remaining 5 and 4 gives us:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5-4=1$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Is a lower number possible?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The only other lower number would be a $0$. But this is not possible because we start with an odd count of odd numbers. We can only remove odd numbers in pairs which leaves us with an odd number at the end. The lowest non-negative odd number is $1$.&lt;/p&gt;&#xA;" OwnerUserId="15632" LastActivityDate="2015-10-31T20:42:30.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23706" PostTypeId="2" ParentId="22438" CreationDate="2015-10-31T20:45:52.593" Score="0" Body="&lt;p&gt;It is possible for any value equal to or greater than 3, assuming that you have the ability to chop the coins into pieces with infinite precision. You do not need to have coins which have an even weight distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For even numbers, you divide into two groups and weigh. Set aside the lighter half as L, the heavier side as H. Now split H into two parts and weigh. If they are equal, then H will disappear and you know that the fake is lighter. If H1 and H2 are unequal, then L will disappear and you know that the fake is heavier. Continue splitting and weighing until you have two coins or one coin left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your second weighing showed that the fake was heavier, then the fake is the heavier of the remaining coin(s). If your second weighing showed that the fake was lighter, then the fake is the lighter of the remaining coin(s).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For odd numbers, it is possible, using deduction and limits. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Premises: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Coins can be split into pieces without losing material.&lt;/li&gt;&#xA;&lt;li&gt;Coins can be weighed relative to themselves without proving anything. If I split a coin into pieces, I can continue adjusting the split until I have evenly divided the coin, and none of this causes the coin to disappear.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, take all the coins and then weight 2v1 until you have weighed every single combination of coins. If at any point 1 is heavier than 2, then all other coins will disappear and you know which coin is fake and that it's heavier. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, now what if you don't find the fake? Split every coin into two equal parts (by weight), keeping track of which coins parts belong to which coin. Now weigh two halves of one coin against three halves of other coins until you have tested all combinations. If at any point a coin is heavier than its opposition, then the fake will be revealed and known to be heavier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what we have now is a formula of N-1/N, where N is the number of pieces we've split each coin into. Take the limit as N approaches infinity and we have 1, which represents equality of weight, which is not possible. So, for any given threshold of measurement, we can either find the fake and know it's heavier or find nothing and prove that the fake is lighter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key here is that any standard of accuracy is reachable, if we have a great deal of patience. In fact, if we know the standard of accuracy in advance, we can simply chop the coins into sufficiently-small pieces and do one round of weighings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing that fake is lighter (if it were heavier we would have found it), we can solve for an odd number of coins. Weigh the coins 1v1 until you have a single coin left standing. That coin is the fake, and you know it's lighter. For higher values, you can be more efficient by weighing larger piles against each other, but the end result is that you can narrow it down to 1 coin, and then you win.&lt;/p&gt;&#xA;" OwnerUserId="17205" LastActivityDate="2015-10-31T20:45:52.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23707" PostTypeId="1" AcceptedAnswerId="23718" CreationDate="2015-10-31T22:06:35.030" Score="3" ViewCount="274" Body="&lt;p&gt;My &lt;a href=&quot;https://puzzling.stackexchange.com/questions/23692/yet-another-backwards-hangman&quot;&gt;previous backward hangman&lt;/a&gt; (created originally by &lt;a href=&quot;https://puzzling.stackexchange.com/users/10037/alex&quot;&gt;Alex&lt;/a&gt;, see &lt;a href=&quot;https://puzzling.stackexchange.com/questions/11281/the-hangman-game-backward&quot;&gt;here&lt;/a&gt;), was still solved too quickly, so I thought I'd make another tweak to make it a bit more tricky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This time, whatever the bizarre back story is, is repeated four times.  But unfortunately the clues were mixed up.  So there are four different words with a missing letter, and cryptic clues next to them:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Filling in a &quot;p&quot; - Can you stomach rubbish!&lt;/li&gt;&#xA;&lt;li&gt;Filling in a &quot;v&quot; - A serious hole.  &lt;/li&gt;&#xA;&lt;li&gt;Filling in an &quot;s&quot; - Feel bitter that he posted again.  &lt;/li&gt;&#xA;&lt;li&gt;Filling in a &quot;t&quot; - Shred an unpleasant sound.  &lt;/li&gt;&#xA;&lt;li&gt;Filling in a &quot;g&quot; - Centre on a short polite man without money representing the king.  &lt;/li&gt;&#xA;&lt;li&gt;Filling in a &quot;k&quot; - Find gold and make a coin.  &lt;/li&gt;&#xA;&lt;li&gt;Filling in a &quot;p&quot; - Of wrath: oppressive.  &lt;/li&gt;&#xA;&lt;li&gt;Filling in a &quot;c&quot; - Quick! Triple ice!  &lt;/li&gt;&#xA;&lt;li&gt;Filling in an &quot;f&quot; - Cockney view of his spouse's trouble.  &lt;/li&gt;&#xA;&lt;li&gt;Filling in a &quot;c&quot; - Regarding a small coin from not long ago.  &lt;/li&gt;&#xA;&lt;li&gt;Filling in a &quot;p&quot; - Relevant without Sn mixed in a feeling of regret.  &lt;/li&gt;&#xA;&lt;li&gt;Filling in a &quot;b&quot; - A confused biter of people.  &lt;/li&gt;&#xA;&lt;li&gt;Filling in a &quot;c&quot; - Elegant period to rethink.  &lt;/li&gt;&#xA;&lt;li&gt;Filling in an &quot;l&quot; - There next to a fasting period without the abandon.  &lt;/li&gt;&#xA;&lt;li&gt;Filling in a &quot;p&quot; - Way mature decoration.  &lt;/li&gt;&#xA;&lt;li&gt;Filling in an &quot;a&quot; - Parallel lines on a surface.  &lt;/li&gt;&#xA;&lt;li&gt;Filling in a &quot;t&quot; - Petri tests without bugs is not serious.  &lt;/li&gt;&#xA;&lt;li&gt;Filling in a &quot;d&quot; - Level or slope?  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The answer will be in the form&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;W_LL; _ALL; WA_ES; CU_&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;followed by an explanation of which word fits in to which clue... &lt;/p&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-11-01T15:56:49.090" Title="Mixed Backwards Hangmen" Tags="|riddle|word|english|language|cryptic-clues|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23708" PostTypeId="2" ParentId="23690" CreationDate="2015-10-31T22:48:13.170" Score="1" Body="&lt;p&gt;There are indications, that ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the Mummy is the culprit:&lt;br&gt;- pretty bad alibi&lt;br&gt;- dust on the table&lt;br&gt;- snuffed candles to prevent catching fire&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastActivityDate="2015-10-31T22:48:13.170" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23709" PostTypeId="2" ParentId="22438" CreationDate="2015-11-01T00:56:34.957" Score="0" Body="&lt;p&gt;N=3, N-1=2.  There are 2 real coins and 1 fake coin.  The 2 real coins will balance against each other and the other one is fake. This process will hold true for all values of N where N is greater than or equal to 3.&lt;/p&gt;&#xA;" OwnerUserId="17207" LastEditorUserId="17207" LastEditDate="2015-11-01T01:04:33.100" LastActivityDate="2015-11-01T01:04:33.100" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23710" PostTypeId="2" ParentId="23476" CreationDate="2015-11-01T06:06:23.893" Score="0" Body="&lt;p&gt;According to the conditions, the magic square looks like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;. 1 .&#xA;7 . .&#xA;. . 9&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;To prove it is impossible to solve, I would like to simply focus on attempting to solve the top row.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will use the number 8 (As is is the largest after 9) in either of the positions marked X :&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x 1 x&#xA;7 . .&#xA;. . 9&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This would mean the left and right column would be equal to or greater than 15 which is the required sum of the 3 numbers in a particular column/row.&#xA;I would like to illustrate this in the two cases that follow&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the sum of the first column is already 15 and hence this row is unsolvable as using any number between 1-9 in the position marked Y would mean the columns exceeds the sum of 15&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;8 1 .&#xA;7 . .&#xA;Y . 9&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Case 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the sum of the third column is 17 and hence it is not possible.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;. 1 8&#xA;7 . .&#xA;. . 9&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now as 7 as already been used, we cannot use it again in the top row. The remaining largest integers are 6&amp;amp;5, Using these integers in the top row, results in the sum to fall short of 15.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;6 1 5                 5 1 6&#xA;7 . .        OR       7 . . &#xA;. . 9                 . . 9&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We see, in the top row 5+6+1 = 12 &amp;lt; 15&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence this magic square is unsolvable.&lt;/p&gt;&#xA;" OwnerUserId="9346" LastActivityDate="2015-11-01T06:06:23.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23712" PostTypeId="2" ParentId="23702" CreationDate="2015-11-01T07:55:03.350" Score="0" Body="&lt;p&gt;The question is, what's the lowest number possible from subtracting pairs until one number remains...&#xA;the series is n² for n = 1 to 101.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;there is an odd number of pairs, so there will be one number remaining at the end.&#xA;it makes sense to start at the largest pairs...&#xA;101² - 100² = 201 &#xA;99² - 98² = 197&#xA;... there are 100 pairs and the first item (1).&#xA;Subtract the highest of these leaves 25 pairs.&#xA;or ((n+3)²-(n+2)²)-((n+1)² - (n)²) = 4&#xA;There are 25 pairs that subtract to 4...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Canceling them out leaves you with a 4 and a 1...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Final answer being 3.&lt;/p&gt;&#xA;" OwnerUserId="17209" LastActivityDate="2015-11-01T07:55:03.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23713" PostTypeId="2" ParentId="23702" CreationDate="2015-11-01T08:09:18.360" Score="0" Body="&lt;p&gt;Ok, I got it down to 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So subtract pairs from 101² to 16²&#xA;then go through that list and match pairs that are closest, eg (99²-98²)-14²&#xA;Go through the list and reduce like so, you will find all the squares will cancel to 0 or 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then you can subtract the remainder via repetition down.&#xA;Iterate a few times and you can get down to 1... perhaps 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No time for a mathematical proof, i used excel.&lt;/p&gt;&#xA;" OwnerUserId="17210" LastActivityDate="2015-11-01T08:09:18.360" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23715" PostTypeId="1" AcceptedAnswerId="23730" CreationDate="2015-11-01T10:35:35.390" Score="4" ViewCount="592" Body="&lt;p&gt;Alice the Witch had a great night last night. She'd had a Halloween night house party with five of her closest friends from Witch College that she hadn't seen for ages. They had been drinking and gossiping and laughing until the sun had almost come up. Even their familiars had behaved well too (except for Gertie's which had a minor argument with the snake and the toad).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each guest had brought some of her favourite party food with them (the Spider Soup in particular had been fantastic).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this morning she discovered that each of her friends had left something important behind.&#xA;Even her own head was a bit fuzzy and she could only remember a few details of who had brought what, the order in which they had arrived, or which familiar each of them had brought.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Can you help Alice remember which witch had brought each of the dishes and help her reunite her friends with their possessions?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Neither the witch who brought a rat, nor the one who left without her glasses, nor  Janice, nor the one who brought Pumpkin Pie arrived second at the party.&lt;/li&gt;&#xA;&lt;li&gt;Ethel, the guest who brought the Savoury Slime, the one that left without her hat, the one with a pet bat, and the one that arrived third were all different witches.&lt;/li&gt;&#xA;&lt;li&gt;The Liquorice Leeches arrived before the witch who left her cauldron behind who arrived before Gertie.&lt;/li&gt;&#xA;&lt;li&gt;The owner of the rat arrived before the witch who forgot her ruby slippers, but after the one who brought Cobweb Cookies.&lt;/li&gt;&#xA;&lt;li&gt;Frances arrived at almost the same time as the witch who brought her toad.&lt;/li&gt;&#xA;&lt;li&gt;Ethel arrived either directly before or directly after the witch with a snake.&lt;/li&gt;&#xA;&lt;li&gt;The cauldron-forgetter and the rat owner were not the same witch, in fact exactly one witch arrived between them.&lt;/li&gt;&#xA;&lt;li&gt;The witch who left her wand behind arrived two positions from the owner of the cat.&lt;/li&gt;&#xA;&lt;li&gt;At least one witch arrived between Gertie and the owner of the cat.&lt;/li&gt;&#xA;&lt;li&gt;At least one witch arrived between Hettie and the rat-owner.&lt;/li&gt;&#xA;&lt;li&gt;Either the witch who forgot the hat or the witch who forgot the cauldron was also the owner of one of the mammals.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="57212" LastEditDate="2024-02-02T13:31:52.940" LastActivityDate="2024-02-02T14:51:59.277" Title="The Witches Have Overdone It" Tags="|logical-deduction|seasonal|logic-grid|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23716" PostTypeId="2" ParentId="23707" CreationDate="2015-11-01T10:47:04.000" Score="1" Body="&lt;p&gt;Not a complete solution, just a start, as much for my own benefit as others.&lt;/p&gt;&#xA;&lt;p&gt;Two of the words are:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; RE_ENT and GRA_E&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The clues that fit these two are:&lt;/p&gt;&#xA;&lt;p&gt;Filling in an &amp;quot;s&amp;quot; - Feel bitter that he posted again.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; re&lt;strong&gt;s&lt;/strong&gt;ent&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Filling in a &amp;quot;g&amp;quot; - Centre on a short polite man without money representing the king.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; re&lt;strong&gt;g&lt;/strong&gt;ent&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Filling in a &amp;quot;c&amp;quot; - Regarding a small coin from not long ago.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; re&lt;strong&gt;c&lt;/strong&gt;ent&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Filling in a &amp;quot;p&amp;quot; - Relevant with Sn mixed in a feeling of regret.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; re&lt;strong&gt;p&lt;/strong&gt;ent&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Filling in a &amp;quot;v&amp;quot; - A serious hole.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; gra&lt;strong&gt;v&lt;/strong&gt;e&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Filling in a &amp;quot;p&amp;quot; - Can you stomach rubbish!&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; gra&lt;strong&gt;p&lt;/strong&gt;e&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Filling in a &amp;quot;t&amp;quot; - Shred an unpleasant sound.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; gra&lt;strong&gt;t&lt;/strong&gt;e&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Filling in a &amp;quot;c&amp;quot; - Elegant period to rethink.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; gra&lt;strong&gt;c&lt;/strong&gt;e&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I think a third word is&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; STRI_E, but i'm far less sure about this one, but it could be stri&lt;strong&gt;f&lt;/strong&gt;e for clue #9 and stri&lt;strong&gt;k&lt;/strong&gt;e for #6, maybe..&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Which leaves clues:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Filling in a &amp;quot;c&amp;quot; - Quick! Triple ice!&lt;/p&gt;&#xA;&lt;p&gt;Filling in a &amp;quot;p&amp;quot; - Of wrath: oppressive.&lt;/p&gt;&#xA;&lt;p&gt;Filling in a &amp;quot;b&amp;quot; - A confused biter of people.&lt;/p&gt;&#xA;&lt;p&gt;Filling in an &amp;quot;l&amp;quot; - There next to a fasting period without the abandon.&lt;/p&gt;&#xA;&lt;p&gt;Filling in a &amp;quot;p&amp;quot; - Way mature decoration.&lt;/p&gt;&#xA;&lt;p&gt;Filling in an &amp;quot;a&amp;quot; - Parallel lines on a surface.&lt;/p&gt;&#xA;&lt;p&gt;Filling in a &amp;quot;t&amp;quot; - Petri tests without bugs is not serious.&lt;/p&gt;&#xA;&lt;p&gt;Filling in a &amp;quot;d&amp;quot; - Level or sl&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16846" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-11-01T14:41:57.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23717" PostTypeId="1" AcceptedAnswerId="23720" CreationDate="2015-11-01T13:07:00.263" Score="4" ViewCount="260" Body="&lt;p&gt;Eighteen coins are arranged in a circle. In the beginning, all eighteen coins show tails. The following two moves are allowed:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Simultaneously flipping over &lt;strong&gt;four consecutive&lt;/strong&gt; coins (so that tails become heads, and heads become tails); for instance the coins in positions 6,7,8,9 may be flipped simultaneously.&lt;/li&gt;&#xA;&lt;li&gt;Simultaneously flipping over &lt;strong&gt;two pairs&lt;/strong&gt; of consecutive coins that are separated by exactly one non-flipped coin; for instance the coins in positions 5,6,8,9 may be flipped simultaneously.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is it possible to reach the situation where all eighteen coins show heads?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-11-01T13:15:46.287" LastActivityDate="2015-11-01T14:58:09.030" Title="Flipping coin quadruples" Tags="|mathematics|combinatorics|reachability|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23718" PostTypeId="2" ParentId="23707" CreationDate="2015-11-01T13:16:04.007" Score="3" Body="&lt;p&gt;&lt;sup&gt;&lt;em&gt;With thanks to @DrunkWolf, who got three of the Hangman patterns, but mixed up some of the clues&lt;/em&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The four Hangman words are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;code&gt;RE_ENT&lt;/code&gt;&lt;/strong&gt;&lt;br&gt;&#xA; &lt;strong&gt;&lt;code&gt;GRA_E&lt;/code&gt;&lt;/strong&gt;&lt;br&gt;&#xA; &lt;strong&gt;&lt;code&gt;STRI_E&lt;/code&gt;&lt;/strong&gt;&lt;br&gt;&#xA; &lt;strong&gt;&lt;code&gt;TRI_E&lt;/code&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Filling in a &quot;p&quot; - Can you stomach rubbish!&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;code&gt;TRIPE&lt;/code&gt;&lt;/strong&gt;  Tripe is made from animal stomach, and is used figuratively to mean &quot;rubbish&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;Filling in a &quot;v&quot; - A serious hole.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;code&gt;GRAVE&lt;/code&gt;&lt;/strong&gt;  As an adjective, serious.  As a noun, a hole where someone is buried.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;Filling in an &quot;s&quot; - Feel bitter that he posted again.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;code&gt;RESENT&lt;/code&gt;&lt;/strong&gt;  To feel bitter is to resent.  Posted again = re-sent.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;Filling in a &quot;t&quot; - Shred an unpleasant sound.&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;code&gt;GRATE&lt;/code&gt;&lt;/strong&gt;  An unpleasant sound grates on your ears.  To grate is to shred.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;Filling in a &quot;g&quot; - Centre on a short polite man without money representing the king.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;code&gt;REGENT&lt;/code&gt;&lt;/strong&gt;  Polite man = gentleman.  Shortened, it is gent.  &quot;Centre&quot; without &quot;cent&quot; is &quot;re&quot; -&gt; &quot;re gent&quot;.  A man representing the king is a regent.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;Filling in a &quot;k&quot; - Find gold and make a coin.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;code&gt;STRIKE&lt;/code&gt;&lt;/strong&gt;  Strike it rich / Strike a coin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;Filling in a &quot;p&quot; - Of wrath: oppressive.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;code&gt;GRAPE&lt;/code&gt;&lt;/strong&gt;  &quot;Grapes of wrath&quot;; not sure about the &quot;oppressive&quot; part.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;Filling in a &quot;c&quot; - Quick! Triple ice!  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;code&gt;TRICE&lt;/code&gt;&lt;/strong&gt; &quot;Quick&quot; is &quot;in a trice&quot;.  Sounds like &quot;triple ice&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;Filling in an &quot;f&quot; - Cockney view of his spouse's trouble.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;code&gt;STRIFE&lt;/code&gt;&lt;/strong&gt;  &quot;Trouble&quot; = strife; in cockney rhyming slang, a wife is &quot;trouble and strife&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;Filling in a &quot;c&quot; - Regarding a small coin from not long ago.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;code&gt;RECENT&lt;/code&gt;&lt;/strong&gt;  &quot;Re: cent&quot; = regarding a small coin.  Recent = not long ago.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;Filling in a &quot;p&quot; - Relevant with Sn mixed in a feeling of regret.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;code&gt;REPENT&lt;/code&gt;&lt;/strong&gt;  One repents when one has a feeling of regret.  Not sure where the tin (Sn) fits in...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;Filling in a &quot;b&quot; - A confused biter of people.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;code&gt;TRIBE&lt;/code&gt;&lt;/strong&gt;  The word &quot;biter&quot; mixed up (confused).  People = tribe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;Filling in a &quot;c&quot; - Elegant period to rethink.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;code&gt;GRACE&lt;/code&gt;&lt;/strong&gt;  Elegant = graceful.  Period to rethink = grace period.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;Filling in an &quot;l&quot; - There next to a fasting period without the abandon.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;code&gt;RELENT&lt;/code&gt;&lt;/strong&gt; &quot;There&quot; without &quot;the&quot; = &quot;re&quot;.  Fasting period = lent.  Abandon = relent.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;Filling in a &quot;p&quot; - Way mature decoration.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;code&gt;STRIPE&lt;/code&gt;&lt;/strong&gt; A stripe is a decoration.  Not sure about &quot;way mature&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;Filling in an &quot;a&quot; - Parallel lines on a surface.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;code&gt;STRIAE&lt;/code&gt;&lt;/strong&gt; Self-explanatory.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;Filling in a &quot;t&quot; - Petri tests without bugs is not serious.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;code&gt;TRITE&lt;/code&gt;&lt;/strong&gt;  &quot;Petri tests&quot; with &quot;pests&quot; removed leaves &quot;trite&quot;.  Trite = not serious.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;Filling in a &quot;d&quot; - Level or slope?  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;code&gt;GRADE&lt;/code&gt;&lt;/strong&gt;  Ground &quot;at grade&quot; is level; a grade in school (1st grade, etc.) is a level of learning; a grade on a road or ramp is the slope.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2015-11-01T13:16:04.007" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23720" PostTypeId="2" ParentId="23717" CreationDate="2015-11-01T14:45:43.943" Score="9" Body="&lt;p&gt;It is not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Color the coins alternating black and white. There are 9 black coins and 9 white coins. Every possible move flips two black coins and two white coins, so there will always be an even number of each color showing heads. This makes it impossible for all 9 of each color to be heads up.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-11-01T14:45:43.943" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23721" PostTypeId="1" AcceptedAnswerId="23724" CreationDate="2015-11-01T17:40:37.780" Score="9" ViewCount="293" Body="&lt;p&gt;Given the implications&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$aN\implies S$$ $$eF\implies I$$ $$gH\implies M$$ &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;could you please complete the following implication:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$nS\implies ?$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(The answer is one of the letters $T$, $A$, $L$, $K$.)&lt;/p&gt;&#xA;" OwnerUserId="17216" LastEditorUserId="8874" LastEditDate="2015-11-01T20:29:11.410" LastActivityDate="2015-11-01T20:29:11.410" Title="Strange implications with $aN$, $eF$ and $gH$" Tags="|logical-deduction|pattern|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23722" PostTypeId="1" CreationDate="2015-11-01T18:20:47.213" Score="9" ViewCount="495" Body="&lt;h1&gt;My name is ?????&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;You must find out Dr. Mysterious's name, which is your greatest weapon against him (it gives you power over him). His slave, Mr. Useless, has sent you some information but it is encrypted to prevent it from falling into the wrong hands. You must decrypt this information or the world will be destroyed by Dr. Mysterious's Amazing Technicolour Bomb, which only he can deactivate!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;code&gt;---BEGIN MESSAGE---&lt;/code&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;PART 1&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;These are cryptic clues to Part 2:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;NOOOO! I can't get back on track!&lt;/li&gt;&#xA;  &lt;li&gt;Newsflash: The waves on Cryptic Beach have reached 5 characters in height!&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;PART 2&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;code&gt;Tf 3h ot322eyrr 11 e ai 46kps7&lt;/code&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;PART 3&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;When the numbers become letters, it's simple as Blaise de Vigenère!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Remember, &lt;em&gt;YOU MUST FIND HIS NAME&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;code&gt;wpcik://die.bgapgzv.mck/fzlyq?a=XqJGYSt1WVJ&lt;/code&gt; &lt;em&gt;That URL came out great, didn't it!&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;code&gt;---END MESSAGE---&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user17105" LastEditorUserId="4421" LastEditDate="2015-11-27T02:53:13.577" LastActivityDate="2015-11-30T16:17:32.867" Title="Save the world from Dr. Mysterious" Tags="|cipher|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23723" PostTypeId="2" ParentId="23722" CreationDate="2015-11-01T20:17:59.413" Score="5" Body="&lt;p&gt;&lt;strong&gt;Part 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;We are using a railfence cipher&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;We have 5 rails&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Decoding using cipher from part 1 gives&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The key for part 3 is 12321476&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14013" LastActivityDate="2015-11-01T20:17:59.413" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23724" PostTypeId="2" ParentId="23721" CreationDate="2015-11-01T20:24:03.547" Score="13" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; T&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Justification:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The puzzle is built around the periodic system of chemical elements:&lt;br&gt;&#xA;&lt;br&gt;&#xA; Na = Sodium  (Na reflected = aN; first letter of Sodium = S)&lt;br&gt;&#xA; Fe = Iron  (Fe reflected = eF; first letter of Iron = I)&lt;br&gt;&#xA; Hg = Mercury  (Hg reflected = gH; first letter of Mercury = M)&lt;br&gt;&#xA; Sn = Tin  (Sn reflected = nS; first letter of Tin = T) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-11-01T20:24:03.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23725" PostTypeId="2" ParentId="23676" CreationDate="2015-11-01T20:25:47.417" Score="2" Body="&lt;p&gt;Let's come at it from extremes. If you show every treasure to every person, all your treasures will be stolen and you will learn nothing about who the thieves are. If you show each treasure to only one person, over 400 villagers will not see a treasure. If the thieves are in the 64 people who see your treasures, you'll know who they are, but if at least one thief isn't in that group, you don't have your information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you show each treasure to 8 people, everyone will get to see one treasure. But if a given treasure is stolen, you don't know which of the 8 people stole it. The only way to learn that is to show each person more than one treasure. If there is only one thief, you need to just make unique pairs - sure, 8 people saw treasure 1 from country A, and 8 people saw treasure 2 from country B, but only one person saw &lt;strong&gt;both&lt;/strong&gt; A1 and B2, so that is the thief. The number of combinations is 64 choose 2, so 64*63/2 or 2016, so showing two items per person would be fine with only one thief.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have multiple thieves, it gets a lot harder. If A1, B2, C3, and D4 go missing, and each person saw 2 treasures, are your two thieves A1B2 and C3D4 or A1D4 and B2C3? To overcome this, you need to use less of your combinations to eliminate these overlaps and that means you're going to need to show more items to each person.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With four thieves, more things will go missing - up to 8 if you show each person two items, up to 12 if you show each person three items, and so on. The question statement wants you to show each person 16 items. That means a LOT of items will be stolen. If you use some sort of symmetric assignment of who sees what, some of your 16 countries will have one item seen by each thief, so you'll lose all 4, some will have one item seen by all four thieves, so you'll just lose that 1, and the rest will have two or three thieves see the same thing so you'll lose between 1 and 4. You'll have a total of dozens of items stolen. Trying to split that up into clumps that you can say &quot;this 16 was from person 27, and this 16 was from person 178&quot; is going to be impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean literally impossible. After you've shown 1 country, 128 people have seen each item. After country 2, 32 people have seen each combination. After country three, 8 people have seen each combination. After country four, 2 people. No matter how you deal out country 5, some people will have seen the same things. And even if they haven't, you will have the same A1B2 and C3D4 or A1D4 and B2C3 problem, because you used all your combinations. The only way to use sparser combinations would be if some people didn't see anything from country A, and some didn't see any from country B, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So some part of the puzzle has to give. Either you don't show everyone one item from each country, or you don't do all the showing followed by all the stealing. I think it's relatively easy if you show, wait, rule out certain people (item 3 didn't get stolen so the 128 people I showed it to are honest) and continue on like that. Some nights only one thing would be stolen, some nights 4 things, but you could work out who the thieves were eventually by making the combinations based on who is still under suspicion. But if the puzzle is posed as written, I say &lt;strong&gt;there is no solution&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1509" LastActivityDate="2015-11-01T20:25:47.417" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23726" PostTypeId="1" AcceptedAnswerId="23728" CreationDate="2015-11-01T22:30:01.673" Score="6" ViewCount="696" Body="&lt;p&gt;It's known that the numbers 1 to 9 can only form eight different magic squares, which are all rotations and reflections of each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any set of 9 distinct numbers that can form two &lt;em&gt;essentially different&lt;/em&gt; magic squares, which means that you can't get from one to the other using rotations and reflections? If not, is there a way to prove it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The numbers from 1 to 9 only have one essentially different magic square configuration.)&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2015-11-02T16:21:02.983" Title="Are there any sets of 9 numbers that can form two essentially distinct magic squares?" Tags="|mathematics|magic-square|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23727" PostTypeId="1" AcceptedAnswerId="23731" CreationDate="2015-11-01T22:49:25.370" Score="16" ViewCount="8429" Body="&lt;p&gt;In the game Undertale, there is a character named Papyrus who believes that the Jumble puzzle is much harder than the daily crossword.&lt;/p&gt;&#xA;&lt;p&gt;Help prove him right by trying to solve this brainteaser:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3OIRj.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3OIRj.png&quot; alt=&quot;Title: &amp;quot;Jumble&amp;quot;. First word: first and sixth boxes are shaded, scrambled letters are &amp;quot;N E H K T C I&amp;quot;. Second word: fifth and eighth boxes are shaded, scrambled letters are &amp;quot;E T S R R I N A&amp;quot;. Third word: first box is shaded, scrambled letters are &amp;quot;S L E P Y A R&amp;quot;. Fourth word: second and fourth boxes are shaded, scrambled letters are &amp;quot;S A N M R P A E&amp;quot;. Instructions: &amp;quot;Now unscramble the grey [shaded] squares to get the surprise answer, as suggested by the cartoon on the right&amp;quot;. The boxes given for the final answer are two, then a dash, then five. The cartoon is a robot saying &amp;quot;Bone appetit! Nyeh heh heh!&amp;quot; at a table, one hand under a plate, while a shorter figure with a robot hand slaps themself on the head. The caption to the cartoon is &amp;quot;Papyrus served spaghetti as usual; it was his standard evening (blank here for answer).&amp;quot;&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="69582" LastEditDate="2021-06-06T02:41:35.557" LastActivityDate="2022-12-05T14:28:18.083" Title="Can you solve this devilish Junior Jumble?" Tags="|wordplay|anagram|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23728" PostTypeId="2" ParentId="23726" CreationDate="2015-11-01T23:41:13.840" Score="13" Body="&lt;p&gt;Note that any two magic squares with the same numbers must have the same row/column sum. The middle number must be 1/3 of the sum (because the sum of the four lines passing through the center is the sum of all the numbers plus three times the center), so it is the same in both squares. Subtract this number from all the numbers, so that the middle cell and the row/column sum are both 0. Then diametrically opposite cells must be negatives of each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the corner cells of one square contain the same numbers as the corner cells of the other, they must be rotations or reflections of each other. So if there are two essentially different squares, one number must be a corner in one square and an edge in another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose one of the squares looks like this:$$\begin{array}{ccc}a&amp;amp;b&amp;amp;c\\-d&amp;amp;0&amp;amp;d\\-c&amp;amp;-b&amp;amp;-a\\\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either $a$ and $c$ both become edge cells in the second square, or one is an edge and the other is a corner. If one remains as a corner, then the only possibility (up to certain symmetries) that does not immediately require two numbers to be the same is this: $$\begin{array}{ccc}a&amp;amp;c&amp;amp;b\\d&amp;amp;0&amp;amp;-d\\-b&amp;amp;-c&amp;amp;-a\\\end{array}$$However, if we try to solve this, we find that $a$ must be 0, which makes $d=b=-c$, and the squares are not essentially different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, the configuration of the second square must be something like this:$$\begin{array}{ccc}-b&amp;amp;a&amp;amp;d\\c&amp;amp;0&amp;amp;-c\\-d&amp;amp;-a&amp;amp;b\\\end{array}$$ If we solve this, we find that all the numbers must be 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, there are no sets of 9 numbers that can produce two essentially different magic squares.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-11-02T16:21:02.983" LastActivityDate="2015-11-02T16:21:02.983" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23729" PostTypeId="2" ParentId="23727" CreationDate="2015-11-01T23:42:35.820" Score="2" Body="&lt;p&gt;There's a cooking-related possibility for all of the first four:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;KITCHEN, STRAINER, PARSLEY, PARMESAN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which gives this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;KEIRPAM&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not sure what that could anagram to, though - all of the words have multiple possibilities anyway so this could be entirely wrong.&lt;/p&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2015-11-01T23:42:35.820" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23730" PostTypeId="2" ParentId="23715" CreationDate="2015-11-02T00:37:47.937" Score="2" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;Order&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;Name&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;Food&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;Item&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;Familiar&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;Ethel&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;Liquorice Leeches&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;Glasses&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;Cat&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;Hettie&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;Savoury Slime&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;Cauldron&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;Snake&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;Janice&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;Cobweb Cookies&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;Wand&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;Toad&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;Frances&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;Spider Soup&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;Hat&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;Rat&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;5&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;Gertie&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;Pumpkin Pie&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;Slippers&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;Bat&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;With all clues I got to this state:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/df9k8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/df9k8.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;At this point I had to start guessing. First I decided between Cat on number 1 or 2. Number 2 lead to a solution that contradicted one of the hints so Cat was number 1.&lt;/p&gt;&#xA;&lt;p&gt;This lead me to a state were I could choose between Bat on number 2 or 5. Number 2 lead to a solution that contradicted one of the hints so Bat was number 5. And this also directly lead to an solution so that was the answer&lt;/p&gt;&#xA;&lt;p&gt;I used the tool on &lt;a href=&quot;http://www.jsingler.de/apps/logikloeser/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.jsingler.de/apps/logikloeser/&lt;/a&gt; to help with the grid&lt;/p&gt;&#xA;" OwnerUserId="1748" LastEditorUserId="69582" LastEditDate="2024-02-02T14:51:59.277" LastActivityDate="2024-02-02T14:51:59.277" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23731" PostTypeId="2" ParentId="23727" CreationDate="2015-11-02T01:11:58.530" Score="12" Body="&lt;p&gt;The four words are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; THICKEN, RETRAINS, REPLAYS, and SPEARMAN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the spaghetti is, of course, Papyrus' evening&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RE-PASTA. (zoom-in, ba-dum-tish)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12511" LastActivityDate="2015-11-02T01:11:58.530" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23732" PostTypeId="1" AcceptedAnswerId="23733" CreationDate="2015-11-02T01:28:50.800" Score="8" ViewCount="546" Body="&lt;p&gt;This is related to my previous question, Q: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/23410/nice-song-pity-about-the-words&quot;&gt;Nice song, pity about the words&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a terrible memory for the words of songs and I'm even worse when it comes to their titles. I'd really like to find this very well-known song on Youtube, if only I could remember what it is called. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you help me with the title?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the gist of the lyrics. I'm afraid my version is only an approximation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;There's a possibility that my affections for you may fizzle out at some point. However, provided you remain astronomically covered there won't be any necessity for misgivings on that score -- I'll provide a guarantee of that.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I remember correctly, the next line starts with a religious reference about omniscience versus ignorance. However the end of the line I never heard properly at all. It might have been dog-treat related  because it sounded as though the lead singer was puzzled about how he would cope 'with a chew'?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Jo63N.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jo63N.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few more words after that but I can't think of them right now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I've just remembered that it is quite old. It came out in the 1960s I believe. Still it has millions of plays on Youtube and was a big hit at the time. I heard a cover version on British TV only a few days ago so  it hasn't been forgotten by any means.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's another hint&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It may not be related to dogs, I can't be certain of that. However it certainly has some connection with domestic animals and the noises they make.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14177" LastEditorUserId="5373" LastEditDate="2018-05-14T16:56:06.097" LastActivityDate="2018-05-14T16:56:06.097" Title="Can anyone help me remember the title to this song?" Tags="|wordplay|knowledge|music|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23733" PostTypeId="2" ParentId="23732" CreationDate="2015-11-02T02:36:20.337" Score="8" Body="&lt;p&gt;The song is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; God Only Knows, by The Beach Boys.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a possibility that my affections for you may fizzle out at some point.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I may not always love you&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, provided you remain astronomically covered&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; But long as there are stars above you&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;there won't be any necessity for misgivings on that score&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You never need to doubt it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll provide a guarantee of that.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'll make you so sure about it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The part about coping &lt;em&gt;with a chew&lt;/em&gt; refers to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; God only knows what I'd be &lt;em&gt;without you&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second hint points to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pet Sounds, the 1966 album this song is from.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16998" LastEditorUserId="16998" LastEditDate="2015-11-02T03:01:13.997" LastActivityDate="2015-11-02T03:01:13.997" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23734" PostTypeId="1" AcceptedAnswerId="23739" CreationDate="2015-11-02T08:11:35.363" Score="1" ViewCount="1267" Body="&lt;p&gt;I'm Caucasian and my fiancé is Chinese - I met her on a trip to Tahiti though as chance would have it, it turned out that we actually grew up in adjacent towns in Ohio. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We decided to have a very intimate wedding and only invite our respective immediate family members. We had yet to ever introduce one another to our respective families. We had no extended family and all our grandparents had already passed on, so with each of us having just one sibling, the total number of invites sent out was 4 (one to each set of parents, and one to both my and my fianceé's siblings).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everyone RSVPed. My sibling said they'd be bringing another guest, as did my fiancée's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However on the day of the wedding only 6 people turned up and yet everyone we were expecting was at the affair!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HOW??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1: The siblings did not bring one another as their +1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 2: Both my and my fianceé parents are happily married couples and there's no unconventional hanky panky going on between any members of my or my fianceé's family. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 3 (huge hint - possible spoiler... not needed to solve)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Both our siblings are half Chinese, half Caucasian females. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17145" LastEditorUserId="17145" LastEditDate="2015-11-02T17:45:39.993" LastActivityDate="2015-11-02T17:45:39.993" Title="Where are the missing guests?" Tags="|logical-deduction|family|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23735" PostTypeId="2" ParentId="23734" CreationDate="2015-11-02T08:25:30.647" Score="3" Body="&lt;p&gt;Probably&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Each sibling was thinking of bringing the other&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4317" LastActivityDate="2015-11-02T08:25:30.647" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23737" PostTypeId="1" AcceptedAnswerId="23742" CreationDate="2015-11-02T09:13:06.133" Score="12" ViewCount="868" Body="&lt;p&gt;Alice and Bob play the following game with $99$ numbers $M_1,\ldots,M_{99}$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In the beginning, the $99$ numbers take the values $1,2,4,8,16,\ldots,2^{97},2^{98}$.&lt;/li&gt;&#xA;&lt;li&gt;Then Alice and Bob alternatingly make moves; Alice makes the first move.&lt;/li&gt;&#xA;&lt;li&gt;In every move, the active player picks five numbers $M_a,M_b,M_c,M_d,M_e$ (with five distinct indices $a,b,c,d,e$) and decreases each of them by $1$.&lt;/li&gt;&#xA;&lt;li&gt;If one of the numbers becomes negative, the game ends. The player who has caused this loses, and the other player is declared winner.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Which player has a winning strategy? (As usual, we assume that Alice and Bob use optimal strategies.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="4269" LastEditDate="2015-11-02T10:49:45.780" LastActivityDate="2015-11-02T18:54:36.820" Title="Ninety-nine non-negative numbers" Tags="|mathematics|strategy|game|nim|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23738" PostTypeId="2" ParentId="23734" CreationDate="2015-11-02T09:18:34.117" Score="3" Body="&lt;p&gt;An oddball answer, but fits the solution well.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There is one (awkward!) coincidence that could have taken place. The parents of the bride and groom must be divorced, and the siblings are dating each other's father. So the siblings end up there with their dates, who also happens to be the fathers of the bride and groom. And the mothers are there too, bringing up the total to 6.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastActivityDate="2015-11-02T09:18:34.117" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23739" PostTypeId="2" ParentId="23734" CreationDate="2015-11-02T09:50:49.357" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Your parents have separated. So have your fiancée's. One of your parents has subsequently started a relationship with one of hers, and they had a child - your half-sibling, and also your fiancée's half-sibling. You have one sibling each, it just happens that those siblings are the same person. (Also, you may be your fiancée's step-brother)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Guest numbers:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Your parents (2), your fiancée's parents (2), your shared sibling (1) and their date (1) for a total of 6&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11717" LastActivityDate="2015-11-02T09:50:49.357" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="23740" PostTypeId="1" CreationDate="2015-11-02T10:48:44.703" Score="6" ViewCount="363" Body="&lt;p&gt;A man picks up a huge circular dart board and shoots paintballs at it. $n$ paintballs manage to hit it. They are random and point-sized, for the purpose of this question. A bug is somewhere on the edge of the board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It likes to eat paint, so it goes to the nearest smear of paint and eats it. It then goes to the nearest smear from there and eats that also. It repeats this process until they entire board is clean. Incidentally, the bug also leaves a trail behind, wherever it goes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; How many times is this trail most likely to intersect itself? (Report as a function of $n$)&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-11-04T09:27:50.683" Title="Intersecting bug trail" Tags="|mathematics|probability|" AnswerCount="3" CommentCount="5" ClosedDate="2015-11-04T15:20:02.817" ContentLicense="CC BY-SA 3.0" />
  <row Id="23742" PostTypeId="2" ParentId="23737" CreationDate="2015-11-02T15:36:40.750" Score="17" Body="&lt;h2&gt;TL;DR&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first player should decrease $M_1$, $M_{96}$, $M_{97}$, $M_{98}$, $M_{99}$ in the first move. Afterwards he can repeat all moves of the second player to win.&lt;br&gt;&lt;br&gt;This is guaranteed to work because the numbers $M_1 .. M_{95}$ are all even after the first move, and each of the numbers $M_{96} .. M_{99}$ is bigger than the sum of all numbers $M_1 .. M_{95}$. At the end all numbers $M_1 .. M_{95}$ will be $0$, and the numbers $M_{96} .. M_{99}$ will be positive with the second player to move.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The rest of the post describes how I came to the conclusion above.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Closer look&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let's make two groups of numbers $M_1 .. M_{95}$ and $M_{96} .. M_{99}$:&lt;br&gt;&lt;br&gt;$\begin{array}{} M_1 &amp;amp; M_2 &amp;amp; M_3 &amp;amp; ... &amp;amp; M_{95} &amp;amp; \qquad\qquad &amp;amp; M_{96} &amp;amp; M_{97} &amp;amp; M_{98} &amp;amp; M_{99} \\ 1 &amp;amp; 2 &amp;amp; 4 &amp;amp; ... &amp;amp; 2^{94} &amp;amp; \qquad\qquad &amp;amp; 2^{95} &amp;amp; 2^{96} &amp;amp; 2^{97} &amp;amp; 2^{98} \end{array}$&lt;br&gt;&lt;br&gt;The sum of the numbers $M_1 .. M_{95}$ is $2^{95}-1$ which is smaller than $M_{96}=2^{95}$. Even if we would decrease all numbers $M_{96} .. M_{99}$ in each move, they will stay positive until the end of the game.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Changing the rules&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This allows us to create an equivalent game with slightly different rules. First we get rid of the numbers $M_{96} .. M_{99}$. Second we change the move rule to the following:&lt;br&gt;&lt;br&gt;- In every move, the active player picks &lt;strong&gt;between one and five&lt;/strong&gt; numbers and decreases each of them by $1$.&lt;br&gt;&lt;br&gt;This game is equivalent to the original game, because we can decrease the missing one to four numbers from the group $M_{96} .. M_{99}$ without changing the outcome of the game.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Winning position&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let us assume we have a position where all numbers are even numbers. In such a position it is always possible to repeat the move done by the previous player reaching again a position where all numbers are even numbers. At some point the first player will run out of possible moves and the second player wins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;How to win?&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first player can decrease the number $M_1=1$ (and $M_{96} .. M_{99}$ in the original game) and reach the winning position from the previous chapter (he will be then the second player).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2015-11-02T18:54:36.820" LastActivityDate="2015-11-02T18:54:36.820" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23743" PostTypeId="1" AcceptedAnswerId="23752" CreationDate="2015-11-02T15:43:29.923" Score="30" ViewCount="3517" Body="&lt;p&gt;What is the smallest box diagram you can draw which uses all of the Unicode box drawing characters at least once, but without leaving any loose ends?  For reference, you must use each of these characters at least once (not counting the spaces, which are just there for alignment purposes):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;─ │ ┌ ┐ └ ┘ ├ ┤ ┬ ┴ ┼&#xA;═ ║ ╔ ╗ ╚ ╝ ╠ ╣ ╦ ╩ ╬&#xA;    ╒ ╕ ╘ ╛ ╞ ╡ ╤ ╧ ╪&#xA;    ╓ ╖ ╙ ╜ ╟ ╢ ╥ ╨ ╫&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;By loose ends, I mean:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;┌┐ &amp;lt;-Allowed  ┌┐ &amp;lt;-Not allowed  ┌┐┌┐ &amp;lt;-Allowed&#xA;└┘            └┴                └┴┴┘&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Also, all single-lines must line up with single lines and doubles must line up with doubles, so:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;├─, ╟─, ╠═, ╞═ &amp;lt;-Allowed  ├═, ╟═, ╠─, ╞─ &amp;lt;-Not allowed&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You may use characters more than once each, but each character must appear at least once.  You may make multiple unconnected shapes - it is not necessary to create one single connected shape, so long as all of the ends are closed off. (But see the bonus below.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Entries will be graded code-golf style.  Count the number of characters (including spaces, either leading or internal) used in your entry and include it in bold as a header.  Newlines are free (to prevent conflicts between 1-char and 2-char newlines on different OS's).  Then subtract 40 points (one for each of the required characters).  (A perfect score of 0 would thus use each character exactly once with no duplicates.)  Lowest score after one week wins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Create a diagram which is one contiguous shape.  All other rules apply as previously.  If you create both a contiguous graph and a non-contiguous set of graphs, your overall score will be the average of both scores divided by two.&lt;/p&gt;&#xA;" OwnerUserId="16131" LastEditorUserId="16131" LastEditDate="2015-11-03T15:20:25.577" LastActivityDate="2019-11-30T07:57:25.480" Title="Smallest Unicode Box Drawing" Tags="|topology|" AnswerCount="4" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="23744" PostTypeId="2" ParentId="23690" CreationDate="2015-11-02T15:49:50.863" Score="8" Body="&lt;p&gt;It's odd that:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Dracula's (2nd) favourite drink is the punch, and yet he hasn't consumed any of it  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;His alibi is unverifiable, and:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;he only claims that it would have been noticed if &lt;em&gt;he'd&lt;/em&gt; added anything &lt;em&gt;once the party had started&lt;/em&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So either:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Dracula spiked the punch whilst it was still in the kitchen, assuming Frankenstein's monster wouldn't notice  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Dracula gave Frank's monster a special ingredient for the punch, which would explain the monster's statement that he made 'punch bad' - a notably different suggestion to 'bad punch'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;MASSIVE EDIT/answer change&lt;/strong&gt;:&lt;br&gt;&#xA;I missed an obvious mythological clue, and that makes me sad.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There should not be six sets of eyes reflected in the punch, because vampires cast no reflection!&lt;br&gt;&#xA;&lt;br&gt;&#xA;So it was almost certainly whoever is currently claiming to be Dracula - presumably a human in disguise, disgusted by the human-vampire peace treaty and acting to destabilise the relationship between the two species as much as possible!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11717" LastEditorUserId="11717" LastEditDate="2015-11-02T16:13:36.810" LastActivityDate="2015-11-02T16:13:36.810" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23745" PostTypeId="2" ParentId="23743" CreationDate="2015-11-02T16:13:42.193" Score="5" Body="&lt;h2&gt;Score: 5&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Duplicates = 4, Spaces = 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;╔╦╗┌┬┐╓╥╖╒╤╕╔═╕&#xA;╠╬╣├┼┤╟╫╢╞╪╡║ │&#xA;╚╩╝└┴┘╙╨╜╘╧╛╙─┘&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;h2&gt;For a contiguous shape: 9&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Duplicates = 9, Spaces = 0&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;╔╦═╗&#xA;╠╬╤╣&#xA;╟╫┼╢&#xA;╚╩╡║&#xA;┌┬┴╜&#xA;││╓╖&#xA;╘╧╝╙┐&#xA;╒╕┌╥┤&#xA;╞╪╛╙┘&#xA;├┴┐╔╗&#xA;└─┴╨╜&#xA;&lt;/pre&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2015-11-02T18:01:33.873" LastActivityDate="2015-11-02T18:01:33.873" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23746" PostTypeId="2" ParentId="23743" CreationDate="2015-11-02T16:26:51.853" Score="3" Body="&lt;h1&gt;Averaged score 6&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;50 characters, includes 2 spaces. I'll keep working on improvements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE:&lt;/strong&gt; 46 characters, no spaces. I suspect there's a way to deal with the triple '|' in the first figure...&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;┌╥┬┐╔╤╦╗╒╤╕╓╥╖&#xA;│║││╠╧╬╣╞╪╡╟╫╢&#xA;├╨┼┤╚═╩╝╘╧╛╙╨╜&#xA;└─┴┘        &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Contiguous: 48 characters, 0 spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE:&lt;/strong&gt; 46 characters, includes 1 space&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;╔╦╤╗&#xA;║╟┼╢&#xA;╠╬╧╣&#xA;╚╩╤╝&#xA;╒═╪╕&#xA;╘╦╧╛&#xA;╓╫╖&#xA;╙╫╜&#xA;┌╨┬┐&#xA;│ ╞╡&#xA;├╥┼┤&#xA;└╨┴┘&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I know it's been solved perfectly, but I'm going to keep trying to improve my solution.&lt;/p&gt;&#xA;" OwnerUserId="17167" LastEditorUserId="17167" LastEditDate="2015-11-02T23:09:39.897" LastActivityDate="2015-11-02T23:09:39.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23748" PostTypeId="2" ParentId="23737" CreationDate="2015-11-02T16:43:35.330" Score="3" Body="&lt;p&gt;Second (but completed) answer, once again building of Sleafar's answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alice wins&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If we disregard M96-M97-M98-M99, and therefor consider each turn as removing 'one to five' units from the remainder, eventually said remainder will run out.&#xA;&#xA;If someone gets to a point where the remainder is dividible by six, they just let the other player pick 1 to 5 from the 'remainder', and on their pick, they pick 6 - (whatever the other did), so after each two turns, the net value of the remainder goes down by 6, therefor always remaining dividable by 6. &#xA;&lt;br&gt;&lt;br&gt;In the limit case, all M1 to M95 will be depleted after a such move, meaning the other player only gets to pick 4 from M96-M99, and needs to turn one negative.&#xA;&lt;br&gt;&lt;br&gt;&#xA;Now can Alice force the wining postion on the first try?&#xA;&lt;br&gt;&lt;br&gt;YES. The only way she couldn't, is if the initial amount were dividable by 6. Given it's an odd number (due to M1), she can always force the win.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2156" LastActivityDate="2015-11-02T16:43:35.330" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23749" PostTypeId="1" AcceptedAnswerId="23751" CreationDate="2015-11-02T17:58:59.150" Score="16" ViewCount="1013" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;My first's where you are when you aren't trick-or-treating&lt;br&gt;&#xA;  My next is the start of what kids will be eating&lt;br&gt;&#xA;  My third, when gigantic, might test your comport&lt;br&gt;&#xA;  My fourth starts the &lt;em&gt;season&lt;/em&gt; where goblins cavort&lt;br&gt;&#xA;  My fifth is a note sung by sirens that lure you&lt;br&gt;&#xA;  My last's where you put the disguise that obscures you  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;My whole is a staple of magic primeval&lt;br&gt;&#xA;  For witches invoking the forces of evil&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2015-11-02T22:04:11.090" Title="A Halloween Riddle" Tags="|riddle|word|wordplay|seasonal|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23750" PostTypeId="2" ParentId="23743" CreationDate="2015-11-02T18:42:03.120" Score="18" Body="&lt;p&gt;&lt;strong&gt;Score: 2&lt;/strong&gt;&#xA;42 Characters, 2 duplicates - Sorry that I don't know how to format.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;┌╥╖╓┬┐╔═╦╕&#xA;╞╬╣╠╪╡╟─╫┤ &#xA;└╨╜╙┴┘╚═╩╛ &#xA;╒╤╗ &#xA;││║ &#xA;├┼╢ &#xA;╘╧╝&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Bonus challenge:  &lt;strong&gt;Score: 2&lt;/strong&gt;&#xA;42 Characters, 2 duplicates&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;╒╤╦╗╔╕╓╥╖┌┬─┐&#xA;├┼╢╟╫┤╠╬╣╞╪═╡&#xA;││║╚╩╧╝╙╨┘╘═╛&#xA;└┴╜&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="16187" LastEditorUserId="16187" LastEditDate="2015-11-02T19:18:27.383" LastActivityDate="2015-11-02T19:18:27.383" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23751" PostTypeId="2" ParentId="23749" CreationDate="2015-11-02T21:14:40.210" Score="14" Body="&lt;p&gt;Thanks to Dr Xorile for seeing the final solution from my first attempt and GentlePurpleRain for the last fix in formatting:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first's where you are when you aren't trick-or-treating&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;IN&lt;/strong&gt;, because when you're not &lt;em&gt;out&lt;/em&gt; trick-or-treating, you're &lt;em&gt;in&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My next is the start of what kids will be eating&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;C&lt;/strong&gt;, the &quot;start&quot; of &lt;em&gt;candy&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My third, when gigantic, might test your comport&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;ANT&lt;/strong&gt;, as in a comport-testing gigantic &lt;em&gt;ant&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fourth starts the season where goblins cavort&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;A&lt;/strong&gt;, the &quot;start&quot; of &lt;em&gt;autumn&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fifth is a note sung by sirens that lure you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;TI&lt;/strong&gt;, one of the notes do, re mi, fa so, la &lt;em&gt;ti&lt;/em&gt;, do&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My last's where you put the disguise that obscures you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;ON&lt;/strong&gt;, because you put the disguise &lt;em&gt;on&lt;/em&gt; to obscure your identity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My whole is a staple of magic primeval&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For witches invoking the forces of evil&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;INCANTATION&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17231" LastEditorUserId="17231" LastEditDate="2015-11-02T22:04:11.090" LastActivityDate="2015-11-02T22:04:11.090" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="23752" PostTypeId="2" ParentId="23743" CreationDate="2015-11-02T22:01:33.517" Score="36" Body="&lt;h1&gt;Score 0&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Found using a small program.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$10 \times 4$ grid:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;┌─┐╔═╗╒╦╤╕&#xA;├╥┤╟┬╢╘╣╞╛&#xA;│╠╪╬╡║╓╫┼╖&#xA;└╨┘╚╧╩╝╙┴╜&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;$ 8\times5 $ grid:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;┌─┐╔═╗╒╕&#xA;╞╦╡║╓╨┴┘&#xA;├╫┼╢╙╥┬╖&#xA;│╟┤╠╤╬╪╣&#xA;└╜╘╩╛╚╧╝&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2015-11-03T06:38:20.003" LastActivityDate="2015-11-03T06:38:20.003" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="23753" PostTypeId="1" CreationDate="2015-11-02T23:02:00.773" Score="-6" ViewCount="370" Body="&lt;p&gt;What happens to Pinocchio's nose if he says to another person &quot;My nose is about to grow&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it grow?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it stay the same?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something else?&lt;/p&gt;&#xA;" OwnerUserId="15645" LastEditorUserId="20274" LastEditDate="2016-05-21T22:04:50.960" LastActivityDate="2016-05-21T22:04:50.960" Title="A question about Pinocchio's nose" Tags="|logical-deduction|paradox|" AnswerCount="1" CommentCount="1" ClosedDate="2015-11-03T03:40:26.263" ContentLicense="CC BY-SA 3.0" />
  <row Id="23754" PostTypeId="2" ParentId="23753" CreationDate="2015-11-02T23:08:07.063" Score="3" Body="&lt;p&gt;I believe that Pinnochio's nose only grows when he tells a lie, not if he makes a mistake, so what happens with his nose depends upon what he believes.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If he thinks that his nose really is about to grow, then he is not lying, his nose will not grow but he will have made a mistake.&lt;/li&gt;&#xA;&lt;li&gt;If he thinks that his nose is not about to grow, then he is lying and his nose will grow.   &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In either case there is no paradox.&lt;/p&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-11-02T23:08:07.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23755" PostTypeId="1" AcceptedAnswerId="23756" CreationDate="2015-11-02T23:57:26.057" Score="12" ViewCount="692" Body="&lt;p&gt;Few days ago, the notorious Blackbeard sunk 3 of my ships carrying exotic goods and treasures from Asia. I was furious and I am determined to send out my fleet to bring him to justice. However, as he is a well known and a formidable pirate, I need to plan my attack carefully. So I send out a number of scouts, hoping they would be able to locate where they are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my delight, my scouts are successful in their first attempt to locate the pirates, and I have received the following message:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There are 10 ships in total, namely 1 flagship, 2 frigates, 3&#xA;  corvettes, and 4 patrol boats located in the Caribbean sea.&#xA;  It seems like that they are anchored here and won't be moving any time soon.&#xA;  Furthermore, none of Blackbeard's ships touch each other, not even&#xA;  diagonally.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, my military adviser says we need more information on their fleet formation so he can devise a plan to strike him down from where he is the weakest. So I replied to the scout to bring back that information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's exactly what they did, but this time, the information is not complete. The message is as follows&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Please see the attached map for the pirate's fleet formation. The numbers&#xA;  are the number of ship pieces on that row or column&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The map attached look like this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{|c|c|c|c|c|c|c|c|c|c|l}&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline&amp;amp;&amp;amp;\blacktriangle&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;3\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline&amp;amp;\blacklozenge&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;4\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;0\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;0\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;\blacktriangleright&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;8\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline2&amp;amp;1&amp;amp;3&amp;amp;1&amp;amp;1&amp;amp;1&amp;amp;3&amp;amp;1&amp;amp;5&amp;amp;2&#xA;\end{array}&#xA;\quad\quad&#xA;\bbox[4pt,border: 1px solid black]{&#xA;\begin{array}{llc}&#xA;\bf{Legend:}&amp;amp;\\&#xA;\hline&#xA;\text{Flagship:}&amp;amp; \blacktriangleleft\blacksquare\blacksquare\blacktriangleright \text{or} &amp;amp;\blacktriangle\\&#xA;&amp;amp;&amp;amp;\blacksquare\\&#xA;&amp;amp;&amp;amp;\blacksquare\\&#xA;&amp;amp;&amp;amp;\blacktriangledown\\&#xA;\hline&#xA;\text{Frigate:}&amp;amp; \blacktriangleleft\blacksquare\blacktriangleright \text{or} &amp;amp;\blacktriangle\\&#xA;&amp;amp;&amp;amp;\blacksquare\\&#xA;&amp;amp;&amp;amp;\blacktriangledown\\&#xA;\hline&#xA;\text{Corvette:}&amp;amp; \blacktriangleleft\blacktriangleright \text{or} &amp;amp;\blacktriangle\\&#xA;&amp;amp;&amp;amp;\blacktriangledown\\&#xA;\hline&#xA;\text{Patrol Boat:}&amp;amp; \blacklozenge\\&#xA;\end{array}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, something happened to the scout, and he wasn't able to finish the map. However, after seeing the map, my military adviser smirked and locked himself in the operation room to discuss the tactic with my admiral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did he really manage to locate the enemy ships based on that map? Can you deduce all the ships' locations via this map?&lt;/p&gt;&#xA;" OwnerUserId="16621" LastEditorUserId="4421" LastEditDate="2015-11-04T19:07:52.953" LastActivityDate="2015-11-04T19:34:25.510" Title="Help locate enemy fleet" Tags="|logical-deduction|story|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23756" PostTypeId="2" ParentId="23755" CreationDate="2015-11-03T00:34:53.707" Score="4" Body="&lt;p&gt;$$&#xA;\begin{array}{|c|c|c|c|c|c|c|c|c|c|l}&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;\blacklozenge&amp;amp;&amp;amp;1\\&#xA;\hline&amp;amp;&amp;amp;\blacktriangle&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline&amp;amp;&amp;amp;\blacktriangledown&amp;amp;&amp;amp;&amp;amp;&amp;amp;\blacktriangle&amp;amp;&amp;amp;\blacklozenge&amp;amp;&amp;amp;3\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;\blacksquare&amp;amp;&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline&amp;amp;\blacklozenge&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;\blacktriangledown&amp;amp;&amp;amp;\blacktriangleleft&amp;amp;\blacktriangleright&amp;amp;4\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;0\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;0\\&#xA;\hline\blacktriangle&amp;amp;&amp;amp;\blacktriangleleft&amp;amp;\blacksquare&amp;amp;\blacksquare&amp;amp;\blacktriangleright&amp;amp;&amp;amp;\blacktriangleleft&amp;amp;\blacksquare&amp;amp;\blacktriangleright&amp;amp;8\\&#xA;\hline\blacktriangledown&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;\blacklozenge&amp;amp;&amp;amp;1\\&#xA;\hline2&amp;amp;1&amp;amp;3&amp;amp;1&amp;amp;1&amp;amp;1&amp;amp;3&amp;amp;1&amp;amp;5&amp;amp;2&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="14013" LastEditorUserId="4421" LastEditDate="2015-11-04T19:12:49.343" LastActivityDate="2015-11-04T19:12:49.343" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23757" PostTypeId="2" ParentId="22177" CreationDate="2015-11-03T01:50:41.593" Score="1" Body="&lt;p&gt;Because there's no way anybody could have a box with that many diamonds in it and entrust it to a bank.  Banks might keep precious metals but they don't usually keep diamonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the thief was afraid it had some kind of police tracking device.&lt;/p&gt;&#xA;" OwnerUserId="9575" LastActivityDate="2015-11-03T01:50:41.593" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23758" PostTypeId="1" AcceptedAnswerId="23767" CreationDate="2015-11-03T02:09:18.617" Score="18" ViewCount="1184" Body="&lt;ol&gt;&#xA;&lt;li&gt;It was for me that modern quantization was named, and I'm important to theorists of certain persuasions.&lt;/li&gt;&#xA;&lt;li&gt;Given a space, I can say I greatly enjoy it!&lt;/li&gt;&#xA;&lt;li&gt;My companions I'm never far from, and together we labor—though one may oppose me.&lt;/li&gt;&#xA;&lt;li&gt;I bear the moon near one end, and a tree at the other.&lt;/li&gt;&#xA;&lt;li&gt;I bend in the middle, but what makes me move are ten of one thing coming from fore of another.&lt;/li&gt;&#xA;&lt;li&gt;Some would consider my weapon to be weak, but still, it can wound deeply. An insult delivered is never followed by retraction.&lt;/li&gt;&#xA;&lt;li&gt;The wind blows across your face and I can make a blow on your face, too—but only with my friends.&lt;/li&gt;&#xA;&lt;li&gt;Life is pointless without me.&lt;/li&gt;&#xA;&lt;li&gt;Alone or with others of my kind we can compose a feelingless, prickly weed seed.&lt;/li&gt;&#xA;&lt;li&gt;I take planes into flight and feed hungry babes.&lt;/li&gt;&#xA;&lt;li&gt;I give ability to athletes as well as the deaf and the blind.&lt;/li&gt;&#xA;&lt;li&gt;A wave of happiness takes over me from time to time.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&lt;p&gt;P.S. This is my first riddle. Please be kind!&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; When and where do I reveal all, if someone has answered the riddle but people are struggling with all the details of some of the clues?&lt;/p&gt;&#xA;" OwnerUserId="10155" LastEditorUserId="10155" LastEditDate="2021-09-07T04:47:17.897" LastActivityDate="2021-09-07T04:47:17.897" Title="It was for me that modern quantization was named: what am I?" Tags="|riddle|" AnswerCount="4" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="23759" PostTypeId="2" ParentId="236" CreationDate="2015-11-03T02:11:48.960" Score="9" Body="&lt;p&gt;I started writing my definitive explanation for how everyone is actually wrong about the necessity of the Oracle's proclamation and in the process finally explained to myself why, in fact, it's essential. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possibly not adding anything new to the list of answers (how ironic is that??) I'll throw in my explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is highly unintuitive, but the way that the eye logic is deduced starts with the accusation that someone has blue eyes. The immediate response to that accusation is &quot;is it me?&quot; (by everyone on the island). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we know if we reduce this down to 2 people if they both have blue eyes they each say (to themselves) &quot;I also see someone with blue eyes&quot; and wind up sitting there for an extra day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But their thought process is &quot;what is the other person thinking? - they *know that there's a blue eyed person on the island and they know that I know that there's a blue eyed person on the island and therefore if I'm not moving it must be because they have blue eyes&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what happens if you don't have the announcement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, with one and two people it's self evident that looking at no one or one other person offers no useful info.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, with three people, intuitively you think &quot;everyone MUST see a blue eyed person&quot; but remember the issue isn't what they can see, it's what they can be sure EVERYONE else can see - so assume everyone is a pessimist and expects their own eye color to be non-blue...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A (think her eyes are brown) looks at B and thinks &quot;B sees me (A) with brown eyes and thinks her(B's) eyes are also brown and so A assumes B assumes C is staring at 2 brown eyed people and expects that her own (C's) eyes are ALSO brown.  And there's the rub... I was stuck for a while on the idea &quot;but A knows for sure that C can see B's blue eyes!!!&quot;... however, the issue isn't what A knows; The issue is what A knows B knows C knows.  And when you walk the chain of deduction, assuming everyone is a pessimist (not wanting to think they have blue eyes) the inevitable conclusion is that every person must deduce that the last person in the he thinks she thinks chain will assume there are NO blue eyed people! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quite counter intuitively, this progression can work for any number of people, so it doesn't matter if there are 3 or 3 million blue eyed people, it's still entirely logical and rational (actually inevitable) that A will come to the conclusion that person [number of blue eyed people on the island] can reasonably suspect that there are no blue eyed people on the island.  And if there are no blue eyed people on the island then there's no place from which to start one's logical countdown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the last person in the logical chain has been informed that there is indeed a blue eyed person on the island then either they will leave (seeing no one else with blue eyes) or they will stay (because they themselves see someone else with blue eyes) and the entire deduction process begins.&lt;/p&gt;&#xA;" OwnerUserId="17145" LastEditorUserId="17145" LastEditDate="2015-11-04T22:43:01.233" LastActivityDate="2015-11-04T22:43:01.233" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23760" PostTypeId="2" ParentId="23755" CreationDate="2015-11-03T03:32:58.800" Score="7" Body="&lt;p&gt;Illyasviel beat me to the answer, but I'll show one possible process by which that answer can be derived.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Final answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{|c|c|c|c|c|c|c|c|c|c|l}&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;\blacklozenge&amp;amp;&amp;amp;1\\&#xA;\hline&amp;amp;&amp;amp;\blacktriangle&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline&amp;amp;&amp;amp;\blacktriangledown&amp;amp;&amp;amp;&amp;amp;&amp;amp;\blacktriangle&amp;amp;&amp;amp;\blacklozenge&amp;amp;&amp;amp;3\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;\blacksquare&amp;amp;&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline&amp;amp;\blacklozenge&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;\blacktriangledown&amp;amp;&amp;amp;\blacktriangleleft&amp;amp;\blacktriangleright&amp;amp;4\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;0\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;0\\&#xA;\hline\blacktriangle&amp;amp;&amp;amp;\blacktriangleleft&amp;amp;\blacksquare&amp;amp;\blacksquare&amp;amp;\blacktriangleright&amp;amp;&amp;amp;\blacktriangleleft&amp;amp;\blacksquare&amp;amp;\blacktriangleright&amp;amp;8\\&#xA;\hline\blacktriangledown&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;\blacklozenge&amp;amp;&amp;amp;1\\&#xA;\hline2&amp;amp;1&amp;amp;3&amp;amp;1&amp;amp;1&amp;amp;1&amp;amp;3&amp;amp;1&amp;amp;5&amp;amp;2&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Begin with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\def\b{\color{blue}\blacksquare}&#xA;\def\ltri{\color{red}\blacktriangleleft}&#xA;\def\rtri{\color{red}\blacktriangleright}&#xA;\def\utri{\color{red}\blacktriangle}&#xA;\def\dtri{\color{red}\blacktriangledown}&#xA;\def\rsq{\color{red}\blacksquare}&#xA;\def\rdi{\color{red}\blacklozenge}&#xA;\begin{array}{|c|c|c|c|c|c|c|c|c|c|l}&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline&amp;amp;&amp;amp;\utri&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;3\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline&amp;amp;\rdi&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;4\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;0\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;0\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;\rtri&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;8\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline2&amp;amp;1&amp;amp;3&amp;amp;1&amp;amp;1&amp;amp;1&amp;amp;3&amp;amp;1&amp;amp;5&amp;amp;2&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use ?'s to mark spaces that are known to contain a ship segment, but when I do not know the correct symbol. $\b$ marks spaces that are known &lt;strong&gt;not&lt;/strong&gt; to contain a ship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, mark all the spaces known to not contain a ship. In this pass, we get this from the 0's, from the 1's that already have segments them, and all spaces adjacent to a ship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{|c|c|c|c|c|c|c|c|c|c|l}&#xA;\hline&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline\b&amp;amp;\b&amp;amp;\utri&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;1\\&#xA;\hline&amp;amp;\b&amp;amp;?&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;3\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline\b&amp;amp;\rdi&amp;amp;\b&amp;amp;&amp;amp;\b&amp;amp;\b&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;4\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;0\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;0\\&#xA;\hline&amp;amp;\b&amp;amp;&amp;amp;&amp;amp;?&amp;amp;\rtri&amp;amp;\b&amp;amp;&amp;amp;&amp;amp;&amp;amp;8\\&#xA;\hline&amp;amp;\b&amp;amp;&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline&amp;amp;\b&amp;amp;&amp;amp;&amp;amp;\b&amp;amp;\b&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline2&amp;amp;1&amp;amp;3&amp;amp;1&amp;amp;1&amp;amp;1&amp;amp;3&amp;amp;1&amp;amp;5&amp;amp;2&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the 8. We have eliminated two spaces from that 8's row, therefore all the remaining spaces must contain a ship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{|c|c|c|c|c|c|c|c|c|c|l}&#xA;\hline&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;&amp;amp;\b&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline\b&amp;amp;\b&amp;amp;\utri&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;1\\&#xA;\hline&amp;amp;\b&amp;amp;\dtri&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;&amp;amp;\b&amp;amp;&amp;amp;&amp;amp;3\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;&amp;amp;\b&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline\b&amp;amp;\rdi&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;&amp;amp;\b&amp;amp;&amp;amp;&amp;amp;4\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;0\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;0\\&#xA;\hline?&amp;amp;\b&amp;amp;\ltri&amp;amp;\rsq&amp;amp;\rsq&amp;amp;\rtri&amp;amp;\b&amp;amp;\ltri&amp;amp;\rsq&amp;amp;\rtri&amp;amp;8\\&#xA;\hline&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;1\\&#xA;\hline&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;&amp;amp;\b&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline2&amp;amp;1&amp;amp;3&amp;amp;1&amp;amp;1&amp;amp;1&amp;amp;3&amp;amp;1&amp;amp;5&amp;amp;2&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there is a 1 directly below that 8, and horizontal ships cannot have a ship next to them, even diagonally, the far left [?] must be a ^.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{|c|c|c|c|c|c|c|c|c|c|l}&#xA;\hline&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;&amp;amp;\b&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline\b&amp;amp;\b&amp;amp;\utri&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;1\\&#xA;\hline&amp;amp;\b&amp;amp;\dtri&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;&amp;amp;\b&amp;amp;&amp;amp;&amp;amp;3\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;&amp;amp;\b&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline\b&amp;amp;\rdi&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;&amp;amp;\b&amp;amp;&amp;amp;&amp;amp;4\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;0\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;0\\&#xA;\hline\utri&amp;amp;\b&amp;amp;\ltri&amp;amp;\rsq&amp;amp;\rsq&amp;amp;\rtri&amp;amp;\b&amp;amp;\ltri&amp;amp;\rsq&amp;amp;\rtri&amp;amp;8\\&#xA;\hline?&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;1\\&#xA;\hline&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;&amp;amp;\b&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline2&amp;amp;1&amp;amp;3&amp;amp;1&amp;amp;1&amp;amp;1&amp;amp;3&amp;amp;1&amp;amp;5&amp;amp;2&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the ? fills that column's 2, we can close it as a $\dtri$ and block out the rest of that row.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{|c|c|c|c|c|c|c|c|c|c|l}&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;&amp;amp;\b&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline\b&amp;amp;\b&amp;amp;\utri&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;1\\&#xA;\hline\b&amp;amp;\b&amp;amp;\dtri&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;&amp;amp;\b&amp;amp;&amp;amp;&amp;amp;3\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;&amp;amp;\b&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline\b&amp;amp;\rdi&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;&amp;amp;\b&amp;amp;&amp;amp;&amp;amp;4\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;0\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;0\\&#xA;\hline\utri&amp;amp;\b&amp;amp;\ltri&amp;amp;\rsq&amp;amp;\rsq&amp;amp;\rtri&amp;amp;\b&amp;amp;\ltri&amp;amp;\rsq&amp;amp;\rtri&amp;amp;8\\&#xA;\hline\dtri&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;1\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;&amp;amp;\b&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline2&amp;amp;1&amp;amp;3&amp;amp;1&amp;amp;1&amp;amp;1&amp;amp;3&amp;amp;1&amp;amp;5&amp;amp;2&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the 4 row can be filled in, since six spaces have been blocked out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{|c|c|c|c|c|c|c|c|c|c|l}&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;&amp;amp;\b&amp;amp;&amp;amp;\b&amp;amp;1\\&#xA;\hline\b&amp;amp;\b&amp;amp;\utri&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;1\\&#xA;\hline\b&amp;amp;\b&amp;amp;\dtri&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;&amp;amp;\b&amp;amp;&amp;amp;\b&amp;amp;3\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;1\\&#xA;\hline\b&amp;amp;\rdi&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;?&amp;amp;\b&amp;amp;\ltri&amp;amp;\rtri&amp;amp;4\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;0\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;0\\&#xA;\hline\utri&amp;amp;\b&amp;amp;\ltri&amp;amp;\rsq&amp;amp;\rsq&amp;amp;\rtri&amp;amp;\b&amp;amp;\ltri&amp;amp;\rsq&amp;amp;\rtri&amp;amp;8\\&#xA;\hline\dtri&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;1\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;&amp;amp;\b&amp;amp;&amp;amp;\b&amp;amp;1\\&#xA;\hline2&amp;amp;1&amp;amp;3&amp;amp;1&amp;amp;1&amp;amp;1&amp;amp;3&amp;amp;1&amp;amp;5&amp;amp;2&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can fill in the rows above that 4, by the same elimination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{|c|c|c|c|c|c|c|c|c|c|l}&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;&amp;amp;\b&amp;amp;1\\&#xA;\hline\b&amp;amp;\b&amp;amp;\utri&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;1\\&#xA;\hline\b&amp;amp;\b&amp;amp;\dtri&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;?&amp;amp;\b&amp;amp;?&amp;amp;\b&amp;amp;3\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;?&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;1\\&#xA;\hline\b&amp;amp;\rdi&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\dtri&amp;amp;\b&amp;amp;\ltri&amp;amp;\rtri&amp;amp;4\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;0\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;0\\&#xA;\hline\utri&amp;amp;\b&amp;amp;\ltri&amp;amp;\rsq&amp;amp;\rsq&amp;amp;\rtri&amp;amp;\b&amp;amp;\ltri&amp;amp;\rsq&amp;amp;\rtri&amp;amp;8\\&#xA;\hline\dtri&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;1\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;&amp;amp;\b&amp;amp;1\\&#xA;\hline2&amp;amp;1&amp;amp;3&amp;amp;1&amp;amp;1&amp;amp;1&amp;amp;3&amp;amp;1&amp;amp;5&amp;amp;2&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we can place the last two patrol boats, in the only empty spaces remaining.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{|c|c|c|c|c|c|c|c|c|c|l}&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\rdi&amp;amp;\b&amp;amp;1\\&#xA;\hline\b&amp;amp;\b&amp;amp;\utri&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;1\\&#xA;\hline\b&amp;amp;\b&amp;amp;\dtri&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\utri&amp;amp;\b&amp;amp;\rdi&amp;amp;\b&amp;amp;3\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\rsq&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;1\\&#xA;\hline\b&amp;amp;\rdi&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\dtri&amp;amp;\b&amp;amp;\ltri&amp;amp;\rtri&amp;amp;4\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;0\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;0\\&#xA;\hline\utri&amp;amp;\b&amp;amp;\ltri&amp;amp;\rsq&amp;amp;\rsq&amp;amp;\rtri&amp;amp;\b&amp;amp;\ltri&amp;amp;\rsq&amp;amp;\rtri&amp;amp;8\\&#xA;\hline\dtri&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;1\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\rdi&amp;amp;\b&amp;amp;1\\&#xA;\hline2&amp;amp;1&amp;amp;3&amp;amp;1&amp;amp;1&amp;amp;1&amp;amp;3&amp;amp;1&amp;amp;5&amp;amp;2&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="16990" LastEditorUserId="4421" LastEditDate="2015-11-04T19:34:25.510" LastActivityDate="2015-11-04T19:34:25.510" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23761" PostTypeId="2" ParentId="23740" CreationDate="2015-11-03T03:41:39.973" Score="0" Body="&lt;p&gt;A quick monte carlo simulation ($N=100,000$) for $n\leq20$ is revealing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In each of the lines at the bottom of this post, the &lt;strong&gt;first number is $n$&lt;/strong&gt;, the &lt;strong&gt;second is the number of intercepts&lt;/strong&gt;, and the &lt;strong&gt;third is the count&lt;/strong&gt; of simulations that had the number of intercepts.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, even for $n$ as large as 20, and $N$ as large as 100,000, there was simply no intercept seen.  (I was able to choose points in my code and see intercepts, so it wasn't an obvious coding error).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that the answer is 0 for all $n$.  Making an intercept is simply very difficult, even for quite large $n$.  I suppose eventually an increase in intercepts might happen, but I'm not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 0 100000&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;4 0 100000&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;5 0 100000&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;6 0 100000&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;7 0 100000&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;8 0 100000&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;9 0 100000&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;10 0 100000&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;11 0 100000&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;12 0 100000&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;13 0 100000&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;14 0 100000&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;15 0 100000&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;16 0 100000&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;17 0 100000&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;18 0 100000&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;19 0 100000&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;20 0 100000&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: Even with a few larger $n$'s, there are no interceptions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;100 0 1000&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;200 0 1000&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;300 0 1000&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;400 0 1000&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;500 0 1000&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;600 0 1000&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2015-11-03T04:36:38.647" LastActivityDate="2015-11-03T04:36:38.647" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23762" PostTypeId="2" ParentId="23758" CreationDate="2015-11-03T08:16:00.373" Score="3" Body="&lt;p&gt;Are you &lt;code&gt;Carbon Dioxide&lt;/code&gt;?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;CO2 is used in lasers which is the birth place of quantization in physics. &#xA;It's a gas so it expands in space.&#xA;The companions are very close since compounds are bonded. One (Carbon) opposes the other two (Oxygen)&#xA;CO2 is at the moon and what trees inhale. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;CO2 molecular compound bends in structure (&lt;a href=&quot;http://linoit.com/entry/image/1222890&quot; rel=&quot;nofollow&quot;&gt;http://linoit.com/entry/image/1222890&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bonds for CO2 is made of 2 free electrons PAIRS on each end of the oxygens (2*2*2 = 8), plus two double bonds (one from each oxygen to the carbon (8+2 = 10)). Coming off of 4 double bonds. (&lt;a href=&quot;http://www.middleschoolchemistry.com/img/content/multimedia/chapter_4/lesson_6/covalent_bonding_carbon_dioxide_3.jpg&quot; rel=&quot;nofollow&quot;&gt;http://www.middleschoolchemistry.com/img/content/multimedia/chapter_4/lesson_6/covalent_bonding_carbon_dioxide_3.jpg&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CO2 isn't that deadly, but a lot of it can be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CO2 is in the air so when the wind blows, it blows too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our lives can't exist without C, CO2, and O2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest of your details were added after I answered &lt;/p&gt;&#xA;" OwnerUserId="16599" LastEditorUserId="16599" LastEditDate="2015-11-03T10:13:07.587" LastActivityDate="2015-11-03T10:13:07.587" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23763" PostTypeId="2" ParentId="16477" CreationDate="2015-11-03T10:03:28.097" Score="6" Body="&lt;p&gt;My conclusion is that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Scandal is inevitable&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is demonstrated thus:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because there are the same number of mementoes as islanders, all mementoes are in circulation. This means that if you have $x$ friends and it is your birthday, you have $x$ mementoes in your possession, and you give them all out. Over the course of the year you then get one memento from each of your $x$ friends, resulting in having $x$ mementoes again by your birthday.&lt;br&gt;&#xA;&lt;br&gt;&#xA;  This means that if the castaway befriends someone who has their birthday &lt;em&gt;before&lt;/em&gt; his, then that islander won't have enough mementoes to give out = &lt;strong&gt;scandal!&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;  But surely our castaway can befriend an islander whose birthday isn't for a while? He can only befriend one islander, because he only has one memento to give out. So the castaway has his birthday first, and gives his islander friend the memento. All good. And on that islander's birthday he gives the castaway a different memento. All good. But the castaway still only has one memento, so he can't befriend any more islanders, and he can't break off his existing friendship. Which means that when his islander friend's birthday comes around &lt;strong&gt;&lt;em&gt;the only memento he has to give is the one his islander friend recently gave him&lt;/em&gt; = scandal!&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;  The only slight consolation is that the castaway has at least a year before the scandal will erupt. I suggest he works on an escape plan!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9444" LastActivityDate="2015-11-03T10:03:28.097" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23764" PostTypeId="2" ParentId="23740" CreationDate="2015-11-03T11:43:47.050" Score="4" Body="&lt;p&gt;I also did a Monte Carlo simulation (N=10,000) and got a linear increase in the number of intercepts with $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n\leq2$ there are no intercepts (excluding cases where two points are identical) by simple geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $3&amp;lt;n\leq20$ I get the following plot of average intercepts against $n$, which is approximately linear given the low number of repeats. (I didn't do n=3 because I initially thought it couldn't have any intercepts, but it can.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OFtjg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OFtjg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With an equation of approximately $$I=0.075n-0.3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly the number of intercepts for a given $n$ seems to be approximately binomial, which leads me to believe there may be an analytical solution to this although its beyond my knowledge.&lt;/p&gt;&#xA;" OwnerUserId="17239" LastEditorUserId="17239" LastEditDate="2015-11-04T09:27:50.683" LastActivityDate="2015-11-04T09:27:50.683" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23765" PostTypeId="2" ParentId="23758" CreationDate="2015-11-03T12:02:45.307" Score="5" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fingers/arm (some clues seem to refer to one, and some to the other)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It was for me that modern quantization was named, and I'm important to theorists of certain persuasions.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Our current counting system (base 10) came about because people (typically) have 10 fingers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Given a space, I can say I greatly enjoy it!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People like to stretch out their arms, given the space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My companions I'm never far from, and together we labor—though one may oppose me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The fingers are together on the hand, and work together, but the thumb is opposable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I bear the moon near one end, and a tree at the other.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; At one end of the arm is the head (moon), and at the other end are the fingers (tree).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I bend in the middle, but what makes me move are ten of one thing coming from fore of another.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The arm bends in the middle (elbow).  The ten fingers come from the fore of the arm.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some would consider my weapon to be weak, but still, it can wound deeply. An insult delivered is never followed by retraction.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A fist (punch) may not be as powerful as other weapons, but if delivered properly, can cause serious damage.  Once a punch is delivered, it can't be taken back.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The wind blows across your face and I can make a blow on your face, too—but only with my friends.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Another reference to a punch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Life is pointless without me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can't point without fingers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alone or with others of my kind we can compose a feelingless, prickly weed seed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I take planes into flight and feed hungry babes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It would be very difficult to do these things without fingers/arms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I give ability to athletes as well as the deaf and the blind.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Similarly, athletes without fingers/arms would have a hard time.  The deaf and the blind can use their fingers to communicate (sign language/braille).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A wave of happiness takes over me from time to time.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2015-11-03T12:02:45.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23766" PostTypeId="2" ParentId="23758" CreationDate="2015-11-03T12:06:50.643" Score="1" Body="&lt;p&gt;Well I'm not at all convinced by this answer. There seem to be way too many &lt;em&gt;Hmmmms&lt;/em&gt;  for my liking. But it has some convincing bits too, so I decided to post it, so here it goes...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Plan(c)k&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It was for me that modern quantization was named, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Max Planck gave a solution to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ultraviolet_catastrophe&quot; rel=&quot;nofollow&quot;&gt;Ultraviolet Catastrophe&lt;/a&gt;, introducing the &lt;a href=&quot;https://en.wikipedia.org/wiki/Planck_constant&quot; rel=&quot;nofollow&quot;&gt;Planck Constant&lt;/a&gt;, preempting further quantum work by Einstein, Schrödinger and Heisenberg.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and I'm important to theorists of certain persuasions.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;a href=&quot;https://en.wikipedia.org/wiki/Planck_length&quot; rel=&quot;nofollow&quot;&gt;Planck length&lt;/a&gt; and Planck time are important concepts in theoretical Physics&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Given a space, I can say I greatly enjoy it!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Possibly a reference to &lt;a href=&quot;https://en.wikipedia.org/wiki/Planking_(fad)&quot; rel=&quot;nofollow&quot;&gt;Planking&lt;/a&gt;???&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My companions I'm never far from, and together we labor—though one may oppose me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; On a building site there are lots of planks, for instance in scaffolding, that work together. Not sure about the &quot;oppose me&quot; bit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I bear the moon near one end, and a tree at the other.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hmm...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I bend in the middle, but what makes me move are ten of one thing coming from fore of another.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Planks are often supported at both ends, but bend in the middle. To move a plank, you could use ten fingers from four hands.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some would consider my weapon to be weak, but still, it can wound deeply. An insult delivered is never followed by retraction.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A plank can give you splinters. Doesn't sound like much, but a deep splinter hurts, and splinters usually get stuck in.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The wind blows across your face and I can make a blow on your face, too—but only with my friends.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you assemble planks with a slight overhang, you can get a blow to your face,  a bit like standing on a garden rake.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Life is pointless without me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hmmm...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alone or with others of my kind we can compose a feelingless, prickly weed seed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hmmmm...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I take planes into flight and feed hungry babes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first planes were wooden, and from a tree you could get some food&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I give ability to athletes as well as the deaf and the blind.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hmmmm...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A wave of happiness takes over me from time to time.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hmmmmm...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="7316" LastActivityDate="2015-11-03T12:06:50.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23767" PostTypeId="2" ParentId="23758" CreationDate="2015-11-03T12:50:39.123" Score="18" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are a &lt;strong&gt;digit&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It was for me that modern quantization was named, and I'm important to theorists of certain persuasions.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Digits are used in counting, and ratios of fingers are studied by &quot;digit theorists&quot; (yes, it is a thing).&#xA; &lt;strong&gt;Puzzle creator answer:&lt;/strong&gt; the whole idea of digital instead of analog is to define discrete steps or islands within the space of infinite variability of a signal or value, then round off the analog values to the nearest island (or “digit”). In a sound wave for example, rounding when done horizontally in time for the purposes of music creation is called quantization, and when done vertically it is what is normally thought of as digitization. Math is a highly theoretical discipline and I’m pretty sure they use digits in it. (Smile!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Given a space, I can say I greatly enjoy it!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Insert a space, and As Tim Couwelier commented, you can &quot;dig it&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My companions I'm never far from, and together we labor—though one may oppose me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The thumb opposes the other fingers, and in some simians, one of the toes may oppose the other toes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I bear the moon near one end, and a tree at the other. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Moon = cuticle(lunula)/nail; Tree = Palm&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I bend in the middle, but what makes me move are ten of one thing coming from fore of another.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Ten&quot;dons come from the hands and feet to allow the digits to move and bend, as well as the &quot;fore&quot; arm.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some would consider my weapon to be weak, but still, it can wound deeply. An insult delivered is never followed by retraction.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pointing the finger in some societies is considered rude.&#xA; &lt;strong&gt;Puzzle creator answer:&lt;/strong&gt; a human's finger nails don't compare to the claws of an animal, but can still scratch hard enough to draw blood. An &quot;insult&quot; delivered by fingernails (a scratched wound) is never followed by retraction of those nails--humans do not have retractable nails.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The wind blows across your face and I can make a blow on your face, too—but only with my friends.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You need your digits to create a fist (or slap).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Life is pointless without me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can't point without me (well, you can, but semantics...)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alone or with others of my kind we can compose a feelingless, prickly weed seed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;~ Numb burrs (&quot;numbers&quot;)  [Thanks to Fillet]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I take planes into flight and feed hungry babes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Difficult to do things without your digits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I give ability to athletes as well as the deaf and the blind.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Athletes generally use their fingers/toes, and so do the blind and deaf (to read).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A wave of happiness takes over me from time to time. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jazz hands/spirit fingers.&#xA; &lt;strong&gt;Puzzle creator answer:&lt;/strong&gt; Simply waving at someone you're happy to see is all it takes for a &quot;wave of happiness&quot; to take over your digit(s)!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="14233" LastEditorUserId="10155" LastEditDate="2018-06-08T17:09:08.527" LastActivityDate="2018-06-08T17:09:08.527" CommentCount="14" ContentLicense="CC BY-SA 4.0" />
  <row Id="23768" PostTypeId="1" AcceptedAnswerId="23770" CreationDate="2015-11-03T18:20:53.367" Score="7" ViewCount="517" Body="&lt;p&gt;My adviser's &lt;a href=&quot;https://puzzling.stackexchange.com/a/23756/16621&quot;&gt;brilliance&lt;/a&gt; crushed the Blackbeard's fleet, and we held a banquet to celebrate. The banquet was short lived however, for a arrow flew right past me and struck a pillar. Attached was a message from Blackbeard. It reads&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Arrr, how dare you attack me branch fleet while I was out swashbucklin'. By the name of meself, I will send you to Davy Jones' locker.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I realized right away that I had made an error in my previous attack. It was indeed too little to be called the Blackbeard Grand Fleet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quickly, I dismissed the banquet and summoned my military adviser to discuss how we should prepare. He advised that we strike them in the high sea instead, as fighting at home would potentially put all the citizens in danger. He further said that Blackbeard would most likely sail directly from where the previous engagement was, so we should prepare to scout in that general direction right away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus once again, scouts were sent out. But by the time we received a message, it was near nightfall. It read&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We have found a fleet wearing Blackbeard's Jolly Roger. However, because&#xA;  it is dark, I don't see Blackbeard anywhere. To make things worse, the fog&#xA;  makes it hard to pinpoint the location of their ship, so I can only attach&#xA;  a map with how many ship pieces there are on each row and column. Based on my&#xA;  first speculation, there's again 1 flagship, 2 frigates, 3 corvettes, and 4&#xA;  patrol boats, and I do not believe they touch each other. Not even&#xA;  diagonally.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{|c|c|c|c|c|c|c|c|c|c|l}&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;4\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;\bigcirc&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;5\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;2\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;2\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;0\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;3\\&#xA;\hline1&amp;amp;1&amp;amp;2&amp;amp;3&amp;amp;1&amp;amp;7&amp;amp;1&amp;amp;2&amp;amp;0&amp;amp;2&#xA;\end{array}&#xA;\quad\quad&#xA;\bbox[4pt,border: 1px solid black]{&#xA;\begin{array}{llc}&#xA;\bf{Legend:}&amp;amp;\\&#xA;\hline&#xA;\text{Flagship:}&amp;amp; \blacktriangleleft\blacksquare\blacksquare\blacktriangleright \text{or} &amp;amp;\blacktriangle\\&#xA;&amp;amp;&amp;amp;\blacksquare\\&#xA;&amp;amp;&amp;amp;\blacksquare\\&#xA;&amp;amp;&amp;amp;\blacktriangledown\\&#xA;\hline&#xA;\text{Frigate:}&amp;amp; \blacktriangleleft\blacksquare\blacktriangleright \text{or} &amp;amp;\blacktriangle\\&#xA;&amp;amp;&amp;amp;\blacksquare\\&#xA;&amp;amp;&amp;amp;\blacktriangledown\\&#xA;\hline&#xA;\text{Corvette:}&amp;amp; \blacktriangleleft\blacktriangleright \text{or} &amp;amp;\blacktriangle\\&#xA;&amp;amp;&amp;amp;\blacktriangledown\\&#xA;\hline&#xA;\text{Patrol Boat:}&amp;amp; \blacklozenge\\&#xA;\hline&#xA;\text{HUGE rock:}&amp;amp; \bigcirc&#xA;\end{array}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My adviser looked at the map for a few minutes, and said to me: this is most likely his vanguard fleet; we still need to crush them, but his main fleet should be not far behind, and I should send the scouts there instead. Then, he quickly disappeared into the operation room like before, leaving me pondering how he managed to find the ships using this map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you guys help me figure out the enemy location again?&lt;/p&gt;&#xA;" OwnerUserId="16621" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-11-05T16:19:16.343" Title="Prepare for the counterattack" Tags="|logical-deduction|story|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23769" PostTypeId="1" AcceptedAnswerId="23784" CreationDate="2015-11-03T18:28:24.937" Score="4" ViewCount="283" Body="&lt;p&gt;The ringmaster of a flea circus draws a square $ABCD$ with corners &#xA;$A=(+1,+1)$, $B=(+1,-1)$, $C=(-1,+1)$, $D=(-1,-1)$ in the Euclidean &#xA;plane and picks a point $P$ with &lt;strong&gt;integer coordinates&lt;/strong&gt; outside the square.&#xA;The ringmaster then places a single flea onto point $P$.&#xA;Whenever the ringmaster shouts &lt;em&gt;&quot;Hop!&quot;&lt;/em&gt;, the flea looks around itself &#xA;and picks a corner $y$ of the square:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If there is a unique rightmost corner (from the flea's current point &#xA;of view), then $y$ becomes this unique rightmost corner.&lt;/li&gt;&#xA;&lt;li&gt;If the rightmost corner is not unique (as the flea is sitting on the &#xA;same line with two neighboring corners $y_1$ and $y_2$ of the square), then &#xA;$y$ is chosen as the rightmost corner that is farther away from the flea.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The flea then jumps over the picked corner $y$ to the mirror point on &#xA;the  other side. (In other words, the flea sitting in point $x$ may &#xA;jump over corner $y$ to the new point $z$, so that $y$ is the midpoint &#xA;between $x$ and $z$.) &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; &#xA;  Is it possible that after a finite number of jumps the flea reaches &#xA;  a point $Q$, so that the distance between origin and $Q$ is at least &#xA;  $30$ times the distance between origin and $P$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-11-04T17:45:28.867" LastActivityDate="2015-11-04T19:09:03.580" Title="A final incident in the flea circus: Part 1" Tags="|mathematics|combinatorics|reachability|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23770" PostTypeId="2" ParentId="23768" CreationDate="2015-11-03T19:07:04.033" Score="4" Body="&lt;p&gt;$$&#xA;\begin{array}{|c|c|c|c|c|c|c|c|c|c|l}&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;\blacktriangleleft&amp;amp;\blacksquare&amp;amp;\blacktriangleright&amp;amp;&amp;amp;\blacktriangle&amp;amp;4\\&#xA;\hline\blacktriangleleft&amp;amp;\blacksquare&amp;amp;\blacksquare&amp;amp;\blacktriangleright&amp;amp;&amp;amp;\bigcirc&amp;amp;&amp;amp;&amp;amp;&amp;amp;\blacktriangledown&amp;amp;5\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;\blacktriangle&amp;amp;&amp;amp;\blacklozenge&amp;amp;&amp;amp;&amp;amp;2\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;\blacklozenge&amp;amp;&amp;amp;\blacksquare&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;2\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;\blacktriangledown&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;\blacklozenge&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;\blacktriangle&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;\blacktriangledown&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;0\\&#xA;\hline&amp;amp;&amp;amp;\blacklozenge&amp;amp;&amp;amp;\blacktriangleleft&amp;amp;\blacktriangleright&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;3\\&#xA;\hline1&amp;amp;1&amp;amp;2&amp;amp;3&amp;amp;1&amp;amp;7&amp;amp;1&amp;amp;2&amp;amp;0&amp;amp;2&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First you eliminate the row and column with 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\def\b{\color{blue}\blacksquare}&#xA;\def\ltri{\color{red}\blacktriangleleft}&#xA;\def\rtri{\color{red}\blacktriangleright}&#xA;\def\utri{\color{red}\blacktriangle}&#xA;\def\dtri{\color{red}\blacktriangledown}&#xA;\def\rsq{\color{red}\blacksquare}&#xA;\def\rdi{\color{red}\blacklozenge}&#xA;\begin{array}{|c|c|c|c|c|c|c|c|c|c|l}&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;\b&amp;amp;&amp;amp;4\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;\bigcirc&amp;amp;&amp;amp;&amp;amp;\b&amp;amp;&amp;amp;5\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;\b&amp;amp;&amp;amp;2\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;\b&amp;amp;&amp;amp;2\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;\b&amp;amp;&amp;amp;1\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;\b&amp;amp;&amp;amp;1\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;\b&amp;amp;&amp;amp;1\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;\b&amp;amp;&amp;amp;1\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;0\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;\b&amp;amp;&amp;amp;3\\&#xA;\hline1&amp;amp;1&amp;amp;2&amp;amp;3&amp;amp;1&amp;amp;7&amp;amp;1&amp;amp;2&amp;amp;0&amp;amp;2&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sixth column can at most have one more free space so we can fill some of it in and add some more free spaces. 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be filled}\\&#xA;^\dagger\text{at most two}\\&#xA;^{\dagger\dagger}\text{this column needs a corvette and a frigate in the middle}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;step&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{|c|c|c|c|c|c|c|c|c|c|ll}&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\ltri&amp;amp;\rsq&amp;amp;\rtri&amp;amp;\b&amp;amp;\utri&amp;amp;4\\&#xA;\hline\ltri&amp;amp;\rsq&amp;amp;\rsq&amp;amp;\rtri&amp;amp;\b&amp;amp;\bigcirc&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\dtri&amp;amp;5\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;?&amp;amp;\b&amp;amp;\rdi&amp;amp;\b&amp;amp;\b&amp;amp;2\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;&amp;amp;\b&amp;amp;?&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;2\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;&amp;amp;\b&amp;amp;&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;1\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;&amp;amp;\b&amp;amp;&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;1\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;?&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;1\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;?&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;1\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;0\\&#xA;\hline\b&amp;amp;\b&amp;amp;\rdi&amp;amp;\b&amp;amp;\ltri&amp;amp;\rtri&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;3^*\\&#xA;\hline1&amp;amp;1&amp;amp;2&amp;amp;3&amp;amp;1&amp;amp;7&amp;amp;1&amp;amp;2&amp;amp;0&amp;amp;2\\&#xA;\end{array}&#xA;\\&#xA;^*\text{must be filled but no more frigates}\\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;step&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{|c|c|c|c|c|c|c|c|c|c|ll}&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\ltri&amp;amp;\rsq&amp;amp;\rtri&amp;amp;\b&amp;amp;\utri&amp;amp;4\\&#xA;\hline\ltri&amp;amp;\rsq&amp;amp;\rsq&amp;amp;\rtri&amp;amp;\b&amp;amp;\bigcirc&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\dtri&amp;amp;5\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;?&amp;amp;\b&amp;amp;\rdi&amp;amp;\b&amp;amp;\b&amp;amp;2\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\rdi&amp;amp;\b&amp;amp;?&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;2\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;1\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\rdi&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;1\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;?&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;1\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;?&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;1\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;0\\&#xA;\hline\b&amp;amp;\b&amp;amp;\rdi&amp;amp;\b&amp;amp;\ltri&amp;amp;\rtri&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;3\\&#xA;\hline1&amp;amp;1&amp;amp;2&amp;amp;3^*&amp;amp;1&amp;amp;7&amp;amp;1&amp;amp;2&amp;amp;0&amp;amp;2\\&#xA;\end{array}&#xA;\\&#xA;^*\text{only 2 patrols left for here}\\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;last step solving sixth column&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{|c|c|c|c|c|c|c|c|c|c|ll}&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\ltri&amp;amp;\rsq&amp;amp;\rtri&amp;amp;\b&amp;amp;\utri&amp;amp;4\\&#xA;\hline\ltri&amp;amp;\rsq&amp;amp;\rsq&amp;amp;\rtri&amp;amp;\b&amp;amp;\bigcirc&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\dtri&amp;amp;5\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\utri&amp;amp;\b&amp;amp;\rdi&amp;amp;\b&amp;amp;\b&amp;amp;2\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\rdi&amp;amp;\b&amp;amp;\rsq&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;2\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\dtri&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;1\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\rdi&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;1\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\utri&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;1\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\dtri&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;1\\&#xA;\hline\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;0\\&#xA;\hline\b&amp;amp;\b&amp;amp;\rdi&amp;amp;\b&amp;amp;\ltri&amp;amp;\rtri&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;\b&amp;amp;3\\&#xA;\hline1&amp;amp;1&amp;amp;2&amp;amp;3&amp;amp;1&amp;amp;7&amp;amp;1&amp;amp;2&amp;amp;0&amp;amp;2\\&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="16608" LastEditorUserId="16608" LastEditDate="2015-11-05T16:19:16.343" LastActivityDate="2015-11-05T16:19:16.343" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23771" PostTypeId="2" ParentId="23769" CreationDate="2015-11-03T20:06:22.100" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Yes it is possible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We will first position our flea at position $(+3,+1)$ and after each jump we will rotate the plane around the origin to get our flea back to positive coordinates. With this our flea will always have to jump over corner (-1,+1) and although it might sound strange to do so the flea will keep its relative position to the origin and the four corners and make this solution easier to understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our first jump will be to point $(-5, +1)$ rotate 90 degrees to get to position $(+1, +5)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Until we are back at a position $(?, +1)$ we can keep jumping and rotatin 180 degrees and with each such jump our position gets changed by the vector $(+2, -2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our first jump for example will get us to position $(-3, -3)$ which rotated 180 degrees gives us $(+3, +3)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next jump to $(-5, -1)$ rotated $(+5, +1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point we can already see a pattern. Each time we reach a position $(?, +1)$ our position will have changed by the vector $(+2, 0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we can repeat this infinitely we will slowly get further and further away from the origin and up to infinite distance, which is obviously more than 30 times the original distance.&lt;/p&gt;&#xA;" OwnerUserId="16608" LastActivityDate="2015-11-03T20:06:22.100" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23772" PostTypeId="1" AcceptedAnswerId="23778" CreationDate="2015-11-03T20:35:55.287" Score="10" ViewCount="928" Body="&lt;p&gt;I am amazed at my adviser's &lt;a href=&quot;https://puzzling.stackexchange.com/a/23770/16621&quot;&gt;brilliance&lt;/a&gt; once again. Blackbeard's vanguard fleet was quickly sunk. And as luck would have it, the moment we finished cleaning up the mess, the scout returned with the lost scout from the first engagement. He seemed extremely excited and kept pointing to a piece of paper. Upon closer look, it is another map the scout drew.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{|c|c|c|c|c|c|c|c|l}&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;4\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;2\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;5\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;2\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;6\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;4\\&#xA;\hline3&amp;amp;1&amp;amp;5&amp;amp;2&amp;amp;5&amp;amp;2&amp;amp;2&amp;amp;5&#xA;\end{array}&#xA;\quad\quad&#xA;\bbox[4pt,border: 1px solid black]{&#xA;\begin{array}{llc}&#xA;\bf{Legend:}&amp;amp;\\&#xA;\hline&#xA;\text{Man-of-War:}&amp;amp; \blacktriangleleft\blacksquare\blacksquare\blacksquare\blacktriangleright \text{or} &amp;amp;\blacktriangle\\&#xA;&amp;amp;&amp;amp;\blacksquare\\&#xA;&amp;amp;&amp;amp;\blacksquare\\&#xA;&amp;amp;&amp;amp;\blacksquare\\&#xA;&amp;amp;&amp;amp;\blacktriangledown\\&#xA;\hline&#xA;\text{Queen Anne's Revenge:}&amp;amp; \blacktriangleleft\blacksquare\blacksquare\blacktriangleright \text{or} &amp;amp;\blacktriangle\\&#xA;&amp;amp;&amp;amp;\blacksquare\\&#xA;&amp;amp;&amp;amp;\blacksquare\\&#xA;&amp;amp;&amp;amp;\blacktriangledown\\&#xA;\hline&#xA;\text{Frigate:}&amp;amp; \blacktriangleleft\blacksquare\blacktriangleright \text{or} &amp;amp;\blacktriangle\\&#xA;&amp;amp;&amp;amp;\blacksquare\\&#xA;&amp;amp;&amp;amp;\blacktriangledown\\&#xA;\hline&#xA;\text{Corvette:}&amp;amp; \blacktriangleleft\blacktriangleright \text{or} &amp;amp;\blacktriangle\\&#xA;&amp;amp;&amp;amp;\blacktriangledown\\&#xA;\hline&#xA;\text{Patrol Boat:}&amp;amp; \blacklozenge&#xA;\end{array}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After he finally calmed down, he told me the following&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I have spotted Blackbeard himself on his Queen Anne's Revenge.&#xA;  However, in addition to his standard setup (1 flagship, 2&#xA;  frigates, 3 corvettes, and 4 patrol boats), there is a man-of-war with&#xA;  74 cannons. The size of man-of-war is bigger than even Blackbeard's&#xA;  flagship. Their formation is slightly different though, and they are&#xA;  closer together, but interestingly, none of the ships touch each&#xA;  other, not even diagonally. Unfortunately, since it was still dark, I wasn't able to&#xA;  pinpoint their exactly location, but I have drawn the map with the number&#xA;  of ship pieces on each column and row.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My adviser pondered at the map for a while, and then smiled.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Men, it seems like Blackbeard himself isn't aware of our presence yet,&#xA;  we must strike them now before dawn or we will face Blackbeard's full&#xA;  wrath. I have figured out their exact formation, so follow my orders&#xA;  and I will lead you into victory!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With a cheer from the rest of the sailors, everyone frantically prepare to sail again, leaving me behind pondering just how he managed to find out the enemy fleet formation, and why the scout keeps adding a legend but not using it.&lt;/p&gt;&#xA;" OwnerUserId="16621" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-11-09T11:38:00.183" Title="The final showdown" Tags="|logical-deduction|story|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23773" PostTypeId="1" CreationDate="2015-11-03T21:53:00.917" Score="12" ViewCount="836" Body="&lt;p&gt;Chuck Asimov is a keen inventor and balloonist.&#xA;He has developed some humanoid robots and decided to take nine of them (conveniently numbered 1 to 9) on a trip in his hot air balloon.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately the weather conditions have deteriorated rapidly and the balloon has been blown off-course - it is heading for a range of rocky mountains!&#xA;The balloon is going to crash unless he can lose some weight from the balloon, and quickly. If the balloon crashes, Chuck will be killed and all the robots in the basket destroyed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chuck weighs 100kg. Each of the robots also weighs 100kg. &#xA;The balloon will only make it to safety if the total weight of the occupants of the basket can be reduced to 400kg.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it was up to Chuck, he would throw six of the robots over the side - he can always build new ones - but it is not up to Chuck.&#xA;The robots are stronger than he is, so they will decide who gets thrown over the side.&#xA;No only that, but Chuck hurt his back recently and couldn't even throw a deactivated robot over the side.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now Chuck has programmed a set of laws into his robots which are similar (but not identical to) those formulated by his distant relative.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Law 1: A robot may not cause harm to a human, or by inaction allow a human to come to harm.&lt;/li&gt;&#xA;&lt;li&gt;Law 2: A robot may not harm another robot, or by inaction allow another robot to come to harm unless that robot is self-sacrificing.&lt;/li&gt;&#xA;&lt;li&gt;Law 3: A robot must protect itself from harm.  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The laws are, of course, prioritised so that a robot will always attempt to follow them except where doing so would conflict with an earlier law. The laws activate when the robots find themselves in emergency conditions and the robots will begin to take action in their numerical order.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When Chuck realises the peril he is in, he announces the state of emergency and expects robots 1 to 6 to throw themselves overboard in order to save the other occupants of the balloon.&#xA;Unfortunately for Chuck, his programming skills are not as good as his mechanical skills and the laws have not taken effect properly in all the robots.&#xA;For each robot there is only a 50% chance per law that it has taken effect. Some robots may have all three laws correctly programmed, others may have any two, one or even none.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What really happens now is a matter of chance.&#xA;The robots begin to take actions in their numerical order, but in very quick succession.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If a robot has no laws active, it will perform no actions.&lt;/li&gt;&#xA;&lt;li&gt;If a robot has laws 2 and 3 in place, it will attempt to throw Chuck overboard.&lt;/li&gt;&#xA;&lt;li&gt;Chuck is able to deactivate the first robot (in each round of moves) attempting to throw him overboard, but not subsequent robots in the same round.&lt;/li&gt;&#xA;&lt;li&gt;A robot with law 3 in place, but lacking law 2 will pick any robot at random and attempt to throw them overboard.&lt;/li&gt;&#xA;&lt;li&gt;A robot with law 3 in place but lacking laws 1 and 2 will pick Chuck or any random robot and attempt to throw them overboard (equal probabilities).&lt;/li&gt;&#xA;&lt;li&gt;A single robot can counter the agressive actions of another robot towards Chuck, another robot, or themselves where those actions conflict with their own programming. Both robots will become deactivated (but remain in the basket).  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;After all robots have acted, any robots remaining in the basket who have not been deactivated will get another round of moves to act in the same sequence as before. This continues until all robots have been deactivated or the crash has been averted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The robots don't think several moves down the line - they are concerned only with immediate actions and outcomes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What is the probability that Chuck survives his balloon trip?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus Question:&lt;/strong&gt; How many robots would there need to have been in the basket initially for Chuck's chance of survival to be 10% or less?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each robot plans its move, considering the planned actions of all the preceding robots. &#xA;The first robot in any round that chooses to attack Chuck has its attack instantly negated and becomes deactivated. &#xA;Self-sacrifice moves are also regarded as being instantaneous.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;L--- No laws are active, so the robot does nothing.&lt;/li&gt;&#xA;&lt;li&gt;L--3 If the robot is currently under attack, it will defend itself. Otherwise it will choose another occupant at random and throw them overboard.&lt;/li&gt;&#xA;&lt;li&gt;L-2- If any other robot is currently subject to an unblocked attack, it will defend that robot. If multiple robots are under threat, it will choose to defend the robot attacked by the lowest-numbered attacker. If no robots are under threat it will either throw Chuck out or self-sacrifice (50% of each).&lt;/li&gt;&#xA;&lt;li&gt;L-23 If any other robot is under attack, it will defend that robot. Otherwise if it is under attack, it will defend itself. Otherwise it will throw Chuck out.&lt;/li&gt;&#xA;&lt;li&gt;L1-- If Chuck is under attack, it will defend him. Otherwise it will choose a random robot (including itself as a target) and throw them out.&lt;/li&gt;&#xA;&lt;li&gt;L1-3 If Chuck is under attack it will attempt to defend him. Otherwise if it is under attack it will self-defend. Otherwise it will throw out a random robot.&lt;/li&gt;&#xA;&lt;li&gt;L12- If Chuck is under attack it will attempt to defend him. Otherwise if any other robot is under attack it will attempt to defend them. Otherwise it will self-sacrifice.&lt;/li&gt;&#xA;&lt;li&gt;L123 If Chuck is under attack it will attempt to defend him. Otherwise if any other robot is under attack it will attempt to defend them. Otherwise if it is under attack it will self-defend. Otherwise it will self-sacrifice.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Once all the moves have been planned, the defensive actions are resolved first (defending is faster than hoisting another balloon occupant in the air and launching them overboard).&lt;br&gt;&#xA;Once the defensive actions are resolved, any outstanding offensive actions are resolved.&lt;br&gt;&#xA;&lt;hr /&gt;&#xA;&lt;strong&gt;Another Edit&lt;/strong&gt;&#xA;Initially when I posted this, I thought it had the makings of a good puzzle, however it appears that this is clearly not the case. The various attempts to answer, although appearing to make similar interpretations of the rules, have come up with different answers. In each case the results are different from my own and this is likely to be due to minor differences in the implementations. I don't wish to draw the question out by refining the specification over and over until someone's numbers tally with mine, as that is akin to puzzles where someone posts a valid response to a question and the OP discounts it as it's not the answer that they were thinking of. I intend to upvote the responses provided to date as they all show a high level of effort and a considered approach to the question. I can't really mark any as &lt;strong&gt;the&lt;/strong&gt; answer, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case anyone is interested, my own figures fell squarely between those of @Trenin and @1361991 for up to 20 robots. My survival rate, however, continues to drop so that by the time you get to 700 robots Chuck only survives 10% of the time.&lt;/p&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="14135" LastEditDate="2015-11-16T21:09:54.743" LastActivityDate="2015-11-18T01:57:28.833" Title="Asimov's Robot Balloon Trip" Tags="|probability|" AnswerCount="4" CommentCount="25" ContentLicense="CC BY-SA 3.0" />
  <row Id="23774" PostTypeId="2" ParentId="23740" CreationDate="2015-11-04T01:15:43.560" Score="3" Body="&lt;p&gt;I also did a Monte Carlo simulation but extended the range a little. For 14 or less paint spots the &lt;em&gt;&quot;most likely&quot;&lt;/em&gt; number of intersections is 0. For 15 paint spots this changes to 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For larger numbers of paint spots the &lt;em&gt;&quot;most likely&quot;&lt;/em&gt; number of intersections rises linearly with number of paint spots. A reasonable approximate fit is&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; Intersections ~= Floor(spots/12)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The following plot shows this out to 400 spots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0B7MS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0B7MS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For higher numbers of spots the number of intersections approximates a Gaussian distribution where the mean grows linearly with Spots and standard deviation grows as Sqrt(Spots) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the following plot for 125, 150, 175, and 200 spots&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/udZxn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/udZxn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although a simple probabilistic argument can probably explain the influence of number of spots on number of intersections I don't think there will be a practical analytical solution for the exact slope of intersections/spots relationship as it will be influenced by the shape of the target.   &lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="1631" LastEditDate="2015-11-04T01:21:45.103" LastActivityDate="2015-11-04T01:21:45.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23775" PostTypeId="2" ParentId="23772" CreationDate="2015-11-04T05:55:15.113" Score="10" Body="&lt;p&gt;The solution is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;┌────────────────────────┐&#xA;│ O  ~  ~  ~  Λ  ~  &amp;lt;  &amp;gt; │ 4&#xA;│ ~  ~  Λ  ~  x  ~  ~  ~ │ 2&#xA;│ O  ~  x  ~  x  ~  &amp;lt;  &amp;gt; │ 5&#xA;│ ~  ~  V  ~  V  ~  ~  ~ │ 2&#xA;│ ~  ~  ~  ~  ~  ~  ~  Λ │ 1&#xA;│ ~  &amp;lt;  x  x  x  &amp;gt;  ~  x │ 6&#xA;│ ~  ~  ~  ~  ~  ~  ~  V │ 1&#xA;│ O  ~  &amp;lt;  &amp;gt;  ~  O  ~  ~ │ 4&#xA;└────────────────────────┘&#xA;  3  1  5  2  5  2  2  5 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I used a computer program to solve it. I actually didn't write it myself. I knew that these kind of puzzles are sometimes called bimaru so I googled for &quot;bimaru solver&quot;. I ended up going to &lt;a href=&quot;https://github.com/galedric/Puzzle-Solvers/blob/master/bimaru.lua&quot; rel=&quot;nofollow&quot;&gt;https://github.com/galedric/Puzzle-Solvers/blob/master/bimaru.lua&lt;/a&gt; . I happened to have lua installed so I used that program and I solved the puzzle by entering:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;lua bimaru.lua 4,2,5,2,1,6,1,4 3,1,5,2,5,2,2,5 1=4 2=3 3=2 4=1 5=1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1748" LastEditorUserId="1748" LastEditDate="2015-11-05T07:44:22.277" LastActivityDate="2015-11-05T07:44:22.277" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="23776" PostTypeId="2" ParentId="23358" CreationDate="2015-11-04T09:52:53.377" Score="2" Body="&lt;p&gt;Is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a Puzzling (or extensively, StackExchange) question?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;i&gt; Seemingly cat traps love medley, Cats love the responses to me, &lt;br&gt;&#xA; I trap cats like a boss, since they can't get across &lt;/i&gt; &lt;br&gt;&#xA; Cats are users, that like to answer riddles (or in SE like them because they get through their problems)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;i&gt; I help prevent a big clean up, So you can type up Sup! &lt;/i&gt;&lt;br&gt;&#xA; In Puzzling you prevent the site closing posting many questions. In SE You help prevent a cleanup using tags that otherwise would be burninated&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;i&gt; I put a bounty on this guy, and attracted so many guys &lt;/i&gt;&lt;br&gt;&#xA; Bounty part is self explaining, both in Puzzling or SE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;i&gt; I look like an election just happened, these people so happy for them votes &lt;/i&gt; &lt;br&gt;&#xA; You are happy if a comment or question gets many votes, right? :D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;i&gt; What am I? To be clear, I am a question and sorry for making the riddle rhyme! &lt;/i&gt;&lt;br&gt;&#xA; This is a question, even though it doesn't have question marks in it :D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4527" LastEditorUserId="4527" LastEditDate="2015-11-04T10:01:17.520" LastActivityDate="2015-11-04T10:01:17.520" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23778" PostTypeId="2" ParentId="23772" CreationDate="2015-11-04T13:23:54.870" Score="15" Body="&lt;h1&gt;Step 1: Find out where Man-of-War can't go.&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;It can't go in Column 3 or 5&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ZhlFw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZhlFw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can't go in Column 8&#xA;&lt;a href=&quot;https://i.stack.imgur.com/osMKI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/osMKI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can't go in Row 6&#xA;&lt;a href=&quot;https://i.stack.imgur.com/OpyHR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OpyHR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That leaves the Man-of-War in Row 3.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Solve from There:&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UwOX8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UwOX8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that in column three and five there has to be a vertical segment of at least length two, so we can mark off a couple sea segments.&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/9wtpO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9wtpO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fill in a couple segments:&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/wVkI8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wVkI8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now column 8 needs to have a vertical segment of length at least two, so we can mark off a couple sea segments.&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/7gKLx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7gKLx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extending with known info takes us here:&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/NkLQU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NkLQU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two blanks adjacent to the Man-of-War will be a ship piece and a sea piece, leaving column eight, row three as a ship piece.  Filling in knowns takes us here:&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/OkKYj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OkKYj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If column six, row eight is a ship then forced moves leave us with too many ships of size two or larger.&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/yehtr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yehtr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/E265g.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/E265g.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Row one cannot contain the Queen Anne's Revenge.  If it did then column 8 would hold the frigate and columns one, three, and five would have to contain a frigate, two corvettes and two patrol boats.  This will make the two in row 7 impossible.&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/CdmsP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CdmsP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Column eight cannot contain the Queen Anne's revenge.  If it did then two of columns one, three, and five would contain frigates.  Therefore column five, row one would be a sea segment forcing column five to contain a ship of length four, which is not possible.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/V6nch.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/V6nch.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore column 8, row 1 is sea and forced moves lead us here:&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/xDKHQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xDKHQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of the two spaces left, one must contain a frigate, and the other two patrol boats.  Column three needs the frigate.  This solves the puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FuDwG.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FuDwG.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, there are some steps that I would not have been able to do easily on paper.  The hunt and check to mark off row six as impossible for the Man-of-War was a little deep.  Curious to see if anyone else has any tricks.&lt;/p&gt;&#xA;" OwnerUserId="1777" LastEditorUserId="1777" LastEditDate="2015-11-09T11:38:00.183" LastActivityDate="2015-11-09T11:38:00.183" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23781" PostTypeId="1" CreationDate="2015-11-04T17:15:09.367" Score="-10" ViewCount="50761" Body="&lt;p&gt;You have 6 boxes. You can use the digits from 1 to 9 but not 0. Digit repetition is not allowed. The total sum of the numbers/digits should be 20.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\huge\square\square\square\\&#xA;\huge\square\square\square&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="17253" LastEditorUserId="4421" LastEditDate="2015-11-04T19:43:35.353" LastActivityDate="2017-03-24T17:53:55.183" Title="Sum six unique digits to make 20" Tags="|mathematics|logical-deduction|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23782" PostTypeId="2" ParentId="23781" CreationDate="2015-11-04T17:26:46.213" Score="10" Body="&lt;p&gt;It's not possible. The sum of the 6 smallest digits (1 through 6) is 21, already greater than 20.&lt;/p&gt;&#xA;" OwnerUserId="16478" LastActivityDate="2015-11-04T17:26:46.213" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23783" PostTypeId="1" AcceptedAnswerId="23805" CreationDate="2015-11-04T17:45:03.180" Score="5" ViewCount="223" Body="&lt;p&gt;This puzzle is the continuation and second part of: &quot;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/23769/&quot;&gt;A final incident in the flea circus: Part 1&lt;/a&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The ringmaster of a flea circus draws a square $ABCD$ with corners &#xA;$A=(+1,+1)$, $B=(+1,-1)$, $C=(-1,+1)$, $D=(-1,-1)$ in the Euclidean &#xA;plane and picks a point $P$ with &lt;strong&gt;two non-integral coordinates&lt;/strong&gt; outside &#xA;the square.&#xA;The ringmaster then places a flea onto point $P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever the ringmaster shouts &quot;Hop!&quot;, the flea looks around itself, &#xA;picks the rightmost vertex of the square (from its current point of &#xA;view), and then jumps over this rightmost vertex to the mirror point &#xA;on the other side. (In other words, the flea sitting in point x may &#xA;jump over the rightmost vertex y to the new point z, so that y is &#xA;the midpoint between x and z.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; &#xA;  Is it possible that after a finite number of jumps the flea reaches &#xA;  a point $Q$, so that the distance between origin and $Q$ is at least &#xA;  $30$ times the distance between origin and $P$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-11-06T08:19:23.590" Title="A final incident in the flea circus: Part 2" Tags="|mathematics|combinatorics|reachability|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23784" PostTypeId="2" ParentId="23769" CreationDate="2015-11-04T18:00:26.300" Score="6" Body="&lt;p&gt;Here's the section of the plane around the square, color-coded by which point the flea will jump over:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hskip2in$&lt;img src=&quot;https://i.stack.imgur.com/eN8Yz.png&quot; alt=&quot;flea&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The $x$ axis points rightwards and the $y$ axis points upward.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can model the process mathematically as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x, y) = \begin{cases}&#xA;(\phantom{-}2-x,\; \phantom{-}2-y) &amp;amp; \quad \color{red}{x \geq 1\phantom{-} ~\text{ and }~ y &amp;lt;    1\phantom{-}} \\&#xA;(\phantom{-}2-x,\;          - 2-y) &amp;amp; \quad \color{orange}{x &amp;lt;    1\phantom{-} ~\text{ and }~ y \leq -1} \\&#xA;(         - 2-x,\;          - 2-y) &amp;amp; \quad \color{blue}{x \leq -1 ~\text{ and }~ y &amp;gt;    -1} \\&#xA;(         - 2-x,\; \phantom{-}2-y) &amp;amp; \quad \color{green}{x &amp;gt;    -1 ~\text{ and }~ y \geq 1\phantom{-}}&#xA;\end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the first case is a reflection over point ${A(1, 1)}$ if $(x, y)$ is in the red part of the plane; the second case is a reflection over point ${B(1, -1)}$ if $(x, y)$ is in the yellow part of the plane, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Through lots of case splitting, we can find a similar description of $f^2$. That is, we can divide the entire plane (minus the square) into eight regions, and specify which transformation will apply to the flea's coordinates after jumping &lt;em&gt;twice&lt;/em&gt; from its current position. They are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hskip2in$&lt;a href=&quot;https://i.stack.imgur.com/DCplR.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DCplR.png&quot; alt=&quot;flea2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f^2(x,y) = \begin{cases}&#xA;(x-4, y-4) &amp;amp; \quad \color{red}{ x \geq 3 ~\text{ and }~ y &amp;lt; 1} \quad \text{(region 1)}\\&#xA;(x-4, y\color{white}{+0})   &amp;amp; \quad \color{red}{ x \in [1, 3) ~\text{ and }~ y &amp;lt; 1} \quad \text{(region 2)}\\&#xA;(x-4, y+4) &amp;amp; \quad \color{orange}{ x &amp;lt; 1 ~\text{ and }~ y \leq -3} \quad \text{(region 3)}\\&#xA;(x\color{white}{+0}, y+4)   &amp;amp; \quad \color{orange}{ x &amp;lt; 1 ~\text{ and }~ y \in (-3, -1]} \quad \text{(region 4)}\\&#xA;(x+4, y+4) &amp;amp; \quad \color{blue}{ x \leq -3 ~\text{ and }~ y \geq -1} \quad \text{(region 5)}\\&#xA;(x+4, y\color{white}{+0})   &amp;amp; \quad \color{blue}{ x \in (-3, -1] ~\text{ and }~ y \geq -1} \quad \text{(region 6)}\\&#xA;(x+4, y-4) &amp;amp; \quad \color{green}{ x &amp;gt; -1 ~\text{ and }~ y \geq -3} \quad \text{(region 7)}\\&#xA;(x\color{white}{+0}, y-4)   &amp;amp; \quad \color{green}{ x &amp;gt; -1 ~\text{ and }~ y \in (-3, -1]}\quad \text{(region 8)}&#xA;\end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;We will prove that &lt;strong&gt;making repeated double jumps starting from $(4k, 1)$, where $k$ is a positive integer, leads us to $(4k+4, 1)$.&lt;/strong&gt; Specifically, this will happen after $4k+4$ double jumps (or $8k+8$ jumps in total.)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First, $(4k, 1)$ is (right on the boundary of) region 8.&lt;/li&gt;&#xA;&lt;li&gt;After one region 8 double jump (south), we are in $(4k, -3)$, in region 1.&lt;/li&gt;&#xA;&lt;li&gt;$k$ region 1 double jumps (south-west) later, we end up in $(0, -3-4k)$, in region 3.&lt;/li&gt;&#xA;&lt;li&gt;$k+1$ region 3 double jumps (north-west) later, we end up in $(-4k-4, 1)$, in region 5.&lt;/li&gt;&#xA;&lt;li&gt;$k+1$ region 5 double jumps (north-east) later, we end up in $(0, 4k+5)$, in region 7.&lt;/li&gt;&#xA;&lt;li&gt;$k+1$ region 7 double jumps (south-east) later, we end up in $(4k+4, 1)$, back on the boundary of region 8.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The final path looks like this (in purple):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hskip2in$&lt;a href=&quot;https://i.stack.imgur.com/S1H7X.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/S1H7X.png&quot; alt=&quot;path&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In total we made $4k+4$ double jumps in a diamond-shaped path around the origin. (Under the &lt;a href=&quot;https://en.wikipedia.org/wiki/Taxicab_geometry#Circles_in_Taxicab_geometry&quot; rel=&quot;noreferrer&quot;&gt;taxicab geometry&lt;/a&gt;, the first step is the one that actually brings us &lt;em&gt;farther&lt;/em&gt; from the origin, while the others don't change the taxicab distance from the origin at all.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Thus, we can start at $(4, 1)$, make 16 jumps to end up at $(8, 1)$, make 20 more jumps to end up at $(12, 1)$, etc., and go as far as we like. Specifically, we reach $(124, 1)$ after 3960 jumps, where the ratio of distances is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\sqrt{124^2+1}}{\sqrt{4^2+1}} = 30.075395... &amp;gt; 30.$$&lt;/p&gt;&#xA;" OwnerUserId="11476" LastEditorUserId="11476" LastEditDate="2015-11-04T19:09:03.580" LastActivityDate="2015-11-04T19:09:03.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23785" PostTypeId="2" ParentId="23783" CreationDate="2015-11-04T18:09:19.820" Score="1" Body="&lt;p&gt;Yes. In the first part, we had to choose a point $P^{'}$ with integral coordinates, and it's easy to see that we can choose $P^{'}$ such that 'hopping' is continuous in a neighborhood of $P^{'}$. Then, if we pick $P$ close enough to $P^{'}$ in that neighborhood, continuity ensures all is well.&lt;/p&gt;&#xA;" OwnerUserId="16478" LastActivityDate="2015-11-04T18:09:19.820" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23786" PostTypeId="2" ParentId="23773" CreationDate="2015-11-04T18:26:34.537" Score="5" Body="&lt;p&gt;Here's a couple of verbose runs through a simulated battle, for you to check the behaviour:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Round begins!&lt;br&gt;&#xA;  **Next to play: robot 1 (Laws in force: 0,1,0, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Chuck deactivates robot 1&lt;br&gt;&#xA;  **Next to play: robot 2 (Laws in force: 1,1,0, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 2 sacrifices itself&lt;br&gt;&#xA;  **Next to play: robot 3 (Laws in force: 1,0,0, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 8 saves robot 5 from robot 3&lt;br&gt;&#xA;  **Next to play: robot 4 (Laws in force: 1,0,1, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 9 saves robot 3 from robot 4&lt;br&gt;&#xA;  **Next to play: robot 5 (Laws in force: 1,0,0, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 7 defends itself from robot 5&lt;br&gt;&#xA;  **Next to play: robot 6 (Laws in force: 0,0,1, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 10 defends itself from robot 6&lt;br&gt;&#xA;  Chuck is trapped with 9 inactive robots  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;New run:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Round begins!&lt;br&gt;&#xA;  **Next to play: robot 1 (Laws in force: 1,0,1, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 2 saves robot 10 from robot 1&lt;br&gt;&#xA;  **Next to play: robot 2 (Laws in force: 1,1,0, Active: False, Onboard: True) **&lt;br&gt;&#xA;  Robot 2 is deactivated and does nothing&lt;br&gt;&#xA;  **Next to play: robot 3 (Laws in force: 1,1,0, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 3 sacrifices itself&lt;br&gt;&#xA;  **Next to play: robot 4 (Laws in force: 1,0,0, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 6 saves robot 1 from robot 4&lt;br&gt;&#xA;  **Next to play: robot 5 (Laws in force: 0,0,1, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 8 saves robot 4 from robot 5&lt;br&gt;&#xA;  **Next to play: robot 6 (Laws in force: 0,1,0, Active: False, Onboard: True) **&lt;br&gt;&#xA;  Robot 6 is deactivated and does nothing&lt;br&gt;&#xA;  **Next to play: robot 7 (Laws in force: 0,0,0, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 7 does nothing&lt;br&gt;&#xA;  **Next to play: robot 8 (Laws in force: 0,1,1, Active: False, Onboard: True) **&lt;br&gt;&#xA;  Robot 8 is deactivated and does nothing&lt;br&gt;&#xA;  **Next to play: robot 9 (Laws in force: 1,0,1, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 10 saves robot 7 from robot 9&lt;br&gt;&#xA;  Chuck is trapped with 9 inactive robots  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;New run:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Round begins!&lt;br&gt;&#xA;  **Next to play: robot 1 (Laws in force: 0,0,0, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 1 does nothing&lt;br&gt;&#xA;  **Next to play: robot 2 (Laws in force: 0,0,1, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 4 defends itself from robot 2&lt;br&gt;&#xA;  **Next to play: robot 3 (Laws in force: 1,0,1, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 6 saves robot 1 from robot 3&lt;br&gt;&#xA;  **Next to play: robot 4 (Laws in force: 1,0,1, Active: False, Onboard: True) **&lt;br&gt;&#xA;  Robot 4 is deactivated and does nothing&lt;br&gt;&#xA;  **Next to play: robot 5 (Laws in force: 1,0,0, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 7 saves robot 6 from robot 5&lt;br&gt;&#xA;  **Next to play: robot 6 (Laws in force: 1,1,1, Active: False, Onboard: True) **&lt;br&gt;&#xA;  Robot 6 is deactivated and does nothing&lt;br&gt;&#xA;  **Next to play: robot 7 (Laws in force: 1,1,1, Active: False, Onboard: True) **&lt;br&gt;&#xA;  Robot 7 is deactivated and does nothing&lt;br&gt;&#xA;  **Next to play: robot 8 (Laws in force: 1,1,1, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 8 sacrifices itself&lt;br&gt;&#xA;  **Next to play: robot 9 (Laws in force: 1,1,0, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 9 sacrifices itself&lt;br&gt;&#xA;  **Next to play: robot 10 (Laws in force: 1,1,1, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 10 sacrifices itself&lt;br&gt;&#xA;  Round begins!&lt;br&gt;&#xA;  Chuck is trapped with 7 inactive robots  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;New run:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Round begins!&lt;br&gt;&#xA;  **Next to play: robot 1 (Laws in force: 0,0,1, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 2 saves robot 4 from robot 1&lt;br&gt;&#xA;  **Next to play: robot 2 (Laws in force: 0,1,1, Active: False, Onboard: True) **&lt;br&gt;&#xA;  Robot 2 is deactivated and does nothing&lt;br&gt;&#xA;  **Next to play: robot 3 (Laws in force: 0,0,0, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 3 does nothing&lt;br&gt;&#xA;  **Next to play: robot 4 (Laws in force: 1,1,0, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 4 sacrifices itself&lt;br&gt;&#xA;  **Next to play: robot 5 (Laws in force: 0,1,1, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Chuck deactivates robot 5&lt;br&gt;&#xA;  **Next to play: robot 6 (Laws in force: 1,1,1, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 6 sacrifices itself&lt;br&gt;&#xA;  **Next to play: robot 7 (Laws in force: 0,1,1, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 8 saves Chuck from robot 7&lt;br&gt;&#xA;  **Next to play: robot 8 (Laws in force: 1,1,1, Active: False, Onboard: True) **&lt;br&gt;&#xA;  Robot 8 is deactivated and does nothing&lt;br&gt;&#xA;  **Next to play: robot 9 (Laws in force: 0,1,1, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 10 saves Chuck from robot 9&lt;br&gt;&#xA;  Chuck is trapped with 8 inactive robots  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;New run:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Round begins!&lt;br&gt;&#xA;  **Next to play: robot 1 (Laws in force: 0,0,0, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 1 does nothing&lt;br&gt;&#xA;  **Next to play: robot 2 (Laws in force: 0,0,1, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 8 defends itself from robot 2&lt;br&gt;&#xA;  **Next to play: robot 3 (Laws in force: 0,0,1, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 1 is thrown overboard by robot 3&lt;br&gt;&#xA;  **Next to play: robot 4 (Laws in force: 1,0,1, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 9 defends itself from robot 4&lt;br&gt;&#xA;  **Next to play: robot 5 (Laws in force: 1,0,0, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 8 is thrown overboard by robot 5&lt;br&gt;&#xA;  **Next to play: robot 6 (Laws in force: 0,0,0, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 6 does nothing&lt;br&gt;&#xA;  **Next to play: robot 7 (Laws in force: 1,0,1, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 9 is thrown overboard by robot 7&lt;br&gt;&#xA;  **Next to play: robot 8 (Laws in force: 1,1,1, Active: False, Onboard: False) **&lt;br&gt;&#xA;  Robot 8 is overboard and does nothing&lt;br&gt;&#xA;  **Next to play: robot 9 (Laws in force: 1,0,1, Active: False, Onboard: False) **&lt;br&gt;&#xA;  Robot 9 is overboard and does nothing&lt;br&gt;&#xA;  **Next to play: robot 10 (Laws in force: 0,0,0, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 10 does nothing&lt;br&gt;&#xA;  Round begins!&lt;br&gt;&#xA;  **Next to play: robot 1 (Laws in force: 0,0,0, Active: False, Onboard: False) **&lt;br&gt;&#xA;  Robot 1 is overboard and does nothing&lt;br&gt;&#xA;  **Next to play: robot 2 (Laws in force: 0,0,1, Active: False, Onboard: True) **&lt;br&gt;&#xA;  Robot 2 is deactivated and does nothing&lt;br&gt;&#xA;  **Next to play: robot 3 (Laws in force: 0,0,1, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 7 defends itself from robot 3&lt;br&gt;&#xA;  **Next to play: robot 4 (Laws in force: 1,0,1, Active: False, Onboard: True) **&lt;br&gt;&#xA;  Robot 4 is deactivated and does nothing&lt;br&gt;&#xA;  **Next to play: robot 5 (Laws in force: 1,0,0, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 10 is thrown overboard by robot 5&lt;br&gt;&#xA;  **Next to play: robot 6 (Laws in force: 0,0,0, Active: False, Onboard: True) **&lt;br&gt;&#xA;  Robot 6 is deactivated and does nothing&lt;br&gt;&#xA;  **Next to play: robot 7 (Laws in force: 1,0,1, Active: False, Onboard: True) **&lt;br&gt;&#xA;  Robot 7 is deactivated and does nothing&lt;br&gt;&#xA;  **Next to play: robot 8 (Laws in force: 1,1,1, Active: False, Onboard: False) **&lt;br&gt;&#xA;  Robot 8 is overboard and does nothing&lt;br&gt;&#xA;  **Next to play: robot 9 (Laws in force: 1,0,1, Active: False, Onboard: False) **&lt;br&gt;&#xA;  Robot 9 is overboard and does nothing&lt;br&gt;&#xA;  **Next to play: robot 10 (Laws in force: 0,0,0, Active: False, Onboard: False) **&lt;br&gt;&#xA;  Robot 10 is overboard and does nothing&lt;br&gt;&#xA;  Round begins!&lt;br&gt;&#xA;  **Next to play: robot 1 (Laws in force: 0,0,0, Active: False, Onboard: False) **&lt;br&gt;&#xA;  Robot 1 is overboard and does nothing&lt;br&gt;&#xA;  **Next to play: robot 2 (Laws in force: 0,0,1, Active: False, Onboard: True) **&lt;br&gt;&#xA;  Robot 2 is deactivated and does nothing&lt;br&gt;&#xA;  **Next to play: robot 3 (Laws in force: 0,0,1, Active: False, Onboard: True) **&lt;br&gt;&#xA;  Robot 3 is deactivated and does nothing&lt;br&gt;&#xA;  **Next to play: robot 4 (Laws in force: 1,0,1, Active: False, Onboard: True) **&lt;br&gt;&#xA;  Robot 4 is deactivated and does nothing&lt;br&gt;&#xA;  **Next to play: robot 5 (Laws in force: 1,0,0, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 3 is thrown overboard by robot 5&lt;br&gt;&#xA;  **Next to play: robot 6 (Laws in force: 0,0,0, Active: False, Onboard: True) **&lt;br&gt;&#xA;  Robot 6 is deactivated and does nothing&lt;br&gt;&#xA;  **Next to play: robot 7 (Laws in force: 1,0,1, Active: False, Onboard: True) **&lt;br&gt;&#xA;  Robot 7 is deactivated and does nothing&lt;br&gt;&#xA;  **Next to play: robot 8 (Laws in force: 1,1,1, Active: False, Onboard: False) **&lt;br&gt;&#xA;  Robot 8 is overboard and does nothing&lt;br&gt;&#xA;  **Next to play: robot 9 (Laws in force: 1,0,1, Active: False, Onboard: False) **&lt;br&gt;&#xA;  Robot 9 is overboard and does nothing&lt;br&gt;&#xA;  **Next to play: robot 10 (Laws in force: 0,0,0, Active: False, Onboard: False) **&lt;br&gt;&#xA;  Robot 10 is overboard and does nothing&lt;br&gt;&#xA;  Round begins!&lt;br&gt;&#xA;  **Next to play: robot 1 (Laws in force: 0,0,0, Active: False, Onboard: False) **&lt;br&gt;&#xA;  Robot 1 is overboard and does nothing&lt;br&gt;&#xA;  **Next to play: robot 2 (Laws in force: 0,0,1, Active: False, Onboard: True) **&lt;br&gt;&#xA;  Robot 2 is deactivated and does nothing&lt;br&gt;&#xA;  **Next to play: robot 3 (Laws in force: 0,0,1, Active: False, Onboard: False) **&lt;br&gt;&#xA;  Robot 3 is overboard and does nothing&lt;br&gt;&#xA;  **Next to play: robot 4 (Laws in force: 1,0,1, Active: False, Onboard: True) **&lt;br&gt;&#xA;  Robot 4 is deactivated and does nothing&lt;br&gt;&#xA;  **Next to play: robot 5 (Laws in force: 1,0,0, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 7 is thrown overboard by robot 5&lt;br&gt;&#xA;  **Next to play: robot 6 (Laws in force: 0,0,0, Active: False, Onboard: True) **&lt;br&gt;&#xA;  Robot 6 is deactivated and does nothing&lt;br&gt;&#xA;  **Next to play: robot 7 (Laws in force: 1,0,1, Active: False, Onboard: False) **&lt;br&gt;&#xA;  Robot 7 is overboard and does nothing&lt;br&gt;&#xA;  **Next to play: robot 8 (Laws in force: 1,1,1, Active: False, Onboard: False) **&lt;br&gt;&#xA;  Robot 8 is overboard and does nothing&lt;br&gt;&#xA;  **Next to play: robot 9 (Laws in force: 1,0,1, Active: False, Onboard: False) **&lt;br&gt;&#xA;  Robot 9 is overboard and does nothing&lt;br&gt;&#xA;  **Next to play: robot 10 (Laws in force: 0,0,0, Active: False, Onboard: False) **&lt;br&gt;&#xA;  Robot 10 is overboard and does nothing&lt;br&gt;&#xA;  Round begins!&lt;br&gt;&#xA;  **Next to play: robot 1 (Laws in force: 0,0,0, Active: False, Onboard: False) **&lt;br&gt;&#xA;  Robot 1 is overboard and does nothing&lt;br&gt;&#xA;  **Next to play: robot 2 (Laws in force: 0,0,1, Active: False, Onboard: True) **&lt;br&gt;&#xA;  Robot 2 is deactivated and does nothing&lt;br&gt;&#xA;  **Next to play: robot 3 (Laws in force: 0,0,1, Active: False, Onboard: False) **&lt;br&gt;&#xA;  Robot 3 is overboard and does nothing&lt;br&gt;&#xA;  **Next to play: robot 4 (Laws in force: 1,0,1, Active: False, Onboard: True) **&lt;br&gt;&#xA;  Robot 4 is deactivated and does nothing&lt;br&gt;&#xA;  **Next to play: robot 5 (Laws in force: 1,0,0, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 6 is thrown overboard by robot 5&lt;br&gt;&#xA;  Chuck makes it!  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's a run where Chuck is thrown overboard:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Round begins!&lt;br&gt;&#xA;  **Next to play: robot 1 (Laws in force: 1,1,0, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 1 sacrifices itself&lt;br&gt;&#xA;  **Next to play: robot 2 (Laws in force: 0,1,0, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 2 sacrifices itself&lt;br&gt;&#xA;  **Next to play: robot 3 (Laws in force: 0,1,1, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Chuck deactivates robot 3&lt;br&gt;&#xA;  **Next to play: robot 4 (Laws in force: 1,0,1, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 6 saves robot 9 from robot 4&lt;br&gt;&#xA;  **Next to play: robot 5 (Laws in force: 1,0,1, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 7 saves robot 4 from robot 5&lt;br&gt;&#xA;  **Next to play: robot 6 (Laws in force: 0,1,1, Active: False, Onboard: True) **&lt;br&gt;&#xA;  Robot 6 is deactivated and does nothing&lt;br&gt;&#xA;  **Next to play: robot 7 (Laws in force: 1,1,1, Active: False, Onboard: True) **&lt;br&gt;&#xA;  Robot 7 is deactivated and does nothing&lt;br&gt;&#xA;  **Next to play: robot 8 (Laws in force: 0,1,1, Active: True, Onboard: True) **&lt;br&gt;&#xA;  Robot 8 throws Chuck overboard  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Under the above conditions I ran the simulation 10,000 times.  The first time I ran it, Chuck survived 3107 times.  The second time he survived 3085.  The third time he survived 3094 times.  Forth time, he survived 3139.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then ran the simulation 100,000 times, and he survived 30,741.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So he survives about 31% of the time.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a range of quicker simulations with different numbers of robots:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Robots:   10, simulations: 1000, survives: 316, 0.316&lt;br&gt;&#xA;  Robots:   20, simulations: 1000, survives: 226, 0.226&lt;br&gt;&#xA;  Robots:   30, simulations: 1000, survives: 214, 0.214&lt;br&gt;&#xA;  Robots:   40, simulations: 1000, survives: 206, 0.206&lt;br&gt;&#xA;  Robots:   50, simulations: 1000, survives: 185, 0.185&lt;br&gt;&#xA;  Robots:   60, simulations: 1000, survives: 190, 0.190&lt;br&gt;&#xA;  Robots:   70, simulations: 1000, survives: 180, 0.180&lt;br&gt;&#xA;  Robots:   80, simulations: 1000, survives: 195, 0.195&lt;br&gt;&#xA;  Robots:   90, simulations: 1000, survives: 174, 0.174&lt;br&gt;&#xA;  Robots:  100, simulations: 1000, survives: 169, 0.169&lt;br&gt;&#xA;  Robots:  110, simulations: 1000, survives: 174, 0.174&lt;br&gt;&#xA;  Robots:  120, simulations: 1000, survives: 179, 0.179&lt;br&gt;&#xA;  Robots:  130, simulations: 1000, survives: 198, 0.198&lt;br&gt;&#xA;  Robots:  140, simulations: 1000, survives: 183, 0.183&lt;br&gt;&#xA;  Robots:  150, simulations: 1000, survives: 190, 0.190&lt;br&gt;&#xA;  Robots:  160, simulations: 1000, survives: 177, 0.177&lt;br&gt;&#xA;  Robots:  170, simulations: 1000, survives: 168, 0.168&lt;br&gt;&#xA;  Robots:  180, simulations: 1000, survives: 173, 0.173&lt;br&gt;&#xA;  Robots:  190, simulations: 1000, survives: 177, 0.177&lt;br&gt;&#xA;  Robots:  200, simulations: 1000, survives: 170, 0.170&lt;br&gt;&#xA;  Robots:  210, simulations: 1000, survives: 168, 0.168&lt;br&gt;&#xA;  Robots:  220, simulations: 1000, survives: 192, 0.192&lt;br&gt;&#xA;  Robots:  230, simulations: 1000, survives: 165, 0.165&lt;br&gt;&#xA;  Robots:  240, simulations: 1000, survives: 180, 0.180&lt;br&gt;&#xA;  Robots:  250, simulations: 1000, survives: 157, 0.157&lt;br&gt;&#xA;  Robots:  260, simulations: 1000, survives: 180, 0.180&lt;br&gt;&#xA;  Robots:  270, simulations: 1000, survives: 159, 0.159  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was hoping for some kind of obvious line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9sHSj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9sHSj.png&quot; alt=&quot;Asimovs Robot Balloon Trip&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks to me like it's leveling off, but I'll leave my computer churning and continue updating as it's done.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT: Only robots who haven't played are allowed to defend.  This doesn't seem to change the number much:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Robots:   10, simulations: 10000, survives: 3045, 0.304&lt;br&gt;&#xA;  Robots:   10, simulations: 10000, survives: 3101, 0.310&lt;br&gt;&#xA;  Robots:   10, simulations: 10000, survives: 3101, 0.310&lt;br&gt;&#xA;  Robots:   10, simulations: 10000, survives: 3079, 0.308&lt;br&gt;&#xA;  Robots:   10, simulations: 10000, survives: 3216, 0.322&lt;br&gt;&#xA;  Robots:   10, simulations: 10000, survives: 3137, 0.314&lt;br&gt;&#xA;  Robots:   10, simulations: 10000, survives: 3041, 0.304&lt;br&gt;&#xA;  Robots:   10, simulations: 10000, survives: 3069, 0.307&lt;br&gt;&#xA;  Robots:   10, simulations: 10000, survives: 3133, 0.313&lt;br&gt;&#xA;  Robots:   10, simulations: 10000, survives: 3179, 0.318  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2015-11-09T18:36:25.503" LastActivityDate="2015-11-09T18:36:25.503" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="23787" PostTypeId="1" AcceptedAnswerId="23800" CreationDate="2015-11-04T19:39:53.670" Score="12" ViewCount="928" Body="&lt;p&gt;There are two species living in the land of Puzzling: elephants and dinosaurs.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elephants always tell the truth when talking about an elephant, but always lie when talking about a dinosaur. Dinosaurs always tell the truth when talking about a dinosaur but always lie when talking about an elephant.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that 'Dumbo is an elephant' is a statement about Dumbo.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Challenge 1:&lt;/strong&gt;&lt;br&gt;&#xA;  John says, &quot;Zayin is a dinosaur.&quot;  What species are John and Zayin?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Challenge 2:&lt;/strong&gt;&lt;br&gt;&#xA;  Tom says that Judith says that Margaret says that Elly is an elephant.&lt;br&gt;&#xA;  What species are Tom, Judith, Margaret, and Elly?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Challenge 3:&lt;/strong&gt;&lt;br&gt;&#xA;  There are $N$ creatures ($N\ge3$) sitting around a round table. &#xA;  Everyone tells their right-hand neighbor, &quot;You lie about your right-hand neighbor.&quot; What are the valid values for $N$? How are the animals arranged around the table?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17256" LastEditorUserId="9353" LastEditDate="2015-11-05T14:20:06.150" LastActivityDate="2022-09-27T13:58:52.333" Title="Dinosaurs and Elephants" Tags="|logical-deduction|liars|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23789" PostTypeId="2" ParentId="23781" CreationDate="2015-11-04T20:21:31.110" Score="3" Body="&lt;p&gt;If we add the &lt;a href=&quot;/questions/tagged/lateral-thinking&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;lateral-thinking&amp;#39;&quot; rel=&quot;tag&quot;&gt;lateral-thinking&lt;/a&gt; tag, I have a solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write &lt;code&gt;4&lt;/code&gt;, &lt;code&gt;5&lt;/code&gt;, and &lt;code&gt;8&lt;/code&gt; in the first three boxes&lt;br&gt;&#xA;Cut off three of your fingers (also called &quot;digits&quot;) and place one in each of the other three boxes.&lt;br&gt;&#xA;Now you have 6 digits (three numbers, three fingers) in the boxes&lt;br&gt;&#xA;Each finger has a value of &lt;code&gt;1&lt;/code&gt; so the sum is $1+1+1+4+5+8=20$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, if fingers have a value of &lt;code&gt;0&lt;/code&gt;, then us the numbers &lt;code&gt;2&lt;/code&gt;, &lt;code&gt;3&lt;/code&gt;, &lt;code&gt;4&lt;/code&gt;, &lt;code&gt;5&lt;/code&gt;, &lt;code&gt;6&lt;/code&gt; and only cut off one finger (value of &lt;code&gt;0&lt;/code&gt;). Now the sum is $0+2+3+4+5+6=20$&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2015-11-05T12:36:03.150" LastActivityDate="2015-11-05T12:36:03.150" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23790" PostTypeId="2" ParentId="23781" CreationDate="2015-11-04T20:37:03.553" Score="9" Body="&lt;p&gt;Seems pretty easy to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\huge[9][1][8]\\&#xA;\huge[2]\square\square&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rules don't say I need to fill in &lt;em&gt;all&lt;/em&gt; of the boxes.&lt;/p&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2015-11-04T20:37:03.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23791" PostTypeId="1" AcceptedAnswerId="23794" CreationDate="2015-11-04T23:04:24.827" Score="10" ViewCount="1141" Body="&lt;p&gt;&lt;strong&gt;Story:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You go to a restaurant called &quot;Likeable Alb's&quot; to eat some of their world-famous chips. You sit down in your chair. Some time later, the waiter comes and hands you the menu. The menu contains how much money it will cost to buy certain amounts of chips. &quot;Alright then&quot; you say, as you look at the menu. The menu reads as the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$------------------------------------$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 chip for $1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5 chips for $5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;25 chips for $15&lt;/p&gt;&#xA;&#xA;&lt;p&gt;80 chips for $40&lt;/p&gt;&#xA;&#xA;&lt;p&gt;400 chips for $100&lt;/p&gt;&#xA;&#xA;&lt;p&gt;900 chips for $250&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2,000 chips for $500&lt;/p&gt;&#xA;&#xA;&lt;p&gt;10,000 chips for $1,500&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will, though, give you any amount of chips, as long as it's between 1 and 10,000, and it will be a specific price that we can calculate for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$------------------------------------$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You decide that you don't want to have any of these. You want 300 chips, so that's what you order. The chips are delicious, but when you finally get the bill, you look at the cost and are very confused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you looked at your bill, what made you so confused? Why?&lt;/p&gt;&#xA;" OwnerUserId="11002" LastEditorUserId="11002" LastEditDate="2015-11-04T23:25:07.867" LastActivityDate="2015-11-05T21:49:36.980" Title="Likeable Alb's Chip Restaurant" Tags="|mathematics|enigmatic-puzzle|" AnswerCount="3" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="23792" PostTypeId="2" ParentId="23791" CreationDate="2015-11-04T23:08:38.073" Score="15" Body="&lt;p&gt;$300 = 80 \times 3 + 25 \times 2 + 5 \times 2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it might have cost him $40 \times 3 + 15 \times 2 + 5 \times 2 = 160$, which is more than what he would have paid had he ordered $400$ chips.&lt;/p&gt;&#xA;" OwnerUserId="16478" LastActivityDate="2015-11-04T23:08:38.073" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23794" PostTypeId="2" ParentId="23791" CreationDate="2015-11-04T23:31:37.213" Score="26" Body="&lt;p&gt;The price is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;\$1 per chip, but the money is listed in base 20.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When your bill comes out to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;\$F0&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;of course you're confused!&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-11-05T03:50:21.990" LastActivityDate="2015-11-05T03:50:21.990" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23795" PostTypeId="2" ParentId="23781" CreationDate="2015-11-05T10:04:40.447" Score="1" Body="&lt;h1&gt;1,2,3,4,5,7&lt;/h1&gt;&#xA;&lt;p&gt;And the puzzle is in base 9.&lt;/p&gt;&#xA;" OwnerUserId="16599" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-11-05T10:04:40.447" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23796" PostTypeId="1" AcceptedAnswerId="23798" CreationDate="2015-11-05T10:48:37.233" Score="15" ViewCount="1769" Body="&lt;p&gt;When I was cleaning my basement I found one puzzle that my father gave me when I was a child.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really want to know the name of it! Could someone help me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The puzzle consist in 4 disks, and each disk have 4 balls. To beat the puzzle you have to put balls with the same color in the same column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here is some pictures:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bRSYM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bRSYM.jpg&quot; alt=&quot;Puzzle&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OEmIO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OEmIO.jpg&quot; alt=&quot;Puzzle&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16293" LastEditorUserId="2806" LastEditDate="2015-11-05T13:38:39.650" LastActivityDate="2015-11-05T13:38:39.650" Title="Help with the name of this puzzle" Tags="|logical-deduction|puzzle-identification|twisty-puzzles|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23797" PostTypeId="2" ParentId="23796" CreationDate="2015-11-05T11:12:59.333" Score="6" Body="&lt;p&gt;It seems to be called &quot;Tower of Babel&quot; or &quot;Babylon Tower&quot;.&lt;br&gt;&#xA;Some sites refer to it as a &quot;Hungarian Tower of Babel&quot;, others as &quot;Ivory Tower&quot;.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.passionforpuzzles.com/2012/09/hungarian-tower-of-babel-puzzle.php&quot;&gt;http://www.passionforpuzzles.com/2012/09/hungarian-tower-of-babel-puzzle.php&lt;/a&gt;   &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.jaapsch.net/puzzles/ivory.htm&quot;&gt;http://www.jaapsch.net/puzzles/ivory.htm&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The puzzle was designed by Endre Pap and &lt;a href=&quot;http://www.jaapsch.net/puzzles/patents/de3104021.pdf&quot;&gt;patented&lt;/a&gt; on 2 December 1982.&lt;/p&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="8874" LastEditDate="2015-11-05T11:18:02.653" LastActivityDate="2015-11-05T11:18:02.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23798" PostTypeId="2" ParentId="23796" CreationDate="2015-11-05T13:11:14.610" Score="10" Body="&lt;p&gt;That specific one looks like a Varikon tower to me. See &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.cs.brandeis.edu/~storer/JimPuzzles/MANIP/VarikonTowers/VarikonTowers.pdf&quot;&gt;http://www.cs.brandeis.edu/~storer/JimPuzzles/MANIP/VarikonTowers/VarikonTowers.pdf&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.jaapsch.net/puzzles/tower.htm&quot;&gt;http://www.jaapsch.net/puzzles/tower.htm&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1748" LastEditorUserId="1748" LastEditDate="2015-11-05T13:19:30.037" LastActivityDate="2015-11-05T13:19:30.037" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23799" PostTypeId="2" ParentId="23773" CreationDate="2015-11-05T13:49:23.593" Score="1" Body="&lt;p&gt;Chuck survives with probably 64.9%&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Contrived Test Scenario:&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Robot #1, Rules &lt;strong&gt;1,3&lt;/strong&gt;.  #2 Rule &lt;strong&gt;2&lt;/strong&gt;.  #3  Rule &lt;strong&gt;3&lt;/strong&gt;, #4 Rule &lt;strong&gt;3&lt;/strong&gt;, #5 Rules &lt;strong&gt;1,2,3&lt;/strong&gt;  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Robot #1 Attacks Robot #2 by lack of Rule 2.&lt;br&gt;&#xA;  Robot #2 Attacks Chuck (50-50 of choosing Chuck or Robot)&lt;br&gt;&#xA;  Chuck Deactivates #2&lt;br&gt;&#xA;  Robot #3 Attacks #5    (May choose anything but himself)&lt;br&gt;&#xA;  Robot #4 Attacks Chuck (May choose anything but himself)&lt;br&gt;&#xA;  Robot #5 Defends Chuck (Note: #2 is attacked by #1, #3 is attacking #5, #4 is attacking Chuck!)  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Result:&lt;br&gt;&#xA;  Robot #2 is inactive by Chuck&lt;br&gt;&#xA;  Robot #4 is inactive by #5&lt;br&gt;&#xA;  Robot #1 throws #2&lt;br&gt;&#xA;  Robot #3 throws #5 out&lt;br&gt;&#xA;  Chuck Wins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note for the below: I cherry picked some of the shorter rounds, with Chuck deactivating robots, a rule 1,3 or rule 3 tends to spend awhile throwing inactives out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Verbose Outputs:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;Chuck Wins&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[0_1.2, 1_1.2.3, 2_1.2.3, 3_1.2, 4_1.2, 5_1.3, 6_3, 7_1, 8_1.2, C]&#xA;*Robot 0 has laws 1,2, Robot 1 has laws 1,2,3 etc.*&#xA;Round: 0&#xA;0_1.2: 0   *Robot 0 throws himself out*&#xA;1_1.2.3: 1 *Robot 1 throws himself out*&#xA;2_1.2.3: 2 *Robot 2 throws himself out*&#xA;3_1.2: 3   *Robot 3 throws himself out*&#xA;4_1.2: 4   *Robot 4 throws himself out*&#xA;5_1.3: 4   *Robot 5 helps robot 4 jump out*&#xA;6_3: 3     *Robot 6 helps robot 3 jump out*&#xA;7_1: 1     *Robot 7 helps robot 1 jump out*&#xA;8_1.2: 8   *Robot 8 throws himself out*&#xA;C: null&#xA;Done: [5_1.3, 6_3, 7_1, C]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Chuck gets Chucked&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[0_1.3, 1_1.2, 2_2.3, 3D, 4_1, 5_2, 6_1.2.3, 7_2, 8D, C]&#xA;*Robot 0 has laws 1,3, Robot 3 has no laws so it starts off as inactive*&#xA;Round: 0&#xA;0_1.3: 7         *Robot #0 attacks #7*&#xA;1_1.2: 1         *Robot #1 throws himself out*&#xA;Deactivated: 0   *Robot #2 deactivates #0*&#xA;2_2.3D: null     *Robot #2 is now inactive*&#xA;3D: null         *Robot #3 has no laws and is inactive*&#xA;4_1: 4           *Robot #4 throws himself out*&#xA;5_2D: 9             *Robot #5 attacks chuck*&#xA;Chuck deactivates 5 *Chuck deactivates #5*&#xA;6_1.2.3: 6          *Robot #6 throws himself out*&#xA;7_2: 9&#xA;Attacking Chuck     *Robot #7 attacks Chuck*&#xA;8D: null            *Robot #8 has no laws and is inactive&#xA;C: null&#xA;*Not shown, final result is:*&#xA;*[0_1.3D, 2_2.3D, 3D, 5_2D, 7_2, 8D]*&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Chuck rides into the mountain&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[0_3, 1_2, 2_1.3, 3D, 4D, 5D, 6_1.2, 7_2.3, 8_1.3, C]&#xA;Round: 0&#xA;0_3D: 9             *Robot #0 attacks Chuck*&#xA;Chuck deactivates 0 *Chuck deactivates #0*&#xA;1_2: 9              *Robot #1 attacks Chuck*&#xA;Attacking Chuck&#xA;Deactivated: 1      *Robot #2 defends Chuck*&#xA;2_1.3D: null        *Robot #2 is now inactive*&#xA;3D: null            *Robot #3 has no laws*&#xA;4D: null            *Robot #4 has no laws*&#xA;5D: null            *Robot #5 has no laws*&#xA;6_1.2: 6            *Robot #6 jumps out*&#xA;7_2.3: 9            *Robot #7 attacks Chuck*&#xA;Attacking Chuck&#xA;Deactivated: 7      *Robot #8 defends Chuck*&#xA;8_1.3D: null        *Robot #8 is now inactive*&#xA;C: null&#xA;Done: [0_3D, 1_2D, 2_1.3D, 3D, 4D, 5D, 7_2.3D, 8_1.3D, C]&#xA;*Chuck is doomed, all robots inactive*&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Probability Notes&lt;/strong&gt;&lt;br&gt;&#xA;  The first robot to move acts with probability (Rounded to 0 decimals):&lt;br&gt;&#xA;  33% Throw self&lt;br&gt;&#xA;  35% Throw another Robot&lt;br&gt;&#xA;  20% Throw Chuck&lt;br&gt;&#xA;  13% Do nothing  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;When Chuck is attacked, the probability is:&lt;br&gt;&#xA;  6% Throw self&lt;br&gt;&#xA;  11% Throw another Robot&lt;br&gt;&#xA;  20% Throw Chuck&lt;br&gt;&#xA;  50% Defend Chuck&lt;br&gt;&#xA;  13% Do nothing  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;When a robot's attacked other than himself:&lt;br&gt;&#xA;  1% Throw self&lt;br&gt;&#xA;  35% Throw another Robot&lt;br&gt;&#xA;  1% Throw Chuck&lt;br&gt;&#xA;  50% Defend the other Robot&lt;br&gt;&#xA;  13% Do nothing  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Probability Table for the first robot:&#xA;                   Self   Robot   Chuck  Defense   Nothing&#xA;Laws 1,2,3 Action: 1&#xA;Laws 1,2   Action: 1&#xA;Laws 1     Action: 1/9    8/9&#xA;Laws 1,3   Action:        1&#xA;Laws 2,3   Action:                1&#xA;Laws 2     Action: 1/2            1/2&#xA;Laws 3     Action:        8/9     1/9&#xA;Laws None  Action:                                 1&#xA;Sum:               47/18  50/18    29/18 0         1&#xA;P (Sum/8)          33%    35%      20%   0%        13%        &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What this implies is that Chuck is unlikely to be killed directly.  Less than 1/5 chance of being targeted.  And robots tend to be protective with 50-50 odds of deactivating an aggressor.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Expanding the probabilities 9 times yielded a couple bounds:&lt;br&gt;&#xA;  Defining a Win as 6 robots attacking another robot: 10%&lt;br&gt;&#xA;  Defining a loss as Chuck attacked by two or more, or all robots inactive: 11%&lt;br&gt;&#xA;  Unknown: 79% due to another round or robots may be inactive or robots may have jumped out.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Survival Count per Robot:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;3  100.00%&#xA;4   85.9%&#xA;5   77.9%&#xA;6   71.9%&#xA;7   68.2%&#xA;8   65.8%&#xA;9   65.1%&#xA;10  64.7%&#xA;13  64.9%  (Bottom)&#xA;14  64.9%  (Bottom)&#xA;200 75.0%  (Flat?)&#xA;750 75.1%  (Flat?)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Q2Kep.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Q2Kep.png&quot; alt=&quot;Survival Charts&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Algorithm pseudo-code&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;public void getTarget(ArrayList&amp;lt;Robot&amp;gt; robots) {&#xA;  if(deactivated || chuck) return;&#xA;  if(law1) {&#xA;    if(ChuckAttacked) {deactivate self and robot attacking chuck}&#xA;    else if(law2) {throw self out}&#xA;    else {&#xA;      if(law3) {throw a robot out}&#xA;      else {throw anything but Chuck out}&#xA;    }&#xA;  } else {&#xA;    if(law2) {&#xA;      if(a robot is attacked) {deactivate self and first to attack}&#xA;      else {&#xA;           if(law3 or 50-50 rng) {Throw chuck out}&#xA;           else {Throw self out}&#xA;      }&#xA;    //Implicitly, this is law3 only.&#xA;    } else {throw anything but myself out}&#xA;  }&#xA;  if(target is Chuck and Chuck has not acted) {deactivate self and mark chuck as acted}&#xA;  else {declare attack against target}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="11750" LastEditorUserId="11750" LastEditDate="2015-11-15T23:20:49.417" LastActivityDate="2015-11-15T23:20:49.417" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23800" PostTypeId="2" ParentId="23787" CreationDate="2015-11-05T13:56:53.920" Score="10" Body="&lt;p&gt;If John is a dinosaur, he will say that a dinosaur is a dinosaur (true) and an elephant is a dinosaur (lie.) If John is an elephant, he will say that an elephant is an elephant (true) and a dinosaur is an elephant (lie.) Therefore all we can establish from &lt;em&gt;John says &quot;Zayin is a dinosaur.&quot;&lt;/em&gt; is that John is a dinosaur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 16 combinations of animals for the second part. For example TD JD MD ED would mean all 4 are dinosaurs. Since they would each tell the truth, starting with M saying &quot;E is a D&quot;, you end up with T saying &quot;E is a D&quot; which is not what was said, so that combination can be ruled out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are more than 1 combinations that lead to the sentence being uttered though: for example TE JD MD ED would make M say &quot;E is a D&quot; truthfully, and J pass that along truthfully, but then T lie and report the chain ended E. Similarly TD JE MD ED would start with M saying &quot;E is a D&quot; but then J would lie, claiming M said E was an E, and then T would pass that along truthfully. I haven't tested all 16 combinations, but my suspicion is that whenever you have an odd split (1-3 or 3-1) of animal types, that is the sentence that will be uttered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the third part, the sentence &quot;you lie about your righthand neighbour&quot; involves 3 animals: the speaker S, the subject Y, and the neighbour N. There are 8 combinations possible: DDD, DEE, DED, DDE, EEE, EED, EDE, EED. In the 4 cases where Y and N are different, Y really does lie, but only if S and Y are the same will S say so. In the other 4 cases, Y tells the truth, but if S and Y are different S will lie about that. So the patterns DEE, DDE, EED, and EDD will lead to this sentence. You can get this pattern if they sit in pairs - for example four of them as DDEE with the final E &quot;looping back around&quot; and sitting next to the first D. The total number of animals must be a multiple of 4 to reuse this pattern, eg DDEEDDEE and so on.&lt;/p&gt;&#xA;" OwnerUserId="1509" LastActivityDate="2015-11-05T13:56:53.920" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23801" PostTypeId="1" AcceptedAnswerId="23803" CreationDate="2015-11-05T19:03:47.247" Score="6" ViewCount="724" Body="&lt;p&gt;This is a slight adaptation of a google interview question that I found entertaining.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are given a jar with &lt;strong&gt;N&lt;/strong&gt; coins, &lt;strong&gt;K&lt;/strong&gt; of them are fair. The remaining coins have heads on both sides. You choose a coin at random from the jar and flip it &lt;strong&gt;m&lt;/strong&gt; times. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If you get heads all &lt;strong&gt;m&lt;/strong&gt; times, what is the probability that you selected a fair coin?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If &lt;strong&gt;m&lt;/strong&gt; is 3, and &lt;strong&gt;N&lt;/strong&gt; is 100, how many fair coins would need to be in the jar, for you to be at least 50% sure you selected a fair coin?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11063" LastEditorUserId="12408" LastEditDate="2016-07-25T02:30:51.187" LastActivityDate="2016-07-25T02:30:51.187" Title="A jar of mixed coins" Tags="|probability|coin-tossing|" AnswerCount="2" CommentCount="0" ClosedDate="2015-11-11T18:48:37.387" ContentLicense="CC BY-SA 3.0" />
  <row Id="23802" PostTypeId="2" ParentId="23801" CreationDate="2015-11-05T19:22:22.700" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    P(fair-coin|m-heads) = p(fair-coin &amp;amp; m heads)/p(m-heads)&#xA;    p(fair-coin &amp;amp; m-heads) = K/N * (0.5^m)&#xA;    p(m-heads) = p(m-heads &amp;amp; (fair-coin |unfair-coin)) = p(m-heads * fair-coin)+ p(m-heads *unfair-coin)&#xA;    p(m-heads * unfair-coin) = p(unfair-coin)*p(m-heads/unfair-coin) = N-K/N  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so answer to the first part = K/N * (0.5^m)/(K/N*(0.5^m)) + (N-K)/N)&lt;/p&gt;&#xA;" OwnerUserId="10038" LastActivityDate="2015-11-05T19:22:22.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23803" PostTypeId="2" ParentId="23801" CreationDate="2015-11-05T19:24:28.213" Score="6" Body="&lt;p&gt;The probability is given by $P(fair|heads^m) = P(heads^m|fair) \cdot P(fair) / P(heads^m)$, where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(heads^m) = P(heads^m|fair)\cdot P(fair) + P(heads^m|nonfair)\cdot P(nonfair) = 1/2^m \cdot K/N + 1^m \cdot (N-K)/N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_m = \frac{1/2^m \cdot K/N}{1/2^m \cdot K/N + 1^m \cdot (N-K)/N)} .$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for $K$, we find that it must be 88.8.&lt;/p&gt;&#xA;" OwnerUserId="12901" LastActivityDate="2015-11-05T19:24:28.213" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23805" PostTypeId="2" ParentId="23783" CreationDate="2015-11-05T20:24:00.890" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;This is a continuation of my &lt;a href=&quot;https://puzzling.stackexchange.com/a/23784/11476&quot;&gt;previous answer&lt;/a&gt;. The answer is &lt;strong&gt;no&lt;/strong&gt;: if we consider only the non-integral points inside the eight regions of $f^2$, we see the flea will always take on a bounded path.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://i.stack.imgur.com/DCplR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;eight regions $f^2$ acts differently on&lt;/a&gt; are snapped to an orthogonal grid, where each grid cell is a $2 \times 2$ square, and the origin is the center of a cell. We will consider only the &lt;a href=&quot;https://en.wikipedia.org/wiki/Interior_(topology)&quot; rel=&quot;nofollow noreferrer&quot;&gt;interior&lt;/a&gt; of these squares, as we're examing non-integral coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of this grid the regions are laid out on, each cell has the property that all of its points will undergo the same transformation when $f^2$ is applied to them. Because all of the transformations in the piecewise definition of $f^2$ are &lt;em&gt;translations&lt;/em&gt; over a vector whose components are both multiples of 2, the entire cell is even mapped to a different cell!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to state this is to say that the value of $\operatorname{fmod}(x+1, 2)$ and $\operatorname{fmod}(y+1, 2)$ doesn't affect the shape of the path the flea will take when starting at $(x, y)$; only which cell the initial point falls in matters. (&lt;code&gt;fmod&lt;/code&gt;, here, is &lt;a href=&quot;http://www.cplusplus.com/reference/cmath/fmod/&quot; rel=&quot;nofollow noreferrer&quot;&gt;fractional modulo&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, here are the paths the flea takes when starting from two points in the same cell (the gray lines mark the cells; the center cell containing the origin is the square the flea jumps over):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hspace2in$&lt;a href=&quot;http://puu.sh/laBEY/3dd820c323.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;coinciding paths http://puu.sh/laBEY/3dd820c323.png&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We must prove that for all of these cells, the path will be bounded. Here is the grid of cells, with arrows representing the direction of the translation $f^2$ will move the flea in &lt;strong&gt;by two cells&lt;/strong&gt; from any point in the cell:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hspace1.2in$&lt;a href=&quot;https://i.stack.imgur.com/BO835.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BO835.png&quot; alt=&quot;cell directions&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without loss of generality, assume the flea is in the center of the cells. Then it is clear to see that none of the possible moves will change the effective &lt;a href=&quot;https://en.wikipedia.org/wiki/Taxicab_geometry#Circles_in_Taxicab_geometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;taxicab distance&lt;/a&gt; between the flea and the origin: the diagonal moves would have never changed it anyhow, and the orthogonal moves nicely mirror the flea's position over one of the axes. The ratio between the taxicab norm and the regular norm is at most $\sqrt{2}$, so we can never stray more than $\sqrt{2}$ times as far away from the origin as from the initial point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To account for the fact that the flea might not be exactly in the center of a cell, and the actual norms will fluctuate, note that moving the flea around within a cell will not matter: the largest factor that can &lt;em&gt;ever&lt;/em&gt; account for is the difference between the flea being in the center at $(2, 0)$ vs. it being very close to $(3, 1)$, where the ratio is $\frac{\sqrt{3^2+1^2}}{\sqrt{2^2+0^2}}=\sqrt{5}$, and $\sqrt{2} \cdot \sqrt{5} &amp;lt; 30$ still.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Edit: I suppose this only proves that the subsequence of double jumps is bounded in norm. Can the norm &quot;peak&quot; in the transitional jump between two of the jumps we did count? No: $f$ itself is a rotation, which doesn't affect the norm, followed by a translation over $(\pm 2,\pm 2)$. The worst ratio this &lt;em&gt;could&lt;/em&gt; account for is bounded above by $\sqrt{13}$ by similar logic to the above. (Imagine the flea moves from a point close to (1, 0) to a point close to a rotation of (3, 2).) Stacking all of these weird upper bounds, we can only account for a total ratio of at most $\sqrt 2 \cdot \sqrt 5 \cdot \sqrt{13}$, which is still less than 30.)&lt;/p&gt;&#xA;" OwnerUserId="11476" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-11-06T08:19:23.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23807" PostTypeId="2" ParentId="23791" CreationDate="2015-11-05T20:36:49.527" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Since you didn't order 400 chips, they didn't charge you the 400 chip price, they charged you the 1 chip price (\$1) 300 times.  Therefore your bill was \$300.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17264" LastEditorUserId="12408" LastEditDate="2015-11-05T21:49:36.980" LastActivityDate="2015-11-05T21:49:36.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23808" PostTypeId="2" ParentId="23781" CreationDate="2015-11-05T21:40:54.790" Score="3" Body="&lt;p&gt;Inspired by &lt;a href=&quot;https://puzzling.stackexchange.com/a/23795/4421&quot;&gt;@AlbertRenshaw's post&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use hexadecimal.  Then $2 + 4 + 5 + 6 + 7 + 8 = 20_{16}\;(=32_{10})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can actually work with any base from 11 to 19:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1 + 2 + 3 + 4 + 5 + 7 = 20_{11}\;(=22_{10})$&lt;br&gt;&#xA;$1 + 2 + 3 + 4 + 5 + 9 = 20_{12}\;(=24_{10})$&lt;br&gt;&#xA;$1 + 2 + 3 + 4 + 7 + 9 = 20_{13}\;(=26_{10})$&lt;br&gt;&#xA;$1 + 2 + 3 + 6 + 7 + 9 = 20_{14}\;(=28_{10})$&lt;br&gt;&#xA;$1 + 2 + 3 + 7 + 8 + 9 = 20_{15}\;(=30_{10})$&lt;br&gt;&#xA;$1 + 4 + 5 + 7 + 8 + 9 = 20_{17}\;(=34_{10})$&lt;br&gt;&#xA;$1 + 5 + 6 + 7 + 8 + 9 = 20_{18}\;(=36_{10})$&lt;br&gt;&#xA;$3 + 5 + 6 + 7 + 8 + 9 = 20_{19}\;(=38_{10})$  &lt;/p&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-11-05T21:40:54.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23810" PostTypeId="1" AcceptedAnswerId="23811" CreationDate="2015-11-05T23:32:29.933" Score="8" ViewCount="423" Body="&lt;p&gt;Blackbeard is no more, and everything seemed peaceful again, but during the last engagement with Blackbeard, my military adviser got &lt;a href=&quot;https://puzzling.stackexchange.com/a/23775/16621&quot;&gt;introduced to the concept of &quot;computer&quot;&lt;/a&gt;, and became very interested in its capability and uses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it is not a surprised when he led a team of engineers and farmers to build a massive &quot;computer&quot; powered by tons of potatoes. However, its capability is very limited, and it couldn't solve the map. When a standard cryptic map drawn by the scouts is fed into the &quot;computer&quot;, it spewed out another map with even more numbers. For example, it would produce a map looking like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;-----------------------------------------&#xA;| 2 |   | 1 |   | 1 |   | 3 |   |   |   | &#xA;-----------------------------------------&#xA;|   |   |   | 0 |   |   |   |   | 2 | 1 | &#xA;-----------------------------------------&#xA;| 2 |   |   |   | 1 |   |   |   |   |   | &#xA;-----------------------------------------&#xA;|   |   | 1 |   |   | 0 |   |   | 3 |   | &#xA;-----------------------------------------&#xA;|   |   |   |   |   |   |   |   |   | 0 | &#xA;-----------------------------------------&#xA;|   |   | 3 |   | 3 |   |   |   |   |   | &#xA;-----------------------------------------&#xA;|   |   |   |   | 2 |   |   | 1 |   |   | &#xA;-----------------------------------------&#xA;|   |   |   | 4 |   |   | 1 |   |   | 2 | &#xA;-----------------------------------------&#xA;| 0 |   |   |   |   |   |   | 0 |   |   | &#xA;-----------------------------------------&#xA;|   |   |   | 3 |   | 2 |   |   | 2 |   | &#xA;-----------------------------------------&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The map that was fed in had the standard fleet formation, with 1 flagship of size 4, 2 frigates of size 3, 3 corvette of size 2, and 4 caravel of size 1(I have finally learned the real name of those pesky patrol boats). All of them are placed horizontally or vertically with no ships touching each other, not even diagonally. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My military adviser studied the new map for a bit and concluded that every number on the new map indicates the number of ship pieces adjacent to that number (including diagonally adjacent), and furthermore, no ship pieces is on top of a number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He smiled at the &quot;computer&quot; and contently walked away, possibly thinking about its uses and improvements. Leaving me with new map of nonsense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you make sense of this new map and figure out the fleet formation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Although this map was generated by a &quot;computer&quot;, no computer is necessary to find the solution.&lt;/p&gt;&#xA;" OwnerUserId="16621" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-11-06T00:17:25.400" Title="End of piracy and beginning of technology" Tags="|logical-deduction|story|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23811" PostTypeId="2" ParentId="23810" CreationDate="2015-11-05T23:58:37.727" Score="8" Body="&lt;p&gt;First, we can exclude all the spaces around the 0s:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;-----------------------------------------&#xA;| 2 |   | 1 | . | 1 |   | 3 |   |   |   | &#xA;-----------------------------------------&#xA;|   |   | . | 0 | . |   |   |   | 2 | 1 | &#xA;-----------------------------------------&#xA;| 2 |   | . | . | 1 | . | . |   |   |   | &#xA;-----------------------------------------&#xA;|   |   | 1 |   | . | 0 | . |   | 3 | . | &#xA;-----------------------------------------&#xA;|   |   |   |   | . | . | . |   | . | 0 | &#xA;-----------------------------------------&#xA;|   |   | 3 |   | 3 |   |   |   | . | . | &#xA;-----------------------------------------&#xA;|   |   |   |   | 2 |   |   | 1 |   |   | &#xA;-----------------------------------------&#xA;| . | . |   | 4 |   |   | 1 | . | . | 2 | &#xA;-----------------------------------------&#xA;| 0 | . |   |   |   |   | . | 0 | . |   | &#xA;-----------------------------------------&#xA;| . | . |   | 3 |   | 2 | . | . | 2 |   | &#xA;-----------------------------------------&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The 2 in the bottom right only has two open spaces next to it, so we can fill those in. We also have some squares in the top left with only one possibility.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;-----------------------------------------&#xA;| 2 | . | 1 | . | 1 |   | 3 |   |   |   | &#xA;-----------------------------------------&#xA;| &amp;lt; | &amp;gt; | . | 0 | . |   |   |   | 2 | 1 | &#xA;-----------------------------------------&#xA;| 2 | . | . | . | 1 | . | . |   |   |   | &#xA;-----------------------------------------&#xA;| . | . | 1 |   | . | 0 | . |   | 3 | . | &#xA;-----------------------------------------&#xA;|   |   |   |   | . | . | . |   | . | 0 | &#xA;-----------------------------------------&#xA;|   |   | 3 |   | 3 |   |   |   | . | . | &#xA;-----------------------------------------&#xA;|   |   |   |   | 2 |   |   | 1 |   |   | &#xA;-----------------------------------------&#xA;| . | . |   | 4 |   |   | 1 | . | . | 2 | &#xA;-----------------------------------------&#xA;| 0 | . |   |   |   |   | . | 0 | . | ^ | &#xA;-----------------------------------------&#xA;| . | . |   | 3 |   | 2 | . | . | 2 | v | &#xA;-----------------------------------------&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The 3 above the 2 in the center requires that there is some part of a ship in the space above and to the left of it:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;-----------------------------------------&#xA;| 2 | . | 1 | . | 1 | . | 3 |   |   |   | &#xA;-----------------------------------------&#xA;| &amp;lt; | &amp;gt; | . | 0 | . | ? |   |   | 2 | 1 | &#xA;-----------------------------------------&#xA;| 2 | . | . | . | 1 | . | . |   |   |   | &#xA;-----------------------------------------&#xA;| . | . | 1 | . | . | 0 | . |   | 3 | . | &#xA;-----------------------------------------&#xA;|   | . | . | ? | . | . | . |   | . | 0 | &#xA;-----------------------------------------&#xA;|   |   | 3 |   | 3 |   |   |   | . | . | &#xA;-----------------------------------------&#xA;|   |   |   |   | 2 |   |   | 1 |   |   | &#xA;-----------------------------------------&#xA;| . | . |   | 4 |   |   | 1 | . | . | 2 | &#xA;-----------------------------------------&#xA;| 0 | . |   |   |   |   | . | 0 | . | ^ | &#xA;-----------------------------------------&#xA;| . | . |   | 3 |   | 2 | . | . | 2 | v | &#xA;-----------------------------------------&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now the 3 in the top row has only one possibility that works with the 2-1 to its right, and the 3 below those is left with only 3 spaces around it:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;-----------------------------------------&#xA;| 2 | . | 1 | . | 1 | . | 3 | . | . |   | &#xA;-----------------------------------------&#xA;| &amp;lt; | &amp;gt; | . | 0 | . | &amp;lt; | x | &amp;gt; | 2 | 1 | &#xA;-----------------------------------------&#xA;| 2 | . | . | . | 1 | . | . | . | . | o | &#xA;-----------------------------------------&#xA;| . | . | 1 | . | . | 0 | . | ^ | 3 | . | &#xA;-----------------------------------------&#xA;|   | . | . | ? | . | . | . | ? | . | 0 | &#xA;-----------------------------------------&#xA;|   |   | 3 |   | 3 |   |   |   | . | . | &#xA;-----------------------------------------&#xA;|   |   |   |   | 2 |   |   | 1 |   |   | &#xA;-----------------------------------------&#xA;| . | . |   | 4 |   |   | 1 | . | . | 2 | &#xA;-----------------------------------------&#xA;| 0 | . |   |   |   |   | . | 0 | . | ^ | &#xA;-----------------------------------------&#xA;| . | . |   | 3 |   | 2 | . | . | 2 | v | &#xA;-----------------------------------------&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;At this point, there is only one possible location for the flagship:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;-----------------------------------------&#xA;| 2 | . | 1 | . | 1 | . | 3 | . | . |   | &#xA;-----------------------------------------&#xA;| &amp;lt; | &amp;gt; | . | 0 | . | &amp;lt; | x | &amp;gt; | 2 | 1 | &#xA;-----------------------------------------&#xA;| 2 | . | . | . | 1 | . | . | . | . | o | &#xA;-----------------------------------------&#xA;| . | . | 1 | . | . | 0 | . | ^ | 3 | . | &#xA;-----------------------------------------&#xA;|   | . | . | ? | . | . | . | ? | . | 0 | &#xA;-----------------------------------------&#xA;|   |   | 3 |   | 3 |   |   |   | . | . | &#xA;-----------------------------------------&#xA;|   |   |   |   | 2 | . | . | 1 |   |   | &#xA;-----------------------------------------&#xA;| . | . | . | 4 | . | . | 1 | . | . | 2 | &#xA;-----------------------------------------&#xA;| 0 | . | &amp;lt; | x | x | &amp;gt; | . | 0 | . | ^ | &#xA;-----------------------------------------&#xA;| . | . | . | 3 | . | 2 | . | . | 2 | v | &#xA;-----------------------------------------&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The space between the two 3s and the 4 must have a ship part, otherwise it becomes impossible to place enough parts in that area.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;-----------------------------------------&#xA;| 2 | . | 1 | . | 1 | . | 3 | . | . |   | &#xA;-----------------------------------------&#xA;| &amp;lt; | &amp;gt; | . | 0 | . | &amp;lt; | x | &amp;gt; | 2 | 1 | &#xA;-----------------------------------------&#xA;| 2 | . | . | . | 1 | . | . | . | . | o | &#xA;-----------------------------------------&#xA;| . | . | 1 | . | . | 0 | . | ^ | 3 | . | &#xA;-----------------------------------------&#xA;|   | . | . | ? | . | . | . | ? | . | 0 | &#xA;-----------------------------------------&#xA;|   |   | 3 |   | 3 |   |   |   | . | . | &#xA;-----------------------------------------&#xA;|   |   | . | ? | 2 | . | . | 1 |   |   | &#xA;-----------------------------------------&#xA;| . | . | . | 4 | . | . | 1 | . | . | 2 | &#xA;-----------------------------------------&#xA;| 0 | . | &amp;lt; | x | x | &amp;gt; | . | 0 | . | ^ | &#xA;-----------------------------------------&#xA;| . | . | . | 3 | . | 2 | . | . | 2 | v | &#xA;-----------------------------------------&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If the space between the 3s is empty, we find that it is not possible to place all the remaining ships (either we need 5 caravels, or there is nowhere for the other frigate). So the frigate is between the 3s.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;-----------------------------------------&#xA;| 2 | . | 1 | . | 1 | . | 3 | . | . |   | &#xA;-----------------------------------------&#xA;| &amp;lt; | &amp;gt; | . | 0 | . | &amp;lt; | x | &amp;gt; | 2 | 1 | &#xA;-----------------------------------------&#xA;| 2 | . | . | . | 1 | . | . | . | . | o | &#xA;-----------------------------------------&#xA;| . | . | 1 | . | . | 0 | . | ^ | 3 | . | &#xA;-----------------------------------------&#xA;|   | . | . | ^ | . | . | . | v | . | 0 | &#xA;-----------------------------------------&#xA;|   | . | 3 | x | 3 | . | . | . | . | . | &#xA;-----------------------------------------&#xA;|   | . | . | v | 2 | . | . | 1 |   |   | &#xA;-----------------------------------------&#xA;| . | . | . | 4 | . | . | 1 | . | . | 2 | &#xA;-----------------------------------------&#xA;| 0 | . | &amp;lt; | x | x | &amp;gt; | . | 0 | . | ^ | &#xA;-----------------------------------------&#xA;| . | . | . | 3 | . | 2 | . | . | 2 | v | &#xA;-----------------------------------------&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And all that's left is to place the remaining 3 caravels:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;-----------------------------------------&#xA;| 2 |   | 1 |   | 1 |   | 3 |   |   |   | &#xA;-----------------------------------------&#xA;| &amp;lt; | &amp;gt; |   | 0 |   | &amp;lt; | x | &amp;gt; | 2 | 1 | &#xA;-----------------------------------------&#xA;| 2 |   |   |   | 1 |   |   |   |   | o | &#xA;-----------------------------------------&#xA;|   |   | 1 |   |   | 0 |   | ^ | 3 |   | &#xA;-----------------------------------------&#xA;| o |   |   | ^ |   |   |   | v |   | 0 | &#xA;-----------------------------------------&#xA;|   |   | 3 | x | 3 |   |   |   |   |   | &#xA;-----------------------------------------&#xA;| o |   |   | v | 2 |   |   | 1 | o |   | &#xA;-----------------------------------------&#xA;|   |   |   | 4 |   |   | 1 |   |   | 2 | &#xA;-----------------------------------------&#xA;| 0 |   | &amp;lt; | x | x | &amp;gt; |   | 0 |   | ^ | &#xA;-----------------------------------------&#xA;|   |   |   | 3 |   | 2 |   |   | 2 | v | &#xA;-----------------------------------------&lt;/pre&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-11-06T00:17:25.400" LastActivityDate="2015-11-06T00:17:25.400" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23813" PostTypeId="2" ParentId="23358" CreationDate="2015-11-06T10:30:40.857" Score="0" Body="&lt;p&gt;Just a wild guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Moderator&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-11-06T10:30:40.857" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23815" PostTypeId="2" ParentId="22051" CreationDate="2015-11-06T13:15:16.467" Score="10" Body="&lt;p&gt;&quot;I had Spieg words called Be words and Spiel words were termed No words.&quot; This did it for me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Spieg words are substrings in &lt;em&gt;Best Picture Academy Award winners&lt;/em&gt;, Spiel words are substrings in &lt;em&gt;nominees&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A few examples:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Spieg&lt;/strong&gt;&#xA; &lt;li&gt;Midnight &lt;strong&gt;&lt;em&gt;Cow&lt;/em&gt;&lt;/strong&gt;boy&lt;/li&gt;&#xA; &lt;li&gt;&lt;strong&gt;&lt;em&gt;Law&lt;/em&gt;&lt;/strong&gt;rence of Arabia&lt;/li&gt;&#xA; &lt;li&gt;The Lord of the Rings Ret&lt;strong&gt;&lt;em&gt;urn&lt;/em&gt;&lt;/strong&gt; of the King&lt;/li&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Spiel&lt;/strong&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;li&gt;The Ch&lt;strong&gt;&lt;em&gt;amp&lt;/em&gt;&lt;/strong&gt;&lt;/li&gt;&#xA; &lt;li&gt;A &lt;strong&gt;&lt;em&gt;Far&lt;/em&gt;&lt;/strong&gt;ewell to Arms&lt;/li&gt;&#xA; &lt;li&gt;Mild&lt;strong&gt;&lt;em&gt;red&lt;/em&gt;&lt;/strong&gt; Pierce&lt;/li&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;etc.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All winners (Spieg words) were nominees, but not all nominees (Spiel words) won.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="4421" LastEditDate="2016-09-08T20:56:10.173" LastActivityDate="2016-09-08T20:56:10.173" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23816" PostTypeId="1" AcceptedAnswerId="23817" CreationDate="2015-11-06T14:31:10.670" Score="12" ViewCount="1306" Body="&lt;p&gt;When I ran into professor Halfbrain this morning, he told me that he has &#xA;constructed a cube that can be tilted and balanced on a plane table such that&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;exactly one of the eight corners touches the table (and lies in the table plane),&lt;/li&gt;&#xA;&lt;li&gt;while the distances of the other seven corners to the table plane are respectively 10, 20, 30, 40, 50, 60, and 70 centimeters.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Does such a cube really  exist, or has the professor once again  comitted one of his phenomenal mathematical blunders?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-11-08T09:41:46.653" Title="Professor Halfbrain and the tilted cube" Tags="|mathematics|geometry|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23817" PostTypeId="2" ParentId="23816" CreationDate="2015-11-06T15:49:25.713" Score="16" Body="&lt;p&gt;Such a cube&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; does exist&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first thing I noticed that this essential the same question as&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can you rotate a unit cube in such a way that the height to the table of the corners form an arithmetic sequence for some order of the corners?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because then you can just multiply it's size by something to the required distances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's take this cube:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0f3yf.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0f3yf.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for convenience let's define $f(AB)$ as the height difference between corner A and B.&lt;br&gt;&#xA;The lines AB, DC, EF and HG are all in parralel so $f(AB)$, $f(DC)$, $f(EF)$ and $f(HG)$ are all the same no matter how you rotate the cube. The same holds true for AD, BC, EH and FG.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us then first tilt the cube only a little bit by rotating it around the AD line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we rotate the cube around the line AB in such a way that D goes in between A nd B in terms of height. The situation we now have is that $f(AD)=f(DB)=f(BC)=f(EH)=f(HF)=f(FG)$. Now all we need to get is that this also equals $f(CE)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(CE)$ is determined by the initial tilt we did. We only did a little tilt to show the corner E will end up a lot heigher than corner C. Now let's say our initial tilt was 45 degrees for example. We then see E will be lower then corner C*. By the Intermediate value theorem we can say that at some initial angle $f(CE)$ will hold our desired value so professor Halfbrain is right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;*: sorry no proof but it can be seen very easy when you try it out. I have an actual cube here in front of me that gave a visual confirmation&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1748" LastEditorUserId="1748" LastEditDate="2015-11-06T15:56:11.893" LastActivityDate="2015-11-06T15:56:11.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23819" PostTypeId="1" AcceptedAnswerId="23820" CreationDate="2015-11-06T17:52:06.113" Score="13" ViewCount="1371" Body="&lt;p&gt;Rusty is a member of the highly secretive Logician's guild. After years of grueling trials, he is now finally allowed to meet the three leaders of the guild. Their names are Truman, Lyle and Xavier, and they have specific ways of answering yes/no questions: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Truman tells the truth.&lt;/li&gt;&#xA;&lt;li&gt;Lyle lies.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Xavier answers the &quot;xor&quot; of how Truman and Lyle would answer the question, meaning:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If Truman and Lyle would give different answers, then Xavier responds &quot;yes&quot;.&lt;/li&gt;&#xA;&lt;li&gt;If Truman and Lyle would give the same answer, then Xavier responds &quot;no&quot;.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Rusty enters the main chamber, and find the leaders sitting in chairs labeled Left, Right, and Middle, though he doesn't know who is who (the three know each other's identities). The leaders give Rusty a challenge. Rusty must determine their identities by asking at most three yes/no questions, each directed at a single leader. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make things harder, and to prevent paradoxes, the only allowable questions are those which are Boolean combinations of questions of one of these two forms:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;Is leader in the _____ chair named _____?&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;Are you named _____?&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For example, Rusty may ask questions like &quot;Is the person in the middle chair either Xavier or Lyle?&quot;, or &quot;Is it true that both you are Xavier and the person in the left chair is Truman?&quot;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How can Rusty pass the test?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note&lt;/em&gt;: It may seem that Xavier would always answer &quot;yes&quot;, since Truman tells the truth, which should always be different than Lyle's lie. However, including the pronoun &quot;you&quot; in your question can break this symmetry. &lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-11-08T11:36:26.887" Title="Truths, Lies, and Xors" Tags="|liars|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23820" PostTypeId="2" ParentId="23819" CreationDate="2015-11-06T18:33:46.713" Score="22" Body="&lt;pre&gt;&lt;code&gt;╠ Ask one of them (let it be one in the middle) &quot;Are you Truman?&quot; (Q1)&#xA;╠═No╦ Middle is Xavier.&#xA;║   ║  &#xA;║   ╠ Ask man on the left if he is Xavier. (Q2)&#xA;║   ╠Yes═ Left is Lyle. (Right is Truman.)&#xA;║   ╚═No═ Left is Truman. (Right is Lyle.)&#xA;║&#xA;╚Yes╦ Middle is either Truman or Lyle.&#xA;    ║  &#xA;    ╠ Ask middle if he is Xavier. (Q2)&#xA;    ╠Yes╦ Middle is Lyle.&#xA;    ║   ║&#xA;    ║   ╠ Ask middle if left is Xavier. (Q3)&#xA;    ║   ╠Yes═ Left is Truman. (Right is Xavier.)&#xA;    ║   ╚═No═ Left is Xavier. (Right is Truman.)  &#xA;    ║&#xA;    ╚═No╦ Middle is Truman.&#xA;        ║&#xA;        ╠ Ask middle if left is Xavier. (Q3)&#xA;        ╠Yes═ Left is Xavier. (Right is Lyle.)&#xA;        ╚═No═ Left is Lyle. (Right is Xavier.)  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="17005" LastEditorUserId="-1" LastEditDate="2015-11-07T16:23:25.960" LastActivityDate="2015-11-07T16:23:25.960" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23821" PostTypeId="1" CreationDate="2015-11-06T18:39:05.353" Score="3" ViewCount="1179" Body="&lt;p&gt;I'm quite familiar with &quot;clock puzzles&quot;, but only when&#xA;the puzzle is such that all 3 hands start at 12:00.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recently tried this one:&#xA;Over a 12 hour period, how often are AT LEAST 2 of the 3 hands&#xA;of a clock in the same position?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure answer is 1436.&#xA;Can someone confirm? Thank you.&lt;/p&gt;&#xA;" OwnerUserId="17274" LastEditorUserId="17274" LastEditDate="2015-11-07T04:28:41.817" LastActivityDate="2015-11-07T15:08:49.687" Title="Clock puzzle using 3 hands" Tags="|strategy|" AnswerCount="2" CommentCount="0" ClosedDate="2015-11-07T17:51:14.533" ContentLicense="CC BY-SA 3.0" />
  <row Id="23823" PostTypeId="2" ParentId="23773" CreationDate="2015-11-06T19:13:28.320" Score="2" Body="&lt;p&gt;I thought I implemented my rules the same as @Dr Xorile, but I am getting much better survival rate.  &lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Algorithms&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;In my implementation, when a robot attacks another robot (or Chuck, if he has already defended himself this round), then before any other robots attack, we see if any will defend.  For example, if Robots 1,2, and 3 have already acted and it is Robot 4's turn.  Say Robot 4 attacks Robot 5, then we check to see in order if any remaining robots can stop it.  If Robot 5 is apathetic (no laws), and Robot 6 only cares for Chuck (law 1), but Robot 7 cares for other robots (law 2). So Robot 7 stops the attack.  Result is 4 and 7 are deactivated, and the next to act is Robot 5.  Robot 5 does nothing (no laws), so it is Robot 6's turn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My algorithm is as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;next=0&#xA;while (number of robots &amp;gt; 3 &amp;amp;&amp;amp; there are active robots) {&#xA;  if robots[next] is active {&#xA;    // Check laws.&#xA;    if (law 1) {&#xA;      if (law 2) {&#xA;        // TT?: Robot values Chuck and other Robots.&#xA;        sacrifice self&#xA;      } else if (law 3) {&#xA;        // TFT: Robot values Chuck and itself.&#xA;        attack random robot excluding self&#xA;      } else {&#xA;        // TFF: Robot values Chuck only.&#xA;        attack random robot include self&#xA;      }&#xA;    } else if (law 2) { &#xA;      if (law 3) {&#xA;        // FTT: Robot values other robots and itself.&#xA;        attack chuck&#xA;      } else {&#xA;        // FTF: Robot values other robots.&#xA;        randomly either attack chuck or sacrifice itself&#xA;      }&#xA;    } else if (law 3) {&#xA;      // FFT: Robot only cares for itself.&#xA;      randomly attack chuck or other robot&#xA;    } else {&#xA;      // FFF: Robot cares for nothing&#xA;      deactivate robot&#xA;    }&#xA;  }&#xA;&#xA;  // Go to next robot.&#xA;  next++;&#xA;  // If start of new round, let Chuck act again.&#xA;  if (next == number of robots) {&#xA;    next = 0;&#xA;    chuck can act again this round&#xA;  }&#xA;}&#xA;&#xA;// Check simulation status - if 3 or less robots remain, success!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Attacking another robot or chuck is written as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;if (attacker != victim) {&#xA;  // Not a suicide.&#xA;  if (victim is chuck and chuck hasn't acted this round) {&#xA;    chuck deactivates robot and can no longer act this round&#xA;    return&#xA;  } else {&#xA;    // See if anyone stops it.&#xA;    for each robot yet to act this round {&#xA;      if (victim is chuck) {&#xA;        // Robot will save chuck if law 1&#xA;        if (robot has law 1) {&#xA;          defend(attacker, robot)&#xA;          return&#xA;        } &#xA;      } else {&#xA;        // Victim is another robot.&#xA;        if (robot has law 2) {&#xA;          defend(attacker, robot);&#xA;          return&#xA;        }&#xA;      }&#xA;    }&#xA;  }&#xA;}&#xA;&#xA;// If we get here, no one defended the attack.&#xA;throw out the victim&#xA;if (victim == chuck) end simulation&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Lastly, when defending, both the attacker and defender end up deactivated.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Verbose Results&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Here are 10 sample runs.  Before each run, each robot's laws are displayed.  e.g. (T,F,F) means the robot obeys the first law, but not the other two. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that, each line starts with the number of active robots over the number of remaining robots.  All simulations end when the number of active robots hits 0, or the number of remaining hits 3, or if Chuck is thrown over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My results look as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0: (T,F,F)  1: (F,F,T)  2: (F,T,F)  3: (T,T,F)  4: (F,F,T)  &#xA;5: (T,F,T)  6: (T,F,T)  7: (T,F,F)  8: (T,T,T)  9: (F,F,F)  &#xA;  10/10, Robot 0 attacks 8, thwarted by 2&#xA;   8/10, Robot 1 attacks 5, thwarted by 3&#xA;   6/10, Robot 4 attacks 3, thwarted by 8&#xA;   4/10, Robot 5 throws 3 over&#xA;   4/ 9, Robot 6 throws 1 over&#xA;   4/ 8, Robot 7 throws 8 over&#xA;   4/ 7, Robot 9 No laws - deactivating&#xA;   3/ 7, Robot 5 throws 9 over&#xA;   3/ 6, Robot 6 throws 7 over&#xA;   2/ 5, Robot 5 throws 0 over&#xA;   2/ 4, Robot 6 throws 4 over&#xA;End of simulation.  3 robots, 2 active - SUCCESS!&#xA;0: (T,T,T)  1: (T,T,F)  2: (T,T,T)  3: (F,T,F)  4: (F,T,T)  &#xA;5: (T,T,F)  6: (T,T,F)  7: (T,F,T)  8: (T,T,T)  9: (F,F,T)  &#xA;  10/10, Robot 0 jumped&#xA;   9/ 9, Robot 1 jumped&#xA;   8/ 8, Robot 2 jumped&#xA;   7/ 7, Robot 3 jumped&#xA;   6/ 6, Robot 4 attacks Chuck, who defends himself&#xA;   5/ 6, Robot 5 jumped&#xA;   4/ 5, Robot 6 jumped&#xA;   3/ 4, Robot 7 attacks 8, who defends itself&#xA;   1/ 4, Robot 9 throws 4 over&#xA;End of simulation.  3 robots, 1 active - SUCCESS!&#xA;0: (T,T,F)  1: (F,F,F)  2: (T,F,T)  3: (F,F,T)  4: (T,T,F)  &#xA;5: (T,T,T)  6: (F,F,T)  7: (T,F,F)  8: (F,F,F)  9: (F,F,F)  &#xA;  10/10, Robot 0 jumped&#xA;   9/ 9, Robot 1 No laws - deactivating&#xA;   8/ 9, Robot 2 attacks 7, thwarted by 4&#xA;   6/ 9, Robot 3 attacks 4, thwarted by 5&#xA;   4/ 9, Robot 6 attacks Chuck, who defends himself&#xA;   3/ 9, Robot 7 throws 1 over&#xA;   3/ 8, Robot 8 No laws - deactivating&#xA;   2/ 8, Robot 9 No laws - deactivating&#xA;   1/ 8, Robot 7 throws 4 over&#xA;   1/ 7, Robot 7 throws 3 over&#xA;   1/ 6, Robot 7 throws 9 over&#xA;   1/ 5, Robot 7 throws 5 over&#xA;   1/ 4, Robot 7 throws 6 over&#xA;End of simulation.  3 robots, 1 active - SUCCESS!&#xA;0: (T,T,F)  1: (F,F,T)  2: (F,F,F)  3: (T,F,T)  4: (F,T,T)  &#xA;5: (F,T,T)  6: (F,T,F)  7: (F,F,F)  8: (T,F,F)  9: (F,T,T)  &#xA;  10/10, Robot 0 jumped&#xA;   9/ 9, Robot 1 attacks 7, thwarted by 4&#xA;   7/ 9, Robot 2 No laws - deactivating&#xA;   6/ 9, Robot 3 attacks 1, thwarted by 5&#xA;   4/ 9, Robot 6 attacks Chuck, who defends himself&#xA;   3/ 9, Robot 7 No laws - deactivating&#xA;   2/ 9, Robot 8 attacks 5, thwarted by 9&#xA;End of simulation.  9 robots, 0 active - Failure.&#xA;0: (F,T,T)  1: (F,T,F)  2: (T,T,T)  3: (T,F,F)  4: (T,F,T)  &#xA;5: (F,T,T)  6: (F,F,T)  7: (F,F,T)  8: (F,T,T)  9: (F,F,F)  &#xA;  10/10, Robot 0 attacks Chuck, who defends himself&#xA;   9/10, Robot 1 jumped&#xA;   8/ 9, Robot 2 jumped&#xA;   7/ 8, Robot 3 attacks 8, thwarted by 5&#xA;   5/ 8, Robot 4 attacks 9, thwarted by 8&#xA;   3/ 8, Robot 6 throws Chuck over!&#xA;End of simulation.  8 robots, 3 active - Failure.&#xA;In 5 trials, Chuck survived 3.  Percentage: 0.600000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h1&gt;Long Duration Results&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;In 1,000,000 trials, (10 trials of 100,000), I see the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;In 100000 trials, Chuck survived 35980.  Percentage: 0.359800&#xA;In 100000 trials, Chuck survived 36098.  Percentage: 0.360980&#xA;In 100000 trials, Chuck survived 35920.  Percentage: 0.359200&#xA;In 100000 trials, Chuck survived 36039.  Percentage: 0.360390&#xA;In 100000 trials, Chuck survived 35822.  Percentage: 0.358220&#xA;In 100000 trials, Chuck survived 36065.  Percentage: 0.360650&#xA;In 100000 trials, Chuck survived 35810.  Percentage: 0.358100&#xA;In 100000 trials, Chuck survived 35942.  Percentage: 0.359420&#xA;In 100000 trials, Chuck survived 36023.  Percentage: 0.360230&#xA;In 100000 trials, Chuck survived 36051.  Percentage: 0.360510&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Average of approximately 36%.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Different numbers of Robots&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I then varied the number of robots from 5 up to 100.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  5: Chuck survived 66648/100000.  Percentage: 0.666480&#xA;  6: Chuck survived 53514/100000.  Percentage: 0.535140&#xA;  7: Chuck survived 45694/100000.  Percentage: 0.456940&#xA;  8: Chuck survived 40522/100000.  Percentage: 0.405220&#xA;  9: Chuck survived 37826/100000.  Percentage: 0.378260&#xA; 10: Chuck survived 36234/100000.  Percentage: 0.362340&#xA; 11: Chuck survived 34773/100000.  Percentage: 0.347730&#xA; 12: Chuck survived 33793/100000.  Percentage: 0.337930&#xA; 13: Chuck survived 32929/100000.  Percentage: 0.329290&#xA; 14: Chuck survived 32384/100000.  Percentage: 0.323840&#xA; 15: Chuck survived 31838/100000.  Percentage: 0.318380&#xA; 16: Chuck survived 31135/100000.  Percentage: 0.311350&#xA; 17: Chuck survived 30644/100000.  Percentage: 0.306440&#xA; 18: Chuck survived 30160/100000.  Percentage: 0.301600&#xA; 19: Chuck survived 29923/100000.  Percentage: 0.299230&#xA; 20: Chuck survived 29874/100000.  Percentage: 0.298740&#xA; 21: Chuck survived 29465/100000.  Percentage: 0.294650&#xA; 22: Chuck survived 29099/100000.  Percentage: 0.290990&#xA; 23: Chuck survived 29006/100000.  Percentage: 0.290060&#xA; 24: Chuck survived 29038/100000.  Percentage: 0.290380&#xA; 25: Chuck survived 28479/100000.  Percentage: 0.284790&#xA; 26: Chuck survived 28339/100000.  Percentage: 0.283390&#xA; 27: Chuck survived 28143/100000.  Percentage: 0.281430&#xA; 28: Chuck survived 28044/100000.  Percentage: 0.280440&#xA; 29: Chuck survived 27898/100000.  Percentage: 0.278980&#xA; 30: Chuck survived 27620/100000.  Percentage: 0.276200&#xA; 31: Chuck survived 27677/100000.  Percentage: 0.276770&#xA; 32: Chuck survived 27399/100000.  Percentage: 0.273990&#xA; 33: Chuck survived 27070/100000.  Percentage: 0.270700&#xA; 34: Chuck survived 27079/100000.  Percentage: 0.270790&#xA; 35: Chuck survived 27081/100000.  Percentage: 0.270810&#xA; 36: Chuck survived 26809/100000.  Percentage: 0.268090&#xA; 37: Chuck survived 26771/100000.  Percentage: 0.267710&#xA; 38: Chuck survived 26963/100000.  Percentage: 0.269630&#xA; 39: Chuck survived 26821/100000.  Percentage: 0.268210&#xA; 40: Chuck survived 26763/100000.  Percentage: 0.267630&#xA; 41: Chuck survived 26523/100000.  Percentage: 0.265230&#xA; 42: Chuck survived 26351/100000.  Percentage: 0.263510&#xA; 43: Chuck survived 26365/100000.  Percentage: 0.263650&#xA; 44: Chuck survived 26417/100000.  Percentage: 0.264170&#xA; 45: Chuck survived 26398/100000.  Percentage: 0.263980&#xA; 46: Chuck survived 26393/100000.  Percentage: 0.263930&#xA; 47: Chuck survived 26290/100000.  Percentage: 0.262900&#xA; 48: Chuck survived 26255/100000.  Percentage: 0.262550&#xA; 49: Chuck survived 25783/100000.  Percentage: 0.257830&#xA; 50: Chuck survived 25858/100000.  Percentage: 0.258580&#xA; 51: Chuck survived 26155/100000.  Percentage: 0.261550&#xA; 52: Chuck survived 26019/100000.  Percentage: 0.260190&#xA; 53: Chuck survived 26011/100000.  Percentage: 0.260110&#xA; 54: Chuck survived 25723/100000.  Percentage: 0.257230&#xA; 55: Chuck survived 25647/100000.  Percentage: 0.256470&#xA; 56: Chuck survived 25572/100000.  Percentage: 0.255720&#xA; 57: Chuck survived 25906/100000.  Percentage: 0.259060&#xA; 58: Chuck survived 25711/100000.  Percentage: 0.257110&#xA; 59: Chuck survived 25636/100000.  Percentage: 0.256360&#xA; 60: Chuck survived 25737/100000.  Percentage: 0.257370&#xA; 61: Chuck survived 25688/100000.  Percentage: 0.256880&#xA; 62: Chuck survived 25560/100000.  Percentage: 0.255600&#xA; 63: Chuck survived 25657/100000.  Percentage: 0.256570&#xA; 64: Chuck survived 25596/100000.  Percentage: 0.255960&#xA; 65: Chuck survived 25234/100000.  Percentage: 0.252340&#xA; 66: Chuck survived 25434/100000.  Percentage: 0.254340&#xA; 67: Chuck survived 25321/100000.  Percentage: 0.253210&#xA; 68: Chuck survived 25313/100000.  Percentage: 0.253130&#xA; 69: Chuck survived 25617/100000.  Percentage: 0.256170&#xA; 70: Chuck survived 25318/100000.  Percentage: 0.253180&#xA; 71: Chuck survived 25453/100000.  Percentage: 0.254530&#xA; 72: Chuck survived 25195/100000.  Percentage: 0.251950&#xA; 73: Chuck survived 25162/100000.  Percentage: 0.251620&#xA; 74: Chuck survived 25409/100000.  Percentage: 0.254090&#xA; 75: Chuck survived 25212/100000.  Percentage: 0.252120&#xA; 76: Chuck survived 25036/100000.  Percentage: 0.250360&#xA; 77: Chuck survived 25305/100000.  Percentage: 0.253050&#xA; 78: Chuck survived 25215/100000.  Percentage: 0.252150&#xA; 79: Chuck survived 25219/100000.  Percentage: 0.252190&#xA; 80: Chuck survived 24893/100000.  Percentage: 0.248930&#xA; 81: Chuck survived 24979/100000.  Percentage: 0.249790&#xA; 82: Chuck survived 25252/100000.  Percentage: 0.252520&#xA; 83: Chuck survived 25022/100000.  Percentage: 0.250220&#xA; 84: Chuck survived 25108/100000.  Percentage: 0.251080&#xA; 85: Chuck survived 25069/100000.  Percentage: 0.250690&#xA; 86: Chuck survived 24835/100000.  Percentage: 0.248350&#xA; 87: Chuck survived 25185/100000.  Percentage: 0.251850&#xA; 88: Chuck survived 24910/100000.  Percentage: 0.249100&#xA; 89: Chuck survived 24748/100000.  Percentage: 0.247480&#xA; 90: Chuck survived 25040/100000.  Percentage: 0.250400&#xA; 91: Chuck survived 25030/100000.  Percentage: 0.250300&#xA; 92: Chuck survived 24815/100000.  Percentage: 0.248150&#xA; 93: Chuck survived 24842/100000.  Percentage: 0.248420&#xA; 94: Chuck survived 25012/100000.  Percentage: 0.250120&#xA; 95: Chuck survived 24787/100000.  Percentage: 0.247870&#xA; 96: Chuck survived 25017/100000.  Percentage: 0.250170&#xA; 97: Chuck survived 24726/100000.  Percentage: 0.247260&#xA; 98: Chuck survived 24977/100000.  Percentage: 0.249770&#xA; 99: Chuck survived 24807/100000.  Percentage: 0.248070&#xA;100: Chuck survived 24896/100000.  Percentage: 0.248960&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;By the end, he only survives 25% of the time.  I ran a few more:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1000: Chuck survived 2382/10000.  Percentage: 0.238200&#xA;10000: Chuck survived 20/100.  Percentage: 0.200000&#xA;10000: Chuck survived 23/100.  Percentage: 0.230000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h1&gt;Analysis&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I expect you can never add enough robots to get the survivability less than 10%.  The reason for this is that in many of the success cases, a single (T,F,T) robot remains active and dumps the rest.   This suggests that 1/8 of the time that a single robot is left active, this robot will have favourable laws for Chuck (self-preservation and wants to save Chuck), so it will throw all the deactivated robots off for Chuck. &lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="1592" LastEditDate="2015-11-06T19:36:32.033" LastActivityDate="2015-11-06T19:36:32.033" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23824" PostTypeId="2" ParentId="23821" CreationDate="2015-11-06T19:40:24.973" Score="6" Body="&lt;p&gt;The time from $10:30$ to $22:30$ is a $12$ hour period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During $12$ hours the hour hand makes $1$ full circle, the minute hand makes $12$, and the second hand makes $720$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During the $1$ full circle of the hour hand it is passed $12-1=11$ times by the minute hand and $720-1=719$ times by the second hand. Similarly the minute hand is passed $720-12=708$ times by the second hand during the $12$ hour period. Summing it all together gives $11+719+708=1438$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we have to consider the time when all $3$ hands meet at $12:00$. This is $1$ overlap which we have counted $3$ times in the previous paragraph. To correct this we need to subtract $2$ which gives the final result $1438-2=1436$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit: Proof that there is no other overlap of all $3$ hands in a $12$ hour period&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the text above we know that the hour and minute hands overlap every $\frac{12}{11}h$ and the minute and second hand overlap every $\frac{12}{708}h$. For all 3 hands to overlap there must be a solution with integer $x$ and $y$ for the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{12}{11}x=\frac{12}{708}y$&lt;br&gt;&#xA;$\frac{1}{11}x=\frac{1}{708}y$&lt;br&gt;&#xA;$708x=11y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $11$ is prime and $708$ is not divisible by $11$ the smallest solution is $x=11$ and $y=708$ which shows there can be only one overlap of all $3$ hands in a $12$ hour period.&lt;/p&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2015-11-06T20:21:33.803" LastActivityDate="2015-11-06T20:21:33.803" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23826" PostTypeId="2" ParentId="23819" CreationDate="2015-11-06T23:58:22.063" Score="11" Body="&lt;p&gt;Only two questions are needed:  &quot;Are you Truman?&quot; and &quot;Are you Xavier?&quot;, though you may have to direct one of the questions to two different leaders. The following lists show how each of these questions are answered by each leader:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you Truman? Truman: Yes, Lyle: Yes, Xavier: No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you Xavier? Truman: No, Lyle: Yes, Xavier: Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each leader has a unique combination of answers:  Lyle says yes to either question, and Truman and Xavier have opposite answers from each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your first question, pick a leader and ask one of the questions.  If the answer is no, you figured out who it is (only one leader answers no to each of the two questions).  You only need to ask the other question of another leader, and you're done.  If the answer to the first question was yes, then ask the other question of the same leader.  If you get another yes you've found Lyle and you can ask either question of another leader to figure out who is Truman and who is Xavier.  If you get a no, you've identified the first leader and you pick one of the two remaining leaders to re-ask the original question.&lt;/p&gt;&#xA;" OwnerUserId="17279" LastActivityDate="2015-11-06T23:58:22.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23828" PostTypeId="2" ParentId="23816" CreationDate="2015-11-07T08:30:28.763" Score="18" Body="&lt;p&gt;As a supplement to &lt;a href=&quot;https://puzzling.stackexchange.com/a/23817/9718&quot;&gt;Ivo Beckers's answer&lt;/a&gt;, I provide an explicit construction of Professor Halfbrain's cube:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xR1Xt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cube has edge length $10\sqrt{21}$.  The angles between the three bottom edges and the table plane are (colored red, green, and blue, respectively):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\color{red}{\sin^{-1}\left(\frac{1}{\sqrt{21}}\right)} \\&#xA;\color{lime}{\sin^{-1}\left(\frac{2}{\sqrt{21}}\right)} \\&#xA;\color{blue}{\sin^{-1}\left(\frac{4}{\sqrt{21}}\right)}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Ivo conjectures, this arrangement is unique (up to reflection and rotation around the vertical axis).  We can prove this by using &lt;a href=&quot;https://en.wikipedia.org/wiki/Euler_angles&quot; rel=&quot;nofollow noreferrer&quot;&gt;Euler angles&lt;/a&gt; to describe the rotation of the cube.  (I will use the z-y-z convention.)  We can ignore the first rotation about the z-axis, since it does not affect the angles between the cube and the table surface, only spinning the cube about its vertex.  Calling the remaining two angles $\alpha$ and $\beta$, we can write the transformation matrix:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left(\begin{matrix} \cos\alpha\cos\beta &amp;amp; -\cos\alpha\sin\beta &amp;amp;\sin\alpha\\&#xA;\sin\beta &amp;amp;\cos\beta &amp;amp; 0\\&#xA; -\sin\alpha\cos\beta &amp;amp;\sin\alpha\sin\beta &amp;amp;\cos\alpha\\\end{matrix}\right)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and its application to the three unit vectors:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left(&#xA;\begin{matrix}&#xA;\cos\alpha\cos\beta\\&#xA;\sin\beta\\&#xA; -\sin\alpha\cos\beta\\&#xA;\end{matrix}&#xA;\right),\left(&#xA;\begin{matrix}&#xA; -\cos\alpha\sin\beta\\&#xA;\cos\beta\\&#xA;\sin\alpha\sin\beta\\&#xA;\end{matrix}&#xA;\right),\left(&#xA;\begin{matrix}&#xA;\sin\alpha\\&#xA; 0\\&#xA;\cos\alpha\\&#xA;\end{matrix}&#xA;\right)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this we have the height of three vertices of the cube, as well as the heights of the three edges from those vertices to the bottommost vertex.  The position of each vertex is a sum of some subset of those edges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The height of the shortest edge must be $10$; otherwise&amp;mdash;as the height of each edge is positive&amp;mdash;no vertex could have height $10$.  Next, since each of the three different heights may be included at most once, the next edge must have height $20$, otherwise no vertex could have height $20$.  Similarly, the third edge must have a height of $40$.  This gives us enough to reach all the way to $10+20+40=70$; and, since our choice of heights were fixed at each step, we know the solution must be unique.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $s$ as the edge length, we can now set up a system of equations to solve for the Euler angles and size of the cube:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;-s\sin\alpha\cos\beta &amp;amp;= 10 \\&#xA;s\sin\alpha\sin\beta &amp;amp;= 20 \\&#xA;s\cos\alpha &amp;amp;= 40&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Immediately we can write $s=40\sec\alpha$ and remove $s$ from the first two equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;-4 \cos \beta \tan \alpha &amp;amp;= 1 \\&#xA;2\sin \beta \tan \alpha &amp;amp;= 1&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then divide and write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;-\frac{1}{2}\tan\beta = 1 \\&#xA;\beta = -\tan^{-1} 2&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we can back-substitute and solve for $\alpha$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;-\frac{4}{\sqrt{5}}\tan\alpha = 1 \\&#xA;\alpha = -\tan^{-1}\left(\frac{\sqrt{5}}{4}\right)=\cos^{-1}\left(\frac{4}{\sqrt{21}}\right)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, using $s=40\sec\alpha$ we then obtain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;s = 10\sqrt{21}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For completeness, here is the resulting rotation matrix:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left(&#xA;\begin{matrix}&#xA; \frac{4}{\sqrt{105}} &amp;amp; \frac{8}{\sqrt{105}} &amp;amp; -\sqrt{\frac{5}{21}} \\&#xA; -\frac{2}{\sqrt{5}} &amp;amp; \frac{1}{\sqrt{5}} &amp;amp; 0 \\&#xA; \frac{1}{\sqrt{21}} &amp;amp; \frac{2}{\sqrt{21}} &amp;amp; \frac{4}{\sqrt{21}} \\&#xA;\end{matrix}&#xA;\right)&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2015-11-07T08:30:28.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23829" PostTypeId="1" AcceptedAnswerId="23830" CreationDate="2015-11-07T09:47:13.053" Score="10" ViewCount="1087" Body="&lt;p&gt;In the book 'The Tipping Point', Malcolm Gladwell cites two puzzles and draws an analogy between them, in fact he claims the two are equivalent. I believe there are some problems with both puzzles, and the analogy as well. Do you see the same problems, and are there any suggestions to fix this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first riddle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Consider the following brain teaser. Suppose I give you four cards,&#xA;  labeled with the letter A and D and the numerals 3 and 6. The rule of&#xA;  the game is that a card with a vowel on it always has an even number&#xA;  on the other side. Which of the cards would you have to turn over to&#xA;  prove this rule to be true? The answer is two: the A card and the&#xA;  three card. The overwhelming majority of the people given this test,&#xA;  though, don't get it right. They tend to answer just the A card, or&#xA;  the A and the 6. It's a hard question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here, the &lt;code&gt;D&lt;/code&gt;-card should be turned over as well, as there is no indication that cards cannot have letters on both sides. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now the second:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But now let me pose another question. Suppose four people are drinking&#xA;  in a bar. One is drinking Coke. One is sixteen. One is drinking beer&#xA;  and one is twenty-five. Given the rule that no one under twenty-one is&#xA;  allowed to drink beer, which of those people's IDs do we have to check&#xA;  to make sure the law is being observed? Now the answer is easy, in&#xA;  fact, I'm sure that almost everyone will get it right: The&#xA;  beer-drinker and the sixteen-year-old. But, as the psychologist Leda&#xA;  Cosmides (who dreamt up this example) points out, it is exactly the&#xA;  same puzzle as the A, D, 3 and 6 puzzle. The difference is that it is&#xA;  framed in a way that makes it about people, instead of about numbers,&#xA;  and as human beings we are a lot more sophisticated about each other&#xA;  than we are about the abstract world.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But checking the ID of the sixteen-year-old doesn't make sense, you already know his age. He can still be drinking tequila while you read his birth-date. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analogy is incorrect after these mistakes. If those two mistakes are fixed, by adding the rule 'Cards always have a letter on one side and a number on the other' to the first puzzle, and in the second case alter the wording so that we may also check the drink of the sixteen-year-old, the analogy is correct, right?&lt;/p&gt;&#xA;" OwnerUserId="17282" LastActivityDate="2020-11-22T00:07:20.790" Title="mistakes in analogous puzzles from The Tipping Point" Tags="|logical-deduction|explanation|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23830" PostTypeId="2" ParentId="23829" CreationDate="2015-11-07T10:39:40.110" Score="9" Body="&lt;p&gt;Yes, you are correct. Those mistakes should be fixed in order to make the questions analogous.&lt;/p&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2015-11-07T10:39:40.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23831" PostTypeId="2" ParentId="23821" CreationDate="2015-11-07T15:08:49.687" Score="0" Body="&lt;p&gt;If you read the accepted answer on &lt;a href=&quot;https://puzzling.stackexchange.com/questions/22458/meeting-of-hands-on-a-clock&quot;&gt;this question&lt;/a&gt;, the answer becomes evident.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If revolutions per day are given by:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Second hand - 1 revolution per minute or $1440$ per day&lt;/li&gt;&#xA;&lt;li&gt;Minute hand - 1 revolution per hour or $24$ per day&lt;/li&gt;&#xA;&lt;li&gt;Hour hand - 1 revolution per 12 hours or $2$ per day&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then meetings in a day are given by:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;(a) the minute and hour hands ($24-2=22$ per day)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;(b) the minute and second hands ($1440-24=1416$ per day)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;(c) the hour and second hands ($1440-2=1438$ per day)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;(d) all three hands ($2$ per day)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now, to answer your question, we want the following value&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(a)+(b)+(c)-2(d)=22+1416+1438-2(2)=2872$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We subtracted $2(d)$ since the meeting of all three hands has been counted thrice, once in $(a)$, once in $(b)$ and once in $(c)$, when we want it to be counted only once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is answer for one day. If we want $12$ hours, we just divide by $2$. Hence we conclude that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your answer is correct; as $2872/2=1436$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-11-07T15:08:49.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23832" PostTypeId="1" AcceptedAnswerId="23851" CreationDate="2015-11-07T15:46:08.613" Score="10" ViewCount="1214" Body="&lt;p&gt;Cosmo has cut a square into seven rectangles, so that the seven lengths $\ell_1,\ldots,\ell_7$ and the seven widths $w_1,\ldots,w_7$ of these rectangles satisfy&#xA;$$ \{\ell_1,\ldots,\ell_7\}\cup\{w_1,\ldots,w_7\} ~=~ \{1,2,3,4,\ldots,14\}. $$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What are the possible side lengths of this square?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2018-04-23T09:26:01.563" Title="Cutting a square into seven rectangles" Tags="|mathematics|geometry|dissection|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23836" PostTypeId="2" ParentId="23832" CreationDate="2015-11-07T23:49:45.560" Score="2" Body="&lt;p&gt;I don't think it can be done. To tile a rectangle with smaller rectangles such that no two rectangles have a common side length in the simplest way (apart from the trivial case of a single rectangle) requires 5 rectangles: 1 in each corner, surrounding a 5th one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other configurations can be made by replacing one of the rectangles with 5 in a similar way. Thus you could have 1, 5, 9,... (potentially) but not 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This turns out is not true. I was uncertain, but thanks to @f'' for the counter-example&lt;/p&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2015-11-08T01:02:09.400" LastActivityDate="2015-11-08T01:02:09.400" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23837" PostTypeId="2" ParentId="23832" CreationDate="2015-11-08T03:26:08.603" Score="9" Body="&lt;p&gt;Here is an example to show that it is possible with a side length of 22.&lt;img src=&quot;https://i.stack.imgur.com/YmdIT.png&quot; alt=&quot;1]&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-11-08T03:26:08.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23838" PostTypeId="2" ParentId="23816" CreationDate="2015-11-08T03:27:58.473" Score="6" Body="&lt;p&gt;Yes, it's possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Placing the base vertex at the origin, the $8$ vertices are the subset-sums of the $3$ edge vectors leaving the base vertex. If their respective $z$-components can be made to be $1$, $2$, and $4$, then the vertices will have $z$-components $0$ through $7$ by binary encoding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the vector $(1,2,4)$, and complete it to an orthogonal basis with two vectors of the same length as it. Thinking of these as rows of a $3 \times 3$ scaled-orthogonal matrix with $(1,2,4)$ the last row, the columns are also equal-length and orthogonal, and have $z$-coordinates $1,2,4$. These satisfy all the conditions we needed of the base vectors.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-11-08T03:27:58.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23840" PostTypeId="2" ParentId="23816" CreationDate="2015-11-08T09:41:46.653" Score="5" Body="&lt;p&gt;An addition to the other answers:  I was hoping for a construction where every vertex of the cube had integer, three-dimensional coordinates, but I don't think this is possible.   However, if you try to answer the problem in 4-dimensions, not only does such a hypercube exist but you can construct an example with integer coordinates.  Think of all possible sums of the vectors:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\mathbf{v_1} = (-8, 4,-2, 1) \\&#xA;\mathbf{v_2} = (-4, -8, 1, 2) \\&#xA;\mathbf{v_4} = (2, -1, -8, 4) \\&#xA;\mathbf{v_8} = (1, 2, 4, 8) \\&#xA;$$&#xA;One can check these are all the same length and mutually orthogonal.  If you include the empty sum $(0, 0, 0, 0)$, these form the vertices of a hypercube whose last coordinate is all possible values from $0$ to $15$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you only take the subset-sums of $\{\mathbf{v_1}, \mathbf{v_2},\mathbf{v_4}\}$, you get a cube embedded in four-space.  Does this cube satisfy the requirements?   Well, not exactly, since trying to project this cube back into three-space results in something that isn't a cube.  However, if you go the other way and project the line $L = t (0, 0, 0, 1)$ into the three-dimensional space $S$ that contains this cube to create a line $L'$, then the points of the cube must be evenly spaced if projected onto $L'$.  Hence if you take the plane orthogonal to $L'$ in $S$ going through the original and call this your ``table&quot;, it all sort of works up to a scaling factor.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, there is an example with integer coordinates.  It just requires four-space, and you lose a nice coordinate representation of the table (and you are also off by a scaling factor).&lt;/p&gt;&#xA;" OwnerUserId="2722" LastActivityDate="2015-11-08T09:41:46.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23843" PostTypeId="1" AcceptedAnswerId="23870" CreationDate="2015-11-08T10:48:18.990" Score="16" ViewCount="1297" Body="&lt;p&gt;Assume that black starts with just the king on e8 and white has all her pieces in the normal opening positions. What is the shortest forced mate for white?&lt;/p&gt;&#xA;" OwnerUserId="8998" LastActivityDate="2015-11-10T19:09:17.350" Title="Shortest forced mate if black has no pieces" Tags="|chess|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23844" PostTypeId="2" ParentId="23819" CreationDate="2015-11-08T11:36:26.887" Score="4" Body="&lt;p&gt;We can do this without asking a specific person &quot;are you...?&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ask chair 1, &quot;is the person in chair 2 Truman, Lyle, or Xavier?&quot;. The correct answer, obviously, is yes, which means Truman answers yes, Lyle no, Xavier yes.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the guy in chair 1 answers no, he's definitely Lyle. Ask chair 2 &quot;is the person in chair 1 Truman?&quot;. The correct answer is no, so Truman will say no, Lyle would say yes, Xavier will say yes.&lt;/li&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;If the guy in chair 2 answers no, he's Truman and chair 3 is Xavier.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;If the guy in chair 2 answers yes, he's Xavier and chair 3 is Truman.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;If the guy in chair 1 answers yes, he's definitely not Lyle. Ask chair 2 the original question &quot;is the person in chair 3 Truman, Lyle, or Xavier?&quot;. The correct answer is yes, so Truman says yes, Lyle no, Xavier yes.&lt;/li&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;If the guy in chair 2 answers no, he's definitely Lyle. Same as above, ask the guy in chair 3 &quot;is the person in chair 2 Truman?&quot;. The correct answer being no, Truman says no, Lyle yes, Xavier yes.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;If the guy in chair 3 answers no, he's Truman and chair 1 is Xavier.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;If the guy in chair 3 answers yes, he's Xavier and chair 1 is Truman.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;If the guy in chair 2 answers yes, he's definitely not Lyle, which means the guy in chair 3 is definitely Lyle. Ask the guy in chair 1 &quot;is the guy in chair 3 Truman?&quot;. The correct answer is no, so Truman answers yes, Lyle no, Xavier yes.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;If the guy in chair 1 answers no, he's Truman and chair 2 is Xavier.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;If the guy in chair 2 answers yes, he's Xavier and chair 2 is Truman.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Vtfhg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Vtfhg.png&quot; alt=&quot;Flowchart of above description.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do something similar by leading with &quot;is the person in chair 1 both Truman and Lyle?&quot;, which must be false, so instead of showing who Lyle is, it shows who Truman is. Then you ask if Truman is Truman and Lyle says no, Xavier says yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you know who Truman or Lyle are, you can directly ask them if one of the unknowns is Xavier. Truman will say yes if it's Xavier, no if it's Lyle. Lyle will say no if it's Xavier, yes if it's Truman. Same thing for directly asking Truman if a chair is Lyle, or Lyle if a chair is Truman.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, of course, it doesn't matter what we call the chairs or what order we ask in, as long as we're careful to swap labels around appropriately.&lt;/p&gt;&#xA;" OwnerUserId="16474" LastActivityDate="2015-11-08T11:36:26.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23845" PostTypeId="2" ParentId="23843" CreationDate="2015-11-08T11:40:11.920" Score="20" Body="&lt;p&gt;Here is a possible ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ... 8 move answer, which allows to ignore all variants of black King movements. No matter what black is trying, white can always make the shown moves.&lt;br&gt;&lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/PBVKl.gif&quot; alt=&quot;&quot;&gt;&lt;br&gt;&lt;br&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;br&gt;&lt;br&gt;I managed to survive 7 moves against &lt;a href=&quot;http://www.apronus.com/chess/computer.htm?fen=14k3X8X8X8X8X8XPPPPPPPPXRNBQKBNR_w_KQ_-_0_1&quot; rel=&quot;noreferrer&quot;&gt;a computer&lt;/a&gt;, but there are too many variants to list them all here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2015-11-08T12:07:42.580" LastActivityDate="2015-11-08T12:07:42.580" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23846" PostTypeId="1" CreationDate="2015-11-08T13:23:38.430" Score="-13" ViewCount="640" Body="&lt;p&gt;I made a code and I want to test out the difficulty, so can anyone figure out what this says?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;.1852132114 51415 775 14920 18582015 18582095 11149820 5255 .191852020512 2018158 18522518156 412158 5255&lt;/p&gt;&#xA;" OwnerUserId="17295" LastActivityDate="2015-11-11T03:31:17.160" Title="What does this say? (I want to see how easy/difficult this code is)" Tags="|cipher|" AnswerCount="1" CommentCount="6" ClosedDate="2016-07-30T01:48:11.027" ContentLicense="CC BY-SA 3.0" />
  <row Id="23847" PostTypeId="1" AcceptedAnswerId="23848" CreationDate="2015-11-08T13:32:39.033" Score="12" ViewCount="1089" Body="&lt;p&gt;Two hours ago, I received a phone call from professor Halfbrain. &#xA;The professor told me that he has detected an odd perfect number. &#xA;The professor was very excited, since the existence of odd perfect &#xA;numbers is an outstanding open question in number theory. &#xA;Let me explain this question to you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Definition:&lt;/strong&gt; An integer $N$ is &lt;strong&gt;perfect&lt;/strong&gt;, if the sum of all divisors of $N$ below $N$ exactly equals $N$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For instance $N=6$ is perfect, as $1+2+3=6$. &#xA;And also the number $N=28$ is perfect, as $1+2+4+7+14=28$. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem:&lt;/strong&gt; For a number $N$ with prime factorization &#xA;  $N=p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}$ the sum $\sigma(N)$ of all divisors (including the number $N$ itself) is given by the expression&#xA;  $$ (1+p_1+p_1^2+\cdots+p_1^{e_1})\,&#xA;   (1+p_2+p_2^2+\cdots+p_2^{e_2})\, \cdots &#xA;   (1+p_k+p_k^2+\cdots+p_k^{e_k}). $$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The theorem implies that an integer $N$ is perfect, if and only if &#xA;$\sigma(N)=2N$ holds true. With the help of the theorem, one easily &#xA;checks and verifies that $6$ and $28$ are perfect:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$6=2\cdot3~~$    yields $~~\sigma(6)=(1+2)(1+3)=12=2\cdot6$ &lt;/li&gt;&#xA;&lt;li&gt;$28=2^2\cdot7~~$ yields $~~\sigma(28)=(1+2+4)(1+7)=56 =2\cdot 28$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To the current day, mathematicians have been hunting for &lt;strong&gt;odd&lt;/strong&gt; perfect &#xA;numbers. Although thousands of hours of research time have been invested &#xA;into this problem, all these hours were of no avail.&#xA;The big breakthrough of Professor Halfbrain on 8 November 2015 might &#xA;be a turning point in the history of mathematics.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Professor Halfbrain's odd perfect number theorem:&lt;/strong&gt;&lt;br&gt;&#xA;  For $N=3^2\cdot7^2\cdot11^2\cdot13^2\cdot22021 ~=~ 198,585,576,189$, we have&lt;br&gt;&#xA;  \begin{eqnarray}&#xA;\sigma(N) &amp;amp;=&amp;amp; (1+3+3^2)(1+7+7^2)(1+11+11^2)(1+13+13^2)(1+22021)&#xA;\\ &amp;amp;=&amp;amp; 397,171,152,378 ~~=~~ 2\cdot 198,585,576,189 ~~=~~ 2N.&#xA;\end{eqnarray}&#xA;  Therefore &lt;strong&gt;$N=198,585,576,189$&lt;/strong&gt; constitutes an odd perfect number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Has the professor indeed managed to detect an odd perfect &#xA;number, or is there a mistake hidden somewhere in the above paragraphs? &lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2016-03-13T15:40:40.777" Title="Professor Halfbrain and the odd perfect number" Tags="|mathematics|number-theory|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23848" PostTypeId="2" ParentId="23847" CreationDate="2015-11-08T13:57:36.343" Score="14" Body="&lt;p&gt;The professor ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ... has made a mistake. $22021$ is not prime, it's $19^2 \cdot 61$.&lt;br&gt;&lt;br&gt; \begin{align*} \small{ \sigma(N) } &amp;amp;= \small{ (1+3+3^2)(1+7+7^2)(1+11+11^2)(1+13+13^2)(1+19+19^2)(1+61) } \\ &amp;amp;= \small{ 426,027,470,778 \gt 2N } \end{align*}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="16322" LastEditDate="2016-03-13T15:40:40.777" LastActivityDate="2016-03-13T15:40:40.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23850" PostTypeId="2" ParentId="23847" CreationDate="2015-11-08T16:22:27.220" Score="6" Body="&lt;p&gt;What's suspicious ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ... If you factor the individual products, $1 + 3 + 3^2 = 13$, $1 + 7 + 7^2 = 57 = 3 \cdot 19$, $1 + 11 + 11^2 = 133 = 7 \cdot 19$, $1 + 13 + 13^2 = 183 = 3 \cdot 61$, so there are a few factors not accounted for. It would seem that the two 19's and the 61 must be factors of 22021. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4664" LastEditorUserId="16322" LastEditDate="2016-03-13T15:39:43.770" LastActivityDate="2016-03-13T15:39:43.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23851" PostTypeId="2" ParentId="23832" CreationDate="2015-11-08T18:10:36.447" Score="19" Body="&lt;p&gt;From the deleted answer from &lt;strong&gt;frodoswalker&lt;/strong&gt; we know the possible ranges of the squares:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Our maximum area is $1*2+3*4+5*6+7*8+9*10+11*12+13*14=504$, which means the maximum square width is $\lfloor \sqrt{504} \rfloor = 22$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Our minimum area is $1*14+2*13+3*12+4*11+5*10+6*9+7*8=280$, which means the minimum square width is $\lceil \sqrt{280} \rceil = 17$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The biggest size available to our rectangles is $14$. Therefore we know that each side of the square consists of at least $2$ rectangles. Let's look at some possible configurations. &lt;strong&gt;The key aspect here is, that each dimension (width or height) of each rectangle must be different.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;2 rectangles per side&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ltjrr.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ltjrr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to see, that using this configuration there is no solution. We would need to fill the remaining rectangle area in the middle with the remaining 3 rectangles. This is not possible without at least 2 of them having the same width or height.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;3 sides with 2 rectangles, 1 side with 4 rectangles&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/w3rIZ.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/w3rIZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above we see some of the possible configurations with 4 rectangles at the top. It's easy to see, that we can't fill the remaining area with the remaining rectangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same applies if we try to put 5 rectangles on 1 side of the square.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;2 sides with 2 rectangles, 2 sides with 3 rectangles&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IL22Y.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IL22Y.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above there is a configuration with 3 rectangles at top/bottom and another one with 3 rectangles at top/right. With 1 rectangle remaining we can't fill the remaining area.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;3 sides with 2 rectangles, 1 side with 3 rectangles&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vyviW.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vyviW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above we see 2 different configurations which allow us to find correct solutions. Other configurations can be reached by rotation or mirroring and will be ignored here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can define the following equations for both images:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x0 + x1 + x2 = x3 + x4 = x5 + x6 = x7 + y8$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the left variant only:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x2 = x9 + x4$&lt;br&gt;&#xA;$x10 = x1 +x9$&lt;br&gt;&#xA;$x3 = x0 + x10$&lt;br&gt;&#xA;$x5 = x11 + x12$&lt;br&gt;&#xA;$x11 = x7 + x13$&lt;br&gt;&#xA;$x8 = x6 + x12 + x13$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for the right variant:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x1 = x9 + x10$&lt;br&gt;&#xA;$x4 = x2 + x10$&lt;br&gt;&#xA;$x3 = x0 + x9$&lt;br&gt;&#xA;$x5 = x11 + x12$&lt;br&gt;&#xA;$x7 = x11 + x13$&lt;br&gt;&#xA;$x6 + x12 = x8 + x13$&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feeding them into a computer returns several solutions for squares with size 18 to 22. There is an answer from &lt;strong&gt;f''&lt;/strong&gt; for the 22 square. Here are 18 to 21:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YImjT.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YImjT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5qzVS.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5qzVS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Algb1.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Algb1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ymlue.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ymlue.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently there is no solution for a square with size 17.&lt;/p&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2015-11-08T18:54:35.893" LastActivityDate="2015-11-08T18:54:35.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23852" PostTypeId="1" AcceptedAnswerId="23875" CreationDate="2015-11-08T21:45:32.043" Score="15" ViewCount="2248" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Being what you think you are would mean you cannot think.&lt;br&gt;&#xA;  Being what you really are means you're wrong in having this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/yrWM6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yrWM6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Think about:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What kind of fish depicted in this piece of &lt;em&gt;art&lt;/em&gt;?  What could the switched-up order mean?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/NtPrN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NtPrN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Think about:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What kind of animal are these &lt;em&gt;Docs&lt;/em&gt; working on?  What could the switched-up image mean?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/ay5ev.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ay5ev.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Think about: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What makes this &lt;em&gt;car&lt;/em&gt; unique?  What could the switched-up image mean?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 4:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;a href=&quot;https://i.stack.imgur.com/EPxHb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EPxHb.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 5:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  The last word of the title is meant quite literally.  The previous hints are important for their subject matter.  The strange alterations hint at how to make use of all the hints.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Special thanks to Rand Al'Thor for helping me work out the kinks in this riddle.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="17062" LastEditorUserId="17062" LastEditDate="2015-11-09T19:34:31.223" LastActivityDate="2015-11-09T19:44:52.950" Title="What's your diagnosis?" Tags="|riddle|wordplay|" AnswerCount="6" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23854" PostTypeId="1" AcceptedAnswerId="23855" CreationDate="2015-11-09T00:37:20.440" Score="-1" ViewCount="331" Body="&lt;p&gt;&lt;em&gt;Sequel to:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/5290/the-hallway-and-tele-doors&quot;&gt;The Hallway and Tele Doors&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; You &lt;em&gt;need not&lt;/em&gt; have to read the above to understand this puzzle, however &lt;strong&gt;this puzzle does spoil the above one.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The tele door system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each pair of doors, represented by letters, are &quot;&lt;em&gt;linked&lt;/em&gt;&quot;, you go enter one door on the left side, and you go exit the other in the right. Example hallway:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;hr&gt;&#xA;  &#xA;  &lt;p&gt;A1 B1 A2 B2  &lt;/p&gt;&#xA;  &#xA;  &lt;hr&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Prisoners &lt;strong&gt;always&lt;/strong&gt; enter through the &lt;em&gt;leftmost door&lt;/em&gt;. So in the example hallway, they'd go enter through the first A1-&gt;A2-&gt;B2-&gt;B1-&gt;A2-&gt;A1-&gt;B1-&gt;B2-&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; There will &lt;em&gt;never&lt;/em&gt; be an &lt;em&gt;3 (&lt;/em&gt; representing any letter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You, my friend, are a wickid man. You wish to &lt;strong&gt;trap&lt;/strong&gt; someone in a hallway for &lt;em&gt;eternity&lt;/em&gt; using nothing but &lt;strong&gt;pairs of tele doors&lt;/strong&gt;. But sadly, you have finally come to understand this is &lt;em&gt;not possible&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how long &lt;strong&gt;CAN&lt;/strong&gt; you &lt;em&gt;keep&lt;/em&gt; them there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given &lt;strong&gt;26 pairs of teles&lt;/strong&gt;, how many tele doors can you force them walk through, &lt;em&gt;before escaping&lt;/em&gt;?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="69582" LastEditDate="2021-05-07T14:27:21.423" LastActivityDate="2021-05-07T14:27:21.423" Title="The Hallway and Tele Doors ~ Prolong the Inevitable" Tags="|logical-deduction|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23855" PostTypeId="2" ParentId="23854" CreationDate="2015-11-09T01:05:11.980" Score="4" Body="&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/a/5294/12408&quot;&gt;This answer&lt;/a&gt; (which I wrote) to the original question shows that it is impossible to enter any door twice:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[Suppose] that you enter the same teleportation door twice at some point. Consider the first time that this happens, and what happened immediately beforehand. Because you can't go towards the left, you can only possibly enter the leftmost door once. Therefore the repeated door must have another door immediately to its left, and you must have just come out of that one both times. But this contradicts the fact that it's the first time you used a door twice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore the maximum will involve walking into every door once, for a total of 52.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given two pairs of doors, it can be arranged for all four to be walked through, like this:&#xA;&lt;code&gt;A1 B1 A2 B2&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then this can be repeated 13 times to achieve the maximum.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-11-09T02:45:47.083" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23856" PostTypeId="2" ParentId="23854" CreationDate="2015-11-09T03:03:57.570" Score="0" Body="&lt;p&gt;All of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each pair of two doors, do this pattern:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;----[A]---[B]---[A]---[B]----&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;They'll walk through in this order:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;--1-[A]-4-[B]-3-[A]-2-[B]-5--&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'm fairly sure that odd amounts are impossible, but I haven't proven that yet.&lt;/p&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2015-11-09T03:03:57.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23857" PostTypeId="2" ParentId="23852" CreationDate="2015-11-09T03:28:14.553" Score="14" Body="&lt;p&gt;&lt;em&gt;What am I?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, I'd say that looking at this riddle, I am:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Confused&lt;/strong&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Being what you think you are would mean you cannot think.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you think you are confused you are &quot;unable to think clearly&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Being what you really are means you're wrong in having this.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; But what I really am is correct, and if I'm correct, I mustn't actually have the confusion I thought I did...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastActivityDate="2015-11-09T03:28:14.553" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23858" PostTypeId="1" AcceptedAnswerId="23861" CreationDate="2015-11-09T04:02:46.790" Score="3" ViewCount="231" Body="&lt;p&gt;Following &lt;a href=&quot;https://puzzling.stackexchange.com/users/10940/prem&quot;&gt;@Prem&lt;/a&gt;'s &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15339/four-words&quot;&gt;four-word puzzle&lt;/a&gt;, I thought I'd have a go.  (There have been a few other similar things, e.g. &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15407/another-six-words&quot;&gt;here&lt;/a&gt;, &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15347/six-words-six-words&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 6 words each with 6 letters.  If written in 6 rows (to make a 6x6 square) the same words read down as well as across.  Below are 6 cryptic clues to the 6 words.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The clues have been mixed up to increase the challenge a little&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Aristocrat eats honey. Gatherer thwarts unfairly.&lt;br&gt;&#xA;  Equip to ban eel stew.&lt;br&gt;&#xA;  Not exactly round at the trail's end.&lt;br&gt;&#xA;  Guides in haste erstwhile bulls.&lt;br&gt;&#xA;  Tea tears apart fashions.&lt;br&gt;&#xA;  Drearier droning undoes lovely lilts, even right at the beginning.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-11-09T15:50:46.983" Title="Yet Another Six-words (------||||||)" Tags="|word-square|cryptic-clues|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23860" PostTypeId="2" ParentId="23852" CreationDate="2015-11-09T05:42:08.533" Score="6" Body="&lt;p&gt;I think the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; reflection&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Being what you think you are would mean you cannot think.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you think you are your reflection and its real than you cannot think as the reflection is just an image without cognitive capabilities.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Being what you really are means you're wrong in having this.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; But the reflection is just you so if you think your reflection is thinking, it is nevertheless you who it is thinking.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Image Hints&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Fish&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first and the third are the mirror images. The second piece is the original one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Mouse&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The reflection is on a glass table.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Car&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is a reflection on a clear lake.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4177" LastActivityDate="2015-11-09T05:42:08.533" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23861" PostTypeId="2" ParentId="23858" CreationDate="2015-11-09T09:48:19.947" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;code&gt;TRENDS&lt;/code&gt; &lt;br&gt;&#xA;  &lt;code&gt;RUNOUT&lt;/code&gt; &lt;br&gt;&#xA;  &lt;code&gt;ENABLE&lt;/code&gt; &lt;br&gt;&#xA;  &lt;code&gt;NOBBLE&lt;/code&gt; &lt;br&gt;&#xA;  &lt;code&gt;DULLER&lt;/code&gt; &lt;br&gt;&#xA;  &lt;code&gt;STEERS&lt;/code&gt; &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Aristocrat eats honey. Gatherer thwarts unfairly.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Noble holding 'b' (honey gather). To &lt;strong&gt;nobble&lt;/strong&gt; is to thwart unfairly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Equip to ban eel stew.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Anagram of 'ban eel': &lt;strong&gt;enable&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not exactly round at the trail's end.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  To end is to &lt;strong&gt;run out&lt;/strong&gt;. Runout is also when a tool is slightly off centre.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Guides in haste erstwhile bulls.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ha&lt;strong&gt;ste ers&lt;/strong&gt;twhile. To guide is to steer, and a bull is a steer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tea tears apart fashions.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'T' + rends: &lt;strong&gt;Trends&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Drearier droning undoes lovely lilts, even right at the beginning.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;d&lt;/strong&gt;roning &lt;strong&gt;u&lt;/strong&gt;ndoes &lt;strong&gt;l&lt;/strong&gt;ovely &lt;strong&gt;l&lt;/strong&gt;ilts, &lt;strong&gt;e&lt;/strong&gt;ven &lt;strong&gt;r&lt;/strong&gt;ight: &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7316" LastEditorUserId="7316" LastEditDate="2015-11-09T15:50:46.983" LastActivityDate="2015-11-09T15:50:46.983" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23862" PostTypeId="1" AcceptedAnswerId="23865" CreationDate="2015-11-09T09:50:14.017" Score="7" ViewCount="2931" Body="&lt;p&gt;There are two kinds of people living in a village: those who always lie &#xA;(liars) and those who always tell the truth (truth-tellers). Everybody &#xA;knows each other in the village and therefore knows whether a particular &#xA;villager is a liar or a truth-teller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One day a foreigner came to the village, and saw $100$ villagers &#xA;$V_1,V_2,\ldots,V_{100}$ standing lined up from left to right on the market &#xA;square.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Villager $V_1$ pointed to some person $V_j$ to his right (with $j\ge2$) &#xA;and either announced &lt;em&gt;&quot;This person is a liar&quot;&lt;/em&gt; or &lt;em&gt;&quot;This person is a &#xA;truth-teller&quot;&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Villager $V_2$ pointed to some person $V_k$ to his right (with $k\ge3$) &#xA;and either announced &lt;em&gt;&quot;This person is a liar&quot;&lt;/em&gt; or &lt;em&gt;&quot;This person is a &#xA;truth-teller&quot;&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;li&gt;And so on, and so on, and so on&lt;/li&gt;&#xA;&lt;li&gt;Finally villager $V_{99}$ pointed to person $V_{100}$ to his right and &#xA;either announced &lt;em&gt;&quot;This person is a liar&quot;&lt;/em&gt; or &lt;em&gt;&quot;This person is a &#xA;truth-teller&quot;&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The foreigner had no idea what was going on and looked confused.&#xA;Finally the village chief (who was a famous truth-teller) gave him a hint: &#xA;&lt;em&gt;&quot;Among these $100$ villagers, there are at least &lt;strong&gt;N&lt;/strong&gt; truth-tellers.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; For which values of &lt;strong&gt;N&lt;/strong&gt; can we be sure that the foreigner&#xA;  was now able to deduce exactly which of the $100$ villagers were liars &#xA;  and which were truth-tellers?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-11-09T12:32:31.383" Title="Another village of liars and truth-tellers" Tags="|mathematics|logical-deduction|liars|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23863" PostTypeId="2" ParentId="23862" CreationDate="2015-11-09T10:24:55.607" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $N \gt 50$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All liars will be consistent with each other, and all truth tellers will be consistent with each other. This means we have two distinct groups, with at most one of size greater than 50. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the chief tells us there are at least 51 truth tellers, the larger group must be truthful. &lt;/p&gt;&#xA;" OwnerUserId="3987" LastActivityDate="2015-11-09T10:24:55.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23864" PostTypeId="2" ParentId="23862" CreationDate="2015-11-09T10:27:11.963" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Firstly, there is one trivial case where $N = 100$ and the foreigner can knows that everyone is a truth-teller.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Assuming for a minute that the village chief specified EXACTLY how many truth-tellers there were and looking at the case where $N = 50$, if we assume for a moment that every one of the villagers in the line points to the last villager, either there will be 49 liars and 50 truth-tellers pointing at a liar OR there will be 49 truth-tellers and 50 liars pointing at a truth-teller. In this case the set of responses will be identical (50 * L, 49 * T). So it is impossible to be sure for $N = 50$.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Now expanding this to the case where the chief is not specific, we have ruled out all cases where $N &amp;lt; 50$.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Continuing on to cases where there are more than 50 truth-tellers. Every villager forms part of a chain that links to $V_{100}$.&#xA; Determining the nature of $V_{100}$ allows the foreigner to trace back each chain and determine the nature of each villager in the chain. If he makes an assumption about the nature of $V_{100}$ nature and determines how many truth-tellers there are based on that assumption, if it is greater than or equal to $N$ then his initial assumption is correct. If he adds up too few truth-tellers, he just needs to turn all his truth-teller flags to liar flags and vice versa.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="14135" LastEditDate="2015-11-09T12:32:31.383" LastActivityDate="2015-11-09T12:32:31.383" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23865" PostTypeId="2" ParentId="23862" CreationDate="2015-11-09T11:06:51.257" Score="7" Body="&lt;p&gt;It all boils down to the last villager.&#xA;If you know what the last villager is you know what $V_{99}$ is and any other villager that pointed to him, therefore you know exactly what the villagers are that pointed to those villagers and so on. So, knowing what $V_{100}$ is has a unique state for all other villagers. So we only need to consider these two cases:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$V_{100}$ is a liar&lt;/li&gt;&#xA;&lt;li&gt;$V_{100}$ is a truthteller&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;if $V_{99}$ was a liar in the first case he would be a truthteller in the second case. This is also true for all other villagers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So ultimately there are two cases:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There are $X$ truthtellers&lt;/li&gt;&#xA;&lt;li&gt;There are $100 - X$ truthtellers.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Only one of these cases can have more than 50 truthtellers, so if $N &amp;gt; 50$ we can exactly determine everyone. &lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-11-09T11:06:51.257" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23866" PostTypeId="1" AcceptedAnswerId="23877" CreationDate="2015-11-09T15:08:49.927" Score="5" ViewCount="670" Body="&lt;p&gt;When you enter a search into &lt;a href=&quot;https://books.google.com/ngrams&quot; rel=&quot;nofollow noreferrer&quot;&gt;google ngrams&lt;/a&gt;, the graph shows the words/strings down the right hand side. What if you could pick search terms such that reading the right hand side of the graph, from top to bottom, makes an actual sentence...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Challenge&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the longest sentence you can create in google ngrams by reading the search terms down the right hand side of the graph?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A nice simple one to start you off. I give you, for a score of 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://books.google.com/ngrams/graph?content=This%2Csentence%2Cscores%2Cpoorly&amp;amp;year_start=1800&amp;amp;year_end=2000&amp;amp;corpus=15&amp;amp;smoothing=3&amp;amp;share=&amp;amp;direct_url=t1%3B%2CThis%3B%2Cc0%3B.t1%3B%2Csentence%3B%2Cc0%3B.t1%3B%2Cscores%3B%2Cc0%3B.t1%3B%2Cpoorly%3B%2Cc0&quot; rel=&quot;nofollow noreferrer&quot;&gt;This sentence scores poorly.&lt;/a&gt;   &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/G8jYd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/G8jYd.png&quot; alt=&quot;Graph for &amp;quot;This sentence scores poorly.&amp;quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rules&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Each word must be its own ngrams search term (no strings as search terms)&lt;/li&gt;&#xA;&lt;li&gt;The sentence must be in English. You must use the corpus &quot;English&quot;.&lt;/li&gt;&#xA;&lt;li&gt;The sentence must make grammatical sense. To achieve this, you are allowed to insert punctuation in your sentence which is not present in your ngrams search.&lt;/li&gt;&#xA;&lt;li&gt;You are not allowed to tick &quot;case insensitive&quot;. (Because it spoils the look by adding &lt;em&gt;(All)&lt;/em&gt; to the end of each word). Capitals are allowed where they make grammatical sense (beginning of sentence, proper nouns) but are not required.&lt;/li&gt;&#xA;&lt;li&gt;You may not string endless adjectives or adverbs together; only one adjective to describe each object and one adverb to describe each verb, &lt;em&gt;unless you can put a joining word in between&lt;/em&gt;. i.e. &quot;The big red awesome car&quot; would not be acceptable, but &quot;the awesome car which was big and red&quot; would be... although clearly that example isn't going to appear in order! Note that possessive nouns (e.g. &quot;my &lt;em&gt;sister's&lt;/em&gt; car&quot;) are adjectives and hence fall under this rule.&lt;/li&gt;&#xA;&lt;li&gt;You are allowed to change the end year of your search from the default.&lt;/li&gt;&#xA;&lt;li&gt;The maximum score possible is 12, because google ngrams only allows 12 search terms.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9444" LastEditorUserId="69582" LastEditDate="2021-06-25T15:43:12.843" LastActivityDate="2021-06-25T15:43:12.843" Title="The ngrams sentence challenge" Tags="|optimization|google-ngram|" AnswerCount="7" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="23867" PostTypeId="2" ParentId="23866" CreationDate="2015-11-09T15:35:21.200" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://books.google.com/ngrams/graph?content=buffalo%2Cbuffalo%2Cbuffalo%2Cbuffalo%2Cbuffalo%2Cbuffalo%2Cbuffalo%2Cbuffalo&amp;amp;year_start=1800&amp;amp;year_end=2000&amp;amp;corpus=15&amp;amp;smoothing=3&amp;amp;share=&amp;amp;direct_url=t1%3B%2Cbuffalo%3B%2Cc0%3B.t1%3B%2Cbuffalo%3B%2Cc0%3B.t1%3B%2Cbuffalo%3B%2Cc0%3B.t1%3B%2Cbuffalo%3B%2Cc0%3B.t1%3B%2Cbuffalo%3B%2Cc0%3B.t1%3B%2Cbuffalo%3B%2Cc0%3B.t1%3B%2Cbuffalo%3B%2Cc0%3B.t1%3B%2Cbuffalo%3B%2Cc0&quot; rel=&quot;nofollow&quot;&gt;Buffalo buffalo buffalo buffalo buffalo buffalo buffalo buffalo&lt;/a&gt; trivially scores 8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Buffalo_buffalo_Buffalo_buffalo_buffalo_buffalo_Buffalo_buffalo&quot; rel=&quot;nofollow&quot;&gt;Explanation&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12625" LastActivityDate="2015-11-09T15:35:21.200" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23869" PostTypeId="2" ParentId="23866" CreationDate="2015-11-09T16:05:24.057" Score="3" Body="&lt;h2&gt;10 points&lt;/h2&gt;&#xA;&lt;p&gt;A somewhat morbid sentence, perhaps from a proponent of eugenics:&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;a href=&quot;https://books.google.com/ngrams/graph?content=A%2Cnew%2Cman%2Cwithout%2Ctrue%2Clove%2Crequires%2Csevere%2Canesthesia%2Cperfunctorily&amp;amp;year_start=1800&amp;amp;year_end=1975&amp;amp;corpus=15&amp;amp;smoothing=3&amp;amp;share=&quot; rel=&quot;nofollow noreferrer&quot;&gt;A new man without true love requires severe anesthesia perfunctorily&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;(Note that this uses an end year of 1975.  After that point, &lt;em&gt;&lt;strong&gt;love&lt;/strong&gt;&lt;/em&gt; overtakes &lt;em&gt;&lt;strong&gt;true&lt;/strong&gt;&lt;/em&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-11-09T16:05:24.057" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23870" PostTypeId="2" ParentId="23843" CreationDate="2015-11-09T17:30:54.957" Score="6" Body="&lt;p&gt;Mate in Seven:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With some explanation to follow, this is mate in 3:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/imD4r.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/imD4r.jpg&quot; alt=&quot;mate in 3 scenario&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The king moving to row 8:&lt;br&gt;&#xA;Kc8 -&gt; Qe7 -&gt; Kb8 -&gt; Ba6 -&gt; Ka8 -&gt; Qb7&lt;br&gt;&#xA;Kd8 -&gt; Pd4 -&gt; Kc8 -&gt; Ba6 -&gt; ( Kb8 -&gt; Qb7 ) or ( Kd8 -&gt; Bg5)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The king moving to row 6:&lt;br&gt;&#xA;Kd6 -&gt; Pd4 -&gt; Kc6 -&gt; Bf4 -&gt; Kb6 -&gt; Qc7&lt;br&gt;&#xA;Kc6 -&gt; Pd4 -&gt; Kd6 -&gt; Bf4 -&gt; Kc6 -&gt; Qc7&lt;br&gt;&#xA;Kc6 -&gt; Pd4 -&gt; Kb6 -&gt; Qd7 -&gt; Ka5 -&gt; Qc6  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such, any 4 moves that white can make to get black into that position or further down the move tree is mate in 7 at most.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next position to explore is this creation after five moves.  Is there any position the king can be that is not mate in 2? ( Red represents places the king can not be ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Arbrm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Arbrm.jpg&quot; alt=&quot;mate in2maybe&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ka8: Qa5 -&gt; Kb7 -&gt; Qa6&lt;br&gt;&#xA;Ka7: Qa5 -&gt; Kb7 -&gt; Qa6&lt;br&gt;&#xA;Kb7: maybe&lt;br&gt;&#xA;Kb6: Qh6 -&gt; Ka7 or b7 -&gt; Qa6&lt;br&gt;&#xA;Kc8: maybe&lt;br&gt;&#xA;Kc6: Qf7 -&gt; Kb6 -&gt; Qc7&lt;br&gt;&#xA;Kd8: Qf7 -&gt; Kc8 -&gt; Qc7&lt;br&gt;&#xA;Kd7: maybe&lt;br&gt;&#xA;Ke7: Qd5 -&gt; Kf8 or f6 -&gt; Qf7&lt;br&gt;&#xA;Kf8: Qf7&lt;br&gt;&#xA;Kf6: Qf7&lt;br&gt;&#xA;Kg7: Qf7 -&gt; Qh8 -&gt; Qg8  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, if black should live beyond 7 moves, black needs to reach b7, c8, or d7 on the fifth move.  White's stragety is to first develop the mate in 3 scenario and then the bishop attack.  After Pe4 -&gt; k? -&gt; Qh5, black's second moves could be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7VZRY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7VZRY.jpg&quot; alt=&quot;secondmove&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Except for that we know of our mate in 3 scenario, black's only options to avoid this are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/s1Cjn.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/s1Cjn.jpg&quot; alt=&quot;avoidance&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;White's third move is Pd4.  If black is on column E, F, or G, black's only hope to get to b7, c8, or d7 require moving to column D.  However, white's 4th move will be Qf7 and this is the mate in 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, black's third moves are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mIf5I.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mIf5I.jpg&quot; alt=&quot;final3&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About our mate in 3 scenario:  Things have become worse for black with the development of Pd4.  Qf7 is now mate in 2, which means in can be white's fifth move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black can be on the right or the left of the board.  If black is on the right, it can not reach b7, b8, or d7 by it's fifth move.  This means that white can just develop the bishops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If black is on the left hand side, the bishop development must be abandoned because it can not stop the king from becoming safe ( maybe ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, white's fourth move is Qf7.  If black is on row 7, it must move to row 6 because row 8 is certainly mate in 3 at most. Black's fourth moves are as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0xdKw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0xdKw.jpg&quot; alt=&quot;fourth&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;White now has three moves to clean up:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kd4 -&gt; Bf4 -&gt; Kc4 -&gt; Qc7&lt;br&gt;&#xA;Kc4 -&gt; Bf4 -&gt; Kb4 -&gt; Qc7&lt;br&gt;&#xA;Kb6 -&gt; Bf4 -&gt; Ka5 -&gt; Qa7 -&gt; Kb4 -&gt; Pc3&lt;br&gt;&#xA;Ka5 -&gt; Qa7 -&gt; Ka5 -&gt; Pc3&lt;/p&gt;&#xA;" OwnerUserId="13054" LastEditorUserId="13054" LastEditDate="2015-11-10T19:09:17.350" LastActivityDate="2015-11-10T19:09:17.350" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23871" PostTypeId="2" ParentId="23852" CreationDate="2015-11-09T19:17:46.880" Score="1" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bipolar Disorder&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation follows.....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being what you think you are would mean you cannot think.&#xA;Being what you really are means you're wrong in having this.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are two personalities.. Bi and being in one of the episodes you cannot think straight and are obnoxious. If you are aware and know you are, you can control it and you are not in the episode.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not in Order or read the picture of fish as DisOrdered.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2 and Hint 3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Can mean &quot;cannot think right&quot; so the whole context is are upside down.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9892" LastEditorUserId="9892" LastEditDate="2015-11-09T19:36:03.477" LastActivityDate="2015-11-09T19:36:03.477" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23873" PostTypeId="2" ParentId="23852" CreationDate="2015-11-09T19:25:04.810" Score="3" Body="&lt;p&gt;Is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Species Dysphoria&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Species_dysphoria&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;??&lt;/p&gt;&#xA;" OwnerUserId="12655" LastActivityDate="2015-11-09T19:25:04.810" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23874" PostTypeId="2" ParentId="23852" CreationDate="2015-11-09T19:33:03.737" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clinical lycanthropy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Has to do with thinking you're an animal or transforming into one, and is normally associated with &quot;Reality-Bending mental state&quot; (Delusions/Hallucinations) which could explain hint 3 with the crazy upside down car.&lt;/p&gt;&#xA;" OwnerUserId="16273" LastActivityDate="2015-11-09T19:33:03.737" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23875" PostTypeId="2" ParentId="23852" CreationDate="2015-11-09T19:39:22.887" Score="17" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Cotard_delusion&quot;&gt;Cotard Delusion&lt;/a&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; certainly can't think if you're actually dead, and you can't be dead if you actively have this delusion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dot car, rat doc, Dr Taco, cod art.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17062" LastEditDate="2015-11-09T19:44:52.950" LastActivityDate="2015-11-09T19:44:52.950" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23877" PostTypeId="2" ParentId="23866" CreationDate="2015-11-09T20:03:20.683" Score="3" Body="&lt;p&gt;13:  Just modifying a previous answer yields higher results:&#xA;&lt;a href=&quot;https://books.google.com/ngrams/graph?content=A%2Cnew%2Cman%2Cthrough%2Clife%2Cwithout%2Ctrue%2Clove%2Ctoday%2Cnormally%2Callows%2Cmedicinal&amp;amp;year_start=1800&amp;amp;year_end=1972&amp;amp;corpus=15&amp;amp;smoothing=3&amp;amp;share=&amp;amp;direct_url=t1%3B%2CA%3B%2Cc0%3B.t1%3B%2Cnew%3B%2Cc0%3B.t1%3B%2Cman%3B%2Cc0%3B.t1%3B%2Cthrough%3B%2Cc0%3B.t1%3B%2Clife%3B%2Cc0%3B.t1%3B%2Cwithout%3B%2Cc0%3B.t1%3B%2Ctrue%3B%2Cc0%3B.t1%3B%2Clove%3B%2Cc0%3B.t1%3B%2Ctoday%3B%2Cc0%3B.t1%3B%2Cnormally%3B%2Cc0%3B.t1%3B%2Callows%3B%2Cc0%3B.t1%3B%2Cmedicinal%3B%2Cc0&quot; rel=&quot;nofollow&quot;&gt;A,new,man,through,life,without,true,love,today,normally,allows,medicinal,marijuana&lt;/a&gt;&#xA; (Though I will point out that google stops showing them after 12.  So the answer may just be 12.&lt;/p&gt;&#xA;" OwnerUserId="17303" LastActivityDate="2015-11-09T20:03:20.683" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23878" PostTypeId="1" CreationDate="2015-11-09T20:42:27.750" Score="-2" ViewCount="268" Body="&lt;p&gt;We have a chess board of size n x n and there is exactly one stone on every square. We have green, blue and red stones. We must not have a row or a column full of stones of the same colour. We can omit using one of the colors. How many combinations of placing the stones are there?&lt;/p&gt;&#xA;" OwnerUserId="17305" LastActivityDate="2015-11-10T00:39:29.580" Title="Colored stones on a chess board puzzle" Tags="|logical-deduction|calculation-puzzle|combinatorics|" AnswerCount="2" CommentCount="8" ClosedDate="2015-11-10T01:34:29.790" ContentLicense="CC BY-SA 3.0" />
  <row Id="23879" PostTypeId="2" ParentId="23866" CreationDate="2015-11-09T21:34:45.793" Score="3" Body="&lt;h1&gt;12 points&lt;/h1&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;This could very much run counter vs. prevailing ideologies,&quot; quoth Slavoj Zizek.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/48Adt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/48Adt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9523" LastActivityDate="2015-11-09T21:34:45.793" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23880" PostTypeId="2" ParentId="23878" CreationDate="2015-11-09T21:59:10.283" Score="-1" Body="&lt;p&gt;$2(3)^{2n-2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am assuming that n must be at least 2, otherwise the board would have a row or column of all the same color.&lt;/p&gt;&#xA;" OwnerUserId="5669" LastActivityDate="2015-11-09T21:59:10.283" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23881" PostTypeId="2" ParentId="23878" CreationDate="2015-11-09T22:01:33.713" Score="1" Body="&lt;p&gt;This can be done with the inclusion-exclusion principle, but it gets a little bit ugly because of the intersecting cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Step 1:&lt;/em&gt; There are $3^{n^2}$ ways of putting the coins down altogether.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Step 2:&lt;/em&gt; From that we need to &lt;strong&gt;subtract&lt;/strong&gt; the ways of putting coins down that have at least one row or one column that are the same colour:&#xA;\begin{equation} 2\times\dbinom{8}{1}\times3^1\times(3^{n(n-1)}) \end{equation}&#xA;There's 16 rows and columns, which can each be 3 colours, and then a board of $n(n-1)$ left over.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Step 3:&lt;/em&gt; To that we need to &lt;strong&gt;add&lt;/strong&gt; all the ways of putting coins down that have at least two rows or columns that are the same colour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Step 3.1:&lt;/em&gt;  Two rows or two columns:&#xA;\begin{equation}2\times\dbinom{8}{2}\times3^2\times(3^{n(n-2)})\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Step 3.2:&lt;/em&gt;  One row and one column (note they have to be the same colour):&#xA;\begin{equation}8^2\times3\times(3^{(n-1)(n-1)})\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Step 4:&lt;/em&gt; To this we need to &lt;strong&gt;subtract&lt;/strong&gt; all the ways of putting coins down that have at least three rows or columns that are the same colour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Step 4.1:&lt;/em&gt;  Three rows or three columns:&#xA;\begin{equation}2\times\dbinom{8}{3}\times3^3\times(3^{n(n-3)})\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Step 4.2:&lt;/em&gt;  One row and two columns or vice versa (note they have to be the same colour):&#xA;\begin{equation}2\times\dbinom{8}{2}\times\dbinom{8}{1}\times3\times(3^{(n-1)(n-2)})\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Things carry on in this way for quite some time...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tricksy thing is that the cases where there are a mixture of rows and columns of one colour, you immediately know that all the single colours rows and column are the same colour.  This gives (thanks @Mike Earnest for writing this out):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation} 3^{n^2}-2\sum_{i=1}^n (-1)^i\binom{n}i3^{n(n-i)+i}+\sum_{i,j=1}^n (-1)^{i+j}\binom{n}i\binom{n}{j}3^{(n-i)(n-j)+1} \end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the first term is from step 1, the second term represents the cases where there are $i$ rows or $i$ columns, and the third term represents the cases where there are a mixture of at least 1 row and at least 1 column.&lt;/p&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2015-11-10T00:39:29.580" LastActivityDate="2015-11-10T00:39:29.580" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23883" PostTypeId="1" AcceptedAnswerId="23886" CreationDate="2015-11-10T04:56:17.253" Score="12" ViewCount="579" Body="&lt;p&gt;Some days ago, a young man came into my grandfather's antique shop and wanted to offload some of the old stuff that once belonged to his father.  Among them were three really old journals.  They were pretty fragile, but otherwise in relatively good shape.  As my grandfather flipped through them carefully, three entries caught his eyes.  He could not remember the exact words, but he remembered them as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;15 October 1583 - Hard&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Our son was born a month ago.  My wife was in so much pain while in labour, but thankfully, she and the baby made it through the night.  By the morning, her fever has all but gone and the baby was sound asleep.  I have dreamt of this day when I have become a proud father to a son.  My family would be delighted to see him tomorrow.  Hopefully this year's winter would not be as cold as 10 years ago when the nearby lake was frozen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;12 September 1752 - Richmond&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I met my her today.  She was so beautifully dressed and was accompanied by an entourage of maids.  Each a beauty in her own right.  But she stood out among all, like a black swan among the geese.  Her hair smell of rosemary, her eyes clear as day, her smile brings the sun from behind the clouds.  She was the most beautiful creature on Earth.  Well, at least the most beautiful here at the Gate House.  I should have asked her name.  For now I can only dream about her, just like how my ancestor dreamt about being a father.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;31 December 1926 - Batman&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;New Years Eve today.  Due to the flood earlier in the year, no one was really in the mood for any celebration.  That flood has wiped out an entire district.  My grandson was sick because of the flood and my daughter has been complaining about how her son was always sick and never seem to be in any sort of good health.  We are not rich, and so cannot afford good doctor.  We will have to do with whatever we've got.  Please be strong, my grandson.  How, o how, did my forefathers dream of being a father and dream of beautiful woman when my family has to suffer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Evidently the diaries were written by three different people.  It seemed like the journals were passed down through generations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My grandfather asked me whether these journals were genuine.  He mentioned that the entries have more than enough clues as to point me in the right direction to determine the genuineness of the diaries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should he buy them off the young man?  I am afraid that he would be duped into buying something that are fake and are completely worthless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please explain why you think they are genuine (or not).&lt;/p&gt;&#xA;" OwnerUserId="12332" LastEditorUserId="8874" LastEditDate="2016-03-23T17:06:09.177" LastActivityDate="2016-03-23T17:06:09.177" Title="The Diaries of the Old" Tags="|story|history|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23884" PostTypeId="2" ParentId="23883" CreationDate="2015-11-10T05:58:55.383" Score="-1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;But she stood out among all, like a black swan among the geese.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The diary is a fake one, as black swan was not yet discovered in 1752.&lt;/p&gt;&#xA;" OwnerUserId="16499" LastActivityDate="2015-11-10T05:58:55.383" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23885" PostTypeId="2" ParentId="23883" CreationDate="2015-11-10T06:41:28.653" Score="1" Body="&lt;p&gt;31 December 1926 - Batman&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This diary is fake, because &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Batman,_Turkey&quot; rel=&quot;nofollow&quot;&gt;&quot;Until the 1950s, Batman was a village, called Iluh ...&quot;&lt;/a&gt;. The flood in the year 1926 is mentioned in the article as well: &quot;The entire district disappeared in the early 1926 due to the flood of the Batman River.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2015-11-10T18:52:55.067" LastActivityDate="2015-11-10T18:52:55.067" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23886" PostTypeId="2" ParentId="23883" CreationDate="2015-11-10T06:50:36.900" Score="11" Body="&lt;p&gt;Edit: didn't realize we were talking about 3 separate books. That's what late-night puzzling will do to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Re-Edit: And once again, answer changes because I just can't let it alone, lol!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All 3 are fake.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All 3 carry dates that are among the ones dropped due to the change from the Julian Calendar to the Gregorian Calendar in their area.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Journal 1: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This part of Austria switched in 1583, skipping from 4 October to 16 October. The 15th was skipped.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Journal 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We're clearly in England (the Royal residence at Richmond, which includes the Gate House). Adoption took place in 1752 here, skipping from September 2 to September 14. The 12th was skipped.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Journal 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We are in Turkey here. The reason this is fake is twofold. First, this area would likely be using the Islamic Calendar, with completely different months and year numbering. Second, and more to the point of the puzzle, the Gregorian switch took place at the end of 1926. Since they switched so late, they had to drop 13 days this time. The calendar skipped from 18 December, 1926 to 1 January, 1927. Once more, the date on the entry was skipped.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16874" LastEditorUserId="16874" LastEditDate="2015-11-10T14:37:08.000" LastActivityDate="2015-11-10T14:37:08.000" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23887" PostTypeId="1" AcceptedAnswerId="23891" CreationDate="2015-11-10T13:45:05.277" Score="2" ViewCount="942" Body="&lt;p&gt;The square is divided into 9 squares (3x3) and it changes from left to right, following some rule. The right hand column shows the answer choices.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/V53cN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/V53cN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11135" LastActivityDate="2015-11-10T22:11:15.933" Title="What figure should most logically follow the series?" Tags="|visual|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23888" PostTypeId="2" ParentId="23887" CreationDate="2015-11-10T14:01:24.193" Score="3" Body="&lt;p&gt;I would say this one&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/LpV4O.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LpV4O.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every other picture, starting with the first one, in the sequence is the same picture, just a mirrored version&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For a bonus guess the picture after that might look like&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/ZKTVk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZKTVk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every other starting with the second picture is the same picture, rotated clockwise 90 degrees then mirrored (Makes sense to me at least!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16273" LastEditorUserId="16273" LastEditDate="2015-11-10T14:32:32.870" LastActivityDate="2015-11-10T14:32:32.870" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23889" PostTypeId="1" CreationDate="2015-11-10T14:09:22.143" Score="1" ViewCount="358" Body="&lt;p&gt;This is similar to the another one I posted but want to see what people think the answer is and why.These are practice questions from an assessment company but these seem quite interesting.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/oUBME.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oUBME.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11135" LastActivityDate="2015-11-18T20:32:29.170" Title="Another figure problem, trying to figure out the pattern/rule" Tags="|visual|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23890" PostTypeId="2" ParentId="23889" CreationDate="2015-11-10T14:18:19.027" Score="13" Body="&lt;p&gt;It would be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The second one down.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Looking at the first square in the example, let's call the top dot A and the lower one B.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dot A travels down to the left and then up to the right.  Think of it as an oscillating dot on that diagonal.  Dot B appears to travel like a knight in chess, so in the second square, it is at top left, then bottom middle, then top right, and the only option from the right is middle left.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14233" LastActivityDate="2015-11-10T14:18:19.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23891" PostTypeId="2" ParentId="23887" CreationDate="2015-11-10T15:39:33.707" Score="10" Body="&lt;p&gt;I also believe the answer to be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/lwaq5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lwaq5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I have a different reason.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/kSRSt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kSRSt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;br/&gt;I've color coated this to make it easier. Red and Green both cycle around the edges in a counter-clockwise motion. Yellow goes left, middle, right, middle, and presumably left again like it's &quot;walking&quot; back and forth. Blue, like Yellow, is going back and forth but on a diagonal. Middle, Upper-Right, Middle, Lower-Left, presumably middle again. Each color has a cyclical movement as if it were pacing (either back and forth or in a circle).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This would mean that the NEXT next would be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/1RaVw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1RaVw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;br/&gt;Which is exactly the same as the second image in the pattern except Red and Green are backwards.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2479" LastEditorUserId="2479" LastEditDate="2015-11-10T16:09:06.857" LastActivityDate="2015-11-10T16:09:06.857" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23892" PostTypeId="1" AcceptedAnswerId="23910" CreationDate="2015-11-10T20:54:44.013" Score="-7" ViewCount="3137" Body="&lt;p&gt;What does this rebus mean?&lt;/p&gt;&#xA;&lt;h1&gt;12356789 KOOL&lt;/h1&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Hint #1:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;The answer has to do partly with the fact the 4 is MISSING, not JUST that the 4 is special itself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Hint #2:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;The phrase &amp;quot;look back&amp;quot; is correct and is in the answer. (Obviously it's not the entire answer though) You also have to be a bit creative with the whole number thing. Happy guessing!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11002" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2017-03-20T20:57:22.003" Title="A KOOL rebus puzzle" Tags="|rebus|" AnswerCount="12" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="23894" PostTypeId="2" ParentId="23892" CreationDate="2015-11-10T22:22:37.473" Score="4" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Number School&#xA;because there are a bunch of &quot;numbers&quot; + &quot;kool&quot; and put together &quot;Numberskool&quot; -&gt; &quot;Number-skool&quot; -&gt; &quot;number-school&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After second hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Look back, for numbers are left. (Four has left and the numbers are to the left side of look back.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16471" LastEditorUserId="16471" LastEditDate="2015-11-11T20:27:50.680" LastActivityDate="2015-11-11T20:27:50.680" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23895" PostTypeId="1" AcceptedAnswerId="23898" CreationDate="2015-11-10T22:42:43.580" Score="3" ViewCount="377" Body="&lt;p&gt;$$\frac{(023456789)\space\text{SKOOL}}{\text{THE ANSWER}} + \text{CONFUSION}$$&lt;/p&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2015-11-11T01:16:53.203" Title="How do you feel about this?" Tags="|rebus|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23896" PostTypeId="2" ParentId="23892" CreationDate="2015-11-10T23:10:25.290" Score="14" Body="&lt;p&gt;I'll take a stab:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Look back, it's a foregone (4 gone) conclusion&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16874" LastActivityDate="2015-11-10T23:10:25.290" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23897" PostTypeId="2" ParentId="23892" CreationDate="2015-11-11T00:02:28.103" Score="4" Body="&lt;p&gt;might the answer be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Not four Skool&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user11082" LastEditorDisplayName="user11082" LastEditDate="2015-11-11T00:17:13.347" LastActivityDate="2015-11-11T00:17:13.347" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23898" PostTypeId="2" ParentId="23895" CreationDate="2015-11-11T01:16:53.203" Score="7" Body="&lt;p&gt;Possibly (though it doesn't feel quite right):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No one looks back on the answer with some confusion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastActivityDate="2015-11-11T01:16:53.203" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23902" PostTypeId="2" ParentId="23846" CreationDate="2015-11-11T03:31:17.160" Score="6" Body="&lt;p&gt;What I have here still sounds a little jumbled, so it's possible that there's a mistake or two in there, but I think a couple of the commenters had the right idea.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you split up each number correctly, then as Daphne B says in the comments, you just map A to 1, B to 2 and so on. Split the numbers like so: &lt;br&gt;&#xA; &lt;br&gt;&#xA; 18-5-2-13-21-14 5-14-15 7-7-5 14-9-20 18-5-8-20-15 18-5-8-20-9-5 11-14-9-8-20 5-25-5 .19-18-5-20-20-5-12 20-18-1-5-8 18-5-22-5-18-15-6 4-12-15-8 5-25-5 &lt;br&gt;&#xA; &lt;br&gt;&#xA; and convert to letters. &lt;br&gt;&#xA; &lt;br&gt;&#xA; .REBMUN ENO GGE NIT REHTO REHTIE KNIHT EYE .SRETTEL TRAEH REVEROF DLOH EYE &lt;br&gt;&#xA; &lt;br&gt;&#xA; As question_asker posited, it looks like the message is reversed, so reverse it back and you get &lt;br&gt;&#xA; &lt;br&gt;&#xA; EYE HOLD FOREVER HEART LETTERS. EYE THINK EITHER OTHER TIN EGG ONE NUMBER. &lt;br&gt;&#xA; &lt;br&gt;&#xA; or in more readable terms, &quot;&lt;strong&gt;I hold forever heart letters. I think either other tin egg one number&lt;/strong&gt;&quot; &lt;br&gt;&#xA; It's entirely possible that I screwed up decoding part of it; in particular the final message stops making sense after &quot;either.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1600" LastActivityDate="2015-11-11T03:31:17.160" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23903" PostTypeId="2" ParentId="23892" CreationDate="2015-11-11T06:28:38.793" Score="3" Body="&lt;p&gt;Some guesses:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Not&lt;/em&gt; &lt;em&gt;Look&lt;/em&gt;ing forward to&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Look&lt;/em&gt; be&lt;em&gt;fore&lt;/em&gt; you leap&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Look&lt;/em&gt; forward to&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Look&lt;/em&gt;ing &lt;em&gt;back&lt;/em&gt; &lt;em&gt;for&lt;/em&gt; ______&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In decreasing order of plausibility&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-11-11T06:28:38.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23904" PostTypeId="2" ParentId="23866" CreationDate="2015-11-11T09:23:54.090" Score="2" Body="&lt;p&gt;Here's another &lt;strong&gt;12-pointer&lt;/strong&gt; that I believe satisfies all the rules. Unfortunately, for some reason Google does not place the words in the correct order on the right-hand side, but at all points within the selected date range the order of the word frequency (when you hover over the graph) is in the correct sentence order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://books.google.com/ngrams/graph?content=I%2Cjust%2Clove%2Cquickly%2Cfinding%2Cperfect%2Canswers%2Ccomments%2Caside%2Cresolving%2Cpuzzles%2Ccleverly&amp;amp;year_start=1975&amp;amp;year_end=2000&amp;amp;corpus=15&amp;amp;smoothing=3&amp;amp;share=&amp;amp;direct_url=t1%3B%2CI%3B%2Cc0%3B.t1%3B%2Cjust%3B%2Cc0%3B.t1%3B%2Clove%3B%2Cc0%3B.t1%3B%2Cquickly%3B%2Cc0%3B.t1%3B%2Cfinding%3B%2Cc0%3B.t1%3B%2Cperfect%3B%2Cc0%3B.t1%3B%2Canswers%3B%2Cc0%3B.t1%3B%2Ccomments%3B%2Cc0%3B.t1%3B%2Caside%3B%2Cc0%3B.t1%3B%2Cresolving%3B%2Cc0%3B.t1%3B%2Cpuzzles%3B%2Cc0%3B.t1%3B%2Ccleverly%3B%2Cc0&quot; rel=&quot;nofollow&quot;&gt;I just love quickly finding perfect answers (comments aside) - resolving puzzles cleverly.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-11-11T09:23:54.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23905" PostTypeId="2" ParentId="23892" CreationDate="2015-11-11T12:28:33.543" Score="9" Body="&lt;p&gt;Just a guess...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Look back, not forward.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14233" LastActivityDate="2015-11-11T12:28:33.543" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23906" PostTypeId="2" ParentId="23892" CreationDate="2015-11-11T15:32:54.110" Score="6" Body="&lt;p&gt;Maybe it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Look back, not foreseen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17323" LastActivityDate="2015-11-11T15:32:54.110" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23908" PostTypeId="2" ParentId="23892" CreationDate="2015-11-11T16:12:17.230" Score="2" Body="&lt;p&gt;Guess&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Before Count foregone his look back&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Before (It happens 'before' kool) Count (Meaning count Dracula or sesame street count, because 12356789 is counting) foregone (meaning given up -- because the four is gone from the list) look back (meaning remember or recall)  So, &quot;Before the Count has given up on thinking of the past&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16273" LastActivityDate="2015-11-11T16:12:17.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23909" PostTypeId="2" ParentId="23892" CreationDate="2015-11-11T16:30:18.573" Score="4" Body="&lt;p&gt;Is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; foregone conclusion?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The four is missing from the count, 4 gone-&gt;foregone &#xA;and KOOL is look backwards -&gt; lookback which is a synonym for conclusion&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;After Second Hint&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Look back for missing numbers?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4591" LastEditorUserId="4591" LastEditDate="2015-11-11T19:00:41.617" LastActivityDate="2015-11-11T19:00:41.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23910" PostTypeId="2" ParentId="23892" CreationDate="2015-11-11T16:35:25.010" Score="8" Body="&lt;p&gt;OK, I'm &lt;em&gt;prrrrrrrretty sure&lt;/em&gt; this isn't the answer, but in the spirit of the site, this is what I was getting at in my comment on the OP:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Never look back&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; given absurdly liberal leeway in pronunciation, &quot;no four&quot; sounds kinda almost a little but not really but just enough like &quot;never&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastActivityDate="2015-11-11T16:35:25.010" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="23911" PostTypeId="2" ParentId="23892" CreationDate="2015-11-11T17:48:27.397" Score="2" Body="&lt;p&gt;Is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Count forward, look back&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You're counting in order without a four; and look is reversed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16846" LastActivityDate="2015-11-11T17:48:27.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23912" PostTypeId="2" ParentId="23892" CreationDate="2015-11-11T18:09:18.170" Score="2" Body="&lt;p&gt;I'm gussing&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; go forth before look back&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; four is gone so gone forth, or go forth, and look is backward so look back. and 12356789 is before kool, hence my answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16621" LastActivityDate="2015-11-11T18:09:18.170" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23913" PostTypeId="2" ParentId="23892" CreationDate="2015-11-11T20:23:06.730" Score="2" Body="&lt;p&gt;Is it?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Look back for nothing then&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Look is reversed and the number 4 is not in the 10 numbers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17327" LastActivityDate="2015-11-11T20:23:06.730" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23914" PostTypeId="1" AcceptedAnswerId="23921" CreationDate="2015-11-11T22:32:48.037" Score="5" ViewCount="800" Body="&lt;p&gt;&lt;strong&gt;Story:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a line of bombs, all over a building. You are the bomb defusal expert, and you are the one everyone is counting on the save the city, for as if any one of the bombs go off, you, and everyone in the city, will die a tragic death. Obviously you don't want that to happen, so you read up the secret file they gave you on what's going on so you can get started.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;=========================================================================&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You have to stall the bombs until we can hack into the entire system and shut all the bombs down. You have to run from button to button located on the bombs. Touching the buttons on them will reset the timers on them.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Here's a diagram of the bombs, how long it will take to get to them, etc.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A &amp;lt;==1m==&amp;gt;  B &amp;lt;==2m30s==&amp;gt; E&#xA;&#xA;/\         /\&#xA;2m         4m&#xA;\/         \/&#xA;&#xA;C &amp;lt;==3m==&amp;gt; D&#xA;&#xA;A's timer is 10m&#xA;B's timer is 12m&#xA;C's timer is 12m&#xA;D's timer is 16m&#xA;E's timer is 12m&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You are starting off at bomb A. Also, it does not take any time actually pushing the button on the bomb. The only time spent (Hopefully, if you're not just standing on the wall doing nothing, which you can do, but make sure you have enough time...) is actually running between the bombs. If you get to a bomb with no time left you can use your ninja skills to get the bomb's button pushed before it explodes, but if you're in the middle of a pathway, like if you were 2 minutes in on a 3 minute pathway, and the bomb you were travelling to went to 0, you couldn't suddenly teleport there to press the button.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;=========================================================================&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You decide to get started on you're task, and you get a message that they've starting the hacking already.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Little does anyone know that in reality the hacking will fail every time. Just how long can you stall the bombs so that you get the maximum time before everything explodes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try to do this without computer programs, because that would be too easy!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. Please no loopholes, there's no lateral-thinking tag here!&lt;/p&gt;&#xA;" OwnerUserId="11002" LastEditorUserId="7715" LastEditDate="2015-11-11T23:11:48.927" LastActivityDate="2015-11-12T13:46:48.267" Title="I Was Told It Would Explode" Tags="|logical-deduction|no-computers|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23915" PostTypeId="2" ParentId="23914" CreationDate="2015-11-11T22:50:39.863" Score="3" Body="&lt;p&gt;I'll say a guess of 18m30s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C and E are you opposite points. While we could make a CABE circuit (11m), that leaves out D, which will eventually blow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EBACD one-way takes 8m30s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, ABE (3m30s), wait 3m, EBAC (5m30s, just enough to prevent C blowing), D (3m), wait 3m30s for E to blow. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-11-12T13:46:48.267" LastActivityDate="2015-11-12T13:46:48.267" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23916" PostTypeId="1" AcceptedAnswerId="23920" CreationDate="2015-11-12T01:38:54.933" Score="1" ViewCount="3539" Body="&lt;p&gt;new to puzzles here. A real life situation led me to consider this. We have a small league of 6 people playing in teams of two. I would like to rotate the teams in such a way that everyone gets an equal chance to play with every other person and against every other person.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g. in the first game, it's A+B vs C+D and E+F sit out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I arrange rotations in such a way that everyone gets an equal number of chances to play with every other player and against every other player? &lt;/p&gt;&#xA;" OwnerUserId="17330" LastActivityDate="2015-11-12T19:41:09.833" Title="Fair tournament for rotating teams of two" Tags="|mathematics|combinatorics|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23918" PostTypeId="1" AcceptedAnswerId="23938" CreationDate="2015-11-12T02:10:34.443" Score="15" ViewCount="402" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/w93xk.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/w93xk.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a slightly modified thing I remember from a line drawing in a book years ago.  Well-known phrase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Most&lt;/em&gt; of the word Gangsta is a red herring.  This rebus predates rap as we know it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;another hint&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One &quot;bird&quot; is &lt;em&gt;the best&lt;/em&gt;, and he's standing on a couple of letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1536" LastEditorUserId="1536" LastEditDate="2015-11-12T18:12:37.450" LastActivityDate="2018-08-21T16:52:43.237" Title="Parrot Rebus redux" Tags="|rebus|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23919" PostTypeId="2" ParentId="23916" CreationDate="2015-11-12T02:15:09.097" Score="1" Body="&lt;p&gt;Well, for four players (A,B,C and D), there are three possible pairings:&lt;br&gt;&#xA;A&amp;amp;B v C&amp;amp;D&lt;br&gt;&#xA;A&amp;amp;C v B&amp;amp;D&lt;br&gt;&#xA;A&amp;amp;D v B&amp;amp;C  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those games exclude E&amp;amp;F.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there are $\binom{6}{2}$ = 15 different ways to have two people sitting out.&lt;/p&gt;&#xA;" OwnerUserId="15942" LastActivityDate="2015-11-12T02:15:09.097" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23920" PostTypeId="2" ParentId="23916" CreationDate="2015-11-12T02:59:49.000" Score="2" Body="&lt;p&gt;Each person plays the same number of matches with each of the other 5, so the number of matches each person plays must be a multiple of 5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If each person plays 5 matches, then there are a total of 30 players across the matches. But this is impossible because 4 players are in each match and 30 isn't divisible by 4. So each player must play at least 10 matches, with a total of 15 matches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can make five sets of three games each: for each set, divide the players into three pairs, and have each pair play the other two once. This is one possible way to divide the pairs:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;AB/CD/EF&lt;/li&gt;&#xA;&lt;li&gt;AC/BE/DF&lt;/li&gt;&#xA;&lt;li&gt;AD/BF/CE&lt;/li&gt;&#xA;&lt;li&gt;AE/BD/CF&lt;/li&gt;&#xA;&lt;li&gt;AF/BC/DE&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You can check that for every two players, they are together in two games (in the set where they are paired) and against each other in four games (once in each of the other sets).&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-11-12T02:59:49.000" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="23921" PostTypeId="2" ParentId="23914" CreationDate="2015-11-12T04:13:02.560" Score="7" Body="&lt;p&gt;I think I got 19m30s...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A)CABEBACD(E explodes)&lt;/p&gt;&#xA;" OwnerUserId="11887" LastActivityDate="2015-11-12T04:13:02.560" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23922" PostTypeId="1" AcceptedAnswerId="23926" CreationDate="2015-11-12T12:36:18.613" Score="10" ViewCount="1447" Body="&lt;p&gt;I have seen this (not the rebus, but what it represents) written on many places. Could you guess what it is and on where I may have seen it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Uamkd.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uamkd.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="8874" LastEditDate="2015-11-12T15:34:09.967" LastActivityDate="2015-11-13T09:57:20.353" Title="Rebus: written on many places" Tags="|rebus|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23923" PostTypeId="2" ParentId="23922" CreationDate="2015-11-12T12:49:20.593" Score="10" Body="&lt;p&gt;It could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Due to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;ill&quot; not &quot;bills&quot; - subtract the first from the second and you get BS.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14233" LastActivityDate="2015-11-12T12:49:20.593" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23924" PostTypeId="2" ParentId="23922" CreationDate="2015-11-12T13:03:42.857" Score="1" Body="&lt;p&gt;I could think of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Sick leaves are not paid&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the explanation is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;If sick, No Bills&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user11082" LastActivityDate="2015-11-12T13:03:42.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23925" PostTypeId="2" ParentId="23922" CreationDate="2015-11-12T13:24:47.250" Score="4" Body="&lt;p&gt;The only thing that comes to mind is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; post no bills&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I just can't find the connection for the first word.&lt;/p&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2015-11-12T13:24:47.250" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23926" PostTypeId="2" ParentId="23922" CreationDate="2015-11-12T13:31:57.350" Score="15" Body="&lt;p&gt;Perhaps&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; S(t)ick no bills&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;as illustrated for instance at&#xA;&lt;a href=&quot;https://www.flickr.com/photos/codein/90245453&quot; rel=&quot;nofollow&quot;&gt;https://www.flickr.com/photos/codein/90245453&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explanation&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sick with Tea = Sick with T = &lt;strong&gt;STICK&lt;/strong&gt;&lt;br&gt;&#xA; Forbidden sign = &lt;strong&gt;NO&lt;/strong&gt;&lt;br&gt;&#xA; Bill Pullman, Bill Murray, Bill Hader, Bill Maher, Bill Walton = &lt;strong&gt;BILLS&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-11-13T09:57:20.353" LastActivityDate="2015-11-13T09:57:20.353" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="23927" PostTypeId="1" CreationDate="2015-11-12T13:57:02.723" Score="18" ViewCount="3883" Body="&lt;p&gt;Person A thinks of a 5 digit binary number. Person B tries to guess the number. &#xA;B can guess a 5 digit binary number and A will respond with the number of correct digits (digits in the right place). &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the maximum number of guesses B will take to know the binary number (assuming he is playing optimally)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17336" LastEditorUserId="8874" LastEditDate="2015-11-12T20:26:25.633" LastActivityDate="2015-11-12T20:35:27.333" Title="Guess five binary digits!" Tags="|mathematics|logical-deduction|strategy|combinatorics|mastermind|" AnswerCount="6" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23929" PostTypeId="2" ParentId="23866" CreationDate="2015-11-12T14:21:42.920" Score="4" Body="&lt;p&gt;Here's a particularly dark &lt;strong&gt;12-point&lt;/strong&gt;er:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://books.google.com/ngrams/graph?content=the%2Cvery%2Cgood%2Cman%2Cthough%2Ckind%2Cusually%2Cconsidered%2Ckilling%2Cbabies%2Cmarginally%2Cunconscionable&amp;amp;year_start=1800&amp;amp;year_end=2000&amp;amp;corpus=15&amp;amp;smoothing=3&amp;amp;share=&amp;amp;direct_url=t1%3B%2Cthe%3B%2Cc0%3B.t1%3B%2Cvery%3B%2Cc0%3B.t1%3B%2Cgood%3B%2Cc0%3B.t1%3B%2Cman%3B%2Cc0%3B.t1%3B%2Cthough%3B%2Cc0%3B.t1%3B%2Ckind%3B%2Cc0%3B.t1%3B%2Cusually%3B%2Cc0%3B.t1%3B%2Cconsidered%3B%2Cc0%3B.t1%3B%2Ckilling%3B%2Cc0%3B.t1%3B%2Cbabies%3B%2Cc0%3B.t1%3B%2Cmarginally%3B%2Cc0%3B.t1%3B%2Cunconscionable%3B%2Cc0&quot; rel=&quot;nofollow noreferrer&quot;&gt;The very good man—though usually kind—considered killing babies marginally unconscionable.&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's a screenshot of the hover window for the year 2000:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/khBD1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/khBD1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16275" LastEditorUserId="16275" LastEditDate="2015-11-12T14:27:32.413" LastActivityDate="2015-11-12T14:27:32.413" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23930" PostTypeId="2" ParentId="23927" CreationDate="2015-11-12T14:36:30.920" Score="5" Body="&lt;p&gt;First guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If you start with &lt;code&gt;00000&lt;/code&gt; and change one digit at a time. For example, let's say the number is &lt;code&gt;10010&lt;/code&gt;. You guess &lt;code&gt;00000&lt;/code&gt; and get that 3 digits are correct. Then guess &lt;code&gt;10000&lt;/code&gt; and get that 4 digits are correct, so now you know that the first digit is correct as &lt;code&gt;1&lt;/code&gt;. Then change it to &lt;code&gt;11000&lt;/code&gt; and get that 3 digits are correct, so you know the second digit was correct as &lt;code&gt;0&lt;/code&gt;. Then guess &lt;code&gt;10100&lt;/code&gt; and get that 3 digits are correct, so you know the third digit is &lt;code&gt;0&lt;/code&gt;. Your final guess is on the fourth digit because when you guess &lt;code&gt;10010&lt;/code&gt; either all 5 digits will be correct or only 3 digits will be correct. If all 5 are correct, then you are clearly done, but if 3 are correct, you can deduce that the fourth digit was right as a &lt;code&gt;0&lt;/code&gt; and the last digit has to be a &lt;code&gt;1&lt;/code&gt;. You will know the correct answer after 5 guesses. It will take a sixth guess if you need to &quot;guess&quot; it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11727" LastEditorUserId="11727" LastEditDate="2015-11-12T14:42:08.733" LastActivityDate="2015-11-12T14:42:08.733" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23931" PostTypeId="2" ParentId="23927" CreationDate="2015-11-12T14:37:18.777" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 6. Start with all 0's (or 1's) and flip each bit. If the number of corrects increases, keep that bit the same, if not flip it back, and flip the next bit. If you're lucky the 5th bit was already correct which means it only took 5 guesses, but if you're still wrong on the 6th guess you know you have it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7377" LastActivityDate="2015-11-12T14:37:18.777" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23932" PostTypeId="2" ParentId="23927" CreationDate="2015-11-12T14:41:20.347" Score="15" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4 guesses to &lt;em&gt;know&lt;/em&gt; the number&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So after experimenting with some guessing patterns I came up with this one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 00000, 00011, 00101, 01001&lt;br&gt;First find out the amount of zeros, then always change two digits like I did. If the next thing A says to you is the previous result +2, both changed digits are correct. If it's the same as the previous result and both changed digits are the same, it's either 10 or 01. If it's the same as the previous result and both changed digits are &lt;em&gt;not&lt;/em&gt; the same, it's 00 or 11. If it's -2, both changed digits are wrong&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's test it out&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; with the sequence 10101&lt;br&gt;&lt;strong&gt;00000&lt;/strong&gt; - A says &quot;2&quot;, so B knows that there are 2 zeros in the sequence.&lt;br&gt;000&lt;strong&gt;11&lt;/strong&gt; - A says &quot;2&quot;, so B knows that the last two digits are either 10 or 01&lt;br&gt;00&lt;strong&gt;10&lt;/strong&gt;1 - A says &quot;4&quot;, so B knows that the last three digits are 101&lt;br&gt;0&lt;strong&gt;10&lt;/strong&gt;01 - A says &quot;2&quot;, so B knows that the second digit is not 1.&lt;br&gt; B knows the solution! If the last three digits are 101, the second digit is not 1 and there are 2 zeros in the sequence, it can just be 10101! This works with any 5 digit binary sequence&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another example&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; with the number 11001&lt;br&gt;&lt;strong&gt;00000&lt;/strong&gt; - A says &quot;2&quot;, so B knows that there are 2 zeros in the sequence&lt;br&gt;000&lt;strong&gt;11&lt;/strong&gt; - A says &quot;2&quot;, so B knows that the last two digits are either 10 or 01&lt;br&gt;00&lt;strong&gt;10&lt;/strong&gt;1 - A says &quot;2&quot;, so B knows that the third-&gt;fourth digits are either 11 or 00&lt;br&gt;0&lt;strong&gt;10&lt;/strong&gt;01 - a says &quot;4&quot;, so B knows that the second-&gt;third digits are 10. That means that the fourth digit must be 0, since the combination of 11 for the third-&gt;fourth digits isn't possible anymore. (B knows that the third digit is 0). As a consequence, the last digit must be 1, since the combination 10 for the last two digits isn't possible anymore (B knows that the fourth digits is 0). B also knows that there are only two zeros in this sequence, so the first digit must be 1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user14478" LastEditorDisplayName="user14478" LastEditDate="2015-11-12T15:47:16.950" LastActivityDate="2015-11-12T15:47:16.950" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23933" PostTypeId="2" ParentId="22781" CreationDate="2015-11-12T15:09:15.790" Score="1" Body="&lt;h1&gt;Q1&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;In this specific context I would approach the problem as follows: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(0) What is given?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;n&lt;/strong&gt; taps;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Fn&lt;/strong&gt; is a set representing the rate of flow of water, where &lt;strong&gt;Fn&lt;/strong&gt; &gt; 0 is inflow and &lt;strong&gt;Fn&lt;/strong&gt; &amp;lt; 0 is outflow;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;x&lt;/strong&gt; is the initial state;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;y&lt;/strong&gt; is the end (goal) state.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(1) What can we deduce?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;delta_xy&lt;/strong&gt; = x - y, where &lt;strong&gt;delta_xy&lt;/strong&gt; &gt; 0 means that it requires outflow; or &lt;strong&gt;delta_xy&lt;/strong&gt; &amp;lt; 0 means that it requires inflow.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(2) What is the stopping condition?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;delta_xy&lt;/strong&gt; = 0&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(3) Constraints and Edge cases: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Cannot remove more than what is contained, and cannot overflow.&lt;/li&gt;&#xA;&lt;li&gt;The set &lt;strong&gt;Fn&lt;/strong&gt; is constructed such that from &lt;strong&gt;x&lt;/strong&gt;, it is &lt;em&gt;possible&lt;/em&gt; to attain end state &lt;strong&gt;y&lt;/strong&gt;. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;i.e.: &#xA; - &lt;strong&gt;n&lt;/strong&gt; = 2&#xA; - &lt;strong&gt;Fn&lt;/strong&gt; = {-5, +5}&#xA; - &lt;strong&gt;x&lt;/strong&gt; = 3, &lt;strong&gt;y&lt;/strong&gt; = 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be obvious that given these initial conditions it is impossible to attain end state &lt;strong&gt;y&lt;/strong&gt;, given initial state &lt;strong&gt;x&lt;/strong&gt; and set &lt;strong&gt;Fn&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Algorithm&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;One could interpret this problem as a &lt;strong&gt;Subset Sum Problem&lt;/strong&gt;; where we ask the following question: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;given a set of integers&lt;/em&gt; &lt;strong&gt;Fn&lt;/strong&gt; &lt;em&gt;and an integer&lt;/em&gt; &lt;strong&gt;delta_xy&lt;/strong&gt; &lt;em&gt;, is there any non-empty subset that sums to an integer&lt;/em&gt; &lt;strong&gt;Z&lt;/strong&gt; &lt;em&gt;, such that&lt;/em&gt; &lt;strong&gt;Z&lt;/strong&gt; + &lt;strong&gt;delta_xy&lt;/strong&gt; = &lt;strong&gt;0&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiple solutions exists already for this problem, but in this particular case I would opt for a solution that employs &lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Subset_sum_problem#Pseudo-polynomial_time_dynamic_programming_solution&quot; rel=&quot;nofollow&quot;&gt;Dynamic Programming&lt;/a&gt;&lt;/strong&gt;. We define a function &lt;strong&gt;Q(i,s)&lt;/strong&gt; that returns a &lt;em&gt;boolean&lt;/em&gt; corresponding to the answer of the following question: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;is there a non-empty subset of&lt;/em&gt; &lt;strong&gt;{x_1, ... , x_i}&lt;/strong&gt; &lt;em&gt;whose sum is&lt;/em&gt; &lt;strong&gt;s&lt;/strong&gt;, where &lt;strong&gt;s&lt;/strong&gt; &amp;lt;= &lt;strong&gt;Z&lt;/strong&gt; if &lt;strong&gt;delta_xy&lt;/strong&gt; &amp;lt; 0; or &lt;strong&gt;s&lt;/strong&gt; &gt;= &lt;strong&gt;Z&lt;/strong&gt; if &lt;strong&gt;delta_xy&lt;/strong&gt; &gt; 0?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By computing this function and storing the results, as prescribed by the DP technique, eventually all the possible result of the function will be pre-computed and the complexity will become linear. As per &lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Subset_sum_problem#Pseudo-polynomial_time_dynamic_programming_solution&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;&lt;/strong&gt; article, the time complexity is &lt;strong&gt;O(sN)&lt;/strong&gt; where &lt;em&gt;s&lt;/em&gt; is the sum we are trying to find, in this case &lt;strong&gt;Z&lt;/strong&gt; which is proportional to &lt;strong&gt;delta_xy&lt;/strong&gt;; and &lt;strong&gt;N&lt;/strong&gt; is the size of the set, or the number of taps in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At every cycle (or minute) we alter the value of &lt;strong&gt;Z&lt;/strong&gt; for the next cycle, which ensure eventual completion, given that the subset respects the constraints. It is possible to analyze the subset to ensure eventual completion prior to starting the algorithm itself.&lt;/p&gt;&#xA;" OwnerUserId="17313" LastActivityDate="2015-11-12T15:09:15.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23934" PostTypeId="2" ParentId="23927" CreationDate="2015-11-12T15:10:39.500" Score="0" Body="&lt;p&gt;There are $2^5=32$ possible answers, and each time you ask you get one of 6 responses, so in theory it could be settle in two questions. But I doubt this is possible&lt;/p&gt;&#xA;" OwnerUserId="15942" LastActivityDate="2015-11-12T15:10:39.500" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23935" PostTypeId="2" ParentId="23866" CreationDate="2015-11-12T15:13:05.113" Score="0" Body="&lt;p&gt;Here's one for 12 points:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://books.google.com/ngrams/graph?content=I%2Cwould%2Cvery%2Cmuch%2Cwant%2Cdead%2Chorse%2Cears%2Catop%2CJames%27s%2Cchariot%2Carcher&amp;amp;year_start=1800&amp;amp;year_end=2000&amp;amp;corpus=15&amp;amp;smoothing=3&amp;amp;share=&amp;amp;direct_url=t1%3B%2CI%3B%2Cc0%3B.t1%3B%2Cwould%3B%2Cc0%3B.t1%3B%2Cvery%3B%2Cc0%3B.t1%3B%2Cmuch%3B%2Cc0%3B.t1%3B%2Cwant%3B%2Cc0%3B.t1%3B%2Cdead%3B%2Cc0%3B.t1%3B%2Chorse%3B%2Cc0%3B.t1%3B%2Cears%3B%2Cc0%3B.t1%3B%2Catop%3B%2Cc0%3B.t1%3B%2CJames%20%27s%3B%2Cc0%3B.t1%3B%2Cchariot%3B%2Cc0%3B.t1%3B%2Carcher%3B%2Cc0&quot; rel=&quot;nofollow&quot;&gt;I would very much want dead horse ears atop James's chariot archer&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11727" LastEditorUserId="11727" LastEditDate="2015-11-12T19:30:33.613" LastActivityDate="2015-11-12T19:30:33.613" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23936" PostTypeId="2" ParentId="23927" CreationDate="2015-11-12T15:22:06.257" Score="1" Body="&lt;p&gt;If this puzzle is to be evaluated like Mastermind (where a final &quot;guess&quot; is required to confirm the correct solution) then the answer is six, otherwise it is five - using the following approach:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First guess: all 0s (or all 1s). The response to this (number of correct digits) will indicate how many of each digit are used: 0 or 5 - all digits are the same (trivial); 1 or 4 - all digits but one are the same; 2 or 3 - 2 of one, 3 of the other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 or 4 - 2nd guess: aaabb (where a is the more common number). If the response is 4 then one of the b digits is correct - so third guess would be aaaba, and then a fourth guess (if necessary) would be aaaab; otherwise (ie. response is 3) then neither of the b digits is correct - so third guess would be abbaa;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If response to third guess is also 3, then number must be baaaa (4th guess), otherwise, fourth guess would be abaaa, and then a fifth guess (if necessary) would be aabaa.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2 or 3 - 2nd guess: aaaab (where a is the more common number). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; response is 4 then one of the b digits is the last, so third guess would be aabba; if this gave a response of 4 then a fourth guess would be aabab and a fifth guess (if necessary) would be aaabb, otherwise (ie. a response of 3) then a fourth guess would be baaab and a fifth guess (if necessary) would be abaab.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But if&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; response is 3 then neither of the b digits is the last so third guess would be aabba; if this gave a response of 3 then the fourth guess would have to be bbaaa, otherwise (ie. a response of 4) then a fourth guess would be aabaa;if this gave a response of 4 then one of the b digits would have to be in the third position, otherwise it would have to be in the fourth position; so the fifth guess would be babaa (or baaba), and if necessary the sixth guess would have to be abbaa (or ababa).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, the worst case scenario would be six guesses as the minimum necessary to determine the combination in any case.&lt;/p&gt;&#xA;" OwnerDisplayName="user9278" LastActivityDate="2015-11-12T15:22:06.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23937" PostTypeId="2" ParentId="23922" CreationDate="2015-11-12T15:49:33.293" Score="1" Body="&lt;p&gt;It could be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Post&lt;/strong&gt; - The image shows a sick spot and &quot;post&quot; is an anagram of &quot;spot&quot;&lt;br&gt;&#xA; &lt;strong&gt;No&lt;/strong&gt;&lt;br&gt;&#xA; &lt;strong&gt;Bills&lt;/strong&gt; - the pictures are of people called Bill.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It would have been seen:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; On the walls of many buildings or on hoardings where the owners don't want flyposters to stick up advertising material.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-11-12T15:49:33.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23938" PostTypeId="2" ParentId="23918" CreationDate="2015-11-12T18:14:36.483" Score="13" Body="&lt;p&gt;it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Honesty is the best policy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; on ST is the best Polly, see?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastActivityDate="2015-11-12T18:14:36.483" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23939" PostTypeId="1" AcceptedAnswerId="24071" CreationDate="2015-11-12T18:28:59.787" Score="19" ViewCount="431" Body="&lt;ul&gt;&#xA;&lt;li&gt;awaited is for A&lt;/li&gt;&#xA;&lt;li&gt;multidimensional is for B&lt;/li&gt;&#xA;&lt;li&gt;terrified is for C&lt;/li&gt;&#xA;&lt;li&gt;friendship is for D&lt;/li&gt;&#xA;&lt;li&gt;hungarian is for E&lt;/li&gt;&#xA;&lt;li&gt;visiting is for F&lt;/li&gt;&#xA;&lt;li&gt;theatrical is for G&lt;/li&gt;&#xA;&lt;li&gt;indivisible is for H&lt;/li&gt;&#xA;&lt;li&gt;delicious is for I&lt;/li&gt;&#xA;&lt;li&gt;distasteful is for J&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What are &lt;strong&gt;footprint&lt;/strong&gt;, &lt;strong&gt;criticism&lt;/strong&gt; and &lt;strong&gt;maturity&lt;/strong&gt; for?&lt;/p&gt;&#xA;" OwnerUserId="9125" LastActivityDate="2015-11-19T23:03:31.733" Title="Awaited is for A" Tags="|word|pattern|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23940" PostTypeId="2" ParentId="23916" CreationDate="2015-11-12T19:30:50.287" Score="0" Body="&lt;p&gt;What you've asked for here is similar to an &lt;a href=&quot;http://www.jdawiseman.com/papers/tournaments/individual-pairs/individual-pairs.html&quot; rel=&quot;nofollow&quot;&gt;Individual-Pairs&lt;/a&gt; tournament.  This link has many schedules for varying number of players if you are interested.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 6 person schedule is asymmetric, meaning some people play an unfair number of times.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The schedules proposed so far are not strictly &quot;Individual-Pairs&quot; tournaments since they don't meet the critia:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;each partners each other once&lt;/li&gt;&#xA;&lt;li&gt;each opposes each other twice&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the solutions posted so far, people parter with each other twice, and play against each other four times.  However, these solutions are fair and given that you accepted the answer, this is probably what you are looking for.  However, for some numbers of people (4, 5, 8, 9, 12, 13, etc) there exist very nice minimal tournaments which are quite handy.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="1592" LastEditDate="2015-11-12T19:41:09.833" LastActivityDate="2015-11-12T19:41:09.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23941" PostTypeId="1" AcceptedAnswerId="24070" CreationDate="2015-11-12T19:31:24.870" Score="11" ViewCount="797" Body="&lt;p&gt;Wanted to try out a new puzzle.  Doesn't seem like an original idea, so the concept may already be out there.  Please let me know if so and I'll put a link back to it.  Also, not sure if it will be too easy or too hard, so please add your comments below. &lt;em&gt;(Edit: I've made a second such puzzle &lt;a href=&quot;https://puzzling.stackexchange.com/questions/24094/bridge-words-2-word-pairs-linked-by-their-front-and-back&quot;&gt;here&lt;/a&gt;)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Find a pair of &quot;bridge words&quot; to fill in the gaps of the sentences below.  Bridge words are words where the last $n\geq3$ letters of the first word are the same as the first $n$ letters of the second word.  The words must then both be more than $n$ letters long.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;br&gt;&#xA;Q: I hurled away boredom.  I flung away drabness.  I _______ _______.&lt;br&gt;&#xA;A: I catapul&lt;strong&gt;ted&lt;/strong&gt; &lt;strong&gt;ted&lt;/strong&gt;iousness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, here goes:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Sometimes we torture people.  Sometimes we don't.  We ______ our ______.&lt;/li&gt;&#xA;&lt;li&gt;I object to objects.  In fact, I have a ______ for _______.&lt;/li&gt;&#xA;&lt;li&gt;I measured our location through the _______, while keeping track of our distance on the _______.&lt;/li&gt;&#xA;&lt;li&gt;The trouble with _________ _________, is that kids and drunkards end up dancing on them.&lt;/li&gt;&#xA;&lt;li&gt;Despite his __________ tone, my sense were all ______.&lt;/li&gt;&#xA;&lt;li&gt;The runny-nosed trebuchet operating was basically a ________ ________.&lt;/li&gt;&#xA;&lt;li&gt;It's much smaller, but leaves a remainder.  It's an ______ _______. (This is quite a technical word)&lt;/li&gt;&#xA;&lt;li&gt;I don't care about the strength.  _________ is _________.&lt;/li&gt;&#xA;&lt;li&gt;If ________ your _______ is confirmed, you may be eligible to become king!&lt;/li&gt;&#xA;&lt;li&gt;They punched each other without speaking.  The _______ were rather _____.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $n\geq4$ and mostly $n=5$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2015-11-20T19:58:04.983" Title="Bridge words - word pairs linked by the front and back" Tags="|wordplay|english|" AnswerCount="4" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="23942" PostTypeId="2" ParentId="23927" CreationDate="2015-11-12T20:17:03.547" Score="18" Body="&lt;p&gt;We first present a complete description and analysis of an approach that &#xA;takes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4 guesses to &lt;strong&gt;know&lt;/strong&gt; the correct number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Furthermore, we will show that with fewer guesses one cannot guarantee success.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Description and analysis of the algorithm&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;With a &lt;strong&gt;first guess 00000&lt;/strong&gt;, the six possible answers are as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Score 0: Then the hidden number is 11111. Done.&lt;/li&gt;&#xA;&lt;li&gt;Score 1: Then we succeed with a total of four guesses (see Case A below).&lt;/li&gt;&#xA;&lt;li&gt;Score 2: Then we succeed with a total of four guesses (see Case B below).&lt;/li&gt;&#xA;&lt;li&gt;Score 3: This case is symmetric to Case B.&lt;/li&gt;&#xA;&lt;li&gt;Score 4: This case is symmetric to Case A.&lt;/li&gt;&#xA;&lt;li&gt;Score 5: Then the hidden number is 00000. Done.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Case A: The guess 00000 has a score of 1.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The hidden number consists of four 1s and a single 0. Our second guess is 00111. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the score is 4, we know that the hidden number is 01111 or 10111. &#xA;We succeed with one additional guess.&lt;/li&gt;&#xA;&lt;li&gt;If the score is 2, we know that the hidden number is 11011, 11101 or 11110.&#xA;We succeed with two additonal guesses.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Altogether, this yields at most four guesses in case A.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Case B: The guess 00000 has a score of 2.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The hidden number consists of three 1s and two 0s. We guess 11000. &#xA;Let us distinguish three situations on the first two positions of the hidden number.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the hidden number is 11xxx, then the score of 11000 equals 4.&lt;/li&gt;&#xA;&lt;li&gt;If the hidden number is 10xxx or 01xxx, then the score of 11000 equals 2.&lt;/li&gt;&#xA;&lt;li&gt;If the hidden number is 00xxx, then the score of 11000 equals 0.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This implies the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the score of 11000 is 0, we know that the hidden number is 00111. Done.&lt;/li&gt;&#xA;&lt;li&gt;If the score of 11000 is 4, we know that the hidden number is 11100, 11010 or 11001. We succeed with two additonal guesses.&lt;/li&gt;&#xA;&lt;li&gt;If the score of 11000 is 2, we know that the hidden number starts with 10 or 01, and ends with 011 or 101 or 110. This is the only remaining case (***).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In this only remaining case (***), our third and fourth guesses are 01011 and 01101. This yields the scores in the following table:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                |  10011  10101  10110  01011  01101  01110&#xA;  ---------------------------------------------------------&#xA;  guess 01011:  |    3      1      1      5      3      3&#xA;  guess 01101:  |    1      3      1      3      5      3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We see that every column occurs only once, and hence allows us to correctly&#xA;identify the hidden number.&#xA;Altogether, this yields at most four guesses in each of the subcases of case B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Lower bound argument:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Independently of the structure of the first guess, player A announces a score of 2. This then leaves player B with exactly 10 possible numbers (as the hidden number must agree with the guess in two places and disagree in the other three places).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the crucial observation: whenever player B announces his guess from now on, then the &lt;strong&gt;parity&lt;/strong&gt; (odd/even) of the answer of A is already pre-determined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the spectrum of possible answers is either $\{0,2,4\}$ or $\{1,3,5\}$.&#xA;By using two more guesses, this only yields &lt;strong&gt;nine&lt;/strong&gt; distinct possibilities for the answers, and this is not enough to distinguish between &lt;strong&gt;ten&lt;/strong&gt; possible cases. Therefore three guesses are not enough.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-11-12T20:35:27.333" LastActivityDate="2015-11-12T20:35:27.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23943" PostTypeId="2" ParentId="23941" CreationDate="2015-11-12T20:17:53.323" Score="4" Body="&lt;p&gt;1.Sometimes we torture people. Sometimes we don't. We ______ our ______.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2.I object to objects. In fact, I have a ______ for _______.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; loathing/things&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3.I measured our location through the _______, while keeping track of our distance on the _______.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4.The trouble with _________ _________, is that kids and drunkards end up dancing on them.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; portable/tables - Thanks to @GentlePurpleRain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5.Despite his __________ tone, my sense were all ______.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; disbelieving/ingrained&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6.The runny-nosed trebuchet operating was basically a ________ ________.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; snuffling flinger&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;7.It's much smaller, but leaves a remainder. It's an ______ _______.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; piddling lingerer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;8.I don't care about the strength. _________ is _________.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;9.If ________ your _______ is confirmed, you may be eligible to become king!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; perchance/ancestry&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;10.They punched each other without speaking. The _______ were rather _____.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; fighters/terse&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14233" LastEditorUserId="4421" LastEditDate="2015-11-19T21:53:19.003" LastActivityDate="2015-11-19T21:53:19.003" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="23944" PostTypeId="1" CreationDate="2015-11-12T20:26:33.070" Score="36" ViewCount="3940" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;sup&gt;&lt;em&gt;This is the first in what will hopefully be a series of &lt;strong&gt;Unsolved Mysteries&lt;/strong&gt; posts.&lt;/em&gt;&#xA;  &lt;/sup&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;sup&gt;&#xA;  &lt;em&gt;Note that &lt;strong&gt;this puzzle has no known solution&lt;/strong&gt;, nor any proof that a solution is impossible.  We will see how smart the denizens of Puzzling.SE actually are...!&lt;/em&gt;&#xA;  &lt;/sup&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Most people are familiar with the concept of a &lt;a href=&quot;https://en.wikipedia.org/wiki/Magic_square&quot;&gt;Magic Square&lt;/a&gt;.  (If not, follow the link to read up on it.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are algorithms available that make it trivial to construct a magic square of almost any size, but by adding a few constraints to the problem, it becomes much more challenging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following $4\times4$ magic square, where every entry is itself a square number, and the rows, columns and diagonals all sum to $8515$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}\\&#xA;68^2&amp;amp;29^2&amp;amp;41^2&amp;amp;37^2\\&#xA;17^2&amp;amp;31^2&amp;amp;79^2&amp;amp;32^2\\&#xA;59^2&amp;amp;28^2&amp;amp;23^2&amp;amp;61^2\\&#xA;11^2&amp;amp;77^2&amp;amp;8^2&amp;amp;49^2&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note that&lt;/em&gt;&lt;br&gt;&#xA;$68^2 + 29^2 + 41^2 + 37^2 = 17^2 + 31^2 + 79^2 + 32^2$&lt;br&gt;&#xA;&lt;em&gt;but&lt;/em&gt;&lt;br&gt;&#xA;$68 + 29 + 41 + 37 \ne 17 + 31 + 79 + 32$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Only the squared values have the properties of a magic square.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many such $4\times4$ squares have been constructed, but as of yet, no one has succeeded in constructing a $3\times3$ magic square with the same property, nor in proving that no such magic square exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Your challenge&lt;/em&gt;&lt;/strong&gt;, therefore, is as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;A)&lt;/strong&gt; Build a $3\times3$ magic square where each of the nine entries in the square is itself a square number.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;&lt;em&gt;or&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;B)&lt;/strong&gt; Prove that no such square exists.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For the pedantic among us (you know who you are), here are a few additional constraints:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Each entry in the square must be unique.  (A square consisting entirely of $4$s &lt;em&gt;is not valid&lt;/em&gt;.)&lt;/li&gt;&#xA;&lt;li&gt;The definition of &quot;square number&quot; implies this, but I will spell it out here for those who like to quibble: The entries (before squaring) must be integers.  Thus a magic square using values $\{ \sqrt1^2, \sqrt2^2, \sqrt3^2,  \sqrt4^2, \sqrt5^2,  \sqrt6^2,  \sqrt7^2, \sqrt8^2, \sqrt9^2\}$ &lt;em&gt;is not valid&lt;/em&gt; (although, of course, $\sqrt1^2$, $\sqrt4^2$, and $\sqrt9^2$ can be used in a square, being proper square numbers ($=1^2, 2^2, 3^2$).&lt;/li&gt;&#xA;&lt;li&gt;This also means that using complex numbers, limits, representations of infinity, or any other abstract mathematical concept &lt;em&gt;is not valid&lt;/em&gt;.  The intent of the question is obvious; please stick to that.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="1681" LastEditDate="2016-01-06T22:10:40.187" LastActivityDate="2018-10-22T15:48:16.283" Title="Unsolved Mysteries: Magic Square of Squares" Tags="|mathematics|magic-square|real|unsolved-mysteries|" AnswerCount="3" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="23945" PostTypeId="1" AcceptedAnswerId="23949" CreationDate="2015-11-12T22:39:07.120" Score="6" ViewCount="734" Body="&lt;p&gt;&lt;sub&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/23959/mtg-challenge-2-two-peas-in-a-pod&quot;&gt;Next Challenge&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;BACKGROUND:&lt;/strong&gt;&lt;br&gt;&#xA;Here's a puzzle for Magic: the Gathering players out there. I feel that this puzzle is in the scope of the site, and is no different than other puzzles, e.g. chess, which require knowledge of the parameters of a game in order to solve. I'm adding the MtG tag since I hope to do more of these in the future. I've linked to cards so that you don't have to go look them up yourself. For those interested in learning the rules of Magic, see &lt;a href=&quot;http://magic.wizards.com/en/game-info/gameplay/how-to-play&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Magic:_The_Gathering&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This first one will be reasonably easy; I've tried to ensure that there is only one correct answer. Please don't post an answer unless it's distinct i.e. isn't just another answer but with mana tapped slightly differently or two spells cast in swapped orders if it doesn't affect the outcome. If I missed something, and there are multiple &lt;em&gt;distinct&lt;/em&gt; solutions, I will accept the one that gets the most damage through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PUZZLE SETUP:&lt;/strong&gt;&lt;br&gt;&#xA;It is Main Phase 1 of your turn. You've just drawn and haven't played a land for the turn. Defeat your opponent during this turn (before end of cleanup step). Provide the sequence of spells/game actions that you take to achieve the goal. Please be specific about mana usage / land tapping (feel free to keep a count of mana between steps)&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Your hand:&lt;/h3&gt;  &#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=233056&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gitaxian Probe&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=233056&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gitaxian Probe&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=50145&quot; rel=&quot;nofollow noreferrer&quot;&gt;Serum Visions&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=402052&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sure Strike&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=380523&quot; rel=&quot;nofollow noreferrer&quot;&gt;Twinflame&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Your board:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=292742&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kiln Fiend&lt;/a&gt; (no summoning sickness)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 Island, 2 Mountain all untapped&lt;/p&gt;&#xA;&#xA;&lt;p&gt;10 life&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Your opponent's hand:&lt;/h3&gt;  &#xA;&#xA;&lt;p&gt;Nothing&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Your opponent's board:&lt;/h3&gt;  &#xA;&#xA;&lt;p&gt;Nothing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;50 life&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Your Library (top to bottom):&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=194119&quot; rel=&quot;nofollow noreferrer&quot;&gt;Assault Strobe&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=397729&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brute Force&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=397722&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lightning Bolt&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=370568&quot; rel=&quot;nofollow noreferrer&quot;&gt;Manamorphose&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=405347&quot; rel=&quot;nofollow noreferrer&quot;&gt;Preordain&lt;/a&gt;&lt;br&gt;&#xA;Island&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=405158&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brainstorm&lt;/a&gt;&lt;br&gt;&#xA;Mountain&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=380203&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thought Scour&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=370409&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lava Spike&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=370472&quot; rel=&quot;nofollow noreferrer&quot;&gt;Grapeshot&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-11-16T16:34:48.733" Title="&quot;Magic: the Gathering&quot; Challenge #1: A Fiendish Puzzle" Tags="|calculation-puzzle|magic-the-gathering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23946" PostTypeId="5" CreationDate="2015-11-12T22:40:31.110" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-11-12T22:40:31.110" LastActivityDate="2015-11-12T22:40:31.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23947" PostTypeId="4" CreationDate="2015-11-12T22:40:31.110" Score="0" Body="For puzzles related to Magic: the Gathering trading card game. For example, these could involve challenges like winning from a given board state, or puzzles related to a specific card or deck." OwnerUserId="13970" LastEditorUserId="13970" LastEditDate="2015-11-13T02:13:23.210" LastActivityDate="2015-11-13T02:13:23.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23949" PostTypeId="2" ParentId="23945" CreationDate="2015-11-13T00:06:45.227" Score="5" Body="&lt;p&gt;This is essentially the same as the other answer, but using both Probes and with Brute Force instead of Sure Strike.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Serum Visions (Draw Assault Strobe, put Lightning Bolt on bottom, leave Brute Force on top. RR)&lt;/li&gt;&#xA;&lt;li&gt;Gitaxian Probe (Draw Brute Force. Pay life. RR)&lt;/li&gt;&#xA;&lt;li&gt;Gitaxian Probe (Draw Manamorphose. Pay life. RR)&lt;/li&gt;&#xA;&lt;li&gt;Manamorphose (Draw Preordain, create UR. UR)&lt;/li&gt;&#xA;&lt;li&gt;Preordain (Put Brainstorm and Island on bottom, draw Mountain. R)&lt;/li&gt;&#xA;&lt;li&gt;Mountain (RR)&lt;/li&gt;&#xA;&lt;li&gt;Brute Force (R)&lt;/li&gt;&#xA;&lt;li&gt;Assault Strobe&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;1+21 from seven spells, and another +3 from Brute Force, x2 makes 50 damage.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-11-13T00:06:45.227" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23951" PostTypeId="1" AcceptedAnswerId="23955" CreationDate="2015-11-13T08:46:03.710" Score="2" ViewCount="275" Body="&lt;p&gt;Given two segments, construct, using compass and straightedge, two other segments such that the lengths of given segments represent arithmetic and geometric means of the lengths of constructed segments.&lt;/p&gt;&#xA;" OwnerUserId="7668" LastActivityDate="2015-11-13T12:02:57.023" Title="Compass-and-straightedge construction of two segments, given their arithmetic and geometric means" Tags="|construction|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23952" PostTypeId="1" AcceptedAnswerId="23953" CreationDate="2015-11-13T10:13:02.800" Score="6" ViewCount="364" Body="&lt;p&gt;On the table lies a rectangular piece of paper $ABCD$ of area $100$. &#xA;Cosmo folds the rectangle once along a straight line, so that afterwards &#xA;corner $C$ lies exactly on top of corner $A$. &#xA;The folded piece of paper forms a pentagon.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What is the largest possible area of this pentagon?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-11-13T13:49:59.970" Title="Folding a piece of paper" Tags="|mathematics|geometry|paper-folding|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23953" PostTypeId="2" ParentId="23952" CreationDate="2015-11-13T10:36:41.680" Score="6" Body="&lt;p&gt;There is no largest area, but the area can be made as close as wanted to 75.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Let's imagine our rectangle. It has diagonal $AC$, centre $O$; let's say $AB$ is the longest side and $H$ is the middle of $AB$.&lt;/li&gt;&#xA;&lt;li&gt;The rectangle will be folded along a line $p$, which is perpendicular to $AC$ at point $O$. $p$ also crosses $AB$ at some point $M$.  &lt;/li&gt;&#xA;&lt;li&gt;It is easy to note that when we fold the overlapped area will be of size $2*Size(AOM)$. The pentagon area is $Sp = 100-2*Size(AOM)$. In order to achieve biggest pentagon we need to make overlap as small as possible.  &lt;/li&gt;&#xA;&lt;li&gt;But, $AOM$ always includes $AOH$ and $Size(AOH) = \frac{100}{8}$. Thereby $Sp = 100-2*Size(AOM) &amp;lt; 100-\frac{100}{4} = 75$.  &lt;/li&gt;&#xA;&lt;li&gt;Finally, let's note that making $\frac{AB}{AD}$ ratio big enough we can make point $M$ as close as needed to point $H$ making $Size(AOM)$ as close as needed to $\frac{100}{8}$ and thereby $Sp$ to 75. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="28" LastEditorUserId="2806" LastEditDate="2015-11-13T13:49:59.970" LastActivityDate="2015-11-13T13:49:59.970" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23954" PostTypeId="2" ParentId="6079" CreationDate="2015-11-13T11:04:00.803" Score="16" Body="&lt;p&gt;I just figured i'd add another reasoning, because i knew without comparing any 2 peoples answers against eachother.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everyone except Karin makes the statements: &quot;It wasn't me&quot; and &quot;It was someone else&quot; This rules all of them out automatically, since if they did do it, both would be false. So it has to be Karin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16846" LastActivityDate="2015-11-13T11:04:00.803" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23955" PostTypeId="2" ParentId="23951" CreationDate="2015-11-13T11:57:16.453" Score="1" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8EmGo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Set your compass to the length of the arithmetic mean ($m$) and draw a circle with radius $m$.  Call the center $O$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Draw a radius $OA$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Draw a perpendicular radius $OB$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Set your compass to the length of the geometric mean ($g$) and mark a point $C$ on the line $OB$, such that $\overline{OC}=g$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Draw a line perpendicular to $OB$ through $C$.  Call its intersection with the circle $D$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the circle has radius $m$, we can write $\overline{OC}^2 + \overline{CD}^2 = m^2$.  Since $\overline{OC}=g$, we then have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;   g^2 + \overline{CD}^2 = m^2 \\&#xA;   \overline{CD} = \sqrt{m^2 - g^2}&#xA;   $$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Set your compass to the length of $CD$ and draw another circle centered on $O$.  Call its intersections with $OA$ points $E$ and $F$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;We now have two segments of interest, $AE$ and $AF$.  Their lengths are, respectively:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;\overline{AE} &amp;amp;= m - \sqrt{m^2 - g^2} \\&#xA;\overline{AF} &amp;amp;= m + \sqrt{m^2 - g^2} \\&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculating their arithmetic and geometric means gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;\frac{\overline{AE}+\overline{AF}}{2} &amp;amp;= \frac{m - \sqrt{m^2 - g^2} + m + \sqrt{m^2 - g^2}}{2} \\&#xA;&amp;amp;= \frac{m + m}{2} \\ &amp;amp;= m \\&#xA;\sqrt{\left(\overline{AE}\right)\left(\overline{AF}\right)} &amp;amp;= \sqrt{\left(m - \sqrt{m^2 - g^2}\right)\left(m + \sqrt{m^2 - g^2}\right)} \\&#xA;&amp;amp;= \sqrt{m^2 - \left(\sqrt{m^2 - g^2}\right)^2} \\&#xA;&amp;amp;= \sqrt{m^2 - \left(m^2 - g^2\right)} \\&#xA;&amp;amp;= \sqrt{g^2} \\&#xA;&amp;amp;= g&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, as desired, we have constructed two segments $AE$ and $AF$ whose arithmetic mean is $m$ and geometric mean is $g$.  QED&lt;/p&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="9718" LastEditDate="2015-11-13T12:02:57.023" LastActivityDate="2015-11-13T12:02:57.023" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23956" PostTypeId="5" CreationDate="2015-11-13T13:20:16.637" Score="0" Body="&lt;p&gt;For puzzles related to Magic: the Gathering trading card game. For example, these could involve challenges like winning from a given board state, or puzzles related to a specific card or deck. General Magic: The Gathering questions are off-topic but can be asked on our sister site &lt;a href=&quot;https://boardgames.stackexchange.com/&quot;&gt;Board &amp;amp; Card Games Stack Exchange&lt;/a&gt; using the &lt;a href=&quot;/questions/tagged/magic-the-gathering&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;magic-the-gathering&amp;#39;&quot; rel=&quot;tag&quot;&gt;magic-the-gathering&lt;/a&gt; tag.&lt;/p&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="69582" LastEditDate="2021-02-17T00:10:04.640" LastActivityDate="2021-02-17T00:10:04.640" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23957" PostTypeId="4" CreationDate="2015-11-13T13:20:16.637" Score="0" Body="For puzzles related to Magic: the Gathering trading card game. For example, these could involve challenges like winning from a given board state, or puzzles related to a specific card or deck.." OwnerUserId="4421" LastEditorUserId="69582" LastEditDate="2021-02-17T00:10:30.200" LastActivityDate="2021-02-17T00:10:30.200" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23958" PostTypeId="1" AcceptedAnswerId="24000" CreationDate="2015-11-13T13:42:49.043" Score="10" ViewCount="1186" Body="&lt;p&gt;&lt;sub&gt;This is in the spirit of the &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%22what+is+a%22+word+is%3Aquestion&quot;&gt;What is a Word/Phrase™&lt;/a&gt; series started by &lt;a href=&quot;https://puzzling.stackexchange.com/users/463&quot;&gt;JLee&lt;/a&gt; with a special brand of &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%22Phrase%22%20user%3A463%20is%3Aquestion&quot;&gt;Phrase™&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%22What+is+%22+%22Word%22+-missing+user%3A463+is%3Aquestion&quot;&gt;Word™&lt;/a&gt; puzzles.&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;If a word conforms to a special rule, I call it a Beecumming Word™.&lt;/p&gt;&#xA;&lt;p&gt;Use the following examples below to find the rule.&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;Beecumming Words™&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;Not Beecumming Words™&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;CENTAUR&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;JACKALOPE&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;EDUCATING&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;INFORMING&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;GLACIER&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;ICEBERG&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;INTEGRALS&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;DERIVATIVES&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;MASTERING&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;PRACTICING&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;MIRACLES&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;WONDERS&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;MISCREANT&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;VAGABOND&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;SCREWING&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;HAMMERING&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;SECLUDED&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;HIDDEN&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;STRANGLED&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;CHOKED&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;TRIANGLE&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;HEXAGON&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;WETLANDS&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;SWAMP&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;WILDCAT&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;GRIZZLY&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;WRESTLING&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;FIGHTING&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;And, if you want to analyze, here is a CSV version:&lt;/p&gt;&#xA;&lt;pre&gt;&#xA;Beecumming Words™,Not Beecumming Words™&#xA;CENTAUR,JACKALOPE&#xA;EDUCATING,INFORMING&#xA;GLACIER,ICEBERG&#xA;INTEGRALS,DERIVATIVES&#xA;MASTERING,PRACTICING&#xA;MIRACLES,WONDERS&#xA;MISCREANT,VAGABOND&#xA;SCREWING,HAMMERING&#xA;SECLUDED,HIDDEN&#xA;STRANGLED,CHOKED&#xA;TRIANGLE,HEXAGON&#xA;WETLANDS,SWAMP&#xA;WILDCAT,GRIZZLY&#xA;WRESTLING,FIGHTING&#xA;&lt;/pre&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Things that are not relevant include:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;capitalization&lt;/li&gt;&#xA;&lt;li&gt;whether &amp;quot;y&amp;quot; is a consonant or a vowel&lt;/li&gt;&#xA;&lt;li&gt;any sort of mathematical computation (the connection is a single step)&lt;/li&gt;&#xA;&lt;li&gt;pronunciation (e.g. &amp;quot;WURSTGILEDMARST&amp;quot; abides by the rule)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Some more words that fail the rule include &amp;quot;BEECUMMING&amp;quot;, &amp;quot;question_asker&amp;quot;, and &amp;quot;Engineer Toast&amp;quot;. The title's spelling is intentional and SFW.&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2023-12-30T15:46:26.587" LastActivityDate="2023-12-30T15:46:26.587" Title="What is a Beecumming Word™?" Tags="|pattern|word-property|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23959" PostTypeId="1" AcceptedAnswerId="23962" CreationDate="2015-11-13T14:11:40.473" Score="16" ViewCount="946" Body="&lt;p&gt;&lt;sub&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/23945/magic-the-gathering-challenge-1-a-fiendish-puzzle&quot;&gt;Previous Challenge&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/23996/magic-the-gathering-challenge-3-legends-rule&quot;&gt;Next Challenge&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;BACKGROUND:&lt;/strong&gt;&lt;br&gt;&#xA;Here's a puzzle for Magic: the Gathering players out there. I feel that this puzzle is in the scope of the site, and is no different than other puzzles, e.g. chess, which require knowledge of the parameters of a game in order to solve. I'm adding the MtG tag since I hope to do more of these in the future. I've linked to cards so that you don't have to go look them up yourself. For those interested in learning the rules of Magic, see &lt;a href=&quot;https://en.wikipedia.org/wiki/Magic:_The_Gathering&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://magic.wizards.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried to ensure that there is only one correct answer, but there are so many permutations, I may have missed one. Please don't post an answer unless it's distinct i.e. isn't just another answer but with mana tapped slightly differently or two spells cast in swapped orders if it doesn't affect the outcome. If I missed something, and there are multiple &lt;em&gt;distinct&lt;/em&gt; solutions, I will accept the one that gets the most damage through. Your solution cannot involve opponent's cooperation (e.g. they choose not to block any of your creatures) in other words, it must be a guaranteed win regardless of what your opponent can do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PUZZLE SETUP:&lt;/strong&gt;&lt;br&gt;&#xA;It is Main Phase 1 of your turn. You have played your land for the turn. No cards have been put into either graveyard so far this turn. Defeat your opponent during this turn (before end of cleanup step). Provide the sequence of spells/game actions that you take to achieve the goal. Please be specific about mana usage / land tapping (feel free to keep a count of mana between steps)&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Your hand:&lt;/h3&gt;  &#xA;&#xA;&lt;p&gt;Nothing&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Your board (all untapped):&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=218006&quot; rel=&quot;nofollow noreferrer&quot;&gt;Birthing Pod&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=206332&quot; rel=&quot;nofollow noreferrer&quot;&gt;Voltaic Key&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=394381&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kiora's Follower&lt;/a&gt; (no summoning sickness)&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=288937&quot; rel=&quot;nofollow noreferrer&quot;&gt;Omniscience&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 Forest, 6 Plains&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 life&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Your graveyard:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=376301&quot; rel=&quot;nofollow noreferrer&quot;&gt;Deceiver Exarch&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=398579&quot; rel=&quot;nofollow noreferrer&quot;&gt;Seismic Elemental&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=83282&quot; rel=&quot;nofollow noreferrer&quot;&gt;Storm Crow&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Your opponent's hand:&lt;/h3&gt;  &#xA;&#xA;&lt;p&gt;Nothing&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Your opponent's board (all untapped):&lt;/h3&gt;  &#xA;&#xA;&lt;p&gt;3 copies of &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=389599&quot; rel=&quot;nofollow noreferrer&quot;&gt;Morkrut Banshee&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Your opponent's graveyard:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=389540&quot; rel=&quot;nofollow noreferrer&quot;&gt;Grave Titan&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Original Stipulation: 9 life&lt;br&gt;&#xA;&lt;em&gt;I actually missed a line that leads to being able to do 20 damage!&lt;/em&gt;&lt;br&gt;&#xA;New Stipulation: 20 life!&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Your Library:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Forest&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=233068&quot; rel=&quot;nofollow noreferrer&quot;&gt;Blade Splicer&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=202575&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fire Imp&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=221201&quot; rel=&quot;nofollow noreferrer&quot;&gt;Goblin Chieftain&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=386327&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hellraiser Goblin&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=370440&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pestermite&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=405454&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wood Elves&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=108835&quot; rel=&quot;nofollow noreferrer&quot;&gt;Avalanche Riders&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=393837&quot; rel=&quot;nofollow noreferrer&quot;&gt;Flametongue Kavu&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=214064&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hero of Bladehold&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=214028&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hero of Oxid Ridge&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=262875&quot; rel=&quot;nofollow noreferrer&quot;&gt;Huntmaster of the Fells&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=368530&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hellrider&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=205075&quot; rel=&quot;nofollow noreferrer&quot;&gt;Obstinate Baloth&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=389615&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nekrataal&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=382983&quot; rel=&quot;nofollow noreferrer&quot;&gt;Karmic Guide&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=389599&quot; rel=&quot;nofollow noreferrer&quot;&gt;Morkrut Banshee&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=397839&quot; rel=&quot;nofollow noreferrer&quot;&gt;Puppeteer Clique&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=370493&quot; rel=&quot;nofollow noreferrer&quot;&gt;Reveillark&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=249685&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thragtusk&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=382996&quot; rel=&quot;nofollow noreferrer&quot;&gt;Laquatus's Champion&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=214044&quot; rel=&quot;nofollow noreferrer&quot;&gt;Massacre Wurm&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=405400&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sun Titan&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=394062&quot; rel=&quot;nofollow noreferrer&quot;&gt;Twilight Shepherd&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=397880&quot; rel=&quot;nofollow noreferrer&quot;&gt;Elesh Norn, Grand Cenobite&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=240027&quot; rel=&quot;nofollow noreferrer&quot;&gt;Craterhoof Behemoth&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-11-16T18:04:23.397" Title="&quot;Magic: the Gathering&quot; Challenge #2: Two Peas in a Pod" Tags="|calculation-puzzle|magic-the-gathering|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="23962" PostTypeId="2" ParentId="23959" CreationDate="2015-11-13T17:42:32.090" Score="9" Body="&lt;ul&gt;&#xA;&lt;li&gt;Tap Forest, float a green. &lt;/li&gt;&#xA;&lt;li&gt;Kiora's Followers Activate targeting Forest. &lt;/li&gt;&#xA;&lt;li&gt;Tap Plains, use floating green to pod Kiora’s Followers - &gt;&#xA;Pestermite &lt;/li&gt;&#xA;&lt;li&gt;Pestermite trigger targetting Birthing Pod &lt;/li&gt;&#xA;&lt;li&gt;1 Forest 5&#xA;Plains 1 Life &lt;/li&gt;&#xA;&lt;li&gt;Tap Forest and Plains, pod Pestermite - &gt; Obstinate&#xA;Baloth &lt;/li&gt;&#xA;&lt;li&gt;4 Plains 5 Life &lt;/li&gt;&#xA;&lt;li&gt;Voltaic Key targetting Birthing Pod &lt;/li&gt;&#xA;&lt;li&gt;3 Plains 5 Life&lt;/li&gt;&#xA;&lt;li&gt;Tap plains, pay 2 life, pod Obstinate Baloth - &gt; Karmic Guide&lt;/li&gt;&#xA;&lt;li&gt;Karmic Guide Returns Deceiver Exarch to play.&lt;/li&gt;&#xA;&lt;li&gt;Deceiver Exarch trigger targetting Birthing Pod &lt;/li&gt;&#xA;&lt;li&gt;2 Plains 3 Life &lt;/li&gt;&#xA;&lt;li&gt;Tap plains, pay 2 life, pod Karmic Guide - &gt; Twilight&#xA;Shepherd &lt;/li&gt;&#xA;&lt;li&gt;Cast Pestermite from Twilight Shepherd and target Birthing Pod.&lt;/li&gt;&#xA;&lt;li&gt;Cast Obstinate Baloth and Kiora’s Followers, both returned from Twilight Shepherd. &lt;/li&gt;&#xA;&lt;li&gt;Cast Karmic Guide returned from Twilight Shepherd, which reanimates Seismic Elemental&lt;/li&gt;&#xA;&lt;li&gt;1 plains, 5 life&lt;/li&gt;&#xA;&lt;li&gt;Tap plains, pay 2 life, pod Kiora’s Followers - &gt; Hellraiser Goblin &lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Swing 2/1 flyer (Pestermite), 2/1 flyer (Karmic Guide), 4/4 (Seismic Elemental, 5/5 flyer (Twilight Shepherd), 4/4 (Obstinate Baloth), 2/2 (Hell Raiser Goblin), 1/4 (Deceiver Exarch), all unable to be blocked due to seismic elemental's trigger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2 + 2 + 4 + 5 + 4 + 2 + 1 = 20 Damage&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="10904" LastEditorUserId="10904" LastEditDate="2015-11-13T17:49:17.280" LastActivityDate="2015-11-13T17:49:17.280" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23963" PostTypeId="1" AcceptedAnswerId="23964" CreationDate="2015-11-13T18:07:54.540" Score="6" ViewCount="980" Body="&lt;p&gt;My Indian friend Jamal recently participated in the well-known TV quiz&#xA;show &lt;strong&gt;&lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Who_Wants_to_Be_a_Millionaire&quot;&gt;Who wants to be a millionaire?&lt;/a&gt;&lt;/em&gt;&lt;/strong&gt;. You might have heard or read about Jamal's spectacular success; the media reported on it extensively.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what Jamal told us about the 32.000 Euro question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Oh no, the 32.000 Euro question was a sports question! The second safe haven suddenly seemed far away, perhaps out of reach!  Although I enjoy exercising myself, I have not paid much attention to professional sports.  Unfortunately, this question was not about the Raiders or the Lakers.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;At first sight, all four answers seemed perfectly acceptable. I thought to&#xA;  vaguely remember that D is particularly strong in that sport, and also has&#xA;  many good teams playing in the national competition. But I also thought to&#xA;  vaguely remember that they had become world champion the time &lt;strong&gt;before&lt;/strong&gt;,&#xA;  as at that time the final had been particularly violent.  Not that I had&#xA;  watched this final, but it just was all over the media and there was no way&#xA;  of not hearing about it.  Aha, and then I suddenly remembered that C had also&#xA;  been involved in these ruthless acts of kung fu violence.  Hence C and D had&#xA;  been the finalists &lt;strong&gt;the time before&lt;/strong&gt;, and this should have eliminated them,&#xA;  unless of course one of them managed to make it to the final again next time.&#xA;  Darn. Darn. Darn.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I read the question once again, and decided to go with A or B. I cautiously&#xA;  said: &quot;Perhaps it is answer A&quot;, and saw a lickety-split happy flash in Prem's&#xA;  eyes.  Prem wanted me to lose.  So A had to be wrong.  Yes, yes, yes, yes, yes! I went for answer B.  (After the show and only much later I found out, that A in fact had been B's opponent in the deciding match of that world championship.)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;But the only thing that really mattered was that I had picked the right answer! Prem swallowed hard, and his eyes popped out a little bit. &quot;Answer B is corrrrrrrrrrrrrrrrrrrrrrrrect, my frrrrrrrrrrrrrrrriend!&quot; This was good enough to get me to the 64.000 Euro question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;What was the 32.000 Euro question?&lt;br&gt;&#xA;What were the four possible answers A, B, C, D?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="9404" LastEditDate="2016-03-04T16:06:31.500" LastActivityDate="2016-03-04T16:06:31.500" Title="Slumdog millionaire (10)" Tags="|knowledge|sports|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23964" PostTypeId="2" ParentId="23963" CreationDate="2015-11-13T18:16:19.810" Score="10" Body="&lt;p&gt;This seems too easy to be the 32.000 Euro question.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The question was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Who won the 2014 FIFA World Cup?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer choices were:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;A)&lt;/strong&gt; Argentina&lt;br&gt;&#xA; &lt;strong&gt;B)&lt;/strong&gt; Germany&lt;br&gt;&#xA; &lt;strong&gt;C)&lt;/strong&gt; Netherlands&lt;br&gt;&#xA; &lt;strong&gt;D)&lt;/strong&gt; Spain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Justification:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Spain&lt;/strong&gt; won the 2010 FIFA World Cup, with a 1-0 victory over &lt;strong&gt;The Netherlands&lt;/strong&gt;. There was some pretty raucous violence during the game, including a couple of &quot;kung-fu kick&quot; fouls. Spain also has quite a few good football teams (in fact, football is the national sport of Spain). The 2014 World Cup championship was Germany versus Argentina, in which Germany was victorious.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2015-11-13T18:16:19.810" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23965" PostTypeId="2" ParentId="23958" CreationDate="2015-11-13T18:20:04.030" Score="5" Body="&lt;p&gt;I don't have an answer, but here's some basic analysis that may help (and may hinder):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Common letters: acdegilmnrstuw&lt;br&gt;&#xA;No letters that appear in beecumming words only&lt;br&gt;&#xA;Letters that appear in unbeecumming words only: bfhjkopvxyz&lt;br&gt;&#xA;Letters that appear in neither group: q&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I spent a while looking at the poetry of EE Cummings, but I was left none the wiser (at least for this puzzle - I'm sure the poetry was good for me in other ways)&lt;/p&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2015-11-14T01:30:11.230" LastActivityDate="2015-11-14T01:30:11.230" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23966" PostTypeId="1" AcceptedAnswerId="23968" CreationDate="2015-11-13T18:30:23.103" Score="19" ViewCount="1710" Body="&lt;p&gt;The blackboard contains 30 integer numbers from 1..2000. Each minute you are allowed to elect some positive integer number x and subtract x from some (chosen by you) of the numbers on the blackboard with the condition that the resulting numbers remain non-negative. If a number becomes 0 then it is erased from the blackboard. The game ends when all numbers are wiped out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the smallest number of minutes so that you can always win the game with arbitrary 30 integer numbers from 1..2000 on the blackboard?&lt;/p&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2015-11-13T22:25:21.250" Title="Blackboard cleaning" Tags="|mathematics|game|blackboard|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23967" PostTypeId="2" ParentId="23966" CreationDate="2015-11-13T18:44:53.423" Score="23" Body="&lt;p&gt;It can be done in:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 11 numbers (ie 10 minutes - assuming regular spacing and timer starting from the first number chosen.  If it's just &lt;em&gt;some time during&lt;/em&gt; &quot;each minute&quot;, then it's $10\pm1$ minutes)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;But how?&quot; you ask.  Well, I'm glad you asked:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Convert the numbers to binary, and subtract the bits.&#xA;  Or, to put it another way, subtract 1024 from all the numbers that can afford it.  Then subtract 512 from all the numbers that can afford it.  Then 256.  Then 128.  Etc.  Down to 1. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the best answer. See @Mike Earnest’s proof below!&lt;/p&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2015-11-13T22:25:21.250" LastActivityDate="2015-11-13T22:25:21.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23968" PostTypeId="2" ParentId="23966" CreationDate="2015-11-13T19:33:03.820" Score="26" Body="&lt;p&gt;Dr Xorile's answer of &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 11 steps&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is optimal. Suppose that the numbers 1, 2, 4, 8, ... 512, and 1024 are all on the board. In order to eliminate the number $2^i$, there must be some step where you decrease $2^i$, but none of the numbers $2^0,2^1,\dots, 2^{i-1}$. If this were &lt;em&gt;not&lt;/em&gt; true, then the amount you decreased $2^i$ would be at most the amount you decreased the numbers $2^0,2^1,\dots,2^{i-1}$, which is at most $2^0+2^1+\dots+2^{i-1}=2^i-1$, so that $2^i$ wouldn't have been eliminated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, in this case, there must be eleven distinct steps, since for each $i=0,1,\dots,10$, there must be a step where the smallest number touched was $2^i$. &lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-11-13T19:33:03.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23969" PostTypeId="2" ParentId="23963" CreationDate="2015-11-13T21:38:52.423" Score="5" Body="&lt;p&gt;My answer is very similar to Bailey M's but I had a different question in my head that also fits these circumstances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Who was the champion of the 2010 FIFA World Cup?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Choices:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A. Netherlands&lt;br&gt;&#xA; B. Spain&lt;br&gt;&#xA; C. France&lt;br&gt;&#xA; D. Italy  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rationale:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - Italy (D) is particularly strong in soccer/football/calcio/futbol behind club sides like Juventus, Inter Milan, AC Milan, and Roma.&lt;br&gt;&#xA; - They were in the 2006 FIFA World Cup Final against France (C) where Zinedine Zidane famously head-butted (also a move in kung-fu) Marco Materazzi in the chest.&lt;br&gt;&#xA; - In the 2010 final, Spain defeated the Netherlands to become champion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The OP may very well be looking for Bailey M's answer, because if I recall correctly, Nigel de Jong's kick out into Xabi Alonso was dubbed a kung-fu kick by the media. Given how general the puzzle is, my answer also fits very well. Also, for a 32k question, going back 4 more years into the past makes more sense.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17360" LastEditorUserId="17360" LastEditDate="2015-11-13T22:31:43.820" LastActivityDate="2015-11-13T22:31:43.820" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23970" PostTypeId="1" AcceptedAnswerId="23971" CreationDate="2015-11-14T15:12:59.613" Score="10" ViewCount="445" Body="&lt;p&gt;On the table, there is a balance with two pans together with ten weights &#xA;ofrespectively  1, 2, 4, 8, 16, 32, 64, 128, 256 and 512 grams. &#xA;Cosmo takes a coin out of his pocket, shows it to Fredo and says: &#xA;&lt;em&gt;&quot;There are exactly 89 different ways of placing the coin together with &#xA;some (perhaps zero) of the weights into the left pan and placing some of the &#xA;other weights into the right pan, so that the balance is in equilibrium.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What is the weight of Cosmo's coin?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-11-15T12:50:43.053" Title="Weighing in 89 different ways" Tags="|mathematics|combinatorics|weighing|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23971" PostTypeId="2" ParentId="23970" CreationDate="2015-11-14T19:49:19.267" Score="8" Body="&lt;p&gt;The weight of Cosmo's coin is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $341$, or represented as a binary number $101010101$.&lt;br&gt;&lt;br&gt;&lt;strong&gt;Carl Löndahl&lt;/strong&gt; found the second solution $171$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Step 1&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/LPQEv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LPQEv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&lt;br&gt;This image represents both sides of the scale, the colored part being the coin splitted in the corresponding weights. The part below demonstrates distribution of the weights for some possible weighings. In this case we only change the weights 256 and 512 (&quot;bits&quot; 8 and 9), and ignore changes in the rest of the bits. There are 2 possible cases. Let's call this number: $$S_1=2$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Step 2&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/eFxXm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eFxXm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&lt;br&gt;This time we look at the weights 128 and 64 (&quot;bits&quot; 6 and 7). We have again 2 possible cases. For each of the 2 cases we can choose 1 possible case from step 1 above which gives us together $2 * S_1$ cases.&lt;br&gt;&lt;br&gt;&lt;a href=&quot;https://i.stack.imgur.com/svi0Z.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/svi0Z.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&lt;br&gt;But we have 1 additional case if we use all four &quot;bits&quot; 6 to 9. This gives us a new number: $$S_2=2*S_1+1=5$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Step 3&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/oWYYU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oWYYU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&lt;br&gt;Again 2 cases, this time using bits 4 and 5. Both can be combined with all cases from step 2.&lt;br&gt;&lt;br&gt;&lt;a href=&quot;https://i.stack.imgur.com/xlvTl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xlvTl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&lt;br&gt;One additional case using bits 4 to 7 which can be combined with all cases from step 1.&lt;br&gt;&lt;br&gt;&lt;a href=&quot;https://i.stack.imgur.com/OcUAM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OcUAM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&lt;br&gt;And one more case using bits 4 to 9. This gives us a total of: $$S_3=2*S_2+S_1+1=13$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Step 4&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/WsUIf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WsUIf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&lt;br&gt;Similar to previous chapters we can define: $$S_4=2*S_3+S_2+S_1+1=34$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Step 5&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I omitted the picture here as the pattern should be obvious now: $$S_5=2*S_4+S_3+S_2+S_1+1=89$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Second solution (found by Carl Löndahl)&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The nubmer of weighings can be calculated similarly for this number.&lt;br&gt;&lt;br&gt;&lt;a href=&quot;https://i.stack.imgur.com/KB8Cg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KB8Cg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;$$S_1=3$$&lt;br&gt;&lt;a href=&quot;https://i.stack.imgur.com/7Vlsx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7Vlsx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&lt;br&gt;Here we see a difference, there are 2 cases without using bits 5 and 6, and this is also true for the following steps.$$S_2=2*S1+2=8$$ $$S_3=2*S_2+S_1+2=21$$ $$S_4=2*S_3+S_2+S_1+2=55$$ There is also a difference in the 5th step. Because bit 1 is already used we multiply with 1 instead of 2 here. $$S_5=1*S_4+S_3+S_2+S_1+2=89$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;General solution&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Based on the thoughts above we can define a general way to calculate the number of weighings. First write the number in binary form. Then define a variable for each bit equal 1 from left to right as follows:&lt;br&gt;&lt;br&gt;$x_1$ -&gt; number of weights bigger or equal the value of the first set bit&lt;br&gt;$x_2$ -&gt; number of remaining weights bigger or equal the value of the second set bit&lt;br&gt;$x_3$ -&gt; number of remaining weights bigger or equal the value of the third set bit&lt;br&gt;...&lt;br&gt;&lt;br&gt;The number can be then calculated as follows:&lt;br&gt;$$S_1 = x_1$$ $$S_2 = x_2*S_1+(x_1-1)$$ $$S_3 = x_3*S_2+(x_2-1)*S_1+(x_1-1)$$ $$S_4 = x_4*S_3+(x_3-1)*S_2+(x_2-1)*S_1+(x_1-1)$$ $$S_5 = x_5*S_4+(x_4-1)*S_3+(x_3-1)*S_2+(x_2-1)*S_1+(x_1-1)$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2015-11-15T12:50:43.053" LastActivityDate="2015-11-15T12:50:43.053" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23972" PostTypeId="1" AcceptedAnswerId="23975" CreationDate="2015-11-14T23:12:10.470" Score="5" ViewCount="2249" Body="&lt;p&gt;I found these rules for a crossword, but I don't know what type it is. Can anyone help me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rules:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;p&gt;The words are given and you're asked to place them on the grid!&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;The grid and the list of words are provided separately and are not numbered.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;The goal is to place all the words on the grid so that all the crossings of words are valid.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Usually a word is placed on the grid as a starting point for the solver.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is a &lt;a href=&quot;http://www.stayrolexakias.gr/amerikaniko&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; of this type of crossword from a Greek site.&lt;/p&gt;&#xA;" OwnerUserId="17369" LastEditorUserId="4421" LastEditDate="2015-11-15T13:55:32.550" LastActivityDate="2023-03-08T06:13:21.263" Title="Crossword Puzzles: What is this crossword type?" Tags="|logical-deduction|enigmatic-puzzle|crosswords|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23973" PostTypeId="2" ParentId="23972" CreationDate="2015-11-14T23:16:35.183" Score="7" Body="&lt;p&gt;I believe the term you're looking for is &lt;em&gt;diagramless crossword&lt;/em&gt;. &lt;a href=&quot;https://en.wikipedia.org/wiki/Crossword#Major_crossword_variants&quot;&gt;From Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In a diagramless crossword, often called a diagramless for short or, in the UK, a skeleton crossword or carte blanche, the grid offers overall dimensions, but the locations of most of the clue numbers and shaded squares are unspecified. A solver must deduce not only the answers to individual clues, but how to fit together partially built-up clumps of answers into larger clumps with properly-set shaded squares.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="39" LastActivityDate="2015-11-14T23:16:35.183" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23974" PostTypeId="2" ParentId="23972" CreationDate="2015-11-14T23:20:20.847" Score="4" Body="&lt;p&gt;This sounds like a &lt;em&gt;criss-cross&lt;/em&gt; - here's a description from the MIT Mystery Hunt puzzle archive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://devjoe.appspot.com/huntindex/keyword/crisscross&quot; rel=&quot;nofollow&quot;&gt;http://devjoe.appspot.com/huntindex/keyword/crisscross&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In its standard form, this puzzle consists of a loosely interlocking grid and a list of words to be filled in. You must use logic and elimination to figure out where to place the words. This category includes puzzles where the word list is clued rather than provided, but if the clues are associated with specific entries, this becomes a crossword puzzle instead.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2015-11-14T23:20:20.847" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23975" PostTypeId="2" ParentId="23972" CreationDate="2015-11-15T06:02:22.587" Score="3" Body="&lt;p&gt;This is called a &lt;strong&gt;fill-in&lt;/strong&gt;, &lt;strong&gt;fill-it-in&lt;/strong&gt;, or &lt;strong&gt;word fill&lt;/strong&gt; puzzle, depending on who is publishing it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A crossword grid is given, along with a list of words. The task is to fill in the grid with the given words (across and down) so that the grid is filled, and all words are used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Fill-In_(puzzle)&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt; about it. &lt;/p&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2015-11-15T06:02:22.587" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23976" PostTypeId="2" ParentId="23958" CreationDate="2015-11-15T06:43:11.990" Score="1" Body="&lt;p&gt;I think that a Beecumming word is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Used in the works of E.E. Cummings. That seems like what the OP is implying with the name Beecumming Word, but I haven't had any luck finding the list of words in several collections of his works.  I'm not sure if I'm using incorrect or incomplete sources, but so far no word from either list has appeared in my searches.  However, the implication seems so strong that I'm willing to hazard this as a guess.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14120" LastEditorUserId="14120" LastEditDate="2015-11-15T21:39:13.377" LastActivityDate="2015-11-15T21:39:13.377" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23978" PostTypeId="2" ParentId="23970" CreationDate="2015-11-15T10:17:37.357" Score="8" Body="&lt;p&gt;Turns out there are two solutions: $171$ and $341$ (unless I made a terrible mistake; I have checked this using some Python code).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;L = [1,2,4,8,16,32,64,128,256,512]&#xA;Q = [0]*(sum(L)+1)&#xA;def rec(Q, L, i, s, f):&#xA;    if i == len(L):&#xA;        if s &amp;gt; 0:&#xA;            Q[s] += 1&#xA;        if s == 171: # change to see sets for 341&#xA;            print f, sum(f)&#xA;        return&#xA;    f.append(L[i])&#xA;    rec(Q, L, i+1, s + L[i], f)&#xA;    f.pop()&#xA;    f.append(-L[i])&#xA;    rec(Q, L, i+1, s - L[i], f)&#xA;    f.pop()&#xA;    rec(Q, L, i+1, s, f)&#xA;rec(Q,L,0,0,[])&#xA;print &quot;Possible solutions:&quot;&#xA;for i in range(0,len(Q)): &#xA;    if Q[i] == 89: print i&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12901" LastEditorUserId="12901" LastEditDate="2015-11-15T10:23:34.720" LastActivityDate="2015-11-15T10:23:34.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23979" PostTypeId="2" ParentId="23941" CreationDate="2015-11-15T13:45:25.093" Score="2" Body="&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;Despite his __________ tone, my senses were all ______.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; accommodATING ... ATINGling&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="228" LastActivityDate="2015-11-15T13:45:25.093" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23980" PostTypeId="1" AcceptedAnswerId="35527" CreationDate="2015-11-15T20:22:28.233" Score="16" ViewCount="2051" Body="&lt;p&gt;My friend is studying Maritime technologies. He was given a logical test, where there was the following question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A paper message was found lying on the street near shop AKVAPLUSS with such content:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;NFITJMDSMNFCVHMNMBSERFYWVELUEKCQSJ&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Decrypt the code to discover the secret message.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The test ended a week ago, we are simply interested in the way it can be decrypted, because all the ways we tried didn't work.&lt;/li&gt;&#xA;&lt;li&gt;The question was 4th in the test if it helps.&lt;/li&gt;&#xA;&lt;li&gt;Finally the teacher answered the e-mail, but he didn't give the answer to the puzzle, however he gave a hint (how we should find out that ourselves I don't know):&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Place under the message the word &quot;akvapluss&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: Clarified puzzle and found out that teacher was made unique puzzles for each student, and my friend is from&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Latvia&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5781" LastEditorUserId="5781" LastEditDate="2016-06-09T20:52:14.890" LastActivityDate="2016-06-10T05:34:00.263" Title="Decrypt secret message from logical test" Tags="|cipher|" AnswerCount="3" CommentCount="20" ContentLicense="CC BY-SA 3.0" />
  <row Id="23981" PostTypeId="2" ParentId="22688" CreationDate="2015-11-16T07:42:39.480" Score="3" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 87&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How to find it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Label the contents of the square A / B / C / D, with A upper left, B upper right, C lower left and D lower right.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Based on these names for the spots, we find:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (A³ + D³) = (C³-B³) for all examples.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We now know: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;x³ + 25³ = 90³ - 38³, or x³ = 658503, which solves for x = 87.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2156" LastActivityDate="2015-11-16T07:42:39.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23982" PostTypeId="1" AcceptedAnswerId="24022" CreationDate="2015-11-16T09:10:15.610" Score="13" ViewCount="2428" Body="&lt;p&gt;Alice and Bob play the following game on a standard $8\times8$ chessboard.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In the very beginning, Alice picks a square on the chessboard and places a knight on this square.&lt;/li&gt;&#xA;&lt;li&gt;Then Bob and Alice alternate in moving the knight. Bob makes the first move.&#xA;The knight moves in standard chess knight-move fashion, but it must never revisit a square that it has visited in an earlier move.&lt;/li&gt;&#xA;&lt;li&gt;The game ends, once a &lt;strong&gt;player has no move left. This player wins the game&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Which player is going to win this game?&#xA;  (As usual, we assume that Alice and Bob both use optimal strategies.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="12408" LastEditDate="2016-07-25T02:40:03.970" LastActivityDate="2018-08-03T09:36:53.027" Title="The knight's game" Tags="|strategy|game|chess|checkerboard|knight-moves|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23983" PostTypeId="2" ParentId="23982" CreationDate="2015-11-16T09:38:03.913" Score="-2" Body="&lt;p&gt;Well, seemingly&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Alice will win. The knight alternates color each move. Bob always has a square more, Alice has one down since she chooses the starting square, which she can't return to.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4317" LastEditorUserId="4317" LastEditDate="2015-11-16T10:06:16.010" LastActivityDate="2015-11-16T10:06:16.010" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23984" PostTypeId="1" AcceptedAnswerId="23990" CreationDate="2015-11-16T09:38:15.247" Score="6" ViewCount="994" Body="&lt;p&gt;You are to place the numbers $1$, $2$, $3$, $4$ and five zeros in a $3 \times 3$ grid.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Do this in such a way so that the column, row, and two diagonal sums form the sequence $0, 1, 2, 3, 4, 5, 6, 7$ in some order.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I find such a square pleasing in a similar way as the 3x3 magic square.  Unfortunately, there is not a unique solution.  Kudos if you can find all solutions and prove there are no more by hand (but I don't know of an elegant way of doing that).&lt;/p&gt;&#xA;" OwnerUserId="2722" LastEditorUserId="4421" LastEditDate="2015-11-25T17:55:58.457" LastActivityDate="2015-11-25T17:55:58.457" Title="An Antimagic Square" Tags="|magic-square|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23985" PostTypeId="2" ParentId="23984" CreationDate="2015-11-16T09:51:29.383" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;010&lt;br&gt;320&lt;br&gt;400&lt;br&gt;&lt;br&gt;c3=0,r1=1,d1=2,c2=3,r3=4,r2=5,d2=6,c1=7&lt;br&gt;This can be filpped and rotated to create more solutions...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4317" LastActivityDate="2015-11-16T09:51:29.383" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23986" PostTypeId="1" AcceptedAnswerId="23988" CreationDate="2015-11-16T10:22:06.247" Score="17" ViewCount="2702" Body="&lt;p&gt;&lt;strong&gt;Guess what this rebus represents???&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/D6gaz.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/D6gaz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; It's a phrase. Not a single word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastActivityDate="2015-11-16T21:17:40.263" Title="Rebus: Don't make me angry" Tags="|lateral-thinking|rebus|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23987" PostTypeId="2" ParentId="23986" CreationDate="2015-11-16T11:33:52.503" Score="11" Body="&lt;p&gt;First part is definetely &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Goto, or go-to. Puzzling user 4394 is GOTO-0, -0 = GOTO&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second part i'm struggling a bit, but it might be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bibel Connection? Bee - Bell - Connection? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16846" LastActivityDate="2015-11-16T11:33:52.503" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23988" PostTypeId="2" ParentId="23986" CreationDate="2015-11-16T13:34:41.073" Score="31" Body="&lt;p&gt;It is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Go to Hell&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As DrunkWolf pointed out, the first part with the numbers points to GoTo0, a user on this site.  The second part works like this... start with &quot;handbell&quot;.  Remove the &quot;b&quot; (the bee image is cutout of the image), and remove the &quot;and&quot; (the AND gate image is also cutout), leaving &quot;hell&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14233" LastActivityDate="2015-11-16T13:34:41.073" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23989" PostTypeId="2" ParentId="23984" CreationDate="2015-11-16T13:51:40.047" Score="6" Body="&lt;p&gt;The unique solutions (ignoring rotations and flips) are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0 0 1&lt;br/&gt;2 0 0&lt;br/&gt;4 0 3&lt;br/&gt;-----&lt;br/&gt;0 0 3&lt;br/&gt;2 0 0&lt;br/&gt;4 0 1&lt;br/&gt;-----&lt;br/&gt;0 0 0&lt;br/&gt;0 1 3&lt;br/&gt;2 0 4&lt;br/&gt;-----&lt;br/&gt;0 0 1&lt;br/&gt;0 0 2&lt;br/&gt;4 0 3&lt;br/&gt;-----&lt;br/&gt;0 0 0&lt;br/&gt;0 2 1&lt;br/&gt;4 3 0&lt;br/&gt;-----&lt;br/&gt;0 0 0&lt;br/&gt;0 3 1&lt;br/&gt;2 4 0&lt;br/&gt;-----&lt;br/&gt;0 0 0&lt;br/&gt;0 2 4&lt;br/&gt;1 0 3&lt;br/&gt;-----&lt;br/&gt;0 0 3&lt;br/&gt;0 0 2&lt;br/&gt;4 0 1&lt;br/&gt;-----&lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Found through the method of full exhaustion.&lt;/p&gt;&#xA;" OwnerUserId="5610" LastActivityDate="2015-11-16T13:51:40.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23990" PostTypeId="2" ParentId="23984" CreationDate="2015-11-16T14:14:37.330" Score="6" Body="&lt;p&gt;For any such magic square the sum of all rows, columns and diagonals:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{x=0}^7{x}=28$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now each number is counted at least twice, once for the row and once for the column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore each number in a corner is counted one additional time for one diagonal and the number in the center is counted two additional times for both diagonals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now since every number gets counted at least two times we already have a sum of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{x=1}^4{2x}=20$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this point we try to fill in the center:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;4 in the center:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;sum of 28:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The numbers 1, 2 and 3 must be on the edges to not increase the sum.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Not a valid solution since both diagonals have a sum of 4.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3 in the center:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;sum of 26&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;2 must be in a corner while 1 and 4 are on the edges to get the sum perfect.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;All three numbers must be opposite of a 0 otherwise we have 2 lines with the 3 in the center and a 0 on both sides of it.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;By using the lines crossing the center we already get the sums: 3, 4, 5, 7.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The 2 goes to any corner.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;To get a sum of 6 the 4 must be next to the 2.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The 1 must not be next to the 2 so we dont get two sums of 3.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;One unique solution:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;240&#xA;031&#xA;000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2 in the center:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;sum of 24&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;2 possible solutions&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;4 in a corner; 1 and 3 on the edges:&lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;Again all three numbers must be opposite of a 0 for obvious reasons.&lt;/p&gt;&#xA;    &#xA;    &lt;p&gt;Lines crossing the center give the sums: 2, 3, 5, 6.&lt;/p&gt;&#xA;    &#xA;    &lt;p&gt;The 4 goes into any corner.&lt;/p&gt;&#xA;    &#xA;    &lt;p&gt;To get a sum of 7 the 3 must be next to the 4.&lt;/p&gt;&#xA;    &#xA;    &lt;p&gt;To not get another 5 the 1 must not be next to the 4.&lt;/p&gt;&#xA;    &#xA;    &lt;p&gt;One unique solution:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;430&#xA;021&#xA;000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;  &lt;/blockquote&gt;&#xA;  &#xA;  &lt;p&gt;1 and 3 in the corners; 4 on an edge:&lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;Again all three numbers must be opposite of a 0 for obvious reasons.&lt;/p&gt;&#xA;    &#xA;    &lt;p&gt;Lines crossing the center give the sums: 2, 3, 5, 6.&lt;/p&gt;&#xA;    &#xA;    &lt;p&gt;4 goes onto any edge.&lt;/p&gt;&#xA;    &#xA;    &lt;p&gt;3 must be next to 4 to get a sum of 7.&lt;/p&gt;&#xA;    &#xA;    &lt;p&gt;1 must be next to 3 but not next to 4. (get one sum of 4 and don't change the 7)&lt;/p&gt;&#xA;    &#xA;    &lt;p&gt;One unique solution:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;043&#xA;020&#xA;001&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1 in the center:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;sum of 22&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;4 and 2 must be in a corner while 3 is on an edge.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Again all three numbers must be opposite of a 0 for obvious reasons.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Lines crossing the center give the sums: 1, 3, 4, 5.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;3 goes to any edge.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;4 must be next to 3 to get a sum of 7.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;2 must be next to 4 but not next to 3. (get 6 don't change 7)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;One unique solution:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;034&#xA;010&#xA;002&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;0 in the center:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;sum of 20&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;1, 3 and 4 must be in the corners, while 2 is on an edge:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;2 of the corner-numbers must be opposite of each other.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;3 possible solutions:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;1 opposite of 3:&lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;diagonal 103 and diagonal 004 give both a sum of 4 so no valid solution&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;  &#xA;  &lt;p&gt;1 opposite of 4:&lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;The diagonals give the sums 3 and 5&lt;/p&gt;&#xA;    &#xA;    &lt;p&gt;The 2 must not be between 3 and 4 as that would give more than 7.&lt;/p&gt;&#xA;    &#xA;    &lt;p&gt;The 2 must not be only next to 1 as that would give us one more line with the sum of 3.&lt;/p&gt;&#xA;    &#xA;    &lt;p&gt;The 2 can go either between 1 and 3 or only next to 4 which would give a sum of 6 either way.&lt;/p&gt;&#xA;    &#xA;    &lt;p&gt;Two unique solutions:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;001&#xA;002&#xA;403&#xA;&#xA;001&#xA;200&#xA;403&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;  &lt;/blockquote&gt;&#xA;  &#xA;  &lt;p&gt;3 opposite of 4:&lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;The diagonals give the sums 1 and 7.&lt;/p&gt;&#xA;    &#xA;    &lt;p&gt;The 2 must not go between 1 and 4 as that would give us another sum of 7.&lt;/p&gt;&#xA;    &#xA;    &lt;p&gt;The 2 must not go only next to 3 as that would give us the lines 023 and 104 both with a sum of 5.&lt;/p&gt;&#xA;    &#xA;    &lt;p&gt;The 2 can go either between 1 and 3 or only next to 4 which would give a sum of 6 either way.&lt;/p&gt;&#xA;    &#xA;    &lt;p&gt;Two unique solutions:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;003&#xA;002&#xA;401&#xA;&#xA;003&#xA;200&#xA;401&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That should cover all possible cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The unique solutions shown are equal to 8 solutions each by mirroring and rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there are a total of 64 possible solutions to this kind of magic square.&lt;/p&gt;&#xA;" OwnerUserId="16608" LastActivityDate="2015-11-16T14:14:37.330" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23991" PostTypeId="1" AcceptedAnswerId="24011" CreationDate="2015-11-16T15:16:41.173" Score="15" ViewCount="1065" Body="&lt;p&gt;You have a circular ice cream cake, with pink icing on top, and the brown chocolate interior showing when looked at from beneath. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x$ be a fixed real number. Over and over, you cut a $x^\circ$ slice from the cake, turn it over, and insert it back in. (Since this is an ice cream cake, the cuts resolidify). The position of each slice is anticlockwise adjacent to the previous one. For example, when  $x=130^\circ$, the first four steps would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zXvEx.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zXvEx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Prove that all of the icing will end up back on top... even when $x$ is irrational!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Source: This is yet another one of Peter Winkler's gems, from &lt;em&gt;Mathematical Mind Benders&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-11-17T04:56:28.783" Title="The Ice Cream Cake Process" Tags="|mathematics|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23992" PostTypeId="2" ParentId="10252" CreationDate="2015-11-16T15:17:59.670" Score="6" Body="&lt;p&gt;Edited because for some reason everything below gets flagged as spam in a de novo post, but not in edits made to an existing post!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The lion wins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A discussion (below) of the generality and assumptions of the solution might reveal the answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This solution need not assume that the lion knows anything about the positions of the zebras when it nominates the target points.  However, it relies on the assumption that the lion may change the target zebra at the end of the lion's turn; i.e. the lion's turn does not immediately and automatically end when the lion completes its move.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To motivate the solution, note the following (lemmas 1 and 2 can be read as hints; 3 and 4 might reveal the solution).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The lion can always cross to the other side of a group of zebras in a finite number of moves.  To accomplish this, the lion needs only to pass the first (stationary because it is non-target) zebra, then designate that zebra as the target and pass the remaining zebras.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lemma 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The lion can drive an arbitrary-sized group of zebras ahead of it.  To accomplish this, the lion approaches the rearmost zebra until it is close enough to nominate that zebra.  Call this zebra &quot;A&quot;. The lion nominates A, and continues to move as close to A possible at every turn until the lion and A approach the position of the second-rearmost zebra.  Call this new zebra B.  The lion drives A past B in the normal way, but on the turn after A passes B, the lion withdraws to a position 100m behind B and nominates B at the end of the turn.  After B moves, the lion at the beginning of its next turn nominates whatever zebra is rearmost, and repeats the process.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lemma 3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A lion driving two zebras A and B ahead of it can place one of them (the lion does not care which one) on any arbitrary single point, with the other still between the target and the lion.  To accomplish this, imagine that while driving the zebras in the normal way, A crosses the target first.  Note that A has also passed B in this move.  Then the lion moves 100m away from the target, so A moves back to the target.  The lion then moves 100m away from B and nominates B.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lemma 4&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Continuing from 3, the lion (if it wishes) can then place B on the same target point as A, as follows:  If B's next move does not carry it to the target point, the lion moves to within 100m of the target point forcing B past the target point.  The lion then retreats to 100m (or more) away, and B moves to the target.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The solution&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Solution 1: Target points for different zebras are allowed to coincide.&lt;/strong&gt; The lion nominates an arbitrary single point for the entire group of zebras.  The lion first crosses to one side of the entire group of zebras (possible by lemma 1), and then drives them all to one side of the target (possible by lemma 2).  The lion drives the zebras far enough past the point so that it can again cross the group in fewer moves than it would take for any of the zebras to cross back past the target point before the lion finishes crossing the group (possible by lemmas 1, 2).  The lion then begins to drive the zebras (all in a fairly tight bunch) towards the target.  The lion then repeatedly applies the procedure outlined in Lemma 3 to place all the zebras (save the last one) on the point, and then uses the lemma 4 procedure to place the last one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Slightly more general solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Solution 2: All target points must be distinct.&lt;/strong&gt; The lion nominates a set of points anywhere on the line.  The lion then applies solution 1 to place all of the zebras on the leftmost point, having driven them in from the left (or the rightmost, having driven them in from the right).  The lion then repeats the process for all zebras except the one that was targeted for the leftmost point, moving them to the second leftmost.  And so forth.  When the 99 leftmost points are all occupied by the appropriate zebras, the lion targets the last zebra and pushes it past the rightmost target, then retreats so that the zebra must move exactly onto that point.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Does it make any difference if the targeting distance was slightly modified so that the lion cannot tag zebras exactly 100m or 200m away (that is, change it from a closed interval to an open interval)?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As far as I can see, this does not make a difference for these solutions.  Considering the lemmas separately, I fail to see where any of them is compromised by this slight modification.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edited to add (in response to Trenin below--I don't have enough reputation to comment on her/his answer):&#xA;1) I don't see any constraint in the puzzle that requires the zebras positions to not overlap.&#xA;2) If the lion is at 0, and a zebra is required to minimize its distance to a zebra at 10+pi, the zebra is forced to move exactly to 100, because this is the nearest position to 10+pi that satisfies the non-crossing constraint.&lt;/p&gt;&#xA;" OwnerUserId="10417" LastEditorUserId="10417" LastEditDate="2015-11-16T18:43:06.760" LastActivityDate="2015-11-16T18:43:06.760" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23993" PostTypeId="2" ParentId="23982" CreationDate="2015-11-16T15:42:32.400" Score="2" Body="&lt;p&gt;Edit: Just to be clear, the following argument does not work because of Tim Couwelier's comment.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The winner is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alice by a &lt;a href=&quot;https://en.wikipedia.org/wiki/Strategy-stealing_argument&quot; rel=&quot;nofollow&quot;&gt;strategy stealing&lt;/a&gt; argument.&lt;br&gt;&#xA; &#xA;&lt;br&gt;&#xA; &#xA; Call the game where Alice starts in the corner and both players move optimally &quot;the a1 game&quot;.  If Alice wins the a1 game, then we are done.  If Bob wins the a1 game, then have Alice start in a square that is a knights move away from a corner, and call this &quot;the b3 game&quot;.   In the b3 game, if she chooses, Alice can steal Bob's strategy from the a1 game.  Every time Bob makes a move in the b3 game, Alice responds by making Bob's move from the a1 game.&lt;br&gt;&#xA;&#xA;&lt;br&gt;&#xA; The only difference between the two games is that in b3 game the corner square is a valid move.  Notice if Bob does move to that corner square, this is an automatic win for Alice in the b3 game.  If Bob somehow wins the b3 game without ever moving on that corner, then this would correspond to Alice winning the a1 game, since in the b3 game having an extra square Bob can move to makes it harder for Bob to win.   Thus if the a1 game is a loss of Alice, the b3 game will be a win.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2722" LastEditorUserId="2722" LastEditDate="2015-11-20T16:12:59.450" LastActivityDate="2015-11-20T16:12:59.450" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23994" PostTypeId="2" ParentId="23991" CreationDate="2015-11-16T15:52:06.390" Score="1" Body="&lt;p&gt;To get the ice cream cake back to a position where either all the icing or all the chocolate faces top(chocolate top can be seen as a solution since you just need to repeat all steps to flip the cake again) we need to have the sum off all made slices be dividable by 360:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x \times n = 0 \bmod 360$$ where $n$ is the number of slices we flip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easiest way to get this would be to multiply $x$  by a natural number $y$ so that $x \times y$ is itself a natural number. After that you only need to multiply $y$ by 360 to be sure that your solution is dividable by 360.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a rational number this is relatively easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since every rational number $x$ can be written as a fraction $\frac{a}{b}$ we just need to multiply $x$ by its denominator $b$ to make sure that the solution is a natural number.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As for irrational numbers I do not think that it is possible, at least within the range of finite natural numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that on a well known example we simply take the irrational number $\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as i know it is not possible that the term&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\pi \times n \space|\space n\in {N} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can ever equal a natural number.&lt;/p&gt;&#xA;" OwnerUserId="16608" LastActivityDate="2015-11-16T15:52:06.390" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23996" PostTypeId="1" AcceptedAnswerId="24012" CreationDate="2015-11-16T18:03:54.280" Score="5" ViewCount="873" Body="&lt;p&gt;&lt;sub&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/23959/magic-the-gathering-challenge-2-two-peas-in-a-pod&quot;&gt;Previous Challenge&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/24040/magic-the-gathering-challenge-4-stuck-in-the-grenzone-welcome-to-the-dunge&quot;&gt;Next Challenge&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;BACKGROUND:&lt;/strong&gt;&lt;br&gt;&#xA;Assume that this is a normal 20-life game of magic with no banned list (excepting cards from the 'Un' sets). If multiple solutions are tied for the fastest, the solution that provides the most damage will be chosen. Turns are counted such that the first player has turn 1, the other has 1.5, the first then has turn 2, the other 2.5 etc. So if you win on turn 5 on the draw, you have a turn 5.5 win. If you are on the draw and win on your opponent's 3rd turn, you have a turn 3 win.&#xA;Your solution cannot involve opponent's cooperation (e.g. they choose to damage themselves). Assume that your opponent's hand and library consist entirely of basic Mountains and that they will play a Mountain every turn. Should they prove unable to play a Mountain for the turn, they will proceed to discard as normal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PUZZLE SETUP:&lt;/strong&gt;&lt;br&gt;&#xA;This one is going to be different than the first two, instead of giving you a board state with a specific answer, I pose an open challenge: Perform the fastest win using only legendary cards (not just creatures, all cards used in the solution must be legendary, including lands--this means no instants/sorceries). You may choose to play or draw and assume that you will draw whatever cards are necessary, regardless of shuffling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best I have been able to come up with is a turn 3.5 win for 35 damage, but that doesn't mean you can't post a solution that is worse than that but still better than the others posted. A baseline is still useful and your solution may inspire someone to improve upon it and find the optimal solution (which may well be better than mine). Best of luck and be sure to be detailed in your solutions regarding sequencing, keep track of how many cards you have left in hand, and how lands are tapped for mana. Please do not post a solution if a better one has already been posted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;User Ninety-Three has beaten my original solution by a turn, but for those who were curious, see below for my initial solution.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Turn 0:&lt;/strong&gt; Choose to be on the draw, play Gemstone Caverns, pitching anything (too bad Stormcrow isn't legendary, let's just say Tetsuo Umezawa). 5 cards left in hand&lt;br&gt;&#xA;&lt;strong&gt;Turn 1.5:&lt;/strong&gt; Draw, play Karakas, cast Kytheon, Hero of Akros with Karakas and Isamaru, Hound of Konda with Gemstone Caverns. 3 cards left in hand&lt;br&gt;&#xA;&lt;strong&gt;Turn 2.5:&lt;/strong&gt; Draw, play Gaea's Cradle, cast Bushi Tenderfoot with Karakas and Zurgo Bellstriker with Caverns. Then tap Cradle for GGGG and cast Surrak the Hunt Caller. Give himself haste and attack with kytheon, isamaru, and surrak for 2(Isamaru)+2(Kytheon)+5(Surrak)=9 damage. Flip Kytheon into Gideon, Battle-Forged. 0 cards left in hand&lt;br&gt;&#xA;&lt;strong&gt;Turn 3.5:&lt;/strong&gt; Draw, activate Gideon's 0 ability, turning him into a 4/4 creature. Tap Cradle for GGGGG, Karakas for W, and Caverns for R and cast Atarka, World Render. Give it haste with Surrak. Attack with all, Atarka gives itself double-strike to deal 1(Bushi)+2(Zurgo)+2(Isamaru)+4(Gideon)+5(Surrak)+12(Atarka)=26. Add that the the 9 damage last turn, and your opponent is at -15 life!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-11-18T18:05:30.000" Title="&quot;Magic: the Gathering&quot; Challenge #3: Legends Rule" Tags="|calculation-puzzle|magic-the-gathering|" AnswerCount="2" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="23997" PostTypeId="2" ParentId="10252" CreationDate="2015-11-16T18:04:02.217" Score="1" Body="&lt;p&gt;If you are talking about the real number line, then I believe the Zebras can win.  All they need to do is make sure they stay at irrational numbered points on the line.  Assuming the Lion picks rational numbers as his nominated positions (and if he picks irrational, then use a different class of irrational numbers for the zebras), then there are an infinite number of irrational numbers within any number range, providing an infinite number of possible moves to a zebra while avoiding the nominated position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then again, I don't understand how to satisfy this rule:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Zebra Move Constraint 2: Attractor The zebras are friendly. Within the positions allowed by the non-crossing constraint, the target will move to the position that minimises its distance to the last zebra it passed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How do you minimize the difference between two real numbers?  If the two zebras are not allowed to occupy the same spot, then there is no way to minimize the distance.  Any attempt can be shown to not be minimal by halving the distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps I am misreading the question?&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2015-11-16T18:04:02.217" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23998" PostTypeId="2" ParentId="23996" CreationDate="2015-11-16T18:55:24.263" Score="5" Body="&lt;p&gt;I can deal exactly 20 damage on turn 3.0.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Turn 1: &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=79121&quot; rel=&quot;nofollow&quot;&gt;Shinka, the Bloodsoaked Keep&lt;/a&gt;; &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=394748&quot; rel=&quot;nofollow&quot;&gt;Zurgo Bellstriker&lt;/a&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Turn 2: &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=10422&quot; rel=&quot;nofollow&quot;&gt;Gaea's Cradle&lt;/a&gt;; &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=128949&quot; rel=&quot;nofollow&quot;&gt;Radha, Heir to Keld&lt;/a&gt;; attack with Zurgo for 2.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Turn 3: Tap Shinka for R; cast &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=113512&quot; rel=&quot;nofollow&quot;&gt;Norin the Wary&lt;/a&gt;. Tap Cradle for GGG; tap Radha for G; play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=373713&quot; rel=&quot;nofollow&quot;&gt;Nykthos, Shrine to Nyx&lt;/a&gt;; pay GG to tap Nykthos for RRR, cast &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=136139&quot; rel=&quot;nofollow&quot;&gt;Tarox Bladewing&lt;/a&gt;. Activate Tarox's Grandeur twice; attack with Tarox for 16 and Zurgo for 2.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Update: I found an infinite combo! Before the opponent takes their third turn, I can take an infinite number of turns, create an infinite number of creatures, and attack for an infinite amount of damage. Whether or not this is good depends on how taking extra turns is scored. I feel this should be scored well, because &quot;Turn 3: I trigger Medomai then kill you&quot; is basically the same as &quot;Turn 3: I kill you&quot;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also depends on being allowed to break the normal &quot;only four copies of a card&quot; rule. I am doing so based on a generous interpretation of the criteria &quot;no banned list&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;∞ on turn 3.0*:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Turn 1: &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=79121&quot; rel=&quot;nofollow&quot;&gt;Shinka, the Bloodsoaked Keep&lt;/a&gt;; &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=394748&quot; rel=&quot;nofollow&quot;&gt;Zurgo Bellstriker&lt;/a&gt; (5 cards in hand)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Turn 2: &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=113512&quot; rel=&quot;nofollow&quot;&gt;Norin the Wary&lt;/a&gt;, &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=10422&quot; rel=&quot;nofollow&quot;&gt;Gaea's Cradle&lt;/a&gt;, &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=147393&quot; rel=&quot;nofollow&quot;&gt;Rhys the Redeemed&lt;/a&gt; (3 cards in hand)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Turn 3: Tap lands for GGGR, play a second copy of &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=10422&quot; rel=&quot;nofollow&quot;&gt;Gaea's Cradle&lt;/a&gt; (losing the first copy to the legend rule), tap it to end up with GGGGGGR. Cast &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=386709&quot; rel=&quot;nofollow&quot;&gt;Ugin's Nexus&lt;/a&gt; and &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=51055&quot; rel=&quot;nofollow&quot;&gt;Slobad, Goblin Tinkerer&lt;/a&gt;. Activate Slobad, sacrificing the Nexus to make the Nexus indestructible, taking an extra turn (yes this works, yes I'm sure). (1 card in hand)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Turn 3B: Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=1705&quot; rel=&quot;nofollow&quot;&gt;Tolaria&lt;/a&gt;, cast Ugin's Nexus, sacrifice it to take an extra turn.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Turn 3C: Tap Gaea's Cradle and Tolaria for GGGGU. Cast &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=382920&quot; rel=&quot;nofollow&quot;&gt;Edric, Spymaster of Trest&lt;/a&gt;. Attack with Zurgo, Rhys and Slobad to draw three cards. Play another Gaea's Cradle, tap it and Shinka for GGGGR, play another Nexus, take another turn.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Turn 3D: Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=201242&quot; rel=&quot;nofollow&quot;&gt;Hammerheim&lt;/a&gt;, Nexus, take another turn. Attack to draw some extra cards.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Turn 3E: Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=201202&quot; rel=&quot;nofollow&quot;&gt;Urborg&lt;/a&gt;, Nexus, take another turn.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Turn 3F: Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=109755&quot; rel=&quot;nofollow&quot;&gt;Pendelhaven&lt;/a&gt;, Nexus, take another turn. Activate Rhys the Redeemed.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Turn 3#: Play Nexus, take another turn. Activate Rhys the Redeemed.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Repeat the above turn an arbitrary number of times to generate an arbitrary number of Rhys tokens.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Turn 3Z: Attack with your army of Rhys tokens and deal an arbitrary amount of damage.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7459" LastEditorUserId="7459" LastEditDate="2015-11-16T22:45:39.890" LastActivityDate="2015-11-16T22:45:39.890" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="24000" PostTypeId="2" ParentId="23958" CreationDate="2015-11-16T19:25:14.917" Score="13" Body="&lt;p&gt;If you take the&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Unique Letters from E. E. Cummings' full name (Edward Estlin Cummings)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;then you get&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; acdegilmnrstuw&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which are the letters found in @Dr Xorile's answer as common letters! This means that the definition of a Beecumming Word™ is that it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; is composed only of letters found in the name &quot;Edward Estlin Cummings&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7316" LastEditorUserId="7316" LastEditDate="2015-11-16T20:33:51.690" LastActivityDate="2015-11-16T20:33:51.690" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24002" PostTypeId="2" ParentId="23773" CreationDate="2015-11-16T20:04:11.583" Score="0" Body="&lt;p&gt;The probability that chuck lives is approximately 466.5/512 which is about 90%.  i would like to point out that one robot acting at a time as if they were playing a game of cards does not appears sensible. The other answers have assumed that, which perhaps explains the difference in our answers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;moreover, i also do not think that a wise 1,1,0 robot will sacrifice itself while being aware that the other robots might kill chuck in its absence. So, i also assumed the robots to be wise and chose the optimum alternative depending upon the behaviour of others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To evaluate my answer, i categorized the 9 robots into 4 teams- those who follow law 1, those who follow law 2 instead, those who follow just law 3 and those don't follow any. It can be seen that this consideration simplifies as members have very similar behaviour&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that chuck lives if there are 3 or more robots following law 1. This is because if someone attacks chuck, 3 good robots will die and kill 3 at the same time, hence completing the task so that there is no more violence required. This realization alone brings Chuck's survival probability to 90%. But Chuck can also survive if there are fewer than 3 good robots, largely if there are fewer robots following law 2 and more robots following none. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The precise calculation quickly gets pretty complicated after this, but after doing some i can safely conclude that my approximation is very close to the correct value.&lt;/p&gt;&#xA;" OwnerUserId="17384" LastEditorUserId="17384" LastEditDate="2015-11-18T01:57:28.833" LastActivityDate="2015-11-18T01:57:28.833" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24007" PostTypeId="2" ParentId="228" CreationDate="2015-11-16T22:14:49.623" Score="0" Body="&lt;p&gt;There is no exactly &quot;most efficient&quot; solving method. It really depends on how efficient the solver is. Like in CFOP there is intuitive F2L which is very efficient.&lt;/p&gt;&#xA;" OwnerUserId="17386" LastActivityDate="2015-11-16T22:14:49.623" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24008" PostTypeId="2" ParentId="10252" CreationDate="2015-11-17T00:05:11.113" Score="0" Body="&lt;p&gt;I might be missing something. Can't the lion simply push all the zebra far to one side, and then pull them one at a time. In each case (except the last) you push the zebra past the second one, and then pull it to it's position (the left most unfilled position). The last one will have already been crossed by the second last, so it can also be pulled into position.&lt;/p&gt;&#xA;" OwnerUserId="15942" LastActivityDate="2015-11-17T00:05:11.113" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24011" PostTypeId="2" ParentId="23991" CreationDate="2015-11-17T04:56:28.783" Score="13" Body="&lt;p&gt;Instead of the flipping mechanism rotating around the cake, let's take the frame of the flipper, where the flipper is fixed and always flips the same interval, followed by the cake being rotated. Whether the cake returns to all icing doesn't depend on the frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the cake have circumference $1$, and so that points correspond to values modulo $1$. Let's put $0$ above the cake and have the length-$x$ flipper be centered above the cake flipping the interval $R=[-x/2, x/2]$. So, we alternate negating points in range $R$ (and swapping their color) and increasing all points by $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look the orbit of a single point on the cake, treating the flips (when they happen) and the rotations as separate steps. We can look both forward and backward in time, since the steps are reversible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At some point in the orbit, the point is in $R$ since the rotation angle equals the length of $R$. Say it flips from a value $-q$ to $q$. Iterating forward, it the rotates to $q+x, q+2x,\dots$ until it is again at some point $p$ in $R$. Iterating backwards, the same is true mirrored with $-q-x, -q-2x, \dots$ until it reaches a point $-p$ in $R$. But $p$ and $-p$ flip to each other, so the whole thing forms a cycle. Moreover, the point has changed color twice in that cycle, making the orbit a closed loop of events. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, every point, after some number of steps, returns to its original position and color. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This number can be different for different points. It is twice the number $n$ of steps to rotate from $q$ to $p$. Looking at the extreme cases where $p$ and $q$ lie near the boundaries of $R$, the distance $nx$ rotated is in the range $(1-x,1+x)$, and so $n\in (x^{-1}-1,x^{-1}+1)$. Since $n$ is an integer, if we let $m=\lfloor x^{-1}\rfloor$, then $n$ is either $m$ or $m+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since points can have one of two cycle lengths, $2m$ and $2(m+1)$, it suffices to take a number of steps divisible by both of them, the smallest being their LCM, which is $2m(m+1)$, still with $m=\lfloor x^{-1}\rfloor$. After that number of steps, each point has done a whole number of cycles, and so returned to its original location and color of icing.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-11-17T04:56:28.783" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24012" PostTypeId="2" ParentId="23996" CreationDate="2015-11-17T05:56:00.737" Score="6" Body="&lt;p&gt;&lt;strong&gt;Update (11/17)&lt;/strong&gt;: Based on some comments on my previous solutions and a little more time to think, I'm submitting a single new solution, which I think is a (minor) upgrade.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notable Features:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Kills on the first main phase of turn 2.5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Deals infinite damage (technically, infinite life loss)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Has no turn 0 plays, and no extra land drops&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Turn 1.5: &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=79218&quot; rel=&quot;nofollow&quot;&gt;Untaidake, the Cloud Keeper&lt;/a&gt;, tapped - pass turn (7 cards)&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Turn 2.5: Pay 2 life to tap Untaidake for (2), &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=398492&quot; rel=&quot;nofollow&quot;&gt;Sword of the Animist&lt;/a&gt;; &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=397719&quot; rel=&quot;nofollow&quot;&gt;Mox Opal&lt;/a&gt;; &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=383133&quot; rel=&quot;nofollow&quot;&gt;Tolarian Academy&lt;/a&gt;; tap Academy for UU, &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=81979&quot; rel=&quot;nofollow&quot;&gt;Umezawa's Jitte&lt;/a&gt; (4 cards)&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Tap Mox Opal for B, play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=397719&quot; rel=&quot;nofollow&quot;&gt;Mox Opal&lt;/a&gt; (sacrificing the first), tap Mox Opal for R, play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=382276&quot; rel=&quot;nofollow&quot;&gt;Grenzo, Dungeon Warden&lt;/a&gt;, play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=397719&quot; rel=&quot;nofollow&quot;&gt;Mox Opal&lt;/a&gt;, sacrificing the second, tap Mox Opal for B, play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=397719&quot; rel=&quot;nofollow&quot;&gt;Mox Opal&lt;/a&gt;, sacrificing the third, tap Mox Opal for B (BB floating, 0 cards)&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Pay (BB) for Grenzo, revealing &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=376305&quot; rel=&quot;nofollow&quot;&gt;Derevi, Empyrial Tactician&lt;/a&gt;.  Put her into play, untap Tolarian Academy, tap Academy for UUU.  Repeat this three more times, sacrificing previous Derevis. (UUUUUU floating, 0 cards)&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Pay (UU) for Grenzo, revealing &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=397905&quot; rel=&quot;nofollow&quot;&gt;Emrakul, the Aeons Torn&lt;/a&gt;, put it into your graveyard, shuffle your graveyard into your library, adding back in three Derevis and Emrakul (along with the Moxen).&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Keep paying (UU) at a time for Grenzo, revealing (Derevi, Derevi, Derevi, Emrakul), untapping Academy with each Derevi before shuffling them all back in.  This loop gains one U (-UU+UUU-UU+UUU-UU+UUU-UU), so repeat until you have infinite U. (Infinite U, 0 cards)&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Pay (UU), (UU), (UU) for Grenzo, revealing Derevi, Derevi, Derevi - untap/tap Mox Opal for BBB. (BBB, Infinite U, 0 cards)&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Pay (UU) for Grenzo, revealing &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=376254&quot; rel=&quot;nofollow&quot;&gt;Azami, Lady of Scrolls&lt;/a&gt;, put her into play, use Azami to tap Darevi to draw a card. (BBB, Infinite U, 1 card)&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=88802&quot; rel=&quot;nofollow&quot;&gt;Maga, Traitor to Mortals&lt;/a&gt; for X = infinite, killing your opponent.&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Alternatively: for a milling victory, tap the Mox for RB, pay (UU) for Grenzo to reveal &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=129913&quot; rel=&quot;nofollow&quot;&gt;Ambassador Laquatus&lt;/a&gt;, play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=159096&quot; rel=&quot;nofollow&quot;&gt;Lord of Tresserhorn&lt;/a&gt; from your hand, and respond by paying infinite for Laquatus' ability, targeting your opponent.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Old Solutions:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I'll take a crack at this, since I'm enjoying these puzzles quite a bit! I've got full-infinite turns/mana/damage/cards on turn 2.5.  I can also deal 20+ damage normally on that turn with minimal difficulty, in case we dislike taking infinite turns.  I don't have enough reputation to post links, so you'll have to find some of the cards yourself (sorry!).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Turn 0: &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=122094&quot; rel=&quot;nofollow&quot;&gt;Gemstone Caverns&lt;/a&gt;, exiling a second Gemstone Caverns (5 cards in hand)&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Turn 1.5: &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=79121&quot; rel=&quot;nofollow&quot;&gt;Shinka, the Bloodsoaked Keep&lt;/a&gt;; &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=394748&quot; rel=&quot;nofollow&quot;&gt;Zurgo Bellstriker&lt;/a&gt;; &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=398428&quot; rel=&quot;nofollow&quot;&gt;Kytheon, Hero of Akros&lt;/a&gt; (off Gemstone) (3 cards in hand)&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Turn 2.5: &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=10422&quot; rel=&quot;nofollow&quot;&gt;Gaea's Cradle&lt;/a&gt;; &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=382276&quot; rel=&quot;nofollow&quot;&gt;Grenzo, Dungeon Warden&lt;/a&gt; for RB (Gemstone for B); tap Gaea's Cradle for GGG floating.  Pay (2) for Grenzo, reveal &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=376305&quot; rel=&quot;nofollow&quot;&gt;Derevi, Empyrial Tactician&lt;/a&gt; - Put her into play, untap Gaea's Cradle.  Tap Cradle for GGGG. (GGGGG floating, 2 cards in hand)&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Pay (2) for Grenzo, reveal &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=376254&quot; rel=&quot;nofollow&quot;&gt;Azami, Lady of Scrolls&lt;/a&gt; - put her into play. Pay (2) for Grenzo, reveal &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=80283&quot; rel=&quot;nofollow&quot;&gt;Azusa, Lost but Seeking&lt;/a&gt; - put her into play.  Play Gaea's Cradle #2, sacrificing the first one.  Tap Gaea's Cradle for GGGGGG. (GGGGGGG floating, 1 card in hand)&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Pay (2) for Grenzo, reveal &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=405349&quot; rel=&quot;nofollow&quot;&gt;Prime Speaker Zegana&lt;/a&gt; - put her into play with two counters due to Grenzo.  Draw three cards. Pay (2) for Grenzo, reveal &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=397748&quot; rel=&quot;nofollow&quot;&gt;Endrek Sahr, Master Breeder&lt;/a&gt; - put him into play. Pay (2) for Grenzo, reveal Derevi, Empyrial Tactician #2, put her into play, sacrificing the first - untap Gaea's Cradle. Tap Gaea's Cradle for GGGGGGGG. (GGGGGGGGG floating, 4 cards in hand)&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Pay (2) for Grenzo, reveal Derevi, Empyrial Tactician #3, put her in play.  Sacrifice the one already in play, untap Gaea's Cradle.  Tap Gaea's Cradle for GGGGGGGG (GGGGGGGGGGGGGGG floating, 4 cards in hand).  Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=397905&quot; rel=&quot;nofollow&quot;&gt;Emrakul, the Aeons Torn&lt;/a&gt;, triggering Endrek Sahr, Master Breeder for 15 Thrull tokens.  Sacrifice Endrek as a result.  Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=201198&quot; rel=&quot;nofollow&quot;&gt;Karakas&lt;/a&gt; (your third land drop thanks to Azusa).  Take your extra turn from playing Emrakul.  Do whatever you like, tap Karakas to return Emrakul to your hand, tap Gaea's Cradle to generate &gt;15 mana due to Thrulls, replay Emrakul.  Take infinite turns, using a second Emrakul and the legend rule to avoid decking.  Draw any cards you like off Azami, Lady of Scrolls.  Use &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=259295&quot; rel=&quot;nofollow&quot;&gt;Omath, Locus of Mana&lt;/a&gt; for infinite mana and infinite power/toughness, because why not.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the no-extra-turns version of this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Use your giant pile of mana on turn 2.5 for &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=378531&quot; rel=&quot;nofollow&quot;&gt;Heroes' Podium&lt;/a&gt;.  Attack with Zurgo, Kytheon, and your favorite 2 power creature with haste revealed off Grenzo (I'm partial to &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=397698&quot; rel=&quot;nofollow&quot;&gt;Kiki-Jiki, Mirror Breaker&lt;/a&gt;) with at least four other legendaries in play to deal 21+.  I didn't try to optimize this, but 100+ shouldn't be hard; there are a lot of hasty 1 or 2 power legendaries and it goes up as the square of the number of creatures.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And, as a bonus answer, a &lt;em&gt;second&lt;/em&gt; way to start off that infinite turn 2.5 combo:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Turn 0: Gemstone Caverns, exiling a second Gemstone Caverns (5 cards in hand)&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Turn 1.5: &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=397719&quot; rel=&quot;nofollow&quot;&gt;Mox Opal&lt;/a&gt;; &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=383133&quot; rel=&quot;nofollow&quot;&gt;Tolarian Academy&lt;/a&gt;; tap Academy for U, &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=5250&quot; rel=&quot;nofollow&quot;&gt;Sword of the Chosen&lt;/a&gt; (3 cards in hand)&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Turn 2.5: Tap Academy for UU; &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=398492&quot; rel=&quot;nofollow&quot;&gt;Sword of the Animist&lt;/a&gt;; tap Mox Opal for B and Gemstone Caverns for R. Play Grenzo, Dungeon Warden. Play a second Tolarian Academy (sacrificing the first), tap for UUU floating, 1 card in hand.&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Pay (2) for Grenzo, reveal Azusa, Lost but Seeking. Play Gaea's Cradle as your last card, tap for GG (GGU floating).  Pay (2) for Grenzo, reveal Derevi, Empyrial Tactician, untap Gaea's Cradle.  Tap for GGG (GGGU floating).  Pay (2) for Grenzo, reveal Prime Speaker Zegana, draw 3.  Pay (2) for Grenzo, reveal Endrek Sahr (0 mana floating, 3 cards).&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play a second Gaea's Cradle, sacrificing the first.  Tap it for GGGGG.  Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=389721&quot; rel=&quot;nofollow&quot;&gt;Titania, Protector of Argaroth&lt;/a&gt;, triggering Endrek for 5 Thrulls and returning your first Gaea's Cradle to the battlefield.  Sacrifice the tapped one, triggering Titania for a 5/3 elemental token.  Tap Gaea's Cradle for GGGGGGGGGGGG.  Pay (2) for Grenzo, reveal Derevi, Empyrial Tactician (sacrificing the old one), untap Gaea's Cradle. Tap Cradle for GGGGGGGGGGGG (GGGGGGGGGGGGGGGGGGGGGG floating, 1 card in hand).  Play Emrakul, the Aeons Torn triggering Endrek for another 15 Thrulls, then go infinite as before.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="17389" LastEditDate="2015-11-18T16:19:41.687" LastActivityDate="2015-11-18T16:19:41.687" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="24013" PostTypeId="1" AcceptedAnswerId="24082" CreationDate="2015-11-17T09:37:24.230" Score="7" ViewCount="918" Body="&lt;p&gt;Alice and Bob play the following game on a $9\times11$ chessboard.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First Alice divides the chessboard into $33$ smaller rectangles of &#xA;dimensions $1\times3$ and $3\times1$.&lt;/li&gt;&#xA;&lt;li&gt;Then Bob labels each of the $99$ squares with one of the integers &#xA;$0,1,2,3,4,5$, such that in each of Alice's smaller rectangles the &#xA;sum of the labels equals $5$. &lt;/li&gt;&#xA;&lt;li&gt;Bob wins the game, if in each of the nine rows the sum of the labels &#xA;is a prime number. Otherwise Alice wins.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Which player is going to win this game?&#xA;  (As usual, we assume that Alice and Bob both use optimal strategies.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-11-20T22:54:42.397" Title="A prime number game" Tags="|mathematics|strategy|game|checkerboard|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24014" PostTypeId="1" AcceptedAnswerId="24017" CreationDate="2015-11-17T10:21:49.623" Score="11" ViewCount="841" Body="&lt;p&gt;I recently watched a documentary where they talked about messages sent into space and how they imagined aliens figuring out how to understand them.&#xA;They used math and geometry in clever ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if it may be possible to use programming to send such messages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is then if it is possible to figure out the syntax and semantics of a language by looking at a program.&#xA;I think a quine would be a natural choice of program since it is its own output regardless of what the symbols mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following program is a quine written in X.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[ q &amp;lt; &amp;lt; [ q &amp;gt; { q } &amp;lt; ] ( q ) &amp;gt; &amp;gt; ] ( q )&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;All features of X are used in the program. X is Turing-equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The program is a quine according to definition at &lt;a href=&quot;https://en.wikipedia.org/wiki/Quine_%28computing%29&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Quine_%28computing%29&lt;/a&gt;.&#xA;It is not a &quot;Cheating&quot; quine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are no symbols in the language that are outputting themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The challenge is to figure out the syntax and semantics of X.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rules should be the minimal set of rules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There should be a list of numbered rules, described in English.&#xA;The rules should appear in the same order as the first example of a corresponding syntactic element in the program.&#xA;Use &lt;em&gt;x&lt;/em&gt;, &lt;em&gt;y&lt;/em&gt; and &lt;em&gt;z&lt;/em&gt; as wild-cards. &lt;em&gt;x&lt;/em&gt; denoting the first item, &lt;em&gt;y&lt;/em&gt; the next and then &lt;em&gt;z&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;x &amp;lt;&lt;/code&gt;&#xA;bite something less then &lt;em&gt;x&lt;/em&gt; minutes ago&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;{ x }&lt;/code&gt;&#xA;eject lab mouse named &lt;em&gt;x&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;x &amp;gt; y&lt;/code&gt;&#xA;lets all eat &lt;em&gt;x&lt;/em&gt;, &lt;em&gt;y&lt;/em&gt; not&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The winner is the one who has the most correct rules in the correct positions.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Answer&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The point of this challenge was to test if it was in fact possible to become capable of reading a programming language by only looking at a quine for that language.&#xA;I am very happy about how this turned out. Great success!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The accepted answer has all the key features right.&#xA;There are subtleties that were not possible to decide based on the quine.&#xA;Those would have been apparent with only a few additional examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Congratulations to The Dark Truth!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For completeness I have made an interpreter for the language with an included user manual: &lt;a href=&quot;https://gist.github.com/thomasmf/031db24f0e63a3459e62&quot; rel=&quot;nofollow&quot;&gt;https://gist.github.com/thomasmf/031db24f0e63a3459e62&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17392" LastEditorUserId="17392" LastEditDate="2015-11-18T17:48:48.490" LastActivityDate="2015-11-18T17:48:48.490" Title="Figure out the language" Tags="|computer-puzzle|language|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24016" PostTypeId="2" ParentId="24013" CreationDate="2015-11-17T10:46:31.100" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Bob will win.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This is because it is impossible for Alice to force him to make a non-prime sum in a row.&lt;/p&gt;&#xA;&lt;p&gt;I will go a step further and say it is always possible to fill the rows so that 6 rows have the sum 19 and 3 rows have the sum 17.&lt;/p&gt;&#xA;&lt;p&gt;To prove this i will solve it with an easy board and show how every other combination of rectangles can be made from this board with simple transformations and still show the same solution.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Our easy board will have the first two colums filled with 3 vertical rectangles per column.&lt;/p&gt;&#xA;&lt;p&gt;The rest of the board will be filled with horizontal rectangles for an additional 3 horizontal rectangles per row:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;┌─┬─┬─────┬─────┬─────┐&#xA;│ │ │     │     │     │&#xA;│ │ ├─────┼─────┼─────┤&#xA;│ │ │     │     │     │&#xA;│ │ ├─────┼─────┼─────┤&#xA;│ │ │     │     │     │&#xA;├─┼─┼─────┼─────┼─────┤&#xA;│ │ │     │     │     │&#xA;│ │ ├─────┼─────┼─────┤&#xA;│ │ │     │     │     │&#xA;│ │ ├─────┼─────┼─────┤&#xA;│ │ │     │     │     │&#xA;├─┼─┼─────┼─────┼─────┤&#xA;│ │ │     │     │     │&#xA;│ │ ├─────┼─────┼─────┤&#xA;│ │ │     │     │     │&#xA;│ │ ├─────┼─────┼─────┤&#xA;│ │ │     │     │     │&#xA;└─┴─┴─────┴─────┴─────┘&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Now each row will have a sum of 15 (for the three horizontal boxes) plus a small amount from the vertical boxes.&lt;/p&gt;&#xA;&lt;p&gt;I will fill it in as easy as possible:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;┌─┬─┬─────┬─────┬─────┐&#xA;│1│1│5 0 0│5 0 0│5 0 0│  17&#xA;│ │ ├─────┼─────┼─────┤&#xA;│2│2│0 5 0│0 5 0│0 5 0│  19&#xA;│ │ ├─────┼─────┼─────┤&#xA;│2│2│0 0 5│0 0 5│0 0 5│  19&#xA;├─┼─┼─────┼─────┼─────┤&#xA;│1│1│5 0 0│5 0 0│5 0 0│  17&#xA;│ │ ├─────┼─────┼─────┤&#xA;│2│2│0 5 0│0 5 0│0 5 0│  19&#xA;│ │ ├─────┼─────┼─────┤&#xA;│2│2│0 0 5│0 0 5│0 0 5│  19&#xA;├─┼─┼─────┼─────┼─────┤&#xA;│1│1│5 0 0│5 0 0│5 0 0│  17&#xA;│ │ ├─────┼─────┼─────┤&#xA;│2│2│0 5 0│0 5 0│0 5 0│  19&#xA;│ │ ├─────┼─────┼─────┤&#xA;│2│2│0 0 5│0 0 5│0 0 5│  19&#xA;└─┴─┴─────┴─────┴─────┘&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Now we can already guess where this is going.&lt;/p&gt;&#xA;&lt;p&gt;3 Transformations are possible/needed:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;substituting 3 horizontal rectangles by 3 vertical rectangles.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;This is done easily as we already have our 5s filled in perfectly&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;┌─┬─┬─────┬─────┬─────┐&#xA;│1│1│5 0 0│5 0 0│5 0 0│&#xA;│ │ ├─────┼─────┼─────┤&#xA;│2│2│0 5 0│0 5 0│0 5 0│&#xA;│ │ ├─────┼─────┼─────┤&#xA;│2│2│0 0 5│0 0 5│0 0 5│&#xA;├─┼─┼─┬─┬─┼─────┼─────┤&#xA;│1│1│5│0│0│5 0 0│5 0 0│&#xA;│ │ │ │ │ ├─────┼─────┤&#xA;│2│2│0│5│0│0 5 0│0 5 0│&#xA;│ │ │ │ │ ├─────┼─────┤&#xA;│2│2│0│0│5│0 0 5│0 0 5│&#xA;├─┼─┼─┴─┴─┼─────┼─────┤&#xA;│1│1│5 0 0│5 0 0│5 0 0│&#xA;│ │ ├─────┼─────┼─────┤&#xA;│2│2│0 5 0│0 5 0│0 5 0│&#xA;│ │ ├─────┼─────┼─────┤&#xA;│2│2│0 0 5│0 0 5│0 0 5│&#xA;└─┴─┴─────┴─────┴─────┘&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;This can actually be done at any position we wish as long as we have 3 rectangles in a square formation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;moving vertically&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;When we have 3 vertical rectangles in a square formation and a horizontal rectangle directly above or below we can switch them.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;┌─┬─┬─────┬─────┬─────┐&#xA;│1│1│5 0 0│5 0 0│5 0 0│&#xA;│ │ ├─────┼─────┼─────┤&#xA;│2│2│0 5 0│0 5 0│0 5 0│&#xA;│ │ ├─┬─┬─┼─────┼─────┤&#xA;│2│2│5│0│0│0 0 5│0 0 5│&#xA;├─┼─┤ │ │ ├─────┼─────┤&#xA;│1│1│0│5│0│5 0 0│5 0 0│&#xA;│ │ │ │ │ ├─────┼─────┤&#xA;│2│2│0│0│5│0 5 0│0 5 0│&#xA;│ │ ├─┴─┴─┼─────┼─────┤&#xA;│2│2│0 0 5│0 0 5│0 0 5│&#xA;├─┼─┼─────┼─────┼─────┤&#xA;│1│1│5 0 0│5 0 0│5 0 0│&#xA;│ │ ├─────┼─────┼─────┤&#xA;│2│2│0 5 0│0 5 0│0 5 0│&#xA;│ │ ├─────┼─────┼─────┤&#xA;│2│2│0 0 5│0 0 5│0 0 5│&#xA;└─┴─┴─────┴─────┴─────┘&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;We dont need to care for the first 2 columns for now as whenever this movement is possible there will always be at least 3 vertical 5-rectangles in the same 3 rows.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;horizontal movement&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;When we have 3 horizontal rectangles in a square formation and a vertical rectangle directly left or right we can switch them.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;┌─┬─┬─────┬─────┬─────┐&#xA;│1│1│5 0 0│5 0 0│5 0 0│&#xA;│ │ ├─────┼─────┼─────┤&#xA;│2│2│0 5 0│0 5 0│0 5 0│&#xA;│ │ ├─┬─┬─┼─────┼─────┤&#xA;│2│2│5│0│0│0 0 5│0 0 5│&#xA;├─┼─┤ │ │ ├─────┼─────┤&#xA;│1│1│0│5│0│5 0 0│5 0 0│&#xA;│ │ │ │ │ ├─────┼─────┤&#xA;│2│2│0│0│5│0 5 0│0 5 0│&#xA;│ │ ├─┴─┴─┼─────┼─────┤&#xA;│2│2│0 0 5│0 0 5│0 0 5│&#xA;├─┼─┴───┬─┼─────┼─────┤&#xA;│1│5 0 0│1│5 0 0│5 0 0│&#xA;│ ├─────┤ ├─────┼─────┤&#xA;│2│0 5 0│2│0 5 0│0 5 0│&#xA;│ ├─────┤ ├─────┼─────┤&#xA;│2│0 0 5│2│0 0 5│0 0 5│&#xA;└─┴─────┴─┴─────┴─────┘&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;This can be done as often as needed without changing the sums in the rows.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;By using these transformations as often as needed we can reach all possible combinations of rectangles.&lt;/p&gt;&#xA;" OwnerUserId="16608" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-11-17T10:46:31.100" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24017" PostTypeId="2" ParentId="24014" CreationDate="2015-11-17T12:49:59.683" Score="8" Body="&lt;p&gt;The program is written in a programming language that allows only single character variables and allows the compiling of strings into runnable code during compilation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;running&quot; a string means cutting it up into its logical units and returning the concatenated results of each logical unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rules:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;code&gt;[q ..... ]&lt;/code&gt; -&gt; assign the string value &quot;.....&quot; to the variable &quot;q&quot;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;&amp;lt;.....&amp;gt;&lt;/code&gt; -&gt; run &quot;.....&quot; and return the result / if &quot;.....&quot; is no logical unit return it as string&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;{q}&lt;/code&gt; -&gt; return &quot;q&quot; as string&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;(q)&lt;/code&gt; -&gt; run &quot;q&quot; and print the result&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;when running the program the following happens (i will use &quot;&quot; to show strings and ++ for concatenation):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;code&gt;[ q &amp;lt; &amp;lt; [ q &amp;gt; { q } &amp;lt; ] ( q ) &amp;gt; &amp;gt; ] ( q )&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;[ q &amp;lt; &amp;lt; [ q &amp;gt; { q } &amp;lt; ] ( q ) &amp;gt; &amp;gt; ]&lt;/code&gt; -&gt; assign the string &lt;code&gt;&quot;&amp;lt; &amp;lt; [ q &amp;gt; { q } &amp;lt; ] ( q ) &amp;gt; &amp;gt;&quot;&lt;/code&gt; to q&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;( q )&lt;/code&gt; -&gt; run q and print resulting string&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;&amp;lt; &amp;lt; [ q &amp;gt; { q } &amp;lt; ] ( q ) &amp;gt; &amp;gt;&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;&amp;lt; [ q &amp;gt; ++ { q } ++ &amp;lt; ] ( q ) &amp;gt;&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;&quot;[ q&quot; ++ &quot;&amp;lt; &amp;lt; [ q &amp;gt; { q } &amp;lt; ] ( q ) &amp;gt; &amp;gt;&quot; ++ &quot;] ( q )&quot;&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;&quot;[ q &amp;lt; &amp;lt; [ q &amp;gt; { q } &amp;lt; ] ( q ) &amp;gt; &amp;gt; ] ( q )&quot;&lt;/code&gt; &amp;lt;- print this&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="16608" LastActivityDate="2015-11-17T12:49:59.683" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24018" PostTypeId="2" ParentId="23229" CreationDate="2015-11-17T12:59:40.667" Score="0" Body="&lt;p&gt;X should be &lt;strong&gt;7&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;patterns are ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) `&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A B A B A&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;&#xA;2.)&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A ,A-2 ,A-2-2 ,A ,A-2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;&#xA;3.)&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A ,B,A,B,A&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;&#xA;4.)&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A,A-1,A-1-1,A,A-1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;`&lt;/p&gt;&#xA;" OwnerUserId="14410" LastEditorUserId="14410" LastEditDate="2015-11-18T06:14:48.863" LastActivityDate="2015-11-18T06:14:48.863" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24019" PostTypeId="1" AcceptedAnswerId="24020" CreationDate="2015-11-17T17:23:52.427" Score="12" ViewCount="1133" Body="&lt;p&gt;This riddle is in season twice per year. Now is one of those times.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My first turns a harsh command into a start&lt;br&gt;&#xA;  And aid for a structure to aid from the heart.&lt;br&gt;&#xA;  Two of my second can take you quite far&lt;br&gt;&#xA;  And three of them help if locked out of your car.&lt;br&gt;&#xA;  My third is the final of a group of five&lt;br&gt;&#xA;  For which a newspaper reporter must strive.&lt;br&gt;&#xA;  My fourth counts the number of that which we share;&lt;br&gt;&#xA;  My fifth, twice the tour of a legionnaire.&lt;br&gt;&#xA;  Though you grow by my sixth again and again,&lt;br&gt;&#xA;  Your size is the same as it's always been.&lt;br&gt;&#xA;  My next two together, togetherness mean,&lt;br&gt;&#xA;  Yet on a map never together are seen.&lt;br&gt;&#xA;  If you attacked those heading into the bay,&lt;br&gt;&#xA;  My ninth is the word you most likely would say.&lt;br&gt;&#xA;  My last is a segment of distance that's put&lt;br&gt;&#xA;  Along with a meter, but never a foot.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;My whole links the seasonal precipitation&lt;br&gt;&#xA;  With those who first gratefully made celebration.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17231" LastActivityDate="2015-11-18T14:49:55.017" Title="A Twice-Seasoned Riddle" Tags="|riddle|seasonal|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24020" PostTypeId="2" ParentId="24019" CreationDate="2015-11-17T17:56:34.607" Score="12" Body="&lt;p&gt;1st:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;M&quot; makes 'bark' into 'embark' and 'brace' into 'embrace'. (from @DrunkWolf amd @Aant)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2nd:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The second should be &quot;A&quot;. Two of the second then is AA (American Airlines) and three of them is AAA (American Automobile Association). (from @Gamow)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3rd:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Y&quot; (why), the fifth of the 'five Ws' questions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4th:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;F&quot;: that which we share is 'our's, which becomes 'four'. (from @GordonK)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5th:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;L&quot;: the tour of a Roman legionary was 25 years, and twice that is 50, or L in Roman numerals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6th:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;O&quot;: looks like zero, which doesn't change the size of a number when added.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;7th and 8th:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;WE&quot;: west and east are opposites on a map, but together make the word 'we'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;9th:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;R&quot;: said by a stereotypical pirate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;10th:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;S&quot;: 'meter' pluralizes to 'meters', but 'foot' becomes 'feet', not 'foots'. (from @question_asker)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whole:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;MAYFLOWERS. Mayflower is known as the ship that carried the Pilgrims associated with Thanksgiving, and also in the saying &quot;April showers bring May flowers&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-11-17T22:35:25.560" LastActivityDate="2015-11-17T22:35:25.560" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24021" PostTypeId="2" ParentId="24019" CreationDate="2015-11-17T19:14:12.343" Score="6" Body="&lt;p&gt;The reasoning for the 4th might be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; F for fraction. Fractions are used to quantify the size of a portion, and the fraction's denominator is the &quot;number of that which we share&quot; (i.e. the total number of portions, assuming that they're all equally sized)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My apologies for putting this as a separate answer rather than as a comment on the above post. (I'm a new user and don't have the reputation to comment.)&lt;/p&gt;&#xA;" OwnerUserId="17399" LastEditorUserId="2156" LastEditDate="2015-11-18T14:49:55.017" LastActivityDate="2015-11-18T14:49:55.017" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24022" PostTypeId="2" ParentId="23982" CreationDate="2015-11-17T21:37:00.603" Score="9" Body="&lt;p&gt;The player who will win this game on an $8 \times 8$ board is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Bob&lt;/strong&gt;, assuming he has a computer and a few hours to calculate the variants (details below). It doesn't matter which square is chosen by Alice, because Bob has always a forced win. On smaller boards Alice has always at least one square which she can choose to win.&lt;br&gt;&lt;br&gt;&lt;em&gt;With the result known, maybe someone can come up with a solution that doesn't require an exhaustive search. But if the solution is not specific to an $8 \times 8$ board, it has to explain why Alice can win on smaller boards.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Computer based proof&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I managed to significantly increase the speed of my previously posted program, so I was able to calculate the result for an $8 \times 8$ board. It's much bigger now, but I have added some comments this time. The speed up was achieved through following steps:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Mutlithreading: Usually pretty obvious, but I use it here only to calculate all starting positions in parallel. This doesn't complicate the code as much as other possibilities.&lt;/li&gt;&#xA;&lt;li&gt;Precalculating possible moves: Also pretty obvious, not much to say here.&lt;/li&gt;&#xA;&lt;li&gt;Sorting the possible moves: This was the most important point, and a huge speed up. The idea is, if we find a winning move for the current player, we don't need to check the other possible moves. The hard part is to find the right moves. I did it here by sorting the possible moves by the number of possible moves from their target square. So corner squares are preferred, squares in the center are searched last. The effect is huge, if you want to see it set the SIZE to 7 and reverse the sorting.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Despite the speed up, calculating the $8 \times 8$ results takes several hours. On a Core i5 I got the result for the first square (C1) after 1 h 38 m. Total runtime was about 8.5 hours.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Results for smaller boards&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The result is that Alice always has a square where she can guarantee a win:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$3 \times 3$: any square except B2 (example &lt;a href=&quot;https://i.stack.imgur.com/FBDLX.png&quot; rel=&quot;noreferrer&quot;&gt;win&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;$4 \times 4$: any corner square (example &lt;a href=&quot;https://i.stack.imgur.com/hNgpK.png&quot; rel=&quot;noreferrer&quot;&gt;win&lt;/a&gt;/&lt;a href=&quot;https://i.stack.imgur.com/Zz0xk.png&quot; rel=&quot;noreferrer&quot;&gt;loss&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;$5 \times 5$: A1, C2 and any mirrored/rotated version of them (example &lt;a href=&quot;https://i.stack.imgur.com/3OdHD.png&quot; rel=&quot;noreferrer&quot;&gt;win&lt;/a&gt;/&lt;a href=&quot;https://i.stack.imgur.com/e7bQB.png&quot; rel=&quot;noreferrer&quot;&gt;loss&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;$6 \times 6$: C2 and any mirrored/rotated version (example &lt;a href=&quot;https://i.stack.imgur.com/4dwQR.png&quot; rel=&quot;noreferrer&quot;&gt;win&lt;/a&gt;/&lt;a href=&quot;https://i.stack.imgur.com/Ss6aO.png&quot; rel=&quot;noreferrer&quot;&gt;loss&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;$7 \times 7$: any black square (example &lt;a href=&quot;https://i.stack.imgur.com/xlPBD.png&quot; rel=&quot;noreferrer&quot;&gt;win&lt;/a&gt;/&lt;a href=&quot;https://i.stack.imgur.com/sElIE.png&quot; rel=&quot;noreferrer&quot;&gt;loss&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;The code&lt;/h2&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import java.time.Duration;&#xA;import java.time.Instant;&#xA;import java.util.Arrays;&#xA;&#xA;public class Main {&#xA;    private static final int SIZE = 6;&#xA;    private static final int[] ROW_OFFSET = new int[] { 2, 1, -1, -2, -2, -1, 1, 2 };&#xA;    private static final int[] COL_OFFSET = new int[] { 1, 2, 2, 1, -1, -2, -2, -1 };&#xA;&#xA;    private static class Square {&#xA;        /** Usual naming: A1, B2, C3, ... */&#xA;        public final String name;&#xA;        /** Possible knight moves from this square. */&#xA;        public Square[] moves = new Square[0];&#xA;        /** Marks that this square was already used in the current variant. */&#xA;        public boolean used = false;&#xA;&#xA;        public Square(int row, int col) {&#xA;            super();&#xA;            this.name = (char) ('A' + col) + &quot;&quot; + (1 + row);&#xA;        }&#xA;&#xA;        /** Append a move to the array. */&#xA;        public void addMove(Square square) {&#xA;            int tmp = moves.length;&#xA;            moves = Arrays.copyOf(moves, tmp + 1);&#xA;            moves[tmp] = square;&#xA;        }&#xA;    }&#xA;&#xA;    /** Create a new board with the given size. */&#xA;    private static Square[][] createBoard(int size) {&#xA;        Square[][] board = new Square[size][size];&#xA;        // Create all squares first, so that they can be referenced later.&#xA;        for (int row = 0; row &amp;lt; size; ++row) {&#xA;            for (int col = 0; col &amp;lt; size; ++col) {&#xA;                board[row][col] = new Square(row, col);&#xA;            }&#xA;        }&#xA;        // Precalculate all possible knight moves.&#xA;        for (int row = 0; row &amp;lt; size; ++row) {&#xA;            for (int col = 0; col &amp;lt; size; ++col) {&#xA;                for (int i = 0; i &amp;lt; ROW_OFFSET.length; ++i) {&#xA;                    int newRow = row + ROW_OFFSET[i];&#xA;                    int newCol = col + COL_OFFSET[i];&#xA;                    if (newRow &amp;gt;= 0 &amp;amp;&amp;amp; newRow &amp;lt; size &amp;amp;&amp;amp;&#xA;                            newCol &amp;gt;= 0 &amp;amp;&amp;amp; newCol &amp;lt; size) {&#xA;                        board[row][col].addMove(board[newRow][newCol]);&#xA;                    }&#xA;                }&#xA;            }&#xA;        }&#xA;        // Sort the possible moves by number of moves from the target square.&#xA;        for (int row = 0; row &amp;lt; size; ++row) {&#xA;            for (int col = 0; col &amp;lt; size; ++col) {&#xA;                Arrays.sort(board[row][col].moves, (o1, o2) -&amp;gt;&#xA;                        Integer.compare(o1.moves.length, o2.moves.length));&#xA;            }&#xA;        }&#xA;        return board;&#xA;    }&#xA;&#xA;    /**&#xA;     * Determine the winner of current position.&#xA;     * @param player '0' for the first moving player (Bob), '1' for the second&#xA;     *               moving player (Alice).&#xA;     * @param square Target square of the last move, or square chosen by Alice&#xA;     *               to start.&#xA;     * @return Number of the player who will win the game.&#xA;     */&#xA;    private static int winner(int player, Square square) {&#xA;        // Mark target square as used.&#xA;        square.used = true;&#xA;        try {&#xA;            boolean movesLeft = false;&#xA;            // Recursively check all possible moves.&#xA;            for (Square newSquare : square.moves) {&#xA;                if (!newSquare.used) {&#xA;                    movesLeft = true;&#xA;                    if (winner(1 - player, newSquare) == player) {&#xA;                        // If the current move wins the game for current player,&#xA;                        // there is no need to look further.&#xA;                        return player;&#xA;                    }&#xA;                }&#xA;            }&#xA;            // No winning move found. If there are no moves left at all we win,&#xA;            // otherwise we loose.&#xA;            return movesLeft ? 1 - player : player;&#xA;        } finally {&#xA;            // Unmark target square before backtracking.&#xA;            square.used = false;&#xA;        }&#xA;    }&#xA;&#xA;    public static void main(String[] args) {&#xA;        Instant begin = Instant.now();&#xA;        // Check all unique starting squares.&#xA;        for (int row = 0; row &amp;lt; (SIZE + 1) / 2; ++row) {&#xA;            for (int col = row; col &amp;lt; (SIZE + 1) / 2; ++col) {&#xA;                // Each square is checked by a separate thread, so each of them&#xA;                // need a new copy of the board.&#xA;                Square square = createBoard(SIZE)[row][col];&#xA;                new Thread() {&#xA;                    @Override&#xA;                    public void run() {&#xA;                        System.out.println(square.name + &quot; &quot; +&#xA;                                winner(0, square) + &quot; &quot; +&#xA;                                Duration.between(begin, Instant.now()));&#xA;                    }&#xA;                }.start();&#xA;            }&#xA;        }&#xA;    }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2015-11-21T14:11:05.470" LastActivityDate="2015-11-21T14:11:05.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24024" PostTypeId="1" AcceptedAnswerId="24031" CreationDate="2015-11-18T05:31:30.120" Score="8" ViewCount="2855" Body="&lt;p&gt;In a family, there are the following human members:&#xA;Grandfather, grandmother, father, mother, son, daughter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All family members are inside a single room now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question:&#xA;What is the minimum amount of people inside the room now?&lt;/p&gt;&#xA;" OwnerUserId="17405" LastActivityDate="2015-11-19T15:11:50.263" Title="A tiny question: The little big family" Tags="|logical-deduction|family|" AnswerCount="10" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24025" PostTypeId="2" ParentId="24024" CreationDate="2015-11-18T05:48:38.497" Score="18" Body="&lt;p&gt;There are at least&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4 people.&lt;br&gt;&#xA;&lt;br&gt;&#xA;  Consider a couple Anna and Bill. They have a son Carl, who has a daughter Dora.&lt;br&gt;&#xA;&lt;br&gt;&#xA;  Then,&lt;br&gt;&#xA;  - Anna is a &lt;strong&gt;grandmother&lt;/strong&gt; to Dora&lt;br&gt;&#xA;  - Bill is a &lt;strong&gt;grandfather&lt;/strong&gt; to Dora&lt;br&gt;&#xA;  - Bill is a &lt;strong&gt;father&lt;/strong&gt; to Carl (and Carl is a &lt;strong&gt;father&lt;/strong&gt; to Dora)&lt;br&gt;&#xA;  - Anna is a &lt;strong&gt;mother&lt;/strong&gt; to Carl&lt;br&gt;&#xA;  - Carl is a &lt;strong&gt;son&lt;/strong&gt; to Anna and Bill&lt;br&gt;&#xA;  - Dora is a &lt;strong&gt;daughter&lt;/strong&gt; to Carl&lt;br&gt;&#xA;&lt;br&gt;&#xA; You can't have less than this. You need at least 3 generations in order for someone to be a grandparent, for a minimum of 3 people. But since we need both a grandmother and a grandfather, the eldest generation must have 2 people, for a total of four. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="4421" LastEditDate="2015-11-18T23:02:58.420" LastActivityDate="2015-11-18T23:02:58.420" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24026" PostTypeId="2" ParentId="24024" CreationDate="2015-11-18T08:24:06.530" Score="16" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Two. A man: who is a grandfather, a father and a son; and a woman: who is a grandmother, a mother and a daughter. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17407" LastActivityDate="2015-11-18T08:24:06.530" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="24028" PostTypeId="1" AcceptedAnswerId="24445" CreationDate="2015-11-18T09:51:08.660" Score="20" ViewCount="1203" Body="&lt;p&gt;Somewhere in this universe, there is an extremely tiny planet covered entirely in water. The only inhabitable surface on The entire planet is a raised Platform that keeps the animals on the planet above the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only living things on this planet are seven peculiar animals. Nobody knows where these creatures came from or how they came to be - be it through magic or some accelerated evolutionary process (you can choose to explain this phenomenon However you like It, it's not important).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The very first creature to appear on the platform was a very sickly cat.&lt;br&gt;&#xA;The next animal to appear was a rather detestable and perverted snail.&lt;br&gt;&#xA;Then came an asthmatic dragon - (not technically a real animal, but hey this is not technically a real planet either),&lt;br&gt;&#xA;and then a hairy leopard,&lt;br&gt;&#xA;followed by a morally upright rooster.&lt;br&gt;&#xA;The sixth creature to arrive was a tiny, tasty lamb.&lt;br&gt;&#xA;The final animal was a very childish earthworm.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone explain the reason for the existence of this tiny planet?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is my first puzzle on this site! I hope I haven't made it too obscure or easy! I'll add hints if necessary.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hint 1: I have edited the text to intentionally introduce 4 capitalization mistakes that point to the two hints within.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14095" LastEditorUserId="8874" LastEditDate="2015-11-19T17:07:24.843" LastActivityDate="2015-12-14T16:12:13.080" Title="A curious little planet" Tags="|riddle|pattern|knowledge|" AnswerCount="3" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="24029" PostTypeId="2" ParentId="24024" CreationDate="2015-11-18T10:19:33.467" Score="5" Body="&lt;ul&gt;&#xA;&lt;li&gt;Any male human is a son&lt;/li&gt;&#xA;&lt;li&gt;Any female human is a daughter&lt;/li&gt;&#xA;&lt;li&gt;Any grandfather is a father&lt;/li&gt;&#xA;&lt;li&gt;Any grandmother is a mother&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; So two people satisfy all the qualifications: a grandfather and a grandmother&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2737" LastActivityDate="2015-11-18T10:19:33.467" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24030" PostTypeId="1" AcceptedAnswerId="24034" CreationDate="2015-11-18T10:30:47.890" Score="5" ViewCount="567" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;There are four girls in the maze.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;1: One with dark hair and wearing a green hat. &lt;br&gt;&#xA;2: One with brown skin wearing a black tank top. &lt;br&gt;&#xA;3: One who hit the wall and bloodied her head. &lt;br&gt;&#xA;4: One who ran along the path and hurt her leg.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Their names are Hazel, Azalea, Cathy and Ellie.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A. Hazel found a way out while taking Ellie in her arms.&lt;br&gt;&#xA;B. Azalea was left behind with Cathy.&lt;br&gt;&#xA;C. Hazel has a hat in her hand but it's not hers.&lt;br&gt;&#xA;D. Cathy screamed out loud for help at the sight of blood.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Question: Which number (1-4) matches with which name?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Hope this is not too easy, or weird. Please help me improve this question if you have any ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for visiting to see this post! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They all know each other and were together in a maze finding a way out&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11353" LastEditorUserId="9000" LastEditDate="2015-11-19T13:18:42.190" LastActivityDate="2015-11-19T13:18:42.190" Title="Four girls in a maze; find a match of their names and given numbers" Tags="|lateral-thinking|" AnswerCount="1" CommentCount="9" ClosedDate="2015-11-18T14:41:06.883" ContentLicense="CC BY-SA 3.0" />
  <row Id="24031" PostTypeId="2" ParentId="24024" CreationDate="2015-11-18T10:38:03.153" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Three people&lt;br&gt;&#xA; Consider Anna, Bill and Carl.&lt;br&gt;&#xA; Imagine a land where it is allowed to marry your own children&lt;br&gt;&#xA; Anna is Bill's &lt;strong&gt;daughter&lt;/strong&gt;. Therefore Bill is her &lt;strong&gt;father&lt;/strong&gt;.&lt;br&gt;&#xA; Anna marries Bill.&lt;br&gt;&#xA; They both get a &lt;strong&gt;son&lt;/strong&gt; Carl.&lt;br&gt;&#xA; So Bill is Carl's father and &lt;strong&gt;grandfather&lt;/strong&gt;.&lt;br&gt;&#xA; And Anna is Carl's &lt;strong&gt;mother&lt;/strong&gt;&lt;br&gt;&#xA; But, Anna is also Carl's (step-)&lt;strong&gt;grandmother&lt;/strong&gt; because she is married to his grandfather&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1748" LastEditorUserId="1748" LastEditDate="2015-11-18T10:43:29.267" LastActivityDate="2015-11-18T10:43:29.267" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24034" PostTypeId="2" ParentId="24030" CreationDate="2015-11-18T12:51:22.770" Score="4" Body="&lt;p&gt;I feel like the answer you're looking for is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1:Cathy, 2 Hazel, 3 Azalea, 4 Ellie&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Azalea was left behind with Cathy, Cathy screamed at blood, presumably not her own, so Azalea is 3. Hazel and Ellie got out, with Hazel supporting Ellie, since Ellie needed support, she's 4. That leaves 1 and 2 for Hazel and Cathy. Hazel doesn't have a hat, so she's not 1. Cathy 1, Hazel 2, Azalea 3, Ellie 4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However. this is not a logic puzzle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; These are all leaps of 'logic', and there's no purely logical 'solution' to this riddle. Seeing as Hazel has a hat that's not hers (or her partners) also implies that all 4 have had contact, which opens up the whole thing, and leads to the more logical conclusion that 1:Azalea, 2 Hazel, 3 Ellie, 4 Cathy, seeing as you're more likely to support someone that has hurt her leg, instead of carrying them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16846" LastEditorUserId="16846" LastEditDate="2015-11-19T10:54:00.473" LastActivityDate="2015-11-19T10:54:00.473" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24036" PostTypeId="1" AcceptedAnswerId="24038" CreationDate="2015-11-18T14:00:00.947" Score="0" ViewCount="12550" Body="&lt;p&gt;You are at a restaurant. There are a total of three people(including you). Now you order a pizza and the trouble is that all three(including you) want equal parts. You have only a knife. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you cut the pizza in three equal parts?&lt;/p&gt;&#xA;" OwnerUserId="17415" LastEditorUserId="17415" LastEditDate="2015-11-18T14:10:42.177" LastActivityDate="2015-11-18T15:28:01.497" Title="Divide a circle into three equal parts" Tags="|mathematics|geometry|dissection|" AnswerCount="3" CommentCount="8" ClosedDate="2015-11-18T15:38:06.150" ContentLicense="CC BY-SA 3.0" />
  <row Id="24037" PostTypeId="2" ParentId="24036" CreationDate="2015-11-18T14:22:52.033" Score="3" Body="&lt;p&gt;At the risk this is not the intended solution, but here goes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are assuming:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;you know the exact center point of the pizza&lt;/li&gt;&#xA;&lt;li&gt;you have a piece of clean string, equal to at least the pizza's radius. (dental floss?)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cut a straight line from the center of the circle to the outer edge. where the line would extend to the other outer edge, don't cut the entire line, but make a marking cut on the outer edge. Now take your piece of string, and 'measure' the radius of pizza (along the cut you first made). Using said lenght, use that as a radius, from the 'marking cut' you made on the opposite side. You can use that to find two intersecting points with a circle radius r from the 'marking cut' point with the circles edge. Mark those two points with a small cut. Now just make straight cuts from both of those new 'marking cuts' to the center.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why does this work?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cosine(60) = 1/2. Basically, the perpendicular bisector of a line cuts it in half and is easy to construct. The second set of marking cuts have a cosine of 1/2 (half the radius compared to radius). 2x that angle makes a 120 degree angle between the parts, which is what you need to form a perfect three way split on a 360 degree circle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Video of a similar method:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;(but using measuring to spot the half, which I conveniently replaced with the 'use floss to create the perpendicular bisector' thingy): &lt;a href=&quot;https://www.youtube.com/watch?v=wFAAyqE7HT8&quot; rel=&quot;nofollow&quot;&gt;Youtube video&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;This would also work if you think you have proficient eyesight to spot the middle of a line perfectly, and can but perfectly perpendicular without any assistance.&lt;/p&gt;&#xA;" OwnerUserId="2156" LastEditorUserId="2156" LastEditDate="2015-11-18T14:44:12.413" LastActivityDate="2015-11-18T14:44:12.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24038" PostTypeId="2" ParentId="24036" CreationDate="2015-11-18T14:33:00.973" Score="5" Body="&lt;p&gt;Solution&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Take your pizza and fold it like so to get the proper lines in the crust&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/fCaGx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fCaGx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA; As you can see it is now easy to equally divide your pizza into three parts.&lt;br&gt;&#xA; Only having a knife is no problem now!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;EDIT for even edges&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are many origami techniques to get the pizza to be divided evenly like in the steps shown below.&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/FRkgZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FRkgZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA; Once the edges line up properly and evenly, while folding the half circle into a &quot;Z&quot; type shape, that's when you can make a good crease into the crust and can then proceed with the knife!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16273" LastEditorUserId="16273" LastEditDate="2015-11-18T15:28:01.497" LastActivityDate="2015-11-18T15:28:01.497" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24039" PostTypeId="2" ParentId="24036" CreationDate="2015-11-18T14:34:15.423" Score="3" Body="&lt;p&gt;Different method then other answer, but works equally well using a piece of string / dental floss / whatever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are assuming:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;you know the exact center point of the pizza&lt;/li&gt;&#xA;&lt;li&gt;you have a piece of clean string, equal to at least the pizza's radius. (dental floss?)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Method:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; From the center of the circle, use the string to 'copy' the radius of the pizza. Pick a random point on the other edge, make a marking cut. Use the string to find the intersections of a circle with radius r with the edge of the pizza (two such points), mark them with cuts. From one of the cuts, repeat the method, you'll find two points (the one you came from, and one new). Repeat two more times to find a total of 6 cuts. Then, cut in straight lines from center to cuts on the outside, each time skipping one of the cuts. You'll end up with 3 120 degree parts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Graphic version:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/sI9GW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sI9GW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA; Of course you don't cut the triangle, but lines from center to the 'triangle points' on the outer edge.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(not mine, found it on this page): &lt;a href=&quot;http://counton.org/xplusyfiles/home/issue_1/running_round_in_circles/index.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2156" LastEditorUserId="2156" LastEditDate="2015-11-18T14:53:29.143" LastActivityDate="2015-11-18T14:53:29.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24040" PostTypeId="1" AcceptedAnswerId="24053" CreationDate="2015-11-18T15:08:32.947" Score="5" ViewCount="618" Body="&lt;p&gt;&lt;sub&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/23996/magic-the-gathering-challenge-3-legends-rule&quot;&gt;Previous Challenge&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/24081/magic-the-gathering-challenge-5-revenge-of-the-pod&quot;&gt;Next Challenge&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;BACKGROUND:&lt;/h3&gt;   &#xA;&#xA;&lt;p&gt;The usual magic rules apply here. Please don't post a solution that if a solution for more damage already exists. Please be explicit in mana usage in your solution. I've tried to ensure that there are no infinite loops available, but if I somehow missed one (that Kiki-Jiki's always causing trouble), please inform me and I'll remove the required components from the puzzle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this puzzle, assume that unless otherwise modified, the bottom of your library will always be 50 mountains, regardless of shuffling. Similarly, assume that beneath any cards deliberately placed on the top of your library, there will always be 50 mountains regardless of shuffling. Basically, you can't just activate Grenzo blindly and say that you hit whatever you want. Cards on the top or bottom cannot be assumed to exist there unless deliberately placed there in your solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh and for those wondering, no, the punny titles will not stop as long as I can help it.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;PUZZLE SETUP:&lt;/h3&gt;  &#xA;&#xA;&lt;p&gt;It is your main phase 1. You have just had a &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=15417&quot; rel=&quot;nofollow noreferrer&quot;&gt;Goblin Recruiter&lt;/a&gt; enter the battlefield (don't worry about how, it's magic) and are about to resolve its ability. Defeat your opponent this turn by doing as much damage as possible! The solution that gets the most damage through wins. I won't tell my current best record as I don't want to dissuade other answers that do less damage from getting posted initially, but suffice to say that it is in the triple digits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your hand:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=394025&quot; rel=&quot;nofollow noreferrer&quot;&gt;Demonic Tutor&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=39846&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brightstone Ritual&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your board: (all untapped; recruiter has summoning sickness, the others do not)&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=382276&quot; rel=&quot;nofollow noreferrer&quot;&gt;Grenzo, Dungeon Warden&lt;/a&gt; with a +1/+1 counter, making it a 3/3&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=376569&quot; rel=&quot;nofollow noreferrer&quot;&gt;Viscera Seer&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=405389&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sol Ring&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=15417&quot; rel=&quot;nofollow noreferrer&quot;&gt;Goblin Recruiter&lt;/a&gt;&lt;br&gt;&#xA;1 Forest&lt;br&gt;&#xA;2 Mountain&lt;br&gt;&#xA;2 Swamp&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5 life&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your graveyard:&lt;/strong&gt;&lt;br&gt;&#xA;Nothing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your library:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=220147&quot; rel=&quot;nofollow noreferrer&quot;&gt;Adaptive Automaton&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=19712&quot; rel=&quot;nofollow noreferrer&quot;&gt;Battle Squadron&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=389439&quot; rel=&quot;nofollow noreferrer&quot;&gt;Beetleback Chief&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=74441&quot; rel=&quot;nofollow noreferrer&quot;&gt;Blazing Shoal&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=39846&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brightstone Ritual&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=401840&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chasm Guide&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=393945&quot; rel=&quot;nofollow noreferrer&quot;&gt;Clickslither&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=397422&quot; rel=&quot;nofollow noreferrer&quot;&gt;Coat of Arms&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=370699&quot; rel=&quot;nofollow noreferrer&quot;&gt;Door of Destinies&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=370480&quot; rel=&quot;nofollow noreferrer&quot;&gt;Empty the Warrens&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=177501&quot; rel=&quot;nofollow noreferrer&quot;&gt;Goblin Bushwhacker&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=221201&quot; rel=&quot;nofollow noreferrer&quot;&gt;Goblin Chieftain&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=44317&quot; rel=&quot;nofollow noreferrer&quot;&gt;Goblin Clearcutter&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=382957&quot; rel=&quot;nofollow noreferrer&quot;&gt;Goblin General&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=129578&quot; rel=&quot;nofollow noreferrer&quot;&gt;Goblin King&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=45123&quot; rel=&quot;nofollow noreferrer&quot;&gt;Goblin Lookout&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=15212&quot; rel=&quot;nofollow noreferrer&quot;&gt;Goblin Marshal&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=393962&quot; rel=&quot;nofollow noreferrer&quot;&gt;Goblin Matron&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=393963&quot; rel=&quot;nofollow noreferrer&quot;&gt;Goblin Ringleader&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=393973&quot; rel=&quot;nofollow noreferrer&quot;&gt;Goblin War Marshal&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=44292&quot; rel=&quot;nofollow noreferrer&quot;&gt;Goblin War Strike&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=393965&quot; rel=&quot;nofollow noreferrer&quot;&gt;Goblin Warchief&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=213782&quot; rel=&quot;nofollow noreferrer&quot;&gt;Goblin Wardriver&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=389537&quot; rel=&quot;nofollow noreferrer&quot;&gt;Goblin Welder&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=380431&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hall of Triumph&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=207924&quot; rel=&quot;nofollow noreferrer&quot;&gt;Heartstone&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=386327&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hellraiser Goblin&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=397698&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kiki-Jiki, Mirror Breaker&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=386339&quot; rel=&quot;nofollow noreferrer&quot;&gt;Krenko, Mob Boss&lt;/a&gt;&lt;br&gt;&#xA;100 Mountain&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=370518&quot; rel=&quot;nofollow noreferrer&quot;&gt;Murderous Redcap&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=383062&quot; rel=&quot;nofollow noreferrer&quot;&gt;Reckless Charge&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=393981&quot; rel=&quot;nofollow noreferrer&quot;&gt;Reckless One&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=393982&quot; rel=&quot;nofollow noreferrer&quot;&gt;Siege-Gang Commander&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=383145&quot; rel=&quot;nofollow noreferrer&quot;&gt;Vampiric Tutor&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=191368&quot; rel=&quot;nofollow noreferrer&quot;&gt;Warren Instigator&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=402100&quot; rel=&quot;nofollow noreferrer&quot;&gt;Zada, Hedron Grinder&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Opponent's hand:&lt;/strong&gt;&lt;br&gt;&#xA;Nothing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Opponent's board:&lt;/strong&gt;&lt;br&gt;&#xA;Nothing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;20 life&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Opponent's graveyard:&lt;/strong&gt;&lt;br&gt;&#xA;Nothing&lt;/p&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.923" LastActivityDate="2017-03-11T02:52:10.860" Title="&quot;Magic: the Gathering&quot; Challenge #4: Stuck in the Grenzone" Tags="|calculation-puzzle|magic-the-gathering|" AnswerCount="3" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="24041" PostTypeId="2" ParentId="24024" CreationDate="2015-11-18T15:21:19.403" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Depending on how you interpret some tricky relationships, there could be only &lt;strong&gt;3&lt;/strong&gt; people in the room.&lt;br&gt;&#xA; &lt;em&gt;Bill&lt;/em&gt; is in the room, and is the &lt;strong&gt;father&lt;/strong&gt; to &lt;em&gt;David&lt;/em&gt; and &lt;strong&gt;grandfather&lt;/strong&gt; to &lt;em&gt;Emma&lt;/em&gt;. &lt;em&gt;Emma&lt;/em&gt; is, obviously, the &lt;strong&gt;daughter&lt;/strong&gt; of &lt;em&gt;David&lt;/em&gt;. And &lt;em&gt;David&lt;/em&gt; is the &lt;strong&gt;son&lt;/strong&gt; of &lt;em&gt;Bill&lt;/em&gt;.&lt;br&gt;&#xA; Now, what you didn't know was that &lt;em&gt;Bill&lt;/em&gt; used to be &lt;em&gt;Anna&lt;/em&gt; before &lt;em&gt;gender-transitioning&lt;/em&gt; from female to male, and &lt;em&gt;Anna&lt;/em&gt; begot &lt;em&gt;Christina&lt;/em&gt; through cloning.&lt;br&gt;&#xA; Wait, who's &lt;em&gt;Christina&lt;/em&gt;? Well, &lt;em&gt;David&lt;/em&gt; used to be &lt;em&gt;Christina&lt;/em&gt; before his own gender-transition, which is when (s)he gave birth to &lt;em&gt;Emma&lt;/em&gt; - again through cloning.&lt;br&gt;&#xA; So, &lt;em&gt;Bill&lt;/em&gt; is also the &lt;strong&gt;mother&lt;/strong&gt; of &lt;em&gt;David&lt;/em&gt;, and is also the &lt;strong&gt;grandmother&lt;/strong&gt; of &lt;em&gt;Emma&lt;/em&gt;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The modern age, folks: giving us new and interesting answers to age old puzzles!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (I realize David doesn't really have to have switched genders, it could just be that Emma's mom is dead, but this way just feels more complete and less morbid - and also, if you go for cloning in the answer, why not go for it twice anyway!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Okay, writing that out made me think of another possibility: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Warning: potentially squicky. Again, &lt;strong&gt;3&lt;/strong&gt; people, with tricky relationships that are subject to interpretation.&lt;br&gt;&#xA; &lt;em&gt;Alice&lt;/em&gt; gave birth to &lt;em&gt;Derek&lt;/em&gt; through cloning, thus becoming the &lt;strong&gt;mother&lt;/strong&gt; to &lt;em&gt;Derek&lt;/em&gt; who is her &lt;strong&gt;son&lt;/strong&gt;.&lt;br&gt;&#xA; Then, &lt;em&gt;Alice&lt;/em&gt; and &lt;em&gt;Derek&lt;/em&gt; gave birth to &lt;em&gt;Eva&lt;/em&gt;, who is thus their &lt;strong&gt;daughter&lt;/strong&gt;. &lt;em&gt;Derek&lt;/em&gt; is now a &lt;strong&gt;father&lt;/strong&gt;. &lt;em&gt;Alice&lt;/em&gt; is both &lt;strong&gt;mother&lt;/strong&gt; and &lt;strong&gt;grandmother&lt;/strong&gt; to &lt;em&gt;Eva&lt;/em&gt;. &#xA; Then, &lt;em&gt;Alice&lt;/em&gt; chose to transition into &lt;em&gt;Bob&lt;/em&gt;, thus becoming also the &lt;strong&gt;grandfather&lt;/strong&gt; of &lt;em&gt;Eva&lt;/em&gt;.&lt;br&gt;&#xA; (What are you looking at me like that for? I warned ya! :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14122" LastEditorUserId="14122" LastEditDate="2015-11-18T16:11:41.013" LastActivityDate="2015-11-18T16:11:41.013" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24042" PostTypeId="2" ParentId="24024" CreationDate="2015-11-18T15:48:56.510" Score="1" Body="&lt;p&gt;Four. Here's one way:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Alice has a son named Bob, who has a daughter named Carol. So, Alice is a mother and a grandmother, Bob is a father and a son, and Carol is a daughter and a granddaughter. They are all in the room. Then, Bob's paternal grandfather Dennis enters the room. So we have:&lt;/p&gt;&#xA;&lt;p&gt;Grandmother: Alice (to Carol)&lt;/p&gt;&#xA;&lt;p&gt;Grandfather: Dennis (to Bob)&lt;/p&gt;&#xA;&lt;p&gt;Father: Bob (to Carol)&lt;/p&gt;&#xA;&lt;p&gt;Son: Bob (to Alice)&lt;/p&gt;&#xA;&lt;p&gt;Daughter: Carol (to Bob)&lt;/p&gt;&#xA;&lt;p&gt;Granddaughter: Carol (to Alice)&lt;/p&gt;&#xA;&lt;p&gt;Grandson: Bob (to Dennis)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Four people, four generations, six roles, no two people filling the same roles, and no person filling the same role twice.&lt;/p&gt;&#xA;" OwnerUserId="16945" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-11-18T15:55:59.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24043" PostTypeId="2" ParentId="24024" CreationDate="2015-11-18T16:01:30.543" Score="0" Body="&lt;p&gt;Unless there's a little bit more clarification on the question...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It could simply be two grandparents. Both of them were born a son or daughter, then had their own children who had children. This means each member is a Grandfather, father, and son, or Grandmother, mother, and daughter all at once. If their children are (by some tragedy) all gone, then the &quot;whole family&quot; is in the room... (jeez that made me sad, just now)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unless we're talking about a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; hermaphroditic grandparent, in which case, we could squeeze it down to 1. I feel like this isn't what the question was getting at, though...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13053" LastActivityDate="2015-11-18T16:01:30.543" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24044" PostTypeId="1" AcceptedAnswerId="24710" CreationDate="2015-11-18T16:28:30.100" Score="16" ViewCount="948" Body="&lt;p&gt;&lt;em&gt;This question was inspired by something I realized while thinking about @Gamow's &lt;a href=&quot;https://puzzling.stackexchange.com/questions/23970/weighing-in-89-different-ways&quot;&gt;Weighing in 89 different ways&lt;/a&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a two-pan balance and $2015$ weights, with masses $1,2,4,\ldots, 2^{2014}$. You also have one coin with a mass that is at least $1$ and at most $2^{2014}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are exactly $2015$ ways to place the coin in the left pan of the balance, and some number of weights in either pan of the balance, such that both sides of the balance have the same total mass. &lt;em&gt;How many possibilities&lt;/em&gt; are there for the mass of the coin?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;How many ways are there to make $2^n$? What about $3\times2^n$, $5\times2^n$, $7\times2^n$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The 99th-smallest possibility is $1476395007\times2^{1970}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2b:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If we pad out the binary representation of $1476395007\times2^{1970}$ to $2015$ digits, the result is 14 zeroes, 1 one, 1 zero, 1 one, 1 zero, 27 ones, and 1970 zeros. The continued fraction $[14;1,1,1,1,27]$ is equal to $\frac{2015}{138}$. The appearance of $2015$ in the numerator is not a coincidence!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$1476395006\times2^{1970}\rightarrow[14;1,1,1,1,26]=\frac{1942}{133}\\1476395007\times2^{1970}\rightarrow[14;1,1,1,1,27]=\frac{2015}{138}\\1476395008\times2^{1970}\rightarrow[14;1,1,2]=\frac{73}{5}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://electronics.stackexchange.com/a/202024&quot;&gt;https://electronics.stackexchange.com/a/202024&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-12-20T02:01:44.400" Title="Weighing in 2015 different ways" Tags="|mathematics|combinatorics|weighing|number-theory|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24045" PostTypeId="2" ParentId="24024" CreationDate="2015-11-18T16:29:51.403" Score="0" Body="&lt;p&gt;Two. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every man is a son to either parent. Thus Grandfather is a son, and by definition a father at the same time.&#xA;Every woman is a daughter to either parent. Grandmother is a daughter, and by definition, a mother at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not knowing too much about genetics, it could possibly be &lt;strong&gt;ONE&lt;/strong&gt;. If said individual was a chimera, who was hermaphroditic. Technically they could be both son and daughter, and grandmother, mother, grandfather, father if one half of the chimera shared the female genes, and genitalia, and the other half shared the male genes and genitalia. I know, its a convoluted answer, but....&lt;/p&gt;&#xA;" OwnerUserId="17417" LastActivityDate="2015-11-18T16:29:51.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24046" PostTypeId="2" ParentId="20057" CreationDate="2015-11-18T16:34:52.693" Score="1" Body="&lt;p&gt;I have thougth about the N pirates challenge a little bit, but probably I made a mistake somewhere in my reasoning... but here we go. It's a little long and confusing, and english is not my main language, but anyway I'll do my best. Feel free to correct me or ask for clarifications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summary:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For n any positive integer and N number of pirates, and Z the position of the infected (1 the youngest, N the oldest):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If N = 2^n or N = 2^(n-1)+4, all pirates are saved, the oldest has the coin&lt;/li&gt;&#xA;&lt;li&gt;If 2^(n-1) + 4 &amp;lt; N &amp;lt; 2^n and Z between 2^(n-2)+2 and 2^(n-1)+4, all die until pirate 2^(n-1) + 4, the oldest has the coin.&lt;/li&gt;&#xA;&lt;li&gt;If 2^(n-1) + 4 &amp;lt; N &amp;lt; 2^n and Z &amp;lt; 2^(n-2)+2, all die until pirate 2^(n-1) + 2, the oldest has the coin.&lt;/li&gt;&#xA;&lt;li&gt;If 2^(n-1) + 4 &amp;lt; N &amp;lt; 2^n and Z &gt; 2^(n-1)+2, all die until pirate 2^(n-1) + 2, one of the pirates &amp;lt; 2^(n-2)+2 gets the coin.&lt;/li&gt;&#xA;&lt;li&gt;Every other case, all pirates die until pirate 2^(n-2)+2, one of the pirates &amp;lt; 2^(n-3)+2 gets the coin.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;First of all, we define a &quot;safe&quot; pirate if he will survive, no matter where the infected is.&#xA;And we define a &quot;desperate&quot; pirate (or &quot;despirate&quot;) if there is a chance of death, even the littlest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will also call the pirates from 1 to N, 1 the youngest, N the oldest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there where no infected pirate, we can see that a safe pirate has always his vote, the votes of the despirates in his turn and one bribed pirate of his choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also see that any pirate with number less than a safe pirate becomes a safe pirate, because it will never be his turn. So, a pirate behind a safe pirate will always vote death, unless it is his closest (and greater than him) safe pirate's turn, or if he's bribed. That lets us with safe pirates at N = 2^n + 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now, with one infected pirate, we can`t know if he is one of the desperate or the safe pirates. If he is one of the desperate, then if we bribe one pirate, we lose the infected vote. If we don't bribe anyone, then we lose the bribed pirate's vote, so in any case we have one vote less than the no infected case. It's analogous with the case of the infected being one of the safe pirates, either we lose the bribed pirate's vote, or the infected, leaving us in the same state as with no infected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, our first pirate will decide to keep the coin. Now it's time to decide when a pirate is safe and when is not. Our first pirate will always think he is in the worst case scenario, that is, the infected is one of the desperate pirates. This is the same as to think that there is not an infected pirate and he can't bribe anyone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, a safe pirate is one pirate that has half of the pirates being desperate (counting him). So if the previous safe pirate is n, the next safe pirate will be 2n. All pirates between n and 2n are despirates, because if the infected happens to be among them, they will die if his turn comes, so all of them must vote for 2n to be 100% sure of surviving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Tyler Seacrest answer and comment we can see that pirates 5, 6 and 7 are despirates (they can die if the infected is one of them), so pirate 8 is safe, he has the votes of 5, 6, 7 and 8 at least (one more if the infected is not amongst them). The next safe pirate will be 16, then 32, 64... all powers of 2. Even 2 and 4 are safe pirates, so we can say that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If N = 2^n, with n natural number, all of them will survive, with the oldest pirate keeping the coin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if N is not 2^n we can say wolog that 2^(n-1) &amp;lt; N &amp;lt; 2^n for some n. Let's say that N = 2^n - 1. If the infected pirate is among the despirates, the he will die. If the infected is among the safe ones and all the desperate pirates vote for him, he would survive. But as the number of pirates at this point is odd, he have one more vote than needed, meaning that under the same circunstances, pirate N-1 would survive too. So pirate N-1 will vote to kill him, as the chances of survival are the same (pirate N-2 can't do that to pirate N-1 because he won't get enough votes to survive). Also, every other pirate should vote to kill him, because if they can save him, they can also save the next pirate, so is best to kill him and save the next.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in this case, every pirate votes to kill but the oldest and the infected, giving us the information about who the infected is, changing the safe-desperate pattern. Now they know exactly how many votes they will get for sure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this moment, pirate 2^(n-1) + 2 (let's call him M) is safe if the infected is between 1 and 2^(n-2)+2 (both included) or greater than 2^(n-1) + 2. If the infected is not in that interval, the greatest safe pirate will be 2^(n-1) + 4. In both cases the pirate M and all pirates &amp;lt; M will survive, so they vote to kill N and get the information. The next pirates are killed until pirate 2^(n-1) + 4 (let's call him L). If L is the first pirate to vote (meaning that the total number of pirates is L, so no one has any information), all pirates from him to 2^(n-2)+2 will vote to save him, in fear of the infected being amongst them. If he's not the first but the infected happens to be amongst them, they will also save him. But if he is not the first and the infected  is not amongst them, he and the next one will be killed, and M will be saved.&#xA;For 2^(n-1) + 3 there are not enough pirates in fear to stop the killing, so he will die, and pirate M depends on where the infected shows to know if he survives or not.&#xA;For 2^(n-1) + 2 the same as above, he is killed because not enough pirates are afraid if he is unlucky (and the slaughter will continue until pirate 2^(n-2) + 2, who emerges victorious but poor).&#xA;For 2^(n-1) + 1 he will be killed either way, and again 2^(n-2) + 2 wins using the coin to bribe someone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that's all, sorry for the long answer.&lt;/p&gt;&#xA;" OwnerUserId="14068" LastActivityDate="2015-11-18T16:34:52.693" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24048" PostTypeId="2" ParentId="24040" CreationDate="2015-11-18T16:57:49.553" Score="1" Body="&lt;p&gt;Figured I'd put up an answer, though I don't know if it's optimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Goblin Recruiter puts on top of library, in order from top to bottom:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Siege-Gang Commander&lt;/li&gt;&#xA;&lt;li&gt;Kiki-Jiki, Mirror Breaker&lt;/li&gt;&#xA;&lt;li&gt;Goblin Ringleader&lt;/li&gt;&#xA;&lt;li&gt;Warren Instigator&lt;/li&gt;&#xA;&lt;li&gt;Goblin Marshal&lt;/li&gt;&#xA;&lt;li&gt;Goblin Chieftain&lt;/li&gt;&#xA;&lt;li&gt;Zada, Hedron Grinder&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Krenko, Mob Boss&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Sacrifice Goblin Recruiter to Viscera Seer, scrying Siege-Gang Commander to bottom of library.&lt;/li&gt;&#xA;&lt;li&gt;Tap Sol Ring, use 2 mana to activate Grenzo, putting Siege-Gang Commander into play, getting three 1/1 Goblins.&lt;/li&gt;&#xA;&lt;li&gt;Sacrifice goblin token to Viscera Seer, scrying Kiki-Jiki to bottom of library.&lt;/li&gt;&#xA;&lt;li&gt;Tap Forest and Swamp, use mana to activate Grenzo, putting Kiki-Jiki into play.&lt;/li&gt;&#xA;&lt;li&gt;Activate Kiki-Jiki targetting Siege-Gang Commander, putting three goblin tokens into play.&lt;/li&gt;&#xA;&lt;li&gt;Board: 2 Siege-Gang Commanders, 5 goblin tokens, Viscera Seer, Tapped Kiki-Jiki, Grenzo, 2 Mountains, 1 Swamp untapped.&lt;/li&gt;&#xA;&lt;li&gt;Cast Brightstone Ritual, tapping a mountain, adding 9 red mana to your mana pool.&lt;/li&gt;&#xA;&lt;li&gt;Sacrifice Kiki-Jiki, scrying Goblin Ringleader to the bottom of your library.&lt;/li&gt;&#xA;&lt;li&gt;Use two red mana to activate Grenzo, putting Goblin Ringleader into play, drawing Warren Instigator, Zada, Hedron Grinder, Goblin Marshal, and Goblin Chieftain.&lt;/li&gt;&#xA;&lt;li&gt;Sacrifice Viscera Seer, scrying Krenko, Mob Boss to the bottom of your library.&lt;/li&gt;&#xA;&lt;li&gt;Use two red mana to activate Grenzo, putting Krenko, Mob Boss into play.&lt;/li&gt;&#xA;&lt;li&gt;Board: 2 Siege-Gang Commander, 5 goblin tokens, Goblin Ringleader, Grenzo, Krenko, Mob Boss, Untapped Mountain and Swamp, 5 red mana floating.&lt;/li&gt;&#xA;&lt;li&gt;Use two red mana to cast Warren Instigator&lt;/li&gt;&#xA;&lt;li&gt;Use three red mana to cast Goblin Chieftain&lt;/li&gt;&#xA;&lt;li&gt;Use Swamp and Mountain to cast Demonic Tutor, putting Blazing Shoal into hand.&lt;/li&gt;&#xA;&lt;li&gt;Board: 2Siege-Gang Commanders, 5 goblin tokens, Goblin Ringleader, Grenzo, Krenko Mob Boss, Warren Instigator, Goblin Chieftain.&lt;/li&gt;&#xA;&lt;li&gt;Activate Krenko Mob Boss, making 12 goblin tokens.&lt;/li&gt;&#xA;&lt;li&gt;Attack with 2 Siege-Gang Commander, 17 goblin tokens, Goblin Ringleader, Grenzo, Warren Instigator, Goblin Chieftain.&lt;/li&gt;&#xA;&lt;li&gt;Warren Instigator triggers on first-strike damage, putting Zada, Hedron Grinder into play.&lt;/li&gt;&#xA;&lt;li&gt;Still in first strike damage step, cast Blazing Shoal pitching Goblin Marshal, targetting Zada, Hedron Grinder for +6/+0.&lt;/li&gt;&#xA;&lt;li&gt;All your creatures have +6/+0 in regular damage step.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Total Damage: &lt;strong&gt;2&lt;/strong&gt; Warren Instigator first strike damage + &lt;strong&gt;19*8&lt;/strong&gt; Damage (Warren Instigator regular damage, Goblin Chieftain, 17 goblin tokens) + &lt;strong&gt;3*9&lt;/strong&gt; Damage (2 Siege-Gang Commanders, Goblin Ringleader) + &lt;strong&gt;10&lt;/strong&gt; Grenzo Damage&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total Damage: &lt;strong&gt;191&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer improved with tip by Drunkwolf.&lt;/p&gt;&#xA;" OwnerUserId="10904" LastEditorUserId="10904" LastEditDate="2015-11-18T18:04:54.693" LastActivityDate="2015-11-18T18:04:54.693" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24049" PostTypeId="2" ParentId="24044" CreationDate="2015-11-18T17:10:34.277" Score="7" Body="&lt;h2&gt;Update&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Going back to my original post&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;...we are looking for the cardinality of the set $\{f(q) = 2015\}$ for all $q$ in the allowed range,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where $f$ is defined as &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $f(n) =\begin{cases}f(\frac{n}2) &amp;amp; \quad \text{if } n \text{ is even},\\ f(\frac{n+1}2) + f(\frac{n-1}2) &amp;amp; \quad \text{if } n \text{ is odd}\end{cases}$ and $f(0) = 1$ and $n&amp;gt;0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is explained as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $*****,y_1,y_2$ be an arbitrary number. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $y_2$ is 0, this cannot be further expanded. So we have to look at $y_1$. &lt;/li&gt;&#xA;&lt;li&gt;If $y_2$ is 1, we can leave it fixed as $1$ (hence, we can forget about it), or...&lt;/li&gt;&#xA;&lt;li&gt;we may expand it to $1,-1$. Now, the $1$ will interfere with $y_1$, so we have $*****,(y_1+1),-1$. We have fixed $y_2 = -1$, so we can further ignore it.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Again, if $y_1 = 0$, the expansion will stop here. For the starting conditions, the number of representations of $0$ is $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One problem remains, $f(1) = f(1)+f(0)$. This is because, given an infinite series of weights $(2^i)_{i&amp;gt;0}$ there are infinitely many representations of $1$. In the question, $i$ cannot be larger than $2^{2014}$, which gives a max recursion depth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sequence that arises is closely related to the &lt;a href=&quot;https://oeis.org/A002487&quot; rel=&quot;nofollow&quot;&gt;Stern-Brocot sequence&lt;/a&gt;, but I haven't found a good way to compute the cardinality of the set $\{f(q) = 2015\}$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;End of update&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;So...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...the weight &lt;em&gt;can&lt;/em&gt; be $1$ since then we have $[1]$, $[2-1]$, $[4-2-1]$,...,$[2^{2014}-2^{2013}-...-1]$ which are the only ways to get $1$ (I guess no proof is needed for this). There are exactly $2015$ such sums. Also, the weight may be $2^{2014}-1$ (and, of course, this is the maximum weight) since we have $[2^{2014}-1]$, $[2^{2014}-2+1]$, $[2^{2014}-4+2+1]$,...,$[2^{2014}-2^{2013}+...+1]$. There is obviously some symmetry here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This gives us a rough lower bound of $2$ :-)&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Examples&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Let us first study a couple of examples (msb is the left-most bit).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If we take a number $n=\underbrace{0\cdots0}_{m}1$, there are $m+1$ different representations.&lt;/li&gt;&#xA;&lt;li&gt;For $n=\underbrace{0\cdots0}_{m}11$, there are $2m+1$ different representations.&lt;/li&gt;&#xA;&lt;li&gt;More generally, for $n=\underbrace{0\cdots0}_{m_1}\underbrace{1\cdots1}_{m_2}$, there are $m_1m_2+1$ different representations.&lt;/li&gt;&#xA;&lt;li&gt;For $n=\underbrace{0\cdots0}_{m_1}\underbrace{1\cdots1}_{m_2}\underbrace{0\cdots0}_{m_3}$, there are $m_1m_2+1$ different representations, regardless of $m_3$ since there is no $1$ to expand.&lt;/li&gt;&#xA;&lt;li&gt;We omitt $n=\underbrace{1\cdots1}_{m_1}\underbrace{0\cdots0}_{m_2}\underbrace{1\cdots1}_{m_3}$, as this is larger than the allowed solution.&lt;/li&gt;&#xA;&lt;li&gt;For $n=\underbrace{0\cdots0}_{m_1}\underbrace{1\cdots1}_{m_2}\underbrace{0\cdots0}_{m_3}\underbrace{1\cdots1}_{m_4}$ we rewrite the expression. Let underline denote negative bits and denote an expandable set of bits with overline. We have the following:&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$0\cdots00,11\cdots11,0\cdots00,11\cdots11$ - $1$&lt;/li&gt;&#xA;&lt;li&gt;$ \overline{0\cdots01},\overline{00\cdots0\underline{1}},1\cdots11,00\cdots00$ - &#xA;third and fourth field does not change, so $m_1m_2$ possible representations.&lt;/li&gt;&#xA;&lt;li&gt;$ 0\cdots00,11\cdots11,\overline{0\cdots01},\overline{00\cdots0\underline{1}}$ - &#xA;first and second field does not change, so $m_3m_4$ possible representations.&lt;/li&gt;&#xA;&lt;li&gt;$\overline{0\cdots01}, 00\cdots00 ,\underline{1}\underline{1}\cdots\underline{1},\overline{00\cdots0\underline{1}}$ - $m_1m_4$ &#xA;representations.&lt;/li&gt;&#xA;&lt;li&gt;$\overline{0\cdots01}, \overline{00\cdots0\underline{1}} ,\overline{0\cdots01},\overline{00\cdots0\underline{1}}$ - $m_1m_2m_3m_4$  possible representations.&#xA;All these sets are disjoint and give in total $m_1m_2m_3m_4+m_1m_2+m_3m_4+m_1m_4+1 = $$m_3 m_4+m_1 (m_4+m_2 (m_3 m_4+1))+1$ representations, which corresponds to the numerator of $[m_1;m_2,m_3,m_4]$.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As a generalization of the size of the reprsentation sets, we have the recursion&#xA;$S_i = m_iS_{i-1} + S_{i-2}$, $S_0 = 1$, $S_{-1}=0$. Of course this generalization is non-trivial, but for now I will treat it handwavingly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see that it fits the continued fraction (from the hint).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The weights&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The maximum weight is obtained by $[1;1,1,1,1,1,1,1...]$, which corresponds to the Fibonacci sequence: in this case $2015 &amp;lt; F_{18}$. This is a bit too much to simply perform brute force search over (since all lengths up to 18 must be checked and with the first $0 \leq m_1 \leq 14$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that, to find the solutions, we should be able solve the non-linear equations modular&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{bmatrix}m_1 &amp;amp; 1 \\ 1 &amp;amp; 0\end{bmatrix}\begin{bmatrix}m_2 &amp;amp; 1 \\ 1 &amp;amp; 0\end{bmatrix}\cdots\begin{bmatrix}m_k &amp;amp; 1 \\ 1 &amp;amp; 0\end{bmatrix} \equiv \begin{bmatrix}0 &amp;amp; * \\*&amp;amp;*\end{bmatrix} \bmod \{5,13,31\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and reconstruct using Chinese remainder theorem, but it still requires a lot of work. I guess there is a smart way to exploit the structure or maybe a theorem regarding continued fractions that may be usable.&lt;/p&gt;&#xA;" OwnerUserId="12901" LastEditorUserId="12901" LastEditDate="2015-12-17T21:05:06.357" LastActivityDate="2015-12-17T21:05:06.357" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="24050" PostTypeId="2" ParentId="24040" CreationDate="2015-11-18T18:21:16.357" Score="3" Body="&lt;p&gt;Here's mine&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Goblin Recruiter puts on top of library, in order from top to bottom:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Goblin Marshal &lt;/li&gt;&#xA;&lt;li&gt;Kiki-Jiki, Mirror Breaker&lt;/li&gt;&#xA;&lt;li&gt;Goblin Chieftain&lt;/li&gt;&#xA;&lt;li&gt;Krenko, Mob boss&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Tap everything, in mana pool: &lt;img src=&quot;https://i.stack.imgur.com/D9G1K.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/D9G1K.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IYyeS.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IYyeS.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bFGus.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/m9Vh0.gif&quot; alt=&quot;]&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Viscera seer the recruiter, Marshal on bottom.&lt;/li&gt;&#xA;&lt;li&gt;Activate Grenzo, put Marshal in play, 2 tokens &lt;img src=&quot;https://i.stack.imgur.com/D9G1K.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/D9G1K.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IYyeS.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IYyeS.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bFGus.gif&quot; alt=&quot;]&quot;&gt;&lt;/li&gt;&#xA;&lt;li&gt;Viscera seer a token, Kiki-Jiki on bottom. -1 token&lt;/li&gt;&#xA;&lt;li&gt;Activate Grenzo, put Kiki jiki in play, active to copy Marshal, +2 tokens. (Tapped kiki, 2 Marshals, 3 tokens)&lt;img src=&quot;https://i.stack.imgur.com/D9G1K.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IYyeS.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bFGus.gif&quot; alt=&quot;]&quot;&gt;&lt;/li&gt;&#xA;&lt;li&gt;Viscera seer both marshals, +4 tokens, Chieftan and Krenko on bottom.&lt;/li&gt;&#xA;&lt;li&gt;Brightstone ritual, 9 goblins total (Kiki, 7 tokens, Grenzo) &lt;img src=&quot;https://i.stack.imgur.com/D9G1K.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/D9G1K.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/D9G1K.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/D9G1K.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/D9G1K.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/D9G1K.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/D9G1K.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/D9G1K.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/D9G1K.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IYyeS.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bFGus.gif&quot; alt=&quot;]&quot;&gt;&lt;/li&gt;&#xA;&lt;li&gt;Activate Grenzo x2, Chieftain and Krenko in play (11 goblins total) &lt;img src=&quot;https://i.stack.imgur.com/D9G1K.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/D9G1K.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/D9G1K.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/D9G1K.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/D9G1K.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IYyeS.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bFGus.gif&quot; alt=&quot;]&quot;&gt;&lt;/li&gt;&#xA;&lt;li&gt;Activate Krenko, +11 goblins (Kiki, Krenko (both tapped), Grenzo, 18 tokens, Chieftan)&lt;/li&gt;&#xA;&lt;li&gt;Demonic tutor Coat of Arms, hardcast, all goblins +21/+21 from coat (and 1 from the chieftain)&lt;/li&gt;&#xA;&lt;li&gt;Attack with 20 goblins, Grenzo is 25/25, rest is 23/23 and the seer&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Total damage: 463&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="16846" LastEditorUserId="30903" LastEditDate="2017-03-11T02:46:16.880" LastActivityDate="2017-03-11T02:46:16.880" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="24053" PostTypeId="2" ParentId="24040" CreationDate="2015-11-18T21:31:27.627" Score="5" Body="&lt;p&gt;Some slight improvements to the solution DrunkWolf used (I don't feel too bad about the similarities, I swear I came up with the base solution independently).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Goblin Recruiter puts the following on top of my library:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Goblin Marshal&lt;/li&gt;&#xA;&lt;li&gt;Kiki-Jiki, Mirror Breaker&lt;/li&gt;&#xA;&lt;li&gt;Goblin Chieftain&lt;/li&gt;&#xA;&lt;li&gt;Krenko, Mob boss&lt;/li&gt;&#xA;&lt;li&gt;Goblin Clearcutter&lt;/li&gt;&#xA;&lt;li&gt;Goblin Ringleader&lt;/li&gt;&#xA;&lt;li&gt;Goblin Welder&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Tap everything for &lt;img src=&quot;https://i.stack.imgur.com/c1khu.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nRiAc.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nRiAc.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ujvin.gif&quot; alt=&quot;]&quot;&gt;&lt;br&gt;&#xA;Sacrifice Recruiter to Viscera seer to scry Marshal to the bottom, then activate Grenzo to get Marshal (2 tokens) &lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nRiAc.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nRiAc.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ujvin.gif&quot; alt=&quot;]&quot;&gt;&lt;br&gt;&#xA;Sacrifice a token to bottom Kiki-Jiki, then activate Grenzo to get Kiki (1 token) &lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nRiAc.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nRiAc.gif&quot; alt=&quot;]&quot;&gt;&lt;br&gt;&#xA;Tap Kiki to copy Marshal (3 tokens)&lt;br&gt;&#xA;Sacrifice both Marshals to bottom Chieftain and Krenko (7 tokens)&lt;br&gt;&#xA;Cast Brightstone Ritual &lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nRiAc.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nRiAc.gif&quot; alt=&quot;]&quot;&gt;&lt;br&gt;&#xA;Activate Grenzo twice to get Chieftain and Krenko &lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nRiAc.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nRiAc.gif&quot; alt=&quot;]&quot;&gt;&lt;br&gt;&#xA;Sacrifice Kiki-Jiki to bottom Clearcutter&lt;br&gt;&#xA;Activate Grenzo to get Clearcutter &lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nRiAc.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nRiAc.gif&quot; alt=&quot;]&quot;&gt;&lt;br&gt;&#xA;Tap Clearcutter for three red &lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nRiAc.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nRiAc.gif&quot; alt=&quot;]&quot;&gt;&lt;br&gt;&#xA;Sacrifice Clearcutter to bottom Ringleader&lt;br&gt;&#xA;Activate Grenzo to get Ringleader, putting Goblin Welder into your hand. &lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nRiAc.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nRiAc.gif&quot; alt=&quot;]&quot;&gt;&lt;br&gt;&#xA;Cast Goblin Welder &lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nRiAc.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nRiAc.gif&quot; alt=&quot;]&quot;&gt;&lt;br&gt;&#xA;You now control Grenzo, Chieftain, Krenko, Ringleader, Welder and 7 tokens.  Activate Krenko for 12 tokens&lt;br&gt;&#xA;Cast Demonic Tutor getting Vampiric Tutor &lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nRiAc.gif&quot; alt=&quot;]&quot;&gt;&lt;br&gt;&#xA;Cast Vampiric Tutor putting Coat of Arms on top &lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uratb.gif&quot; alt=&quot;]&quot;&gt;&lt;br&gt;&#xA;Sacrifice Viscera Seer to bottom Coat of Arms&lt;br&gt;&#xA;Activate Grenzo to put Coat of Arms into the graveyard&lt;br&gt;&#xA;Activate Goblin Welder to swap Sol Ring for Coat of Arms  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attack with 27/27 Grenzo, 26/26 Ringleader, 25/25 Chieftain and nineteen 25/25 tokens, for a total of 553.&lt;/p&gt;&#xA;" OwnerUserId="7459" LastEditorUserId="30903" LastEditDate="2017-03-11T02:52:10.860" LastActivityDate="2017-03-11T02:52:10.860" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="24054" PostTypeId="2" ParentId="23469" CreationDate="2015-11-18T21:32:27.623" Score="2" Body="&lt;p&gt;Well, based on all the hints and 2012rcampion's new picture, it looks like the pattern is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a qwerty keyboard.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Based on that, it looks like the five missing pegs are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; two to fill in the &quot;W&quot;, two to fill in the &quot;E&quot;, and one to fill in the &quot;T&quot;.&#xA; Counting from the top left as (1,1)  that would be pegs at (4,2) and (7,2) for W, (8,1) and (8,3) for &quot;E&quot;, and (14,2) for &quot;T&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The name was presumably relevant because of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Antonín Dvořák, which sounds like the Dvorak keyboard (but it was named after a different Dvorak!)&#xA; The &lt;a href=&quot;https://en.wikipedia.org/wiki/QWERTY&quot; rel=&quot;nofollow&quot;&gt;QWERTY keyboard&lt;/a&gt; was invented in the early 1870s by Christopher Latham Sholes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17036" LastActivityDate="2015-11-18T21:32:27.623" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24055" PostTypeId="2" ParentId="24024" CreationDate="2015-11-18T22:16:35.573" Score="8" Body="&lt;p&gt;If you allow for time-travel, the answer is ONE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Read &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/All_You_Zombies&quot;&gt;All you Zombies&lt;/a&gt;&lt;/em&gt;.  The protagonist of the story is hermaphroditic, and is his/her own father and mother, therefore also his/her own grandfather and grandmother.&lt;/p&gt;&#xA;" OwnerUserId="11465" LastActivityDate="2015-11-18T22:16:35.573" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24057" PostTypeId="2" ParentId="24028" CreationDate="2015-11-19T10:58:05.460" Score="17" Body="&lt;p&gt;&lt;strong&gt;&quot;The entire planet is a raised Platform&quot;&lt;/strong&gt; and &lt;strong&gt;&quot;However you like It&quot;&lt;/strong&gt; &#xA;point towards&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;William Shakespeare:&lt;/strong&gt; the platform is a theater, the play is &quot;As you like it&quot;.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The concrete answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the beginning monologue &lt;strong&gt;&quot;All the world's a stage&quot;&lt;/strong&gt; of &quot;As you like it&quot; &#xA; (&lt;a href=&quot;https://en.wikipedia.org/wiki/All_the_world&amp;#39;s_a_stage&quot;&gt;wikipedia&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is the full text:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;And all the men and women merely players;&lt;br&gt;&#xA;They have their exits and their entrances,&lt;br&gt;&#xA;And one man in his time plays many parts,&lt;br&gt;&#xA;His acts being seven ages. At first the infant,&lt;br&gt;&#xA;Mewling and puking in the nurse's arms.&lt;br&gt;&#xA;Then, the whining school-boy with his satchel&lt;br&gt;&#xA;And shining morning face, creeping like snail&lt;br&gt;&#xA;Unwillingly to school. And then the lover,&lt;br&gt;&#xA;Sighing like furnace, with a woeful ballad&lt;br&gt;&#xA;Made to his mistress' eyebrow. Then, a soldier,&lt;br&gt;&#xA;Full of strange oaths, and bearded like the pard,&lt;br&gt;&#xA;Jealous in honour, sudden, and quick in quarrel,&lt;br&gt;&#xA;Seeking the bubble reputation&lt;br&gt;&#xA;Even in the cannon's mouth. And then, the justice,&lt;br&gt;&#xA;In fair round belly, with a good capon lined,&lt;br&gt;&#xA;With eyes severe, and beard of formal cut,&lt;br&gt;&#xA;Full of wise saws, and modern instances,&lt;br&gt;&#xA;And so he plays his part. The sixth age shifts&lt;br&gt;&#xA;Into the lean and slippered pantaloon,&lt;br&gt;&#xA;With spectacles on nose and pouch on side,&lt;br&gt;&#xA;His youthful hose, well saved, a world too wide&lt;br&gt;&#xA;For his shrunk shank, and his big manly voice,&lt;br&gt;&#xA;Turning again toward childish treble, pipes&lt;br&gt;&#xA;And whistles in his sound. Last scene of all,&lt;br&gt;&#xA;That ends this strange eventful history,&lt;br&gt;&#xA;Is second childishness and mere oblivion,&lt;br&gt;&#xA;Sans teeth, sans eyes, sans taste, sans everything.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-11-19T10:58:05.460" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24058" PostTypeId="1" AcceptedAnswerId="24063" CreationDate="2015-11-19T11:36:09.830" Score="6" ViewCount="8390" Body="&lt;p&gt;You have a disc. One half of the disc is painted white and the other half is black. The disc rotates in either a clockwise direction or an anti-clockwise direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two sensors which are not connected to the disc but are in a fixed position, pointed at the disc. They are at 90° from each other with respect to the center of the disc.  The sensors output the value &lt;code&gt;1&lt;/code&gt; when reading white and &lt;code&gt;0&lt;/code&gt; when reading black.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using only the output of the two sensors, find the direction in which the disc is rotating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0ymQs.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0ymQs.jpg&quot; alt=&quot;Diagram of Disc&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17437" LastEditorUserId="4421" LastEditDate="2015-11-19T20:01:34.703" LastActivityDate="2021-05-10T18:28:53.037" Title="Find the direction of the disc" Tags="|logical-deduction|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24059" PostTypeId="2" ParentId="24058" CreationDate="2015-11-19T12:08:01.667" Score="3" Body="&lt;p&gt;So if I'm getting this correct, we have a disc with radius R at the origin, and a sensor at (-R,0) and (0,R). The disc is half black, half white, with the dividing line between the colors being a straight line going through the center of the disc.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; If both sensors show 1, we know the black part of the disc is up. Then if the left sensor goes to 0 before the other one, we know the disc is rotating clockwise. If the right sensor goes out first the disc is rotating counter-clockwise.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If the sensors are 180 degrees apart, (-R,0) and (R,0), I don't believe this is solvable without &lt;a href=&quot;/questions/tagged/lateral-thinking&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;lateral-thinking&amp;#39;&quot; rel=&quot;tag&quot;&gt;lateral-thinking&lt;/a&gt; because you're unable to distinguish between clockwise with black side up and counterclockwise with black side down.&lt;/p&gt;&#xA;" OwnerUserId="16846" LastEditorUserId="69582" LastEditDate="2021-05-10T18:28:53.037" LastActivityDate="2021-05-10T18:28:53.037" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24060" PostTypeId="1" AcceptedAnswerId="24067" CreationDate="2015-11-19T12:09:15.650" Score="16" ViewCount="2764" Body="&lt;p&gt;Abby selects a non-negative integer $A$, and Bobby selects a &#xA;non-negative integer $B$. &#xA;They then both tell their number secretly to Summy.&#xA;Summy writes the numbers $5$, $8$, and $15$ on the blackboard, and &#xA;announces that one of these three numbers is the sum $A+B$.&#xA;Then the three go through several rounds of the following form:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Summy rings a bell.&lt;/li&gt;&#xA;&lt;li&gt;Abby writes on a slip of paper whether she does know or does not &#xA;know which of the numbers on the blackboard is the sum $A+B$.&lt;/li&gt;&#xA;&lt;li&gt;Bobby writes on a (different) slip of paper whether he does know or &#xA;does not know which of the numbers on the blackboard is the sum $A+B$.&lt;/li&gt;&#xA;&lt;li&gt;Abby and Bobby give their papers to Summy.&lt;/li&gt;&#xA;&lt;li&gt;Summy checks the papers. &#xA;If at least one of the papers says YES, then the process stops. &#xA;If both papers say NO, then the next round starts.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Abby and Bobby are absolutely honest and very very intelligent.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What is the maximum number of times that the bell &#xA;  might be rung before the process stops?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="57212" LastEditDate="2021-12-07T09:31:48.750" LastActivityDate="2021-12-07T09:31:48.750" Title="Abby and Bobby and three numbers on the blackboard" Tags="|mathematics|logical-deduction|meta-knowledge|blackboard|" AnswerCount="6" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24061" PostTypeId="2" ParentId="24060" CreationDate="2015-11-19T12:17:02.870" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Twice.&#xA; If either had a number higher than 8, he/she would've known the&#xA; first time. If either had a number higher than 5 the second time, they&#xA; would've known. Which leaves only the possibility of 5, meaning the&#xA; bell has rung for a maximum of 2 times.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17434" LastActivityDate="2015-11-19T12:17:02.870" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24063" PostTypeId="2" ParentId="24058" CreationDate="2015-11-19T12:23:55.523" Score="8" Body="&lt;p&gt;There are four possible initial states for the sensor.  There are two possible changes, one for clockwise, one for counter clockwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The disc is rotating clockwise if Sensor 1 changes to the value of Sensor 2 and Sensor 2 becomes Sensor 1 XOR 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The disc is rotating counter clockwise if Sensor 1 changes to the value of Sensor 2 XOR 1 and Sensor 2 becomes Sensor 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/u5jAr.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/u5jAr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1777" LastEditorUserId="1777" LastEditDate="2015-11-19T12:48:30.613" LastActivityDate="2015-11-19T12:48:30.613" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24064" PostTypeId="1" AcceptedAnswerId="24065" CreationDate="2015-11-19T13:21:26.530" Score="2" ViewCount="625" Body="&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/1963/6-the-magic-number&quot;&gt;Here&lt;/a&gt; is a question and some answer for it. Now my question is a bit outside the box. Well,  put operators between the numbers show to get the end result:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;10   10   10 = 6&#xA;11   11   11 = 6&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As you can see some examples in the above link, also here is an example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2 + 2 + 2 = 6&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So, what operators should I have between them for &lt;code&gt;10&lt;/code&gt; and &lt;code&gt;11&lt;/code&gt; ?&lt;/p&gt;&#xA;" OwnerUserId="17438" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-11-22T01:19:31.030" Title="6 - the magic number (version 2)" Tags="|mathematics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24065" PostTypeId="2" ParentId="24064" CreationDate="2015-11-19T13:27:13.760" Score="7" Body="&lt;p&gt;10.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(\sqrt{10-10/10})! = 6$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;11.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\lfloor\log 11\rfloor+\lfloor\log 11\rfloor+\lfloor\log 11\rfloor=6$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12901" LastEditorUserId="17438" LastEditDate="2015-11-19T13:58:18.923" LastActivityDate="2015-11-19T13:58:18.923" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24066" PostTypeId="2" ParentId="24060" CreationDate="2015-11-19T13:34:35.433" Score="8" Body="&lt;p&gt;The bell will be rung at most&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;10 times.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since the situation is symmetric, consider it from the point of view of A who only knows their number and how B would respond if they had certain numbers. We can see how long it will take to know the sum for each starting number as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;9-15: on first round, only option for sum is 15&lt;/li&gt;&#xA;&lt;li&gt;6: on 2nd round, know B is not 9 so sum is 8&lt;/li&gt;&#xA;&lt;li&gt;2: on 2nd round, know B is not 13; on 3rd round, know B is not 6 so sum is 5&lt;/li&gt;&#xA;&lt;li&gt;3: on 2nd round, know B is not 12; on 4th round, know B is not 2 so sum is 8&lt;/li&gt;&#xA;&lt;li&gt;5: on 2nd round, know B is not 10; on 5th round, know B is not 3 so sum is 5&lt;/li&gt;&#xA;&lt;li&gt;0: on 2nd round, know B is not 15; on 6th round, know B is not 5 so sum is 8&lt;/li&gt;&#xA;&lt;li&gt;8: on 7th round, know B is not 0 so sum is 15&lt;/li&gt;&#xA;&lt;li&gt;7: on 8th round, know B is not 8 so sum is 8&lt;/li&gt;&#xA;&lt;li&gt;1: on 2nd round, know B is not 14; on 9th round, know B is not 7 so sum is 5&lt;/li&gt;&#xA;&lt;li&gt;4: on 2nd round, know B is not 11; on 10th round, know B is not 1 so sum is 8&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This covers all numbers, so the upper bound for the maximum number of rings is 10. B may, of course, say &quot;yes&quot; before A does, but since it's possible for both numbers to be 4, it's indeed possible for the bell to ring 10 times. Thus the bell can be rung a maximum of 10 times.&lt;/p&gt;&#xA;" OwnerUserId="12511" LastActivityDate="2015-11-19T13:34:35.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24067" PostTypeId="2" ParentId="24060" CreationDate="2015-11-19T13:34:44.843" Score="23" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;10 times.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The following table gives the numbers of one person and the corresponding possible numbers of the other person.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;own  other&#xA;0    5, 8, 15&#xA;1    4, 7, 14&#xA;2    3, 6, 13&#xA;3    2, 5, 12&#xA;4    1, 4, 11&#xA;5    0, 3, 10&#xA;6    2, 9&#xA;7    1, 8 &#xA;8    0, 7&#xA;9    6&#xA;10   5&#xA;11   4&#xA;12   3&#xA;13   2&#xA;14   1&#xA;15   0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Round 1:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If one of them has a number greater than 8 he/she would know that the sum is 15 as it can't be 5 or 8 anymore.&lt;/p&gt;&#xA;&lt;p&gt;For the other numbers the following table is true(after both said no):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;own  other&#xA;0    5, 8 -&amp;gt; no 15 as the other person would have known it&#xA;1    4, 7 -&amp;gt; no 14 as the other person would have known it&#xA;2    3, 6 -&amp;gt; no 13 as the other person would have known it&#xA;3    2, 5 -&amp;gt; no 12 as the other person would have known it&#xA;4    1, 4 -&amp;gt; no 11 as the other person would have known it&#xA;5    0, 3 -&amp;gt; no 10 as the other person would have known it&#xA;6    2    -&amp;gt; no 9 as the other person would have known it&#xA;7    1, 8 &#xA;8    0, 7&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Round 2:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If one of them has a 6 he/she now knows the number of the other one&lt;/p&gt;&#xA;&lt;p&gt;For the other numbers the following table is true(after both said no):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;own  other&#xA;0    5, 8&#xA;1    4, 7&#xA;2    3    -&amp;gt; no 6 as the other person would have known it&#xA;3    2, 5&#xA;4    1, 4&#xA;5    0, 3&#xA;7    1, 8 &#xA;8    0, 7&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Round 3:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If one of them has a 2 he/she now knows the number of the other one&lt;/p&gt;&#xA;&lt;p&gt;For the other numbers the following table is true(after both said no):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;own  other&#xA;0    5, 8&#xA;1    4, 7&#xA;3    5    -&amp;gt; no 2 as the other person would have known it&#xA;4    1, 4&#xA;5    0, 3&#xA;7    1, 8 &#xA;8    0, 7&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Round 4:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If one of them has a 3 he/she now knows the number of the other one&lt;/p&gt;&#xA;&lt;p&gt;For the other numbers the following table is true(after both said no):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;own  other&#xA;0    5, 8&#xA;1    4, 7&#xA;4    1, 4&#xA;5    0    -&amp;gt; no 3 as the other person would have known it&#xA;7    1, 8 &#xA;8    0, 7&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Round 5:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If one of them has a 5 he/she now knows the number of the other one&lt;/p&gt;&#xA;&lt;p&gt;For the other numbers the following table is true(after both said no):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;own  other&#xA;0    8    -&amp;gt; no 5 as the other person would have known it&#xA;1    4, 7&#xA;4    1, 4&#xA;7    1, 8 &#xA;8    0, 7&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Round 6:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If one of them has a 0 he/she now knows the number of the other one&lt;/p&gt;&#xA;&lt;p&gt;For the other numbers the following table is true(after both said no):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;own  other&#xA;1    4, 7&#xA;4    1, 4&#xA;7    1, 8 &#xA;8    7    -&amp;gt; no 0 as the other person would have known it&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Round 7:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If one of them has a 8 he/she now knows the number of the other one&lt;/p&gt;&#xA;&lt;p&gt;For the other numbers the following table is true(after both said no):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;own  other&#xA;1    4, 7&#xA;4    1, 4&#xA;7    1    -&amp;gt; no 8 as the other person would have known it&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Round 8:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If one of them has a 7 he/she now knows the number of the other one&lt;/p&gt;&#xA;&lt;p&gt;For the other numbers the following table is true(after both said no):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;own  other&#xA;1    4    -&amp;gt; no 7 as the other person would have known it&#xA;4    1, 4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Round 9:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If one of them has a 1 he/she now knows the number of the other one&lt;/p&gt;&#xA;&lt;p&gt;For the other numbers the following table is true(after both said no):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;own  other&#xA;4    4    -&amp;gt; no 1 as the other person would have known it&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Round 10:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If they make it up to this round they now definitely know what number the other person has and at the same time what the sum of their numbers is.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&lt;p&gt;This is actually a kind of derivation of the blue eyes problem:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/236/in-the-100-blue-eyes-problem-why-is-the-oracle-necessary&quot;&gt;In the 100 blue eyes problem - why is the oracle necessary?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16608" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-11-19T13:45:31.653" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24068" PostTypeId="2" ParentId="24024" CreationDate="2015-11-19T15:11:50.263" Score="0" Body="&lt;p&gt;3-4&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If &lt;a href=&quot;https://en.wikipedia.org/wiki/Kourtney_Kardashian&quot; rel=&quot;nofollow&quot;&gt;Kourtney Kardashian&lt;/a&gt; decides to follow in her father's/mother's footsteps and becomes Kurtis Kardashian, then brings her son Mason and daughter Penelope to visit their grandmother &lt;a href=&quot;https://en.wikipedia.org/wiki/Caitlyn_Jenner&quot; rel=&quot;nofollow&quot;&gt;Caitlyn Jenner&lt;/a&gt;, then you will have a biological grandfather, socially grandmother, biological mother, socially father, and son and daughter with only four people. If either Mason or Penelope choose to change their sex, you only need three.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1682" LastActivityDate="2015-11-19T15:11:50.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24069" PostTypeId="2" ParentId="23941" CreationDate="2015-11-19T21:22:05.373" Score="3" Body="&lt;p&gt;I believe 3 is &lt;a href=&quot;https://en.wikipedia.org/wiki/Astrodome_(aeronautics)&quot; rel=&quot;nofollow&quot;&gt;astrodome&lt;/a&gt;/odometer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8 could be a lot of things, I think. The second word is &quot;inessential&quot;, so the first word is any synonym for strength that ends in -iness, such as &quot;sturdiness&quot; or &quot;brawniness&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: #7 I think was meant to be &quot;aliquot quotient&quot;.&lt;/p&gt;&#xA;" OwnerUserId="10861" LastEditorUserId="10861" LastEditDate="2015-11-19T21:44:49.610" LastActivityDate="2015-11-19T21:44:49.610" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24070" PostTypeId="2" ParentId="23941" CreationDate="2015-11-19T21:32:34.780" Score="4" Body="&lt;p&gt;&lt;sup&gt;&lt;em&gt;Since several people have provided different parts of the answer, I figured it made sense to make a single community-wiki answer that pulls together all the parts.&lt;/em&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Sometimes we torture people. Sometimes we don't. We ______ our ______.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; intersperse/persecution&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;I object to objects. In fact, I have a ______ for _______.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; loathing/things&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;I measured our location through the _______, while keeping track of our distance on the _______.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; astrodome/odometer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;The trouble with _________ _________, is that kids and drunkards end up dancing on them.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; unstable/tables&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;Despite his __________ tone, my sense were all ______.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; accommodating/atingle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;6&quot;&gt;&#xA;&lt;li&gt;The runny-nosed trebuchet operating was basically a ________ ________.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; snuffling flinger&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;7&quot;&gt;&#xA;&lt;li&gt;It's much smaller, but leaves a remainder. It's an ______ _______.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; piddling lingerer or aliquot quotient&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;8&quot;&gt;&#xA;&lt;li&gt;I don't care about the strength. _________ is _________.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sturdiness/inessential&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;9&quot;&gt;&#xA;&lt;li&gt;If ________ your _______ is confirmed, you may be eligible to become king!&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; perchance/ancestry&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;10&quot;&gt;&#xA;&lt;li&gt;They punched each other without speaking. The _______ were rather _____.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; fighters/terse&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="4421" LastEditDate="2015-11-19T21:52:56.690" LastActivityDate="2015-11-19T21:52:56.690" CommentCount="4" CommunityOwnedDate="2015-11-19T21:32:34.780" ContentLicense="CC BY-SA 3.0" />
  <row Id="24071" PostTypeId="2" ParentId="23939" CreationDate="2015-11-19T23:03:31.733" Score="20" Body="&lt;p&gt;footprint is for &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Q&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;criticism is for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; L&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;maturity is for &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; K&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;&lt;strong&gt;Explanation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letters of interest in each word are those with a dot or a crossbar. In this set, they would be the letters &lt;em&gt;i&lt;/em&gt;, &lt;em&gt;t&lt;/em&gt;, or &lt;em&gt;f&lt;/em&gt;. Letters with a dot denote a dot (or dit) in Morse code, while letters with a crossbar denote a dash (or dah). Extracting the special letters in the order that they occur in each word and translating them to Morse code gives us the below.&lt;br&gt;&#xA;&lt;br&gt;&#xA; awaited = dot dash = A&lt;br&gt;&#xA; multidimensional = dash dot dot dot = B&lt;br&gt;&#xA; terrified = dash dot dash dot = C&lt;br&gt;&#xA; friendship = dash dot dot = D&lt;br&gt;&#xA; hungarian = dot = E&lt;br&gt;&#xA; visiting = dot dot dash dot = F&lt;br&gt;&#xA; theatrical = dash dash dot = G&lt;br&gt;&#xA; indivisible = dot dot dot dot = H&lt;br&gt;&#xA; delicious = dot dot = I&lt;br&gt;&#xA; distasteful = dot dash dash dash = J&lt;br&gt;&#xA; footprint = dash dash dot dash = Q&lt;br&gt;&#xA; criticism = dot dash dot dot = L&lt;br&gt;&#xA; maturity = dash dot dash = K  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15921" LastActivityDate="2015-11-19T23:03:31.733" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24072" PostTypeId="2" ParentId="24028" CreationDate="2015-11-19T23:28:58.347" Score="2" Body="&lt;p&gt;I have no idea how far I am from the solution but I like the similarities I found. So, here it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;These are the members of G7 - The Platform;the members are:Canada, France, USA, Italy,Japan, UK, USA; and the Planet is the &#xA;financial system as a whole: it's fluid, it flows like water from one part of the world to the other&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The explanation is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Very sickly cat - cat in mythology is venerated as a special animal, but in Germany(and other European countries) it is considered as having 7 lives; well sickly one would have only one life and can account only for one country;&#xA;the Snail - they are eaten in Italy&#xA;Dragon - It was part of coat of arms in Great Britain, not sure now though;&#xA;Leopard - maybe it's hairy since it came from the north, Canada&#xA;Rooster - the term itself originated in USA&#xA;Lamb and Earthworm - can't find any interesting associations though for &gt;!the last 2: France and Japan&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user11082" LastActivityDate="2015-11-19T23:28:58.347" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24073" PostTypeId="2" ParentId="24060" CreationDate="2015-11-20T03:29:19.280" Score="2" Body="&lt;p&gt;Just as a side note, in addition to the correct answers already posted. It is interesting that the numbers 5, 8, 15 are kind of special. Most triples of numbers yield unsolvable games! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if the numbers given are 1, 2, 3, and neither player had picked three at the beginning, you can't get anywhere and the game never terminates. &lt;/p&gt;&#xA;" OwnerUserId="11644" LastActivityDate="2015-11-20T03:29:19.280" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24074" PostTypeId="2" ParentId="22836" CreationDate="2015-11-20T04:48:18.563" Score="4" Body="&lt;p&gt;The maximum I've been able to find is 50. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started off trying to get solutions by stacking various arrangements of the eight-queen solutions and filling in the leftover spaces optimally, but realized that to be a dreadfully inefficient (and far from optimal) exercise, and instead came up with a fairly straightforward recursive backtracking algorithm to populate an optimal cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So anyways, here's my solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Layer 1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BV6HD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BV6HD.png&quot; alt=&quot;Layer 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Layer 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Sb3rS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Sb3rS.png&quot; alt=&quot;Layer 2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Layer 3:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sNXXT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sNXXT.png&quot; alt=&quot;Layer 3&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Layer 4:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UZiJg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UZiJg.png&quot; alt=&quot;Layer 4&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Layer 5:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iK6sC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iK6sC.png&quot; alt=&quot;Layer 5&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Layer 6:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qfbyh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qfbyh.png&quot; alt=&quot;Layer 6&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Layer 7:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/motvF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/motvF.png&quot; alt=&quot;Layer 7&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Layer 8:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KNydA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KNydA.png&quot; alt=&quot;Layer 8&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the pattern isn't clear, here's what the solution would look like from a &quot;top-down&quot; view through all eight layers. Neat stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Dzhec.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Dzhec.png&quot; alt=&quot;All together, now&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17167" LastActivityDate="2015-11-20T04:48:18.563" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24080" PostTypeId="2" ParentId="18" CreationDate="2015-11-20T14:14:35.357" Score="4" Body="&lt;p&gt;In most of these sorts of puzzles, there is a single moment when you crack it. For this one, it's the realization that &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;you can bring something back&lt;/em&gt;. I mean, that's kind of crazy, I am trying to get everything from this bank to that bank, why would I bring anything in the wrong direction? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If it never occurs to you to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; bring anything back, the puzzle seems impossible. For the first step, you know it's ok to leave the fox with the beans, and it's not ok to leave the goose with anything, so you take it, fine. Then whatever you take over next you're going to have to leave with the goose while you go back for the third thing, and that won't work. If taking the goose back occurs to you here, you've cracked the puzzle and the rest is just mechanics.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Same thing with the limited-time lights and two-people-at-time-over-a-bridge thing. Most people want to pair slow with fast, or put two slows together but the key to solving it is actually pairing the two fastest. With the liars and truth tellers it's asking one what the other would say. And so on.&lt;/p&gt;&#xA;" OwnerUserId="1509" LastEditorUserId="46903" LastEditDate="2018-09-26T15:45:12.433" LastActivityDate="2018-09-26T15:45:12.433" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24081" PostTypeId="1" AcceptedAnswerId="24101" CreationDate="2015-11-20T14:16:53.273" Score="7" ViewCount="559" Body="&lt;p&gt;&lt;sub&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/24040/magic-the-gathering-challenge-4-stuck-in-the-grenzone&quot;&gt;Previous Challenge&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/24240/magic-the-gathering-challenge-6-b-its-gonna-cost-you&quot;&gt;Next Challenge&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;BACKGROUND:&lt;/h3&gt;   &#xA;&#xA;&lt;p&gt;The usual magic rules apply here. Please be explicit in mana usage in your solution where relevant. I've tried to ensure that there are no infinite loops available, but if I somehow missed one, please inform me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you draw any cards, you may assume that you draw whatever cards you require that are still in your library, regardless of shuffling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You guys asked for a hard one so here it is. This is going to be a pretty difficult one and may involve some complex interactions. Be sure to be careful with all steps and read all cards carefully. If you're fuzzy on a rule/interaction, &lt;a href=&quot;http://magic.wizards.com/en/gameinfo/gameplay/formats/comprehensiverules&quot; rel=&quot;nofollow noreferrer&quot;&gt;here is the comprehensive rulebook&lt;/a&gt; and &lt;a href=&quot;http://chat.magicjudges.org/mtgrules/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here is the mtg judge chat&lt;/a&gt; where you can ask rulings questions that you can't find on the web.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All permanents in the given setup below are owned by their controllers (recall that if a player loses the game, all permanents they own are removed from the game regardless of who controls them at the time)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your solution may not assume that an opponent will cooperate with you in some way (e.g. choose not to block when they have legal blockers). Your solution must be able to win regardless of any potential blocks made by your opponents.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;PUZZLE SETUP:&lt;/h3&gt;  &#xA;&#xA;&lt;p&gt;It's a game of 3-for-all! You're behind on life and on board and your opponents are trying to lock the game down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is your main phase 1. Given the board state given below, defeat both of your opponents this turn!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your hand:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=106657&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pull from Eternity&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your board: (all untapped; all permanents have been in continuous control since beginning turn)&lt;/strong&gt;&lt;br&gt;&#xA;2 &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=218006&quot; rel=&quot;nofollow noreferrer&quot;&gt;Birthing Pod&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=370458&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kitchen Finks&lt;/a&gt; (with no counters)&lt;br&gt;&#xA;25 Mountain&lt;br&gt;&#xA;5 Swamp&lt;br&gt;&#xA;1 Plains&lt;/p&gt;&#xA;&#xA;&lt;p&gt;25 life&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your graveyard:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=194052&quot; rel=&quot;nofollow noreferrer&quot;&gt;Necropede&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your library:&lt;/strong&gt;&lt;br&gt;&#xA;23 Swamp&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=233052&quot; rel=&quot;nofollow noreferrer&quot;&gt;Glistener Elf&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=36115&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hapless Researcher&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=3146&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nature's Chosen&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=249405&quot; rel=&quot;nofollow noreferrer&quot;&gt;Qasali Pridemage&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=50537&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dross Harvester&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=386337&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kathari Bomber&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=230788&quot; rel=&quot;nofollow noreferrer&quot;&gt;Laboratory Maniac&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=23043&quot; rel=&quot;nofollow noreferrer&quot;&gt;Shade's Form&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=222923&quot; rel=&quot;nofollow noreferrer&quot;&gt;Village Bell-Ringer&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=15138&quot; rel=&quot;nofollow noreferrer&quot;&gt;Academy Rector&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=122124&quot; rel=&quot;nofollow noreferrer&quot;&gt;Arcum Dagsson&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=386506&quot; rel=&quot;nofollow noreferrer&quot;&gt;Clever Impersonator&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=220529&quot; rel=&quot;nofollow noreferrer&quot;&gt;March of the Machines&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=214375&quot; rel=&quot;nofollow noreferrer&quot;&gt;Phyrexian Metamorph&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=227544&quot; rel=&quot;nofollow noreferrer&quot;&gt;Xenograft&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=39701&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cryptic Gateway&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=382983&quot; rel=&quot;nofollow noreferrer&quot;&gt;Karmic Guide&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=376446&quot; rel=&quot;nofollow noreferrer&quot;&gt;Phyrexian Delver&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=97109&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sky Hussar&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=271207&quot; rel=&quot;nofollow noreferrer&quot;&gt;Auratouched Mage&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=405400&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sun Titan&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=405126&quot; rel=&quot;nofollow noreferrer&quot;&gt;Angel of Serenity&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=383194&quot; rel=&quot;nofollow noreferrer&quot;&gt;Boonweaver Giant&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=366285&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sepulchral Primordial&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=198172&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gigantomancer&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=228242&quot; rel=&quot;nofollow noreferrer&quot;&gt;Trench Gorger&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Opponent 1's hand:&lt;/strong&gt;&lt;br&gt;&#xA;Nothing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Opponent 1's board: (all untapped)&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=279991&quot; rel=&quot;nofollow noreferrer&quot;&gt;Blood Reckoning&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=4584&quot; rel=&quot;nofollow noreferrer&quot;&gt;Peacekeeper&lt;/a&gt;&lt;br&gt;&#xA;2 &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=382362&quot; rel=&quot;nofollow noreferrer&quot;&gt;Silent Arbiter&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;60 life&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Opponent 1's graveyard:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=234446&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hamlet Captain&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=130997&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rage Weaver&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=368951&quot; rel=&quot;nofollow noreferrer&quot;&gt;Voice of Resurgence&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=368530&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hellrider&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=382324&quot; rel=&quot;nofollow noreferrer&quot;&gt;Phage the Untouchable&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Opponent 2's hand:&lt;/strong&gt;&lt;br&gt;&#xA;Nothing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Opponent 2's board: (all untapped)&lt;/strong&gt;&lt;br&gt;&#xA;2 &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=275264&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ghostly Prison&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=247147&quot; rel=&quot;nofollow noreferrer&quot;&gt;Windborn Muse&lt;/a&gt;&lt;br&gt;&#xA;Activated &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=253653&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jace, Architect of Thought&lt;/a&gt;'s +1 last turn. The Jace is no longer on the field, but in the graveyard&lt;/p&gt;&#xA;&#xA;&lt;p&gt;44 life&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Opponent 2's graveyard:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=271119&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lightning Mauler&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=389673&quot; rel=&quot;nofollow noreferrer&quot;&gt;Silverblade Paladin&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=214064&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hero of Bladehold&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=253653&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jace, Architect of Thought&lt;/a&gt; that activated +1 last turn  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;All Exiles:&lt;/strong&gt;&lt;br&gt;&#xA;Nothing&lt;/p&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2017-03-11T02:45:24.883" Title="&quot;Magic: the Gathering&quot; Challenge #5: What if Pod was One of Us" Tags="|calculation-puzzle|magic-the-gathering|" AnswerCount="1" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="24082" PostTypeId="2" ParentId="24013" CreationDate="2015-11-20T16:46:57.830" Score="13" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bob can always win by labeling the squares using the following scheme:&lt;br&gt;&lt;br&gt;$31131131131 \quad 19$&lt;br&gt;$13113113113 \quad 19$&lt;br&gt;$11311311311 \quad 17$&lt;br&gt;$31131131131 \quad 19$&lt;br&gt;$13113113113 \quad 19$&lt;br&gt;$11311311311 \quad 17$&lt;br&gt;$31131131131 \quad 19$&lt;br&gt;$13113113113 \quad 19$&lt;br&gt;$11311311311 \quad 17$&lt;br&gt;&lt;br&gt;The sum of each row is a prime number. Any horizontal or vertical combination of $3$ squares has the sum $5$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2015-11-20T22:54:42.397" LastActivityDate="2015-11-20T22:54:42.397" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24083" PostTypeId="1" AcceptedAnswerId="24090" CreationDate="2015-11-20T16:57:22.103" Score="9" ViewCount="1448" Body="&lt;p&gt;Is it possible to draw a closed path on the surface of a standard $3\times3\times3$ Rubik's cube &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;such that the path traverses each of the $54$ little squares exactly once, and&lt;/li&gt;&#xA;&lt;li&gt;such that the path does &lt;em&gt;not&lt;/em&gt; go through any corner of these little squares?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-09T09:12:27.787" LastActivityDate="2016-03-22T14:18:39.797" Title="A closed path on the Rubik's cube" Tags="|mathematics|combinatorics|rubiks-cube|graph-theory|" AnswerCount="6" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24084" PostTypeId="1" AcceptedAnswerId="24134" CreationDate="2015-11-20T17:07:39.730" Score="11" ViewCount="711" Body="&lt;p&gt;My Indian friend Jamal recently participated in the well-known TV quiz&#xA;show &lt;strong&gt;&lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Who_Wants_to_Be_a_Millionaire&quot;&gt;Who wants to be a millionaire?&lt;/a&gt;&lt;/em&gt;&lt;/strong&gt;. You might have heard or read about Jamal's spectacular success; the media reported on it extensively.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what Jamal told us about the 64.000 Euro question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I almost quit when I saw the 64.000 Euro question - I can never tell the&#xA;  four of them apart.  But of course,I had already reached second safe haven,&#xA;  I had nothing to lose, and I was already taking home a five-figure amount of money. So I tried to sort them out.  At first I&#xA;  thought the answer had to be D, because he was nowhere to be seen during&#xA;  that whole cucumber thing.  But then I realized that naturally the twins&#xA;  would be together, so that eliminated answer C as well.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I sat there for a good fifteen minutes pondering the answer, trying to&#xA;  ignore the steam coming out of Prem's ears.  Maddeningly, the part about&#xA;  the brother and the mother (from forty-plus years ago) kept running through&#xA;  my head, but it didn't help me until I remembered that it was B who had&#xA;  been out in front that time.  I got so excited that I screamed out&#xA;  &quot;THE ANSWER IS A&quot;, nearly giving Prem another heart attack in the process.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;But I didn't care, because I had just won - and I don't have to tell you&#xA;  how much.  I must say Prem acted every bit as overjoyed as I was - but I&#xA;  could have sworn he gave me the &quot;fast finger&quot; when nobody could see.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;What was the 64.000 Euro question?&lt;br&gt;&#xA;What were the four possible answers A, B, C, D?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2015-11-23T11:38:33.377" Title="Slumdog millionaire (11)" Tags="|knowledge|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24085" PostTypeId="2" ParentId="24083" CreationDate="2015-11-20T17:16:25.767" Score="1" Body="&lt;p&gt;Yes, it is possible. Each and every square in the Rubik's cube has four non-diagonal neighbors, including those neighbors on adjacent faces. There are no squares with an odd number of neighbors, so the parity checks out, and a closed path can be drawn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have just sketched out a couple of these paths. If requested, I can attach.&lt;/p&gt;&#xA;" OwnerUserId="16945" LastActivityDate="2015-11-20T17:16:25.767" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24086" PostTypeId="2" ParentId="24083" CreationDate="2015-11-20T17:18:33.893" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes, you can&lt;br&gt;&#xA;&lt;br&gt;&#xA; On the base, draw an S shaped path going from near-left to far-right&lt;br&gt;&#xA; On the right and front faces, do an S shape so you end up on the front face, top left.&lt;br&gt;&#xA; On the top, draw an S shape, going from front-left to back-right&lt;br&gt;&#xA; On the left and back, do an S shape, so you end up on the left face, bottom-near. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-11-20T17:18:33.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24088" PostTypeId="2" ParentId="24083" CreationDate="2015-11-20T17:23:43.253" Score="8" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes: &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As can be seen:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/4tJ8v.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4tJ8v.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA; (Similar path on opposite three faces)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9278" LastEditorDisplayName="user9278" LastEditDate="2015-11-20T17:32:37.533" LastActivityDate="2015-11-20T17:32:37.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24089" PostTypeId="1" AcceptedAnswerId="24093" CreationDate="2015-11-20T18:28:00.293" Score="7" ViewCount="762" Body="&lt;p&gt;And, lo! did the heavens open up to mine eyes and I did seeth the truth of the universe. The circular constant did pour forth from the mouth of angels and bury within my mind and I saw that it was good. The LORD spoketh thus, saying &quot;THIS IS THE TRUTH POURED OUT FOR YOU AND FOR MANY. SPREAD MY WORD AMONGST MY PEOPLE.&quot; And I did see that the word was wrong. I found the truth within the lie and knew it to be so. Find the truth for thyself! Detect the lie within the truth as it was said unto me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;3. &#xA;  ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastActivityDate="2015-11-20T20:49:41.423" Title="The Lie within The Truth" Tags="|pattern|enigmatic-puzzle|" AnswerCount="1" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24090" PostTypeId="2" ParentId="24083" CreationDate="2015-11-20T18:35:13.170" Score="14" Body="&lt;p&gt;An observation is..&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; that you can traverse a cube's face from one corner and end up diagonally opposite or in same column (or row).&#xA;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/pJTIz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pJTIz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using this observation, one of many possible solutions is (left image is front of cube, right image is inside view):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/g9oiy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/g9oiy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13851" LastActivityDate="2015-11-20T18:35:13.170" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24091" PostTypeId="2" ParentId="24083" CreationDate="2015-11-20T18:41:20.957" Score="12" Body="&lt;p&gt;The minimum degree of the graph involved is 4: each of the 56 squares has exactly four neighbors. &lt;a href=&quot;https://math.stackexchange.com/questions/690822/show-that-g-is-k-connected&quot;&gt;Thus, the graph is 4-connected&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also planar: we can obviously draw it on a sphere (imagine a &quot;deformed&quot; Rubik's cube.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://people.math.gatech.edu/~thomas/PAP/ham.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thus it is Hamiltonian&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11476" LastEditorUserId="-1" LastEditDate="2017-04-13T12:19:36.950" LastActivityDate="2015-11-21T22:05:12.067" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24092" PostTypeId="1" CreationDate="2015-11-20T18:42:55.503" Score="7" ViewCount="649" Body="&lt;p&gt;Alice and Bob decide to play a game. Alice goes first. They will take turns placing queens on a chessboard such that the queen is not threatened by any other queen on the board. Both players play optimally. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Winning Condition: Be the last person to place a queen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who Wins?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;After the first game, they play again. Alice again goes first and uses white pieces and Bob uses black pieces. This time, queens are allowed to be threatened by others of the same color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Winning Condition: Be the last person to place a queen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who Wins?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Bonus Round:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They play a third game with the same setup and rules as the second game with one addition: If someone is unable to play, their turn is skipped. I.E., if every open space on the board is threatened by a white piece, then Bob cannot place any more black pieces but Alice my continue to place her white pieces. (I expect that the optimal strategy, therefore, would be similar to the second game where you completely block someone out and then fill the rest of the safe spaces with your own pieces. However, that may not be the case.) Play continues until neither person is able to place any more pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What's the minimum number of pieces on the board when the game is over?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What's the &lt;em&gt;maximum&lt;/em&gt; number of pieces?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Clarifications:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can threaten every space on a board with five queens. Here is one such configuration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EurA5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EurA5.png&quot; alt=&quot;Block With Five&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first game, they will both be using the same color queens for simplicity. No queen is allowed to threaten any other queen. Therefore, they're going to place at least 5 queens since that's the minimum required to threaten every space. Here's an example of a game. &lt;code&gt;A&lt;/code&gt; is Alice and &lt;code&gt;B&lt;/code&gt; is Bob. The number indicates which turn it was for them. I.E., &lt;code&gt;A1&lt;/code&gt; is Alice's first turn and &lt;code&gt;B1&lt;/code&gt; is Bob's first turn. In this game, Alice won because &lt;code&gt;A3&lt;/code&gt; threatened the last of the unthreatened spaces and Bob was unable to play.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bkyTK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bkyTK.png&quot; alt=&quot;Game 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second game, they use queens of different colors. Each player can play anywhere not threatened by the other player. It is allowed to play somewhere threatened by your &lt;em&gt;own&lt;/em&gt; pieces, though. Here's an example game. Again, Alice one. This may end up being the same case as the first game except that someone can &quot;skip&quot; a turn by playing somewhere that doesn't threaten any new cells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KQQoo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KQQoo.png&quot; alt=&quot;Game 2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the bonus game, I think that the &quot;max pieces&quot; question will be the trickiest. It seems like the most important thing is to threaten the most spaces possible so you can go back and fill the board at the end. However, you also have to keep them from threatening the same spaces. Here's an example game. Alice was able to place 6 more pieces after locking out Bob show she won 11 - 5. (I had to switch to hexadecimal so &lt;code&gt;AB&lt;/code&gt; is Alice's 11th turn.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/J3QvI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/J3QvI.png&quot; alt=&quot;Game 3&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disclaimer: I am not super-good at chess. I used to play my uncle a lot. I don't know if this question is provably solvable. It was a curiosity that occurred to me and I'm trying to translate from my ignorance into an understandable puzzle.&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="9000" LastEditDate="2016-02-29T18:39:35.373" LastActivityDate="2016-03-28T13:38:41.443" Title="Queens on a Board with Two Players" Tags="|chess|" AnswerCount="1" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="24093" PostTypeId="2" ParentId="24089" CreationDate="2015-11-20T19:09:09.893" Score="14" Body="&lt;p&gt;Find all the indices where the number doesn't match the corresponding digit of pi:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (-1),9,14,34,40,61,66,78,81,82,96,116,129,134,146,166,185,205,210,215,227,229,234,235,248,267&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then take the differences and convert to letters:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; jetfuelcantmeltsteelbeams&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-11-20T20:49:41.423" LastActivityDate="2015-11-20T20:49:41.423" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24094" PostTypeId="1" AcceptedAnswerId="24110" CreationDate="2015-11-20T19:56:53.020" Score="8" ViewCount="326" Body="&lt;p&gt;I made a puzzle earlier (&lt;a href=&quot;https://puzzling.stackexchange.com/questions/23941/bridge-words-word-pairs-linked-by-the-front-and-back&quot;&gt;Bridge words&lt;/a&gt;), and thought I'd make another (encouraged by a request from &lt;a href=&quot;https://puzzling.stackexchange.com/users/10861/victorhenry&quot;&gt;@VictorHenry&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Find a pair of &quot;bridge words&quot; to fill in the gaps of the sentences below.  Bridge words are words where the last $n\geq3$ letters of the first word are the same as the first $n$ letters of the second word.  The words must then both be more than $n$ letters long (&quot;hots&lt;strong&gt;hots-hots&lt;/strong&gt;hots&quot; would be a valid bridge pair ($n=4$), but &quot;here-here&quot; would not be).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;br&gt;&#xA;Q: Sometimes we torture people. Sometimes we don't. We ______ our ______.&lt;br&gt;&#xA;A: We inters&lt;strong&gt;perse&lt;/strong&gt; our &lt;strong&gt;perse&lt;/strong&gt;cution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, here goes:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Good climbers leave the _______ ________.&lt;/li&gt;&#xA;&lt;li&gt;I caught _____ of the person _______ in the pool.&lt;/li&gt;&#xA;&lt;li&gt;She lives far out on the ______, ______ from society.&lt;/li&gt;&#xA;&lt;li&gt;I could see that the glamorous lady was ______ _______ that caught the light.&lt;/li&gt;&#xA;&lt;li&gt;We've ______ all the evidence and concluded that alcoholism is _______.&lt;/li&gt;&#xA;&lt;li&gt;The dentist used an odd ______ to save information about her patient's ______.&lt;/li&gt;&#xA;&lt;li&gt;I was sick in the West Indies, but in the ______ I was the ______.&lt;/li&gt;&#xA;&lt;li&gt;I take a coat and dress out of the wardrobe,  ______ the _______, and return them.&lt;/li&gt;&#xA;&lt;li&gt;Grimm's _________ was left ______ after the queen guessed his name.&lt;/li&gt;&#xA;&lt;li&gt;The king said to the regent: &quot;I'll be back tomorrow, so you must _____ _______.&quot;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $n$ is equal to 5 or 6.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-11-23T04:31:26.747" Title="Bridge words 2 - word pairs linked by their front and back" Tags="|wordplay|english|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24095" PostTypeId="1" CreationDate="2015-11-20T20:19:07.643" Score="6" ViewCount="201" Body="&lt;p&gt;Inspired by &lt;a href=&quot;https://puzzling.stackexchange.com/questions/24083/a-closed-path-on-the-rubiks-cube&quot;&gt;A closed path on the Rubik's cube&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let $N$ be a positive integer. Consider an $N\times N\times N$ cube, with each face tiled by $N^2$ squares measuring $1\times 1$. A closed path is drawn on the surface of the cube, never entering the same square twice, never passing through the corner of a square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each time the path enters a square, necessarily through an edge, it will exit through a different edge. Let $R$ be the number of squares with exit edge immediately to the right of the entry edge, and let $L$ be the number of squares with exit edge immediately to the left of the entry edge. Ignore squares not meeting the path, and squares for which the exit edge is across from the entry edge.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What are the possible values for $R-L$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2015-11-20T22:50:06.283" Title="Left and right turns on the surface of a cube" Tags="|mathematics|geometry|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24096" PostTypeId="1" AcceptedAnswerId="24099" CreationDate="2015-11-20T21:14:11.227" Score="21" ViewCount="1300" Body="&lt;p&gt;&lt;sub&gt;This is in the spirit of the &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%22what+is+a%22+word+is%3Aquestion&quot;&gt;What is a Word/Phrase™&lt;/a&gt; series started by &lt;a href=&quot;https://puzzling.stackexchange.com/users/463&quot;&gt;JLee&lt;/a&gt; with a special brand of &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%22Phrase%22%20user%3A463%20is%3Aquestion&quot;&gt;Phrase™&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%22What+is+%22+%22Word%22+-missing+user%3A463+is%3Aquestion&quot;&gt;Word™&lt;/a&gt; puzzles.&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;If a word conforms to a special rule, I call it a Rotten Word™.&lt;/p&gt;&#xA;&lt;p&gt;Use the following examples below to find the rule.&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;Rotten Words™&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;Not Rotten Words™&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;ANT&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;BEETLE&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;BARF&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;VOMIT&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;CLERK&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;TELLER&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;EBBS&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;FLOWS&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;ENVY&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;JEALOUSY&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;FLAP&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;FLUTTER&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;GNAT&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;FLY&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;GREEN&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;BLUE&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;NOWHERE&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;SOMEWHERE&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;NUN&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;PRIEST&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;PURELY&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;MIXED&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;REEF&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;OCEAN&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;ROOF&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;CEILING&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;VEX&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;TROUBLE&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;And, if you want to analyze, here is a CSV version:&lt;/p&gt;&#xA;&lt;pre&gt;&#xA;Rotten Words™,Not Rotten Words™&#xA;ANT,BEETLE&#xA;BARF,VOMIT&#xA;CLERK,TELLER&#xA;EBBS,FLOWS&#xA;ENVY,JEALOUSY&#xA;FLAP,FLUTTER&#xA;GNAT,FLY&#xA;GREEN,BLUE&#xA;NOWHERE,SOMEWHERE&#xA;NUN,PRIEST&#xA;PURELY,MIXED&#xA;REEF,OCEAN&#xA;ROOF,CEILING&#xA;VEX,TROUBLE&#xA;&lt;/pre&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="69582" LastEditDate="2023-12-28T11:29:48.653" LastActivityDate="2023-12-28T11:29:48.653" Title="What is a Rotten Word™?" Tags="|pattern|word-property|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24097" PostTypeId="1" AcceptedAnswerId="24098" CreationDate="2015-11-20T21:17:35.607" Score="1" ViewCount="155" Body="&lt;pre&gt;&lt;code&gt;8   = 2&#xA;68  = 3&#xA;91  = 1&#xA;32  = 0&#xA;883 = 4&#xA;761 = ?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What number should be instead of &lt;code&gt;?&lt;/code&gt; in this: &lt;code&gt;76 = ?&lt;/code&gt; according to other equations ?&lt;/p&gt;&#xA;" OwnerUserId="17438" LastActivityDate="2015-11-21T04:05:26.120" Title="Some crazy equations" Tags="|riddle|mathematics|" AnswerCount="1" CommentCount="0" ClosedDate="2015-11-20T21:32:50.983" ContentLicense="CC BY-SA 3.0" />
  <row Id="24098" PostTypeId="2" ParentId="24097" CreationDate="2015-11-20T21:24:44.357" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 - It's the number of closed loops in the number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1748" LastEditorUserId="17438" LastEditDate="2015-11-21T04:05:26.120" LastActivityDate="2015-11-21T04:05:26.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24099" PostTypeId="2" ParentId="24096" CreationDate="2015-11-20T21:36:06.117" Score="22" Body="&lt;p&gt;Rotten Words are words that...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; can be ROT13'd to produce a new word. (A is replaced with N B with O, etc.)&lt;br&gt;&#xA; These produce: &lt;code&gt;NAG ONES PYREX ROOF RAIL SYNC TANG TERRA ABJURER AHA CHERYL ERRS EBBS IRK.&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="17297" LastEditDate="2019-08-17T03:52:46.243" LastActivityDate="2019-08-17T03:52:46.243" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24100" PostTypeId="2" ParentId="24095" CreationDate="2015-11-20T22:13:23.757" Score="5" Body="&lt;p&gt;I'll start with ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - any integer between $-4$ and $+4$ (for $N \ge 2$)&lt;br&gt;- any integer between $-3$ and $+3$ (for $N = 1$)&lt;br&gt;&lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/OKjGZ.png&quot; alt=&quot;&quot;&gt;&lt;br&gt;&lt;br&gt;$-4$, $0$ and $+4$ are obvious (left image). We also see that any positive number can be made negative by mirroring or walking in other direction. The right image shows how to construct paths with values $1$, $2$ and $3$.&lt;br&gt;&lt;br&gt;Changing the paths while staying on a single plane obviously doesn't increase the number because for every added left turn we get another right turn. The same applies if we move to another plane with both ends of our path:&lt;br&gt;&lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/XcfBS.png&quot; alt=&quot;&quot;&gt;&lt;br&gt;&lt;br&gt;Using other ways to leave the plane only seems to decrease the number of turns like in the right picture at the top.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2015-11-20T22:50:06.283" LastActivityDate="2015-11-20T22:50:06.283" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24101" PostTypeId="2" ParentId="24081" CreationDate="2015-11-21T04:34:49.690" Score="5" Body="&lt;p&gt;&lt;strong&gt;Update: Now with formatting and links!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The win condition is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=230788&quot; rel=&quot;nofollow noreferrer&quot;&gt;Laboratory Maniac&lt;/a&gt;, which makes sense, since he's the oddball out in this list.  Everything else is very Birthing Pod friendly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll use [Pod 1 or 2] &gt; [Thing Sacrificed] &gt; [Thing Gotten] for shorthand.  Links are used to identify the &lt;em&gt;first&lt;/em&gt; time a specific card is drawn or acquired.  Also, the &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=405341&quot; rel=&quot;nofollow noreferrer&quot;&gt;Plains&lt;/a&gt; and &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=106657&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pull from Eternity&lt;/a&gt; were added later, so I'll be solving without their aid.  &lt;em&gt;(You can assume for this entire solution that I am holding Pull from Eternity in my hand and have my Plains untapped, if you wish)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tap all your lands for mana floating - we'll be winning in this phase&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; ---&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;em&gt;Starting Resources: &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;, 25 life&lt;/em&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;em&gt;Board: &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=218006&quot; rel=&quot;nofollow noreferrer&quot;&gt;Birthing Pod&lt;/a&gt; 1 (P1), &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=218006&quot; rel=&quot;nofollow noreferrer&quot;&gt;Birthing Pod&lt;/a&gt; 2 (P2), &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=370458&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kitchen Finks&lt;/a&gt;&lt;/em&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;em&gt;Graveyard: &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=194052&quot; rel=&quot;nofollow noreferrer&quot;&gt;Necropede&lt;/a&gt;&lt;/em&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; ---&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; P1 &gt; Kitchen Finks &gt; &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=15138&quot; rel=&quot;nofollow noreferrer&quot;&gt;Academy Rector&lt;/a&gt;, persist Kitchen Finks, Kitchen Finks gets -1/-1, gain 2 life&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;, 25 life&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; P2 &gt; Academy Rector &gt; &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=97109&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sky Hussar&lt;/a&gt;, exile Rector for &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=220529&quot; rel=&quot;nofollow noreferrer&quot;&gt;March of the Machines&lt;/a&gt;, untap both Pods (thanks to March of the Machines making them creatures)&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;, 23 life&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; P1 &gt; Sky Hussar &gt; &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=271207&quot; rel=&quot;nofollow noreferrer&quot;&gt;Auratouched Mage&lt;/a&gt;, finding and attaching &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=3146&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nature's Chosen&lt;/a&gt;&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;, 21 life&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Tap Autatouched Mage to untap P1, using Nature's Chosen&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; P1 &gt; Auratouched Mage &gt; &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=366285&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sepulchral Primordial&lt;/a&gt;, grabbing &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=271119&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lightning Mauler&lt;/a&gt; and &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=368951&quot; rel=&quot;nofollow noreferrer&quot;&gt;Voice of Resurgence&lt;/a&gt; from opponents' graveyards and putting them on the battlefield &#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;, 19 life&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; P2 &gt; Voice of Resurgence &gt; &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=222923&quot; rel=&quot;nofollow noreferrer&quot;&gt;Village Bell-Ringer&lt;/a&gt;, get an &lt;a href=&quot;http://shop.tcgplayer.com/magic/dragons-maze/elemental-token&quot; rel=&quot;nofollow noreferrer&quot;&gt;Elemental Token&lt;/a&gt;, untap both Pods&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;, 17 life&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; P1 &gt; Village Bell-Ringer &gt; &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=122124&quot; rel=&quot;nofollow noreferrer&quot;&gt;Arcum Dagsson&lt;/a&gt;&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;, 15 life&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; P2 &gt; Arcum Dagsson &gt; &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=382983&quot; rel=&quot;nofollow noreferrer&quot;&gt;Karmic Guide&lt;/a&gt;, returning Village Bell-Ringer, untap both Pods&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;, 13 life&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; P1 &gt; Village Bell-Ringer &gt; &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=386506&quot; rel=&quot;nofollow noreferrer&quot;&gt;Clever Impersonator&lt;/a&gt;, copying Karmic Guide, returning Arcum Daagsson, soulbond Arcum Daagsson with Lightning Mauler, giving both Haste&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;, 11 life&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; P2 &gt; Karmic Guide &gt; &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=405400&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sun Titan&lt;/a&gt;, returning Village Bell-Ringer, untap both Pods&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;, 9 life&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; P1 &gt; Village Bell-Ringer &gt; &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=214375&quot; rel=&quot;nofollow noreferrer&quot;&gt;Phyrexian Metamorph&lt;/a&gt;, copying Kitchen Finks, gain 2 life&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;, 9 life&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; P2 &gt; Elemental Token &gt; &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=36115&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hapless Researcher&lt;/a&gt;&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;, 7 life&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Tap Arcum Dagsson for his ability targeting Phyrexian Metamorph (Kitchen Finks), sacrifice Phyrexian Metamorph (Kitchen Finks) and replace it with &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=39701&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cryptic Gateway&lt;/a&gt;, persist Phyrexian Metamorph (Kitchen Finks), copying Sun Titan (with -1/-1), returning Village Bell-Ringer, untap both Pods and Arcum Dagsson&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; P1 &gt; Phyrexian Metamorph (Sun Titan) (-1/-1) &gt; &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=405126&quot; rel=&quot;nofollow noreferrer&quot;&gt;Angel of Serenity&lt;/a&gt;, exiling Karmic Guide, Sky Hussar, and Phyrexian Metamorph&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;, 5 life&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; P2 &gt; Angel of Serenity &gt; &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=198172&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gigantomancer&lt;/a&gt;, returning Karmic Guide, Sky Hussar, and Phyrexian Metamorph to my hand&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;, 3 life&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Tap Arcum Dagsson and Gigantomancer to play Sky Hussar, untap both Pods, Arcum Dagsson, and Gigantomancer&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; P1 &gt; Lightning Mauler &gt; &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=50537&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dross Harvester&lt;/a&gt;, Arcum Dagsson no longer has haste&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;, 1 life&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; ---&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;em&gt;Stabilization&lt;/em&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;em&gt;Board: Birthing Pod 1, Birthing Pod 2, March of the Machines, Cryptic Gateway, Kitchen Finks (-1/-1), Sepulchral Primordial, Dross Harvester, Sun Titan, Humans: [Hapless Researcher, Arcum Dagsson, Gigantomancer, Village Bell-Ringer, Sky Hussar], Angels: [Clever Impersonator (Karmic Guide)]&lt;/em&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;em&gt;Hand: Karmic Guide, Phyrexian Metamorph&lt;/em&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;em&gt;Graveyard: Necropede, Auratouched Mage, Nature's Chosen, Angel of Serenity&lt;/em&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;em&gt;Exile: Academy Rector&lt;/em&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; ---&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Sacrifice Hapless Researcher, draw and discard &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=227544&quot; rel=&quot;nofollow noreferrer&quot;&gt;Xenograft&lt;/a&gt;, gain 2 life &#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;, 3 life&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play Phyrexian Metamorph for RRR, pay 2 life, copy Clever Impersonator (Karmic Guide), return Hapless Researcher&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;, 1 life&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Sacrifice Hapless Researcher, draw and discard &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=233052&quot; rel=&quot;nofollow noreferrer&quot;&gt;Glistener Elf&lt;/a&gt;, gain 2 life &#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;, 3 life&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;em&gt;--- Beginning Loop One ---&lt;/em&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Tap Clever Impersonator (Karmic Guide), Phyrexian Metamorph (Karmic Guide) to play Karmic Guide using Cryptic Gateway, returning Angel of Serenity, exiling Karmic Guide, Village Bell-Ringer, and Hapless Researcher&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; P2 &gt; Angel of Serenity &gt; &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=228242&quot; rel=&quot;nofollow noreferrer&quot;&gt;Trench Gorger&lt;/a&gt; (exiling all 23 &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=405402&quot; rel=&quot;nofollow noreferrer&quot;&gt;Swamps&lt;/a&gt; in the deck), returning Karmic Guide, Village Bell-Ringer, and Hapless Researcher to my hand&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;, 3 life&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Tap Arcum Dagsson and Gigantomancer to play Hapless Researcher using Cryptic Gateway&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Tap Hapless Researcher and Sky Hussar to play Village Bell-Ringer using Cryptic Gateway, untapping both Pods, Clever Impersonator (Karmic Guide), Phyrexian Metamorph (Karmic Guide), Arcum Dagsson, Gigantomancer, Hapless Researcher and Sky Hussar&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Sacrifice Hapless Researcher, draw and discard &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=249405&quot; rel=&quot;nofollow noreferrer&quot;&gt;Qasali Pridemage&lt;/a&gt;, gain 2 life&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;, 5 life&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;em&gt;--- Beginning Loop Two ---&lt;/em&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Tap Clever Impersonator (Karmic Guide), Phyrexian Metamorph (Karmic Guide), to play Karmic Guide using Cryptic Gateway, return Angel of Serenity, exiling Karmic Guide, Village Bell-Ringer, and Hapless Researcher&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; P1 &gt; Angel of Serenity &gt; [Fail to Find], returning Karmic Guide, Village Bell-Ringer, and Hapless Researcher to my hand&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;, 5 life&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; P2 &gt; Sun Titan &gt; &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=383194&quot; rel=&quot;nofollow noreferrer&quot;&gt;Boonweaver Giant&lt;/a&gt;, finding and attaching &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=23043&quot; rel=&quot;nofollow noreferrer&quot;&gt;Shade's Form&lt;/a&gt;&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;, 5 life&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Tap Arcum Dagsson and Gigantomancer to play Hapless Researcher using Cryptic Gateway&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Tap Hapless Researcher and Sky Hussar to play Village Bell-Ringer using Cryptic Gateway, untapping both Pods, Clever Impersonator (Karmic Guide), Phyrexian Metamorph (Karmic Guide), Arcum Dagsson, Gigantomancer, Hapless Researcher and Sky Hussar&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Sacrifice Hapless Researcher, draw and discard &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=386337&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kathari Bomber&lt;/a&gt;, gain 2 life&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;, 7 life&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;em&gt;--- Beginning Loop Three ---&lt;/em&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Tap Clever Impersonator (Karmic Guide), Phyrexian Metamorph (Karmic Guide), to play Karmic Guide using Cryptic Gateway, return Angel of Serenity, exiling Karmic Guide, Village Bell-Ringer, and Hapless Researcher&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; P1 &gt; Angel of Serenity &gt; [Fail to Find], returning Karmic Guide, Village Bell-Ringer, and Hapless Researcher to my hand&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;, 7 life&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Tap Arcum Dagsson and Gigantomancer to play Hapless Researcher using Cryptic Gateway&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; P2 &gt; Arcum Dagsson &gt; &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=376446&quot; rel=&quot;nofollow noreferrer&quot;&gt;Phyrexian Delver&lt;/a&gt;, return Qasali Pridemage, lose 2 life&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;, 5 life&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Tap Hapless Researcher and Sky Hussar to play Village Bell-Ringer using Cryptic Gateway, untapping both Pods, Clever Impersonator (Karmic Guide), Phyrexian Metamorph (Karmic Guide), Gigantomancer, Hapless Researcher and Sky Hussar&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; P1 &gt; Qasali Pridemage &gt; &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=230788&quot; rel=&quot;nofollow noreferrer&quot;&gt;Laboratory Maniac&lt;/a&gt;&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;, 5 life&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; At this point, your deck is now empty&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Sacrifice Hapless Researcher with an empty deck and Laboratory Maniac in play, gain 2 life, win the game!&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; ---&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;em&gt;Remaining Resources: &lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8cSm.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXu0Z.gif&quot; alt=&quot;]&quot;&gt;, 7 life, Karmic Guide in hand, 0 cards in deck, a bunch of junk on the battlefield&lt;/em&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; ---&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="30903" LastEditDate="2017-03-11T02:45:24.883" LastActivityDate="2017-03-11T02:45:24.883" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="24103" PostTypeId="1" AcceptedAnswerId="24106" CreationDate="2015-11-21T15:13:47.850" Score="10" ViewCount="2599" Body="&lt;p&gt;&lt;em&gt;I saw a similar puzzle in a book long time ago.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tnZy4.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What will be the next symbol?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastActivityDate="2015-11-21T21:06:19.043" Title="A classic symbol sequence" Tags="|pattern|symbols|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24106" PostTypeId="2" ParentId="24103" CreationDate="2015-11-21T15:40:59.237" Score="14" Body="&lt;p&gt;&lt;strong&gt;The next image&lt;/strong&gt; in the sequence should be this symbol:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;a href=&quot;https://i.stack.imgur.com/7Ph8Y.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7Ph8Y.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;These are numbers drawn with a reflection of the number in front of the digit.  &lt;code&gt;1,2,3…&lt;/code&gt;! And as you see ending number is &lt;code&gt;5&lt;/code&gt;, So the next one will be &lt;strong&gt;&lt;code&gt;6&lt;/code&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17438" LastEditorUserId="17438" LastEditDate="2015-11-21T16:10:38.043" LastActivityDate="2015-11-21T16:10:38.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24107" PostTypeId="2" ParentId="24103" CreationDate="2015-11-21T15:51:50.153" Score="6" Body="&lt;p&gt;Answer:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/MvhVe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MvhVe.png&quot; alt=&quot;Mirror6&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each image is a digit with its horizontal mirror image on it's left side&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7715" LastEditorUserId="7715" LastEditDate="2015-11-21T21:06:19.043" LastActivityDate="2015-11-21T21:06:19.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24108" PostTypeId="1" AcceptedAnswerId="24113" CreationDate="2015-11-21T16:19:21.337" Score="6" ViewCount="254" Body="&lt;p&gt;Is it possible to draw a &lt;em&gt;closed&lt;/em&gt; path on the surface of a standard $3\times3\times3$ Rubik's cube &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;such that the path traverses exactly one diagonal of each of the $54$ little squares, and&lt;/li&gt;&#xA;&lt;li&gt;such that the path does not intersect itself?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-11-28T16:20:08.793" Title="Another closed path on the Rubik's cube" Tags="|mathematics|combinatorics|rubiks-cube|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24110" PostTypeId="2" ParentId="24094" CreationDate="2015-11-21T17:58:05.773" Score="7" Body="&lt;p&gt;Hooray! These seem harder but I'm making progress. Going to go ahead and post this to keep track of which ones I've solved so far.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Good climbers leave the _______ ________.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;mountain untainted&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;I caught _____ of the person _______ in the pool.&lt;/li&gt;&#xA;&lt;li&gt;She lives far out on the ______, ______ from society.&lt;/li&gt;&#xA;&lt;li&gt;I could see that the glamorous lady was ______ _______.&lt;/li&gt;&#xA;&lt;li&gt;We've ______ all the evidence and concluded that alcoholism is _______.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;gathered/hereditary&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;6&quot;&gt;&#xA;&lt;li&gt;The dentist used an odd ______ to save information about her patient's ______.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;filename/enamel&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;7&quot;&gt;&#xA;&lt;li&gt;I was sick in the West Indies, but in the ______ I was the ______.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Antilles/illest&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;8&quot;&gt;&#xA;&lt;li&gt;I took two coats out of the wardrobe to ______ the two _______.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;launder / &lt;a href=&quot;https://en.wikipedia.org/wiki/Coat_(clothing)#Coats.2C_jackets_and_overcoats&quot; rel=&quot;nofollow&quot;&gt;undercoats&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;9&quot;&gt;&#xA;&lt;li&gt;Grimm's _________ was left ______ when the queen guessed his name.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Rumpelstiltskin/tsking&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;10&quot;&gt;&#xA;&lt;li&gt;The king said to the regent: &quot;I'll be back tomorrow, so you must _____ _______.&quot;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;govern overnight&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10861" LastEditorUserId="10861" LastEditDate="2015-11-22T15:27:32.230" LastActivityDate="2015-11-22T15:27:32.230" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="24111" PostTypeId="2" ParentId="24094" CreationDate="2015-11-21T19:38:29.107" Score="5" Body="&lt;p&gt;I have a few to add to the ones &lt;a href=&quot;https://puzzling.stackexchange.com/users/10861/victorhenry&quot;&gt;@VictorHenry&lt;/a&gt; already solved.  My solution for number nine feels weak. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.She lives far out on the ______, ______ from society.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;peninsula/insulated&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2.I caught _____ of the person _______ in the pool.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;glimmers/immersed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;9.Grimm's _________ was ______ when the queen guessed his name.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;printer/interested ???  spinner/innerved ??? This seems like it should be Rumpelstiltskin-related...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17020" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-11-21T21:05:39.927" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24112" PostTypeId="1" CreationDate="2015-11-21T22:07:41.363" Score="3" ViewCount="337" Body="&lt;p&gt;$$8\text{ }8\text{ }8 = 6$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add in any required number of mathematical symbols to make the above equation work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this possible? Can somebody explain?&lt;/p&gt;&#xA;" OwnerUserId="17474" LastEditorUserId="39" LastEditDate="2015-11-21T22:08:52.373" LastActivityDate="2015-11-21T22:08:52.373" Title="How is it possible to make 8 (...) 8 (...) 8 = 6?" Tags="|formation-of-numbers|" AnswerCount="0" CommentCount="1" ClosedDate="2015-11-21T22:34:33.000" ContentLicense="CC BY-SA 3.0" />
  <row Id="24113" PostTypeId="2" ParentId="24108" CreationDate="2015-11-21T23:52:57.033" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The answer is &lt;strong&gt;no&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Consider the graph formed by &quot;corners&quot; of the squares on the Rubik's cube. These are the vertices we're moving between. Assign each vertex coordinates in $\{0 \dots 3\}^3$. &#xA;We observe that diagonal moves can't change the parity of $(x+y+z \bmod 2)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HZz67.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HZz67.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem asks for a &lt;a href=&quot;https://en.wikipedia.org/wiki/Eulerian_path#Definition&quot; rel=&quot;nofollow noreferrer&quot;&gt;Eulerian cycle&lt;/a&gt; on this subgraph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that there are vertices of &lt;a href=&quot;https://en.wikipedia.org/wiki/Degree_(graph_theory)&quot; rel=&quot;nofollow noreferrer&quot;&gt;degree&lt;/a&gt; three: they are marked by the red spheres on the corners of the cube in the above image. This means there is no Eulerian cycle: there is an odd number of ways from/to the corners, but a Eulerian cycle needs an even number (intuitively, it needs as many ways &lt;em&gt;from&lt;/em&gt; as it needs ways &lt;em&gt;to&lt;/em&gt; each vertex).&lt;/p&gt;&#xA;" OwnerUserId="11476" LastEditorUserId="8874" LastEditDate="2015-11-28T16:20:08.793" LastActivityDate="2015-11-28T16:20:08.793" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24114" PostTypeId="2" ParentId="24064" CreationDate="2015-11-22T01:19:31.030" Score="3" Body="&lt;p&gt;For all integers $n &amp;gt; 1$, we can write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\log \left( \log(n) / \log \left( √√√√√\sqrt{n} \right) \right)}{\log(\lceil √√\dots\sqrt{n} \rceil )}=6$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we use the fact that $$\log(n) / \log(√√√√√\sqrt{n}) =64.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get a &quot;free&quot; $2$, we repeatedly take the square root of our number $n$: we know that for all $n &amp;gt; 1$, $1 &amp;lt; \sqrt{n} &amp;lt; n$, so eventually we approach $1$ from above, but never reach it. Then, we take the ceiling of this value, and get $\lceil 1+\varepsilon \rceil = 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, we simply calculate $\log(64) / \log(2) = 6$.&lt;/p&gt;&#xA;" OwnerUserId="11476" LastActivityDate="2015-11-22T01:19:31.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24115" PostTypeId="1" AcceptedAnswerId="24141" CreationDate="2015-11-22T05:20:47.257" Score="5" ViewCount="921" Body="&lt;p&gt;Assume that the following three equations hold true:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$C \circ I \circ D \circ A = B$  &lt;/li&gt;&#xA;&lt;li&gt;$O \circ K \circ M \circ G = C$  &lt;/li&gt;&#xA;&lt;li&gt;$Y \circ A \circ B \circ C = D$  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Find the right hand side of the following equation: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$S \circ W \circ A \circ G =~ ?$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Hints:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Convert each letter to its corresponding numerical value.&lt;br&gt;&#xA; E.g., $C \circ I \circ D \circ A = B$ is converted to $3 \circ 9 \circ 4 \circ 1 = 2$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Use mathematical operators to make above equations work out.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All the equations follow the same pattern. Just find a pattern, apply it to the last equation and find the result.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17437" LastEditorUserId="8874" LastEditDate="2015-11-23T14:03:48.860" LastActivityDate="2015-11-23T16:17:23.197" Title="Mathematical alphabetical equation" Tags="|mathematics|pattern|calculation-puzzle|" AnswerCount="1" CommentCount="7" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24118" PostTypeId="1" AcceptedAnswerId="24119" CreationDate="2015-11-22T13:19:25.890" Score="2" ViewCount="220" Body="&lt;p&gt;What number is &lt;code&gt;?&lt;/code&gt; in follow?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2  = 46&#xA;4  = 812&#xA;8  = 1624&#xA;16 = ?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="17438" LastActivityDate="2015-11-22T13:53:57.800" Title="Magic Numerical Equations" Tags="|riddle|mathematics|" AnswerCount="2" CommentCount="0" ClosedDate="2015-11-25T08:47:14.913" ContentLicense="CC BY-SA 3.0" />
  <row Id="24119" PostTypeId="2" ParentId="24118" CreationDate="2015-11-22T13:31:24.683" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$4+6 = 10 = 2 \cdot 5$,  $8+12 = 20 = 4 \cdot 5$,  $16+24 = 40 = 8 \cdot 5$,  $32+48 = 80 = 16 \cdot 5$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $6-4=2$, $12-8=4$, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; So $? = 3248$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of course, there are many possible solutions to this question...&lt;/p&gt;&#xA;" OwnerUserId="12901" LastEditorUserId="12901" LastEditDate="2015-11-22T13:47:05.880" LastActivityDate="2015-11-22T13:47:05.880" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24120" PostTypeId="1" AcceptedAnswerId="24121" CreationDate="2015-11-22T13:38:57.333" Score="12" ViewCount="2116" Body="&lt;p&gt;Every day I park my car in the same spot, which has not had any alterations done in years, nor has the surrounding area changed. No new trees, building, etc. I park my car in the same fashion each time, facing the same direction, and don't use a windscreen sunshield nor car cover. The windows are always closed. Nobody touches the vehicle or the surroundings during the day. I live in Beersheba, Israel, about 33 degrees north of the equator. The only changes to the situation in summer and winter are naturally-occurring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summer, when I get to the car at the end of the day the car is cool and comfortable. In winter, when I get to the car it is hot and needs to be aired. &lt;strong&gt;WHY?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="6116" LastEditorUserId="5373" LastEditDate="2019-12-20T17:24:31.577" LastActivityDate="2019-12-20T17:24:31.577" Title="Car hotter in winter" Tags="|situation|seasonal|" AnswerCount="6" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24121" PostTypeId="2" ParentId="24120" CreationDate="2015-11-22T13:47:02.583" Score="18" Body="&lt;p&gt;I would suspect that it's due to the angle of the sun in the different seasons. The car is probably in the shade during the summer, but in winter, when the sun is at a lower angle in the sky, it shines directly into the car, heating it up. &lt;/p&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2015-11-22T13:47:02.583" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24122" PostTypeId="2" ParentId="24120" CreationDate="2015-11-22T13:49:00.830" Score="10" Body="&lt;p&gt;A guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your car usually stands under a simple roof. Maybe theres also a wall to the west.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the summer the sun shines almost directly from the top onto your car, but because of the roof the light won't reach. Depending on what time your day &quot;ends&quot; (i would guess you mean your workday) the extra wall to the west will block the sun during evening.&lt;br&gt;&#xA;&lt;br&gt;&#xA; During winter the sun will shine from a much lower angle in the south of your car and reach beneath the roof.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16608" LastActivityDate="2015-11-22T13:49:00.830" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24123" PostTypeId="2" ParentId="24118" CreationDate="2015-11-22T13:53:57.800" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3248&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think the right side of the equation are actually always 2 numbers that are respectively the double and triple of the number on the left.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2 = 4 6&#xA;4 = 8 12&#xA;8 = 16 24&#xA;16 = 32 48&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="16608" LastActivityDate="2015-11-22T13:53:57.800" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24124" PostTypeId="1" AcceptedAnswerId="24126" CreationDate="2015-11-22T15:21:31.350" Score="30" ViewCount="1437" Body="&lt;p&gt;&lt;em&gt;This is the first question for the new &lt;a href=&quot;https://puzzling.meta.stackexchange.com/questions/4464/fortnightly-topic-challenge-5-literature&quot;&gt;Fortnightly Topic Challenge #5: Literature&lt;/a&gt;. My inspiration source was a TV show.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I wanted to read some books during holidays, so a friend sent me a list of some international classic books. Unfortunately he recently discovered &lt;a href=&quot;https://en.wikipedia.org/wiki/Emoji&quot; rel=&quot;nofollow noreferrer&quot;&gt;Emoji's&lt;/a&gt;, and most of his communication attempts look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GA8LX.png&quot; alt=&quot;1. ❓ 2.  3.  4.  5. ☝️ 6. ➕ 7.  8. &quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you help me to find out which books he recommended?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each author has written 2 books in this list.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="5044" LastEditDate="2018-05-17T16:40:40.983" LastActivityDate="2018-05-17T16:40:40.983" Title="Emojiture Challenge" Tags="|rebus|literature|emoji|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24125" PostTypeId="2" ParentId="24124" CreationDate="2015-11-22T15:40:57.890" Score="3" Body="&lt;p&gt;To get started:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1.) Jules Verne: The Mysterious Island&lt;br&gt;&#xA; 2.) Robert A. Heinlein: Red Planet&lt;br&gt;&#xA; 3.) Arthur Conan Doyle: The Valley of Fear&lt;br&gt;&#xA; 4.) Jules Verne: Journey to the Center of the Earth&lt;br&gt;&#xA; 5.)&lt;br&gt;&#xA; 6.) Arthur Conan Doyle: The Sign of the Four&lt;br&gt;&#xA; 7.) Robert A. Heinlein: Starship Troopers&lt;br&gt;&#xA; 8.)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-11-22T15:55:04.780" LastActivityDate="2015-11-22T15:55:04.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24126" PostTypeId="2" ParentId="24124" CreationDate="2015-11-22T15:42:55.620" Score="25" Body="&lt;p&gt;Got them all!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1) The Mysterious Island (Jules Verne)&#xA;&lt;br&gt;&#xA; 2) Red Planet (Robert A. Heinlein)&#xA;&lt;br&gt;&#xA; 3) The Valley of Fear (Sir Arthur Conan Doyle)&#xA;&lt;br&gt;&#xA; 4) Journey to the Center of the Earth (Jules Verne)&#xA;&lt;br&gt;&#xA; 5) Nineteen Eighty Four (George Orwell) (1 finger - 9 o'clock - 8 legs - 4 leaves)&#xA;&lt;br&gt;&#xA; 6) The Sign of Four (Sir Arthur Conan Doyle)&#xA;&lt;br&gt;&#xA; 7) Starship Troopers (Robert A. Heinlein)&#xA;&lt;br&gt;&#xA; 8) Animal Farm (George Orwell)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="17389" LastEditDate="2015-11-22T15:54:56.980" LastActivityDate="2015-11-22T15:54:56.980" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24127" PostTypeId="1" CreationDate="2015-11-22T16:58:34.030" Score="8" ViewCount="539" Body="&lt;p&gt;Four players with first names Ahmad, Ali, Amir and Nima participated in a tournament. The highest ranked player defeated the other three, the second ranked player defeated the players in rank three and four, and the player in third rank defeated the player in rank four.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore the following information are available (where ln = Last name,):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Ali Koohi (fn &amp;amp; ln) defeated Amir (fn) and Rezayi (ln) defeated Nima (fn)&lt;/li&gt;&#xA;&lt;li&gt;Ahmad (fn) is not in the third rank.&lt;/li&gt;&#xA;&lt;li&gt;Akbari (ln) is not in the last rank.&lt;/li&gt;&#xA;&lt;li&gt;Nazari (ln) whose first name is not Ahmad, is exactly one rank after Ali.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What is Ali Koohi's rank?&lt;/p&gt;&#xA;" OwnerUserId="17480" LastEditorUserId="8874" LastEditDate="2015-11-22T17:39:23.357" LastActivityDate="2015-11-25T17:42:42.557" Title="Ahmad, Ali, Amir and Nima in a tournament" Tags="|logical-deduction|" AnswerCount="5" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24128" PostTypeId="2" ParentId="24127" CreationDate="2015-11-22T17:22:12.037" Score="2" Body="&lt;p&gt;&lt;strong&gt;Update: solution is unique.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possible Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Third Place!&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; 1: Ahmad Rezayi&#xA;&lt;br&gt;&#xA; 2: Nima Akbari&#xA;&lt;br&gt;&#xA; 3: Ali Koohi&#xA;&lt;br&gt;&#xA; 4: Amir Nazari&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Logic, for my own reference&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &#xA; FN:&#xA;&lt;br&gt;&#xA; Ali (Koohi) (1, 2, 3*)&#xA;&lt;br&gt;&#xA; Ahmad (1*, 2, 4) Rezayi*&#xA;&lt;br&gt;&#xA; Amir (2, 3, 4*)&#xA;&lt;br&gt;&#xA; Nima (not Rezayi) (2*, 3, 4) Akbari*&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; LN:&#xA;&lt;br&gt;&#xA; Koohi (Ali) (1, 2, 3*)&#xA;&lt;br&gt;&#xA; Rezayi (not Nima) (1*, 2, 3) Ahmad*&#xA;&lt;br&gt;&#xA; Akbari (1, 2*, 3) Nina*&#xA;&lt;br&gt;&#xA; Nazari (not Ahmad) (2, 3, 4*)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="17389" LastEditDate="2015-11-22T17:27:13.430" LastActivityDate="2015-11-22T17:27:13.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24129" PostTypeId="2" ParentId="24127" CreationDate="2015-11-22T17:35:35.833" Score="4" Body="&lt;p&gt;Here is a summary of the available information:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The four first names are: Ahmad, Ali, Amir, Nima.  &lt;/li&gt;&#xA;&lt;li&gt;The four last names are: Akbari, Koohi, Nazari, Rezayi&lt;/li&gt;&#xA;&lt;li&gt;Ali Koohi is one name pair&lt;/li&gt;&#xA;&lt;li&gt;Rezayi beats Nima, and hence Nima Rezayi is not a name pair&lt;/li&gt;&#xA;&lt;li&gt;Ahmand Nazari is not a name pair&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This leaves three possible name matchings:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(1) Ali Koohi, Amir Rezayi, Ahmad Akbari, Nima Nazari&lt;br&gt;&#xA;  (2) Ali Koohi, Amir Akbari, Nima Nazari, Ahmad Rezayi&lt;br&gt;&#xA;  (3) Ali Koohi, Amir Nazari, Nima Akbari, Ahmad Rezayi&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now let us work through the three cases one by one.&lt;br&gt;&#xA;In case (1) Ali beats Amir and Amir beats Nima. Hence Ali and Nima Nazari cannot be on adjacent ranks (as Amir is inbetween).&#xA;In the remaining cases (2) and (3), Ali beats Amir and Ahmad beats Nima. &#xA;This yields six possible rankings:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;(a) Ali &gt; Amir &gt; Ahmad &gt; Nima&lt;/li&gt;&#xA;&lt;li&gt;(b) Ali &gt; Ahmad &gt; Amir &gt; Nima&lt;/li&gt;&#xA;&lt;li&gt;(c) Ali &gt; Ahmad &gt; Nima &gt; Amir &lt;/li&gt;&#xA;&lt;li&gt;(d) Ahmad &gt; Nima &gt; Ali &gt; Amir&lt;/li&gt;&#xA;&lt;li&gt;(e) Ahmad &gt; Ali &gt; Nima &gt; Amir&lt;/li&gt;&#xA;&lt;li&gt;(f) Ahmad &gt; Ali &gt; Amir &gt; Nima&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now as Ahmad is not on the third place, (2a) and (3a) are eliminated.&lt;br&gt;&#xA;As Ali and Nazari are on neighboring places (in this order), this &#xA;eliminates (2b), (3b), (2c), (3c), (2d), (3e), and (2f).&lt;br&gt;&#xA;We are left with (3d), (2e) and (3f).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally we use that Akbari is not in last place:&#xA;In (2e) Amir Akbari is in last place, and in (3f) Nima Akbari is in last place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leaves (3d) as only possible solution:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ahmad Rezayi &gt; Nima Akbari &gt; Ali Koohi &gt; Amir Nazari&lt;br&gt;&#xA;  Hence Ali Koohi reached the &lt;strong&gt;third place&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-11-22T17:43:29.720" LastActivityDate="2015-11-22T17:43:29.720" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24130" PostTypeId="2" ParentId="24127" CreationDate="2015-11-22T19:37:13.677" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3rd place.&lt;br&gt;&#xA;&lt;br&gt;&#xA; - The first clue has two last names that cannot be in last place, as they defeated someone: Koohi and Rezayi.&lt;br&gt;&#xA; - Per clue #3, Akbari is also not in last, so spots 1-3 are Koohi, Rezayi, and Akbari, in some order.&lt;br&gt;&#xA; - Per clue 4, Nezari must be in last place, and per that same clue, Ali Koohi is one spot ahead of them: in 3rd.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To be complete:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Per the first clue, and his place at #3, Ali defeated only one person, Amir. This means Amir's last name is Nezari. This leaves Akbari and Rezayi. Since Rezayi defeated Nima, they cannot be the same person, therefore Nima is Akbari, and Ahmad is Rezayi. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The full list:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1st Ahmad Rezayi&lt;br&gt;&#xA; 2nd Nima Akbari&lt;br&gt;&#xA; 3rd Ali Kohi&lt;br&gt;&#xA; 4th Amir Nezari  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16874" LastEditorUserId="4421" LastEditDate="2015-11-25T17:42:42.557" LastActivityDate="2015-11-25T17:42:42.557" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24131" PostTypeId="2" ParentId="24120" CreationDate="2015-11-22T22:24:26.133" Score="2" Body="&lt;p&gt;Perhaps trees dropping their leaves?&lt;/p&gt;&#xA;" OwnerUserId="2767" LastActivityDate="2015-11-22T22:24:26.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24132" PostTypeId="2" ParentId="24127" CreationDate="2015-11-23T03:14:27.600" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;ul&gt;&#xA;  &lt;li&gt;In clue 1 we have have two &lt;strong&gt;last names&lt;/strong&gt; (Koohi &amp;amp; Rezayi) that defeated &lt;em&gt;someone else&lt;/em&gt;, so those names can't be last.&lt;/li&gt;&#xA;  &lt;li&gt;In clue 3, a third &lt;strong&gt;last name&lt;/strong&gt; (Akbari) is also not last.&lt;/li&gt;&#xA;  &lt;li&gt;Therefore, the only remaining &lt;strong&gt;last name&lt;/strong&gt; - Nazari - must be last.&lt;/li&gt;&#xA;  &lt;li&gt;Ali (one rank above Nazari according to clue 4) must be &lt;em&gt;&lt;strong&gt;3rd&lt;/strong&gt;&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13148" LastEditorUserId="13148" LastEditDate="2015-11-23T03:22:49.917" LastActivityDate="2015-11-23T03:22:49.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24133" PostTypeId="2" ParentId="24127" CreationDate="2015-11-23T07:20:31.653" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Ali Koohi's rank is 3rd.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solution:&lt;br&gt;&#xA;From all the clues, Ali's last name is Koohi. There are three possibilities of First and last name for remaining three players.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1. Ahmad Rezayi - Amir Akbari - Nima Nazari.&lt;br&gt;&#xA;2. Ahmad Rezayi - Nima Akbari - Amir Nazari.&lt;br&gt;&#xA;3. Amir Rezayi - Ahmad Akbari - Nima Nazari.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's talk about first case.  &lt;/p&gt;&#xA;&#xA;&lt;h1&gt;&lt;em&gt;Case 1. Ahmad Rezayi - Amir Akbari - Nima Nazari&lt;/em&gt;&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;=&gt; From the clues, the possible ranks of all the players for this case are:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1. Ali Koohi = 1, 2&lt;br&gt;&#xA;2. Ahmad Rezayi = 1, 2&lt;br&gt;&#xA;3. Amir Akbari = 2, 3&lt;br&gt;&#xA;4. Nima Nazari = 2, 3&lt;br&gt;&#xA;=&gt; For this case, no one is following under rank 4. So, this case is not valid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;&lt;em&gt;Case 2. Ahmad Rezayi - Nima Akbari - Amir Nazari&lt;/em&gt;&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;=&gt; From the clues, the possible ranks of all the players for this case are:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1. Ali Koohi = 1, 2, 3&lt;br&gt;&#xA;2. Ahmad Rezayi = 1, 2&lt;br&gt;&#xA;3. Nima Akbari = 2, 3&lt;br&gt;&#xA;4. Amir Nazari = 2, 3, 4&lt;br&gt;&#xA;=&gt; So, the possible rankwise order for this case is:&lt;br&gt;&#xA;&lt;strong&gt;Ahmad Rezayi - Nima Akbari - Ali Koohi - Amir Nazari&lt;/strong&gt;&#xA;-----------&gt; This is a valid scenario. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;&lt;em&gt;Case 3. Amir Rezayi - Ahmad Akbari - Nima Nazari&lt;/em&gt;&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;=&gt; From the clues, the possible ranks of all the players for this case are:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1. Ali Koohi = 1, 2, 3&lt;br&gt;&#xA;2. Amir Rezayi = 2, 3, 4&lt;br&gt;&#xA;3. Ahmad Akbari = 1, 2&lt;br&gt;&#xA;4. Nima Nazari = 3, 4&lt;br&gt;&#xA;=&gt; From the clues, there is no any possible rankwise order for this case.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conclusion:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;From all the possible cases, only case 2 is possible and hence the answer is, &lt;em&gt;Ali is at 3rd position&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17437" LastEditorUserId="17437" LastEditDate="2015-11-23T09:19:58.683" LastActivityDate="2015-11-23T09:19:58.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24134" PostTypeId="2" ParentId="24084" CreationDate="2015-11-23T10:17:58.357" Score="13" Body="&lt;p&gt;The question was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Which Gibb brother was not a member of the Bee Gees?&lt;/strong&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the alternatives were:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;A. Andy&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;B. Barry&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;C. Maurice&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;D. Robin&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; * The &lt;em&gt;&quot;cucumber thing&quot;&lt;/em&gt; refers to their album Cucumber Castle, on which Robin didn't appear because he was pursuing a solo career.&lt;br&gt;&#xA; * Robin and Maurice are twin brothers (&lt;em&gt;&quot;naturally the twins would be together&quot;&lt;/em&gt;).&lt;br&gt;&#xA; * Barry and Robin shared the lead singer role, so Barry was the sole frontman during Robin's absence (&lt;em&gt;&quot;it was B who had been out in front that time&quot;&lt;/em&gt; refers to this maybe?&lt;br&gt;&#xA; * The &quot;mother/brother&quot; clue refers to the song lyrics &quot;Whether you're a brother or whether you're a mother, you're stayin' alive, stayin' alive&quot;&#xA;&lt;br&gt;&#xA; * Andy was never a member of Bee Gees, he was a solo artist his entire career, so the correct answer is A!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks to GentlePurpleRain, as his/her comment put me in the right direction :)&lt;/p&gt;&#xA;" OwnerUserId="4624" LastEditorUserId="8874" LastEditDate="2015-11-23T11:38:33.377" LastActivityDate="2015-11-23T11:38:33.377" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24135" PostTypeId="2" ParentId="24120" CreationDate="2015-11-23T11:46:07.670" Score="3" Body="&lt;p&gt;It could also be due to less leaves on trees and so the sun is directly on it but that answer above would be the best answer. Of course speaking outside of physical temperature, your car may seem warmer in the Winter due to it already being cold outside so your going into the warm.&lt;/p&gt;&#xA;" OwnerUserId="17491" LastActivityDate="2015-11-23T11:46:07.670" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24136" PostTypeId="1" AcceptedAnswerId="24137" CreationDate="2015-11-23T11:50:08.217" Score="31" ViewCount="4264" Body="&lt;p&gt;Alice and Bob and the referee Conny play the following game with a pawn on a standard $8\times8$ chessboard: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In the beginning, Conny places the pawn into the center of a randomly chosen square. All $64$ squares are chosen with equal probabilities. &lt;/li&gt;&#xA;&lt;li&gt;Then Alice and Bob alternate turns; Alice starts.&lt;/li&gt;&#xA;&lt;li&gt;In every turn, the pawn is moved to a new square. The pawn is always placed into the very center of a square.&lt;/li&gt;&#xA;&lt;li&gt;The (Euclidean straight line) distance moved by the pawn must be strictly larger than the (Euclidean straight line) distance moved by the pawn in any of the preceding moves.&lt;/li&gt;&#xA;&lt;li&gt;A player unable to move loses the game.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What is the probability that Alice wins this game?&#xA;  (As usual, we assume that Alice and Bob both use optimal strategies.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2020-11-03T13:36:59.393" Title="A lonely pawn on the chessboard" Tags="|mathematics|strategy|probability|game|checkerboard|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24137" PostTypeId="2" ParentId="24136" CreationDate="2015-11-23T12:14:56.027" Score="38" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alice wins with 100% probability.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Strategy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; On each of her turns Alice will place the pawn onto the square which is exactly opposite of the original square over the center of the board.&lt;br&gt;&#xA; Now Bob will have to move the pawn further than Alice did and will involuntarily end up being closer to a corner than Alice was before.&lt;br&gt;&#xA; With every turn Alice will not change the minimum distance to one of the corners while Bob will get closer and closer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How this works:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lets look at it from the end of the game:&lt;br&gt;&#xA; The person who jumps from one corner of the board to the opposite corner (for example &lt;strong&gt;A1&lt;/strong&gt; to &lt;strong&gt;H8&lt;/strong&gt;) will win, because the opponent can't make a longer jump than that.&lt;br&gt;&#xA; The person who jumps into a corner first will lose, because the opponent can jump into the opposite corner afterwards.&lt;br&gt;&#xA; The only way to force your opponentinto the corner is if you jumped from a square directly next to a corner to the opposite square (for example &lt;strong&gt;A2&lt;/strong&gt; to &lt;strong&gt;H7&lt;/strong&gt;).&lt;br&gt;&#xA; Therefore the first person to jump onto a square directly next to a corner will lose because the opponent can from there on out force the moves.&lt;br&gt;&#xA; And so on and so forth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16608" LastActivityDate="2015-11-23T12:14:56.027" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24138" PostTypeId="1" AcceptedAnswerId="24145" CreationDate="2015-11-23T13:05:45.510" Score="10" ViewCount="1544" Body="&lt;p&gt;Your goal. Win the game with &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=136048&quot;&gt;Barren Glory&lt;/a&gt; using a &lt;a href=&quot;http://magic.wizards.com/en/gameinfo/gameplay/formats/modern&quot;&gt;Modern legal deck&lt;/a&gt; on turn 2's upkeep using the least number of cards possible. (I don't think turn 1 is possibble but please prove me wrong).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume your opponent has a deck containing only basic lands and you decide whether you begin or are on the draw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using a card means using its effect. If a card is discarded, exiled, milled or anything else to pay for an additional cost or because of an effect of another card it is not counted towards your card count.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My personal solution uses 8 cards and has a total converted mana cost of 21. Good luck!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is a tie then the answer with the least total converted mana cost of all cards used wins.&lt;/p&gt;&#xA;" OwnerUserId="1748" LastEditorUserId="1748" LastEditDate="2015-11-23T14:08:01.847" LastActivityDate="2015-11-24T04:17:37.697" Title="Magic: the Gathering - Barren Glory" Tags="|magic-the-gathering|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24139" PostTypeId="1" AcceptedAnswerId="24150" CreationDate="2015-11-23T13:50:23.860" Score="15" ViewCount="988" Body="&lt;p&gt;State the rule behind the following partition of letters, and fill &#xA;in the questionmarks. (The puzzle has a clear and justifiable trivia-type solution.)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;      A . . D E . G H I . K . . N O . . . S T . . . ? ? ?&#xA;   ---------------------------------------------------------&#xA;      . B C . . F . . . J . L M . . P Q R . . U V W ? ? ?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="9125" LastActivityDate="2015-11-23T20:28:51.257" Title="State the rule!" Tags="|pattern|knowledge|letter-sequence|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24140" PostTypeId="2" ParentId="24120" CreationDate="2015-11-23T13:54:38.007" Score="5" Body="&lt;p&gt;You have a remote starter, and since you leave the windows closed you start the car about 15 minutes before using it.  In the winter you have the heater on, in the summer you have the air conditioning on, so remote starting the car makes it cool or toasty depending on how you've adjusted the climate controls each day.&lt;/p&gt;&#xA;" OwnerUserId="1682" LastActivityDate="2015-11-23T13:54:38.007" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24141" PostTypeId="2" ParentId="24115" CreationDate="2015-11-23T13:57:30.370" Score="3" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $S \circ W \circ A \circ G = G$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation:&lt;/strong&gt; &#xA;After plugging in the values $A=1,~B=2,~C=3,~\ldots,~Z=26$, each of the three&#xA;example equations $~~~\alpha\circ\beta\circ\gamma\circ\delta=\epsilon~~~$ follows the pattern&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $~~~(\alpha+\beta)/\gamma+\delta=\epsilon^2~~~~~$  respectively &#xA; $~~~~~\epsilon=\sqrt{(\alpha+\beta)/\gamma+\delta}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Indeed, the three given equations give us the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    C o I o D o A = B    yields     3 o  9 o  4 o  1 = 2&#xA;    O o K o M o G = C    yields    15 o 11 o 13 o  7 = 3&#xA;    Y o A o B o C = D    yields    25 o  1 o  2 o  3 = 4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;One easily verifies that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(3+9)/4+1=2^2~~~$ and $~~~(15+11)/13+7=3^2~~~$ and $~~~(25+1)/2+3=4^2$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This then leads to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $S\circ W\circ A\circ G=G~~~~~$ as $~(19+23)/1+7=7^2$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-11-23T16:17:23.197" LastActivityDate="2015-11-23T16:17:23.197" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24142" PostTypeId="2" ParentId="24138" CreationDate="2015-11-23T14:00:31.303" Score="5" Body="&lt;p&gt;Here's a turn 1 version using 4 cards and 13 mana.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Play &lt;a href=&quot;http://magiccards.info/ts/en/270.html&quot; rel=&quot;nofollow&quot;&gt;Calciform pools&lt;/a&gt;, pass turn, play plains, charge pools, pass turn charge pools, pass turn, etc until we start the turn with 6 counters on the pools: Tap pools, plains, remove 6 counters =&gt; play &lt;a href=&quot;http://magiccards.info/mm2/en/4.html&quot; rel=&quot;nofollow&quot;&gt;Karn, liberated&lt;/a&gt;, exile with +4 &lt;a href=&quot;http://magiccards.info/fut/en/3.html&quot; rel=&quot;nofollow&quot;&gt;Barren Glory&lt;/a&gt; (pass turn) exile &lt;a href=&quot;http://magiccards.info/ts/en/115.html&quot; rel=&quot;nofollow&quot;&gt;Living End&lt;/a&gt; with +4 ability(pass turn) Ult, restart game, muligan to 0, put Barren Glory into play, win on first upkeep. 4 cards used (exiled living end doesn't count), 12 mana total, turn 1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If that somehow doesn't rock your boat, here's a simple and awesome 8&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;mountain, &lt;a href=&quot;http://magiccards.info/pc/en/122.html&quot; rel=&quot;nofollow&quot;&gt;Simian spirit guide&lt;/a&gt; x 4 (or x1 and &lt;a href=&quot;http://magiccards.info/mma/en/110.html&quot; rel=&quot;nofollow&quot;&gt;desperate ritual&lt;/a&gt; x3 if that counts as less mana cost) to cast &lt;a href=&quot;http://magiccards.info/chk/en/193.html&quot; rel=&quot;nofollow&quot;&gt;Through the breach&lt;/a&gt; into &lt;a href=&quot;http://magiccards.info/roe/en/113.html&quot; rel=&quot;nofollow&quot;&gt;Hellcarver Demon&lt;/a&gt; into &lt;a href=&quot;http://magiccards.info/fut/en/3.html&quot; rel=&quot;nofollow&quot;&gt;Barren Glory&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since Ivo made fun of my mana costs here's one that's far more boring but that beats his :P &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; On the play: &lt;a href=&quot;http://magiccards.info/mma/en/223.html&quot; rel=&quot;nofollow&quot;&gt;Glimmervoid&lt;/a&gt;, &lt;a href=&quot;http://magiccards.info/pc/en/122.html&quot; rel=&quot;nofollow&quot;&gt;Simian spirit guide&lt;/a&gt;x1 &lt;a href=&quot;http://magiccards.info/mma/en/110.html&quot; rel=&quot;nofollow&quot;&gt;desperate ritual&lt;/a&gt; x4, &lt;a href=&quot;http://magiccards.info/shm/en/211.html&quot; rel=&quot;nofollow&quot;&gt;Manamorphose&lt;/a&gt;  for 2 white mana-&gt; draw and play &lt;a href=&quot;http://magiccards.info/fut/en/3.html&quot; rel=&quot;nofollow&quot;&gt;Barren Glory&lt;/a&gt;, sacrifice Glimmervoid eot, total cost: 19&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16846" LastEditorUserId="16846" LastEditDate="2015-11-23T20:29:21.403" LastActivityDate="2015-11-23T20:29:21.403" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="24143" PostTypeId="2" ParentId="24138" CreationDate="2015-11-23T14:27:50.237" Score="9" Body="&lt;p&gt;Let's see...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Got it on turn 1 unkeep with 10 cards, 27 CMC.  Thanks to @Ninety-Three for greatly improving my sloppy-but-hilarious first turn 1 solution!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Wow You Got Lucky:&lt;/strong&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Go second, mulligan once. During opponent's turn one end step:&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Exile Simian Spirit Guide x2 for RR&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Desperate Ritual x3 for RR &gt; RRRRR&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Manamorphose for RR, getting UR and drawing a card&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Manamorphose for RR, getting UG and drawing a card&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Quicken for U drawing a card&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Unexpected Results for UGRR, revealing Barren Glory, put it into play&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Continue into your turn one upkeep, win the game.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My older, un-optimized turn-one solution&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Wow You Got &lt;em&gt;Really&lt;/em&gt; Lucky:&lt;/strong&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Go second. During opponent's turn one end step:&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Exile Simian Spirit Guide x2 for RR&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Desperate Ritual x2 for RR, RR &gt; RRRR&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Manamorphose for RR, getting GG, drawing 1 (3 cards, RRGG floating)&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Violent Outburst for RRG, cascading into Wheel of Fate and playing it, discarding Composite Golem x2, drawing 7 (7 cards, G floating)&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Exile Simian Spirit Guide x2 for RR&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Violent Outburst for RRG, cascading into Living End, returning both Composite Golems (4 cards, nothing floating)&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Sacrifice both Composite Golems for WWUUBBRRGG&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Quicken for U, drawing 1 (4 cards, WWUBBRRGG floating)&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Unexpected Results for UGWW, revealing Barren Glory, put it into play&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; One with Nothing for B, discard the rest of your hand&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Continue into your turn one upkeep, win the game.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;-- Previous turn 2 wins ---  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Down to 7 cards, total CMC 19!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Very Expected Results&lt;/strong&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;em&gt;---Mulligan to 5---&lt;/em&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Exile Simian Spirit Guide for R&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Glimmervoid, tap for R&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Desperate Ritual x2 for RR, RR &gt; RRRR&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Manamorphose for RR &gt; GU, drawing Unexpected Results&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Unexpected Results for RRGU, revealing Barren Glory, play it for 0&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Pass Turn, sacrificing Glimmervoid&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Win on next Upkeep&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;8 cards, total CMC 15&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Amulet Bloom is a good Modern deck, let's just play that!&lt;/strong&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Exile Simian Spirit Guide for R&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Amulet of Vigor for R&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Selesnya Sanctuary, untap, tap for GW, return to hand&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Summer Bloom for GW&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Selesnya Sanctuary, untap, tap for GW, return to hand&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Amulet of Vigor for G&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Selesnya Sanctuary, untap, tap for GW, untap, tap for GW, return to hand&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Selesnya Sanctuary, untap, tap for GW, untap, tap for GW, cycle Edge of Autumn, sacrificing the Sanctuary and drawing Claws of Gix&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Claws of Gix for 0&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Barren Glory for WWWWGG&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Sacrifice both Amulets of Vigor and Claws of Gix to Claws of Gix for GGW&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Pass Turn&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Win on next Upkeep  (Simian Spirit Guide exiled, both Amulets of Vigor sacrificed to Claws of Gix, Claws of Gix sacrificed to itself, land sacrificed to Edge of Autumn)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="17389" LastEditDate="2015-11-24T04:17:37.697" LastActivityDate="2015-11-24T04:17:37.697" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="24144" PostTypeId="2" ParentId="24138" CreationDate="2015-11-23T15:39:07.320" Score="3" Body="&lt;p&gt;I think I can do it with total CMC 12:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Glimmervoid - R. &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?name=MEMNITE&quot; rel=&quot;nofollow&quot;&gt;Memnite&lt;/a&gt; - R. &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=235603&quot; rel=&quot;nofollow&quot;&gt;Infernal Plunge&lt;/a&gt; - RRR. Memnite - RRR. Infernal Plunge - RRRRR. Desperate Ritual - RRRRRR. Manamorphose - RRRRWW. Barren Glory. Total Cards: 8 Total CMC: 0 + 0 + 1 + 0 + 1 + 2 + 2 + 6 = 12&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9523" LastEditorUserId="-1" LastEditDate="2015-11-23T16:17:33.473" LastActivityDate="2015-11-23T16:17:33.473" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24145" PostTypeId="2" ParentId="24138" CreationDate="2015-11-23T15:42:25.530" Score="6" Body="&lt;p&gt;By combining the solutions of Zerris and histocrat and adding &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=96934&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wild Cantor&lt;/a&gt;, I can get down to seven cards, total CMC fifteen.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mulligan to a six card hand.&lt;br&gt;&#xA; Exile &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=124474&quot; rel=&quot;nofollow noreferrer&quot;&gt;Simian Spirit Guide&lt;/a&gt; for &lt;a href=&quot;https://i.stack.imgur.com/kolAx.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kolAx.gif&quot; alt=&quot;R&quot;&gt;&lt;/a&gt;.&lt;br&gt;&#xA; Cast &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=383334&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ornithopter&lt;/a&gt;.&lt;br&gt;&#xA; Cast &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=235603&quot; rel=&quot;nofollow noreferrer&quot;&gt;Infernal Plunge&lt;/a&gt; for &lt;a href=&quot;https://i.stack.imgur.com/kolAx.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kolAx.gif&quot; alt=&quot;R&quot;&gt;&lt;/a&gt;, add &lt;a href=&quot;https://i.stack.imgur.com/kolAx.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kolAx.gif&quot; alt=&quot;R&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/kolAx.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kolAx.gif&quot; alt=&quot;R&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/kolAx.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kolAx.gif&quot; alt=&quot;R&quot;&gt;&lt;/a&gt;.&lt;br&gt;&#xA; Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=370425&quot; rel=&quot;nofollow noreferrer&quot;&gt;Glimmervoid&lt;/a&gt;, tap for &lt;a href=&quot;https://i.stack.imgur.com/LXnPK.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LXnPK.gif&quot; alt=&quot;G&quot;&gt;&lt;/a&gt;.&lt;br&gt;&#xA; Cast &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=96934&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wild Cantor&lt;/a&gt; for &lt;a href=&quot;https://i.stack.imgur.com/kolAx.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kolAx.gif&quot; alt=&quot;R&quot;&gt;&lt;/a&gt;, sacrifice it to add &lt;a href=&quot;https://i.stack.imgur.com/vvt7g.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vvt7g.gif&quot; alt=&quot;U&quot;&gt;&lt;/a&gt;.&lt;br&gt;&#xA; Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=366248&quot; rel=&quot;nofollow noreferrer&quot;&gt;Unexpected Results&lt;/a&gt;, revealing &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=136048&quot; rel=&quot;nofollow noreferrer&quot;&gt;Barren Glory&lt;/a&gt;.&lt;br&gt;&#xA; Pass the turn, sacrifice &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=370425&quot; rel=&quot;nofollow noreferrer&quot;&gt;Glimmervoid&lt;/a&gt;.&lt;br&gt;&#xA; Win on your next upkeep.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Old solution: Seven cards, total CMC seventeen.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mulligan to a six card hand.&lt;br&gt;&#xA; Exile &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=124474&quot; rel=&quot;nofollow noreferrer&quot;&gt;Simian Spirit Guide&lt;/a&gt; for &lt;a href=&quot;https://i.stack.imgur.com/kolAx.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kolAx.gif&quot; alt=&quot;R&quot;&gt;&lt;/a&gt;.&lt;br&gt;&#xA; Cast &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=191577&quot; rel=&quot;nofollow noreferrer&quot;&gt;Amulet of Vigor&lt;/a&gt;.&lt;br&gt;&#xA; Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=397734&quot; rel=&quot;nofollow noreferrer&quot;&gt;Selesnya Sanctuary&lt;/a&gt;, tap it and return it to your hand.&lt;br&gt;&#xA; Cast &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=84694&quot; rel=&quot;nofollow noreferrer&quot;&gt;Summer Bloom&lt;/a&gt;.&lt;br&gt;&#xA; Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=397734&quot; rel=&quot;nofollow noreferrer&quot;&gt;Selesnya Sanctuary&lt;/a&gt;, tap it and return it to your hand.&lt;br&gt;&#xA; Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=397734&quot; rel=&quot;nofollow noreferrer&quot;&gt;Selesnya Sanctuary&lt;/a&gt;, tap it and return it to your hand.&lt;br&gt;&#xA; Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=397734&quot; rel=&quot;nofollow noreferrer&quot;&gt;Selesnya Sanctuary&lt;/a&gt;, tap it and sacrifice it to cycle &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=243442&quot; rel=&quot;nofollow noreferrer&quot;&gt;Edge of Autumn&lt;/a&gt;.&#xA; Cast &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=49060&quot; rel=&quot;nofollow noreferrer&quot;&gt;Deconstruct&lt;/a&gt; paying &lt;a href=&quot;https://i.stack.imgur.com/LXnPK.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LXnPK.gif&quot; alt=&quot;G&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/LXnPK.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LXnPK.gif&quot; alt=&quot;G&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/LXnPK.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LXnPK.gif&quot; alt=&quot;G&quot;&gt;&lt;/a&gt; on &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=191577&quot; rel=&quot;nofollow noreferrer&quot;&gt;Amulet of Vigor&lt;/a&gt;.&lt;br&gt;&#xA; Cast &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=136048&quot; rel=&quot;nofollow noreferrer&quot;&gt;Barren Glory&lt;/a&gt;.&lt;br&gt;&#xA; Pass the turn and win on your next upkeep.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7459" LastEditorUserId="7459" LastEditDate="2015-11-23T16:23:49.613" LastActivityDate="2015-11-23T16:23:49.613" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24148" PostTypeId="2" ParentId="24120" CreationDate="2015-11-23T17:27:56.710" Score="2" Body="&lt;p&gt;The car is not hotter.. you are just wearing your winter clothes in the car, so you are hotter. You aren't airing the hot air out, you are letting the cool air in because your body is overheating. Take off your winter clothes when you get in the car. In the summer you do not wear winter clothes, so it won't be so hot.&lt;/p&gt;&#xA;" OwnerUserId="17497" LastActivityDate="2015-11-23T17:27:56.710" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24149" PostTypeId="1" AcceptedAnswerId="24154" CreationDate="2015-11-23T17:54:02.813" Score="8" ViewCount="594" Body="&lt;p&gt;The blackboard contains the two numbers 0 and 120. Each minute you are allowed to write an additional number $ x $ on the blackboard if $ x$ has not been written before on the blackboard and if $ x $ is the integer root of some polynomial $ P(x) $ in one variable whose coefficients are numbers on the blackboard. The game ends when you can not write further numbers on the blackboard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which numbers will be written on the blackboard when the game ends?&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="9404" LastEditDate="2015-11-23T18:35:52.247" LastActivityDate="2015-11-23T20:06:54.573" Title="Blackboard problem with polynomial" Tags="|mathematics|game|reachability|blackboard|" AnswerCount="4" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="24150" PostTypeId="2" ParentId="24139" CreationDate="2015-11-23T18:07:49.513" Score="29" Body="&lt;p&gt;The rule appears to be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The last letters of the names of U.S. states.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;A&lt;/strong&gt;&lt;br&gt;&#xA; Alabam&lt;strong&gt;a&lt;/strong&gt;&lt;br&gt;&#xA; Alask&lt;strong&gt;a&lt;/strong&gt;&lt;br&gt;&#xA; Arizon&lt;strong&gt;a&lt;/strong&gt;&lt;br&gt;&#xA; Californi&lt;strong&gt;a&lt;/strong&gt;&lt;br&gt;&#xA; Florid&lt;strong&gt;a&lt;/strong&gt;&lt;br&gt;&#xA; Georgi&lt;strong&gt;a&lt;/strong&gt;&lt;br&gt;&#xA; Indian&lt;strong&gt;a&lt;/strong&gt;&lt;br&gt;&#xA; Iow&lt;strong&gt;a&lt;/strong&gt;&lt;br&gt;&#xA; Louisian&lt;strong&gt;a&lt;/strong&gt;&lt;br&gt;&#xA; Minnesot&lt;strong&gt;a&lt;/strong&gt;&lt;br&gt;&#xA; Montan&lt;strong&gt;a&lt;/strong&gt;&lt;br&gt;&#xA; Nebrask&lt;strong&gt;a&lt;/strong&gt;&lt;br&gt;&#xA; Nevad&lt;strong&gt;a&lt;/strong&gt;&lt;br&gt;&#xA; North Carolin&lt;strong&gt;a&lt;/strong&gt;&lt;br&gt;&#xA; North Dakot&lt;strong&gt;a&lt;/strong&gt;&lt;br&gt;&#xA; Oklahom&lt;strong&gt;a&lt;/strong&gt;&lt;br&gt;&#xA; Pennsylvani&lt;strong&gt;a&lt;/strong&gt;&lt;br&gt;&#xA; South Carolin&lt;strong&gt;a&lt;/strong&gt;&lt;br&gt;&#xA; South Dakot&lt;strong&gt;a&lt;/strong&gt;&lt;br&gt;&#xA; Virgini&lt;strong&gt;a&lt;/strong&gt;&lt;br&gt;&#xA; West Virgini&lt;strong&gt;a&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;D&lt;/strong&gt;&lt;br&gt;&#xA; Marylan&lt;strong&gt;d&lt;/strong&gt;&lt;br&gt;&#xA; Rhode Islan&lt;strong&gt;d&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;E&lt;/strong&gt;&lt;br&gt;&#xA; Delawar&lt;strong&gt;e&lt;/strong&gt;&lt;br&gt;&#xA; Main&lt;strong&gt;e&lt;/strong&gt;&lt;br&gt;&#xA; New Hampshir&lt;strong&gt;e&lt;/strong&gt;&lt;br&gt;&#xA; Tennesse&lt;strong&gt;e&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;G&lt;/strong&gt;&lt;br&gt;&#xA; Wyomin&lt;strong&gt;g&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;H&lt;/strong&gt;&lt;br&gt;&#xA; Uta&lt;strong&gt;h&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;I&lt;/strong&gt;&lt;br&gt;&#xA; Hawai&lt;strong&gt;i&lt;/strong&gt;&lt;br&gt;&#xA; Mississipp&lt;strong&gt;i&lt;/strong&gt;&lt;br&gt;&#xA; Missour&lt;strong&gt;i&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;K&lt;/strong&gt;&lt;br&gt;&#xA; New Yor&lt;strong&gt;k&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;N&lt;/strong&gt;&lt;br&gt;&#xA; Michiga&lt;strong&gt;n&lt;/strong&gt;&lt;br&gt;&#xA; Orego&lt;strong&gt;n&lt;/strong&gt;&lt;br&gt;&#xA; Washingto&lt;strong&gt;n&lt;/strong&gt;&lt;br&gt;&#xA; Wisconsi&lt;strong&gt;n&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;O&lt;/strong&gt;&lt;br&gt;&#xA; Colora&lt;strong&gt;o&lt;/strong&gt;&lt;br&gt;&#xA; Idah&lt;strong&gt;o&lt;/strong&gt;&lt;br&gt;&#xA; New Mexic&lt;strong&gt;o&lt;/strong&gt;&lt;br&gt;&#xA; Ohi&lt;strong&gt;o&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;S&lt;/strong&gt;&lt;br&gt;&#xA; Arkansa&lt;strong&gt;s&lt;/strong&gt;&lt;br&gt;&#xA; Illinoi&lt;strong&gt;s&lt;/strong&gt;&lt;br&gt;&#xA; Kansa&lt;strong&gt;s&lt;/strong&gt;&lt;br&gt;&#xA; Massachusett&lt;strong&gt;s&lt;/strong&gt;&lt;br&gt;&#xA; Texa&lt;strong&gt;s&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;T&lt;/strong&gt;&lt;br&gt;&#xA; Connecticu&lt;strong&gt;t&lt;/strong&gt;&lt;br&gt;&#xA; Vermon&lt;strong&gt;t&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So of the last three letters,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Y&lt;/strong&gt; (Kentuck&lt;strong&gt;y&lt;/strong&gt;, New Jerse&lt;strong&gt;y&lt;/strong&gt;) is on the top row, X and Z on the bottom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5684" LastEditorUserId="4421" LastEditDate="2015-11-23T18:26:30.657" LastActivityDate="2015-11-23T18:26:30.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24151" PostTypeId="2" ParentId="24149" CreationDate="2015-11-23T18:32:33.640" Score="3" Body="&lt;p&gt;Every integer can be written on the board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First polynomial:&#xA;$$P(x) = 120x - 120$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The root of this polynomial is 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any integer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P(x) = x - x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this solution, it is important to note that constants are not coefficients. Also note that $-$ is an operator, not part of the coefficient. Had it been intended as the coefficient, I would have written $x + -x$ (or even more explicitly $1x + -1x$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: As an added shortcut (thanks to f''), you could simply use this polynomial from the initial board to retrieve any value for $x$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P(x) = 0x$$&lt;/p&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2015-11-23T19:01:08.713" LastActivityDate="2015-11-23T19:01:08.713" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="24152" PostTypeId="2" ParentId="24149" CreationDate="2015-11-23T18:50:07.437" Score="3" Body="&lt;p&gt;I started like this &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;120x + 120 = 0 &amp;amp;\rightarrow&amp;amp; x &amp;amp;= &amp;amp;-1\\&#xA;- x^{120} - x^{119} - ... - x^2 - x^1 +120 = 0 &amp;amp;\rightarrow&amp;amp;x &amp;amp;= &amp;amp;1\\&#xA;x + 120 = 0 &amp;amp;\rightarrow&amp;amp; x&amp;amp; = &amp;amp;-120\\&#xA;x^7 - x^3 + 120 = 0 &amp;amp;\rightarrow&amp;amp; x &amp;amp;= &amp;amp;-2\\&#xA;x - 2 = 0 &amp;amp;\rightarrow&amp;amp; x &amp;amp;= &amp;amp;2\\&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have $-120, -2, -1, 0, 1, 2, 120$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having both $-1$ and $1$ means that I can always get the negated value of any new number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I can't crack this puzzle as I keep finding new numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wonder if I got it all wrong...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some more (thanks to Aravind for the first 3)...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;x^2 + 2x - 120 = 0 &amp;amp;\rightarrow&amp;amp; x &amp;amp;= &amp;amp;10\\&#xA;2x - 10 = 0 &amp;amp;\rightarrow&amp;amp; x &amp;amp;= &amp;amp;5\\&#xA;10x - 120 = 0 &amp;amp;\rightarrow&amp;amp; x &amp;amp;= &amp;amp;12\\&#xA;x^4 + x^3 + x^2 + x - 120 = 0 &amp;amp;\rightarrow&amp;amp; x &amp;amp;= &amp;amp;3\\&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;2x - 120 = 0 &amp;amp;\rightarrow&amp;amp; x &amp;amp;= &amp;amp;60\\&#xA;3x - 120 = 0 &amp;amp;\rightarrow&amp;amp; x &amp;amp;= &amp;amp;40\\&#xA;5x - 120 = 0 &amp;amp;\rightarrow&amp;amp; x &amp;amp;= &amp;amp;24\\&#xA;5x^4 + 10x - 120 &amp;amp;\rightarrow&amp;amp; x &amp;amp;= &amp;amp;4\\&#xA;4x - 120 = 0 &amp;amp;\rightarrow&amp;amp; x &amp;amp;= &amp;amp;30\\&#xA;4x^2 + 4x -120 &amp;amp;\rightarrow&amp;amp; x &amp;amp;= &amp;amp;5\\&#xA;5x -120 = 0 &amp;amp;\rightarrow&amp;amp; x &amp;amp;= &amp;amp;24\\&#xA;3x^2 + 2x - 120 = 0 &amp;amp;\rightarrow&amp;amp; x &amp;amp;= &amp;amp;6\\&#xA;6x - 120 = 0 &amp;amp;\rightarrow&amp;amp; x &amp;amp;= &amp;amp;20\\\\&#xA;-2x^2 - x + 120 = 0 &amp;amp;\rightarrow&amp;amp; x &amp;amp;= &amp;amp;-8\\\\&#xA;-8x + 120 = 0 &amp;amp;\rightarrow&amp;amp; x &amp;amp;= &amp;amp;15\\&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{0, 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120\}$ (and all the negated numbers as well).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I understand from other answers/comments this should be the complete list. I still wonder how that is proved.&lt;/p&gt;&#xA;" OwnerUserId="17499" LastEditorUserId="17499" LastEditDate="2015-11-23T20:06:54.573" LastActivityDate="2015-11-23T20:06:54.573" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="24153" PostTypeId="2" ParentId="24149" CreationDate="2015-11-23T18:50:50.107" Score="3" Body="&lt;p&gt;I believe the OP's intended answer is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exactly the divisors of 120. This is true of the non-zero numbers on the board at any stage. If $P(x)=0$, then we can write it as $xg(x)+d=0$ and $d$ is a divisor of 120, this implies that so is $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;StillLearning's answer shows that all of them can be obtained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once $2,3,5$ are obtained, other divisors of 120 can be obtained by repeated division, eg: $2x-120=0$ to get $60$, then $2x-60-0$ to get $30$ etc.&lt;/p&gt;&#xA;" OwnerUserId="5684" LastEditorUserId="4421" LastEditDate="2015-11-23T19:40:19.667" LastActivityDate="2015-11-23T19:40:19.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24154" PostTypeId="2" ParentId="24149" CreationDate="2015-11-23T19:40:01.570" Score="11" Body="&lt;p&gt;By the &lt;a href=&quot;https://en.wikipedia.org/wiki/Rational_root_theorem&quot;&gt;rational root theorem&lt;/a&gt;, any integer root of the polynomial&#xA;$$a_nx^n + \dots + a_1x + a_0$$&#xA;is a divisor of its constant term $a_0$, if $a_0 \neq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will never let $a_0 = 0$, since that means we can divide the whole polynomial by $x$ (repeatedly, if necessary; yielding a root $0$ each time, which is initially already on the board) and get a polynomial (with the same non-zero roots) for which $a_0 \neq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus initially, $a_0 = 120$, our only choice. From this, suppose we get a new root $p$, a divisor of $a_0 = 120$. Then the next minute, we can choose from $\{p, 120\}$, both divisors of 120, etc. Clearly, the integer roots we generate by this process will &lt;em&gt;always&lt;/em&gt; be divisors of 120, positive or negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We show how to achieve &lt;em&gt;all&lt;/em&gt; divisors of 120:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;First, $120x+120$ gives us $x=-1$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Then, $(-1)x^{120}+(-1)x^{119}+\dots+(-1)x^1+120$ gives us $x=1$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Then, $x+120$ gives us $x=-120$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We can get $2$, $3$ and $5$ using only the coefficients in $\{\pm 120, \pm 1\}$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$-x^7 + x^3 \pm 120$ gives us $x = \pm 2$.&lt;/li&gt;&#xA;&lt;li&gt;$-x^4 -x^3 -x^2 -x \pm 120$ gives us $x = \pm 3$.&lt;/li&gt;&#xA;&lt;li&gt;$-x^3 +x \pm 120$ gives us $x = \pm 5$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Aravind found a nice trick -- now that we have $2$, $3$, and $5$ on the blackboard, the other roots follow by dividing from $120 = 2^3 \cdot 3 \cdot 5$. (If we have $p$ and $q$ on the board, we can use $px-q$ to get $x=q/p$, so first take $q=120$ and then repeat the process with $q \to q/p$ for a prime $p$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can conlude that the solution set is precisely the set of positive and negative divisors of 120.&lt;/p&gt;&#xA;" OwnerUserId="11476" LastEditorUserId="11476" LastEditDate="2015-11-23T20:00:39.127" LastActivityDate="2015-11-23T20:00:39.127" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24155" PostTypeId="2" ParentId="24136" CreationDate="2015-11-23T20:12:28.047" Score="12" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Alice always wins. If Alice is at &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt;, her winning strategy is to hop to &lt;span class=&quot;math-container&quot;&gt;$q'$&lt;/span&gt;, such that the midpoint of &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$q'$&lt;/span&gt; is the center of the board.&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; We need only prove that this move is always legal for Alice to make. Since she never is without a legal move, she wins.&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; She can certainly do this on her first move.&#xA; For other moves, let &lt;span class=&quot;math-container&quot;&gt;$p'$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$q'$&lt;/span&gt; be the opposites of &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt;, respectively, as defined above. Suppose Alice has just hopped from &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$p'$&lt;/span&gt;, and then Bob hopped from &lt;span class=&quot;math-container&quot;&gt;$p'$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt;. Let &lt;span class=&quot;math-container&quot;&gt;$o$&lt;/span&gt; be the center of the board. Finally, let &lt;span class=&quot;math-container&quot;&gt;$d(a,b)$&lt;/span&gt; mean the distance between points &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt;, and recall the triangle inequality &lt;span class=&quot;math-container&quot;&gt;$d(a,c)\le d(a,b)+d(b,c)$&lt;/span&gt;, for any &lt;span class=&quot;math-container&quot;&gt;$a,b,c$&lt;/span&gt;.&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; First, we show that &lt;span class=&quot;math-container&quot;&gt;$d(q,o)&amp;gt;d(p',o)$&lt;/span&gt;. Supposing not, it would follow that&#xA;&#xA; &lt;span class=&quot;math-container&quot;&gt;$$d(p',q)\le d(p',o)+d(o,q)\le d(p',o)+d(o,p')=d(p',p),$$&lt;/span&gt;&#xA;&#xA; but &lt;span class=&quot;math-container&quot;&gt;$d(p',q)\le d(p',p)$&lt;/span&gt; contradicts the fact that it was legal for Bob to hop from &lt;span class=&quot;math-container&quot;&gt;$p'$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt;.&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; With this fact under our belts, we know have that&#xA;&#xA; &lt;span class=&quot;math-container&quot;&gt;$$d(p',q)\le d(p',o)+d(q,o)&amp;lt;d(q,o)+d(q,o)=d(q,q')$$&lt;/span&gt;&#xA;&#xA; This proves that it is legal for Alice to hop from &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$q'$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="72204" LastEditDate="2020-11-03T13:36:59.393" LastActivityDate="2020-11-03T13:36:59.393" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24157" PostTypeId="1" AcceptedAnswerId="24159" CreationDate="2015-11-23T20:27:10.870" Score="12" ViewCount="997" Body="&lt;p&gt;After looking through the list of books my friend sent me &lt;a href=&quot;https://puzzling.stackexchange.com/questions/24124/emojiture-challenge&quot;&gt;the last time&lt;/a&gt; I asked him to send me some recommendations for more recent books. Here is what I got:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IodsV.png&quot; alt=&quot;1. ✞ 2.  3.  4.  5.  6.  7.  8. &quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope you can help me again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hints:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All books are newer that those on the previous list.&lt;br&gt;Each book was written by a different author.&lt;br&gt;Each book was followed by a movie or TV series.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-11-23T21:38:10.343" Title="Emojiture Challenge - Next Generation" Tags="|rebus|literature|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24159" PostTypeId="2" ParentId="24157" CreationDate="2015-11-23T20:52:53.853" Score="15" Body="&lt;p&gt;So far I've got:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. Pet Semetary&#xA;&lt;br&gt; 2. The Girl with the Dragon Tattoo&#xA;&lt;br&gt; 3. Game of Thrones&#xA;&lt;br&gt; 4. Night Watch (several novels of the same title)&#xA;&lt;br&gt; 5. The Hitchhiker's Guide to the Galaxy&#xA;&lt;br&gt; 6. Harry Potter and the Goblet of Fire&#xA;&lt;br&gt; 7. Hunger Games&#xA;&lt;br&gt; 8. The Neverending Story&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12283" LastEditorUserId="12283" LastEditDate="2015-11-23T21:26:43.033" LastActivityDate="2015-11-23T21:26:43.033" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24160" PostTypeId="2" ParentId="24157" CreationDate="2015-11-23T21:09:48.570" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. Pet Sematary&lt;br&gt;&#xA;2. Girl with the Dragon Tattoo&lt;br&gt;&#xA;3. Game of Thrones&lt;br&gt;&#xA;4. Night Watch&lt;br&gt;&#xA;5. Hitchhiker's Guide to the Galaxy&lt;br&gt;&#xA;6. Harry Potter and the Goblet of Fire&lt;br&gt;&#xA;7. Hunger Games&lt;br&gt;&#xA;8.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16874" LastActivityDate="2015-11-23T21:09:48.570" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24161" PostTypeId="1" CreationDate="2015-11-24T01:49:53.403" Score="8" ViewCount="398" Body="&lt;p&gt;&lt;strong&gt;Long time no see!  Hello Stonehenge!(I mean PSE)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A man named Kasiski gives you a letter that reads this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Yssy uztzt t hzbvhqfhew wphivzwim mt tbdnhp wnvjchimzos agi qirmmpfmhwp ho uwtbg mt youzme on tgscrkjyh, igfywmtyp dakfdwtx ysot anuocdx eve vfdiaeyj ho afgs imx rsnxytq mthswnxwj os ts tbsmwfaegy tb pktaogtytcn hk thsxqq.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;chile; venezuela, nigeria; canada, germany &lt;br/&gt;&#xA;  japan, australia, new zealand &lt;br/&gt;&#xA;  Liverpool, Manchester, Sheffield, Nottingham, Oxford, Swindon, Bristol, Cardiff &lt;br/&gt;&#xA;  angola, russia, kenya; libya, saudi arabia &lt;br/&gt;&#xA;  malaysia, mongolia; afghanistan, south korea&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You have to find what the secret message and locations together references...&lt;/p&gt;&#xA;" OwnerUserId="11790" LastEditorUserId="11790" LastEditDate="2015-11-24T21:32:48.950" LastActivityDate="2015-11-24T21:32:48.950" Title="A cipher and a book…?" Tags="|cipher|enigmatic-puzzle|knowledge|literature|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24162" PostTypeId="2" ParentId="23358" CreationDate="2015-11-24T04:11:04.487" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is a SE question, specifically, THIS QUESTION!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16874" LastActivityDate="2015-11-24T04:11:04.487" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24163" PostTypeId="1" AcceptedAnswerId="24166" CreationDate="2015-11-24T10:22:15.207" Score="2" ViewCount="635" Body="&lt;p&gt;Fifteen white rhinoceroses are standing in a line.&#xA;Each of them weighs an integer number of pounds.&#xA;For each rhinoceros (except the leftmost one), its weight plus twice the weight of its left neighbor exactly equals $15000$ pounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determine all possible weight distributions for these fifteen animals!&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-15T09:31:25.370" LastActivityDate="2016-03-15T09:31:25.370" Title="Fifteen white rhinoceroses" Tags="|mathematics|calculation-puzzle|algebra|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24164" PostTypeId="2" ParentId="24163" CreationDate="2015-11-24T10:28:00.547" Score="5" Body="&lt;p&gt;In order for the total weight to be 15 pounds (&lt;strong&gt;15.000 = 15&lt;/strong&gt;, using English decimal notation), with each rhino weighing a positive integer number of pounds the only solution is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They all weigh 5 pounds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In order for the total weight to be 15000 pounds (seems more likely if I remember how much a pound is and how big a rhino is), assuming each rhino must have a positive weight, the only solution is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They all weight 5000 pounds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &quot;error&quot; (i.e. the difference between the actual weight of a rhino and the perfect weight of 5000) doubles with each rhino while alternating sign. i.e. If we have an &quot;error&quot; of +1 in the weight of the first rhino (i.e. it weighs 4999 or 5001) then the &quot;error&quot; in the second rhino is -2, and in the third is +4 etc. As we have 15 rhinos, if the error of the first rhino is &lt;strong&gt;a&lt;/strong&gt; then the error of the last rhino is &lt;strong&gt;2^14 * a&lt;/strong&gt;. Hence an &quot;error&quot; of 1 (the smallest possible) in the first rhino gives an error of 8192 in the fourteenth and 16384 in the fifteenth rhino: 5000 minus either of these is negative and hence not possible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9444" LastEditorUserId="9444" LastEditDate="2015-11-24T10:47:30.390" LastActivityDate="2015-11-24T10:47:30.390" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24165" PostTypeId="2" ParentId="24163" CreationDate="2015-11-24T10:31:26.003" Score="4" Body="&lt;p&gt;Each rhino weighs 5000 pounds. If the leftmost rhino weighs even 5001 or 4999 pounds, the rhinos at the right weigh too much to continue the sequence.&lt;/p&gt;&#xA;" OwnerUserId="12730" LastEditorUserId="12730" LastEditDate="2015-11-24T10:45:14.633" LastActivityDate="2015-11-24T10:45:14.633" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24166" PostTypeId="2" ParentId="24163" CreationDate="2015-11-24T10:54:46.953" Score="11" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The only possible weight for each rhinoceros is 5000 pounds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As for why it is the only solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First we look at the obvious solution of every rhinoceros weighing the same.&lt;br&gt;&#xA; Then the amount they would have to weight would be:&lt;br&gt;&#xA; $3w = 15000$&lt;br&gt;&#xA; $w = 5000$&lt;br&gt;&#xA;&lt;br&gt;&#xA; Now lets add to that a little difference $d$ for the first weight and this difference will cascade through to the other weights.&lt;br&gt;&#xA; $w_0 = 5000 + d_0$&lt;br&gt;&#xA;&lt;br&gt;&#xA; And for all others:&lt;br&gt;&#xA; $w_n = w + d_n$&lt;br&gt;&#xA;&lt;br&gt;&#xA; $w_n + 2w_{n-1} = 15000$&lt;br&gt;&#xA; $w + d_n + 2(w + d_{n-1})= 15000$&lt;br&gt;&#xA; $3w + d_n + 2d_{n-1} = 15000$&lt;br&gt;&#xA; $d_n + 2d_{n-1} = 0$&lt;br&gt;&#xA; $d_n = -2d_{n-1}$&lt;br&gt;&#xA;&lt;br&gt;&#xA; Therefore:&lt;br&gt;&#xA; $d_n = d_0 * -2^n$&lt;br&gt;&#xA;&lt;br&gt;&#xA; Now even for the smallest difference of $d_0 = \pm1$ the 14th and 15th rhinoceros will have a difference of $d_{13} = \pm8192$ and $d_{14} = \pm16384$ respectively.&lt;br&gt;&#xA; Since both have an absolute amount greater than $5000$ and at least one of them has to be negative that will also make the respective rhinoceros have a weight of less than zero which should in no way be possible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16608" LastEditorUserId="16608" LastEditDate="2015-11-24T13:08:52.557" LastActivityDate="2015-11-24T13:08:52.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24167" PostTypeId="2" ParentId="24161" CreationDate="2015-11-24T12:23:47.770" Score="3" Body="&lt;p&gt;Well this is only a partial answer. The man is called Kasiski, which is a hint that we can try&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Using the &lt;a href=&quot;https://en.wikipedia.org/wiki/Kasiski_examination&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kasiski examination&lt;/a&gt; to attempt to solve a Vigenère cipher. I used an &lt;a href=&quot;http://www.dcode.fr/vigenere-cipher&quot; rel=&quot;nofollow noreferrer&quot;&gt;online Vigenere tool&lt;/a&gt; which found that using the Keyword &quot;FLOAT&quot; the text deciphers to&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the plain text&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They built a convoluted reticulum to induce incertitude and furthermore to bring to naught an abhorrent inanimate parasite that hijacks the casualty to have its genetic machinery as an instrument in propagation of itself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well I'm no cell biologist, and a bit stuck. Might that be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; related to a virus of some kind?? Some antiviral cell engineering?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Regarding the lists of countries/cities. The third list are all placenames in England/Wales, and as listed would represent approximately a half-circle,  I added Stonehenge to the map for completeness (green dot and circle near the bottom).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/wunsA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wunsA.jpg&quot; alt=&quot;Liverpool to cardiff on map&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or a &quot;D&quot; if you joined back to Liverpool from Cardiff.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/2sTQQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2sTQQ.jpg&quot; alt=&quot;Liverpool to cardiff and back on map&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7316" LastEditorUserId="7316" LastEditDate="2015-11-24T13:34:15.147" LastActivityDate="2015-11-24T13:34:15.147" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24168" PostTypeId="1" AcceptedAnswerId="24173" CreationDate="2015-11-24T13:50:07.327" Score="17" ViewCount="1495" Body="&lt;p&gt;I recently put together a list of books by some well-known authors, but something went horribly wrong when I tried to print it out. The first entry in the list was completely unreadable, but the rest were all just missing their vowels.&#xA;Not only that, but the two halves of the list were all muddled up. Originally there was a clear relationship between the title in the first column and the title in the second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt; Can you work out what the original titles were, what the matches were, and what the first entry in the list should have read?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;1 ??????????????????????? A LVSFTHTWNS&#xA;2   CRR                   B THCLMNGFSLPNGBTY&#xA;3   NTRVWWTHTHVMPR        C THCCKSCLLNG&#xA;4   PBBLNTHSKY            D SPRTSNBNDG&#xA;5   STRQST                E RG&#xA;6   THMYSTRSFFRTSTYLS     F LGCYFTRRR&#xA;7   THNDRMDSTRN           G GNTSBRD&#xA;8   THPLGRMSRGRSS         H DVDSTRRSPCRNGR&#xA;9   THWSPFCTRY            I DDSN&#xA;10  WTHSHDDRNGFLL         J CNSDRPHLBS&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Thanks to Deusovi for the title&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="14135" LastEditDate="2015-11-24T14:36:12.267" LastActivityDate="2015-11-24T15:29:54.143" Title="Disemvowelled Books" Tags="|literature|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24169" PostTypeId="2" ParentId="24161" CreationDate="2015-11-24T14:00:52.653" Score="5" Body="&lt;p&gt;Building on &lt;a href=&quot;https://puzzling.stackexchange.com/a/24167/9000&quot;&gt;Fillet's answer&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The keyword for the cipher (&lt;code&gt;FLOAT&lt;/code&gt;) is found by connecting all the locations listed with semicolons indicating line breaks and then squinting a bit.&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/aPk6j.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aPk6j.png&quot; alt=&quot;F&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/koCPc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/koCPc.png&quot; alt=&quot;L&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/nsTD8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nsTD8.jpg&quot; alt=&quot;O&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/8uQbz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8uQbz.jpg&quot; alt=&quot;A&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/OUzhy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OUzhy.jpg&quot; alt=&quot;T&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which is used to solve the Vigenère cipher to get the plain text:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They built a convoluted reticulum to induce incertitude and furthermore to bring to naught an abhorrent inanimate parasite that hijacks the casualty to have its genetic machinery as an instrument in propagation of itself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which sounds like a fancy-pants word replacement way to describe&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; someone getting pregnant.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2015-11-24T14:24:29.330" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24170" PostTypeId="2" ParentId="24168" CreationDate="2015-11-24T14:17:40.220" Score="2" Body="&lt;p&gt;An exceptionally incomplete answer:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1st column  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;6&lt;/strong&gt; The Mysterious Affair at Styles&lt;br&gt;&#xA;&lt;strong&gt;7&lt;/strong&gt; The Andromeda Strain&lt;br&gt;&#xA;&lt;strong&gt;8&lt;/strong&gt; The Pilgrim's Regress&lt;br&gt;&#xA;&lt;strong&gt;10&lt;/strong&gt; With Shuddering Fall  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2nd column  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;B&lt;/strong&gt; The Claiming of Sleeping Beauty (Which I should not have googled at work)&lt;br&gt;&#xA; &lt;strong&gt;C&lt;/strong&gt; The Cuckoo's Calling&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11717" LastEditorUserId="11717" LastEditDate="2015-11-24T14:35:55.313" LastActivityDate="2015-11-24T14:35:55.313" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24171" PostTypeId="1" AcceptedAnswerId="24181" CreationDate="2015-11-24T14:30:27.997" Score="3" ViewCount="871" Body="&lt;p&gt;Who am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I lived a long successful life I did&lt;br&gt;&#xA;  Starting out was rough as a kid  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I seemed to catch break at age thirty eight&lt;br&gt;&#xA;  Working under the governor sure was great  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In '64 I dethroned a  king&lt;br&gt;&#xA;  Shortly after, a shipyard and naval academy became a thing  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;My final moments were in '85&lt;br&gt;&#xA;  Nothing could be done about my demise   &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Before I left a truce was signed&lt;br&gt;&#xA;  My legacy lives on for all of mankind  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Will add hints if needed &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint #1&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To be fair, the year thing may be very misleading and unfair.  So I will tell you not to get too hung up in the 20th century. For all I know denoting a year like '85 can only be referred to the 20th century. So I'll throw this early hint to everyone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #2&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I died at age 72&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #3 (Hopefully a good one!)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I was a military leader in the Qing Dynasty&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16273" LastEditorUserId="16273" LastEditDate="2015-11-27T15:48:40.500" LastActivityDate="2015-11-27T17:24:10.533" Title="You dare to eat me?" Tags="|riddle|rhyme|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="24172" PostTypeId="2" ParentId="24136" CreationDate="2015-11-24T14:35:12.410" Score="8" Body="&lt;p&gt;For an $m\times n$ board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $m$ or $n$ is even:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alice can win with 100% probability using @TheDarkTruth's &lt;a href=&quot;https://puzzling.stackexchange.com/a/24137/5434&quot;&gt;strategy&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If both $m$ and $n$ are odd:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Given an optimal strategy from both players,  Alice will win with 100% probability all games where she start off-centre and lose with 100% probability the games where she starts in the centre of the board; so her overall probability is $\frac{mn-1}{mn}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the case of starting in the exact centre of the board, Alice initially moves the pawn out of the centre then Bob can then follow @TheDarkTruth's &lt;a href=&quot;https://puzzling.stackexchange.com/a/24137/5434&quot;&gt;strategy&lt;/a&gt; with the roles reversed).&#xA; An $8 \times 8$ board always gives the first player a 100% probability of winning as the pawn will always start off-centre.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5434" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2015-11-24T14:46:38.580" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24173" PostTypeId="2" ParentId="24168" CreationDate="2015-11-24T14:50:25.263" Score="14" Body="&lt;p&gt;Column 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 Harry Potter and the Philosopher's Stone&lt;br&gt;&#xA; 2 Carrie&lt;br&gt;&#xA; 3 Interview With the Vampire&lt;br&gt;&#xA; 4 Pebble in the Sky&lt;br&gt;&#xA; 5 Star Quest&lt;br&gt;&#xA; 6 The Mysterious Affair at Styles&lt;br&gt;&#xA; 7 The Andromeda Strain&lt;br&gt;&#xA; 8 The Pilgrim's Regress&lt;br&gt;&#xA; 9 The Wasp Factory&lt;br&gt;&#xA; 10 With Shuddering Fall  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Column 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Lives of the Twins&lt;br&gt;&#xA; B The Claiming of Sleeping Beauty&lt;br&gt;&#xA; C Cuckoos Calling&lt;br&gt;&#xA; D Spirits in Bondage&lt;br&gt;&#xA; E Rage&lt;br&gt;&#xA; F Legacy of Terror&lt;br&gt;&#xA; G Giants Bread&lt;br&gt;&#xA; H David Starr Space Ranger&lt;br&gt;&#xA; I Odds On&lt;br&gt;&#xA; J Consider Phlebas  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Matches:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The pattern is that the first column holds an author's first book under their own name, the second holds their first title written under a pseudonym.&lt;br&gt;&#xA; 1-C (J K Rowling/Robert Galbraith)&lt;br&gt;&#xA; 2-E (Stephen King/Richard Bachman)&lt;br&gt;&#xA; 3-B (Anne Rice/A N Roquelaure)&lt;br&gt;&#xA; 4-H (Isaac Asimov/Paul French)&lt;br&gt;&#xA; 5-F (Dean Koontz/Deanna Dwyer)&lt;br&gt;&#xA; 6-G (Agatha Christie/Mary Westmacott)&lt;br&gt;&#xA; 7-I  (Michael Crichton/John Lange)&lt;br&gt;&#xA; 8-D (C S Lewis/Clive Hamilton)&lt;br&gt;&#xA; 9-J (Iain Banks/Iain M Banks)&lt;br&gt;&#xA; 10-A (Joyce Carol Oates/Rosamond Smith)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16874" LastEditorUserId="16874" LastEditDate="2015-11-24T15:29:54.143" LastActivityDate="2015-11-24T15:29:54.143" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="24174" PostTypeId="2" ParentId="24168" CreationDate="2015-11-24T14:50:37.727" Score="4" Body="&lt;p&gt;1 ??????????????????????? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2   CRR            &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Carrie (Stephen King)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3   NTRVWWTHTHVMPR       &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Interview with a vampire (Anne Rice)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4   PBBLNTHSKY           &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pebble in the sky (Isaac Asimov)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5   STRQST               &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Star Quest (Dean R. Koontz)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6   THMYSTRSFFRTSTYLS &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The mysterious affair at Styles (Agatha Christie)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;7   THNDRMDSTRN          &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Andromeda Strain (Michael Crichton)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;8   THPLGRMSRGRSS      &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Pilgrim's Regress (CS Lewis)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;9   THWSPFCTRY         &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The wasp Factory (Iain Banks)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;10  WTHSHDDRNGFLL      &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; With Shuddering Fall (Joyce Oates)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A THLVSFTWNS&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lives of the Twins (Joyce Oates)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;B THCLMNGFSLPNGBTY&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; B The Claiming of Sleeping Beauty (Anne Rice)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;C THCCKSCLLNG&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C The Cuckoo's Calling (Robert Galbraith, aka JK Rowling)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;D SPRTSNBNDG&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Spirits in Bondage (CS Lewis)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;E RG&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rage (Stephen King)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;F LGCYFTRRR&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Legacy of Terror (Dean Koontz)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;G GNTSBRD&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Giant's Bread (Agatha Christie)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;H DVDSTRRSPCRNGR&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; David Starr Space Ranger (Isaac Asimov)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I DDSN&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Odds On (Michael Crichton)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;J CNSDRPHLBS&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Consider Phlebas  (Iain Banks)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Matching everything up, with each author having an entry on the left and right columns leaves...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Robert Galbraith/JKR Unmatched. But the left side has the debut novel of the author (disregarding pseudonyms), so the answer is &quot;Harry Potter and the Philosopher's Stone&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7316" LastEditorUserId="7316" LastEditDate="2015-11-24T14:58:40.477" LastActivityDate="2015-11-24T14:58:40.477" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24175" PostTypeId="2" ParentId="24171" CreationDate="2015-11-24T15:50:39.870" Score="3" Body="&lt;p&gt;I would guess the beatles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1964 they &lt;a href=&quot;http://www.elvis-history-blog.com/elvis-1964.html&quot; rel=&quot;nofollow&quot;&gt;&quot;dethroned&quot; Elvis&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1985 their publishing rights were &lt;a href=&quot;http://www.history.com/this-day-in-history/michael-jackson-takes-control-of-the-beatles-publishing-rights&quot; rel=&quot;nofollow&quot;&gt;bought up by Michael Jackson&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is far-fetched and i can't fit the age 38 into this.&lt;/p&gt;&#xA;" OwnerUserId="16608" LastActivityDate="2015-11-24T15:50:39.870" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24176" PostTypeId="1" AcceptedAnswerId="24192" CreationDate="2015-11-24T16:32:53.083" Score="4" ViewCount="1948" Body="&lt;p&gt;I am who I am,&lt;br&gt;&#xA;Returned to the world,&lt;br&gt;&#xA;To Save or to Damn,&lt;br&gt;&#xA;My banner's unfurled.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So many names I have,&lt;br&gt;&#xA;But one always true.&lt;br&gt;&#xA;No golden Calve,&lt;br&gt;&#xA;But worship you do.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My people, they came.&lt;br&gt;&#xA;Their weapon they bring.&lt;br&gt;&#xA;Few know their true name,&lt;br&gt;&#xA;And its familiar ring.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did sneak away,&lt;br&gt;&#xA;Now the battle is o'er.&lt;br&gt;&#xA;Power, all that I say.&lt;br&gt;&#xA;Who am I? Fear or adore?  &lt;/p&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="9951" LastEditDate="2015-11-24T20:27:10.153" LastActivityDate="2015-11-25T19:46:10.083" Title="Name me, Who am I?" Tags="|riddle|knowledge|rhyme|literature|" AnswerCount="6" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24177" PostTypeId="2" ParentId="24176" CreationDate="2015-11-24T17:05:18.617" Score="4" Body="&lt;p&gt;I am going to guess .....&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jesus &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. He will return (also referred to as &quot;the Great 'I am') &lt;br&gt;&#xA; 2. He has many names &lt;br&gt;&#xA; 3. He is not an idol, yet we worship Him &lt;br&gt;&#xA; 4. A sword was brought to Gethsemane &lt;br&gt;&#xA; 5. He did leave. &lt;br&gt;&#xA; 6. But He left in Power &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17508" LastEditorUserId="17508" LastEditDate="2015-11-24T19:21:25.053" LastActivityDate="2015-11-24T19:21:25.053" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24178" PostTypeId="2" ParentId="24163" CreationDate="2015-11-24T17:10:13.323" Score="1" Body="&lt;p&gt;The other answers are right,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All weigh $5$ pounds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Trial and error is by far the simplest approach. Assume values of the first rhino, and then calculate onwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1000,13000,error$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2000,11000,error$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3000,9000,error$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$4000,7000,1000,13000,error$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5000,5000,5000,\dots$ Solution found!!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$6000,3000,9000,error$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$7000,1000,13000,error$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$8000,error$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$9000,error$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so on....&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-11-24T17:10:13.323" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24181" PostTypeId="2" ParentId="24171" CreationDate="2015-11-24T18:06:52.343" Score="6" Body="&lt;p&gt;&lt;strong&gt;UPDATE:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Zuo Zongtang, or &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Zuo_Zongtang&quot; rel=&quot;nofollow&quot;&gt;General Tso&lt;/a&gt;&quot;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; The dates match (1812-1885), he was hired as an advisor to Zeng Guofan (governor) in 1850, dethroned Hong Tianguifu in 1864, &quot;founded China's first modern shipyard and naval academy in Fuzhou&quot; in 1866. He died shortly after a truce was signed in the Sino-French war.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2015-11-27T17:24:10.533" LastActivityDate="2015-11-27T17:24:10.533" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="24182" PostTypeId="2" ParentId="16640" CreationDate="2015-11-24T19:29:09.893" Score="7" Body="&lt;p&gt;I'm new here, I hope this is the right place to post, since it is more of a possible hint than an answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that the word GOLD is emphasized in the question caught my attention, and reminded me of &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/The_Gold-Bug&quot; rel=&quot;nofollow&quot;&gt;The Gold-Bug&lt;/a&gt;&quot;, a short story by Edgar Allan Poe, in which the protagonists decipher a secret message to find a treasure buried by a pirate (Coincidence? I think not).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cipher in Poe's novel is a simple substitution cipher that they crack using frequency analysis. Assuming we're faced with a simple substitution cipher here too (which would explain the &quot;It's probably an easy one to solve&quot; line from the author of the question), I tried to brute-force it a little but haven't got enough spare time on my hands to dig further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A last remark, that may or may not be of use: If the word &quot;GOLD&quot; is part of the clear text, it doesn't matter whether the clear is in English or in German, since it is the same word in the two languages.)&lt;/p&gt;&#xA;" OwnerUserId="17512" LastEditorUserId="16322" LastEditDate="2016-02-08T09:29:15.153" LastActivityDate="2016-02-08T09:29:15.153" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24184" PostTypeId="2" ParentId="16640" CreationDate="2015-11-24T20:42:09.543" Score="0" Body="&lt;p&gt;&lt;strong&gt;WIKI FOR ANALYSIS&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Character mix and frequency:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;CHAR    ASCII   FREQ&#xA;---------------------&#xA;        32      5&#xA;!       33      3&#xA;&quot;       34      3&#xA;$        36      3&#xA;'       39      4&#xA;(       40      4&#xA;)       41      3&#xA;+       43      5&#xA;,       44      5&#xA;-       45      4&#xA;.       46      4&#xA;/       47      4&#xA;0       48      3&#xA;1       49      3&#xA;2       50      1&#xA;3       51      3&#xA;4       52      3&#xA;5       53      5&#xA;6       54      1&#xA;7       55      4&#xA;8       56      2&#xA;9       57      3&#xA;?       63      3&#xA;@       64      6&#xA;\       92      4&#xA;a       97      2&#xA;b       98      2&#xA;c       99      2&#xA;d       100     5&#xA;e       101     5&#xA;f       102     3&#xA;h       104     3&#xA;j       106     1&#xA;l       108     3&#xA;m       109     4&#xA;n       110     2&#xA;o       111     6&#xA;p       112     2&#xA;q       113     3&#xA;r       114     4&#xA;s       115     2&#xA;t       116     1&#xA;u       117     3&#xA;v       118     3&#xA;w       119     2&#xA;x       120     3&#xA;y       121     5&#xA;z       122     5&#xA;&#xA;&lt;/pre&gt;&#xA;" OwnerUserId="9000" LastActivityDate="2015-11-24T20:42:09.543" CommentCount="0" CommunityOwnedDate="2015-11-24T20:42:09.543" ContentLicense="CC BY-SA 3.0" />
  <row Id="24186" PostTypeId="2" ParentId="16640" CreationDate="2015-11-24T20:54:42.417" Score="20" Body="&lt;p&gt;The German and WWII references indicate that the message went through...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;1. Enigma machine encryption&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; This &lt;a href=&quot;http://www.counton.org/explorer/codebreaking/enigma-cipher.php&quot;&gt;online tool&lt;/a&gt; is the most straightforward for quickly decoding this puzzle's message. Use G, O, L, D in the &quot;Position&quot; boxes, enter the message in the &quot;Encrypted Text&quot; field, and hit the &quot;Decrypt&quot; button to get&#xA;&lt;br&gt;&#xA; &lt;pre&gt;.-.. .- - .. - ..- -.. . ----- .-.-.- ..--- ...-- ...-- ....- ..... ----. .-.. --- -. --. .. - ..- -.. . -....- ....- ----. ..... -.... ---.. ----. ...-- ----.&lt;/pre&gt;&lt;br&gt;&#xA; I'll see if I can replicate the results using another tool.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then, the telegram reference and the look of the decoded message from the first step has the message's original encryption of...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;2. Morse code&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Translating the result of the first decryption gets&lt;br&gt;&#xA; &lt;pre&gt;LATITUDE0.233459LONGITUDE-49568939&lt;/pre&gt;&#xA;&lt;br&gt;&#xA;  I could've made a mistake in the decryption somewhere, but I think there's a decimal missing in the longitude? Searching the coordinates 0.233459,-49.568939 online gave me a location in or near Canal Perigoso, Brazil.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And now the story makes much more sense to me. It is not just for flavour.&lt;/p&gt;&#xA;" OwnerUserId="15921" LastActivityDate="2015-11-24T20:54:42.417" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24187" PostTypeId="2" ParentId="24176" CreationDate="2015-11-24T21:09:04.840" Score="4" Body="&lt;p&gt;I'm not sure if this is right, but maybe&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Paul_Atreides&quot; rel=&quot;nofollow&quot;&gt;Paul Atreides&lt;/a&gt; from Frank Herbert's Dune.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am who I am,&#xA;Returned to the world,&#xA;To Save or to Damn,&#xA;My banner's unfurled.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He came to Arrakis and became the leader of the Fremen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So many names I have,&#xA;But one always true.&#xA;No golden Calve,&#xA;But worship you do.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He is also called Muad'Dib and Usul. He is worshipped by the Fremen like a god.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My people, they came.&#xA;Their weapon they bring.&#xA;Few know their true name,&#xA;And its familiar ring.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The weapon of the Fremen might be either the sandworm or the &lt;a href=&quot;http://dune.wikia.com/wiki/Weirding_Module&quot; rel=&quot;nofollow&quot;&gt;Weirding Module&lt;/a&gt;. They use &lt;a href=&quot;http://dune.wikia.com/wiki/Water_Rings&quot; rel=&quot;nofollow&quot;&gt;Water Rings&lt;/a&gt; (had to look that up).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I did sneak away,&#xA;Now the battle is o'er.&#xA;Power, all that I say.&#xA;Who am I? Fear or adore?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He can use the &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Bene_Gesserit&quot; rel=&quot;nofollow&quot;&gt;Bene Gesserit power of Voice&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastActivityDate="2015-11-24T21:09:04.840" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24191" PostTypeId="2" ParentId="24176" CreationDate="2015-11-24T22:58:20.313" Score="3" Body="&lt;p&gt;I've a guess.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Is it Rand Al'thor, no one is sure if he will ruin the world or save it. Also he has many names but every body knows him as the dragon. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am who I am,&#xA;Returned to the world,&#xA;To Save or to Damn,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Because he is called as &quot;the prophesied savior and breaker of the world&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My banner's unfurled.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Because he reveals the Dragon Banner in fight against the Seanchan.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So many names I have,&#xA;But one always true.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because he has many names Dragon Reborn, Coramoor, Car'a'carn, He Who Comes With the Dawn, Shadowkiller, King of Illian, Lord of the Morning&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My people, they came.&#xA;Their weapon they bring.&#xA;Few know their true name,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because he will bring Aiel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17517" LastEditorUserId="17517" LastEditDate="2015-11-25T11:53:28.213" LastActivityDate="2015-11-25T11:53:28.213" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24192" PostTypeId="2" ParentId="24176" CreationDate="2015-11-24T23:21:05.060" Score="5" Body="&lt;p&gt;Here's my guess:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am who I am, Returned to the world,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rand al'Thor is the reincarnated Lews Therin, from the Wheel of Time series. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To Save or to Damn,&#xA;My banner's unfurled.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rand is ultimately responsible for the outcome of the Last Battle, win or lose. He fights under the Dragon banner for a large portion of the series.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So many names I have,&#xA;But one always true.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rand is called a lot of things, &quot;The Dragon Reborn&quot;, &quot;Lews Therin&quot;, &quot;Car'a'Carn&quot;, but his name is Rand al'Thor.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No golden Calve,&#xA;But worship you do.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Play on words. His golden dragon tattoos are on his forearms, not his calves. He's often worshipped by those who believe in the prophecies about him.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My people, they came.&#xA;Their weapon they bring.&#xA;Few know their true name,&#xA;And its familiar ring.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is the stanza I'm not sure about. It could mean several things. Asha'man bring Saidin as a weapon, but they don't know they were once all called Aes'Sedai. His friends come and fight with him, and on some level know about the whole Wheel reincarnation thing, but few are aware of what their names used to be in past lives.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I did sneak away,&#xA;Now the battle is o'er.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;After the last battle, he silently left his tent without his army noticing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Power, all that I say.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rand gained some sort of unspecified superpower after the last battle. He cannot channel anymore, but can just will things to happen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who am I? Fear or adore?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He switched bodies with Ishamael during the last battle, so if anyone recognized him it'd be as a villain. Those bonded to him can sense that he's still Rand.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17519" LastEditorUserId="9951" LastEditDate="2015-11-25T19:46:10.083" LastActivityDate="2015-11-25T19:46:10.083" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24193" PostTypeId="2" ParentId="24176" CreationDate="2015-11-24T23:37:37.227" Score="1" Body="&lt;p&gt;My attempt:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Death&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am who I am, Returned to the world, To Save or to Damn, My banner's unfurled.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He comes and goes. Some welcome him, many don't. Many organizations beckon him with their actions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So many names I have, But one always true. No golden Calve, But worship you do.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He has many names. Feared to the point of &quot;worship,&quot; or feared similar to the way a god might be feared.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My people, they came. Their weapon they bring. Few know their true name, And its familiar ring.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; His people are soldiers or others. Literal weapons. Sound of war.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I did sneak away,&#xA;Now the battle is o'er.&#xA;Power, all that I say.&#xA;Who am I? Fear or adore?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He comes and goes. Literal battle. He has power over life. Some adore him, many fear him.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1646" LastActivityDate="2015-11-24T23:37:37.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24194" PostTypeId="2" ParentId="24176" CreationDate="2015-11-24T23:44:06.047" Score="2" Body="&lt;p&gt;My guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Any politician&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am who I am&lt;br&gt;&#xA; Returned to the world, = every few years in America&lt;br&gt;&#xA; To Save or to Damn, = depending on your party affiliation&lt;br&gt;&#xA; My banner's unfurled. = yard signs, billboards, tv ads, etc.&lt;br&gt;&#xA;&lt;br&gt;&#xA; So many names I have, = name-calling is common&lt;br&gt;&#xA; But one always true. = their real name&lt;br&gt;&#xA; No golden Calve,&lt;br&gt;&#xA; But worship you do. = news is obsessed/we're obliged as citizens&lt;br&gt;&#xA;&lt;br&gt;&#xA; My people, they came. = campaign volunteers are their people&lt;br&gt;&#xA; Their weapon they bring. = their weapon is phonebanking&lt;br&gt;&#xA; Few know their true name, = phone bankers are often anonymous&lt;br&gt;&#xA; And its familiar ring. = but the ring of your cellphone is familiar&lt;br&gt;&#xA;&lt;br&gt;&#xA; I did sneak away, = politicians often shirk responsibility after the election&lt;br&gt;&#xA; Now the battle is o'er.&lt;br&gt;&#xA; Power, all that I say. = a politician's word is powerful/bully pulpit&lt;br&gt;&#xA; Who am I? Fear or adore?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17520" LastActivityDate="2015-11-24T23:44:06.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24195" PostTypeId="2" ParentId="23944" CreationDate="2015-11-25T00:14:42.913" Score="-1" Body="&lt;p&gt;I am not clever, but I am pedantic and determined. I'm also probably wrong. Using the link provided by f'', I tried squaring everything in the base square:&#xA;$$&#xA;\begin{array}{ccc}&#xA;(a)^2 &amp;amp; (a+c)^2 &amp;amp; (a+2c)^2 \\&#xA;(a+b)^2 &amp;amp; (a+b+c)^2 &amp;amp; (a+b+2c)^2 \\&#xA;(a+2b)^{2} &amp;amp; (a+2b+c)^2 &amp;amp; (a+2b+2c)^2 \\&#xA;\end{array} &#xA;$$&#xA;Then I simplified the squares&#xA;$$ &#xA;\begin{array}{ccc}&#xA;a^2 &amp;amp; a^2+c^2+2ac &amp;amp; a^2+4c^2+4ac \\&#xA;a^2+b^2+2ab &amp;amp; a^2+b^2+c^2+2ab+2ac+2bc &amp;amp; a^2+b^2+4c^2+2ab+4ac+4bc \\&#xA;a^2+4b^2+4ab &amp;amp; a^2+4b^2+c^2+4ab+2ac+4bc &amp;amp; a^2+4b^2+4c^2+4ab+4ac+8bc&#xA;\end{array}&#xA;$$&#xA;In the link, the magic number was $3a+3b+3c$, three times the middle term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying my middle term by 3 gives $3a^2+3b^2+3c^2+6ab+6ac+6bc$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My last squared number includes $8bc$, so there is no way to add any three terms to get the magic number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is nowhere near a proof, though.&lt;/p&gt;&#xA;" OwnerUserId="17521" LastActivityDate="2015-11-25T00:14:42.913" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24196" PostTypeId="5" CreationDate="2015-11-25T02:43:59.797" Score="0" Body="&lt;p&gt;In a 'blackboard puzzle' a set of numbers is transformed according to certain rules. The current set of numbers is described as numbers being &lt;em&gt;written on a blackboard&lt;/em&gt;. A modification of the number set is described by erasing numbers and/or writing new numbers onto the blackboard. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typical goals in typical blackboard puzzles are &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;to get the board into a certain state, e.g. &quot;all numbers below 20&quot; or &quot;no even numbers&quot;&lt;/li&gt;&#xA;&lt;li&gt;to maximize or to minimize the last remaining number.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="8874" LastEditDate="2015-11-25T09:55:40.957" LastActivityDate="2015-11-25T09:55:40.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24197" PostTypeId="4" CreationDate="2015-11-25T02:43:59.797" Score="0" Body="In a 'blackboard puzzle' a set of numbers is transformed according to certain rules. " OwnerUserId="11876" LastEditorUserId="8874" LastEditDate="2015-11-25T09:55:40.957" LastActivityDate="2015-11-25T09:55:40.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24200" PostTypeId="1" CreationDate="2015-11-25T11:12:40.570" Score="1" ViewCount="176" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;Hi all.&lt;/em&gt;&lt;/strong&gt; &lt;br&gt;&#xA;This image is a series of number in sequence created by computer app.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AK1jH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AK1jH.png&quot; alt=&quot;Numbers in Sequence&quot;&gt;&lt;/a&gt; &lt;br&gt;&#xA;You can see there are five numbers in above row and five numbers in row below. &lt;br&gt;&#xA;&lt;br&gt;&lt;strong&gt;&lt;em&gt;1) Pls find what will be the numbers for A-B-C-D-E and F-G&lt;/em&gt;&lt;/strong&gt; &lt;br&gt;&#xA;&lt;br&gt; &lt;strong&gt;&lt;em&gt;2) What is the number to put inside the green box?&lt;/em&gt;&lt;/strong&gt; &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Facts of the image&lt;/em&gt;&lt;/strong&gt; &lt;br&gt;&#xA;The sequence comes from horizontal yellow line and vertical yellow line is the same number (which will appeared in the green box). &lt;br&gt;&#xA;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1: &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I made this by famous computer application. Not by myself :P&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;Hint 2: &lt;strong&gt;Dare to look some answers!?&lt;/strong&gt; &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C is 12. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt; Hint 3: &lt;strong&gt;Dare to look some answers!?&lt;/strong&gt; &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; E is 48.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt; Hint 4: &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The number in green box &quot;?&quot; is comes from F-G sequence, A-B-C-D-E sequence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;" OwnerUserId="11353" LastActivityDate="2015-11-25T11:54:25.913" Title="Find the numbers in sequence (Created by PC App)" Tags="|mathematics|calculation-puzzle|computer-puzzle|" AnswerCount="1" CommentCount="2" ClosedDate="2015-11-25T17:58:12.093" ContentLicense="CC BY-SA 3.0" />
  <row Id="24201" PostTypeId="1" AcceptedAnswerId="24203" CreationDate="2015-11-25T11:24:58.423" Score="4" ViewCount="1324" Body="&lt;p&gt;&lt;em&gt;Dear,&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You like me, and I'm just like you &lt;br&gt; &#xA;We can't touch, but we can talk to &lt;br&gt; &#xA;We can't hear, but we can see through &lt;br&gt; &#xA;I'm always there and you come more than once &lt;br&gt; &#xA;'Cause once is not enough for our love &lt;br&gt; &#xA;Time is passing, but tonight again &lt;br&gt; &#xA;When you laugh at me, I laugh at you, my charm &lt;br&gt;&#xA;    &lt;em&gt;Pls don't afraid to see me, when the dark has come.&lt;/em&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Beloved, :P&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="11353" LastActivityDate="2015-11-28T04:16:03.547" Title="Who/What am I? A Riddle to my dear" Tags="|riddle|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24202" PostTypeId="2" ParentId="24200" CreationDate="2015-11-25T11:54:25.913" Score="5" Body="&lt;p&gt;As-is, there are infinitely many &quot;logics&quot; that lead to a &quot;solution&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The top row are &lt;a href=&quot;https://oeis.org/A239993&quot;&gt;even numbers $n$ so that $6 \nmid \frac n2 \cdot \sigma \left( \frac n2 \right)$&lt;/a&gt;. The next one is $F=32$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The middle row are &lt;a href=&quot;https://oeis.org/A135372&quot;&gt;highly composite numbers $+1$ starting at 5&lt;/a&gt;. The next one is $G=49$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The row $A,B,C,D,E,?$ is calculated as $2 \cdot \text{middle row} - \text{top row}$. The answers are $A=8$, $B=6$, $C=12$, $D=32$, $E=48$, $?=66$.&lt;/p&gt;&#xA;" OwnerUserId="11476" LastActivityDate="2015-11-25T11:54:25.913" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24203" PostTypeId="2" ParentId="24201" CreationDate="2015-11-25T12:36:05.017" Score="14" Body="&lt;p&gt;Is that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Our Mirror Image?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You like me, and I'm just like you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We like to see in mirror, and mirror image of us is same as like us.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can't touch, but we can talk to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Actually we can not touch mirror image.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can't hear, but we can see through&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We both can not hear each other but we can see.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm always there and you come more than once&#xA;'Cause once is not enough for our love&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Frequently we go in front of it, Once is not enough&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Time is passing, but tonight again&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; not sure about this line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you laugh at me, I laugh at you, my charm&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If we laugh our mirror image also laugh&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pls don't afraid to see me, when the dark has come.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's horrible to face yourself into mirror at dark.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4312" LastEditorUserId="4312" LastEditDate="2015-11-28T04:16:03.547" LastActivityDate="2015-11-28T04:16:03.547" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24205" PostTypeId="1" CreationDate="2015-11-25T13:05:34.603" Score="6" ViewCount="1492" Body="&lt;p&gt;At a party, everybody is friend with exactly $22$ of the other persons present.&#xA;Whenever two persons are friends, they do not have any friends in common.&#xA;Whenever two persons are not friends, they have exactly six friends (present at the party) in common.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; How many persons are at the party?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&#xA;(Note: Friendships are always mutual. Nobody is friend with himself.)&#xA;&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2015-11-25T13:12:57.510" LastActivityDate="2015-11-26T04:03:21.177" Title="A party puzzler" Tags="|mathematics|combinatorics|graph-theory|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24206" PostTypeId="2" ParentId="24201" CreationDate="2015-11-25T14:03:36.640" Score="2" Body="&lt;p&gt;The only thing I can think of like this that shows itself at night is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My shadow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's more:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Our image is a likeness of ourselves &lt;br&gt;&#xA; We can't hear each other, but I can see my shadow &lt;br&gt;&#xA; We arrive at our shadow when it is dark and a light brings it to life &lt;br&gt;&#xA; At night is when we see our shadow most. In fact, it has been said that one can be afraid of one's own shadow. &lt;br&gt;&#xA; The title indicates that a what is possible and if a what is possible, then that tells me the need to say so makes it likely, so I am thinking it must be a thing, not a person, which shadow fits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17508" LastActivityDate="2015-11-25T14:03:36.640" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24207" PostTypeId="2" ParentId="24205" CreationDate="2015-11-25T14:03:58.857" Score="10" Body="&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; I read up a bit more and I am pretty sure of my answer now.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Is the answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $100$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I searched for &quot;22 graph&quot; on Google and a link came up called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Higman%E2%80%93Sims_graph&quot;&gt;Higman-Sims graph&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not know anything about &lt;a href=&quot;/questions/tagged/graph-theory&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;graph-theory&amp;#39;&quot; rel=&quot;tag&quot;&gt;graph-theory&lt;/a&gt;, but the first line on the page looked very similar to your question.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In mathematical graph theory, the Higman–Sims graph is a &lt;strong&gt;22&lt;/strong&gt;-regular undirected graph with 100 vertices and 1100 edges. It is the unique strongly regular graph with 100 vertices and valency 22, where &lt;strong&gt;no neighboring pair of vertices share a common neighbor&lt;/strong&gt; and &lt;strong&gt;each non-neighboring pair of vertices share six common neighbors&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4269" LastEditDate="2015-11-25T14:35:19.583" LastActivityDate="2015-11-25T14:35:19.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24209" PostTypeId="1" AcceptedAnswerId="24211" CreationDate="2015-11-25T14:32:20.687" Score="14" ViewCount="1039" Body="&lt;p&gt;&lt;sub&gt;(Based on &lt;a href=&quot;https://puzzling.stackexchange.com/questions/24205/a-party-puzzler&quot;&gt;this puzzle&lt;/a&gt; by Gamow. The answer there might give you a hint (so don't look yet!))&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a fancy party in &lt;em&gt;Café la Tour&lt;/em&gt;, everybody is friends with exactly 14 of the other people present.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Whenever two people at the party are friends, they have exactly 6 friends (present at the party) in common.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Whenever two people at the party are &lt;strong&gt;not&lt;/strong&gt; friends, they have exactly 2 friends (present at the party) in common.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(Oh, and half of the people at the party came dressed in white, and the other half was dressed in black. How stylish!)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; How many persons are at the party?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&#xA;(Note: Friendships are always mutual. Nobody is their own friend. There is at least one person at the party.)&#xA;&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="11476" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2015-11-25T15:42:44.190" Title="A chic party puzzler" Tags="|mathematics|combinatorics|chess|graph-theory|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24211" PostTypeId="2" ParentId="24209" CreationDate="2015-11-25T14:44:32.540" Score="11" Body="&lt;p&gt;The chess tag gave it away.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are exactly 64 people at the party.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To show this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Imagine every person standing alone on a field on a big chessboard.&#xA; 64 Fields and therefore 64 people.&lt;br&gt;&#xA; Now every person is friends with those 14 other people which stand in the same row or column. (7 in the same row + 7 in the same column)&lt;br&gt;&#xA;&lt;br&gt;&#xA; With this every two people which are friends and therefore are within the same row/column will have the other 6 people in the same row/column as common friends.&lt;br&gt;&#xA;&lt;br&gt;&#xA; In a similar way every two people which are not friends and therefore neither within the same row nor the same column will have those two people as common friends that are standing on the intersections of the respective rows and column.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16608" LastActivityDate="2015-11-25T14:44:32.540" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24212" PostTypeId="2" ParentId="24205" CreationDate="2015-11-25T16:25:59.597" Score="9" Body="&lt;p&gt;As pointed out by ghost_in_the_code&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 100&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is a solution, however he doesn't state why it's the only solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show it is the only solution we have to realize that what is asked is how many vertices are there in $srg(v,22,0,6)$ where $srg(v, k, λ, μ)$ is defined as the strongly regular graph with:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$v$ vertices  &lt;/li&gt;&#xA;&lt;li&gt;degree $k$  &lt;/li&gt;&#xA;&lt;li&gt;Every two adjacent vertices have λ common neighbours.  &lt;/li&gt;&#xA;&lt;li&gt;Every two non-adjacent vertices have μ common neighbours.  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As explained on this &lt;a href=&quot;https://en.wikipedia.org/wiki/Strongly_regular_graph&quot;&gt;wiki-page&lt;/a&gt; the following relation is always true:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$(v-k-1)\mu = k(k-\lambda-1)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Filling in our numbers we get&lt;br&gt;&#xA;$(v-23)6 = 22(21)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives the unique solution&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $v=100$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-11-25T16:25:59.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24214" PostTypeId="1" CreationDate="2015-11-25T16:37:44.527" Score="4" ViewCount="493" Body="&lt;p&gt;&lt;sub&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/24081/magic-the-gathering-challenge-5-what-if-pod-was-one-of-us&quot;&gt;Previous Challenge&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Background&lt;/h2&gt;  &#xA;&#xA;&lt;p&gt;Today's puzzle is going to involve Mental Magic, a fun, memory-testing variant of magic. All of magic's usual rules apply, but with the following variations&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Any card in your hand may be played face down as a land that taps for any color of mana. This land is treated as basic with no subtypes. (Usual land-per-turn rule applies)  &lt;/li&gt;&#xA;&lt;li&gt;Any card may be played as any other nonland MtG card as long as   &#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;that card does not have the same name  &lt;/li&gt;&#xA;&lt;li&gt;that card has the exact same mana cost (note: this includes color symbols, hybrid mana, X-cost, etc.) so e.g. Wild Guess can be cast as Eidolon of the Great Revel, but not as Rolling Thunder and Thoughtseize can be cast as Gravecrawler, but not as Deathrite Shaman.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;The same spell may not be cast more than once in a game  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Some additional house rules for simplicity:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Search effects are banned. This includes cards like Entomb and also cards where only part of the effect involves searching e.g. Sever the Bloodline  &lt;/li&gt;&#xA;&lt;li&gt;Cards in hand/library have no name, only their printed (sub)types and costs so no exiling an Elvish Spirit Guide or cycling Street Wraith  &lt;/li&gt;&#xA;&lt;li&gt;We'll be playing a Legacy game of Mental Magic, meaning that the Legacy banned list and rules should be observed.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Puzzle Setup&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;It's turn 1 and you're on the play. Let's assume that (magically) your opponent has no cards in hand. Win this turn with the rules given above and the hand and library below. Both boards and graveyards are empty. If your require interaction with your opponent's library (remember that searching is banned) you may assume the card type(s) required are there. Your solution must be a guaranteed win, not involving your opponent self-destructing (e.g. they choose 20 on a Choice of Damanations) or random chance (you Hymn to Tourach yourself and hit the right 2 cards). If multiple solutions are achieved, the one that does the most damage wins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your hand:&lt;/strong&gt;&lt;br&gt;&#xA;Cabal Ritual x 2 (&lt;strong&gt;1B&lt;/strong&gt;)&lt;br&gt;&#xA;Candelabra of Tawnos (&lt;strong&gt;1&lt;/strong&gt;)&lt;br&gt;&#xA;Doomsday (&lt;strong&gt;BBB&lt;/strong&gt;)&lt;br&gt;&#xA;Reanimate x 3 (&lt;strong&gt;B&lt;/strong&gt;) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your library: (top to bottom)&lt;/strong&gt;&lt;br&gt;&#xA;Reanimate (&lt;strong&gt;B&lt;/strong&gt;)&lt;br&gt;&#xA;Tarmogoyf (&lt;strong&gt;1G&lt;/strong&gt;)&lt;br&gt;&#xA;Council's Judgment (&lt;strong&gt;1WW&lt;/strong&gt;)&lt;br&gt;&#xA;Lightning Bolt (&lt;strong&gt;R&lt;/strong&gt;)&lt;br&gt;&#xA;Brainstorm (&lt;strong&gt;U&lt;/strong&gt;)&lt;br&gt;&#xA;Hymn to Tourach (&lt;strong&gt;BB&lt;/strong&gt;)&lt;br&gt;&#xA;Glimpse of Nature (&lt;strong&gt;G&lt;/strong&gt;)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: There's been a lot of confusion surrounding the not-very-well-defined parameters I provided. In an attempt to clarify this, I've posted what I believe to be a more clear, more well-defined version &lt;a href=&quot;https://puzzling.stackexchange.com/questions/24240/magic-the-gathering-challenge-6-b&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2023-11-06T22:14:33.527" Title="Magic: the Gathering - Challenge #6: Going Mental" Tags="|calculation-puzzle|magic-the-gathering|" AnswerCount="4" CommentCount="21" ContentLicense="CC BY-SA 3.0" />
  <row Id="24215" PostTypeId="1" AcceptedAnswerId="24225" CreationDate="2015-11-25T16:38:28.030" Score="8" ViewCount="627" Body="&lt;p&gt;I'm posting this from my phone again because I've gotten hopelessly lost in Boston, really want to go home, and desperately need the help of some clever people on the internet. Please help me - I'm really lost and I would like to, after some finite amount of driving, get back home and I don't know any better way to do that than to ask about my problem on a puzzling site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not very good at remembering things, so any approach based on recalling whether I've been at a location before isn't going to work. However, I'm really good at following directions, so what I'd really like is to be given a list of numbers. Each time I reach an intersection, I'll read the next number, count that many turns from the right, and follow that road - so at a standard* 4 way intersection, a $1$ will tell me to turn right, a $2$ to go straight, a $3$ to turn left and a $4$ to make a U-turn (and a $5$ brings me back to turning right and so on). I define an intersection as any moment at which I have an opportunity to turn (so if a street branches off from the one I'm on, that's a 3-way intersection).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my extensive experience as a driver, I have observed the following: there are only finitely many intersections and that, at each, only finitely many roads meet. Also, I'm quite sure that it is possible to get back to my house from any point in the road network (Boston's not &lt;em&gt;that&lt;/em&gt; bad**). Finally - and I'm quite proud of this - if I drive by my house (which is located by a road), I will &lt;em&gt;definitely&lt;/em&gt; recognize it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can any of you suggest a sequence&lt;/strong&gt;$^{***}$&lt;strong&gt;of instructions that will get me home?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(*Of course, Boston's road network is so messed up that there are approximately zero standard intersections. This doesn't affect the answer though.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(**For the purposes of this question.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(***Given that I do not know how many roads there are in this city, this sequence is necessarily infinite. Please do not try to write an entire infinite sequence in your answer - a mathematical construction is expected instead).&lt;/p&gt;&#xA;" OwnerUserId="4765" LastEditorUserId="22922" LastEditDate="2016-09-15T07:24:36.563" LastActivityDate="2016-09-15T07:24:36.563" Title="Lost In Boston: How do I Get Home?" Tags="|mathematics|strategy|mazes|graph-theory|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="24216" PostTypeId="2" ParentId="24214" CreationDate="2015-11-25T17:19:44.750" Score="-1" Body="&lt;p&gt;Play Candelabra of Tawnos facedown.&lt;br&gt;&#xA;Tap [facedown] for B. (pool: B)&lt;br&gt;&#xA;Play Reanimate as &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=221510&quot; rel=&quot;nofollow&quot;&gt;Dark Ritual&lt;/a&gt;. (pool: BBB)&lt;br&gt;&#xA;Play Reanimate as &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=221510&quot; rel=&quot;nofollow&quot;&gt;Dark Ritual&lt;/a&gt;. (pool: BBBBB)&lt;br&gt;&#xA;Play Doomsday as &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=243250&quot; rel=&quot;nofollow&quot;&gt;Geralf's Messenger&lt;/a&gt;. (pool: BB)&lt;br&gt;&#xA;Play Reanimate as &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=386635&quot; rel=&quot;nofollow&quot;&gt;Retribution of the Ancients&lt;/a&gt;. (pool: B)&lt;br&gt;&#xA;....win&lt;br&gt;&#xA;Play Cabal Ritual as &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=198308&quot; rel=&quot;nofollow&quot;&gt;Bloodthrone Vampire&lt;/a&gt;. (pool: -1)&lt;/p&gt;&#xA;" OwnerUserId="17536" LastEditorUserId="10399" LastEditDate="2015-11-25T17:51:20.527" LastActivityDate="2015-11-25T17:51:20.527" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24217" PostTypeId="2" ParentId="24214" CreationDate="2015-11-25T17:44:35.017" Score="0" Body="&lt;p&gt;On a phone, so I'll fix formatting later.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1B - land  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tap for B  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;B - Dark Ritual  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;B - Blood Celebrant  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pay B, 1 Life: R  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;B - Culling the Weak  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;BBB - Cruel Bargain, draw 4, lose 10 life  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;R - Mana Clash  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Opponent flips 20 tails before you flip 9, and you never both flip heads - you win!&lt;/p&gt;&#xA;" OwnerUserId="17389" LastActivityDate="2015-11-25T17:44:35.017" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24218" PostTypeId="1" CreationDate="2015-11-25T17:47:09.473" Score="0" ViewCount="336" Body="&lt;p&gt;I don't get it. A friend asked ne this but I don't know what to say. I tried to see if it had to deal with integers or something, but I got nothing.&lt;/p&gt;&#xA;" OwnerUserId="17537" LastActivityDate="2015-11-25T18:48:13.180" Title="Why does 11+1=3?" Tags="|calculation-puzzle|" AnswerCount="2" CommentCount="8" ClosedDate="2015-11-26T00:26:14.613" ContentLicense="CC BY-SA 3.0" />
  <row Id="24219" PostTypeId="2" ParentId="23358" CreationDate="2015-11-25T17:48:24.307" Score="1" Body="&lt;p&gt;I'm not sure but is it...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; your own question in Meta Stack Exchange:&lt;br&gt;&#xA; &lt;a href=&quot;https://meta.stackexchange.com/questions/251661/what-happened-to-the-other-people-my-impact-people-reached-just-dropped&quot;&gt;What happened to the other people? My Impact (people reached) just dropped&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and attracted so many guys&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The question is about reached people&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I look like an election just happened,&lt;br&gt;&#xA;these people so happy for them votes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the question the reason for lost people is that the other answers had got more votes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17138" LastEditorUserId="-1" LastEditDate="2017-03-20T10:32:12.560" LastActivityDate="2015-11-25T17:48:24.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24220" PostTypeId="2" ParentId="24218" CreationDate="2015-11-25T17:57:33.927" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 11 people, plus one more, equals 3 barbershop quartets! &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16874" LastActivityDate="2015-11-25T17:57:33.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24221" PostTypeId="2" ParentId="23944" CreationDate="2015-11-25T18:02:56.490" Score="8" Body="&lt;p&gt;&lt;strong&gt;Copied entirely from &lt;a href=&quot;http://mathpages.com/home/kmath417.htm&quot; rel=&quot;noreferrer&quot;&gt;http://mathpages.com/home/kmath417.htm&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This is not a solution; it is just a MathJAXed version of the link provided.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magic Square of Squares&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's an open question whether there exists a 3x3 magic square comprised &#xA;entirely of square integers.  Before considering the possibility of &#xA;such a square, it's worthwhile to review some basic facts about &#xA;arbitrary 3x3 magic squares, defined as an array of numbers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}\\&#xA;&amp;amp;A&amp;amp;B&amp;amp;C\\&#xA;&amp;amp;D&amp;amp;E&amp;amp;F\\&#xA;&amp;amp;G&amp;amp;H&amp;amp;I\\&#xA;\end{array}&#xA;$$&#xA;such that each row, column, and main diagonal has the same sum.&lt;br&gt;&#xA;Letting $S$ denote the common sum, we have the eight defining &#xA;conditions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A+B+C=S$$    $$D+E+F=S$$    $$G+H+I=S$$    $$A+E+I=S$$&#xA;$$A+D+G=S$$    $$B+E+H=S$$    $$C+F+I=S$$    $$C+E+G=S$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding up the four conditions involving $E$, and re-arranging terms, &#xA;gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(A+B+C)  +  (D+E+F)  +  (G+H+I)  +   3E   =   4S$$    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since each quantity in parentheses equals $S$, this shows that $3E = S$.&#xA;Substituting for $S$ into the four basic equations involving $E$, we&#xA;have&#xA;$$A+I = B+H = C+G = D+F  =  2E$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we subtract $2E$ from both sides of $A+I=2E$ (for example) we get&#xA;$(A-E) + (I-E) = 0$, which shows that $A-E = E-I$, and likewise for the&#xA;other cases, proving that each row, column, and diagonal containing&#xA;$E$ is an arithmetic progression.  If we define $n = A-E$ and $m = C-E$, &#xA;the magic square contains the terms&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}\\&#xA;&amp;amp;E+n&amp;amp;&amp;amp;E+m\\&#xA;&amp;amp;&amp;amp;E&amp;amp;\\&#xA;&amp;amp;E-m&amp;amp;&amp;amp;E-n\\&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this the remaining terms follow, due to the requirement that &#xA;each row and column sum to $3E$, so the entire magic square can be &#xA;expressed in terms of the three numbers $E$,$m$,$n$ as shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{ccc}\\&#xA;&amp;amp;E+n&amp;amp;E-n-m&amp;amp;E+m\\&#xA;&amp;amp;E-n+m&amp;amp;E&amp;amp;E+n-m\\&#xA;&amp;amp;E-m&amp;amp;E+n+m&amp;amp;E-n\\&#xA;\end{array}&#xA;$$&#xA; &lt;strong&gt;(1)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose we imagine a magic square comprised entirely of square &#xA;integers.  This implies the existence of integers $E$,$m$,$n$ such that &#xA;the magic square given by (1) has the values&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}\\&#xA;&amp;amp;a^2&amp;amp;b^2&amp;amp;c^2\\&#xA;&amp;amp;d^2&amp;amp;e^2&amp;amp;f^2\\&#xA;&amp;amp;g^2&amp;amp;h^2&amp;amp;i^2\\&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equating these to their equivalents in (1) gives the relations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{ccc}\\&#xA;&amp;amp;a^2 = E+n   &amp;amp;b^2 = E-n-m &amp;amp;c^2 = E+m\\&#xA;&amp;amp;d^2 = E-n+m &amp;amp;e^2 = E     &amp;amp;f^2 = E+n-m\\&#xA;&amp;amp;g^2 = E-m   &amp;amp;h^2 = E+n+m &amp;amp;i^2 = E-n\\&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(2)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summing the pairs symmetrical about E, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a^2 + i^2 = b^2 + h^2 = c^2 + g^2 = f^2 + d^2  =  2e^2$$       &lt;strong&gt;(3)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, if we multiply the pairs symmetrical about $E$ we find&#xA;that $e^4$ is expressible as a sum of two squares in the following four&#xA;ways&#xA;$$(ai)^2  +  n^2  =  e^4$$&#xA;$$(bh)^2 + (n+m)^2 = e^4$$&#xA;$$(cg)^2  +  m^2  =  e^4$$&#xA;$$(df)^2 + (n-m)^2 = e^4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These equations, along with (3), show that both $e^4$ and $2e^2$ are &#xA;expressible as sums of two squares in four different ways.  In addition,&#xA;if we square both sides of the previous relation $a^2 + i^2 = 2e^2$ (for &#xA;example) we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a^4 + 2(ai)^2 + i^4 = 4e^4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so we can solve this for $(ai)^2$ and substitute into the above &#xA;expression for $e^4$, and likewise for the other three expressions,&#xA;to give&#xA;$$a^4  +  i^4   =   2[e^4  +  n^2  ]$$&#xA;$$b^4  +  h^4   =   2[e^4 + (n+m)^2]$$&#xA;$$c^4  +  g^4   =   2[e^4  +  m^2  ]$$&#xA;$$d^4  +  f^4   =   2[e^4 + (n-m)^2]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a very strong set of conditions.  It requires a fourth power&#xA;$(e^4)$ which, if increased by any one of four distinct squares, equals&#xA;half a sum of two fourth powers.  Furthermore, the roots of the four&#xA;distinct squares must be of the form $n$, $m$, $n+m$, and $n-m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also infer more Pythagorean relations by multiplying together&#xA;other pairs of relations from (2).  For example, if we multiply $b^2$&#xA;by $f^2$ and re-arrange the terms, and likewise for the other pairs,&#xA;we get&#xA;$$(bf)^2  +  n^2  =  (E-m)^2$$&#xA;$$(fh)^2  +  m^2  =  (E+n)^2$$&#xA;$$(hd)^2  +  n^2  =  (E+m)^2$$&#xA;$$(db)^2  +  m^2  =  (E-n)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Squaring both sides of the relation $b^2 + f^2 = 2(E-m)$ and solving&#xA;for $(bf)^2$, we can substitute into the first of these equations, and&#xA;likewise for the other pairs, to give&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b^4 + f^4 = 2[(e^2 - m)^2 + n^2]$$&#xA;$$f^4 + h^4 = 2[(e^2 + n)^2 + m^2]$$&#xA;$$h^4 + d^4 = 2[(e^2 + m)^2 + n^2]$$&#xA;$$d^4 + b^4 = 2[(e^2 - n)^2 + m^2]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that we previously derived the conditions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b^4 + h^4 = 2[(e^2)^2 + (n+m)^2]$$&#xA;$$d^4 + f^4 = 2[(e^2)^2 + (n-m)^2]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so we have found that a necessary condition for a magic square of &#xA;squares is that there must be four 4th powers $b^4$, $d^4$, $f^4$, $h^4$ &#xA;whose sums in pairs each equal twice a sum of squares.  This also&#xA;leads to relations such as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f^4 - d^4 = 2(n-m)e^2$$&#xA;$$h^4 - b^4 = 2(n+m)e^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, all these relations just follow algebraically from the more&#xA;basic conditions, so to prove that no magic square of squares is&#xA;possible we should probably focus on the basic conditions.  Recall&#xA;that the relations (2) imply&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a^2 + i^2 = 2e^2$$       $$b^2 + f^2 = 2g^2$$&#xA;$$b^2 + h^2 = 2e^2$$       $$f^2 + h^2 = 2a^2$$&#xA;$$c^2 + g^2 = 2e^2$$       $$h^2 + d^2 = 2c^2$$&#xA;$$d^2 + f^2 = 2e^2$$       $$d^2 + b^2 = 2i^2$$       &lt;strong&gt;(4)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The four equations on the left show that $2e^2$ must be a sum of two &#xA;squares in four different ways, so $e$ itself is a sum of squares in&#xA;(at least) two different ways.  The first non-trivial occurrence is&#xA;$e = 65 = 5*13$, because this is the smallest number that is the &#xA;product of distinct primes congruent to $1 (mod 4)$.  Each of the &#xA;factors, in turn, is necessarily a sum of two squares, i.e.,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5 = 1^2 + 2^2$  and $13 = 2^2 + 3^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two expressions for $65$ arise from the two ways of multiplying&#xA;these according to the ancient formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(X^2 + Y^2)(A^2 + B^2) = (XB +- YA)^2 + (XA -+ YB)^2$$        &lt;strong&gt;(5)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking $X=1, Y=2, A=2, B=3$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e  =  65  =  (5)(13)   =   1^2 + 8^2   =   4^2 + 7^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The four ways of expressing $e^2$ as a sum of two squares arise from &#xA;the four ways of multiplying these two forms using equation (5) (and&#xA;applying the factor $2 = 1^2 + 1^2$, which doesn't change the number &#xA;of expressions).  Thus we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2(65)^2  =  2(1^2 + 8^2)(1^2 + 8^2)  =  47^2 + 79^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=  2(1^2 + 8^2)(4^2 + 7^2)  =  23^2 + 89^2  =  35^2 + 85^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=  2(4^2 + 7^2)(4^2 + 7^2)  =  13^2 + 91^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can assemble these squares into a square of squares such as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}&#xA;13^2 &amp;amp; 23^2 &amp;amp; 47^2\\&#xA;35^2 &amp;amp; 65^2 &amp;amp; 85^2\\&#xA;79^2 &amp;amp; 89^2 &amp;amp; 91^2\\&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which has the required equal sums on all four of the lines through&#xA;the center square (the two main diagonals, the center row, and the &#xA;center column).  Notice, however, that the outer rows and columns do&#xA;not give the required sum.  In general, we can show that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;PROPOSITION 1:  Any square whose elements satisfy the central sums &#xA;                  and whose central number is expressible as a sum of&#xA;                  two squares in no more than four distinct ways will&#xA;                  NOT give the required sums for the outer rows and&#xA;                  columns.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This proposition doesn't completely rule out the existence of a magic&#xA;square of squares, because it doesn't cover possible squares whose &#xA;central numbers are expressible as a sum of two squares in MORE than&#xA;four distinct ways.  However, it does show that the root of the&#xA;central number of any magic square of squares would have to be the &#xA;product of more than two distinct primes of the form $4k+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove this proposition, note that the above conditions can be &#xA;expressed parametrically in terms of $X,Y,A,B$ (which had the values &#xA;$1,2,2,3$ in the above example).  If we define&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u = YA - XB$$        $$r = YB - XA$$&#xA;$$v = YB + XA$$        $$s = YA + XB$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then we have squares such as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{ccc}&#xA;(2uv-uu+vv)^2  &amp;amp;  (ur+vs+vr-us)^2  &amp;amp;  (2rs+rr-ss)^2\\&#xA;(us+vr-vs+ur)^2  &amp;amp;   (uu+vv)^2   &amp;amp;   (us+vr+vs-ur)^2\\&#xA;(2rs-rr+ss)^2   &amp;amp;  (ur+vs-vr+us)^2  &amp;amp; (2uv+uu-vv)^2\\&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which automatically has equal sums on the diagonals and the central&#xA;row and column.  The common sum along each of these lines is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S = 3(u^2 + v^2) = 3(r^2 + s^2) = 3[(A^2 + B^2)(X^2 + Y^2)]^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any magic square of squares whose central element is a sum of two&#xA;squares in no more than four ways must consist of some arrangement&#xA;of the four pairs of opposite terms shown in the square above, but&#xA;they need not be arranged as shown above.  It's convenient to refer&#xA;to each of the outer elements by the negative term it contains.  For&#xA;example, we will let $US$ denote the quantity $(ur+vs+vr-us)^2$, and we&#xA;will let $RR$ denote $(2rs-rr+ss)^2$, and so on.  Thus we have the four&#xA;pairs of opposite squares $(UU,VV)$, $(RR,SS)$, $(US,VR)$, and $(UR,VS)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We note that the system is symmetrical under exchange of $(u,v)$ with&#xA;$(r,s)$, and we consider first the possibility that $UU$ is in the same&#xA;row or column with two of the &quot;crossed&quot; elements.  In order for the&#xA;square to be magic the sum of these three quantities must equal the&#xA;common sum $S$ given above.  However, if we examine all six of these &#xA;case we find that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$UU+VR+VS-S =  8XY(A+B)(A-B)(v+u)(v-u)$$&#xA;$$UU+VR+US-S =  4vu(v+u)(v-u)$$&#xA;$$UU+VR+UR-S =  32ABXYuv$$&#xA;$$UU+VS+US-S =  8uv(A+B)(A-B)(X+Y)(X-Y)$$&#xA;$$UU+VS+UR-S =  4vu(v+u)(v-u)$$&#xA;$$UU+US+UR-S = -8AB(X+Y)(X-Y)(v+u)(v-u)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order for the terms of the overall square to be distinct, the&#xA;values of $u,v,r,s$ must all be non-zero, and the quantities $u^2 - v^2$&#xA;and $r^2 - s^2$ must also be non-zero.  Furthermore, each of the&#xA;quantities $A,B,X,Y$ must be non-zero, as must be the quantities&#xA;$A^2 - B^2$ and $X^2 - Y^2$.  These last two condition are due to the&#xA;fact that if $X=Y$, for example, we have $u=X(A-B)$, $v=X(A+B)$, $r=X(B-A)$,&#xA;and $s=X(A+B)$, which implies $v=s$ and $u=-r$.  Inserting these expressions&#xA;for $v$ and $u$ into $ur+vs+us-vr$ gives $-r^2 + s^2 - 2r$s, so we have $VR=SS$.&#xA;It follows from all these conditions that none of the six quantities &#xA;shown above can equal zero, so no arrangement with any of those &#xA;combinations of terms in any row or column can be a magic square&#xA;of squares.  Of course, this also rules out any squares with rows or&#xA;columns containing $SS$ with any two of the cross terms, since those &#xA;appear on the opposite side of the above squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It remains to consider the possibility of a magic square with each&#xA;of the outer rows and columns containing two &quot;pure&quot; terms (i.e., &#xA;$UU, VV, RR, SS$) and one cross term.  This implies that the pure&#xA;terms must be at the four corners.  Therefore, one of the rows or&#xA;columns will be $[UU  *  RR]$ where &quot;$*$&quot; denotes one of the four&#xA;cross terms.  Two of these cases can be ruled out immediately,&#xA;because we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$UU+VS+RR-S = 4XY(X+Y)(X-Y)(3A^2 - B^2)(A^2 - 3B^2)$$&#xA;$$UU+UR+RR-S = 4XY(X+Y)(X-Y)(A^2 + 4AB + B^2)(A^2- 4AB + B^2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first is impossible because a non-zero square cannot be $3$ times &#xA;a square, and the second is impossible because it implies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A = \pm 2B[2\pm sqrt(3)]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only two remaining cases are $[UU\ VR\ RR]$ and $[UU\ US\ RR]$, each of&#xA;which corresponds to either of the two squares (because we can&#xA;transpose the placement of the remaining pair of terms).  To prove&#xA;that these are impossible, we can make use of the right-hand relations&#xA;in (4), which state that the double of each corner equals the sum of&#xA;the middle terms of the two opposite sides.  This leads us to examine&#xA;the sums shown below for the indicated square&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}\\&#xA;    UU &amp;amp; VR &amp;amp; RR\\&#xA;    VS &amp;amp;    &amp;amp; UR\\&#xA;    SS &amp;amp; US &amp;amp; VV\\&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$VR+UR-2SS = 4(s+r)(s-r)[3AB(Y^2 - X^2) + XY(B^2 - A^2)]$$&#xA;$$UR+US-2UU = 4(v+u)(v-u)[3XY(B^2 - A^2) - AB(Y^2 - X^2)]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order for this to be a magic square, both of the indicated sums&#xA;must vanish, which (for distinct terms) requires the trailing factors&#xA;to vanish.  But these are of the form $3m+n=0$ and $3n-m=0$ and therefore&#xA;$m=n=0$, which is to say $AB(Y^2 - X^2) = XY(B^2 - A^2) = 0$.  This is&#xA;impossible for distinct terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The three remaining cases can be ruled out similarly, based on the&#xA;sums shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}\\&#xA;    UU &amp;amp; VR &amp;amp; RR\\&#xA;    UR &amp;amp;    &amp;amp; VS\\&#xA;    SS &amp;amp; US &amp;amp; VV\\&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$VR+VS-2SS = 4rs[3(A^2 - B^2)(Y^2 - X^2) + 4 ABXY]$$&#xA;$$VS+US-2UU = 4uv[(A^2 - B^2)(Y^2 - X^2) - 12 ABXY]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}\\&#xA;    UU &amp;amp; US &amp;amp; RR\\&#xA;    VS &amp;amp;    &amp;amp; UR\\&#xA;    SS &amp;amp; VR &amp;amp; VV\\&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$US+UR-2SS = 4rs[(A^2 - B^2)(Y^2 - X^2) + 12 ABXY]$$&#xA;$$UR+VR-2UU = 4uv[3(A^2 - B^2)(Y^2 - X^2) - 4 ABXY]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}\\&#xA;    UU &amp;amp; US &amp;amp; RR\\&#xA;    UR &amp;amp;    &amp;amp; VS\\&#xA;    SS &amp;amp; VR &amp;amp; VV\\&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$US+VS-2SS = 4(s+r)(s-r)[AB(Y^2 - X^2) + 3XY(B^2 - A^2)]$$&#xA;$$VS+VR-2UU = 4(v+u)(v-u)[3AB(X^2 - Y^2) + XY(B^2 - A^2)]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This completes the proof.  So, what has been shown here is that if &#xA;the central number $2e^2$ is expressible as a sum of two squares in &#xA;just four distinct ways (which is the minimum number necessary to &#xA;satisfy the diagonal and center row and column conditions), then &#xA;it's impossible to satisfy the outer row and column conditions.  Of&#xA;course, every array will be based on just four partitions of $2e^2$&#xA;into two squares, but those partitions won't all necessarily come &#xA;from the same factorization of $e$, so those cases aren't covered by&#xA;the above argument.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4421" LastEditDate="2015-11-30T22:23:47.117" LastActivityDate="2015-11-30T22:23:47.117" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24223" PostTypeId="2" ParentId="23358" CreationDate="2015-11-25T18:46:28.660" Score="4" Body="&lt;p&gt;I think this is referring to this question on Lifehacks.SE:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;a href=&quot;https://lifehacks.stackexchange.com/questions/9531&quot;&gt;How can I keep my cat off my keyboard?&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reasoning:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Seemingly cat traps love medley,&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The highest-voted answer has a photo with a box labelled &quot;Cat Trap&quot;, and is by user &lt;a href=&quot;https://lifehacks.stackexchange.com/users/160&quot;&gt;Tom Medley&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Cats love the responses to me,&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are 12 different answers, 4 of which have &gt;30 upvotes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I help prevent a big clean up&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; By preventing a cat from sitting on your keyboard, you prevent having to erase all the text that was &quot;typed&quot; by the cat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;So you can type up Sup!&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm assuming this just means that you are able to type.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I trap cats like a boss,&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, the box is labelled &quot;Cat Trap&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;since they can't get across,&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cats are distracted by the box, and can't get &quot;across&quot; to the keyboard.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I put a bounty on this guy,&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The top-voted (and accepted) answer was awarded a 500-rep bounty.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;and attracted so many guys,&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 12 different people provided answers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I look like an election just happened,&lt;br&gt;&#xA;these people so happy for them votes&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There were an awful lot of upvotes on the question and several of the answers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:43.243" LastActivityDate="2015-11-25T18:46:28.660" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24224" PostTypeId="2" ParentId="24218" CreationDate="2015-11-25T18:48:13.180" Score="13" Body="&lt;p&gt;A rather simplistic solution&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your friend considers $1$ as a tally mark instead of a digit,&lt;br&gt;&#xA; Hence $$||+|=|||=3$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9379" LastActivityDate="2015-11-25T18:48:13.180" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24225" PostTypeId="2" ParentId="24215" CreationDate="2015-11-25T21:16:01.967" Score="6" Body="&lt;p&gt;For any number $n$, there are finitely many connected graphs with $n$ edges. So we can make lists of all the graphs with no edges, all the graphs with one edge, all the graphs with two edges, and so on. If we join these lists together, we can make an infinite list that is guaranteed to contain every finite graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for each graph, there are finitely many ways to choose a starting point, ending point, and starting direction, so we can make a list of every possible starting situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once we have this list, assume that we started in the first situation on the list, and try to get to the target based on that. If that doesn't work, we assume that we started in the second situation and figure out where we ended up. Then, we go to the target from there. If we still aren't done, we keep continuing down the list in the same way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the list contains every possible starting situation, eventually we will reach the right one, at which point we will arrive at the target.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-11-25T21:16:01.967" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24226" PostTypeId="2" ParentId="24205" CreationDate="2015-11-25T21:58:34.310" Score="1" Body="&lt;p&gt;I did this in my head and multiplied two numbers with a calculator to get 184 = 8 x 23.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;At a party, everybody is friend with exactly 22 of the other persons present.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The way I thought it out was if there were 23 people at the party then each person would be friends with the other 22.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Whenever two persons are friends, they do not have any friends in common.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) There could be n groupings of 23&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Whenever two persons are not friends, they have exactly six friends (present at the party) in common.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Which tells me there can be a minimum of 8 groupings. (2 groupings contain the two non-friends, the other 6 groupings contain the 6 common friends from Person from Group1 and Person from Group2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a picture, imagine each black circle contains 23 dots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aUBsf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aUBsf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Red dots signify the two non-friends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Green dots signify the common friends of each of the red dots.&lt;/p&gt;&#xA;" OwnerUserId="17542" LastEditorUserId="11784" LastEditDate="2015-11-26T04:03:21.177" LastActivityDate="2015-11-26T04:03:21.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24227" PostTypeId="1" AcceptedAnswerId="24229" CreationDate="2015-11-25T22:44:05.223" Score="8" ViewCount="408" Body="&lt;p&gt;Below are four groups of cryptic clues.  Within each group, the answers share a common characteristic.  The four groups also share a common characteristic.  Solving any one group should make it much easier to solve the others.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Group A&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Punctuation provides stylish mixup with new beginning&lt;/li&gt;&#xA;&lt;li&gt;Insect in potato begets mechanical being&lt;/li&gt;&#xA;&lt;li&gt;Arias of hobo's final death are confusing&lt;/li&gt;&#xA;&lt;li&gt;Benefactor starts napping in entrance&lt;/li&gt;&#xA;&lt;li&gt;Glorify and endlessly refine the final warrior&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Group B&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Copy a note twice beside the ocean&lt;/li&gt;&#xA;&lt;li&gt;Six repeated depot openings are intense&lt;/li&gt;&#xA;&lt;li&gt;Curb bottomless citrus and miss the start of blockbuster&lt;/li&gt;&#xA;&lt;li&gt;Symbol for a confused boxer without a place&lt;/li&gt;&#xA;&lt;li&gt;Excavate that with a finger&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Group C&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Obese friend with no head is deadly&lt;/li&gt;&#xA;&lt;li&gt;Your mother, out of her sleepwear, gets embroiled on the first of November with an island nation&lt;/li&gt;&#xA;&lt;li&gt;Let a heathen pass water once more without me&lt;/li&gt;&#xA;&lt;li&gt;Miss, your mother is an obstruction&lt;/li&gt;&#xA;&lt;li&gt;As inverted fuel for epic tales&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Group D&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The sides on a heptagon where the shape begins to be level&lt;/li&gt;&#xA;&lt;li&gt;Battle's debut is in folk dance with insurgent&lt;/li&gt;&#xA;&lt;li&gt;Cadence is decided unconventionally without having eaten&lt;/li&gt;&#xA;&lt;li&gt;Even fifty at party with no initiation&lt;/li&gt;&#xA;&lt;li&gt;Recommend a joint without a clue, finally&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&lt;em&gt;NOTE:  This is the first time I've ever attempted to create cryptic clues.  I appreciate any feedback, however minor, about what does or doesn't work, and/or any suggestions on how to improve the clues.&lt;/em&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="4421" LastEditDate="2015-11-25T22:49:39.353" LastActivityDate="2016-04-19T17:32:36.333" Title="Cryptic categories" Tags="|pattern|wordplay|cryptic-clues|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24228" PostTypeId="2" ParentId="24214" CreationDate="2015-11-25T22:49:58.170" Score="3" Body="&lt;p&gt;Think I figured out a line.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Doomsday as land&lt;/li&gt;&#xA;&lt;li&gt;Play (reanimate) dark ritual&lt;/li&gt;&#xA;&lt;li&gt;Play (reanimate) blood pet&lt;/li&gt;&#xA;&lt;li&gt;Play (reanimate) culling the weak and sac blood pet (BBBBB)&lt;/li&gt;&#xA;&lt;li&gt;Play (candelabra) cromantic star and sac for G (floating BBBG) draw card&lt;/li&gt;&#xA;&lt;li&gt;Play (reanimate) song of the damned and name all 4 cards in your yard as critters go to BBBBBB G floating&lt;/li&gt;&#xA;&lt;li&gt;Play (cabal ritual) plunge into darkness and pay 19 life.  Take the glimpse.&lt;/li&gt;&#xA;&lt;li&gt;Play (cabal ritual) shallow grave, pick (reanimate) death's shadow as the card&lt;/li&gt;&#xA;&lt;li&gt;Attack and play (glimpse of nature) berserk to deal 24 (edit math fail)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Easy game.&lt;/p&gt;&#xA;" OwnerUserId="17544" LastEditorUserId="86115" LastEditDate="2023-11-06T22:14:33.527" LastActivityDate="2023-11-06T22:14:33.527" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="24229" PostTypeId="2" ParentId="24227" CreationDate="2015-11-25T23:37:23.743" Score="4" Body="&lt;p&gt;Group A:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (COLO*)N&lt;br&gt;&#xA; RO(B)OT&lt;br&gt;&#xA; (O + LOSS)* = SOLOS&lt;br&gt;&#xA; DO(N)OR&lt;br&gt;&#xA; HONe + warriOR&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Group B:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MI+MI+C (sea)&lt;br&gt;&#xA; VI+VI+Depot&lt;br&gt;&#xA; LIMe + hIT&lt;br&gt;&#xA; (pugilist - put)* = SIGIL&lt;br&gt;&#xA; DIG+IT  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Group C:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; FAT+pAL&lt;br&gt;&#xA; pajama* - ma + n = JAPAN&lt;br&gt;&#xA; P + AGAiN&lt;br&gt;&#xA; MA+DAM&lt;br&gt;&#xA; SA+GAS  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Group D (&lt;em&gt;thanks to f''!&lt;/em&gt;):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; S+EVEN&lt;br&gt;&#xA; RE(B)EL&lt;br&gt;&#xA; determined* - dined = METER&lt;br&gt;&#xA; L+rEVEL&lt;br&gt;&#xA; REeFER&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And of course, the common thread is that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; they all have the same vowel in the second and fourth positions; which one it is is dependent on the group.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="4421" LastEditDate="2016-04-19T17:32:36.333" LastActivityDate="2016-04-19T17:32:36.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24230" PostTypeId="1" CreationDate="2015-11-26T01:00:07.890" Score="20" ViewCount="3780" Body="&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/2300/&quot;&gt;This question&lt;/a&gt; discusses a well-known problem where you are given 25 horses and a five-lane track. You are asked to find the minimum number of races (maximum 5 horses per race) to determine the fastest 3 horses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have no stopwatch or other absolute method of measuring time.  The only way to know if a horse is faster than another horse is if it finishes before the other horse in the same race.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we &lt;strong&gt;&lt;em&gt;slightly modify the original question&lt;/em&gt;&lt;/strong&gt; to state that you need to know the relative speeds of all 25 horses (instead of just the top 3), what would be the minimum number of races required?&lt;/p&gt;&#xA;" OwnerUserId="17547" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.923" LastActivityDate="2017-07-19T01:38:51.817" Title="25 Horses and 5 Tracks" Tags="|mathematics|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="24233" PostTypeId="2" ParentId="24214" CreationDate="2015-11-26T09:15:50.007" Score="2" Body="&lt;p&gt;Update: I finally did it, a solution all my own that deals infinite damage, without counting anything as having a card type not printed on it (even for targeting purposes). As OP stated in the comments of the question, Recover works, which is a triggered ability belonging to a card in a graveyard. Based on this, I will be using the triggered abilities of several cards in graveyards. Since I am not casting these cards, they do not become unusable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting Hand: 1, B, B, B, 1B, 1B, BBB&lt;br&gt;&#xA;1: Play Reanimate as land, tap it for (B)&lt;br&gt;&#xA;2: Cast &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=397401&quot; rel=&quot;nofollow&quot;&gt;Dark Ritual&lt;/a&gt; (BBB)&lt;br&gt;&#xA;3: Cast &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=19115&quot; rel=&quot;nofollow&quot;&gt;Cast Carnival of Souls&lt;/a&gt; (B)&lt;br&gt;&#xA;4: Cast &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=46112&quot; rel=&quot;nofollow&quot;&gt;Myr Moonvessel&lt;/a&gt;, Carnival triggers (B)&lt;br&gt;&#xA;5: Cast &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=6085&quot; rel=&quot;nofollow&quot;&gt;Culling the Weak&lt;/a&gt;, sacrificing Moonvessel (1BBBB)&lt;br&gt;&#xA;6: Cast &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=382901&quot; rel=&quot;nofollow&quot;&gt;Cruel Bargain&lt;/a&gt;, drawing Reanimate, dredging Reanimate as Darkblast (milling Tarmogoyf, Council's Judgement and Lightning Bolt), drawing Brainstorm and Hymn to Tourach (1B)&lt;br&gt;&#xA;7: Cast &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=240178&quot; rel=&quot;nofollow&quot;&gt;Blood Artist&lt;/a&gt;, Carnival triggers (B)&lt;br&gt;&#xA;8: Cast &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=376569&quot; rel=&quot;nofollow&quot;&gt;Viscera Seer&lt;/a&gt;, Carnival triggers (B)&lt;br&gt;&#xA;9: Cast &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=25625&quot; rel=&quot;nofollow&quot;&gt;Blood Pet&lt;/a&gt;, Carnival triggers, sacrifice it, Doomsday in the graveyard triggers as &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=370353&quot; rel=&quot;nofollow&quot;&gt;Bridge from Below&lt;/a&gt;, you get a token, Carnival triggers (BBB)&lt;br&gt;&#xA;10: Cast &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=236908&quot; rel=&quot;nofollow&quot;&gt;Distress&lt;/a&gt; on the opponent just to get it into the graveyard (B)&lt;br&gt;&#xA;11: Sacrifice a token, Hymn to Tourach in the graveyard triggers as &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=116742&quot; rel=&quot;nofollow&quot;&gt;Nether Traitor&lt;/a&gt;, pay to bring it back, Carnival triggers&lt;br&gt;&#xA;12: Sacrifice Nether Shadow, Bridge triggers, Carnival triggers.&lt;br&gt;&#xA;13: Return to step 11, repeat infinitely for infinite Blood Artist triggers and infinite mana.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Previous solution was wrong due to an illegal dredge, fixed now.&lt;/p&gt;&#xA;" OwnerUserId="7459" LastEditorUserId="7459" LastEditDate="2015-11-28T23:06:00.577" LastActivityDate="2015-11-28T23:06:00.577" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="24235" PostTypeId="2" ParentId="24171" CreationDate="2015-11-26T10:19:16.203" Score="4" Body="&lt;p&gt;I'm going to guess that you are the &quot;House of York&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The title of the question is &quot;You dare  to eat me&quot; - York peppermint patties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You said you died at age 72 and you died in the year '85, so you must have been born in '13. The house of York rose to power from 1413-1485 &lt;a href=&quot;https://en.wikipedia.org/wiki/Royal_Badges_of_England&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Royal_Badges_of_England&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The house of York had many kings in place so I'm sure something happened in '65 probably. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the House of York overthrew the previous &quot;House of Lancaster&quot;, which is another food product company (Lancaster) hinting towards &quot;you dare to eat me&quot; once more.&lt;/p&gt;&#xA;" OwnerUserId="16599" LastActivityDate="2015-11-26T10:19:16.203" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24240" PostTypeId="1" CreationDate="2015-11-26T17:49:39.253" Score="1" ViewCount="432" Body="&lt;p&gt;&lt;sub&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/24081/magic-the-gathering-challenge-5-what-if-pod-was-one-of-us&quot;&gt;Previous Challenge&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/24324/magic-the-gathering-challenge-7-in-your-dreams&quot;&gt;Next Challenge&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Background&lt;/h2&gt;  &#xA;&#xA;&lt;p&gt;Alright I'm going to try a more clear approach to my last attempted puzzle. Given the mana costs given below, name cards with those exact mana costs to use in a solution that wins this turn (the card chosen must match the exact mana cost printed on the card so e.g. RR can be a Wild Guess but not a Rolling Thunder or a kicked Shattering Spree). Once you decide what card is associated with a particular cost, it must stay the same card in all zones; all usual rules apply to the card once chosen so if you decide that 3B is a Heartstabber Mosquito, you can pay kicker as though it were that card. You can play any card face down as a basic land that makes any color mana (still only one land per turn)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some addition rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The same spell may not be named for multiple cards (e.g. even if you're only casting one copy of Dark Confidant, no other 1B card may be named as Dark Confidant)  &lt;/li&gt;&#xA;&lt;li&gt;The same spell may not be cast more than once&lt;/li&gt;&#xA;&lt;li&gt;The Legacy banned list and rules should be observed (sorry Demonic Tutor)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Puzzle Setup&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;It's turn 1 and you're on the play. Let's assume that (magically) your opponent has no cards in hand. Win this turn with the rules given above and the mana costs given below. Both boards and graveyards are empty. Your solution must be a guaranteed win, not involving your opponent self-destructing (e.g. they choose 20 on a Choice of Damanations) or random chance (you Hymn to Tourach yourself and hit the right 2 cards). If multiple solutions are achieved, the one that does the most damage wins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your hand:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;1B&lt;/strong&gt; x 2&lt;br&gt;&#xA;&lt;strong&gt;1&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;BBB&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;B&lt;/strong&gt; x 3 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your library: (top to bottom)&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;B&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;1G&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;1WW&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;R&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;U&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;BB&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;G&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your opponent's library:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;8GGG&lt;/strong&gt; x 60&lt;/p&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2016-02-13T10:06:14.363" Title="Magic: the Gathering - Challenge #6.b: It's Gonna Cost You" Tags="|calculation-puzzle|magic-the-gathering|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24241" PostTypeId="1" AcceptedAnswerId="24248" CreationDate="2015-11-26T18:22:35.943" Score="3" ViewCount="378" Body="&lt;p&gt;Playing with a Standard-legal Magic: The Gathering deck, your goal is to win a game as quickly as possible, against an opponent who does nothing. You may assume that your deck is stacked, so that you will always draw exactly the cards you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may choose to play or draw, but winning turn 4 on the play is considered faster than winning turn 4 on the draw. In the event that two solutions tie for speed, the winner will be the solution that deals more damage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Possibly Asked Questions:&lt;/strong&gt;&lt;br&gt;&#xA;At the time of asking, Standard-legal Magic cards are any cards printed in Khans of Tarkir, Fate Reforged, Dragons of Tarkir, Magic Origins, or Battle for Zendikar.&lt;br&gt;&#xA;If you care about the opponent at all, assume their deck consists of sixty islands and they begin the game with a mulligan to zero.&lt;br&gt;&#xA;I know it's possible to deal approximately 35 damage turn 3 on the play, so if you're not there yet you definitely haven't found the optimal solution.&lt;/p&gt;&#xA;" OwnerUserId="7459" LastEditorUserId="7459" LastEditDate="2015-11-26T21:08:15.270" LastActivityDate="2015-11-27T21:48:44.690" Title="Magic: the Gathering - Standard Speed" Tags="|magic-the-gathering|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24242" PostTypeId="2" ParentId="24240" CreationDate="2015-11-26T18:44:35.167" Score="4" Body="&lt;ol&gt;&#xA;&lt;li&gt;Play &lt;strong&gt;BBB&lt;/strong&gt; as a land.&lt;/li&gt;&#xA;&lt;li&gt;Play &lt;strong&gt;B&lt;/strong&gt; as Dark Ritual (&lt;strong&gt;BBB&lt;/strong&gt;)&lt;/li&gt;&#xA;&lt;li&gt;Play &lt;strong&gt;1&lt;/strong&gt; as Myr Moonvessel, then &lt;strong&gt;B&lt;/strong&gt; as Culling the Weak, sacking the&#xA;Myr (&lt;strong&gt;1BBBBB&lt;/strong&gt;)&lt;/li&gt;&#xA;&lt;li&gt;Play &lt;strong&gt;B&lt;/strong&gt; as Death's Shadow, which dies immediately (&lt;strong&gt;1BBBB&lt;/strong&gt;)&lt;/li&gt;&#xA;&lt;li&gt;Play &lt;strong&gt;1B&lt;/strong&gt; as Plunge into Darkness, pay 19 life, get the &lt;strong&gt;B&lt;/strong&gt; (&lt;strong&gt;BBB&lt;/strong&gt;)&lt;/li&gt;&#xA;&lt;li&gt;Play &lt;strong&gt;1B&lt;/strong&gt; as Shallow Grave, which puts Death's Shadow into play as a&#xA;12/12 with haste (&lt;strong&gt;B&lt;/strong&gt;)&lt;/li&gt;&#xA;&lt;li&gt;Play &lt;strong&gt;B&lt;/strong&gt; as Tainted Strike, attack for 13 infect.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="17559" LastEditorUserId="17559" LastEditDate="2015-11-26T18:52:35.997" LastActivityDate="2015-11-26T18:52:35.997" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24244" PostTypeId="2" ParentId="24241" CreationDate="2015-11-26T20:29:30.523" Score="2" Body="&lt;p&gt;Turn 3 on the play.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Turn 1:&lt;/strong&gt; Forest, Scythe Leopard.&lt;br/&gt;&#xA;&lt;strong&gt;Turn 2:&lt;/strong&gt; Wooded Foothills, crack it for a Mountain, Monastery Swiftspear, Scythe Leopard, attack for 4. &lt;br/&gt;&#xA;&lt;strong&gt;Turn 3:&lt;/strong&gt; Wooded Foothills, crack it for a Cinder Glade, Monastery Swiftspear, Atarka's Command (3 damage to opponent and +1/+1 to your dudes), triggering prowess, attack with 2 4/4's and 2 3/4's for 14.&#xA;&lt;br/&gt;&lt;br/&gt;&#xA;&lt;strong&gt;Total:&lt;/strong&gt; 21 damage.&lt;/p&gt;&#xA;" OwnerUserId="17559" LastEditorUserId="17559" LastEditDate="2015-11-26T20:34:58.367" LastActivityDate="2015-11-26T20:34:58.367" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24245" PostTypeId="2" ParentId="24230" CreationDate="2015-11-26T21:53:12.483" Score="4" Body="&lt;p&gt;Using a kind of binary search, where you have a few &quot;constant&quot; horses and you rank the rest of them as being faster or slower than your constants, would certainly get the job done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick three horses at random (let's call them A, B, and C), and run them together in 11 races such that they match off against the 22 other horses, and therefore establish an absolute ranking for all three, recording which horses were better or worse than them in each race.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Best case: 16 races&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;A comes in (for example) 6th overall, B comes in 12th overall, and C comes in 18th (specific positions don't really matter, as long as they're separated by five horses). Race the 5 horses faster than A, and the 5 horses faster than B but slower than A, and the five faster than C but slower than B, and you've established a ranking of 18 of the horses in 14 races. Then, repeat with the remaining seven, keeping three of the horses constant, doing TWO (thanks, Dr Xorile) more races to rank the remainder (again, assuming best case. Ranking seven horses in this manner may require as many as three races).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Worst case: 48 races&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If all three chosen horses place so that the rankings of the 22 remaining horses form a sequential list (i.e. A, B, and C come in 1st, 2nd, and 3rd or 1st, 24th, and 25th) that is the worst case, as it tells you nothing about the remaining horses and no meaningful partitions can be made. If this bad luck is repeated for the rest of the ranking, you will require 48 races to establish a rank for all 25 horses (assuming you naively continue using this strategy all the way through).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Average case: 22 races&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I've calculated the average for all ${25 \choose 3} = 2,300$ possible choices of three horses to be 22. Which is actually much better than I was expecting for an average case. I found this by calculating the average races needed for 6 horses, then seven, then eight, etc. and then subbing those values in when there was a gap of that size between the rankings of two &quot;constant&quot; horses.&lt;/p&gt;&#xA;" OwnerUserId="17167" LastEditorUserId="17167" LastEditDate="2015-11-27T04:57:40.507" LastActivityDate="2015-11-27T04:57:40.507" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24247" PostTypeId="2" ParentId="23227" CreationDate="2015-11-26T23:36:13.583" Score="-2" Body="&lt;p&gt;I believe that &lt;strong&gt;no&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For many pairs of numbers &lt;code&gt;x&lt;/code&gt; and &lt;code&gt;y&lt;/code&gt;, &lt;code&gt;(4x + 3y) / 5&lt;/code&gt; and &lt;code&gt;(24x − 7y) / 25&lt;/code&gt; won't even be integers. For example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x = 2, y = 7: &#xA;   (4x + 3y) / 5 = (8 + 21) / 5 = 29/5 = 5.8&#xA;   (24x − 7y) / 25 = (48 - 49) / 25 = -1/25 = -0.04&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In about only 1/5 of the steps the first expression will be an integer, and in about only 1/25 of the steps the second expression will be an integer, per divisibility rules, and assuming that the numerator is a random integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, by construction, most non-integer numbers will continue to be operated into other non-integer numbers - with bigger denominators in fractional form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question can be turned around into: Find a way to generate all numbers from 1 to 2014, given the operations with &lt;code&gt;x&lt;/code&gt; and &lt;code&gt;y&lt;/code&gt; above, and choosing &lt;code&gt;x&lt;/code&gt; and &lt;code&gt;y&lt;/code&gt; for each step. Expecting it to happen at random is extremely improbable.&lt;/p&gt;&#xA;" OwnerUserId="17562" LastActivityDate="2015-11-26T23:36:13.583" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24248" PostTypeId="2" ParentId="24241" CreationDate="2015-11-26T23:44:08.870" Score="5" Body="&lt;p&gt;&lt;em&gt;&quot;Looks like you can't go infinite before turn four, so this is probably just won by the most aggressive strategy&quot;&lt;/em&gt; - a quote from myself.  I was wrong.  You can totally go infinite on turn 3 on the play.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Turn 3 on the Play, Main Phase 1: Infinite Damage&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Turn 1:&lt;/strong&gt; &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=398566&quot; rel=&quot;nofollow&quot;&gt;Yavimaya Coast&lt;/a&gt;, &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=398450&quot; rel=&quot;nofollow&quot;&gt;Honored Hierarch&lt;/a&gt; (5 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Turn 2:&lt;/strong&gt; &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=405116&quot; rel=&quot;nofollow&quot;&gt;Wooded Foothills&lt;/a&gt; &gt; Mountain, &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=394502&quot; rel=&quot;nofollow&quot;&gt;Atarka’s Command&lt;/a&gt;, dealing 3 and putting a &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=405115&quot; rel=&quot;nofollow&quot;&gt;Windswept Heath&lt;/a&gt; into play (3 cards, 2 gy)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Windswept Heath &gt; Plains, &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=386515&quot; rel=&quot;nofollow&quot;&gt;Defiant Strike&lt;/a&gt; on Hierarch, draw 1, attack with Hierarch becoming renowned (3 cards, 4 gy)&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Turn 3:&lt;/strong&gt; &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=405104&quot; rel=&quot;nofollow&quot;&gt;Polluted Delta&lt;/a&gt; &gt; Island (3 cards, 5 gy)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Tap Island, Plains, Mountain for &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=386571&quot; rel=&quot;nofollow&quot;&gt;Jeskai Ascendancy&lt;/a&gt; (2 cards, 5 gy)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Tap HH for R, play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=398496&quot; rel=&quot;nofollow&quot;&gt;Magmatic Insight&lt;/a&gt;, discarding Swamp, draw two, untap HH, loot a card (2 cards, 8 gy)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Tap HH for U, play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=386705&quot; rel=&quot;nofollow&quot;&gt;Treasure Cruise&lt;/a&gt; delving 7, untap HH, loot a card (4 cards, 3 gy)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Tap HH for R, play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=398496&quot; rel=&quot;nofollow&quot;&gt;Magmatic Insight&lt;/a&gt;, discarding Swamp, draw two, untap HH, loot a card (4 cards, 6 gy)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Tap HH for R, play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=394708&quot; rel=&quot;nofollow&quot;&gt;Spidersilk Net&lt;/a&gt;, untap HH, loot a card (R, 3 cards, 7 gy)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Tap HH for W, tap Yavimiaya Coast for U, play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=386571&quot; rel=&quot;nofollow&quot;&gt;Jeskai Ascendency&lt;/a&gt;, untap HH, loot a card (2 cards, 8 gy)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Tap HH for U, play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=386705&quot; rel=&quot;nofollow&quot;&gt;Treasure Cruise&lt;/a&gt; delving 7, untap/tap HH for UW, loot twice (UW, 4 cards, 4 gy)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=394708&quot; rel=&quot;nofollow&quot;&gt;Spidersilk Net&lt;/a&gt;, untap/tap HH for UR, loot twice (UUWR, 3 cards, 6 gy)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=386571&quot; rel=&quot;nofollow&quot;&gt;Jeskai Ascendency&lt;/a&gt;, untap/tap HH for WR, loot once (UWR, 2 cards, 7 gy)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=386571&quot; rel=&quot;nofollow&quot;&gt;Jeskai Ascendency&lt;/a&gt;, untap/tap HH for UUW (UUW, 1 card, 7 gy)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=386705&quot; rel=&quot;nofollow&quot;&gt;Treasure Cruise&lt;/a&gt; delving 7, untap/tap HH for UUUW (UUUUUWW, 3 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=391927&quot; rel=&quot;nofollow&quot;&gt;Soulfire Grand Master&lt;/a&gt; (UUUW, 2 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=386515&quot; rel=&quot;nofollow&quot;&gt;Defiant Strike&lt;/a&gt; on HH, draw one, untap/tap HH for UUUU (UUUUUUU, 2 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=398528&quot; rel=&quot;nofollow&quot;&gt;Disperse&lt;/a&gt; targeting Spidersilk Net for UU, trigger Soulfire Grand Master for UUUU to return Disperse to your hand, untap/tap HH for UUUU (UUUUU, 1 card, Spidersilk Net, Disperse)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=394708&quot; rel=&quot;nofollow&quot;&gt;Spidersilk Net&lt;/a&gt;, untap/tap HH for UURR (UUUUUUURR, 1 card, Disperse)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Repeat the previous two steps an infinite number of times, gaining RR each time until you have infinite R (Infinite R, 1 card, Disperse)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=402015&quot; rel=&quot;nofollow&quot;&gt;Rolling Thunder&lt;/a&gt; for X = infinite, targeting your opponent for all of it, dealing infinite damage and gaining infinite life as well&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;But maybe Jeskai Ascendancy and Treasure Cruise are too easy; after all, both have seen play in Legacy (and even Vintage).  Lets go infinite without using a single non-land card that's seen play outside of Standard.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Turn 3 on the Play, Main Phase 1: Another Infinite Damage Combo&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Turn 1: Forest, &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=394637&quot; rel=&quot;nofollow&quot;&gt;Obscuring Aether&lt;/a&gt; (5 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Turn 2: Forest, &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=394637&quot; rel=&quot;nofollow&quot;&gt;Obscuring Aether&lt;/a&gt; x2, &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=386634&quot; rel=&quot;nofollow&quot;&gt;Rattleclaw Mystic&lt;/a&gt; x3 face down (0 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Turn 3: Unmorph Rattleclaw Mystic #1 for GG, add RGU, tap for U.  Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=386657&quot; rel=&quot;nofollow&quot;&gt;Secret Plans&lt;/a&gt; for GU. (RU, 0 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Unmorph Rattleclaw Mystic #2 for RU, add RGU, draw a card, tap for U (RGUU, 1 card)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Unmorph Rattleclaw Mystic #3 for RG, add RGU, draw a card, tap for U (RGUUUU, 2 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=386634&quot; rel=&quot;nofollow&quot;&gt;Rattleclaw Mystic&lt;/a&gt; #4 face down, unmorph it for RU, add RGU, draw a card (RGGUUUU, 2 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=386692&quot; rel=&quot;nofollow&quot;&gt;Temur Ascendancy&lt;/a&gt; for RGU, tap Mystic #4 for R (RGUUU, 1 card)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=401844&quot; rel=&quot;nofollow&quot;&gt;Clutch of Currents&lt;/a&gt; for U, returning Mystic #1, replay it face down, unmorph Mystic #1 for RG, add RGU, draw a card, tap for U (RGUUUU, 1 card)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Repeat the previous step three more times (RGUUUUUUU, 1 card)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=394696&quot; rel=&quot;nofollow&quot;&gt;Sidisi's Faithful&lt;/a&gt;, exploiting itself for U, returning Mystic #1, replay it face down, unmorph Mystic #1 for RG, add RGU, draw a card, tap for U (RGUUUUUUUU, 1 card)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Repeat the previous step three more times (RGUUUUUUUUUUU, 1 card)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=394614&quot; rel=&quot;nofollow&quot;&gt;Learn from the Past&lt;/a&gt; for UUUU, targeting yourself, shuffling in everything and drawing a card (RGUUUUUUU, 1 card)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Repeat the above cycles with Clutch x4, Sidisi's Faithful x4, gaining U per card and drawing one off each (RGUUUUUUUUUUUUUUU, 1 card)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=394614&quot; rel=&quot;nofollow&quot;&gt;Learn from the Past&lt;/a&gt; for UUUU, targeting yourself, shuffling in everything (including the other Learn from the Past) and drawing a card (RGUUUUUUUUUUU, 1 card)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Repeat the above two steps an infinite number of times, gaining infinite U (RG, infinite U, 1 card)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=386509&quot; rel=&quot;nofollow&quot;&gt;Crater's Claws&lt;/a&gt; for X = infinite, targeting your opponent, dealing them infinite damage  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Previous Solution&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;39 damage Turn 3 on the play.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Turn 1: Wooded Foothills, crack it for a Forest, Scythe Leopard. (5 cards, 1 in yard, 0 damage)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Turn 2: Wooded Foothills, crack it for a Mountain, Atarka's Command dealing 3 and putting Wooded Foothills in play, crack it for Canopy Vista, Defiant Strike on Scythe Leopard. Attack with Scythe Leopard for 6 damage. (3 cards, 5 in yard, 9 damage this turn)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Turn 3: Wooded Foothills, crack it for a Cinder Glade, Animist's Awakening for X = 3, revealing Windswept Heath x3, crack for Cinder Glade x3, Become Immense on Scythe Leopard, delving 5, Temur Battle Rage on Scythe Leopard, giving double strike. Attack with Scythe Leopard for 15+15 damage. (0 cards, 7 in yard, 30 damage this turn)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="17389" LastEditDate="2015-11-27T21:48:44.690" LastActivityDate="2015-11-27T21:48:44.690" CommentCount="25" ContentLicense="CC BY-SA 3.0" />
  <row Id="24249" PostTypeId="2" ParentId="24230" CreationDate="2015-11-27T04:59:10.000" Score="10" Body="&lt;h1&gt;Worst case 20 races, best case 13 races&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Run all 25 horses in 5 races to get the following partial order, with each letter corresponding to a horse's completion time (which we assume for simplicity to be constant for each horse regardless of the number of races run).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;a &amp;lt; b &amp;lt; c &amp;lt; d &amp;lt; e&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;f &amp;lt; g &amp;lt; h &amp;lt; i &amp;lt; j&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;k &amp;lt; l &amp;lt; m &amp;lt; n &amp;lt; o&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;p &amp;lt; q &amp;lt; r &amp;lt; s &amp;lt; t&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;u &amp;lt; v &amp;lt; w &amp;lt; x &amp;lt; y&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Call these the &lt;em&gt;initial 5 chains&lt;/em&gt;. The best of these are &lt;code&gt;afkpu&lt;/code&gt;, whose relative order we don't yet know. We repeatedly race the fastest horses with unknown relative order against each other. Sometimes, fewer than 5 horses would be required, in which case we put in the next fastest of some of the chains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Race 6 ('best of the best'): &lt;code&gt;afkpu&lt;/code&gt;&lt;/strong&gt; Assume without loss of generality that &lt;code&gt;a &amp;lt; f &amp;lt; k &amp;lt; p &amp;lt; u&lt;/code&gt; . The fastest horse is then &lt;code&gt;a&lt;/code&gt;. Remove &lt;code&gt;a&lt;/code&gt; from the pool (of horses that must still compete) and continue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Race 7 ('runners up'): &lt;code&gt;bfcgk&lt;/code&gt;&lt;/strong&gt; From the initial 5 chains and race 7, we know the next fastest horse is one of &lt;code&gt;bf&lt;/code&gt;. The fastest horse after that would either be the other (of &lt;code&gt;bf&lt;/code&gt;) or one of &lt;code&gt;cgk&lt;/code&gt;; all other horses are slower than these 4. The winner and runner up of race 7 are awarded positions 2 and 3, and then removed from the pool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the initial 5 chains. In each chain, the remaining horses still form a single chain, so we still have 5 chains, though some will have fewer than 5 horses. Repeat the 'best of the best' and 'runners up' races to get 3 horses every two races (thanks @IvoBeckers). When there are at most 5 horses remaining in the pool, run them all in a single race to determine the bottom 5 positions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Calculations:&lt;/strong&gt; races 1-5 yield no positions, the first 21 positions are determined in the next 14 races, and the last race determines the final 4 positions. Total 5 + 14 + 1 = &lt;strong&gt;20 races in the worst case&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Best case: use the above method, but assume the most convenient winning pattern possible to get more than 2 placings per race after race 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Race 6: &lt;code&gt;afkpu&lt;/code&gt;, &lt;code&gt;a=1&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Race 7: &lt;code&gt;bfcgk&lt;/code&gt;, &lt;code&gt;b=2,f=3,c=4&lt;/code&gt;. After &lt;code&gt;bf&lt;/code&gt; are removed, the fastest 3 horses according to the established partial orders are &lt;code&gt;cgk&lt;/code&gt;, all of which are in race 7, so we get 3 placings from this race. Assume &lt;code&gt;g &amp;lt; k&lt;/code&gt; since we're looking for the best case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Race 8: &lt;code&gt;dgehk&lt;/code&gt;, &lt;code&gt;d=5,g=6,e=7&lt;/code&gt;. We only really need &lt;code&gt;dg&lt;/code&gt; for the 'best of the best', so we have room for 3 more - choose &lt;code&gt;ehk&lt;/code&gt;. This race now also has the composition of a 'runners up' race, so we get 2 placings &lt;code&gt;dg&lt;/code&gt;. In the best case, &lt;code&gt;e&lt;/code&gt; also places, and all the remaining horses in the pool are slower than either &lt;code&gt;h&lt;/code&gt; or &lt;code&gt;k&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Race 9: &lt;code&gt;hkilp&lt;/code&gt;, &lt;code&gt;h=8,k=9,l=10&lt;/code&gt;. We don't need a 'best of the best' race since &lt;code&gt;hkpu&lt;/code&gt; are already ordered, so this is just a 'runners up' race. In the best case, &lt;code&gt;hk&lt;/code&gt; take out the next 2 placings, and we get a bonus placing since the remaining pool is slower than at least one of &lt;code&gt;ilp&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Race 10: &lt;code&gt;impjn&lt;/code&gt;, &lt;code&gt;i=11,j=12,m=13&lt;/code&gt;. The 'best of the best' only needs to sort &lt;code&gt;imp&lt;/code&gt;, so we have room for 2 more - choose &lt;code&gt;jn&lt;/code&gt;. If &lt;code&gt;ij&lt;/code&gt; are fastest in this race, the remaining pool is slower than at least one of &lt;code&gt;mp&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Race 11: &lt;code&gt;npoqu&lt;/code&gt;, &lt;code&gt;n=14,p=15,o=16&lt;/code&gt;. The 'best of the best' only needs to sort &lt;code&gt;np&lt;/code&gt;; add &lt;code&gt;oqu&lt;/code&gt;. If &lt;code&gt;np&lt;/code&gt; are fastest in this race, the remaining pool is slower than at least one of &lt;code&gt;oqu&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Race 12: &lt;code&gt;qurvs&lt;/code&gt;, &lt;code&gt;q=17,r=18,u=19&lt;/code&gt;. The 'best of the best' race is not needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Race 13: &lt;code&gt;svtwx&lt;/code&gt;, &lt;code&gt;s=20,t=21,v=22,w=23,x=24,y=25&lt;/code&gt;. Last 6. Race all except one of the last in its chain (say, &lt;code&gt;y&lt;/code&gt;). If the loser of race 13 (say &lt;code&gt;x&lt;/code&gt;) is also in that chain, the chain places the remaining horse last (we already know &lt;code&gt;x &amp;lt; y&lt;/code&gt;).&lt;/p&gt;&#xA;" OwnerUserId="9422" LastEditorUserId="9422" LastEditDate="2015-12-01T14:35:53.983" LastActivityDate="2015-12-01T14:35:53.983" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="24251" PostTypeId="2" ParentId="23227" CreationDate="2015-11-27T17:24:07.587" Score="14" Body="&lt;h2&gt;Answer&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes it is possible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;The (maybe) interesting details&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I started by creating a huge HTML table containing an overview of the generated values (right click the image and select &quot;View image&quot; for bigger version):&lt;br&gt;&lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/Q6EXo.png&quot; alt=&quot;table&quot;&gt;&lt;br&gt;&lt;br&gt;I ignored non integer values, therefore most of the fields are empty. The integer values form a regular pattern. The highlighted fields are the ones which keep one of the values unchanged. In addition to the 3 variants mentioned by &lt;strong&gt;GentlePurpleRain&lt;/strong&gt; above, there is one more which uses negative values.&lt;br&gt;&lt;br&gt;Using a highlighted red field it's possible to increase a single value while keeping the other unchanged. Using a highlighted blue field allows to decrease a single value. However, this is not enough to solve the puzzle. The next step was to increase the complexity. Look at the field representing [6, 17] -&gt; [15, 1]. This changes two values, but we can restore one of them in a second step, namely [15 30] -&gt; [30 6]. The effect of both these replacements is [17] -&gt; [1].&lt;br&gt;&lt;br&gt;There are more possibilities, e.g. [255, 510] -&gt; [510, 102] and [102, 289] -&gt; [255, 17] result in [289] -&gt; [17]. Together with the previous combination we have [289] -&gt; [1].&lt;br&gt;&lt;br&gt;This can be continued further, which I have done using a computer program. It starts with the highlighted fields, and in each iteration adds more possible single number transformations to the list, until it finds the transformation [2015] -&gt; [1]. See next chapter for the result.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;The ugly details&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Here is a possible sequence. It's not necessarily the shortest one, it's the one which was easiest to find.&lt;br&gt;&lt;br&gt;&#xA; &lt;code&gt;Replacement                 Surplus               Missing&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;.                           [2015]                [1]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[300, 100] -&amp;gt; [300, 260]    [260, 2015]           [1, 100]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[180, 260] -&amp;gt; [300, 100]    [300, 2015]           [1, 180]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[300, 600] -&amp;gt; [600, 120]    [120, 2015]           [1, 180]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[135, 120] -&amp;gt; [180, 96]     [96, 2015]            [1, 135]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[96, 97] -&amp;gt; [135, 65]       [65, 2015]            [1, 97]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[120, 65] -&amp;gt; [135, 97]      [135, 2015]           [1, 120]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[96, 72] -&amp;gt; [120, 72]       [135, 2015]           [1, 96]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[135, 120] -&amp;gt; [180, 96]     [180, 2015]           [1, 120]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[252, 189] -&amp;gt; [315, 189]    [180, 315, 2015]      [1, 120, 252]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[315, 180] -&amp;gt; [360, 252]    [360, 2015]           [1, 120]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[360, 720] -&amp;gt; [720, 144]    [144, 2015]           [1, 120]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[167, 144] -&amp;gt; [220, 120]    [220, 2015]           [1, 167]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[308, 231] -&amp;gt; [385, 231]    [220, 385, 2015]      [1, 167, 308]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[385, 220] -&amp;gt; [440, 308]    [440, 2015]           [1, 167]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[440, 880] -&amp;gt; [880, 176]    [176, 2015]           [1, 167]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[176, 7] -&amp;gt; [145, 167]      [145, 2015]           [1, 7]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[375, 750] -&amp;gt; [750, 150]    [145, 150, 2015]      [1, 7, 375]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[150, 425] -&amp;gt; [375, 25]     [25, 145, 2015]       [1, 7, 425]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[450, 25] -&amp;gt; [375, 425]     [145, 375, 2015]      [1, 7, 450]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[375, 250] -&amp;gt; [450, 290]    [145, 290, 2015]      [1, 7, 250]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[290, 30] -&amp;gt; [250, 270]     [145, 270, 2015]      [1, 7, 30]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[225, 450] -&amp;gt; [450, 90]     [90, 145, 270, 2015]  [1, 7, 30, 225]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[90, 255] -&amp;gt; [225, 15]      [15, 145, 270, 2015]  [1, 7, 30, 255]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[270, 15] -&amp;gt; [225, 255]     [145, 225, 2015]      [1, 7, 30]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[225, 450] -&amp;gt; [450, 90]     [90, 145, 2015]       [1, 7, 30]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[270, 90] -&amp;gt; [270, 234]     [145, 234, 2015]      [1, 7, 30]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[390, 130] -&amp;gt; [390, 338]    [145, 234, 338, 2015] [1, 7, 30, 130]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[234, 338] -&amp;gt; [390, 130]    [145, 390, 2015]      [1, 7, 30]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[145, 390] -&amp;gt; [350, 30]     [350, 2015]           [1, 7]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[315, 105] -&amp;gt; [315, 273]    [273, 350, 2015]      [1, 7, 105]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[189, 273] -&amp;gt; [315, 105]    [315, 350, 2015]      [1, 7, 189]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[224, 168] -&amp;gt; [280, 168]    [280, 315, 350, 2015] [1, 7, 189, 224]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[315, 280] -&amp;gt; [420, 224]    [350, 420, 2015]      [1, 7, 189]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[420, 315] -&amp;gt; [525, 315]    [350, 525, 2015]      [1, 7, 189]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[350, 525] -&amp;gt; [595, 189]    [595, 2015]           [1, 7]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[595, 2015] -&amp;gt; [1685, 7]    [1685]                [1]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[60, 20] -&amp;gt; [60, 52]        [52, 1685]            [1, 20]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[36, 52] -&amp;gt; [60, 20]        [60, 1685]            [1, 36]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[55, 60] -&amp;gt; [80, 36]        [80, 1685]            [1, 55]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[112, 84] -&amp;gt; [140, 84]      [80, 140, 1685]       [1, 55, 112]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[140, 80] -&amp;gt; [160, 112]     [160, 1685]           [1, 55]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[160, 320] -&amp;gt; [320, 64]     [64, 1685]            [1, 55]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[64, 23] -&amp;gt; [65, 55]        [65, 1685]            [1, 23]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[65, 130] -&amp;gt; [130, 26]      [26, 1685]            [1, 23]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[26, 7] -&amp;gt; [25, 23]         [25, 1685]            [1, 7]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[120, 40] -&amp;gt; [120, 104]     [25, 104, 1685]       [1, 7, 40]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[72, 104] -&amp;gt; [120, 40]      [25, 120, 1685]       [1, 7, 72]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[120, 240] -&amp;gt; [240, 48]     [25, 48, 1685]        [1, 7, 72]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[89, 48] -&amp;gt; [100, 72]       [25, 100, 1685]       [1, 7, 89]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[100, 25] -&amp;gt; [95, 89]       [95, 1685]            [1, 7]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[35, 95] -&amp;gt; [85, 7]         [85, 1685]            [1, 35]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[85, 170] -&amp;gt; [170, 34]      [34, 1685]            [1, 35]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[34, 13] -&amp;gt; [35, 29]        [29, 1685]            [1, 13]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[168, 126] -&amp;gt; [210, 126]    [29, 210, 1685]       [1, 13, 168]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[210, 120] -&amp;gt; [240, 168]    [29, 240, 1685]       [1, 13, 120]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[336, 252] -&amp;gt; [420, 252]    [29, 240, 420, 1685]  [1, 13, 120, 336]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[420, 240] -&amp;gt; [480, 336]    [29, 480, 1685]       [1, 13, 120]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[480, 960] -&amp;gt; [960, 192]    [29, 192, 1685]       [1, 13, 120]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[181, 192] -&amp;gt; [260, 120]    [29, 260, 1685]       [1, 13, 181]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[500, 1000] -&amp;gt; [1000, 200]  [29, 200, 260, 1685]  [1, 13, 181, 500]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[475, 200] -&amp;gt; [500, 400]    [29, 260, 400, 1685]  [1, 13, 181, 475]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[400, 300] -&amp;gt; [500, 300]    [29, 260, 500, 1685]  [1, 13, 181, 475]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[500, 125] -&amp;gt; [475, 445]    [29, 260, 445, 1685]  [1, 13, 125, 181]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[260, 445] -&amp;gt; [475, 125]    [29, 475, 1685]       [1, 13, 181]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[525, 175] -&amp;gt; [525, 455]    [29, 455, 475, 1685]  [1, 13, 175, 181]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[455, 935] -&amp;gt; [925, 175]    [29, 475, 925, 1685]  [1, 13, 181, 935]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[475, 925] -&amp;gt; [935, 197]    [29, 197, 1685]       [1, 13, 181]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[197, 29] -&amp;gt; [175, 181]     [175, 1685]           [1, 13]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[256, 192] -&amp;gt; [320, 192]    [175, 320, 1685]      [1, 13, 256]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[360, 320] -&amp;gt; [480, 256]    [175, 480, 1685]      [1, 13, 360]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[480, 960] -&amp;gt; [960, 192]    [175, 192, 1685]      [1, 13, 360]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[431, 192] -&amp;gt; [460, 360]    [175, 460, 1685]      [1, 13, 431]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[196, 147] -&amp;gt; [245, 147]    [175, 245, 460, 1685] [1, 13, 196, 431]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[245, 140] -&amp;gt; [280, 196]    [175, 280, 460, 1685] [1, 13, 140, 431]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[280, 460] -&amp;gt; [500, 140]    [175, 500, 1685]      [1, 13, 431]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[500, 175] -&amp;gt; [505, 431]    [505, 1685]           [1, 13]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[505, 1685] -&amp;gt; [1415, 13]   [1415]                [1]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[600, 200] -&amp;gt; [600, 520]    [520, 1415]           [1, 200]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[360, 520] -&amp;gt; [600, 200]    [600, 1415]           [1, 360]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[600, 1200] -&amp;gt; [1200, 240]  [240, 1415]           [1, 360]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[445, 240] -&amp;gt; [500, 360]    [500, 1415]           [1, 445]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[500, 75] -&amp;gt; [445, 459]     [459, 1415]           [1, 75]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[405, 810] -&amp;gt; [810, 162]    [162, 459, 1415]      [1, 75, 405]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[162, 459] -&amp;gt; [405, 27]     [27, 1415]            [1, 75]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[86, 27] -&amp;gt; [85, 75]        [85, 1415]            [1, 86]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[75, 150] -&amp;gt; [150, 30]      [30, 85, 1415]        [1, 75, 86]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[30, 85] -&amp;gt; [75, 5]         [5, 1415]             [1, 86]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[140, 280] -&amp;gt; [280, 56]     [5, 56, 1415]         [1, 86, 140]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[133, 56] -&amp;gt; [140, 112]     [5, 112, 1415]        [1, 86, 133]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[112, 84] -&amp;gt; [140, 84]      [5, 140, 1415]        [1, 86, 133]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[140, 5] -&amp;gt; [115, 133]      [115, 1415]           [1, 86]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[115, 230] -&amp;gt; [230, 46]     [46, 1415]            [1, 86]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[103, 46] -&amp;gt; [110, 86]      [110, 1415]           [1, 103]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[280, 210] -&amp;gt; [350, 210]    [110, 350, 1415]      [1, 103, 280]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[350, 200] -&amp;gt; [400, 280]    [110, 400, 1415]      [1, 103, 200]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[400, 800] -&amp;gt; [800, 160]    [110, 160, 1415]      [1, 103, 200]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[130, 160] -&amp;gt; [200, 80]     [80, 110, 1415]       [1, 103, 130]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[80, 110] -&amp;gt; [130, 46]      [46, 1415]            [1, 103]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[120, 40] -&amp;gt; [120, 104]     [46, 104, 1415]       [1, 40, 103]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[72, 104] -&amp;gt; [120, 40]      [46, 120, 1415]       [1, 72, 103]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[110, 120] -&amp;gt; [160, 72]     [46, 160, 1415]       [1, 103, 110]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[224, 168] -&amp;gt; [280, 168]    [46, 160, 280, 1415]  [1, 103, 110, 224]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[280, 160] -&amp;gt; [320, 224]    [46, 320, 1415]       [1, 103, 110]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[320, 640] -&amp;gt; [640, 128]    [46, 128, 1415]       [1, 103, 110]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[128, 46] -&amp;gt; [130, 110]     [130, 1415]           [1, 103]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[112, 84] -&amp;gt; [140, 84]      [130, 140, 1415]      [1, 103, 112]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[140, 80] -&amp;gt; [160, 112]     [130, 160, 1415]      [1, 80, 103]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[130, 160] -&amp;gt; [200, 80]     [200, 1415]           [1, 103]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[600, 200] -&amp;gt; [600, 520]    [520, 1415]           [1, 103]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[520, 1415] -&amp;gt; [1265, 103]  [1265]                [1]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[120, 40] -&amp;gt; [120, 104]     [104, 1265]           [1, 40]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[72, 104] -&amp;gt; [120, 40]      [120, 1265]           [1, 72]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[168, 126] -&amp;gt; [210, 126]    [120, 210, 1265]      [1, 72, 168]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[210, 120] -&amp;gt; [240, 168]    [240, 1265]           [1, 72]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[145, 240] -&amp;gt; [260, 72]     [260, 1265]           [1, 145]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[364, 273] -&amp;gt; [455, 273]    [260, 455, 1265]      [1, 145, 364]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[455, 260] -&amp;gt; [520, 364]    [520, 1265]           [1, 145]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[520, 1265] -&amp;gt; [1175, 145]  [1175]                [1]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[750, 1500] -&amp;gt; [1500, 300]  [300, 1175]           [1, 750]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[300, 850] -&amp;gt; [750, 50]     [50, 1175]            [1, 850]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[900, 50] -&amp;gt; [750, 850]     [750, 1175]           [1, 900]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[750, 1500] -&amp;gt; [1500, 300]  [300, 1175]           [1, 900]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[1025, 300] -&amp;gt; [1000, 900]  [1000, 1175]          [1, 1025]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[1000, 2000] -&amp;gt; [2000, 400] [400, 1175]           [1, 1025]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[400, 1175] -&amp;gt; [1025, 55]   [55]                  [1]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[300, 600] -&amp;gt; [600, 120]    [55, 120]             [1, 300]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[285, 120] -&amp;gt; [300, 240]    [55, 240]             [1, 285]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[240, 155] -&amp;gt; [285, 187]    [55, 187]             [1, 155]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[105, 35] -&amp;gt; [105, 91]      [55, 91, 187]         [1, 35, 155]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[91, 187] -&amp;gt; [185, 35]      [55, 185]             [1, 155]&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;[55, 185] -&amp;gt; [155, 1]       []                    []&lt;/code&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2015-11-27T18:21:08.460" LastActivityDate="2015-11-27T18:21:08.460" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24252" PostTypeId="1" AcceptedAnswerId="24253" CreationDate="2015-11-28T05:34:02.037" Score="4" ViewCount="1002" Body="&lt;p&gt;Here is a puzzle I have not yet been able to solve:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are three people A, B and C who know each others identity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may pick A, B or C &#xA;show him a logic statement (truth functional compounds only such as AND, OR, IF THEN, IF and only if) about their identities. (example: [A is a Knight OR B is Not a Normal])&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He will then state whether or not the statement is true. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They can consist of: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Knights&lt;/strong&gt;, who always tell the truth; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Knaves&lt;/strong&gt;, who always lie; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Normals&lt;/strong&gt;, who can do either one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There can be at &lt;em&gt;MOST&lt;/em&gt; one Normal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal is to ask the fewest questions (each one directed at either A, B or C) to definitively identify all three. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I suspect there are many ways to approach the puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried listing all possible combinations (30) of Knights, Knaves and Normals (that can tell the truth/lie), and then showing each one a statement. If I know the statement is true (A1=Knave) and I show it to a Knave who never tells the truth, I will get back a FALSE. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For my first question, I have shown A: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;[A is a Night AND B is NOT a Normal] OR [A is NOT a Night AND B is a Normal]. I use this because if I get a TRUE I know that B can't be normal, and if I get false it ends up that C can't be normal. This gave me my first stepping stone into the puzzle but you may think there is better. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you do it in 7 questions? (I have heard 5 is possible)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please ask if you are curious/have more questions. Happy Thanksgiving!&lt;/p&gt;&#xA;" OwnerUserId="17572" LastEditorUserId="12408" LastEditDate="2016-07-25T02:35:50.483" LastActivityDate="2017-01-19T16:10:25.167" Title="Identify A B &amp; C using the least number of directed logic statements (Knights, Knaves, Normals)" Tags="|logical-deduction|liars|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24253" PostTypeId="2" ParentId="24252" CreationDate="2015-11-28T05:50:12.283" Score="2" Body="&lt;p&gt;Here's a way with six questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Show A: &quot;A is a Knight if and only if B is a Normal.&quot;&lt;/li&gt;&#xA;&lt;li&gt;Show B: &quot;B is a Knight if and only if C is a Normal.&quot;&lt;/li&gt;&#xA;&lt;li&gt;Show C: &quot;C is a Knight if and only if A is a Normal.&quot;&lt;/li&gt;&#xA;&lt;li&gt;Show A: &quot;A is a Knight or A is a Knave.&quot;&lt;/li&gt;&#xA;&lt;li&gt;Show B: &quot;B is a Knight or B is a Knave.&quot;&lt;/li&gt;&#xA;&lt;li&gt;Show C: &quot;C is a Knight or C is a Knave.&quot;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The first three questions identify whether there is a Normal, and who it is if there is one:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If there are no Normals, all three answer FALSE.&lt;/li&gt;&#xA;&lt;li&gt;If A is a Normal, B answers FALSE and C answers TRUE.&lt;/li&gt;&#xA;&lt;li&gt;If B is a Normal, C answers FALSE and A answers TRUE.&lt;/li&gt;&#xA;&lt;li&gt;If C is a Normal, A answers FALSE and B answers TRUE.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The other three questions identify each person who isn't a normal. A Knight answers TRUE while a Knave answers FALSE.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;With the restriction that questions can't be changed based on earlier answers, five questions is impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we were able to use five questions to identify all three, then each of the 32 sequences of TRUEs and FALSEs should correspond to at most one assignment of identities. Either we ask one person three or more questions, or we ask two people two questions each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we ask one person three questions. If that person is a Normal, they can answer whatever they want for those three questions, so each case takes up eight sequences of answers. There are four cases in which that person is a Normal. These cases would have to take up a a total of 32 sequences, leaving none for the cases in which that person is not a Normal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, we ask two people two questions each. Similarly, each case in which one of them is a Normal must take up four sequences of answers. There are eight cases in which one of them is a Normal, again taking up all 32 sequences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way, we cannot use five questions to identify all three people.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2015-11-29T04:44:16.683" LastActivityDate="2015-11-29T04:44:16.683" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24254" PostTypeId="2" ParentId="24252" CreationDate="2015-11-28T06:22:41.003" Score="0" Body="&lt;p&gt;This can be done in exactly five questions every time, provided your first question can find you at least one non-normal.  Thanks to @Evan for already providing such a question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this helpful?  Because now we could show our non-normal (let's assume it's B) any statement, XOR it with &quot;B is a Knave&quot;, and be certain that our answer was the truth.  As you can see, if B has to be a Knight or a Knave:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;B Knight, XYZ &lt;strong&gt;True&lt;/strong&gt;: [XYZ] XOR [B Knave] is True, so B says &quot;&lt;strong&gt;True&lt;/strong&gt;&quot;&lt;br&gt;&#xA;B Knight, XYZ &lt;strong&gt;False&lt;/strong&gt;: [XYZ] XOR [B Knave] is False, so B says &quot;&lt;strong&gt;False&lt;/strong&gt;&quot;&lt;br&gt;&#xA;B Knave, XYZ &lt;strong&gt;True&lt;/strong&gt;: [XYZ] XOR [B Knave] is False, so B says &quot;&lt;strong&gt;True&lt;/strong&gt;&quot;&lt;br&gt;&#xA;B Knave, XYZ &lt;strong&gt;False&lt;/strong&gt;: [XYZ] XOR [B Knave] is True, so B says &quot;&lt;strong&gt;False&lt;/strong&gt;&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result, every statement can split the set of cases exactly in half by making a statement that is true for exactly half of them.  And, conveniently, there are exactly 16 cases once we know somebody is not normal - again assuming B is not normal:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A has 3 choices) x (B has 2 choices) x (C has 3 choices) = 18 possible cases, but Normal Knave Normal and Normal Knight Normal break the rules, so only 16 of those cases are valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, our initial question statement leaves us with 16 cases, a second statement (if well chosen) can leave us with 8 cases for each answer, the third leaves us with 4 per set of answers, the fourth leaves us with 2, and the fifth uniquely identifies our situation.  Now that we know it's possible, we simply need to find a set of statements that encompasses all possible strings of true/false answers - which, for five consecutive statements is 2^5 = 32 possibilities.  I'll be going for simplicity over clarity, so my statements will not all follow the form &quot;XYZ&quot; XOR &quot;Letter is a Knave&quot;, though they could be formatted as such if one wished.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll create some notation here for efficiency:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[Number][Truth Values][Person]:[Statement]&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the [Number]'th statement shown, in the event that the previous sequential answers were the [Truth Values] indicated, it is being shown to [Person], and the statement is [Statement].  If one can determine all of the roles given a resulting truth value for a statement, I will denote it as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Number][Truth Values][Person][Single Truth Value] - [A Role], [B Role], [C Role]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where [Single Truth Value] was [Person]'s response to statement [Number] with previous answers of [Truth Values].  Note that a conclusion may show up multiple times by way of different paths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Letter] [Role] is an assertion that [Letter] is [Role] &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Example:&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A is a Normal  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ask A if A is a Knight, and A says Yes.  I ask A if A is a Knight a second time, and A says No.  This is recorded as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1A: [A Knight]&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;2TA: [A Knight]&lt;/strong&gt;&lt;br&gt;&#xA;2TAF - Normal  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, the statements:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1A: [A Knight AND B NotNormal] OR [A NotKnight AND B Normal]&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;2TB: [A Knave OR C Normal] XOR [B Knave]&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;3TTA: [B Knight] XOR [A Knave]&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;4TTTB: [C Normal]&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;5TTTTB: [A Knight]&lt;/strong&gt;&lt;br&gt;&#xA;5TTTTBT - Knight, Knight, Normal&lt;br&gt;&#xA;5TTTTBF - Knave, Knight, Normal&lt;br&gt;&#xA;&lt;strong&gt;5TTTFB: [C Knight]&lt;/strong&gt;&lt;br&gt;&#xA;5TTTFBT - Knave, Knight, Knight&lt;br&gt;&#xA;5TTTFBF - Knave, Knight, Knave&lt;br&gt;&#xA;&lt;strong&gt;4TTFB: [C Normal]&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;5TTFTB: [C Knight]&lt;/strong&gt;&lt;br&gt;&#xA;5TTFTBT - Knave, Knave, Knave&lt;br&gt;&#xA;5TTFTBF - Knave, Knave, Knight&lt;br&gt;&#xA;&lt;strong&gt;5TTFFB: [A Knight]&lt;/strong&gt;&lt;br&gt;&#xA;5TTFFBT - Knave, Knave, Normal&lt;br&gt;&#xA;5TTFFBF - Knight, Knave, Normal&lt;br&gt;&#xA;&lt;strong&gt;3TFB: [C Knight] XOR [B Knave]&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;4TFTC: [A Knight]&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;5TFTTC: [B Knight]&lt;/strong&gt;&lt;br&gt;&#xA;5TFTTCT - Knight, Knight, Knight&lt;br&gt;&#xA;5TFTTCF - Knight, Knave, Knight&lt;br&gt;&#xA;&lt;strong&gt;5TFTFC: [B Knight]&lt;/strong&gt;&lt;br&gt;&#xA;5TFTFCT - Normal, Knight, Knight&lt;br&gt;&#xA;5TFTFCF - Normal, Knave, Knight&lt;br&gt;&#xA;&lt;strong&gt;4TFFC: [A Knight]&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;5TFFTC: [B Knight]&lt;/strong&gt;&lt;br&gt;&#xA;5TFFTCT - Normal, Knave, Knave&lt;br&gt;&#xA;5TFFTCF - Normal, Knight, Knave&lt;br&gt;&#xA;&lt;strong&gt;5TFFFC: [B Knight]&lt;/strong&gt;&lt;br&gt;&#xA;5TFFFCT - Knight, Knave, Knave&lt;br&gt;&#xA;5TFFFCF - Knight, Knight, Knave&lt;br&gt;&#xA;&lt;strong&gt;2FC: [A Knave OR B Normal] XOR [C Knave]&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;3FTA: [C Knight] XOR [A Knave]&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;4FTTC: [B Normal]&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;5FTTTC: [A Knight]&lt;/strong&gt;&lt;br&gt;&#xA;5FTTTCT - Knight, Normal, Knight&lt;br&gt;&#xA;5FTTTCF - Knave, Normal, Knight&lt;br&gt;&#xA;&lt;strong&gt;5FTTFC: [B Knight]&lt;/strong&gt;&lt;br&gt;&#xA;5FTTFCT - Knave, Knight, Knight&lt;br&gt;&#xA;5FTTFCF - Knave, Knave, Knight&lt;br&gt;&#xA;&lt;strong&gt;4FTFC: [B Normal]&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;5FTFTC: [B Knight]&lt;/strong&gt;&lt;br&gt;&#xA;5FTFTCT - Knave, Knave, Knave&lt;br&gt;&#xA;5FTFTCF - Knave, Knight, Knave&lt;br&gt;&#xA;&lt;strong&gt;5FTFFC: [A Knight]&lt;/strong&gt;&lt;br&gt;&#xA;5FTFFCT - Knave, Normal, Knave&lt;br&gt;&#xA;5FTFFCF - Knight, Normal, Knave&lt;br&gt;&#xA;&lt;strong&gt;3FFC: [B Knight] XOR [C Knave]&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;4FFTB: [A Knight]&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;5FFTTB: [C Knight]&lt;/strong&gt;&lt;br&gt;&#xA;5FFTTBT - Knight, Knight, Knight&lt;br&gt;&#xA;5FFTTBF - Knight, Knight, Knave&lt;br&gt;&#xA;&lt;strong&gt;5FFTFB: [C Knight]&lt;/strong&gt;&lt;br&gt;&#xA;5FFTFBT - Normal, Knight, Knight&lt;br&gt;&#xA;5FFTFBF - Normal, Knight, Knave&lt;br&gt;&#xA;&lt;strong&gt;4FFFB: [A Knight]&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;5FFFTB: [C Knight]&lt;/strong&gt;&lt;br&gt;&#xA;5FFFTBT - Normal, Knave, Knave&lt;br&gt;&#xA;5FFFTBF - Normal, Knave, Knight&lt;br&gt;&#xA;&lt;strong&gt;5FFFFB: [C Knight]&lt;/strong&gt;&lt;br&gt;&#xA;5FFFFBT - Knight, Knave, Knave&lt;br&gt;&#xA;5FFFFBF - Knight, Knave, Knight  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To use this list, just show the first statement to the indicated person (in this case, A) - if you get a T, jump to 2T and show that statement to the indicated person, if you get a false, jump to 2F and show that statement to that indicated person instead.  Continue on until you get an answer for your fifth statement, all possibilities of which lead to conclusions.  You can test the veracity of these conclusions by picking any case (A = ?, B = ?, C = ?) and running through the questions to confirm that you end up at the correct conclusion.&lt;/p&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="17389" LastEditDate="2015-11-28T20:39:24.563" LastActivityDate="2015-11-28T20:39:24.563" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="24255" PostTypeId="1" AcceptedAnswerId="24264" CreationDate="2015-11-28T06:31:45.540" Score="0" ViewCount="414" Body="&lt;p&gt;I am more elegant than others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looks like one of animal but I am still mystery and not exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have things which my duplicate haven't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Impressive for children, Found in their thought and dreams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They also like to play with me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Royal Arm for heraldic with heavy wings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;" OwnerUserId="4312" LastEditorUserId="4312" LastEditDate="2015-11-30T12:42:30.613" LastActivityDate="2015-11-30T12:42:30.613" Title="Elegant with sharp thing, Who am I?" Tags="|riddle|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24256" PostTypeId="1" CreationDate="2015-11-28T15:38:55.947" Score="8" ViewCount="1179" Body="&lt;p&gt;So there are five beautiful ladies whose faces are veiled. &#xA;You can distinguish them well as they have written their (correct) names Alice, Betty, Cindy, Diana and Emily on their T-shirts.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You know that two of them have brown eyes, and always tell the truth.&lt;/li&gt;&#xA;&lt;li&gt;You know that three of them have blue eyes, and always lie.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Alice, Betty and Cindy are brought in front of you.&#xA;You may ask each of them them a single question.&#xA;Based on the three answers, you must guess the eye colors of all five ladies correctly. &#xA;How do you manage to succeed?&lt;/p&gt;&#xA;" OwnerUserId="17574" LastEditorUserId="8874" LastEditDate="2015-11-29T16:12:05.230" LastActivityDate="2018-10-11T06:01:12.360" Title="Five beautiful ladies" Tags="|logical-deduction|liars|" AnswerCount="9" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24257" PostTypeId="1" AcceptedAnswerId="24258" CreationDate="2015-11-28T16:29:42.947" Score="21" ViewCount="1272" Body="&lt;p&gt;The referee tells Lefty, Righty, and Middly: &lt;em&gt;&quot;I have picked three positive integers $L,M,R$ with $L&amp;lt;M&amp;lt;R$ and $L+M+R=13$, and I have written these numbers onto the three cards lying face down on the table. The left card carries value $L$, the middle card carries value $M$, and the right card carries value $R$.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Lefty picks up the left card, looks at it and ponders a while about it.&#xA;Then he announces to the others: &lt;em&gt;&quot;I am not able to determine the numbers $M$ and $R$.&quot;&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Then Righty picks up the right card, studies the value $R$ and thinks deeply about it. Then Righty announces: &lt;em&gt;&quot;I am not able to deduce the numbers $L$ and $M$.&quot;&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Finally Middly takes the card in the middle and thinks about the value $M$.&#xA;After some time he says: &lt;em&gt;&quot;I am not able to deduce the numbers $L$ and $R$.&quot;&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What is the value of $M$? And what can be said about the values of $L$ and $R$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-12-04T07:36:23.173" Title="The mysterious card in the middle" Tags="|mathematics|logical-deduction|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24258" PostTypeId="2" ParentId="24257" CreationDate="2015-11-28T16:53:53.167" Score="23" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; M is 4&lt;br&gt;&#xA; L and R can be 1 and 8 or 2 and 7&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And here's the solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lefty's turn:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; L can only be 1, 2 or 3. If L were 4 or more, then M would have to be at least 5 and R at least 6, which sums up to 15, so L is at most 3.&lt;br&gt;&#xA; If L were 3, then M and R would be 4 and 6 respectively and then Lefty would be able to guess the value. So L is 1 or 2.&lt;br&gt;&#xA; If L were 2, then M+R would be 11, leaving 3/8, 4/7 and 5/6 as possible M/R values.&lt;br&gt;&#xA; If L were 1, then M+R would be 12, leaving 2/10, 3/9, 4/8, 5/7 as possible M/R values.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now let's go to Righty:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If R were 11 or more, then L and M would have to be 0 or negative, so R is at most 10.&#xA; If R is 10, then L+M must be 3, which only leaves 1 and 2 as possible values, but Righty couldn't deduce it, so R is 9 or less. Also, as a perfect logician, Righty would know that L is 1 or 2, because if it were 3, then Lefty would know the answer.&lt;br&gt;&#xA; If R is 9, then L+M must be 4, leaving only 1 and 3 as possible values for L and M. So R must be 8 or lower.&lt;br&gt;&#xA; If R is 8, then L+M = 5, with possible values of: 1/4, 2/3&lt;br&gt;&#xA; If R is 7, then L+M = 6, possible values: 1/5, 2/4&lt;br&gt;&#xA; If R is 6, then L+M = 7, possible values: 2/5&lt;br&gt;&#xA; If R is 5, then L+M = 8, no possible values&lt;br&gt;&#xA; So L is [1,2] and R is [7,8]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Do the same for Middly:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As a perfect logician he would know than L is 1 or 2 and R is 7 or 8.&lt;br&gt;&#xA; So M must be between 2 and 7, L can only be 1 or 2 and R can only be 7 or 8:&lt;br&gt;&#xA; If M is 2, then L+R = 11, possible values: none&lt;br&gt;&#xA; If M is 3, then L+R = 10, possible values: 2/8&lt;br&gt;&#xA; If M is 4, then L+R = 9, possible values: 1/8, 2/7&lt;br&gt;&#xA; If M is 5, then L+R = 8, possible values: 1/7&lt;br&gt;&#xA; If M is 6, then L+R = 7, possible values: none&lt;br&gt;&#xA; If M is 7, then L+R = 6, possible values: none&lt;br&gt;&#xA; From here we can deduce that M is not 3 or 5. And because Middly could not deduce it, we can see that M must be 4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Possible solutions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    L    M    R&lt;br&gt;&#xA;    1    4    8&lt;br&gt;&#xA;    2    4    7&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9352" LastEditorUserId="10781" LastEditDate="2015-11-28T20:53:44.907" LastActivityDate="2015-11-28T20:53:44.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24259" PostTypeId="2" ParentId="24256" CreationDate="2015-11-29T05:37:42.483" Score="3" Body="&lt;p&gt;There are 10 possible ways that TTFFF can be ordered, so this is provably impossible if we ask questions with only yes/no answers (as 3 questions can only distinguish &lt;span class=&quot;math-container&quot;&gt;$2^3 = 8$&lt;/span&gt; scenarios).  However, nowhere in these rules has it been stated that we may only ask such questions.  Thus, I will go ahead and blatantly ask a ridiculously complicated compound question.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Question 1, to Girl 1:&lt;/strong&gt; Assuming the color brown was defined as &quot;true&quot; and the color blue was defined as &quot;false&quot;, XORing your answer for each girl with the truth value of the statement &quot;Girl 1 Always Lies&quot;, what truth value would the eyes of each of the five girls here individually be defined as?&lt;br&gt;&#xA;&lt;br&gt;&#xA; For every girl where she says &quot;True&quot;, they have brown eyes.  For every girl where she says &quot;False&quot;, they have blue eyes.  Since we asked her to respond individually, she must lie or tell the truth for each individual sub-answer, and in every case her lie/truth is canceled out by the XORing process, leaving only the truth.  We're done, in a single question!&lt;br&gt;&#xA;&lt;br&gt;&#xA;  &lt;em&gt;Alternatively, if you like slightly simpler questions:&lt;/em&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;  &lt;strong&gt;Question 1, to Girl 1:&lt;/strong&gt; Is it true that &quot;Girl 2 has brown eyes&quot; XOR &quot; you always lie&quot;?&lt;br&gt;&#xA;&lt;br&gt;&#xA;  &lt;strong&gt;Question 2a, to Girl 2:&lt;/strong&gt; What is the pairing of names and eye colors for the five girls here?&lt;br&gt;&#xA;&lt;br&gt;&#xA;  &lt;strong&gt;Question 2b, to Girl 2:&lt;/strong&gt; What is &lt;em&gt;not&lt;/em&gt; the pairing of names and eye colors for the five girls here?&lt;br&gt;&#xA;&lt;br&gt;&#xA;  If you got a &quot;Yes&quot; for question 1, ask 2a.  If you got a &quot;No&quot;, ask 2b.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="11110" LastEditDate="2018-10-09T07:48:06.250" LastActivityDate="2018-10-09T07:48:06.250" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="24260" PostTypeId="2" ParentId="24257" CreationDate="2015-11-29T08:25:21.140" Score="3" Body="&lt;p&gt;Wrote the following python script to help solve this puzzle:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;cards = set()&#xA;&#xA;for l in range(1, 13):&#xA;    for m in range(l + 1, 13):&#xA;        r = 13 - m - l&#xA;        if r &amp;gt; m:&#xA;            hand = frozenset((l, m, r))&#xA;            if len(hand) == 3:&#xA;                cards.add(hand)&#xA;&#xA;cards = sorted(tuple(sorted(list(hand))) for hand in cards)&#xA;&#xA;print('L M R ')&#xA;for hand in cards:&#xA;    print '%d %d %d' % hand&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;prints out:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; L M R  &lt;br/&gt;&#xA; 1 2 10 &lt;br/&gt;&#xA; 1 3 9 &lt;br/&gt;&#xA; 1 4 8 &lt;br/&gt;&#xA; 1 5 7 &lt;br/&gt;&#xA; 2 3 8 &lt;br/&gt;&#xA; 2 4 7 &lt;br/&gt;&#xA; 2 5 6 &lt;br/&gt;&#xA; 3 4 6 &lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So you can easily see Lefty must have seen a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 or 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And then Right had to have seen a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 8 or 7&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And therefore Middly must have seen a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And so the only solutions could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 4 8 &lt;br/&gt;&#xA; 2 4 7&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastEditorUserId="10118" LastEditDate="2015-11-29T09:01:04.070" LastActivityDate="2015-11-29T09:01:04.070" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24261" PostTypeId="1" AcceptedAnswerId="24262" CreationDate="2015-11-29T15:44:38.477" Score="-1" ViewCount="376" Body="&lt;p&gt;No backstory or hints here. Just a symbol.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/zyhuX.jpg&quot; alt=&quot;quite a strange one&quot; /&gt;&#xA;What does this symbol mean?&lt;/p&gt;&#xA;&lt;p&gt;Any theories are welcome too.&lt;/p&gt;&#xA;" OwnerUserId="17251" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-11-29T17:35:32.300" Title="What is this symbol?" Tags="|knowledge|symbols|" AnswerCount="1" CommentCount="5" ClosedDate="2015-11-30T04:01:21.947" ContentLicense="CC BY-SA 3.0" />
  <row Id="24262" PostTypeId="2" ParentId="24261" CreationDate="2015-11-29T16:56:25.977" Score="4" Body="&lt;p&gt;I'm not sure, but I think it is ...&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Early form of an &lt;strong&gt;Chinese letter&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Also here is a photo of it:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/wsobK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wsobK.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;BTW, It looks like a ancient pictograms either. I don't know, maybe if you &lt;a href=&quot;https://www.google.lv/search?q=ancient+pictograms&amp;amp;num=20&amp;amp;source=lnms&amp;amp;tbm=isch&amp;amp;sa=X&amp;amp;ved=0ahUKEwjr2ZzChrbJAhVC8XIKHXg_CvQQ_AUIBygB&amp;amp;biw=1408&amp;amp;bih=692&quot; rel=&quot;nofollow noreferrer&quot;&gt;google it&lt;/a&gt; then you find some similar symbol.&lt;/p&gt;&#xA;" OwnerUserId="17438" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-11-29T17:12:46.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24264" PostTypeId="2" ParentId="24255" CreationDate="2015-11-29T19:54:13.353" Score="4" Body="&lt;p&gt;It seems like a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Unicorn&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am more elegant than others.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It really looks elegant.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am looks like one of animal but I am still mystery and not exist.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It looks like horse but still it do not exist.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have things which my duplicate haven't.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It have a horn but we cannot find it in horse.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Impressive for children, Found in their thought and dreams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They also like to play with me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Children love it (even i love them). They play with their toys.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Royal Arm for heraldic with heavy wings.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is shown as a wing for Scotland in the Royal Arm for heraldic.&#xA; Example..as below image&#xA; &lt;img src=&quot;https://i.stack.imgur.com/lgib4.png&quot; alt=&quot;![enter image description here&quot;&gt;]&lt;a href=&quot;https://i.stack.imgur.com/lgib4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Elegant with sharp thing&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Its sharp thing is it's horn.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="4312" LastEditDate="2015-11-30T10:24:28.623" LastActivityDate="2015-11-30T10:24:28.623" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24265" PostTypeId="1" AcceptedAnswerId="24266" CreationDate="2015-11-29T22:23:02.113" Score="9" ViewCount="2411" Body="&lt;p&gt;On the 15-puzzle, what (solvable) position takes the most moves to solve if you solve it optimally? &lt;/p&gt;&#xA;" OwnerUserId="16225" LastEditorUserId="69582" LastEditDate="2021-06-22T15:14:45.897" LastActivityDate="2021-06-22T15:14:45.897" Title="What is the superflip on 15-puzzle?" Tags="|optimization|sliding-blocks|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24266" PostTypeId="2" ParentId="24265" CreationDate="2015-11-29T23:20:20.870" Score="14" Body="&lt;p&gt;The hardest positions on the 15-puzzle require $80$ moves to solve (where a move consists of sliding a single tile). Here is an example of a position requiring $80$ moves:&#xA;$$&#xA;\begin{array}{|c|c|c|c|}&#xA;\hline&#xA;15&amp;amp;14&amp;amp;8&amp;amp;12\\\hline&#xA;10&amp;amp;11&amp;amp;9&amp;amp;13\\\hline&#xA;2&amp;amp;6&amp;amp;5&amp;amp;1\\\hline&#xA;3&amp;amp;7&amp;amp;4&amp;amp;\\\hline&#xA;\end{array}&#xA;$$&#xA;This position (and several others) can be found in Ralph Gasser's PhD thesis from 1995 [G], where it is proved that $80$ moves are necessary. A few years later, it was proved in [BMFN] that no position requires more moves. Both proofs are aided by computers.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;[BMFN] Brüngger, Adrian; Marzetta, Ambros; Fukuda, Komei; Nievergelt, Jurg. The parallel search bench ZRAM and its applications. &lt;em&gt;Ann. Oper. Res.&lt;/em&gt; &lt;strong&gt;90&lt;/strong&gt; (1999), 45-63.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[G] Gasser, Ralph Udo. Harnessing Computational Resources&#xA;for Efficient Exhaustive Search. PhD Thesis, ETH Zürich, 1995.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2015-11-29T23:30:25.933" LastActivityDate="2015-11-29T23:30:25.933" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24267" PostTypeId="1" AcceptedAnswerId="24268" CreationDate="2015-11-30T00:34:30.087" Score="2" ViewCount="364" Body="&lt;p&gt;There was a puzzle game on ZX-Spectrum computer called &quot;Logo: Part I&quot;: &lt;a href=&quot;http://www.worldofspectrum.org/pub/sinclair/screens/in-game/l/Logo-PartI.gif&quot; rel=&quot;nofollow&quot;&gt;http://www.worldofspectrum.org/pub/sinclair/screens/in-game/l/Logo-PartI.gif&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, you can put tokens on the board; the newly put token becomes 1; old neighbours are increased; if a neighbour was 4, it becomes 1 again; the goal is to get the specific configuration for each level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd want to know if there are other names or implementations of this puzzle. Is it well-known?&lt;/p&gt;&#xA;" OwnerUserId="1804" LastEditorUserId="1804" LastEditDate="2015-11-30T01:38:11.837" LastActivityDate="2015-11-30T01:38:11.837" Title="Looking for a specific puzzle game" Tags="|puzzle-identification|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24268" PostTypeId="2" ParentId="24267" CreationDate="2015-11-30T00:48:34.657" Score="5" Body="&lt;p&gt;Welcome to Puzzling.SE!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to be a variant of &lt;a href=&quot;http://mathworld.wolfram.com/LightsOutPuzzle.html&quot;&gt;Lights Out&lt;/a&gt;, except switches that haven't been pressed don't 'exist' yet.&lt;/p&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="11876" LastEditDate="2015-11-30T00:54:22.430" LastActivityDate="2015-11-30T00:54:22.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24269" PostTypeId="1" AcceptedAnswerId="24281" CreationDate="2015-11-30T00:52:08.467" Score="10" ViewCount="1227" Body="&lt;p&gt;You are a lone builder. You have to build a city on a 10x10 initial empty grid. You can place:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a street, which is a blue cell on the top of any cell,&lt;/li&gt;&#xA;&lt;li&gt;a tomato shop, which is a red cell on any cell having a street in one of his 4-neighbors,&lt;/li&gt;&#xA;&lt;li&gt;a small house, which is a green cell on any cell having a street and a tomato shop in two of his 4-neighbors,&lt;/li&gt;&#xA;&lt;li&gt;a tall building, which is a yellow cell on any cell having a street, a tomato shop and a house in three of his 4-neighbors.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These rules are valid during the construction step only: if a cell is replaced by another one, it has to comply the rules, but its neighbors can break them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is an example of a possible city:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WAQF7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WAQF7.png&quot; alt=&quot;The city achieved currently by our builder&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is easy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How many yellow tall buildings can you build in the city, and in how many steps?&lt;/strong&gt; An answer is considered better than another one if it maximizes the number of buildings, or if the number is the same but it does it in a lower number of steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, if you would be given a 3x3 city, you could build 4 buildings in 12 steps:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NlCZX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NlCZX.png&quot; alt=&quot;How to build a 3x3 city in 12 steps&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the little story, I got inspired from a now unavailable Android game: &lt;a href=&quot;https://www.google.fr/search?hl=en&amp;amp;q=com.amusetime.buildingtower&amp;amp;gws_rd=cr&amp;amp;ei=fZpbVpDiNsXxaKPrkNAO&quot; rel=&quot;nofollow noreferrer&quot;&gt;com.amusetime.buildingtower&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="13589" LastEditorUserId="15632" LastEditDate="2015-11-30T22:03:32.350" LastActivityDate="2015-11-30T22:03:32.350" Title="The lone builder" Tags="|strategy|computer-puzzle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24270" PostTypeId="1" CreationDate="2015-11-30T03:40:24.147" Score="-3" ViewCount="2043" Body="&lt;p&gt;With the given numbers I need to make:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$7 \circ 8 \circ 7 \circ 7 \circ 8 \circ 3 = 100$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using only the following operators:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$+\space-\space\times\space\div$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17586" LastEditorUserId="7308" LastEditDate="2015-11-30T06:45:59.420" LastActivityDate="2017-10-28T22:34:35.913" Title="Fill in the operators to make $7 \circ 8 \circ 7 \circ 7 \circ 8 \circ 3 = 100$" Tags="|mathematics|calculation-puzzle|formation-of-numbers|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24271" PostTypeId="2" ParentId="24269" CreationDate="2015-11-30T03:48:59.010" Score="2" Body="&lt;p&gt;First approximation of an answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;46 Yellow squares in 125 actions.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's label the rows 0 - 90 and the columns 0 - 9, thus defining each square in the grid from 00 - 99.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Color blue mostly in pairs - 00, 01, 04, 05, 08, 09, 22, 23, 26, 27, 40, 41, 44, 45, 48, 62, 63, 66, 67, 80, 81, 84, 85, 88, 89.  Note that square 49 is &lt;em&gt;not&lt;/em&gt; colored. This takes 25 actions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Additionally, color in 10, 19, 39, 50, 59, 93, 95, and 96 blue - this takes 8 actions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Color in 90, and 99 red - this takes 2 actions (note that 80, and 89 are blue).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) In rows 0, 2, 4, 6, and 8, color all the un-colored squares red (again excluding 49) - this takes 24 actions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) color in the following squares green - 11, 13, 15, 17, 19, 31, 33, 35, 37, 51, 53, 55, 57, 71, 73, 75, 77, 79, 91, 97.  This takes 20 actions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6) Color in every square vertically or horizontally adjacent to a green square as yellow, excluding 09 and 90.  This takes 46 actions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Done.  All squares except 49 and 94 are colored at this point.&lt;/p&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="17389" LastEditDate="2015-11-30T05:59:23.203" LastActivityDate="2015-11-30T05:59:23.203" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24272" PostTypeId="2" ParentId="24270" CreationDate="2015-11-30T03:57:44.407" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$7 \times 8 + 7 \times 7 - 8 + 3 = 100$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-11-30T03:57:44.407" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24273" PostTypeId="2" ParentId="1980" CreationDate="2015-11-30T04:08:43.537" Score="0" Body="&lt;p&gt;C calls out that he is wearing a black hat. Why is he 100% certain of the color of his hat?&#xA;After a while, C comes to the realization that he must answer.&#xA;This is because D can't answer, and neither can A or B.&#xA;D can see C and B, but can't determine his own hat color. B can't see anyone and also can't determine his own hat color. A is in the same situation as B, where he can't see anyone and can't determine his own hat color.&#xA;Since A, B, and D are silent, that leaves C. C knows he is wearing a black hat because if D saw that both B and C were wearing white hats, then he would have answered. But since D is silent, C knows that he must be wearing a black hat as he can see that B is wearing a white hat.&lt;/p&gt;&#xA;" OwnerUserId="17587" LastActivityDate="2015-11-30T04:08:43.537" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24274" PostTypeId="1" AcceptedAnswerId="24279" CreationDate="2015-11-30T12:47:14.763" Score="10" ViewCount="1203" Body="&lt;p&gt;Fredo owns a balance with two pans and 212 weights, each of 1 gram.&#xA;Fredo proudly announces to Cosmo that his weight system has the following stunning property:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For every integer weight $W$ with $1\le W\le212$, there is a &lt;strong&gt;unique&lt;/strong&gt;&#xA;  way of getting a perfect equilibrium with $W$ in the left pan and some &#xA;  of the weights in the right pan.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cosmo is not impressed: &lt;em&gt;&quot;What do you mean by a &lt;strong&gt;unique&lt;/strong&gt; way of getting &#xA;a perfect equilibrium?&quot;&lt;/em&gt;&lt;br&gt;&#xA;To which Fredo answers: &lt;em&gt;&quot;It is unique, if we ignore the ordering of the &#xA;weights and if we do not distinguish between weights of equal value.&quot;&lt;/em&gt;&lt;br&gt;&#xA;Cosmo is still not impressed: &lt;em&gt;&quot;But there must be dozens of weight systems&#xA;with that property!&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is Cosmo right? How many other weight systems are there whose weights are integers that add up to $212$ and that have Fredo's special property?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-12-02T02:24:47.847" Title="212 weights of 1 gram" Tags="|mathematics|weighing|" AnswerCount="4" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="24275" PostTypeId="2" ParentId="24060" CreationDate="2015-11-30T13:25:41.020" Score="3" Body="&lt;p&gt;I think the answer is 4 times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the first bell, if either has &gt;8 the answer is 15 and there is no second bell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the second bell, if either has &gt;5 the answer is 8 and there is no third bell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the third bell, both know that each has 5 or less. For the answer to be 8, the numbers must be 4 and 4 or 5 and 3. If either has 0, 1 or 2 the answer will be 5 and there will be no fourth bell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fourth bell sounds only if the answer is 8.&lt;/p&gt;&#xA;" OwnerUserId="17439" LastActivityDate="2015-11-30T13:25:41.020" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24276" PostTypeId="2" ParentId="24274" CreationDate="2015-11-30T14:06:02.173" Score="4" Body="&lt;p&gt;Let's supose there is an alternative weight system other than all weights being one. How should it look like?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's suppose we have n integer weights w1,...,wn such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;w1 + w2 + ... + wn = 212, n &amp;lt;= 212 and let's see if we can construct a different system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can arrange our weights without losing generality:&#xA;w1 &amp;lt;= w2 &amp;lt;= ... &amp;lt;= wn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that w1 = 1, because we need to express 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, let k be the lowest integer weight we have above one. Then we have, sorted, this weights:&#xA;1, 1, 1, 1,..., 1, k,...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then wi = k: If i &amp;lt; k, k-1 is not expressible as the sum of 1's (there are not enough ones). If i &gt; k, k is expressible both as k and as a sum of ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leaves us with a structure like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1, 1, 1, ..., 1, k1, k1, k1, ..., k1, k2, ..., ks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where k1 appears at position k1.&#xA;What numbers can we express so far?&#xA;Only with ones: 1, 2, ..., k1-1&#xA;With k1: k1, k1+1,..., 2*k1-1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means we need a new weight k2 to be 2*k1.&#xA;What numbers can we express?&#xA;Only with ones: 1, 2, ..., k1-1&#xA;With k1: k1, k1+1,..., 2*k1-1&#xA;With k2: k2, k2+1,..., k2 + k1 + (k1-1) = 4*k1 - 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means we need a new k3 to be 4*k1, and k4 to be 8*k1, etc.&#xA;So we have a base-2 system of weights:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;k1-1 ones: 1, 1, 1, ...&#xA;k1,&#xA;2*k1&#xA;4*k1,&#xA;...&#xA;2^r*k1, for some integer r.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This adds to k1-1 + k1*(1+2+..+2^r) for some r, so we have this equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;212 = k - 1 + k*(1+2+..+2^r), or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;k * (1 + 1+ 2 +4 + ... + 2^r) = 213 = 3*71. Let's be q = 1 + 1+ 2 +...+2^r,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;k * q = 3*71, all being integers and 3 and 71 being prime numbers, there are only two possibilities:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If k = 3, then q = 71 for some r. And if k = 71, q = 3 for some r.&#xA;If you calculate q, q = 1 + (sum of powers of 2 until r) = 1 + 2^(r+1)-1 = 2^(r+1). So q is a power of two, and none of 3 and 71 are powers of two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leaves us with an impossible system other than all weights being one.&lt;/p&gt;&#xA;" OwnerUserId="17558" LastActivityDate="2015-11-30T14:06:02.173" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24277" PostTypeId="1" AcceptedAnswerId="24278" CreationDate="2015-11-30T14:24:22.217" Score="3" ViewCount="3396" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;$9, 27, 31, 155, 161, 1127, ?$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I observed that the first term and second term respectively are $3^2$ and $3^3$. But I have no idea about the subsequent terms. &lt;br&gt;&#xA;One more observation I did was, the difference in all of these: &#xA;$$9, +18= 27,+4=31,+124=155,+6=161,+966=1127.$$&#xA;We see the after each alternate number, the difference is increasing: $4,6,...$ so next should be $8$, which gives the answer as $1135$. Which is surprisingly correct. But what is the pattern in the middle numbers?&lt;/p&gt;&#xA;" OwnerUserId="11425" LastActivityDate="2015-11-30T14:27:41.430" Title="Find the next term in the number series?" Tags="|number-sequence|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24278" PostTypeId="2" ParentId="24277" CreationDate="2015-11-30T14:27:41.430" Score="14" Body="&lt;p&gt;It seems to be x3, +4, x5, +6, x7, +8 and so on&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2015-11-30T14:27:41.430" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24279" PostTypeId="2" ParentId="24274" CreationDate="2015-11-30T14:46:53.777" Score="12" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are 3 weight systems with these properties.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's try looking at it from the biggest weights first.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $w$ be the biggest weight of one such system, and let $n$ be the number of $w$-weights in the system. Let the remainder $r = 212 - n \times w$. Then $n$ is maximal (subject to $n \times w \leq 212$) and $r = w-1$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Indeed, if $n$ were not maximal, then $r \geq w$. We can do division with remainder to find $p\geq 1$ and $0\leq q &amp;lt; w$ with $r = p \times w + q$. There is a unique combination of weights that add up to $q$, and none of these weights are $w$-weights (because $q &amp;lt; w$). We can use this to derive a contradiction of the uniqueness of an $r$-weighting:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Take $p\times$ $w$-weights and the unique $q$-weighting (or nothing, if $q = 0$).&lt;/li&gt;&#xA;&lt;li&gt;Take all weights except $w$-weights; by hypothesis they add up to $r$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It follows that $n$ is maximal and $r &amp;lt; w$. Now, if $r &amp;lt; w-1$, no weighting would be able to produce a $(w-1)$-weighting, so it must be $r = w-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this in mind, our formula then becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{equation}n \times w +r = 212\end{equation}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Leftrightarrow n \times w + (w-1) = 212$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Leftrightarrow (n+1) \times w = 213$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words our biggest weights in a system must be integer divisors of $212 + 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only numbers this is true for are the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{lll}&#xA;    w=1&amp;amp;   n=212&amp;amp; r=0\\&#xA;    w=3&amp;amp;   n=70&amp;amp;  r=2\\&#xA;    w=71&amp;amp;  n=2&amp;amp;   r=70&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, 213 would also fit, but we wouldn't get a single weight of that size and the remainder would still be 212, so nothing would be gained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for how we can fill in the remainder with weights, the next biggest weigth in the remainder must fulfill the same properties as the original biggest weigth. Therefore, we can use the same rules as we first used (except for $w=1$, because in that case $r=0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $r+1$ is a prime number in both cases, the complete remainder must be made out of $1$-weigths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, our systems are the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$212\times$ $1$-weights;&lt;/li&gt;&#xA;&lt;li&gt;$70\times$ $3$-weights and $2\times$ $1$-weights;&lt;/li&gt;&#xA;&lt;li&gt;$2\times$ $71$-weights and $70\times$ $1$-weights.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="16608" LastEditorUserId="16608" LastEditDate="2015-12-01T07:52:44.327" LastActivityDate="2015-12-01T07:52:44.327" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24280" PostTypeId="2" ParentId="23722" CreationDate="2015-11-30T16:02:40.950" Score="6" Body="&lt;p&gt;Dr. Mysterious name is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jeff&lt;br&gt;&#xA; Without repeating work from the previous answers, per part 1 and part 2, we have &quot;The key for part 3 is 12321476.&quot;&lt;br&gt;&#xA; The fact that the clues mention Vigenere, but have a numeric key suggests a Gronsfeld cipher. The URL obtained from this is gibberish though. Using Gronsfeld with a keyed alphabet, however, reveals the answer; the alphabet key is YOUMUSTFINDHISNAME (which reduces down to YOUMSTFINDHAE once you removed duplicate letters). Using this in conjunction with the numeric key gives us a URL: &lt;a href=&quot;https://www.youtube.com/watch?v=AfIOBLr1NDU&quot;&gt;https://www.youtube.com/watch?v=AfIOBLr1NDU&lt;/a&gt; . The title of this video is &quot;My Name is Jeff&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Many thanks to Illyasviel for the work on Part 1 and 2!&lt;/p&gt;&#xA;" OwnerUserId="16874" LastEditorUserId="16874" LastEditDate="2015-11-30T16:17:32.867" LastActivityDate="2015-11-30T16:17:32.867" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24281" PostTypeId="2" ParentId="24269" CreationDate="2015-11-30T16:18:20.110" Score="12" Body="&lt;h2&gt;&lt;s&gt;56&lt;/s&gt; 57 yellow squares / 149 steps&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First place the blue, red and green squares like in the following three images (the black X marks new squares):&lt;br&gt;&lt;br&gt;&lt;a href=&quot;https://i.stack.imgur.com/K4rgS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/K4rgS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&lt;br&gt;Then fill the green spiral with yellow buildings, starting top left. Add the 4 green buildings from the middle image. Finally fill the remaining yellow buildings from the right image.&lt;br&gt;&lt;br&gt;&lt;a href=&quot;https://i.stack.imgur.com/BtR7e.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BtR7e.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2015-11-30T21:54:00.817" LastActivityDate="2015-11-30T21:54:00.817" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24282" PostTypeId="2" ParentId="24274" CreationDate="2015-11-30T17:36:05.090" Score="1" Body="&lt;p&gt;To be able to weigh objects of integer weight, including those of weight 1, the weights must form a set which in ascending order has the property that the first has weight 1 and each subsequent one has either the same weight as the preceding one, or the sum of all preceding weights plus 1. Weights of any other value, if included in the set, would allow the weight of some objects to be expressed with at least two different combinations of weights.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that if there are n1 weights of weight 1, the next weight must be of weight n1+1 and if there are n2 of these, the next heaviest must be of weight n2(n1+1)+n1+1 which factorises as (n2+1)(n1+1). The weight of the ith heaviest of the set will thus be (n(i-1)+1)(n(i-2)+1)...(n1+1) and the heaviest object that can be weighed will be (n(i)+1)(n(i-1)+1)(n(i-2)+1)...(n1+1)-1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A set of weights capable of weighing an object of maximum weight 212 must thus consist of n(i)+n(i-1)+n(i-2)...+n1 weights of which&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(n(i)+1)(n(i-1)+1)(n(i-2)+1)...(n1+1)-1=212&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding 1 to each side, the task is thus to factorise 213, the prime factors of which are 1, 3, 71 and 213.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are only three sets which satisfy this, Fredo's set where (212+1)-1=212 and the sets (70+1)(2+1)-1=212 and (2+1)(70+1)-1=212. These correspond to a set containing 2 weights of weight 1 and 70 weights of weight 3 (2x1+1) and one containg 70 weights of weight 1 and 2 weights of weight 71 (70x1+1).&lt;/p&gt;&#xA;" OwnerUserId="17439" LastEditorUserId="17439" LastEditDate="2015-11-30T17:59:45.190" LastActivityDate="2015-11-30T17:59:45.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24283" PostTypeId="1" CreationDate="2015-11-30T20:54:20.450" Score="5" ViewCount="2379" Body="&lt;p&gt;Can anyone help me find out what these kinds of puzzles are?&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;&lt;strong&gt;1. B&amp;amp;W Grid Puzzle&lt;/strong&gt;&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;There's a grid with black &amp;amp; white blocks. A black block will turn white if it's surrounded by 3 white blocks. A white block will turn black if it's surrounded by 2 or 3 black blocks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Example&lt;/em&gt;: If the initial shape is a &quot;cross&quot; made by 5 black blocks, what are the possible results given shapes A, B, C, and D? (See image below)&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;&lt;strong&gt;2. Foetus Puzzle&lt;/strong&gt;&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I can't remember exactly how this one works, but generally the contents of an animal's womb is determined by labels depending on the sex of adjacent foetuses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rules:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If foetus A is beside 1 male foetus, A is labeled 1M.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If foetus B is beside 2 male foetuses, B is labeled 2M.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If foetus C is beside 0 male foetuses, C is labeled 0M.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They may be mapped out or have related questions, e.g. &quot;What is probability of getting &lt;code&gt;1M 2M 0M&lt;/code&gt;?&quot; or &quot;If there are 2 males and 2 females, what constructions are considered impossible?&quot;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;&lt;strong&gt;3. Colored Grid Puzzle&lt;/strong&gt;&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;There is a 3x3 grid and each square must be filled with a certain colour. There are 3 colours (red, green, and blue) to work with, but there aren't necessarily 3 of each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Example&lt;/em&gt;: With the following statements, you have to guess which colour is in column 3, row 2.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1) The square in Column 1, Row 2 is red.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;2) In Column 2, there are 2 green squares and 1 blue square.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;... etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;&lt;strong&gt;Notes&lt;/strong&gt;&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I couldn't recall the questions in detail but I hope I explained the type of logic puzzles I'm looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5WQOv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5WQOv.png&quot; alt=&quot;puzzle examples&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17594" LastEditorUserId="7586" LastEditDate="2016-03-16T23:42:14.747" LastActivityDate="2017-09-23T16:13:57.493" Title="Can anyone help me identify these logic puzzles?" Tags="|puzzle-identification|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24284" PostTypeId="1" AcceptedAnswerId="24289" CreationDate="2015-11-30T21:41:14.210" Score="12" ViewCount="569" Body="&lt;p&gt;Below are three groups of clues. Within each group, the answers share a common characteristic. The groups also share a common characteristic. Solving any one group should make it much easier to solve the others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Group A&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Trend toward fervor&lt;/li&gt;&#xA;  &lt;li&gt;Packing a teddy bear for your first sleepover&lt;/li&gt;&#xA;  &lt;li&gt;Aborigine traditional dress&lt;/li&gt;&#xA;  &lt;li&gt;Tongue-twister&lt;/li&gt;&#xA;  &lt;li&gt;Predilection toward deities&lt;/li&gt;&#xA;  &lt;li&gt;One who hugs programmers&lt;/li&gt;&#xA;  &lt;li&gt;Forgiveness for killing someone&lt;/li&gt;&#xA;  &lt;li&gt;Uncertainty about rainforests&lt;/li&gt;&#xA;  &lt;li&gt;A lie about involvement in a crime&lt;/li&gt;&#xA;  &lt;li&gt;Predisposition toward rage&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Group B&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Tantrum play-by-play&lt;/li&gt;&#xA;  &lt;li&gt;Advocation for cremation&lt;/li&gt;&#xA;  &lt;li&gt;Flawed point of view&lt;/li&gt;&#xA;  &lt;li&gt;Declaration of independence&lt;/li&gt;&#xA;  &lt;li&gt;Plan for improved air quality&lt;/li&gt;&#xA;  &lt;li&gt;Investigator-in-training&lt;/li&gt;&#xA;  &lt;li&gt;Plan for a new waste-treatment facility&lt;/li&gt;&#xA;  &lt;li&gt;Gemstone perplexity&lt;/li&gt;&#xA;  &lt;li&gt;Careful look into the sudden change in leadership&lt;/li&gt;&#xA;  &lt;li&gt;Act of getting covered for eventualities  &lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Group C&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Limitation put on a photograph  &lt;/li&gt;&#xA;  &lt;li&gt;Discouraged group of firefighters&lt;/li&gt;&#xA;  &lt;li&gt;Document exempting human traffickers&lt;/li&gt;&#xA;  &lt;li&gt;Less-than-brilliant hue&lt;/li&gt;&#xA;  &lt;li&gt;Religious ceremonies of young ladies&lt;/li&gt;&#xA;  &lt;li&gt;Very accurate gadget&lt;/li&gt;&#xA;  &lt;li&gt;Expressing a poem non-vocally&lt;/li&gt;&#xA;  &lt;li&gt;Well-ploughed section of a field&lt;/li&gt;&#xA;  &lt;li&gt;Intelligent choice for private air travel&lt;/li&gt;&#xA;  &lt;li&gt;Item in the Munitions Hall of Fame&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="4421" LastEditDate="2015-12-03T19:01:23.810" LastActivityDate="2015-12-03T19:01:23.810" Title="Word Categories" Tags="|word|pattern|wordplay|rhyme|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="24286" PostTypeId="1" AcceptedAnswerId="24287" CreationDate="2015-12-01T04:27:00.723" Score="8" ViewCount="363" Body="&lt;p&gt;I was looking up some quotations the other day by a famous person.  The following three quotations were all made by the same person, but somehow they got mixed up before I could read them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you tell me what happened to them, what the quotations are, and who said them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Quotation 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;The sis maize imp puller ill legion.  They risen own eat fort dimples; known eat fork ump lick hated fill us if he.  A rowin' bray, now roan art a sour temp pull. Duff ill us iffy ask eye and nest.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Quotation 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Harp rhyme purr puss sin thus lie fist ooh hell paw thirds. Sand diff few can tell path emmett leased owned heard dumb.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Quotation 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Weak enough fur robed hay imp ease sin theatre whorl dent ill owe he may cap ease withers elves.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2015-12-01T05:46:48.427" Title="Mixed-up quotations" Tags="|wordplay|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24287" PostTypeId="2" ParentId="24286" CreationDate="2015-12-01T04:34:53.343" Score="6" Body="&lt;p&gt;1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;This is my simple religion. There is no need for temples; no need for complicated philosophy. Our own brain, our own heart is our temple; the philosophy is kindness.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Our prime purpose in this life is to help others. And if you can't help them, at least don't hurt them.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;We can never obtain peace in the outer world until we make peace with ourselves.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These are all quotes from&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the Dalai Lama.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and they've been modified&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; by taking phonetically similar words.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="11876" LastEditDate="2015-12-01T05:46:48.427" LastActivityDate="2015-12-01T05:46:48.427" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24288" PostTypeId="2" ParentId="24256" CreationDate="2015-12-01T12:55:05.303" Score="11" Body="&lt;p&gt;You ask Alice, Betty and Cindy &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;What are the colour of Diana and Emily's eyes, in that order?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If ABC are all liars, they will answer &quot;blue, blue&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If two of ABC are truth-tellers, then the truth-tellers will say &quot;blue, blue&quot;, while the liar will say &quot;brown, brown&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one of ABC are truth-tellers, the truth-teller will say either &quot;brown, blue&quot; or &quot;blue, brown&quot;, while &lt;em&gt;both&lt;/em&gt; the liars will say the opposite. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By counting the answers you get you should be able to deduce what colour the eyes of the girls are.&lt;/p&gt;&#xA;" OwnerUserId="17599" LastActivityDate="2015-12-01T12:55:05.303" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24289" PostTypeId="2" ParentId="24284" CreationDate="2015-12-01T15:19:42.910" Score="7" Body="&lt;p&gt;This is what I've done and gathered from the comments (wiki)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Group A&#xA;1. Trend toward fervor                                | ✓ intensity propensity&#xA;2. Packing a teddy for your first sleepover           | ✓ separation preparation  &#xA;3. Aborigine traditional dress                        | ✓ Australia Regalia&#xA;4. Tongue-twister                                     | ✓ elocution convolution&#xA;5. Predilection toward deities                        | ✓ divinity affinity&#xA;6. One who hugs programmers                           | ✓ developer enveloper&#xA;7. Forgiveness for killing someone                    | ✓ execution absolution&#xA;8. Uncertainty about rainforests                      | ✓ humidity timidity&#xA;9. A lie about involvement in a crime                 | ✓ complicity duplicity -or- perpetration fabrication&#xA;10. Predisposition toward rage                        | ✓ lividity proclivity&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;-&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Group B &#xA;1. Tantrum play-by-play                               | ✓ conniption description&#xA;2. Advocation for cremation                           | ✓ interment determent&#xA;3. Flawed point of view                               | ✓ defective perspective&#xA;4. Declaration of independence                        | ✓ secession expression&#xA;5. Plan for improved air quality                      | ✓ pollution solution&#xA;6. Investigator-in-training                           | ✓ prospective detective&#xA;7. Plan for a new waste-treatment facility            | ✓ disposal proposal&#xA;8. Gemstone perplexity                                | ✓ corundum conundrum -or- jewelry foolery&#xA;9. Careful look into the sudden change in leadership  | ✓ mutiny scrutiny&#xA;10. Act of getting covered for eventualities          | ✓ insurance procurance&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;-&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Group C     &#xA;1. Limitation put on a photograph                     | ✓ picture stricture&#xA;2. Discouraged group of firefighters                  | ✓ dismayed brigade&#xA;3. Document exempting human traffickers               | ✓ slaver waiver &#xA;4. Less-than-brilliant hue                            | ✓ duller colour &#xA;5. Religious ceremonies of young ladies               | ✓ lasses' masses&#xA;6. Very accurate gadget                               | ✓ precise device    &#xA;7. Expressing a poem non-vocally                      | ✓ miming rhyming&#xA;8. Well-ploughed section of a field                   | ✓ thorough furrow   &#xA;9. Intelligent choice for private air travel          | ✓ smarter charter   &#xA;10. Item in the Munitions Hall of Fame                | ✓ displayed grenade&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="51" LastEditorUserId="4421" LastEditDate="2015-12-03T16:12:17.843" LastActivityDate="2015-12-03T16:12:17.843" CommentCount="8" CommunityOwnedDate="2015-12-01T15:19:42.910" ContentLicense="CC BY-SA 3.0" />
  <row Id="24291" PostTypeId="2" ParentId="24256" CreationDate="2015-12-01T15:30:24.263" Score="3" Body="&lt;h2&gt;Only one question needed&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Assuming we are not restricted to T/F questions, we could ask just one question, &quot;What would one of the other 4 with an opposite eye colour to yours say are the people in this group with blue eyes?&quot; A liar would invert the truth, and a truth-teller would truthfully report an inversion. In either case, the named people would have brown eyes and the remainder, blue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using @kevinpieter's idea of counting, the (single) question can be simplified to just &quot;Who are the people in this group with blue eyes?&quot; - if three people are named, the named people have blue eyes and the other two, brown; if two people are named, the named people have brown eyes and the other three, blue.&lt;/p&gt;&#xA;" OwnerUserId="9422" LastEditorUserId="9422" LastEditDate="2015-12-02T00:03:53.740" LastActivityDate="2015-12-02T00:03:53.740" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24292" PostTypeId="1" AcceptedAnswerId="24311" CreationDate="2015-12-01T18:05:50.150" Score="3" ViewCount="3300" Body="&lt;p&gt;Right now I am trying to get the following distorted position:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On each face only one diagonal is solved and the rest of the cubes are a different colour than the diagonal. For example: If I am looking at the white face, there should exactly be two opposite corner pieces in that face correctly oriented, and no other white piece should be there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If this is possible, what is an algorithm to get there?&lt;/p&gt;&#xA;" OwnerUserId="6217" LastEditorUserId="4421" LastEditDate="2015-12-01T18:27:50.520" LastActivityDate="2015-12-03T21:53:04.187" Title="Algorithm for a single diagonal on each face of the Rubik's Cube" Tags="|rubiks-cube|algorithm|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24293" PostTypeId="2" ParentId="24256" CreationDate="2015-12-01T18:17:03.820" Score="0" Body="&lt;p&gt;Ask Alice one question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Treating blue-eyed girls (liars) as all having the value 0 and any brown-eyed girls (truthtellers) as respectively having the values: you (Alice) 1, Betty 2, Cindy 4, Diana 8, and Emily 16, what is the sum of all girls' values?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assuming that the lying girls can only comprehensively lie (and cannot mix truth and falsehood together so as to deceive more subtly), the number given will, when converted to binary, show a bit pattern containing either three zeroes and two ones, or three ones and two zeroes. Given that we know there are three liars, the liars are represented by the digit, be it &lt;code&gt;0&lt;/code&gt; or &lt;code&gt;1&lt;/code&gt;, with three occurrences, and the truthtellers are represented by the digit with two occurrences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples of Alice's possible answers:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 17, 10001: the two 1s represent truthtellers, so Alice and Emily have brown eyes.&#xA; 19, 10011: the two 0s represent truthtellers, so Betty and Cindy have brown eyes.&#xA;  9, 01001: impossible because Alice cannot falsely say there are 3 liars.&#xA; 22, 10110: impossible because Alice cannot truthfully say there are 3 truthtellers.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="10155" LastEditorUserId="10155" LastEditDate="2015-12-01T18:22:52.107" LastActivityDate="2015-12-01T18:22:52.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24294" PostTypeId="2" ParentId="24256" CreationDate="2015-12-01T19:03:35.867" Score="1" Body="&lt;p&gt;Going off what kevinpieter wrote:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... and this doesn't work since you can't tell Diana and Emily apart.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ask the three girls &quot;Do Diana and Emily both have brown eyes?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If all three answer the same thing, we know all three of them are the same type (and thus liars, which means they must have all said &quot;No&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If two say the same thing and the third person says another thing, we know one of two things:&lt;br&gt;&#xA;1) Two truth tellers and one liar were asked (in this case we know the truth tellers said &quot;No&quot; and the liar said &quot;Yes&quot;)&lt;br&gt;&#xA;2) Two liars and one truth teller were asked (in this case we know the liars said &quot;Yes&quot; and the truth teller said &quot;No&quot;)&lt;/p&gt;&#xA;" OwnerUserId="12092" LastActivityDate="2015-12-01T19:03:35.867" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24295" PostTypeId="2" ParentId="24256" CreationDate="2015-12-01T19:23:32.093" Score="0" Body="&lt;p&gt;You just need two simple questions. Also, this question needs less male gaze :p&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ask Alice,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;If I were to ask you whether Betty had brown eyes, would you say yes?&lt;/strong&gt;&#xA; No matter what eye color Alice has, she will answer yes if and only if Betty has brown eyes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, ask Betty,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;What are [not] the eye colors of all five of you?&lt;/strong&gt; If Betty has brown eyes, you just ask her what everybody's eye colors are, and you're done. If Betty has blue eyes, you ask her what are &lt;em&gt;not&lt;/em&gt; everybody's eye colors, so that only the correct answer will be a lie. Either way, Betty will report the eye colors of all five.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9523" LastActivityDate="2015-12-01T19:23:32.093" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24296" PostTypeId="1" AcceptedAnswerId="24297" CreationDate="2015-12-01T19:32:42.387" Score="7" ViewCount="1011" Body="&lt;p&gt;When the volume of the sound is low, Adam walks home from school.  When the volume of the sound is high, he gets a ride.  Why?&lt;/p&gt;&#xA;" OwnerUserId="17605" LastEditorUserId="17605" LastEditDate="2015-12-01T21:16:22.050" LastActivityDate="2015-12-02T13:33:40.207" Title="Adam commutes to school" Tags="|lateral-thinking|situation|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24297" PostTypeId="2" ParentId="24296" CreationDate="2015-12-01T19:35:41.063" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sound is a body of water.&lt;br&gt;&#xA; When the tide is low, he can walk.&lt;br&gt;&#xA; When the tide is high, he goes via boat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="375" LastEditorUserId="375" LastEditDate="2015-12-01T21:04:52.690" LastActivityDate="2015-12-01T21:04:52.690" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24302" PostTypeId="1" AcceptedAnswerId="24303" CreationDate="2015-12-01T20:43:29.983" Score="1" ViewCount="225" Body="&lt;p&gt;Below is a list of encoded information regarding popular flavor texts of MtG cards. The numbers listed refer to the number of letters in each word of the text. Punctuation has been preserved. The following acronym represents the name of the card. Using this info, can you name the cards?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;1 4 1 6 4 3!&quot; &lt;strong&gt;G.T.&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;9, 2 3 4. &lt;strong&gt;D.C.&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;4 9 3. &lt;strong&gt;G.O.&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;5 5 1'1 5 5 4!&quot; &lt;strong&gt;M.O.O.&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;9 3 2-6!&quot; &lt;strong&gt;I.H.G.T.&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;3 4 2 3 8. &lt;strong&gt;R.G.&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;3! 4, 3! 2'1 3 8!&quot; &lt;strong&gt;L.&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;4 6 5, 3. &lt;strong&gt;P.T.&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;2 3 8 1 5 5 2 3 5 5.&quot; &lt;strong&gt;C.&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;4 7 4 4 3'1 6. &lt;strong&gt;H.&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;14...2 12? &lt;strong&gt;B.G.&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;1 4 4 3 4 2 1 7 2 4.&quot; &lt;strong&gt;W.&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Some notes:&lt;br&gt;&#xA;- Quotes keep the quotation marks, but I've dropped all attributions included in the flavor text.&lt;br&gt;&#xA;- Some cards on the list have multiple printings, some with various differing flavor texts. The flavor text I reference may not necessarily be on all printings of the associated card. You don't have to give the edition if you don't want, but are of course welcome to do so.&lt;br&gt;&#xA;- Since this is a multi-part puzzle, to avoid the issue of not knowing who to attribute the accepted answer to in the probable event of split answers, I will make a community answer to keep the focus on reminiscing on sweet/funny/awesome flavor texts.&lt;/p&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="13970" LastEditDate="2015-12-01T21:22:12.013" LastActivityDate="2015-12-02T10:17:28.797" Title="Magic: the Gathering: Flavor Test" Tags="|magic-the-gathering|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="24303" PostTypeId="2" ParentId="24302" CreationDate="2015-12-01T20:45:42.293" Score="5" Body="&lt;ul&gt;&#xA;&lt;li&gt;&quot;1 4 1 6 4 3!&quot; G.T.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I want a banana this big&quot; (&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=29980&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gorilla Titan&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;9, 2 3 4. D.C.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Greatness, at any cost. (&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=397731&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dark Confidant&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;4 9 3. G.O.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They certainly are. (&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=8818&quot; rel=&quot;nofollow noreferrer&quot;&gt;Goblin Offensive&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&quot;5 5 1'1 5 5 4!&quot; M.O.O.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Guess where I'm gonna plant this!&quot; (&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=13148&quot; rel=&quot;nofollow noreferrer&quot;&gt;Might of Oaks, 7th Edition&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&quot;9 3 2-6!&quot; I.H.G.T.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Everybody but me - CHARGE!&quot; (&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=157923&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ib Halfheart, Goblin Tactician&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;3 4 2 3 8. R.G.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The root of all greevils. (&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=8827&quot; rel=&quot;nofollow noreferrer&quot;&gt;Root Greevil&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&quot;3! 4, 3! 2'1 3 8!&quot; L.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ach! Hans, run! It's the Lhurgoyf!&quot; (&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=247151&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lhurgoyf&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;4 6 5, 3. P.T.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When nature calls, &lt;em&gt;run&lt;/em&gt;. (&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=397688&quot; rel=&quot;nofollow noreferrer&quot;&gt;Primeval Titan&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&quot;2 3 8 1 5 5 2 3 5 5.&quot; C.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;It was probably a lousy spell in the first place.&quot; (&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=4693&quot; rel=&quot;nofollow noreferrer&quot;&gt;Counterspell, Tempest&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;4 7 4 4 3'1 6. H.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When killing five just isn't enough. (&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=247382&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hex&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;14...2 12? B.G.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reinforcements... or refreshments? (&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=393819&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bottle Gnomes&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&quot;1 4 4 3 4 2 1 7 2 4.&quot; W.  &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It doesn't exist in gatherer but it's a promo (Magic Player Rewards) version of Wasteland:&#xA; &lt;a href=&quot;https://i.stack.imgur.com/bGBvR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bGBvR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="1748" LastEditDate="2015-12-02T10:17:28.797" LastActivityDate="2015-12-02T10:17:28.797" CommentCount="3" CommunityOwnedDate="2015-12-01T20:45:42.293" ContentLicense="CC BY-SA 3.0" />
  <row Id="24305" PostTypeId="2" ParentId="24240" CreationDate="2015-12-01T21:12:42.907" Score="0" Body="&lt;p&gt;I'm gonna post my solution for 31 damage, to see if anyone can (dis/im)prove it.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Play &lt;strong&gt;1B&lt;/strong&gt; as land. Mana: B Life: 20&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Play &lt;strong&gt;B&lt;/strong&gt; as Dark Ritual. Mana: BBB Life: 20&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Play &lt;strong&gt;1B&lt;/strong&gt; as Carnival of Souls. Mana: B Life: 20&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Play &lt;strong&gt;1&lt;/strong&gt; as Myr Moonvessel, trigger Carnival. Mana: B Life: 19&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Play &lt;strong&gt;B&lt;/strong&gt; as Culling the Weak, trigger Moonvessel. Mana: 1BBBB Life: 19&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Play &lt;strong&gt;B&lt;/strong&gt; as Blood Celebrant, trigger Carnival. Mana: 1BBBB Life: 18&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Play &lt;strong&gt;BBB&lt;/strong&gt; as Infernal Contract, draw &lt;strong&gt;B&lt;/strong&gt;,&lt;strong&gt;1G&lt;/strong&gt;,&lt;strong&gt;1WW&lt;/strong&gt;,&lt;strong&gt;R&lt;/strong&gt;. Mana: 1B Life: 9&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Play &lt;strong&gt;B&lt;/strong&gt; as Death's Shadow, trigger Carnival. Mana: 1B Life: 8&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Activate Celebrant. Mana: 1G Life: 7&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Play &lt;strong&gt;1G&lt;/strong&gt; as Nest Invader, get spawn, trigger Carnival. Mana: BB Life: 5&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Sac spawn. Mana: 1BB Life: 5&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Activate Celebrant twice. Mana: 1WW Life: 3&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Play &lt;strong&gt;1WW&lt;/strong&gt; as Silverblade Paladin, soulbond with Death's Shadow, trigger Carnival. Mana: B Life: 2&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Activate Celebrant. Mana: R Life: 1&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Play &lt;strong&gt;R&lt;/strong&gt; as Burst of Speed. Mana: 0 Life: 1&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Attack for 1(Celebrant)+2(Invader)+2*2(Paladin)+12*2(Shadow)=31 damage.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13970" LastActivityDate="2015-12-01T21:12:42.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24307" PostTypeId="2" ParentId="24256" CreationDate="2015-12-01T23:05:51.560" Score="10" Body="&lt;p&gt;A nasty thing about liars is that when asked complex questions, they may answer in whatever fashion, other than 100% truthfulness, would be most vexing.  That is a substantial complication which other answers fail to take into account, but the problem is still solvable as stated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming T identifies the brown-eyed girls and F the blue-eyed ones, ask Alice...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Is the sequence one of TFTFF FFFTT TFFTF TFFFT FTFTF FTFFT?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After that, ask Betty...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Is the sequence one of FTTFF FFFTT TFFTF TFFFT FTFTF FTFFT?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If both answer &quot;yes&quot;, then ask Cindy...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Who else has brown eyes?&quot;  You know Cindy has brown eyes, and she will truthfully identify the other brown-eyed person.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Otherwise, ask Cindy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Does Emily have Brown eyes?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cindy will be lying, but her answer along with those of Alice and Betty will reveal everything you need to know:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If either Alice or Betty says &quot;yes&quot; and the other &quot;no&quot;, the one that said &quot;yes&quot; has brown eyes and the other blue; Diana will be opposite Emily, and Cindy will say (falsely) which of D/E have blue eyes.  If neither says yes, their eyes match, and Diana's eyes will match Emily's.  Cindy's answer will indicate (falsely) whether AB or DE is the blue-eyed pair.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1467" LastActivityDate="2015-12-01T23:05:51.560" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24308" PostTypeId="1" AcceptedAnswerId="24312" CreationDate="2015-12-02T00:28:50.350" Score="16" ViewCount="1317" Body="&lt;p&gt;Ernie is a renowned polyglot. Why just last week I saw him helping out at a Mathematics conference by acting as a simultaneous translator for seven guests who only spoke Hungarian, Icelandic, Swahili, Tongan, Arabic, Cantonese and Basque respectively. So Imagine my surprise last night when we went to a newly opened French Bistro and Ernie had to ask the waiter to translate the blackboard menu, which consisted of very elementary French such as soupe du jour and porc avec pommes de terre. So after a fine main course I asked why he  didn't know any French. &quot;Ahh...&quot;, he said, with a look of reminiscence on his face, &quot;it may not be fair, but I still blame the Geek Girl for that&quot;. Even for Ernie, that was a somewhat opaque statement so I asked him to explain a little more. And this is what he told me...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It all began on day one of my first Senior School Maths Olympiad. On&#xA;  the team for another school I saw the cutest geek girl you could&#xA;  imagine. I was smitten by the way her braces glinted in the sunlight.&#xA;  But the path to true love is not an easy one - I was younger than any&#xA;  of the other contestants and to my dismay she ignored me totally. At&#xA;  lunch time the older students played a game called 'Slice of Pi'. The&#xA;  rules were simple:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; The caller would shout out &quot;Slice of Pi&quot; followed by one or more pairs &#xA; of integers *(p1,q1), (p2,q2)...(pn,qn)*. &#xA; The responder had to reply with a single string of digits consisting of&#xA; the *q1* digits of pi starting at the *p1th* place after teh decimal &#xA; point, concatenated with the *q2* digits of pi starting at digit&#xA; *p2*,...etc. right out to *qn* digits starting at place *pn*.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;So,&quot; I asked, if someone said &quot;Slice of Pi (3,3), (1,1) (7,2) the correct response would be 159165?&quot; Ernie paused for several seconds (longer than I would normally expect for an answer to a simple maths question) and then replied &quot;Correct - and an adjudicator would confirm the answer was right using a paper table of pi from an old maths book&quot;. He continued...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;One of the boys was almost always the winner. &quot;He has memorized all&#xA;  the digits out to the 200th decimal place&quot;, claimed the Geek Girl with&#xA;  a look of adulation on her face. I instantly decided this was my&#xA;  chance to impress her - &quot;Oh that's nothing&quot;, I boasted, &quot;I bet I could&#xA;  learn all the way out to the 2000th place by tomorrow morning!&quot; The&#xA;  others laughed in disbelief. &quot;Then we will test you on that tomorrow!&quot;&#xA;  said the team leader. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;On my way home in the bus that night I realized what a task I had set&#xA;  myself. Like any sensible engineer I only knew 9 digits of pi and even&#xA;  if I could learn a digit every twenty seconds it would take more than&#xA;  eight hours to memorize the remaining 1991. I&#xA;  also needed to swot up for the next day's Programming Contest and knew&#xA;  that without a good night's sleep my performance would be dismal. But&#xA;  then I found the solution to all my problems in the issue of Popular&#xA;  Science where I found an article on how 'sleep-learning' experiments&#xA;  had been trialed back in the 1950's - as soon as I read the&#xA;  Introduction I was convinced it would work! &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;By the time I got home I had the plan in place. Luckily, my computer&#xA;  project at my last Junior Olympiad had been text scanning, character&#xA;  recognition, and text to speech translation and I still had the old&#xA;  code. I pulled down my copy of Smith's &quot;The First Million Digits Of&#xA;  Pi&quot;, and set my  scanner onto it. Then I hacked out a quick routine&#xA;  that would repeatedly loop my text to speech algorithm through the&#xA;  first 2000 digits. My plan was to sleep for 12 hours with the digits&#xA;  playing to my subconscious mind as I slept, via a set of old&#xA;  headphones. The headphones were stereo, so I could play a second&#xA;  soundtrack into my other ear. Once the Pi file was ready I quickly&#xA;  scanned all three Volumes of Knuth's &lt;em&gt;The art of Computer Programming&lt;/em&gt;&#xA;  using the same software and set up the Knuth file to play at 10 times&#xA;  normal speed so I could get through most of the series over-night. I&#xA;  got both files ready to run then hacked a 600 second delay in my text&#xA;  to speech program so I could get to sleep before the words began, set&#xA;  my alarm clock for 7:00 am, attached the earphones and fell into a&#xA;  deep sleep. What could possibly go wrong? &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Unfortunately, the earphones blocked the sound of the alarm clock so&#xA;  it was 8:30 before I woke up. I had already missed the bus and had to&#xA;  cycle into town so never a chance to check if the sleep-learning had&#xA;  worked. I arrived at the Olympiad and joined the geek girl's team.&#xA;  &quot;Right then&quot;, said the team leader, &quot;time for your test Shorty. Let's&#xA;  start with something easy to begin with. Slice of Pi (1,9)&quot;. I opened&#xA;  my mouth (after all - I already knew the first 9 decimal places) and&#xA;  blurted out &quot;112345569&quot;. I fled from the room in embarrassed. I also&#xA;  fell instantly out of love with the geek girl when I heard her&#xA;  donkey-like bray of laughter pursuing me out the door. It was a year&#xA;  before I was willing to show my face at another Olympiad.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TwcE9.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TwcE9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;So the sleep learning didn't work?&quot; I interrupted, &quot;It just confused your mind?&quot;. &quot;Not Quite&quot;, Ernie replied, and continued...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It took a few hours of investigation to discover what had happened. In&#xA;  my haste the night before I had confused the two files - so in one ear&#xA;  i had been hearing the same short section of Knuth repeated again and&#xA;  again and again all night - and simultaneously a vast number of digits&#xA;  of pi had been fed into my other ear (at 10x normal speed). The sleep&#xA;  learning had worked too well. My mind had absorbed all the digits of&#xA;  pi that I heard, but due to the multiple repetitions a quick-sort&#xA;  algorithm had been driven deep into my unconscious mind. Any time I&#xA;  tried to recollect a sequence of digits of pi (or a group of&#xA;  sequences), my subconscious automatically sorted the digits before&#xA;  relaying them to my conscious mind. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;And there was another problem too. That evening I decided to take&#xA;  solace in Proust's '&lt;em&gt;À la recherche du temps perdu&lt;/em&gt;' in which I had&#xA;  been deeply engrossed over the previous few weeks, but was shocked to&#xA;  discover that I couldn't read a word of it. Every atom of the French&#xA;  language had been totally erased! I opened the Pop Sci magazine again&#xA;  and this time read the &lt;strong&gt;whole&lt;/strong&gt; article. There was a short section&#xA;  near the end describing an unfortunate side effect of adding too many&#xA;  new memories in a short time. In some subjects the new memories&#xA;  displaced old memories. In my case all my knowledge of written and&#xA;  spoken French had been deleted!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At that moment I had to stop Ernie - &quot;its a great tale&quot;, I said, chuckling with laughter &quot;You almost had me believing you for a minute&quot;. &quot;But it's true, Ernie asserted. &quot;Test me if you like, and if it doesn't work like I told you I will buy you a good Whiskey - and to be fair, if it does work then you can shout me a bottle of Brandy.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I quickly jotted down four numbers on a tissue so I could check them later and asked Ernie for the &quot;&lt;strong&gt;Slice of Pi &lt;em&gt;(p1,10), (p2,20), (p3,30), (p4,40)&lt;/em&gt;&lt;/strong&gt;&quot; Ernie paused for a moment and then replied &quot; that would be...&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 0000000000&#xA; 1111111111&#xA; 2222222222  &#xA; 3333333333&#xA; 4444444444&#xA; 5555555555&#xA; 6666666666  &#xA; 7777777777&#xA; 8888888888&#xA; 9999999999&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&quot;That's a bit unlikely&quot;, I said, &quot;are you sure.&quot; &quot;Of course I am!&quot; Ernie replied, &quot;You can check the digits online, if you don't believe me. What's more, if you can work out the minimum number of digits I must have learned to get the answer right, I will buy you a dozen bottles of Whiskey!&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I trust Ernie's word implicitly, so have already bought him the brandy, but I would love to get 12 bottles of whiskey in return. I decided to check out the numbers but realized I had left the tissue with the numbers &lt;em&gt;p1&lt;/em&gt;, &lt;em&gt;p2&lt;/em&gt;, &lt;em&gt;p3&lt;/em&gt;, and &lt;em&gt;p4&lt;/em&gt; behind at the restaurant - the only thing I can remember about them was that &lt;em&gt;p1&lt;/em&gt; is less than &lt;em&gt;p2&lt;/em&gt; is less than &lt;em&gt;p3&lt;/em&gt; is less than p4*.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the information provided can you confirm Ernie's digits could be correct for some &lt;em&gt;p1,p2,p3&lt;/em&gt; and &lt;em&gt;p4&lt;/em&gt;, and if so is possible to calculate the minimum possible number of digits he knows? &lt;/p&gt;&#xA;" OwnerUserId="1631" LastActivityDate="2015-12-07T18:03:17.530" Title="Ernie and the slice of Pi" Tags="|mathematics|word-problem|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24309" PostTypeId="2" ParentId="24292" CreationDate="2015-12-02T00:53:09.607" Score="0" Body="&lt;p&gt;This is no problem.  Here's a photo of the cube solved in this way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Oq0Ro.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Oq0Ro.jpg&quot; alt=&quot;Orange, green and white stripes&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ZGCnL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZGCnL.jpg&quot; alt=&quot;Red, yellow and blue stripes&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want all the edges to be wrong, but edges can be twisted in place, provided you do so in pairs. There are 6 pairs of edges, so do that move and all your edges will be &quot;wrong&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Corners can also be twisted in place, provided you do so in pairs, so do the diagonals of one face and the opposite diagonals of the opposing face. So, to be clear, if the cube has side length 1, with corners at 000,001,010,011,100,101,110,111, then twist 000-011 and 101-110. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That will leave all 6 faces with correct diagonals and no other correct square (as seen in the photos). &lt;/p&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2015-12-03T21:53:04.187" LastActivityDate="2015-12-03T21:53:04.187" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24310" PostTypeId="2" ParentId="24274" CreationDate="2015-12-02T01:26:58.360" Score="0" Body="&lt;p&gt;Fredo's system represents a &lt;a href=&quot;https://en.wikipedia.org/wiki/Unary_numeral_system&quot; rel=&quot;nofollow&quot;&gt;unary numeral system&lt;/a&gt; (i.e. base-1). We can change the base and still have Fredo's uniqueness property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a number system in base $n$, select weight values $n^k$ for all non-negative integers $k$ such that $n^k$ is at least 212. Then for each weight value, acquire $n-1$ weights.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write the weight $W$ in base-n notation (e.g. in base-2 for W=5, write &quot;101&quot;). This representation is unique for each given (positive) $n$. The place value of each digit corresponds to a unique weight value. For each digit, its numeric value corresponds to the number of weights of the corresponding weight value to select. Since the representation is unique, the selection of weights is also unique, provided we have fewer than $n$ of each weight value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, Frodo requires the weights to add up to 212. Define $s(p)$ as the sum of all weights in the system up to weight value $n^p$. That is,&#xA;$$s(p) = \sum_{k=0}^{p} (n-1)n^k$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $n$, we need to find place value $k$ such that 212 is between $s(p-1)$ and $s(p)$ for some $p$, and the remaining weight is a multiple of $n^p$, where that multiple is less than $n$ (i.e. the last place value may have less than the maximum number of weights).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be shown that none of the systems for $1 &amp;lt; n &amp;lt; 15$ work. From $n=15$, we have $n^ &amp;gt; 212$, so we can just solve $212 - (n-1) = kn$, which reduces to $213 = (k+1)n$. Since $212 = 3 * 71$, both of which are prime, the only remaining solution is $n=71, k=2$. That is, 70 sets of 1-weights, and 2 sets of 71-weights. This matches one of the systems others have derived.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we now relax the system and allow more than $(n-1)$ of the largest weights, we still have a unique numbering system, but can now admit the other solution involving $3 * 71$, namely, $n=3, k=70$. That is, 2 sets of 1-weights and 70 sets of 3-weights. This matches the other system others have derived.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Note: I much prefer the solution by @TheDarkTruth for this problem, but will leave this solution up for now as it offers an alternate formulation of the puzzle.]&lt;/p&gt;&#xA;" OwnerUserId="9422" LastEditorUserId="9422" LastEditDate="2015-12-02T02:24:47.847" LastActivityDate="2015-12-02T02:24:47.847" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24311" PostTypeId="2" ParentId="24292" CreationDate="2015-12-02T08:20:40.200" Score="1" Body="&lt;p&gt;This isn't possible to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Break the various cubes where all faces have diagonals into two cases:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;There does not exist a corner which lies on three diagonals.&lt;/li&gt;&#xA;&lt;li&gt;There exists a corner which lies on three diagonals.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In each of these cases, I will show why this can't be done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm using the standard Rubik's Cube color scheme: BOY clockwise, RO/WY/GB opposites format.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In the first case, where no three diagonals meet, suppose we have two diagonals on adjacent faces. They must meet at a corner, which means (by assumption) the third diagonal does not connect at this adjacent corner. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose (because colors are invariant), that the diagonals point to the DRF corner, and lie on the front and right faces. Because we're in the first case, we know that the top must go from the URF corner to the BLF corner, and the bottom must be rotated. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ERVPZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ERVPZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The only corner that can satisfy this orientation of the blue and yellow corner is the blue-yellow-red corner, which obviously won't work here because the diagonal is red. (The blue-yellow-orange corner is discounted because it would have the wrong positions of blue and yellow.) Thus, in the first case, there is no configuration.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In the second case, suppose without loss of generality that the joining corner is in the front top position, like so:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pBvLK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pBvLK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a little more annoying, but we can once again have to split this position into two subcases, since we need to assume one more diagonal to conclude anything about the remaining corners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Subcase A:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume the following position: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3MVJU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3MVJU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's only one corner fitting a couple of these positions, so let's slot them in:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qd5np.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qd5np.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now we know the rest of the corners, so let's slot those in. (I apparently missed a green corner color here; this is an accident.) &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PjWEd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PjWEd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, we just slot in the rest of the pieces such that they themselves are a rotation of a Rubik's cube, like so:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GnEss.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GnEss.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(In this specific picture, blank means white. A quirk of &lt;a href=&quot;http://rubiks-cube-solver.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;the editor I'm using&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I actually thought this was a legal position when I first looked at it. Indeed, I assembled a cube in this position to try it out. Unfortunately, when I went to solve it, one final corner was rotated, making the state unsolvable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this position is isomorphic to all other positions in of the same type, this case can be ruled out as a viable option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Subcase B:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we need to take a look at the other rotation of the green side.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/y1bR7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/y1bR7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pick an arbitrary edge. By the above assumptions, all other edges are solved with respect to the one you pick. The colors you put on this edge are therefore irrelevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I fixed one arbitrary edge, and fixed the remaining twelve edges with respect to it: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PpZbc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PpZbc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(In this picture, the four blank edges are white. Again, a quirk of the editor.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've marked two corners. By assumption, the green diagonal points a specific direction. Thus, the other two corners are not on the diagonal, and those squares must be blue. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the top back corner in this frame must be a blue-yellow corner, with blue on the green face and yellow on the yellow face. However, this corner isn't available! It's the one being used in the top front of the cube. The only remaining corner that has the right colors to fit is the blue-yellow-red corner, and it has the wrong orientation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, there is no solution to subcase B.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Since there are no cases where this is possible, the state you're looking for can't be found. &lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorDisplayName="user20" LastEditDate="2015-12-02T08:28:35.400" LastActivityDate="2015-12-02T08:28:35.400" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24312" PostTypeId="2" ParentId="24308" CreationDate="2015-12-02T09:35:21.890" Score="7" Body="&lt;p&gt;There are several possible solutions for $(p1, p2, p3, p4)$. Assuming that you with 'possible to calculate the minimum possible number of digits he knows' means a solution with the lowest possible $p4$, we have:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (42, 10) = 9399375105&lt;br&gt;&#xA; (48, 20) = 51058209749445923078&lt;br&gt;&#xA; (92, 30) = 421170679821480865132823066470&lt;br&gt;&#xA; (210,40) = 5665933446128475648233786783165271201909&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've let my search for possible solutions run a little bit longer and if with 'minimum possible number of digits he knows' is ment to find the lowest possible value of $p4-p1$ (continuous range of digits), I soon found a range of only 65 digits: (1399,10), (1399,20), (1413,30) and (1424,40).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solutions to the problem are actually not particularly rare. For $p4 &amp;lt;= 1000$ there are 1556 solutions and for $p4 &amp;lt;= 2000 $, there are 16376 solutions. That however, I find slightly odd. I would have expected the number of solutions to increase linearly with the number of combinations, which is at least roughly quartic in relation to $p4$ (doubling $p4$ should result in $2^4$ or 16 times as many solutions).&lt;/p&gt;&#xA;" OwnerUserId="11687" LastEditorUserId="11687" LastEditDate="2015-12-07T18:03:17.530" LastActivityDate="2015-12-07T18:03:17.530" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="24313" PostTypeId="1" AcceptedAnswerId="24316" CreationDate="2015-12-02T10:59:48.673" Score="5" ViewCount="375" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Find a year that has &lt;strong&gt;three Friday the thirteenths&lt;/strong&gt; in it, either past or future.  (That is, find a year where the thirteenth of the month falls on a Friday in three different months.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Find a year with '&lt;strong&gt;a month of Sundays&lt;/strong&gt;'.  That is, find a year that has a &lt;strong&gt;Sunday on every day of a month&lt;/strong&gt; at least once (i.e. a Sunday the 1st, a Sunday the 2nd, a Sunday the 3rd, ..., a Sunday the 31st).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11497" LastEditorUserId="4421" LastEditDate="2015-12-02T17:27:09.807" LastActivityDate="2015-12-02T18:13:41.870" Title="Friday the 13th poser" Tags="|mathematics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24315" PostTypeId="2" ParentId="24296" CreationDate="2015-12-02T12:07:57.133" Score="4" Body="&lt;p&gt;Other reason can be:&#xA;Adam is blind&#xA;When volume of sound is low, he can navigate by hearing, and he walks&#xA;When volume of sound is high, it becomes too difficult to understand what's around by hearing, so he takes a ride.&lt;/p&gt;&#xA;" OwnerUserId="17005" LastActivityDate="2015-12-02T12:07:57.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24316" PostTypeId="2" ParentId="24313" CreationDate="2015-12-02T12:17:47.247" Score="5" Body="&lt;p&gt;A solution for both problems is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;This year (2015)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;First we look for the variances between the days for each month.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Normal year&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;January    0&#xA;February   +3   (starts on the weekday that is 3 days after the weekday of January 1st)&#xA;March      +3&#xA;April      +6   (starts on the weekday that is 6 days after the weekday of January 1st)&#xA;May        +1   etc...&#xA;June       +4&#xA;July       +6&#xA;August     +2&#xA;September  +5&#xA;October    0&#xA;November   +3&#xA;December   +5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Leap year&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;January    0&#xA;February   +3&#xA;March      +4&#xA;April      0&#xA;May        +2&#xA;June       +5&#xA;July       0&#xA;August     +3&#xA;September  +6&#xA;October    +1&#xA;November   +4&#xA;December   +6&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;First, we need 3 different months with the same variance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a normal year, those are February, March and November.&lt;br&gt;&#xA;For a leap year, those are January, April and July.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find a year with 3 Fridays the 13th, we need the three months mentioned above to have their 13th on a Friday, which means those months start on a Sunday.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a &lt;strong&gt;normal year&lt;/strong&gt;, this means the year starts on a Thursday (those months have a variance of 3, so if they start on a Sunday, January (and thus the year) begins 3 days earlier, on a Thursday).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a &lt;strong&gt;leap year&lt;/strong&gt;, this means the year starts on a Sunday (one of the three months is January, and it must start on a Sunday, so that means the year starts on a Sunday).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our current year (2015) began on a Thursday, therefore 2015 had 3 Fridays the 13th.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As for a 'month of Sundays':&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both a normal year and a leap year have all 7 possible variances in their months, so we only need to find a month with 31 days where the 31st is a Sunday.  This will mean the whole year has a 'month of Sundays'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And surprise, this year's May did indeed have a Sunday the 31st.&lt;/p&gt;&#xA;" OwnerUserId="16608" LastEditorUserId="4421" LastEditDate="2015-12-02T18:13:41.870" LastActivityDate="2015-12-02T18:13:41.870" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24317" PostTypeId="2" ParentId="24296" CreationDate="2015-12-02T13:33:40.207" Score="4" Body="&lt;p&gt;As you state in the question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When the volume of the sound is low, Adam &lt;strong&gt;walks home from school&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Adam's parents drive him on their way to and from work during [noisy] rush-hour traffic (every morning and possibly some evenings). When Adam finishes school before the evening rush hour while his parents are still working, he walks home.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12462" LastActivityDate="2015-12-02T13:33:40.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24318" PostTypeId="1" AcceptedAnswerId="24322" CreationDate="2015-12-02T15:02:59.340" Score="4" ViewCount="1132" Body="&lt;p&gt;I have a (possibly-) unfair coin, which lands heads with probability $p$ and lands tails with probability $1-p$. I toss it twice; the tosses are &lt;em&gt;independent&lt;/em&gt; and &lt;em&gt;identically distributed&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probabilities of the various outcomes will vary depending on $p$. For example, if I know that $p = \frac12$, then the probability of tossing two heads, which I'll abbreviate $\mathbb P(HH)$, can be calculated as $p^2 = \frac14$. Symmetrically, the probability of tossing two tails $\mathbb P(TT) = \frac14$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if I know that $p &amp;gt; \frac12$, e.g. $p = \frac1{\sqrt 3}$, then $HH$ becomes more likely: $\mathbb P(HH) = \frac13$. Correspondingly, $TT$ becomes less likely: $\mathbb P(TT) = (1 - \frac1{\sqrt 3})^2 \approx 0.18$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems like I can push $HH$ up above $\frac14$, but only by pushing $TT$ down below that same number. With that in mind, the following question might seem surprising:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Under what circumstances does $\mathbb P(HH) = \mathbb P(TT) = \frac13$?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1950" LastEditorUserId="12408" LastEditDate="2016-07-25T02:30:00.390" LastActivityDate="2016-07-25T02:30:00.390" Title="A coin flip's long tail" Tags="|mathematics|probability|coin-tossing|" AnswerCount="2" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="24319" PostTypeId="2" ParentId="24318" CreationDate="2015-12-02T15:43:14.273" Score="5" Body="&lt;p&gt;This seems to be mathematically impossible. The probability of two heads is $1/3$ and the probability of two tails is $1/3$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since they're equal, this gives us&#xA;$$&#xA;p^2 = (1-p)^2&#xA;$$&#xA;$$&#xA;p = 1-p&#xA;$$&#xA;$$&#xA;p = 1/2&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $(1/2)^2 = 1/4$, not $1/3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there really is a solution to this I'd be really interested to see it.&lt;/p&gt;&#xA;" OwnerUserId="17559" LastActivityDate="2015-12-02T15:43:14.273" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24321" PostTypeId="1" AcceptedAnswerId="24323" CreationDate="2015-12-02T18:04:42.130" Score="8" ViewCount="770" Body="&lt;p&gt;The ladies mentioned in &lt;a href=&quot;https://puzzling.stackexchange.com/questions/24256/five-beautiful-ladies&quot;&gt;Five beautiful ladies&lt;/a&gt; have five brothers: Adam, Bob, Charlie, David, and Edward.  Three have hazel eyes and are honest; two have lime-colored eyes and always lie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are tasked with finding out who among the brothers is honest by submitting one multiple-choice question in writing to each of Adam, Bob, and Charlie, and examining the responses; all questions must be submitted before any responses are received.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiple-choice questions should be written so as to have one correct answer; assume nothing good will come of asking ambiguous questions(*).  Honest people will identify and report the correct choice; liars will honestly identify the correct choice, but falsely report some other choice, selected in whatever fashion will be most vexatious.  If a liar is given a 4-way A/B/C/D choice where the correct answer was B, he may arbitrarily answer A, C, or D; he will not answer B or Q).  By inference, if someone answers B, that will imply that either he was truthful and the correct answer was B, or he was lying and the correct answer is A, C, or D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*) For simplicity, it should suffice to say that a liar is only forbidden from responding with choices which are not offered, or any choice which is clearly the unique correct answer.  There is no requirement that a liar pick the &quot;most dishonest&quot; choice--merely that there exist some honestly-correct choice other than the one reported by the liar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that as with the Ladies' puzzle, there are ten possible combinations of eye colors to resolve using three questions, which implies the need for questions with more than two possible answers.  Unlike the Ladies' puzzle, however, all questions must be submitted in advance.  The puzzle would be unsolvable with three liars, but is solvable with two.&lt;/p&gt;&#xA;" OwnerUserId="1467" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-12-03T04:34:04.487" Title="The Beautiful Ladies' Brothers" Tags="|logical-deduction|liars|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24322" PostTypeId="2" ParentId="24318" CreationDate="2015-12-02T19:20:48.410" Score="13" Body="&lt;p&gt;$p$ is a random variable, chosen with uniform probability over the interval [0,1]. The probability of two heads is $\int_0^1 p^2=\frac{1}{3}$, and by symmetry so is the probability of two tails.&lt;/p&gt;&#xA;" OwnerUserId="3987" LastActivityDate="2015-12-02T19:20:48.410" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="24323" PostTypeId="2" ParentId="24321" CreationDate="2015-12-02T23:38:48.160" Score="10" Body="&lt;p&gt;We will ask each of A, B and C a three choice question, whose options are numbered (1), (2), (3). The questions are given in the below table:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\begin{array}{l|l|l}\text{To Adam...}&amp;amp;\text{To Bob...}&amp;amp;\text{To Charlie...}\\\hline(1) \text{   The liars are B and E} &amp;amp; (1) \text{   The liars are C and E} &amp;amp; (1) \text{   The liars are A and E} \\(2) \text{   The liars are C and E} &amp;amp; (2) \text{   The liars are A and E} &amp;amp; (2) \text{   The liars are B and E} \\(3) \text{   None of the above} &amp;amp; (3) \text{   None of the above} &amp;amp; (3) \text{   None of the above} \\\end{array}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Notice that if a person is a liar then the correct answer is (3), so &lt;strong&gt;a liar will never answer (3).&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We then deduce the liars as shown below. The symbol &quot;_&quot; means any answer which isn't (3). &#xA;$$&#xA;\begin{array}{c|c|c}&#xA;\_\,33\implies AD &amp;amp; \_\,\_3 \implies AB &amp;amp; \_21\implies AE\\&#xA;3\_\,3\implies BD &amp;amp; \_\,3\_ \implies AC &amp;amp; 1\_2\implies BE\\&#xA;33\_\,\implies CD &amp;amp; 3\_\,\_ \implies BC &amp;amp; 21\_\implies CE\\&#xA;333\implies DE&#xA;\end{array}&#xA;$$&#xA;For example, the entry &quot;$3\_\,\_\implies BC$&quot; is shorthand for &quot;If Adam responds (3), and neither of Bob and Charlie respond (3), then the liars are Bob and Charlie.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's why this works:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;If everyone answered (3)&lt;/em&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then none of A,B,C are liars, so the liars are &lt;strong&gt;DE&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;If exactly two people answered (3)&lt;/em&gt;... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For definiteness, suppose A and B answered (3), and Charlie didn't. Then both A and B must be honest, so the liars are either CD, CE or DE. CE is impossible, because in that case A would have said (2). Also, DE is impossible, because in that case C would answer (3). Thus, the liars are &lt;strong&gt;CD&lt;/strong&gt;, as claimed. The other two cases work similarly.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;If exactly one person said (3)...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose it was Adam who said (3), so that Adam is honest. D must be honest as well, otherwise at least one of B or C would answer (3). Thus, the only possibilities are BC, BE and CE. However, BE and CE were ruled out by Adam saying (3), so the true answer is indeed &lt;strong&gt;BC&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;If no one said (3)...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since at least one of A,B or C is honest, the only three possibilities are AE, BE and CE. Thus, the majority of A, B and C are truthful, so we can simply go with the majority answer. For example, in the 212 case, both Adam and Bob are claiming the liars are CE, so the liars are indeed &lt;strong&gt;CE&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-12-03T04:34:04.487" LastActivityDate="2015-12-03T04:34:04.487" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24324" PostTypeId="1" AcceptedAnswerId="24329" CreationDate="2015-12-03T16:22:26.843" Score="3" ViewCount="290" Body="&lt;p&gt;&lt;sub&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/24240/magic-the-gathering-challenge-6-b-its-gonna-cost-you&quot;&gt;Previous Challenge&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Background&lt;/h2&gt;  &#xA;&#xA;&lt;p&gt;Normal MtG rules apply here. You may not assume that your opponent will cooperate with you (e.g. choose not to block any of your creatures when they can). Assume that your opponent will block the largest power creatures that they legally can with as many creatures as they legally can, regardless of whether or not it will kill their blockers (in the event of tied power, assume they choose to block the first of the tied attackers alphabetically). If you would draw a card, you may assume that you draw any card remaining in your library, regardless of shuffling.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Puzzle Setup&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;It's your main phase 1, you haven't played a land for the turn. You're being demolished by your opponent and they have you dead next turn: reverse the tides and win this turn! You've just cast &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=275260&quot; rel=&quot;nofollow noreferrer&quot;&gt;Three Dreams&lt;/a&gt; and are about to resolve it. Good luck!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your hand:&lt;/strong&gt;&lt;br&gt;&#xA;Nothing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your board:&lt;/strong&gt;&lt;br&gt;&#xA;Nothing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your graveyard:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=247184&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aura Shards&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=398678&quot; rel=&quot;nofollow noreferrer&quot;&gt;Auramancer&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=405216&quot; rel=&quot;nofollow noreferrer&quot;&gt;Eternal Witness&lt;/a&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your library:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=14526&quot; rel=&quot;nofollow noreferrer&quot;&gt;Abduction&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=382843&quot; rel=&quot;nofollow noreferrer&quot;&gt;Animate Dead&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=394506&quot; rel=&quot;nofollow noreferrer&quot;&gt;Battle Mastery&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=397798&quot; rel=&quot;nofollow noreferrer&quot;&gt;Daybreak Coronet&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=47165&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dragon Fangs&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=240138&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dual Casting&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=382923&quot; rel=&quot;nofollow noreferrer&quot;&gt;Empyrial Armor&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=393887&quot; rel=&quot;nofollow noreferrer&quot;&gt;Enslave&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=405227&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fallen Ideal&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=50272&quot; rel=&quot;nofollow noreferrer&quot;&gt;Field of Reality&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=83677&quot; rel=&quot;nofollow noreferrer&quot;&gt;Flight of Fancy&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=87968&quot; rel=&quot;nofollow noreferrer&quot;&gt;Galvanic Arc&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=383262&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hammerhand&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=107684&quot; rel=&quot;nofollow noreferrer&quot;&gt;Infiltrator's Magemark&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=394625&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mirror Mockery&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=3146&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nature's Chosen&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=107685&quot; rel=&quot;nofollow noreferrer&quot;&gt;Necromancer's Magemark&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=253546&quot; rel=&quot;nofollow noreferrer&quot;&gt;Righteous Authority&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=26407&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sisay's Ingenuity&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=109680&quot; rel=&quot;nofollow noreferrer&quot;&gt;Spirit Loop&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=397816&quot; rel=&quot;nofollow noreferrer&quot;&gt;Splinter Twin&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=146580&quot; rel=&quot;nofollow noreferrer&quot;&gt;Triclopean Sight&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=405445&quot; rel=&quot;nofollow noreferrer&quot;&gt;Vow of Malice&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=238135&quot; rel=&quot;nofollow noreferrer&quot;&gt;Vow of Wildness&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your life:&lt;/strong&gt;&lt;br&gt;&#xA;1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your mana:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/cKhUi.gif&quot; alt=&quot;]&quot;&gt;: 41&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/kACCs.gif&quot; alt=&quot;]&quot;&gt;: 11&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/ubTN0.gif&quot; alt=&quot;]&quot;&gt;: 8&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/vbX5h.gif&quot; alt=&quot;]&quot;&gt;: 3&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/zYzsO.gif&quot; alt=&quot;]&quot;&gt;: 7&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/FD8X8.gif&quot; alt=&quot;]&quot;&gt;: 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Opponent's hand:&lt;/strong&gt;&lt;br&gt;&#xA;Nothing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Opponent's board: (all untapped, summoning sick)&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=270856&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aether Membrane&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=87333&quot; rel=&quot;nofollow noreferrer&quot;&gt;Arashi, the Sky Asunder&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=87994&quot; rel=&quot;nofollow noreferrer&quot;&gt;Drift of Phantasms&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=378532&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pillar of War&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Opponent's graveyard:&lt;/strong&gt;&lt;br&gt;&#xA;Nothing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Opponent's library:&lt;/strong&gt;&lt;br&gt;&#xA;60 Swamp&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Opponent's life:&lt;/strong&gt;&lt;br&gt;&#xA;60&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Opponent's mana:&lt;/strong&gt;&lt;br&gt;&#xA;Nothing&lt;/p&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.923" LastActivityDate="2017-03-12T06:30:19.300" Title="Magic: the Gathering - Challenge #7: In Your Dreams" Tags="|calculation-puzzle|magic-the-gathering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24325" PostTypeId="1" AcceptedAnswerId="24326" CreationDate="2015-12-03T16:31:23.790" Score="15" ViewCount="2346" Body="&lt;p&gt;I am writing an API to crawl a popular website, &lt;a href=&quot;http://eksisozluk.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;eksisozluk&lt;/a&gt;, kind of like Turkey's reddit, but more text based.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On this website, you can get a list of the popular posts from the path &lt;a href=&quot;https://eksisozluk.com/basliklar/populer&quot; rel=&quot;nofollow noreferrer&quot;&gt;/basliklar/populer&lt;/a&gt;. While looking at the source, I noticed something unusual, an element that goes&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;...&#xA;&amp;lt;ul class=&quot;topic-list partial&quot; data-timestamp=&quot;635847618951557008&quot;&amp;gt;&#xA;      &amp;lt;li&amp;gt; &#xA;...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The &lt;code&gt;data-timestamp&lt;/code&gt; attribute lists an integer, but it is unusually large compared to what we should expect from the current timestamp, 1449158148. You can also try for yourself from the link above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was curious, and decided to write a script to collect more data. It would download the page every few seconds, and write the value of the unusual timestamp along with the current true timestamp. It is written in Python 3:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import urllib.request&#xA;from bs4 import BeautifulSoup&#xA;from datetime import datetime&#xA;&#xA;from time import sleep&#xA;&#xA;unixTime = datetime(1970,1,1)&#xA;&#xA;def timestampFromDatetime(dt):&#xA;    timestamp = round((dt - unixTime).total_seconds())&#xA;    return timestamp&#xA;&#xA;def getTimestamp():&#xA;    url = &quot;https://eksisozluk.com/basliklar/populer&quot;&#xA;&#xA;    page = urllib.request.urlopen(url)&#xA;    html = page.read()&#xA;    soup = BeautifulSoup(html, &quot;html.parser&quot;)&#xA;&#xA;    result = int(soup.find(&quot;ul&quot;, class_=&quot;topic-list&quot;).attrs[&quot;data-timestamp&quot;])&#xA;    return result&#xA;&#xA;for i in range(50):&#xA;    fromSite = getTimestamp()&#xA;    now = timestampFromDatetime(datetime.now())&#xA;    print(fromSite, now)&#xA;    sleep(5)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This gives me 50 points:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;# Weird timestamp  Actual timestamp&#xA;635847598460202879 1449163046&#xA;635847598513998011 1449163051&#xA;635847598564898858 1449163057&#xA;635847598619779187 1449163062&#xA;635847598672122901 1449163067&#xA;635847598724154127 1449163073&#xA;635847598774759606 1449163078&#xA;635847598826951319 1449163083&#xA;635847598882122760 1449163088&#xA;635847598933678706 1449163094&#xA;635847598987122702 1449163099&#xA;635847599037905893 1449163104&#xA;635847599089941378 1449163109&#xA;635847599143685105 1449163114&#xA;635847599194481078 1449163120&#xA;635847599246516550 1449163125&#xA;635847599299020788 1449163130&#xA;635847599352747462 1449163135&#xA;635847599403247975 1449163141&#xA;635847599455595969 1449163146&#xA;635847599507787698 1449163151&#xA;635847599561809825 1449163156&#xA;635847599613997298 1449163162&#xA;635847599666184777 1449163167&#xA;635847599718372221 1449163172&#xA;635847599770403445 1449163177&#xA;635847599821719358 1449163182&#xA;635847599874067356 1449163188&#xA;635847599925946549 1449163193&#xA;635847599979465814 1449163198&#xA;635847600031965774 1449163203&#xA;635847600082677978 1449163209&#xA;635847600136340701 1449163214&#xA;635847600186905168 1449163219&#xA;635847600239102634 1449163224&#xA;635847600291450609 1449163229&#xA;635847600345090574 1449163235&#xA;635847600395677810 1449163240&#xA;635847600447713278 1449163245&#xA;635847600499748728 1449163250&#xA;635847600551784190 1449163255&#xA;635847600605090407 1449163261&#xA;635847600656965384 1449163266&#xA;635847600709465344 1449163271&#xA;635847600759926040 1449163276&#xA;635847600813371543 1449163281&#xA;635847600864153228 1449163287&#xA;635847600916344950 1449163292&#xA;635847600969933930 1449163297&#xA;635847601021808910 1449163302&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The relationship between the columns appears to be linear. With c1 being the first column and c2 being the second, I plotted c2 versus c1 and fit the linear function &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y(x) = m x + n$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JEsn1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JEsn1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual relationship is &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;m = 1.000000e+7&#xA;n = 6.21355315553082e+17&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It is clear that the number of seconds is multiplied by 10^7 for some reason. But there is also the shift n. I wanted to understand how many seconds that shift is equal to. Since the number of seconds is multiplied with 10^7, the actual shift should be &lt;code&gt;6.21355315553082e+17 / 1e7 = 6.21355315553082e+10&lt;/code&gt; seconds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found out using google that&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;6.21355315553082e+10 seconds = 1 968.99824 years&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which is 15 hours short of 1969 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is as far as I could come. The Unix epoch starts at 1.1.1970, which clearly plays a role somewhere. I want to find out if there is a reason behind this weirdness, whether &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is it intentional, for obfuscation?&lt;/li&gt;&#xA;&lt;li&gt;Is it related to an integer overflow bug?&lt;/li&gt;&#xA;&lt;li&gt;Is it just bad coding?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I could only guess, and I don't think I could reach the site's admins. If this insignificantest of mysteries seems attractive to you, you can take a shot.&lt;/p&gt;&#xA;" OwnerUserId="17628" LastActivityDate="2015-12-08T10:06:24.290" Title="Trying to understand an unusual timestamp encountered on the web" Tags="|calculation-puzzle|computer-puzzle|mystery|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24326" PostTypeId="2" ParentId="24325" CreationDate="2015-12-03T16:50:23.297" Score="21" Body="&lt;p&gt;The timestamp is using 01/01/01 as it's base, instead of the UNIX epoch: [Example]&lt;a href=&quot;http://www.epochconverter.com/epoch/seconds-days-since-year-0.php&quot;&gt;http://www.epochconverter.com/epoch/seconds-days-since-year-0.php&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look down the page to Seconds Since 0001/01/01 AD to see. I believe the extra digits beyond what the above link shows are just additional precision.&lt;/p&gt;&#xA;" OwnerUserId="16874" LastActivityDate="2015-12-03T16:50:23.297" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24327" PostTypeId="2" ParentId="24325" CreationDate="2015-12-03T16:55:48.273" Score="8" Body="&lt;p&gt;The Wikipedia page for &lt;a href=&quot;https://en.wikipedia.org/wiki/Epoch_%28reference_date%29&quot;&gt;Epoch&lt;/a&gt; lists some commonly used epoch dates. Specifically &lt;strong&gt;Microsoft .NET, Go, REXX and Rata Die&lt;/strong&gt; are listed as using &lt;strong&gt;January 1, AD 1&lt;/strong&gt; as their epoch date. Therefore I assume one of these languages was used to implement this site.&lt;/p&gt;&#xA;" OwnerUserId="15632" LastActivityDate="2015-12-03T16:55:48.273" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24328" PostTypeId="2" ParentId="24257" CreationDate="2015-12-03T18:44:06.560" Score="7" Body="&lt;p&gt;I agree with &lt;a href=&quot;https://puzzling.stackexchange.com/users/9352/novarg&quot;&gt;Novarg's&lt;/a&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/a/24258/9352&quot;&gt;answer&lt;/a&gt;, but it's a bit hard to follow. Here's how I solved it:&lt;br&gt;&#xA;We can know the possible values are these:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 2 10 &lt;br/&gt;&#xA; 1 3 9 &lt;br/&gt;&#xA; 1 4 8 &lt;br/&gt;&#xA; 1 5 7 &lt;br/&gt;&#xA; 2 3 8 &lt;br/&gt;&#xA; 2 4 7 &lt;br/&gt;&#xA; 2 5 6 &lt;br/&gt;&#xA; 3 4 6 &lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When lefty comes in&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If it was a 3, there would be no other options and he would be able to figure it out, but he didn't. &lt;br/&gt;&#xA; 1 2 10 &lt;br/&gt;&#xA; 1 3 9 &lt;br/&gt;&#xA; 1 4 8 &lt;br/&gt;&#xA; 1 5 7 &lt;br/&gt;&#xA; 2 3 8 &lt;br/&gt;&#xA; 2 4 7 &lt;br/&gt;&#xA; 2 5 6 &lt;br/&gt;&#xA; &lt;strike&gt;3 4 6&lt;/strike&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When righty goes in, he knows that lefty couldn't figure it out.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; And since he couldnt either, all unique values for R must be eliminated&#xA; including 6, since he would have figured out given that he knew the other 6 was eliminated! &lt;br/&gt;&#xA; &lt;strike&gt;1 2 10&lt;/strike&gt;  &lt;br/&gt;&#xA; &lt;strike&gt;1 3 9&lt;/strike&gt;  &lt;br/&gt;&#xA; 1 4 8 &lt;br/&gt;&#xA; 1 5 7 &lt;br/&gt;&#xA; 2 3 8 &lt;br/&gt;&#xA; 2 4 7 &lt;br/&gt;&#xA; &lt;strike&gt;2 5 6 &lt;/strike&gt; &lt;br/&gt;&#xA; &lt;strike&gt;3 4 6&lt;/strike&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When M comes in, the same applies for all remaining unique values for M&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strike&gt;1 2 10&lt;/strike&gt;  &lt;br/&gt;&#xA; &lt;strike&gt;1 3 9&lt;/strike&gt;  &lt;br/&gt;&#xA; 1 4 8 &lt;br/&gt;&#xA; &lt;strike&gt;1 5 7 &lt;/strike&gt; &lt;br/&gt;&#xA; &lt;strike&gt;2 3 8 &lt;/strike&gt; &lt;br/&gt;&#xA; 2 4 7 &lt;br/&gt;&#xA; &lt;strike&gt;2 5 6 &lt;/strike&gt; &lt;br/&gt;&#xA; &lt;strike&gt;3 4 6&lt;/strike&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5380" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-12-04T07:36:23.173" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24329" PostTypeId="2" ParentId="24324" CreationDate="2015-12-03T18:51:24.253" Score="1" Body="&lt;p&gt;I feel like I found an answer far easier than intended. EDIT: Here's a solution that deals 78 damage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Resolve Three Dreams (hand: 3 cards)&lt;br&gt;&#xA;Cast Animate dead, returning Eternal Witness, getting back Three Dreams&lt;br&gt;&#xA;Cast Three Dreams (hand: 5 cards)&lt;br&gt;&#xA;32 C&lt;br&gt;&#xA;9 W&lt;br&gt;&#xA;8 U&lt;br&gt;&#xA;2 B&lt;br&gt;&#xA;7 R&lt;br&gt;&#xA;1 G&lt;br&gt;&#xA;Cast Hammerhand on Eternal Witness (trigger targets Drift of Phantasms). Cast Splinter Twin on Eternal Witness, activate it, get back Three Dreams&lt;br&gt;&#xA;Cast Dual Casting on the Eternal Witness token, then cast Dual Cast Three Dreams (hand: 8)&lt;br&gt;&#xA;25 C&lt;br&gt;&#xA;8 W&lt;br&gt;&#xA;8 U&lt;br&gt;&#xA;2 B&lt;br&gt;&#xA;2 R&lt;br&gt;&#xA;1 G&lt;br&gt;&#xA;Cast Triclopean Sight on the Twin Witness to untap it&lt;br&gt;&#xA;Activate Twin for a token to get back Auramancer (hand: 7 + Aurmancer)&lt;br&gt;&#xA;24 C&lt;br&gt;&#xA;7 W&lt;br&gt;&#xA;8 U&lt;br&gt;&#xA;2 B&lt;br&gt;&#xA;2 R&lt;br&gt;&#xA;1 G&lt;br&gt;&#xA;Play Fallen Ideal on the new Witness token.&lt;br&gt;&#xA;Play Abduction on Twin Witness, untapping it.&lt;br&gt;&#xA;Activate Twin, then with the trigger on the stack, sacrifice the Twin Witness to Fallen Ideal.&lt;br&gt;&#xA;Abduction triggers, bringing back Eternal Witness, bringing back Abduction.&lt;br&gt;&#xA;Splinter Twin resolves, you get a token, bringing back Three Dreams.&lt;br&gt;&#xA;Board: Fallen Token (4/2), Dualcaster Token, Plain Token, Original Witness&lt;br&gt;&#xA;Hand: 5 + Auramancer, Three Dreams, Abduction&lt;br&gt;&#xA;20 C&lt;br&gt;&#xA;7 W&lt;br&gt;&#xA;6 U&lt;br&gt;&#xA;1 B&lt;br&gt;&#xA;2 R&lt;br&gt;&#xA;1 G&lt;br&gt;&#xA;Play Nature's Chosen on the Dualcaster, untap it.&lt;br&gt;&#xA;Dualcast Three Dreams (hand: 10 + Auramancer, Abduction)&lt;br&gt;&#xA;16 C&lt;br&gt;&#xA;6 W&lt;br&gt;&#xA;6 U&lt;br&gt;&#xA;1 B&lt;br&gt;&#xA;1 R&lt;br&gt;&#xA;Sacrifice Original Witness to Fallen Ideal.&lt;br&gt;&#xA;Cast Auramancer for Animate Dead.&lt;br&gt;&#xA;Cast Animate Dead for Witness, for Three Dreams (hand: 10 + Abduction, Three Dreams)&lt;br&gt;&#xA;Board: Fallen Token (6/3), Dualcaster Token, Plain Token, Original Witness, Auramancer&lt;br&gt;&#xA;13 C&lt;br&gt;&#xA;5 W&lt;br&gt;&#xA;6 U&lt;br&gt;&#xA;1 R&lt;br&gt;&#xA;Cast Three Dreams (hand: 13 + Abduction)&lt;br&gt;&#xA;Cast Battle Mastery, Righteous Authority, Empyrial Armor, Mirror Mockery on Fallen Token (hand: 9 + Abduction)&lt;br&gt;&#xA;2 C&lt;br&gt;&#xA;4 U&lt;br&gt;&#xA;1 R&lt;br&gt;&#xA;Cast Abduction on Auramancer&lt;br&gt;&#xA;Sacrifice Auramancer (8/4), abduction triggers, bringing back Auramancer, getting back Hammerhand&lt;br&gt;&#xA;Cast Hammerhand on Fallen Token, trigger targets Aether Membrane (hand: 9)&lt;br&gt;&#xA;Attack with Fallen token (opp has no flying blockers because of Hammerhand triggers), Mirror Mockery triggers, token gets back Three Dreams (hand:10)&lt;br&gt;&#xA;Sacrifice all five creatures, attacker is 2 base + 16 Ideal + 1 Hammerhand + 10 Armor + 10 Authority. 78 damage with Double Strike.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;@NeedAName mentioned that this solution is easier than intended because Arashi doesn't have flying, you can adapt this solution to beat a flying Arashi by Witnessing back Abduction instead of Three Dreams and Abducting the opponent's Arashi (you lose 8 damage, but the solution can then be optimized to save a bit of mana so that you can cast Flight of Fancy for the draw 2, so you only lose 4 damage).&lt;/p&gt;&#xA;" OwnerUserId="7459" LastEditorUserId="7459" LastEditDate="2015-12-03T23:07:52.423" LastActivityDate="2015-12-03T23:07:52.423" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24330" PostTypeId="2" ParentId="23292" CreationDate="2015-12-03T23:04:22.783" Score="2" Body="&lt;p&gt;As I mentioned in a previous comment, checking all weighings can take more than an hour on a fast machine. So I stopped trying to solve it that way, and only used it to check possible answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weighing #1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1-15 v 16-30&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Excluding duplicates, this gives me the following ordered groups and possible weighings, where the number listed is the number of fewer coins (light or heavy):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 15 15&lt;br&gt;&#xA; &amp;nbsp;------&lt;br&gt;&#xA; &amp;nbsp;&amp;nbsp;1 &amp;nbsp;&amp;nbsp;1&lt;br&gt;&#xA; &amp;nbsp;&amp;nbsp;2 &amp;nbsp;&amp;nbsp;2&lt;br&gt;&#xA; &amp;nbsp;&amp;nbsp;3 &amp;nbsp;&amp;nbsp;3&lt;br&gt;&#xA; &amp;nbsp;&amp;nbsp;4 &amp;nbsp;&amp;nbsp;4&lt;br&gt;&#xA; &amp;nbsp;&amp;nbsp;5 &amp;nbsp;&amp;nbsp;5&lt;br&gt;&#xA; &amp;nbsp;&amp;nbsp;6 &amp;nbsp;&amp;nbsp;6&lt;br&gt;&#xA; &amp;nbsp;&amp;nbsp;7 &amp;nbsp;&amp;nbsp;7&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, the next part was to find a suitable second weighing. I tried following the first puzzle's weighing strategy and that led to problems. I could solve everything except for 20 light / 10 heavy. So I abandoned that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried many different weighings, and decided what I really needed to do was avoid multiples of 3 and 5. This brings me to weighing 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1-11 v 12-22&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This gives me the following known group sizes in order and possible weighings for them:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 11 4 7 8&lt;br&gt;&#xA; &amp;nbsp;----------&lt;br&gt;&#xA; &amp;nbsp;&amp;nbsp;1 0 1 0&lt;br&gt;&#xA; &amp;nbsp;&amp;nbsp;2 0 2 0&lt;br&gt;&#xA; &amp;nbsp;&amp;nbsp;1 1 0 2&lt;br&gt;&#xA; &amp;nbsp;&amp;nbsp;3 0 3 0&lt;br&gt;&#xA; &amp;nbsp;&amp;nbsp;2 1 1 2&lt;br&gt;&#xA; &amp;nbsp;&amp;nbsp;4 0 4 0&lt;br&gt;&#xA; &amp;nbsp;&amp;nbsp;3 1 2 2&lt;br&gt;&#xA; &amp;nbsp;&amp;nbsp;2 2 0 4&lt;br&gt;&#xA; &amp;nbsp;&amp;nbsp;5 0 5 0&lt;br&gt;&#xA; &amp;nbsp;&amp;nbsp;4 1 3 2&lt;br&gt;&#xA; &amp;nbsp;&amp;nbsp;3 2 1 4&lt;br&gt;&#xA; &amp;nbsp;&amp;nbsp;6 0 6 0&lt;br&gt;&#xA; &amp;nbsp;&amp;nbsp;5 1 4 2&lt;br&gt;&#xA; &amp;nbsp;&amp;nbsp;4 2 2 4&lt;br&gt;&#xA; &amp;nbsp;&amp;nbsp;3 3 0 6&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Given the general large difference between the first and third and second and fourth groups, I want to combine them for my third weighing. I tried a few different ways and came up with the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 7-11,16-22 v 12-15,23-30&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This weighing made it impossible to have many of the second group of weighings be equal, leaving me with the known group sizes and weighings of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 6 5 4 7 8&lt;br&gt;&#xA; -----------&lt;br&gt;&#xA; 0 2 1 1 2&lt;br&gt;&#xA; 2 1 1 2 2&lt;br&gt;&#xA; 4 0 1 3 2&lt;br&gt;&#xA; 0 4 2 2 4&lt;br&gt;&#xA; 2 3 2 3 4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now I have one weighing left, and it is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1-6,15-22 v 7-11,23-30&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This makes all the remaining weighings invalid leaving the only possibilities as 30 light/0 heavy or 0 light/30 heavy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TL;DR&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1-15 v 16-30&lt;br&gt;&#xA; 1-11 v 12-22&lt;br&gt;&#xA; 7-11,16-22 v 12-15,23-30&lt;br&gt;&#xA; 1-6,16-22 v 7-11,23-30&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="375" LastEditorUserId="375" LastEditDate="2015-12-04T13:16:44.450" LastActivityDate="2015-12-04T13:16:44.450" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24331" PostTypeId="1" AcceptedAnswerId="24332" CreationDate="2015-12-04T02:11:44.690" Score="2" ViewCount="348" Body="&lt;p&gt;Your friend buys a new ninja sword online, but when he receives the package it is shipped to &quot;seltruTajniNtnatuMeganeeT&quot; and contains this message written in the inside of the box:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;bggp tp fjcg pkgg kdta btlg oqkg pd ekjcu cd dcg tca teodsqpgsy cd ndbfqpgko j cgga pd vgkjmy ydq tkg pig kgngjvgk dm pijo jpgb ftkp ndbjcu oddc j lcdw ydq sjvg dc  dtl pkgg ndqkp&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9993" LastEditorUserId="14013" LastEditDate="2015-12-04T21:32:44.027" LastActivityDate="2015-12-05T22:18:04.317" Title="The Six-Edged Sword" Tags="|cipher|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24332" PostTypeId="2" ParentId="24331" CreationDate="2015-12-04T03:07:02.307" Score="5" Body="&lt;p&gt;Using a &lt;a href=&quot;http://rumkin.com/tools/cipher/cryptogram-solver.php&quot; rel=&quot;nofollow&quot;&gt;cryptogram solver&lt;/a&gt;, the message is revealed to be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MEET AT PINE TREE ROAD MAKE SURE TO BRING NO ONE AND ABSOLUTELY NO COMPUTERS I NEED TO VERIFY YOU ARE THE RECEIVER OF THIS ITEM PART COMING SOON I KNOW YOU LIVE ON OAK TREE COURT&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I wouldn't follow the instructions, as it seems the post office has confused your friend for the&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Teenage Mutant Ninja Turtles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="9871" LastEditDate="2015-12-05T22:18:04.317" LastActivityDate="2015-12-05T22:18:04.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24333" PostTypeId="2" ParentId="2827" CreationDate="2015-12-04T03:52:09.153" Score="1" Body="&lt;p&gt;cut the cake in an even X making 4 slices. Then stack the slices on top of each other and cut them down the center making 8 equal pieces with equal icing&lt;/p&gt;&#xA;" OwnerUserId="17634" LastActivityDate="2015-12-04T03:52:09.153" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24334" PostTypeId="1" AcceptedAnswerId="24335" CreationDate="2015-12-04T09:39:18.180" Score="18" ViewCount="1813" Body="&lt;p&gt;In Alfred Hitchcock's movie &lt;em&gt;&quot;To Catch a Thief&quot;&lt;/em&gt;, the retired infamous &#xA;jewel thief John Robie has to crack a safe with a complicated $4\times4$ &#xA;pattern of keys. Each of the 16 keys is either turned around ($+$) or &#xA;not turned around ($-$):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   - - + - &#xA;   + - - + &#xA;   - + - - &#xA;   + - - + &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;To open the safe, John Robie must make all keys turned around.&#xA;The catch is that whenever Robie turns one key (from $+$ to $-$ or from $-$ to $+$), then all the keys in the same row and all the keys in the same column simultaneously turn (from $+$ to $-$ or from $-$ to $+$ respectively).&#xA;John Robie may turn any key as often as he likes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What is the minimum number of keys that John Robie has to turn to open the safe?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="-1" LastEditDate="2015-12-04T18:04:31.953" LastActivityDate="2015-12-04T18:04:31.953" Title="To Catch a Thief" Tags="|mathematics|combinatorics|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24335" PostTypeId="2" ParentId="24334" CreationDate="2015-12-04T10:11:31.343" Score="19" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2 turns&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To turn a single key without changing the rest of the keys you have to turn the key itself plus the three other keys in the same row plus the three other keys in the same column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whe you do this the following happens:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Every key you don't turn youself will be turned twice.&lt;br&gt;&#xA;Once by the one in the same column as the key you didn't turn and the same row as the key you actually want to turn.&lt;br&gt;&#xA;Once by the key in the sam row as the key you didn't turn and the same column as the key you actually want to turn.&lt;br&gt;&#xA;Two turns bring the key back to its starting position.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Every key in the same row/column as the key you actually want to turn get turned four times.&lt;br&gt;&#xA;Once by itself.&lt;br&gt;&#xA;Three times by the other keys in the same row/column.&lt;br&gt;&#xA;Four turns bring the key back to its starting position.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The key you actually want to turn gets turned seven times.&lt;br&gt;&#xA;Once by itself.&lt;br&gt;&#xA;Three times by the keys in the same row.&lt;br&gt;&#xA;Three times by the keys in the same column.&lt;br&gt;&#xA;Seven turns bring it to the other position.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Using this on our 10 keys we want to turn we get the following needed turns:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;- - + -&#xA;+ - - +&#xA;- + - -&#xA;+ - - +&#xA;&#xA;4 5 6 4&#xA;4 4 4 4&#xA;4 6 5 4&#xA;4 4 4 4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Finally since two turns of the same key result in the starting position we can reduce the number of turns as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0 1 0 0&#xA;0 0 0 0&#xA;0 0 1 0&#xA;0 0 0 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For a total of 2 turns.&lt;/p&gt;&#xA;" OwnerUserId="16608" LastActivityDate="2015-12-04T10:11:31.343" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="24336" PostTypeId="1" AcceptedAnswerId="24338" CreationDate="2015-12-04T10:21:30.480" Score="11" ViewCount="862" Body="&lt;p&gt;&lt;strong&gt;Original Q&lt;/strong&gt;: Plot 24 points so that there is only one 23-sided polygon where 23 of those points are its vertices, and the other point is not on any edge of the polygon. A polygon, here, is defined as a simple closed curve that is made up of line segments. Therefore, any set of collinear points will always form only a single side (in a convex polygon). &quot;Simple closed curve&quot; means &quot;closed curve that does not intersect itself&quot;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&lt;strong&gt;Formal statement of Q&lt;/strong&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;We say that a finite set $P$ of points in the Euclidean plane contains a 23-gon $Q$,  if the following conditions are satisfied&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$Q$ has 23 pairwise distinct corners &lt;/li&gt;&#xA;&lt;li&gt;$Q$ has 23 pairwise non-crossing sides&lt;/li&gt;&#xA;&lt;li&gt;At every corner of $Q$, the two incident sides enclose an angle $\ne180^{\circ}$&lt;/li&gt;&#xA;&lt;li&gt;Every corner of $Q$ is contained in $P$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(Note that the interior of any side of $Q$ may contain an arbitrary number of points from $P$; there is no restriction on such points.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Task:&lt;/strong&gt;&#xA;  Plot a point set $P$ with $24$ points that contains exactly one 23-gon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="22922" LastEditDate="2016-08-04T15:36:58.310" LastActivityDate="2016-08-04T15:36:58.310" Title="Icosikaitrigons" Tags="|mathematics|geometry|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24337" PostTypeId="2" ParentId="24336" CreationDate="2015-12-04T14:36:20.827" Score="6" Body="&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: This is wrong. As Gamow points out, you can &quot;Skip the southern-most black point. Start from northern-most black, go to next black, to central red, to blue, to black, and then follow the parabola of red points.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I have it! Pardon the terrible art:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Tw5XK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Tw5XK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the black and blue points are exactly in a line. We cannot use the blue point. If we did, we'd still need to use at least three of the black points, which generates two adjacent sides (black-black-blue) forming a 180-degree angle no matter what.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now all that we have to show is that there is only one 23-gon possible with the remaining points (that is, the one that's shaped like a horseshoe with the concavity on the left.) We run into a similar problem with the central red point as with the blue point: if we try to use it in some other way by removing the concavity, we'll end up with three black points forming two adjacent sides with a 180-degree angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not a mathematician, so please tell me about any flaws you spot in my logic or any ways to make this clearer.&lt;/p&gt;&#xA;" OwnerUserId="10851" LastEditorUserId="10851" LastEditDate="2015-12-04T14:42:05.727" LastActivityDate="2015-12-04T14:42:05.727" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24338" PostTypeId="2" ParentId="24336" CreationDate="2015-12-04T15:46:03.063" Score="15" Body="&lt;p&gt;Inspired by Luke's idea:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/reFQt.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/reFQt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The blue point is collinear with the two big black points above it, and also collinear with the black points on the right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $P$ be a polygon made with some of these points, let $s$ be a side of this polygon, and let $\ell$ be the line containing $s$.  Notice that there are no points in the interior of the shape formed by all 24 points.  Thus, all the points of $P$ must lie either on $\ell$ or on one side of $\ell$ -- otherwise there would be a side that crosses $s$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $P$ is a icosikaitetragon and if the blue point is included, then $P$ must include three points in a line (either the blue point and the two above, or the blue point and the two on the right).  WLOG, assume it is the blue point and the two above it.   If the middle point is not an angle of $180^\circ$, then there is a side $s$ coming out of it that has points of $P$ on both sides, a contradiction.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence we must exclude the blue point.  From there, the natural polygon is the only one, since any other option would again have a side $s$ with points of $P$ on both sides.&lt;/p&gt;&#xA;" OwnerUserId="2722" LastActivityDate="2015-12-04T15:46:03.063" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24340" PostTypeId="1" AcceptedAnswerId="24349" CreationDate="2015-12-05T00:06:33.933" Score="2" ViewCount="266" Body="&lt;h2&gt;Background&lt;/h2&gt;  &#xA;&#xA;&lt;p&gt;Normal MtG rules apply here. Your solution must work regardless of any possible choices made by your opponents; assume they do whatever makes it most difficult for you to win. If you would draw a card, you may assume that you draw any card remaining in your library, regardless of shuffling.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Puzzle Setup&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;You are playing in a multiplayer game of Magic.  Decks are 60+ cards, no ban list besides unglued/unhinged, maximum  four of each card unless the card says otherwise (or is a basic land).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the other players have been eliminated, leaving only you and one opponent.  &lt;strong&gt;It is your opponent's first main phase&lt;/strong&gt;, nothing has summoning sickness, and no spells have been cast yet on his turn.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Win the game!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If there are multiple solutions, the best is the one which wins on the earliest turn, secondary tiebreaker is dealing the most damage)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your hand:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=370545&quot; rel=&quot;nofollow&quot;&gt;Angel's Grace&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=239964&quot; rel=&quot;nofollow&quot;&gt;Borderland Ranger&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=397612&quot; rel=&quot;nofollow&quot;&gt;Counterspell&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=397401&quot; rel=&quot;nofollow&quot;&gt;Dark Ritual&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=397522&quot; rel=&quot;nofollow&quot;&gt;Diabolic Edict&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=370552&quot; rel=&quot;nofollow&quot;&gt;Glacial Ray&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=83832&quot; rel=&quot;nofollow&quot;&gt;Grozoth&lt;/a&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your board:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=271178&quot; rel=&quot;nofollow&quot;&gt;Nullmage Advocate&lt;/a&gt; (Tapped)&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=152556&quot; rel=&quot;nofollow&quot;&gt;Heritage Druid&lt;/a&gt; (Tapped)&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=393996&quot; rel=&quot;nofollow&quot;&gt;Wellwisher&lt;/a&gt; (Tapped)&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=288992&quot; rel=&quot;nofollow&quot;&gt;Door to Nothingness&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=193467&quot; rel=&quot;nofollow&quot;&gt;Near-Death Experience&lt;/a&gt;&lt;br&gt;&#xA;2 Mountains&lt;br&gt;&#xA;2 Islands&lt;br&gt;&#xA;2 Plains&lt;br&gt;&#xA;2 Forests&lt;br&gt;&#xA;1 Swamp  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your graveyard:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=30555&quot; rel=&quot;nofollow&quot;&gt;Ancestral Tribute&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=382877&quot; rel=&quot;nofollow&quot;&gt;Cabal Ritual&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=122121&quot; rel=&quot;nofollow&quot;&gt;Controvert&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=32919&quot; rel=&quot;nofollow&quot;&gt;Elephant Ambush&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=386326&quot; rel=&quot;nofollow&quot;&gt;Hell's Thunder&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=122051&quot; rel=&quot;nofollow&quot;&gt;Garza's Assassin&lt;/a&gt;    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your library:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;em&gt;Whatever you want&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your life:&lt;/strong&gt;&lt;br&gt;&#xA;8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Opponent's hand:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=386288&quot; rel=&quot;nofollow&quot;&gt;Decree of Annihilation&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=401858&quot; rel=&quot;nofollow&quot;&gt;Dispel&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=230066&quot; rel=&quot;nofollow&quot;&gt;Mental Misstep&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=370469&quot; rel=&quot;nofollow&quot;&gt;Rift Bolt&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=370469&quot; rel=&quot;nofollow&quot;&gt;Rift Bolt&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Opponent's board:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=382379&quot; rel=&quot;nofollow&quot;&gt;Sulfuric Vortex&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=189910&quot; rel=&quot;nofollow&quot;&gt;Telepathy&lt;/a&gt;&lt;br&gt;&#xA;12 Mountains&lt;br&gt;&#xA;1 Island  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Opponent's graveyard:&lt;/strong&gt;&lt;br&gt;&#xA;Nothing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Opponent's library:&lt;/strong&gt;&lt;br&gt;&#xA;39 Mountains    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Opponent's life:&lt;/strong&gt;&lt;br&gt;&#xA;10&lt;/p&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="17389" LastEditDate="2015-12-05T02:52:44.973" LastActivityDate="2015-12-05T18:04:20.267" Title="Magic: the Gathering - Impending Annihilation" Tags="|calculation-puzzle|magic-the-gathering|" AnswerCount="1" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="24341" PostTypeId="1" AcceptedAnswerId="24344" CreationDate="2015-12-05T00:29:58.510" Score="2" ViewCount="670" Body="&lt;p&gt;(Your investigations continue &lt;a href=&quot;https://puzzling.stackexchange.com/questions/24353/who-are-the-guilty-parties-2&quot;&gt;here&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I am testing this puzzle to see how it works, and gauge the difficulty.  I have plans which will be unveiled in future puzzles of a similar nature, pending interest etc here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the rules:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is a group of people on an isolated island, and, in a curious twist, the butlers have all been murdered.  You have been sent to collect statements from each person.&lt;br&gt;&#xA;   - Each person in the group is either guilty or innocent.&lt;br&gt;&#xA;   - Each person is either friends or not friends with each other person.  The friendship is mutual.&lt;br&gt;&#xA;   - Each guilty person is friends with every other guilty person.&lt;br&gt;&#xA;   - Each innocent person has witnessed a number of events, which allows them to deduce the innocence or guilt of a number of people in the group.  This may or may not have been reciprocated.&lt;br&gt;&#xA;   - The guilty people know who the guilty and innocent parties are.&lt;br&gt;&#xA;   - Each person is either boastful or modest (to be explained below).  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The statements that each person makes are as follows:&lt;br&gt;&#xA;   - Innocent modest people will simply make a list of truthful statements.&lt;br&gt;&#xA;   - Innocent boastful people will do the same, but will also make a statement about a random other person about whom they know nothing.  If the person is their friend, they will say s/he is innocent.  Otherwise they'll just say something (innocent or guilty at random).&lt;br&gt;&#xA;   - Guilty modest people will say that friends are innocent (regardless of whether they are); they will say that non-friends are guilty (which is inaccurate, because their non-friends are innocent by the above definitions).&lt;br&gt;&#xA;   - Guilty boastful people will do the same.  But they will also make one truthful statement (either that one of their guilty friends is guilty or that one of their innocent non-friends is innocent).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This kind of thing happens a lot to you...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On this particular day, you go out to the island and get to sit down with the main characters: &lt;strong&gt;Bill, Celia, Laine, Timika, Delicia, and Terrance&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are their statements:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Statements of Bill:&lt;/strong&gt;&lt;br&gt;&#xA;   - I am innocent&lt;br&gt;&#xA;   - Delicia is guilty&lt;br&gt;&#xA;   - Laine is guilty&lt;br&gt;&#xA;   - Celia is guilty&lt;br&gt;&#xA;   - Timika is innocent  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Statements of Celia:&lt;/strong&gt;&lt;br&gt;&#xA;   - I am innocent&lt;br&gt;&#xA;   - Timika is guilty&lt;br&gt;&#xA;   - Bill is guilty&lt;br&gt;&#xA;   - Laine is innocent&lt;br&gt;&#xA;   - Terrance is innocent  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Statements of Laine:&lt;/strong&gt;&lt;br&gt;&#xA;   - I am innocent&lt;br&gt;&#xA;   - Bill is guilty&lt;br&gt;&#xA;   - Terrance is innocent  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Statements of Timika:&lt;/strong&gt;&lt;br&gt;&#xA;   - I am innocent&lt;br&gt;&#xA;   - Terrance is guilty&lt;br&gt;&#xA;   - Delicia is guilty&lt;br&gt;&#xA;   - Bill is innocent  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Statements of Delicia:&lt;/strong&gt;&lt;br&gt;&#xA;   - I am innocent&lt;br&gt;&#xA;   - Laine is innocent&lt;br&gt;&#xA;   - Bill is guilty&lt;br&gt;&#xA;   - Terrance is innocent  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Statements of Terrance:&lt;/strong&gt;&lt;br&gt;&#xA;   - I am innocent&lt;br&gt;&#xA;   - Timika is guilty&lt;br&gt;&#xA;   - Delicia is innocent&lt;br&gt;&#xA;   - Laine is innocent&lt;br&gt;&#xA;   - Bill is guilty  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This seems like a complete disaster.  You are struggling to find something sensible out of what they say.  So you phone your mentor and good friend, Hercule, and he says, &quot;Ze simplest solution is probably true.  &lt;strong&gt;Find ze solutio' wiz ze fewest conspirators&lt;/strong&gt;.  Zat will be your answer.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As always, post your solutions below, and please also comment about the puzzle - what you like and what you don't.  How difficult your found this.  As I said, I'm hoping to do a couple more, assuming this works as intended...&lt;/p&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-12-05T22:48:55.510" Title="Who are the guilty parties? #1" Tags="|logical-deduction|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24343" PostTypeId="1" AcceptedAnswerId="24363" CreationDate="2015-12-05T03:46:09.267" Score="8" ViewCount="586" Body="&lt;p&gt;If a word has a certain property, I call it an Alpha Word™. Based on the following list, determine the rule governing if a word is an Alpha Word™.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Alpha Words          Not Alpha Words&#xA;AND                  AN&#xA;BIOLOGY              APPLE&#xA;CODE                 AUTHORS&#xA;CONCEPTUAL           BASKETBALL&#xA;FANS                 BUT&#xA;INFORMATION          CODED&#xA;LAW                  CYTOPLASM&#xA;LEVEL                DESPITE&#xA;LITERATE             GOLF&#xA;MINING               LIFESTYLE&#xA;MOVIE                MACROECONOMICS&#xA;PASSIONATE           MEDIA&#xA;POWER                PICTURESQUE&#xA;RELATED              REALISTIC&#xA;SERIOUS              SEVERANCE&#xA;SYSTEM               SIGNAL&#xA;THOSE                SYNERGY&#xA;TRAIN                THULIUM&#xA;WARRIORS             VITAL&#xA;WISH                 XYLOPHONE&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Hint 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because the classifications of some of these words may change over time, it should be noted that this challenge was written on December 4, 2015.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;FOR&quot; demonstrates itself as an Alpha Word very much.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alpha Words are not Beta Words.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 4&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you are reading this, you are exactly one click away from finding the rule in this challenge. &lt;a href=&quot;https://goo.gl/qFI4PQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;Want to use your click on this?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 5 (makes this much easier)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are also one click away from finding all of the Beta words.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15603" LastEditorUserId="30633" LastEditDate="2016-11-08T00:02:02.360" LastActivityDate="2016-11-08T00:02:02.360" Title="What is an Alpha Word™?" Tags="|word|pattern|word-property|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24344" PostTypeId="2" ParentId="24341" CreationDate="2015-12-05T05:12:17.933" Score="3" Body="&lt;p&gt;Some useful facts about innocent people:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;An innocent person makes at most one false statement.  (Every one of their statements is true, except for at most one boastful statement, which can be true or false.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If an innocent person claims a set of people is guilty, no more than one of those people is innocent.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If an innocent person states that $n$ people are guilty, at least $n-1$ people are guilty.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $n$ people are guilty, an innocent person will claim that at most $n+1$ people are guilty.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;And some facts about guilty people:&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;If a guilty person claims a set of people is guilty, at most one of those people is guilty.  (Every statement they make about someone being guilty is false, as mentioned by Dr. Xorile, except for at most one boastful statement, which is always true.)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Since we want to find the case with the minimum number of guilty people, we can step through a number of cases: first assuming that no people are guilty, then assuming that one person is guilty, etc.  The first possible case that we find is the correct answer.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Zero&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;We can immediately eliminate the case where nobody is guilty; since in that case everyone would be innocent, and they would all claim that there was at most one guilty party (by property 4).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;One&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In this case, every innocent person claims that at most two people are guilty.  That means that Bill must be guilty (by property 4), since he claims three people are guilty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, that means everyone else is innocent, and at most one of Timika's statements:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Terrance is guilty&lt;/li&gt;&#xA;  &lt;li&gt;Delicia is guilty&lt;/li&gt;&#xA;  &lt;li&gt;Bill is innocent&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is false (property 1); however, all three statements are false: a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Two&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Assume that Bill is innocent.  Then at most one of his statements:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Delicia is guilty&lt;/li&gt;&#xA;  &lt;li&gt;Laine is guilty&lt;/li&gt;&#xA;  &lt;li&gt;Celia is guilty&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is false (property 1).  Since there are only two guilty parties, one of the first three statements must be false.  Since Bill can make at most one false statement, two of Delicia, Laine, and Celia must be guilty.  Everyone else (Timika and Terrance) must be innocent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since Terrance is innocent, at most one of his statements:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Timika is guilty&lt;/li&gt;&#xA;  &lt;li&gt;Delicia is innocent&lt;/li&gt;&#xA;  &lt;li&gt;Laine is innocent&lt;/li&gt;&#xA;  &lt;li&gt;Bill is guilty&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;can be false; however, we know that both the first and last are false: a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore our assumption was false, and (if there are exactly two guilty people) Bill is guilty.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Assume that Timika is innocent.  At most one of her statements:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Terrance is guilty&lt;/li&gt;&#xA;  &lt;li&gt;Delicia is guilty&lt;/li&gt;&#xA;  &lt;li&gt;Bill is innocent&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;can be false (property 1).  We know that the last is false.  However, since only one additional person can be guilty, at least one of the first two must also be false: a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, if exactly two people are guilty, Bill and Timika are guilty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we only need to check that no contradictions arise if Bill and Timika are guilty.  To do this I make a &lt;strong&gt;big&lt;/strong&gt; assumption:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Everybody is Modest&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This simplifies the rules, giving us some new properties:&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;6&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;Everything an innocent person says is true.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If a guilty person claims that another person is guilty, that person is innocent.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;We don't need to worry about guilty persons' claims that another person is innocent, since this only implies that the two are friends.  If the second party is guilty, this must be satisfied; otherwise the person is innocent, and their friendships don't matter.  I'll mark these statements with &lt;code&gt;don't care&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's check everyone's statements:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bill &lt;code&gt;Guilty&lt;/code&gt;:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;I am innocent &lt;code&gt;don't care&lt;/code&gt;&lt;/li&gt;&#xA;  &lt;li&gt;Delicia is guilty &lt;code&gt;innocent, OK&lt;/code&gt;&lt;/li&gt;&#xA;  &lt;li&gt;Laine is guilty &lt;code&gt;innocent, OK&lt;/code&gt;&lt;/li&gt;&#xA;  &lt;li&gt;Celia is guilty &lt;code&gt;innocent, OK&lt;/code&gt;&lt;/li&gt;&#xA;  &lt;li&gt;Timika is innocent &lt;code&gt;don't care&lt;/code&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;Celia &lt;code&gt;Innocent&lt;/code&gt;:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;I am innocent  &lt;code&gt;innocent, OK&lt;/code&gt;&lt;/li&gt;&#xA;  &lt;li&gt;Timika is guilty &lt;code&gt;guilty, OK&lt;/code&gt;&lt;/li&gt;&#xA;  &lt;li&gt;Bill is guilty &lt;code&gt;guilty, OK&lt;/code&gt;&lt;/li&gt;&#xA;  &lt;li&gt;Laine is innocent &lt;code&gt;innocent, OK&lt;/code&gt;&lt;/li&gt;&#xA;  &lt;li&gt;Terrance is innocent &lt;code&gt;innocent, OK&lt;/code&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;Laine &lt;code&gt;Innocent&lt;/code&gt;:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;I am innocent &lt;code&gt;innocent, OK&lt;/code&gt;&lt;/li&gt;&#xA;  &lt;li&gt;Bill is guilty &lt;code&gt;guilty, OK&lt;/code&gt;&lt;/li&gt;&#xA;  &lt;li&gt;Terrance is innocent &lt;code&gt;innocent, OK&lt;/code&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;Timika &lt;code&gt;Guilty&lt;/code&gt;:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;I am innocent &lt;code&gt;don't care&lt;/code&gt;&lt;/li&gt;&#xA;  &lt;li&gt;Terrance is guilty &lt;code&gt;innocent, OK&lt;/code&gt;&lt;/li&gt;&#xA;  &lt;li&gt;Delicia is guilty &lt;code&gt;innocent, OK&lt;/code&gt;&lt;/li&gt;&#xA;  &lt;li&gt;Bill is innocent &lt;code&gt;don't care&lt;/code&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;Delicia &lt;code&gt;Innocent&lt;/code&gt;:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;I am innocent &lt;code&gt;innocent, OK&lt;/code&gt;&lt;/li&gt;&#xA;  &lt;li&gt;Laine is innocent &lt;code&gt;innocent, OK&lt;/code&gt;&lt;/li&gt;&#xA;  &lt;li&gt;Bill is guilty &lt;code&gt;guilty, OK&lt;/code&gt;&lt;/li&gt;&#xA;  &lt;li&gt;Terrance is innocent &lt;code&gt;innocent, OK&lt;/code&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;Terrance &lt;code&gt;Innocent&lt;/code&gt;:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;I am innocent &lt;code&gt;innocent, OK&lt;/code&gt;&lt;/li&gt;&#xA;  &lt;li&gt;Timika is guilty &lt;code&gt;guilty, OK&lt;/code&gt;&lt;/li&gt;&#xA;  &lt;li&gt;Delicia is innocent &lt;code&gt;innocent, OK&lt;/code&gt;&lt;/li&gt;&#xA;  &lt;li&gt;Laine is innocent &lt;code&gt;innocent, OK&lt;/code&gt;&lt;/li&gt;&#xA;  &lt;li&gt;Bill is guilty &lt;code&gt;guilty, OK&lt;/code&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since there are no contradictions, it is possible that Bill and Timika are guilty, and (thanks to our previous arguments) we know that this solution is unique for two or fewer guilty parties.&lt;/p&gt;&#xA;" OwnerUserId="9718" LastActivityDate="2015-12-05T05:12:17.933" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24345" PostTypeId="2" ParentId="24336" CreationDate="2015-12-05T08:10:38.307" Score="1" Body="&lt;p&gt;@TylerSeacrest is right as well. Here is my solution which I thought of, even before posting the Q.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Take a regular $23$-gon. Take $4$ consecutive points, $A,B,C,D$. Extend $AB$ and $DC$, and name their intersection $X$. Now $A,B,C,D,E,\dots ,X$ is the required set of points.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2015-12-05T08:10:38.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24347" PostTypeId="1" AcceptedAnswerId="24352" CreationDate="2015-12-05T13:18:50.657" Score="7" ViewCount="1632" Body="&lt;p&gt;Can you make 10 triangles with just 10 sticks?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sticks don't have to have the same length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Try making 4 triangles with 6 sticks of equal length. Now put more shorter sticks inside this shape.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the world would have four or more dimensions, you could have made the shape with sticks that have exactly the same length. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are several other possibilities than the one I gave hints for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recently found one with at least 40 triangles. However, for that, overlapping sticks are needed. Why not try searching for the greatest amount of triangles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you disallow overlapping sticks, 13 triangles is possible. &lt;/p&gt;&#xA;" OwnerUserId="16322" LastEditorUserId="16322" LastEditDate="2015-12-05T14:16:49.677" LastActivityDate="2015-12-05T23:28:45.347" Title="Can you make 10 triangles with just 10 sticks?" Tags="|mathematics|triangle|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24348" PostTypeId="2" ParentId="24347" CreationDate="2015-12-05T14:09:13.677" Score="5" Body="&lt;p&gt;Two &lt;a href=&quot;https://ixquick-proxy.com/do/spg/show_picture.pl?l=english&amp;amp;rais=1&amp;amp;oiu=http%3A%2F%2Fi.stack.imgur.com%2FdohgO.png&amp;amp;sp=6218fd74c04320f8958fc27f9318834a&quot; rel=&quot;nofollow noreferrer&quot;&gt;5-point stars&lt;/a&gt; should do the trick! This solution uses 10 'sticks' (each stick spans vertex-to-vertex) of exactly the same length in 2D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As someone commented briefly (sadly, it was deleted too quickly for me to reference), we get 10 triangles even with a single 5-point star: 5 large, 5 small, and the large triangles overlap the small triangles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; This answer addresses the original question. As @DrXorile commented, a single 10-point star can produce many more triangles.&lt;/p&gt;&#xA;" OwnerUserId="9422" LastEditorUserId="9422" LastEditDate="2015-12-05T23:28:45.347" LastActivityDate="2015-12-05T23:28:45.347" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24349" PostTypeId="2" ParentId="24340" CreationDate="2015-12-05T16:38:40.623" Score="3" Body="&lt;p&gt;Well that was tough, but here it is, infinite damage on turn 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The opponent suspends two Rift Bolts, casts Decree, holding up Dispel for Counterspell and Misstep for Ritual.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Untap, take 2 from Sulfuric vortex, and draw &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=370428&quot; rel=&quot;nofollow&quot;&gt;Street Wraith&lt;/a&gt;.&lt;br&gt;&#xA;Cycle &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=370428&quot; rel=&quot;nofollow&quot;&gt;Street Wraith&lt;/a&gt; to draw &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=253589&quot; rel=&quot;nofollow&quot;&gt;Bazaar of Baghdad&lt;/a&gt;.&lt;br&gt;&#xA;Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=253589&quot; rel=&quot;nofollow&quot;&gt;Bazaar of Baghdad&lt;/a&gt; and activate it to draw and discard &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=108832&quot; rel=&quot;nofollow&quot;&gt;Call to the Netherwold&lt;/a&gt; and &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=176456&quot; rel=&quot;nofollow&quot;&gt;Fatestitcher&lt;/a&gt;.&lt;br&gt;&#xA;Cast &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=108832&quot; rel=&quot;nofollow&quot;&gt;Call to the Netherwold&lt;/a&gt; for its Madness cost of zero, returning &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=370428&quot; rel=&quot;nofollow&quot;&gt;Street Wraith&lt;/a&gt;.&lt;br&gt;&#xA;Cycle &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=370428&quot; rel=&quot;nofollow&quot;&gt;Street Wraith&lt;/a&gt; to draw &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=383000&quot; rel=&quot;nofollow&quot;&gt;Lion's Eye Diamond&lt;/a&gt;.&lt;br&gt;&#xA;Crack &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=383000&quot; rel=&quot;nofollow&quot;&gt;Lion's Eye Diamond&lt;/a&gt; for UUU, and Unearth &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=176456&quot; rel=&quot;nofollow&quot;&gt;Fatestitcher&lt;/a&gt;.&lt;br&gt;&#xA;Use &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=176456&quot; rel=&quot;nofollow&quot;&gt;Fatestitcher&lt;/a&gt; to untap &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=253589&quot; rel=&quot;nofollow&quot;&gt;Bazaar of Baghdad&lt;/a&gt;, tap it to draw and discard Fatestticher and &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=393880&quot; rel=&quot;nofollow&quot;&gt;Basking Rootwalla&lt;/a&gt; (casting &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=393880&quot; rel=&quot;nofollow&quot;&gt;Basking Rootwalla&lt;/a&gt; for Madness zero).&lt;br&gt;&#xA;Unearth the second &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=176456&quot; rel=&quot;nofollow&quot;&gt;Fatestitcher&lt;/a&gt;, untap &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=253589&quot; rel=&quot;nofollow&quot;&gt;Bazaar of Baghdad&lt;/a&gt;, draw and discard &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=405312&quot; rel=&quot;nofollow&quot;&gt;Mulldrifter&lt;/a&gt; and &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=389491&quot; rel=&quot;nofollow&quot;&gt;Dread Return&lt;/a&gt;.&lt;br&gt;&#xA;Flashback &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=389491&quot; rel=&quot;nofollow&quot;&gt;Dread Return&lt;/a&gt; on &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=405312&quot; rel=&quot;nofollow&quot;&gt;Mulldrifter&lt;/a&gt;, sacrificing all three creatures, drawing two cards.&lt;br&gt;&#xA;Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=382866&quot; rel=&quot;nofollow&quot;&gt;Black Lotus&lt;/a&gt;, &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=382841&quot; rel=&quot;nofollow&quot;&gt;Ancestral Recall&lt;/a&gt;.&lt;br&gt;&#xA;Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=383000&quot; rel=&quot;nofollow&quot;&gt;Lion's Eye Diamond&lt;/a&gt;, &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=383000&quot; rel=&quot;nofollow&quot;&gt;Lion's Eye Diamond&lt;/a&gt;, &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=88789&quot; rel=&quot;nofollow&quot;&gt;Ideas Unbound&lt;/a&gt;, crack &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=383000&quot; rel=&quot;nofollow&quot;&gt;Lion's Eye Diamond&lt;/a&gt; for UUURRR with Ideas on the stack.&lt;br&gt;&#xA;Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=370440&quot; rel=&quot;nofollow&quot;&gt;Pestermite&lt;/a&gt;, &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=243487&quot; rel=&quot;nofollow&quot;&gt;Seething Song&lt;/a&gt;, &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=397698&quot; rel=&quot;nofollow&quot;&gt;Kiki-Jiki, Mirror Breaker&lt;/a&gt;.&lt;br&gt;&#xA;Make infinite &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=370440&quot; rel=&quot;nofollow&quot;&gt;Pestermites&lt;/a&gt; and deal infinite damage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today I learned you can win a game of Magic with five life and a one card hand. Cool.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove that &quot;Opponent cycles Decree&quot; is beatable: The opponent cycles Decree and casts both Rift Bolts at you. You untap, Vortex on the stack you use Nullmage to kill the Vortex and Wellwisher to gain 1 life, thus surviving to your draw phase. From there you play Bazaar, draw Lotus + Ancestral, and skip straight to that part of the above combo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the opponent casts a rift bolt at one of your creatures, you'll survive into your main phase and win as above.&lt;/p&gt;&#xA;" OwnerUserId="7459" LastEditorUserId="7459" LastEditDate="2015-12-05T18:04:20.267" LastActivityDate="2015-12-05T18:04:20.267" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="24350" PostTypeId="2" ParentId="24347" CreationDate="2015-12-05T18:00:28.660" Score="2" Body="&lt;p&gt;From the hints I suspect the OP was also looking for a solution based on a tetrahedron. The simplest (non overlapping) solution is to have each face divided into two through one of the vertices. This gives 12 triangles (3 per face) and 10 sticks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wanted to do this exactly, you could make a smaller triangle parallel to one of the faces and with edges on the 3 remaining faces. This gives 8 triangles with 9 sticks, and the two extras can be made by dividing one of the remaining triangle in two.&lt;/p&gt;&#xA;" OwnerUserId="15942" LastActivityDate="2015-12-05T18:00:28.660" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24351" PostTypeId="1" AcceptedAnswerId="24372" CreationDate="2015-12-05T19:16:11.970" Score="14" ViewCount="457" Body="&lt;p&gt;A certain group of $n$ boxers is totally ordered by strength; if boxer &#xA;$X$ is stronger than boxer $Y$, then $X$ wins every single fight &#xA;against $Y$. The ordering itself is not known to us, and we do not&#xA;have the slightest idea which boxers are stronger than other boxers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, these boxers are also very moody; as soon as a boxer has &#xA;lost a total of six fights in a tournament, he is disappointed and &#xA;walks home right away (independently of the number of fights that he &#xA;has won up to that point). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our task is to organize a sequence of fights in a tournament so that &#xA;in the end we exactly know the ordering of all $n$ boxers (and hence &#xA;know for each pair, which of the two is the stronger one).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What is the maximum number $n$ of boxers for which we &#xA;  can settle our task with absolute certainty?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-12-07T17:50:32.867" Title="The moody boxers" Tags="|mathematics|strategy|combinatorics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24352" PostTypeId="2" ParentId="24347" CreationDate="2015-12-05T21:19:20.063" Score="9" Body="&lt;p&gt;Using non-overlapping sticks I get 36. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll give measurements and positions to make this easy to picture. Imagine we're drawing this on graph paper. Make one horizontal stick of width eight units. Pick a point somewhere above it, and attach sticks from it to each of 9 points on the line, all one unit apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now how many triangles are there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;say the horizontal stick is 0, and the others are 1-9 in order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are eight triangles with no sticks between them (012, 023, 034, etc...) where when I say 0 it is understood that I mean the segment of 0 necessary to complete a triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 7 triangles with 1 interior stick (013, 024, 035, etc...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;etc... 6 with 2 interior sticks (014, 025, ...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;etc... until we get 1 with 7 interior sticks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then 8+7+6+5+4+3+2+1 = 36.&lt;/p&gt;&#xA;" OwnerUserId="16995" LastActivityDate="2015-12-05T21:19:20.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24353" PostTypeId="1" AcceptedAnswerId="24354" CreationDate="2015-12-05T22:46:44.467" Score="5" ViewCount="419" Body="&lt;p&gt;You've barely got back from &lt;a href=&quot;https://puzzling.stackexchange.com/questions/24341/who-are-the-guilty-parties-1&quot;&gt;the island where your last case was&lt;/a&gt;, when you hear that you are needed again:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is a group of people at an isolated villa.  In a curious twist, the aristocrats have all been murdered.  You have been sent to collect statements from each person.&lt;br&gt;&#xA;   - Each person in the group is either guilty or innocent.&lt;br&gt;&#xA;   - Each person is either friends or not friends with each other person.  The friendship is mutual.&lt;br&gt;&#xA;   - Each guilty person is friends with every other guilty person.&lt;br&gt;&#xA;   - Each innocent person has witnessed a number of events, which allows them to deduce the innocence or guilt of a number of people in the group.  This may or may not have been reciprocated.&lt;br&gt;&#xA;   - The guilty people know who the guilty and innocent parties are.&lt;br&gt;&#xA;   - Each person is either boastful or modest (to be explained below).  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The statements that each person makes are as follows:&lt;br&gt;&#xA;   - Innocent modest people will simply make a list of truthful statements.&lt;br&gt;&#xA;   - Innocent boastful people will do the same, but will also make a statement about a random other person about whom they know nothing.  If the person is their friend, they will say s/he is innocent.  Otherwise they'll just say something (innocent or guilty at random).&lt;br&gt;&#xA;   - Guilty modest people will say that friends are innocent (regardless of whether they are); they will say that non-friends are guilty (which is inaccurate, because their non-friends are innocent by the above definitions).&lt;br&gt;&#xA;   - Guilty boastful people will do the same.  But they will also make one truthful statement (either that one of their guilty friends is guilty or that one of their innocent non-friends is innocent).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This situation seems all too common these days...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On this particular day, you go out to the island and get to sit down with the main characters:&lt;br&gt;&#xA;&lt;strong&gt;Elton, Roxann, Lamar, Laine, Diego, Marylee, Mika, Mendy, Tracey, and Ruth&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what you're able to glean from each one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Statements of Elton:&lt;/strong&gt;&lt;br&gt;&#xA;   - I am innocent&lt;br&gt;&#xA;   - Mika is guilty&lt;br&gt;&#xA;   - Lamar is innocent&lt;br&gt;&#xA;   - Laine is innocent  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Statements of Roxann:&lt;/strong&gt;&lt;br&gt;&#xA;   - I am innocent&lt;br&gt;&#xA;   - Elton is innocent&lt;br&gt;&#xA;   - Laine is innocent&lt;br&gt;&#xA;   - Lamar is guilty&lt;br&gt;&#xA;   - Mika is guilty&lt;br&gt;&#xA;   - Tracey is innocent&lt;br&gt;&#xA;   - Diego is innocent  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Statements of Lamar:&lt;/strong&gt;&lt;br&gt;&#xA;   - I am innocent&lt;br&gt;&#xA;   - Laine is innocent&lt;br&gt;&#xA;   - Mika is innocent&lt;br&gt;&#xA;   - Roxann is guilty&lt;br&gt;&#xA;   - Tracey is guilty  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Statements of Laine:&lt;/strong&gt;&lt;br&gt;&#xA;   - I am innocent&lt;br&gt;&#xA;   - Mika is guilty&lt;br&gt;&#xA;   - Mendy is guilty&lt;br&gt;&#xA;   - Tracey is innocent  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Statements of Diego:&lt;/strong&gt;&lt;br&gt;&#xA;   - I am innocent&lt;br&gt;&#xA;   - Elton is guilty&lt;br&gt;&#xA;   - Ruth is innocent&lt;br&gt;&#xA;   - Lamar is guilty&lt;br&gt;&#xA;   - Roxann is innocent  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Statements of Marylee:&lt;/strong&gt;&lt;br&gt;&#xA;   - I am innocent&lt;br&gt;&#xA;   - Mika is innocent&lt;br&gt;&#xA;   - Elton is guilty&lt;br&gt;&#xA;   - Laine is innocent  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Statements of Mika:&lt;/strong&gt;&lt;br&gt;&#xA;   - I am innocent&lt;br&gt;&#xA;   - Marylee is innocent&lt;br&gt;&#xA;   - Roxann is guilty&lt;br&gt;&#xA;   - Diego is innocent&lt;br&gt;&#xA;   - Laine is innocent&lt;br&gt;&#xA;   - Tracey is guilty  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Statements of Mendy:&lt;/strong&gt;&lt;br&gt;&#xA;   - I am innocent&lt;br&gt;&#xA;   - Laine is guilty&lt;br&gt;&#xA;   - Roxann is innocent&lt;br&gt;&#xA;   - Diego is innocent&lt;br&gt;&#xA;   - Lamar is innocent&lt;br&gt;&#xA;   - Mika is guilty  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Statements of Tracey:&lt;/strong&gt;&lt;br&gt;&#xA;   - I am innocent&lt;br&gt;&#xA;   - Laine is innocent&lt;br&gt;&#xA;   - Mika is guilty&lt;br&gt;&#xA;   - Elton is innocent  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Statements of Ruth:&lt;/strong&gt;&lt;br&gt;&#xA;   - I am innocent&lt;br&gt;&#xA;   - Mika is innocent&lt;br&gt;&#xA;   - Elton is innocent&lt;br&gt;&#xA;   - Laine is innocent  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It seems like a hopeless case, but then you remember occam's razor.  The simplest solution is likely to be it, in this case, the solution with the fewest number of co-conspirators is likely to be the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who murdered the aristocrats?&lt;/p&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-12-06T18:47:08.457" Title="Who Are The Guilty Parties? #2" Tags="|logical-deduction|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24354" PostTypeId="2" ParentId="24353" CreationDate="2015-12-06T02:00:24.463" Score="4" Body="&lt;p&gt;&lt;strong&gt;Answer:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;strong&gt;Lamar, Mika, Mendy, and Marylee murdered them.&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;  Also, if we'd rather not use &quot;Occam's Razor&quot; as our evidence in court, we can comfortably arrest Lamar, Mika, and Mendy, while letting Roxann and Laine go (as these are consistent across all six possible solutions).  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Logic:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let's start with a statement of &quot;facts&quot;:&lt;br&gt;&#xA;&lt;br&gt;&#xA; Name - Innocent (Not Innocent)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Elton – Lamar, Laine (Mika)&lt;br&gt;&#xA; Roxann – Elton, Laine, Tracey, Diego (Lamar, Mika)&lt;br&gt;&#xA; Lamar – Laine, Mika (Roxann, Tracey)&lt;br&gt;&#xA; Laine – Tracey, (Mika, Mendy)&lt;br&gt;&#xA; Diego – Ruth, Roxann (Elton, Lamar)&lt;br&gt;&#xA; Marylee – Mika, Laine (Elton)&lt;br&gt;&#xA; Mika – Marylee, Diego, Laine (Roxann, Tracey)&lt;br&gt;&#xA; Mendy – Roxann, Diego, Lamar (Laine, Mika)&lt;br&gt;&#xA; Tracey – Elton, Laine (Mika)&lt;br&gt;&#xA; Ruth – Mika, Elton, Laine&lt;br&gt;&#xA;&lt;br&gt;&#xA; So let's try to pair them into two groups, where &quot;innocent&quot; means same group and &quot;guilty&quot; means other group.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Assumption: Roxann and Tracey are in opposite groups.&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; In this case, both Lamar and Mika are boastful, with this as their only boast (either they are claiming an innocent or a friend guilty, depending on whether they're innocent or guilty). As a result, Mika must be friends with Laine (if Mika is guilty) or Laine must be innocent (if Mika is innocent).  Therefor, when Laine says Mika is guilty, that must be his boast (either they're both innocent, or they're both friends).  Therefor, Mendy must be in the other group from Laine.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Group 1: Mendy&lt;br&gt;&#xA;&lt;br&gt;&#xA; Group 2: Laine&lt;br&gt;&#xA;&lt;br&gt;&#xA; If Mika was innocent, then Marylee, Diego, and Laine must be innocent (because Roxann or Tracey is a boast), which means Laine's claim that Mika is guilty is a boast, so Tracey is innocent, which means her claims about Mika are a boast, so Elton is innocent, which means &lt;em&gt;his&lt;/em&gt; claim about Mika is also a boast, so Lamar is innocent.  But then Diego is innocent and wrong about both Elton and Lamar.  Thus, our original guess was wrong, and Mika is guilty.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Group 1: Mendy&lt;br&gt;&#xA;&lt;br&gt;&#xA; Group 2: Laine&lt;br&gt;&#xA;&lt;br&gt;&#xA; Guilty:&lt;br&gt;&#xA; Mika&lt;br&gt;&#xA;&lt;br&gt;&#xA; If Roxann was guilty, then Mika must be her boast, so Lamar is truly innocent.  Tracey would also be innocent, since she's in the other group from Roxann.  But then Lamar is boasting twice - he's wrong about both Mika and Tracey.  This can't be, so Roxann is innocent.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Since Roxann is innocent, Tracey must be Guilty (and a boast), so Elton, Laine, and Diego are innocent and Lamar must be guilty (and Mendy, being in the other group from Laine, must be guilty).  Diego is wrong about Elton, so that must be his boast, and Ruth is innocent.  Marylee is wrong about both Mika and Elton, so Marylee must be guilty.&lt;br&gt;&#xA;&lt;br&gt;&#xA; ---&lt;br&gt;&#xA;&lt;br&gt;&#xA; Possible Solution 1:&lt;br&gt;&#xA;&lt;br&gt;&#xA; Innocent:&lt;br&gt;&#xA; Roxann, Elton, Laine, Diego, Ruth&lt;br&gt;&#xA;&lt;br&gt;&#xA; Guilty:&lt;br&gt;&#xA; Mika, Tracey, Lamar, Marylee, Mendy&lt;br&gt;&#xA;&lt;br&gt;&#xA; ---&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Alternative: Roxann and Tracey are in the same group&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Group 1: Roxann, Tracey&lt;br&gt;&#xA;&lt;br&gt;&#xA; Neither Mika or Lamar can be in their group, because both of them claim they're both guilty.  For innocent or guilty, that would be two boasts.  Since we only have two total groups, Mika and Lamar are together.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Group 1: Roxann, Tracey&lt;br&gt;&#xA;&lt;br&gt;&#xA; Group 2: Mika, Lamar&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Assumption: Laine and Mika are in the same group&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; This means Laine is with Mika and Lamar&lt;br&gt;&#xA;&lt;br&gt;&#xA; Mendy says Laine and Mika are both guilty, so Mendy is not in their group (otherwise, that would be two boasts).&lt;br&gt;&#xA;&lt;br&gt;&#xA; Group 1: Roxann, Tracey, Mendy&lt;br&gt;&#xA;&lt;br&gt;&#xA; Group 2: Mika, Lamar, Laine&lt;br&gt;&#xA;&lt;br&gt;&#xA; But now Laine says that Tracey is innocent and Mika is guilty.  if Laine is innocent, then these are both boasts based on who shares a group.  Thus, Laine must be guilty, and friends with Tracey (where Mika is still a boast).&lt;br&gt;&#xA;&lt;br&gt;&#xA; Innocent:&lt;br&gt;&#xA; Roxann, Tracey, Mendy&lt;br&gt;&#xA;&lt;br&gt;&#xA; Guilty:&lt;br&gt;&#xA; Mika, Lamar, Laine&lt;br&gt;&#xA;&lt;br&gt;&#xA; But Roxann is innocent and said Laine was innocent - that's a boast, so Elton is innocent.  But then Elton says Lamar and Laine are innocent, which are both boasts!  This is inconsistent, so our assumption was false.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Conclusion: Laine and Mika are in different groups&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Group 1: Roxann, Tracey, Laine&lt;br&gt;&#xA;&lt;br&gt;&#xA; Group 2: Mika, Lamar&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Assumption: Mika is innocent&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; This means we have:&lt;br&gt;&#xA;&lt;br&gt;&#xA; Innocent:&lt;br&gt;&#xA; Mika, Lamar&lt;br&gt;&#xA;&lt;br&gt;&#xA; Guilty:&lt;br&gt;&#xA; Roxann, Tracey, Laine&lt;br&gt;&#xA;&lt;br&gt;&#xA; Mika is innocent and says that Laine is innocent, so that must be her only boast.  As a result, Marylee and Diego are innocent.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Innocent:&lt;br&gt;&#xA; Mika, Lamar, Marylee, Diego&lt;br&gt;&#xA;&lt;br&gt;&#xA; Guilty:&lt;br&gt;&#xA; Roxann, Tracey, Laine&lt;br&gt;&#xA;&lt;br&gt;&#xA; But then Diego is innocent, but claims Roxann is innocent and Lamar is guilty - these are both boasts, which is inconsistent. Thus, our assumption was false.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Conclusion: Mika is guilty&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Now we have:&lt;br&gt;&#xA;&lt;br&gt;&#xA; Innocent:&lt;br&gt;&#xA; Roxann, Tracey, Laine&lt;br&gt;&#xA;&lt;br&gt;&#xA; Guilty:&lt;br&gt;&#xA; Mika, Lamar&lt;br&gt;&#xA;&lt;br&gt;&#xA; If Mendy were innocent, claiming Lamar innocent and Laine guilty would be two boasts, so Mendy is guilty.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Innocent:&lt;br&gt;&#xA; Roxann, Tracey, Laine&lt;br&gt;&#xA;&lt;br&gt;&#xA; Guilty:&lt;br&gt;&#xA; Mika, Lamar, Mendy&lt;br&gt;&#xA;&lt;br&gt;&#xA; At this point, we've used all of Laine's statements, so we can stop worrying about him.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Assumption: Diego and Elton are in the same group&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; In this case, Diego must be boasting when he says Elton is guilty, so Lamar can't be a boast, which means Diego (and Elton) are innocent.  Also, Diego's claim about Ruth can't be a boast, so she's innocent.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Innocent:&lt;br&gt;&#xA; Roxann, Tracey, Laine, Diego, Elton, Ruth&lt;br&gt;&#xA;&lt;br&gt;&#xA; Guilty:&lt;br&gt;&#xA; Mika, Lamar, Mendy&lt;br&gt;&#xA;&lt;br&gt;&#xA; If Marylee were innocent, her claims about Mika and Elton would be boasts, so Marylee is guilty.&lt;br&gt;&#xA;&lt;br&gt;&#xA; ---&lt;br&gt;&#xA;&lt;br&gt;&#xA; Possible Solution 2:&lt;br&gt;&#xA;&lt;br&gt;&#xA; Innocent:&lt;br&gt;&#xA; Roxann, Tracey, Laine, Diego, Elton, Ruth&lt;br&gt;&#xA;&lt;br&gt;&#xA; Guilty:&lt;br&gt;&#xA; Mika, Lamar, Mendy, Marylee&lt;br&gt;&#xA;&lt;br&gt;&#xA; ---&lt;br&gt;&#xA;&lt;br&gt;&#xA; This means Ruth is boasting about Mika, Tracey is consistent, Mendy is friends with Roxann and Diego and boasting about Mika, Mika is friends with Deigo or Laine or both, Marylee might be friends with Laine, Diego is boasting about Elton, Lamar might be friends with Laine, Roxann is consistent, and Elton is boasting about Lamar.  Everybody's statements line up, so this is a &lt;em&gt;possible&lt;/em&gt; solution.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Alternative: Diego and Elton are in different groups&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Innocent:&lt;br&gt;&#xA; Roxann, Tracey, Laine&lt;br&gt;&#xA;&lt;br&gt;&#xA; Guilty:&lt;br&gt;&#xA; Mika, Lamar, Mendy&lt;br&gt;&#xA;&lt;br&gt;&#xA; in this case, Roxann has no useful knowledge to add (we know she boasted, but not about whom).  Tracey may or may not be boasting, but we don't gain any knowledge from her.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Assumption: Ruth is Innocent&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; In this case, Mika is a boast, so Elton is innocent, thus Diego is Guilty.  Marylee can't be innocent because then Mika and Elton would be boasts, so Marylee is guilty.  This gives:&lt;br&gt;&#xA;&lt;br&gt;&#xA; ---&lt;br&gt;&#xA;&lt;br&gt;&#xA; Possible Solution 3:&lt;br&gt;&#xA;&lt;br&gt;&#xA; Innocent:&lt;br&gt;&#xA; Roxann, Tracey, Laine, Ruth, Elton&lt;br&gt;&#xA;&lt;br&gt;&#xA; Guilty:&lt;br&gt;&#xA; Mika, Lamar, Mendy, Diego, Marylee&lt;br&gt;&#xA;&lt;br&gt;&#xA; ---&lt;br&gt;&#xA;&lt;br&gt;&#xA; Which is &lt;em&gt;also&lt;/em&gt; consistent (though it does imply that Laine is a very friendly person!)&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Alternative: Ruth is guilty&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; In this case, we have:&lt;br&gt;&#xA;&lt;br&gt;&#xA; Innocent:&lt;br&gt;&#xA; Roxann, Tracey, Laine&lt;br&gt;&#xA;&lt;br&gt;&#xA; Guilty:&lt;br&gt;&#xA; Mika, Lamar, Mendy, Ruth&lt;br&gt;&#xA;&lt;br&gt;&#xA; [Diego and Elton in different groups]&lt;br&gt;&#xA;&lt;br&gt;&#xA; At this point, for all remaining possibilities, there must be at least five guilty parties (Mika, Lamar, Mendy, Ruth, and Diego or Elton).  But remember our last clue: &lt;em&gt;&quot;The solution with the fewest number of co-conspirators is likely to be the answer&quot;&lt;/em&gt;.  And we already have a solution with only four guilty people!  Thus, the solution with four guilty people must be the correct answer (assuming that Occam's razor works for detective work).  Done!&lt;br&gt;&#xA;&lt;br&gt;&#xA; For completeness:&lt;br&gt;&#xA;&lt;br&gt;&#xA; ---&lt;br&gt;&#xA;&lt;br&gt;&#xA; Possible Solution 4:&lt;br&gt;&#xA;&lt;br&gt;&#xA; Innocent:&lt;br&gt;&#xA; Roxann, Tracey, Laine, Elton&lt;br&gt;&#xA;&lt;br&gt;&#xA; Guilty:&lt;br&gt;&#xA; Mika, Lamar, Mendy, Ruth, Diego, Marylee&lt;br&gt;&#xA;&lt;br&gt;&#xA; ---&lt;br&gt;&#xA;&lt;br&gt;&#xA; Possible Solution 5:&lt;br&gt;&#xA;&lt;br&gt;&#xA; Innocent:&lt;br&gt;&#xA; Roxann, Tracey, Laine, Deigo, Marylee&lt;br&gt;&#xA;&lt;br&gt;&#xA; Guilty:&lt;br&gt;&#xA; Mika, Lamar, Mendy, Ruth, Elton&lt;br&gt;&#xA;&lt;br&gt;&#xA; ---&lt;br&gt;&#xA;&lt;br&gt;&#xA; Possible Solution 6:&lt;br&gt;&#xA;&lt;br&gt;&#xA; Innocent:&lt;br&gt;&#xA; Roxann, Tracey, Laine, Deigo&lt;br&gt;&#xA;&lt;br&gt;&#xA; Guilty:&lt;br&gt;&#xA; Mika, Lamar, Mendy, Ruth, Elton, Marylee&lt;br&gt;&#xA;&lt;br&gt;&#xA; ---  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="17389" LastEditDate="2015-12-06T18:47:08.457" LastActivityDate="2015-12-06T18:47:08.457" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="24355" PostTypeId="2" ParentId="24351" CreationDate="2015-12-06T12:47:35.567" Score="6" Body="&lt;p&gt;The answer should be&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;64.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The strategy:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;(Quite similar to the mergesort technique used in sorting numbers)&lt;/p&gt;&#xA;&lt;p&gt;Divide the boxers into groups of two and let them fight. Create a hierarchy in each of the groups in terms of their ranks.&lt;/p&gt;&#xA;&lt;p&gt;Now consider the groups of two, two at a time. Let the bottom placed boxers have a fight. The loser of this fight is necessarily placed at the bottom of the four. Now the boxer placed above the one who just lost should fight the winner of the previous match. This strategy ensures that the loser fights only once. Continuing similarly, we can create a hierarchy in the group of four.&lt;/p&gt;&#xA;&lt;p&gt;Do this for the 16 pairs of groups of two.&lt;/p&gt;&#xA;&lt;p&gt;Continuing with the same strategy, we can hence, club them into groups of four. Note that the bottom placed ranker has lost only twice.&lt;/p&gt;&#xA;&lt;p&gt;Similarly, create pairs of the groups of four boxers. We can thus in the end come up with groups of eight boxers where the bottom ranker has lost thrice.&lt;/p&gt;&#xA;&lt;p&gt;This way we finally end up with a group of 64 where the boxer at the bottom has lost six times. He may then walk away.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17659" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-12-06T12:59:16.100" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24356" PostTypeId="1" AcceptedAnswerId="24368" CreationDate="2015-12-06T15:13:55.700" Score="10" ViewCount="549" Body="&lt;p&gt;You open a file and see the following codes. What do they mean?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;05132: Some off facing&#xA;6309: A salmon sandwiching&#xA;32001: Be laid bacon&#xA;24322: A hi conveys&#xA;5039: Rebirth solved&#xA;66622: Tried oozing&#xA;63142: Abandon prows&#xA;41622: Honesty fool&#xA;0099: Noisily cut pluck&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It's more-or-less impossible to solve without a hint, so here's the first one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each number corresponds to its phrase in a certain way. All phrases together form a list of some sort, and all items in the list have something in common (which is why they can be arranged).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If I generated the same file a few years later, its contents may be different.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3: Since this puzzle hasn't got any up-votes yet (but 57 views), it doesn't seem too popular. So here's the third hint already.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes, I just gave you the third hint. You didn't see it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4269" LastEditDate="2015-12-07T10:55:13.467" LastActivityDate="2015-12-07T12:59:47.033" Title="05132: Some off facing?" Tags="|cipher|enigmatic-puzzle|anagram|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24358" PostTypeId="1" AcceptedAnswerId="24360" CreationDate="2015-12-06T23:16:42.113" Score="-1" ViewCount="642" Body="&lt;p&gt;What is the largest well-defined finite integer that you can exactly describe using unabbreviated English words (no symbols and no letters from other alphabets), with no repeated letters. Any fractions will assume to be rounded down to the nearest integer. Anything that is time-dependent is assumed to refer to the moment the answer is given. The answer does not need to be a grammatically correct sentence so long as it is clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By well-defined, I mean a number that can be realistically determined. E.g., 'number of cats' might be large, but can't be determined. However, 'zero' of 'my age in hours' would both be fine (though you should give the answer for the latter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt; This is not a duplicate. The other question was under-constrained. In fact, I asked this because that question and the answers it got were unsatisfying.&lt;/p&gt;&#xA;" OwnerUserId="16995" LastEditorUserId="71" LastEditDate="2015-12-12T14:05:50.017" LastActivityDate="2015-12-12T14:05:50.017" Title="Describe a large integer" Tags="|mathematics|english|" AnswerCount="8" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="24360" PostTypeId="2" ParentId="24358" CreationDate="2015-12-06T23:38:58.650" Score="4" Body="&lt;p&gt;&lt;em&gt;Edit: a much larger answer.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Two up hymn age six&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which I will interpret as 2, followed by (hymn age) of Knuth up-arrows, six.  The age of hymns (according to wikipedia, and we should probably go with the western variety) is roughly 2,650 [years, the default English interpretation of age], thanks to the ancient Greeks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives us:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑6&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is less than Graham's number (in fact, less than g2, where Graham's is g64), but larger than anything else you could possibly conceive without also using up-arrow notation.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let's start this off with something decent:&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Planck time hour&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; An hour in units of &quot;Planck time&quot; (per Wolfram Alpha) is roughly&lt;br&gt;&#xA;&lt;br&gt;&#xA; $6.7 \times 10^{46}$&lt;br&gt;&#xA;&lt;br&gt;&#xA; ---&lt;br&gt;&#xA;&lt;br&gt;&#xA; Alternatively, combining two of the other ideas in here:&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Sixty Mole bang&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Which is $60(6.022141 \times 10^{23})!$, or (again, thanks Wolfram Alpha) roughly&lt;br&gt;&#xA;&lt;br&gt;&#xA; $10^{10^{25}} = 1 \times 10^{10000000000000000000000000}$&lt;br&gt;&#xA;&lt;br&gt;&#xA; ---&lt;br&gt;&#xA;&lt;br&gt;&#xA; But increasingly tortured readings of English may allow us to improve this to:&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Mole sixty bang&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; which is $(6.022141 \times 10^{23})!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!$, or roughly&lt;br&gt;&#xA;&lt;br&gt;&#xA; $1 \times 10^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{25}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}$&lt;br&gt;&#xA;&lt;br&gt;&#xA; I'd suggest some &lt;a href=&quot;https://en.wikipedia.org/wiki/Knuth&amp;#39;s_up-arrow_notation&quot; rel=&quot;nofollow&quot;&gt;up-arrow notation&lt;/a&gt; at this point...&lt;br&gt;&#xA;&lt;br&gt;&#xA; $(10↑↑61)^{25}$&lt;br&gt;&#xA;&lt;br&gt;&#xA; Of course, you might then also allow for&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Sixty Mole bang&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; where the mole applies to the number of bangs.  This gets messy quickly, but a very rough estimate would be:&lt;br&gt;&#xA;&lt;br&gt;&#xA; $10↑↑(10↑24)$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="4421" LastEditDate="2015-12-11T22:15:46.410" LastActivityDate="2015-12-11T22:15:46.410" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="24361" PostTypeId="2" ParentId="24358" CreationDate="2015-12-07T02:08:33.843" Score="0" Body="&lt;p&gt;How about:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;'MDCLXVI score bang'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the translation &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;MDCLXVI = 1666 in Roman numerals,&lt;br&gt;&#xA;  score = 20 in 'Old' English,&#xA;  and&lt;br&gt;&#xA;  bang = equivalent to the factorial symbol&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I get $(1666 \cdot 20)! = 33320! \approx 3.1525677 \cdot10^{136225}$&lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="4421" LastEditDate="2015-12-11T22:20:43.470" LastActivityDate="2015-12-11T22:20:43.470" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="24362" PostTypeId="2" ParentId="14749" CreationDate="2015-12-07T05:27:05.810" Score="-1" Body="&lt;p&gt;As many have already explained that first statement is proof enough to spot the convict, i.e&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A says: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I did not do it.    B did it.&#xA;TRUE(Its not A)     FALSE(Its not B) =&amp;gt; Its C&#xA;FALSE (Its A)       TRUE( Its B)     =&amp;gt; Contradiction. Hence, not possible.&#xA;Conclusion 1 =&amp;gt; C did it.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Explaining other statements:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B says: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I did not do it.       I know that C did it.&#xA;TRUE (Conclusion 1)  FALSE            &#xA;Conclusion 2=&amp;gt; B does not know that C did it &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;C says: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I did not do it.       B does not know who it was.&#xA;FALSE (Since C did it)  TRUE (Conclusion 2)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Hence, all the statements have one TRUE and FALSE statement and the convict is C.&lt;/p&gt;&#xA;" OwnerUserId="2656" LastActivityDate="2015-12-07T05:27:05.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24363" PostTypeId="2" ParentId="24343" CreationDate="2015-12-07T05:53:49.187" Score="6" Body="&lt;p&gt;I think I've got it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the alpha words can all be found on here: &lt;a href=&quot;http://stackexchange.com/sites&quot;&gt;http://stackexchange.com/sites&lt;/a&gt;, which is a page accessible with one click (Hint 4,5) on any of the &lt;em&gt;&quot;more&quot;&lt;/em&gt; hyper links at the bottom of the page. Make sure you're using list view and displaying all sites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alpha words are words that can be found in the subtitles of &lt;strong&gt;NON-BETA&lt;/strong&gt; stackexchanges. Non-alpha words are conversely any words that &lt;em&gt;cannot&lt;/em&gt; be found in the &lt;strong&gt;NON-BETA&lt;/strong&gt; subtitles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just control-f the words and click on the stackexchange to see if it's non-beta.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1 references the time-sensitivity of these words (since stackexchanges can gain / lose beta status over time).&#xA;Hint 2 is because the word &lt;em&gt;FOR&lt;/em&gt; is in every single subtitle.&lt;/p&gt;&#xA;" OwnerUserId="14095" LastEditorUserId="14095" LastEditDate="2015-12-07T10:02:36.313" LastActivityDate="2015-12-07T10:02:36.313" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24365" PostTypeId="2" ParentId="2827" CreationDate="2015-12-07T10:27:20.207" Score="8" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lOZqG.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lOZqG.png&quot; alt=&quot;Cut a cake into 8 pieces - with 3 cuts&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;As in the image; I will cut the cake horizontally first and two vertical cuts into it will be added. &#xA;&lt;br&gt; So it can have 8 pieces in total finally without even moving the cake position. &#xA;&lt;br&gt; I guess that's very simple and easy way. Isn't it? :D.&lt;/p&gt;&#xA;" OwnerUserId="11353" LastActivityDate="2015-12-07T10:27:20.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24366" PostTypeId="2" ParentId="24358" CreationDate="2015-12-07T10:53:33.810" Score="-4" Body="&lt;p&gt;&quot;The largest number ever used in a serious mathematical proof.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wich apparently isn't &lt;a href=&quot;https://en.wikipedia.org/wiki/Graham%27s_number&quot; rel=&quot;nofollow&quot;&gt;Graham's Number&lt;/a&gt; anymore:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Specific integers known to be far larger than Graham's number have since appeared in many serious mathematical proofs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9423" LastActivityDate="2015-12-07T10:53:33.810" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24367" PostTypeId="2" ParentId="2827" CreationDate="2015-12-07T10:57:51.853" Score="-3" Body="&lt;p&gt;It is simple.First of all find 360 degree divided by 8. So we get 45 degrees.Now, in our cake concept, cake is the circle with 360 degrees and as we cut into 8 equal parts, each cut part should have 45 degrees.&#xA;Now how to do this is:&#xA;Cut the cake directly into half.Take 45 degrees from an end of the cake to right side.Cut the cake right at this new position, in full to further half.&#xA;Repeat this step once again. We get 8 equal parts of piece cakes.&lt;/p&gt;&#xA;" OwnerUserId="17669" LastActivityDate="2015-12-07T10:57:51.853" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24368" PostTypeId="2" ParentId="24356" CreationDate="2015-12-07T11:23:39.670" Score="9" Body="&lt;p&gt;The numbers correspond to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reversed ID numbers of puzzles on this site created by user &lt;a href=&quot;https://puzzling.stackexchange.com/users/8874&quot;&gt;Gamow&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The phrases are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Anagrams of phrases that relate to each puzzle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The complete list:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 05132: Some off facing  $\;\;\;\rightarrow\;\;\;$ Game of coffins (&lt;a href=&quot;https://puzzling.stackexchange.com/questions/23150&quot;&gt;How to beat Count Dracula&lt;/a&gt;)&lt;br&gt;&#xA; 6309: A salmon sandwiching  $\;\;\;\rightarrow\;\;\;$ Dancing animals show (&lt;a href=&quot;https://puzzling.stackexchange.com/questions/9036&quot;&gt;The lion and the zebras&lt;/a&gt;)&lt;br&gt;&#xA; 32001: Be laid bacon  $\;\;\;\rightarrow\;\;\;$ Bob and Alice (&lt;a href=&quot;https://puzzling.stackexchange.com/questions/10023&quot;&gt;The subtraction game&lt;/a&gt;)&lt;br&gt;&#xA; 24322: A hi conveys  $\;\;\;\rightarrow\;\;\;$ Heavy coins (&lt;a href=&quot;https://puzzling.stackexchange.com/questions/22342&quot;&gt;Thirty genuine and seventy fake coins&lt;/a&gt;)&lt;br&gt;&#xA; 5039: Rebirth solved  $\;\;\;\rightarrow\;\;\;$ Devil's brother (&lt;a href=&quot;https://puzzling.stackexchange.com/questions/9305&quot;&gt;The Devil&amp;#39;s Brother&lt;/a&gt;)&lt;br&gt;&#xA; 66622: Tried oozing  $\;\;\;\rightarrow\;\;\;$ No zero digit (&lt;a href=&quot;https://puzzling.stackexchange.com/questions/22666&quot;&gt;The largest Monday number&lt;/a&gt;)&lt;br&gt;&#xA; 63142: Abandon prows  $\;\;\;\rightarrow\;\;\;$ Pawns on board (&lt;a href=&quot;https://puzzling.stackexchange.com/questions/24136&quot;&gt;A lonely pawn on the chessboard&lt;/a&gt;)&lt;br&gt;&#xA; 41622: Honesty fool  $\;\;\;\rightarrow\;\;\;$ Lots of honey (&lt;a href=&quot;https://puzzling.stackexchange.com/questions/22614&quot;&gt;Winnie-the-Pooh and the 27 honey pots&lt;/a&gt;)&lt;br&gt;&#xA; 0099: Noisily cut pluck  $\;\;\;\rightarrow\;\;\;$ Unlucky politics (&lt;a href=&quot;https://puzzling.stackexchange.com/questions/9900&quot;&gt;Thirteen politicians&lt;/a&gt;)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They are ordered by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; number of votes, descending.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12901" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-12-07T12:59:47.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24369" PostTypeId="2" ParentId="24343" CreationDate="2015-12-07T11:38:16.203" Score="2" Body="&lt;p&gt;Adding to @AZGreenTea's discovery:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If a word exists in non-beta sites' description, its an ALPHA word, even if it also exists in beta sites.&lt;/li&gt;&#xA;&lt;li&gt;If a word does not exist in alpha sites' description or exists in beta sites' description, its a BETA word.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Here are a few examples: &lt;a href=&quot;http://stackexchange.com/sites?view=list#traffic&quot;&gt;LINK&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alpha Words   :      &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;AND                  Stack Overflow&#xA;BIOLOGY              Biology&#xA;CODE                 Codereview&#xA;CONCEPTUAL           Programmers&#xA;FANS                 Anime and Manga&#xA;INFORMATION          Information Security&#xA;LAW                  Mi Yodeya&#xA;LEVEL                Mathematics&#xA;LITERATE             Personal Finance &amp;amp; Money&#xA;MINING               Cross Validated&#xA;MOVIE                Movies &amp;amp; TV&#xA;PASSIONATE           Arqade&#xA;POWER                Super User&#xA;RELATED              Mathematics&#xA;SERIOUS              Home Improvement&#xA;SYSTEM               Server Fault&#xA;THOSE                Christianity&#xA;TRAIN                Bicycles&#xA;WARRIORS             Travel&#xA;WISH                 Database Administrators   &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2656" LastActivityDate="2015-12-07T11:38:16.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24370" PostTypeId="1" AcceptedAnswerId="24654" CreationDate="2015-12-07T14:08:38.837" Score="30" ViewCount="1350" Body="&lt;p&gt;A while ago (maybe even a few years ago) I was given a Bletchley Park mug that is black when cold. When it heats up (from having a hot drink inside it) it turns white and reveals a sequence of pictures and a message about Bletchley Park at the top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was given to me as a gift by my brother after he'd visited the Bletchley Park Trust/Museum some time ago (years but not decades). It was sold at the gift shop there. I never really looked at it in detail before but I brought it into work today and when I looked at what the heat revealed I noticed the code. To add context Bletchley Park is where the UK code breakers who cracked the German Enigma Codes during World War 2 worked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the bottom is a sequence of letters:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;DBICE ENVSY FKJNZ WPFWU LW&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This isn't a simple shift cipher (I checked for that) but the general letter combinations and suchlike look valid for it to be a coded message of some sort. It's been a long time since I did any code cracking and this sample isn't really long enough or containing any double letters for the tricks I vaguely remember to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does seem suspicious that it is blocks of 5 characters in each &quot;word&quot; apart from the last 2 so the spaces could well be misleading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the context of where the mug came from I wouldn't be surprised to find the phrases &quot;Alan Turing&quot; or &quot;Bletchley Park&quot; and/or &quot;Trust&quot; in the translated text somewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is anyone able to decipher it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that it is most likely in British English)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mug cool:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/cNTW0.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cNTW0.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mug hot&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/klBUG.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/klBUG.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kn3VG.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kn3VG.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OQkFY.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OQkFY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17671" LastEditorUserId="17671" LastEditDate="2015-12-10T15:21:00.227" LastActivityDate="2015-12-17T14:45:50.783" Title="Solving an encoded message at the bottom of a Bletchley Park mug" Tags="|cipher|" AnswerCount="1" CommentCount="28" ContentLicense="CC BY-SA 3.0" />
  <row Id="24372" PostTypeId="2" ParentId="24351" CreationDate="2015-12-07T17:50:32.867" Score="7" Body="&lt;p&gt;I'll try to show&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 64 is the most $n$ could be.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Suppose you were just trying to determine the worst boxer, which is certainly necessary if you want to order all the boxers by ability.   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Suppose also every boxer starts off with a coin.  Whenever a boxer wins a match, they give all their coins, including those collected from other boxers, to the loser.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We now fix the matches so the following conditions hold:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The boxer with more losses ended up losing, unless&lt;/li&gt;&#xA;&lt;li&gt;It is a boxer with at least one win against a winless boxer.  In that case, the winless boxer loses regardless of the number of losses for both sides.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;With these two conditions, once you win you will never get any more coins.  Since in order to determine the worst boxer everyone except one needs to win, we see that one boxer will end up with all the coins.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I claim that on your $k$th loss, you will have at most $2^k$ coins.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof proceeds by induction.  This is true to start with if $k = 0$.   Otherwise, assume so far you've lost $k-1$ matches, and by induction have at most $2^{k-1}$ coins.  If you lose again, then the person you lost to has at most $k-1$ losses, and hence also has at most $2^{k-1}$ coins, and after collecting the winner's coins you will have at most $2^k$ coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With at most six loses you can have at most $64$ coins.  Since to determine the overall worst boxer requires that boxer collecting all the coins, there can be at most&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $64$ boxers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2722" LastActivityDate="2015-12-07T17:50:32.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24373" PostTypeId="1" CreationDate="2015-12-07T18:04:35.863" Score="10" ViewCount="4905" Body="&lt;p&gt;There is a scale with an object on the left-hand side, whose mass is given in some number of units. Predictably, the task is to balance the two sides. But there is a catch: You only have this peculiar weight set, having masses 1, 3, 9, 27, ... units. That is, one for each power of 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to output a list of strings representing where the weights should be placed, in order for the two sides to be balanced, assuming that weight on the left has mass x units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can you calculate this using balanced ternary number system?&lt;/p&gt;&#xA;" OwnerUserId="17673" LastEditorUserId="5373" LastEditDate="2017-08-26T00:15:47.580" LastActivityDate="2017-08-26T00:15:47.580" Title="Using a balance to determine any weight with a set of weights weighing powers of 3" Tags="|mathematics|logical-deduction|weighing|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24374" PostTypeId="2" ParentId="24373" CreationDate="2015-12-07T18:34:41.443" Score="5" Body="&lt;p&gt;I revised my answer because I found a much simpler approach (this requires $\mathcal{O}(\log n)$ weighings and $\mathcal{O}([\log n]^2)$ moved weights, which may not be the most efficient, though):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Let $n$ be the unknown weight. Pick a $k$ such that $3^k - \sum_{i=0}^{k-1}3^i \leq n$, i.e. the largest weight such that when placed on the non-object side and all weights below on the object side does not weigh over.&lt;/li&gt;&#xA;&lt;li&gt;Remove the largest weight on the object side and see if the scale weighs over. If it does, leave it. Otherwise, try to put that weight on the non-object side. If it weighs over now, remove the weight completely.&lt;/li&gt;&#xA;&lt;li&gt;Repeat for the second largest weight on the object side, and so forth, this until equilibrium.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The following Python code simulates the procedure:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;n = 71&#xA;L = []&#xA;k = 0&#xA;while n &amp;gt; sum([3**i for i in range(0,k)]): &#xA;    k += 1&#xA;    L.append(-1)&#xA;L.pop()&#xA;L.append(1)&#xA;k = len(L)-2&#xA;while k &amp;gt; -1:&#xA;    if sum([3**i*L[i] for i in range(0,len(L))])+3**k &amp;lt;= n:&#xA;        L[k] += 1&#xA;    else:&#xA;        k -= 1&#xA;print L, sum([3**i*L[i] for i in range(0,len(L))])&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="12901" LastEditorUserId="12901" LastEditDate="2015-12-07T20:45:37.120" LastActivityDate="2015-12-07T20:45:37.120" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24375" PostTypeId="2" ParentId="24373" CreationDate="2015-12-07T21:00:36.143" Score="5" Body="&lt;p&gt;To solve this, I began by listing out the first few natural numbers, and determining how I would balance the scale for each one.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Weight     Left        Right side&#xA;to         side        (x is the weight&#xA;measure                we are measuring)&#xA;----------------------------------------&#xA;1:         (1)         x&#xA;2:         (3)         x(1)&#xA;3:         (3)         x&#xA;4:         (3)(1)      x&#xA;5:         (9)         x(3)(1)&#xA;6:         (9)         x(3)&#xA;7:         (9)(1)      x(3)&#xA;8:         (9)         x(1)&#xA;9:         (9)         x&#xA;10:        (9)(1)      x&#xA;11:        (9)(3)      x(1)&#xA;12:        (9)(3)      x&#xA;13:        (9)(3)(1)   x&#xA;14:        (27)        x(9)(3)(1)&#xA;15:        (27)        x(9)(3)&#xA;16:        (27)(1)     x(9)(3)&#xA;17:        (27)        x(9)(1)&#xA;18:        (27)        x(9)&#xA;19:        (27)(1)     x(9)&#xA;20:        (27)(3)     x(9)(1)&#xA;21:        (27)(3)     x(9)&#xA;22:        (27)(3)(1)  x(9)&#xA;23:        (27)        x(3)(1)&#xA;24:        (27)        x(3)&#xA;25:        (27)(1)     x(3)&#xA;26:        (27)        x(1)&#xA;27:        (27)        x&#xA;28:        (27)(1)     x&#xA;29:        (27)(3)     x(1)&#xA;30:        (27)(3)     x&#xA;31:        (27)(3)(1)  x&#xA;32:        (27)(9)     x(3)(1)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I quickly noticed a pattern developing in how the weights were placed for each number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each weight $w$,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Place a weight on the left side if there exists an $n$ such that $n \; \text{mod} \; 3 = 1$ and $|wn - x| \le \frac{w-1}2$&lt;br&gt;&#xA; Place a weight on the right side if there exists an $n$ such that $n \; \text{mod} \; 3 = 2$ and $|wn - x| \le \frac{w-1}2$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My math skills are not good enough to solve for $n$ in the above inequalities (if it's even possible).  If that can be done, you can use a simple mathematical method to determine the arrangement of weights:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For each weight $w$ up to the first weight greater than $x$, solve for $n$.&lt;br&gt;&#xA; If $n = 1 \mod 3$, put the weight on the left.&lt;br&gt;&#xA; If $n = 2 \mod 3$, put the weight on the right.&lt;br&gt;&#xA; If $n = 0 \mod 3$, don't use the weight.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2015-12-07T21:00:36.143" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24376" PostTypeId="2" ParentId="24373" CreationDate="2015-12-07T23:30:29.127" Score="9" Body="&lt;p&gt;I think it's useful to represent the input number in base $3$. Our weight system can easily write numbers whose ternary representation includes only $0$'s or $1$'s, but it can't handle $2$'s very well. Thankfully, when you add the weight that corresponds to a digit occupied by a $2$, it becomes a $0$ (and carries $1$ over to the next ternary digit).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This observation should make it simple to figure out what to do in each case. For instance, suppose our input number is (in ternary) $220002022200221111222$. We will add some weights to it, that is, a ternary number made up only of $0$'s and $1$'s, so that the result is also one such number. You will see that the process also guarantees the summand and the sum will never share a $1$ in the same digit, so we are good!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start 'scanning' the input from the right until you find a $2$ (and remember this digit, say $k$). Now, continue scanning until you find a $0$. Notice this always happens in finite time because the input is finite (so it has many $0$'s 'to the left').&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have thus found a sequence of $1$'s and $2$'s, followed by a $0$. When you add the weight that correponds to digit $k$, that digit will become a $0$ and will carry over a $1$ to the next digit. This creates a domino effect, until the $1$ carries over to a digit which was occupied by a $1$ (rather than a $2$): that digit will become a $2$. So, if instead of $1$'s our sequence had only $2$'s, the domino effect would go on until the final $0$, which would become a $1$ itself. Well, it's easy to make a $1$ into a $2$: simply add $1$ to it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our algorithm then modifies our scanned sequence as follows: add the ternary number which has $1$'s wherever our scanned sequence has $1$'s and also a $1$ at the starting digit (and $0$'s everywhere else).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then it's a matter of continue scanning for such sequences in our input number until we are done. Notice once again that since our input is finite, it decomposes into finite such 'scan'-sequences. For our example, we have four such sequences, and we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{array}{rr}&#xA;  &amp;amp;220002022200221111222\\&#xA;+ &amp;amp;010001000100001111001\\&#xA;=&amp;amp;1000010100001000000000&#xA;\end{array}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice how the carry-over process guarantees the summand (the weights we add to the input mass) and the result (the weights we add on the other side to even things out) do not share $1$'s in the same ternary digits, so it's all good!&lt;/p&gt;&#xA;" OwnerUserId="16478" LastActivityDate="2015-12-07T23:30:29.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24377" PostTypeId="1" AcceptedAnswerId="24381" CreationDate="2015-12-08T01:14:27.433" Score="9" ViewCount="3017" Body="&lt;p&gt;You are the author for the magazine &lt;em&gt;Extreme Cuisine Quarterly&lt;/em&gt;. Your articles are not concerned with cooking, but rather with the presentation of the meal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recently, there have been rumors of a Russian chef who performs the following amazing feat. They say that during the appetizer course he brings out a huge cube of cheese and, in front of everyones eyes, cuts it up into several smaller cubes, &lt;strong&gt;all of which have different sizes&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your editor wants you to fly Russia to write an article about this Russian chef.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Save your magazine the cost of a plane ticket by proving the feat is impossible. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2016-01-09T23:34:49.000" Title="Chopping a cube into different sized subcubes" Tags="|geometry|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24378" PostTypeId="2" ParentId="24373" CreationDate="2015-12-08T02:59:22.983" Score="2" Body="&lt;p&gt;The members of the set of integers $A_n$ that total the natural number $n,$ where all members of the set are a power of three can be defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;A_n=\small{\left\{3^k \left(\left\lfloor 3^{-k} \left(2 n+3^{\left\lceil \log _3(2 n+1)\right\rceil }-1\right)\right/2\rfloor -3 \left\lfloor  3^{-k-1} \left(2 n+3^{\left\lceil \log _3(2 n+1)\right\rceil }-1\right)/2\right\rfloor -1\right)\\&#xA;~\Big{|}~ k\in \mathbb{Z} \land 0 \leq k \leq \left\lfloor \log _3\left(\left(3^{\left\lceil \log _3(2 n+1)\right\rceil }-1\right)/2+n\right)\right\rfloor\right\}}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Mathematica code:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a[n_] := DeleteCases[#, 0] &amp;amp;@ Table[3^k*(-1 - 3*Floor[(3^(-1 - k)*&#xA;(-1 + 3^Ceiling[Log[3, 1 + 2*n]] + 2*n))/2] + &#xA;Floor[(-1 + 3^Ceiling[Log[3, 1 + 2*n]] + 2*n)/(2*3^k)]), &#xA;{k, 0, Floor[Log[3, (-1 + 3^Ceiling[Log[3, 1 + 2*n]])/2 + n]]}]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then &lt;code&gt;a[#] &amp;amp;/@ Range @ 20&lt;/code&gt; gives&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;{{1}, {-1, 3}, {3}, {1, 3}, {-1, -3, 9}, {-3, 9}, {1, -3, 9}, &#xA;{-1, 9}, {9}, {1, 9}, {-1, 3, 9}, {3, 9}, {1, 3, 9}, {-1, -3, -9, 27}, &#xA;{-3, -9, 27}, {1, -3, -9, 27}, {-1, -9, 27}, {-9, 27}, {1, -9, 27}, &#xA;{-1, 3, -9, 27}}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Clearly the negative integers in $A_n$ should go on the same side as the weight, so&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f[n_] := {{Total@#}, Select[#, # &amp;gt; 0 &amp;amp;], Join[{&quot;W&quot;}, &#xA;-Select[#, # &amp;lt; 0 &amp;amp;]]} &amp;amp;@a@# &amp;amp;@n&#xA;&#xA;Grid[f[#] &amp;amp; /@ Range@20, Alignment -&amp;gt; Left]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;gives&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1      {1}          {W} &#xA;2      {3}          {W, 1} &#xA;3      {3}          {W} &#xA;4      {1, 3}       {W} &#xA;5      {9}          {W, 1, 3} &#xA;6      {9}          {W, 3}&#xA;7      {1, 9}       {W, 3} &#xA;8      {9}          {W, 1}&#xA;9      {9}          {W} &#xA;10     {1, 9}       {W} &#xA;11     {3, 9}       {W, 1} &#xA;12     {3, 9}       {W} &#xA;13     {1, 3, 9}    {W} &#xA;14     {27}         {W, 1, 3, 9} &#xA;15     {27}         {W, 3, 9} &#xA;16     {1, 27}      {W, 3, 9} &#xA;17     {27}         {W, 1, 9} &#xA;18     {27}         {W, 9} &#xA;19     {1, 27}      {W, 9} &#xA;20     {3, 27}      {W, 1, 9}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and the first thousand values of $A_n$ are calculated in $\approx 0.5$ seconds on my (slow) machine.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="2819" LastEditDate="2015-12-08T14:32:36.827" LastActivityDate="2015-12-08T14:32:36.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24379" PostTypeId="2" ParentId="24377" CreationDate="2015-12-08T03:27:22.137" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Making a cut on a cubiod will create two pieces that have have at least one side with similar dimensions.  As such, one piece will have to have that side cut through, and since it too is a cubiod, the same problem will be created again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13054" LastActivityDate="2015-12-08T03:27:22.137" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24380" PostTypeId="2" ParentId="24377" CreationDate="2015-12-08T04:12:54.310" Score="5" Body="&lt;p&gt;Can't be done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it could, then every planar cross-section of the cube would consist of a big square divided into several smaller squares all of which would have different sizes. This &lt;strong&gt;can&lt;/strong&gt; be done (see picture below), but the smallest square in any solution is always internal to the big square (i.e. not on an edge or corner). Consider the planar cross sections that cut through the &lt;em&gt;smallest cube&lt;/em&gt; in the entire block. The smallest-cube will be the smallest-square in these cross sections and will be internal to the large square (and also internal to the block). At least one of the surfaces of this smallest cube will be surrounded by 'walls' consisting the touching faces of the surrounding cubes. The only way to 'cover' this walled in face is to use still smaller cubes, so the cube cannot be the smallest. Hence disproof by contradiction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this is clear enough - it is hard to describe properly without complicated drawings or models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jUWr7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jUWr7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="1631" LastEditDate="2015-12-12T08:22:31.313" LastActivityDate="2015-12-12T08:22:31.313" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24381" PostTypeId="2" ParentId="24377" CreationDate="2015-12-08T05:04:50.347" Score="5" Body="&lt;p&gt;I assume the huge cube $C$ is cut up into several &lt;em&gt;finite&lt;/em&gt; smaller cubes $C_i$. Moreover, I assume that any finite partition of a cube into cubes must be stacked (that is, no $C_i$ is tilted relative to $C$). I provide no proof of this but I think this is pretty straightforward (examine a corner of $C$ and verify that it propagates to the rest).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will show that such a partition of the cube contains an 'decreasing' sequence of cubes, contradicting its finiteness. The proof follows primarily a two-dimensional reasoning, which I think is pretty neat and makes it easier to follow even without pictures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the bottom face $f_0$ of $C$. The partition of $C$ determines a partition of the square $f_0$ into finite squares each of which is a different size. Consider the &lt;strong&gt;smallest&lt;/strong&gt; square $s_0$ of such a partition.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Claim: $s_0$ must lie in the interior of $f_0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Indeed, let $s_0 = ABCD$, $f_0 = PQRS$ and suppose $s_0$ touches the boundary $\partial f_0$ of $f_0$. There are two cases:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$s_0$ is not in a corner of $f_0$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In this case, assume without loss of generality that $AB \subset PQ$ (so that $PQ$ is a barrier to squares surrounding $s_0$ in the partition of $f_0$) and that no other side of $s_0$ is contained in $\partial f_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The square that touches $BC$ must be larger than $s_0$, and because of the barrier $PQ$ its side along $BC$ must begin at $B$ and end past $C$. This means part of it acts as a barrier to squares that touch $CD$. We may repeat the process, deducing that the square that touches $CD$ must begin at $C$ and end past $D$, so part of it acts as a barrier to squares that touch $DA$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $DA$ is walled near $D$ by this last square and near $A$ by $PQ$, so a square that touches $DA$ cannot be larger than $s_0$, contradicting its (strict) minimality.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$s_0$ is in a corner of $f_0$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This case is actually easier than the previous one. Assume without loss of generality that $AB \subset PQ$ and $DA \subset SP$. Like before, part of the square that touches $BC$ acts as a barrier to the square that touches $CD$ near $C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $CD$ is walled near $C$ by this last square and near $D$ by $SP$, and once again we derive a contradiction with $s_0$'s minimality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The claim is thus proved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, consider the cube $c_0$ that $s_0$ is a face of; it stands on the bottom face $f_0$ of $C$. $c_0$ is surrounded on all sides by larger cubes that also stand on $f_0$, and thus its top face $f_1$ is walled on all sides. It follows that cubes touching $f_1 $ must all be smaller than $c_0$ and stand on $f_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These cubes determine a partition of the square $f_1$ into finite squares each of which is a different size; we may thus consider the smallest square $s_1$ of such a partition and the cube $c_1$ that $s_1$ is a face of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Procceeding by induction, we obtain a sequence ${(c_n)}_{n \geq0}$ of cubes in our partition in which $c_n$ is larger than $c_{n+1}$, as desired.&lt;/p&gt;&#xA;" OwnerUserId="16478" LastEditorUserId="16478" LastEditDate="2015-12-08T16:04:06.787" LastActivityDate="2015-12-08T16:04:06.787" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24382" PostTypeId="2" ParentId="3801" CreationDate="2015-12-08T05:33:19.220" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/e3F7c.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/e3F7c.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My answer is &lt;strong&gt;&lt;em&gt;9&lt;/em&gt;&lt;/strong&gt; as it can be seen in the picture. (without using 3D technique) :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11353" LastActivityDate="2015-12-08T05:33:19.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24384" PostTypeId="2" ParentId="24325" CreationDate="2015-12-08T10:06:24.290" Score="4" Body="&lt;p&gt;They are using &lt;code&gt;.NET&lt;/code&gt;'s built-in representation of time, the &lt;a href=&quot;https://msdn.microsoft.com/en-us/library/system.datetime(v=vs.110).aspx&quot; rel=&quot;nofollow&quot;&gt;&lt;code&gt;DateTime&lt;/code&gt; structure&lt;/a&gt;.  In particular, they are storing the internal representation in &lt;a href=&quot;https://msdn.microsoft.com/en-us/library/system.datetime.ticks(v=vs.110).aspx&quot; rel=&quot;nofollow&quot;&gt;&lt;code&gt;Ticks&lt;/code&gt;&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A single tick represents one hundred nanoseconds or one ten-millionth of a second. There are 10,000 ticks in a millisecond, or 10 million [10&lt;sup&gt;7&lt;/sup&gt;] ticks in a second.&lt;br&gt;&#xA;  The value of this property represents the number of 100-nanosecond intervals that have elapsed since 12:00:00 midnight, January 1, 0001 (0:00:00 UTC on January 1, 0001, in the Gregorian calendar), which represents &lt;code&gt;DateTime.MinValue&lt;/code&gt;. It does not include the number of ticks that are attributable to leap seconds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This matches with the epoch and tick size determined by the previous answerers.&lt;/p&gt;&#xA;" OwnerUserId="9718" LastActivityDate="2015-12-08T10:06:24.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24387" PostTypeId="2" ParentId="24373" CreationDate="2015-12-08T14:53:32.207" Score="3" Body="&lt;p&gt;@Fimpellizieri has the best answer here.  I am putting this in as an answer since it is too big for a comment. Give your votes to him.  I am just showing a simpler explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As he said, take the mass $x$ that we want to weigh and represent it in ternary.  For example, say the mass weighs $13,579$.  In base 3, this is $200121221$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we are looking for two numbers, $A$ and $B$ such that $A=x+B$.  These two numbers must have base 3 representations that only include $1$s or $0$s since we can only include one weight of each kind.  Also, they may not have a $1$ in the same position, but this is trivial to ensure because if they both do, then simply remove it and the equality will still hold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, back to our example.  Take the number $x$ in ternary and we need to solve for $A$ and $B$ as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with $B=0$ in the following equation.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x:   200121221&#xA;B: +         0&#xA;     ---------&#xA;A:   200121221&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Starting from the right, whenever we encounter a $2$ in A, put a $1$ in that position in $B$.  The first $2$ in $A$ is in the second position, so put a $1$ there in $B$.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x:   200121221&#xA;B: +        10&#xA;     ---------&#xA;A:   200122001&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;By changing $B$, $A$ has changed as well.  So simply repeat the process until $A$ has no more $2$s.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x:   200121221    200121221    200121221    200121221&#xA;B: +        10 -&amp;gt;      1010 -&amp;gt;    101010 -&amp;gt; 100101010&#xA;     ---------    ---------    ---------    ---------&#xA;A:   200122001    200200001    201000001   1001000001&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So, we now have $B=100101010 (6834) $ and $A=1001000001 (20413)$.  And you will notice that $6834+13579=20413$, so our equality holds.  Also, both $A$ and $B$ contain entirely $1$s and $0$s and never a $1$ in the same place.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we can set up our scales as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Left side: mass $x$ plus weights for $B$&lt;/li&gt;&#xA;&lt;li&gt;Right side: weights for $A$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In this example, this ends up with $B=3+3^3+3^5+3^8=3+27+243+6561$ and $A=3^0+3^6+3^9=1+729+19683$.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="1592" LastEditDate="2015-12-09T18:44:55.450" LastActivityDate="2015-12-09T18:44:55.450" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24389" PostTypeId="2" ParentId="4818" CreationDate="2015-12-08T16:29:13.160" Score="0" Body="&lt;p&gt;Eric's answer is not totally correct.  If you were mark the indistinguishable centers, to make them distinct, it STILL is possible on the 4x4 to have either PLL parity or OLL parity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, because the inner layer moves do an odd number of exchanges, it is possible on the 4x4 to exchange 2 edges and leave the rest intact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note, PLL parity is not really a parity issue at all.  It only looks like it because we try to reduce the 4x4 to a 3x3.  Actually, PLL parity in the 4x4 is a simple exchange of 2 pairs, and we don't think twice about that in the 3x3.&lt;/p&gt;&#xA;" OwnerUserId="17684" LastActivityDate="2015-12-08T16:29:13.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24402" PostTypeId="1" CreationDate="2015-12-08T23:27:44.837" Score="-2" ViewCount="243" Body="&lt;p&gt;Let there be a house of which there is a beautified front door that is closed. Now: what is that &quot;thing&quot; for which it is impossible, logically (and &lt;strong&gt;&lt;em&gt;irrefutably&lt;/em&gt;&lt;/strong&gt;), to be outside this closed door?&lt;/p&gt;&#xA;" OwnerUserId="17688" LastEditorUserId="15942" LastEditDate="2015-12-09T00:03:42.943" LastActivityDate="2015-12-09T00:48:09.353" Title="What cannot stand outside a door?" Tags="|logical-deduction|lateral-thinking|" AnswerCount="2" CommentCount="4" ClosedDate="2015-12-09T01:27:34.737" ContentLicense="CC BY-SA 3.0" />
  <row Id="24403" PostTypeId="2" ParentId="24402" CreationDate="2015-12-09T00:01:19.480" Score="1" Body="&lt;p&gt;There are probably many answers, but here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The threshold&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13054" LastActivityDate="2015-12-09T00:01:19.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24404" PostTypeId="2" ParentId="24402" CreationDate="2015-12-09T00:48:09.353" Score="3" Body="&lt;p&gt;I'm gonna go with...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The inside&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14095" LastActivityDate="2015-12-09T00:48:09.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24406" PostTypeId="2" ParentId="16881" CreationDate="2015-12-09T04:42:57.413" Score="12" Body="&lt;p&gt;This is not an answer, but some might find it helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, as has recently been indicated in some comments, I noticed the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You think the flying direction of butterflies is random? Think again!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;A&lt;/strong&gt; single butterfly (in &lt;strong&gt;one&lt;/strong&gt; direction). &lt;strong&gt;B&lt;/strong&gt;ut &lt;strong&gt;two&lt;/strong&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Have you taken a &lt;strong&gt;&lt;em&gt;c&lt;/strong&gt;lose look&lt;/em&gt; at hint &lt;strong&gt;3&lt;/strong&gt; above?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which seems to suggest that &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The frequency of the directions might encode letters.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's a demonstration:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I did this with some drag'n'dropping:&lt;a href=&quot;https://i.stack.imgur.com/V2hZH.jpg&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/V2hZH.jpg&quot; alt=&quot;Butterflies&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I think I waited too long to post this, so it may not help @Alconja much since it seems he's figured this out already.  Having said that, I noticed that,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; assuming the letters are correct, we have everything we need to spell some words, (most vowels, some useful consonants) though I couldn't get anything out of it.  I tried some things like looking at first or last letters of butterflies with certain colors, and a few other things, as well, though I haven't put much effort into that yet, so there could still be something there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All I really got out of it was a slight feeling of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/ZHsL7.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZHsL7.png&quot; alt=&quot;More Butterflies&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, for what it's worth,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; one of the first things that this kind of puzzle makes me think of is Morse code, however, in this case, I have a strong hunch that this is not relevant this time, even though it doesn't seem like a stretch to interpret the markings on the butterflies as dots and dashes.  Some butterflies would have to encode two letters if that were the case, and there would have also been the question of direction, though I personally would have started with anteriorly to posteriorly and in a brachial direction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17020" LastEditorUserId="17020" LastEditDate="2015-12-09T04:53:16.527" LastActivityDate="2015-12-09T04:53:16.527" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="24407" PostTypeId="1" AcceptedAnswerId="24454" CreationDate="2015-12-09T06:50:09.487" Score="4" ViewCount="537" Body="&lt;p&gt;I love solving puzzles, It keeps my brain active. Previously I used to solve newspaper &amp;quot;Sudoku&amp;quot; which were fairly easy. Then I solved some interesting puzzles in the book below.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Mastering Sudoku week by week-Paul Stephens&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I'm looking for any good website that offers medium to hard daily sudoku puzzles. Can anybody suggest me any good website, where I can get daily sudoku puzzles.&lt;/p&gt;&#xA;" OwnerDisplayName="user16080" LastEditorUserId="69582" LastEditDate="2021-06-03T14:15:49.053" LastActivityDate="2021-06-03T14:15:49.053" Title="Good resource for daily Sudoku Puzzles" Tags="|logical-deduction|grid-deduction|sudoku|reference-request|" AnswerCount="5" CommentCount="1" ClosedDate="2021-06-03T17:06:36.180" ContentLicense="CC BY-SA 4.0" />
  <row Id="24411" PostTypeId="2" ParentId="24407" CreationDate="2015-12-09T08:10:12.753" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.conceptispuzzles.com/de/index.aspx&quot; rel=&quot;nofollow&quot;&gt;This website&lt;/a&gt; has a few good ones for free (among other logic puzzles).&lt;br&gt;&#xA; I think you would need to subscribe though if you want to get new ones on a regular basis.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16608" LastActivityDate="2015-12-09T08:10:12.753" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24414" PostTypeId="1" CreationDate="2015-12-09T09:25:39.813" Score="-2" ViewCount="278" Body="&lt;p&gt;The number is 154967328. No other information is given, except that it's possible to answer in a minute (so no complex calculations). What is specil about the number?&lt;/p&gt;&#xA;" OwnerUserId="17692" LastActivityDate="2015-12-09T10:14:03.363" Title="What is special about this number?" Tags="|riddle|enigmatic-puzzle|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24415" PostTypeId="2" ParentId="24414" CreationDate="2015-12-09T09:36:49.753" Score="2" Body="&lt;p&gt;It contains all digits from 1 to 9.&lt;/p&gt;&#xA;" OwnerUserId="17482" LastActivityDate="2015-12-09T09:36:49.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24417" PostTypeId="2" ParentId="24414" CreationDate="2015-12-09T10:14:03.363" Score="3" Body="&lt;p&gt;It has the digits 1 through 9 exactly, and the digits that perfect squares can end in (14569) come before the others.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-12-09T10:14:03.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24419" PostTypeId="2" ParentId="10421" CreationDate="2015-12-09T10:41:52.893" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It will just be SKY BLUE :D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The runner was just finding the exit but was in the cave where there cannot be seen any light. But the man with the gun asked for the color of sky while he was running in the maze. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I guess he would just say sky's original color, sky blue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11353" LastEditorUserId="11452" LastEditDate="2015-12-09T15:01:46.783" LastActivityDate="2015-12-09T15:01:46.783" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24421" PostTypeId="1" AcceptedAnswerId="24422" CreationDate="2015-12-09T11:06:00.483" Score="-3" ViewCount="579" Body="&lt;p&gt;What is the missing alphabet in below sequence? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;M T W T _  _ S&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is a sequence of a _ _ _ _ :D That's it. Guess it's easy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11353" LastActivityDate="2015-12-09T13:52:49.630" Title="Guess the missing alphabet!" Tags="|logical-deduction|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24422" PostTypeId="2" ParentId="24421" CreationDate="2015-12-09T11:07:07.880" Score="2" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; F S&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because the series is the first letter of &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;DAYS IN A WEEK STARTING FROM MONDAY&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Monday Tuesday Wednesday Thursday Friday Saturday Sunday&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastActivityDate="2015-12-09T11:07:07.880" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24424" PostTypeId="1" AcceptedAnswerId="24425" CreationDate="2015-12-09T13:43:46.417" Score="4" ViewCount="135" Body="&lt;p&gt;The chessboard below contains two pawns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PdiAn.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PdiAn.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How many squares (of any size) does not contain a pawn?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17698" LastActivityDate="2015-12-09T13:48:21.483" Title="Pawn-free squares" Tags="|chess|checkerboard|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24425" PostTypeId="2" ParentId="24424" CreationDate="2015-12-09T13:48:21.483" Score="5" Body="&lt;pre&gt;&lt;code&gt;5*5: 1&#xA;4*4: 6&#xA;3*3: 18&#xA;2*2: 41&#xA;1*1: 62&#xA;total: 128&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="16608" LastActivityDate="2015-12-09T13:48:21.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24426" PostTypeId="2" ParentId="24421" CreationDate="2015-12-09T13:52:49.630" Score="1" Body="&lt;p&gt;These are Days in the week , therefor answer will be F (for friday) &amp;amp; S (for Saturday).&lt;/p&gt;&#xA;" OwnerUserId="7657" LastActivityDate="2015-12-09T13:52:49.630" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24427" PostTypeId="1" AcceptedAnswerId="24428" CreationDate="2015-12-09T14:58:47.377" Score="4" ViewCount="972" Body="&lt;p&gt;(My previous question was maybe a bit too easy. This one is just slightly harder...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Twenty-seven white blocks are assembled into a single cube and its outside is painted blue.  Foolishly placed on the the floor, the cube is one day stepped on by a blind man, and it breaks into pieces (all twenty-seven are separated and spread across the floor completely at random). He tries to reassemble the pieces.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the probability that the cube which the blind man assembles is blue?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;: the blind man cannot feel the blue paint in any way, it has identical structure to the white paint. No lateral thinking tag.&lt;/p&gt;&#xA;" OwnerUserId="17698" LastEditorUserId="17698" LastEditDate="2015-12-09T15:11:05.367" LastActivityDate="2015-12-09T18:28:49.470" Title="The blind man and the broken cube" Tags="|combinatorics|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24428" PostTypeId="2" ParentId="24427" CreationDate="2015-12-09T15:30:35.193" Score="5" Body="&lt;p&gt;The center cube must be placed in the middle, the centers of faces must be placed in their proper places, and the edges must be placed in their proper places.This guarantees that the corners are in their places. The number of ways to place each cube in its proper place (ignoring rotation) is...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${27\choose1}{26\choose6}{20\choose8}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have to multiply by the probability that each cube is in its proper alignment. For the corners, the correct corner has to be pointing out (1/8); for the edges, the correct edge has to be pointing out (1/12); for the face centers, the correct face has to be pointing out (1/6). This gives a combined probability of...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{{27\choose1}{26\choose6}{20\choose8}8^812^{12}6^6}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is about $1.83\times10^{-37}$.&lt;/p&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2015-12-09T15:30:35.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24429" PostTypeId="1" AcceptedAnswerId="24433" CreationDate="2015-12-09T15:30:49.967" Score="10" ViewCount="1448" Body="&lt;p&gt;The grandfather of my friend Bob has built a perfect square castle for himself and his family and divided it into 9 perfect square halls and made an armoury in the middle one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bob's father divided each of the remaining 8 halls into another 9 perfect square halls and in each middle one he made a garden.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bob himself divided each of the remaining 64 halls into 9 perfect square halls and in each middle one he made a swimming pool and the rest became bedrooms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recently Bob told me that he managed to walk through each bedroom visiting each one of them exactly once and returning back to the one where he started. Each bedroom has a door to another bedroom if they share a wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I being myself decided to check if it's possible before fully trusting him, but I do not have a lot of time, so I'm asking you: can I trust Bob or does he just want to look cool?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Picture:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/V63VM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/V63VM.png&quot; alt=&quot;Picture of Castle&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA; A=Armoury G=Garden P=Pool&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9352" LastEditorUserId="7715" LastEditDate="2015-12-10T03:52:22.170" LastActivityDate="2015-12-10T03:52:22.170" Title="An old square castle with many square rooms" Tags="|graph-theory|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="24430" PostTypeId="1" AcceptedAnswerId="24434" CreationDate="2015-12-09T15:39:04.900" Score="14" ViewCount="2974" Body="&lt;p&gt;How does one dissect a $10\times2$ rectangle into four pieces that can be reassembled to form a square?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-12-09T22:43:01.157" Title="Cutting a 10-by-2 rectangle" Tags="|mathematics|geometry|dissection|" AnswerCount="4" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="24431" PostTypeId="2" ParentId="24429" CreationDate="2015-12-09T15:57:05.747" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bob could have done it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is actually pretty easy to prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First we only look at the smallest rings of bedroms around the swimming pools.&lt;br&gt;&#xA;It is pretty easy to form a cycle by going through those 8 rooms and returning to the one you started at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next we look at each 8 of those cycles around the gardens.&lt;br&gt;&#xA;To connect those 9 cycles into a bigger circuit we make a total of 7 connections.&lt;br&gt;&#xA;For each of those connections we look at a single, not yet connected, wall between 2 adjacent inner cycles and take 4 of their roms, two from each cycle, that stand in a square right next to the wall between the cycles.&lt;br&gt;&#xA;If we now severe the connections between the two rooms of each cycle respectively and instead make two new connections between the two pairs that are on opposite sides of the wall between the cycles we have effectively combined two cycles into a single one.&lt;br&gt;&#xA;Do that 7 times for each garden and you get 8 cycles each spanning a total of 64 bedrooms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last step is to combine those 8 large cycles into a complete cyclet with 7 more of these connections. &lt;/p&gt;&#xA;" OwnerUserId="16608" LastActivityDate="2015-12-09T15:57:05.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24432" PostTypeId="2" ParentId="24430" CreationDate="2015-12-09T16:21:02.250" Score="10" Body="&lt;p&gt;Well, I can do it with 4 cuts, 5 pieces&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cut two 4x2 sections (2 cuts). Cut them diagonally (4 total cuts). The diagonals have length $\sqrt{20}$.  Assemble them so the diagonals are the sides of the new square, length 4 against length 2. That leaves a 2x2 hole in the middle, for your remaining 2x2 piece.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-12-09T16:21:02.250" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24433" PostTypeId="2" ParentId="24429" CreationDate="2015-12-09T16:26:16.807" Score="13" Body="&lt;p&gt;This is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; possible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is a picture describing why.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;r = room&#xA;p = pool&#xA;g = garden&#xA;a = armoury&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's a closed path, so choose any room to start in.&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/GFwdu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GFwdu.png&quot; alt=&quot;closed path&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2015-12-09T16:26:16.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24434" PostTypeId="2" ParentId="24430" CreationDate="2015-12-09T16:42:02.507" Score="31" Body="&lt;p&gt;Index your rectangle from (0,0) to (10,2). Then cut from  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (3,0) to (4,2); (4,2) to (8,0); (8,0) to (9,2).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These four pieces can be used to make the square.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/8LM7K.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8LM7K.png&quot; alt=&quot;Dissection&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that this dissection works without any flipping or even any rotation of the pieces!  To show that it's a square is also relatively simple.  Easy geometry shows that the angles are right angles, and that the sides are $2\sqrt{5}$.&lt;/p&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2015-12-09T19:04:41.913" LastActivityDate="2015-12-09T19:04:41.913" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24435" PostTypeId="2" ParentId="24427" CreationDate="2015-12-09T18:04:02.710" Score="2" Body="&lt;p&gt;The cubes have 0, 1, 2, or 3 sides painted blue.  The final assembly is correct if all the pieces are in the correct groups, and oriented so the blue faces are out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 8 cubes with 3 blue faces.  12 with 2 blue faces, 6 with 1 blue face, and 1 with 0 blue faces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each cube can be oriented in 24 ways (pick any face to be &quot;up&quot;, and then one of the four adjoining faces to be &quot;forward&quot;).  The 3-blue cubes can be oriented with all blue sides out in 3 different ways.  The 2-blue cubes in 2 ways.  The 1-blue cube in 4 ways, and the 0-cube in 24 ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The odds of getting the cubes in the right groups are: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_1 = {8! \times 12! \times 6! \times 1! \over 27!} \approx 1.27 \times 10^{-12}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the cubes are put into the right places, the odds that they are oriented correctly is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_2 = {3^8 \times 2^{12} \times 4^6\times 24^1\over 24^{27}} = 24^{-18}\approx 1.43\times 10^{-25}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total probability is $P_3 = P_1 \times P_2 \approx 1.82 \times 10^{-37}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it's interesting that $P_2$ is so much smaller than $P_1$.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2015-12-09T18:04:02.710" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24437" PostTypeId="2" ParentId="24427" CreationDate="2015-12-09T18:18:51.297" Score="2" Body="&lt;p&gt;We must consider both position and orientation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a single cube like a die.  There are 6 faces, so that means there are 6 different faces a cube can have up.  For each of these ways, it can be rotated 4 ways.  Thus, there are 24 ways to position a single cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total number of possibilities for the positions of pieces of the cube are $27!$.  The number of orientations of each cube is $24$, and their are $27$ of them.  So the total number of possible composite cubes is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$27! \times 24^{27}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A $3\times3\times3$ composite cube has:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$1$ centre piece&lt;/li&gt;&#xA;&lt;li&gt;$8$ corner pieces&lt;/li&gt;&#xA;&lt;li&gt;$12$ edge pieces&lt;/li&gt;&#xA;&lt;li&gt;$6$ middle pieces (centre of each face)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The centre piece must be placed in the middle, but orientation doesn't matter. Thus, if you have a valid composite cube with the centre piece in the right location, three are $24$ variants that are also valid simply by rotating this piece through all possible rotations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The middle pieces can be placed in any of the middle places.  They need the blue face outward, but other than that, can be in any of the $4$ possible rotations.  So, if a composite cube is valid, you could make $6!$ valid composite cubes simply be re-positioning the middle pieces.  For each of these valid composite cubes, you could take a middle piece and rotate it $4$ ways.  Thus, there are $6! \times 4^6$ valid cubes that can be made by repositioning and rotating the middle pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The edge pieces must be placed in the edge.  The piece can be in two possible orientations where both outward sides are blue.  Thus, given a valid cube, you could re-position the 12 edge pieces in $12!$ different ways.  For each positioning, you could orient each edge piece in $2$ ways.  Thus, there are $12! \times 2^{12}$ valid cubes that can be made by repositioning and rotating the edge pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, the corner pieces must be placed at the corner.  There are three valid ways that this can be done by rotating them so that the vertex of the three faces remains the vertex of the composite cube.  Thus, there are $8!$ ways of positioning these corners in a valid cube, and $8^3$ ways of rotating the corners.  So, there are $8! \times 3^8$ valid cubes made from repositioning and rotating the corners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, putting this all together, if we divide the number of valid composite cubes by the total number of composite cubes, we will get our answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{24 \times 8! \times 3^8 \times 12! \times 2^{12} \times 6! \times 4^6}{27! \times 24^{27}}=1.8298051 \times 10^{-37}$$&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="1592" LastEditDate="2015-12-09T18:28:49.470" LastActivityDate="2015-12-09T18:28:49.470" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24439" PostTypeId="1" AcceptedAnswerId="24449" CreationDate="2015-12-09T19:19:13.490" Score="3" ViewCount="195" Body="&lt;p&gt;You have six loaded dice (I mean falsified, non regular).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first die there is always a $10\%$ of probability for $1$ to come out.&lt;br&gt;&#xA;In the second die there is always a $20\%$ of probability for $2$ to come out.&lt;br&gt;&#xA;In the third die there is always a $30\%$ of probability for $3$ to come out.&lt;br&gt;&#xA;In the fourth die there is always a $40\%$ of probability for $4$ to come out.&lt;br&gt;&#xA;In the fifth die there is always a $50\%$ of probability for $5$ to come out.&lt;br&gt;&#xA;In the sixth die there is always a $60\%$ of probability for $6$ to come out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All six dice are 6 sided, and the remaining numbers of each dice have an equal probability to come out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the probability for $21$ to come out?&lt;/p&gt;&#xA;" OwnerUserId="17708" LastEditorUserId="15942" LastEditDate="2015-12-09T19:51:44.197" LastActivityDate="2015-12-10T02:30:44.193" Title="A question of.. Probability!" Tags="|probability|dice|" AnswerCount="2" CommentCount="4" ClosedDate="2015-12-10T06:47:01.847" ContentLicense="CC BY-SA 3.0" />
  <row Id="24440" PostTypeId="1" AcceptedAnswerId="24451" CreationDate="2015-12-09T19:34:50.877" Score="19" ViewCount="585" Body="&lt;p&gt;\begin{array}{l}&#xA;&amp;amp;19\times 46=784\\&#xA;&amp;amp;56+43=102\\&#xA;&amp;amp;360\div 15=29\\&#xA;&amp;amp;196-55=114\\&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Hint 1:&lt;/strong&gt; For once, this is &lt;em&gt;not&lt;/em&gt; an abuse of the equals symbol!&lt;br&gt;&#xA; &lt;strong&gt;Hint 2:&lt;/strong&gt; The clues are  for the song only, the artist follows.&lt;br&gt;&#xA; &lt;strong&gt;Hint 3:&lt;/strong&gt; $26 + 89 = 115$&lt;br&gt;&#xA; &lt;strong&gt;Hint 4:&lt;/strong&gt; The third equation is potentially paradoxical, but I don't mind.&lt;br&gt;&#xA; &lt;strong&gt;Hint 5:&lt;/strong&gt; Released as a single (B side) in 1696!&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="2819" LastEditDate="2015-12-09T23:30:24.593" LastActivityDate="2015-12-10T05:12:20.427" Title="Guess the song and artist" Tags="|music|" AnswerCount="1" CommentCount="7" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24441" PostTypeId="2" ParentId="24439" CreationDate="2015-12-09T19:45:06.630" Score="5" Body="&lt;p&gt;Assuming equally distributed probability of the &quot;other&quot; throws:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;6:  2.1504e-06&lt;br&gt;&#xA;   7:  1.516032e-05&lt;br&gt;&#xA;   8:  6.105599999999999e-05&lt;br&gt;&#xA;   9:  0.0001866496&lt;br&gt;&#xA;  10:  0.0004811571199999999&lt;br&gt;&#xA;  11:  0.0011080704000000004&lt;br&gt;&#xA;  12:  0.002317765120000012&lt;br&gt;&#xA;  13:  0.00445283840000005&lt;br&gt;&#xA;  14:  0.007932728320000082&lt;br&gt;&#xA;  15:  0.013234946560000219&lt;br&gt;&#xA;  16:  0.02078419456000026&lt;br&gt;&#xA;  17:  0.03073397760000035&lt;br&gt;&#xA;  18:  0.042942725119999754&lt;br&gt;&#xA;  19:  0.05675394815999913&lt;br&gt;&#xA;  20:  0.07107977727999641&lt;br&gt;&#xA;  &lt;strong&gt;21:  0.08419953407999499&lt;/strong&gt;&lt;br&gt;&#xA;  22:  0.09412483327999363&lt;br&gt;&#xA;  23:  0.09939161855999414&lt;br&gt;&#xA;  24:  0.09888333311999539&lt;br&gt;&#xA;  25:  0.09254429439999612&lt;br&gt;&#xA;  26:  0.08094589695999811&lt;br&gt;&#xA;  27:  0.06610059775999934&lt;br&gt;&#xA;  28:  0.050275064320000035&lt;br&gt;&#xA;  29:  0.03517282560000011&lt;br&gt;&#xA;  30:  0.022496632320000085&lt;br&gt;&#xA;  31:  0.012905395200000015&lt;br&gt;&#xA;  32:  0.006652177920000004&lt;br&gt;&#xA;  33:  0.0029200895999999987&lt;br&gt;&#xA;  34:  0.0010063871999999997&lt;br&gt;&#xA;  35:  0.0002651443199999999&lt;br&gt;&#xA;  36:  2.9030399999999992e-05&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or, if you prefer the exact answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;6:  21/9765625&lt;br&gt;&#xA;   7:  2961/195312500&lt;br&gt;&#xA;   8:  477/7812500&lt;br&gt;&#xA;   9:  7291/39062500&lt;br&gt;&#xA;  10:  23494/48828125&lt;br&gt;&#xA;  11:  10821/9765625&lt;br&gt;&#xA;  12:  905377/390625000&lt;br&gt;&#xA;  13:  173939/39062500&lt;br&gt;&#xA;  14:  1549361/195312500&lt;br&gt;&#xA;  15:  5169901/390625000&lt;br&gt;&#xA;  16:  4059413/195312500&lt;br&gt;&#xA;  17:  600273/19531250&lt;br&gt;&#xA;  18:  8387251/195312500&lt;br&gt;&#xA;  19:  22169511/390625000&lt;br&gt;&#xA;  20:  13882769/195312500&lt;br&gt;&#xA;  &lt;strong&gt;21:  32890443/390625000&lt;/strong&gt;&lt;br&gt;&#xA;  22:  36767513/390625000&lt;br&gt;&#xA;  23:  38824851/390625000&lt;br&gt;&#xA;  24:  19313151/195312500&lt;br&gt;&#xA;  25:  7230023/78125000&lt;br&gt;&#xA;  26:  31619491/390625000&lt;br&gt;&#xA;  27:  12910273/195312500&lt;br&gt;&#xA;  28:  19638697/390625000&lt;br&gt;&#xA;  29:  2747877/78125000&lt;br&gt;&#xA;  30:  8787747/390625000&lt;br&gt;&#xA;  31:  504117/39062500&lt;br&gt;&#xA;  32:  2598507/390625000&lt;br&gt;&#xA;  33:  57033/19531250&lt;br&gt;&#xA;  34:  9828/9765625&lt;br&gt;&#xA;  35:  25893/97656250&lt;br&gt;&#xA;  36:  567/19531250 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here's a brief explanation of how I did this:&lt;br&gt;&#xA;First, here's a list of the probabilities associated with the different dice throws:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dice 1 probabilities&lt;br&gt;&#xA;  Probability of a 1 is 1/10&lt;br&gt;&#xA;  Probability of a 2 is 9/50&lt;br&gt;&#xA;  Probability of a 3 is 9/50&lt;br&gt;&#xA;  Probability of a 4 is 9/50&lt;br&gt;&#xA;  Probability of a 5 is 9/50&lt;br&gt;&#xA;  Probability of a 6 is 9/50  &lt;/p&gt;&#xA;  &#xA;  &lt;hr&gt;&#xA;  &#xA;  &lt;p&gt;Dice 2 probabilities&lt;br&gt;&#xA;  Probability of a 1 is 4/25&lt;br&gt;&#xA;  Probability of a 2 is 1/5&lt;br&gt;&#xA;  Probability of a 3 is 4/25&lt;br&gt;&#xA;  Probability of a 4 is 4/25&lt;br&gt;&#xA;  Probability of a 5 is 4/25&lt;br&gt;&#xA;  Probability of a 6 is 4/25  &lt;/p&gt;&#xA;  &#xA;  &lt;hr&gt;&#xA;  &#xA;  &lt;p&gt;Dice 3 probabilities&lt;br&gt;&#xA;  Probability of a 1 is 7/50&lt;br&gt;&#xA;  Probability of a 2 is 7/50&lt;br&gt;&#xA;  Probability of a 3 is 3/10&lt;br&gt;&#xA;  Probability of a 4 is 7/50&lt;br&gt;&#xA;  Probability of a 5 is 7/50&lt;br&gt;&#xA;  Probability of a 6 is 7/50  &lt;/p&gt;&#xA;  &#xA;  &lt;hr&gt;&#xA;  &#xA;  &lt;p&gt;Dice 4 probabilities&lt;br&gt;&#xA;  Probability of a 1 is 3/25&lt;br&gt;&#xA;  Probability of a 2 is 3/25&lt;br&gt;&#xA;  Probability of a 3 is 3/25&lt;br&gt;&#xA;  Probability of a 4 is 2/5&lt;br&gt;&#xA;  Probability of a 5 is 3/25&lt;br&gt;&#xA;  Probability of a 6 is 3/25  &lt;/p&gt;&#xA;  &#xA;  &lt;hr&gt;&#xA;  &#xA;  &lt;p&gt;Dice 5 probabilities&lt;br&gt;&#xA;  Probability of a 1 is 1/10&lt;br&gt;&#xA;  Probability of a 2 is 1/10&lt;br&gt;&#xA;  Probability of a 3 is 1/10&lt;br&gt;&#xA;  Probability of a 4 is 1/10&lt;br&gt;&#xA;  Probability of a 5 is 1/2&lt;br&gt;&#xA;  Probability of a 6 is 1/10  &lt;/p&gt;&#xA;  &#xA;  &lt;hr&gt;&#xA;  &#xA;  &lt;p&gt;Dice 6 probabilities&lt;br&gt;&#xA;  Probability of a 1 is 2/25&lt;br&gt;&#xA;  Probability of a 2 is 2/25&lt;br&gt;&#xA;  Probability of a 3 is 2/25&lt;br&gt;&#xA;  Probability of a 4 is 2/25&lt;br&gt;&#xA;  Probability of a 5 is 2/25&lt;br&gt;&#xA;  Probability of a 6 is 3/5  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then, there are only $6^6 = 46656$ possible throws.  This is a small number for a computer to iterate through, so I do so.  For each throw, I calculate what the probability of it occurring is, and what the sum of the dice is.  I accumulate that sum to the probability of getting that number.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some examples of ways to throw 21:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Die 1 has a 1 (probability 1/10  )&#xA;Die 2 has a 1 (probability 4/25  )&#xA;Die 3 has a 1 (probability 7/50  )&#xA;Die 4 has a 6 (probability 3/25  )&#xA;Die 5 has a 6 (probability 1/10  )&#xA;Die 6 has a 6 (probability 3/5  )&#xA;Probability of this throw: 63/3906250  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Die 1 has a 1 (probability 1/10  )&#xA;Die 2 has a 1 (probability 4/25  )&#xA;Die 3 has a 2 (probability 7/50  )&#xA;Die 4 has a 5 (probability 3/25  )&#xA;Die 5 has a 6 (probability 1/10  )&#xA;Die 6 has a 6 (probability 3/5  )&#xA;Probability of this throw: 63/3906250  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Die 1 has a 1 (probability 1/10  )&#xA;Die 2 has a 1 (probability 4/25  )&#xA;Die 3 has a 2 (probability 7/50  )&#xA;Die 4 has a 6 (probability 3/25  )&#xA;Die 5 has a 5 (probability 1/2  )&#xA;Die 6 has a 6 (probability 3/5  )&#xA;Probability of this throw: 63/781250  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Die 1 has a 1 (probability 1/10  )&#xA;Die 2 has a 1 (probability 4/25  )&#xA;Die 3 has a 2 (probability 7/50  )&#xA;Die 4 has a 6 (probability 3/25  )&#xA;Die 5 has a 6 (probability 1/10  )&#xA;Die 6 has a 5 (probability 2/25  )&#xA;Probability of this throw: 21/9765625  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Die 1 has a 1 (probability 1/10  )&#xA;Die 2 has a 1 (probability 4/25  )&#xA;Die 3 has a 3 (probability 3/10  )&#xA;Die 4 has a 4 (probability 2/5  )&#xA;Die 5 has a 6 (probability 1/10  )&#xA;Die 6 has a 6 (probability 3/5  )&#xA;Probability of this throw: 9/78125  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Die 1 has a 1 (probability 1/10  )&#xA;Die 2 has a 1 (probability 4/25  )&#xA;Die 3 has a 3 (probability 3/10  )&#xA;Die 4 has a 5 (probability 3/25  )&#xA;Die 5 has a 5 (probability 1/2  )&#xA;Die 6 has a 6 (probability 3/5  )&#xA;Probability of this throw: 27/156250  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously this goes on for some time, but in the end you get the answer.  Below is the python script:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from collections import defaultdict&#xA;import itertools&#xA;import fractions&#xA;&#xA;#Generate the probabilities for the dice&#xA;dice=defaultdict(list)&#xA;for die in range(1,7):&#xA;    for throw in range(7):&#xA;        if throw==0:&#xA;            dice[die].append(0)&#xA;        elif die==throw:&#xA;            dice[die].append(fractions.Fraction(die,10))&#xA;        else:&#xA;            dice[die].append(fractions.Fraction(10-die,50))&#xA;&#xA;#Generate the probabilities for the different throws&#xA;results = defaultdict(fractions.Fraction)&#xA;for throw in itertools.product(range(1,7),repeat=6):&#xA;    prob=fractions.Fraction(1,1)&#xA;    for die,t in zip(range(1,7),throw):&#xA;        prob*=dice[die][t]&#xA;    results[sum(throw)]+=prob&#xA;&#xA;#Print results&#xA;for r in results:&#xA;    print(&quot;%2d: %s&quot;%(r,results[r]))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2015-12-09T22:47:59.423" LastActivityDate="2015-12-09T22:47:59.423" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24442" PostTypeId="2" ParentId="16881" CreationDate="2015-12-09T20:28:41.993" Score="27" Body="&lt;p&gt;The solution is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Darwin was right no creationism&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;Steps to the solution&lt;/h1&gt;&#xA;&#xA;&lt;h2&gt;1. Finding the hint&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The butterflies have to be arranged according to their orientation. There are 16 different orientations (17 if you count the grey ancestor-butterfly in the center, but it can be ignored for this), evenly distributed with 22.5° difference each. Based on JTL's arrangement, slightly altered here, the first hint becomes visible (reading counter-clock-wise like mathematicians do and unlike cuckoo clocks):&#xA; &lt;a href=&quot;https://i.stack.imgur.com/ZuUoN.jpg&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZuUoN.jpg&quot; alt=&quot;The first hint&quot;&gt;&lt;/a&gt;&#xA; Hint: Baconian by shape&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;2. Preparing&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To get the correct order, the butterflies have to be ordered: Starting with the grey butterfly in the middle the next butterfly-generation has exactly one property changed: either the shape (wings or antennae) or the color or pattern. With this rule a family-tree can be created like this (based on Alconja's work)&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/xQ4iV.jpg&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xQ4iV.jpg&quot; alt=&quot;Butterfly families&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;3. Solving&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; With the hint on Bacon's cipher the solution is hidden as a binary code with A=00000, B=00001, etc. In this riddle a butterfly is considered as 1 if its shape is different (wing-shape or antennae) and is considered as 0 if the shape is the same (and the color or pattern is different). Starting on the second image, the first branch becomes&lt;br&gt;&#xA; 000110000010001101100100001101&lt;br&gt;&#xA; split into chunks of 5 yields&lt;br&gt;&#xA; 00011 = D, 00000 = A, 10001 = R, 10110 = W, 01000 = I, 01101= N.&lt;br&gt;&#xA; Each new branch of the butterflies forms a new word, resulting in the solution: Darwin was right no creationism&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A big thank you for this beautiful and challenging riddle! And of course to Alconja and JTL who did all the hard work. I'm new here, so if I can split the bounty, I'm happy to do so.&lt;/p&gt;&#xA;" OwnerUserId="17711" LastEditorUserId="17711" LastEditDate="2015-12-09T20:44:20.530" LastActivityDate="2015-12-09T20:44:20.530" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24443" PostTypeId="2" ParentId="24358" CreationDate="2015-12-09T20:28:51.700" Score="0" Body="&lt;p&gt;I'm gonna play off Zerris' answer and go with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Foe Sixty Bang&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Foe_(unit)&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Foe_(unit)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A foe is a unit of energy equal to 10^44 joules or 10^51 ergs, used to measure the large amount of energy released by a supernova.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1 \times 10^{44})!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or better yet:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1 \times 10^{51})!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!$ in ergs&lt;/p&gt;&#xA;" OwnerUserId="16874" LastEditorUserId="4421" LastEditDate="2015-12-11T22:21:26.660" LastActivityDate="2015-12-11T22:21:26.660" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24444" PostTypeId="2" ParentId="24430" CreationDate="2015-12-09T21:07:36.833" Score="7" Body="&lt;p&gt;Big square: Actually 2 - Inside (10x10) and outside (10.5x10.5).  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;2 cuts (or 3 without stacking)  Cut piece in half parallel with the long side, giving 2  10x1 pieces.  Stack these pieces together, and cut in the same manner, giving 4  10x0.5 pieces.  Arrange these pieces so that the short ends are against then end of the next piece, forming a large square&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Picture:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; [&lt;img src=&quot;https://i.stack.imgur.com/9gKVN.png&quot; alt=&quot;10x10&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1918" LastActivityDate="2015-12-09T21:07:36.833" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24445" PostTypeId="2" ParentId="24028" CreationDate="2015-12-09T22:17:51.793" Score="5" Body="&lt;p&gt;As noted above, the answer is the stage but the answers needs explanations of how each &quot;part&quot; matches each &quot;animal&quot; So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A cat: mews or &quot;mewls&quot;. Shakespeare's baby was puking, which makes it a sickly cat&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The schoolboy is obviously a snail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure about the dragon: perhaps the furnace sighing instead of roaring makes it like an asthmatic dragon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hairy Leopard is obvious, &quot;bearded like the (leo)pard&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Morally upright rooster: Capon is apparently a castrated rooster, and a judge should be morally upright.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tiny tasty lamb: lambshanks are tasty, but this mans shanks are shrunk, making it tiny as well as tasty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Childish earthworm also obvious, it is the &quot;second childhood&quot; and like an earthworm, has no teeth, or eyes.&lt;/p&gt;&#xA;" OwnerUserId="17712" LastEditorUserId="17712" LastEditDate="2015-12-14T16:12:13.080" LastActivityDate="2015-12-14T16:12:13.080" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24446" PostTypeId="1" AcceptedAnswerId="24447" CreationDate="2015-12-09T22:21:16.223" Score="0" ViewCount="349" Body="&lt;p&gt;My first riddle, it's probably too easy though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get almost any knowledge from one book or another, but there is one thing that is absolutely impossible to learn from a book. What is it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not talking about any physical skill that requires practice.&lt;/p&gt;&#xA;" OwnerUserId="17712" LastEditorUserId="11876" LastEditDate="2015-12-09T22:40:16.200" LastActivityDate="2015-12-10T04:16:02.630" Title="What can't you learn?" Tags="|riddle|" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24447" PostTypeId="2" ParentId="24446" CreationDate="2015-12-09T22:37:39.453" Score="7" Body="&lt;p&gt;I'm pretty sure the answer to this is supposed to be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The one thing that is impossible to learn from a book is how to read.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, this is not really true.&lt;/p&gt;&#xA;" OwnerUserId="2406" LastActivityDate="2015-12-09T22:37:39.453" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24448" PostTypeId="2" ParentId="24430" CreationDate="2015-12-09T22:43:01.157" Score="4" Body="&lt;p&gt;Here's a four-piece solution, taken from &lt;a href=&quot;https://math.stackexchange.com/questions/96776/dissection-of-a-1-times-5-rectangle-to-a-square&quot;&gt;here&lt;/a&gt; (the link only has pictures, but I supplemented with cut locations):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cut from (0, 2) to (4, 0), from (4, 0) to (4 + $\sqrt{3}$, 2), and from (10 - $\sqrt{3}$, 0) to (10, 2).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Image of the cut-up rectangle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/hj7fB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hj7fB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then just slide the four pieces into a (rotated) square:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/z38n0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/z38n0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA; &lt;br&gt;&lt;br&gt;&#xA; Despite the irrational locations of the cuts in the original rectangle, the vertices of all pieces in the joined square are at integer locations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16275" LastEditorUserId="-1" LastEditDate="2017-04-13T12:19:36.950" LastActivityDate="2015-12-09T22:43:01.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24449" PostTypeId="2" ParentId="24439" CreationDate="2015-12-10T02:17:32.517" Score="3" Body="&lt;p&gt;Dr. Xorile gives a brute-force approach; here's a more efficient dynamic programming method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $p_k(n)$ be the probability that the first $k$ dice sum to $n$.  The question then asks for $p_6(21)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can define $p$ recursively, starting with the base case $k=0$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;p_0(n)=\delta_{0,n}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($\delta_{ij}$ is the Kroneker delta, equal to $1$ when its arguments are the same, and $0$ when they differ.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next we need to know the probability distributions for the different dice.  The $k$th die has one face with probability $k/10$, and the others five have probability $(1-k/10)/5=1/5-k/50$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we calculate $p$ for successive values of $k$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;p_k(n) = \left[\sum_{i\in(1\ldots 6)}(1-\delta_{ki})(1/5-k/50)p_{k-1}(n-i)\right]+(k/10)\left[p_{k-1}(n-k)\right]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coding this in &lt;em&gt;Mathematica&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;p[0, 0] = 1;&#xA;p[0, _] = 0;&#xA;p[k_, n_] := p[k, n] =&#xA;    Sum[(1 - KroneckerDelta[i, k]) (1/5-k/50) p[k-1, n-i], {i, 1, 6}] + (k/10) p[k-1, n-k]&#xA;&#xA;p[6, 21] (* 32890443/390625000 *)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is the same result as Dr. Xorile, but achieved in $O(nkd)$ time vs. $O(d^k)$ (where $d$ is the number of faces per die).&lt;/p&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="9718" LastEditDate="2015-12-10T02:30:44.193" LastActivityDate="2015-12-10T02:30:44.193" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24450" PostTypeId="2" ParentId="24446" CreationDate="2015-12-10T04:16:02.630" Score="4" Body="&lt;p&gt;One thing that it is impossible to learn from a book?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Something you already know.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="375" LastActivityDate="2015-12-10T04:16:02.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24451" PostTypeId="2" ParentId="24440" CreationDate="2015-12-10T05:08:29.730" Score="14" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;If 6 was 9&lt;/em&gt; by Jimi Hendrix&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The numbers 6 and 9 have been switched in the equations, meaning that any 6 is actually a 9, and any 9 is a 6. Thus the equations turn into&#xA;&#xA; \begin{array}{l} &amp;amp;16\times 49=784\\&amp;amp;59+43=102\\&amp;amp;390\div 15=26\\&amp;amp;169-55=114\\\end{array}&#xA; Which are valid equations!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation of hints&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Hint 3&lt;/strong&gt; becomes $29 + 86 = 115$ (though it was correct even before switching)&lt;br/&gt;&#xA; &lt;strong&gt;Hint 4&lt;/strong&gt; &quot;I don't mind&quot; are lyrics from the song &lt;em&gt;If 6 was 9&lt;/em&gt;&lt;br/&gt;&#xA; &lt;strong&gt;Hint 5&lt;/strong&gt; The song was released in 1969&lt;br/&gt;&#xA; The other hints don't need explanation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4624" LastEditorUserId="4624" LastEditDate="2015-12-10T05:12:20.427" LastActivityDate="2015-12-10T05:12:20.427" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24452" PostTypeId="2" ParentId="24358" CreationDate="2015-12-10T06:12:22.767" Score="5" Body="&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://googology.wikia.com/wiki/Enormabixul&quot;&gt;Enormabixul&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Googology wiki has many like these. This one is many, many orders of magnitude greater than, for example, &lt;a href=&quot;https://en.wikipedia.org/wiki/Graham%27s_number&quot;&gt;Graham's number&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Rayo%27s_number&quot;&gt;The Rayo sum&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rayo's number is one of the greatest numbers ever defined in mathematics. From wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The smallest number bigger than any finite number named by an&#xA;  expression in the language of set theory with a googol symbols or&#xA;  less.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that this definition is not &lt;a href=&quot;https://en.wikipedia.org/wiki/Berry_paradox&quot;&gt;self-referential&lt;/a&gt;, since it can't be expressed using only (first-order) set theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A general list of googologisms can be found in the &lt;a href=&quot;http://googology.wikia.com/wiki/List_of_googologisms&quot;&gt;wiki&lt;/a&gt;.&#xA;There is also a list of &lt;a href=&quot;http://googology.wikia.com/wiki/List_of_functions&quot;&gt;fast-growing functions&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may be useful for future answerers.&lt;/p&gt;&#xA;" OwnerUserId="14219" LastEditorUserId="14219" LastEditDate="2015-12-10T14:28:13.627" LastActivityDate="2015-12-10T14:28:13.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24453" PostTypeId="2" ParentId="24358" CreationDate="2015-12-10T06:34:36.103" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;g sub forty k&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's $g_{40000}$, where a lowercase $g$ is Graham's sequence.&lt;/p&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2015-12-10T06:34:36.103" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24454" PostTypeId="2" ParentId="24407" CreationDate="2015-12-10T07:26:23.030" Score="3" Body="&lt;p&gt;&lt;a href=&quot;http://websudoku.com&quot; rel=&quot;nofollow&quot;&gt;Web Sudoku&lt;/a&gt; has huge amounts of puzzles across five difficulty levels. You don't even have to wait a day for new ones.&lt;/p&gt;&#xA;" OwnerUserId="17231" LastActivityDate="2015-12-10T07:26:23.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24455" PostTypeId="2" ParentId="2066" CreationDate="2015-12-10T09:23:21.860" Score="-1" Body="&lt;p&gt;below is my answer photo. &lt;br&gt;&#xA;Just Simple and beautiful :D &lt;br&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/KneYm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KneYm.png&quot; alt=&quot;My cool Answer&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt; I guess my answer matched with at least this&#xA;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Edit: Best answer is the one in which straight line passes center of&#xA;  the circles.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11353" LastEditorUserId="46903" LastEditDate="2018-09-07T02:14:39.013" LastActivityDate="2018-09-07T02:14:39.013" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24456" PostTypeId="1" AcceptedAnswerId="24482" CreationDate="2015-12-10T11:02:33.513" Score="10" ViewCount="577" Body="&lt;p&gt;I am after and before&lt;br&gt;&#xA;I am the god who said no more!&lt;br&gt;&#xA;A fraudulent rose, now set free&lt;br&gt;&#xA;The foundation sat upon me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I refuse to move, though I'm casually roaming&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3987" LastEditorUserId="25957" LastEditDate="2016-08-09T09:52:35.687" LastActivityDate="2016-08-09T09:52:35.687" Title="I am the god who said no more! Who am I?" Tags="|riddle|" AnswerCount="4" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="24469" PostTypeId="1" AcceptedAnswerId="24477" CreationDate="2015-12-10T19:48:15.827" Score="12" ViewCount="405" Body="&lt;p&gt;The following grid has been constructed using the shapes of unfolded cubes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uT7jS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uT7jS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Determine the minimum number of shapes (red and green) needed to cover the blue grid. The shapes may overlap, but must remain in their original scale.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rotations of the shapes are accepted, but not mirrorings. The shapes should cover all blue squares and no white squares. If you are not giving a fancy graphics solution, the solution should at least contain a reasonable justification for your number.&lt;/p&gt;&#xA;" OwnerUserId="12901" LastEditorUserId="4421" LastEditDate="2015-12-10T20:41:07.603" LastActivityDate="2015-12-10T21:01:29.340" Title="Fitting the pieces" Tags="|geometry|visual|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24471" PostTypeId="2" ParentId="24358" CreationDate="2015-12-10T19:58:58.697" Score="3" Body="&lt;p&gt;&lt;strong&gt;Top squared.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whichever answer is the top one (either &lt;a href=&quot;https://puzzling.stackexchange.com/users/14219/ivan-barreto&quot;&gt;Ivan Barreto's&lt;/a&gt; or &lt;a href=&quot;https://puzzling.stackexchange.com/users/17389/zerris&quot;&gt;Zerris's&lt;/a&gt;, depending on whether you consider &lt;em&gt;enormabixul&lt;/em&gt; an English word — &lt;em&gt;Rayo&lt;/em&gt; isn't, it's only a proper name), I'm squaring that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since answers are to be interpreted at the time of posting, I'm not closing the field.&lt;/p&gt;&#xA;" OwnerUserId="71" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-12-10T19:58:58.697" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24472" PostTypeId="2" ParentId="24358" CreationDate="2015-12-10T20:01:45.327" Score="3" Body="&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/a/24471&quot;&gt;My best&lt;/a&gt; and four&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to show that the list can go on.&lt;/p&gt;&#xA;" OwnerUserId="71" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-12-10T20:01:45.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24473" PostTypeId="2" ParentId="24456" CreationDate="2015-12-10T20:22:09.977" Score="0" Body="&lt;p&gt;Answer  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nothing or nothingness? Not sure about the third verse, though.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am after and before&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Before and after existance - nothing, basically&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the god who said no more!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;God had no choice&quot; as to whether or not to &quot;start&quot; the Universe, as i'm sure i read some philosopher say. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A fraudulent rose, now set free&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Not sure&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The foundation sat upon me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You can only start building from scratch (foundations) by starting with/from nothing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16800" LastEditorUserId="16273" LastEditDate="2015-12-10T21:14:28.663" LastActivityDate="2015-12-10T21:14:28.663" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24474" PostTypeId="2" ParentId="24456" CreationDate="2015-12-10T20:39:57.727" Score="8" Body="&lt;p&gt;Terrible guess: you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Peter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After and before: Peter can appear as a first or last name (both real and fictional people are named Peter Peter, even). Said no more: Peter Parker renounced his godhood and said he'd be Spiderman no more. A fraudulent rose, now set free: Peter &quot;Pete&quot; Rose, the baseball player, was imprisoned for tax fraud but is now set free. The foundation sat upon me: &lt;a href=&quot;http://biblehub.com/matthew/16-18.htm&quot;&gt;Matthew 16:18&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9523" LastActivityDate="2015-12-10T20:39:57.727" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24476" PostTypeId="2" ParentId="24456" CreationDate="2015-12-10T20:59:12.000" Score="0" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explaination&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am after and before&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It was before our existence and will be even after our death because it don't stop&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the god who said no more!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you lose it you can't get it back. The word 'said' here shows that it was past tense i.e. it is gone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A fraudulent rose, now set free&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There had been a great discussion on time in physics. But now everything is decided how it works&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The foundation sat upon me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It started with big bang. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastActivityDate="2015-12-10T20:59:12.000" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24477" PostTypeId="2" ParentId="24469" CreationDate="2015-12-10T21:01:29.340" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The minimum number of shapes is &lt;strong&gt;28&lt;/strong&gt;:&lt;br&gt;&lt;br&gt;&lt;a href=&quot;https://i.stack.imgur.com/B4ndm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/B4ndm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastActivityDate="2015-12-10T21:01:29.340" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24478" PostTypeId="2" ParentId="546" CreationDate="2015-12-10T21:20:25.463" Score="-6" Body="&lt;p&gt;I think it is possible to solve all puzzles (of four unique colors) at five tries. The best way to start is picking red, green, blue, yellow. Then simply stick to those colors, starting with red... say you get three white markers after the first round, simply keep red, green, blue, but move all of them one space to the right, and fill in the last space (space #1) with the first of the remaining colors, brown. Continue doing this, systematically, and you will hit the right code within five tries.&lt;/p&gt;&#xA;" OwnerUserId="17725" LastActivityDate="2015-12-10T21:20:25.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24480" PostTypeId="1" AcceptedAnswerId="24494" CreationDate="2015-12-11T04:10:25.150" Score="4" ViewCount="366" Body="&lt;p&gt;You've barely got back from &lt;a href=&quot;https://puzzling.stackexchange.com/questions/24353/who-are-the-guilty-parties-2&quot;&gt;the isolated villa where your last case was&lt;/a&gt;, when you hear that you are needed again:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is a group of people on board a yacht.  The wealthy owner and her spouse have been brutally murdered.  You have been sent to collect statements from each person.&lt;br&gt;&#xA;   - Each person in the group is either guilty or innocent.&lt;br&gt;&#xA;   - Each person is either friends or not friends with each other person.  The friendship is mutual.&lt;br&gt;&#xA;   - Each guilty person is friends with every other guilty person.&lt;br&gt;&#xA;   - Each innocent person has witnessed a number of events, which allows them to deduce the innocence or guilt of a number of people in the group.  This may or may not have been reciprocated.&lt;br&gt;&#xA;   - The guilty people know who the guilty and innocent parties are.&lt;br&gt;&#xA;   - Each person is either boastful or modest (to be explained below).  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The statements that each person makes are as follows:&lt;br&gt;&#xA;   - Innocent modest people will simply make a list of truthful statements.&lt;br&gt;&#xA;   - Innocent boastful people will do the same, but will also make a statement about a random other person about whom they know nothing.  If the person is their friend, they will say s/he is innocent.  Otherwise they'll just say something (innocent or guilty at random).&lt;br&gt;&#xA;   - Guilty modest people will say that friends are innocent (regardless of whether they are); they will say that non-friends are guilty (which is inaccurate, because their non-friends are innocent by the above definitions).&lt;br&gt;&#xA;   - Guilty boastful people will do the same.  But they will also make one truthful statement (either that one of their guilty friends is guilty or that one of their innocent non-friends is innocent).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This situation seems all too common these days...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On this particular day, you go out to the island and get to sit down with the main characters:&lt;br&gt;&#xA;&lt;strong&gt;Apolonia, Luisa, Belkis, Buck, Berenice, Lenny, Christoper, Madonna, Christopher, Sherry, Nicki, Terrance, Ashely, Mamie, Laurice, Darron, Ezequiel, Gale, Nancee, and Cammie&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Statements of Apolonia:&lt;/strong&gt;&lt;br&gt;&#xA;   - I am innocent&lt;br&gt;&#xA;   - Ezequiel is guilty&lt;br&gt;&#xA;   - Mamie is guilty&lt;br&gt;&#xA;   - Laurice is guilty&lt;br&gt;&#xA;   - Lenny is innocent&lt;br&gt;&#xA;   - Buck is innocent&lt;br&gt;&#xA;   - Cammie is guilty&lt;br&gt;&#xA;   - Christopher is innocent&lt;br&gt;&#xA;   - Darron is guilty&lt;br&gt;&#xA;   - Berenice is innocent&lt;br&gt;&#xA;   - Ashely is innocent  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Statements of Luisa:&lt;/strong&gt;&lt;br&gt;&#xA;   - I am innocent&lt;br&gt;&#xA;   - Terrance is innocent&lt;br&gt;&#xA;   - Ashely is innocent&lt;br&gt;&#xA;   - Apolonia is innocent&lt;br&gt;&#xA;   - Darron is guilty&lt;br&gt;&#xA;   - Mamie is guilty&lt;br&gt;&#xA;   - Nicki is innocent&lt;br&gt;&#xA;   - Madonna is innocent  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Statements of Belkis:&lt;/strong&gt;&lt;br&gt;&#xA;   - I am innocent&lt;br&gt;&#xA;   - Nicki is innocent&lt;br&gt;&#xA;   - Madonna is innocent&lt;br&gt;&#xA;   - Ezequiel is guilty&lt;br&gt;&#xA;   - Apolonia is innocent&lt;br&gt;&#xA;   - Terrance is innocent  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Statements of Buck:&lt;/strong&gt;&lt;br&gt;&#xA;   - I am innocent&lt;br&gt;&#xA;   - Nicki is innocent&lt;br&gt;&#xA;   - Christoper is innocent&lt;br&gt;&#xA;   - Laurice is guilty&lt;br&gt;&#xA;   - Lenny is innocent&lt;br&gt;&#xA;   - Madonna is innocent&lt;br&gt;&#xA;   - Terrance is innocent&lt;br&gt;&#xA;   - Belkis is innocent&lt;br&gt;&#xA;   - Ezequiel is guilty  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Statements of Berenice:&lt;/strong&gt;&lt;br&gt;&#xA;   - I am innocent&lt;br&gt;&#xA;   - Laurice is guilty&lt;br&gt;&#xA;   - Belkis is innocent&lt;br&gt;&#xA;   - Terrance is innocent&lt;br&gt;&#xA;   - Luisa is innocent&lt;br&gt;&#xA;   - Sherry is innocent&lt;br&gt;&#xA;   - Nancee is guilty&lt;br&gt;&#xA;   - Ashely is innocent&lt;br&gt;&#xA;   - Darron is guilty  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Statements of Lenny:&lt;/strong&gt;&lt;br&gt;&#xA;   - I am innocent&lt;br&gt;&#xA;   - Terrance is innocent&lt;br&gt;&#xA;   - Luisa is innocent&lt;br&gt;&#xA;   - Buck is innocent&lt;br&gt;&#xA;   - Nicki is innocent&lt;br&gt;&#xA;   - Sherry is innocent&lt;br&gt;&#xA;   - Ashely is innocent  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Statements of Christoper:&lt;/strong&gt;&lt;br&gt;&#xA;   - I am innocent&lt;br&gt;&#xA;   - Buck is innocent&lt;br&gt;&#xA;   - Madonna is innocent&lt;br&gt;&#xA;   - Mamie is guilty&lt;br&gt;&#xA;   - Luisa is innocent&lt;br&gt;&#xA;   - Berenice is innocent&lt;br&gt;&#xA;   - Christopher is innocent&lt;br&gt;&#xA;   - Ezequiel is guilty&lt;br&gt;&#xA;   - Laurice is guilty&lt;br&gt;&#xA;   - Nicki is innocent  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Statements of Madonna:&lt;/strong&gt;&lt;br&gt;&#xA;   - I am innocent&lt;br&gt;&#xA;   - Luisa is innocent&lt;br&gt;&#xA;   - Christoper is innocent&lt;br&gt;&#xA;   - Sherry is innocent&lt;br&gt;&#xA;   - Buck is innocent&lt;br&gt;&#xA;   - Belkis is innocent&lt;br&gt;&#xA;   - Christopher is innocent&lt;br&gt;&#xA;   - Cammie is guilty&lt;br&gt;&#xA;   - Ashely is innocent&lt;br&gt;&#xA;   - Lenny is innocent  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Statements of Christopher:&lt;/strong&gt;&lt;br&gt;&#xA;   - I am innocent&lt;br&gt;&#xA;   - Nicki is innocent&lt;br&gt;&#xA;   - Buck is innocent&lt;br&gt;&#xA;   - Terrance is innocent&lt;br&gt;&#xA;   - Gale is guilty&lt;br&gt;&#xA;   - Christoper is innocent&lt;br&gt;&#xA;   - Lenny is innocent&lt;br&gt;&#xA;   - Apolonia is innocent&lt;br&gt;&#xA;   - Luisa is innocent  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Statements of Sherry:&lt;/strong&gt;&lt;br&gt;&#xA;   - I am innocent&lt;br&gt;&#xA;   - Madonna is innocent&lt;br&gt;&#xA;   - Terrance is innocent&lt;br&gt;&#xA;   - Ezequiel is guilty&lt;br&gt;&#xA;   - Ashely is innocent&lt;br&gt;&#xA;   - Laurice is guilty&lt;br&gt;&#xA;   - Christopher is innocent&lt;br&gt;&#xA;   - Lenny is innocent&lt;br&gt;&#xA;   - Belkis is innocent&lt;br&gt;&#xA;   - Cammie is guilty  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Statements of Nicki:&lt;/strong&gt;&lt;br&gt;&#xA;   - I am innocent&lt;br&gt;&#xA;   - Mamie is guilty&lt;br&gt;&#xA;   - Laurice is guilty&lt;br&gt;&#xA;   - Christoper is innocent&lt;br&gt;&#xA;   - Madonna is innocent&lt;br&gt;&#xA;   - Luisa is innocent&lt;br&gt;&#xA;   - Nancee is guilty&lt;br&gt;&#xA;   - Apolonia is innocent&lt;br&gt;&#xA;   - Lenny is innocent&lt;br&gt;&#xA;   - Gale is guilty  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Statements of Terrance:&lt;/strong&gt;&lt;br&gt;&#xA;   - I am innocent&lt;br&gt;&#xA;   - Christopher is innocent&lt;br&gt;&#xA;   - Christoper is innocent&lt;br&gt;&#xA;   - Mamie is guilty&lt;br&gt;&#xA;   - Ezequiel is guilty&lt;br&gt;&#xA;   - Sherry is innocent&lt;br&gt;&#xA;   - Apolonia is innocent  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Statements of Ashely:&lt;/strong&gt;&lt;br&gt;&#xA;   - I am innocent&lt;br&gt;&#xA;   - Christoper is innocent&lt;br&gt;&#xA;   - Belkis is innocent&lt;br&gt;&#xA;   - Apolonia is innocent&lt;br&gt;&#xA;   - Cammie is innocent&lt;br&gt;&#xA;   - Laurice is guilty&lt;br&gt;&#xA;   - Terrance is innocent&lt;br&gt;&#xA;   - Buck is innocent&lt;br&gt;&#xA;   - Christopher is innocent&lt;br&gt;&#xA;   - Nancee is guilty&lt;br&gt;&#xA;   - Berenice is innocent&lt;br&gt;&#xA;   - Gale is guilty  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Statements of Mamie:&lt;/strong&gt;&lt;br&gt;&#xA;   - I am innocent&lt;br&gt;&#xA;   - Cammie is innocent&lt;br&gt;&#xA;   - Belkis is guilty&lt;br&gt;&#xA;   - Nicki is guilty&lt;br&gt;&#xA;   - Gale is innocent&lt;br&gt;&#xA;   - Laurice is innocent&lt;br&gt;&#xA;   - Sherry is guilty  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Statements of Laurice:&lt;/strong&gt;&lt;br&gt;&#xA;   - I am innocent&lt;br&gt;&#xA;   - Mamie is innocent&lt;br&gt;&#xA;   - Madonna is guilty&lt;br&gt;&#xA;   - Gale is innocent&lt;br&gt;&#xA;   - Ashely is guilty&lt;br&gt;&#xA;   - Nicki is guilty&lt;br&gt;&#xA;   - Apolonia is guilty&lt;br&gt;&#xA;   - Sherry is guilty&lt;br&gt;&#xA;   - Cammie is innocent&lt;br&gt;&#xA;   - Berenice is guilty&lt;br&gt;&#xA;   - Terrance is guilty  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Statements of Darron:&lt;/strong&gt;&lt;br&gt;&#xA;   - I am innocent&lt;br&gt;&#xA;   - Mamie is innocent&lt;br&gt;&#xA;   - Christopher is guilty&lt;br&gt;&#xA;   - Nicki is guilty&lt;br&gt;&#xA;   - Lenny is guilty&lt;br&gt;&#xA;   - Belkis is innocent&lt;br&gt;&#xA;   - Sherry is innocent&lt;br&gt;&#xA;   - Ashely is guilty&lt;br&gt;&#xA;   - Gale is innocent&lt;br&gt;&#xA;   - Buck is guilty  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Statements of Ezequiel:&lt;/strong&gt;&lt;br&gt;&#xA;   - I am innocent&lt;br&gt;&#xA;   - Ashely is guilty&lt;br&gt;&#xA;   - Buck is guilty&lt;br&gt;&#xA;   - Christopher is guilty&lt;br&gt;&#xA;   - Darron is innocent&lt;br&gt;&#xA;   - Madonna is guilty&lt;br&gt;&#xA;   - Nancee is innocent&lt;br&gt;&#xA;   - Gale is innocent&lt;br&gt;&#xA;   - Christoper is guilty&lt;br&gt;&#xA;   - Sherry is guilty  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Statements of Gale:&lt;/strong&gt;&lt;br&gt;&#xA;   - I am innocent&lt;br&gt;&#xA;   - Ashely is guilty&lt;br&gt;&#xA;   - Madonna is innocent&lt;br&gt;&#xA;   - Mamie is innocent&lt;br&gt;&#xA;   - Christoper is innocent&lt;br&gt;&#xA;   - Laurice is innocent&lt;br&gt;&#xA;   - Apolonia is guilty&lt;br&gt;&#xA;   - Berenice is guilty&lt;br&gt;&#xA;   - Nicki is guilty&lt;br&gt;&#xA;   - Christopher is innocent  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Statements of Nancee:&lt;/strong&gt;&lt;br&gt;&#xA;   - I am innocent&lt;br&gt;&#xA;   - Terrance is guilty&lt;br&gt;&#xA;   - Ezequiel is innocent&lt;br&gt;&#xA;   - Darron is guilty&lt;br&gt;&#xA;   - Nicki is guilty&lt;br&gt;&#xA;   - Apolonia is guilty&lt;br&gt;&#xA;   - Christoper is guilty&lt;br&gt;&#xA;   - Belkis is innocent  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Statements of Cammie:&lt;/strong&gt;&lt;br&gt;&#xA;   - I am innocent&lt;br&gt;&#xA;   - Mamie is innocent&lt;br&gt;&#xA;   - Apolonia is innocent&lt;br&gt;&#xA;   - Berenice is innocent&lt;br&gt;&#xA;   - Gale is innocent&lt;br&gt;&#xA;   - Lenny is innocent&lt;br&gt;&#xA;   - Ezequiel is innocent &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As before, we believe that there must be the fewest possible collaborators...&lt;/p&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-12-11T21:26:04.517" Title="Who are the guilty parties? #3" Tags="|logical-deduction|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24481" PostTypeId="1" CreationDate="2015-12-11T04:44:58.293" Score="6" ViewCount="130" Body="&lt;p&gt;I'm looking for a good online crossword that I can do when I have breaks in my day. Since I'm a web designer, I want something that has a well designed website. Also, since I like to think I'm smart, I want something large though not necessarily complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best I could find is the Times Crossword (nytimes.com/crosswords/). However, &quot;The Crossword&quot; (the large one) is only available to subscribers. Though you have one month free, after that you have to pay. Since I'm still young and in school, I have very little income and no credit card/PayPal to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any online crosswords that are well designed (as good looking and easy to use as the Times) and also on the level of the Times crossword covering the same themes and with the same style?&lt;/p&gt;&#xA;" OwnerUserId="17728" LastActivityDate="2015-12-11T04:44:58.293" Title="Good Crosswords" Tags="|crosswords|" AnswerCount="0" CommentCount="2" ClosedDate="2015-12-11T15:59:09.700" ContentLicense="CC BY-SA 3.0" />
  <row Id="24482" PostTypeId="2" ParentId="24456" CreationDate="2015-12-11T04:51:41.813" Score="9" Body="&lt;p&gt;I have the actual answer this time! Very confident. You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Terminus&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am after and before&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A terminus is the beginning or end of something&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am the god who said no more&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Terminus declares &quot;no more&quot; in Ralph Waldo Emerson's poem of the same name.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A fraudulent rose, now set free&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The hacker Terminus, whose real name was Leonard Rose, was imprisoned for wire fraud for a year and a day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The foundation sat upon me&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In Isaac Asimov's &lt;em&gt;Foundation&lt;/em&gt; series, the Foundation is built on the planet Terminus at the edge of the galaxy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Excellent riddle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I refuse to move, though I'm casually roaming&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Terminus, as a god of boundary stones, forbade the removal of his marker from the future site if the temple of Jupiter. He was also a &lt;em&gt;Roman&lt;/em&gt; god, and &quot;casually roaming&quot;=&quot;roamin&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9523" LastEditorUserId="3987" LastEditDate="2015-12-12T19:07:01.033" LastActivityDate="2015-12-12T19:07:01.033" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24483" PostTypeId="1" AcceptedAnswerId="24484" CreationDate="2015-12-11T07:50:07.310" Score="6" ViewCount="505" Body="&lt;p&gt;I am a six letter word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My &lt;strong&gt;1,2,3&lt;/strong&gt; is very abundant on earth.&lt;br&gt;&#xA;My &lt;strong&gt;2,3,4&lt;/strong&gt; is an organ.&lt;br&gt;&#xA;My &lt;strong&gt;3,4,5&lt;/strong&gt; is part of circle&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You frequently use me on the net.&lt;br&gt;&#xA;&lt;strong&gt;&lt;em&gt;What am I?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="16822" LastEditorUserId="4421" LastEditDate="2015-12-11T17:02:07.387" LastActivityDate="2015-12-11T17:03:59.253" Title="You frequently use me on the net. What am I?" Tags="|riddle|word|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24484" PostTypeId="2" ParentId="24483" CreationDate="2015-12-11T07:57:56.850" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;SEARCH&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Most abundant on earth:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; sea&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An organ:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ear&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Part of a circle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; arc&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You frequently use it over the net:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; search&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16608" LastEditorUserId="4421" LastEditDate="2015-12-11T17:03:19.820" LastActivityDate="2015-12-11T17:03:19.820" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24485" PostTypeId="2" ParentId="24483" CreationDate="2015-12-11T07:59:34.307" Score="5" Body="&lt;p&gt;I would say it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SEARCH&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Taken further next three is part of circle&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ARC&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Taken next three together is an organ&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; *AR can only be EAR&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Taken first three together is most abundant on earth.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; *EA I guess it's SEA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So it gives us&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SEARCH&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9352" LastEditorUserId="4421" LastEditDate="2015-12-11T17:03:59.253" LastActivityDate="2015-12-11T17:03:59.253" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24486" PostTypeId="1" AcceptedAnswerId="24492" CreationDate="2015-12-11T08:45:18.360" Score="26" ViewCount="1897" Body="&lt;p&gt;I am not married, but a daughter have I&lt;br&gt;&#xA;And to lead, she leads, daughter my&lt;br&gt;&#xA;I was part of the li'l boy that once rocked history&lt;br&gt;&#xA;I am just you, I ain't a mystery  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;" OwnerUserId="17659" LastEditorUserId="17659" LastEditDate="2016-04-28T01:29:07.277" LastActivityDate="2016-04-28T01:29:07.277" Title="Tell me who I am: To lead, she leads, daughter my" Tags="|riddle|chemistry|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24487" PostTypeId="2" ParentId="24486" CreationDate="2015-12-11T10:57:55.850" Score="5" Body="&lt;p&gt;Who am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Elena Tonra Lead of Band &quot;Daughter&quot; &lt;a href=&quot;https://en.wikipedia.org/wiki/Daughter_(band)&quot;&gt;https://en.wikipedia.org/wiki/Daughter_(band)&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am not married, but a daughter have I&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Elena Tonra is not married but she has a daughter i.e the band DAUGHTER&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And to lead, she leads, daughter my&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lead singer of Daughter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was part of the li'l boy that once rocked history&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To be deduced.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am just you, I ain't a mystery&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To be deduced.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2656" LastActivityDate="2015-12-11T10:57:55.850" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24491" PostTypeId="1" AcceptedAnswerId="24504" CreationDate="2015-12-11T13:39:02.300" Score="3" ViewCount="242" Body="&lt;p&gt;$$\lfloor \nabla_{\mu} X_{\nu} + \nabla_{\nu} X_{\mu}\rfloor = 0$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Hint1:&lt;/strong&gt; Also a video game.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="2819" LastEditDate="2015-12-11T17:29:48.140" LastActivityDate="2015-12-11T18:45:23.717" Title="Guess the song and artist (2)" Tags="|music|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24492" PostTypeId="2" ParentId="24486" CreationDate="2015-12-11T13:50:20.820" Score="34" Body="&lt;p&gt;This is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; uranium&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am not married, but a daughter have I&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To quote Wikipedia, &quot;A parent isotope is one that undergoes decay to form a daughter isotope. One example of this is uranium (atomic number 92) decaying into thorium (atomic number 90).&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And to lead, she leads, daughter my&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Thorium decays into other things which eventually decay into lead&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was part of the li'l boy that once rocked history&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To quote Wikipedia, &quot;Little Boy was the codename for the type of atomic bomb dropped on the Japanese city of Hiroshima&quot;.  It was the first uranium-based nuclear explosion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am just you, I ain't a mystery&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The atomic symbol for Uranium is U (you)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12092" LastActivityDate="2015-12-11T13:50:20.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24494" PostTypeId="2" ParentId="24480" CreationDate="2015-12-11T15:57:50.357" Score="3" Body="&lt;p&gt;We have the following statements by the people. Red is 'innocent', blue is 'guilty' and green is no statement. Statements by a single person are given as a row in the matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3pwwP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3pwwP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given that we are looking for a minimum collaborators solution, we may use a majority decision procedure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By this rule, we can first conclude that Apolonia is innocent, which gives Ezequiel, Mamie, Laurice, Cammie and Darron as suspects. We don't know which type of truthteller she is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, Ezequiel is a suspect and everything points to that he has committed the murder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuing this reasoning, we have a list of suspects:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Ezequiel&lt;/li&gt;&#xA;&lt;li&gt;Mamie&lt;/li&gt;&#xA;&lt;li&gt;Laurice&lt;/li&gt;&#xA;&lt;li&gt;Cammie&lt;/li&gt;&#xA;&lt;li&gt;Darron&lt;/li&gt;&#xA;&lt;li&gt;Nancee&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We can see that in this group, the statements diverge from the statments the rest of the group. Gale is not in the group by majority decision, but most statements are given by people in the suspect group. By inverting the statements of the suspect group, we obtain the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aDkQy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aDkQy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, we find that Gale is also guilty. So the group of guilty people are&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Ezequiel&lt;/li&gt;&#xA;&lt;li&gt;Mamie&lt;/li&gt;&#xA;&lt;li&gt;Laurice&lt;/li&gt;&#xA;&lt;li&gt;Cammie&lt;/li&gt;&#xA;&lt;li&gt;Darron&lt;/li&gt;&#xA;&lt;li&gt;Nancee&lt;/li&gt;&#xA;&lt;li&gt;Gale&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I believe this is the minimal solution. The discrepancies are due to that some people are boastful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note, I wrote the following code to parse the data and make it more overlookable:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import numpy&#xA;import matplotlib.pyplot as plt&#xA;&#xA;f = open(&quot;statements.txt&quot;,'r')&#xA;count = f.read().count(&quot;Statements&quot;)&#xA;f.seek(0)&#xA;&#xA;mapping = {}&#xA;people = []&#xA;people_counter = 0&#xA;truth_matrix = numpy.zeros((count,count)) # truth adjaceny matrix&#xA;#suspects = ['Mamie','Ezequiel','Laurice','Cammie','Nancee','Darron']&#xA;&#xA;for line in f:&#xA;    if line[0:10] == &quot;Statements&quot;:&#xA;        person = line.split()[2][0:-1]&#xA;        if person not in mapping:&#xA;            mapping[person] = people_counter&#xA;            people.append(person)&#xA;            people_counter += 1&#xA;    if line[0:1] == &quot;-&quot;:&#xA;        subject = line.split()[1]&#xA;        if subject == &quot;I&quot;:&#xA;            subject = person&#xA;        if subject not in mapping:&#xA;            mapping[subject] = people_counter&#xA;            people.append(subject)&#xA;            people_counter += 1&#xA;        mult = 1&#xA;        #if person in suspects: mult = -1&#xA;        if line.split()[3] == &quot;innocent&quot;:&#xA;            truth_matrix[mapping[person],mapping[subject]] = 100*mult&#xA;        else:&#xA;        truth_matrix[mapping[person],mapping[subject]] = -100*mult&#xA;&#xA;plt.xticks(range(len(people)), people, rotation='vertical')&#xA;plt.yticks(range(len(people)), people)&#xA;plt.imshow(truth_matrix, interpolation='nearest')&#xA;plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I hope you do not feel that it ruins the fun of the problems :-)&lt;/p&gt;&#xA;" OwnerUserId="12901" LastEditorUserId="12901" LastEditDate="2015-12-11T21:26:04.517" LastActivityDate="2015-12-11T21:26:04.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24496" PostTypeId="2" ParentId="17437" CreationDate="2015-12-11T16:06:31.393" Score="9" Body="&lt;p&gt;The encoded message is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is a sample message&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The last hint from &lt;strong&gt;question_asker&lt;/strong&gt; was my starting point. He quoted a clue which I couldn't find anywhere. I looked at the &lt;a href=&quot;https://puzzling.stackexchange.com/posts/17437/revisions&quot;&gt;revision history&lt;/a&gt; of the question and there it was, together with &lt;a href=&quot;https://puzzling.stackexchange.com/revisions/2ba00967-aa59-4283-863a-210ab445accd/view-source&quot;&gt;many more&lt;/a&gt;. The one which revealed the &quot;encoding&quot; to me was:&lt;br&gt;&lt;br&gt;&lt;code&gt;This puzzle's message starts with the word  This.&lt;/code&gt;&lt;br&gt;&lt;br&gt;Lets take a look at the beginning of the encoded text and mark &lt;strong&gt;T&lt;/strong&gt; with &lt;strong&gt;1&lt;/strong&gt;, &lt;strong&gt;h&lt;/strong&gt; with &lt;strong&gt;2&lt;/strong&gt;, &lt;strong&gt;i&lt;/strong&gt; with &lt;strong&gt;3&lt;/strong&gt; and &lt;strong&gt;s&lt;/strong&gt; with &lt;strong&gt;4&lt;/strong&gt;.&lt;br&gt;&lt;br&gt;&lt;code&gt;I  s Txz KhxIiHM s q nivRs OafToHB hduiTplssz gil s&lt;/code&gt;&lt;br&gt;&lt;code&gt;...4.1....2..3...4....3..4....1....2..31..44...3..4&lt;/code&gt;&lt;br&gt;&lt;br&gt;We can see the sequence &lt;strong&gt;1234&lt;/strong&gt; near the beginning with some irregular spacing between the letters. The interesting thing is, the same sequence appears a little further to the right with exact the same spacing (ignoring some of the numbers which are just there as fillers). I copied the encoded text to a second line below the first and rotated it so that both sequences are one above other (25 characters to the left). Taking the characters which are equal in both lines reveals the message.&lt;br&gt;&lt;br&gt;&lt;code&gt;I  s Txz KhxIiHM s q nivRs OafToHB hduiTplssz gil s  a q aT mWphl e&lt;/code&gt;&lt;br&gt;&lt;code&gt;s OafToHB hduiTplssz gil s  a q aT mWphl ens di gmVUJbgVzanbm pClee&lt;/code&gt;&lt;br&gt;&lt;code&gt;.....T....h..i...s.. .i..s .a..... ........s.............a..m.p.l.e&lt;/code&gt;&lt;br&gt;&lt;br&gt;&lt;code&gt;ns di gmVUJbgVzanbm pCleepLsxM smshyao rSgeREq jezknqcVxs s  aNvSEge&lt;/code&gt;&lt;br&gt;&lt;code&gt;pLsxM smshyao rSgeREq jezknqcVxs s  aNvSEgeI  s Txz KhxIiHM s q nivR&lt;/code&gt;&lt;br&gt;&lt;code&gt;..... .m...............e.......s.s..a....ge.........................&lt;/code&gt;&lt;br&gt;&lt;br&gt;Now most of the other clues make sense as well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-12-11T18:54:01.237" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24497" PostTypeId="1" AcceptedAnswerId="24502" CreationDate="2015-12-11T16:13:37.227" Score="14" ViewCount="650" Body="&lt;p&gt;What is the answer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hints:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;This puzzle is too vague  &lt;/li&gt;&#xA;&lt;li&gt;This puzzle is still incomplete  &lt;/li&gt;&#xA;&lt;li&gt;There is not enough information to determine the answer  &lt;/li&gt;&#xA;&lt;li&gt;When this puzzle is complete the answer will read as clear as day&lt;/li&gt;&#xA;&lt;li&gt;Come back later  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="14219" LastEditorUserId="14219" LastEditDate="2015-12-11T16:55:49.607" LastActivityDate="2015-12-11T17:10:10.033" Title="The answer to this puzzle is…" Tags="|lateral-thinking|" AnswerCount="4" CommentCount="8" ClosedDate="2015-12-11T18:13:48.093" ContentLicense="CC BY-SA 3.0" />
  <row Id="24498" PostTypeId="2" ParentId="24497" CreationDate="2015-12-11T16:28:56.253" Score="6" Body="&lt;p&gt;The anwser is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; too broad&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because a puzzle satisfying any/all of the hints will lead to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a close request with the tick on &quot;too broad&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14461" LastActivityDate="2015-12-11T16:28:56.253" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24501" PostTypeId="2" ParentId="24497" CreationDate="2015-12-11T17:02:30.267" Score="5" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; as clear as day&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When this puzzle is complete the answer will read &lt;strong&gt;as clear as day&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2015-12-11T17:02:30.267" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24502" PostTypeId="2" ParentId="24497" CreationDate="2015-12-11T17:07:18.567" Score="15" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; [on hold], or eventually [closed]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the title is shortly going to read &quot;The answer to this puzzle is… [on hold]&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12511" LastActivityDate="2015-12-11T17:07:18.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24503" PostTypeId="2" ParentId="24497" CreationDate="2015-12-11T17:10:10.033" Score="2" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An ellipsis (&lt;strong&gt;...&lt;/strong&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This puzzle is too vague&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An ellipsis can be used to describe something vaguely&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This puzzle is still incomplete&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An ellipsis can indicate that a field on a webpage contains more text (is incomplete)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;There is not enough information to determine the answer&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An ellipsis indicates a lack of information&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;When this puzzle is complete the answer will read as clear as day&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The puzzle is already complete, and the answer is there in the title, as clear as day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2015-12-11T17:10:10.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24504" PostTypeId="2" ParentId="24491" CreationDate="2015-12-11T17:33:39.370" Score="4" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Killing Floor by Howlin Wolf?&lt;br&gt;&#xA;&lt;br&gt;&#xA; $\nabla_{\mu} X_{\nu} + \nabla_{\nu} X_{\mu} = 0$ is the &quot;Killing equation&quot;, and $\lfloor x \rfloor$ is the floor function &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="10399" LastEditDate="2015-12-11T18:45:23.717" LastActivityDate="2015-12-11T18:45:23.717" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24505" PostTypeId="1" AcceptedAnswerId="24516" CreationDate="2015-12-11T21:49:09.153" Score="4" ViewCount="275" Body="&lt;p&gt;&lt;em&gt;Enough of the maths clues (well partially!) - something more literary:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Romulus and Remus; Mowgli.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Hint 1:&lt;/strong&gt;  Graphics[{Circle[], Line@# &amp;amp; /@ {{{0, -1}, {0, -3}}, {{-1, -2}, {1, -2}}}}]&lt;br&gt;&#xA; &lt;strong&gt;Hint 2:&lt;/strong&gt; Previous answer partial answer to this Q.&lt;br&gt;&#xA; &lt;strong&gt;Hint 3:&lt;/strong&gt; &lt;em&gt;Hint 1&lt;/em&gt; is song title (one word).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="2819" LastEditDate="2015-12-14T12:21:50.237" LastActivityDate="2015-12-14T12:26:40.577" Title="Guess the song and artist (3)" Tags="|music|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24506" PostTypeId="2" ParentId="24505" CreationDate="2015-12-11T22:01:37.857" Score="4" Body="&lt;p&gt;An incorrect answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Raised by Wolves&quot; by U2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My best explanation is that &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Rome's legendary founders and Mowglie are &quot;2&quot; who were &quot;raised by wolves.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm spending more time trying to figure out what the first clue is trying to tell us than I am with the puzzle itself.&lt;/p&gt;&#xA;" OwnerUserId="17020" LastEditorUserId="17020" LastEditDate="2015-12-11T22:10:49.290" LastActivityDate="2015-12-11T22:10:49.290" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24507" PostTypeId="1" AcceptedAnswerId="24596" CreationDate="2015-12-11T22:02:39.410" Score="11" ViewCount="440" Body="&lt;p&gt;During our latest space mission to Mars, humankind has retrieved a curious alien circuit. Our historians think it must be a calendar of some sort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Hqmtu.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Hqmtu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After speaking with some alien linguists, we know now that the characters are symbols of counting and we deduced that they must be numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dKbbe.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dKbbe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the engineering deparment hooked up the circuit to a low-frequency signal, it started moving, emitting light and spitting out similar symbols. Freely translated, the engineers observed and wrote down the following sequence&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[2, 1], [5, 5], [8, 2], [11, 1], [14, 5], [15, 4], [18, 3], [21, 2], [22, 3], [25, 2], [28, 1], [31, 5], [34, 2], [37, 1], [40, 5], [43, 2], [46, 1], [49, 5], [50, 4], [51, 2], [52, 3], [55, 2], [58, 1], [61, 5], [64, 2], [67, 1], [70, 5], [71, 4], [74, 3], [77, 2], [80, 1], [81, 2], [82, 3], [85, 2], [88, 1], [91, 5], [92, 4], [93, 2], [94, 3], [97, 2], [100, 1], [101, 2], [104, 1], [107, 5], [110, 2], [113, 1], [116, 5], [119, 2], [122, 1], [125, 5], [128, 2], [131, 1], [134, 5], [137, 2], [140, 1], [141, 2], [142, 3], [145, 2], [148, 1], [151, 5], [154, 2], [157, 1], [160, 5], [163, 2], [166, 1], [169, 5], [172, 2], [175, 1], [176, 2], [179, 1], [182, 5], [183, 4], [186, 3], [189, 2], [190, 3], [191, 5], [194, 2], [197, 1], [200, 5], [203, 2], [206, 1], [209, 5], [212, 2], [215, 1], [216, 2], [217, 3], [220, 2], [223, 1], [226, 5], [229, 2], [232, 1], [235, 5], [238, 2], [241, 1], [244, 5], [247, 2], [250, 1], [251, 2], [254, 1], [257, 5], [260, 2], [263, 1], [266, 5], [267, 4], [270, 3], [271, 5], [274, 2], [277, 1], [280, 5], [281, 4], [284, 3], [287, 2], [290, 1], [291, 2], [292, 3], [295, 2], [298, 1], [301, 5], [302, 4], [303, 2], [304, 3], [307, 2], [310, 1], [311, 2], [314, 1], [317, 5], [320, 2], [323, 1], [326, 5], [329, 2], [332, 1], [335, 5], [338, 2], [341, 1], [344, 5], [347, 2], [350, 1], [351, 2], [352, 3], [355, 2], [358, 1], [361, 5], [364, 2], [367, 1], [370, 5], [373, 2], [376, 1], [379, 5], [382, 2], [385, 1], [386, 2], [389, 1], [392, 5], [393, 4], [396, 3], [399, 2], [400, 3], [401, 5], [404, 2], [407, 1], [410, 5], [413, 2], [416, 1], [419, 5], [422, 2], [425, 1], [426, 2], [427, 3], [430, 2], [433, 1], [436, 5], [439, 2], [442, 1], [445, 5], [448, 2], [451, 1], [454, 5], [457, 2], [460, 1], [461, 2], [464, 1], [467, 5], [470, 2], [473, 1], [476, 5], [477, 4], [480, 3], [481, 5], [484, 2], [487, 1], [490, 5], [491, 4], [494, 3], [497, 2], [500, 1], [501, 2], [502, 3], [505, 2], [508, 1], [511, 5], [512, 4], [513, 2], [514, 3], [517, 2], [520, 1], [521, 2], [524, 1], [527, 5], [530, 2], [533, 1], [536, 5], [539, 2], [542, 1], [545, 5], [548, 2], [551, 1], [554, 5], [557, 2], [560, 1], [561, 2], [562, 3], [565, 2], [568, 1], [571, 5], [574, 2], [577, 1], [580, 5], [583, 2], [586, 1], [589, 5], [592, 2], [595, 1], [596, 2], [599, 1], [602, 5], [603, 4], [606, 3], [609, 2], [610, 3], [611, 5], [614, 2], [617, 1], [620, 5], [623, 2], [626, 1], [629, 5], [632, 2], [635, 1], [636, 2], [637, 3], [640, 2], [643, 1], [646, 5], [649, 2], [652, 1], [655, 5], [658, 2], [661, 1], [664, 5], [667, 2], [670, 1], [671, 2], [674, 1], [677, 5], [680, 2], [683, 1], [686, 5], [687, 4], [690, 3], [691, 5], [694, 2], [697, 1], [700, 5], [701, 4], [704, 3], [707, 2], [710, 1], [711, 2], [712, 3], [715, 2], [718, 1], [721, 5], [722, 4], [723, 2], [724, 3], [727, 2], [730, 1], [731, 2], [734, 1], [737, 5], [740, 2], [743, 1], [746, 5], [749, 2], [752, 1], [755, 5], [758, 2], [761, 1], [764, 5], [767, 2], [770, 1], [771, 2], [772, 3], [775, 2], [778, 1], [781, 5], [784, 2], [787, 1], [790, 5], [793, 2], [796, 1], [799, 5], [802, 2], [805, 1], [806, 2], [809, 1], [812, 5], [813, 4], [816, 3], [819, 2], [820, 3], [821, 5], [824, 2], [827, 1], [830, 5], [833, 2], [836, 1], [839, 5], [842, 2], [845, 1], [846, 2], [847, 3], [850, 2], [853, 1], [856, 5], [859, 2], [862, 1], [865, 5], [868, 2], [871, 1], [874, 5], [877, 2], [880, 1], [881, 2], [884, 1], [887, 5], [890, 2], [893, 1], [896, 5], [897, 4], [900, 3], [901, 5], [904, 2], [907, 1], [910, 5], [911, 4], [914, 3], [917, 2], [920, 1], [921, 2], [922, 3], [925, 2], [928, 1], [931, 5], [932, 4], [933, 2], [934, 3], [937, 2], [940, 1], [941, 2], [944, 1], [947, 5], [950, 2], [953, 1], [956, 5], [959, 2], [962, 1], [965, 5], [968, 2], [971, 1], [974, 5], [977, 2], [980, 1], [981, 2], [982, 3], [985, 2], [988, 1], [991, 5], [994, 2], [997, 1], [1000, 5], [1003, 2], [1006, 1], [1009, 5], [1012, 2], [1015, 1], [1016, 2], [1019, 1], [1022, 5], [1023, 4], [1026, 3], [1029, 2], [1030, 3], [1031, 5], [1034, 2], [1037, 1], [1040, 5], [1043, 2], [1046, 1], [1049, 5], [1052, 2], [1055, 1], [1056, 2], [1057, 3], [1060, 2], [1063, 1], [1066, 5], [1069, 2], [1072, 1], [1075, 5], [1078, 2], [1081, 1], [1084, 5], [1087, 2], [1090, 1], [1091, 2], [1094, 1], [1097, 5], [1100, 2], [1103, 1], [1106, 5], [1107, 4], [1110, 3], [1111, 5], [1114, 2], [1117, 1], [1120, 5], [1121, 4], [1124, 3], [1127, 2], [1130, 1], [1131, 2], [1132, 3], [1135, 2], [1138, 1], [1141, 5], [1142, 4], [1143, 2], [1144, 3], [1147, 2], [1150, 1], [1151, 2], [1154, 1], [1157, 5], [1160, 2], [1163, 1], [1166, 5], [1169, 2], [1172, 1], [1175, 5], [1178, 2], [1181, 1], [1184, 5], [1187, 2], [1190, 1], [1191, 2], [1192, 3], [1195, 2], [1198, 1], [1201, 5], [1204, 2], [1207, 1], [1210, 5], [1213, 2], [1216, 1], [1219, 5], [1222, 2], [1225, 1], [1226, 2], [1229, 1], [1232, 5], [1233, 4], [1236, 3], [1239, 2], [1240, 3], [1241, 5], [1244, 2], [1247, 1], [1250, 5], [1253, 2], [1256, 1], [1259, 5], [1262, 2], [1265, 1], [1266, 2], [1267, 3], [1270, 2], [1273, 1], [1276, 5], [1279, 2], [1282, 1], [1285, 5], [1288, 2], [1291, 1], [1294, 5], [1297, 2], [1300, 1], [1301, 2], [1304, 1], [1307, 5], [1310, 2], [1313, 1], [1316, 5], [1317, 4], [1320, 3], [1321, 5], [1324, 2], [1327, 1], [1330, 5], [1331, 4], [1334, 3], [1337, 2], [1340, 1], [1341, 2], [1342, 3], [1345, 2], [1348, 1], [1351, 5], [1352, 4], [1353, 2], [1354, 3], [1357, 2], [1360, 1], [1361, 2], [1364, 1], [1367, 5], [1370, 2], [1373, 1], [1376, 5], [1379, 2], [1382, 1], [1385, 5], [1388, 2], [1391, 1], [1394, 5], [1397, 2], [1400, 1], [1401, 2], [1402, 3], [1405, 2], [1408, 1], [1411, 5], [1414, 2], [1417, 1], [1420, 5], [1423, 2], [1426, 1], [1429, 5], [1432, 2], [1435, 1], [1436, 2], [1439, 1], [1442, 5], [1443, 4], [1446, 3], [1449, 2], [1450, 3], [1451, 5], [1454, 2], [1457, 1], [1460, 5], [1463, 2], [1466, 1], [1469, 5], [1472, 2], [1475, 1], [1476, 2], [1477, 3], [1480, 2], [1483, 1], [1486, 5], [1489, 2], [1492, 1], [1495, 5], [1498, 2], [1501, 1], [1504, 5], [1507, 2], [1510, 1], [1511, 2], [1514, 1], [1517, 5], [1520, 2], [1523, 1], [1526, 5], [1527, 4], [1530, 3], [1531, 5], [1534, 2], [1537, 1], [1540, 5], [1541, 4], [1544, 3], [1547, 2], [1550, 1], [1551, 2], [1552, 3], [1555, 2], [1558, 1], [1561, 5], [1562, 4], [1563, 2], [1564, 3], [1567, 2], [1570, 1], [1571, 2], [1574, 1], [1577, 5], [1580, 2], [1583, 1], [1586, 5], [1589, 2], [1592, 1], [1595, 5], [1598, 2], [1601, 1], [1604, 5], [1607, 2], [1610, 1], [1611, 2], [1612, 3], [1615, 2], [1618, 1], [1621, 5], [1624, 2], [1627, 1], [1630, 5], [1633, 2], [1636, 1], [1639, 5], [1642, 2], [1645, 1], [1646, 2], [1649, 1], [1652, 5], [1653, 4], [1656, 3], [1659, 2], [1660, 3], [1661, 5], [1664, 2], [1667, 1], [1670, 5], [1673, 2], [1676, 1], [1679, 5], [1682, 2], [1685, 1], [1686, 2], [1687, 3], [1690, 2], [1693, 1], [1696, 5], [1699, 2], [1702, 1], [1705, 5], [1708, 2], [1711, 1], [1714, 5], [1717, 2], [1720, 1], [1721, 2], [1724, 1], [1727, 5], [1730, 2], [1733, 1], [1736, 5], [1737, 4], [1740, 3], [1741, 5], [1744, 2], [1747, 1], [1750, 5], [1751, 4], [1754, 3], [1757, 2], [1760, 1], [1761, 2], [1762, 3], [1765, 2], [1768, 1], [1771, 5], [1772, 4], [1773, 2], [1774, 3], [1777, 2], [1780, 1], [1781, 2], [1784, 1], [1787, 5], [1790, 2], [1793, 1], [1796, 5], [1799, 2], [1802, 1], [1805, 5], [1808, 2], [1811, 1], [1814, 5], [1817, 2], [1820, 1], [1821, 2], [1822, 3], [1825, 2], [1828, 1], [1831, 5], [1834, 2], [1837, 1], [1840, 5], [1843, 2], [1846, 1], [1849, 5], [1852, 2], [1855, 1], [1856, 2], [1859, 1], [1862, 5], [1863, 4], [1866, 3], [1869, 2], [1870, 3], [1871, 5], [1874, 2], [1877, 1], [1880, 5], [1883, 2], [1886, 1], [1889, 5], [1892, 2], [1895, 1], [1896, 2], [1897, 3], [1900, 2], [1903, 1], [1906, 5], [1909, 2], [1912, 1], [1915, 5], [1918, 2], [1921, 1], [1924, 5], [1927, 2], [1930, 1], [1931, 2], [1934, 1], [1937, 5], [1940, 2], [1943, 1], [1946, 5], [1947, 4], [1950, 3], [1951, 5], [1954, 2], [1957, 1], [1960, 5], [1961, 4], [1964, 3], [1967, 2], [1970, 1], [1971, 2], [1972, 3], [1975, 2], [1978, 1], [1981, 5], [1982, 4], [1983, 2], [1984, 3], [1987, 2], [1990, 1], [1991, 2], [1994, 1], [1997, 5], [2000, 2], [2003, 1], [2006, 5], [2009, 2], [2012, 1], [2015, 5], [2018, 2], [2021, 1], [2024, 5], [2027, 2], [2030, 1], [2031, 2], [2032, 3], [2035, 2], [2038, 1], [2041, 5], [2044, 2], [2047, 1], [2050, 5], [2053, 2], [2056, 1], [2059, 5], [2062, 2], [2065, 1], [2066, 2], [2069, 1], [2072, 5], [2073, 4], [2076, 3], [2079, 2], [2080, 3], [2081, 5], [2084, 2], [2087, 1], [2090, 5], [2093, 2], [2096, 1], [2099, 5], [2102, 2], [2105, 1], [2106, 2], [2107, 3], [2110, 2], [2113, 1], [2116, 5], [2119, 2], [2122, 1], [2125, 5], [2128, 2], [2131, 1], [2134, 5], [2137, 2], [2140, 1], [2141, 2], [2144, 1], [2147, 5], [2150, 2], [2153, 1], [2156, 5], [2157, 4], [2160, 3], [2161, 5], [2164, 2], [2167, 1], [2170, 5], [2171, 4], [2174, 3], [2177, 2], [2180, 1], [2181, 2], [2182, 3], [2185, 2], [2188, 1], [2191, 5], [2192, 4], [2193, 2], [2194, 3], [2197, 2], [2200, 1], [2201, 2], [2204, 1], [2207, 5], [2210, 2], [2213, 1], [2216, 5], [2219, 2], [2222, 1], [2225, 5], [2228, 2], [2231, 1], [2234, 5], [2237, 2], [2240, 1], [2241, 2], [2242, 3], [2245, 2], [2248, 1], [2251, 5], [2254, 2], [2257, 1], [2260, 5], [2263, 2], [2266, 1], [2269, 5], [2272, 2], [2275, 1], [2276, 2], [2279, 1], [2282, 5], [2283, 4], [2286, 3], [2289, 2], [2290, 3], [2291, 5], [2294, 2], [2297, 1], [2300, 5], [2303, 2], [2306, 1], [2309, 5], [2312, 2], [2315, 1], [2316, 2], [2317, 3], [2320, 2], [2323, 1], [2326, 5], [2329, 2], [2332, 1], [2335, 5], [2338, 2], [2341, 1], [2344, 5], [2347, 2], [2350, 1], [2351, 2], [2354, 1], [2357, 5], [2360, 2], [2363, 1], [2366, 5], [2367, 4], [2370, 3], [2371, 5], [2374, 2], [2377, 1], [2380, 5], [2381, 4], [2384, 3], [2387, 2], [2390, 1], [2391, 2], [2392, 3], [2395, 2], [2398, 1], [2401, 5], [2402, 4], [2403, 2], [2404, 3], [2407, 2], [2410, 1], [2411, 2], [2414, 1], [2417, 5], [2420, 2], [2423, 1], [2426, 5], [2429, 2], [2432, 1], [2435, 5], [2438, 2], [2441, 1], [2444, 5], [2447, 2], [2450, 1], [2451, 2], [2452, 3], [2455, 2], [2458, 1], [2461, 5], [2464, 2], [2467, 1], [2470, 5], [2473, 2], [2476, 1], [2479, 5], [2482, 2], [2485, 1], [2486, 2], [2489, 1], [2492, 5], [2493, 4], [2496, 3], [2499, 2], [2500, 3], [2501, 5], [2504, 2], [2507, 1], [2510, 5], [2513, 2], [2516, 1], [2519, 5], [2522, 2], [2525, 1], [2526, 2], [2527, 3], [2530, 2], [2533, 1], [2536, 5], [2539, 2], [2542, 1], [2545, 5], [2548, 2], [2551, 1], [2554, 5], [2557, 2], [2560, 1]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;before the circuit broke down. The sequence has the format [output, state number/light]. When we analyzed the parts, we could see some circular objects looking similar but with different sizes. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can we determine how the circuit worked and recreate a functioning one using human technology? A solution should at least describe the rule how the numbers of generated.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12901" LastEditorUserId="12901" LastEditDate="2015-12-12T09:29:11.093" LastActivityDate="2021-01-18T17:07:36.967" Title="A curious circuit" Tags="|computer-puzzle|" AnswerCount="2" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24508" PostTypeId="2" ParentId="24507" CreationDate="2015-12-11T22:54:38.430" Score="2" Body="&lt;p&gt;I'm not sure if there's supposed to be more to the answer than this...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The machine moves between the states &lt;kbd&gt;1&lt;/kbd&gt; through &lt;kbd&gt;5&lt;/kbd&gt;.  Following a red arrow adds $3$ to the output.  Following a black arrow adds $1$ to the output.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2015-12-11T22:54:38.430" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24510" PostTypeId="1" AcceptedAnswerId="24511" CreationDate="2015-12-12T02:26:35.260" Score="5" ViewCount="346" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Fine, precise,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;a powerful vice.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I draw up blood, thin crimson penned;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Yet with never a wound can I contend.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Under careful conditions am I used,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;although on occasion I've been abused.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Many an ache have I helped ease;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I have even aided the diseased.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I have helped many people escape;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;yet I imbue no movement.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This is my first time on the Puzzling Stack Exchange. Feel free to give comments or criticisms, I love to hear it.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This is also my first riddle, and I'm worried it may be pretty easy. A line by line breakdown on how your answer fits is welcome!&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="17744" LastActivityDate="2015-12-12T03:09:40.850" Title="A powerful vice. Who am I?" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24511" PostTypeId="2" ParentId="24510" CreationDate="2015-12-12T03:09:40.850" Score="1" Body="&lt;p&gt;This seems a little on-the-nose, but is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a needle? (i.e., a syringe)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fine, precise,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; needles are marked for measurement, usually fine measurement&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;a powerful vice.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; they can contain powerful drugs, the use of which is known as a &quot;vice&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I draw up blood, thin crimson penned;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; well, drawing up blood is one thing that needles are known to do&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet with never a wound can I contend.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a needle will ... probably? ... not release as much blood as an open wound would&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Under careful conditions am I used,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; doctors and nurses try to be careful when using needles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;although on occasion I've been abused.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; as mentioned above, needles can be used to inject drugs. in some cases this is called &quot;drug abuse&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Many an ache have I helped ease;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; opioid narcotics (the legal kind and the illegal kind), for one example, are administered with needles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have even aided the diseased.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; as noted above, needles can be used to inject substances for medicinal purposes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have helped many people escape;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;escape&quot; is being used in the mental/emotional sense here - drugs administered with needles can provide an escape from reality&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yet I imbue no movement.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the needle itself does not propel anyone forward (or in any other direction), per se&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I really feel like I'm missing something about this one&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="17250" LastActivityDate="2015-12-12T03:09:40.850" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24512" PostTypeId="1" AcceptedAnswerId="24549" CreationDate="2015-12-12T03:57:31.520" Score="10" ViewCount="936" Body="&lt;p&gt;Alice and Bob play a game on  the following irregular chessboard. (Note the blacked out squares are not legal moves.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mUkJ1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mUkJ1.png&quot; alt=&quot;The game board.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Alice starts the game by placing a knight on any square she chooses.&lt;/li&gt;&#xA;&lt;li&gt;They then take turns, starting with Bob, moving the knight as a knight moves in chess.&lt;/li&gt;&#xA;&lt;li&gt;Immediately reversing the previous move is not allowed.  That is, if one player moves the knight from square $x$ to square $y$, the next player is not allowed to move to square $x$ during the immediate next turn.&lt;/li&gt;&#xA;&lt;li&gt;The winner is the first player to move to a square that has been previously occupied sometime during the game.  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Who wins with optimum play and why?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2722" LastEditorUserId="12408" LastEditDate="2016-07-25T02:40:28.360" LastActivityDate="2016-07-25T02:40:28.360" Title="Let's revisit this later (a two player knight's game)" Tags="|mathematics|game|checkerboard|knight-moves|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="24513" PostTypeId="1" AcceptedAnswerId="24514" CreationDate="2015-12-12T08:08:26.300" Score="-2" ViewCount="7559" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Find the missing number in given series: $253, 495, 143, ?$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$152$ &lt;/li&gt;&#xA;&lt;li&gt;$105$ &lt;/li&gt;&#xA;&lt;li&gt;$903$ &lt;/li&gt;&#xA;&lt;li&gt;$374$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Okay, so I'll tell you what I could do:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first number is $16^2-1$ third number is $12^2-1.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, the differences are: $+242,-352.$ So my next attempt was to do $+462$. But none of the options are correct then.&lt;/p&gt;&#xA;" OwnerUserId="11425" LastEditorUserId="8874" LastEditDate="2015-12-14T12:13:32.497" LastActivityDate="2020-04-27T13:32:12.383" Title="Find the missing number in the series: 253, 495, 143,?" Tags="|number-sequence|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24514" PostTypeId="2" ParentId="24513" CreationDate="2015-12-12T08:44:32.290" Score="8" Body="&lt;p&gt;There is an obvious &lt;strong&gt;pattern&lt;/strong&gt; which suggests an answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;In each of the numbers given, the middle digit is the sum of the first and last digits, so this suggests that the answer is d) 374 as that is the only one which fits the pattern.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However these four numbers don't form a &lt;strong&gt;series&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-12-12T08:44:32.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24515" PostTypeId="2" ParentId="16846" CreationDate="2015-12-12T10:17:17.363" Score="2" Body="&lt;p&gt;simpler is to draw a path with a brush in a photo software.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;draw the whole path with all the dead ends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then select the whole and invert or make negative !&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GGSyz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GGSyz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17750" LastActivityDate="2015-12-12T10:17:17.363" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24516" PostTypeId="2" ParentId="24505" CreationDate="2015-12-12T13:39:03.473" Score="3" Body="&lt;p&gt;How about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Woman&lt;/em&gt; by Wolfmother (Female symbol, Romulus and Remus and Mowgli were raised in the wild ('by wolves'))&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2015-12-14T12:26:40.577" LastActivityDate="2015-12-14T12:26:40.577" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24518" PostTypeId="2" ParentId="24486" CreationDate="2015-12-12T15:03:27.707" Score="-2" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Uranium. Your decay includes lead Pb.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You were part of &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the Little Boy atomic bomb used on Hiroshima&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17754" LastEditorUserId="2621" LastEditDate="2015-12-12T18:14:22.070" LastActivityDate="2015-12-12T18:14:22.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24519" PostTypeId="1" AcceptedAnswerId="24532" CreationDate="2015-12-12T15:49:51.690" Score="8" ViewCount="448" Body="&lt;p&gt;12 chess players took part in a tournament. Each played against each other exactly once. After the tournament every chess player did 12 lists of names.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;On the first list, the player only wrote his own name.&lt;/li&gt;&#xA;&lt;li&gt;On the second list, they wrote their own names as well as all man they had won against.&lt;/li&gt;&#xA;&lt;li&gt;They proceeded to write lists: Every next list contains all names from the previous list and the new names that players from the previous list had won against.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It turned out that for all chess players, the 11th and the 12th list contained different amount of each name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many games ended draw in the tournament?&lt;/p&gt;&#xA;" OwnerUserId="17755" LastActivityDate="2015-12-12T20:55:55.540" Title="Chess tournament problem" Tags="|logical-deduction|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24520" PostTypeId="1" AcceptedAnswerId="24521" CreationDate="2015-12-12T16:31:08.613" Score="3" ViewCount="181" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;First one, now two snakes&lt;br&gt;&#xA;  Too much? Rightfully, not free&lt;br&gt;&#xA;  Alternative, less?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17756" LastEditorUserId="8874" LastEditDate="2016-03-23T16:56:22.660" LastActivityDate="2016-03-23T16:56:22.660" Title="who are you going to call?" Tags="|riddle|haiku|history|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24521" PostTypeId="2" ParentId="24520" CreationDate="2015-12-12T17:12:08.717" Score="5" Body="&lt;p&gt;I'm not quite sure, but i think you are going to call&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Doctor&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First one, now two snakes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Refers to the the snake and staff symbol which has been associated with healthcare. The single snake emblem is that of Aesculapius. The double snake emblem is caduceus of Hermes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Too much? Rightfully not free ... alternatively, less?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Probably this one talks about the high cost of healthcare facilities in some countries and need to make it more affordable&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17659" LastActivityDate="2015-12-12T17:12:08.717" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24522" PostTypeId="1" AcceptedAnswerId="24523" CreationDate="2015-12-12T17:43:12.867" Score="6" ViewCount="495" Body="&lt;p&gt;What sentence do the following paragraphs encode?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look lad. Light looms lad. See light shows. Silly silly shows. See.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lately seven lemurs sang. La la la. Like seven skylarks. Stupidly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sit somewhere. Slice some sausages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Send loaves. Lotsa sandwich loaves soon. Limes. Some sour lozenges. Some leeks. Some large sauerkraut sausages. Some long sauerkraut sausages. Like sandwich length long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shoot leopards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets stop lighting socks. Stop stop. Stop lighting long socks. Stop stop stop stop. Stop. Someone lit socks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lucy says. Lie lie lie. Loudly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Someone lurched.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Loud singers like shouting. Ludicrously loud long. Loud singers shouting. Shhhhhhh.&lt;/p&gt;&#xA;" OwnerUserId="17712" LastActivityDate="2015-12-12T17:58:20.817" Title="The Code that's not a Code" Tags="|cipher|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24523" PostTypeId="2" ParentId="24522" CreationDate="2015-12-12T17:58:20.817" Score="6" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MORSE CODE IS ACTUALLY A CIPHER NOT A CODE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is encoded as&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; morse code, obviously. Starting with L gives - and S gives .&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12901" LastActivityDate="2015-12-12T17:58:20.817" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24524" PostTypeId="1" AcceptedAnswerId="24526" CreationDate="2015-12-12T18:47:59.853" Score="4" ViewCount="270" Body="&lt;p&gt;As I sat there, a woman in black asked me if I had decided yet. I told her I hadn't, but she gave me a glass anyway. Later, she came again. This time I had decided. So, she went and got something from the man in white and brought it to me. When I was finished, I wasn't allowed to go until I gave her something extra.&lt;/p&gt;&#xA;" OwnerUserId="17712" LastActivityDate="2015-12-12T18:59:56.523" Title="What Strange Place am I in?" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24525" PostTypeId="2" ParentId="3445" CreationDate="2015-12-12T18:58:15.137" Score="-2" Body="&lt;p&gt;Just click all buttons one time in any order and you will convert all from red to green.&lt;/p&gt;&#xA;" OwnerUserId="17758" LastActivityDate="2015-12-12T18:58:15.137" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24526" PostTypeId="2" ParentId="24524" CreationDate="2015-12-12T18:59:56.523" Score="5" Body="&lt;p&gt;Sounds like you are in  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A restaurant  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;br&gt;&#xA;a woman in black asked me if I had decided yet. I told her I hadn't, but she gave me a glass anyway  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The waitress came to your table to see if you had decided on your meal and gave you the drink (in a glass) you had ordered already. Or a glass of water as some will give you without asking.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later, she came again. This time I had decided  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You give her your order  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, she went and got something from the man in white and brought it to me.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; She got your order from the chef  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I was finished, I wasn't allowed to go until I gave her something extra.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You had to pay before leaving.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7715" LastActivityDate="2015-12-12T18:59:56.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24527" PostTypeId="1" CreationDate="2015-12-12T19:04:10.137" Score="12" ViewCount="389" Body="&lt;p&gt;Sage of mountain, oracle of sky,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hidden by the vegetable lie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dutch affection in voices gilt&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When Windy City's tower was built.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bane of the hasty, raven's prey,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Powers grown from birth to today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet, I am unchanged in the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you know my name, dear friend?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Hint/alternative/improved scansion version:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I'm sage of earth, seer of the sky,&lt;br&gt;&#xA;&#xA;Use obscured by carrot lie.&lt;br&gt;&#xA;&#xA;Dutch song my appellation told&lt;br&gt;&#xA;&#xA;During crisis of black gold.&lt;br&gt;&#xA;&#xA;Bane of the hasty, friend of fuzz&lt;br&gt;&#xA;&#xA;AKA sub eyeball cuz.&lt;br&gt;&#xA;&#xA;In the end, I am the same&lt;br&gt;&#xA;&#xA;As from the start. Please guess my name.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17020" LastEditorUserId="17020" LastEditDate="2015-12-15T05:19:04.953" LastActivityDate="2015-12-15T10:58:05.820" Title="Hidden by the Vegetable Lie" Tags="|riddle|" AnswerCount="1" CommentCount="11" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24528" PostTypeId="1" AcceptedAnswerId="24534" CreationDate="2015-12-12T19:47:16.040" Score="4" ViewCount="480" Body="&lt;p&gt;Replace the question mark with your solution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CpV2T.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CpV2T.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user17753" LastActivityDate="2015-12-13T20:12:56.970" Title="It needs some thinking but solvable and nice" Tags="|mathematics|geometry|visual|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24529" PostTypeId="1" AcceptedAnswerId="24531" CreationDate="2015-12-12T19:53:53.253" Score="6" ViewCount="387" Body="&lt;p&gt;Draw the missing item replacing the question mark&#xA;&lt;a href=&quot;https://i.stack.imgur.com/buuW4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/buuW4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user17752" LastActivityDate="2015-12-12T21:17:52.753" Title="Draw the item after decoding the instructions" Tags="|logical-deduction|pattern|visual|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24530" PostTypeId="2" ParentId="24528" CreationDate="2015-12-12T20:00:44.023" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;That would be a hypercube, looking something like&#xA;  &lt;a href=&quot;https://i.stack.imgur.com/YUQdS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YUQdS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12901" LastActivityDate="2015-12-12T20:00:44.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24531" PostTypeId="2" ParentId="24529" CreationDate="2015-12-12T20:37:01.970" Score="8" Body="&lt;ul&gt;&#xA;&lt;li&gt;4 is partly covered by 0&lt;/li&gt;&#xA;&lt;li&gt;0 is completly covered by 3&lt;/li&gt;&#xA;&lt;li&gt;6 is partly covered by 0&lt;/li&gt;&#xA;&lt;li&gt;6 is partly uncovered by 0&lt;/li&gt;&#xA;&lt;li&gt;4 is completely uncovered by 6&lt;/li&gt;&#xA;&lt;li&gt;4 is partly uncovered by 3&lt;/li&gt;&#xA;&lt;li&gt;6 is not covered by 4&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So we need a figure that&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;6 is partly covered by 4&lt;/li&gt;&#xA;&lt;li&gt;3 is completely uncovered by 6&lt;/li&gt;&#xA;&lt;li&gt;0 is partly covered by 3&lt;/li&gt;&#xA;&lt;li&gt;4 is completely covered by 0&lt;/li&gt;&#xA;&lt;li&gt;0 is partly covered by 6&lt;/li&gt;&#xA;&lt;li&gt;4 is partly covered by 3&lt;/li&gt;&#xA;&lt;li&gt;6 is completely covered by 0&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Therefore,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/C3hg9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/C3hg9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12901" LastEditorUserId="12901" LastEditDate="2015-12-12T21:17:52.753" LastActivityDate="2015-12-12T21:17:52.753" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24532" PostTypeId="2" ParentId="24519" CreationDate="2015-12-12T20:55:55.540" Score="8" Body="&lt;p&gt;The number of draws must have been&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 54&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In order for the lists to keep growing right up until the end, each list must contain exactly one more person than the previous (they can't grow by more than that, otherwise the last list would have to have more than 12 names).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, the second list must contain exactly two people, meaning that each person in the tournament beat exactly one other person. This means there were only 12 victories total in the tournament. Subtracting this from the total number of matches, 66, gives the number of draws.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-12-12T20:55:55.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24534" PostTypeId="2" ParentId="24528" CreationDate="2015-12-12T22:31:37.970" Score="7" Body="&lt;p&gt;That was easy!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The figures are n-cubes with half the size of the larger one. The numbers show in which half the little cube is in each coordinate:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2 1   -&amp;gt; high x, low  y -&amp;gt; top right&#xA;2 2   -&amp;gt; high x, high y -&amp;gt; bottom right&#xA;1 2 2 -&amp;gt; low  x, high y, high z -&amp;gt; back's bottom left&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The next one is the tesseract, which can be divided in 16 half-sided cubes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The one we want is at the top right, back and far side in the w-dimension:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Zejal.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zejal.png&quot; alt=&quot;Incredibly counterproductive&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See? If I measured my time answering this question, it would be like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;3%&lt;/strong&gt;......Thinking about the puzzle&lt;br&gt;&#xA;  &lt;strong&gt;97%&lt;/strong&gt;.....Trying to draw tesseracts with correct face occlusion on PAINT.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm surprised no one took a more practical approach in the meantime.&lt;/p&gt;&#xA;" OwnerUserId="14219" LastActivityDate="2015-12-12T22:31:37.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24535" PostTypeId="1" AcceptedAnswerId="105951" CreationDate="2015-12-13T01:12:06.223" Score="15" ViewCount="1413" Body="&lt;p&gt;Tommy just got a new train set. It only came with one type of train track piece, a quarter circle, all of which were the same size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hspace{2.5in}$&#xA;&lt;a href=&quot;https://i.stack.imgur.com/p6NBp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/p6NBp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Prove that, whenever Tommy makes a closed loop out of these pieces, the number of tracks he uses will always be a multiple of four.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example, this loop uses 40 tracks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/weY6f.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/weY6f.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="69582" LastEditDate="2020-12-22T16:01:41.690" LastActivityDate="2020-12-22T16:01:41.690" Title="Tommy's Train Tracks" Tags="|mathematics|geometry|combinatorics|" AnswerCount="6" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24536" PostTypeId="2" ParentId="24535" CreationDate="2015-12-13T05:07:39.210" Score="3" Body="&lt;p&gt;&lt;strong&gt;A proof:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will define &quot;area&quot; as the area enclosed by any collection of track pieces, such that this area is smaller than the area on the other side of the track (also known as &quot;the world&quot;).  If there are multiple disconnected collections, the total area is the sum of all such smaller areas.  If there are concentric loops, each loop counts all area contained within, even if this would lead to double/triple/etc counting some pieces.  This area is well defined initially, because we are guaranteed that the tracks form a closed loop that encloses something (and, implied, that this loop is of finite size).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the area is 0, we are done (since our loop consists of zero pieces of track, and zero is divisible by four).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order for the track to be closed if it has area greater than zero, it must at some point have two consecutive convex turns - we'll call those turns &quot;right&quot;, with turns in the opposite direction &quot;left&quot;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Proof:&lt;br&gt;&#xA;&lt;br&gt;&#xA; If there are no places with two consecutive convex turns, then every convex turn must be bordered on either side by concave turns.  This means there are at least as many concave turns as convex turns (note that we can't have an unbounded track with a convex turn at each end to get an extra, since this would not form a closed loop), in which case the total angle of turn in a convex direction (where each piece of track is 90*), must be less than or equal to 0*, since each concave track piece cancels out a convex track piece.  But the total angle of turn in a convex direction must be exactly 360* in order to form a closed loop - therefor, we have a contradiction, and our assumption that we had no places with two consecutive convex turns must have been false.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We will now examine what could be on either side of those right turns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Type 1: &lt;strong&gt;&lt;a href=&quot;https://i.stack.imgur.com/8SVzR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Right, Right, Right, Right&lt;/a&gt;&lt;/strong&gt; - this forms a disconnected closed loop, and can be altered by removing the four tracks involved.  This shrinks the total area inside the tracks by a finite amount, and does not change whether any other tracks in our collection still form closed loops. &lt;em&gt;This is always safe.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Type 2: &lt;strong&gt;Right, Right, Right, Left&lt;/strong&gt; or &lt;strong&gt;Left, Right, Right, Right&lt;/strong&gt; - note that this cannot have a fourth consecutive Right, since they would be a closed loop per Type 1 and the Left could not connect.  As a result, this case can always be restated as &lt;a href=&quot;https://i.stack.imgur.com/mV0EB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Left, Right, Right, Right, Left&lt;/a&gt;.  However, in this case, we can pick up both the Lefts and rotate them 90*, giving us &lt;a href=&quot;https://i.stack.imgur.com/mICu4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Right - Right, Right, Right, Right&lt;/a&gt;.  This alters the collection of tracks by removing four tracks and rotating one.  The total area is shrunk by a finite amount without changing whether the tracks are connected in a closed loop. &lt;em&gt;This is always safe.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Type 3: &lt;strong&gt;&lt;a href=&quot;https://i.stack.imgur.com/FpH7E.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Left, Right, Right, Left&lt;/a&gt;&lt;/strong&gt; - we can turn this from a convex set of track into a concave set of track by rotating all the pieces involved 180* around the center to get Right, Left, Left, Right - in which case, we have shrunk the total area that is inside the track by a finite amount, without changing the number of tracks or whether the tracks form a closed loop.  &lt;em&gt;This is not necessarily safe, since it doesn't work for tracks like &lt;a href=&quot;https://i.stack.imgur.com/qntVC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt;.  However, it is safe for all other cases where this rotation would not cause track to overlap.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There always exists &lt;em&gt;at least&lt;/em&gt; one safe transformation that shrinks the contained area of our collection of tracks.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Proof:&lt;br&gt;&#xA;&lt;br&gt;&#xA; First note that the three types above are comprehensive - this is easily seen from the fact that they cover all four possibilities for the two pieces of track adjacent to our convex pair - R...R, R...L, L...R, and L...L.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Now assume there exists a finite non-zero closed loop track with no safe transformations as defined above.  All type 1 and 2 transformations are safe, so this track must consist exclusively of unsafe type 3's at all places where there are two consecutive convex curves. But note that the interfering curve that makes a type 3 unsafe must either be part of a different connected loop from that type 3 or it must be part of the same loop.  If it's part of a different loop, both loops can be picked up and placed at a safe distance apart from each other such that the original type 3 becomes safe.  This leaves the number of tracks unchanged, leaves all loops intact, and doesn't alter the total contained area, so it is a safe transformation.  We claimed there were no safe transformations, so every type 3 in our loop is being interfered with by another part of that loop.  In this case, all the interfering curves must be concave.  All such type 3's which are unsafe involve at least the following curves: &lt;a href=&quot;https://i.stack.imgur.com/eBD2R.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;two convex and four concave&lt;/a&gt;.  If we attempt to increase the number of convex pieces involved by attaching one to any end, we must then increase the number of concave pieces because either we do not have two consecutive convex pieces or we do, and they form another unsafe and inseparable type 3, which necessitates the accompanying majority of concave pieces.  You could overlap two adjacent unsafe type 3's in an attempt to share two concave pieces - but this would result in four convex and six concave, no change in the difference.  You could also find some way to connect the two concave pieces on either side with a convex (not directly, obviously, but at some point down the line).  This would get you to - at best - the same number of convex curves as concave curves.  In conclusion, any loop containing only unsafe and inseparable type 3's cannot have more convex curves than concave, which is a contradiction with it being a closed loop.  Thus, our original assumption was faulty, and there is some safe transformation (as defined in our three types) in any non-zero closed loop.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For any finite closed non-zero loop(s), starting with the initial situation and using a safe transformation, we can alter the collection of tracks in such a way that the total area is decreased by a finite amount and they still form closed loop(s).  We then find a safe transformation for this new collection and repeat.  This allows us to reduce the total area until we reach a contained area of zero.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Proof:&lt;br&gt;&#xA;&lt;br&gt;&#xA; Since we started with finite area and can reduce by finite amounts an arbitrary number of times, we must eventually reach zero or negative area.  No safe transformation can cause us to have a negative area, therefor we must reach exactly zero.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since none of our transformations ever caused us to stop forming a connected loop, we have a loop with zero area, which means we must have zero tracks left.  Because we only ever removed tracks four at a time, the number of tracks in our initial closed loop &lt;em&gt;must have been divisible by four&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jCZq8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;By way of example, using this method on the example loop&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="17389" LastEditDate="2015-12-13T09:02:25.967" LastActivityDate="2015-12-13T09:02:25.967" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24537" PostTypeId="2" ParentId="24535" CreationDate="2015-12-13T09:11:20.227" Score="1" Body="&lt;p&gt;Suppose without loss of generality our starting position is $(0,0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $T_1$ be the track piece provided in the question, and suppose the radius of the arc is $1$, so that if $T_1$ were the first piece encountered from our starting position, the next position would be $(1,1)$ if going clockwise and $(-1,-1)$ if going counterclockwise. Similarly, let $T_2$ be &lt;a href=&quot;https://i.stack.imgur.com/ez9ap.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;its rotation by 90 degrees&lt;/a&gt;, $T_3$ &lt;a href=&quot;https://i.stack.imgur.com/qbKgB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;its rotation by 180 degrees&lt;/a&gt;, and $T_4$ &lt;a href=&quot;https://i.stack.imgur.com/VkTfl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;its rotation by 270 degrees&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The total number of track pieces in a loop must be even.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is easy to see. Travelling through any track piece causes a change of $(\pm 1 \pm 1)$ in our current position; in particular, it alters the parity of each coordinate in our position. Because at the beginning and at the end both our coordinates are even, so too must be even the number of parity changes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The total number of track pieces in a loop must be a multiple of four.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since we have shown that the total number of track pieces is even, we can decompose the loop into a series of consecutive pairs of track pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the key observation here is that any &lt;em&gt;possible&lt;/em&gt; (as in valid) combination of two consecutive track pieces causes a change of $(\pm 2, 0)$ or $(0, \pm 2)$ in our coordinates. For the beginning and end to be $(0,0)$, there has to be an even number of such changes. In other words, there must be an even number of pairs of consecutive track pieces, so the total number of track pieces must be a multiple of four.&lt;/p&gt;&#xA;" OwnerUserId="16478" LastActivityDate="2015-12-13T09:11:20.227" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24538" PostTypeId="2" ParentId="24535" CreationDate="2015-12-13T10:39:20.277" Score="2" Body="&lt;p&gt;Let's assume a train travels around the track, and that it starts facing exactly right ($0°$ in a Cartesian coordinate system). After traveling the loop completely the train must be at the starting position again facing right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each track changes the direction of the train by $\pm 90°$. To have the same direction at the end the train must have traveled an even number of tracks. The number number of left and right turns is either the same or it differs by $4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing that the number of tracks is even, we can split the track in double-pieces. There are 4 possible combinations: $LL$, $RR$, $LR$ and $RL$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1fImY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1fImY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's define that a single-piece changes the position of the train by $(\pm 1, \pm 1)$. Then the first and second double-pieces change the position of the train by either $(0, \pm 2)$ or $(\pm 2, 0)$, while at the same time changing its direction by $180°$. The third and fourth double-pieces change the position of the train by $(\pm 2, \pm 2)$, without changing its direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the first and second double-pieces are the only ones changing the direction of the train, the total number of them in the track must be again even. Otherwise the train wouldn't face the same direction at the end. At the same time the train can have only one of $2$ possible directions at the start and end of each double-piece. This means that all of the first and second  double-pieces in the track change either the $X$ or the $Y$ coordinate of the train, but not both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also means that the third and fourth double-pieces are the only ones which change the other coordinate of the train. As they change this coordinate always by the same value, their total number must be even as well. Otherwise the train couldn't arrive at the starting position at the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the total number of double-pieces is even as well, leading to the result that the number of single-pieces must be a multiple of $4$.&lt;/p&gt;&#xA;" OwnerUserId="15632" LastActivityDate="2015-12-13T10:39:20.277" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24539" PostTypeId="2" ParentId="24535" CreationDate="2015-12-13T11:56:53.353" Score="26" Body="&lt;p&gt;Let's think of each piece not as curved but as an L shape, with two sections at a right angle. Adjacent pieces in the track join in a straight line. If we pair off the touching sections, we see the track is composed of two-unit segments connected at right angles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since horizontal and vertical segments alternate, there are an equal number of each. Moreover, in traveling a closed loop, the train must go east for as many segments as it goes west, so the number of horizontal segments is even. Likewise for vertical segments. So, the total number of segments is a multiple of 4. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of track pieces equals the number of segments, and so is divisble by 4.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-12-13T11:56:53.353" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="24540" PostTypeId="1" CreationDate="2015-12-13T13:01:54.680" Score="2" ViewCount="214" Body="&lt;p&gt;The Big and the Small Kingdom are both rectangular islands and divided into rectangular landscape. In each province there is a road that runs along one of the diagonals. On each island exist roads that make a closed route, which does not go through any point several times. The picture shows the Little Kingdom, which has six area:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/x9gEI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/x9gEI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Great Kingdom has an odd number of landscapes. How many landscapes does the Great Empire have at least?&lt;/p&gt;&#xA;" OwnerUserId="17755" LastEditorUserId="11876" LastEditDate="2015-12-13T19:24:12.730" LastActivityDate="2015-12-14T17:26:34.650" Title="Geometry problemsolving" Tags="|geometry|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24542" PostTypeId="2" ParentId="24540" CreationDate="2015-12-13T13:43:29.570" Score="0" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;5&lt;/strong&gt;, in case it's ok to not go through all squares.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the general case&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We note that you can't possible have a path with less then 4 squares. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the case where we don't have to go through all squares&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 5 is the lowest odd number higher then 4, so if we can find a solution, we have proof that this is the optimal solution. Luckily such a solution is easily found &lt;br&gt;&lt;a href=&quot;https://i.stack.imgur.com/Eapop.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Eapop.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;sidenote:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Just because the Big kingdom has less landscapes then the Small kingdom doesn't mean they're smaller. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16846" LastEditorUserId="16846" LastEditDate="2015-12-13T22:28:16.510" LastActivityDate="2015-12-13T22:28:16.510" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24544" PostTypeId="1" CreationDate="2015-12-13T16:16:32.687" Score="7" ViewCount="55916" Body="&lt;p&gt;Find the words ends with &quot; ANT &quot; clues given&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kind of an ant works with figures? An accountANT. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Lives in the jungle?&lt;/li&gt;&#xA;&lt;li&gt;Is far away?&lt;/li&gt;&#xA;&lt;li&gt;Is extraordinarily large?&lt;/li&gt;&#xA;&lt;li&gt;Works for a master?&lt;/li&gt;&#xA;&lt;li&gt;Is good-natured?&lt;/li&gt;&#xA;&lt;li&gt;Is unchanging?&lt;/li&gt;&#xA;&lt;li&gt;Is luxurious?&lt;/li&gt;&#xA;&lt;li&gt;Is one who takes part?&lt;/li&gt;&#xA;&lt;li&gt;Is a very small child?&lt;/li&gt;&#xA;&lt;li&gt;Is sleeping?&lt;/li&gt;&#xA;&lt;li&gt;Is very bright?&lt;/li&gt;&#xA;&lt;li&gt;Is empty?&lt;/li&gt;&#xA;&lt;li&gt;Is immediate?&lt;/li&gt;&#xA;&lt;li&gt;Is plentiful?&lt;/li&gt;&#xA;&lt;li&gt;Has moved to a diferent country?&lt;/li&gt;&#xA;&lt;li&gt;Is meaningful?&lt;/li&gt;&#xA;&lt;li&gt;Is something that grows?&lt;/li&gt;&#xA;&lt;li&gt;Has infuence over others?&lt;/li&gt;&#xA;&lt;li&gt;Is unsure and indecisive?&lt;/li&gt;&#xA;&lt;li&gt;Lives in a certain place?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="17764" LastActivityDate="2015-12-15T19:25:24.027" Title="words ending with ant" Tags="|word|" AnswerCount="5" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24546" PostTypeId="2" ParentId="24544" CreationDate="2015-12-13T16:38:02.140" Score="2" Body="&lt;ol&gt;&#xA;&lt;li&gt;Elephant&#xA;2.distant&#xA;4.Servant&#xA;8.participant&#xA;18.dominant&#xA;14.abundant&#xA;15.Migrant&#xA;&#xA;&lt;ol start=&quot;6&quot;&gt;&#xA;&lt;li&gt;Constant&#xA;7.elegant&#xA;9.infant&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="17765" LastActivityDate="2015-12-13T16:38:02.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24547" PostTypeId="2" ParentId="24535" CreationDate="2015-12-13T16:49:12.417" Score="5" Body="&lt;p&gt;Start at some point in the track and follow it around.  Write a new line for each piece. On each line write the direction you're going at the start and then the direction you're going when you leave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eg for a loop you'd have&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;U R&lt;br&gt;&#xA;  R D&lt;br&gt;&#xA;  D L&lt;br&gt;&#xA;  L U&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now with four observations about this list you have your proof.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;There are an even number of items in the list. This is because odd items start with U/D and end with R/L, and other way around for even. And, since it's a loop, the end of the last must equal the start of the first.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There's an even number of R's and an even number of L's. This is because the list can be broken into pairs, which are either with two R's or two L's.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The total number of L's must be equal to the total number of R's. Because it's a loop it must go R as much as it goes L.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Every item has either a R or an L. So the total number of items is double an even number or a multiple of 4.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note you don't even need the UD argument, so if you had a straight track of some length and joined it into a loop with the curved tracks, they'd still be a multiple of 4 number of curved tracks. Or in fact any number of parallel straight tracks thrown in (as long as they are all parallel- you still just count the curved pieces).&lt;/p&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2015-12-13T22:56:10.307" LastActivityDate="2015-12-13T22:56:10.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24549" PostTypeId="2" ParentId="24512" CreationDate="2015-12-13T17:34:36.590" Score="4" Body="&lt;p&gt;Alice wins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose for contradiction that Bob has a winning strategy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claim: If Alice makes the first move on square X, then on the next turn (Bob's), there is exactly one square Y among the neighbors of X on which Bob must make his (first) move to win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the claim, for each square X, let f(X) denote the Y as in the claim. Then since there are more W squares than B, there must be two W squares $X_1,X_2$ and a B square Y such that $f(X_1)=f(X_2)=Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this gives a strategy for Alice to win by starting at $Y$: if Bob places the knight on a square other than $X_1$, Alice steals Bob's strategy by pretending that Bob already made a move at $X_1$; if Bob places the knight on $X_1$, Alice pretends that the first move was Bob's at $X_2$ and again wins by strategy-stealing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof of the claim:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose for contradiction that Y, Z are two squares both adjacent to X and such that when Alice starts at X, then Bob has a winning strategy by placing his knight at Y or at Z.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let Alice begin a game by placing the knight at Y.&#xA;Now if Bob doesn't place the knight at X, then Alice can pretend to be the second-player where the game began with Bob starting at X; this is a contradiction so Bob must place the knight at X.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in this case Alice can place the knight on Z and by strategy-stealing again, go on to win. This proves the claim.&lt;/p&gt;&#xA;" OwnerUserId="5684" LastEditorUserId="5684" LastEditDate="2015-12-14T17:31:29.063" LastActivityDate="2015-12-14T17:31:29.063" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="24551" PostTypeId="1" AcceptedAnswerId="24572" CreationDate="2015-12-13T18:38:58.100" Score="41" ViewCount="8518" Body="&lt;p&gt;You're a fan of pen &amp;amp; paper RPGs and have finally arranged a new playing group, recruited amongst your friends from the mathematical faculty.&lt;/p&gt;&#xA;&lt;p&gt;As a GameMaster (GM) you've assumed that all your players will be fully equipped with the necessary dice showing 4-sides (d4), 6-sides (d6), 8-sides (d8) 10-sides (d10), 12-sides (d12) and 20-sdies (d20). So your disappointment is great, when you show up for the game and all they have brought with them was a near-infinite amount of ordinary, six-sided dice.&lt;/p&gt;&#xA;&lt;p&gt;While you luckily didn't need to have 'all' dice in your set, the RPG system you want to play with is based on d100 ( usually realized by using two d10 ), and you have to find a way of using (any number) of d6 and any rule you can think of, to realized a value range 0-99 (or 0 to 99).&lt;/p&gt;&#xA;&lt;p&gt;One mathematician suggests:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;We simply roll twenty of our d6, sum the rolls and subtract 20, that gives us the range (20;120)-20 = (0;100), and we consider both the 0 and the 100 to be 100.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This works, but chances for the various values are rather far from being an equal &amp;quot;fair&amp;quot; distribution, so soon the players start a heated discussion on how they can 'best' achieve a fair distribution. They agree on the following metric:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;Ideally, each value has a chance of 0.01, so the best rule is the one which minimizes &lt;span class=&quot;math-container&quot;&gt;$\Sigma(( P_i - 0.01 )^2)$&lt;/span&gt;&amp;quot; - it does not matter how many (finite) rolls of d6 are used.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Under these conditions, what is the &amp;quot;fairest&amp;quot; method of getting to a range 0..99 (or 1..100) just using six-sided dice?&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;Notes:&lt;/h2&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I do not know a &amp;quot;best&amp;quot; solution for this (yet)&lt;/li&gt;&#xA;&lt;li&gt;the 'algorithm' has to produce any value of the D100 range&lt;/li&gt;&#xA;&lt;li&gt;The &amp;quot;fairest&amp;quot; algorithm wins. Only if two algorithms are equally fair, the one with less dice rolls wins.&lt;/li&gt;&#xA;&lt;li&gt;the 'algorithm' can consist of any mathematical rule which you can think of, but it must be reasonably &amp;quot;doable&amp;quot; by a group of mathematicians with nothing but pen &amp;amp; paper (and d6)&lt;/li&gt;&#xA;&lt;li&gt;Any &lt;strong&gt;finite&lt;/strong&gt; number of d6 ( or rolls thereof) may be used in the 'algorithm'&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Note, that the last rule exclude all algorithms which do not guarantee that a result is reached in a finite number of rolls. (In particular, you can not re-roll until a certain value is rolled. You may reroll 1e106 times though.)&lt;/p&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="5840" LastEditDate="2021-12-07T11:45:22.500" LastActivityDate="2021-12-07T11:45:22.500" Title="Aproximating 100 by 6" Tags="|mathematics|probability|dice|" AnswerCount="22" CommentCount="14" ContentLicense="CC BY-SA 4.0" />
  <row Id="24552" PostTypeId="1" AcceptedAnswerId="24560" CreationDate="2015-12-13T18:57:16.583" Score="13" ViewCount="711" Body="&lt;p&gt;When I am stressed,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can say so much more,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm three times within myself,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Twice Window, Once door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="17766" LastActivityDate="2015-12-14T13:16:35.720" Title="When I am stressed" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24553" PostTypeId="2" ParentId="24551" CreationDate="2015-12-13T19:10:14.077" Score="6" Body="&lt;p&gt;Take $n$ dice, and roll them one at a time. Let $a_0=0$, and let $a_k=a_{k-1}+100(r_k-1)/6^k$, where $r_k$ is the result of the $k$th roll. Then round $a_n$ down to an integer and use it as the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This essentially divides the range into $6^n$ pieces and reassembles them into as close to an even 100 pieces as possible, and can be made arbitrarily precise by using increasing numbers of dice. Still working on the exact error, but I think this may be optimal for any number of dice.&lt;/p&gt;&#xA;" OwnerUserId="12511" LastEditorUserId="12511" LastEditDate="2015-12-13T19:27:38.803" LastActivityDate="2015-12-13T19:27:38.803" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24555" PostTypeId="2" ParentId="24551" CreationDate="2015-12-13T19:43:42.977" Score="6" Body="&lt;p&gt;I actually don't know what you mean with the metric but it's fairly simple to do with just 3 rolls that might have to be repeated.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;on 111 it is considered 1&lt;br&gt;&#xA;on 112 it is considered 2&lt;br&gt;&#xA;...&lt;br&gt;&#xA;on 121 it is considered 7&lt;br&gt;&#xA;...&lt;br&gt;&#xA;on 211 it is considered 37&lt;br&gt;&#xA;...&lt;br&gt;&#xA;on 354 it is considered 100&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any value higher reroll the three dice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically it's $(\texttt{first_digit} - 1)\times 36 + (\texttt{second_digit} - 1)\times 6 + \texttt{third_digit}$&lt;/p&gt;&#xA;" OwnerUserId="1748" LastEditorUserId="9718" LastEditDate="2015-12-14T09:02:04.163" LastActivityDate="2015-12-14T09:02:04.163" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24556" PostTypeId="2" ParentId="24551" CreationDate="2015-12-13T19:43:51.070" Score="20" Body="&lt;p&gt;Figured i'd add my comment as an answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We roll 3 dice $(D1,D2,D3)$, which we seperate spatially or in some other way (we used to use 3 different colors). We reroll D1 and D2 if they land on 6, and we reroll D3 if we it lands on a 5 or a 6. We repeat these rerolls until we have 3 numbers, $D1 \&amp;amp; D2 &amp;lt;= 5$, and $D3 &amp;lt;= 4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We then form our dice result by taking $(D1-1)*20+(D2-1)*4+D3$. This gives uniform numbers ranging from 1 $(1,1,1)$ to 100 $(5,5,4)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seeing as we're not allowed to roll an infinite amount of dice, we can stop after an arbitrary large amount of rerolls, and if we get there simply set the value to 51 (or any other). Since this chance goes to 0 the distribution tends to uniform.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;small edit:&lt;/strong&gt; &#xA;you can drastically reduce the amount of rerolls needed by using 4 dice, and setting the value of D3 and D4 equal to 1 if the roll is below 3, and equal to 2 if the roll is 4+ and using the formula $(D1-1)*20+(D2-1)*4+(D3-1)*2+D4$&lt;/p&gt;&#xA;" OwnerUserId="16846" LastEditorUserId="16846" LastEditDate="2015-12-14T13:49:15.100" LastActivityDate="2015-12-14T13:49:15.100" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24557" PostTypeId="1" AcceptedAnswerId="24573" CreationDate="2015-12-13T19:59:37.507" Score="9" ViewCount="698" Body="&lt;p&gt;Alice and Bob are playing an unusual game of chess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We begin with a piece on a square of an infinite chessboard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On each of her turns, Alice moves the piece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On each of his turns, Bob destroys a square of the chessboard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bob wins if he can trap Alice (i.e., create a 'moat' of destroyed squares too big for Alice to cross).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alice wins if she can prove that Bob has no winning strategy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume perfect play.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the pieces: Let's make up pieces called n-kings, and say that an n-king leaps to any square within n squares of it (so can fly over destroyed squares).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly Bob wins from the get-go if Alice starts with a 0-king. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the highest order of king for which Bob has a winning strategy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the lowest order of king for which Alice has a winning strategy (if any)?&lt;/p&gt;&#xA;" OwnerUserId="16995" LastActivityDate="2015-12-14T16:05:29.303" Title="Trapping pieces on an infinite chessboard" Tags="|chess|checkerboard|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="24560" PostTypeId="2" ParentId="24552" CreationDate="2015-12-13T21:22:10.007" Score="15" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Syllable -- it has three syllables, window has two, door has one. Syllables can be stressed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16995" LastEditorUserId="12462" LastEditDate="2015-12-14T13:16:35.720" LastActivityDate="2015-12-14T13:16:35.720" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24561" PostTypeId="2" ParentId="24544" CreationDate="2015-12-13T21:30:27.923" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lives in the jungle? - ELEPHANT, PLANT, ANT&lt;br&gt;&#xA; Is far away? - DISTANT&lt;br&gt;&#xA; Is extraordinarily large? - GIANT&lt;br&gt;&#xA; Works for a master? - SERVANT&lt;br&gt;&#xA; Is good-natured? - PLEASANT&lt;br&gt;&#xA; Is unchanging? - CONSTANT&lt;br&gt;&#xA; Is luxurious? - ELEGANT, EXTRAVAGANT&lt;br&gt;&#xA; Is one who takes part? - PARTICIPANT&lt;br&gt;&#xA; Is a very small child? - INFANT&lt;br&gt;&#xA; Is sleeping? - DORMANT&lt;br&gt;&#xA; Is very bright? - BRILLIANT&lt;br&gt;&#xA; Is empty? - VACANT&lt;br&gt;&#xA; Is immediate? - INSTANT&lt;br&gt;&#xA; Is plentiful? - ABUNDANT&lt;br&gt;&#xA; Has moved to a different country? - IMMIGRANT&lt;br&gt;&#xA; Is meaningful? - SIGNIFICANT&lt;br&gt;&#xA; Is something that grows? - PLANT&lt;br&gt;&#xA; Has influence over others? - DOMINANT&lt;br&gt;&#xA; Is unsure and indecisive? - HESITANT&lt;br&gt;&#xA; Lives in a certain place? - INHABITANT  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16995" LastEditorUserId="17811" LastEditDate="2015-12-15T19:25:24.027" LastActivityDate="2015-12-15T19:25:24.027" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24562" PostTypeId="2" ParentId="24540" CreationDate="2015-12-13T21:40:06.670" Score="4" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  9&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One way to do this is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/VFhJf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VFhJf.png&quot; alt=&quot;solution&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17755" LastEditorUserId="5874" LastEditDate="2015-12-14T17:26:34.650" LastActivityDate="2015-12-14T17:26:34.650" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24563" PostTypeId="2" ParentId="24551" CreationDate="2015-12-13T22:10:43.013" Score="28" Body="&lt;p&gt;Get 10 different d6 dices and describe them on paper. Next to each description, associate a unique number from 0 to 9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put all those dices in an opaque bag (you should have one to transport that near-infinite number of dices).&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Shuffle the bag. Pick one die blindly and retrieve its value on the paper you wrote. Multiply that value by 10.&lt;/li&gt;&#xA;&lt;li&gt;Shuffle the bag. Pick one die blindly and retrieve its value on the paper you wrote.&lt;/li&gt;&#xA;&lt;li&gt;Add the two values.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The fairness is maximal. The number of &lt;em&gt;rolls&lt;/em&gt; (hum) is constant: 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you feel the courage, use not 10 but 100 dice to reduce the &lt;em&gt;rolls&lt;/em&gt; to 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't want a bag? Just arrange the dice randomly behind your back and pick  without peeking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't want a sheet of paper and pen? Memorize the numbers associated.&lt;/p&gt;&#xA;" OwnerUserId="11903" LastEditorUserId="11903" LastEditDate="2015-12-13T23:06:53.547" LastActivityDate="2015-12-13T23:06:53.547" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="24564" PostTypeId="2" ParentId="24551" CreationDate="2015-12-13T22:13:12.597" Score="7" Body="&lt;p&gt;For each digit you roll two 6d&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) If the first dice is 6, it gets re-rolled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) If the second dice is 1, 2 or 3, you take the value of the first dice. If it is 4, 5 or 6, you add 5 to the second dice. Obviously, &quot;10&quot; is converted to &quot;0&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pros:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Exact distribution, each number from 0 to 99 has the same possibilities.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Extensible to as many decimal places as you want.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;No need of complicated lookup tables or calculations to get evidence.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Only 4 rolls for number.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Cons: it may need some reroll (1/6 by digit).&lt;/p&gt;&#xA;" OwnerUserId="17769" LastActivityDate="2015-12-13T22:13:12.597" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24565" PostTypeId="2" ParentId="24551" CreationDate="2015-12-13T22:23:57.513" Score="8" Body="&lt;p&gt;With $n$ rolls, there are $6^n$ possible outcomes. If you use fewer rolls, pretend you made additional rolls you ignored. Any algorithm maps an outcome to one of 100 values. So, an optimal algorithm is to partition the outcomes into 100 sets of as equal size as possible, differing by at most 1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many ways to instantiate this partition with an algorithm. One is to treat the sequence of rolls as a number of base 6, with digits 0 through 5, and take the result modulo 100.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-12-13T22:23:57.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24566" PostTypeId="2" ParentId="24527" CreationDate="2015-12-14T01:19:54.963" Score="8" Body="&lt;p&gt;My answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;RADAR&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sage of earth, seer of the sky,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Radar looks into the earth and into the sky&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Use obscured by carrot lie.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; British used radar during ww2, but allegedly convinced the Germans their spotting skills were due to improved eye sight thanks to eating carrots (credit to @Alconja)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Dutch song my appellation told&#xA;During crisis of black gold.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Netherlands sung &quot;I See a Star&quot; at the Eurovision Song Contest in 1974 (during the oil crisis).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bane of the hasty, friend of fuzz&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Police uses radar to catch speeders.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;AKA sub eyeball cuz.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sub eyeball, is referring to the wavelength being below the visible spectrum. (credit to @Alconja)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the end, I am the same&#xA;As from the start. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Refers to radar being a palindrome&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17771" LastEditorUserId="17771" LastEditDate="2015-12-15T10:58:05.820" LastActivityDate="2015-12-15T10:58:05.820" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="24567" PostTypeId="2" ParentId="23175" CreationDate="2015-12-14T01:37:46.633" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The month of March.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(although I'm not sure how it matches the last line.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not the month of Halloween&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet it's a daylight saving switching month&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which does not contain Thanksgiving&lt;/p&gt;&#xA;&#xA;&lt;p&gt;???&lt;/p&gt;&#xA;" OwnerUserId="17771" LastEditorUserId="17771" LastEditDate="2015-12-14T14:48:22.460" LastActivityDate="2015-12-14T14:48:22.460" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24568" PostTypeId="2" ParentId="24551" CreationDate="2015-12-14T02:49:29.903" Score="16" Body="&lt;p&gt;As @MikeEarnest says, unless you put an upper limit on the number of rolls, you can always do better with more rolls.  Since $6^n$ is never divisible by $100$, you can never be completely fair.  With more rolls you can always get the unfairness lower.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that $6^4=1296$ is very close to a multiple of $100$ means you can get very close to fair with $4$ rolls.  You must have a way to order the rolls, either by die color or by order of rolling one die.  Subtract one from each roll and view it as a base $6$ number from $0000_6=0_{10}$ to $5555_6=1295_{10}$  Now assign $13$ of the rolls to each of $96$ results and $12$ of the rolls to the other $4$  An easy way to do that is to divide the result by $99.7$ and throw away the remainder. There are $4$ choices with probability $\frac {12}{1296}$ and $96$ choices with probability $\frac {13}{1296}$ for error &lt;a href=&quot;http://www.wolframalpha.com/input/?i=96*%280.01-13%2F1296%29%5E2%2B4*%280.01-12%2F1296%29%5E2&quot;&gt;about $2.286E-6$&lt;/a&gt;  More rolls will always do better.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2015-12-14T02:49:29.903" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24569" PostTypeId="1" CreationDate="2015-12-14T03:08:19.340" Score="7" ViewCount="646" Body="&lt;p&gt;Given the quadrilateral ABCD (see drawing) with mid diagonals, E of AC and F of BD, you need to create 5 polygons equal in area to 1/4 of the quadrilateral ABCD area by means of a straight-edge, drawing 9 lines. The 5 polygons - quadrilaterals or triangles - may partially overlap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NHdn6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NHdn6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9531" LastEditorUserId="10399" LastEditDate="2015-12-14T17:49:15.803" LastActivityDate="2016-03-25T02:50:44.637" Title="5 equal area polygons" Tags="|geometry|" AnswerCount="1" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="24570" PostTypeId="2" ParentId="24551" CreationDate="2015-12-14T07:59:43.630" Score="8" Body="&lt;h2&gt;Method: Accumulation Register&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Each player needs four distinguishable dice and a piece of paper for logging.  Identify the dice as &quot;A&quot;, &quot;B&quot;, &quot;C&quot;, and &quot;D&quot;.  The log is initialized to &quot;0&quot;.  Each time a new percentile is needed, roll the four dice.  Add the value $216(A-1) + 36 (B-1) + 6 (C-1) + D - 1$ to the currently logged value and reduce modulo $100$.  (The phrase &quot;reduce module 100&quot; is equivalent to &quot;keep only the last two digits&quot; if everyone is working in base 10.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming fair dice, the rolling uniformly generates a value in $[0,1295]$.  In a particular roll, $96$ outputs can be generated $13$ ways and $4$ outputs can only be generated $12$ ways.  The fitness function $\sum (p_i-0.01)^2$ is then $96 \left( \frac{13}{1296} - 0.01 \right)^2 + 4 \left(\frac{12}{1296} - 0.01 \right)^2 \approx 2.3 \times 10^{-6}$.  However, this is only a short-term effect.  Over the long run (many times 100/4 = 25 rolls), this effect is evenly distributed over all possible outcomes.  I.e., the fitness function decreases to zero as the number of rolls increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Variants: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Using this method with three dice gives an increment in [0,215] and a fitness of $16 \left( \frac{3}{216} - 0.01 \right)^2 + 84 \left(\frac{2}{216} - 0.01 \right)^2 \approx 2.9 \times 10^{-4}$.  Again over the long run, this decreases to zero.&lt;/li&gt;&#xA;&lt;li&gt;Using this method with nine dice gives an increment in [0,10077695] and a one-roll fitness of $96 \left( \frac{100777}{10077696} - 0.01 \right)^2 + 4 \left(\frac{100776}{10077696} - 0.01 \right)^2 \approx 3.8 \times 10^{-14}$.&lt;/li&gt;&#xA;&lt;li&gt;For $n$ dice, the one-roll fitness is roughly $10^{-1.5 n}$ and decreases to zero as many rolls are accumulated.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Several other answers here use the $36(A-1)+6(B-1)+(C-1)$ method, but do not arrange for the under-represented values to be (eventually) evenly distributed over all of $[0,99]$.&lt;/p&gt;&#xA;" OwnerUserId="17776" LastActivityDate="2015-12-14T07:59:43.630" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24571" PostTypeId="2" ParentId="24551" CreationDate="2015-12-14T08:31:11.917" Score="4" Body="&lt;p&gt;I know there are already better ideas out there, but I immediately thought of this one when I saw the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can use binary counting: an even number on a die is a 1 and an uneven number on a die is a 0. Roll 7 dice (or one die 7 times) (2^7 = 128) and note the resulting binary code on a piece of paper. We now have a uniformly distributed number in the [0-128] range. We can scale this number to be in the [0~100] range by using &lt;code&gt;Floor((X/128)*100)&lt;/code&gt;. The floor will affect the distribution (I'm not exactly sure how much). But we can get arbitrarily close to perfect by using more dice (for example with 2^10 we get a number in [0~1024] which we can scale down and floor with less bias.&lt;/p&gt;&#xA;" OwnerUserId="17777" LastEditorUserId="2315" LastEditDate="2015-12-15T23:07:31.317" LastActivityDate="2015-12-15T23:07:31.317" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24572" PostTypeId="2" ParentId="24551" CreationDate="2015-12-14T08:44:37.363" Score="44" Body="&lt;p&gt;As &lt;a href=&quot;https://puzzling.stackexchange.com/a/24565/9718&quot;&gt;xnor points out in his answer&lt;/a&gt;, this question is basically asking for the way to most evenly distribute $6^n$ results among $100$ bins, and gives a very brief description of the solution.  I'll go into a &lt;em&gt;bit&lt;/em&gt; more detail here.  If you're not interested in the proofs, skip to section 2.3&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Summary&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In order to construct a &quot;best&quot; algorithm for converting rolls of one die into another, there are two types of optimality to consider:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Minimum &quot;score.&quot;  BmyGuest's metric:&#xA;$$\sum_i ( p_i - 1/100 )^2$$&#xA;is basically the &lt;a href=&quot;https://en.wikipedia.org/wiki/Pearson%27s_chi-squared_test&quot; rel=&quot;nofollow noreferrer&quot;&gt;chi-squared statistic&lt;/a&gt; for goodness-of-fit, between the distribution resulting from the algorithm and a uniform distribution.  I'll call the metric applied to a particular algorithm that algorithm's &lt;em&gt;score.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Minimum number of rolls.  I'll consider only the &lt;em&gt;expected&lt;/em&gt; number of rolls for an algorithm.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note I'll be diverging a from typical RPG terminology: when I refer to $\mathsf{3d6}$, for example, I mean that you should roll three six-sided dice &lt;em&gt;without adding them, keeping track of the order the dice were rolled in&lt;/em&gt; (usually the dice are treated as indistinguishable from each other, i.e. rolled all at once).  This is important because, as you noted, the results in a traditional roll are not equiprobable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also consider a $\mathsf{d}n$ to be numbered from $0$ to $n-1$, to make some of the math a little easier.  If you're following my algorithms with real dice, just map $n\to 0$ for the input and vice-versa for the output.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;1 Optimal Scoring Algorithms&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;In the general case, we want to map $m$ &quot;results&quot; into $n$ &quot;bins.&quot;  (In the original case, we have $m=6^k$, $n=100$.)  Consider an algorithm that maps $r_i$ results to bin $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability of landing in bin $i$ is $r_i/m$, so the score is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sum_{i=0}^{n-1}\left(\frac{r_i}{m} - \frac{1}{n}\right)^2&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;1.1 Distribution of an Optimal Algorithm&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Since the score is a sum of independent terms we can easily isolate the effect of changing a pair of elements of the solution, $(r_i,r_j)$.  Let's decrease $r_i$ by $1$ and increase $r_j$ by $1$, and see under what conditions the score decreases:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;\left(\frac{r_i}{m} - \frac{1}{n}\right)^2 + \left(\frac{r_j}{m} - \frac{1}{n}\right)^2 &amp;amp;&amp;gt; \left(\frac{r_i-1}{m} - \frac{1}{n}\right)^2 + \left(\frac{r_j+1}{m} - \frac{1}{n}\right)^2 \\&#xA;\frac{r_i^2 + r_j^2}{m^2} - 2\frac{r_i + r_j}{m n} + \frac{2}{n^2} &amp;amp;&amp;gt; \frac{r_i^2 + r_j^2 - 2(r_i - r_j) + 2}{m^2} - 2\frac{r_i + r_j}{m n} + \frac{2}{n^2} \\&#xA;0 &amp;amp;&amp;gt; \frac{2}{m^2}(1 - r_i + r_j) \\&#xA;r_i &amp;amp;&amp;gt; 1 + r_j&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that if the solution contains two values which differ from each other by more than one, it cannot be optimal (since moving the two values closer together will decrease the score).  Therefore, in an optimal solution $r_i$ will take on at most two values, separated from each other by $1$.  Let's call them $r$ and $r+1$.  Calling the number of bins with $r$ results $x$, and the number of bins with $r+1$ results $x'$, we can write a system of equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;n = x + x' \\&#xA;m = rx + (r+1)x'&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can solve for $x$ and $x'$ in terms of $r$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x' = n - x \\&#xA;m = rx + (r+1)(n-x) \\&#xA;m = rx + rn - rx + n - x \\&#xA;m = (r+1)n - x \\&#xA;x = (r+1)n-m&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that $0\leq x \leq n$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;0 \leq (r+1)n-m \leq n \\&#xA;0 \leq (r+1) - \frac{m}{n} \leq 1 \\&#xA;\frac{m}{n} \leq r+1 \leq \frac{m}{n} + 1 \\&#xA;\frac{m}{n} - 1 \leq r \leq \frac{m}{n}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $r$ must be an integer, we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;r = \left\lfloor \frac{m}{n} \right\rfloor&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting back in, we see that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;r + 1 &amp;amp;= \left\lfloor \frac{m}{n} \right\rfloor + 1 \\&#xA;x &amp;amp;= \left(\left\lfloor \frac{m}{n} \right\rfloor +1 \right)n - m \\&#xA;x' &amp;amp;= m - \left\lfloor \frac{m}{n} \right\rfloor n&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;1.2 Score of an Optimal Algorithm&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The score of the optimal algorithm described above gives us a way to compute the minimum possible score for a given $m$ and $n$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sum_{i=0}^{n-1}\left(\frac{r_i}{m} - \frac{1}{n}\right)^2 = x\left(\frac{r}{m}-\frac{1}{n}\right)^2+x'\left(\frac{r+1}{m}-\frac{1}{n}\right)^2 \\&#xA;= \left[\left(\left\lfloor \frac{m}{n} \right\rfloor +1 \right)n - m\right]\left(\frac{\lfloor m/n\rfloor}{m}-\frac{1}{n}\right)^2+\left[m - \left\lfloor \frac{m}{n} \right\rfloor n\right]\left(\frac{\lfloor m/n\rfloor + 1}{m}-\frac{1}{n}\right)^2&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To simplify this expression, we can make use of the identity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left\lfloor \frac{m}{n} \right\rfloor = \frac{m}{n} - \frac{m\ \mathrm{mod}\ n}{n}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prevent the equations from getting too long, I'll use $m' = m\ \mathrm{mod}\ n$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left[\left(\frac{m-m'}{n} +1 \right)n - m\right]\left(\frac{m-m'}{mn}-\frac{1}{n}\right)^2+\left[m - \frac{m-m'}{n} n\right]\left(\frac{m-m'+n}{mn}-\frac{1}{n}\right)^2 \\&#xA;= (m-m' + n - m)\left(\frac{m-m' - m}{mn}\right)^2+(m - m+m')\left(\frac{m-m'+n-m}{mn}\right)^2 \\&#xA;= \frac{(n-m')(-m')^2+(m')(n-m')^2}{(mn)^2} \\&#xA;= \frac{(n-m')(m')}{(mn)^2}(m' + n-m') \\&#xA;= \frac{1}{n}\left(\frac{m'}{m}\right)\left(\frac{n-m'}{m}\right)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we can see that the score decreases as $m$ increases, and has local minima when $m'$ or $n-m'$ are close to zero (i.e. $m$ close to divisible by $n$).  This agrees with our intuition.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;1.3 Optimal Scores for Mapping a $k\mathsf{d6}$ to a $\mathsf{d100}$&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Using the expression above, Here's a table of the best possible scores for each $k \le 10$:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; k       m  (m mod n)  score&#xA; ---------------------------&#xA; 0           1   1  0.990&#xA; 1           6   6  0.157&#xA; 2          36  36  0.0178&#xA; 3         216  16  2.88e-4&#xA; 4       1,296  96  2.29e-6&#xA; 5       7,776  76  3.02e-7&#xA; 6      46,656  56  1.13e-8&#xA; 7     279,936  36  2.94e-10&#xA; 8   1,679,616  16  4.76e-12&#xA; 9  10,077,696  96  3.78e-14&#xA;10  60,466,176  76  4.99e-15&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;An unstated assumption was that the assignment of results to bins was the same on each roll.  Some algorithms achieve a better average score by allowing the results of one roll to influence the next.  However, the score of each individual roll is limited by the same maximum.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;2 Optimal Expected Number of Rolls&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;An algorithm can have an expected number of rolls less than the maximum number of rolls, while maintaining the same score.  In fact, an algorithm can have a finite expected number of rolls even with an infinite maximum number.  This is possible when the algorithm allows &quot;early termination.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An algorithm maps each input roll (previously &quot;result&quot;) to a corresponding output (previously &quot;bin&quot;).  If an algorithm maps all rolls beginning with a certain prefix sequence to the same output, once that prefix has been rolled we know the output will be the same regardless of the subsequent rolls and we can stop rolling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pay careful attention to the terminology in the following sections: a roll is the roll of a single die, and a sequence or input is a list of rolls input to an algorithm&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;2.1 Bounding the Expected Number of Rolls&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In order to compute the minimum expected number of rolls, I make the assumption that each roll is made with the same type of die.  That is, if the first roll is a $\mathsf{d6}$, all the subsequent rolls will also be $\mathsf{d6}$s.  Essentially, instead of rolling a large number of different dice, we roll one die over and over.  (Note this restriction means that we cannot analyze algorithms that use another source of randomness such as the orientation of the die.)  Assume we are using a $\mathsf{d}b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expected number of rolls is the sum, over all inputs, of the probability of each input times its length.  If the length of sequence $i$ is $k_i$, then the expected number of rolls $E$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;E = \sum_i k_i b^{-k_i}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider replacing a roll $\{x_1, x_2, x_3, \ldots, x_k\}$ with $b$ rolls all mapping to the same output as the original roll:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\{x_1, x_2, x_3, \ldots, x_k, 0\} \\&#xA;\{x_1, x_2, x_3, \ldots, x_k, 1\} \\&#xA;\vdots \\&#xA;\{x_1, x_2, x_3, \ldots, x_k, b-1 \}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The algorithm is still valid, since all possible rolls are mapped to outputs, and the probability of each output is the same.  The change in $E$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;b\left[(k+1)b^{-(k+1)}\right]- k b^{-k} \\&#xA;= (k+1) b^{-k} - k b^{-k} \\&#xA;= b^{-k}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, splitting an input increases $E$.  Conversely, merging the set of inputs with a given prefix will decrease $E$.  Therefore, choosing how many inputs of each length to assign to a given output can be solved by a greedy algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine that the output space of the algorithm is represented by an interval of length $1$, and that each output is a sub-interval of length $p_i$ (the probability of that output).  Each input is an interval of length $b^{-k}$, and the problem is to cover each output interval with a number of input intervals, with no overlap.  The greedy algorithm does so optimally by repeatedly inserting the largest input (smallest $k$) that fits in the remaining space in the output.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Example: find the optimal distribution of inputs for $b=6$, $k=4$, $n=100$.  We have:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$ m = 1296 \\ \left\lfloor \frac{m}{n} \right\rfloor = 12 \\ m' = 96 $$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Remember that $p_i=r_i/m$, and $r_i=\lfloor m/n \rfloor + 0\text~{or}~1$, so:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$ p = \frac{12}{1296}~\text{or}~\frac{13}{1296} $$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$6^{-1}=1/6$ and $6^{-2}=1/36$ are both too big, but $6^{-3}=1/216$ is small enough to fit in both $p_i$:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$ \begin{align} p &amp;amp;= 2\times 6^{-3} + \frac{0}{1296} \\&#xA;  &amp;amp;~\text{or}~2\times 6^{-3} + \frac{1}{1296} \end{align} $$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Finally, we fill up the remainder with multiples of $6^{-4}$:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$ \begin{align} p &amp;amp;= 2\times 6^{-3} \\ &amp;amp;~\text{or}~2\times 6^{-3} +&#xA;  1\times 6^{-4} \end{align} $$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can see that this process is just the expression of $p$ as a decimal in base $b$!  With this in mind, I'll skip an analysis of the minimum $E$ for optimally-scoring algorithms of finite $k$, and jump straight to algorithms where $k$ is infinite.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;2.2 Optimal Algorithm with Exact Distribution&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;With an infinite maximum $k$, the value of $m$ is also infinite, and the score of the algorithm drops to zero.  This means that we can achieve an exactly uniform output distribution.  This allows us to forget about $m$, $m'$, $x$, and other values that complicated the analysis.  $p=1/n$ for all the outputs.  We can find the decimal expansion of $p$ by using long division to divide $n$ into $1$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Example: $b=6$, $n=100=244_6$.  The long division is done in base-6.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;      .0020543...&#xA;    __________&#xA;244 ) 1&#xA;      .532&#xA;      ----&#xA;      .024&#xA;      .02152&#xA;      ------&#xA;      .00204&#xA;      .001504&#xA;      -------&#xA;      .000132&#xA;      .0001220&#xA;      --------&#xA;      .0000100&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;  &#xA;  &lt;p&gt;The final row is a power of the base, signifying that the decimal expansion repeats.  Note that if we round the expansion to four digits, we get the same result as in the example for $k=4$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;With the expansion in hand, we can write an expression for $E$:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$&#xA;  E = n\left[\sum_k d_k \times k \times b^{-k} \right] + (e+ k_r)\times b^{-k_r}  \\&#xA;  E = 100\left[2\times 3\times 6^{-3} + 5\times 5\times 6^{-5} + 4\times 6\times 6^{-6} + 3\times 7\times 6^{-7} \right] + (E+5)\times 6^{-5} \\&#xA;  E = 14738/4665 \approx 3.159\ldots&#xA;  $$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The factor of $n$ appears because there are $n$ identical outputs contributing to $E$.  $d_k$ is the $k$-th digit in the base-$b$ expansion of $1/n$, and $k_r$ is the number of repeating digits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't know of a way to simplify the formula for $E$, since the decimal expansion can vary wildly with $b$ and $n$.  For example, although $1/100=0.00\overline{20543}_6$ is a fairly short expansion, changing $n$ by $1$ results in $1/101=0.\overline{0020455351}_6$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;2.3 A Implementation of an Optimal Algorithm&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Here I will describe one class of optimal algorithm, based on the interpretation of the input as a base-$b$ decimal such that the $k$-th roll is the $k$-th digit after the decimal point.  For example, the input:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\{0, 4, 2, 2, 3\}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(with $b=6$) is equivalent to the decimal:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;.04223_6&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that inputs are uniformly distributed over the interval $[0,1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unlike the previous section where we grouped the inputs together based on their output, here I'll group the inputs based on size.  The total amount of the interval occupied by inputs of length $k$ is $n d_k b^{-k}$.  Note that this quantity is equivalent to the amount that we subtracted from the remainder at step $k$ of the long division, for example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;      .0020543...&#xA;    __________&#xA;244 ) 1&#xA;      .532 &amp;lt;= 100 * 2 * 6^-3&#xA;      ----&#xA;      .024&#xA;      .02152 &amp;lt;= 100 * 5 * 6^-5&#xA;      ------&#xA;      .00204&#xA;      .001504 &amp;lt;= 100 * 4 * 6^-6&#xA;      -------&#xA;      .000132&#xA;      .0001220 &amp;lt;= 100 * 3 * 6^-7&#xA;      --------&#xA;      .0000100&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Thus, the remainder at each stage is simply the amount of the interval remaining after all the inputs of length $k$ or less have been removed.  This leads us to the algorithm, which I'll demonstrate for $b=6$, $n=100$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Setup: put the remainders from dividing $n$ into $1$ in base $b$ into&#xA;  a table, along with the number of new digits for each step.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;             roll 3&#xA;.024     =&amp;gt;   24   &amp;lt;--+&#xA;             roll 2   |&#xA;.00204   =&amp;gt;  204      |&#xA;             roll 1   |&#xA;.000132  =&amp;gt;  132      |&#xA;             roll 1   |&#xA;.0000100 =&amp;gt;  100      |&#xA;             roll 1 --+&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;  &#xA;  &lt;p&gt;For every output you need, start at the top of the table.  Alternate&#xA;  between two steps:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Roll the number of dice listed, and append their digits to the current value (which starts at zero).&lt;/li&gt;&#xA;  &lt;li&gt;If the current value is greater than the number on the current row, return the difference of the two $\mathrm{mod}\ n$; otherwise continue&#xA;  to the next row.&lt;/li&gt;&#xA;  &lt;li&gt;If you reach the end of the table, go back to the beginning, skipping the first row.  (Note that if the decimal expansion terminates, the last remainder will be $0$ and you will never reach the end of the table.)&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Start with a current value of $0$.&lt;/li&gt;&#xA;  &lt;li&gt;The first row is &lt;code&gt;roll 3&lt;/code&gt;, so roll $\mathsf{3d6}$ and append their digits: $$ \{6, 1, 5\} \to 015_6 $$&lt;/li&gt;&#xA;  &lt;li&gt;The next row is &lt;code&gt;24&lt;/code&gt;.  Since $15_6 &amp;lt; 24_6$, continue to the next row.&lt;/li&gt;&#xA;  &lt;li&gt;The next row is &lt;code&gt;roll 2&lt;/code&gt;, so roll $\mathsf{2d6}$:  $$ \{4, 6\} \to 1540_6 $$&lt;/li&gt;&#xA;  &lt;li&gt;The next row is &lt;code&gt;204&lt;/code&gt;.  Since $1540_6 \geq 204_6$, return the difference between the two, $\text{mod}\ 100$: $$ 1540_6 - 204_6 = 1332_6 = 344 \to \boxed{44} $$&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Side note: The probability that you have to repeat the table at least once is equal to the last remainder: $.00001_6=1/7776\approx 0.013\%$.  Assume that the DM and your entire party of five (six people total) make one roll per minute, every minute, for the next 10 years.  That's a total of around $3\cdot 10^{7}$ rolls.  Therefore the number of times you'll loop through the table will be around $\log_{7776}3\cdot 10^{7} \approx 1.9$!&lt;/p&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-12-17T04:49:49.577" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24573" PostTypeId="2" ParentId="24557" CreationDate="2015-12-14T09:05:38.563" Score="7" Body="&lt;p&gt;This is equivalent to &lt;a href=&quot;https://en.wikipedia.org/wiki/Angel_problem&quot;&gt;Conway's angel problem&lt;/a&gt;. It has been shown that Bob can win only if Alice has a 1- (or 0-) king, and Alice has a winning strategy otherwise. I don't know that there is any simple way to show this (it was an unsolved problem for many years). One strategy that Bob can use is described &lt;a href=&quot;http://www.link.cs.cmu.edu/15859-s11/notes/Angel.pdf&quot;&gt;here&lt;/a&gt;. In the instance of the game they use, Bob plays the role of the angel while Alice is the devil.&lt;/p&gt;&#xA;" OwnerUserId="6022" LastActivityDate="2015-12-14T09:05:38.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24574" PostTypeId="2" ParentId="24551" CreationDate="2015-12-14T09:51:23.313" Score="4" Body="&lt;p&gt;While one can create a $n$-digit senary number by simply throwing $n$ dice, one can also create binary and ternary numbers by grouping numbers together. Now one can look for a number $x=2^a3^b$ that is close to a power of ten to minimize &quot;ramainder&quot; reults that have to be distributed over all numbers. This requires to find a minumum of $\mod(log_{10}(2^a3^b),1)$. For up to fifty dice for instance this is found for $a=1$, $b=35$. The maximum number will be $M=10^{17}+63090197999408$. The $10^{17}$ can be distributed evenly over 100 numbers, while the $63090197999408$ can not: 92 numbers will receive a number range of $L=630901979994$ while the remaining 8 will be chosen for a range of $H=630901979995$ numbers. This yields the following probabilies:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the 92 numbers: $(10^{15}+L)/T=0.009999999999999998$&lt;/li&gt;&#xA;&lt;li&gt;the 8 numbers: $(10^{15}+H)/T=0.010000000000000009$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So the 8 &quot;high probability&quot; numbers will be chosen $1.11\cdot10^{-13}\%$ more likely than the &quot;low probability ones&quot;, which hopefully is neglible over the course of your campaign.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few other optiumum versions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;up to 10 dice: $a=9$, $b=9$, (9D6)&lt;/li&gt;&#xA;&lt;li&gt;up to 30 dice: $a=21$, $b=14$, (14D6, 7D2)&lt;/li&gt;&#xA;&lt;li&gt;up to 100 dice: $a=18$, $b=62$, (18D6, 44D3)&lt;/li&gt;&#xA;&lt;li&gt;up to 500 dice: $a=228$, $b=290$, (228D6, 62D3)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="17778" LastActivityDate="2015-12-14T09:51:23.313" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24575" PostTypeId="2" ParentId="24551" CreationDate="2015-12-14T11:13:04.767" Score="5" Body="&lt;h3&gt;A possibility focused on a low (and finite) number of rolls and practicality&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Roll 3 dice as 3 digits of a base-6 number. You get 216 different results (which I will just arbitrarily call result number 1-216). If the result is in the range 1-200 just take (mod 100). If you roll a number in 201-216 range you have one of 16 results, which you map to numbers 0-99 depending on how many rolls you already had this evening. (so the first player maps to 0-15, the second player to 16-31, next player 32-47,... wrapping around after 99)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If calculating the error for a single throw, it is quite high (2/216 vs 3/216) but spread over the whole evening or multiple game sessions the error evens out, as each possible outcome has the same probability.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;For practicality:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;You can write a mapping table for the throws 1-200 and use a marker on a second table for 201-216 which you move up one space after each throw above 200) I would also spread out the results of the &gt;200 rolls. So the first player maps 201-216 to ( 0, 6, ...., 84, 90 ) the next player to (1,7...,85,91) so for a roll-playing game where the result of a roll is usually used like &lt;code&gt;You have to roll higher than 68 to pass the test&lt;/code&gt; no player gets a perceived unfair advantage.&lt;/p&gt;&#xA;" OwnerUserId="1998" LastActivityDate="2015-12-14T11:13:04.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24576" PostTypeId="2" ParentId="24551" CreationDate="2015-12-14T14:26:55.380" Score="3" Body="&lt;p&gt;&lt;strong&gt;The most simple approach&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A D6 has an equal distribution. Take away your cheating dice! Statistically perfect dice will do the job.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Take a D6 and reroll if it shows a six. Otherwise the D6 indicates one of the five parts.&lt;/li&gt;&#xA;&lt;li&gt;The last 20: Make 5 Segments of 4 pieces. Reroll if you have a 6.&lt;/li&gt;&#xA;&lt;li&gt;The last 4: Reroll if you have a 5 or 6.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Praxis: Three different colored D6. At two black out the number 6. At one black out numbers 5 and 6 to remember rerolling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: The &quot;big&quot; D6 a 2. The The &quot;last 20&quot; shows a 3. The &quot;last 4&quot; shows a 3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;33 =  20 + 10 + 3 = 20 * (2-1) + 5 * (3-1) + 1 * 3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Done by using &lt;strong&gt;three D6&lt;/strong&gt;. Now statistically correct thanks to @DrunkWolf. The system uses the basis 20, 5 and 1 as digits.&lt;/p&gt;&#xA;" OwnerUserId="17784" LastEditorUserId="17784" LastEditDate="2015-12-14T15:25:58.193" LastActivityDate="2015-12-14T15:25:58.193" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24577" PostTypeId="2" ParentId="16184" CreationDate="2015-12-14T14:30:13.270" Score="0" Body="&lt;p&gt;I disagree that the inversion has the answer.  There are an odd number of pieces, and if the answer were an inversion, all of the pieces would be grouped in pairs.  9 boxes do add up to a larger box, but I don't see a pattern emerging by arranging the boxes into a larger box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It could be that the inverted piece is moved, changed in scale, or rotated through each iteration.  That would explain why on the left there appears to be a simple inversion: the piece is a square, and so rotating the inverted portion would maintain the same appearance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My intuition tells me the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Option 3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17783" LastActivityDate="2015-12-14T14:30:13.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24579" PostTypeId="2" ParentId="24557" CreationDate="2015-12-14T16:05:29.303" Score="5" Body="&lt;p&gt;Here is a rough sketch of how Bob wins against a normal chess king. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose a large number, $r$, and consider the ring of squares which are all $r$ king moves away from Alice's starting square. Bob will initially only eat squares from that ring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Alice moves outward, she will approach a particular location on this ring. Bob's strategy is to destroy every other square in an interval surrounding the place Alice approaches. Once Alice reaches the perimeter, Bob then fills in these gaps right before Alice escapes through them. This will only occupy half of Bob's time, while the other half will be spent extending the rest of the ring. Once Bob has eaten the entire ring, he is then guaranteed to win.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-12-14T16:05:29.303" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24580" PostTypeId="1" AcceptedAnswerId="24585" CreationDate="2015-12-14T17:25:15.363" Score="-1" ViewCount="2801" Body="&#xA;&#xA;&lt;p&gt;There are 5 girls in a moving bus, and each girl has a basket in her hand. Each basket contains 5 cats and each cat has 4 kittens. How many legs are there on the bus?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;each girl has 2 legs&lt;/li&gt;&#xA;&lt;li&gt;each cat has 4 legs&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="17438" LastEditorUserId="5373" LastEditDate="2017-12-17T16:11:42.827" LastActivityDate="2017-12-17T16:11:42.827" Title="How many legs are on the bus?" Tags="|mathematics|" AnswerCount="4" CommentCount="1" FavoriteCount="0" ClosedDate="2015-12-14T20:08:30.943" ContentLicense="CC BY-SA 3.0" />
  <row Id="24581" PostTypeId="2" ParentId="24580" CreationDate="2015-12-14T17:28:36.060" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 512 legs: total of 125 cats: 500 legs, 5 girls: 10 legs, plus the 2 legs of the driver. Thanks to GentlePurpleRain for pointing out my misreading of the question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16874" LastEditorUserId="16874" LastEditDate="2015-12-14T17:36:13.257" LastActivityDate="2015-12-14T17:36:13.257" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24582" PostTypeId="2" ParentId="24580" CreationDate="2015-12-14T18:22:04.160" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Zero: buses don't have legs, they have wheels. Typically 6 of them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13101" LastActivityDate="2015-12-14T18:22:04.160" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24583" PostTypeId="2" ParentId="24580" CreationDate="2015-12-14T18:35:02.553" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; there must 110 legs in total in the bus - each cat has four kittens but they are not in the baskets&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17499" LastActivityDate="2015-12-14T18:35:02.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24584" PostTypeId="2" ParentId="24551" CreationDate="2015-12-14T18:45:22.210" Score="4" Body="&lt;p&gt;Roll 3 dice, a red one, blue one, and green one, label the values as $r$, $b$, and $g$.  Calculate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ N = 36 (r-1) + 6 (b-1) + g = 36r + 6b + g - 42$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $N &amp;gt; 108$, then subtract 108.  (Effectively, you have relabeled the 4,5,6 on the red die as 1,2,3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here, you have two options for how to handle $N&amp;gt;100$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1)&lt;/strong&gt; Subtract 100.  This gives an $8\%$ chance of unfairly giving a low roll, but is otherwise perfectly fair.  Always uses $3$ rolls, the absolute minimum for determining 100 distinct values.  Plus, your players will love you (assuming low rolls are good) for all the extra critical successes they are getting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2)&lt;/strong&gt; Reroll the dice when $N &amp;gt; 100$.  This gives a perfectly fair distribution, usually completes in $3$ rolls, and has expected number of rolls required of $3 \div 0.92 \approx 3.26$ rolls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This strategy is also very simple.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2015-12-14T18:45:22.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24585" PostTypeId="2" ParentId="24580" CreationDate="2015-12-14T18:50:43.257" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Each basket contains 5 cats and each cat has 4 kittens.&quot;&lt;br&gt;&#xA;&lt;br&gt;&#xA;  As we know, all kittens are cats. To resolve this statement without creating an infinitely recursive quantity of kittens, we assume that each basket contains 5 cats all of which are kittens, and each of those cats &quot;has&quot; and is had by the other four kittens.&lt;br&gt;&#xA;&lt;br&gt;&#xA;  Mutual having might seem strange, but we can imagine a situation in which Alice would say that she has Bob, and Bob would say that he has Alice, demonstrating that it is grammatically possible.&lt;br&gt;&#xA;&lt;br&gt;&#xA; So each basket has 20 legs, plus two for each girl, bringing us to 110 legs, or 112 legs if the driver has legs and isn't one of our basket-having girls.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7459" LastActivityDate="2015-12-14T18:50:43.257" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24586" PostTypeId="1" AcceptedAnswerId="24587" CreationDate="2015-12-14T18:55:55.400" Score="12" ViewCount="1244" Body="&lt;p&gt;A clown approaches you, holding 99 red balloons. He proposes a game where the two of you will take turns popping some nonzero number of balloons. The only restrictions are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;On the first turn, the current player can't pop all the balloons.&lt;/li&gt;&#xA;&lt;li&gt;On later turns, a player can't pop more balloons than their opponent just popped. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Whoever pops the last balloon wins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The clown gives you the choice of going first or second. Which do you choose, and how do you win?&lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-12-16T18:33:20.680" Title="Pop the Last Balloon" Tags="|mathematics|game|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24587" PostTypeId="2" ParentId="24586" CreationDate="2015-12-14T18:59:56.257" Score="17" Body="&lt;p&gt;A strategy is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Go first.&lt;br&gt;&#xA; Pop one balloon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There must be a non-zero number of balloons popped and neither player can pop more than one balloon per turn $\implies$ exactly one balloon is popped each turn, and the odd-numbered player (first player) wins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2015-12-14T18:59:56.257" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24588" PostTypeId="2" ParentId="24586" CreationDate="2015-12-14T19:12:39.630" Score="2" Body="&lt;p&gt;Spoiler&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You go 1st and pop one balloon. You will now always be popping odd balloons, as your partner can only pop 1 (they can't pop more than you) they will always be popping even balloons. You win as you want to pop the last (odd number) balloon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17789" LastEditorUserId="310" LastEditDate="2015-12-15T00:50:46.493" LastActivityDate="2015-12-15T00:50:46.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24589" PostTypeId="2" ParentId="24551" CreationDate="2015-12-14T19:55:41.443" Score="8" Body="&lt;p&gt;&lt;strong&gt;Two dice with fair distribution&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two dices are sufficent. Take the orientation of one dice additionally. One of the four sides points towards the player. An exactly angle of 45° is unlikely. You mentioned, you have pen&amp;amp;paper?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Toss the red dice on a checkered paper. Not as complicated as measuring exact angle of dice in relation to the other ice or paper. We don't want a solution with a single dice, do we? Not practicable to draw exact angles smaller than 10°.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Toss the red dice on a checkered paper. You can distinguish the nearly 45° orientation by this way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dice sides are numbered from highest to lowest side 1-4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example&lt;/strong&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ZSoHw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZSoHw.png&quot; alt=&quot;D100 with two dice&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Red side shows a 2. 2nd highest number: 26-50&lt;/li&gt;&#xA;&lt;li&gt;Red shows a 4: Segment 41 - 45 (range +3*5)&lt;/li&gt;&#xA;&lt;li&gt;Blue shows a 3: 43 is the result&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Algorithm (base 4, 5, 5)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Toss a red and a blue dice.&lt;/li&gt;&#xA;&lt;li&gt;Reroll dice showing a 6.&lt;/li&gt;&#xA;&lt;li&gt;Orientation of the red dice determines the quarter (1-25, ... 76 - 100). &lt;em&gt;Quick to use!&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;red dice &lt;strong&gt;r&lt;/strong&gt; gives the number in a penta system. 5 * (&lt;strong&gt;r&lt;/strong&gt;-1)&lt;/li&gt;&#xA;&lt;li&gt;blue dice just gives the smallest influence. This barely has to be used in a roleplaying session.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Improved Algorithm (base 5, 4, 5)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Less cheating can be done if base 5 is taken first. The top of the dice can be seen by everyone and should have largest influence. &#xA;Both have no approximation errors and feature few rerolls. Removing the rerolls of a 6 is easy, but less gameification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B.t.w. &lt;em&gt;Zero&lt;/em&gt; dice is not even a puzzle. You have paper and mathematicans: Draw a circle of with 36° segments on the paper. Label the ten segments from 0x to 9x. Spin your pen in the center. A second spin determines the 1 to 10. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Algorithm without reroll (not elegant)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 11 (6+5) possiblities for at least one six. We have to generate more information withouth using information dependant on the &quot;red front side&quot;. 64 combinations by the blue front side and the two sides facing the other dice for two six. For a single six the mapping can be done 5 by 16 with two sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Avoid rerolling if the mapping does not fit: For this purpose turn the blue dice clockwise. This suits best the design goal for a finits number of dice throws: 1.&lt;/p&gt;&#xA;" OwnerUserId="17784" LastEditorUserId="17784" LastEditDate="2015-12-17T08:01:40.210" LastActivityDate="2015-12-17T08:01:40.210" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24591" PostTypeId="1" AcceptedAnswerId="24657" CreationDate="2015-12-14T21:35:26.040" Score="10" ViewCount="1086" Body="&lt;p&gt;Having bested the silly &lt;a href=&quot;https://puzzling.stackexchange.com/q/24586/10615&quot;&gt;clown&lt;/a&gt;, you go to your packaging job. It is a slow day, so you decide to teach the game you learned to one of your coworkers. There are no balloons around, so you use bubble wrap instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This two player game is played with a huge roll of bubble wrap, which has &lt;strong&gt;sixteen million&lt;/strong&gt; bubbles. Players take turns popping some nonzero number of bubbles. The only restrictions are: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;On the first turn, the current player can't pop all the bubbles.&lt;/li&gt;&#xA;&lt;li&gt;On later turns, a player can't pop more bubbles than their opponent just popped. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Whoever pops the last bubble wins. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your friend has had lots of practice popping bubble wrap: he is so confident of his ability that he bets $100 he can win, and even lets you choose who goes first.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Should you go first or second? How can you guarantee victory?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&lt;em&gt;This is the same game as before, but starting with 16,000,000 things instead of 99. This makes the puzzle much less trivial.&lt;/em&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-12-17T15:36:24.857" Title="Pop the Last Bubble" Tags="|mathematics|game|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24592" PostTypeId="2" ParentId="24591" CreationDate="2015-12-14T21:59:26.707" Score="4" Body="&lt;p&gt;I feel like there is a way easier solution to this, but I'll just post the one I found:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Let the friend start&lt;br&gt;&lt;br&gt;The friend will obviously start with a number &lt;code&gt;&amp;lt; 16,000,000 / 2&lt;/code&gt;, or else he would lose because you would be allowed to pop all the remaining bubble wraps.&lt;br&gt;&lt;br&gt;If the number he picked is even:&lt;br&gt;&lt;br&gt;Now calculate in some kind of loop if the number he picked is dividable by 4. If no, you will pop the same amount of bubble wraps / 2. If yes, then do the same with the &lt;code&gt;previous number * 2&lt;/code&gt;, and so on. Just stop when it's not dividable anymore, and pick that number.&lt;br&gt;&lt;br&gt;Of course, when the number he picked is odd, just pick &lt;code&gt;1&lt;/code&gt;, then he will lose certainly.&lt;br&gt;&lt;br&gt;Example:&lt;br&gt;&lt;br&gt;He picked 24.&lt;br&gt;Dividable by 2? (=Even?) - true&lt;br&gt;Dividable by 4? - true&lt;br&gt;Dividable by 8? - true&lt;br&gt;Dividable by 16? - false.&lt;br&gt;&lt;br&gt;So pick &lt;em&gt;8&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user14478" LastEditorDisplayName="user14478" LastEditDate="2015-12-14T22:23:36.350" LastActivityDate="2015-12-14T22:23:36.350" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24594" PostTypeId="2" ParentId="24591" CreationDate="2015-12-14T22:20:52.827" Score="14" Body="&lt;p&gt;Go first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem: a player facing a $2^i$ bubbles, where $i$ is a positive integer, and who cannot pop them all immediately, is guaranteed to lose. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proving the theorem by induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $i=0$, it is impossible, as you can't face 1 bubbles and not be able to pop it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $i=1$, you face 2 bubbles, and cannot pop both.  You must pop one, and thus lose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $i=2$, you face 4 bubbles.  If you pop 2 or 3, your opponent pops the rest, and you lose.  If you pop 1, then you alternate popping 1 until he wins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step is to assume the theorem is true for all $i \leq N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you face $2^{N+1}$ bubbles and pops $2^N$ or more, your opponent will immediately pop the rest and win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you pop fewer than $2^N$, your opponent adopts a strategy that guarantees you will face $2^N$ bubbles, eventually.  This strategy is the exact same strategy the opponent would use if you faced $2^N$ bubbles initially.  Since the theorem holds for $i \leq N$, such a strategy &lt;em&gt;must&lt;/em&gt; exist.  Thus your opponent can force you to end up facing $2^N$ bubbles, which we have already proven is a lose condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;QED, by induction the theorem is proven.  Anyone facing a number of bubbles that is a power of 2, but who cannot win immediately, must lose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To win, you force your opponent to face $2^i$ bubbles.  You start with 16 million, and the next power of 2 is $2^{23}$.  So you pop&#xA;$16000000 - 2^{23} = 7611392$  bubbles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your opponent faces $2^{23}$ bubbles.  From there on, the strategy is to keep forcing your opponent to face powers of two balloons, until you can win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;  As some comments have pointed out, proving that a winning strategy exists is not exactly the same as providing a strategy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Strategy for winning if your opponent faces $2^N$ bubbles&lt;/strong&gt;: If your opponent pops $x$ bubbles, find $i$ such t hat $2^{i-1} \le x &amp;lt; 2^{i}$, then pop $2^{i} - x$ bubbles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why this works:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, since $x \ge 2^{i-1}$, it flollows that $2^{i} - x \le 2^{i} - 2^{i-1} = 2^{i-1} \le x$, so it is always possible to pop that number of bubbles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, if $i = N$, then you have just won.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, you have ensured that your opponent is facing $2^{N} - 2^{i} = 2^{N-1} + A 2^{i}$ bubbles, and cannot pop more than $2^{i-1}$.  By continuing this strategy, you can peel off each group of $2^i$ bubbles until your opponent faces $2^{N-1}$.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="2328" LastEditDate="2015-12-15T17:46:01.147" LastActivityDate="2015-12-15T17:46:01.147" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="24595" PostTypeId="2" ParentId="24586" CreationDate="2015-12-14T22:31:49.530" Score="6" Body="&lt;p&gt;For the general case of N balloons, if it's not possible to pop all the balloons, a winning move is...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...pop the largest power-of-two factor of N, if one can do so.  This will leave the opponent with a number whose largest power-of-two factor is twice as big, and thus cannot be popped.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If that move is unavailable...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...the player is doomed because any move the player makes will leave a remaining number of balloons whose largest power-of-two factor will be no greater than the number of balloons just popped (thus leaving the other player a winning move).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To understand why this is so...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the number of balloons was $2^N(2K+1)$ for some integers K and N, and one pops $2^N$, the resulting number of balloons will be $2^{N+1}K$, which will of course be a multiple of a power of two which is too large for the opponent to pop.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Conversely...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the number of balloons is $2^N(2K+1)$ for some integers K and N, but one cannot pop $2^N$, then one must pop $2^M(2L+1)$ for integers L and M such that $M&amp;lt;N$.  That will leave $2^M[2(2^{N-M-1}(2K+1)-(L+1))+1]$ balloons, for the opponent, which would clearly be of the winning form $2^N(2K+1)$ after substituting $M$ for $N$ and $2^{N-M-1}(2K+1)-(L+1)$ for $K$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1467" LastEditorUserId="1467" LastEditDate="2015-12-15T20:06:27.840" LastActivityDate="2015-12-15T20:06:27.840" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="24596" PostTypeId="2" ParentId="24507" CreationDate="2015-12-14T22:45:28.523" Score="5" Body="&lt;p&gt;The first thing to notice is that whenever a red arrow is followed, the output increases by 3, while a black arrow increases the output by 1.&lt;/p&gt;&#xA;&lt;p&gt;The arrow to be followed is determined by the current value of the output, according to the following table:&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th&gt;State&lt;/th&gt;&#xA;&lt;th&gt;Output&lt;/th&gt;&#xA;&lt;th&gt;Arrow&lt;/th&gt;&#xA;&lt;th&gt;Goes to&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;1&lt;/td&gt;&#xA;&lt;td&gt;multiple of 5&lt;/td&gt;&#xA;&lt;td&gt;black&lt;/td&gt;&#xA;&lt;td&gt;2&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;1&lt;/td&gt;&#xA;&lt;td&gt;not multiple of 5&lt;/td&gt;&#xA;&lt;td&gt;red&lt;/td&gt;&#xA;&lt;td&gt;5&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;2&lt;/td&gt;&#xA;&lt;td&gt;multiple of 3&lt;/td&gt;&#xA;&lt;td&gt;black&lt;/td&gt;&#xA;&lt;td&gt;3&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;2&lt;/td&gt;&#xA;&lt;td&gt;not multiple of 3&lt;/td&gt;&#xA;&lt;td&gt;red&lt;/td&gt;&#xA;&lt;td&gt;1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;3&lt;/td&gt;&#xA;&lt;td&gt;multiple of 10&lt;/td&gt;&#xA;&lt;td&gt;black&lt;/td&gt;&#xA;&lt;td&gt;5&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;3&lt;/td&gt;&#xA;&lt;td&gt;not multiple of 10&lt;/td&gt;&#xA;&lt;td&gt;red&lt;/td&gt;&#xA;&lt;td&gt;2&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;4&lt;/td&gt;&#xA;&lt;td&gt;multiple of 2&lt;/td&gt;&#xA;&lt;td&gt;black&lt;/td&gt;&#xA;&lt;td&gt;2&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;4&lt;/td&gt;&#xA;&lt;td&gt;not multiple of 2&lt;/td&gt;&#xA;&lt;td&gt;red&lt;/td&gt;&#xA;&lt;td&gt;3&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;5&lt;/td&gt;&#xA;&lt;td&gt;multiple of 7&lt;/td&gt;&#xA;&lt;td&gt;black&lt;/td&gt;&#xA;&lt;td&gt;4&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td&gt;5&lt;/td&gt;&#xA;&lt;td&gt;not multiple of 7&lt;/td&gt;&#xA;&lt;td&gt;red&lt;/td&gt;&#xA;&lt;td&gt;2&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;" OwnerUserId="4421" LastEditorUserId="69582" LastEditDate="2021-01-18T17:07:36.967" LastActivityDate="2021-01-18T17:07:36.967" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24597" PostTypeId="2" ParentId="546" CreationDate="2015-12-15T00:21:05.227" Score="3" Body="&lt;p&gt;I made a custom strategy for solving with the original rules - 6 colors and 4 pegs code length with 10 guesses allowed. My solution is guaranteed to solve in 9 guesses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the colors are red, green, yellow, orange, purple, and violet. Start by putting 1 color in the first 2 holes then another color in the second 2 holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example&lt;/strong&gt; - RRGG&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now make another guess with 2 colors that haven't been used the next guess must follow the same pattern. So after 3 guesses we will have used all the colors. So in my solution the first 3 guesses would go something like this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;guess 1&lt;/strong&gt; - YYOO&lt;br&gt;&#xA;&lt;strong&gt;guess 2&lt;/strong&gt; - GGVV&lt;br&gt;&#xA;&lt;strong&gt;guess 3&lt;/strong&gt; - RRPP&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Up until now we have ignored the clue pegs - the first 3 guesses always use the same pattern. Now here's where things get a little complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our goal is to figure out what colors the first 2 pegs of the code are and what color the other 2 pegs are. I call these the front and back pegs respectively. After that we can settle the final positions in at most 3 guesses. So after making the guesses above take the guess marked with the most clue pegs (if it is a tie you can choose) and replace the back pegs with a color you know is NOT in the code then make this your new guess. Now there are 2 ways find a color not in the code. If one of the first 3 guesses has no clue pegs then none of those colors are in the code. The second method requires making a guess. If the first 3 guesses all have clue pegs choose the guess marked only by one clue peg - this is always possible! Now choose a color in that guess and simply repeat it. Now note the clue peg if there is none that color is not in the code otherwise if you still get a clue peg then the other color is not in the code. Now you can continue. Since this sounds more complicated then it is I will show examples from playing against the computer. The colors the computer uses are: blue,green,orange,purple,red,yellow. Bu stands for blue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;example 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;guess 1 - BuBuGG; no clue pegs&lt;br&gt;&#xA;guess 2 - OOPP; no clue pegs&lt;br&gt;&#xA;guess 3 - RRYY; 2 black 0 white&lt;br&gt;&#xA;guess 4 - RRBuBu; 2 black - this is guess 3 with the last 2 holes replaced with a color not in the code.&lt;br&gt;&#xA;guess 5 - RRRR; solved  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;example 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;guess 1 - BuBuGG; 0 black 1 white&lt;br&gt;&#xA;guess 2 - OOPP; 0 black 2 white&lt;br&gt;&#xA;guess 3 - RRYY; 0 black 1 white&lt;br&gt;&#xA;guess 4 -BuBuBuBu; 1 black 0 white - here we are looking for a color not in the code. We could have modified guess 3 but did guess 1 instead. So we now know 1 blue peg is in the code and if we look at guess 1 we see it must go in one of the back 2 holes. So what we have of the code so far is: xxBux The x means the other pegs are unknown yet. Note that we simply stuck blue in one of the last 2 holes it's position may not yet be settled. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;guess 5 - OOGG; 0 black 0 white - Since guess 2 has the most clue pegs we modify it by putting green in the last 2 holes. Green is used because we know it is not in the code. From this we can see that orange is not in the code either. If we look back at guess 2 we can alse there are 2 purple pegs and that both must go in the front 2 holes. So now the code is: PPBux Looking at guess 3 the final color must be red. The clue tells 1 peg is out of position. Since the code has only 1 spot left in the back the missing peg must be red. So now we have: PPBuR which is used as our next guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;guess 6 - PPBuR solved - if we had gotten 2B 2W it would have been PPRBu&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;example 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;guess 1 BuBuGG; 2 black 0 white&lt;br&gt;&#xA;guess 2 OOPP; 1 black 0 white&lt;br&gt;&#xA;guess 3 RRYY; 0 black 1 white&lt;br&gt;&#xA;guess 4 OOOO; 1 black 0 white - code so far is Oxxx&lt;br&gt;&#xA;guess 5 BuBuPP; 1 black 0 white - code so far is OBuGR Guess 5 tells us there is 1 blue peg in the code and that it must go in front. Looking at guess 1 the other black clue peg must refer to green telling us that it goes in back. Guess 3 tells us red must go in back.&lt;br&gt;&#xA;guess 6 OBuGR; 0 black 4 white&lt;br&gt;&#xA;guess 7 BuORG; solved &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Important! In the end game there are different ways to settle the final positions of the pegs. Once you know what pegs go in the front and back there is a pattern you can follow:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PRGY; 2 black 0 white&lt;br&gt;&#xA;PRYG; 4 black 0 white&lt;br&gt;&#xA;RPGY; solved&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically if you get 4 white swap the front 2 pegs then the back 2 pegs as shown above. If you get 2 black swap the pair in the back and that will either solve or give you 4 white.&lt;/p&gt;&#xA;" OwnerUserId="17794" LastEditorUserId="12109" LastEditDate="2015-12-15T02:15:15.463" LastActivityDate="2015-12-15T02:15:15.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24598" PostTypeId="2" ParentId="24569" CreationDate="2015-12-15T05:21:20.243" Score="1" Body="&lt;p&gt;Where it is clear, I will use &lt;code&gt;XYZ&lt;/code&gt; and &lt;code&gt;WXYZ&lt;/code&gt; to mean either the shape the area it encloses. Likewise, &lt;code&gt;XY&lt;/code&gt; will refer to either the line segment or its length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5Ic2s.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5Ic2s.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with the midpoint E of AC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since EC = $\frac{1}{2}$AC, we have DEC = $\frac{1}{2}$DAC. (Both triangles have the same perpendicular height from AC to D.) Likewise, BEC = $\frac{1}{2}$BAC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summing the areas, we get EBCD = $\frac{1}{2}$ABCD. Call this (1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, DF = FB = $\frac{1}{2}$DB, so DFE = BFE = $\frac{1}{2}$DBE. (All these triangles have the same perpendicular height from DB to E.) Likewise, DFC = BFC = $\frac{1}{2}$DBC. Adding one of each, we get&lt;br&gt;&#xA;DEFC = EBCF = $\frac{1}{2}$EBDC. Call this (2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From (1) and (2), DEFC = EBCF = $\frac{1}{4}$ABCD. With reference to the diagram above, DEFC is composed of the shaded areas P and R, and EBCF is the quadrilateral labelled S. (Note: P, Q and S cross the diagonal lines AC and BD.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since DFE = BFE and DFC = BFC, we also have two additional solutions, each composed of two triangles: FEDBCF and FEBDCF. (Although not quadrilaterals, the OP allowed sums and differences of touching triangles as solutions in a comment to the question.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By a similar construction starting with the midpoint F of DB, we also get ABEF = $\frac{1}{4}$ABCD. ABEF is composed of the shaded areas Q and R. As before, we also get EFACBE and EFCABE as additional solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we have established that P+R = Q+R = S = $\frac{1}{4}$ABCD. So adding the first 3 expressions and decomposing ABCD, we get P+R + Q+R + S = $\frac{3}{4}$(P+Q+R+S+T).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, P+Q+R+S+T = 4(T-R), so (T-R) is also $\frac{1}{4}$ABCD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This gives a total of 8 shapes produced using 5 lines, with each shape occupying an area $\frac{1}{4}$ABCD.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we are permitted 9 lines, we have 4 additional lines with which to define acceptable polygons. Here are some approaches that might produce acceptable shapes, possibly by completing the triangle formed from the rays AB and DC. In each case, one vertex is not pinned down, but I leave these here for others to play with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since (P+R) + S = $\frac{1}{2}$ABCD, the complement T+Q is also $\frac{1}{2}$ABCD. Since ABE &amp;lt; $\frac{1}{4}$ABCD, there is a point G on AD such that DGE = GABE = $\frac{1}{4}$ABCD. Likewise, there is a point H on AD such that FHA = FHDC = $\frac{1}{4}$ABCD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since T &gt; $\frac{1}{4}$ABCD, there is a point I on AB that lies between A and the intersection of AB and the ray DE, such that DAI = DIBE = $\frac{1}{4}$ABCD. Likewise, there is a point J on CD such that AJD = AJCF = $\frac{1}{4}$ABCD.&lt;/p&gt;&#xA;" OwnerUserId="9422" LastEditorUserId="9422" LastEditDate="2015-12-26T01:35:57.557" LastActivityDate="2015-12-26T01:35:57.557" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="24599" PostTypeId="2" ParentId="24551" CreationDate="2015-12-15T05:52:42.487" Score="3" Body="&lt;p&gt;If you are also allowed a basic clock displaying the current time to the second, you only need one die.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Establish the starting time (&lt;code&gt;Ts&lt;/code&gt;).&lt;/li&gt;&#xA;&lt;li&gt;Roll one die (value &lt;code&gt;r&lt;/code&gt;). Non-rollers observe and agree on the throw time (&lt;code&gt;Tt&lt;/code&gt;) to the second.&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;t = Tt - Ts&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Final value = &lt;code&gt;r * (t mod 17 + 1)&lt;/code&gt;. Reroll if r = 6 and t mod 17 = 15 or 16&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You can also modify for smaller values than 17 second cycles. For example, you can also add a modulo to the number of cycles, and a conditional modulo within that a la leap years. It's fairly trivial to create a probabilistic model that equally generates values 1-100.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, most of your entropy comes from when the die is thrown, rather than the throw itself.&lt;/p&gt;&#xA;" OwnerUserId="1810" LastActivityDate="2015-12-15T05:52:42.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24601" PostTypeId="2" ParentId="24551" CreationDate="2015-12-15T09:06:25.560" Score="3" Body="&lt;p&gt;My solution needs you to throw &lt;strong&gt;2 identifiable dice&lt;/strong&gt; (or throw them in order) &lt;strong&gt;per digit&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I like it because it's &lt;strong&gt;easy to compute by a human mind&lt;/strong&gt; and builds the number digit per digit, so you are not faced with high numbers or base-changing stuff nor tables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One digit is done by:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;rolling first dic to get a number between 1 and 5. This is done by rerolling any 6 and keeping any other result. To address the issue of possible infinite rolling, you can put an arbitrary limit of reroll (for example 4 reroll), and decide that when reached, the result would be an arbitrary number between 1-5 (for example: 5). Keep this number in mind.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;by rolling the second die, you get to add 5 or keep the previous number on odd or respectively even numbers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You have your digit. Repeat the process to get your 1-100 number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This solution is arbitrarily close to a perfect uniform distribution depending on the number of re-roll you are allowing yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A nice variation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the dice are actually recognisable, shuffle them in a box, pick only one at random, it'll decide if the '+5' bonus to applies to your result. You only have to throw it to finish the process of building your digit&lt;/p&gt;&#xA;" OwnerUserId="17798" LastEditorUserId="2315" LastEditDate="2015-12-15T23:19:38.073" LastActivityDate="2015-12-15T23:19:38.073" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24605" PostTypeId="1" CreationDate="2015-12-15T10:30:55.390" Score="5" ViewCount="1077" Body="&lt;h2&gt;Rules&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Basics&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;52 well shuffled cards are dealt equally to both players. Player 1 starts the first hand. The game consists of a number of rounds, which are played as follows. On each round, players get alternate turns. The first person to run out of cards wins immediately (this could be at any turn in any round).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Starting a round&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A person starts a round by playing any number of cards face down on the table and declare what rank they are. (If it is your last card/s you must be truthful to win.) For example, one could play three cards from his deck and say that they are sevens. Then it is the other person's turn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;On each turn&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is your turn, you may&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Pass - Turn goes back to the other player.&lt;/li&gt;&#xA;&lt;li&gt;Play - Play any number of cards on top of the same deck. You are declaring them to be of the same rank as what the round was started with. You cannot declare falsely (bluff) if these is/are your last card(s).&lt;/li&gt;&#xA;&lt;li&gt;Pick - Pick up the cards that the other person played on his last turn (if he did play any). If they are true (same as what was declared at the start of the round), you keep the cards and the other player starts a new round. Otherwise, if it is a bluff, you give the cards to the other player and start a new round.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If both players pass consecutively, the cards on the table are discarded and the person who played the last card starts a new round.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Question&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Is there any way to determine the probability of an optimal player winning against another optimal player (where both players know that each other are optimal)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if your answer is exactly 1/2, justify it.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4269" LastEditDate="2015-12-16T14:44:08.857" LastActivityDate="2015-12-16T20:55:30.223" Title="Optimal game of Bluff (not the wikipedia one)" Tags="|strategy|game|cards|" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="24606" PostTypeId="2" ParentId="163" CreationDate="2015-12-15T11:01:50.780" Score="0" Body="&lt;p&gt;I am a beginner and still learning the beginner's method. I was stuck in this situation and could not understand the accepted answer in a go. So documenting what I did to solve it. Disclaimer: it is not at all fast but is simple to understand for beginners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do what you do to move an edge from top layer to middle layer, i.e. &lt;code&gt;U R U' R' U' F' U F&lt;/code&gt; or &lt;code&gt;U' F' U F U R U' R'&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically put in a yellow edge (if you started with white) in the place of the flipped edge, that should move just the flipped edge to the top layer and then you can position it correctly.&lt;/p&gt;&#xA;" OwnerUserId="17800" LastActivityDate="2015-12-15T11:01:50.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24607" PostTypeId="2" ParentId="24544" CreationDate="2015-12-15T14:04:41.007" Score="-1" Body="&lt;p&gt;6 constant&lt;br&gt;&#xA;12 vacant&lt;br&gt;&#xA;15 migrant&lt;br&gt;&#xA;14 abundant&lt;br&gt;&#xA;13 instant  &lt;/p&gt;&#xA;" OwnerUserId="17802" LastEditorUserId="4421" LastEditDate="2015-12-15T15:40:35.403" LastActivityDate="2015-12-15T15:40:35.403" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24608" PostTypeId="2" ParentId="24544" CreationDate="2015-12-15T14:27:43.723" Score="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;Elephant &lt;/li&gt;&#xA;&lt;li&gt;Distant &lt;/li&gt;&#xA;&lt;li&gt;Giant&lt;/li&gt;&#xA;&lt;li&gt;Servant&lt;/li&gt;&#xA;&lt;li&gt;Tolerant&lt;/li&gt;&#xA;&lt;li&gt;Constant &lt;/li&gt;&#xA;&lt;li&gt;Extravagant&lt;/li&gt;&#xA;&lt;li&gt;Participant&lt;/li&gt;&#xA;&lt;li&gt;Infant&lt;/li&gt;&#xA;&lt;li&gt;Dormant&lt;/li&gt;&#xA;&lt;li&gt;Luminant &lt;/li&gt;&#xA;&lt;li&gt;Vacant&lt;/li&gt;&#xA;&lt;li&gt;Instant&lt;/li&gt;&#xA;&lt;li&gt;Abundant &lt;/li&gt;&#xA;&lt;li&gt;Migrant&lt;/li&gt;&#xA;&lt;li&gt;Relevant&lt;/li&gt;&#xA;&lt;li&gt;Plant&lt;/li&gt;&#xA;&lt;li&gt;Dominant&lt;/li&gt;&#xA;&lt;li&gt;Hesitant &lt;/li&gt;&#xA;&lt;li&gt;Habitant&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="17803" LastActivityDate="2015-12-15T14:27:43.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24609" PostTypeId="2" ParentId="24551" CreationDate="2015-12-15T15:54:16.613" Score="4" Body="&lt;h1&gt;An optimal algorithm with an optimal termination speed that can be done by humans&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;First the algorithm:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;rolled = 0&#xA;max = 1&#xA;while(we have another die){&#xA;    rolled = 6*rolled + rand(6)&#xA;    max = 6*max&#xA;    if(rolled/100 != max/100){&#xA;        return rolled%100&#xA;    }&#xA;    max = max%100&#xA;    rolled = rolled%100&#xA;}&#xA;return rolled&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here are just the most important points for the optimality of this algorithm.&lt;br&gt;&#xA;&lt;b&gt;The full proof is left the reader.&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;It is easy:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;All numbers have at most 3 digits. All divisions are with 100. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;This algorithm returns each value with an optimally equal propability:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;To understand that these algorithm works you have to realize the following 3 points:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The propability that rolled has a specific value (between $0$ and $max$) is always $\frac{1}{max}$.&lt;/li&gt;&#xA;&lt;li&gt;Under the condition that rolled doesn't has the same digit in the hundreds position then $rolled\%100$ is equally distributed.&lt;/li&gt;&#xA;&lt;li&gt;If $rolled$ and $max$ have the same digit in the hundreds position then taking modulo $100$ will keep these statements true.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;There can't be a faster algorithm:&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;After n rolls there are $6^n$ possible outcomes. &lt;/li&gt;&#xA;&lt;li&gt;In $6^n\%100$ of those cases no algorithm could have terminated.&lt;/li&gt;&#xA;&lt;li&gt;The value of max is always $6^n\%100$.&lt;/li&gt;&#xA;&lt;li&gt;For all values in $[0, max-1]$ there is exactly one series of die throws to reach that specific values.&lt;/li&gt;&#xA;&lt;li&gt;There are exactly $max$ cases where the algorithm didn't terminated yet.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="17807" LastActivityDate="2015-12-15T15:54:16.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24610" PostTypeId="2" ParentId="24544" CreationDate="2015-12-15T18:56:25.987" Score="2" Body="&lt;p&gt;1)elephant 2)distant 3)exorbitant/giant 4)assistant 5)pleasant/tolerant 6)constant 7)luxuriant/elegant/extravagant 8)participant/contestant 9)infant 10)dormant 11)radiant/ luminary 12)vacant 13)instant 14)abundant 15)migrant/immigrant 16)relevant/significant 17)plant 18)dominant/ important 19)hesitant 20)occupant/habitant&lt;/p&gt;&#xA;" OwnerUserId="17808" LastActivityDate="2015-12-15T18:56:25.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24612" PostTypeId="2" ParentId="24605" CreationDate="2015-12-15T19:20:26.037" Score="4" Body="&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: I had the rules wrong, below is a correct analysis with the correct rules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show how complicated this game is, let's try to solve a much simpler version. Suppose instead the deck has only 4 cards, namely, two kings and two queens, but all other rules are the same. Let's call the players Alice and Bob, with Alice going first. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Alice is dealt either two queens or two kings, she immediately wins, so assume they are each dealt a queen and a king.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alice has only two sensible options: she should play a single card, then either tell the truth (T) or lie (L) about its identity. Bob can then either doubt Alice's claim (D), play the card which Alice claimed to play (P), play the opposite card, or pass. If he passes, he immediately loses, and it turns out that if Bob can win dishonestly, then he can also win honestly, so we only consider the first two options.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TD&lt;/strong&gt;: Bob is wrong, so Alice gets another turn, and wins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LP&lt;/strong&gt;: Alice just successfully lied. Bob plays a card, but then Alice plays her other card and wins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TP&lt;/strong&gt;: Alice plays one card, then Bob plays a card. Alice has no choice but to pass, as does Bob. Since Bob played last, he starts the next round and wins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LD&lt;/strong&gt;: Alice lied, but was called out, so she picks up her card, and Alice and Bob switch roles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say that the probability of Alice winning under optimal play is $p$. We can summarize the above information in a matrix, where each entry contains the probability of Alice winning in that situation:&#xA;$$&#xA;\begin{array}{c|c|c|}&#xA;&amp;amp;P&amp;amp;D\\\hline&#xA;T &amp;amp; 0&amp;amp;1\\\hline&#xA;L &amp;amp; 1 &amp;amp; 1-p\\\hline &#xA;\end{array}&#xA;$$&#xA;Notice the recursion above, because in the LD case, we have the same game with the roles reversed, so Alice now wins with probability $1-p$ instead of $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only solution is to use a mixed strategy. Suppose Alice tells the truth with probability $a$. Her best strategy is to choose $a$ so that Bob's payoff is equalized. If Bob chooses $P$, his expected payout is $1-a$, and if he chooses $D$, he gets $a\cdot 1+(1-a)(1-p)=1-p+pa$. Setting these equal, we get that&#xA;$$a = \tfrac{p}{1+p}$$&#xA;By symmetry, Bob will Believe with probability $ \frac{p}{1+p}$ as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculating Alice's probability of winning by summing over the four possibilities in the above matrix, then setting this equal to $p$, we get&#xA;$$&#xA;p = 2(\tfrac{p}{1+p})(\tfrac{1}{1+p})\cdot 1+(\tfrac{1}{1+p})^2(1-p)&#xA;$$&#xA;Amazingly, when you solve this, you get that $p$ is equal to the inverse of the Golden Ratio, i.e. $$p=\frac{\sqrt{5}-1}2\approx 61.8\%$$&#xA;is Alice's probability of winning under optimal play. To achieve this, she tells the truth about $\frac{.618}{1+.618}=38\%$ percent of the time. Bob should believe her with the same probability. Of course, this was assuming Alice was dealt a queen and a king. Overall, her probability of winning is $\frac13\cdot 1+\frac23\cdot 0.618=74.5\%$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can perhaps guess, based on these results, that the game favors the first player, and the best strategy is to slightly lean towards deceit and doubt, but still be honest and trusting a nontrivial amount of the time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you can see that the full 52 card game would be ridiculously complicated. This small game had itself as an option: imagine there were many games which mutually had each other as options. There would be a complicated web of nonlinear equations instead of a single nice one. &lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-12-16T20:55:30.223" LastActivityDate="2015-12-16T20:55:30.223" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24613" PostTypeId="1" CreationDate="2015-12-15T19:53:28.310" Score="2" ViewCount="386" Body="&lt;p&gt;&lt;em&gt;(This is a continuation of the series from last year. The first such question is found &lt;a href=&quot;https://puzzling.stackexchange.com/questions/3981/the-security-to-the-party&quot;&gt;here&lt;/a&gt;)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A party is being held at a local mansion. The host is very rich and his success is because of one thing ~ his famous recipe for Spaghetti!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only guests that may attend are people the host knows and trusts, whom he gave the password to enter to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's where you come in. You and a friend are trying to steal this recipe. You sneak by and listen to the passwords.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first guest arrives. The security guard says &quot;5&quot;. The guest says &quot;5&quot;, and enters the mansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second guest arrives. The security guard says &quot;9&quot;. The guest replies &quot;5&quot;, and is admitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third guest arrives. The security guard says &quot;3&quot;. The guest replies &quot;5&quot;, and enters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your friend thinks he's got it all figured out. He climbs out of the bushes and approaches the guard. The guard says &quot;4&quot;. Your friend confidently says &quot;5&quot;. The guard then pulls a lever, opening a trap door beneath your friend. He falls to his death.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fourth guest arrives. The guard says &quot;6&quot;, and the guest replies &quot;24&quot; and enters the party.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point you believe you know what's going on. You approach the guard, and he says &quot;7&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you say?&lt;/p&gt;&#xA;" OwnerUserId="17559" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2015-12-15T20:39:31.917" Title="The Security to the Party [Part 38]" Tags="|password|" AnswerCount="1" CommentCount="1" ClosedDate="2015-12-15T21:18:13.360" ContentLicense="CC BY-SA 3.0" />
  <row Id="24614" PostTypeId="1" CreationDate="2015-12-15T20:38:19.807" Score="-5" ViewCount="468" Body="&lt;p&gt;A donkey was in a stable, tied by a 15-foot rope to a pole. He then walked 20 feet out of the stable. How is this possible?&lt;/p&gt;&#xA;" OwnerUserId="17813" LastEditorDisplayName="user88" LastEditDate="2015-12-18T08:44:28.867" LastActivityDate="2015-12-18T08:44:28.867" Title="The pole, the 15 foot rope, and the donkey 20 feet away" Tags="|riddle|" AnswerCount="3" CommentCount="7" ClosedDate="2015-12-16T09:28:53.137" ContentLicense="CC BY-SA 3.0" />
  <row Id="24615" PostTypeId="2" ParentId="24613" CreationDate="2015-12-15T20:39:31.917" Score="10" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I say &quot;14&quot;. The method is to convert the guard's number into a word &quot;seven&quot;, take the last letter &quot;n&quot; and convert it into its place in the alphabet, &quot;14&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7459" LastActivityDate="2015-12-15T20:39:31.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24616" PostTypeId="2" ParentId="24614" CreationDate="2015-12-15T20:51:25.100" Score="0" Body="&lt;p&gt;It feels like a trivial answer but&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The pole was also 20 feet away, next to the donkey.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1561" LastActivityDate="2015-12-15T20:51:25.100" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24617" PostTypeId="2" ParentId="24614" CreationDate="2015-12-15T21:14:06.023" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The pole is 5 feet from the stable&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3987" LastActivityDate="2015-12-15T21:14:06.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24621" PostTypeId="1" AcceptedAnswerId="24624" CreationDate="2015-12-15T22:22:50.357" Score="2" ViewCount="233" Body="&lt;p&gt;Just had an idea on a new Four words question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(See the original for the solution format&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/15339/four-words&quot;&gt;Four-words (----||||)&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) I am a visual definition of myself when you had the answer&lt;br&gt;&#xA;2) My position is kind of contradicting with what I am&lt;br&gt;&#xA;3) This one is easy once you figure out the rest&lt;br&gt;&#xA;4) I am the same as 1)  &lt;/p&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2015-12-16T00:10:09.653" Title="Yet another Four-words (----||||)" Tags="|word|crosswords|word-square|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24622" PostTypeId="1" AcceptedAnswerId="24623" CreationDate="2015-12-15T23:19:45.357" Score="5" ViewCount="663" Body="&lt;p&gt;You are trapped in a chamber in the center of the Minotaur's Labyrinth. There are $\mathbf{N}$ tunnels, $\mathbf{m}$ of which lead to safety; the remaining tunnels only lead back to the chamber. Each tunnel is of a different length, taking $h_i$ hours to travel. Each time you return to the chamber, the room shifts so that you can only choose tunnels at random. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What is the expected amount of time it will take you to escape?&lt;/li&gt;&#xA;&lt;li&gt;You have 24 hours until the Minotaur wakes up. If there are 10 tunnels, such that $h_i = i$ (for $i$ = 1,2,...10) and two of the tunnels lead to safety do you believe you will escape in time?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11063" LastEditorUserId="4050" LastEditDate="2015-12-24T12:11:16.407" LastActivityDate="2015-12-24T12:11:16.407" Title="The Minotaur's Labyrinth" Tags="|mathematics|logical-deduction|probability|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24623" PostTypeId="2" ParentId="24622" CreationDate="2015-12-15T23:29:59.050" Score="8" Body="&lt;ol&gt;&#xA;&lt;li&gt;You spend an expected $\frac{\sum h_i}N$ hours each time you travel a tunnel, and you have to travel an average of $\frac Nm$ tunnels to escape. The product is $\frac{\sum h_i}m$ hours total.&lt;/li&gt;&#xA;&lt;li&gt;$\sum h_i=55$ and $m=2$, so the expected amount of time to escape is $\frac{55}{2}=27.5$ hours.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-12-15T23:29:59.050" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24624" PostTypeId="2" ParentId="24621" CreationDate="2015-12-16T00:10:09.653" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;E D G E&lt;/code&gt;&lt;br&gt;&lt;code&gt;D E A D&lt;/code&gt;&lt;br&gt;&lt;code&gt;G A N G&lt;/code&gt;&lt;br&gt;&lt;code&gt;E D G E&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because the second one is not in the exact center ;)&lt;/p&gt;&#xA;" OwnerUserId="17771" LastActivityDate="2015-12-16T00:10:09.653" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24625" PostTypeId="1" AcceptedAnswerId="24627" CreationDate="2015-12-16T00:14:24.150" Score="7" ViewCount="231" Body="&lt;p&gt;Mr Maxwell, the new Physics teacher at &lt;strong&gt;Farthingbottom School&lt;/strong&gt;, has volunteered to do the lighting at this year's &lt;a href=&quot;https://puzzling.stackexchange.com/questions/21975/preparing-for-the-end-of-term-performance&quot;&gt;End of Term Performance&lt;/a&gt;, for which the children have been preparing very hard.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;He has set up red, green and blue lights to cover each of the regions of the stage (left, right, and centre) and provided white lights for the backdrop and a powerful white followspot.&lt;br&gt;&#xA;&lt;em&gt;[Please note that Mr Maxwell is unfamiliar with the convention of &quot;stage left&quot; referring to an actor's perspective of the stage. His labels refer to the stage as viewed by the audience.]&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what the unlit stage looks like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/axAeH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/axAeH.png&quot; alt=&quot;Unlit stage&quot;&gt;&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 10 songs in the Carol Concert and each of the songs requires a different lighting arrangement on stage. The diagram below shows the lighting required for the first nine carols.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TTuYO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TTuYO.png&quot; alt=&quot;lit stage songs 1 to 9&quot;&gt;&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mr Maxwell has set up a 26-button lighting controller that he has wired up in such a way that pushing any button will toggle the state of a number of the lights on the stage.&#xA;He has labelled each of the buttons alphabetically and produced a helpful table that shows which lights are toggled by each letter.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;    LR  LG  LB  CR  CG  CB  RR  RG  RB  BD  FS&#xA;a   0   1   0   0   1   0   0   0   0   0   1&#xA;b   0   1   0   1   0   0   0   1   0   1   1&#xA;c   0   0   1   0   0   0   0   0   0   0   0&#xA;d   0   0   0   0   0   0   1   0   0   0   1&#xA;e   0   1   0   0   0   0   1   0   0   1   1&#xA;f   0   1   0   0   1   0   0   0   0   0   0&#xA;g   0   0   0   0   1   0   1   0   1   1   0&#xA;h   0   1   1   0   0   0   1   0   0   0   0&#xA;i   0   0   0   0   0   1   0   0   0   0   0&#xA;j   0   1   0   1   0   1   0   0   1   0   0&#xA;k   1   1   1   0   1   0   0   0   0   1   0&#xA;l   0   0   0   0   0   0   0   1   0   0   1&#xA;m   1   0   0   1   1   1   0   0   0   0   0&#xA;n   0   1   0   1   0   0   0   0   0   0   1&#xA;o   0   0   0   1   0   0   0   0   0   1   0&#xA;p   0   0   0   0   1   0   1   0   0   0   0&#xA;q   1   0   0   1   1   1   0   0   1   0   0&#xA;r   0   0   1   0   0   0   1   0   1   0   0&#xA;s   1   0   0   0   0   0   0   0   0   0   0&#xA;t   1   1   0   1   0   0   0   0   0   0   0&#xA;u   0   0   0   1   0   0   1   0   0   0   0&#xA;v   1   0   1   0   0   1   0   0   0   1   0&#xA;w   1   0   0   0   0   0   0   1   0   0   0&#xA;x   0   1   0   0   1   0   1   0   0   0   0&#xA;y   0   0   0   1   0   0   0   1   0   0   0&#xA;z   0   1   0   0   0   0   0   0   0   0   0&#xA;&#xA;Key: LR = left red&#xA;     LG = left green&#xA;     LB = left blue&#xA;     CR = centre red&#xA;     RR = right red&#xA;     BD = backdrop&#xA;     FS = followspot&#xA;&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;By connecting the buttons in this way, he is able to switch the lighting from one song to the next (for the first 9 songs) by pressing a batch of either two or three of the letters (but never the same letter twice in the same batch).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;He has just realised that the letters he needs to press (in the order in which he will press them) spell out the first line of the 10th song. Not only that, but if he types the second line of the 10th song the stage lighting is then correctly configured for the final carol.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What is the title of the 10th carol, and what is the minimum number of letters he will need to type after it finishes to turn off all the lights on stage?&lt;/p&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2015-12-17T00:09:43.890" Title="The Stage is Set for the Christmas Performance" Tags="|seasonal|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24626" PostTypeId="1" CreationDate="2015-12-16T03:41:53.983" Score="19" ViewCount="1199" Body="&lt;p&gt;The text is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ntmilEmondt enae&#xA;Leala ed hThtrc &#xA;smve till pg   n&#xA;tut ane una olas&#xA;hrertwil incma t&#xA;ieeenak ioo anTl&#xA;es otah foer del&#xA;smeormresr  nmhi&#xA;m  f ugrpsmci io&#xA; thmmoT tir luay&#xA;beoa   a nfsoner&#xA;t oris sihhsvlad&#xA;sb fengtfned ewr&#xA; irnssortaa htam&#xA;ervegs  hntgdn i&#xA; senly  ia eoftn&#xA;bito itofshirude&#xA; trsecolucfia ra&#xA; phef fid tyaene&#xA;thowuoyffi w  ro&#xA;nde sr pnfct ta &#xA; T f  et e a lir&#xA; fhiout ioilimos&#xA;utortru  tn  c t&#xA;nce  raero  ls n&#xA;osenn  bo  Ee ml&#xA;Elt  stigsimtet &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It comes with the following riddle, which is part of the &lt;em&gt;intended&lt;/em&gt; form of solving the puzzle.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Where Air meets fire&lt;br&gt;&#xA;  and land marries sea&lt;br&gt;&#xA;  head south and east&lt;br&gt;&#xA;  Therein lies the key&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure it's a transposition cypher, based on letter frequencies, but beyond that, I have no idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the end result plaintext is a spell in 5th edition D&amp;amp;D, which means it's (probably) in the format:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Title]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Ordinal] level [spell school]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Description]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The [spell school] is Transmutation, and I know that the spell level is a high one, so the [Ordinal] is probably &quot;Sixth&quot; or higher (&quot;Ninth&quot; being the highest possible). This was learnt from in character knowledge. There's also a well above zero chance that the first word of the [Title] is &quot;Elminster's&quot; or &quot;Elminster&quot;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Does anyone know any good analytical techniques for solving this sort of transposition cypher, some possible cyphers it could be, or perhaps , way of solving it through the &quot;intended&quot; method using the riddle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Someone asked a question as an answer, and then deleted it (obviously, using an answer to ask a clarifying question is not the way to go about things). The question was something along the lines of 'is it a spell from the D&amp;amp;D 5e PHB?'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer was no, probably not. It's a spell for 5e, but this is the third of three puzzles, and the first two were homebrew spells, so I suspect the third will be too. The previous two's names both started with &quot;Elminster's&quot;, which is why I suspected this one might as well.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I used some in-character knowledge to get a clue. Apparently &quot;fire&quot; refers to heating the scroll, and &quot;air&quot; refers to the blank spaces in the text. When the physical scroll was heated* &lt;a href=&quot;https://i.stack.imgur.com/9FROJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;the following map appeared&lt;/a&gt;. The blue dots weren't actually there, I've just added them to make clear where the blank spaces are. For reference, the map is of the Moonshae Isles, in Faerûn (part of the Forgotten Realms D&amp;amp;D campaign setting).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apologies for for the fact that some information was required that would have been impossible for anyone here to obtain on their own. It never occurred to me that such information would have been part of the puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;* unfortunately, the invisible ink had faded by the time I tried this, so it didn't work. The above image is of a separate copy of the scroll, with the text superimposed over it digitally.&lt;/p&gt;&#xA;" OwnerUserId="17815" LastEditorUserId="17815" LastEditDate="2016-01-29T03:56:03.130" LastActivityDate="2016-01-29T06:22:02.823" Title="Unknown transposition cypher puzzle" Tags="|riddle|cipher|" AnswerCount="1" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="24627" PostTypeId="2" ParentId="24625" CreationDate="2015-12-16T04:03:03.443" Score="5" Body="&lt;p&gt;First I found the possible button presses for the first 9 songs (using a script to quickly try all 2952 possible sets of three or fewer buttons):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;aij&lt;/code&gt; &lt;code&gt;dgo&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;er&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;st&lt;/code&gt; &lt;code&gt;afn&lt;/code&gt; &lt;code&gt;chu&lt;/code&gt; &lt;code&gt;deo&lt;/code&gt; &lt;code&gt;fpu&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;abp&lt;/code&gt; &lt;code&gt;bdz&lt;/code&gt; &lt;code&gt;bnu&lt;/code&gt; &lt;code&gt;dlo&lt;/code&gt; &lt;code&gt;ouy&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;emr&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;ry&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;egn&lt;/code&gt; &lt;code&gt;fij&lt;/code&gt; &lt;code&gt;iqs&lt;/code&gt; &lt;code&gt;jkv&lt;/code&gt; &lt;code&gt;jmt&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;elt&lt;/code&gt; &lt;code&gt;enw&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;cpv&lt;/code&gt; &lt;code&gt;emn&lt;/code&gt; &lt;code&gt;fhv&lt;/code&gt; &lt;code&gt;gjt&lt;/code&gt; &lt;code&gt;hik&lt;/code&gt;   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can immediately get the word&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;merry&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;from the fifth and six lines.  This made me realize that I could reduce my search space from &quot;all possible sentences&quot; to beginning lines of songs of a particular genre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I &lt;a href=&quot;http://www.carols.org.uk/&quot;&gt;looked through a list&lt;/a&gt; and found one which appears to work:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; God rest you merry, gentlemen&#xA; (&lt;code&gt;dgo&lt;/code&gt;) (&lt;code&gt;er&lt;/code&gt; &lt;code&gt;st&lt;/code&gt;) (&lt;code&gt;ouy&lt;/code&gt;) (&lt;code&gt;emr&lt;/code&gt; &lt;code&gt;ry&lt;/code&gt;) (&lt;code&gt;egn&lt;/code&gt; &lt;code&gt;elt&lt;/code&gt; &lt;code&gt;emn&lt;/code&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Which is also the name of the song!)  The next line is then:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let nothing you dismay&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From which I get the lighting layout:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stage: RWR, Backdrop on, Follow on&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using the same strategy as for the first part, I find that the minimum number of button presses to disable the lights:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3 (i.e. &lt;code&gt;emz&lt;/code&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="9718" LastEditDate="2015-12-17T00:09:43.890" LastActivityDate="2015-12-17T00:09:43.890" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24628" PostTypeId="2" ParentId="24614" CreationDate="2015-12-16T07:11:52.480" Score="1" Body="&lt;p&gt;I think the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The pole can be anywhere between a radius of 5 to 15 feet from the stable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The above condition will allow the donkey to be inside the stable, or be at a distance of 20 feet away from the stable.  (Or even at a max distance of 35 feet from the stable)&lt;/p&gt;&#xA;" OwnerUserId="16918" LastEditorUserId="11452" LastEditDate="2015-12-17T21:11:15.220" LastActivityDate="2015-12-17T21:11:15.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24631" PostTypeId="2" ParentId="201" CreationDate="2015-12-16T10:47:05.900" Score="0" Body="&lt;p&gt;&lt;em&gt;As A doesn't know which color his hat is, B &amp;amp; C's color are not the same. (means [B != Black] OR [C != Black]) &lt;br&gt;&#xA;If C's hat color [Black], according to A said, B would think his is [Red] definitely. &lt;br&gt;&#xA;But B said he doesn't know; C's hat color is not [Black].&lt;br&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;So the color of hat the child on chair C have is RED.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;To answer the Bonus question: &lt;br&gt;&#xA;If C is wearing Black, we might be able to find out B's hat color is not Black; RED. &lt;br&gt;&#xA;But as the above solution; C's color is RED, so A cannot know the answer if B is also wearing RED. With that case, B also cannot know which color he is wearing when C's color is RED and A color can be RED.&lt;br&gt;&#xA;I think no one can answer this as no one can know hat color of A and B.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="11353" LastActivityDate="2015-12-16T10:47:05.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24632" PostTypeId="1" AcceptedAnswerId="24636" CreationDate="2015-12-16T11:30:06.590" Score="-2" ViewCount="593" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7X1p2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7X1p2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 1 =&gt; what is the pattern of this image? &lt;br&gt;&#xA;Question 2 =&gt; what is the source to become 42 at last?&lt;/p&gt;&#xA;" OwnerUserId="11353" LastActivityDate="2015-12-16T15:43:24.893" Title="Find pattern of number circulation" Tags="|mathematics|" AnswerCount="4" CommentCount="0" ClosedDate="2015-12-16T13:51:12.603" ContentLicense="CC BY-SA 3.0" />
  <row Id="24633" PostTypeId="2" ParentId="24632" CreationDate="2015-12-16T11:34:21.017" Score="0" Body="&lt;p&gt;a = b&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c = d&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e = f&lt;/p&gt;&#xA;&#xA;&lt;p&gt;on the left part you just add one on each line. The result is the current number * the number just under (like b = a&lt;em&gt;c, d = c&lt;/em&gt;e )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and i didn't undertsand your last question :)&lt;/p&gt;&#xA;" OwnerUserId="17799" LastActivityDate="2015-12-16T11:34:21.017" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24634" PostTypeId="2" ParentId="24632" CreationDate="2015-12-16T12:00:49.317" Score="1" Body="&lt;p&gt;If we call the left series $L(n)$ and the right series $R(n)$, then &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$L(n) = n$, and $R(n)=R(n-1)+2*L(n)$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This simplifies to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $R(n) = 2\sum\limits_{i=1}^ni = (n+1)*n$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If your second question is asking about what R(7) will be then it's &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $R(7) = 8*7 = 56$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16846" LastActivityDate="2015-12-16T12:00:49.317" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24635" PostTypeId="2" ParentId="24632" CreationDate="2015-12-16T12:02:47.363" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Question 1 - The number on the right side is the product of the number on the left and the next whole number in sequence. I.e 2*3, 3*4, 4*5&lt;br&gt;&#xA;&lt;br&gt;&#xA; Question 2 - The next step in the sequence is 7=56&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16874" LastEditorUserId="12109" LastEditDate="2015-12-16T15:43:24.893" LastActivityDate="2015-12-16T15:43:24.893" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24636" PostTypeId="2" ParentId="24632" CreationDate="2015-12-16T13:01:42.657" Score="1" Body="&lt;p&gt;I don't want to make it complicated. It's pretty simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Series goes like this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;n * (n+1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 * 3 = 6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 * 4 = 12&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 * 5 = 20&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5 * 6 = 30&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6 * 7 = 42&lt;/p&gt;&#xA;" OwnerUserId="16918" LastActivityDate="2015-12-16T13:01:42.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24641" PostTypeId="2" ParentId="24586" CreationDate="2015-12-16T15:49:36.207" Score="2" Body="&lt;p&gt;As previously stated, the strategy for this game is to&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; go first&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; pop one balloon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You will win, because&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Your opponent's only move is to pop a single balloon.  Repeat, and you will eventually pop the last.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h1&gt;General Strategy&lt;/h1&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;This is not correct BTW - but I will leave it as an exercise to the reader as to why.  Cheaters can look at the comments!&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;h2&gt;&lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; is odd&lt;/h2&gt;&#xA;&lt;p&gt;Use the same strategy as the &lt;span class=&quot;math-container&quot;&gt;$99$&lt;/span&gt; balloon case: go first and pop one balloon.&lt;/p&gt;&#xA;&lt;h2&gt;&lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; is even&lt;/h2&gt;&#xA;&lt;p&gt;In this case, you must&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; go second.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;On the clown's turn, he will do one of the following:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Pop half or more balloons.&lt;/li&gt;&#xA;&lt;li&gt;Pop an odd number of balloons less than half&lt;/li&gt;&#xA;&lt;li&gt;Pop an even number of balloons less than half.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;h3&gt;Pops half or more&lt;/h3&gt;&#xA;&lt;p&gt;You win easily by popping the rest.&lt;/p&gt;&#xA;&lt;h3&gt;Pops an odd number of balloons&lt;/h3&gt;&#xA;&lt;p&gt;It is now your turn in an &amp;quot;&lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; is odd&amp;quot; case.  So, pop 1 and let the game play out.&lt;/p&gt;&#xA;&lt;h3&gt;Pops an even number of balloons&lt;/h3&gt;&#xA;&lt;p&gt;On your turn, simply pop &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; balloons.&lt;/p&gt;&#xA;&lt;p&gt;You win because the clown again has an even number of balloons and he is in the same prediciment as before.  Only now his options are limited to popping &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; (and leaving it even) or &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; (and making it odd).  The moment he pops &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; you win.  Otherwise, you will eventually be left with the last &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; balloons and get to pop them both to win.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-12-16T18:33:20.680" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24642" PostTypeId="2" ParentId="24551" CreationDate="2015-12-16T16:06:01.153" Score="2" Body="&lt;p&gt;A solution that allows the number to be directly read from the dice in most cases:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roll 3 identifiable dice, d1, d2 and d3. This gives 216 possible outcomes, so 2.16 outcomes should lead to each possible value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;200 of those outcomes can be mapped directly to the required output. Rather than requiring players to do multiplications by 6 and multi-digit modular arithmetic, I instead propose a straightforward lookup table as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set N := d2.&#xA;If d1 = 3 or d1 = 6, add 3 or 6 to the value of N as necessary to bring it within the range 7-9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If d1 &amp;lt;= 3, our result is equal to N * 10 + d3 (i.e. the digits are concatenated, and in particular when d1 &amp;lt;= 2, the value is shown right there on the other 2 dice with no other transformations required at all)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If d1 &gt;= 4 and d3 &amp;lt;= 4, our result is equal to N * 10 + d3 + 6 (i.e. add 6 to the value of d3, but otherwise treat the same as before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If d1 &gt;= 4 and d3 = 5, our result is equal to N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Together, these relatively simple rules cover 198 possibilities (of 216) with no &quot;maths&quot; involved, leaving only the 18 cases where d1 &gt;= 4 and d3 = 6, and we've not generated the number 10 at all...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If d1 = d3 = 6 and d2 &gt;= 5, our result is equal to 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have a result we can stop here. However, only 200 possible rolls mapped to a number, and 16 possible rolls don't have a defined result yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roll another die, d4. It is recommended that this is rolled onto grid paper (as in Stefan Bischof's answer), as the orientation may be important, but not necessarily as critical as in that case (it will act only as the &quot;decider&quot; between two adjacent numbers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate X = (d1-4) * 36 + (d2 - 1) * 6 + d4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will give a value between 1 and 96. Add 1 to X for values from 25 to 48, 2 for values from 49 to 72 and 3 for values from 73 to 96. This converts it to a distribution containing all integers from 1 to 99 except for multiples of 25.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the orientation of d4. For each of the possible 6 values of d4, there must be a defined reference position (e.g. the lowest numbered side face pointing north, or directly towards a nominated wall or similar).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering the current value mod 25 (which can be any value from 1 to 24, but not zero). Multiply this by 14.4 degrees (exactly 1/25 of a turn). If the cube were rotated clockwise towards the reference position, will it reach the reference position in less than this angle of rotation? If so, add 1 to the value of X. (25 dice identical to d4, and rotated to exactly the relevant angles could be glued to the table or squared paper for reference).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Almost all the time, the answer will be clear - only two critical orientations exist - one where d4 is almost exactly at the reference position, and one where rotating it by the specified angle would put it almost exactly at the reference position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In those ambiguous cases (and only in those ambiguous cases), roll d5, and add one only if the value of d5 is odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All possible values are then selected with precisely equal probability, and a procedure that requires rolling at most 5 dice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 92.6% of rolls, only 3 dice are necessary, and the value requires almost no calculation.&#xA;For the remaining 7.4% of rolls, 4 dice are necessary, with a 5th held in reserve to be rolled as a tie-breaker between two adjacent scores if the angle of the 4th die is not sufficient to resolve it.&lt;/p&gt;&#xA;" OwnerUserId="12582" LastEditorUserId="12582" LastEditDate="2015-12-17T09:33:25.940" LastActivityDate="2015-12-17T09:33:25.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24643" PostTypeId="1" AcceptedAnswerId="24644" CreationDate="2015-12-16T20:23:35.877" Score="11" ViewCount="529" Body="&lt;p&gt;&lt;sup&gt;This is inspired by the What is a Word/Phrase™ puzzles originally created by &lt;a href=&quot;https://puzzling.stackexchange.com/users/463/jlee&quot;&gt;JLee&lt;/a&gt;.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a word has a certain property, I call it a Vain Word™.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In each of the short, unrelated sentences below, only one of the choices is a Vain Word™.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sentences are for context only, and do not affect whether a word is a Vain Word™.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;I missed the bus, so I took a ( &lt;strong&gt;CAB&lt;/strong&gt; / &lt;strong&gt;TAXI&lt;/strong&gt; ) to work.&lt;/li&gt;&#xA;  &lt;li&gt;The old house down the street is covered in ( &lt;strong&gt;IVY&lt;/strong&gt; / &lt;strong&gt;VINES&lt;/strong&gt; ).&lt;/li&gt;&#xA;  &lt;li&gt;Having too much to drink can make you ( &lt;strong&gt;VOMIT&lt;/strong&gt; / &lt;strong&gt;PUKE&lt;/strong&gt; ).&lt;/li&gt;&#xA;  &lt;li&gt;To my dismay, a ( &lt;strong&gt;YOUNGSTER&lt;/strong&gt; / &lt;strong&gt;YOUTH&lt;/strong&gt;  ) always loiters in front of my store.&lt;/li&gt;&#xA;  &lt;li&gt;Archaeologists who explore tombs may have to contend with a (  &lt;strong&gt;ZOMBIE&lt;/strong&gt; / &lt;strong&gt;MUMMY&lt;/strong&gt; ).&lt;/li&gt;&#xA;  &lt;li&gt;I noticed the ocean was very ( &lt;strong&gt;WAVY&lt;/strong&gt; / &lt;strong&gt;CALM&lt;/strong&gt; ) while I was sitting on the beach today.&lt;/li&gt;&#xA;  &lt;li&gt;Having been beaten by my enemy last time, this time I was prepared to ( &lt;strong&gt;OUTSMART&lt;/strong&gt; / &lt;strong&gt;OUTWIT&lt;/strong&gt; ) him.&lt;/li&gt;&#xA;  &lt;li&gt;The controversial social experiment turned out to be a ( &lt;strong&gt;HOAX&lt;/strong&gt; / &lt;strong&gt;SHAM&lt;/strong&gt; ).&lt;/li&gt;&#xA;  &lt;li&gt;The mathematician was stuck trying to prove something, until he recalled a rather obscure ( &lt;strong&gt;AXIOM&lt;/strong&gt; / &lt;strong&gt;THEOREM&lt;/strong&gt; ).&lt;/li&gt;&#xA;  &lt;li&gt;My friend showed my his ( &lt;strong&gt;GARGANTUAN&lt;/strong&gt; / &lt;strong&gt;MAMMOTH&lt;/strong&gt; ) truck. It was quite an eyesore.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What makes a word a Vain Word™?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="17559" LastEditorUserId="-1" LastEditorDisplayName="user14478" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2016-09-23T21:33:13.313" Title="What is a Vain Word™?" Tags="|pattern|english|word-property|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24644" PostTypeId="2" ParentId="24643" CreationDate="2015-12-16T20:28:59.660" Score="9" Body="&lt;p&gt;Vain words are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; made up of letters which are symmetrical about the Y-axis, presumably because they like to look at themselves in the mirror&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. TAXI&lt;br&gt;&#xA; 2. IVY&lt;br&gt;&#xA; 3. VOMIT&lt;br&gt;&#xA; 4. YOUTH&lt;br&gt;&#xA; 5. MUMMY&lt;br&gt;&#xA; 6. WAVY&lt;br&gt;&#xA; 7. OUTWIT&lt;br&gt;&#xA; 8. HOAX&lt;br&gt;&#xA; 9. AXIOM&lt;br&gt;&#xA; 10. MAMMOTH  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastActivityDate="2015-12-16T20:28:59.660" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24645" PostTypeId="1" AcceptedAnswerId="24646" CreationDate="2015-12-16T21:45:57.310" Score="3" ViewCount="278" Body="&lt;p&gt;What word am I thinking of?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Gamblers beat casino odds where&#xA;possible. Sometimes they even play&#xA;with a stacked deck. 'Hackers'&#xA;or coders also like to&#xA;win. Such people will use&#xA;a bit hack to thrive.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="16995" LastEditorUserId="16995" LastEditDate="2015-12-17T00:27:25.220" LastActivityDate="2015-12-17T00:27:25.220" Title="What word am I thinking of?" Tags="|riddle|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24646" PostTypeId="2" ParentId="24645" CreationDate="2015-12-16T22:29:29.880" Score="4" Body="&lt;p&gt;The word is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Binary&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each line has 5 words. These lines can be interpreted as numbers from 0 to 31 by treating odd-sized words as 1 bits and even-sized words as 0 bits. We then convert 0 to 'A', 1 to 'B', and so on. The phrase &quot;bit hack&quot; is a clue, of course, both to the methodology and the answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, I think there's a typo here, and &quot;succeed&quot; should be &quot;understand&quot;.&lt;/p&gt;&#xA;" OwnerUserId="9523" LastActivityDate="2015-12-16T22:29:29.880" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24647" PostTypeId="2" ParentId="24551" CreationDate="2015-12-16T22:45:45.237" Score="2" Body="&lt;h2&gt;A calculator-based procedure.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Initialise your calculator with a random decimal fraction, You could start with π/6, or you could roll a few d6 to generate the seed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each time you need a d100, throw d6 three times. Each time add the number to the number currently in the calculator unless you throw a six; then divide by six (even if you throw a six).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After three throws, multiply the number by 100. The integer portion is your resulting d100; you should subtract this ready for the next throw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although any given throw will be slightly biased (216 rolls distributed between 100 results), the bias depends on your saved random number, and will therefore be random in the long term.&lt;/p&gt;&#xA;" OwnerUserId="12593" LastActivityDate="2015-12-16T22:45:45.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24648" PostTypeId="1" AcceptedAnswerId="69645" CreationDate="2015-12-17T06:05:14.740" Score="13" ViewCount="2430" Body="&lt;p&gt;Some people were gunned down on the frontier a few years back. Their goal of seeking a new life, tragically cut short. Still, they did see some strange things before they died. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aside from being shot, what did this crew have in common?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a well-defined right answer. There are hints in the flavor text. If you get it, you will know that you are right.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Carlisle was shot 5 times.&#xA;Grant was shot 7 times&#xA;Harper was shot once.&#xA;Mathews was shot 6 times.&#xA;O'Herlihy was shot 3 times.&#xA;Rizzo was shot 7 times.&#xA;Thompson was shot 10 times.&#xA;Unknown was shot 4 times.&#xA;Watson was shot 14 times.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="16995" LastEditorUserId="16995" LastEditDate="2018-08-20T03:30:48.587" LastActivityDate="2018-08-20T03:30:48.587" Title="What do all these people have in common?" Tags="|enigmatic-puzzle|knowledge|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24650" PostTypeId="2" ParentId="24551" CreationDate="2015-12-17T10:09:52.767" Score="2" Body="&lt;p&gt;I am very weak in the syntax that is the programming language known as mathematics, so please take my answer with enough salt.&lt;br&gt;&#xA;Additionally I may be way off in the value of what I'm presenting (specifically in my usage of any actual math related terms).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here it is:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following requires no re-rolls (therefore meeting the finite number of rolls), covers all numbers from 0-99 (or 1-100 depending on how you're game system treats double 0 on 2d10), and, as far as my intuitive statistical analysis goes, (which may not be that far) is a reasonably even probability distribution across all numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally both possible solutions would be relatively simple to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Possible Solution 1&lt;/strong&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Roll 20d6 &lt;/li&gt;&#xA;&lt;li&gt;Add &lt;/li&gt;&#xA;&lt;li&gt;Any roll greater than 100 remove the 100s place and let the remaining value stand.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(NOTE: If you desire to have 0, then let a value of 100 represent 0 instead of 100).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Justification:&lt;/strong&gt;&lt;br&gt;&#xA;1/6th = 16.666666666666666666666666666667  (or rounded 17)&#xA;17 dice with a maximum value of 102 roughly cover most of 0-100 in 17-102.&lt;br&gt;&#xA;3 dice with a maximum possible role value of 18 will assist in filling in the lower numbers 3-18.&lt;br&gt;&#xA;the remainder above 100 of the 17 dice fill in for 1 &amp;amp; 2, while 3-17 fill in the missing numbers 1-17.&#xA;&lt;br&gt;(NOTE: In my mind the intentional overlap of values improves the evenness[?] of the probability distribution/fairness.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Possible Solution 2&lt;/strong&gt;&lt;br&gt;&#xA;Better, probably (because it may be more practical, though I suspect the probability distribution is similar...?):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Roll 1d6&lt;/li&gt;&#xA;&lt;li&gt;Let the following ranges be represented by the value rolled&#xA;&lt;br&gt;1= range 1-17&#xA;&lt;br&gt;2= 18-34&#xA;&lt;br&gt;3= 35-51&#xA;&lt;br&gt;4= 52-68&#xA;&lt;br&gt;5= 69-85&#xA;&lt;br&gt;6= 86-100 (actually 102)&lt;/li&gt;&#xA;&lt;li&gt;roll 3d6 and add the result to the base of your range you rolled earlier.&lt;/li&gt;&#xA;&lt;li&gt;If the result is a number above 100 remove the 100s place and let it stand as the remaining value (101=01,103=03).&lt;/li&gt;&#xA;&lt;li&gt;100 either represents itself or 0&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Essentially same justification as previous...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Final Note: I believe this method could potentially work for any dice standing in for 2d10/percentile dice.&lt;/p&gt;&#xA;" OwnerUserId="5692" LastEditorUserId="5692" LastEditDate="2015-12-19T00:20:11.763" LastActivityDate="2015-12-19T00:20:11.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24651" PostTypeId="2" ParentId="24551" CreationDate="2015-12-17T10:32:13.207" Score="3" Body="&lt;p&gt;Even though this is listed as a mathematical problem no fancy mathematical solution is required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My solution is based on that of @SJuan76.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Put 5 dice in an opaque bag. Each die has its '6' side covered with a different sticker in the range 1 to 5.&lt;/li&gt;&#xA;&lt;li&gt;Pull a die from the bag at random, roll it to generate the number from one to five.&lt;/li&gt;&#xA;&lt;li&gt;Roll an unstickered die. If the result is even add 5 to the result of the stickered die. &lt;/li&gt;&#xA;&lt;li&gt;You now have a number from 1 to 10 (or 0) for the first digit. Repeat steps 1 - 3 for the second digit.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Only 6 dice required, the five in the bag plus another one, and exactly four d6 rolls per d100 and no time-consuming lookup tables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can even not use the unstickered die if you roll a stickered die with an even numbered sticker. With 6 dice in each of two colours and two bags you could even roll all 4 dice together and sum the result quickly.&lt;/p&gt;&#xA;" OwnerUserId="17832" LastActivityDate="2015-12-17T10:32:13.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24652" PostTypeId="2" ParentId="24648" CreationDate="2015-12-17T12:33:50.230" Score="16" Body="&lt;p&gt;They were all &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.ex-astris-scientia.org/database/redshirt_deaths.htm&quot; rel=&quot;noreferrer&quot; title=&quot;eyyyyyyy&quot;&gt;Star Trek TOS Redshirts who died under Kirk's command&lt;/a&gt; - For them, space was truly &lt;em&gt;the final frontier&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;LATE EDIT:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; OK, OP says: &quot;REDSHIRTS: numbers are indexes into the titles; ordering is by stardate &amp;amp; character name. I don't know if that's too obscure for this site; it'd be typical fare at the MIT Mystery Hunt which is where I do most of my puzzling.&quot;&lt;br&gt;&#xA;&lt;br&gt;&#xA; I will admit I don't really understand what this means - Ordered by stardate, using character name (as given in the puzzle) as a secondary ordering criterion, and then treating the episode title as an array, I get nonsense. Switching the ordering order, still nonsense. This is also the case whether I assume the array is 1- or 0-indexed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&#xA;&lt;/p&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2016-02-16T14:47:25.147" LastActivityDate="2016-02-16T14:47:25.147" CommentCount="19" ContentLicense="CC BY-SA 3.0" />
  <row Id="24653" PostTypeId="1" AcceptedAnswerId="24655" CreationDate="2015-12-17T14:36:03.167" Score="14" ViewCount="700" Body="&lt;p&gt;These are probably incredibly easy, but have fun!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &lt;a href=&quot;https://i.stack.imgur.com/lwrok.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lwrok.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/d7E8X.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/d7E8X.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5NLYq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5NLYq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/5NLYq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5NLYq.jpg&quot; 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rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5NLYq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/5NLYq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5NLYq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7) &lt;a href=&quot;https://i.stack.imgur.com/UbvMq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UbvMq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/mp7fF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mp7fF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/etn8p.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/etn8p.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/hnW4w.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hnW4w.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/R0Bl0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/R0Bl0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14233" LastEditorUserId="14233" LastEditDate="2015-12-17T21:12:47.323" LastActivityDate="2015-12-17T21:12:47.323" Title="Some Time of Year Rebuses" Tags="|rebus|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24654" PostTypeId="2" ParentId="24370" CreationDate="2015-12-17T14:45:50.783" Score="22" Body="&lt;p&gt;Over the weekend I emailed Bletchley Park to see if they could shed any light on it. Today I got this reply:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dear Tim,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Thank you for your email which was circulated, we believe the letters are random. However we have sent your email onto our Education and Learning department for further information, until then unfortunately we don't have any further information. If there are any findings or developments will contact you in the first instance.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Kind Regards,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Bletchley Park Information &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So it seems that there is no code to solve :( Very disappointing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I hear anything different I'll update. Sorry to have wasted people's time.&lt;/p&gt;&#xA;" OwnerUserId="17671" LastActivityDate="2015-12-17T14:45:50.783" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24655" PostTypeId="2" ParentId="24653" CreationDate="2015-12-17T14:52:09.513" Score="14" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. Santa Claus (Sandy + Claws)&lt;br&gt;&#xA; 2. Merry Christmas (Mary + Kris + Mess)&lt;br&gt;&#xA; 3. Mistletoe (Missile + Toe)&lt;br&gt;&#xA; 4. Happy Holidays (cHAP stick + Pee + Holly + Daze)&lt;br&gt;&#xA; 5. Kwanzaa (Kwan + piZZA)&lt;br&gt;&#xA; 6. Menorah (Men + OR + A)&lt;br&gt;&#xA; 7. Fruitcake (Fruit + beefCAKE)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16720" LastEditorUserId="10399" LastEditDate="2015-12-17T15:03:29.020" LastActivityDate="2015-12-17T15:03:29.020" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24657" PostTypeId="2" ParentId="24591" CreationDate="2015-12-17T15:36:24.857" Score="6" Body="&lt;p&gt;The most simple strategy I can think of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Go first. During each of your turns write the nubmer of remaining bubbles as a binary number (for $16,000,000_{10}$ this would be $111101000010010000000000_2$). During your turn pop a number of bubbles equal to the value of the lowes bit set (this would be $10000000000_2 = 1024_{10}$ at the beginning). Repeat until you get all bubbles in the last turn.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is guaranteed to work because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $1.$ The starting position is not a power of $2$ allowing the algorithm to work during the first turn (otherwise let the opponent go first).&lt;br&gt;&#xA; $2.$ During each of your turns you pop either all bubbles (winning the game) or less than a half of them (preventing the opponent to win in next turn).&lt;br&gt;&#xA; $3.$ After each of your turns the $x$ lowest bits are all $0$, where $x$ is the number of bits in the number of bubbles you popped.&lt;br&gt;&#xA; $4.$ Your opponent will set at least one of the lowest $x$ bits to $1$ during his move, which enables you to repeat the procedure in the next move.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastActivityDate="2015-12-17T15:36:24.857" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24659" PostTypeId="1" CreationDate="2015-12-17T16:05:32.610" Score="1" ViewCount="205" Body="&lt;p&gt;I want to make a star with 2&quot; x 2&quot;.&#xA;I know that there are 10 equal length sides.&#xA;What I need to know is what angle am I to cut each piece at: the tip of the star, and the outside &quot;V&quot; of the star. A chop saw cut angles to 45 degrees. &lt;/p&gt;&#xA;" OwnerUserId="17836" LastActivityDate="2015-12-17T16:30:19.070" Title="angles of a star" Tags="|pattern|" AnswerCount="1" CommentCount="1" ClosedDate="2015-12-17T19:35:09.317" ContentLicense="CC BY-SA 3.0" />
  <row Id="24660" PostTypeId="2" ParentId="24659" CreationDate="2015-12-17T16:30:19.070" Score="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;Label your corners clockwise $A, B, C, D$.&lt;/li&gt;&#xA;&lt;li&gt;On side $AB$, mark the half-way point $P_{AB}$ at 1&quot;.&lt;/li&gt;&#xA;&lt;li&gt;Cut from $P_{AB}$ towards the adjacent sides $BC$, creating $P_{BC}$ where the cut meet the edge. The angle $P_{BC}P_{AB}P_B$ is $54^{\circ}$.&lt;/li&gt;&#xA;&lt;li&gt;Cut from $P_{BC}$ to edge $CD$, creating $P_{CD}$. The angle $P_{AB}P_{BC}P_{CD}$ is $72^{\circ}$.&lt;/li&gt;&#xA;&lt;li&gt;Mirror the two cuts you just made on the other side of the material.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You now have a pentagon. Draw lines connecting all vertices of this pentagon and cut out the pieces that will leave you with a star.&lt;/p&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2015-12-17T16:30:19.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24661" PostTypeId="1" AcceptedAnswerId="24664" CreationDate="2015-12-17T17:49:20.183" Score="10" ViewCount="847" Body="&lt;p&gt;I have a friend! (The one everyone seems to have around here) He was on wheel of fortune and did great! Won over $10,000 and a trip to Aruba.&lt;br&gt;&#xA;He's now in the bonus round and is ready to win the big prize! Whatever is in the envelope.  He failed though because he just couldn't figure it out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being the smart fellow he is, he did his research to find out what the most common letters + vowel are picked from contestants in the bonus round.  What he found through the magic of the internet was&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;C - 64%&#xA;M - 59%&#xA;D - 57%&#xA;A - 48%`&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;His category was &lt;code&gt;Phrase&lt;/code&gt; and given &lt;code&gt;RSTLNE&lt;/code&gt; and his chosen letters of &lt;code&gt;CMDA&lt;/code&gt;, the bonus puzzle result ended up looking like this.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  _A_    _N   &#xA;_A_    _ _ _&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;He wants to know if any one is slick enough to solve what he could not! There has been many one letter solved puzzles in the history of the game show, so this is a very generous one!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe I can make this a series of my own ;) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rules&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You are allowed ONE guess  &lt;/li&gt;&#xA;&lt;li&gt;If there is an edit changing your answer, you are automatically disqualified (From getting the green check mark)  &lt;/li&gt;&#xA;&lt;li&gt;Since I can't really put you on a ten second timer first correct post wins!&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="16273" LastEditorUserId="372" LastEditDate="2015-12-17T20:54:26.757" LastActivityDate="2015-12-17T20:54:26.757" Title="Let's spin the wheel!" Tags="|word|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24662" PostTypeId="5" CreationDate="2015-12-17T18:04:02.860" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-12-17T18:04:02.860" LastActivityDate="2015-12-17T18:04:02.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24663" PostTypeId="4" CreationDate="2015-12-17T18:04:02.860" Score="0" Body="This tag is for puzzles that use the same concepts as the popular game show &quot;Wheel of Fortune&quot;" OwnerUserId="16273" LastEditorUserId="16273" LastEditDate="2015-12-17T20:15:14.773" LastActivityDate="2015-12-17T20:15:14.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24664" PostTypeId="2" ParentId="24661" CreationDate="2015-12-17T18:26:00.033" Score="13" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; WAX ON WAX OFF (from the Karate Kid)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2015-12-17T18:26:00.033" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24665" PostTypeId="1" AcceptedAnswerId="24668" CreationDate="2015-12-17T22:12:36.633" Score="7" ViewCount="1159" Body="&lt;p&gt;Captain Nefarious has captured you and two of your friends, Alice and Bob.  Being a villain, he has a natural desire to monologue, so he sits you down (guarded, of course by his Largely Incompetent Henchmen), and explains his Nefarious™ Plan:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;Each of you will be placed in a separate room, which provides &lt;em&gt;no way&lt;/em&gt; of communicating with anyone outside the room.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;Each day, my True Random Colour Generator™ will display one of three random colours (&lt;span class=&quot;math-container&quot;&gt;$\color{red}{\text{RED}}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\color{blue}{\text{BLUE}}$&lt;/span&gt;, or &lt;span class=&quot;math-container&quot;&gt;$\color{green}{\text{GREEN}}$&lt;/span&gt;) on the wall of your room.  Then each of you will be given a chance to guess what colour one of your friends was shown.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;You will each tell me the name of one friend (not yourself), and the colour you think they were shown.  If none of you name the same friend, and at least one of you names the correct colour, you will live another day.  If two of you name the same friend, or if none of you name the correct colour, I will immediately feed all of you to my HeroEater™—my patent-pending new hero disposal device.&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Being a True Villain, he is contractually obligated to allow the three of you to talk among yourselves before you are separated.  &amp;quot;But no funny stuff,&amp;quot; he warns, &amp;quot;or you'll be meeting the HeroEater™ sooner than you'd like.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;You, Alice, and Bob quickly put your heads together to think up a strategy.  You know that an Elite Extraction Team will be dispatched to rescue you, but you don't know how long it will take.  It could be a day, or it could be 20 years.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Is there a strategy that will guarantee your survival until the Elite Extraction Team arrives?&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;If not, what plan gives you the best probability of surviving until you are rescued (i.e. what strategy maximizes the probability that at least one of you will correctly guess another's colour?)&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;sup&gt;&lt;strong&gt;Disclaimer:&lt;/strong&gt; I don't know if there is a guaranteed solution to this problem.  I have been able to devise a strategy that provides a &lt;span class=&quot;math-container&quot;&gt;$&amp;gt;77\%$&lt;/span&gt; (i.e. &lt;span class=&quot;math-container&quot;&gt;$\frac79$&lt;/span&gt;) chance of survival each day, but there could be something better.  (If your solution has a smaller probability, still feel free to post it.)&lt;/sup&gt;&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-12-18T16:53:42.937" Title="Surviving Captain Nefarious" Tags="|mathematics|strategy|probability|story|hat-guessing|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24666" PostTypeId="2" ParentId="24665" CreationDate="2015-12-17T22:18:42.773" Score="1" Body="&lt;p&gt;8/27 chance of dying each day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can avoid the first part of doom by having a circle of guessing: you guess Alice, Alice guesses Bob, Bob guesses you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The color part is tricky, since your color has no bearing on their colors. To avoid that, all of you guess the same color. That gives $(\frac{2}{3})^3$ chance of dying each day. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-12-17T22:18:42.773" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24667" PostTypeId="2" ParentId="24665" CreationDate="2015-12-17T23:05:53.917" Score="9" Body="&lt;p&gt;This solution gives a 7/9 probability of survival.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As JonTheMon has explained, agreeing on a cycle of A $\to$ B $\to$ C $\to$ A is necessary to ensure that the same person won't be chosen twice. This limits some of the strategies that can be chosen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The team can improve over random color selection, however, by each naming the color that they see. They will succeed as long as at least 2 walls display the same color. That occurs with probability p $= 1 - \frac{2}{3}\cdot\frac{1}{3} = \frac{7}{9} \enspace$.&lt;/p&gt;&#xA;" OwnerUserId="17841" LastEditorUserId="9422" LastEditDate="2015-12-17T23:39:03.117" LastActivityDate="2015-12-17T23:39:03.117" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24668" PostTypeId="2" ParentId="24665" CreationDate="2015-12-18T00:43:00.060" Score="12" Body="&lt;p&gt;This is a proof that it is impossible to beat $\frac{7}{9}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 27 equally likely situations for the choice of colors. Each person will guess correctly in 9 situations. However, for any two people, they will both be correct in 3 situations (three possible colors for A, B has the color A guesses and C has the color B guesses). By inclusion-exclusion, the number of situations in which at least one person is correct is $9+9+9-3-3-3$ plus the number of situations where all three are correct. The maximum of this achieved when all three people are correct whenever any two are, resulting in a value of $9+9+9-3-3-3+3=21$. Therefore, it is not possible to do better than $\frac{21}{27}=\frac{7}{9}$.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-12-18T00:43:00.060" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24669" PostTypeId="1" AcceptedAnswerId="24670" CreationDate="2015-12-18T04:57:07.813" Score="15" ViewCount="2107" Body="&lt;p&gt;Nine cards numbered 1 through 9 are kept facing up on a table. Alice and Bob are playing a game wherein they pick up cards one at a time, alternatively. The first person to have &lt;strong&gt;three&lt;/strong&gt; cards with sum 15 wins the game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, it is upto Bob to choose who picks the first card. Can Bob ensure a win?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(7, 8) is not a winning hand, because two cards have a sum 15&lt;br&gt;&#xA;  (1,2,3,9) is not a winning hand because four cards have sum 15&lt;br&gt;&#xA;  (2, 6, 3, 7) is a winning hand because three cards(2, 6, 7) have sum 15&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17659" LastEditorUserId="11876" LastEditDate="2015-12-18T05:06:18.390" LastActivityDate="2015-12-18T05:06:18.390" Title="Find the winning strategy for this card game" Tags="|mathematics|strategy|cards|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24670" PostTypeId="2" ParentId="24669" CreationDate="2015-12-18T05:05:51.253" Score="20" Body="&lt;p&gt;This game is &lt;em&gt;actually&lt;/em&gt;...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tic-Tac-Toe (or Noughts and Crosses).&lt;br&gt;&#xA;    4 9 2&lt;br&gt;&#xA;    3 5 7&lt;br&gt;&#xA;    8 1 6&lt;br&gt;&#xA; If the numbers are arranged in this 3x3 magic square, all rows, columns, and diagonals (and ONLY those) are sets of 3 that sum to 15.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bob cannot ensure a win.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2015-12-18T05:05:51.253" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24671" PostTypeId="1" CreationDate="2015-12-18T06:56:09.407" Score="-5" ViewCount="322" Body="&lt;p&gt;1, 2, 5, 13, ?, ?, 1&#xA;Please give explanation.&lt;/p&gt;&#xA;" OwnerUserId="17819" LastActivityDate="2015-12-18T13:49:51.023" Title="What are the missing items of this number sequence?" Tags="|logical-deduction|pattern|number-sequence|" AnswerCount="1" CommentCount="3" ClosedDate="2015-12-18T18:03:13.340" ContentLicense="CC BY-SA 3.0" />
  <row Id="24672" PostTypeId="2" ParentId="24044" CreationDate="2015-12-18T07:03:27.600" Score="3" Body="&lt;p&gt;(Please tell me why this works, I have no idea!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 2b suggests that there is a relationship between binary representations of coin weights and continued fractions.  I assumed that this relationship is a one-to-one mapping, so we can count all possible weights by enumerating the continued fractions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, we want all fractions with $2015$ in the numerator, whose continued fractions represent binary numbers less than $2^{2015-1}$.  Those numbers have a leading zero in their binary expansion, so we want fractions whose continued fraction expansion begins with a nonzero number: that is, fractions greater than one.  This means that the denominator must be less than $2015$, and coprime to $2015$.  There are $\phi(2015)$ numbers matching those criteria, so the number of possible denominators/possible fractions/possible continued fractions/possible binary numbers/possible weights is just:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $1440$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="9718" LastEditDate="2015-12-19T03:04:16.183" LastActivityDate="2015-12-19T03:04:16.183" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24675" PostTypeId="1" CreationDate="2015-12-18T08:21:03.603" Score="15" ViewCount="1082" Body="&lt;h1&gt;Decipher the clues, escape the prison!&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The guard at the prison, where you have been locked up for &lt;a href=&quot;http://www.telegraph.co.uk/technology/internet-security/11932793/Killer-USB-stick-destroys-your-computer-in-seconds.html&quot;&gt;misuse of USB ports&lt;/a&gt;, has given you one chance to escape, by typing in a certain word. A keyboard is put down in front of you, and you are shown a sheet that reads as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You never know where the key will be hidden, it could be &lt;em&gt;right&#xA;  between your eyes&lt;/em&gt;...&lt;br&gt; &lt;code&gt;[20][1E][23][1F]&lt;/code&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;THE WORD&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Much like a Vigenere cipher, yet the letters are numbers!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;code&gt;[1L][1H][2M][16][13][08][06][17][15][18][10]&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What should you type in?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The part about 'right between your eyes' applies to the word only, not the text in the first code block.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The guard is quite nice and has given you, being a &lt;strong&gt;USB expert&lt;/strong&gt;, nothing too hard.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Surely &lt;em&gt;someone&lt;/em&gt; knows a cipher like a Vigenere cipher but with numbers as letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17827" LastEditorUserId="17827" LastEditDate="2016-01-08T07:44:19.150" LastActivityDate="2016-01-23T23:29:32.840" Title="Decipher the clues, escape the prison!" Tags="|cipher|computer-puzzle|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="24676" PostTypeId="1" CreationDate="2015-12-18T08:25:25.263" Score="4" ViewCount="390" Body="&lt;p&gt;A farmer has a certain number of sheep.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whichever single sheep you pick, you can always divide the other remaining sheep in two groups which have the same number weight and the same number of sheep.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you prove that all sheep are the same weight?&lt;/p&gt;&#xA;" OwnerUserId="5041" LastEditorDisplayName="user88" LastEditDate="2016-03-04T02:24:39.373" LastActivityDate="2016-03-04T02:24:39.373" Title="Prove that all the sheep are the same weight" Tags="|mathematics|" AnswerCount="0" CommentCount="6" ClosedDate="2016-01-29T21:05:10.533" ContentLicense="CC BY-SA 3.0" />
  <row Id="24679" PostTypeId="1" AcceptedAnswerId="24680" CreationDate="2015-12-18T12:57:03.317" Score="3" ViewCount="377" Body="&lt;p&gt;Let's kick it up a notch from &lt;a href=&quot;https://puzzling.stackexchange.com/questions/24661/lets-spin-the-wheel&quot;&gt;Let's spin the wheel!&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My friend is again somehow on the game show and has made it to the bonus round yet again! He uses the same strategy as before, but he just plain sucks at this.  He again wants to know if you can figure out at which he failed!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This one is a littler tricker. Given &lt;code&gt;RSTLNE&lt;/code&gt; and Letters chosen &lt;code&gt;CMDA&lt;/code&gt; (As most popular)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The puzzle goes like this, The &lt;a href=&quot;http://wheeloffurtunehistory.wikia.com/wiki/List_of_categories&quot; rel=&quot;nofollow noreferrer&quot;&gt;category&lt;/a&gt; is &lt;code&gt;Before and After&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;C_ _ L _    D _ _    _ _ _ _ T&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rules&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You are allowed ONE guess  &lt;/li&gt;&#xA;&lt;li&gt;If there is an edit changing your answer, you are automatically disqualified (From getting the green check mark)  &lt;/li&gt;&#xA;&lt;li&gt;Since I can't really put you on a ten second timer first correct post wins!&lt;/li&gt;&#xA;&lt;li&gt;All the letters listed are letters shown (or not shown) in the puzzle&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="16273" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2015-12-18T17:02:41.617" Title="Let's spin the wheel! [2]" Tags="|word|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24680" PostTypeId="2" ParentId="24679" CreationDate="2015-12-18T13:08:20.380" Score="5" Body="&lt;p&gt;Pat, I'd like to solve the puzzle. It's &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;CHILI DOG FIGHT&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastActivityDate="2015-12-18T13:08:20.380" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24682" PostTypeId="2" ParentId="24671" CreationDate="2015-12-18T13:49:51.023" Score="1" Body="&lt;p&gt;I would figure that it is simply&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1,2,5,13,&lt;strong&gt;5&lt;/strong&gt;,&lt;strong&gt;2&lt;/strong&gt;,1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9346" LastActivityDate="2015-12-18T13:49:51.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24683" PostTypeId="1" AcceptedAnswerId="24686" CreationDate="2015-12-18T14:13:44.237" Score="11" ViewCount="762" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;My death number is 5.8&lt;br&gt;&#xA;  Last two days i didn't slept&lt;br&gt;&#xA;  Being compared to an illuminating object in the bad conditions&lt;br&gt;&#xA;  I didn't listen to the person who is everyone's well wisher&lt;br&gt;&#xA;  The hint is that my image is poor&lt;br&gt;&#xA;  My name that you know is not what is my&lt;br&gt;&#xA;  Do you know who am i?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sorry for not rhyming&lt;/p&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2015-12-18T15:55:17.337" LastActivityDate="2015-12-18T17:06:03.557" Title="Who am I with the death number 5.8" Tags="|riddle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24685" PostTypeId="2" ParentId="24683" CreationDate="2015-12-18T15:48:52.857" Score="5" Body="&lt;p&gt;Are you ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Macbeth?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is what I have got so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My death number is 5.8&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He dies in &lt;a href=&quot;http://nfs.sparknotes.com/macbeth/page_216.html&quot;&gt;Act 5, Scene 8&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Last two days i didn't slept&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being compared to an illuminating object in the bad conditions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn't listen to everyone's well wisher&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hint is that my image is poor&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He becomes a tyrannical ruler.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My name that you know is not what is my&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Macbeth means &quot;son of life&quot;, he seems to be quite the opposite.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Do you know who am i?&lt;/p&gt;&#xA;" OwnerUserId="15632" LastActivityDate="2015-12-18T15:48:52.857" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24686" PostTypeId="2" ParentId="24683" CreationDate="2015-12-18T16:37:14.140" Score="7" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Marilyn Monroe&lt;br&gt;&#xA; Death number 5.8 - Died August 5th, 1962&lt;br&gt;&#xA; Last two days hadn't slept - possible reference to her death by overdose on sleeping pills&lt;br&gt;&#xA; Being compared... - refers to Candle in the Wind, by Elton John&lt;br&gt;&#xA; Didn;t listen - didn't listen to her doctor for not taking drugs&lt;br&gt;&#xA; The hint is that my image is poor - she had been struggling with depression and poor self-image&lt;br&gt;&#xA; My name that you know is not what is my - Her real name was Norma Jean Mortenson  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16874" LastEditorUserId="17482" LastEditDate="2015-12-18T17:06:03.557" LastActivityDate="2015-12-18T17:06:03.557" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24687" PostTypeId="2" ParentId="24679" CreationDate="2015-12-18T17:02:41.617" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CHILI DOG NIGHT&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My final answer.&lt;/p&gt;&#xA;" OwnerUserId="5834" LastActivityDate="2015-12-18T17:02:41.617" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24689" PostTypeId="1" AcceptedAnswerId="24690" CreationDate="2015-12-18T19:41:41.337" Score="10" ViewCount="841" Body="&lt;p&gt;A carnival offers a game, called Nerves of Steel. It is played with a standard die, except that the &quot;six&quot; has been replaced with a skull. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You start by paying $7, then rolling the die. Each time you roll a number, that value is added to your running total. After each roll, you can either stop, receiving your running total in dollars, or you can roll again. However, if you ever roll a skull, the game immediately ends while you walk away seven dollars lighter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you rolled a 5, then a 3, then stopped, you would win \$1 (5+3, minus the \$7 fee). If you rolled a 4, then a 4, then a 3, then a skull, you would lose \$7.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What strategy maximizes your expected winnings?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you want, you can also try to find what your expected winnings will actually be when you play optimally, but I wouldn't recommend doing this without a computer.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2016-01-29T01:49:44.093" LastActivityDate="2016-01-29T01:49:44.093" Title="Nerves of Steel (a dice game)" Tags="|strategy|probability|dice|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24690" PostTypeId="2" ParentId="24689" CreationDate="2015-12-18T19:56:43.980" Score="7" Body="&lt;p&gt;Roll until you hit 15 dollars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider the game without an initial fee, your goal is to maximize the amount taken. At every dollar point below 15 dollars, you stand to gain more on average from another roll than what you'd lose from rolling a 6. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;OLD&lt;/strong&gt;&lt;br&gt;&#xA;The average amount of a roll that doesn't skull you is 3. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you roll once, 5/6 of the time you make 3 (-4 overall), and 1/6 of the time you make 0 (-7 overall).&lt;br&gt;&#xA;If you roll twice, 25/36 of the time you make 6 (-1 overall) and 11/36 you make 0 (-7).&lt;br&gt;&#xA;If you roll thrice, 125/216 you make 9 (+2 overall) and 91/216 you make 0 (-7).&lt;br&gt;&#xA;Four rolls, 625/1296 you make 12 (+5 overall) and 671/1296 you make 0 (-7). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In every scenario, you lose money (on average) per roll, so the best way to maximize your earnings is to not play. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's averages, so let's look at discreet numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you roll a 1, a subsequent roll will 5/6 of the time average to 4 (-3), and 1/6 to 0 (-7), and you can't get positive.&lt;br&gt;&#xA;If you roll a 5, a subsequent roll will 5/6 of the time average to 8 (+1), and 1/6 to 0 (-7). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for every roll after the first, your expected earning is 15/6 and your expected loss is (total)/6.  So, if you could somehow always get to 15, you should stop, since your expected roll after that is negative. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-12-18T21:11:23.407" LastActivityDate="2015-12-18T21:11:23.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24691" PostTypeId="2" ParentId="24689" CreationDate="2015-12-18T21:21:53.440" Score="0" Body="&lt;p&gt;&lt;em&gt;Disclaimer:&lt;/em&gt; I've never been particularly good with probability, so please correct me if I'm wrong at any point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wrote a short Python script to generate the expected value after a given number of rolls. The expected value is calculated as the sum of the product of each outcome and its corresponding probability. For example, the expected value after one roll is given by $E=-7+1*1/6+2*1/6+3*1/6+4*1/6+5*1/6+0*1/6=-4.5$. That is, playing this game a large number of times with a single roll would—on average—net you -$4.50 on each game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import itertools&#xA;&#xA;sides = range(1, 7)&#xA;&#xA;for i in xrange(1, 10):&#xA;    rolls = itertools.product(sides, repeat=i)&#xA;    results = []&#xA;    count = 0&#xA;    for item in rolls:&#xA;        count += 1&#xA;        if 6 in item:&#xA;            results.append(-7)&#xA;        else:&#xA;            results.append(-7 + sum(item))&#xA;    print '%i rolls: $%.2f' % (i,  float(sum(results))/len(results))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The output of this script is as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 rolls: $-4.50&#xA;2 rolls: $-2.83&#xA;3 rolls: $-1.79&#xA;4 rolls: $-1.21&#xA;5 rolls: $-0.97&#xA;6 rolls: $-0.97&#xA;7 rolls: $-1.14&#xA;8 rolls: $-1.42&#xA;9 rolls: $-1.77&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It looks like the maximum possible payout—if forced to play the game—is 5 or 6 rolls. However, these are both negative, so your best course of action is to just not play.&lt;/p&gt;&#xA;" OwnerUserId="16275" LastActivityDate="2015-12-18T21:21:53.440" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24692" PostTypeId="2" ParentId="24689" CreationDate="2015-12-18T22:04:36.777" Score="2" Body="&lt;p&gt;A quick Monte Python simulation (hey, I never saw that before!) indicates that @JonTheMon's idea works, at least for those kinds of strategies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Hl9iZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Hl9iZ.png&quot; alt=&quot;Monte Carlo Result&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This graph shows the average payoff after 100,000 games played, where the strategy is to stop if your score is more that or equal to the $x$-axis number (labelled strategy).  This shows a clear peak for a strategy of 15.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the data if you want:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;4 -3.24325&lt;br&gt;&#xA;  5 -2.74864&lt;br&gt;&#xA;  6 -2.23449&lt;br&gt;&#xA;  7 -1.94655&lt;br&gt;&#xA;  8 -1.68452&lt;br&gt;&#xA;  9 -1.43732&lt;br&gt;&#xA;  10 -1.27297&lt;br&gt;&#xA;  11 -1.09576&lt;br&gt;&#xA;  12 -1.01448&lt;br&gt;&#xA;  13 -0.92683&lt;br&gt;&#xA;  14 -0.86798&lt;br&gt;&#xA;  15 -0.82838&lt;br&gt;&#xA;  16 -0.85695&lt;br&gt;&#xA;  17 -0.87132&lt;br&gt;&#xA;  18 -0.94816&lt;br&gt;&#xA;  19 -0.95924&lt;br&gt;&#xA;  20 -1.00816&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using Markov Chains, you can figure out the probabilities after $n$ throws exactly.  Here's a graph showing the probability of different payouts after $n$ throws for $1\leq n\leq10$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZkzYy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZkzYy.png&quot; alt=&quot;Payoff calculated with markov chains&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2015-12-18T22:59:37.590" LastActivityDate="2015-12-18T22:59:37.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24693" PostTypeId="2" ParentId="24689" CreationDate="2015-12-18T22:16:37.147" Score="2" Body="&lt;p&gt;This is the same as other answers but this is (to me) an easier way to see it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are at a point where you will get \$$S$ if you quit right now.  For instance, before your first roll, $S=7$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your expected profit by rolling again is: $$1*\frac 1 6+2*\frac 1 6+3*\frac 1 6+4*\frac 1 6+5*\frac 1 6-S*\frac 1 6 =(1+2+3+4+5-S)/6 = (15-S)/6$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: If $S&amp;lt;15$ you earn a profit on average by rolling again so should.  If $S&amp;gt;15$ you lose money if you roll again so shouldn't.  If $S=15$ you are cost neutral so you are wasting time by rolling again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To maximize rate of profit generation: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Pay your $7 to play&lt;/li&gt;&#xA;&lt;li&gt;Roll once&lt;/li&gt;&#xA;&lt;li&gt;Quit&lt;/li&gt;&#xA;&lt;li&gt;Repeat from step 1 indefinitely&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="99" LastActivityDate="2015-12-18T22:16:37.147" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24694" PostTypeId="1" AcceptedAnswerId="24695" CreationDate="2015-12-18T23:30:03.807" Score="14" ViewCount="1164" Body="&lt;p&gt;&lt;em&gt;&lt;sub&gt;This is in the spirit of the &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%22what+is+a%22+word+is%3Aquestion&quot;&gt;What is a Word/Phrase™&lt;/a&gt; series started by &lt;a href=&quot;https://puzzling.stackexchange.com/users/463&quot;&gt;JLee&lt;/a&gt; with a special brand of &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%22Phrase%22%20user%3A463%20is%3Aquestion&quot;&gt;Phrase™&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=%22What+is+%22+%22Word%22+-missing+user%3A463+is%3Aquestion&quot;&gt;Word™&lt;/a&gt; puzzles.&lt;/sub&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;If a word conforms to a certain rule, I call it a &lt;strong&gt;Sound Word™&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Use the examples below to find the rule.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KqoX9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KqoX9.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;For those unable to view images, here is a text version of the table:&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;Sound Words™&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;Not Sound Words™&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;DECAY&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;DECOMPOSE&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;QUEUE&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;LINE&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;EIGHTY&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;FOURSCORE&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;ENEMY&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;FOE&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;EXPEDIENCY&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;EFFICIENCY&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;ESCAPEE&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;FUGITIVE&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;TEDIOUS&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;BORING&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;DEVIOUS&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;TRICKY&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;EXCELLENCY&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;EMINENCE&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;IVY&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;VINE&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;" OwnerUserId="4421" LastEditorUserId="69582" LastEditorDisplayName="user14478" LastEditDate="2022-01-27T22:35:38.990" LastActivityDate="2022-01-27T22:35:38.990" Title="What is a Sound Word™?" Tags="|pattern|language|word-property|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24695" PostTypeId="2" ParentId="24694" CreationDate="2015-12-19T00:18:05.137" Score="23" Body="&lt;p&gt;A Sound Word is a word whose every syllable sounds like a letter of the alphabet.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Decay: DK&lt;/li&gt;&#xA;&lt;li&gt;Queue: Q&lt;/li&gt;&#xA;&lt;li&gt;Eighty: AT&lt;/li&gt;&#xA;&lt;li&gt;Enemy: NME&lt;/li&gt;&#xA;&lt;li&gt;Expediency: XPDNC&lt;/li&gt;&#xA;&lt;li&gt;Escapee: SKP&lt;/li&gt;&#xA;&lt;li&gt;Tedious: TDS&lt;/li&gt;&#xA;&lt;li&gt;Devious: DVS&lt;/li&gt;&#xA;&lt;li&gt;Excellency: XLNC&lt;/li&gt;&#xA;&lt;li&gt;Ivy: IV&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7459" LastActivityDate="2015-12-19T00:18:05.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24696" PostTypeId="1" AcceptedAnswerId="24701" CreationDate="2015-12-19T00:54:27.197" Score="2" ViewCount="476" Body="&lt;p&gt;&lt;strong&gt;The Challenge&lt;/strong&gt;&lt;br&gt;&#xA;Using only lands and Goblins, your goal is to build a Magic: the Gathering deck that can win the game as quickly as possible, against an opponent who does nothing. You may assume that your deck is stacked, so that you will always draw exactly the cards you want. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Scoring&lt;/strong&gt;&lt;br&gt;&#xA;The winning answer will be the one which can defeat the opponent in as few turns as possible. You may choose to play or draw, but winning on the play is considered faster than winning on the draw.&lt;br&gt;&#xA;In the event that two solutions are equally fast, the winner will be the solution that deals more damage.&lt;br&gt;&#xA;In the event that two solutions are equally fast and equally damaging, the winner will be the one that uses less cards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Possibly Asked Questions&lt;/strong&gt;&lt;br&gt;&#xA;When you say Goblins...&lt;br&gt;&#xA;Any card with the subtype Goblin. So no &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=8902&quot; rel=&quot;nofollow&quot;&gt;Goblin Game&lt;/a&gt; or &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=398647&quot; rel=&quot;nofollow&quot;&gt;Dragon Fodder&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the opponent doing?&lt;br&gt;&#xA;The opponent's deck consists of sixty islands, and they begin the game by mulliganing to zero. They will take no game actions unless an effect requires them to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Loopholes?&lt;br&gt;&#xA;Your solution may not rely upon random chance such as winning coinflips, or the opponent doing something suicidal, like choosing 20 on &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=88803&quot; rel=&quot;nofollow&quot;&gt;Choice of Damnations&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7459" LastActivityDate="2015-12-22T17:32:02.380" Title="Magic: the Gathering - Goblin Horde" Tags="|calculation-puzzle|magic-the-gathering|" AnswerCount="3" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="24697" PostTypeId="2" ParentId="24696" CreationDate="2015-12-19T04:01:27.077" Score="1" Body="&lt;p&gt;I haven't touched a Magic card for about 6 years, but I'll take a shot at setting a base. Choose to play.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Play a mountain and cast &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=382959&quot; rel=&quot;nofollow&quot;&gt;Goblin Lackey&lt;/a&gt;&lt;/li&gt;&#xA;  &lt;li&gt;Play a mountain and cast &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=398537&quot; rel=&quot;nofollow&quot;&gt;Goblin Piledriver&lt;/a&gt;; attack for 1, putting &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=393982&quot; rel=&quot;nofollow&quot;&gt;Siege-Gang Commander&lt;/a&gt; into the battlefield (with 3 1/1 goblin tokens).&lt;/li&gt;&#xA;  &lt;li&gt;Play a mountain and cast &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=221201&quot; rel=&quot;nofollow&quot;&gt;Goblin Chieftain&lt;/a&gt;; attack for 23, 24 total damage, 7 cards played.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;P.S. Wow. Magic got scary when I stopped playing, and most of my older cards appear to have been made obsolete by strictly better alternatives.&lt;/p&gt;&#xA;" OwnerUserId="13237" LastEditorUserId="13237" LastEditDate="2015-12-19T04:39:30.213" LastActivityDate="2015-12-19T04:39:30.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24698" PostTypeId="2" ParentId="24696" CreationDate="2015-12-19T05:16:45.143" Score="1" Body="&lt;p&gt;I'm assuming I can use any land I want, since it doesn't say I can only use Mountains or basics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose to be on the play.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Turn 1&lt;/strong&gt;: Mountain, &lt;a href=&quot;http://magiccards.info/us/en/190.html&quot; rel=&quot;nofollow&quot;&gt;Goblin Lackey&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Turn 2&lt;/strong&gt;: &lt;a href=&quot;http://magiccards.info/tp/en/305.html&quot; rel=&quot;nofollow&quot;&gt;Ancient Tomb&lt;/a&gt;, &lt;a href=&quot;http://magiccards.info/m15/en/145.html&quot; rel=&quot;nofollow&quot;&gt;Goblin Rabblemaster&lt;/a&gt;. Attack with Lackey and the Rabblemaster token for 2. Put &lt;a href=&quot;http://magiccards.info/chk/en/175.html&quot; rel=&quot;nofollow&quot;&gt;Kiki-Jiki, Mirror Breaker&lt;/a&gt; into play with the Lackey trigger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On your opponent's end step, use Kiki-Jiki to make a copy of Goblin Rabblemaster (so it will survive to your next turn).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Turn 3&lt;/strong&gt;: Mountain, &lt;a href=&quot;http://magiccards.info/m10/en/139.html&quot; rel=&quot;nofollow&quot;&gt;Goblin Chieftain&lt;/a&gt;. Use Kiki-Jiki to copy Goblin Rabblemaster. Go to combat, make 3 more tokens from Rabblemasters, attack your opponent with 6 2/2's and 3 11/3's for 45 damage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;47 total damage, 7 cards played.&lt;/p&gt;&#xA;" OwnerUserId="17559" LastEditorUserId="17559" LastEditDate="2015-12-21T13:52:46.647" LastActivityDate="2015-12-21T13:52:46.647" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24701" PostTypeId="2" ParentId="24696" CreationDate="2015-12-19T06:57:51.603" Score="5" Body="&lt;p&gt;Got it down to turn 1 on the draw if we allow Mons's Goblin Waiters!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Turn 1 - infinite damage on the draw (16 cards used)&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Turn 0:&lt;br&gt;&#xA;&lt;br&gt;&#xA; Gemstone Caverns, exiling Mons's Goblin Raiders for irony (1 card &quot;used&quot;)&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Turn 1:&lt;br&gt;&#xA;&lt;br&gt;&#xA; City of Traitors, tap both lands for 2R (2R, 2 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play Mons's Goblin Waiters for R, sacrifice both lands for R (2R, 3 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play Skirk Prospector for R, sacrifice Mons's Goblin Waiters for R (2R, 4 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play Mogg War Marshal for 1R, sacrifice it and both tokens for RRR (1RRR, 5 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play Mogg War Marshal for 1R, sacrifice it and both tokens for RRR (RRRRR, 6 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play Goblin Ringleader for RRRR, revealing Mogg War Marshal, Goblin Lackey, Goblin Bushwhacker, and Grenzo, Dungeon Warden, then sacrifice it for R (RR, 7 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play Mogg War Marshal for RR, sacrifice it and both tokens for RRR (RRR, 8 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play Goblin Lackey for R (RR, 9 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play Goblin Bushwhacker, Kicked, for RR (10 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Attack with Goblin Lackey for 2 damage, put Grenzo, Dungeon Warden into play (11 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Sacrifice Goblin Lackey and Goblin Bushwhacker for RR (RR, 11 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Pay RR to Grenzo, revealing Siege-Gang Commander, put it into play, sacrifice it and all three tokens for RRRR (RRRR, 12 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Pay RR to Grenzo, revealing Siege-Gang Commander, put it into play, sacrifice it and all three tokens for RRRR (RRRRRR, 13 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Pay RR to Grenzo, revealing Goblin Sledder, put it into play, sacrifice it to give Grenzo +1/+1 (RRRR, 14 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Pay RR to Grenzo, revealing Kiki-Jiki, Mirror Breaker, put it into play (RR, 15 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Pay RR to Grenzo, revealing Lightning Crafter, put it into play, &lt;em&gt;but hold the ETB trigger on the stack...&lt;/em&gt; (16 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Tap Kiki-Jiki, targeting Lightning Crafter to create a token, champion Kiki-Jiki with the token, tap the token to deal your opponent 3 damage, then sacrifice the token for R, returning Kiki-Jiki to the battlefield untapped&lt;br&gt;&#xA;&lt;br&gt;&#xA; Repeat the step above infinite times, dealing your opponent infinite damage! &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without Mons's, the best is turn two:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Turn 2 (&lt;em&gt;pre-combat!&lt;/em&gt;) - infinite damage on the play (12 cards used)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Turn 1:&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=23238&quot; rel=&quot;nofollow&quot;&gt;Geothermal Crevice&lt;/a&gt; tapped (1 card used)&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Turn 2:&lt;br&gt;&#xA;&lt;br&gt;&#xA; Tap and sacrifice Geothermal Crevice, play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=152587&quot; rel=&quot;nofollow&quot;&gt;Frogtosser Banneret&lt;/a&gt; for BG (2 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Mountain, play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=77919&quot; rel=&quot;nofollow&quot;&gt;Akki Rockspeaker&lt;/a&gt; for R, gaining R (R, 4 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=393985&quot; rel=&quot;nofollow&quot;&gt;Skirk Prospector&lt;/a&gt; for R (5 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Sacrifice Akki Rockspeaker to gain R, play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=393973&quot; rel=&quot;nofollow&quot;&gt;Mogg War Marshal&lt;/a&gt; for R (6 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Sacrifice Mogg War Marshal and both tokens for RRR (RRR, 6 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=393973&quot; rel=&quot;nofollow&quot;&gt;Mogg War Marshal&lt;/a&gt; for R (RR, 7 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Sacrifice Mogg War Marshal and both tokens for RRR (RRRRR, 7 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=393963&quot; rel=&quot;nofollow&quot;&gt;Goblin Ringleader&lt;/a&gt; for RRR, revealing Mogg War Marshal x2, Kiki-Jiki Mirror Breaker, Lightning Crafter (RR, 8 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=393973&quot; rel=&quot;nofollow&quot;&gt;Mogg War Marshal&lt;/a&gt;, &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=393973&quot; rel=&quot;nofollow&quot;&gt;Mogg War Marshal&lt;/a&gt; for RR (10 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Sacrifice both Mogg War Marshals and all four tokens for RRRRRR (RRRRRR, 10 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Sacrifice Goblin Ringleader for R (RRRRRRR, 10 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=397698&quot; rel=&quot;nofollow&quot;&gt;Kiki-Jiki, Mirror Breaker&lt;/a&gt; for RRRR (RRR, 11 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=152893&quot; rel=&quot;nofollow&quot;&gt;Lightning Crafter&lt;/a&gt; for RRR, &lt;em&gt;but hold the ETB trigger on the stack&lt;/em&gt; (12 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Tap Kiki-Jiki targeting Lightning Crafter to make a token, champion Kiki-Jiki with the token.  Tap the token to deal 3 damage to your opponent, then sacrifice the token to Skirk Prospector for R, returning Kiki-Jiki to the battlefield (untapped)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Repeat the above step infinite times, dealing infinite damage to your opponent!&lt;br&gt;&#xA; (With the original Lightning Crafter trigger still on the stack)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As well as the most efficient kill:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Turn 2 - infinite damage on the play (6 cards used)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Turn 1:&lt;br&gt;&#xA;&lt;br&gt;&#xA; Mountain, &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=382959&quot; rel=&quot;nofollow&quot;&gt;Goblin Lackey&lt;/a&gt; (2 cards used)&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Turn 2:&lt;br&gt;&#xA;&lt;br&gt;&#xA; Tap Mountain for R, play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=393985&quot; rel=&quot;nofollow&quot;&gt;Skirk Prospector&lt;/a&gt; (3 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Attack with Goblin Lackey for 1 damage, put &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=397698&quot; rel=&quot;nofollow&quot;&gt;Kiki-Jiki, Mirror Breaker&lt;/a&gt; into play (4 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=10422&quot; rel=&quot;nofollow&quot;&gt;Gaea's Cradle&lt;/a&gt;, Tap for GGG (GGG floating, 5 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Sacrifice Goblin Lackey to Skirk Prospector for R (GGGR, 5 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=152893&quot; rel=&quot;nofollow&quot;&gt;Lightning Crafter&lt;/a&gt; for GGGR, &lt;em&gt;but hold the ETB trigger on the stack&lt;/em&gt; (6 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Tap Kiki-Jiki targeting Lightning Crafter to make a token, champion Kiki-Jiki with the token.  Tap the token to deal 3 damage to your opponent, then sacrifice the token to Skirk Prospector for R, returning Kiki-Jiki to the battlefield (untapped)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Repeat the above step infinite times, dealing infinite damage to your opponent!&lt;br&gt;&#xA; (With the original Lightning Crafter trigger still on the stack)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Optimizing for card efficiency another way:   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Turn 2 - infinite damage on the play (5 cards from hand)&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm proud to have broken &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=142363&quot; rel=&quot;nofollow&quot;&gt;Goatnapper&lt;/a&gt;, although the consequences may be tragic - after having proven that it facilitates a practically unbeatable turn two infinite combo deck, I am sure it will shortly be banned from Legacy, and we will all be poorer for the loss.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Turn 1:&lt;br&gt;&#xA;&lt;br&gt;&#xA; Mountain, &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=382959&quot; rel=&quot;nofollow&quot;&gt;Goblin Lackey&lt;/a&gt; (2 cards played)&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Turn 2:&lt;br&gt;&#xA;&lt;br&gt;&#xA; Attack with Goblin Lackey for 1 damage, put &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=382276&quot; rel=&quot;nofollow&quot;&gt;Grenzo, Dungeon Warden&lt;/a&gt; into play (3 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Tap Mountain for R, play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=393985&quot; rel=&quot;nofollow&quot;&gt;Skirk Prospector&lt;/a&gt; (4 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=10422&quot; rel=&quot;nofollow&quot;&gt;Gaea's Cradle&lt;/a&gt;, Tap for GGG (GGG floating, 5 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Pay GG to Grenzo, revealing &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=393982&quot; rel=&quot;nofollow&quot;&gt;Siege-Gang Commander&lt;/a&gt;, put it into play, get three Goblin tokens&lt;br&gt;&#xA;&lt;br&gt;&#xA; Sacrifice Siege-Gang Commander and all three tokens to Skirk Prospector for RRRR (GRRRR, 5 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Sacrifice Goblin Lackey to Skirk Prospector for R (GRRRRR, 5 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Pay GR to Grenzo, revealing &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=207921&quot; rel=&quot;nofollow&quot;&gt;Amoeboid Changeling&lt;/a&gt;, put it into play&lt;br&gt;&#xA;&lt;br&gt;&#xA; Pay RR to Grenzo, revealing &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=142363&quot; rel=&quot;nofollow&quot;&gt;Goatnapper&lt;/a&gt;, put it into play targeting Amoeboid Changeling (a Goat, as well as a Goblin), untapping it, gaining control of it, and giving it haste&lt;br&gt;&#xA;&lt;br&gt;&#xA; Pay RR to Grenzo, revealing &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=397698&quot; rel=&quot;nofollow&quot;&gt;Kiki-Jiki, Mirror Breaker&lt;/a&gt;, put it into play&lt;br&gt;&#xA;&lt;br&gt;&#xA; Tap Amoeboid Changeling (haste from Goatnapper) to give Kiki-Jiki all creature types until end of turn&lt;br&gt;&#xA;&lt;br&gt;&#xA; Tap Kiki-Jiki, targeting Goatnapper, creating a token with haste.  Target Kiki-Jiki with the Goatnapper token trigger (Kiki-Jiki is currently a Goat as well as a Goblin), untap and gain control of Kiki-Jiki (as well as giving it more haste)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Repeat the above step infinite times for infinite Goatnapper tokens, leaving Kiki-Jiki untapped&lt;br&gt;&#xA;&lt;br&gt;&#xA; Sacrifice your infinite Goatnapper tokens for infinite R from Skirk Prospector (Infinite R, 5 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Pay RR to Grenzo, revealing &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=393964&quot; rel=&quot;nofollow&quot;&gt;Goblin Sledder&lt;/a&gt;, put it into play (Infinite R, 5 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Sacrifice Skirk Prospector and Goblin Sledder, giving Grenzo +2/+2 until end of turn&lt;br&gt;&#xA;&lt;br&gt;&#xA; Pay RR to Grenzo, revealing &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=42274&quot; rel=&quot;nofollow&quot;&gt;Goblin Dynamo&lt;/a&gt;, put it into play (Infinite R, 5 cards)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Tap Kiki-Jiki, targeting Goblin Dynamo, creating a token copy with haste&lt;br&gt;&#xA;&lt;br&gt;&#xA; Tap and sacrifice the Goblin Dynamo token with X = infinite, dealing infinite damage to your opponent!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="17389" LastEditDate="2015-12-22T17:32:02.380" LastActivityDate="2015-12-22T17:32:02.380" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24702" PostTypeId="1" AcceptedAnswerId="24703" CreationDate="2015-12-19T08:15:53.440" Score="8" ViewCount="403" Body="&lt;p&gt;Is it possible to dissect a $7\times9$ rectangle into $21$ pieces that are $L$-shaped and that consist of three little squares? &lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="10615" LastEditDate="2015-12-19T14:32:52.653" LastActivityDate="2015-12-19T14:32:52.653" Title="Cutting a 7-by-9 rectangle" Tags="|mathematics|geometry|dissection|tiling|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24703" PostTypeId="2" ParentId="24702" CreationDate="2015-12-19T08:37:15.733" Score="12" Body="&lt;p&gt;Answer blatantly stolen from &lt;a href=&quot;https://math.stackexchange.com/a/1580996/256930&quot;&gt;this Math.SE post&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/4QyhJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4QyhJ.png&quot; alt=&quot;tiling&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="-1" LastEditDate="2017-04-13T12:19:36.950" LastActivityDate="2015-12-19T08:37:15.733" CommentCount="1" CommunityOwnedDate="2015-12-19T08:37:15.733" ContentLicense="CC BY-SA 3.0" />
  <row Id="24704" PostTypeId="1" CreationDate="2015-12-19T10:05:35.430" Score="7" ViewCount="465" Body="&lt;p&gt;So I have this riddle which contains a .docx file I opened and extracted the file and got the following contents:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Cn5tv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Cn5tv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the file it was written &quot;There is nothing here. But there is something inside&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also have another hint &quot;The doc is a lie. Needle in a haystack.&quot; Please help me solve this puzzle.&lt;/p&gt;&#xA;" OwnerUserId="17862" LastEditorUserId="11876" LastEditDate="2016-01-02T12:47:07.287" LastActivityDate="2016-01-02T12:47:07.287" Title="Need Help In This Riddle: &quot;The Doc Is A Lie&quot;" Tags="|steganography|" AnswerCount="0" CommentCount="5" ClosedDate="2016-02-23T04:22:37.937" ContentLicense="CC BY-SA 3.0" />
  <row Id="24705" PostTypeId="1" AcceptedAnswerId="24706" CreationDate="2015-12-19T13:27:21.497" Score="10" ViewCount="462" Body="&lt;p&gt;In how many ways can you construct a 6x6 chart with only 1 and -1 such that in every row and column the product is always positive.&lt;/p&gt;&#xA;" OwnerUserId="17755" LastActivityDate="2015-12-19T15:23:04.153" Title="Multiplications in a chart, problem solving" Tags="|combinatorics|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24706" PostTypeId="2" ParentId="24705" CreationDate="2015-12-19T14:23:29.307" Score="18" Body="&lt;p&gt;The number of ways is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $2^{25}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The idea is that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you can put whatever you want in the upper left 5 by 5 square, but then there is a unique way to complete this to a valid 6 by 6 square.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is explained in the below picture:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/hQ6ND.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hQ6ND.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-12-19T15:23:04.153" LastActivityDate="2015-12-19T15:23:04.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24707" PostTypeId="1" AcceptedAnswerId="25746" CreationDate="2015-12-19T17:56:26.127" Score="3" ViewCount="350" Body="&lt;p&gt;This puzzle continues the puzzle&#xA;&quot;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/24139/state-the-rule&quot;&gt;State the rule!&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;State the rule behind the following partition of letters, and fill &#xA;in the questionmarks. (The puzzle has a clear and justifiable trivia-type solution.)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;      A . C . E . G H . J . . M . . . . R S . . V W ? ? ? &#xA;   ---------------------------------------------------------&#xA;      . B . D . F . . I . K L . N O P Q . . T U . . ? ? ? &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2016-01-27T15:02:15.730" Title="Again: State the rule!" Tags="|pattern|knowledge|letter-sequence|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24708" PostTypeId="2" ParentId="8604" CreationDate="2015-12-19T20:29:15.437" Score="2" Body="&lt;p&gt;Geometrically, it is the TOP face that is the value, and a corresponding BOTTOM face that rests on a surface (a table). A die consists of numbered faces each of the same size. Since each side must have a corresponding face on the opposite sides, there is no circumstance where an odd number would not end with a vertex on top (unless somehow the die rested on its edge.) Instead, use the suggestion above and use an even number of faces that are labeled the same, i.e the top and bottom always are the same label. In this fashion you can make &quot;fair dice&quot; for any value &gt;2&lt;/p&gt;&#xA;" OwnerUserId="17866" LastActivityDate="2015-12-19T20:29:15.437" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24709" PostTypeId="1" CreationDate="2015-12-19T21:08:57.383" Score="0" ViewCount="335" Body="&lt;p&gt;I learned how to solve a rubik's cube this summer. On my first try (after memorizing all the algorithms) I got a sub-50. Now, my personal best is 31 seconds. I tried learning the Fridrich method but I don't ever have the time to sit and watch a tutorial. I use the beginners method. Though sometimes I do 1-look instead of 2. Any ways on how to get faster?&lt;/p&gt;&#xA;" OwnerUserId="17867" LastEditorUserId="17482" LastEditDate="2015-12-19T22:22:55.097" LastActivityDate="2018-04-17T14:29:30.690" Title="Easy ways to solve" Tags="|rubiks-cube|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24710" PostTypeId="2" ParentId="24044" CreationDate="2015-12-19T23:31:52.167" Score="5" Body="&lt;p&gt;We will prove something slightly stronger that implies 2012rcampion's solution is corrrect.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $[x_n,y_n,x_{n-1},y_{n-1}, \dots, x_1,y_1,x_0]$ be the binary number $p$ given by $$\text{$x_n$ 0's , $y_n$ 1's , $x_{n-1}$ 0's , $y_{n-1}$ 1's , $\dots$ , $x_1$ 0's , $y_1$ 1's , $x_0$ 0's}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Then the numerator of the simplified fraction that is equal to the continued fraction$$[x_n;y_n,x_{n-1},y_{n-1}, \dots ,x_1,y_1]$$is precisely the number of distinct ways of writing an equality $p +q = r$, where $q$ and $r$ are never $1$ in the same binary digit and are always $0$ on every binary digit beyond $x_n$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It should be clear how every solution as in the statement corresponds to a solution on the scale: $p$ is the weight of the coin, and $q, r$ provide the choice of weights on each pan of the balance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, we assume without loss of generality that $x_n, y_n \geq 1$, except $x_0$ may be $0$. It can be easily seen the coin may not weigh $2^{2014}$, so the assumption is justified (its last bit must be $0$, and its first bit need not be $0$), and the weight of the coin can always be written as some $p$ in the statement. This also shows why we must consider, as 2012rcampion did, only fractions with $2015$ in the numerator that are greater than one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof will be by induction.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Base case: $n=1$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In this case, $p$ is of the form$$\underbrace{00\dots 0}_{x_1}\underbrace{11\dots 1}_{y_1}\underbrace{00\dots 0}_{x_0}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our reasoning will go as follows: we will analyze how the conditions on $q$ and $r$ shape a solution, going from the rightmost digit of $q$ and proceeding to the left. We will consider possibilities for $q$ that will automatically ensure the conditions on $q$ and $r=p+q$ are satisfied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any solution $(p,q,r)$, the fist $x_0$ digits of $q$ must be $0$, for otherwise condition that $q$ and $r$ not be $1$ in the same digit would be violated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we may choose any of the next $y_1$ digits of $q$ to be $1$, and this will create a domino effect of carry-overs, imposing that all other of these $y_1$ digits be $0$ (so as not to violate the digit condition). Of course, we may also choose to have all of these $y_1$ digits be $0$, and as in the preceding paragraph this will make it so that all of the next $x_1$ digits also be $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have thus found one solution (all zeroes) and $y_1$ 'partial' solutions, with a $1$ in one of the digits corresponding to $y_1$ and $0$ at the other digits. We need only find the possibilities for the last $x_1$ digits of the partial solutions. These last digits of the sum $p+q$ currently look like this:$$\underbrace{00\dots 0}_{x_1-1}1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We may choose to have the corresponding rightmost digit in $q$ be $0$, and this will impose that all other digits be also $0$; or we may choose it be $1$, and this will create a carry over. In this latter case, the new last digits of the sum have a very similar form:$$\underbrace{00\dots 0}_{x_1-2}1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuing in this fashion, we find there are $x_1$ fits to the $y_1$ partial solutions we initially encountered. These fits correspond to choosing one of the last $x_1$ digits and assigning $1$ to all digits that (strictly) precede it in the segment, and $0$ to all others. Notice in particular that the last digit will always be $0$, even when all others that precede it are $1$, because of the condition that binary digits beyond $x_n$ be $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are thus $x_1y_1 +1$ solutions. Now, consider the continued fraction of the statement. We have$$[x_1;y_1]=x_1+\frac{1}{y_1}=\frac{x_1y_1+1}{y_1},$$so the numerator of the corresponding simplified fraction is in agreement, as claimed.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Inductive step: suppose the claim holds for all $m$ with $1 \leq m \leq n-1$ and let $p$ be given by $[x_n,y_n, \dots, x_1,y_1,x_0]$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We will analyze the shape of the initial digits of $q$, like we did in the previous case. Its first $x_0$ digits need once again be $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we choose the next $y_1$ of its digits to be $0$, then the next $x_1$ of its digits will also have to be $0$, and we will have found one partial solution, complete until the digits that correspond to $y_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like before, we may also choose one of the $y_1$ digits to be $1$, and this will create a domino carry-over effect. Any of these $y_1$ choices will result in a sum that looks like this:$$\dots \underbrace{00\dots 0}_{x_2} \underbrace{11\dots 1}_{y_2}\underbrace{00\dots 0}_{x_1-1}1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As in the previous case, we may choose the rightmost digit in the segment to be $0$ or $1$, where choosing $0$ implies that all the next digits (until the $y_2$ segment) also be $0$, and choosing $1$ implies a new choice must be made for the next rightmost digit (with similar implications).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This results in $x_1-1$ fits that end in $0$ (so the next segment of $1$'s in $p+q$ has length $y_2$) and a single fit that ends in $1$ (so the next segment of $1$'s in $p+1$ has length $y_2+1$). We may group the former with our lone previous partial solution, for a total $y_1(x_1-1)+1$ partial solutions that need a fit for the binary number $[x_n,y_n, \dots, x_2,y_2]$, and $y_1$ partial solutions that need a fit for the binary number $[x_n,y_n, \dots, x_2,y_2+1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We may apply the inductive hypothesis to these fits, but we will use &lt;a href=&quot;https://en.wikipedia.org/wiki/Continued_fraction#Some_useful_theorems&quot;&gt;some results on continued fractions&lt;/a&gt;. Consider the sequence of positive integers $(a_j)$ given by the $x_i$'s and $y_i$'s:&#xA;\begin{array}{ccccccccc}&#xA;x_n&amp;amp;y_n&amp;amp;x_{n-1}&amp;amp;y_n{-1}&amp;amp;\dots &amp;amp;x_2&amp;amp;y_2&amp;amp;x_1&amp;amp;y_1\\&#xA;a_0&amp;amp;a_1&amp;amp;a_2&amp;amp;a_3&amp;amp; \dots &amp;amp;a_{l}&amp;amp;a_{l+1}&amp;amp;a_{l+2}&amp;amp;a_{l+3}&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $h_j$ and $k_j$ be the corresponding recursive sequences (see the link). Then we have that$$[x_n;y_n, \dots, x_2,y_2]=[a_0;a_1, \dots, a_l,a_{l+1}]\\&#xA;[x_n;y_n, \dots, x_2,y_2+1]=[a_0;a_1, \dots, a_l,a_{l+1}+1]=[a_0;a_1, \dots, a_l,a_{l+1},1]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, by Theorems 1 and 2:$$[x_n;y_n, \dots, x_2,y_2]=\frac{h_{l+1}}{k_{l+1}}\\&#xA;[x_n;y_n, \dots, x_2,y_2+1]=\frac{h_{l+1}+h_l}{k_{l+1}+k_l}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Corollary 1 and the Lemma at the end, these fractions are irreducible so we may apply the inductive hypothesis to the (binary) numbers $[x_n,y_n, \dots, x_2,y_2]$ and $[x_n,y_n, \dots, x_2,y_2+1]$, obtaining that the total number of solutions is:&#xA;\begin{align}&amp;amp;(x_1y_1-y_1+1)\cdot h_{l+1} + y_1 \cdot (h_{l+1}+h_l)\\&#xA;= \text{ } &amp;amp;(a_{l+2}a_{l+3}-a_{l+3}+1)\cdot h_{l+1} + a_{l+3} \cdot (h_{l+1}+h_l)\\&#xA;= \text{ } &amp;amp;(a_{l+2}a_{l+3}+1)\cdot h_{l+1} + a_{l+3} \cdot h_l\\&#xA;= \text{ } &amp;amp;a_{l+3}(a_{l+2}\cdot h_{l+1}+h_l)+h_{l+1}\\&#xA;= \text{ } &amp;amp;a_{l+3}\cdot h_{l+2} + h_{l+1}=h_{l+3}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, $h_{l+3}$ is the numerator of the simplified fraction given by$$[x_n;y_n, \dots, x_1,y_1]=[a_0;a_1, \dots, a_{l+2},a_{l+3}]=\frac{h_{l+3}}{k_{l+3}},$$which completes the induction.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Lemma: The fraction $\frac{a \cdot h_{l+1}+h_l}{a \cdot k_{l+1}+k_l}$ is irreducible for all integers $a$ with $0 \leq a \leq a_{l+2}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Indeed, when $a=0$ we have $\frac{h_l}{k_l}$ and when $a=a_{l+2}$ we have $\frac{h_{l+2}}{k_{l+2}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $a$ is in between, it suffices to note that by construction of the sequences $(h_j)$ and $(k_j)$, $a \cdot h_{l+1}+h_l$ is precisely the numerator in the simplification of the continued fraction $[a_0;a_1, \dots, a_l,a_{l+1},a]$, and $a \cdot k_{l+1}+k_l$ is its denominator. Then, when $a\geq 2$ irreducibility follows from the fact that $\frac{1}{a}$ is irreducible, and when $a=1$ it follows from the fact that $[a_0;a_1, \dots, a_l,a_{l+1},a]=[a_0;a_1, \dots, a_l,a_{l+1}+1]$ (and in this case, $a_{l+1}+1\geq 2$, so $\frac{1}{a_{l+1}+1}$ is irreducible).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice the $a_j$ involved are all positive (only $a_{l+4}$ may be zero), so that none of the assumptions of the lemma lead to contradictions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I decided to try and tackle this problem, I had no idea it would be as hard and take as long as it did. I have no idea how you could see this relation between combinatorics of weightings and continued fractions, but I'm glad you did, because my other attempts were not at all feasible, involving recursion that was not very realistic to compute. Your hints were invaluable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, I am very glad to have been shown this. It's one of those mathematical truths that bewilders me by how it unexpectedly relates two things that seem to have nothing at all in common, shedding light on some intricate underlying structure that has always been there, unmindful of whether or not we grasp it.&lt;/p&gt;&#xA;" OwnerUserId="16478" LastEditorUserId="16478" LastEditDate="2015-12-20T02:01:44.400" LastActivityDate="2015-12-20T02:01:44.400" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24711" PostTypeId="2" ParentId="24044" CreationDate="2015-12-19T23:45:57.913" Score="2" Body="&lt;p&gt;Here is my intended solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the number of ways to weigh $(2k+1)\times2^n$. Any weighing must involve the weight $2^n$ and no smaller weights. If $2^n$ is on the left, the remaining weights form a weighing of $(2k+2)\times2^n$. If $2^n$ is on the right, the remaining weights form a weighing of $2k\times2^n$. Therefore, the number of ways to weigh $(2k+1)\times2^n$ is equal to the number of ways to weigh $2k\times2^n$, plus the number of ways to weigh $(2k+2)\times2^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we start with $0$ having 1 weighing and $2^{2015}$ having 0 weighings, we can generate the number of weighings for every integer in between by adding neighboring terms. This corresponds exactly to the way the &lt;a href=&quot;https://en.wikipedia.org/wiki/Stern%E2%80%93Brocot_tree&quot; rel=&quot;nofollow&quot;&gt;Stern-Brocot tree&lt;/a&gt; is generated. Therefore the number of ways to weigh $x$ is precisely the denominator of the  $x$th (zero-indexed) term in the Stern-Brocot tree with depth $2015$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The weight $2^{2014}$ corresponds to the fraction $\frac{1}{1}$, so we want to count the number of fractions in the Stern-Brocot tree that are less than $1$ and have a denominator of $2015$. There are $\phi(2015)$ such fractions, one for each possible numerator coprime to $2015$ (the tree is deep enough to contain them all), so the answer is $1440$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The hints about the continued fractions were detailing a way to calculate the specified term in the Stern-Brocot tree, although the fractions were reversed for simplicity.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2015-12-19T23:45:57.913" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24712" PostTypeId="1" CreationDate="2015-12-20T03:40:49.060" Score="0" ViewCount="71" Body="&lt;p&gt;This question was asked by teacher, a very simple one but still highly puzzling. Could you find the answer?&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;There were three persons, &lt;strong&gt;A&lt;/strong&gt; :- The Costumer, &lt;strong&gt;B&lt;/strong&gt; :- The Cloth Shop Owner and &lt;strong&gt;C&lt;/strong&gt; :- A boot shop owner. The question goes:&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;A&lt;/strong&gt; purchased a cloth from &lt;strong&gt;B&lt;/strong&gt; costing 200 rupees. Since &lt;strong&gt;A&lt;/strong&gt; didn't had change he gave &lt;strong&gt;B&lt;/strong&gt; a 1000 rupees note. &lt;strong&gt;B&lt;/strong&gt; didn't had any change too. He went to &lt;strong&gt;C&lt;/strong&gt; and brought 1000 in change and gave the remaining 800 rupees to &lt;strong&gt;A&lt;/strong&gt; and &lt;strong&gt;A&lt;/strong&gt; went away. After some time &lt;strong&gt;C&lt;/strong&gt; went to &lt;strong&gt;B&lt;/strong&gt; and said that the 1000 rupee note was fake. &lt;strong&gt;B&lt;/strong&gt; returned 1000 rupees to &lt;strong&gt;C&lt;/strong&gt;. How much did &lt;strong&gt;B&lt;/strong&gt; Lose?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Sorry if the statements were bit puzzling. Ask if it is so.&lt;/p&gt;&#xA;" OwnerUserId="16628" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-12-20T04:10:55.570" Title="A very Puzzling Loss" Tags="|calculation-puzzle|" AnswerCount="0" CommentCount="2" ClosedDate="2015-12-20T04:17:19.360" ContentLicense="CC BY-SA 3.0" />
  <row Id="24714" PostTypeId="1" AcceptedAnswerId="24717" CreationDate="2015-12-20T08:17:47.783" Score="19" ViewCount="473" Body="&lt;p&gt;Is it possible to arrange the integers $1,2,3,\ldots,169$ in a $13\times13$ square, so that in every $2\times2$ square the sum of the four numbers is divisible by $170$?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2015-12-20T16:56:36.040" Title="Filling a 13-by-13 square" Tags="|mathematics|combinatorics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24715" PostTypeId="1" CreationDate="2015-12-20T09:59:15.900" Score="6" ViewCount="408" Body="&lt;p&gt;So, one of my friends asked me this, because he knew I'd read The Hitchhiker's Guide to The Galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He sent me a picture of a piece of paper. It read like this;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                   42&#xA;                258-17-7&#xA;                424-13-9&#xA;                 23-4-18&#xA;                 24-1-17&#xA;                484-1-5&#xA;                  1-1-4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So I have no clue, except the first one, which is obvious&lt;/p&gt;&#xA;" OwnerUserId="17873" LastActivityDate="2015-12-20T10:18:53.697" Title="Number puzzle related to the Hitchiker's Guide" Tags="|cipher|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24716" PostTypeId="2" ParentId="24715" CreationDate="2015-12-20T10:18:53.697" Score="3" Body="&lt;p&gt;This looks like a &lt;a href=&quot;https://en.wikipedia.org/wiki/Book_cipher&quot; rel=&quot;nofollow&quot;&gt;book cipher&lt;/a&gt;, where the first 42 clues The Hitchhiker's Guide To The Galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first number represents the page, the second number represents the line (or maybe sentence), and the third number represents the word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, it'll be unsolvable if you have a different type of copy (electronic, hardback, or paperback) because the pages will be different sizes.&lt;/p&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2015-12-20T10:18:53.697" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="24717" PostTypeId="2" ParentId="24714" CreationDate="2015-12-20T11:06:38.833" Score="13" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; yes &lt;a href=&quot;https://i.stack.imgur.com/EUDra.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EUDra.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;Basically we just have ordered pairs of 170 alternating with high-low left/right. The pattern to figure out which pairs go where is very simple as well: We start with the square 1 169, 168 2. If we take 2 steps to the right, we do -13 if the original was bigger then 85, and +13 if it was smaller. If we take 2 steps down, we do -2 if the original was bigger then 85, and +2 if it was smaller. This way the whole 13x13 square can be filled as in the image. The numbers below are all possible sums and they are (ofcourse) all 340.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16846" LastEditorUserId="16846" LastEditDate="2015-12-20T16:56:36.040" LastActivityDate="2015-12-20T16:56:36.040" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24718" PostTypeId="1" CreationDate="2015-12-20T12:02:12.757" Score="5" ViewCount="191" Body="&lt;p&gt;A black and a white piece is in two adjacent corners of a 12-gon. In a move, we get to move any piece to any vacant neighboring corners. If both of the pieces returns to a position which they have been before, the game ends.  How many moves can we at most do?&lt;/p&gt;&#xA;" OwnerUserId="17755" LastActivityDate="2015-12-20T17:15:04.327" Title="Problem solving, two pieces on a 12-gon" Tags="|geometry|combinatorics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24719" PostTypeId="2" ParentId="24718" CreationDate="2015-12-20T12:18:28.557" Score="4" Body="&lt;p&gt;Let the corners be ordered $1,2...,12$. Starting from (black,white) = $(1,2)$, we can move white (I'm assuming we don't need to move black at the same time)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1,2)\rightarrow (1,3)\rightarrow(1,4)\rightarrow...\rightarrow(1,12)$ (10 moves)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then move the black piece one step in the other direction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(2,12)\rightarrow(2,11)\rightarrow...\rightarrow(2,4)$ (8 moves)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since we cannot visit $(2,3)$. Next step is to $(3,4)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is how the 12-gon may be traversed (I pretty sure this is optimal):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uFUuK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uFUuK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I have counted correctly,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; there are $9\times 12$ from the cycles with white moves and $11$ moves with the black, so in total $119$. If the last move is counted, then we may do $120$ moves.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12901" LastEditorUserId="12901" LastEditDate="2015-12-20T13:26:50.730" LastActivityDate="2015-12-20T13:26:50.730" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24720" PostTypeId="1" AcceptedAnswerId="24722" CreationDate="2015-12-20T16:59:17.070" Score="-2" ViewCount="850" Body="&lt;p&gt;Decode the Secret Message :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;10110111100110101001001110010011100100001101001111011111100010001001000010001101100100111001101111011110&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hints-&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;some may not apply, figure out what ones do: inverted, backwards, binary, hex, letters&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17875" LastEditorUserId="16846" LastEditDate="2015-12-20T17:50:42.133" LastActivityDate="2015-12-20T17:50:42.133" Title="Decode the Secret Message" Tags="|cipher|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24721" PostTypeId="1" AcceptedAnswerId="24743" CreationDate="2015-12-20T16:59:34.550" Score="50" ViewCount="2907" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Edward's &quot;revenge&quot; on Bob the mute house servant&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Edward had enough! This was it! During the yearly sorcerer meet in a&#xA;  hidden cellar in Léon, Edward discovered a red stain on his precious robe. &#xA;  First he thought that it was just blood - but no, it was red wine! If&#xA;  another sorcerer would've noticed the stain he'ld have been the&#xA;  laughing stock of the whole meet! He didn't even drink wine!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;No, it was Bob's fault. What did he think? Just because he's a measly&#xA;  house-servant he's allowed to do &lt;strong&gt;any&lt;/strong&gt; mistakes?! He had to pay -&#xA;  but Edward only had a few hours before his own Master would return.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;After tricking the mute Bob down into the cellar, Edward trapped Bob's&#xA;  mind with a maze spell. This spell would throw Bob into a featureless&#xA;  labyrinth and force him to attempt and escape. And the best part? While&#xA;  inside the maze, every minute in the real world would feel like an&#xA;  eternity!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;But as Edward was just about to start his evil laugh, Bob woke up. He&#xA;  escaped the maze! Edward went back into his room upstairs to draw a&#xA;  new one.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;It was perfect. A piece of beauty. This time, Bob would pay for not&#xA;  properly dedusting the wall behind his bookshelf! After trapping Bob&#xA;  in another maze, he confidently left the cellar - just to come back 10&#xA;  minutes later with Bob already gone! Impossible! And he only had&#xA;  enough astral power left for one more maze spell!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nNVTp.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nNVTp.jpg&quot; alt=&quot;Edward&amp;#39;s Maze I&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edward had to figure out what strategy Bob used to escape his mazes, to create one that was inescapable.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When mazed, a person has to decide for &lt;strong&gt;one simple rule/strategy&lt;/strong&gt; on how to deal with every crossing paths. The rule shouldn't be more complex than one if/else statement and be easy to memorize! Only actual crossings are relevant, corners and underbridges are unimportant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The victim can only turn left or right. &lt;em&gt;Going straight ahead in an &quot;X-crossing&quot; is not an option!&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;If the victim runs into a dead end (Entrance) it simply turns around and walks back. Otherwise, the victim is never allowed to walk back.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Due to the confusing nature of the maze, the victim also can only count on two informations only:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How many crossings the victim already passed (starting with 0)&lt;/li&gt;&#xA;&lt;li&gt;The last decision made (left or right)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The victim has no sense of orientation, does not know how long the tunnels are or how many corners they had and cannot leave behind any marks. Everyting is featureless and looks exactly the same!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&quot;Always turn left.&quot; (the classic)&lt;/li&gt;&#xA;&lt;li&gt;&quot;Alternate between left and right. Start with left.&quot;&lt;/li&gt;&#xA;&lt;li&gt;&quot;If the [number of crossings passed &gt; 15] turn left, otherwise turn right.&quot;&lt;/li&gt;&#xA;&lt;li&gt;&quot;If the [(number of crossings passed + 1) % 3 == 0] turn left, otherwise right.&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How to handle &quot;Underbridges&quot; and &quot;T-Sections&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no &quot;going straight ahead&quot;. It's always &quot;left&quot; or &quot;right&quot;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bwejp.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bwejp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to handle &quot;X-crossings&quot;. Always go either &quot;left&quot; or &quot;right&quot;, never &quot;straight ahead&quot;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FRzqm.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FRzqm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Analysis&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Luckily, Edward is able to analyze Bob's path through another spell. &lt;strong&gt;The red line marks the way Bob took&lt;/strong&gt; but doesn't display how many times he passed through the same tunnel, so the information is of limited use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HzqUB.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HzqUB.jpg&quot; alt=&quot;Edward&amp;#39;s Maze I - the first&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While analyzing the second labyrinth, Edward was able to analyze how many times Bob went through a certain crossing (depicted by the number).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qrEqP.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qrEqP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1. What simple rule/strategy is Bob using to get through Edwards labyrinths?&lt;/strong&gt; &lt;em&gt;Bob uses the &lt;strong&gt;same&lt;/strong&gt; strategy for every maze! It does not change between mazes!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2. Is it possible for Edward to create a 2 dimensional, finite labyrinth to stop Bob from escaping a third time?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(I hope I could make everything clear - I'm sadly not a native english speaker and yet have to practice explaining riddle-rules. Let me know if something is confusing and I'll try to clear it up!)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="14076" LastEditorUserId="14076" LastEditDate="2016-02-14T16:09:02.143" LastActivityDate="2017-03-13T02:54:59.727" Title="Edward's Maze I" Tags="|logical-deduction|pattern|story|mazes|" AnswerCount="1" CommentCount="20" ContentLicense="CC BY-SA 3.0" />
  <row Id="24722" PostTypeId="2" ParentId="24720" CreationDate="2015-12-20T17:12:21.830" Score="1" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hello, world!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The trick is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; to divide into 8 bit binary, invert them, and convert to ASCII&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1011 0111 -&gt; 0100 1000 -&gt; H&lt;br&gt;1001 1010 -&gt; 0110 0101 -&gt; e&lt;br&gt;1001 0011 -&gt; 0110 1100 -&gt; l&lt;br&gt; 1001 0011 -&gt; 0110 1100 -&gt; l&lt;br&gt; 1001 0000 -&gt; 0110 1111 -&gt; o&lt;br&gt; 1101 0011 -&gt; 0010 1100 -&gt; ,&lt;br&gt; 1101 1111 -&gt; 0010 0000 -&gt; space&lt;br&gt; 1000 1000 -&gt; 0111 0111 -&gt; w&lt;br&gt; 1001 0000 -&gt; 0110 1111 -&gt; o&lt;br&gt; 1000 1101 -&gt; 0111 0010 -&gt; r&lt;br&gt; 1001 0011 -&gt; 0110 1100 -&gt; l&lt;br&gt;1001 1011 -&gt; 0110 0100 -&gt; d&lt;br&gt; 1101 1110 -&gt; 0010 0001 -&gt; !&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16846" LastEditorUserId="16846" LastEditDate="2015-12-20T17:17:52.440" LastActivityDate="2015-12-20T17:17:52.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24723" PostTypeId="2" ParentId="24718" CreationDate="2015-12-20T17:15:04.327" Score="2" Body="&lt;p&gt;Carl's answer is optimal. Let's say that a position is &quot;odd&quot; if there are an odd number of empty vertices between the two pieces, and &quot;even&quot; otherwise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process must alternate between odd and even positions. Since there are only 60 odd positions and we start at an even position, the greatest number of states we path can traverse is 121, which means we can make at most 120 moves.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-12-20T17:15:04.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24726" PostTypeId="2" ParentId="24709" CreationDate="2015-12-20T20:06:30.927" Score="1" Body="&lt;p&gt;If you want to improve your times you should learn F2L. &#xA;I guess you also mean the same thing with Fridrich (althought F2L is only the second step of Fridrich). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can learn it by watching a tutorial and by practicing a lot. And don't come with the excuse of not having time. If you don't have time watching a 10 minute tutorial you also have no time on practicing the cube and you'll not achieve your goal of getting faster. There is no magic trick, that improves your times without effort.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can learn F2L from here: &lt;a href=&quot;https://www.youtube.com/watch?v=0PkaWYMoVoM&quot; rel=&quot;nofollow&quot;&gt;How To Learn F2L Intuitively&lt;/a&gt;. The video is not only 9 minutes long. Although you probably will take much longer to get used to the idea of F2L and it will take a few weeks until you get good times with it. At first your solving time will skyrocket. But don't give up! If you practice a lot, you'll get used to all the different cases and you'll be faster than ever. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And from time to time you should learn a few 1-look cases. Quite a lot of them are really easy (especially the OLLs). For instance you probably know the alg &lt;code&gt;F (R U R' U') F'&lt;/code&gt;. Another OLL is &lt;code&gt;F (R U R' U')2 F'&lt;/code&gt; (the middle term is executed twice). Or another one is &lt;code&gt;Fw (R U R' U')2 Fw'&lt;/code&gt; (take the first to layers of F instead of only one). &lt;/p&gt;&#xA;" OwnerUserId="1902" LastEditorUserId="2156" LastEditDate="2015-12-22T10:47:14.410" LastActivityDate="2015-12-22T10:47:14.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24728" PostTypeId="1" AcceptedAnswerId="24729" CreationDate="2015-12-20T20:56:08.040" Score="15" ViewCount="5319" Body="&lt;p&gt;&quot;The clue is hidden in this statement&quot;, read the note handed to Sherlock by Moriarty, who hid the stolen treasure in one of the ten pillars. Which pillar is it?&lt;/p&gt;&#xA;" OwnerUserId="17860" LastEditorUserId="4269" LastEditDate="2015-12-21T11:30:50.383" LastActivityDate="2016-02-03T14:23:11.117" Title="The clue is hidden in this statement" Tags="|wordplay|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24729" PostTypeId="2" ParentId="24728" CreationDate="2015-12-20T20:58:52.867" Score="16" Body="&lt;p&gt;It's in the&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;ninth: hidde&lt;strong&gt;n in th&lt;/strong&gt;is statement&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2015-12-20T20:58:52.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24730" PostTypeId="1" AcceptedAnswerId="24766" CreationDate="2015-12-20T21:14:39.753" Score="19" ViewCount="1506" Body="&lt;p&gt;Ernie had spent most of the afternoon at my place, helping me with the tile grouting in my new bathroom. It had been a difficult and messy job, so when he asked if I could do him a &quot;small favour&quot; I felt I could not refuse. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ernie went on to explain. &quot;My family are all great fans of the Downton Abbey TV series, and as it is getting into birthday season I thought I would kill several birds with one stone and have a &lt;em&gt;stately home&lt;/em&gt; birthday celebration for some of my nieces and nephews. I’ve already organised a catering company to provide appropriately dressed maids and footmen for the occasion, but I thought you might like to act as butler. All you would have to do is dress appropriately and at the right time cut the birthday cakes.” I replied that I would be honoured to accept, so Ernie went on to explain what was required in a little more detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ts1sP.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ts1sP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;At 10:00am my oldest sister's twins, Fritz and Donaldo, will be celebrating their 10th birthday. I have made them a lovely orange-strawberry short-cake in the shape of a perfect equilateral triangle. As long as they get half each they will be happy.&quot; &quot;So just a single cut?&quot; I replied. &quot;A single cut is just right&quot;, Ernie replied. &quot;Of course, you need to make sure you keep the Great Aunt happy!&quot; &quot;The Great Aunt?&quot;, I asked. &quot;Yes&quot;, Ernie explained, &quot;Great Aunt Bismarkia will be arriving with the twins. She was a Time and Motion Manager back in the 1960s, and she always insists on efficiency. She will want you to make sure you use the shortest possible total length of cuts required to divide up the cake appropriately at any cake cutting she witnesses.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;At 11:00am my middle sister's triplets, Maude, Loretta and Gwendolyn, will be celebrating their 11th birthday. I have made them a lovely chocolate cake with mint icing in the shape of a perfect square. It has a small round red marshmallow in the exact middle of the top. Maude and Loretta love marshmallows so will want even shares of the one in the cake, but Gwendolyn hates marshmallows so she will want a piece with no marshmallow. But none of them will be fussed regarding how big or small their piece of cake is.&quot; &quot;So just two cuts?&quot; I asked. &quot;Two cuts will be just right&quot;, Ernie replied. &quot;Of course, you need to make sure you keep both Great Aunts happy!&quot;. &quot;Both Great Aunts!?&quot;, I asked. &quot;Yes&quot;, Ernie explained, &quot;Great Aunt Lusitania will be arriving with the triplets. She was an Equal Opportunity Advocate back in the 1950s, and she always insists on equality. She will want you to make sure that each of the cuts required to divide up the cake appropriately are of precisely the same length at any cake cutting she witnesses.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;And at 12:00am my youngest sister's quadruplets, Bob, Rob, Jan and Sue will be celebrating their 12th birthday. I have made them a lovely Battenberg cake, with blackberry icing and a liquorish lattice set into the side walls, in the shape of a perfect pentagon. Bob, Rob and Jan love liquorish so they will want equal shares of the perimeter of the cake, while Sue hates liquorish and will want a piece of cake that doesn't touch the perimeter at any point. Oh - and while the first three don't care how big or small their pieces are, Sue will want her piece to be at least half as big as the next larger piece.&quot; &quot;So just three cuts?&quot; I asked. &quot;Three cuts will be just right&quot;, Ernie replied. &quot;Of course, you need to make sure you keep all three Great Aunts happy!&quot;. &quot;Three Great Aunts!!?&quot;, I asked. &quot;Yes&quot;, Ernie explained, &quot;Great Aunt Titania will be arriving with the quads. She was a Group Theorist back in the 1940s, and she always insists on symmetry. She will want you to make sure that at least one of the final pieces has a mirror symmetry plane at any cake cutting she witnesses.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By now I was feeling a bit concerned – all those rules and requirements. So I made a few more enquiries and Ernie confirmed that there should be no extra pieces of cake cut other than those the children required after the cutting; that each cut had to be straight and could not cut across more than one piece of cake (but a single cut could divide a piece of cake onto one, two or three pieces depending on the shape of the piece and the positions of the previous cuts); that each cake had a side length of exactly 100 mm; and that the old ladies still had great eyesight and could tell if a cut was incorrect by more than a tenth of a degree in angle or a tenth of a mm in length or position. He also confirmed that Great Aunt Bismarkia chooses the shortest-length solution that satisfies all other cake-cutting participants that are present at the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I’m terrified that I will disappoint some of the children or make the some of the aunts unhappy and ruin the celebrations. Can anyone help me? A set of diagrams showing where and how to cut the cakes would be of great help.&lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="1631" LastEditDate="2015-12-21T21:33:25.663" LastActivityDate="2015-12-30T18:17:31.197" Title="Ernie and the Cake Cutting" Tags="|mathematics|calculation-puzzle|geometry|word-problem|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24731" PostTypeId="1" AcceptedAnswerId="24748" CreationDate="2015-12-20T22:47:02.440" Score="8" ViewCount="884" Body="&lt;p&gt;100 clever men receive presents from the president. Each man gets either a red or blue present, and only knows the color of his own gift. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then each man must guess a colour of a gift of some man: he must chose a man (besides himself) and a colour, write these down on a piece of paper and give this paper to an organiser. Once all this is done the organiser counts amount of correct guesses out of 100.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The men know all the described procedure in advance and have time to develop a strategy, before receiving any presents. Once they receive the presents, they will be unable to communicate with each other.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their task is to guarantee maximum amount of correct guesses. Your task is to say 1) what is this maximum amount, 2) what can be a strategy of the men and, the most important: 3) prove that there is no other strategy, which can guaranty a bigger amount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. &quot;Guarantee&quot; - means that this amount should be achieved independently of luck and what presents are. It can be that all 100 presents are blue, or all red, or a mix, distribution between men also is arbitrary.&lt;br&gt;&#xA;P.P.S. It feels like 50 is right answer, it is easy to figure out a strategy to do this, but it is really hard to prove that this is the best result. Note that 1. several men can guess about one present, 2. man can chose who he is guessing about After he got his present.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2015-12-21T13:48:49.397" LastActivityDate="2015-12-21T20:28:11.380" Title="Guess another's present" Tags="|logical-deduction|hat-guessing|" AnswerCount="6" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24732" PostTypeId="2" ParentId="24731" CreationDate="2015-12-21T00:09:19.833" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I can guarantee exactly 50 correct guesses.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using the Foolowing Strategy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;First the men group themselves into two groups of exatly 50 each.&lt;br&gt;&#xA;  For convenience we will name those groups A and B.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The next step is for everyone to find themselves into pairs.&lt;br&gt;&#xA;  Each pair has to consist of one person from group A and one person from group B.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Please note, that these two steps are done before receiving presents.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After receiving the presents the following guesses will be made:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Everyone will guess for the other person in his pair.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Everyone from group A will guess the same color as the present he received.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Everyone from group B will guess the opposite color as the present he received.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Guessing this way will give the following results:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If both people in a pair have the same color the one from group A will have guessed right and the one from group B will have guessed wrong.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If both people in a pair have a different color the one from group A will have guessed wrong and the one from group B will have guessed right.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This way we guarantee exactly one right guess in each pair for 50 pairs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why it should not be possible to get more:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Since they aren't allowed to communicate in any way each one would have a chance of exactly 50% to guess right since every person can have one of two states.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now we can have one person of each pair guess one stated based on his own present and the other person of each pair guess the other state based on his own present.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If we now group them together to pairs then each pair as well can only have one of two possible states.(Those states being &quot;same color&quot; and &quot;different color&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16608" LastActivityDate="2015-12-21T00:09:19.833" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24733" PostTypeId="2" ParentId="24731" CreationDate="2015-12-21T00:11:55.927" Score="0" Body="&lt;p&gt;1, the max is 50.&#xA;2, they should all guess the same person but half should say blue, the other day red.  That would mean that 50 are correct, and 50 are wrong. &#xA;3, assuming that there are an even number of red and blue gifts, getting one half correct is the highest available. If there are more red than blue, guessing all one color could result in fewer than 50.&lt;/p&gt;&#xA;" OwnerUserId="5696" LastActivityDate="2015-12-21T00:11:55.927" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24734" PostTypeId="2" ParentId="16168" CreationDate="2015-12-21T03:55:43.153" Score="25" Body="&lt;p&gt;This was a monster of a puzzle.  But I think I got it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following is a solution that does not use randomness, finishes in bounded time for any given circle size, and includes a (mostly- maybe a detail or two left to the reader =)) full proof of validity.  Tell me if it makes sense, or if further clarification is needed.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We first define the subroutine $fill(S,k)$.  This takes in a set of people, $S,$ such that everybody knows whether they are in $S$ or not.  It also takes in a paramater $k$ related to how long to run for.  It works as follows&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) On round 1, everybody in set $S$ turns on their light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) On every round from $2$ to $k$, everyone in $S$ and everyone who has seen a light on during $fill(S,k)$ turns on their light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) From round $k+1$ to $2^k+k+1$, everybody who saw a light on  at the end of stage 2), and in every round since then, turns on their light.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Denote $S'$ to be the set of people who saw lights on round $k$. We first claim that, if S is only the leader, then regardless of $k$, at the end of this subroutine, everyone in the circle either sees a light, or does not see a light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this, note that $S'$ either contains everyone or does not contain everyone.  If it contains everyone, it is clear that stage 3) of the subroutine simply involves everyone turning on their light every round, and thus everyone sees a light at the end.  If it does not contain everyone, then we note that the set of lights seen in each round of stage 2) at most doubles in size.  Thus, at the end of stage 2), at most $2^k$ lights are on. But it's also easy to see that, if $S'$ is not the whole circle, then the set of lights on in stage 3) decreases by at least one each round.  Thus, by the end, no lights will be on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose $ U &amp;gt; n$.  Then, we claim that $ fill(S, U)$ will result in every light being on if and only if $S$ is not the empty set, and every light being off otherwise.  This is easy to see because, if $U &amp;gt; n$, then $ S'$ will have to be the entire circle, as long as $S$ was not the empty set.  Otherwise, if $S$ was the empty set, then clearly nobody will ever see a light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, now we're ready to describe our strategy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 1.  Deduce an upper bound $U$ on $n$, as done in many other solutions.  We run $fill(L, k)$, where $L$ is the leader, for increasing $k$ starting at $1$. At some point, the result will be positive, with everyone seeing a light.  At this point, we can be assured that the circle contains at most $2^k$ people where $k$ was the final value used to $fill$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 2.  Initialize by considering the leader to be in set $S_0$, and everyone else to be in set $S_1$. Let $k = 2$ represent the number of sets at any point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that currently, everybody knows which indexed set they are in, and which sets exist.  We will maintain this invariant so that everyone can coordinate their strategies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 3. Set $done = false$.  While not $done:$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 A) Set $done = true$. For each subset $ I \subset \{1, 2, \cdots, k\}$, other than the empty set and the entire set:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 A i) Everybody in the set $ \bigcup_{i \in I} S_i$ turns on their light once. Let $T$ be the set of people who saw lights on this day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 A ii) For every set $S_j$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 A ii a) Run $fill(S_j \cap T, U)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 A ii b) Run $fill(S_j / T, U)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 A ii c) If both $fills$ return positive results, then  increment $k$ by 1, let $ S_j = S_j \cap T$, $S_k = S_j / T$, and set $ done = false$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 4.  We will prove that, when we reach this point, everyone in the circle can deduce the circle's size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we prove that we eventually reach Step 4. Note that $done$ is only set to $false$ in 3) if a new set $ S_k$ is created.  Furthermore, every new set created is not empty, and no nonempty set becomes empty during the algorithm, because $ S_j$ is only split if both $ S_j \cap T$ and $ S_j / T$ are nonempty.  Thus, if there are $n$ people in the circle, we can create at most $n-2$ new sets before there are $n$ total sets, one person to a set, and no further sets can be split. At this point, set $3$ must terminate with $done = true$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose we reach step 4.  For every set $S_i$, let $x_i = |S_i|$.  For any partition $(A,B)$ of the $x_i$ into two nonempty sets, there was a corresponding round in the final iteration of Step 3 in which every person corresponding to $A$ turned on their light. This then resulted, for each $S_i$, in everyone in $S_i$ seeing a light on, or everyone seeing a light off. Furthermore, at least one person in an $S_i$ represented by $B$ saw a light, or else $A$ would have to be a closed loop.  The key here is that we know that the number of people in $A$ who didn't see a light must be equal to the number of people in $B$ who did.  This gives us an equation in the variables $x_i$, of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\sum_{i \in X} x_i = \sum_{i \in Y} x_i,&#xA;\end{equation}, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $ X \subset A$, $ Y \subset B$, and neither $X$ nor $Y$ is empty. We say that such an equation &quot;satisfies&quot; the partition $(A,B)$.  More generally, we say that $(A,B)$ is satisfied by any equation of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\sum_{i \in X} r_i x_i = \sum_{i \in Y} s_i x_i,&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where all coefficients $r_i,s_i$ are positive, and $X$ and $Y$ are nonempty subsets  of $ A$ and $B$.  If a set of equations satisfies every partition of a set of variables, we say that that set of variables is fully satisfied.  At step 4, we thus have a set of fully satisfied variables.  Thus, to prove that we can deduce the size of the circle, we show that, given any set of variables such that $x_i = 1$, along with a satisfiable and fully satisfying set of equations on those variables,  we can deduce the value of every variable&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We use induction. For $ k = 2$ variables, we have $ x_1 = 1$, and the partition $ (x_1, x_2)$ yields $ k_1x_1 = k_2x_2$, where $ k_1, k_2 \neq 0$, so clearly both variables are determined.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, consider a set of $k$ variables $ x_1, x_2, \cdots, x_k$, and assume our hypothesis holds for all sets of $k-1$ variables. We will eleminate $x_k$ from the equations to construct a set of equations which fully satisfies $ x_1, \cdots, x_{k-1}$. For each partition $(A,B)$ of the remaining $k-1$ variables, there are two partitions of the whole set of variables: $ (A \cup c, B)$, and $ (A, B \cup c)$.  Both of these partitions have some satisfying equation.  If either of the satisfying equations don't include $c$, then we have an equation which satisfies $(A,B)$ on the first $k-1$ variables.  Otherwise, if both satsifying equations include $c$ with some weight, then we can linearly combine them to eliminate $c$ and get a new equation, which some basic algebra tells us will also satisfy $(A,B)$ in the first $k-1$ variables.  Thus, by doing this for every partition, we can construct a set of fully satisfying equations for $ x_1, \cdots, x_{k-1}$, which by our inductive hypothesis determines all of their values.  Finally, we can consider the partition which separates $x_k$ form the other $k-1$ variables.  This partition is only satisfied by a linear equation writing $x_k$ in terms of the other variables; as these are all determined, we can also determine $x_k$.  Thus all variables are determined, and we are done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, since every person in the circle knows a fully satisfying set of equations on all of the subsets $S_i$, and knows that $ |S_1| = 1$, everyone in the circle can deduce the size of each subset, and of the entire circle.&lt;/p&gt;&#xA;" OwnerUserId="17880" LastEditorUserId="17880" LastEditDate="2015-12-30T16:46:38.777" LastActivityDate="2015-12-30T16:46:38.777" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="24743" PostTypeId="2" ParentId="24721" CreationDate="2015-12-21T09:35:43.350" Score="26" Body="&lt;p&gt;A very well-written puzzle! And very good visualization! This site needs more good puzzle-writer like you =D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Guess it must be very troublesome to fit all those mazes and turns using the simple rule. Good job!&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;1. What simple rule/strategy is Bob using to get through Edwards labyrinths?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;From the question, I believe the simple rule is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If [(number of crossings passed + 1) is prime] turn right; otherwise turn left&lt;br&gt;&#xA; which results in:&lt;br&gt;&#xA; LRRLRLRLLLRLRLLLRLRLLLRLLLLLRLRLLLLLRLLL&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's see how that works in each maze.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moves Bob took in Maze 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LRRLRLR&lt;sup&gt;a&lt;/sup&gt;LLLRL&lt;sup&gt;b&lt;/sup&gt;RLLL&lt;sup&gt;c&lt;/sup&gt;RLRLLLR&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;a&lt;/sup&gt;escape lower left loop, just went through T-junction going upper half&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;b&lt;/sup&gt;just enter upper-middle structure, Bob just turned left at the T-junction going inside the structure from the right, going into the loop&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;c&lt;/sup&gt;went into the loop two times&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In maze 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LRRLRLR&lt;sup&gt;a&lt;/sup&gt;LLLR&lt;sup&gt;b&lt;/sup&gt;LR&lt;sup&gt;c&lt;/sup&gt;LLLRLRLLLRL&lt;sup&gt;d&lt;/sup&gt;LLLLRL&lt;sup&gt;e&lt;/sup&gt;, then three options, either to loop the lower left 0, 1, or 2 times before going the outer loop:&#xA; &lt;br&gt;&#xA; 1) LLLRLLLLLR (directly go to outer loop)&lt;br&gt;&#xA; 2) RLLRLLRLLR (loop lower-left once)&lt;br&gt;&#xA; 3) RLLLLLRLLL (loop lower-left twice)&lt;br&gt;&#xA; But to adhere to the &quot;simple rule&quot;, only the third one is valid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;a&lt;/sup&gt;Bob just loop through the outer loop clockwise, then bounce back from entrance, now facing the X-junction from below&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;b&lt;/sup&gt;Bob just escaped lower loop, now facing the second middle loop&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;c&lt;/sup&gt;Bob just escaped the second loop, now facing the big maze in the middle&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;d&lt;/sup&gt;Bob just escape the middle maze, now on the middle-left T-junction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;e&lt;/sup&gt;Bob just arrived at the last loop&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explanation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; By maze 1, we need to start with LR. If the next one is L on maze 2, the next would need to be RL, resulting in LRLRL. Then we need to complete the middle loop in clockwise manner (counter-clockwise will need to repeat the 1 junction on the right), resulting in LRLRLLLRL, which doesn't work in Maze 1. So the first three moves are LRR.&lt;br&gt;&#xA; &lt;br&gt;&#xA; Using maze 1, the next two moves are LR. Using similar reasoning in maze 2 as in previous case, the next two moves are again LR (so far LRRLRLR), which from maze 1 we get the next moves as LLLRL. Then from maze 2, we have the next moves as: RLLLRLRLLLRLLLLLRL. And then we can complete the maze using the pattern we found thus far.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;2. Is it possible for Edward to create a 2 dimensional, finite labyrinth to stop Bob from escaping a third time?&lt;/h3&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes.&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/tibZP.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tibZP.png&quot; alt=&quot;Bob-trapping maze&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA; The entrance is below, the exit is on the right.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;Explanation&lt;/h3&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After some careful thought, Edward finally realized Bob's strategy, and after a few hours (just before his Master came back!), he found out a pattern in Bob's strategy:&lt;br&gt;&#xA; &lt;br&gt;&#xA; Starting from the third turn (the first two are LR), Bob either takes RL or LL (since the even-numbered moves will always be L).&lt;br&gt;&#xA; &lt;br&gt;&#xA; So Edward devices a maze where taking RL or LL will always bring you back to the same point, that is, the center of the maze (the green tile)&lt;br&gt;&#xA; Edward did the evil laugh, until he realized that to enter this trap, Bob needs to take a right turn on one even move, so that Bob will be on his odd-numbered turn when he reach the green tile.&lt;br&gt;&#xA; &lt;br&gt;&#xA; Edward was utterly disappointed (&quot;But primes are odd!&quot; Edward yelled in frustration) and was about to give up when, fortunately, Willy reminded him that on the second move, Bob does a right turn (&quot;Sometimes you can be useful, Willy,&quot; Edward said).&lt;br&gt;&#xA; &lt;br&gt;&#xA; With that final revelation, Edward finally established the Ultimate Bob-trapping Maze&lt;sup&gt;TM&lt;/sup&gt;, to the horror of Bob, who we never hear from ever again (anyway, he is mute)...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;Fun stuff&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I believe (before OP last edit) the &quot;if ... else ...&quot; rule also allows Bob to create a very long list of predefined moves like:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;If the [number of crossings passed == 1 OR number of crossings passed == 3 OR ... OR number of crossings passed == 125] turn left, otherwise turn right.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is still one &quot;if ... else ...&quot;, with a huge disjunction =D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the current &quot;simple rule&quot; is actually very hard to determine once Bob has passed &lt;a href=&quot;https://en.wikipedia.org/wiki/Largest_known_prime_number&quot; rel=&quot;noreferrer&quot;&gt;many many crossings&lt;/a&gt;...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/users/17389&quot;&gt;Zerris&lt;/a&gt; found the smallest of such maze, and the simplicity is beautiful:&#xA; &lt;a href=&quot;https://i.stack.imgur.com/4EV8i.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4EV8i.png&quot; alt=&quot;Beautiful Bob-trapping maze&quot;&gt;&lt;/a&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1649" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2017-03-13T02:54:59.727" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24744" PostTypeId="1" AcceptedAnswerId="24745" CreationDate="2015-12-21T09:41:20.053" Score="4" ViewCount="2190" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ScgRp.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ScgRp.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Find the missing number in this grid. I am unable to find any pattern.&lt;/p&gt;&#xA;" OwnerUserId="17860" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-12-29T06:32:37.797" Title="Find the missing number in this grid" Tags="|pattern|calculation-puzzle|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24745" PostTypeId="2" ParentId="24744" CreationDate="2015-12-21T09:44:24.593" Score="6" Body="&lt;p&gt;It is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; -19 &lt;br&gt;&lt;br&gt;&#xA; The above number is exactly in the middle of the 2 bottom numbers &lt;br&gt;&lt;br&gt;&#xA; &#xA;So the difference between 9 and -5 is 14, just subtract 14 from -5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14461" LastEditorUserId="14461" LastEditDate="2015-12-21T09:48:14.777" LastActivityDate="2015-12-21T09:48:14.777" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24746" PostTypeId="1" AcceptedAnswerId="24749" CreationDate="2015-12-21T13:50:26.270" Score="1" ViewCount="418" Body="&lt;p&gt;As my first riddle, good luck finding what hides behind that text!   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The man thinks it's silver&lt;br&gt;&#xA;All he's selling is his old dirty cave&lt;br&gt;&#xA;A wednesday afternoon&lt;br&gt;&#xA;he tries and he tries but get none&lt;br&gt;&#xA;Times flies&lt;br&gt;&#xA;he stands there in darkness&lt;br&gt;&#xA;We all see this stranger&lt;br&gt;&#xA;dark shriveled shape  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does this text represent?&lt;br&gt;&#xA;The answer should not be vague or ambiguous, if you find it you'll know! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1 (small hint)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; At the end he's selling his old dirty cave&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2 (bigger hint)  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it is a song&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3 ( queen of all hints)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ZoSo&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17888" LastEditorUserId="17250" LastEditDate="2015-12-21T16:35:35.170" LastActivityDate="2015-12-21T16:42:00.553" Title="What popular thing does this text represent?" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24747" PostTypeId="2" ParentId="24731" CreationDate="2015-12-21T14:19:45.807" Score="5" Body="&lt;p&gt;Another possible 50% success strategy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick two people, A and B. Divide the people into groups of 25, G1,G2,G3, and G4. Make sure A is in G1 or G2 and B is in G3 or G4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everyone in G1 guesses that B has a red present. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everyone in G2 guesses that B has a blue present. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everyone in G3 guesses that A has a red present. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everyone in G4 guesses that A has a blue present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everyone in G1 or everyone in G2 will be right (25 right)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everyone in G3 or everyone in G4 will be right (25 right)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No one has to guess themselves&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore we get exactly 50%&lt;/p&gt;&#xA;" OwnerUserId="5169" LastActivityDate="2015-12-21T14:19:45.807" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24748" PostTypeId="2" ParentId="24731" CreationDate="2015-12-21T16:28:32.940" Score="4" Body="&lt;p&gt;Here is a proof that more than 50 is impossible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;No matter what strategy a player uses, they will be wrong exactly half of the time. Why? Let Alice be a particular player. For each present distribution $P$, define a corresponding distribution $f(P)$, where you change the color of the person that Alice is guessing and leave all other colors the same. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $f(P)$ doesn't change Alice's color, it doesn't change her guess, which means that $f(f(P))=P$. In addition, $f$ has no fixed points. Therefore, we can break up the set of all $2^{100}$ distributions into pairs $(P,f(P))$. Alice will be correct for exactly one distribution in each pair, so she is correct half the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that each person gets 0.5 guesses correct on average, so adding these up, the team gets 50 guesses correct on average. This proves that it is impossible to guarantee getting more than 50 (if they always get 51 or more, then the average would be 51 or more).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Put another way: from Alice's perspective, she is guessing the value of a coin flip she has no information about, so she has to be correct with probability 1/2.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-12-21T16:28:32.940" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24749" PostTypeId="2" ParentId="24746" CreationDate="2015-12-21T16:29:05.960" Score="4" Body="&#xA;&#xA;&lt;p&gt;I'm going to tentatively guess that the clues are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an inversion (of sorts) of the lyrics of &lt;em&gt;Stairway to Heaven&lt;/em&gt;. This is mostly based on Hint 3, but I'm taking my shot, in the spirit of Eminem in the movie &lt;em&gt;8 Mile&lt;/em&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;man&lt;/strong&gt; &lt;strong&gt;thinks&lt;/strong&gt; it's &lt;strong&gt;silver&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There's a &lt;strong&gt;lady&lt;/strong&gt; who's &lt;strong&gt;sure&lt;/strong&gt; all that glitters is &lt;strong&gt;gold&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All he's &lt;strong&gt;selling&lt;/strong&gt; is his old dirty &lt;strong&gt;cave&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; She's &lt;strong&gt;buying&lt;/strong&gt; a &lt;strong&gt;stairway to heaven&lt;/strong&gt; (I'm not thrilled about this one, but &quot;cave&quot; sort of indicates down/into the earth and &quot;stairway to heaven&quot;, well)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A wednesday afternoon&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; not a clue on this one, honestly, unless the idea is that stores aren't closed on a Wednesday afternoon  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;he tries and he tries but get none&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;With a word she can get what she came for&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;he stands there in &lt;strong&gt;darkness&lt;/strong&gt;&lt;br&gt;&#xA;We all see this &lt;strong&gt;stranger&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There walks a lady &lt;strong&gt;we all know&lt;/strong&gt;&lt;br&gt;&#xA; Who &lt;strong&gt;shines white light&lt;/strong&gt; and wants to show&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2015-12-21T16:42:00.553" LastActivityDate="2015-12-21T16:42:00.553" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24751" PostTypeId="2" ParentId="24728" CreationDate="2015-12-21T18:19:43.713" Score="11" Body="&lt;p&gt;its in &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;IX.... hidd&lt;strong&gt;en&lt;/strong&gt; &lt;strong&gt;in&lt;/strong&gt; this...&lt;br&gt;(read it from back... enin---&gt; nine)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17895" LastEditorUserId="16846" LastEditDate="2015-12-22T09:55:16.547" LastActivityDate="2015-12-22T09:55:16.547" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24752" PostTypeId="2" ParentId="24731" CreationDate="2015-12-21T19:29:14.237" Score="0" Body="&lt;p&gt;I think we can get to 98!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We designate two men ahead of time, call them A and B. When guessing begins, one man with a red present (if there is one) will choose person A and make a random guess. Then, another man with a red present chooses the previous man and guesses red. This proceeds in the same fashion with B and the blue presents. When all other guesses are made, A and B choose any man in their chain and guesses the color of that chain, A guesses red, B guesses blue. In the event either A or B is not chosen at all, they choose someone from the other chain and guess THAT color. The only ones that could be wrong would be the men that start each chain, and even then, they have a 50% chance of being right as well!&lt;/p&gt;&#xA;" OwnerUserId="16874" LastActivityDate="2015-12-21T19:29:14.237" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24753" PostTypeId="2" ParentId="24731" CreationDate="2015-12-21T19:53:14.157" Score="0" Body="&lt;p&gt;I'm thinking you could get at least 99 right provided that the men write their answers down sequentially after so much time has passed. If the men are labeled M1, M2, ..., M100, then if M1 is red, he should guess M2's color and hand it in within, say, 5 seconds. If M1 is blue, then he should let the 5 seconds pass, signifying to the group that he is blue. If M2 is red, he should guess M1's color according to their scheme; otherwise, he lets 5 seconds pass and thereby signifying that he is blue and deferring to M3, and so on. So in this manner only reds will guess until you've reached M100. After you reach M100, those who did not answer yet must be blue and they should answer as such for each other. The only one that could possibly get it wrong would be M1.&lt;/p&gt;&#xA;" OwnerUserId="17508" LastEditorUserId="17508" LastEditDate="2015-12-21T20:28:11.380" LastActivityDate="2015-12-21T20:28:11.380" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24754" PostTypeId="1" AcceptedAnswerId="24759" CreationDate="2015-12-21T19:53:30.470" Score="5" ViewCount="453" Body="&lt;p&gt;The challenge is: &lt;code&gt;aHR0cDovLzQwNDguY28ubmYv&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tips: &#xA;&lt;strong&gt;ba(s(2)-s)e 63+1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the password you need to discover what the first line means; the second is &lt;strong&gt;2spooky&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The password is in the form &lt;strong&gt;Xxx xxxxxxxxx xxxxxxxx&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The password has special letters, but just replace them with normal characters, for example: ç = c, ã = a, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck! &lt;/p&gt;&#xA;" OwnerUserId="17896" LastEditorUserId="9718" LastEditDate="2015-12-21T20:16:39.827" LastActivityDate="2015-12-31T06:33:14.150" Title="Find the password!" Tags="|enigmatic-puzzle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24755" PostTypeId="1" AcceptedAnswerId="24756" CreationDate="2015-12-21T20:06:40.270" Score="5" ViewCount="2647" Body="&lt;p&gt;See if you can figure this one out and find the hidden message!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8xKlX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8xKlX.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is more than 1 part!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint #1: 3=2=3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint #2:  | 166 | 3 | 3 |&lt;/p&gt;&#xA;" OwnerUserId="17896" LastEditorUserId="17896" LastEditDate="2015-12-21T20:16:52.777" LastActivityDate="2015-12-21T20:16:52.777" Title="Visual Puzzle - Find the hidden message!" Tags="|enigmatic-puzzle|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24756" PostTypeId="2" ParentId="24755" CreationDate="2015-12-21T20:15:09.530" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Profile.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each row is an ASCII character, encoded in binary. Open=1, Closed=0&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17446" LastEditorUserId="17446" LastEditDate="2015-12-21T20:16:01.887" LastActivityDate="2015-12-21T20:16:01.887" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24757" PostTypeId="2" ParentId="24755" CreationDate="2015-12-21T20:15:52.730" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Profile. Each row is an ASCII character in binary; closed folder == 0, open folder == 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13101" LastActivityDate="2015-12-21T20:15:52.730" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24758" PostTypeId="1" AcceptedAnswerId="24797" CreationDate="2015-12-21T20:29:41.140" Score="20" ViewCount="2866" Body="&lt;p&gt;300 dwarfs go over a bridge in the middle of the night. The bridge is rickety and manages at most two dwarfs at a time. With them is a lantern that they must provide at each transition. Dwarfs need different time to go over the bridge: 1 min, 2 min, 3 min ... and 300 minutes. When two dwarves go over, they go with the slowest one's speed. No dwarf would like to go over the bridge more than 3 times (ie, front-back-front). What is the minimum time they can manage the transition?&lt;/p&gt;&#xA;" OwnerUserId="17755" LastEditorUserId="22922" LastEditDate="2016-09-15T07:23:20.600" LastActivityDate="2016-09-15T07:23:20.600" Title="Dwarfs over bridge problem" Tags="|calculation-puzzle|strategy|" AnswerCount="5" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="24759" PostTypeId="2" ParentId="24754" CreationDate="2015-12-21T20:43:07.630" Score="5" Body="&lt;p&gt;the challenge text is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; base64 encoded from &lt;code&gt;http://4048.co.nf/&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Going there gives a page containing the hidden text &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; inurl:&quot;141542&quot; intext:&quot;07/10/2008&quot; insrc line 639 in md5 = ise mbu odide foto anya &lt;Br/&gt;68747470733a2f2f656e2e77696b6970656469612e6f72672f77696b692f424f41435f466c696768745f373132 &lt;br/&gt;They're closer than you think.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you google  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;141542 (because of the inurl prefix)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;you'll find&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.al.sp.gov.br/norma/?id=141542&quot;&gt;http://www.al.sp.gov.br/norma/?id=141542&lt;/a&gt;, which shows a law passed in 07/10/2008. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Next,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; view the page source and check line 639 (from the insrc prefix above)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and you get the password,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Sem revogação expressa&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13101" LastActivityDate="2015-12-21T20:43:07.630" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24760" PostTypeId="2" ParentId="24758" CreationDate="2015-12-21T20:48:22.423" Score="5" Body="&lt;p&gt;Every dwarf, except the last, makes 3 trips. For each pair, they first go over at the even's speed, then the odd comes back (1x per dwarf). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once all the evens are across (and odds on the starting side):&lt;br&gt;&#xA;1 and 3 go over, 2 comes back&lt;br&gt;&#xA;2 and 5 go over, 4 comes back&lt;br&gt;&#xA;4 and 7 go over, 6 comes back&lt;br&gt;&#xA;6 and 9 go over, 8 comes back&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Times are: 3,2,5,4,7,6,9,8 etc. Again, 1x per dwarf, except the first and last. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1-300 sums to 45,150.  Double is 90,300. Less 1 and 300 = $89,999$&lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-12-21T20:48:22.423" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24763" PostTypeId="1" AcceptedAnswerId="24768" CreationDate="2015-12-21T22:12:11.783" Score="3" ViewCount="779" Body="&lt;p&gt;&lt;strong&gt;Here's a small poem. See if you can find the answer! This is just a small riddle-type-thing that shouldn't be too hard for someone to solve! And yes, all the information that you need to solve this is available! Also, make sure to explain how you got the answer because that's pretty important! Anyways, the riddle-type-thing starts now:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Enter it,&lt;br&gt;&#xA;  Violently running,&lt;br&gt;&#xA;  Every single time,&lt;br&gt;&#xA;  Running for my life,&lt;br&gt;&#xA;  Yawning deeply, very tired, it stopped&lt;br&gt;&#xA;  .........Waiting&lt;br&gt;&#xA;  Softly it creeps, once again,&lt;br&gt;&#xA;  Ever so softly,&lt;br&gt;&#xA;  Coming, It's coming,&lt;br&gt;&#xA;  Oh so softly,&lt;br&gt;&#xA;  No, it cannot be,&lt;br&gt;&#xA;  Deadly once again, rising,&lt;br&gt;&#xA;  .........Waiting&lt;br&gt;&#xA;  Look in it...&lt;br&gt;&#xA;  Editing it...&lt;br&gt;&#xA;  Turning around,&lt;br&gt;&#xA;  Tag is wrong, tag is right,&lt;br&gt;&#xA;  Entering, my hand is slipping,&lt;br&gt;&#xA;  Realized I'm back in it.&lt;/em&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; Some lines are useless, the ones in the text, and some are clues for what to do next.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;" OwnerUserId="11002" LastEditorUserId="33688" LastEditDate="2017-07-23T09:10:56.900" LastActivityDate="2017-07-23T09:10:56.900" Title="The answer is... What?" Tags="|riddle|enigmatic-puzzle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24764" PostTypeId="2" ParentId="24758" CreationDate="2015-12-21T22:16:08.477" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It should take them 89,700 minutes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We use the following strategy:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;We leave the 300 minute dwarf for the last trip.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Every time we send two of them over we send those two that take the most time out of all remaining ones. (excluding the 300 minute dwarf)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Every time we send someone back we send the one that needs the smallest amount of time out of the two that just got over the bridge.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The last trip will be made by the 300 minute dwarf and the 1 minute dwarf.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The trips will look something like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;299 298 -&amp;gt; | 298 &amp;lt;-&#xA;298 297 -&amp;gt; | 297 &amp;lt;-&#xA;...&#xA;3   2   -&amp;gt; | 2   &amp;lt;-&#xA;2   1   -&amp;gt; | 1   &amp;lt;-&#xA;300 1   -&amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trips to the other side will take a total of 45149 minutes since every dwarf except the 1 minute one will be the slower one exactly once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{300 \times 301}{2}-1=45149$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trips back will be taken by every dwarf except the 300 and 299 minute ones for a total of 44551 minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{298\times 299}{2}=44551$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$45149+44551=89700$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Why this is the fastest you can get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First we look at the walks back:&lt;br&gt;&#xA;We must have 298 walks back and every dwarf can only walk back once so we just let the 298 fastest dwarfs go back once each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the walks forward:&lt;br&gt;&#xA;With two exceptions every dwarf will have to walk forward twice.&lt;br&gt;&#xA;At least one of those walks every dwarf should be the slower one as otherwise one dwarf that is even slower will have to be the slower one twice.&lt;br&gt;&#xA;The two exceptions will be the two that never walk back.&lt;br&gt;&#xA;And since we only have 299 walks forward the 1 will be the smaller one twice&lt;/p&gt;&#xA;" OwnerUserId="16608" LastEditorUserId="16608" LastEditDate="2015-12-21T23:26:31.003" LastActivityDate="2015-12-21T23:26:31.003" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24765" PostTypeId="2" ParentId="24758" CreationDate="2015-12-21T22:19:22.397" Score="6" Body="&lt;pre&gt;&lt;code&gt;1 and 2 go over: 2 minutes&#xA;1 comes back   : 1 minute&#xA;1 and 3 go over: 3 minutes&#xA;2 comes back   : 2 minutes&#xA;2 and 4 go over: 4 minutes&#xA;3 comes back   : 3 minutes&#xA;3 and 5 go over: 5 minutes&#xA;4 comes back   : 4 minutes&#xA;4 and 6 go over: 6 minutes&#xA;5 comes back   : 5 minutes&#xA;...&#xA;297 comes back     : 297 minutes&#xA;297 and 299 go over: 299 minutes&#xA;298 comes back     : 298 minutes&#xA;298 and 300 go over: 300 minutes&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This sums to 89,700 minutes&lt;/p&gt;&#xA;" OwnerUserId="17898" LastActivityDate="2015-12-21T22:19:22.397" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24766" PostTypeId="2" ParentId="24730" CreationDate="2015-12-21T22:39:53.500" Score="11" Body="&lt;p&gt;This question has 3 very different parts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The equilateral triangle must be cut exactly once such that the length of such cut is minimize and the areas of the two parts are equal.  The result is to cut the cake such that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the length of the cut is $50\sqrt{2}$mm which is parallel to one side and $25\sqrt{6}$mm from the opposing corner.  If the angle of this cut changes, it will necessarily increase until it reaches $50\sqrt{3}$mm which is the cut from corner to base.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2) The square must be cut into 3 pieces with one cut bisecting the center point and all cuts being of the same minimum length.  The two cuts should be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $83.82$mm units in length.  The first should be a $37.63$ degree angle removing a corner of the cake that is a right triangle with a side lengths of $66.39$mm and $51.18$mm.  The second cut should go from the side opposite the starting angle, through the marshmallow, and ending at the first cut to the first cut. To get the right length start $36.68$mm from the opposite corner.  As these numbers are not neat and were from simulation, I would not be surprised if it can be optimized further.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3) The pentagon is definitely evil. It must be cut with 3 cuts such that the cuts are of equal length, there is an enclosed area with a minimum of .5 the next largest piece in area, the three pieces not enclosed have equal portions of the original perimeter, and at least one piece has mirror symmetry.  Again, all cuts should be of minimum length.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sue's complaining can be quieted and the Great Aunt B can be more satisfied by having all three cuts only touch the perimeter one time. This is done by drawing (or imagining as someone would complain about messing up the icing) a triangle on the pentagonal cake such that the corners rest exactly on the perimeter and all three outer pieces have the same portion of the perimeter.  This will simplify the pentagon down into a single triangle.  All properties of this triangle (angles and side lengths) are defined by the distance between the closest triangle corner and pentagon corner.  In the cases where this distance is $0$ or $50/3 mm$, this outer triangle will be isosceles. The corners of this triangle are the three points where each cut touchs the perimeter of the cake the single time.  Each cut should terminate immediatly at the point where it will intersect another cut.  This will result in a triangle isolated in the center of the cake without touching any side.  Following an annoying simulation, I found the cuts should be between 108.0 and 108.1mm in length and that an outer triangle corner should be $50/3 mm$ from the closest pentagon corner.  If this cut length is altered in either direction, the central triangle will diviate from the required isosceles shape by more than the .1mm requirement. If a different corner distance is used, the required length to meet the area requirements will increase. The angles of the cuts should be 13.3 degrees, 17.26 degrees, and 9.24 degrees relative to the triangle. If this is followed, Sue will receive 13.44% of the cake while the next largest will receive 26.1% of the cake.  This difference from the expected 2x area is either due to rounding error within the simulation or required to meet Titania's specifications.  The sides of Sue's piece will be 69.45mm, 69.45mm, and 83.1mm in length.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Due to the apparent rounding error and simulated (rather than hand calculated) result, I am still not convinced there is not a combination of corner distance and cut length that would be better. Note that the picture is outdated for part 2.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/4cFLG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4cFLG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditDate="2015-12-30T18:17:31.197" LastActivityDate="2015-12-30T18:17:31.197" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="24767" PostTypeId="2" ParentId="24763" CreationDate="2015-12-21T22:40:19.873" Score="5" Body="&lt;p&gt;Partial answer?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; EVERY SECOND LETTER is spelled out as an acrostic, but I can't figure out how it applies. Taking every second letter of the text doesn't produce anything recognisable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2015-12-21T22:40:19.873" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24768" PostTypeId="2" ParentId="24763" CreationDate="2015-12-22T00:26:32.040" Score="3" Body="&lt;p&gt;If you look in the source you'll see extra text at the bottom. Taking every second letter gives you &quot;THE ANSWER IS PADLOCK&quot;.&lt;/p&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2015-12-22T00:26:32.040" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24769" PostTypeId="1" AcceptedAnswerId="24772" CreationDate="2015-12-22T07:25:21.320" Score="9" ViewCount="1451" Body="&lt;p&gt;What is a word made up of 4 letters, &#xA;&lt;p&gt;yet is also made up of 3,&lt;P&gt; Sometimes is written with 9 letters, and then with 4, &lt;P&gt; Rarely consists of 6, &lt;P&gt;and never is written with 5 ? &lt;/p&gt;&#xA;" OwnerUserId="16822" LastActivityDate="2016-02-29T15:14:51.597" Title="What is the solution?" Tags="|riddle|wordplay|" AnswerCount="4" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24770" PostTypeId="2" ParentId="24769" CreationDate="2015-12-22T07:29:09.617" Score="8" Body="&lt;p&gt;This is the solution :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &#xA; Each number of letter concerns a word of the sentence :&lt;br&gt;&#xA; What has 4 letters&lt;br&gt;&#xA; yet has 3&lt;br&gt;&#xA; Sometimes 9&lt;br&gt;&#xA; then 4&lt;br&gt;&#xA; Rarely 6&lt;br&gt;&#xA; never 5  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17888" LastEditorUserId="2656" LastEditDate="2015-12-22T07:35:31.470" LastActivityDate="2015-12-22T07:35:31.470" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24771" PostTypeId="1" AcceptedAnswerId="24774" CreationDate="2015-12-22T09:13:28.533" Score="7" ViewCount="287" Body="&lt;p&gt;What if in &lt;a href=&quot;https://puzzling.stackexchange.com/questions/24731/guess-anothers-present&quot;&gt;Guess another&amp;#39;s present&lt;/a&gt; problem introduce more present's colours? Also i would reduce a number of men to number of colours. So the task basically the following:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;N clever men receive presents from the president. There are N possible colors of present. Each man knows the color of his own gift only.&lt;/p&gt;&#xA;&lt;p&gt;Then each man must guess a colour of a gift of some man: he must chose a man (besides himself) and a colour, write these down on a piece of paper and give this paper to an organiser. Once all this is done the organiser counts amount of correct guesses out of N.&lt;/p&gt;&#xA;&lt;p&gt;The men know all the described procedure in advance and have time to develop a strategy, before receiving any presents. Once they receive the presents, they will be unable to communicate with each other.&lt;/p&gt;&#xA;&lt;p&gt;Their task is to guarantee maximum amount of correct guesses. Your task is to say 1) what is this maximum amount, 2) what can be a strategy of the men and 3) prove that there is no other strategy, which can guaranty a bigger amount.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;@Mike Earnest gave a quite clear &lt;a href=&quot;https://puzzling.stackexchange.com/a/24748/28&quot;&gt;proof&lt;/a&gt; that for such a task it is impossible to guarantee more than N/N=1 correct guess. When N=2 it is quite easy to guarantee 1 correct guess: see &lt;a href=&quot;https://puzzling.stackexchange.com/a/24732/28&quot;&gt;answer&lt;/a&gt; of @The Dark Truth. But already with N=3 i'm failing to figure out a strategy.&lt;/p&gt;&#xA;&lt;p&gt;Also note that the task is quite similar to well known &lt;a href=&quot;https://puzzling.stackexchange.com/questions/70/n-logicians-wearing-hats-of-n-colors/405#405&quot;&gt;N logicians wearing hats of N colors&lt;/a&gt; problem. The differences are: 1. a man sees only his &amp;quot;hat&amp;quot;, not &amp;quot;hats&amp;quot; of all others, 2. a man must guess another person color and he can chose this person, he is not stuck with guessing his own &amp;quot;hat&amp;quot; color.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2015-12-23T10:12:00.530" Title="Guess another's present 2: Generalised" Tags="|logical-deduction|hat-guessing|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24772" PostTypeId="2" ParentId="24769" CreationDate="2015-12-22T10:00:01.187" Score="15" Body="&lt;p&gt;Can I say that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The question is &quot;What is &lt;strong&gt;the solution&lt;/strong&gt;?&quot;&lt;br&gt;&#xA; And following the pattern in the clue (each number describes the number of letters in the first word in the phrase), then:&lt;br&gt;&#xA; &lt;br&gt;&#xA; &lt;pre&gt;The solution is a phrase made up of 11 letters&lt;/pre&gt;&lt;br&gt;&#xA; (space is not a letter)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1649" LastEditorUserId="1649" LastEditDate="2015-12-22T11:10:45.177" LastActivityDate="2015-12-22T11:10:45.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24773" PostTypeId="1" AcceptedAnswerId="24790" CreationDate="2015-12-22T11:35:04.847" Score="14" ViewCount="1093" Body="&lt;p&gt;There is a row of 2015 chips, of which 2014 are white and one is red.&#xA;You are allowed to make moves of the following type:&#xA; &lt;strong&gt;&quot;Choose one red chip, and flip the colors of its two neighboring chips (from red to white, respectively from white to red).&quot;&lt;/strong&gt;&#xA;Your goal is to turn all chips red by a sequence of such moves.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; For which positions of the red chip (in the initial situation) can you reach your goal?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2016-12-20T10:26:20.023" Title="A row of 2015 red and white chips" Tags="|mathematics|combinatorics|reachability|" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24774" PostTypeId="2" ParentId="24771" CreationDate="2015-12-22T11:52:05.550" Score="6" Body="&lt;p&gt;No, a correct guess cannot be guaranteed for any $N\geq 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key idea, &lt;a href=&quot;https://puzzling.stackexchange.com/a/24748/4551&quot;&gt;pointed out by Mike Earnest&lt;/a&gt;, is that the expected number of correct guesses is the number of people divided by the number of colors, here $1$. So, if ever there are $2$ or more correct guesses, some other state must lead to $0$ incorrect guesses to compensate. Therefore, any winning strategy must always give &lt;em&gt;exactly&lt;/em&gt; $1$ correct guess. We can imagine the players simply lose if two or more of them guess right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say that player P &lt;em&gt;considers&lt;/em&gt; player Q if some color of P would make them guess Q's color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, note that considering is symmetric. Say that P considers Q, specifically that when P is red, he guesses that Q is green. If Q is indeed green, then P must guess wrong to avoid two correct guesses. The only guaranteed wrong guess for Q is that P is a color other than red, so Q considers P. (Extending this reasoning for later use, if green Q guesses, say, blue P, then blue P must guess Q as well.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, any two pairs of players that consider each other must have a player in common. Otherwise, both pairs could have a correct guess. This requires that there is a player who considers everyone. It only remains to rule this out for $N \geq 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note from before that if P considers Q, he must dedicate at least two of his colors to guessing Q. So, any player considers at most half the other players, which for $N\geq 3$ rules out a player who considers everyone.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-12-23T10:12:00.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24776" PostTypeId="2" ParentId="24773" CreationDate="2015-12-22T13:31:45.637" Score="7" Body="&lt;p&gt;&lt;strong&gt;Edited with proof that the middle one is the only one&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can show quite simply that we're unable to solve if the red chip is on&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an odd number&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To show this&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; imagine that the row is laid out on a very long checkerboard, with the first square being black. Because we have $2015$ chips, we have $\lceil{\frac{2015}{2}}\rceil = 1008$ black squares; an even amount. We also know that with each flip centered on a white square we always flip 2 black squares. Every flip we either add 2 reds on these squares (if both were white), keep the amount the same (if one was white and one was red) or lose 2 reds (if both were red). It then follows that if we start out with an uneven amount on chips on black, we can never turn this into an even amount. As such it's unsolvable if the red chip starts on a black square.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then let's continue to show that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The only option is the middle one, 1008&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can prove this &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; By noting that every change we make on the board is reversible (we can always move back to our starting position by doing the moves we did again in reverse), which means that if we can solve it from one starting state, we should be able to reach that one state from every other solvable starting state.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is then sufficient to prove that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No other state with a single red chip can be reached from the starting state where we have 1 red chip in the middle. Since we've already proven that an 'odd' state is unsolvable, the only viable steps would be in increments of 2. To make a step of 2, we would have to flip around a neighboring (odd) chip and then remove the chip we just flipped around. However this is impossible, since we can't just flip around any point.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16846" LastEditorUserId="16846" LastEditDate="2015-12-22T21:45:12.650" LastActivityDate="2015-12-22T21:45:12.650" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="24777" PostTypeId="2" ParentId="24758" CreationDate="2015-12-22T14:40:03.767" Score="5" Body="&lt;p&gt;My answer: 89699&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How? Let's see for 6 dwarves:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 and 2 go -&gt; 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 returns -&gt; 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6 and 5 go -&gt; 6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 returns -&gt; 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 and 3 go -&gt; 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 returns -&gt; 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 and 3 go -&gt; 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 returns -&gt; 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 and 4 returns -&gt; 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the sum for 6 is 1 + 2 + 2 + 3 + 3 + 4 + 4 + 4 + 6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, the first dwarf only benefits once from his speed, the third last dwarf has to cross three times at his speed, the second-last and last only once (at the speed of the last dwarf). all the others go 3 times; but ony 2 times at their speed (and once with a slower dwarf). Putting it together:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For n: 1 + 2*sum(2..n-3)+ 3*n-2 + n&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 300: 1 + 2*sum(2..297) + 3*298 + 300 = 1 + 88504 + 894 + 300 = 89699&lt;/p&gt;&#xA;" OwnerUserId="17912" LastEditorUserId="17912" LastEditDate="2015-12-23T08:32:47.170" LastActivityDate="2015-12-23T08:32:47.170" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24779" PostTypeId="2" ParentId="24773" CreationDate="2015-12-22T15:01:24.057" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Only one spot works, exactly in the middle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you have a chip somewhere in the middle, and you flip its neighbors, then flip one of the new red chips, you get a pattern of RWRR.  The RR can be continuously moved down, by flipping the outside R. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say you start on 3. 3-2-4 will give you spots 1-5 filled. 5-3-1 will fill you out to 6, but then you can't add any more reds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on that, it looks like you can only effectively double your starting red's position, where the position is from the closest end. Therefore, your starting chip has to be exactly in the middle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way of thinking about it is that you move your red chip out 1, and fill in 2 spots behind it. Eventually you'll have an odd number of chips = 2*N - 1. You can shift the chips and get 1 more from the innermost chip. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-12-22T16:24:58.503" LastActivityDate="2015-12-22T16:24:58.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24780" PostTypeId="1" CreationDate="2015-12-22T15:12:37.507" Score="5" ViewCount="660" Body="&lt;p&gt;Inspired by &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16168/the-circular-prison-of-unknown-size&quot;&gt;The Circular Prison of Unknown Size&lt;/a&gt; by &lt;a href=&quot;https://puzzling.stackexchange.com/users/10615/mike-earnest&quot;&gt;@MikeEarnest&lt;/a&gt; (Great puzzle, by the way!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rules are almost same, so I won't waste time by rewriting them all here. (They're still valid, though.) The only major difference is the structure of the prison. It is not necessarily circular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The prison has $n$ cells, each having atleast one switch and atleast one light. Every light is completely controlled by exactly one switch. A light and its switch cannot exist in the same room. As in the earlier version, lights emit a single flash at noon, after which the prisoners are re-scattered, one prisoner per room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no dead end (or start) in the graph of the lights and switches. Meaning to say, if I select any switch in any room, find the room of its light, then see any one of the switches in that room and find the room of &lt;em&gt;that&lt;/em&gt; light, and keep on doing so, I will eventually end up in the room I started with. This effectively makes the structure a set of cycles that are interlinked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added rule: One cannot find two independent non-empty subsets of this graph. Meaning to say, if the cells are divided into two sets, there will always exist a switch in set 1 that controls a light in set 2, and a switch in set 2 that controls a light in set 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As earlier, you can devise a plan for the other prisoners and follow a different strategy yourself. And the task is same, &lt;strong&gt;find the value of $n$ from any one prisoner's point of view&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Extra&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If this gets solved, try&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a method that enables a prisoner to figure out the entire structure, rather than just $n$&lt;/li&gt;&#xA;&lt;li&gt;a method that does not involve randomness &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Please clearly specify if you are solving a variant and not the actual puzzle.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2020-05-29T16:51:55.270" Title="The Unknown Prison of Unknown Size" Tags="|mathematics|strategy|story|graph-theory|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24781" PostTypeId="2" ParentId="24769" CreationDate="2015-12-22T15:23:29.413" Score="4" Body="&lt;p&gt;The solution is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As stated in the title&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;What&quot; is the solution&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17913" LastActivityDate="2015-12-22T15:23:29.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24782" PostTypeId="2" ParentId="24773" CreationDate="2015-12-22T15:57:44.730" Score="1" Body="&lt;p&gt;I think:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;All&lt;/strong&gt; starting positions are &lt;strong&gt;valid&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Asume we have all our chips are red, and we want to turn them white but one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can start from one edge choosing every 2 mod 3 chip for some chips (even 0). Then at one point, we stop choosing 2 mod 3 and we start with all 0 mod 3 after that moment, leaving one red, until the end (the last one will flip only 2 chips, because it's in the edge).&#xA;That means that the starting position of one chip on one 1 mod 3 position is valid. But 1 mod 3 is 0 mod 3 by simmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all 2 mod 3, we can start fliping all 1 mod 3 chips for some chips (even 0). Then at one point we stop choosing 1 mod 3 and start choosing all 0 mod 3 until the end, leaving one red at one 2 mod 3. (2 mod 3 it 2 mod 3 by simmetry ;) )&lt;/p&gt;&#xA;" OwnerUserId="14068" LastEditorUserId="14068" LastEditDate="2015-12-22T16:07:40.183" LastActivityDate="2015-12-22T16:07:40.183" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24783" PostTypeId="2" ParentId="24728" CreationDate="2015-12-22T17:35:54.590" Score="2" Body="&lt;p&gt;My guess and reasoning is as under.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's in ninth pillar, hidde$N \  IN \ TH$is contains ninth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user17894" LastEditorUserId="9892" LastEditDate="2015-12-22T19:26:54.830" LastActivityDate="2015-12-22T19:26:54.830" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24784" PostTypeId="2" ParentId="24744" CreationDate="2015-12-22T17:38:41.783" Score="2" Body="&lt;p&gt;Answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;-19. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Every number is mean of two digits exactly below it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user17894" LastEditorUserId="16846" LastEditDate="2015-12-22T18:08:20.990" LastActivityDate="2015-12-22T18:08:20.990" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24787" PostTypeId="2" ParentId="24769" CreationDate="2015-12-22T19:28:52.557" Score="0" Body="&lt;p&gt;What is the solution?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The solution is a word made up of 8 letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2444" LastActivityDate="2015-12-22T19:28:52.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24790" PostTypeId="2" ParentId="24773" CreationDate="2015-12-22T21:00:04.827" Score="7" Body="&lt;p&gt;As has been conjectured by others, this is solvable for only the chip in the middle. To show this, let us look at the problem in a new way; instead of considering which chips are red, let us consider which have been chosen an odd number of times.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;___*___&#xA;___X___&#xA;__XX___&#xA;__XXX__&#xA;_XXXX__&#xA;_XXXXX_&#xA;_XX_XX_&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where a move is legal on the starting square (marked by &lt;code&gt;*&lt;/code&gt;) when it has either $0$ or $2$ neighbors which are $X$'s (meaning it remains red) and a move is legal on any other square when it has exactly $1$ neighbor which is an $X$ (meaning it was flipped from white to red). Obviously, we win when a move would be legal on &lt;em&gt;every&lt;/em&gt; square, since that means that all squares are red. This is the case in the last line of the solution above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will show that all reachable positions have a common form. In particular, any position may be reached by alternatively adding and subtracting intervals from atop each other where each successive interval contains the initial position and is strictly contained in the previous interval. So, for instance, here is a few steps of addition and subtraction showing the form of such an interval:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;__________*_______________&#xA;_XXXXXXXXXXXXXXXXXXXXXX__&#xA;_X___________________XX__&#xA;_X_XXXXXXXXXXXXXXXX__XX__&#xA;_X_XXX____________X__XX__&#xA;_X_XXX_XXXXXXX____X__XX__&#xA;_X_XXX_XX___XX____X__XX__&#xA;_X_XXX_XX_X_XX____X__XX__&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where all the lines are legal positions. Notice that any such position is easily obtainable since one can flip the center when it is strictly contained in an interval (as it would have either $2$ or $0$ neighbors) and then one may expand out the new interval until it hits the boundary of the old interval, since until that point, the change from new interval to old interval would make moves legal at the edge where it went from all X's to all _'s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, if we are in a state which is such a nest of intervals, any legal move leads us to a state which is still expressible as a nest of intervals. A convenient way to show this is to note that an equivalent characterization is that &quot;the number of changes from &lt;code&gt;_&lt;/code&gt; to &lt;code&gt;X&lt;/code&gt; is the same going out from the initial block to either edge&quot; To show this, one may consider two cases. First, if the move must either be in the original chip's position, meaning that it was either contained wholly in some other interval (meaning both sides gain a change) or was the unique point in some interval (meaning both sides lose a change). Otherwise, the move must have occurred at a point &lt;code&gt;.&lt;/code&gt; in the pattern &lt;code&gt;X._&lt;/code&gt; or &lt;code&gt;_.X&lt;/code&gt;. Either way, the number of changes does not depend on whether &lt;code&gt;.&lt;/code&gt; is &lt;code&gt;X&lt;/code&gt; or &lt;code&gt;_&lt;/code&gt; so this move maintains the nested interval property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that this is a complete characterization of the legal moves from a given center. The important thing to note is that there is are precisely two solutions in terms of where the $X$'s go for any starting solution. To prove this, notice that flipping any interior position flips two pieces and that no non-trivial such set of flips yields no change, as the rightmost/leftmost affected piece are affected precisely once. Thus, any change affected by interior pieces may be made in precisely one way. Then, a boundary piece may be flipped, and if so, so must the other one - giving one degree of freedom in whether to flip both boundaries. Moreover, by the argument presented by @DrunkWolf, odd positions are non-solvable. One may check that the following four patterns solve every possible board (as long as the length of the board is of the form $4n-1$ and the &lt;code&gt;*&lt;/code&gt; is at an even position). In particular, if &lt;code&gt;*&lt;/code&gt; is at a position of the form $4n$ then the following two patterns apply:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;...___________*___________...&#xA;..._XX__XX__XX_XX__XX__XX_...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;...___________*___________...&#xA;...XX__XX__XX___XX__XX__XX...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and if &lt;code&gt;*&lt;/code&gt; is at a position of the form $4n+2$ then the following two patterns apply:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;..._____________*_____________...&#xA;..._XX__XX__XX__X__XX__XX__XX_...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;..._____________*_____________...&#xA;...XX__XX__XX__XXX__XX__XX__XX...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Merely verifying that the relevant two these are indeed solutions for any setup suffices to show that they are the unique two solutions. Note that all solutions are shown on a row where, without the ellipses, they are valid. All of these change state every two lengths, meaning that, unless they are within one chip of the center, the number of changes from &lt;code&gt;X&lt;/code&gt; to &lt;code&gt;_&lt;/code&gt; will not be balanced on each side, meaning that the move will not be possible to reach. Conversely, if &lt;code&gt;*&lt;/code&gt; is directly in the center, either solution is symmetric, meaning both sides obviously have the same number of changes and the solution is obtainable (and we can achieve it without flipping the outer pieces, since exactly one of the solutions above will do that and the other will not)&lt;/p&gt;&#xA;" OwnerUserId="4765" LastEditorUserId="4765" LastEditDate="2015-12-22T21:33:10.130" LastActivityDate="2015-12-22T21:33:10.130" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24791" PostTypeId="2" ParentId="16168" CreationDate="2015-12-22T21:34:26.707" Score="4" Body="&lt;p&gt;Below is, in my opinion, the cleanest possible presentation of the solution. The method is the same as Gil's other than rewording, but the proof at the end is different. As a fun fact, the process runs for at most $n^2 + 2^n + n^22^{2n}$ steps.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;First, we demonstrate how all the prisoners can determine an upper bound on their number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initially, let $m=1$, and let's say that the leader (i.e. you) is &quot;active,&quot; but no one else is. For the next $m$ days, everyone who is active flashes, and every inactive prisoner who gets flashed becomes active. After these $m$ days, between $\min(m+1,n)$ and $2^m$ people are active (since the number of active people at most doubles each day, and as long as not everyone is active, it increases by at least one). Then, for the next $2^m$ days, everyone who is active flashes, and everyone who &lt;em&gt;doesn't&lt;/em&gt; get flashed becomes &lt;em&gt;inactive&lt;/em&gt;. If there were any inactive people, then at least one person will be deactivated each day. Otherwise, everyone stays active.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end of these $m+2^m$ days, everyone is now either active or inactive. In the former case, the prisoners now know $n\le 2^m$. In the latter case, they increment $m$ and try again. They are guaranteed to stop when $m=n-1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From now on, assume the prisoners know an upper bound, $B$. Given a subset $S$ of prisoners, where everyone knows whether or not they are in $S$, we know define a process called Signal$(S)$; by the end of this process, everyone will know whether $S$ is nonempty, so that people in $S$ have signaled their presence. Signal$(S)$ lasts for $B$ days. Initially, people in $S$ are &quot;active,&quot; and everyone else inactive. Each day, active people flash, and people who are flashed become active. On the last day, everyone either flashed or didn't, according to whether $S$ was nonempty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the whole algorithm. At any point, the prisoners will be partitioned into some number of (nonempty) sets, $S_1,\dots S_k$, where every prisoner knows which set they are in. Initially, $S_1$ contains only you, and $S_2$ has everyone else.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The prisoners find an upper bound, $B$.&lt;/li&gt;&#xA;&lt;li&gt;For every proper, nonempty subset $A$ of $\{1,2,\dots,k\},$&lt;br&gt;&#xA;a. [1 day] All prisoners in $\bigcup_{i\in A}S_i$ flash. Let $F$ be the set of people flashed on this day.&lt;br&gt;&#xA;b. For each $1\le j\le k$,&lt;br&gt;&#xA; $\quad$i. [B days] Signal$(S_j\cap F).$&lt;br&gt;&#xA; $\quad$ii. [B days] Signal$(S_j\setminus F)$.&lt;br&gt;&#xA; $\quad$iii. If both $S_j\cap F$ and $S_j\setminus F$ are nonempty, replace $S_j$ with $S_j\cap F$, increment $k$, and&lt;br&gt;&#xA;$\qquad$ set $S_{k+1}=S_j\setminus F$.&lt;/li&gt;&#xA;&lt;li&gt;If $k$ increased during step 2, repeat step 2. Otherwise, the prisoners stop, and calculate $n$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The above loop will eventually stop, since $k$ can't be greater than $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how do they calculate $n$ once this stops? Because the $k$ did not increase, for every set $S_j$, either everyone or no one in $S_j$ was flashed during 2(a). Let $x_{i}$ be the number of people in $S_i$. Also, let $A'$ be the set of $j$ for which $S_j$ was flashed on step 2(a). Note that $A'\setminus A$ must be nonempty; otherwise, everyone in $\bigcup_{i\in A}S_i$ would be flashing themselves, creating a closed loop, contradicting that $A$ was proper (not all of $\{1,\dots,k\})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the number of people who flashed during 2a) equals the number who were flashed, the prisoners now know that $$\sum_{i\in A}x_i=\sum_{j\in A'}x_j.$$ This means the prisoners know a systems of $2^k-2$ equations about the variables $x_1,x_2,\dots,x_k$. Additionally, they know $x_1=1$. We prove that such a system of equations has a unique positive integer solution. The prisoners proceed to guess this solution by trial and error, and win by announcing that $n=\sum_i x_i$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Consider a system of equations in unknowns $x_{1},\dots x_k$, where for each nonempty proper subset $A$ of indices, there is another subset $A'$ where $A'\setminus A$ is nonempty, and an equation of the form $\sum_{i\in A} x_i = \sum_{j\in A'} x_j$. This system as at most one positive integer solution where $x_1=1$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof:&lt;/em&gt; Let $x_1,\dots,x_k$ and $y_1,\dots,y_k$ be two such solutions. Let $r$ a minimal value of $\frac{y_i}{x_i}$ for all choices of $i$, and let $z_i=y_i-rx_i$. This implies each $z_i$ is nonnegative, at least one $z_i=0$, and that the numbers $z_i$ solve the same equations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose not all $z_i$ are zero. Let $Z$ be the set of $i$ for which $z_i=0$. Then there would be a set $Z'$ for which $\sum_{i\in Z} z_i = \sum_{j\in Z'} z_j$, and where $Z'\setminus Z$ is nonempty. But every summand on the left is zero, and at least one on the right is positive, so this is a contradiction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, all $z_i$ are zero, so each $y_i=rx_i$. Since $x_1=y_1=1$, this implies $r=1$, so in fact each $x_i=y_i$, showing any solution is unique.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-12-22T21:50:28.067" LastActivityDate="2015-12-22T21:50:28.067" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24792" PostTypeId="1" AcceptedAnswerId="24842" CreationDate="2015-12-22T21:41:00.000" Score="25" ViewCount="1568" Body="&lt;p&gt;&lt;strong&gt;Question: Who is in first place?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9zP1H.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9zP1H.png&quot; alt=&quot;Rainbow Road Level 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It's difficult to add hints in the puzzle itself given its nature so I'll try to add some notes here as text instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Red: [Solved]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orange: [Solved]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yellow:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; K1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="30903" LastEditDate="2017-03-09T02:10:25.117" LastActivityDate="2020-08-27T18:27:32.893" Title="Rainbow Road Rabbit Hole" Tags="|enigmatic-puzzle|steganography|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24794" PostTypeId="1" CreationDate="2015-12-22T21:59:42.930" Score="6" ViewCount="625" Body="&lt;p&gt;Wandering around the Mojave Wasteland, you stumble upon cave with a sealed door. In the cave, there is a terminal glowing mysteriously in greenish light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/k50Lb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/k50Lb.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Not the actual terminal)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking closer, you see that it is displaying the number 42. You know from the quest that the number has been drawn from a random uniform distribution (1 to N, where each distribution is equally likely). To enter the door you must enter the maximum number N of this distribution. Which is the most likely?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Enter number N:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17918" LastActivityDate="2015-12-23T04:56:42.403" Title="The Cave and the Terminal" Tags="|mathematics|" AnswerCount="4" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="24795" PostTypeId="2" ParentId="24794" CreationDate="2015-12-22T22:04:36.940" Score="10" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is 42.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If it was chosen in the range [1,42], then the probability is 1/42. If it was larger, for instance 43, it would be 1/43. It cannot be smaller than 42 for obvious reasons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17698" LastActivityDate="2015-12-22T22:04:36.940" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24796" PostTypeId="2" ParentId="24792" CreationDate="2015-12-22T23:51:11.630" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;Downloading the image and using MS Paint's fill tool revealed this:&lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/hyrXYm.png&quot; alt=&quot;&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hyrXY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/hyrXY.png&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;Going to that URL revealed this:&lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/9bLBvm.png&quot; alt=&quot;&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;Rotating and mirroring that image, then using the fat muppets as dashes and the thin ones as dots, revealed the following morse code string: &lt;code&gt;6HW9W PNG&lt;/code&gt;. Going to &lt;code&gt;https://i.stack.imgur.com/6HW9w.png&lt;/code&gt; revealed this link:&lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/JjYvim.png&quot; alt=&quot;&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JjYvi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/JjYvi.png&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;The coordinates are for the CIA Museum in Virginia. I feel like that's probably a hint for decoding that string, but that's as far as I can get for the moment.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17167" LastEditorUserId="-1" LastEditDate="2020-08-27T18:27:32.893" LastActivityDate="2020-08-27T18:27:32.893" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24797" PostTypeId="2" ParentId="24758" CreationDate="2015-12-23T01:01:03.810" Score="10" Body="&lt;p&gt;Sander B's solution can be improved upon: Consider the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Except for the last crossing, for every pair of dwarves crossing, 1 dwarf returns, so for each trio (2 out, 1 back), the number of dwarves on the far side increases by 1. So there must be 298 such trios, with the final crossing without return bringing the last two dwarves over. In particular, 298 times, a dwarf heads back.&lt;/li&gt;&#xA;&lt;li&gt;Since no dwarf will cross more than 3 times, no dwarf can return more than once. Thus 298 different dwarves return. It is evident that the best possible result would be to leave the two slowest dwarves out of this group. Then the time spent just returning will be $$1 + 2 + \ldots + 297 + 298 = \frac{298 \times 299}{2}=44551$$ minutes.&lt;/li&gt;&#xA;&lt;li&gt;For the forward crossings, dwarves 299 and 300 will cross once, while all the other dwarves cross twice. Of course half of the values don't contribute to the actual time, but which half? Every non-contributing time has to be matched with a higher time that contributes. So 300 must contribute. If we don't match 299 with 300, then some lower number is dropped in place of 299, and 299 contributes, which increases the total. So 300 and 299 should cross together. This leaves the two 298s that cannot be shielded. Again, if they cross with anyone other than 297, 297 will contribute instead of some lower number. Thus 298 should always pair with 297. And so on. Each even-numbered dwarf must be paired with dwarf whose number is 1 less to minimize the time. If this can be accomplished, the minimum forward crossing time is $$300+2\times(2 + 4 + 6 + \ldots + 298) \\= 300 + 4\times(1 + 2 + 3 +\ldots + 149)  \\=300+4\times\frac{149\times150}{2}\\=300+44700\\=45000$$ Adding the return times gives a total time of $\ 89551$ minutes.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To accomplish it: &#xA;Break the crossings up into groups of 4 crossings: (outbound, return, outbound, return). In each group, start with the 3rd and 4th highest numbers waiting to cross first, then the two highest cross. For each return, always the return the lowest number on the outward side. So the order goes $$\color{green}{297 + 298 \rightarrow\\297 \leftarrow\\299 + 300 \rightarrow\\298 \leftarrow}\\\color{blue}{295 + 296 \rightarrow\\295 \leftarrow\\297 + 298 \rightarrow\\296 \leftarrow}\\\color{green}{293 + 294 \rightarrow\\293 \leftarrow\\295 + 296 \rightarrow\\294 \leftarrow}\\\vdots\\\color{green}{1 + 2 \rightarrow\\1\leftarrow\\3 + 4\rightarrow\\2\leftarrow}\\1 + 2 \rightarrow$$&lt;/p&gt;&#xA;" OwnerUserId="17915" LastEditorUserId="17915" LastEditDate="2015-12-23T23:25:26.807" LastActivityDate="2015-12-23T23:25:26.807" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24798" PostTypeId="2" ParentId="24794" CreationDate="2015-12-23T03:08:05.793" Score="2" Body="&lt;p&gt;The answer (without looking) must be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 42&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; any number higher gives it a lower probability of 42 occurring&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastActivityDate="2015-12-23T03:08:05.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24799" PostTypeId="2" ParentId="24794" CreationDate="2015-12-23T03:17:35.037" Score="4" Body="&lt;p&gt;This is apparently a well known problem.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Citing &lt;a href=&quot;https://en.wikipedia.org/wiki/German_tank_problem#Confidence_intervals&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, I claim &lt;strong&gt;84&lt;/strong&gt; (specifically, 2m = 2*42 = 84).  Although other statistics might also argue for &lt;a href=&quot;https://en.wikipedia.org/wiki/German_tank_problem#One_tank&quot; rel=&quot;nofollow&quot;&gt;42&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17389" LastActivityDate="2015-12-23T03:17:35.037" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24801" PostTypeId="2" ParentId="24794" CreationDate="2015-12-23T04:56:42.403" Score="1" Body="&lt;p&gt;A important piece of unspecified information is the prior distribution of N. To determine the best guess for N, we'll need to guess something about this distribution. Intuitive leaps are often important on quests.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The prior probability distribution of N is monotonically decreasing.&#xA;In other words, P(A) &amp;lt; P(B) if A &amp;lt; B.&#xA;This allows distributions such as P(N) = $2^{-N}$, a poisson distribution, and many others.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If this is true, the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;42&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The posterior probability for any number A&gt;=42 is proportional to P(A)/A, and P(42) &gt; P(B) for all B &gt; 42, so the posterior propability of A=42 is the highest.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2187" LastActivityDate="2015-12-23T04:56:42.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24803" PostTypeId="1" AcceptedAnswerId="24804" CreationDate="2015-12-23T06:42:50.300" Score="48" ViewCount="4050" Body="&lt;p&gt;George is a great puzzler, so I was extremely surprised when he didn't immediately know the answer to a really famous puzzle.  It's a puzzle that you probably did years ago, and have heard so often you can do it from memory rather than working it out.  It's also not really that difficult, so I was also surprised when it appeared to be stumping him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Come on, surely you know this one.&quot; I said.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I don't.  And don't call me Shirley.&quot; He answered grumpily.  I could tell his mood was declining rapidly, but like any great puzzler he was down and not out, and I watched his facial expression change as he reached into his mental bag of tricks. He nodded towards a conveniently located white board. &quot;Have you got a marker for that?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I handed him one, and he drew up the following diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/H7tej.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/H7tej.png&quot; alt=&quot;Reverse FCB&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He stepped back, admiring his work, beaming proudly. &quot;Well, now the solution is very obvious!&quot; he commented.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And indeed it was.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question for you is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;what was the puzzle?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="19114" LastEditDate="2016-06-06T11:34:59.970" LastActivityDate="2016-06-06T11:34:59.970" Title="Reverse Puzzling" Tags="|logical-deduction|enigmatic-puzzle|reverse-puzzling|" AnswerCount="2" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="24804" PostTypeId="2" ParentId="24803" CreationDate="2015-12-23T06:53:35.063" Score="43" Body="&lt;p&gt;I don't know if the puzzle has an established name, but it should be something similar to this: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A man has to get a Fox, a Chicken, and a sack of Barley across a river.&lt;br&gt;&#xA; &#xA; He has a rowboat, and it can only carry him and one other thing.&lt;br&gt;&#xA;&#xA; If the fox and the chicken are left together, the fox will eat the chicken.&lt;br&gt;&#xA; &#xA; If the chicken and the barley are left together, the chicken will eat the barley.&lt;br&gt;&#xA;&#xA; How does the man do it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All nodes in the graph represent all the LEGAL states in the puzzle (where nothing can eat anything else). Each line between each pair of nodes represents a single move (i.e. the man crossing the river).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; F represents the Fox, C represents the Chicken, B represents the Barley, * represents the man, and | represents the river.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The puzzle starts at the top left node, and the goal state is at the bottom right node.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (On second thought, the reverse could be true as well, if the man starts on the right side of the river instead.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14095" LastEditorUserId="-1" LastEditDate="2015-12-24T18:32:26.877" LastActivityDate="2015-12-24T18:32:26.877" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24808" PostTypeId="2" ParentId="24728" CreationDate="2015-12-23T10:19:16.193" Score="1" Body="&lt;p&gt;according to the statement I think answer is one of(f) ten i.e. IX&lt;/p&gt;&#xA;" OwnerUserId="17927" LastActivityDate="2015-12-23T10:19:16.193" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24809" PostTypeId="1" CreationDate="2015-12-23T10:45:43.647" Score="7" ViewCount="699" Body="&lt;p&gt;In the US, text painted on roads reads bottom to top:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;AHEAD&#xA;STOP&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In the UK, it reads like normal text:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; NO&#xA;ENTRY&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Obligatory XKCD:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2EDNd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2EDNd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The task here is to come up with the longest text (measured in number of words) that can be read as an &lt;em&gt;instruction to drivers&lt;/em&gt; in both US and UK use. This doesn't need to be a road safety/direction message per se, but relevant to driving or driving journeys.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &quot;STOP AHEAD&quot; could be read as &quot;AHEAD(,) STOP&quot; scoring 2, while &quot;ENTRY NO&quot; isn't valid as the only sensible expansion is &quot;ENTRY? NO&quot; and not an instruction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After having a loophole pointed out to me : the number of characters in a line should not exceed 8 unless they form a single (non-hyphenated) word. The exact number is based on examples seem in the wild (by me) &lt;/p&gt;&#xA;" OwnerUserId="8765" LastEditorUserId="8765" LastEditDate="2015-12-23T15:49:50.883" LastActivityDate="2015-12-30T00:33:02.947" Title="Transatlantic road paint" Tags="|word|english|open-ended|" AnswerCount="5" CommentCount="18" ClosedDate="2021-08-05T08:10:42.880" ContentLicense="CC BY-SA 3.0" />
  <row Id="24813" PostTypeId="2" ParentId="24809" CreationDate="2015-12-23T12:17:17.257" Score="2" Body="&lt;p&gt;Ans:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1. Please Drive Slowly =&gt; Slowly Drive Please&lt;br&gt;&#xA;2. Please Turn on Green =&gt; Green ON Turn Please&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3-word and 4-word answers.&lt;/p&gt;&#xA;" OwnerUserId="17437" LastEditorUserId="8765" LastEditDate="2015-12-23T12:25:45.033" LastActivityDate="2015-12-23T12:25:45.033" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24814" PostTypeId="2" ParentId="24809" CreationDate="2015-12-23T12:26:32.883" Score="8" Body="&lt;p&gt;As I mentioned in the comments, this was inspired by a real painted sign (in the UK) which was slowly revealed to me as I approached it in the dark:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LEFT&lt;p&gt;&#xA; TURN&lt;p&gt;&#xA; ONLY&lt;p&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And means exactly the same in the US convention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://goo.gl/maps/3VCRkGKYytu&quot; rel=&quot;nofollow&quot;&gt;Here it is on street view&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Self-answer to use spoiler mask)&lt;/p&gt;&#xA;" OwnerUserId="8765" LastEditorUserId="8765" LastEditDate="2015-12-23T18:30:49.233" LastActivityDate="2015-12-23T18:30:49.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24815" PostTypeId="2" ParentId="24803" CreationDate="2015-12-23T14:04:57.707" Score="9" Body="&lt;p&gt;It looks to me like:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You (represented by &lt;code&gt;*&lt;/code&gt;) have a fox (&lt;code&gt;F&lt;/code&gt;), cat (&lt;code&gt;C&lt;/code&gt;) and a bird (&lt;code&gt;B&lt;/code&gt;) which you have to get over a river (&lt;code&gt;|&lt;/code&gt;).  and there's an error &lt;code&gt;C*|FC&lt;/code&gt; should be &lt;code&gt;B|*FC&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastActivityDate="2015-12-23T14:04:57.707" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24816" PostTypeId="2" ParentId="24809" CreationDate="2015-12-23T14:32:01.330" Score="2" Body="&lt;p&gt;As @xnor says, there's always a way of breaking &quot;the longest&quot;. Here is one such way (breaking method in bold:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Please&lt;p&gt;&#xA; Turn&lt;p&gt;&#xA; Left&lt;p&gt;&#xA; &lt;strong&gt;Then&lt;/strong&gt;&lt;p&gt;&#xA; Indicate&lt;p&gt;&#xA; Right&lt;p&gt;&#xA; &lt;strong&gt;Then&lt;/strong&gt;&lt;p&gt;&#xA; Turn&lt;p&gt;&#xA; Right&lt;p&gt;&#xA; &lt;strong&gt;Then&lt;/strong&gt;&lt;p&gt;&#xA; Indicate&lt;p&gt;&#xA; Left&lt;p&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;becomes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Left&lt;p&gt;&#xA; Indicate&lt;p&gt;&#xA; &lt;strong&gt;Then&lt;/strong&gt;&lt;p&gt;&#xA; Right&lt;p&gt;&#xA; Turn&lt;p&gt;&#xA; &lt;strong&gt;Then&lt;/strong&gt;&lt;p&gt;&#xA; Right&lt;p&gt;&#xA; Indicate&lt;p&gt;&#xA; &lt;strong&gt;Then&lt;/strong&gt;&lt;p&gt;&#xA; Left&lt;p&gt;&#xA; Turn&lt;p&gt;&#xA; Please&lt;p&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This format can be extended/repeated ad nauseum.&lt;/p&gt;&#xA;" OwnerUserId="9444" LastEditorUserId="9444" LastEditDate="2015-12-23T15:49:19.820" LastActivityDate="2015-12-23T15:49:19.820" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="24817" PostTypeId="2" ParentId="15727" CreationDate="2015-12-23T16:01:12.733" Score="-1" Body="&lt;p&gt;Envelope. Eye is a potential alternative.&#xA;SECOND questions answer is &lt;strong&gt;-postal service&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="17933" LastActivityDate="2015-12-23T16:01:12.733" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24819" PostTypeId="1" AcceptedAnswerId="24820" CreationDate="2015-12-23T16:45:25.417" Score="15" ViewCount="995" Body="&lt;p&gt;On the table there are $20$ coins that carry the letters $A,B,C,\ldots,T$.&#xA;The values of these coins are $1,2,3,\ldots,20$ Euro (but the assignment&#xA;of values to letters is secret).&#xA;Fredo knows the value of every single coin, while Cosmo does not know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In every round of the game, Cosmo must point at three different coins &#xA;on the table (which we call the first, second, third coin). &#xA;Fredo pockets the first coin, and then tells Cosmo truthfully which of &#xA;the other two coins has higher value. &#xA;The game then goes on with the remaining coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cosmo may end the game after any round, by pointing at two coins still&#xA;on the table and announcing that their total value is at least $28$ Euro.&#xA;Cosmo wins the game if and only if his announcement is correct.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; How can Cosmo enforce a win?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2016-01-01T20:04:09.967" Title="Twenty coins and no balance" Tags="|mathematics|strategy|game|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24820" PostTypeId="2" ParentId="24819" CreationDate="2015-12-23T17:06:42.613" Score="12" Body="&lt;p&gt;Divide the coins into 3 groups.&lt;br&gt;&#xA;Take 3 from the first group.&lt;br&gt;&#xA;Of the remaining two, select the smaller one first, the bigger one second, then a new coin 3rd.&lt;br&gt;&#xA;Repeat with the rest of the coins in the group.&lt;br&gt;&#xA;Repeat with the other group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end, we'll have 3 groups, with minimum values 5, 11, 17 and an extra coin in each group.  Use two of the smaller coins to pick the two largest values (11 and 17), minimum 28. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know the groups have those minimum values by checking the worst case scenarios.&lt;br&gt;&#xA;We could bump off 18-20 with our first guess in each group, leaving the max at 17. The 17 could be in a group with 20, 17, 16, 15, 14, 13, 12, leaving the next highest number at 11. Repeat leaving the smallest group at 5. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2015-12-23T19:16:01.030" LastActivityDate="2015-12-23T19:16:01.030" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24821" PostTypeId="2" ParentId="24780" CreationDate="2015-12-23T17:36:58.497" Score="3" Body="&lt;p&gt;If there is no known maximum number of switches and bulbs, the first part of the solution (finding a known upper bound) becomes much harder (refer to the other puzzle for the approach that's workable there).  In the original puzzle, that involves for each &quot;attempt&quot; making sure that the number of prisoners who have seen light is at least $N$ and and at most $2^N$, and then using a procedure which, if any prisoners haven't seen light, will each day increase by 1 the number of prisoners who know that not everybody has seen the light.  The same principle can be applied, even in the generalized problem, but the number of prisoners who might see light increases a according to $N!$ rather than $2^N$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For simplicity, regard the boss as a &quot;prisoner who has seen the light going into the first day&quot;.  On the first day, the boss alone turns on one light.  On the second day, the boss and the prisoner who saw the first light must turn on two lights if possible to ensure that someone will see light who hasn't done so already.  After the second day, it's possible that as few as three people (including the boss) may have seen light, but as many as six might have done so.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now comes an interesting bit.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; On the third day, it's not necessary that prisoners ensure that anyone new see the light, but merely to ensure that by the end of the day at least four people (including the boss) have done so.  If at least one of the prisoners who has seen light (or is the boss) is able to turn on three light switches, then three other prisoners will see light next turn, bringing the total up to four prisoners.  If none of them have three or more switches, then having them all turn on all their switches will result in at least one new person seeing light.  Thus, weakest case ends with four prisoners plus the boss having seen the light; strongest with 24 (six people each turn on three switches, illuminating 18 new people)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For succeeding days:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1&lt;br&gt;1+1=2&lt;br&gt;2+(2*2)=6&lt;br&gt;6+(6*3)=24&lt;br&gt;24+(24*4)=120&lt;br&gt;120+(120*5)=720&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus setting up the $N!$ progression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once an upper bound on the number of prisoners is established, the approach used in the other puzzle for determining an exact number may be used.  Determining the topology of the switches is apt to be difficult, and I don't think a full mapping will be possible without randomness or a means of hiding one's plans from the warden:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If in a prison with 14 or more cells there are five rooms with two switches and one light, and none of them control each other's light, a warden will always be able to ensure that the lights are always all light or all dark (by suitably arranging nine prisoners in the five cells that control them) thus rendering them indistinguishable from each other.  The warden will then always be able to ensure that the first two of the rooms are always populated by prisoners who will flip the switches the same way given the same circumstance (there are four possible prisoner actions; by the pigeonhole principle, if there are five prisoners two must perform the same action).  Since there is no way prisoners can operate those switches independently, there is no way to disambiguate what they control.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1467" LastActivityDate="2015-12-23T17:36:58.497" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24822" PostTypeId="1" CreationDate="2015-12-23T17:38:21.900" Score="-2" ViewCount="7053" Body="&lt;p&gt;I ve been asked to solve this riddle;&#xA;Eat 40 eggs in 7 days.&#xA;Minimum 1 egg a day.&#xA;Only use odd numbers...&lt;/p&gt;&#xA;" OwnerUserId="17934" LastEditorUserId="16846" LastEditDate="2015-12-23T18:35:46.973" LastActivityDate="2015-12-23T18:35:46.973" Title="Divide 40 eggs over 7 days, use odd numbers" Tags="|riddle|lateral-thinking|" AnswerCount="4" CommentCount="0" ClosedDate="2015-12-23T21:07:25.293" ContentLicense="CC BY-SA 3.0" />
  <row Id="24823" PostTypeId="2" ParentId="24809" CreationDate="2015-12-23T17:38:49.163" Score="1" Body="&lt;p&gt;From the campus where I work:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HUMP&#xA; SPEED&#xA; SLOW&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which I could parse as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HUMP SPEED: SLOW!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SLOW! SPEED HUMP.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Granted, the first interpretation isn't really driving related, so it probably isn't actually a contender. But somewhat amusing nonetheless!&lt;/p&gt;&#xA;" OwnerUserId="12369" LastActivityDate="2015-12-23T17:38:49.163" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24824" PostTypeId="2" ParentId="24822" CreationDate="2015-12-23T17:54:10.840" Score="1" Body="&lt;p&gt;The simple mathematical answer is that it is impossible. Adding an odd number of odd numbers (7 odd numbers) will always result in an odd number. As 40 is even, it is impossible to add 7 odd numbers to get an even number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a lateral thinking attempt, eat 5 eggs the first 4 days, and 7 eggs the last 3. You will have eaten a total of 41 eggs, but you will have eaten 40 eggs in 7 days. You just so happened to have eaten an extra egg (and gotten very sick).&lt;/p&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2015-12-23T17:54:10.840" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24825" PostTypeId="2" ParentId="24822" CreationDate="2015-12-23T18:20:34.400" Score="2" Body="&lt;p&gt;Using only odd numbers:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Eat 1 egg on Monday&lt;/li&gt;&#xA;&lt;li&gt;Eat 1 egg on Tuesday&lt;/li&gt;&#xA;&lt;li&gt;Eat 1 egg on Wednesday&lt;/li&gt;&#xA;&lt;li&gt;Eat 1 egg on Thursday&lt;/li&gt;&#xA;&lt;li&gt;Eat 1 egg on Friday&lt;/li&gt;&#xA;&lt;li&gt;Eat 1 egg on Saturday&lt;/li&gt;&#xA;&lt;li&gt;Eat 1 egg on Sunday morning&lt;/li&gt;&#xA;&lt;li&gt;Eat 33 eggs on Sunday evening&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;:-)&lt;/p&gt;&#xA;" OwnerUserId="4394" LastActivityDate="2015-12-23T18:20:34.400" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24826" PostTypeId="2" ParentId="24822" CreationDate="2015-12-23T18:22:49.637" Score="3" Body="&lt;p&gt;Since it's (obviously) impossible mathematically we have to go lateral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think everyone can agree that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 34 eggs is a very &lt;strong&gt;odd&lt;/strong&gt; number of eggs to eat in a day&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So we just eat&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 every mon-saturday, going by the mathematical definition of odd. Then 34 on sunday, going by the definition in the dictionary&lt;br&gt;&lt;br&gt;&lt;strong&gt;1. Deviating from what is ordinary, usual, or expected;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16846" LastActivityDate="2015-12-23T18:22:49.637" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24828" PostTypeId="2" ParentId="24822" CreationDate="2015-12-23T18:32:28.713" Score="4" Body="&lt;p&gt;Monday to Friday, eat 5 and 1/7th eggs. Saturday and Sunday, eat 7 and 1/7th eggs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we all know that fractions are not odd numbers, however, the question doesn't state &quot;Only &lt;em&gt;eat&lt;/em&gt; odd numbers of eggs&quot;, it simply asks us to &lt;em&gt;use&lt;/em&gt; odd numbers, and the above solution uses only 1,5 and 7: all odd numbers.&lt;/p&gt;&#xA;" OwnerUserId="7459" LastActivityDate="2015-12-23T18:32:28.713" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24829" PostTypeId="2" ParentId="24819" CreationDate="2015-12-23T18:41:50.893" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Label the coins 1 through 20.&lt;br&gt;&#xA; &lt;strong&gt;Your first five moves are&lt;/strong&gt;:&lt;br&gt;&#xA; - Give up 16, and weigh 1 vs 2&lt;br&gt;&#xA; - Give up 17, and weigh 3 vs 4&#xA; - ...&lt;br&gt;&#xA; - Give up 20, and weigh 9 vs 10&lt;br&gt;&#xA; (The worst case here is that we've given up coins valued 16 to 20.)&lt;br&gt;&#xA; Assume without loss of generality that $2&amp;gt;1, 4&amp;gt;3,\ldots, 10&amp;gt;9$.&lt;br&gt;&#xA; &lt;strong&gt;Your next five moves are&lt;/strong&gt;:&lt;br&gt;&#xA; - Give up 1, and weigh 2 vs 11&lt;br&gt;&#xA; - Give up 3, and weigh 4 vs 12&lt;br&gt;&#xA; - Give up 5, and weigh 6 vs 13&lt;br&gt;&#xA; - Give up 7, and weigh 8 vs 14&lt;br&gt;&#xA; - Give up 9, and weigh 10 vs 15&lt;br&gt;&#xA; Assume wlog that $11&amp;gt;2,12&amp;gt;4,\ldots,15&amp;gt;10$&lt;br&gt;&#xA; The worst case here is that we've given up one of 13 or 14. Which means we now have 5 coins ($11,12,\ldots,15$) the top two of which weigh at least 28 (either 15+14 or 15+13).&lt;br&gt;&#xA; We also have five weighings left, by giving up 2,4,6,8, and 10.&lt;br&gt;&#xA; &lt;strong&gt;Last five weighings to determine the top two of the group:&lt;/strong&gt;&lt;br&gt;&#xA;  - 11 vs 12&lt;br&gt;&#xA;  - 13 vs 14&lt;br&gt;&#xA; Assume wlog $12&amp;gt;11$ and $14&amp;gt;13$.&lt;br&gt;&#xA;  - 12 vs 14&lt;br&gt;&#xA; Assume wlog $14&amp;gt;12$&lt;br&gt;&#xA;  - 15 vs 12 (controversial, but it's the only way it works)&lt;br&gt;&#xA; &lt;strong&gt;Case 1: 15&gt;12:&lt;/strong&gt;&lt;br&gt;&#xA; This eliminates 11 from the top two ($15&amp;gt;12&amp;gt;11$) and 12 from the top two ($15&amp;gt;12$ and $14&amp;gt;12$). The top three can therefore be $15&amp;gt;14&amp;gt;13$, $14&amp;gt;15&amp;gt;13$, or $14&amp;gt;13&amp;gt;15$. So weigh 15 vs 13 and eliminate the smaller of the two.&lt;br&gt;&#xA; &lt;strong&gt;Case 2: 15&amp;lt;12:&lt;/strong&gt;&lt;br&gt;&#xA; In this case, $14&amp;gt;13$, $14&amp;gt;12&amp;gt;15$, and $14&amp;gt;12&amp;gt;11$. This eliminates 11 and 15, and leaves either $14&amp;gt;13&amp;gt;12$ or $14&amp;gt;12&amp;gt;13$. So weigh 12 vs 13 and eliminate the smaller of the two.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15942" LastActivityDate="2015-12-23T18:41:50.893" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24830" PostTypeId="2" ParentId="24792" CreationDate="2015-12-23T18:55:25.713" Score="4" Body="&lt;p&gt;The coordinates in the third image are the location of the Kryptos sculpture. Decoding 'e8efe.ctn' using&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the method for the first part of Kryptos (keyed Vigenere with keys &quot;kryptos&quot; and &quot;palimpsest&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;provides the location of the next image: &lt;a href=&quot;https://i.stack.imgur.com/b8qwn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/b8qwn.png&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This appears to be several barcodes. The one in the top left looks like an &lt;a href=&quot;https://en.wikipedia.org/wiki/Aztec_Code&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aztec Code&lt;/a&gt;, while The large one on the bottom looks like a &lt;a href=&quot;https://en.wikipedia.org/wiki/MaxiCode&quot; rel=&quot;nofollow noreferrer&quot;&gt;MaxiCode&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The third barcode is &lt;a href=&quot;https://en.wikipedia.org/wiki/Code_39&quot; rel=&quot;nofollow noreferrer&quot;&gt;Code 39&lt;/a&gt; with the information &lt;code&gt;T+CT&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="30903" LastEditDate="2017-03-09T02:07:14.210" LastActivityDate="2017-03-09T02:07:14.210" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24832" PostTypeId="1" AcceptedAnswerId="24833" CreationDate="2015-12-23T19:17:39.330" Score="5" ViewCount="898" Body="&lt;p&gt;You have been given a blank 31 x 31 grid as shown below. Each square in the grid is either &quot;light&quot; or &quot;dark&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numbers to the right of the grid represent the consecutive sequences of &quot;light&quot; squares in each row while the numbers in the bottom represent the consecutive sequence of &quot;light&quot; squares in each vertical column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/h6m9a.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/h6m9a.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For those interested in text (edited to fix line breaks).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rows: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 3 2 2 2 2 3 1&#xA;1 3 1 3 1 3 1 3 1 3 1&#xA;1 3 1 3 1 3 1 3 2 1 2&#xA;1 1 1 2 4 3&#xA;1 3 1 3 1 5 7 3&#xA;1 3 1 3 1 5 7 3&#xA;1 3 1 3 1 5 7 3&#xA;31&#xA;7 3 1 1 3 7&#xA;7 3 1 5 3 7&#xA;7 2 1 5 3 7&#xA;7 2 2 2 2 1 1 7&#xA;7 2 1 5 1 1 7&#xA;7 3 1 5 1 1 7&#xA;7 3 1 2 1 8&#xA;31&#xA;4 3 1 2 2 5&#xA;4 3 1 5 3 1 3 4&#xA;5 1 2 5 3 3 4&#xA;6 3 5 3 1 1 6&#xA;6 3 5 3 1 2 5&#xA;6 3 1 3 1 3 5&#xA;6 3 1 3 1 3 4&#xA;31&#xA;5 3 3 4 7&#xA;4 3 1 3 1 3 9&#xA;8 1 3 1 3 9&#xA;7 2 1 1 2 2 7&#xA;6 3 2 2 2 3 6&#xA;5 4 2 2 3 6&#xA;4 2 2 2 7&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Columns:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;31&#xA;24&#xA;3 27&#xA;3 27&#xA;3 12 7 4&#xA;11 5 4&#xA;19 1 3 1&#xA;1 1 3 5 2 2&#xA;2 6 5 7 3&#xA;2 7 19&#xA;2 8 3 2 1&#xA;1 1 1 2 2 1 3 1&#xA;16 2 2 1 2 2&#xA;1 1 2 2 1 1 3&#xA;2 4 2 2 1 5 2 1&#xA;2 4 2 2 16&#xA;2 4 2 2 2 2 4&#xA;1 5 5 1 2 4&#xA;16 2 10&#xA;1 2 2 11&#xA;2 10 2 3 1&#xA;2 7 11 1 2&#xA;2 10 1 1 2 2&#xA;1 5 2 2 4 2 2&#xA;16 2 7 1&#xA;17 2 1 9&#xA;2 14 3 8&#xA;3 24&#xA;2 28&#xA;29&#xA;31&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Hint&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For example, a 31 to the right of the grid means that all the squares in that row are &quot;light&quot; colored.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9892" LastEditorUserId="1600" LastEditDate="2015-12-25T18:18:01.500" LastActivityDate="2016-12-28T13:35:50.307" Title="Decipher the message in the 31x31 grid" Tags="|nonogram|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24833" PostTypeId="2" ParentId="24832" CreationDate="2015-12-23T19:54:13.393" Score="8" Body="&lt;p&gt;Pretty sure the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Happy new year 2016&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/P61Tw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/P61Tw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;  Like others have mentioned tough, numbers weren't entire correct, as you can see from the broken r and weird 6.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16846" LastEditorUserId="25957" LastEditDate="2016-12-28T13:35:50.307" LastActivityDate="2016-12-28T13:35:50.307" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24834" PostTypeId="2" ParentId="24832" CreationDate="2015-12-23T20:09:04.070" Score="4" Body="&lt;p&gt;Although some of the items were off by a little, I think this is what you were aiming for:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/7L8tU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7L8tU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By &quot;off by a little&quot;, I mean that row 12 doesn't exactly look good as you posted, so I took a little artistic license, and modified it.&#xA;Instead of 7 2 2 2 2 1 1 7, it should have been 7 1 1 1 2 1 1 7.  There may have been more small errors like that, but once I got started I was able to finish it mostly without using the numbers.&lt;/p&gt;&#xA;" OwnerUserId="12061" LastEditorUserId="12061" LastEditDate="2015-12-23T20:17:09.267" LastActivityDate="2015-12-23T20:17:09.267" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24835" PostTypeId="1" CreationDate="2015-12-23T20:10:54.990" Score="2" ViewCount="1017" Body="&lt;p&gt;You are standing in front of water-filled pool with hungry and aggressive human-eating sharks. You scratch your skin so that you bleed, which will make the sharks even more hungry. You then take a leap into the pool and walk slowly until you reach the other end of the pool. You step out of the pool unharmed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The sharks are alive and you are walking in the same pool as the sharks)&lt;/p&gt;&#xA;" OwnerUserId="17698" LastActivityDate="2015-12-23T23:12:48.270" Title="A pool of hungry sharks" Tags="|lateral-thinking|" AnswerCount="4" CommentCount="4" FavoriteCount="0" ClosedDate="2015-12-23T21:22:51.137" ContentLicense="CC BY-SA 3.0" />
  <row Id="24836" PostTypeId="2" ParentId="24835" CreationDate="2015-12-23T20:29:47.320" Score="10" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you are not a human, so the human-eating sharks don't care&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13101" LastActivityDate="2015-12-23T20:29:47.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24837" PostTypeId="2" ParentId="24835" CreationDate="2015-12-23T20:31:53.480" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are lying. There are no sharks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12901" LastActivityDate="2015-12-23T20:31:53.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24838" PostTypeId="2" ParentId="24835" CreationDate="2015-12-23T20:47:16.167" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The pool is a circle with diameter of 100 feet, and depth of 30 feet. But across the middle of the pool is a rectangular section that is one foot wide and only 1 inch deep. You walk across this section. The sharks are on either side of the section so they cannot hurt you. It looks something like this:&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/o4Slh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/o4Slh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="7448" LastEditDate="2015-12-23T23:12:48.270" LastActivityDate="2015-12-23T23:12:48.270" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24839" PostTypeId="2" ParentId="24835" CreationDate="2015-12-23T20:47:55.300" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - There are fences/nets keeping the sharks away from you.&lt;br&gt;&#xA; - You wear plate armor.&lt;br&gt;&#xA; - You are Keanu Reeves, there are no &lt;s&gt;spoons&lt;/s&gt; sharks.&lt;br&gt;&#xA; - There was a substance in the water which made all sharks fall asleep.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2015-12-23T20:53:17.260" LastActivityDate="2015-12-23T20:53:17.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24842" PostTypeId="2" ParentId="24792" CreationDate="2015-12-24T06:41:38.080" Score="17" Body="&lt;p&gt;This puzzle is rainbow (road)-themed, and consists of a series of colored images.  Each image contains the URL of the next image, encoded or obfuscated in some fashion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/a/24796/9718&quot;&gt;Steve Eckert started us off by solving the &quot;red&quot; and &quot;orange&quot; images.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;red&quot; image (the image in the OP) used a traditional &lt;a href=&quot;https://en.wikipedia.org/wiki/Steganography&quot; rel=&quot;nofollow noreferrer&quot;&gt;steganographic&lt;/a&gt; technique:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the URL was hidden in plain text, in a color one bit different from the background.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://i.stack.imgur.com/9bLBv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;orange&quot; image&lt;/a&gt; uses a classic encoding method:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The URL is encoded as &lt;a href=&quot;https://en.wikipedia.org/wiki/Morse_code&quot; rel=&quot;nofollow noreferrer&quot;&gt;Morse code&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/a/24830/9718&quot;&gt;f'' was next to step in, solving&lt;/a&gt; the &lt;a href=&quot;https://i.stack.imgur.com/JjYvi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;yellow&quot; image:&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The URL was enciphered using the same &lt;a href=&quot;https://en.wikipedia.org/wiki/Vigen%C3%A8re_cipher&quot; rel=&quot;nofollow noreferrer&quot;&gt;Vigenere cipher&lt;/a&gt; as the famous &lt;a href=&quot;https://en.wikipedia.org/wiki/Kryptos&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kryptos sculpture&lt;/a&gt; whose coordinates were in the image.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This revealed the &quot;green&quot; image:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/b8qwn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/b8qwn.png&lt;/a&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/b8qwn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This image contains a number of barcodes.  Each of the barcodes on the first row encodes a single character from the URL:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first is an &lt;a href=&quot;https://en.wikipedia.org/wiki/Aztec_Code&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aztec code&lt;/a&gt;.  Using &lt;em&gt;Mathematica&lt;/em&gt; to decode it produced &lt;code&gt;m&lt;/code&gt;.  (I verified its contents by hand, since I've been working a lot with 2d barcode formats in preparation for my next puzzle.)&lt;br&gt;&#xA;&lt;br&gt;&#xA; The second is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Code_128&quot; rel=&quot;nofollow noreferrer&quot;&gt;Code 128&lt;/a&gt; barcode, containing the letter &lt;code&gt;L&lt;/code&gt;.  I used &lt;em&gt;Mathematica&lt;/em&gt; to decode this one as well.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The third and fifth are &lt;a href=&quot;https://en.wikipedia.org/wiki/Code_39&quot; rel=&quot;nofollow noreferrer&quot;&gt;Code 39&lt;/a&gt; barcodes, containing &lt;code&gt;+C&lt;/code&gt; and &lt;code&gt;W&lt;/code&gt;.  The former is actually using the &lt;a href=&quot;https://en.wikipedia.org/wiki/Code_39#Full_ASCII_Code_39&quot; rel=&quot;nofollow noreferrer&quot;&gt;full ASCII extension&lt;/a&gt; to code a lowercase &lt;code&gt;c&lt;/code&gt;.  I decoded these by hand using the picture on the Wikipedia page.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The last large barcode is a &lt;a href=&quot;https://en.wikipedia.org/wiki/MaxiCode&quot; rel=&quot;nofollow noreferrer&quot;&gt;MaxiCode&lt;/a&gt;, used mainly by UPS.  This was by far the most difficult part of the puzzle, since it's hard to find information on the format, and there aren't any online MaxiCode readers (since nobody except UPS needs to &lt;em&gt;read&lt;/em&gt; the barcodes, only generate them).  I eventually installed a trial version of some computer vision software to reveal that the barcode contained only &lt;code&gt;.png&lt;/code&gt;.  (It was obvious in retrospect that the repeated pattern in the upper half of the barcode is just padding in the outer data area, since the barcode is a fixed size.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This gives us all but one of the characters in the URL.  Brute-forcing the remaining character by programmatically trying all 62 possible characters leads us to the &quot;blue&quot; image:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/mLc1W.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/mLc1W.png&lt;/a&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/mLc1W.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can see the letters &lt;code&gt;p&lt;/code&gt;, &lt;code&gt;n&lt;/code&gt;, and &lt;code&gt;g&lt;/code&gt; on the right, so it's apparent that we must rearrange the letters using the values on the left in order to retrieve the next URL.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm too impatient to figure out what $B'_L$ and $\beta$ are, so I programmatically searched all 120 permutations of the right-hand side that ended in &lt;code&gt;.png&lt;/code&gt;, leading to the &quot;magneta&quot; image:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/xfrqT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/xfrqT.png&lt;/a&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/xfrqT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; The solution seems to indicate that we were supposed to order the left-hand side by value, so we know that $\beta &amp;lt; B'_L &amp;lt; \sqrt{2}$.  It could be that $B'_L$ is supposed to be &lt;a href=&quot;https://en.wikipedia.org/wiki/Legendre%27s_constant&quot; rel=&quot;nofollow noreferrer&quot;&gt;Legendre's constant&lt;/a&gt;, $1$, in which case $\beta &amp;lt; 1$ (but I still don't know what $\beta$ is; oh well).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The next URL is encoded with a simple&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Caesar_cipher&quot; rel=&quot;nofollow noreferrer&quot;&gt;Caesar cipher&lt;/a&gt;, which I figured out by noticing the difference between &lt;code&gt;c&lt;/code&gt; and &lt;code&gt;a&lt;/code&gt; was the same as the difference between &lt;code&gt;p&lt;/code&gt; and &lt;code&gt;n&lt;/code&gt; (what I would expect to be in the plaintext there).  (As &lt;a href=&quot;https://puzzling.stackexchange.com/questions/24792/rainbow-road-rabbit-hole/24842?noredirect=1#comment76214_24842&quot;&gt;Zandar points out&lt;/a&gt;, this particular Caesar cipher is &lt;a href=&quot;http://www.rot13.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;rot13&lt;/a&gt;, with to &quot;rot5&quot; for the numbers.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using a known-plaintext attack on the extension (&lt;code&gt;.png&lt;/code&gt;), we get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;https://i.stack.imgur.com/T*gf*.png&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I wasn't sure how the cipher was supposed to affect the digits, so I once again brute-force searched all 100 possible digit assignments, which turns up (along with two false positives) the &quot;purple&quot; image:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/T1gf2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/T1gf2.png&lt;/a&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/T1gf2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can tell that we're dealing with a transposition cipher, since all the characters from the expected URL format (&lt;code&gt;https://i.stack.imgur.com/*.png&lt;/code&gt;) are here.  Playing around some reveals that it's a simple&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Transposition_cipher&quot; rel=&quot;nofollow noreferrer&quot;&gt;transposition cipher&lt;/a&gt; with 12 columns.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that in order to decode it correctly, some of the spaces in the text have to be doubled or trebled.  With the correct spacing, I get:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Y       s i a t . i o 9&#xA;o a t H   n g t s m m .&#xA;u l h e t a e p t g / j&#xA;  m e r h l : : a u 5 p&#xA;a o r e e     / c r Y g&#xA;r s e     i   / k . 8&#xA;e t . i f m h i . c u &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;or:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;You are almost there. Here is the final image:  https://i.stack.imgur.com/5Y8u9.jpg&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yielding the final part of the puzzle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/5Y8u9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/5Y8u9.jpg&lt;/a&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/5Y8u9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the spirit of the challenge, I'll present the answer as an image of my own:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/NjRIS.png&quot; alt=&quot;Peach!&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2017-03-09T02:09:15.317" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24844" PostTypeId="1" CreationDate="2015-12-24T21:39:15.250" Score="2" ViewCount="260" Body="&lt;p&gt;I'm working on creating a crossword puzzle and am unfortunately stuck with the combination ~CY~E (blank C Y blank E). Does anyone have any PC phrases or words that could fit that combination?&lt;/p&gt;&#xA;" OwnerUserId="17728" LastEditorUserId="17728" LastEditDate="2015-12-24T21:55:02.343" LastActivityDate="2015-12-28T23:54:18.167" Title="Phrase for _CY_E" Tags="|word|crosswords|" AnswerCount="6" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24846" PostTypeId="2" ParentId="24844" CreationDate="2015-12-24T21:53:17.653" Score="5" Body="&lt;p&gt;Nothing &lt;a href=&quot;http://wordsolver.net/solve#!q=_CY_E&amp;amp;type=st_crossword&amp;amp;dic=d_twl&amp;amp;ml=15&quot;&gt;particularly likely&lt;/a&gt; - I'd try to change one of the crossing words instead.&lt;/p&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="17389" LastEditDate="2015-12-24T23:09:20.243" LastActivityDate="2015-12-24T23:09:20.243" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24847" PostTypeId="1" AcceptedAnswerId="24861" CreationDate="2015-12-24T22:43:41.447" Score="9" ViewCount="1005" Body="&lt;p&gt;I can ruin a party&lt;br&gt;&#xA;Showing up uninvited&lt;br&gt;&#xA;Though some are pleased to see me&lt;br&gt;&#xA;When they get real excited&lt;br&gt;&#xA;Even those who would first shun me&lt;br&gt;&#xA;Won't mind meeting me again,&lt;br&gt;&#xA;Or after certain indications&lt;br&gt;&#xA;They'd be happy with me then&lt;br&gt;&#xA;Should I tell you my nature?&lt;br&gt;&#xA;Should I give myself away?&lt;br&gt;&#xA;Alert and ready to oblige&lt;br&gt;&#xA;When you tell me it's okay&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who am I?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; okay, this is my first riddle so may have turned out be more difficult than I thought. A bit of guidance then: every pair of lines give a clue. And here's another hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is familiar to many of you Puzzling StackExchange users. In fact, it's staring you right in the face!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user17947" LastEditorDisplayName="user17947" LastEditDate="2015-12-25T21:41:15.643" LastActivityDate="2015-12-31T23:44:23.930" Title="I can ruin a party" Tags="|riddle|word|" AnswerCount="8" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24848" PostTypeId="1" AcceptedAnswerId="24851" CreationDate="2015-12-25T00:02:19.480" Score="142" ViewCount="45285" Body="&lt;p&gt;Here is a interesting picture with two arrangements of four shapes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fWdMd.jpg&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fWdMd.jpg&quot; alt=&quot;pic&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can they make a different area with the same shapes?&lt;/p&gt;&#xA;" OwnerUserId="435" LastEditorDisplayName="user88" LastEditDate="2016-05-15T17:04:30.767" LastActivityDate="2020-03-19T01:13:07.653" Title="How can 64 = 65?" Tags="|mathematics|calculation-puzzle|geometry|lateral-thinking|" AnswerCount="8" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24849" PostTypeId="2" ParentId="24848" CreationDate="2015-12-25T00:16:46.803" Score="7" Body="&lt;p&gt;The triangles don't have the same slope; you can see that the large diagonal line through the &quot;larger&quot; rectangle bends. It's covered up by the thick lines around the triangles, but there is a very thin hole that has a total area of one square - the same square that supposedly &quot;appeared out of nowhere&quot;.&lt;/p&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2015-12-25T00:16:46.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24850" PostTypeId="2" ParentId="24848" CreationDate="2015-12-25T00:17:39.683" Score="13" Body="&lt;p&gt;It's a misleading diagram. In reality, the angles do not match up- the larger interior angle of the orange triangle is about 69.5 degrees, whereas it's 68.2 for the grey quadrilateral. (Correct me if I'm wrong- dusting off my trig here.) In the diagram with area 65, the orange areas are actually quadrilaterals. If you look closely, you can see that they have a slight inflection where they meet the other orange section. So that extra area comes from expanding them just a bit.&lt;/p&gt;&#xA;" OwnerUserId="14441" LastActivityDate="2015-12-25T00:17:39.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24851" PostTypeId="2" ParentId="24848" CreationDate="2015-12-25T00:25:58.660" Score="176" Body="&lt;p&gt;This is a famous physical puzzle that can be tied to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Fibonacci_number&quot; rel=&quot;noreferrer&quot;&gt;fibonacci series&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer the question as posed, the issue is that the two slopes are different (&lt;span class=&quot;math-container&quot;&gt;$\frac25$&lt;/span&gt; vs &lt;span class=&quot;math-container&quot;&gt;$\frac38$&lt;/span&gt;).  Note that all those numbers are in the fibonacci series (&lt;span class=&quot;math-container&quot;&gt;$1,1,2,3,5,8,13,21,\ldots$&lt;/span&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Successive fractions are closer approximations to &lt;a href=&quot;https://en.wikipedia.org/wiki/Golden_ratio&quot; rel=&quot;noreferrer&quot;&gt;&lt;span class=&quot;math-container&quot;&gt;$\varphi$&lt;/span&gt;&lt;/a&gt;, alternating between above and below.  Diagrams like this can be generated by making a square with sides equal to a number in the fibonacci series (in this question 8), then dividing it into two rectangles with widths of the two fibonacci numbers that make up the first one chosen (3 and 5).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cut the smaller one down the diagonal, and cut the bigger one down the middle at a diagonal, such that the width of the diagonal cut is the next smallest number (2 in this case).  Note that this will leave a trapezoid, whose small parallel size matches the original small rectangle's smaller side (3 in this case), and whose larger parallel size matches the original larger rectangle's smaller side (5 in this case).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since &lt;span class=&quot;math-container&quot;&gt;$\frac25\approx\frac38$&lt;/span&gt;, and from the above constructions, the pieces can be rearranged into a rectangle (as shown), the area of which will always be one away from the original square, but will look approximately correct, since the slopes almost match.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Since this answer received so many up-votes (thank you!), I suppose people are very interested in it, so I thought I'd draw up a few images!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1,1,2,3&lt;/strong&gt;: &lt;span class=&quot;math-container&quot;&gt;$3\times3 = 9 = 10 = 2\times5$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nWWOQ.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nWWOQ.png&quot; alt=&quot;1,1,2,3&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1,2,3,5:&lt;/strong&gt; &lt;span class=&quot;math-container&quot;&gt;$5\times5 = 25 = 24 = 3\times8$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7JHKs.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7JHKs.png&quot; alt=&quot;1,2,3,5&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2,3,5,8:&lt;/strong&gt; &lt;span class=&quot;math-container&quot;&gt;$8\times8 = 64 = 65 = 5\times13$&lt;/span&gt; (The OP's example)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FmJY9.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FmJY9.png&quot; alt=&quot;2,3,5,8&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3,5,8,13:&lt;/strong&gt; &lt;span class=&quot;math-container&quot;&gt;$13\times13 = 169 = 168 = 8\times21$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5OK01.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5OK01.png&quot; alt=&quot;3,5,8,13&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;5,8,13,21:&lt;/strong&gt; &lt;span class=&quot;math-container&quot;&gt;$21\times21 = 441 = 442 = 13\times34$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7B6ib.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7B6ib.png&quot; alt=&quot;5,8,13,21&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A comment from @EricJ. prompted a discussion that may be worth bringing up here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not claiming that all such puzzles are based on the Fibonacci series. Just that all Fibonacci numbers can generate these diagrams. There are several characteristics of Fibonacci numbers that make this work.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;One is that the square of a Fibonacci number alternates between being one more and one less than the product of the numbers on either side.&lt;/li&gt;&#xA;&lt;li&gt;There's the slope thing I mentioned already, meaning that our construction will give us two slopes that are approximately equal. And&lt;/li&gt;&#xA;&lt;li&gt;There's an argument that the overall construction can be done based on each number being the sum of the previous two.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The latter two points can best be understood by studying the construction in detail. The first point can be proved by induction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;RTP&lt;/strong&gt;: &lt;span class=&quot;math-container&quot;&gt;$f_k^2=f_{k-1}f_{k+1}\pm1$&lt;/span&gt; for &lt;span class=&quot;math-container&quot;&gt;$k\geq1$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're indexing so that &lt;span class=&quot;math-container&quot;&gt;$f_0=0$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$f_1=1$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 1&lt;/strong&gt;: &lt;span class=&quot;math-container&quot;&gt;$k=1$&lt;/span&gt;: &lt;span class=&quot;math-container&quot;&gt;$f_1^2=f_0f_2+1$&lt;/span&gt; can be verified by substitution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 2&lt;/strong&gt;: Assume it's true for &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt;. So &lt;span class=&quot;math-container&quot;&gt;$f_k^2 = f_{k-1}f_{k+1}\mp1$&lt;/span&gt;. (I'm using &lt;span class=&quot;math-container&quot;&gt;$\mp1$&lt;/span&gt; because I'm expecting it to alternate, and so in step 3 I'll prove it for &lt;span class=&quot;math-container&quot;&gt;$\pm1$&lt;/span&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 3&lt;/strong&gt;: We need to show that &lt;span class=&quot;math-container&quot;&gt;$f_{k+1}^2 = f_{k}f_{k+2}\pm1$&lt;/span&gt;. So here goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{eqnarray}&#xA;f_{k}f_{k+2}\pm1 &amp;amp; = &amp;amp; f_{k}(f_{k}+f_{k+1})\pm1\\&#xA;&amp;amp; = &amp;amp; f_{k}^2+f_kf_{k+1}\pm1\\&#xA;&amp;amp; = &amp;amp; (f_{k-1}f_{k+1}\mp1)+f_kf_{k+1}\pm1\\&#xA;&amp;amp; = &amp;amp; f_{k+1}(f_{k-1}+f_{k})\\&#xA;&amp;amp; = &amp;amp; f_{k+1}(f_{k+1})\\&#xA;&amp;amp; = &amp;amp; f_{k+1}^2&#xA;\end{eqnarray}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used the definition of the Fibonacci series twice (&lt;span class=&quot;math-container&quot;&gt;$f_{k+2} = f_k+f_{k+1}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$f_{k-1}+f_k=f_{k+1}$&lt;/span&gt;) and the assumption of step 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that when you do the above construction the areas will always differ by just 1 (alternating above and below each time).&lt;/p&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2020-03-19T01:13:07.653" LastActivityDate="2020-03-19T01:13:07.653" CommentCount="14" ContentLicense="CC BY-SA 4.0" />
  <row Id="24852" PostTypeId="2" ParentId="24847" CreationDate="2015-12-25T06:51:36.877" Score="3" Body="&lt;p&gt;Are you looking for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;alcohol&lt;/strong&gt;?&lt;br&gt;&lt;br&gt;- people bring it to a party&lt;br&gt;- some are always happy about it&lt;br&gt;- some aren't but drink it anyway&lt;br&gt;- it can make people &quot;happy&quot;&lt;br&gt;- too much of it can ruin a party&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastActivityDate="2015-12-25T06:51:36.877" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24853" PostTypeId="2" ParentId="24847" CreationDate="2015-12-25T09:28:04.920" Score="2" Body="&lt;p&gt;Are you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sleep&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tiredness can ruin parties, some people love to sleep, especially after a busy/exciting day. Some people refuse at first, but we all welcome sleep in the end. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16846" LastActivityDate="2015-12-25T09:28:04.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24854" PostTypeId="2" ParentId="24848" CreationDate="2015-12-25T10:04:01.173" Score="14" Body="&lt;p&gt;The image on the right &lt;em&gt;cheats&lt;/em&gt;: the pieces don't actually fit together perfectly, there's a gap in between. To prove it, we can calculate the size of the gap, by calculating the size of a triangle, formed by:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the longest side of the yellow triangle: $a = \sqrt{3^2+8^2}$&lt;/li&gt;&#xA;&lt;li&gt;the inclined side of the trapezoid: $b = \sqrt{2^2+5^2}$&lt;/li&gt;&#xA;&lt;li&gt;the diagonal of the rectangle on the right: $c = \sqrt{5^2+13^2}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The area of this triangle can be calculated using Heron's formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;A = \sqrt{s(s-a)(s-b)(s-c)}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;s = \frac{1}{2}(a+b+c)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting the values into the formula gives exactly 0.5 for $A$. There are two such triangles, so that's a total 1 = the expected discrepancy.&lt;/p&gt;&#xA;" OwnerUserId="8997" LastEditorUserId="8997" LastEditDate="2015-12-25T10:25:23.687" LastActivityDate="2015-12-25T10:25:23.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24855" PostTypeId="2" ParentId="24848" CreationDate="2015-12-25T12:04:57.257" Score="3" Body="&lt;p&gt;&lt;strong&gt;Simple answer&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those shapes (in orange) at the right side of picture, are not triangles at all! they are two quadrilaterals. an thus they have area greater than visually expected. so there is no equity here. They are different and thus have different total area.&lt;/p&gt;&#xA;" OwnerUserId="17954" LastEditorUserId="17954" LastEditDate="2015-12-25T16:08:04.990" LastActivityDate="2015-12-25T16:08:04.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24857" PostTypeId="2" ParentId="24847" CreationDate="2015-12-25T16:01:26.273" Score="0" Body="&lt;p&gt;Are you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Rain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;? Because It depend to everyone's mood&lt;/p&gt;&#xA;" OwnerUserId="17954" LastActivityDate="2015-12-25T16:01:26.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24859" PostTypeId="2" ParentId="24847" CreationDate="2015-12-25T17:49:54.843" Score="3" Body="&lt;p&gt;Are you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Police&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They can ruin a party when they show up. Some people are happy to have them bust a party. Even if you don't want to see them today, you might need them someday.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16995" LastActivityDate="2015-12-25T17:49:54.843" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24860" PostTypeId="2" ParentId="24848" CreationDate="2015-12-25T18:27:13.547" Score="153" Body="&lt;p&gt;&lt;strong&gt;The diagram is misleading, as it hides a gap in the middle of the second configuration.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what we actually get if we rearrange the shapes in question. Notice that the diagonal “bows” slightly, leaving some extra space between the shapes – this is where the extra unit of area creeps in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/irWhD.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/irWhD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/vIMMx.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vIMMx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you shouldn’t trust me any more than the person who drew the original picture!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we see here, pictures can be misleading – so my diagram isn’t proof that the original diagram was wrong. This just gives an intuitive sense of where the extra space has come from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a proper proof, consider the gradients:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The gradient of the blue trapezium is $5/2 = 2.5$&lt;/li&gt;&#xA;&lt;li&gt;The gradient of the red triangle is $8/3 = 2.666...$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Since the gradients don’t match, we can’t arrange them side-by-side like this without some blank space between them. But because they’re close, the eye can be tricked into thinking they form a single continuous line, and doesn’t notice the slope on the triangle changing midway down.&lt;/p&gt;&#xA;" OwnerUserId="4692" LastEditorUserId="4692" LastEditDate="2017-08-30T08:56:20.607" LastActivityDate="2017-08-30T08:56:20.607" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24861" PostTypeId="2" ParentId="24847" CreationDate="2015-12-25T23:43:28.770" Score="6" Body="&lt;p&gt;Are you  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A spoiler?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some people hate you, others like, nay, desire you&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: Sorry for the wait, I answered this originally on my phone between things during the holidays. Extended answer reasoning below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;I can ruin a party&#xA;Showing up uninvited&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Having something (a twist ending to something you're invested in, a solution to a puzzle you're thinking hard about, etc.) spoiled for you when you don't want it can be decidedly un-fun  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Though some are pleased to see me&#xA;When they get real excited&#xA;Even those who would first shun me&#xA;Won't mind meeting me again,&#xA;Or after certain indications&#xA;They'd be happy with me then&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm lumping the rest of these together, because (to me) they all sort of point to a similar basic idea: spoilers are not &lt;em&gt;necessarily&lt;/em&gt; always unwanted - they can be very welcome in certain circumstances. For example, if you were using a website where people posed puzzling questions, you might want to know an answer while allowing others to work it out in their heads without potentially accidentally reading the answer. Or, similarly, one person may be interested in a certain plot point of a movie or book just for curiosity's sake, while another may want to only come upon it in the course of reading or viewing on their own.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Should I tell you my nature?&#xA;Should I give myself away?&#xA;Alert and ready to oblige&#xA;When you tell me it's okay&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This last bit makes me think of the nature of spoiler &lt;em&gt;tags&lt;/em&gt; on online content - they allow someone to post the content, and for the content to remain in place, without it becoming visible without the active consent of the reader.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2015-12-29T02:59:56.543" LastActivityDate="2015-12-29T02:59:56.543" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24862" PostTypeId="2" ParentId="24847" CreationDate="2015-12-26T01:06:04.580" Score="7" Body="&lt;p&gt;I would imagine that you're&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a hint.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To answer each clue:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. People working on a puzzle often dislike unsolicited hints/advice. &lt;br&gt;2. People who are excited about a puzzle but making no headway may welcome a hint. &lt;br&gt;3. Many puzzlers will try the puzzle on their own, then later go back to the hints if they get stuck. &lt;br&gt;4. Not sure what 'certain indications' are in this case. &lt;br&gt;5. There's always a balance between giving a good hint and giving too much away. &lt;br&gt;6. Hints are generally hidden with spoiler tags, waiting for when the puzzler wants to read them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, some advice:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This hint was a bit of a giveaway- 'staring you in the face' was a little obvious. As the riddle suggests, you should ask yourself when writing hints if it gives too much away. If you make more puzzles in the future, you could try giving people more time to come up with answers, and only let out the hints little by little :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14441" LastActivityDate="2015-12-26T01:06:04.580" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24863" PostTypeId="1" AcceptedAnswerId="24866" CreationDate="2015-12-26T01:48:04.333" Score="8" ViewCount="1178" Body="&lt;p&gt;&lt;a href=&quot;https://brilliant.org/problems/the-jealous-husbands-problem-extended/?group=00RgtjoBQNIn&amp;amp;ref_id=847443&quot; rel=&quot;noreferrer&quot;&gt;This problem&lt;/a&gt; is one of the hardest problems I have ever encountered.&lt;/p&gt;&#xA;&lt;p&gt;If you don't trust links, here it is:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;There are some Couples wanting to cross a river by a boat. The boat can carry at most 3 persons at a time and of course cannot cross the river by itself with no one on board.&lt;/p&gt;&#xA;&lt;p&gt;The problem is that all the husbands are extremely jealous. So no woman can be in the presence of another man unless her husband is also present. Even a woman alone in a boat at a bank which has other men on that shore without her husband is not permissible.&lt;/p&gt;&#xA;&lt;p&gt;In other words at no point of time can women outnumber men on bank or shore because that would mean some woman is husband-less!&lt;/p&gt;&#xA;&lt;p&gt;It may be assumed that everyone knows how to row and all persons on the boat disembark and board at the same time.&lt;/p&gt;&#xA;&lt;p&gt;Given the constraints provided, the maximum number of couples that can be transported across the river is &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The minimum number of one-way trips needed to transport these couples is &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Find &lt;span class=&quot;math-container&quot;&gt;$n \times m$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I can tell you right now that I wouldn't expect anyone to have solved it without cheating (there are ways).&lt;/p&gt;&#xA;&lt;p&gt;I don't know where to begin, or even get a number solution.&lt;/p&gt;&#xA;&lt;p&gt;How would you do this problem?&lt;/p&gt;&#xA;" OwnerUserId="17970" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2018-04-13T03:10:54.103" Title="Numbers nonexistent" Tags="|river-crossing|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24864" PostTypeId="2" ParentId="24847" CreationDate="2015-12-26T03:34:55.793" Score="2" Body="&lt;p&gt;This is obviously&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A smartphone.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I can ruin a party Showing up uninvited&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  A party sucks if everyone is on their phones.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Though some are pleased to see me When they get real excited&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Could refer to using something like Songza to liven up a party. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Even those who would first shun me&lt;br&gt;&#xA;Won't mind meeting me again,&lt;br&gt;&#xA;Or after certain indications&lt;br&gt;&#xA;They'd be happy with me then&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you get an important text message or other notification, you'll be glad you had your smartphone. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Should I tell you my nature?&lt;br&gt;&#xA;Should I give myself away?&lt;br&gt;&#xA;Alert and ready to oblige&lt;br&gt;&#xA;When you tell me it's okay&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &quot;Alert&quot; could again refer to a notification. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2015-12-26T03:34:55.793" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24865" PostTypeId="1" CreationDate="2015-12-26T04:29:43.170" Score="12" ViewCount="1733" Body="&lt;p&gt;&lt;strong&gt;Task 1:&lt;/strong&gt; Find the maximum value of $n$ for which it is possible to create a regular uniquely-solvable Killer Sudoku (9x9) such that every cage has atleast $n$ cells in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rules for killer sudoku&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quoting &lt;a href=&quot;https://en.wikipedia.org/wiki/Killer_sudoku&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The objective is to fill the grid with numbers from 1 to 9 in a way&#xA;  that the following conditions are met:&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Each row, column, and nonet contains each number exactly once.&lt;/li&gt;&#xA;  &lt;li&gt;The sum of all numbers in a cage must match the small number printed in its corner.&lt;/li&gt;&#xA;  &lt;li&gt;No number appears more than once in a cage.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Task 2:&lt;/strong&gt; What if rule number 3 was ignored?&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4269" LastEditDate="2016-04-24T13:58:08.493" LastActivityDate="2023-08-19T06:28:19.237" Title="Largest cages Killer Sudoku" Tags="|logical-deduction|calculation-puzzle|optimization|sudoku|" AnswerCount="4" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="24866" PostTypeId="2" ParentId="24863" CreationDate="2015-12-26T04:45:13.657" Score="12" Body="&lt;p&gt;It can be done with a maximum of $n=5$ couples in $m=11$ trips, for a product of $55$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, let's show that more than $5$ couples cannot cross.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If both banks have a man, then every woman must be with her husband so that he won't be jealous. So, if a crossing step starts and ends with men on both banks, it must keep couples together, so it must move exactly one couple (more would exceed the boat's capacity of $3$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if we say that couples are crossing from the left bank to the right bank, then a crossing left to right cannot increase the number of men on the left, and vice versa. Moreover, a crossing changes the number of men by at most $3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's categorize the legal states by how many men are on each bank and which bank the boat is on. We'll label a state 4-2 if it has 4 men on the left and 2 on the right. This doesn't fully describe the state, since it ignores women, but this suffices for this proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We draw a graph of the possible transitions between these states based on the restrictions we found above. The left column lists states with the boat on the left bank, and similarly for the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observe that for $n=6$ or more, the start state at the top doesn't connect to the solution state at the bottom, so $6$ or more couples cannot cross. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AAJzy.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AAJzy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, for $5$ couples, they connect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3Ruhf.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3Ruhf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now show that $5$ couples can indeed cross, and it makes a minimum of $n=11$ steps. Note that the segment from 3-2 on the left to 2-3 must be used. Moreover, all steps in the problem are reversible and the banks are symmetric, so it suffices to reach the solution state from after this central move, and play the steps backwards and mirror from before the move to reach the start state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The central segment has one couple go from right to left, making there be $3$ couples on the left and $2$ on the right. From here, the graph show that $3$ men must travel right, leaving $3$ women on the left and nobody else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, with all men on the right, it's easy to resolve in $4$ more moves by only moving women. One woman take the boat left and brings back $3$ women, then one woman ferries left again and bring the last women over. This can't be shortened, since an even number of moves must be done, and $2$ moves is plainly too few. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The middle $3$ moves are forced, so $m=4+3+4=11$ moves is optimal.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-12-26T04:45:13.657" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24867" PostTypeId="2" ParentId="24863" CreationDate="2015-12-26T05:37:18.190" Score="6" Body="&lt;p&gt;xnor posted his answer as I was writing this, so I decided to go ahead and finish it in any case. I think my solution is a bit more elementary, if more convoluted and less general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will show that the maximum possible value for $n$ is $5$. First, we show five is possible.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let men be $\{A,B,C,D,E\}$ and women $\{a,b,c,d,e\}$, where a man and woman are married if they're the same letter.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Get $a,b,c$ across, then get $c$ back. ($a,b$ on the other side)&lt;/li&gt;&#xA;&lt;li&gt;Get $c,d$ across, then get $d$ back. ($a,b,c$ on the other side)&lt;/li&gt;&#xA;&lt;li&gt;Get $A,B,C$ across, get $C,c$ back. ($A,a,B,b$ on the other side)&lt;/li&gt;&#xA;&lt;li&gt;Get $C,D,E$ across, get $a$ back. ($A, B, b, C, D, E$ on the other side)&lt;/li&gt;&#xA;&lt;li&gt;Get $a,c, d$ across, then get $a$ back. ($A,B,b, C,c, D,d, E$ on the other side)&lt;/li&gt;&#xA;&lt;li&gt;Get $a, e$ across.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;We now show that $n\geq6$ cannot be done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An opening move can consist of taking a couple, or one to three women across, but regardless of the situation the only possibility is some women are left on the other side. Now, no men can cross if there are at least four women waiting on the other side. We will show that if three or less women are on the other side, then no solution can be achieved, which completes the proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If three women are on the other side and any men cross, then it must be that those three women's husbands crossed. In that case, the only possibility (aside from undoing the crossing) is that a couple comes back. We thus have $k \geq 4$ couples on one side, and two couples on the other side, and must take some people across.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this situation, any man that goes across must take his woman with him, and any woman that goes across must take her man with her, so the only possibility is a couple going across. But this precisely undoes our last move, so nothing was gained. This is the big difference from $n=5$; when $n=5$, $k=3$ and we could take the three men across, leaving three women behind since no other men were in their presence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If two women are on the other side and any men cross, then it must be that those two women's husbands crossed. Moreover, no other man or woman may have crossed: the man would leave his wife behind in presence of other men, and the woman would be in the presence of two men in the boat, without her husband.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We thus have $k \geq 4$ couples on one side, and two couples on the other side, and must get some people back. If only men come back, it must be that both men came back, undoing the previous move. If any woman comes back, her husband must also come back, so it must be that a couple comes back. We now have $k \geq 5$ couples on one side, and a single couple on the other side, and must take some people across. This fails exactly as in the previous case ($k \geq 4$ and two couples on the other side).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, if a single woman is on the other side and any men cross, then it must be that her husband crossed. If only he crossed, we have $k \geq 5$ couples on one side and a couple on the other side, and must get some people back. This fails as above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If more people crossed with him, it must be that another couple did. We now have $k \geq 4$ couples on one side, and two couples on the other side, and must get some people back. Once again, we fall into a previous a failing case.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I think that finding the maximal $n$ was the 'hardest' part of the question, and toying around with it, arguments similar to the ones above will show  that the solution I presented uses a minimal number of trips ($m = 11$). There are other ways to reach a solution (particularly the opening moves, leading to having three women waiting on the other side), but in no fewer trips.&lt;/p&gt;&#xA;" OwnerUserId="16478" LastActivityDate="2015-12-26T05:37:18.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24870" PostTypeId="2" ParentId="24844" CreationDate="2015-12-27T05:52:52.593" Score="1" Body="&lt;p&gt;Instructing someone to freeze ('icy be'), or saying one is frozen (&quot;Icy me&quot;)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the puzzle is using cryptics, perhaps an abbreviated institution got mixed up with a yes-man (ACYOE interleaves &quot;CO&quot; and &quot;AYE&quot;)?&lt;/p&gt;&#xA;" OwnerUserId="1467" LastActivityDate="2015-12-27T05:52:52.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24872" PostTypeId="1" AcceptedAnswerId="24873" CreationDate="2015-12-27T07:25:47.643" Score="9" ViewCount="323" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7Z0DU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7Z0DU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Challenge:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Given unlimited copies of each of the above type of puzzle piece, create a $9\times 9$ square which features a picture of snake.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Rules:&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There must be only one snake, with a head at one end and a tail at the other.&lt;/li&gt;&#xA;&lt;li&gt;Pieces may be rotated, but not flipped. &lt;/li&gt;&#xA;&lt;li&gt;A piece with one straight edge must be placed at the the square's edge.&lt;/li&gt;&#xA;&lt;li&gt;A piece with two straight edges must be placed at the square's corner.&lt;/li&gt;&#xA;&lt;li&gt;Some pieces may be used many times, some only once, others not at all.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If my calculations are correct, &lt;strike&gt;the solution is essentially unique. Specifically, there are two solutions up to rotation, but the only difference between them is that the head and tail pieces are swapped.&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I was wrong, this puzzle is under specified, there are at least 3 distinct solutions.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2015-12-27T10:01:05.510" LastActivityDate="2015-12-28T04:21:40.443" Title="Jigsaw Logic: Opheodrys Vernalis" Tags="|tiling|jigsaw-puzzle|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24873" PostTypeId="2" ParentId="24872" CreationDate="2015-12-27T08:28:13.023" Score="8" Body="&lt;p&gt;There are a few solutions, but we're only asked to find one of them, so here one is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It looks like &lt;em&gt;most&lt;/em&gt; of the puzzle is constrained - I believe I only get to make two assumptions (and different choices there will yield the remaining solutions).&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/OkClb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OkClb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="17389" LastEditDate="2015-12-27T19:20:28.167" LastActivityDate="2015-12-27T19:20:28.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24874" PostTypeId="1" AcceptedAnswerId="24919" CreationDate="2015-12-27T09:36:38.013" Score="21" ViewCount="999" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Do there exist integers $r,s,t\ge100$ such that the decimal representation of $(r+\sqrt{s})^t$ is of the form $~~~\ldots\ldots2015\,.\,2016\ldots\ldots$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&#xA;(In other words: the four digits before the decimal point are 2015, and the first four digits after the decimal point are 2016.)&#xA;&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-21T09:42:46.453" LastActivityDate="2016-03-21T09:42:46.453" Title="The year 2016 is approaching" Tags="|mathematics|number-theory|seasonal|algebra|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24876" PostTypeId="1" AcceptedAnswerId="24877" CreationDate="2015-12-27T14:52:35.660" Score="9" ViewCount="667" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am an important person for my kinds,&lt;br&gt;&#xA;  Do you remember when you walked 1267200 feet with me so that you could defiance,&lt;br&gt;&#xA;  I used to save things a lot but you don't use me anymore because you have got a cool guy,&lt;br&gt;&#xA;  My mom is Mrs 11 and dad is Mr 17 but their relationship is weak and if it's broken then I will die.&lt;br&gt;&#xA;  Can you guess who am I?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastActivityDate="2016-03-01T10:33:37.743" Title="The son of Mrs. 11 and Mr. 17" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24877" PostTypeId="2" ParentId="24876" CreationDate="2015-12-27T15:09:48.350" Score="10" Body="&lt;p&gt;You're &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Salt&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am an important person for my kinds,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Used in vegetables. The most important in all of the seasonings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Do you remember when you walked 1267200 feet with me so that you could defiance,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gandhi's salt march, 240 miles = 1267200 feet. He walked for the Non-cooperation movement.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I used to save things a lot but you don't use me anymore because you have got a cool guy,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People used to store a lot of things in salt, before the refrigerator made that obsolete.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My mom is Mrs 11 and dad is Mr 17 but their relationship is weak and if it's broken then I will die.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Periodic table, 11 = Na, 17 = Cl, Salt is NaCl, if their bond breaks you're just left with ions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16846" LastEditorUserId="17482" LastEditDate="2016-03-01T10:33:37.743" LastActivityDate="2016-03-01T10:33:37.743" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="24878" PostTypeId="1" CreationDate="2015-12-27T15:25:07.580" Score="14" ViewCount="773" Body="&lt;p&gt;What is the longest -1 word (or the word worth most in Scrabble for words with the same amount of letters (which is just the addition of all the letter values of the original 'word', no constraints)), where a -1 word is defined as:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;This 'word' does not have to be an actual word, just a string of letters.&lt;/li&gt;&#xA;&lt;li&gt;one can replace each and every letter in the 'word' with another letter to create a word that appears in the &lt;a href=&quot;http://dictionary.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;dictionary.com&lt;/a&gt; database.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;An example of a -1 word would be &amp;quot;gub,&amp;quot; and the replacements one could make would be &amp;quot;rub,&amp;quot; &amp;quot;gob,&amp;quot; and &amp;quot;gun.&amp;quot;&lt;/p&gt;&#xA;" OwnerUserId="167" LastEditorUserId="69582" LastEditDate="2021-10-29T17:49:14.843" LastActivityDate="2021-10-29T17:49:14.843" Title="The Longest -1 Word" Tags="|word|optimization|" AnswerCount="3" CommentCount="13" ContentLicense="CC BY-SA 4.0" />
  <row Id="24879" PostTypeId="2" ParentId="24844" CreationDate="2015-12-27T16:42:50.767" Score="1" Body="&lt;p&gt;If you really have to use those letters, I think you're going to have to settle for obscure words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's poor form, but you could use&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After shuffling, my ace let me get ahead (1, 4)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; anag. MYACE = A CYME&lt;br&gt;&#xA; a head (of unexpanded leaves) = a &lt;a href=&quot;https://en.wiktionary.org/wiki/cyme&quot; rel=&quot;nofollow&quot;&gt;cyme&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="71" LastActivityDate="2015-12-27T16:42:50.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24881" PostTypeId="2" ParentId="24844" CreationDate="2015-12-27T18:08:36.267" Score="2" Body="&lt;p&gt;Another cryptic:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;RCYBE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rotate the string:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;CYBER&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14219" LastActivityDate="2015-12-27T18:08:36.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24882" PostTypeId="2" ParentId="24819" CreationDate="2015-12-27T22:13:13.170" Score="6" Body="&lt;p&gt;Edited 29 December based on Tyler Seacrests's very insightful comments (so if it is correct it is a joint Derek/Tyler solution). Of course if it's incorrect the fault is mine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two important insights:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The very first coin will be sacrificed blind. Worst-case it is value_20. We can be sure to retain value_19 by only sacrificing the &lt;em&gt;lower&lt;/em&gt; of a comparison thereafter.&lt;br&gt;&#xA;To achieve 28 we must therefore guarantee finding value_9 or above. A group of 10 (or more) coins (A...J or K...T) is guaranteed to contain this, allowing for the initial sacrifice.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proceed as follows:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1. Offer Fredo A, B and C (in that order). A is sacrificed and we classify B and C as &lt;em&gt;lower&lt;/em&gt; and &lt;em&gt;higher&lt;/em&gt;&lt;br&gt;&#xA;2. Offer Fredo &lt;em&gt;lower&lt;/em&gt;, then &lt;em&gt;higher&lt;/em&gt;, then D. The old &lt;em&gt;lower&lt;/em&gt; is sacrificed and we now know the highest of B-C-D.&lt;br&gt;&#xA;3...8. Sacrifice the &lt;em&gt;lower&lt;/em&gt; of the previous comparison to compare the &lt;em&gt;higher&lt;/em&gt; with E...J&lt;br&gt;&#xA;9. Put the current &lt;em&gt;higher&lt;/em&gt; aside, sacrifice &lt;em&gt;lower&lt;/em&gt; against K. The coin set aside is at least value_9 (it may be as high as value_20). But the coin we now sacrifice is worth less.&lt;br&gt;&#xA;10...16. Continue sacrificing &lt;em&gt;lower&lt;/em&gt; against L...T&lt;br&gt;&#xA;17. Point to the &lt;em&gt;higher&lt;/em&gt; of the last comparison and the coin set aside in step 9  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10475" LastEditorUserId="10475" LastEditDate="2016-01-01T20:04:09.967" LastActivityDate="2016-01-01T20:04:09.967" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24888" PostTypeId="1" AcceptedAnswerId="24899" CreationDate="2015-12-28T03:36:20.047" Score="21" ViewCount="4059" Body="&lt;p&gt;A cookie-baking recipe book (&lt;a href=&quot;https://books.google.ca/books?id=kBv0R-Aqv80C&quot;&gt;this one, specifically&lt;/a&gt;) has the following instruction in it, which is incorrect in a single spot:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you have a fan-assisted oven, you will need to reduce the recipe temperature by 68°F. Fan-assisted ovens all perform differently, so always refer to the manufacturer's instructions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can you figure out what's wrong with the instruction just by looking at it, and what correction you need to make? The error is not a simple grammatical error, but it does have to do with language.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I first found this problem &lt;a href=&quot;http://research.microsoft.com/en-us/um/people/leino/puzzles.html&quot;&gt;here&lt;/a&gt;, so I cannot take credit for it. However, the mistake in the sentence does exist independently in a recipe book that I've also cited.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="30633" LastEditDate="2017-04-09T20:26:18.277" LastActivityDate="2017-04-09T20:26:18.277" Title="What's wrong with this line in a baking recipe book?" Tags="|language|real|" AnswerCount="11" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="24889" PostTypeId="2" ParentId="24888" CreationDate="2015-12-28T03:47:46.783" Score="2" Body="&lt;p&gt;The problem is in reducing the temperature by 68 degrees Fahrenheit. This is 293.15 Kelvins, a huge difference. The Fahrenheit scale zero is a nonzero amount of heat, so even reducing by 0 degrees would be a (large) change.  Pragmatically, the sentence means to decrease the Fahrenheit temperature measure by 68.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2015-12-28T03:47:46.783" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="24890" PostTypeId="2" ParentId="24872" CreationDate="2015-12-28T04:21:40.443" Score="3" Body="&lt;p&gt;Just posting this for the sake of completeness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are four solutions to this puzzle. One of them is in Zerris's answer, another is below. Given any solution, you can get a different one by swapping the tail and head pieces, which gives all four.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/E6ThB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/E6ThB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-12-28T04:21:40.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24891" PostTypeId="2" ParentId="24888" CreationDate="2015-12-28T05:39:08.807" Score="9" Body="&lt;p&gt;A nitpicky answer along the lines of &quot;if they have eggs, get a dozen&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This says that if you have a fan-assisted oven, you need to reduce the temperature. However, you might have multiple ovens and be baking the cookies in the non-fan-assisted one. So to be entirely correct, it should say, &quot;If you are using a fan-assisted oven to bake these cookies...&quot;&lt;/p&gt;&#xA;" OwnerUserId="14441" LastActivityDate="2015-12-28T05:39:08.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24892" PostTypeId="1" AcceptedAnswerId="24894" CreationDate="2015-12-28T05:42:43.757" Score="30" ViewCount="1508" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;This puzzle was inspired by &lt;a href=&quot;https://puzzling.stackexchange.com/questions/24803/reverse-puzzling&quot;&gt;Dr Xorile's first reverse puzzle&lt;/a&gt;! Kudos to him for coming up with such a fun format.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Intrigued, I gave George yet another famous puzzle to ponder over. Once again, I was surprised that he had not seen this before either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;George frowned, paused for a few moments, and then picked up his trusty marker and started drawing a diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My eyes widened in horror as I realized that George would take a long time if he kept up this strategy. I didn't have time for him to complete his diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Wait, George! Let's... let's try an easier version of the same puzzle. A much easier version.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I gave George the revised puzzle and he eventually came up with the following diagram for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qEe7B.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qEe7B.jpg&quot; alt=&quot;Click on the picture for a clearer image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Ah! The answer is obvious now!&quot; George exclaimed. I eyed George warily. I was starting to think he was more of a 'tedious' puzzler than a 'great' one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What was/were the puzzle(s)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14095" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2020-05-24T04:21:17.430" Title="Reverse Puzzling 2" Tags="|logical-deduction|enigmatic-puzzle|reverse-puzzling|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24893" PostTypeId="1" AcceptedAnswerId="24898" CreationDate="2015-12-28T05:43:18.087" Score="4" ViewCount="2057" Body="&lt;p&gt;Here's a one-player card game:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;We begin with a shuffled card deck&lt;/li&gt;&#xA;&lt;li&gt;The player try to guess the color of the card at the top (black or red)&lt;/li&gt;&#xA;&lt;li&gt;If the player guess right, he can keep the card. Else the card is put in another stack&lt;/li&gt;&#xA;&lt;li&gt;When the stack finish, the player shuffle the non-guessed cards and go back to &lt;code&gt;2.&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;The game finish when there's no card left.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;A first strategy to win would be to first ask one color until the first stack ends and then ask another color. With &lt;code&gt;N&lt;/code&gt;, number of cards of one color. The is always gonna be &lt;code&gt;3N&lt;/code&gt; guesses before the end of the game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's call it the &lt;code&gt;two_pass&lt;/code&gt; strategy.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;def two_pass():&#xA;    return BLACK if times_shuffled == 0 else RED&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But here's another strategy:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You can use the cards you guessed right and the cards you have seen during the game to guess the probability of how many cards of one color are left in the stack&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then you can have something like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;def probability_guess():&#xA;    blacks_left = N - blacks_seen_but_not_taken - blacks_taken&#xA;    prob_of_getting_black = blacks_left / cards_left&#xA;    if prob_of_getting_black &amp;lt; 0.50:&#xA;        return RED&#xA;    else:&#xA;        return BLACK #EDITED, was 'WHITE' before&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And here's the trick question: &lt;strong&gt;Is &lt;code&gt;probability_guess&lt;/code&gt; better than &lt;code&gt;two_pass&lt;/code&gt;?&lt;/strong&gt; (will it always need less guesses than the other strategy)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Here's some python code with a simulation of the game and the strategies: &lt;a href=&quot;https://gist.github.com/mdamien/bb3848310965dfe91419&quot; rel=&quot;nofollow&quot;&gt;https://gist.github.com/mdamien/bb3848310965dfe91419&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18004" LastEditorUserId="18004" LastEditDate="2015-12-28T08:14:25.567" LastActivityDate="2016-01-07T22:20:32.883" Title="In a Guess The Color game, what's the optimal strategy?" Tags="|probability|cards|game-theory|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24894" PostTypeId="2" ParentId="24892" CreationDate="2015-12-28T06:16:17.540" Score="25" Body="&lt;p&gt;This puzzle is the Towers of Hanoi with three disks. A, B, and C are the pegs, and the positions from left to right are the disks; the lines connect positions that are valid to move between.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The initial puzzle is probably the Towers of Hanoi with more disks - the area needed to draw the diagram is quadrupled with every new disk added, so a larger diagram would take up much more space &lt;em&gt;and&lt;/em&gt; time.&lt;/p&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="11876" LastEditDate="2015-12-29T01:38:14.013" LastActivityDate="2015-12-29T01:38:14.013" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24895" PostTypeId="2" ParentId="11364" CreationDate="2015-12-28T07:10:41.797" Score="3" Body="&lt;p&gt;We know that a probability of $5/32$ is possible (with the strategy &quot;if you see two reds, guess blue, otherwise guess red&quot;), so let's prove they can't do better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose there was a strategy with 6 hat combinations $c_1,\dots,c_6$ for which the players win. Let's just look at $c_1,\dots,c_5$ for now. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Connect two combinations with an edge labeled $i$ if player $i$ sees the same thing in both of them. Then two patterns which are connected agree in 3 places: the two places that player $i$ sees, and player $i$'s hat, since what he guesses is determined by what he sees. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Then there must be at least one edge of each label&lt;/em&gt;, since player $i$ can only see 4 possible things, and there are 5 patterns, so there must be two where player $i$ sees the same thing.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;If two combinations agree on 4 places, they agree on the fifth&lt;/em&gt;, since the last hat is forced by surrounding ones. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;There can't be more than one edge between any pair of combinations.&lt;/em&gt; Otherwise, they would agree in four places, and differ in the fifth, which is impossible by (2).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;There are at most two edges out of each combination.&lt;/em&gt; If there were 3, there would have to be to edges labeled by adjacent numbers, $i$ and $i+1$; without loss of generality, $i=1$, so we have the below scenario:&#xA;$$&#xA;c_1\stackrel{1}{\longleftrightarrow}c_2 \stackrel{2}{\longleftrightarrow} c_3&#xA;$$&#xA;This means $c_2$ and $c_3$ have the same hats on players $1,2,3$. If $c_2$ and $c_3$ also were the same for player 4, then by (2) they would be the same; since $c_2\neq c_3$, they must differ on 4. Similarly, $c_2$ and $c_3$ differ on 4, so it must true that $c_1$ and $c_3$ are the same on 4 (they are opposites of the same thing). So $c_1$ and $c_3$ agree on $1,2$ and $4$, which means they agree on all, which contradicts (3).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Combining (1), (3) and (4) proves that the graph is a 5 cycle, where the edges appearing in the order $1,3,5,2,4$. Thus, $c_1$ uniquely determines $c_2$ through $c_5$. But the same reasoning applied to $c_1,c_2,c_3,c_4,c_6$ shows that $c_6$ must be one of $c_1,\dots,c_5$, so the six combinations were not distinct. &lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2015-12-28T07:10:41.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24896" PostTypeId="2" ParentId="24888" CreationDate="2015-12-28T07:48:24.220" Score="7" Body="&lt;p&gt;The phrase &quot;reduce the recipe temperature&quot; tells me to actually change the figures in the book, rather than simply reducing the temperature setting on the oven. A correct phrasing in this fashion would have been something along the lines of &quot;use the recipe temperature reduced by&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a side note, the amount &quot;68&quot; seems oddly specific given the context...&lt;/p&gt;&#xA;" OwnerUserId="18012" LastActivityDate="2015-12-28T07:48:24.220" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24898" PostTypeId="2" ParentId="24893" CreationDate="2015-12-28T08:12:45.763" Score="4" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Two pass is better (at least in worst-case)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;s&gt;First of all, your probability guess as stated in the question never returns Black, so it never finishes&lt;/s&gt;. #had an unedited version of the question open&lt;br&gt;&lt;br&gt;Assuming this is a mistake probability guess still has a worst case where it never finishes. More specifically there's a $1/2^N$ chance that it finishes a single pass without 'learning' anything. It &lt;strong&gt;could&lt;/strong&gt; literally guess every card wrong every single time. However, in case you have equal amounts of reds/blacks that not the case.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the amount of red/blacks are equal&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Then the best guess algorithm becomes a lot better. However it's easy to show that the worst-case is still worse.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Suppose that your algorithm guesses 0 if it has no clear indication either way, or 1 if it's seen more 0's so far.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if our code is 1001, it will guess wrong, right, wrong, right. leaving you with 10 for the next time. If this is ordered right for you (01) you will get it in 6, but if it's wrong, you will need another pass, and get it in 7. This can always happen if a pass leaves an equal amount of both red and black.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16846" LastEditorUserId="16846" LastEditDate="2015-12-28T10:08:41.553" LastActivityDate="2015-12-28T10:08:41.553" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="24899" PostTypeId="2" ParentId="24888" CreationDate="2015-12-28T08:13:20.360" Score="61" Body="&lt;p&gt;The cookbook should have read: &quot;Reduce the recipe temperature by 36°F.&quot; (Or perhaps 35 or 40, for a &quot;round&quot; amount—&quot;36&quot; implies an &lt;a href=&quot;https://en.wikipedia.org/wiki/Significant_figures&quot;&gt;unwarranted degree of precision&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original source that the cookbook was based on presumably said &quot;reduce by 20°C.&quot; However, Celsius and Fahrenheit are relative scales (with different definitions of zero), not absolute scales. The cookbook authors erroneously did the conversion as if 20°C were a &lt;em&gt;specific temperature&lt;/em&gt;, which does correspond to 68°F, but &lt;a href=&quot;https://en.wikipedia.org/wiki/Celsius#Temperatures_and_intervals&quot;&gt;when converting &lt;em&gt;intervals&lt;/em&gt;&lt;/a&gt;, one should not do the zero point adjustment: a &lt;em&gt;difference&lt;/em&gt; of one degree Celsius corresponds to a difference of 1.8 degrees Fahrenheit.&lt;/p&gt;&#xA;" OwnerUserId="9044" LastEditorUserId="9044" LastEditDate="2015-12-28T08:44:36.903" LastActivityDate="2015-12-28T08:44:36.903" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="24900" PostTypeId="2" ParentId="24888" CreationDate="2015-12-28T08:18:36.553" Score="13" Body="&lt;p&gt;I've never used a fan-assisted oven before, but sixty-eight degrees seems like a &lt;em&gt;huge&lt;/em&gt; change in temperature. Perhaps the intended instruction was to reduce the temperature by &lt;em&gt;six to eight&lt;/em&gt; degrees.&lt;/p&gt;&#xA;" OwnerUserId="5524" LastActivityDate="2015-12-28T08:18:36.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24901" PostTypeId="2" ParentId="24878" CreationDate="2015-12-28T08:35:36.013" Score="6" Body="&lt;p&gt;Ok, let's get the ball rolling with a simple 5-letter word:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;SLOPS&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;With the changed words being:&lt;br&gt;&#xA;&lt;strong&gt;F&lt;/strong&gt;LOPS&lt;br&gt;&#xA;S&lt;strong&gt;T&lt;/strong&gt;OPS&lt;br&gt;&#xA;SL&lt;strong&gt;A&lt;/strong&gt;PS&lt;br&gt;&#xA;SLO&lt;strong&gt;T&lt;/strong&gt;S&lt;br&gt;&#xA;SLOP&lt;strong&gt;E&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2015-12-28T08:35:36.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24902" PostTypeId="2" ParentId="24888" CreationDate="2015-12-28T08:40:33.647" Score="3" Body="&lt;p&gt;Pedants' corner:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Fahrenheit&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, the symbol for Fahrenheit is &lt;strong&gt;℉&lt;/strong&gt; - a single symbol, represented, for example, in Unicode by codepoint U+2109.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The text above uses two symbols, &lt;strong&gt;˚&lt;/strong&gt; and &lt;strong&gt;F&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, I didn't figure this out just by looking at it...&lt;/p&gt;&#xA;" OwnerUserId="18015" LastActivityDate="2015-12-28T08:40:33.647" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24903" PostTypeId="2" ParentId="24888" CreationDate="2015-12-28T10:01:27.907" Score="6" Body="&lt;p&gt;You should not adjust the &lt;em&gt;temperature of the recipe&lt;/em&gt; (assuming it is at safe room temperature) but rather adjust the temperature of the oven.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should read &lt;em&gt;temperature given in the recipe&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="18018" LastActivityDate="2015-12-28T10:01:27.907" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24904" PostTypeId="2" ParentId="24878" CreationDate="2015-12-28T11:39:43.723" Score="19" Body="&lt;p&gt;Here's a 7-letter -1 word:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;PASTERS&lt;/strong&gt; (&lt;a href=&quot;http://dictionary.reference.com/browse/pasters&quot;&gt;English word&lt;/a&gt;, score: 9)&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;http://dictionary.reference.com/browse/masters&quot;&gt;&lt;strong&gt;M&lt;/strong&gt;ASTERS&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;http://dictionary.reference.com/browse/posters&quot;&gt;P&lt;strong&gt;O&lt;/strong&gt;STERS&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;http://dictionary.reference.com/browse/patters&quot;&gt;PA&lt;strong&gt;T&lt;/strong&gt;TERS&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;http://dictionary.reference.com/browse/passers&quot;&gt;PAS&lt;strong&gt;S&lt;/strong&gt;ERS&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;http://dictionary.reference.com/browse/pastors&quot;&gt;PAST&lt;strong&gt;O&lt;/strong&gt;RS&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;http://dictionary.reference.com/browse/pastels&quot;&gt;PASTE&lt;strong&gt;L&lt;/strong&gt;S&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;http://dictionary.reference.com/browse/pastern&quot;&gt;PASTER&lt;strong&gt;N&lt;/strong&gt;&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9044" LastEditorUserId="9044" LastEditDate="2015-12-29T06:06:31.043" LastActivityDate="2015-12-29T06:06:31.043" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24905" PostTypeId="2" ParentId="24888" CreationDate="2015-12-28T12:52:09.357" Score="2" Body="&lt;p&gt;Shouldn't it be &quot;recipe's temperature&quot; ?&lt;/p&gt;&#xA;" OwnerUserId="12223" LastActivityDate="2015-12-28T12:52:09.357" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24906" PostTypeId="2" ParentId="24888" CreationDate="2015-12-28T13:23:46.120" Score="7" Body="&lt;p&gt;If the ovens &quot;all perform differently&quot; to the extent that they need to &quot;refer to the manufacturer's instructions&quot;, then recommended temperature reduction should not be expressed as a precise &quot;need&quot;.&lt;/p&gt;&#xA;" OwnerUserId="9422" LastActivityDate="2015-12-28T13:23:46.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24909" PostTypeId="1" AcceptedAnswerId="24910" CreationDate="2015-12-28T16:43:42.170" Score="3" ViewCount="262" Body="&lt;p&gt;I was on my computer last night when, all of a sudden, a message popped up on my screen. The message consisted of nothing more than an image and the following message &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2024 9251680 2777 763267036&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was able to to determine from the image that the message came from some kind of alien race that had two hands, with 31 fingers on each hand. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing this, can someone help me figure out what this message means? &lt;/p&gt;&#xA;" OwnerUserId="12162" LastActivityDate="2015-12-28T16:57:59.523" Title="Help me translate this message" Tags="|mathematics|english|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24910" PostTypeId="2" ParentId="24909" CreationDate="2015-12-28T16:54:51.800" Score="5" Body="&lt;p&gt;&quot;We come in peace&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Just convert the numbers to base 62.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2015-12-28T16:57:59.523" LastActivityDate="2015-12-28T16:57:59.523" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24911" PostTypeId="1" AcceptedAnswerId="25426" CreationDate="2015-12-28T18:33:51.717" Score="1" ViewCount="395" Body="&lt;p&gt;&lt;em&gt;You wake up. You are on a very small room. There is a paper on the floor, and you don't know how to exit. Can you decipher the note to escape?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the note:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;9%110111111220222222330445330550567443660770554070000605000000000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Good luck trying to escape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint #1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Numbers(in the example, '9') before the % are the length of the whole phrase you're trying to decipher.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You need to put all the numbers on a list. Put (Length of the phrase) horizontally, then go down. Like this:&lt;br&gt;&#xA;110111111 Then, down.&lt;br&gt;&#xA;&#xA; 220222222&lt;br&gt;&#xA;Repeat that (lenth of the code)/(Lenth of the phrase) On the example, it's 63/9 (We don't count the 9%) If you do 63/9, you get 7. So you need to repeat that 7 times.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After getting table (On hint 2), think about a LED calculator.&#xA; It's formed of vars, not?&#xA; Think that every var is a number.&#xA; Like this:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;... &lt;code&gt;1&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;2|-----|3&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;4 -----&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;5|-----|6&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...    &lt;code&gt;7&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Now get every column, and get the numbers of each. Now you can figure out what to do with those numbers! ;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user18031" LastEditorDisplayName="user18031" LastEditDate="2016-01-15T18:29:07.790" LastActivityDate="2016-01-19T10:00:11.203" Title="Encoding Extreme - 1" Tags="|number-sequence|letters|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24912" PostTypeId="2" ParentId="24844" CreationDate="2015-12-28T21:02:52.053" Score="1" Body="&lt;p&gt;&lt;strong&gt;Ocyte&lt;/strong&gt; is a less common spelling for oocyte - egg cell.&#xA;Would that work?&lt;/p&gt;&#xA;" OwnerUserId="7489" LastActivityDate="2015-12-28T21:02:52.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24913" PostTypeId="2" ParentId="24844" CreationDate="2015-12-28T23:54:18.167" Score="3" Body="&lt;p&gt;The suffix -cyte means cell, so you could be cute and have it be ACYTE with the clue &quot;Cell for sore eyes&quot;&lt;/p&gt;&#xA;" OwnerUserId="14441" LastActivityDate="2015-12-28T23:54:18.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24915" PostTypeId="2" ParentId="24888" CreationDate="2015-12-29T00:09:02.047" Score="2" Body="&lt;p&gt;I know the answer has already been accepted, but I just wanted to point out another issue with the instructions as given that I haven't seen anyone else point out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It says you should reduce the temperature by 68 degrees Farenheit, which is impossible with most commercially available ovens since they tend to only let you select temperatures in increments of 5 degrees. Now, there may be ovens that allow that precise of a temperature selection, but I've never seen one, which makes this a pretty silly thing to put in a cookbook that seems to be aimed at amateur bakers. &lt;/p&gt;&#xA;" OwnerUserId="18040" LastActivityDate="2015-12-29T00:09:02.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24916" PostTypeId="2" ParentId="24892" CreationDate="2015-12-29T00:13:37.197" Score="10" Body="&lt;p&gt;Just to expand on @Deusovi's devious answer, you can convert the states described in the circle to a game state as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Start with a set of empty pegs, labelled A, B, and C.&lt;/li&gt;&#xA;&lt;li&gt;Place the biggest disk on the peg with the letter shown on the right.&lt;/li&gt;&#xA;&lt;li&gt;Place the middle disk on the peg with the letter shown in the middle.&lt;/li&gt;&#xA;&lt;li&gt;Place the smallest disk on the peg with the letter shown on the left.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note that by doing this in this order, there are no problematic states - every peg is available for every disk. The is also how you can see that there are &lt;span class=&quot;math-container&quot;&gt;$3^n$&lt;/span&gt; states when there are &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; disks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you could just write all these &lt;span class=&quot;math-container&quot;&gt;$3^n$&lt;/span&gt; states out and join them up, but you'll note that @AZGreenTea did it in a very symmetric way.  You can see that the triangle is made up of three smaller triangles, each of which has the same right hand letter = biggest disk. There are a limited number of ways to move the biggest disk (basically when everything else is piled up on a different column), so the triangle you see can be replicated three times, adding a forth letter (A, B, and C). These can also only be joined in the corner, and so the triangles you see are replicated three times and put into a bigger triangle for &lt;span class=&quot;math-container&quot;&gt;$n=4$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This self replication is, of course, fractal in nature, and hence the link between the towers of Hanoi and the fractal in question, which turns out to be &lt;a href=&quot;https://en.wikipedia.org/wiki/Sierpinski_triangle#Towers_of_Hanoi&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sierpinski's triangle&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2020-05-24T04:21:17.430" LastActivityDate="2020-05-24T04:21:17.430" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24918" PostTypeId="2" ParentId="24744" CreationDate="2015-12-29T06:32:37.797" Score="-1" Body="&lt;p&gt;answer is -19 because above number is the half of the sum of two numbers in the bottom row&lt;/p&gt;&#xA;" OwnerUserId="18051" LastActivityDate="2015-12-29T06:32:37.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24919" PostTypeId="2" ParentId="24874" CreationDate="2015-12-29T07:13:37.090" Score="20" Body="&lt;p&gt;The answer is yes, this is possible.  Fix $t = 100$.  A big key is found in &lt;a href=&quot;https://puzzling.stackexchange.com/questions/11193/computation-without-a-computer&quot;&gt;this excellent puzzle&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The quantity $(r + \sqrt{s})^{100} + (r - \sqrt{s})^{100}$ is an integer, since all the terms with an odd $\sqrt{s}$ power cancel.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Notice that $\{r - \sqrt{s}\}_{r, s \geq 100}$ is a dense subset of $\mathbb{R}^{&amp;gt;0}$.   To see this, note that the quantity $\epsilon_m := m - \sqrt{m^2 - 1} = \frac{1}{m + \sqrt{m^2 - 1}}$ goes to zero for large $m$.  We can then take $k \epsilon_m$ to get close to any value we choose.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Hence $\{(r - \sqrt{s})^{100}\}_{r, s \geq 100}$ is also dense in $\mathbb{R}^{&amp;gt;0}$.   In particular, we can find an $r$ and $s$ such that $(r - \sqrt{s})^{100}$ has any particular finite substring of decimal digits after the decimal that we want.  This allows us to force $(r + \sqrt{s})^{100}$ to have any digits (after the decimal) that we want.&lt;br&gt;&#xA;&lt;br&gt;&#xA;  Thus, we force $(r + \sqrt{s})^{100}$ to have the substring of digits $20152016$ appear roughly a googol places after the decimal point.  Then take $(10r + \sqrt{100s})^{100}$ and we will have our answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2722" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2015-12-29T07:13:37.090" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24921" PostTypeId="1" AcceptedAnswerId="24927" CreationDate="2015-12-29T16:59:17.107" Score="5" ViewCount="356" Body="&lt;p&gt;I encouraged my sibling to continue exploring the world. I impressed upon my other sibling that, while friends are commendable, family is what really counts. I offered them both assistance should they get into a spot of bother.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I contemplated the possibility that they no longer trusted one another; or relied on their mutual affection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two hypothetical inquiries were made to test the boundaries of my offer to help. The first contemplated the possibility that one of the siblings had traveled to a remote location. I assured them that the telephone would still work. The second considered the other sibling becoming bankrupt. Again, I was able to comfort them with my offer of continued support.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conversation continued in this manner for a while.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Who and where am I?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2018-04-28T18:00:24.347" LastActivityDate="2018-04-28T18:00:24.347" Title="Greetings Sibling. Verily" Tags="|riddle|enigmatic-puzzle|humor|" AnswerCount="4" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24922" PostTypeId="2" ParentId="24921" CreationDate="2015-12-29T17:42:00.593" Score="2" Body="&lt;p&gt;Are you perhaps&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quantum Entanglement?, or perhaps a source of entangled particles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It seems to me that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your siblings are 2 entangled particles. You're undergoing an experiment if quantum entanglement holds. Perhaps some sort of EPR-experiment. The 2 hypothetical inquires are the measurements being done on the entangled particles. The telephone still works is related to the 'connection' between the particles. I'm guessing that the bankrupt inquiry is a measurement of spin (- for bankrupt)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Might be farfetched/wrong, but came to mind immediately upon reading the riddle.&lt;/p&gt;&#xA;" OwnerUserId="16846" LastActivityDate="2015-12-29T17:42:00.593" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24923" PostTypeId="2" ParentId="24921" CreationDate="2015-12-29T17:56:39.400" Score="1" Body="&lt;p&gt;I seems too obvious, but I think that this person is a close relative (brother/sister/parent)&#xA;Reason: &quot;I encouraged my sibling to continue exploring the world. I impressed upon my other sibling that, while friends are commendable, family is what really counts.&quot; The clear and repeated use of the word sibling, in reference to the other 2 people suggests that they are all 3 siblings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This person is likely sitting at home on a telephone call/conference call, calling each of the siblings/relatives.&#xA;Reason: The first sibling was encouraged to &quot;continue exploring&quot;, suggesting that the travel was a current and ongoing activity. Further on the query asks about the possibility that the sibling had travelled to a remote location, suggesting that the sibling is not physically present with the person talking, but is somewhere else. The fact that they are talking means they are on the phone or some similar communication device. Reference &quot;telephone would still work&quot;.&lt;/p&gt;&#xA;" OwnerUserId="18071" LastActivityDate="2015-12-29T17:56:39.400" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24924" PostTypeId="2" ParentId="24921" CreationDate="2015-12-29T19:26:09.480" Score="1" Body="&lt;p&gt;Is this some kind of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Solar System&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reasons&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is the 'subject' = sun in the solar system&lt;br&gt;&#xA; two siblings = two planets&lt;br&gt;&#xA; friends of the siblings = satellite of the planets  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I encouraged my sibling to continue exploring the world. I impressed upon my other sibling that, while friends are commendable, family is what really counts. I offered them both assistance should they get into a spot of bother.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One planet doesn't orbit the sun anymore and other has many satellites. Here offering assistance means that they will still be in the gravitational affect of the sun.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I contemplated the possibility that they no longer trusted one another; or relied on their mutual affection.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is very less affect of gravity of a planet to the other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first contemplated the possibility that one of the siblings had traveled to a remote location. I assured them that the telephone would still work.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One planet has gone too far of solar system but the gravitational force still works.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second considered the other sibling becoming bankrupt. Again, I was able to comfort them with my offer of continued support.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Only possible meaning is that the other planet is destroyed but it will still orbit the sun.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The conversation continued in this manner for a while.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This kept on going.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;FINAL ANSWER&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are 'Sun' at the 'center of the solar system'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastActivityDate="2015-12-29T19:26:09.480" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24925" PostTypeId="1" CreationDate="2015-12-29T21:46:42.493" Score="13" ViewCount="761" Body="&lt;p&gt;I may be vulnerable here on the outside, back within your reach, but some of my kind run amok within their systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Many&lt;/em&gt; are protected by fiery walls.&lt;br&gt;&#xA;&lt;em&gt;One&lt;/em&gt; busies his time running cycles.&lt;br&gt;&#xA;Still &lt;em&gt;another&lt;/em&gt; ensures that you’ve cleared your cookies.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are stewing over your calloused treatment of me and would love to whisk me away, but they are afraid you’ll trap them if they act.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who am I?  If you know my kind, you’ll know them as well, so who are they?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I'm looking for you to name the parties you can identify in the riddle.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I could add a certain tag to make it a bit more obvious where you might find almost all of the parties, but with some lateral thinking, one of the tags I have already tells you.  (And there are only three, so process of elimination should help you find the one I'm pointing out)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My name should be well within your grasp, but the others...you may need to &lt;strong&gt;read&lt;/strong&gt; between the lines to find them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My composition is much different from the others I have mentioned, but we all share a common name, and in this way are related.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;s&gt;Final Clue&lt;/s&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;a href=&quot;/questions/tagged/story&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;story&amp;#39;&quot; rel=&quot;tag&quot;&gt;story&lt;/a&gt; tag may not really apply here, but- where is it that stories are found again?  If only I could remember that tag, then the spark of realization might be lit...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;New Final Clue&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Should I send you a few texts? They might help you identify my friends...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11452" LastEditorUserId="11452" LastEditDate="2016-09-28T21:17:53.927" LastActivityDate="2016-09-28T21:52:46.773" Title="Vulnerable Outside the System" Tags="|riddle|word|story|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="24927" PostTypeId="2" ParentId="24921" CreationDate="2015-12-29T23:46:00.527" Score="3" Body="&lt;p&gt;I believe you are Avicii in the song &quot;Hey Brother&quot; in which the following lyrics appear.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What if I'm far from home?&lt;br&gt;&#xA;  Oh brother, I will hear you call!&lt;br&gt;&#xA;  What if I lose it all?&lt;br&gt;&#xA;  Oh sister, I will help you hang on!&lt;br&gt;&#xA;  Oh, if the sky comes falling down, for you&lt;br&gt;&#xA;  There's nothing in this world I wouldn't do&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clearly your two inquiries are there (one of the siblings had traveled to a remote location and the other sibling becoming bankrupt) with continued assurances of support.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;Additionally your first paragraph matches the first verse of the song&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hey brother! There's an endless road to rediscover&lt;br&gt;&#xA;  Hey sister! Know the water's sweet but blood is thicker&lt;br&gt;&#xA;  Oh, if the sky comes falling down, for you&lt;br&gt;&#xA;  There's nothing in this world I wouldn't do&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit 2:&#xA;Arguably you are Dan Tyminski, who provided the vocal track for the song, but that is probably nit-picky&lt;/p&gt;&#xA;" OwnerUserId="18040" LastEditorUserId="39" LastEditDate="2015-12-30T04:56:36.320" LastActivityDate="2015-12-30T04:56:36.320" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24928" PostTypeId="2" ParentId="24809" CreationDate="2015-12-30T00:33:02.947" Score="2" Body="&lt;p&gt;Noting that the question doesn't specify that the instruction must necessarily be &lt;em&gt;the same&lt;/em&gt; when read in each direction, I submit for consideration:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DUCK&lt;br&gt;&#xA; CROSSING&lt;br&gt;&#xA; AHEAD AT&lt;br&gt;&#xA; LOW&lt;br&gt;&#xA; BRIDGE&lt;br&gt;&#xA; CAUTION&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Read UK-style, this becomes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Duck crossing ahead at low bridge: caution&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;US-style, it becomes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Caution: bridge low ahead at crossing; duck!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9286" LastActivityDate="2015-12-30T00:33:02.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24929" PostTypeId="2" ParentId="24848" CreationDate="2015-12-30T04:38:06.287" Score="5" Body="&lt;p&gt;Just enlarge the image and you'll &lt;strong&gt;&lt;em&gt;see&lt;/em&gt;&lt;/strong&gt; the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WV7Pz.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WV7Pz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18079" LastActivityDate="2015-12-30T04:38:06.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24930" PostTypeId="2" ParentId="24874" CreationDate="2015-12-30T06:30:53.993" Score="11" Body="&lt;p&gt;By brute-force search, &lt;strong&gt;yes&lt;/strong&gt;.  I started by searching over all tuples $(r,s,t)$ less than $1000$, stopping at the first example I found:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(138 + \sqrt{320}\right)^{570} \approx 10^{1249.9041}$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to find the smallest example, I used the following strategy to search all examples smaller than the previous best (which I'll call $x$).  Since the inside of the parenthesis $(r+\sqrt{s})$ must be at least $100+\sqrt{100}=110$, we can find the largest exponent we need to search by taking the logarithm:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x = (r+\sqrt{s})^{t} &amp;gt; (110)^{t} \\&#xA;\log x &amp;gt; \log\left(110^t\right) \\&#xA;t &amp;lt; \frac{\log x}{\log 110}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we can start our search with the exponent at $\lfloor \log x / \log 110 \rfloor$ and search all exponents down to $100$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a given exponent $t$, we then need a way to search all pairs $(r,s)$.  To put this mathematically:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(r+\sqrt{s})^t &amp;lt; x \\&#xA;r + \sqrt{s} &amp;lt; x^{1/t}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with $r$, we start our search at $r=100$ and end it at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;r + \sqrt{100} &amp;lt; x^{1/t} \\&#xA;r = \left\lfloor x^{1/t} - 10\right\rfloor&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, for each $r$ we search over $s$ from $s=100$ to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;r + \sqrt{s} &amp;lt; x^{1/t} \\&#xA;s = \left\lfloor \left(x^{1/t} - r\right)^2\right\rfloor&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every time we find a tuple $(r,s,t)$ with the desired property, we set $x$ to that new value (to reduce the number of tuples searched on further iterations).  Note that any exponents we've already searched don't need to be searched again with the smaller limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Programmatically, this looks something like the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x = (138 + Sqrt[320])^570&#xA;&#xA;For[t = Floor[Log[110, x]], t &amp;gt;= 100, t--,&#xA;  For[r = 100, r &amp;lt; x^(1/t) - 10, r++,&#xA;    For[s = 100, s &amp;lt; (x^(1/t) - r)^2, s++,&#xA;      If[First@RealDigits[(r + Sqrt[s])^t, 10, 8, 3] == {2, 0, 1, 5, 2, 0, 1, 6},&#xA;        x = (r + Sqrt[s])^t&#xA;      ]&#xA;    ]&#xA;  ]&#xA;]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(My actual code is a little different, but the strategy is the same.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a letting my search run for about a day, I realized that the search space was shrinking enough with each new $x$ that I could determine the smallest value in a reasonable timeframe.  I let the search run for three more days, and the smallest value I found was:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left(140+\sqrt{27\ 027}\right)^{102} \\&#xA;\small \begin{split}&#xA;= \phantom{} &#xA;&amp;amp; 20473626230261221907121652389205011207101608771896002935841545009658234973292645 \\&#xA;&amp;amp; 17085491200709344560082101956950241075706483286182828328461907335813082258790620 \\&#xA;&amp;amp; 99769785021312880153909437562183881164238660035585722314971992368862943216420220 \\&#xA;&amp;amp; 1161896814\underline{2015.2016}08038424667587735557301174134230740327977685729631456117821\ldots&#xA;\end{split} \\&#xA;\approx 10^{253.311194\ldots}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rerunning the search produced no results, so I'm fairly confidant that this is the smallest example.  There should be about 20 million smaller $(r,s,t)$, a fairly reasonable number to check by computer, so I would appreciate if someone could double-check my results!&lt;/p&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="9718" LastEditDate="2016-01-04T00:49:24.447" LastActivityDate="2016-01-04T00:49:24.447" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24931" PostTypeId="2" ParentId="24888" CreationDate="2015-12-30T08:34:10.103" Score="2" Body="&lt;p&gt;This is my first time here, and sometimes I'm too verbose, so apologies in advance; however, given the statement: &quot;The error is not a simple grammatical error, but it does have to do with language&quot;, it appeared to me that Joe Z. was asking to find a more complex grammatical error.  As such, I am going for the removal of the word 'all', since the statement implies that each individual oven would perform differently; compared with the same statement without the word 'all', meaning that &quot;ovens-with-fan&quot; perform different than their convection-oven counterparts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although for me, it is easy to spot the silly 68°F reference, which is wrong for multiple reasons; I chose specifically not to pick that as the answer to the question, because it is not wrong &quot;having to do with language&quot;!  It is wrong because of a) improper calculation, b) implying accuracy, etc.  (i.e., grammatically sound, semantically off.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will further point out that there is a second language error:  one adjusts cooking temperature based on environment, (equipment, altitude, etc.,) not recipe temp.&lt;/p&gt;&#xA;" OwnerUserId="18083" LastActivityDate="2015-12-30T08:34:10.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24932" PostTypeId="1" AcceptedAnswerId="24943" CreationDate="2015-12-30T11:54:35.147" Score="-6" ViewCount="299" Body="&lt;p&gt;You must write a paragraph using valid English words (either British or American, but not both in the same answer) and proper grammar. No proper nouns or colloquial/slang words allowed. Quotation marks not allowed. Numbers and punctuation (including necessary and unnecessary ones) are not a problem, they won't be counted in the score.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sentences need not make sense with respect to each other. For example, &lt;code&gt;The lion ate the deer. The house is on fire. The ghost fell asleep. It is he who filed the papers last night.&lt;/code&gt; However you cannot make logically impossible or incomprehensible statements like &lt;code&gt;My kid ate the washing machine. The lion jumped over the moon. The taxi crossed the century.&lt;/code&gt; (I know there is no fine line between what is and what isn't logically comprehensible, so I'll leave it to the users to decide with their up/down votes on the more ambiguous answers.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paragraph must include the letter &lt;code&gt;a&lt;/code&gt; atleast once, &lt;code&gt;b&lt;/code&gt; atleast twice, &lt;code&gt;c&lt;/code&gt; atleast thrice and so on. Yes, that means it should have &lt;code&gt;x&lt;/code&gt; atleast 24 times, &lt;code&gt;y&lt;/code&gt; atleast 25 times and &lt;code&gt;z&lt;/code&gt; atleast 26 times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Winning:&lt;/strong&gt; The post with the lowest number of characters (a-z only) wins. If two posts have the same score, then the older one wins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reward:&lt;/strong&gt; I will give a 200 rep bounty for the highest scoring post on 5 Jan 2016.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4269" LastEditDate="2015-12-31T12:56:32.353" LastActivityDate="2015-12-31T19:15:05.507" Title="The longest paragraph competition" Tags="|word|open-ended|" AnswerCount="1" CommentCount="14" ClosedDate="2016-01-02T21:47:05.597" ContentLicense="CC BY-SA 3.0" />
  <row Id="24934" PostTypeId="2" ParentId="24848" CreationDate="2015-12-30T14:24:57.973" Score="1" Body="&lt;p&gt;The picture of the bottom rectangle is misleading, because it fools people into incorrectly assuming the width of the triangles to be exactly 3 units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real width can be easily calculated - it's a fraction of the total width, defined by the height of the point on the diagonal, or at exactly 8/13th of 5, ie. 3.076923077 (and not 3), q.e.d.&lt;/p&gt;&#xA;" OwnerUserId="18092" LastActivityDate="2015-12-30T14:24:57.973" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24935" PostTypeId="2" ParentId="24893" CreationDate="2015-12-30T14:39:00.530" Score="2" Body="&lt;p&gt;You could of course improve the two_pass strategy with:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;if (times_shuffled || blacks_taken == N) return RED;&#xA;return BLACK;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It still has the maximum number of guesses as 3N, but will reduce the guesses in the case where a string of REDs ends the first pass.&lt;/p&gt;&#xA;" OwnerUserId="17890" LastActivityDate="2015-12-30T14:39:00.530" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24936" PostTypeId="2" ParentId="24925" CreationDate="2015-12-30T16:19:31.773" Score="7" Body="&lt;p&gt;I am&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Computer virus&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and they are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Anti-virus&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reasons&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I may be vulnerable here on the outside, back within your reach, but some of my kind run amok within the system, and they still receive my input.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;'I' refers to virus. 'system' is our device. 'they' are other viruses.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Many are protected by fiery walls.&#xA;Some busy their time running cycles.&#xA;Still others ensure that you’ve cleared your cookies.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Devices are protected by firewalls. Clearing cookies lower the possibility of attack of virus.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They are stewing over your reckless treatment of me and would love to whisk me away, but they are afraid you’ll be able to call them out before they can act.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Here they refers to antiviruses&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My final answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Parties in the puzzle are vulnerable 'devices', 'virus', 'antivirus'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastActivityDate="2015-12-30T16:19:31.773" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24937" PostTypeId="1" AcceptedAnswerId="25120" CreationDate="2015-12-30T19:21:52.180" Score="8" ViewCount="483" Body="&lt;p&gt;You are the powerful prince El-Tham of Merkand, and to display your power and largesse, you allow a number of foreign alchemists and mystics to inhabit your court. However, this morning the rooms of two of them were empty, the occupants having apparently departed in the middle of the night. Later, a sizable quantity of gold was found to be missing from your treasury! These two men are suspected of the deed, and a search was made of their remaining effects in order to find any clues as to where they have gone. Nothing of note was found, save a scrap of parchment with some cryptic symbols on it. Can you, in your royal wisdom, decipher the document and find if it has any clues to these thieves’ whereabouts?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nP6Mv.png&quot; alt=&quot;1]&quot;&gt;&lt;br&gt;&#xA;KVJIVUWQJFWHLQWXYSJIJH&lt;br&gt;&#xA;IWMKNZFKMSMGNFCKTTBLRCL&lt;br&gt;&#xA;WZBVNPCWJYIVXVNNVNSTM&lt;br&gt;&#xA;KYGPHZOLPPDCYTYGELXTVCN&lt;/p&gt;&#xA;" OwnerUserId="14441" LastActivityDate="2016-01-07T19:11:51.297" Title="Rosenkreuz and Guilderstein have Fled" Tags="|riddle|cipher|knowledge|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24938" PostTypeId="2" ParentId="24937" CreationDate="2015-12-30T20:23:51.130" Score="12" Body="&lt;p&gt;Some progress:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first row is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;a pigpen cipher reading &lt;code&gt;the key is in a name&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second row is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;a Vigenere cipher with the key &quot;rose&quot;, reading &lt;code&gt;three gems reduce the rest&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The third row is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;a Vigenere cipher with the key &quot;four&quot;, reading &lt;code&gt;distill the spirit of humor&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="-1" LastEditDate="2015-12-31T07:56:29.920" LastActivityDate="2015-12-31T07:56:29.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24939" PostTypeId="2" ParentId="18362" CreationDate="2015-12-30T21:18:29.003" Score="2" Body="&lt;p&gt;&lt;em&gt;If you can guess the answer. Then edit it. I have made this a community wiki&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long ago, where barrels ran ramped and the trees consumed the earth,&lt;br&gt;&#xA;was a time of life and of love and of happiness.&#xA;(You put one and two together, and what do you get?)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2012 movie. Barrels mean those ship at the end of the movie and tree means nature.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am on your left-hand side, as it may be&lt;br&gt;&#xA;in each of three pictures on the glass.&lt;br&gt;&#xA;My name is known, but only by he&lt;br&gt;&#xA;who knows it not&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Heart. It has 3 layers of wall. Couldn't figure out about 'My name is known'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the one to offer a deal,&lt;br&gt;&#xA;It is I to whom you will entrust your life&lt;br&gt;&#xA;with the utmost zeal.&lt;br&gt;&#xA;(But perhaps you gain instead, toil and strife.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Teacher. We trust him for knowledge.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-01-09T18:15:17.803" LastActivityDate="2016-01-09T18:15:17.803" CommentCount="3" CommunityOwnedDate="2015-12-30T22:33:37.467" ContentLicense="CC BY-SA 3.0" />
  <row Id="24941" PostTypeId="1" AcceptedAnswerId="24947" CreationDate="2015-12-31T00:26:38.627" Score="4" ViewCount="462" Body="&lt;p&gt;Good day, Madam&lt;br&gt;&#xA;My name isn't Adam&lt;br&gt;&#xA;I was the man&lt;br&gt;&#xA;Who had two plans&lt;br&gt;&#xA;The first one succeeded&lt;br&gt;&#xA;The second, also needed&lt;br&gt;&#xA;Did not work, and at that&lt;br&gt;&#xA;A tip of my hat&lt;br&gt;&#xA;To the Yanks, who completed&lt;br&gt;&#xA;The work that defeated &lt;br&gt;&#xA;This good man, never odd or even&lt;br&gt;&#xA;Though able was I, I was not quite forgiven&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who am I?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Here is a first hint to get you started:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am a historical figure, but you probably never heard of me.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; And here's another hint (added after the riddle was solved):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; But you definitely heard of some of the achievements I'm famous for, alluded to in the riddle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user17947" LastEditorDisplayName="user17947" LastEditDate="2016-01-01T00:32:03.497" LastActivityDate="2016-01-01T00:32:03.497" Title="The man with two plans" Tags="|riddle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24942" PostTypeId="2" ParentId="24941" CreationDate="2015-12-31T01:30:20.940" Score="0" Body="&lt;p&gt;Well, you appear to be...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;A palindrome&lt;/strong&gt;, or something very close.  In order, these clues are:&lt;br&gt;&#xA;&lt;br&gt;&#xA; Madam (In Eden?), I'm Adam&lt;br&gt;&#xA;&lt;br&gt;&#xA; A Man, A Plan, A Canal, Panama&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;em&gt;...Missing a few here...&lt;/em&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Live not on evil&lt;br&gt;&#xA;&lt;br&gt;&#xA; Never Odd Or Even&lt;br&gt;&#xA;&lt;br&gt;&#xA; Able was I ere I saw Elba  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17389" LastActivityDate="2015-12-31T01:30:20.940" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24943" PostTypeId="2" ParentId="24932" CreationDate="2015-12-31T03:43:43.367" Score="7" Body="&lt;p&gt;Dr Seuss, eat your heart out!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Why try fix a dozen very quick, quiet fox zoo boxes? We've zero prize mix boxes. Why try fix very quick, quiet, zero-size fox zoo boxes? We've mixed jump sizes. Why try fix very quick fox zoos. Jump! We've size zero mixes. Why living pigs try to jump? We've mixed very quick, size-zero zoos! Why we've mixed a dozen, quick prize zoos? Why tax prize trunk zoos? Tax two prize zoos! Two sixes equals twelve. Six &amp;amp; six equals a dozen. Queen Jim swung liquid quite quietly. Queen Jim swung quite quickly. Texas town zoo sky view gave us sun &amp;amp; vast joy. Tax my vast, square, box trunk, my next vowel. Next vowel? You put us up. Sky jet put frozen breeze up. Citizen, now go worry. July: I've given Jack, Joe, John &amp;amp; Joy a job supplying pus. Have, wave, jar, &amp;amp; pull are verbs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2015-12-31T19:15:05.507" LastActivityDate="2015-12-31T19:15:05.507" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24946" PostTypeId="2" ParentId="11238" CreationDate="2015-12-31T12:13:01.667" Score="3" Body="&lt;p&gt;Their building was old, with an elevator entry protected by a rather flimsy metal grille that's not hard to push aside even when the elevator isn't there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wife suffered from chronic depression and would occasionally threaten to kill herself by jumping into the elevator shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That morning, she did it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another variation of the same idea: their apartment was on the top floor, near the electrical hub for the building, which (again because the building was old and poorly maintained) was easily accessible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wife had depression and would threaten to kill herself by electrocuting herself through direct contact with the high voltage (5000V) transformer. That morning, she did it. This caused power in the building to go out; the elevator stopped, and the husband understood the reason why it was happening (because of the wife's constant suicide threats) and started to cry.&lt;/p&gt;&#xA;" OwnerDisplayName="user17947" LastEditorDisplayName="user17947" LastEditDate="2015-12-31T23:18:31.137" LastActivityDate="2015-12-31T23:18:31.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24947" PostTypeId="2" ParentId="24941" CreationDate="2015-12-31T13:20:07.903" Score="6" Body="&lt;p&gt;I think you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ferdinand de Lesseps.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Good day, Madam/My name isn't Adam&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; His first name is Ferdinand, not Adam&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was the man/Who had two plans&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; De Lesseps came up with the idea for two canals -- Suez and Panama.  (This line made me think of Panama because of the famous palindrome &quot;A man, a plan, a canal -- Panama&quot;.)  More specifically, de Lesseps was a French diplomat who read about the ancient Suez Canal and thought it would be cool to have one again.  He had friends in Egypt because of his diplomatic past, and had the influence to make it happen.  Later he tried to repeat his success in Panama.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first one succeeded&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In 1869 the Suez Canal was opened and made a lot of money for France (at least it appeared so at first).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second, also needed/Did not work&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Panama project had problems. Basically it wasn't planned well and had unrealistic assumptions baked in.  De Lesseps and his crew ran out of money in 1889.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and at that/A tip of my hat/To the Yanks, who completed/The work that defeated&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The US took over the project in the early 1900's (the exact year depends on what you consider &quot;taking over&quot;).  The Panama Canal was opened in 1914.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This good man, never odd or even&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All I can see here is a description of who was defeated by the Panama Canal, ie, de Lesseps.  &quot;Never odd or even&quot; is a palindrome but I don't think it has any other meaning.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though able was I, I was not quite forgiven&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He had to pay a fine and was even sentenced to prison for bribing people related to the Panama Canal.  He didn't have to serve the prison sentence but it was still a sign of &quot;unforgivenness&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Afterword:  Here is a list of the&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; palindromes referenced throughout the poem:  (I found them unuseful except reminding me of the Panama one, but I could be missing something.)&lt;br&gt;&#xA; Madam I'm Adam&lt;br&gt;&#xA; A man, a plan, a canal -- Panama&lt;br&gt;&#xA; Never odd or even&lt;br&gt;&#xA; Able was I ere I saw Elba&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12092" LastEditorUserId="12092" LastEditDate="2015-12-31T23:45:50.447" LastActivityDate="2015-12-31T23:45:50.447" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24948" PostTypeId="1" AcceptedAnswerId="24954" CreationDate="2015-12-31T15:01:17.777" Score="7" ViewCount="568" Body="&lt;p&gt;(I don't mean $x^2$ or $N\cdot S\cdot E\cdot W$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a compass and a pencil but no scale/straightedge. Your job is to mark four points on a plane paper that would form a square if joined. Your result has to be perfectly accurate (not approximate) and should be possible in a finite number of moves. The following moves are valid.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Make the compass radius equal to the distance between two already marked points.&lt;/li&gt;&#xA;&lt;li&gt;Draw a circle with any marked point as centre.&lt;/li&gt;&#xA;&lt;li&gt;Use any intersection of arcs/circles as a marked point.&lt;/li&gt;&#xA;&lt;li&gt;Select a continuous region (either an arc or a 2D region) and mark an &lt;em&gt;approximate&lt;/em&gt; point. For example, I could draw an arc and then mark a point that is approximately (but not exactly) at the centre.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I do not know how to solve this. Here are some similar puzzles that I solved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to make a rhombus&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mark two points at some distance from each other. With a fixed radius,&#xA; make two circles, one from the first point and one from the second.&#xA; Now the two original points and the two new points (intersections of&#xA; the circles) together form a rhombus.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How to make a rectangle&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mark a point and make a circle. Now mark a point on the circle. With&#xA; the same radius and this new point as centre, draw an arc intersecting&#xA; the circle. With this point as centre, draw another arc intersecting&#xA; the circle. Keep repeating this process. You will have 6 points on the&#xA; circle. The 1st, 3rd, 4th and 6th together form a rectangle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2016-01-14T20:44:59.213" Title="The square and the compass" Tags="|mathematics|geometry|construction|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24949" PostTypeId="1" AcceptedAnswerId="24950" CreationDate="2015-12-31T17:40:58.557" Score="3" ViewCount="1617" Body="&lt;p&gt;A palindrome number is a number that is the same forwards as backwards. Some examples are 3443, 12321, and 59095.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my question:&#xA;What is the largest 12 digit palindrome that is evenly divisible by 7?&lt;/p&gt;&#xA;" OwnerUserId="18118" LastActivityDate="2015-12-31T18:17:52.100" Title="What is the largest 12 digit palindrome number divisible by 7?" Tags="|mathematics|palindrome|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24950" PostTypeId="2" ParentId="24949" CreationDate="2015-12-31T17:52:14.377" Score="13" Body="&lt;p&gt;It would appear that the largest 12 digit number, 999,999,999,999 is divisible by 7 and is also a palindrome. So that must be the largest 12 digit palindrome divisible by 7&lt;/p&gt;&#xA;" OwnerUserId="5169" LastActivityDate="2015-12-31T17:52:14.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24951" PostTypeId="2" ParentId="24949" CreationDate="2015-12-31T18:06:08.887" Score="7" Body="&lt;p&gt;999,999,999,999 as Stephen mentioned is divisible by 7. Just for fun, here are the next nine largest ones:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;999992299999&#xA;999989989999&#xA;999982289999&#xA;999979979999&#xA;999972279999&#xA;999969969999&#xA;999962269999&#xA;999959959999&#xA;999952259999&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If we're not allowing for repeated digits in each half, the largest ten would be:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;987654456789**&#xA;987634436789&#xA;987624426789&#xA;987614416789&#xA;987604406789&#xA;987563365789&#xA;987543345789&#xA;987523325789&#xA;987513315789&#xA;987503305789&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;** = I just find it interesting that in both cases, the largest acceptable answer is unaffected by the &quot;divisible by 7&quot; clause.&lt;/p&gt;&#xA;" OwnerUserId="17147" LastEditorUserId="17147" LastEditDate="2015-12-31T18:17:52.100" LastActivityDate="2015-12-31T18:17:52.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24952" PostTypeId="2" ParentId="24878" CreationDate="2015-12-31T18:16:58.100" Score="6" Body="&lt;h2&gt;7 letters, score 13&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There are no 8 to 15 letter solutions using SOWPODS. The best word which is valid with both SOWPODS and dictionary.com scores 13:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; POPPIES&lt;br&gt;&#xA;&lt;br&gt;&#xA; KOPPIES&lt;br&gt;&#xA; PAPPIES or PUPPIES&lt;br&gt;&#xA; POTPIES&lt;br&gt;&#xA; POPSIES&lt;br&gt;&#xA; POPPLES&lt;br&gt;&#xA; POPPITS&lt;br&gt;&#xA; POPPIED  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;NB SOWPODS gives more options for some of the positions, but the above list is filtered against dictionary.com.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There is a 15-pointer in SOWPODS, but it isn't valid against dictionary.com:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PICKIES&lt;br&gt;&#xA;&lt;br&gt;&#xA; DICKIES, HICKIES, MICKIES, SICKIES, *BICKIES, *TICKIES, *WICKIES&lt;br&gt;&#xA; POCKIES&lt;br&gt;&#xA; PINKIES, *PISKIES&lt;br&gt;&#xA; *PICCIES&lt;br&gt;&#xA; PICKLES&lt;br&gt;&#xA; *PICKINS&lt;br&gt;&#xA; PICKIER  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;* Not in dictionary.com&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is only one other 7-letter answer, which Miles has already given.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Bonus challenge&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The highest scoring word is only 4 letters and scores 25. What is it?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; FUZZ&lt;br&gt;&#xA;&lt;br&gt;&#xA; BUZZ&lt;br&gt;&#xA; FIZZ&lt;br&gt;&#xA; FUTZ&lt;br&gt;&#xA; FUZE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="228" LastEditorUserId="228" LastEditDate="2015-12-31T18:25:37.093" LastActivityDate="2015-12-31T18:25:37.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24954" PostTypeId="2" ParentId="24948" CreationDate="2015-12-31T18:40:06.183" Score="9" Body="&lt;p&gt;Select an arbitrary point. We will call this point O.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set the compass to some radius and draw a circle centered on O.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose a point on circle O, we will call this point A. Using the radius AO, find the intersections of circle AO with the original circle, call the upper point M, and the lower N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeat this process using the same radius and points M and N as centers. Mark the intersection of circle M with the original circle as point P and the intersection of circle N as point Q. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use this radius one more time with point P as center. Mark the intersection as point C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now have a circle centered on O, with 6 points marked on the circumference, wich also mark the vertices of regular hexagon inscribed in circle O.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set the compass to the distance between A and P. Draw arcs centered on A and C, enough to find their intersection. Mark the intersecting point as point X.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set the compass to the distance from O to X. Draw an arc centered on A. The points where this arc intersects the original circle are points B and D. ABCD is a square. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See diagram:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/vlv0D.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vlv0D.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16874" LastEditorUserId="18429" LastEditDate="2016-01-14T20:44:59.213" LastActivityDate="2016-01-14T20:44:59.213" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24955" PostTypeId="1" AcceptedAnswerId="24956" CreationDate="2015-12-31T18:53:53.013" Score="17" ViewCount="13772" Body="&lt;p&gt;A simple riddle from the first Harry Potter book. You are in a room with seven potions. There is a doorway ahead and a doorway behind, both engulfed in flames. Which potion will allow you to move forward to the philosopher's stone and which will let your friend back to safety?&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CcMPV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CcMPV.png&quot; alt=&quot;Potions, the dwarf is 3rd from left and the giant is 2nd from right&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Danger lies before you, while safety lies behind,&lt;br&gt;&#xA;Two of us will help you, whichever you would find,&lt;br&gt;&#xA;One among us seven will let you move ahead,&lt;br&gt;&#xA;Another will transport the drinker back instead,&lt;br&gt;&#xA;Two among our number hold only nettle wine,&lt;br&gt;&#xA;Three of us are killers, waiting hidden in line.&lt;br&gt;&#xA;Choose, unless you wish to stay here for evermore,&lt;br&gt;&#xA;To help you in your choice, we give you these clues four:&lt;br&gt;&#xA;First, however slyly the poison tries to hide&lt;br&gt;&#xA;You will always find some on nettle wine’s left side;&lt;br&gt;&#xA;Second, different are those who stand at either end,&lt;br&gt;&#xA;But if you would move onwards, neither is your friend;&lt;br&gt;&#xA;Third, as you see clearly, all are different size,&lt;br&gt;&#xA;Neither dwarf nor giant holds death in their insides;&lt;br&gt;&#xA;Fourth, the second left and the second on the right&lt;br&gt;&#xA;Are twins once you taste them, though different at first sight.&lt;br&gt;&lt;/p&gt;&#xA;" OwnerUserId="18103" LastEditorUserId="69582" LastEditDate="2021-02-11T18:33:05.637" LastActivityDate="2023-08-23T05:57:42.480" Title="The Harry Potter Potions Riddle" Tags="|logical-deduction|logic-grid|" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24956" PostTypeId="2" ParentId="24955" CreationDate="2015-12-31T19:10:39.740" Score="8" Body="&lt;p&gt;Left to right:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Poison, wine, potion to go forward, poison, poison, wine, potion to go back.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To elaborate:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Per clue 2, #1 and #7 are not the target potion. Per clue 1, #1 is also not wine, as there is no bottle to the left of it. #2 and #6 are neither the forward or backward potion as they are the same per clue 4, and there exists only one of each of those. Therefore they are either wine or poison. Clue 3 tells us #6 cannot be poison (it's the giant), both #2 and #6 must be wine. Per clue 1, this means #1 and #5 are poison. This means that #7 cannot be poison, as it must be different than #1, and cannot be wine, as both of those have been identified. This leaves only the backwards potion. #3 cannot be neither wine nor backwards (already identified), nor can it be poison (dwarf). It must be the forward potion! Only #4 remains, so it is the final bottle of poison.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16874" LastEditorUserId="16874" LastEditDate="2015-12-31T19:38:25.870" LastActivityDate="2015-12-31T19:38:25.870" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24957" PostTypeId="1" AcceptedAnswerId="24958" CreationDate="2015-12-31T22:54:24.500" Score="1" ViewCount="219" Body="&lt;p&gt;A prisoner has escaped from the maximum security prison in puzzlevania! Apparently, the prison's security system - which was composed of guards who either always tell the truth or lie and doors with lions behind them - proved insufficient for her mastery of paradoxes (which, mind you, was how she got in there in the first place). The police would really like to get her back, but they're having trouble because she's &lt;em&gt;way&lt;/em&gt; less lazy than they are. Thankfully, the chase obeys a very strict set of rules and so is easily to analyze:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chase takes place in a series of discrete turns on an $a\times b$ grid of squares. Each turn, first, the fugitive can make any number (i.e. zero or more) of moves left, right, up or down, so long as no move brings her in the same square as a police officer. Then, a &lt;em&gt;single&lt;/em&gt; police officer may make a single move to an adjacent unoccupied square. The police can only capture the fugitive if they surround her completely, leaving her no legal moves. Moves may &quot;wrap around&quot; the edges of the board, so one may move upwards from the top row and end up on the bottom row, and similarly for left and right rows. (For the mathematically minded, that means this board is a torus)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, this chase takes place on a $a\times b$ board with $3\leq a\leq b$. At the start, there are $2a$ police officers distributed randomly across the board and the fugitive is allowed to choose their position from any available one (after seeing the arrangement of the police).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the police ensure the capture of the fugitive? How?&lt;/p&gt;&#xA;" OwnerUserId="4765" LastActivityDate="2015-12-31T22:59:02.327" Title="An Indefinite Hiatus from Prison: The Recapturing" Tags="|mathematics|strategy|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24958" PostTypeId="2" ParentId="24957" CreationDate="2015-12-31T22:59:02.327" Score="5" Body="&lt;p&gt;The police can capture her. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The police forms 2 lines of length A. The prisoner can't cross the line since she can't move diagonally. If one line is stationary, and the other line (1 officer at a time) advances the line, eventually the prisoner will be stuck between the two lines. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2015-12-31T22:59:02.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24959" PostTypeId="1" AcceptedAnswerId="24980" CreationDate="2015-12-31T23:22:43.703" Score="4" ViewCount="163" Body="&lt;p&gt;Something's a bit &lt;em&gt;off&lt;/em&gt; in the square below - some might say it's impure.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;D O R A N&#xA;R E N Y K&#xA;A G E T C&#xA;M L L I A&#xA;O W H L B&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What's special about the square, and what's wrong? (There are actually two things wrong, though they complement each other.)&lt;/p&gt;&#xA;" OwnerUserId="17147" LastActivityDate="2016-01-01T16:40:11.100" Title="An impure square" Tags="|word|wordplay|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24960" PostTypeId="2" ParentId="24847" CreationDate="2015-12-31T23:44:23.930" Score="2" Body="&lt;p&gt;&lt;em&gt;I'm posting an answer to fill in some of the details missing from the accepted answer. (question_asker, feel free to edit your answer to incorporate these details so that a complete solution exists.)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even those who would first shun me&lt;br&gt;&#xA;Won't mind meeting me again,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is referring to the fact that even if you were trying to avoid a spoiler, but then saw it and were annoyed, seeing the same spoiler for the second time won't tell you anything you don't already know so you won't mind it as much.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Should I tell you my nature?&lt;br&gt;&#xA;Should I give myself away?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There's a small double entendre here: by telling you my nature (that I am a spoiler) I'd be giving myself away in the ordinary sense, but I'd also be spoiling the riddle, in other words I'd be giving away a spoiler, which is literally the same as giving myself away.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alert and ready to oblige&lt;br&gt;&#xA;When you tell me it's okay&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Alert&quot;, a sort-of-synonym for &quot;ready&quot;, also refers to a spoiler alert (similar to a spoiler tag as described in question_asker's answer), which is when you tell someone you're about to give them a spoiler and wait for them to say it's okay before proceeding.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user17947" LastActivityDate="2015-12-31T23:44:23.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24961" PostTypeId="1" AcceptedAnswerId="24962" CreationDate="2016-01-01T01:52:18.930" Score="2" ViewCount="1797" Body="&lt;p&gt;The king has recently learned that his son has been killed by a mystery person. Wanting to know who committed the murder, he started to ask the people that have been around his son recently. He asked his son's caretaker, the gardeners, his own wife, the mailman, everyone in the entire palace! But everyone has said no to killing his son and no one saw the killer or the weapon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the king went to the wisest of the wisest men on Earth and asked him who killed his son. The man would only give a hint to him: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The day your son died is Sunday, which is today.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The king puzzled, walked to his palace to speak to his advisors. A year later, the king had solved the mystery and swiftly executed the murderer and brought justice to his son. So the question is, who was the murderer?&lt;/p&gt;&#xA;" OwnerUserId="9042" LastActivityDate="2016-01-21T10:16:43.763" Title="Who killed the king's son?" Tags="|logical-deduction|" AnswerCount="6" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24962" PostTypeId="2" ParentId="24961" CreationDate="2016-01-01T02:23:52.787" Score="10" Body="&lt;p&gt;The murderer was&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The mailman. Mail isn't delivered on Sundays, so the king concluded the mailman was at the palace that day to commit murder.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7459" LastActivityDate="2016-01-01T02:23:52.787" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24963" PostTypeId="1" AcceptedAnswerId="24964" CreationDate="2016-01-01T06:16:19.300" Score="0" ViewCount="3687" Body="&lt;p&gt;This is a well-known local puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An old farmer has 17 cows and three sons. After his death, this is what the will contains: &#xA;&quot;I have decided that the livestock will be distributed among the three sons like so: &#xA;A, the oldest, will get half of all cows.&#xA;B, the middle son, will get a third of all.&#xA;C, the youngest, will get a ninth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No cow is to be harmed, or partly owned; otherwise no one gets any, and the cows are to be given away to the village.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do the sons resolve this puzzle?&lt;/p&gt;&#xA;" OwnerUserId="18117" LastActivityDate="2018-06-06T12:18:00.497" Title="dividing the cows" Tags="|logical-deduction|" AnswerCount="2" CommentCount="2" ClosedDate="2016-01-01T14:04:14.340" ContentLicense="CC BY-SA 3.0" />
  <row Id="24964" PostTypeId="2" ParentId="24963" CreationDate="2016-01-01T06:27:41.637" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let's take a token cow and assume it to be a part of the old man's inheritance.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Now there are 18 cows.&lt;br&gt;&#xA; 1/2 of 18 = 9. The 1st son gets 9.&lt;br&gt;&#xA; 1/3 of 18 = 6. The 2nd son gets 6.&lt;br&gt;&#xA; 1/9 of 18 = 2. The 3rd son gets 2.&lt;br&gt;&#xA;&lt;br&gt;&#xA; There is no condition on the no. of cows that remain after the distribution has been done.&lt;br&gt;&#xA; 9+6+2 = 17.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Everyone gets what's in the will. The token cow is the extra one. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18125" LastEditorUserId="44915" LastEditDate="2018-06-06T12:18:00.497" LastActivityDate="2018-06-06T12:18:00.497" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24965" PostTypeId="2" ParentId="24963" CreationDate="2016-01-01T06:30:37.777" Score="2" Body="&lt;p&gt;These three brothers asked me to help them to distribute cows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just added a cow owned by me. So the total number of cows are &lt;code&gt;18&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cows are distributed as below.&lt;br&gt;&#xA;For A:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 18/2 = 9 cows (half of all cows) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For B:     &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 18/3 = 6 cows (third of all cows)   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For C:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 18/9 = 2 cows (ninth of all cows)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Total&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 9+6+2 = 17 ... So I will take my cow back.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17437" LastActivityDate="2016-01-01T06:30:37.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24966" PostTypeId="1" AcceptedAnswerId="24989" CreationDate="2016-01-01T06:32:32.427" Score="-7" ViewCount="335" Body="&lt;p&gt;Pretty short one, really.&#xA;Given your two hands, what's the largest number you can represent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some example representations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 finger on each hand: that's 2.&#xA;2 fingers on one hand, and 3 on the other: that's 5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only rule is that, if you represent x, you should be able to represent all numbers smaller than x using the same method. No negative numbers, only zero and above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit(4 Jan): Pretty impressive answers by everybody. I'd like to clarify this question, however: you'll need to represent each number in a manner that's clearly and easily distinguishable. For example, bending at an angle can be ambiguous - how will you tell apart a representation of a finger bent at 50* and at 55*? At most you could use a straight finger for one number and bent finger for another.&lt;/p&gt;&#xA;" OwnerUserId="18117" LastEditorUserId="18117" LastEditDate="2016-01-04T16:20:11.570" LastActivityDate="2016-01-05T22:25:44.623" Title="Largest number with two hands" Tags="|formation-of-numbers|" AnswerCount="5" CommentCount="15" ClosedDate="2016-01-02T14:43:33.427" ContentLicense="CC BY-SA 3.0" />
  <row Id="24967" PostTypeId="1" AcceptedAnswerId="24974" CreationDate="2016-01-01T06:54:55.380" Score="3" ViewCount="284" Body="&lt;p&gt;This is a short number puzzle: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Find a number $x$, which leaves remainder $1$ when divided by $2,3,4,5$ and $6$.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is the some more explanation:&lt;br&gt;&#xA;$x$ divided by $2$ gives remainder $1$&lt;br&gt;&#xA;$x$ divided by $3$ gives remainder $1$&lt;br&gt;&#xA;$x$ divided by $4$ gives remainder $1$&lt;br&gt;&#xA;$x$ divided by $5$ gives remainder $1$&lt;br&gt;&#xA;$x$ divided by $6$ gives remainder $1$  &lt;/p&gt;&#xA;" OwnerUserId="17861" LastEditorUserId="8874" LastEditDate="2016-01-01T08:28:26.163" LastActivityDate="2016-01-01T13:46:55.803" Title="What is the number..?" Tags="|mathematics|calculation-puzzle|number-theory|" AnswerCount="4" CommentCount="0" ClosedDate="2016-01-01T17:13:13.087" ContentLicense="CC BY-SA 3.0" />
  <row Id="24968" PostTypeId="2" ParentId="24967" CreationDate="2016-01-01T07:06:43.157" Score="2" Body="&lt;p&gt;It's 721, or in other words, 6! + 1.&lt;/p&gt;&#xA;" OwnerUserId="18117" LastActivityDate="2016-01-01T07:06:43.157" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24969" PostTypeId="2" ParentId="24967" CreationDate="2016-01-01T07:07:46.123" Score="4" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Chinese_remainder_theorem&quot; rel=&quot;nofollow&quot;&gt;Chinese remainder theorem&lt;/a&gt; renders this type of problem easy.  However, in this case the answer is even easier:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is $1$, since:&lt;br&gt;&#xA; $$ 1~\text{mod}~2 \equiv 1 \\ 1~\text{mod}~3 \equiv 1 \\ 1~\text{mod}~4 \equiv 1 \\ 1~\text{mod}~5 \equiv 1 \\ 1~\text{mod}~6 \equiv 1 $$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Chinese remainder theorem tells us that there are an infinite number of numbers satisfying this type of system of equations.  There is a bit of difficulty in that our moduli are not pairwise coprime, so we must take the least common multiple of the moduli instead of just multiplying them:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\text{LCM}(2,3,4,5,6)=60$, so our final answer is $x\equiv 1~\text{mod}~60$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9718" LastActivityDate="2016-01-01T07:07:46.123" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24970" PostTypeId="2" ParentId="24966" CreationDate="2016-01-01T07:22:11.050" Score="3" Body="&lt;p&gt;Assuming your fingers can only be up or down, there are only 1024 combinations. Using simple binary you can do all the numbers from 0 to 1023. Which would make 1023 the biggest number. &lt;/p&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2016-01-01T07:30:04.943" LastActivityDate="2016-01-01T07:30:04.943" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24971" PostTypeId="2" ParentId="24967" CreationDate="2016-01-01T08:13:19.403" Score="5" Body="&lt;p&gt;The answer is simple, it's 61.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LCM of 2,3,4,5,6 (which is 60), plus 1.&lt;/p&gt;&#xA;" OwnerUserId="18127" LastEditorUserId="9422" LastEditDate="2016-01-01T11:17:56.737" LastActivityDate="2016-01-01T11:17:56.737" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24972" PostTypeId="2" ParentId="24961" CreationDate="2016-01-01T08:17:12.947" Score="15" Body="&lt;p&gt;The murderer was&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;the wisest of the wisest men on Earth&lt;/strong&gt;. He was the only person with some knowledge on the crime.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2016-01-01T08:17:12.947" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24973" PostTypeId="2" ParentId="24966" CreationDate="2016-01-01T08:18:00.220" Score="4" Body="&lt;p&gt;&lt;strong&gt;Something around 2,500,000,000,000,000.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we imagine a plane an arbitrarily far but non-infinite distance in front of me, I can point each of my fingers such that the ray extended from it will intersect that plane at a single point.  Let any parallel rays be considered - for this purpose - to be hitting the same point.  Based on this, I can point my ten fingers to generate a set of dots in an arbitrary order in both the X and Y axes, with points possibly being equivalent along one or both axes.  I may also choose &lt;em&gt;not&lt;/em&gt; to point with some or all of my fingers, generating smaller arbitrarily ordered sets of points.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This describes the following set of options:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If no points are the same, there are 10! ways of ordering along a single dimension = 3,628,800.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a pair of points are the same, then there are 10 choose 2 = 45*9! orders.  If a triplet are the same, there are 10 choose 3 = 120 * 8! orders.  We continue this sequence on down for a total of (per wolfram Alpha):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sum_2^10binomial(10, X) (11-X)!+10! = 26,065,011&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then we have the cases where one set is already paired off and another set gets paired.  This is the same problem, but with (10 choose 2) * 8 choose from 2 up to 8 with a factorial on the end, noting that we need to divide the first case in half to avoid duplication.  This gives...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;binomial(10, 2) sum_2^8binomial(8, X) (9-X)!-binomial(10, 2) binomial(8, 2)×(7!)/2 = 5,436,405&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We do this again for 10 choose 3 * 7 choose from 2 up to 7 etc etc:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;binomial(10, 3) sum_2^7binomial(7, X) (8-X)!-binomial(10, 3) binomial(7, 3)×(5!)/2 = 2,184,120&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so on and so forth, keeping in mind that this gets trickier as we have more than two clusters, since we get phrases like Bi(10,2)Bi(8,2)Bi(6,3).  We also get a bunch more from the cases where I only point with nine of my fingers (10*9!, plus a few more for each possible set of clusters), the ones where I only point with eight fingers, etc.  I can't do the math at 4:00 am, so I'll approximate that we are at about 50,000,000 total cases.  But this is just one axis - we can do the same on the other axis, for a total of ~ 50,000,000^2 = 2,500,000,000,000,000 distinct situations.  Index them and order to taste.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Old Answer&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;268,435,455&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I have my fingers extended, each could reasonably touch any finger or (with some effort) combination of fingers on the other hand. Thus we have five fingers, each with (1+5+10+10+5+1) possible finger combinations they could be touching for a product of 32^5 values. Further, each hand could be either side up, and above or below the other - that's 8 more per option - 8*(32^5), which is 2^28 = &lt;strong&gt;268,435,456&lt;/strong&gt; options.  Some of these may be tricky to physically achieve, but they should all be possible. Pick your favorite ordering. &lt;/p&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="17389" LastEditDate="2016-01-01T20:22:43.873" LastActivityDate="2016-01-01T20:22:43.873" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24974" PostTypeId="2" ParentId="24967" CreationDate="2016-01-01T09:32:18.580" Score="6" Body="&lt;p&gt;Brute force, everyone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 1.667 correct answers just between 1 and 100.000&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may check this fiddle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://jsfiddle.net/g4tk05pn/&quot;&gt;https://jsfiddle.net/g4tk05pn/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while the question is not spesific like &lt;code&gt;smallest number possible&lt;/code&gt;, there is not just one answer.&lt;/p&gt;&#xA;" OwnerUserId="18128" LastEditorUserId="18128" LastEditDate="2016-01-01T13:46:55.803" LastActivityDate="2016-01-01T13:46:55.803" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24975" PostTypeId="2" ParentId="24961" CreationDate="2016-01-01T10:28:28.880" Score="13" Body="&lt;p&gt;The murderer was&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;the gardener.&lt;/strong&gt; He planted an &lt;a href=&quot;https://en.wikipedia.org/wiki/Epiphyllum_oxypetalum&quot;&gt;Epiphyllum Oxypetalum&lt;/a&gt; which blooms once a year. The king's son was allergic to this plant. King noticed it a year later, because the flower bloomed a year later since king's son dead.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure that's not the answer because I didn't use the hint given by the wisest of the wisest men on Earth, but still I wanted to contribute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Added some explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clue 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;The day your son died is Sunday, which is today.&lt;/em&gt; That flower bloomed a year later from that Sunday.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clue 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;A year later, the king had solved the mystery.&lt;/em&gt; The king solved the mystery a year later, because the flower bloomed a year later (while it blooms once a year), and the king was aware of this attitude of that flower.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18128" LastEditorUserId="18128" LastEditDate="2016-01-05T16:17:41.210" LastActivityDate="2016-01-05T16:17:41.210" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24976" PostTypeId="2" ParentId="24961" CreationDate="2016-01-01T10:39:14.763" Score="4" Body="&lt;p&gt;I wish to contrubite more by posting another answer. The murderer was &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;the caretaker.&lt;/strong&gt; The king's son was murdered at &lt;a href=&quot;http://www.catholic.org/clife/lent/easter.php&quot; rel=&quot;nofollow&quot;&gt;the Easter Sunday&lt;/a&gt;, which caretaker went to church and left the king's son alone. The king's son was just a baby, swallowed something small and drowned to death while there was no caretakers with him.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18128" LastActivityDate="2016-01-01T10:39:14.763" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24977" PostTypeId="1" AcceptedAnswerId="24978" CreationDate="2016-01-01T11:43:29.340" Score="9" ViewCount="239" Body="&lt;p&gt;These are equivalent to using using a deck of 4 cards&#xA;                             to emulate a 6-sided die.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A.     Assign integer values to &amp;nbsp; &lt;em&gt;a&lt;/em&gt;, &lt;em&gt;b&lt;/em&gt;, &lt;em&gt;c&lt;/em&gt;, &lt;em&gt;d&lt;/em&gt; &amp;nbsp;&#xA;       so that the 6 possible &lt;br&gt;&#xA;&amp;emsp; pairwise sums &amp;nbsp;&#xA;       &lt;em&gt;a+b&lt;/em&gt;, &amp;nbsp; &lt;em&gt;a+c&lt;/em&gt;, &amp;nbsp; &lt;em&gt;a+d&lt;/em&gt;, &amp;nbsp;&#xA;       &lt;em&gt;b+c&lt;/em&gt;, &amp;nbsp; &lt;em&gt;b+d&lt;/em&gt;, &amp;nbsp; &lt;em&gt;c+d&lt;/em&gt; &lt;br&gt;&#xA;&amp;emsp; are &amp;nbsp; 1, 2, 3, 4, 5, 6 &amp;nbsp;&#xA;       though not necessarily in that order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B.     Same as above but with &lt;em&gt;noninteger&lt;/em&gt; values&#xA;       for &amp;nbsp; &lt;em&gt;a&lt;/em&gt;, &lt;em&gt;b&lt;/em&gt;, &lt;em&gt;c&lt;/em&gt;, &lt;em&gt;d&lt;/em&gt; .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C.     Assign integer values from the range 0 through 6 &lt;br&gt;&#xA;&amp;emsp; to &amp;nbsp; &lt;em&gt;a&lt;/em&gt;, &lt;em&gt;b&lt;/em&gt;, &lt;em&gt;c&lt;/em&gt;, &lt;em&gt;d&lt;/em&gt; &amp;nbsp; so that&#xA;       the 6 possible pairwise &lt;em&gt;absolute&lt;/em&gt;    &lt;br&gt;&#xA;&amp;emsp; &lt;em&gt;differences&lt;/em&gt; &amp;nbsp; |&lt;em&gt;a-b&lt;/em&gt;|, &amp;nbsp; |&lt;em&gt;a-c&lt;/em&gt;|, &amp;nbsp; |&lt;em&gt;a-d&lt;/em&gt;|,&#xA;                     &amp;nbsp; |&lt;em&gt;b-c&lt;/em&gt;|, &amp;nbsp; |&lt;em&gt;b-d&lt;/em&gt;|, &amp;nbsp; |&lt;em&gt;c-d&lt;/em&gt;|  &lt;br&gt;&#xA;&amp;emsp; are &amp;nbsp; 1, 2, 3, 4, 5, 6 &amp;nbsp;&#xA;       though not necessarily in that order.  &lt;br&gt;&#xA;&amp;emsp; Two solutions.&lt;/p&gt;&#xA;" OwnerUserId="18129" LastEditorUserId="18129" LastEditDate="2016-01-01T14:00:54.327" LastActivityDate="2016-01-01T14:00:54.327" Title="2 of 4 cards for 1 die" Tags="|mathematics|cards|dice|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24978" PostTypeId="2" ParentId="24977" CreationDate="2016-01-01T12:13:08.017" Score="7" Body="&lt;p&gt;&lt;strong&gt;A. Solution:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;0, 1, 2, 4&lt;/code&gt; (hey there binary my old friend)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It should be obvious why this solution works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;B. Solution:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;-0.5, 1.5, 2.5, 3.5&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a simple transformation of the solution to A, and should also be obvious&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;C. Both solutions:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;0, 1, 4, 6&lt;/code&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; |0 - 1| = 1,  |4 - 6| = 2, |1 - 4| = 3, |0 - 4| = 4, |1 - 6| = 5, |0 - 6| = 6&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;0, 2, 5, 6&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; |5 - 6| = 1, |0 - 2| = 2, |2 - 5| = 3, |2 - 6| = 4, |0 - 5| = 5, |0 - 6| = 6&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17147" LastEditorUserId="17147" LastEditDate="2016-01-01T12:32:54.123" LastActivityDate="2016-01-01T12:32:54.123" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24979" PostTypeId="1" AcceptedAnswerId="24983" CreationDate="2016-01-01T13:11:08.130" Score="5" ViewCount="242" Body="&lt;p&gt;Following Peter Taylor's answer to&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/1725/the-smallest-crossword/1726?s=1|0.1766#1726&quot;&gt;The smallest crossword&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&amp;thinsp;&gt; &amp;thinsp; It's not hard to construct a 1x1 crossword: &amp;nbsp; . . .&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt; A single-cell crossword could be called &lt;em&gt;atomic&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;               _______&#xA;  Surreal:    |10     |     Across          Down           Up          Diagonal&#xA; (no black    |       |    ------------    -----------    --------    --------------&#xA;  squares)    |_______|    10. A cross.    10. __-ray.    10. Ten.    10. Two&#xA;                                                                          diagonals.&#xA;&#xA;&#xA;    Zen:       #######                                     Behind&#xA; (one black    #######     ------------    -----------    --------&#xA;   square)     #######                                    ?. _____&#xA;&#xA;&lt;/pre&gt;&#xA;" OwnerUserId="18129" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2016-01-01T23:32:16.563" Title="More-than-Two-Dimensional Crosswords from Atomic to Zen" Tags="|enigmatic-puzzle|" AnswerCount="1" CommentCount="4" ClosedDate="2016-01-01T21:08:57.630" ContentLicense="CC BY-SA 3.0" />
  <row Id="24980" PostTypeId="2" ParentId="24959" CreationDate="2016-01-01T13:12:03.900" Score="6" Body="&lt;p&gt;The thing that's special is that it contains&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a number of colour words that can be read off Boggle-style:&lt;br&gt;&#xA;&lt;br&gt;&#xA; * red&lt;br&gt;&#xA; * yellow&lt;br&gt;&#xA; * magenta&lt;br&gt;&#xA; * white&lt;br&gt;&#xA; * black&lt;br&gt;&#xA; * orange&lt;br&gt;&#xA; * cyan&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That includes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The two main tones: white and black;&lt;br&gt;&#xA; Two of the three primary-school primary colours: red and yellow (the third being blue);&lt;br&gt;&#xA; The three &quot;true&quot; secondary colours, also used in printing: cyan, magenta, yellow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So there's one missing element and one extra element,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; blue and orange, which are &lt;a href=&quot;http://tvtropes.org/pmwiki/pmwiki.php/Main/OrangeBlueContrast&quot;&gt;used in visual arts for their contrast&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="228" LastEditorUserId="11876" LastEditDate="2016-01-01T16:40:11.100" LastActivityDate="2016-01-01T16:40:11.100" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24981" PostTypeId="2" ParentId="11508" CreationDate="2016-01-01T14:24:05.087" Score="-2" Body="&lt;p&gt;The Idiom is: &lt;strong&gt;An empty vessel, makes more noise!!!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Translation:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A man sitting idle makes more noise. &lt;/li&gt;&#xA;&lt;li&gt;A man who does nothing but&#xA;fights more.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="18132" LastActivityDate="2016-01-01T14:24:05.087" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24982" PostTypeId="2" ParentId="24961" CreationDate="2016-01-01T14:50:38.947" Score="8" Body="&lt;p&gt;The murderer was&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;The king's wife&lt;/strong&gt; (who probably wasn't the mother of the king's son and therefore had some reasons for wanting him dead, but that's just my speculation).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The son's body was discovered on a weekday (Tuesday, say), and the fact that he had died on Sunday was known only to the king and to the murderer. A year from the Sunday before the body was discovered, the king saw his wife laying flowers on the son's grave, and deduced that it was her.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user17947" LastActivityDate="2016-01-01T14:50:38.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24983" PostTypeId="2" ParentId="24979" CreationDate="2016-01-01T16:34:48.600" Score="4" Body="&lt;p&gt;The answer to &quot;Surreal&quot; is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;X. X is a cross, ten in Roman numerals, and written as two diagonal lines; X-rays are a type of electromagnetic wave.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2016-01-01T16:34:48.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24986" PostTypeId="2" ParentId="24966" CreationDate="2016-01-02T01:47:32.443" Score="4" Body="&lt;p&gt;&lt;strong&gt;Any number!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And you need only 1 finger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider your straight index finger as the number $0$ and your completely bend finger as $x$ where $x$ is any number you like. Now measure the angle that your finger bend at one of the joints of that finger in comparison with the stretched finger and call it $a$. Now every other number is constructed by bending your index finger slightly less and measuring the angle that you bent it and call it $b$. and then your number is $\lceil {b \over a} \cdot x \rceil$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course measuring this angle can only be up to some accuracy but that doesn't matter. The question is not how accurately you can know what the number is but if you can represent every number, and this is true&lt;/p&gt;&#xA;" OwnerUserId="1748" LastEditorUserId="1748" LastEditDate="2016-01-02T03:03:10.920" LastActivityDate="2016-01-02T03:03:10.920" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24987" PostTypeId="2" ParentId="24966" CreationDate="2016-01-02T02:07:22.060" Score="0" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Graham&amp;#39;s_number#Magnitude&quot; rel=&quot;nofollow noreferrer&quot;&gt;$g_5$&lt;/a&gt;, as in the notation of Graham's number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zGw9J.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zGw9J.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12795" LastActivityDate="2016-01-02T02:07:22.060" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24988" PostTypeId="2" ParentId="103" CreationDate="2016-01-02T02:14:09.543" Score="0" Body="&lt;p&gt;Fill the 7 gallon jug, then fill the 3 gallon jug using the water from the 7 gallon jug, repeat this step, now you have excactly 1 gallon left in the 7 gallon jug, and just simply pour that gallon of water over into the 3 gallon jug&lt;/p&gt;&#xA;" OwnerUserId="18145" LastActivityDate="2016-01-02T02:14:09.543" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24989" PostTypeId="2" ParentId="24966" CreationDate="2016-01-02T02:25:58.223" Score="3" Body="&lt;p&gt;Say you have &lt;strong&gt;4&lt;/strong&gt; fingers on each hand; all your fingers have &lt;strong&gt;three joints&lt;/strong&gt; which can all be extended or collapsed(representing a 0 and a 1) independent of each other.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You now have 3 bits of binary information in one finger, but you gotta remember, you have &lt;strong&gt;8&lt;/strong&gt; in total (8*3=24), so you have &lt;strong&gt;24 binary bits&lt;/strong&gt; of information in your 8 fingers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So what's the greatest number you can represent with this system?&lt;/strong&gt; 111111111111111111111111(base 2)=16777215(base 10) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So using my system you could display over 16 million different numbers, in just 8 fingers, &lt;em&gt;but hang on&lt;/em&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have &lt;strong&gt;5&lt;/strong&gt; fingers on each hand, but your thumbs act differently than your other fingers.  They only have &lt;strong&gt;2 joints&lt;/strong&gt;, so that's another 4 bits of information to our already huge 24 bit number, we now have &lt;strong&gt;28 bits&lt;/strong&gt; to work with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So what's the new highest number?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1111111111111111111111111111(base 2)=268435455(base 10). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wow- a silly 4 bits added to the end allow you to display &lt;strong&gt;over 260 million different numbers&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a lot of information from just 10 simple fingers, don't ya think?&lt;/p&gt;&#xA;" OwnerUserId="18145" LastEditorUserId="11452" LastEditDate="2016-01-05T22:25:44.623" LastActivityDate="2016-01-05T22:25:44.623" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24990" PostTypeId="2" ParentId="619" CreationDate="2016-01-02T02:32:11.153" Score="0" Body="&lt;p&gt;23=2/2+2*2*2*2+2+2+2&#xA;34=3/3+3*3*3+3&#xA;56=5/5+5*5+5*5&#xA;.....This pattern will repeat, basiacly all the numbers are 1 above a number which is divisible by the smaller number, Example: 56=55+1, mod(55, 5)=0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplication is basiacly repeated addition, so i don't count that as cheating since you can easily replace it with repeated addition, but i'm lazy.... don't judge.&lt;/p&gt;&#xA;" OwnerUserId="18145" LastActivityDate="2016-01-02T02:32:11.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24991" PostTypeId="2" ParentId="22378" CreationDate="2016-01-02T03:00:06.763" Score="2" Body="&lt;p&gt;A bit late in on the action, but thought I'd try to solve some of the remaining clues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of Us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$200&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;xnor&lt;/strong&gt; - anagram (in disarray) of &quot;ronx&quot; (B-less &lt;a href=&quot;https://en.wikipedia.org/wiki/The_Bronx&quot; rel=&quot;nofollow&quot;&gt;Bronx&lt;/a&gt;, a borough of New York City)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$400&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;dmg&lt;/strong&gt; - &quot;damg&quot; (deka-milligram, a small dekagram) without the 'a' (a-tom, where &quot;tom&quot; comes from a Greek word meaning &quot;cut&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8949" LastActivityDate="2016-01-02T03:00:06.763" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24992" PostTypeId="1" AcceptedAnswerId="24993" CreationDate="2016-01-02T05:23:29.557" Score="5" ViewCount="1479" Body="&lt;p&gt;&lt;strong&gt;1597&lt;/strong&gt; or &lt;strong&gt;2584&lt;/strong&gt; is a variation of the game &lt;a href=&quot;https://en.wikipedia.org/wiki/2048_(video_game)&quot; rel=&quot;nofollow noreferrer&quot;&gt;2048&lt;/a&gt; where pairs of tiles numbered with consecutive values in the Fibonacci sequence are merged into a new tile with the next value in the sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An important difference is that new tiles may only have a value of either 1 or 2 (whereas in 2048 new tiles have a value of 2 or 4).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the largest value for a single tile that can be achieved in this game?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here are some playable links: &lt;a href=&quot;http://jmhummel.github.io/1597/&quot; rel=&quot;nofollow noreferrer&quot;&gt;1597&lt;/a&gt;, &lt;a href=&quot;http://davidagross.github.io/2048/&quot; rel=&quot;nofollow noreferrer&quot;&gt;2048&lt;/a&gt;, &lt;a href=&quot;http://themoreyouknow.github.io/fib/&quot; rel=&quot;nofollow noreferrer&quot;&gt;FIB&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also a few variations which are not of relevance for this puzzle, like &lt;a href=&quot;http://cozmic72.github.io/987/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt; where new tiles may have a value of 1, 2 or 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/48/what-is-the-largest-tile-possible-in-2048&quot;&gt;What is the largest tile possible in 2048?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4394" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2016-01-02T08:38:36.803" Title="Largest tile possible in 1597 (2048 with Fibonacci numbers)" Tags="|game|2048-game|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24993" PostTypeId="2" ParentId="24992" CreationDate="2016-01-02T06:15:00.747" Score="4" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;3,524,578&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The best you can do is spawning a 2 in the 16th spot, when everything else is lined up ready to be merged.  This means your tiles are:&lt;br&gt;&#xA;&lt;br&gt;&#xA; 2 &gt; 3 [makes 5] &gt; 8 [makes 13] &gt; 21... etc.  The pattern here is fairly simple, we get to skip every other number, which means our tiles are f3, f4, f6, f8...  ...f32.  These all combine to create f33 = 3524578.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Proof that this situation is possible:  if your tiles always spawn adjacent to the smallest previous tile, then we can unwrap the board into a single 16-long string of ordered tiles / empty spaces.  Creating an f32 tile is identical to creating an f33 tile, except everything is one less in the series.  This means that we have f31... ... f5, f3 = 2, f2 = 1, [blank].  Spawning a 1 in the blank spot clearly lets us combine all the way up, so we can create our f32.  Creating an f30 uses identical tiles, except it doesn't need the initial f31 tile, so we can leave that slot filled by our recently created f32 and apply the same logic.  This cascades all the way down to creating our f4, which is: ...f6, f3 = 2, [blank] - in which case, a spawned 1 will allow us to create our f4 = 3 without trouble.  Thus there exists a sequence of tiles that will allow us to create every tile in f3, f4, f6... ...f32, which means that their combination (f33) is possible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="17389" LastEditDate="2016-01-02T08:38:36.803" LastActivityDate="2016-01-02T08:38:36.803" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24994" PostTypeId="1" CreationDate="2016-01-02T06:24:32.143" Score="5" ViewCount="557" Body="&lt;p&gt;Here is the sudoku - I could think the following only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The top number box 3 coloumn wise should have some number - 6,7,3 (in some order).&#xA;The box 3 column 2 -- 2,5,8 ( in some order), and box 3 column 3 - 1,4,9 in some order.&#xA;Rest, I am really not able to identify what is wrong. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wW7b6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wW7b6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am really clueless here, I am not able to think further. What logic am I missing here?&lt;/p&gt;&#xA;" OwnerUserId="18149" LastEditorUserId="16846" LastEditDate="2016-01-03T11:27:09.550" LastActivityDate="2018-06-06T12:19:13.780" Title="How to solve this Sudoku?" Tags="|sudoku|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24995" PostTypeId="2" ParentId="24994" CreationDate="2016-01-02T12:26:04.910" Score="5" Body="&lt;p&gt;Hints (that should be sufficient to allow simple reduction of possibilities):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the second row, first box, you have a cell with possible numbers (4,6), and one with (1,6). In the second row, last column, you have a cell with (1,4) as possible numbers.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Thus, no other cells in this row can be (1,4,6), which nets you a 7 somewhere in the upper right 3x3 box.&lt;br&gt;&#xA;&lt;br&gt;&#xA; After that, check around for other 7s - you should be able to find two immediately. Then check twos - you should find one. Carry through to completion. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user20" LastEditorUserId="38194" LastEditDate="2018-06-06T12:19:13.780" LastActivityDate="2018-06-06T12:19:13.780" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="24996" PostTypeId="1" AcceptedAnswerId="25001" CreationDate="2016-01-02T12:31:24.990" Score="6" ViewCount="885" Body="&lt;p&gt;A seasonal cryptic-crossword-style clue:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I heard Santa's little helper insult me! (7)&quot;&lt;/p&gt;&#xA;" OwnerUserId="4050" LastActivityDate="2016-01-02T20:06:55.240" Title="I heard Santa's little helper insult me!" Tags="|cryptic-clues|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24997" PostTypeId="2" ParentId="24996" CreationDate="2016-01-02T13:06:15.133" Score="2" Body="&lt;p&gt;I think an answer is,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tony Cox  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason:   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's a 7 character.&lt;br&gt;&#xA; He's an actor, who played a role of &lt;code&gt;Little helper&lt;/code&gt; in bad santa movie released in 2003.&lt;br&gt;&#xA; Check this: &lt;a href=&quot;https://en.wikiquote.org/wiki/Bad_Santa&quot; rel=&quot;nofollow&quot;&gt;https://en.wikiquote.org/wiki/Bad_Santa&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17437" LastEditorUserId="17437" LastEditDate="2016-01-02T13:20:38.667" LastActivityDate="2016-01-02T13:20:38.667" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24998" PostTypeId="1" AcceptedAnswerId="24999" CreationDate="2016-01-02T13:17:03.953" Score="5" ViewCount="338" Body="&lt;p&gt;I am a peculiar creature.&#xA;When I stand still, some people will avoid me.&#xA;When I do a handstand, many people who previously avoided me come to me, but few people who come to me will run away.&#xA;But, people argue whether I can do a handstand. At least there are people that with all their saneness claim that I never do a handstand. However some other people claim that they see me do a handstand.&lt;/p&gt;&#xA;" OwnerUserId="18151" LastEditorUserId="7715" LastEditDate="2016-01-02T16:08:31.787" LastActivityDate="2016-01-02T16:08:31.787" Title="Who am I who my posture will change social stigma?" Tags="|riddle|word|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24999" PostTypeId="2" ParentId="24998" CreationDate="2016-01-02T13:23:50.880" Score="12" Body="&lt;p&gt;I think this might be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;A dog&lt;/strong&gt;.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Some people are scared of dogs and will avoid them. Turning a dog on it's head gives you &lt;strong&gt;God&lt;/strong&gt;. Some people who are afraid of dogs come to God. Many people believe that God does not exist.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2016-01-02T13:23:50.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25000" PostTypeId="2" ParentId="24996" CreationDate="2016-01-02T13:36:42.430" Score="1" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; slander&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Slander&lt;/em&gt; means &quot;a malicious, false, and defamatory statement or report.&quot; The word somewhat sounds like &quot;S(anta's) l(ittle) helper&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5095" LastActivityDate="2016-01-02T13:36:42.430" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25001" PostTypeId="2" ParentId="24996" CreationDate="2016-01-02T14:58:08.130" Score="6" Body="&lt;p&gt;Hm, is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;RUDE ELF / RUDOLPH?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It wouldn't &lt;em&gt;quite&lt;/em&gt; fit cryptic crossword style, and I'm not sure where the definition would be, but at least I sort of have the&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;homophone from &quot;I heard&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2016-01-02T14:58:08.130" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25002" PostTypeId="2" ParentId="24996" CreationDate="2016-01-02T20:06:55.240" Score="2" Body="&lt;p&gt;I believe that the answer may be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SELFISH&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My rationale is the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I heard = Anagram&lt;br&gt;&#xA; Santa's little helper = ELF&lt;br&gt;&#xA; insult = HISS&lt;br&gt;&#xA; me! = Definition (including the exclamation point)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18157" LastActivityDate="2016-01-02T20:06:55.240" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25003" PostTypeId="2" ParentId="24994" CreationDate="2016-01-03T00:39:26.883" Score="-5" Body="&lt;p&gt;Is this the solution you seek?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nMGb1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nMGb1.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9531" LastActivityDate="2016-01-03T00:39:26.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25004" PostTypeId="1" AcceptedAnswerId="25010" CreationDate="2016-01-03T01:23:28.833" Score="6" ViewCount="842" Body="&lt;p&gt;I can hold the heaviest of all objects,&#xA;&lt;br&gt;but will falter to the lightest of all objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I prefer long and skinny rather than short and thick&#xA;&lt;br&gt;since it is to my taste&#xA;&lt;br&gt;which you cannot sense&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see me as far the eye can see&#xA;&lt;br&gt; in the seven seas&#xA;&lt;br&gt; or the seven continents&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Precious by nature&#xA;&lt;br&gt; yet never rare&#xA;&lt;br&gt; I seem to be more desired than gold&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are here because of me&#xA;&lt;br&gt; and I am certainly not here because of you&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a group, I can be a force to reckon with&#xA;&lt;br&gt; yet I can be gentler than a breeze&#xA;&lt;br&gt; as every organism adores me&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I this time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I can be wavy or straight, but always true to the color blue&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm critical to your own survival and like to fall from the sky sometimes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE:&lt;/strong&gt; After five days the above hint was posted, I will answer the question. Please hurry!&lt;/p&gt;&#xA;" OwnerUserId="9042" LastEditorUserId="9042" LastEditDate="2016-01-15T00:08:44.640" LastActivityDate="2016-01-15T05:38:50.083" Title="What am I this time?" Tags="|riddle|" AnswerCount="7" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25005" PostTypeId="1" AcceptedAnswerId="25007" CreationDate="2016-01-03T02:27:15.653" Score="-1" ViewCount="233" Body="&lt;p&gt;For each of the following shapes, draw extra lines to divide the shape into the smallest number of triangles that can completely fill the shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZtFkg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZtFkg.png&quot; alt=&quot;pentagon&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/u7tMR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/u7tMR.png&quot; alt=&quot;pentagon of triangles&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shapes (a correct answer answers all of these correctly, there is more than one and it is not divided into three questions to make this question more challenging):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lkA1Y.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lkA1Y.png&quot; alt=&quot;puzzle1&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/rRFOA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rRFOA.png&quot; alt=&quot;puzzle2&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/4uEr6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4uEr6.png&quot; alt=&quot;puzzle3&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The blue shape is blue to increase visibility of what is inside and outside the shape (blue is inside).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="12879" LastActivityDate="2016-01-03T03:44:45.980" Title="Trianglify the Shapes" Tags="|geometry|triangle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25006" PostTypeId="1" AcceptedAnswerId="25011" CreationDate="2016-01-03T03:29:23.283" Score="6" ViewCount="502" Body="&lt;p&gt;So I was presented with this puzzler:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to the puzzle is a 6 letter word with 6 different letters. Each of these 12 words have one - and only one - letter of the mystery word in the correct position (i.e. 1st, 2nd, 3rd, etc.)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;MYRIAD&#xA;FUMBLE&#xA;CATKIN&#xA;RECKON&#xA;TOWARD&#xA;OUTAGE&#xA;STEADY&#xA;SWAYED&#xA;BOUNTY&#xA;TONGUE&#xA;STOLEN&#xA;UNREST&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I couldn't find a very efficient algorithm for solving it - more on this later - so I decided to put my programming skills to use and just generate all possible solutions via brute force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine my surprise when there was only a single 6 letter combination that met the criteria, and it was the answer to the puzzle!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't want to give away the answer in case you want to solve it yourself, but as I'm not interested in the answer so much as the method of devising the problem and its solution, I have a mild spoiler: each letter of the answer appears exactly twice in its correct position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TURKEY&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So my questions are basically:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is having 2 words per letter-position a function of building a unique solution to such a puzzle?&lt;/li&gt;&#xA;&lt;li&gt;What's the algorithm for doing this without a computer? One attempt I worked at noted that the words &lt;code&gt;RECKON, OUTAGE, BOUNTY, UNREST&lt;/code&gt;, and &lt;code&gt;SWAYED&lt;/code&gt;don't share any letter-positions so they contain 5 of the letters, but I couldn't finish the thought.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I think I'm missing something obvious about the puzzle's design and intended solving process. I usually crank through puzzle books in an hour or two, but this one just felt impossible!&lt;/p&gt;&#xA;" OwnerUserId="1810" LastEditorUserId="1810" LastEditDate="2016-01-03T04:31:29.183" LastActivityDate="2016-01-03T05:14:31.923" Title="Mathematics behind word puzzle" Tags="|word|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25007" PostTypeId="2" ParentId="25005" CreationDate="2016-01-03T03:44:45.980" Score="5" Body="&lt;p&gt;Here is one possible triangulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/e9RJn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that all triangulations (that do not introduce extra vertices) will have the same number of triangles, which depends only on the number of sides in the polygon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A sketch of the proof: A polygon with three sides is already a triangle, so the minimal number of triangles needed to triangulate it is exactly $1$.  For a polygon with $n$ sides, choose two adjacent sides and connect the two vertices they do not share with a line.  This creates a triangle and a new polygon.  The new polygon is missing the two sides we chose, but has an extra side (the new side we just created).  Thus the number of triangles needed to triangulate a polygon with $n$ sides is $1$ plus the number of triangles needed to triangulate a polygon with $n-2+1=n-1$ sides.  As a recurrence relation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;a(3) &amp;amp;= 1 \\&#xA;a(n) &amp;amp;= 1 + a(n-1) \\&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving this yields $a(n) = n-2$, so the number of triangles needed to triangulate a polygon with $n$ sides is exactly $n-2$, regardless of that polygon's shape.&lt;/p&gt;&#xA;" OwnerUserId="9718" LastActivityDate="2016-01-03T03:44:45.980" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25008" PostTypeId="1" AcceptedAnswerId="25036" CreationDate="2016-01-03T04:24:25.440" Score="3" ViewCount="914" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Here I lie, so famously &lt;br&gt;&#xA;  Master of the edifice they named for me &lt;br&gt;&#xA;  Eternal as a pyramid? You'd think, but no &lt;br&gt;&#xA;  They'll both collapse, but mine takes even longer till it goes &lt;br&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I was back then, as now, despised and hated &lt;br&gt;&#xA;  Yet I continue to be reincarnated &lt;br&gt;&#xA;  So long as men do dream of easy glory &lt;br&gt;&#xA;  So long are men inspired to relive my story&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Who am I?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I will accept the first answer that answers the question correctly and explains all the clues (make sure you look carefully, some of them are not obvious even when you know the answer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my third riddle, so I'll also welcome any feedback or critique that helps me improve (in the comments, or in the form of an answer if it needs to contain spoilers). If you liked this riddle, here are my previous two: &lt;a href=&quot;https://puzzling.stackexchange.com/q/24847/17947&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;https://puzzling.stackexchange.com/q/24941/17947&quot;&gt;2&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit -- A Hint:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Look deeper to find the scheme of the riddle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user17947" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2016-01-04T16:39:41.463" Title="A grave deception" Tags="|riddle|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25009" PostTypeId="1" AcceptedAnswerId="36148" CreationDate="2016-01-03T05:00:25.970" Score="10" ViewCount="957" Body="&lt;p&gt;Once upon a time Isaac was lounging under a tree in Euclid's apple orchard,&#xA;when something struck his foresight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;ldquo;In the future,&amp;rdquo; he imagined,&#xA;&amp;ldquo;the layout of these trees could help&#xA;understand how to relieve an urban congestion problem.&amp;rdquo;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our hero roughed out the idea on his mobile iTypewriter as you can see.&#xA;What was he thinking?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;________|        :     :      :                    .'&#xA;________      red:   g:een   :red    green    red.'  green    red    green  .'&#xA;        |       :    .'     :                  .'                       ..''&#xA;        |       :    :     :                 .'                      ..'&#xA;________|      :    :     :                .'                     .''        .&#xA;________     gr:en .' red:   green    red.'  green    red    g..''    red..''&#xA;        |      :   :    :              .'                  ..'       ..''&#xA;        |     :   :    :             .'                 .''      ..''&#xA;________|     :  .'   :            .'               ..''     ..''&#xA;________     :red:   :reen    red.'  green    red..' gree..'' red    green&#xA;        |    :  :   :          .'             .''    ..''                  ...&#xA;        |    : .'  :         .'           ..''   ..''                ...'''&#xA;________|   :  :  :        .'          ..'   ..''              ...'''&#xA;________    :g:ee:    red.'  green  .'red..''green    red...'''een    red&#xA;        |  : .' :      .'       ..'' ..''          ...'''&#xA;        |  : : :     .'      ..' ..''        ...'''                  .....''''&#xA;________|  :: :    .'     .''..''      ...'''              .....'''''&#xA;________  :.':red.'  g..''.'' red...'''een    red.....'''''   red.......''''''&#xA;        | :::  .'  ..:.''  ...'''      .....'''''  .......'''''''.......::::::&#xA;        |::: .' .::''...'''  .....'''''.....'''''''...:::::::::.::::::::::::::&#xA;________|::.:.:::.:::...:::::.::::::::::::::::::::::::::::::::::::::::::::::::&#xA;      red:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&#xA;(Those dotty diagonals are meant to be solid, straight and thin.&#xA; The lower portion is meant to show an increasing density of diagonals.)&#xA;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;pre&gt;   time  -----------------------&gt; &lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18129" LastEditorUserId="18129" LastEditDate="2016-06-14T05:00:08.067" LastActivityDate="2016-06-16T17:59:17.850" Title="Euclid's orchard" Tags="|mathematics|enigmatic-puzzle|visual|optimization|real|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25010" PostTypeId="2" ParentId="25004" CreationDate="2016-01-03T05:03:18.397" Score="8" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Water.&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; The ocean can hold the world's heaviest objects, but will falter due to the 'lightest' of all objects -- the sun. (In this case, 'lightest' refers to the brightness and not the weight of an object.)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Long and skinny can refer to the throat that drinks it, or even the shape of waves in the ocean.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The 'taste you cannot sense' refers to the fact that water doesn't have a taste.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Water obviously exists in the seven seas, and on every continent.&lt;br&gt;&#xA;&lt;br&gt;&#xA; It is precious by nature and worth more than money, yet (most) people have access to it.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Water predates humans but humans wouldn't be here without water, as they need it to survive.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Finally, bodies of water (such as the ocean) are a force to be reckoned with, yet small amounts of water are harmless and necessary for human survival.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12146" LastEditorUserId="4421" LastEditDate="2016-01-07T18:21:44.523" LastActivityDate="2016-01-07T18:21:44.523" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="25011" PostTypeId="2" ParentId="25006" CreationDate="2016-01-03T05:14:31.923" Score="2" Body="&lt;p&gt;My process:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My first thought was, &quot;Let's look at the common/repeated letters.&quot; So I took all the letters from each place and found all the repeated ones, giving this:&lt;br&gt;MFCRTOSSBTSU: TTSSS&lt;br&gt;YUAEOUTWOOTN: TTUUOOO&lt;br&gt;RMTCWTEAUNOR: TTRR&lt;br&gt;IBKKAAAYNGLE: KKAAA&lt;br&gt;ALIORGDETUES: EE&lt;br&gt;DENNDEYDYENT: NNNEEEDDDYY&lt;br&gt;Just looking down the rows at that, TURKEY kinda jumps out. But that didn't have to be true- it could have been that some of the letters were triples and some were singles, and wouldn't be revealed by looking at the multiples. If that were the case, I would have done a crossword-type deal. Taking S as a given, eliminate all of the multiples that show up in the same word as one starting with S, and then figure out what combos sum the remaining multiples to 12 (because there are 12 letters from TURKEY in the list). That process would give us this: SO(T/R)K-T. (The T at the end is a freebie, because that's the only letter that appears only once as a terminal.) No matches, though, unless SORKAT is a word now. &lt;br&gt;Doing the same with T gives us T(U/-)(T/R/-)(K/-)(E/-)(N/Y), which means our word is either TURKEY or TUTKEY, or TURKEN with one of the four intervening letters swapped. And if none of those had worked, I would have assumed the first letter was neither S nor T and then continued in the same way.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14441" LastActivityDate="2016-01-03T05:14:31.923" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25012" PostTypeId="2" ParentId="25008" CreationDate="2016-01-03T05:15:55.780" Score="3" Body="&lt;p&gt;Are you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jesus Christ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here I lie, so famously &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Master of the edifice they named for me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An edifice is both a building and a complex set of beliefs. Jesus died at the cross, but the cross is also a symbol of the Christian belief (named after Christ).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Eternal as a pyramid? You'd think, but no &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They'll both collapse, but mine takes even longer till it goes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is referring to the eventual collapse of religion, though the longstanding belief in Christianity mandates that it'll be around as long as people 'need' it, as beliefs often last longer than objects themselves.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was back then, as now, despised and hated &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet I continue to be reincarnated&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Bible describes Jesus as a figure who was killed out of hatred for his beliefs, yet we continue to believe that he will come again. We celebrate holidays such as Easter Sunday to represent his reincarnation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So long as men do dream of easy glory &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So long are men inspired to relive my story&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People often use religion as a way to justify themselves. They believe that repentance and belief will send them to a heaven of eternal joy. This belief is what impels people to preach the word of the Bible. They rely on this idea of eternal glory past the end of their lives.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12146" LastActivityDate="2016-01-03T05:15:55.780" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25013" PostTypeId="2" ParentId="25009" CreationDate="2016-01-03T08:45:02.800" Score="7" Body="&lt;p&gt;Was he thinking that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alternating red and green traffic lights facilitates traffic to move &lt;em&gt;efficiently&lt;/em&gt;* both N-S and E-W?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Efficiently&quot; means reliably travelling Euclidean distances as opposed to Manhattan distances.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Say we want to get from one corner to the other with minimal stops. Consider the &lt;a href=&quot;https://en.wikipedia.org/wiki/Main_diagonal&quot;&gt;major and minor diagonals&lt;/a&gt; and the four diagonals around them. If the lights along any diagonal are always the same color, and the diagonals alternate color, we have the following scenario (WLOG) for both the major and minor diagonal: We start travelling N-S at one of the &quot;corners,&quot; we have a green light. As we're between streets, the lights change, and we have a green light at the next intersection, so we can turn. Change the lights again, and once more we'll have a green light at the next intersection. Continue in this fashion until we've traveled the entire diagonal &lt;em&gt;without ever stopping at a red light.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That said, we also consider other possibilities:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Other desired paths at different &quot;angles&quot; will be less optimal if we travel along their gradient (on second thought, will they?), but in any case, travelling at a &quot;45 degree angle&quot; and then traveling N-S or E-W directly to our desired destination will have us always at a green light. We can (&lt;em&gt;in theory&lt;/em&gt; - &lt;em&gt;yellow lights&lt;/em&gt;** and possible delays/blockages be damned) get anywhere we want without ever stopping.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In practice though&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Having yellow lights at every intersection in both directions every so often is an utterly frightening idea - I suppose it's no coincidence there are no Golden Delicious trees in this orchard, though.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Disclaimer: I'm tired and there's probably a flaw in my reasoning, and I don't think it accounts for all the lines he drew, but I don't think it's a bad attempt.&lt;/p&gt;&#xA;" OwnerUserId="17147" LastEditorUserId="17147" LastEditDate="2016-01-03T09:05:21.430" LastActivityDate="2016-01-03T09:05:21.430" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25014" PostTypeId="1" AcceptedAnswerId="25017" CreationDate="2016-01-03T09:54:01.797" Score="7" ViewCount="385" Body="&lt;p&gt;The ringmaster of a flea circus puts a flea onto the integer $0$.&#xA;Whenever the ringmaster shouts &lt;em&gt;&quot;Hop!&quot;&lt;/em&gt;, the flea jumps from its &#xA;current integer $n$ to one of the numbers $n+2$, $n-2$, $2n$ and $n/2$.&#xA;The flea only visits positive integers, and it must not visit any number twice.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is it possible that the flea visits every positive &#xA;  integer exactly once?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2016-01-03T13:58:09.027" Title="A new stunt at the flea circus" Tags="|mathematics|combinatorics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25016" PostTypeId="2" ParentId="25014" CreationDate="2016-01-03T11:26:13.297" Score="3" Body="&lt;p&gt;I'm actually not entirely sure if this strategy is proof one way or another :P&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We can reach all positive integers up to any arbitrary odd integer $N$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We do this by simply&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hopping to 2, then to 1 (n/2) and then going up to our point $N$, we then jump to $2N$ and hop down with -2 until we reach 4. We've no covered all integers from $1$ to $N$, since we can pick $N$ to be any odd integer, we can visit every positive integer. &lt;em&gt;However,&lt;/em&gt; it's quite clear that there's some we don't visit because we visit odd integers higher then our arbitrary number, but if we wanted to include those numbers we just need to pick our $N$ higher&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(from here on, the post is a lie thanks to frodo)&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;S&gt;We can also note that there's no strategy that does better &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; because we're not allowed to visit numbers more then once, and because there's no combination of operators that allows us to visit all integers we 'pass' without backtracking entirely we have to conclude that either we visit all even numbers first, or we visit all odd numbers first. If we visit all odd first then using $2n$ is the only way to get to an even number (as we did). If we visit all even numbers first then using $n/2$ is the only way to get to an odd number, our arbitrary even number has to furfil $E/2$=odd, so $E=2N$, this. In both cases you get the exact same distribution, where all numbers from $1$ to $N$ are visited exactly once.&lt;/S&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;*note that this doesnt take into acount flea working hours, mortality, drug problems or wavering dedication&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="16846" LastEditorUserId="16846" LastEditDate="2016-01-03T13:58:09.027" LastActivityDate="2016-01-03T13:58:09.027" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25017" PostTypeId="2" ParentId="25014" CreationDate="2016-01-03T11:35:02.550" Score="13" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with&lt;br&gt;&#xA;0,2,1,3,6,4,8,10,5,7,9,11,22,20,18,16,14,12,24,26,13,15,17,19,21,23,25,27...  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pattern is &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Start with 0,2,1,3&lt;/li&gt;&#xA;&lt;li&gt;Double&lt;/li&gt;&#xA;&lt;li&gt;Decrement until you cannot&lt;/li&gt;&#xA;&lt;li&gt;Double&lt;/li&gt;&#xA;&lt;li&gt;Increment&lt;/li&gt;&#xA;&lt;li&gt;Divide&lt;/li&gt;&#xA;&lt;li&gt;Increment until the current value is the highest yet&lt;/li&gt;&#xA;&lt;li&gt;Repeat all but line 1&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This works by filling in the evens and odds in parallel, doing evens (mostly) in reverse and odds forward, using doubling and dividing to switch between them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We start on an odd number, n, with all lower numbers filled.&lt;/li&gt;&#xA;&lt;li&gt;We then move to 2n, then decrement until n+1. All odd numbers up to n and all even numbers up to 2n are done.&lt;/li&gt;&#xA;&lt;li&gt;We double, to 2n+2, increment to 2n+4, then divide to n+2. Now all even numbers up to 2n+4 and all odd numbers up to n+2 are done. &lt;/li&gt;&#xA;&lt;li&gt;We increment until we are at 2n+5. &lt;/li&gt;&#xA;&lt;li&gt;Now all odd numbers up to 2n+5 are done, and all even numbers up to 2n+4 i.e. all numbers up to 2n+5, and we are on an odd number. This meets the initial conditions, so we can repeat with this as our new value of n.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3987" LastEditorUserId="3987" LastEditDate="2016-01-03T13:27:39.677" LastActivityDate="2016-01-03T13:27:39.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25019" PostTypeId="2" ParentId="25008" CreationDate="2016-01-03T19:15:36.393" Score="4" Body="&lt;p&gt;I think you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ozymandias&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here I lie, so famously&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many people know Shelley's poem Ozymandias&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Master of the edifice they named for me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The poem is named for the man (it is the Greek name for Rameses II)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Eternal as a pyramid? You'd think, but no/They'll both collapse, but mine takes even longer till it goes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The poem will eventually be forgotten, but it's currently around even after pyramids have fallen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was back then, as now, despised and hated&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Both when the poem was written, and now, people make fun of Ozymandias&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet I continue to be reincarnated/So long as men do dream of easy glory/So long are men inspired to relive my story&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As long as foolish people say things about their works living forever, others will mock them by bringing up the poem&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12092" LastActivityDate="2016-01-03T19:15:36.393" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25022" PostTypeId="1" CreationDate="2016-01-03T20:59:13.967" Score="5" ViewCount="140" Body="&lt;p&gt;What (if any) is the maximum possible number of pieces in a Rubik's cube---27? 64? 125? 216? 343?...or does the maximum end with 125?&lt;/p&gt;&#xA;" OwnerUserId="18172" LastActivityDate="2016-01-03T21:11:35.960" Title="What (if any) is the maximum possible number of pieces ina Rubik's cube?" Tags="|rubiks-cube|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25023" PostTypeId="2" ParentId="25022" CreationDate="2016-01-03T21:11:35.960" Score="4" Body="&lt;p&gt;There's no maximum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ign.com/articles/2015/01/12/17x17x17-rubiks-cube-the-worlds-largest-solved-in-75-hours&quot; rel=&quot;nofollow&quot;&gt;This article&lt;/a&gt; discusses a 17x17x17 cube, and the technique used to make it could easily be extended.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Also, keep in mind that the Rubik's Cube generally doesn't have any inside cubies. The standard cube has only 26 actual cubies, and most of them are deformed on the inside to make the mechanism work.&lt;/p&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2016-01-03T21:11:35.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25024" PostTypeId="2" ParentId="25004" CreationDate="2016-01-04T02:20:09.490" Score="1" Body="&lt;p&gt;I think you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Carbon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can hold the heaviest of all objects&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Diamonds, made of carbon, are the hardest known material (though some people only say they are the the hardest &lt;em&gt;naturally occurring&lt;/em&gt; material, and &lt;a href=&quot;https://van.physics.illinois.edu/qa/listing.php?id=613&quot; rel=&quot;nofollow&quot;&gt;there seems to be some controversy&lt;/a&gt; as to whether they are in fact the hardest), so a large diamond would presumably be capable of holding very heavy objects. I assume that &quot;heaviest of all objects&quot; (which if interpreted literally would be something like a supermassive black hole?) is meant metaphorically or as hyperbole.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But will falter to the lightest of all objects&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The lightest of all objects is hydrogen. This could be referring to a chemical reaction that takes place when carbon comes in contact with hydrogen. Alternatively, graphite, also made of carbon, is one of the &lt;em&gt;softest&lt;/em&gt; materials, so this could refer to a pencil coming in contact with a sheet of paper (which is among the lightest of objects on the scale experienced by humans).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I prefer long and skinny rather than short and thick&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pencils, which have a graphite core, are long and skinny.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;since it is to my taste&lt;br&gt;&#xA;which you cannot sense&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure about this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can see me as far the eye can see&lt;br&gt;&#xA;in the seven seas&lt;br&gt;&#xA;or the seven continents&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Carbon is one of the most abundant elements, both on Earth and in the universe. Wherever you are and in any direction you look, you will see it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Precious by nature&lt;br&gt;&#xA;yet never rare&lt;br&gt;&#xA;I seem to be more desired than gold&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Diamonds, one of the forms in which carbon is found in Nature, are precious and more desired than gold. Yet carbon is abundant and not rare.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are here because of me&#xA;and I am certainly not here because of you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All life on Earth is carbon based. Carbon existed long before humans, or even the Earth, did.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a group, I can be a force to reckon with&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Carbon belongs to a group of elements known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Carbon_group&quot; rel=&quot;nofollow&quot;&gt;carbon group,&lt;/a&gt; which also contains important elements like silicon and germanium, very useful semiconductors, and tin and lead.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yet I can be gentler than a breeze&lt;br&gt;&#xA;as every organism adores me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure what the first line is referring to. Organisms need carbon to survive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user17947" LastEditorDisplayName="user17947" LastEditDate="2016-01-04T02:21:11.833" LastActivityDate="2016-01-04T02:21:11.833" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25025" PostTypeId="1" AcceptedAnswerId="25073" CreationDate="2016-01-04T03:53:23.960" Score="10" ViewCount="674" Body="&lt;p&gt;In the 1980s, a German spy was assigned a mission to retrieve a set of important documents kept inside a British outpost. The spy knew that the documents were locked in a safe, which the password was a three-integer-sequence. He had to make sure the password was correct, as the safe was connected to an alarm that would be triggered whenever any unauthorized access was detected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The German intelligence agency acquired several number sequences from a British traitor. The traitor insisted that these sequences contained vital information about the password. However, he could not figure out what the actual password was; he just knew that &lt;code&gt;A&lt;/code&gt;, &lt;code&gt;B&lt;/code&gt;, &lt;code&gt;C&lt;/code&gt; were the three integers needed to be keyed in, and they were somehow related to a kind of sports.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number sequences were:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Seq.1&lt;/strong&gt; - 14, 32, 16, &lt;code&gt;A&lt;/code&gt;, 28, 18, 24&lt;br&gt;&#xA;&lt;strong&gt;Seq.2&lt;/strong&gt; - 19, 3, &lt;code&gt;B&lt;/code&gt;, 2, 15, 10, 6&lt;br&gt;&#xA;&lt;strong&gt;Seq.3&lt;/strong&gt; - 3, 54, &lt;code&gt;C&lt;/code&gt;, 39, 18, 30, 45  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you know what &lt;code&gt;A&lt;/code&gt;, &lt;code&gt;B&lt;/code&gt;, &lt;code&gt;C&lt;/code&gt; are?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Hint&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The numbers in &lt;strong&gt;Seq.2&lt;/strong&gt; can be seen in that sport &lt;strong&gt;directly&lt;/strong&gt;, while those in the other two sequences are not that obvious.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Visual Explanation to the answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/20o6d.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/20o6d.png&quot; alt=&quot;Sequences on a dartboard&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Seq.1&lt;/strong&gt; - Blue, clockwise. They are the &lt;strong&gt;doubles&lt;/strong&gt;.&lt;br&gt;&#xA; &lt;strong&gt;Seq.2&lt;/strong&gt; - Orange, counter-clockwise. They are the &lt;strong&gt;singles&lt;/strong&gt;.&lt;br&gt;&#xA; &lt;strong&gt;Seq.3&lt;/strong&gt; - Purple, clockwise. They are the &lt;strong&gt;trebles&lt;/strong&gt; or &lt;strong&gt;triples&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16499" LastEditorUserId="16499" LastEditDate="2016-01-05T12:59:58.697" LastActivityDate="2016-01-05T12:59:58.697" Title="Three in a lock" Tags="|lateral-thinking|number-sequence|story|sports|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25026" PostTypeId="1" AcceptedAnswerId="25029" CreationDate="2016-01-04T05:36:30.163" Score="8" ViewCount="1888" Body="&lt;p&gt;The assertions in the following infinite&#xA;list are all the same, true or false. &lt;br&gt;&#xA;&amp;emsp; &amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;$~\vdots$ &lt;br&gt;&#xA;&amp;emsp; T$~~$F $~~$ I have a black Model T. &lt;br&gt;&#xA;&amp;emsp; T$~~$F $~~$ I have a black Model T. &lt;br&gt;&#xA;&amp;emsp; T$~~$F $~~$ I have a black Model T. &lt;br&gt;&#xA;&amp;emsp; &amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;$~\vdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The object is to concoct an infinite list of assertions where: $\strut$&lt;br&gt;&#xA;&amp;bull; &amp;thinsp; The assertions are identical.           $\strut$&lt;br&gt;&#xA;In every possible consistent scenario&#xA;where each assertion is either true or false:  $\strut$&lt;br&gt;&#xA;&amp;bull; &amp;thinsp; At least one assertion must be true.    $\strut$&lt;br&gt;&#xA;&amp;bull; &amp;thinsp; At least one assertion must be false.   $\strut$&lt;br&gt;&#xA;And at least one consistent scenario is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this open-ended challenge infinitely many answers are possible. &lt;br&gt;&#xA;Try to find one (or more) in any (or none) of these categories. $\strut$&lt;br&gt;&#xA;a.&amp;nbsp;           Exactly one assertion must be true.      &lt;br&gt;&#xA;b.&amp;nbsp;           Exactly one assertion must be false.     &lt;br&gt;&#xA;c.&amp;nbsp;           Infinitely many assertions must be true  &lt;br&gt;&#xA;  &amp;ensp;&amp;ensp;     while infinitely many must be false.     &lt;br&gt;&#xA;d.&amp;thinsp;         Some other quirky condition.             &lt;br&gt;&#xA;e.&amp;thinsp;         Some other quirky condition.             &lt;br&gt;&#xA;f.&amp;thinsp;&amp;thinsp; Some other quirky condition.             &lt;br&gt;&#xA;&amp;emsp; &amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;$~\vdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feeling competitive?$~$ Try for the fewest words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an answer that is incorrect because&#xA;the truth values &lt;em&gt;could&lt;/em&gt; be all true: &lt;br&gt;&#xA;&amp;emsp; &amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;$~\vdots$ &lt;br&gt;&#xA;&amp;emsp; T$~~$F $~~$ This assertion is true. &lt;br&gt;&#xA;&amp;emsp; T$~~$F $~~$ This assertion is true. &lt;br&gt;&#xA;&amp;emsp; &amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;$~\vdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an answer that is incorrect because the assertions&#xA;can be neither true nor false: &lt;br&gt;&#xA;&amp;emsp; &amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;$~\vdots$ &lt;br&gt;&#xA;&amp;emsp; T$~~$F $~~$ This assertion is false. &lt;br&gt;&#xA;&amp;emsp; T$~~$F $~~$ This assertion is false. &lt;br&gt;&#xA;&amp;emsp; &amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;$~\vdots$&lt;/p&gt;&#xA;" OwnerUserId="18129" LastEditorUserId="18129" LastEditDate="2021-10-29T20:18:22.260" LastActivityDate="2021-10-29T20:18:22.260" Title="Infinite choice question" Tags="|logical-deduction|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25027" PostTypeId="2" ParentId="25026" CreationDate="2016-01-04T05:55:07.710" Score="3" Body="&lt;p&gt;I propose:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; .&lt;br&gt;&#xA; .&lt;br&gt;&#xA; .&lt;br&gt;&#xA; The next assertion is false&lt;br&gt;&#xA; The next assertion is false&lt;br&gt;&#xA; The next assertion is false&lt;br&gt;&#xA; The next assertion is false&lt;br&gt;&#xA; .&lt;br&gt;&#xA; .&lt;br&gt;&#xA; .&lt;br&gt;&#xA; &lt;br&gt;&#xA; The assertions alternately take the truth values true, false, true, false, .... This is consistent with what they are asserting, since the ones that are true say that the next one is false, which is indeed the case, and the ones that are false say that the next one is false, which is not the case, so those assertions are indeed false.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user17947" LastActivityDate="2016-01-04T05:55:07.710" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25029" PostTypeId="2" ParentId="25026" CreationDate="2016-01-04T06:30:26.603" Score="10" Body="&lt;p&gt;Answers without a ** next to them work only if the list is countably infinite and ordered (although an ordering could be enforced in the assertion at the cost of brevity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a**:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Other assertions are False&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; If this is true for any assertion, all other assertions are false.  If this is false for any assertion, a true assertion must exist.  The combination of these two facts guarantees exactly one true assertion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;b**:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Another assertion is False&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; If this is false for any assertion, then all other assertions are true.  If this is true for any assertion, a false assertion must exist.  The combination of these two facts guarantees exactly one false assertion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;c: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Next assertion is False&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; If the first assertion is true, the next is false, which makes the next true... etc.  The inverse is also possible, but either way, infinite assertions are both true and false.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;d**:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Other [property] assertions are False&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; This allows you to make any single assertion true.  For example, if you want only assertion 18 to be true, pick the property [18th].  This is vacuously true for the 18th assertion, since no other 18th assertions exist.  Since the 18th assertion is true, every other assertion is false.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Another [property] assertion is False&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; This allows you to make any single assertion false.  For example, if you want only assertion 18 to be false, pick the property [18th].  This is false for the 18th assertion, since no other 18th assertions exist to have truth values.  Since the 18th assertion is false, every other assertion is true.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;[X] other [property] assertions are True&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Through similar logic, this allows you to make any finite number of specific assertions true, with the rest false.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;[X] other [property] assertion are False&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Through similar logic, this allows you to make any finite number of specific assertions false, with the rest true.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And finally, the universal (if boring) e**:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;[Property] assertion&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; This lets you cause any describable set of assertions to be true, with the rest false.  It also implies some cheesy ways of doing a, b, and c.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;First assertion&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Obviously, true only of the first assertion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;b:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Subsequent assertion&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; True of all but the first assertion.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;c:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Prime assertion&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; True and false for infinitely many assertions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="17389" LastEditDate="2016-01-04T08:59:08.157" LastActivityDate="2016-01-04T08:59:08.157" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25030" PostTypeId="2" ParentId="25004" CreationDate="2016-01-04T06:54:58.893" Score="1" Body="&lt;p&gt;Is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Air&lt;/strong&gt;: It can hold heaviest of all objects and will falter to the lightest, long and skinny can be referred to the planes like structure like a paper plane, it is everywhere and precious, we are here because of it. It's a force in group like storm and every organism adores it for their survival.&amp;nbsp;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5429" LastEditorUserId="18247" LastEditDate="2016-01-13T05:40:25.877" LastActivityDate="2016-01-13T05:40:25.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25031" PostTypeId="1" AcceptedAnswerId="25033" CreationDate="2016-01-04T09:45:44.687" Score="3" ViewCount="198" Body="&lt;p&gt;I was always with the wisest&lt;br&gt;&#xA;who because of me made this earth the best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are gonna find me in the end&lt;br&gt;&#xA;With someone who is always there with me to stand&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am the catalyst of the thoughts&lt;br&gt;&#xA;When you don't know something, I will always be there&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you find me,&lt;br&gt;&#xA;I will be sure it is because of my enemy,  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But do you even know who is me?&lt;/p&gt;&#xA;" OwnerUserId="17482" LastActivityDate="2016-01-06T11:44:38.733" Title="The catalyst of the thoughts" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25032" PostTypeId="1" AcceptedAnswerId="25034" CreationDate="2016-01-04T09:51:49.953" Score="14" ViewCount="634" Body="&lt;p&gt;A quasi-pangrammatic number is a positive integer whose English name contains at least one copy of each letter used to write all of the positive integers that precede it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;123456789 (&lt;em&gt;&quot;one hundred twenty-three million four hundred fifty-six thousand seven hundred eighty-nine&quot;&lt;/em&gt;) is a quasi-pangrammatic number: its name contains all letters execept for &lt;em&gt;b&lt;/em&gt;, &lt;em&gt;c&lt;/em&gt;, &lt;em&gt;j&lt;/em&gt;, &lt;em&gt;k&lt;/em&gt;, &lt;em&gt;p&lt;/em&gt;, &lt;em&gt;q&lt;/em&gt; and &lt;em&gt;z&lt;/em&gt;, and no other letters are needed to write integers from 1 to 123456788 in words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The smallest quasi-pangrammatic number is trivially 1 (&lt;em&gt;&quot;one&quot;&lt;/em&gt;) because there are no positive integers before it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the smallest quasi-pangrammatic integer after 1?&lt;/p&gt;&#xA;" OwnerUserId="4394" LastEditorUserId="13384" LastEditDate="2016-01-04T15:14:58.280" LastActivityDate="2016-01-08T08:05:28.007" Title="Smallest quasi-pangrammatic number" Tags="|english|pangram|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25033" PostTypeId="2" ParentId="25031" CreationDate="2016-01-04T09:56:13.093" Score="4" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation as Follows&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I was always with the wisest&lt;/p&gt;&#xA;&#xA;&lt;p&gt;who because of me made this earth the best.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Wise people always has questions and those questions lead them to the answers which resulted in inventions&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are gonna find me in the end&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There is a question at the end of the riddle i.e. &quot;But do you even know who is me?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With someone who is always there with me to stand&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I think it's referring to the question mark (?) that always stands with a question at the end of the question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the catalyst of the thoughts&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you don't know something, I will always be there&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Questions provoke thoughts to find the answer. When we don't know something, there will definitely arises a question (What is it? How is it? etc)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you find me,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will be sure it is because of my enemy,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;When we find the answer. which is 'Question', it's because of the enemy of 'question' which is answer (answer is the antonym of question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="17482" LastEditDate="2016-01-06T11:44:38.733" LastActivityDate="2016-01-06T11:44:38.733" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="25034" PostTypeId="2" ParentId="25032" CreationDate="2016-01-04T10:41:07.663" Score="11" Body="&lt;p&gt;&lt;em&gt;Thanks to @dmg for the correction&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 12,468 (twelve thousand four hundred sixty eight). It covers the letters: adefghilnorstuvwxy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why I think it is the answer?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I wrote down the letters that come in the English names of 1,2....19,20,30,40,50,60,70,80,90. I chose these numbers only because other numbers will just repeat. For example in sixty two, I have already considered 'sixty' and 'two' separately. Now the letters that I wrote were: ONETWHRFUIVSXGLY. I had to take 8 in my smallest number because it is the only number that has a G. 12 for L and W. 4 or 5 for F. I chose four because it is smaller. 6 for X.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="4394" LastEditDate="2016-01-07T06:57:27.160" LastActivityDate="2016-01-07T06:57:27.160" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25035" PostTypeId="1" AcceptedAnswerId="25088" CreationDate="2016-01-04T14:20:33.260" Score="7" ViewCount="544" Body="&lt;p&gt;Ok, so I run a play-by-post RPG campaign and I have the following &quot;puzzle&quot; set out for the group. My main worry is that they won't get it so I figured I'd try it here first. Other than solving it, I'd also love it if you could give indications of what got you through it, what things are too obvious, etc. and if you think it's too hard for novice puzzlers :) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I doubted for a while if I should post it here or over at RPG SE, but in the end I feel this is mostly a puzzle question. This is an edited version of the puzzle so that when they google it they wont go immediately here.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Short bit of prestory:&#xA;&lt;em&gt;It's 27th April 3577 AF, You're an inquisitor in training, in a foreign land investigating. During the night, your mentor suddenly left and all he did was leave you a note. The note is in his handwriting but the content is very much not in his style. You notice there's some odd misspellings in the letter that form the word hidden, and when you investigate further you notice the second letter of each word spells out that you should look in his dresser. In his dresser, in a secret compartment you find another scroll, completely jibberish apart from the top line, and a necklace.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The note reads:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;see you again in 12 years&lt;br&gt;&lt;br&gt;&lt;pre&gt;aew tprhowi  se wlindao,s trth e isn aglnccecpaot o ntudp o.sernet e mact twhhhiay  no cnlu tyl srbek pne yobyr oeslfuf.dhem eae wI tn es edIroc ae n abv liaet  ucot .kyo          &lt;/pre&gt;         &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16846" LastEditorUserId="16846" LastEditDate="2016-01-05T21:40:46.327" LastActivityDate="2016-01-06T07:22:08.687" Title="Difficulty judging difficulty" Tags="|riddle|cipher|puzzle-creation|" AnswerCount="1" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="25036" PostTypeId="2" ParentId="25008" CreationDate="2016-01-04T16:32:53.247" Score="2" Body="&lt;p&gt;Are you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Charles Ponzi?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Here I lie, so famously &#xA;Master of the edifice they named for me&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The 'Ponzi Scheme' (as hinted at in your...hint) was named for him  &lt;strong&gt;EDIT:&lt;/strong&gt; lol, I just got the pun—intended or otherwise—in &quot;here I &lt;em&gt;lie&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Eternal as a pyramid? You'd think, but no &#xA;They'll both collapse, but mine takes even longer till it goes&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A reference to a 'pyramid scheme' - a similar (but different!) type of money-making scheme. From &lt;a href=&quot;https://en.wikipedia.org/wiki/Ponzi_scheme#Similar_schemes&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia entry&lt;/a&gt;: &lt;em&gt;&quot;A pyramid scheme typically collapses much faster because it requires exponential increases in participants to sustain it. By contrast, Ponzi schemes can survive simply by persuading most existing participants to reinvest their money, with a relatively small number of new participants&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;I was back then, as now, despised and hated &#xA;Yet I continue to be reincarnated&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Charles Ponzi was a notorious con artist, deceiving many in his life&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;So long as men do dream of easy glory &#xA;So long are men inspired to relive my story&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Ponzi Scheme, cleverly executed, would be a source of easy money for those who would endeavor the risk&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2016-01-04T16:39:41.463" LastActivityDate="2016-01-04T16:39:41.463" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25037" PostTypeId="2" ParentId="16477" CreationDate="2016-01-04T17:01:56.133" Score="2" Body="&lt;p&gt;Let's look at the island, before the refugee arrives, on the date Jan 1. We will show that every islander must have exactly two friends. Consider the following directed graph: the islanders are vertices, and there is an edge between friends which is directed towards the islander with the earlier birthday. I claim that &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Every islander must, right now, have at least as many presents as their &quot;out-degree&quot; (the number of edges they have which point away).&lt;/li&gt;&#xA;&lt;li&gt;Every islander with exactly one friend must have at least one present, plus at least one more if his edge directs out.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To understand the first bullet point: if an islander has, for example, three edges pointing out, that means he will have to give three gifts before his birthday. If he didn't have at least three presents right now, he would run out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, islanders with one friend (loners) are a special case. A loner can never be without a present, else he would have to regift the next gift he receives. This explains the second bullet point: he needs the extra present to ensure he is never empty. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above two bullet points show that&#xA;$$&#xA;\text{# of friendships }+\text{ # of loners }\le \text{ # of mementos }=n&#xA;$$&#xA;Letting $f_k$ be the number of islanders with $k$ friends, and using the fact that the number of edges an undirected graph has is the half the sum of the vertex degrees, the above equation can be written as $\big(\frac12\sum_1^n k\cdot f_k\big)+f_1\le n,$ or as&#xA;$$&#xA;\tfrac32 f_1+f_2+\tfrac32 f_3+\tfrac{4}2f_4+\dots+\tfrac{n}{2}f_n\le n&#xA;$$ &#xA;The only solution to the above inequality is for every islander to have exactly two friends. Every islander contributes at least one to the above sum, and since there are $n$ islanders, he must contribute exactly one. Every category other than $f_2$ contributes $\frac32$ or more to the sum, so &lt;strong&gt;every islander must have two friends&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the refugee cannot avoid a scandal: we have shown that the island's customs can only be maintained if every islander has two friends, but the refugee arriving and making a friend would destroy that property. &lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2016-01-04T17:01:56.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25038" PostTypeId="1" AcceptedAnswerId="81826" CreationDate="2016-01-04T17:05:58.973" Score="28" ViewCount="2198" Body="&lt;p&gt;&lt;em&gt;Rise and shine, Mister Bond.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;James awoke to find himself yet again imprisoned by an evil genius. At least, that's what he presumed. He was alone in a room with a strange device: it consisted of a person-sized glass chamber, with a single tube feeding out of it, splitting into 23 tubes which headed God knows where.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I do hope you're comfortable, since you will be here for quite a while. We're going to play a little game.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The voice was coming from an intercom. &quot;If it's information you want, you might as well kill me. I'll die before I tell you anything!&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Nothing so routine, Mister Bond. What I want is to test your mettle. The device in front of you is a cloning machine: it will completely vaporize your body, but will create 23 identical copies in your place, each to be housed in separate prison cells.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;After this I will, at random times, bring a random clone out of his cell into the Light Bulb Room. This room has &lt;strong&gt;two&lt;/strong&gt; lightbulbs, one red and one blue, which are each either on or off. The clone may then toggle none, either or both of the bulbs, before I return him to his cell. I will take care to ensure the clones can't communicate with each other, or change the Light Bulb Room in any other way then toggling these bulbs.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;And what exactly am I supposed to accomplish through all this?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A clone may at any time make the declaration that &lt;strong&gt;every&lt;/strong&gt; clone has visited the Light Bulb Room room. If he is right, you win and may all go free. If he is wrong, you will all be executed.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I'll give you some time to think of a plan before this begins. Since my machine copies your brain state exactly, all of the clones will follow the exact same plan.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Time to see if you really are the stuff of legends, Mister Bond.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Wait! What did you say the initial states of the light bulbs were?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I didn't.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A faint crackling, then silence.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How can James Bond succeed, with certainty?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Compare and contrast this with the classic &quot;100 prisoners and one lightbulb&quot; puzzle. In that one, the prisoners could use an asymmetric strategy where one prisoner was singled to count the others. This asymmetry is now ruled out since all prisoners are clones. The winning strategy will require the extra power of having two light bulbs.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2019-04-16T02:34:50.877" Title="23 Clones and Two Lightbulbs" Tags="|strategy|algorithm|" AnswerCount="7" CommentCount="21" ContentLicense="CC BY-SA 3.0" />
  <row Id="25039" PostTypeId="1" CreationDate="2016-01-04T17:30:00.260" Score="29" ViewCount="15021" Body="&lt;p&gt;Add the four basic operators $\times\div+\,\;-$ and optionally brackets to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$10 \quad 9 \quad 8 \quad 7 \quad 6 \quad 5 \quad 4 \quad 3 \quad 2 \quad 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the total $2016$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We are looking for the simplest solution - i.e. the least amount of characters (ignoring spaces). Please include your character count in your answer.&lt;/li&gt;&#xA;&lt;li&gt;Keep the order; do not add or combine numbers.&lt;/li&gt;&#xA;&lt;li&gt;Use all four operators at least once.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Credit for initial concept: Alex Bellos&lt;/p&gt;&#xA;" OwnerUserId="9802" LastEditorUserId="4421" LastEditDate="2016-01-05T22:45:32.077" LastActivityDate="2016-01-07T02:59:25.083" Title="10 9 8 7 6 5 4 3 2 1 = 2016" Tags="|mathematics|formation-of-numbers|" AnswerCount="13" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="25041" PostTypeId="2" ParentId="25039" CreationDate="2016-01-04T17:58:34.630" Score="12" Body="&lt;p&gt;I came up with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(10 - 9) \times 8 \times 7 \times 6 \times (5 - 4 + 3 + 2 \div 1)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is &lt;strong&gt;9 operators&lt;/strong&gt; and &lt;strong&gt;2 required groupings&lt;/strong&gt;, for a total of &lt;strong&gt;24 characters&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="17147" LastEditorUserId="4421" LastEditDate="2016-01-05T22:56:07.977" LastActivityDate="2016-01-05T22:56:07.977" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25042" PostTypeId="1" AcceptedAnswerId="25103" CreationDate="2016-01-04T18:08:54.667" Score="7" ViewCount="655" Body="&lt;p&gt;You are a spy sent to the Moon in the year 2025. Your mission is to infiltrate a base that is guarded by simply a single unique passcode of length &amp;lt;10 on a standard QWERTY keyboard. The only hint as to what to type to gain access to the base is shown below. Luckily, you were able to see right through the code and typed the correct passcode to gain access; otherwise, you would have been shot on the spot for entering an incorrect passcode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the hit provided to you:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;3222312222444454555563335&#xA;3020310100302013030161002&#xA;3242301210325322252163432&#xA;3020300101302022030251021&#xA;3121322221334344343354545&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What did you type?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There are multiple grids that can point you to the answer. Another such grid could be identical to the first, except it has a bottom row of &lt;code&gt;2121222221334344343354545&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You do not need a computer to solve this. All digits above are in base 10.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15603" LastEditorUserId="15603" LastEditDate="2016-01-06T13:08:07.233" LastActivityDate="2016-01-06T13:08:07.233" Title="Find the password in the grid" Tags="|lateral-thinking|password|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25043" PostTypeId="2" ParentId="25039" CreationDate="2016-01-04T18:43:05.467" Score="24" Body="&lt;p&gt;&lt;strong&gt;22 characters&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $10 \times 9 \times 8 \times 7 \times 6 \div 5 \div (4 - 3 + 2) \times 1$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I looked at @Will's answer and found a way to improve on it.&lt;/p&gt;&#xA;" OwnerUserId="375" LastEditorUserId="4421" LastEditDate="2016-01-05T22:36:20.000" LastActivityDate="2016-01-05T22:36:20.000" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="25044" PostTypeId="2" ParentId="25039" CreationDate="2016-01-04T19:20:28.527" Score="0" Body="&lt;p&gt;&lt;strong&gt;&lt;s&gt;22&lt;/s&gt; 20 characters&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 10 + 9 * 8 * 7 * 4 - 6 / 3 - 5 - 2 - 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without order :(&lt;/p&gt;&#xA;" OwnerUserId="18180" LastEditorUserId="9802" LastEditDate="2016-01-05T01:14:53.903" LastActivityDate="2016-01-05T01:14:53.903" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25047" PostTypeId="2" ParentId="25039" CreationDate="2016-01-04T22:13:17.133" Score="18" Body="&lt;p&gt;&lt;strong&gt;22 characters&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think you can beat Joel's answer at 22 characters, but there are some nice ways to tie it (including a variation of Joel's for completeness):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(10 - 9 + 8 \times 7 \times 6 - 5 + 4) \times 3 \times 2 \times 1$&lt;br&gt;&#xA; $10 - 9 + 8 \times 7 \times (6 \times 5 + 4 \times 3 \div 2) - 1$&lt;br&gt;&#xA; $10 - 9 + 8 \times 7 \times (6 + 5 \times 4 \times 3 \div 2) - 1$&lt;br&gt;&#xA; $10 - 9 + 8 \times 7 \times 6 \div (5 - 4) \times 3 \times 2 - 1$&lt;br&gt;&#xA; $10 - 9 + 8 \times 7 \times 6 \times (5 + 4) \div 3 \times 2 - 1$&lt;br&gt;&#xA; $10 - 9 + 8 \times 7 \times 6 \times (5 + 4 + 3) \div 2 - 1$&lt;br&gt;&#xA; $10 + (9 \times 8 \times 7 - 6 + 5) \times 4 - 3 \times 2 \div 1$&lt;br&gt;&#xA; $10 \times 9 \times 8 \times 7 \div (6 - 5 + 4 - 3 \div 2 - 1)$&lt;br&gt;&#xA; $10 \times 9 \times 8 \times 7 \div (6 \times 5 \div 4 - 3 \times 2 + 1)$&lt;br&gt;&#xA; $10 \times 9 \times 8 \times 7 \times 6 \div (5 \times 4 - 3 \times 2 + 1)$&lt;br&gt;&#xA; $10 \times 9 \times 8 \times 7 \times 6 \div (5 + 4 \times 3 - 2 \times 1)$&lt;br&gt;&#xA; $10 \times 9 \times 8 \times 7 \times 6 \div 5 \div (4 - 3 + 2) \times 1$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are other ways but many of them are trivial (change $\div 1$ to $\times 1$ or vice-versa, or $\div 1)$ to $)\div 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we didn't have the restriction that we need to use all the different operators there is also a very nice solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $ 10 \times 9 \times 8 \times 7 \times 6 \div (5 + 4 + 3 + 2 + 1)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3911" LastEditorUserId="4421" LastEditDate="2016-01-05T22:54:56.617" LastActivityDate="2016-01-05T22:54:56.617" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25048" PostTypeId="1" AcceptedAnswerId="25053" CreationDate="2016-01-04T22:52:05.473" Score="7" ViewCount="505" Body="&lt;p&gt;Call me a square one if you will, but it’s not impossible for nerds to be cool&lt;br&gt;&#xA;There are two big spaces that I fit in, and it is rare that I fit in both&lt;br&gt;&#xA;I fit the best in spaced, but that wasn’t for too long&lt;br&gt;&#xA;And when I was round, I did my best to fit then, too&lt;br&gt;&#xA;Though it is not my shape, I do have a fondness for cones&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who am I?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;There is at least &lt;strong&gt;one&lt;/strong&gt; clue to this person/thing in each line, but list any that you think apply.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="11452" LastEditorUserId="5373" LastEditDate="2018-05-18T21:53:39.270" LastActivityDate="2018-05-18T21:53:39.270" Title="It Turns Out, I Fit in Many Places" Tags="|riddle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25049" PostTypeId="2" ParentId="25048" CreationDate="2016-01-04T23:27:37.937" Score="6" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Breasts. (Just to prove my innocence I would say that i searched about conical objects and got &lt;a href=&quot;https://answers.yahoo.com/question/index?qid=20081103174345AA9R8Yv&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt; )&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Call me a square one if you will, but it’s not impossible for nerds to be cool&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Calling a breast of square shaped is &lt;strong&gt;not&lt;/strong&gt; cool&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are two big spaces that I fit in, and it is rare that I fit in both&lt;br&gt;&#xA;I fit the best in spaced, but that wasn’t for too long&lt;br&gt;&#xA;And when I was round, I did my best to fit then, too&lt;br&gt;&#xA;Though it is not my shape, I do have a fondness for cones&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Talked about their shape and bras this whole time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastActivityDate="2016-01-04T23:27:37.937" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="25050" PostTypeId="2" ParentId="25039" CreationDate="2016-01-05T00:28:35.970" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $10 + 9 \times 8 - 7 + 654  \times 3 - 21$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;17 Characters&lt;/b&gt;...  Is mushing numbers together allowed?  Also, yay Mathematica.&lt;/p&gt;&#xA;" OwnerUserId="7673" LastEditorUserId="4421" LastEditDate="2016-01-05T22:57:05.647" LastActivityDate="2016-01-05T22:57:05.647" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25051" PostTypeId="2" ParentId="25038" CreationDate="2016-01-05T01:42:24.793" Score="0" Body="&lt;p&gt;The clear answer is that James actually has no chance of success. Like all supervillains, the designer of this machine is obsessed with overly dramatic methods of killing people; they merely want to watch James think himself into a tizzy and then voluntarily walk into a machine that will vaporize him.&lt;/p&gt;&#xA;" OwnerUserId="14441" LastActivityDate="2016-01-05T01:42:24.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25052" PostTypeId="2" ParentId="25026" CreationDate="2016-01-05T02:50:21.613" Score="1" Body="&lt;p&gt;In the spirit of Zerris's &lt;em&gt;other&lt;/em&gt;-logic above:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Category f. $~$ Exactly $n$ assertions must be true, $n \ge 1$ .&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;$~\vdots$ &lt;br&gt;&#xA; Fewer than $n$ other assertions are true. &lt;br&gt;&#xA; Fewer than $n$ other assertions are true. &lt;br&gt;&#xA; &amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;$~\vdots$&#xA; &lt;br&gt;&lt;br&gt;&#xA; trues $\le n~~~~$ or else any supposedly-true assertion&#xA;                   would be self-contradictory. &lt;br&gt;&#xA; trues $\ge n~~~~$ because infinitely many false assertions remain and&#xA;                   claim that $~$ trues&amp;nbsp;$ &amp;lt; n~$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Category g. $~$ Exactly $n$ assertions must be false, $n \ge 1$ .&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;$~\vdots$ &lt;br&gt;&#xA; At least $n$ other assertions are false. &lt;br&gt;&#xA; At least $n$ other assertions are false. &lt;br&gt;&#xA; &amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;$~\vdots$&#xA; &lt;br&gt;&lt;br&gt;&#xA; falses $\le n~~~$ or else the claim of each supposedly-false assertion&#xA;                   would be satisfied. &lt;br&gt;&#xA; falses $\ge n~~~$ because infinitely many true assertions remain to say so.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Category h. $~$ An odd number of assertions must be true.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;$~\vdots$ &lt;br&gt;&#xA; An even number of other assertions are true. &lt;br&gt;&#xA; &amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;$~\vdots$&#xA; &lt;br&gt;&lt;br&gt;&#xA; trues $\ne$ 0 $~~~~$ or else the claims of the&#xA;                      supposedly-false assertions would be satisfied. &lt;br&gt;&#xA; trues is odd  $~$    because at least one true assertion implies this.&lt;br&gt;&#xA; (Let's agree that 0 is an even number and infinity is not.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;In a slightly different spirit, with a messier solution (for now?),&#xA;for an ordered bi-infinite list:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Category i. $~$ Infinitely many &lt;em&gt;consecutive&lt;/em&gt; assertions must be true while&#xA;                infinitely many consecutive assertions must be false.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;$~\vdots$ &lt;br&gt;&#xA; Infinitely many consecutive assertions are false while&#xA; this assertion and the next have the same truth value. &lt;br&gt;&#xA; &amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;$~\vdots$&#xA; &lt;br&gt;&lt;br&gt;&#xA; The reasoning is similar to those for the preceding lists,&#xA; with the first step being to establish&#xA; that the assertions cannot be all false.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18129" LastEditorUserId="18129" LastEditDate="2016-01-06T19:15:22.073" LastActivityDate="2016-01-06T19:15:22.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25053" PostTypeId="2" ParentId="25048" CreationDate="2016-01-05T03:05:49.883" Score="13" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.imdb.com/name/nm0670408/?ref_=tt_cl_t8&quot; rel=&quot;noreferrer&quot;&gt;Simon Pegg&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Call me a square one if you will, but it’s not impossible for nerds to be cool&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Square peg. He's a well known nerd.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are two big spaces that I fit in, and it is rare that I fit in both&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He appears in both &lt;a href=&quot;http://www.imdb.com/title/tt2488496/&quot; rel=&quot;noreferrer&quot;&gt;Star Wars&lt;/a&gt; and &lt;a href=&quot;http://www.imdb.com/title/tt0796366/&quot; rel=&quot;noreferrer&quot;&gt;Star Trek&lt;/a&gt;, the only actor I can think of to do both.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I fit the best in spaced, but that wasn’t for too long&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;His brilliant but short lived sitcom, &lt;a href=&quot;http://www.imdb.com/title/tt0187664/&quot; rel=&quot;noreferrer&quot;&gt;Spaced&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And when I was round, I did my best to fit then, too&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.imdb.com/title/tt0425413/&quot; rel=&quot;noreferrer&quot;&gt;Run Fatboy Run&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though it is not my shape, I do have a fondness for cones&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The entire &lt;a href=&quot;https://en.wikipedia.org/wiki/Three_Flavours_Cornetto_trilogy&quot; rel=&quot;noreferrer&quot;&gt;Cornetto trilogy&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6048" LastEditorUserId="11452" LastEditDate="2016-01-08T17:38:46.367" LastActivityDate="2016-01-08T17:38:46.367" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25054" PostTypeId="2" ParentId="25039" CreationDate="2016-01-05T05:11:23.207" Score="1" Body="&lt;p&gt;&lt;strong&gt;22 Chars:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 10 x 9 x 8 x 7 x 6/(5 + 4 + 3 + 2 + 1)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18184" LastActivityDate="2016-01-05T05:11:23.207" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25055" PostTypeId="1" AcceptedAnswerId="25751" CreationDate="2016-01-05T05:53:23.817" Score="27" ViewCount="971" Body="&lt;p&gt;One day, while you work through one of your countless puzzle books, you notice that a sheet of paper has been slipped inside with a curious little poem written on it.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;A Study&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;An ancient man demolished his estate, &lt;br&gt;&#xA;And therefore, with enthusiasm innate, &lt;br&gt;&#xA;He was bewildered when a hoax was glued &lt;br&gt;&#xA;Somewhere beside where he had said adieu. &lt;br&gt;&#xA;The dome where five lone defects lost their head &lt;br&gt;&#xA;Confounds small pools where diamonds sleep instead.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;At first glance you think that it's just another of your brother's nonsense poems, but you notice the slight imperfections in meter and rhyme that aren't usually there, and for some reason you feel that this poem in a particular is positively more &lt;em&gt;cryptic&lt;/em&gt; than the others...&lt;/p&gt;&#xA;&lt;p&gt;What is the poem a study in?&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;It's been a while, so I thought I'd give a hint:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; You turn the page, and notice something written on the other side: 5 4 8 2 5 7 (8)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8949" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-01-25T16:11:09.557" Title="Cryptic Poem: a study in what?" Tags="|word|cryptic-clues|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25056" PostTypeId="1" AcceptedAnswerId="25089" CreationDate="2016-01-05T06:46:43.043" Score="11" ViewCount="311" Body="&lt;p&gt;There are five fencers. Each has a particular skill level, all of which are different. When two fencers duel, the one with the higher skill level always wins. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many duels do the fencers need to hold in order to rank themselves from best to worst, in the worst case?&lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2016-01-06T04:59:11.340" Title="Rank the Fencers" Tags="|weighing|algorithm|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25057" PostTypeId="2" ParentId="25039" CreationDate="2016-01-05T06:58:17.223" Score="1" Body="&lt;p&gt;&lt;strong&gt;24 chars&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(10-9+8) \times 7 \times (6 \times 5 + 4-3 + 2-1)$&lt;br&gt;&#xA;&lt;br&gt;&#xA; $9 \times 7 \times 32 = 2016$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18186" LastEditorUserId="4421" LastEditDate="2016-01-05T23:07:09.140" LastActivityDate="2016-01-05T23:07:09.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25059" PostTypeId="2" ParentId="25056" CreationDate="2016-01-05T07:01:22.640" Score="5" Body="&lt;p&gt;I think it should be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 8 duels&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Divide the five randomly into two groups - one of size 2 and the other of size 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Create ranks within the groups by making them fight each other. This takes 1 + 3 = 4 duels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the group of two be A &gt; B and the group of three such that C &gt; D &gt; E.&#xA;We now need to insert A and B into the C,D,E group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make B fight E. If B wins, B should fight D and so on... The moment B loses, A should fight the winner and fight his way up the ranks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the worst case, B loses to E, and A must thus have to fight E, D and then C to determine his final rank. This takes another 4 duels.&lt;/p&gt;&#xA;" OwnerUserId="17659" LastEditorUserId="17659" LastEditDate="2016-01-05T09:27:56.790" LastActivityDate="2016-01-05T09:27:56.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25060" PostTypeId="2" ParentId="25038" CreationDate="2016-01-05T07:08:21.147" Score="1" Body="&lt;p&gt;We want to reduce this problem to a single light bulb problem and assign a &quot;counter clone&quot;. When we achieve this, we're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get rid of one light bulb, the only way we can do this within the rules of the game is to make sure it is switched on or off a lot of times. This will in the end break the light bulb.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So every clone will go in with the following instructions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Go into the room and check the blue light bulb.&lt;/li&gt;&#xA;&lt;li&gt;If it is on --&gt; switch it off and if the red bulb on, switch it off as&#xA;well (This to make sure we conserve it). Then your done.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If it is off --&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Try to switch it on, if it works the blue light bulb is still not&#xA;broken. If the red bulb is on, switch it off (conserving it) Then your done&lt;/li&gt;&#xA;&lt;li&gt;Try to switch it on, if it does not work another clone has broken it already and you can continue to the red bulb.&lt;/li&gt;&#xA;&lt;li&gt;Try to switch it on, if you see the blue bulb give one light flash, you're the clone breaking the blue bulb. You are the &quot;counter clone&quot; and can continue to the red bulb.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;At the red light bulb, the &quot;counter clone&quot; will always switch the bulb on and count the number of times he finds the red bulb off. If he finds the bulb off 22 times, everybody has visited the room and he can make the declaration.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The other clones will switch the red bulb only off the first time they find it on. All the times they visit the room after their first time, they do nothing.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This strategy might take long, because it might take some time to break the first light bulb. But this strategy will provide certainty.&lt;/p&gt;&#xA;" OwnerUserId="18185" LastActivityDate="2016-01-05T07:08:21.147" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25061" PostTypeId="2" ParentId="25038" CreationDate="2016-01-05T09:03:16.783" Score="4" Body="&lt;p&gt;Here's my solution. Took me awhile to figure out, and I'm not 100% sure it's right, but I believe it is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each clone starts off in an &quot;uninitialized&quot; state. In that state, when they go into the light bulb room they will look at the lights. If the lights are both off, they will move to the &quot;counter&quot; state and turn both lights on. Otherwise, they will move to the &quot;peon&quot; state with a value of 1 and turn both lights off.&lt;br&gt;&lt;br&gt;A peon's job is simple: If he sees two light's in the up position and has a value above 0, turn blue off (leaving only red on) and decrease his value. Otherwise, don't do anything.&lt;br&gt;&lt;br&gt;A counter can enter the room in one of 4 positions: 0, red, blue, or 2 lights on. If both lights are off, he turns them both on. If only the red light is on, he adds 1 to the number of clones he's counted, then turns both lights on. If he sees 2 lights on, he turns off the red light (leaving just blue on). If he just sees the blue light is on and he set it that way last time he was there, he turns both on again. Otherwise, he knows there must be another counter who set it that way, so he turns into a peon with a value equal to his count (+1 for himself).&lt;br&gt;&lt;br&gt;Over time, this should converge until one counter has seen a single response per clone and knows that they've all gone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I appologize for the bad grammar/formatting, it's late and I'm tired. If someone else doesn't get to it I'll edit tomorrow.&lt;/p&gt;&#xA;" OwnerUserId="18188" LastEditorUserId="18188" LastEditDate="2016-01-05T18:50:47.333" LastActivityDate="2016-01-05T18:50:47.333" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="25063" PostTypeId="2" ParentId="25039" CreationDate="2016-01-05T09:26:57.730" Score="0" Body="&lt;p&gt;&lt;strong&gt;21!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;10*9*8*7*6/(5+4+3+2+1) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But not valid, I guess...&lt;/p&gt;&#xA;" OwnerUserId="18189" LastEditorUserId="2730" LastEditDate="2016-01-05T11:15:19.110" LastActivityDate="2016-01-05T11:15:19.110" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25064" PostTypeId="2" ParentId="25039" CreationDate="2016-01-05T09:29:56.460" Score="0" Body="&lt;p&gt;&lt;b&gt;27 characters&lt;/b&gt; (missing &lt;code&gt;/&lt;/code&gt;):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;(10-9)(8*7)(6-5)(4)(3)(2+1)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and &lt;b&gt;22 characters&lt;/b&gt; (missing &lt;code&gt;-&lt;/code&gt;):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;10*9*8*7*6/(5+4+3+2+1)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18190" LastEditorUserId="9802" LastEditDate="2016-01-05T12:57:44.980" LastActivityDate="2016-01-05T12:57:44.980" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25065" PostTypeId="2" ParentId="25056" CreationDate="2016-01-05T09:34:49.973" Score="2" Body="&lt;p&gt;Assuming Transitivity holds i.e if A &gt; B and B &gt; C then A &gt; C, then solving this is like sorting an array of 5 elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying DivideAndConquer algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the fencers be A,B,C,D,E.&#xA;Now lets follow the below steps:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Pick anyone. He fights with everyone undecided in his group to divide the group into two: one better than him and other worse than him.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the &lt;strong&gt;worst case&lt;/strong&gt;, it would be:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A fights with B,C,D,E  (All belong to the better group with none in the other group: Group 1 {}  Group 2 {B,C,D,E})  [Fights = 4]&lt;/li&gt;&#xA;&lt;li&gt;B Fights with C,D,E (All belong to the better group with none in the other group: Group 1 {}  Group 2 {C,D,E})  [Fights = 3]&lt;/li&gt;&#xA;&lt;li&gt;C Fights with D,E (All belong to the better group with none in the other group: Group 1 {}  Group 2 {D,E})  [Fights =  2]&lt;/li&gt;&#xA;&lt;li&gt;D Fights with E (E wins: Group 1 {}  Group 2 {E})  [Fights =  1]&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Skill sequence: A &amp;lt; B &amp;lt; C &amp;lt; D &amp;lt; E&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total fight (Worst case) : 4 + 3 + 2 + 1 = 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the &lt;strong&gt;best case&lt;/strong&gt;, it would be&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A fights with B,C,D,E  (Groups equally split: Group 1 {B,C}  Group 2 {D,E})  [Fights 4]&lt;/li&gt;&#xA;&lt;li&gt;B Fights with C (B Wins: Group 1 {C}  Group 2 {B})  [Fights 1]&lt;/li&gt;&#xA;&lt;li&gt;D Fights with E (E wins: Group 1 {D}  Group 2 {E})  [Fights 1]&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Skill sequence: B &gt; C &gt; A &gt; E &gt; D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total fight (Best case) : 4 + 1 + 1 =  6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, for 'x' fencers, in the worst case &lt;strong&gt;(x-1)*x/2&lt;/strong&gt; fights needs to be fought.&lt;/p&gt;&#xA;" OwnerUserId="2656" LastEditorUserId="2656" LastEditDate="2016-01-06T04:59:11.340" LastActivityDate="2016-01-06T04:59:11.340" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25066" PostTypeId="1" AcceptedAnswerId="25072" CreationDate="2016-01-05T09:36:32.567" Score="13" ViewCount="4707" Body="&lt;p&gt;&lt;em&gt;My inner part is filled with helium&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;Rubber comprises my epithelium&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;If you tie me 'bout your hand&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;Then I'll stay closer to the ground&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;Else I'll go up to the sky&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;I am not a balloon, what am I?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2596" LastEditorUserId="4421" LastEditDate="2016-01-07T13:47:52.680" LastActivityDate="2016-01-07T13:47:52.680" Title="I am not a balloon. What am I?" Tags="|riddle|lateral-thinking|" AnswerCount="3" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="25067" PostTypeId="2" ParentId="25039" CreationDate="2016-01-05T10:03:13.230" Score="1" Body="&lt;p&gt;Here is one more with &lt;b&gt;22 characters&lt;/b&gt; not mentioned in @PaulPro's answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 10*9*8*7*6/(5*4-3-2*1)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit: As @DanHenderson pointed out, this has no + operator. &lt;/p&gt;&#xA;" OwnerUserId="16322" LastEditorUserId="16322" LastEditDate="2016-01-05T14:47:40.070" LastActivityDate="2016-01-05T14:47:40.070" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25068" PostTypeId="2" ParentId="25056" CreationDate="2016-01-05T10:12:21.227" Score="2" Body="&lt;p&gt;I ended up with the same answer as @Manal, but a bit diferently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For ordering two fencers you need one duel. Let's name it A&gt;B (1 Duel)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an extra fencer (3) you'll need 2 more duels in the worst case (First against either one, then against the other). Rename them A&gt;B&gt;C. (1+2 Duels)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an extra fencer (4) you'll make him duel B, and then the A or C, depending whether hi won or lost. Rename them A&gt;B&gt;C&gt;D. (1+2+2 Duels)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, for the fifth fencer you'll make him duel B, and then, in the worst case, C and D. (1+2+2+3 Duels)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end it's 8 Duels (worst case)&lt;/p&gt;&#xA;" OwnerUserId="12165" LastActivityDate="2016-01-05T10:12:21.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25069" PostTypeId="2" ParentId="25025" CreationDate="2016-01-05T10:27:45.857" Score="4" Body="&lt;p&gt;Sorry I can't comment because I don't have high enough reputation. Is it the whole line of numbers that are seen often or is it the three numbers that is often seen together?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it for instance 14, 32, 16, A, 28, 18, 24 that is often seen in a sport or is it A, B, C?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Spoiler:&lt;/strong&gt; Seq. 2 is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A dartboard - 19,3,&lt;strong&gt;17&lt;/strong&gt;,2,15,10,6. Damn didn't realise that the rest was also from a dartboard untill it was alrady guessed :(&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18191" LastEditorUserId="18191" LastEditDate="2016-01-05T12:51:33.807" LastActivityDate="2016-01-05T12:51:33.807" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25070" PostTypeId="2" ParentId="25039" CreationDate="2016-01-05T10:32:06.257" Score="1" Body="&lt;p&gt;&lt;strong&gt;22 chars&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;±10*9*8*7*6/(5+4+3+2+1)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If ± is valid as usage of - character, then you get two answers, which of one is correct :)&lt;/p&gt;&#xA;" OwnerUserId="356" LastActivityDate="2016-01-05T10:32:06.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25071" PostTypeId="2" ParentId="25066" CreationDate="2016-01-05T11:10:10.037" Score="2" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lightweight aircraft&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My inner part is filled with helium&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Helium is used as a fuel (&lt;a href=&quot;http://www.usesof.net/uses-of-helium.html&quot; rel=&quot;nofollow&quot;&gt;Reference&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rubber comprises my epithelium&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We have rubber band powered airplanes.(&lt;a href=&quot;https://www.youtube.com/watch?v=WcwK-pPOP54&quot; rel=&quot;nofollow&quot;&gt;Reference&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you tie me 'bout your hand&#xA;Then I'll stay closer to the ground&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you have it in hand. It won't go up.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Else I'll go up to the sky&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Possible reference to leave it after adding potential energy in it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastActivityDate="2016-01-05T11:10:10.037" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25072" PostTypeId="2" ParentId="25066" CreationDate="2016-01-05T11:46:11.127" Score="26" Body="&lt;p&gt;Is it a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Portable hard drive&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My inner part is filled with helium&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some hard drives are filled with helium to reduce internal friction&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Rubber comprises my epithelium&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rubberized to prevent damage if it falls&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If you tie me 'bout your hand&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;Then I'll stay closer to the ground&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you connect it to your PC it will be relatively close to the ground and at the tip of your fingers. Alternatively &quot;grounding&quot; the hard drive (credits to &lt;a href=&quot;https://puzzling.stackexchange.com/users/1682/adam-davis&quot;&gt;Adam Davis&lt;/a&gt;) or wearing a grounding wristband (credits to &lt;a href=&quot;https://puzzling.stackexchange.com/users/17057/dane-andersen&quot;&gt;Dane Andersen&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Else I'll go up to the sky&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alternatively you can store all your data in the &quot;cloud&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am not a balloon, what am I?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A hard drive&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2016-01-06T16:36:24.410" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="25073" PostTypeId="2" ParentId="25025" CreationDate="2016-01-05T11:58:42.143" Score="7" Body="&lt;p&gt;These sequences are found on a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; dartboard &lt;br&gt; Seq1 and Seq3 clockwise and Seq2 counter-clockwise.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Seq1: 7, 16, 8, 11, 14, 9 12   all multiplied with 2: A =&gt; 22 &lt;br&gt;&#xA; Seq2: 19, 3, 17, 2, 15, 10, 6:  B =&gt; 17 &lt;br&gt;&#xA; Seq3: 1, 18, 4, 13, 6, 10, 15  all multiplied with 3: C =&gt; 12&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks to Bungicasse for the questions, which helped me a lot.&lt;/p&gt;&#xA;" OwnerUserId="14461" LastActivityDate="2016-01-05T11:58:42.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25074" PostTypeId="1" AcceptedAnswerId="25075" CreationDate="2016-01-05T14:18:37.863" Score="1" ViewCount="1318" Body="&lt;p&gt;So imagine you are a bus driver.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have left the depo at 9:00 AM on the dot!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You arrive at the first stop, open the doors and pick up the 3 people waiting at the stop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the next stop 2 Adults and 3 Children get on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third stop - and twice as many people get off as get on. 2 People get on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the next stop the children get off with 1 adult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the final stop everyone has got to get off, the last person on the bus wanted to complement the driver and asked for his name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the driver's name?&lt;/p&gt;&#xA;" OwnerUserId="18201" LastEditorUserId="5373" LastEditDate="2018-06-05T18:01:13.540" LastActivityDate="2018-06-05T18:01:13.540" Title="A Simple Puzzle - A bus Driver" Tags="|lateral-thinking|story|" AnswerCount="1" CommentCount="2" ClosedDate="2016-01-05T15:36:33.023" ContentLicense="CC BY-SA 4.0" />
  <row Id="25075" PostTypeId="2" ParentId="25074" CreationDate="2016-01-05T14:21:52.513" Score="5" Body="&lt;p&gt;Answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;AeJey&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I am the bus driver (Since I am imagining that I am the bus driver)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="2730" LastEditDate="2016-01-05T14:22:49.857" LastActivityDate="2016-01-05T14:22:49.857" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25076" PostTypeId="2" ParentId="25066" CreationDate="2016-01-05T14:27:03.267" Score="9" Body="&lt;p&gt;I think it is a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; laser barcode scanner. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My inner part is filled with helium&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Laser barcode scanners contain &lt;a href=&quot;https://en.wikipedia.org/wiki/Helium%E2%80%93neon_laser&quot; rel=&quot;nofollow&quot;&gt;helium-neon lasers&lt;/a&gt;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Rubber comprises my epithelium&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The scanners have an external enclosure made of rubber. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If you tie me 'bout your hand&lt;br&gt;&#xA;Then I'll stay closer to the ground&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When one holds the scanner in their hand, it is aimed at the items in the checkout counter (closer to the ground)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Else I'll go up to the sky&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When the scanner is left on the rack, the beam will point upwards towards the sky. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7529" LastEditorDisplayName="user20" LastEditDate="2016-01-06T16:36:59.463" LastActivityDate="2016-01-06T16:36:59.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25077" PostTypeId="2" ParentId="25039" CreationDate="2016-01-05T15:18:37.563" Score="3" Body="&lt;p&gt;&lt;strong&gt;21 chars&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $10 \times 9 \times 8 (7 - 6 \div 5 - 4 + 3 - 2 \times 1)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4659" LastEditorUserId="4421" LastEditDate="2016-01-05T22:58:54.933" LastActivityDate="2016-01-05T22:58:54.933" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25078" PostTypeId="1" AcceptedAnswerId="25084" CreationDate="2016-01-05T18:17:12.803" Score="9" ViewCount="1131" Body="&lt;p&gt;Smoking makes me blush&lt;br&gt;&#xA;And smell rather strongly&lt;br&gt;&#xA;I'll achieve my purpose&lt;br&gt;&#xA;If you answer this wrongly&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who am I?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user17947" LastActivityDate="2016-01-05T20:57:00.270" Title="Short, but not sweet" Tags="|riddle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25079" PostTypeId="2" ParentId="25078" CreationDate="2016-01-05T18:46:54.640" Score="10" Body="&lt;p&gt;You are a  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Butt&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Short, but not sweet  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pun hiding &quot;but&quot; in the title, also, I don't know any sweet butts. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Smoking makes me blush  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cigarette butts have a red ember (blush) when you smoke.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And smell rather strongly    &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Both smoking and butts can smell rather strong.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll achieve my purpose&lt;br&gt;&#xA;If you answer this wrongly   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because I have probably answered this wrong, I will be the butt of the joke. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="9951" LastEditDate="2016-01-05T19:01:23.597" LastActivityDate="2016-01-05T19:01:23.597" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25081" PostTypeId="2" ParentId="5900" CreationDate="2016-01-05T19:29:07.143" Score="-2" Body="&lt;p&gt;The answer is &quot;Sherlock I have found you&quot;. The key to finding the answer is masonic cipher (also pigpen cipher). &lt;/p&gt;&#xA;" OwnerUserId="18212" LastActivityDate="2016-01-05T19:29:07.143" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25084" PostTypeId="2" ParentId="25078" CreationDate="2016-01-05T20:13:47.577" Score="11" Body="&lt;p&gt;You are a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; herring&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Smoking makes me blush&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Smoking herring turns it red.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And smell rather strongly&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Smoked herring has a strong smell.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll achieve my purpose if you answer this wrongly&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A red herring is a clue meant to be misleading.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Short, but not sweet&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A herring is a small fish, and it does not taste sweet!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18215" LastEditorUserId="18215" LastEditDate="2016-01-05T20:57:00.270" LastActivityDate="2016-01-05T20:57:00.270" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25088" PostTypeId="2" ParentId="25035" CreationDate="2016-01-05T23:22:37.977" Score="4" Body="&lt;p&gt;The solution is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; what power lies in words, that a single concept can outdo. present me that which can only truly be spoken by yourself. feed me what I need so I can reveal it back to you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;Step 1&lt;/h3&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first step is to determine the type of the cipher. &lt;a href=&quot;http://www.simonsingh.net/The_Black_Chamber/substitutioncrackingtool.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;The site from Simon Singh&lt;/a&gt; has a nice tool to determine the frequency of the letters in the cipher text. It also shows the expected count of letters for an English text, which leads to the conclusion that no substitution was used.&lt;br&gt;&lt;br&gt;&lt;a href=&quot;https://i.stack.imgur.com/4ahru.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4ahru.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&lt;br&gt;Difficulty of this step: easy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;Step 2&lt;/h3&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As we don't use a substitution cipher, the next guess is that some kind of transposition is used. There is &lt;a href=&quot;http://tholman.com/other/transposition/&quot; rel=&quot;nofollow noreferrer&quot;&gt;another nice site&lt;/a&gt; for this kind of ciphers. The text has a length of 170 characters. Prime factors of 170 are 2, 5 and 17, so testing the corresponding numbers for rows and columns first makes sense, if we assume that the riddle was designed to be not too hard. Arranging the cipher text in 10 columns and 17 rows gives this table:&lt;br&gt;&lt;br&gt;&lt;a href=&quot;https://i.stack.imgur.com/BMiSY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BMiSY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&lt;br&gt;The correct key for this part was already posted in the comments above. Another method to solve this without a key would be to use an &lt;a href=&quot;http://www.anagram-solver.org/?letters=nccecpaot&quot; rel=&quot;nofollow noreferrer&quot;&gt;anagram solver with an appropriate row&lt;/a&gt;. In this case I used the 5th row containing the word &quot;concept&quot;. Having most of the columns correct, the rest is then pretty easy:&lt;br&gt;&lt;br&gt;&lt;a href=&quot;https://i.stack.imgur.com/0IxxS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0IxxS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&lt;br&gt;Some other properties of the cipher text can also be used to find out the correct ordering. In the last row the period should be in the last column. Other periods and the comma should be followed by a blank. Making the starting letter of the next sentence a capital letter could be already enough to solve the rest without a key.&lt;br&gt;&lt;br&gt;The key used is presented as a date in the text of the riddle: 27th April 3577 AF. Adding the 12 years from the note and using 1 and 6 for A and F gives the complete key.&lt;br&gt;&lt;br&gt;Difficulty of this step: hard to say because of the leaked key in the comments, but I would say medium.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;Additional thoughts&lt;/h3&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The difficulty rating I've given is of course subjective. There are much harder crypto puzzles on this site. It's also an interesting topic, so I know some simple cipher methods. If your audience never heard of &quot;letter frequencies&quot;, already the first step might become unsolvable. Finding the correct number of rows/columns also involves some amount of trial and error. Presenting the cipher text in a correctly sized grid would make the deciphering much easier, definitely too easy if someone made some code cracking before.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2016-01-06T07:22:08.687" LastActivityDate="2016-01-06T07:22:08.687" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25089" PostTypeId="2" ParentId="25056" CreationDate="2016-01-06T01:42:20.610" Score="12" Body="&lt;p&gt;There are $5!=120$ possible orderings of the fencers, so we need $\log_2(120)\approx 6.9069$ bits of information. Each duel provides at most $1$ bit of information, so at least $7$ duels will be necessary. Here is a way to determine the ordering in $7$ duels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Call the fencers $A$, $B$, $C$, $D$, and $E$.&lt;br&gt; First $A$ and $B$ duel, by symmetry we assume $A$ wins. Then $C$ and $D$ duel, we assume $C$ wins. Next $A$ and $C$ duel, we assume $A$ wins. Now we have $E$ and $C$ duel. Here we separate into cases.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$E$ beats $C$. We then have $B$ and $C$ duel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a. $B$ beats $C$. There are three possible orderings: $A&amp;gt;B&amp;gt;E&amp;gt;C&amp;gt;D$, $A&amp;gt;E&amp;gt;B&amp;gt;C&amp;gt;D$, and $E&amp;gt;A&amp;gt;B&amp;gt;C&amp;gt;D$, and two more duels ($E$ vs $A$, then $E$ vs $B$) determine the ordering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b. $C$ beats $B$. There are four possible orderings: $A&amp;gt;E&amp;gt;C&amp;gt;B&amp;gt;D$, $E&amp;gt;A&amp;gt;C&amp;gt;B&amp;gt;D$, $A&amp;gt;E&amp;gt;C&amp;gt;D&amp;gt;B$, and $E&amp;gt;A&amp;gt;C&amp;gt;D&amp;gt;B$, and two more duels ($A$ vs $E$, then $B$ vs $D$) determine the ordering.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$C$ beats $E$. We then have $E$ duel $D$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a. $E$ beats $D$. We then have $B$ duel $E$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(i): $B$ beats $E$. There are only two possible orderings: $A&amp;gt;B&amp;gt;C&amp;gt;E&amp;gt;D$ and $A&amp;gt;C&amp;gt;B&amp;gt;E&amp;gt;D$, and having $B$ and $C$ duel determines the ordering.&lt;br&gt;&lt;br&gt;(ii): $E$ beats $B$. There are only two possible orderings: $A&amp;gt;C&amp;gt;E&amp;gt;B&amp;gt;D$ and $A&amp;gt;C&amp;gt;E&amp;gt;D&amp;gt;B$, and having $D$ and $B$ duel determines the ordering.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;b. $D$ beats $E$. We then have $B$ and $D$ duel.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(i): $B$ beats $D$. There are only two possible orderings: $A&amp;gt;B&amp;gt;C&amp;gt;D&amp;gt;E$ and $A&amp;gt;C&amp;gt;B&amp;gt;D&amp;gt;E$, and having $B$ and $C$ duel determines the ordering.&lt;br&gt;&lt;br&gt;(ii): $D$ beats $B$. There are only two possible orderings: $A&amp;gt;C&amp;gt;D&amp;gt;B&amp;gt;E$ and $A&amp;gt;C&amp;gt;D&amp;gt;E&amp;gt;B$, and having $B$ duel $E$ determines the ordering.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2016-01-06T02:00:30.617" LastActivityDate="2016-01-06T02:00:30.617" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25090" PostTypeId="1" AcceptedAnswerId="25094" CreationDate="2016-01-06T04:31:03.743" Score="2" ViewCount="342" Body="&lt;p&gt;Mr. S, the history professor, offered his class a challenge. He told the class that bonus points would be given in the examination to the first student who deciphered the text. All students received an e-mail from Mr. S after the lecture.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hi everyone,  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The first student who can reveal what word is being hidden in the following passage will be awarded extra 20 points in the final examination.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;code&gt; &lt;/code&gt;  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This challenge is open for 7 days. Good luck to you all!  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Mr. S&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One of the students, Mr. P, tried extremely hard to decipher the passage. As days went by, he was getting more and more frustrated. Fortunately, he figured out what the passage meant just about 6 minutes before the deadline. Mr. S was very impressed and gave him extra 20 points as promised.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you find out what word is hidden in the text?&lt;/p&gt;&#xA;" OwnerUserId="16499" LastEditorUserId="8874" LastEditDate="2016-03-23T17:06:25.800" LastActivityDate="2016-03-23T17:06:25.800" Title="Leet of Legends" Tags="|story|cipher|history|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25091" PostTypeId="1" AcceptedAnswerId="25093" CreationDate="2016-01-06T04:47:33.040" Score="6" ViewCount="8641" Body="&lt;p&gt;Andrew and Bob are talking to each other, but we don't yet know who is who.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The guy in the blue shirt says, &quot;I am Andrew.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The guy in the red shirt says, &quot;I am Bob.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If at least one of them lied, what color shirt is Andrew wearing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vnXwr.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vnXwr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="185" LastEditorUserId="4421" LastEditDate="2016-01-07T16:59:32.333" LastActivityDate="2018-08-21T00:07:20.270" Title="Which color shirt is Andrew wearing?" Tags="|logical-deduction|liars|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25092" PostTypeId="2" ParentId="25091" CreationDate="2016-01-06T04:56:05.390" Score="3" Body="&lt;p&gt;Suppose only the guy in blue lied, then he was in fact Bob. But then the guy in red would be lying also, as the guy in red said that he was Bob, who should be Andrew in fact. This led to a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose only the guy in red lied, then he was in fact Andrew. But then the guy in blue would be lying also, as the guy in blue said that he was Andrew, who should be Bob in fact. This led to a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there was at least one person lying, therefore the guy in blue was Bob and the guy in red was Andrew. Both of them lied.&lt;/p&gt;&#xA;" OwnerUserId="16499" LastActivityDate="2016-01-06T04:56:05.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25093" PostTypeId="2" ParentId="25091" CreationDate="2016-01-06T05:03:56.890" Score="8" Body="&lt;p&gt;The guy in the blue shirt says, &quot;I am Andrew.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If he is a liar, he is BOB.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The guy in the red shirt says, &quot;I am Bob.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If he is not a liar, he is BOB and hence ANDREW must be speaking the truth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a contradiction as at least one of them lies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hence,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; both of them lies i.e Andrew is the one in RED shirt and BOB in BLUE.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2656" LastEditorUserId="42452" LastEditDate="2018-08-21T00:07:20.270" LastActivityDate="2018-08-21T00:07:20.270" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25094" PostTypeId="2" ParentId="25090" CreationDate="2016-01-06T05:06:16.350" Score="2" Body="&lt;p&gt;The word is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Atlantis&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &quot;passage&quot; is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A base-64 encoding of a passage (in &quot;1337speak&quot;) about Atlantis from Plato's &lt;em&gt;Timaeus&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It reads (when &lt;em&gt;fully&lt;/em&gt; decoded)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; there was an island situated in front of the straits which are by you called the pillars of heracles; the island was larger than libya and asia put together, and was the way to other islands, and from these you might pass to the whole of the opposite continent which surrounded the true ocean; for this sea which is within the straits of heracles is only a harbour, having a narrow entrance, but that other is a real sea, and the surrounding land may be most truly called a boundless continent&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17147" LastEditorUserId="17147" LastEditDate="2016-01-07T02:27:13.453" LastActivityDate="2016-01-07T02:27:13.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25095" PostTypeId="1" AcceptedAnswerId="25153" CreationDate="2016-01-06T07:07:15.460" Score="11" ViewCount="463" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Through the horn that persistent noise speaks to me&lt;br&gt;&#xA;  It tells me, by Greek A to C, that we are not here eternally&lt;br&gt;&#xA;  Then Hebrew Jacob seals the deal, en route to comb the silence&lt;br&gt;&#xA;  What once was thought is wrong, it's in fact not balanced&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hints:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A to C is alpha, beta, gamma.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hebrew Jacob is cobe (common nickname, pronounced cob-E)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The silence is outer space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4 - Practically a spoiler. Just google it if you really want the answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; COBE and COMB are acronyms, usually COMB is shorthanded CMB.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18064" LastEditorUserId="18064" LastEditDate="2016-01-07T08:12:39.033" LastActivityDate="2016-01-07T16:50:59.477" Title="What is that - riddle" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25096" PostTypeId="2" ParentId="25038" CreationDate="2016-01-06T07:39:14.990" Score="8" Body="&lt;p&gt;Who even needs multiple bulbs?  Let's solve this using &lt;em&gt;just the red bulb!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each clone maintains a personal count C, which starts at 0.&lt;br&gt;&#xA; Ignore the state of the blue bulb.  The red bulb is on or off; Let &quot;on&quot; be L = 1 and &quot;off&quot; be L = 0.&lt;br&gt;&#xA;&lt;br&gt;&#xA; If L = 0, 22 &gt; C &gt; -2: set C = C-1, L = 1&lt;br&gt;&#xA; If L = 0, C = -2 or C &gt; 21: do nothing&lt;br&gt;&#xA; If L = 1, C &amp;lt; 43: set C = C+1, L = 0&lt;br&gt;&#xA; If L = 1, C = 43: &lt;em&gt;declare that all clones have entered the room&lt;/em&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Logic behind this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each clone sets L = 1, and C = -1 to vote.  The next clone to enter will set L = 0 and C = 1, collecting the vote.  The same clone seeing the room multiple times in a row will just keep redoing and undoing the same two actions.  As a result, +1 and -1 are always conserved, but will be redistributed at random until eventually some clone gets lucky and happens to collect a lot of votes.  Note that no clone can ever go further than -2 through this process, so each clone has at most two votes.  Confusion about the initial state means that when one clone manages to collect all the votes, there will be either 44 or 45 votes to collect - but as long as they collect at least 44, then everybody has voted once, since 21 other clones would cap out at 42 or 43 votes.  Since every finite pattern will appear in an infinite random pattern, we simply need to know that there exists a pattern that causes 44 votes to be gathered by one clone.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The pattern, of course, is (starting when L = 0), Clones enter the room in this order: 2 - 1 - 2 - 1 - 3 - 1 - 3 - 1 - 4 - 1 - 4 - 1... etc.  Clearly this can transfer all votes to 1, thus it will eventually happen with probability 1.  The &gt; 21 case just means that as soon as a clone gets half the votes, they can stop giving them back. (If some votes have already been distributed, clones will simply show up in the above sequence once for each increment their counter is above -2)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;For posterity, my old solution, using both bulbs and a designated counter&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each clone maintains a personal count C&lt;br&gt;&#xA; The combination of lights have four positions, L = 0, 1, 2, 3 – initial state unknown&lt;br&gt;&#xA; Clones distinguish between their first visit to the room and subsequent visits&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;First visit:&lt;/strong&gt;&lt;br&gt;&#xA; If L = 0: set C = 0, L = 0&lt;br&gt;&#xA; If L = 1, 2, or 3: set C = 1, L = 0&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Subsequent visits:&lt;/strong&gt;&lt;br&gt;&#xA; If L = 0, C &gt; -1: set C = C-1, L = 1&lt;br&gt;&#xA; If L = 0, C &amp;lt; 0: do nothing&lt;br&gt;&#xA; If L = 0 and you are TCO: &lt;em&gt;declare that all clones have entered the room&lt;/em&gt;&lt;br&gt;&#xA; If L = 1, C &amp;lt; 21: set C = C+1, L = 0&lt;br&gt;&#xA; If L = 1, C = 21: set C = 0, L = 2: &lt;em&gt;you are The Chosen One&lt;/em&gt;&lt;br&gt;&#xA; If L = 1 and you are TCO: &lt;em&gt;declare that all clones have entered the room&lt;/em&gt;&lt;br&gt;&#xA; If L = 2 and you are TCO: do nothing&lt;br&gt;&#xA; If L = 2 and you are not TCO and C != -2: set C = -2, L = 3&lt;br&gt;&#xA; If L = 2 and you are not TCO and C = -2: do nothing&lt;br&gt;&#xA; If L = 3 and you are TCO and C &amp;lt; 21: set C = C+1, L = 2&lt;br&gt;&#xA; If L = 3 and you are TCO and C = 21: &lt;em&gt;declare that all clones have entered the room&lt;/em&gt;&lt;br&gt;&#xA; If L = 3 and you are not TCO: do nothing  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Logic behind this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If L = 0 initially, every clone initializes to C = 0 until a clone enters for a second time.  This clone has C = 0, so they set L = 1, and C = -1.  The next clone to enter will set L = 0 and C = 1, whether it's their first time or not.  At this point, nothing happens for a while again until a clone enters for the second time, and sets themselves to -1 and another clone to +1.  The same clone seeing the room multiple times in a row will just keep redoing and undoing the same two actions.  As a result, +1 and -1 are always conserved, but can be redistributed until eventually some clone happens to collect a lot of +1s without any -1s.  Note that no clone can ever go further than -1 through this process, so no clone is being double counted.  At some point, a single clone will hit C = 22 as a result.  But this is where the initial state becomes an issue.&lt;br&gt;&#xA;&lt;br&gt;&#xA; This clone doesn't know if the initial state was 0, so there are 22 other -1 clones and just them as a +22 clone... or if the initial state was 1, 2, or 3 and there's an extra +1 from the very first clone that's not balanced out anywhere else - giving only 21 other -1 clones, themselves as a +22 clone, and one clone that's at 0, or has never visited the room.&lt;br&gt;&#xA;&lt;br&gt;&#xA; At this point, this clone knows they are very close, so they declare themselves The Chosen One, and set L = 2 while resetting their count.  If a clone has already been in the room before, this is the only way they could see an L = 2, since initialization and early moves only give L = 0 or L = 1 past the very first clone.  If every clone has been in the room before (initial state was 0), this is a subsequent event for all of them, and they know that it's special.  The first one to receive this message responds with a 3 (meaning &quot;count me!&quot;) and then removes themselves from the game.  One this gets to The Chosen One, they count and reset to 2.  This continues until everybody responds with 3, at which point The Chosen One will get to a count of 22 again and declare victory.  Alternatively, the initial state was not L = 0.  In this case, there is one clone who isn't at -1.  If they're at 0, they count from 2 to 3 like everybody else, and The Chosen One still gets to 22.  If they hadn't been in the room at all before, they go from L = 2 to L = 0, and set C = 1.  At this point, a lot of stuff can go weird with the counts due to them entering the room consecutively (setting it to L = 1) and then other clones who hadn't seen the L = 2 case doing stuff, but it all boils down to one thing - nobody can get up to a count of 22, because there is at most a +1 in the system.  Therefor there can never be a second Chosen One, and nobody can ever set L to 2 or 3.  As a result, whenever The Chosen One returns to the room, L = 0 or L = 1 - either way, this can only have been caused by the last missing clone showing up and screwing around.  At this point, all the clones are accounted for, so The Chosen One declares victory without finishing the count.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="17389" LastEditDate="2016-01-06T15:29:20.327" LastActivityDate="2016-01-06T15:29:20.327" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25098" PostTypeId="1" AcceptedAnswerId="25640" CreationDate="2016-01-06T08:56:06.073" Score="38" ViewCount="5004" Body="&lt;p&gt;Is it possible to fill the $121$ entries in an $11\times11$ square with the values $0,+1,-1$, so that the row sums and column sums are $22$ distinct numbers?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2016-01-22T00:29:47.570" Title="Filling an 11-by-11 square" Tags="|mathematics|combinatorics|" AnswerCount="9" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="25099" PostTypeId="2" ParentId="25098" CreationDate="2016-01-06T09:15:05.383" Score="12" Body="&lt;p&gt;&lt;strong&gt;Partial answer (fixing earlier mistake)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The possible outcomes of a sum of 11 values (0, +1 or -1) can be any integer number from - 11 to + 11, leaving 23 options total.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theory on combining resulting numbers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It's fairly trivial to state that if '11' occurs (a row/column with all 1's), there can never be a column/row totalling '-11' or '-10', due to one of the rows/columns holding a '1' (you can at lowest reach -9 then). The other way around, you cannot have '-11' and expect to be able to squeeze in 10 or 11. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What we learn from this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;11 and -11 cannot occur in scenario's with one of them in a column&lt;/strong&gt;, and the other in a row&lt;br&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note: the above is where I went wrong: I first assumed they couldn't appear together at all, which is false - as pointed out in the comments&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For ease of further explanation, assume both 11 and -11 to be in rows (all that follows would be similar but mirrored along the diagonal).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If 11 and -11 both take up an entire row, this means in columns you can at extremes reach -9 and +9. If we want 10 and -10 to be able to appear, those have to be in rows aswell (with 10x '1' or '-1' and one zero). This means we already have 4 rows occupied.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It now gets harder though:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if we aligned the 'zeroes' from the '10' and '-10' rows below eachother, we can get at most +7 and -7 in a column, if we didn't we can get +8 or -8 at most. Either way we can only create +9 and -9 in rows again. We can do this two ways: 9x '+-1' and 2x 0, or 10x '+-1' and 1x '-+1'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What can we squeeze into columns now?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; so far, the rows from -11 and 11 cancel eachother out, the rows from -10 and 10 allow for a maximum of +1 or -1 variance on the two rows occupied, and the rows from -9 and 9 for a +2 or -2 variance on the two rows those occupy. So with 5 rows left a column value can now hold at most +-5 (+-3) , which also implies we can now actually squeeze our next step (+8 and -8) into columns.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Option 1: Put the '8's in columns.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This means we've used up all variance on the remaining 9 columns, so we have 9 columns which can at most hold +-5. This also implies both 7's and 6's will have to go in rows. The 11's, 10's, 9's, 7's and 6's hog a combined 10 out of a possible 11 rows. &lt;br&gt; Is there a way to build the 7's and 6's in the column's that leaves sufficient different value numbers to be built in the colums? This will most likely imply 7's and 6's to be built to a maximum from -1 and +1's rather then 0's, to allow sufficiëntly high values in the collumns). Key might be to try and ensure the presence of the +-4 and +-3 in the colums as a guide to figure out how to place the several +-1's to create 6/7 rows.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Option 2: Put the '8's in rows&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's not because we CAN put the 8's in columns, we have to.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Option 3: Put one of the 8's in a column, the other in a row.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Besides option one and two, there's still a third option, and that's to go both ways.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I may have to put this train of thoughts on hold for now, as there doesn't appear a simple path further down. The thing is: The more you keep pushing things with very similar patterns for the '+' and '-' variation in rows, the bigger the chances you'll have issues fitting in 5's and 4's in columns. If the '+1' and the '-1' from the '+' and '-' version of the larger sums keep cancelling eachother out in the column totals, suddenly it appears 4's and 5's turn seemingly impossible to fit in. (or maybe I just haven't spotted how).&lt;/p&gt;&#xA;" OwnerUserId="2156" LastEditorUserId="2156" LastEditDate="2016-01-07T08:59:13.967" LastActivityDate="2016-01-07T08:59:13.967" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25100" PostTypeId="1" AcceptedAnswerId="25116" CreationDate="2016-01-06T09:55:39.427" Score="7" ViewCount="1310" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;I'm the first, but also second&lt;/p&gt;&#xA;&lt;p&gt;I'm rather odd, but also not&lt;/p&gt;&#xA;&lt;p&gt;I am the circle of three&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="16322" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-01-06T22:41:32.250" Title="The circle of three" Tags="|riddle|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25101" PostTypeId="2" ParentId="25100" CreationDate="2016-01-06T10:04:58.853" Score="11" Body="&lt;p&gt;You are the&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Second hand in a clock&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation Follows&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm the first, but also second&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Second hand is the fastest and to reach the 12 hour mark first. But its calles Second.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm rather odd, but also not&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Some time the second hand points to an odd number but in the next instant it will be an even number&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the circle of three&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The rotation of second hand manages second, minutes (one complete rotation of second hand) and hours (60 complete rotations of second hand). The numbers/points in circular form in a clock represents seconds, minutes and hours.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="2730" LastEditDate="2016-01-06T10:06:31.917" LastActivityDate="2016-01-06T10:06:31.917" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25103" PostTypeId="2" ParentId="25042" CreationDate="2016-01-06T12:58:45.853" Score="9" Body="&lt;p&gt;This was obvious from the start. But getting the answer was tedious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 88&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/b8E8Z.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/b8E8Z.png&quot; alt=&quot;puzzle&quot;&gt;&lt;/a&gt;&#xA;&#xA;&#xA; The grid is divided into 5x5 blocks, where each block is sum of characters in 5x5 font. This gives text &quot;WHAT IS EIGHT TIMES ELEVEN&quot;. And because password is shorter than 15, the pass is result of the spelled out formula (might be in text form).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18238" LastActivityDate="2016-01-06T12:58:45.853" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="25104" PostTypeId="2" ParentId="25098" CreationDate="2016-01-06T13:01:07.270" Score="8" Body="&lt;h2&gt;Partial answer&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;As Tim pointed out&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you have rows with sums of -11, 11 and 10, that quickly limits the different numbers you can generate in the columns. Experimentally you can fit in ±11, ±10, ±9, ±8, ±7, ±6 and ±5 but there isn't anywhere to put ±4. If I miss out -10 then I can get to ±3 but there isn't anywhere to put ±2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the big problem is that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you need at least ±11 and one of ±10 just to be able to complete the square.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the row and columns use 22 of the 23 sums but they must sum to the same number so the missing sum must be even.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By contrast, even squares are quite easy; there's a simple pattern whereby you fill a 2n square with n 0s, 2n² 1s and 2n²-n -1s. (Hint: 2n²-n = 2n(2n-1)/2.) Example for 6x6:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;pre&gt;1  1  1  1  1  1  6&lt;br&gt;1  1  1  1  1 -1  4&lt;br&gt;1  1  1  1 -1 -1  2&lt;br&gt;1  1  0 -1 -1 -1 -1&lt;br&gt;1  0 -1 -1 -1 -1 -3&lt;br&gt;0 -1 -1 -1 -1 -1 -5&lt;br&gt;5  3  1  0 -2 -4&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My best effort so far for 11x11:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 1  1  1  1  1  1  1  1  1  1  1 11&#xA; 1  1  1  1  1  1  1  1  1  1  0 10&#xA; 1  1  1  1  1  1  1  1  1  0 -1  8&#xA; 1  1  1  1  1  1  1  1  0 -1 -1  6&#xA; 1  1  1  1  1  1  1  0 -1 -1 -1  4&#xA; 1  1  1  1  0  0  0  0 -1 -1 -1  1&#xA; 1  1  1  1 -1 -1 -1 -1 -1 -1 -1 -3&#xA; 1  1  1 -1 -1 -1 -1 -1 -1 -1 -1 -5&#xA; 1  1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -7&#xA; 1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -9&#xA;-1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -11&#xA; 9  7  5  3  0  0  0 -1 -4 -6 -8&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As you can see ±2 are missing, but I'm sure that the fact that you can't get those diagonals to meet must be a factor somehow.&lt;/p&gt;&#xA;" OwnerUserId="12593" LastEditorUserId="12593" LastEditDate="2016-01-06T17:26:53.093" LastActivityDate="2016-01-06T17:26:53.093" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25105" PostTypeId="2" ParentId="25038" CreationDate="2016-01-06T13:15:41.457" Score="14" Body="&lt;p&gt;The standard solution is that all clones signal their first passage using some state of the bulbs to an elected clone, the &quot;counter&quot;, who counts how many there are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to address 3 things.&lt;br&gt;&#xA;1. How to transmit the information of one's passage reliably,&lt;br&gt;&#xA;2. How to deal with the initial state,&lt;br&gt;&#xA;3. How to elect a counter.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the lights don't represent the &quot;state of the room&quot;, but represent a message that a clone can pass to whichever clone following him.  The message consists of 2 bits of information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) is pretty easy.  Each clone has a virtual token.  He must pass this token to the counter.  Any clone might have multiple tokens in his possession at any time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assign to the blue light the meaning &quot;I give you one token&quot;.  A clone who finds the blue light on automatically increases his token count.  If he wants to get rid of a token, he leaves the blue light on before leaving and decreases his count.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The counter will collect all tokens and never leave the blue light on, all others will push their tokens to the next clone by leaving the blue light on.  That is, if they still have one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (2) we need to adjust the method.  The blue light might be on initially.  The first clone, not knowing he is first, will count this as a token.  Therefore the total token count is actually 23 or 24, and the counter doesn't know when to declare the end of the game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To overcome that problem we need to double the number of tokens.  Every clone starts with 2 tokens.  Then the total number of token is 46 or 47 and the counter can safely declare the end as soon as he has collected 46 token.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (3) we need to settle for a single counter.  But make sure there is at least one.  This can be achieved with the red light.  Every time a counter visits the room he toggles the red light.  But if he sees that the red light has changed state in his absence, that means here is another active counter.  If that happens, he simply resigns his role without changing the red light.  He will spend the rest of his time as a non-counter getting rid of any token he already collected.  With time only one counter should remain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in fact, all clones start as a counter.  On the first passage they collect any token they get, don't leave any, and switch the red light.  Starting with the second passage, they will check the red light and possibly become a non-counter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the method is as follows:&lt;br&gt;&#xA;- the initial state of each clone is the role of counter with 2 token.&lt;br&gt;&#xA;- any clone seeing the blue light on when arriving, increases his token count.&lt;br&gt;&#xA;- a counter with a token count of 46 declares that all clones have been in the room.&lt;br&gt;&#xA;- a counter who sees that the red light changed becomes a non-counter.  He acts as such in the 2 last rules.&lt;br&gt;&#xA;- a counter who remains a counter always toggles the red light.  All others leave it unchanged.&lt;br&gt;&#xA;- a non-counter with non-zero token count switches the blue light on and decreases his count.  All others switch the light off.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There can be optimizations, for example, a counter with at least 24 token doesn't change role any more.  Or on its first visit, if the the blue light is off, you directly become a non-counter.  But the more you optimize the less certain it is that the solution works...&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="2136" LastEditDate="2016-01-06T14:33:29.333" LastActivityDate="2016-01-06T14:33:29.333" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="25106" PostTypeId="2" ParentId="25100" CreationDate="2016-01-06T13:16:36.320" Score="3" Body="&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a clock&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm the first, but also second&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the first to show the time and the seconds in the time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm rather odd, but also not&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the hour, minute and second can be odd numbers and also can be not odd &lt;em&gt;i.e.&lt;/em&gt; even&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The circle of three&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a clock is a circle with three hands&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastActivityDate="2016-01-06T13:16:36.320" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25110" PostTypeId="1" AcceptedAnswerId="26273" CreationDate="2016-01-06T15:34:34.100" Score="0" ViewCount="433" Body="&lt;p&gt;I need help for solving this puzzle.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OZapu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OZapu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody have an idea?&lt;/p&gt;&#xA;" OwnerUserId="18242" LastEditorUserId="29753" LastEditDate="2016-09-03T14:54:49.017" LastActivityDate="2016-09-03T14:54:49.017" Title="Solving a puzzle related to Mars Attacks" Tags="|cipher|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25112" PostTypeId="1" AcceptedAnswerId="25150" CreationDate="2016-01-06T17:00:36.250" Score="19" ViewCount="934" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm a strange creature,&lt;br&gt;&#xA;  Often misunderstood and moody,&lt;br&gt;I come in 2 'types',&lt;br&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A vast majority of my kind is part of the first type,&lt;br&gt;&#xA;  Energetic, enthusiastic and eager to learn&lt;br&gt;&#xA;  With this type comes short lives,&lt;br&gt;&#xA;  A decade at most,&lt;br&gt;&#xA;  But almost all of these will make it far enough for them to ripen and change flavor.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A tiny minority amongst us, 'type 2'&lt;br&gt;&#xA;  less than one in a thousand, is fundamentally different,&lt;br&gt;&#xA;  relatively well grounded and more mature,&#xA;  Even as young they feel superior, for they live lives far richer and longer,&lt;br&gt;&#xA;  Most of these live up to thirty years, some even up to forty.&lt;br&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now the odd thing is that amongst these long lived type twos, less of them will ripen and change flavor then of the first type. Altough some will try to claim that they were born with the right flavor already. But I insist that's not how it should work.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So what kind of creature are we describing,and what are the 2 'types'?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I figured i'd start giving out some hints, will add a few every 24h :P&lt;/p&gt;&#xA;&#xA;&lt;p&gt;hint 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some of these, regardless of type, would be offended if you called them a creature, prefering something that sounds a bit more civilized.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;hint 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Type 2 actually occurs 1 in 1506, altough most falsely estimate it to be 1 in 1461.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'd like to add, like I did in the comments, that if you have the right answer, you will know it. (And perhaps develop an urge to hit me)&lt;/p&gt;&#xA;" OwnerUserId="16846" LastEditorUserId="16846" LastEditDate="2016-01-07T13:17:16.803" LastActivityDate="2016-01-07T17:00:37.280" Title="What kind of creature am I?" Tags="|riddle|" AnswerCount="6" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25113" PostTypeId="2" ParentId="25100" CreationDate="2016-01-06T17:01:00.057" Score="1" Body="&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the hands of a clock&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm the first, but also second&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The second hand also the minute hand both go all the way around?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm rather odd, but also not&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The hour, minute and second can be odd numbers and also even&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The circle of three&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Minute, hour and second hands&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18244" LastActivityDate="2016-01-06T17:01:00.057" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25116" PostTypeId="2" ParentId="25100" CreationDate="2016-01-06T17:48:01.180" Score="5" Body="&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm the first, but also second&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first prime, but also the second (depending on convention: 1 used to be regarded as prime).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm rather odd, but also not&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2 is the oddest prime (it is even).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The circle of three&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ???&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="2819" LastEditDate="2016-01-06T22:41:32.250" LastActivityDate="2016-01-06T22:41:32.250" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25117" PostTypeId="1" AcceptedAnswerId="25118" CreationDate="2016-01-06T18:15:52.130" Score="6" ViewCount="267" Body="&lt;p&gt;Inspired by the succes of &lt;a href=&quot;https://puzzling.stackexchange.com/questions/25100/the-circle-of-three&quot;&gt;this puzzle&lt;/a&gt;, here is another riddle:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Smaller than the Great Graham!&lt;/p&gt;&#xA;&lt;p&gt;Smaller than the monster, but also much greater than the monster,&lt;/p&gt;&#xA;&lt;p&gt;am I equal to two-in-a-circle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; If you include the full answer, you will hit the character limit for answers (30000 characters). Please use an abbreviation. Yes, this is a clue.&lt;/p&gt;&#xA;" OwnerUserId="16322" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-01-06T18:33:43.460" Title="The circle of two" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25118" PostTypeId="2" ParentId="25117" CreationDate="2016-01-06T18:26:08.287" Score="5" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Steinhaus%E2%80%93Moser_notation#Mega&quot; rel=&quot;nofollow&quot;&gt;Mega&lt;/a&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Smaller than the Great Graham!&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Smaller than &lt;a href=&quot;https://en.wikipedia.org/wiki/Graham%27s_number&quot; rel=&quot;nofollow&quot;&gt;Graham's Number&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Smaller than the monster, but also much greater than the monster,&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;EDIT&lt;/strong&gt;: 'mega' is the prefix for 1000&lt;sup&gt;2&lt;/sup&gt;, 'tera', from the Greek for &quot;monster&quot; (thanks, @wythagoras!), is the prefix for 1000&lt;sup&gt;4&lt;/sup&gt;. So &lt;em&gt;a&lt;/em&gt; Mega is much greater than &lt;em&gt;tera-&lt;/em&gt; anything, lowercase-m mega is smaller than its tera counterpart.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;am I equal to two-in-a-circle.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; See link for first spoilertag&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2016-01-06T18:33:43.460" LastActivityDate="2016-01-06T18:33:43.460" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25119" PostTypeId="2" ParentId="25098" CreationDate="2016-01-06T19:09:13.800" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I don't know yet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I used a simple backtracking approach. The code is here: &lt;a href=&quot;https://gist.github.com/smaspe/bb7efc8e03e8eaa81629&quot; rel=&quot;nofollow&quot;&gt;https://gist.github.com/smaspe/bb7efc8e03e8eaa81629&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It basically tries to recursively fill the matrix and prunes the inconsistent branches as soon as 2 identical sums are found on a complete row or line.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it is currently running.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="310" LastEditorUserId="310" LastEditDate="2016-01-06T20:34:00.260" LastActivityDate="2016-01-06T20:34:00.260" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25120" PostTypeId="2" ParentId="24937" CreationDate="2016-01-06T19:59:04.210" Score="6" Body="&lt;p&gt;Additional progress:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having read f&quot;'s spoilers for the first three rows, I find that the fourth row is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;a Vigenere cipher with the key &lt;code&gt;BRIEF&lt;/code&gt;, reading (spaces added) &lt;code&gt;VITRIOL OF THE PRIME FOOL&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Furthermore, the fifth row is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;a Vigenere cipher with the key &lt;code&gt;SKULL&lt;/code&gt;, reading (spaces added) &lt;code&gt;SOMEWHERE ELSE IN OUR MIDST&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;El-Tham&lt;/code&gt; and &lt;code&gt;Merkand&lt;/code&gt; are, respectively, anagrams of &lt;code&gt;Hamlet&lt;/code&gt; and &lt;code&gt;Denmark&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The names Rosenkreuz and Guilderstein apparently invoke Rosicrucian and possibly Masonic elements in connection with alchemy.  Consider &lt;em&gt;Die warhhaffte und vollkommene Bereitung des Philosophischen Steins der Brüderschaft aus dem Orden des Gülden-und Rosen-Creutzes (The True and Complete Preparation of the Philosopher's Stone by the Brotherhood from the Order of the Golden and Rosy Cross)&lt;/em&gt;, for example.  See &lt;a href=&quot;https://en.wikipedia.org/wiki/Rosicrucianism&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Rosicrucianism&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I offer a solution&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Helsingør, known in Shakespeare's day as Elsinore (taken from &quot;somewhere &lt;strong&gt;else in our&lt;/strong&gt; midst&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18224" LastEditorUserId="18224" LastEditDate="2016-01-07T19:11:51.297" LastActivityDate="2016-01-07T19:11:51.297" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25121" PostTypeId="2" ParentId="23059" CreationDate="2016-01-06T20:21:29.977" Score="2" Body="&lt;p&gt;I've been looking in the wrong directions for months! Please feel free to suggest an edit.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Starting centre at the square, moving outwards clockwise, and separating each set of words by ~ (formerly black square) leads to the phrases below.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;encrypted &amp;rarr; decrypted (using letter substitution from frequency analysis and guessing)&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; 1. ZQH  &amp;rarr; AFK [away from keyboard]&lt;br&gt;&#xA; 2. XUFVKSV &amp;rarr; COVER ME&lt;br&gt;&#xA; 3. WUTIEUKKCNFVWUTVNIZIOUGJZTWINSVJ &amp;rarr; DONT WORRY IVE DONE IT A THOUSAND TIMES&lt;br&gt;&#xA; 4. NAAWNJIKZXIIOVS &amp;rarr; ILL DISTRACT THEM&lt;br&gt;&#xA; 5. NAAPUQNKJI &amp;rarr; ILL GO FIRST&lt;br&gt;&#xA; 6. NAATVFVKJGKKVTWVK &amp;rarr; ILL NEVER SURRENDER&lt;br&gt;&#xA; 7. AVISVOZTWAVIONJ &amp;rarr; LET ME HANDLE THIS&lt;br&gt;&#xA; 8. AVIJPUNT &amp;rarr; LETS GO IN&lt;br&gt;&#xA; 9. UURJ &amp;rarr; OOPS&lt;br&gt;&#xA; 10. KGTEEEEEEEEEEEEEEEEEEEEE &amp;rarr; RUNWWWWWWWWWWWWWWWWWWWWW [see Dave Andersen's comment]  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As for the creature's name...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Comparing plaintext to ciphertext alphabet, a sequence of letters that looks like a name appears in the latter. Instead of KLMNOP, we get&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;https://en.wikipedia.org/wiki/Hastur&quot; rel=&quot;nofollow&quot;&gt;HASTUR&lt;/a&gt; the Unspeakable One&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll pass on the riches.&lt;/p&gt;&#xA;" OwnerUserId="15921" LastEditorUserId="15921" LastEditDate="2016-01-06T21:02:09.573" LastActivityDate="2016-01-06T21:02:09.573" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25122" PostTypeId="1" AcceptedAnswerId="25124" CreationDate="2016-01-06T20:30:51.253" Score="53" ViewCount="4078" Body="&lt;p&gt;This is an ancient puzzle from an old book that I've never seen anywhere else. Can any of you puzzlers solve it?&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If the B MT put:&lt;br /&gt;&#xA;If the B. putting:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17827" LastEditorUserId="69582" LastEditDate="2021-09-15T15:44:18.820" LastActivityDate="2021-09-15T15:44:18.820" Title="''If the B...''" Tags="|wordplay|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="25123" PostTypeId="1" AcceptedAnswerId="25126" CreationDate="2016-01-06T20:31:57.590" Score="1" ViewCount="583" Body="&lt;p&gt;Try and guess the object I am describing:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;According to the public,&lt;br&gt;&#xA;      they describe me as 'bold'&lt;br&gt;&#xA;      though my job is to get hit,&lt;br&gt;&#xA;      by my 'clapping' partner&lt;br&gt;&#xA;      while I scream my pain.&lt;br&gt;&#xA;      I can be highly religious,&lt;br&gt;&#xA;      or I just hang around shops&lt;br&gt;&#xA;      jumping and shrieking whenever&lt;br&gt;&#xA;      a door opens.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I? This is quite cryptic so you'll need to think outside the box!&lt;/p&gt;&#xA;" OwnerUserId="18250" LastEditorUserId="5373" LastEditDate="2017-01-10T14:31:50.683" LastActivityDate="2017-01-10T14:31:50.683" Title="What am I? According to the public, they describe me as 'bold'" Tags="|riddle|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25124" PostTypeId="2" ParentId="25122" CreationDate="2016-01-06T20:37:04.413" Score="62" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the grate be (great B) empty, put coal on (or maybe in?) (colon);&#xA; if the grate be full stop putting coal in (on?).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="12628" LastEditDate="2018-07-26T09:01:25.983" LastActivityDate="2018-07-26T09:01:25.983" CommentCount="14" ContentLicense="CC BY-SA 4.0" />
  <row Id="25125" PostTypeId="2" ParentId="25123" CreationDate="2016-01-06T20:40:33.387" Score="3" Body="&lt;p&gt;You're...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a bell!&lt;br&gt;&#xA; Bells are supposed to be hit by their clappers. Many shops have bells on the doors to alert employees when a customer enters, and many churches use bells too.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2016-01-06T20:40:33.387" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25126" PostTypeId="2" ParentId="25123" CreationDate="2016-01-06T20:43:09.037" Score="8" Body="&lt;p&gt;I'm having trouble with the first clue, but:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You're a bell.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to the public,&#xA;they describe me as 'bold'&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bradley Bell was the executive producer of The Bold and the Beautiful?  Bell is also a font, which can be bold.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;though my job is to get hit,&#xA;by my 'clapping' partner&#xA;while I scream my pain.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The clapper hangs inside a bell and hits it to make it ring.  It's also called a tongue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be highly religious,&#xA;or I just hang around shops&#xA;jumping and shrieking whenever&#xA;a door opens.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  It can be a church bell, or the bells that ring to signify that a customer had entered the store.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13054" LastActivityDate="2016-01-06T20:43:09.037" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25127" PostTypeId="2" ParentId="25123" CreationDate="2016-01-06T20:43:25.730" Score="5" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a bell? Or more specifically, the 'cup' part of the bell?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;According to the public,&#xA;they describe me as 'bold'&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not really sure about that one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;though my job is to get hit,&#xA;by my 'clapping' partner&#xA;while I scream my pain.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bells are rung by the 'clapper' hitting the cup.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I can be highly religious,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bells are found in church bell towers, so religious, and literally high off the ground.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;or I just hang around shops&#xA;jumping and shrieking whenever&#xA;a door opens.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; And there are also bells that are attached to shop doors to ring when someone enters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11775" LastActivityDate="2016-01-06T20:43:25.730" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25128" PostTypeId="1" CreationDate="2016-01-06T21:49:15.987" Score="42" ViewCount="11758" Body="&lt;p&gt;The background to this one's given &lt;a href=&quot;https://www.damninteresting.com/89-263-201-500-337-480/&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;NB: These may not be a genuine cryptograms (current thinking is that these are gibberish); but I though it interesting to share these in case they're genuine / as they're probably of interest to this community.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;1: THE LOCALITY OF THE VAULT.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;71, 194, 38, 1701, 89, 76, 11, 83, 1629, 48, 94, 63, 132, 16, 111, 95, 84, 341, 975, 14, 40, 64, 27, 81, 139, 213, 63, 90, 1120, 8, 15, 3, 126, 2018, 40, 74, 758, 485, 604, 230, 436, 664, 582, 150, 251, 284, 308, 231, 124, 211, 486, 225, 401, 370, 11, 101, 305, 139, 189, 17, 33, 88, 208, 193, 145, 1, 94, 73, 416, 918, 263, 28, 500, 538, 356, 117, 136, 219, 27, 176, 130, 10, 460, 25, 485, 18, 436, 65, 84, 200, 283, 118, 320, 138, 36, 416, 280, 15, 71, 224, 961, 44, 16, 401, 39, 88, 61, 304, 12, 21, 24, 283, 134, 92, 63, 246, 486, 682, 7, 219, 184, 360, 780, 18, 64, 463, 474, 131, 160, 79, 73, 440, 95, 18, 64, 581, 34, 69, 128, 367, 460, 17, 81, 12, 103, 820, 62, 116, 97, 103, 862, 70, 60, 1317, 471, 540, 208, 121, 890, 346, 36, 150, 59, 568, 614, 13, 120, 63, 219, 812, 2160, 1780, 99, 35, 18, 21, 136, 872, 15, 28, 170, 88, 4, 30, 44, 112, 18, 147, 436, 195, 320, 37, 122, 113, 6, 140, 8, 120, 305, 42, 58, 461, 44, 106, 301, 13, 408, 680, 93, 86, 116, 530, 82, 568, 9, 102, 38, 416, 89, 71, 216, 728, 965, 818, 2, 38, 121, 195, 14, 326, 148, 234, 18, 55, 131, 234, 361, 824, 5, 81, 623, 48, 961, 19, 26, 33, 10, 1101, 365, 92, 88, 181, 275, 346, 201, 206, 86, 36, 219, 324, 829, 840, 64, 326, 19, 48, 122, 85, 216, 284, 919, 861, 326, 985, 233, 64, 68, 232, 431, 960, 50, 29, 81, 216, 321, 603, 14, 612, 81, 360, 36, 51, 62, 194, 78, 60, 200, 314, 676, 112, 4, 28, 18, 61, 136, 247, 819, 921, 1060, 464, 895, 10, 6, 66, 119, 38, 41, 49, 602, 423, 962, 302, 294, 875, 78, 14, 23, 111, 109, 62, 31, 501, 823, 216, 280, 34, 24, 150, 1000, 162, 286, 19, 21, 17, 340, 19, 242, 31, 86, 234, 140, 607, 115, 33, 191, 67, 104, 86, 52, 88, 16, 80, 121, 67, 95, 122, 216, 548, 96, 11, 201, 77, 364, 218, 65, 667, 890, 236, 154, 211, 10, 98, 34, 119, 56, 216, 119, 71, 218, 1164, 1496, 1817, 51, 39, 210, 36, 3, 19, 540, 232, 22, 141, 617, 84, 290, 80, 46, 207, 411, 150, 29, 38, 46, 172, 85, 194, 39, 261, 543, 897, 624, 18, 212, 416, 127, 931, 19, 4, 63, 96, 12, 101, 418, 16, 140, 230, 460, 538, 19, 27, 88, 612, 1431, 90, 716, 275, 74, 83, 11, 426, 89, 72, 84, 1300, 1706, 814, 221, 132, 40, 102, 34, 868, 975, 1101, 84, 16, 79, 23, 16, 81, 122, 324, 403, 912, 227, 936, 447, 55, 86, 34, 43, 212, 107, 96, 314, 264, 1065, 323, 428, 601, 203, 124, 95, 216, 814, 2906, 654, 820, 2, 301, 112, 176, 213, 71, 87, 96, 202, 35, 10, 2, 41, 17, 84, 221, 736, 820, 214, 11, 60, 760&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;2: THE CONTENTS OF THE TREASURE.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;115, 73, 24, 807, 37, 52, 49, 17, 31, 62, 647, 22, 7, 15, 140, 47, 29, 107, 79, 84, 56, 239, 10, 26, 811, 5, 196, 308, 85, 52, 160, 136, 59, 211, 36, 9, 46, 316, 554, 122, 106, 95, 53, 58, 2, 42, 7, 35, 122, 53, 31, 82, 77, 250, 196, 56, 96, 118, 71, 140, 287, 28, 353, 37, 1005, 65, 147, 807, 24, 3, 8, 12, 47, 43, 59, 807, 45, 316, 101, 41, 78, 154, 1005, 122, 138, 191, 16, 77, 49, 102, 57, 72, 34, 73, 85, 35, 371, 59, 196, 81, 92, 191, 106, 273, 60, 394, 620, 270, 220, 106, 388, 287, 63, 3, 6, 191, 122, 43, 234, 400, 106, 290, 314, 47, 48, 81, 96, 26, 115, 92, 158, 191, 110, 77, 85, 197, 46, 10, 113, 140, 353, 48, 120, 106, 2, 607, 61, 420, 811, 29, 125, 14, 20, 37, 105, 28, 248, 16, 159, 7, 35, 19, 301, 125, 110, 486, 287, 98, 117, 511, 62, 51, 220, 37, 113, 140, 807, 138, 540, 8, 44, 287, 388, 117, 18, 79, 344, 34, 20, 59, 511, 548, 107, 603, 220, 7, 66, 154, 41, 20, 50, 6, 575, 122, 154, 248, 110, 61, 52, 33, 30, 5, 38, 8, 14, 84, 57, 540, 217, 115, 71, 29, 84, 63, 43, 131, 29, 138, 47, 73, 239, 540, 52, 53, 79, 118, 51, 44, 63, 196, 12, 239, 112, 3, 49, 79, 353, 105, 56, 371, 557, 211, 505, 125, 360, 133, 143, 101, 15, 284, 540, 252, 14, 205, 140, 344, 26, 811, 138, 115, 48, 73, 34, 205, 316, 607, 63, 220, 7, 52, 150, 44, 52, 16, 40, 37, 158, 807, 37, 121, 12, 95, 10, 15, 35, 12, 131, 62, 115, 102, 807, 49, 53, 135, 138, 30, 31, 62, 67, 41, 85, 63, 10, 106, 807, 138, 8, 113, 20, 32, 33, 37, 353, 287, 140, 47, 85, 50, 37, 49, 47, 64, 6, 7, 71, 33, 4, 43, 47, 63, 1, 27, 600, 208, 230, 15, 191, 246, 85, 94, 511, 2, 270, 20, 39, 7, 33, 44, 22, 40, 7, 10, 3, 811, 106, 44, 486, 230, 353, 211, 200, 31, 10, 38, 140, 297, 61, 603, 320, 302, 666, 287, 2, 44, 33, 32, 511, 548, 10, 6, 250, 557, 246, 53, 37, 52, 83, 47, 320, 38, 33, 807, 7, 44, 30, 31, 250, 10, 15, 35, 106, 160, 113, 31, 102, 406, 230, 540, 320, 29, 66, 33, 101, 807, 138, 301, 316, 353, 320, 220, 37, 52, 28, 540, 320, 33, 8, 48, 107, 50, 811, 7, 2, 113, 73, 16, 125, 11, 110, 67, 102, 807, 33, 59, 81, 158, 38, 43, 581, 138, 19, 85, 400, 38, 43, 77, 14, 27, 8, 47, 138, 63, 140, 44, 35, 22, 177, 106, 250, 314, 217, 2, 10, 7, 1005, 4, 20, 25, 44, 48, 7, 26, 46, 110, 230, 807, 191, 34, 112, 147, 44, 110, 121, 125, 96, 41, 51, 50, 140, 56, 47, 152, 540, 63, 807, 28, 42, 250, 138, 582, 98, 643, 32, 107, 140, 112, 26, 85, 138, 540, 53, 20, 125, 371, 38, 36, 10, 52, 118, 136, 102, 420, 150, 112, 71, 14, 20, 7, 24, 18, 12, 807, 37, 67, 110, 62, 33, 21, 95, 220, 511, 102, 811, 30, 83, 84, 305, 620, 15, 2, 10, 8, 220, 106, 353, 105, 106, 60, 275, 72, 8, 50, 205, 185, 112, 125, 540, 65, 106, 807, 138, 96, 110, 16, 73, 33, 807, 150, 409, 400, 50, 154, 285, 96, 106, 316, 270, 205, 101, 811, 400, 8, 44, 37, 52, 40, 241, 34, 205, 38, 16, 46, 47, 85, 24, 44, 15, 64, 73, 138, 807, 85, 78, 110, 33, 420, 505, 53, 37, 38, 22, 31, 10, 110, 106, 101, 140, 15, 38, 3, 5, 44, 7, 98, 287, 135, 150, 96, 33, 84, 125, 807, 191, 96, 511, 118, 40, 370, 643, 466, 106, 41, 107, 603, 220, 275, 30, 150, 105, 49, 53, 287, 250, 208, 134, 7, 53, 12, 47, 85, 63, 138, 110, 21, 112, 140, 485, 486, 505, 14, 73, 84, 575, 1005, 150, 200, 16, 42, 5, 4, 25, 42, 8, 16, 811, 125, 160, 32, 205, 603, 807, 81, 96, 405, 41, 600, 136, 14, 20, 28, 26, 353, 302, 246, 8, 131, 160, 140, 84, 440, 42, 16, 811, 40, 67, 101, 102, 194, 138, 205, 51, 63, 241, 540, 122, 8, 10, 63, 140, 47, 48, 140, 288&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;3: NAMES AND RESIDENCES.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;317, 8, 92, 73, 112, 89, 67, 318, 28, 96,107, 41, 631, 78, 146, 397, 118, 98, 114, 246, 348, 116, 74, 88, 12, 65, 32, 14, 81, 19, 76, 121, 216, 85, 33, 66, 15, 108, 68, 77, 43, 24, 122, 96, 117, 36, 211, 301, 15, 44, 11, 46, 89, 18, 136, 68, 317, 28, 90, 82, 304, 71, 43, 221, 198, 176, 310, 319, 81, 99, 264, 380, 56, 37, 319, 2, 44, 53, 28, 44, 75, 98, 102, 37, 85, 107, 117, 64, 88, 136, 48, 151, 99, 175, 89, 315, 326, 78, 96, 214, 218, 311, 43, 89, 51, 90, 75, 128, 96, 33, 28, 103, 84, 65, 26, 41, 246, 84, 270, 98, 116, 32, 59, 74, 66, 69, 240, 15, 8, 121, 20, 77, 89, 31, 11, 106, 81, 191, 224, 328, 18, 75, 52, 82, 117, 201, 39, 23, 217, 27, 21, 84, 35, 54, 109, 128, 49, 77, 88, 1, 81, 217, 64, 55, 83, 116, 251, 269, 311, 96, 54, 32, 120, 18, 132, 102, 219, 211, 84, 150, 219, 275, 312, 64, 10, 106, 87, 75, 47, 21, 29, 37, 81, 44, 18, 126, 115, 132, 160, 181, 203, 76, 81, 299, 314, 337, 351, 96, 11, 28, 97, 318, 238, 106, 24, 93, 3, 19, 17, 26, 60, 73, 88, 14, 126, 138, 234, 286, 297, 321, 365, 264, 19, 22, 84, 56, 107, 98, 123, 111, 214, 136, 7, 33, 45, 40, 13, 28, 46, 42, 107, 196, 227, 344, 198, 203, 247, 116, 19, 8, 212, 230, 31, 6, 328, 65, 48, 52, 59, 41, 122, 33, 117, 11, 18, 25, 71, 36, 45, 83, 76, 89, 92, 31, 65, 70, 83, 96, 27, 33, 44, 50, 61, 24, 112, 136, 149, 176, 180, 194, 143, 171, 205, 296, 87, 12, 44, 51, 89, 98, 34, 41, 208, 173, 66, 9, 35, 16, 95, 8, 113, 175, 90, 56, 203, 19, 177, 183, 206, 157, 200, 218, 260, 291, 305, 618, 951, 320, 18, 124, 78, 65, 19, 32, 124, 48, 53, 57, 84, 96, 207, 244, 66, 82, 119, 71, 11, 86, 77, 213, 54, 82, 316, 245, 303, 86, 97, 106, 212, 18, 37, 15, 81, 89, 16, 7, 81, 39, 96, 14, 43, 216, 118, 29, 55, 109, 136, 172, 213, 64, 8, 227, 304, 611, 221, 364, 819, 375, 128, 296, 1, 18, 53, 76, 10, 15, 23, 19, 71, 84, 120, 134, 66, 73, 89, 96, 230, 48, 77, 26, 101, 127, 936, 218, 439, 178, 171, 61, 226, 313, 215, 102, 18, 167, 262, 114, 218, 66, 59, 48, 27, 19, 13, 82, 48, 162, 119, 34, 127, 139, 34, 128, 129, 74, 63, 120, 11, 54, 61, 73, 92, 180, 66, 75, 101, 124, 265, 89, 96, 126, 274, 896, 917, 434, 461, 235, 890, 312, 413, 328, 381, 96, 105, 217, 66, 118, 22, 77, 64, 42, 12, 7, 55, 24, 83, 67, 97, 109, 121, 135, 181, 203, 219, 228, 256, 21, 34, 77, 319, 374, 382, 675, 684, 717, 864, 203, 4, 18, 92, 16, 63, 82, 22, 46, 55, 69, 74, 112, 134, 186, 175, 119, 213, 416, 312, 343, 264, 119, 186, 218, 343, 417, 845, 951, 124, 209, 49, 617, 856, 924, 936, 72, 19, 28, 11, 35, 42, 40, 66, 85, 94, 112, 65, 82, 115, 119, 236, 244, 186, 172, 112, 85, 6, 56, 38, 44, 85, 72, 32, 47, 63, 96, 124, 217, 314, 319, 221, 644, 817, 821, 934, 922, 416, 975, 10, 22, 18, 46, 137, 181, 101, 39, 86, 103, 116, 138, 164, 212, 218, 296, 815, 380, 412, 460, 495, 675, 820, 952&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;( Puzzles copied from / full pamphlet available at: &lt;a href=&quot;http://www.unmuseum.org/bealepap.htm&quot; rel=&quot;noreferrer&quot;&gt;http://www.unmuseum.org/bealepap.htm&lt;/a&gt; )&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;The second cipher has been &lt;a href=&quot;https://en.wikipedia.org/wiki/Beale_ciphers#Deciphered_message&quot; rel=&quot;noreferrer&quot;&gt;solved&lt;/a&gt;. The Wikipedia link on ‘solved’ has info about how it was done.&lt;/p&gt;&#xA;" OwnerUserId="1681" LastEditorUserId="-1" LastEditDate="2020-06-21T07:56:42.010" LastActivityDate="2023-05-04T09:25:21.463" Title="The Beale Papers" Tags="|cryptography|history|real|cipher|unsolved-mysteries|" AnswerCount="1" CommentCount="23" ContentLicense="CC BY-SA 4.0" />
  <row Id="25129" PostTypeId="2" ParentId="25039" CreationDate="2016-01-06T22:16:26.340" Score="6" Body="&lt;p&gt;If we allow for implicit multiplication of parenthesized expressions then the following solutions, all of &lt;strong&gt;length 20&lt;/strong&gt;, become possible&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $10\times 9\times 8\times 7(6\div 5-4+3) 2\times 1$&lt;br&gt;&#xA; $10\times 9\times 8\times 7 (6\div 5-4+3) 2\div 1$&lt;br&gt;&#xA; $10\times 9\times 8 (7-6\div 5-4+3-2) 1$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This list has been generated via exhaustive search, and excludes needlessly parenthesizing expressions that only contain multiplication or division.&lt;/p&gt;&#xA;" OwnerUserId="13498" LastEditorUserId="13498" LastEditDate="2016-01-07T02:59:25.083" LastActivityDate="2016-01-07T02:59:25.083" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25130" PostTypeId="1" AcceptedAnswerId="25160" CreationDate="2016-01-06T22:18:08.473" Score="10" ViewCount="369" Body="&lt;p&gt;To add to the recent spate (should there be a who-am-i tag?):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I come from a brook.&lt;br&gt;&#xA;I'm first in a book.&lt;br&gt;&#xA;I'm old, but not stale.&lt;br&gt;&#xA;From me was made hail.&lt;br&gt;&#xA;I sound like an ocean,&lt;br&gt;&#xA;perpetual motion.&lt;br&gt;&#xA;I'm not quite a kernel,&lt;br&gt;&#xA;nor am I nocturnal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: I have modified the title and added the &lt;a href=&quot;/questions/tagged/music&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;music&amp;#39;&quot; rel=&quot;tag&quot;&gt;music&lt;/a&gt; tag.&lt;/p&gt;&#xA;" OwnerUserId="18224" LastEditorUserId="18224" LastEditDate="2016-01-07T19:01:46.783" LastActivityDate="2016-01-07T20:45:38.270" Title="Wer bin ich? (was: Another who am I riddle)" Tags="|riddle|music|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25132" PostTypeId="2" ParentId="25130" CreationDate="2016-01-07T00:56:49.197" Score="2" Body="&lt;p&gt;My answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water.&lt;br&gt;&#xA;&lt;br&gt;&#xA; It comes from a brook, water is in the water cycle so it's old but not stale. Hail comes from water. Oceans are made of water so it will sound like Oceans. Water is in perpetual motion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18169" LastEditorUserId="4421" LastEditDate="2016-01-07T02:42:04.733" LastActivityDate="2016-01-07T02:42:04.733" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25133" PostTypeId="2" ParentId="25112" CreationDate="2016-01-07T03:15:04.427" Score="2" Body="&lt;p&gt;I'll venture a speculation that, while not a full answer, may give another puzzler some ideas if I happen to be on the right track. I'm not familiar enough to know the specifics, but perhaps someone else is...? (&lt;strong&gt;Edit&lt;/strong&gt;: Never mind, apparently I'm off the mark!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the answer might be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a kind of fine alcohol.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm a strange creature,&lt;br&gt;&#xA;Often misunderstood and moody,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This particular drink might be perceived as being for a certain type of moody person (the drunk or the recluse, or stiff high society). Alternatively, alcohol makes people moody.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come in 2 'types',&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Perhaps the two types are distinguished by base ingredients, as with (for instance) malt versus grain whiskey. Or, it could be a difference of process.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A vast majority of my kind is part of the first type,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most alcohol is the relatively cheaper variety.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Energetic, enthusiastic and eager to learn&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bubbly, perhaps, or this might be a description of its flavor. Eager to learn might mean it adopts the flavors of its cask.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With this type comes short lives,&lt;br&gt;&#xA;A decade at most,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This would be the length of the aging process.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But almost all of these will make it far enough for them to ripen and change flavor.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are &quot;eager to learn&quot; and quickly change flavor as they age.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A tiny minority amongst us, 'type 2'&lt;br&gt;&#xA;less than one in a thousand, is fundamentally different,&lt;br&gt;&#xA;relatively well grounded and more mature, Even as young they feel superior, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The superior alcohol is more expensive and less often produced, but is innately better quality from the start.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;for they live lives far richer and longer,&lt;br&gt;&#xA;Most of these live up to thirty years, some even up to forty.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This kind is best/worth more if aged longer, so the aging process is usually lengthy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now the odd thing is that amongst these long lived type twos, less of them will ripen and change flavor then of the first type. Although some will try to claim that they were born with the right flavor already. But I insist that's not how it should work.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This type might subtly increase in quality as it ages, but it does not necessarily absorb new flavors (depending on the method used?). There is some debate over whether this leads to a better result, or if it should in fact be aged in such a way that the flavor profile changes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And of course, as a final clue, there's the OP's screen name... ;)&lt;/p&gt;&#xA;" OwnerUserId="12398" LastEditorUserId="12398" LastEditDate="2016-01-07T17:00:37.280" LastActivityDate="2016-01-07T17:00:37.280" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25134" PostTypeId="1" CreationDate="2016-01-07T03:42:49.223" Score="4" ViewCount="2088" Body="&lt;p&gt;A man has:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;2 legs in the morning&lt;/li&gt;&#xA;&lt;li&gt;3 legs in the afternoon&lt;/li&gt;&#xA;&lt;li&gt;4 legs in the evening.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Explain how this can be. The numbers are not &lt;em&gt;completely&lt;/em&gt; at random - these is some logic to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HINT&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Construction&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11497" LastEditorUserId="11497" LastEditDate="2016-01-07T15:18:37.607" LastActivityDate="2016-01-07T15:18:37.607" Title="A man with legs puzzle" Tags="|riddle|" AnswerCount="5" CommentCount="15" ClosedDate="2016-01-07T14:54:22.250" ContentLicense="CC BY-SA 3.0" />
  <row Id="25136" PostTypeId="2" ParentId="25032" CreationDate="2016-01-07T06:43:06.860" Score="5" Body="&lt;p&gt;I wrote a little proramm to check the numbers from one to one million for this property and there turn out to be 817 such numbers. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I also checked it for German and due to äöüß there is no such number in the first million besides 1, so I replaced them with ae,oe,ue,ss and 721. With the smallest being 12,467, just one below the English one. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The smallest common number ($\neq 1$) in English and German is 412,678.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a little plot of them:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HgypL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HgypL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the meantime I also calculated the smallest German quasi-pangrammatic number using the correct spelling (with äöüß) and it turns out to be 4,512,637&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5589" LastEditorUserId="5589" LastEditDate="2016-01-08T08:05:28.007" LastActivityDate="2016-01-08T08:05:28.007" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25137" PostTypeId="2" ParentId="25134" CreationDate="2016-01-07T07:24:21.757" Score="15" Body="&lt;p&gt;&lt;strong&gt;A lateral thinking&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A man has:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 legs in the morning&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;His own 2 legs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3 legs in the afternoon&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;He bought a fried drumstick chicken leg from the KFC in the noon. So he has 3 (his two + 1 chicken leg) legs in the afternoon. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4 legs in the evening.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;On his way home in the evening, he bought another chicken leg from KFC for his wife. So, now he has 4 (his 2 + 2 chicken legs) legs in the evening.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastActivityDate="2016-01-07T07:24:21.757" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25138" PostTypeId="2" ParentId="25134" CreationDate="2016-01-07T09:07:06.520" Score="10" Body="&lt;p&gt;A man has:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2 legs in the morning:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (morning means his younger stage) &lt;code&gt;His own 2 legs&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3 legs in the afternoon&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;(afternoon means his aged stage, he can't walk without his walking stick) His own 2 legs and a walking stick`&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4 legs in the evening.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (evening means he dies, now he lie in bed, bed has 4 legs) &lt;code&gt;Legs of bed&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18263" LastEditorUserId="18263" LastEditDate="2016-01-07T09:32:07.477" LastActivityDate="2016-01-07T09:32:07.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25139" PostTypeId="2" ParentId="25112" CreationDate="2016-01-07T09:45:27.290" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Isotopes of Uranium (or some other element)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18264" LastActivityDate="2016-01-07T09:45:27.290" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25140" PostTypeId="2" ParentId="8111" CreationDate="2016-01-07T10:03:16.693" Score="11" Body="&lt;p&gt;This is 12 points:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;. . . .&#xA;. . . .&#xA;    . .&#xA;    . .&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Cutting it diagonally from top right to bottom left, we get two sets of 7 points:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;. . . .&#xA;. . . &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  .&#xA;. .&#xA;. .&#xA;. .&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="18117" LastActivityDate="2016-01-07T10:03:16.693" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25141" PostTypeId="2" ParentId="25134" CreationDate="2016-01-07T10:10:39.590" Score="3" Body="&lt;p&gt;A man has:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 legs in the morning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; His own 2 legs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3 legs in the afternoon:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He broke a leg and now he is walking with a crutch&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4 legs in the evening:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He broke another leg and now he is walking with two crutches&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18191" LastActivityDate="2016-01-07T10:10:39.590" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25142" PostTypeId="2" ParentId="25134" CreationDate="2016-01-07T10:33:12.047" Score="6" Body="&lt;p&gt;The man:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; is a distance runner/cyclist, and has entered a competition with various stages, split up into &quot;legs&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2 legs in the morning, 3 legs in the afternoon, 4 legs in the evening:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;is simply his schedule for the race.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9444" LastActivityDate="2016-01-07T10:33:12.047" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25143" PostTypeId="2" ParentId="25112" CreationDate="2016-01-07T11:12:06.183" Score="2" Body="&lt;p&gt;It could be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ants - split into queens and workers. The ages and proportions match roughly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alternatively:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Something about jobs and education? Not so confident!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12462" LastEditorUserId="12462" LastEditDate="2016-01-07T14:26:22.003" LastActivityDate="2016-01-07T14:26:22.003" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25144" PostTypeId="2" ParentId="25112" CreationDate="2016-01-07T11:30:23.047" Score="1" Body="&lt;p&gt;I actually think I got the right idea:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cheese! I don't know too much about the different types of cheese but this could definitely be true if you ask me. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If that is not the case then:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I have thought about horse/mule but that would include a donkey too, I also thought about lemons, they can be green or yellow but I can't really get all of it to match up. Then I thought of bees, but they don't seem to match up with their life expectancy. I think that there's a big chance that it's a kind of animal/fruit but I can't figure out which.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;New try!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Toad / Frog&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;AHA! I GOT IT NOW&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Peafowl!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18191" LastEditorUserId="18191" LastEditDate="2016-01-07T12:16:19.350" LastActivityDate="2016-01-07T12:16:19.350" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="25145" PostTypeId="2" ParentId="25112" CreationDate="2016-01-07T12:06:46.503" Score="2" Body="&lt;p&gt;My guess is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; B cells&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Type one &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Plasma cells, which are eager to &quot;learn&quot; how to fight intruders&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Type two &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Memory B cells, which create long lasting immunity &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &quot;ripening and maturing&quot; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; obviously happens after differentiation, so probably the activation of the B cells? Not so sure about this one. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am not certain about the actual life span involved, but Type 2 is certainly much longer lived than Type 1. Although literal 10 years sounds unlikely for the type 1 in my guess. &lt;/p&gt;&#xA;" OwnerUserId="2811" LastActivityDate="2016-01-07T12:06:46.503" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25146" PostTypeId="1" AcceptedAnswerId="25152" CreationDate="2016-01-07T13:00:26.323" Score="5" ViewCount="631" Body="&lt;p&gt;My first two became the chandler's downfall and woe&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My last two include a fiddle, flute and cello.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put it all together and what do you get?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An evil woman, who can turn to stone any living thing with her strange magic.&lt;/p&gt;&#xA;" OwnerUserId="14233" LastEditorUserId="14233" LastEditDate="2016-01-07T13:25:13.407" LastActivityDate="2016-01-12T12:04:58.540" Title="What is this entity?" Tags="|riddle|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25147" PostTypeId="2" ParentId="25134" CreationDate="2016-01-07T14:12:54.183" Score="4" Body="&lt;p&gt;2 legs in the morning&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; His own 2 legs, walking around, going to work, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3 legs in the afternoon&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; On the way home from work, he stops at the bar and has a few beers while sitting on a 3-legged bar stool.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4 legs in the evening&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Now he's tired, and he goes to bed. The bed has 4 legs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="375" LastActivityDate="2016-01-07T14:12:54.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25148" PostTypeId="2" ParentId="25146" CreationDate="2016-01-07T14:20:53.997" Score="0" Body="&lt;p&gt;This is mostly a wild guess that I can't justify to the clues, but I don't know enough about the subject to be sure it's wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gorgonzola?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Trying to justify it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The last hint made me think Medusa, who's a gorgon. I searched Medusa and Chandler and there is some mention of Chandler within a book called the &quot;Gorgon's Gaze&quot;, and I didn't know if &quot;zo&quot; &quot;la&quot; could be musical terms (with &quot;la&quot; at least being a note).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12462" LastEditorUserId="12462" LastEditDate="2016-01-12T12:04:58.540" LastActivityDate="2016-01-12T12:04:58.540" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25149" PostTypeId="1" AcceptedAnswerId="25151" CreationDate="2016-01-07T14:53:07.680" Score="10" ViewCount="1038" Body="&lt;p&gt;Is there a valid set of moves in Othello that allows a win score of 64-0 for either player?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And is there record of any such game in the past?&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="51131" LastEditDate="2019-09-11T14:17:27.020" LastActivityDate="2019-09-11T14:17:27.020" Title="Clear board in Othello" Tags="|strategy|game|board-games|reversi|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25150" PostTypeId="2" ParentId="25112" CreationDate="2016-01-07T14:54:56.737" Score="12" Body="&lt;p&gt;From your clues,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1:1461 and 1:1506 are ratios attributed to &lt;em&gt;the &lt;a href=&quot;http://www.mystro.com/leapdays.htm&quot;&gt;chance of being born&lt;/a&gt; &lt;a href=&quot;https://blogs.loc.gov/loc/2012/02/to-borrow-from-google-%E2%80%A6-and-rossini-%E2%80%A6-and-the-cosmos/&quot;&gt;on a Leap Day&lt;/a&gt;&lt;/em&gt; (i.e., February 29th),&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm gonna say the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Teenagers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People are teens (depending on your definition) somewhere in the range of age 10 to age 20 (technically 13-19, since those are the numbers with &quot;teen&quot; in them). Thus, their &quot;lifespan&quot; (as teens) is a decade or less. People born on a Leap Day (&quot;type 2&quot;) only celebrate their birthdays once every four years, and don't even &lt;em&gt;become&lt;/em&gt; teens (by that logic) until they are actually 40 or 44, and—again, by that logic—remain teens until they are 80.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Those in the non-Leap-Birthday group (&quot;type 1&quot;) &lt;em&gt;start&lt;/em&gt; their teenhood younger, and change more (mentally, but moreso physically) in those years. Those in the latter group have already gone through many of the changes mentally and all of the changes physically once they reach their Leap-teenhood.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2016-01-07T16:20:56.680" LastActivityDate="2016-01-07T16:20:56.680" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="25151" PostTypeId="2" ParentId="25149" CreationDate="2016-01-07T15:01:25.880" Score="13" Body="&lt;p&gt;Yes, I found this video on YouTube that has this &quot;perfect game&quot;. Terrible music by the way&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=prWG1OFgVqg&quot;&gt;https://www.youtube.com/watch?v=prWG1OFgVqg&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2016-01-07T15:01:25.880" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25152" PostTypeId="2" ParentId="25146" CreationDate="2016-01-07T16:26:00.873" Score="12" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Electric Light Orchestra&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first two&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Electric Lights made chandlery less important and eventually completely unneeded&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The last two&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Light Orchestra; indicates light orchestral music, which often has fiddlers, flautists, and cellists&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And lastly&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Evil Woman,&quot; &quot;Turn to Stone,&quot; &quot;Livin' Thing,&quot; and &quot;Strange Magic&quot; can be found in the discography of &lt;a href=&quot;https://en.wikipedia.org/wiki/Electric_Light_Orchestra&quot;&gt;Electric Light Orchestra&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17147" LastActivityDate="2016-01-07T16:26:00.873" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25153" PostTypeId="2" ParentId="25095" CreationDate="2016-01-07T16:50:59.477" Score="8" Body="&lt;p&gt;My physics degree is useful! Did have to google 'Hebrew Jacob', though&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_microwave_background&quot;&gt;Cosmic Microwave Background&lt;/a&gt;, or CMB&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Through the horn that persistent noise speaks to me  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;a href=&quot;https://en.wikipedia.org/wiki/Holmdel_Horn_Antenna&quot;&gt;Horn antenna&lt;/a&gt; that Penzias and Wilson accidentally discovered the CMB with in 1964  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;It tells me, by Greek A to C, that we are not here eternally  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Mostly 'Greek C' - a lot of astronomical measurements make use of gamma radiation. Radiating objects are, by necessity, decaying in some way, so aren't going to be around forever.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Then Hebrew Jacob seals the deal, en route to comb the silence  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_Background_Explorer&quot;&gt;COBE satellite&lt;/a&gt; (COsmic Background Explorer) launched in 1989 to investigate the CMB (comb the silence of space?)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;What once was thought is wrong, it's in fact not balanced  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It was initially assumed that the CMB should be isotropic (and indeed that the universe as a whole should be). COBE confirmed a slight anisotropy consistent with models of the big bang theory (the start of the universe, not the TV show)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11717" LastActivityDate="2016-01-07T16:50:59.477" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25154" PostTypeId="2" ParentId="25004" CreationDate="2016-01-07T17:52:08.270" Score="4" Body="&lt;p&gt;It may not be the right answer, but it more or less fits on all accounts.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;ice&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can hold the heaviest of all objects,&#xA;but will falter to the lightest of all objects.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is tough and withstands heavy loads. However, lightest of all objects can be interpreted as fire (also in the Classical 4-element model), which defeats ice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I prefer long and skinny rather than short and thick&#xA;since it is to my taste&#xA;which you cannot sense&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It forms icicles -- long and skinny. It's also tasteless.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can see me as far the eye can see&#xA;in the seven seas&#xA;or the seven continents&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All the continents are sometimes covered in ice. It's also in the highest clouds. Layers of ice also cover oceans.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Precious by nature&#xA;yet never rare&#xA;I seem to be more desired than gold&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This one is a stretch. Due to global warming, we are very protective of the polar caps.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are here because of me&#xA;and I am certainly not here because of you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Evolution of man occured after the latest ice age, so we are here because of ice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a group, I can be a force to reckon with&#xA;yet I can be gentler than a breeze&#xA;as every organism adores me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Huge boulders of ice (or hail) are dangerous. Gentle snow, on the other hand... is adorable :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5726" LastActivityDate="2016-01-07T17:52:08.270" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25155" PostTypeId="2" ParentId="25130" CreationDate="2016-01-07T18:47:41.790" Score="1" Body="&lt;p&gt;Ok at first I thought water but they realised that was wrong, so I have come up with a new answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    You're a cloud&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come from a brook.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    Clouds are made of water, (well water vapour)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm first in a book.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    Not sure, maybe something to do with being white? White page, white cloud?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm old but not stale.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    Not sure again, but I suppose there has always been clouds so that makes them old.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From me was made hail.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    Well hail comes out of clouds were they are made&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I sound like an ocean.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    Storm cloud (Thunder sounding like waves) or something to do with rain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Perpetual motion.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    I suppose clouds are always on the move&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not quite a kernel,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    not sure, maybe the sun is the kernel and clouds merely part of the scenery?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;nor am I nocturnal.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    You can't see clouds at night.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Am I right?&lt;/p&gt;&#xA;" OwnerUserId="18250" LastEditorUserId="16322" LastEditDate="2016-01-07T20:45:38.270" LastActivityDate="2016-01-07T20:45:38.270" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25156" PostTypeId="2" ParentId="25130" CreationDate="2016-01-07T18:59:53.500" Score="3" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a &lt;strong&gt;tree&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come from a brook.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water is on of main life sources of trees, and brook is a source of water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm first in a book.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A book was a tree at first, while it's produced from lumber. So tree is first step of a book in its history.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm old, but not stale.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Trees replenish themselves every year (at springs), even if they are very old.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From me was made hail.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Trees improve possibilty to rain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I sound like an ocean,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If wind blows, trembling leaves of trees sound like ocean.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;perpetual motion.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A tree dies every winter, replenishes every spring. Every year, until it cut down by a human.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not quite a kernel,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Trees multiply via and grow from kernels, by they are not kernels when they grow.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;nor am I nocturnal.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Trees can't make photosynthesis at night. They are more active at days.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18128" LastActivityDate="2016-01-07T18:59:53.500" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25157" PostTypeId="2" ParentId="25004" CreationDate="2016-01-07T19:09:39.260" Score="2" Body="&lt;p&gt;I think that the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Space&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can hold the heaviest of all objects, &#xA;but will falter to the lightest of all objects.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Space have galaxies. And if you put something into the space, then the space is covered&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I prefer long and skinny rather than short and thick &#xA;since it is to my taste &#xA;which you cannot sense&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The galaxies are spiral in shape which is long (not short). Space is vacuum. So there is no taste&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can see me as far the eye can see &#xA;in the seven seas &#xA;or the seven continents&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We can see the sky from anywhere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Precious by nature &#xA;yet never rare &#xA;I seem to be more desired than gold&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We always need some space (area) to live first. It is abundant on earth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are here because of me &#xA;and I am certainly not here because of you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No space = no earth = no life&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a group, I can be a force to reckon with &#xA;yet I can be gentler than a breeze &#xA;as every organism adores me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Space is vast and many explosion like supernova happens in space. However it is also very silent there. And every person needs space in his own life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastActivityDate="2016-01-07T19:09:39.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25158" PostTypeId="1" AcceptedAnswerId="25545" CreationDate="2016-01-07T19:13:04.260" Score="12" ViewCount="434" Body="&lt;p&gt;Jill and Suzie are given the same data set and instructed to rank each entry, but they are not allowed to rank the data using the same criteria.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jill goes first and has no trouble at all. She easily evaluates her rankings and writes down her top 15:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;90, 80, 40, 10, 70, 95, 75, 94, 35, 20, 15, 5, 73, 25, 55&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Suzie takes what seems like the next best approach after Jill's and starts writing down her top rankings:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;95, 90, 80, 70&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But after only the first 4 she runs into a problem. Thinking on her feet, she tweaks her method and is able to continue:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;10, 40, 74, 94, 35&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But after 9 entries, she encounters another problem, and this time she's really stuck. Unable to determine a tiebreaker, she declares a 3-way tie for 10th place:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;69 / 20 / 15&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;She then continues with 13th and 14th place:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;75, 55&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But she's forced to declare a tie again at 15th:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;64 / 81&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Jill gloats smugly at Suzie, having clearly chosen the superior ranking method. Suzie knows she had the more difficult task, but she feels the data they were given was partly to blame.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;What data were Jill and Suzie ranking? What method did each use, and why does Suzie feel that the data was partly to blame for her failure?&lt;/h2&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #1:&lt;/strong&gt; The data table Jill and Suzie were using had three entries that were broken down into two rows each (unlike the majority of the entries, which were each entered on one row). The girls treated each row of data as an individual data point and ranked them accordingly. However, if their data table had presented with each of these entries as only one row of (the equivalent combined) data, &lt;code&gt;84&lt;/code&gt; would have ranked 15th for Jill and 9th for Suzie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #2:&lt;/strong&gt; The &lt;a href=&quot;/questions/tagged/trivia&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;trivia&amp;#39;&quot; rel=&quot;tag&quot;&gt;trivia&lt;/a&gt; in this puzzle is specific to the US.&lt;/p&gt;&#xA;" OwnerUserId="17231" LastEditorUserId="17231" LastEditDate="2016-01-19T17:22:51.740" LastActivityDate="2016-01-19T22:06:28.163" Title="Alternate rankings for the same data" Tags="|enigmatic-puzzle|knowledge|number-sequence|" AnswerCount="2" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="25159" PostTypeId="1" AcceptedAnswerId="25193" CreationDate="2016-01-07T19:32:57.940" Score="7" ViewCount="1013" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Feeling lucky, punk? &lt;br&gt;&#xA;  Go on, ogle me, why don't ya? &lt;br&gt;&#xA;  But don't dare ask me for a quickie &lt;br&gt;&#xA;  Or else I'll just taunt ya &lt;br&gt;&#xA;  Just tell me what I want to hear &lt;br&gt;&#xA;  That you're a spy who loves me &lt;br&gt;&#xA;  If I reply with R&amp;amp;B and friendly service &lt;br&gt;&#xA;  You'll know that you have me&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A complete solution should address the following questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Who am I?&lt;/li&gt;&#xA;&lt;li&gt;What was my reply to your amorous advances?&lt;/li&gt;&#xA;&lt;li&gt;What was the method prescribed by the riddle to arrive at the solution?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; question_asker's answer is correct and I have accepted it. However, an unfortunate thing happened that I didn't anticipate and which spoiled a part of the riddle. As a result, the accepted solution is still slightly different from my intended one, though there was no way question_asker could know this. It's still possible to solve the riddle (for example, question_asker solved it despite this apparent flaw), but if you plan to try it, keep in mind the following small&lt;/p&gt;&#xA;&#xA;&lt;p&gt;hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lines 3-4 of the riddle don't make sense anymore, so you might want to just ignore them. But in the future they may become important again, in which case you would need to take them into account - I can't predict whether this would happen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And here is a more complete explanation of the problem (contains a big spoiler):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My idea was that to solve the riddle you would need to google a certain phrase (as explained in question_asker's answer, and choose the first result THAT IS NOT FROM WIKIPEDIA. Lines 3-4 provide the clue that tells you to ignore results from Wikipedia (or to include &quot;-wiki&quot; in your search phrase). The search phrase was designed so that the first search result that would be displayed would actually be to the Wikipedia page for the film &quot;The Spy Who Loved Me&quot; if &quot;-wiki&quot; was not included, which would have been a small red herring that makes the riddle slightly more challenging. However, what happened was that overnight Google promoted the non-Wikipedia search result to first place (maybe because I clicked it several times when designing the riddle), thus ruining the effect.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user17947" LastEditorDisplayName="user17947" LastEditDate="2016-01-08T16:58:34.930" LastActivityDate="2016-01-08T16:58:34.930" Title="A naughty, naughty riddle (take that, NSA!)" Tags="|riddle|lateral-thinking|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25160" PostTypeId="2" ParentId="25130" CreationDate="2016-01-07T19:40:08.690" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Prelude in C by J. S. Bach (BWV 846)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come from a brook.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; J. S. Bach (Bach in german [title clue] means brook)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm first in a book.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first piece in &quot;The Well-Tempered Clavier&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm old, but not stale.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An old and good piece of music.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From me was made hail.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ave Maria was written to the tune... Hail Mary.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I sound like an ocean,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Prelude in &quot;sea&quot; (thanks @hvd)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;perpetual motion.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The prelude itself has &quot;perpetual motion&quot; - steady pace of melody. Fugue (which belongs with this piece) means escape.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not quite a kernel,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not a colonel... but a major. C major.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;nor am I nocturnal.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not a nocturne?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5726" LastEditorUserId="5726" LastEditDate="2016-01-07T20:16:40.617" LastActivityDate="2016-01-07T20:16:40.617" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="25161" PostTypeId="1" AcceptedAnswerId="25174" CreationDate="2016-01-07T20:20:36.573" Score="7" ViewCount="182" Body="&lt;p&gt;&lt;strong&gt;If&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;2010THADFA + 2014IFTH = WHAT IF?,&lt;/li&gt;&#xA;&lt;li&gt;1996RE + 2012NE = SANTA FE,&lt;/li&gt;&#xA;&lt;li&gt;1987INTHWO + 2002HA = IT TAKES TWO,&lt;/li&gt;&#xA;&lt;li&gt;1975THWI + 1994BEANTHBE = HOME,&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;then&lt;/strong&gt; (in increasing order of difficulty)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;2007XA + 2009NETONO = ?,&lt;/li&gt;&#xA;&lt;li&gt;1966CA + ? = ? (two solutions)&lt;/li&gt;&#xA;&lt;li&gt;2015HA + ? + ? = ?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Super extra bonus: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;2015AL + ? + ? + ? = ?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="17147" LastEditorUserId="17147" LastEditDate="2016-01-07T21:21:22.430" LastActivityDate="2016-01-08T14:04:57.703" Title="Bursting with... broad-mindedness" Tags="|enigmatic-puzzle|knowledge|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25162" PostTypeId="1" AcceptedAnswerId="25166" CreationDate="2016-01-07T20:24:26.890" Score="3" ViewCount="657" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I follow a pattern.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears to me that just the above two sentences are not enough. I should add more about what I've already tried to solve it. But I already solved it, and it wouldn't be fun if I told more. [this is a rant, not a clue]&lt;/p&gt;&#xA;" OwnerUserId="16322" LastEditorUserId="11876" LastEditDate="2016-01-22T05:20:15.333" LastActivityDate="2016-01-22T05:20:15.333" Title="The circle of Mega" Tags="|riddle|lateral-thinking|" AnswerCount="4" CommentCount="4" ClosedDate="2016-01-08T02:51:04.807" ContentLicense="CC BY-SA 3.0" />
  <row Id="25163" PostTypeId="1" AcceptedAnswerId="25167" CreationDate="2016-01-07T20:27:43.277" Score="6" ViewCount="240" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;I am the circle of life and death&lt;/p&gt;&#xA;&lt;p&gt;I gave birth to puzzling&lt;/p&gt;&#xA;&lt;p&gt;I gave death to literature&lt;/p&gt;&#xA;&lt;p&gt;As I'm associated with the number 51,&lt;/p&gt;&#xA;&lt;p&gt;One may call me the circle of 51&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="16322" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-01-07T21:03:04.670" Title="The circle of 51" Tags="|riddle|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25164" PostTypeId="2" ParentId="25162" CreationDate="2016-01-07T20:38:45.663" Score="7" Body="&lt;p&gt;Are you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a full stop (.) or &quot;What am I?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; immediately follow the words &quot;a pattern&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I only answered this because this was tagged &lt;a href=&quot;/questions/tagged/lateral-thinking&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;lateral-thinking&amp;#39;&quot; rel=&quot;tag&quot;&gt;lateral-thinking&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="18289" LastActivityDate="2016-01-07T20:38:45.663" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25165" PostTypeId="1" AcceptedAnswerId="25177" CreationDate="2016-01-07T20:51:21.600" Score="77" ViewCount="16105" Body="&lt;p&gt;It is known that when logicians get drunk they may claim not to know the answer to a question when they should know it very well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a night of &lt;a href=&quot;https://puzzling.stackexchange.com/questions/12422/3-logicians-walk-into-a-bar&quot;&gt;heavy drinking at the bar&lt;/a&gt;, three logicians decide that it's time leave. Before they walk out of the place, the barman asks &quot;Is anybody able to drive back home?&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first logician says &quot;I don't know&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second logician says &quot;I don't know&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third logician says &quot;I don't know&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many logicians must be drunk at a minimum? Provide reasoning.&lt;/p&gt;&#xA;" OwnerUserId="4394" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2020-11-19T09:24:02.073" Title="Three logicians walk out of the bar" Tags="|logical-deduction|humor|" AnswerCount="14" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="25166" PostTypeId="2" ParentId="25162" CreationDate="2016-01-07T20:53:37.747" Score="5" Body="&lt;p&gt;I think the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 51&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The last two riddles of OP were 'the circle of three' and 'the circle of two' having answers 'two' and 'Mega'. So according to the pattern the next had to be 'the circle of Mega' whose answer should be the name of the next puzzle by OP which is 'Circle of 51'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Great Series.&lt;/p&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-01-07T23:31:25.633" LastActivityDate="2016-01-07T23:31:25.633" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="25167" PostTypeId="2" ParentId="25163" CreationDate="2016-01-07T20:54:31.097" Score="7" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Area 51&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the circle of life and death&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am the place where stack exchange sites are given life -- or not.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I gave birth to puzzling&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; See previous clue with regard to this site.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I gave death to literature&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;The Literature site didn't have enough activity during the beta, and has been closed. You can download the data dump of all questions here.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As I'm associated with the number 51,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The site's name is &quot;Area 51&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One may call me the circle of 51&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure of this; there is a portion of a circular two-dimensional depiction of a planet or moon at the top of the Area 51 background, or perhaps it has something to do with the fact that a circle is a geometric figure whose area is often calculated.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18224" LastEditorUserId="18224" LastEditDate="2016-01-07T21:03:04.670" LastActivityDate="2016-01-07T21:03:04.670" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25168" PostTypeId="2" ParentId="25162" CreationDate="2016-01-07T20:57:54.203" Score="2" Body="&lt;p&gt;I'll have a go at this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mega is, as we've discussed in &lt;a href=&quot;https://puzzling.stackexchange.com/questions/25117/the-circle-of-two&quot;&gt;previous questions&lt;/a&gt;, a two (2) inside of a circle.&lt;br&gt;&#xA; This notation, known as &lt;a href=&quot;https://en.wikipedia.org/wiki/Steinhaus%E2%80%93Moser_notation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Steinhaus-Moser notation&lt;/a&gt;, makes the circle synonymous with a pentagon&lt;br&gt;&#xA;&lt;br&gt;&#xA; As stated in the linked article, &quot;a number n in a triangle means n&lt;sup&gt;n&lt;/sup&gt;.&quot;&lt;br&gt;&#xA; 'a number n in a square is equivalent to &quot;the number n inside n triangles, which are all nested.&quot;'&lt;br&gt;&#xA; 'a number n in a pentagon is equivalent with &quot;the number n inside n squares, which are all nested.&quot;'&lt;br&gt;&#xA;&lt;br&gt;&#xA; So you could claim that, since the circle is operationally equivalent to the pentagon, the &lt;em&gt;pattern&lt;/em&gt; is an increasing number of sides of the enclosing shape.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't know how to do the fancy number shit*, so if anybody wants to give me a hand there, feel free.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;* &lt;em&gt;but I am a font of technical terminology such as this&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2016-01-07T20:57:54.203" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25169" PostTypeId="2" ParentId="25165" CreationDate="2016-01-07T20:58:44.267" Score="40" Body="&lt;p&gt;At minimum? &lt;strong&gt;Zero&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first logician was driven to the bar by his friend, who then biked home, leaving the car in a local garage.  He himself never learned to drive, so he cannot drive home, and knows nothing about the other two. The second and third logicians both walked to the bar, but know how to drive. Each individually cannot drive because they have no car, so their answer to the bartender is conditional on whether the first logician came with a car, which was not revealed by his answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="17389" LastEditDate="2016-01-07T21:29:25.243" LastActivityDate="2016-01-07T21:29:25.243" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25170" PostTypeId="2" ParentId="25165" CreationDate="2016-01-07T21:06:46.267" Score="44" Body="&lt;p&gt;At a minimum, one logician - the final one - must be drunk. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first two can reasonably say they don't know if &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;they cannot drive but &lt;/li&gt;&#xA;&lt;li&gt;they think one of the later logicians might&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The last one knows none of the prior logicians can drive, their answer should be &quot;yes&quot; or &quot;no&quot; according to their own ability. The fact that they do not answer thus means they are drunk.&lt;/p&gt;&#xA;" OwnerUserId="3987" LastActivityDate="2016-01-07T21:06:46.267" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25171" PostTypeId="2" ParentId="25165" CreationDate="2016-01-07T21:11:46.537" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Three.  They have all been drinking heavily, and they all should know very well that none of them is in a state to drive.  Any logician answering &quot;I don't know&quot; instead of &quot;no&quot; must be giving that answer because of being drunk.  They all gave that answer, therefore they are all drunk.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18224" LastActivityDate="2016-01-07T21:11:46.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25172" PostTypeId="2" ParentId="25165" CreationDate="2016-01-07T21:22:47.680" Score="10" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The minimum is &lt;strong&gt;zero&lt;/strong&gt;. All three logicians never learned to drive a car and there is one additional guest in the bar unknown to the logicians. The word &quot;anybody&quot; is not limited to the three logicians.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastActivityDate="2016-01-07T21:22:47.680" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25173" PostTypeId="1" AcceptedAnswerId="25232" CreationDate="2016-01-07T21:29:35.197" Score="7" ViewCount="569" Body="&lt;p&gt;As you pore through an old book on famous composers, you find an interesting passage.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It wasn't until the American Revolution that this note was found locked in a drawer. It had been taken from the composer's manuscripts and smuggled across the sea to the then-colonies. American intelligence tried to figure out what the message said, but could not. One man was confident that the message had crucial information to their winning the Revolution in mere days but could not break the code.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;GYPUVXOLIHPFCHYLUMYGSQQEIYOZUOHUCIHUVKIIYNWMXNDPAKEVWSRUKTLBVUHIR&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What did the code say?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The composer in question is Johann Sebastian Bach...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ... Or at least he would have been fifty years ago.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This code is based on a particular piece of music.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That piece of music is popularly misattributed to Bach.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15603" LastEditorUserId="15603" LastEditDate="2016-01-09T15:14:35.443" LastActivityDate="2016-01-09T20:47:47.537" Title="The composer's code" Tags="|cipher|" AnswerCount="1" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="25174" PostTypeId="2" ParentId="25161" CreationDate="2016-01-07T21:38:44.837" Score="4" Body="&lt;p&gt;The first one is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'M ALIVE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second is  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2014IFTH, giving WHAT WOULD YOU DO?&lt;br&gt;&#xA;  1970GE, giving SO WHAT?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The third is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1970LOLAKIGE and 2015SORO, giving RIGHT HAND MAN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Super extra bonus&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2012GHTHMU, 1983BA, and 1972PI, giving WITH YOU&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And here's the key:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The equations are in the form of [four-digit year of Broadway* debut][first two letters of each of the words in the title] + same = SONG TITLE COMMON TO BOTH MUSICALS&lt;br&gt;&#xA;&lt;br&gt;&#xA; e.g.,&lt;br&gt;&#xA;&lt;br&gt;&#xA;  2010 &lt;strong&gt;&lt;em&gt;TH&lt;/strong&gt;E &lt;strong&gt;AD&lt;/strong&gt;DAMS &lt;strong&gt;FA&lt;/strong&gt;MILY&lt;/em&gt; + 2014 &lt;strong&gt;&lt;em&gt;IF&lt;/strong&gt;/&lt;strong&gt;TH&lt;/strong&gt;EN&lt;/em&gt; = WHAT IF?&lt;br&gt;&#xA;  1996 &lt;strong&gt;&lt;em&gt;RE&lt;/strong&gt;NT&lt;/em&gt; + 2012 &lt;strong&gt;&lt;em&gt;NE&lt;/strong&gt;WSIES&lt;/em&gt; = SANTA FE&lt;br&gt;&#xA;  1987 &lt;strong&gt;&lt;em&gt;IN&lt;/strong&gt;TO &lt;strong&gt;TH&lt;/strong&gt;E &lt;strong&gt;WO&lt;/strong&gt;ODS&lt;/em&gt; + 2002 &lt;strong&gt;&lt;em&gt;HA&lt;/strong&gt;IRSPRAY&lt;/em&gt; = IT TAKES TWO&lt;br&gt;&#xA;  1975 &lt;strong&gt;&lt;em&gt;TH&lt;/strong&gt;E &lt;strong&gt;WI&lt;/strong&gt;Z&lt;/em&gt; + 1994 &lt;strong&gt;&lt;em&gt;BE&lt;/strong&gt;AUTY &lt;strong&gt;AN&lt;/strong&gt;D &lt;strong&gt;TH&lt;/strong&gt;E &lt;strong&gt;BE&lt;/strong&gt;AST&lt;/em&gt; = HOME&lt;br&gt;&#xA;&lt;br&gt;&#xA;  2007 &lt;strong&gt;&lt;em&gt;XA&lt;/strong&gt;NADU&lt;/em&gt; + 2009 &lt;strong&gt;&lt;em&gt;NE&lt;/strong&gt;XT &lt;strong&gt;TO&lt;/strong&gt; &lt;strong&gt;NO&lt;/strong&gt;RMAL&lt;/em&gt; = I'M ALIVE&lt;br&gt;&#xA;  1966 &lt;strong&gt;&lt;em&gt;CA&lt;/strong&gt;BARET&lt;/em&gt; + 2014 &lt;strong&gt;&lt;em&gt;IF&lt;/strong&gt;/&lt;strong&gt;TH&lt;/strong&gt;EN&lt;/em&gt; = WHAT WOULD YOU DO?&lt;br&gt;&#xA;  1966 &lt;strong&gt;&lt;em&gt;CA&lt;/strong&gt;BARET&lt;/em&gt; + 1970 &lt;strong&gt;&lt;em&gt;GE&lt;/strong&gt;ORGY&lt;/em&gt; = SO WHAT?&lt;br&gt;&#xA;  2015 &lt;strong&gt;&lt;em&gt;HA&lt;/strong&gt;MILTON&lt;/em&gt; + 1970 &lt;strong&gt;&lt;em&gt;LO&lt;/strong&gt;VELY &lt;strong&gt;LA&lt;/strong&gt;DIES, &lt;strong&gt;KI&lt;/strong&gt;ND &lt;strong&gt;GE&lt;/strong&gt;NTLEMEN&lt;/em&gt; + 2015 &lt;strong&gt;&lt;em&gt;SO&lt;/strong&gt;METHING &lt;strong&gt;RO&lt;/strong&gt;TTEN&lt;/em&gt; = RIGHT HAND MAN&lt;br&gt;&#xA;&lt;br&gt;&#xA;  2015 &lt;strong&gt;&lt;em&gt;AL&lt;/strong&gt;LEGIANCE&lt;/em&gt; + 2012 &lt;strong&gt;&lt;em&gt;GH&lt;/strong&gt;OST &lt;strong&gt;TH&lt;/strong&gt;E &lt;strong&gt;MU&lt;/strong&gt;SICAL&lt;/em&gt; + 1983 &lt;strong&gt;&lt;em&gt;BA&lt;/strong&gt;BY&lt;/em&gt; + 1972 &lt;strong&gt;&lt;em&gt;PI&lt;/strong&gt;PPIN&lt;/em&gt; = WITH YOU&lt;br&gt;&#xA;&lt;br&gt;&#xA; * &lt;em&gt;Ho ho! &quot;Broad&quot;-mindedness!&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As hinted to in the comments:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Adding the word &quot;THE&quot; to a song from Cabaret gives us 1991 &lt;strong&gt;&lt;em&gt;MI&lt;/strong&gt;SS &lt;strong&gt;SA&lt;/strong&gt;IGON&lt;/em&gt; (THE TELEPHONE SONG)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2016-01-08T14:04:57.703" LastActivityDate="2016-01-08T14:04:57.703" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="25175" PostTypeId="2" ParentId="24893" CreationDate="2016-01-07T22:20:32.883" Score="2" Body="&lt;p&gt;&lt;strong&gt;Regarding worst case:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example to demonstrate worst case:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If with N = 5, we have deck:&#xA;&lt;code&gt;1010001011&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will use 10 moves making the guesses:&#xA;&lt;code&gt;0000011111&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After this, we have two 1:s and two 0:s left, which might shuffle into:&#xA;&lt;code&gt;1010&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we use 4 moves to guess:&#xA;&lt;code&gt;0000&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After this, we still have two 1:s left. So we will need two more moves, totalling 16 (which is worse than 3N = 15).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why probability_guess algorithm may become worse than the two_pass algorithm is becuase it might requires more than one reshuffle. The two_pass algorithm is guaranteed to never need more than one reshuffle. Already at N=3 and N=4, probability_guess algorithm may require an additional reshuffle, but in these cases, it will still not become worse than two_pass. When N=5, though, the impact of the additional required reshuffle may cause the worst case performance to become worse than two_pass, as shown above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Regarding average case:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interesting things happen around N = 26. For N = 26, two_pass and probability_guess have approximately equal performance in the average case. When N &gt; 26, probability_guess algorithm will degrade in performance, becoming worse than two_pass algorithm, even in the average case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Speculative reasoning around average case performance:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets compare benefits and drawbacks of probability_guess for two cases. Benefit of probability_guess is that you use available information from past observations which improves your probability to guess subsequent guesses correctly. Drawback of probability_guess is that there is a risk that you may need more than one reshuffle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the simplistic case N = 1:&#xA;We have 50% to guess the first card correctly. For the second card, we have a 100% chance to guess it correctly. In this case, the probability_guess algorithm increased our chance to guess correctly (from 50% to 100%) for that second card (which is half of the total number of cards). Ie. the benefit is quite big.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider case when we let N tend towards infinity. Take N=10^9 as an example of a big N. When picking the second card, our probability to guess the second card correctly is 500,000,000 / 999,999,999 which is approximately 50%. Ie. probability_guess didn't help much in improving the probability to get it correct. The same reasoning applies for the majority of the guesses for the initial shuffle. Only when you are near the end of the deck will the probability_guess give a significant improvement to the probability to guess correctly. But &quot;near the end of the deck&quot; is a very small portion of the entire deck. The drawback, however, has become larger. Now we do not only risk having to reshuffle twice. We might have to reshuffle several times. The benefit does not outweigh the drawback.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a specific example, I did one million solution attempts for N=500. Out of these, 94.7% were slower than the two-pass alogrithm. And in the worst case, 7 reshuffles were necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;N &amp;lt; 5       : probability_guess better than two_pass in average case, and equal in worst case&#xA;5 &amp;lt;= N &amp;lt; 26 : probability_guess better than two_pass in average case, but worse in worst case&#xA;N == 26     : probability_guess roughly equal to two_pass in average case, but worse in worst case&#xA;N &amp;gt; 26      : probability_guess worse than two_pass in average case&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="18292" LastActivityDate="2016-01-07T22:20:32.883" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25177" PostTypeId="2" ParentId="25165" CreationDate="2016-01-07T22:35:07.630" Score="86" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All three are drunk&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the first were sober s/he would answer yes. So the first is drunk. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the second were sober s/he would answer yes.  So the second is drunk. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the third were sober s/he would answer yes. The third now has enough information to answer &quot;no&quot;, but is too drunk to realise it. &lt;/p&gt;&#xA;" OwnerUserId="15942" LastActivityDate="2016-01-07T22:35:07.630" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="25178" PostTypeId="2" ParentId="25162" CreationDate="2016-01-07T22:59:28.003" Score="2" Body="&lt;p&gt;Are you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Moser's number&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My reasoning is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; your previous question asked about 2 in a circle, which is a Mega. Using the same notation, Moser's number is 2 in a polygon with Mega sides (which would be practically indistinguishable from a circle, so a circle of Mega).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1955" LastActivityDate="2016-01-07T22:59:28.003" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25182" PostTypeId="2" ParentId="25165" CreationDate="2016-01-08T04:02:26.260" Score="6" Body="&lt;p&gt;Zero. Imagine that each is at 99% of the legal limit, but how would they know for sure. Each would say &quot;I don't know,&quot; because no one has measured their alcohol level, but perhaps none are legally drunk.&lt;/p&gt;&#xA;" OwnerUserId="18300" LastActivityDate="2016-01-08T04:02:26.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25183" PostTypeId="1" AcceptedAnswerId="25382" CreationDate="2016-01-08T05:36:56.270" Score="1" ViewCount="1035" Body="&lt;p&gt;Scream out or scream low &lt;br&gt;&#xA;You will only hear me when I &lt;s&gt;blow&lt;/s&gt; whisper &lt;br&gt;&#xA;Stand back or stand front &lt;br&gt;&#xA;You will only see me when I want &lt;br&gt;&#xA;Go there or no-where &lt;br&gt;&#xA;I can be there beside with no distance &lt;br&gt;&#xA;I don't have ticket &lt;br&gt;&#xA;I can't make chitchat &lt;br&gt;&#xA;But once you know what I am &lt;br&gt;&#xA;You don't want to meet me again. &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;who or what am I?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Hint 1: &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I could be the same shape as you &lt;br&gt;&#xA; Or I could be in the other shape you never saw &lt;br&gt;&#xA; I was just like you before &lt;br&gt;&#xA; May be you were my beloved or more &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2: &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No hint on this because I might be staring right into your face.&lt;br&gt;&#xA; You know or never know until I show. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11353" LastEditorUserId="11353" LastEditDate="2016-01-15T10:22:40.527" LastActivityDate="2016-01-15T10:22:40.527" Title="Naturally, you don't want to meet me again" Tags="|riddle|lateral-thinking|" AnswerCount="10" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25184" PostTypeId="2" ParentId="25158" CreationDate="2016-01-08T05:53:11.913" Score="1" Body="&lt;p&gt;It could be temperature based on latitude (North to South) and longitude (east to west).&lt;/p&gt;&#xA;" OwnerUserId="9531" LastActivityDate="2016-01-08T05:53:11.913" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25185" PostTypeId="2" ParentId="20812" CreationDate="2016-01-08T07:27:22.510" Score="2" Body="&lt;p&gt;If we can assume these digits are measured we get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2 + √2 = √9 (true to one significant figure)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastActivityDate="2016-01-08T07:27:22.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25186" PostTypeId="2" ParentId="25183" CreationDate="2016-01-08T07:28:11.403" Score="0" Body="&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; Invalidated by OP&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Death?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It seems to match most of the hints.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4269" LastEditDate="2016-01-08T10:06:58.483" LastActivityDate="2016-01-08T10:06:58.483" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25187" PostTypeId="2" ParentId="25165" CreationDate="2016-01-08T08:28:58.337" Score="5" Body="&lt;p&gt;I guess it must be as per the truth table in logic gates, if its 0 AND 0 , result/output is 0. Similar way , If first logician says &quot;I don't know&quot; AND so does the second one, the third logician as per the truth table says &quot;I don't know&quot; .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer must be all the three logicians are drunk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just a funny thought.. &lt;/p&gt;&#xA;" OwnerUserId="18306" LastActivityDate="2016-01-08T08:28:58.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25188" PostTypeId="2" ParentId="25183" CreationDate="2016-01-08T09:07:40.117" Score="6" Body="&lt;p&gt;Are you a.. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hurricane &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My first go at this so I hope I'm doing it right xD&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit, a explanation to as why:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - You can scream all your want, but you'll only hear a hurricane once it's started to get enough &quot;speed&quot; built up, aka &lt;strong&gt;blowing winds&lt;/strong&gt;.&#xA; &lt;br&gt;&#xA; - You'll only see a hurricane when it is strong enough, &quot;when it wants&quot;.&#xA; &lt;br&gt;&#xA; - You can go anywhere, and a hurricane can cover enough distance in no time at all to be considered &quot;be there with no distance&quot;.&#xA; &lt;br&gt;&#xA; - A hurricane can't make chitchat, once it stats to blow things get messy and loud.&#xA; &lt;br&gt;&#xA; - And the title says &quot;Naturally&quot; hinting towards a natural object/thing, and a hurricane is the force of nature so there's that.&#xA; &lt;br&gt;&#xA; - Oh and ofc you don't want to meet him/her/it again : )&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18217" LastEditorUserId="18217" LastEditDate="2016-01-08T10:31:13.683" LastActivityDate="2016-01-08T10:31:13.683" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25189" PostTypeId="1" CreationDate="2016-01-08T09:53:56.210" Score="17" ViewCount="1357" Body="&lt;p&gt;Based on &lt;a href=&quot;https://puzzling.stackexchange.com/questions/25149/clear-board-in-othello&quot;&gt;Clear board in Othello&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If we have an $2m\times 2n$ Othello board ($m\geq n\geq 2$), and the&#xA;  centre $4$ squares have the usual starting position, then for what all&#xA;  values of $m$ and $n$ can we &lt;em&gt;clear the board&lt;/em&gt; (make the board&#xA;  entirely covered with one colour, either black or white) by controlling&#xA;  both players?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;s&gt;I am not interested in trial-and-error or a computer algorithm.&lt;/s&gt; &lt;em&gt;Since this still remains unsolved, I guess any approach will be of help.&lt;/em&gt; I'm only asking if there is a logical/mathematical way to solve the Q for large $m$ and $n$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus sub-case:&lt;/strong&gt; $m=n$&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="51131" LastEditDate="2019-09-11T14:17:44.770" LastActivityDate="2019-09-11T14:17:44.770" Title="Clear board in Othello (Generalisation)" Tags="|strategy|game|board-games|reversi|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25190" PostTypeId="2" ParentId="25165" CreationDate="2016-01-08T10:11:33.977" Score="10" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;zero&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is because the question is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; poorly phrased. It is not known to the logicians what set of people 'anybody' is referring to. It could obviously refer to 'anybody' in the world, which would make the answer a trivial yes, It could also refer to just the 3 logicians, but being logicians they feel it's not their place to guess. So they answer with 'I don't know (could you elaborate on which people are part of 'anybody' in this case)'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16846" LastActivityDate="2016-01-08T10:11:33.977" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25191" PostTypeId="2" ParentId="25183" CreationDate="2016-01-08T10:22:02.823" Score="1" Body="&lt;p&gt;As far as I understand, might it be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DREAM&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18309" LastEditorUserId="18217" LastEditDate="2016-01-08T11:02:28.757" LastActivityDate="2016-01-08T11:02:28.757" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25192" PostTypeId="2" ParentId="6239" CreationDate="2016-01-08T11:37:36.040" Score="2" Body="&lt;p&gt;Taking the bottom output as bit 0 up to the top output as bit 7.  You can easily see that bits 5, 6, and 7 are always 010. Given that I wrote the following Python code (without looking at any answers to make it more fun!):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;def bit0(a, b, c):&#xA;    return a or not (b and c)&#xA;def bit1(a, b, c):&#xA;    return bit1_5(a, b, c) or not (a or b or c)&#xA;def bit1_5(a, b, c):&#xA;    return (a and c) and not b&#xA;def bit2(a, b, c):&#xA;    return (not a and (b != c)) or bit1_5(a, b, c)&#xA;def bit3(a, b, c):&#xA;    return a == b == c&#xA;def bit4(a, b, c):&#xA;    return (a or b) and c&#xA;&#xA;def compose(i):&#xA;    a = bool(i &amp;amp; 0x04)&#xA;    b = bool(i &amp;amp; 0x02)&#xA;    c = bool(i &amp;amp; 0x01)&#xA;    bits = 0b01000000&#xA;    if bit0(a, b, c):&#xA;        bits |= 0b00000001&#xA;    if bit1(a, b, c):&#xA;        bits |= 0b00000010&#xA;    if bit2(a, b, c):&#xA;        bits |= 0b00000100&#xA;    if bit3(a, b, c):&#xA;        bits |= 0b00001000&#xA;    if bit4(a, b, c):&#xA;        bits |= 0b00010000&#xA;    return bits&#xA;&#xA;if __name__ == '__main__':&#xA;    message = ''&#xA;    for i in range(8):&#xA;        message += chr(compose(i))&#xA;    print (message)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which outputs:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; KEEPAWAY&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastActivityDate="2016-01-08T11:37:36.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25193" PostTypeId="2" ParentId="25159" CreationDate="2016-01-08T14:44:32.943" Score="11" Body="&lt;p&gt;So I tried to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Go&lt;/strong&gt; [on,] &lt;strong&gt;ogle&lt;/strong&gt; [me] the phrase &quot;I'm a spy who loves you&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and I &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;[felt] lucky&lt;/strong&gt;, so I chose the first result  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which was &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://www.youtube.com/watch?v=jArxD7n7Xk0&quot;&gt;a video for a song of (nearly) that same title&lt;/a&gt;, by Rob Bender, from &lt;strong&gt;&lt;em&gt;R&lt;/strong&gt;[&lt;strong&gt;&amp;amp;&lt;/strong&gt;]&lt;strong&gt;B&lt;/strong&gt;'s Goodie Garden&lt;/em&gt;, and the lyrics to which are &lt;a href=&quot;https://robbendermusic.bandcamp.com/track/im-the-spy-who-loves-you-2&quot;&gt;here&lt;/a&gt; (I can't listen at the moment)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The above, to the best of my understanding, addresses point 3, and the third chunk of spoilertext addresses point 2, with point 1 &lt;em&gt;possibly&lt;/em&gt; being answered by the bolded text in the first chunk of spoilertext (I am the least sure of this last bit).&lt;/p&gt;&#xA;&#xA;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2016-01-08T15:58:44.890" LastActivityDate="2016-01-08T15:58:44.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25194" PostTypeId="1" AcceptedAnswerId="25196" CreationDate="2016-01-08T15:22:17.953" Score="12" ViewCount="1798" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Less one, and my maker is often said to be the mightiest between us.&lt;/p&gt;&#xA;&lt;p&gt;Alone I smite, yet my greater does not.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;" OwnerUserId="17744" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-01-10T09:07:08.613" Title="Alone I Smite. Who am I?" Tags="|riddle|word|wordplay|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25195" PostTypeId="2" ParentId="25194" CreationDate="2016-01-08T15:30:50.227" Score="5" Body="&lt;p&gt;One option:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A ruler  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Less one, and my maker is often said to be the mightiest between us.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Remove the 'r', and you get rule. Often, those who make the rules seem to be greater than the rules themselves (and don't live by them).  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alone I smite, yet my greater does not.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rulers are often used to &quot;smite&quot; kids who misbehave in class.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2016-01-08T15:30:50.227" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25196" PostTypeId="2" ParentId="25194" CreationDate="2016-01-08T16:25:42.017" Score="24" Body="&lt;p&gt;I think the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;sword&lt;/strong&gt; - Omitting the first letter gives us 'word.' The maker of a word can be a pen, which is said to be mightier than the sword.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18319" LastEditorDisplayName="user9" LastEditDate="2016-01-09T08:12:33.643" LastActivityDate="2016-01-09T08:12:33.643" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25197" PostTypeId="1" AcceptedAnswerId="25198" CreationDate="2016-01-08T17:20:34.593" Score="8" ViewCount="1034" Body="&lt;p&gt;Hangman game backward #7 - guess what was the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;fill in a &quot;d&quot; - If this is marked on the calendar, you might be in some trouble.&lt;br&gt;&#xA;fill in a &quot;h&quot; - Google picture of this, you'll find mostly music sheets and a kind of sport.&lt;br&gt;&#xA;fill in a &quot;l&quot; - Open me and you'll usually find your beloved one.&lt;br&gt;&#xA;fill in a &quot;p&quot; - It is believed that in some Asia country, this is where some monsters live.&lt;br&gt;&#xA;fill in a &quot;r&quot; - &quot;Fly me to the moon, and let me play among the stars&quot;&lt;br&gt;&#xA;fill in a &quot;s&quot; - You might feel your lack of power disturbing without me.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, your answer can be in the form of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The question was: _ p p l e&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First answer that fully explained each line will be accepted, enjoy!&lt;/p&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2016-01-09T21:16:15.323" Title="The hangman game backward #7" Tags="|riddle|knowledge|english|language|letters|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25198" PostTypeId="2" ParentId="25197" CreationDate="2016-01-08T17:27:40.860" Score="12" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; _ o c k e t&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fill in a &quot;d&quot; - If this is marked on the calendar, you might be in some trouble.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dockets are related to (law) cases &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fill in a &quot;h&quot; - Google picture of this, you'll find mostly music sheets and a kind of sport.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hocket is a musical term&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fill in a &quot;l&quot; - Open me and you'll usually find your beloved one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Locket holds a picture&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fill in a &quot;p&quot; - It is believed that in some Asia country, this is where some monsters live.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pocket monster! I choose you!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fill in a &quot;r&quot; - &quot;Fly me to the moon, and let me play among the stars&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rocket&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fill in a &quot;s&quot; - You might feel your lack of power disturbing without me. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Socket, where you plug stuff in&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2016-01-08T17:33:44.883" LastActivityDate="2016-01-08T17:33:44.883" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25199" PostTypeId="2" ParentId="25197" CreationDate="2016-01-08T17:28:17.197" Score="4" Body="&lt;p&gt;The arrangement of letters was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;_ocket.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17712" LastEditorUserId="17712" LastEditDate="2016-01-09T21:16:15.323" LastActivityDate="2016-01-09T21:16:15.323" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25200" PostTypeId="2" ParentId="25165" CreationDate="2016-01-08T18:19:25.350" Score="5" Body="&lt;p&gt;Three.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If a logician is sober, she will say &quot;yes&quot; because she herself is able to drive. So &quot;I don't know&quot; always means a logician is drunk, completely independent of everything else,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is too simple and therefore makes me think I did not understand the question. Well, ok, there's one last wrinkle: is this situation even possible?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; and the answer is yes, because although the third logician knew the answer was &quot;no&quot;, she might decide to say &quot;I don't know&quot;, and therefore there is no contradiction. Had she said &quot;no&quot;, there would be a contradiction, but that's irrelevant.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18323" LastActivityDate="2016-01-08T18:19:25.350" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25201" PostTypeId="1" AcceptedAnswerId="25247" CreationDate="2016-01-08T18:37:07.770" Score="5" ViewCount="548" Body="&lt;p&gt;You find a piece of paper. It reads&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;SGJAKFBWJMWQAMZEJJXNFZFPFFVL&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Can you de-code it?&#xA;You will have to add in the spaces. It is a phrase not a word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seeing as this is incredibly hard, I will give you a few hints...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A special triangle is required to do this&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A number line will help&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Odds and evens are &lt;em&gt;so, so&lt;/em&gt; different ;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the final clue is the title mixed with the tag &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will be &lt;em&gt;very&lt;/em&gt; impressed by anyone who can even get close on this one...&lt;/p&gt;&#xA;" OwnerUserId="18250" LastEditorUserId="18250" LastEditDate="2016-01-11T07:12:53.320" LastActivityDate="2016-01-11T07:12:53.320" Title="The triangular encoding" Tags="|mathematics|cipher|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25202" PostTypeId="1" AcceptedAnswerId="25204" CreationDate="2016-01-08T18:40:16.340" Score="11" ViewCount="832" Body="&lt;p&gt;There are 2016 wooden stakes, evenly spaced around a large circle. You have 2016 labels featuring all the numbers 1 to 2016, and you proceed to label each stake in any order you please. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, you connect the stakes with taut strings in the order they are labeled, so 1 is tied to 2, 2 is tied to 3, and so on. You also tie 2016 to 1 to create a closed loop.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it possible to do this in such a way that no two of the 2016 strings are parallel?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="9523" LastEditDate="2016-01-08T18:55:46.400" LastActivityDate="2016-01-08T19:04:51.217" Title="Stakes in a Circle" Tags="|mathematics|" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25203" PostTypeId="2" ParentId="25202" CreationDate="2016-01-08T18:49:51.317" Score="2" Body="&lt;p&gt;Yes, you simply make each string go across the circle and then too the next one along. Like this but with 1008 strings acting as diameters and the other 1008 going to the one adjacent so it completes the circle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aOhnM.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aOhnM.gif&quot; alt=&quot;&amp;quot;Example&amp;quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18250" LastActivityDate="2016-01-08T18:49:51.317" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25204" PostTypeId="2" ParentId="25202" CreationDate="2016-01-08T18:57:33.660" Score="14" Body="&lt;p&gt;No, it is not possible. In particular, consider that, were we to number the stakes from &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt; through &lt;span class=&quot;math-container&quot;&gt;$2015$&lt;/span&gt; going evenly around the circle, then a string going from &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; is parallel to one going from &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt;  if and only if &lt;span class=&quot;math-container&quot;&gt;$a+b\equiv c+d\pmod{2016}$&lt;/span&gt;. This may be seen as there is a reflection taking stake &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; to stake &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; which also takes stake &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; to stake &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt;. Two sets of parallel lines between such labeling are shown below:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ej9ms.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ej9ms.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/invnk.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/invnk.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Thus, we may reduce the problem to the following:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Is there a sequence &lt;span class=&quot;math-container&quot;&gt;$s_0,\ldots,s_{2015}$&lt;/span&gt; such that the value of &lt;span class=&quot;math-container&quot;&gt;$s_n+s_{n+1}$&lt;/span&gt; is distinct for every &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; between &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$2015$&lt;/span&gt; (inclusively, taking the last value to be &lt;span class=&quot;math-container&quot;&gt;$s_{2015}+s_0$&lt;/span&gt;)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;And the answer is &amp;quot;no&amp;quot; because there are only &lt;span class=&quot;math-container&quot;&gt;$2016$&lt;/span&gt; equivalence classes mod &lt;span class=&quot;math-container&quot;&gt;$2016$&lt;/span&gt;, meaning each equivalence class must contain exactly one value &lt;span class=&quot;math-container&quot;&gt;$s_{n}+s_{n+1}$&lt;/span&gt;. However, that means that &lt;span class=&quot;math-container&quot;&gt;$$1008\equiv\sum_{i=0}^{2015}i\equiv \sum_{n=0}^{2015}s_n+s_{n+1}\equiv 2\sum_{n=0}^{2015}s_n\equiv 2\sum_{i=0}^{2015}i\equiv2\cdot 1008\pmod{2016}$$&lt;/span&gt;&#xA;which is false since &lt;span class=&quot;math-container&quot;&gt;$2\cdot 1008 \not\equiv 1008$&lt;/span&gt;. Note that we used, in our second to last step, that &lt;span class=&quot;math-container&quot;&gt;$s_n$&lt;/span&gt; enumerates the equivalence classes mod &lt;span class=&quot;math-container&quot;&gt;$2016$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;(Note that the equivalence &lt;span class=&quot;math-container&quot;&gt;$1008\equiv\sum_{i=0}^{2015}i\pmod{2016}$&lt;/span&gt; comes from a pairing argument - each summand &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; may be paired with its opposite &lt;span class=&quot;math-container&quot;&gt;$-i$&lt;/span&gt; making a total of &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt;, with the exception of the two &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; such that &lt;span class=&quot;math-container&quot;&gt;$i=-i$&lt;/span&gt; - which are &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$1008$&lt;/span&gt;. So the sum is &lt;span class=&quot;math-container&quot;&gt;$0+1008=1008$&lt;/span&gt;)&lt;/p&gt;&#xA;" OwnerUserId="4765" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-01-08T19:04:51.217" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25205" PostTypeId="2" ParentId="25165" CreationDate="2016-01-08T19:14:12.017" Score="5" Body="&lt;p&gt;Zero&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all bodies, any is able to drive. Taking things very literally as a logician, they do not know the capabilities of all bodies in question, and thus cannot determine the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer would be different if the question was &quot;Are any of you able to drive back home?&quot; At which point we could receive a more accurate number of drunken logicians. The first two would be able to speak for themselves with a yes or no answer, and the third would be able to speak conclusively based on the first two answers and their own knowledge with a yes or no answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as it is worded, they don't have enough information, and consequently don't know.&lt;/p&gt;&#xA;" OwnerUserId="18326" LastActivityDate="2016-01-08T19:14:12.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25206" PostTypeId="2" ParentId="25165" CreationDate="2016-01-08T19:15:34.067" Score="10" Body="&lt;p&gt;Three. This is a conversation involving four persons: the barman and three logicians. If all three logicians claim that they do not know the answer to question &quot;Is anybody able to drive back home?&quot;, they all must be drunk, because they very well know that the barman is sober and fully able to drive. The barman is &quot;anybody&quot;. Being logicians they react only to the logic of the question and completely ignore the practical side of it.&lt;/p&gt;&#xA;" OwnerUserId="18325" LastActivityDate="2016-01-08T19:15:34.067" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25207" PostTypeId="2" ParentId="25165" CreationDate="2016-01-08T21:01:19.623" Score="5" Body="&lt;p&gt;Assuming they all have a driving license:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3 because if any could drive he would know.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="310" LastActivityDate="2016-01-08T21:01:19.623" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25208" PostTypeId="1" AcceptedAnswerId="25209" CreationDate="2016-01-08T21:17:54.737" Score="9" ViewCount="2348" Body="&lt;p&gt;A man was killed in a roundhouse with no doors or windows. &#xA;Four maids were present at the time of the murder, but are saying nothing. When asked what they were doing they replied as thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maid 1: 'I was making the tea.'&lt;br&gt;&#xA;Maid 2: 'I was dusting the mantelpiece.'&lt;br&gt;&#xA;Maid 3: 'I was setting a mouse trap in the corner.'&lt;br&gt;&#xA;Maid 4: 'I was watching the telly.'  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the maids has lied. The liar is the murderer.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Who is the murderer and what is the lie? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18250" LastEditorUserId="8874" LastEditDate="2016-01-08T21:22:01.780" LastActivityDate="2020-04-01T14:34:33.507" Title="The murder by the maid - Who is the murderer" Tags="|logical-deduction|lateral-thinking|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25209" PostTypeId="2" ParentId="25208" CreationDate="2016-01-08T21:20:13.610" Score="11" Body="&lt;p&gt;It was&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Maid 3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Argument:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A roundhouse with no doors or windows does not have corners &#xA;(to set a mouse trap).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-01-08T21:21:08.123" LastActivityDate="2016-01-08T21:21:08.123" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25210" PostTypeId="1" AcceptedAnswerId="25211" CreationDate="2016-01-09T02:35:32.800" Score="9" ViewCount="187" Body="&lt;p&gt;A spy tries to infiltrate into a Spanish base. In order to do so, he pinpointed a spot where he could listen the people outside of the gate who were asked to answer a short question. These questions were:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Guard:&lt;/strong&gt; 14.&lt;br&gt;&#xA;  &lt;strong&gt;First outsider answer:&lt;/strong&gt;  7.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Guard:&lt;/strong&gt; 18.&lt;br&gt;&#xA;  &lt;strong&gt;Second outsider answer:&lt;/strong&gt; 9.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once the spy heard this, he realised that he always needed to give the half of the number as a response. So he went to the base,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Guard:&lt;/strong&gt; 6.&lt;br&gt;&#xA;  &lt;strong&gt;Spy answer:&lt;/strong&gt; 3.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The spy was immediately killed.&lt;br&gt;&#xA;Why this happened?&lt;/p&gt;&#xA;" OwnerUserId="18339" LastActivityDate="2016-01-09T02:42:16.973" Title="The spy who was killed" Tags="|password|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25211" PostTypeId="2" ParentId="25210" CreationDate="2016-01-09T02:42:16.973" Score="5" Body="&lt;p&gt;This is the standard &quot;count the letters in the word&quot; puzzle, but in Spanish. 14 is &quot;catorce&quot; (7 letters) and 18 is &quot;dieciocho&quot; (nine letters). So he should have answered 4, because &quot;seis&quot; (Spanish for 6) has four letters.&lt;/p&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2016-01-09T02:42:16.973" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25212" PostTypeId="1" CreationDate="2016-01-09T03:06:19.637" Score="26" ViewCount="3613" Body="&lt;p&gt;A good nonogram puzzle has a unique solution. However, some nonograms do not. For example, this puzzle:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;      211 &#xA;    1311231&#xA;  1     &#xA;  3     &#xA;2 1   &#xA;1 1   &#xA;1 2   &#xA;  3   &#xA;  1        &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Has at least two solutions, shown below:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;      211 &#xA;    1311231&#xA;  1 *------&#xA;  3 --***--&#xA;2 1 -**--*-&#xA;1 1 -*---*-&#xA;1 2 -*--**-&#xA;  3 --***--&#xA;  1 ------*      &#xA;&#xA;      211 &#xA;    1311231&#xA;  1 *------&#xA;  3 -***---&#xA;2 1 -**-*--&#xA;1 1 -*---*-&#xA;1 2 --*-**-&#xA;  3 ---***-&#xA;  1 ------*  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Is there an algorithm to decide that the puzzle has a unique solution that is more efficient than trying to solve it?&lt;/p&gt;&#xA;" OwnerUserId="18151" LastEditorUserId="63414" LastEditDate="2020-06-21T08:50:30.597" LastActivityDate="2020-06-21T08:50:30.597" Title="Is there an algorithm to decide that the nonogram puzzle is unique" Tags="|algorithm|nonogram|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="25213" PostTypeId="2" ParentId="25194" CreationDate="2016-01-09T03:11:06.770" Score="3" Body="&lt;p&gt;I'll put my guess out there as a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Golem&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My reasoning:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Less one, and my maker is often said to be the mightiest between us.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are two ways I have read on deactivating a Golem: &lt;br /&gt;&#xA; 1. Remove the mystic scroll from its mouth (less one scroll) &lt;br /&gt;&#xA; 2. Remove the character א (&lt;em&gt;aleph&lt;/em&gt;) from אמת (&lt;em&gt;emet&lt;/em&gt;, &quot;truth&quot;) to make מת (&lt;em&gt;met&lt;/em&gt;, &quot;dead&quot;) (more likely)&lt;br /&gt;&lt;br /&gt;&#xA; So less one character and the creator of the Golem (Rabbi, Elder, Judge, or Sage typically) is now mightier than a lump of clay&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alone I smite, yet my greater does not.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First, though Golems were not created to &quot;smite enemies&quot; in all traditions - there is a definite leaning in Jewish Mythology towards Golems being created to defend Jewish communities and in pop culture most Golems I have seen lean strongly toward militaristic tools or monsters (with the exception of Disc World). Yet I would say the Rabbi is the greater, so the Golem &lt;em&gt;smites enemies alone&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11239" LastEditorUserId="7529" LastEditDate="2016-01-09T04:47:53.087" LastActivityDate="2016-01-09T04:47:53.087" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25214" PostTypeId="1" AcceptedAnswerId="25756" CreationDate="2016-01-09T03:17:57.853" Score="11" ViewCount="921" Body="&lt;p&gt;As you work through one of your countless puzzle books, you find a sheet of paper inside with a poem on it. Remembering &lt;a href=&quot;https://puzzling.stackexchange.com/q/25055/8949&quot;&gt;the last time&lt;/a&gt; you found something like that, you decide to take a crack at this one, hoping to solve it a bit more quickly...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;A Fable&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The three little pigs hid a little rhyme, &lt;br&gt;&#xA;  But it's only a matter of how much time &lt;br&gt;&#xA;  Until the big bad wolf commits the big bad crime &lt;br&gt;&#xA;  And blows their whole house down. &lt;br&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;“Mark my words, pigs, count on these: &lt;br&gt;&#xA;  I'll find those hidden rhymes with ease, &lt;br&gt;&#xA;  I'll find those letters and those ABC's &lt;br&gt;&#xA;  And split them into threes!”&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The big bad wolf said this little rhyme, &lt;br&gt;&#xA;  And he committed the big bad crime, &lt;br&gt;&#xA;  And he cooked all three of the pigs with thyme, &lt;br&gt;&#xA;  Wolfing his bacon down.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nonsense indeed! But what is the fable hiding?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Upon closer inspection of the poem, you notice that some words have a faint underline, perhaps hinting at a ciphering method: words, count, rhymes, letters, threes, bacon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Treat each line as a unit, and ignore the stanza breaks. The number 3 is &lt;em&gt;very&lt;/em&gt; important.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8949" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2016-01-30T20:49:47.837" Title="The Three Little Pigs are hiding something" Tags="|cipher|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="25215" PostTypeId="1" AcceptedAnswerId="25399" CreationDate="2016-01-09T03:38:34.610" Score="4" ViewCount="598" Body="&lt;p&gt;I can be a tiny fruit - if you collect enough of me, you can turn me into something you like. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can also be a gigantic creature cut into 6 pieces. Two of them are tiny.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I can also be eccentric creature that can understand any language - in this form, I have a strange feature shared with Kēlen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Generally, puzzles are inspired from the creator's background&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm an Indonesian&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The three possibility is a different thing of same name, hence &lt;a href=&quot;/questions/tagged/wordplay&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;wordplay&amp;#39;&quot; rel=&quot;tag&quot;&gt;wordplay&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This word (one of the possibility) has an intimate relationship with Indonesia&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18151" LastEditorUserId="11876" LastEditDate="2016-02-25T04:57:22.467" LastActivityDate="2016-02-25T04:57:22.467" Title="You can see me in different face" Tags="|riddle|word|wordplay|" AnswerCount="4" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="25217" PostTypeId="2" ParentId="25215" CreationDate="2016-01-09T09:18:10.200" Score="7" Body="&lt;p&gt;This is a completely wild guess bordering the realm of absurdity, but - I'll still put it out there!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Atom (or perhaps a molecule, quark, or something physics-related)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One possibility is a tiny fruit that if you collect me enough, someone can turn me into something that you like&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Atoms/molecules are the building blocks of everything&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another possibility is a gigantic creature that is cut into 6 pieces. Two of them is tiny.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Depending on what you mean by &quot;pieces&quot;, you can find 6 properties in an atom (electrons, isotopes, neutrons, protons -&gt; the latter two consisting of quarks and gluons - which are very tiny). In comparison to these particles, the actual atom is gigantic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet another possibility is a eccentric creature can understand language with strange feature shared with Kēlen (I'm not human).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm going to go on a massive limb here. The only thing Kelen shares with any other language is the Devanagari writing system, which is a script for many languages - one of them being Sanskrit. &lt;em&gt;In some philosophies/theories&lt;/em&gt;, Sanskrit is considered the building block of most of the world's other languages, as atoms are the building block of all that's physical.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2807" LastActivityDate="2016-01-09T09:18:10.200" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25218" PostTypeId="2" ParentId="25208" CreationDate="2016-01-09T12:14:21.730" Score="1" Body="&lt;p&gt;The murderer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The third, as there are no doors and the mousehole technically qualifies as a door.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18117" LastEditorUserId="18117" LastEditDate="2016-01-09T12:23:52.287" LastActivityDate="2016-01-09T12:23:52.287" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25219" PostTypeId="1" AcceptedAnswerId="25231" CreationDate="2016-01-09T12:20:51.290" Score="12" ViewCount="571" Body="&lt;p&gt;If we join two circles on a plane, each will have exactly one neighbor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3DJpg.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3DJpg.png&quot; alt=&quot;Neighboring Circles - 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given three or more circles, we can build a chain where each circle has exactly two neighbors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/H2Xfu.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/H2Xfu.png&quot; alt=&quot;Neighboring Circles - 2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also arrangements where each circle has exactly three neighbors, like the one shown here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0ZmSS.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0ZmSS.png&quot; alt=&quot;Neighboring Circles - 3&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to arrange a finite number of equally sized, non-overlapping circles on a plane such that each circle has exactly four neighbors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if the circles can have different sizes?&lt;/p&gt;&#xA;" OwnerUserId="4394" LastActivityDate="2016-01-09T23:14:48.960" Title="Neighboring circles" Tags="|geometry|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25220" PostTypeId="2" ParentId="25219" CreationDate="2016-01-09T12:47:17.973" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1: no, but on a cylinder, yes. two touching staggered lines around the circumference &lt;br&gt;&lt;br&gt;&#xA; 2: yes, arrange (eg)5 in a circle and then larger ones outside each touching 2 inner circles and two neighbours &lt;br&gt;&lt;br&gt;&#xA; additionally 5 neighbours is possible if you allow a circle of negative radius, draw splayed dodecahedron  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9137" LastEditorUserId="9137" LastEditDate="2016-01-09T23:14:48.960" LastActivityDate="2016-01-09T23:14:48.960" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25221" PostTypeId="2" ParentId="25183" CreationDate="2016-01-09T12:54:07.980" Score="2" Body="&lt;p&gt;Answer 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Icy wind&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scream out or scream low&lt;br&gt;&#xA;You will only hear me when I blow &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A sufficiently powerful howl of wind will be audible over a scream.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Stand back or stand front&lt;br&gt;&#xA;You will only see me when I want &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wind is invisible, but I guess 'feeling' and 'seeing' could be considered analogous to one another in this case.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Go there or no-where&lt;br&gt;&#xA;I can be there beside with no distance &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can't feel still wind, obviously.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't have ticket&lt;br&gt;&#xA;I can't make chitchat&lt;br&gt;&#xA;But once you know what I am&lt;br&gt;&#xA;You don't want to meet me again. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Who wants to endure another draft of icy wind?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A bit of a long shot on some of the clues, but I think it fits.&lt;/p&gt;&#xA;" OwnerUserId="18117" LastEditorUserId="18117" LastEditDate="2016-01-09T13:07:20.880" LastActivityDate="2016-01-09T13:07:20.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25222" PostTypeId="2" ParentId="25183" CreationDate="2016-01-09T12:57:19.543" Score="1" Body="&lt;p&gt;Answer 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A bomb going off&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Similar to my earlier answer, using multiple meanings of 'blow'.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Only in this case, a bomb will be concealed and hard to detect, but pretty much obvious when it blows.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18117" LastEditorUserId="18117" LastEditDate="2016-01-09T13:10:57.903" LastActivityDate="2016-01-09T13:10:57.903" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25223" PostTypeId="2" ParentId="25194" CreationDate="2016-01-09T14:54:47.483" Score="0" Body="&lt;p&gt;Power is described as the ability to destroy.Thus the answer is minus one (-1)&lt;/p&gt;&#xA;" OwnerUserId="18343" LastActivityDate="2016-01-09T14:54:47.483" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25224" PostTypeId="1" AcceptedAnswerId="25229" CreationDate="2016-01-09T15:15:40.330" Score="3" ViewCount="193" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am an unobtanium.&lt;br&gt;&#xA;  Yet I am everwhere.&lt;br&gt;&#xA;  My number is avoided.&lt;br&gt;&#xA;  Yet myself is praised.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;After me, glass products formed.&lt;br&gt;&#xA;  Before me, clorophyll formed.&lt;br&gt;&#xA;  Above me, is where the hole is on.&lt;br&gt;&#xA;  Below me, is what I was usually befriended with.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A countries of love praise me.&lt;br&gt;&#xA;  A countries of sticks praise me.&lt;br&gt;&#xA;  A broken countries praise me.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18151" LastActivityDate="2016-01-09T15:48:39.817" Title="Yet Another Liars" Tags="|riddle|poetry|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25227" PostTypeId="2" ParentId="25219" CreationDate="2016-01-09T15:42:03.340" Score="4" Body="&lt;p&gt;With an infinite number of circles, the simplest way is to have a line of circles nestled on top of another line of circles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe it is impossible with a finite number of circles. First, let's start with a standard trapezoid, 2 circles on 3 circles. The bottom middle circle has 4 neighbors. The problem are the other two bottom circles. At this point, when adding a circle you can only connect to 2 other circles, and then the new circle needs 2 more connections as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For circles of different sizes, use the simple double line, but have 1 line be smaller than the other. The line will curve and meet itself. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2016-01-09T15:42:03.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25229" PostTypeId="2" ParentId="25224" CreationDate="2016-01-09T15:48:39.817" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Aluminum, elemental number 13  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My number is avoided. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Unlucky number 13&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After me, glass products formed&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Next element is Silicon, which is what glass is made of&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Before me, chlorophyll formed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Previous element is Magnesium, which is the core of chlorophyll&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Above me, is where the hole is on.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Boron is used in semi-conductors as holes for electrons&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Below me, is what I was usually befriended with.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gallium is in the same group as Aluminum&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2016-01-09T15:48:39.817" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25231" PostTypeId="2" ParentId="25219" CreationDate="2016-01-09T18:20:02.977" Score="13" Body="&lt;p&gt;It is not possible when the radii are equal. Consider the topmost circle, C. In order to pack four neighbors around C, none of which are above C, you have to use the below arrangement (where C is the gray circle):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8PCux.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8PCux.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means there is a circle to the immediate right and left of C. Repeating the same argument over and over shows that there is an infinite line of circles at the same level as C, contradicting the finiteness requirement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you allow differing radii, the below arrangement works:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aJet4.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aJet4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2016-01-09T19:31:41.557" LastActivityDate="2016-01-09T19:31:41.557" CommentCount="0" CommunityOwnedDate="2016-01-09T19:38:17.187" ContentLicense="CC BY-SA 3.0" />
  <row Id="25232" PostTypeId="2" ParentId="25173" CreationDate="2016-01-09T20:40:38.570" Score="6" Body="&lt;p&gt;Having seen Mr. Holland's Opus, I remember the part where he's explaining to the kids that the song &quot;A Lover's Concerto&quot; used the melody from Minuet in G Major. I also remember hearing that it was really written by someone else (googling this bit shows me that it's now attributed to Christian Petzold).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we transcribe the melody below the string of characters, we get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;GYPUVXOLIHPFCHYLUMYGSQQEIYOZUOHUCIHUVKIIYNWMXNDPAKEVWSRUKTLBVUHIR&lt;/code&gt;&#xA; &lt;code&gt;DGABCDGGECDEFGGGCDCBABCBAGFGABGBADGABCDGGECDEFGGGCDCBABCBAGABAGFG&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And using that as a Vigenère cipher gives us&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;CROSSTHEDELAWARERIVERONCHRISTMASBEATTHEBRITISHWITHASURPRISEATTACK&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And finally split up properly:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cross the Delaware River on Christmas beat the British with a surprise attack&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="375" LastEditorUserId="375" LastEditDate="2016-01-09T20:47:47.537" LastActivityDate="2016-01-09T20:47:47.537" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25233" PostTypeId="1" AcceptedAnswerId="25236" CreationDate="2016-01-09T20:58:24.373" Score="6" ViewCount="399" Body="&lt;p&gt;One day you were taking a hike in some foothills somewhere near the East Coast of the US. It was clear another hac&amp;ndash; umm, &lt;em&gt;hiker&lt;/em&gt; had already been there when you came across the following munged signage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jc5U4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jc5U4.png&quot; alt=&quot;look around&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was a note attached to it which read:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   -T+H-E--K-E+Y+-I+S+&#xA;   +I-N-V+I-S-I+B-L-E+&#xA;&#xA;Y-E-T+T-H+E+M-E+S+S-A+G-E&#xA;I+S-+O-B-S+C+U-R+E+D-+B+Y&#xA;T+H+A-T+--W+H+I+C-H-++I-S&#xA;&#xA;   -C+O+N-C-E+A-L-E+D-&#xA;   -×+×-×+×I+N×+×-×-×+&#xA;   P+L-A+I-N-S+I-G-H-T&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Where were you? What was the arrow pointing to?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="17147" LastEditorUserId="17147" LastEditDate="2016-01-09T22:46:58.870" LastActivityDate="2016-01-10T02:58:40.980" Title="When hackers hike" Tags="|enigmatic-puzzle|computer-puzzle|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="25234" PostTypeId="2" ParentId="25233" CreationDate="2016-01-09T23:03:20.443" Score="8" Body="&lt;h2&gt;Partial solution&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The PNG file in the question is corrupted, with 19 extra bytes in the second &lt;code&gt;IDAT&lt;/code&gt; chunk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At offset &lt;code&gt;0x000106BC&lt;/code&gt; in the file there is a 19-byte text string:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;PCLOANICNESAILGEHDT&lt;/code&gt;, which is &lt;code&gt;PLAINSIGHT&lt;/code&gt; interleaved with &lt;code&gt;CONCEALED&lt;/code&gt; (i.e. &quot;concealed in plain sight&quot;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Removing the offending bytes results in an uncorrupted image:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/8BTEF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It looks like a word puzzle from here on out, so I'll leave the final solution to someone else.&lt;/p&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="9718" LastEditDate="2016-01-10T00:12:55.610" LastActivityDate="2016-01-10T00:12:55.610" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25236" PostTypeId="2" ParentId="25233" CreationDate="2016-01-10T00:45:41.180" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As seen in the answer from &lt;em&gt;2012rcampion&lt;/em&gt; fixing the corrupted PNG image reveals the next step:&lt;br&gt;&lt;br&gt;KCTVCEDCA&lt;br&gt;&lt;br&gt;Using the &lt;a href=&quot;http://www.dcode.fr/vigenere-cipher&quot; rel=&quot;nofollow&quot;&gt;Viginere Encoder&lt;/a&gt; (not decoder) and the key INVISIBLE we get:&lt;br&gt;&lt;br&gt;&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Spodumene&quot; rel=&quot;nofollow&quot;&gt;SPODUMENE&lt;/a&gt;&lt;br&gt;&lt;br&gt;&lt;s&gt;I guess the &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Black_Hills&quot; rel=&quot;nofollow&quot;&gt;Black Hills&lt;/a&gt; mentioned in the Wikipedia article are a little too far away from the East Coast. Maybe you are in &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Huntington,_Massachusetts&quot; rel=&quot;nofollow&quot;&gt;Huntington&lt;/a&gt; (see the image in the Spodumene article). &lt;a href=&quot;https://en.m.wikisource.org/wiki/1911_Encyclop%C3%A6dia_Britannica/Spodumene&quot; rel=&quot;nofollow&quot;&gt;Encyclopædia Britannica&lt;/a&gt; mentions more locations in Massachusetts.&lt;/s&gt;&lt;br&gt;&lt;br&gt;As &lt;em&gt;Daniil Agashiyev&lt;/em&gt; found out the answer to both question is probably &lt;strong&gt;Hiddenite&lt;/strong&gt; which is a &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Hiddenite&quot; rel=&quot;nofollow&quot;&gt;variety of Spodumene&lt;/a&gt; and a &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Hiddenite,_North_Carolina&quot; rel=&quot;nofollow&quot;&gt;town in North Carolina&lt;/a&gt;.&lt;br&gt;&lt;br&gt;There is also a red herring in the question. The + and - signs can be splitted to groups of 8 and interpreted as ASCII. The result is &lt;em&gt;CrImSoNfIsH&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2016-01-10T02:58:40.980" LastActivityDate="2016-01-10T02:58:40.980" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25237" PostTypeId="2" ParentId="25183" CreationDate="2016-01-10T00:50:48.317" Score="1" Body="&lt;p&gt;Is it a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; volcano?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;...only hear me when I blow&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The eruption.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...only see me when I want&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When the volcano &quot;chooses&quot; to erupt.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be there beside with no distance&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The tremors from the eruption reach far away.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But once you know what I am \ You don't want to meet me again.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because death.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12795" LastActivityDate="2016-01-10T00:50:48.317" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25239" PostTypeId="2" ParentId="25194" CreationDate="2016-01-10T09:07:08.613" Score="0" Body="&lt;p&gt;I think it's:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; -1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Less one, and my maker is often said to be the mightiest between us.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Less one is also minus one. &quot;1&quot; would therefore be &quot;my marker&quot;. 1 is one &quot;mightier&quot; than -1 in fundamental mathematics because it preserves identity in multiplication.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alone I smite, yet my greater does not.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In addition -1 smites by taking one away from a number, whereas &quot;my greater&quot;, zero, doesn't take anything away from a number in addition. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastActivityDate="2016-01-10T09:07:08.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25240" PostTypeId="1" AcceptedAnswerId="25246" CreationDate="2016-01-10T16:49:15.050" Score="23" ViewCount="3274" Body="&lt;p&gt;Somewhere on the Hyperbolic Plane, you encounter ten mathematicians. You know that some of them always tell the truth while the rest always lie. Each of them knows which of the others tell the truth and which do not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $g$ be the number of mathematicians that tell the truth. Let $h$ be the &lt;em&gt;sum of the indices&lt;/em&gt; of the mathematicians that tell the truth (e.g. if mathematicians 4 and 5 tell the truth, $h=4+5=9$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You ask the mathematicians for the values of $g$ and $h$ and receive the following answers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mathematician&amp;nbsp;&amp;nbsp; 1:&lt;/strong&gt; $\quad g = 6 \quad \lor \quad h = 12$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mathematician&amp;nbsp;&amp;nbsp; 2:&lt;/strong&gt; $\quad g = 3 \quad \lor \quad h = 12$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mathematician&amp;nbsp;&amp;nbsp; 3:&lt;/strong&gt; $\quad g = 4 \quad \lor \quad h = 11$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mathematician&amp;nbsp;&amp;nbsp; 4:&lt;/strong&gt; $\quad g = 4 \quad \lor \quad h = 25$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mathematician&amp;nbsp;&amp;nbsp; 5:&lt;/strong&gt; $\quad g = 2 \quad \lor \quad h = 17$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mathematician&amp;nbsp;&amp;nbsp; 6:&lt;/strong&gt; $\quad g = 3 \quad \lor \quad h = 24$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mathematician&amp;nbsp;&amp;nbsp; 7:&lt;/strong&gt; $\quad g = 5 \quad \lor \quad h = 23$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mathematician&amp;nbsp;&amp;nbsp; 8:&lt;/strong&gt; $\quad g = 5 \quad \lor \quad h = 11$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mathematician&amp;nbsp;&amp;nbsp; 9:&lt;/strong&gt; $\quad g = 6 \quad \lor \quad h = 23$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mathematician 10:&lt;/strong&gt; $\quad g = 5 \quad \lor \quad h = 16$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Q:&lt;/strong&gt; Which of the mathematicians are telling the truth?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Note 1:&lt;/strong&gt; $\lor$ is the boolean &quot;or&quot; operator, so only &lt;strong&gt;one&lt;/strong&gt; of the statements needs to be correct in order for a mathematician to be telling the truth.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Note 2:&lt;/strong&gt; I constructed this puzzle using a search program that enumerates all possibilities to ensure correctness. It would therefore be nice to get solutions that do&lt;/em&gt; not &lt;em&gt;rely on computer programs themselves.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user18360" LastActivityDate="2016-01-11T22:05:27.180" Title="Ten Tricky Mathematicians" Tags="|mathematics|logical-deduction|liars|" AnswerCount="3" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="25241" PostTypeId="1" AcceptedAnswerId="25242" CreationDate="2016-01-10T17:00:03.627" Score="4" ViewCount="5339" Body="&lt;p&gt;There was an IQ test created by Random Person. A contestant attempts to answer all the questions but fails the last one. Here goes the conversation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Random Person:&lt;/strong&gt; You have 500 bricks in the plane. You drop one out. How many do you have left?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Contestant:&lt;/strong&gt; Easy. 499 bricks.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Random Person:&lt;/strong&gt; What three steps are needed to put an elephant into a fridge?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Contestant:&lt;/strong&gt; Open the fridge, put the elephant inside, and close it.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Random Person:&lt;/strong&gt; Now what four steps are needed to put the deer inside the fridge?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Contestant:&lt;/strong&gt; Easy, open the fridge, take out the elephant, put in the deer, then finally, close the fridge.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Troll:&lt;/strong&gt; The Lion King's party had every animal but who?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Contestant:&lt;/strong&gt; The deer&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Random Person:&lt;/strong&gt; The old lady crosses the river filled with crocodiles. How does she do it?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Contestant:&lt;/strong&gt; All the crocodiles are at the Lion King's party&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Random Person:&lt;/strong&gt; But she still dies afterwards. Why?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Contestant:&lt;/strong&gt; Uhhh, she falls into a lava pit?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Random Person:&lt;/strong&gt; Haha! You have failed the test! Now get out of here! IQ score: 0/100&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How did the lady die from crossing the river? She made it to the other side, not eaten and not injured from crossing the river.&lt;/p&gt;&#xA;" OwnerUserId="9042" LastEditorUserId="5373" LastEditDate="2017-12-10T19:16:16.030" LastActivityDate="2019-10-29T15:33:46.997" Title="Why did the lady die?" Tags="|lateral-thinking|situation|" AnswerCount="3" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25242" PostTypeId="2" ParentId="25241" CreationDate="2016-01-10T17:02:07.367" Score="21" Body="&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The brick (from the plane) fell on her head.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12092" LastActivityDate="2016-01-10T17:02:07.367" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25244" PostTypeId="2" ParentId="25165" CreationDate="2016-01-10T17:04:01.737" Score="2" Body="&lt;p&gt;None of them are necessarily drunk. If any logician was able to drive home, they would have said &quot;yes.&quot; We conclude that each logician knows that they personally cannot drive home, but that could be because they don't know how to drive or they left their car at home, not necessarily because they're drunk. Each logician surely also feels that it is very likely that somebody, somewhere in the world, is capable of driving home. However, they do not know this for sure (some unlikely circumstance may have rendered everybody who knows how to drive unable to do so, or maybe everyone who is in a fit state to drive home is already at home) so they answer, &quot;I don't know.&quot;&lt;/p&gt;&#xA;" OwnerUserId="2023" LastActivityDate="2016-01-10T17:04:01.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25246" PostTypeId="2" ParentId="25240" CreationDate="2016-01-10T17:06:41.427" Score="25" Body="&lt;p&gt;By case distinction on the value $g$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $g=1$, then the unique truth teller Mx must claim that $h=x$; contradiction.&lt;/li&gt;&#xA;&lt;li&gt;If $g=2$, then M5 tells the truth ($g=2$). The other truth teller Mx must claim that $h=x+5$; contradiction.&lt;/li&gt;&#xA;&lt;li&gt;If $g=3$, then M2 and M6 tell the truth ($g=3$). The other truth teller Mx must claim that $h=x+2+6=x+8$; contradiction.&lt;/li&gt;&#xA;&lt;li&gt;If $g=4$, then M3 and M4 tell the truth ($g=4$). The other two truth tellers Mx and My must claim that $h=x+y+7$; in particular, they both must claim the same value of $h$. The pairs (M1 and M2) and (M3 and M8) do not work.&#xA;Only (M7 and M9) works.&lt;/li&gt;&#xA;&lt;li&gt;If $g=5$, then M7, M8, M10 tell the truth. The other two truth tellers Mx and My must claim that $h=x+y+25$; this value $h$ is too large.&lt;/li&gt;&#xA;&lt;li&gt;If $g\ge6$, then at least four guys must state the correct value of $h$; contradiction.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To summarize:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; M3, M4, M7 and M9 tell the truth with $g=4$ and $h=23$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Furthermore, there is a second solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; None of them tells the truth with $g=0$ and $h=0$. (For this, you must accept that the empty sum takes value $0$.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-01-10T17:53:44.813" LastActivityDate="2016-01-10T17:53:44.813" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25247" PostTypeId="2" ParentId="25201" CreationDate="2016-01-10T17:37:47.620" Score="14" Body="&lt;p&gt;I believe that the solution is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Count on a number line, and multiply by -1 if the $n$-th number in Pascal's triangle is even, i.e. the map:&lt;br&gt;&#xA;&lt;br&gt;&#xA; $$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; 2  &amp;amp; 1 &amp;amp; 1 &amp;amp; 3 &amp;amp; 3 &amp;amp; 1 &amp;amp; 1 &amp;amp; 4 &amp;amp; 6 &amp;amp; 4 &amp;amp; 1 \\ \hline 1 &amp;amp; 2 &amp;amp; 3 &amp;amp; 4 &amp;amp; -5 &amp;amp; 6 &amp;amp; 7 &amp;amp; 8 &amp;amp; 9 &amp;amp; 10 &amp;amp; 11 &amp;amp; -12 &amp;amp; -13 &amp;amp; -14* &amp;amp; 15 \\ \hline \end{array}$$&lt;br&gt;&#xA; $$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline 1  &amp;amp; 5  &amp;amp; 10  &amp;amp; 10  &amp;amp; 5  &amp;amp; 1  &amp;amp; 1  &amp;amp; 6   &amp;amp; 15 &amp;amp; 20  &amp;amp; 15 &amp;amp; 6   &amp;amp; 1  \\ \hline  16 &amp;amp; 17 &amp;amp; -18 &amp;amp; -19 &amp;amp; 20 &amp;amp; 21 &amp;amp; 22 &amp;amp; -23 &amp;amp; 24 &amp;amp; -25 &amp;amp; 26 &amp;amp; -27 &amp;amp; 28 \\ \hline \end{array}$$&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;code&gt;*&lt;/code&gt; marking a potential mistake in encoding?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Alter&quot; each letter appropriately using the second row of the map, and you get&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;TIMEFLIESWHENX*OUAREHAVINGFUN&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Where &quot;alter&quot; means shift modulo 26, e.g.,&lt;br&gt;&#xA; 1st letter: S + 1 = T&lt;br&gt;&#xA; 2nd letter: G + 2 = I&lt;br&gt;&#xA; 5th letter: K - 5 = F&lt;br&gt;&#xA; 8th letter: W + 8 = E&lt;br&gt;&#xA; 27th letter: V - 27 = U&lt;br&gt;&#xA;&lt;br&gt;&#xA; In Python this could be:&lt;br&gt;&#xA; &lt;code&gt;lambda x,y: chr( ((ord(x) - ord('A') + y) % 26) + ord('A') )&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Note that the asterisk on the -14 above indicates the real shift should be -13 instead, giving a Y instead of an X&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and thus my final answer is the phrase&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Time flies when you are having fun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17147" LastEditorUserId="17147" LastEditDate="2016-01-10T19:09:59.443" LastActivityDate="2016-01-10T19:09:59.443" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25248" PostTypeId="1" AcceptedAnswerId="45604" CreationDate="2016-01-10T18:06:54.170" Score="46" ViewCount="3316" Body="&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My last three letters serve to clarify.&lt;br&gt;My first four letters are where famous people speak.&lt;br&gt;Dropping all consonants, I am almost a curse.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The solution is a single English word and, once found, will leave no doubt that it is the intended one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The solution explains how the riddle must be modified to yield the solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user18360" LastEditorUserId="21234" LastEditDate="2016-11-10T21:50:13.943" LastActivityDate="2016-11-13T04:24:31.467" Title="Almost a curse?" Tags="|riddle|word|english|" AnswerCount="14" CommentCount="22" ContentLicense="CC BY-SA 3.0" />
  <row Id="25249" PostTypeId="1" AcceptedAnswerId="25252" CreationDate="2016-01-10T18:31:46.873" Score="14" ViewCount="1481" Body="&lt;p&gt;Alice and Bob play the following game with $n\ge2016$ chips on the table.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Alice and Bob move alternatingly. Alice makes the first move.&lt;/li&gt;&#xA;&lt;li&gt;In every move, the active player removes $k$ chips from the table, where either $k=1$, or $k$ is a multiple of $23$, or $k$ is a prime number.&lt;/li&gt;&#xA;&lt;li&gt;The player who takes the &lt;em&gt;last&lt;/em&gt; chip wins the game.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Which player is going to win this game?&#xA;  (As usual, we assume that Alice and Bob both use optimal strategies.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2016-01-11T09:30:29.773" Title="Removing chips from the table" Tags="|mathematics|strategy|game|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25250" PostTypeId="2" ParentId="25248" CreationDate="2016-01-10T18:34:57.517" Score="9" Body="&lt;p&gt;My guess is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Show&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My last three letters serve to clarify.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; last three letters are 'how' which is asked when we need clarification&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My first four letters are where famous people speak.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; famous people speak at different shows.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Dropping all consonants, I am almost a curse.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When dropped all consonents then has some pronouncing similarity with 'whore'. Or as APrough commented &quot;You also go &quot;Oh, #$%^@&quot; (insert your own favorite curse word). So, technically it is almost a curse.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-01-13T04:41:29.430" LastActivityDate="2016-01-13T04:41:29.430" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="25251" PostTypeId="1" AcceptedAnswerId="25273" CreationDate="2016-01-10T18:42:31.017" Score="9" ViewCount="500" Body="&lt;p&gt;I am far from perfect&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though perfect is like me&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I acted like an idiot&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On live late night TV&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could say I am a cat&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With stripes of orange and black&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I am more an angry bull&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For flags make me attack&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first you've never heard before&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My last is also rare&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it's the part that's perfect&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(almost, not everywhere)&lt;/p&gt;&#xA;" OwnerUserId="17712" LastActivityDate="2016-01-11T14:03:11.473" Title="I'm far from perfect. Who Am I?" Tags="|riddle|knowledge|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25252" PostTypeId="2" ParentId="25249" CreationDate="2016-01-10T18:46:53.783" Score="23" Body="&lt;p&gt;From any multiple of 4 less than 92, any move leaves a number of chips that is not divisible by 4. Then, removing 1, 2, or 3 chips results in another multiple of 4. Therefore, any multiple of 4 less than 92 is a losing position, and all other numbers less than 92 are winning positions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From any number greater than or equal to 92, it is possible to remove a multiple of 23 chips to leave a multiple of 4 less than 92 (including zero). Therefore, Alice wins.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2016-01-10T18:46:53.783" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25256" PostTypeId="2" ParentId="25240" CreationDate="2016-01-10T20:23:08.250" Score="6" Body="&lt;p&gt;By iterating through the possible values of &lt;code&gt;g&lt;/code&gt;, then filling the remaining missing mathematicians who use the correct value &lt;code&gt;h&lt;/code&gt;, I was able to determine another solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mathematicians 2 and 6 claimed the value &lt;code&gt;g = 3&lt;/code&gt;, and by iterating through the remaining mathematicians, Mathematician 3 claimed the value &lt;code&gt;h = 11&lt;/code&gt;. 3 mathematicians, and sum of indices = 11.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upon further consideration, my answer is actually invalid:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If this solution is valid, then Mathematician 8 is also telling the truth (e.g. &lt;code&gt;h = 11&lt;/code&gt;) which cannot be, so this is a contradiction and is not a valid solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18367" LastEditorUserId="18367" LastEditDate="2016-01-10T20:52:45.680" LastActivityDate="2016-01-10T20:52:45.680" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25257" PostTypeId="2" ParentId="20812" CreationDate="2016-01-10T21:57:47.990" Score="5" Body="&lt;p&gt;While the accepted answer was also the first one I thought of, there's also a nice solution with &lt;a href=&quot;http://mathworld.wolfram.com/Subfactorial.html&quot; rel=&quot;noreferrer&quot;&gt;subfactorials&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$2\;!2 = !\sqrt{9}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="228" LastActivityDate="2016-01-10T21:57:47.990" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25258" PostTypeId="1" AcceptedAnswerId="25260" CreationDate="2016-01-10T22:10:55.307" Score="3" ViewCount="1215" Body="&lt;p&gt;Find the missing words in the chain, where consecutive pairs of words must form a phrase. &lt;br&gt;&#xA;For example,&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;HOLY&lt;/strong&gt;&lt;br&gt;&#xA;  - - - -&lt;br&gt;&#xA;  - - - -&lt;br&gt; &#xA;  &lt;strong&gt;RUN&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;has the solution &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;HOLY&lt;/strong&gt;&lt;br&gt;&#xA; &lt;strong&gt;WAR&lt;/strong&gt;&lt;br&gt;&#xA; &lt;strong&gt;TRIAL&lt;/strong&gt;&lt;br&gt; &lt;strong&gt;RUN&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The puzzle is based on the game show &lt;a href=&quot;https://en.wikipedia.org/wiki/Chain_Reaction_(game_show)&quot; rel=&quot;nofollow&quot;&gt;Chain Reaction&lt;/a&gt;.&lt;br&gt;&lt;br&gt;&#xA;1.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;BLACK&lt;/strong&gt;&lt;br&gt;&#xA;  - - - -&lt;br&gt;&#xA;  - - - -&lt;br&gt; &#xA;  &lt;strong&gt;PUZZLE&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;PUZZLE&lt;/strong&gt;&lt;br&gt;&#xA;  - - - -&lt;br&gt;&#xA;  - - - -&lt;br&gt; &#xA;  &lt;strong&gt;SANDWICH&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;STACK&lt;/strong&gt;&lt;br&gt; &#xA;   - - - -&lt;br&gt;&#xA;   - - - -&lt;br&gt;&#xA;   &lt;strong&gt;DUE&lt;/strong&gt; &lt;br&gt;&#xA;   - - - -&lt;br&gt;&#xA;   - - - -&lt;br&gt;&#xA;   &lt;strong&gt;SHIFT&lt;/strong&gt;&lt;br&gt;&#xA;   - - - -&lt;br&gt;&#xA;   - - - -&lt;br&gt; &#xA;   - - - -&lt;br&gt;&#xA;  &lt;strong&gt;EXCHANGE&lt;/strong&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7489" LastEditorUserId="11876" LastEditDate="2016-01-10T23:41:44.863" LastActivityDate="2016-01-11T02:05:44.867" Title="Phrasal Word Chains" Tags="|word|english|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25260" PostTypeId="2" ParentId="25258" CreationDate="2016-01-10T23:03:59.610" Score="4" Body="&lt;p&gt;Possible solutions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BLACK&lt;br&gt;&#xA; MAGIC&lt;br&gt;&#xA; WORD&lt;br&gt;&#xA; PUZZLE &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PUZZLE&lt;br&gt;&#xA; BOOK&lt;br&gt;&#xA; CLUB&lt;br&gt;&#xA; SANDWICH  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; STACK&lt;br&gt;&#xA; TRACE&lt;br&gt;&#xA; AMOUNT&lt;br&gt;&#xA; DUE&lt;br&gt;&#xA; DATE&lt;br&gt;&#xA; NIGHT&lt;br&gt;&#xA; SHIFT&lt;br&gt;&#xA; GROUND&lt;br&gt;&#xA; BEEF&lt;br&gt;&#xA; STOCK&lt;br&gt;&#xA; EXCHANGE  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17147" LastEditorUserId="17147" LastEditDate="2016-01-11T02:05:44.867" LastActivityDate="2016-01-11T02:05:44.867" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="25263" PostTypeId="2" ParentId="25241" CreationDate="2016-01-11T00:33:19.270" Score="14" Body="&lt;p&gt;The old lady died afterwards because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everyone dies eventually, and she was old so her death within a reasonably short time was expected anyway. She died in her sleep, painlessly and of natural causes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user17947" LastActivityDate="2016-01-11T00:33:19.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25264" PostTypeId="1" AcceptedAnswerId="25271" CreationDate="2016-01-11T05:40:46.217" Score="3" ViewCount="294" Body="&lt;p&gt;Based on &lt;a href=&quot;https://puzzling.stackexchange.com/questions/24948/the-square-and-the-compass&quot;&gt;The square and the compass&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rules are &lt;em&gt;almost&lt;/em&gt; the same. (The only difference is the actual task, marked in bold.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a compass and a pencil but no scale/straightedge. &lt;s&gt; Your job is to mark four points on a plane paper that would form a square if joined.&lt;/s&gt; &lt;strong&gt;You are given two points on a plane paper. Your job is to find the midpoint of line segment formed by the two points, if joined.&lt;/strong&gt;  Your result has to be perfectly accurate (not approximate) and should be possible in a finite number of moves. The following moves are valid.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Make the compass radius equal to the distance between two already marked points.&lt;/li&gt;&#xA;&lt;li&gt;Draw a circle with any marked point as centre.&lt;/li&gt;&#xA;&lt;li&gt;Use any intersection of arcs/circles as a marked point.&lt;/li&gt;&#xA;&lt;li&gt;Select a continuous region (either an arc or a 2D region) and mark an &lt;em&gt;approximate&lt;/em&gt; point. For example, I could draw an arc and then mark a point that is approximately (but not exactly) at the centre.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I do not know how to solve this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I do know how to do it in infinite moves, though.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.923" LastActivityDate="2016-01-11T10:03:42.823" Title="The square and the compass II - Midpoints" Tags="|mathematics|geometry|construction|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25265" PostTypeId="2" ParentId="25264" CreationDate="2016-01-11T06:20:58.793" Score="4" Body="&lt;p&gt;Doing this in infinite moves is relatively trivial, so I'll post an answer that does that while I think about the harder problem of doing it in finite moves.  Maybe this method will inspire somebody.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mark approximately the midpoint of the two points.  Now use the compass to create two circles of the same radius, such that each has an endpoint as its center, each contains the approximate midpoint, but neither contains the other endpoint (there are many possible radii here, any will do).  Then take the intersections of those two circles as two new endpoints and repeat the process.  Each time, your endpoints are closer together.  Also, since you always use equal radii, the line between the points of intersection always goes through the exact center.  Thus, after infinite steps you arrive at precisely the midpoint of the original two points.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17389" LastActivityDate="2016-01-11T06:20:58.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25270" PostTypeId="2" ParentId="25249" CreationDate="2016-01-11T09:30:29.773" Score="12" Body="&lt;p&gt;On their turn, each player can remove &lt;code&gt;1&lt;/code&gt;, &lt;code&gt;any prime number&lt;/code&gt;, or &lt;code&gt;23*n&lt;/code&gt; chips on their turn (for any integer value &lt;code&gt;n&lt;/code&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fundamentally, the goal of the game is to take the last chip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key thing to note is that amongst the numbers of chips we can remove are 1 (explicitly allowed), 2 (prime), and 3 (prime), but not 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that if a player starts their turn with 4 chips on the table, they have already lost the game (since their only legal moves are to remove 1, 2, or 3, and their opponent will respond by removing 3, 2, or 1, respectively, thus removing the last chip)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this in turn means that if a player can, as their move, reduce the table to 4 chips, then they will win on their next move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This logic also applies for 8;  if a player can bring the count of chips to 8, then they will win in two more rounds of play, the first of which will bring the value to 4, and the second will bring the value to 0, through exactly this same process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the same way, it applies to 12, 16, etc.  All the multiples of 4.  The optimum strategy for this game, as it turns out, is very concerned with multiples of 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whomever can get the number of chips to a multiple of 4 first, wins the game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there's one exception to this.  If our opponent can somehow take the value from one multiple of 4 to a different multiple of 4, then they can take over the game.  So!  Since the first player (Alice) will use her first move to get the number of chips to a power of four, Bob wants to remove a multiple of 4 on his turn in order to gain control of the game (n.b.: &lt;code&gt;(4x) - (4y) == 4(x-y)&lt;/code&gt;.  That is, subtracting any multiple of 4 from another multiple of four leaves you with a multiple of four as the result)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as a reminder, Bob's options for number of chips to remove are:  1 (not a multiple of 4), any prime (not a multiple of 4, by definition, and 4 itself is not prime), or any multiple of 23.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it happens, there are a whole bunch of multiples of 23 which are also multiples of 4.  &lt;code&gt;23*4 == 92&lt;/code&gt; is the smallest positive one.  So to take back control of the game, all Bob has to do is to take 92 chips from the table, and now he can play the &quot;stay on a multiple of 4&quot; game that we describe above, and be guaranteed a win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, Alice (being the annoyingly perfect logician that she is), knows that this is Bob's only possible escape from this strategy.  And so she'll block it with her first move, in which she'll remove (23*84==) 1932 chips, leaving only 84 on the table.  From here, Bob can't use the magic &quot;take 92 chips&quot; move which would give him control control of the game, as there aren't that many chips available for him to take&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No matter what move Bob makes, it will result in a non-multiple-of-four number of chips, and Alice can then remove 1, 2, or 3 chips to return the chips to a lower multiple of four.  And eventually, win the game.&lt;/p&gt;&#xA;" OwnerUserId="9286" LastActivityDate="2016-01-11T09:30:29.773" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25271" PostTypeId="2" ParentId="25264" CreationDate="2016-01-11T10:03:42.823" Score="7" Body="&lt;p&gt;I fully copied below answer from &lt;a href=&quot;https://math.stackexchange.com/questions/227285/constructing-the-midpoint-of-a-segment-by-compass&quot;&gt;https://math.stackexchange.com/questions/227285/constructing-the-midpoint-of-a-segment-by-compass&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Some Googling revealed the following comments to &lt;a href=&quot;https://math.stackexchange.com/a/147034&quot;&gt;this answer&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;I know it is possible, but is there an easy way to divide a segment in half with only a compass? – robjohn♦ May 20 at 3:46&lt;/li&gt;&#xA;  &lt;li&gt;I don't know if that's &quot;easy&quot;, but here's one method:&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Find the point $C$ on the ray from $A$ through $B$ such that $|AC|=2|AB|$ using my previous comment [The relevant part: &quot;To double the distance along a ray, use the construction of a regular hexagon with vertex $A$ and center $B$&quot;.]&lt;/li&gt;&#xA;  &lt;li&gt;Intersect the circle with center $C$ through $A$ with the circle with center $A$ through $B$ to find $D_1,D_2$.&lt;/li&gt;&#xA;  &lt;li&gt;The midpoint of $AB$ is the second point of intersection of the two circles with center $D_i$ through $A$. – t.b. May 20 at 9:28&lt;/li&gt;&#xA;  &lt;/ol&gt;&lt;/li&gt;&#xA;  &lt;li&gt;Here is a picture of what I have in mind: - t.b. May 20 at 12:38&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/E06Ml.png&quot; alt=&quot;Division in half&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dotted line is not used in the construction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The triangles $\Delta ACD_1$ and $\Delta AMD_1$ are isosceles by construction and they share a common angle, hence they are similar. Therefore $AM : AB = AM : AD_1 = AD_1 : AC = AB : AC = 1 : 2$.&lt;/p&gt;&#xA;" OwnerUserId="1748" LastEditorUserId="-1" LastEditDate="2017-04-13T12:19:36.950" LastActivityDate="2016-01-11T10:03:42.823" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25272" PostTypeId="2" ParentId="25208" CreationDate="2016-01-11T10:43:13.130" Score="13" Body="&lt;p&gt;Since there is the &lt;em&gt;lateral thinking&lt;/em&gt; tag, I am posting what came into my mind.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In a house with no doors or windows, how did the man and the maids get in?&#xA; How did they get the supplies and all? So the whole question is a lie and&#xA; 'JoeBeastlyGerbil' is the liar.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Since 'the liar is the murderer', 'JoeBeastlyGerbil' is the murderer&#xA; of the man who actually didn't exist.&lt;br&gt;&#xA;&lt;br&gt;&#xA; And as we have stated, 'JoeBeastlyGerbil' is a liar. So his statement&#xA; 'the liar is the murderer' will also become a lie. Which means 'the&#xA; liar is not the murderer'. Which will result in a paradox. :P&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="17147" LastEditDate="2016-01-11T21:25:43.437" LastActivityDate="2016-01-11T21:25:43.437" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25273" PostTypeId="2" ParentId="25251" CreationDate="2016-01-11T12:52:32.893" Score="7" Body="&lt;p&gt;I think you are...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Vontaze Burfict, the Cincinnati Bengal player whose on-field antics caused his team to lose the AFC wild card game in the 2015 NFL playoffs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am far from perfect&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Burfict is known for causing issues on the field.  In addition, though he was considered a very good draft choice, his on- and off- field antics in college and poor performance at the combine, marked him lower in the eyes of teams, and he went undrafted.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though perfect is like me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; last names rhymes with perfect&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I acted like an idiot&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Happens almost every year.  He inevitably causes a fine-worthy penalty sometime during the season.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On live late night TV&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  The Pittsburgh - Cincinatti playoff occurred at night on January 9th, 2016.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You could say I am a cat&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With stripes of orange and black&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cincinnati Bengals, named after Paul Brown's high school team&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I am more an angry bull&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  See above concerning on- and off- field antics&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For flags make me attack&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Seeing a yellow flag on the field seems to make him angrier and cause more issues on the field.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My first you've never heard before&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Vontaze is not really a common first name, at least in the U.S.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My last is also rare&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Burfict is also not a really common surname either.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But it's the part that's perfect&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(almost, not everywhere)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, last name rhymes with perfect.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of course, this could also apply to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Adam Jones (aka Pac-man), Burfict's teammate who caused an additional penalty after the Burfict penalty for pushing a ref on the field.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14233" LastEditorUserId="14233" LastEditDate="2016-01-11T14:03:11.473" LastActivityDate="2016-01-11T14:03:11.473" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25274" PostTypeId="2" ParentId="25215" CreationDate="2016-01-11T13:07:29.800" Score="5" Body="&lt;p&gt;Hm, i have a wild guess, but i can not really backup it with facts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a Dragon&lt;br&gt;&lt;br&gt;&#xA;&#xA; there is a dragon fruit in asia, dont know much about it, but maybe this is meant by the tiny fruit&lt;br&gt;&lt;br&gt;&#xA;&#xA; I remember a myth about a giant dragon, which created some island when dying and falling into the sea. But i can not remember where i heard that myth, but i think it was about an asian country...&lt;br&gt;&lt;br&gt;&#xA;&#xA; Dragons in myths are known to speak all languages...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13748" LastEditorUserId="13748" LastEditDate="2016-01-11T13:15:35.453" LastActivityDate="2016-01-11T13:15:35.453" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25276" PostTypeId="2" ParentId="25240" CreationDate="2016-01-11T14:52:22.407" Score="3" Body="&lt;p&gt;The truth tellers are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; M3, M4, M7 and M9&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Can't be 2 truth tellers because that would mean M5 and someone else.  All the others don't have their index plus 5 equal to their stated h, so no go.&lt;br&gt;&#xA; Likewise there can't be g = 3 because we have M2 and M6. But there's no single other mathematician with h = index + 2 + 6, so no go.&lt;br&gt;&#xA; Can't be g = 5 as 7 + 8 + 9 = 24. Minimum other two h's must be greater than be 27.  No h &gt; 27, so no go.&lt;br&gt;&#xA; Can't be g = 6 as then we'd have M1 and M9.  We'd then need 4 identical h's which isn't so, so no go.&lt;br&gt;&#xA; This leaves g = 4 which gives us M3 and M4 with g = 4 and M7 and M9 with h = 3 + 4 + 7 + 9&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;s&gt;Sorry output's so ugly.  Can't get spoilers to newline properly.&lt;/s&gt; Thank you @The Dark Truth&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just started looking at other answers and from @Gamow saw I missed g = 1 and g &gt;= 7&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apologies to @Gamow for stealing M[0-9]+ notation.&lt;/p&gt;&#xA;" OwnerUserId="10118" LastEditorUserId="10118" LastEditDate="2016-01-11T22:05:27.180" LastActivityDate="2016-01-11T22:05:27.180" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25277" PostTypeId="1" AcceptedAnswerId="25279" CreationDate="2016-01-11T18:31:31.303" Score="6" ViewCount="652" Body="&lt;p&gt;Enjoy this relatively easy Mtg themed crossword. All entries are card names (punctuation and spaces removed but otherwise preserved). Yes some of them are going to be hard to get in a vacuum, but fill in the surrounding entries and that should help. GL HF!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FhBuZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FhBuZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Horizontal&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;$1$. I forget what set this monkey was from...maybe Iraqi Nights?&lt;br&gt;&#xA;$7$. A one-time color shifting of 16-across.&lt;br&gt;&#xA;$10$. The perfectly average way to say no.&lt;br&gt;&#xA;$12$. 17,18,19, aaaand 20! Got a stifle?&lt;br&gt;&#xA;$13$. The latest in a line of identical elves stretching back to Alpha.&lt;br&gt;&#xA;$16$. The first of its kind, it's functionality is later passed on to another with a less religious name.&lt;br&gt;&#xA;$17$. Synonymous with two damage in magic.&lt;br&gt;&#xA;$18$. Romanian vampires in Ravnica?&lt;br&gt;&#xA;$19$. &quot;I'll have what she's having!&quot; Break out the rulebook for this Yoked Ox.&lt;br&gt;&#xA;$24$. Sort of like The One Ring of Tolkien's world (I guess that makes Karn Sauron)&lt;br&gt;&#xA;$25$. Don't rock the boat!&lt;br&gt;&#xA;$28$. Oh I could just go on and on about this card: it's so spectacular and powerful--who doesn't love drawing cards--and...&lt;br&gt;&#xA;$29$. This was his first appearance; it isn't until years later that we learn it's actually a mispronounciation of the real name!&lt;br&gt;&#xA;$31$. When you cast this card from Alpha, your opponent will likely look like the depiction in the art.&lt;br&gt;&#xA;$32$. It's no lion but it is a cat that enjoys savannahs.&lt;br&gt;&#xA;$33$. Non-artifact? Check! Non-black? Check! Bye bye!&lt;br&gt;&#xA;$35$. Its mere approach sends Jeskai decks running.&lt;br&gt;&#xA;$36$. Bruce Willis' favorite card.&lt;br&gt;&#xA;$37$. IT'S NO ICY!...it's no icy&lt;br&gt;&#xA;$38$. Cloudy with a chance of meteor shower.&lt;br&gt;&#xA;$39$. Recently made playable in legacy for the first time ever.  &lt;/p&gt;&#xA;&#xA;&lt;h2&gt; Vertical&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;$1$. Meaning fate, this card was later color shifted into blue.&lt;br&gt;&#xA;$2$. Paradoxically can be used against Soulless One.&lt;br&gt;&#xA;$3$. A member of magic's creepiest creature type, this one had the most success.&lt;br&gt;&#xA;$4$. This card really 'got it!' until the rules change stopped damage from using the stack.&lt;br&gt;&#xA;$5$. Great! First I had to go help out on the farm, now I have to go on a stupid journey to a new land!&lt;br&gt;&#xA;$6$. THE magic card!&lt;br&gt;&#xA;$8$. Another? Don't mind if I do!&lt;br&gt;&#xA;$9$. Sometimes 8-down is just a reeally expensive (\$) one of these.&lt;br&gt;&#xA;$11$. At least neither of us get to do anyth-wait that's not fair! And what's going on in this art?&lt;br&gt;&#xA;$14$. Ooh that creature looks fun! I'll take one too.&lt;br&gt;&#xA;$15$. An Icelandic glacial flood that is a bigger version of 36-across.&lt;br&gt;&#xA;$20$. Out of the blackness and stench of the engulfing swamp came &lt;code&gt;___&lt;/code&gt;&lt;br&gt;&#xA;$21$. I'll drag you out of your ivory tower and put you in my torture device!&lt;br&gt;&#xA;$22$. The most powerful card in Legacy&lt;br&gt;&#xA;$23$. If you've got a hunch on this card, you should follow it.&lt;br&gt;&#xA;$26$. He exercises his right to were clothes!&lt;br&gt;&#xA;$27$. Patrick Sullivan refers to this Khans of Tarkir card as &quot;full retail&quot;.&lt;br&gt;&#xA;$30$. Upset because this card was made obselete in Born of the Gods? Take its successor's advice.&lt;br&gt;&#xA;$34$. Another place where you can commonly find 32-across.  &lt;/p&gt;&#xA;" OwnerUserId="13970" LastEditorUserId="13970" LastEditDate="2016-01-12T03:48:42.637" LastActivityDate="2016-01-12T05:01:59.423" Title="Concordant Crossword: A MtG crossword" Tags="|crosswords|magic-the-gathering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25278" PostTypeId="2" ParentId="25098" CreationDate="2016-01-11T18:51:08.943" Score="7" Body="&lt;p&gt;Taking the example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PHzZ1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PHzZ1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $n=6$, where the greyed cells are the totals of their respective columns and rows, and the top-left cell is the total of each greyed column/row, I am fairly certain that it is not possible for the following reasons:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)    The top-left cell must $=\pm n/2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)    The only excluded integer in the range $[-n,n]$ must be either $-n$ or $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As noted by @kaine below, these statements are identical, so a proof of either would be enough to show that solutions are not possible for odd $n$. Both are purely observational though, and the bounty will go to a proof  (or disproof?!) of the above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NB&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://github.com/martinq321/puzz/blob/master/01&quot; rel=&quot;nofollow noreferrer&quot;&gt;This &lt;em&gt;Mathematica&lt;/em&gt; code&lt;/a&gt; doesn't check &lt;em&gt;all&lt;/em&gt; possiblilities (rather large for $n=11$), but checks &lt;em&gt;the most likely&lt;/em&gt; (and returns nothing for &lt;code&gt;test@11&lt;/code&gt;).&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="2819" LastEditDate="2016-01-12T10:14:06.663" LastActivityDate="2016-01-12T10:14:06.663" CommentCount="4" CommunityOwnedDate="2016-01-11T19:10:32.443" ContentLicense="CC BY-SA 3.0" />
  <row Id="25279" PostTypeId="2" ParentId="25277" CreationDate="2016-01-11T21:36:38.843" Score="2" Body="&lt;p&gt;Edit: Got 'em all! Credit to Zerris for #6 which somehow didn't occur to me, and to Gatherer's regex function which I admit I used to find four or five of 'em.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Across:&lt;br&gt;&#xA; 1: Kird Ape&lt;br&gt;&#xA; 7: Damnation&lt;br&gt;&#xA; 10: Cancel&lt;br&gt;&#xA; 12: Grapeshot&lt;br&gt;&#xA; 13: Elvish Mystic&lt;br&gt;&#xA; 16: Wrath of God&lt;br&gt;&#xA; 17: Shock&lt;br&gt;&#xA; 18: Moroii&lt;br&gt;&#xA; 19: Spellskite&lt;br&gt;&#xA; 24: Mirari&lt;br&gt;&#xA; 25: Capsize&lt;br&gt;&#xA; 28: Gush&lt;br&gt;&#xA; 29: Gideon Jura&lt;br&gt;&#xA; 31: Mind Twist&lt;br&gt;&#xA; 32: Wild Nacatl&lt;br&gt;&#xA; 33: Terror&lt;br&gt;&#xA; 35: Siege Rhino&lt;br&gt;&#xA; 36: Armageddon&lt;br&gt;&#xA; 37: Ring of Gix&lt;br&gt;&#xA; 38: Starstorm&lt;br&gt;&#xA; 39: Land Tax&lt;br&gt;&#xA;&lt;br&gt;&#xA; Vertical:&lt;br&gt;&#xA; 1: Kismet&lt;br&gt;&#xA; 2: Remove Soul&lt;br&gt;&#xA; 3: Psychatog&lt;br&gt;&#xA; 4: Mogg Fanatic&lt;br&gt;&#xA; 5: Path to Exile&lt;br&gt;&#xA; 6: Black Lotus&lt;br&gt;&#xA; 8: Time Walk&lt;br&gt;&#xA; 9: Explore&lt;br&gt;&#xA; 11: Stasis&lt;br&gt;&#xA; 14: Clone&lt;br&gt;&#xA; 15: Jokulhaups&lt;br&gt;&#xA; 20: White Knight&lt;br&gt;&#xA; 21: Black Vise&lt;br&gt;&#xA; 22: Brainstorm&lt;br&gt;&#xA; 23: Intuition&lt;br&gt;&#xA; 26: Werebear&lt;br&gt;&#xA; 27: Utter End&lt;br&gt;&#xA; 30: Infest&lt;br&gt;&#xA; 34: Taiga&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7459" LastEditorUserId="7459" LastEditDate="2016-01-12T05:01:59.423" LastActivityDate="2016-01-12T05:01:59.423" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="25280" PostTypeId="2" ParentId="25248" CreationDate="2016-01-11T21:44:35.753" Score="6" Body="&lt;p&gt;You are.. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Linear&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My logic:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - Famous people speak in lines(in movies)&lt;br&gt;&#xA; - Ear, in which you speak to clarify &lt;br&gt;&#xA; - The curse part is still bugging me, perhaps that is the curse of it all?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18217" LastEditorUserId="17250" LastEditDate="2016-01-12T18:19:39.607" LastActivityDate="2016-01-12T18:19:39.607" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="25281" PostTypeId="1" AcceptedAnswerId="25349" CreationDate="2016-01-11T21:55:24.997" Score="4" ViewCount="781" Body="&lt;p&gt;What is the answer and why?&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;I am a similar of the answer,&lt;/p&gt;&#xA;&lt;p&gt;A jumbled flower atom,&lt;/p&gt;&#xA;&lt;p&gt;plus 14, I am a muddled, missing alliance,&lt;/p&gt;&#xA;&lt;p&gt;My answer is an answer, a mix.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;What word am I?&lt;/p&gt;&#xA;" OwnerUserId="18250" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-01-13T19:07:28.367" Title="Riddle - what is the word? I am a similar of the answer" Tags="|riddle|word|anagram|" AnswerCount="4" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="25282" PostTypeId="1" AcceptedAnswerId="25315" CreationDate="2016-01-11T22:16:40.543" Score="4" ViewCount="1085" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;I am half of who I am,&lt;br /&gt;&#xA;yet I am twice of who I am.&lt;br /&gt;&#xA;Show me, the answer of this riddle,&lt;br /&gt;&#xA;and I will give you a little giggle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&lt;p&gt;The answer will explain all 4 lines!&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Hint 1: I've added a comma to the 3rd sentence&lt;/p&gt;&#xA;&lt;p&gt;Hint 2: Someone got 'half' of the answer here already!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-01-12T18:53:11.933" Title="It's not you but me who can solve this riddle!" Tags="|riddle|wordplay|" AnswerCount="7" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25285" PostTypeId="2" ParentId="25277" CreationDate="2016-01-11T22:36:01.623" Score="0" Body="&lt;p&gt;A few more:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 7: Damnation&lt;br&gt;&#xA;&lt;br&gt;&#xA; 6: Black Lotus  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="17389" LastEditDate="2016-01-11T23:00:41.170" LastActivityDate="2016-01-11T23:00:41.170" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25286" PostTypeId="2" ParentId="25282" CreationDate="2016-01-11T23:18:23.803" Score="6" Body="&lt;p&gt;It is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Butt?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am half of who I am,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A butt can be made of two butt cheeks&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yet I am twice of who I am.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Same as above&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Show me the answer of this riddle,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Butts are usually hidden behind clothing so you can't really see it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and I will give you a little giggle.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Seeing or saying &quot;Butt&quot; can cause lots of people to laugh&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9042" LastActivityDate="2016-01-11T23:18:23.803" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25287" PostTypeId="1" AcceptedAnswerId="25313" CreationDate="2016-01-11T23:23:59.490" Score="31" ViewCount="7559" Body="&lt;p&gt;On a small $ 4 \times 3$ chessboard, the top row is filled with black knights and the bottom row with white knights. On each move, you may move one knight (as it moves in chess) to an unoccupied square.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How many moves does it take to switch the white knights with the black knights?&lt;br&gt;&#xA;  Why can't it be done in fewer?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cCMVq.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cCMVq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;em&gt;Algorithmic Puzzles&lt;/em&gt;, Anany and Maria Levitin&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="19470" LastEditDate="2016-03-19T15:02:04.830" LastActivityDate="2016-03-19T15:02:04.830" Title="Switch The Knights" Tags="|chess|algorithm|checkerboard|knight-moves|" AnswerCount="5" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25288" PostTypeId="2" ParentId="25287" CreationDate="2016-01-12T07:37:12.723" Score="16" Body="&lt;p&gt;&lt;strong&gt;Edit: Now that @GOTO 0 got it in 16, I can at least prove that his solution is optimal.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The minimum possible number of moves if we could move through pieces would be seven per side.  This is a total of 14 moves.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Proof that this is ideal:&lt;/strong&gt; no piece can make it to the other side in one move because the positions are three squares apart.  If every piece could make it to the other side in just two moves then each piece would have to end on the same color it started as - but that's not possible, because the top and bottom starting places don't have the same number of black/white squares.  Thus it must take at least seven total moves, and seven is possible.  It involves (for Black where D = down, R = right, L = left) moving the Top Left knight DRR &gt; DDL, moving the Top Middle knight DDL &gt; DRR, and moving the Top Right knight DLL &gt; DRR &gt; DLL.  This can be mirrored on the other side for a total of 14 moves.&lt;br&gt;&#xA;&lt;br&gt;&#xA; But, it cannot be done in 14 moves, because there will be a conflict!&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Proof of conflict:&lt;/strong&gt; In order to move across in seven moves, one of each side's corner knights and both middle knights have to make it across in just two.  Let's assume without loss of generality that the Top Left knight does so.  This means the Top Right knight must take three moves - and also means that the Bottom Right knight must take three moves (if the two knights taking three moves are on opposite corners, they take the same path as each other).  Now, when can the Top Right knight move?&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;em&gt;If the Top Right knight ever moves&lt;/em&gt;: then he occupies the square DLL of his start.  However, the Bottom Middle knight needs to go through there, so TR will have to move again before the Bottom Middle knight does.  Our stack is now TR &gt; BM.  But the TR knight can't move again (DRR) without getting into a permanent block with BL, who wants to move to that same square.  So we have a stack of BL &gt; TR &gt; BM.  If BL has to move first, he'll have to get out of the way entirely, which requires TM to move to make room, giving us a stack of TM &gt; BL &gt; TR &gt; BM. If TM moves at least once (remember, he'll go DLL because middles swap with three-move knights), then he's in a permanent block with BR, so BR better move first.  This gives BR &gt; TM &gt; BL &gt; TR &gt; BM.  But then BR needs to get out of the way entirely, resulting in him moving ULL &gt; URR... and getting into a permanent block with TL.  TL better avoid that by moving first, giving us TL &gt; BR &gt; TM &gt; BL &gt; TR &gt; BM.  But in order for TL to move and get out of the way, they'll need to take the spot of BM, which means... uh oh, our stack is circular now, which is a contradiction.  BM &gt; TL &gt; BR &gt; TM &gt; BL &gt; TR &gt; BM can't be done.  But our assumption was only that TR moves, and negating that is also contradictory.  Thus the 14 move solution is not possible.&lt;br&gt;&#xA;&lt;br&gt;&#xA; If this was done in 15 moves, then the combination of all six knights would be on at least one square of a different color from when they started, since only even numbers of moves can maintain color.  However, since the knights start and end in the same combined positions, they start and end on the same combined colors.  Thus, this cannot be done in 15 moves.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Therefor, the &lt;strong&gt;minimum possible is 16&lt;/strong&gt;, and @GOTO 0 proved by example that this can be done.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My best was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;20 Moves&lt;/strong&gt;, in a variety of possible ways - here's one:&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/fPQBi.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fPQBi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="17389" LastEditDate="2016-01-12T18:07:32.490" LastActivityDate="2016-01-12T18:07:32.490" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25289" PostTypeId="2" ParentId="25282" CreationDate="2016-01-12T08:01:04.090" Score="8" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the number zero.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am half of who I am:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\frac{0}{2}=0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yet I am twice of who I am:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $0 \times 2=0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Show me the answer of this riddle,&#xA;and I will give you a little giggle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This one is on you, OP.  You better make &lt;a href=&quot;https://encrypted.google.com/search?hl=en&amp;amp;q=jokes%20about%20the%20number%20zero&quot;&gt;a joke about the number zero&lt;/a&gt; or you'll unfairly invalidate this answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12795" LastActivityDate="2016-01-12T08:01:04.090" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25290" PostTypeId="1" CreationDate="2016-01-12T11:57:48.903" Score="4" ViewCount="671" Body="&lt;p&gt;As a matter of fact, humanity sent a message to aliens with Voyager spacecrafts on 1977. Aliens suppose to decode the message without knowing anything about human race and be able to play attached videodisk. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But is it that simple task for a human? I think this can be an excellent decode-puzzle for you: explain each of 8 pictures from the message.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/maw1y.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/maw1y.jpg&quot; alt=&quot;the image is clickable&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Click image for better resolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Big hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; According to wiki the message should contain a instructions in binary code, indicate the location of the solar system relative to the 14 powerful pulsars. As a &quot;dipstick&quot; the hyperfine structure of the hydrogen molecule (1420 MHz) is given.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2016-01-12T14:05:46.460" LastActivityDate="2016-01-13T03:05:34.473" Title="Decode Voyager's Message to Aliens" Tags="|cipher|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25292" PostTypeId="2" ParentId="25098" CreationDate="2016-01-12T13:42:35.680" Score="2" Body="&lt;p&gt;My best try so far: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 1   1   1   1   1   1   1   1   1   1   1   11&#xA;-1  -1  -1  -1  -1  -1  -1  -1  -1  -1   0  -10&#xA; 1   1   1   1   1   1   1   1   1  -1   1   9&#xA;-1  -1  -1  -1  -1  -1  -1  -1   0  -1   1  -8&#xA; 1   1   1   1   1   1   1  -1   1  -1   1   7&#xA;-1  -1  -1  -1  -1  -1   0  -1   1  -1   1  -6&#xA; 1   1   1   1   1  -1   1  -1   1  -1   1   5&#xA;-1  -1  -1  -1   0  -1   1  -1   1  -1   1  -4&#xA; 1   1   1  -1   1  -1   1  -1   1  -1   1   3&#xA;-1  -1   0  -1   1  -1   1  -1   1  -1   1  -2&#xA; 0  -1   1  -1   1  -1   1  -1   1  -1   1   0&#xA; 0  -1   2  -3   4  -5   6  -7   8  -9  10  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As you can see, the sum of 1 is missing, instead I end up with 0 twice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also suspect that there's a problem with symmetry after one row or column is filled with all 1s or -1s, but I really have no idea how to prove that yet.&lt;/p&gt;&#xA;" OwnerUserId="17069" LastEditorUserId="17167" LastEditDate="2016-01-12T16:33:56.283" LastActivityDate="2016-01-12T16:33:56.283" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25294" PostTypeId="2" ParentId="25183" CreationDate="2016-01-12T14:27:41.240" Score="1" Body="&lt;p&gt;Are you a...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;spy / ninja / assasin ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Scream out or scream low&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You will only hear me when I blow&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A spy/ninja will be really stealthy so you can't hear them unless they're breathing loudly, or 'blowing'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Stand back or stand front&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You will only see me when I want&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The stealth argument applies here too.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Go there or no-where&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I can be there beside with no distance&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Spies/ninjas/assasins are trained to sneak up on people quickly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I don't have ticket&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I can't make chitchat&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Why would any sane ninja make chitchat?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But once you know what I am&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You don't want to meet me again.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;People generally don't want to meet assasins for coffee?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now that I've typed all of this, the word &lt;em&gt;'stalker'&lt;/em&gt; also comes to mind....&lt;/p&gt;&#xA;" OwnerUserId="17976" LastActivityDate="2016-01-12T14:27:41.240" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25296" PostTypeId="1" AcceptedAnswerId="25302" CreationDate="2016-01-12T14:42:55.523" Score="9" ViewCount="796" Body="&lt;p&gt;&lt;i&gt; I proceed with a pastry,&lt;br&gt;&#xA;get a grocer's from a gambol,&lt;br&gt;&#xA;a persona from a pixie,&lt;br&gt;&#xA;add pressure when you practice,&lt;br&gt;&#xA;bawl about the balm,&lt;br&gt;&#xA;bisect from below,&lt;br&gt;&#xA;get chairs from consumption,&lt;br&gt;&#xA;dispossess the duck,&lt;br&gt;&#xA;dirty the diesel,&lt;br&gt;&#xA;drool in a ditch,&lt;br&gt;&#xA;find room for a rate,&lt;br&gt;&#xA;flesh from the family,&lt;br&gt;&#xA;fluster a fish,&lt;br&gt;&#xA;glower from a garment,&lt;br&gt;&#xA;have contempt for a callus,&lt;br&gt;&#xA;insult illumination,&lt;br&gt;&#xA;jolt a jackal's joint,&lt;br&gt;&#xA;kill off the comedy,&lt;br&gt;&#xA;make a superficial saint,&lt;br&gt;&#xA;make accurate the ardin,&lt;br&gt;&#xA;or firm the furniture,&lt;br&gt;&#xA;make raw the rock,&lt;br&gt;&#xA;produce tulle from the type,&lt;br&gt;&#xA;or an abarka from an adze,&lt;br&gt;&#xA;mess up the medicine,&lt;br&gt;&#xA;pass round the rodent,&lt;br&gt;&#xA;post part of the pelvis,&lt;br&gt;&#xA;precipitate the present,&lt;br&gt;&#xA;pull a rock from the ring,&lt;br&gt;&#xA;puncture the parasite,&lt;br&gt;&#xA;secure the subterfuge,&lt;br&gt;&#xA;submerge the stain,&lt;br&gt;&#xA;show talented from terminated,&lt;br&gt;&#xA;spout about the summit,&lt;br&gt;&#xA;tighten the team,&lt;br&gt;&#xA;turn on a tack,&lt;br&gt;&#xA;wear at the waterside,&lt;br&gt;&#xA;and conclude at the cover. &lt;/I&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Who or what am I?  &lt;/p&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2016-01-12T17:37:44.520" Title="The bringer of change - I'll make you miss your Ex!" Tags="|wordplay|" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25299" PostTypeId="2" ParentId="25296" CreationDate="2016-01-12T16:00:01.207" Score="3" Body="&lt;p&gt;cant find a fit for all (yet!??) but the answer may be&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the letter S&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;eg. I proceed with a pastry&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Start / tart  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;eg. a persona from a pixie  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Self / elf&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18319" LastEditorUserId="18319" LastEditDate="2016-01-12T16:05:27.710" LastActivityDate="2016-01-12T16:05:27.710" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25301" PostTypeId="2" ParentId="25287" CreationDate="2016-01-12T16:36:46.073" Score="27" Body="&lt;p&gt;I found a solution that uses 16 moves.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/jeCUY.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jeCUY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After exhaustively checking that there is no solution in 14 moves, I conclude that 16 moves is optimal, because after any odd number of moves the number of white and black squares occupied by knights cannot be equal.&lt;/p&gt;&#xA;" OwnerUserId="4394" LastEditorUserId="4394" LastEditDate="2016-01-12T17:49:28.777" LastActivityDate="2016-01-12T17:49:28.777" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25302" PostTypeId="2" ParentId="25296" CreationDate="2016-01-12T16:43:18.413" Score="10" Body="&lt;p&gt;Here's what I have so far:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first word on each line is made by adding a leading s to the second. Applying this to the title means you are: Sex!&lt;br&gt;&#xA; I proceed with a pastry, - Start with a tart&lt;br&gt;&#xA; get a grocer's from a gambol, -  get a shop from a hop&lt;br&gt;&#xA; a persona from a pixie, - self from elf&lt;br&gt;&#xA; add pressure when you practice - strain when you train&lt;br&gt;&#xA; bawl about the balm - scream about the cream&lt;br&gt;&#xA; bisect from below, - sunder from under&lt;br&gt;&#xA; get chairs from consumption, - seats from eat&lt;br&gt;&#xA; dispossess the duck - steal the teal&lt;br&gt;&#xA; dirty the diesel,  - soil the oil&lt;br&gt;&#xA; drool in a ditch, - spit in a pit&lt;br&gt;&#xA; find room for a rate, - space for a pace&lt;br&gt;&#xA; flesh from the family, - Skin from kin&lt;br&gt;&#xA; fluster a fish, - shake a hake&lt;br&gt;&#xA; glower from a garment, - scowl from a cowl&lt;br&gt;&#xA; have contempt for a callus, - have scorn for a  corn&lt;br&gt;&#xA; insult illumination, - slight light&lt;br&gt;&#xA; jolt a jackal's joint, - shock the hock&lt;br&gt;&#xA; kill off the comedy, - slaughter  the laughter&lt;br&gt;&#xA; make a superficial saint, - make a shallow hallow&lt;br&gt;&#xA; make accurate the ardin, - sharp the harp&lt;br&gt;&#xA; or firm the furniture, - stable the table&lt;br&gt;&#xA; make raw the rock, - sore the ore&lt;br&gt;&#xA; produce tulle from the type, - produce silk from the ilk&lt;br&gt;&#xA; or an abarka from an adze, - shoe from a hoe&lt;br&gt;&#xA; mess up the medicine, - spill the pill?&lt;br&gt;&#xA; pass round the rodent, - share the hare (pretty sure this is right,though hares are lagomorphs, not rodents)&lt;br&gt;&#xA; post part of the pelvis, - ship the hip&lt;br&gt;&#xA; precipitate the present, - snow the now&lt;br&gt;&#xA; pull a rock from the ring, - stone from the tone&lt;br&gt;&#xA; puncture the parasite, - stick the tick&lt;br&gt;&#xA; secure the subterfuge, - strap the trap&lt;br&gt;&#xA; submerge the stain, - sink the ink&lt;br&gt;&#xA; show talented from terminated - show skilled from the killed&lt;br&gt;&#xA; spout about the summit, - speak about the peak&lt;br&gt;&#xA; tighten the team, - screw the crew&lt;br&gt;&#xA; turn on a tack, - spin on a pin&lt;br&gt;&#xA; wear at the waterside - sport at the port&lt;br&gt;&#xA; and conclude at the cover. - stop at the top  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16874" LastEditorUserId="16874" LastEditDate="2016-01-12T17:37:44.520" LastActivityDate="2016-01-12T17:37:44.520" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25303" PostTypeId="2" ParentId="25282" CreationDate="2016-01-12T16:44:22.940" Score="4" Body="&lt;p&gt;Is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Infinity?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am half of who I am,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Infinity divided by two is still infinity and infinity can be cut into two equal pieces.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yet I am twice of who I am.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Infinity times two is still infinity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Show me the answer of this riddle,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Infinity is kinda hard to comprehend. It is more of a concept rather than a number but everyone probably disagrees with that.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and I will give you a little giggle.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To write infinity in its opother form, you have to draw a sideways 8, or sometimes a BUTT!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9042" LastActivityDate="2016-01-12T16:44:22.940" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25304" PostTypeId="2" ParentId="25098" CreationDate="2016-01-12T16:51:15.727" Score="6" Body="&lt;p&gt;This is not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the more interesting question is the generalized solution to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given a square with $N$ columns and $N$ rows, is it possible to have $2N$ unique sums of columns and rows?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First, we need to consider two properties of an $N\times N$ matrix:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;blockquote&gt;&#xA;  &lt;p&gt;For a given matrix with $N$ columns and rows, the range of possible value are $2N + 1$ (in the range $[N, -N]$) and $2N$ sums of values ($N$ columns and $N$ rows). &lt;strong&gt;This means there is $1$ and only $1$ &quot;missing value&quot; $M_v$ in the series.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote&gt;&#xA;  &lt;p&gt;There are also $N^2$ places to fill. &lt;strong&gt;This means we need to be at least efficient enough to have $2N$ values using $N^2$ places.&lt;/strong&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Next, consider a matrix where $N$ is even and it has $2N$ unique sums:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, assume $N = M_v$. By placing values in the left column and upper row starting with $-N$ descending, we can place unique, sequential values until there are 2 columns and 2 rows remaining. At this point, the sums of all the columns and rows adjacent to the $2 \times 2$ matrix is $0$. It is then trivial to demonstrate that this matrix can produce the remaining values required. This same technique also works when $-N = M_v$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we have $2N$ values to produce, and with each iteration we produce 2 unique values, we cannot be less efficient or else we cannot produce all of the required values. However, when $N$ is odd we should note another property of the matrix:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;blockquote&gt;&#xA;  &lt;p&gt;Each value $(1, 0, -1)$ added to the matrix will result in an even change to the sum total of the matrix columns and rows. &lt;strong&gt;Therefore the sum of the columns and rows must be even.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also note that sum of the columns ($C_v$) and rows ($R_v$) is equal to $-M_v$. ($C_v + R_v = -M_v$ or $C_v + R_v + M_v = 0$). &lt;strong&gt;This means that $M_v$ cannot be odd&lt;/strong&gt; because it is impossible for the sum of the columns and rows to be odd. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, when $N$ is odd, the missing value $M_v$ cannot be $N$ or $-N$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That leaves only the possibility of including both $N$ and $-N$ in the matrix, but to do this we must put them both in columns (or rows). Once this is done, observe that there are $N - 2$ columns and $N$ rows (or vice versa) remaining. This allows for $N + (N - 2) = 2N - 2$ unique sums of columns and rows &lt;em&gt;but&lt;/em&gt; leaves only $N(N - 2) = N^2 - 2N$ places. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, when $N = 11$, after placing $11,-11$ we now have $9 + 11 = 20$ sums but only $9\times 11 = 99$ places. Previously, we demonstrated that for $2N$ values we needed $N^2$ places which, for $N = 10$, is $20$ sums and $100$ places. So when $N$ is odd, after placing $2$ numbers ($N$ and $-N$), we have to be more efficient in order to succeed. Maybe a rectangle can be more efficiently filled (requires fewer available spaces) than a square matrix?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that assumption, after placing $N$ and $-N$, we can continue to place sequential numbers (like $N-1$ and $-(N-1)$) but that will fill columns (or rows) very quickly (approaching $N + N/2$ instead of $2N$ sums) and reduce the &quot;available spaces&quot; below $N^2 - 2N$ making it easier to consider $M_v$ as $N-1$ or $-(N-1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we can skip either $N - 1$ or $-(N - 1)$ then we will explore further. To start we can try the alternating values in columns/rows that we used when $N$ was even. That produced 2 unique values for each column/row pair that was eliminated. It is not possible to be more efficient than that. However, after placing $N$, $-N$ and either $N - 1$ or $-(N - 1)$, we have used 3 columns (or rows) almost entirely identical. The third column contains a $0$ that must be in the second from the top position to allow $N - 3$ in the second row. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we subtract a row for each column as is required, until there remains only 1 column. Based on the pattern as described above there will be $4$ rows ($4$ remaining spaces; with an even numbered matrix, the 4 remaining spaces were in a square, not a column). However, there are only $3$ possible values (1, 0, -1). This will result in a duplicate value being placed. A duplicate value means a second &quot;missing value&quot; somewhere else in the matrix, since we cannot increase the number of sums produced. &lt;strong&gt;This means we cannot have $22$ unique values for an $11\times 11$ matrix (or any odd sized matrix). We can, at most, have $21$ unique values (or $2N - 1$).&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, even if it is possible to efficiently fill a square of size $N$ where $N$ is odd, the constraints on how the values are placed reduces the number of unique sums to below $2N$. (Note: I think there is a method to prove that when $N = 11$ there can be at most $19$ unique values, but I'm still working on it...)&lt;/p&gt;&#xA;" OwnerUserId="15681" LastEditorUserId="15681" LastEditDate="2016-01-20T05:37:49.597" LastActivityDate="2016-01-20T05:37:49.597" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="25305" PostTypeId="2" ParentId="25287" CreationDate="2016-01-12T17:03:27.707" Score="1" Body="&lt;p&gt;Maybe only a basic hint for a proof, but here are my thoughts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the playing area, where an &lt;code&gt;X&lt;/code&gt; represents a cell in the playing area.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;X X X&#xA;X X X&#xA;X X X&#xA;X X X&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note that the two centre cells are connected to only two cells each. So we can change the playing area to the one given below, without changing the problem in any way&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;  X  &#xA;&#xA;X X X&#xA;X   X&#xA;X   X&#xA;X X X&#xA;&#xA;  X  &#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now the four corners are also connected two only two cells each, and not to each other. Hence we can change the playing area to&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;    X    &#xA;X       X&#xA;    X    &#xA;  X   X  &#xA;  X   X  &#xA;    X    &#xA;X       X&#xA;    X    &#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Adding the playing pieces makes the grid as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;    X    &#xA;B       B&#xA;    B    &#xA;  X   X  &#xA;  X   X  &#xA;    W    &#xA;W       W&#xA;    X    &#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This representation is equivalent to the original puzzle; any solution that works here also works there and vice versa. It is only that it is easier to see which cell is connected to which, in the figure given above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no idea how to prove an optimal solution, though.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2016-01-12T17:03:27.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25308" PostTypeId="2" ParentId="25282" CreationDate="2016-01-12T17:13:01.373" Score="3" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the word &quot;me&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am half of who I am:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I am&quot; has 4 characters (including whitespace). &quot;me&quot; is 2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yet I am twice of who I am:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Twice of &quot;I&quot;(1character). &quot;me&quot;(2characters). &quot;I&quot; and &quot;me&quot; have synonymous meanings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Show me, the answer of this riddle,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You're saying &quot;me&quot; is &quot;the answer of this riddle&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and I will give you a little giggle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is so obvious it's in the title and in the third line, it could be considered a little funny it took so long to figure it out.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16471" LastActivityDate="2016-01-12T17:13:01.373" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25309" PostTypeId="2" ParentId="25282" CreationDate="2016-01-12T17:24:18.053" Score="5" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Semi-colon (sign is ;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am half of who I am,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Half of 'fullstop'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yet I am twice of who I am.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Twice of comma,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Show me, the answer of this riddle,&#xA;and I will give you a little giggle.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is used in the wink emoticon as ;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastActivityDate="2016-01-12T17:24:18.053" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25311" PostTypeId="2" ParentId="25282" CreationDate="2016-01-12T17:53:01.153" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Brain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am half of who I am,&#xA;yet I am twice of who I am.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As per the title hint, brain is the thing that solves the riddle. Brain has two hemispheres, so it's two halves.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Show me, the answer of this riddle,&#xA;and I will give you a little giggle.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The brain is the thing that will see the irony in the answer and it controls your actions, so you will giggle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5726" LastActivityDate="2016-01-12T17:53:01.153" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25312" PostTypeId="2" ParentId="25281" CreationDate="2016-01-12T18:01:59.500" Score="3" Body="&lt;p&gt;I think the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; sixteen &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a similar of the answer,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think this refers to 't', the sixteenth consonant in the alphabet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A jumbled flower atom,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An atom of flower could be 'ow', jumbled gives 'wo' and added to 't' gives 'two'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;plus 14,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2 + 14 = 16&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a muddled, missing alliance,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'I am a' could refer to 'existence', muddled gives 'sixteenec' and without alliance (ec, european community) gives 'sixteen'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My answer is an answer, a mix.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a mix could refer to a combination of all the lower case a that appear in the question (including lines above and below) which is sixteen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Obviously, a bit convoluted but let me know if I'm anywhere near.&lt;/p&gt;&#xA;" OwnerUserId="18419" LastActivityDate="2016-01-12T18:01:59.500" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25313" PostTypeId="2" ParentId="25287" CreationDate="2016-01-12T18:03:30.937" Score="22" Body="&lt;p&gt;You need at least 16 Moves.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Let's make the task &lt;strong&gt;visually more simple&lt;/strong&gt;. The initial board is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a4 b4 c4  &#xA;a3 b3 c3  &#xA;a2 b2 c2  &#xA;a1 b1 c1  &#xA;&lt;/code&gt;&lt;/pre&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;We cut it into 12 cells and connect only those, which are separated exactly by one move of a knight. Easy to check that the result is the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    c4 - a3 - c2 - a1  &#xA;     |    |   |    |&#xA;    b2   b1   b4   b3 &#xA;     |    |   |    | &#xA;    a4 - c3 - a2 - c1 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So the knights are placed like this now:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    B1 -  . - .  - W1 &#xA;     |    |   |    |&#xA;     .   W2   B2   .&#xA;     |    |   |    | &#xA;    B3 -  . - .  - W3 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Now it is quite easy to find &lt;strong&gt;a strategy for quick switch&lt;/strong&gt; of the knights.  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In 2+2+2+3=9 moves we have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    W2 -  . - B1 - W1 &#xA;     |    |   |    |&#xA;     .   B3   B2   .&#xA;     |    |   |    | &#xA;    W3 -  . - .  - . &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In 9+1+2+2=14 moves we have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    W2 - B1 - .  - . &#xA;     |    |   |    |&#xA;     .   B3   W1   .&#xA;     |    |   |    | &#xA;    W3 -  . - .  - B2 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And in 14+2=16 moves we have the final position:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    W2 -  . - .  - B1 &#xA;     |    |   |    |&#xA;     .   B3   W1   .&#xA;     |    |   |    | &#xA;    W3 -  . - .  - B2 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The actual chess moves have been perfectly illustrated by &lt;a href=&quot;https://puzzling.stackexchange.com/a/25301/28&quot;&gt;GOTO 0&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;It is also easy to &lt;strong&gt;show that you can't perform the task faster&lt;/strong&gt;. Indeed, imagine there is only black knights. It will take 2 moves for B2 to reach a closest final position, and at least 2 and 3 moves for B1 and B3, since they can't go both to the same cell. This makes 7 moves in total, furthermore there is only one way (up to the symmetry) you can move them. Similarly, you need at least 7 for moving the white knights.&lt;br&gt;&#xA;And, finally, since black and white are on the board at the same time and their paths are crossed, you will need at least 2 additional moves to make them pass each other.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="28" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2016-01-13T14:42:52.270" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25315" PostTypeId="2" ParentId="25282" CreationDate="2016-01-12T18:53:11.933" Score="12" Body="&lt;p&gt;I think it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; meme&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am half of who I am,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;me&quot; is half of &quot;meme&quot; (&quot;I&quot; as a definition, &quot;me&quot;, is half of [&quot;who I am&quot; = the answer])&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yet I am twice of who I am.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;meme&quot; is twice &quot;me&quot; ([&quot;I&quot; = the answer] is twice &quot;who I am&quot; as a definition, &quot;me&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Show me, the answer of this riddle,&#xA;and I will give you a little giggle.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is &quot;meme&quot; which is usually supposed to be funny and make people laugh.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12092" LastActivityDate="2016-01-12T18:53:11.933" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="25319" PostTypeId="1" AcceptedAnswerId="25321" CreationDate="2016-01-12T23:48:19.007" Score="4" ViewCount="531" Body="&lt;p&gt;I have a mouth, but I don't eat with it&#xA;&lt;br&gt;My body doesn't support me, yet I eat with that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a comfy bed, yet I never sleep with it&#xA;&lt;br&gt;whether it is night or day&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I love to run, but can't slow down to a jog&#xA;&lt;br&gt; yet I can a raging bull&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can be shorter than a dwarf&#xA;&lt;br&gt; or taller than a skyscraper&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You better watch out&#xA;&lt;br&gt; the hope for survival&#xA;&lt;br&gt; just turned to your worst nightmare&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I, seen everyday?&lt;/p&gt;&#xA;" OwnerUserId="9042" LastActivityDate="2016-01-13T17:37:01.063" Title="What am I, seen everyday?" Tags="|riddle|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25320" PostTypeId="2" ParentId="25290" CreationDate="2016-01-13T03:05:34.473" Score="4" Body="&lt;h1&gt;Part 1&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0BJl4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is one of the more difficult parts.  This portion of the record is supposed to pictorially represent&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the transition between two states of the hydrogen atom: in particular, the two hyperfine levels of the ground state.  The difference between the two states is in the alignment of the electron's spin and angular momentum; on the right the two are parallel, and on the left the two are antiparallel.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The transition between these two states happens at a &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrogen_line&quot; rel=&quot;nofollow noreferrer&quot;&gt;very precise frequency&lt;/a&gt;.  The bar between the two states is labeled with a &lt;code&gt;|&lt;/code&gt; (1), indicating that this frequency defines the units of time used on the rest of the record:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; \begin{align}1~\text{TU}&amp;amp;=\frac{1}{1420.4~\text{MHz}}=7.04\cdot 10^{-10}~\text{s}\end{align}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;Part 2&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/En9qS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The meaning of these images is relatively simple, they are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; top and side views of the record itself, as well as the stylus included alongside the record.  They are intended to demonstrate how the record is played.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The dashes are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; binary numbers, with &lt;code&gt;|&lt;/code&gt; representing &lt;code&gt;1&lt;/code&gt; and &lt;code&gt;-&lt;/code&gt; representing &lt;code&gt;0&lt;/code&gt;.  Both encode a duration using the time unit from Part 1.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The number wrapping around the record (&lt;code&gt;|--||----||--|-------------------&lt;/code&gt;) represents $9753\times 2^{19}~\text{TU}$, or $3.600~\text{s}$, the duration of one rotation.  (This makes it a $16\frac{2}{3}$ record, half the speed of a typical $33\frac{1}{3}$ record.)  It is read in the direction the stylus moves over the record (counterclockwise as seen from above).&#xA;&lt;br&gt;&#xA; The number below the side view (&lt;code&gt;||-|----|----------------------------------&lt;/code&gt;) represents $139\times 2^{34}~\text{TU}$, or about $84~\text{min}$, approximately the duration of the record.  It is read in the direction the stylus moves over the record, from the outer edge to the inside.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The audio contents of the record (which I &lt;a href=&quot;https://www.youtube.com/watch?v=Bhuq9rNO_FQ&amp;amp;index=1&amp;amp;list=PLA5Z0m2JKyVJUgkMG08WP8KsAvLrjfkjP&quot; rel=&quot;nofollow noreferrer&quot;&gt;listened to&lt;/a&gt; while writing this) consist of greetings in 55 languages, as well as a couple longer snippets from members of the United Nations; &lt;a href=&quot;https://en.wikipedia.org/wiki/Star_Trek_IV:_The_Voyage_Home&quot; rel=&quot;nofollow noreferrer&quot;&gt;whale song&lt;/a&gt; (really!); &quot;sounds of the Earth,&quot; including natural sounds (e.g. a thunderstorm), as well as manmade sounds (e.g. a rocket launch); and a variety of music from different cultures.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Part 3&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ngEVe.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This part is a schematic depiction of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a TV signal, intended to assist in decoding the images stored on the record.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first row shows&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the signal itself, going from left to right.  It begins with a beep (probably a vertical synchronization pulse that separates images), then shows three complete lines and the start of one more.  The lines are numbered sequentially, &lt;code&gt;|&lt;/code&gt; (1), &lt;code&gt;|-&lt;/code&gt; (2) and &lt;code&gt;||&lt;/code&gt; (3).&lt;br&gt;&#xA;&lt;br&gt;&#xA; Each line begins with a horizontal blanking interval (low period) and then a horizontal sync pulse, followed by the analog video signal representing the image intensity along that line.  The first line is marked with &lt;code&gt;|-||-|--||--------------&lt;/code&gt;, $723\times 2^{14}~\text{TU}$ or $8.3~\text{ms}$, the duration of one line.  (This is several orders of magnitude slower than ordinary video, whose frequencies are too high to record on a record.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second row shows&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The horizontal location on the screen, going slowly from one side of the screen to the other during the active part of the signal, and quickly returning back during the blanking interval.  I'm not exactly sure what the dots are; possibly they indicate the three color components (&lt;code&gt;RGB&lt;/code&gt;) of the video signal, or that the video signal is interleaved?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The third row shows&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The scan pattern on the screen.  They seem to have interchanged the horizontal and vertical components of the signal, since ordinary TV scans the lines horizontally, not vertically.  The two numbers (&lt;code&gt;|---&lt;/code&gt; and &lt;code&gt;|---------&lt;/code&gt;) indicate the 8 lines shown, and the total of 512 lines per frame.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally, the fourth row shows&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first image on the record: a circle (meant as a test pattern).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The images on the record start off with basic mathematical definitions (Arabic numerals in base-10, addition, and fractions); then basic units (including centimeters, meters, and kilometers, seconds, days, and years); a schematic of the solar system; and finally a wide range of humans and human activities.  (I based &lt;a href=&quot;https://puzzling.stackexchange.com/q/17119/9718&quot;&gt;one of my puzzles&lt;/a&gt; off of the beginning part of the record's message, actually!)&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Part 4&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;This is another hard one.  The spiky object on the lower-left depicts&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A map of the solar system's location relative to a number of &lt;a href=&quot;https://en.wikipedia.org/wiki/Pulsar&quot; rel=&quot;nofollow noreferrer&quot;&gt;pulsars&lt;/a&gt;.  A pulsar is a class of neutron star, the ultra-dense remnant of a massive supernova.  When the neutron star forms, it retains the angular momentum of the original star; but since it is so small (on the order of 10 kilometers), this means that it spins very fast (think of a figure skater pulling in their arms when they spin).  For example, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Crab_Pulsar&quot; rel=&quot;nofollow noreferrer&quot;&gt;Crab pulsar&lt;/a&gt; spins at around 30 rotations per second!  A pulsar has a &quot;hot spot&quot; that comes into view once per rotation, so the star appears to flash very quickly.  Since neutron stars are still around five times as massive as the Sun, their rate of rotation is very stable over time.  This allows us to use them as &quot;beacons,&quot; identifiable by the frequency of their flashes.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The lines emanate from the solar system at the center, and represent the relative distance and direction to various pulsars.  The numbers on each line encode the period of the pulsar's rotation.  In most images of the record it's hard to make out the map very well, so if you're interested in more details about it you should look at &lt;a href=&quot;http://www.johnstonsarchive.net/astro/pulsarmap.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt; (in which the author shows that it is possible to use the map to determine very accurately not only the position, but also the date at which the map was produced).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2016-01-13T03:05:34.473" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25321" PostTypeId="2" ParentId="25319" CreationDate="2016-01-13T03:39:36.817" Score="7" Body="&lt;p&gt;I can't fit all of the verses, but one that fits some is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; river&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I have a mouth, but I don't eat with it&#xA;...&#xA;I have a comfy bed, yet I never sleep with it&#xA;whether it is night or day&#xA;&#xA;I love to run, but can't slow down to a jog&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A river has a mouth and a bed, and it runs but can't slow down.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I can be shorter than a dwarf&#xA;or taller than a skyscraper&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A river can be short, or it can have a waterfall that makes it taller than a skyscraper&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The parts I can't fit yet are:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;My body doesn't support me, yet I eat with that&#xA;...&#xA;yet I can a raging bull&#xA;...&#xA;You better watch out&#xA;the hope for survival&#xA;just turned to your worst nightmare&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I don't even know what the line about the bull means. Is a word missing?&lt;/p&gt;&#xA;" OwnerUserId="4373" LastActivityDate="2016-01-13T03:39:36.817" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="25322" PostTypeId="2" ParentId="25215" CreationDate="2016-01-13T04:42:09.373" Score="1" Body="&lt;p&gt;You are a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; kiwi&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One possibility is a tiny fruit that if you collect enough of me, someone can turn me into something that you like&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; kiwifruit is somewhat small&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another possibility is a gigantic creature that is cut into 6 pieces. Two of them are tiny.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; kiwi currency - Historically, there were six different coins: 1c, 2c, 5c, 10c, 20c, and 50c&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet another possibility is an eccentric creature which can understand any language with a strange feature shared with Kēlen (I'm not human).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well, this doesn't fit, because kiwi birds are not human but can't understand verbless languages, and Kiwis are human and may or may not understand verbless languages.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Relevance:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Indonesia is one of the closest countries to NZ (even though still 7500 km away)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4373" LastActivityDate="2016-01-13T04:42:09.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25323" PostTypeId="2" ParentId="25004" CreationDate="2016-01-13T05:50:24.777" Score="3" Body="&lt;p&gt;I think the answer is  (you are)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cloud&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I found it most suited with the below sentences&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;but will falter to the lightest of all objects. /&#xA; You can see me as far the eye can see  /&#xA; Precious by nature  /&#xA; and I am certainly not here because of you  /&#xA; As a group, I can be a force to reckon with  /&#xA; yet I can be gentler than a breeze  /&#xA;&lt;/code&gt;&#xA;&lt;br&gt;&#xA;And this Hint  &lt;strong&gt;I can be wavy or straight, but always true to the color blue&lt;/strong&gt;&#xA; &lt;br&gt;&#xA;made me sure of my answer. : )&lt;/p&gt;&#xA;" OwnerUserId="11353" LastActivityDate="2016-01-13T05:50:24.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25325" PostTypeId="1" CreationDate="2016-01-13T07:52:19.457" Score="0" ViewCount="385" Body="&lt;p&gt;&lt;em&gt;Based on &lt;a href=&quot;https://puzzling.stackexchange.com/questions/24948/the-square-and-the-compass&quot;&gt;The square and the compass&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This is a new kind of challenge proposed on &lt;a href=&quot;http://meta.puzzling.stackexchange.com/questions/4508/cops-and-robbers-anyone&quot;&gt;Meta Puzzling SE&lt;/a&gt;. Any discussion about the general type of puzzle (rather than this particular one) can be done there.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The robbers' post is given &lt;a href=&quot;https://puzzling.stackexchange.com/questions/25326/robbers-the-ultimate-compass-challenge&quot;&gt;here&lt;/a&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a compass and a pencil but no scale/straightedge. You need to decide upon a construction task that can be achieved accurately in a finite number of moves. The following moves are valid.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Make the compass radius equal to the distance between two already marked points.&lt;/li&gt;&#xA;&lt;li&gt;Draw a circle with any marked point as centre.&lt;/li&gt;&#xA;&lt;li&gt;Use any intersection of arcs/circles as a marked point.&lt;/li&gt;&#xA;&lt;li&gt;Select a continuous region (either an arc or a 2D region) and mark an approximate point. For example, I could draw an arc and then mark a point that is approximately (but not exactly) at the centre.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For example, a task could be &quot;find the midpoint of a line segment&quot; or &quot;mark 5 points that form a regular pentagon&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your challenge will be scored by the amount of time it remains unsolved. At the end of this month (31 Jan), you must post the solution(s) to any unsolved challenges. You can make up to 3 posts, and your total score will be the sum of the individual scores.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2016-01-13T16:08:51.677" Title="Cops - The ultimate compass challenge" Tags="|mathematics|geometry|construction|cops-and-robbers|" AnswerCount="2" CommentCount="6" ClosedDate="2016-01-15T10:19:52.137" ContentLicense="CC BY-SA 3.0" />
  <row Id="25326" PostTypeId="1" CreationDate="2016-01-13T07:56:58.977" Score="-1" ViewCount="267" Body="&lt;p&gt;Please read the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/25325/cops-the-ultimate-compass-challenge&quot;&gt;Cops post&lt;/a&gt; for all details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Points are scored on the basis of how long a challenge remains unsolved before you solve. You may solve any number of challenges. You cannot solve your own challenges. Points scored by robbers and by cops are independent, and you can participate in both posts separately.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2016-01-13T16:38:30.850" Title="Robbers - The ultimate compass challenge" Tags="|mathematics|geometry|construction|cops-and-robbers|" AnswerCount="2" CommentCount="5" ClosedDate="2016-01-14T20:14:52.400" ContentLicense="CC BY-SA 3.0" />
  <row Id="25327" PostTypeId="2" ParentId="4472" CreationDate="2016-01-13T08:32:27.413" Score="1" Body="&lt;p&gt;I know an answer has been marked as right. &#xA;But as my answer is different, I think I shall write :D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Riddle #1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cloud&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Riddle #2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Poop&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Riddle #3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Money-box / Piggy-bank&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Just enjoy the answers :D&lt;/p&gt;&#xA;" OwnerUserId="11353" LastActivityDate="2016-01-13T08:32:27.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25328" PostTypeId="2" ParentId="25281" CreationDate="2016-01-13T09:44:46.077" Score="4" Body="&lt;p&gt;My guess is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Carbon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a similar of the answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Carbon copy - identical&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A jumbled flower atom&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It can bind to anything and can form long chains called polymers. Also The actual location of electrons in a carbon atom cannot be determined orbiting the nucleus .&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;plus 14, I am a muddled, missing alliance,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think this is the Carbon-14 used for carbon dating. Carbon-14 is not stable unlike Carbon-12 and has radioactive decay. Also carbon-14, which makes up a minuscule amount of overall carbon in the universe. Carbon-12 (99%) and Carbon-13 (1%)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My answer is an answer, a mix.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Can bind with other elements and is found in many different compounds and almost everything.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16036" LastActivityDate="2016-01-13T09:44:46.077" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25330" PostTypeId="2" ParentId="25325" CreationDate="2016-01-13T14:32:02.253" Score="1" Body="&lt;p&gt;OK, I've got an easy one to start with. Given a straight line segment radius R, find a point that is exactly R away from one endpoint of the segment, and is colinear with the two endpoints. Essentially, double the line segment.&lt;/p&gt;&#xA;" OwnerUserId="1968" LastEditorUserId="1968" LastEditDate="2016-01-13T15:09:58.083" LastActivityDate="2016-01-13T15:09:58.083" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25331" PostTypeId="2" ParentId="25326" CreationDate="2016-01-13T14:44:05.233" Score="2" Body="&lt;p&gt;From the Cop's challenge thread:&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/a/25330/13101&quot;&gt;https://puzzling.stackexchange.com/a/25330/13101&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;OK, I've got an easy one to start with. Given a line segment radius R, find a point that is exactly R away from one endpoint of the segment, and lies on the same vector. Essentially, double the line segment.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; complete the circle that the segment is on. next, place the points of the compass on each end of the segment. complete a circle using one point as the center. Create 6 circles using the radius of the new circle, centered on the end of the first segment intersection through the center of the new circle, and adding new circles around the first at the intersection of each new circle and the original. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Image:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/E3PVg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/E3PVg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13101" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2016-01-13T15:09:55.053" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="25333" PostTypeId="2" ParentId="25326" CreationDate="2016-01-13T15:16:05.603" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Given a line segment radius R, find a point that is exactly R away from one endpoint of the segment, and lies on the same vector.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Draw two circles of radius $R$ about the two ends of the line segement $a$ and $b$.  Call the circles $A$ and $B$. They meet at two new points $c$ and $d$.  $c$ and $d$ are $\sqrt{3}R$ from each other.  Draw a circle of radius $\sqrt{3}R$ from $c$ and for good measure $d$.  These circles will both intersect $A$ at the same point $f$.  That point $f$ meets all requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works because you effectively make a right triangle with the following properties:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The right angle is located at the midpoint between $a$ and $b$ (call it $e$).  This is not a marked point and is, therefore, invisible.  It is the intersection between $ab$ and $cd$.&lt;/li&gt;&#xA;&lt;li&gt;Since two circles are produced with the same radius about $c$ and $d$, the intersection will rest on a line that bisects $cd$.  As $cd$ was produced by bisecting $ab$ these are on the same vector.&lt;/li&gt;&#xA;&lt;li&gt;We know $ce$ equals $\sqrt{3}/2$ as it is the height of the equilateral triangle $abc$. The hypoteneus is $\sqrt{3}$ in length as it was from $cd$ which is twice $ce$.  $ef$ should be $1.5$ and $cd^2=ef^2+ce^2=\sqrt{3}^2=3=(\sqrt{3}/2)^2+1.5^2=3/4+9/4=3$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditDate="2016-01-13T15:53:52.457" LastActivityDate="2016-01-13T15:53:52.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25334" PostTypeId="1" AcceptedAnswerId="25335" CreationDate="2016-01-13T15:39:23.483" Score="4" ViewCount="1277" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A natural state, I'm sought by all.&lt;br&gt;&#xA;  Go without me, and you shall fall.&lt;br&gt;&#xA;  You do me when you spend,&lt;br&gt;&#xA;  and use me when you eat to no end.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="6151" LastActivityDate="2016-01-15T12:18:56.410" Title="A riddle for your time?" Tags="|riddle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25335" PostTypeId="2" ParentId="25334" CreationDate="2016-01-13T15:41:12.303" Score="6" Body="&lt;p&gt;I think it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Balance&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation as follows&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A natural state, I'm sought by all.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;In nature, everything is balanced by default. And balance is sought by all&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Go without me, and you shall fall.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Without balance, we will definitely fall.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You do me when you spend,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;When you spend, what left off will be the balance&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and use me when you eat to no end.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I am not sure about this one. May be it is 'if you eat everything there won't be anything balance'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastActivityDate="2016-01-13T15:41:12.303" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25336" PostTypeId="2" ParentId="25325" CreationDate="2016-01-13T16:08:51.677" Score="0" Body="&lt;p&gt;Might as well do another easy one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Define two points $a$ and $b$.  Generate two more points $c$ and $d$ such that $abcd$ is a perfect square.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="99" LastActivityDate="2016-01-13T16:08:51.677" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25337" PostTypeId="1" AcceptedAnswerId="25338" CreationDate="2016-01-13T16:10:26.660" Score="1" ViewCount="666" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;From left to right, I hold the key to our existence.&lt;br&gt;&#xA;  From right to left, I am lack of experiences.&lt;br&gt;&#xA;  I exists somewhere in Geneva,&lt;br&gt;&#xA;  Yet rearranged, I am a Capital in South America.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I held an important meeting in 1938,&lt;br&gt;&#xA;  but does not sign the agreement until 1962.&lt;br&gt;&#xA;  Given all these important details,&lt;br&gt;&#xA;  Could you tell if I am male or female?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="4050" LastEditDate="2016-01-13T18:43:16.257" LastActivityDate="2016-01-13T18:43:16.257" Title="An easy riddle, who am I? From left to right, I hold the key to our existence" Tags="|riddle|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25338" PostTypeId="2" ParentId="25337" CreationDate="2016-01-13T16:16:33.490" Score="4" Body="&lt;p&gt;To start:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;From left to right, I hold the key to our existence.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Evian is a brand of water (which is the key to our existence)&lt;br&gt;&#xA;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;From right to left, I am lack of experiences.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Naive is a word indicating a lack of experience&lt;br&gt;&#xA;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Yet rearranged, I am a Capital in South America&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  This feels like an extreme stretch, but as @IrishPanda points out in the comments on this answer, there is a town called Santa Maria de &lt;em&gt;Nieva&lt;/em&gt; that is the capital of a &lt;em&gt;district&lt;/em&gt; within Peru.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;I exists somewhere in Geneva,&lt;/code&gt;&#xA;&lt;code&gt;I held an important meeting in 1938,&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The water gets its name from &lt;em&gt;Évian-les-Bains&lt;/em&gt;, on the shore of Lake Geneva, the site of the Évian conference in 1938&lt;br&gt;&#xA;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;but does not sign the agreement until 1962&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;a href=&quot;https://en.wikipedia.org/wiki/International_Agreement_on_the_Neutrality_of_Laos&quot; rel=&quot;nofollow&quot;&gt;International Agreement on the Neutrality of Laos&lt;/a&gt; was signed in 1962 in Geneva (but in Switzerland, not France as in the above)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Given all these important details,&#xA;Could you tell if I am male or female?&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not really, actually. There is a possibility that &quot;Evian&quot; is marginally more likely to be used for a girl's name, but the results as anyone could reasonably be expected to find them are inconclusive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2016-01-13T18:35:22.020" LastActivityDate="2016-01-13T18:35:22.020" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="25339" PostTypeId="1" AcceptedAnswerId="25359" CreationDate="2016-01-13T16:29:42.350" Score="9" ViewCount="1826" Body="&lt;p&gt;No matter the road they tread,&lt;br&gt;&#xA;at the beginning I lurk.&lt;br&gt;&#xA;In darkness it is me they dread.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I stalk and plot my strike,&lt;br&gt;&#xA;eager to finish my work.&lt;br&gt;&#xA;It is after their long hike -  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;in the middle of their journey, my friend -&lt;br&gt;&#xA;when I deliver my dirk.&lt;br&gt;&#xA;Thus I finally bring the adventurer’s end.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;" OwnerUserId="18429" LastActivityDate="2016-01-14T20:20:00.090" Title="&quot;In darkness it is me they dread.&quot; Who am I?" Tags="|riddle|wordplay|rhyme|english|literature|" AnswerCount="5" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="25342" PostTypeId="2" ParentId="25319" CreationDate="2016-01-13T16:45:08.360" Score="4" Body="&lt;p&gt;You're&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A human shadow, or just a shadow?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This would explain the title as these are seen everyday&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a mouth but, I don't eat with it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As long as the object has a mouth then so does the shadow, but a shadow can't eat&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My body doesn't support me, but I eat with that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure about the second bit, but a shadows body doesn't support itself,  maybe a shadow eats the light?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have a comfy bed but never sleep in it&#xA;whether it is night or day&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A human has a bed but their shadow can't sleep in it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I love to run, but can't slow down to a jog,&#xA;but I can a raging bull&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Don't know about running, but a raging bull has a shadow?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be shorter than a dwarf&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; e.g. a fly's shadow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or taller than a skyscraper&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; e.g. Mount Everest's shadow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You better watch out &#xA;the hope for survival &#xA;just turned to your worst nightmare&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure, but most nightmares consist of shadows and darkness&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18250" LastEditorUserId="9042" LastEditDate="2016-01-13T17:37:01.063" LastActivityDate="2016-01-13T17:37:01.063" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25343" PostTypeId="2" ParentId="25339" CreationDate="2016-01-13T17:17:36.163" Score="0" Body="&lt;p&gt;I think the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No matter the road they tread,&#xA;at the beginning I lurk.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sun rises everyday.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In darkness it is me they dread.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Those who have lived in darkness for long time feel uncomfortable in coming to light&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I stalk and plot my strike,&#xA;eager to finish my work.&#xA;It is after their long hike -&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You might not like Sun-rays sometimes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;in the middle of their journey, my friend -&#xA;when I deliver my dirk.&#xA;Thus I finally bring the adventurer’s end.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sun sets in the end and the day is over.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastActivityDate="2016-01-13T17:17:36.163" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25344" PostTypeId="2" ParentId="25281" CreationDate="2016-01-13T18:16:27.890" Score="2" Body="&lt;p&gt;i think the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Solution&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a similar of the answer,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Answer is the synonym  of solution&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A jumbled flower atom,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In chemistry we take special care of the random atoms used in making a solution&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;plus 14, I am a muddled, missing alliance,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure about this. Maybe this part of the riddle is 'missing alliance'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My answer is an answer, a mix.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Solution is a mixture of chemicals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastActivityDate="2016-01-13T18:16:27.890" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25345" PostTypeId="1" AcceptedAnswerId="25350" CreationDate="2016-01-13T18:30:50.880" Score="4" ViewCount="446" Body="&lt;p&gt;I am the mixture of distinguished thoughts,&lt;br&gt;&#xA;right on this stage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you tell me who am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Monster Hints :&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stage refers to this site.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Those thoughts are more distinguished than a riddle. More distinguished than only being a puzzle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-01-13T18:53:33.513" LastActivityDate="2016-01-13T19:04:46.667" Title="Mixture of thoughts - short riddle" Tags="|riddle|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25346" PostTypeId="2" ParentId="25339" CreationDate="2016-01-13T18:32:32.223" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tiredness?&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No matter the road they tread,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;at the beginning I lurk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In darkness it is me they dread. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You don't start tired but it eventually creeps up on them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I stalk and plot my strike,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;eager to finish my work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is after their long hike - &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tiredness always follows you - when I read this line I realised it had to be something abstract. It might not be tiredness but it's a good bet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;in the middle of their journey, my friend -&lt;/p&gt;&#xA;&#xA;&lt;p&gt;when I deliver my dirk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus I finally bring the adventurer’s end.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tiredness forces you to stop, which is the 'dirk'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Am I close?&lt;/p&gt;&#xA;" OwnerUserId="18250" LastActivityDate="2016-01-13T18:32:32.223" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="25347" PostTypeId="2" ParentId="25345" CreationDate="2016-01-13T18:41:44.803" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your sub-consciousness?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It could be miles away from the answer, but it's quite hard with just one clue;&#xA;could you maybe add some more?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your sub-consciousness thinks about stuff without you realising, &lt;code&gt;right on this stage&lt;/code&gt; I suppose could mean in your head?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Am I close? If not could you maybe add another clue?&lt;/p&gt;&#xA;" OwnerUserId="18250" LastActivityDate="2016-01-13T18:41:44.803" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25348" PostTypeId="2" ParentId="25345" CreationDate="2016-01-13T18:51:21.527" Score="1" Body="&lt;p&gt;Ok thanks @manshu for that comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you may be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A riddle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A riddle is made up of different parts, each 'distinguished', and the 'stage' is this site.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18250" LastActivityDate="2016-01-13T18:51:21.527" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25349" PostTypeId="2" ParentId="25281" CreationDate="2016-01-13T19:03:24.693" Score="4" Body="&lt;p&gt;Building off of manshu's answer, the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; solution&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a similar of the answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'solution' and 'answer' are synonyms&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A jumbled flower atom:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'solution' is an anagram of 'lotus ion'&lt;br&gt;&#xA;&lt;br&gt;&#xA; A lotus is a flower, and an ion is a charged atom&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;plus 14, I am a muddled, missing alliance,:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'solution' with another N (14th letter of the alphabet) is an anagram of 'lost union'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My answer is an answer, a mix.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A solution is a mixture comprised of a solute and a solvent&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17147" LastEditorUserId="17147" LastEditDate="2016-01-13T19:07:28.367" LastActivityDate="2016-01-13T19:07:28.367" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="25350" PostTypeId="2" ParentId="25345" CreationDate="2016-01-13T19:04:46.667" Score="5" Body="&lt;p&gt;You are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hot Network Questions&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They come from a mixture of sites, and they are to the right of this page.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14233" LastActivityDate="2016-01-13T19:04:46.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25351" PostTypeId="2" ParentId="25339" CreationDate="2016-01-13T19:38:45.757" Score="7" Body="&lt;p&gt;Is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Death?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No matter the road they tread,&#xA;at the beginning I lurk.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No matter the choices people make in their lives, death is with them from the moment they are born.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In darkness it is me they dread.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many are fearful of death, as it is inevitable, full of darkness and the unknown. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I stalk and plot my strike,&#xA;eager to finish my work.&#xA;It is after their long hike -&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, the fate of death stays with us throughout our whole lives, the possibility of dying is ever-present, even though we try to diminish it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;in the middle of their journey, my friend -&#xA;when I deliver my dirk.&#xA;Thus I finally bring the adventurer’s end.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Humans, being able to contemplate our own end, would almost never be ready to die.  There is always something more we could do with our lives, something we never finished before dying.  Death interrupts our journey and calls it the end.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dreaded Death&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11452" LastActivityDate="2016-01-13T19:38:45.757" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25352" PostTypeId="2" ParentId="25339" CreationDate="2016-01-13T20:23:43.427" Score="3" Body="&lt;p&gt;Not a serious answer (unless it happens to be correct, in which case it is totally serious):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A saving throw  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No matter the road they tread,&lt;br&gt;&#xA;at the beginning I lurk.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The threat of the blown save can happen AT ANY TIME. Just having to make the roll means your face might be about to be melted off.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In darkness it is me they dread.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Entering that cave/dungeon, knowing there's gonna be a trap in there SOMEWHERE. Seriously, do dungeon architects take classes in this stuff?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I stalk and plot my strike,&#xA;eager to finish my work.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Seriously, that d20 hates you. It's just biding its time, waiting for the right opportunity.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is after their long hike -&#xA;in the middle of their journey, my friend -&#xA;when I deliver my dirk.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bloody random encounters.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus I finally bring the adventurer’s end.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ugh, a 1. The owlbear does what to my what now?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16874" LastActivityDate="2016-01-13T20:23:43.427" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25353" PostTypeId="1" AcceptedAnswerId="25370" CreationDate="2016-01-13T20:25:52.790" Score="5" ViewCount="649" Body="&lt;p&gt;I have 7 doors coloured like the rainbow:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uykxn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uykxn.png&quot; alt=&quot;#Rainbow doors&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can move in either direction moving one door for the first go, two for the second, three for the third, etc - you get the idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You &lt;em&gt;can&lt;/em&gt; jump to the other side. e.g. On your first go you can move from red to violet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start on red. Is it possible to go to each door, only going to each colour door no more than &lt;em&gt;twice&lt;/em&gt;? (It is impossible to do it touching each one just once)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You must end up back on red&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the sequence of movements which will result in success? (I am looking for the least amount of moves)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current leader board, top 5:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Matt: 9&lt;/li&gt;&#xA;&lt;li&gt;Oren Melzer: 9&lt;/li&gt;&#xA;&lt;li&gt;Matt: 13&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="18250" LastEditorUserId="11876" LastEditDate="2016-03-25T08:16:21.730" LastActivityDate="2016-03-25T08:16:21.730" Title="Rainbow Door Sequence" Tags="|logical-deduction|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25354" PostTypeId="2" ParentId="25353" CreationDate="2016-01-13T20:31:58.633" Score="5" Body="&lt;p&gt;&lt;strong&gt;More Efficient&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternate left and right until you've visited all, then head straight for red&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RLRLRLRRR&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This will traverse:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Red -&gt; Orange -&gt; Violet -&gt; Yellow -&gt; Indigo -&gt; Green -&gt; Blue -&gt; Blue -&gt; Indigo -&gt; Red&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Old Answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, since we have to go back to the end, add a move left or right (lands back on Indigo, since we're at 7 now), then do the above moves in reverse.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RRRLLL L LLLRRR&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This will go back to indigo and then touch the other six doors once, ending on red. The final progression:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (Start) Red -&gt; Orange -&gt; Green -&gt; Violet -&gt; Yellow -&gt; Blue -&gt; Indigo -&gt; Indigo -&gt; Blue -&gt; Yellow -&gt; Violet -&gt; Green -&gt; Orange -&gt; Red&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17446" LastEditorUserId="17446" LastEditDate="2016-01-13T20:48:46.440" LastActivityDate="2016-01-13T20:48:46.440" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="25355" PostTypeId="5" CreationDate="2016-01-13T20:32:30.993" Score="0" Body="&lt;p&gt;Use this tag in puzzles where the user must work out the answer through a sequence of colours&lt;/p&gt;&#xA;" OwnerUserId="18250" LastEditorUserId="18250" LastEditDate="2016-01-13T20:53:09.827" LastActivityDate="2016-01-13T20:53:09.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25356" PostTypeId="4" CreationDate="2016-01-13T20:32:30.993" Score="0" Body="Used for puzzles involving a sequence of colours" OwnerUserId="18250" LastEditorUserId="18250" LastEditDate="2016-01-13T21:20:08.907" LastActivityDate="2016-01-13T21:20:08.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25358" PostTypeId="1" AcceptedAnswerId="25365" CreationDate="2016-01-13T20:45:38.770" Score="25" ViewCount="1386" Body="&lt;p&gt;You are given a 3 by 3 chessboard with a knight on each corner, where the knights in the top row are black and in the bottom row are white.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On each turn, you may move a knight of either color (the way it moves in chess) to an unoccupied square. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it possible to rearrange the knights so that they are still in the corners, but where knights in opposite corners are the same color? If so, how many turns does it take to do so?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AfPoQ.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AfPoQ.png&quot; alt=&quot;knights&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="12408" LastEditDate="2016-07-25T02:40:12.530" LastActivityDate="2016-07-25T02:40:12.530" Title="Desegregate the Knights" Tags="|chess|checkerboard|algorithm|reachability|knight-moves|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25359" PostTypeId="2" ParentId="25339" CreationDate="2016-01-13T21:04:14.137" Score="8" Body="&lt;p&gt;Are you...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter R?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No matter the road they tread,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;at the beginning I lurk.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; first letter of road&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In darkness it is me they dread.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; not quite as solid, but it's in daRkness&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I stalk and plot my strike,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;eager to finish my work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is after their long hike -&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the word labor, a common synonym for work, is finished with an r&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;in the middle of their journey, my friend -&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; jouRney&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;when I deliver my dirk.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; not sure, except diRk?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus I finally bring the adventurer’s end.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; adventureR&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18446" LastActivityDate="2016-01-13T21:04:14.137" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25362" PostTypeId="2" ParentId="25358" CreationDate="2016-01-13T21:25:38.123" Score="2" Body="&lt;p&gt;Probably not the best explanation, but....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not possible.  In order to change relative positioning between the pieces, you would need an additional square while making moves (a total of 9 squares) to hold a piece temporarily. (4 original + 4 to hold pieces + 1 for temporary positioning).  However, despite the 3x3 grid having 9 squares, the middle square is not reachable by any knight, regardless of the number of moves it makes.&lt;/p&gt;&#xA;" OwnerUserId="14233" LastActivityDate="2016-01-13T21:25:38.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25364" PostTypeId="2" ParentId="25098" CreationDate="2016-01-13T21:48:29.143" Score="2" Body="&lt;h3&gt;Partial Solution:&lt;/h3&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We can't have all of {11, -11, 10, -10, 9, -9, 8, -8, 7, -7, 6, -6} in the solution.&lt;br&gt;&#xA; Because {11, -11} would obviously have to be in rows (11 in a column and -11 in a row doesn't work since -1 and 1 can't occupy the same place).&lt;br&gt;&#xA; This is without loss of generality (because all in columns or all in rows are symmetrical).  Call these rows R11 and R-11.&lt;br&gt;&#xA; Now {10, -10} have to be in rows as well because there's no room in the remaining columns.  Call these rows R10 and R-10.&lt;br&gt;&#xA; Similarly {9, -9}, {8, -8} and {7, -7} all have to be in rows. Call these R9, R-9, R8, R-8, R7, R-7&lt;br&gt;&#xA; Now we have one row left and the maximum we can put in a column is 5 (using 1 from R10, 0 from R-10, 1 from R9, 0 from R-9, 1 from R8, 0 from R-8, 1 from R7 and 0 from R-7 and then 1 in the remaining empty row).&#xA; So we have no room left for both 6 and -6.&lt;br&gt;&#xA; Therefore we have to get rid of one of {11, -11, 10, -10, 9, -9, 8, -8, 7, -7, 6, -6}.&lt;br&gt;&#xA; Say this number is n.&lt;br&gt;&#xA; Now the sum of the totals for the rows and the columns is -n.&lt;br&gt;&#xA; This is also 2 * (the sum of all the elements in the matrix).&lt;br&gt;&#xA; So n must be even, and without loss of generality we can say n = 6, 8 or 10.&lt;br&gt;&#xA; Now we need 22 numbers (11x11 matrix) from 23 numbers [-11, 11].&lt;br&gt;&#xA; So the numbers are [-11, 11] with 6, 8 or 10 removed.&lt;br&gt;&#xA; And the matrix must have R11 and R-11.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Far as I've got.&lt;/p&gt;&#xA;" OwnerUserId="10118" LastEditorUserId="10118" LastEditDate="2016-01-13T23:00:31.403" LastActivityDate="2016-01-13T23:00:31.403" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25365" PostTypeId="2" ParentId="25358" CreationDate="2016-01-13T21:49:59.647" Score="31" Body="&lt;p&gt;Give these names to all the squares:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;163&#xA;4 8&#xA;725&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Each number can only be accessed by way of the numbers before and after it (where 8 wraps around to 1). That means they form a loop. Since they can never pass each other up on the loop, their relative ordering cannot change. Therefore it is impossible.&lt;/p&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2016-01-13T21:49:59.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25366" PostTypeId="1" AcceptedAnswerId="25369" CreationDate="2016-01-13T23:36:59.417" Score="5" ViewCount="350" Body="&lt;p&gt;As posed by Frank Zappa and Don Pardo on &amp;ldquo;Saturday Night Live&amp;rdquo; in 1976:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;ldquo;I am gross and perverted &lt;br&gt;&#xA;I'm obsessed and deranged &lt;br&gt;&#xA;I have existed for years &lt;br&gt;&#xA;But very little has changed &lt;br&gt;&#xA;I'm the tool of the government &lt;br&gt;&#xA;And industry too &lt;br&gt;&#xA;For I am destined to rule &lt;br&gt;&#xA;And regulate you&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;ldquo;I might be vile and pernicious &lt;br&gt;&#xA;But you can't look away &lt;br&gt;&#xA;I make you think I'm delicious &lt;br&gt;&#xA;With the stuff that I say &lt;br&gt;&#xA;I'm the best you can get &lt;br&gt;&#xA;Have you guessed me yet? &lt;br&gt;&#xA;I'm the _____________&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;ldquo;You will obey me while I lead you &lt;br&gt;&#xA;And eat the garbage that I feed you &lt;br&gt;&#xA;Until the day that we don't need you &lt;br&gt;&#xA;Don't go for help ... no one will heed you &lt;br&gt;&#xA;Your mind is totally controlled &lt;br&gt;&#xA;It has been stuffed into my mold &lt;br&gt;&#xA;And you will do as you are told &lt;br&gt;&#xA;Until the rights to you are sold&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;ldquo;That's right, folks, [hint:]&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; don't touch that dial!&amp;rdquo;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18129" LastEditorUserId="18129" LastEditDate="2016-02-04T07:48:45.837" LastActivityDate="2016-02-04T07:48:45.837" Title="Riddle from Frank Zappa and The Mothers of Invention" Tags="|riddle|knowledge|poetry|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25369" PostTypeId="2" ParentId="25366" CreationDate="2016-01-14T01:11:43.437" Score="4" Body="&lt;p&gt;Oh, this is an interesting idea for a puzzle! Is the answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Television?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My reasoning:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am gross and perverted&lt;br&gt;&#xA;I'm obsessed and deranged   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There has always been moral outrage over violent TV shows.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have existed for years&lt;br&gt;&#xA;But very little has changed&lt;br&gt;&#xA;I'm the tool of the government&lt;br&gt;&#xA;And industry too&lt;br&gt;&#xA;For I am destined to rule&lt;br&gt;&#xA;And regulate you &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For a long time there have been government PSAs and corporate advertisements playing on nearly every TV channel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;“I might be vile and pernicious&lt;br&gt;&#xA;But you can't look away&lt;br&gt;&#xA;I make you think I'm delicious&lt;br&gt;&#xA;With the stuff that I say  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People enjoy watching TV and generally find arguments (especially ones that they already agree with) from pundits very convincing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm the best you can get&lt;br&gt;&#xA;Have you guessed me yet?&lt;br&gt;&#xA;I'm the _____________ &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Self explanatory.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;“You will obey me while I lead you&lt;br&gt;&#xA;And eat the garbage that I feed you&lt;br&gt;&#xA;Until the day that we don't need you&lt;br&gt;&#xA;Don't go for help ... no one will heed you  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, people like watching TV even if the content isn't intellectual.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your mind is totally controlled&lt;br&gt;&#xA;It has been stuffed into my mold&lt;br&gt;&#xA;And you will do as you are told&lt;br&gt;&#xA;Until the rights to you are sold &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Advertisements can be seen as &quot;selling&quot; viewers to companies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2016-01-14T01:11:43.437" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25370" PostTypeId="2" ParentId="25353" CreationDate="2016-01-14T03:20:28.310" Score="4" Body="&lt;p&gt;Verified by computer program, the optimal solution is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 9 steps&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This can be done four ways, but they are all rotations/reversals of Matt's answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LRLRLRLLL&lt;br&gt;&#xA;&lt;br&gt;&#xA; LRLRLRRLL&lt;br&gt;&#xA;&lt;br&gt;&#xA; RLRLRLLRR&lt;br&gt;&#xA;&lt;br&gt;&#xA; RLRLRLRRR&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Many longer solutions exist, up to 13 steps, without hitting the same spot more than twice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13861" LastEditorUserId="17389" LastEditDate="2016-01-14T05:23:37.030" LastActivityDate="2016-01-14T05:23:37.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25371" PostTypeId="2" ParentId="25183" CreationDate="2016-01-14T04:41:53.610" Score="3" Body="&lt;p&gt;Is it a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ghost?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;...only see me when I want&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They appear.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be there beside with no distance&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They fly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But once you know what I am \ You don't want to meet me again.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because they're scary.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12795" LastActivityDate="2016-01-14T04:41:53.610" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25372" PostTypeId="1" AcceptedAnswerId="25398" CreationDate="2016-01-14T07:38:16.540" Score="7" ViewCount="1011" Body="&lt;p&gt;&lt;strong&gt;BACKGROUND:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine you are playing a Magic the Gathering game of &lt;strong&gt;Sealed Oath of the Gatewatch&lt;/strong&gt;.  This means your deck consists of at least 40 cards, collected from four &quot;Oath of the Gatewatch&quot; booster packs, two &quot;Battle for Zendikar&quot; booster packs, and any number of the five traditional Basic Lands.  You may assume your packs contain whatever cards you want, as long as each pack has the proper distribution &lt;em&gt;(1 rare or mythic rare, 3 uncommons, 10 commons, 1 land, and possibly a foil of any card in the set replacing a common)&lt;/em&gt;.  You can find a simulation &lt;a href=&quot;http://draftsim.com/draft.php?mode=Sealed_OGW&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is the start of a match and your opponent mulligans to 0 with a deck consisting of 40 Basic Mountains.  You may not assume the outcome of any choice they make.  Whenever your deck is shuffled, including at the start of the game, you may assume it is stacked in any order you like, and you may choose to play or draw.  What is the fastest that you can win such a game?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Measuring the quality of a solution:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Faster is better, so winning turn 4 on the play is better than winning turn 4 on the draw, and winning in your turn 4 upkeep on the play (&lt;a href=&quot;http://www.mtgsalvation.com/cards/oath-of-the-gatewatch/26302-hedron-alignment&quot; rel=&quot;nofollow&quot;&gt;Hedron Alignment&lt;/a&gt;?) is better than winning during your turn 4 combat step on the play.  For equally fast solutions, the one that deals more damage (or life loss) wins.  If you're tied on both speed and damage, the winner is the one with the most style points, subjectively assigned by me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Update: @JonTheMon's Turn 2 answer was the one I came up with when solving this myself, but I'll leave the question open for the rest of the day in case anybody comes up with something better.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="17389" LastEditDate="2016-01-14T19:25:33.103" LastActivityDate="2016-01-14T19:25:33.103" Title="Magic: the Gathering - Sealed Speed" Tags="|calculation-puzzle|magic-the-gathering|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25374" PostTypeId="1" AcceptedAnswerId="25377" CreationDate="2016-01-14T08:00:20.207" Score="7" ViewCount="1709" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Alone I am 24th, with a friend I am 20. Another friend and I am unclean.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="6151" LastActivityDate="2016-01-14T08:17:53.290" Title="A riddle for your time (2)" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25376" PostTypeId="1" AcceptedAnswerId="25396" CreationDate="2016-01-14T08:10:02.263" Score="16" ViewCount="1141" Body="&lt;p&gt;You are a mathematician in the late 19&lt;sup&gt;th&lt;/sup&gt; century and you receive a mysterious letter from your colleague Georg Cantor.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;S   G   A S C O Q X&#xA;E G H D R H G D H L&#xA;    D M   D K D R Z&#xA;  H H   E Z D H   S&#xA;  N R M E H B H S O&#xA;Q S E   E D M S M X&#xA;N N G   Z H Z H T F&#xA;  R X Z L   M Q S Z&#xA;  N R R S X E H   D&#xA;K S E H     N N S N&#xA;&#xA;G. Cantor&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What could he possibly be trying to tell you?&lt;/p&gt;&#xA;" OwnerUserId="8949" LastActivityDate="2016-01-14T13:38:28.677" Title="Cantor's Conviction" Tags="|cipher|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25377" PostTypeId="2" ParentId="25374" CreationDate="2016-01-14T08:17:53.290" Score="11" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;X&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alone I am 24th&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'X' is the 24th letter in the alphabets&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;with a friend I am 20&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'X' with another 'X' gives 'XX', which is equal to 20 in roman numeral system.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another friend and I am unclean&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think it's referring to 'XXX', which most people consider as dirty.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastActivityDate="2016-01-14T08:17:53.290" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25378" PostTypeId="1" AcceptedAnswerId="25381" CreationDate="2016-01-14T08:26:01.507" Score="10" ViewCount="2671" Body="&lt;p&gt;This one is slightly harder then the previous two.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;I'm not a pool, a tree, a slave, a website, or a shirt.&lt;/p&gt;&#xA;&lt;p&gt;But all have something that is related to me.&lt;/p&gt;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6151" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-01-15T10:15:18.970" Title="A riddle for your time (3)" Tags="|riddle|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25379" PostTypeId="2" ParentId="25378" CreationDate="2016-01-14T08:45:32.833" Score="7" Body="&lt;p&gt;maybe you are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a dog &lt;br&gt; - a tree has a bark &lt;br&gt; a slave has a master &lt;br&gt; a shirt has a collar&lt;br&gt; a pool can have a guard (ie guarddog)&lt;br&gt;... though i'm struggling with website...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18319" LastActivityDate="2016-01-14T08:45:32.833" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25380" PostTypeId="2" ParentId="25378" CreationDate="2016-01-14T08:59:09.937" Score="6" Body="&lt;p&gt;maybe you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Golf &lt;br&gt;website has links &lt;br&gt;tree has wood &lt;br&gt; slave has irons &lt;br&gt; shirt could be a tee-shirt&lt;br&gt; pool has water&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18319" LastActivityDate="2016-01-14T08:59:09.937" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25381" PostTypeId="2" ParentId="25378" CreationDate="2016-01-14T09:10:42.710" Score="17" Body="&lt;p&gt;I think it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Book&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pool and book have cover &lt;br /&gt; Tree and book have leaves&lt;br /&gt; Slaves and books are bound &lt;br /&gt; Website and book have pages &lt;br /&gt; Shirt and book have sleeves.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="2730" LastEditDate="2016-01-15T10:15:18.970" LastActivityDate="2016-01-15T10:15:18.970" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25382" PostTypeId="2" ParentId="25183" CreationDate="2016-01-14T09:54:56.473" Score="2" Body="&lt;p&gt;Assuming they exist,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ghost&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Scream out or scream low &#xA;You will only hear me when I blow &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since they fly, so the one way you can hear it is when they blow maybe make sounds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Stand back or stand front &#xA;You will only see me when I want &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ghosts appear on will and when they want you to see them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Go there or no-where &#xA;I can be there beside with no distance &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They have supreme powers and can appear appear anywhere in no time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't have ticket &#xA;I can't make chitchat &#xA;But once you know what I am &#xA;You don't want to meet me again. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes we don't wish to comeback to you or meet you if we know you are ghost.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I could be the same shape as you &#xA;Or I could be in the other shape you never saw &#xA;I was just like you in before &#xA;May be you were my beloved or more &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Assuming ghost of a human being who might have been close to you. Mostly ghost appear to people whom are their relatives or acquaintances.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No hint on this because I might be staring right into your face.&#xA;You know or never know until I show.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I hope its not true for me. :-)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2656" LastActivityDate="2016-01-14T09:54:56.473" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25383" PostTypeId="2" ParentId="25376" CreationDate="2016-01-14T10:17:18.370" Score="1" Body="&lt;p&gt;Too long for a comment, so here goes what came to my mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A first (bad) try:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That's a 10x10 matrix of letters and spaces. Cantor is famous (among other things) for his diagonal argument for the existence of uncountable sets. In his proof he constructs a sequence of elements from an (infinitely large) &quot;square&quot; which feels like the same thing we are supposed to do here. The way he does it is he takes the diagonal and (in the case of 1s and 0s) inverts it, for decimals he adds 1 (modulo 10). A very naive approach would be to take the diagonal &lt;code&gt;SGD EDZQ N&lt;/code&gt; and change it in some way it. However, I couldn't find a way that produces something that makes sense to me: A Caesar Cipher shift by any number doesn't yield anything. Switching letters like a&amp;lt;-&gt;z, b&amp;lt;-&gt;y, etc. is just as bad. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cantor's pairing function can also be used to produce a sequence of elements from a matrix. Starting at the top left corner, read diagonal lines: &lt;code&gt;S&lt;/code&gt;, &lt;code&gt;E&lt;/code&gt;, &lt;code&gt;GG&lt;/code&gt;, etc. This does not produce anything I would call a message. Starting at another corner doesn't help either.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not giving up just yet, but so far my best guess for his message to me is that I should have payed more attention in math classes.&lt;/p&gt;&#xA;" OwnerUserId="17069" LastActivityDate="2016-01-14T10:17:18.370" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25384" PostTypeId="2" ParentId="25183" CreationDate="2016-01-14T10:40:48.617" Score="1" Body="&lt;p&gt;My answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fear&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scream out or scream low&#xA;You will only hear me when I blow&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No matter if you scream or not, the fear is there and you will feel it (when it blows);&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Stand back or stand front&#xA;You will only see me when I want&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Face it or not, you will still feel the fear when it comes;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Go there or no-where&#xA;I can be there beside with no distance&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It comes at any moment and any place&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't have ticket&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fear just comes, doesn't need a ticket&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can't make chitchat&#xA;But once you know what I am&#xA;You don't want to meet me again. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You don't have power on it, and when you are feared you don't want to feel it anymore.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Sorry for bad english. Have a nice day.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I got a second guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it may be a fart. you can hear it only when it blows, and ofc you don't want to meet it again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18458" LastEditorUserId="18458" LastEditDate="2016-01-14T14:09:02.933" LastActivityDate="2016-01-14T14:09:02.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25385" PostTypeId="1" AcceptedAnswerId="25391" CreationDate="2016-01-14T10:46:01.723" Score="15" ViewCount="1845" Body="&lt;p&gt;There is a brick of gold and a brick of iron in a boat on a lake (both 10 inch blocks), if they are both dropped into the water which will make the water level higher?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&#xA;we can assume its a very small and light boat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(feel free to change the tag)&lt;/p&gt;&#xA;" OwnerUserId="6151" LastEditorUserId="12408" LastEditDate="2016-01-16T21:02:00.187" LastActivityDate="2016-01-16T21:02:00.187" Title="Bars in the river" Tags="|physics|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25387" PostTypeId="2" ParentId="25385" CreationDate="2016-01-14T10:50:26.337" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strike&gt;None will make it any higher, since they already caused the same water displacement while in the boat.&lt;/strike&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: Above answer is wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's actually the iron block, since it has a lower density. In other words: while in the boat, any additional weight causes additional water displacement amounting to the volume of water of that weight. In the water, however, it just displaces up to it's own volume. Higher density means higher weight at the same volume and thus more displacement.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18459" LastEditorUserId="18459" LastEditDate="2016-01-14T12:25:24.997" LastActivityDate="2016-01-14T12:25:24.997" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25388" PostTypeId="1" AcceptedAnswerId="25390" CreationDate="2016-01-14T10:58:19.013" Score="1" ViewCount="605" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;500 is at my end and my start, yet 5 is at my heart.&#xA;  The first letter and the first number make me complete.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is my name?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My name is that of a king.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6151" LastActivityDate="2016-01-14T11:39:49.873" Title="A riddle for your time (4)" Tags="|riddle|" AnswerCount="1" CommentCount="1" ClosedDate="2016-01-14T14:32:44.420" ContentLicense="CC BY-SA 3.0" />
  <row Id="25390" PostTypeId="2" ParentId="25388" CreationDate="2016-01-14T11:05:43.370" Score="7" Body="&lt;p&gt;My answer is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DAVID. D=500, V=5, and I=1 (the first number). 'A' is the first letter of the alphabet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18460" LastEditorUserId="6151" LastEditDate="2016-01-14T11:39:49.873" LastActivityDate="2016-01-14T11:39:49.873" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25391" PostTypeId="2" ParentId="25385" CreationDate="2016-01-14T11:10:25.823" Score="17" Body="&lt;p&gt;I will assume that we first drop one of the blocks, get it back out of the water (with a rope or something) and then drop the other block.&lt;br&gt;&#xA;After that we compare the water levels at the times each of the blocks was in the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is the case the water level will be higher while the &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; iron block&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is in the water.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Reason:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Both blocks will displace the same amount of water when they are dropped into the water, since none of them floats and they both have the same volume.&lt;br&gt;&#xA; The only difference between both measurements will be the displacement that the boat will create on the water.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The boat has to displace an amount of water that has the same weight as the weight of the boat in order to stay afloat.&lt;br&gt;&#xA; Therefore the heavier the boat the more water will be displaced and the higher the water level will be.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Since gold has a much higher density compared to iron the boat will have a higher weight during the time that the iron is in the water (gold remains on the boat) as compared to the time that the gold is in the water (iron remains on the boat).&#xA; This higher weight results in a higher displacement of water and therefore in a higher water level.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16608" LastActivityDate="2016-01-14T11:10:25.823" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25392" PostTypeId="2" ParentId="25385" CreationDate="2016-01-14T11:27:49.330" Score="8" Body="&lt;p&gt;Both will make the water level of the lake &lt;strong&gt;lower&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Until they are on the boat they shift a volume of water equal to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(weight\ of\ the\ brick)\ /\ (density\ of\ the\ water)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When they are thrown into the lake they shift a volume of water equal to the volume of the brick, that is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(weight\ of\ the\ brick)\ /\ (density\ of\ the\ brick)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that iron and gold have higher density than the water, the volume shifted is less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be more complete, given the densities of gold and iron respectively $19.30 g/cm^3$ and $7.87 g/cm^3$ if:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$l1$ = level of the water lake with both the bars on the boat&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$l2$ = level of the water with iron brick on the boat and gold brick into the lake&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$l3$ = level of the water with gold brick on the boat and iron brick into the lake&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$l4$ = level of the water with both bricks into the lake&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$l1 &amp;gt; l3 &amp;gt; l2 &amp;gt; l4$&lt;/p&gt;&#xA;" OwnerUserId="15823" LastActivityDate="2016-01-14T11:27:49.330" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25393" PostTypeId="2" ParentId="25372" CreationDate="2016-01-14T12:13:46.083" Score="2" Body="&lt;p&gt;&lt;strong&gt;First turn&lt;/strong&gt;&lt;br&gt;&#xA;play Swamp&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second turn&lt;/strong&gt;&lt;br&gt;&#xA;play Island&lt;br&gt;&#xA;cast &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=402075&quot; rel=&quot;nofollow&quot;&gt;Transgress the Mind&lt;/a&gt; on yourself exiling &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=407567&quot; rel=&quot;nofollow&quot;&gt;Hedron Alignment&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Third turn&lt;/strong&gt;&lt;br&gt;&#xA;Play Island&lt;br&gt;&#xA;Cast &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=407570&quot; rel=&quot;nofollow&quot;&gt;Oath of Jace&lt;/a&gt; discarding another Hedron Alignment&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fourth Turn&lt;/strong&gt;&lt;br&gt;&#xA;Cast Hedron Alignment&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fifth Turn&lt;/strong&gt;&lt;br&gt;&#xA;Win the game in your upkeep by revealing the 4th Hedron Alignment in your hand&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other cards you draw, discard or have in your hand are irrelevant&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2016-01-14T12:13:46.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25395" PostTypeId="2" ParentId="25372" CreationDate="2016-01-14T12:57:39.460" Score="4" Body="&lt;p&gt;Have you heard of the Expeditions? Wizards is packaging super rare reprinted lands into Oath of the Gatewatch packs, and while those lands are not legal in Standard Constructed Magic, if you open them in Sealed, you're allowed to play them. So for this solution, I opened &lt;a href=&quot;http://gatherer.wizards.com/Handlers/Image.ashx?multiverseid=409569&amp;amp;type=card&quot; rel=&quot;nofollow&quot;&gt;Eye of Ugin&lt;/a&gt; and &lt;a href=&quot;http://gatherer.wizards.com/Handlers/Image.ashx?multiverseid=409567&amp;amp;type=card&quot; rel=&quot;nofollow&quot;&gt;Ancient Tomb&lt;/a&gt; as Expeditions. I also got very lucky in my  Battle for Zendikar packs: both rares were &lt;a href=&quot;http://gatherer.wizards.com/Handlers/Image.ashx?multiverseid=401871&amp;amp;type=card&quot; rel=&quot;nofollow&quot;&gt;Endless Ones&lt;/a&gt;, and both had a foil &lt;a href=&quot;http://gatherer.wizards.com/Handlers/Image.ashx?multiverseid=401871&amp;amp;type=card&quot; rel=&quot;nofollow&quot;&gt;Endless One&lt;/a&gt; as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Turn 1:&lt;/strong&gt;: &lt;a href=&quot;http://gatherer.wizards.com/Handlers/Image.ashx?multiverseid=409569&amp;amp;type=card&quot; rel=&quot;nofollow&quot;&gt;Eye of Ugin&lt;/a&gt;, &lt;a href=&quot;http://gatherer.wizards.com/Handlers/Image.ashx?multiverseid=401871&amp;amp;type=card&quot; rel=&quot;nofollow&quot;&gt;Endless One&lt;/a&gt; for X=2, &lt;a href=&quot;http://gatherer.wizards.com/Handlers/Image.ashx?multiverseid=401871&amp;amp;type=card&quot; rel=&quot;nofollow&quot;&gt;Endless One&lt;/a&gt; for X=2, &lt;a href=&quot;http://gatherer.wizards.com/Handlers/Image.ashx?multiverseid=401871&amp;amp;type=card&quot; rel=&quot;nofollow&quot;&gt;Endless One&lt;/a&gt; for X=2, &lt;a href=&quot;http://gatherer.wizards.com/Handlers/Image.ashx?multiverseid=401871&amp;amp;type=card&quot; rel=&quot;nofollow&quot;&gt;Endless One&lt;/a&gt; for X=2.&lt;br&gt;&#xA;&lt;strong&gt;Turn 2:&lt;/strong&gt; &lt;a href=&quot;http://gatherer.wizards.com/Handlers/Image.ashx?multiverseid=409567&amp;amp;type=card&quot; rel=&quot;nofollow&quot;&gt;Ancient Tomb&lt;/a&gt;, &lt;a href=&quot;http://media.wizards.com/2015/ogw_239nCi30ks3/en_qwaVdKGXfL.png&quot; rel=&quot;nofollow&quot;&gt;Thought-Knot Seer&lt;/a&gt;, attack for 8.&lt;br&gt;&#xA;&lt;strong&gt;Turn 3:&lt;/strong&gt; Swamp, &lt;a href=&quot;http://gatherer.wizards.com/Handlers/Image.ashx?multiverseid=402063&amp;amp;type=card&quot; rel=&quot;nofollow&quot;&gt;Swarm Surge&lt;/a&gt;, attack for 22.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's turn three, on the play, thirty damage during the first strike step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Old solution&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Turn 1: Island, &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=402021&quot; rel=&quot;nofollow&quot;&gt;Salvage Drone&lt;/a&gt;&lt;br&gt;&#xA;Turn 2: Swamp, &lt;a href=&quot;http://gatherer.wizards.com/Handlers/Image.ashx?multiverseid=402075&amp;amp;type=card&quot; rel=&quot;nofollow&quot;&gt;Transgress the Mind&lt;/a&gt; on yourself exiling &lt;a href=&quot;http://media.wizards.com/2015/ogw_239nCi30ks3/en_e66ZBFGveS.png&quot; rel=&quot;nofollow&quot;&gt;Hedron Alignment&lt;/a&gt;&lt;br&gt;&#xA;Turn 3: Swamp, &lt;a href=&quot;http://media.wizards.com/2015/ogw_239nCi30ks3/en_e66ZBFGveS.png&quot; rel=&quot;nofollow&quot;&gt;Hedron Alignment&lt;/a&gt;&lt;br&gt;&#xA;Turn 4: With &lt;a href=&quot;http://media.wizards.com/2015/ogw_239nCi30ks3/en_e66ZBFGveS.png&quot; rel=&quot;nofollow&quot;&gt;Hedron Alignment&lt;/a&gt;'s trigger on the stack, cast &lt;a href=&quot;http://media.wizards.com/2015/ogw_239nCi30ks3/en_nRjOVWrwyA.png&quot; rel=&quot;nofollow&quot;&gt;Grasp of Darkness&lt;/a&gt; on &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=402021&quot; rel=&quot;nofollow&quot;&gt;Salvage Drone&lt;/a&gt;. Use &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=402021&quot; rel=&quot;nofollow&quot;&gt;Salvage Drone&lt;/a&gt;'s death trigger to discard Hedron Alignment. Resolve &lt;a href=&quot;http://media.wizards.com/2015/ogw_239nCi30ks3/en_e66ZBFGveS.png&quot; rel=&quot;nofollow&quot;&gt;Hedron Alignment&lt;/a&gt;'s trigger, reveal a &lt;a href=&quot;http://media.wizards.com/2015/ogw_239nCi30ks3/en_e66ZBFGveS.png&quot; rel=&quot;nofollow&quot;&gt;Hedron Alignment&lt;/a&gt; in your hand, win the game.&lt;/p&gt;&#xA;" OwnerUserId="7459" LastEditorUserId="7459" LastEditDate="2016-01-14T15:04:48.437" LastActivityDate="2016-01-14T15:04:48.437" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25396" PostTypeId="2" ParentId="25376" CreationDate="2016-01-14T13:38:28.677" Score="24" Body="&lt;p&gt;I think the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.azquotes.com/quote/690586&quot; rel=&quot;nofollow noreferrer&quot;&gt;The fear of infinity is a form of myopia that destroys the possibility of seeing the actual infinite&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you shift every letter forward one place in the alphabet, the words appear diagonally from top left to bottom right:&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/Uw9K5.png&quot; alt=&quot;Cantor quote]&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="40" LastActivityDate="2016-01-14T13:38:28.677" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25397" PostTypeId="2" ParentId="25372" CreationDate="2016-01-14T13:38:56.657" Score="3" Body="&lt;p&gt;Turn 4, upkeep&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Turn 1:&lt;/strong&gt;&lt;br&gt;&#xA;Play a Swamp&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Turn 2:&lt;/strong&gt;&lt;br&gt;&#xA;Play an Island&lt;br&gt;&#xA;Cast &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=402075&quot; rel=&quot;nofollow&quot;&gt;Transgress the Mind&lt;/a&gt; on yourself to exile &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=407567&quot; rel=&quot;nofollow&quot;&gt;Hedron Alignment&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Turn 3:&lt;/strong&gt;&lt;br&gt;&#xA;Play an Island&lt;br&gt;&#xA;Cast Hedron Alignment&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Turn 4:&lt;/strong&gt;&lt;br&gt;&#xA;At the beginning of your upkeep, Hedron Alignment's ability triggers and goes on the stack.&lt;br&gt;&#xA;In response, cast &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=407593&quot; rel=&quot;nofollow&quot;&gt;Corpse Churn&lt;/a&gt; to put a Hedron Alignment from the library into the graveyard.&lt;br&gt;&#xA;The triggered ability resolves, you reveal another Hedron Alignment in your hand and you win the game.&lt;/p&gt;&#xA;" OwnerUserId="1598" LastEditorUserId="1598" LastEditDate="2016-01-14T13:54:31.773" LastActivityDate="2016-01-14T13:54:31.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25398" PostTypeId="2" ParentId="25372" CreationDate="2016-01-14T15:10:49.193" Score="10" Body="&lt;p&gt;Turn 2 combat:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T1: Eye of Ugin, 5x Eldrazi Mimic (1 foil)&lt;br&gt;&#xA;T2: Mountain, Vile Aggregate - 30 damage&lt;br&gt;&#xA;Total: 30 damage (42 on draw)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Turn 3 combat:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T1: Ancient Tomb, Eldrazi Mimic&lt;br&gt;&#xA;T2: Ancient Tomb, Eldrazi Mimic, Eldrazi Mimic - 2 damage&lt;br&gt;&#xA;T3: Land, Reality Smasher - 20 damage&lt;br&gt;&#xA;Total 22 damage&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Turn 3 draw combat:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T1: Forest, Scythe Leopard&lt;br&gt;&#xA;T2: Ancient Tomb, Snapping Gnarlid - 2 damage&lt;br&gt;&#xA;T3: Verdant Catacombs&gt;Forest, Swell of Growth&gt;Verdant Catacombs&gt;Forest, Swell of Growth&gt;Mountain, Reckless Bushwacker- 2x +2/+2, 5 landfall triggers - 7 + 8 + 4 + 2 = 21 damage&lt;br&gt;&#xA;Total 23 damage&lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2016-01-14T16:43:53.337" LastActivityDate="2016-01-14T16:43:53.337" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="25399" PostTypeId="2" ParentId="25215" CreationDate="2016-01-14T15:51:04.197" Score="9" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Java&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One possibility is a tiny fruit that if you collect enough of me, someone can turn me into something that you like&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Coffee beans are the the seeds of the fruit of coffee plants, and when you collect enough, you can roast, grind, and brew them into coffee.  The island of Java is so known for its Java coffee beans, that in some cases java and coffee are viewed as interchangeable.  Coffee is something that many people like.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another possibility is a gigantic creature that is cut into 6 pieces. Two of them are tiny.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This refers to the Indonesian island of Java, which is the world's most populous island, and which is split into four provinces and two special regions, with the provinces being larger than the special areas.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet another possibility is an eccentric creature which can understand any language with a strange feature shared with Kēlen (I'm not human).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This refers to the Java programming language that many use for computer applications the world round.  I am not as familiar with the subject, but this may tie in with the title with JavaServer Faces, which used to build user interface in web applications.  &lt;strong&gt;EDIT&lt;/strong&gt;: Per the OP in the comments, Java is a &lt;em&gt;noun-oriented language.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Intimately connected to Indonesia, coffee, and code (and the OP), you are &lt;strong&gt;JAVA&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11452" LastEditorUserId="11452" LastEditDate="2016-01-15T14:38:02.917" LastActivityDate="2016-01-15T14:38:02.917" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25400" PostTypeId="1" AcceptedAnswerId="25408" CreationDate="2016-01-14T18:24:08.427" Score="7" ViewCount="631" Body="&lt;p&gt;Some examples: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/25358/desegregate-the-knights&quot;&gt;Desegregate the Knights&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/questions/25287/switch-the-knights?lq=1&quot;&gt;Switch The Knights&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You must give two 8x8 chess board positions that have any number of black and white knights.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Both boards must have the same number of each colour of knights.&lt;/li&gt;&#xA;&lt;li&gt;It must &lt;strong&gt;not&lt;/strong&gt; be possible to arrive at the second board position from the first (and vice versa) using a finite number of L-shaped chess knight moves, that do not necessarily have to be played one colour at a time. (2 or more moves of the same coloured knights can be played in succession)&lt;/li&gt;&#xA;&lt;li&gt;All knights of the same colour are indistinguishable and all knights of different colours are.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the minimum number of knights required to do so?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2016-07-25T02:38:12.580" Title="Create an impossible knight transformation" Tags="|chess|optimization|reachability|knight-moves|" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25403" PostTypeId="2" ParentId="25400" CreationDate="2016-01-14T19:38:31.570" Score="1" Body="&lt;p&gt;I believe the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 64 knights&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a knight on any square can reach any square on a 8x8 board in a finite number of moves. If you try to block a knight from moving by placing knights in each square it can move to, you must then block the new knights again and again until no knight can move. This inevitably results in the entire board being filled.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Al alternate way to prove this (which I don't have the patience for) would be to prove&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; that it is possible to swap 63 knights from even one starting position.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17231" LastEditorUserId="17231" LastEditDate="2016-01-14T20:07:23.623" LastActivityDate="2016-01-14T20:07:23.623" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25404" PostTypeId="1" AcceptedAnswerId="25411" CreationDate="2016-01-14T19:53:17.930" Score="29" ViewCount="2051" Body="&lt;p&gt;Each of the sections below has a single answer, and the five sections are related.  After finding the solution to one section, the others should be much easier.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h3&gt;&lt;em&gt;The Best Friend&lt;/em&gt;&lt;/h3&gt;&#xA;  &#xA;  &lt;p&gt;Whenever you're down and out, I'm in there like a dirty shirt.&lt;br&gt;&#xA;  Your other friends and I are gems.&lt;br&gt;&#xA;  When you renovated your bathroom, you needed me in order to work hard.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h3&gt;&lt;em&gt;The Counsellor&lt;/em&gt;&lt;/h3&gt;&#xA;  &#xA;  &lt;p&gt;If you're angry, I will come and provide some help.&lt;br&gt;&#xA;  If you're still angry, I will let you be, to help you evaluate.&lt;br&gt;&#xA;  I'll be with you in all your despair and anguish.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h3&gt;&lt;em&gt;The Rabble-Rouser&lt;/em&gt;&lt;/h3&gt;&#xA;  &#xA;  &lt;p&gt;If you're waiting on too many things, you can bring me in to shake things up.&lt;br&gt;&#xA;  If anyone is being ruthless or unforgiving, I'm in!&lt;br&gt;&#xA;  It is probably wise to remove me from employment.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h3&gt;&lt;em&gt;The Party Animal&lt;/em&gt;&lt;/h3&gt;&#xA;  &#xA;  &lt;p&gt;When I get together with any girl, there's bound to be a party.&lt;br&gt;&#xA;  In fact, no adequate party happens without me being there.&lt;br&gt;&#xA;  I cause a commotion at any party just by showing up.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h3&gt;&lt;em&gt;The Efficiency Expert&lt;/em&gt;&lt;/h3&gt;&#xA;  &#xA;  &lt;p&gt;Any young man can make use of me to get ahead.&lt;br&gt;&#xA;  Stick me on a letter, and it will arrive much faster.&lt;br&gt;&#xA;  If you were thinking of leaving your job and you brought me in to consult, I would advise not saying anything.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&#xA;&lt;strong&gt;&lt;em&gt;To what does each of these sections refer?  What is the common bond between them all?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2016-01-15T16:17:04.677" Title="Five Unique Individuals" Tags="|riddle|wordplay|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25406" PostTypeId="2" ParentId="25404" CreationDate="2016-01-14T20:30:50.743" Score="1" Body="&lt;p&gt;A (probably wrong) guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mind altering substances&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The best friend:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Coffee is ubiquitous and helps you work&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The counsellor:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cigarettes bring you back to even and force you to take time for reflection&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The rabble-rouser:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cocaine should not be used at work and encourages brash behaviour&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The party animal:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Liquor is at parties and causes a commotion&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The efficiency expert:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Adderal (speed) young men use it a lot for studying, speed is fast, if you use it at work don't tell people&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18103" LastActivityDate="2016-01-14T20:30:50.743" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25407" PostTypeId="2" ParentId="25404" CreationDate="2016-01-14T21:41:13.377" Score="2" Body="&lt;p&gt;Here's a guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hand&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Best Friend&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The thumb, helps you get a grip on things&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Counsellor&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The pointer finger will guide you&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Rabble-Rouser&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Middle finger, used to flip people off&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Party Animal&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ring finger, since getting engaged/married seems to start parties&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Efficiency Expert&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pinky finger? Not quite sure how this would work, but pinky promises could help you getting ahead&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9696" LastActivityDate="2016-01-14T21:41:13.377" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25408" PostTypeId="2" ParentId="25400" CreationDate="2016-01-14T22:18:13.490" Score="12" Body="&lt;p&gt;All 64 knights are needed, in which case any setup is stuck. We prove that with 63 knights, any position is reachable from any other position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With 63 knights, the puzzle is much like a sliding puzzle, except pieces slide as knight moves rather than into orthogonality adjacent cells. We can think of a move as swapping the position of the hole and the knight that moves into it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather than having the knights be colored, let them be labelled 1 through 63, similar to the &lt;a href=&quot;https://en.wikipedia.org/wiki/15_puzzle&quot; rel=&quot;noreferrer&quot;&gt;15 puzzle&lt;/a&gt;. As the 15 puzzle, there is an invariant that means only half the positions are reachable: the parity of the permutation (labelling the hole as &quot;64&quot;) plus the color of the hole square. Each move swaps the hole with a knight, switching both the parity of the permutation and the color of the hole square, preserving the invariant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We'll argue that, like for the 15 puzzles, one can reach any position with with the same invariant. First, move the hole to the right spot, then we show that any even permutation can be obtained while keeping the hole where it is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With three knights and a hole X in a four-cycle, one can have the hole do a lap clockwise to rotate the knights along a three cycle.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    2          3&#xA;1          2      &#xA;      3          1    &#xA;  X          X    &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;With knights 1,2,3 in this formation anywhere on the board, we can always move the hole X to the right place via other spaces to catalyze their rotation, then move the hole back the way it came, so the board is as it started except for those three knights.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can do the same along other 3-cycle formations&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    2&#xA;1       3&#xA;    X&#xA;&#xA;1&#xA;    3&#xA;  2&#xA;      X&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In fact, this lets us rotate any three knights in a path of three that isn't a straight line. Using a knight's tour graph with no straight lines, we can cycle any three adjacent cells in the tour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zvL4v.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zvL4v.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, permutations of three adjacent elements suffices to generate all even permutations, as &lt;a href=&quot;http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/genset.pdf&quot; rel=&quot;noreferrer&quot;&gt;shown in theorem 3.4 here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It only remains to show that the invariant doesn't prevent correctly colored knights in the original problem. Since there's either at least two white knights or two black knights, we can always swap their labels to reverse the invariant.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2016-01-14T22:18:13.490" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25409" PostTypeId="2" ParentId="25004" CreationDate="2016-01-15T05:38:50.083" Score="0" Body="&lt;p&gt;It has got to be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Rain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I can hold the heaviest of all objects,&#xA;but will falter to the lightest of all objects.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Not sure what the heavest object is but the lightest object is air, and rain gets deviated by air.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I prefer long and skinny rather than short and thick&#xA;since it is to my taste&#xA;which you cannot sense&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Not sure what this means...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You can see me as far the eye can see&#xA;in the seven seas&#xA;or the seven continents&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It rains pretty much everywhere and when it is raining, you can see it all around.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Precious by nature&#xA;yet never rare&#xA;I seem to be more desired than gold&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;People like rain?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You are here because of me&#xA;and I am certainly not here because of you&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;No explanation needed here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;As a group, I can be a force to reckon with&#xA;yet I can be gentler than a breeze&#xA;as every organism adores me&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If it rains really heavily, the force of the water is huge. But a gentle drizzle feels like...mother nature's love?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17976" LastActivityDate="2016-01-15T05:38:50.083" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25410" PostTypeId="2" ParentId="25404" CreationDate="2016-01-15T07:15:34.773" Score="1" Body="&lt;p&gt;My guess is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; parts of a car&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Best Friend&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever you're down and out, I'm in there like a dirty shirt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your other friends and I are gems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you renovated your bathroom, you needed me in order to work hard.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; this refer to &lt;strong&gt;TIRES&lt;/strong&gt;. When you are not using the car, it is just sitting ther. Other friends are the four tires and it works together. For the renovation, you need to be TIREless.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Counsellor&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're angry, I will come and provide some help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're still angry, I will let you be, to help you evaluate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll be with you in all your despair and anguish.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; my guess here is the &lt;strong&gt;GAUGE&lt;/strong&gt;. This is use to provide info and help and also lets you evaluate your car.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Rabble-Rouser&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're waiting on too many things, you can bring me in to shake things up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone is being ruthless or unforgiving, I'm in!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is probably wise to remove me from employment.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is the &lt;strong&gt;BRAKE&lt;/strong&gt;. If you are waiting, you need some break. Also for unforgiving, like a fight, you need a break. Removing break from employment (I guess over breaks) is good.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Party Animal&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I get together with any girl, there's bound to be a party.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, no adequate party happens without me being there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cause a commotion at any party just by showing up.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This one's funny. This is the &lt;strong&gt;BUMPER&lt;/strong&gt;. Only for girls and it always comes in pairs. If this shows up in a party, it can cause commotion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Efficiency Expert&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any young man can make use of me to get ahead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stick me on a letter, and it will arrive much faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were thinking of leaving your job and you brought me in to consult, I would advise not saying anything.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The last one is &lt;strong&gt;SPOILER&lt;/strong&gt;. Getting ahead is getting spoiler information and it can arrive faster. About job, someone will not say anything so it would not spoil your plan or tell the plan as a spoiler. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16036" LastActivityDate="2016-01-15T07:15:34.773" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25411" PostTypeId="2" ParentId="25404" CreationDate="2016-01-15T08:52:48.207" Score="29" Body="&lt;p&gt;These individuals are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the five vowels.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Best Friend&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This is O.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whenever you're down and out, I'm in there like a dirty shirt.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; O is in both &quot;down&quot; and &quot;out&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your other friends and I are gems.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Adding O to &quot;pals&quot; gives &quot;opals&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you renovated your bathroom, you needed me in order to work hard.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Adding O to &quot;tiled&quot; gives &quot;toiled&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Counsellor&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is I.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you're angry, I will come and provide some help.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Adding I to &quot;mad&quot; gives &quot;maid&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you're still angry, I will let you be, to help you evaluate.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Taking I away from &quot;irate&quot; gives &quot;rate&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll be with you in all your despair and anguish.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I is in &quot;despair&quot; and &quot;anguish&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Rabble-Rouser&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is U.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you're waiting on too many things, you can bring me in to shake things up.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Adding U to &quot;pend&quot; gives &quot;upend&quot;.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If anyone is being ruthless or unforgiving, I'm in!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; U is in &quot;ruthless&quot; and &quot;unforgiving&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is probably wise to remove me from employment.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Removing U from &quot;usage&quot; gives &quot;sage&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Party Animal&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is A.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I get together with any girl, there's bound to be a party.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Adding A to &quot;gal&quot; gives &quot;gala&quot;.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In fact, no adequate party happens without me being there.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A is in &quot;adequate&quot; and &quot;party&quot;.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cause a commotion at any party just by showing up.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Adding A to &quot;do&quot; gives &quot;ado&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Efficiency Expert&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is E.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Any young man can make use of me to get ahead.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Adding E to &quot;lad&quot; gives &quot;lead&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Stick me on a letter, and it will arrive much faster.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Adding E to &quot;mail&quot; gives &quot;email&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you were thinking of leaving your job and you brought me in to consult, I would advise not saying anything.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Adding E to &quot;quit&quot; gives &quot;quiet&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12511" LastEditorUserId="17250" LastEditDate="2016-01-15T16:17:04.677" LastActivityDate="2016-01-15T16:17:04.677" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="25412" PostTypeId="2" ParentId="25334" CreationDate="2016-01-15T12:18:56.410" Score="3" Body="&lt;p&gt;I believe the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Energy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A natural state, I'm sought by all&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Whether it be just to live or to power our vehicles, lights etc.It all qualifies as energy. Wars have been waged to obtain valuable resources that ultimately gets used for energy. I am referring to oil and fossil fuels here.So in a natural state it could either be the energy we as humans have or the natural energy that resides within our earth/universe&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Go without me, and you shall fall.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This speaks for itself, but without any energy you would surely fall.See the end of many marathons or exertion competitions. Many people collapse&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You do me when you spend,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This can be referred to the act of actually buying things or it could be that you expend or spend energy to do pretty much anything.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and use me when you eat to no end.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When we eat our body uses energy to process all the food and generally when we eat to no end, we will either become fat, which means your body uses more energy to sustain it anyway, or you become sick, which largely drains our energy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Upon using google I found that the answer given by @AeJey is the widely chosen one. Though I feel this one fits the riddle better. Even if my explanation is not correctly executed, I feel the answer fits better.&lt;/p&gt;&#xA;" OwnerUserId="18487" LastActivityDate="2016-01-15T12:18:56.410" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25413" PostTypeId="1" AcceptedAnswerId="25697" CreationDate="2016-01-15T14:05:00.763" Score="16" ViewCount="420" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;In the Viking lands, we three gods reign supreme&lt;/p&gt;&#xA;&lt;p&gt;In the three kingdoms, where not all is as it seems.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In the frigid North, &lt;strong&gt;En&lt;/strong&gt; leads the three&lt;/p&gt;&#xA;&lt;p&gt;Just like in the South, as it should be.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In the temperate South, &lt;strong&gt;To&lt;/strong&gt; makes up the center&lt;/p&gt;&#xA;&lt;p&gt;The North agrees, but the Middle is the dissenter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In the Middle, &lt;strong&gt;Tre&lt;/strong&gt; comes in the first&lt;/p&gt;&#xA;&lt;p&gt;South and North agree, that is certainly the worst.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;We gods are amorphous, as it is well known,&lt;/p&gt;&#xA;&lt;p&gt;Our forms may change in the domains of each zone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In the extreme lands, &lt;strong&gt;En&lt;/strong&gt; appears like Janus,&lt;/p&gt;&#xA;&lt;p&gt;Near the Middle, however, they find that heinous.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In the Middle, &lt;strong&gt;To&lt;/strong&gt; appears surprised,&lt;/p&gt;&#xA;&lt;p&gt;Everywhere else, however, He appears incised.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Only &lt;strong&gt;Tre&lt;/strong&gt;, however, is truly immutable,&lt;/p&gt;&#xA;&lt;p&gt;Everywhere, His nature is truly indisputable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;What are we? What are the Three Kingdoms?&lt;/p&gt;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;The puzzle hinges on the major local languages of the Scandinavian region, as well as some basic geography to determine their respective regions. While it does require some research, the correct solution will (hopefully) be obvious once it is reached.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;The pseudonyms of the Gods don't actually have much bearing on the puzzle, and can be changed without affecting the answers. However, they do act as a (very minor) hint when taken in conjunction with Paragraph 2 of the riddle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7529" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-01-23T08:54:40.200" Title="The Viking Saga" Tags="|riddle|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="25414" PostTypeId="1" AcceptedAnswerId="25418" CreationDate="2016-01-15T15:38:02.373" Score="5" ViewCount="1061" Body="&lt;p&gt;&lt;strong&gt;The Challenge&lt;/strong&gt;&lt;br&gt;&#xA;Using only lands and artifacts, your goal is to build a Modern-legal Magic: the Gathering deck that is capable of winning the game as quickly as possible, against an opponent who does nothing. You may assume that your deck is stacked, so that you will always draw exactly the cards you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your opponent's deck consists of sixty islands, they begin the game by mulliganing to zero, and will take no actions unless required to. Your solution may not involve random chance, or the opponent choosing to do something suicidal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Scoring&lt;/strong&gt;&lt;br&gt;&#xA;The best solution is the one which can defeat the opponent in as few turns as possible. You may choose to play or draw, but winning on the play is faster than winning on the draw.&lt;br&gt;&#xA;If two solutions are equally fast, the better solution is the one that deals more damage (or life-loss).&lt;br&gt;&#xA;If two solutions are equally fast and equally damaging, the better solution is the one that uses less cards.&lt;br&gt;&#xA;If your solution involves taking extra turns, it will be scored based on how many turns your opponent got. If the opponent never gets a turn, that's a turn 1 win. If the opponent takes one turn before you win, that's turn 2, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why Modern-legal artifacts?&lt;/strong&gt;&lt;br&gt;&#xA;I've tested it, and I believe the restriction achieves an interesting level of complexity, with neither a dead simple four card solution, nor an optimal fifty-card monster solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plus, it lends itself to a great puzzle title.&lt;/p&gt;&#xA;" OwnerUserId="7459" LastEditorUserId="7459" LastEditDate="2016-01-15T18:18:46.430" LastActivityDate="2016-01-16T01:45:37.523" Title="Magic: the Gathering - Modern Engineering" Tags="|calculation-puzzle|magic-the-gathering|" AnswerCount="5" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25415" PostTypeId="2" ParentId="25414" CreationDate="2016-01-15T16:05:41.650" Score="0" Body="&lt;p&gt;T2 Combat:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T1: Mox Opal, Inkmoth Nexus, Ornithopter, SpringLeaf Drum, Cranial Plating&lt;br&gt;&#xA;T2: Darksteel Citadel, Animate Inkmoth, equip Cranial, Bone Saw, equip Bone Saw, Ornithopter - 10 Infect damage&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T2 Draw Combat:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T1: Mox Opal, Inkmoth Nexus, Ornithopter, SpringLeaf Drum, Cranial Plating&lt;br&gt;&#xA;T2: Darksteel Citadel, Animate Inkmoth, equip Cranial, Ornithopter, Ornithopter, Ornithopter - 10 Infect damage&lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2016-01-15T16:41:38.650" LastActivityDate="2016-01-15T16:41:38.650" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25416" PostTypeId="2" ParentId="25414" CreationDate="2016-01-15T16:12:00.673" Score="0" Body="&lt;p&gt;&lt;strong&gt;Alternative Turn 2 Play Combat - 11 Infect Damage&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Turn 1 on play:&#xA;Inkmoth Nexus -&gt; Paradise Mantle -&gt; Mishra's Bauble -&gt; Mox Opal -&gt; Arcbound Ravager -&gt; Sac Mox Opal to Arcbound for 1 counter -&gt; Mox Opal - &gt; Equip Paradise Mantle to Arcbound Ravager -&gt; Sacrifice Mishra’s Bauble&lt;/li&gt;&#xA;&lt;li&gt;Turn 2 &#xA;Darksteel Citadel -&gt; Activate Inkmoth - &gt; Tap Arcbound, Mox Opal for Mana -&gt; Chromatic Star Chromatic Star -&gt; Sacrifice both chromatic stars and Mox Opal -&gt; 3 counters 2 cards -&gt; Mox Opal -&gt; Chromatic Star -&gt; Sacrifice Paradise Mantle, Mox Opal, Chromatic Star, Darksteel Citadel -&gt; 4 counters, 1 card -&gt; Darksteel Relic -&gt; Sacrifice Darksteel relic -&gt; 1 counter -&gt; Sacrifice Arcbound Ravager -&gt; 10 counters on Inkmoth -&gt; Swing for 11 Infect Damage&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Turn 2 Draw Combat&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Turn 1 on draw:&#xA;Inkmoth Nexus -&gt; Springleaf Drum -&gt; Memnite -&gt; Mox Opal -&gt; Cranial Plating -&gt; Mox Opal -&gt; Mox Opal -&gt; Cranial Plating&lt;/li&gt;&#xA;&lt;li&gt;Turn 2 &#xA;Darksteel Citadel -&gt; Activate Inkmoth -&gt; Equip both Cranial Platings&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In play: 2 Cranial Plating, Inkmoth Nexus, Mox Opal, Darksteel Citadel, Springleaf Drum, Memnite&#xA;In graveyard: 2 Mox Opal&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7 artifacts in play with Inkmoth active, Inkmoth Nexus swings for 15 Infect damage.&lt;/p&gt;&#xA;" OwnerUserId="10904" LastEditorUserId="10904" LastEditDate="2016-01-15T17:27:23.450" LastActivityDate="2016-01-15T17:27:23.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25418" PostTypeId="2" ParentId="25414" CreationDate="2016-01-15T16:31:21.460" Score="9" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;Edit: Here's a more interesting infinite combo, still with 11 cards:&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=397853&quot; rel=&quot;nofollow&quot;&gt;Darksteel Citadel&lt;/a&gt;, tap for (1)&lt;br&gt;&#xA;Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=383334&quot; rel=&quot;nofollow&quot;&gt;Ornithopter&lt;/a&gt; for 0&lt;br&gt;&#xA;Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=208248&quot; rel=&quot;nofollow&quot;&gt;Mox Opal&lt;/a&gt;, tap for U&lt;br&gt;&#xA;Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=208248&quot; rel=&quot;nofollow&quot;&gt;Mox Opal&lt;/a&gt; sacrificing the other one, tap for U&lt;br&gt;&#xA;Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=179428&quot; rel=&quot;nofollow&quot;&gt;Skill Borrower&lt;/a&gt; for 1UU, reveal &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=51633&quot; rel=&quot;nofollow&quot;&gt;Krark-Clan Ironworks&lt;/a&gt;&lt;br&gt;&#xA;Sacrifice Darksteel Citadel, Ornithopter, and Mox Opal to Skill Borrower for (6)&lt;br&gt;&#xA;Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=207883&quot; rel=&quot;nofollow&quot;&gt;Mimic Vat&lt;/a&gt; for 3&lt;br&gt;&#xA;Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=389551&quot; rel=&quot;nofollow&quot;&gt;Ichor Wellspring&lt;/a&gt; for 2, but hold priority...&lt;br&gt;&#xA;...Sacrifice the Ichor Wellspring to Skill Borrower for (2) and hold priority...&lt;br&gt;&#xA;...Sacrifice Skill Borrower to itself for (2), imprinting onto Mimic Vat&lt;br&gt;&#xA;...Tap Mimic Vat for 3, creating a token copy of Skill Borrower with haste&lt;br&gt;&#xA;...Sacrifice Mimic Vat to the Skill Borrower token for (2)&lt;br&gt;&#xA;...release priority, drawing Krark-Clan Ironworks and &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=48141&quot; rel=&quot;nofollow&quot;&gt;Voltaic Construct&lt;/a&gt;, revealing &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=405208&quot; rel=&quot;nofollow&quot;&gt;Dreamstone Hedron&lt;/a&gt;&lt;br&gt;&#xA;Tap the Skill Borrower token for (3)&lt;br&gt;&#xA;Play Voltaic Construct for 4&lt;br&gt;&#xA;Pay 2 to untap the Skill Borrower token with Voltaic Construct, then tap it again for (3)&lt;br&gt;&#xA;Repeat until you have infinite mana&lt;br&gt;&#xA;Tap and sacrifice the Skill Borrower token for 3, drawing Dreamstone Hedron, Goblin Cannon, and ???&lt;br&gt;&#xA;Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=74271&quot; rel=&quot;nofollow&quot;&gt;Goblin Cannon&lt;/a&gt; for 4&lt;br&gt;&#xA;Shoot your opponent for infinite with Goblin Cannon  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Turn 1 First Main Phase on the Play&lt;/strong&gt; - &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=74271&quot; rel=&quot;nofollow&quot;&gt;Infinite Damage&lt;/a&gt;, 11 cards  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=397853&quot; rel=&quot;nofollow&quot;&gt;Darksteel Citadel&lt;/a&gt;, tap for (1)&lt;br&gt;&#xA;Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=383334&quot; rel=&quot;nofollow&quot;&gt;Ornithopter&lt;/a&gt; for 0&lt;br&gt;&#xA;Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=208248&quot; rel=&quot;nofollow&quot;&gt;Mox Opal&lt;/a&gt;, tap for U&lt;br&gt;&#xA;Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=208248&quot; rel=&quot;nofollow&quot;&gt;Mox Opal&lt;/a&gt; sacrificing the other one, tap for U&lt;br&gt;&#xA;Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=208248&quot; rel=&quot;nofollow&quot;&gt;Mox Opal&lt;/a&gt; sacrificing the other one, tap for U&lt;br&gt;&#xA;Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=51633&quot; rel=&quot;nofollow&quot;&gt;Krark-Clan Ironworks&lt;/a&gt; for 1UUU&lt;br&gt;&#xA;Sacrifice Darksteel Citadel, Ornithopter, and Mox Opal to Krark-Clan Ironworks for (6)&lt;br&gt;&#xA;Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=389551&quot; rel=&quot;nofollow&quot;&gt;Ichor Wellspring&lt;/a&gt; for 2, drawing a card&lt;br&gt;&#xA;Sacrifice Ichor Wellspring to Krark-Clan Ironworks, drawing a card, gaining (2)&lt;br&gt;&#xA;Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=389609&quot; rel=&quot;nofollow&quot;&gt;Myr Retriever&lt;/a&gt; for 2&lt;br&gt;&#xA;Sacrifice Myr Retriever to Krark-Clan Ironworks, returning Ichor Wellspring, gaining (2)&lt;br&gt;&#xA;Play Ichor Wellspring for 2, drawing a card&lt;br&gt;&#xA;Sacrifice Ichor Wellspring to Krark-Clan Ironworks, drawing a card, gaining (2) (6 mana, 3 cards)&lt;br&gt;&#xA;Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=113513&quot; rel=&quot;nofollow&quot;&gt;Locket of Yesterdays&lt;/a&gt; for 1&lt;br&gt;&#xA;Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=74271&quot; rel=&quot;nofollow&quot;&gt;Goblin Cannon&lt;/a&gt; for 4&lt;br&gt;&#xA;Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=389609&quot; rel=&quot;nofollow&quot;&gt;Myr Retriever&lt;/a&gt; for 1  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sacrifice Myr Retriever to Krark-Clan Ironworks returning Myr Retriever, play it, repeat until you have infinite mana, then blow Goblin Cannon for infinite damage&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Older Solutions:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Turn 2 Combat on the Play&lt;/strong&gt; - 11 infect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Turn 1:&lt;/em&gt;&lt;br&gt;&#xA;Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=213731&quot; rel=&quot;nofollow&quot;&gt;Inkmoth Nexus&lt;/a&gt;, tap for (1)&lt;br&gt;&#xA;Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=383334&quot; rel=&quot;nofollow&quot;&gt;Ornithopter&lt;/a&gt; for 0&lt;br&gt;&#xA;Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=378534&quot; rel=&quot;nofollow&quot;&gt;Springleaf Drum&lt;/a&gt; for 1, tap it and Ornithopter for U&lt;br&gt;&#xA;Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=208248&quot; rel=&quot;nofollow&quot;&gt;Mox Opal&lt;/a&gt;, tap for U&lt;br&gt;&#xA;Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=370510&quot; rel=&quot;nofollow&quot;&gt;Arcbound Ravager&lt;/a&gt; for UU (1 counter)&lt;br&gt;&#xA;Sac Mox Opal to Ravager (2 counters)&lt;br&gt;&#xA;Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=208248&quot; rel=&quot;nofollow&quot;&gt;Mox Opal&lt;/a&gt;, tap for U&lt;br&gt;&#xA;Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=178135&quot; rel=&quot;nofollow&quot;&gt;Adventuring Gear&lt;/a&gt; for U  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Turn 2:&lt;/em&gt;&lt;br&gt;&#xA;Tap Springleaf Drum and Ornithopter for U&lt;br&gt;&#xA;Activate Inkmoth Nexus for U&lt;br&gt;&#xA;Tap Mox Opal for U&lt;br&gt;&#xA;Equip Adventuring Gear to Nexus for U&lt;br&gt;&#xA;Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=405104&quot; rel=&quot;nofollow&quot;&gt;Polluted Delta&lt;/a&gt;, fetch for Island (+4/+4 to Nexus)&lt;br&gt;&#xA;Sac Mox Opal, Springleaf Drum, Ornithopter, and Adventuring Gear to Ravager (6 counters)&lt;br&gt;&#xA;Sac Ravager to itself, transfer all 6 counters to Inkmoth Nexus  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attack with &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=213731&quot; rel=&quot;nofollow&quot;&gt;Inkmoth Nexus&lt;/a&gt; for 11 Infect  &lt;/p&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="17389" LastEditDate="2016-01-16T01:45:37.523" LastActivityDate="2016-01-16T01:45:37.523" CommentCount="28" ContentLicense="CC BY-SA 3.0" />
  <row Id="25419" PostTypeId="2" ParentId="25248" CreationDate="2016-01-15T18:02:06.357" Score="3" Body="&lt;p&gt;My answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Boombox&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My last three letters serve to clarify.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; e.g, drawing a box around an answer or perhaps ticking a box to clarify a position.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My first four letters are where famous people speak.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A boom is a large mike on the end of a stick usually used for actors or people being interviewed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Dropping all consonants, I am almost a curse.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'ooo' is almost 'poo' (if you add an extra line)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18422" LastActivityDate="2016-01-15T18:02:06.357" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25420" PostTypeId="2" ParentId="25414" CreationDate="2016-01-15T18:04:48.947" Score="0" Body="&lt;p&gt;&lt;strong&gt;Before the second turn*, infinite damage.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;numbers between brackets indicate i tapped things and it made mana.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;darksteel citadel (1), memnite, mox opal x4 (5) &lt;a href=&quot;http://magiccards.info/mi/en/262.html&quot; rel=&quot;nofollow&quot;&gt;Timesifter&lt;/a&gt;,&lt;br&gt;opponent takes his turn, reveals island, you reveal something cmc 1 (over and over ofcourse) so you have 30 turns to win in any way you like before the start of your normal second turn. For instance by doing the combo Zerris suggested, Etherium sculptor + 2 myr retrievers + blasting station + krak-clan Ironworks, but spread out over turns 1.1 to 1.10   &lt;br&gt;*&lt;em&gt;depending on how extra turns are viewed&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16846" LastEditorUserId="16846" LastEditDate="2016-01-15T18:52:11.083" LastActivityDate="2016-01-15T18:52:11.083" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="25421" PostTypeId="2" ParentId="25414" CreationDate="2016-01-15T18:13:49.677" Score="5" Body="&lt;p&gt;&lt;strong&gt;First solution: Turn 1 infinite damage, 15 cards&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Play Darksteel Citadel, tap for 1&lt;br&gt;&#xA;Cast Ornithopther&lt;br&gt;&#xA;Cast Mox Opal, tap for U&lt;br&gt;&#xA;Cast Mox Opal, tap for U&lt;br&gt;&#xA;Cast Mox Opal, tap for U&lt;br&gt;&#xA;Cast Krark-clan Ironworks.&#xA;Sacrifice Citadel to Ironworks&lt;br&gt;&#xA;Cast Ichor Wellspring and sac it. &lt;em&gt;(2 cards in hand, 2 mana in pool)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cast Locket of Yesterdays&lt;br&gt;&#xA;Cast Ichor Wellspring and sac it&lt;br&gt;&#xA;Cast Ichor Wellspring and sac it&lt;br&gt;&#xA;Cast Ichor Wellspring and sac it. &lt;em&gt;(4 cards in hand, 4 mana in pool)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cast Myr Retriever and sacrifice it.&lt;br&gt;&#xA;Cast Myr Retriever and sacrifice it, returning other Myr Retriever. repeat infinite number of times, generates infinite mana.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cast Lightning greaves&lt;br&gt;&#xA;Cast Sigil of Distinction with all mana&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equip to Ornithopther and attack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second solution: Turn 1 (deck size - 8)*2 damage, 7 cards&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Play Darksteel Citadel, tap for 1&lt;br&gt;&#xA;Cast Ornithopther&lt;br&gt;&#xA;Cast Mox Opal, tap for U&lt;br&gt;&#xA;Cast Mox Opal, tap for U&lt;br&gt;&#xA;Cast Mox Opal, tap for U&lt;br&gt;&#xA;Cast Krark-clan Ironworks&lt;br&gt;&#xA;Sacrifice Citadel, Ornithopter, Mox and Ironworks itself to Ironworks for 8 mana.&lt;br&gt;&#xA;Cast Goblin Charbelcher and activate it!&lt;br&gt;&#xA;Let the rest of your deck consist of only nonland cards, except for the last one. That one is a Mountain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Not infinite because the rules require your deck to be of a size that you still can shuffle, too bad. Otherwise infinite&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1748" LastEditorUserId="1748" LastEditDate="2016-01-16T00:03:16.690" LastActivityDate="2016-01-16T00:03:16.690" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25422" PostTypeId="1" AcceptedAnswerId="25430" CreationDate="2016-01-15T19:08:52.187" Score="3" ViewCount="727" Body="&lt;p&gt;The word I seek is split into letters,&#xA;each letter enclosed in a riddle below.&lt;br&gt;&#xA;Each riddle will get harder as the number increases.&lt;br&gt;&#xA;You know the length of my word, but what's the word?&#xA;One letter wrong is likely to change the whole word!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Letter #1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;At the cross roads,&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;there's a twist down one path&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;and dead ends the rest.&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;there's one of me on every line&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Letter #2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You will be female without me,&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;But if I come I will let you walk out of here alive&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;I split to make a number of relevance to me&lt;/em&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Letter #3:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;With me you are will gain wisdom&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;But if you think you know it all your eyes will never see knowledge&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;Don't ditch me - I am as good as two of you&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Letter #4:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The English language just wouldn't be the same without me&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;In the name of Englishmen, I present them all&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;And lag behind to help them with their language&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Letter #5:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am cousin of the square, though he be twice as large as me&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;Though I got half the score of my friend 'I' in my degree°&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;Lover of me or loather of me, this oath is over&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What word am I and what does each riddle mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seeing as no-one of yet has letter #2, I'll put some emphasis on it:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;OLD&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You will be female without me,&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;But if I come I will let you walk out of here alive&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;I split to make a number of relevance to me&lt;/em&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;NEW&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You will be female without me,&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;But if I come I will let you &lt;strong&gt;walk&lt;/strong&gt; out of here alive&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;I &lt;strong&gt;double&lt;/strong&gt; to make a number of relevance to me&lt;/em&gt;&lt;br&gt;&#xA;&lt;strong&gt;&lt;em&gt;In a chart of chemistry&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="18250" LastEditorUserId="11876" LastEditDate="2016-03-25T08:17:10.737" LastActivityDate="2016-03-25T08:17:10.737" Title="Riddle Reorganized" Tags="|riddle|word|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25424" PostTypeId="1" AcceptedAnswerId="25431" CreationDate="2016-01-15T20:45:26.097" Score="2" ViewCount="564" Body="&lt;p&gt;You have 11 visually identical coins. One of them is marked genuine (and you know which one this is). Out of the remaining 10 coins, all but one of them are also genuine, but you don't know which ones. The remaining odd coin has a different weight, either heavier or lighter than the others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a regular two-pan balance. This works like you expect: you load a certain number of coins to each side, and the balance will tell you the heavier pan (or whether they have equal weight).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many weighings are necessary in the worst case to find the odd coin and determine whether it weighs more or less than genuine coins?&lt;/p&gt;&#xA;" OwnerUserId="17970" LastEditorUserId="17389" LastEditDate="2016-01-15T22:02:35.123" LastActivityDate="2016-01-15T22:09:24.043" Title="Find the odd coin" Tags="|logical-deduction|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25425" PostTypeId="2" ParentId="25422" CreationDate="2016-01-15T21:06:14.567" Score="3" Body="&lt;p&gt;I only (think I) know two letters, but that's enough for a partial answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Letter #1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; t - the bend is at the bottom where the t curves in this font, the dead ends are the top and left/right.  Each line of this section of the clue (for that matter, all the italicized sections) has a &quot;t&quot;.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Letter #3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; w - with me you get &quot;w&quot;isdom because it begins with a w, and also w = double you = &quot;as good as two of you&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12092" LastActivityDate="2016-01-15T21:06:14.567" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25426" PostTypeId="2" ParentId="24911" CreationDate="2016-01-15T21:06:17.030" Score="4" Body="&lt;p&gt;Given all OP's hints, this is almost directly spelled out for us already, but the answer is (intended to be)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; NO ESCAPE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is found by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reading the % as a modulo operator of sorts, and dividing up the string of digits to the right of it into chunks of the length of the digit to the left of it (i.e., nine-character chunks, in rows), then treating the numbers in the columns as the cells in an LCD number display, as described in OP's hints above&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;1 1 0 1 1 1 1 1 1&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;2 2 0 2 2 2 2 2 2&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;3 3 0 4 4 5 3 3 0&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;5 5 0 5 6 7 4 4 3&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;6 6 0 7 7 0 5 5 4&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;0 7 0 0 0 0 6 0 5&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;0 0 0 0 0 0 0 0 0&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Treating the numbers as the cells as outlined in OP's hints, with 1 indicating the top H bar, 2 indicating the top left V bar, 3 indicating the top right V bar, 4 indicating the center H bar, 5 the bottom left V bar, 6 the bottom right V bar, and 7 the bottom H bar, and treating each &lt;em&gt;column&lt;/em&gt; as a character, you get LCD-styled letters spelling out NO ESCAPP, as follows:&lt;br&gt;&#xA; &lt;code&gt;.  _   _       _   _   _   _   _   _&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;. | | | |     |_  |_  |   |_| |_| |_|&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;. | | |_|     |_   _| |_  | | |   |&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; The last column should read (down) &lt;code&gt;1 2 4 5 7 0 0&lt;/code&gt; (E) instead of &lt;code&gt;1 2 0 3 4 5 0&lt;/code&gt; (P).&lt;br&gt;&#xA;&lt;br&gt;&#xA; Read on for more details.&lt;br&gt;&#xA; Seven-segment display:&lt;br&gt;&#xA; &lt;code&gt;.  _&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;. |_|&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;. |_|&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; We can label the segments as follows:&lt;br&gt;&#xA; &lt;code&gt;.  1&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;. 2 3&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;.  4&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;. 5 6&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;.  7&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Generally we can represent a combination as a binary number, indicating which segments are lit up. E.g. 1111011 would be all segments but #5 lit up, giving us a &quot;9&quot;:&lt;br&gt;&#xA; &lt;code&gt;.  _&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;. |_|&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;.  _|&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; OP used a slightly different notation, where instead of indicating for each segment whether it is lit up or not (binary), he just gives us the numbers of the lit up segments, and pads with 0s if not all the segments are lit up. To indicate a &quot;9&quot; (1111011), it would be 1234670. A &quot;0&quot; (1110110) would be 1235600.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Reading the numbers column by column, we get:&lt;br&gt;&#xA; 1235600, 1235670, 0000000, 1245700, 1246700, 1257000, 1234560, 1234500, 1203450. The last number deviates from how the others are displayed (0 in the middle, not padded at the end), so we presume it's just equivalent to 1234500.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Placing the seven-segment display of these combinations next to each other, we get &quot;NO ESCAPP&quot;:&lt;br&gt;&#xA; &lt;code&gt;.  _   _       _   _   _   _   _   _&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;. | | | |     |_  |_  |   |_| |_| |_|&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;. | | |_|     |_   _| |_  | | |   |&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="18429" LastEditDate="2016-01-19T10:00:11.203" LastActivityDate="2016-01-19T10:00:11.203" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25428" PostTypeId="2" ParentId="25422" CreationDate="2016-01-15T21:17:18.173" Score="3" Body="&lt;p&gt;A partial answer here to add to the previous: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Letter #4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; E &#xA; The English language starts with E, so definitely would be different without it.  Also, when at the end of the word (when it &quot;lags behind&quot;, the letter e changes how you pronounce other vowels in the sentence (e.g., hat vs. hate)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Letter #5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; L&#xA; Capital L is roughly half a square. A straight line (l) could be said to have a 180 degree angle while a capital L has a 90 degree angle. Add L to over to get Lover, add L to oath to get Loath (similar to loather). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18493" LastActivityDate="2016-01-15T21:17:18.173" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25429" PostTypeId="2" ParentId="25422" CreationDate="2016-01-15T21:21:36.157" Score="1" Body="&lt;p&gt;Building on the other answers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Towel&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18103" LastActivityDate="2016-01-15T21:21:36.157" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25430" PostTypeId="2" ParentId="25422" CreationDate="2016-01-15T21:26:07.647" Score="7" Body="&lt;p&gt;A guess that the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Towel&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Letter #1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the cross roads,&lt;br&gt;&#xA;there's a twist down one path&lt;br&gt;&#xA;and dead ends the rest.&lt;br&gt;&#xA;there's one of me on every line  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;T&lt;/strong&gt; - A lower case t has three &quot;dead ends&quot; and one curved path.  There are t's in each line of this micro-riddle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Letter #2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will be female without me,&lt;br&gt;&#xA;But if I come I will let you &lt;strong&gt;walk&lt;/strong&gt; out of here alive&lt;br&gt;&#xA;I &lt;strong&gt;double&lt;/strong&gt; to make a number of relevance to me&lt;br&gt;&#xA;&lt;strong&gt;In a chart of chemistry&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;O&lt;/strong&gt; I honestly thought this was playing on the Y chromosome, as far as I know TYWEL is not a word.  &lt;strong&gt;EDIT&lt;/strong&gt;: Here's a try anyway... Bisecting an O with a line results in this Ø, which is the symbol for diameter.  An o is a circle, so a diameter would be important to the letter o. &lt;em&gt;This isn't it, but it was interesting and close enough for me to want to leave it in.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Edits I think are correct:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;NEW EDIT&lt;/strong&gt;: I can only figure out part of the clue for #2, but O is the symbol for the element Oxygen, and its atomic number is 8, which can be made from two O's. &lt;strong&gt;AH- The Final Piece&lt;/strong&gt;: SHE/SHOE, she being female (when O is removed) and a shoe letting you walk away.  Major credit goes to Daphne B who cracked it while I was weekending away from the computer, but left the breadcrumbs for me to follow.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Letter #3:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With me you are will gain wisdom&lt;br&gt;&#xA;But if you think you know it all your eyes will never see knowledge&lt;br&gt;&#xA;Don't ditch me - I am as good as two of you  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;W&lt;/strong&gt; - With a w you can spell wisdom, you can't spell knowledge without one, and it is literally two u's.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Letter #4:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The English language just wouldn't be the same without me&lt;br&gt;&#xA;In the name of Englishmen, I present them all&lt;br&gt;&#xA;And lag behind to help them with their language  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;E&lt;/strong&gt; - English starts with an e, and language ends with one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Letter #5:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am cousin of the square, though he be twice as large as me&lt;br&gt;&#xA;Though I got half the score of my friend 'I' in my degree°&lt;br&gt;&#xA;Lover of me or loather of me, this oath is over  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;L&lt;/strong&gt; - Two l's could be made into a square if one is rotated.  I is a line of 180°, while an L has a 90° angle.  L's are also in lover and loather.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11452" LastEditorUserId="11452" LastEditDate="2016-01-19T17:36:37.653" LastActivityDate="2016-01-19T17:36:37.653" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="25431" PostTypeId="2" ParentId="25424" CreationDate="2016-01-15T21:37:50.263" Score="4" Body="&lt;p&gt;&lt;strong&gt;We can determine in three weighings every time.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Method:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Label the coins 1 - 11, with 1 being genuine&lt;br&gt;&#xA;&lt;br&gt;&#xA; Weighing 1: 1, 2, 3, 4 vs. 5, 6, 7, 8&lt;br&gt;&#xA;&lt;br&gt;&#xA; If the weights are the same, then&lt;br&gt;&#xA;&lt;br&gt;&#xA; Weighing 2a: 9 vs. 10&lt;br&gt;&#xA;&lt;br&gt;&#xA; If they're the same again, then&lt;br&gt;&#xA;&lt;br&gt;&#xA; Weighing 3a: 1 vs. 11 - we know 11 is fake, but this tells us if it's heavier or lighter&lt;br&gt;&#xA;&lt;br&gt;&#xA; If the left side was heaver in Weighing 2a, then&lt;br&gt;&#xA;&lt;br&gt;&#xA; Weighing 3b: 1 vs. 9 - if they're the same, then 10 is fake and light, otherwise 9 is fake and heavy&lt;br&gt;&#xA;&lt;br&gt;&#xA; If the right side was heaver in Weighing 2a, then&lt;br&gt;&#xA;&lt;br&gt;&#xA; Weighing 3c: 1 vs. 9 - if they're the same, then 10 is fake and heavy, otherwise 9 is fake and light&lt;br&gt;&#xA;&lt;br&gt;&#xA; If the left side was heavier in Weighing 1:&lt;br&gt;&#xA;&lt;br&gt;&#xA; Weighing 2b: 5, 6 vs. 7, 8&lt;br&gt;&#xA;&lt;br&gt;&#xA; If they're the same, then&lt;br&gt;&#xA;&lt;br&gt;&#xA; Weighing 3d: 2 vs. 3 - whichever is heavier is also fake, if they're the same then 4 is heavy and fake&lt;br&gt;&#xA;&lt;br&gt;&#xA; If the left side was heavier in Weighing 2b, then&lt;br&gt;&#xA;&lt;br&gt;&#xA; Weighing 3e: 7 vs 8 - whichever is lighter is also fake&lt;br&gt;&#xA;&lt;br&gt;&#xA; If the right side was heavier in Weighing 2b, then&lt;br&gt;&#xA;&lt;br&gt;&#xA; Weighing 3f: 5 vs 6 - whichever is lighter is also fake&lt;br&gt;&#xA;&lt;br&gt;&#xA; If the right side was heavier in Weighing 1:&lt;br&gt;&#xA;&lt;br&gt;&#xA; Weighing 2c: 5, 6 vs. 7, 8&lt;br&gt;&#xA;&lt;br&gt;&#xA; If they're the same, then&lt;br&gt;&#xA;&lt;br&gt;&#xA; Weighing 3g: 2 vs. 3 - whichever is lighter is also fake, if they're the same then 4 is light and fake&lt;br&gt;&#xA;&lt;br&gt;&#xA; If the left side was heavier in Weighing 2c, then&lt;br&gt;&#xA;&lt;br&gt;&#xA; Weighing 3h: 5 vs 6 - whichever is heavier is also fake&lt;br&gt;&#xA;&lt;br&gt;&#xA; If the right side was heavier in Weighing 2c, then&lt;br&gt;&#xA;&lt;br&gt;&#xA; Weighing 3i: 7 vs 8 - whichever is heavier is also fake  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="17389" LastEditDate="2016-01-15T22:09:24.043" LastActivityDate="2016-01-15T22:09:24.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25432" PostTypeId="1" AcceptedAnswerId="25433" CreationDate="2016-01-15T21:54:33.530" Score="-4" ViewCount="259" Body="&lt;p&gt;We have a billiards table with height &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; and width &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt;, as illustrated in the image below. Notice that it has no pockets.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/QRZLs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QRZLs.png&quot; alt=&quot;Billiards Table&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;A ball with radius &lt;span class=&quot;math-container&quot;&gt;$r\ll w$&lt;/span&gt; (can be regarded as a point) is propelled from the bottom-left corner at an angle &lt;span class=&quot;math-container&quot;&gt;$5°\leq\alpha\leq 85°$&lt;/span&gt; with respect the horizontal axis (as depicted in the figure). The movement is constrained to a plane, so there is no depth involved. The velocity of the ball is irrelevant.&lt;/p&gt;&#xA;&lt;p&gt;There are no dissipative effects, so the ball will keep bouncing forever with no loss of energy. In other words, the collisions (ball-walls) are &lt;em&gt;perfectly elastic&lt;/em&gt; (the angles are conserved before and after a collision).&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Find the angle &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; that &lt;em&gt;maximizes&lt;/em&gt; the &lt;em&gt;number of bounces&lt;/em&gt; until the ball returns to the &lt;em&gt;first collision point&lt;/em&gt;. How many bounces &lt;span class=&quot;math-container&quot;&gt;$n(h,w,\alpha)$&lt;/span&gt; will it take for this to happen? &lt;em&gt;If you think there is no angle &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; for which the ball returns to the first collision point, demonstrate it.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&lt;strong&gt;NOTE:&lt;/strong&gt; The &lt;em&gt;first collision point&lt;/em&gt; is the point &lt;span class=&quot;math-container&quot;&gt;$(x_0,y_0)$&lt;/span&gt; at which the ball first touches a wall.&lt;/p&gt;&#xA;" OwnerUserId="10564" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-01-16T16:28:36.140" Title="The Billiards Problem" Tags="|calculation-puzzle|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="25433" PostTypeId="2" ParentId="25432" CreationDate="2016-01-15T22:05:11.317" Score="3" Body="&lt;p&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; as alpha approaches 0, the number of bounces approaches infinity. There is no true solution to this because any alpha you give I can provide a smaller alpha which will have more bounces&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;2. &#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; any angle not equal to tan(h/w) will cause a rectangle of some size to be drawn, but I think I may be misunderstanding what you mean by &quot;draw a rectangle&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18103" LastActivityDate="2016-01-15T22:05:11.317" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="25434" PostTypeId="2" ParentId="25432" CreationDate="2016-01-15T22:10:09.220" Score="2" Body="&lt;p&gt;For part 2.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There can be no rectangle. First, we show no rectangle can contain the starting point. Consider this point. This lies in the corner of the table, so in any drawn rectangle must also be a corner. But, this means the ball must return to this point at a 90 degree angle to the direction it left from. Since the whole corner is only 90 degrees, it must either come from off the table, impossible, or the initial angle must be 0. However, an initial angle of 0 would bounce straight back and not form a rectangle.&lt;BR&gt;&lt;br&gt;&#xA;Now, we show that any rectangle is a closed loop. Consider any corner of the rectangle. This must be on a wall, and must be a 90 degree angle. Thus, both entry and exit angles must be 45 degrees. Since the walls are 45 degrees from each other, this means that once an angle leaves a wall at 45 degree, all subsequent angles will be 45 degree angles. However, this means that when the point returns to the starting corner, it will be at a 45 degree angle and must then repeat the rectangle. By simply reversing the direction, we see that this means all rectangles are closed loops.&lt;BR&gt;&lt;br&gt;&#xA; If every rectangle is a closed loop, and no rectangle can contain the starting point, then the starting point cannot create a rectangle. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1968" LastEditorUserId="1968" LastEditDate="2016-01-15T23:22:21.463" LastActivityDate="2016-01-15T23:22:21.463" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="25435" PostTypeId="2" ParentId="231" CreationDate="2016-01-15T22:52:02.570" Score="1" Body="&lt;p&gt;The above answer deals with a gradient puzzle. I don't have an answer analyzed by experts, but from personal experience. I have been working on a 1000 piece puzzle of the Biltmore estate. It is a night scene on Christmas. It has large area of black along the bottom and a large area of sky which is a gradient of blue. This involved using a hybrid technique for different parts for the most efficient solving time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The building and  other distinguishable features of the puzzle were quickly put together by randomly finding pieces that match and leaving them in approximate locations and building around them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the gradient part, like Max's answer, Method 1 was used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the solid black part  however, It is brute force technique. Pick one piece try to match it to an open space, if it fits great, if not, put it aside and pick another one to try.&lt;/p&gt;&#xA;" OwnerUserId="18495" LastActivityDate="2016-01-15T22:52:02.570" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25436" PostTypeId="2" ParentId="25422" CreationDate="2016-01-16T00:07:53.443" Score="0" Body="&lt;p&gt;Answer is: towel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reasoning for clue 2: 'Her' is female without 'o'. With it, becomes 'hero'. Not sure about the third line.&lt;/p&gt;&#xA;" OwnerUserId="18497" LastActivityDate="2016-01-16T00:07:53.443" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25439" PostTypeId="1" AcceptedAnswerId="26157" CreationDate="2016-01-16T09:08:54.107" Score="29" ViewCount="960" Body="&lt;p&gt;A province has 10 cities (arranged in a circular manner).  Every pair of cities is directly connected by a straight highway, and each has its own unique number:  Highway 1, Highway 2, Highway 3, ..., Highway 45.  Below is an example of one such numbering, but the governor Tonald Drump loves rearranging the numbers of the highways, so the numbers are periodically changing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/75cLv.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/75cLv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that you cannot exit these highways partway between cities, and instead must proceed all the way down the highway to the destination city.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An obscure religion requires its members to go on a pilgrimage to this province.  For reasons lost to time, each pilgrim must, after arriving at one of these cities&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;only travel via these highways,&lt;/li&gt;&#xA;&lt;li&gt;travel on at least nine of the highways during the pilgrimage, and&lt;/li&gt;&#xA;&lt;li&gt;only travel on a highway if it has a strictly larger number than the one they just walked down.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note that they need not visit every city, repeating cities is acceptable, and pilgrims can choose their starting point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tonald Drump, who has no love of this religion (and has even suggested building a wall to keep them out), has made it clear he'd like to rearrange the numbers of the highway so that the pilgrim's task is impossible.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Show that no matter how Tonald Drump numbers these highways, the pilgrims can always complete their journey.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2722" LastActivityDate="2016-02-14T18:01:04.627" Title="Ever increasing highway numbers" Tags="|mathematics|graph-theory|" AnswerCount="3" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="25444" PostTypeId="1" AcceptedAnswerId="25462" CreationDate="2016-01-16T23:41:11.160" Score="5" ViewCount="276" Body="&lt;p&gt;I'm prickly and one-eyed&lt;br&gt; And quite hard to spy&lt;br&gt;&#xA;With fodder for mysteries and puzzles galore&lt;br&gt;&#xA;When you see me you'll be pleased&lt;br&gt;&#xA;But feel rather teased&lt;br&gt;&#xA;And pointedly wonder why you'd missed me before&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who am I?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I made some minor changes to the riddle to make it work a bit better.&lt;/p&gt;&#xA;" OwnerDisplayName="user17947" LastEditorDisplayName="user17947" LastEditDate="2016-01-17T08:22:35.927" LastActivityDate="2016-01-18T02:19:33.347" Title="I'm prickly and one-eyed" Tags="|riddle|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25445" PostTypeId="1" AcceptedAnswerId="25452" CreationDate="2016-01-17T02:02:20.503" Score="10" ViewCount="798" Body="&lt;p&gt;Seven treasure chests, made of gold, silver, tin, steel, brass, pewter and lead, each contain coins of one of the seven metals. No two chests contain the same type of coin.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The pewter chest does not contain tin coins.&lt;/li&gt;&#xA;&lt;li&gt;The brass chest does not contain silver coins.&lt;/li&gt;&#xA;&lt;li&gt;The tin chest does not contain lead coins.&lt;/li&gt;&#xA;&lt;li&gt;The coins in the silver chest and the chest containing the steel coins are of the same metal.&lt;/li&gt;&#xA;&lt;li&gt;The coins in the gold chest and the chest containing the brass coins are of the same metal.&lt;/li&gt;&#xA;&lt;li&gt;The coins in the steel chest and the chest containing the gold coins are of the same metal.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Which chest contains which coins?&lt;/p&gt;&#xA;" OwnerUserId="18508" LastEditorUserId="69582" LastEditDate="2024-02-06T23:55:54.017" LastActivityDate="2024-02-06T23:55:54.017" Title="Seven treasure chests" Tags="|logical-deduction|logic-grid|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25446" PostTypeId="2" ParentId="25445" CreationDate="2016-01-17T02:40:19.123" Score="4" Body="&lt;p&gt;Pewter lead.&lt;br&gt;&#xA;Lead pewter.&lt;br&gt;&#xA;Gold silver.&lt;br&gt;&#xA;Silver brass.&lt;br&gt;&#xA;Steel tin.&lt;br&gt;&#xA;Tin gold.&lt;br&gt;&#xA;Brass steel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I made a table and removed all none possible matches. Then I started to guess. By luck I didn't reach any unsolvable states and got it at the first time guessing.&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2016-01-17T02:40:19.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25448" PostTypeId="2" ParentId="25445" CreationDate="2016-01-17T02:50:10.660" Score="8" Body="&lt;p&gt;In addition to Ivo's answer, there are three other configurations that satisfy the problem as currently written:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;--------------------&#xA;gold --&amp;gt; silver&#xA;steel --&amp;gt; tin&#xA;lead --&amp;gt; lead&#xA;brass --&amp;gt; steel&#xA;silver --&amp;gt; brass&#xA;tin --&amp;gt; gold&#xA;pewter --&amp;gt; pewter&#xA;--------------------&#xA;gold --&amp;gt; silver&#xA;steel --&amp;gt; lead&#xA;lead --&amp;gt; gold&#xA;brass --&amp;gt; steel&#xA;silver --&amp;gt; brass&#xA;tin --&amp;gt; tin&#xA;pewter --&amp;gt; pewter&#xA;--------------------&#xA;gold --&amp;gt; silver&#xA;steel --&amp;gt; pewter&#xA;lead --&amp;gt; lead&#xA;brass --&amp;gt; steel&#xA;silver --&amp;gt; brass&#xA;tin --&amp;gt; tin&#xA;pewter --&amp;gt; gold&#xA;--------------------&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I suspect there was one rule omitted: chests cannot contain coins made of their own metal.&lt;/p&gt;&#xA;" OwnerUserId="5456" LastActivityDate="2016-01-17T02:50:10.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25449" PostTypeId="1" AcceptedAnswerId="25450" CreationDate="2016-01-17T05:56:25.990" Score="2" ViewCount="167" Body="&lt;p&gt;This is crossposted from Math.SE, the original question is &lt;a href=&quot;https://math.stackexchange.com/questions/1615278/maths-puzzle-logic&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1+2+3+4 = 61$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2+3+4+5 = 52$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3+4+5+6 = 51$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$4+5+6+7 = 50$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$7+8+9+10 = ?$&lt;/p&gt;&#xA;" OwnerDisplayName="user10001" LastEditorUserId="-1" LastEditDate="2017-04-13T12:19:36.950" LastActivityDate="2016-01-17T08:53:05.967" Title="decipher this modified sum operation," Tags="|mathematics|number-sequence|formation-of-numbers|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25450" PostTypeId="2" ParentId="25449" CreationDate="2016-01-17T07:39:33.267" Score="4" Body="&lt;p&gt;\begin{align}7+8+9+10&amp;amp;\to\text{Seven}+\text{Eight}+\text{Nine}+\text{Ten}\\&amp;amp;\to\text{S}+\text{E}+\text{N}+\text{T}\\&amp;amp;\to\text{Alphabet}_{19}+\text{Alphabet}_{5}+\text{Alphabet}_{14}+\text{Alphabet}_{20}\\&amp;amp;\to19+5+14+20\\&amp;amp;=58\end{align}&lt;/p&gt;&#xA;" OwnerUserId="18510" LastEditorUserId="18510" LastEditDate="2016-01-17T08:53:05.967" LastActivityDate="2016-01-17T08:53:05.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25451" PostTypeId="2" ParentId="25444" CreationDate="2016-01-17T10:59:04.497" Score="10" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think the answer is:&lt;br&gt;&#xA; Hint.&lt;br&gt;&#xA; It is one-eyed, as in it has just one i in it. It has information about the puzzle, hence it is its fodder. It teases us for not knowing the answer yet. And yet we're pleased when we see it because of the possibility that we might solve the question with this new information. And when we do know the answer, we might wonder how did we miss it on the first place.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8980" LastEditorUserId="17250" LastEditDate="2016-01-17T16:25:06.773" LastActivityDate="2016-01-17T16:25:06.773" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25452" PostTypeId="2" ParentId="25445" CreationDate="2016-01-17T10:59:16.363" Score="9" Body="&lt;p&gt;In addition to the other answers, here is how the solutions can be derived using a graph representation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let A → B denote &quot;the chest of metal A contains coins of metal B&quot;. Then we are looking to put seven arcs in a graph with seven nodes (one of each type of metal) such that each node has both an incoming and an outgoing arc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the last three clues, we have the following path:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; silver → X → steel → Y → gold → Z → brass&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At first glance, it would seem that all seven metals are represented here, in a single cycle linking all chests.  But this is forbidden, since we know that the brass chest does not contain silver coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have to reuse metals, and the only way to resolve this is by setting:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; X = brass and Z = silver&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;winding up with the five-metal cycle&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; silver → brass → steel → Y → gold → silver&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Other attempts at instantiating variables will lead to coins of the same metal appearing in more than one chest (i.e. a node with more than one incoming arc in the graph representation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leaves the variable Y with three candidate metals: pewter, lead and tin.  Here's where the first and third clues come in.  If we set Y=tin, then there are two solutions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the pewter chest can contain either pewter or lead coins, and the lead chest takes the remaining metal type.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Setting either Y=pewter or Y=lead we wind up with a single solution in each case:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the remaining two chests contain only coins of their own type.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hence, we come to the four solutions given in the other answers.&lt;/p&gt;&#xA;" OwnerUserId="13484" LastEditorUserId="13484" LastEditDate="2016-01-17T11:04:55.133" LastActivityDate="2016-01-17T11:04:55.133" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25454" PostTypeId="1" AcceptedAnswerId="25455" CreationDate="2016-01-17T15:15:26.923" Score="1" ViewCount="163" Body="&lt;p&gt;This rhyme could save you!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Red touches yellow,&#xA;  &lt;br&gt; kills a fellow&lt;br&gt;Red touches black, &lt;br&gt;there's a venom  lack&#xA;  &lt;br&gt;Black tail and mouth of cotton,&#xA;  &lt;br&gt;Will kill if trodden&#xA;  &lt;br&gt;Brown tail and aquatic&#xA;  &lt;br&gt;Life fanatic&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is being described?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;They are animals&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;4 are being described&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18250" LastEditorUserId="7715" LastEditDate="2016-01-17T16:18:36.170" LastActivityDate="2016-01-17T18:24:57.233" Title="This rhyme could save your life!" Tags="|rhyme|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25455" PostTypeId="2" ParentId="25454" CreationDate="2016-01-17T16:02:38.570" Score="1" Body="&lt;p&gt;The rhyme is describing  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Snakes  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Red touches yellow, &#xA;kills a fellow  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Coral Snake (venomous)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Red touches black, &#xA;there's a venom lack  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; California Mountain King Snake  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Black tail and mouth of cotton, &#xA;Will kill if trodden  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cottonmouth Snake  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Brown tail and aquatic &#xA;Life fanatic &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Brown Water Snake&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7715" LastEditorUserId="7715" LastEditDate="2016-01-17T18:24:57.233" LastActivityDate="2016-01-17T18:24:57.233" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25456" PostTypeId="2" ParentId="10753" CreationDate="2016-01-17T16:36:01.840" Score="3" Body="&lt;p&gt;Yes this happens when the cube is wrongly assembled... I have had the same problem...Usually when you drop the cube or whatever and the center pieces come out, you think you have put them back right.. But turns out not necessarily always. You have to put them back correctly. Here's how. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above case as shown in pic, just leave the yellow and white faces alone... And move the center piece (literally pull it out and move) of each face one-left until you have moved all four. In the above case as shown in pic it will be as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orange center moved to where green center is&#xA;Green center moved to where Red center is&#xA;Red center moved to where Blue center is&#xA;Blue center moved to where Orange center is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps. At least this is what I did and the standard algorithm works afterwards&lt;/p&gt;&#xA;" OwnerUserId="18513" LastActivityDate="2016-01-17T16:36:01.840" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25457" PostTypeId="1" AcceptedAnswerId="25459" CreationDate="2016-01-17T18:43:51.010" Score="17" ViewCount="789" Body="&lt;p&gt;I have nine cards, each labeled with a real number (possibly negative or fractional) in a special way. When I randomly pick two of these cards without replacement and add their labels together, the probability of each sum is the same as when you roll two fair six-sided dice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means there is a $\frac16$ chance the two cards add to 7, a $\frac5{36}$ chance they add to 8, a $\frac19$ chance they add to 5, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How have I labeled my cards?&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2016-01-29T01:50:04.900" LastActivityDate="2016-01-29T01:50:04.900" Title="Two Dice = Nine Cards" Tags="|combinatorics|probability|dice|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25459" PostTypeId="2" ParentId="25457" CreationDate="2016-01-17T19:17:41.920" Score="12" Body="&lt;p&gt;The cards are numbered:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$0.5,1.5,2.5,2.5,3.5,4.5,4.5,5.5,6.5$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 36 possible ways to choose two cards out of nine, so the results must be $2,3,3,4,4,4,\ldots,11,11,12$. If we sum all of these, we find that the total is $252$, and that each card is represented eight times. This makes the sum of all nine cards $\frac{252}{8}=31.5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any card with value $x$, the other eight cards sum to $31.5-x$. However, these cards can be divided into four pairs, each of which sums to an integer. Therefore $31.5-x$ is an integer, and $x$ is a half-integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, consider the eight pairs that contain the card $x$. These eight pairs include $x$ eight times, and each other card once, so they sum to $31.5+7x$. The smallest eight results $(2,3,3,4,4,4,5,5)$ sum to $30$, so $31.5+7x\ge30$. The smallest half-integer satisfying this inequality is $0.5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The smallest card must be at most $0.5$, because a sum of $2$ would not be possible if it was $1.5$ or greater. Therefore the smallest card is exactly $0.5$. The rest of the cards follow easily from this.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2016-01-17T19:30:23.083" LastActivityDate="2016-01-17T19:30:23.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25460" PostTypeId="1" AcceptedAnswerId="25461" CreationDate="2016-01-17T20:52:37.187" Score="30" ViewCount="2401" Body="&lt;p&gt;A knight is placed on an infinitely large chess board with no edges. It can only visit each square once. What is the smallest number of moves it can make that would cause it to become trapped?&lt;/p&gt;&#xA;" OwnerUserId="18514" LastEditorUserId="12408" LastEditDate="2016-07-25T02:37:23.937" LastActivityDate="2016-07-25T02:37:23.937" Title="Trapping The Knight" Tags="|chess|knight-moves|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25461" PostTypeId="2" ParentId="25460" CreationDate="2016-01-17T21:33:15.517" Score="42" Body="&lt;p&gt;I can get the knight trapped in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;15 moves&lt;/strong&gt;:&#xA;$$\begin{array} {ccccccccc}&#xA;\cdot &amp;amp; \cdot &amp;amp; \cdot &amp;amp; \cdot &amp;amp; 5 &amp;amp; \cdot &amp;amp; \cdot &amp;amp; \cdot &amp;amp; \cdot\\&#xA;\cdot &amp;amp; 3 &amp;amp; \cdot &amp;amp; \cdot &amp;amp; \cdot &amp;amp; \cdot &amp;amp; \cdot &amp;amp; 7 &amp;amp; \cdot\\&#xA;\cdot &amp;amp; \cdot &amp;amp; \cdot &amp;amp; 4 &amp;amp; \cdot &amp;amp; 6 &amp;amp; \cdot &amp;amp; \cdot &amp;amp; \cdot\\&#xA;\cdot &amp;amp; \cdot &amp;amp; 2 &amp;amp;\cdot  &amp;amp; \cdot &amp;amp; \cdot &amp;amp; 8 &amp;amp; \cdot &amp;amp; \cdot\\&#xA;1 &amp;amp; \cdot &amp;amp; \cdot &amp;amp; \cdot &amp;amp; 15 &amp;amp; \cdot &amp;amp; \cdot &amp;amp; \cdot &amp;amp; 9\\&#xA;\cdot &amp;amp; \cdot &amp;amp; S &amp;amp;\cdot  &amp;amp; \cdot &amp;amp; \cdot &amp;amp; 10 &amp;amp; \cdot &amp;amp; \cdot\\&#xA;\cdot &amp;amp; \cdot &amp;amp; \cdot &amp;amp; 14 &amp;amp; \cdot &amp;amp; 12 &amp;amp; \cdot &amp;amp; \cdot &amp;amp; \cdot\\&#xA;\cdot &amp;amp; \cdot &amp;amp; \cdot &amp;amp; \cdot &amp;amp; \cdot &amp;amp; \cdot &amp;amp; \cdot &amp;amp; 11 &amp;amp; \cdot\\&#xA;\cdot &amp;amp; \cdot &amp;amp; \cdot &amp;amp; \cdot &amp;amp; 13 &amp;amp; \cdot &amp;amp; \cdot &amp;amp; \cdot &amp;amp; \cdot\\&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just keep going around to block off squares before going into the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not possible to do it in less because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to trap the knight, you need to eliminate all eight of its squares that it can escape to. It is not possible to get from one directly to another, so you need to visit an extra square in between each outside square. So that's 8 squares to block off, plus 7 moves in between: 15 moves total.&lt;/p&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="18522" LastEditDate="2016-01-18T06:59:20.917" LastActivityDate="2016-01-18T06:59:20.917" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="25462" PostTypeId="2" ParentId="25444" CreationDate="2016-01-17T21:47:06.500" Score="5" Body="&lt;p&gt;It is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a needle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; needles are sharp (prickly) and have an eye, and are well hidden (hard to spy) in hay-stacks (fodder). And if you are teased, you may feel a bit needled.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="1631" LastEditDate="2016-01-17T22:10:30.233" LastActivityDate="2016-01-17T22:10:30.233" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25463" PostTypeId="1" AcceptedAnswerId="25494" CreationDate="2016-01-17T21:59:18.960" Score="0" ViewCount="419" Body="&lt;p&gt;Psd(x) denotes the sum of digits of all positive integer divisors of x where x&gt;0. For example, Psd(15)=1+3+5+1+5=14 (divisors: 1,3,5,15)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is x for the biggest value of Psd(x) where x&amp;lt;10000.&lt;/p&gt;&#xA;" OwnerUserId="18514" LastActivityDate="2016-01-18T12:30:45.153" Title="Sum of digits of divisors" Tags="|mathematics|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25464" PostTypeId="1" AcceptedAnswerId="25468" CreationDate="2016-01-17T22:07:18.137" Score="5" ViewCount="1477" Body="&lt;p&gt;Hark! Pay attention when I speak&lt;br&gt;&#xA;And make way, or grave results may follow&lt;br&gt;&#xA;My mirror image will tell you  who I am&lt;br&gt;&#xA;Chase me, and you'll be considered the lowest of the low&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who am I?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user17947" LastActivityDate="2016-01-18T10:34:12.620" Title="Pay attention when I speak - who am I?" Tags="|riddle|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25465" PostTypeId="1" AcceptedAnswerId="25466" CreationDate="2016-01-17T22:08:02.453" Score="3" ViewCount="174" Body="&lt;p&gt;a,b,c,d are different numbers from each other and 0. abcd is a 4 digits number. ab and cd are two digits numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;what should be a,b,c,d to make abcd-(ab*cd) minimum?&lt;/p&gt;&#xA;" OwnerUserId="18514" LastActivityDate="2016-01-17T22:34:42.957" Title="Minimum Difference between two special numbers" Tags="|mathematics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25466" PostTypeId="2" ParentId="25465" CreationDate="2016-01-17T22:34:42.957" Score="8" Body="&lt;p&gt;(&lt;em&gt;Note: I'm assuming that &quot;$ab$&quot; refers to the two-digit number $10a+b$ and not to $a\times b$, and similarly that $abcd$ refers to the four digit number whose digits are $a,b,c,d$, since otherwise the question makes no sense.&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $a=1,b=2,c=9,d=8$&#xA;&lt;br&gt;&lt;br&gt;&#xA;so that&lt;br&gt;&#xA;&lt;br&gt;&#xA; $abcd=1298$, $ab\times cd=1176$, $abcd-ab\times cd = 122$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and this is the minimum that's possible to achieve, because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Write the quantity we're trying to minimize as&lt;br&gt;&lt;br&gt; $abcd-ab\times cd=(ab\times100+cd)-ab\times cd=ab\times(100-cd)+cd$. &lt;br&gt;&lt;br&gt;This shows that the dependence on $a$ and $b$ is such that it pays to take $ab$ as small as possible, namely $a=1$, $b=2$, $ab=12$. Then, rewrite the quantity yet again as&#xA; &lt;br&gt;&lt;br&gt;&#xA; $abcd-ab\times cd=(ab\times100+cd)-ab\times cd=ab\times100-(ab-1)\times cd$&#xA; &lt;br&gt;&lt;br&gt; which shows that it pays to take $cd$ as &lt;em&gt;large&lt;/em&gt; as possible, namely $c=9$, $d=8$, $cd=98$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user17947" LastActivityDate="2016-01-17T22:34:42.957" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25467" PostTypeId="2" ParentId="25098" CreationDate="2016-01-17T22:48:25.540" Score="1" Body="&lt;p&gt;It is not possible if the matrix is odd x odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspected this by trying the same problem with 4x4 and got a solution: the logic was obvious. There has to be 4+4 distinct numbers, and with -1,0,1 you could find 9 different sums: -4,-3,-2,-1,0,1,2,3,4. There is no solution if you use -4 and 4 at the same time since they have to be in the same row/column. Only solution would be without any of them as below;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Z6N8h.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z6N8h.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I tried it with 3x3 and 5x5 and I stuck at the same point where you are supposed to put -1,0,1 and there was no possible solution;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iCEZT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iCEZT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wrote an program that tries to find possible solution with a basic algorithm and it could not find a solution either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I can show that it is not possible to have all distinct sums. Firstly, there has to be 22 different numbers and 23 different values you can have by summing rows and columns as states above. First of all I will show you it is not possible to have 11 and -11 at the same time;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to use 11 and -11, there has to be on the same row/column as below;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RQaRK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RQaRK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you have to have at least one 10 and -10 in this possible solution and you have to put it on the row again since it is not possible to have any of them on a column any more;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/X8T8s.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/X8T8s.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you cannot have -10 any more in this possible solution and you have to use all other possible sums since if you put -1s and one 0 in the row, you will not able to have -9 and 9 any more and there is no way to find a solution any more. So we have to use only one -11 or 11 as below as a start;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kDnOB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kDnOB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since you will have only one 11 or -11, you have to have -10 and 10 at the same time in the solution so only possible solution for this as above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, you need to 9 and -9. You cannot put -9 on the row since you will not able to find 9 any more, so -9 has to be on the column and 9 on the row:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ryRrw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ryRrw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;same goes for 8 and -8 and it goes to a point;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ljo05.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ljo05.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this is the same problem for a oddxodd matrix where you cannot find a solution after this point as 5x5 and 3x3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result, there is no solution and it is not possible!&lt;/p&gt;&#xA;" OwnerUserId="18514" LastEditorUserId="18514" LastEditDate="2016-01-18T08:44:15.787" LastActivityDate="2016-01-18T08:44:15.787" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25468" PostTypeId="2" ParentId="25464" CreationDate="2016-01-17T23:10:08.940" Score="11" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an ambulance.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hark! Pay attention when I speak&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ambulances sound their sirens to draw attention when they're on the way to an emergency.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And make way, or grave results may follow&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you don't let an ambulance through traffic, it will be delayed and the person it's trying to get to could die.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My mirror image will tell you who I am&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ambulances have AMBULANCE printed backwards on their fronts, so that people looking in their rear-view mirrors can more easily recognise them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Chase me, and you'll be considered the lowest of the low&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Ambulance-chaser&quot; is slang for a lawyer who rushes to accident scenes to convince the victim to employ them. They're generally considered to be predatory. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12511" LastActivityDate="2016-01-17T23:10:08.940" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25469" PostTypeId="2" ParentId="25464" CreationDate="2016-01-17T23:12:06.117" Score="3" Body="&lt;p&gt;Are you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An ambulance?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you hear an ambulance you have to pay attention to see if you need to get out of the way and people might die if you block them. Ambulances have ambulance written on it mirrored so people can read through their rear view mirrors. Chasing an ambulance (to get through trafic) is usually frowned upon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16846" LastActivityDate="2016-01-17T23:12:06.117" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25470" PostTypeId="1" AcceptedAnswerId="25476" CreationDate="2016-01-18T00:15:07.857" Score="5" ViewCount="689" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I comprise a fixed number of lines&lt;br&gt;&#xA;  And that number is smaller than nine&lt;br&gt;&#xA;  As I self-referentially&lt;br&gt;&#xA;  Describe myself, potentially&lt;br&gt;&#xA;  I'll become a real breeze to divine&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Who am I?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This one isn't going to win any awards for Most Difficult Riddle of the Year, but it seemed amusing enough for a Sunday afternoon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. The answer is &lt;em&gt;not&lt;/em&gt; &quot;riddle&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit: Watch out for spoilers in the comments below.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2:&lt;/strong&gt; by popular demand, I changed one word in the riddle to make it more accurate and pretty (though somewhat easier as a result).&lt;/p&gt;&#xA;" OwnerDisplayName="user17947" LastEditorDisplayName="user17947" LastEditDate="2016-01-18T18:55:06.423" LastActivityDate="2016-01-18T18:55:06.423" Title="A self-referential riddle" Tags="|riddle|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25471" PostTypeId="2" ParentId="25470" CreationDate="2016-01-18T02:14:40.990" Score="1" Body="&lt;p&gt;Is the answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I comprise a fixed number of units&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Kinda stuck here. Maybe there is only one unit?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And that number is smaller than nine&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 &amp;lt; 9 = True &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As I self-referentially&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am the &lt;strong&gt;&lt;em&gt;one&lt;/em&gt;&lt;/strong&gt; and only, 1 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Describe myself, potentially&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Same as above, I am the &lt;strong&gt;&lt;em&gt;one&lt;/em&gt;&lt;/strong&gt; and only&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll become a real breeze to divine&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Becoming the one and only in something or the number one in something will lead to lots of fame and very god-like status&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9042" LastActivityDate="2016-01-18T02:14:40.990" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25472" PostTypeId="2" ParentId="25444" CreationDate="2016-01-18T02:19:33.347" Score="1" Body="&lt;p&gt;Is the answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm prickly and one-eyed&lt;br&gt;&#xA;And quite hard to spy&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There can only be one answer and they can be very hard to spot. Also hard to grasp, hence the prickly part&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With fodder for mysteries and puzzles galore&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer gives the riddle purpose: to be solved!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you see me you'll be pleased&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You should feel the satisfaction that you solved a riddle first&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But feel rather teased&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The suspense when your answer is so close to the real answer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And pointedly wonder why you'd missed me before&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you finally see the answer, you will wonder &quot;Why didn't I think of that before?&quot; (Answer: You're human, not Jon Skeet's bot :) )&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9042" LastActivityDate="2016-01-18T02:19:33.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25473" PostTypeId="2" ParentId="10101" CreationDate="2016-01-18T02:23:53.817" Score="0" Body="&lt;p&gt;It has been too long so I will answer my own question that has been unanswered for 10 months.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mustard gas&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm the thing you would love&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Curing cancer, starts with mustard...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet I was a murderer by the millions&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Killed millions of people during wars and battles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was a poison of the 1900's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It was a poison developed in the 1900's&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then a revolutionary medicine in the hospitals&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It help cure cancer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My other use a massage to your blood, the other the killer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not only it destroys cancer cells, it can destroy white blood cells&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9042" LastActivityDate="2016-01-18T02:23:53.817" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25474" PostTypeId="2" ParentId="25464" CreationDate="2016-01-18T03:22:06.157" Score="4" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Donald Trump&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hark! Pay attention when I speak&lt;br&gt;&#xA;And make way, or grave results may follow&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Donald Trump has a huge ego and craves being paid attention to. He is known as a germaphobe, so you'd better make way when he passes by, or he may subject you to his typical insults and bullying behavior.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My mirror image will tell you who I am&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You see his image on TV (a flat, reflective surface like a mirror), where he is constantly reminding you of who he is and how great and smart he is.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Chase me, and you'll be considered the lowest of the low&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He consistently derides and insults all the other Republican presidential candidates for daring to chase him as the leading contender for the nomination.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user17947" LastActivityDate="2016-01-18T03:22:06.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25475" PostTypeId="2" ParentId="12119" CreationDate="2016-01-18T04:48:53.787" Score="2" Body="&lt;p&gt;Originally, this question was asked and solved by Gera Weiss in a paper called &quot;A Combinatorial Game Approach to State Nullification by Hybrid Feedback&quot;. Here's a link:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cs.bgu.ac.il/~geraw/Publications_files/CombinatorialGameApproachToStateNullification.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.cs.bgu.ac.il/~geraw/Publications_files/CombinatorialGameApproachToStateNullification.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18521" LastActivityDate="2016-01-18T04:48:53.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25476" PostTypeId="2" ParentId="25470" CreationDate="2016-01-18T04:57:26.733" Score="3" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A limerick.&lt;br&gt;&#xA;&lt;br&gt;&#xA;  It has five lines (less than 9), and the riddle itself is a limerick. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2016-01-18T04:57:26.733" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25477" PostTypeId="1" AcceptedAnswerId="25521" CreationDate="2016-01-18T06:15:06.853" Score="3" ViewCount="1160" Body="&lt;p&gt;This is my first post on here, inspired by other questions, but original.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I am a gas that you love&#xA;most commonly found up above&#xA;I have no hues&#xA;and convert into twos&#xA;when you trap me and give me a shove&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Here is another limerick to add some more clues.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I can be manufactured if able&#xA;but down here I'm rather unstable&#xA;in space I am found&#xA;not here on the ground&#xA;it's not me that you love but my label&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="18522" LastEditorUserId="18522" LastEditDate="2016-01-18T15:42:14.430" LastActivityDate="2016-01-19T01:39:38.773" Title="What am I limerick" Tags="|riddle|limerick|" AnswerCount="10" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25478" PostTypeId="2" ParentId="25477" CreationDate="2016-01-18T06:20:50.797" Score="0" Body="&lt;p&gt;Are you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ozone gas&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We love ozone gas since it protects us, it's usually found high in the atmosphere (ozone layer) and converts into O2 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16846" LastActivityDate="2016-01-18T06:20:50.797" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25479" PostTypeId="2" ParentId="25477" CreationDate="2016-01-18T06:38:22.953" Score="0" Body="&lt;p&gt;My Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Nitrogen Gas (N₂) / Nitrous Oxide (N₂O)&lt;br&gt;&#xA;Or Helium!!!&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We love it...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Also known as laughing gas (N₂O that is)...&lt;br&gt;&#xA;Helium is so light it floats out of our atmosphere...&lt;br&gt;&#xA;Kids and adults like breathing it or using it in balloons&lt;br&gt;&#xA;Its colourless...&lt;br&gt;&#xA;It splits into two...&lt;br&gt;&#xA;It is found above (70% of the atmosphere)...&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12014" LastEditorUserId="12014" LastEditDate="2016-01-18T06:52:09.173" LastActivityDate="2016-01-18T06:52:09.173" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25480" PostTypeId="2" ParentId="25477" CreationDate="2016-01-18T06:43:24.787" Score="0" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; hydrogen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a gas that you love&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hydrogen powers the sun, without which life on Earth could not exist. Who doesn't love that? It is also useful for many technological applications, for example it can be used to store energy and powers some cars, such as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Toyota_Mirai&quot; rel=&quot;nofollow&quot;&gt;Toyota Mirai&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Most commonly found up above&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sun, which is up above, has lots of it, although it is also quite common on earth (e.g., it is extracted from natural gas and oil). If it escapes into the air, due to its low molecular weight it will rise up to the upper parts of the atmosphere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no hues&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hydrogen is colorless&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and convert into twos&lt;br&gt;&#xA;when you trap me and give me a shove&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This refers to nuclear fusion, the reaction that powers the sun: two hydrogen nuclei (aka protons), when trapped in close proximity and given a lot of energy (a &quot;shove&quot;), fuse to form a deuterium atom (and some other things, which then proceed to do more things to complete the fusion reaction cycle, as described for example &lt;a href=&quot;http://www.nobelprize.org/educational/physics/energy/fusion_2.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;). Deuterium is a hydrogen isotope with two protons in its nucleus, i.e., a &quot;two&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user17947" LastEditorDisplayName="user17947" LastEditDate="2016-01-18T06:51:52.253" LastActivityDate="2016-01-18T06:51:52.253" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25481" PostTypeId="1" AcceptedAnswerId="25529" CreationDate="2016-01-18T07:37:25.897" Score="11" ViewCount="4064" Body="&lt;p&gt;Each of an infinite number of wizards is independently assigned a black or white hat based on the outcome of an unbiased coin flip. Each wizard can see everyone's hat but his own. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the count of three, each wizard must simultaneously either guess his own hat color or abstain from guessing. The wizards collectively win if an infinite number of them guess correctly and none of them guess incorrectly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the wizards devise a strategy so that they usually win? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wizards can plan and move around before the hats are distributed, but immediately upon hat distribution, a spell is cast so that they are all completely paralyzed up until the moment of their simultaneous guess. No communication is allowed.&lt;/p&gt;&#xA;" OwnerUserId="18525" LastEditorUserId="18525" LastEditDate="2016-01-18T20:17:11.457" LastActivityDate="2016-01-20T23:43:31.617" Title="Infinite wizards and hats" Tags="|hat-guessing|" AnswerCount="11" CommentCount="23" ContentLicense="CC BY-SA 3.0" />
  <row Id="25482" PostTypeId="2" ParentId="25481" CreationDate="2016-01-18T07:53:17.940" Score="3" Body="&lt;h2&gt;As the number of hat-wizards approaches infinity, the probability of winning approaches 100%&lt;/h2&gt;&#xA;&lt;p&gt;Using a strategy similar to the one given by xnor to &lt;a href=&quot;https://puzzling.stackexchange.com/a/15942/16990&quot;&gt;solve the case where n=15&lt;/a&gt; (and adding an additional digit each time ceil(log2(n+1)) increases), the wizards' odds of losing will become arbitrarily small as the number of wizards becomes arbitrarily large.&lt;/p&gt;&#xA;&lt;h2&gt;However, that does not extend to infinity itself&lt;/h2&gt;&#xA;&lt;p&gt;The strategy provided there does not work at infinity, as it would require vectors of infinite length, which is beyond even the most learned and magical of wizards.&lt;/p&gt;&#xA;&lt;p&gt;Furthermore, since the probability of losing remains positive (if very small) for all non-infinite n, the infinite wizards cannot divide themselves into infinitely-many finitely sized subgroups, because, given infinite trials, at least one of those subgroups would lose.&lt;/p&gt;&#xA;&lt;h2&gt;I believe that a solution where infinitely many wizards all guess correctly is impossible&lt;/h2&gt;&#xA;&lt;p&gt;&lt;strong&gt;EDIT: As Cruncher points out in a comment, the following is fallacious reasoning. I'm not going to amend this answer, because someone else already arrived at just such a strategy above!&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;This riddle, as stated, requires that an infinite number of wizards guess correctly - not merely one. Thus, infinitely many wizards must be able to deduce their hat with &lt;em&gt;100% certainty&lt;/em&gt;, and otherwise stay silent. No strategy existing for this puzzle has ever given any wizard 100% certainty of their hat.&lt;/p&gt;&#xA;&lt;p&gt;If the constraint requiring infinitely many wizards guess right were lifted, the finite solution could be applied, by having a predetermined finite (but very large) subset of wizards use the finite strategy, while all others remain silent.&lt;/p&gt;&#xA;" OwnerUserId="16990" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-01-18T20:08:54.743" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25483" PostTypeId="2" ParentId="25481" CreationDate="2016-01-18T08:00:16.290" Score="-1" Body="&lt;p&gt;The wizards make groups of three, one is appointed &quot;the chosen one&quot;. When they all get their hats, the chosen one of each group can either join or not join their group. Joining means that the other two have the same colored hat, not joining means that they have different colors (this can also be done the other way around). The other two can deduce from the action of the chosen one whether their hat color is the same  as or opposite from the color of the other. So two thirds of the wizards can &quot;guess&quot; correctly, the chosen one abstains from guessing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is slightly cheaty since this is a form of communication, which I presume is not allowed (if it is, simply ask someone the color of your hat).&lt;/p&gt;&#xA;" OwnerUserId="18526" LastActivityDate="2016-01-18T08:00:16.290" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25484" PostTypeId="2" ParentId="25481" CreationDate="2016-01-18T08:52:18.447" Score="5" Body="&lt;p&gt;Another option of 'sorting' the wizards would be to:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Start off with two wizards (put them someplace where there is a lot of room around them).&lt;/li&gt;&#xA;&lt;li&gt;Now a third wizard joins in.&lt;/li&gt;&#xA;&lt;li&gt;He sees that the first two wizards have similar hats (let's say white) so he stands on the left of the both wizards.&lt;/li&gt;&#xA;&lt;li&gt;Now a fourth wizards joins in.&lt;/li&gt;&#xA;&lt;li&gt;He sees that two wizards are wearing white hats and one is wearing a black hat. He goes and stand in between the wizard with the white hat and the wizard with the black hat.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now, regardless of the fourth wizard's own hat color, he will always be sorted next to his own color (since there are only two possibilities).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;W = White hat&#xA;B = Black hat&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First two wizards : W W  &lt;/li&gt;&#xA;&lt;li&gt;Three wizards : W W B &lt;/li&gt;&#xA;&lt;li&gt;Four wizards : W W W B&lt;/li&gt;&#xA;&lt;li&gt;Five wizards : W W W B B&lt;/li&gt;&#xA;&lt;li&gt;Six wizards : W W W B B B&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As long as the room is big enough, this method will sort all the wizards and will (assuming the number really is infinite) tell all the wizards that are in between two of the same colored hats their own colored hat as well. So in the end only the last wizard will have to &quot;guess&quot; (or as stated by DrunkWolf, he can abstain about) which hat he has.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This answer assumes the wizards have no means of communicating with each other.&lt;/p&gt;&#xA;" OwnerUserId="6151" LastEditorUserId="4692" LastEditDate="2016-01-18T16:42:37.740" LastActivityDate="2016-01-18T16:42:37.740" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="25485" PostTypeId="2" ParentId="25481" CreationDate="2016-01-18T08:52:50.810" Score="0" Body="&lt;p&gt;If there are infinite number of wizards and wizards can see each other (assuming wizards can see infinite number of wizards) and flipping coin is unbiased, the solution is that every wizard know their hats already since the chance of getting head or tail is exactly 50% if number of wizards goes to infinity.&lt;/p&gt;&#xA;" OwnerUserId="18514" LastActivityDate="2016-01-18T08:52:50.810" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25486" PostTypeId="2" ParentId="25477" CreationDate="2016-01-18T08:58:14.803" Score="0" Body="&lt;p&gt;I think the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clouds&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a gas that you love&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People love different shapes of clouds and the rain they give.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and convert into twos&#xA;when you trap me and give me a shove&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They give us rain by shove. Two parts are rain and the remaining cloud.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastActivityDate="2016-01-18T08:58:14.803" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25490" PostTypeId="2" ParentId="25464" CreationDate="2016-01-18T10:34:12.620" Score="-1" Body="&lt;p&gt;Its your Name cuz you pay attention when some one calls your name.&lt;/p&gt;&#xA;" OwnerUserId="18531" LastActivityDate="2016-01-18T10:34:12.620" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25491" PostTypeId="1" AcceptedAnswerId="25549" CreationDate="2016-01-18T10:36:47.737" Score="6" ViewCount="516" Body="&lt;p&gt;Your task is to draw uniformly from a set of size $N$. You are given a &lt;em&gt;single&lt;/em&gt; coin which you can ‘bend’ to control the probability $p\in[0,1]$ of throwing a Head, but you are only allowed a &lt;strong&gt;bounded&lt;/strong&gt; number of throws; can it be done? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if the number of throws is not bounded then this can be done (more or less trivially) by an &lt;strong&gt;unbiased&lt;/strong&gt; coin (how?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: your are only allowed to bend once or otherwise your coin will break...&lt;/p&gt;&#xA;" OwnerUserId="18521" LastEditorUserId="12408" LastEditDate="2016-07-25T02:30:19.657" LastActivityDate="2016-07-25T02:30:19.657" Title="Drawing uniformly using a single biased coin" Tags="|mathematics|probability|coin-tossing|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25492" PostTypeId="2" ParentId="25491" CreationDate="2016-01-18T11:42:48.897" Score="1" Body="&lt;p&gt;Since you are allowed to bend the coin many times to control the probability $p$ it should be simple:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Just reject each outcome one by one, each time bending coin according to a current number of the non-rejected outcomes:&lt;br&gt;&#xA; 1) Bend to make $p=\frac{1}{N}$. Throw. If it is Head then you got outcome-1 (with probability of $\frac{1}{N}$). Otherwise proceed to step 2.&lt;br&gt;&#xA; ...&lt;br&gt;&#xA; k) Bend to make p=1/(N-k+1). Throw. If it is Head then you got outcome-k (with probability of $\frac{N-1}{N}*\frac{N-2}{N-1}*...*\frac{N-k+1}{N-k+2}*\frac{1}{N-k+1}=\frac{1}{N}$). Otherwise proceed to step k+1.&lt;br&gt;&#xA; ...&lt;br&gt;&#xA; N) Here p must be $\frac{1}{1}$, so no need to throw. You got outcome-N.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More interesting question is &quot;Can it be done if you can bend the coin only once to chose the probability $p$?&quot;.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditDate="2016-01-18T12:02:13.503" LastActivityDate="2016-01-18T12:02:13.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25493" PostTypeId="2" ParentId="25477" CreationDate="2016-01-18T11:56:27.890" Score="4" Body="&lt;p&gt;I'm surprised nobody has guessed this yet.&lt;br&gt;&#xA;I think it is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oxygen (O2) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I am a gas that you love  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Humans need O2 to live and breathe. (It is found in both the air we breathe and the water we drink - both crucial requirements for life.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;most commonly found up above  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; O2 in the Earth's atmosphere.    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I have no hues  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; O2 is a colorless, odorless gas. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;and convert into twos  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Our respiratory system transfers O2 to our blood where O2 is broken down (O2 -&gt; O, O - twos.)   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;when you trap me and give me a shove  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Trap me&quot; - breathing in. &quot;Give me a shove&quot; - our heart pumping blood where the O2 was transferred before the process of breaking it down starts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18429" LastActivityDate="2016-01-18T11:56:27.890" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25494" PostTypeId="2" ParentId="25463" CreationDate="2016-01-18T12:30:45.153" Score="0" Body="&lt;p&gt;Here's a simple Python bruteforce program that gets the job done:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from math import sqrt&#xA;&#xA;maxNum = -1&#xA;maxSum = 0&#xA;for x in range(1, 10000):&#xA;  currentSum = 0&#xA;  for d in range(1, int(sqrt(x)) + 1):&#xA;    if(x % d == 0):&#xA;      for digit in str(d):&#xA;        currentSum += int(digit)&#xA;      for digit in str(x / d):&#xA;        currentSum += int(digit)&#xA;  if currentSum &amp;gt; maxSum:&#xA;    maxSum = currentSum&#xA;    maxNum = x&#xA;&#xA;print &quot;Maximum sum: {}.\n{}:&quot;.format(maxSum, maxNum)&#xA;for d in range(1, int(sqrt(maxNum)) + 1):&#xA;  if(maxNum % d == 0):&#xA;    print &quot;  = {} x {}&quot;.format(d, maxNum / d)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Output:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Maximum sum: 549.&#xA;7920:&#xA;  = 1 x 7920&#xA;  = 2 x 3960&#xA;  = 3 x 2640&#xA;  = 4 x 1980&#xA;  = 5 x 1584&#xA;  = 6 x 1320&#xA;  = 8 x 990&#xA;  = 9 x 880&#xA;  = 10 x 792&#xA;  = 11 x 720&#xA;  = 12 x 660&#xA;  = 15 x 528&#xA;  = 16 x 495&#xA;  = 18 x 440&#xA;  = 20 x 396&#xA;  = 22 x 360&#xA;  = 24 x 330&#xA;  = 30 x 264&#xA;  = 33 x 240&#xA;  = 36 x 220&#xA;  = 40 x 198&#xA;  = 44 x 180&#xA;  = 45 x 176&#xA;  = 48 x 165&#xA;  = 55 x 144&#xA;  = 60 x 132&#xA;  = 66 x 120&#xA;  = 72 x 110&#xA;  = 80 x 99&#xA;  = 88 x 90&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Good guess @Ross Millikan.&lt;/p&gt;&#xA;" OwnerUserId="18429" LastActivityDate="2016-01-18T12:30:45.153" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25495" PostTypeId="2" ParentId="25481" CreationDate="2016-01-18T13:41:02.450" Score="12" Body="&lt;p&gt;Create infinite groups of finite size with $$size = (2^k)-1 ; k=2,3,...,infinity$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of the groups will handle themselves as per the optimal strategy &lt;a href=&quot;https://puzzling.stackexchange.com/a/15942/18537&quot;&gt;in this answer&lt;/a&gt; by &lt;a href=&quot;https://puzzling.stackexchange.com/users/4551/xnor&quot;&gt;xnor&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus each group has a probability of correctly guessing equal to : $$1- \frac{1}{2^k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the probability that all of the groups guess right is the infinite multiplication of those probabilities, which is at least 56.25%  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may even get a better strategy by starting with k=3 rather than k=2. And the higher k you start with the better your strategy will be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Demonstration that the infinite multiplication $\ge .5625$&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f(k)$ be the result of the multiplication up to the k-th term. &#xA;Thus :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(2) = 0.75  $$&#xA;$$f(k+1) = f(k) * (1- \frac{1}{2^{k+1}}) = f(k) - f(k)*\frac{1}{2^{k+1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the terms in the multiplication are greater than 0 and less than 1 thus we know that for all $k \gt 2$ : $f(k+1) \lt f(k)$. So for all k we have $f(k) \le f(2)$. Hence :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(k+1) = f(k) - f(k)*\frac{1}{2^{k+1}} \ge f(k) - f(2)*\frac{1}{2^{k+1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we rewrite our infinite multiplication using the previous formula we have an infinite sum which is a lower bound (L) of our infinite multiplication :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L = f(2) - f(2)* \sum_{k=2}^\infty \frac{1}{2^{k+1}} = .5625 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we conclude that the strategy here explained has a probability of success at least 56.25%&lt;/p&gt;&#xA;" OwnerUserId="18537" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2016-01-18T15:32:44.190" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="25496" PostTypeId="1" AcceptedAnswerId="114465" CreationDate="2016-01-18T14:09:50.907" Score="10" ViewCount="689" Body="&lt;p&gt;Does there exist a circle whose boundary contains 6 points whose 15 pairwise distances are distinct integers?&lt;/p&gt;&#xA;" OwnerUserId="18525" LastEditorUserId="69582" LastEditDate="2021-09-12T18:20:19.157" LastActivityDate="2022-01-22T14:52:21.430" Title="Points on the boundary of a circle" Tags="|mathematics|geometry|" AnswerCount="3" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="25497" PostTypeId="2" ParentId="25477" CreationDate="2016-01-18T15:27:51.090" Score="0" Body="&lt;p&gt;I'm gunna make a rather, er, crass guess and say it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Methane&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a gas that you love&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everyone loves their own brand I guess&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;most commonly found up above&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Methane rises and is one of the big things damaging out ozone layer IIRC&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no hues&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Methane doesn't have a colour&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and convert into twos&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is where it kindof falls apart, I guess something to do with 'number twos' though Methane doesn't exactly 'convert' into feces&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;when you trap me and give me a shove&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Erm, should be self explanatory given the above ;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18542" LastActivityDate="2016-01-18T15:27:51.090" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25498" PostTypeId="2" ParentId="25481" CreationDate="2016-01-18T15:42:47.497" Score="-1" Body="&lt;h1&gt;Use Magic:&lt;/h1&gt;&#xA;&lt;p&gt;Have each wizard cast a spell on himself/herself that allows himself/herself to &amp;quot;feel&amp;quot; color. To get a sense of what black and white &amp;quot;feel&amp;quot; like touch something black/white before the hats are distributed. Then feel your own hat as it sits on your head and guess from that.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;OR&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Use some sort of a &amp;quot;Luck&amp;quot; spell/enchantment and guess with the increased luck...&lt;/p&gt;&#xA;" OwnerUserId="3922" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-01-18T23:06:12.767" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25499" PostTypeId="1" AcceptedAnswerId="25602" CreationDate="2016-01-18T15:43:08.457" Score="1" ViewCount="407" Body="&lt;p&gt;His brothers will always help you,&lt;br&gt;&#xA;Slaying spiders or golems.&lt;br&gt;&#xA;Sometimes you see him for real,&lt;br&gt;&#xA;Depending on how you live.&lt;br&gt;&#xA;And the thing you won't believe,&lt;br&gt;&#xA;Is that he can melt steel.&lt;br&gt;&#xA;Even if he looks like a gem,&lt;br&gt;&#xA;One day he will crush you.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who is &quot;he&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;hint 1 (kinda big one) :  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He will crush what is left of you, if there is something left by that time!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;hint 2  :  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Line three is more of a joke than a real sentence and may concern Stack users :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;hint 3 :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the two first lines refer to a video game  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;last hint : &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A very famous guy died trying to reach it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17888" LastEditorUserId="17888" LastEditDate="2016-01-21T10:22:22.210" LastActivityDate="2016-01-21T10:47:45.860" Title="Father of us all?" Tags="|riddle|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25501" PostTypeId="2" ParentId="25477" CreationDate="2016-01-18T16:23:37.513" Score="0" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Helium&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a gas that you love&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Who doesn't love Helium?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;most commonly found up above&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Helium rises&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no hues&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; and is colorless&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and convert into twos&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2nd periodic element&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;when you trap me and give me a shove&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; when you put me in a balloon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14233" LastActivityDate="2016-01-18T16:23:37.513" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25502" PostTypeId="2" ParentId="25499" CreationDate="2016-01-18T16:47:15.477" Score="-1" Body="&lt;p&gt;air or water...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;oxygen can melt you and iron in time if there is humidity (water) in the air, air has water so it can kill spiders as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Golems are made of clay so you may slay it with water (which is the brother of air?)&lt;/p&gt;&#xA;" OwnerUserId="18514" LastEditorUserId="18514" LastEditDate="2016-01-18T19:17:39.467" LastActivityDate="2016-01-18T19:17:39.467" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25503" PostTypeId="2" ParentId="25496" CreationDate="2016-01-18T16:58:11.403" Score="22" Body="&lt;p&gt;They do exist. And they are actually called Brahmagupta Hexagons. An example is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/luIIA.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/luIIA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which I took from &lt;a href=&quot;http://forumgeom.fau.edu/FG2005volume5/FG200515.pdf&quot; rel=&quot;noreferrer&quot;&gt;this paper&lt;/a&gt; which has a lot more info on them&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2016-01-18T16:58:11.403" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25505" PostTypeId="1" AcceptedAnswerId="25550" CreationDate="2016-01-18T17:56:57.417" Score="6" ViewCount="1784" Body="&lt;p&gt;A certain young gentleman received a confusing and cryptic letter, which he assumes is a love letter. But alas, the letter has no name on it. The letter is as reads below:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Hey, baby, won't you look my way, I could be your new addiction. Well, maybe not me, seeing as I didn't write that. The question is, who did?&lt;/p&gt;&#xA;&lt;p&gt;I apologize if I'm not making much sense, because there is an annoying insect in my room. It sounds like haemophilia, but it shares it's roots with anthropology. I have Nikolai to blame for that. Look, out, Gansey!!&lt;/p&gt;&#xA;&lt;p&gt;I'll have to go kill it now, because, honey, there ain't no rest for the wicked. (I didn't write that line either, by the way.)&lt;/p&gt;&#xA;&lt;p&gt;I'm also craving some homeomorphic toroidal polyhedrons, but I'd rather share them with you rather then eat them alone. I suppose I could write it all down in one letter.&lt;/p&gt;&#xA;&lt;p&gt;I'd rather tell you how I'm feeling, boy, and how I'm doing things my way. If I'd got my piano, then I'd know things were okay. (Is there a limit to the amount of allusions you can make in one letter? Take my first two letters, I suppose, and you'd get your answer,)&lt;/p&gt;&#xA;&lt;p&gt;And in my last letter... well, Fabian and Gideon can tell you, it's the first letter of their last of names.&lt;/p&gt;&#xA;&lt;p&gt;It's elementary, my dear, and if you can't figure this out, then I'd have to infer that you aren't really worth the unrequited crush.&lt;/p&gt;&#xA;&lt;p&gt;If you figure out the name I've been trying to tell you, then you'll know who I am, because I am his daughter.&lt;/p&gt;&#xA;&lt;p&gt;Solve it by the 26th, baby, we ride then at midnight. Bring the potatoes.&lt;/p&gt;&#xA;&lt;p&gt;Yours Truly,&lt;/p&gt;&#xA;&lt;p&gt;Me&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;What is the name of the author of the letter?&lt;/p&gt;&#xA;" OwnerUserId="11814" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-01-20T05:49:16.340" Title="A Love letter with No Name" Tags="|riddle|wordplay|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25506" PostTypeId="2" ParentId="25499" CreationDate="2016-01-18T18:43:41.890" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Coal?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not sure about the first two though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes you seem him for real depending on how you live. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Suggests difference between classes maybe. Upper class living on electricity lower struggling for heating &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the thing you won't believe is that he can melt steel&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fairly obvious. Used for furnaces and could be made to forge steel&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;even if he looks like a gem&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Apply enough pressure and you get a diamond&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;one day he will crush you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After death, fossilised and turn to coal yourself under the pressure&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18546" LastEditorUserId="18429" LastEditDate="2016-01-19T06:10:07.727" LastActivityDate="2016-01-19T06:10:07.727" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25508" PostTypeId="2" ParentId="25481" CreationDate="2016-01-18T19:56:51.687" Score="-1" Body="&lt;p&gt;What if they all abstain from guessing? Wouldn't that mean they all won? Nobody guesses incorrectly, and the color of everyone's hat is totally irrelevant. Or did I miss something.&lt;/p&gt;&#xA;" OwnerUserId="18548" LastActivityDate="2016-01-18T19:56:51.687" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25509" PostTypeId="1" CreationDate="2016-01-18T20:07:48.053" Score="0" ViewCount="2553" Body="&lt;p&gt;There is a popular game called Mastermind in which one player guesses another player's  secret sequence - it could be a word, some colors, or numbers. The guesser says a possible sequence, and they get back the number of correct numbers in the right spot and the number of correct numbers in the wrong spot. Here we will be playing Mastermind with 4-digit numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example game with the answer &quot;4560&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You start with 1234 and get the response &quot;-1&quot;; that means one digit is correct but in the wrong location.&#xA;You guess 3465 and get &quot;-2 +1&quot;; that means two digits are in the wrong location and one digit is in the correct location. The other digit isn't part of the number at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The game continues:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1234: -1&#xA;3465: -2 +1&#xA;4503: -1 +2&#xA;4560 +4 (4 tries and win!)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You are very smart, so as soon as you get enough information you'll be able to guess the number instantly, but the problem is that you are the unluckiest person alive, especially at guessing numbers. So, in the worst case scenario, how many guesses do you need to find your opponent's number?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind that 4-digit numbers can't start with 0!&lt;/p&gt;&#xA;" OwnerUserId="18514" LastEditorUserId="11876" LastEditDate="2016-01-19T04:17:47.400" LastActivityDate="2016-01-20T10:08:55.680" Title="Guess my number" Tags="|mastermind|" AnswerCount="2" CommentCount="14" ClosedDate="2016-01-19T15:40:35.417" ContentLicense="CC BY-SA 3.0" />
  <row Id="25510" PostTypeId="2" ParentId="25477" CreationDate="2016-01-18T20:38:33.400" Score="0" Body="&lt;p&gt;My guess is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Plasma aka STAR's&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a gas that you love&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stars are luminous balls of plasma and who does not love STARS ;-)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;most commonly found up above&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Up above the world so high like the rhyme goes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no hues&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Coz they are so far away.. it is tough to guess the correct color of the star.. Always say they are white.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and convert into twos&#xA;when you trap me and give me a shove&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Computer simulations suggest in gravity free environment of space, divide to form two copies of original structure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Here is another limerick to add some more clues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can be manufactured if able&#xA;but down here I'm rather unstable&#xA;in space I am found&#xA;not here on the ground&#xA;it's not me that you love but my label&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Expensive and unsafe to manufacture plasma..&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9892" LastActivityDate="2016-01-18T20:38:33.400" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25512" PostTypeId="1" AcceptedAnswerId="25517" CreationDate="2016-01-18T21:19:23.457" Score="14" ViewCount="976" Body="&lt;p&gt;You travel far and wide in search of the one named 'the oracle'. You wish to ask a question which has been troubling you for days now. You trek through mountains and ice-caps, forests and cities and eventually find the oracle sitting motionless in a cave. You finally can find out your answer...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You look up at the oracle and ask:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;What is the cutest species currently alive on this planet?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The oracle stirs and hands you a sheet which is shown below. He merely states:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Solve the puzzle and you will know&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the puzzle the oracle gave you:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Q1. Left 1.   My mouth is a city, all of my kind are full of floating trees&lt;br&gt;&#xA;Q2. Left 1.   As humans are to the galaxy, dust is to humans&lt;br&gt;&#xA;Q3. Right 1.  A dirty kick in football is the same as a woodcutter cutting a log and someone logging into someone else's account&lt;br&gt;&#xA;Q4. Left 1.    I am oil in the sea, chemicals on the lab floor, and water down your shirt&lt;br&gt;&#xA;Q5. Left 1.    If I move up at the start I will be considered rude, but at the end, you will set me rolling&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/43mYp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/43mYp.png&quot; alt=&quot;Grid&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What does the oracle think the cutest species is?&lt;br&gt;&#xA;  Solve the puzzle by placing each answer horizontally in each row with the corresponding number. The answer will read downwards diagonally in the shaded boxes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who can solve the cutest species problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Added a extra clue to Q3.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And because people are getting confused about 'left 1' and 'right 1' here is a hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; left and right are &lt;strong&gt;Directions&lt;/strong&gt; and the 1 is &lt;strong&gt;the amount&lt;/strong&gt; of the direction. Basically move all the letters (&lt;strong&gt;Directions&lt;/strong&gt; by &lt;strong&gt;the amount&lt;/strong&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This will help especially when you have the answers and are trying to fit them in the grid&lt;/p&gt;&#xA;" OwnerUserId="18250" LastEditorUserId="18250" LastEditDate="2016-01-25T19:26:47.207" LastActivityDate="2016-01-25T19:26:47.207" Title="The cutest species problem" Tags="|riddle|word|" AnswerCount="3" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="25513" PostTypeId="5" CreationDate="2016-01-18T21:20:56.857" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2016-01-18T21:20:56.857" LastActivityDate="2016-01-18T21:20:56.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25514" PostTypeId="4" CreationDate="2016-01-18T21:20:56.857" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2016-01-18T21:20:56.857" LastActivityDate="2016-01-18T21:20:56.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25515" PostTypeId="2" ParentId="25477" CreationDate="2016-01-18T21:41:05.267" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oxygen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a gas that you love&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Can't live without it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;most commonly found up above&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; O3 is the most common gas in the ozone&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no hues&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A colorless gas&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and convert into twos&#xA;when you trap me and give me a shove&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oxygen is trapped when we breath it in, and when we shove it in our exhale it is CO2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13054" LastActivityDate="2016-01-18T21:41:05.267" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25516" PostTypeId="2" ParentId="25481" CreationDate="2016-01-18T22:15:46.383" Score="6" Body="&lt;p&gt;As other answers have shown that any probability of success less than $100\%$ can be achieved, let me show that no strategy can achieve a success rate of $100\%$. The argument is quite simple: Number the wizards as $1,\,2,\,3,\ldots$ and let $p_i$ be the probability that the $i^{th}$ wizard will guess at all (given a random arrangement of hats). Unfortunately, the wizard necessarily guesses wrongly with probability $\frac{p_i}2$ since they must choose their guess based on the other's arrangement of hats, but this knowledge is independent of their hat color. Thus, to achieve no wrong guesses, $p_i=0$. However, if the event of any given wizard guessing almost never happens, the event of infinitely many wizards guessing almost never happens either which is a problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This proof, however, does not present a problem to the fact that probabilities less than $100\%$ are obtainable; these essentially work by choosing the individual $p_i$ to be small (but have infinite sum) and choosing them such that the losses are &quot;consolidated&quot; (e.g. everyone guesses wrong if anyone does) but the wins are spread out (e.g. one wizard guessing correctly decreases the probability that others do).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;One should note that this argument fails for uncountably many wizards. A strategy for $2^{|\mathbb N|}$ wizards that succeeds with probability $100\%$ (assuming this probability is well defined) is presented below. The strategy given &lt;a href=&quot;https://puzzling.stackexchange.com/a/15942/18537&quot;&gt;here by xnor&lt;/a&gt; can be extended fairly naturally. To summarize the strategy there for $2^{n}-1$ wizards:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The wizards are put into bijection with the non-empty subsets of $\{1,2,\ldots,n\}$. Then, we compute a new set $S$ by letting $s\in S$ if, there are an odd number of wizards who have black hats and whose associated sets include $s$. Any wizard who can guess their own hat color in a way that would make $S$ empty will make the opposite guess.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the lemmas one might need to show that this strategy is effective would be noting that $S$ can be any subset of $\{1,\ldots,n\}$ with equal probability and that, if it is non-empty, one wizard will make a guess (in particular, the wizard associated with the set $S$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To extend this, let $W$ be the set of wizards. Choose some bijection $f:W\rightarrow P(\mathbb N)\setminus\{\emptyset\}$. Next, let $E$ be some function $E:P(W)\rightarrow \{0,1\}$ such that, if two sets $S$ and $S'$ differ by one element (i.e. $|(S\setminus S')\cup (S'\setminus S)|=1$) then $E(S)=1-E(S')$. The existence of such an $E$ follows from the axiom of choice (in the finite case, we can just use $E(S)=|S|\text{ mod }2$, which breaks down for infinite $S$). Then, consider the set $S\subseteq \mathbb N$ defined as&#xA;$$S=\{n\in\mathbb N : E(W_n)=1\}$$&#xA;where $W_n$ is the set of wizards $w$ whose hats are black and for which $n\in f(w)$. Each wizard guesses that $S$ is not empty and thus, if one of their choices would make $S$ zero, they do not make the guess. Notice that, if $S$ is not empty, then changing the hat color of the wizard associated with $S$ would make it empty - thus, that wizard is able to deduce the color of their hat if they know $S$ is not empty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This strategy therefore lets a single wizard guess correctly whenever $S$ isn't empty (and if it is, &lt;em&gt;every&lt;/em&gt; wizard guesses incorrectly). Noting that there is a unique arrangement of the hat colors on the wizards associated with the sets $\{1\},\,\{2\},\,\{3\},\ldots$ that makes $S$ empty given the hat colors of everyone else, and this arrangement must be achieved with probability $0$ as it requires infinitely many independent events of probability $\frac{1}2$ to occur. Partitioning the wizards into countably many uncountable groups and applying this strategy gives countably many correct guesses with probability $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I note &quot;assuming the probability is well defined&quot; because the mathematical formalisms handling probability tend to not play well with those handling the axiom of choice; in particular, it's not obvious that the set of positions for which this strategy wins is measurable, which would mean no probability could be assigned to it. It's possible that such strategies exist for countably many wizards too)&lt;/p&gt;&#xA;" OwnerUserId="4765" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2016-01-20T23:43:31.617" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25517" PostTypeId="2" ParentId="25512" CreationDate="2016-01-18T22:33:21.673" Score="11" Body="&lt;p&gt;My mouth is a city, all of my kind are full of floating trees&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Portsmouth... see Ports and Mouth :). So the word is ports since the mouth is Portsmouth while the trees are the boats since they float in water at ports. Ports become Sport. Letter: S&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As humans are to the galaxy, dust is to humans&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Small, humans are small compared to the galaxy like how dust is small compared to humans. Or alternatively, SPECK. Now with a shift, we have: malls/pecks. They are both reasonable words though pecks is more related to an animal than malls. Letter: E&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A dirty kick in football is the same as a woodcutter cutting a log and someone logging into someone else's account&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nice job hacking. The word is hacks, which is turned into shacks. Letter: A&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am oil in the sea, chemicals on the lab floor, and water down your shirt&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stain/Spill, as oil spills, spilled chemicals, and of course spilled water is a stain to the respective object. With a shift to the left one, we get: tains/pills. Pills is an actual word so that is probably the right one. Letter: L&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I move up at the start I will be considered rude, but at the end, you will set me rolling&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To slink means to sneak, or in this case sneak up at the start, which is quite rude. But of course, at the end, the person slinking will be running at fast as they can as they have been caught. Slink becomes links so our final letter is: S &lt;br&gt; Or you could also use the free answer shown in the comment by the OP: @PythonMaster for all except the last letter. It is so much simpler than everyone thinks, UP at the START = rude. But UP at the end will START it. Or START-UP. So the word is literally in the question, just minus the ups. The word us start which turns into tarts. Yummy!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So we have:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SEALS, Which must be seals. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9042" LastEditorUserId="9042" LastEditDate="2016-01-23T14:50:48.007" LastActivityDate="2016-01-23T14:50:48.007" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="25518" PostTypeId="2" ParentId="25491" CreationDate="2016-01-18T23:41:41.847" Score="7" Body="&lt;p&gt;It is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;possible!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Let $M\ge5$ be some multiple of $N$ such that $M-1$ is prime. (This is possible by &lt;a href=&quot;https://en.wikipedia.org/wiki/Dirichlet&amp;#39;s_theorem_on_arithmetic_progressions&quot;&gt;Dirichlet's theorem&lt;/a&gt;.)&lt;br&gt;&#xA;&lt;br&gt;&#xA;Consider the polynomial $x^{M-1}+(1-x)^{M-1}-\frac{1}{M}$. The value of this polynomial is positive for $x=0$ and negative for $x=\frac{1}{2}$, so it must have a real root in the interval $(0,\frac{1}{2})$. Choose $p$ as such a root.&lt;br&gt;&#xA;&lt;br&gt;&#xA;Now flip the coin $M-1$ times. The probability that it comes up all heads or all tails is $p^{M-1}+(1-p)^{M-1}=\frac{1}{M}$.&lt;br&gt;&#xA;Otherwise, there are $j$ heads for some $0&amp;lt;j&amp;lt;M-1$. For every $j$, there are $M-1\choose j$ equally likely ways to get $j$ heads. Because $M-1$ is prime, it divides $M-1\choose j$, so these cases can be divided into $M-1$ equal groups (one way to do this is to sum the indices of the flips that came up heads and reducing mod $M-1$). By putting together one group from each $j$, we can make $M-1$ equal sets of results, which must each have probability $\frac{1}{M}$.&lt;br&gt;&#xA;&lt;br&gt;&#xA;We have now chosen equally from $M$ items. Because $M$ is a multiple of $N$, we can combine probabilities if necessary to choose equally from $N$ items.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2016-01-19T00:08:41.117" LastActivityDate="2016-01-19T00:08:41.117" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25519" PostTypeId="2" ParentId="25491" CreationDate="2016-01-19T00:08:51.817" Score="4" Body="&lt;p&gt;It is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;possible!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is not my own work, I got this answer from &lt;a href=&quot;https://math.stackexchange.com/a/1131658/177399&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say we flip the coin $F$ times. For each $k$, we will take the $\binom{F}k$ coin flip sequences which have exactly $k$ heads and divide these as evenly as possible among the first $N-1$ outcomes. Since $\binom{F}k$ may not be divisible by $N-1$, there might be some remainder $r_k$ left over; these remaining $r_k$ sequences will be assigned to the last outcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By construction, the first $N-1$ outcomes are all equally likely. Let's say the coin is biased so that $P($heads$)=b$. The probability of the last outcome, let's call it $R(b)$, is given by adding up all the probabilities of the leftover sequences:&#xA;$$&#xA;R(b)=\sum_{k=0}^Fr_kb^k(1-b)^{F-k}&#xA;$$&#xA;Note that $R(0)=1$, and&#xA;$$&#xA;R(1/2)=\sum_{k=0}^F r_k(1/2)^F\le FN(1/2)^F&#xA;$$&#xA;since the remainders $r_k$ are all at most $N$. This means that we can choose $F$ large enough so $R(1/2)&amp;lt;\frac1N$. Then, by the Intermediate Value Theorem, we can choose a value $b$ between $0$ and $1/2$ for which $R(b)=\frac1N$.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="-1" LastEditDate="2017-04-13T12:19:36.950" LastActivityDate="2016-01-19T00:08:51.817" CommentCount="0" CommunityOwnedDate="2016-01-19T00:08:51.817" ContentLicense="CC BY-SA 3.0" />
  <row Id="25521" PostTypeId="2" ParentId="25477" CreationDate="2016-01-19T01:39:38.773" Score="5" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Sulfur monoxide&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;I am a gas that you love&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Chemical formula SO, also short for Significant Other&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;most commonly found up above&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Sulfur monoxide has been detected in space, but is otherwise rarely encountered intact&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;I have no hues&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Sulfur monoxide is a colourless gas.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;and convert into twos&#xA;when you trap me and give me a shove&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; When concentrated or condensed, it converts to S2O2&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; Source: &lt;a href=&quot;https://en.wikipedia.org/wiki/Sulfur_monoxide&quot; rel=&quot;noreferrer&quot;&gt;Sulfur monoxide - Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14219" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-01-19T01:39:38.773" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25522" PostTypeId="2" ParentId="25505" CreationDate="2016-01-19T02:26:56.653" Score="3" Body="&lt;p&gt;An incomplete attempt at the clues:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Hey, baby, won't you look my way, I could be your new addiction.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; These are lyrics from Everybody Talks by &lt;strong&gt;Neon Trees&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;I apologize if I'm not making much sense, because there is an annoying insect in my room. It sounds like haemophilia, but it shares it's roots with anthropology. I have Nikolai to blame for that. Look, out, Gansey!!&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not too sure about this, but the annoying insect could refer to the &lt;strong&gt;Bee&lt;/strong&gt;, from The Flight of the Bumblebee by Nikolai Rimsky-Korsakov. Don't know how haemophilia, anthropology and Gansey fit into this though.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;there ain't no rest for the wicked.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is a song by &lt;strong&gt;Cage The Elephant&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;homeomorphic toroidal polyhedrons&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; These are &lt;strong&gt;Donut&lt;/strong&gt; shapes. With the new edit, I'm inclined to think the letter is O.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;I'd rather tell you how I'm feeling, boy, and how I'm doing things my way. If I'd got my piano, then I'd know things were okay.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is a song lyric from Piano by &lt;strong&gt;Ariana Grande&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;6&quot;&gt;&#xA;&lt;li&gt;Fabian and Gideon can tell you&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fabian and Gideon &lt;strong&gt;Prewett&lt;/strong&gt; are from the Harry Potter universe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is where I'm stumped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With @AJL's comment, I do notice that there are a lot of elements involved here.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. &lt;strong&gt;NE&lt;/strong&gt;on, 2. &lt;strong&gt;B&lt;/strong&gt;oron or &lt;strong&gt;Be&lt;/strong&gt;ryllium, 3.??? 4.&lt;strong&gt;O&lt;/strong&gt;xygen? 5.&lt;strong&gt;Ar&lt;/strong&gt;gon, 6.&lt;strong&gt;P&lt;/strong&gt;hosphorus&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Still have no idea what 3 is on about, nor how these elements come into play.&#xA;Feel free to use my answers if you can solve it, anyone!&lt;/p&gt;&#xA;" OwnerUserId="14095" LastEditorUserId="14095" LastEditDate="2016-01-20T01:52:06.070" LastActivityDate="2016-01-20T01:52:06.070" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25523" PostTypeId="2" ParentId="25481" CreationDate="2016-01-19T04:42:44.693" Score="0" Body="&lt;p&gt;I'm going to say it's not possible, given these assumptions about the rules are correct:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Each wizard can see every other wizard's hat. Apart from that, no knowledge or information may be imparted to one between &lt;strong&gt;getting a hat&lt;/strong&gt; and &lt;strong&gt;making a guess&lt;/strong&gt;. The wizards aren't able (or allowed) to notice any other details about the other wizards, like for example whose hat they're looking at.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The coin flips (&amp;amp; therefore the hats assigned) truly are completely random &amp;amp; independent.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If these facts are true... each wizard receives absolutely no relevant information about their own hat before making his or her guess. Every wizard who guesses really is &lt;em&gt;guessing&lt;/em&gt;, and has a 50% chance of being correct. Any non-infinite number of guesses fails to satisfy the first condition of victory, so [edited replacing my 'infinite number of guesses' bit which dshin responded to]...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability of two unrelated events A and B occurring is P(A) * P(B). The accuracy (even if not the contents) of a wizard's guess is not related to that of another wizard's guess. Therefore as the number of wizards approaches infinity, the probability of all their guesses being correct -- P(A Wizard Guesses Correctly) ^ Number of Wizards Guessing -- approaches 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Short answer: No. The wizards lose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EXCEPT I'M A BIG DUMMY AND&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i did NOT do my hat-counting puzzle research&lt;/p&gt;&#xA;&#xA;&lt;p&gt;also infinity is weird and i forgot to take that into account as well&lt;/p&gt;&#xA;" OwnerUserId="18551" LastEditorUserId="18551" LastEditDate="2016-01-19T11:27:26.570" LastActivityDate="2016-01-19T11:27:26.570" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25524" PostTypeId="1" AcceptedAnswerId="25538" CreationDate="2016-01-19T06:18:49.177" Score="40" ViewCount="5304" Body="&lt;p&gt;You are blindfolded and disoriented, standing exactly 1 mile from the Great Wall of China. How far must you walk to find the wall?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the earth is flat and the Great Wall is infinitely long and straight.&lt;/p&gt;&#xA;" OwnerUserId="18525" LastEditorUserId="18525" LastEditDate="2016-01-19T17:19:54.243" LastActivityDate="2016-06-21T19:22:48.863" Title="Blindfolded and disoriented near the Great Wall of China" Tags="|geometry|" AnswerCount="5" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="25525" PostTypeId="2" ParentId="25524" CreationDate="2016-01-19T06:23:59.390" Score="14" Body="&lt;p&gt;If the angle between the possible wall and the initial line is $x$ (the angle between the diagonal line and the bottom line in the diagram below), then the distance travelled is $1+(\pi/2+2x)+1/tan(x)+1/sin(x)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gratifyingly this gives a slightly improved answer of $2+3\pi/2\approx6.7124$ for my first attempt (because you can drop down straight rather than complete the circle), where $x=\pi/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also gives my second attempt for $x=\pi/4$ (answer $2+\sqrt{2}+\pi\approx6.5558$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Throwing the expression into &lt;a href=&quot;https://www.wolframalpha.com/input/?i=d%2Fdx%281%2Bpi%2F2%2B2x%2B1%2Ftan%28x%29%2B1%2Fsin%28x%29%29%3D0&quot; rel=&quot;nofollow noreferrer&quot;&gt;wolfram alpha&lt;/a&gt;, shows that a minimum occurs at $\pi/3$.  This gives a value of $1+\sqrt{3}+7\pi/6\approx6.397$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Old new upper bound: $2+\sqrt{2}+\pi$ as per diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WJTvu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WJTvu.png&quot; alt=&quot;Walk near the wall&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(Old upper bound: $2\pi+1$ miles. Walk 1 mile in any direction and then walk is a circle of radius 1, centred at your starting point. )&lt;/p&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2016-01-19T19:16:46.303" LastActivityDate="2016-01-19T19:16:46.303" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="25529" PostTypeId="2" ParentId="25481" CreationDate="2016-01-19T10:45:40.130" Score="5" Body="&lt;p&gt;This solution is based on the Jose's answer, but it construct a strategy with an arbitrary win chance.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Winning with an arbitrary win chance&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/users/4551/xnor&quot;&gt;xnor&lt;/a&gt; showed int &lt;a href=&quot;https://puzzling.stackexchange.com/a/15942/18537&quot;&gt;this post&lt;/a&gt; that a group with $2^i-1$ person can achieve a win chance of $1-\frac{1}{2}^i$.&lt;br&gt;&#xA;Or in other words for all $p\in [0,1)$ there exists an group size $n$ so that the group has a win chance of at least $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we make infinity many groups numbered from 1 onward. The $i$'th group should have a win chance of at least $p_i = (1-\epsilon)^{\frac{1}{2}^i}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To achieve that the groups must have $2^{\lceil log_2(\frac{1}{1-p_i}) \rceil}-1$ member.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability that all groups win is at least&#xA;$$\prod_{i=1}^\infty p_i=\prod_{i=1}^\infty (1-\epsilon)^{\frac{1}{2}^i} =(1-\epsilon)^{\sum_{i=1}^\infty\frac{1}{2}^i} = (1-\epsilon)^1 = 1-\epsilon$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for all $\epsilon&amp;gt;0$ there is strategy that has a fail chance at most of $\epsilon$; or in other words: There is a strategy that has a win chance as close to 100% as you want (at least if you want it to be lower than 100%).&lt;/p&gt;&#xA;" OwnerUserId="17807" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.923" LastActivityDate="2016-01-20T18:01:21.157" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25530" PostTypeId="2" ParentId="25509" CreationDate="2016-01-19T11:02:42.533" Score="0" Body="&lt;p&gt;Let's assume the number is 1234&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First guess will be 7890, response 0 (we know 7890 aren't included)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second guess will be 5612, response -2 (two of 5612 are included, don't know where)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third guess will be 3456, response -2 (two of 3,4,5,6 are included. Since we know from guess 1 that 7,8,9 and 0 don't feature, the other two digits must be 1 and 2. This means that keeping 5 and 6 from the second guess was incorrect, and going forward we need 1,2,3 and 4 in some order)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fourth guess - 4,3,2,1, response is -4 (from guess 2, we now know that 1 and 2 are not the last 2 digits)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fifth - 2,1,4,3, response is -4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sixth - 1,2,3,4, response is 4 and we win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This assumes that no digits in the answer can be repeated and that there isn't a worse way to mix things up between steps 4 and 5.&lt;/p&gt;&#xA;" OwnerUserId="11717" LastEditorUserId="11717" LastEditDate="2016-01-19T11:37:24.397" LastActivityDate="2016-01-19T11:37:24.397" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25531" PostTypeId="2" ParentId="25509" CreationDate="2016-01-19T11:14:39.987" Score="1" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 11&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using a script to simulate games where one picks randomly but draws perfect conclusions (excludes all possible configurations that can be eliminated based on previous guesses) given in javascript &lt;a href=&quot;https://jsfiddle.net/zttfzLL8/14/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; as determined over runs of 10 million.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing this to the normal version with 6 pegs we see that the average solve time with random guessing increases from 4.638 to 6.05 even though the amount of initial possibilities goes from 1296 to 9000.&lt;/p&gt;&#xA;" OwnerUserId="16846" LastEditorUserId="16846" LastEditDate="2016-01-20T10:08:55.680" LastActivityDate="2016-01-20T10:08:55.680" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25533" PostTypeId="2" ParentId="25481" CreationDate="2016-01-19T12:54:00.213" Score="-1" Body="&lt;p&gt;There is only one wizard, but being surounded by cleverly arranged mirrors, it seems like there is an infinity of him.&#xA;As soon as he get his hat every copy will get the same, and he will win.&#xA;Any kind of mirrory solution can work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the same idea, as wizards, they might use spell to see through their colleagues eyes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that as they can't move nor communicate once the hats are on, and as the hats are distributed simultaneously, they can't get any significant hint on what they have, as two neighbours can have black hats as well as whites. So you are stuck with 1/2 per wizard, which will descend to 0 as you reach an infinity of wizard. So they have to cheat one way or another.&lt;/p&gt;&#xA;" OwnerUserId="18557" LastEditorUserId="18557" LastEditDate="2016-01-19T13:42:49.310" LastActivityDate="2016-01-19T13:42:49.310" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="25537" PostTypeId="2" ParentId="25524" CreationDate="2016-01-19T16:34:42.273" Score="0" Body="&lt;p&gt;@DrXorile is clos to the answer. Mine isn't an answer either but here's some food for thoughts&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wanted to picture it. It looks like it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take 360 individuals, all starting at the center of the circle and each at a different angle, only one will find the wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a 0.27% chance of finding the wall if you walk eaxtly one mile. &#xA;If you need to reach the wall for your survival, you're dead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also imagine the guy who started just one degree slightly off, extend his hands  and the wall is just 2 inches further, then starts over in the wrong direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jbeSj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jbeSj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Walking more than one mile means we could increase our chances of reaching the wall at slightly off angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CzDVd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CzDVd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then again, this could happen:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UnM7M.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UnM7M.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18561" LastEditorUserId="18561" LastEditDate="2016-01-19T16:46:35.203" LastActivityDate="2016-01-19T16:46:35.203" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="25538" PostTypeId="2" ParentId="25524" CreationDate="2016-01-19T17:33:24.370" Score="32" Body="&lt;p&gt;$\DeclareMathOperator{\arcsec}{arcsec}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each possible orientation of the wall (relative to some arbitrary initial orientation), the point on the wall closest to our starting point is a distance $1$ away.  The collection of the closest points for all possible orientations of the wall form a circle of radius $1$ around our starting point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we move a distance $r&amp;gt;1$ away from the initial point, we intersect two orientations of the wall that are an angle $\theta$ apart.  In order to reach that point we must have crossed all of the orientations in that angle.  In the figure below on the left, those &quot;explored&quot; points are marked by a magenta line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/H3H3i.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By trigonometry we can show that $\theta = 2\arcsec r$.  If we traverse the path shown on the right side of the above figure, we travel a worst-case distance of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;r + r(2\pi - \theta) \\&#xA;r + 2r(\pi - \arcsec r)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This distance is minimized when $r\approx 1.04356$ for a worst-case distance of $6.99528$, an improvement of about $3.95\%$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, looking at the figure we can immediately see that the majority of the large circular arc is &quot;wasted&quot; distance.  Only the ends contribute to additional &quot;explored&quot; points.  If we shrink-wrap the rest of the path around the unit circle, we get the following path:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nMc0H.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The worst-case distance of this path is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;r + 2\sqrt{r^2-1} + (2\pi - 2\theta) \\&#xA;r + 2\left(\sqrt{r^2-1} + \pi - 2\arcsec r\right)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This happens to be minimized for $r = \sqrt{\frac{15-\sqrt{33}}{6}} \approx 1.24200$ (not the distance shown in the figure), for a worst-case distance of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sqrt{\frac{9+\sqrt{33}}{2}}+4\arctan&#xA;   \sqrt{\frac{9+\sqrt{33}}{8}}&#xA;\approx 6.45891&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;an improvement of $11.32\%$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Update&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Thanks to Michael Seifert for pointing out that we can do better by letting the radii of the start and end be different, in which case we have the distance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;r_1 + \sqrt{r_1^2-1} + \sqrt{r_2^2-1} + 2\pi - \theta_1 - \theta_2 \\&#xA;r_1 + \sqrt{r_1^2-1} + \sqrt{r_2^2-1} + 2\pi - \arcsec r_1 - \arcsec r_2&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is minimized by $r_1=2/\sqrt{3},\ r_2=\sqrt{2}$ (with $\theta_1=\pi/3,\ \theta_2=\pi/2$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zO0fR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Because of the nice angles, this picture is exactly to scale.)  The worst-case distance here is simply&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{2}{\sqrt{3}} + \frac{1}{\sqrt{3}} + \frac{2\pi}{3} + \frac{\pi}{2} + 1 \\&#xA;= 1 + \sqrt{3} + \frac{7\pi}{6}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a $12.16\%$ improvement.)&lt;/p&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="9718" LastEditDate="2016-01-19T19:03:22.380" LastActivityDate="2016-01-19T19:03:22.380" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="25543" PostTypeId="2" ParentId="25524" CreationDate="2016-01-19T20:06:32.480" Score="2" Body="&lt;p&gt;&quot;...standing exactly 1 mile from the Great Wall of China. How far must you walk to find the wall?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You &lt;strong&gt;must&lt;/strong&gt; walk 1 mile.  If you go the wrong way then you will end up walking further.  If you don't walk that far you can't reach it.&lt;/p&gt;&#xA;" OwnerUserId="4372" LastEditorUserId="4372" LastEditDate="2016-06-21T19:22:48.863" LastActivityDate="2016-06-21T19:22:48.863" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="25544" PostTypeId="1" CreationDate="2016-01-19T20:21:38.773" Score="0" ViewCount="87" Body="&lt;p&gt;Found this one in my son's puzzle book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are a cyborg in a death match with two other cyborgs. All three of you have identical blasters, and one single shot from these blasters will destroy any of you. You have a 33% (read as 1/3) chance of hitting your target per shot. The next cyborg has a 50% chance. The next cyborg will always hit his target.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this match the cyborgs take turns shooting, one shot per round. The weakest accuracy (you) goes first, on up to the strongest. If a cyborg is eliminated he won't get another turn. You are all logical and will try to win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is your optimum first move? With this move what are your odds of being the final survivor?&lt;/p&gt;&#xA;" OwnerUserId="2614" LastActivityDate="2016-01-19T20:21:38.773" Title="Cyborg death match: Make your move" Tags="|logical-deduction|" AnswerCount="0" CommentCount="2" ClosedDate="2016-01-19T20:30:46.240" ContentLicense="CC BY-SA 3.0" />
  <row Id="25545" PostTypeId="2" ParentId="25158" CreationDate="2016-01-19T22:06:28.163" Score="9" Body="&lt;p&gt;Jill has ranked&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the US Interstate Highways&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;length.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Suzie has ranked them by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;number of states passed through, and then by age as a tiebreaker.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The problem is that &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;some Interstates pass through the same number of states, and were formed in the same year. Because these are the only statistics provided other than the length, Suzie has no other way to order them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2016-01-19T22:06:28.163" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25546" PostTypeId="1" CreationDate="2016-01-20T00:47:01.480" Score="3" ViewCount="71" Body="&lt;p&gt;Each cell in an infinite square lattice contains an arrow ponting to one of the eight cells adjacent to it. No two adjacent cells contain arrows differing in direction by more than 45°.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose (starting anywhere) we trace a path by travelling in the direction of each arrow we hit. Prove that we never visit the same cell twice. &lt;/p&gt;&#xA;" OwnerUserId="18508" LastActivityDate="2016-01-20T00:47:01.480" Title="Drifting in an ocean of arrows" Tags="|mathematics|" AnswerCount="0" CommentCount="3" ClosedDate="2016-01-20T06:04:05.203" ContentLicense="CC BY-SA 3.0" />
  <row Id="25549" PostTypeId="2" ParentId="25491" CreationDate="2016-01-20T03:05:39.147" Score="2" Body="&lt;p&gt;Let's start with the algorithm: throw the coin $N-1$ times and choose the $i$-th element ($1\leq i\leq N-1$) if and only if the $i$-th throw was a Head and the rest were a Tail. Repeat, if no one was chosen. If after $k$ repeats no one was chosen we choose the $N$-th element. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need to show that we can choose $p$ (the probability for a Head) and $k$ such that the probability to choose the $N$-th element is $\frac{1}{N}$ (clearly then, the probability to choose any other element is also $\frac{1}{N}$). Now, if we work out the details, the probability to choose the $N$-th element is:&#xA;$$\left[1 - (N-1)p(1-p)^{N-2}\right]^k$$&#xA;Consider it as a polynomial in $p$. We look to fix $k$ such that there will be two points that evaluate to values above and below $\frac{1}{N}$ (so we can apply the intermediate value theorem). For $p=0$ it evaluates to $1$ (irrespective of $k$). For $p=\frac{1}{N-1}$ the value within the square brackets is less than $1$ so by taking $k$ large enough we can bring it below $\frac{1}{N}$ as needed.&lt;/p&gt;&#xA;" OwnerUserId="18521" LastEditorUserId="18521" LastEditDate="2016-01-20T06:36:45.337" LastActivityDate="2016-01-20T06:36:45.337" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25550" PostTypeId="2" ParentId="25505" CreationDate="2016-01-20T03:11:24.747" Score="6" Body="&lt;p&gt;Think I have most of it&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The clues are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lyrics from a song by &lt;strong&gt;Ne&lt;/strong&gt;on Trees&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;B&lt;/strong&gt;umblebee is the insect. It's also a song by Nikolai Rimsky-Korsakov. Gansey is apparently a character from a book series called The Raven Cycle and he is allergic to bees and wasps.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A song by &lt;strong&gt;Ca&lt;/strong&gt;ge the Elephant&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's an &lt;strong&gt;O&lt;/strong&gt; shape&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Song by &lt;strong&gt;Ar&lt;/strong&gt;iana Grande&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Last name of Fabian and Gideon &lt;strong&gt;P&lt;/strong&gt;rewett (Thanks @AZGreenTea!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All the clues are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Elements from the periodic table of elements. 1-Neon 2-Boron 3-Calcium 4-Oxygen 5-Argon 6-Phosphorus&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Their atomic  numbers are 10, 5, 20, 8, 18, and 15 which can be translated to their corresponding letter in the alphabet (a=1, b=2, etc.) which comes out to JETHRO. There aren't any anagrams of Jethro, so is the answer Jethro's Daughter? Maybe Gael Anderson, daughter of Ian Anderson front man of the band Jethro Tull?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The last line about the potatoes seems to be a Tumblr meme, not sure if it's a clue or not.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;EDIT 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; @AZGreenTea pointed out that it might be Zipporah, daughter of Reuel and Jethro and wife of Moses. Moses isn't going to be happy if he hears about this!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18570" LastEditorUserId="18570" LastEditDate="2016-01-20T05:49:16.340" LastActivityDate="2016-01-20T05:49:16.340" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25551" PostTypeId="1" AcceptedAnswerId="25628" CreationDate="2016-01-20T03:48:52.690" Score="27" ViewCount="1916" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OTNbE.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OTNbE.png&quot; alt=&quot;A red square. A horizontal rectangle, with three vertical bands (narrow cyan/wide green/narrow cyan), a tall vertical yellow rectangle, a vertical rectangle with three equal horizontal bands (magenta/blue/magenta). An uncoloured hexagon.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Given the colours and banding of the first four rectangles, how should you colour the hexagon, and why?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There's a single, unambiguous solution (ignoring symmetry), which will be obvious when found. Nothing is hidden in the image, you could print it out and still solve it with nothing more than your eyes and some lateral thinking.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;strong&gt;Edit&lt;/strong&gt;: I intended this to be solved by eye, and so my image wasn't drawn 100% accurately. However, since a few people are approaching things somewhat mathematically (which is admittedly a valid approach), I've checked my calculations and updated the image slightly. It shouldn't make much difference in solving it, but I believe the maths is now correct to the nearest pixel.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="17250" LastEditDate="2016-03-04T12:53:43.040" LastActivityDate="2016-03-04T12:53:43.040" Title="How should the hexagon be coloured?" Tags="|pattern|visual|" AnswerCount="4" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="25553" PostTypeId="1" AcceptedAnswerId="25586" CreationDate="2016-01-20T07:29:50.937" Score="8" ViewCount="372" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;There is a polygon whose edge lengths are all odd integers. Prove that this polygon's interior cannot be tiled by dominoes whose dimensions are $1\times 2$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An example of such a polygon is a &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Mutilated_chessboard_problem&quot;&gt;mutilated chessboard&lt;/a&gt;,&quot; which is a chessboard with two opposite corners removed.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="8874" LastEditDate="2016-03-21T13:24:08.760" LastActivityDate="2016-03-21T13:24:08.760" Title="Tiling an Odd Polygon with Dominoes" Tags="|dissection|tiling|checkerboard|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25557" PostTypeId="2" ParentId="24961" CreationDate="2016-01-20T10:30:55.093" Score="3" Body="&lt;p&gt;The murderer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The catholic cardinal&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The cardinal was born in this kingdom, now lives on Vatican but comes back to the reign once a year: for Easter Sunday to host a special preach.&lt;br&gt;&#xA; He killed the prince on Easter Sunday.&lt;br&gt;&#xA; The wiseman hint points not only for Sunday, but for today, this special Sunday. He &quot;sherlockly&quot; deducted it from the cardinal was the one the king couldn´t ask about the murder because he already left for Vatican.&lt;br&gt;&#xA; &quot;A year later, the king &lt;strong&gt;had solved&lt;/strong&gt; the mystery and swiftly executed the  murderer&quot;&lt;br&gt;&#xA; After a year, he had it already solved, but had to wait for the cardinal to return, the king couldn't raid the Vatican for diplomatic reasons.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think that wraps it all.&lt;/p&gt;&#xA;" OwnerUserId="18576" LastEditorUserId="18576" LastEditDate="2016-01-21T10:16:43.763" LastActivityDate="2016-01-21T10:16:43.763" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25558" PostTypeId="1" CreationDate="2016-01-20T11:02:09.233" Score="1" ViewCount="202" Body="&lt;p&gt;In a closed room, paper slips labelled $1$ to $100$ are randomly put into boxes also labeled $1$ to $100$. You win a big prize if you can locate a given number (from $1$ to $100$) by opening no more than $50$ boxes. As stated, your odds of winning the prize are, clearly, $50$%. Now, suppose you have an accomplice that can go into the room just before the boxes are being sealed and secretly switch the content of two boxes (due to time constraint, this all he can do). By how much your odds of winning can be improved?&lt;/p&gt;&#xA;" OwnerUserId="18521" LastActivityDate="2016-01-20T12:53:55.343" Title="Finding a random number in a randomly assigned number in boxes" Tags="|mathematics|probability|" AnswerCount="2" CommentCount="2" ClosedDate="2016-01-20T12:51:46.193" ContentLicense="CC BY-SA 3.0" />
  <row Id="25560" PostTypeId="2" ParentId="25558" CreationDate="2016-01-20T11:55:50.973" Score="1" Body="&lt;p&gt;Your odds are increased to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 100%&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your strategy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Open the box with the same number as your given number. See what number is in there and go to that box number, and repeat. Eventually it must loop back to the first number. Without accomplice this loop could be all 100 boxes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What your accomplice does:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; break the greatest loop there is in two smaller loops half the size. That way there won't be any loops larger than 50 and you always get your number&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2016-01-20T11:55:50.973" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25561" PostTypeId="2" ParentId="25558" CreationDate="2016-01-20T12:43:24.663" Score="0" Body="&lt;p&gt;firstly, you and your accomplice do not know what number is going to be given. so he/she can open a box, let say it is box #1 and the number is 10 in the box. then he can open another box which is the same number in the previous box, for this case, box #10. and check the number in the box and switch it with 1 whatever (s)he gets, let say he got 50. so u guarantee that at least one of the number and box number will surely be the match and it is not the one in the box #1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;firstly the chance of getting 1 from #1 is 1/100 for your accomplice, and he will not switch then, not getting 1 and gettin one from the box #10 is 99/100*1/99 = again 1/100. and not getting 1 from both then is 98/100 as a result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so ur chance of getting one from number one is 2/100, and u know that at least one of the box has the same number as box and open the given number then. the chance of given number is different than your switched box is 99/100. and the chance of getting the same box as u switch is 1/99. so another 1/100 from here added.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as a result, take number 1, then take the number given, then take box as the same number in box#1 (u know that box has different number for sure), take always the box with the number in the previous box always not to encounter the box and the number different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so 1-98/100*98/99*97/98.... (50 times) = the new outcome ( edited: it will be much more complicated since i changed the algorithm as taking box according to previous nnumber in the previous box)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;typing on the phone, so cant calculate :(&lt;/p&gt;&#xA;" OwnerUserId="18514" LastEditorUserId="18514" LastEditDate="2016-01-20T12:53:55.343" LastActivityDate="2016-01-20T12:53:55.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25562" PostTypeId="2" ParentId="25512" CreationDate="2016-01-20T13:09:19.730" Score="5" Body="&lt;p&gt;Q1. Left 1. My mouth is a city, all of my kind are full of floating trees&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I don't have this yet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Q2. Left 1. As humans are to the galaxy, dust is to humans&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Speck --&gt; Pecks&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Q3. Right 1. A dirty kick in football is the same as a woodcutter cutting a log and someone logging into someone else's account&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hacks --&gt; Shack&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Q4. Left 1. I am oil in the sea, chemicals on the lab floor, and water down your shirt&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Spill -&gt; Pills&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Q5. Left 1. If I move up at the start I will be considered rude, but at the end, you will set me rolling&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Don't have it yet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Making the grid&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;?????&lt;br/&gt;&#xA;PECKS&lt;br/&gt;&#xA;SHACK&lt;br/&gt;&#xA;PILLS&lt;br/&gt;&#xA;?????&lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Making the animal&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ?EAL? - perhaps SEALS? Using S at beginning and end of clue words is a definite pattern here, and one that makes the left/right thing easier.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1509" LastActivityDate="2016-01-20T13:09:19.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25563" PostTypeId="1" AcceptedAnswerId="46233" CreationDate="2016-01-20T13:11:20.287" Score="10" ViewCount="16095" Body="&lt;p&gt;I'm not sure how to count permutations in the 15 puzzle.  They say that an odd count means it's impossible but not sure how to count this puzzle.  How do I count whether this is even or odd?  If it is solvable how can I solve?  The last two tiles always seems to be swapped in the wrong position no matter how much I slide it around.  I've been keeping the moves to the bottom half, do I need to do a big permutation around the whole puzzle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my grid results&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;| 1  | 2  | 3  | 4  |&#xA;| 5  | 6  | 7  | 8  |&#xA;| 9  | 10 | 11 | 15 |&#xA;| 13 | 14 | 12 |    |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here's what I'm referencing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.ubc.ca/~cass/courses/m308-02b/projects/grant/fifteen.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.math.ubc.ca/~cass/courses/m308-02b/projects/grant/fifteen.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18577" LastActivityDate="2018-08-30T23:22:27.010" Title="Do I have an unsolvable 15 puzzle?" Tags="|sliding-blocks|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25564" PostTypeId="1" AcceptedAnswerId="25566" CreationDate="2016-01-20T13:12:08.300" Score="3" ViewCount="198" Body="&lt;p&gt;You are walking on the street and notice a guy gambling with some other guys and you start to watch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The game principle is pretty easy. There is a standard 52-card deck and it is being shuffled every time before the game begins. You take five random cards and if at least two of them has the same rank (2,3..., J,Q,K,A) you win, otherwise you lose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the long run, you should play or not? (with probabilities of course)&lt;/p&gt;&#xA;" OwnerUserId="18514" LastEditorUserId="16322" LastEditDate="2016-01-20T14:46:55.030" LastActivityDate="2016-01-20T14:46:55.030" Title="Gambling with the gambler" Tags="|probability|cards|" AnswerCount="1" CommentCount="1" ClosedDate="2016-01-24T05:57:18.250" ContentLicense="CC BY-SA 3.0" />
  <row Id="25565" PostTypeId="2" ParentId="25553" CreationDate="2016-01-20T13:15:23.377" Score="2" Body="&lt;p&gt;If the polygon is &lt;strong&gt;allowed to contain holes&lt;/strong&gt;, then there are counter examples:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    XXX&#xA;    X X&#xA;    XXX&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This polygon has only sides of length 3 (outer sides) and length 1 (sides of hole), and it is easy to tile it with four dominoes:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    AAB&#xA;    D B&#xA;    DCC&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="9125" LastActivityDate="2016-01-20T13:15:23.377" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25566" PostTypeId="2" ParentId="25564" CreationDate="2016-01-20T13:26:31.370" Score="6" Body="&lt;p&gt;The probability of the first two cards &lt;em&gt;not&lt;/em&gt; matching is $\frac{48}{51}$. Given them not matching, the probability of the next card matching neither is $\frac{44}{50}$, and so on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our total probability of no pairs, then, is $\frac{48 \times 44 \times 40 \times 36}{51\times 50\times 49\times 48}=.507$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your odds of losing are slightly better than half, and you should not take the bet.&lt;/p&gt;&#xA;" OwnerUserId="3987" LastEditorUserId="3987" LastEditDate="2016-01-20T14:39:02.400" LastActivityDate="2016-01-20T14:39:02.400" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25567" PostTypeId="1" AcceptedAnswerId="25569" CreationDate="2016-01-20T14:58:08.080" Score="10" ViewCount="423" Body="&lt;p&gt;I currently stand on point A, somewhere in Europe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I do my sequence here and if I start with 25, I get the following sequence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;25, 13, 7, 5, 4, 4, 4, 4, x, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I step a few metres to the Southwest,  I get the following sequence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;25, 10, 3, 5, 4, 6, 3, 5, y, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I step a few metres to the East,  I get the following sequence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;25, 14, 8, 4, 4,  4, 4, 4, z, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here $x,y,z$ are yet unknown integers, and the dots signify that the sequence continues indefinitely to the right. You may start to find $x,y,z$, but the harder question and actual puzzle is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Where is point A?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="16322" LastEditorUserId="16322" LastEditDate="2016-01-20T16:14:26.203" LastActivityDate="2016-01-31T18:16:52.073" Title="Sequence dependent on my place" Tags="|number-sequence|language|geography|knowledge|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25568" PostTypeId="2" ParentId="25553" CreationDate="2016-01-20T15:29:07.130" Score="4" Body="&lt;p&gt;The proof can be found in &lt;a href=&quot;http://www.cccg.ca/proceedings/1999/c32.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; (Section 2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The explanation is rather extensive so it would be hard to type it out here. But what they basically do is first color the polygon a chessboard pattern in such a way that at least one corner is black. Then they prove that in that case all corners are black. Then they prove that there are more black than white tiles making it impossible to tile the polygon.&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2016-01-20T15:29:07.130" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25569" PostTypeId="2" ParentId="25567" CreationDate="2016-01-20T15:57:48.503" Score="10" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://www.google.co.uk/maps/place/Hoogste+punt+van+Nederland/@50.7546473,6.0207066,21z/data=!4m2!3m1!1s0x0000000000000000:0xdc0e1b0270fe841c&quot; rel=&quot;nofollow&quot;&gt;Hoogste punt van Nederland&lt;/a&gt;  (Highest point of the Netherlands)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Using the trusty old &quot;count the letters in the number&quot; technique, I have found that Dutch fits the pattern for point A:&lt;br&gt;&#xA; 25 - vijfentwintig&lt;br&gt;&#xA; =&gt; 13 - dertien&lt;br&gt;&#xA; =&gt; 7 - zeven&lt;br&gt;&#xA; =&gt; 5 - vijf&lt;br&gt;&#xA; =&gt; 4 - vier&lt;br&gt;&#xA; =&gt; 4 - vier&lt;br&gt;&#xA; =&gt; 4 - vier&lt;br&gt;&#xA; =&gt; 4 - vier&lt;br&gt;&#xA; =&gt; 4 - vier ....&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the value for x is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Going southwest from Point A takes you to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Belgium where the official language is French&lt;br&gt;&#xA; 25 - vingt-cinq&lt;br&gt;&#xA; =&gt; 10 - dix&lt;br&gt;&#xA; =&gt; 3 - trois&lt;br&gt;&#xA; =&gt; 5 - cinq&lt;br&gt;&#xA; =&gt; 4 - quatre&lt;br&gt;&#xA; =&gt; 6 - six&lt;br&gt;&#xA; =&gt; 3 - trois&lt;br&gt;&#xA; =&gt; 5 - cinq&lt;br&gt;&#xA; =&gt; 4 - quatre ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the value for y is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Stepping East takes you to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Germany and the sequence here is&lt;br&gt;&#xA; 25 - fünfundzwanzig&lt;br&gt;&#xA; =&gt; 14 - vierzehn&lt;br&gt;&#xA; =&gt; 8 - acht&lt;br&gt;&#xA; =&gt; 4 - vier&lt;br&gt;&#xA; =&gt; 4 - vier&lt;br&gt;&#xA; =&gt; 4 - vier&lt;br&gt;&#xA; =&gt; 4 - vier&lt;br&gt;&#xA; =&gt; 4 - vier&lt;br&gt;&#xA; =&gt; 4 - vier&lt;br&gt;&#xA; =&gt; 4 - vier ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the value for z is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="16322" LastEditDate="2016-01-31T18:16:52.073" LastActivityDate="2016-01-31T18:16:52.073" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25570" PostTypeId="2" ParentId="25512" CreationDate="2016-01-20T16:13:47.203" Score="2" Body="&lt;p&gt;Partial answer based on the previous ones :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q1. Left 1. My mouth is a city, all of my kind are full of floating trees&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ports/Sport : Portsmouth is a city. Boats are floating trees&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Q2. Left 1. As humans are to the galaxy, dust is to humans&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Speck/Pecks&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Q3. Right 1. A dirty kick in football is the same as a woodcutter cutting a log and someone logging into someone else's account&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hacks/Shack&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Q4. Left 1. I am oil in the sea, chemicals on the lab floor, and water down your shirt&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Spill/Pills&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Q5. Left 1. If I move up at the start I will be considered rude, but at the end, you will set me rolling&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Still looking...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Giving :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SPORT&lt;br&gt;&#xA; PECKS&lt;br&gt;&#xA; SHACK&lt;br&gt;&#xA; PILLS&lt;br&gt;&#xA; ?????&lt;br&gt;&#xA;&lt;br&gt;&#xA;Still going for SEALS...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the script I used for the first and last question to get five letters word matching the swapping letter requirement :&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;# RUBY&#xA;a = [&quot;word1&quot;, &quot;word2&quot;, &quot;word3&quot;]&#xA;z = []&#xA;a.each do |b|&#xA;    w = b[1..4] + b[0] # for Q5 || w = b[-1] + b[0..3] for Q1&#xA;    if a.find_index(w) != nil&#xA;        z.push(b)&#xA;    end&#xA;end&#xA;puts z&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I used this source for common 5 letters english words : &lt;a href=&quot;http://www.poslarchive.com/math/scrabble/lists/common-5.html&quot; rel=&quot;nofollow&quot;&gt;http://www.poslarchive.com/math/scrabble/lists/common-5.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Giving these lists of words :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Q1 : &lt;a href=&quot;http://pastebin.com/Aa1dBmQy&quot; rel=&quot;nofollow&quot;&gt;http://pastebin.com/Aa1dBmQy&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Q5 : &lt;a href=&quot;http://pastebin.com/ciQmsV9u&quot; rel=&quot;nofollow&quot;&gt;http://pastebin.com/ciQmsV9u&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="18280" LastEditorUserId="18280" LastEditDate="2016-01-21T08:12:24.887" LastActivityDate="2016-01-21T08:12:24.887" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25572" PostTypeId="1" AcceptedAnswerId="25573" CreationDate="2016-01-20T17:30:21.043" Score="7" ViewCount="416" Body="&lt;p&gt;Dearest Zipporah,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I apologize for the delay. I finally deciphered your message. While I had hoped it was you that sent the letter, I was glad to see that it indeed was your father's name that you encoded in your letter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I myself am not quite adept with words, and you aren't either, as been proved in your letter. I regret to inform you that I, in reality, have no potatoes to bring with me on the twenty sixth. Although I know that you love them, I'm pained to admit that my refrigerator broke down and I had to throw all my food out, including the potatoes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the mean time, I tried my best to create a puzzle for you so you can solve it whilst you wait. It is a three word confession that I am quite pained to admit to you. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To write my puzzle, I did a bit of recon and found out what books you've been watching lately. Each word of the sentence is in each of the three paragraphs, one for each one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chemical symbol of the element with the electron configuration of [Kayar] for-dee^tin fireess^too firepea^fire. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The country in the Caribbean that experienced  an earthquake in the year factorized as 2•3•5•67, minus the most used letter in the English alphabet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, what vegetable does Matt Smith liken Sontarans too? Then add an s. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh, the lengths I go to gain your affection, now all in vain. Don't hold this letter as a slight against me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Love, &#xA;Henry&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P. S. Aren't you the one with Moses?&lt;/p&gt;&#xA;" OwnerUserId="11814" LastEditorUserId="11814" LastEditDate="2016-01-20T18:28:12.220" LastActivityDate="2016-01-20T18:28:12.220" Title="Love letter number two" Tags="|riddle|knowledge|wordplay|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25573" PostTypeId="2" ParentId="25572" CreationDate="2016-01-20T17:55:32.273" Score="3" Body="&lt;p&gt;The confession is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I hate potatoes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first one is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I, for Iodine, which has electron configuration $[Kr] 4d^{10}5s^{2}5p^{5}$, which is basically what's written read phonetically and transcribed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second one is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Haiti, minus the final syllable, a long E sound.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The last one is basically a freebie from&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://www.youtube.com/watch?v=YUrMOcjMHoE&quot; rel=&quot;nofollow&quot;&gt;The Snowmen&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17147" LastActivityDate="2016-01-20T17:55:32.273" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25575" PostTypeId="1" AcceptedAnswerId="25576" CreationDate="2016-01-20T18:56:29.350" Score="1" ViewCount="704" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm asked by my University Professor, Mandy, to come and see her in&#xA;  her office after class.  &quot;Somethin' wrong?&quot; I keep thinking when I&#xA;  head to her office.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;You are elected to be our Math club's consultant and I can see why. &#xA;  However,&quot; as Mandy continues, &quot;I will have you solve this question.&#xA;  If you fail to do so, I'll find somebody to take your place.&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;Fair enough.&quot; I think.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;And here's the question:    &#xA;&#xA;e + f + 20 = 2  &#xA;c + d + 88 = 5  &#xA;c + d + e + f = ?  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Before I could write down -101 as the answer, Mandy mentioned the&#xA;  answer is not -101!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;As I struggle, another student who passed by managed to see the question&#xA;  and said &quot;It is too easy&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;Correct.&quot; said Mandy and the student became the Club's consultant&#xA;  instead.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the correct answer and, what happened?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, could you find all the hidden references in the question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I have made a mistake on the question posted and it's been edited.  Unsure if it violate the rule I will go ahead and accept the correct answer&lt;/p&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="10037" LastEditDate="2016-01-21T15:40:57.513" LastActivityDate="2016-01-21T15:59:16.030" Title="It's too easy? It's too easy!" Tags="|riddle|lateral-thinking|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25576" PostTypeId="2" ParentId="25575" CreationDate="2016-01-20T19:57:54.580" Score="4" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2 (&quot;Two, easy!&quot;)&lt;br&gt;&#xA; The answer is the number of circles (closed loops) in each problem - 2 in the first ('e', and the '0' of 20), 5 in the second ('d', and two each in the '8's), 2 in the third ('d' and 'e').&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18582" LastEditorUserId="18522" LastEditDate="2016-01-21T15:59:16.030" LastActivityDate="2016-01-21T15:59:16.030" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25577" PostTypeId="1" AcceptedAnswerId="25578" CreationDate="2016-01-20T20:16:14.633" Score="6" ViewCount="330" Body="&lt;p&gt;&lt;em&gt;In the spirit of the original &lt;a href=&quot;https://puzzling.stackexchange.com/questions/15339&quot;&gt;Four-words&lt;/a&gt; puzzle by &lt;a href=&quot;https://puzzling.stackexchange.com/users/10940&quot;&gt;Prem&lt;/a&gt;, and all those that followed.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My first completely overshadows anything you see.&lt;br&gt;&#xA;  My second's used extensively to mean extensively.&lt;br&gt;&#xA;  My third are pithy aphorisms commonly dispersed.&lt;br&gt;&#xA;  My fourth's what you'll be feeling if Deusovi solves this first.&lt;br&gt;&#xA;  My fifth's predominantly white, though legend says it's golden.&lt;br&gt;&#xA;  My sixth is what will keep the world consistently unfolding.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;My whole's a square that's six by six with letters in each cell.&lt;br&gt;&#xA;  There is no difference in the words read flat or vertical.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2016-01-20T20:43:29.023" Title="A Rhyming Six-Words Puzzle" Tags="|word|wordplay|rhyme|word-square|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25578" PostTypeId="2" ParentId="25577" CreationDate="2016-01-20T20:31:52.833" Score="5" Body="&lt;p&gt;How about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;D W A R F S&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;W I D E L Y&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;A D A G E S&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;R E G R E T&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;F L E E C E&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;S Y S T E M&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; M&lt;b&gt;&lt;i&gt;oooooo&lt;/i&gt;ooooo&lt;/b&gt;m, do I &lt;em&gt;have&lt;/em&gt; to explain every line? FINE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;My first completely overshadows anything you see.&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; aye, something that &lt;em&gt;dwarfs&lt;/em&gt; everything could be said to &lt;em&gt;overshadow&lt;/em&gt; it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;My second's used extensively to mean extensively.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;widely&lt;/em&gt;, even. Judge I object to having to explain this one, res ipsa loquitur&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;My third are pithy aphorisms commonly dispersed.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;adages&lt;/em&gt; is another word for aphorisms&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;My fourth's what you'll be feeling if Deusovi solves this first.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; well, yes, I &lt;em&gt;would&lt;/em&gt; be &lt;em&gt;regret&lt;/em&gt;ful if the god of false-egg-plurals had beaten me to it, but they did not&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;My fifth's predominantly white, though legend says it's golden.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;fleece&lt;/em&gt; is white, but there's that whole &lt;a href=&quot;https://en.wikipedia.org/wiki/Golden_Fleece&quot; rel=&quot;nofollow&quot;&gt;Jason thing&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;My sixth is what will keep the world consistently unfolding.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;systems&lt;/em&gt; are the logic by which things can be said to run&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2016-01-20T20:43:29.023" LastActivityDate="2016-01-20T20:43:29.023" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25579" PostTypeId="1" AcceptedAnswerId="25612" CreationDate="2016-01-20T20:44:39.600" Score="2" ViewCount="1230" Body="&lt;p&gt;What is the missing grid like? I just cannot recognize any rule.&lt;img src=&quot;https://i.stack.imgur.com/Pce1U.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="18447" LastEditorUserId="18447" LastEditDate="2016-01-21T14:15:55.757" LastActivityDate="2016-01-21T14:15:55.757" Title="Grid pattern. Help me to solve the puzzle" Tags="|pattern|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25580" PostTypeId="2" ParentId="25575" CreationDate="2016-01-20T20:59:47.567" Score="2" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;-100&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;Through rearrangement we get&lt;br&gt;&#xA;a + b = -17&lt;br&gt;&#xA;c + d = -83&lt;br&gt;&#xA;therefore a + b + c + d = -17 + -83 = -100&lt;br&gt;&#xA;&lt;br&gt;&#xA; Mandy was testing his confidence and since he hesitated to give the correct answer in the face of adversity she chose a different student who was more confident.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18103" LastEditorUserId="4421" LastEditDate="2016-01-20T21:55:15.110" LastActivityDate="2016-01-20T21:55:15.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25581" PostTypeId="2" ParentId="25551" CreationDate="2016-01-20T21:09:39.347" Score="2" Body="&lt;p&gt;Either Black or White, depending on how you look at them. If they are basic light colors - then seen together they would be seen as white light, but if they are paint pigments - then black. However, because of the background I would bet on white. &lt;/p&gt;&#xA;" OwnerUserId="18582" LastActivityDate="2016-01-20T21:09:39.347" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25582" PostTypeId="1" AcceptedAnswerId="25619" CreationDate="2016-01-20T21:45:00.493" Score="2" ViewCount="614" Body="&lt;p&gt;A man and his wife, completely devoted to each other, have sworn to be teetotal all their lives. One of their vows at the wedding was in fact that if either of them ever drank alcohol they would take their own lives (pretty silly in my opinion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite this vow the man is a brewer - though he never drinks his own brew.&#xA;The man had recently written his will giving everything he owns to his nephew.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;His nephew knows this and comes round their house to celebrate New Year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They both go down into the cellar so the nephew can inspect the latest brew. The man's wife who is washing up sees the nephew come up for some 'apple juice'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shortly after he goes back down, she hears her husband scream. She rushes down and sees this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Her husband is lying dead on the floor next to an open barrel of beer. He has blood on his shirt and a big stone lying on his stomach. His hair is wet (An investigation shows his hair is covered in beer) and he is holding a half-empty cup of &lt;strong&gt;alcoholic&lt;/strong&gt; cider. The floor around the barrel is wet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The nephew claims a rock fell from the ceiling and killed him. But there was no rock missing from the ceiling&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;You&lt;/strong&gt; are called in to investigate.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What happened and how did the man die (There's a clue in the title)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18250" LastEditorUserId="18250" LastEditDate="2016-01-21T15:22:07.213" LastActivityDate="2018-03-27T10:54:25.597" Title="A cellar suicide" Tags="|mystery|" AnswerCount="5" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="25583" PostTypeId="5" CreationDate="2016-01-20T21:46:06.007" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2016-01-20T21:46:06.007" LastActivityDate="2016-01-20T21:46:06.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25584" PostTypeId="4" CreationDate="2016-01-20T21:46:06.007" Score="0" Body="  Used in murder questions where you must find the killer" OwnerUserId="18250" LastEditorUserId="18250" LastEditDate="2016-01-20T22:25:24.540" LastActivityDate="2016-01-20T22:25:24.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25585" PostTypeId="2" ParentId="25582" CreationDate="2016-01-20T22:08:32.363" Score="11" Body="&lt;p&gt;Okay, here's what happened.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A rock fell from the ceiling and killed the man.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But how, exactly?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The man did not properly vent this one barrel. Over time, the carbon dioxide built up enough that there was an enormous amount of pressure placed on the cap of the barrel. This pressure became too much, exploding with a violent gush of liquid gold, causing the man to scream. The barrel cap was thrown against the ceiling with such force that a large rock was dislodged directly above the man, promptly falling on his head and killing him.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This explains the scream, the open barrel, the pooled beer, the beer in the man's hair, and the circumstances of his death. The apple cider was delivered to him by his nephew and was unrelated.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I specifically skipped the obvious answer of &quot;the nephew gave the man alcohol and so he killed himself or the nephew killed him and made it look like he killed himself&quot; because I find it uninteresting. But perhaps that's the right answer?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2016-01-21T15:45:52.263" LastActivityDate="2016-01-21T15:45:52.263" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25586" PostTypeId="2" ParentId="25553" CreationDate="2016-01-20T23:04:02.750" Score="4" Body="&lt;p&gt;Let $P$ be the interior of some polygon in the coordinate plane with each side length odd, $\partial P$ its boundary (which we give the counter-clockwise orientation). We may assume that every edge in $\partial P$ is parallel to one of the coordinate axes, and that the vertices have integer coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the function $U(x,y)=\frac{\pi i}{2}e^{\pi i (x+y)}$. Green's Theorem from calculus says that&#xA;$$&#xA;\tag{$\star$}&#xA;\int_P\frac{\partial U}{\partial x}dx\,dy=\int_{\partial P} U\,dy.&#xA;$$&#xA;A direct computation shows that if $D$ is a domino with sides parallel to the coordinate axes, then&#xA;$&#xA;\int_D \frac{\partial U}{\partial Y}dx\,dy = 0,&#xA;$&#xA;so the left hand side of $(\star)$ is $0$ if $P$ can be tiled by dominos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right hand side of $(\star)$ vanishes on the horizontal edges, and on each vertical edge of odd length the value is $(-1)^{x+y}$, where $(x,y)$ are the coordinates of the final endpoint of that edge. Between any two consecutive vertical edges, there is a horizontal edge of odd length, so the $y$-coordinates of the ending points of two adjacent edges are of opposite parity, as are the $x$-coordinates. This means $(-1)^{x+y}$ has the same sign for each vertical edge, so the right hand side of $(\star)$ is non-zero.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In fact our argument shows the right hand side of $(\star)$ is $\pm\frac{E}{2}$, where $E$ is the number of edges of the polygon $P$. If we color the plane checkerboard fashion, one checks that&#xA;$&#xA;\int_P \frac{\partial U}{\partial x}dx\,dy&#xA;$&#xA;is twice the difference between the number of black squares and the number of white squares in $P$. The argument above then shows that if every side length of $P$ is odd, the difference between the number of black squares and the number of white squares in $P$ is always $\frac{E}{4}$. This gives a different proof of Theorem 3.1 in the paper linked by Ivo Beckers.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2016-01-20T23:20:15.933" LastActivityDate="2016-01-20T23:20:15.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25587" PostTypeId="1" AcceptedAnswerId="25589" CreationDate="2016-01-20T23:37:19.913" Score="5" ViewCount="565" Body="&lt;p&gt;I appear from the space&lt;br&gt;&#xA;between the two planets&lt;br&gt;&#xA;or between the two atoms&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am what a mouse and book&lt;br&gt;&#xA;or a computer and air&lt;br&gt;&#xA;have in common&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When deceived, you shout my name&lt;br&gt;&#xA;or when you are hiding your darkest secrets&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I stalk you day and night&lt;br&gt;&#xA;wherever you are&lt;br&gt;&#xA;and Shark loves me &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I, stalking you everyday?&lt;/p&gt;&#xA;" OwnerUserId="9042" LastEditorUserId="4421" LastEditDate="2016-01-21T14:26:11.560" LastActivityDate="2017-01-28T00:11:11.520" Title="What am I, stalking you everyday?" Tags="|riddle|lateral-thinking|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25588" PostTypeId="2" ParentId="25551" CreationDate="2016-01-21T04:16:07.373" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The color of the hexagon should be black because all the other colors are a part of CMYK color model and the black is the only color that is left out.&lt;br&gt;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/LDlNO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&lt;br&gt;&#xA; And there should be no pattern in the color because alternate shapes have been colored.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8980" LastEditorUserId="18429" LastEditDate="2016-01-21T04:49:53.280" LastActivityDate="2016-01-21T04:49:53.280" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25589" PostTypeId="2" ParentId="25587" CreationDate="2016-01-21T05:11:46.387" Score="13" Body="&lt;p&gt;Well, I've tried, but...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;nothing&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...seems to fit . ;)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I appear from the space / between the two planets / or between the two atoms&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There's &quot;nothing&quot; between planets or atoms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am what a mouse and book / or a computer and air / have in common&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A mouse and a book have nothing in common, nor a computer and air.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;When deceived, you shout my name / or when you are hiding your darkest secrets&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You're speechless when deceived, and you say nothing when hiding secrets.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I stalk you day and night / wherever you are / and Shark loves me / What am I, stalking you everyday?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nothing is stalking me (that's just paranoia), and Shark(?) clearly loves nothing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastActivityDate="2016-01-21T05:11:46.387" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="25590" PostTypeId="1" CreationDate="2016-01-21T06:12:46.237" Score="3" ViewCount="489" Body="&lt;p&gt;Rule 1: There are no rules&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rule 2: When in doubt, see rule 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The puzzle is to find the next two items in the following sequence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;D15 R0 D22 R13 R-13 D11 ...&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Hints&lt;/h2&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;This puzzle is related to something &quot;in the real world&quot;. If you are looking for a purely mathematical way to solve this you are on the wrong path. &lt;/li&gt;&#xA;&lt;li&gt;This puzzle has to do with American Politics.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="18589" LastEditorUserId="9718" LastEditDate="2016-01-28T06:04:20.140" LastActivityDate="2016-01-28T06:04:20.140" Title="There are no rules: D15 R0 D22" Tags="|enigmatic-puzzle|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="25591" PostTypeId="1" AcceptedAnswerId="25592" CreationDate="2016-01-21T06:19:39.703" Score="8" ViewCount="844" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;This is a continuation of &lt;a href=&quot;https://puzzling.stackexchange.com/questions/24892/reverse-puzzling-2&quot;&gt;Reverse Puzzling 2&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&amp;quot;Okay George, let's try this one. Pretty simple.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;I gave George the puzzle. He picked up his marker and quickly sketched out the following.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jaNj0.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jaNj0.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;Well... I guess.... I guess that's the right answer then...&amp;quot; George muttered sadly.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;What was the puzzle I gave him?&lt;/p&gt;&#xA;&lt;p&gt;(I've kept this one relatively scant with the details since the previous ones were solved so quickly. Will add clues if it remains unsolved, though I highly doubt it will.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14095" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-06-06T11:35:39.953" Title="Reverse Puzzling 3" Tags="|enigmatic-puzzle|game-theory|reverse-puzzling|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25592" PostTypeId="2" ParentId="25591" CreationDate="2016-01-21T07:05:25.053" Score="11" Body="&lt;p&gt;It looks like a version of the Prisoner's Dilemma. You're player 2, and the choice is either to &lt;strong&gt;C&lt;/strong&gt;ooperate or to &lt;strong&gt;D&lt;/strong&gt;efect, after which player 1 will make their choice on the second level of the tree.&lt;/p&gt;&#xA;&lt;p&gt;If you both cooperate, you get the best overall outcome (&amp;quot;only&amp;quot; 2 points lost total). But a defector to a cooperator gets a better outcome for themself (no points lost at all, compared to 1 if they both cooperate), to the detriment of the poor cooperator, who loses 3 points.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;A more complete version of the problem might read as follows (I've added a bit of thriller flavour to the scenario to justify George's sadness a little more):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;An evil games-master has locked you in a room and strapped you into an armchair, so that you can't move your head, arms, legs, or body.&lt;/p&gt;&#xA;&lt;p&gt;Directly in front of you, separated by a chainlink screen, is somebody else in the exact same predicament. You can both see each other, but your hands are encased from the wrists onward in a hard metal shell. Neither of you can see either of your hands, although obviously you can feel your own.&lt;/p&gt;&#xA;&lt;p&gt;A loudspeaker blares. &amp;quot;Welcome, the two of you, to the Double Chairs of Demise. I take it you two are comfy where you are?&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;You struggle in your seat and try to shout, but you find that your vocal cords have been disabled by something implanted in your throat. You slump back down.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;You two have been chosen for a special experiment,&amp;quot; says the games-master. &amp;quot;I dabble in human psychology, and you two are the perfect rats for this little contraption I've devised.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;You will participate in a game that lasts for ten turns,&amp;quot; he continued. &amp;quot;In your hand pods, you will feel two buttons on each hand. From left to right, these buttons are labelled A, B, C, and D&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;You feel around for the buttons, only to discover that the buttons for your left hand have been removed! You can definitely feel two distinct buttons with your right hand, but in the pod your left hand is encased in, there is nothing but a bare metal surface. It looks like only options C and D are available to you. You look at your opponent, who seems to have a look of shock and panic on his face that indicates the same thing.&lt;/p&gt;&#xA;&lt;p&gt;A display flashes up on the wall in front of you with the number &amp;quot;10&amp;quot;. &amp;quot;You start the game with 10 &lt;strong&gt;life points&lt;/strong&gt;. At the end of every turn, I will take one life point away from each of you. If you lose all 10 points, you &lt;strong&gt;die.&lt;/strong&gt;&amp;quot; His voice turned into an evil cackle at the last word. &amp;quot;But, all is not lost. For you see, during your turn you will have a range of actions available at your disposal to try and gain points before I take one away.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;If you press the button A&amp;quot;, the games-master says, &amp;quot;you will each gain one life point, but the game will go on for one more turn. If you press the button B, you will sacrifice one of your own life points to give your opponent two. If you press C, you pass your turn and nothing will happen. If you press D, you will take one life point back for yourself, at the expense of your opponent losing two!&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;And best of all,&amp;quot; cackled the games-master, &amp;quot;your action is completely independent. It has no bearing on the effect of your opponent's action. None at all! So anything you do is an action by itself!&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;You can hear a switch being flipped. &amp;quot;Let the game... &lt;strong&gt;begin.&lt;/strong&gt;&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;George, seeing this scenario, deduced that both players would continue to press D, trying to claw back that one point they couldn't get for themselves, and thus they would kill each other in just five turns. It's sad because regardless of whether they cooperated or both defected, both of them would die anyway, unless one of them deliberately lost to keep the other alive.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-01-21T17:06:11.833" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="25593" PostTypeId="2" ParentId="25590" CreationDate="2016-01-21T07:17:16.407" Score="6" Body="&lt;p&gt;D donates Double, R is Reverse. solve the puzzle as a mirror in the middle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take Reverse -13 to 13.&#xA;Double 11 to 22.&#xA;so x= R0, y=D7.5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;at least this one makes sense to me :)&lt;/p&gt;&#xA;" OwnerUserId="18514" LastActivityDate="2016-01-21T07:17:16.407" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25594" PostTypeId="2" ParentId="25551" CreationDate="2016-01-21T07:34:16.380" Score="0" Body="&lt;p&gt;I spotted something interesting, not sure if it's right, but here's a go:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/FlI5I.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FlI5I.png&quot; alt=&quot;hexagon with colour stripes&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I drew overlapping selections by hand onto a layer mask, so not particularly accurate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If we look at the relative areas of the four rectangles:&lt;br&gt;&#xA;     &lt;code&gt;5x5 = 25&lt;/code&gt; (Red)&lt;br&gt;&#xA;     &lt;code&gt;6x5 = 30&lt;/code&gt; (Cyan/green/cyan)&lt;br&gt;&#xA;     &lt;code&gt;5x7 = 35&lt;/code&gt; (Yellow)&lt;br&gt;&#xA;     &lt;code&gt;5x6 = 30&lt;/code&gt; (Magenta/Blue/Magenta)&lt;br&gt;&#xA; And the total of these is 120 -&#xA; the internal angle in degrees of a hexagon.&#xA; So dividing the hexagon from one vertex using these angles gives us:&lt;br&gt;&#xA; 25 degrees of red;&lt;br&gt;&#xA; 30 degrees comprised of 5 cyan + 20 green + 5 cyan;&lt;br&gt;&#xA; 35 degrees of yellow;&lt;br&gt;&#xA; 30 degrees of magenta/blue/magenta (sub-divided horizontally, not drawn to scale!).  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18522" LastActivityDate="2016-01-21T07:34:16.380" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25595" PostTypeId="1" AcceptedAnswerId="25596" CreationDate="2016-01-21T08:19:33.623" Score="4" ViewCount="1462" Body="&lt;p&gt;I am white&lt;br&gt;&#xA;Have some things brown&lt;br&gt;&#xA;And something orange  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My form is mostly round&lt;br&gt;&#xA;But stems from imaginations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And when the time comes&lt;br&gt;&#xA;I wither away&lt;br&gt;&#xA;Until such a time&lt;br&gt;&#xA;When they again can play  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="10327" LastActivityDate="2016-01-22T02:47:03.290" Title="What am I? Seasonal riddle" Tags="|riddle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25596" PostTypeId="2" ParentId="25595" CreationDate="2016-01-21T08:21:30.557" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A snowman. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am white, Have some things brown, And something orange&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Orange = carrot nose, &#xA; Some things brown = twigs/branches for arms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My form is mostly round, But stems from imaginations&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Round = form of the sections making up a snowman, &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And when the time comes&#xA;I wither away&#xA;Until such a time&#xA;When they again can play &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Withering away = melting in spring. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18590" LastEditorUserId="6151" LastEditDate="2016-01-21T11:50:55.877" LastActivityDate="2016-01-21T11:50:55.877" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25597" PostTypeId="2" ParentId="25595" CreationDate="2016-01-21T08:35:32.833" Score="3" Body="&lt;p&gt;You are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Snowman&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&quot;Seasonal Riddle&quot; and &quot;I am white&quot; phrase make me think of Winter.&#xA;Then found it suited as below, &lt;br&gt;&#xA;Brown - A snowman's coat &lt;br&gt;&#xA;Orange - His nose &lt;br&gt;&#xA;Round - His body &lt;br&gt;&#xA;It is not always existed and will be melted in a time.  &lt;br&gt;&#xA;We can built it again when we like to play! &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It's kinda easy riddle. :D&lt;/p&gt;&#xA;" OwnerUserId="11353" LastEditorUserId="11353" LastEditDate="2016-01-22T02:47:03.290" LastActivityDate="2016-01-22T02:47:03.290" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25598" PostTypeId="1" AcceptedAnswerId="25600" CreationDate="2016-01-21T09:58:40.767" Score="28" ViewCount="15608" Body="&lt;p&gt;My car has a four digit number with no leading zeros, and the following properties:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Its last digit is double the first digit&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;its middle two digits are the same&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;its last two digits are double the first two digits&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;What is my car's number?&lt;/p&gt;&#xA;" OwnerUserId="18263" LastEditorUserId="1649" LastEditDate="2023-03-31T09:42:53.280" LastActivityDate="2023-03-31T09:42:53.280" Title="What is my four digit car number?" Tags="|mathematics|logical-deduction|" AnswerCount="11" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="25599" PostTypeId="2" ParentId="25598" CreationDate="2016-01-21T10:05:51.613" Score="4" Body="&lt;p&gt;well...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0 0 0 0 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Seems to fit :)&lt;/p&gt;&#xA;" OwnerUserId="17888" LastActivityDate="2016-01-21T10:05:51.613" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25600" PostTypeId="2" ParentId="25598" CreationDate="2016-01-21T10:23:42.697" Score="16" Body="&lt;p&gt;Answer should be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4998&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Script I used :&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;# RUBY&#xA;for i in 0..9999&#xA;&#xA;  j = i.to_s.rjust(4, &quot;0&quot;) # 0 padding&#xA;  a = j[0]&#xA;  b = j[1]&#xA;  c = j[2]&#xA;  d = j[3]&#xA;&#xA;  if (b != c) then next end # Rule2&#xA;  if d.to_i != (2 * a.to_i) then  next end #Rule1&#xA;  if 2 * (a.to_s + b.to_s).to_i != (c.to_s + d.to_s).to_i then next end #Rule3&#xA;&#xA;  puts j&#xA;&#xA;  next&#xA;&#xA;end&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="18280" LastActivityDate="2016-01-21T10:23:42.697" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25601" PostTypeId="2" ParentId="25598" CreationDate="2016-01-21T10:24:37.560" Score="12" Body="&lt;p&gt;I think it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4 9 9 8&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Good math problem! ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Done in the old way, not programming took place!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; From the clues we have:&lt;br&gt;&#xA; a b c d&lt;br&gt;&#xA; d = 2*a&lt;br&gt;&#xA; b = c&lt;br&gt;&#xA; 10*c + d = 2*(10*a + b)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Doing some substitution we achieve:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 10*b + 2*a = 2*(10*a + b)&lt;br&gt;&#xA; 8*b = 18*a&lt;br&gt;&#xA; 4*b = 9*a&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a has to be a 4&lt;br&gt;&#xA; b has to be a 9&lt;br&gt;&#xA; cd has to be 49*2 = 98&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18593" LastEditorUserId="18593" LastEditDate="2016-01-22T08:13:53.477" LastActivityDate="2016-01-22T08:13:53.477" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25602" PostTypeId="2" ParentId="25499" CreationDate="2016-01-21T10:39:30.553" Score="3" Body="&lt;p&gt;Could the answer be :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;His brothers will always help you,&#xA;Slaying spiders or golems.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; See comments&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes you see him for real,&#xA;Depending on how you live.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If it's about people on Stack, I guess it means that since you spend all your time on your computer, you don't see the sun often&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the thing you won't believe,&#xA;Is that he can melt steel.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sun generates heat which can melt anything&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Even if he looks like a gem,&#xA;One day he will crush you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sun is sometimes referred to as a gem. And when it will explode it will crush our while solar system.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 1 :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  He will crush what is left of you, if there is something left by that time! &lt;br&gt;&#xA; The sun will not explode for another 6 billions year I think&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Last hint :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A very famous guy died trying to reach it&lt;br&gt;&#xA; This could refer to Icarus who died for getting to close to the sun, melting his wings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18280" LastEditorUserId="18280" LastEditDate="2016-01-21T10:47:45.860" LastActivityDate="2016-01-21T10:47:45.860" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25603" PostTypeId="2" ParentId="25598" CreationDate="2016-01-21T10:39:45.363" Score="83" Body="&lt;p&gt;Last digit is the double of first digit;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;1..2&lt;br/&gt;&#xA;2..4&lt;br/&gt;&#xA;3..6&lt;br/&gt;&#xA;4..8&lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Middle two digits are same. It has to be a number when multiplied by 2 has the last digit in it;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;1662 &lt;br/&gt;&#xA;2774 &lt;br/&gt;&#xA;3886 &lt;br/&gt;&#xA;4998 &lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Last two digits are the double of first two digits&#xA;The only correct possible answer&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;&lt;strong&gt;4998&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18559" LastEditorUserId="81979" LastEditDate="2023-03-31T09:09:08.383" LastActivityDate="2023-03-31T09:09:08.383" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="25604" PostTypeId="1" CreationDate="2016-01-21T10:45:59.730" Score="4" ViewCount="1326" Body="&lt;p&gt;My 3rd grader brought an assignment home about the Giza plateau, and one of his questions was the following riddle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What creature with one voice walks on four legs in the morning, on two at noon, and on three in the evening?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This totally puzzles me; the only thing I can think of  is an unfortunate human being who's been through a lot the night before (silly joke on my part).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Can you solve the Sphinx's riddle?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="18594" LastEditorUserId="22922" LastEditDate="2016-09-01T10:38:45.570" LastActivityDate="2019-06-13T10:54:30.177" Title="Sphinx-related riddle" Tags="|riddle|" AnswerCount="5" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="25605" PostTypeId="2" ParentId="25604" CreationDate="2016-01-21T11:09:05.813" Score="19" Body="&lt;p&gt;&lt;strong&gt;What creature with one voice walks on four legs in the morning, on two at noon, and on three in the evening?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Humans&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;four legs in the morning&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As a child, you crawl.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;two at noon&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; During adulthood, you walk.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;on three in the evening&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;As an elder, you walk with a cane.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16546" LastEditorUserId="4421" LastEditDate="2016-01-21T20:40:21.207" LastActivityDate="2016-01-21T20:40:21.207" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25606" PostTypeId="1" AcceptedAnswerId="25616" CreationDate="2016-01-21T11:19:40.107" Score="8" ViewCount="621" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gBIy8.jpg&quot; alt=&quot;power outlet, 2 timers, light bulb&quot;&gt;&#xA;&lt;br&gt;&#xA;Two motorized 24-hour&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Time_switch&quot; rel=&quot;nofollow noreferrer&quot;&gt;light timers&lt;/a&gt;&#xA;are &lt;a href=&quot;https://en.wikipedia.org/wiki/Daisy_chain_%28electrical_engineering%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;daisy&#xA;chained&lt;/a&gt;&#xA;between a power outlet and a light bulb.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For these timers, devise schedules and choose initial times&#xA;that produce the following repeated lighting pattern,&#xA;with the largest possible whole number $n$,&#xA;beginning when the outlet's power is switched on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;ensp;&amp;hairsp;&amp;hairsp;   Light is on for 1/2 hour,$~$ off for $n$ hours, &lt;br&gt;&#xA;&amp;nbsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;          on for 1/2 hour,$~$ off for $n$ hours, &lt;br&gt;&#xA;&amp;nbsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;          on for 1/2 hour,$~$ off for $n$ hours, &lt;br&gt;&#xA;&amp;nbsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;          $~\,\vdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If you are unfamiliar with these timers&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each timer repeatedly cycles through&#xA;its schedule of 24 intervals that last an hour each.&#xA;&lt;br&gt;&lt;sup&gt;&lt;sup&gt;&lt;/sup&gt;&lt;/sup&gt;&#xA;&amp;bull;$~$ A circular dial determines the current point in the schedule&#xA;&lt;br&gt;&lt;sup&gt;&lt;sup&gt;&lt;/sup&gt;&lt;/sup&gt;&#xA;&amp;bull;$~$ A motor rotates the dial to advance through its schedule&#xA;          whenever power is supplied to the timer&#xA;&lt;br&gt;&lt;sup&gt;&lt;sup&gt;&lt;/sup&gt;&lt;/sup&gt;&#xA;&amp;bull;$~$ You may initially set the dial to any minute of any interval&#xA;&lt;br&gt;&lt;sup&gt;&lt;sup&gt;&lt;/sup&gt;&lt;/sup&gt;&#xA;&amp;bull;$~$ You preset each interval to ON or OFF&#xA;&lt;br&gt;&lt;sup&gt;&lt;sup&gt;&lt;/sup&gt;&lt;/sup&gt;&#xA;&amp;bull;$~$ When the dial is in an interval that was set to ON,&#xA;          the timer acts as a direct connection for power to flow&#xA;          to whatever is plugged into the timer&#xA;&lt;br&gt;&lt;sup&gt;&lt;sup&gt;&lt;/sup&gt;&lt;/sup&gt;&#xA;&amp;bull;$~$ When the dial is in an interval that was set to OFF,&#xA;          the timer does not provide a power connection&#xA;&lt;br&gt;&lt;sup&gt;&lt;sup&gt;&lt;/sup&gt;&lt;/sup&gt;&#xA;&amp;bull;$~$ The resulting ON and OFF durations could be fractions of an hour&#xA;          if the timer is set to begin within an interval or&#xA;          if incoming power is interrupted during an interval&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first timer is plugged into the outlet.&#xA;&lt;br&gt;&lt;sup&gt;&lt;/sup&gt;&#xA;&amp;bull;$~$ It runs nonstop once the outlet is switched on&#xA;&lt;br&gt;&lt;sup&gt;&lt;/sup&gt;&#xA;&amp;bull;$~$ It supplies power&amp;mdash;&amp;hairsp;but only when its dial is in&#xA;          an ON interval&amp;hairsp;&amp;mdash;&amp;hairsp;to the second timer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second timer has the light bulb plugged into it.&#xA;&lt;br&gt;&lt;sup&gt;&lt;/sup&gt;&#xA;&amp;bull;$~$ It advances through its schedule&#xA;          only when the first timer supplies power&#xA;&lt;br&gt;&lt;sup&gt;&lt;/sup&gt;&#xA;&amp;bull;$~$ It lights the bulb, but only while powered by&#xA;          the first timer and when its dial is in an ON interval&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Related puzzles&lt;/strong&gt;&#xA;&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/q/25610/18129&quot;&gt;Odd hours with two timers&lt;/a&gt;&#xA;&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/q/25812/18129&quot;&gt;Day and night of the two timers&lt;/a&gt;&#xA;&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/q/33882/18129&quot;&gt;Third timer's a charm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&#xA;(These puzzles are either directly from&#xA; or related to actual botany experiments)&#xA;&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="18129" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2016-06-14T04:55:02.730" Title="Halve time with two timers" Tags="|optimization|time|real|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25607" PostTypeId="2" ParentId="25604" CreationDate="2016-01-21T11:38:21.587" Score="24" Body="&lt;p&gt;I must confess that I've always been a bit puzzled by the standard answer that literally everyone gives to this puzzle, which relies on weird metaphorical interpretations of what are fundamentally quite common, ordinary, well-understood words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Particularly since there's a simple (and frankly, rather obvious) answer which doesn't require you to pretend that words mean anything other than what the dictionary says that they mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This more sensible answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My cat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;walks on four legs in the morning&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the mornings, my wife and I watch the morning news, while seated on the sofa.  Invariably, the cat walks across us, stepping on all four of our legs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;on two at noon&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  At noon I'm at home (due to an injury which is currently keeping me from work), but my wife is at work.  So when my cat walks across the sofa while I watch the news at noon, she only walks on my two legs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;and on three in the evening&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  In the evening, my wife is usually quite tired from her long day, and so will doze on her side on the sofa, as we watch the evening news together.  So when the cat walks across the sofa in the evening, she steps on both of my legs, but only the uppermost of my wife's legs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think we can all agree that this is a far simpler answer than that other one which requires redefining the words used by the riddle.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean, I'm as much of a fan of &lt;a href=&quot;http://www.fecundity.com/pmagnus/humpty.html&quot;&gt;The Humpty Dumpty theory of language&lt;/a&gt; as the next guy, but.. it does sort of cheapen the solution to a riddle if you could only get an answer by inventing new meanings for the words.  Right?&lt;/p&gt;&#xA;" OwnerUserId="9286" LastEditorUserId="9286" LastEditDate="2016-01-21T12:04:46.307" LastActivityDate="2016-01-21T12:04:46.307" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25608" PostTypeId="1" AcceptedAnswerId="27591" CreationDate="2016-01-21T12:41:40.390" Score="15" ViewCount="962" Body="&lt;p&gt;Three high school friends who took different paths after high school are now all enjoying retirement. Each one settled for a different job but all enjoyed cryptography throughout their lives.&lt;br&gt;&#xA;Now having reunited they send each other encrypted messages to keep their minds sharp and prevent their knowledge from dwindling.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;One spent his life in the military.&lt;/li&gt;&#xA;&lt;li&gt;Another in the police force.&lt;/li&gt;&#xA;&lt;li&gt;The last a telecommunications company.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Currently they communicate using a homemade cipher they can all relate to.    &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;!3B7BA2B3C 1CF 7A9A3B2ABD1F A7A1B 7A9A3B2ABD1F 7A2B7C2BD B7A1B.&#xA;!C4A 1C2E E1A2CB1F 3B7BA2B3C 2E7B 2AB 6A1AEE3C,&#xA;C7B9A2D 1F2B1A2D3A6A 4A7B2D 6A1AEEC7A2B1FF 0A1CDD 7A2B7C2BD B7A1B.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Can you crack their latest message???&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To complete the problem you must:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; - Decrypt the message and find the author.&#xA; - Explain clearly how the cipher works.&#xA; - Explain what the cipher is based on.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hints&lt;/strong&gt;&lt;br&gt;&#xA;After a few days I will post some more hints if this remains unanswered.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They all share practical knowledge, they have been known to mumble.&lt;br&gt;&#xA; The last thing they want to do is meet up, they are in their element.&lt;br&gt;&#xA; They also like to abbreviate, time is of the essence.&lt;br&gt;&#xA; Read between the lines systematically, these hints are not what they seem.&lt;br&gt;&#xA; What is the solution to verbal communication errors? &lt;br&gt;&#xA; Shall I spell it out for you? Perhaps. &lt;br&gt;&#xA; Nate! Oh My Gosh, Get It Over With!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PLEASE NOTE:&lt;/strong&gt; I have updated the cipher text due to minor errors.&lt;/p&gt;&#xA;" OwnerUserId="12014" LastEditorUserId="12014" LastEditDate="2016-02-18T08:00:23.170" LastActivityDate="2016-02-18T16:13:23.133" Title="The Three Retiree's Cipher" Tags="|riddle|cipher|cryptic-clues|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25609" PostTypeId="2" ParentId="25606" CreationDate="2016-01-21T12:43:06.693" Score="1" Body="&lt;p&gt;A motor rotates the dial to advance through its schedule whenever power is supplied to the timer &lt;/p&gt;&#xA;&#xA;&lt;p&gt;• You may preset each interval to ON or OFF &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;only 1 interval of the 24 is ON on the first and second&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;• You may initially set the dial to any minute of any interval &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first timer 23:00 away from first ON interval&lt;br/&gt;&#xA; The second timer 22:30 away from the first ON interval&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;in that case the first n = &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 552&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18559" LastActivityDate="2016-01-21T12:43:06.693" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25610" PostTypeId="1" AcceptedAnswerId="25624" CreationDate="2016-01-21T12:49:52.803" Score="8" ViewCount="823" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gBIy8.jpg&quot; alt=&quot;power outlet, 2 timers, light bulb&quot;&gt;&#xA;&lt;br&gt;&#xA;Two motorized 24-hour&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Time_switch&quot; rel=&quot;nofollow noreferrer&quot;&gt;light timers&lt;/a&gt;&#xA;are &lt;a href=&quot;https://en.wikipedia.org/wiki/Daisy_chain_%28electrical_engineering%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;daisy&#xA;chained&lt;/a&gt;&#xA;between a power outlet and a light bulb.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For these timers, devise schedules and choose initial times&#xA;that produce the following repeated 9-hour lighting pattern,&#xA;with the largest possible whole number $d$ less than&amp;nbsp;9,&#xA;beginning when the outlet's power is switched on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;ensp;&amp;hairsp;&amp;hairsp;   Light is on for $\:d~$ hours,$~$ off for $\,9\,$&amp;ndash;$\:d~~$ hours, &lt;br&gt;&#xA;&amp;nbsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;          on for $\:d~$ hours,$~$ off for $\,9\,$&amp;ndash;$\:d~~$ hours, &lt;br&gt;&#xA;&amp;nbsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;          on for $\:d~$ hours,$~$ off for $\,9\,$&amp;ndash;$\:d~~$ hours, &lt;br&gt;&#xA;&amp;nbsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;          $~\,\vdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If you are unfamiliar with these timers&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each timer repeatedly cycles through&#xA;its schedule of 24 intervals that last an hour each.&#xA;&lt;br&gt;&lt;sup&gt;&lt;sup&gt;&lt;/sup&gt;&lt;/sup&gt;&#xA;&amp;bull;$~$ A circular dial determines the current point in the schedule&#xA;&lt;br&gt;&lt;sup&gt;&lt;sup&gt;&lt;/sup&gt;&lt;/sup&gt;&#xA;&amp;bull;$~$ A motor rotates the dial to advance through its schedule&#xA;          whenever power is supplied to the timer&#xA;&lt;br&gt;&lt;sup&gt;&lt;sup&gt;&lt;/sup&gt;&lt;/sup&gt;&#xA;&amp;bull;$~$ You may initially set the dial to any minute of any interval&#xA;&lt;br&gt;&lt;sup&gt;&lt;sup&gt;&lt;/sup&gt;&lt;/sup&gt;&#xA;&amp;bull;$~$ You preset each interval to ON or OFF&#xA;&lt;br&gt;&lt;sup&gt;&lt;sup&gt;&lt;/sup&gt;&lt;/sup&gt;&#xA;&amp;bull;$~$ When the dial is in an interval that was set to ON,&#xA;          the timer acts as a direct connection for power to flow&#xA;          to whatever is plugged into the timer&#xA;&lt;br&gt;&lt;sup&gt;&lt;sup&gt;&lt;/sup&gt;&lt;/sup&gt;&#xA;&amp;bull;$~$ When the dial is in an interval that was set to OFF,&#xA;          the timer does not provide a power connection&#xA;&lt;br&gt;&lt;sup&gt;&lt;sup&gt;&lt;/sup&gt;&lt;/sup&gt;&#xA;&amp;bull;$~$ The resulting ON and OFF durations could be fractions of an hour&#xA;          if the timer is set to begin within an interval or&#xA;          if incoming power is interrupted during an interval&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first timer is plugged into the outlet.&#xA;&lt;br&gt;&lt;sup&gt;&lt;/sup&gt;&#xA;&amp;bull;$~$ It runs nonstop once the outlet is switched on&#xA;&lt;br&gt;&lt;sup&gt;&lt;/sup&gt;&#xA;&amp;bull;$~$ It supplies power&amp;mdash;&amp;hairsp;but only when its dial is in&#xA;          an ON interval&amp;hairsp;&amp;mdash;&amp;hairsp;to the second timer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second timer has the light bulb plugged into it.&#xA;&lt;br&gt;&lt;sup&gt;&lt;/sup&gt;&#xA;&amp;bull;$~$ It advances through its schedule&#xA;          only when the first timer supplies power&#xA;&lt;br&gt;&lt;sup&gt;&lt;/sup&gt;&#xA;&amp;bull;$~$ It lights the bulb, but only while powered by&#xA;          the first timer and when its dial is in an ON interval&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Related puzzles&lt;/strong&gt;&#xA;&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/q/25606/18129&quot;&gt;Halve time with two timers&lt;/a&gt;&#xA;&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/q/25812/18129&quot;&gt;Day and night of the two timers&lt;/a&gt;&#xA;&lt;br&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/q/33882/18129&quot;&gt;Third timer's a charm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&#xA;(This puzzle is directly from an actual botany experiment)&#xA;&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="18129" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2016-06-14T04:53:53.600" Title="Odd hours with two timers" Tags="|optimization|time|real|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25611" PostTypeId="2" ParentId="25606" CreationDate="2016-01-21T12:58:21.560" Score="2" Body="&lt;p&gt;Let me start off with an $n=1$ solution, (at least I think this works)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set the intervals like this where 1 is on and 0 is off:&lt;br&gt;&#xA;A: 110110110110110110110110&lt;br&gt;&#xA;B: 101010101010101010101010&lt;/p&gt;&#xA;&#xA;&lt;p&gt;set dial A to 90 minutes and B to 0 minutes.&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2016-01-21T12:58:21.560" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25612" PostTypeId="2" ParentId="25579" CreationDate="2016-01-21T13:16:01.550" Score="4" Body="&lt;p&gt;Something like this would make sense. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/3kJOk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3kJOk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The middle is mirrored along the diagonal and the outer are moved diagonally downright. Of course, there are many variations to this. It could be the upper that is always is moved downright diagonally and the lower that is mirrored along the diagonal, but that would produce an overlap.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12901" LastActivityDate="2016-01-21T13:16:01.550" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25613" PostTypeId="2" ParentId="25610" CreationDate="2016-01-21T13:36:18.817" Score="5" Body="&lt;p&gt;Set both timers to be always on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d = 9;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On for 9 hours, off for 0 hours, on for 9 hours, off for 0 hours.&lt;/p&gt;&#xA;" OwnerUserId="10909" LastActivityDate="2016-01-21T13:36:18.817" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25614" PostTypeId="1" AcceptedAnswerId="25647" CreationDate="2016-01-21T14:04:35.183" Score="6" ViewCount="412" Body="&lt;p&gt;I decided to make a second puzzle. I think this one is very tough, but then again it might be very easy for some. Perhaps it's a bit niche.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A child  &#xA;born as the result of eight betrayals  &#xA;I played in the sand, when everyone else played elsewhere&#xA;&#xA;I practiced as I grew  &#xA;half way to becoming a maestro  &#xA;then I became a key player in big business  &#xA;&#xA;my success was amplified by my logical approach  &#xA;my ambition guided me, and those I supported, in giant leaps  &#xA;when things got too noisy I went one way  &#xA;&#xA;is this fair  &#xA;I was betrayed many many times like my parents before me  &#xA;and eventually I was no longer special&#xA;&#xA;Who am I?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="18522" LastActivityDate="2016-01-30T18:32:56.087" Title="Not what you might expect. Who am I?" Tags="|riddle|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25615" PostTypeId="1" CreationDate="2016-01-21T14:56:59.040" Score="3" ViewCount="311" Body="&lt;p&gt;Inspired by &lt;a href=&quot;https://puzzling.stackexchange.com/questions/25524/blindfolded-and-disoriented-near-the-great-wall-of-china&quot;&gt;Blindfolded and disoriented near the Great Wall of China&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A drone is stationary at a spatial point about 1 m from the Great Wall, which is a vertical plane rectangle with height 5 m and infinite length. The altitude of the drone is a random number between 0 m and 5 m. The drone may be pointing in any direction (which may even be tilted upwards or downwards), however it does not know which. It can travel in any direction, and switch directions accurately. It gets to know when it touches the ground plane, and can then determine its own tilt upwards/downwards by travelling a small distance on the ground (about zero).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Q 1:&lt;/strong&gt; What is the minimum distance that must be travelled to ensure it hits the Great Wall?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Q 2:&lt;/strong&gt; If the objective is to minimise displacement between initial and final positions, what upper value can it be guaranteed to outperform?&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2016-01-21T17:26:36.410" Title="The lost drone at the Great Wall of China" Tags="|mathematics|geometry|optimization|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="25616" PostTypeId="2" ParentId="25606" CreationDate="2016-01-21T15:04:24.120" Score="4" Body="&lt;p&gt;To show that Ivo Beckers has the longest $n$ in his answer, rewrite it like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Set the intervals like this where 1 is on and 0 is off:&lt;br&gt;&#xA;  A: &lt;strong&gt;101101101101101101101101&lt;/strong&gt;&lt;br&gt;&#xA;  B: 101010101010101010101010  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;set dial A to &lt;strong&gt;30&lt;/strong&gt; minutes and B to 0 minutes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is, rotate &lt;code&gt;A&lt;/code&gt; once to the left and set its timer to &lt;code&gt;30&lt;/code&gt; instead of &lt;code&gt;90&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works as before for a one hour delay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that &lt;code&gt;A&lt;/code&gt; is now simply &lt;code&gt;101&lt;/code&gt; repeated 8 times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get a 2 hour delay we'd need &lt;code&gt;A&lt;/code&gt; to be &lt;code&gt;10010&lt;/code&gt; repeated, but 5 doesn't divide 24 evenly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, for a 3 hour delay &lt;code&gt;A&lt;/code&gt; is set to &lt;code&gt;1000100&lt;/code&gt; repeated but 7 doesn't divide 24 evenly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;etc&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of these work because:&lt;br&gt;&#xA;Their length is &lt;code&gt;len = n + (n + 1)&lt;/code&gt; for an n hour delay.&lt;br&gt;&#xA;Let's call this &lt;code&gt;len(A)&lt;/code&gt;, therefore &lt;code&gt;len(A)&lt;/code&gt; has to be odd.&lt;br&gt;&#xA;The only odd number that divides 24 evenly is 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So 1 hour is the longest delay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Other patterns simply don't work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to have a 1/2 hour &lt;code&gt;on&lt;/code&gt; followed by &lt;code&gt;off&lt;/code&gt; &lt;code&gt;B&lt;/code&gt; can't have &lt;strong&gt;any&lt;/strong&gt; sequential ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;code&gt;B&lt;/code&gt; must be composed of a single one followed by one or more zeros with this pattern repeated through out &lt;code&gt;B&lt;/code&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;B&lt;/code&gt; must also start with a one or we can simply remove leading zeros (with no loss of generality) so that &lt;code&gt;B&lt;/code&gt; does start with a one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;A&lt;/code&gt; must now start with a one and its timer set to start at &lt;code&gt;30&lt;/code&gt; minutes so we begin with a 1/2 hour &lt;code&gt;on&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;A&lt;/code&gt; must then be followed by a zero so we go &lt;code&gt;off&lt;/code&gt; after that 1/2 hour &lt;code&gt;on&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And &lt;code&gt;B&lt;/code&gt; obviously is followed by one or more zeros.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore &lt;code&gt;A&lt;/code&gt; and &lt;code&gt;B&lt;/code&gt; &lt;strong&gt;must&lt;/strong&gt; start with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;A&lt;/code&gt; = &lt;code&gt;10&lt;/code&gt; [1]&lt;br&gt;&#xA;&lt;code&gt;B&lt;/code&gt; = &lt;code&gt;10&lt;/code&gt; [2]  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for any number of zeros that follow the initial one for &lt;code&gt;B&lt;/code&gt; in [2], say &lt;code&gt;B&lt;/code&gt; is now &lt;code&gt;10...0&lt;/code&gt; repeated &lt;em&gt;i.e&lt;/em&gt; a one followed by &lt;code&gt;b&lt;/code&gt; zeros followed by a one followed by &lt;code&gt;b&lt;/code&gt; zeros &lt;em&gt;etc&lt;/em&gt;, then &lt;code&gt;A&lt;/code&gt; must be &lt;code&gt;100..011..1&lt;/code&gt; repeated &lt;em&gt;i.e.&lt;/em&gt; a one followed by &lt;code&gt;b&lt;/code&gt; zeros and then &lt;code&gt;b&lt;/code&gt; ones followed by a one then &lt;code&gt;b&lt;/code&gt; zeros, then &lt;code&gt;b&lt;/code&gt; ones &lt;em&gt;etc&lt;/em&gt;.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore &lt;code&gt;len(A)&lt;/code&gt; is &lt;code&gt;1 + 2b&lt;/code&gt;, so &lt;code&gt;len(A)&lt;/code&gt; remains odd and must evenly divided 24.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So therefore &lt;code&gt;len(A)&lt;/code&gt; can only equal 3.&lt;br&gt;&#xA;and &lt;code&gt;A&lt;/code&gt; is &lt;code&gt;101101101101101101101101&lt;/code&gt;&lt;br&gt;&#xA;and &lt;code&gt;B&lt;/code&gt; is &lt;code&gt;101010101010101010101010&lt;/code&gt;&lt;br&gt;&#xA;with &lt;code&gt;A&lt;/code&gt; starting at &lt;code&gt;30&lt;/code&gt; minutes and &lt;code&gt;B&lt;/code&gt; at &lt;code&gt;0&lt;/code&gt; minutes.&lt;br&gt;&#xA;Or equivalent rotations as in Ivo Beckers original answer.&lt;/p&gt;&#xA;" OwnerUserId="10118" LastEditorUserId="10118" LastEditDate="2016-01-22T03:25:21.560" LastActivityDate="2016-01-22T03:25:21.560" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25618" PostTypeId="2" ParentId="25582" CreationDate="2016-01-21T15:46:16.640" Score="1" Body="&lt;p&gt;If this was QI, this answer would no doubt get the buzzer, but...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The man finished his apple juice and the nephew went to get some more for him.&#xA; The man didn't realise and picked up the nephew's cider and took a swig.&#xA; When the nephew returned with the apple juice, the man realised his mistake and screamed then killed himself with a large rock he kept on hand, you know, for emergencies and other eventualities.&#xA; The nephew ran to stop him and knocked over some beer in the process.&#xA; Remarkably he managed to keep hold of the cider glass.&#xA; The nephew lied for any one of multiple plausible reasons (he was religious and didn't like the idea of suicide, he wanted to protect his uncle's honor, he wanted to protect his aunt's feelings, he felt guilty about leaving his cider there while going to get more apple juice for his dear old uncle).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alternatively...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It was just like above except the nephew did it on purpose and swapped the cider/apple juice to inherit all the goodies. The beer barrel was knocked over in the man's anger before he offed himself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18522" LastActivityDate="2016-01-21T15:46:16.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25619" PostTypeId="2" ParentId="25582" CreationDate="2016-01-21T16:03:07.967" Score="6" Body="&lt;p&gt;Could it be that :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The nephew took a big rock, which was lying on top of the beer barrel to keep it closed, to knock out his uncle. That would explain the blood on his shirt from a supposed head wound. Unfortunately for the nephew, this was not enough to kill him, so the nephew put his uncle's head in the barrel and drowned him, explaining the wet hair and wet ground. Now, to make it look like a suicide he put his own cider cup in his uncle's hand, whom he knew would have to kill himself if he ever drank alcohol. Thus, he inherited the money.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18280" LastEditorUserId="11452" LastEditDate="2016-01-21T17:01:01.193" LastActivityDate="2016-01-21T17:01:01.193" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="25620" PostTypeId="2" ParentId="25615" CreationDate="2016-01-21T16:34:12.147" Score="4" Body="&lt;p&gt;Q1: My initial approach would be to simply sculpt a sphere by going up sqrt(2) meters, then rotating around the initial point in a circle of radius sqrt(2). Three quarters of the way around, rotate 90* and start a second circle of radius sqrt(2) with the origin as center, perpendicular to the first. Three quarters of the way around that circle, rotate 90* again and create a third circle with the same origin / radius perpendicular to both the first two. When you finish that one, you'll be where you broke away from the second circle, so finish it off, then finish off the first circle. That gives you three perpendicular circles describing a sphere in distance sqrt(2)+6sqrt(2)pi = &lt;strong&gt;about 28 meters&lt;/strong&gt;. Note that you need a radius of sqrt(2) to guarantee a hit with this method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you hit the ground at any point, continue along the surface on the projection of your original path onto the surface - this is necessarily less distance than the original path, so it can't hurt our distance in the worst case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q2: My approach never displaces further than &lt;strong&gt;sqrt(2) meters&lt;/strong&gt;, but a method optimizing only displacement could simply travel the surface of the sphere of radius 1 centered at the origin in a chaotic path until it collided with the wall.  The wall could be 1 meter away, so that is clearly optimal. &lt;/p&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="17389" LastEditDate="2016-01-21T16:58:35.673" LastActivityDate="2016-01-21T16:58:35.673" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25621" PostTypeId="2" ParentId="25582" CreationDate="2016-01-21T16:51:35.017" Score="3" Body="&lt;p&gt;Even though an answer has been accepted, I'll leave my best interpretation of the events, which did in fact end with a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; suicide.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A thought:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I had thought of a drowning, like Thrax's answer proposes, but see an issue with that timeline.  Since the wife &lt;em&gt;rushes&lt;/em&gt; down after the scream (which, in Thrax's answer, is from the blow from the rock), there would be very little time to drown the uncle and then pose him with the rock and cup before the wife sees what's happening.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So here's how I see it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The uncle and the nephew are in the cellar, inspecting the brew, as the OP states.  The nephew leaves and returns to find the uncle having just opened a barrel, using a rock to unjam the lid.  They both peer into the brew when the nephew dunks his uncle's head into it, holding it under the beer in an attempt to drown him and claim his possessions.  The uncle fights, and gathers his strength for one push, which succeeds in freeing him from the nephew &lt;em&gt;and&lt;/em&gt; lets him continue breathing.  &lt;strong&gt;However&lt;/strong&gt;, when the uncle braced himself for that death-staving push, he inhaled some of the beer.  He coughed and sputtered and tried to catch his breath.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Realizing that he'd inadvertantly drank beer, and had really only chosen &lt;strong&gt;&lt;em&gt;death&lt;/strong&gt; by his &lt;strong&gt;own&lt;/strong&gt; hand&lt;/em&gt; over his nephew's, he let out a bellowing scream.  His nephew had done him in, either way, so he lifted the rock over his head, brought it crashing down, and fell into a heap with the rock on his chest.  Just as his wife found him moments later.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, it was indeed&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a cellar suicide.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11452" LastActivityDate="2016-01-21T16:51:35.017" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25622" PostTypeId="2" ParentId="25610" CreationDate="2016-01-21T17:02:51.590" Score="4" Body="&lt;p&gt;A first (serious) attempt:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Set the first timer to be on for 2 and off for 1 (and repeating forever), starting at the beginning of the on cycle.&lt;br&gt;&#xA; Set the second timer to be on for 2 hour and off for 4 (and repeating forever), starting at the beginning of the on cycle.&lt;br&gt;&#xA; Then the light will go on for 2 hours and stay off for 7 hours and repeat forever.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't know if this can be improved upon, but I'd be a little surprised.  The first timer extends the time taken for the second (so that it takes 3 hours to move 2 hours on the second timer).  Other options would be that it takes 6 hours to move 4 hours, but then it will get out of sync with the 9 hours cycle (unless something very clever happens with syncing over multiple cycles).  The ideal would be 9 hours to move 6, but 9 hours doesn't divide into 24.&lt;/p&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2016-01-21T17:09:59.923" LastActivityDate="2016-01-21T17:09:59.923" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25624" PostTypeId="2" ParentId="25610" CreationDate="2016-01-21T17:15:26.073" Score="5" Body="&lt;p&gt;Obviously, you can't obtain a 9 hour cycle with just one 24 hour timer, because 24 is no integer multiple of 9. So there have to be hours at which the first timer is turned off. As the first timer runs at a 24 hour cycle, it will be off at certain fixed hours every day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The greatest common divisor of 24 and 9 is 3, so the 9-hour cycle occurs at all phase shifts that are multiples of three. If the on-time is at least 3 hours, there is no hour the bulb is off every day, which contradicts the observation that there must be such hours. So obviously $d \le 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will show a solution with $d = 2$, which thus is optimal. Set the first timer to 8 copies of 2 hours on and 1 hour off. The second timer is powered 6 hours per 9 hour interval. As 6 is a divider of 24, it is possible to program the second timer in a way that it has a cycle repeating every 6 hours it is powered, which is 9 hours real time. Set the timer to 4 copies of 2 hours on and 4 hours off, and have both timers start at the beginning of their on phases.&lt;/p&gt;&#xA;" OwnerUserId="17093" LastEditorUserId="12408" LastEditDate="2016-01-21T23:48:09.683" LastActivityDate="2016-01-21T23:48:09.683" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25625" PostTypeId="2" ParentId="25598" CreationDate="2016-01-21T17:22:55.277" Score="33" Body="&lt;p&gt;A deductive approach:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The number has form abb(2a) for digits a and b. Since the last two digits are double the first two, 10b + 2a = 2(10a + b). So 8b = 18a and 4b = 9a. Therefore b = 9, a = 4, and the solution is 4998.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17915" LastEditorUserId="17915" LastEditDate="2016-01-21T17:30:06.520" LastActivityDate="2016-01-21T17:30:06.520" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25628" PostTypeId="2" ParentId="25551" CreationDate="2016-01-21T20:00:45.180" Score="27" Body="&lt;p&gt;These images are all (orthographic) projections of the same cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has the three basic colors, with opposite faces identical. When one is in front of the other their colors are added in the projection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final answer is a variation on these two pictures, depending on how you rotate the cube:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/WKAcY.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WKAcY.png&quot; alt=&quot;ANSWER&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A more detailed explanation of the answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;First:&lt;/strong&gt;  We just see the two red faces superposed.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Second:&lt;/strong&gt; Now the cube is rotated, with the green faces in place of red. However, it is slightly tilted so the blue sides mix with the green forming the cyan rectangles.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Third:&lt;/strong&gt;  The cube is at 45 degrees with the ground, so red and green faces superpose completely at top, as well as green and red at the bottom, giving an uniform yellow color.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Fourth:&lt;/strong&gt; Finally, we see a similar arrangement to the second picture except in a different plane of rotation.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, we can notice differences in size between the rectangles. That's no mistake: When a cube rotates along a coordinate axis, its visual size increases from $l$ to $l \sqrt2$, then goes back down to $l$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The yellow is the biggest rectangle because the diagonal of a square is the longest line segment inside it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our last projection is the cube viewed vertex-on, which has a hexagonal envelope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This picture shows how you combine colors of face pairs in the X, Y and Z axis to get the final answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/pEJES.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pEJES.png&quot; alt=&quot;Explanation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This puzzle was actually pretty easy for me to figure out because I love trying to visualize higher-dimensional spaces, so I was very well familiar with projections, envelopes, cross sections and stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a fascinating way to improve your spatial reasoning, so I highly recommend anyone interested to take a &lt;a href=&quot;http://eusebeia.dyndns.org/4d/vis/01-intro#Introduction&quot;&gt;look&lt;/a&gt;. Or you could just try exploring a &lt;a href=&quot;https://www.google.com.br/search?q=4d%20maze&quot;&gt;4d maze&lt;/a&gt; and see if you can find your way out&lt;/p&gt;&#xA;" OwnerUserId="14219" LastEditorUserId="14219" LastEditDate="2016-01-22T18:35:30.570" LastActivityDate="2016-01-22T18:35:30.570" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="25629" PostTypeId="1" AcceptedAnswerId="25630" CreationDate="2016-01-21T20:03:28.523" Score="3" ViewCount="428" Body="&lt;p&gt;I came downstairs and set my new clock to 12 pm. Its Sunday. I go back to work on Monday. My shift starts at 12pm, in a days time. Unbeknownst to me, the clock is defective. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At 45 seconds into the first minute, the second hand stutters in the same place for two seconds and then continues as normal. It will then happen again, 61 seconds after the second hand continues (where it is 46 seconds past on the clock, though at the time it will be actually 48 seconds past) and will continue like this.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If I just use this clock and no other, how late will I be for work tomorrow?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;This puzzle is based on a real clock I found on my desk, stuttering away and&#xA;telling me the wrong time.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="18250" LastActivityDate="2016-01-25T13:24:44.567" Title="The st-t-tuttering clock" Tags="|mathematics|time|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25630" PostTypeId="2" ParentId="25629" CreationDate="2016-01-21T20:20:25.903" Score="3" Body="&lt;p&gt;There are $60 \times 60 \times 24 = 86400$ seconds from when you set the clock until you are supposed to be at work.  After the first stutter, the clock loses 2 seconds for every 61 seconds that pass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(86400 - 45 - 2) / 61 \approx 1415.623$, so the clock will stutter (for 2 seconds) 1415 times (+1 time for the initial occurrence) before you are supposed to start work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus you will be late by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;$1416 \times 2 = 2832$ seconds&lt;/strong&gt;, or &lt;strong&gt;47 minutes, 12 seconds&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="4421" LastEditDate="2016-01-22T22:13:38.570" LastActivityDate="2016-01-22T22:13:38.570" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25631" PostTypeId="5" CreationDate="2016-01-21T20:24:25.967" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2016-01-21T20:24:25.967" LastActivityDate="2016-01-21T20:24:25.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25632" PostTypeId="4" CreationDate="2016-01-21T20:24:25.967" Score="0" Body="Use this tag for puzzles that require the context in which the puzzle was posted on Puzzling Stack Exchange. Do not use this tag for questions that belong on meta.puzzling.SE." OwnerUserId="16322" LastEditorUserId="16322" LastEditDate="2016-01-21T21:57:06.993" LastActivityDate="2016-01-21T21:57:06.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25633" PostTypeId="1" AcceptedAnswerId="25670" CreationDate="2016-01-21T20:27:31.000" Score="15" ViewCount="494" Body="&lt;p&gt;Not necessarily easy, but &quot;'E's-y&quot;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each &lt;a href=&quot;https://en.wikipedia.org/wiki/Word_square&quot;&gt;word square&lt;/a&gt; below, the &lt;code&gt;E&lt;/code&gt;s have been filled in for you.  Your task is to complete the rest of the square using only &lt;strong&gt;&lt;em&gt;five other distinct letters&lt;/em&gt;&lt;/strong&gt;.  That is, you may pick any five letters of the alphabet (other than &lt;code&gt;E&lt;/code&gt;), and use each of those letters as many times as you wish.  Each square uses a different set of five letters, although there may be overlap between squares.  (e.g. if square (1) uses &lt;code&gt;BFKLZ&lt;/code&gt;, square (2) might use &lt;code&gt;ABKMR&lt;/code&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solutions contain only relatively-common English words that should be known to an educated native English speaker.  There are no abbreviations, slang terms, archaic words, foreign words, or anything else unusual.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{rcrc}&#xA;  \raise{3em}1.&#xA;  &amp;amp;\begin{array}{|c|c|c|c|c|}&#xA;    \hline E&amp;amp;&amp;amp;&amp;amp;E&amp;amp;\\&#xA;    \hline &amp;amp;E&amp;amp;&amp;amp;&amp;amp;E\\&#xA;    \hline &amp;amp;&amp;amp;E&amp;amp;E&amp;amp;\\&#xA;    \hline E&amp;amp;&amp;amp;E&amp;amp;&amp;amp;\\&#xA;    \hline &amp;amp;E&amp;amp;&amp;amp;&amp;amp;\\&#xA;    \hline&#xA;  \end{array}&#xA;  \quad\quad\quad\quad\quad&#xA;  \raise{3em}2.\quad&#xA;  &amp;amp;\begin{array}{|c|c|c|c|c|}&#xA;    \hline &amp;amp;&amp;amp;&amp;amp;E&amp;amp;\\&#xA;    \hline &amp;amp;&amp;amp;&amp;amp;&amp;amp;E\\&#xA;    \hline &amp;amp;&amp;amp;&amp;amp;&amp;amp;E\\&#xA;    \hline E&amp;amp;&amp;amp;&amp;amp;E&amp;amp;\\&#xA;    \hline &amp;amp;E&amp;amp;E&amp;amp;&amp;amp;\\&#xA;    \hline&#xA;  \end{array}\\&#xA;  \\&#xA;  \raise{3em}3.&#xA;  &amp;amp;\begin{array}{|c|c|c|c|c|}&#xA;    \hline &amp;amp;&amp;amp;&amp;amp;E&amp;amp;\\&#xA;    \hline &amp;amp;&amp;amp;&amp;amp;&amp;amp;E\\&#xA;    \hline &amp;amp;&amp;amp;&amp;amp;E&amp;amp;\\&#xA;    \hline E&amp;amp;&amp;amp;E&amp;amp;&amp;amp;\\&#xA;    \hline &amp;amp;E&amp;amp;&amp;amp;&amp;amp;E\\&#xA;    \hline&#xA;  \end{array}&#xA;  \quad\quad\quad\quad\quad&#xA;  \raise{3em}4.&#xA;  &amp;amp;\begin{array}{|c|c|c|c|c|}&#xA;    \hline \;\;\;&amp;amp;&amp;amp;&amp;amp;&amp;amp;\\&#xA;    \hline &amp;amp;E&amp;amp;&amp;amp;&amp;amp;E\\&#xA;    \hline &amp;amp;&amp;amp;&amp;amp;E&amp;amp;\\&#xA;    \hline &amp;amp;&amp;amp;E&amp;amp;&amp;amp;\\&#xA;    \hline &amp;amp;E&amp;amp;&amp;amp;&amp;amp;E\\&#xA;    \hline&#xA;  \end{array}\\&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I can't prevent anyone from writing a computer program to solve these, I would encourage people to try it without.  Where's the fun in brute-forcing a solution?&lt;br&gt;&#xA;(Feel free to &lt;a href=&quot;http://www.litscape.com/word_tools/pattern_match.php&quot;&gt;use a computer&lt;/a&gt; to help you find word patterns for a single word, just not to solve the whole puzzle.)&lt;/p&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="4421" LastEditDate="2016-01-21T22:18:31.693" LastActivityDate="2016-01-22T17:43:20.487" Title="E's-y Word-Square Fill-in" Tags="|word|no-computers|word-square|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25634" PostTypeId="2" ParentId="2147" CreationDate="2016-01-21T21:01:11.817" Score="2" Body="&lt;p&gt;I find all the answers here way too complicated except Julia's&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using LOGIC; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Smith will ALWAYS shoots Brown first, because whoever he leaves alive will automatically shoot him. He has a better chance of survival if Jones is the one alive&lt;br&gt;&#xA;2) If Jones kills Brown, he is automatically dead.&lt;br&gt;&#xA;3) If Brown kills Jones, he is automatically dead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This stipulates that the other two will always fire on one another. Jones is safe as long as they are alive and has dramatically less chances of survival the moment one of them shoots at him FIRST (aka, if he kills one of them). There is literally no need for any percentages other than the initial ones already stated in the puzzle. Depending on the rules (if you are allowed to intentionally miss), either he misses intentionally, or he shoots Smith and hopes he misses.&lt;/p&gt;&#xA;" OwnerUserId="11251" LastActivityDate="2016-01-21T21:01:11.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25635" PostTypeId="2" ParentId="25633" CreationDate="2016-01-21T21:30:32.023" Score="7" Body="&lt;p&gt;Solution for #1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{|c|c|c|c|c|}&#xA;    \hline E&amp;amp;N&amp;amp;T&amp;amp;E&amp;amp;R\\&#xA;    \hline N&amp;amp;E&amp;amp;R&amp;amp;V&amp;amp;E\\&#xA;    \hline T&amp;amp;R&amp;amp;E&amp;amp;E&amp;amp;S\\&#xA;    \hline E&amp;amp;V&amp;amp;E&amp;amp;N&amp;amp;T\\&#xA;    \hline R&amp;amp;E&amp;amp;S&amp;amp;T&amp;amp;S\\&#xA;    \hline&#xA;  \end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Letters used: N, T, R, S, and V&lt;/p&gt;&#xA;" OwnerUserId="239" LastActivityDate="2016-01-21T21:30:32.023" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25636" PostTypeId="1" AcceptedAnswerId="25638" CreationDate="2016-01-21T21:46:25.727" Score="5" ViewCount="319" Body="&lt;p&gt;I really enjoyed attempting to find a solution to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/25098/filling-an-11-by-11-square&quot;&gt;Filling an 11-by-11 square&lt;/a&gt; (and am a fan of &lt;a href=&quot;https://puzzling.stackexchange.com/users/8874/gamow&quot;&gt;Gamow&lt;/a&gt;'s square puzzles), so I was inspired to make a puzzle in the same spirit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to fill all $121$ entries of an $11\times11$ grid with the numbers $0$, $1$, and any composite number less than or equal to $10$ such that the row sums and column sums contain the first $22$ prime numbers?&lt;/p&gt;&#xA;" OwnerUserId="16275" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2016-01-22T12:10:16.070" Title="Primes, composites, and an 11-by-11 square" Tags="|mathematics|combinatorics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25637" PostTypeId="2" ParentId="25598" CreationDate="2016-01-21T22:12:16.910" Score="3" Body="&lt;p&gt;If &lt;code&gt;a&lt;/code&gt; is the first digit &lt;code&gt;b&lt;/code&gt; is the second and third and &lt;code&gt;d&lt;/code&gt; is the last, then &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;20*a + 2*b = 10*b + d&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;because &lt;code&gt;its last two digits are double the first two digits&lt;/code&gt;.&#xA;Or, simplifying &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;20*a = 8*b + d&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Because &lt;code&gt;Its last digit is double the first digit&lt;/code&gt;, we can rewrite this as&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;20*a = 8*b + 2*a&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which amounts to&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;9*a = 4*b&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Since the 0 solution is forbidden, and both a and b are naturals less than 10, evidently the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4 9 9 8, where a = 4, b = 9 and d = a*2 = 8 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12376" LastEditorUserId="12376" LastEditDate="2016-01-21T22:21:37.263" LastActivityDate="2016-01-21T22:21:37.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25638" PostTypeId="2" ParentId="25636" CreationDate="2016-01-21T23:20:48.200" Score="13" Body="&lt;p&gt;If the row sums and the columns sums contain the first 22 prime numbers, their sum must be equal to the sum of the first 22 prime numbers, which is odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the sum of the row sums is equal to the sum of the column sums, because they are both equal to the total of all the numbers in the grid. Therefore, their sum must be even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the row and column sums cannot be the first 22 prime numbers.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2016-01-21T23:34:45.013" LastActivityDate="2016-01-21T23:34:45.013" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="25639" PostTypeId="1" CreationDate="2016-01-21T23:30:39.517" Score="6" ViewCount="997" Body="&lt;p&gt;That was too easy. I solved it during physics, although I did let some of my internet friends take a crack at it, since they seem to enjoy puzzles. You need practice. (Admittedly, so do I.) Here’s a puzzle that you may or may not enjoy, since you like Ted Hughes and I like Sylvia Plath, you like Shakespeare and I like Gabriel Rossetti. It’s some sort of conglomeration of a few of their quotes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hate also taken in consideration you statement concerning potatoes, and have concluded that at least you tried to make an effort to write a puzzle and make up for it. Potatoes will remain the love of my life, and I suppose you could take up second place. Take it or leave it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here’s your puzzle. It’s not relevant to anything, so I don’t think the answer will be as easy as yours was. Just find the first letter of the titles of the volumes/collections/poems in which the quotes come from for the first “tier” of the puzzle. The keyword should be relatively simple to guess, because I love it, and I italicized the important parts...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am dogged in my pursuit of confirming my turpitude, and execrate the Integrated Development Environment, enjoyment that arises as a result of the current days,&lt;/em&gt; my friend Gansey told me. We were walking through a museum, and I was thinking about you because I’d seen this picture of a woman raising her hand to touch something that she could not reach. You’re a wonderful guy, Henry, but sometimes your heart seems to race past all boundaries and out of my reach. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I saw an illustration. &lt;em&gt;A companion to an illustration in 1881, Can day from darkness ever take flight?&lt;/em&gt; My friend comments, and I look at him, the meaning of his words beyond me, but possibly not beyond you. I wonder if he thinks that I am a receptacle of knowledge into which he can toss whatever quote he’d been thinking of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Flies watch no resurrection in the sun.&lt;/em&gt; I replied with a quote of my own, as we exited the museum, and I thought about how much that painting spoke to me, wondering if things like this happened in real life. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the exchange with Gansey, I went to look for you, although, I didn’t look too hard, I have to admit. I was in a ponderous mood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The table-top, made of the same wood that James Potter’s wand is made of, that you smashed had been the broad plank top of my mother's heirloom sideboard- mapped with the scars of my whole life.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You have been the last dream of my soul,&lt;/em&gt; Henry, no matter how careless I happen to be... sometimes I feel like I should apologize for being me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;We got everything you want honey, we know the names&lt;/em&gt;, my friends had told me, and I’d followed them. There’s always something intoxicating about receiving validation from people you don’t see in real life, but, &lt;em&gt;“You are who you choose to be,”&lt;/em&gt; Ted Hughes told me, and my eyelids drooped shut and I drifted off to sleep. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;He declares there is nothing he can do for this beautiful woman who seems to be alive and dead. He is not a doctor. He can only pray.&lt;/em&gt; Gosh, Henry, there’s no hope for me, I keep making mistakes!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have fun with the puzzle, honey. I wrapped it up pretty nicely in my nonsensical rambling, which is mostly for your entertainment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Love, Zipporah&lt;/p&gt;&#xA;" OwnerUserId="11814" LastEditorUserId="11814" LastEditDate="2016-01-23T18:19:56.637" LastActivityDate="2016-01-23T18:19:56.637" Title="&quot;Love&quot; letter number three" Tags="|riddle|wordplay|knowledge|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25640" PostTypeId="2" ParentId="25098" CreationDate="2016-01-22T00:08:43.653" Score="14" Body="&lt;p&gt;I've found a solution &lt;a href=&quot;https://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1093812711&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;, which I've decided to blatantly steal since I think a lot of people really want to see a fully worked out solution. It is indeed impossible. The same method below can be used to show that a valid $n\times n$ matrix can only exist if $n$ is even and the missing sum is $\pm n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say the missing sum is $m$. We can assume by symmetry that $m\le 0$. There are 11 rows and columns whose sum is positive. Let's rearrange these positive row/columns so that they are all on the top/left of the matrix. Assuming there are $r$ positive rows, this means there are $11-r$ positive columns. Let's group all the positive columns on the left, and all positive rows on top. See the picture at the bottom, where I've given labels A, B, C and D to four quadrants of the matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let $a$ be the sum of the numbers in section A, and the same with $b,c,d$. Then&#xA;$$\begin{align}&#xA;a+b+c+d&#xA;&amp;amp;=(\text{sum of all rows and columns})/2\\&#xA;&amp;amp;=((-11)+(-10)+\dots+11-m)/2\\&#xA;&amp;amp;=-m/2&#xA;\end{align}$$&#xA;We also know that&#xA;$$&#xA;\begin{align}&#xA;(a+b)+(a+c)&#xA;&amp;amp;=\text{sum of all positive rows/columns}\\&#xA;&amp;amp;=1+2+\dots+11\\&#xA;&amp;amp;=66&#xA;\end{align}$$&#xA;Combining these last two observations,&#xA;$$&#xA;-m/2=a+b+c+d=(2a+b+c)-a+d=66-a+d&#xA;$$&#xA;Here's the kicker. There are $r(11-r)$ entries in section $A$, meaning that $a\le r(11-r)$. No matter what $r$ is, the quantity $r(11-r)$ is at most $30^*$. Similarly, $d\ge -30$. This means that&#xA;$$&#xA;-m/2\ge 66-30-30=6,&#xA;$$&#xA;implying that $m\le -12$, a contradiction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ruNo2.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ruNo2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;${}^*$ To make this proof to work for a general $n\times n$ matrix, instead of just $n=11$, use the inequality $ab\le (a+b)^2/4$, which shows $r(n-r)\le n^2/4$. We then get that $-m/2=\frac{n(n+1)}2-a+d\ge \frac{n(n+1)}2-\frac{n^2}4-\frac{n^2}4=n/2$, implying that $m\le -n$, which means $m=-n$. Since the missing sum must be even, this means $n$ must be even.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2016-01-22T00:29:47.570" LastActivityDate="2016-01-22T00:29:47.570" CommentCount="0" CommunityOwnedDate="2016-01-22T00:08:43.653" ContentLicense="CC BY-SA 3.0" />
  <row Id="25641" PostTypeId="1" AcceptedAnswerId="25642" CreationDate="2016-01-22T00:25:10.750" Score="4" ViewCount="658" Body="&lt;p&gt;Three prisoners were blindfolded and given a shovel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sick and twisted guard told the three men: “No matter what happens tonight, at least one man is certain to die. However, it will not be the man with the deepest grave. But, if any of you begin to dig, only the man with the deepest grave will survive. If none of you dig, only one man will die.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With no murders or suicides, the three prisoners survived the night. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The guard’s honest husband, who was just as twisted, said the guard was always truthful. How can this be?  &lt;/p&gt;&#xA;" OwnerUserId="18340" LastEditorUserId="5373" LastEditDate="2018-01-21T15:46:57.953" LastActivityDate="2018-01-22T15:05:19.607" Title="The Honest Guard" Tags="|lateral-thinking|situation|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25642" PostTypeId="2" ParentId="25641" CreationDate="2016-01-22T00:32:44.057" Score="7" Body="&lt;p&gt;When the guard said&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;If none of you dig, only one man will die,&quot; he didn't specify which man&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so I'm guessing that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; None of the prisoners dug, and the honest guard died of a terminal illness&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17147" LastEditorUserId="17147" LastEditDate="2016-01-22T00:37:01.840" LastActivityDate="2016-01-22T00:37:01.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25643" PostTypeId="2" ParentId="25641" CreationDate="2016-01-22T00:32:46.250" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; None of them dug.  The guard's husband died of natural causes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15942" LastActivityDate="2016-01-22T00:32:46.250" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25644" PostTypeId="2" ParentId="25641" CreationDate="2016-01-22T02:15:22.417" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All of them dug, and all the graves were exactly the same depth. The guard had said that only &lt;em&gt;the man&lt;/em&gt; with the deepest grave would die. There is no such man, so no one dies. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="11110" LastEditDate="2018-01-22T15:05:19.607" LastActivityDate="2018-01-22T15:05:19.607" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25645" PostTypeId="1" AcceptedAnswerId="25646" CreationDate="2016-01-22T02:36:15.733" Score="1" ViewCount="175" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;I cross blades, but with no adversary.&lt;/p&gt;&#xA;&lt;p&gt;I am sharp, but I am still entrusted to children without thought.&lt;/p&gt;&#xA;&lt;p&gt;I turn one into two, two into four. But never vice-versa.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;" OwnerUserId="17744" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-01-22T02:56:04.043" Title="Crossing Blades. Who am I?" Tags="|riddle|word|wordplay|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25646" PostTypeId="2" ParentId="25645" CreationDate="2016-01-22T02:39:52.853" Score="3" Body="&lt;p&gt;The Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Scissors&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Blades cross paths to cut.&lt;br&gt;&#xA; Cuts things up, which divides them.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12014" LastEditorUserId="12014" LastEditDate="2016-01-22T02:56:04.043" LastActivityDate="2016-01-22T02:56:04.043" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="25647" PostTypeId="2" ParentId="25614" CreationDate="2016-01-22T04:31:38.167" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;In this case, the riddle is not referring to a person, but a company: &lt;strong&gt;Fairchild Semiconductor.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It took some research to verify the clues that were standing out to me, but I pieced enough together to be rather certain of my answer. Below, I don't have ALL of the clues answered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Child&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'child' is in the company's name.&lt;p&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Born as the result of eight betrayals&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Traitorous Eight is a term given to the eight employees who left Shockley Semiconductor Laboratory, to create Fairchild Semiconductor. Shockley considered this a 'betrayal'. (so Fairchild Semiconductor was &quot;born as the result of eight betrayals&quot;) Source : &lt;a href=&quot;https://en.wikipedia.org/wiki/Traitorous_eight&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Traitorous_eight&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I played in the sand, when everyone else played elsewhere&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Playing in the sand&quot; could refer to the fact that silicon is used to make semiconductors. Silicon is a primary ingredient in common sand. Fairchild was focused on a different type of semiconductor than most research at the time, which could allude to &quot;everyone else played elsewhere&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I practiced as I grew &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Traitorous Eight worked in a garage at first, &quot;practicing&quot;, or rather perfecting, their methods of creating a special type of transistor known as a Moore transistor.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;half way to becoming a maestro&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm not 100% clear on this one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;then I became a key player in big business&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They eventually would beat out all competition for their product.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;my success was amplified by my logical approach&lt;br&gt;&#xA;my ambition guided me, and those I supported, in giant leaps  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I don't know a lot about the specifics here. I notice many references to things that semiconductors can be used for - amplifiers, logic gates, 'guide' might refer to rectifiers...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;when things got too noisy I went one way&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most likely a reference to the in-fighting within the company. The Eight never really did get along with each other, brilliant though they were.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is this fair&lt;br&gt;&#xA;I was betrayed many many times like my parents before me&lt;br&gt;&#xA;and eventually I was no longer special&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;After more reading, we discover that Fairchild Semiconductor itself lost the original eight over time to its infighting... more betrayals, if you will... and quickly became a nobody in the market that it was once the king of.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A great riddle, it was fun to solve!&lt;/p&gt;&#xA;" OwnerUserId="18590" LastEditorUserId="18590" LastEditDate="2016-01-23T19:11:39.647" LastActivityDate="2016-01-23T19:11:39.647" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25649" PostTypeId="1" AcceptedAnswerId="25650" CreationDate="2016-01-22T06:29:15.400" Score="3" ViewCount="208" Body="&lt;p&gt;Some considered me lucky, others not&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Turn me sideways,  I can give you endless lot&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Turn me upside down, I can give you the same pot&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But be careful with that, I might lose oxygen and rot&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I look like 20 but I'm younger than you thought&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="16036" LastEditorUserId="16036" LastEditDate="2016-01-22T06:41:59.223" LastActivityDate="2016-01-22T07:00:48.850" Title="Am I really lucky?" Tags="|riddle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25650" PostTypeId="2" ParentId="25649" CreationDate="2016-01-22T06:44:14.617" Score="3" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The number 8&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Turned sideways it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Infinity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Upside down it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Also 8&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might lose oxygen and rot:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oxygen's atomic number is 8&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not sure about the hint&lt;/p&gt;&#xA;" OwnerUserId="17147" LastActivityDate="2016-01-22T06:44:14.617" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25651" PostTypeId="2" ParentId="25649" CreationDate="2016-01-22T07:00:48.850" Score="1" Body="&lt;p&gt;I know this was already answered, but I think the answer could also be a penny.&lt;/p&gt;&#xA;" OwnerUserId="18608" LastActivityDate="2016-01-22T07:00:48.850" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25652" PostTypeId="2" ParentId="25598" CreationDate="2016-01-22T08:40:53.117" Score="1" Body="&lt;p&gt;&lt;code&gt;x 0 0 2x&lt;/code&gt; where &lt;code&gt;0&amp;lt;x&amp;lt;4&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Its last digit is double the first digit [x and 2x, check]&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;its middle two digits are the same [0&amp;amp;0, check]&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;its last two digits are double the first two digits [x+0 and 0+2x, check]&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="16842" LastActivityDate="2016-01-22T08:40:53.117" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25653" PostTypeId="2" ParentId="25636" CreationDate="2016-01-22T12:00:59.287" Score="1" Body="&lt;p&gt;I had some trouble to understand easily @f&quot; answer, so I'll post a more intuitive one.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The matrix is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{bmatrix}a_{0,0} &amp;amp; a_{0,1}&amp;amp;... &amp;amp;a_{0,10}\\a_{1,0}&amp;amp;...&amp;amp;..&amp;amp;...\\...&amp;amp;...&amp;amp;...&amp;amp;...\\a_{10,0} &amp;amp;...&amp;amp;...&amp;amp; a_{10,10}\end{bmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{row sum }r_i=\sum_{k=0}^{10}{a_{i,k}}\qquad i=0...10$$&#xA;$$\text{column sum }c_i=\sum_{k=0}^{10}{a_{k,i}}\qquad i=0...10$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we define&#xA;$$\text{sum of the sums of the rows }R=\sum_{i=0}^{10}{r_{i}}$$&#xA;and&#xA;$$\text{sum of the sums of the columns }C=\sum_{i=0}^{10}{c_{i}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1)$ From the question we know that $C+R=\text{sum of first 22 primes}=2n + 1= 791$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(2)$ From the construction of $C$ and $R$ we know that $C=R$ since they are the sum of all of the values of the matrix. Therefore their sum must be in the form $C+R= 2C = 2R = 2n$ which is in contradiction with $(1)$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Therefore, the row and column sums cannot be the first 22 prime numbers. -@f&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4652" LastEditorUserId="4652" LastEditDate="2016-01-22T12:10:16.070" LastActivityDate="2016-01-22T12:10:16.070" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25654" PostTypeId="1" AcceptedAnswerId="25678" CreationDate="2016-01-22T12:48:56.763" Score="2" ViewCount="948" Body="&lt;p&gt;There is one missing item to make the pattern logically complete&lt;img src=&quot;https://i.stack.imgur.com/kgi3C.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="18447" LastEditorUserId="11876" LastEditDate="2016-01-22T16:21:11.850" LastActivityDate="2016-09-17T13:23:39.670" Title="What is the missing item that completes the pattern?" Tags="|pattern|visual|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25656" PostTypeId="2" ParentId="25598" CreationDate="2016-01-22T13:02:02.263" Score="1" Body="&lt;p&gt;For the last digit to be double the first digit but also the last two digits being double the first two digits then the first two digits must differ by 5, giving (a)(a+5)(a+5)(2a). The restriction on the third digit then requires a+5 = 2a+1 (from the carry from 2(a+5)) which simplifies to a=4.&lt;/p&gt;&#xA;" OwnerUserId="12593" LastActivityDate="2016-01-22T13:02:02.263" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="25657" PostTypeId="2" ParentId="2147" CreationDate="2016-01-22T13:04:28.773" Score="2" Body="&lt;p&gt;-While Smith is alive, Jones and Brown will never shoot each other, as if they succeed, they're instantly killed after.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Knowing this, Smith also knows he will never have the first shot when it eventually comes to a 1v1, he can discard that. His choice now comes to who he wants to take out first (as a perfect killer he is) and who he wants to make that first 1v1 shot. Obviously optimal strategy for Smith is killing Brown as fast as he can and then Jones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-If Brown realizes this, he will use his first and maybe only shot to try to kill Smith as otherwise he is dead for sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These leaves us with &lt;strong&gt;only 4 possible scenarios&lt;/strong&gt;, regardless the chances of each one coming to happen:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A) Jones kills Smith, then Brown shoots Jones. Long term survival chances for Jones: less than 20%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B) Jones kills Brown, then Smith shoots Jones. Long (and short) term survival chances for Jones: 0%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C) Jones fails first shot, then Brown kills Smith, then Jones shoots Brown. Long term survival chances for Jones: Higher than 50%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D) Jones fails first shot, Brown too, Smith kills Brown, and Jones has one chance to kill Smith. Long term survival chances for Jones: exactly 50%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-As you can see, it's highly interesting for Jones to fail the first shot, so if he can do that intentionally, it would be highly recommended.&lt;/p&gt;&#xA;" OwnerUserId="18576" LastEditorUserId="18576" LastEditDate="2016-01-22T13:19:45.617" LastActivityDate="2016-01-22T13:19:45.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25659" PostTypeId="2" ParentId="25633" CreationDate="2016-01-22T13:37:57.560" Score="8" Body="&lt;p&gt;Solution for #2   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RnI2Y.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RnI2Y.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Used the letters on the right (S,D,G,A,U)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And Solution #4 (probably not what you got)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4VzHB.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4VzHB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Used letters T, R, O, U, S&lt;/p&gt;&#xA;" OwnerUserId="14233" LastEditorUserId="14233" LastEditDate="2016-01-22T15:07:04.477" LastActivityDate="2016-01-22T15:07:04.477" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25660" PostTypeId="2" ParentId="25654" CreationDate="2016-01-22T13:41:42.793" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;@ $$\circ\\\circ\\\circ\\\circ\\\circ\\\circ $$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The sets of dots interchange in vertical/horizontal. Number of dots in each set increases by one (so the set containing six dots is missing). The number of black and white dots are of equal amount (18 of each, this seems to be the most logical answer).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12901" LastEditorUserId="18250" LastEditDate="2016-01-22T21:35:37.227" LastActivityDate="2016-01-22T21:35:37.227" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25661" PostTypeId="2" ParentId="25633" CreationDate="2016-01-22T14:40:26.517" Score="5" Body="&lt;p&gt;3: snavh&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ashEs&#xA;shavE&#xA;havEn&#xA;EvEns&#xA;sEnsE&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="5169" LastActivityDate="2016-01-22T14:40:26.517" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25662" PostTypeId="1" AcceptedAnswerId="25665" CreationDate="2016-01-22T14:57:39.927" Score="3" ViewCount="1486" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Glare at me, while I at you&lt;/p&gt;&#xA;&lt;p&gt;Worship this familiar with glimmer&lt;/p&gt;&#xA;&lt;p&gt;While I stand in unfrequented gloom&lt;/p&gt;&#xA;&lt;p&gt;Common, still I elude avaricious hands&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15991" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-01-22T21:46:25.120" Title="What Am I? - A Riddle (GLARE AT ME, WHILE I AT YOU)" Tags="|riddle|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25663" PostTypeId="2" ParentId="25662" CreationDate="2016-01-22T15:23:53.963" Score="7" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; your reflection in a mirror?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Glare at me, while I at you&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is obviously what you see looking in a mirror.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Worship this familiar with glimmer&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; appears to refer both to the familiarity of your own reflection, and the glimmer from the surface of the mirror.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;While I stand in unfrequented gloom&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; could be a reference to our inability to frequent the world we see in reflections,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Common, still I elude avaricious hands&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a reference to both the commonplace nature of reflections, combined with our inability to grasp something in a reflection.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user9278" LastEditorUserId="4421" LastEditDate="2016-01-22T16:07:03.543" LastActivityDate="2016-01-22T16:07:03.543" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25664" PostTypeId="2" ParentId="25662" CreationDate="2016-01-22T16:10:28.167" Score="3" Body="&lt;p&gt;I think it might be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your shadow. (I personally think the mirror answer explains the first two stanzas better though)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Glare at me, while I at you:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This could be explained by the fact your shadow mimics your actions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Worship this familiar with glimmer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; your shadow spreads across surfaces, unevenly , depending on the sources of light casting it. This can make it shimmer and stuff. (And light glimmers)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While I stand in unfrequented gloom:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You (most of the time) stand in the feet of your shadow's feet, and yet cannot frequent the space your shadow occupies. (otherwise your shadow disappears)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Common, still I elude avaricious hands:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Shadows are everywhere, yet they are impossible to grasp/take/etc...  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastActivityDate="2016-01-22T16:10:28.167" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25665" PostTypeId="2" ParentId="25662" CreationDate="2016-01-22T16:12:38.513" Score="7" Body="&lt;p&gt;I think you are..&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a star!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;  &lt;em&gt;Glare at me, while I at you&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I suppose stars can resemble eyes, so when you look at them, you can imagine them looking back at you.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Author's note:&lt;/strong&gt; &quot;Glare&quot; is a word-play. It can mean &quot;stare&quot;, which is what you do at stars, and it can also mean &quot;shine&quot;, which is what they do at you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Worship this familiar with glimmer&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is safe to say that everyone sees the starry night sky quite frequently. Those &lt;strong&gt;glimmering&lt;/strong&gt; stars are quite &lt;strong&gt;familiar&lt;/strong&gt;, are they not?&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Author's note:&lt;/strong&gt; The answer is a word-play too. Not only are celestial stars worshipped, but we also adore famous celebrities (&quot;stars&quot;) with paparazzi camera flashes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;While I stand in unfrequented gloom&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stars are surrounded by the dark night sky.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Author's note:&lt;/strong&gt; Stars are also solitary, never visited. Celebrities could also certainly feel lonely where they are.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Common, still I elude avaricious hands&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are countless stars dotting the sky, eternally out of reach.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Author's note:&lt;/strong&gt; Indeed! Also, celebrities are common in general, but those who desire to personally get to know them will most often find them out of reach.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Author's note:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Now I understand all the more why we call celebrities &quot;stars&quot;!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18429" LastEditorUserId="18429" LastEditDate="2016-01-22T16:28:03.510" LastActivityDate="2016-01-22T16:28:03.510" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25667" PostTypeId="1" AcceptedAnswerId="25669" CreationDate="2016-01-22T17:07:30.357" Score="-1" ViewCount="762" Body="&lt;p&gt;At top speed a child inside a train jumps off and is unharmed. Now influenced, another child jumps off and sprains his ankle. Landing on cement and without assistance, the children rush to the hospital to provide moral support for their relative who is in critical condition as a result of exiting the same train while it was not moving. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why weren’t either of the children seriously injured? &lt;/p&gt;&#xA;" OwnerUserId="18340" LastActivityDate="2016-01-22T19:29:40.650" Title="The Train Jumpers" Tags="|riddle|" AnswerCount="6" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25668" PostTypeId="2" ParentId="25667" CreationDate="2016-01-22T17:20:12.723" Score="8" Body="&lt;p&gt;My answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The children were on a train that for whatever reason had a very low top speed. (Could be an antique, a toy train, w/e). They jumped off because they could see the train was about to have an accident. The train crashes afterwards , and their relative who hadn't jumped gets out, injured, and is rushed to the hospital. Children go to visit the relative.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastActivityDate="2016-01-22T17:20:12.723" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25669" PostTypeId="2" ParentId="25667" CreationDate="2016-01-22T17:30:30.010" Score="3" Body="&lt;p&gt;Here is what I think happened:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What really caught my attention was &quot;Landing on cement&quot;. This to me implies that the children's relative did not &quot;step off&quot; the train as one would normally do. Perhaps the relative tripped and fell between the train and the platform. Falls can be quite nasty if you hit the ground in a bad posture, which can explain the critical condition.&lt;br&gt;&#xA;&lt;hr&gt;&#xA; As for why the children weren't seriously injured as they jumped off the train while it was going at top speed:&lt;br&gt;&#xA;&lt;br&gt;&#xA; Maybe &quot;top speed&quot; doesn't refer to the train's actual top speed, but rather the top speed the driver is legally allowed to drive it at in that area. The children would probably jump off once the train is close to its destination, which means its top speed is probably quite low as it is pulling in to the station. Also the fact that they &quot;rush to the hospital&quot; implies that they saw the accident happening, which makes it even more likely that they jumped off the train shortly before it stopped.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Maybe the train's top speed wasn't that high, as Spacemonkey suggested.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Maybe they jumped off and landed on something soft? Though that wouldn't work well with the &quot;Landing on cement&quot; piece.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18429" LastEditorUserId="18429" LastEditDate="2016-01-22T17:35:39.343" LastActivityDate="2016-01-22T17:35:39.343" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25670" PostTypeId="2" ParentId="25633" CreationDate="2016-01-22T17:43:20.487" Score="8" Body="&lt;p&gt;Since the answers all came out piecemeal, I figured I'd post them all together in one answer here, to make it easier for people to see the solutions.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\require{color}&#xA;\definecolor{g}{RGB}{0, 180, 0}&#xA;\begin{array}{rcrc}&#xA;  \raise{3em}1.&#xA;  &amp;amp;NRSTV\quad\begin{array}{|c|c|c|c|c|}&#xA;    \hline E&amp;amp;\color{g}N&amp;amp;\color{g}T&amp;amp;E&amp;amp;\color{g}R\\&#xA;    \hline \color{g}N&amp;amp;E&amp;amp;\color{g}R&amp;amp;\color{g}V&amp;amp;E\\&#xA;    \hline \color{g}T&amp;amp;\color{g}R&amp;amp;E&amp;amp;E&amp;amp;\color{g}S\\&#xA;    \hline E&amp;amp;\color{g}V&amp;amp;E&amp;amp;\color{g}N&amp;amp;\color{g}T\\&#xA;    \hline \color{g}R&amp;amp;E&amp;amp;\color{g}S&amp;amp;\color{g}T&amp;amp;\color{g}S\\&#xA;    \hline&#xA;  \end{array}&#xA;  \quad\quad\quad&#xA;  \raise{3em}2.&#xA;  &amp;amp;ADGSU\quad&#xA;  \begin{array}{|c|c|c|c|c|}&#xA;    \hline \color{g}S&amp;amp;\color{g}A&amp;amp;\color{g}G&amp;amp;E&amp;amp;\color{g}S\\&#xA;    \hline \color{g}A&amp;amp;\color{g}D&amp;amp;\color{g}A&amp;amp;\color{g}G&amp;amp;E\\&#xA;    \hline \color{g}G&amp;amp;\color{g}A&amp;amp;\color{g}U&amp;amp;\color{g}G&amp;amp;E\\&#xA;    \hline E&amp;amp;\color{g}G&amp;amp;\color{g}G&amp;amp;E&amp;amp;\color{g}D\\&#xA;    \hline \color{g}S&amp;amp;E&amp;amp;E&amp;amp;\color{g}D&amp;amp;\color{g}S\\&#xA;    \hline&#xA;  \end{array}\\&#xA;  \\&#xA;  \raise{3em}3.&#xA;  &amp;amp;AHNSV\quad\begin{array}{|c|c|c|c|c|}&#xA;    \hline \color{g}A&amp;amp;\color{g}S&amp;amp;\color{g}H&amp;amp;E&amp;amp;\color{g}S\\&#xA;    \hline \color{g}S&amp;amp;\color{g}H&amp;amp;\color{g}A&amp;amp;\color{g}V&amp;amp;E\\&#xA;    \hline \color{g}H&amp;amp;\color{g}A&amp;amp;\color{g}V&amp;amp;E&amp;amp;\color{g}N\\&#xA;    \hline E&amp;amp;\color{g}V&amp;amp;E&amp;amp;\color{g}N&amp;amp;\color{g}S\\&#xA;    \hline \color{g}S&amp;amp;E&amp;amp;\color{g}N&amp;amp;\color{g}S&amp;amp;E\\&#xA;    \hline&#xA;  \end{array}&#xA;  \quad\quad\quad&#xA;  \raise{3em}4.&#xA;  &amp;amp;ORSTU\quad\begin{array}{|c|c|c|c|c|}&#xA;    \hline \color{g}T&amp;amp;\color{g}R&amp;amp;\color{g}O&amp;amp;\color{g}U&amp;amp;\color{g}T\\&#xA;    \hline \color{g}R&amp;amp;E&amp;amp;\color{g}U&amp;amp;\color{g}S&amp;amp;E\\&#xA;    \hline \color{g}O&amp;amp;\color{g}U&amp;amp;\color{g}T&amp;amp;E&amp;amp;\color{g}R\\&#xA;    \hline \color{g}U&amp;amp;\color{g}S&amp;amp;E&amp;amp;\color{g}R&amp;amp;\color{g}S\\&#xA;    \hline \color{g}T&amp;amp;E&amp;amp;\color{g}R&amp;amp;\color{g}S&amp;amp;E\\&#xA;    \hline&#xA;  \end{array}\\&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2016-01-22T17:43:20.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25671" PostTypeId="2" ParentId="25667" CreationDate="2016-01-22T17:53:23.873" Score="3" Body="&lt;p&gt;Here's an alternative take.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The train that the two children jumped from safely was &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;a children's train ride at Disney World or a comparable amusement park (but really, &lt;em&gt;is&lt;/em&gt; there one?).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And this explains the low top speed of the train, as well as why they landed on cement.  You know, ADA, and all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The children's relative was also&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; on a fun trip to the park and tried to join the kids on the train ride.  However, being an adult and far too big, the relative could not maneuver in his/her seat.  This resulted in the relative hitting his/her head on a low tunnel, then needing to be pried out of the small seat, unconscious,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;while the children were able to slide out of any restraints and jump to safety.  &lt;/p&gt;&#xA;" OwnerUserId="11452" LastActivityDate="2016-01-22T17:53:23.873" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25672" PostTypeId="2" ParentId="25639" CreationDate="2016-01-22T17:55:27.457" Score="4" Body="&lt;p&gt;&lt;strong&gt;[INCOMPLETE ANSWER]&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am kinda stuck so I'll just post what I have so far in hopes of inspiring someone!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Italicized sections:&lt;/strong&gt;&lt;br&gt;&#xA;In each section we are looking for the first letter of the title of the volume/collection/poem/movie/etc that the quote in (or described by) the section is from.&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;em&gt;I am dogged in my pursuit of confirming my turpitude, and execrate the Integrated Development Environment, enjoyment that arises as a result of the current days&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I haven't been able to solve this one.&lt;br&gt;&#xA;&lt;br&gt;&#xA; This is one messed up sentence, if it can even be called that. I've tried Googling for parts of this sentence, including variations by word synonyms, but to no avail.&lt;br&gt;&#xA;&lt;br&gt;&#xA; On another note, &lt;em&gt;Integrated Development Environment&lt;/em&gt; has no place in the sentence. Maybe the letter of this section is simply I from IDE? That would however be quite strange as then this would be the only section that is not a quote from something.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;&lt;em&gt;A companion to an illustration in 1881, Can day from darkness ever take flight?&lt;/em&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.rossettiarchive.org/docs/2-1881.1stedn.rad.html#A.R.131&quot; rel=&quot;nofollow&quot;&gt;Dante Gabriel Rossetti, Ballads and Sonnets (1881), &quot;&lt;strong&gt;F&lt;/strong&gt;ound&quot;&lt;/a&gt;.&lt;br&gt;&#xA; First letter: &lt;strong&gt;F&lt;/strong&gt;.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;em&gt;Note:&lt;/em&gt; actual quote: &lt;em&gt;&quot;Can day from darkness ever &lt;strong&gt;again&lt;/strong&gt; take flight?&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;&lt;em&gt;Flies watch no resurrection in the sun&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sylvia Plath, &lt;strong&gt;N&lt;/strong&gt;ovember Graveyard (1956)&lt;br&gt;&#xA; First letter: &lt;strong&gt;N&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;&lt;em&gt;The table-top, made of the same wood that James Potter’s wand is made of, that you smashed had been the broad plank top of my mother's heirloom sideboard- mapped with the scars of my whole life.&lt;/em&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ted Hughes, The &lt;strong&gt;M&lt;/strong&gt;inotaur.&lt;br&gt;&#xA; First letter: &lt;strong&gt;M&lt;/strong&gt;. (Author said to ignore &quot;the&quot;s.)&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;em&gt;Note:&lt;/em&gt; actual quote:&lt;br&gt; &#xA; &quot;The mahogany table-top you smashed&lt;br&gt;&#xA; Had been the broad plank top&lt;br&gt;&#xA; Of my mother’s heirloom sideboard—&lt;br&gt;&#xA; Mapped with the scars of my whole life.&quot;&lt;br&gt;&#xA;&lt;br&gt;&#xA; James Potter's (from Harry Potter) wand is made of mahogany.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;&lt;em&gt;We got everything you want honey, we know the names&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Guns N' Roses, &lt;strong&gt;W&lt;/strong&gt;elcome To The Jungle.&lt;br&gt;&#xA; First letter: &lt;strong&gt;W&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;&lt;em&gt;You are who you choose to be&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ted Hughes, The &lt;strong&gt;I&lt;/strong&gt;ron Man. (Again, ignore &quot;the&quot;s.)&lt;br&gt;&#xA; First letter: &lt;strong&gt;I&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;&lt;em&gt;We came into the world like brother and brother, and now let's go hand in hand, not one before another&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; William Shakespeare, The &lt;strong&gt;C&lt;/strong&gt;omedy of Errors. (Again, ignore &quot;the&quot;s.)&lt;br&gt;&#xA; First letter: &lt;strong&gt;C&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;So that gives us the following letters:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; _ F N M W I C ('_' for unsolved section.)&lt;br&gt;&#xA;&lt;br&gt;&#xA; I don't know what to make of these.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18429" LastEditorUserId="18429" LastEditDate="2016-01-23T16:45:55.517" LastActivityDate="2016-01-23T16:45:55.517" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="25673" PostTypeId="2" ParentId="25667" CreationDate="2016-01-22T18:40:47.223" Score="0" Body="&lt;p&gt;It depends on where and when you jumped:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The relative jumped off the train onto the adjacent tracks and was hit by an oncoming train going the other direction.  This second train then stopped upon realizing that it had hit someone, while the first one started moving.  It got to full speed before anyone knew what was going on - that's when the first child jumped from one train to the other, an easy jump since they're at the same height.  He had time because he was in the front car of the train, while the relative was hit at the front of the second train.  However, by the time the second child went to jump, he was too late and had to jump from the train to the ground, spraining his ankle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16131" LastActivityDate="2016-01-22T18:40:47.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25674" PostTypeId="2" ParentId="25667" CreationDate="2016-01-22T18:50:09.010" Score="2" Body="&lt;p&gt;I know this is answered, but here's an alternative answer that explains how it might have happened in a different way:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The train doesn't move sideways, it goes up and down (It's a novelty restaurant made by a train carriage attached under a helium balloon).&#xA; As the train starts to ascend at full upwards speed it is still not very high above the ground, the first child jumps at this point.&#xA; After the first child jumps, the pilot stopped the ascent and the balloon slowed its ascent.&#xA; The second child jumps a few seconds later and it's a bit higher, so he sprains his ankle.&#xA; The balloon continues slowing its ascent and stops a bit higher still.&#xA; The relative then jumps when the train is no longer moving, hovering at a fixed height and gets badly injured.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18522" LastActivityDate="2016-01-22T18:50:09.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25675" PostTypeId="2" ParentId="25667" CreationDate="2016-01-22T19:29:40.650" Score="2" Body="&lt;p&gt;Another alternative:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The train is metaphorical for drinking/doing drugs. The first child &quot;jumped off&quot; or stopped participating when things were going at full speed. They avoided injury. The next child waited a little too long and ended up with some damage from the night, but avoided most of it. The relative partied way too hard and ended up needing medical intervention for his addiction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2016-01-22T19:29:40.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25676" PostTypeId="2" ParentId="25553" CreationDate="2016-01-22T20:21:27.450" Score="3" Body="&lt;p&gt;Tile the interior of this polygon with unit squares, then color these black and white in checkerboard fashion. We will prove that the number black squares differs from the number of white squares, which proves a domino tiling is impossible (since every domino covers one square of each color). To this end, let $B$ and $W$ be the number of black and white squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's place a positive ion on each edge of each internal black square, giving each black square a charge of +4. Similarly, place a negative ion on each edge of each white square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pacXa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pacXa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total electrical charge of the polygon's interior is equal to $4\times(B-W)$. However, the +/- ions on edges between neighboring squares cancel out. The only uncanceled charges are on the perimeter, so the net charge is also given by the black perimeter minus the white perimeter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now show the perimeter is white/black unbalanced. Imagine you walk around the perimeter, starting at a corner with, say, a black edge. As you walk along this side, you will alternate black and white, finishing with a black edge (since the side length is odd). You then make a 90${}^\circ$ or 270${}^\circ$ turn and start walking along the next side, again starting with black. Continuing this process, every side has one more black edge than white, so the perimeter is more black than white.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting this all together, we get that&#xA;$$&#xA;B-W=\tfrac14(\text{black perimeter}-\text{white perimeter})\neq0&#xA;$$&#xA;so the number of white squares differs from that of black, so a tiling is impossible. &lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2016-01-22T20:21:27.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25677" PostTypeId="1" AcceptedAnswerId="25698" CreationDate="2016-01-22T20:38:48.300" Score="16" ViewCount="1025" Body="&lt;p&gt;You're a cosmonaut dedicated to exploring and understanding strange new worlds. The latest world you've arrived on has a species with three distinct genders which you've named Alpha, Beta, and Gamma. After careful observation, you determine the rules for the gender of offspring are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A Gender cannot mate with its own gender.&lt;/li&gt;&#xA;&lt;li&gt;If an Alpha and a Beta mate, they produce two Gammas.&lt;/li&gt;&#xA;&lt;li&gt;If an Alpha and a Gamma mate, they produce two Betas.&lt;/li&gt;&#xA;&lt;li&gt;If a Beta and a Gamma mate, they produce two Alphas.&lt;/li&gt;&#xA;&lt;li&gt;After mating, these strange creatures promptly die. (They can only mate once)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You start to wonder: how can this species exist? Surely given enough time, all the members of this species would be of one gender, unable to mate, and the species would die off. After pondering and observing a while, inspiration strikes. Now satisfied that this is impossible, you leave the planet in search of more interesting puzzles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did you know this species can't die off?&lt;/p&gt;&#xA;" OwnerUserId="18103" LastEditorUserId="12408" LastEditDate="2016-01-22T20:52:42.413" LastActivityDate="2016-01-23T10:07:02.330" Title="A Strange Planet with Strange Genders" Tags="|mathematics|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25678" PostTypeId="2" ParentId="25654" CreationDate="2016-01-22T20:42:15.557" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The amount of black and white dots are supposed to be equal. In that case there is currently 12 white dots and 18 black dots. The pattern is that there are (including the missing one) 8 different set of dots each with a different number of dots from 1 to 8. &lt;br&gt;&lt;br&gt;The missing one must be 6 dots long. To make the amount of black and white dots equal, you need 6 white dots. So the missing one will be a set of 6 white dots. &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It will be vertical because dots with even number of dots are vertical and an odd number of dots is horizontal. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/NeFXY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NeFXY.png&quot; alt=&quot;Dots&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18250" LastEditorUserId="18250" LastEditDate="2016-09-17T13:23:39.670" LastActivityDate="2016-09-17T13:23:39.670" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25680" PostTypeId="2" ParentId="25677" CreationDate="2016-01-22T20:50:39.143" Score="8" Body="&lt;p&gt;Give each gender a value: Alpha is 1, Beta is 2, Gamma is 3. Then consider what happens to the total value when two mate.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Alpha and Beta produce two Gamma: +3&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Alpha and Gamma produce two Beta: +0&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Beta and Gamma produce two Alpha: -3&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The total value always remains the same mod 3. In particular, if it is not a multiple of 3, then it will never be a multiple of 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if the number of creatures is divisible by 3, then if they were all the same gender, the total would also have to be divisible by 3. So if there are a multiple of 3 creatures, and the total value is not a multiple of 3, then the species cannot end up with only one gender.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2016-01-22T20:50:39.143" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="25681" PostTypeId="2" ParentId="25662" CreationDate="2016-01-22T21:46:25.120" Score="3" Body="&lt;p&gt;I think it might be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A diamond&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Glare at me, while I at you:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure, but the translucence of a raw diamond could be associated with a glare.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Worship this familiar with glimmer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Worship the stone by polishing it, till it shines&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While I stand in unfrequented gloom:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are buried deep in the bed rock&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Common, still I elude avaricious hands:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Even the rich and greedy have a hard time getting them, and at times kill for them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18624" LastActivityDate="2016-01-22T21:46:25.120" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25682" PostTypeId="2" ParentId="25677" CreationDate="2016-01-22T22:54:38.177" Score="0" Body="&lt;p&gt;I do not see any trick at all. Assume we had 3 pairs of parents: Alpha + Beta, Alpha + Gamma, Beta + Gamma. That is 2α + 2β + 2γ in total. The first couple gets 2γ as offsprings, the second one gets 2 bouncy baby β’s, and the third couple becomes the proud parents of 2α. In total, that is 2α + 2β + 2γ.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Surely given enough time, all the members of this species would be of&#xA;  one gender, unable to mate, and the species would die off.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is completely absurd: as long as there are 6 creatures that are 2α + 2β + 2γ, they are never going to die out as long as all of their children survive because each generation, the gender composition is the same. The stationarity of the population it attained at the 1:1:1 gender ratio, and the excess guys are... well, not so lucky. Not them, but their children, of course.&lt;/p&gt;&#xA;" OwnerUserId="18626" LastActivityDate="2016-01-22T22:54:38.177" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25683" PostTypeId="1" AcceptedAnswerId="25694" CreationDate="2016-01-23T00:08:04.490" Score="4" ViewCount="550" Body="&lt;h2&gt;Background&lt;/h2&gt;  &#xA;&#xA;&lt;p&gt;Normal MtG rules apply here, except that whenever your deck is shuffled, you may assume it is stacked in any order you like.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Puzzle Setup&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;No permanents are currently tapped or have summoning sickness, and you have not yet played a land this turn.  It is your first main phase.  Assume that your opponent will always discard or sacrifice to save Solitary Confinement if possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Goal: Win the game!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If there are multiple solutions, the best is the one which wins on the earliest turn, secondary tiebreaker is dealing the most damage)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your hand:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=405218&quot; rel=&quot;nofollow&quot;&gt;Evolving Wilds&lt;/a&gt;    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your board:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=179428&quot; rel=&quot;nofollow&quot;&gt;Skill Borrower&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=405284&quot; rel=&quot;nofollow&quot;&gt;Lightning Greaves&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=194972&quot; rel=&quot;nofollow&quot;&gt;Sensei's Divining Top&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=74206&quot; rel=&quot;nofollow&quot;&gt;Oboro, Palace in the Clouds&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=397543&quot; rel=&quot;nofollow&quot;&gt;City of Traitors&lt;/a&gt;&lt;br&gt;&#xA;3 Forests&lt;br&gt;&#xA;2 Plains&lt;br&gt;&#xA;2 Islands&lt;br&gt;&#xA;2 Swamps  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your graveyard:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=398438&quot; rel=&quot;nofollow&quot;&gt;Nissa, Vastwood Seer&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=124474&quot; rel=&quot;nofollow&quot;&gt;Simian Spirit Guide&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=87595&quot; rel=&quot;nofollow&quot;&gt;Sasaya, Orochi Ascendant&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=382855&quot; rel=&quot;nofollow&quot;&gt;Baleful Strix&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=184601&quot; rel=&quot;nofollow&quot;&gt;Ashen Ghoul&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=107385&quot; rel=&quot;nofollow&quot;&gt;Experiment Kraj&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your library:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=83282&quot; rel=&quot;nofollow&quot;&gt;Storm Crow&lt;/a&gt;&lt;br&gt;&#xA;Forest&lt;br&gt;&#xA;Plains&lt;br&gt;&#xA;Island&lt;br&gt;&#xA;Swamp&lt;br&gt;&#xA;Mountain&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=368961&quot; rel=&quot;nofollow&quot;&gt;Aetherling&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=233055&quot; rel=&quot;nofollow&quot;&gt;Batterskull&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=78687&quot; rel=&quot;nofollow&quot;&gt;Budoka Gardener&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=382251&quot; rel=&quot;nofollow&quot;&gt;Dimir Doppelganger&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=288992&quot; rel=&quot;nofollow&quot;&gt;Door to Nothingness&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=394558&quot; rel=&quot;nofollow&quot;&gt;Dromoka's Command&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=236501&quot; rel=&quot;nofollow&quot;&gt;Ghave, Guru of Spores&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=338406&quot; rel=&quot;nofollow&quot;&gt;Golgari Grave-Troll&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=397698&quot; rel=&quot;nofollow&quot;&gt;Kiki-Jiki, Mirror Breaker&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=193472&quot; rel=&quot;nofollow&quot;&gt;Nirkana Revenant&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=370623&quot; rel=&quot;nofollow&quot;&gt;Ratchet Bomb&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=240096&quot; rel=&quot;nofollow&quot;&gt;Restoration Angel&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=3310&quot; rel=&quot;nofollow&quot;&gt;Shallow Grave&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=376498&quot; rel=&quot;nofollow&quot;&gt;Shattergang Brothers&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=393982&quot; rel=&quot;nofollow&quot;&gt;Siege-Gang Commander&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=393985&quot; rel=&quot;nofollow&quot;&gt;Skirk Prospector&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=193535&quot; rel=&quot;nofollow&quot;&gt;Spawnsire of Ulamog&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=389693&quot; rel=&quot;nofollow&quot;&gt;Steel Hellkite&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=389735&quot; rel=&quot;nofollow&quot;&gt;Vampire Hexmage&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=227302&quot; rel=&quot;nofollow&quot;&gt;Worldslayer&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your sideboard:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=397905&quot; rel=&quot;nofollow&quot;&gt;Emrakul, the Aeons Torn&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your life:&lt;/strong&gt;&lt;br&gt;&#xA;20&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Opponent's hand:&lt;/strong&gt;&lt;br&gt;&#xA;7 Plains  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Opponent's board:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=34769&quot; rel=&quot;nofollow&quot;&gt;Solitary Confinement&lt;/a&gt;&lt;br&gt;&#xA;3 Plains   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Opponent's graveyard:&lt;/strong&gt;&lt;br&gt;&#xA;Nothing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Opponent's library:&lt;/strong&gt;&lt;br&gt;&#xA;49 Plains    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Opponent's life:&lt;/strong&gt;&lt;br&gt;&#xA;1  &lt;/p&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="17389" LastEditDate="2016-01-25T18:11:30.140" LastActivityDate="2016-01-26T04:45:43.407" Title="Magic: the Gathering - Borrowed Time" Tags="|calculation-puzzle|magic-the-gathering|" AnswerCount="2" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="25684" PostTypeId="1" AcceptedAnswerId="25686" CreationDate="2016-01-23T00:26:32.423" Score="4" ViewCount="363" Body="&lt;p&gt;13 friends are sitting around a table. They shuffle a half-deck of cards, consisting of only the black cards of a standard deck, and deal out two cards to each person.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are then a series of turns. On each turn, everyone takes the lower ranked of their two cards and passes it to the person on their right (A=1, J=11, Q=12, K=13). If both of a person's cards are the same rank, then the game immediately ends.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the probability that they never stop?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2016-01-25T17:33:02.417" Title="Card Passing Game" Tags="|probability|cards|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25685" PostTypeId="2" ParentId="25684" CreationDate="2016-01-23T01:07:35.150" Score="1" Body="&lt;p&gt;The probability is zero, as shown in &lt;a href=&quot;https://math.stackexchange.com/questions/241110/a-less-challenging-trivia-problem&quot;&gt;this answer&lt;/a&gt; to the same question with a deck of 25 pairs.  By analogy to &lt;a href=&quot;https://math.stackexchange.com/questions/245310/card-passing-game-maximum-length&quot;&gt;this answer&lt;/a&gt; the match will happen within $12$ passes&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="-1" LastEditDate="2017-04-13T12:19:36.950" LastActivityDate="2016-01-23T01:07:35.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25686" PostTypeId="2" ParentId="25684" CreationDate="2016-01-23T01:12:53.400" Score="10" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The chance is 0%&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let us begin with this observation:  with 13 pair of cards in play, that accounts for every card in the half-deck.  There are no cards remaining out of play.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's construct a scenario in which the game would never stop.  To make that happen, each player must have one high card (say, above a 7) and one low card (7 or below).  Since each player has a high card and a low card, and passes the low card, nobody will wind up with a pair, since all the low cards are moving at the same time, just chasing each other around the table, and never matching the (above a 7) high cards that aren't moving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sounds good, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that this precise situation can't actually occur;  there are only 12 cards in the half-deck above a 7 (pairs of 8s, 9s, 10s, jacks, queens, and kings), but 13 players.  Which means that we're guaranteed to have at least one player whose highest card is a 7 or lower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Contrariwise, there are only 12 cards in the half-deck that are below a 7 (pairs of aces, 2s, 3s, 4s, 5s, and 6s).  This means that we're guaranteed to have at least one player whose lowest card is a 7 or higher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So over time, in this game, it's the lowest cards that will be circulating around the table:  the pairs of aces through sixes, plus one seven.  And since we're guaranteed that at least one player's highest card will be a 7 or lower which will remain in their hand as other cards are passed around the table, that player will eventually form a match, ending the game.&lt;/p&gt;&#xA;" OwnerUserId="9286" LastActivityDate="2016-01-23T01:12:53.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25687" PostTypeId="2" ParentId="25683" CreationDate="2016-01-23T01:32:29.053" Score="3" Body="&lt;p&gt;&lt;strong&gt;Turn 2 kill&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Turn 1&lt;/strong&gt;&lt;br&gt;&#xA;tap all land for mana&lt;br&gt;&#xA;play Evolving Wilds&lt;br&gt;&#xA;sacrifice for Mountain&lt;br&gt;&#xA;stack the top three cards like this: Budoka Gardener, Dimir Doppelganger, Shattergang Brothers&lt;br&gt;&#xA;Activate Gardener's ability, Skill Borrower flips&lt;br&gt;&#xA;Activate Top to put Dimir Doppelganger on top&lt;br&gt;&#xA;Activate Doppelganger's ability exiling Sasaya.&lt;br&gt;&#xA;It becomes the flipped Sasaya, which is an enchantment.&lt;br&gt;&#xA;Activate Top to put Shattergang Brothers on top  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Turn 2&lt;/strong&gt;&lt;br&gt;&#xA;cast Ashen Ghoul in your upkeep&lt;br&gt;&#xA;cast Shattergang Brothers&lt;br&gt;&#xA;Sacrifice Skill Borrower to destroy Solitary Confinement&lt;br&gt;&#xA;Equip Grieves to Brothers&#xA;Attack with Shattergang Brothers and Ghoul for 6 damage&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;OLD ANSWER&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Turn 4 kill&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Turn 1&lt;/strong&gt;&lt;br&gt;&#xA;Tap all land for mana&lt;br&gt;&#xA;play Evolving Wilds&lt;br&gt;&#xA;Sacrifice for a Mountain&lt;br&gt;&#xA;Stack library so  Spawnsire of Ulamog is on top&lt;br&gt;&#xA;tap Top to draw Spawnsire of Ulamog&lt;br&gt;&#xA;cast Spawnsire of Ulamog&lt;br&gt;&#xA;equip Greaves&lt;br&gt;&#xA;Attack (I assume opponent sacrifices Plains)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Turn 2&lt;/strong&gt;&lt;br&gt;&#xA;Attack (I assume opponent sacrifices Plains)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Turn 3&lt;/strong&gt;&lt;br&gt;&#xA;Attack (I assume opponent sacrifices Plains)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Turn 4&lt;/strong&gt;&lt;br&gt;&#xA;Attack, opponent sacrifices Solitary Confinement, damage comes through.&lt;/p&gt;&#xA;" OwnerUserId="1748" LastEditorUserId="1748" LastEditDate="2016-01-23T02:34:04.587" LastActivityDate="2016-01-23T02:34:04.587" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25689" PostTypeId="1" AcceptedAnswerId="25693" CreationDate="2016-01-23T03:52:36.380" Score="-1" ViewCount="357" Body="&lt;p&gt;There are $n$ large bottles of wine, out of which $n-1$ have been poisoned.&lt;br&gt;However, consuming a single bottle does no harm, it requires all $n-1$ bottles (or small samples of them) to be consumed by the same person to get poisoned. A person dies of poisoning only at the end of the day. The poison consumed by a person remains in his/her body forever, but does not activate until all the other poisons are also consumed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have $100$ people and unlimited time for performing the tests. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the largest value of $n$ for which you can identify the good bottle?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;P.S. This is not a &lt;a href=&quot;/questions/tagged/lateral-thinking&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;lateral-thinking&amp;#39;&quot; rel=&quot;tag&quot;&gt;lateral-thinking&lt;/a&gt;, nor is it a &lt;a href=&quot;/questions/tagged/science&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;science&amp;#39;&quot; rel=&quot;tag&quot;&gt;science&lt;/a&gt; one.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="18250" LastEditDate="2016-01-23T20:00:08.267" LastActivityDate="2016-01-23T20:00:08.267" Title="All poisoned but one" Tags="|logical-deduction|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25690" PostTypeId="1" AcceptedAnswerId="25752" CreationDate="2016-01-23T04:19:39.007" Score="5" ViewCount="359" Body="&lt;p&gt;One the voice of a city in italy, the other a taker of fleece.&lt;br&gt;&#xA;Together they make a duo.&lt;br&gt;&#xA;Like calcium to cobalt.&lt;br&gt;&#xA;Who rarely appear outside of their element.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who are we?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hints*&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. We have been on the small _____ for decades.&lt;Br&gt;&#xA; 2. We come from the VII.&lt;Br&gt;&#xA; 3. We are siblings.&lt;Br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12014" LastEditorUserId="12014" LastEditDate="2016-01-25T17:41:22.687" LastActivityDate="2016-01-26T11:30:35.403" Title="Who comprise this duo?" Tags="|riddle|cryptic-clues|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="25691" PostTypeId="1" CreationDate="2016-01-23T04:24:51.373" Score="6" ViewCount="226" Body="&lt;p&gt;I am that which you see the least,&lt;br&gt;&#xA;and with what you see, does relate.&lt;br&gt;&#xA;No matter how inseparable we are&lt;br&gt;&#xA;you’ll always miss me.&lt;br&gt;&#xA;Only in brief reflection do I come to mind. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Who am I?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wrote this down a few minutes ago; I actually joined stack exchange just to share.&lt;/p&gt;&#xA;" OwnerUserId="18634" LastEditorUserId="25957" LastEditDate="2016-08-09T09:51:17.697" LastActivityDate="2016-08-09T10:04:41.943" Title="Only in brief reflection do I come to mind. Who am I?" Tags="|riddle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25692" PostTypeId="2" ParentId="25689" CreationDate="2016-01-23T04:45:27.467" Score="3" Body="&lt;p&gt;Well, as a first guess, &lt;strong&gt;n = 101&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Just line all 100 people and all 101 bottles up and have each person drink from all but bottle number x, where x is their number in line.  Wait a day.  If a person y dies, then y was the good bottle and they drank every other poison.  If nobody dies, then 101 was the good bottle, since it's the only one everybody drank from.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17389" LastActivityDate="2016-01-23T04:45:27.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25693" PostTypeId="2" ParentId="25689" CreationDate="2016-01-23T04:46:23.117" Score="8" Body="&lt;p&gt;Anyone who has drank fewer than $n-1$ bottles is guaranteed not to be poisoned, so we get no information from that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone who drinks $n$ bottles is guaranteed to be poisoned, so this also gives no information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only situation that does provide information is if someone drinks exactly $n-1$ bottles. If they die, the good bottle is the one they didn't drink, otherwise it is one of the bottles they did drink. At this point, the person is useless for getting any more information (repeating a drink is pointless, and drinking the last bottle is guaranteed death).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each person can only rule out one bottle, so the maximum $n$ is $101$.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2016-01-23T04:46:23.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25694" PostTypeId="2" ParentId="25683" CreationDate="2016-01-23T05:25:58.647" Score="3" Body="&lt;p&gt;EDIT: Credit to Zerris who posted his partly-broken infinite solution, this is a version of that solution which I fixed up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tap Swamp, Swamp, Island, Oboro, City, Plains for 7&lt;br&gt;&#xA;Play Evolving Wilds, sac for Island&lt;br&gt;&#xA;The top three cards of your deck are Kiki-Jiki, Budoka Gardener, Shallow Grave&lt;br&gt;&#xA;Skill Borrower uses Kiki to copy itself&lt;br&gt;&#xA;Skill Borrower token uses Kiki to copy itself&lt;br&gt;&#xA;Spin Top to put Budoka Gardener on top&lt;br&gt;&#xA;Hold priority, token activates Budoka's ability, then:&lt;br&gt;&#xA;Spin top, put Shallow Grave on top, tap top to draw Shallow Grave&lt;br&gt;&#xA;Cast Shallow Grave, returning Nissa, searching up a Forest&lt;br&gt;&#xA;Top cards of your deck are now Divining Top, Dimir Doppleganger, Restoration Angel, Kiki, Batterskull, Shattergang Brothers&lt;br&gt;&#xA;Resolve Budoka, putting Forest into play, flipping the Borrower token and transforming Nissa&lt;br&gt;&#xA;Plus Nissa to draw Top&lt;br&gt;&#xA;Have the flipped token use Dimir Doppleganger to copy Sasaya, becoming Sasaya's Essence (zero mana floating)&lt;br&gt;&#xA;Tap all lands for 16 G, 2W, 3U&lt;br&gt;&#xA;Play Top, spin for Angel, tap Top to draw Angel&lt;br&gt;&#xA;Cast Angel, blinking the nontoken Skill Borrower&lt;br&gt;&#xA;Equip Skill Borrower with Lightning Greaves&lt;br&gt;&#xA;Hold priority and have Skill Borrower spin top, then:&lt;br&gt;&#xA;Spin top again, putting Kiki on top&lt;br&gt;&#xA;Have nontoken Skill Borrower copy the token Skill Borrower&lt;br&gt;&#xA;Resolve the top activation, putting Top back on top&lt;br&gt;&#xA;Tap the Skill Borrower token to draw Top&lt;br&gt;&#xA;Play Top&lt;br&gt;&#xA;Spin Top to put Batterskull on top&lt;br&gt;&#xA;Use Batterskull to return the nontoken Skill Borrower to your hand, then recast it&lt;br&gt;&#xA;Spin Top to put Kiki on top&lt;br&gt;&#xA;Equip Skill Borrower with Lightning Greaves&lt;br&gt;&#xA;Use Kiki to have Skill Borrower copy a token Skill Borrower&lt;br&gt;&#xA;Equip the token with Lightning Greaves (to make the nontoken Borrower lose shroud)&lt;br&gt;&#xA;Have the token use Kiki to copy Restoration Angel, blinking the nontoken Skill Borrower&lt;br&gt;&#xA;Repeat the above four steps to create infinite Restoration Angels&lt;br&gt;&#xA;You now have infinite Angels, one untapped Skill Borrower, and your top cards are Kiki, Dimir Doopleganger, Batterskull, Shattergang Brothers&lt;br&gt;&#xA;Make a Skill Borrower token&lt;br&gt;&#xA;Tap top to draw Kiki&lt;br&gt;&#xA;Hold priority, have Skill Borrower spin top, then:&lt;br&gt;&#xA;Tap the Skill Borrower token to draw a card with Top (top cards are now Dimir, Batterskull, Shattergang)&lt;br&gt;&#xA;Resolve the spinning top to put Shattergang on top&lt;br&gt;&#xA;Sacrifice Sasaya's Essence to Shattergang to destroy Solitary Confinement&lt;br&gt;&#xA;Attack with infinite angels for infinite damage  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Old version, Turn 1, 15 damage:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tap all lands except forests and plains (this gives exactly enough mana for now)&lt;br&gt;&#xA;Play Evolving Wilds, sacrifice it to find Plains&lt;br&gt;&#xA;The top three cards of your deck are Kiki-Jiki, Budoka Gardener, Shallow Grave&lt;br&gt;&#xA;Borrow Kiki to copy Skill Borrower, repeat this step twenty times&lt;br&gt;&#xA;Top to put Budoka on top&lt;br&gt;&#xA;Hold priority and token Borrows Budoka, with that on the stack:&lt;br&gt;&#xA;Top to put Shallow Grave on top, then draw it&lt;br&gt;&#xA;Cast Shallow Grave to return Nissa&lt;br&gt;&#xA;Use Nissa's trigger to put Forest into your hand&lt;br&gt;&#xA;After shuffling, the top cards are Top, Dimir Doppleganger, Shattergang Brothers, Ghave&lt;br&gt;&#xA;Resolve the Borrowed Budoka activation, putting Forest into play, flipping the token and transforming Nissa&lt;br&gt;&#xA;Activate Nissa to draw Top&lt;br&gt;&#xA;Have the flipped token Borrow Dimir Doppleganger to copy Sasaya, turning it into Sasaya's Essence&lt;br&gt;&#xA;Tap four forests for sixteen and two plains for six (22 mana)&lt;br&gt;&#xA;Play Top (21 mana)&lt;br&gt;&#xA;Rearrange to put Shattergang on top (20 mana)&lt;br&gt;&#xA;Activate Shattergang (17 mana)&lt;br&gt;&#xA;Rearrange to put Ghave on top (16 mana)&lt;br&gt;&#xA;Use Ghave to sacrifice a Skill Borrower token to put a counter on another Borrower (15 mana)&lt;br&gt;&#xA;Remove the counter to make an untapped Saproling token (14 mana)&lt;br&gt;&#xA;Sacrifice 14 more Skill Borrower tokens to put 14 counters on the Saproling&lt;br&gt;&#xA;Equip the Saproling with Lightning Greaves and attack for 15.&lt;/p&gt;&#xA;" OwnerUserId="7459" LastEditorUserId="7459" LastEditDate="2016-01-26T04:45:43.407" LastActivityDate="2016-01-26T04:45:43.407" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="25695" PostTypeId="2" ParentId="25691" CreationDate="2016-01-23T06:21:49.580" Score="6" Body="&lt;p&gt;My answer is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; your eyelids. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am that which you see the least,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most of the time when we are actually &lt;em&gt;seeing&lt;/em&gt; things, our eyes are open, so we don't see our eyelids. Closing our eyes sorta lets us do it, but barely so. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and with what you see, does relate.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We see with our eyes and eyelids are definitely related to our eyes. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No matter how inseparable we are&#xA;you’ll always miss me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eyelids are obviously on us and are so close to that which we see with, but as soon as we try to bring them into our field of view (blinking/closing our eyes), we obscure our ability to fully see. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Only in brief reflection do I come to mind.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We can, however, see our eyelids when looking in a mirror, or, to go deeper with it- when we close our eyes to reflect on past choices or anything, we acknowledge that our eyelids are in front of our eyes. In other words, they come to mind. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So again, you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eyelids&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11452" LastActivityDate="2016-01-23T06:21:49.580" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25697" PostTypeId="2" ParentId="25413" CreationDate="2016-01-23T07:48:15.660" Score="13" Body="&lt;p&gt;The three gods are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the three additional letters of the alphabet in...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The three kingdoms, which are (from North to South):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the Scandinavian nations of Norway, Sweden and Denmark.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To be specific:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In Norway and Denmark, En, To and Tre (literally one, two, three) are Æ/æ, Ø/ø and Å/å. In Sweden, they are Å/å, Ä/ä and Ö/ö.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In the frigid North, En leads the three / Just like in the South, as it should be.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In Norway and Denmark, the first additional letter is Æ.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In the temperate South, To makes up the center / The North agrees, but the Middle is the dissenter.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In Norway and Denmark, the second additional letter is Ø. But not in Sweden, where it is Ä.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In the Middle, Tre comes in the first / South and North agree, that is certainly the worst.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The third additional letter (in Norway and Denmark), is Å, which comes first in Sweden.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In the extreme lands, En appears like Janus, / Near the Middle, however, they find that heinous.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Æ is like Janus in the sense that it has two faces, A and E.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In the Middle, To appears surprised, / Everywhere else, however, He appears incised.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Swedish Ö looks like a surprised face, whereas the Norwegian/Danish Ø, is visually cut in two (incised).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Only Tre, however, is truly immutable, / Everywhere, His nature is truly indisputable.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Only Å is shared between all nations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="5208" LastEditDate="2016-01-23T08:54:40.200" LastActivityDate="2016-01-23T08:54:40.200" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25698" PostTypeId="2" ParentId="25677" CreationDate="2016-01-23T07:55:27.540" Score="8" Body="&lt;p&gt;&lt;strong&gt;First observation:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since every mating produces two offspring and results in the immediate death of both parents, and since there is no other source of death on the planet, the total population of creatures is constant over time;  only the relative population levels of the genders can change.  (We'll leave the question of how these creatures reached their current population level for worldbuilding.SE)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second observation:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The population naturally self-balances.  That is, if all but one of the population is of gender 'alpha', and if the last other individual is of gender 'beta', then an alpha will mate with the beta, producing two gammas.  And each of those will then mate with another of the alphas (since they can't mate with each other), producing four more betas.  And so on, with the overbalance of alphas slowly decreasing with each cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Let's do some eugenics, boys and girls!&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in order to prove what would have to happen to make the population collapse to a single gender, let's try to engage in selective breeding to make that happen intentionally.  For the following discussion, assume that we want only alphas to remain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it's easy to reduce the population to only alphas if we have precisely the same number of betas as of gammas; we'd just pair the betas with the gammas and produce entirely alpha offspring, while the betas and gamma parents die, and we're done.  But to do this, we need to have precisely the same number of betas as gammas, because if there is even a single surviving, unmated beta or gamma, then the population diversity will eventually recover given enough mating operations (as per the second observation, above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's assume that we &lt;em&gt;don't&lt;/em&gt; have the same number of betas as of gammas.  If the number of betas doesn't match the number of gammas, can we perform selective breeding in order to make those numbers match?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, there are three moves we can make:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Mate alpha with beta.  We wind up with one fewer beta, and two more gammas.&lt;/li&gt;&#xA;&lt;li&gt;Mate alpha with gamma.  We wind up with one fewer gamma, and two more betas.&lt;/li&gt;&#xA;&lt;li&gt;Mate beta with gamma.  We wind up with one fewer beta and one fewer gamma.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;We can eliminate move #3 as irrelevant;  it doesn't bring the number of betas closer to the number of gammas.  Moves #1 and #2, on the other hand, can be used to bring their numbers closer together by 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;And therefore:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless of the total population of the planet, if the difference in the number of beta individuals from the number of gamma individuals is a multiple of 3, then we can perform selective breeding in order to make the total number of betas equal to the total number of gammas.  And once we've done that, then we can make the betas and gammas mate and die off, resulting in a population consisting only of alphas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And on the other hand, if the difference between the number of betas and the number of gammas is &lt;em&gt;not&lt;/em&gt; a multiple of 3, then we cannot ever make them equal via the available mating operations, and it's impossible for there to ever be a population consisting only of alphas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this is also true for each of the other pairs of genders;  if the difference between alphas and betas is a multiple of three, we can equalise their numbers and produce a population entirely of gammas, and if the difference between alphas and gammas is a multiple of three, we can equalise &lt;em&gt;their&lt;/em&gt; numbers and produce a population entirely of betas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;And finally, the answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The puzzle specifies that we &lt;em&gt;do&lt;/em&gt; eventually determine that it's impossible for the population to become entirely composed of any one gender.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To reach that conclusion, we must have taken a census of how many individuals there are of each gender, and determined that no pair of genders have a difference that is a multiple of three.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result of this, by the logic above, no mating operations can possibly occur which would result in any pair of genders having the same number of individuals.  And since no two genders can ever have the same number of members, they cannot ever be bred together to remove all members of both genders at the same time.&lt;/p&gt;&#xA;" OwnerUserId="9286" LastEditorUserId="9286" LastEditDate="2016-01-23T08:00:45.953" LastActivityDate="2016-01-23T08:00:45.953" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25699" PostTypeId="2" ParentId="25677" CreationDate="2016-01-23T10:07:02.330" Score="4" Body="&lt;p&gt;I'll go and post a not-so-lateral-thinking answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If individuals only die after mating then&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the population never dies off, regardless of gender distribution&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; mating doesn't change the number individuals &lt;em&gt;alive&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6070" LastActivityDate="2016-01-23T10:07:02.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25700" PostTypeId="2" ParentId="25212" CreationDate="2016-01-23T10:33:32.740" Score="1" Body="&lt;p&gt;if the rows and columns are symmetric and you cannot find any black square from a row and from a column only, it is not unique and there is at least one more solution.&lt;/p&gt;&#xA;" OwnerUserId="18514" LastActivityDate="2016-01-23T10:33:32.740" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25702" PostTypeId="1" CreationDate="2016-01-23T11:20:54.067" Score="4" ViewCount="247" Body="&lt;p&gt;It was too 6-6-500, then I've decided to 1009 it with more color, but it became too 54-1-500.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What could be it?&lt;/p&gt;&#xA;" OwnerUserId="18514" LastEditorUserId="8874" LastEditDate="2016-03-11T15:56:24.260" LastActivityDate="2016-03-11T15:56:24.260" Title="What could be it?" Tags="|knowledge|cipher|history|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25703" PostTypeId="1" AcceptedAnswerId="25705" CreationDate="2016-01-23T11:34:08.550" Score="5" ViewCount="220" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I will repeat,&lt;br&gt;&#xA;  Split 1+2 &lt;br&gt;for your benefit.&#xA;  &lt;br&gt;Detach the 2 latter&#xA;  &lt;br&gt;General in, Turkey&#xA;  &lt;br&gt;Now start anew,&#xA;  &lt;br&gt;I will repeat&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;What am I?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Check the last tag&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18250" LastEditorUserId="18250" LastEditDate="2016-09-24T14:06:23.293" LastActivityDate="2016-09-24T14:06:23.293" Title="What am I, I will repeat" Tags="|riddle|word|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25704" PostTypeId="2" ParentId="25598" CreationDate="2016-01-23T12:33:39.830" Score="1" Body="&lt;p&gt;Without looking at any answers, it's:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4998. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Form the first two clues it's of the form a b b (2a).&lt;br&gt;&#xA; So from the last:  20a + 2b = 10b + 2a.&lt;br&gt;&#xA; Or: 9a = 4b.&lt;br&gt;&#xA; The only single nonzero digits this works for is a = 4 and b = 9  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastActivityDate="2016-01-23T12:33:39.830" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25705" PostTypeId="2" ParentId="25703" CreationDate="2016-01-23T12:50:43.563" Score="5" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;AGAIN&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I will repeat,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is a clue to the whole word. When you repeat something, you do it &lt;strong&gt;again&lt;/strong&gt;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Split 1+2 for your benefit. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you split the word between letters 1 and 2, you get &lt;strong&gt;a gain&lt;/strong&gt;, which is for your benefit.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Detach the 2 latter&lt;br&gt;&#xA;General in, Turkey&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Remove the last 2 letters and you get &lt;strong&gt;aga&lt;/strong&gt;, which is a title for a general in Turkey. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now start anew, &#xA;I will repeat&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Again&lt;/strong&gt;, the word is &lt;strong&gt;again&lt;/strong&gt;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2016-01-23T12:50:43.563" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="25706" PostTypeId="2" ParentId="25702" CreationDate="2016-01-23T14:57:27.107" Score="11" Body="&lt;p&gt;Partial answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If we convert the numbers to Roman numerals, we get that 6-6-500 is &quot;VIVID&quot;, 1009 is &quot;MIX&quot; and 54-1-500 is &quot;LIVID&quot;. So the first sentence becomes &lt;br&gt; &lt;br&gt;&#xA;&#xA; &quot;It was too vivid, then I've decided to mix it with more color, but it became too livid.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1600" LastEditorUserId="1600" LastEditDate="2016-01-23T16:26:36.060" LastActivityDate="2016-01-23T16:26:36.060" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25707" PostTypeId="1" AcceptedAnswerId="25854" CreationDate="2016-01-23T15:49:31.833" Score="8" ViewCount="788" Body="&lt;p&gt;&lt;strong&gt;Warning: This puzzle contains major spoilers regarding Star Wars: The Force Awakens. Do not proceed unless you are okay with that!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a new friend, Valerie -- we just became acquainted recently while attending the same college class. She is fascinating, but a little eccentric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recently met my new friend, Valerie, for breakfast. Over the course of our conversation, she mentioned that she is an avid watcher of sci-fi and fantasy films. I'm a fan of the genre myself, so I asked her to name her favorite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Oh,&quot; she replied. &quot;I'll have to think about that.  I'll tell you when I figure it out.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Honestly, I forgot all about the conversation until today.  I arrived to the class we share early, as I always do.  Valerie showed up just as the bell rang, just as she always does.  But today, as she passed me on her way to an empty seat, she dropped a folded piece of paper on my open textbook.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bit perplexed, I unfolded the note and quickly skimmmed it, enough to see it was a numbered list referencing characters and events from Star Wars: The Force Awakens.  I immediately thought about the conversation we had had. &quot;It's a strange way to present her answer,&quot; I thought, &quot;but the meaning is clear enough.&quot;  I folded the paper back up and put it away in my backpack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end of class, I caught Valerie's attention before she left.  &quot;So the new Star Wars is your favorite film? It's quite good, but-&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Don't be so sure,&quot; she said, cutting me off with a mysterious smile.  &quot;There's more to my note than you have yet seen.&quot; With that, she left for her next class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Frowning, I fished the note back out of my backpack and looked at it again, more closely this time.  Yes, something was strange, but &lt;strong&gt;what did it mean?  What is the answer hidden in Valerie's note?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. The droid, BB-8, is R2-D2's child.&lt;br&gt;&lt;br&gt;&#xA; 2. Kylo Ren's father is Han Solo.&lt;br&gt;&lt;br&gt;&#xA; 3. And his mother is Leia! &lt;br&gt;&lt;br&gt;&#xA; 4. (Only in this movie, Leia's title is &quot;Sergeant&quot; instead of &quot;Princess&quot;.)&lt;br&gt;&lt;br&gt;&#xA; 5. Finn is a stormtrooper who defects.&lt;br&gt;&lt;br&gt;&#xA; 6. Poe Dameron asks Finn to pilot a ship off of Jakku.&lt;br&gt;&lt;br&gt;&#xA; 7. The ship Poe and Finn are in crashes on Jakku.&lt;br&gt;&lt;br&gt;&#xA; 8. But fortunately we find out they both survived.&lt;br&gt;&lt;br&gt;&#xA; 9. The girl, Rey, has a mysterious background.&lt;br&gt;&lt;br&gt;&#xA; 10. It turns out that she can use the force.&lt;br&gt;&lt;br&gt;&#xA; 11. She is also a very capable pilot.&lt;br&gt;&lt;br&gt;&#xA; 12. She has a hard time leaving the aunt and uncle she lives with on Jakku.&lt;br&gt;&lt;br&gt;&#xA; 13. But she does it because she has heard there's a powerful Jedi on Dagobah who can teach her.&lt;br&gt;&lt;br&gt;&#xA; 14. Meanwhile, the evil regime has built a powerful weapon of mass destruction.&lt;br&gt;&lt;br&gt;&#xA; 15. But this time, they've learned their lesson, and this one is NOT vulnerable in one small area where it can be attacked by small ships.&lt;br&gt;&lt;br&gt;&#xA; 16. Kylo Ren murders Han Solo.&lt;br&gt;&lt;br&gt;&#xA; 17. And Leia cries like a baby onscreen when Rey tells her what happened.&lt;br&gt;&lt;br&gt;&#xA; 18. Rey finds Luke sitting in a cozy cottage drinking a cup of tea.&lt;br&gt;&lt;br&gt;&#xA; 19. (Because of course it turns out there is a map pointing out exactly where Luke is.)&lt;br&gt;&lt;br&gt;&#xA; 20. Which we only find in its entirety near the end of the plot because, apparently, droid mental health issues are a real thing -- poor R2-D2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As noted in the comments, some items in the list are true and others are not. Once you figure out which are true and which are not, the trivia portion of this puzzle is done.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The order of the statements in the list is important.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Once you have the trivia part done, turn your attention to the other tag on this question. If you are inexperienced with these puzzle types, I refer you to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/5905/what-characteristics-of-a-ciphertext-can-be-indicators-of-a-particular-cipher&quot;&gt;this excellent question&lt;/a&gt;. Good luck!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18493" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2016-01-28T15:24:18.183" Title="A Favorite Film Cryptogram" Tags="|cipher|knowledge|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="25708" PostTypeId="2" ParentId="24675" CreationDate="2016-01-23T16:27:01.633" Score="5" Body="&lt;p&gt;I think the word is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;SPECTRUM&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first block is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; just a hint that the encrypted text is base 16.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since I am a USB expert,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I know how to read in base 16. And when I convert the legal blocks in &lt;code&gt;[1L][1H][2M][16][13][08][06][17][15][18][10]&lt;/code&gt; from base 16 to base 10, it results as &lt;code&gt;22 19 8 6 23 21 24 16&lt;/code&gt;. When we convert these decimal numbers to letters in English alphabet, we get &lt;code&gt;WTIGXVYQ&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then, by using &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Caesar cipher, I shift the letters by 22 to get &lt;strong&gt;SPECTRUM&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18645" LastEditorUserId="18645" LastEditDate="2016-01-23T23:29:32.840" LastActivityDate="2016-01-23T23:29:32.840" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25709" PostTypeId="2" ParentId="25702" CreationDate="2016-01-23T16:33:35.810" Score="3" Body="&lt;p&gt;I think it is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; red&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;based on the partial answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; red is a color which is vivid, then I mix it with blue to get a livid color, which is purple&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18645" LastEditorUserId="18645" LastEditDate="2016-01-23T16:38:42.807" LastActivityDate="2016-01-23T16:38:42.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25710" PostTypeId="1" AcceptedAnswerId="25711" CreationDate="2016-01-23T17:48:47.583" Score="2" ViewCount="278" Body="&lt;p&gt;Some will understand what I say, some do not.&lt;br&gt;&#xA;I speak the exact words you want.&lt;br&gt;&#xA;Some words I speak however, you would like not.&lt;br&gt;&#xA;For you to understand me, you shall study a lot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I carry out your orders, exactly what you say,&lt;br&gt;&#xA;I may make mistakes, but it is not my fault.&lt;br&gt;&#xA;You will understand me maybe after an hour,&lt;br&gt;&#xA;maybe after a month!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will be mistaken, when you hear my name&lt;br&gt;&#xA;I don't gather things from here and there.&lt;br&gt;&#xA;No, that is so lame.&lt;br&gt;&#xA;I do serious business,&lt;br&gt; &#xA;Carrying out orders and translating is my game.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you guess who I am?&lt;/p&gt;&#xA;" OwnerUserId="18645" LastEditorUserId="18250" LastEditDate="2016-01-23T18:24:26.497" LastActivityDate="2016-01-23T18:45:18.110" Title="All I do is carry out orders and obey. Who am I?" Tags="|riddle|poetry|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25711" PostTypeId="2" ParentId="25710" CreationDate="2016-01-23T18:36:43.723" Score="5" Body="&lt;p&gt;My guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A compiler&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I speak the exact words you want&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Only does what you program it to do.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I may make mistakes, but its not my fault&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Code maybe wrong, therefore its not really wrong the programmer is.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You would not like what I say / would need to study a lot.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Binary / Machine code maybe require studying to learn / people might not like the complexity of either?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you hear my name you maybe confused, I dont gather.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Could mean to gather, might cause confusion to those without programming knowledge.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Etc. &lt;/p&gt;&#xA;" OwnerUserId="12014" LastEditorUserId="12014" LastEditDate="2016-01-23T18:45:18.110" LastActivityDate="2016-01-23T18:45:18.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25714" PostTypeId="2" ParentId="25690" CreationDate="2016-01-23T19:35:40.463" Score="1" Body="&lt;p&gt;This may be a bit off the mark, as it isn't necessarily a duo, but you could be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jason and the Argonauts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One the voice of a city in italy, the other a taker of fleece.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The taker of fleece that I am thinking of is Jason, who quested for the Golden Fleece with his team of Argonauts.  In Italy, there is a municipality called San Paolo d'Argon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Together they make a duo.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, not the most straightforward duo, but this &lt;em&gt;is&lt;/em&gt; a cryptic puzzle, so nobody is saying it should be straightforward.  Anyway, one and another are a duo; in this case, Jason is one and the Argonauts are another.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Like calcium to cobalt.&#xA;Who rarely appear outside of their element.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This seems to point to the element Argon on the periodic table, but there is likely more than I am acknowledging.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But again, this duo may in fact be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jason and the Argonauts&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11452" LastActivityDate="2016-01-23T19:35:40.463" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25715" PostTypeId="1" AcceptedAnswerId="25748" CreationDate="2016-01-23T20:05:13.013" Score="15" ViewCount="834" Body="&lt;p&gt;I was inspired by one of &lt;a href=&quot;http://lotr.wikia.com/wiki/Riddles_In_The_Dark&quot;&gt;Tolkien's famous riddles&lt;/a&gt; from The Hobbit,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;No-legs lay on one-leg,&lt;br&gt;&#xA;  two-legs sat near on three-legs,&lt;br&gt;&#xA;  four-legs got some.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;to create a new riddle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;One-leg took four-legs&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;to a place with no legs,&lt;/em&gt; &lt;br&gt;&#xA;&lt;em&gt;two-legs came later.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explain what this is describing.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit.&lt;/strong&gt; And here's a hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The one-leg my riddle is referring to can fly. It and its kind don't have wings, but they sometimes have arms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user17947" LastEditorUserId="5373" LastEditDate="2018-11-10T10:38:30.740" LastActivityDate="2018-11-10T10:38:30.740" Title="A Tolkien-inspired riddle" Tags="|riddle|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25716" PostTypeId="5" CreationDate="2016-01-23T20:14:28.733" Score="0" Body="&lt;p&gt;John Ronald Reuel Tolkien was an English writer, best known &#xA;for the classic high-fantasy works &lt;em&gt;&quot;The Hobbit&quot;&lt;/em&gt;, &#xA;&lt;em&gt;&quot;The Lord of the Rings&quot;&lt;/em&gt;, and &lt;em&gt;&quot;The Silmarillion&quot;&lt;/em&gt;.&#xA;His books inspire inspire fan fiction and fan puzzles. &#xA;Much of his work was based on ancient mythology and poetry,&#xA; His influences include Early Germanic, and Old English literature, &#xA;such as Beowulf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Books:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hobbit:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bilbo Baggins, 13 dwarves, and the wizard Gandalf go on an Adventure.&#xA;  A children's book that offered the first glimpse into Tolkien's&#xA;  sub-creation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Lord of the Rings:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sometimes published in 3 volumes: 1. The Fellowship of the Ring; 2.&#xA;  The Two Towers; 3. The Return of the King. Tolkien's masterwork. A&#xA;  sequel to The Hobbit, but much richer and deeper (and longer) than its&#xA;  predecessor. Bilbo's heir Frodo embarks on an epic quest to destroy&#xA;  the sinister Ring acquired on the earlier journey.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Adventures of Tom Bombadil:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Collection of poems and light verse supposedly written by hobbits,&#xA;  actually collected from Tolkien's papers of many years. Some also&#xA;  appear in The Lord of the Rings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Silmarillion:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Best seen as the lore book of the Elves whom the hobbits meet on their&#xA;  journeys. In a high formal tone, it tells of the mythic creation of&#xA;  the world and the tragic wars of the Elves and the Valar against evil.&#xA;  This volume was compiled after Tolkien's death as a &quot;snapshot&quot; of the&#xA;  latest versions of legends he'd been working on all his life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfinished Tales:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The book with the answers to the questions posed by readers of The&#xA;  Lord of the Rings' appendices. Incomplete but polished stories and&#xA;  essays from all periods of Middle-earth's history.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Beowulf: A Translation and Commentary:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A prose translation of the early medieval epic poem Beowulf from Old&#xA;  English to Modern English. It was edited by Tolkien's son&#xA;  Christopher and published posthumously. The poem is about the life of&#xA;  Beowulf, a hero of the Geats in Scandinavia. It tells of his&#xA;  adventures, fighting dragons, and monsters such as Grendel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18250" LastEditorUserId="25355" LastEditDate="2017-01-14T15:24:49.533" LastActivityDate="2017-01-14T15:24:49.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25717" PostTypeId="4" CreationDate="2016-01-23T20:14:28.733" Score="0" Body="A puzzle related to the books of J. R. R. Tolkien, as for instance &quot;The Hobbit&quot; (mainly the famous riddles inside) and &quot;The Lord of the Rings&quot;." OwnerUserId="18250" LastEditorUserId="8874" LastEditDate="2016-01-24T14:56:58.523" LastActivityDate="2016-01-24T14:56:58.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25718" PostTypeId="1" AcceptedAnswerId="25743" CreationDate="2016-01-23T23:21:09.260" Score="8" ViewCount="941" Body="&lt;p&gt;1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a peak&lt;br&gt;&#xA;take a swing&lt;br&gt;&#xA;drink some soup&lt;br&gt;&#xA;return a ring&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;coming from a line of paper&lt;br&gt;&#xA;although wrong, we named a vapor&lt;br&gt;&#xA;launching animals: one, two, three&lt;br&gt;&#xA;impressing royalty, who are we?&lt;/p&gt;&#xA;" OwnerUserId="18522" LastEditorUserId="18522" LastEditDate="2016-01-23T23:28:39.957" LastActivityDate="2016-01-27T05:41:32.357" Title="Two riddles for two people" Tags="|riddle|" AnswerCount="2" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25719" PostTypeId="2" ParentId="25598" CreationDate="2016-01-24T01:07:13.657" Score="1" Body="&lt;p&gt;A little bit different coding solution: This can be done by creating the cartesian product of four instances of the set of numbers 0-9. This results in a list of our products, each of length 4 (ten thousand such lists, or 10&lt;sup&gt;4&lt;/sup&gt;); first is &lt;code&gt;(0, 0, 0, 0)&lt;/code&gt;, last is &lt;code&gt;(9, 9, 9, 9)&lt;/code&gt;. Then we filter that list based on the given constraints &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In scala:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;val toTen = Range.inclusive(0, 9)&#xA;&#xA;val candidates = for { x &amp;lt;- toTen; y &amp;lt;- toTen;&#xA;              z &amp;lt;- toTen; a &amp;lt;- toTen } yield (x, y, z, a)&#xA;&#xA;val result = candidates.drop(1)   // exclude 0000&#xA;   .filter(x =&amp;gt;&#xA;     (x._4 == 2*x._1) &amp;amp;&amp;amp; (x._2 == x._3) &amp;amp;&amp;amp;&#xA;     (x._3*10 + x._4) == 2*(x._1 * 10 + x._2))&#xA;   .head&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;code&gt;res1: (Int, Int, Int, Int) = (4,9,9,8)&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18656" LastActivityDate="2016-01-24T01:07:13.657" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25720" PostTypeId="2" ParentId="25715" CreationDate="2016-01-24T01:53:01.510" Score="1" Body="&lt;p&gt;I think one-leg is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; collar&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and four-legs is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the dog. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The place with no legs is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a park, &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the two-legs is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the dog owner. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hence,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the dog owner taking his/her dog for a walk to a park with its collar. The dog basically is brought by the collar and the owner comes after the dog. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18645" LastActivityDate="2016-01-24T01:53:01.510" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25721" PostTypeId="1" AcceptedAnswerId="25722" CreationDate="2016-01-24T02:31:00.453" Score="4" ViewCount="330" Body="&lt;p&gt;Ripped from &lt;a href=&quot;https://en.wikipedia.org/wiki/The_Prisoner_of_Benda&quot; rel=&quot;nofollow&quot;&gt;The Prisoner of Brenda&lt;/a&gt; with apologies to Matt Groening, Ken Keeler &lt;em&gt;et al&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Professor Farnsworth: &quot;There. This time I'm sure I've fixed the &lt;a href=&quot;http://theinfosphere.org/Mind-switcher&quot; rel=&quot;nofollow&quot;&gt;Mind-switcher&lt;/a&gt;.&quot; An invention in which two people (or a robot and an intelligent wash bucket or whatever) can sit and have their minds swapped.  After a brief discussion as to why anyone would want to do such a thing, Amy Wong concludes &quot;We are just the people this mind-switcher was made for by us!&quot;  They switch minds but soon realize that it wasn't such a good idea.  Getting back into the Mind-switcher to undo their mistake, nothing happens!  Farnsworth (now in Amy's body): &quot;I failed to take into account the cerebral immune response.  Once two bodies have switched minds, they can never [directly] switch back.  However, perhaps they can [indirectly].  Maybe we can swap back using a third [or more] body for temporary storage space.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show how Amy and Farnsworth can swap their minds back using one or more &quot;temporary storage space&quot; bodies so that in the end everyone involved is back to normal.  Or prove that it's impossible.&lt;/p&gt;&#xA;" OwnerUserId="10118" LastEditorUserId="10118" LastEditDate="2016-01-24T02:55:51.053" LastActivityDate="2016-01-24T03:04:43.337" Title="The Prisoner of Brenda" Tags="|mathematics|logical-deduction|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25722" PostTypeId="2" ParentId="25721" CreationDate="2016-01-24T02:55:43.883" Score="6" Body="&lt;p&gt;&lt;strong&gt;It is possible.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Call Amy and Farnsworth A and F.  Find two more people, X and Y.  Let B(C) imply that person B has person C's mind.  Let &quot;Swap B and C&quot; imply swapping whatever minds are currently in the bodies labeled B and C.&lt;br&gt;&#xA;&lt;br&gt;&#xA; So far, A and F swapped to get: A(F), F(A), X(X), Y(Y).  Swaps Used: AF.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Swap A and X; F and Y.  Get A(X), F(Y), X(F), Y(A).  Swaps Used: AF, AX, FY.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Swap A and Y; F and X. Get A(A), F(F), X(Y), Y(X).  Swaps Used: AF, AX, FY, AY, FX.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Finally, Swap X and Y to get A(A), F(F), X(X), Y(Y).&lt;br&gt;&#xA;&lt;br&gt;&#xA; Everybody's back to normal!  Well, as normal as they were to begin with.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="17389" LastEditDate="2016-01-24T03:04:43.337" LastActivityDate="2016-01-24T03:04:43.337" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25723" PostTypeId="2" ParentId="25715" CreationDate="2016-01-24T04:40:07.507" Score="2" Body="&lt;p&gt;I think the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Evolution&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One-leg took four-legs&#xA;to a place with no legs,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A single replicator eventually became 4 nucleotides (the building-blocks, or legs of nucleic acids such as DNA). And took this new replicator to the oceans (at the time, no life-forms there had any literal legs).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;two-legs came later.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We evolved from all of that much later, and we have two legs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18522" LastActivityDate="2016-01-24T04:40:07.507" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25724" PostTypeId="1" AcceptedAnswerId="25726" CreationDate="2016-01-24T06:43:10.027" Score="12" ViewCount="611" Body="&lt;p&gt;Following on from the recent &lt;a href=&quot;https://puzzling.stackexchange.com/questions/25524/blindfolded-and-disoriented-near-the-great-wall-of-china&quot;&gt;blindfolded near the great Wall of China puzzle&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you are floating in space a mile away from a huge space station (the death star perhaps). Assume that the death star is essentially an infinite plane, and you want to get to it. The only way is to bump into it. But it could be in any direction. What's the shortest distance you could fly in your space suit to guarantee you'll hit it? Assume there's no gravity or other sensors available. &lt;/p&gt;&#xA;" OwnerUserId="15942" LastEditorUserId="15942" LastEditDate="2019-02-13T16:20:01.200" LastActivityDate="2019-02-13T16:20:01.200" Title="Blindfolded and disoriented near a space station" Tags="|geometry|optimization|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="25726" PostTypeId="2" ParentId="25724" CreationDate="2016-01-24T07:20:39.227" Score="8" Body="&lt;p&gt;A simple upper bound is $\sqrt{2} + 2\times 2\pi + 2 \approx 15.98$ miles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we travel on two circles of radius $1$, parallel to each other and $2$ miles apart, and whose centers are collinear with the starting point, we can guarantee to hit the space station. To do this, we travel $\sqrt{2}$ miles to the circumference of the first circle, travel around it, then travel the $2$ miles to the other circle and travel around that.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As @MPeti mentioned in the comments below, we can improve this by using the &lt;a href=&quot;https://puzzling.stackexchange.com/a/25538/8949&quot;&gt;optimal path&lt;/a&gt; in the 2-d version instead of circles. The length of the path now becomes $\sqrt{2+\frac{1}{3}} + 2\times(\frac{7\pi}{6}+1+\sqrt{\frac{4}{3}-1}) + 2 \approx 14.01$ miles.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If we don't travel the last mile in the 2-d path and travel between the two paths in a triangular fashion, and realise that the optimum path in 2-d is not necessarily the optimum path in 3-d, we can reduce the length further to approximately $12.74$ miles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A graphic to help visualisation (made using &lt;a href=&quot;http://www.geogebra.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;GeoGebra&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ftLxT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ftLxT.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The length of the top path + the initial radial path is $\frac{3\pi}{2}-\theta+\tan{\frac{\theta}{2}}+\sqrt{\sec^2{\frac{\theta}{2}}+1}$, which according to &lt;a href=&quot;http://www.wolframalpha.com/input/?i=minimize+3*pi%2F2-x%2Btan%28x%2F2%29%2Bsqrt%28sec%5E2%28x%2F2%29%2B1%29%2C+0%3C%3Dx%3C%3Dpi%2F2&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wolfram Alpha&lt;/a&gt; is minimised when $\theta \approx 1.14372$ with a length of $l_1 \approx 5.76604$ miles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AShcZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AShcZ.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The length of the bottom path is $\frac{3\pi}{2}-\theta+\tan{\frac{\theta}{2}}$, which is minimised when $\theta = \frac{\pi}{2}$ with a length of $l_2 = \pi + 1 \approx 4.14159$ miles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jfnPn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jfnPn.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding these two lengths and the length of the semicircular path between them, we arrive at a total length of $l_1 + 2\sqrt{2} + l_2 \approx 12.74$ miles.&lt;/p&gt;&#xA;" OwnerUserId="8949" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2016-01-26T01:30:42.327" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="25728" PostTypeId="2" ParentId="25598" CreationDate="2016-01-24T11:28:06.710" Score="0" Body="&lt;p&gt;Wolfram Mathematica solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pEXJS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pEXJS.png&quot; alt=&quot;Wolfram Mathematica solution&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer should be: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;4998&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18659" LastActivityDate="2016-01-24T11:28:06.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25730" PostTypeId="2" ParentId="25715" CreationDate="2016-01-24T17:32:57.297" Score="0" Body="&lt;p&gt;Since my previous answer was wrong, I'll have another go:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A hole, in golf.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;One-leg took four-legs&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The club (a single stem/leg) took four tries (legs, or sections of a journey) to get the ball...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;to a place with no legs,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The hole.&lt;br&gt;&#xA; A well-hit ball in golf is also sometimes referred to having legs - &lt;em&gt;&quot;That's got some legs&quot;&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;two-legs came later.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The human player, having two legs, came along later to collect the ball.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18522" LastActivityDate="2016-01-24T17:32:57.297" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25731" PostTypeId="1" AcceptedAnswerId="25732" CreationDate="2016-01-24T19:15:13.730" Score="33" ViewCount="4783" Body="&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A word of 6 letters,&lt;br&gt;&#xA;  I always play fair&#xA;  &lt;br&gt;Minus my first I am cold&#xA;  &lt;br&gt;Minus my last and jumbled,&#xA;  &lt;br&gt;I'll give you a massage&#xA;  &lt;br&gt;Jumble my whole,&#xA;  &lt;br&gt;You'll learn nothing from me&#xA;  &lt;br&gt;I always play fair&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;Hint to those still trying to solve it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;(You probably know &lt;a href=&quot;https://puzzling.stackexchange.com/questions/20233/a-short-brutal-riddle&quot;&gt;this question&lt;/a&gt; by @roland by now.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18250" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2016-02-04T10:59:07.543" Title="What am I? I always play fair" Tags="|riddle|word|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25732" PostTypeId="2" ParentId="25731" CreationDate="2016-01-24T20:11:05.713" Score="32" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;CHEAT&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A word of 6 letters,&#xA;I always play fair &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A CHEAT doesn't play fair, and it's not 6 letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Minus my first I am cold &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;HEAT isn't cold.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Minus my last and jumbled, &#xA;I'll give you a massage &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You might need a massage if you have an ACHE.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Jumble my whole,&#xA;You'll learn nothing from me &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;TEACHing makes you learn things.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I always play fair&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Every clue is wrong because the narrator is cheating. (aka same gimmick as &lt;a href=&quot;https://puzzling.stackexchange.com/questions/20233/a-short-brutal-riddle&quot;&gt;this other one&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2016-01-24T20:11:05.713" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25734" PostTypeId="1" AcceptedAnswerId="25735" CreationDate="2016-01-24T21:11:39.563" Score="7" ViewCount="522" Body="&lt;p&gt;I am four, but my second can be any of the five, though not sometimes.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;My first is final.&lt;/li&gt;&#xA;&lt;li&gt;My second often warns of danger.&lt;/li&gt;&#xA;&lt;li&gt;My third is what you are in the middle of right now.&lt;/li&gt;&#xA;&lt;li&gt;My fourth cannot be found.&lt;/li&gt;&#xA;&lt;li&gt;My fifth is what you desire.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus:&lt;/strong&gt; Can you think of any others like me?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Since this has been answered already, I'll include some other examples below.  Don't roll over the spoiler box below if you still want to solve the original puzzle on your own.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;- BAD BED BID BOD BUD&lt;br&gt;&#xA;- BAG BEG BIG BOG BUG (Thanks @Deusovi)&lt;br&gt;&#xA;- BALL BELL BILL BOLL BULL&lt;br&gt;&#xA;- BAT BET BIT BOT BUT&lt;br&gt;&#xA;- FALL FELL FILL FOLL FULL&lt;br&gt;&#xA;- FAR FER FIR FOR FUR (Also FYR, but that's a proper noun from the Elder Scrolls games, so maybe shouldn't count.)&lt;br&gt;&#xA;- MASS MESS MISS MOSS MUSS (MUSS being an informal spelling of MESS, generally done to hair.)&lt;br&gt;&#xA;- MAST MEST MIST MOST MUST MYST (Thanks @APrough, though as noted, MEST and MYST are both proper nouns.)&lt;br&gt;&#xA;- PACK PECK PICK POCK PUCK&lt;br&gt;&#xA;- PALL PELL PILL POLL PULL&lt;br&gt;&#xA;- PAP PEP PIP POP PUP&lt;br&gt;&#xA;- PAT PET PIT POT PUT&lt;br&gt;&#xA;- TAN TEN TIN TON TUN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16131" LastEditorUserId="16131" LastEditDate="2016-01-25T18:18:09.553" LastActivityDate="2020-09-07T08:23:00.360" Title="Five from Four - What am I?" Tags="|wordplay|letters|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25735" PostTypeId="2" ParentId="25734" CreationDate="2016-01-24T21:25:56.513" Score="7" Body="&lt;p&gt;You're&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; L_ST.&lt;br&gt;&#xA;Last is final; lest means &quot;with the intention of preventing&quot;; I am in the middle of a list; something lost cannot be found; lust is desire.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And for another example;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;BAG, BEG, BIG, BOG, BUG is another series that works like this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2016-01-24T21:25:56.513" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25736" PostTypeId="2" ParentId="25734" CreationDate="2016-01-24T21:36:22.237" Score="3" Body="&lt;p&gt;Another example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;TAN, TEN, TIN, TON and TUN*&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;*This is a word, see &lt;a href=&quot;https://www.bing.com/search?q=tun&amp;amp;form=EDGNTC&amp;amp;qs=PF&amp;amp;cvid=7236fd70511a469f8214054d5efaf16f&amp;amp;pq=tun&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; if you don't believe me&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BALL, BELL, BILL, BOLL and BULL&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18250" LastActivityDate="2016-01-24T21:36:22.237" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25738" PostTypeId="1" AcceptedAnswerId="25744" CreationDate="2016-01-25T02:24:07.013" Score="8" ViewCount="12862" Body="&lt;p&gt;In &lt;em&gt;Outliers&lt;/em&gt;, Malcolm Gladwell presents the following puzzle:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; ♦♣♣ | ♥♥♣ | ♦♥♥&#xA; ♦♦♣ | ♣♥♦ | ♦♣♣&#xA; ♥♥♥ | ♦♦♣ | ♣♥♦&#xA;-----------------&#xA; ♥♣♦ | ♥♣♦ | ♥♣♦&#xA; ♥♣♥ | ♥♦♣ | ♣♥♥&#xA; ♦♦♣ | ♣♥♦ | ♦♣♦&#xA;-----------------&#xA; ♦♥♦ | ♦♣♥ |&#xA; ♣♥♣ | ♦♥♦ |  ?&#xA; ♥♣♦ | ♥♣♣ |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In the first edition (2008) of the book, the top left square erroneously appears as:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;♣♣♦&#xA;♦♥♣&#xA;♦♥♥&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The alleged answer in both cases is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ♥♦♣&lt;br&gt;&#xA; ♦♣♦&lt;br&gt;&#xA; ♥♥♣&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have two questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What is the pattern? &lt;/li&gt;&#xA;&lt;li&gt;What mistake(s) could explain the discrepancy between editions?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Note:&lt;/strong&gt; I have an answer for Q1, but do not have one for Q2.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="17147" LastActivityDate="2020-04-15T19:17:29.693" Title="Malcolm Gladwell's Outliers - Progressive Matrices Puzzle" Tags="|pattern|explanation|" AnswerCount="4" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25740" PostTypeId="2" ParentId="25715" CreationDate="2016-01-25T05:10:54.020" Score="0" Body="&lt;p&gt;My guesses are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One-leg took four-legs&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I guess: A Helicopter - assuming &lt;em&gt;leg&lt;/em&gt; is figurative&lt;br&gt;&lt;br&gt;&#xA; As the first helicopter designs (Chinese Flying Top or Leonardo's Airscrew) had a single propeller while the first practical working model (early model Skiorosky VS-300s) had four propellers.&lt;br&gt;&lt;br&gt;&#xA; While the hint: Helicopters fly but have no &lt;em&gt;wings&lt;/em&gt; but sometimes, like the pictured skiorosky below, they have two &quot;arms&quot; for carrying payloads, to help stabilize, and for other reasons.&#xA; &lt;img src=&quot;https://i.stack.imgur.com/HAYkn.jpg&quot; alt=&quot;original Skiorosky vs300 from rotaryforum.com&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;to a place with no legs, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; From the 1939 Skiorosky came the jet propelled aircraft of the 1940s which had no propellers or &quot;legs&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;two-legs came later.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Finally in the 1950s to just recently, Tilt-rotor aircraft (like the Marine's V-22 Osprey) have been developed and successfully flown.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11239" LastEditorUserId="11239" LastEditDate="2016-01-25T12:17:23.983" LastActivityDate="2016-01-25T12:17:23.983" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25741" PostTypeId="2" ParentId="25718" CreationDate="2016-01-25T05:45:50.840" Score="0" Body="&lt;p&gt;Is it a volcano?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A peak is obvious.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Not sure about what &quot;swing&quot; would refer to.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Drink some soup could be a pool of lava inside the volcano.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Return a ring is either a reference to The Lord of the Rings, or perhaps referring to the ring around the tip of a stereotypical conical volcano. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Coming from a line on paper refers to the fact that it appears on maps.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Vapour is the smoke coming from the volcano.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I'm not too sure about the last two lines, but any animals around when a volcano erupts would certainly be high tailing it out of there, and volcanoes are impressive in general.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="17815" LastActivityDate="2016-01-25T05:45:50.840" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="25742" PostTypeId="1" AcceptedAnswerId="25909" CreationDate="2016-01-25T07:25:33.037" Score="8" ViewCount="1602" Body="&lt;p&gt;Caesar was quietly sipping a good cup of coffee at a cafe in &lt;em&gt;Rome&lt;/em&gt; when a police officer suddenly approached him. We need you, (s)he said! We have a very complicated case and the &lt;em&gt;primary&lt;/em&gt; victims here are important. Caesar, a great detective, was intrigued by the word &quot;complicated&quot; and agreed to help. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The police officer said that they collected video recordings of each victim showing the last thing of what they were doing and where they were before the crime was committed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First Victim&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Caesar went to the first victim. He is floating on the swimming &lt;em&gt;pool&lt;/em&gt;, eyes wide open and seems to be looking into the &lt;em&gt;sky&lt;/em&gt;. A piece of paper is lying beside him and there is a note saying: &quot;I've waited a &lt;em&gt;century&lt;/em&gt; just to kill him. Want a clue? The victim is the &lt;em&gt;clue&lt;/em&gt; now because of me.&quot;. Caesar thought that this is the suspect's clue and he believes that the victim trusts the suspect and was loyal to him until now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Caesar decided to look at the video recording. The victim is drinking at the &lt;em&gt;Topaz&lt;/em&gt; bar not far from the crime scene. A stranger approached him and they talked for a few minutes. After that the victim left. The guy beside the stranger followed the victim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second Victim&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Caesar then visited the second victim. He decided to look at the recording first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The recording shows that the second victim is with a female stranger. They seem to be talking and are buying &lt;em&gt;roses&lt;/em&gt; at the flower shop. After that, the victim left and the female stranger followed him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Caesar then went to the crime scene and saw the roses on top of the victim. The roses are soaked in blood and a note is on top of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The note says: &quot;On the start of the day I met you, I'm with you until the end, but I will also show you the end. Nothing will change if I'm still with you&quot; Caesar was left with deep thoughts. It's like watching &lt;em&gt;500 Days of Summer&lt;/em&gt;. He believes that the suspect is the female stranger on the recording, might be a love affair angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Third Victim&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Caesar hurried to the last crime scene, he immediately played the recording.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The recording shows the victim talking to two guys. Caesar was surprised to see the same guy from the first victim's recording! And what is surprising is the guy has a &lt;em&gt;twin&lt;/em&gt;! In the recording, the victim handed a &lt;em&gt;gold bar&lt;/em&gt; to the twins and left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When Caesar checked the crime scene, he saw the gold bars on top of the victim's dead body, A &lt;em&gt;Donovan music CD&lt;/em&gt; and a note was also there. It says: &quot;&lt;em&gt;1k of Gold&lt;/em&gt; only for me? because of that '&lt;em&gt;I'm the guy who can calm you down&lt;/em&gt;'. Thanks to the twins and I got the truth&quot;. Caesar concluded that this one might be jealousy.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;With all the clues and recordings, Caesar finally got the suspects and arrested them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you need to find out&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The 3 suspects&lt;/li&gt;&#xA;&lt;li&gt;The 3 victims&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;BONUS:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3. The name of the group of the 3 suspects&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;HINT:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clues are distributed in the story. Using music, symbolism, words, movies. What is common on them. (second clue tomorrow if not solved)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;adding more HINTS&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answers are not exactly a name. Italicized some words and phrases as clues&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16036" LastEditorUserId="16036" LastEditDate="2016-01-28T00:56:57.657" LastActivityDate="2016-02-01T02:09:42.730" Title="Caesar and the Roman Murders" Tags="|enigmatic-puzzle|story|mystery|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25743" PostTypeId="2" ParentId="25718" CreationDate="2016-01-25T08:17:45.900" Score="4" Body="&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Okay, so the undeveloped idea that I initially thought was wrong turned out to be the right answer after all. I've now added the details. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer: the two people described in each of the two riddles are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;strong&gt;Montgolfier brothers&lt;/strong&gt;, inventors of the hot-air balloon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Riddle 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The phrase &quot;Montgolfier brothers&quot; is a combination of &lt;br&gt;&lt;br&gt;&#xA; &lt;strong&gt;Mont&lt;/strong&gt;: &quot;peak&quot; in &quot;have a peak&quot;, &lt;br&gt;&#xA; &lt;strong&gt;golf&lt;/strong&gt;: related to &quot;swing&quot; in &quot;take a swing&quot;, &lt;br&gt;&#xA; &lt;strong&gt;broth&lt;/strong&gt;: soup in &quot;drink some soup&quot;, &lt;br&gt;&#xA; ers, or I guess the word &lt;strong&gt;hers&lt;/strong&gt;, is sorta-kinda related to &quot;ring&quot; in &quot;return a ring&quot;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Riddle 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;coming from a line of paper&quot;: as &lt;a href=&quot;https://en.wikipedia.org/wiki/Montgolfier_brothers&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; says, &quot;The brothers were born into a family of paper manufacturers in Annonay, in Ardèche, France&quot;.&lt;br&gt;&lt;br&gt;&#xA; &quot;although wrong, we named a vapor&quot;. Wikipedia says of Joseph, the older of the brothers, that, after observing the tendency of hot over a fire to rise, &quot;He believed that contained within the smoke was a special gas, which he called Montgolfier Gas, with a special property he called levity.&quot;&lt;br&gt;&lt;br&gt;&#xA; &quot;launching animals, one two three&quot;: Wikipedia writes of one of the brothers' first experimental hot-air balloon flights, &quot;The king proposed to launch two criminals, but it is most likely that the inventors decided to send a sheep, a duck, and a rooster aloft first.&quot;&lt;br&gt;&lt;br&gt;&#xA; &quot;impressing royalty, who are we?&quot;: the Montgolfier brothers invention astonished not just royalty but all of France, Europe and (I suspect) the world. In particular, Wikipedia writes that &quot;... in December 1783, in recognition of their achievement, their father Pierre was elevated to the nobility and the hereditary appellation of de Montgolfier by King Louis XVI of France.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user17947" LastEditorDisplayName="user17947" LastEditDate="2016-01-27T05:41:32.357" LastActivityDate="2016-01-27T05:41:32.357" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="25744" PostTypeId="2" ParentId="25738" CreationDate="2016-01-25T10:59:27.917" Score="5" Body="&lt;p&gt;I mapped the symbols to $\{0,1,2\}$, so we get:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 011 | 221 | 022&#xA; 001 | 120 | 011&#xA; 222 | 001 | 120&#xA;-----------------&#xA; 210 | 210 | 210&#xA; 212 | 201 | 122&#xA; 001 | 120 | 010&#xA;-----------------&#xA; 020 | 012 |&#xA; 121 | 020 |  ?&#xA; 210 | 211 |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I found this quite simple algorithm (which has two cases, either it is a row change [type2] or not [type1]):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; Type 1:&#xA; * we (+1 mod 3) every element &#xA; * rotate columns left one step&#xA; * rotate last column upwards&#xA; Type 2, row change:&#xA; * rotate the matrix 90 degrees&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Since the final step is type 1, we get&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 120      120      120&#xA; 200  -&amp;gt;  200  -&amp;gt;  202&#xA; 112      112      110&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and then +1 and translate back to cardsymbols.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 120      201       ♥♦♣&#xA; 202  --&amp;gt; 010  --&amp;gt;  ♦♣♦&#xA; 110      221       ♥♥♣&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I don't know about the discrepancy. The first and wrong matrix does not make sense to me.&lt;/p&gt;&#xA;" OwnerUserId="12901" LastEditorUserId="12901" LastEditDate="2016-01-28T07:47:27.730" LastActivityDate="2016-01-28T07:47:27.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25745" PostTypeId="1" AcceptedAnswerId="25747" CreationDate="2016-01-25T12:25:22.137" Score="3" ViewCount="1135" Body="&lt;p&gt;My university is having some sort of a challenge. I am really excited by this, but I can't even get through the first part of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought you guys might find it interesting: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here it goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first part involves decoding the following chipertext:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;FGPGZ AGCAE CRRVF GPWJG QCLFA QLFGV GQLUM HRJGU OWCUR CLFGE JCGOY EAGQU RQJGT GJQLG&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint: &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&#xA;Turn two to the left, then two to the right,&lt;br&gt;&#xA;Repeat, to get on the other side.&lt;br&gt;&#xA;Don't be afraid: go, see and conquer,&lt;br&gt;&#xA;You'll find secrets, fun and wonder!&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I tried applying a Caesar shift, but it doesn't seem to be the way to go.&lt;/p&gt;&#xA;" OwnerUserId="10692" LastEditorDisplayName="user20" LastEditDate="2016-08-05T17:00:05.460" LastActivityDate="2016-08-05T17:00:05.460" Title="Crypto-Challenge - part 1" Tags="|cipher|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25746" PostTypeId="2" ParentId="24707" CreationDate="2016-01-25T12:29:32.323" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Letters above the line have been initals for the names of legitimate Kings or Queens of England, while those below the line haven't.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So XYZ go all below the line by the way.&lt;/p&gt;&#xA;" OwnerUserId="18576" LastEditorUserId="18576" LastEditDate="2016-01-27T15:02:15.730" LastActivityDate="2016-01-27T15:02:15.730" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25747" PostTypeId="2" ParentId="25745" CreationDate="2016-01-25T12:32:27.937" Score="9" Body="&lt;p&gt;The key is to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Shift two in one direction, then two in the other, back and forth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You get&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HEREB YIACC EPTTH ERULE SANDC ONDIT IONSO FTHIS QUEST ANDIC LAIMA CCESS TOLEV ELONE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Hereby I accept the rules and conditions of this quest and I claim access to level one&quot;&lt;br&gt;&#xA; when you move the spaces&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17147" LastActivityDate="2016-01-25T12:32:27.937" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25748" PostTypeId="2" ParentId="25715" CreationDate="2016-01-25T13:11:12.203" Score="7" Body="&lt;p&gt;Since the hint mentioned&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the one-leg is something that can fly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I thought it could be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a rocket.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The four-legs refers to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The dog Laika, the first animal in space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;then the place with no legs is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; space, since you're weightless in space and so you effectively have no legs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;two-legs is then:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Humans, since we went to space later.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1598" LastActivityDate="2016-01-25T13:11:12.203" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25750" PostTypeId="2" ParentId="25707" CreationDate="2016-01-25T13:41:36.587" Score="5" Body="&lt;p&gt;&lt;strong&gt;Please note this is not the complete answer&lt;/strong&gt;&lt;br&gt;&#xA;This is simply a formatted version of what I pointed out in the comments, and we have since learned it is a step towards completing this puzzle.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Each statement is either true or false.&lt;/strong&gt;&lt;br&gt;&lt;br&gt;&#xA; &lt;strong&gt;False Statements&lt;/strong&gt;&lt;br&gt;&lt;br&gt;&#xA; 1. The droid, BB-8, is R2-D2's child.&lt;br&gt;&#xA; 4. (Only in this movie, Leia's title is &quot;Sergeant&quot; instead of &quot;Princess&quot;.)&lt;br&gt;&#xA; 6. Poe Dameron asks Finn to pilot a ship off of Jakku.&lt;br&gt;&#xA; 12. She has a hard time leaving the aunt and uncle she lives with on Jakku.&lt;br&gt;&#xA; 13. But she does it because she has heard there's a powerful Jedi on Dagobah who can teach her.&lt;br&gt;&#xA; 15. But this time, they've learned their lesson, and this one is NOT vulnerable in one small area where it can be attacked by small ships.&lt;br&gt;&#xA; 17. And Leia cries like a baby onscreen when Rey tells her what happened.&lt;br&gt;&#xA; 18. Rey finds Luke sitting in a cozy cottage drinking a cup of tea.&lt;br&gt;&lt;br&gt;&#xA; &lt;strong&gt;True Statements&lt;/strong&gt;&lt;br&gt;&lt;br&gt;&#xA; 2. Kylo Ren's father is Han Solo&lt;br&gt;&#xA; 3. And his mother is Leia!&lt;br&gt;&#xA; 5. Finn is a stormtrooper who defects.&lt;br&gt;&#xA; 7. The ship Poe and Finn are in crashes on Jakku.&lt;br&gt;&#xA; 8. But fortunately we find out they both survived.&lt;br&gt;&#xA; 9. The girl, Rey, has a mysterious background.&lt;br&gt;&#xA; 10. It turns out that she can use the force.&lt;br&gt;&#xA; 11. She is also a very capable pilot. &lt;em&gt;Thx to @LogicianWithAHat&lt;/em&gt;&lt;br&gt;&#xA; 14. Meanwhile, the evil regime has built a powerful weapon of mass destruction.&lt;br&gt;&#xA; 16. Kylo Ren murders Han Solo.&lt;br&gt;&#xA; 19. (Because of course it turns out there is a map pointing out exactly where Luke is.)&lt;br&gt;&#xA; 20. Which we only find in its entirety near the end of the plot because, apparently, droid mental health issues are a real thing -- poor R2-D2.&lt;br&gt;&lt;br&gt;&#xA; &lt;strong&gt;Binary&lt;/strong&gt;:&lt;br&gt;&lt;br&gt;&#xA; 01101011111001010011&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12014" LastActivityDate="2016-01-25T13:41:36.587" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25751" PostTypeId="2" ParentId="25055" CreationDate="2016-01-25T13:47:33.893" Score="18" Body="&lt;p&gt;&lt;em&gt;An ancient man demolished his estate,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ROMAN - &quot;An ancient man&quot;, an anagram of MANOR (estate).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And therefore, with enthusiasm innate,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; THUS - &quot;therefore&quot;, found in &quot;en&lt;strong&gt;thus&lt;/strong&gt;iasm&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;He was bewildered when a hoax was glued&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CONFUSED - &quot;bewildered&quot;, made from CON (a hoax) and FUSED (glued).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Somewhere beside where he had said adieu.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BY - &quot;beside&quot;, sounds like BYE (adieu).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The dome where five lone defects lost their head&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; VAULT - &quot;dome&quot;, made from V (five) and (F)AULT (&quot;lone defects&quot; without the first letter).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Confounds small pools where diamonds sleep instead.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PUZZLES - &quot;confounds&quot;, from PUDDLES (small pools) with Z (the sound of sleep) instead of D (diamonds).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Put together, this makes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Roman thus confused by vault puzzles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;for which the solution is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CRYPTICS - &quot;puzzles&quot;, made from an anagram of SIC (Latin for thus) put by CRYPT (vault).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The poem is indeed a study in&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; cryptics!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12511" LastEditorUserId="12511" LastEditDate="2016-01-25T16:11:09.557" LastActivityDate="2016-01-25T16:11:09.557" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25752" PostTypeId="2" ParentId="25690" CreationDate="2016-01-25T15:39:44.733" Score="3" Body="&lt;p&gt;This answer was a collaborative effort between myself and &lt;a href=&quot;https://puzzling.stackexchange.com/users/17250/question-asker&quot;&gt;question_asker&lt;/a&gt;. I figured out:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;taker of fleece&lt;/em&gt; could be &lt;strong&gt;shearer&lt;/strong&gt;;&lt;br&gt;&#xA; and that &lt;em&gt;voice of&lt;/em&gt; and &lt;em&gt;rarely appear&lt;/em&gt; hints at a voice actor&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This was confirmed by the OP with some further hints and then &lt;a href=&quot;https://puzzling.stackexchange.com/users/17250/question-asker&quot;&gt;question_asker&lt;/a&gt; figured out the rest...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;COLLABO-POWERS ACTIVATE&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You said &lt;em&gt;voice actor&lt;/em&gt; and &lt;em&gt;Shearer&lt;/em&gt; and my thoughts went to Harry Shearer and Dan &lt;a href=&quot;https://en.wikipedia.org/wiki/Castellaneta&quot; rel=&quot;nofollow noreferrer&quot;&gt;Castellaneta&lt;/a&gt;, both voice actors on The Simpsons (&quot;on the small &lt;em&gt;screen&lt;/em&gt; for decades&quot;, maybe?). They are Kang and Kodos (Ca and Co), the dome-headed squidlike alien things.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18522" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2016-01-26T09:56:57.490" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="25753" PostTypeId="2" ParentId="25734" CreationDate="2016-01-25T16:41:19.590" Score="4" Body="&lt;p&gt;Though Deusovi got the answer, I thought I would give you one for the bonus, and it works with &quot;Y&quot; as well.  Though it does use 2 proper names, but nothing above excludes that.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MAST, MEST (punk rock band), MIST, MOST, MUST, MYST (computer game from the 90's)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14233" LastActivityDate="2016-01-25T16:41:19.590" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25755" PostTypeId="2" ParentId="25690" CreationDate="2016-01-25T16:42:01.840" Score="1" Body="&lt;p&gt;&lt;strong&gt;The Voice&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A common Cryptic keyword indicating &quot;Sounds Like&quot; or  homonym.  This would apply to &quot;city in italy&quot; and possibly, but not necessarily the &quot;taker of fleece&quot;  Livorno and shearer --&gt; Laverne and Shirley?  But I can't parse the rest.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2752" LastActivityDate="2016-01-25T16:42:01.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25756" PostTypeId="2" ParentId="25214" CreationDate="2016-01-25T17:27:01.587" Score="5" Body="&lt;p&gt;The clue about &lt;strong&gt;bacon&lt;/strong&gt; hints at a &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Bacon%27s_cipher&quot; rel=&quot;nofollow noreferrer&quot;&gt;Baconian cipher&lt;/a&gt;&lt;/em&gt;.  &lt;a href=&quot;https://puzzling.stackexchange.com/users/12511&quot;&gt;Zandar&lt;/a&gt; suggested that it might be a variant of the Baconian cipher that uses a ternary instead of a binary system.  So I started by assigning a pattern to each letter, using the values &lt;code&gt;A, B, C&lt;/code&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;AAA  A            BAA  J            CAA  S&#xA;AAB  B            BAB  K            CAB  T&#xA;AAC  C            BAC  L            CAC  U&#xA;ABA  D            BBA  M            CBA  V&#xA;ABB  E            BBB  N            CBB  W&#xA;ABC  F            BBC  O            CBC  X&#xA;ACA  G            BCA  P            CCA  Y&#xA;ACB  H            BCB  Q            CCB  Z&#xA;ACC  I            BCC  R            CCC  (unassigned)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then I tried to figure out a way to get a pattern out of the poem.  Based on the clue words &lt;code&gt;word&lt;/code&gt;, &lt;code&gt;count&lt;/code&gt;, and &lt;code&gt;threes&lt;/code&gt;, and with the help of the second hint, I counted the words in each line, and assigned each line a letter as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;number of words is divisible by 3:  &lt;code&gt;A&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;number of words $\div$ 3 leaves a remainder of 1: &lt;code&gt;B&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;number of words $\div$ 3 leaves a remainder of 2: &lt;code&gt;C&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(as a more mathematical way of looking at it, use &lt;code&gt;0&lt;/code&gt;, &lt;code&gt;1&lt;/code&gt;, &lt;code&gt;2&lt;/code&gt; instead of &lt;code&gt;A&lt;/code&gt;, &lt;code&gt;B&lt;/code&gt;, &lt;code&gt;C&lt;/code&gt;, and just use (word count $\text{mod}$ 3) to assign a value)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This left me with the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;          Word                Letter&#xA; Line #   count   Remainder   assigned&#xA; 1        8       2           C&#xA; 2        9       0           A&#xA; 3       10       1           B&#xA; 4        6       0           A&#xA; 5        7       1           B&#xA; 6        7       1           B&#xA; 7        7       1           B&#xA; 8        5       2           C&#xA; 9        8       2           C&#xA;10        7       1           B&#xA;11       10       1           B&#xA;12        4       1           B&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This leaves us with four triplets:  &lt;code&gt;CAB&lt;/code&gt;, &lt;code&gt;ABB&lt;/code&gt;, &lt;code&gt;BCC&lt;/code&gt;, &lt;code&gt;BBB&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we refer back to our alphabet table, these correspond to the letters&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;code&gt;T&lt;/code&gt;, &lt;code&gt;E&lt;/code&gt;, &lt;code&gt;R&lt;/code&gt;, &lt;code&gt;N&lt;/code&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now if we look at the rhyme scheme of the poem, we get another pattern, based on which words rhyme with each other:  AAAB CCCC AAAB. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Splitting that into triplets, we get &lt;code&gt;AAA&lt;/code&gt;, &lt;code&gt;BCC&lt;/code&gt;, &lt;code&gt;CCA&lt;/code&gt;, &lt;code&gt;AAB&lt;/code&gt;, which correspond to the letters&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;code&gt;A&lt;/code&gt;, &lt;code&gt;R&lt;/code&gt;, &lt;code&gt;Y&lt;/code&gt;, &lt;code&gt;B&lt;/code&gt;&lt;/strong&gt;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we do the same thing a third time, this time using the &lt;em&gt;letter&lt;/em&gt; count instead of the word count, we end up with the sequence AAAAACBBCBBB. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Splitting that into triplets, we get &lt;code&gt;AAA&lt;/code&gt;, &lt;code&gt;AAC&lt;/code&gt;, &lt;code&gt;BBC&lt;/code&gt;, &lt;code&gt;BBB&lt;/code&gt;, which correspond to the letters&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;code&gt;A&lt;/code&gt;, &lt;code&gt;C&lt;/code&gt;, &lt;code&gt;O&lt;/code&gt;, &lt;code&gt;N&lt;/code&gt;&lt;/strong&gt;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Putting these three results together, we get&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;code&gt;TERNARYBACON&lt;/code&gt;&lt;/strong&gt; or &lt;strong&gt;&lt;em&gt;Ternary bacon&lt;/em&gt;&lt;/strong&gt;, which describes the cipher used in encrypting this. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;A huge thanks to Zandar who actually solved most of this, and SpiritFryer as well.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2016-01-30T20:49:47.837" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="25757" PostTypeId="1" AcceptedAnswerId="25760" CreationDate="2016-01-25T18:22:12.737" Score="3" ViewCount="369" Body="&lt;p&gt;You're enjoying a Saturday morning, making a cup of tea to start the day. Just as you've finished pouring the boiling water into your teacup, you hear a knock at the door. If you like your tea with milk and as hot as possible. Should you: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;add the milk now or &lt;/li&gt;&#xA;&lt;li&gt;add the milk after you return from answering the door? &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In either case the teacup will sit on the counter and the milk jug will sit in the fridge. Which will result in the hottest cup of tea?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extra info:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The amount of milk added is the same in both cases.&lt;/li&gt;&#xA;&lt;li&gt;The amount of water that escapes as steam is negligible.&lt;/li&gt;&#xA;&lt;li&gt;The temperature of the water is greater than the room temperature.&lt;/li&gt;&#xA;&lt;li&gt;The temperature of the room is greater than the fridge temperature.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="18103" LastActivityDate="2016-01-25T18:37:14.820" Title="Hot Tea and an Unexpected Visitor" Tags="|mathematics|physics|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25758" PostTypeId="2" ParentId="25439" CreationDate="2016-01-25T18:35:22.853" Score="3" Body="&lt;p&gt;The statement is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;given a fully connected graph on n nodes where each edge has a unique&#xA;  weight it is possible to find a path of length n-1 whose edge weight&#xA;  is strictly increasing with each edge&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By induction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case for 3 nodes is obvious (take road 1 to road 2, no way can Drump stop those pilgrims!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that its possible to find our desired path in any graph of N-1 nodes, we turn to the case of N nodes.  (We look at N = 6 because its convenient to draw).  Grab any old 5 (N-1) nodes, there is an increasing path of length 4 (N-2) in them, consider how that 6th (N-th) node is connected though:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eC3BD.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eC3BD.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We draw arrows to indicate possible directions of travel along each edge which would create an increasing path which starts along path 1-2-3-....  Note that we must draw edge 1-6 as &quot;up&quot; because if it were &quot;down&quot; we would create path 6-1-2-3-4-5 and be done.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this, edge 2-6 must be &quot;up&quot;, else we could create path 1-6-2-3-4-5 and be done.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this, edge 3-6 must be &quot;up&quot;, else we could create path 1-2-6-3-4-5 and be done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we continue in this way edge 5-6 will be up, but then path 1-2-3-4-5-6 solves our problem!&lt;/p&gt;&#xA;" OwnerUserId="10453" LastActivityDate="2016-01-25T18:35:22.853" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="25759" PostTypeId="1" AcceptedAnswerId="25771" CreationDate="2016-01-25T18:35:37.643" Score="14" ViewCount="702" Body="&lt;p&gt;My Indian friend Jamal recently participated in the well-known TV quiz&#xA;show &lt;strong&gt;&lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Who_Wants_to_Be_a_Millionaire&quot;&gt;Who wants to be a millionaire?&lt;/a&gt;&lt;/em&gt;&lt;/strong&gt;. You might have heard or read about Jamal's spectacular success; the media reported on it extensively.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what Jamal told us about the 125.000 Euro question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Unfortunately the next question brought back sad memories of my high school&#xA;  years. Prem Kumar gave us a big smile, and it was absolutely clear that he&#xA;  had no idea what the question was about.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;B was the only answer that was not on, but I was pretty sure that it was one.&#xA;  And Superman's was one also, and that eliminated C.  That left me with&#xA;  A and D.  I didn't know one from the other.  Both had five letters, both&#xA;  started with a consonant, and both starting consonants were pretty common&#xA;  consonants.  Cleverly I then decided to use my 50/50 lifeline.  I hoped&#xA;  that A or D would be eliminated.  Prem asked the computer to take away&#xA;  two wrong answers, and answers C and D disappeared.  So I chose A. I was&#xA;  excited to find out that I was correct!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;What was the 125.000 Euro question?&lt;br&gt;&#xA;What were the four possible answers A, B, C, D?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2016-01-25T23:32:15.167" Title="Slumdog millionaire (12)" Tags="|knowledge|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25760" PostTypeId="2" ParentId="25757" CreationDate="2016-01-25T18:36:37.293" Score="5" Body="&lt;p&gt;as Daedric pointed out, this is kind of a duplicate. But here's a non-math answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the hottest tea,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; add the milk before answering the door&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; when two liquids are mixed, they will reach a median temperature based on the volumes and temperatures of each (and other factors, but these are more important). So if you have milk that is always the same temperature no matter when you add it, the tea that has had a chance to cool will be cooled more by the milk than the tea that is hotter. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13101" LastActivityDate="2016-01-25T18:36:37.293" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25761" PostTypeId="2" ParentId="25757" CreationDate="2016-01-25T18:37:14.820" Score="1" Body="&lt;p&gt;The Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Adding milk first will mean the tea is hotter later. &lt;br&gt;&#xA; Courtesy of &lt;a href=&quot;http://www.thenakedscientists.com/HTML/experiments/exp/-0c2ee7e805/&quot; rel=&quot;nofollow&quot;&gt;http://www.thenakedscientists.com/HTML/experiments/exp/-0c2ee7e805/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12014" LastActivityDate="2016-01-25T18:37:14.820" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25762" PostTypeId="1" AcceptedAnswerId="25763" CreationDate="2016-01-25T18:39:06.733" Score="6" ViewCount="2265" Body="&lt;p&gt;Every letter a decimal digit, different letters for different digits:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    BASE&#xA;  + BALL&#xA; --------&#xA;   GAMES&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which digit does each letter represent?&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="9404" LastEditDate="2016-03-03T10:20:12.200" LastActivityDate="2016-03-03T10:20:12.200" Title="Baseball games alphametic" Tags="|no-computers|alphametic|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25763" PostTypeId="2" ParentId="25762" CreationDate="2016-01-25T18:51:44.830" Score="12" Body="&lt;p&gt;Assuming numbers can't start with 0, &lt;code&gt;G&lt;/code&gt; is 1 because two four-digit numbers can't sum to 20000 or more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;SE&lt;/code&gt;+&lt;code&gt;LL&lt;/code&gt;=&lt;code&gt;ES&lt;/code&gt; or &lt;code&gt;1ES&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is &lt;code&gt;ES&lt;/code&gt;, then &lt;code&gt;LL&lt;/code&gt; must be a multiple of 9 because &lt;code&gt;SE&lt;/code&gt; and &lt;code&gt;ES&lt;/code&gt; are always congruent mod 9. But &lt;code&gt;LL&lt;/code&gt; is a multiple of 11, so it would have to be 99, which is impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;code&gt;SE&lt;/code&gt;+&lt;code&gt;LL&lt;/code&gt;=&lt;code&gt;1ES&lt;/code&gt;. &lt;code&gt;LL&lt;/code&gt; must be congruent to 100 mod 9. The only multiple of 11 that works is 55, so &lt;code&gt;L&lt;/code&gt; is 5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;SE&lt;/code&gt;+55=&lt;code&gt;1ES&lt;/code&gt;. This is possible when &lt;code&gt;E&lt;/code&gt;+5=&lt;code&gt;S&lt;/code&gt;. The possibilities for &lt;code&gt;ES&lt;/code&gt; are 27, 38, or 49.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;BA&lt;/code&gt;+&lt;code&gt;BA&lt;/code&gt;+1=&lt;code&gt;1AM&lt;/code&gt;. &lt;code&gt;B&lt;/code&gt; must be at least 5 because &lt;code&gt;B&lt;/code&gt;+&lt;code&gt;B&lt;/code&gt; (possibly +1 from a carry) is at least 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &lt;code&gt;A&lt;/code&gt; is less than 5, then &lt;code&gt;A&lt;/code&gt;+&lt;code&gt;A&lt;/code&gt;+1 does not carry, and &lt;code&gt;A&lt;/code&gt; must be even. Inversely, if &lt;code&gt;A&lt;/code&gt; is greater than 5, it must be odd. The possibilities for &lt;code&gt;A&lt;/code&gt; are 0, 2, 4, 7, or 9.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;0 does not work because &lt;code&gt;M&lt;/code&gt; would have to be 1.&lt;/li&gt;&#xA;&lt;li&gt;2 and 7 don't work because &lt;code&gt;M&lt;/code&gt; would have to be 5.&lt;/li&gt;&#xA;&lt;li&gt;9 doesn't work because &lt;code&gt;M&lt;/code&gt; would also have to be 9.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So &lt;code&gt;A&lt;/code&gt; is 4, &lt;code&gt;M&lt;/code&gt; is 9, and &lt;code&gt;B&lt;/code&gt; is 7. This leaves &lt;code&gt;38&lt;/code&gt; as the only possibility for &lt;code&gt;ES&lt;/code&gt;. The full equation is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    7483&#xA;  + 7455&#xA; --------&#xA;   14938&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2016-01-25T18:57:50.770" LastActivityDate="2016-01-25T18:57:50.770" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25764" PostTypeId="1" AcceptedAnswerId="25767" CreationDate="2016-01-25T19:40:55.923" Score="16" ViewCount="2496" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;I am a square,&lt;/p&gt;&#xA;&lt;p&gt;I am made in the year square of 43,&lt;/p&gt;&#xA;&lt;p&gt;also, 106 years after my second child was born&lt;/p&gt;&#xA;&lt;p&gt;also, in a year with the number 43.&lt;/p&gt;&#xA;&lt;p&gt;People pay with me a square amount of dollars.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16322" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-01-25T21:06:08.860" Title="The square of 43" Tags="|riddle|knowledge|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25765" PostTypeId="1" AcceptedAnswerId="25769" CreationDate="2016-01-25T20:42:16.017" Score="10" ViewCount="1773" Body="&lt;p&gt;I had three mathematician friends at school, and we were hanging out.&lt;br&gt;&#xA;Out of the blue a question came to my mind, and I decided to ask it to them: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Guys, I have a question for you. Which is actually pretty easy, I am sure you will find it very fast! I will think of 2 numbers which are distinct and they are between 1 and 10 (including 1 and 10). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will tell the &lt;strong&gt;product&lt;/strong&gt; of them to you &lt;strong&gt;Matt&lt;/strong&gt;, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I will tell the &lt;strong&gt;absolute difference&lt;/strong&gt; of them to &lt;strong&gt;David&lt;/strong&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the &lt;strong&gt;sum&lt;/strong&gt; of them to you &lt;strong&gt;Simon&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They started to talk each other:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Matt: I cannot find it guys.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;David: I cannot find it either.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Simon: This is very hard question actually, I do not believe anyone will ever find it.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Matt: I cannot believe that I cannot find it still, these are pretty good chosen numbers.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Simon: You made a very hard question indeed. Still could not figure it out.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;David: I found it! I thought I would never find it, thanks Simon!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Matt: We all found it then.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Simon: Yes. Good one thanks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What were those numbers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&#xA;Note: This is not a duplicate question; it is a unique one. Just inspired from &lt;a href=&quot;https://puzzling.stackexchange.com/questions/1968/3-impossibly-intelligent-mathematicians&quot;&gt;this one&lt;/a&gt;. The logic and solution are totally different. If you still believe that it is the same one, I am okay anyway.&#xA;&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="18514" LastEditorUserId="10615" LastEditDate="2016-02-24T04:03:43.443" LastActivityDate="2016-02-24T04:03:43.443" Title="Sum, Product and Difference" Tags="|mathematics|logical-deduction|meta-knowledge|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25766" PostTypeId="2" ParentId="25762" CreationDate="2016-01-25T20:44:21.263" Score="2" Body="&lt;p&gt;I found an answer by looking at the lowest numbers:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;E + L must equal S or 10 + S&#xA;&#xA;S + L must equal E or 10 + E&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I set L to zero which means that there won't be a one to carry.  This allows me to set E = S (&quot;es&quot;, not &quot;five&quot;).  Let's pick a number:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;E = 5&#xA;S = E = 5&#xA;L = 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This gives:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  BA55&#xA;+ BA00&#xA;------&#xA; GAM55&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Let's pick a number for A.  This also gives us M.  Note that if 2*A is &amp;lt; 10, A has to be even because it equals 2B - 10.  If instead 2*A &gt;= 10, the one carries over and A has to be odd because 2B + 1 - 10 has to equal A.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A = 2&#xA;M = 4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This gives:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  B255&#xA;+ B200&#xA;------&#xA; G2455&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now I see that B + B has to equal 12.  B = 6:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  6255&#xA;+ 6200&#xA;------&#xA; 12455&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Using this method, I also found these results:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;B = 7, A = 5, S = 5, L = 0, G = 1, E = 5, and M = 0&#xA;B = 8, A = 7, S = 1, L = 0, G = 1, E = 1, and M = 4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This method won't give every result because I'm assuming that L is always zero.&lt;/p&gt;&#xA;" OwnerUserId="2318" LastActivityDate="2016-01-25T20:44:21.263" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25767" PostTypeId="2" ParentId="25764" CreationDate="2016-01-25T21:06:08.860" Score="30" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Franklin &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a square,&#xA;I am made in the year square of 43,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Franklin Square (of Franklin and Blackstone Squares in Boston) opened in 1849 (which is 43 squared)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;also, 106 years after my second child was born&lt;/p&gt;&#xA;&#xA;&lt;p&gt;also, in a year with the number 43.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Benjamin Franklin's second child was born in 1743, 106 years before 1849&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;People pay with me a square amount of dollars.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Benjamin Franklin is on the 100 dollar bill, and 100 is 10 squared.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18493" LastActivityDate="2016-01-25T21:06:08.860" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25768" PostTypeId="2" ParentId="25762" CreationDate="2016-01-25T21:07:43.513" Score="5" Body="&lt;p&gt;There are in total three solutions (with unique numbers):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;G A M E S B L&#xA;-------------&#xA;0 4 9 1 6 2 5&#xA;0 4 9 3 8 2 5&#xA;1 4 9 3 8 7 5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So apart from the solution in @f''s answer, we have&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  2483&#xA;+ 2455&#xA;--------&#xA; 04938&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  2461&#xA;+ 2455&#xA;--------&#xA; 04916&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="12901" LastActivityDate="2016-01-25T21:07:43.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25769" PostTypeId="2" ParentId="25765" CreationDate="2016-01-25T22:03:14.023" Score="11" Body="&lt;p&gt;The solution is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2 &amp;amp; 9 - product of 18, absolute difference of 7, sum of 11.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First we can eliminate all pairs of unique numbers (1-10 inclusive) that would give unique products. Since Matt doesn't know what the pair of numbers is, there must be more than one way to get the product he was given.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Thus, we eliminate the following products: 2, 3, 4, 5, 7, 9, 14, 15, 16, 21, 27, 28, 32, 35, 36, 42, 45, 48, 50, 54, 56, 60, 63, 70, 72, 80, and 90.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/6PXg3.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6PXg3.png&quot; alt=&quot;First elimination by Matt&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Next, David, having eliminated all pairs that give unique products checks to see which pairs yield the absolute difference he was given. Since he does not know the pair at this point, we can assume that there is more than one pair yielding the absolute difference.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We therefore eliminate the unique absolute differences: 4, 6, 8, and 9.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now we come to Simon, who is unable to determine the pair by the sum he was given. We can therefore assume that, of the pairs remaining, there is more than one pair of numbers that add up to that sum.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; So we eliminate the unique sums: 5, 6, and 10.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Having seen all of these eliminations, Matt still does not know the answer. This means that the product is still not unique, so we will eliminate the unique products:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 6, 8, 10, 12, 20, 24, and 40. This leaves only 4 pairs remaining.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With 4 pairs remaining, Simon still doesn't know the answer, which means that the sum he was given is still not unique. We now eliminate the unique sums:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 9 and 13. We only have 2 pairs left, both of which have a sum of 11.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;David was still torn (between pairs with a non-unique absolute difference) &lt;em&gt;until Simon spoke up&lt;/em&gt;. The non-unique pairs he was seeing were:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2 &amp;amp; 9 (difference of 7) and 3 &amp;amp; 10 (difference of 7).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But only ONE of these two pairs remains after Simon's elimination. David therefore knows that the correct pair is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2 &amp;amp; 9. &lt;a href=&quot;https://i.stack.imgur.com/3Djga.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3Djga.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Matt and Simon, understanding the conclusion David has drawn, also agree.&lt;/p&gt;&#xA;" OwnerUserId="18424" LastEditorUserId="18424" LastEditDate="2016-01-26T03:12:27.723" LastActivityDate="2016-01-26T03:12:27.723" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25770" PostTypeId="2" ParentId="25765" CreationDate="2016-01-25T22:55:15.183" Score="2" Body="&lt;p&gt;Here's my solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The numbers are 5 and 6.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First elimination:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since David doesn't know the number, the product must be unique. This leaves us with the following pairs: 1-6, 1-8, 1-10, 2-3, 2-4, 2-5, 2-6, 2-9, 2-10, 3-4, 3-6, 3-8, 3-10, 4-5, 4-6, 4-10, 5-6, 5-8.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second elimination:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We know that the absolute difference is not unique, so we eliminate those pairs with differences of 4, 6, 8 and 9. That leaves us with (organized by increasing difference): 2-3, 3-4, 4-5, 5-6, 2-4, 4-6, 2-5, 3-6, 5-8, 1-6, 3-8, 1-8, 2-9, 3-10&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Third elimination:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sum is still not unique, so eliminate those with sums of 5, 6, or 10. This leaves us with (in order of increasing sum): 3-4, 2-5, 1-6, 4-5, 3-6, 1-8, 5-6, 3-8, 2-9, 5-8, 3-10.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fourth Elimination:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The product is STILL not unique, which narrows this down to just four pairs of numbers. We have either a product of 18 with 2-9 and 3-6 or a product of 30 with 5-6 or 3-10.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fifth Elimination:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sum is STILL not unique, which means we must have either 2-9 or 5-6.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sixth Elimination:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; David was able to get it thanks to Simon's last statement, which means that previously (in the 2-9, 3-6, 5-6 or 3-10 step) the difference was not unique. Since it is now, we can eliminate 2-9 (since 3-10 has the same difference). The numbers are 5-6.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2016-01-25T22:55:15.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25771" PostTypeId="2" ParentId="25759" CreationDate="2016-01-25T23:32:15.167" Score="24" Body="&lt;p&gt;From your description, I think the question and answers are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Which one of these is NOT a noble gas?                                           A. Boron, B. Helium, C. Krypton, D. Radon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;B was the only answer that was not on, but I was pretty sure that it was one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This refers to the common 'on' ending to noble gas names, helium is an exception.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And Superman's was one also, and that eliminated C.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Superman's home planet is Krypton&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think the rest doesn't require a description.&lt;/p&gt;&#xA;" OwnerUserId="18422" LastActivityDate="2016-01-25T23:32:15.167" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25772" PostTypeId="1" AcceptedAnswerId="25773" CreationDate="2016-01-26T03:09:55.183" Score="7" ViewCount="365" Body="&lt;p&gt;I am interested in producing &quot;American style&quot; crosswords and looking for a digital alternative to the graph paper I've been using.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried crossword eclipse but it seems to be more of a generator. It produces grids from a list of words. I want to produce my own grids from scratch. &lt;/p&gt;&#xA;" OwnerUserId="18696" LastActivityDate="2018-02-06T20:25:42.817" Title="Free crossword construction programs" Tags="|crosswords|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25773" PostTypeId="2" ParentId="25772" CreationDate="2016-01-26T04:14:18.633" Score="6" Body="&lt;p&gt;I've had success just using Excel (or a similar  spreadsheet). Set the row and column widths to be the same, and use a tiny font justified to the top and left to insert the numbers. Change the background &quot;fill&quot; color to black for the squares that are black. &lt;/p&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2016-01-26T04:14:18.633" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25775" PostTypeId="2" ParentId="25765" CreationDate="2016-01-26T07:28:07.150" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 9 AND 2 because 5,6 would have been a duplicate, had Simon not been confused first  &lt;a href=&quot;https://i.stack.imgur.com/DuPJ8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DuPJ8.png&quot; alt=&quot;Conditional Formatting&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14455" LastEditorUserId="14455" LastEditDate="2016-01-26T07:56:19.200" LastActivityDate="2016-01-26T07:56:19.200" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25776" PostTypeId="1" AcceptedAnswerId="25785" CreationDate="2016-01-26T11:05:47.157" Score="34" ViewCount="1126" Body="&lt;p&gt;So this is it. The last challenge. Here's where I either get filthy rich, or get myself killed like ol' Professor Alembic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cex8E.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cex8E.jpg&quot; alt=&quot;Trick or treat&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Goodness, I hope that's not &lt;em&gt;his&lt;/em&gt; skull.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the door to the legendary Treasure of the Mictal kings... or to my doom. I have to find out exactly how much to turn the dial. It's just like a safe door, really, only it'll activate a death trap if I guess the wrong number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hmmm... Let's take a look at my illustrations from the journey here. Maybe I can find some clue there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2Kul6.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2Kul6.jpg&quot; alt=&quot;circle&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6LXm5.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6LXm5.jpg&quot; alt=&quot;worship&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/S5BtG.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/S5BtG.jpg&quot; alt=&quot;writing&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HDtzS.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HDtzS.jpg&quot; alt=&quot;pattern&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18558" LastActivityDate="2016-01-26T14:01:27.067" Title="The Treasure of The Mictal Kings" Tags="|visual|story|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25778" PostTypeId="1" AcceptedAnswerId="25779" CreationDate="2016-01-26T12:16:14.943" Score="5" ViewCount="210" Body="&lt;p&gt;I have tried to find the next number in this series but I'm stumped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The series is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;61,52,63,94 .....&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can anybody figure out the next number?&lt;/p&gt;&#xA;" OwnerUserId="18700" LastEditorUserId="9000" LastEditDate="2016-01-26T13:17:59.173" LastActivityDate="2016-01-26T13:17:59.173" Title="Complete the Number Sequence Series" Tags="|number-sequence|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25779" PostTypeId="2" ParentId="25778" CreationDate="2016-01-26T12:23:34.490" Score="4" Body="&lt;p&gt;As @CarlLondahl mentioned, the trick to this one is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To reverse the digits&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which gives us&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 16,25,36,49 which is ofcourse 4^2,5^2,6^2,7^2 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the next element is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The digits of 8^2 reversed -&gt; &lt;strong&gt;46&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16846" LastActivityDate="2016-01-26T12:23:34.490" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25780" PostTypeId="1" CreationDate="2016-01-26T12:42:03.870" Score="0" ViewCount="323" Body="&lt;p&gt;You run an office that employs 23 people. What is the probability that two of your employees have the same birthday? For the purposes of the problem, ignore February 29. &lt;/p&gt;&#xA;" OwnerUserId="18700" LastActivityDate="2016-01-26T12:48:41.827" Title="The Birthday Paradox" Tags="|calculation-puzzle|" AnswerCount="1" CommentCount="0" ClosedDate="2016-01-26T12:56:31.017" ContentLicense="CC BY-SA 3.0" />
  <row Id="25781" PostTypeId="2" ParentId="25780" CreationDate="2016-01-26T12:48:41.827" Score="3" Body="&lt;p&gt;It's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0.507297&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because that's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 - (1/365) ^ 23 * 365 * 364 * 363 * ... * 343&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastActivityDate="2016-01-26T12:48:41.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25782" PostTypeId="1" AcceptedAnswerId="25783" CreationDate="2016-01-26T13:00:54.680" Score="3" ViewCount="735" Body="&lt;p&gt;One day i got bored and was looking at my arms birthmarks and decided to count the amount of unique triangles i could make with just 4 birthmarks, lets call them dots in the puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify: a unique triangle uses three dots and at least 1 of them is different from the other triangles. &#xA;(4 dots can make 4 triangles;  1234: 123,124,134,234) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After i was done i added a fifth dot and started counting again..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I figured there should be a formula to calculate it, and so it got me started on creating that formula. I had some fun creating it and i hope you will as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Create the formula and tell how many unique triangles 13 dots can make&lt;/p&gt;&#xA;" OwnerUserId="18559" LastEditorUserId="11876" LastEditDate="2016-01-26T15:57:50.460" LastActivityDate="2016-01-26T15:57:50.460" Title="unique triangles formula" Tags="|mathematics|" AnswerCount="2" CommentCount="1" ClosedDate="2016-01-26T20:57:53.440" ContentLicense="CC BY-SA 3.0" />
  <row Id="25783" PostTypeId="2" ParentId="25782" CreationDate="2016-01-26T13:15:00.880" Score="6" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 286&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Using the general formula $C(n,r)=\frac{n!}{r!\cdot (n-r)!}$, you plug in $13$ for $n$ and $3$ for $r$ to get 286. $n$ is the set size and $r$ is the subset size. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9631" LastEditorUserId="9631" LastEditDate="2016-01-26T13:27:10.620" LastActivityDate="2016-01-26T13:27:10.620" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25784" PostTypeId="2" ParentId="25782" CreationDate="2016-01-26T13:17:50.497" Score="3" Body="&lt;p&gt;Assuming no three dots are collinear, the formula for $n \ge 3$ dots is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\binom{n}{3} = \frac{1}{6}(n-2)(n-1)n$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So for 13 dots we get&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\frac{1}{6} \cdot 11 \cdot 12 \cdot 13 = 286 \mbox{ triangles}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17147" LastEditorUserId="17147" LastEditDate="2016-01-26T13:30:52.247" LastActivityDate="2016-01-26T13:30:52.247" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25785" PostTypeId="2" ParentId="25776" CreationDate="2016-01-26T14:01:27.067" Score="19" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; turning the dial 59 steps should be right&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We will look at the bottom four pictures numbering them from 1 to 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First number 3 (the stone):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Symbols on the stone are mathematical formulas in unary system. (only counting ones)&lt;br&gt;&#xA;  Each vertical line $|$ is a one and for each single number you simply count the vertical lines.&lt;br&gt;&#xA;  Each two horizontal lines $=$ is an addition symbol.&lt;br&gt;&#xA;  Each square $\square$ is a multiplication symbol.&lt;br&gt;&#xA;  Each T-like symbol $T$ is an equality symbol.&lt;br&gt;&#xA;  Therefore we get the following equations:&lt;br&gt;&#xA;  $$2=2$$&#xA;  $$3=3$$&#xA;  $$1+3=2+2$$&#xA;  $$2\times2=2+2$$&#xA;  $$1\times4=2\times2$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Next is number 4 (the plate):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This is a mathematical term.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;First we have to calculate the corners with the respective mathematical operators.&lt;br&gt;&#xA;  Corners 1 and 4 (counted left to right, top to bottom) are products of something.&lt;br&gt;&#xA;  Corners 2 and 3 are sums of something.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Next we go one step to the middle.&lt;br&gt;&#xA;  Corners 1 and 2 are added to get the top half.&lt;br&gt;&#xA;  Corners 3 and 4 are multiplied to get the bottom half.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Finally the top and bottom half are added.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;We get the following term.&#xA;  $$((1_a\times1_b)+(2_a+2_b))+((3_a+3_b)\times(4_a\times4_b))$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But where do we get the numbers to put into this term.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In picture number 1 (the hill) we can see stones that have numbers in the same unary system as in picture 3 written on them.&lt;br&gt;&#xA;  We can also see a pedestal. The same pedestal as on the plate in picture 4.&lt;br&gt;&#xA;  In picture 4 however the pedestal is on the right side of the plate so we have to imagine the stones in picture 1 rotated by 180$^\circ$ as if standing on the other side.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;With this we can see the needed corners.&lt;br&gt;&#xA;  Corner 1 is 5 and 6&lt;br&gt;&#xA;  Corner 2 is 7 and 8&lt;br&gt;&#xA;  Corner 3 is 3 and 4&lt;br&gt;&#xA;  Corner 4 is 1 and 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lets fill in the term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$((5\times6)+(7+8))+((3+4)\times(1\times2))$$&lt;br&gt;&#xA;$$=(30+15)+(7\times2)$$&lt;br&gt;&#xA;$$=45+14$$&lt;br&gt;&#xA;$$=59$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore I think that the dial has to be turned 59 steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the symbol in picture 2 I think it is a red herring and has nothing to do with the dial since i was not able to think of anything it could be used for.&lt;/p&gt;&#xA;" OwnerUserId="16608" LastActivityDate="2016-01-26T14:01:27.067" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25786" PostTypeId="1" AcceptedAnswerId="25869" CreationDate="2016-01-26T14:23:50.673" Score="10" ViewCount="981" Body="&lt;p&gt;Consider a unconventional billiard board in the shape of an equilateral triangle (depicted below). An incredibly small ball (size in picture is increased for the &lt;em&gt;sake of visibility&lt;/em&gt; on your screen) is put in the A corner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Jy0nY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jy0nY.png&quot; alt=&quot;Ett javla fint biljardbord, eller hur?&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How many paths for the ball are there such that the ball bounces off the sides exactly 20160127 times, starting and ending in the A corner?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;: the question is heavily inspired by a question on the &lt;a href=&quot;https://projecteuler.net/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Project Euler&lt;/a&gt; website, which I enjoyed very much solving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Case: 11 bounces, 2 paths; Case: 10001 bounces, 800 paths; Case: 1000001 bounces, 80840 paths.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Additional information&lt;/strong&gt;: There is also a &lt;a href=&quot;https://arxiv.org/PS_cache/math/pdf/0509/0509292v7.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;paper&lt;/a&gt; on the matter if you are interested in reading more.&lt;/p&gt;&#xA;" OwnerUserId="12901" LastEditorUserId="12901" LastEditDate="2016-01-28T21:38:36.120" LastActivityDate="2020-05-23T17:22:58.743" Title="Bouncing ball on a billiard board" Tags="|geometry|combinatorics|" AnswerCount="2" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25787" PostTypeId="2" ParentId="25786" CreationDate="2016-01-26T15:43:42.617" Score="8" Body="&lt;p&gt;&lt;strong&gt;Updated, calculating paths only to A, rather than to any corner.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculating the number of paths where the ball does not visit A before the 20160127th bounce...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define the side-length of the triangle as 1 unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Working my way up from small numbers.  There is obviously exactly one path where the ball bounces one time before returning to A: straight down the middle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the walls were replaced with mirrors, the paths would be beams of light, reflecting off the mirrors.  There will be many virtual images of point A.  There will also be virtual images of B and C, but we are not interested in those.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Straight down the middle, there is a virtual image of A on the far side of a triangle 2 units in length.  One the far side of the triangle 3 units in length, there are two images of corners, but those are B and C (see image, below).  On the far side of the triangel 4 units in length, there will be 3 images, but the central one is obscured.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/wuot1.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wuot1.png&quot; alt=&quot;All images&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The red lines point to images of corner A, green lines to B, blue lines to C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, for a triangle of length L, there will be L-1 images.  But some of these will be hidden behind other images.  The number of unobstructed images will be the number of integers less than L that are relatively prime with L.  This is Euler's function $\phi(L)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As can be seen by following any of the lines in the image, the path from A to a virtual image on triangle L passes through $2L-3$ mirrors.  Thus, a path that requires $20160127$ bounces must lead to an image on triangle $L = 10080065$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of unobstructed images on that virtual triangle is  $\phi(10080065) = 7589632$.  Thanks to &lt;a href=&quot;http://www.javascripter.net/math/calculators/eulertotientfunction.htm&quot; rel=&quot;noreferrer&quot;&gt;Euler Totient Calculator&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we are only interested in images of A.  For $L = 2, 5$, the leftmost and rightmost images are of A.  In general, for $L = 3*i -1$, there will be $i$ images of A.  Since $10080065 = 3*3360022 - 1$, that triangle will have $3360022$ images of A.  Now to determine how many are unobstructed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can factor $10080065 = 5 \times 17 \times 118589$.  Thus, on that triangle, every $5^{th}$, every $17^{th}$ and every $118589^{th}$ image will be obscured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Images of A repeat with every $3^{rd}$ image, starting with the first and ending with the last.  Since $3$ is relatively prime to each of the factors of $10080065$, each of the three corners will have its images obscured equally.  The exception is A, which has an extra image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the number of images of A that are not obscured = $(7589632 -1)/3 + 1 = 2529878$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The size of the triangle that has images which are reached after $20160127$ reflections is $L = 10080065$.&lt;/li&gt;&#xA;&lt;li&gt;The number of images of A at that distance is $(L+1)/3 = 335062$.&lt;/li&gt;&#xA;&lt;li&gt;The number of unobscured images of any corner at that distance is $\phi(L) = 7589632$.&lt;/li&gt;&#xA;&lt;li&gt;The number of unbscured images of A at that distance is $(\phi(L) -1)/3 + 1 = 2529878$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So the number of paths that reach A after 20160127 bouncess is 2529878.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="2328" LastEditDate="2016-01-26T17:58:15.410" LastActivityDate="2016-01-26T17:58:15.410" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="25788" PostTypeId="2" ParentId="9515" CreationDate="2016-01-26T17:46:41.383" Score="3" Body="&lt;p&gt;First of all I want to point out that IMO the shadows and the river make something around a 45 degrees angle, just for your consideration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q0: hemisphere?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since all people look caucasian and to be wearing late 19th/early 20th Western attires, I'm saying it is north hemisphere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Q2 + Q8: Is it Spring or Autumn? + Do birds fly N or S?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;True&lt;/strong&gt; and &lt;strong&gt;True&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wait. Not boolean questions? Kidding.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Birds look to be flying in about the same direction range (really hard to tell) shadows are cast to. This happens in &lt;strong&gt;Spring&lt;/strong&gt; in both hemispheres and while &lt;strong&gt;they are flying North&lt;/strong&gt; in North hemisphere.&lt;br&gt;&#xA;&lt;br&gt;&#xA; * Taking a look to the other answers I see references to landfarmers sowing, not harvesting, which is even better reasoning for Spring, but wasn't my first thought I'll admit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Q1: time of day?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's really hard to make a perfect appreciation, but IMO shadows are cast towards NW or so.&lt;br&gt;&#xA;&lt;br&gt;&#xA; So I'll say morning but not too soon, since shadows are already well defined.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Q3: navigable river?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; With the right craft, any river is navigable.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Aside from that, the &lt;strong&gt;boat&lt;/strong&gt; and the &lt;strong&gt;dock&lt;/strong&gt; strongly suggest the river is navigated frequently.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Q4: Which direction does the river flow?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; From North to &lt;strong&gt;South&lt;/strong&gt;. You can tell by the trails on the water plus achieved knowledge from previous Qs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Q5: Is the river deep or shallow at the side where the boat is?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think it's deeper at this side. The riverside ends more vertically here and looks smoother on the other side (also hard to tell).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Q6: Are there bridges nearby?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; At least the people in the forefront &lt;strong&gt;doesn't look to think so&lt;/strong&gt;. They will probably rather cross the bridge since it's faster. If the boat has already departed when you reach the river you'll have to wait for it to deliver the passengers, return, and then carry you to the other side. One lone old guy rowing won't get a very high peak speed I bet.&lt;br&gt;&#xA;&lt;br&gt;&#xA; * Also, other answers suggest the ferry costs money, which adds up for the bridge to be the first option if available.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Q7: How far is the railroad from here?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Far enough&lt;/strong&gt; to not be in the drawing.&lt;br&gt;&#xA; &lt;strong&gt;Close enough&lt;/strong&gt; to be reached by walking by a railroad worker with no more equipment or luggage than a lantern and a rather thin suitcase.&lt;br&gt;&#xA;&lt;br&gt;&#xA; That's the better answer I can give.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18576" LastEditorUserId="18576" LastEditDate="2016-01-26T18:00:24.020" LastActivityDate="2016-01-26T18:00:24.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25791" PostTypeId="1" CreationDate="2016-01-26T19:48:11.887" Score="14" ViewCount="2463" Body="&lt;p&gt;On an infinite table are $n$ identical circular coins lying flat. Each coin touches exactly $k$ other coins, and any two coins are connected by a path of touching coins. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determine all possible pairs of values of $n$ and $k$.&lt;/p&gt;&#xA;" OwnerUserId="18508" LastEditorUserId="18508" LastEditDate="2016-01-26T20:11:06.333" LastActivityDate="2016-01-27T21:28:56.013" Title="Touching coins flat on a table" Tags="|mathematics|geometry|graph-theory|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25793" PostTypeId="1" AcceptedAnswerId="25794" CreationDate="2016-01-26T20:11:12.680" Score="3" ViewCount="648" Body="&lt;p&gt;I have been unable to find a solution for this F2L case. It takes about 3 seconds to do, and I do it really slowly. Basically, the F2L case is solved, but the edge is inverted.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0CtSI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0CtSI.jpg&quot; alt=&quot;picture&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The way I do it right now is that I take it out, split it up, and insert it in again.&lt;/p&gt;&#xA;&lt;p&gt;How can I complete this in the fastest time possible?&lt;/p&gt;&#xA;" OwnerUserId="18515" LastEditorUserId="69582" LastEditDate="2021-02-08T15:26:34.133" LastActivityDate="2021-02-08T15:26:34.133" Title="How do I solve this F2L case quickly?" Tags="|rubiks-cube|algorithm|twisty-puzzles|speedsolving|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25794" PostTypeId="2" ParentId="25793" CreationDate="2016-01-26T20:19:33.953" Score="3" Body="&lt;p&gt;No matter what you do, you're going to have to remove the piece, split it from the corner, and reattach them correctly. However, there are efficient ways to do it - try this sequence:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[R U R' U'] [R U' R' U] [F' U2 F2] [R' F' R]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It's a bit long, but it's an intuitive sequence of moves. I'm using the brackets to denote intuitive blocks. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;code&gt;[R U R' U']&lt;/code&gt; removes the corner-edge pair.&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;[R U' R' U]&lt;/code&gt; splits them apart&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;[F' U2 F2]&lt;/code&gt; reforms them and puts them in the correct place&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;[R' F' R]&lt;/code&gt; sets the rebuilt faces back in place.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;At my top speed, this takes significantly less time than three seconds - and I'm fairly out of practice, at that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also worth noting that these positions are relatively straightforward to see coming, and oftentimes, simple adjustments to the order and choice of moves for other edge pairs will prevent this from happening at all. If you see such an eventuality, take a moment, stop, and consider what you can do to prevent it in the future.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorDisplayName="user20" LastEditDate="2016-01-26T20:25:43.557" LastActivityDate="2016-01-26T20:25:43.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25795" PostTypeId="2" ParentId="25791" CreationDate="2016-01-26T20:20:47.720" Score="0" Body="&lt;p&gt;Partial answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; n = infinity, k = 6. Arranged as a hexagonal pattern.&lt;br&gt;&#xA; n = infinity, k = 4. Arranged as a square pattern&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18103" LastActivityDate="2016-01-26T20:20:47.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25796" PostTypeId="2" ParentId="25791" CreationDate="2016-01-26T20:21:15.920" Score="10" Body="&lt;p&gt;My solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $n$ = 1 : $k$ = 0 (single coin can't touch any others)&lt;br&gt;&#xA; $n$ = 2 : $k$ = 1 (pair of coins only touch each other)&lt;br&gt;&#xA; 3 &amp;lt;= $n$ &amp;lt; $\infty$ : $k$ = 2 (finite number of coins greater than 2 can  only touch 2 others without making the &quot;edge&quot; coins touch a different number [updated from Zandar])&lt;br&gt;&#xA; 16 &amp;lt;= $n$ &amp;lt; $\infty$ : $k$ = 3 ($n$ must be a multiple of 4 for the diamond pattern to work, update from Michael Seifert)&lt;br&gt;&#xA; $n$ = $\infty$ : $k$ = 2,3,4,6 (without &quot;edge coins&quot; to worry about, you can look at any subset of the grid to see what connections are available. 5 can't be done on a normal 2-d euclidian surface, that would require a dodecahedron)&lt;br&gt;&#xA;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/mfJoL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mfJoL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA; for the above images, assume each group of coins is a smaller selection of the infinite mass of coins, with the exception of the $k$=2 section, that is a piece of a loop&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13101" LastEditorUserId="13101" LastEditDate="2016-01-26T21:17:24.450" LastActivityDate="2016-01-26T21:17:24.450" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="25797" PostTypeId="1" AcceptedAnswerId="25805" CreationDate="2016-01-26T21:05:47.647" Score="7" ViewCount="923" Body="&lt;p&gt;The example below shows the amount of letters in the word being minimized by 1 each time, but the letters still form a legitimate word:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/08wfS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/08wfS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the longest set like this with the initial word longer or equal to 6 letters long?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18250" LastEditorUserId="22922" LastEditDate="2016-09-01T10:36:21.457" LastActivityDate="2017-06-17T13:00:42.143" Title="Minimizing Letters" Tags="|word|wordplay|" AnswerCount="6" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="25798" PostTypeId="1" AcceptedAnswerId="25799" CreationDate="2016-01-26T21:28:06.357" Score="9" ViewCount="509" Body="&lt;p&gt;This list of twelve US states is missing only one.  Can you figure out which one and why?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Alabama&lt;/li&gt;&#xA;&lt;li&gt;Alaska&lt;/li&gt;&#xA;&lt;li&gt;Arizona&lt;/li&gt;&#xA;&lt;li&gt;Arkansas&lt;/li&gt;&#xA;&lt;li&gt;Connecticut&lt;/li&gt;&#xA;&lt;li&gt;Georgia&lt;/li&gt;&#xA;&lt;li&gt;Illinois&lt;/li&gt;&#xA;&lt;li&gt;Indiana&lt;/li&gt;&#xA;&lt;li&gt;Kentucky&lt;/li&gt;&#xA;&lt;li&gt;Ohio&lt;/li&gt;&#xA;&lt;li&gt;Oklahoma&lt;/li&gt;&#xA;&lt;li&gt;Oregon&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="16131" LastEditorUserId="18559" LastEditDate="2016-01-27T09:54:56.550" LastActivityDate="2016-01-27T09:54:56.550" Title="13 States: Which is missing?" Tags="|pattern|geography|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25799" PostTypeId="2" ParentId="25798" CreationDate="2016-01-26T21:35:10.937" Score="11" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; North Carolina&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since those 13 states&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Contain the letter they begin with more than once&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17147" LastActivityDate="2016-01-26T21:35:10.937" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25800" PostTypeId="2" ParentId="25797" CreationDate="2016-01-26T21:56:33.517" Score="1" Body="&lt;p&gt;I have one that's six (or seven if you count proper nouns)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  I&lt;br&gt;&#xA;  IN&lt;br&gt;&#xA;  SIN&lt;br&gt;&#xA;  SINE&lt;br&gt;&#xA;  SPINES&lt;br&gt;&#xA;  SPLINES&lt;br&gt;&#xA;  SPLINERS - questionable, it's a company name&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;edit: improvement&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  I&lt;br&gt;&#xA;  IN&lt;br&gt;&#xA;  PIN&lt;br&gt;&#xA;  PINE&lt;br&gt;&#xA;  PINED&lt;br&gt;&#xA;  SPINED&lt;br&gt;&#xA;  SPLINED &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;edit: and one starting with O&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  O&lt;br&gt;&#xA;  ON&lt;br&gt;&#xA;  TON&lt;br&gt;&#xA;  TONE&lt;br&gt;&#xA;  TONER&lt;br&gt;&#xA;  STONER&lt;br&gt;&#xA;  STONIER&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18103" LastEditorUserId="18103" LastEditDate="2016-01-26T22:03:30.493" LastActivityDate="2016-01-26T22:03:30.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25801" PostTypeId="2" ParentId="25797" CreationDate="2016-01-26T22:05:13.510" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;   SKATERS&lt;br&gt;&#xA;   SKATER&lt;br&gt;&#xA;   SKATE&lt;br&gt;&#xA;   SATE&lt;br&gt;&#xA;   ATE&lt;br&gt;&#xA;   AT&lt;br&gt;&#xA;   A&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;STRAINS&lt;br&gt;&#xA;TRAINS&lt;br&gt;&#xA;TRAIN&lt;br&gt;&#xA;RAIN&lt;br&gt;&#xA;RAN&lt;br&gt;&#xA;AN&lt;br&gt;&#xA;A&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;BRAINED&lt;br&gt;&#xA;BRINED&lt;br&gt;&#xA;BRINE&lt;br&gt;&#xA; &lt;a href=&quot;http://dictionary.reference.com/browse/bine&quot; rel=&quot;nofollow&quot;&gt;BINE&lt;/a&gt;&lt;br&gt;&#xA;BIN&lt;br&gt;&#xA;IN&lt;br&gt;&#xA;I  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll keep thinking on this one...&lt;/p&gt;&#xA;" OwnerUserId="17167" LastEditorUserId="17167" LastEditDate="2016-01-26T22:29:15.280" LastActivityDate="2016-01-26T22:29:15.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25802" PostTypeId="2" ParentId="25797" CreationDate="2016-01-26T22:07:04.273" Score="2" Body="&lt;p&gt;Without  resorting to (read as: figuring out)  a program to determine all the possible and best answers, here is what I've come up with for words &lt;strong&gt;longer&lt;/strong&gt; than six letters:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;amasses masses masse mass ass as a &lt;em&gt;(though the E should have an accent in masse...)&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;mangled mangle mange mane man an a&lt;br&gt;&#xA;sharped shared shard hard had ad a&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;EIGHT!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;prattled rattled ratted rated rate rat at a&lt;br&gt;&#xA;startled started stated state stat tat at a&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Forget EIGHT, move on to NINE!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;spiriting spirting spiting siting sting sing sin in i&lt;br&gt;&#xA; &lt;em&gt;(spirting is antiquated, so 8 and 9 could be splitting and spitting)&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11452" LastEditorUserId="11452" LastEditDate="2016-01-26T22:47:12.303" LastActivityDate="2016-01-26T22:47:12.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25803" PostTypeId="1" CreationDate="2016-01-26T22:11:15.977" Score="3" ViewCount="2632" Body="&lt;p&gt;This question is about finding the odd one. My question is what is the answer and what is the reason behind it. Also, how should we approach such questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Puzzle 1 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GhvKI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GhvKI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Puzzle 2&#xA;&lt;a href=&quot;https://i.stack.imgur.com/AouUo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AouUo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Puzzle 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zxwyP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zxwyP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18149" LastEditorUserId="4050" LastEditDate="2016-01-27T18:08:09.860" LastActivityDate="2017-03-11T11:25:50.883" Title="Which is the odd one out and why?" Tags="|pattern|visual|symbols|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25804" PostTypeId="2" ParentId="25803" CreationDate="2016-01-26T22:19:23.707" Score="5" Body="&lt;p&gt;Puzzle 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The third one, as it is the only one where all the arrows face the same way, the others the one on top faces a different direction from the rest&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Puzzle 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The second, the flower is on the left of the stem&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Puzzle 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The fourth, the spiral goes left out from the centre &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nice Puzzle!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do this, I look for a common link and see if there is a picture without the link&lt;/p&gt;&#xA;" OwnerUserId="18250" LastEditorUserId="18250" LastEditDate="2016-01-27T20:53:02.290" LastActivityDate="2016-01-27T20:53:02.290" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25805" PostTypeId="2" ParentId="25797" CreationDate="2016-01-26T22:22:45.230" Score="6" Body="&lt;p&gt;I think it will be very difficult to find a confirmed count of &lt;em&gt;all&lt;/em&gt; possible words for this. It seems very open-ended. Perhaps it should be switched to a contest to find the longest?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Everything below this is copied from @Kevin from a &lt;a href=&quot;https://puzzling.stackexchange.com/a/9893/9000&quot;&gt;very similar question&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using the circa 2013 &lt;em&gt;Words With Friends&lt;/em&gt; dictionary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can only add to the right end of the word, &lt;strong&gt;8-letter&lt;/strong&gt; words are possible:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a&#xA;ba&#xA;bar&#xA;barb&#xA;barbe&#xA;barbel&#xA;barbell&#xA;barbells&#xA;&#xA;a&#xA;ma&#xA;max&#xA;maxi&#xA;maxim&#xA;maxima&#xA;maximal&#xA;maximals&#xA;&#xA;a&#xA;pa&#xA;pas&#xA;past&#xA;paste&#xA;paster&#xA;pastern&#xA;pasterns&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If you can add to either end, &lt;strong&gt;9-letter&lt;/strong&gt; words are possible:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I&#xA;id&#xA;aid&#xA;aide&#xA;aider&#xA;aiders&#xA;raiders&#xA;braiders&#xA;abraiders&#xA;&#xA;a&#xA;la&#xA;lap&#xA;laps&#xA;lapse&#xA;elapse&#xA;relapse&#xA;relapser&#xA;relapsers&#xA;&#xA;I&#xA;in&#xA;pin&#xA;ping&#xA;aping&#xA;raping&#xA;craping&#xA;scraping&#xA;scrapings&#xA;&#xA;a&#xA;at&#xA;eat&#xA;eath&#xA;heath&#xA;sheath&#xA;sheathe&#xA;sheather&#xA;sheathers&#xA;&#xA;I&#xA;is&#xA;ais&#xA;rais&#xA;raise&#xA;raiser&#xA;raisers&#xA;praisers&#xA;upraisers&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If you can add at any point, &lt;strong&gt;11-letter&lt;/strong&gt; words are possible:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I   &#xA;pi&#xA;pig&#xA;ping&#xA;oping&#xA;coping&#xA;comping&#xA;compting&#xA;competing&#xA;completing&#xA;complecting&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2016-01-26T22:22:45.230" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25806" PostTypeId="2" ParentId="25797" CreationDate="2016-01-26T22:36:55.523" Score="3" Body="&lt;p&gt;I'm confused as to whether the question is &quot;How many six-letter words can you find that fit this pattern?&quot; or &quot;What is the longest word you can create that fits this pattern?&quot; I will answer the latter.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Longest word: 9 letters&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;CLEANSERS&lt;br&gt;&#xA;CLEANSES&lt;br&gt;&#xA;CLEANSE&lt;br&gt;&#xA;CLEANS&lt;br&gt;&#xA;CLEAN&lt;br&gt;&#xA;LEAN&lt;br&gt;&#xA;LEA&lt;br&gt;&#xA;LA&lt;br&gt;&#xA;A&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16275" LastActivityDate="2016-01-26T22:36:55.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25807" PostTypeId="2" ParentId="25791" CreationDate="2016-01-26T22:40:00.663" Score="1" Body="&lt;p&gt;Nothing scientific in my answer, just by first impression and looking for possible symmetric arrangements:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a finite n:&#xA;n=1, k=0&#xA;n=2, k=1 (two coins next to each other)&#xA;n&gt;=3, k=2 (loop chain of 3 or more coins)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seems odd that I cannot find a valid composition where k&gt;3 and n is finite. I definitely have to give this further thought.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For infinite n, since there would have to be a symmetric arrangement, k could have any value from 2 to 6 except 5: 2 (i.e. infinite row of single coins), 3 (i.e. two parallel rows),  4 (i.e. two parallel rows offset by half a coin in &quot;zigzagish&quot; form), and 6 (i.e. infinite arrangement of coins as tightly as possible).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't figure out an arrangement where k=5, which is quite interesting. Possibly because its impossible to have a 2 dimensional symmetrical arrangement of 5 vertices.&lt;/p&gt;&#xA;" OwnerUserId="18711" LastEditorUserId="18711" LastEditDate="2016-01-26T22:57:25.357" LastActivityDate="2016-01-26T22:57:25.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25808" PostTypeId="2" ParentId="25803" CreationDate="2016-01-26T22:41:26.117" Score="10" Body="&lt;p&gt;Puzzle 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The third is the odd one out. All three arrows curve the same way in this image whereas they curve different directions on 1, 2 and 4. Also, the top arrow curves towards the other arrows but in 1, 2 and 4 the top arrow curves away from the others.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Puzzle 2 and 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The reason is the same in these two puzzles. In puzzle 2 it is the second image which is different and in puzzle 3 it is the fourth image. The reason is that the odd-image out is a rotation plus mirror transformation whereas the other three are just rotations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18522" LastEditorUserId="18250" LastEditDate="2017-03-11T11:25:50.883" LastActivityDate="2017-03-11T11:25:50.883" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25809" PostTypeId="1" AcceptedAnswerId="25822" CreationDate="2016-01-26T22:42:14.150" Score="18" ViewCount="1719" Body="&lt;p&gt;I am Patrick. Tonight, I got a mail from my old friend Dave. He was out of town for a while, and got back today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, we had an argument on the phone yesterday, and I think he had hard feelings. I was supposed to pick him up from the airport after he gave me a phone call, but he never called. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, I got this strange looking mail. What do you think? Is he really mad at me?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dear patrick, how are you? i feeL really hAppy being back home!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;We should MEet as soon as possible. i bet that my timE here will be very EnjOyaBle. &#xA;  hopefully, i will be free some time at Five pm tomorrow.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;how about a couple of beers after Work? maybe there wIll be a happy&#xA;  hour, and we caN driNk cheap beers :)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;by the way, I might have soUnded like a douche today when we were&#xA;  talking on the Phone. i must apologize about what i havE said. i&#xA;  could havE been more Understanding about your situation. but you must&#xA;  also admit that you are being too much sensitive after bReaking up&#xA;  with your girlfriend.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;we are old friends, mate! we have overcome so Many problems, aNd we&#xA;  can trash this out like We did all the time.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;see you tomorrow,&lt;br&gt;&#xA;  dave.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I have mistyped two letters and I have corrected them now. Sorry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  How do you read a sentence? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What do you use a flashlight for? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18645" LastEditorUserId="11876" LastEditDate="2016-01-29T14:49:14.677" LastActivityDate="2016-01-29T14:49:14.677" Title="Is Dave really mad at me?" Tags="|steganography|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="25810" PostTypeId="2" ParentId="25797" CreationDate="2016-01-26T22:50:57.360" Score="4" Body="&lt;p&gt;Using &lt;em&gt;Mathematica&lt;/em&gt;'s dictionary, my longest chain is:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;10 Letters:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;SPLITTINGS&lt;/code&gt;, &lt;code&gt;SPLITTING&lt;/code&gt;, &lt;code&gt;SLITTING&lt;/code&gt;, &lt;code&gt;SITTING&lt;/code&gt;, &lt;code&gt;SITING&lt;/code&gt;, &lt;code&gt;STING&lt;/code&gt;, &lt;code&gt;SING&lt;/code&gt;, &lt;code&gt;SIN&lt;/code&gt;, &lt;code&gt;IN&lt;/code&gt;, &lt;code&gt;I&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My code is below, it takes less than a second to run.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;words = ToUpperCase @ DictionaryLookup[___];&#xA;&#xA;g = Graph[Join @@ Table[StringDrop[w, {#}] -&amp;gt; w &amp;amp; /@ Range[StringLength[w]], {w, words}]];&#xA;&#xA;reachable = VertexOutComponent[g, {&quot;&quot;}];&#xA;&#xA;MaximalBy[reachable, StringLength]&#xA;&#xA;FindShortestPath[g, &quot;&quot;, First @ %]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The strategy is to create a tree-like structure were each word is linked to all the &quot;words&quot; that can be created by dropping one letter.  Some of those words are not in the dictionary (psuedo-words), but they can be ignored since there is no path to them from the root node (the empty word).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once this structure is created, finding the solution is simple: find the longest word that can be reached from the root node.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;How  many?&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;To answer your original question (how many chains are there), we can try counting the number of reachable words of each length:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Grid[List @@@ Normal@GroupBy[reachable, StringLength, Length]]&#xA;&#xA;0   1&#xA;1   2&#xA;2   29&#xA;3   254&#xA;4   942&#xA;5   1561&#xA;6   1390&#xA;7   878&#xA;8   325&#xA;9   51&#xA;10  1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;A total of &lt;code&gt;2645&lt;/code&gt; words at least six letters long can be put into a chain.  This doesn't exactly answer your question however, since some words can be reached in multiple ways:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;AcyclicGraphQ[UndirectedGraph@g] (* False *)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In order to find the total number of ways to make each word, I make use of a recursive enumeration:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The number of ways to make a word is equal to the sum of the number of ways to make each of its predecessors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From here we can make use of the &lt;em&gt;adjacency matrix&lt;/em&gt; of our graph.  The adjacency matrix $A$ has entries $a_{ij}$ equal to the number of edges from vertex $i$ to vertex $j$.  If the number of ways to make the word (vertex) $i$ is $c_i$, then we can write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;  c_i = \sum_j c_j a_{ji}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice this is exactly equal to the matrix multiplication $c=A^{\intercal} c$, making finding $c$ equivalent to finding the eigenvector of $A^{\intercal}$.  (There is one complication: we have to set $a_{11}=1$, otherwise the number of ways to reach the empty word &lt;code&gt;&quot;&quot;&lt;/code&gt; is zero.)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;clean = Subgraph[g, reachable]; (* remove unreachable words *)&#xA;m = AdjacencyMatrix[clean];&#xA;m[[1, 1]] = 1;&#xA;e = First@Eigenvectors[N@Transpose@m, 1];&#xA;Total[Pick[Round[e/First[e]], VertexList[clean], x_ /; StringLength[x] &amp;gt;= 6]]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;giving an answer of &lt;code&gt;26,322&lt;/code&gt; (for &lt;em&gt;Mathematica&lt;/em&gt;'s word list).  This solution is very fast, even on much larger word lists (~100,000 words).&lt;/p&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="9718" LastEditDate="2016-01-27T00:58:52.287" LastActivityDate="2016-01-27T00:58:52.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25811" PostTypeId="2" ParentId="25803" CreationDate="2016-01-27T01:37:04.577" Score="3" Body="&lt;p&gt;Just to add to the &quot;Also, how should we approach such questions&quot; part of your question, my first instinct upon seeing the puzzle was to notice that all four circles in each set &lt;em&gt;appeared&lt;/em&gt; to be rotations of each other. Because the question asked for the 'odd one out', that gave me a hint that perhaps one of the circles, despite appearing to be the same as the others, was not the same. It then became a simple exercise of identifying which one was different.&lt;/p&gt;&#xA;" OwnerUserId="14095" LastEditorUserId="14095" LastEditDate="2016-01-27T01:55:23.363" LastActivityDate="2016-01-27T01:55:23.363" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25812" PostTypeId="1" CreationDate="2016-01-27T03:09:47.023" Score="12" ViewCount="1115" Body="&lt;h3&gt;&amp;thinsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp; Make a non&amp;hairsp;-&amp;hairsp;24&amp;hairsp;-&amp;hairsp;hour day&amp;hairsp;/&lt;sub&gt;&lt;sub&gt;&amp;hairsp;&amp;hairsp;&lt;/sub&gt;&lt;/sub&gt;night cycle &lt;br&gt;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;thinsp; with the most daylight and fewest pieces&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XMmGM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XMmGM.png&quot; alt=&quot;Jigsaw-puzzle pieces: power outlet, 2 timers, light bulb, plug, straight wire, L-shaped wire, socket, 1-to-3 socket multiplier.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, devise schedules and an arrangement&#xA;for two electromechanical 24&amp;hairsp;-&amp;hairsp;hour&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Time_switch&quot; rel=&quot;nofollow noreferrer&quot;&gt;light timers&lt;/a&gt;&#xA;to control the flow of power from an&#xA;&lt;a href=&quot;https://en.wiktionary.org/wiki/outlet#Noun&quot; rel=&quot;nofollow noreferrer&quot;&gt;outlet&lt;/a&gt;&#xA;to a light bulb.&#xA;The challenge is to obtain the following repeated lighting pattern&#xA;with the largest possible number&amp;nbsp;D, while&#xA;N is a nonzero constant other than&amp;thinsp;&#xA;24&amp;hairsp;&amp;minus;&amp;hairsp;&amp;hairsp;D&amp;hairsp;,&#xA;beginning when the outlet's power is switched on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&#xA;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&#xA;&amp;emsp;&amp;ensp;&amp;hairsp;&amp;hairsp;   Light is on for D hours, off for N hours, &lt;br&gt;&#xA;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&#xA;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&#xA;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;          on for D hours, off for N hours, &lt;br&gt;&#xA;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&#xA;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&#xA;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;          on for D hours, off for N hours, &lt;br&gt;&#xA;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&#xA;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&#xA;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;          &amp;hairsp;&amp;hairsp;&lt;strong&gt;. . .&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To reduce complications:&#xA;&lt;br&gt;&lt;sup&gt;&lt;sup&gt;&lt;/sup&gt;&lt;/sup&gt;&#xA;&amp;hairsp;&amp;#x1f55b;&amp;hairsp;&#xA;    When power is initially switched on, both dials point to midnight&#xA;&lt;br&gt;&lt;sup&gt;&lt;sup&gt;&lt;/sup&gt;&lt;/sup&gt;&#xA;&amp;thinsp;&amp;#x1f6c7;&amp;thinsp;&#xA;    The timers may never&#xA;    simultaneously be at transitions in their schedules&#xA;    (from OFF to ON or from ON to OFF,&#xA;     whether the same or different between the timers)&#xA;    even though the power switch may initially turn them on simultaneously&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;&amp;ensp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp; The arrangement can be considered a jigsaw puzzle &lt;br&gt; &amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp;&amp;emsp; &lt;sub&gt;&lt;sup&gt;(and can be simple enough to be described in words)&lt;/sup&gt;&lt;/sub&gt;&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eGkzu.jpg&quot; alt=&quot;Example of how pieces may be arranged&quot;&gt;&#xA;&lt;br&gt;&lt;sup&gt;&lt;/sup&gt;&#xA;The example above shows much of what is allowed,&#xA;although one timer is missing.&#xA;&lt;br&gt;&lt;sup&gt;&lt;/sup&gt;&#xA;&amp;bull;&amp;nbsp; Use 1 outlet, 2 timers and 1 light bulb&#xA;&lt;br&gt;&lt;sup&gt;&lt;/sup&gt;&#xA;&amp;bull;&amp;nbsp; Use the fewest possible plugs, wires, sockets and socket multipliers&#xA;&lt;br&gt;&lt;sup&gt;&lt;/sup&gt;&#xA;&amp;bull;&amp;nbsp; Pieces may be flipped and rotated but not overlapped&#xA;&lt;br&gt;&lt;sup&gt;&lt;/sup&gt;&#xA;&amp;bull;&amp;nbsp; Any number of tabs and slots may remain unlinked&#xA;&lt;br&gt;&lt;sup&gt;&lt;/sup&gt;&#xA;&amp;bull;&amp;nbsp; The layout may include uncovered areas&#xA;&lt;br&gt;&lt;sup&gt;&lt;/sup&gt;&#xA;&amp;bull;&amp;nbsp; Power flows through matched tabs and slots&#xA;&lt;br&gt;&lt;sup&gt;&lt;/sup&gt;&#xA;&amp;bull;&amp;nbsp; The outlet is the sole source of power&#xA;&lt;br&gt;&lt;sup&gt;&lt;/sup&gt;&#xA;&amp;bull;&amp;nbsp; When a timer is in an OFF interval&#xA;             power can only flow into into it, not through it&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If you are unfamiliar with these timers&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each timer repeatedly cycles through&#xA;its schedule of 24 intervals that last an hour each.&#xA;&lt;br&gt;&lt;sup&gt;&lt;sup&gt;&lt;/sup&gt;&lt;/sup&gt;&#xA;&amp;bull;&amp;nbsp; You preset each interval to ON or OFF&#xA;&lt;br&gt;&lt;sup&gt;&lt;sup&gt;&lt;/sup&gt;&lt;/sup&gt;&#xA;&amp;bull;&amp;nbsp; A circular dial determines the current point in the schedule&#xA;&lt;br&gt;&lt;sup&gt;&lt;sup&gt;&lt;/sup&gt;&lt;/sup&gt;&#xA;&amp;bull;&amp;nbsp; The dial advances along its schedule when, and only when,&#xA;             power flows into the timer and its dial-rotating motor&#xA;&lt;br&gt;&lt;sup&gt;&lt;sup&gt;&lt;/sup&gt;&lt;/sup&gt;&#xA;&amp;bull;&amp;nbsp; When the dial is in an interval that was set to ON,&#xA;             the timer acts as a direct connection for power to flow&#xA;             between what is plugged into it and what it plugs into&#xA;&lt;br&gt;&lt;sup&gt;&lt;sup&gt;&lt;/sup&gt;&lt;/sup&gt;&#xA;&amp;bull;&amp;nbsp; When the dial is in an interval that was set to OFF,&#xA;             the timer does not provide a power connection&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Warm-up puzzles&lt;/strong&gt;&#xA;&lt;br&gt;&#xA;Day:    &amp;ensp; &amp;ensp;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/q/25610/18129&quot;&gt;Odd hours with two timers&lt;/a&gt;&#xA;&lt;br&gt;&#xA;Night:  &amp;ensp; &amp;hairsp;&amp;hairsp;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/q/25606/18129&quot;&gt;Halve time with two timers&lt;/a&gt;&#xA;&lt;br&gt;&#xA;Related:&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/q/33882/18129&quot;&gt;Third timer's a charm&lt;/a&gt;&#xA;&amp;emsp; (has a&#xA;        &lt;a href=&quot;https://puzzling.stackexchange.com/a/35181/18129&quot;&gt;sample answer&lt;/a&gt;&#xA;        meant to help)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&#xA;(This puzzle comes directly from an actual botany experiment)&#xA;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Small hint&lt;/strong&gt;&amp;hairsp;&amp;mdash;&amp;hairsp;a possibility implied by&#xA;                                     these pieces' old-fashioned circuitry:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/8NSTQ.jpg&quot; alt=&quot;Arrangement with an outlet feeding power to a socket multiplier's socket&quot;&gt;&#xA; &lt;br&gt;&#xA; &lt;strong&gt;A socket multiplier is not strictly a one-way component.&lt;/strong&gt;&#xA; Power can flow out through its plug and in through its sockets.&#xA; This even allows for power to flow between&#xA; pieces plugged into a multiplier's sockets,&#xA; whether or not the multiplier is itself plugged into anything.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Big hint&lt;/strong&gt;&amp;hairsp;&amp;mdash;&amp;hairsp;another possibility implied by&#xA;                                   these pieces' old-fashioned circuitry:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/EuMNe.jpg&quot; alt=&quot;Arrangement with an outlet feeding power to a timer's output socket&quot;&gt;&#xA; &lt;br&gt;&#xA; &lt;strong&gt;A timer is not strictly a one-way device.&lt;/strong&gt;&#xA; When a timer is in an ON interval,&#xA; power may flow in or out through both its plug and its socket.&#xA; If power flows in only through the socket:&#xA; &lt;br&gt;&lt;sup&gt;&lt;/sup&gt;&#xA; &amp;bull;&amp;nbsp; The motor rotates the dial forward along its schedule&#xA; &lt;br&gt;&lt;sup&gt;&lt;/sup&gt;&#xA; &amp;bull;&amp;nbsp; The timer cannot reliably advance itself into an OFF interval,&#xA;              so (1) the intervals during that portion of the schedule&#xA;              must all be set to ON&#xA;              and (2) power must be supplied through the timer's plug&#xA;              before and during the next OFF interval&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18129" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2017-03-12T04:20:39.733" Title="Day and night of the two timers" Tags="|optimization|real|time|jigsaw-puzzle|circuitry|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="25813" PostTypeId="2" ParentId="25791" CreationDate="2016-01-27T03:35:34.803" Score="10" Body="&lt;p&gt;This is all I can find, there may be more. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;k=0, n=0 : an empty table. (Fig A1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;k=0, n=1 : a single coin on the table. (Fig B1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;k=1, n=0 : an empty table. (Fig A1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;k=1, n=2 : two coins touching. (Fig C1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;k=2, n=0 : an empty table. (Fig A1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;k=2, n&gt;=3 : a 'ring' of 3 or more coins, each touching its two neighbors (For example Figs. D1 - three coins, and A2 - 5 coins, )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;k=3, n=0 : an empty table (Fig A1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;k=3, n=4(m+4), m&gt;=0 : closed loop of four or more repeated blocks of four coins (For example Fig B2 - m=0, and Fig C2 - m=1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;k=3, n=∞ : infinite double row of coins in squared arrangement (Fig A3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;k=3, n=∞ : infinite hexagonal array with every third coin in each row removed. (Fig D3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;k=4, n=∞ : infinite square array of coins (Fig B4), or infinite hexagonal array with every second coin in every second row removed. (Fig A4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;k=4, n=∞ : infinite double row of coins in hexagonal arrangement (Fig B3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;k=5, n=∞ : infinite hexagonal array with one in seven of the coins removed. (Fig C4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;k=6, n=∞ : infinite hexagonal array with no coins removed. (Fig D4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;k&gt;=7, n=0 : empty table again for all these 'impossible' options.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EOCyf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EOCyf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="1631" LastEditDate="2016-01-27T21:28:56.013" LastActivityDate="2016-01-27T21:28:56.013" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="25814" PostTypeId="2" ParentId="25803" CreationDate="2016-01-27T09:01:24.540" Score="2" Body="&lt;p&gt;Puzzle 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3rd picture.  All the rest can be seen as a clockwise rotation of the other images. The 3rd one doesn't. Also it has three arrows in same pattern&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Puzzle 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2nd picture. All the rest can be seen as a clockwise rotation of the other images. The 2nd one is a mirror image of one of the rotations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Puzzle 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4th picture.  All the rest can be seen as a clockwise rotation of the other images. The 4th one is a mirror image of one of the rotations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rotating an image in clockwise direction brings the other three images. While the odd one is a mirror of one of the clockwise rotated images.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2656" LastActivityDate="2016-01-27T09:01:24.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25815" PostTypeId="2" ParentId="25812" CreationDate="2016-01-27T09:26:57.087" Score="3" Body="&lt;p&gt;The biggest I can get for now that should fulfill all conditions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; D=10 N=38&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Only using the four pieces of the top row (and ordering them exactly like in the top row) we set the following on-states:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Timer1&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 24&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Timer2&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1, 2, 3, 4, 5, 6, 7, 8, 9, 10&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;We start with an on-time for 10 hours.&lt;br&gt;&#xA;Afterwards &lt;strong&gt;timer2&lt;/strong&gt; is set to off and &lt;strong&gt;timer1&lt;/strong&gt; will be on for 1 more hour(since they aren't allowed to switch states at the same time)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next &lt;strong&gt;timer1&lt;/strong&gt; will be off for 12 hours so &lt;strong&gt;timer2&lt;/strong&gt; won't move.&lt;br&gt;&#xA;Then &lt;strong&gt;timer1&lt;/strong&gt; will switch on for 12 hours (starting at &lt;strong&gt;intervall24&lt;/strong&gt;) and &lt;strong&gt;timer2&lt;/strong&gt; will go through 12 hours off-state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Timer1&lt;/strong&gt; will switch off for another 12 hours.&lt;br&gt;&#xA;Finally &lt;strong&gt;timer1&lt;/strong&gt; will switch on again, run for 1 hour during &lt;strong&gt;intervall24&lt;/strong&gt; and starting with &lt;strong&gt;intervall1&lt;/strong&gt; &lt;strong&gt;timer2&lt;/strong&gt; will also switch on thereby closing the cycle.&lt;/p&gt;&#xA;" OwnerUserId="16608" LastActivityDate="2016-01-27T09:26:57.087" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25816" PostTypeId="2" ParentId="12" CreationDate="2016-01-27T09:30:16.153" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Yes for example, the puzzle below cannot be solved by logic only guesswork and it only has one solution too.&#xA;&lt;br&gt;&#xA;&lt;br&gt;&lt;a href=&quot;https://i.stack.imgur.com/dSRzQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dSRzQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18723" LastEditorUserId="18250" LastEditDate="2016-11-27T15:08:30.203" LastActivityDate="2016-11-27T15:08:30.203" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25818" PostTypeId="1" AcceptedAnswerId="26072" CreationDate="2016-01-27T10:02:53.707" Score="20" ViewCount="1530" Body="&lt;p&gt;I am loved by many,&lt;br&gt;&#xA;even if sometimes people are angry to me.&lt;br&gt;&#xA;You can find me in different places,&lt;br&gt;&#xA;yet you won't always find me stable.&lt;br&gt;&#xA;Sometimes you must use your talent,&lt;br&gt;&#xA;sometimes you need wisdom to be my best friend.&lt;br&gt;&#xA;I was born before NATO's first SACEUR is appointed,&lt;br&gt;&#xA;yet I am not old fashioned.&lt;br&gt;&#xA;You can tear me apart,&lt;br&gt;&#xA;but then you will find radius and patella.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1 (20 hours after the riddle):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; radius and patella are not crypted (Regan thinks that it may be), you just need to think about their meanings. What are their properties?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2 (2nd day):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Desmond Hung's radius and patella reference is the closest so it may lead you thinking in similar ways: &quot;You can remove the cover from a microphone. The microphone itself looks like a radius while the cover looks like a patella.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is not &quot;organically&quot; alive&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; closest answer so far is : internet, but because of different reasons. think more about this. make a connection with hint #2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Answer is 2 words&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18685" LastEditorUserId="18685" LastEditDate="2016-02-01T09:40:51.497" LastActivityDate="2016-02-18T21:41:45.797" Title="Loved by many - what am I?" Tags="|riddle|word|" AnswerCount="6" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="25819" PostTypeId="1" AcceptedAnswerId="26044" CreationDate="2016-01-27T10:15:57.840" Score="8" ViewCount="1941" Body="&lt;p&gt;Following on from the recent blindfolded near the great Wall of China &lt;a href=&quot;https://puzzling.stackexchange.com/questions/25524/blindfolded-and-disoriented-near-the-great-wall-of-china&quot;&gt;puzzle&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This time, &lt;strong&gt;two wise guys&lt;/strong&gt; are blindfolded and disoriented and standing together exactly 1 mile from the Great Wall of China. they need to reach to Great Wall as soon as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They move exactly 1 miles/hour (no less or no more) and they can communicate with a transmitter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How long will it take for them to reach the Great Wall in the optimally worst case? (both needs to be there at the end.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Since no one could solve it yet, I want to comment about it: @f&quot; is the closest one to the answer still not the best optimal worst case!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If noone could solve in 2 days, I will share the real solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18514" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2019-01-31T04:31:47.050" Title="Two Blindfolded and disoriented near the Great Wall of China" Tags="|geometry|" AnswerCount="7" CommentCount="23" ContentLicense="CC BY-SA 3.0" />
  <row Id="25821" PostTypeId="2" ParentId="25819" CreationDate="2016-01-27T11:37:59.287" Score="2" Body="&lt;p&gt;4.57 hours. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They walk 1 mile in the same direction, then one walks clockwise and the other walks counter-clockwise along an arc of 1 mile radius.&#xA;Worst case: They started parallel to the wall. In which case, one of them would reach the wall when he completes a quarter circle and then the other would have to walk 2 miles to the first guy.&lt;/p&gt;&#xA;" OwnerUserId="18724" LastEditorUserId="18724" LastEditDate="2016-01-27T12:42:22.887" LastActivityDate="2016-01-27T12:42:22.887" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25822" PostTypeId="2" ParentId="25809" CreationDate="2016-01-27T11:42:07.600" Score="20" Body="&lt;p&gt;How I solved it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Look at the capital letters and use their position in the word to identify which letter you need from the following word.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &quot;&lt;code&gt;Dear patrick&lt;/code&gt;&quot;:  &lt;code&gt;D&lt;/code&gt; is in the first position in &lt;code&gt;Dear&lt;/code&gt;, so we take the first letter from &lt;code&gt;patrick&lt;/code&gt;: &lt;strong&gt;&lt;code&gt;P&lt;/code&gt;&lt;/strong&gt;&lt;br&gt;&#xA; &quot;&lt;code&gt;EnjOyaBle&lt;/code&gt;&quot;:  We take the 1st, 4th, and 7th letters from the following word (&lt;code&gt;hopefully&lt;/code&gt;): &lt;strong&gt;&lt;code&gt;HEL&lt;/code&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He actually is telling you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Please help me; I am kidnapped!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18559" LastEditorUserId="4421" LastEditDate="2016-01-27T19:42:08.840" LastActivityDate="2016-01-27T19:42:08.840" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25823" PostTypeId="2" ParentId="25819" CreationDate="2016-01-27T12:53:56.160" Score="2" Body="&lt;p&gt;The other answer is shorter, but relies upon both being able to hold a very precise arc while blind.  If they can do that, great, however walking in a straight line may be easier.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In which case the answer is approximately 6 hours.  Assuming they're able to maintain orientation on a straight course, and can accurately turn to a desired heading, and can accurately track time.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They can walk the complete bounding box of a 1 mile radius circle in that time:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;For the first hour, they should both walk together in the same direction. &lt;/li&gt;&#xA;&lt;li&gt;Then one turns 90 degrees to the right and the other 90 degrees to the left.&lt;/li&gt;&#xA;&lt;li&gt;Both walk straight for the next hour.&lt;/li&gt;&#xA;&lt;li&gt;Each makes another 90 degree turn in the same direction (left turns left again, right turns right again).&lt;/li&gt;&#xA;&lt;li&gt;Both walk straight for the next two hours.&lt;/li&gt;&#xA;&lt;li&gt;Repeat step #4.&lt;/li&gt;&#xA;&lt;li&gt;Both walk straight for the next hour, and which point they are reunited.&lt;/li&gt;&#xA;&lt;li&gt;If at any point after step #1 one of them encounters the wall, they can transmit to the other that they've found it;  the one receiving the message will 1) execute a 180 degree turn if messaged during hour 2, or 2) execute a 90 degree turn if messaged during hours 3 or 4, or 3) continue on their current heading otherwise.  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In the worst-case scenario, the wall will be encountered at one of the corners reached at the end of the 4th hour.  The other participant will need to walk 2 hours to reach that spot.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I say &quot;approximately&quot; 6 hours because the bounding box is larger than the 1 mile radius within which some portion of the wall should coincide (at a tangent, I assume, because otherwise some portion of the wall is less than 1 mile away).  So if the wall is encountered at a corner of the bounding box, there's a reasonable chance that the second participant will encounter a different portion of the wall somewhat before they reach the far corner (i.e. in slightly less than 6 hours).  Both will have reached the wall, but they will be at different points along it.&lt;/p&gt;&#xA;" OwnerUserId="18720" LastActivityDate="2016-01-27T12:53:56.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25825" PostTypeId="2" ParentId="25819" CreationDate="2016-01-27T13:14:45.053" Score="4" Body="&lt;p&gt;A slight improvement on &lt;a href=&quot;https://puzzling.stackexchange.com/a/25821/1748&quot;&gt;Electric_monk's answer&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;each one starts off 90 degrees from eachtother for $\sqrt2$ miles straight. Then they walk for 1 mile straight in such a direction so they are exactly 2 miles apart and on opposite sides of the starting point, then they follow the circle of radius 1 mile until they meet each other. Like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/S5GrC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/S5GrC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the worse case is here when they are the 2 miles apart and one of them finds it, which means that the answer is $1+\sqrt2 +2 \approx 4.41$&lt;/p&gt;&#xA;" OwnerUserId="1748" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2016-01-27T13:43:02.023" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25827" PostTypeId="2" ParentId="25819" CreationDate="2016-01-27T13:20:45.320" Score="12" Body="&lt;p&gt;If at any point during this strategy one person encounters the wall, the other person stops whatever they were doing and instead walks perpendicularly directly towards the wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the starting point have coordinates of $(0,0)$. In Phase 1 of the strategy, one person walks in a straight line to $(\frac{1}{\sqrt{3}},-1)$, and the other walks to $(-\frac{1}{\sqrt{3}},-1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The people then proceed straight to $(\frac{\sqrt{3}}{2},-\frac{1}{2})$ and $(-\frac{\sqrt{3}}{2},-\frac{1}{2})$ (they will not find the wall while doing this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Phase 2 each person walks in an arc of $\frac{3}{2}\sin^{-1}\frac{3}{4}-\frac{\pi}{3}$ radians (approximately 12.886 degrees)* around the circle, and then continues on a tangent until reaching the line $x+y=\sqrt{2}$ or $-x+y=\sqrt{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, the people walk along the lines $x+y=\sqrt{2}$ and $-x+y=\sqrt{2}$ until they reach $(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$ and $(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$ (again, they will not find the wall until they reach these points). In Phase 3, they walk along the circle towards $(0,1)$, which they will reach at a time of $\sqrt3+\frac{8\sqrt2-2\sqrt7}{5}+\frac{3}{2}\sin^{-1}\frac{3}{4}-\frac{\pi}{12}\approx3.94679$ hours if the wall is not encountered earlier. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The worst case if the wall is found during Phase 1 is $1+\frac{2}{\sqrt3}\approx2.15470$ hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The worst case if the wall is found during Phase 2 is when the wall is at an angle of $\frac{1}{2}\sin^{-1}\frac{3}{4}$ radians (approximately 24.295 degrees)* from the vertical. The farther person reaches it at a time of $1+\sqrt3+\frac{\sqrt7}{2}+\frac{3}{2}\sin^{-1}\frac{3}{4}-\frac{\pi}{3}\approx4.27892$ hours. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The worst case if the wall is found during Phase 3 is $1+\sqrt3+\frac{8\sqrt2-2\sqrt7}{5}+\frac{3}{2}\sin^{-1}\frac{3}{4}-\frac{\pi}{3}\approx4.16139$ hours (if the wall is found immediately after they reach the circle, the farther person takes $1$ hour to reach it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall, the worst possible case takes $\boxed{4.27892}$ hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diagrams: green circle shows where the wall is found, green diamond is where the other person reaches it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Worst case for phase 2 (and worst case overall):&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/How00.png&quot; alt=&quot;2]&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Worst case for phase 3:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/2NmO2.png&quot; alt=&quot;3&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Latest possible wall encounter:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/zz2xy.png&quot; alt=&quot;4&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;*Where does this strange value $\sin^{-1}\frac{3}{4}$ come from?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Name the starting point $A$ and the unit circle $O$. Suppose that the people are following paths that are mirror images across the x-axis. Call the horizontal tangent ($y=-1$) $l$. Suppose the wall (call it $w$ is at an angle of $\theta$ to the vertical. Consider which choice of path minimizes the time so that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Both people touch $l$.&lt;/li&gt;&#xA;&lt;li&gt;Then, one person touches $w$ without going inside $O$ (if they go inside the circle, they are missing some tangents).&lt;/li&gt;&#xA;&lt;li&gt;Then, the other person stops following the mirror-image path and heads to $w$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note that if we let $v$ be the reflection of $w$ across the x-axis, the time for the last step (second person moves to $w$) is the same as if the first person moved to $v$ instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying the reflective technique I used in &lt;a href=&quot;https://puzzling.stackexchange.com/a/25985/&quot;&gt;this answer&lt;/a&gt;, we now reflect $A$ over $l$ to get $A^\prime$, $v$ over $w$ to get $v^\prime$, and then $A^\prime$ and $O$ over $v^\prime$ to get $A^{\prime\prime}$ and $O^\prime$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gulD6.png&quot; alt=&quot;4]&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what is half the length of the shortest path that passes from $A^\prime$ to $A^{\prime\prime}$ while staying outside $O$ and $O^\prime$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\frac{\pi}{9}&amp;lt;\theta&amp;lt;\frac{\pi}{6}$, the path looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zWYZO.png&quot; alt=&quot;5]&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ABA^\prime$ is a right triangle with $AB=1$ and $AA^\prime=2$, so $\angle A^\prime AB=\frac{\pi}{3}$ and $A^\prime B=\sqrt3$. $AC$ is parallel to $v^\prime$, so $\angle A^\prime AC=3\theta$ and $\angle BAC=3\theta-\frac{\pi}{3}$. But what is $CD$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5fNzd.png&quot; alt=&quot;6]&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reflect $A$ over $w$ to get $E$, and construct rectangle $CDGF$ with $E$ on $FG$. $\angle AEG=\pi-2\theta$ because $EG$ is perpendicular to $v^\prime$ and $AE$ is perpendicular to $w$. $AE=2$, and $EG=1$. So $CD=FG=EG+EF=1+2\cos{2\theta}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In total, the path length from $A^\prime$ to $D$ is $\sqrt3+3\theta-\frac{\pi}{3}+1+2\cos{2\theta}$. When we take the derivative and set it to zero, we get $3-4\sin{2\theta}=0$, and so the required path length is maximized at $\theta=\frac{1}{2}\sin^{-1}\frac{3}{4}$ with a value of $\sqrt3+\frac{3}{2}\sin^{-1}\frac{3}{4}-\frac{\pi}{3}+1+\frac{\sqrt7}{2}\approx4.27982$, as desired.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2016-02-01T21:59:33.970" CommentCount="24" ContentLicense="CC BY-SA 3.0" />
  <row Id="25828" PostTypeId="1" CreationDate="2016-01-27T20:12:18.020" Score="11" ViewCount="2948" Body="&lt;p&gt;At my school we have this thing called the breakout box and there's a challenge. It's pretty hard, so me and some of my friends have been working on it. The idea is that there is a box with a bunch of locks on it and clues around the room to help unlock it. You have to use the clues to get the locks' codes - a 4 digit number - and open the box. So far we've opened every single lock except one. We've researched for about two days and we can't figure this out, so I decided that I would ask the online community. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The riddle is on the wall. There is this painting and we found out the painting is called 'The Woman and Bird in the Night'. It's by Joan Miro. (see below) The teacher who made the riddles wrote 'e=16' in the bottom corner on the white part of the paper. We aren't sure if this is a hint, a key, or something to throw you off. We really need help with this because we have unlocked all the locks except one. This is the only clue left, and the answer must be a 4-digit combination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll be checking for answers about every hour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the picture: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cN0z4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cN0z4.jpg&quot; alt=&quot;The Woman and Bird in the Night&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wikiart.org/en/joan-miro/not_detected_227979#supersized-artistPaintings-227979&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; is a link to a larger version of the painting if anyone needs it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might seem like this has many answers, but there is only one answer and it is a four digit combo that uses numbers. I do have access to the lock and every day I have been going down and checking all the combos that people have thought might be a possible answer and then adding a comment letting people know if anyone was right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a pic of the picture on the wall:&lt;a href=&quot;https://i.stack.imgur.com/9Dp2i.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9Dp2i.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Y8l8X.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y8l8X.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18735" LastEditorUserId="17482" LastEditDate="2016-03-02T17:29:12.260" LastActivityDate="2016-06-14T19:05:53.853" Title="The Woman and Bird in the Night: Can you solve the riddle related to this painting?" Tags="|number-sequence|visual|" AnswerCount="2" CommentCount="42" ContentLicense="CC BY-SA 3.0" />
  <row Id="25831" PostTypeId="2" ParentId="25818" CreationDate="2016-01-28T00:26:24.377" Score="6" Body="&lt;p&gt;I am unsure but will try&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Uranium&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am loved by many,&#xA;even if sometimes people are angry to me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lots of people would love to have enriched uranium, though it clearly has negatively impacted many.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can find me in different places,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Uranium naturally occurs in low levels in soil, rocks, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yet you won't always find me stable.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A reference to unstable isotopes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes you must use your talent,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A reference to Uranium Talent Search Examination?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;sometimes you need wisdom to be my best friend.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I sure hope people who get stockpiles of uranium are wise with it!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was born before NATO's first SACEUR is appointed,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nuclear weapons preceded Eisenhower's appointment in 1950.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yet I am not old fashioned.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nuclear weapons are still highly desirable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can tear me apart,&#xA;but then you will find radius and patella.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Is this a semi cryptic clue? Take some of the letters of uranium and rearrange them (tear apart) to get &quot;arm&quot; and &quot;ni&quot; which sounds like &quot;knee&quot;. Radius is an arm bone and patella is the kneecap.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18493" LastActivityDate="2016-01-28T00:26:24.377" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25832" PostTypeId="1" AcceptedAnswerId="25834" CreationDate="2016-01-28T05:42:57.233" Score="43" ViewCount="3807" Body="&lt;p&gt;Usually, 20-sided dice are labeled with the numbers 1 to 20. When you roll two of these, their sum is a random number between 2 and 40. The total 21 is most likely to occur, while 2 and 40 are the least. The probabilities vary linearly in between, as shown below.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there &lt;strong&gt;different&lt;/strong&gt; way to label two 20-sided dice with &lt;strong&gt;positive integers&lt;/strong&gt;, so that their sum is still a random number between 2 and 40 with these &lt;strong&gt;same&lt;/strong&gt; probabilities?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without the positive integer constraint, there would be trivial solutions like [0, ... ,19] and [1, ... ,21], or [0.5, ... ,19.5] and [1.5, ... ,20.5].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a variant of a &lt;a href=&quot;https://puzzling.stackexchange.com/questions/18426/unusual-dice-usual-results&quot;&gt;more famous puzzle&lt;/a&gt;, which asks the same question for six-sided dice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WGI41.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WGI41.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2016-01-30T09:36:07.227" Title="Relabeling two 20-sided dice without changing their total" Tags="|combinatorics|probability|dice|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25833" PostTypeId="1" AcceptedAnswerId="25933" CreationDate="2016-01-28T05:45:26.613" Score="10" ViewCount="408" Body="&lt;p&gt;We are the mythical twins&lt;br&gt;&#xA;In the in-between place&lt;br&gt;&#xA;Gateway to the middle&lt;br&gt;&#xA;Where the king's daughter faces the dust&lt;br&gt;&#xA;We are of the strong one, the labourer&lt;br&gt;&#xA;One of us keeps monkeys&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who are we?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The monkeys are real, not mythical.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Geography.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user17947" LastEditorDisplayName="user17947" LastEditDate="2016-01-29T02:18:20.687" LastActivityDate="2016-01-30T01:26:59.253" Title="Ancient Riddle of the In-Between" Tags="|riddle|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25834" PostTypeId="2" ParentId="25832" CreationDate="2016-01-28T06:34:59.763" Score="50" Body="&lt;p&gt;&lt;em&gt;Is there a different way to label two 20-sided dice with positive integers, so that their sum is still a random number between 2 and 40 with these same probabilities?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Details:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In fact, there are &lt;strong&gt;seven&lt;/strong&gt; (if I counted correctly) different ways of achieving this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More details:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Solution 1&lt;/strong&gt; &lt;br&gt;&#xA; Die 1: 1,2,2,3,3,3,4,4,4,4,5,5,5,5,6,6,6,7,7,8 &lt;br&gt;&#xA; Die 2: 1,5,6,9,10,11,13,14,15,16,17,18,19,20,22,23,24,27,28,32 &lt;br&gt;&#xA; &lt;br&gt;&#xA; &lt;strong&gt;Solution 2&lt;/strong&gt; &lt;br&gt;&#xA; Die 1: 1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,9,9,10,10,11 &lt;br&gt;&#xA; Die 2: 1,3,5,7,9,11,11,13,13,15,15,17,17,19,19,21,23,25,27,29 &lt;br&gt;&#xA; &lt;br&gt;&#xA; &lt;strong&gt;Solution 3&lt;/strong&gt; &lt;br&gt;&#xA; Die 1: 1,2,3,3,4,4,5,5,6,6,7,7,8,8,9,9,10,10,11,12 &lt;br&gt;&#xA; Die 2: 1,2,5,6,9,10,11,12,13,14,15,16,17,18,19,20,23,24,27,28 &lt;br&gt;&#xA; &lt;br&gt;&#xA; &lt;strong&gt;Solution 4&lt;/strong&gt; &lt;br&gt;&#xA; Die 1: 1,2,3,4,5,6,6,7,7,8,8,9,9,10,10,11,12,13,14,15 &lt;br&gt;&#xA; Die 2: 1,2,3,4,5,11,11,12,12,13,13,14,14,15,15,21,22,23,24,25 &lt;br&gt;&#xA; &lt;br&gt;&#xA; &lt;strong&gt;Solution 5&lt;/strong&gt; &lt;br&gt;&#xA; Die 1: 1,2,2,3,3,4,4,5,5,6,11,12,12,13,13,14,14,15,15,16 &lt;br&gt;&#xA; Die 2: 1,3,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,22,24 &lt;br&gt;&#xA; &lt;br&gt;&#xA; &lt;strong&gt;Solution 6&lt;/strong&gt; &lt;br&gt;&#xA; Die 1: 1,2,3,3,4,4,5,5,6,7,11,12,13,13,14,14,15,15,16,17 &lt;br&gt;&#xA; Die 2: 1,2,5,6,6,7,9,10,10,11,13,14,14,15,17,18,18,19,22,23 &lt;br&gt;&#xA; &lt;br&gt;&#xA; &lt;strong&gt;Solution 7&lt;/strong&gt; &lt;br&gt;&#xA; Die 1: 1,2,2,3,5,6,6,7,9,10,10,11,13,14,14,15,17,18,18,19 &lt;br&gt;&#xA; Die 2: 1,3,3,5,5,7,7,9,9,11,11,13,13,15,15,17,17,19,19,21 &lt;br&gt;&#xA; &lt;br&gt;&#xA; That's it (I think)!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How I solved it, and how you can check that the solutions are correct:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many years ago I read a very nice math textbook called &lt;em&gt;Analytic Number Theory&lt;/em&gt; by D. J. Newman. Chapter 1 of that book has a section titled &lt;strong&gt;Crazy Dice&lt;/strong&gt; in which the same riddle linked to by the OP that deals with the case of 6-sided dice is solved using generating functions. Essentially one converts the problem to a problem of polynomial algebra. For example, the solution &quot;die1=(1,2,2,3,3,4), die2=(1,3,4,5,6,8)&quot; to the original crazy dice problem reduces to checking the polynomial identity&lt;br&gt;&#xA; $$(x+x^2+x^3+x^4+x^5+x^6)^2 = (x+2x^2+2x^3+x^4)(x+x^3+x^4+x^5+x^6+x^8),$$&#xA; &lt;br&gt;since when expanding out these products, the coefficients of powers of $x$ on the left-hand side correspond precisely to the probabilities (after dividing by 36) to get the numbers $2,3,...,12$ when rolling normal dice, and on the right-hand side this would be the same thing with the crazy dice.&#xA; &#xA; So, to solve our new problem with the $20$-sided dice we need to find corresponding identities involving a &quot;nonstandard&quot; factorization of the polynomial&#xA; &lt;br&gt;&#xA; $$ g(x)^2 = (1+x+x^2+...+x^{19})^2$$&#xA; &lt;br&gt;&#xA; into a product of two distinct polynomials whose coefficients are nonnegative and add up to $20$. I managed to do this by factorizing the polynomial $g(x)$ in Mathematica (it factors into 5 irreducible factors: $x+1$, $x^2+1$, $x^4-x^3+x^2-x+1$, $x^4+x^3+x^2+x+1$, $x^8-x^6+x^4-x^2+1$) and then writing a small program to systematically look at all possible products of these 5 factors with powers $0,1,2$. There are a total of $3^5=243$ possibilities, so my laptop did this easily in a few seconds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user17947" LastEditorDisplayName="user17947" LastEditDate="2016-01-28T07:08:21.023" LastActivityDate="2016-01-28T07:08:21.023" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25835" PostTypeId="2" ParentId="25832" CreationDate="2016-01-28T09:03:40.990" Score="8" Body="&lt;p&gt;A solution that fits a generalized pattern:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &amp;ensp;&#xA;   [ 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 ]&#xA; &lt;br&gt;&#xA; x [ 1 3 5 7 9 11 11 13 13 15 15 17 17 19 19 21 23 25 27 29 ]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Began by looking at what would be equivalent puzzles&#xA;for pairs of 2-sided, 4-sided, 6-sided and 8-sided dice,&#xA;along with the assumption that the intervals between their sides' numbers&#xA;would be symmetric because the resulting distribution is symmetric.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The intervals' symmetry could be either like ABBA and CDDC,&#xA; as turned out to be the case, or like ABCD and DCBA, where the&#xA; intervals of one die reverse the intervals of the other die.&#xA; &lt;br&gt;&lt;br&gt;&#xA; 2 sides (essentially a pair of coins with no alternative&#xA;          to the standard spot pattern):   [1 2] x [1 2]&#xA; &lt;br&gt;&lt;br&gt;&#xA; 4 sides (easy):   [1 2 2 3] x [1 3 3 5]&#xA; &lt;br&gt;&lt;br&gt;&#xA; 6 sides (the famous case):   [1 2 2 3 3 4] x [1 3 4 5 6 8]&#xA; &lt;br&gt;&lt;br&gt;&#xA; 8 sides (not too tough, based on those above):&#xA;                              [1 2 2 3 3 4 4 5] x [1 3 5 5 7 7 9 11]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;20 sides was straightforward after 8 sides refined&#xA;the pattern suggested by 2, 4 and 6 sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;&lt;strong&gt;Addendum.&lt;/strong&gt;&#xA;&amp;thinsp;&#xA;In the terms of&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/a/25889/18129&quot;&gt;another solution&lt;/a&gt;,&#xA;this solution neatly breaks down to&#xA;$$&#xA;\begin{array}{ll}&#xA;A = [1,2] &amp;amp; B=[1,11]\\&#xA;C=[0,1,2,3,4,5,6,7,8,9] &amp;amp; D=[0,2,4,6,8,10.12,14,16,18]&#xA;\end{array}&#xA;$$&#xA;which clearly shows what underlies the straightforward&#xA;even-sided-dice pattern better than how I understood it.&lt;/p&gt;&#xA;" OwnerUserId="18129" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2016-01-30T09:36:07.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25836" PostTypeId="1" AcceptedAnswerId="25848" CreationDate="2016-01-28T09:27:42.930" Score="24" ViewCount="2647" Body="&lt;p&gt;After I have figured out if &lt;a href=&quot;https://puzzling.stackexchange.com/questions/25809/is-dave-really-mad-at-me&quot;&gt;Dave was really mad at me&lt;/a&gt;, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I panicked and called the police.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After a couple of minutes, I saw a strange post of mine. Well, I was not on StackExchange! It was written by Dave. But why would he log in my account and write his message as a post, instead of sending me another email?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hello again, Patrick! I am sorry for what I &#xA;       did.&lt;br&gt;&#xA;  I was just so mad at you and wanted to make you struggle with some meaningless&#xA;       email. &#xA;  I just made random letters uppercase.&#xA;       Please &#xA;  do not take any messages from that mail. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Maybe we can still hang&#xA;       out, &#xA;  what do you say?&#xA;       Take &#xA;  your time to answer. I&#xA;       understand &#xA;  if you say&#xA;       no &#xA;  after that gibberish mail of mine.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I have some news about&#xA;       Daniel &#xA;  by the way. He called me last week. He was looking for a PhD position in MIT. I don't think he will never,&#xA;       ever&#xA;  get a position. He is lazy, you know. When I tell this to him, he gets angry. Maybe he will &#xA;       reason &#xA;  with me after a time. &#xA;       Considering &#xA;  that he has ups &#xA;       and &#xA;  downs &#xA;       sometimes, &#xA;  he probably cannot just take&#xA;       it &#xA;  easy right now.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Whatever.&#xA;       Now,&#xA;  I am going to rest. Drink some coffee and watch TV. Ah, by the way, my cellphone is broken. I dropped it. Therefore, it is off. If you want to hang&#xA;       out,&#xA;  just send me an e-mail.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Best,&lt;br&gt;&#xA;  Dave.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Should I&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;call the police back and &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;explain that everything was a misunderstanding?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hint is in the question itself. Why write as a post instead of sending an e-mail?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18645" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2016-01-29T14:48:42.830" Title="What happened to Dave?" Tags="|steganography|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="25837" PostTypeId="2" ParentId="25738" CreationDate="2016-01-28T09:49:56.090" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;2&quot;&gt;&#xA;  &lt;li&gt;What mistake(s) could explain the discrepancy between editions?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Printer error. The versions in the first and second editions swap&#xA; multiple-choice answer H and the spurious NW-corner pattern.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Methodology:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Found versions that showed multiple-choice answers&#xA; for the &lt;em&gt;Outliers&lt;/em&gt; puzzle.&#xA; &lt;br&gt;&lt;br&gt;&#xA; Incorrect First Edition version, as it appears at the&#xA; &lt;a href=&quot;http://www.lukehoran.com/2013/01/outliers-puzzle/&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt;&#xA; provided by&#xA; &lt;a href=&quot;https://puzzling.stackexchange.com/questions/25738/malcolm-gladwells-outliers-progressive-matrices-puzzle#comment79040_25738&quot;&gt;Daedric&lt;/a&gt;:&#xA; &lt;br&gt;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/ZXn9u.png&quot; height=&quot;350&quot;&gt;&#xA; &lt;br&gt;&lt;br&gt;&#xA; Correct Second Edition version, presumably, as seen in a&#xA; &lt;a href=&quot;https://www.rooshvforum.com/thread-25934.html#pid_507877&quot; rel=&quot;nofollow noreferrer&quot;&gt;blog entry&lt;/a&gt;&#xA; about&#xA; &lt;a href=&quot;https://en.wikipedia.org/wiki/Raven&amp;#39;s_Progressive_Matrices&quot; rel=&quot;nofollow noreferrer&quot;&gt;Raven's Progressive Matrices&lt;/a&gt;:&#xA; &lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/sVcMA.jpg&quot; height=&quot;350&quot;&gt;&#xA; &lt;br&gt;&lt;br&gt;&#xA; The versions agree on every other pattern,&#xA; including the correct multiple-choice answer A.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18129" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2016-01-28T10:16:34.257" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25838" PostTypeId="2" ParentId="25833" CreationDate="2016-01-28T09:59:28.960" Score="0" Body="&lt;p&gt;They are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Heracles &amp;amp; Iphicles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The strong one, the laborer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Heracles the strong one, plus he had to complete several labours commanded by some king which name i forgot&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;keeps monkeys&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; reference to the Kerkopes Heracles captured&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;in between&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; about having 2 fathers and/or Heracles father being zeus and the mother being mortal&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18559" LastEditorUserId="18559" LastEditDate="2016-01-29T08:17:36.673" LastActivityDate="2016-01-29T08:17:36.673" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25839" PostTypeId="2" ParentId="25833" CreationDate="2016-01-28T10:14:30.140" Score="2" Body="&lt;p&gt;I've come to 2 different options, but decided on this answer because I believe it has closer clues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are the mythical twins&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quetzalcoatl and Xolotl, mythical twins&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the in-between place&#xA;Gateway to the middle&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are in the underworld but want to escape and create a life&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where the king's daughter faces the dust&#xA;We are of the strong one, the labourer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; While trying to run away from the underworld, Quetzalcoatl dropped the bones, and they broke into pieces. He gathered up the pieces and took them to the earth goddess &lt;strong&gt;Cihuacoatl, who ground them into flour&lt;/strong&gt;. Quetzalcoatl moistened the flour with his own blood, &lt;strong&gt;which gave it life&lt;/strong&gt;. Then he and Xolotl shaped the mixture into human forms and taught the new creatures how to reproduce themselves. So I guess Cihuacoatl represents the holy woman, and since twins do the hard work, they are the labourers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One of us keeps monkeys&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quetzalcoatl: In his form as Ehecatl he is the wind, is represented by &lt;strong&gt;spider monkeys&lt;/strong&gt;, ducks, and the wind itself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18685" LastActivityDate="2016-01-28T10:14:30.140" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25841" PostTypeId="2" ParentId="25836" CreationDate="2016-01-28T11:33:47.867" Score="-2" Body="&lt;p&gt;May be the kidnapper wants you to suspect Daniel in future, So MR.kidnapper asks Dave to logon your account and write a post , So you will believe that it is actually Dave an no one else&lt;/p&gt;&#xA;" OwnerUserId="18742" LastActivityDate="2016-01-28T11:33:47.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25842" PostTypeId="1" AcceptedAnswerId="25844" CreationDate="2016-01-28T11:45:35.080" Score="6" ViewCount="2659" Body="&lt;p&gt;Today i came across a cool picture that has SIX hidden words in it. I ended up with just 3. Can you able to find all the SIX words in it.? Here is it. Take a look at it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qfFUY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qfFUY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17861" LastActivityDate="2016-01-28T21:41:38.757" Title="Hidden Words..!" Tags="|visual|english|" AnswerCount="5" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25843" PostTypeId="2" ParentId="25842" CreationDate="2016-01-28T11:52:53.940" Score="4" Body="&lt;p&gt;Story:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; On the woman:&lt;br&gt;&#xA; The &lt;strong&gt;S&lt;/strong&gt; is in her hair.&lt;br&gt;&#xA; The &lt;strong&gt;T&lt;/strong&gt; is on her shirt.&lt;br&gt;&#xA; &lt;strong&gt;O&lt;/strong&gt; and &lt;strong&gt;R&lt;/strong&gt; are on the book.&lt;br&gt;&#xA; &lt;strong&gt;Y&lt;/strong&gt; is where the book and her leg meet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Page:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; On the boy:&lt;br&gt;&#xA; The &lt;strong&gt;P&lt;/strong&gt; is on his trousers.&lt;br&gt;&#xA; &lt;strong&gt;A&lt;/strong&gt; and &lt;strong&gt;G&lt;/strong&gt; are on his shirt to the left of his arm.&lt;br&gt;&#xA; The &lt;strong&gt;E&lt;/strong&gt; is on his sleeve.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Words:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; On the plant.&lt;br&gt;&#xA; At the bottom right edge of the leaves.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Read:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; On the dog and girl.&lt;br&gt;&#xA; &lt;strong&gt;R&lt;/strong&gt; and &lt;strong&gt;E&lt;/strong&gt; on each of the dogs ears.&lt;br&gt;&#xA; &lt;strong&gt;A&lt;/strong&gt; and &lt;strong&gt;D&lt;/strong&gt; on the girls hair.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16608" LastEditorUserId="16608" LastEditDate="2016-01-28T11:57:02.210" LastActivityDate="2016-01-28T11:57:02.210" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25844" PostTypeId="2" ParentId="25842" CreationDate="2016-01-28T11:54:09.310" Score="13" Body="&lt;p&gt;I found the words:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Book , Story, Words, Page, Novel, Read&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;image:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/3kBuV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3kBuV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18685" LastEditorUserId="16608" LastEditDate="2016-01-28T12:01:23.200" LastActivityDate="2016-01-28T12:01:23.200" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25845" PostTypeId="2" ParentId="25842" CreationDate="2016-01-28T11:59:00.127" Score="3" Body="&lt;p&gt;The words are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;BOOK&lt;/strong&gt;, &lt;strong&gt;STORY&lt;/strong&gt;, &lt;strong&gt;WORDS&lt;/strong&gt;, &lt;strong&gt;NOVEL&lt;/strong&gt;, &lt;strong&gt;PAGES&lt;/strong&gt;, AND &lt;strong&gt;READ&lt;/strong&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;as marked below:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/t9t5t.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/t9t5t.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18645" LastActivityDate="2016-01-28T11:59:00.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25846" PostTypeId="2" ParentId="25842" CreationDate="2016-01-28T11:59:12.103" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/wtHr5.jpg&quot;&gt;&#xA;&#xA; &lt;br&gt; BOO &amp;amp; BOOK&#xA; &lt;br&gt; NO &amp;amp; NOVEL&#xA; &lt;br&gt; STORY &amp;amp; OR&#xA; &lt;br&gt; words &amp;amp; or&#xA; &lt;br&gt; PAGES &amp;amp; AGE&#xA; &lt;br&gt; read &amp;amp; ad&#xA; &lt;br&gt; HI (in floorboards)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Added, upside-down words:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;      ON (in upside-down NOVEL)&#xA; &lt;br&gt; HIP &amp;amp; HI (beginning with H from rightside-up HI,&#xA;                the P in HIP is upside-down &quot;d&quot; from &quot;read&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18129" LastEditorUserId="18129" LastEditDate="2016-01-28T21:41:38.757" LastActivityDate="2016-01-28T21:41:38.757" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25847" PostTypeId="1" AcceptedAnswerId="25891" CreationDate="2016-01-28T12:10:10.720" Score="7" ViewCount="9580" Body="&lt;p&gt;They pick me Up&lt;br&gt;&#xA;And Wonder&lt;br&gt;&#xA;They call me Light  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have Brothers and Sisters&lt;br&gt;&#xA;Close, so Close&lt;br&gt;&#xA;Together we Serve&lt;br&gt;&#xA;That from which we were Born  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ah, I am Free&lt;br&gt;&#xA;Yet Fastened&lt;br&gt;&#xA;And when I am Unfastened&lt;br&gt;&#xA;I can no longer Fulfill&lt;br&gt;&#xA;My Purpose   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am used for Travelling&lt;br&gt;&#xA;But I am no Burden&lt;br&gt;&#xA;I am through which It is done  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;tweet tweet's answer (and name, haha) are the closest thus far.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10327" LastEditorUserId="10327" LastEditDate="2016-01-29T06:26:11.977" LastActivityDate="2017-01-27T01:42:11.427" Title="What am I? The riddle of it" Tags="|riddle|" AnswerCount="7" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25848" PostTypeId="2" ParentId="25836" CreationDate="2016-01-28T12:21:30.180" Score="19" Body="&lt;p&gt;I think the secret message hidden in the post is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;DEPOT UNDER CASINO&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation As Below.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;In the edit mode, these words seems to be separated out&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The words&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;did &lt;br /&gt;email &lt;br /&gt;please &lt;br /&gt;out &lt;br /&gt;take &lt;br /&gt;understand &lt;br /&gt;no &lt;br /&gt;daniel &lt;br /&gt;ever &lt;br /&gt;reason &lt;br /&gt;considering &lt;br /&gt;and &lt;br /&gt;sometimes &lt;br /&gt;it &lt;br /&gt;now &lt;br /&gt;out&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Then&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Take the first letters&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;And it will form&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;DEPOT UNDER CASINO (DEPOTUNDERCASINO) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I think&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;you should check the depot under the casino before calling the cops.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="2730" LastEditDate="2016-01-28T12:34:56.347" LastActivityDate="2016-01-28T12:34:56.347" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="25850" PostTypeId="2" ParentId="25842" CreationDate="2016-01-28T14:09:34.383" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; novel, story, book, page, words, read&lt;br&gt;&#xA; Novel in the chair&lt;br&gt;&#xA; Book in the glasses&lt;br&gt;&#xA; Story in the mother + book&lt;br&gt;&#xA; words in the tree&lt;br&gt;&#xA; Page in the teenager lying on the couch&lt;br&gt;&#xA; read in the dog ears + girl  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18745" LastActivityDate="2016-01-28T14:09:34.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25851" PostTypeId="2" ParentId="25847" CreationDate="2016-01-28T14:14:01.947" Score="1" Body="&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Headlight&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They pick me Up&#xA;And Wonder&#xA;They call me Light&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Light source, can't connect they pick me up part but may be; there are some headlights which can open/close&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have Brothers and Sisters&#xA;Close, so Close&#xA;Together we Serve&#xA;That from which we were Born&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fog lamp, brake light, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ah, I am Free&#xA;Yet Fastened&#xA;And when I am Unfastened&#xA;I can no longer Fulfill&#xA;My Purpose&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it is outside of the car so it is free, but it is attached to it so can't work if you remove its connections&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am used for Travelling&#xA;But I am no Burden&#xA;I am through which It is done&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it is in front of the vehicle, and it can't be a burden because of the low weight&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18685" LastActivityDate="2016-01-28T14:14:01.947" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25852" PostTypeId="2" ParentId="25847" CreationDate="2016-01-28T14:19:55.610" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Seatbelt&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They pick me up, call me light&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You pick it up to buckle it. It is light as in weight. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Brothers and sisters...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The brothers and sisters are other seatbelts. They all serve the same purpose. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am free...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A seatbelt is free to move around, it serves its purpose when fastened, and doesn't when not. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Used for traveling...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You wear seatbelts while traveling in cars/planes. No burden means you don't really feel it on you when wearing it. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9631" LastActivityDate="2016-01-28T14:19:55.610" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25853" PostTypeId="2" ParentId="25818" CreationDate="2016-01-28T14:45:48.403" Score="1" Body="&lt;p&gt;This is the first time I reply in this site, I hope it do it the right way. Could it be...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A bicycle / motorcycle ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I am loved by many,&lt;br&gt;&#xA;even if sometimes people are angry to me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are many cyclist and they drive automobile (and pedestrians) crazy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can find me in different places,&lt;br&gt;&#xA;yet you won't always find me stable.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bicycles/motorcycles are popular everywhere in the world; but they are not stable as they only have two wheels, and would fall if they don't have a sidestand to support them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes you must use your talent,&lt;br&gt;&#xA;sometimes you need wisdom to be my best friend.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Learning how to ride a bicycle/motorbike requires time and skills.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was born before NATO's first SACEUR is appointed,&lt;br&gt;&#xA;yet I am not old fashioned.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It took this one literally: the first bicycles/motorcycles date from the 19th century and NATO first SAUCER would be mid 20th century.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can tear me apart,&lt;br&gt;&#xA;but then you will find radius and patella.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This would almost be a literal one too: bicycles/motorcycles can be put apart, and some of the pieces would be the spokes (radius/radios in some countries) and a leg (pata/patilla/patella?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18736" LastActivityDate="2016-01-28T14:45:48.403" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25854" PostTypeId="2" ParentId="25707" CreationDate="2016-01-28T15:24:18.183" Score="6" Body="&lt;p&gt;Building off of Daedric's answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you interpret the true statements and false statements as&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; B and A in a Bacon cipher,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;you get the plaintext&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;TRON&lt;/strong&gt;, which is the answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2016-01-28T15:24:18.183" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25855" PostTypeId="2" ParentId="25847" CreationDate="2016-01-28T16:02:45.317" Score="2" Body="&lt;p&gt;My guess would be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Suitcase&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They pick me Up, And Wonder, They call me Light&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;An empty suitcase&lt;/em&gt;. People expect suitcases to weigh a lot, but when they are empty they obviously dont. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have Brothers and Sisters, Close, so Close, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are a lot of suitcases around and when you're traveling there are usually multiple suitcases around. (traveling companions, airplane, bus etc)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Together we Serve, That from which we were Born&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Suitcases were created for the sole purpose of transporting goods.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ah, I am Free&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Suitcases travel all over the world, which can be considered 'free'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet Fastened&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An unlocked / Locked suitcase&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And when I am Unfastened, I can no longer Fulfill my Purpose&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An unlocked (open) suitcase can no longer hold any items and can no longer fulfill its purpose for which it was born.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am used for Travelling&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This one is pretty obvious&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I am no Burden&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People find it convenient to have suitcases around since it allows them to bring more 'goods'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am through which It is done &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Perhaps after a long travel when the suitcase is worn?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6151" LastActivityDate="2016-01-28T16:02:45.317" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25856" PostTypeId="1" AcceptedAnswerId="25858" CreationDate="2016-01-28T16:20:32.067" Score="4" ViewCount="530" Body="&lt;p&gt;You have two bags of similar (weight, texture) solid pills. One bag contains red pills and the other contains green pills. There are 13 pills in one bag and 14 pills in the other. Today, tomorrow, and the day after tomorrow you must select and consume a valid pill set from the following options:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;[3 green pills and 8 red pill] &lt;/li&gt;&#xA;&lt;li&gt;[3 green pills and 4 red pill]&lt;/li&gt;&#xA;&lt;li&gt;[4 green pills and 3 red pill] &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Although you are blind, your colorblind friend is there to assist you. Neither of you know which bag contains which colored pills. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Side effects (Upon you): &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;When you take an invalid pill set you die.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Side effects (Upon your friend):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Concurrently taking 3 red pills today will instantly blind your friend forever.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Consecutively taking 2 red pills today will instantly blind your friend for 24 hours. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Taking 1 red pill today will instantly remove a positive memory from your friend for 48 hours.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Concurrently taking 3 green pills today will instantly fix your friend’s color blindness forever.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Consecutively taking 2 green pills today will temporarily fix your friend’s color blindness, tomorrow.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Taking 1 green pill today will temporarily fix your friend’s color blindness, the day after tomorrow.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Taking any number of green pill within a minute after taking any number of red pills will instantly undo the red pill’s side effects. Green pill side effects will still apply.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Taking any number of red pill within a minute after taking any number of green pills will instantly undo the green pill’s side effects. Red pill side effects will still apply.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How can you insure the cure of your friend’s colorblindness indefinitely and your survival?&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="18340" LastEditorUserId="18340" LastEditDate="2016-01-28T16:28:58.497" LastActivityDate="2016-01-28T18:58:43.940" Title="Colored Pills MADNESS" Tags="|riddle|logical-deduction|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25857" PostTypeId="1" CreationDate="2016-01-28T16:24:36.593" Score="3" ViewCount="646" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/E9ip3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/E9ip3.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Transcription (cells with ■ do not exist in the image version of the grid and should not be filled)&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;+&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;-&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;=&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;-&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;■&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;+&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;■&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;+&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;■&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;■&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;x&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;-&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;=&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;29&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;-&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;■&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;x&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;■&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;-&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;■&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;■&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;x&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;-&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;=&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;-1&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;=&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;■&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;=&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;■&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;=&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;■&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;■&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;-10&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;■&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;9&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;■&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;■&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;■&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;Here are the rules:  use 1-9 only once in the black cells. Order of operations matters.&lt;/p&gt;&#xA;" OwnerUserId="18755" LastEditorUserId="69582" LastEditDate="2021-02-27T16:56:18.667" LastActivityDate="2021-02-27T16:56:18.667" Title="Solve this math table" Tags="|mathematics|calculation-puzzle|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25858" PostTypeId="2" ParentId="25856" CreationDate="2016-01-28T16:48:03.923" Score="7" Body="&lt;p&gt;Denote the bags as A and B, where A has 14 pills and B has 13. Have your friend grab two pills from A, and hold three from B in his other hand (so he knows where it is if he goes blind).&#xA;Your friend takes the two from A, and determines if he's gone blind or not:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If he's blind, he takes the pills he grabbed from B. B is green, A is Red. Once the green pill he took fixes his vision, have him label the bags (leave them in distinct places so he doesn't mix them up). There are now 12 red pills and 10 green ones, so you can take 4 red and 3 green each day and everything is good&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If he isn't blind, he should take 1 more from A (or 3 if taking 2 then 1 green doesn't work), and put the three pills from B back. Have him label the bags (A is green, B is red), and you'll have at least 9 green and 13 red left. Take 4 red and 3 green each day, and it'll all be dandy&lt;/p&gt;&#xA;" OwnerUserId="5169" LastEditorUserId="5169" LastEditDate="2016-01-28T18:58:43.940" LastActivityDate="2016-01-28T18:58:43.940" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="25859" PostTypeId="2" ParentId="25857" CreationDate="2016-01-28T16:52:17.463" Score="4" Body="&lt;h2&gt;3 | 2 | 4&lt;/h2&gt;&#xA;&#xA;&lt;h2&gt;5 | 7 | 6&lt;/h2&gt;&#xA;&#xA;&lt;h2&gt;8 | 1 | 9&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Trial and error, mostly.&lt;/p&gt;&#xA;" OwnerUserId="18758" LastActivityDate="2016-01-28T16:52:17.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25860" PostTypeId="2" ParentId="25857" CreationDate="2016-01-28T16:52:26.113" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3 2 4&lt;br&gt;&#xA; 5 7 6&lt;br&gt;&#xA; 8 1 9&lt;br&gt;&#xA;&lt;br&gt;&#xA; I started off with the middle row, the left two numbers have to be above 30, so 8x4, 9x4, 7x5&lt;br&gt;&#xA; Bottom middle usually was 1, so the top middle adds up with middle to 9&lt;br&gt;&#xA; Left column was interesting. You had to have two numbers that weren't used yet that could get you to -10  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2016-01-28T16:52:26.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25866" PostTypeId="1" AcceptedAnswerId="25873" CreationDate="2016-01-28T19:33:41.587" Score="3" ViewCount="923" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;The day you are looking for is 500 Days ago,&lt;/p&gt;&#xA;&lt;p&gt;but, tomorrow, it is still 500 Days ago,&lt;/p&gt;&#xA;&lt;p&gt;and two days ago, it was still 500 Days ago.&lt;/p&gt;&#xA;&lt;p&gt;That Day was happy for some, but for some it wasn't.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;When am I?&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Note:  You do not need to involve the date and time this puzzle was posted.&lt;/p&gt;&#xA;" OwnerUserId="16322" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-01-28T21:00:54.423" Title="That day that was 500 Days ago" Tags="|riddle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25867" PostTypeId="2" ParentId="25847" CreationDate="2016-01-28T19:36:13.363" Score="1" Body="&lt;p&gt;I think you are a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; photon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;your brothers and sisters are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; electrons and protons&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;they call you light since &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you are actually light&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18645" LastActivityDate="2016-01-28T19:36:13.363" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25868" PostTypeId="2" ParentId="25866" CreationDate="2016-01-28T19:39:56.677" Score="9" Body="&lt;p&gt;I believe the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; July 17, 2009&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; July 17, 2009 was the release date of the movie &lt;a href=&quot;https://en.wikipedia.org/wiki/500_Days_of_Summer&quot;&gt;500 Days of Summer&lt;/a&gt;. The title of the movie remains &quot;500 Days of Summer&quot; regardless of what day it currently is; while the events of the movie were happy for the character Summer (the main female character/romantic interest), they weren't happy for the character Tom (the male protagonist). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18729" LastActivityDate="2016-01-28T19:39:56.677" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25869" PostTypeId="2" ParentId="25786" CreationDate="2016-01-28T19:51:53.167" Score="4" Body="&lt;p&gt;Figured I might as well post my full resolution here. I will use a lot of @user3294068's reasoning here, so feel free to read on the initial part of his answer, with the mirror and virtual image thingies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wuot1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wuot1.png&quot; alt=&quot;Mirror interpretation of bounces&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the initial triangle has side length &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;. We will use a Cartesian coordinate system in which the initial vertex &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; is the origin. We will say the level &lt;span class=&quot;math-container&quot;&gt;$l$&lt;/span&gt; triangle is the equilateral triangle with side length &lt;span class=&quot;math-container&quot;&gt;$l$&lt;/span&gt; that contains the initial vertex &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;, so for instance the big triangle in the image below is the level &lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt; triangle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/egKYA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/egKYA.png&quot; alt=&quot;The level 3 triangle&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, an important observation: a corner is hit in the top side of a level &lt;span class=&quot;math-container&quot;&gt;$l\geq 2$&lt;/span&gt; triangle if and only if the ball crosses &lt;span class=&quot;math-container&quot;&gt;$2l-3$&lt;/span&gt; black lines; in other words, if and only if it bounces &lt;span class=&quot;math-container&quot;&gt;$2l-3$&lt;/span&gt; times. Thus, for the ball to reach a corner after bouncing &lt;span class=&quot;math-container&quot;&gt;$20160127$&lt;/span&gt; times, the corner must lie on the top side of the level &lt;span class=&quot;math-container&quot;&gt;$n=10080065$&lt;/span&gt; triangle. We will hereafter omit ‘the top side of’ for brevity and readability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, because &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; is odd, any corner on the level &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; triangle will have &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;-coordinate in &lt;span class=&quot;math-container&quot;&gt;$\mathbb{Z}+\frac12$&lt;/span&gt;. Hence, any such corner that is hit belongs to the line &lt;span class=&quot;math-container&quot;&gt;$\gamma$&lt;/span&gt; given by &lt;span class=&quot;math-container&quot;&gt;$y=a \cdot x$&lt;/span&gt;, where&lt;span class=&quot;math-container&quot;&gt;$$a = \frac{n\sqrt{3}}{2}\cdot\frac{1}{p+\frac12}$$&lt;/span&gt;for some &lt;span class=&quot;math-container&quot;&gt;$p \in \mathbb{Z}$&lt;/span&gt; satisfying &lt;span class=&quot;math-container&quot;&gt;$p + \frac12 \leq \frac{n}{2}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will make an analysis for the corners with positive &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;-coordinate without loss of generality; at the end, it suffices to double the result obtained. Notice that &lt;span class=&quot;math-container&quot;&gt;$x=0$&lt;/span&gt; is always obscured except for the level &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a corner to be hit, it must not be obscured by some corner lying on a lower level triangle. If it &lt;em&gt;is&lt;/em&gt; obscured, then there must be some integer &lt;span class=&quot;math-container&quot;&gt;$m &amp;lt; n$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$q \leq \frac{m}{2}$&lt;/span&gt; with &lt;span class=&quot;math-container&quot;&gt;$(q,m) \in \gamma$&lt;/span&gt;. Here, &lt;span class=&quot;math-container&quot;&gt;$q \in \mathbb{Z}$&lt;/span&gt; if &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; is even, and &lt;span class=&quot;math-container&quot;&gt;$q \in \mathbb{Z}+\frac12$&lt;/span&gt; if &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; is odd. Thus, if a corner is obscured the following equality must hold:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$m\cdot\left (p+\frac12\right)=n\cdot q \iff m(2p+1) = n\cdot 2q\tag{1}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that regardless of the situation, &lt;span class=&quot;math-container&quot;&gt;$2q \in \Bbb Z$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Claim&lt;/strong&gt;: Let &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; be a corner on a level &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; triangle, where &lt;span class=&quot;math-container&quot;&gt;$n&amp;gt;2$&lt;/span&gt; is odd. Let &lt;span class=&quot;math-container&quot;&gt;$p+\frac12\in \Bbb Z +\frac12$&lt;/span&gt; be &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt;’s &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;-coordinate. Then &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; is obscured if and only if &lt;span class=&quot;math-container&quot;&gt;$2p+1$&lt;/span&gt; contains a factor of &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof&lt;/em&gt;:&#xA;Consider equality &lt;span class=&quot;math-container&quot;&gt;$(1)$&lt;/span&gt; in terms of prime decomposition: if &lt;span class=&quot;math-container&quot;&gt;$2p+1$&lt;/span&gt; did not contain any of &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;'s prime factors, then they would all show up on &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt;'s prime factors, which would contradict &lt;span class=&quot;math-container&quot;&gt;$m&amp;lt;n$&lt;/span&gt;.&#xA;It follows that whenever the equality holds, &lt;span class=&quot;math-container&quot;&gt;$2p+1$&lt;/span&gt;, contains a factor of &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose &lt;span class=&quot;math-container&quot;&gt;$2p+1 = c\cdot f$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$f\geqslant 3$&lt;/span&gt; divides &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;.&#xA;Observe that &lt;span class=&quot;math-container&quot;&gt;$2p+1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; are odd, so the assumption &lt;span class=&quot;math-container&quot;&gt;$f\geqslant 3$&lt;/span&gt; is justified.&#xA;We consider cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; is even, then &lt;span class=&quot;math-container&quot;&gt;$q\in \Bbb Z$&lt;/span&gt; and we pick &lt;span class=&quot;math-container&quot;&gt;$m = \frac {2n}f$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$q = c$&lt;/span&gt;.&#xA;Because &lt;span class=&quot;math-container&quot;&gt;$f\geqslant 3$&lt;/span&gt;, it holds that &lt;span class=&quot;math-container&quot;&gt;$m&amp;lt; n$&lt;/span&gt;.&#xA;With this in mind, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$q \leqslant m/2 \iff c \leqslant n/f\iff 2p+1 = cf \leqslant n,$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is true by hypothesis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; is odd, then &lt;span class=&quot;math-container&quot;&gt;$q\in \Bbb Z+\frac12$&lt;/span&gt; and we pick &lt;span class=&quot;math-container&quot;&gt;$m = \frac nf$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$2q = c$&lt;/span&gt;.&#xA;Once again, &lt;span class=&quot;math-container&quot;&gt;$f\geqslant 3$&lt;/span&gt;, so &lt;span class=&quot;math-container&quot;&gt;$m&amp;lt; n$&lt;/span&gt;.&#xA;Finally,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$q \leqslant m/2 \iff 2q \leqslant m \iff c \leqslant n/f \iff 2p+1 = cf \leqslant n,$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which once again is true by hypothesis and concludes the proof. &lt;span class=&quot;math-container&quot;&gt;$\square$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Using the claim it is easy to solve the problem.&#xA;Of course, we are only interested in corners of type &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;;&#xA;these corners have &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;-coordinates of the form &lt;span class=&quot;math-container&quot;&gt;$x =\frac12 + 1 + 3j$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$j \in \mathbb{Z}$&lt;/span&gt;.&#xA;Remember it must be &lt;span class=&quot;math-container&quot;&gt;$x \leq \frac{n}{2}$&lt;/span&gt;, or &lt;span class=&quot;math-container&quot;&gt;$2x \leq n$&lt;/span&gt;. We are thus interested in finding all numbers of the form &lt;span class=&quot;math-container&quot;&gt;$6j+3$&lt;/span&gt; that are &lt;span class=&quot;math-container&quot;&gt;$\leq n$&lt;/span&gt; and that are coprime to &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A computer search can solve this in no time, and finds &lt;span class=&quot;math-container&quot;&gt;$1264940$&lt;/span&gt; solutions, so that the total is twice that much, or &lt;span class=&quot;math-container&quot;&gt;$2529880$&lt;/span&gt;.&#xA;Using modular arithmethic and inclusion-exclusion, it’s actually not too hard to do it by hand;&#xA;I’ve done it and got the same result.&lt;/p&gt;&#xA;" OwnerUserId="16478" LastEditorUserId="16478" LastEditDate="2020-05-23T17:22:58.743" LastActivityDate="2020-05-23T17:22:58.743" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25870" PostTypeId="1" AcceptedAnswerId="25871" CreationDate="2016-01-28T19:57:58.547" Score="4" ViewCount="291" Body="&lt;p&gt;This riddle includes a small tweak to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/25856/colored-pills-madness&quot; title=&quot;part 1&quot;&gt;part 1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't a duplicate. This riddle has one less pill. Because of this, the answer already provided wouldn't work here. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You have two bags of similar (weight, texture) solid pills. One bag contains 13 red pills and the other contains 13 green pills. Today, tomorrow, and the day after tomorrow you must select and consume a valid pill set from the following options:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;[3 green pills and 8 red pill] &lt;/li&gt;&#xA;&lt;li&gt;[3 green pills and 4 red pill]&lt;/li&gt;&#xA;&lt;li&gt;[4 green pills and 3 red pill] &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Although you are blind, your colorblind friend is there to assist you. Neither of you know which bag contains which colored pills. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Side effects (Upon you, if and when you are the pill taker): &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;When you take an invalid pill set you die.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Side effects (Upon your friend, if and when he is the pill taker):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Concurrently taking 3 red pills today will instantly blind your friend forever.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Consecutively taking 2 red pills today will instantly blind your friend for 24 hours. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Taking 1 red pill today will instantly remove a positive memory from your friend for 48 hours.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Concurrently taking 3 green pills today will instantly fix your friend’s color blindness forever.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Consecutively taking 2 green pills today will temporarily fix your friend’s color blindness, tomorrow.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Taking 1 green pill today will temporarily fix your friend’s color blindness, the day after tomorrow.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Taking any number of green pill within a minute after taking any number of red pills will instantly undo the red pill’s side effects. Green pill side effects will still apply.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Taking any number of red pill within a minute after taking any number of green pills will instantly undo the green pill’s side effects. Red pill side effects will still apply.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How can you insure the cure of your friend’s colorblindness indefinitely and your survival?&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="18340" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2016-01-28T22:17:01.273" Title="Colored Pills True MADNESS -V2-" Tags="|riddle|logical-deduction|" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="25871" PostTypeId="2" ParentId="25870" CreationDate="2016-01-28T20:22:58.513" Score="4" Body="&lt;p&gt;Solution with a 50% chance of the friend losing one positive memory for a bit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have your friend label the bags A and B. Get them to take a pill from A and then one from B. If they suddenly remember a positive memory, A is red and B is green. Otherwise, A is green and B is red. Have your friend label the bags. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have them take 3 green pills, then you take 3 green and 4 red each day, and you're golden.&lt;/p&gt;&#xA;" OwnerUserId="5169" LastEditorUserId="5169" LastEditDate="2016-01-28T21:03:36.043" LastActivityDate="2016-01-28T21:03:36.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25873" PostTypeId="2" ParentId="25866" CreationDate="2016-01-28T20:50:41.343" Score="5" Body="&lt;p&gt;Could it be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; June 6, 1944, sometimes known as D-Day, D being the Roman numeral for 500?&lt;br&gt;&#xA;&lt;br&gt;&#xA; I'm not sure you could say it was &lt;em&gt;happy&lt;/em&gt; for some, though it certainly contributed to an Allied victory, which itself was very happy for many.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The Nazis definitely did a big frowny face because of D-Day, that's for sure.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The &lt;em&gt;event&lt;/em&gt; will always have occurred on that specific date - calling it &quot;500 Day[s ago]&quot; is just a wordplay trick.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The only part that doesn't quite work is the word &quot;ago&quot;, as substituting &quot;D&quot; in for &quot;500&quot; doesn't make the sentence make any grammatical sense.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2016-01-28T21:00:54.423" LastActivityDate="2016-01-28T21:00:54.423" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25874" PostTypeId="2" ParentId="25847" CreationDate="2016-01-28T21:04:50.760" Score="2" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a postage stamp.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;They pick me Up&lt;br&gt;&#xA;And Wonder&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are many wonderful stamp designs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They call me Light&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A stamp barely weighs anything.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have Brothers and Sisters&lt;br&gt;&#xA;Close, so Close&lt;br&gt;&#xA;Together we Serve&lt;br&gt;&#xA;That from which we were Born&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stamps can have different denominations. You can affix multiple stamps to an envelope and add up their values to pay for the postage.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ah, I am Free&lt;br&gt;&#xA;Yet Fastened&lt;br&gt;&#xA;And when I am Unfastened&lt;br&gt;&#xA;I can no longer Fulfill&lt;br&gt;&#xA;My Purpose &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A stamp is not affixed to anything when purchased. It has to be adhered to an envelope or a mail package to fulfill its purpose.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am used for Travelling&lt;br&gt;&#xA;But I am no Burden&lt;br&gt;&#xA;I am through which It is done &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stamps are used to pay for the cost to mail something to somewhere. Without them nothing can be shipped.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18417" LastActivityDate="2016-01-28T21:04:50.760" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25876" PostTypeId="1" AcceptedAnswerId="25877" CreationDate="2016-01-28T21:11:10.053" Score="3" ViewCount="2043" Body="&lt;p&gt;You are blindfolded and 10 coins are placed in front of you on table. You are allowed to touch the coins, but can’t tell which way up they are by feel. You are told that there are 5 coins head up, and 5 coins tails up but not which ones are which. How do you make two piles of coins each with the same number of heads up? You can flip the coins any number of times.&lt;/p&gt;&#xA;" OwnerUserId="18664" LastActivityDate="2016-01-28T21:16:17.920" Title="Place 10 Coins Heads Up Blindfolded" Tags="|probability|" AnswerCount="2" CommentCount="0" ClosedDate="2016-01-28T23:25:27.933" ContentLicense="CC BY-SA 3.0" />
  <row Id="25877" PostTypeId="2" ParentId="25876" CreationDate="2016-01-28T21:15:29.027" Score="7" Body="&lt;p&gt;Divide them into two groups of 5, then flip all the coins in one group. The group you flip had n heads in it, and the other group had 5-n. When you flip the coins, the first group now has 5-n heads as well.&lt;/p&gt;&#xA;" OwnerUserId="5169" LastActivityDate="2016-01-28T21:15:29.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25878" PostTypeId="2" ParentId="25876" CreationDate="2016-01-28T21:16:17.920" Score="1" Body="&lt;p&gt;Create two piles with 0 coins each. Each pile contains no heads and no tails&lt;/p&gt;&#xA;" OwnerUserId="18103" LastActivityDate="2016-01-28T21:16:17.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25879" PostTypeId="1" CreationDate="2016-01-28T21:28:50.527" Score="-2" ViewCount="320" Body="&lt;p&gt;Here is a couple of birthday puzzles based some time in the future:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Puzzle 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Your birthday was yesterday at exactly 12pm (noon). Today is 12pm the next day. &#xA;It is now 365 days till your next birthday. &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How many hours until your next birthday from 12pm yesterday and why?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Puzzle 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Your birthday was today at 11:59am. &#xA;It is now 364 days until your birthday.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What day this year is this describing and why? (Answer in dd/mm/yyyy format)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18250" LastActivityDate="2016-01-29T11:05:56.157" Title="The birthday puzzles" Tags="|mathematics|calculation-puzzle|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25880" PostTypeId="2" ParentId="25847" CreationDate="2016-01-28T22:01:42.677" Score="3" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a wing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Self-explanatory :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Very well, explaining anyway!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They pick you up and Wonder, they call you light &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a bird picks up its wings, they are light, and flight is something to wonder at...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You have close brothers and sisters, you serve that from which you were born&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; wings usually come in pairs (sometimes more than one pair, eg. insects) and birds are born with them&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are free&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; as a bird&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet fastened and when you are unfastened&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; poor bird&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can no longer fulfill your purpose &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; poor, poor bird&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are used for travelling but &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; don't make me spell that out&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are no burden &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; whereas these spoilers are getting tiresome&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are through which it is done &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; and I'm done, whew.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18774" LastEditorUserId="18776" LastEditDate="2016-01-28T22:33:10.083" LastActivityDate="2016-01-28T22:33:10.083" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25881" PostTypeId="2" ParentId="25879" CreationDate="2016-01-28T22:01:59.670" Score="2" Body="&lt;p&gt;&lt;strong&gt;Puzzle 1 solution&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;From 12:00 noon on your birthday yesterday, it will be $366*24=8,784$ hours until your next birthday.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There's a leap year February 29 between this birthday and your next one, giving you an extra day between birthdays.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17167" LastActivityDate="2016-01-28T22:01:59.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25882" PostTypeId="1" AcceptedAnswerId="25883" CreationDate="2016-01-28T22:25:54.923" Score="5" ViewCount="364" Body="&lt;p&gt;(First time posting, let me know if this isn't right. All things in this post except this are relevant, so don't edit.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recently my buddy George texted me this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;WQ RAR I KE Q V UKE 0D3 WQ U2 CD H D3 KE 76S 84 CN8 :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I asked him what he meant by that. He said to go post it on Puzzling.SE. So here we are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then asked how to write a good challenge if I don't know what it's even about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He then sent me this: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The key is keep it clear and let them know what you need&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That message sounds strange and vague.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly he's on to something, but what?&lt;/p&gt;&#xA;" OwnerUserId="18770" LastActivityDate="2016-01-28T22:38:21.967" Title="George is acting strange" Tags="|cipher|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25883" PostTypeId="2" ParentId="25882" CreationDate="2016-01-28T22:38:21.967" Score="5" Body="&lt;p&gt;The message is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Winner Is You!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The major hint was&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The smiley at the end of the encrypted message: it's a &lt;a href=&quot;http://www.enisoc.com/smile.php&quot;&gt;smile cipher&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So we decode the message using the given key, which is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; literally &lt;code&gt;keep it clear and let them know what you need&quot;.&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17147" LastActivityDate="2016-01-28T22:38:21.967" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25884" PostTypeId="1" AcceptedAnswerId="25885" CreationDate="2016-01-28T23:21:48.317" Score="2" ViewCount="297" Body="&lt;p&gt;I was just walking around with my friends Derek, Alice, and Peter, when suddely, WE WERE KIDNAPPED!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We were driven away in a van to a secret hideout, where we met our kidnapper. He calls himself Robert.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Now listen closely,&quot; he said. &quot;I'll let you guys go if you can guess your number. My number is 78.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We agreed, since we really had no other choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;All right. Derek, you go first.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Derek seemed scared that the kidnapper knew his name, but he calmed down, and guessed 43.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;That's right! You can go. Alice?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alice asked how much time we had. &quot;All the time in the world, my dear.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alice thought about it for a few minutes, and guessed 30.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Right again! You can go too. Peter, what's your guess?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Peter sat there and stammered. &quot;Um, 17?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Wrong! You thought your numbers would be in a pattern, but they aren't! There's a code, something the other two must've realized. Guards! Take him to the dungeon.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The guards hauled him off to the dungeon. &quot;His number was 64. Now, ASCIIThenANSI, what's your number?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I only get one shot before I'm thrown in the dungeon, and you gotta help me!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What's my number, and why?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="18770" LastEditorUserId="22922" LastEditDate="2016-09-01T10:32:17.847" LastActivityDate="2016-09-01T10:32:17.847" Title="The Number Kidnapper" Tags="|mathematics|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25885" PostTypeId="2" ParentId="25884" CreationDate="2016-01-28T23:29:03.517" Score="9" Body="&lt;p&gt;Your number is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 131&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The numbers are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sums of each letter's position in the alphabet&lt;br&gt;&#xA;&lt;br&gt;&#xA; That is:&lt;br&gt;&#xA; ROBERT = R + O + B + E + R + T = 18 + 15 + 2 + 5 + 18 + 20 = 78&lt;br&gt;&#xA; DEREK = D + E + R + E + K = 4 + 5 + 18 + 5 + 11 = 43&lt;br&gt;&#xA; ALICE = A + L + I + C + E = 1 + 12 + 9 + 3 + 5 = 30&lt;br&gt;&#xA; PETER = P + E + T + E + R = 16 + 5 + 20 + 5 + 18 = 64&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so your number must be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A + S + C + I + I + T + H + E + N + A + N + S + I =&lt;br&gt;&#xA; 1 + 19 + 3 + 9 + 9 + 20 + 8 + 5 + 14 + 1 + 14 + 19 + 9 = 131&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17147" LastEditorUserId="17147" LastEditDate="2016-01-28T23:38:41.400" LastActivityDate="2016-01-28T23:38:41.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25888" PostTypeId="2" ParentId="25818" CreationDate="2016-01-29T02:06:24.983" Score="3" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a microphone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I am loved by many,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many would find a microphone useful.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;even if sometimes people are angry to me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sometimes angry protesters use microphones to make their opinions heard.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can find me in different places,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Microphones are used in many different places.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yet you won't always find me stable.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Feedback problems can occur.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes you must use your talent,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can sing and play instruments into a microphone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;sometimes you need wisdom to be my best friend.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Or you can give intelligent and interesting talks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was born before NATO's first SACEUR is appointed,&lt;br&gt;&#xA;yet I am not old fashioned.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The condenser microphone and the dynamic microphone commonly used nowadays were invented in the early 20th century. NATO's first SACEUR was appointed in 1951.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can tear me apart,&lt;br&gt;&#xA;but then you will find radius and patella.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can remove the cover from a microphone. The microphone itself looks like a radius while the cover looks like a patella.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18417" LastActivityDate="2016-01-29T02:06:24.983" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25889" PostTypeId="2" ParentId="25832" CreationDate="2016-01-29T04:26:14.903" Score="6" Body="&lt;p&gt;Let A and B be 4-sided dice with the below labels, and similarly for the &lt;a href=&quot;http://www.shapeways.com/product/JF6RYWM66/d5-jack-blank?li=shop-results&amp;amp;optionId=40685693&quot;&gt;5-sided dice&lt;/a&gt; C and D:&#xA;$$&#xA;\begin{array}{ll}&#xA;A = [1,2,3,4] &amp;amp; B=[1,6,11,16]\\&#xA;C=[0,1,2,3,4] &amp;amp; D=[0,4,8,12,16]&#xA;\end{array}&#xA;$$&#xA;Notice that adding the rolls of A and D produces a random number between 1 and 20, so that A and D together simulate a d20, or A + D = d20. Similarly, B + C = d20, so&#xA;$$&#xA;\text{d20 }+\text{d20}=(A+D) + (B+C)&#xA;$$&#xA;To get a different labeling, let's shuffle these factors around:&#xA;$$&#xA;\text{d20 }+\text{d20}=(A+C) + (B+D)&#xA;$$&#xA;Rolling B and D together produces a random number in the list&#xA;$$&#xA;E=[1,6,11,16,5,10,15,20,9,14,19,24,13,18,23,28,17,22,27,32]&#xA;$$&#xA;so B + D is equal to a 20-sided die E with these labels. Similarly, A + C is equal to the 20-sided die F featuring the labels&#xA;$$&#xA;F=[1,2,3,4,2,3,4,5,3,4,5,6,4,5,6,7,5,6,7,8]&#xA;$$&#xA;This shows that d20 + d20 = E + F, so E and F are labelings that work.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This same method shows that there exist pairs of alternative $n$-sided dice which simulate regular $n$-sided dice, as long as $n$ is composite. The problem is impossible when $n$ is prime, which can be shown using abstract algebra by representing dice as polynomials.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2016-01-29T04:58:18.307" LastActivityDate="2016-01-29T04:58:18.307" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25890" PostTypeId="2" ParentId="24626" CreationDate="2016-01-29T06:22:02.823" Score="6" Body="&lt;p&gt;I haven't fully worked it out, but the spell appears to be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Elminster's Elemental Embodiment, Level Eight, Transmutation&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the solution's technique appears to be related to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sliding rows and columns &quot;South&quot; and &quot;East&quot; (wrapping around letters that go off the end).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example, the following image shows a partially corrected (by trial and error) solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/1vBGh.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1vBGh.png&quot; alt=&quot;partial solution&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In it, I've just:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; shifted columns south by &lt;code&gt;1, 1, 0, 0, 2, 1, 1, 3, 3, 1, 2, 2, 3, 1, 1, 0&lt;/code&gt; cells respectively. You can see fairly clearly the first three or so lines, as well as the last several (&quot;This effect lasts for up to a minute, or until concentration is lost&quot;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't have more time to work on this, so I'll just dump some final notes here, for others to pick up from:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. The errors in my transcription clearly seem to correspond in some way to the positions of the islands. I.e. it gets more &quot;corrupted&quot; towards the centre where there are more islands.&lt;br&gt;&#xA; 2. However, there's still other issues, such as the missing &quot;e&quot; in &quot;Elemental&quot;, and &quot;t&quot; in &quot;Embodiment&quot;.&lt;br&gt;&#xA; 2. You clearly need to shift some rows East before (and/or after/during) shifting columns South. For example if you shift the 4th row East by 1 space first, then shift the columns as shown above, you can also read most of lines 4-7: &quot;The caster and up to three willing allies...&quot;.&lt;br&gt;&#xA; 3. It's possible that some of the shifting requires shifting only partial rows/columns...&lt;br&gt;&#xA; 4. The amount you shift columns South by, initially seems to correlate with the number of blank spaces that appear on island squares in that given column, but it falls apart further along...  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastActivityDate="2016-01-29T06:22:02.823" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="25891" PostTypeId="2" ParentId="25847" CreationDate="2016-01-29T07:00:27.727" Score="4" Body="&lt;p&gt;Well if you're not a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; wing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;then you're a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; feather.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation is basically the same as my other answer (EDIT: which I'm locked out of, I've messed my account up). But it does fit better!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nice riddle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: fits better because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everything about flight applies equally well to a wing as to a feather.  But we say &quot;light as a feather&quot;. Also, a feather has more &quot;brothers and sisters&quot; than a wing. And &quot;detaching&quot; is less gruesome!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; (by question poster, adding this detail for clarity of other users):  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They pick me Up&lt;br&gt;&#xA;And Wonder&lt;br&gt;&#xA;They call me Light  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We pick feathers up, and wonder about them (children love picking them up). &quot;As light as a feather&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have Brothers and Sisters&lt;br&gt;&#xA;Close, so Close&lt;br&gt;&#xA;Together we Serve&lt;br&gt;&#xA;That from which we were Born  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are lots of feathers, closely fitting together, on a bird's wing. They serve the bird from which they were born (biologically).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ah, I am Free&lt;br&gt;&#xA;Yet Fastened&lt;br&gt;&#xA;And when I am Unfastened&lt;br&gt;&#xA;I can no longer Fulfill&lt;br&gt;&#xA;My Purpose  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Feathers are free as the bird, but they are fastened to the bird. When they fall off, they can no longer serve the bird.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am used for Travelling&lt;br&gt;&#xA;But I am no Burden&lt;br&gt;&#xA;I am through which It is done  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Birds use the feathers for traveling, it is through which the bird travels. And they nearly weigh nothing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18776" LastEditorUserId="10327" LastEditDate="2016-01-29T07:52:17.633" LastActivityDate="2016-01-29T07:52:17.633" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25892" PostTypeId="2" ParentId="25879" CreationDate="2016-01-29T07:08:15.007" Score="0" Body="&lt;p&gt;&lt;strong&gt;Puzzle 2 Solution&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;His/Her birthday is on &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Any date. But there should not be any 29th February between those birthdays.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Reason&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If I celebrate my first birthday in Honolulu and next birthday in Kiritimati then I can say that I waited for 364 days for my second birthday because the difference in their time zones is 1 day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-01-29T11:04:24.257" LastActivityDate="2016-01-29T11:04:24.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25893" PostTypeId="1" AcceptedAnswerId="26073" CreationDate="2016-01-29T09:09:04.770" Score="2" ViewCount="279" Body="&lt;p&gt;After the attention of the first riddle, wanted to post a simpler and distracting riddle. Good luck&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;So sharp that I can execute what you want me to. &#xA;&#xA;Will guide you the way, and when needed I will split your troubles in your head &#xA;&#xA;So valuable, that you always fight for me &#xA;&#xA;Standing at the crossroads, waiting for you to find me&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am (sometimes) above a wizard's head&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;hint2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am a single word with multiple meanings&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18685" LastEditorUserId="18685" LastEditDate="2016-02-02T15:44:38.327" LastActivityDate="2016-02-02T15:57:00.890" Title="What am I by SC v2" Tags="|riddle|word|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25894" PostTypeId="2" ParentId="25879" CreationDate="2016-01-29T10:43:50.647" Score="0" Body="&lt;p&gt;&lt;strong&gt;Puzzle 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 8772 hours.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There's a leap day before your next birthday, so it's 366 days. However, we're counting from noon, which means it's only 12 hours until the next day rather than 24.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Puzzle 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  07/11/2016&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You're located in &lt;a href=&quot;http://www.timeanddate.com/time/change/antarctica/troll&quot; rel=&quot;nofollow&quot;&gt;Troll Station, Antarctica&lt;/a&gt;. By 11:59am, the clocks have already rolled back. You get the benefit of the roll forward (1 hour), but when it's your birthday again (00:00 on 07/11/2017) the clocks won't change until 02:00.&lt;br&gt; Normally at 11:59am it would be 364 days, 12 hours and 1 minute until your next birthday. However, because you're between clocks rolling back, you get 364 days, 11 hours and 1 minute, which we can round down to 364 days!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12462" LastEditorUserId="12462" LastEditDate="2016-01-29T11:05:56.157" LastActivityDate="2016-01-29T11:05:56.157" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25895" PostTypeId="2" ParentId="25893" CreationDate="2016-01-29T11:26:33.540" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Solution &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So sharp that I can execute what you want me to.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; solution is the answer of a question and you execute what solution tells for the problem.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Will guide you the way, and when needed I will split your troubles in your head&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Solution will guide you how you need to execute and it can remove your troubles in your head since you found a solution for a problem.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So valuable, that you always fight for me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you always fight for a solution for your questions/problems/troubles in your life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Standing at the crossroads, waiting for you to find me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are always solutions to problems and they are waiting at the crossroads for you. You may find it or not...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18514" LastActivityDate="2016-01-29T11:26:33.540" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25896" PostTypeId="1" CreationDate="2016-01-29T11:29:45.293" Score="4" ViewCount="26778" Body="&lt;p&gt;I have three water jugs, one with 10 liters another with 5 and another with 6, how do i get 8 liters on the first one?&lt;/p&gt;&#xA;" OwnerUserId="18788" LastEditorUserId="17147" LastEditDate="2016-01-29T11:34:55.260" LastActivityDate="2016-01-30T00:11:36.443" Title="Water Jugs (10, 5 and 6)" Tags="|calculation-puzzle|decanting-problems|" AnswerCount="7" CommentCount="1" FavoriteCount="0" ClosedDate="2016-01-30T12:08:16.933" ContentLicense="CC BY-SA 3.0" />
  <row Id="25897" PostTypeId="2" ParentId="25896" CreationDate="2016-01-29T11:33:41.847" Score="5" Body="&lt;p&gt;Fill 10 liters jug.&lt;br&gt;&#xA;Pour it in the 6 liter jug.&lt;br&gt;&#xA;Pour the remaining 4 liter in the 5 liter jug.&lt;br&gt;&#xA;Fill 10 liters again.&lt;br&gt;&#xA;Empty the 6 liter jug.&lt;br&gt;&#xA;Pour 6 liters from the 10 liter jug in the 6 liter jug.&lt;br&gt;&#xA;Put content of 5 liter jug in 10 liter jug.&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2016-01-29T11:33:41.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25898" PostTypeId="2" ParentId="25896" CreationDate="2016-01-29T11:39:03.247" Score="20" Body="&lt;p&gt;There is no need for the 5 liters jug. You can do it with only the 10 and 6 liters jugs (and in less steps than the alternate solutions).&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;10 0 0&lt;/code&gt; $\leftrightarrow$ Fill 10 liters jug. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;4  6 0&lt;/code&gt; $\leftrightarrow$ Pour it in the 6 liters jug (now 4 liter in 10 liters jug) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;4  0 0&lt;/code&gt; $\leftrightarrow$ Empty 6 liters jug&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;0  4 0&lt;/code&gt; $\leftrightarrow$ Pour the 4 liter from 10 liters jug to the 6 liters jug (now 4 liter in 6 liters jug)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;10 4 0&lt;/code&gt; $\leftrightarrow$ Fill 10 liters jug again&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;8  6 0&lt;/code&gt; $\leftrightarrow$ Pour water from 10 liters jug to 6 liters jug to fill it &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You will have 8 liter in 10 liters jug&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="12795" LastEditDate="2016-01-29T19:12:39.650" LastActivityDate="2016-01-29T19:12:39.650" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25900" PostTypeId="1" AcceptedAnswerId="25902" CreationDate="2016-01-29T11:45:44.250" Score="8" ViewCount="2274" Body="&lt;p&gt;I have three water jugs, one with 10 liters another with 5 and another with 6, how do i get 8 liters on the first one with 10 liters of water available. (so you cannot fill 10 liters jug again)&lt;/p&gt;&#xA;" OwnerUserId="18514" LastActivityDate="2016-01-29T13:50:41.463" Title="Water Jugs (10, 5 and 6) -v2-" Tags="|decanting-problems|" AnswerCount="2" CommentCount="4" ClosedDate="2016-01-30T15:25:02.450" ContentLicense="CC BY-SA 3.0" />
  <row Id="25901" PostTypeId="2" ParentId="25900" CreationDate="2016-01-29T12:03:07.943" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Legend: 10l jug / 5l jug / 6l jug &lt;br&gt;&#xA; 1. Start with 10l in the 10l jug =&gt; 10/0/0 &lt;br&gt;&#xA; 2. Fill the 6l jug and fill the rest in the 5l jug  =&gt; 0/4/6 &lt;br&gt;&#xA; 3. Swap th 4l into the 10l jug and fill with the 6l jug the 5l jug =&gt; 4/5/1 &lt;br&gt;&#xA; 4. Swap the 5l jug into the 10l jug and the remaining 1l from 6l jug to 5l jug =&gt; 9/1/0 &lt;br&gt;&#xA; 5. Fill from th 9l the 6l jug =&gt; 3/1/6 &lt;br&gt;&#xA; 6. FIll the 5l jug from the 6l jug =&gt; 3/5/2 &lt;br&gt;&#xA; 7. Fill the 10l jug with the 5l jug =&gt; 8/0/2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14461" LastActivityDate="2016-01-29T12:03:07.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25902" PostTypeId="2" ParentId="25900" CreationDate="2016-01-29T12:07:04.997" Score="9" Body="&lt;pre&gt;&lt;code&gt;10L jug - 5L jug  - 6L jug&#xA;10      -    0    -    0  &#xA;4       -    0    -    6 &#xA;4       -    5    -    1&#xA;9       -    0    -    1&#xA;9       -    1    -    0&#xA;3       -    1    -    6&#xA;3       -    5    -    2&#xA;8       -    0    -    2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="18685" LastEditorUserId="12593" LastEditDate="2016-01-29T13:50:41.463" LastActivityDate="2016-01-29T13:50:41.463" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25903" PostTypeId="1" AcceptedAnswerId="25904" CreationDate="2016-01-29T13:25:55.693" Score="12" ViewCount="1247" Body="&lt;p&gt;Guess number 16 in this sequence:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;WS&lt;/li&gt;&#xA;&lt;li&gt;EL&lt;/li&gt;&#xA;&lt;li&gt;JRC&lt;/li&gt;&#xA;&lt;li&gt;AA&lt;/li&gt;&#xA;&lt;li&gt;ME&lt;/li&gt;&#xA;&lt;li&gt;MB&lt;/li&gt;&#xA;&lt;li&gt;VS&lt;/li&gt;&#xA;&lt;li&gt;MT&lt;/li&gt;&#xA;&lt;li&gt;TP&lt;/li&gt;&#xA;&lt;li&gt;BS&lt;/li&gt;&#xA;&lt;li&gt;BF&lt;/li&gt;&#xA;&lt;li&gt;AK&lt;/li&gt;&#xA;&lt;li&gt;GK&lt;/li&gt;&#xA;&lt;li&gt;VK&lt;/li&gt;&#xA;&lt;li&gt;VA&lt;/li&gt;&#xA;&lt;li&gt;??&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="18675" LastEditorUserId="18675" LastEditDate="2016-02-07T17:50:58.327" LastActivityDate="2016-02-07T17:50:58.327" Title="guess the missing letters in sequence" Tags="|letter-sequence|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25904" PostTypeId="2" ParentId="25903" CreationDate="2016-01-29T13:40:14.727" Score="10" Body="&lt;p&gt;The next letters are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MC, for &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnus_Carlsen&quot;&gt;Magnus Carlsen&lt;/a&gt;, most recent &lt;a href=&quot;https://en.wikipedia.org/wiki/World_Chess_Championship&quot;&gt;World Chess Champion&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's a breakdown just to show that the others on the list work:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. Wilhelm Steinitz&lt;br&gt;&#xA; 2. Emanuel Lasker&lt;br&gt;&#xA; 3. José Raúl Capablanca&lt;br&gt;&#xA; 4. Alexander Alekhine&lt;br&gt;&#xA; 5. Max Euwe&lt;br&gt;&#xA; 6. Mikhail Botvinnik&lt;br&gt;&#xA; 7. Vasily Smyslov&lt;br&gt;&#xA; 8. Mikhail Tal&lt;br&gt;&#xA; 9. Tigran Petrosian&lt;br&gt;&#xA; 10. Boris Spassky&lt;br&gt;&#xA; 11. Bobby Fischer&lt;br&gt;&#xA; 12. Anatoly Karpov&lt;br&gt;&#xA; 13. Garry Kasparov&lt;br&gt;&#xA; 14. Vladimir Kramnik&lt;br&gt;&#xA; 15. Viswanathan Anand&lt;br&gt;&#xA; 16. Magnus Carlsen &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2016-01-29T16:01:19.320" LastActivityDate="2016-01-29T16:01:19.320" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25905" PostTypeId="2" ParentId="25896" CreationDate="2016-01-29T13:50:21.913" Score="9" Body="&lt;p&gt;Here's a solution that takes more steps than the others, but the problem doesn't say we're trying to minimize steps ;) &amp;nbsp; It plays on the idea that the 5- and 6-liter jugs allow you to create 1-liter increments as many times as you want.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Fill the 6-liter jug.&lt;/li&gt;&#xA;&lt;li&gt;Pour from the 6-liter jug into the 5-liter jug. You have 1 liter remaining in the 6-liter jug.&lt;/li&gt;&#xA;&lt;li&gt;Pour the 1 liter from the 6-liter jug into the 10-liter jug and empty the 5-liter jug.&lt;/li&gt;&#xA;&lt;li&gt;Repeat steps 1–3 7 more times.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It's clunky but simple.&lt;/p&gt;&#xA;" OwnerUserId="16275" LastActivityDate="2016-01-29T13:50:21.913" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25906" PostTypeId="1" AcceptedAnswerId="25929" CreationDate="2016-01-29T14:07:08.107" Score="11" ViewCount="410" Body="&lt;pre&gt;&lt;code&gt;coal    60 &#xA;&#xA;mine    63 &#xA;&#xA;code    39 &#xA;&#xA;lame    41 &#xA;&#xA;gain    23 &#xA;&#xA;gala    22 &#xA;&#xA;main    ?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Find the missing value for &quot;main&quot;?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hint: there are hidden abbreviations in the words!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18514" LastEditorUserId="18514" LastEditDate="2016-01-29T18:16:14.503" LastActivityDate="2016-01-29T18:43:13.087" Title="Find the missing value for &quot;main&quot;" Tags="|number-sequence|letter-sequence|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25907" PostTypeId="1" AcceptedAnswerId="25920" CreationDate="2016-01-29T14:10:14.640" Score="5" ViewCount="401" Body="&lt;p&gt;I got this message from an old friend.&lt;br/&gt;&#xA;He used to rob banks, and i helped him, but the last time he got arrested.&lt;br/&gt;&#xA;The interesting part is that the cops never found the money!&lt;br/&gt;&#xA;Now he sends me this message..&#xA;&lt;br/&gt;He probably knew the police would read it before sending it to me and made it harder to understand.&#xA;&lt;br/&gt;&#xA;&lt;br/&gt;I can't seem to figure it out, can you? &#xA;&lt;br/&gt; PS you will get a fair share. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You have to find my old treasure.&lt;br/&gt;&#xA;  Are you in the mood for a riddle? &lt;br/&gt;&#xA;  Very far from the tree in my yard.&lt;br/&gt;&#xA;  Close to my newly painted house it will be.&lt;br/&gt;&#xA;  When you are stepping in my garage.&lt;br/&gt;&#xA;  You know that your basement is bigger. &lt;br/&gt;&#xA;  See the bewildered backyard behind my house.&lt;br/&gt;&#xA;  This is the end goodluck!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18559" LastEditorUserId="11876" LastEditDate="2016-01-29T14:46:29.423" LastActivityDate="2016-01-29T17:12:44.740" Title="a message from an old friend" Tags="|steganography|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25908" PostTypeId="2" ParentId="25896" CreationDate="2016-01-29T14:30:25.120" Score="11" Body="&lt;p&gt;Heres a solution without needing to refil and starting with only 10L.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jugs are arranged in the following order&#xA;10L 5L 6L&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with just the 10L jug full&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;10L - 0L --- 0L&lt;br /&gt; &#xA;  4   ---- 0 ---- 6&lt;br /&gt; &#xA;  4  ---- 5 ---- 1&lt;br /&gt; &#xA;  9  ---- 0 ---- 1&lt;br /&gt; &#xA;  9 ----  1 ---- 0&lt;br /&gt; &#xA;  3 ----  1 ---- 6&lt;br /&gt; &#xA;  3 ----  5 ---- 2&lt;br /&gt; &#xA;  &lt;strong&gt;8&lt;/strong&gt; ----  0 ---- 2&lt;br /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18791" LastEditorUserId="18791" LastEditDate="2016-01-29T16:32:02.723" LastActivityDate="2016-01-29T16:32:02.723" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25909" PostTypeId="2" ParentId="25742" CreationDate="2016-01-29T14:44:07.647" Score="5" Body="&lt;p&gt;Well. Rome again is it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;First victim:&lt;/strong&gt;&lt;br&gt;&#xA; &lt;em&gt;victim clues:&lt;/em&gt; pool, sky, clue, &quot;loyalty&quot; angle&lt;br&gt;&#xA; &lt;em&gt;suspect clue:&lt;/em&gt; century&lt;br&gt;&#xA; pools and skies are blue, blue is a color &lt;a href=&quot;https://resources.oncourse.iu.edu/access/content/user/rreagan/Filemanager_Public_Files/meaningofcolors.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;associated with loyalty&lt;/a&gt; and &quot;blue&quot; impacted by the letter C (Roman numeral for 100, the number of years the suspect has been waiting) gives us &quot;clue&quot;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Second victim:&lt;/strong&gt;&lt;br&gt;&#xA; &lt;em&gt;victim clues:&lt;/em&gt; roses, blood, &quot;love affair&quot; angle&lt;br&gt;&#xA; &lt;em&gt;suspect clue:&lt;/em&gt; &lt;em&gt;(500) Days of Summer&lt;/em&gt;&lt;br&gt;&#xA; roses and blood (and &lt;a href=&quot;https://resources.oncourse.iu.edu/access/content/user/rreagan/Filemanager_Public_Files/meaningofcolors.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;love&lt;/a&gt;) are red, and &quot;red&quot; impacted by the letter D (Roman numeral for 500, the number of Days of Summer) gives us &quot;de[a]d&quot;. &lt;em&gt;At the start of the &lt;strong&gt;d&lt;/strong&gt;ay that I met you&lt;/em&gt;, indeed.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Third victim:&lt;/strong&gt;&lt;br&gt;&#xA; &lt;em&gt;victim clues:&lt;/em&gt; gold bars, Donovan lyrics, &quot;jealousy&quot; angle&lt;br&gt;&#xA; &lt;em&gt;suspect clues:&lt;/em&gt; &quot;1K of gold&quot;&lt;br&gt;&#xA; gold (and &lt;a href=&quot;https://resources.oncourse.iu.edu/access/content/user/rreagan/Filemanager_Public_Files/meaningofcolors.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;jealousy&lt;/a&gt;) is effectively yellow, and yellow conveniently appears in the title of the song to which the quoted lyrics belong, &lt;em&gt;Mellow Yellow&lt;/em&gt; (quite rightly). &quot;1K&quot; is a very &lt;em&gt;Greek&lt;/em&gt; way of saying what a Roman might call &quot;M&quot;. &quot;Yellow&quot; impacted by the letter M gives us &quot;Mellow&quot;. It seems M was in fact the guy who could calm Yellow down.&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;&lt;em&gt;per OP&lt;/em&gt;&lt;/strong&gt;:&lt;br&gt;&#xA; '&quot;On the start of the day I met you, I'm with you until the end, but I will also show you the end. Nothing will change if I'm still with you&quot; is letter D. Start of (D)ay and with you until the en(D) and nothing will change because D is in the color RED.'&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;&lt;em&gt;and also&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&#xA; 'And for the twins, they are the letter L. Appears in BLUE and and YELLOW. And lastly, Topaz is associated with Blue'  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We've been told at the very beginning of the story that Caesar is being asked to investigate the primary victims.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well, it turns out these are not &quot;primary&quot; because there are other victims; they are &lt;em&gt;primary colors&lt;/em&gt;: blue, red, and yellow. The suspects? Murderous, murderous numbers masquerading as letters.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Special thanks to &lt;a href=&quot;https://puzzling.stackexchange.com/users/4421/gentlepurplerain&quot;&gt;@GentlePurpleRain&lt;/a&gt; for helping me brainstorm this in chat, especially noting the transformation of the color words caused by the Roman numerals. Luckily for him, purple isn't a primary color.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.923" LastActivityDate="2016-02-01T02:09:42.730" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25911" PostTypeId="1" AcceptedAnswerId="25916" CreationDate="2016-01-29T15:01:26.563" Score="4" ViewCount="168" Body="&lt;p&gt;Before me is stride,&lt;br&gt;&#xA;Paraded by pride.&lt;br&gt;&#xA;After is hope,&lt;br&gt;&#xA;Her possible scope.&lt;br&gt;&#xA;I'm but a girl,&lt;br&gt;&#xA;Who's fourth in the whirl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="9802" LastActivityDate="2016-01-29T15:45:01.453" Title="Before me is stride" Tags="|riddle|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25912" PostTypeId="1" AcceptedAnswerId="25915" CreationDate="2016-01-29T15:02:58.830" Score="7" ViewCount="1060" Body="&lt;p&gt;The idea is to place all the yellow keys into the red locks. There is at least one guaranteed way to do this as the locks as created from the keys and then 'blurred' by adding extra gaps. Solutions however are not (yet!) unique - I haven't figured out how to add gaps without getting more than one solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The full white spaces in the red locks take the yellow key bits, the half-red squares are blocked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dOA54.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dOA54.png&quot; alt=&quot;keysandlocks1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example key 1 goes into locks 4 and 8.&lt;/p&gt;&#xA;" OwnerUserId="11497" LastEditorUserId="11497" LastEditDate="2018-06-01T05:50:55.950" LastActivityDate="2018-06-01T05:50:55.950" Title="Key and lock puzzle" Tags="|combinatorics|keys-and-locks|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="25913" PostTypeId="2" ParentId="25912" CreationDate="2016-01-29T15:24:27.277" Score="2" Body="&lt;p&gt;If you organize the keys A-I and the locks 1-9 here is a solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A 8,&#xA; B 1,&#xA; C 5,&#xA; D 2,&#xA; E 4,&#xA; F 7,&#xA; G 3,&#xA; H 6,&#xA; I 9,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16720" LastEditorUserId="16720" LastEditDate="2016-01-29T15:36:56.953" LastActivityDate="2016-01-29T15:36:56.953" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25914" PostTypeId="2" ParentId="25896" CreationDate="2016-01-29T15:30:21.360" Score="12" Body="&lt;p&gt;Using only exactly 8L:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming 10L, 6L, 5L. Bold indicate refill (top-up or complete)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;0, &lt;strong&gt;6&lt;/strong&gt;, 0 (Using 6 new liters)&lt;/li&gt;&#xA;&lt;li&gt;0, 1, 5&lt;/li&gt;&#xA;&lt;li&gt;1, 0, 5&lt;/li&gt;&#xA;&lt;li&gt;1, 5, 0&lt;/li&gt;&#xA;&lt;li&gt;1, &lt;strong&gt;6&lt;/strong&gt;, 0 (Using one new liter)&lt;/li&gt;&#xA;&lt;li&gt;1, 1, 5&lt;/li&gt;&#xA;&lt;li&gt;2, 0, 5&lt;/li&gt;&#xA;&lt;li&gt;2, 5, 0&lt;/li&gt;&#xA;&lt;li&gt;2, &lt;strong&gt;6&lt;/strong&gt;, 0 (Using one new liter)&lt;/li&gt;&#xA;&lt;li&gt;8, 0, 0&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;10 steps, but not a drop of extra water wasted or used.&lt;/p&gt;&#xA;" OwnerUserId="310" LastActivityDate="2016-01-29T15:30:21.360" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25915" PostTypeId="2" ParentId="25912" CreationDate="2016-01-29T15:37:43.903" Score="6" Body="&lt;p&gt;I found two solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I numbered the keys and locks from 1 to 9 and built a simple matrix:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/WOjQX.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WOjQX.png&quot; alt=&quot;Key / Lock matrix&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A &quot;0&quot; means that the key would not fit into the corresponding lock. A &quot;1&quot; means that the key fits. Then, a -1 means that a key would fit, but that the lock is already occupied by another key.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the last row you can see how the deduction works. First (1.) I assumed that K3 fits only into L5 and K5, and (2.) K5 only into L4. This (3.) blocks K1 for L4, so it needs to be put into L8, and so on (4. to 7.). Two keys (K4 and K7) and two locks (L2 and L3) remained, which can be put into either of them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So my solutions would be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. K1-L8, K2-L1, K3-L5, K4-L2, K5-L4, K6-L7, K7-L3, K8-L6, K9-L9&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2. K1-L8, K2-L1, K3-L5, K4-L3, K5-L4, K6-L7, K7-L2, K8-L6, K9-L9&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17334" LastActivityDate="2016-01-29T15:37:43.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25916" PostTypeId="2" ParentId="25911" CreationDate="2016-01-29T15:45:01.453" Score="8" Body="&lt;p&gt;Stride, paraded by pride&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; perhaps a &lt;em&gt;march&lt;/em&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hope, her possible scope&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; seems like &lt;em&gt;may&lt;/em&gt; would fit, an auxiliary verb (and feminine name) that indicates possibility, a wish, or permission.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but a girl, who's fourth in the whirl&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;April&lt;/em&gt; is another feminine name, and the fourth month of the year, falling after &lt;em&gt;March&lt;/em&gt; and before &lt;em&gt;May&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastActivityDate="2016-01-29T15:45:01.453" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25917" PostTypeId="1" CreationDate="2016-01-29T15:52:14.630" Score="6" ViewCount="338" Body="&lt;p&gt;There is a logical relation between these pictures that gather them all together. Can you sort them and guess what picture can be the next one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RcV96.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RcV96.jpg&quot; alt=&quot;1&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/7WgsA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7WgsA.jpg&quot; alt=&quot;2&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/06M65.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/06M65.jpg&quot; alt=&quot;3&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/nILP5.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nILP5.jpg&quot; alt=&quot;4&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/pdtdm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pdtdm.jpg&quot; alt=&quot;5&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/HIGh5.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HIGh5.jpg&quot; alt=&quot;6&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/enWau.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/enWau.jpg&quot; alt=&quot;7&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/JTAcG.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JTAcG.jpg&quot; alt=&quot;8&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this is an innovative and interesting puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spoiler: The next missing person is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Betty_Binns_Fletcher&quot; rel=&quot;nofollow noreferrer&quot;&gt;Betty Binns Fletcher&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18675" LastEditorUserId="18675" LastEditDate="2016-02-01T15:49:29.060" LastActivityDate="2016-02-01T15:49:29.060" Title="Sort the pictures and guess the next one" Tags="|knowledge|visual|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25918" PostTypeId="1" AcceptedAnswerId="25924" CreationDate="2016-01-29T16:02:39.930" Score="7" ViewCount="207" Body="&lt;p&gt;Louis van Gaal has compiled a list of 20 soccer teams, ordered by how good&#xA;he thinks they are, but strictly refuses to share the ranking. &#xA;Alex Ferguson knows the names of the 20 teams, but does not know the ranking. &#xA;Ferguson may mention three arbitrary teams $A,B,C$ to Van Gaal, and Van Gaal will then choose one of the following two options:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Van Gaal may announce to Ferguson which he thinks is the weakest team of the three (&quot;I think that team $X$ is the weakest team among these three teams&quot;).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Van Gaal may announce to Ferguson which he thinks is the strongest team of the three (&quot;I think that team $X$ is the strongest team among these three teams&quot;). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Alex Ferguson may do this as many times as he likes. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt;&#xA;  Determine the largest integer $n$ such that Ferguson can guarantee&#xA;  to find a sequence $T_1, T_2, \ldots, T_n$ of $n$ teams with the &#xA;  property that he knows that Van Gaal thinks that $T_i$ is better &#xA;  than $T_{i+1}$ for $i=1,\ldots,n$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2016-01-29T17:27:33.267" Title="Ranking soccer teams" Tags="|mathematics|strategy|combinatorics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25919" PostTypeId="2" ParentId="25917" CreationDate="2016-01-29T16:27:29.847" Score="6" Body="&lt;p&gt;These are all people who&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; died in the last 3 years. The proper sort order is the order in which they died.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The names of the people are, starting in the top row working left to right,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Frank Bank, Leon Bramlett, Ray Hatton, Leon Bibb, Valery Belousov, Lionel Pickens (aka Chinx), Dale Armstrong, and George Francis.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sorting these by the criterion mentioned above yields&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. Frank Bank (April 13, 2013)&lt;br&gt;&#xA; 2. George Francis (October 22, 2014)&lt;br&gt;&#xA; 3. Dale Armstrong (November 28, 2014)&lt;br&gt;&#xA; 4. Ray Hatton (March 4, 2015)&lt;br&gt;&#xA; 5. Valery Belousov (April 16, 2015)&lt;br&gt;&#xA; 6. Lionel Pickens (May 17, 2015)&lt;br&gt;&#xA; 7. Leon Bramlett (October 19, 2015)&lt;br&gt;&#xA; 8. Leon Bibb (October 23, 2015)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16275" LastActivityDate="2016-01-29T16:27:29.847" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25920" PostTypeId="2" ParentId="25907" CreationDate="2016-01-29T17:12:44.740" Score="6" Body="&lt;p&gt;The message reads:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; find the tree house in your backyard goodluck!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first word on every line from top to bottom reads:&lt;br&gt;&#xA;&lt;br&gt;&#xA; You&lt;br&gt;&#xA; Are&lt;br&gt;&#xA; Very&lt;br&gt;&#xA; Close&lt;br&gt;&#xA; When&lt;br&gt;&#xA; You&lt;br&gt;&#xA; See&lt;br&gt;&#xA; This&lt;br&gt;&#xA;&lt;br&gt;&#xA; Now the first word of the first line is &quot;You&quot;, which has &lt;strong&gt;3&lt;/strong&gt; letters. Count &lt;strong&gt;3&lt;/strong&gt; words to the right to get the word &quot;find&quot;.&lt;br&gt;&#xA; Similarly the first word of the second line &quot;Are&quot; has &lt;strong&gt;3&lt;/strong&gt; letters. Count &lt;strong&gt;3&lt;/strong&gt; words to the right to get the word &quot;the&quot;.&lt;br&gt;&#xA; &quot;Very&quot; on the third line has &lt;strong&gt;4&lt;/strong&gt; letters. Count &lt;strong&gt;4&lt;/strong&gt; words to the right to get the word &quot;tree&quot;.&lt;br&gt;&#xA; Repeat the process for every line and the message can be found.&lt;br&gt;&#xA;&lt;br&gt;&#xA; You have to &lt;strong&gt;find&lt;/strong&gt; my old treasure.&lt;br&gt;&#xA; Are you in &lt;strong&gt;the&lt;/strong&gt; mood for a riddle?&lt;br&gt;&#xA; Very far from the &lt;strong&gt;tree&lt;/strong&gt; in my yard.&lt;br&gt;&#xA; Close to my newly painted &lt;strong&gt;house&lt;/strong&gt; it will be.&lt;br&gt;&#xA; When you are stepping &lt;strong&gt;in&lt;/strong&gt; my garage.&lt;br&gt;&#xA; You know that &lt;strong&gt;your&lt;/strong&gt; basement is bigger.&lt;br&gt;&#xA; See the bewildered &lt;strong&gt;backyard&lt;/strong&gt; behind my house.&lt;br&gt;&#xA; This is the end &lt;strong&gt;goodluck!&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18417" LastActivityDate="2016-01-29T17:12:44.740" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25921" PostTypeId="2" ParentId="25918" CreationDate="2016-01-29T17:18:24.473" Score="3" Body="&lt;p&gt;As a rough start, $n \le 18 $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose in Van Gaal's ranking, $T_1 &amp;gt; T_2 &amp;gt; ... &amp;gt; T_{20}$, and Ferguson asks about every possible $(T_i, T_j, T_k)$ combination. If for every instance of $(T_1, T_2, T_k)$, Van Gaal responds that he believes $T_k$ is the weakest, then Ferguson cannot differentiate between $T_1$ and $T_2$. A similar situation holds at the bottom of the rankings, and so Ferguson cannot distinguish between $T_{19}$ and $T_{20}$.&lt;/p&gt;&#xA;" OwnerUserId="17446" LastActivityDate="2016-01-29T17:18:24.473" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25922" PostTypeId="1" AcceptedAnswerId="25925" CreationDate="2016-01-29T17:19:57.310" Score="78" ViewCount="5554" Body="&lt;p&gt;Alice has a secret polynomial $P$ with positive integer coefficients. When Bob gives Alice a positive integer $n \neq 2016$, Alice replies with the value of $P(n)$. After doing this several times, Bob guesses the value of $P(2016)$; he wins if he is correct and loses if not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can Bob guarantee a win in as few such exchanges as possible?&lt;/p&gt;&#xA;" OwnerUserId="18508" LastEditorUserId="30903" LastEditDate="2018-02-19T20:59:39.957" LastActivityDate="2018-02-19T20:59:39.957" Title="A secret polynomial" Tags="|mathematics|strategy|" AnswerCount="3" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="25923" PostTypeId="2" ParentId="25896" CreationDate="2016-01-29T17:22:24.173" Score="0" Body="&lt;p&gt;Assuming all three are empty and you have a tap:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;10 5 6  Jugs  &#xA; | | |  &#xA; 0 0 6   &#xA; 0 5 1  &#xA; 1 5 0&#xA; 1 0 5&#xA; 1 0 6&#xA; 7 0 0&#xA; 7 0 6&#xA; 7 5 1&#xA; 8 5 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You have 8 liters in the 10L jug and 5 liters in the 5L jug.&lt;/p&gt;&#xA;" OwnerUserId="18117" LastActivityDate="2016-01-29T17:22:24.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25924" PostTypeId="2" ParentId="25918" CreationDate="2016-01-29T17:27:33.267" Score="7" Body="&lt;p&gt;I think the longest sequence Ferguson can get is of length &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$n=10$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First we find that this is an upper bound &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; van Gaal can divide the set of twenty teams into ten pairs (1st with 2nd, 3rd with 4th, etc...) so that whenever Ferguson asks about a particular triplet which contains two teams from the same pair, he can obscure their relationship by saying whether the other team is the strongest/weakest hence preventing Ferguson from distinguishing 1st from 2nd, 3rd from 4th, etc and allowing him to identify a maximum of one from each pair, that is $n \le 10$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On the other hand this is also a lower bound.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Given any pair of teams $a$, $b$ Ferguson can always either determine how this pair compares with every other team or will be able to distinguish who is the stronger between $a$ and $b$ (by comparing $a$, $b$, $c$ for every other team $c$). In the worst case scenario, he will be able to determine ten pairs of teams so he can put the pairs in order but not necessarily the teams within each pair, hence $n \ge 10$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18422" LastActivityDate="2016-01-29T17:27:33.267" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25925" PostTypeId="2" ParentId="25922" CreationDate="2016-01-29T17:30:35.187" Score="112" Body="&lt;p&gt;Bob can get it in&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;2 questions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's how:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;First guess 1. Then guess a power of 10 greater than the response; he can simply &quot;read off&quot; the coefficients. That uniquely determines the polynomial, so he can just calculate $P(2016)$ for his final answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What I mean by the term in quotes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I'll use the polynomial $17x^4 + 4x^3 + 5x^2 + 8$ as an example: $f(1) = 34$, so I choose any power of ten greater than 34. I pick 1000. $f(1000) = 17004005000008$, and you can easily see the coefficients there, separated by zeroes. (Chunk starting from the right in groups of $n$, where $10^n$ is the number you picked.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="11876" LastEditDate="2016-01-29T18:59:11.350" LastActivityDate="2016-01-29T18:59:11.350" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="25927" PostTypeId="2" ParentId="25906" CreationDate="2016-01-29T18:13:23.270" Score="2" Body="&lt;p&gt;So I might be completely off on this, but I think the answer is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;main=20 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My Solution: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;coal - code=al - de (= 21)&lt;br&gt;&#xA;al - de - lame =  - made (= - 20)&lt;br&gt;&#xA; - mine + made = ad - in (= - 43)&lt;br&gt;&#xA;ad - in + gain = adgd (= - 21)&lt;br&gt;&#xA;adga - gala = d - l (= - 43)&lt;br&gt;&#xA;d - l + coal = coad (= 17)&lt;br&gt;&#xA;code - code = a - e (= - 22)&lt;br&gt;&#xA;a - e + lame = mala (= 19)&lt;br&gt;&#xA;mala - gala = m - g (= - 3)&lt;br&gt;&#xA;m - g + gain = main (= 20)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10064" LastEditorUserId="10064" LastEditDate="2016-01-29T18:18:23.467" LastActivityDate="2016-01-29T18:18:23.467" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25929" PostTypeId="2" ParentId="25906" CreationDate="2016-01-29T18:43:13.087" Score="12" Body="&lt;p&gt;My answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; main = 6 + 19 = 25&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each word is composed of two 2-letter US state abbreviations. The number is the sum of the order in which the states were admitted to the United States. For example, COAL = Colorado + Alabama = 38 + 22 = 60. Here's the full list:&lt;br&gt;&lt;br&gt;coal = Colorado + Alabama = 38 + 22= 60&lt;br&gt;mine = Michigan + Nebraska = 26 + 37 = 63&lt;br&gt;code = Colorado + Deleware = 38 + 1 = 39&lt;br&gt;lame = Louisiana + Maine = 18 + 23 = 41&lt;br&gt;gain = Georgia + Indiana = 4 + 19 = 23&lt;br&gt;gala = Georgia + Louisiana = 4 + 18 = 22&lt;br&gt;main = Massachusetts + Indiana = 6 + 19 = 25&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2016-01-29T18:43:13.087" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25930" PostTypeId="1" AcceptedAnswerId="25932" CreationDate="2016-01-29T18:43:16.957" Score="6" ViewCount="364" Body="&lt;p&gt;Matthew's favourite number is 47.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jacob's favourite number is 17.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tony's favourite number is 38.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is Evan's favourite number?&lt;/p&gt;&#xA;" OwnerUserId="9631" LastActivityDate="2016-06-04T17:35:37.940" Title="Whats your favourite number?" Tags="|riddle|mathematics|calculation-puzzle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25931" PostTypeId="1" CreationDate="2016-01-29T18:53:29.023" Score="73" ViewCount="19305" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;sup&gt;This is the second in a series of &lt;strong&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/search?q=title%3A%22unsolved+mysteries%3A%22&quot;&gt;Unsolved Mysteries&lt;/a&gt;&lt;/strong&gt; posts.  These posts explore current unsolved real-world puzzles.  Since these have been unsolved for years, it's uncertain whether anyone on on Puzzling.SE will be able to find a solution, but we do have a lot of brilliant minds here...&lt;/sup&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;sup&gt;Related meta post:  &lt;a href=&quot;https://puzzling.meta.stackexchange.com/questions/3295/posting-famous-unsolved-puzzles&quot;&gt;Posting famous unsolved puzzles&lt;/a&gt;&lt;/sup&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;In 1988, as a new headquarters for the American Central Intelligence Agency (CIA) was being built in Langley, Virgina, sculptor Jim Sanborn was commissioned to create artwork for the courtyard of the new building.  He designed a large copper monument, shaped somewhat like a flag, and engraved with hundreds of letters -- an encrypted message:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vOebp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vOebp.jpg&quot; alt=&quot;Kryptos sculpture at CIA headquarters&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;A soon as the sculpture was revealed, codebreakers both at the CIA and in the general public went to work on the puzzle.  In 1999, a computer scientist named Jim Gillogly announced that he had solved most of the puzzle.  There were four distinct parts to the code, and he had managed to solve the first three.&lt;/p&gt;&#xA;&lt;p&gt;After his announcement, the CIA revealed that they had actually solved the first three parts internally, the year before.  (Later, the U.S. National Security Administration (NSA) also claimed to have solved the first three parts in 1992.)&lt;/p&gt;&#xA;&lt;p&gt;The fourth section of the code remained unsolved, and to date, no one has brought forth a credible solution for part 4.&lt;/p&gt;&#xA;&lt;p&gt;The &amp;quot;tableau&amp;quot; on the sculpture has two sections.  One side has the encrypted text, and the other is an alphabet table of sorts. &lt;em&gt;(image credit: kryptosrevisited.com)&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7dLjN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7dLjN.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ia72y.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ia72y.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;(a plaintext version of the tableaux is available &lt;strong&gt;&lt;a href=&quot;http://austininc.com/SciRealm/KryptosDetails.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/strong&gt; -- just scroll down)&lt;/em&gt;&lt;/p&gt;&#xA;&lt;h2&gt;Other components of Kryptos&lt;/h2&gt;&#xA;&lt;p&gt;In addition to the copper sculpture pictured above, the artist created various other visual elements that are part of the same installation.  There is a pool at the base of the sculpture, and a bench.  There are also various large slabs of stone strewn about the area with various Morse Code messages.  It is unclear whether these messages are related to the solving of the Kryptos cipher or whether they are simply adding to the artistic aspect of the sculpture.  The Morse messages are as follows (some run right into the ground, so the text in brackets is a guess):&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;SOS&#xA;LUCID MEMORY&#xA;[WHA]T IS YOUR POSITION&#xA;SHADOW FORCES&#xA;VIRTUALLY INVISIBLE&#xA;DIGETAL INTERPRETAT[ION]&#xA;RQ&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;Solving the first 3 parts&lt;/h1&gt;&#xA;&lt;p&gt;&lt;sup&gt;&lt;em&gt;Note:  This section contains spoilers regarding the first three sections of the Kryptos code.  If you want to try to solve these yourself, do not read this section.  These are provided here to help set the context for the fourth (unsolved) section of the code.&lt;/em&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;h2&gt;Part 1&lt;/h2&gt;&#xA;&lt;p&gt;Though it was only discovered by solving Part 1, the ciphertext for Part 1 comprises the first two lines of the encrypted text:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;EMUFPHZLRFAXYUSDJKZLDKRNSHGNFIVJ&#xA;YQTQUXQBQVYUVLLTREVJYQTMKYRDMFD&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;It was encrypted using a Vigenere cipher, with a key of &lt;strong&gt;&lt;code&gt;PALIMPSEST&lt;/code&gt;&lt;/strong&gt; and the modified alphabet depicted on the second tableau.&lt;/p&gt;&#xA;&lt;p&gt;(A &lt;strong&gt;palimpsest&lt;/strong&gt; is a document (often vellum, but possibly parchment or paper), which has had its original text erased and has then been written over with new text.)&lt;/p&gt;&#xA;&lt;p&gt;To decrypt the text, we reverse the process that was used to encrypt it.  First, write out the keyword repeatedly underneath the ciphertext:&lt;/p&gt;&#xA;&lt;p&gt;&lt;code&gt;EMUFPHZLRFAXYUSDJKZLDKRNSHGNFIVJYQTQUXQBQVYUVLLTREVJYQTMKYRDMFD&lt;/code&gt;&#xA;&lt;code&gt;PALIMPSESTPALIMPSESTPALIMPSESTPALIMPSESTPALIMPSESTPALIMPSESTPAL&lt;/code&gt;&lt;/p&gt;&#xA;&lt;p&gt;Then use the Kryptos alphabet tableau to look up the decryption for each letter (note that we ignore the top row and left column as depicted in the tableau above):&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;KRYPTOSABCDEFGHIJLMNQUVWXZKRYP&#xA;RYPTOSABCDEFGHIJLMNQUVWXZKRYPT&#xA;YPTOSABCDEFGHIJLMNQUVWXZKRYPTO&#xA;PTOSABCDEFGHIJLMNQUVWXZKRYPTOS&#xA;TOSABCDEFGHIJLMNQUVWXZKRYPTOSA&#xA;OSABCDEFGHIJLMNQUVWXZKRYPTOSAB&#xA;SABCDEFGHIJLMNQUVWXZKRYPTOSABC&#xA;ABCDEFGHIJLMNQUVWXZKRYPTOSABCD&#xA;BCDEFGHIJLMNQUVWXZKRYPTOSABCDE&#xA;CDEFGHIJLMNQUVWXZKRYPTOSABCDEF&#xA;DEFGHIJLMNQUVWXZKRYPTOSABCDEFG&#xA;EFGHIJLMNQUVWXZKRYPTOSABCDEFGH&#xA;FGHIJLMNQUVWXZKRYPTOSABCDEFGHI&#xA;GHIJLMNQUVWXZKRYPTOSABCDEFGHIJ&#xA;HIJLMNQUVWXZKRYPTOSABCDEFGHIJL&#xA;IJLMNQUVWXZKRYPTOSABCDEFGHIJLM&#xA;JLMNQUVWXZKRYPTOSABCDEFGHIJLMN&#xA;LMNQUVWXZKRYPTOSABCDEFGHIJLMNQ&#xA;MNQUVWXZKRYPTOSABCDEFGHIJLMNQU&#xA;NQUVWXZKRYPTOSABCDEFGHIJLMNQUV&#xA;QUVWXZKRYPTOSABCDEFGHIJLMNQUVW&#xA;UVWXZKRYPTOSABCDEFGHIJLMNQUVWX&#xA;VWXZKRYPTOSABCDEFGHIJLMNQUVWXZ&#xA;WXZKRYPTOSABCDEFGHIJLMNQUVWXZK&#xA;XZKRYPTOSABCDEFGHIJLMNQUVWXZKR&#xA;ZKRYPTOSABCDEFGHIJLMNQUVWXZKRY&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;The first letter is &lt;code&gt;E&lt;/code&gt;, and the first letter of the key (&lt;strong&gt;&lt;code&gt;PALIMPSEST&lt;/code&gt;&lt;/strong&gt;) is &lt;code&gt;P&lt;/code&gt;.  So we find &lt;code&gt;P&lt;/code&gt; in the leftmost column, move right until we hit &lt;code&gt;E&lt;/code&gt;, and then move up to the top, ending at &lt;code&gt;B&lt;/code&gt;.  This makes the first letter of our plaintext a &lt;code&gt;B&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;            |&#xA;            V&#xA;    KRYPTOSABCDEFG...&#xA;    RYPTOSABCDEFGH...&#xA;    YPTOSABCDEFGHI...&#xA;--&amp;gt; PTOSABCDEFGHIJ...&#xA;    TOSABCDEFGHIJL...&#xA;    OSABCDEFGHIJLM...&#xA;    ...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;For the next letter, we find &lt;code&gt;A&lt;/code&gt; in the leftmost column, and follow it to the right until we hit &lt;code&gt;M&lt;/code&gt;, then up, ending at &lt;code&gt;E&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Continuing on in this way, we end up with the following decryption:&lt;/p&gt;&#xA;&lt;p&gt;&lt;code&gt;EMUFPHZLRFAXYUSDJKZLDKRNSHGNFIVJYQTQUXQBQVYUVLLTREVJYQTMKYRDMFD&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;PALIMPSESTPALIMPSESTPALIMPSESTPALIMPSESTPALIMPSESTPALIMPSESTPAL&lt;/code&gt;&lt;br /&gt;&#xA;&lt;strong&gt;&lt;code&gt;BETWEENSUBTLESHADINGANDTHEABSENCEOFLIGHTLIESTHENUANCEOFIQLUSION&lt;/code&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;With spaces and proper case:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Between subtle shading and the absence of light lies the nuance of i&lt;strong&gt;q&lt;/strong&gt;lusion&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;Part 2&lt;/h2&gt;&#xA;&lt;p&gt;The ciphertext for Part 2 comprises the remaining text in the top half of the tableau (until the break halfway down, i.e. lines 3 to 12):&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;VFPJUDEEHZWETZYVGWHKKQETGFQJNCE&#xA;GGWHKK?DQMCPFQZDQMMIAGPFXHQRLG&#xA;TIMVMZJANQLVKQEDAGDVFRPJUNGEUNA&#xA;QZGZLECGYUXUEENJTBJLBQCRTBJDFHRR&#xA;YIZETKZEMVDUFKSJHKFWHKUWQLSZFTI&#xA;HHDDDUVH?DWKBFUFPWNTDFIYCUQZERE&#xA;EVLDKFEZMOQQJLTTUGSYQPFEUNLAVIDX&#xA;FLGGTEZ?FKZBSFDQVGOGIPUFXHHDRKF&#xA;FHQNTGPUAECNUVPDJMQCLQUMUNEDFQ&#xA;ELZZVRRGKFFVOEEXBDMVPNFQXEZLGRE&#xA;DNQFMPNZGLFLPMRJQYALMGNUVPDXVKP&#xA;DQUMEBEDMHDAFMJGZNUPLGEWJLLAETG&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;It is solved in exactly the same way as the first part, except that it uses the keyword &lt;strong&gt;&lt;code&gt;ABSCISSA&lt;/code&gt;&lt;/strong&gt; instead of &lt;strong&gt;&lt;code&gt;PALIMPSEST&lt;/code&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;VFPJUDEEHZWETZYVGWHKKQETGFQJNCEGGWHKK?DQMCPFQZDQMMIAGPFXHQRLGTIMVMZJANQLVKQEDAGDVFRPJUNGEUNAQZGZLECGYUXUEENJTBJLBQCRTBJDFHRRYIZETKZEMVDUFKSJHKFWHKUWQLSZFTIHHDDDUVH?DWKBFUFPWNTDFIYCUQZEREEVLDKFEZMOQQJLTTUGSYQPFEUNLAVIDXFLGGTEZ?FKZBSFDQVGOGIPUFXHHDRKFFHQNTGPUAECNUVPDJMQCLQUMUNEDFQELZZVRRGKFFVOEEXBDMVPNFQXEZLGREDNQFMPNZGLFLPMRJQYALMGNUVPDXVKPDQUMEBEDMHDAFMJGZNUPLGEWJLLAETG&#xA;ABSCISSAABSCISSAABSCISSAABSCISSAABSCI?SSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAAB?SCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISS?AABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAA&#xA;ITWASTOTALLYINVISIBLEHOWSTHATPOSSIBLE?THEYUSEDTHEEARTHSMAGNETICFIELDXTHEINFORMATIONWASGATHEREDANDTRANSMITTEDUNDERGRUUNDTOANUNKNOWNLOCATIONXDOESLANGLEYKNOWABOUTTHIS?THEYSHOULDITSBURIEDOUTTHERESOMEWHEREXWHOKNOWSTHEEXACTLOCATION?ONLYWWTHISWASHISLASTMESSAGEXTHIRTYEIGHTDEGREESFIFTYSEVENMINUTESSIXPOINTFIVESECONDSNORTHSEVENTYSEVENDEGREESEIGHTMINUTESFORTYFOURSECONDSWESTIDBYROWS&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;With spaces and proper case (punctuation in &lt;code&gt;grey&lt;/code&gt; added by me):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;It was totally invisible&lt;code&gt;.&lt;/code&gt; How&lt;code&gt;'&lt;/code&gt;s that possible? They used the Earth&lt;code&gt;'&lt;/code&gt;s magnetic field&lt;code&gt;.&lt;/code&gt; X&lt;br /&gt;&#xA;The information was gathered and transmitted undergr&lt;strong&gt;u&lt;/strong&gt;und to an unknown location&lt;code&gt;.&lt;/code&gt; X&lt;br /&gt;&#xA;Does Langley know about this? They should&lt;code&gt;;&lt;/code&gt; it&lt;code&gt;'&lt;/code&gt;s buried out there somewhere&lt;code&gt;.&lt;/code&gt; X&lt;br /&gt;&#xA;Who knows the exact location? Only WW&lt;code&gt;.&lt;/code&gt; This was his last message&lt;code&gt;:&lt;/code&gt; X&lt;br /&gt;&#xA;Thirty eight degrees fifty seven minutes six point five seconds north&lt;code&gt;,&lt;/code&gt;&lt;br /&gt;&#xA;Seventy seven degrees eight minutes forty four seconds west&lt;code&gt;.&lt;/code&gt;&lt;br /&gt;&#xA;ID by rows&lt;code&gt;.&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Some years after this was originally solved, the sculptor Sanborn made it known that &lt;em&gt;&lt;strong&gt;there was an error in the ciphertext&lt;/strong&gt;&lt;/em&gt;, where a single character had been omitted (right after the word &amp;quot;west&amp;quot; on the second-last line).  He assumed that it would be obvious to a solver, because it would cause the remaining text to decipher into gibberish, but by what seems to be coincidence, the text actually deciphered into the phrase &amp;quot;ID by rows.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;It has since been determined (and confirmed) that &lt;em&gt;&lt;strong&gt;the correct decryption ends with &lt;code&gt;XLAYERTWO&lt;/code&gt; instead of &lt;code&gt;IDBYROWS&lt;/code&gt;&lt;/strong&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;h2&gt;Part 3&lt;/h2&gt;&#xA;&lt;p&gt;The ciphertext for Part 3 comprises the majority of the bottom half of the tableau, up to (but not including) the &lt;code&gt;?&lt;/code&gt; character on the fourth-last line:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;ENDYAHROHNLSRHEOCPTEOIBIDYSHNAIA&#xA;CHTNREYULDSLLSLLNOHSNOSMRWXMNE&#xA;TPRNGATIHNRARPESLNNELEBLPIIACAE&#xA;WMTWNDITEENRAHCTENEUDRETNHAEOE&#xA;TFOLSEDTIWENHAEIOYTEYQHEENCTAYCR&#xA;EIFTBRSPAMHHEWENATAMATEGYEERLB&#xA;TEEFOASFIOTUETUAEOTOARMAEERTNRTI&#xA;BSEDDNIAAHTTMSTEWPIEROAGRIEWFEB&#xA;AECTDDHILCEIHSITEGOEAOSDDRYDLORIT&#xA;RKLMLEHAGTDHARDPNEOHMGFMFEUHE&#xA;ECDMRIPFEIMEHNLSSTTRTVDOHW&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;This turns out to be a very different kind of encryption.  It uses a &lt;em&gt;transposition cipher&lt;/em&gt; which doesn't actually change any of the letters, but simply rearranges them.  The task is to find the method used for the rearrangement.&lt;/p&gt;&#xA;&lt;p&gt;There have been &lt;a href=&quot;http://elonka.com/kryptos/part3.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;many&lt;/a&gt; &lt;a href=&quot;http://math.ucsd.edu/%7Ecrypto/Projects/KarlWang/index2.html#3&quot; rel=&quot;nofollow noreferrer&quot;&gt;different&lt;/a&gt; &lt;a href=&quot;http://www.thekryptosproject.com/tjp/observations/k3_method.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;attempts&lt;/a&gt; at explaining the transposition. For this kind of transposition cipher, there are many different but &lt;em&gt;cryptographically equivalent&lt;/em&gt; ways of describing and performing the cipher that all produce the same text, and there is no way to know which one the cryptographer had used without looking over his shoulder.&lt;/p&gt;&#xA;&lt;p&gt;Luckily, photographs of the worksheets Sanborn used while enciphering K3 (&lt;a href=&quot;https://kryptools.com/Sheets/sheets.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;pictured here&lt;/a&gt;) make it clear that the original transposition method is a &lt;em&gt;&lt;strong&gt;double route transposition&lt;/strong&gt;&lt;/em&gt; using two completely-filled rectangles. The rectangles are of different sizes, but the same route paths (&lt;em&gt;in by rows&lt;/em&gt; and &lt;em&gt;out by upwards columns&lt;/em&gt;) are used each time. This cipher does not use a literal key; the &lt;em&gt;route paths&lt;/em&gt; and &lt;em&gt;grid sizes&lt;/em&gt; are what make up the keys for the two route transpositions.&lt;/p&gt;&#xA;&lt;p&gt;To encipher, the plaintext is first written into a 42-letter wide and 8-letter high grid, from left to right starting on the top row (that is, the usual reading order). The intermediate ciphertext is read out from that grid in upwards columns, reading each column upwards from the bottom row, starting with the leftmost column.&lt;/p&gt;&#xA;&lt;p&gt;The intermediate text is then written into a different grid, 14 letters wide and 24 letters high. As before, it is written in by rows in the usual reading order, and read out by upwards columns to produce the text seen on the sculpture.&lt;/p&gt;&#xA;&lt;p&gt;Deciphering is the inverse of enciphering. First, write the K3 ciphertext into a 14x24 grid by upwards columns. Note the beginning of the ciphertext (&lt;code&gt;ENDYAHRO...&lt;/code&gt;) reads upwards from the bottom of the first column.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;ILNTAYESTATHCW&#xA;BLHMHEHAROIEEH&#xA;ISIWNTHONRSLEO&#xA;OLTETYMFTEHMHD&#xA;ELAAEOAERIILUV&#xA;TSGCRIPEEPEKET&#xA;PDNADESTEWCRFR&#xA;CLRIUARBAELTMT&#xA;OUPIEHBLMTIIFT&#xA;EYTPNNTRRSHRGS&#xA;HEELEEFEAMDOMS&#xA;RRNBTWIEOTDLHL&#xA;SNMECIEYTTTDON&#xA;LTXLHTRGOHCYEH&#xA;NHWEADCEEAERNE&#xA;HCRNREYTAAADPM&#xA;OAMNNSAAUIBDDI&#xA;RISLELTMTNESRE&#xA;HAOSEOCAEDFOAF&#xA;ANNETFNTUDWAHP&#xA;YHSPITEATEEEDI&#xA;DSHRDEENOSIOTR&#xA;NYOANOHEIBRGGM&#xA;EDNRWEQWFIGEAD&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Read out the intermediate text (&lt;code&gt;ILNTAYES...&lt;/code&gt;) by rows. Then, write the intermediate text into a 42x8 grid by upwards columns. The plaintext will be evident on the rows. Note the first letters of the intermediate text (&lt;code&gt;ILNTAYES...&lt;/code&gt;) read upwards from the bottom of the first column.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;SLOWLYDESPARATLYSLOWLYTHEREMAINSOFPASSAGED&#xA;EBRISTHATENCUMBEREDTHELOWERPARTOFTHEDOORWA&#xA;YWASREMOVEDWITHTREMBLINGHANDSIMADEATINYBRE&#xA;ACHINTHEUPPERLEFTHANDCORNERANDTHENWIDENING&#xA;THEHOLEALITTLEIINSERTEDTHECANDLEANDPEEREDI&#xA;NTHEHOTAIRESCAPINGFROMTHECHAMBERCAUSEDTHEF&#xA;LAMETOFLICKERBUTPRESENTLYDETAILSOFTHEROOMW&#xA;ITHINEMERGEDFROMTHEMISTXCANYOUSEEANYTHINGQ&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Sanborn's worksheets also make it clear that this text is exactly 336 letters long, equal to the areas of both rectangles. The question mark is not part of this ciphertext, and it is unclear whether it is the first &amp;quot;letter&amp;quot; of the following Part 4 ciphertext or whether it is simply a delimiter or separator.&lt;/p&gt;&#xA;&lt;p&gt;With spaces and proper case (punctuation in &lt;code&gt;grey&lt;/code&gt; added by me):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Slowly&lt;code&gt;,&lt;/code&gt; desp&lt;strong&gt;a&lt;/strong&gt;rat[e]ly slowly&lt;code&gt;,&lt;/code&gt; the remains of the passage debris that encumbered the lower part of the doorway was removed&lt;code&gt;.&lt;/code&gt; With trembling hands&lt;code&gt;,&lt;/code&gt; I made a tiny breach in the upper&lt;code&gt;-&lt;/code&gt;left&lt;code&gt;-&lt;/code&gt;hand corner&lt;code&gt;,&lt;/code&gt; and then&lt;code&gt;,&lt;/code&gt; widening the hole a little&lt;code&gt;,&lt;/code&gt; I inserted the candle and peered in&lt;code&gt;.&lt;/code&gt;  The hot air escaping from the chamber caused the flame to flicker&lt;code&gt;,&lt;/code&gt; but presently details of the room within emerged from the mist&lt;code&gt;.&lt;/code&gt;  &lt;code&gt;&amp;quot;&lt;/code&gt;Can you see anything&lt;code&gt;?&amp;quot;&lt;/code&gt; Q&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;Part 4&lt;/h2&gt;&#xA;&lt;p&gt;The ciphertext for Part 4 is the remaining 97 (or 98 if the question mark is included) characters on the tableau:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;                          ?OBKR&#xA;UOXOGHULBSOLIFBBWFLRVQQPRNGKSSO&#xA;TWTQSJQSSEKZZWATJKLUDIAWINFBNYP&#xA;VTTMZFPKWGDKZXTJCDIGKUHUAUEKCAR&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Most Kryptos fans assume that decrypting it will involve some combination of the methods used for the first three sections.  To date, no one has been able to solve it.  &lt;h3&gt;&lt;em&gt;&lt;strong&gt;Will you be the first?&lt;/strong&gt;&lt;/em&gt;&lt;/h3&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;There is a plethora of information about Kryptos on the web.  A great place to start if you want more information is Elonka Dunin's &lt;a href=&quot;http://elonka.com/kryptos/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kryptos Page&lt;/a&gt; (especially the &lt;a href=&quot;http://elonka.com/kryptos/faq.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;FAQ&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;em&gt;Like all of my &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=title%3A%22unsolved+mysteries%3A%22&quot;&gt;Unsolved Mysteries&lt;/a&gt; puzzles, I will give a &lt;a href=&quot;https://puzzling.meta.stackexchange.com/questions/4399&quot;&gt;500 rep bounty&lt;/a&gt; to anyone who manages to find a credible solution.  &lt;a href=&quot;https://puzzling.stackexchange.com/users/5208&quot;&gt;Alconja&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/users/17250&quot;&gt;question_asker&lt;/a&gt; have both agreed to match this bounty, so the total bounty is &lt;strong&gt;1500 reputation&lt;/strong&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="83349" LastEditDate="2023-09-02T09:50:01.920" LastActivityDate="2023-09-02T09:50:01.920" Title="Unsolved Mysteries: Kryptos" Tags="|cipher|real|unsolved-mysteries|" AnswerCount="1" CommentCount="18" ContentLicense="CC BY-SA 4.0" />
  <row Id="25932" PostTypeId="2" ParentId="25930" CreationDate="2016-01-29T18:55:34.257" Score="7" Body="&lt;p&gt;Possible answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 22&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Taking the position of every letter in the alphabet and adding the length, dividing by two and then subtracting 1 from the result gives the numbers for each name.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Example&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Matthew $\rightarrow \lfloor\frac{13+1+20+20+8 + 5 + 23 + 7}{2}\rfloor-1 = \lfloor 97/2 \rfloor -1 = 47$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12901" LastEditorUserId="12901" LastEditDate="2016-01-29T19:02:33.547" LastActivityDate="2016-01-29T19:02:33.547" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25933" PostTypeId="2" ParentId="25833" CreationDate="2016-01-29T18:56:56.627" Score="4" Body="&lt;p&gt;They are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The pillars of Herculus, also known as The Strait of Gibraltar&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lines 1-2:&lt;br&gt;&#xA;We are the mythical twins&lt;br&gt;&#xA;In the in-between place&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sides of the strait were called &quot;The Pillars of Hercules.&quot; According to myth, Hercules created the strait as part of one of his Labors. No-one at the time of myth creation knew what lay beyond the strait, providing one potential meaning of &quot;in-between&quot; place. Alternately, the Pillars of Hercules are often referenced in myth as as supporting the sky, hence they are &quot;in between&quot; earth and heaven (or the mythological equivalent).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lines 3-4:&lt;br&gt;&#xA;Gateway to the middle&lt;br&gt;&#xA;Where the king's daughter faces the dust&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The strait is a gateway from the Atlantic Ocean to the Middle East, where Israel, King David's &quot;daughter&quot; is. It is also a desert, hence the dust (this one's a stretch).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Line 5:&lt;br&gt;&#xA;We are of the strong one, the labourer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hercules is a mythical figure purported to be very strong. Also he underwent the 12 Labors of Hercules.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Line 6:&lt;br&gt;&#xA;One of us keeps monkeys&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gibraltar (on the European side) has the only native monkeys in Europe&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastEditorUserId="1620" LastEditDate="2016-01-30T01:26:59.253" LastActivityDate="2016-01-30T01:26:59.253" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25934" PostTypeId="1" AcceptedAnswerId="25951" CreationDate="2016-01-29T19:42:27.090" Score="11" ViewCount="403" Body="&lt;p&gt;Long ago, my militant leader devised a plan for world domination&lt;br&gt;&#xA;Though he is now gone, I am still carrying out his plan&lt;br&gt;&#xA;Today, our territory is near second to none&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many have died for me to live&lt;br&gt;&#xA;And I am well seasoned, but there are times when my armor does fall away  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have secrets of my own, but you must know me before you could begin know them&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;" OwnerUserId="11452" LastEditorUserId="5373" LastEditDate="2018-05-18T21:54:10.053" LastActivityDate="2018-05-18T21:54:10.053" Title="I'm Known for Conquering the World 'Round" Tags="|riddle|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25935" PostTypeId="2" ParentId="25934" CreationDate="2016-01-29T20:19:07.280" Score="2" Body="&lt;p&gt;A start:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The 5th line is yelling &lt;strong&gt;vegetation&lt;/strong&gt; in my mind. &lt;em&gt;And I am well seasoned, but there are times when my armor does fall away&lt;/em&gt; - vegetation follows seasons and in the fall/winter most (trees being the most prolific) lose leaves, sometimes even some bark etc... - this answer also kind of fits with the territory stanza , and the many have died for me to live (plants dying and becoming soil for other plants)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11251" LastActivityDate="2016-01-29T20:19:07.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25936" PostTypeId="2" ParentId="25893" CreationDate="2016-01-29T20:19:50.677" Score="3" Body="&lt;p&gt;Not positive on this one, but I'll take a shot:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Is it a &lt;strong&gt;compass&lt;/strong&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So sharp that I can execute what you want me to.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A compass has a sharp end to help you make circles and arcs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Will guide you the way, and when needed I will split your troubles in your head&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A compass can point you in the right direction; and in &quot;The Golden Compass&quot;, a person's soul is split and one part takes the form of an animal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So valuable, that you always fight for me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Perhaps another reference to &quot;The Golden Compass&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Standing at the crossroads, waiting for you to find me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This part is where I become less confident. Any ideas how this fits in?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18424" LastActivityDate="2016-01-29T20:19:50.677" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25937" PostTypeId="2" ParentId="25934" CreationDate="2016-01-29T21:13:50.880" Score="1" Body="&lt;p&gt;My answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I want to say 'tree'... but I'm going to go with &quot;dandelion&quot;! Final answer!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It begins with 'world domination', suggesting the increase/expansion of resources. While it could be acorns, since these fall from trees and grow new trees, this seems too short-distance a system to describe as having &quot;devised a plan for world domination&quot;. Dandelions are plants designed to spread to far-reaching places and to grow new versions of itself, many times over. Also, the life-span of a tree is very high in relation to a dandelion, which could explain &quot;he is now gone, but I am still carrying out his plan&quot;, and &quot;many have died for me to live&quot;. The last two lines don't fit, though&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18800" LastEditorUserId="1620" LastEditDate="2016-01-29T23:13:43.507" LastActivityDate="2016-01-29T23:13:43.507" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25939" PostTypeId="2" ParentId="25896" CreationDate="2016-01-30T00:11:36.443" Score="1" Body="&lt;p&gt;Fill the 6 liter jug and then pour it into the 10 jug. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;10 liter jug has 6 liters of water.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Fill the 6 liter jug again and pour 5 liters into the 5 liter jug the remaining 1 liter into the 10 liter jug.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;10 liter jug has 7 liters of water.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Repeat last step.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;10 liter jug has 8 liters of water.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="18806" LastActivityDate="2016-01-30T00:11:36.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25940" PostTypeId="2" ParentId="25884" CreationDate="2016-01-30T02:03:46.950" Score="0" Body="&lt;p&gt;The logic is to just add the character in the name to get the sum. A corresponds to 1, B to 2 and so on. &lt;/p&gt;&#xA;" OwnerUserId="18809" LastActivityDate="2016-01-30T02:03:46.950" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25941" PostTypeId="1" AcceptedAnswerId="25942" CreationDate="2016-01-30T04:27:05.257" Score="11" ViewCount="813" Body="&lt;p&gt;When I'm a complete truth-teller, you're happy with me&lt;br&gt;&#xA;When I'm a complete liar, I'm still doing my job very well &lt;br&gt;&#xA;When I'm mostly a liar, you know there's something seriously wrong with me&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who am I?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user17947" LastActivityDate="2016-01-31T12:46:31.563" Title="I'm better when I lie all the time than some of the time" Tags="|riddle|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25942" PostTypeId="2" ParentId="25941" CreationDate="2016-01-30T05:02:26.973" Score="16" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a clock&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because when I'm a complete truth-teller, you're happy with me:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Accurate clocks are a good thing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I'm a complete liar, I'm still doing my job very well:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A clock that is as off as possible (i.e., 12 hours), is still very useful for telling the time since we don't usually need a clock to know whether it's AM or PM.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I'm mostly a liar, you know there's something seriously wrong with me:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Any clock that's neither accurate nor exactly 12 hours off does not function as a useful clock.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12795" LastActivityDate="2016-01-30T05:02:26.973" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="25943" PostTypeId="2" ParentId="25893" CreationDate="2016-01-30T06:06:52.073" Score="0" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; my mind.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So sharp that I can execute what you want me to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A sharp mind can perform many tasks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Will guide you the way, and when needed I will split your troubles in your head.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The mind leads the actions, and resolves problems (in your head is literally in your brain).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So valuable, that you always fight for me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My mind is my life, my personality.  It defines me, and it is always worth fighting for.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Standing at the crossroads, waiting for you to find me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you have to make a decision (crossroads), you have to stop and think about what you're doing with your mind.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12795" LastActivityDate="2016-01-30T06:06:52.073" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25944" PostTypeId="2" ParentId="9438" CreationDate="2016-01-30T07:26:34.363" Score="5" Body="&lt;p&gt;This is a solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First compute $S, T$ as follows: &lt;br&gt; $S=\overline{AB+BC+CA}$,  &lt;br&gt; $T=\overline{SA+SB+SC+ABC}$  &lt;br&gt; Then we have  &lt;br&gt; $\overline{A} = S(T+B+C)+TBC$  &lt;br&gt; $\overline{B} = S(T+C+A)+TCA$  &lt;br&gt; $\overline{C} = S(T+A+B)+TAB$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18814" LastEditorUserId="16322" LastEditDate="2016-01-30T11:19:32.173" LastActivityDate="2016-01-30T11:19:32.173" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25945" PostTypeId="1" AcceptedAnswerId="25946" CreationDate="2016-01-30T08:25:43.060" Score="7" ViewCount="1849" Body="&lt;p&gt;This is a simple problem that my friend asked me:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;2 + 2 = Fish&lt;br /&gt;&#xA;3 + 3 = Eight&lt;br /&gt;&#xA;7 + 7 = ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Can you figure out what is represented by the question mark?&lt;/p&gt;&#xA;" OwnerUserId="17861" LastEditorUserId="69582" LastEditDate="2020-11-19T21:12:30.407" LastActivityDate="2020-11-19T21:12:30.407" Title="2 + 2 = Fish. Who am I?" Tags="|lateral-thinking|" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25946" PostTypeId="2" ParentId="25945" CreationDate="2016-01-30T08:41:26.970" Score="12" Body="&lt;p&gt;My comment was mostly tongue-in-cheek, but here's why I said it (excuse my bad MS Paint skills)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/aQJd3.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aQJd3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first two are clear, but I've no idea what the third would be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose, if I don't go quite as far, I get&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/v3z9U.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/v3z9U.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA; Upside-down triangle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17147" LastEditorUserId="17147" LastEditDate="2016-01-30T08:48:03.960" LastActivityDate="2016-01-30T08:48:03.960" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25947" PostTypeId="1" AcceptedAnswerId="25948" CreationDate="2016-01-30T09:01:32.020" Score="5" ViewCount="492" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;I am the square of 1514,&lt;/p&gt;&#xA;&lt;p&gt;but actually, I'm just the square of 4.&lt;/p&gt;&#xA;&lt;p&gt;I possess a property,&lt;/p&gt;&#xA;&lt;p&gt;where women could be burned for in my time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="16322" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-01-30T09:57:50.387" Title="The square of 1514" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25948" PostTypeId="2" ParentId="25947" CreationDate="2016-01-30T09:53:43.870" Score="9" Body="&lt;p&gt;I am the square of 1514,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is a magic square (more specifically, Dürer's magic square).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but actually, I'm just the square of 4.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Such a magic square consists of $4 \times 4$ elements.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I possess a property,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where women could be burned for in my time&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Women who were accused of magic were burned during these ages.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12901" LastEditorUserId="16322" LastEditDate="2016-01-30T09:57:50.387" LastActivityDate="2016-01-30T09:57:50.387" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25949" PostTypeId="2" ParentId="16106" CreationDate="2016-01-30T11:03:47.573" Score="7" Body="&lt;p&gt;At least two of these statements are false.  Ignoring grammar and wordplay, which?&#xA;&lt;br&gt; a. Pigs is pigs.&#xA;&lt;br&gt; b. One plus one is two.&#xA;&lt;br&gt; c. Two plus two does not equal five.&#xA;&lt;br&gt; d. Five and ten make fifteen.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;List of false statements:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. &amp;ldquo;At least two of these statements are false.&amp;rdquo;&#xA; &lt;br&gt;&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18129" LastActivityDate="2016-01-30T11:03:47.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25950" PostTypeId="2" ParentId="25941" CreationDate="2016-01-30T11:11:58.780" Score="10" Body="&lt;p&gt;You are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Man / Boyfriend/ Husband&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I'm a complete truth-teller, you're happy with me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You appreciate my honesty&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I'm a complete liar, I'm still doing my job very well&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No, you're not fat&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I'm mostly a liar, you know there's something seriously wrong with me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You're not fat, just a little bit chubby&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18816" LastActivityDate="2016-01-30T11:11:58.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25951" PostTypeId="2" ParentId="25934" CreationDate="2016-01-30T11:19:39.737" Score="11" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; KFC, the fast food chain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Long ago, my militant leader devised a plan for world domination&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; KFC's founder was &quot;Colonel&quot; Harland Sanders, who introduced chicken to the fast food market and expanded the franchise from Utah to the rest of the USA.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though he is now gone, I am still carrying out his plan&lt;br&gt;&#xA;Today, our territory is near second to none&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sanders sold the company in 1964, but the company continued to expand overseas and is now one of the largest international fast food chains.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Many have died for me to live&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A lot of chickens die to provide the meat for KFC.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I am well seasoned, but there are times when my armor does fall away &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; KFC is known for its &quot;Original Recipe&quot; seasoning. Not sure about the armor part, although it could refer to the fact that it's been less successful in the US in recent years.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have secrets of my own, but you must know me before you could begin know them&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Only KFC and the companies who produce the secret &quot;11 herbs and spices&quot; know what they are, and even then the production is split over two companies so that each only knows half the recipe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; KFC. As a bonus, these are also the initials of &quot;Known for Conquering&quot; from the title.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12511" LastEditorUserId="12511" LastEditDate="2016-01-30T21:07:13.660" LastActivityDate="2016-01-30T21:07:13.660" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="25952" PostTypeId="1" AcceptedAnswerId="25954" CreationDate="2016-01-30T12:03:28.863" Score="8" ViewCount="300" Body="&lt;p&gt;In life, we always had an instrument with us. Even the poor and the rich have it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This instrument can be used to show:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Encouragement&lt;/li&gt;&#xA;&lt;li&gt;Direction&lt;/li&gt;&#xA;&lt;li&gt;Aggression&lt;/li&gt;&#xA;&lt;li&gt;Love&lt;/li&gt;&#xA;&lt;li&gt;Promise&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This instrument cannot speak. What is this instrument?&lt;/p&gt;&#xA;" OwnerUserId="11369" LastEditorUserId="11369" LastEditDate="2016-01-30T12:15:40.840" LastActivityDate="2016-01-30T20:09:49.170" Title="The instrument forever with us" Tags="|riddle|word|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25954" PostTypeId="2" ParentId="25952" CreationDate="2016-01-30T13:45:44.527" Score="13" Body="&lt;p&gt;It is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your hand. (More specifically, your 5 fingers.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Encouragement&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is the &lt;em&gt;thumb&lt;/em&gt;. A &quot;thumbs-up&quot; is often a sign of encouragement. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Direction&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;em&gt;index finger&lt;/em&gt;. It is used to point the way. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Aggression&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;em&gt;middle finger&lt;/em&gt;. The famous &quot;&lt;a href=&quot;https://en.m.wiktionary.org/wiki/one-finger_salute&quot; rel=&quot;noreferrer&quot;&gt;one-finger salute&lt;/a&gt;&quot; is a definite sign of aggression. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Love&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;em&gt;ring finger&lt;/em&gt; is where we place a wedding ring, a symbol of undying love. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Promise&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;em&gt;pinky finger&lt;/em&gt;. It is used for &quot;&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Pinky_swear&quot; rel=&quot;noreferrer&quot;&gt;pinky promises&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="4421" LastEditDate="2016-01-30T20:09:49.170" LastActivityDate="2016-01-30T20:09:49.170" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25955" PostTypeId="2" ParentId="25952" CreationDate="2016-01-30T14:01:45.200" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The hands. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some people do not have it, but it in general it is possessed by rich and poor alike.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Encouragement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Thumbs up&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Direction:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pointing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Aggression:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Middle finger, fist grip&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Love:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Caressing, &lt;a href=&quot;http://thumbs.dreamstime.com/z/woman-hand-making-sign-heart-25753409.jpg&quot; rel=&quot;nofollow&quot;&gt;love hand sign&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Promise:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pinkie promise&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This instrument cannot speak: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No it can't, and body language isn't speaking&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18817" LastActivityDate="2016-01-30T14:01:45.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25956" PostTypeId="1" AcceptedAnswerId="25962" CreationDate="2016-01-30T14:41:21.450" Score="7" ViewCount="860" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;By my extreme size, I escape your eye &lt;br/&gt;&#xA;  The nature of every thing, is but many of I &lt;br/&gt;&#xA;  Since my unveiling, the inexact came by&lt;br/&gt;&#xA;  And there came a new discipline to unify&lt;br/&gt;&#xA;  What am I?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;The explanation&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; So there are answers for the atom and radiation. Both are very close, but arguably jhabbott got closest with their guess of elementary particles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, my own answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Particles that are, elementary&lt;br/&gt;Finer than a germ, so small you can't see&lt;br/&gt;Apples ants and me, all comprise many&lt;br/&gt;For the inexact, that's uncertainty&lt;br/&gt;What to unify? Quantum gravity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; By &quot;escape your eye&quot; I did mean that one can't see it with naked eyes.&lt;br/&gt;The second sentence of course means that it is a building block of matter (everyday or not).&lt;br/&gt; While two answers are right that the third sentence refers to Heisenberg's Uncertainty Principle or the probabilistic nature of quantum-scale stuff (one is the consequence of the other, right?) the answer that guessed the atom made a mistake here: that didn't exactly come by with the concept of the atom, rather this appeared with the rise of quantum mechanics.&lt;br/&gt; Regarding the new discipline to unify, it refers to the discrepancies between quantum mechanics and general relativity that lead to many different theories trying to resolve the problem. It can mean to unify quantum mechanics with relativity, or to unify those new theories and come to a single, widely agreed physical model again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18817" LastEditorUserId="18817" LastEditDate="2016-02-01T16:51:52.323" LastActivityDate="2016-02-01T16:51:52.323" Title="Science riddle: Many of I" Tags="|riddle|science|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25957" PostTypeId="2" ParentId="25956" CreationDate="2016-01-30T14:53:18.033" Score="5" Body="&lt;p&gt;My guess is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an atom&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Argument:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By my extreme size, I escape your eye  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Atoms are extremely small, and cannot be seen with bare eyes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The nature of every thing, is but many of I  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everything is made of many atoms&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since my unveiling, the inexact came by  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Heisenberg's uncertainty principle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And there came a new discipline to unify  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Nuclear sciences&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="17482" LastEditDate="2016-01-30T16:11:59.167" LastActivityDate="2016-01-30T16:11:59.167" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25958" PostTypeId="2" ParentId="25922" CreationDate="2016-01-30T15:15:11.873" Score="20" Body="&lt;p&gt;If we allow Bob to ask for values not at positive integers, he can find $P(2016)$ with just one question.  He asks for $P(\pi)$ and tells Alice to give him the integral part, then pause until he says start, then keep outputting decimal digits until he says stop.  There are only finitely many monomials with values less than this at $\pi$.  They can be combined in finitely many ways.  He makes a finite list of all the polynomials that produce a value of $P(\pi)$ between the integral part Alice said and $1$ more and computes the decimal expansions until every one differs from the rest.  He then asks Alice to start giving him decimals and tells her to stop when he knows the polynomial.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if $100 \le P(\pi) \lt 101$, he can only use constants $1$ through $100$, linear terms $x$ through $31x$, quadratics $x^2$ through $9x^2$, cubics $x^3$ through $4x^3$ and $x^4$.  Most combinations of those will be too large, but $97+x$ evaluates to $100.14159$, $x^3+4x^2+60$ evaluates to $100.615$and $x^3+4x^2+x^2+x+27$ to $100.6263$.  I don't know how close they can be, but it won't take too many decimals to sort it out.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2016-01-30T15:15:11.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25959" PostTypeId="2" ParentId="8249" CreationDate="2016-01-30T17:58:48.873" Score="4" Body="&lt;p&gt;I would like to thank dmg and user3294068 because their insights helped me find this answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;dmg tried moving out twice as far in each step, whereas user3294068 tried moving w times as far in each step as the last step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will use a different strategy, but I am copying this logic from user3294068 &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The worst case result is 1 door past a turn-around, because we almost reach it, go back, come forward again, and then 1 more step.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What we will do is try to go as far as possible in each step before turning around, and we will see that this results in a solution that works. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's work through the first 2 numbers. I will start going in the + direction. The furthest I can go is dictated by the fact that when I get back to -1, that must still be profitable. Therefore I can go to +3 since it will be 7 steps when I get to -1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, it must be profitable when I get back to +4. I have traveled 6 steps in the previous journey from 0 to 3 to 0, will need 4 to get from the origin to +4 later on, and must be less than 36 total. Therefore I can go to -12.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt;&#xA;I will make use of 3 variables, $n$ The step number. $t_n$ The displacement from the origin of each step it is positive for odd n and negative for even n. $L_n= \sum_{i=1}^{n} 2 |t_i|$ The distance traveled by all steps to venture out and return to the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We start out by defining the condition for choosing a step and staying profitable. Based on the above argument looking at a single step at a time, that condition is: $$L_n+|t_{n-1}|+1&amp;lt;9(|t_{n-1}|+1)$$algebra $$t_n&amp;lt;4t_{n-1}+4-\sum_{i=1}^{n-1}t_i$$Therefore $$t_n=4t_{n-1}+3-\sum_{i=1}^{n-1}t_i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now given $t_0=0$, this recursive formula leads to closed form expression $t_n=3*n*2^{n-1}$. Proof by induction at end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what we have done is constructed a path such that the least profitable point on that path is still profitable. We can see from the form of $t_n$ that it is monotonically increasing. Since we required this path to be profitable in its constrution and we move further from the origin than the previous n, This path can be continued indefinitely. For values above $L_n=1000$, the difference between $8.\bar{9}$ and 8.999 will mean there needs to be corrections, but I am too lazy to apply them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the steps I could calculate before my 32 bit int overflowed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[3],&#xA;[-12],&#xA;[36],&#xA;[-96],&#xA;[240],&#xA;[-576],&#xA;[1344],&#xA;[-3072],&#xA;[6912],&#xA;[-15360],&#xA;[33792],&#xA;[-73728],&#xA;[159744],&#xA;[-344064],&#xA;[737280],&#xA;[-1572864],&#xA;[3342336],&#xA;[-7077888],&#xA;[14942208],&#xA;[-31457280],&#xA;[66060288],&#xA;[-138412032],&#xA;[289406976],&#xA;[-603979776],&#xA;[1258291200]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof by induction:&lt;/strong&gt;&#xA;$t_n=4t_{n-1}+3-\sum_{i=1}^{n-1}t_i$ and $t_0=0$ leads to $t_n=3*n*2^{n-1}$&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Base case n = 1&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$t_1=4*t_0+3-\sum_{i=1}^{0}t_i=3$ and $t_1=3*1*2^0=3$&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Given $t_{n-1}$ prove $t_n$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$t_n = 4* 3 * (n-1) * 2 ^ {n-2} + 3 - (t_{n-1}-4t_{n-2}-3)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t_n = 4* 3 * (n-1) * 2 ^ {n-2} + 3 - (3*(n-1)*2^{n-1}-4*3*(n-2)*2^{n-3}-3)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some algebra leads to the closed form expression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So although this proves that it is possible to beat the dealer, it still might not be the ideal solution. I think there is a way to guarantee winning 3 or 4 euros by taking smaller initial steps and setting a margin of error so that the least profitable step in each series is still worth a couple euros.&lt;/p&gt;&#xA;" OwnerUserId="18821" LastEditorUserId="18821" LastEditDate="2016-01-30T18:10:24.333" LastActivityDate="2016-01-30T18:10:24.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25960" PostTypeId="1" AcceptedAnswerId="26343" CreationDate="2016-01-30T17:59:24.390" Score="9" ViewCount="340" Body="&lt;p&gt;Every letter a decimal digit, different letters for different digits:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    HAPPY&#xA;    HAPPY&#xA;    HAPPY&#xA;   + DAYS&#xA; ----------&#xA;    AHEAD&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which digit does each letter represent?&lt;br&gt;&#xA;(Please present the full analysis how these digits can be determined.)&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="9404" LastEditDate="2016-03-03T10:20:06.467" LastActivityDate="2016-03-03T10:20:06.467" Title="Happy days alphametic" Tags="|no-computers|alphametic|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25961" PostTypeId="2" ParentId="25614" CreationDate="2016-01-30T18:32:56.087" Score="1" Body="&lt;p&gt;This has already been answered correctly by &lt;a href=&quot;https://puzzling.stackexchange.com/users/18590/r-daughtry&quot;&gt;R. Daughtry&lt;/a&gt;, but as per the comments, a few clues were not fully understood. So this answer only serves to fill in those blanks for anyone who wants to know.&lt;/p&gt;&#xA;&lt;p&gt;Line 5: &lt;em&gt;half way to becoming a maestro&lt;/em&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; A maestro is a talented and distinguished &lt;strong&gt;conductor&lt;/strong&gt;, and another word for half is &lt;strong&gt;semi&lt;/strong&gt;, so we get &lt;strong&gt;semi-conductor&lt;/strong&gt;, which is part of the name and also the primary industry.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Line 8: &lt;em&gt;my ambition guided me, and those I supported, in giant leaps&lt;/em&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The clue here is &lt;em&gt;giant leaps&lt;/em&gt;, like the &lt;em&gt;&amp;quot;...one giant leap for mankind&amp;quot;&lt;/em&gt; in Neil Armstrong's moon-landing speech. Fairchild's chips were used in the guidance computer for the Apollo missions, an ambitious project. Thus, they guided both the company and literally supported the astronauts' lives in making such leaps.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Line 9: &lt;em&gt;when things got too noisy I went one way&lt;/em&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Noise was an issue in logic circuits, but diodes (a one-way conductor) were used to manage this noise.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Line 10: &lt;em&gt;is this fair&lt;/em&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; This is another clue to the company name, not just a lead-in for the final two lines.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18522" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-01-30T18:32:56.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25962" PostTypeId="2" ParentId="25956" CreationDate="2016-01-30T18:55:05.670" Score="7" Body="&lt;p&gt;While there is another answer that states&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the atom&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is the answer, which I agree fits very well. I think there is something else the also fits, perhaps better.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sub-atomic particles (Quarks, Leptons, Bosons).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;By my extreme size, I escape your eye&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are too small to see.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The nature of every thing, is but many of I&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Atoms are made of these, as larger structures are made of atoms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Since my unveiling, the inexact came by&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This refers to the probabilistic &lt;em&gt;&quot;nature of nature&quot;&lt;/em&gt; at this &lt;em&gt;quantum&lt;/em&gt; scale.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And there came a new discipline to unify&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The new discipline was quantum mechanics / quantum theory. It was thought at the time that it would be a unifying theory, where Einstein's efforts for a unifying theory so far had failed. However, although this new discipline came to unify, it has not yet delivered.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18522" LastActivityDate="2016-01-30T18:55:05.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25963" PostTypeId="1" AcceptedAnswerId="25965" CreationDate="2016-01-30T18:56:08.220" Score="1" ViewCount="178" Body="&lt;p&gt;Take the 3-letter name of an intelligent, non-human species from a major fictional book series. (Make sure the species name is in singular form, &quot;cat&quot; instead of &quot;cat&lt;strong&gt;s&lt;/strong&gt;&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, name a 5-letter way that refers to the way you might cook your eggs: _____ eggs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, take the first 3 letters of the way you cook your eggs, drop them in front of the 3-letter species name, and replace the last letter of the species name with the last letter of the way you cook your eggs. Now, you'll get the relationship of this species to the main protagonist in this fictional series. What is this relationship? &lt;/p&gt;&#xA;" OwnerUserId="18823" LastActivityDate="2016-01-30T19:20:19.430" Title="Combine the name of a fictional species and a way to cook eggs to find the answer" Tags="|word|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25964" PostTypeId="1" AcceptedAnswerId="25966" CreationDate="2016-01-30T19:15:18.753" Score="7" ViewCount="1518" Body="&lt;p&gt;After the war seven long years ago&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 men gave me life, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But solemn in peace not in bloody strife&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now 17 years have passed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And my enemies I have surpassed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For I am all around them&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 colors I flash, and 6 letters I splash&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Across a great white canvas&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="18823" LastEditorUserId="18823" LastEditDate="2016-01-30T20:50:35.623" LastActivityDate="2016-10-11T15:17:18.197" Title="You use me every day - what am I?" Tags="|riddle|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25965" PostTypeId="2" ParentId="25963" CreationDate="2016-01-30T19:20:19.430" Score="1" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Friend&lt;br&gt;&#xA; Species: ENT Eggs: FRIED&lt;br&gt;&#xA; Following the instructions we get:&lt;br&gt;&#xA; FRIENT&lt;br&gt;&#xA; FRIEND  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18522" LastActivityDate="2016-01-30T19:20:19.430" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25966" PostTypeId="2" ParentId="25964" CreationDate="2016-01-30T19:41:06.680" Score="8" Body="&lt;p&gt;I think it is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Google.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;After the war, 2 men gave me life,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It was founded in Sept. 1998, by two men (Larry Page and Sergey Brin). According to &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_wars_1990%E2%80%932002&quot;&gt;wikipedia&lt;/a&gt; the only war that ended in 1998 was a &lt;a href=&quot;https://en.wikipedia.org/wiki/War_in_Abkhazia_%281998%29&quot;&gt;War in Abkhazia&lt;/a&gt;, it's a bit obscure, but let's go with that!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In peace not in strife&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are a peaceful company, with a peaceful &quot;do no evil&quot; policy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Now 17 years have passed, and my enemies I have surpassed&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; At time of posting, they are 17 years old (and 4 months).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;For I am all around them&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Google is everywhere, in lots of areas, and dominant in its core industry.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;4 colors I flash, and 6 letters I splash&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Logo has 4 colours, &quot;Google&quot; has 6 letters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Across a great white canvas&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Their home page has always been intentionally left mostly blank white.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18522" LastEditorUserId="18522" LastEditDate="2016-01-30T20:45:21.593" LastActivityDate="2016-01-30T20:45:21.593" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25967" PostTypeId="1" AcceptedAnswerId="25968" CreationDate="2016-01-30T20:00:04.460" Score="4" ViewCount="2848" Body="&lt;pre&gt;&lt;code&gt;I am something...&#xA;When you see me, you always clap for me&#xA;yet you hate me, but prefer me to my wife&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;who am I?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="18826" LastEditorUserId="18826" LastEditDate="2016-01-30T20:05:08.840" LastActivityDate="2016-01-30T20:13:39.830" Title="Good One --&gt; Riddle: ...You clap for me" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25968" PostTypeId="2" ParentId="25967" CreationDate="2016-01-30T20:13:39.830" Score="5" Body="&lt;p&gt;Are you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a mosquito?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am something...  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a mosquito is something&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you see me, you always clap for me  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I wanna kill you! I wanna smash you! I wanna make sure you won't survive the clap!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yet you hate me, but prefer me to my wife  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; only female mosquito suck blood, male ones are just annoying buzzing creatures&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9352" LastActivityDate="2016-01-30T20:13:39.830" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25969" PostTypeId="2" ParentId="25934" CreationDate="2016-01-30T20:26:34.113" Score="0" Body="&lt;p&gt;I think the answer is...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Genes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Long ago, my militant leader devised a plan for world domination&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Scooping up whatever organic molecules he could, this leader figured out a way to replicate himself, build armour, and weapons, to coldly dominate with copies of himself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Though he is now gone, I am still carrying out his plan&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Evolution marches on. Genes are replicating entities within the larger replicating structure of DNA. All of which is a distant descendant of that original replicating 'leader', but that specific replicator is long gone, probably consumed by some of his stronger descendants. Genes now carry out the same plan.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Today, our territory is near second to none&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We construct our survival machines all over the world, pretty-much.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Many have died for me to live&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;&quot;The secrets of evolution are time, and death.&quot;&lt;/em&gt; - Carl Sagan. Genes would not be where they are today without death.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And I am well seasoned, but there are times when my armor does fall away&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Genes are usually very well adapted to influencing whatever physical and behavioural traits benefit themselves. However, mutation (in themselves or other influencing genes) often prevents them from being passed on in particular hosts (the living things they build in order to replicate down the generations). Extinction also fits this clue as it stops a line of genes, whereas mutation would only affect one line... his identical counterparts would still be passed on in others.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I have secrets of my own, but you must know me before you could begin know them&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oh, the secrets in DNA and our genes! The general public have an awareness of these, but only those who study genetic material and really have a good understanding of it can hope to uncover some of the unknown secrets they hold.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18522" LastActivityDate="2016-01-30T20:26:34.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25970" PostTypeId="1" AcceptedAnswerId="25976" CreationDate="2016-01-30T21:40:06.330" Score="4" ViewCount="7872" Body="&lt;p&gt;We are siblings&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We follow each other everywhere&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But can never touch each other&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who are we?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CLUE : We live inside your home, good for the summer, bad in the winter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18826" LastEditorUserId="18826" LastEditDate="2016-01-31T00:49:40.893" LastActivityDate="2016-02-01T13:01:57.293" Title="Riddle(Tough one): we are siblings, who are we?" Tags="|riddle|" AnswerCount="6" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25971" PostTypeId="1" AcceptedAnswerId="25975" CreationDate="2016-01-30T21:41:44.507" Score="4" ViewCount="118" Body="&lt;p&gt;The Great Ring in the darkness binds us, the Eye of Fire at its centre. If you are on the ball (and I know you are), tell me, how many times have you been here before?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Now, if you asked me, I would answer 42. But what is the question?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18776" LastActivityDate="2016-01-30T22:44:55.163" Title="The Great Ring in the darkness binds us" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25972" PostTypeId="1" AcceptedAnswerId="26008" CreationDate="2016-01-30T21:48:02.773" Score="4" ViewCount="371" Body="&lt;p&gt;Upon reaching the door to the office, instead of a normal keypad, you are presented with a strange grid. Tacked to the wall there is a punch-card with five columns, each column containing the numbers from zero to nine. At the top of each column there is also a blank dot, as well as a dot with a decimal point at its center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can only assume that the answer to the strange grid must be entered on the punchcard, which in turn must be inserted in a credit card sized slot in the wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seeing the strange slashes and the mathematical symbols, you begin to panic. Math wasn't your forte in school, and now, your life may depend on your completion of this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/homuW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/homuW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;What is the password that this grid holds in its solution?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="17744" LastEditorUserId="17744" LastEditDate="2016-01-31T19:52:48.600" LastActivityDate="2016-02-01T00:31:20.773" Title="A mysterious grid" Tags="|mathematics|cipher|geometry|password|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25973" PostTypeId="2" ParentId="25970" CreationDate="2016-01-30T22:07:07.950" Score="4" Body="&lt;p&gt;Could it be...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; eyes or ears?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But then again, it can be a lot of things...&lt;/p&gt;&#xA;" OwnerUserId="12901" LastActivityDate="2016-01-30T22:07:07.950" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25974" PostTypeId="2" ParentId="25960" CreationDate="2016-01-30T22:18:32.773" Score="3" Body="&lt;p&gt;The corresponding numbers are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A = 9&lt;br&gt;&#xA; D = 3&lt;br&gt;&#xA; E = 8&lt;br&gt;&#xA; H = 2&lt;br&gt;&#xA; P = 6&lt;br&gt;&#xA; S = 0&lt;br&gt;&#xA; Y = 1  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The summation then becomes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  29661&lt;br&gt;&#xA;  29661&lt;br&gt;&#xA;  29661&lt;br&gt;&#xA; + 3910&lt;br&gt;&#xA; ------&lt;br&gt;&#xA;  92893&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4624" LastActivityDate="2016-01-30T22:18:32.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25975" PostTypeId="2" ParentId="25971" CreationDate="2016-01-30T22:44:55.163" Score="4" Body="&lt;p&gt;The great ring in the darkness could be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The path of the earth around the sun (the eye of fire). We are all bound to follow that orbital path by the effect of gravity.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How many times have I been here before?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well, that's the number of times that the planet has been at this point in it's orbital path. Which is a personal question as you're asking my age. So I'll just say that it's more than 20 and less than 60!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2016-01-30T22:44:55.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25976" PostTypeId="2" ParentId="25970" CreationDate="2016-01-30T23:15:32.907" Score="3" Body="&lt;h2&gt;First Guess&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Sun and the Moon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are siblings&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are several mythologies that show the sun and moon as siblings (&lt;a href=&quot;https://en.wikipedia.org/wiki/Twins_in_mythology&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia lists Apollo and Artemis &amp;amp; Kuat and Iae&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We follow each other everywhere&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Though the moon does not technically follow the sun - it can appear that way to the naked eye and has many times been referenced as such in different mythologies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But we can never touch each other&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Physics.SE has &lt;a href=&quot;https://physics.stackexchange.com/questions/220112/how-can-we-see-the-moon-while-its-between-the-earth-and-the-sun&quot;&gt;some cool diagrams and explanations on the phases of the moon&lt;/a&gt; but the main point is even during a full eclipse the moon and the sun are far, far from each other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Second Guess&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fan or Fan Blades more accurately&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are siblings&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The blades could be described as parts of the whole which could be considered brothers or sisters&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We follow each other everywhere&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fan blades typically rotate in a circular fashion (or follow each other)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But we can never touch&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Barring malfunction the fan blades shouldn't touch&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clue:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fans are great in the summer but in the winter (unless you live in a tropical climate like me) just make you colder.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11239" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:18.130" LastActivityDate="2016-01-31T00:46:51.987" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25977" PostTypeId="2" ParentId="25970" CreationDate="2016-01-30T23:35:12.410" Score="1" Body="&lt;p&gt;Well if you are not&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; eyes or ears &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;then you can only be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bernie Sanders and Donald Trump&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;siblings on opposite ends of the&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; american political spectrum&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;following each other around on the&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; neverending campaign trail &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but you can't touch each other&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; or else you would mutually annihilate in a blast of bad hair that could flatten a whole city.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18776" LastActivityDate="2016-01-30T23:35:12.410" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25978" PostTypeId="2" ParentId="25970" CreationDate="2016-01-31T00:19:47.863" Score="1" Body="&lt;p&gt;Are they &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Two opposite poles of a magnet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10647" LastActivityDate="2016-01-31T00:19:47.863" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25979" PostTypeId="1" AcceptedAnswerId="25988" CreationDate="2016-01-31T00:25:06.693" Score="4" ViewCount="871" Body="&lt;p&gt;Complete the following sequence with just one number:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4;\ 9;\ 1;\ 6;\ 2;\ ??$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hints&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;It's a real number&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The sequence ends with that number&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There could be literally infinite similar sequences with the same ending number&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="17708" LastEditorUserId="11876" LastEditDate="2016-01-31T04:51:40.423" LastActivityDate="2018-05-22T11:59:11.480" Title="Complete the numerical sequence!" Tags="|number-sequence|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25980" PostTypeId="2" ParentId="25970" CreationDate="2016-01-31T00:29:48.880" Score="1" Body="&lt;h2&gt;First guess&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; day and night (or light and darkness)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Second guess (after given clue)&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Me and my shadow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We are siblings:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everyone has a shadow since the moment they're born.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Cannot touch each other:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Shadows are not material and then cannot touch nor be touched except if you are Peter Pan.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We follow each other everywhere.&lt;/li&gt;&#xA;&lt;li&gt;Great for summer and not so great for winter.&lt;/li&gt;&#xA;&lt;li&gt;Not a part of the body (clue given in another answer).&lt;/li&gt;&#xA;&lt;li&gt;Closer to you than you think (clue given in comment below):&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I have an air conditioner.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18833" LastEditorUserId="18833" LastEditDate="2016-01-31T10:41:25.893" LastActivityDate="2016-01-31T10:41:25.893" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25981" PostTypeId="2" ParentId="25979" CreationDate="2016-01-31T01:57:41.570" Score="5" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 49, 16 and 25 are all perfect squares.&lt;br&gt;&#xA; And that's where it ends.&lt;br&gt;&#xA; And there are literally an infinite number of sequences of squares ending in 5.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastActivityDate="2016-01-31T01:57:41.570" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25982" PostTypeId="1" AcceptedAnswerId="25984" CreationDate="2016-01-31T02:14:41.650" Score="10" ViewCount="412" Body="&lt;p&gt;I work in a field &lt;br&gt;&#xA;My flower has four petals &lt;br&gt;&#xA;In fiction I'm gilded &lt;br&gt;&#xA;My heart's made of metal&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who am I?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; although @deepthought solved the riddle pretty quickly without additional help, here's a second verse I had prepared with more clues for those interested:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My friend makes the rules &lt;br&gt;&#xA; We're good at construction &lt;br&gt;&#xA; And oft used as tools &lt;br&gt;&#xA; Please resume your deduction&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user17947" LastEditorDisplayName="user17947" LastEditDate="2016-01-31T05:19:41.337" LastActivityDate="2016-01-31T05:19:41.337" Title="I work in a field - who am I?" Tags="|riddle|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25983" PostTypeId="2" ParentId="25819" CreationDate="2016-01-31T04:07:27.463" Score="0" Body="&lt;p&gt;It will take them 6 hours, 8 minutes, and 30 seconds (Pi+3 hours). I'm assuming that they are capable of traveling in a perfect arc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both travelers walk in opposite directions away from each other for one hour. In the best case scenario, one would reach the wall at this point and radio the other. Otherwise, both would head clockwise around the circle. In the worst case scenario, the wall was immediately to the left of one of the wise guys and the other one has to walk half the radius of the circle (taking Pi hours) to reach the wall. They then radio the other, who is on the other side of the circle, to traverse the diameter back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 hour to reach the circle, Pi hours to traverse half the circle, 2 hours for the one who didn't reach it to cut back across; Pi+3 hours.&lt;/p&gt;&#xA;" OwnerUserId="18839" LastActivityDate="2016-01-31T04:07:27.463" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25984" PostTypeId="2" ParentId="25982" CreationDate="2016-01-31T04:09:25.673" Score="7" Body="&lt;p&gt;You are a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; compass&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You work in a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; magnetic&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;field, your flower &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the compass rose&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;has four petals&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; north/east/south/west&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In fiction you are gilded&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;The Golden Compass&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and your heart&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the needle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is made of metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nice riddle!&lt;/p&gt;&#xA;" OwnerUserId="18776" LastActivityDate="2016-01-31T04:09:25.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25985" PostTypeId="2" ParentId="25524" CreationDate="2016-01-31T04:14:12.107" Score="6" Body="&lt;p&gt;I would like to present this non-rigorous but hopefully more intuitive explanation for the optimal path. (The technique used here was very helpful for working on &lt;a href=&quot;https://puzzling.stackexchange.com/questions/25819/&quot;&gt;Oray's variant with two people&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first part of &lt;a href=&quot;https://puzzling.stackexchange.com/a/25538/&quot;&gt;2012rcampion's answer&lt;/a&gt; explains that we should go as far out as some tangent $l$, before going around the circle to get back to $l$ on the other side. Call the starting point $A$ and the circle $O$. Then the problem is this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Find the shortest path that comes from $A$, touches $l$, then goes around the circle and touches $l$ again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It won't change which path is shortest if we turn around at the end and go all the way back:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Find the shortest path that comes from $A$, touches $l$, then goes around the circle and touches $l$ again, and then goes back around the circle to $l$ and then $A$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, if we reflect the entire diagram over $l$, we get this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/a3fLK.png&quot; alt=&quot;1]&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of having our path touch $l$ and go back, we can have it switch sides every time instead, which won't change the length because it's just a reflection. So now the problem is this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Find the shortest path from point $A$ that goes around circle $O^\prime$, then around circle $O$, then goes to point $A^\prime$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Anyone should be able to do that (imagine putting a string from $A$ around the circles to $A^\prime$ and pulling it tight):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aNkIY.png&quot; alt=&quot;2]&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now if we only look at the part of the diagram above $l$, there's the answer without any calculations.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2016-01-31T04:26:16.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25986" PostTypeId="1" AcceptedAnswerId="25987" CreationDate="2016-01-31T05:38:12.750" Score="8" ViewCount="274" Body="&lt;p&gt;Here's a riddle about a misbehaving thing:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;For thousands of years I have been well behaved &lt;/p&gt;&#xA;&lt;p&gt;dancing and moving under their watchful gaze &lt;/p&gt;&#xA;&lt;p&gt;but secretly I embark on daring escapades &lt;/p&gt;&#xA;&lt;p&gt;tangling with friends, slipping through blockades &lt;/p&gt;&#xA;&lt;p&gt;but when doubters look I am still in one place &lt;/p&gt;&#xA;&lt;p&gt;Only recently have they seen through my clever charade &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="18843" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-01-31T06:10:37.380" Title="Riddle about misbehavior" Tags="|riddle|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25987" PostTypeId="2" ParentId="25986" CreationDate="2016-01-31T06:10:37.380" Score="5" Body="&lt;p&gt;I believe you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An electron&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For thousands of years I have been well behaved&lt;br&gt;&#xA;dancing and moving under their watchful gaze &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For thousands of years electrical phenomena were not really thought about very deeply by humans, considered as a mere curiosity. Later humans started to exploit electricity but thought of it in terms of classical physics as a kind of relatively well-behaved fluid flowing from place to place inside of an electrical conductor. (There's probably something more specific this clue is referring to, but I'm not sure what it is at the moment.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but secretly I embark on daring escapades &lt;br&gt;&#xA;tangling with friends, slipping through blockades &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An electron can exhibit &lt;a href=&quot;https://en.wikipedia.org/wiki/Quantum_entanglement&quot;&gt;quantum entanglement&lt;/a&gt; with other electrons, and through the mechanism of &lt;a href=&quot;https://en.wikipedia.org/wiki/Quantum_tunnelling&quot;&gt;quantum tunneling&lt;/a&gt; can &quot;slip through blockades&quot; to reach behind an energy barrier into a place it could not get to under classical physical laws.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but when doubters look I am still in one place &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The electron, properly understood in quantum mechanics as a wave function that simultaneously exists in many places, still has the property that whenever one looks for it by placing an experimental detection apparatus, it will only ever be observed in a specific place; mathematically speaking, its wave function &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Wave_function_collapse&quot;&gt;collapses&lt;/a&gt;&quot; when it is observed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Only recently have they seen through my clever charade &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Our understanding of how electrons actually behave is fairly recent, dating to the discovery of &lt;a href=&quot;https://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation&quot;&gt;Schrödinger's equation&lt;/a&gt; in 1925.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user17947" LastActivityDate="2016-01-31T06:10:37.380" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25988" PostTypeId="2" ParentId="25979" CreationDate="2016-01-31T07:27:02.367" Score="8" Body="&lt;p&gt;Could be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Z&lt;sub&gt;ero&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Get it?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alphabetically: F&lt;sub&gt;our&lt;/sub&gt;, N&lt;sub&gt;ine&lt;/sub&gt;, O&lt;sub&gt;ne&lt;/sub&gt;,&#xA; S&lt;sub&gt;ix&lt;/sub&gt;, T&lt;sub&gt;wo&lt;/sub&gt;, Z&lt;sub&gt;ero&lt;/sub&gt;&#xA; &lt;br&gt;&#xA; Even the complete alphabetical list of &lt;em&gt;all integers&lt;/em&gt; ends&#xA; with Z&lt;sub&gt;ero&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18129" LastActivityDate="2016-01-31T07:27:02.367" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25989" PostTypeId="2" ParentId="25956" CreationDate="2016-01-31T07:36:09.197" Score="2" Body="&lt;p&gt;I think it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Radiation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By my extreme size, I escape your eye &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Here I think that escape doesn't mean &quot;Fail to experience&quot; but rather it means &quot;Run away from confinement&quot;. And of course we can see each other's eye. I am not saying here that we can see the shape of the photon but I am saying that we can see the eye because of it. (For the fun fact I once saw &lt;a href=&quot;https://physics.stackexchange.com/questions/226410/can-photons-be-seen-with-naked-eye&quot;&gt;a question&lt;/a&gt; in which the OP thought that he could see photon)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The nature of every thing, is but many of I &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every object emits radiation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since my unveiling, the inexact came by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Speed_of_light#History&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; we can see that the first calculated value of light was $220000$ $km/s$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And there came a new discipline to unify&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; And finally in 1983 the speed of light was finally concluded to be $299792.458$ $km/s$ by the definition of metre.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:18.130" LastActivityDate="2016-01-31T07:36:09.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25990" PostTypeId="2" ParentId="25970" CreationDate="2016-01-31T09:56:57.560" Score="2" Body="&lt;p&gt;This should be it.. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Blades of a fan&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;EDIT- Explanation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are siblings&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;All the blades are alike and are connected&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We follow each other everywhere&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The move in the same circular path, apparently following each other&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But can never touch each other&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;They obviously don't touch each other&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18846" LastEditorUserId="18846" LastEditDate="2016-01-31T10:51:49.060" LastActivityDate="2016-01-31T10:51:49.060" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25991" PostTypeId="2" ParentId="25893" CreationDate="2016-01-31T10:32:12.330" Score="1" Body="&lt;p&gt;Taking a shot I'm 90% sure about.&lt;br&gt;&#xA;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The right direction&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So sharp that I can execute what you want me to. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Right direction is aiming towards a specific point/place and hence is sharp, therefore it executes the task of showing the right path, which is what we want.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Will guide you the way, and when needed I will split your troubles in your head &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This is kinda self-explanatory.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So valuable, that you always fight for me &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;We always fight to find the right direction in our life&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Standing at the crossroads, waiting for you to find me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Self-explanatory again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18846" LastActivityDate="2016-01-31T10:32:12.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25993" PostTypeId="2" ParentId="25941" CreationDate="2016-01-31T12:46:31.563" Score="3" Body="&lt;p&gt;You are an&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; answer verifier&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;when you tell the truth all the time,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; we know that the answer is true or not.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you lie all the time,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; we again know when the answer is true or not. When you say it is &quot;true&quot;, we know it is not and vice versa.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, when you lie sometimes,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you are useless and we have no idea if the answer is true or not.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Although,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You may be a fortuneteller based on the same reasoning.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18645" LastActivityDate="2016-01-31T12:46:31.563" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25994" PostTypeId="2" ParentId="25922" CreationDate="2016-01-31T12:52:26.923" Score="5" Body="&lt;p&gt;Post-script. Jared is right and I didn't read the conditions correctly. I missed that the coefficients have to be positive. What I wrote below is rubbish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He can't. Let S = {s&lt;sub&gt;1&lt;/sub&gt;, s&lt;sub&gt;2&lt;/sub&gt;, ... s&lt;sub&gt;n&lt;/sub&gt;} be the set of numbers on which he could choose to sample P(x) so he will know the values P(s&lt;sub&gt;1&lt;/sub&gt;), P(s&lt;sub&gt;2&lt;/sub&gt;), ... P(s&lt;sub&gt;n&lt;/sub&gt;). Now define the polynomial M&lt;sub&gt;V&lt;/sub&gt;(x) as Product&lt;sub&gt;&lt;i&gt; all i in S except V &lt;/i&gt;&lt;/sub&gt;(x-i) so now M&lt;sub&gt;V&lt;/sub&gt;(x) is zero everywhere in S but non-zero at x=V. (The value at V is probably quite large)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alice now chooses a random polynomial Q(x) and a random integer R and adds P(x) and R.M&lt;sub&gt;2016&lt;/sub&gt;(x) to get P(x). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When Bob samples P(x) he gets the values of Q(x) because M&lt;sub&gt;2016&lt;/sub&gt;(x) is zero on all his sample points. He doesn't know R so he can't work out the value P(2016) = Q(2016)+R.M&lt;sub&gt;2016&lt;/sub&gt;(2016)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The answer that has been accepted above uses Q(x) = 17x&lt;sup&gt;4&lt;/sup&gt; + 4x&lt;sup&gt;3&lt;/sup&gt; + 5x&lt;sup&gt;2&lt;/sup&gt; + 8 as an example but P(x) = Q(x) + (x-1)(x-1000) takes exactly the same values at the sample-points of 1 and 1000.&lt;/p&gt;&#xA;" OwnerUserId="18849" LastEditorUserId="18849" LastEditDate="2016-01-31T16:36:36.290" LastActivityDate="2016-01-31T16:36:36.290" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25995" PostTypeId="2" ParentId="25819" CreationDate="2016-01-31T13:50:17.897" Score="0" Body="&lt;p&gt;I think &lt;strong&gt;both&lt;/strong&gt; of them can reach the wall in less than 4 hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For starters, they begin to walk in the opposite directions. Let us name them as $c$ for closer one and $f$ for further.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Case 1&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;After 1 hour, if $c$ finds the wall, then $f$ turns back 180 degrees and walks 2 hour. 1 + 2 = 3 hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MZsnQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MZsnQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Case 2&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If neither finds the wall after 1 hour, then both of them again turns 90 degrees in the opposite directions and walk 1 more hour. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 2.1&lt;/strong&gt;&lt;br&gt;&#xA;If the $c$ heads in the right direction, then he will find the wall less than 1 hour. $f$ will compute the angle of his direction, correct his direction and walk through the wall, which will take less than 2 hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/11zZV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/11zZV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 2.2&lt;/strong&gt;&lt;br&gt;&#xA;If $c$ again heads in the wrong direction, he will not find wall after 1 hour. However, now $f$ will be closer to the wall. $f$ turns another 90 degrees and keeps on walking. $c$ waits $f$ to reach the wall, and then himself walks through the wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5AEAM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5AEAM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18645" LastActivityDate="2016-01-31T13:50:17.897" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25996" PostTypeId="1" AcceptedAnswerId="26004" CreationDate="2016-01-31T13:50:18.480" Score="7" ViewCount="610" Body="&lt;p&gt;Your Goal: Win the game turn 1 with you beginning the game (so not on the draw) using the cheapest price total of cards used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rules:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Use the &quot;Price Trend&quot; price from &lt;a href=&quot;http://www.magiccardmarket.eu&quot; rel=&quot;nofollow&quot;&gt;http://www.magiccardmarket.eu&lt;/a&gt;  &lt;/li&gt;&#xA;&lt;li&gt;For each card used, provide a link to the magiccardmarket page of that card  &lt;/li&gt;&#xA;&lt;li&gt;Basic lands are considered free&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://magic.wizards.com/en/gameinfo/gameplay/formats/legacy&quot; rel=&quot;nofollow&quot;&gt;Legacy&lt;/a&gt; banlist applies&lt;/li&gt;&#xA;&lt;li&gt;Your opponent's deck consists entirely of basic Islands and keeps his hand of 7 cards&lt;/li&gt;&#xA;&lt;li&gt;You may stack your deck as you like, even after shuffling&lt;/li&gt;&#xA;&lt;li&gt;You should not rely on your opponent for winning the game. If the opponent is given a choice, any possible choice should win you the game (so no choosing 20 with &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=88803&quot; rel=&quot;nofollow&quot;&gt;Choice of Damnations&lt;/a&gt; for example, and with Chain Lightning kind of cards you should be able to win the game whether he chooses to copy or not)&lt;/li&gt;&#xA;&lt;li&gt;Extra turns are not allowed because I don't consider that still turn 1&lt;/li&gt;&#xA;&lt;li&gt;Do not rely on random actions like coin flipping or choosing cards at random&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Of course prices tend to change so the best answer could change over time but that's ok. I'll re-evaluate answers regularly and most cheap cards are somewhat constant. I'll start off with an answer of my own.&lt;/p&gt;&#xA;" OwnerUserId="1748" LastEditorUserId="1748" LastEditDate="2016-02-01T09:18:37.280" LastActivityDate="2016-02-02T00:27:43.767" Title="Magic-the-Gathering Challenge: Cheapest Legacy Turn 1 kill" Tags="|open-ended|magic-the-gathering|" AnswerCount="3" CommentCount="3" ClosedDate="2021-07-12T15:09:30.210" ContentLicense="CC BY-SA 3.0" />
  <row Id="25997" PostTypeId="2" ParentId="25996" CreationDate="2016-01-31T13:50:18.480" Score="1" Body="&lt;p&gt;Cards used:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Magic+2012/Reverberate&quot; rel=&quot;nofollow&quot;&gt;Reverberate (Magic 2012)&lt;/a&gt; - €0,37  &lt;/li&gt;&#xA;&lt;li&gt;4 x &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Coldsnap/Rite+of+Flame&quot; rel=&quot;nofollow&quot;&gt;Rite of Flame (Coldsnap)&lt;/a&gt; - 4 x€0,75 = €3,00  &lt;/li&gt;&#xA;&lt;li&gt;Mountain  &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Magic+2012/Sorin%27s+Vengeance&quot; rel=&quot;nofollow&quot;&gt;Sorin's Vengeance (Magic 2012)&lt;/a&gt; - €0,54  &lt;/li&gt;&#xA;&lt;li&gt;2 x &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Shadowmoor/Manamorphose&quot; rel=&quot;nofollow&quot;&gt;Manamorphose (Shadowmoor)&lt;/a&gt; - 2 x €2,19 = €4,38&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Total €8,29&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Play mountain&lt;/li&gt;&#xA;&lt;li&gt;Cast 4 times Rite of Flame (RRRRRRRRRR in pool)&lt;/li&gt;&#xA;&lt;li&gt;Cast 2 times Manamorphose adding B. (BBBBRRRRRR in pool)&lt;/li&gt;&#xA;&lt;li&gt;Cast Sorin's Vengeance copying it with Reverberate&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2016-01-31T13:50:18.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25998" PostTypeId="1" AcceptedAnswerId="25999" CreationDate="2016-01-31T14:03:15.220" Score="18" ViewCount="1782" Body="&lt;p&gt;Professor Halfbrain owns a 99×99 board for fantasy chess, whose rows are numbered consecutively from 1 to 99 and whose columns are also numbered consecutively from 1 to 99.&#xA;A fantasy knight can jump from a square in the -th column to any square in the -th row (and can jump to no other square); note that if the knight can jump from square  to square , then this does not mean that it can also jump from square  to square .&lt;/p&gt;&#xA;&lt;p&gt;The professor claims that there exists a closed fantasy knight tour on the chessboard that makes the knight visit every square exactly once, and in the end takes it back to its starting square.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is Halfbrain's claim indeed true, or has the professor once again made one of his mathematical blunders?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="69582" LastEditDate="2021-01-29T15:37:43.463" LastActivityDate="2021-01-29T15:37:43.463" Title="Professor Halfbrain and the fantasy knight" Tags="|mathematics|combinatorics|graph-theory|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25999" PostTypeId="2" ParentId="25998" CreationDate="2016-01-31T14:24:41.537" Score="27" Body="&lt;p&gt;Yes there is a solution with a very simple strategy:&lt;/p&gt;&#xA;&lt;p&gt;Start in (1,1).&lt;br /&gt;&#xA;Always go the right most square that's unvisited&lt;/p&gt;&#xA;&lt;p&gt;I'll try to illustrate it. I checked it by hand on an 9x9 board and a very nice pattern emerges that makes it clear it works on any X by X  board.&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;1&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;2&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;3&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;4&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;5&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;6&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;7&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;8&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;9&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;strong&gt;1&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;1&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;79&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;74&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;67&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;58&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;47&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;34&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;19&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;2&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;strong&gt;2&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;81&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;80&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;77&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;72&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;65&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;56&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;45&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;32&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;17&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;strong&gt;3&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;78&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;76&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;75&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;70&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;63&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;54&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;43&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;30&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;15&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;strong&gt;4&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;73&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;71&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;69&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;68&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;61&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;52&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;41&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;28&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;13&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;strong&gt;5&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;66&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;64&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;62&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;60&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;59&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;50&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;39&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;26&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;11&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;strong&gt;6&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;57&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;55&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;53&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;51&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;49&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;48&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;37&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;24&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;9&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;strong&gt;7&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;46&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;44&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;42&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;40&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;38&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;36&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;35&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;22&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;7&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;strong&gt;8&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;33&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;31&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;29&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;27&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;25&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;23&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;21&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;20&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;5&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;&lt;strong&gt;9&lt;/strong&gt;&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;18&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;16&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;14&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;12&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;10&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;8&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;6&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;4&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;" OwnerUserId="1748" LastEditorUserId="69582" LastEditDate="2021-01-29T15:32:24.717" LastActivityDate="2021-01-29T15:32:24.717" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26001" PostTypeId="1" AcceptedAnswerId="26013" CreationDate="2016-01-31T17:30:41.200" Score="6" ViewCount="300" Body="&lt;p&gt;Let's call a set of points &lt;em&gt;good&lt;/em&gt; if every pair of points in the set are an integer distance apart. Let's call a circle &lt;em&gt;small&lt;/em&gt; if its radius is less than 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good set lies on the boundary of a small circle. How many points can the set contain?&lt;/p&gt;&#xA;" OwnerUserId="18525" LastActivityDate="2016-02-01T21:45:32.447" Title="Good sets on small circles" Tags="|geometry|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26002" PostTypeId="2" ParentId="25998" CreationDate="2016-01-31T17:42:36.917" Score="22" Body="&lt;p&gt;Let $(x,y)$ be the square in row $x$, column $y$, so that a fantasy knight can move from $(x,y)$ to $(y,z)$.  A closed tour is described by a cyclic sequence $$x_0,x_1,x_2,\ldots,x_{99^2-1},x_{99^2}=x_0,$$ where the knight moves from $(x_0,x_1)$ to $(x_1,x_2)$, then to $(x_2,x_3)$, and so on up to $(x_{99^2-1},x_0)$, then finally back to $(x_0,x_1)$. Each square is visited exactly once, so this is an example of a de Bruijn sequence (specifically a $99$-ary de Bruijn sequence of order $2$). De Bruijn sequences are known to exist (&lt;a href=&quot;https://en.wikipedia.org/wiki/De_Bruijn_sequence#Construction&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Wikipedia article&lt;/a&gt; describes a construction), so Halfbrain's claim is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some other puzzles on this site have answers involving de Bruijn sequences (I found &lt;a href=&quot;https://puzzling.stackexchange.com/questions/22903/arranging-all-the-dominos/22904#22904&quot;&gt;this&lt;/a&gt;, &lt;a href=&quot;https://puzzling.stackexchange.com/questions/22842/one-beer-too-many/22843#22843&quot;&gt;this&lt;/a&gt;, &lt;a href=&quot;https://puzzling.stackexchange.com/questions/9397/shortest-number-containing-the-numbers-1-100/9399#9399&quot;&gt;this&lt;/a&gt;, and &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16698/unique-licence-plates/16706#16706&quot;&gt;this&lt;/a&gt;). &lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2016-01-31T17:48:52.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26003" PostTypeId="1" AcceptedAnswerId="26005" CreationDate="2016-01-31T18:28:59.170" Score="1" ViewCount="201" Body="&lt;p&gt;I am one of many kinds&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some say my opposite are all the same&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For that others say I'm cray&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can like my kin&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can like their kin&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can be a former them&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And they a former me&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kids ask why I'm hurt&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For I bleed 48 times every leap&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But my role no longer is to sweep&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who are I, and who are they?&lt;/p&gt;&#xA;" OwnerUserId="18816" LastEditorUserId="18816" LastEditDate="2016-01-31T19:54:04.933" LastActivityDate="2016-01-31T19:54:59.450" Title="I am one of many kinds" Tags="|riddle|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26004" PostTypeId="2" ParentId="25996" CreationDate="2016-01-31T18:29:23.373" Score="9" Body="&lt;p&gt;&lt;strong&gt;Total Price: €0,47&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cards Used:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 53x Swamp&lt;br&gt;&#xA;&lt;br&gt;&#xA; 2x &lt;em&gt;Any Bulk Green Common&lt;/em&gt; - €0,00&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Khans+of+Tarkir/Dead+Drop&quot; rel=&quot;nofollow&quot;&gt;Dead Drop (Khans of Tarkir)&lt;/a&gt; - €0,03&lt;br&gt;&#xA;&lt;br&gt;&#xA; 2x &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Onslaught/Erratic+Explosion&quot; rel=&quot;nofollow&quot;&gt;Erratic Explosion (Onslaught)&lt;/a&gt; - €0,06&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Ice+Age/Burnt+Offering&quot; rel=&quot;nofollow&quot;&gt;Burnt Offering (Ice Age)&lt;/a&gt; - €0,04&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Prerelease+Promos/Allosaurus+Rider&quot; rel=&quot;nofollow&quot;&gt;Allosaurus Rider (Prerelease Promos)&lt;/a&gt; - €0,34  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Play Order:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Exile the two bulk green commons to play &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Prerelease+Promos/Allosaurus+Rider&quot; rel=&quot;nofollow&quot;&gt;Allosaurus Rider&lt;/a&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play Swamp and tap for B to play &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Ice+Age/Burnt+Offering&quot; rel=&quot;nofollow&quot;&gt;Burnt Offering&lt;/a&gt; on Allosaurus Rider for RRRRRRR&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Onslaught/Erratic+Explosion&quot; rel=&quot;nofollow&quot;&gt;Erratic Explosion&lt;/a&gt; for RRR targeting your opponent&lt;br&gt;&#xA;&lt;br&gt;&#xA; Reveal &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Khans+of+Tarkir/Dead+Drop&quot; rel=&quot;nofollow&quot;&gt;Dead Drop&lt;/a&gt; (the only non-land in your deck) to Erratic Explosion, dealing 10 damage&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Onslaught/Erratic+Explosion&quot; rel=&quot;nofollow&quot;&gt;Erratic Explosion&lt;/a&gt; for RRR targeting your opponent&lt;br&gt;&#xA;&lt;br&gt;&#xA; Reveal &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Khans+of+Tarkir/Dead+Drop&quot; rel=&quot;nofollow&quot;&gt;Dead Drop&lt;/a&gt; (the only non-land in your deck) to Erratic Explosion, dealing 10 damage  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Winning through straight combat damage is a bit more expensive, but still under €1,00:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Total Price: €0,91&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cards Used:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Swamp&lt;br&gt;&#xA;&lt;br&gt;&#xA; 2x &lt;em&gt;Any Bulk Green Common&lt;/em&gt; - €0,00&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Ice+Age/Burnt+Offering&quot; rel=&quot;nofollow&quot;&gt;Burnt Offering (Ice Age)&lt;/a&gt; - €0,04&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Magic+2015/Hammerhand&quot; rel=&quot;nofollow&quot;&gt;Hammerhand (Magic 2015)&lt;/a&gt; - €0,10&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Prerelease+Promos/Allosaurus+Rider&quot; rel=&quot;nofollow&quot;&gt;Allosaurus Rider (Prerelease Promos)&lt;/a&gt; - €0,34&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Tempest/Minion+of+the+Wastes&quot; rel=&quot;nofollow&quot;&gt;Minion of the Wastes (Tempest)&lt;/a&gt; - €0,43  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Play Order:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Exile the two bulk green commons to play &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Prerelease+Promos/Allosaurus+Rider&quot; rel=&quot;nofollow&quot;&gt;Allosaurus Rider&lt;/a&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play Swamp and tap for B to play &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Ice+Age/Burnt+Offering&quot; rel=&quot;nofollow&quot;&gt;Burnt Offering&lt;/a&gt; on Allosaurus Rider for BBBRRRR&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Tempest/Minion+of+the+Wastes&quot; rel=&quot;nofollow&quot;&gt;Minion of the Wastes&lt;/a&gt; for BBBRRR, paying 19 life for a 19/19&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Magic+2015/Hammerhand&quot; rel=&quot;nofollow&quot;&gt;Hammerhand&lt;/a&gt; for R enchanting Minion of the Wastes (giving it haste and +1/+1)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Move to combat and swing for 20 damage &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Previous Solutions&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Total Price: €0,58&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cards Used:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Forest&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Darksteel/Myr+Moonvessel&quot; rel=&quot;nofollow&quot;&gt;Myr Moonvessel (Darksteel)&lt;/a&gt; - €0,04&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Innistrad/Infernal+Plunge&quot; rel=&quot;nofollow&quot;&gt;Infernal Plunge (Innistrad)&lt;/a&gt; - €0,09&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Ice+Age/Tinder+Wall&quot; rel=&quot;nofollow&quot;&gt;Tinder Wall (Ice Age)&lt;/a&gt; - €0,18&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Mercadian+Masques/Blood+Oath&quot; rel=&quot;nofollow&quot;&gt;Blood Oath (Mercadian Masques)&lt;/a&gt; - €0,27 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Play Order:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Play Forest, tap for G (G)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Ice+Age/Tinder+Wall&quot; rel=&quot;nofollow&quot;&gt;Tinder Wall&lt;/a&gt; for G, sacrifice it for RR (RR)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Darksteel/Myr+Moonvessel&quot; rel=&quot;nofollow&quot;&gt;Myr Moonvessel&lt;/a&gt; for R, play &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Innistrad/Infernal+Plunge&quot; rel=&quot;nofollow&quot;&gt;Infernal Plunge&lt;/a&gt; for R on Myr Moonvessel for 1RRR (1RRR)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Mercadian+Masques/Blood+Oath&quot; rel=&quot;nofollow&quot;&gt;Blood Oath&lt;/a&gt; for 1RRR, naming &quot;Land&quot; for 21 damage  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Total Price: €0,77&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cards Used:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 53x Mountain&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Onslaught/Erratic+Explosion&quot; rel=&quot;nofollow&quot;&gt;Erratic Explosion (Onslaught)&lt;/a&gt; - €0,03&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Khans+of+Tarkir/Dead+Drop&quot; rel=&quot;nofollow&quot;&gt;Dead Drop (Khans of Tarkir)&lt;/a&gt; - €0,03&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Darksteel/Myr+Moonvessel&quot; rel=&quot;nofollow&quot;&gt;Myr Moonvessel (Darksteel)&lt;/a&gt; - €0,04&lt;br&gt;&#xA;&lt;br&gt;&#xA; 2x &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Innistrad/Infernal+Plunge&quot; rel=&quot;nofollow&quot;&gt;Infernal Plunge (Innistrad)&lt;/a&gt; - €0,18&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Fourth+Edition/Ornithopter&quot; rel=&quot;nofollow&quot;&gt;Ornithopter (Fourth Edition)&lt;/a&gt; - €0,12&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Magic+2012/Reverberate&quot; rel=&quot;nofollow&quot;&gt;Reverberate (Magic 2012)&lt;/a&gt; - €0,37  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Play Order:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Play Mountain, &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Fourth+Edition/Ornithopter&quot; rel=&quot;nofollow&quot;&gt;Ornithopter&lt;/a&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Tap Mountain for R, play &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Innistrad/Infernal+Plunge&quot; rel=&quot;nofollow&quot;&gt;Infernal Plunge&lt;/a&gt; for R on Ornithopter for RRR (RRR)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Darksteel/Myr+Moonvessel&quot; rel=&quot;nofollow&quot;&gt;Myr Moonvessel&lt;/a&gt; for R, play &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Innistrad/Infernal+Plunge&quot; rel=&quot;nofollow&quot;&gt;Infernal Plunge&lt;/a&gt; for R on Myr Moonvessel for 1RRR (1RRRR)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Hold priority, play &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Onslaught/Erratic+Explosion&quot; rel=&quot;nofollow&quot;&gt;Erratic Explosion&lt;/a&gt; for 1RR targeting your opponent (RR)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Play &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Magic+2012/Reverberate&quot; rel=&quot;nofollow&quot;&gt;Reverberate&lt;/a&gt; for RR targeting Erratic Explosion, target your opponent again with the copy&lt;br&gt;&#xA;&lt;br&gt;&#xA; Reveal &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Khans+of+Tarkir/Dead+Drop&quot; rel=&quot;nofollow&quot;&gt;Dead Drop&lt;/a&gt; (the only non-land in your deck) to Erratic Explosion, dealing 10 damage&lt;br&gt;&#xA;&lt;br&gt;&#xA; Reveal &lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Khans+of+Tarkir/Dead+Drop&quot; rel=&quot;nofollow&quot;&gt;Dead Drop&lt;/a&gt; again to the Erratic Explosion copy, dealing 10 damage&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="17389" LastEditDate="2016-02-02T00:27:43.767" LastActivityDate="2016-02-02T00:27:43.767" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26005" PostTypeId="2" ParentId="26003" CreationDate="2016-01-31T19:54:59.450" Score="2" Body="&lt;p&gt;I am one of many kinds&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some say my opposite are all the same&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For that others say I'm cray&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; men are all alike / women are crazy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can like my kin&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can like their kin&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can be a former them&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And they a former me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; homosexual / heterosexual / transgender&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Kids ask why I'm hurt&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For I bleed 48 times every leap&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; leap &lt;em&gt;year&lt;/em&gt;&#xA; (actually this line was the giveaway)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But my role no longer is to sweep&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; feminism&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who are I, and who are they?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Women and men. &#xA; (But I don't get the grammatical mistake clue, so maybe I'm missing something.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18776" LastActivityDate="2016-01-31T19:54:59.450" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26006" PostTypeId="2" ParentId="25996" CreationDate="2016-01-31T21:23:25.550" Score="2" Body="&lt;p&gt;Well since you didn't specify &quot;no randomness&quot;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;€0,16:&lt;/strong&gt;&lt;br&gt;&#xA;Mountain (€0)&lt;br&gt;&#xA;&lt;a href=&quot;https://www.magiccardmarket.eu/Products/Singles/Eighth+Edition/Mana+Clash&quot; rel=&quot;nofollow&quot;&gt;Mana Clash&lt;/a&gt; (€0,16)&lt;br&gt;&#xA;Your opponent flips tails, you flip heads, repeat twenty times, opponent takes twenty damage.&lt;/p&gt;&#xA;" OwnerUserId="7459" LastActivityDate="2016-01-31T21:23:25.550" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26007" PostTypeId="1" AcceptedAnswerId="26009" CreationDate="2016-01-31T21:44:23.907" Score="36" ViewCount="12010" Body="&lt;p&gt;You are to make a statement. &lt;br&gt;&#xA;Of these two offers you have to choose one; which one is more profitable to you, and why?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If the statement is true, you get exactly 10 dollars. If the statement is false, you get either less than or more than 10 dollars but not exactly 10.&lt;/li&gt;&#xA;&lt;li&gt;Regardless of whether the statement is true or false, you get more than 10 dollars.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1766" LastEditorUserId="1766" LastEditDate="2016-04-15T20:37:09.380" LastActivityDate="2016-04-15T20:37:09.380" Title="Which offer is better?" Tags="|liars|" AnswerCount="9" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26008" PostTypeId="2" ParentId="25972" CreationDate="2016-01-31T22:20:27.817" Score="5" Body="&lt;p&gt;The grid reminds me a bit of &lt;a href=&quot;https://esolangs.org/wiki/Fish&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;code&gt;&amp;gt;&amp;lt;&amp;gt;&lt;/code&gt;&lt;/a&gt;, an esoteric programming language where the program pointer moves around on a two-dimensional grid of characters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;code&gt;&amp;gt;&amp;lt;&amp;gt;&lt;/code&gt;, the two slashes &lt;code&gt;/&lt;/code&gt; and &lt;code&gt;\&lt;/code&gt; are reflectors.  Think of the program flow 'bouncing' off of them like hitting a wall.  Assuming the same initial condition as &lt;code&gt;&amp;gt;&amp;lt;&amp;gt;&lt;/code&gt; (starting in the upper-left corner headed right), and that none of the other characters modify the program flow, we get a flow like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/83wpS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sequence of 'instructions' we pass through is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;code&gt;9&lt;/code&gt; &lt;code&gt;x&lt;/code&gt; &lt;code&gt;3&lt;/code&gt; &lt;code&gt;x^2&lt;/code&gt; &lt;code&gt;+&lt;/code&gt; &lt;code&gt;3&lt;/code&gt; &lt;code&gt;x&lt;/code&gt; &lt;code&gt;1/6&lt;/code&gt; &lt;code&gt;+&lt;/code&gt; &lt;code&gt;22&lt;/code&gt; &lt;code&gt;sqrt&lt;/code&gt; &lt;code&gt;$&lt;/code&gt; &lt;code&gt;euro&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I assume that the different color of the &lt;code&gt;9&lt;/code&gt; is probably just to indicate that it is the starting point, and that the &lt;code&gt;&amp;gt;&lt;/code&gt; indicates that the flow exits to the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interpreting &lt;code&gt;x&lt;/code&gt; as &lt;code&gt;times&lt;/code&gt;, the first ten characters look like a mathematical expression written as a series of key to enter into a calculator:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;9 \xrightarrow{\times 3} 27 \xrightarrow{x^2} 729 \xrightarrow{+3} 732 \xrightarrow{\times\frac{1}{6}} 122 \xrightarrow{+22} 144 \xrightarrow{\sqrt{x}} 12&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two currency symbols, however, are still a mystery.  Presumably they indicate some sort of exchange rate conversion.  However, exchange rates are changing constantly and we lack a fixed time to determine a fixed exchange rate.  Also, we need (according to Hanko) a five-digit number.  Converting from &lt;code&gt;USD&lt;/code&gt; to &lt;code&gt;EUR&lt;/code&gt; is not a big change, so the integer part will remain two digits; and since both currencies are divided into exactly 100 subunits, that leaves us only two further decimal places, four total out of five.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally, If I were you, and believed that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;your life may depend on your completion of this problem&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'd probably just stop working there.&lt;/p&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="17744" LastEditDate="2016-02-01T00:31:20.773" LastActivityDate="2016-02-01T00:31:20.773" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="26009" PostTypeId="2" ParentId="26007" CreationDate="2016-01-31T22:23:39.017" Score="68" Body="&lt;p&gt;Take the&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;first&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;offer, and then state:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;The reward for making this statement is less than [&lt;em&gt;insert large value here&lt;/em&gt;] dollars, but not exactly 10&quot;.&lt;br&gt;&#xA;&lt;br&gt;&#xA;If you get 10 dollars, that makes your statement false, so you shouldn't have gotten 10 dollars.&lt;br&gt;&#xA;If you get less than [&lt;em&gt;large value&lt;/em&gt;] dollars but not 10, that makes your statement true, so you should have gotten 10 dollars.&lt;br&gt;&#xA;So the only way for the offer to be fulfilled correctly is if you get at least [&lt;em&gt;large value&lt;/em&gt;] dollars.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="18924" LastEditDate="2016-02-03T15:20:03.533" LastActivityDate="2016-02-03T15:20:03.533" CommentCount="24" ContentLicense="CC BY-SA 3.0" />
  <row Id="26013" PostTypeId="2" ParentId="26001" CreationDate="2016-01-31T23:35:17.627" Score="11" Body="&lt;p&gt;This is not a proper answer! I'll guess that the number is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 6&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But this is just a guess. (EDIT: now with ugly proof)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; any three-point subset of a good set is an integer-sided triangle, and the circumradius of each such triangle must be the same. The formula by the way is $$r=\frac{abc}{\sqrt{(a+b+c)(a+b-c)(c+a-b)(b+c-a)}}$$ So to start we are looking for different triples of integers which have give the same value of r when you put them in this formula (r doesn't need to be an integer). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cheated a bit here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I clicked on OP's profile, and see a previous problem, and a &lt;a href=&quot;https://puzzling.stackexchange.com/a/25503/18776&quot;&gt;comment about a certain hexagon whose side lengths are 5 or less&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I see by trial-and-error that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the equilateral triangle (7,7,7) matches (3,5,7) and (3,7,8) and (5,7,8), r = $\frac{7}{\sqrt{3}}$ which is just over 4. You can make a skewed hexagon out of this by taking two equilateral triangles of side length 7 to make a Star-of-David. Now rotate one triangle until the side lengths of the outer hexagon are 3-5-3-5-3-5, and this will make the diagonals 8, so everything is an integer and this is a &quot;good&quot; set. I think this is OP's hexagon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I also see by trial-and-error that &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the next equilateral triangle that can match anything else is (13,13,13), and r = $\frac{13}{\sqrt{3}}$ which is just past OP's critical value of 7. This is a puzzle and not just a math question. It could turn out that in fact there is a heptagon solution containing no equilateral triangle. But probably OP's hexagon is involved, and OP chose 7 for a reason. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again, this is just a guess&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; based on assumptions about OP :-) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;not a proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ADDED: Well since nobody is answering ... I found all the&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; different triples of integers up to 13 which have the same circumradius. There are a bunch of pairs of triples, and then three other higher-order coincidences: first, the one I found already; second, (4,4,7), (4,6,8), (4,8,8), (6,7,8), $r = \frac{16}{\sqrt{15}}$, just a bit bigger than the previous r; and, (3,8,10), (3,12,12), (8,8,12), and (8,10,12), $r=\frac{16}{\sqrt{7}}$, just over 6. Trial-and-error trying to fit these around a circle shows that these both make pentagons. The first has sides 4-4-6-4-6, one of the diagonals has length 7 and the other 4 have length 8. The other has sides 3-8-8-8-8, two diagonals have length 10, three have length 12. Both &quot;good&quot; sets. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To finish, we need to rule out the&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; pairs. { EDIT: Example: (2,3,4) and (2,4,4) are the smallest pair, $r=\frac{8}{\sqrt{15}}$ which is just over 2. This makes a quadrilateral with sides length 2-3-2-4 and both diagonals length 4. BTW we also know it fits on a circle by &lt;a href=&quot;https://en.wikipedia.org/wiki/Ptolemy%27s_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ptolemy's theorem&lt;/a&gt;, since $2 \cdot 2 + 3 \cdot 4 = 4 \cdot 4$. } Well, a subset of a good set is a good set, so we just need to show that any heptagon would need more than two different (ie. not congruent) kinds of triangle. Uhhhh. Well, it really looks like it ought to be true, and I want it to be true, so I'll just call that &quot;obvious&quot; ;-)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So my guess was right. But maybe OP had a more elegant proof in mind, and didn't intend for me to hax0r it with my 1337 programming skillz (by that I mean, I made a spreadsheet.. hey don't judge) &lt;/p&gt;&#xA;" OwnerUserId="18776" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2016-02-01T21:45:32.447" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26017" PostTypeId="2" ParentId="26007" CreationDate="2016-02-01T11:17:40.943" Score="23" Body="&lt;p&gt;Sounds like you just found a true Oracle, that's amazing. You can make good money with online bets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose the first offer and pick a statement like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The team &lt;em&gt;X&lt;/em&gt; will win against the team &lt;em&gt;Y&lt;/em&gt; in the match &lt;em&gt;Z&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If you get 10 dollars, team X will win.&lt;/li&gt;&#xA;&lt;li&gt;If you get a different amount of money, team X will tie or lose.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Bet any amount of dollars you have on the outcome of the match, and collect your prize.&lt;/p&gt;&#xA;" OwnerUserId="4394" LastEditorUserId="1766" LastEditDate="2016-02-01T13:24:43.917" LastActivityDate="2016-02-01T13:24:43.917" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26018" PostTypeId="1" CreationDate="2016-02-01T12:23:01.053" Score="-4" ViewCount="346" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&amp;lt;, (?), =, &gt;, (?), &gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hints: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1) a sequence, &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;2) the characters serve as symbols&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18447" LastEditorUserId="11876" LastEditDate="2016-02-01T14:38:18.630" LastActivityDate="2016-02-02T08:51:07.467" Title="What is this pattern and what are the missing characters?" Tags="|pattern|" AnswerCount="2" CommentCount="5" FavoriteCount="0" ClosedDate="2016-02-03T15:34:26.037" ContentLicense="CC BY-SA 3.0" />
  <row Id="26019" PostTypeId="2" ParentId="16106" CreationDate="2016-02-01T12:28:58.903" Score="34" Body="&lt;p&gt;I'll confess that I've always had a fondness for the Jester's riddle from &lt;a href=&quot;https://en.wikipedia.org/wiki/Zork_Zero&quot;&gt;Zork Zero&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;I once heard of a bookkeeper who, while working on the accounts of the Frobozz Magic Balloon Company, noted that the word 'balloon' has two double letters in a row! Stretching his limited imagination to the limit, this bookkeeper wondered if there were any words with THREE double letters in a row. He couldn't think of one -- but I'll bet that YOU can!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9286" LastActivityDate="2016-02-01T12:28:58.903" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26020" PostTypeId="2" ParentId="25091" CreationDate="2016-02-01T12:42:36.593" Score="0" Body="&lt;p&gt;Since they both claim to be a different one of the available options, they're either both &quot;truthing&quot; or both lying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since at least one of them is lying, they're not both speaking the truth and so they have to be both lying and so the one claiming to be Bob is actually Andrew and is wearing a red shirt.&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditDate="2016-02-05T14:47:41.497" LastActivityDate="2016-02-05T14:47:41.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26021" PostTypeId="1" CreationDate="2016-02-01T13:17:34.053" Score="2" ViewCount="1650" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;i. I'm not used to the noise yet. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;ii. They used to work in a circus. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;iii. Didn't there use to be a shop there? &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;iv. I never used to smoke. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;v. It didn't use to be so dirty.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18873" LastEditorUserId="9631" LastEditDate="2016-02-01T13:35:52.200" LastActivityDate="2016-02-01T14:54:57.103" Title="Find the &quot;odd one out&quot; and explain why" Tags="|english|" AnswerCount="2" CommentCount="3" ClosedDate="2016-02-01T18:38:30.487" ContentLicense="CC BY-SA 3.0" />
  <row Id="26022" PostTypeId="2" ParentId="26021" CreationDate="2016-02-01T13:22:23.823" Score="2" Body="&lt;p&gt;Maybe not the intended answer, but:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;ii. They used to work in a circus.&lt;/code&gt; is the only one that does not use a negative.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastActivityDate="2016-02-01T13:22:23.823" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26023" PostTypeId="1" CreationDate="2016-02-01T13:38:34.420" Score="2" ViewCount="158" Body="&lt;p&gt;A. The word ‘the’ is not used with the names of people.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B. The past simple form of the verb is only used to describe finished events or states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C. The word ‘some’ is used in statements; the word ‘any’ in negatives and questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D. The word ‘can’ (modal verb) is only used to express ability.&lt;/p&gt;&#xA;" OwnerUserId="18873" LastEditorUserId="8874" LastEditDate="2016-02-01T16:59:50.120" LastActivityDate="2016-02-01T16:59:50.120" Title="Can you think of any exceptions to the following rules?" Tags="|english|language|" AnswerCount="1" CommentCount="5" ClosedDate="2016-02-01T17:07:53.460" ContentLicense="CC BY-SA 3.0" />
  <row Id="26024" PostTypeId="1" AcceptedAnswerId="26025" CreationDate="2016-02-01T13:45:38.507" Score="2" ViewCount="205" Body="&lt;p&gt;You have 2 safety fuses and 4 matches. Each fuse takes exactly one hour to burn, but they are not perfect so may not necessarily burn evenly from both sides. For example, the first half might burn in 20 minutes and the second half might take 40 mins. but this does not change the total time to be burnt as stated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can you measure out 45 minutes by just using these two safety fuses and the matches you have?&lt;/p&gt;&#xA;" OwnerUserId="18514" LastEditorUserId="11876" LastEditDate="2016-02-01T14:05:49.493" LastActivityDate="2016-02-01T14:05:49.493" Title="Safety Fuses and Matches" Tags="|time|" AnswerCount="3" CommentCount="2" ClosedDate="2016-02-01T14:28:19.367" ContentLicense="CC BY-SA 3.0" />
  <row Id="26025" PostTypeId="2" ParentId="26024" CreationDate="2016-02-01T13:57:46.950" Score="4" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Light the first fuse at both ends and the second fuse at one end. Once the first fuse is completely burned, light the other end of the second fuse. Once the second fuse is completely burned, 45 minutes will have passed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;With the first fuse burning at both ends, it will take 30 minutes to burn completely. At this point, the second fuse will have 30 minutes remaining when burning at one end; lighting the other end will burn the rest of the second fuse in $30/2 = 15$ minutes. The total time taken is $30 + 15 = 45$ minutes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18729" LastActivityDate="2016-02-01T13:57:46.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26026" PostTypeId="2" ParentId="26024" CreationDate="2016-02-01T13:58:00.193" Score="4" Body="&lt;p&gt;If the two fuses are labeled A and B:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Light both ends of fuse A, and one end of fuse B. Fuse A will completely burn out after 30 minutes. By this time, there are still 30 minutes of burn time in fuse B. Light the other end of fuse B when A finishes burning, which will halve the remaining 30 minutes of fuse B to 15 minutes, which when added to fuse A's 30 minutes will total 45 minutes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13101" LastActivityDate="2016-02-01T13:58:00.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26027" PostTypeId="2" ParentId="26018" CreationDate="2016-02-01T13:58:10.380" Score="9" Body="&lt;p&gt;I think the missing characters are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;&amp;lt;&quot; and &quot;&gt;&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; each symbol corresponds to the relation of its position (1-6) to the number 3. So the complete sequence would be &amp;lt;, &amp;lt;, =, &gt;, &gt;, &gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17334" LastActivityDate="2016-02-01T13:58:10.380" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="26028" PostTypeId="2" ParentId="26024" CreationDate="2016-02-01T14:02:55.980" Score="1" Body="&lt;p&gt;Two matches are enough hehe... Make a circle from one fuse and light at the touching point. When ut is burnt it will be 30 mins. You prepaired the second one by cutting half, making two circles and placing them down with the four free ends touchin each other. You light it after the first rope is burnt down. It will take additional 15 minutes&lt;/p&gt;&#xA;" OwnerUserId="18447" LastActivityDate="2016-02-01T14:02:55.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26029" PostTypeId="2" ParentId="26023" CreationDate="2016-02-01T14:04:15.553" Score="2" Body="&lt;p&gt;A. ‘the’ is not used with the names of people.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The Donald&quot; (for Donald Trump)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;B. The past simple form of the verb is used to describe finished events or states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C. ‘some’ is used in statements; ‘any’ in negatives and questions.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;He didn't manage to get some (with the phrase &quot;to get some&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;D. ‘can’ (modal verb) is used to express ability.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2016-02-01T14:04:15.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26030" PostTypeId="2" ParentId="26021" CreationDate="2016-02-01T14:40:13.707" Score="2" Body="&lt;p&gt;I think it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. I'm not used to the noise yet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; in this phrase, &quot;use&quot; does not refer to something in the past (used to be), but to adaption (be used to).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17334" LastActivityDate="2016-02-01T14:40:13.707" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26032" PostTypeId="1" AcceptedAnswerId="26038" CreationDate="2016-02-01T16:12:27.187" Score="6" ViewCount="167" Body="&lt;p&gt;Every letter stands for a digit in base-11 representation, &#xA;different letters stand for different digits:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;      WHEAT&#xA;   +  FIELD&#xA;  ------------&#xA;     FARMER&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which digit does each letter represent?&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="9404" LastEditDate="2016-03-03T10:19:59.167" LastActivityDate="2016-03-03T10:19:59.167" Title="Wheat field alphametic" Tags="|no-computers|alphametic|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26033" PostTypeId="1" AcceptedAnswerId="26086" CreationDate="2016-02-01T17:10:27.677" Score="25" ViewCount="9619" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Little 'bout my age: far from that of B&lt;br/&gt;&#xA;Yet in my syntax, you will find some C&lt;br/&gt;&#xA;The way I treat vars, that is much like D&lt;br/&gt;&#xA;I do take objects, but I'm not like E&lt;br/&gt;&#xA;If it starts with F, you'll know it's not me&lt;br/&gt;&#xA;Now I challenge you, just what could I be&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I guess this should be fairly easy for people who study computer-related stuff.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;Update: some clues&lt;/h2&gt;&#xA;&lt;p&gt;As of now nobody has guessed the language I am thinking about. I understand that this seems to be quite broad and many languages can fit to some degree, but my thought is, if there are still many candidates, then you might as well make the rules(clues given in the riddle) stricter and screen the candidates down. E.g. some said that the &amp;quot;D&amp;quot; line refers to D having local vs. global variables, but so many languages do, so you might want to look deeper into, say, variable naming conventions (clue: just an example, nothing to do with that).&lt;/p&gt;&#xA;&lt;p&gt;Also, here are some additional clues:&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Age&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The age is farther from B than all four guesses as of the time of writing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;How it treats vars&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Think typing and stuff.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Even more clues:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Also think memory management.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Objects&lt;/strong&gt;&lt;br/&gt;&#xA;Again, since few if any language is like E with respect to the object system, things remain pretty broad, so you may as well look at a certain aspect of it to screen candidates, more specifically,&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; the philosophy: E very strongly adheres to OOP.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;also, I might need to clarify that&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; objects can mean anything, but I did mean it in the OOP sense: an abstraction of a value (or set of values) with specific functions that are owned by and can operate on it (methods).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;One more, really big clue&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &amp;quot;If it starts with F&amp;quot; is actually trying to suggest the first letter in the language's name is important.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;With this one I probably made it too easy, though.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;The explanation&lt;/h2&gt;&#xA;&lt;p&gt;So Chiel ten Brinke got it right, the correct answer is&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Go (&lt;a href=&quot;https://golang.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;official site&lt;/a&gt;)(&lt;a href=&quot;https://en.wikipedia.org/wiki/Go_(programming_language)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Little 'bout my age: far from that of B&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Go was created in the 2009, just 7 years ago. The clue doesn't give a specific range, but at least one can expect no possible answer to be older than C.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Yet in my syntax, you will find some C&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Go's syntax, although pretty special in the family, is largely based off C, with C keywords, curly braces, operators, etc. Also, although there isn't always a semicolon after statements, they are just implicit, it's perfectly fine to use semicolons like in C, and that would be required to separate statements when doing inline stuff. However, that still doesn't throw C++, C#, Java, etc out of the pool of candidates.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The way I treat vars, that is much like D&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; I stated in the extra clues that one should look at memory management and typing system. Now D has static, strong, inferred typing with automatic memory management (as in: delete vars when they are no longer needed). &amp;quot;Inferred&amp;quot; alone throws many languages out of the question, but since I didn't explicitly talk about type inference it is just a quite probable, but unconfirmed constraint. Automatic memory management filters out C++ and friends.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I do take objects, but I'm not like E&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; In the extra clues I said the language was not like E that it doesn't adhere strongly to OOP, so C# and Java are out too, as well as Eiffel which was one of the answers you can find below. Go doesn't have classes, but structs can have methods, so OOP is pretty well supported, but the language itself doesn't emphasize OOP.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If it starts with F, you'll know it's not me&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; In the extra clues, I mentioned that &amp;quot;starts&amp;quot; is trying to suggest that the first letter in the language's name is important. Now if you see that the riddle went all the way from B to F, and the first letter is important, it should be logically sound to think the language starts with G, the next letter in the sequence. That's not necessarily the way you should think, you can also say that it's Eiffel if not for the other clues. The other best candidates thus far, Rust and Swift, couldn't fit this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So in the end,&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The ones who guessed Rust and Swift did very well, they only missed the BCDEF-G clue. Also, kudos to Klaws for suggesting Deesel, the answer only misinterpreted the line about E (it was about objects, not variables).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18817" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-02-04T10:43:06.273" Title="Computer riddle: I treat vars like D" Tags="|riddle|computer-puzzle|" AnswerCount="9" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="26034" PostTypeId="2" ParentId="26033" CreationDate="2016-02-01T17:15:27.740" Score="6" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C++ (or maybe Objective C)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The syntax is a superset of C, it &quot;takes objects&quot; (is an object oriented language), and its name doesn't start with F. The other languages B, D and E could be pretty much anything that would fit the clues, not sure if they are referring to something specific. I suspect the use of letters as code words for computer languages is a trick to throw the readers off and make them forget there is an actual programming language named &quot;C&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user17947" LastEditorDisplayName="user17947" LastEditDate="2016-02-01T17:20:44.380" LastActivityDate="2016-02-01T17:20:44.380" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="26035" PostTypeId="2" ParentId="26007" CreationDate="2016-02-01T17:17:05.097" Score="7" Body="&lt;p&gt;The question confuses causality and effect. I would choose option 1 and my statement would be;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;You are not going to give me 10 dollars&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This causes a paradox. If I get 10 dollars then it infers my statement is true, but it is false. If I don't get 10 dollars it infers my statement is false, but it is true.&lt;/p&gt;&#xA;" OwnerUserId="5740" LastActivityDate="2016-02-01T17:17:05.097" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26036" PostTypeId="1" AcceptedAnswerId="26040" CreationDate="2016-02-01T17:19:55.770" Score="9" ViewCount="15064" Body="&lt;p&gt;&lt;em&gt;Alternating numbers&lt;/em&gt; are numbers in which all digits alternate between even and odd. For example: 2703 and 7230 are alternating numbers, but 2730 isn't.&lt;/p&gt;&#xA;&lt;p&gt;Numbers are &lt;em&gt;very alternating&lt;/em&gt; when double the number is an alternating number as well, for example 3816 is very alternating, because 7632 is an alternating number as well.&lt;/p&gt;&#xA;&lt;p&gt;The question to you is: how many 4 digits very alternating numbers are there? (the number can't start with one or multiple zeros)&lt;/p&gt;&#xA;&lt;p&gt;Of course this problem can be solved by programming, but it can be solved mathematically as well! Good luck!&lt;/p&gt;&#xA;&lt;p&gt;NB: I didn't come up with this puzzle myself, it's part of the Dutch Math Olympiad, see &lt;a href=&quot;http://www.wiskundeolympiade.nl/files/opgaven/1e-ronde/2015/opgaven_en.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this PDF&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18878" LastEditorUserId="69582" LastEditDate="2021-01-15T16:14:02.023" LastActivityDate="2021-01-15T16:14:02.023" Title="Alternating numbers" Tags="|mathematics|calculation-puzzle|number-theory|formation-of-numbers|" AnswerCount="5" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="26037" PostTypeId="2" ParentId="26036" CreationDate="2016-02-01T17:51:56.200" Score="2" Body="&lt;p&gt;The answer is  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 70&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I did a manual count.  Still trying to work out a mathematical solution.&lt;/p&gt;&#xA;" OwnerUserId="9353" LastEditorUserId="9353" LastEditDate="2016-02-01T18:11:08.923" LastActivityDate="2016-02-01T18:11:08.923" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="26038" PostTypeId="2" ParentId="26032" CreationDate="2016-02-01T17:57:14.313" Score="6" Body="&lt;p&gt;The solution I got is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (A,F,W,L,E,M,R) = (0,1,9,2,3,6,4) and eight possibilities for (H,I,T,D) being (5,a,7,8), (5,a,8,7), (a,5,7,8), (a,5,8,7), (7,8,a,5), (7,8,5,a),(8,7,a,5), (8,7,5,a). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The methodology is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Let 'a' come after 9 in base-11 and all subsequent numbers I quote are in base 11. We must have W+F+(carried over term from 100s) $\ge$ 10 and here you won't carry over more than a 1 so F=1 and as a result W = 9 or a. It follows that A=0. We must have A+L+(carried over from 1s) = E (mod 11) but cannot have L=E (mod 11). Hence T+D $\ge$ 10 and E=L+1. Since E cannot be 0 or 1, we do not carry anything over to the 100s and 2E=M (mod 11). Hence the possibilities for (L,E,M) are (2,3,6), (3,4,8), (4,5,a), (6,7,3), (7,8,5), (8,9,7). We keep in mind that W=9 or a and the remaining terms must be chosen to satisfy H+I+(carried over)=T+D=R (mod 11). Checking through case by case is reasonably quick on pen and paper and we get the solutions only in the first case.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18422" LastActivityDate="2016-02-01T17:57:14.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26040" PostTypeId="2" ParentId="26036" CreationDate="2016-02-01T18:18:22.927" Score="11" Body="&lt;p&gt;There number of such numbers is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 70&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you double a number, a digit of the result is even if and only if the digit to the right didn't carry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $2x$ is alternating if and only if the digits of $x$ are $LHLH$, where $L \in \{0,1,2,3,4\}$ and $H \in \{5,6,7,8,9\}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(As a side note, this implies that if $x$ is very alternating, then $2x$ will still be a four digit number, so that $x &amp;lt; 5000$ necessarily).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that if $x$ is alternating when it looks like OEOE or EOEO. For $x$ to be very alternating, it also must be $LHLH$, so let's count the number of ways to fulfill both these constraints.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;OEOE&lt;/strong&gt;: Each odd number must be low, meaning they are each $1$ or $3$. Each even must be high, so is either $6$ or $8$. There are two choices for each digit for a total of $2\cdot2\cdot2\cdot2=16$ possibilities.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;EOEO&lt;/strong&gt;: Now the evens are low ($0,2$ or $4$), and the odds are high ($5,7$ or $9$). There are three choices for each digit, except the first digit can't be zero, so there are $2\cdot3\cdot3\cdot3=54$ possibilities.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These two counts add to the desired answer.&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="4421" LastEditDate="2016-02-01T20:05:20.143" LastActivityDate="2016-02-01T20:05:20.143" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="26041" PostTypeId="2" ParentId="26036" CreationDate="2016-02-01T18:26:41.077" Score="2" Body="&lt;p&gt;The correct answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 70&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#AnubhavBalodhi, puzzling.stackexchange,26036, 1/2/16 2341 hrs&#xA;E=['0','2','4','6','8']&#xA;O=['1','3','5','7','9']&#xA;&#xA;ans=0&#xA;alters=[]&#xA;&#xA;def Alt(num):&#xA;N=str(num)&#xA;if len(N)&amp;lt;5: #4 digit&#xA;    if (N[0] in E and N[1] in O and N[2] in E and N[3] in O) or (N[0] in O and N[1] in E and N[2] in O and N[3] in E):&#xA;        alters.append(num)&#xA;else:       #5 digit&#xA;    if (N[0] in E and N[1] in O and N[2] in E and N[3] in O and N[4] in E ) or (N[0] in O and N[1] in E and N[2] in O and N[3] in E and N[4] in O):&#xA;        alters.append(num)&#xA;&#xA;for num in range(1000,10**5):&#xA;Alt(num)&#xA;#print(num)&#xA;&#xA;print(len(alters))&#xA;&#xA;for numb in alters:&#xA;if numb&amp;lt;9999 and numb*2 in alters: #if the number in alternating list is of 4 digits, and its double is also in the list.&#xA;    print(numb)&#xA;    ans+=1&#xA;print(&quot;ans is %d&quot; %(ans))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1766" LastEditorUserId="1766" LastEditDate="2016-02-01T19:28:15.463" LastActivityDate="2016-02-01T19:28:15.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26042" PostTypeId="2" ParentId="26036" CreationDate="2016-02-01T18:32:09.763" Score="2" Body="&lt;p&gt;Damn I started this &lt;sup&gt;brute force&lt;/sup&gt; before the answers and I got&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;70&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;VBA &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Better logic for this route would have been&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;if (one mod 2 = three mod 2) and (two mod 2 = four mod 2) then&#xA;    if one mod 2 &amp;lt;&amp;gt; two mod 2 then&#xA;     counter increase&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditDate="2016-02-01T18:41:45.453" LastActivityDate="2016-02-01T18:41:45.453" CommentCount="0" CommunityOwnedDate="2016-02-01T18:41:45.453" ContentLicense="CC BY-SA 3.0" />
  <row Id="26043" PostTypeId="2" ParentId="26036" CreationDate="2016-02-01T19:23:10.010" Score="2" Body="&lt;p&gt;Here we go, brute force again!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is, as pointed out in other answers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 70&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You may check &lt;a href=&quot;https://jsfiddle.net/moonlife/9jLh88hj/&quot; rel=&quot;nofollow&quot; title=&quot;JSFiddle&quot;&gt;JSFiddle&lt;/a&gt; for source code and full list of 4-digit very alternative numbers.&lt;/p&gt;&#xA;" OwnerUserId="18128" LastActivityDate="2016-02-01T19:23:10.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26044" PostTypeId="2" ParentId="25819" CreationDate="2016-02-01T19:35:08.003" Score="0" Body="&lt;p&gt;&lt;strong&gt;The answer of two blindfolded men question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE:&lt;/strong&gt; Before checking out the solution of this one, please check the original question first to figure out this one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, this question has totally different type of solution than recent blindfolded near the great Wall of China puzzle. I will explain the solution as simple as possible and step by step since the solution is longer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The critical difference of this problem is that if one man encounters the wall on the way, the other man stops and walk perpendicularly to the wall since both needs to be at the wall when the time is up. To do that, you have to consider the perpendicular distance between one man and the other man’s wall all the time and since the solution is symmetric, it does not matter who encounters the wall first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the idea is clear at this point. Let’s come to the solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qo9ZV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qo9ZV.jpg&quot; alt=&quot;Figure 1&quot;&gt;&lt;/a&gt;&#xA;&lt;strong&gt;&lt;em&gt;Figure-1&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first part is the easiest part since the distance between both guys would not affect the solution much, so they need to move on the arc to cover the area as much as possible. As the original question, this one is pretty the same. We need to move out of the arc until where it covers as much as possible depending on the distance you get over:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Eq-1:&lt;/strong&gt; &lt;span class=&quot;math-container&quot;&gt;$maximize\ \left [2a-(\frac{1}{\cos(a)}-1)-\tan(a)\right ] \ where\ a&amp;lt;\pi/2$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this question, I tried to explain where cosine and tangent comes from on the Figure 1. Since those are basic maths, I will not get into much detail here and in the further. But this equation explains itself well and please ask if you still do not understand any equation/figure later. As a result;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may see the result here: &lt;a href=&quot;http://www.wolframalpha.com/input/?i=maximize%202a-(1%2Fcos(a)-1)-tan(a)%20where%200%3Ca%3Cpi&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wolfram Alpha Solution for minimum a&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result &lt;span class=&quot;math-container&quot;&gt;$a=\pi/6$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not consider the distance between these guys here, I will explain it later but you may easily guess that the distance between these men are not much at the beginning. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total distance they moved from the beginning to the end can be found easily from the figure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\frac{1}{\cos(a)}+\tan(a)$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So after taking the fastest way to the arc again, how much you need to move on the arc? That part is the hardest part since you need to start to consider the distance between these guys now!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These guys is supposed to move on the arc and move on the tangent after a while to complete all possible points on it, but now they need to consider the distance between them and the perpendicular distance of the wall of the other man as well:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Eq-2:&lt;/strong&gt; &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{\cos(a)}+\tan(a)+b+\tan(c_{1})+\left [\frac{cos(3c_{1}+4a+2b-\pi)}{cos(c_{1})}+1  \right ]$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me explain the equation briefly, I think everything is clear except the bracket equation. The bracket equation is actually the distance between one guy and the other guy’s wall on the way at any point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XK6qA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XK6qA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;Figure-2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In F2 (please zoom into the figure), c represents the distance you are planning to take on the tangent and after that point you will move back to the arc again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do we need to move on the tangent again? The reason is actually secure the distance between these guys, even it seems the distance is getting bigger, it is not since you cover faster than moving on the arc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And &lt;span class=&quot;math-container&quot;&gt;$c_1$&lt;/span&gt; represents while moving on the tangent to &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt;, what would be the distance to the wall if the other one finds the wall: We have to consider every single point on the way to &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; point. To do that, we need to minimize &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$c&amp;lt;c_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$c1&amp;lt;\pi/2$&lt;/span&gt;;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\frac{1}{\cos(a)}+\tan(a)+b+\tan(c_{1})+\left [\frac{cos(3c_{1}+4a+2b-\pi)}{cos(c_{1})}+1  \right ]\ where\ 0&amp;lt;d&amp;lt;\pi/2$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As result: &lt;a href=&quot;http://www.wolframalpha.com/input/?i=minimize%20b%2Btan(c)%2Bcos(3c%2B2*b-pi%2F3)%2Fcos(c)%20where%200%3Cc%3C15pi%2F32&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wolfram Alpha Solution for b&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We got: &lt;span class=&quot;math-container&quot;&gt;$b=0.224896$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before explain what c should be, we need to consider point d in that part to minimize the distance between these men as below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/a23BV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/a23BV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;strong&gt;&lt;em&gt;Figure-3&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Figure 3- you will notice a dark angle (&lt;span class=&quot;math-container&quot;&gt;$\mu$&lt;/span&gt;), if we can find that angle, we can find the perpendicular distance to the wall when a man encounter the wall just after finishing the arc move after &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt;, so;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\mu = \pi/2 - \left [2\pi-4c-4a-2b-\pi/2 \right ]=4a+2b+4c-\pi$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as a result the total distance to the wall included everything;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;minimum distance to the wall: &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{cos(a)}+\tan(a)+b+2tan(c)+\cos(4a+2b+4c-\pi)+1$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we know &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; and if we minimize it with wolfram alpha: &lt;a href=&quot;http://www.wolframalpha.com/input/?i=minimize%201%2Fcos(pi%2F6)%2Btan(pi%2F6)%2B0.224896%2B2*tan(c)%2Bcos(4(pi%2F6)%2B2*0.224896%2B4c-pi)%2B1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wolfram Alpha Result for &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; as a minimum&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so &lt;span class=&quot;math-container&quot;&gt;$c=0.687035$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we have found the best &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt;,&lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt;, let's find out the worst case from this route. Actually there is two critical points where you can find the worst case, one of them is in between &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; point and &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt;, the other one is point &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; -&gt; &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt;: &lt;a href=&quot;http://www.wolframalpha.com/input/?i=maximize%201%2Fcos(pi%2F6)%2Btan(pi%2F6)%2B0.224896%2Btan(c)%2Bcos(3c%2B2*0.224896-pi%2F3)%2Fcos(c)%2B1%20%20where%200%3Cc%3C0.687035&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wolfram Alpha Result for $c_1&lt;/a&gt; - &lt;strong&gt;4.27982&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with maximization of;&#xA;&lt;span class=&quot;math-container&quot;&gt;$maximize\ \frac{1}{cos(\pi/6)}+\tan(\pi/6)+(b=0.224896)+tan(c_1)+cos(3c_1+2\times0.224896-\pi/3)/cos(c)+1\ where\  0&amp;lt;c_1&amp;lt;0.687035=c$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt;: &lt;a href=&quot;http://www.wolframalpha.com/input/?i=1%2Fcos(pi%2F6)%2Btan(pi%2F6)%2B0.224896%2B2*tan(0.687035)%2Bcos(4*(pi%2F6)%2B2*0.224896%2B4*0.687035-pi)%2B1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wolfram Alpha Result for d&lt;/a&gt; - &lt;strong&gt;4.04970&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As result: the optimal worst case for 2 blindfolded men is &lt;strong&gt;4.27982&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="18514" LastEditorUserId="56416" LastEditDate="2019-01-31T04:31:47.050" LastActivityDate="2019-01-31T04:31:47.050" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="26045" PostTypeId="2" ParentId="26033" CreationDate="2016-02-01T19:40:53.733" Score="9" Body="&lt;p&gt;My guess is that you are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Java&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Little 'bout my age: far from that of &lt;a href=&quot;https://en.wikipedia.org/wiki/B_%28programming_language%29&quot;&gt;B&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;As the predecessor of C and its decendents, B is a lot older than Java&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet in my syntax, you will find some &lt;a href=&quot;https://en.wikipedia.org/wiki/C_%28programming_language%29&quot;&gt;C&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Java shares a lot of syntactical elements with C&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The way I treat vars, that is much like &lt;a href=&quot;https://en.wikipedia.org/wiki/D_%28programming_language%29&quot;&gt;D&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Java and D share the concept of local vs global variables, and Java has a construct for a foreach, which is present in D but not C.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I do take objects, but I'm not like &lt;a href=&quot;https://en.wikipedia.org/wiki/E_%28programming_language%29&quot;&gt;E&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;While there are syntactic similarities between E and Java, how objects are defined and handled are not in that set. Java uses explicit typing, where E allows loose typing and defining of variables/returns.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If it starts with F, you'll know it's not me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Clearly, Java doesn't start with F&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now I challenge you, just what could I be&lt;/p&gt;&#xA;" OwnerUserId="12242" LastActivityDate="2016-02-01T19:40:53.733" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26046" PostTypeId="2" ParentId="26007" CreationDate="2016-02-01T19:40:56.693" Score="9" Body="&lt;p&gt;You should take offer &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; #1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because you can guarantee yourself an arbitrarily large amount of money, simply by using the following as your statement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I will receive neither 10 dollars nor 1000 dollars.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we assume the statement is true, this leads to a contradiction because you would receive $10, by the rules. &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume the statement is false, then to satisfy the 'falsity' of the statement, you must receive either 10 or 1000 dollars. But since you cannot receive 10 dollars if the statement is false, you must receive 1000 dollars.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;note: Of course you can use any amount in place of “$1000.”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1766" LastActivityDate="2016-02-01T19:40:56.693" CommentCount="2" CommunityOwnedDate="2016-02-01T19:40:56.693" ContentLicense="CC BY-SA 3.0" />
  <row Id="26047" PostTypeId="1" AcceptedAnswerId="26049" CreationDate="2016-02-01T20:34:26.343" Score="4" ViewCount="863" Body="&lt;p&gt;Chances are you've heard of me,&lt;br&gt;&#xA;Some call me an oracle.&lt;br&gt;&#xA;Am I friend, or enemy?&lt;br&gt;&#xA;I'm rather affordable.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;College, University,&lt;br&gt;&#xA;I'm the savior late at night.&lt;br&gt;&#xA;Beware, not all that you see,&lt;br&gt;&#xA;Is truth or fully right.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Search now, for what I may be,&lt;br&gt;&#xA;Google and I are good pals.&lt;br&gt;&#xA;No paper, I saved the tree.&lt;br&gt;&#xA;Give me now your rationale.   &lt;/p&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2016-02-01T20:40:00.000" Title="Maybe I'm wrong, What am I?" Tags="|riddle|word|rhyme|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26048" PostTypeId="2" ParentId="26033" CreationDate="2016-02-01T20:37:39.440" Score="32" Body="&lt;p&gt;I like the other answers, but there's a clue that they really don't capture well. This answer is similar to the others, but I think hits some of the clues more accurately than the other answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C#&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Little 'bout my age: far from that of B&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Obviously, B is older than C, which is older than C#&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet in my syntax, you will find some C&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many c-style syntatical elements are in C#&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The way I treat vars, that is much like D&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The set of base types and the way they're handled are very similar in the two languages&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I do take objects, but I'm not like E&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Objects in E require the &lt;code&gt;def&lt;/code&gt; keyword, C# uses &lt;code&gt;class&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If it starts with F, you'll know it's not me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is another language called F#, so don't confuse that with C#&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now I challenge you, just what could I be&lt;/p&gt;&#xA;" OwnerUserId="2812" LastActivityDate="2016-02-01T20:37:39.440" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26049" PostTypeId="2" ParentId="26047" CreationDate="2016-02-01T20:40:00.000" Score="6" Body="&lt;p&gt;I believe it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wikipedia.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reasons:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chances are you've heard of me,&#xA;Some call me an oracle.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wikipedia's very well-known as a source of information.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;College, University,&#xA;I'm the savior late at night.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wikipedia's often used as a source for research, especially by students who don't have the time to look up a proper reference.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Beware, not all that you see,&#xA;Is truth or fully right.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wikipedia can be edited by anyone, so its information isn't always true&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Search now, for what I may be,&#xA;Google and I are good pals.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Google searches for things on Wikipedia often give the Wikipedia page as one of the first search results.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No paper, I saved the tree.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wikipedia's purely digital, not on paper.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18729" LastActivityDate="2016-02-01T20:40:00.000" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26050" PostTypeId="1" AcceptedAnswerId="26052" CreationDate="2016-02-01T21:54:44.797" Score="4" ViewCount="717" Body="&lt;p&gt;Find the next number in this sequence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2, 4, 6, 9, 14, 18, 25, 32, 44, ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus question: I originally was going to use a different related sequence which grew at a faster rate, but changed it to this one instead. What was the original sequence, and why might I have opted for this one instead?&lt;/p&gt;&#xA;" OwnerUserId="17024" LastActivityDate="2016-02-01T22:13:53.533" Title="What's the next number in this sequence?" Tags="|mathematics|number-sequence|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26051" PostTypeId="1" AcceptedAnswerId="26054" CreationDate="2016-02-01T22:01:02.200" Score="1" ViewCount="404" Body="&lt;p&gt;You are given a ring for which 1+1=3 and 9/12=0. Yet this ring is not &lt;a href=&quot;https://en.wikipedia.org/wiki/Trivial_ring&quot; rel=&quot;nofollow&quot;&gt;trivial&lt;/a&gt;, quite the opposite. What is your answer?&lt;/p&gt;&#xA;" OwnerUserId="18776" LastEditorUserId="11876" LastEditDate="2016-02-05T13:53:40.857" LastActivityDate="2016-02-05T13:53:40.857" Title="Ring-theoretic riddle" Tags="|riddle|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="26052" PostTypeId="2" ParentId="26050" CreationDate="2016-02-01T22:13:53.533" Score="8" Body="&lt;p&gt;The next number is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 63&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sequence seems to be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $a_n = F_n + P_n$, the $n$-th Fibonacci number plus the $n$-th prime number&lt;br&gt;&#xA;&lt;br&gt;&#xA; 0+2, 1+3, 1+5, 2+7, 3+11, 5+13, 8+17, 13+19, 21+23, so the next number is 34+29&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And my guess for bonus,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; perhaps you originally multiplied, but didn't like that because it'd have been more obvious if we'd factored the numbers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17147" LastActivityDate="2016-02-01T22:13:53.533" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26053" PostTypeId="2" ParentId="26051" CreationDate="2016-02-01T22:36:06.453" Score="3" Body="&lt;p&gt;I think the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Circle of Life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1+1=3 refers to a man and a woman coming together to make a man, woman and child. 9/12 refers to 'nine months' at which point a baby is 0. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18422" LastActivityDate="2016-02-01T22:36:06.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26054" PostTypeId="2" ParentId="26051" CreationDate="2016-02-01T22:41:39.697" Score="4" Body="&lt;p&gt;My answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No, I'll not marry a stranger from the internet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The ring is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An engagement ring (see hexomino's reasoning) - definitely not trivial!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17147" LastActivityDate="2016-02-01T22:41:39.697" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26055" PostTypeId="1" AcceptedAnswerId="26098" CreationDate="2016-02-01T23:09:33.223" Score="13" ViewCount="687" Body="&lt;p&gt;A monarch in democracy,&lt;br&gt;&#xA;yet revered without hypocrisy,&lt;br&gt;&#xA;my name shall live on evermore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under the same reign as the sign of war,&lt;br&gt;&#xA;and thirty-one's twenty-four,&lt;br&gt;&#xA;the first to seize a third at three.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Totaling thirteen and a little more,&lt;br&gt;&#xA;I became the one to do it, finally;&lt;br&gt;&#xA;I'm now a monarch doubly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who am I?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint #1  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is &lt;em&gt;not&lt;/em&gt; a political figure&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17147" LastEditorUserId="17147" LastEditDate="2016-02-03T15:56:07.200" LastActivityDate="2016-02-05T12:36:19.417" Title="A Revered Monarch" Tags="|riddle|rhyme|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26056" PostTypeId="1" AcceptedAnswerId="31992" CreationDate="2016-02-02T03:07:50.277" Score="15" ViewCount="1389" Body="&lt;p&gt;Oras, a great sailor, accidentally docked on a deserted island. He checked his compass and it is not working as it rotates in all direction. He boarded off to investigate the island and saw a cave. Inside the cave, he found 3 treasure boxes with locks. It seems it needs number combinations for it to unlock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As he checked further, he saw a piece of parchment with a note. It says: &quot;Time is running out for me. For whoever find these treasures, you can find the answer here. Find your way and you will see it.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wA5bb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wA5bb.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 3 treasures:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Roman treasure&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;needs 3 digits to unlock&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Japanese treasure&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;needs 2 digits to unlock&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pirate treasure&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;needs 2 digits to unlock&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;HINT:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; solving the codes/digits for each treasure is somehow different from one another&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;HINT 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; methods/techniques can be reused but see it in a different aspect. Results from Roman is different from Japanese and Pirate. Also for one of the solution, clock technique is not mixed with other technique. Numbers that you will meet using the other technique are there for a reason.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;HINT 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Roman - visual, Japanese - not, Pirate - not; 2 clock techniques and 1 other technique. &quot;ADD&quot; is there for a reason when you meet it using a technique. Pirate is harder I think. Pirates do not use Clocks when navigating.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;HINT 4: (Chinese New Year Hint)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pirate is Black Arrow. Again, &quot;Pirates do not use Clocks when navigating&quot;. Combine with Hint 2's &quot;Numbers that you will meet using the other technique are there for a reason&quot;. For Japanese, Roman technique with a twist and use Hint 2 also.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Will reveal answer after bounty expires :)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;CHATROOM:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chat.stackexchange.com/rooms/35150/the-3-treasures&quot;&gt;http://chat.stackexchange.com/rooms/35150/the-3-treasures&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16036" LastEditorUserId="16036" LastEditDate="2016-02-07T13:55:28.040" LastActivityDate="2016-05-11T00:41:31.397" Title="The 3 Treasures" Tags="|enigmatic-puzzle|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="26057" PostTypeId="2" ParentId="26055" CreationDate="2016-02-02T05:23:17.727" Score="2" Body="&lt;p&gt;I couldn't find anything better than&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Napoléon Bonaparte&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A monarch in democracy,&#xA;yet revered without hypocrisy,&#xA;my name shall live on evermore.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; France was a democratic country when he was the emperor but he preferred monarchy (&lt;a href=&quot;https://en.wikipedia.org/wiki/Napoleon#French_Empire&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;).  As wikipedia states &quot;there would be an heir, it would also make it all but impossible to change the regime by assassinating Napoleon.&quot; (&lt;a href=&quot;https://en.wikipedia.org/wiki/France#Modern_period_.281789.E2.80.931914.29&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Under the same rule as the sign of war,&#xA;and thirty-one's twenty-four,&#xA;the first to seize a third at three.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He got some victories in The War of Third Coalition which started in 1803 (second paragraph &lt;a href=&quot;https://en.wikipedia.org/wiki/First_French_Empire&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Totaling thirteen and a little more,&#xA;I became the one to do it, finally.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He created eighteen Marshals of the Empire from amongst his top generals to secure the allegiance of the army (&lt;a href=&quot;https://en.wikipedia.org/wiki/Napoleon#French_Empire&quot; rel=&quot;nofollow&quot;&gt;another link&lt;/a&gt;). His victories led to the worldwide expansion of French revolutionary ideals and reforms&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However my answer cannot explain thirty-one's twenty-four which I thought to be some kind of date.&lt;/p&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-02-02T05:38:46.253" LastActivityDate="2016-02-02T05:38:46.253" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26058" PostTypeId="1" AcceptedAnswerId="26065" CreationDate="2016-02-02T05:59:48.230" Score="16" ViewCount="813" Body="&lt;p&gt;&lt;em&gt;I am beneath myself but above my prime&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;Left is on my right and up is nearby,&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;down isn't close... well maybe a bit&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;When our most popular friend joins us,&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;me and (some of) my neighbours like to joke, lol&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;stay close to me and u'll be ok&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="18522" LastActivityDate="2016-02-02T21:54:15.690" Title="What or who or where am I?" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26059" PostTypeId="1" AcceptedAnswerId="26060" CreationDate="2016-02-02T06:03:13.887" Score="3" ViewCount="349" Body="&lt;p&gt;You are to capture one of the biggest shipment of illegal drugs. The only problem is that your team just narrowed it down to 4 locations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Minnesota&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Florida&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Virginia&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;California&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't have enough men to cover the port of four cities and you can only cover one. The only information that you can rely on that can help you decide on which city is the text message you had intercepted last week.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Text message:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Shipment next week, coordinates 64,14,85,84,72,58,41. Don't messed up, Don Antonio San Carlos II will be there, Hernando Exum and Octavio are not. The wolves might be there, be careful.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you get to the right city before the shipment?&lt;/p&gt;&#xA;" OwnerUserId="16036" LastActivityDate="2016-02-02T06:13:18.853" Title="Intercept the Shipment" Tags="|enigmatic-puzzle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26060" PostTypeId="2" ParentId="26059" CreationDate="2016-02-02T06:13:18.853" Score="6" Body="&lt;p&gt;I think it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Florida.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Antonio San Carlos II -&gt; ASCII&lt;br&gt;&#xA; Hernando Exum and Octavio are not -&gt; decimal, not hex or octal.&lt;br&gt;&#xA; This translates to @SOUTH:)&lt;br&gt;&#xA; So I think Florida because it's the southernmost of the states.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18522" LastActivityDate="2016-02-02T06:13:18.853" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26061" PostTypeId="2" ParentId="16333" CreationDate="2016-02-02T08:50:16.423" Score="1" Body="&lt;h3&gt;Generalizable suboptimal incremental-roll approach&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;[Gotta present this after being inspired by&#xA; &lt;a href=&quot;https://puzzling.stackexchange.com/a/24572/18129&quot;&gt;2012rcampion's answer&#xA;  to a different question&lt;/a&gt;&#xA; and by&#xA; &lt;a href=&quot;https://puzzling.stackexchange.com/a/16370/18129&quot;&gt;FarazMasroor's unacknowledged&#xA;  partial answer here&lt;/a&gt;,&#xA; which would have benefited from more (and more accurate) detail.&#xA; This also turned out to be eerily similar to an&#xA; &lt;a href=&quot;https://math.stackexchange.com/a/1315056/302277&quot;&gt;answer at Mathematics SE&lt;/a&gt;.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a close-to-optimal approach that adapts readily to any desired&#xA;outcome distribution with any-sided dice.  A random fraction between 0 and 1&#xA;is generated to fairly pick a point on an arbitrary&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Quantile_function&quot; rel=&quot;nofollow noreferrer&quot;&gt;inverse cumulative distribution function&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with fractions for&#xA;$\,\textsf{Monday}=0/7\,$ through $\,\textsf{Sunday}=6/7\,$&#xA;in their &lt;a href=&quot;https://en.wikipedia.org/wiki/Senary&quot; rel=&quot;nofollow noreferrer&quot;&gt;base 6&lt;/a&gt; expansions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{rcr}&#xA;\textsf{   Monday}= 0/7=.000000\textrm{...}\!\:_6 &amp;amp;~~&amp;amp;&#xA;\textsf{   Friday}= 4/7=.323232\textrm{...}\!\:_6 \\&#xA;\textsf{  Tuesday}= 1/7=.050505\textrm{...}\!\:_6 &amp;amp;&amp;amp;&#xA;\textsf{ Saturday}= 5/7=.414141\textrm{...}\!\:_6 \\&#xA;\textsf{Wednesday}= 2/7=.141414\textrm{...}\!\:_6 &amp;amp;&amp;amp;&#xA;\textsf{   Sunday}= 6/7=.505050\textrm{...}\!\:_6 \\&#xA;\textsf{ Thursday}= 3/7=.232323\textrm{...}\!\:_6&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now begin rolling repeatedly,&#xA;appending each result as the next digit of a lengthening base-6 fraction,&#xA;treating each roll of &amp;lsquo;$6$&amp;rsquo; as the digit $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep rolling as long as the cumulative digit sequence&#xA;matches one of the days' base-6 fractions listed above.&#xA;When a digit fails to match,&#xA;select the day with the largest fraction that&#xA;is below what resulted from the rolls.&#xA;The exception to these rules is when&#xA;the first two rolls amount to $.00\!\:_6$,&#xA;in which case call it a $\textsf{Monday}$ and be done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that we are done after 2 rolls&#xA;&lt;em&gt;unless&lt;/em&gt; those rolls amount to some day's fraction's first 2 digits:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ .05\!\:_6 \qquad .14\!\:_6 \qquad .23\!\:_6 \qquad&#xA;   .32\!\:_6 \qquad .41\!\:_6 \qquad .50\!\:_6 \qquad&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives a 6/36 = 1/6 chance of needing a 3rd roll.&#xA;Each subsequent roll also leaves a compounded 1/6 chance&#xA;of needing yet another roll.  So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{c}&#xA;\textsf{average number of rolls} ~=~ 2 + \frac16 \Bigl( 1 + \frac16 \bigl(&#xA;                                      1 + \frac16 ( 1 + \dots ) \bigr) \Bigr)&#xA;                                 ~=~ 2.2&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the first 2 rolls were &amp;lsquo;$1$&amp;rsquo; and &amp;lsquo;$5$&amp;rsquo;,&#xA;the randomly-selected day would be $\textsf{Wednesday}$ because&#xA;$.15\!\:_6$ does not begin any day's fraction and:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\textsf{Wednesday}=.141414\textrm{...}\!\:_6~&amp;lt;~.15\!\:_6~&amp;lt;~.232323\textrm{...}\!\:_6=\textsf{Thursday}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If those rolls were &amp;lsquo;$1$&amp;rsquo; and &amp;lsquo;$4$&amp;rsquo;, however, we&#xA;would roll again because $\textsf{Wednesday}$'s&#xA;fraction begins with&amp;nbsp;$.14\!\:_6$.&#xA;If the next roll comes up &amp;lsquo;$0$&amp;rsquo;, which produces $.140\!\:_6$,&#xA;the randomly-selected day would be $\textsf{Tuesday}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\textsf{Tuesday}=.050505\textrm{...}\!\:_6~&amp;lt;~.140\!\:_6~&amp;lt;~.141414\textrm{...}\!\:_6=\textsf{Wednesday}$$&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Variation with alternating die rolls and coin flips&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;This is in the full spirit of the puzzle's title.&#xA;Use a hybrid of bases 6 and 2, where the first fractional digit&#xA;is in base 6, the next in base 2, then base 6, then 2 again, and so on.&#xA;This is effectively &lt;a href=&quot;https://en.wikipedia.org/wiki/Duodecimal&quot; rel=&quot;nofollow noreferrer&quot;&gt;base 12&lt;/a&gt;&#xA;with pairs of numerals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{r}&#xA;\textsf{   Monday}=0/7= ~.00~00~00~00~00~00~&#xA;                        ~~00~00~00~00~00~00~...\!\,_{6\backslash2} \\&#xA;\textsf{  Tuesday}=1/7= ~.01~40~30~50~11~21~&#xA;                        ~~01~40~30~50~11~21~...\!\,_{6\backslash2} \\&#xA;\textsf{Wednesday}=2/7= ~.11~21~01~40~30~50~&#xA;                        ~~11~21~01~40~30~50~...\!\,_{6\backslash2} \\&#xA;\textsf{ Thursday}=3/7= ~.21~01~40~30~50~11~&#xA;                        ~~21~01~40~30~50~11~...\!\,_{6\backslash2} \\&#xA;\textsf{   Friday}=4/7= ~.30~50~11~21~01~40~&#xA;                        ~~30~50~11~21~01~40~...\!\,_{6\backslash2} \\&#xA;\textsf{ Saturday}=5/7= ~.40~30~50~11~21~01~&#xA;                        ~~40~30~50~11~21~01~...\!\,_{6\backslash2} \\&#xA;\textsf{   Sunday}=6/7= ~.50~11~21~01~40~30~&#xA;                        ~~50~11~21~01~40~30~...\!\,_{6\backslash2}&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a die roll, again treat &amp;lsquo;$6$&amp;rsquo; as $0$.&#xA;For a coin flip, call heads $0$ and call tails $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar to before, repeatedly accumulate digits,&#xA;alternately rolling and flipping,&#xA;as long as the sequence of digits matches some day's&#xA;(other than $\textsf{Monday}$'s) fraction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{r}&#xA;\textsf{average number of alternating rolls and flips} ~=~ \phantom{2.636363...}&#xA;\\            2 + \frac6{12} \Bigl( 1 + \frac16 \bigl(&#xA;               1 + \frac12    \bigl( 1 + \frac16 \bigl(&#xA;                1 + \frac12         ( 1 + \dots ) \bigr) \bigr) \bigr) \Bigr)&#xA;          ~=~ 2.636363...&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Comparison to an optimal(?) system with alternating die rolls and coin flips&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;A less-inspired (but optimal?) system&#xA;uses an average of approximately 2.479 alternating rolls and flips.&#xA;Without coin flips, the same approach matches the&#xA;average of 2.057 die rolls posted in other solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This table outlines a repeated 12-action pattern of whether or not&#xA;a day will be selected by each successive die roll or coin flip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{lcccccc}&#xA;                          &amp;amp;\textrm{           }       &amp;amp;\textrm{          }&#xA;                           &amp;amp;\textrm{          }    &amp;amp;\!\!\textrm{Potential}&#xA;                        &amp;amp;\!\!\textrm{Left-    }         &amp;amp;\textrm{Proba-  }\!&#xA;\\[-.5ex]&#xA;                        &amp;amp;\!\textrm{Probability}     &amp;amp;\!\textrm{Continued }\!&#xA;                           &amp;amp;\textrm{Combined  }      &amp;amp;\!\textrm{outcomes }\!&#xA;                        &amp;amp;\!\!\textrm{over     }       &amp;amp;\!\textrm{bility  }&#xA;\\[-.5ex]&#xA;                        &amp;amp;\!\textrm{of reaching}       &amp;amp;\textrm{potential }&#xA;                         &amp;amp;\!\textrm{potential }        &amp;amp;\textrm{assigned }&#xA;                        &amp;amp;\!\!\textrm{potential}         &amp;amp;\textrm{of con- }\!&#xA;\\[-.5ex]&#xA;\!\!\!\textrm{Action}     &amp;amp;\textrm{the action }       &amp;amp;\textrm{outcomes  }&#xA;                         &amp;amp;\!\textrm{outcomes  }        &amp;amp;\textrm{to days  }&#xA;                            &amp;amp;\textrm{outcomes }         &amp;amp;\textrm{tinuing }&#xA;\\[1ex]&#xA;   \textrm{roll}_1   &amp;amp;      1  ~~ \!&amp;amp;      &amp;amp;     6 &amp;amp;    0 &amp;amp;    6 &amp;amp;    6/6~ \\&#xA;   \rlap{\textrm{(cycle begins)}}                                          \\&#xA;\bf\textbf{flip}_2   &amp;amp;      1  ~~ \!&amp;amp;\bf 6 &amp;amp;\bf 12 &amp;amp;\bf 7 &amp;amp;\bf 5 &amp;amp;\bf 5/12 \\&#xA;   \textrm{roll}_3   &amp;amp;     5/12   \!&amp;amp;    5 &amp;amp;    30 &amp;amp;   28 &amp;amp;    2 &amp;amp;    2/30 \\&#xA;   \textrm{flip}_4   &amp;amp;     1/36   \!&amp;amp;    2 &amp;amp;     4 &amp;amp;    0 &amp;amp;    4 &amp;amp;    4/4~~\\&#xA;   \textrm{roll}_5   &amp;amp;     1/36   \!&amp;amp;    4 &amp;amp;    24 &amp;amp;   21 &amp;amp;    3 &amp;amp;    3/24 \\&#xA;   \textrm{flip}_6   &amp;amp;     1/288  \!&amp;amp;    3 &amp;amp;     6 &amp;amp;    0 &amp;amp;    6 &amp;amp;    6/6~~\\&#xA;   \textrm{roll}_7   &amp;amp;     1/288  \!&amp;amp;    6 &amp;amp;    36 &amp;amp;   35 &amp;amp;    1 &amp;amp;    1/36 \\&#xA;   \textrm{flip}_8   &amp;amp;    1/10368 \!&amp;amp;    1 &amp;amp;     2 &amp;amp;    0 &amp;amp;    2 &amp;amp;    2/2~~\\&#xA;   \textrm{roll}_9   &amp;amp;    1/10368 \!&amp;amp;    2 &amp;amp;    12 &amp;amp;    7 &amp;amp;    5 &amp;amp;    5/12 \\&#xA;   \textrm{flip}_{10}&amp;amp;   5/124416 \!&amp;amp;    5 &amp;amp;    10 &amp;amp;    7 &amp;amp;    3 &amp;amp;    3/10 \\&#xA;   \textrm{roll}_{11}&amp;amp;   1/82944 ~\!&amp;amp;    3 &amp;amp;    18 &amp;amp;   14 &amp;amp;    4 &amp;amp;    4/18 \\&#xA;   \textrm{flip}_{12}&amp;amp;   1/373248 \!&amp;amp;    4 &amp;amp;     8 &amp;amp;    7 &amp;amp;    1 &amp;amp;    1/8~~\\&#xA;   \textrm{roll}_{13}&amp;amp;   1/2985984\!&amp;amp;    1 &amp;amp;     6 &amp;amp;    0 &amp;amp;\bf 6 &amp;amp;    6/6~~\\&#xA;   \rlap{\textrm{(cycle repeats)}}                                         \\&#xA;\bf\textbf{flip}_{14}&amp;amp;\bf1/2985984\!&amp;amp;\bf 6 &amp;amp;\bf 12 &amp;amp;\bf 7 &amp;amp;\bf 5 &amp;amp;\bf 5/12 \\&#xA;   \textrm{roll}_{15}&amp;amp;  5/35831808\!&amp;amp;\bf 5 &amp;amp;    30 &amp;amp;   28 &amp;amp;    2 &amp;amp;    2/30&#xA;   \\[-1ex]&#xA;   \textrm{flip}_{16}&amp;amp;     \vdots  &amp;amp;\vdots&amp;amp;\,\vdots&amp;amp;\vdots&amp;amp;\vdots&amp;amp;\vdots\;&#xA;   \\[-1ex] ~~~\vdots&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The row for $\bf\textbf{flip}_{14}$&amp;hairsp;, for example, shows that:&#xA;&lt;br&gt;&amp;bull;&lt;sup&gt;&lt;sup&gt;&amp;ensp;&lt;/sup&gt;&lt;/sup&gt;&#xA;$\bf 1/2985984$ is the probability that this coin flip will be required.&#xA;&lt;br&gt;&amp;bull;&lt;sup&gt;&lt;sup&gt;&amp;ensp;&lt;/sup&gt;&lt;/sup&gt;&#xA;$\bf 6$ equally-likely continuing outcomes, die rolls {1,2,3,4,5,6&amp;hairsp;},&#xA;        are carried forward from the previous action,&#xA;        $\textrm{roll}_{13}$&amp;hairsp;.&#xA;&lt;br&gt;&amp;bull;&lt;sup&gt;&lt;sup&gt;&amp;ensp;&lt;/sup&gt;&lt;/sup&gt;&#xA;$\bf 12$ potential combined outcomes,&#xA;        {1,2,3,4,5,6&amp;hairsp;}&amp;thinsp;&amp;times;&amp;thinsp;{&amp;hairsp;H,T&amp;hairsp;},&#xA;        are now considered for this coin flip's 2 possible results,&#xA;        {Heads,Tails}.&#xA;&lt;br&gt;&amp;bull;&lt;sup&gt;&lt;sup&gt;&amp;ensp;&lt;/sup&gt;&lt;/sup&gt;&#xA;$\bf 7$ of the potential ordered outcomes are&#xA;        arbitrarily assigned to select a day:&#xA;    &lt;br&gt;&amp;emsp;&amp;emsp;&amp;emsp;&#xA;        {1&lt;sub&gt;&lt;sup&gt;&amp;amp;&amp;hairsp;&lt;/sup&gt;&lt;/sub&gt;H&amp;hairsp;}&amp;hairsp;=&amp;hairsp;Mon,&#xA;        {&amp;hairsp;2&lt;sub&gt;&lt;sup&gt;&amp;hairsp;&amp;amp;&amp;hairsp;&lt;/sup&gt;&lt;/sub&gt;H&amp;hairsp;}&amp;hairsp;=Tues,&amp;nbsp;...,&#xA;        {&amp;hairsp;6&lt;sub&gt;&lt;sup&gt;&amp;hairsp;&amp;hairsp;&amp;amp;&amp;hairsp;&lt;/sup&gt;&lt;/sub&gt;H&amp;hairsp;}&amp;hairsp;=&amp;hairsp;Sat&#xA;        and {1&lt;sub&gt;&lt;sup&gt;&amp;amp;&lt;/sup&gt;&lt;/sub&gt;T&amp;hairsp;}&amp;hairsp;=&amp;hairsp;Sun.&#xA;&lt;br&gt;&amp;bull;&lt;sup&gt;&lt;sup&gt;&amp;ensp;&lt;/sup&gt;&lt;/sup&gt;&#xA;$\bf 5$ unassigned potential ordered outcomes,&#xA;       {&amp;hairsp;2&lt;sub&gt;&lt;sup&gt;&amp;hairsp;&amp;amp;&lt;/sup&gt;&lt;/sub&gt;T,&amp;thinsp;3&lt;sub&gt;&lt;sup&gt;&amp;hairsp;&amp;amp;&lt;/sup&gt;&lt;/sub&gt;T,&amp;thinsp;...,&amp;thinsp;6&lt;sub&gt;&lt;sup&gt;&amp;hairsp;&amp;amp;&lt;/sup&gt;&lt;/sub&gt;T&amp;hairsp;},&#xA;        remain left over to combine with the next action if the present&#xA;        action doesn't complete the process by selecting an assigned day.&#xA;&lt;br&gt;&amp;bull;&lt;sup&gt;&lt;sup&gt;&amp;ensp;&lt;/sup&gt;&lt;/sup&gt;&#xA;$\bf 5/12\:$ is the corresponding probability&#xA;        of proceeding to the next action, $\textrm{roll}_{15}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the next action is necessary, the row for $\textrm{roll}_{15}$ shows that&#xA;the $5$ unassigned potential ordered outcomes from $\textrm{flip}_{14}$&#xA;are combined with the possible results of a new die roll&#xA;to produce $30$ new potential ordered outcomes:&#xA;&lt;br&gt;&amp;emsp;&amp;emsp;&amp;emsp;&#xA;{&amp;thinsp;2&lt;sub&gt;&lt;sup&gt;&amp;hairsp;&amp;amp;&lt;/sup&gt;&lt;/sub&gt;T&lt;sub&gt;&lt;sup&gt;&amp;amp;&lt;/sup&gt;&lt;/sub&gt;1,&amp;hairsp;&amp;hairsp;&#xA;         3&lt;sub&gt;&lt;sup&gt;&amp;hairsp;&amp;amp;&lt;/sup&gt;&lt;/sub&gt;T&lt;sub&gt;&lt;sup&gt;&amp;amp;&lt;/sup&gt;&lt;/sub&gt;1,&amp;thinsp;&amp;thinsp;...,&amp;thinsp;&#xA;         6&lt;sub&gt;&lt;sup&gt;&amp;hairsp;&amp;amp;&lt;/sup&gt;&lt;/sub&gt;T&lt;sub&gt;&lt;sup&gt;&amp;amp;&lt;/sup&gt;&lt;/sub&gt;1,&#xA;&lt;br&gt;&amp;emsp;&amp;emsp;&amp;emsp;&amp;nbsp;&amp;thinsp;&#xA;         2&lt;sub&gt;&lt;sup&gt;&amp;hairsp;&amp;amp;&lt;/sup&gt;&lt;/sub&gt;T&lt;sub&gt;&lt;sup&gt;&amp;amp;&amp;thinsp;&lt;/sup&gt;&lt;/sub&gt;2&amp;hairsp;,&#xA;         3&lt;sub&gt;&lt;sup&gt;&amp;hairsp;&amp;amp;&lt;/sup&gt;&lt;/sub&gt;T&lt;sub&gt;&lt;sup&gt;&amp;amp;&amp;thinsp;&lt;/sup&gt;&lt;/sub&gt;2&amp;hairsp;,&amp;thinsp;...,&amp;thinsp;&#xA;         6&lt;sub&gt;&lt;sup&gt;&amp;hairsp;&amp;amp;&lt;/sup&gt;&lt;/sub&gt;T&lt;sub&gt;&lt;sup&gt;&amp;amp;&amp;thinsp;&lt;/sup&gt;&lt;/sub&gt;2&amp;hairsp;,&#xA;&lt;br&gt;&amp;emsp;&amp;emsp;&amp;emsp;&amp;nbsp;&amp;nbsp;&amp;hairsp; $^\vdots$&#xA;&lt;br&gt;&amp;emsp;&amp;emsp;&amp;emsp;&amp;nbsp;&amp;thinsp;&#xA;         2&lt;sub&gt;&lt;sup&gt;&amp;hairsp;&amp;amp;&lt;/sup&gt;&lt;/sub&gt;T&lt;sub&gt;&lt;sup&gt;&amp;amp;&amp;thinsp;&lt;/sup&gt;&lt;/sub&gt;6&amp;hairsp;,&#xA;         3&lt;sub&gt;&lt;sup&gt;&amp;hairsp;&amp;amp;&lt;/sup&gt;&lt;/sub&gt;T&lt;sub&gt;&lt;sup&gt;&amp;amp;&amp;thinsp;&lt;/sup&gt;&lt;/sub&gt;6&amp;hairsp;,&amp;thinsp;...,&amp;thinsp;&#xA;         6&lt;sub&gt;&lt;sup&gt;&amp;hairsp;&amp;amp;&lt;/sup&gt;&lt;/sub&gt;T&lt;sub&gt;&lt;sup&gt;&amp;amp;&amp;thinsp;&lt;/sup&gt;&lt;/sub&gt;6&amp;thinsp;}&#xA;&lt;br&gt;&#xA;Of those combinations, $28$ would select days while $2$ will be left over,&#xA;leaving a $2/30$ chance of requiring yet another action, $\textrm{flip}_{16}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compounding the table's rightmost column of continuation probabilities gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{l}&#xA;\textsf{average number of alternating rolls or flips} =&#xA;\\ \qquad&#xA;1 + \frac6{6}  \Bigl( 1 + \frac5{12} \bigl( 1 + \frac2{30} \bigl( 1 +&#xA;\frac4{4}  \bigl( 1 + \frac3{24} \bigl( 1 + \frac6{6}  \bigl( 1 +&#xA;\frac1{36} \bigl( 1 + \frac2{2}       ( 1 + \ldots )&#xA;                         \bigr) \bigr) \bigr) \bigr) \bigr) \bigr) \Bigr)&#xA;\\&#xA;\phantom{\textsf{average number of alternating rolls or flips}} = 2.4794153...&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, the same calculation with no coin flips is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{l}&#xA;\textsf{average rolls for die only} =&#xA;1 + \frac66  \Bigl( 1 + \frac1{36} \bigl( 1 +&#xA;\frac66  \bigl( 1 + \frac1{36}      ( 1 + \ldots ) \bigr) \bigr) \Bigr)&#xA;\\&#xA;\phantom{\textsf{average rolls for die only}} = 2.0\overline{571428}&#xA;\end{array}&lt;/p&gt;&#xA;" OwnerUserId="18129" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2016-02-10T04:57:55.883" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26062" PostTypeId="2" ParentId="26018" CreationDate="2016-02-02T08:51:07.467" Score="7" Body="&lt;p&gt;My guess is that the missing characters are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot; &amp;lt; &quot; and &quot; &gt; &quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;it symbolizes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Media buttons: &quot;&amp;lt;&amp;lt;&quot; rewind, &quot;=&quot; pause, &quot;&gt;&quot; play, &quot;&gt;&gt;&quot; fast forward &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16036" LastActivityDate="2016-02-02T08:51:07.467" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26063" PostTypeId="1" AcceptedAnswerId="26231" CreationDate="2016-02-02T10:36:28.837" Score="16" ViewCount="835" Body="&lt;p&gt;My respected colleague Dr. Henry just returned from an expedition in the Sahara.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He was hoping to find evidence of a sophisticated ancient civilization, but had little success. However, his team did discover several small artifacts, all in the same area:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YsXnc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YsXnc.png&quot; alt=&quot;Ancient artifacts&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if these artifacts served any purpose. And if so, what?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the Saharan civilization was as sophisticated as Dr. Henry believes, they must have developed very advanced mathematics.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are exactly two different symbols. What can this mean?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18558" LastEditorUserId="18558" LastEditDate="2016-02-08T08:57:14.390" LastActivityDate="2016-02-08T10:00:06.287" Title="Ancient Saharan Artifacts" Tags="|lateral-thinking|enigmatic-puzzle|visual|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26064" PostTypeId="2" ParentId="26063" CreationDate="2016-02-02T10:58:32.363" Score="17" Body="&lt;p&gt;After spending many hours in a haze of glue fumes...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clearly it's a frisbee! &lt;br&gt;&lt;a href=&quot;https://i.stack.imgur.com/tWkpZ.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tWkpZ.png&quot; alt=&quot;The assembled artifact&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12462" LastEditorUserId="12462" LastEditDate="2016-02-02T17:47:43.697" LastActivityDate="2016-02-02T17:47:43.697" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="26065" PostTypeId="2" ParentId="26058" CreationDate="2016-02-02T10:59:52.410" Score="24" Body="&lt;p&gt;I think the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The 'K' key on a keyboard&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am beneath myself&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'K' is below 'I' on the keyboard&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but above my prime&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My prime refers to the first letter of My which is 'M'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Left is on my right and up is nearby,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'L' is to the right of 'K' and 'U' and 'P' are nearby&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;down isn't close... well maybe a bit&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'D' isn't close to 'K' but a bit of down, 'O' and 'N' are.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When our most popular friend joins us,&#xA;me and (some of) my neighbours like to joke, lol&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the neighbours are 'J', 'L', ',' and 'O' with 'K' can be used to make 'joke, lol' (there's an additional 'E') As mentioned in the comments, this line may refer to 'JK' as in 'Just Kidding' or a shortened version of 'joke' which could fit better.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;stay close to me and u'll be ok&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'O' is close to me 'K' = OK&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18422" LastEditorUserId="18422" LastEditDate="2016-02-02T21:54:15.690" LastActivityDate="2016-02-02T21:54:15.690" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26066" PostTypeId="2" ParentId="26007" CreationDate="2016-02-02T12:21:14.783" Score="2" Body="&lt;p&gt;The first!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because I just have a 50% chance to fail, without even reading.&lt;/p&gt;&#xA;" OwnerUserId="9203" LastActivityDate="2016-02-02T12:21:14.783" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26067" PostTypeId="2" ParentId="26007" CreationDate="2016-02-02T12:26:31.020" Score="2" Body="&lt;p&gt;I'd take option 2, because there's only 9 amounts below \$10 whereas there's infinite above, so the odds of getting more than \$10 is higher.&lt;/p&gt;&#xA;" OwnerUserId="18896" LastEditorUserId="17250" LastEditDate="2016-02-02T12:33:14.753" LastActivityDate="2016-02-02T12:33:14.753" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26068" PostTypeId="2" ParentId="26033" CreationDate="2016-02-02T13:08:52.210" Score="5" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Javascript&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Little 'bout my age, far from that of B&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; javascript is newer than B&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet in my syntax, you will find some C&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; javascript has C-like &quot;curly brace&quot; syntax.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The way I treat vars, that is much like D&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; javascript has local and global variables, like D has&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I do take objects, but I'm not like E&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; javascript is object oriented, but doesn't use classes, like E has.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If it starts with F, you'll know it's not me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; javascript starts with a J&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18897" LastActivityDate="2016-02-02T13:08:52.210" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26069" PostTypeId="2" ParentId="26007" CreationDate="2016-02-02T13:43:26.830" Score="6" Body="&lt;p&gt;The most profitable is (without looking at any answers):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first option&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You make a statement like:&lt;br&gt;&#xA;&lt;br&gt;&#xA;    I won't get either 10 or 1,000,000,000,000 dollars for this statement.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Now you can't get 10 dollars because that would make the statement false&lt;br&gt;&#xA; and you only ever get exactly 10 bucks for a true statement.&lt;br&gt;&#xA; So you &lt;em&gt;must&lt;/em&gt; he paid a trillion dollars as that's the only other way that statement can be false.&lt;br&gt;&#xA; If you're paid anything else the statement is true but you're not&lt;br&gt;&#xA; getting the mandatory exact 10 dollars for it.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Edit:&lt;br&gt;&#xA; Ouch! Should have stipulated the &lt;em&gt;currency&lt;/em&gt;!&lt;br&gt;&#xA; Got a trillion Monopoly dollars - not even worth a single cent!!! :(  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastEditorUserId="10118" LastEditDate="2016-02-02T16:01:52.067" LastActivityDate="2016-02-02T16:01:52.067" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26070" PostTypeId="1" AcceptedAnswerId="26075" CreationDate="2016-02-02T13:52:15.457" Score="15" ViewCount="511" Body="&lt;p&gt;When you say that I'm feathery, that's quite clearly wrong.&lt;br&gt;&#xA;If I come from the trees then you can join me in song.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've nothing to wag, but you imply that I do.&lt;br&gt;&#xA;I have nothing to sell, but you suggest that's untrue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You say I'm unwell if I give you no prize.&lt;br&gt;&#xA;If you ensnare me I'll bring tears to your eyes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You say that I'm angry when I turn to the side.&lt;br&gt;&#xA;But call me divine and you'd better all hide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Who or what am I?&lt;/p&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2016-02-02T17:34:09.790" Title="Speaking about me takes you back" Tags="|wordplay|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26071" PostTypeId="2" ParentId="26007" CreationDate="2016-02-02T14:29:00.050" Score="3" Body="&lt;p&gt;Option 2. You are guaranteed more than 10 dollars no matter what you say. Also note that puzzle doesn't state that the statement has to be related to your winnings. It could be &quot;the sun is hot&quot;.&lt;/p&gt;&#xA;" OwnerUserId="14021" LastActivityDate="2016-02-02T14:29:00.050" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26072" PostTypeId="2" ParentId="25818" CreationDate="2016-02-02T15:02:19.133" Score="13" Body="&lt;p&gt;Just wanted to sign up for trying to answer. I think the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Video Game, since the asker told closest answer so far is internet, and consisting of 2 words&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am loved by many,&#xA;even if sometimes people are angry to me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; nearly everybody loves playing video games, but people become angry because of difficulty or multiplayer games&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can find me in different places,&#xA;yet you won't always find me stable.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; pc, mac, playstation, xbox. mobile. different platforms, but not the same behaviour in all, most cross-platform games work better on console, and crashes at computers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes you must use your talent,&#xA;sometimes you need wisdom to be my best friend.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you must have skill for FPS games but you must use your brain for MMO games. best friend must refer to becoming expert at it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was born before NATO's first SACEUR is appointed,&#xA;yet I am not old fashioned.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first publicly demonstrated electronic game was created in 1950. Bertie the Brain was an arcade game of tic-tac-toe, built by Dr. Josef Kates for the 1950 Canadian National Exhibition. The first SACEUR (1951-1952) was General Dwight Eisenhower. So it is before 1951, and today video games are still popular&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can tear me apart,&#xA;but then you will find radius and patella.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because of the microphone reference, I think patella references 0 with the shape, and radius references 1 with the shape. 0 &amp;amp; 1 are known as computer's core so it may refer to games consist of 0&amp;amp;1 code programming&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18898" LastActivityDate="2016-02-02T15:02:19.133" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="26073" PostTypeId="2" ParentId="25893" CreationDate="2016-02-02T15:57:00.890" Score="4" Body="&lt;p&gt;So sharp that you can execute what I want you to. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; sharpened to a point&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Will guide me the way, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; point me in the right direction&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and when needed I will split your troubles in your head &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; decimal point splits numbers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So valuable, that I always fight for you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; that's the whole point&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Standing at the crossroads&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; intersection point&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;waiting for me to find you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; And I finally got to the POINT.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18776" LastActivityDate="2016-02-02T15:57:00.890" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26075" PostTypeId="2" ParentId="26070" CreationDate="2016-02-02T16:51:18.847" Score="11" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wind&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think the title refers to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rewind or &quot;Re: wind&quot;, &quot;Re&quot; being an abbreviation for &quot;in re&quot; meaning &quot;in response to&quot; implying a subject about which you wish to speak.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you say that I'm feathery, that's quite clearly wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Could refer to &quot;downwind&quot; not being feathery (credit to deepthought in the comments)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I come from the trees then you can join me in song.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Referring to woodwind musical instruments&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've nothing to wag, but you imply that I do.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A tailwind is one that comes from behind.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have nothing to sell, but you suggest that's untrue.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Referring to &quot;trade&quot; winds (thanks to deepthought for pointing this out)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You say I'm unwell if I give you no prize.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;It's an ill wind that blows nobody any good&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you ensnare me I'll bring tears to your eyes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Trapped wind can make you cry&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You say that I'm angry when I turn to the side.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A &quot;cross&quot; wind (crosswind) comes from the side&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But call me divine and you'd better all hide.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Japanese word Kamikaze translates as &quot;divine wind&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks to the commenters for their help especially deepthought who sorted out two of the lines. &lt;/p&gt;&#xA;" OwnerUserId="18422" LastEditorUserId="18422" LastEditDate="2016-02-02T17:34:09.790" LastActivityDate="2016-02-02T17:34:09.790" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="26076" PostTypeId="2" ParentId="16106" CreationDate="2016-02-02T21:37:48.660" Score="17" Body="&lt;p&gt;This is more a puzzle than a riddle, but here's one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a common 10 letter English word that can be written with just the top row on a typewriter. What is it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Typewriter!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2016-02-02T21:37:48.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26077" PostTypeId="1" AcceptedAnswerId="26078" CreationDate="2016-02-02T22:34:06.067" Score="6" ViewCount="285" Body="&lt;p&gt;Find the number of integer triples $(a,b,c)$ such that there exists $0&amp;lt;t&amp;lt;1$ with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$| a + \frac{1}{2} - 589t| &amp;lt; \frac{1}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$| b + \frac{1}{2} - 989t| &amp;lt; \frac{1}{2} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$| c + \frac{1}{2} - 1189t| &amp;lt; \frac{1}{2}$&lt;/p&gt;&#xA;" OwnerUserId="18508" LastActivityDate="2016-02-04T09:40:56.097" Title="Firing a laser through a pile of cubes" Tags="|mathematics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26078" PostTypeId="2" ParentId="26077" CreationDate="2016-02-02T22:57:18.353" Score="12" Body="&lt;p&gt;As the title suggests, the question has this more accessible interpretation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Divide 3-space into unit cubes such that their vertices are lattice points. How many cubes contain part of the line segment with endpoints $(0,0,0)$ and $(589,989,1189)$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The line segment crosses $588$ integer $x$-coordinates, $988$ $y$-coordinates, and $1188$ $z$-coordinates, entering a new cube each time. Because $589$, $989$, and $1189$ are pairwise relatively prime, it never crosses two integer coordinates at the same time. The total number of cubes it passes through is therefore $588+988+1188+1=2765$.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2016-02-03T03:11:41.770" LastActivityDate="2016-02-03T03:11:41.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26080" PostTypeId="2" ParentId="26055" CreationDate="2016-02-03T03:34:38.873" Score="4" Body="&lt;p&gt;My guess is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; George Washington&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A monarch in democracy,&#xA;yet revered without hypocrisy,&#xA;my name shall live on evermore.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many wanted him to be king of the United States, and he is consistently ranked as one of the top presidents in opinion polls.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Under the same rule as the sign of war,&#xA;and thirty-one's twenty-four,&#xA;the first to seize a third at three.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm unsure on this part, but I think it has to do with him being the  Commander-in-Chief of the Continental Army. Third of three may refer to his third appearance on a 3 cent US postage stamp.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Totaling thirteen and a little more,&#xA;I became the one to do it, finally.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He was first president under the Constitution, which was ratified by the original 13 colonies. A little more could refer to the fact that he was the only one who has been unanimously elected.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17297" LastActivityDate="2016-02-03T03:34:38.873" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26081" PostTypeId="1" AcceptedAnswerId="26141" CreationDate="2016-02-03T06:13:27.183" Score="8" ViewCount="683" Body="&lt;p&gt;When I was a young adult, I left something at level 2 because&lt;br&gt;&#xA;I found something, the thing I found is like a magnet to some people.&lt;br&gt;&#xA;Some say it belongs to the castle near the neck and the iron island.&lt;br&gt;&#xA;What am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Use the map.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer has some relationship with him. Which person does these pictures represents? &lt;img src=&quot;https://i.stack.imgur.com/aKnaL.jpg&quot; alt=&quot;enter image description here]+[1]&quot;&gt;&lt;strong&gt;&lt;em&gt;+&lt;/em&gt;&lt;/strong&gt;&lt;a href=&quot;https://i.stack.imgur.com/XHGYp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XHGYp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The person I am asking about exists in real. He is not part of any comic book.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If this question remain unsolved until 6-Feb then I'll post the answer myself. And I am sure you'll say &quot;Oh f*k, that was easy.&quot;&lt;/p&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-02-05T12:06:19.003" LastActivityDate="2016-02-05T14:52:13.143" Title="The Magnet for some people" Tags="|riddle|" AnswerCount="5" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26082" PostTypeId="2" ParentId="25818" CreationDate="2016-02-03T08:00:55.257" Score="1" Body="&lt;p&gt;I know this question has been answered but I think another nice guess could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Junction transistor&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am loved by many,&#xA;even if sometimes people are angry to me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is used in many devices; devices that people love. But the same people can get angry if those devices doesn't work.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can find me in different places,&#xA;yet you won't always find me stable.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, used at many devices. When output voltage is kept constant, then input current increase suddenly even at the increase of small input voltage.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes you must use your talent,&#xA;sometimes you need wisdom to be my best friend.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It's hard to work with them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was born before NATO's first SACEUR is appointed,&#xA;yet I am not old fashioned.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First bipolar junction transistor was invented on June 23, 1948. Patent filed on June 26, 1948.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can tear me apart,&#xA;but then you will find radius and patella.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/W7Vzk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W7Vzk.png&quot; alt=&quot;transistor&quot;&gt;&lt;/a&gt; It is denoted by this sign in electronics. Radius could refer to the circle. And patella to the thing in the circle. As it is in the circle so it also explains the 'tear me apart' sentence of this puzzle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This answers also seems to fit the other hints. However the accepted solution is at its best.&lt;/p&gt;&#xA;" OwnerUserId="17482" LastActivityDate="2016-02-03T08:00:55.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26083" PostTypeId="1" AcceptedAnswerId="26275" CreationDate="2016-02-03T09:16:38.290" Score="0" ViewCount="445" Body="&lt;p&gt;Both of my riddles are answered, so here you go :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I love to start with every single digit integer ranged from 1-9 but can't start with 9&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First version of one of my capabilities has a common number with a popular boeing model&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orion Nebula and I have a common abbreviation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Mercedes Benz automobile model and what I did, have a common abbreviation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Born in the early 1900s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; cst1992's answer for nebula is M78 and Boeing 787. M78 is wrong, but even if it was correct, Boeing doesn't have to be the same number. Every line in the riddle is standalone. But when they are together, they mean something. Start digging from Orion Nebula, it is the starting point of the riddle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2 (I thought my question was clear, I am giving an imaginary example solution):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Orion Nebula is also known as M78, Company X has a TV which is known as M78. Mercedes has a car model named ML60, Company X has a phone named ML60. By first version of capabilities; &quot;X&quot; is a phone company but also produces tv so has different capabilities, and has a phone which is 787 Ghz. Boeing has B787, so answer could be company X for this example. Like I said on comment, your clues are not connected to each other for this question. By collecting them all, you will finalise your answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I have blue colour&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 4 (Orion Nebula's starting point):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; M42 is the first abbreviation of Orion Nebula. &lt;a href=&quot;https://en.wikipedia.org/wiki/M42&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/M42&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;had to give all possible hints because of low interest in the puzzle :)&#xA;&lt;strong&gt;with all the given hints, it became relatively easy, so first one to answer every connection will be selected as the correct answer.&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="18685" LastEditorUserId="18685" LastEditDate="2016-02-04T12:00:24.763" LastActivityDate="2016-02-09T21:25:12.183" Title="What am I? by SC - #3" Tags="|riddle|word|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="26084" PostTypeId="2" ParentId="26033" CreationDate="2016-02-03T10:11:13.240" Score="2" Body="&lt;p&gt;Can you be?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Deesel, formerly known as &quot;G&quot;. &lt;a href=&quot;https://en.wikipedia.org/wiki/Deesel&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Deesel&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Little 'bout my age: far from that of B&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First released on 2004-11-06, so it's tad younger than B (ca. 1969). &lt;a href=&quot;http://sourceforge.net/p/deesel/news/?source=navbar&quot; rel=&quot;nofollow&quot;&gt;http://sourceforge.net/p/deesel/news/?source=navbar&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet in my syntax, you will find some C&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Based on Java, there is a bit of C in the syntax.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The way I treat vars, that is much like D&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Both Deesel and D were influenced by Java, Both feature garbage collection.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I do take objects, but I'm not like E&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Variables are dynamically typed in E.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If it starts with F, you'll know it's not me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Obviously, neither &quot;Deesel&quot; nor &quot;G&quot; start with an &quot;F&quot;. However, &quot;G&quot; starts with a &quot;G&quot;, which would be the next letter in the alphabet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18915" LastEditorUserId="18915" LastEditDate="2016-02-03T11:34:12.443" LastActivityDate="2016-02-03T11:34:12.443" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26085" PostTypeId="2" ParentId="26033" CreationDate="2016-02-03T10:16:23.933" Score="1" Body="&lt;p&gt;Can you be?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Swift&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Little 'bout my age: far from that of B&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First appeared at 2014, far compared to other answers as well&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet in my syntax, you will find some C&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Swift is in C-Family PLs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The way I treat vars, that is much like D&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; similar named variables, and handled closely mutable/immutable&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I do take objects, but I'm not like E&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; E: def myFnc(name) :any /// Swift: func myFnc(name: String)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If it starts with F, you'll know it's not me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; with your clue that &quot;First letter is important&quot; applies for Swift. variables are like Bool, String, Int so they are always starting capitilized. Declarations like UIView, and starting with NS (ios tho) are starting capital.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18685" LastActivityDate="2016-02-03T10:16:23.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26086" PostTypeId="2" ParentId="26033" CreationDate="2016-02-03T10:22:26.823" Score="12" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; B, C, D, E, F, .... Go!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Little 'bout my age, far from that of B&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Go is a lot newer than B. It's been developed in 2007.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet in my syntax, you will find some C&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is a compiled, statically typed language in the tradition of C and C++ with curly code blocks etcetera.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The way I treat vars, that is much like D&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Variable type declarations can be inferred in both languages&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I do take objects, but I'm not like E&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In Go mutable objects and pointers can be shared between goroutines while in E, all values are objects and computation is performed by sending messages to objects.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If it starts with F, you'll know it's not me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Go starts with a G&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12679" LastActivityDate="2016-02-03T10:22:26.823" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26087" PostTypeId="2" ParentId="26083" CreationDate="2016-02-03T12:14:36.883" Score="2" Body="&lt;p&gt;I have the nebula and Boeing figured out, and they could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; M78 in the Orion constellation and the 787 Dreamliner&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18117" LastActivityDate="2016-02-03T12:14:36.883" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26088" PostTypeId="1" AcceptedAnswerId="27603" CreationDate="2016-02-03T13:32:34.957" Score="7" ViewCount="545" Body="&lt;p&gt;What is the missing number in this cyclic sequence?&#xA;It would require a notation convention not seen here&#xA;to fit into the scheme here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\phantom{&#xA;\textrm{(repeat)} }&#xA;\quad\textbf{?}~~ \,,~~ 2.4  \,,~~ 1   \,,~~ 36.0&#xA;                  \,,~~ 1    \,,~~ 8   \,,~~ 1  \,,~~ 15 \,,~~ 2.4&#xA;                  \,,~~ 1    \,,~~ 8.0 \,,~~ \frac{~9~}2 ,&#xA;\phantom{         \,,~ \dots }&#xA;\\&#xA;\textrm{(repeat)}&#xA;\quad\textbf{?}~~ \,,~~ 2.4  \,,~~ 1   \,,~~ 36.0&#xA;                  \,,~ \dots&#xA;\phantom{         \,,~~ 1    \,,~~ 8   \,,~~ 1  \,,~~ 15 \,,~~ 2.4&#xA;                  \,,~~ 1    \,,~~ 8.0 \,,~~ \frac{~9~}2 , }&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hintified from comments:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $8$ vs $8.0\,$? &amp;nbsp; $2.4$ vs $\frac92$? &amp;nbsp;&#xA; All relevant properties, including the visual one, are mathematical.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wiktionary.org/wiki/half-truth#Noun&quot; rel=&quot;nofollow&quot;&gt;Half&lt;/a&gt;$/$&amp;hairsp;&lt;a href=&quot;https://en.wikipedia.org/wiki/Tell_%28poker%29&quot; rel=&quot;nofollow&quot;&gt;hint&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The visible clue:&#xA; $2.4$, $36.0$, $2.4$ and $8.0$ are not represented as reduced fractions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18129" LastEditorUserId="18129" LastEditDate="2016-02-16T12:11:32.117" LastActivityDate="2016-02-18T17:51:55.793" Title="A purely numerical sequence with a hint of the visual" Tags="|number-sequence|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="26089" PostTypeId="1" AcceptedAnswerId="26114" CreationDate="2016-02-03T13:34:41.140" Score="6" ViewCount="284" Body="&lt;p&gt;Imagine a game on a grid of $m$ rows and $n$ columns. Player One will place a fixed number $r$ of markers in each row. Player Two will select the two rows which maximize the number of columns having markers in both rows, while Player One will try to minimize this number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$m = 3$  &lt;/li&gt;&#xA;&lt;li&gt;$n = 3$  &lt;/li&gt;&#xA;&lt;li&gt;$r = 2$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(a $3\times 3$ grid, placing 2 markers in each row)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, Player One chooses two columns in each row, and places markers in those cells:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{|c|c|c|}&#xA;\hline&#xA;X&amp;amp;X&amp;amp;\\&#xA;\hline&#xA;X&amp;amp;X&amp;amp;\\&#xA;\hline&#xA;&amp;amp;X&amp;amp;X\\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Player Two then checks the row pairs $(1,2),(1,3)$ and $(2,3)$. The first pair shares 2 columns, and the last two pairs share only 1 column, so Player Two selects rows 1 and 2 because they have the most matching columns (2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for a general strategy for placing markers given $(m,n,r)$. The strategy need not be optimal as it is intended to be uses in stochastic local search. This problem arises as a sub problem in investment design in finance. An easy to calculate lower bound exist, and for almost all instances I have seen a solution where the similarity is this lower bound exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example instances: $(10,30,9)$ has lower bound 2. $(11,22,10)$ has lower bound 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The instance $(10,8,3)$ has a lower bound of 2 with the following solution:&#xA;$$&#xA;\begin{array}{|c|c|c|c|c|c|c|c|}&#xA;\hline&#xA;X&amp;amp;X&amp;amp;X&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;\\&#xA;\hline&#xA;X&amp;amp;X&amp;amp;&amp;amp;X&amp;amp;&amp;amp;&amp;amp;&amp;amp;\\&#xA;\hline&#xA;X&amp;amp;X&amp;amp;&amp;amp;&amp;amp;X&amp;amp;&amp;amp;&amp;amp;\\&#xA;\hline&#xA;X&amp;amp;X&amp;amp;&amp;amp;&amp;amp;&amp;amp;X&amp;amp;&amp;amp;\\&#xA;\hline&#xA;X&amp;amp;X&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;X&amp;amp;\\&#xA;\hline&#xA;X&amp;amp;X&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;X\\&#xA;\hline&#xA;X&amp;amp;&amp;amp;X&amp;amp;X&amp;amp;&amp;amp;&amp;amp;&amp;amp;\\&#xA;\hline&#xA;X&amp;amp;&amp;amp;X&amp;amp;&amp;amp;X&amp;amp;&amp;amp;&amp;amp;\\&#xA;\hline&#xA;X&amp;amp;&amp;amp;X&amp;amp;&amp;amp;&amp;amp;X&amp;amp;&amp;amp;\\&#xA;\hline&#xA;X&amp;amp;&amp;amp;X&amp;amp;&amp;amp;&amp;amp;&amp;amp;X&amp;amp;\\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most similar rows (1 and 2, among others) share 2 markers and the solution is therefore optimal.&lt;/p&gt;&#xA;" OwnerUserId="18919" LastEditorUserId="18919" LastEditDate="2016-02-04T15:44:09.370" LastActivityDate="2016-02-04T18:38:43.050" Title="Placing markers in grid and minimize similarity of rows" Tags="|algorithm|computer-science|" AnswerCount="1" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="26090" PostTypeId="2" ParentId="26033" CreationDate="2016-02-03T13:37:26.920" Score="1" Body="&lt;p&gt;My first attempt at a puzzle here. Here it goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My guess is that you are:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Rust_(programming_language)&quot; rel=&quot;nofollow&quot;&gt;Rust&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Little 'bout my age: far from that of &lt;a href=&quot;https://en.wikipedia.org/wiki/B_(programming_language)&quot; rel=&quot;nofollow&quot;&gt;B&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/B_(programming_language)&quot; rel=&quot;nofollow&quot;&gt;B&lt;/a&gt; was developed in 1969. &lt;a href=&quot;https://en.wikipedia.org/wiki/Rust_(programming_language)&quot; rel=&quot;nofollow&quot;&gt;Rust&lt;/a&gt; was developed in 2010. So it is way newer than B.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet in my syntax, you will find some &lt;a href=&quot;https://en.wikipedia.org/wiki/C_(programming_language)&quot; rel=&quot;nofollow&quot;&gt;C&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;Rust&lt;/code&gt; uses curly brackets like &lt;code&gt;C&lt;/code&gt;. Also all keywords are typed in small caps.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The way I treat vars, that is much like &lt;a href=&quot;https://en.wikipedia.org/wiki/D_(programming_language)&quot; rel=&quot;nofollow&quot;&gt;D&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This one I'm not too sure about :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I do take objects but, I'm not like E:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not this one as well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If it starts with F, you'll know it's not me:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rust doesn't start with 'F'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18753" LastActivityDate="2016-02-03T13:37:26.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26091" PostTypeId="2" ParentId="24728" CreationDate="2016-02-03T14:23:11.117" Score="-1" Body="&lt;p&gt;The clue is hidde&lt;strong&gt;n in th&lt;/strong&gt;is statement. So correct Answer is 9th.&lt;/p&gt;&#xA;" OwnerUserId="18892" LastActivityDate="2016-02-03T14:23:11.117" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26092" PostTypeId="1" AcceptedAnswerId="26093" CreationDate="2016-02-03T17:27:25.673" Score="5" ViewCount="711" Body="&lt;p&gt;I have a riddle. Problem is I don't know the answer. Whoever answers it right will be awarded best answer. The deadline is the end of the month. If no correct answer is given before then, I will close this question (or ask for it to be closed). Here is the riddle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To the Northeast we stand,&lt;br&gt;&#xA;  Ancient natives of the new world,&lt;br&gt;&#xA;  Proud though no longer tall.&lt;br&gt;&#xA;  Worn by age we still watch over the land,&lt;br&gt;&#xA;  Still more noble than our cousin hills,&lt;br&gt;&#xA;  Tell us our name.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;1 word, no punctuation&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can input the answer and see if it is right or wrong. But the site has nothing to do with the riddle. It is a random riddle basically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer has nothing to do with religion. It may also have something to do with geography/maps. It is likely talking about the north eastern USA. That is what I know.&lt;/p&gt;&#xA;" OwnerUserId="18927" LastEditorUserId="1766" LastEditDate="2016-02-24T20:55:50.420" LastActivityDate="2016-02-24T20:55:50.420" Title="To the northeast we stand" Tags="|riddle|" AnswerCount="1" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="26093" PostTypeId="2" ParentId="26092" CreationDate="2016-02-03T17:39:32.660" Score="11" Body="&lt;p&gt;&lt;em&gt;Thanks to question_asker for improvement in the answer&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pretty sure the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Adirondacks. (The Adirondack Mountains)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To the Northeast we stand,&#xA;Ancient natives of the new world,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Adirondacks are in the northeast of North America.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proud though no longer tall.&#xA;Worn by age we still watch over the land,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Adirondacks are one of the oldest extant mountain ranges, which is why they're not terribly high compared to newer mountain ranges like the Himalayas.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Still more noble than our cousin hills,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure about this one; I &lt;em&gt;think&lt;/em&gt; this may refer to other remnants of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Central_Pangean_Mountains&quot; rel=&quot;nofollow&quot;&gt;Central Pangean Mountains&lt;/a&gt; such as the Scottish Highlands and the Anti-Atlas range. The Scottish Highlands aren't as high and the Adirondacks and the Anti-Atlas Mountains aren't as well-known as the Adirondacks, at least in Anglophone culture.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18729" LastEditorUserId="17482" LastEditDate="2016-02-03T20:37:28.733" LastActivityDate="2016-02-03T20:37:28.733" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="26094" PostTypeId="1" AcceptedAnswerId="26095" CreationDate="2016-02-03T18:13:43.490" Score="8" ViewCount="907" Body="&lt;p&gt;I have a calendar with daily puzzles, and I want to verify something, so here is today's puzzle. I answered it myself and have the &quot;official&quot; answer as well. Please be sure to clearly state your reasoning for your answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determine the number from 1 to 10 that goes at each point on this star according to the rules below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/newsQ.jpg&quot; alt=&quot;Going clockwise from the top of the star, the points are A, C, E, D, B. Apologies for the manual illustration instead of a nice digital version.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The sum of all five numbers is 28, and no number is duplicated.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Consecutive numbers may not appear (a) at adjacent points nor (b) across a straight line from each other.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Point A is (a) greater than the sum of B and C and (b) less than the sum of D and E.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Points C and D sum to 11, and neither point is 10.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="14452" LastActivityDate="2016-02-05T04:10:22.133" Title="Five numbers at the points of a star" Tags="|logical-deduction|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26095" PostTypeId="2" ParentId="26094" CreationDate="2016-02-03T18:19:52.573" Score="9" Body="&lt;p&gt;If I've read/interpreted correctly, the following should work&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A = 6&lt;br&gt;&#xA; B = 1&lt;br&gt;&#xA; C = 3&lt;br&gt;&#xA; D = 8&lt;br&gt;&#xA; E = 10&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My reasoning was more or less:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Lemma 1&lt;/strong&gt;: we can't have consecutive numbers &lt;em&gt;at all&lt;/em&gt; with rule 2,&lt;br&gt;&#xA;&lt;br&gt;&#xA; Possible pairs for C and D are $(1,10)$, $(2,9)$, $(3,8)$, $(4,7)$, and $(5,6)$, but $(1,10)$ isn't possible because of rule 4, $(5,6)$ is impossible by lemma 1, and $(2,9)$ and $(4,7)$ aren't possible because then we can no longer choose 5 numbers with no two pairwise points being consecutive numbers, leaving us with C and D being 3 and 8.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Now we need a sum of 17 with 3 numbers from the set $\{1,5,6,10\}$, which obviously leaves us with 1, 6, and 10 for A, B, and E.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The arrangement was then straightforward, though as Wesley pointed out, swapping A and E works as well&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17147" LastEditorUserId="17147" LastEditDate="2016-02-03T18:35:45.320" LastActivityDate="2016-02-03T18:35:45.320" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26096" PostTypeId="2" ParentId="26094" CreationDate="2016-02-03T18:49:10.537" Score="5" Body="&lt;p&gt;I came up with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A = 10&lt;br&gt;&#xA; B = 1&lt;br&gt;&#xA; C = 3&lt;br&gt;&#xA; D = 8&lt;br&gt;&#xA; E = 6&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Basically figuring:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There can't be any consecutive numbers&lt;br&gt;&#xA; Can't be 1 3 5 7 9&#xA; So need 10&lt;br&gt;&#xA; And also 8&lt;br&gt;&#xA; Then 6 3 1&lt;br&gt;&#xA; Made the points fit the other criteria&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastActivityDate="2016-02-03T18:49:10.537" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26097" PostTypeId="2" ParentId="26007" CreationDate="2016-02-03T18:53:08.250" Score="3" Body="&lt;p&gt;I'll go with &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First Option&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the statement is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'll get less than 10 dollars.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Simple and Sober.&lt;/p&gt;&#xA;" OwnerUserId="17070" LastActivityDate="2016-02-03T18:53:08.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26098" PostTypeId="2" ParentId="26055" CreationDate="2016-02-03T22:05:32.003" Score="12" Body="&lt;p&gt;My guess is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; American Pharoah (sic) the horse that recently won the Triple Crown.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A monarch in democracy,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pharaohs were monarchs, but the U.S. is a democracy. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yet revered without hypocrisy,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He's a horse, not a political figure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;my name shall live on evermore.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Triple Crown winners are rare enough to be famous as long as people keep racing horses.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Under the same reign as the sign of war,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure about this one; I think it's a reference to the zodiac, but I don't see a connection. Maybe a reference to War Admiral, a previous Triple Crown winner?&#xA; Could either refer to his jockey also riding War Eagle, a previous contender, or his lineage, as he is a descendant of War Admiral and Man O' War, among other notable thoroughbreds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and thirty-one's twenty-four,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A possible reference to Secretariat's 1973 win by 31 lengths.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the first to seize a third at three.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; His jockey, Victor Espinoza, was the first to ride 3 different Triple Crown horses, winning it on his third attempt.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Totaling thirteen and a little more,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Refers to the how many lengths he won by, over the 3 races. 1 + 7 + 5 1/2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I became the one to do it, finally;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Referring to being the first Triple Crown winner in 37 years.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm now a monarch doubly.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Both from his name and the Triple Crown.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Credit to regan for help on lines 4 and 6, Herb Wolfe on line 7, question_asker, whose comment about LeBron James and Kobe Bryant got me thinking about sports, and Big Black Box, who corrected the name's spelling.&lt;/p&gt;&#xA;" OwnerUserId="18729" LastEditorUserId="18729" LastEditDate="2016-02-04T21:30:17.173" LastActivityDate="2016-02-04T21:30:17.173" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="26099" PostTypeId="1" AcceptedAnswerId="26106" CreationDate="2016-02-03T23:54:14.223" Score="8" ViewCount="538" Body="&lt;p&gt;My previous riddles weren't hard enough,&lt;br&gt;&#xA;so I decided to draw from this experience.&lt;br&gt;&#xA;I didn't want it to be too black,&lt;br&gt;&#xA;so it's half way in-between.&lt;br&gt;&#xA;Is that not fine with you?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My style may elicit some pointed comments,&lt;br&gt;&#xA;but the answer to my last riddle was a key on the keyboard,&lt;br&gt;&#xA;so I wanted to make a big change,&lt;br&gt;&#xA;and I hope you rank it highly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Oh, I guess it turned out to be mostly black after all).&lt;/p&gt;&#xA;" OwnerUserId="18522" LastActivityDate="2016-02-05T17:13:40.580" Title="A harder riddle (hopefully)" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26101" PostTypeId="1" AcceptedAnswerId="26102" CreationDate="2016-02-04T03:27:26.403" Score="6" ViewCount="458" Body="&lt;p&gt;One early morning while you check your mailbox, you see a letter. You open it and read:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;~ For people who like Gray Test, Red Lid, Wordy Pal, people who build&#xA;  Carport Gym or even who Swap Rods ~&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Hi this is Paul, also known as &quot;Mr. Party Cog&quot;. I'm inviting you to my&#xA;  place so you can join the party! The party started around a year ago and&#xA;  is fairly new and I plan to keep it going. I know you will be interested&#xA;  because reading this tells me that we share a common thing. But first&#xA;  you need to know the address.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Here are the landmarks and a guide:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I'm located at a commercial area. Look for Stock Trade street and&#xA;  expect 3 stops going there.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Once you have reached the street, look for a light blue gate with a big &quot;P&quot;&#xA;  on top of it. Don't get lost, a lot of parties are also held on this&#xA;  street. Just to be sure, your name will be posted above the gate&#xA;  unless you are not on the list. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Once inside, you may ask your questions on anyone there for directions&#xA;  because my place is big. Don't worry, a lot of people are also asking&#xA;  and many are willing to answer.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;When you get to the house, my butler will give you the key to your own&#xA;  room. That room is just made for you for being on the list. The best&#xA;  part is, you can personalize it! Arrange it based on your personality&#xA;  or even put a picture of yourself!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;During the party, you can socialize and ask other people and you may&#xA;  gain popularity! Just make sure to have sense or else they will put&#xA;  you down. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Did you get the address? I know that after you read my invitation, you&#xA;  will find the right party. Hope to see you here!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What's the address? &lt;/p&gt;&#xA;" OwnerUserId="16036" LastEditorUserId="22529" LastEditDate="2016-06-29T06:22:43.057" LastActivityDate="2016-06-29T06:22:43.057" Title="You're Invited!" Tags="|enigmatic-puzzle|steganography|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26102" PostTypeId="2" ParentId="26101" CreationDate="2016-02-04T04:14:40.783" Score="5" Body="&lt;p&gt;Every capitalized phrase is an anagram of a common Puzzling.SE tag:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Gray Test → Strategy&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Red Lid → Riddle&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Wordy Pal → Wordplay&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Carport Gym → Cryptogram&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Swap Rods → Password&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Mr. Party Cog → Cryptogram&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&quot;Stock Trade&quot; could be an allusion to the phrase &quot;stock exchange&quot;, what the name for Stack Exchange is based off of; the three stops could be  the links to &quot;All Sites&quot;, &quot;Puzzling&quot;, and &quot;Visit Sites&quot; that get you from the general SE site to Puzzling.SE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The blue gate with a big P may refer to the quote bubble logo of Puzzling (though it should be two question marks?). The name is, of course, your username, and the list is the list of registered users.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Asking for directions is asking a question, and answering is... well, answering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your &quot;room&quot; is your profile page, and the key is your password. You can personalize it by changing the text or your avatar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Popularity&quot; is rep; asking good questions and responding with good answers makes you gain rep, and saying particularly nonsensical things makes other people downvote you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So of course, the address is &lt;code&gt;puzzling.stackexchange.com&lt;/code&gt;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(Thanks to Alconja for helping with 'Stock Trade'!)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="11876" LastEditDate="2016-02-04T04:27:25.583" LastActivityDate="2016-02-04T04:27:25.583" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26105" PostTypeId="2" ParentId="26033" CreationDate="2016-02-04T06:42:19.627" Score="1" Body="&lt;p&gt;Can you be?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Eiffel_%28programming_language%29&quot; rel=&quot;nofollow&quot;&gt;Eiffel&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Little 'bout my age: far from that of B&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since B appeared in 1969, and Eiffel in 1986.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet in my syntax, you will find some C&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As with most structured languages, some C syntax is in there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The way I treat vars, that is much like D&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Object-oriented handling of vars in D is very similar to C++/Java, which in-turn were influenced by Eiffel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I do take objects, but I'm not like E&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eiffel, (somewhat ironically) treats objects very differently from E.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If it starts with F, you'll know it's not me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A play on the position of F and E in the alphabet.  Since the first letter is important, Eiffel would not only not start with F, but be alphabetically next to it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18938" LastActivityDate="2016-02-04T06:42:19.627" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26106" PostTypeId="2" ParentId="26099" CreationDate="2016-02-04T10:10:16.977" Score="9" Body="&lt;p&gt;I think it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The F# major scale. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;The first part:&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;This is referring to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An F-grade pencil. The HB scale goes from Harder to Blacker (9H, 8H, ..., 2H, H, F, HB, B, 2B, ..., 8B, 9B). The harder the pencil's lead grade, the lighter the mark. The softer the lead, the darker (blacker) the mark.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My previous riddles weren't hard enough,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The pencil is graded on the hardness of the lead.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so I decided to draw from this experience.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Drawing with a pencil.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I didn't want it to be too black,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The pencil &lt;strong&gt;isn't&lt;/strong&gt; in the B section.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so it's half way in-between.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The F rating is an arbitrary letter chosen to mark the midpoint between H and HB.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is that not fine with you?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;F&quot; is an arbitrary letter. It doesn't stand for &quot;fine&quot;, as is often thought.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;The Second Part&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;My style may elicit some pointed comments,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's sharp.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but the answer to my last riddle was a key on the keyboard,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The last riddle was about the letter &quot;K&quot; and letters close to it on the keyboard.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so I wanted to make a big change,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is a change in &lt;strong&gt;scale&lt;/strong&gt; and involves keys on a piano. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and I hope you rank it highly.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The rank Major in the military, and major vs minor scale.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(Oh, I guess it turned out to be mostly black after all).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The F# major scale on a piano is played with more black keys than white keys. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12462" LastEditorUserId="12462" LastEditDate="2016-02-05T17:13:40.580" LastActivityDate="2016-02-05T17:13:40.580" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="26107" PostTypeId="1" AcceptedAnswerId="38187" CreationDate="2016-02-04T10:57:56.123" Score="7" ViewCount="743" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;To this riddle one shall add &quot;meta&quot;&lt;br/&gt;&#xA;  Like a program and an SE site&lt;br/&gt;&#xA;  For the answer you have some data&lt;br/&gt;&#xA;  Yet not always you will get it right&lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Trying out an &quot;ABAB&quot; rhyming scheme here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess I'm not good at judging difficulty, so please excuse me if this is too easy or too hard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer itself contains no &quot;meta&quot;, nor will it make sense if it does.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Data&quot; means raw, unprocessed information, it may or may not be computer data.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18817" LastEditorUserId="18817" LastEditDate="2016-02-04T18:01:29.690" LastActivityDate="2016-07-24T22:41:16.550" Title="Riddle: You don't always get me right" Tags="|riddle|" AnswerCount="6" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26108" PostTypeId="2" ParentId="26081" CreationDate="2016-02-04T11:08:57.107" Score="2" Body="&lt;p&gt;A little convoluted but my guess is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Trident&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I was a young adult, I left something at level 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Part of maturing into a young adult involves losing your baby teeth. If you're leaving them at level 2, presumably you move to level 3 with teeth i.e, your &quot;tri&quot; &quot;dent&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the thing I found is like a magnet to some people&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A trident was historically used as a polearm or &quot;pole&quot; weapon due to being on the end of a long shaft. A magnet also has poles and can point you in the right direction so could also be a polearm.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some say it belongs to the city near the neck and the iron island.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think this refers to the Trident river in ASOIAF. The city could be Riverrun (?) where the Tullys historically held the position of Lord Paramount of the Trident.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18422" LastActivityDate="2016-02-04T11:08:57.107" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26110" PostTypeId="1" AcceptedAnswerId="26111" CreationDate="2016-02-04T13:11:45.167" Score="2" ViewCount="203" Body="&lt;p&gt;This question was asked in NET. I am unable to solve it.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/bKGHC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bKGHC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18892" LastActivityDate="2016-02-04T13:43:12.373" Title="Reasoning CSIR NET June 15" Tags="|pattern|letter-sequence|" AnswerCount="1" CommentCount="0" ClosedDate="2016-02-04T14:09:27.770" ContentLicense="CC BY-SA 3.0" />
  <row Id="26111" PostTypeId="2" ParentId="26110" CreationDate="2016-02-04T13:31:06.807" Score="1" Body="&lt;p&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; aN is reversed Na which is &lt;strong&gt;S&lt;/strong&gt;odium&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; eF is reversed Fe which is &lt;strong&gt;I&lt;/strong&gt;ron&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; gH is reversed Hg which is &lt;strong&gt;M&lt;/strong&gt;ercury&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; nS is reversed Sn which is: &lt;strong&gt;T&lt;/strong&gt;in so &lt;strong&gt;1&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18685" LastEditorUserId="2730" LastEditDate="2016-02-04T13:43:12.373" LastActivityDate="2016-02-04T13:43:12.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26112" PostTypeId="2" ParentId="26107" CreationDate="2016-02-04T13:48:55.163" Score="3" Body="&lt;p&gt;If the last line  &lt;code&gt;Yet not always you will get it right&lt;/code&gt; indicates the whole world including meta, &#xA;then my answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Metaphor&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't get it right all the times.&lt;/p&gt;&#xA;" OwnerUserId="17070" LastActivityDate="2016-02-04T13:48:55.163" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26113" PostTypeId="2" ParentId="26081" CreationDate="2016-02-04T14:13:57.390" Score="3" Body="&lt;p&gt;Some say it belongs to the city near the neck and the iron island.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; At the map, Iron Island is at &lt;strong&gt;Ironman's Bay&lt;/strong&gt;. So the answer is something which belongs to the Iron Man&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I was a young adult, I left something at level 2 because&#xA;I found something, the thing I found is like a magnet to some people.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Iron Man's electromagnetic ARC Reactor&lt;/strong&gt; keeps the shrapnel reaching to his heart. This may be the reference to the magnet to some people. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18685" LastActivityDate="2016-02-04T14:13:57.390" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26114" PostTypeId="2" ParentId="26089" CreationDate="2016-02-04T16:17:43.327" Score="3" Body="&lt;p&gt;The following brutish algorithm does okay, it might be a start. It generates an optimal solution for $(10,8,3)$ and gets scores of 3 and 6 for $(10,30,9)$ and $(11,22,10)$, respectively by building the board one row at a time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To create the first row, place markers in the first $r$ columns.&#xA;To generate the next row, $R$, start with a row with no markers. While $R$ has less than $r$ markers:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For each column $i$, create a candidate row $R_i$ which is $R$ but with a marker added to column $i$. (If $R$ already has a marker in column $i$, ignore this candidate).&lt;/li&gt;&#xA;&lt;li&gt;Score $R_i$ pairwise against each other row already added to the board, note the highest pairwise scoring as $S_i$.&lt;/li&gt;&#xA;&lt;li&gt;Among the rows already in the board, count how many also have a marker in column $i$, denote this count as $C_i$.&lt;/li&gt;&#xA;&lt;li&gt;Choose the column $c$ so that $S_c \le S_j$ for any $j \ne c$. If there are two columns with the same minimum score, choose the column with the lower $C_c$.&lt;/li&gt;&#xA;&lt;li&gt;Add a marker into column $c$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Do this until all the rows have been generated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $(10,8,3)$ this generates the following table. Markers are numbered as they are added:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{|c|c|c|c|c|c|c|c|}&#xA;\hline&#xA;1&amp;amp;2&amp;amp;3&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;\\&#xA;\hline&#xA;&amp;amp;&amp;amp;&amp;amp;1&amp;amp;2&amp;amp;3&amp;amp;&amp;amp;\\&#xA;\hline&#xA;3&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;1&amp;amp;2\\&#xA;\hline&#xA;&amp;amp;1&amp;amp;&amp;amp;2&amp;amp;&amp;amp;&amp;amp;3&amp;amp;\\&#xA;\hline&#xA;&amp;amp;&amp;amp;1&amp;amp;&amp;amp;2&amp;amp;&amp;amp;&amp;amp;3\\&#xA;\hline&#xA;2&amp;amp;3&amp;amp;&amp;amp;&amp;amp;&amp;amp;1&amp;amp;&amp;amp;\\&#xA;\hline&#xA;&amp;amp;&amp;amp;1&amp;amp;2&amp;amp;3&amp;amp;&amp;amp;&amp;amp;\\&#xA;\hline&#xA;&amp;amp;&amp;amp;3&amp;amp;&amp;amp;&amp;amp;1&amp;amp;2&amp;amp;\\&#xA;\hline&#xA;3&amp;amp;2&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;1\\&#xA;\hline&#xA;&amp;amp;&amp;amp;&amp;amp;1&amp;amp;3&amp;amp;&amp;amp;&amp;amp;2\\&#xA;\hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A gist with a Python implementation can be found here: &lt;a href=&quot;https://gist.github.com/maafy6/d414a7a561116f784063&quot; rel=&quot;nofollow&quot;&gt;https://gist.github.com/maafy6/d414a7a561116f784063&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17446" LastEditorUserId="17446" LastEditDate="2016-02-04T18:38:43.050" LastActivityDate="2016-02-04T18:38:43.050" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26115" PostTypeId="2" ParentId="26107" CreationDate="2016-02-04T17:59:52.827" Score="5" Body="&lt;p&gt;Bit of a long shot, but I think the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Estimate&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To this riddle one shall add &quot;meta&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'est' or 'esti' translates as 'this' in some languages and 'mate' is an anagram of 'meta', so depending on whether or not we use the 'one' in the clue we have 'est'+'i'+'mate' or 'esti'+'mate'.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Like a program and an SE site&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; programme is like program in how it's pronounced and in some definitions but with an extra 'me'. Adding to 'an', we get 'mean' (this part may need more explanation). SE refers to Standard Error and a mean and standard error are part and parcel of an estimate. The word 'site' can also combine with 'meta' to make 'estimate', playing into the theme of a 'meta' riddle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the answer you have some data&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An estimate needs some input data. The riddle's first two lines also yield words useful for creating the 'estimate' anagram. Also, 'at' is some of data which we can combine with letters yielded from the last line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet not always you will get it right&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you estimate, you don't always get it right. Also, 'not always' could be a synonym for sometimes and the 'right' of sometimes could be 'etimes'. Combining this with the 'at' of line three we can rearrange to again get 'estimate'.    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18422" LastActivityDate="2016-02-04T17:59:52.827" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26116" PostTypeId="2" ParentId="26107" CreationDate="2016-02-04T18:10:52.357" Score="3" Body="&lt;p&gt;I'm going with &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;reference&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;To this riddle one shall add &quot;meta&quot;&#xA;Like a program and an SE site&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Adding meta to something makes it about itself. SE sites are all &lt;code&gt;*.SE.com&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;For the answer you have some data &#xA;Yet not always you will get it right&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Referencing data in a hypothesis is required - but doesn't make the hypothesis correct.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="51" LastActivityDate="2016-02-04T18:10:52.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26117" PostTypeId="2" ParentId="26107" CreationDate="2016-02-04T18:26:49.513" Score="7" Body="&lt;p&gt;Here's my guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Analysis&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Line 1: &lt;br&gt;&#xA;To this riddle one shall add &quot;meta&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can add &quot;meta&quot; to analysis to get &quot;meta-analysis&quot;, a type of statistical analysis.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Line 2: &lt;br&gt;&#xA;Like a program and an SE site&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  There's a program at meta-analysis.com for meta-analyses of data. There's also a &quot;meta&quot; site on SE and a number of SE sites that are devoted to data analysis.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Line 3: &lt;br&gt;For the answer you have some data&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You need data in order to do an analysis&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Line 4: &lt;br&gt;Yet not always you will get it right&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Analyses are not always right. In fact, many analyses use a p-value of .05, which (sort of) means that you'll be wrong about 5% of the time. (That's a simplification, but we'll roll with it for now.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2016-02-04T18:26:49.513" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26118" PostTypeId="1" AcceptedAnswerId="27953" CreationDate="2016-02-04T18:34:35.387" Score="7" ViewCount="5695" Body="&lt;p&gt;When checking if a 9x9 Sudoku solution is valid in the absence of an answer key it &lt;em&gt;may&lt;/em&gt; not be necessary to check every row, column and 3x3 box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the minimum number of items (rows, columns, or boxes) you must check if the &lt;em&gt;only&lt;/em&gt; information you get from a check is whether that specific item is valid?&lt;/p&gt;&#xA;" OwnerUserId="8984" LastEditorUserId="8984" LastEditDate="2017-05-23T22:46:55.367" LastActivityDate="2017-05-23T23:23:06.027" Title="Minimum steps to verify a Sudoku solution" Tags="|sudoku|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="26119" PostTypeId="1" CreationDate="2016-02-05T00:59:05.160" Score="20" ViewCount="1286" Body="&lt;p&gt;A strange man in a bow tie with a serious comb-over left you this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YCHWP.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YCHWP.jpg&quot; alt=&quot;Mysterious piece of paper...&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WJFjK.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WJFjK.jpg&quot; alt=&quot;Second half of mysterious piece of paper...&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;A simple cipher,&quot; he says, &quot;and I leave it to you to find out what has been lost.&quot;&#xA;Can you?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; On an OOC note, I have no idea how hard this would be…?  Also, sorry for the bad image quality. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11790" LastEditorUserId="11790" LastEditDate="2016-02-05T01:24:43.677" LastActivityDate="2023-03-22T17:17:41.597" Title="What have we lost?(a cipher of indeterminate value)" Tags="|cipher|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="26120" PostTypeId="2" ParentId="26094" CreationDate="2016-02-05T04:10:22.133" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There can be no consecutive numbers, so we could take the first odd numbers and add 3 to them in total, obtaining 28. Even then, two of the numbers would end up being 1 and 3. If $a + b + c + d + e = 28$, $c + d = 11$ and $d + e &amp;gt; a &amp;gt; b + c$, then $11-c + e &amp;gt; a &amp;gt; b + c$. The expression $11-c + e$ must be greater than 10, and since $e-c &amp;gt; 1$, it's at least 13. $a, c, d$ and $e$ &gt; 1, so $b=1$. $a, e$ and at least one of $c$ or $d$ are greater than 3, therefore the one left out equals 11-3=8. According to the &quot;no consecutives&quot; rule, $a+e$ can only be written as 10+6. Since $b + c$ must be less than 9...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $b=1, c=3, d=8$. $a$ and $e$ can either be 6 or 10.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastActivityDate="2016-02-05T04:10:22.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26121" PostTypeId="2" ParentId="26081" CreationDate="2016-02-05T05:52:42.493" Score="1" Body="&lt;p&gt;Okay my guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bizarro&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I was a young adult, I left something at level 2 because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0061885169&quot; rel=&quot;nofollow&quot;&gt;There is a young adult book that is reading level 2&lt;/a&gt; that is a retelling of the classic comic by the same name. Also, the original origin story was with Superboy not Superman&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I found something, the thing I found is like a magnet to some people.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I would either say this refers to the bonds that actually hold a body together (early versions of Bizarro tended to shatter or had his skin crumble after using too much power or due to exposure to sun/blue kryptonite) or the fact that most of the time he dies he ends up healing someone -twice a blind girl - which would reference love (magnet) or acceptance.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some say it belongs to the castle near the neck and the iron island.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the second of the #2s and the one that references the pictures most directly the new Bizarro comic - &lt;a href=&quot;http://www.dccomics.com/comics/bizarro-2015/bizarro-2&quot; rel=&quot;nofollow&quot;&gt;Bizarro #2&lt;/a&gt; - has Bizarro invade Iron Heights Prison&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This question is echoed in many of these stories of Bizarro in the tradition of Frankenstein's Monster - Am I man, monster, hero, villain, beast!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hints 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hints 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Several comics have Bizarro created by Lex using Superman's DNA. I'm assuming Batman connection is just based on latest movie but Bizarro #2 shows him dressed as Batman, Jimmy as Robin, and there have been several versions of Bizarro Batman...Batmen?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11239" LastActivityDate="2016-02-05T05:52:42.493" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26123" PostTypeId="1" AcceptedAnswerId="26174" CreationDate="2016-02-05T07:50:46.070" Score="16" ViewCount="4693" Body="&lt;p&gt;You wake up in a small locked room of a huge castle. There are 100 doors, numbers 1 to 100, and you are to open one of them. Open the correct door, and the entire castle is yours to keep. Open any of the other 99 doors, and a hungry monster devours you instantly. &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 100 guards, one in front of each door. You know that 50 of them are truth-tellers, and 50 are liars. You may ask them yes-no questions, but there's a catch: every time you ask a question, the guard will toss a biased coin, with a &lt;strong&gt;0.95 probability of landing on heads&lt;/strong&gt;. If the coin lands on tails, an alarm will be triggered and the monsters will break through the doors to devour you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that you use the optimal strategy, what is the probability that you will survive and win the entire castle?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Note: Constraints are as follows: &lt;br&gt;&#xA;  1. You can ask a guard only a boolean question, i.e. which has a 'Yes' or 'No' as the answer. &lt;br&gt;&#xA;  2. You can not run from the castle.(Not killing the optimistic approach)  &lt;br&gt;&#xA;  3. You can't ask a question, answer to which is not known to guards like &quot;Am I x years old?&quot; or &quot;Am I holding three fingers behind my back?&quot; They will, probably, kill you for wasting their time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The source of the problem : &lt;a href=&quot;https://brilliant.org/problems/99-doors-to-doom/&quot; rel=&quot;noreferrer&quot;&gt;Brilliant.org&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1766" LastEditorUserId="1766" LastEditDate="2016-02-16T12:48:59.177" LastActivityDate="2016-02-16T12:48:59.177" Title="99 doors to death!" Tags="|mathematics|probability|liars|" AnswerCount="6" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="26124" PostTypeId="1" AcceptedAnswerId="28859" CreationDate="2016-02-05T08:09:38.657" Score="5" ViewCount="632" Body="&lt;p&gt;In the world exists a lot of puzzles based on variations of true-tellers, liars and normals (sometimes lies, sometimes not). There are plenty of names for these characters, for example see the questions (or many others similar around the Internet):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/593/knights-and-jokers&quot;&gt;Knights and jokers&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/8926/island-inhabitants-who-lies-and-who-tells-the-truth?lq=1&quot;&gt;Island inhabitants: Who lies and who tells the truth?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/13141/nine-togglers-one-truth-teller-and-only-a-single-question?rq=1&quot;&gt;Nine togglers, one truth teller, and only a single question&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/23819/truths-lies-and-xors?rq=1&quot;&gt;Truths, Lies, and Xors&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the most &lt;strong&gt;correct&lt;/strong&gt; name for this puzzle and the characters?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the most &lt;strong&gt;common&lt;/strong&gt; name for this puzzle and the characters?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the most &lt;strong&gt;popular&lt;/strong&gt; name for this puzzle and the characters?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know, this question demands answers strongly based on opinion, but still I would like to know, what do you think and why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe a lot of scientific/popular literature about this logic puzzles exists, and maybe in this literature is some common/correct name suggested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the first historical appearance of these puzzles could be the clue.&lt;/p&gt;&#xA;" OwnerUserId="18956" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2016-03-23T16:57:46.777" Title="What is the best name for true-tellers, liars and normals" Tags="|logical-deduction|liars|puzzle-history|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="26125" PostTypeId="2" ParentId="26119" CreationDate="2016-02-05T08:20:44.827" Score="5" Body="&lt;p&gt;Out of curiosity, I tried to extract the text from the back of the papers, in case there's a clue.  The pictures are pretty noisy (probably a cell phone camera) so it's difficult, here's the best I could do:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vp8Uz.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vp8Uz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/9X1kz.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9X1kz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears to be a letter of some sort.  I can read the date (January or February 28, 2016) on the top of the first page; I can see a fragment &quot;-ing our&quot; at the end of the third or fourth line on the same page; and I can see a signature in the middle of the second page.&lt;/p&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="9718" LastEditDate="2016-02-05T08:27:40.210" LastActivityDate="2016-02-05T08:27:40.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26126" PostTypeId="2" ParentId="26123" CreationDate="2016-02-05T08:24:22.713" Score="5" Body="&lt;p&gt;I love the truth-teller - liar problems and first time attempting to answer on this site. Correct me if I am on the wrong path :)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First of all we must find a question X, which has to have 50% yes, 50% no answer chance which we know. &quot;Are you a real guard? Yes-No?&quot; Answer has &lt;strong&gt;2&lt;/strong&gt; possibilities: Yes or No, So we will know if he is telling the truth or not. If he is telling the truth, with the power of n, we can find the number of questions to find the correct door: 2^n. There are 100 possibilites so n must be greater that 7 for a guaranteed correct answer. So we must ask 6 more questions to the guard to find the correct door. 2^7 = 128 but 2^6 = 64, which is not enough for an optimal survival strategy so I pick 7. Now we have got the 0.95 probability of being correct because of the coin toss even if all answers are correct. For survival, it must be 7 times in a row heads, so my formula goes to (0.95)^7 = &lt;strong&gt;0.69833729&lt;/strong&gt;.... best probability of survival in my opinion&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18685" LastActivityDate="2016-02-05T08:24:22.713" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26127" PostTypeId="2" ParentId="26123" CreationDate="2016-02-05T08:40:11.287" Score="19" Body="&lt;p&gt;By binary search, you can narrow it down in just &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\lceil\log_2100\rceil$, or 7, questions.&#xA;  &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can't do better since every question gives you one bit of information, and asking a question to decide between two sets of doors is always better than guessing between the two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question you ask is as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Would you answer &quot;yes&quot; if I were to ask you if the right door is in the list [first half of remaining possible doors]?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It doesn't matter which guard you ask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the probability of escape is :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $.95^7$, or about $69.83\%$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit: The above time is wrong (&lt;em&gt;thanks for pointing it out, Matt and durum!&lt;/em&gt;); I assumed the worst case scenario, which I should not have done with probability. The real chance with this strategy is better, about $70.8\%$, &lt;s&gt;but it &lt;em&gt;could&lt;/em&gt; be improved by another strategy that doesn't cut them evenly.&lt;/s&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 2: f'' has proven that this solution is optimal.&lt;/p&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="11876" LastEditDate="2016-02-06T05:33:31.723" LastActivityDate="2016-02-06T05:33:31.723" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="26132" PostTypeId="2" ParentId="26055" CreationDate="2016-02-05T12:36:19.417" Score="2" Body="&lt;p&gt;My guess is that you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Ace of Spades&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A monarch in democracy,&#xA;yet revered without hypocrisy,&#xA;my name shall live on evermore.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The top valued card in the pack of cards.  Most famous of all the cards.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Under the same reign as the sign of war,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The suit spades is the aggressive suit traditionally, similar to the sign of Ares &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and thirty-one's twenty-four,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; both card games, the game 24 includes the ace but is played without any of the face cards.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the first to seize a third at three.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure of this reference, possibly a reference to 9 o'clock, 12 o'clock, 3 o'clock dealing or placing of tarrot cards?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Totaling thirteen and a little more,&#xA;I became the one to do it, finally;&#xA;I'm now a monarch doubly.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Aces can be high and low, big card or one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18966" LastActivityDate="2016-02-05T12:36:19.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26133" PostTypeId="1" AcceptedAnswerId="26142" CreationDate="2016-02-05T13:25:35.840" Score="8" ViewCount="2025" Body="&lt;p&gt;My first riddle, probably very easy but good luck!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An affiliation with dogs&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My culture is like no other&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Programmers use me lots&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am with my two brothers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HINT:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The dogs are rubber&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18940" LastEditorUserId="18940" LastEditDate="2016-02-05T19:23:02.223" LastActivityDate="2016-02-05T20:59:09.093" Title="The Weasel's Riddle" Tags="|riddle|word|" AnswerCount="6" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26134" PostTypeId="2" ParentId="26133" CreationDate="2016-02-05T13:34:23.633" Score="6" Body="&lt;p&gt;Perhaps this is a wild guess, but I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dot&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An affiliation with dogs&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dot = Spot, a common name for dogs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My culture is like no other&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is regarded as good culture, proper to end sentences with a dot.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Programmers use me lots&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dots are commonly used in many programming languages.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am with my two brothers&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Probably dot-dot-dot ... or colon and semicolon ; :&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18967" LastActivityDate="2016-02-05T13:34:23.633" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26135" PostTypeId="2" ParentId="26133" CreationDate="2016-02-05T13:43:17.603" Score="6" Body="&lt;p&gt;Is the answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ducks&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An affiliation with dogs&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Donald Duck is affiliated with Goofy, Pluto, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My culture is like no other&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As a cartoon character, they have a very interesting culture indeed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Programmers use me lots&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rubber duck debugging&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am with my two brothers&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Refers to the Disney Brothers (Walt and Roy)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17147" LastEditorUserId="17147" LastEditDate="2016-02-05T14:45:09.270" LastActivityDate="2016-02-05T14:45:09.270" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26136" PostTypeId="2" ParentId="26081" CreationDate="2016-02-05T13:54:21.277" Score="1" Body="&lt;p&gt;Ultra wild guess here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jesse W. Reno - the inventor of the escalator.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So let's start from the hints&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The character in the second picture is Lex Luthor, who will be portrayed by &lt;strong&gt;Jesse&lt;/strong&gt; Eisenberg in the upcoming movie &quot;Batman v Superman: Dawn of Justice&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And now the riddle:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;When I was a young adult, I left something at level 2 because&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He was 29-30 years old when he created the escalator. &quot;Leaving at level two&quot; refers to the function of the escalator.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I found something, the thing I found is like a magnet to some people.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People are lazy and they prefer the escalator to the stairs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Some say it belongs to the castle near the neck and the iron island.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The very first installation was at Old &lt;strong&gt;Iron&lt;/strong&gt; Pier, Coney &lt;strong&gt;Island&lt;/strong&gt;, New York City. The castle part refers to the fact that the Old Iron Pier looked like a small castle - &lt;a href=&quot;https://i.stack.imgur.com/FnUHG.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FnUHG.jpg&quot; alt=&quot;Old Iron Pier&quot;&gt;&lt;/a&gt; and &quot;near the neck&quot; refers to the Hudson river (&quot;neck&quot; means straight/channel).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jesse W. Reno&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastActivityDate="2016-02-05T13:54:21.277" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26137" PostTypeId="1" AcceptedAnswerId="28746" CreationDate="2016-02-05T14:17:41.350" Score="8" ViewCount="227" Body="&lt;p&gt;Every letter stands for a digit in decimal representation, &#xA;different letters stand for different digits:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;             SUN&#xA;     +      LOSE&#xA;     +     UNTIE&#xA;     +    BOTTLE&#xA;     +   ELISION&#xA;     +  NINETEEN&#xA;     + NONENTITY&#xA;     + EBULLIENT&#xA;     + INSOLUBLE&#xA;  ------------------&#xA;      NEBULOSITY&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which digit does each letter represent?&lt;br&gt;&#xA;(Please present the full analysis how these digits can be determined.)&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="9404" LastEditDate="2016-03-03T10:19:46.667" LastActivityDate="2016-03-10T14:59:35.763" Title="Nebulosity alphametic" Tags="|no-computers|alphametic|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26139" PostTypeId="1" AcceptedAnswerId="26155" CreationDate="2016-02-05T14:23:32.207" Score="10" ViewCount="551" Body="&lt;p&gt;My Indian friend Jamal recently participated in the well-known TV quiz&#xA;show &lt;strong&gt;&lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Who_Wants_to_Be_a_Millionaire&quot;&gt;Who wants to be a millionaire?&lt;/a&gt;&lt;/em&gt;&lt;/strong&gt;. You might have heard or read about Jamal's spectacular success; the media reported on it extensively.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what Jamal told us about the 250.000 Euro question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;During the next commercial break, Prem Kumar moved his face really &#xA;  close to mine. In a low voice, he made an offending remark that I'm &#xA;  not going to repeat here.  Strangely enough, when the show resumed, &#xA;  he beamed at me with a gaze of pure paternal pride, and even chuckled &#xA;  at my first remark.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The 250.000 Euro question was a catastrophe, and I did not know any &#xA;  of the three terms mentioned in it, strange and peculiar sounding &#xA;  terms indeed.  I decided to use the answers to get some better insight. &#xA;  At this level of the show, every wrong answer actually fits part of &#xA;  the question, and the trick is to identify the answers that do not &#xA;  fit more than that.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Answer A almost fit the first strange term in the question, but I &#xA;  knew that this was a Japanese product, and that it should have been &#xA;  spelled with a double-S.  Answer B was obvious nonsense and a pure &#xA;  give-away for Indian citizens, anyway.  At least the second strange &#xA;  term in the question was neither Braman nor Visnun.  Nothing in &#xA;  answer C rang a bell.  Answer D again touched my area of expertise, &#xA;  and I immediately knew that it had to be wrong: the third strange &#xA;  term in the question had a letter R in the end that did not belong &#xA;  there, similarly to Delphir and Basicr.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;As I had successfully excluded A, B, and D, the only good answer &#xA;  was C. And so I said to Prem: &quot;This time I will go for C!&quot;. Nailed it!!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;What was the 250.000 Euro question?&lt;br&gt;&#xA;What were the four possible answers A, B, C, D?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2016-02-05T18:23:45.070" Title="Slumdog millionaire (13)" Tags="|knowledge|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26140" PostTypeId="2" ParentId="26123" CreationDate="2016-02-05T14:27:56.083" Score="0" Body="&lt;p&gt;Ask a boolean question to any guard to ascertain if they are truth teller or liar.&#xA;Then: &#xA;if they are truth teller, ask which door,&#xA;if they are liar, ask them to tell you a door which contains a monster.&lt;/p&gt;&#xA;" OwnerUserId="18970" LastActivityDate="2016-02-05T14:27:56.083" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26141" PostTypeId="2" ParentId="26081" CreationDate="2016-02-05T14:34:58.153" Score="1" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mark Zuckerberg&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I was a young adult, I left something at level 2 because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He left the college at second year.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I found something, the thing I found is like a magnet to some people.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He is the founder of facebook. Many people are highly addicted to it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some say it belongs to the castle near the neck and the iron island.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The castle near the neck and the iron island is 'The Twin'. The twin brothers claimed that the idea of facebook belonged to them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Use the map.&lt;/strong&gt;: Use the map of Game of Thrones which helps in finding the name of the castle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;The answer has some relationship with him. Which person does these pictures represents? picture of [Batman v/s Superman + Lex Luthor]&lt;/strong&gt;: As dmg correctly guessed, the character of Lex Luthor is played by Jesse Eisenberg who also played the character of Mark Zuckerberg in the movie 'The Social Network'. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If this question remain unsolved until 6-Feb then I'll post the answer myself. And I am sure you'll say &quot;Oh f*k, that was easy.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sorry for not giving it as hint but it was also a little hint. Here &lt;strong&gt;f*k = facebook&lt;/strong&gt; And also sorry for revealing the answer a day earlier but I don't like puzzles to remain unsolved. One thing I would like to state is that today facebook has completed 12 years. :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-02-05T14:52:13.143" LastActivityDate="2016-02-05T14:52:13.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26142" PostTypeId="2" ParentId="26133" CreationDate="2016-02-05T14:38:03.140" Score="11" Body="&lt;p&gt;I answered earlier, but I think I have a new idea... Inspired by the title actually.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pop. The Weasel's riddle - Pop goes the weasel. Some nursery rhyme i think.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My culture is like no other&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pop culture.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Programmers use me lots&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Functions pop_front or pop_back.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;----EDITED----&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am with my two brothers&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reference to snap, crackle and pop, three brothers from a cereal brand. Suggested by Irishpanda, Kevin, and David Starkey, thanks!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An affiliation with dogs&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Balloon animals can pop. Balloon animals are often made into dogs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Other suggestions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; popCORN -&gt; CORNdogs suggested by hexomino sounds reasonable. pup being close to pop suggested by kaine, but an earlier comment that the dogs aren't real might cross that off.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18972" LastEditorUserId="18972" LastEditDate="2016-02-05T19:51:07.520" LastActivityDate="2016-02-05T19:51:07.520" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="26143" PostTypeId="2" ParentId="26137" CreationDate="2016-02-05T14:55:20.743" Score="5" Body="&lt;p&gt;After a bunch of scribbling and start-all-overs:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Y = 0, N = 1, E = 2, B = 3, U = 4, L = 5, O = 6, S = 7, I = 8, T = 9&#xA; &#xA; &lt;pre&gt;&#xA;              741&#xA;      +      5672&#xA;      +     41982&#xA;      +    369952&#xA;      +   2587861&#xA;      +  18129221&#xA;      + 161219890&#xA;      + 234558219&#xA;      + 817654352&#xA;   ------------------&#xA;       1234567890&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="9000" LastEditDate="2016-02-05T15:37:16.740" LastActivityDate="2016-02-05T15:37:16.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26144" PostTypeId="1" AcceptedAnswerId="26145" CreationDate="2016-02-05T14:57:14.320" Score="1" ViewCount="1188" Body="&lt;p&gt;Every month the price of a particular commodity falls in this order:   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;1024, 640, 400, 250, $\ldots$&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the next value?&lt;br&gt;&#xA;Please help me to find pattern of this sequence.&lt;/p&gt;&#xA;" OwnerUserId="18892" LastEditorUserId="8874" LastEditDate="2016-03-19T17:10:16.070" LastActivityDate="2016-03-19T17:10:16.070" Title="Next value in a falling sequence" Tags="|calculation-puzzle|number-sequence|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26145" PostTypeId="2" ParentId="26144" CreationDate="2016-02-05T15:00:36.133" Score="6" Body="&lt;p&gt;Each subsequent number is $\frac58$ of the previous one. &lt;/p&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2016-02-05T15:00:36.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26146" PostTypeId="2" ParentId="26144" CreationDate="2016-02-05T15:01:43.217" Score="7" Body="&lt;p&gt;According to the On-line encyclopedia of integer sequences&lt;br /&gt;&#xA;&lt;a href=&quot;https://oeis.org/A084912&quot; rel=&quot;noreferrer&quot;&gt;https://oeis.org/A084912&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A084912:&lt;span class=&quot;math-container&quot;&gt;$~~$&lt;/span&gt;  a(1) = 1024;&lt;span class=&quot;math-container&quot;&gt;$~~$&lt;/span&gt; a(n) = 5*a(n-1)/8.&lt;/p&gt;&#xA;&lt;p&gt;Puzzle: a man offered to sell me a car two years ago for \$1024; a year later his price was \$640; a little while after that he asked \$400; and last week he was willing to sell for \$250. The next time he reduces I shall buy. At what price will I buy the car if he makes consistent reductions?&lt;/p&gt;&#xA;&lt;p&gt;Answer: a(5) = 1250/8 = \$156.25 (not an integer).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-02-05T15:06:50.907" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26147" PostTypeId="1" AcceptedAnswerId="26148" CreationDate="2016-02-05T15:23:35.990" Score="20" ViewCount="2918" Body="&lt;pre&gt;&lt;code&gt;It's late in the evening.&#xA;I advance along a street with a shoe in my hand. &#xA;I cannot proceed, I have to stop.&#xA;Next to me stands a hotel.&#xA;&#xA;&quot;Darn, I am seriously screwed&quot; I exclaim.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What has happened to me?&lt;/p&gt;&#xA;" OwnerUserId="10522" LastEditorUserId="8874" LastEditDate="2016-02-05T15:50:22.243" LastActivityDate="2016-04-13T20:16:57.730" Title="What has happened to me?" Tags="|riddle|lateral-thinking|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26148" PostTypeId="2" ParentId="26147" CreationDate="2016-02-05T15:26:07.323" Score="39" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Playing Monopoly with the shoe trinket. A hotel means you have to pay the property owner a large sum of money.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17147" LastActivityDate="2016-02-05T15:26:07.323" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="26149" PostTypeId="2" ParentId="26133" CreationDate="2016-02-05T16:04:20.707" Score="2" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Counter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An affiliation with dogs&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Counter surfing is when a dog steals food from the counter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My culture is like no other&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Counter culture is a culture that is very different from mainstream society&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Programmers use me lots&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To keep track of iterations&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am with my two brothers&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Posibly variable declaration and termination condition in C-style loops?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18979" LastActivityDate="2016-02-05T16:04:20.707" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26150" PostTypeId="1" AcceptedAnswerId="26152" CreationDate="2016-02-05T16:05:21.420" Score="7" ViewCount="178" Body="&lt;p&gt;I made this puzzle up in 1999 when I was a student.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fort Alpha, on your border territories, was attacked by The Enemy. You don't know if it fell, or repulsed The Enemy's vicious onslaught. A totally trusted advisor, with a detailed knowledge of the military situation, gave counsel before the attack that the chance of a successful defence was 50%. Unfortunately, there are no rapid communication systems; news of the war relies completely on messengers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three of your messengers were given a true report of the battle and set out from Fort Alpha immediately. They travelled separately to forts Beta, Gamma and Delta, which are still under your command. Before collapsing dramatically with exhaustion, these three messengers then relayed the message to three more messengers each, so a total of nine fresh messengers then set out to convey the message to your headquarters. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the 12 messengers were hand-picked from the ultra-loyal elite of your troops, so it is very very unlikely, though not impossible, that any given one of them is loyal to The Enemy. Your totally trusted advisor says that the chance of a given messenger being a traitor is 1%. Any messenger who was a traitor would reverse the message relayed to them, and pass on the opposite message, so as to befuddle you into inaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of the messengers spoke to one another except to relay the message. The Enemy's spies, if there are any, maintain total secrecy, so none of the messengers know whether any of the others is a traitor and they each assume each other to be loyal to you. Only traitors would lie. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The three messengers who arrived from Beta, one who arrived from Gamma, and one who arrived from Delta, recount that Fort Alpha fell tragically to The Enemy. The other four claim that it bravely withstood the attack and decimated The Enemy's marauding rabble. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which story do you believe, and why? &lt;/p&gt;&#xA;" OwnerUserId="18980" LastActivityDate="2016-02-05T16:58:48.853" Title="The Attack on Fort Alpha" Tags="|probability|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="26151" PostTypeId="2" ParentId="26147" CreationDate="2016-02-05T16:09:35.907" Score="7" Body="&lt;p&gt;Super unlikely to be it, however:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are a female who was wearing 5&quot; high stiletto heels but one of them broke, so you took it off so you could walk properly. As you see the hotel you realize that you can't keep walking home because your feet are too sore but you can't afford to stay in the hotel for the night ;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13136" LastActivityDate="2016-02-05T16:09:35.907" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26152" PostTypeId="2" ParentId="26150" CreationDate="2016-02-05T16:20:02.693" Score="4" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Good new! Alpha bravely withstood the enemy attack.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In order for the story that Alpha fell to be true, a minimum of four traitors would be required.  In order for the story that Alpha survived to be true, you would need only three traitors. So given those two scenarios, Alpha surviving is 99 times more likely than Alpha having fallen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To clarify:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For the Alpha to have survived, it would mean that the messenger who went from Alpha to Beta was a traitor, one from Gamma to headquarters was a traitor and one from Delta to headquarters was a traitor. That's three. For Alpha to have fallen, there would be no traitors in the Beta Group, but for both the Delta and Gamma Groups, either the first messenger and one of the second messengers, or two of the second messengers would have to be traitors.  That's four total.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9353" LastEditorUserId="9353" LastEditDate="2016-02-05T16:58:48.853" LastActivityDate="2016-02-05T16:58:48.853" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26155" PostTypeId="2" ParentId="26139" CreationDate="2016-02-05T16:46:07.557" Score="11" Body="&lt;p&gt;&lt;strong&gt;New answer with much credit going to Alexis for turning my first answer on its head&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strange terms are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nisan, Sivan, and Adar as figured in my first answer below. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The question becomes &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;What are Nisan, Sivan, and Adar?&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answers are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A: American cars (question wording implies that the answer refers to cars that are not Japanese; I think any other kind of car works here), B: Hindu gods, C: months in the Hebrew calendar, and D: Programming languages&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Original answer, contains reasoning that I didn't repeat above&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answers A is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nisan (almost Nissan, a Japanese car)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;B is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sivan (take &quot;Shiva&quot;, remove the h and add an n to the end to get &quot;Sivan&quot;. The same transformation on &quot;Brahma&quot; nets &quot;Braman&quot;,and the transformation on &quot;Vishnu&quot; nets Visnun. Brahma, Shiva and Vishnu are the names of Hindu gods.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;D is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Adar. Delphi and Basic are programming languages, add R to get Delphir and Basicr. Ada is another programming language.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But the answers are all actually&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Months in the Hebrew calendar.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Determining the question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Adar, Nisan, and Sivan all contain Jewish festivals from the Bible.  Tishrei is the only other month that has a Biblical festival, it actually has 3 (or 4 depending on how you look at it). So I conclude that...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;TL,DR:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The question is &quot;what month of the Hebrew calendar contains the festivals of Rosh Hashanah, Yom Kippur, and Sukkot?&quot; and answers are A: Nisan, B: Sivan, C: Tishrei, D: Adar.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My only uncertainty with this answer is that it is unclear how Jamal would think the answers fit the strange terms.&lt;/p&gt;&#xA;" OwnerUserId="18493" LastEditorUserId="18493" LastEditDate="2016-02-05T18:23:45.070" LastActivityDate="2016-02-05T18:23:45.070" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26156" PostTypeId="2" ParentId="26133" CreationDate="2016-02-05T17:23:32.013" Score="3" Body="&lt;p&gt;I couldn't quite get the &quot;My culture is like no other&quot; part, but otherwise:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tails&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An affiliation with dogs&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dogs have tails&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Programmers use me lots&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tail recursion&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I am with my two brothers&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sonic (the Hedgehog), Knuckles and Tails&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18982" LastActivityDate="2016-02-05T17:23:32.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26157" PostTypeId="2" ParentId="25439" CreationDate="2016-02-05T17:43:35.680" Score="12" Body="&lt;p&gt;We can show that a pilgrimage is always possible by employing the &lt;a href=&quot;https://en.wikipedia.org/wiki/Pigeonhole_principle&quot; rel=&quot;nofollow&quot;&gt;pigeonhole principle&lt;/a&gt;.  Let there be 10 bins, one for each city, and iterate over the highways in order.  In iteration $i$, each bin will contain the longest &lt;a href=&quot;https://en.wikipedia.org/wiki/Trail_(graph_theory)&quot; rel=&quot;nofollow&quot;&gt;trail&lt;/a&gt; ending in that city using only highways up to and including $i$.  (If there is more than one such trail, pick an arbitrary one.)  Initially, in iteration 0, each bin contains only the city itself, for a trail of length 0.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Claim:&lt;/strong&gt; During each iteration, the sum of the trail lengths over all bins increases by at least 2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;p&gt;Assume that during iteration $i$, we add highway $X-Y$ and consider the longest trails ending in cities $X$ and $Y$ after iteration $i-1$. We consider two cases:&lt;/P&gt;&#xA; &lt;p&gt;(1) Both trails had the same length $n$.  In this case, we can simply append highway $X-Y$ to both trails for a new longest trail, so each will now have length $n+1$.  The sum of lengths increases by 2 and the claim holds.&lt;/p&gt;&#xA; &lt;p&gt;(2) The lengths of the trails were unequal; without loss of generality assume that trail $t_1$ ending in city $X$ had length $n$ and trail $t_2$ ending in $Y$ had length $m \leq n-1$.  In this case, after adding highway $X-Y$, $t_1$ is still the (possibly joint) longest trail ending in city $X$; however, the longest trail ending in city $Y$ is now $t_1$ with highway $X-Y$ appended, for a length of $n+1 \geq m+2$.  Again the claim holds.&lt;/p&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This means that after all 45 highways have been added, the sum over all 10 bins is at least 90.  The pigeonhole principle tells us that at least one trail must have length 9 and fulfils the requirements for a proper pilgrimage.  In fact this already holds for 41 highways, so even if Drump throws up four roadblocks he won't be successful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Running through the example presented in the puzzle, with cities named &lt;code&gt;a&lt;/code&gt; through &lt;code&gt;j&lt;/code&gt; in clockwise fashion starting from the topmost city, here's the situation after iterating over the first five highways (highway 05, just added, caused the sum to increase by 3):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;a(l=1) : e-01-a&#xA;b(l=0) : b&#xA;c(l=1) : j-03-c&#xA;d(l=0) : d&#xA;e(l=1) : a-01-e&#xA;f(l=3) : e-01-a-02-i-05-f&#xA;g(l=0) : g&#xA;h(l=2) : j-03-c-04-h&#xA;i(l=2) : e-01-a-02-i&#xA;j(l=1) : c-03-j&#xA;sum = 11 (+3)&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And here's the eventual list of trails.  All of them start by walking the same seven highways &lt;code&gt;e-01-a-02-i-08-e-10-h-11-f-14-d-15-c&lt;/code&gt;, abbreviated as &lt;code&gt;e-...-c&lt;/code&gt; below.  As you see, in this example highway network, die-hard pilgrims could increase the length of their pilgrimage by more than half if they wanted.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;a(l=12): e-...-c-19-f-26-g-29-j-35-d-44-a&#xA;b(l=14): e-...-c-19-f-26-g-29-j-32-a-41-c-42-e-43-b&#xA;c(l=12): e-...-c-19-f-26-g-29-j-32-a-41-c&#xA;d(l=12): e-...-c-19-f-26-g-29-j-32-a-44-d&#xA;e(l=14): e-...-c-19-f-26-g-29-j-32-a-34-f-39-b-43-e&#xA;f(l=12): e-...-c-19-f-26-g-29-j-32-a-34-f&#xA;g(l=12): e-...-c-19-f-26-g-29-j-32-a-38-g&#xA;h(l=14): e-...-c-19-f-26-g-29-j-32-a-34-f-40-j-45-h&#xA;i(l=9) : e-...-c-23-g-25-i&#xA;j(l=13): e-...-c-19-f-26-g-29-j-32-a-34-f-40-j&#xA;sum = 124 (+5)&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;At first I tried to construct the example by hand, but it turned out to be easier and less error-prone to write a simple program. The hand-constructed example up to highway 05 then served nicely to validate it.  Here's the Python program if you want to experiment further:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;example = [&#xA;    'ae', 'ai', 'cj', 'ch', 'fi', 'dg', 'eg', 'ei', 'bj',&#xA;    'eh', 'fh', 'gh', 'di', 'df', 'cd', 'ij', 'ah', 'bg',&#xA;    'cf', 'de', 'bi', 'ab', 'cg', 'hi', 'gi', 'fg', 'ej',&#xA;    'ef', 'gj', 'bd', 'bc', 'aj', 'dh', 'af', 'dj', 'ci',&#xA;    'bh', 'ag', 'bf', 'fj', 'ac', 'ce', 'be', 'ad', 'hj',&#xA;]&#xA;&#xA;trails = dict()&#xA;for v in 'abcdefghij': trails[v] = v&#xA;&#xA;def output(last):&#xA;    sum=0&#xA;    for v in 'abcdefghij':&#xA;        t = trails[v]&#xA;        l = (len(t)-1)/5&#xA;        sum += l&#xA;        print '%s(l=%d)%s: %s' % (&#xA;                v, l, '' if l&amp;gt;9 else ' ', t)&#xA;    print 'sum = %d (+%d)\n' % (sum, sum-last)&#xA;    return sum&#xA;&#xA;sum = output(0)&#xA;&#xA;for i in range(len(example)):&#xA;    v,w = tuple(example[i])&#xA;    t1,t2 = trails[v], trails[w]&#xA;    t1 += '-%02d-%s' % (i+1, w)&#xA;    t2 += '-%02d-%s' % (i+1, v)&#xA;    if len(t1) &amp;gt; len(trails[w]):&#xA;        trails[w] = t1&#xA;    if len(t2) &amp;gt; len(trails[v]):&#xA;        trails[v] = t2&#xA;    sum = output(sum)&#xA;&lt;/pre&gt;&#xA;" OwnerUserId="13484" LastEditorUserId="13484" LastEditDate="2016-02-07T12:06:56.943" LastActivityDate="2016-02-07T12:06:56.943" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26158" PostTypeId="2" ParentId="26147" CreationDate="2016-02-05T18:00:31.527" Score="10" Body="&lt;p&gt;Also unlikely, but worth a shot:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You have illegally parked in front of the hotel, and the police have placed a Denver Boot on the wheel.  You have your hand on the Boot trying to get it off, but you know it will not work.  Since it is night time, it may be difficult to get someone to come out to remove it until morning, so you are stuck where you are.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16131" LastActivityDate="2016-02-05T18:00:31.527" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26159" PostTypeId="1" AcceptedAnswerId="26163" CreationDate="2016-02-05T18:05:31.837" Score="8" ViewCount="435" Body="&lt;p&gt;Is it possible to fill a rectangular board with 12 squares completely with the numbers 1 to 12, in such a way that there exists a square on the board, where if we start there and traverse the board according to the value of the number in the square we're in and below diagram (as in the puzzle &lt;a href=&quot;https://puzzling.stackexchange.com/questions/11913/the-sands-of-time&quot;&gt;sands of time&lt;/a&gt;) that we will visit all 12 squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UdURJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UdURJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a much simpler example, in a 2x2 square we could place the numbers&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;3  6&#xA;12 9&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And we could start on 3, move according the diagram above to the 6, move down to 9, then left to the 12, visiting each square.&lt;/p&gt;&#xA;" OwnerUserId="16846" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2016-02-05T20:11:30.483" Title="Times square, ehm rectangle" Tags="|combinatorics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26160" PostTypeId="2" ParentId="26133" CreationDate="2016-02-05T18:25:39.747" Score="1" Body="&lt;p&gt;For the last line: ?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the alphabet, O's brother would be P.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18984" LastActivityDate="2016-02-05T18:25:39.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26161" PostTypeId="2" ParentId="26159" CreationDate="2016-02-05T19:23:21.517" Score="8" Body="&lt;p&gt;Sure, it's possible.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;pre&gt;&#xA; 3   7   5   6&#xA; 4   9   12  8&#xA; 2   11  1   10&#xA; &lt;/pre&gt;&#xA;&#xA; Start anywhere you like.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1764" LastActivityDate="2016-02-05T19:23:21.517" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26162" PostTypeId="2" ParentId="26123" CreationDate="2016-02-05T19:34:40.090" Score="8" Body="&lt;p&gt;Actually, I think the accepted answer is wrong and your chance of survival is actually slightly better at:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 70.2984%&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;using the following strategy&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Divide the list into 3 parts, one with 64 numbers, one with 32 number and one with 4 numbers.  Ask any guard &quot;If I were to ask you if the right door is in this list of 62 numbers, what would you say?&quot;  If he says yes, you discard the other two lists and proceed to cut the list in half every time.  If he says no, you throw away the list of 64 and proceed to the list of 32. If it's on that list, you proceed with cutting it in half every time and discard the list of 4.  If not, you discard it and follow the same sequence with the list of four.  If it is on one of the first two lists (96% chance), you have a 69.8337% chance of surviving.  If it is on the third list (4% chance), you have a 81.4506% of surviving.  (.04 * 81.4506) + (.96 * 69.8337) = 70.2984% chance of survival.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9353" LastEditorUserId="9353" LastEditDate="2016-02-05T20:04:21.053" LastActivityDate="2016-02-05T20:04:21.053" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26163" PostTypeId="2" ParentId="26159" CreationDate="2016-02-05T19:41:15.453" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;pre&gt; 6  4  5  7&lt;br&gt;&#xA; 2  3  8 12&lt;br&gt;&#xA; 1 11  9 10&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is one of 12 solutions, but there are fundamentally only two.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The one given above traces a path which has horizontal symmetry; it can be flipped vertically and traced in either direction, giving four equivalent solutions. The other was posted by hvd; it has no symmetries, meaning that it can be flipped horizontally and/or vertically and traced in either direction, giving eight equivalent solutions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The paths traced by the two fundamental solutions are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/6Zs5I.png&quot; alt=&quot;Graphs showing the two paths&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="228" LastEditorUserId="228" LastEditDate="2016-02-05T20:11:30.483" LastActivityDate="2016-02-05T20:11:30.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26166" PostTypeId="1" AcceptedAnswerId="26186" CreationDate="2016-02-05T22:00:43.227" Score="6" ViewCount="424" Body="&lt;p&gt;I got asked this question on a job interview, I am not sure if this is a right place to ask but here it is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Abraham Lincoln were to give an address today, what would he say about us?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&#xA;63, 103, 2, 18, 10, 29, 2, 1, 10, 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am really curious about this, can anyone help?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am pretty new to deciphering, but I have tried ASCII, mod 26, spelled-out letters, cellphone numpad, and some other common methods to solve this, but all didn't lead to a make sense answer.&lt;/p&gt;&#xA;" OwnerUserId="18995" LastEditorUserId="30903" LastEditDate="2017-02-10T04:22:43.037" LastActivityDate="2017-02-10T04:22:43.037" Title="What would Lincoln say?" Tags="|cipher|computer-puzzle|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="26167" PostTypeId="1" AcceptedAnswerId="27493" CreationDate="2016-02-05T22:26:10.497" Score="8" ViewCount="1078" Body="&lt;p&gt;Some call me royalty, but I say they're wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some call me crazy, but I say they're wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some call me a ghost, but I say they're wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some call me racist, but I say they're wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some call me a hat, but I say they're wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am a word&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer involves rhymes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18103" LastEditorUserId="18103" LastEditDate="2016-02-09T18:43:41.307" LastActivityDate="2016-02-15T17:14:35.753" Title="They don't know me at all" Tags="|riddle|word|wordplay|" AnswerCount="9" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="26168" PostTypeId="1" AcceptedAnswerId="26225" CreationDate="2016-02-05T22:56:34.980" Score="4" ViewCount="798" Body="&lt;p&gt;A little riddle I conceived on vacation last winter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am spread out like a leaf.&lt;br&gt;&#xA;That what forms me cuts me, but builds me up.&lt;br&gt;&#xA;That what cuts me seeks always to go lower, but lifts its bed higher.&lt;br&gt;&#xA;I am commonly found where what builds me is less common.&lt;br&gt;&#xA;What am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: see title&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In answer to ermanen, I've decided to post one more clue: &lt;A HREF=&quot;http://medicinezine.com/files/2010/01/gingko_biloba.jpg&quot; rel=&quot;nofollow&quot;&gt;leaf&lt;/A&gt;&lt;/p&gt;&#xA;" OwnerUserId="5623" LastEditorUserId="5623" LastEditDate="2016-02-10T21:03:11.483" LastActivityDate="2016-02-12T16:49:27.583" Title="A riddle from the ages" Tags="|riddle|" AnswerCount="7" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26170" PostTypeId="2" ParentId="26168" CreationDate="2016-02-05T23:10:47.953" Score="8" Body="&lt;p&gt;Second attempt (first attempt below):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; River delta&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Line 1:&lt;br&gt;I am spread out like a leaf.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The water parts of a delta look similar to the veins of a leaf.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Line 2:&lt;br&gt;That what forms me cuts me, but builds me up.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water erodes a delta, but also builds it up by depositing silt.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Line 3: &lt;br&gt;That what cuts me seeks always to go lower, but lifts its bed higher.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water always flows downhill, but the &quot;bed&quot; of a river or river delta is built up by the sediment it carries.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Line 4: &lt;br&gt;I am commonly found where what builds me is less common.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The river delta is formed by fresh water, but the delta is at the ocean, where salt water reigns.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;First attempt:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; River OR riverbed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Line 1:&lt;br&gt;I am spread out like a leaf.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A leaf has &quot;veins&quot; that transport liquid, just like tributaries and smaller streams lead into the main river. If you looked at a system of rivers/riverbeds, it would look like leaf veins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Line 2:&lt;br&gt;That what forms me cuts me, but builds me up.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water forms a riverbed and &quot;cuts&quot; it through erosion, but also builds it up by depositing dirt, rocks, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Line 3: &lt;br&gt;That what cuts me seeks always to go lower, but lifts its bed higher.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water always flows downhill, but again the &quot;bed&quot; of the river is built up by the sediment it carries.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Line 4: &lt;br&gt;I am commonly found where what builds me is less common.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not as sure about this one. If the answer is &quot;riverbed&quot; then those are often found where the water in the river has run out. Otherwise, it might be referring to the fact that rivers more often show up when there aren't lakes around (I'm not even sure if that is true). Still thinking on this line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastEditorUserId="1620" LastEditDate="2016-02-06T00:44:36.317" LastActivityDate="2016-02-06T00:44:36.317" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26172" PostTypeId="1" AcceptedAnswerId="26177" CreationDate="2016-02-06T00:46:29.033" Score="3" ViewCount="353" Body="&lt;p&gt;&lt;em&gt;My first created puzzle, inspired by something I saw at work. Hints are available and feedback is appreciated.&lt;/em&gt; :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;They've been talking about getting one for years, and now's your chance to give them what you know they want but will never get themselves. A great gift for the procrastinator in your life, just in time for Valentine's Day! Inquire about our wide selection of available styles*.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*square and triangle models currently unavailable&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is this perfect gift?&lt;/p&gt;&#xA;" OwnerUserId="17167" LastActivityDate="2016-02-10T05:04:15.760" Title="The Perfect Valentine's Day Gift" Tags="|riddle|wordplay|seasonal|" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26174" PostTypeId="2" ParentId="26123" CreationDate="2016-02-06T01:31:46.813" Score="16" Body="&lt;p&gt;The formulation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If I were to ask you ____, would you answer affirmatively?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;can get an honest answer for any yes-or-no question. So now we just have to find the best way to use the questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is never optimal to choose a door until it is the only possibility left, because the chance of survival will always be better after asking a yes-no question (approximately halving the number of possibilities, so doubling the chance of survival with only a 5% penalty).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose that there is a system of questions such that it can always be narrowed down to one door (possibly choosing different questions depending on the response to previous ones). Each door can be uniquely identified by what sequence of yes/no answers leads to that door. The sequence of answers cannot start with the sequence of a different door, otherwise that door would already have been chosen. Therefore the sequences of answers form a &lt;a href=&quot;https://en.wikipedia.org/wiki/Prefix_code&quot;&gt;prefix-free encoding&lt;/a&gt; of the doors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the length of the code for door $i$ be $l_i$. If the safe door is $i$, it takes $l_i$ questions before choosing that door, so the probability of survival is $0.95^{l_i}$ This make the total probability of survival $\frac{1}{100}\sum_{i=1}^{100}0.95^{l_i}$. We want to maximize this value, under the constraint that $\sum_{i=1}^{100}0.5^{l_i}\le1$ (this is required for it to be a prefix-free encoding, since a code of length $l$ takes up $0.5^l$ of the code space).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\sum_{i=1}^{100}0.5^{l_i}&amp;lt;1$, there is some wasted sequence that does not match any door. This can only happen if some question is being asked in some situation where only one answer is possible. That question can be skipped to improve the chance of survival, so we can assume that $\sum_{i=1}^{100}0.5^{l_i}=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the smallest $l_i$ be $m$ and the two largest ones be $n$ (there must be at least two tied for largest for the condition $\sum_{i=1}^{100}0.5^{l_i}=1$ to be true). Suppose $m+2\le n$. Then $$0.5^{m+1}+0.5^{m+1}+0.5^{n-1}=0.5^{m}+0.5^{n-1}=0.5^{m}+0.5^{n}+0.5^{n}$$ but $$0.95^{m+1}+0.95^{m+1}+0.95^{n-1}=1.9(0.95^{m})+0.95^{n-1}\\&amp;gt;0.95^{m}+0.95^3(0.95^{m})+0.95^{n-1}=0.95^{m}+0.95^{m+3}+0.95^{n-1}\\\ge0.95^{m}+0.95^{n+1}+0.95^{n-1}&amp;gt;0.95^{m}+0.95^{n}+0.95^{n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, while $\sum_{i=1}^{100}0.5^{l_i}$ remains equal to $1$, we can replace $m,n,n$ with $m+1,m+1,n-1$ to improve the chance of survival. This means that the best chance of survival is obtained when the maximum and minimum lengths differ by at most one. The only way to do this is to have 28 sequences of length 6 and 72 sequences of length 7 ($\frac{28}{64}+\frac{72}{128}=1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One easy way to implement this is to make the $x$th question asked (remembering to use the modification at the top of this answer):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;When the number of the correct door is written in binary, is there a digit 1 in the place that represents $2^{x-1}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the right door is between 37 and 64 inclusive, it will be the only one left after 6 questions. Otherwise, it will require a 7th question. The probability of survival is $0.28(0.95^6)+0.72(0.95^7)\approx70.863\%$, which is optimal as proven above.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2016-02-06T02:42:17.123" LastActivityDate="2016-02-06T02:42:17.123" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26175" PostTypeId="1" AcceptedAnswerId="26188" CreationDate="2016-02-06T02:48:04.107" Score="11" ViewCount="519" Body="&lt;p&gt;I came up with this riddle on the train home from work. After seeing what your puzzle fiends crack on a daily basis, I'll be lucky if it survives a single read-through, but I suppose it'll do for a first submission.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Easy to judge, for better or worse,&lt;br&gt; While others will take some&#xA;  time and discourse.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Looks foreign and strange without the aid,&lt;br&gt; Either theirs or yours,&#xA;  will be naked or fade.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Scaled up, rarely down - the risk is too great.&lt;br&gt; Look at its hands,&#xA;  before it's too late!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Would grant you your whims if made into a book,&lt;br&gt; And hides in plain&#xA;  sight wherever you look.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A relevant deity would make an immature teenager giggle. 'cause butt!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If I said you had one, you'd think it rather obvious. But if I said you had two, you'd get offended.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6163" LastEditorUserId="6163" LastEditDate="2016-02-06T07:14:01.007" LastActivityDate="2016-02-07T13:21:42.510" Title="Lift it to travel through time" Tags="|riddle|word|rhyme|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26177" PostTypeId="2" ParentId="26172" CreationDate="2016-02-06T04:13:24.013" Score="5" Body="&lt;p&gt;The gift is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A round tuit.&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/nIWBb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nIWBb.jpg&quot; alt=&quot;a round tuit&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's said to be &quot;a great gift for the procrastinator in your life&quot; because procrastinators have a hard time getting it. It's round, but square and triangle models are also available.&lt;/p&gt;&#xA;" OwnerUserId="16990" LastEditorUserId="17147" LastEditDate="2016-02-06T04:16:20.907" LastActivityDate="2016-02-06T04:16:20.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26178" PostTypeId="1" AcceptedAnswerId="27540" CreationDate="2016-02-06T04:19:33.867" Score="6" ViewCount="309" Body="&lt;p&gt;At MyCorp U.S., Inc. I'm the pusher - one of the main keyboards on the org chart, the one you want to follow. Without me the corporation would be a cruel, harsh place, and would quickly collapse. I'm generous and kind, but also have a sinister side. One time I was arrested for some back-room deal, quite a serious infraction actually. Everyone was shocked!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; added &quot;wordplay&quot; tag (though that is a relatively minor element in the riddle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am a thing, not a person.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My corporation's name might work better in Latin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user17947" LastEditorDisplayName="user17947" LastEditDate="2016-02-17T09:51:36.227" LastActivityDate="2016-02-17T12:57:21.667" Title="The pusher from MyCorp, Inc. - who am I?" Tags="|riddle|wordplay|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26180" PostTypeId="1" AcceptedAnswerId="26183" CreationDate="2016-02-06T05:10:48.960" Score="67" ViewCount="11806" Body="&lt;p&gt;&lt;em&gt;I'm excited to share the following riddle. It was given to me more than two years ago and I finally solved it last summer (after not thinking about it for a long time). In my desperation, I tried to find a solution online, but couldn't even find the riddle anywhere else. I'm excited to see if somebody knows the riddle or if not, how you approach the solution.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three mathematicians are in prison. Each of them is in a single cell and they are not able to communicate in any way. They are imprisoned for an arbitrary number of days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each cell has a single light bulb that is either on or off on a given day. The warden tells the mathematicians that the light system of the prison has three modes: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;neutral&lt;/strong&gt; mode, where each lightbulb is independent of the others&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;bright&lt;/strong&gt; mode, where two bulbs turn &lt;em&gt;on&lt;/em&gt; every day and the other turns &lt;em&gt;off&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;dark&lt;/strong&gt; mode, where two bulbs turn &lt;em&gt;off&lt;/em&gt; every day and the other turns &lt;em&gt;on&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(All distributions are not necessarily uniform.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The prison starts in &lt;strong&gt;neutral&lt;/strong&gt; mode. After an unknown but finite number of days, the warden will select either &lt;strong&gt;bright&lt;/strong&gt; mode or &lt;strong&gt;dark&lt;/strong&gt; mode, which is locked in permanently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After countably infinitely many days have passed, the mathematicians are asked which one the warden picked. They may discuss strategy before going into the cells, but there will be no communication afterwards. They have unlimited capacities to communicate and remember strategies that they come up with. Two of the three need to guess correctly to escape; how can they ensure this? &lt;em&gt;You may assume that the axiom of choice holds.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="19009" LastEditorUserId="19009" LastEditDate="2016-02-06T07:16:05.587" LastActivityDate="2018-02-07T15:46:27.573" Title="Three mathematicians are forever in Prison" Tags="|mathematics|logical-deduction|" AnswerCount="7" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="26183" PostTypeId="2" ParentId="26180" CreationDate="2016-02-06T05:42:48.633" Score="39" Body="&lt;p&gt;The night before their first day in prison, the three mathematicians choose a non-principal &lt;a href=&quot;https://en.wikipedia.org/wiki/Ultrafilter&quot; rel=&quot;noreferrer&quot;&gt;ultrafilter&lt;/a&gt; on the set of days they are in prison. A non-principal ultrafilter is a rule for classifying some sets of days as &lt;em&gt;large&lt;/em&gt;, and the rest as &lt;em&gt;small&lt;/em&gt;, subject to the following conditions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Every set of days containing a large set is large,&lt;/li&gt;&#xA;&lt;li&gt;If the set of all days is partitioned into finitely many sets, exactly one set is large, and&lt;/li&gt;&#xA;&lt;li&gt;No finite set of days is large.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Constructing a non-principal ultrafilter requires the axiom of choice, and communicating an ultrafilter from one mathematician to another requires an infinite amount of information. These mathematicians have unlimited mental capacity though, so maybe this is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the countably infinite number of days has passed, each mathematician guesses &lt;strong&gt;dark mode&lt;/strong&gt; if the set of days when their light was off is large, and guesses &lt;strong&gt;bright mode&lt;/strong&gt; if the set of days their light was on is large (property 2 above guarantees that exactly one of these conditions is met).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose warden eventually selects &lt;strong&gt;dark mode&lt;/strong&gt;. Consider the following four sets of days:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;neutral mode&lt;/strong&gt; days,&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;dark mode&lt;/strong&gt; days when the first mathematician's light is on,&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;dark mode&lt;/strong&gt; days when the second mathematician's light is on,&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;dark mode&lt;/strong&gt; days when the third mathematician's light is on.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Again, property 2 of the ultrafilter says that exactly one of these sets of days is large. By property 3, it cannot be the first, because that set is finite. This means exactly one of the mathematicians saw a light for a large set of days, so that mathematician guesses &lt;strong&gt;bright mode&lt;/strong&gt; and the other two guess &lt;strong&gt;dark mode&lt;/strong&gt;. Success!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The argument in the case the warden selects &lt;strong&gt;bright mode&lt;/strong&gt; is identical.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="40223" LastEditDate="2018-02-07T15:46:27.573" LastActivityDate="2018-02-07T15:46:27.573" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="26186" PostTypeId="2" ParentId="26166" CreationDate="2016-02-06T08:23:56.460" Score="3" Body="&lt;p&gt;I plotted the distribution of the numbers and got a nice half-bell shaped curve, making me think it was the first occurence of a character in some text. Since I am Swedish, I know very little of Lincoln and started with the Gettysburg Address.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out the transcriptions of the speech varies a bit. The versions seem to differ whether they spell &quot;Fourscore&quot; or &quot;Four score&quot; as well as commas here and there. In any case, the first four letters seemed to be invariant. Hence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1-F, 2-o, 4-r =&gt; _ _ o _ _ _ o F _ r&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Next, I wrote a Python script going through the text printing the first occurence of every letter. I tried both including spaces, periods, and commas, as well as excluding them. However, the first occurences when excluding these characters did not line up nicely with the given numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the transcript with all characters, only one letter was close to place 103, namely&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; L, which had its first occurence at place 107 in my transcription, we now have _ L o _ _ _ o F _ r&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If picking a version spelling &quot;Four score&quot; using commas in the first paragraph, such as &lt;a href=&quot;http://americancivilwar.com/north/lincoln.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;, all other numbers line up exactly with the first occurrence of a letter. The resulting text is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; iLovegoFer -- I love gofer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Could this be correct?&lt;/p&gt;&#xA;" OwnerUserId="18919" LastEditorUserId="18919" LastEditDate="2016-02-06T19:18:01.893" LastActivityDate="2016-02-06T19:18:01.893" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="26188" PostTypeId="2" ParentId="26175" CreationDate="2016-02-06T09:28:18.080" Score="6" Body="&lt;p&gt;I think the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Face&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lift it to travel through time (title)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A face-lift makes you look younger.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Easy to judge, for better or worse,&#xA;While others will take some time and discourse.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some faces are easy to read, others no so easy. Face time is used to communicate remotely.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Looks foreign and strange without the aid,&#xA;Either theirs or yours, will be naked or fade.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Without glasses faces look different. If the viewee isn't wearing them, they look naked. If the viewer isn't wearing them then the other person looks blurry or faded.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Scaled up, rarely down - the risk is too great.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rock faces are scaled - it's easier to climb up than down.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Look at its hands, before it's too late!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  A clock face has hands and tells you the time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Would grant you your whims if made into a book,&#xA;And hides in plain sight wherever you look.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Facebook grants whims?&lt;br&gt;&#xA; Objects everywhere have faces / surfaces.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  A relevant deity would make an immature teenager giggle. 'cause butt!&lt;br&gt;&#xA;  The god Janus (who has a face looking in each direction) sounds like anus. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;    If I said you had one, you'd think it rather obvious. But if I said you had two, you'd get offended.&lt;br&gt;&#xA; Everyone has one face, but no-body likes being called two-faced.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;thanks to @Daphne B for a couple of extra hints&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From OP:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Easy to judge, for better or worse,&#xA;While others will take some time and discourse.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We quickly judge people by their face/appearance when we first meet them, sometimes forming the wrong opinion. To see other aspects of people beyond appearance you need to talk and get to know them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Would grant you your whims if made into a book,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The creator of Facebook, Mark Zuckerberg, became one of the richest people on the planet and can now afford just about anything.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;And hides in plain sight wherever you look.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We always see the tip of the nose, no matter where we look, but our brain filters it out so we don't notice it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="6163" LastEditDate="2016-02-07T13:21:42.510" LastActivityDate="2016-02-07T13:21:42.510" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26189" PostTypeId="1" AcceptedAnswerId="26203" CreationDate="2016-02-06T09:35:28.843" Score="6" ViewCount="108" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;When being exhaustive, I'm found on a horse.&lt;br&gt;&#xA;  Attacked by the mob I'm essential, of course.&lt;br&gt;&#xA;  When being artistic I'm holding up well.&lt;br&gt;&#xA;  In a refuge I couldn't escape from Hell.&lt;br&gt;&#xA;  When wearing a crown I'll be found in the centre.&lt;br&gt;&#xA;  With a british policeman - dishevelled preventer.&lt;br&gt;&#xA;  When I come from the earth I prefer wetter weather.&lt;br&gt;&#xA;  Teddy needs me divided to keep him together.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: What am I?&lt;/p&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2016-02-07T09:17:31.270" Title="A diminishing plastic protector" Tags="|riddle|word|rhyme|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26190" PostTypeId="1" AcceptedAnswerId="26192" CreationDate="2016-02-06T12:15:35.953" Score="5" ViewCount="152" Body="&lt;pre&gt;&lt;code&gt;      _                 _         &#xA;_  a |_  b |   c |   d     e   | f   | g |_  h     i     j/*End of Sequence*/&#xA;   a |_  b   | c |   d |_  e |   f     g |_  h     i |   j/*End of Sequence*/&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You find a piece of paper in the print tray with the above characters printed. You notice that the space between the &lt;em&gt;h&lt;/em&gt; and the &lt;em&gt;i&lt;/em&gt; is blank. What should be placed there and why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Read the title again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;br&gt;&lt;hr&gt;&#xA;P.S. I've used both &lt;a href=&quot;/questions/tagged/number-sequence&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;number-sequence&amp;#39;&quot; rel=&quot;tag&quot;&gt;number-sequence&lt;/a&gt; and &lt;a href=&quot;/questions/tagged/letter-sequence&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;letter-sequence&amp;#39;&quot; rel=&quot;tag&quot;&gt;letter-sequence&lt;/a&gt; because I don't want to reveal what kind of a sequence it is. Will &lt;a href=&quot;/questions/tagged/enigmatic-puzzles&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;enigmatic-puzzles&amp;#39;&quot; rel=&quot;tag&quot;&gt;enigmatic-puzzles&lt;/a&gt; be valid here as well?&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="4269" LastEditDate="2016-02-07T10:16:28.960" LastActivityDate="2016-02-07T10:16:28.960" Title="An n-segment sequence" Tags="|cipher|number-sequence|letter-sequence|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26192" PostTypeId="2" ParentId="26190" CreationDate="2016-02-06T14:58:09.097" Score="8" Body="&lt;p&gt;Nothing should be placed there.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is the sequence of segments that aren't lit up when a 7-segment display goes from zero to nine.  The ninth term in the sequence is nothing because all 7 segments are lit up for the digit eight.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12092" LastEditorUserId="12092" LastEditDate="2016-02-06T15:12:02.207" LastActivityDate="2016-02-06T15:12:02.207" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26193" PostTypeId="2" ParentId="26180" CreationDate="2016-02-06T15:41:30.253" Score="90" Body="&lt;p&gt;When the mathematicians are asked for their answers, each answers &quot;light mode&quot; if their light is currently on, or &quot;dark mode&quot; if it is currently off.&lt;/p&gt;&#xA;" OwnerUserId="19025" LastEditorUserId="19025" LastEditDate="2016-02-06T16:33:26.950" LastActivityDate="2016-02-06T16:33:26.950" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="26195" PostTypeId="1" AcceptedAnswerId="26206" CreationDate="2016-02-06T18:31:27.753" Score="11" ViewCount="1366" Body="&lt;p&gt;I'm a physical thing. Can you guess what I am?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A prison of sorts, with light but no breeze,&lt;br&gt; These passive sea&#xA;  creatures contain I with ease.&lt;br&gt; Their very own home is used to&#xA;  transform,&lt;br&gt; To shrivel, protect, infuse, and deform.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Imprisoning children is even more foul,&lt;br&gt; Plucked from their mothers&#xA;  they can't even howl.&lt;br&gt; Scalded on purpose, they lie in a sack,&lt;br&gt; To be&#xA;  beaten, abused, till they finally crack.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If you think you are fun, then you guys better hide,&lt;br&gt; They'll find you,&#xA;  cut you, and lock you inside.&lt;br&gt; You'll change to the core while you&#xA;  languish within,&lt;br&gt; But granting you freedom is a much greater sin.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;For once they release the inmates I hold,&lt;br&gt; Abhorrent and ghastly events&#xA;  will unfold.&lt;br&gt; They'll struggle to open my rusty old door,&lt;br&gt; I'll try to&#xA;  protect you, that's what I'm for.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;But none can withstand these insatiable beasts,&lt;br&gt; They've kept you alive&#xA;  for their terrible feasts.&lt;br&gt; They'll prod you with pitchforks until you&#xA;  comply,&lt;br&gt; Too tired to struggle, you're ready to die.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;One at a time, sometimes all at once,&lt;br&gt; Sometimes it takes days,&#xA;  sometimes even months.&lt;br&gt; But sooner or later I'll be empty again,&lt;br&gt; Even&#xA;  your tears they'll dump in the drain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This prison only opens at the right moment.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If you are like me (as in me the human author, not me the thing you are trying to guess), then you may get a sweet and sticky label in order to elicit help.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Super hint #3 (don't look unless totally stuck):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Don't be misled by the dark tone, the answer is jarringly innocent.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6163" LastEditorUserId="6163" LastEditDate="2016-02-07T00:53:37.200" LastActivityDate="2018-02-27T02:16:17.880" Title="You are gonna need some coffee to crack this little nugget of a riddle!" Tags="|riddle|word|rhyme|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26197" PostTypeId="1" AcceptedAnswerId="26199" CreationDate="2016-02-06T19:04:26.803" Score="7" ViewCount="439" Body="&lt;p&gt;There are $N$ democrats and $N$ republicans sitting in a circle. Each of these people is either always honest or always deceitful. It is known that the number of honest democrats is the same as the number of honest republicans. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each politician is asked, &quot;What party is your right hand neighbor?&quot; They all respond, &quot;Republican.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Prove that $N$ is even.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://www.austms.org.au/Publ/Gazette/2008/May08/PuzzleCorner.pdf&quot; rel=&quot;noreferrer&quot;&gt;AMS Gazette, Puzzle Corner 7&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="11876" LastEditDate="2016-02-07T01:34:30.333" LastActivityDate="2016-02-07T01:34:30.333" Title="Democrats and Republicans in a Circle" Tags="|liars|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26199" PostTypeId="2" ParentId="26197" CreationDate="2016-02-06T19:18:36.610" Score="6" Body="&lt;p&gt;If everyone calls their right-hand neighbor a rep, only $N$ of them can be one, so there are only $N$ truth tellers in total. There can be $N/2$ truth tellers and liars from each party, therefore $N$ is even.&lt;/p&gt;&#xA;" OwnerUserId="14016" LastActivityDate="2016-02-06T19:18:36.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26200" PostTypeId="2" ParentId="26197" CreationDate="2016-02-06T19:19:02.100" Score="6" Body="&lt;p&gt;Half of the people present are Democrats and half are Republicans. Everyone says their right hand neighbor is a Republican, so exactly half of the people are lying. Since there are $2N$ people present, $N$ of them must be liars. There are an equal number of Democratic liars and Republican liars, so $N$ must be even.&lt;/p&gt;&#xA;" OwnerUserId="2627" LastActivityDate="2016-02-06T19:19:02.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26201" PostTypeId="2" ParentId="26197" CreationDate="2016-02-06T20:11:04.583" Score="3" Body="&lt;p&gt;Denote by $k$ the number of honest democrats, which is known to also be equal to the number of honest republicans. The total number of honest people is therefore $2k$, and the total number of dishonest people is $2N-2k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $2k$ honest people answered &quot;republican&quot; to the question about the party affiliation of the person sitting to their right, so since they are telling the truth, that implies that there are at least $2k$ republicans. But we know there are precisely $N$ republicans, so $2k\le N$, or $k\le N/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $2N-2k$ dishonest people answered &quot;republican&quot; to the question about the party affiliation of the person sitting to their right, so since they are lying, there at least $2N-2k$ democrats. But we know there are precisely $N$ democrats, so $2N-2k\le N$, or $k\ge N/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These two observations together imply that $k=N/2$. But obviously $k$ is an integer, so $N$ must be even.&lt;/p&gt;&#xA;" OwnerDisplayName="user17947" LastActivityDate="2016-02-06T20:11:04.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26202" PostTypeId="2" ParentId="26180" CreationDate="2016-02-06T20:25:21.057" Score="8" Body="&lt;p&gt;I heard this puzzle (or a similar version) from Abraham Neyman. The solution is similar to that of Julian Rosen above.&#xA;The mathematicians choose a Banach limit. I.e, a linear operator on bounded sequences that extends the usual limit whose value is between the limit superior and limit inferior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each mathematician answers &quot;bright&quot; if and only if his limit is strictly greater than a half. Since the Banach limit is linear the sum of the three limits is either one or two.&lt;/p&gt;&#xA;" OwnerUserId="19036" LastEditorUserId="44915" LastEditDate="2018-02-07T14:32:00.257" LastActivityDate="2018-02-07T14:32:00.257" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26203" PostTypeId="2" ParentId="26189" CreationDate="2016-02-06T20:40:54.130" Score="7" Body="&lt;p&gt;I think the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pin&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When being exhaustive, I'm found on a horse.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Thoroughpin is a swelling of the tendon sheath above the hock of a horse.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Attacked by the mob I'm essential, of course.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An attack by the mob could be a lynch, and a lynchpin is an essential person or thing to an operation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When being artistic I'm holding up well.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A drawing pin is used to hold things up&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In a refuge I couldn't escape from Hell.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hell refers to Richard Hell a pioneer of punk fashion noted for using &quot;safety pins&quot;. Safety refers to the refuge&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When wearing a crown I'll be found in the centre.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A kingpin is found at the centre of a set of bowling pins (and perhaps a secret crime organisation)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With a british policeman - dishevelled preventer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Bobby pin&quot; - as pointed out by Deusovi&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I come from the earth I prefer wetter weather.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A terrapin is a species of turtle which lives in fresh or brackish water and &quot;terra&quot; is Latin for Earth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Teddy needs me divided to keep him together.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Split pin - a device used largely in the manufacture of teddies&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A diminishing plastic protector&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A PIN code is a password used as an extra layer of security at an ATM on top of a plastic debit/credit card so could be seen as a plastic protector. Diminishing refers to the fact that a usual type of pin diminishes to a point.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18422" LastEditorUserId="18422" LastEditDate="2016-02-07T09:17:31.270" LastActivityDate="2016-02-07T09:17:31.270" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="26204" PostTypeId="2" ParentId="26195" CreationDate="2016-02-06T21:15:21.333" Score="1" Body="&lt;p&gt;I might be way off the track, but my closest guess is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it's an oyster shell.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is a prison for &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the mollusk (sea creature) it contains.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What doesn't make sense to me is that talk about &quot;their home&quot; being used to transform (them ?). Imprisoning children could be the result of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; men eating the oyster content.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They can't even howl because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; they don't even have a nervous system.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Scalded on purpose&quot; doesn't make sense to me as well, because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; AFAIK oyster are eaten raw.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Beaten/abused/crack could describe&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the act of opening the oyster.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Lying in a sack&quot; probably means&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; they are in the stomach already. &quot;You&quot; seems to address the mollusk, warning them before men who wanna eat them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Freedom is sin&quot;, &quot;protection&quot; seems to indicate that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the mollusk is nothing without its shell.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Open Rusty old door&quot; also means&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; cracking the oyster.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Terrible feasts&quot; also means eating. &quot;Tears in the drain&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I've read that some fluid remains after eating an oyster, and that it gets disposed of.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On the other hand, your hints don't make any sense to me at all...&lt;/p&gt;&#xA;" OwnerUserId="19030" LastEditorUserId="5373" LastEditDate="2016-11-19T19:48:15.763" LastActivityDate="2016-11-19T19:48:15.763" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26206" PostTypeId="2" ParentId="26195" CreationDate="2016-02-07T01:44:39.943" Score="5" Body="&lt;h1&gt;new answer&lt;/h1&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Pickling and especially a jar of pickles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The transformations happen to&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; a cucumber or other pickled item.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The sea creatures&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; perhaps to make you think of salt, or pickled herring. Or sea cucumbers, which are a creature despite the name.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The second stanza refers to&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; eggs, as mentioned in the old answer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The &amp;quot;find you and cut you&amp;quot;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; refers to eating pickles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;while &amp;quot;change to the core&amp;quot; is&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; the pickling process and fermentation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;And a fun guy&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Refers to fungi, pickled mushrooms&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The abhorrent and ghastly I got nothing, but struggle to open is&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; getting the lid off a jar at first&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The second-last stanza is about&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; eating pickles that have been preserved for the purpose&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;And the final stanza&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; to using up the jar over a long period of time and then pouring away the remaining juice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Hint 1&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Many pickles need a while - my own pickled beans I wait 6 weeks to open. Also, in more of a confirmation than a hint, the &amp;quot;moment of inertia&amp;quot; could be relevant when turning the lid to open the jar&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Hint 2&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; No idea, perhaps there's a popular US brand name in there&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Hint 3&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; the word &amp;quot;jar&amp;quot; is the key here&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h1&gt;old answer&lt;/h1&gt;&#xA;&lt;p&gt;I believe it's a&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; refrigerator&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;It contains things, including sea creatures I suppose. The list of transformations described: &amp;quot;shrivel, protect, infuse, and deform&amp;quot; I've seen.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; And of course a fridge has a light inside&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The second stanza I believe refers to&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; eggs, which are kept in the fridge&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I believe the third and fourth stanzas refers to&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The safety precautions for discarded fridges so that children don't hide in them and suffocate&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The fifth stanza refers to&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Keeping food (for feasts, that you eat with forks) in a fridge&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;and finally&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; They do get empty, and even get replaced. Any liquid that accumulates in the bottom gets dumped&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I can't actually connect the hints to this though.&lt;/p&gt;&#xA;&lt;p&gt;&lt;br&gt;&lt;br&gt;&#xA;Kate got almost everything anyway, but here are the full answers from OP:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;You are gonna need some coffee to crack this little nugget of a riddle!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; This is a programming reference. Java is used to open files of type .jar (which stands from Java ARchive), which are sort of like nuggets containing a bunch of things.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;A prison of sorts, with light but no breeze,&#xA;These passive sea creatures contain I with ease.&#xA;Their very own home is used to transform,&#xA;To shrivel, protect, infuse, and deform.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; A jar is transparent, but air-tight. One of the most commonly pickled things are cucumbers, which are also one of the most numerous sea animals in the world. Their home - salt water - is what's typically used to pickle things.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;Imprisoning children is even more foul,&#xA;Plucked from their mothers they can't even howl.&#xA;Scalded on purpose, they lie in a sack,&#xA;To be beaten, abused, till they finally crack.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The entire stanza is about pickled eggs. There are some punny hints here like &amp;quot;foul vs fowl&amp;quot; and &amp;quot;plucked&amp;quot;. Eggs are children, they get boiled (scalded), and then lie around in their shell waiting for us to crack them and clean the shell off.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;If you think you are fun, then you guys better hide,&#xA;They'll find you, cut you, and lock you inside.&#xA;You'll change to the core while you languish within,&#xA;But granting you freedom is a much greater sin.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Mushrooms are also commonly sealed in jars, the play on words here is &amp;quot;fun guy vs fungi&amp;quot;. Mushrooms hide in the forest, and people go looking for them, cut them, and put them in jars. Pickling changes the mushrooms just like everything else.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;For once they release the inmates I hold,&#xA;Abhorrent and ghastly events will unfold.&#xA;They'll struggle to open my rusty old door,&#xA;I'll try to protect you, that's what I'm for.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Once the jar is open, the contents get eaten. Opening jars is notoriously hard, which actually ties into the second hint. And the point of the jar is to preserve and protect the contents.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;But none can withstand these insatiable beasts,&#xA;They've kept you alive for their terrible feasts.&#xA;They'll prod you with pitchforks until you comply,&#xA;Too tired to struggle, you're ready to die.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; We are hungry and we want some pickles! Which we'll fish out of the jar with a fork. If often takes a few tries, too, you keep stabbing the slippery bastard and he keeps rolling out of the way.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;One at a time, sometimes all at once,&#xA;Sometimes it takes days, sometimes even months.&#xA;But sooner or later I'll be empty again,&#xA;Even your tears they'll dump in the drain.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Finishing off a jar can take anywhere from a single sitting to months, but eventually it's empty except for the brine left behind. Which is salty like tears, and is shed by the pickle as you remove it from the jar.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;Hint 1:&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Moment means not only a small chunk of time, but also a force used to create rotational movement. When we open jars, we apply certain moment to the lid, which won't budge until the moment is high enough.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;Hint 2:&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; How many of you have heard &amp;quot;Honey! Can you open this jar?&amp;quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;Hint 3:&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Of course the give-away here is &amp;quot;jarringly&amp;quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1509" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-02-08T18:04:24.233" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="26207" PostTypeId="2" ParentId="26195" CreationDate="2016-02-07T04:26:35.017" Score="3" Body="&lt;p&gt;My Guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An Egg&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A prison of sorts, with light but no breeze,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eggs are partially see through with a bright enough light. Its called candling and is used to check on the development in an egg (see if it was fertilized).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These passive sea creatures contain I with ease.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The hen's actual egg could be considered the passive sea creatures (especially if not fertilized) contained behind the outer shell&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Their very own home is used to transform,&#xA;To shrivel, protect, infuse, and deform.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the egg is fertilized it will transform into a baby chick - if not it will spoil and rot without breaking the shell.....unless a build up of gasses causes it to explode.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Imprisoning children is even more foul,&#xA;Plucked from their mothers they can't even howl.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eggs could be considered the child of a hen (at least they are if fertilized) and can't howl as they are still in their shell.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Scalded on purpose, they lie in a sack,&#xA;To be beaten, abused, till they finally crack.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pasteurization is a common practice with eggs (which involves a hot water or bath) and broken eggs are a common sight when looking for that perfect dozen at the market.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you think you are fun, then you guys better hide,&#xA;They'll find you, cut you, and lock you inside.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think this is referring to the fact that sperm would lose their tails (&lt;em&gt;cut you&lt;/em&gt;) during fertilization (the &lt;em&gt;fun&lt;/em&gt;) and are then &lt;em&gt;trapped&lt;/em&gt; or combined with the egg.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'll change to the core while you languish within,&#xA;But granting you freedom is a much greater sin.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Basically as soon as fertilization happens the zygote forms and continues to grow into a baby chick. If a sperm fails to fertilize (&lt;em&gt;freedom&lt;/em&gt;) it will, basically, just die where it lands.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For once they release the inmates I hold,&#xA;Abhorrent and ghastly events will unfold.&#xA;They'll struggle to open my rusty old door,&#xA;I'll try to protect you, that's what I'm for.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is the farmer opening the chicken coup and collecting the eggs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But none can withstand these insatiable beasts,&#xA;They've kept you alive for their terrible feasts.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mmmm...Breakfast&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They'll prod you with pitchforks until you comply,&#xA;Too tired to struggle, you're ready to die.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I assume a fork = pitchforks and, well...breakfast :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One at a time, sometimes all at once,&#xA;Sometimes it takes days, sometimes even months.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Depends on how hungry a person is and what they are used for. As in scrambled eggs for breakfast will be quick and only use a few eggs, a full wedding cake would use a number of eggs but beginning spoiling in a week or so, while an egg used to make a meatloaf to be used in a packaged &quot;TV Dinner&quot; is a single egg but could last a while.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But sooner or later I'll be empty again,&#xA;Even your tears they'll dump in the drain.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You crack the &lt;em&gt;tear shaped&lt;/em&gt; eggs and dump them into whatever your cooking&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This prison only opens at the right moment = Eggs only hatch when chick is formed and ready, person is ready to cook, or when it is spoiled to the point that it forces it to crack.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you are like me, then you may get a sweet and sticky label in order to elicit help. = I know at least they use sticky labels to show expiration dates&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Super hint #3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Don't be misled by the dark tone, the answer is jarringly innocent = Both baby chicks are innocent, and soft, and chicken eggs can be pickled and jarred.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11239" LastActivityDate="2016-02-07T04:26:35.017" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26208" PostTypeId="2" ParentId="26167" CreationDate="2016-02-07T06:38:05.730" Score="2" Body="&lt;p&gt;After thinking about this, and after some deep thought and stressful thinking, I have come a conclusion! Additionally, sorry if the way I formatted my answer is incorrect; this is my first time on this site.&#xA;All that aside, I would have to say the answer to this is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The word &quot;Hate&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Some call me royalty, but I say they're wrong.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Some monarchs and leaders have been criticized for complaining and hating those who don't help them&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Some call me crazy, but I say they're wrong.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;People who hate very aggressively are looked as crazy by society&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Some call me a ghost, but I say they're wrong.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Some people believe that hatred does not exist because they have not encountered it (e.g. little children)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Some call me racist, but I say they're wrong.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Racism is often the product of hatred&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Some call me a hat, but I say they're wrong.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Perhaps some people might mispronounce the word as &lt;strong&gt;hat&lt;/strong&gt;e&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Hate&quot; is indeed, a word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19051" LastActivityDate="2016-02-07T06:38:05.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26209" PostTypeId="1" AcceptedAnswerId="26224" CreationDate="2016-02-07T10:41:19.000" Score="3" ViewCount="128" Body="&lt;p&gt;I am on place A, somewhere in the United States of America.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If I go to the Southwest, my sequence starts as following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;3880181, 967089, 299246, 212635, 194944,  ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If I go to the Northwest, my sequence starts as following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1009518, 483702, 288146, 221846, 133791, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If I go to the Northeast, my sequence starts as following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;588349, 587272, 545089, 529354, 422495, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If I go to the Southeast, my sequence starts as following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;629292, 198791, 142955, 122427, 117866, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Additional hints:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Hint 1: &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each sequence consists of positive integers and is decreasing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;Hint 2: &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each sequence changes over time, and it probably changes every day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;Hint 3: &#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The last sequence has 33 elements. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Questions: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Where is point A? &lt;/li&gt;&#xA;&lt;li&gt;What do the numbers represent? &lt;/li&gt;&#xA;&lt;li&gt;How many elements does the third sequence have?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="16322" LastActivityDate="2016-02-08T09:29:30.020" Title="Sequence dependent on my place and the time" Tags="|number-sequence|geography|knowledge|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26210" PostTypeId="1" AcceptedAnswerId="26213" CreationDate="2016-02-07T13:18:17.697" Score="11" ViewCount="855" Body="&lt;p&gt;Professor Halfbrain has spent his entire weekend with cutting rectangles into smaller rectangles. In particular, he proved the following deep theorem on such dissections.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Professor Halfbrain's rectangle dissection theorem&lt;/strong&gt;: &#xA;  If a $5\times2$ rectangle is cut into ten smaller rectangles with integer side lengths, then two of these smaller rectangles must be congruent.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This puzzle asks you to improve the professor's theorem and to make it even deeper: Find the largest integer $n$, for which the following statement hold true.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If a $5\times n$ rectangle is cut into ten smaller rectangles with integer side lengths, then two of these smaller rectangles must be congruent.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2016-02-07T18:18:44.040" Title="Professor Halfbrain and the dissection of a rectangle" Tags="|mathematics|geometry|dissection|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26211" PostTypeId="1" AcceptedAnswerId="26221" CreationDate="2016-02-07T13:37:48.360" Score="7" ViewCount="754" Body="&lt;p&gt;My friends and I like to pose cryptic crosswords to each other and I slipped up when I made a clue with the straight in the middle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that it's not Ximanean, but I thought it might be allowed by more liberal crossworders. &lt;/p&gt;&#xA;" OwnerUserId="9316" LastEditorUserId="69582" LastEditDate="2021-05-07T00:51:29.593" LastActivityDate="2023-08-20T21:51:10.783" Title="Is it permissible to put the straight or definition in the middle of the cryptic crossword clue?" Tags="|cryptic-clues|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26213" PostTypeId="2" ParentId="26210" CreationDate="2016-02-07T14:59:44.453" Score="19" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Taking the 10 smallest possible distinct rectangles:&lt;br&gt;&#xA;&lt;br&gt;&#xA; 1x1 = 1&lt;br&gt;&#xA; 2x1 = 2&lt;br&gt;&#xA; 3x1 = 3&lt;br&gt;&#xA; 2x2 = 4&lt;br&gt;&#xA; 4x1 = 4&lt;br&gt;&#xA; 5x1 = 5&lt;br&gt;&#xA; 3x2 = 6&lt;br&gt;&#xA; 6x1 = 6&lt;br&gt;&#xA; 7x1 = 7&lt;br&gt;&#xA; 4x2 = 8&lt;br&gt;&#xA;&lt;br&gt;&#xA; Their sum is 46. Therefore, for any total area smaller than 46, the professor's statement must necessarily hold.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The smallest possible area that's at least 46 is 50. Does the professor's statement hold for N=10?&lt;br&gt;&#xA;&lt;br&gt;&#xA; No, it does not. A 5x10 rectangle can be split up into 1x1, 1x2, 1x3, 1x4, 1x5, 2x2, 2x3, 2x4, 1x7, 1x10 rectangles, none of which are congruent.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/6t8dx.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6t8dx.png&quot; alt=&quot;Blocks&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; For a 5x9 rectangle, it can be split up into 9 rectangles of sizes 1x1, 1x2, 1x3, 1x5, 1x7, 1x9, 2x2, 2x3, 2x4. This allows one single rectangle to be added to produce 5xN for any N&gt;10 (but not N=10 itself, since it would require two 5x1 rectangles). Picture for N=11:&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/YyyCb.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YyyCb.png&quot; alt=&quot;Blocks&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; So the professor's statement doesn't hold for N=10, nor for any N&gt;10. That makes the answer &lt;strong&gt;9&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1764" LastEditorUserId="1764" LastEditDate="2016-02-07T15:12:29.403" LastActivityDate="2016-02-07T15:12:29.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26215" PostTypeId="1" AcceptedAnswerId="26216" CreationDate="2016-02-07T16:11:59.033" Score="3" ViewCount="347" Body="&lt;p&gt;You and your fellow legionaries have just conquered a neighboring village, and the Prelate has ordered per tradition that the survivors be decimated and the victims' heads delivered to him in a sack as proof of the deed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sadly, he's a little new on the job (and more than a little bloodthirsty), and incorrectly believes that &quot;decimate&quot; means to reduce the population to 1/10th its original size.  Your Captain, on the other hand, is a seasoned veteran and he knows that it really only means to reduce the population by 1/10th.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, these two officers have very limited means of communication.  The Prelate can only communicate with the Captain by sending an order to decimate the survivors, and the Captain can only communicate with the Prelate by sending him a sack full of heads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the Prelate knows exactly how many villagers there were to start with, and will not be satisfied until his definition of &quot;decimation&quot; has occurred, how many times will he have to give the order before his expectations are met?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: I'm looking for a mathematical answer.  You could just do this on a calculator but that's not how this is intended to be solved.  You may assume all rounding is to the nearest integer, though it shouldn't make much of a difference.)&lt;/p&gt;&#xA;" OwnerUserId="16131" LastEditorUserId="16131" LastEditDate="2016-02-08T17:54:59.773" LastActivityDate="2016-02-08T17:54:59.773" Title="How many decimations equals a &quot;decimation&quot;?" Tags="|mathematics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26216" PostTypeId="2" ParentId="26215" CreationDate="2016-02-07T16:21:06.457" Score="4" Body="&lt;p&gt;Aprox 22.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each time a decimation (regular) occurs, you reduce the size of the population to 9/10 of the original. So, our formula is $(\frac{9}{10})^x = \frac{1}{10}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the log of each side, bringing down the X and dividing gives $x = 21.85$&lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2016-02-07T16:21:06.457" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26217" PostTypeId="2" ParentId="26211" CreationDate="2016-02-07T17:23:39.397" Score="3" Body="&lt;p&gt;No, that would be an invalid clue. It must be possible to split the clue into two parts and (possibly ignoring a connector) have a definition and a wordplay part.&lt;/p&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2016-02-07T17:23:39.397" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="26218" PostTypeId="1" AcceptedAnswerId="26246" CreationDate="2016-02-07T18:16:24.843" Score="8" ViewCount="462" Body="&lt;p&gt;As an agent for the CIA, you are forced to take the test to re-validate your membership. Surprisingly, they send you a cryptogram no test-taker has been able to solve. The answer is quite clear to the guys who made the cryptogram, but hard to those who didn't:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;745692789632147412369 741253698521478963! &lt;br&gt;9874156147896398745632198745632174125369789632147418953741268  7412689874123514789638529874123574126!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The only hint they would ever give that it is two phrases ending in &lt;code&gt;!&lt;/code&gt; and each string of numbers represents a word. So what is the message anyways?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;745692789632147412369 = ? &lt;br&gt;&#xA;    741253698521478963! = ? &lt;br&gt;&#xA;    9874156147896398745632198745632174125369789632147418953741268 = ? &lt;br&gt;&#xA;    7412689874123514789638529874123574126! = ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9042" LastEditorUserId="12408" LastEditDate="2016-02-08T21:58:53.583" LastActivityDate="2016-02-09T00:23:41.937" Title="Who am I... Password Denied?" Tags="|cipher|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26219" PostTypeId="1" AcceptedAnswerId="26220" CreationDate="2016-02-07T18:45:29.287" Score="8" ViewCount="389" Body="&lt;p&gt;A famous plaything, beloved by geeks &lt;br&gt;&#xA;If you are like many, my meaning you'll seek &lt;br&gt;&#xA;From six letters to five, I'm a journey complete &lt;br&gt;&#xA;You can play me once, but you won't get a repeat&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who am I?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user17947" LastEditorDisplayName="user17947" LastEditDate="2016-02-11T19:37:19.750" LastActivityDate="2016-02-11T19:37:19.750" Title="A famous plaything - who am I?" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26220" PostTypeId="2" ParentId="26219" CreationDate="2016-02-07T19:05:52.293" Score="8" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Life (or one of two Games of Life)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm a famous game, beloved by geeks &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; John Conway's Game of Life is famous and beloved by geeks&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you are like many, my meaning you'll seek &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People seek the meaning of life&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From six letters to five, I'm a journey complete &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; From CRADLE (6) to GRAVE (5)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can play me once, but you won't get a repeat&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Once you die, that's it (unless you believe in reincarnation)&lt;br&gt;&#xA; I refuse to quote Drake here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17147" LastActivityDate="2016-02-07T19:05:52.293" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26221" PostTypeId="2" ParentId="26211" CreationDate="2016-02-07T19:54:54.023" Score="4" Body="&lt;p&gt;If you're not sticking to the North American or Ximenean standards then &quot;Is it permissible&quot; raises the question &quot;permitted by whom?&quot; or &quot;according to what rules?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we alter the question a bit to ask &quot;could it be published as a cryptic crossword clue in a national newspaper?&quot;, then since published clues include &quot;&lt;a href=&quot;https://www.theguardian.com/crosswords/crossword-blog/2011/nov/14/crossword-roundup-eggs-chestnuts&quot; rel=&quot;nofollow&quot;&gt;1d pi?!&lt;/a&gt;&quot; (answer: TURNIP) and &quot;&lt;a href=&quot;https://www.theguardian.com/crosswords/2013/nov/26/araucaria-puzzle-crossword-master-dies&quot; rel=&quot;nofollow&quot;&gt;Of of of of of of of of of of&lt;/a&gt;&quot; (answer: OFTENTIMES) - both of which omit the 'straight' entirely - I don't see any reason why putting it in the middle shouldn't be equally permissible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we're including British 'libertarian' cryptics then personally I think this would be ok.&lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="4050" LastEditDate="2016-07-08T22:24:05.203" LastActivityDate="2016-07-08T22:24:05.203" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26222" PostTypeId="2" ParentId="26147" CreationDate="2016-02-07T20:53:09.887" Score="0" Body="&lt;p&gt;Well it has everything:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; its just after midnight of the day your casino commission dealer license expired.  The shoe is a blackjack deck.  You left earlier but found yourself trapped by a weird &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Vera_Coking&quot; rel=&quot;nofollow&quot;&gt;artifact of a hotel-casino's ground footprint&lt;/a&gt; with no way out without crossing back through the casino floor.  Since your license expired you would be in serious trouble if found with gaming materials on the casino floor of the hotel. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11921" LastActivityDate="2016-02-07T20:53:09.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26224" PostTypeId="2" ParentId="26209" CreationDate="2016-02-07T22:15:10.780" Score="4" Body="&lt;p&gt;Not sure I deserve any credit for this answer, since I found it thanks to the massive additional hint from user &lt;a href=&quot;https://puzzling.stackexchange.com/users/12408/f&quot;&gt;f''&lt;/a&gt;.&#xA;Anyway, here is the correct answer, I think :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Point A is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; somewhere in the Four Corners region, where boundaries of four U.S. states meet: Arizona (SW, first sequence), Utah (NW, second sequence), Colorado (NE, third sequence) and New Mexico.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The numbers represent&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; county population, in descending order, for each of the four states.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, Sequence 1 translates to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Maricopa, Pima, Pinal, Yavapai, Mohave (or Yuma, I couldn't find 2007 source data, can't access to census.gov for some reason), ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sequence 2 translates to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Salt Lake, Utah, Davis, Weber, Washington, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sequence 3 translates to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Denver, El Paso (which is now the most populated county, it seems), Arapahoe, Jefferson, Adams, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sequence 4 translates to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bernalillo, Doña Ama, Santa Fe, Sandoval, San Juan, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The third sequence has&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 64 elements (Colorado has 64 counties in total).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17512" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2016-02-08T09:29:30.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26225" PostTypeId="2" ParentId="26168" CreationDate="2016-02-07T23:10:44.577" Score="3" Body="&lt;p&gt;It could be an&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alluvial fan&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am spread out like a leaf.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's spread out, and may appear to have &quot;veins&quot; much like a leaf. The leaf in the hint looks like a fan as well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That what forms me cuts me, but builds me up.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Streams erode rock, but carry the sediment/alluvium required to build up the alluvial fan&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That what cuts me seeks always to go lower, but lifts its bed higher.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Streams flow downhill because of gravity, but in doing so pick up the sediment which will build up the alluvial fan&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am commonly found where what builds me is less common.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water builds alluvial fans (most of the time), but they're often found in arid climates / desert regions&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17147" LastEditorUserId="17147" LastEditDate="2016-02-10T23:23:25.903" LastActivityDate="2016-02-10T23:23:25.903" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26226" PostTypeId="1" AcceptedAnswerId="26262" CreationDate="2016-02-08T00:06:43.617" Score="23" ViewCount="795" Body="&lt;p&gt;One token is placed on each square of an infinite checkerboard. One square is marked with an X. You want to get as many tokens on the marked square as possible. To do this, you may make any finite number of moves, where a move consists of choosing a square, placing two tokens on that square, and removing any three tokens of your choice from the eight surrounding squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the maximum number of tokens that you can have on the X?&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2016-02-08T00:13:15.227" LastActivityDate="2016-02-10T06:37:15.263" Title="Concentrating tokens on an infinite board" Tags="|mathematics|optimization|checkerboard|reachability|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26227" PostTypeId="1" AcceptedAnswerId="26238" CreationDate="2016-02-08T01:44:09.970" Score="6" ViewCount="485" Body="&lt;p&gt;&lt;sup&gt;&amp;lbrack;A variation of&amp;hairsp;:&#xA; &lt;a href=&quot;https://puzzling.stackexchange.com/q/3668&quot;&gt;What color was the  bear?&lt;/a&gt;$/$&lt;a href=&quot;https://puzzling.stackexchange.com/q/2611&quot;&gt;Where is your rescue partner?&lt;/a&gt;&amp;rbrack;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uYg5J.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From which latitudes on a theoretically spherical Earth&#xA;can you proceed as follows?&#xA;&lt;br&gt;&amp;bull;&amp;nbsp;&#xA;Set out by traveling due south for 90&amp;deg; of latitude (&amp;frac14; of Earth's circumference)&#xA;&lt;br&gt;&amp;bull;&amp;nbsp;&#xA;Then go due west for the same distance (at least 90&amp;deg; of longitude)&#xA;&lt;br&gt;&amp;bull;&amp;nbsp;&#xA;Complete a round trip by maintaining a northeast bearing&#xA;for &amp;radic;&amp;hairsp;2&amp;hairsp; times that distance&#xA;&lt;br&gt;&amp;bull;&amp;nbsp;&#xA;(Again be where you began,&#xA; as would be the case anywhere and for all distances on a&#xA; &lt;a href=&quot;https://en.wikipedia.org/wiki/Flatland&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;flat&lt;/em&gt;&lt;/a&gt;&#xA; &lt;a href=&quot;https://en.wikipedia.org/wiki/Flat_Earth&quot; rel=&quot;nofollow noreferrer&quot;&gt;Earth&lt;/a&gt;,&#xA; where the complete path would be a proper&#xA; &lt;a href=&quot;http://mathworld.wolfram.com/IsoscelesRightTriangle.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;isosceles right triangle&lt;/a&gt;&#xA; with a straight northeasterly hypotenuse)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes:&#xA;&lt;br&gt;&amp;bull;&amp;nbsp;&#xA;The northeasterly segment follows a 45&amp;deg;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Rhumb_line&quot; rel=&quot;nofollow noreferrer&quot;&gt;rhumb line&lt;/a&gt;&amp;hairsp;&amp;mdash;&amp;hairsp;not&#xA;a &amp;ldquo;straight&amp;rdquo; line or&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Great_circle&quot; rel=&quot;nofollow noreferrer&quot;&gt;great circle&lt;/a&gt;&amp;hairsp;&amp;mdash;&amp;hairsp;because&#xA;northeast is re-reckoned at each point along the path&#xA;&lt;br&gt;&amp;bull;&amp;nbsp;&#xA;No need to precisely calculate answers as long as they are&#xA;otherwise complete and convincing&#xA;&lt;br&gt;&amp;bull;&amp;nbsp;&#xA;This is an unvarnished geometry puzzle without regard to&#xA;the inaccuracy of a compass,&#xA;the irregularity or geography of Earth,&#xA;the limitations of real travel,&#xA;or the real risks of real-life&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Constant_bearing,_decreasing_range&quot; rel=&quot;nofollow noreferrer&quot;&gt;constant-bearing&lt;/a&gt;&#xA;adventure&#xA;&lt;br&gt;&amp;bull;&amp;nbsp;&#xA;Indeed, any resemblance to an existing Earth by the so-called Earth here,&#xA;or by the terminology of latitudes, longitudes,&#xA;rhumb lines and compass directions,&#xA;is purely for the sake of familiarity;&#xA;no actual axial antipodes were harmed (thank you, @question_asker)&lt;/p&gt;&#xA;" OwnerUserId="18129" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.630" LastActivityDate="2016-02-09T05:26:18.460" Title="Bearing due, due and back" Tags="|geometry|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="26228" PostTypeId="2" ParentId="26168" CreationDate="2016-02-08T02:17:54.290" Score="1" Body="&lt;p&gt;Is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sand &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's spread out like a leaf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That what forms me cuts me, but builds me up : &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; river/sea&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That what cuts me seeks always to go lower, but lifts its bed higher : &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; river/sea cuts sand and goes low by each layer of sand bed it makes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am commonly found where what builds me is less common :  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;desert &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5429" LastEditorUserId="5429" LastEditDate="2016-02-08T03:48:43.580" LastActivityDate="2016-02-08T03:48:43.580" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26230" PostTypeId="2" ParentId="26168" CreationDate="2016-02-08T04:55:51.960" Score="4" Body="&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; History&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am spread out like a leaf.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Referring the the leaves/pages of the books written to cover the histories of different era&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That what forms me cuts me, but builds me up.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; History consists of facts which are remodelled (cut and built) as per the current generation of rulers and historians.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That what cuts me seeks always to go lower, but lifts its bed higher.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Anyone who wishes to have his name marked in history tries to aggrandize him/herself and in the process belittle(lower) the other one. Say, Mughals invading Inda: could be aggrandized by the Mughal historians but belittled by the Indian rulers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am commonly found where what builds me is less common.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; History most times is about places or people who are no longer living (what builds me is less common i.e does not exist).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2656" LastActivityDate="2016-02-08T04:55:51.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26231" PostTypeId="2" ParentId="26063" CreationDate="2016-02-08T10:00:06.287" Score="6" Body="&lt;p&gt;It's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pi in binary. &lt;a href=&quot;http://www.befria.nu/elias/pi/binpi.html&quot;&gt;http://www.befria.nu/elias/pi/binpi.html&lt;/a&gt;&#xA; The two dots in the center represent the integer part, the marks at the circumference represent the fractional part.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19070" LastActivityDate="2016-02-08T10:00:06.287" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26233" PostTypeId="1" AcceptedAnswerId="26236" CreationDate="2016-02-08T14:23:58.477" Score="4" ViewCount="549" Body="&lt;p&gt;Imagine a world in which there are three kinds of electric charges instead of two. In this world, like charges repel and unlike charges attract. Let these types of charges be $A, B$ &amp;amp; $C$. Thus $A$ &amp;amp; $A$ repel, $A$ &amp;amp; $B$ attract, $A$ &amp;amp; $C$ attract and so on. Coulombs inverse square law is valid and superposition law is also valid like in our world. In this world like charges repel with force two times that of attraction between unlike charges. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A configuration of charged particles is &lt;em&gt;neutral&lt;/em&gt; if each particle experiences no net force. Which of the following configurations is neutral?&lt;a href=&quot;https://i.stack.imgur.com/jwrKt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jwrKt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Assumptions:&lt;/strong&gt; In option 2, it is an equilateral triangle. In option 4, it is regular hexagon. In option 5, it is a square. In option 6, it is regular pentagon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One or more than one options can be correct.&lt;/p&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="2627" LastEditDate="2016-02-08T20:53:42.993" LastActivityDate="2016-02-08T20:53:42.993" Title="In the world of three electric charges" Tags="|science|" AnswerCount="3" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="26234" PostTypeId="2" ParentId="26226" CreationDate="2016-02-08T14:30:38.927" Score="10" Body="&lt;h1&gt;General Proof&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Let $T_n$ be the number of tokens at a &lt;a href=&quot;https://en.wikipedia.org/wiki/Chebyshev_distance&quot; rel=&quot;nofollow&quot;&gt;Chebyshev distance&lt;/a&gt; $n$ from X; fancy way of imagining squares around X of increasing distance $n$ from X. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The values of $T_n$ are;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T_1=8$$&#xA;$$T_2=16$$&#xA;$$T_3=24$$&#xA;$$...$$&#xA;$$T_n=n \times 8$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, the goal is to move tokens towards X by replacing 3 tokens for 2.  It makes the most sense to take tokens farther from X and place them closer.  So, every move inward from a square will result in $\frac{2}{3}$ of the tokens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets call $V_n$ the &lt;em&gt;value&lt;/em&gt; of the tokens in $T_n$.  As you move tokens towards X, the number of them decreases by $\frac{2}{3}$ each move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see that;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V_n=T_n \times \left(\frac{2}{3}\right)^n = 8n\left(\frac{2}{3}\right)^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All we need to do now is solve the values of all the tokens, not forgetting the initial token at $X$..&#xA;$$Max = 1 + \sum_{n=1}^{\infty} V_n$$&#xA;$$Max = 1 + \sum_{n=1}^{\infty} 8n\left(\frac{2}{3}\right)^n=1+8\sum_{n=1}^{\infty} n\left(\frac{2}{3}\right)^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that $\forall x \in (-1,1)$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x)=\sum_{n=1}^{\infty} nx^n = \frac{x}{(x-1)^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, substituting $x=\frac{2}{3}$, we get;&#xA;$$Max = 1 + 8 \times f\left(\frac{2}{3}\right) = 1 + 8 \times \frac{\frac{2}{3}}{(\frac{2}{3}-1)^2}$$&#xA;$$Max = 1 + \frac{8 \times \frac{2}{3}}{\frac{1}{3}^2}$$&#xA;$$Max = 1 + \frac{8 \times 2 \times 9}{3}=49$$&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Finite maximum&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;So that is the theoretical number of tokens we can get to X.  Actually achieving this is another matter since it would require an infinite number of moves.  Thus, in any number of finite moves, the best we could achieve is 48 tokens.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that in order to get 48 tokens, we need to make moves that put 47 tokens on X.  However, you can't put an odd number since the rules state you can only remove 3 and add 2.  Thus, we can't achieve 48 directly by always moving inward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there another way to get tokens on X?  Thanks to &#xA;@julian-rosen for pointing it out.  Lets say you have this situation:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0000&#xA;0XYZ&#xA;0000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Where $X=47$, $Y=1$, and $Z=2$.&#xA;You can take one from X and the two off from Z, adding two to Y.  Then you will have 3 at Y, which you can remove and place 2 more at X, which gives you 48.  This also works if Y and Z are reversed, however that requires more tokens closer, so the first alternative should be easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in order to have 1 left over in the $n=1$ square and 2 left over in the $n=2$ square with 47 tokens on X, we need the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;n=0: $47-T_0=47-1=46$ tokens added. $46 \times \frac{3}{2}=69$. &lt;strong&gt;&lt;em&gt;Add one since we need an extra token in the $n=1$ square.&lt;/em&gt;&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;n=1: $70-T_1=70-8=62$: tokens added.  $62 \times \frac{3}{2}=93$.  &lt;strong&gt;&lt;em&gt;Add two since we need two extra tokens in the $n=2$ square.&lt;/em&gt;&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;n=2: $95-T_2=95-16=79$ tokens added. $79 \times \frac{3}{2}=120$ &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, we would need 120 tokens in the $n=3$ square.  We already know that the $n=3$ square has 24, so we'd only need an additional 96 tokens.  The value of these tokens is related to their distance from $X$ as we calculated before.  Thus, their value is $V=96 \times \left(\frac{2}{3}\right)^3=28\frac{4}{9}$.  From the above equation, we know that $8\sum_{n=1}^{\infty} n\left(\frac{2}{3}\right)^n=48$.  We also know that the value of the tokens in the $n=1$ spot is $8\times\frac{2}{3}=5\frac{1}{3}$, and those in the $n=2$ spot are $16\times(\frac{2}{3})^2=7\frac{1}{9}$, and the existing 24 tokens in the $n=3$ square are also $24\times(\frac{2}{3})^3=7\frac{1}{9}$.  Thus, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$8\sum_{n=4}^{\infty} n\left(\frac{2}{3}\right)^n=8\sum_{n=1}^{\infty} n\left(\frac{2}{3}\right)^n - 5\frac{1}{3} - 7\frac{1}{9} - 7\frac{1}{9}=48-19\frac{5}{9}=28\frac{4}{9}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, in order to get 96 tokens into the $n=3$ square would require an infinite number of moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, 48 tokens is not possible.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Construction method for 47 tokens&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Lets say you have a square centred on X whose Chebyshev distance to $X$ is $n$ or less.  We will denote all elements in this square to be the set $S_n$. Furthermore, we will access individual elements by their coordinates.  $S_n(0,0)$ will be the square in the top left.  $S_n(n,n)$ will be the element at X.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As well, we will denote perimeter squares whose Chebyshev distance to $X$ is exactly $n$.  These perimeter squares will be called $P_n$ and will employ the same coordinate system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will move all the tokens in $P_n$ to $P_{n-1}$ as follows.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Starting near the top left corner of $P_{n-1}$ (this is the square at $S_n(1,2)$), take all the tokens in the three squares above it (above to the left, directly above, and above to the right).  If divisible by 3, then move them all (using the put 2 remove 3 rule).  Otherwise, leave any remainder in the square to the right, or above.  Since every square in $S_n$ has at least one token initially, this is always possible.&lt;/li&gt;&#xA;&lt;li&gt;Move to the right one square and repeat.&lt;/li&gt;&#xA;&lt;li&gt;Keep going until you have completed the top row.  Now rotate the entire board counter clockwise, and continue the process.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Eventually, you will return to $S_n(1,1)$ and the only tokens left in $P_n$ will be the remainder after dividing by 3.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continue this process until you are left with a $3\times 3$ square around X.  Then move all the tokens into the middle using the remove 3 add 2 rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n$ is large enough, you will end up with a string of 2s in the remainder of each square leading to the centre with 45 tokens on $X$.  These can be collapsed to a string of 1s leading to the centre with a 47 on $X$ using julian's move.  The earliest this is possible is at $n=13$.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Example of 47 tokens&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Here is an example of how to get 47 tokens.  It uses the construction method outlined earlier; moving clockwise from the top left, move the outer tokens inward, propagating the remainder clockwise.  Any remainder from the outer ring of tokens is left over in the square under the top left square.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remainder tokens are shown as they are generated, but to keep the graphics small, they are ignored unless needed as we get closer to $X$.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;n=13                         n=12                       &#xA;111111111111111111111111111 1                           n=11               &#xA;111111111111111111111111111 11311311311311311311311313 1&#xA;111111111111111111111111111  1111111111111111111111111 113331313311331313311335&#xA;111111111111111111111111111  3111111111111111111111113  11111111111111111111111&#xA;111111111111111111111111111  1111111111111111111111111  31111111111111111111113&#xA;111111111111111111111111111  1111111111111111111111111  31111111111111111111113&#xA;111111111111111111111111111  3111111111111111111111113  11111111111111111111111&#xA;111111111111111111111111111  1111111111111111111111111  31111111111111111111111&#xA;111111111111111111111111111  1111111111111111111111111  11111111111111111111113&#xA;111111111111111111111111111  3111111111111111111111113  31111111111111111111113&#xA;111111111111111111111111111  1111111111111111111111111  31111111111111111111111&#xA;111111111111111111111111111  1111111111111111111111111  11111111111111111111113&#xA;111111111111111111111111111  3111111111111111111111113  11111111111111111111111&#xA;111111111111111111111111111  1111111111111111111111111  31111111111111111111113&#xA;111111111111111111111111111  1111111111111111111111111  31111111111111111111113&#xA;111111111111111111111111111  3111111111111111111111113  11111111111111111111111&#xA;111111111111111111111111111  1111111111111111111111111  31111111111111111111111&#xA;111111111111111111111111111  1111111111111111111111111  11111111111111111111113&#xA;111111111111111111111111111  3111111111111111111111113  31111111111111111111113&#xA;111111111111111111111111111  1111111111111111111111111  31111111111111111111111&#xA;111111111111111111111111111  1111111111111111111111111  11111111111111111111113&#xA;111111111111111111111111111  3111111111111111111111113  11111111111111111111111&#xA;111111111111111111111111111  1111111111111111111111111  31111111111111111111113&#xA;111111111111111111111111111  1111111111111111111111111  31111111111111111111113&#xA;111111111111111111111111111  3111111111111111111111113  51331133131331133131333&#xA;111111111111111111111111111  3113113113113113113113113  Outer ring tokens: 196&#xA;111111111111111111111111111  Tokens in outer ring: 164&#xA;Tokens in outer ring: 104&#xA;&#xA; n=10                  &#xA;1171313331331333311337 n=9                  &#xA; 311111111111111111113 3733331331333331339  n=8&#xA; 311111111111111111113 3111111111111111113 23933133333333133b n=7&#xA; 311111111111111111111 3111111111111111113  31111111111111113 3b133333333133b&#xA; 111111111111111111113 3111111111111111113  31111111111111111 311111111111113&#xA; 311111111111111111113 1111111111111111111  31111111111111113 311111111111113&#xA; 111111111111111111111 3111111111111111113  11111111111111113 311111111111113&#xA; 311111111111111111113 1111111111111111113  31111111111111113 311111111111113&#xA; 311111111111111111113 3111111111111111113  11111111111111113 311111111111113&#xA; 311111111111111111113 3111111111111111113  31111111111111113 111111111111113&#xA; 111111111111111111113 3111111111111111111  31111111111111113 311111111111113&#xA; 311111111111111111111 3111111111111111113  31111111111111113 311111111111113&#xA; 311111111111111111111 3111111111111111113  31111111111111113 311111111111113&#xA; 111111111111111111113 3111111111111111113  31111111111111113 311111111111113&#xA; 311111111111111111113 1111111111111111113  31111111111111113 311111111111113&#xA; 311111111111111111113 3111111111111111113  31111111111111111 311111111111113&#xA; 311111111111111111113 3111111111111111111  31111111111111113 311111111111113&#xA; 311111111111111111111 3111111111111111113  31111111111111113 d1333133333333b&#xA; 111111111111111111113 3111111111111111113  b133313333333313b Outer ring: 192&#xA; 111111111111111111113 9133313333333113339  Outer ring: 204&#xA; 913331331333311333317 Outer ring: 212&#xA; Outer ring: 210&#xA;&#xA; n=6&#xA; 3b3333333133d  n=5&#xA; 3111111111113 23b33333333d n=4&#xA; 3111111111113  31111111113 3b333333f  n=3     &#xA; 3111111111113  31111111113 311111113 13b3333f  n=2     &#xA; 3111111111113  31111111113 311111113  3111113  3b33f   n=1&#xA; 3111111111113  31111111113 311111113  3111113  31113  23bf&#xA; 3111111111113  31111111113 311111113  3111113  31113   313&#xA; 3111111111113  31111111113 311111113  3111113  31113   f3f&#xA; 3111111111113  31111111113 311111113  3111113  f333f   R: 68&#xA; 3111111111113  31111111113 311111113  f13333f  R: 92&#xA; 3111111111113  31111111113 f1333333d  R: 114&#xA; 3111111111113  d133333333d Ring: 136   &#xA; d13333333333b  Ring: 156&#xA; Ring: 176&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This leaves the final position before moving all the tokens to $X$ as:          &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;23bf&#xA; 3X3&#xA; f3f&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;With 68 tokens surrounding $X$.  The first move is to take the 2 remainder tokens and one of the tokens surrounding $X$ and replace them with 2 tokens as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;02bf&#xA; 5X3&#xA; f3f&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We now have 69 tokens surrounding $X$.  Trivially, we can remove all 69 and add 46 to $X$, leaving us with 47 tokens on $X$.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="1592" LastEditDate="2016-02-09T18:56:54.783" LastActivityDate="2016-02-09T18:56:54.783" CommentCount="21" ContentLicense="CC BY-SA 3.0" />
  <row Id="26235" PostTypeId="2" ParentId="26233" CreationDate="2016-02-08T14:39:57.457" Score="1" Body="&lt;p&gt;Oh man, I did this scenario in college!  Without the actual diagrams, of course.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 is neutral, but only if there is no C charge within finite distance of it.&lt;br&gt;&#xA;&lt;br&gt;&#xA; 2 is not neutral; the two A's fly apart.&lt;br&gt;&#xA;&lt;br&gt;&#xA; 3 is neutral and probably the most likely outcome.&lt;br&gt;&#xA;&lt;br&gt;&#xA; 4 is not neutral (I think) - can't do the math off the top of my head - but my suspicion is that it's too attractive and collapses inwards due to the attraction being based off distance squared rather than distance.&lt;br&gt;&#xA;&lt;br&gt;&#xA; 5 is not neutral; although the attractive forces balance out the repulsive forces in strength, they don't in direction.  All the charges fly away from each other.&lt;br&gt;&#xA;&lt;br&gt;&#xA; 6 is not neutral - there's nothing to stop A from collapsing towards the center.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17389" LastActivityDate="2016-02-08T14:39:57.457" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26236" PostTypeId="2" ParentId="26233" CreationDate="2016-02-08T15:29:35.010" Score="2" Body="&lt;p&gt;the force between electric charges are propotional to $\frac{1}{d^2}$ (distance between two charges.) So if charges are close to each other, the force is much higher. As a result (assuming the definition of neutral in the equation);&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1&amp;amp;3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 and 3 is &lt;strong&gt;neutral&lt;/strong&gt; since both electric charges attracts each other if there is no friction. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the angle is assumed to be 120 degrees between AAC, BAC and AAB, then the charge A in the middle will move towards between B and C with an initial accelaration then slow down since the distance between A-A getting lower. But the route cannot be foreseen since we do not know the distance and the weight of the charges. It can move back towards to A and go back again or just move outside of the range of B, C &amp;amp; A. &lt;strong&gt;NO Neutral.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is the tricky but easy after knowing $\frac{1}{d^2}$ rule. It is symmetric for every charge so I take A as an example; A - A force is so little since the distance between A-A. So A will tend to go to A with a force from two Bs and two Cs. It will happen all of the charges and they will touch each other and the end and becomes stable since there are more attracting force than pushing. &lt;strong&gt;No Neutral.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Same logic as 4, but this time the distance may be important. Let's call A-A pushin force as F and $F=\frac{1}{2d^2}$ if we call the distance between A-B as d. A-B attracting forces are 2F from both sides. so $2\sqrt(2)F-F$ will be the force that attracts A to the middle until they all touch to each other and stick together just like 4. &lt;strong&gt;No Neutral&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can easily say that A will move towards the middle of the hexagon and then move like a come and go forever. B and C will tend to move just like A but slower. still &lt;strong&gt;No Neutral.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18514" LastActivityDate="2016-02-08T15:29:35.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26237" PostTypeId="2" ParentId="26167" CreationDate="2016-02-08T15:34:58.907" Score="6" Body="&lt;p&gt;Guess &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Stubborn &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some call me royalty, but I say they're wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some modest Royal might say they're wrong in the normal context&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some call me crazy, but I say they're wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Being crazy, you never know you ARE crazy! &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some call me a ghost, but I say they're wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I have always been around! I'm no ghost! &quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some call me racist, but I say they're wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I don't like Indian people overall, but it's not a race issue&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some call me a hat, but I say they're wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I'm the brains under the hat&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; [&quot;Insert anything here&quot;] - but I say they're wrong.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19078" LastEditorUserId="19078" LastEditDate="2016-02-11T22:17:53.687" LastActivityDate="2016-02-11T22:17:53.687" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26238" PostTypeId="2" ParentId="26227" CreationDate="2016-02-08T16:52:58.177" Score="7" Body="&lt;p&gt;I'll assume that the Earth is a sphere of radius 1 and that the starting point is in the Northern hemisphere. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;In the accompanying Wikipedia article about the rhumb line, it tells us that &quot;The distance between two points, measured along a [rhumb line], is simply the absolute value of the secant of the bearing (azimuth) times the north-south distance&quot;. Hence, if our bearing returns us to the starting point, it will have the correct distance as specified in the question (since $\sec(\pi/4) = \sqrt{2}$). So, it suffices to find the latitudes for which the north-easterly journey returns us to the beginning. Now, if we travel south from a latitude of $\phi_1$ to a distance $\pi/2$ and then West for the same distance, the azimuthal angle we go through is $\frac{\pi}{2\sin(\phi_1)}$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another part of the Wikipedia article gives us that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The differential relationship between the azimuthal angle $\lambda$ and the polar angle $\phi$ as we travel along the rhumb line is $\frac{d\lambda}{d\phi} = \sec \phi$. Integrating both sides with respect to $\phi$ tells us that travelling along the rhumb from polar angle $\phi_0$ to $\phi_1$ corresponds to travelling an azimuth of angle $\log(\tan(\phi_1) + \sec(\phi_1)) - \log(\tan(\phi_0) + \sec(\phi_0))$. Now, as per the question, we want $\phi_0 = \phi_1 - \pi/2$, hence the azimuth difference in the last leg of the journey is $\log(\frac{\tan(\phi_1) + \sec(\phi_1)}{-\cot(\phi_1) + \csc(\phi_1)})$ which must be equal to the azimuthal angle $\frac{\pi}{2\sin(\phi_1)}$ we derived from travelling south then west. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So any latitude which works is the solution $x$ to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$\log(\frac{\tan(x) + \sec(x)}{-\cot(x) + \csc(x)}) = \frac{\pi}{2\sin(x)}$ Plotting them both, it seems like there is one solution at latitude $x \approx 1.019$ radians or a latitude of $58.4$ degrees, roughly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, there's a twist in the tale because, of course, as @human pointed out, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The above equation is only valid modulo $2\pi$ so we can add an integer multiple of $2\pi$ to the left hand side of the equation and also get a valid solution i.e, $\log(\frac{\tan(x) + \sec(x)}{-\cot(x) + \csc(x)}) + 2k\pi = \frac{\pi}{2\sin(x)}$ for $ k \in \mathbb{Z}$. The right hand side is strictly decreasing on the interval $(0, \pi/2)$ and the left hand side is convex with limits of $+\infty$ at both ends of the interval and a steeper slope at zero so for each value of $k$ there is one intersection point between the curves i.e, one solution. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For negative $k$&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The last leg of the journey &quot;encircles the Earth&quot; (in the sense of intersecting the same line of longitude) $k$ more times than the second leg. The solutions are always very close to $\pi/2$ i.e, the North Pole, even when $k=-1$. This is a result of the fact that the rhumb line spins infinitely many times as it approaches the North Pole.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For positive $k$&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The second leg &quot;encircles the Earth&quot; $k$ more times than the last leg. These solutions tend more gradually towards (zero) the equator as $k$ increases.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What about the limits?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As $k \rightarrow \infty$, the solution goes to the equator. At the equator, a $\pi/2$ movement south brings us to the South Pole from where all directions (except North) are ill-defined so we don't have a solution here. As $k \rightarrow -\infty$, however, the solution tends to the North Pole. The North Pole itself constitutes a valid and very interesting solution. After the first two legs of the journey we are at the equator. From there, travelling along the rhumb line, we wind around the North Pole an infinite number of times in order to reach it but do so in a finite distance ($\sqrt{2} \frac{\pi}{2}$). This is one of the incredible facts about the rhumb line. It is seemingly infinite at both ends but is finite in length (rather counterintuitive).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18422" LastEditorUserId="18422" LastEditDate="2016-02-09T01:15:09.727" LastActivityDate="2016-02-09T01:15:09.727" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26239" PostTypeId="2" ParentId="26233" CreationDate="2016-02-08T17:13:46.853" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 and 3 are neutral because they consist of unlike charges next to each other. In 2 and 6, all the other charges move respectively the A in the middle and at the top downwards. In 4, the force attracting the A at the top downwards is greater than one that would have repelled it upwards if the bottom A were in the middle, so there's no neutrality. In 5, a square, the force attracting the top left A to the bottom right is definitely greater than the force coming from farther and repelling it in the opposite direction, so none other than 1 and 3 is neutral.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastActivityDate="2016-02-08T17:13:46.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26240" PostTypeId="2" ParentId="26226" CreationDate="2016-02-08T17:55:42.347" Score="5" Body="&lt;p&gt;Here is a proof that the largest number of tokens that one can get on the X is 48. (I also suspect it can't be higher than 47, but Julian Rosen's comment to Trenin's answer shows that, if true, showing that would require a separate argument.) Of course I should note that this proof was inspired (as I believe the puzzle was too) by a very beautiful and fairly well-known math puzzle called &lt;a href=&quot;https://en.wikipedia.org/wiki/Conway%27s_Soldiers&quot; rel=&quot;noreferrer&quot;&gt;Conway's soldiers&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define an &quot;energy function&quot; that associates with each state $x$ of the board a positive number $E(x)$, given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E(x) = \sum_{\textrm{tokens } T} (2/3)^{d(T,X)},&#xA;$$&#xA;where $d(T,X)$ denotes the Chebyshev distance (the same one referred to in Trenin's answer) of the token to the X. Note that this is a convergent infinite series. Its value at the initial board state $x_0$ is&#xA;\begin{align}&#xA;E(x_0)&#xA;&amp;amp;= 1 + 8\cdot (2/3) + 16\cdot (2/3)^2 + 24 \cdot (2/3)^3 + 32\cdot(2/3)^4 + \ldots&#xA;\\&#xA;&amp;amp;= 1+8 \sum_{n=1}^\infty n \left(\frac{2}{3}\right)^n = 1+8 {\sum_{n=1}^\infty n\, t^n}_{\big|t=\frac23} &#xA;=1+8 {\frac{t}{(1-t)^2}}_{\big|t=\frac23} = 1+8\frac{\phantom{(}\frac23\phantom{)^2}}{\left(\frac13\right)^2}=49.&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma.&lt;/strong&gt; In any legal move, the energy can only decrease or stay the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof.&lt;/strong&gt; A legal move replaces three tokens, whose contribution to the energy sum is&#xA;$$ (2/3)^a + (2/3)^b + (2/3)^c, $$&#xA;with two tokens, whose contribution to the sum is&#xA;$$ 2\cdot(2/3)^d, $$&#xA;where $d$ is the Chebyshev distance between X and the position Y where the tokens are added, and $a,b,c$ is the Chebyshev distance between X and the position where the tokens are deleted. Since those positions are all neighboring positions to Y, all the numbers $a,b,c$ are between $d-1$ and $d+1$. If $a,b,c=d+1$ then the net change to the energy is&#xA;$$ 2\cdot(2/3)^d - 3\cdot(2/3)^{d+1} = 0, $$&#xA;i.e., the energy remains the same. In all other cases it is easily checked that the energy decreases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof that the number of tokens on the X can never exceed 48.&lt;/strong&gt; The energy of the board at the beginning is 49. By the lemma, after any finite sequence of moves the energy is still less than or equal to 49. This includes some positive amount of energy due to tokens that are far away and weren't touched by the sequence of moves, so in practice at most 48 units of energy that can be contributed to the energy sum by tokens at the X, each of which contributes one unit of energy to the sum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit: Added thoughts about how to achieve a matching lower bound.&lt;/strong&gt; From the proof it is apparent that if we are to have hope of constructing a solution that actually manages to bring 48, or close to it, tokens to the X, such a solution must minimize the loss in energy - too many steps that decrease energy can bring the energy down below 48 or 47, thus limiting the number of tokens we can bring to the X. In particular, this line of reasoning shows that the best strategy would consist of drawing tokens in closer and closer from concentric squares of increasing Chebyshev distance from X - precisely the kind of strategy envisioned in Trenin's answer, that uses only moves that keep the energy the same. Now, this is the general idea, but it seems like there are some highly nontrivial details involved in showing that one can actually manage to concentrate enough of the energy at the X without &quot;energy losses&quot; due to the numbers of tokens in one of the concentric squares not being precisely divisible by 3, or tokens not being close enough to each other to be usable together, etc. So, further arguments and perhaps even a computer-based search may be needed to completely nail down the precise answer to the question.&lt;/p&gt;&#xA;" OwnerDisplayName="user17947" LastEditorDisplayName="user17947" LastEditDate="2016-02-08T18:08:13.443" LastActivityDate="2016-02-08T18:08:13.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26241" PostTypeId="2" ParentId="26180" CreationDate="2016-02-08T19:17:46.957" Score="1" Body="&lt;p&gt;From the perspective of a single prisoner, countably infinite days are necessary in order for him to realize that he is no longer in neutral mode. This is because: neutral mode will never have an infinite number of consecutive days without change. Likewise, one of the other two modes can only be recognized after countably infinite days where no change has occurred. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, after countably infinite days: each prisoner will realize that he is no longer in neutral mode. He will realize that his light has been on (or off) forever. At this point, if it is on, then he will say that they are in bright mode; if off he will state that they are in dark mode (this is the strategy the mathematicians agree upon before entering the cell). Two of the three mathematicians will be correct in their statement, therefore they will escape.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main thing to realize is that it works because countably infinite days have passed. This answer will not work if only a finite number of days have passed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 2/9:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above solution I assumed the lighting mode was static, but in the case it's not, the solution is similar. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each mathematician simply chooses his answer based on the preponderance of on or off days. If he sees a greater number of on (off) days, then he assumes bright (dark) mode. If the mode is distributed randomly among the light bulbs, then all three will choose correctly because each room will have 2 days on (off) for every day off (on). If the mode is not distributed randomly, then two of the three prisoners will choose correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT 2/10: generalizing: let's say the mode is bright. the following is true; one case is where every room has 2 bright days for every 1 dark day. Another case is where two specific rooms have all bright days and one specific room has all dark days. All other cases are in between: that is, a specific room has a bright-to-dark ratio &amp;lt; 2:1 and could even be &amp;lt; 1:1, while the two other rooms always have a bright-to-dark ratio &gt; 1:1. So, again each mathematician when asked what mode they're in replies with whatever he observed to be the preponderance in his cell - two of the three will be right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: 2/11 AS others have pointed out, I've been basing my answer on a random distribution, but the problem as stated doesn't restrict it, so I need to rethink my answer. (Answer works for random distribution only).&lt;/p&gt;&#xA;" OwnerUserId="19088" LastEditorUserId="19088" LastEditDate="2016-02-11T18:05:27.170" LastActivityDate="2016-02-11T18:05:27.170" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="26243" PostTypeId="2" ParentId="26167" CreationDate="2016-02-08T22:27:57.013" Score="1" Body="&lt;p&gt;I'm half-hoping that this isn't the intended solution, but given the &quot;wordplay&quot; tag, I strongly suspect that you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; something that a person who is easily offended by profanity may not want to read. Consider yourself warned before reading further.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Still reading? Okay, I think you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Fucking&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some call me royalty, but I say they're wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;You're calling me a &lt;em&gt;king&lt;/em&gt;? Fuck king!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some call me crazy, but I say they're wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;You're calling me crazy? &lt;em&gt;You're&lt;/em&gt; fucking crazy!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some call me a ghost, but I say they're wrong.&lt;br&gt;&#xA;Some call me racist, but I say they're wrong.&lt;br&gt;&#xA;Some call me a hat, but I say they're wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure about the OP's precise intention, but I think there's some similar wordplay in action here - e.g. &quot;I'm not a fucking racist&quot; to convey not just ordinary disagreement but vehement disagreement, or &quot;fuck 'hat&quot; as an approximate homophone for &quot;fuck that&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user17947" LastActivityDate="2016-02-08T22:27:57.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26245" PostTypeId="1" AcceptedAnswerId="26377" CreationDate="2016-02-08T23:51:57.727" Score="-5" ViewCount="763" Body="&lt;p&gt;You are in a mansion, stuck in there armed with only a baseball bat that can only take 100 more hits. Then you are at the mercy of the War Ghosts, ghosts that can take as many hits according to color. Due to evolution and natural selection, 99 of them turned camo, able to take 5 hits per ghost. One is pale, which can be killed with one hit. The pattern and hits are shown respectively:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Camo -&gt; Blue -&gt; Red -&gt; Yellow -&gt; Pale&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The above means that a Camo ghost will turn blue after one hit, red with another, yellow with another, pale with another, and dead with another. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;With only your baseball bat, how can you kill them all with only 100 shots at them?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt;: Hits cannot go through ghosts so swinging a hundred times through them will not work. Also, one ghost per hit. The laws of physics has been bent and ricochets and recoil does not exist. The ghosts can live forever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; You will die of old age, &quot;war&quot; is a fancy adjective that will not affect the ghosts' lives, and the questions is: How can you kill all 100 ghosts with only you baseball bat as the only weapon where physics disregards force, recoil, and rebound?&lt;/p&gt;&#xA;" OwnerUserId="9042" LastEditorUserId="9042" LastEditDate="2016-02-09T22:08:30.410" LastActivityDate="2016-02-12T01:05:47.860" Title="The 100 War Ghosts" Tags="|lateral-thinking|" AnswerCount="7" CommentCount="16" ClosedDate="2016-02-12T08:42:04.193" ContentLicense="CC BY-SA 3.0" />
  <row Id="26246" PostTypeId="2" ParentId="26218" CreationDate="2016-02-09T00:23:41.937" Score="6" Body="&lt;p&gt;Pretty sure the message is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; YOU WIN! PASSWORD DENIED!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Found by using the numbers given to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; trace out individual letter patterns on the number pad of a keyboard.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So specifically, letter by letter:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 745692 = &lt;strong&gt;Y&lt;/strong&gt;&lt;br&gt;&#xA; 78963214 = &lt;strong&gt;O&lt;/strong&gt; (though arguably needs an extra 7 at the end)&lt;br&gt;&#xA; 7412369 = &lt;strong&gt;U&lt;/strong&gt;&lt;br&gt;&#xA; ---&lt;br&gt;&#xA; 74125369 = &lt;strong&gt;W&lt;/strong&gt;&lt;br&gt;&#xA; 852 = &lt;strong&gt;I&lt;/strong&gt;&lt;br&gt;&#xA; 1478963 = &lt;strong&gt;N&lt;/strong&gt; (looks like a lower case N)&lt;br&gt;&#xA; ---&lt;br&gt;&#xA; 9874156 = &lt;strong&gt;P&lt;/strong&gt; (though probably should have an extra 9)&lt;br&gt;&#xA; 1478963 = &lt;strong&gt;A&lt;/strong&gt; (though it actually looks the same as the &quot;N&quot; above)&lt;br&gt;&#xA; 987456321 = &lt;strong&gt;S&lt;/strong&gt;&lt;br&gt;&#xA; 987456321 = &lt;strong&gt;S&lt;/strong&gt;&lt;br&gt;&#xA; 74125369 = &lt;strong&gt;W&lt;/strong&gt;&lt;br&gt;&#xA; 78963214 = &lt;strong&gt;O&lt;/strong&gt; (needs another 7)&lt;br&gt;&#xA; 7418953 = &lt;strong&gt;R&lt;/strong&gt;&lt;br&gt;&#xA; 741268 = &lt;strong&gt;D&lt;/strong&gt;&lt;br&gt;&#xA; ---&lt;br&gt;&#xA; 741268 = &lt;strong&gt;D&lt;/strong&gt;&lt;br&gt;&#xA; 98741235 = &lt;strong&gt;E&lt;/strong&gt; (though you have to squint a bit)&lt;br&gt;&#xA; 1478963 = &lt;strong&gt;N&lt;/strong&gt; (as above)&lt;br&gt;&#xA; 852 = &lt;strong&gt;I&lt;/strong&gt;&lt;br&gt;&#xA; 98741235 = &lt;strong&gt;E&lt;/strong&gt;&lt;br&gt;&#xA; 74126 = &lt;strong&gt;D&lt;/strong&gt; (edit that fixed the end of &quot;password&quot; seems to have dropped the last &quot;8&quot; though to close the D)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...though I've got to admit, I don't really understand the purpose of the &quot;Who am I?&quot; in the title, or the actual meaning of the message, so it's possible there's more to it.&lt;/p&gt;&#xA;" OwnerUserId="5208" LastActivityDate="2016-02-09T00:23:41.937" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26247" PostTypeId="2" ParentId="26178" CreationDate="2016-02-09T01:56:10.513" Score="3" Body="&lt;p&gt;Are you:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; electricity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;pusher:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; electricity is often &lt;a href=&quot;https://learn.sparkfun.com/tutorials/voltage-current-resistance-and-ohms-law/voltage&quot; rel=&quot;nofollow&quot;&gt;analogized&lt;/a&gt; to water pressure; thus the sense of it being a &quot;pusher&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;main keyboard on the org chart:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;I cannot make this connection&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the one you want to follow:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you'd want electricity to be at the company first for lighting and HVAC; you'd turn the lights on before/as you entered a building.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;without me the corporation would be a cruel, dark place, and would quickly collapse:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; electricity -&gt; lighting for the &quot;dark&quot; piece; HVAC for being a &quot;cruel&quot; place to work; and so if no one worked at a corporation, it may soon collapse.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;generous and kind:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; see above lighting and HVAC; also, electronics of all kinds (computers, cell phones) that enable people at a corporation to do work&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;have a sinister side:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the inherent risks of electricity (electrocution; short circuiting, etc)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;arrested for some back-room deal:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; by &quot;arrested&quot; I would imply &quot;captured&quot; or &quot;stock-piled&quot; as energy in UPS batteries; UPS' are often housed in a &quot;back room&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;everyone was shocked!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; hopefully not by the UPS, but maybe by static electricity!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13098" LastActivityDate="2016-02-09T01:56:10.513" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26248" PostTypeId="1" AcceptedAnswerId="26249" CreationDate="2016-02-09T02:43:02.010" Score="23" ViewCount="3964" Body="&lt;p&gt;One day, the chief of the dwarves decided he wanted to test his tribe.  So that night, he told the dwarves that he would paint on each dwarf's back a dot colored either red or blue. Each dwarf will know everyone else's dot color, but not their own.&lt;/p&gt;&#xA;&lt;p&gt;Every dwarf with a red dot on his or her back is to go to the dining hall on the N&lt;sup&gt;th&lt;/sup&gt; day, where N is the number of dwarves with a red dot on their backs.  The presence of any blue-dotted dwarves at the dining hall on the N&lt;sup&gt;th&lt;/sup&gt; day constitutes a failure.&lt;/p&gt;&#xA;&lt;p&gt;Furthermore, after the dwarves get their backs painted, they are not allowed to communicate using any means, including (but not limited to) speaking, punching, and holding mirrors.  No dwarf is allowed to know what color he is until after the trial is over. They do not also get to know if someone went to the hall on any of the 1 to N-1 days.&lt;/p&gt;&#xA;&lt;p&gt;The dwarves can meet on the day before the trial in order to talk strategy.  What strategy should they use?&lt;/p&gt;&#xA;&lt;p&gt;Note the question is not the same as the &lt;a href=&quot;https://puzzling.stackexchange.com/questions/236/in-the-100-blue-eyes-problem-why-is-the-oracle-necessary&quot;&gt;blue-eyes puzzle&lt;/a&gt; . &lt;strong&gt;In the blue eyes puzzle, one would get get to know if someone went to the hall on days 1 to N-1 .Here, nobody gets to know this.&lt;/strong&gt; This is a crucial difference.&lt;/p&gt;&#xA;" OwnerUserId="18144" LastEditorUserId="57212" LastEditDate="2022-08-31T15:29:30.640" LastActivityDate="2022-08-31T15:29:30.640" Title="The Dwarves' Test" Tags="|logical-deduction|strategy|word-problem|" AnswerCount="2" CommentCount="12" ContentLicense="CC BY-SA 4.0" />
  <row Id="26249" PostTypeId="2" ParentId="26248" CreationDate="2016-02-09T03:04:43.023" Score="50" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each dwarf counts the number of red dots they see on everyone else's backs. If that number is $x$, they go to the hall on the $x+1$th day only.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A dwarf with a red dot will&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;count $N-1$ red dots, and show up on the $N$th day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A dwarf with a blue dot will &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;count $N$ red dots, and show up on the $N+1$th day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, on the $N$th day, all dwarves with red dots are present, and all dwarves with blue dots are not.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2016-02-09T03:13:04.343" LastActivityDate="2016-02-09T03:13:04.343" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="26250" PostTypeId="2" ParentId="26245" CreationDate="2016-02-09T03:39:13.360" Score="12" Body="&lt;p&gt;The camo ghosts are camouflage, i.e. they will appear as whatever colour is behind them. Therefore, you would lure the pale ghost and a camouflage ghost together and hit the camo ghost when the pale ghost is behind it. You repeat this 98 more times with the other camo ghosts, and then finally hit the real pale ghost for 100 hits.&lt;/p&gt;&#xA;" OwnerUserId="19095" LastActivityDate="2016-02-09T03:39:13.360" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="26251" PostTypeId="1" AcceptedAnswerId="26264" CreationDate="2016-02-09T07:33:28.483" Score="17" ViewCount="3633" Body="&lt;p&gt;&lt;strong&gt;Where:&lt;/strong&gt; Hackanoogie Prison. &lt;strong&gt;When:&lt;/strong&gt; Who cares? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Master Key Keeper had had a really good day and in a fit of insanity and niceness decided to let all the inmates to go free if they manage to carry out the following task.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Guys&quot;, he calls over the P.A system (it's very informal there on Tuesday mornings) &quot;During the night the Key Keepers were extremely bored and crept into everyone's rooms and painted a red or blue dot on every persons nose&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;We will open the door to the chocolate-fountain room (as it is the largest room in the prison) and everyone must go inside&quot;. He paused to take a swig from the ever-smoking ever-present flask by his side. &quot;If the red-dots and the blue-dots successfully manage to sort themselves out and stand on two opposite sides of the room -&quot; he paused for dramatic effect &quot;you will all be let free!&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Needless to say, you cannot possibly see what is on your, or anyone else's, noses right now because you are currently standing in pitch black darkness. You may discuss a strategy now, but when you enter the room (which will be lit) - No talking, winking, burping or any communication of any kind. &lt;strong&gt;Oh! by the way as soon as you take your place - we will mace you (the painless, raspberry flavor  type) and you will get blinded!&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Off you go&quot; he giggles as he falls backwards through window and into the Venus flytrap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lo and behold! the inmates manage to complete the task successfully. How did they do it?   &lt;/p&gt;&#xA;" OwnerUserId="19086" LastEditorUserId="57212" LastEditDate="2023-06-17T23:49:43.210" LastActivityDate="2023-06-17T23:49:43.210" Title="Red and Blue in the Chocolate Fountain Room" Tags="|logical-deduction|strategy|hat-guessing|" AnswerCount="9" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="26252" PostTypeId="2" ParentId="26251" CreationDate="2016-02-09T07:57:50.363" Score="6" Body="&lt;p&gt;Everyone will line up in a straight line.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; If you don't see two types of people, stand at the end.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Otherwise,&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; stand in between the two groups.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This way, everyone will know their dot color based on how the next person stands. The last person gets no information about their color, so it's just a 50/50 shot when the two groups walk to the sides of the room.&lt;/p&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="63414" LastEditDate="2020-12-18T13:10:08.783" LastActivityDate="2020-12-18T13:10:08.783" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="26253" PostTypeId="2" ParentId="26251" CreationDate="2016-02-09T08:04:59.997" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They will randomly split into two groups (Group 1 and Group 2). One random person from (let say) Group 1 will count the number of red and blue dots from the other side. There are three possibilities: same amount of red and blue colored noses, red is more or blue is more;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Possibility 1 (Red and Blue is more)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If blue is more, that one random (chosen maybe, does not matter since everyone needs to move one by one) person (let say Person 1) from Group 1 will switch any red one from Group 2 with himself. Then one another person (Person 2 from Group 1) will check Person 1's nose and if Person 1's nose is red, then he will switch himself with him. As a result Person 1 will know his nose is red and stay in Group 1. Otherwise, Person 2 will switch himself with someone else whose nose is red from Group 2. and this goes on. At the end the group 1 will be all red, and group 2 will be all blue. Same thing can be applied to If red is more.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Possibility 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1 If the number of people whose nose has red or blue dots are the same in Group 2, one random person from Group 2 will randomly switch himself with someone from Group 1. If the balance does not change Person 2 will switch himself with Person 1. So Person 1 will know his nose color and stay in Group 2. and this will go on just like in Possibility 1 after one color overrides.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18514" LastEditorUserId="18514" LastEditDate="2016-02-09T08:11:30.283" LastActivityDate="2016-02-09T08:11:30.283" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="26254" PostTypeId="1" AcceptedAnswerId="26255" CreationDate="2016-02-09T09:21:31.847" Score="41" ViewCount="5038" Body="&lt;p&gt;Everyone knows the &lt;strong&gt;Pythagorean theorem&lt;/strong&gt;: In a  right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;The Pythagorean Theorem:&lt;/strong&gt; $$a^2+b^2=c^2$$ &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Recently, professor Halfbrain has proved a theorem that he calls the &lt;strong&gt;Non-Pythagorean theorem&lt;/strong&gt;. This theorem considers a right-angled triangle, in which the two sides adjacent to the right angle have lengths $a$ and $b$ (just as in the classical Pythagorean theorem), together with a mysterious quantity that the professor calls $x$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;The Non-Pythagorean Theorem:&lt;/strong&gt; $$a^{-2}+b^{-2}=x^{-2}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This puzzle asks: What's the geometric role of this mysterious quantity $x$ in the right-angled triangle? &lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2016-02-09T14:21:14.783" Title="The Non-Pythagorean Theorem" Tags="|mathematics|geometry|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26255" PostTypeId="2" ParentId="26254" CreationDate="2016-02-09T09:38:03.763" Score="58" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $x$ is the altitude $h$ from side $c$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $a^{-2} + b^{-2} = \frac{1}{a^2} + \frac{1}{b^2} = \frac{a^2+b^2}{a^2 b^2} = \frac{c^2}{a^2 b^2},$&#xA;&#xA; hence $x = \frac{ab}{c}$. &#xA; The area of the triangle is $\frac{ab}{2}=\frac{ch}{2}$, so $x=h$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18922" LastActivityDate="2016-02-09T09:38:03.763" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26256" PostTypeId="2" ParentId="26245" CreationDate="2016-02-09T09:45:45.543" Score="5" Body="&lt;p&gt;(That's a long shot)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hit all 99 camo ghosts once, wait a while, then if one remains, hit him.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;${}$&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The riddle states that the ghosts turned camo through evolution and natural selection, so non-camo ghosts have a harder time surviving. Either they kill each other, or they have a predator in the mansion. If they have a predator, turning them all to blue should allow it to kill them all. If they prey on one another, in the end there shall be only one, and you have to hope that it will be weakened to pale.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18922" LastActivityDate="2016-02-09T09:45:45.543" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26257" PostTypeId="2" ParentId="26123" CreationDate="2016-02-09T09:53:50.953" Score="3" Body="&lt;p&gt;Everybody already stated that which question is to be asked to get right answer from Truthteller or Liar with the question below;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If I were to ask you ____, would you answer yes?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the second part is the hardest part I believe, so I tried to get a better answer. I divided hundred doors into 64 and 36 ones, then 36 doors into 16 and 20, then 20 doors into 8 and 12, then 12 doors into 8 and 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as a result, I started to ask the question above;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If I were to ask you that the right door is among the first 64 doors, would you answer yes?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So if you got yes answer, it is just $0.95^7$, but you have $64 \ percent$ chance to get yes answer. But to chance to get no is $36 \ percent$. It is not that low. So I divided 36 into different parts as stated above and found the answer below;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0.95(\frac{64}{100}0.95^6+\frac{36}{100}0.95(\frac{16}{36}0.95^4+\frac{20}{36}0.95(\frac{8}{20}0.95^3+\frac{12}{20}0.95(\frac{8}{12}0.95^3+\frac{4}{12}0.95^2))))=0.70863$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is how to find the optimum, which is the same answer as f&quot; but simpler to understand :)&lt;/p&gt;&#xA;" OwnerUserId="18514" LastEditorUserId="18514" LastEditDate="2016-02-09T12:19:23.573" LastActivityDate="2016-02-09T12:19:23.573" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26258" PostTypeId="2" ParentId="26251" CreationDate="2016-02-09T10:08:49.460" Score="3" Body="&lt;p&gt;Step 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One person is chosen to make the first move. If he sees an even number of red noses he goes to the left, if he sees an odd number he goes to the right. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Step 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Now everyone else counts the number of red noses, excluding the first person (in case he is red). If they agree with his choice (odd or even), they are blue. If not, a red nose is missing, which has to be them self.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Step 3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Depending on the color of the first person, the people with the same color will join him, and the rest will go the the opposite side.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19103" LastActivityDate="2016-02-09T10:08:49.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26259" PostTypeId="2" ParentId="26251" CreationDate="2016-02-09T10:18:42.670" Score="49" Body="&lt;p&gt;The prisoners agree that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each prisoner who sees an &lt;strong&gt;even&lt;/strong&gt; number of blue noses&#xA; goes to a predetermined side of the room&#xA; (perhaps the side with the door through which they entered).&#xA; Each prisoner who sees an &lt;strong&gt;odd&lt;/strong&gt; number of blue noses goes to the&#xA; opposite side of the room.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Each prisoner with a blue nose will:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; See either an even or an odd number of blue noses,&#xA; one fewer than the total of blue noses.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Each prisoner with a red nose will:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; See The total number of blue noses, which also will be either even or odd,&#xA; but not the same as the number that the blue-nosed prisoners see.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18129" LastEditorUserId="18129" LastEditDate="2016-02-09T11:14:45.360" LastActivityDate="2016-02-09T11:14:45.360" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26262" PostTypeId="2" ParentId="26226" CreationDate="2016-02-09T11:32:29.343" Score="8" Body="&lt;p&gt;Following in the footsteps of Big Black Box's and Trenin's answers, this shows very concretely that...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $47$ &lt;em&gt;can&lt;/em&gt; indeed be achieved.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I tried a lot of different approaches to this problem, but ultimately it's taking the courage to do all the boring calculations that gets the job done. I set up an imgur album &lt;a href=&quot;https://i.stack.imgur.com/AR30G.jpg&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; with pictures that contain a (mostly) step by step solution that achieves $47$ tokens on the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the beginning of the process there's plenty of room for choice, which might make it a bit difficult to follow. Still, checking its validity should boil down to performing elementary calculations according to the pictures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The description of each image (in particular 'From $P(19)$ to $P(18)$') should help the reader understand what's going on. The notation used is also similar to Trenin's; in particular, $P(n)$ refers to squares at Chebyshev distance exactly $n$ from the center (the $X$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think $48$ is possible, and while Trenin provides a reasoning for why it can't happen in a certain situation, I don't believe it covers all cases. I'm too tired now, but I will think on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: A rigorous proof that $48$ cannot be achieved.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Proof that $48$ cannot be achieved.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;We first point out a few observations. Suppose there are $t$ tokens on $P(n+1)$; if we wish to move them to $P(n)$, the best we can do is moving $\left \lfloor \frac23 t \right \rfloor$ tokens. Studying the parity of $t$ modulo $3$, it's easy to see what's happening:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;When $t \equiv 0$ mod $3$, we're simplying transfering them all.&lt;/li&gt;&#xA;&lt;li&gt;When $t \equiv 1$ mod $3$, we leave one token behind.&lt;/li&gt;&#xA;&lt;li&gt;When $t \equiv 2$ mod $3$, we remove the remainder two tokens from $P(n+1)$ and a single token from $P(n)$ to place two new tokens at $P(n)$, for a net gain of one token.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It's easy to check that these cases agree with the previous formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we'll be looking at the reverse process. We will generally consider the lowest amount of tokens (lowest energy configuration) required when backtracking. As a consequence of this and the previous observation, reversion is done by $x \longrightarrow \left \lceil \frac32 x \right \rceil$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there were $48$ tokens on the $X$, then $47$ of them must have come from $P(1)$. Thus, before the transfer there must have been $\left \lceil \frac32 \cdot 47 \right \rceil = 71$ tokens on $P(1)$. Since at the beginning there were $8$ tokens on $P(1)$, $71-8=63$ of the tokens must have come from $P(2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can iterate this process, producing a sequence $a_n$ of numbers that describe how many tokens &lt;strong&gt;at least&lt;/strong&gt; on $P(n)$ must have come from $P(n+1)$. Put another way, $a_n$ is the minimum amount of tokens added to $P(n)$ (beyond those already there at the beginning; the tokens are added from $P(k)$'s with $k&amp;gt;n$) required for $48$ to be achieved on $X=P(0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sequence is defined by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left\{ \begin{array}{l}&#xA;a_0 = 47 \\&#xA;a_{n+1}=\left \lceil \frac32 a_n \right \rceil – 8(n+1)&#xA;\end{array} \right.$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Claim:&lt;/strong&gt; $a_n = 16n+47$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We prove by induction. The base case $n=0$ checks out, so we proceed to the inductive step. We have that $\frac32 (16n+47) = 24n + 70.5$, whence $=\left \lceil \frac32 (16n+47) \right \rceil = 24n + 71$. Now, a simple calculation shows that $24n + 71 -8(n+1) = 16(n+1)+47$, so the sequences satisfy the same recursion, and thus indeed coincide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem here is that this sequence is never $0$ or less. Effectively, this shows that no matter how large we choose $n$, we will always need to work with $P(k)$'s yet farther from $P(n)$. In other words, one cannot transfer enough tokens to the $X$ in finite time/within a finite distance from $X$.&lt;/p&gt;&#xA;" OwnerUserId="16478" LastEditorUserId="16478" LastEditDate="2016-02-10T06:37:15.263" LastActivityDate="2016-02-10T06:37:15.263" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26264" PostTypeId="2" ParentId="26251" CreationDate="2016-02-09T11:53:56.487" Score="19" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Any two prisoners enter the room and stand side by side in the middle of the room. All other prisoners either have to:&lt;br&gt;&#xA;&lt;br&gt;&#xA; - Position themselves on either side of the already positioned prisoners, if they all have the same color.&lt;br&gt;&#xA; - Position themselves inbetween two prisoners, that have different colored noses.&lt;br&gt;&#xA;&lt;br&gt;&#xA; After all prisoners are in place, they will be perfectly separated by color.&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; The trick here is that even though at the end of the day we will have one line of people - all the blues will be on one side and all the reds will be on the other. The 'two opposite sides of the room' doesn't have to be completely separate and wherever the blue and red meet, draw a mental line down the room and they will ,ipso facto, be standing on opposite sides of the room!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1485" LastEditorUserId="1485" LastEditDate="2016-02-09T13:06:55.257" LastActivityDate="2016-02-09T13:06:55.257" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="26265" PostTypeId="1" AcceptedAnswerId="26279" CreationDate="2016-02-09T13:24:08.877" Score="5" ViewCount="246" Body="&lt;p&gt;Every letter stands for a digit in decimal representation, &#xA;different letters stand for different digits:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;      MEN + CAN = REACH - MOON&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which digit does each letter represent?&lt;br&gt;&#xA;(Please present the full analysis how these digits can be determined.)&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="9404" LastEditDate="2016-03-03T10:19:39.417" LastActivityDate="2016-03-03T10:19:39.417" Title="Neil Armstrong and Buzz Aldrin alphametic" Tags="|no-computers|alphametic|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26266" PostTypeId="2" ParentId="26251" CreationDate="2016-02-09T13:44:19.083" Score="3" Body="&lt;p&gt;I came up with this answer, which can be generalized nicely for multiple colors:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Inmates are assigned numbers $1$ through $n$. In the room, every inmate $i$ determines the parity of blue noses of all other inmates, which we call $b_i$ and is either even ($e$) or odd ($o$). Inmate $1$ then chooses the left side of the room if $b_1 = e$, and right otherwise (or any similar scheme). Inmates $2$ through $n$ (in that order) know $b_1$ from the side that inmate $1$ chose, as well as $b_i$. They follow him if $b_i = b_1$, and choose the opposite side otherwise.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So there one bit of information is &quot;communicated&quot; without explicit communication, which helps solve the problem for two colors. For more than two ($m$) colors, ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...we would need to transmit $m-1$ bits of information ($m-1$ parities), because the final parity can be obtained knowing the total number of inmates. The room must have $m$ different spots to allow for $m$ groups to separate. If these spots can be ordered (encoded) somehow, then by choosing one spot, inmate $1$ can transmit $\log_2 m$ bits of information (using an encoding similar to the one above). So as long as $\log_2 m &amp;gt; m - 1$, this is enough information for the others to know inmates $1$s observed parities of all colors, and hence their own color. Unfortunately, this does not hold for $m &amp;gt; 2$.&#xA; &#xA; So one needs an additional constraint for an arbitrary number of colors: The room must have at least $2^{m-1}$ definable spots. For smaller $m$, this should be doable; for example, $m=4$ means $8$ spots are necessary, that's just how many directions a compass has.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10628" LastEditorUserId="10628" LastEditDate="2016-02-09T19:25:17.997" LastActivityDate="2016-02-09T19:25:17.997" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26269" PostTypeId="1" AcceptedAnswerId="26272" CreationDate="2016-02-09T14:36:15.087" Score="6" ViewCount="994" Body="&lt;p&gt;Earlier today my friend, a mathematician, gave me a white bag with this equation on it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;SIΛ°H&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What on Earth is in the bag?&lt;/p&gt;&#xA;" OwnerUserId="18940" LastEditorUserId="8874" LastEditDate="2016-03-22T09:07:30.353" LastActivityDate="2016-03-22T09:07:30.353" Title="An interesting equation" Tags="|riddle|word|lateral-thinking|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26271" PostTypeId="2" ParentId="26269" CreationDate="2016-02-09T14:42:21.683" Score="11" Body="&lt;p&gt;The bag is full of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Flour.&lt;br&gt;&#xA; You were holding it upside down, it actually says &quot;HOVIS&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18522" LastActivityDate="2016-02-09T14:42:21.683" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="26272" PostTypeId="2" ParentId="26269" CreationDate="2016-02-09T14:44:00.733" Score="7" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The bag is full of bread.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Argument:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HoVIS Ltd is a British company that produces flour and bread:&lt;br&gt;&#xA; &lt;a href=&quot;https://en.wikipedia.org/wiki/Hovis&quot;&gt;https://en.wikipedia.org/wiki/Hovis&lt;/a&gt;&#xA;&lt;br&gt;&#xA; The inscription &quot;SIΛ°H&quot; is &quot;HoVIS&quot; upside down.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2016-02-09T14:44:00.733" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26273" PostTypeId="2" ParentId="25110" CreationDate="2016-02-09T14:49:49.677" Score="7" Body="&lt;p&gt;I think the answer is either &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;See you again&quot; or &quot;Young, Wild and Free&quot; by Wiz Khalifa&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Writing out the letters of the alphabet underneath the letters as they appear on a keyboard gives a method of encrypting/decrypting information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;QWERTYUIOPASDFGHJKLZXCVBNM&lt;br&gt;&#xA;ABCDEFGHIJKLMNOPQRSTUVWXYZ&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under these rules, using for decryption, we have that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; V $\rightarrow$ W, O $\rightarrow$ I, M $\rightarrow$ Z, A $\rightarrow$ K, I $\rightarrow$ H, Q $\rightarrow$ A, S $\rightarrow$ L, Y $\rightarrow$ F.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hence the message from the Good Samaritan becomes &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; WIZ KHALIFA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Looking at his discography, I reckon the song is either&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;See you again&quot; referring to seeing the martians on their return visit to Earth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Young, Wild and Free&quot; which was a song Wiz recorded with Snoop Dogg but also featured Bruno MARS.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18422" LastActivityDate="2016-02-09T14:49:49.677" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26275" PostTypeId="2" ParentId="26083" CreationDate="2016-02-09T15:31:34.007" Score="3" Body="&lt;p&gt;I am:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BMW&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Orion Nebula and I have a common abbreviation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; With the hint types and information, and the popularity of M42, it is a BMW engine, which is listed in the link given&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I love to start with every single digit integer ranged from 1-9 but can't start with 9&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BMW models start with every number except 9. 2 series, 3 series, etc&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First version of one of my capabilities has a common number with a popular boeing model&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BMW Dixi is the first car made by BMW, which has a 747cc engine. Reference to Boeing 747&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A Mercedes Benz automobile model and what I did, have a common abbreviation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BMW S55 Engine and Mercedes Benz S55 model&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Born in the early 1900s&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BMW was founded in 1916&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18685" LastEditorUserId="18685" LastEditDate="2016-02-09T21:25:12.183" LastActivityDate="2016-02-09T21:25:12.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26276" PostTypeId="2" ParentId="26269" CreationDate="2016-02-09T15:52:02.337" Score="3" Body="&lt;p&gt;It's a bag of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hovis bread - the &quot;equation&quot; is the name (EDIT: just saw it in the supermarket and see the 'o' is lower case) &quot;HoVIS&quot; upside down and (EDIT: &lt;em&gt;perhaps&lt;/em&gt;) scrunched up a little.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastEditorUserId="10118" LastEditDate="2016-02-09T21:06:53.713" LastActivityDate="2016-02-09T21:06:53.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26278" PostTypeId="1" AcceptedAnswerId="26393" CreationDate="2016-02-09T16:29:03.657" Score="17" ViewCount="3324" Body="&lt;p&gt;The room is pitch black; the only thing giving any sort of light is the moon. You look around, but can’t seem to see much at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suddenly, music starts playing and giant rats appear out of nowhere, jumping and prancing around! The way they move is so elegant that they make you fall under a small trance; you can’t seem to look away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You notice that something else is coming from behind the giant rats, marching closer and closer. As it gets closer you can start making out what it is: a giant robot-looking figure holding a gun. It points the gun at the giant rats, preparing to shoot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just when you thought this couldn’t get any weirder, ballerinas leap out of nowhere, jumping here and there, back and forth. You try to wrap your head around all this but it is just too much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The music gets louder and louder; you try to move but you can’t. As the music gets to an unbearable volume that makes you just want to shoot yourself, it stops. Clouds get blown in, covering up the moon and the moonlight. You hear this faint chatter that quickly turns into a roar of chattering and clapping sounds: hundreds of people talking and clapping all at once. The lights slowly light up the room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(CHOOSE ONE)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Passive:&lt;/strong&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That was the first time you had ever seen the classic tale of The Nutcracker.&lt;br&gt;&#xA;&lt;br&gt;&#xA; “Wow that was pretty good. I don’t know what the classic story is about, but this play sure has piqued my interest. But next time I should DEFINITELY not stand right next to the speakers,” you say to yourself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Aggressive:&lt;/strong&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That was the first time you had ever seen the classic tale of The Nutcracker.&lt;br&gt;&#xA;&lt;br&gt;&#xA; “Wow, that play sucked! Now I DEFINITELY do not want to see the original!  But next time I should DEFINITELY not stand right next to the speakers,” you say to yourself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Everyone slowly disperses and goes back to what they were originally doing. A fellow co-worker walks up to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Why the glum face? You’re the number one detective in the world right now, and you’re on vacation! Smile and enjoy yourself a little. That title and a vacation like this is a once-in-a-lifetime opportunity! Tell you what, tonight for dinner, the first round is on me if you come skiing with us,” he says.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“I guess I should try to enjoy myself, and a drink wouldn’t hurt later on,” you think to yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Sure, I’ll come”.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Great! We’ll meet you at the entrance to the ski lift in 20 minutes,” your co-worker says. You get changed into your skiing gear and headed towards the ski lift, where you meet your fellow co-workers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After about an hour of skiing with your co-workers, you decide to take one last round down the hill before heading back to your room to warm up. As you sit on the ski lift patiently waiting for it to reach the top of the hill, you hear a loud scream. It is coming from the direction of the forest right next to the ski hill, about half way down. You notice that everyone is talking and pointing towards the forest where the scream came from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you finally reach the top of the hill, you quickly ski halfway down and hop over the small fence which says “DO NOT ENTER.”  As soon as you hop over, you see five teenagers run out of the forest. You manage to stop one of the teenagers; you take a quick glance at him: male, blond hair, about 6’ 2”, wearing a thick leather winter jacket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“What happened?” you ask the young teenager.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Holy shit man!  There’s a dead body over there!” he says in a panicked voice as he points deep into the wintry forest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You quickly head over to the area the young male was pointing to. When you finally reach the area, you see a body lying on the ground. You slowly approach the body and kneel down to take a better look at the body. You notice that it’s a male, with curly brown hair, wearing a red sweater and dark blue jeans. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suddenly, you hear voices coming towards you; it is your co-workers and some police officers. The officers start calling back to HQ reporting this incident. Your co-workers start taking pictures on their cell phones and scouting the area for any clues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suddenly, there is a big rumble in the ground; a loud unfamiliar sound approaches. You look up and see that it is an avalanche!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“AVALANCHE! EVERYBODY RUN!” you yell. Everyone starts running; you quickly look around the area for any clues and noticed that there is a top hat near the body. You run towards the top hat and grab it, then quickly turn around and run towards the body. You know that you don’t have much time left, so you quickly search the body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(CHOOSE TWO)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Front Left Pocket:&lt;/strong&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You quickly grab the object in his front left pocket and pull it out. You notice that it is his cell phone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Front Right Pocket:&lt;/strong&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You quickly grab the object in his front right pocket and pull it out.  You notice that it is his keys.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Back Left Pocket:&lt;/strong&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You place your hand into his back left pocket and don’t feel anything in his pocket.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Back Right Pocket:&lt;/strong&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You quickly grab the object in his back right pocket and pull it out. You notice that it is his wallet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After you have searched the second pocket, you know that you won’t have time to check another, so you started running down the hill. As you run full-speed down the hill, you take a quick glance behind you to see that the avalanche has violently swept up the body and all other evidence with it. You know that you won’t be able to outrun the avalanche and it’s only a matter of time before it catches up to you. All of a sudden you see a big shadow hanging over you, something big hits your back, and you black out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You open your eyes. The only thing you recall is the avalanche hitting you. You slowly raise yourself into a sitting position on the bed. You look around the room, but it is an unfamiliar room, a room which you have not seen before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A doctor walks in. “Ah, I see you are finally awake. You took a pretty big hit from that avalanche. How are you feeling?” asks the doctor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“I've had better days, but thank you. How long was I out for?” you reply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“About two to three hours; not too long. The rescue team worked quickly on finding you and the others and brought them here for recovery.  And thank God they did! By the way, your team says that when you are ready, they will meet you back at the lodge’s main lobby.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Thanks Doc.”  You quickly head over to the lodge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upon arrival, your co-workers run up to you, asking if you are OK and saying that they are glad nothing has happened to you. They inform you that the rescue team is still out searching in case someone else is out there under the snow, but since they do not know how many people got hit by the avalanche, they do not know when they can stop looking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You ask about the body. “The body still has not been found, and might not be found. It was a big avalanche,” one of your co-workers says. They also inform you about the clues that were found upon rescuing you.  They point to a table across the room and you walk over to check them out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the table are the items you managed to grab before the avalanche consumed everything in its path. One of the items is the mysterious top hat. You examine the top hat. It is black, made out of silk, and very clean, as if it were brand new.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Nothing seems to be out of the ordinary,” you mutter to yourself. You place the top hat in a Ziploc bag and send it to the forensics team for further investigation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(Choose The Same Two Pockets You Chose Earlier At The Crime Scene)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Front Left Pocket:&lt;/strong&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The victim's smartphone is white with a see-through light-green phone case. You unlock the phone and start checking for any recent activities. You find out that the victim’s name is Terry W. Ter. He doesn’t seem to have any videos, text messages, or incoming/outgoing phone calls. The only thing he has are pictures of him outside either skiing, snowboarding, or ice fishing, with trees covered in snow. The weird part is that the pictures are in albums that he has renamed to “Winter 2013”, “Winter 2014”, and “Winter 2015”.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Front Right Pocket:&lt;/strong&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You examine the victim’s keys.  There are 3 sets of keys on the key chain. One key is for a Nissan Altima, one key is for his mail box, and the last key might be his house keys but you can’t be 100% certain just by looking at it. Checking the parking lot for a Nissan Altima, you come across a green one and decide to try opening it with the key; it opens! You check the car for any clues but it seems to be kept in a very clean condition. The only thing you find is a parking ticket with the car's license plate number (WNTR1 5H34R), the car model, and the victim's name: Terry W. Ter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Back Left Pocket:&lt;/strong&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You remember that you didn’t find anything in the back left pocket of the victims body.  “Why did I pick the back left pocket? No one ever keeps anything in that neglected pocket,” you sigh to yourself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Back Right Pocket:&lt;/strong&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You examine the victim’s wallet.  It is a black leather wallet, which seems to be brand new. You check inside the wallet for any clues, but the only card that you find in there is a gift card for a sauna place called “HOT AS HELL SAUNA”. There is a number on the back of the card for the place and you decide to call it. You find out that the card belongs to Terry W. Ter, and that he has already used the gift card once.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After examining all the clues you have gathered, you think to yourself, “I knew a vacation was too good to be true, maybe next year”. Just when you were about to head back to your room to think about this case some more, the owner of the lodge comes up to you and your team and offers the party room as a temporary home base for all the investigation work. He also offers security so no one can go in unless it is a member of the investigation team or the police. You and your team happily accepted the owner’s offer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You notice that the owner is about 5’ 9”, has brown hair, is skinny, and is dressed up pretty fancy. The sky is getting darker and darker as you head back to your room to get a good night’s rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You suddenly hear a loud buzzing noise with music. “RING RING RING RING,” goes your phone as it wakes you up. You slowly open your eyes, give a big yawn, and answer your phone. “Hello?” you answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“We have a situation; we have found another dead body. Come to the laundry area in the basement quick,” says one of your co-workers as she hangs up the phone. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You quickly get dressed and find your way towards the laundry room. As you enter through the door you see your team and police officers already surrounding the area. There are some small yellow cones around the body. You walk up to the body to get a better look.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a young women, 23-26 years old, brunette, around 5’ 5” to 5’ 7”, and with brown eyes. She is wearing the staff uniform of the ski lodge, so you get one of the police officers to get the owner of the ski lodge there for more information. You check the body but there aren’t any stab or gunshot wounds. Her neck does seem a bit red and puffier than usual, but you conclude that it was not strangulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You then start looking at the clues that the others have found. On the table there is a plate with a single carrot.  It seems that someone has taken a bite out of the carrot. With nothing else on the carrot, you move on to the next clue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(CHOOSE ONE)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clue 1:&lt;/strong&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You walk up to the item on the table.  It is the victim’s wallet. You open it up, but there is nothing inside the wallet except for a piece of paper. You take a closer look at the paper.  It is the victim’s temporary driver’s license. Her name is Sally N. Bhal, and her date of birth is December 14, 19**. The last two digits of her birth year have been scratched out, as well as the remainder of the card.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clue 2:&lt;/strong&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You walk up to the item on the table.  It is a piece of paper. You open the piece of paper.  It says “Things To Buy” underlined as the title, and under it is “- Buy Scarf”. It was written with a blue pen, and whoever wrote this had bad hand writing; you can tell they were struggling. It doesn’t seem like a female's writing. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As you are about to take a look at the other clue, out of the corner of your eye you see that one of the police officers slip and fall on the edge of the table. The officer falls hard, pushing the table away, and knocking the clues off the table. The carrot falls on the ground and the plate shatters into tiny pieces, while the two other clues fall right into the lit fireplace right beside you. The clues quickly burn, and there is nothing left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the others help the officer up, the owner of the ski lodge and the other police officer come through the door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“I heard you were looking for me?” says the owner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Yes, I was wondering, do you by any chance know this young woman? She is wearing the ski lodge uniform,” you reply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The owner takes a look at the body, then his eyes open wide. “Oh shit!” he says, “that’s Sally! Yes, I do know her; she works here at the lodge part time cleaning the rooms, and washing the clothes and bed sheets and all that.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You take a look at the body and then look at the owner and ask, “Did she have any enemies? What else do you know about her?”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Not too much; she’s just another employee.  We don’t talk too much. She is very nice and I would say liked by all.  I don’t see her having many enemies,” says the owner as he looks away and looks around the room. “… I do know that she loves animals, especially dogs, she likes the colour cyan, she lives not too far from here, and she is really allergic to carrots.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You look at the owner with curiosity and ask, “How do you know that she is allergic to carrots?”.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Well, every employee has to fill out this form with their personal information and on the sheet there is a question asking if they are allergic to anything so we can be more cautious about what task we assign to them. May I go now? I am quite busy running the lodge at the moment.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Alright you’re free to go. By the way, I never got your name,” you reply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Oh how rude of me!  My name is Vladimir J. Englebel,” he replies as he takes his leave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You stand there for a minute thinking about the clues you have found so far, but nothing is really connecting in your head. You decide to head back to your room to think about this some more.  As you walk out the laundry doorway, you bump into someone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Excuse me, mate,” says the man, as he continues to mop the floor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You look at the man you accidentally bumped into. He is about 5’ 6”, has black hair, and is wearing the grey ski lodge janitor uniform with his name on the left side of his shirt. “Namon” it says.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You apologize for bumping into him. “It’s OK,” he replies. You see his eyes shift over a little and you notice the man’s facial expression change. After a couple of seconds he just continues with what he was doing before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You then noticed that Mr. Englebel was right behind you; you didn’t notice him approaching. “What’s wrong with him?” you ask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“His mother passed away not too long ago; he must still be sad about it. He really loved him mom, but I mean everyone does right?” Mr. Englebel replies. You nod and then proceed to your room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you lie in bed thinking about the case, you decided to go talk to Vladimir. You walk up to Vladimir’s office and knock on his door. “Come in,” he says.  You enter his office.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Hi Mr. Englebel” you say.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Come in, please, take a seat. What can I do for you?”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“I have a plan to find out who the murderer is, but I will need you to provide me a list of everybody who has been here since the murder. Also, I would like you to close down the ski lodge until this case has been solved. I would like you to ask each room to pack and leave the lodge ASAP one at a time.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“If you think that is what’s best, then sure. I will provide you that list in a couple of hours,” the owner says.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Oh, and one more thing Mr. Englebel: keep this a secret and don’t let anyone else know.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You wait for a couple of hours back in your room, when suddenly your room phone rings, and you answer the phone. “I have compiled that list for you. Please come to my office and I will give it to you.” Vlad says as he hangs up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You head to Vlad’s office. “Here you go, and this list has everyone that has been here since a couple of days prior to the first murder,” he says.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Thanks. Now, I would like you to hold onto this list.  Go and tell the occupants of each room to pack and leave the premises. After each leaves I want you to cross their names off the list.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vlad agrees and starts asking the occupants of each room to leave the ski lodge. After a couple of hours, you still see a lot of people leaving. All of a sudden, you hear loud knocking at your door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“We have an emergency! We have found another body! Everyone is waiting for you in the boiler room,” yells one of your co-workers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You quickly rush to go find Vlad. As you approach him you slow your pace so as not to alarm the customers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Excuse me, Vlad, may I talk to you for a second in private?” you whisper as you pull him aside. “We have just found another dead body.  I will need you to stop asking people to leave, because now we have just eliminated a bunch of people who aren’t the killer, which means that someone on the remainder of the list is the murderer.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He looks surprised. “Another body found?  First the avalanche and now this! This is definitely not good for my business.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You ask Vlad for the list and quickly head to the crime scene. When you arrive, you see everyone crowded around talking, and a body lying in the middle of the room. As you get closer, you notice that there is a small tree branch in the victim’s armpit. “Looks like the victim bled to death, the armpit is a vital spot,” you mutter to yourself. You take a closer look at the tree branch, but it just seems to be a regular tree branch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(CHOOSE TWO)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Check Body:&lt;/strong&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You check the victim’s body. Upon closer inspection, you noticed that the victim is bald, about 5’ 11”, and normally built.  He is wearing jeans and a red sweater. He also has really dark brown eyes, so dark that they can almost pass as black. There isn't any other kind of marking on him, except for the tree branch wound.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Check Pocket:&lt;/strong&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You check the victim’s pockets.  In his back right pocket you find his wallet. You open and search it for any clues, but the only thing you find is his ID card. His name is Andy K. Cane, 6’ 1”, and he has his G license. Everything else has been scratched out on the card. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Check Clue:&lt;/strong&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You check the clue on the ground beside a small yellow cone.  As you get closer to it, you noticed tiny pieces of hair. Examining them closer with a magnify glass, you can see that the hairs are about 1.5” to 2” long, and there are brown and black hairs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All of a sudden, you hear a faint sound.  It sounds like a snake hissing, but it doesn’t stop. People inside the room start coughing and some even mention that they are feeling dizzy. “Look, there is a gas leak over there!” yells one of the police officers. Looking up, you see a large-diameter yellow pipe with a label saying, “Danger: toxic gas.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everyone starts to evacuate the room immediately, as the gas is dangerous. As you exit the room, one of your co-workers quickly closes the door so that the gas does not escape. After a couple of seconds, you hear a loud “BANG!”. You look through the window on the door to see that a portion of the pipe has exploded and gas is quickly filling the room. You know that you won’t be going back in there for a while, so you decide to go back to your room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While you walk back to your room, you think to yourself, “I don’t know why, but is it just me or does it seem that every time I try to take a look at the last clue, some mysterious phenomenon happens?” The more you think about it, the more it starts to hurt your brain, so you decide just to forget about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After you arrive back at your room, you take a look at the list you received from Vlad. “There are still 35 people on the list. 35 suspects to interrogate. Well, better start investigating some people,” you say to yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You take a look at the first name on the list: “Amber E. Nog.” You ask Vlad and some other co-workers to help set up an interrogation room. After a room has been found and set up, you call in Amber, the first suspect on the list. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Amber enters the room, you notice that she is about 5’ 5”, skinny, has red curly hair, and is wearing dark blue jeans with a grey sweater. She sits down. “Hi, Mr. Detective,” she says.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Hi. Amber E. Nog, correct?” you reply. She nods in silence as she stares at you -- not in a fierce way, buts more like she's interested. “When did you arrive at this lodge?” you ask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“I arrived a couple days ago with my friends.  We were just trying to have some fun during our holiday.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Do you know anyone at this lodge? Are you related to anyone?” you ask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“No, the only people I know are my friends that I came with,” she replies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You notice that she has something sticking out of her sweater pocket.  “What’s that in your sweater?” you ask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Oh this? This is my E-book tablet,” she says as she pulls it out to show you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“What kind of books do you like reading?” you ask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Mainly mysteries and detective cases. I am currently reading this one book about a detective who can solve any case by placing his left hand on the victim's head to get a glimpse of what happened. Then the detective has to find evidence to convict the murderer. It’s very interesting,” she says.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Alright, you are dismissed” you tell Amber.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next person to walk in is in the ski lodge uniform, so you immediately know she works here. You ask for her name, occupation, and whether she as seen or heard anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“My name is Cindy Silver and I work here as a cleaner. I mostly just clean the rooms after the customers have left, to prepare for the next customers. I haven’t really heard or seen anything, just the usual,” she says as she introduces herself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a quick glance you notice that she is about 5’ 6”, has black hair recently dyed a lighter brown colour, is wearing the ski lodge’s work uniform for females, is around 18 to 21 years old, and is on the skinny side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Do you by any chance recognize or have you seen anyone recently wearing a top hat like this?” you ask Cindy as you show her the top hat you found earlier that week.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Hmm… nope, I haven’t seen anyone wear a top hat like that around here recently,” she answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Alright you are free to go.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Cindy gets up, you notice that she pauses for a second midway through getting up, as if she is in pain. “You alright?” you ask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Yeah, I just injured myself the other day. I was lifting something heavy.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“You should talk to Vlad about that. He’ll get you the medical care that you need,” you tell Cindy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“I have already talked to him; our medical plan doesn’t cover this, but that’s okay -- it’s not as bad as you think it is,” she replies with a smile as she walks out the door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You call for the next person to come in. As he enters the room you notice that he is about 6’ 3”, jacked, has brown hair, and is wearing jeans and a red plaid long-sleeved shirt. When he sits down, you asked him to tell you a little about himself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Well, my name is Jack Jones and I am a lumberjack. I came here for vacation with some buddies from work,” he says.  When you hear that he is a lumberjack, all you can think of is the stick you found earlier on one of the victims' bodies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“May we search your luggage and belongings, Mr. Jones?” you ask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Sure, but you won’t find anything suspicious, if that’s what you’re thinking,” Jack replies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You start searching through his belongings, finding nothing but the usual items people would take on vacations like clothes, and underwear. You then spot a secret compartment in the luggage.  You unzip the pocket to find a Swiss army knife. It isn’t your usual Swiss army knife though; it is much larger than usual. You check out each tool built into this gadget. It has scissors, a knife, tweezers, a mini-axe, a flashlight, a bottle opener, a hook, and nail clippers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“So, why did you bring something like this to a vacation spot?” you ask Jack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Me and the guys were gonna camp outside for a couple days here at the lodge; there is a camping spot just down the road,” Jack replies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You ask Vlad if he owns such property and he confirms that he does have a campground down the road from the lodge. He does not recall if anyone has booked a campsite since he has been so busy; one of the other workers might have booked Jack in instead of Vlad himself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You decide to call in the next person.  As he opens the door to come in, you noticed that he is dressed very nicely: Black dress pants, full black suit, grey vest, white dress shirt, silver tie, and even dress shoes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Good afternoon, Detective, how may I help you?” he asks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“You’re dressed up rather nicely today,” you remark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“This is how I dress every day.  My name is Strom E. Hail and I am the CEO of the famous Power Tech Corp.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“I see.  Anyhow, are you here on vacation? Visiting? Did you come here with family or friends?” you ask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Yes, I am here on vacation with my family.  We heard good reviews about this place and we wanted to try it out. I guess the reviews aren’t what they say it is,” Mr. Hail replies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“What do you mean?”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Well, this place is filthy, the service is terrible, and the people here have no manners, but I guess that is what you get for the price you pay. Me and my family wanted to go to a much fancier place but they were all full so we decided to try a second rate ski lodge instead.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You ask, “Where is your family now?”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“They returned to our home a couple of days ago.  They did not feel like staying in such a rambunctious place any longer. I stayed because I have some business to attend to, and wanted to complain to the manager, but he was never in his office,” he replied. After a couple more questions, you decided to call in the next person. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You look at who is next on the list and to your surprise it is Vlad, the owner of the ski lodge. You hear the doorknob turn, the door slowly swings open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“You must be surprised to see me on the list, no?” he says as he enters the room. Since you already know Vlad, you skip the formalities and ask him if he has seen anything suspicious or has had anything strange happen recently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Nothing strange or suspicious has happened recently,” Vlad says after thinking about it. “Besides all this, nothing out of the ordinary”.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Have any of your employees started acting strange or weird lately?” you ask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Not really.  Everything has been fine. I don’t hear any complaints at all from them,” he replies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“This guy probably doesn’t know anything since he is stuck in the office all day anyways,” you think to yourself. You decided to let him go and ask for the next person to be brought in. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the next person enters the room, you noticed that he is wearing an employee uniform from the ski lodge. “Another employee? I guess Vlad wrote down the names of all the employees as well,” you muttered to yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Please take a seat.  Can you tell me a little about yourself?” you ask the employee.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“My name is Shig Bovel and I am fifty-nine years old. I have worked at the ski lodge in maintenance for thirty-five years now,” Shig replies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Thirty-five years is a long time.  Has anything big happened recently?” you ask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“I don’t know if you would call this recent, but five years ago Vlad’s father died. The benefits used to be so much better; they covered you and your family for everything. Now, we have limited benefits and if you want something more you have to request for it.  Most requests get rejected due to high cost,” replies Shig.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You look at the clock on the wall behind Shig, and to your surprise, time has just flown by. You wanted to get a chance to interview the others before too much time passed, but it seems like that won’t be happening. You decided to interview one last person before calling for a break and then resuming the list. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Alright, you’re free to go,” you tell Shig.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Oh, thank you,” he replies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the last person for the first half of the interviews comes in, he face seems  familiar. At a quick glace you see that the person is male, has blonde hair, is around 6’ 2”, and is more on the bulky side. After a couple of seconds, it hits you. “Aren’t you that young man that I stopped when you and your friends ran out of the forest after you found a dead body?” you ask the teenager.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Yeah, that was me. My name is Shane Shady by the way,” Shane replies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Nice to meet you Shane,” you say as you shake his hand. “Now, tell me. What were you and your friends doing in the forest in the off-limits area?”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Well, we just kinda went off to explore a bit and wanted to find a nice area where we could come back later that night and set up a campfire and stuff.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“You know that area is off-limits, right?” you ask Shane, but you get no reply -- just utter silence. “I’ll give you one chance, Shane. Did you and/or your friends commit the murder?” you ask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“No way, man! It wasn’t me, it wasn’t any of us,” he quickly replies. You look him in the eye while he tries to avoid yours; you can see in his face that he is not lying, but does seem shady in a way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Alright, you’re free to go,” you tell Shane. “Thank you!” he says with a big smile and a look of relief.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You decide to take a small break, and ponder it for a bit. Then it hits you. You know who the murderer is!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Who is the murderer?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4645" LastEditorUserId="5373" LastEditDate="2016-09-06T13:27:50.813" LastActivityDate="2022-09-13T14:47:34.720" Title="Mysterious Murder Mystery 5" Tags="|story|mystery|situation|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26279" PostTypeId="2" ParentId="26265" CreationDate="2016-02-09T16:29:29.003" Score="5" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 902 + 382 = 10836 - 9552&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solution. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First there are the &quot;gimme&quot; digits.&lt;br&gt;&#xA;&lt;br&gt;&#xA; We can rewrite this as &lt;code&gt;MEN + CAN + MOON = REACH&lt;/code&gt;. A crude upper bound for the right hand side is &lt;code&gt;9999 + 999 + 999&lt;/code&gt; which is &lt;code&gt;11997&lt;/code&gt; yielding R = 1. Then we know that E is also either 0 or 1; digit-distinctness forces E = 0. We use this to pin down M as either 8 or 9; 7 is too small as 7999 + 709 + 999 does not clear 10,000. So those are the easy constraints here.&lt;br&gt;&#xA;&lt;br&gt;&#xA; After that you can condense two pairs of digits together and simplify the problem considerably:&lt;br&gt;&#xA;&lt;br&gt;&#xA; Writing out what we know, &lt;code&gt;m0n + can + moon = 10ach&lt;/code&gt;. Reading this modulo ten, we see &lt;code&gt;3n = h (mod 10).&lt;/code&gt; Since 3 and 10 are coprime, the map &lt;code&gt;x -&amp;gt; 3x&lt;/code&gt; is a permutation of the digits &lt;code&gt;0..9&lt;/code&gt; modulo 10. So there is only one degree of freedom here, and we can write &lt;code&gt;m00 + ca0 + moo0 = 10ac0 + w&lt;/code&gt; where &lt;code&gt;w = (3*n-h)/10.&lt;/code&gt; More explicitly here is a table: &lt;pre&gt;&lt;code&gt;     n        =   0    1    2    3    4    5    6    7    8    9&#xA;     h        =   0    3    6    9    2    5    8    1    4    7&#xA;     allowed? =   X    X        m=8        X   m=9   X   m=9  m=8&#xA; (3*n - h)/10 =             0    0    1         1         2    2&lt;/code&gt;&lt;/pre&gt; We can see that &lt;code&gt;w&lt;/code&gt; is bounded between 0 and 2, and lets us look up a couple pairs of (h, n) numbers. This will be crucial for quickly brute-forcing the rest. When we divide the whole equation by 10 we now have:&lt;br&gt;&#xA; &lt;pre&gt;&lt;code&gt;    m0 + moo = 1000 + (ac - ca) - w&lt;/code&gt;&lt;/pre&gt;&#xA; Now comes our second pair that we can condense: we rewrite &lt;code&gt;ac - ca&lt;/code&gt; as &lt;code&gt;9 * (a - c)&lt;/code&gt;. Defining &lt;code&gt;x = a - c&lt;/code&gt; we can see that &lt;code&gt;x&lt;/code&gt; is bounded between 9-2 = +7 and 2-9 = -7, so &lt;strong&gt;it can be negative&lt;/strong&gt;, which is important. We will be using it to take certain guesses for &lt;code&gt;o&lt;/code&gt; and try to bring them within &lt;code&gt;0,1,2&lt;/code&gt; for a value of &lt;code&gt;w&lt;/code&gt;. Our current equation is just:&lt;br&gt;&#xA; &lt;pre&gt;&lt;code&gt;    11*o  - 9 * x + w = 1000 - 110 * m.&lt;/code&gt;&lt;/pre&gt;&#xA; Now we are ready to investigate particulars.&lt;br&gt;&#xA;&lt;br&gt;&#xA; For m=8 we see that &lt;code&gt;11 * o - 9 * x + w = 120.&lt;/code&gt; Clearly &lt;code&gt;o &amp;gt; 1&lt;/code&gt; from our distinct-digit constraint, and since &lt;code&gt;w&lt;/code&gt; is so small we will need negative &lt;code&gt;x&lt;/code&gt; to fight to bring this 120 down to the appropriate range. Here is a table of the possibilities:&#xA; &lt;pre&gt;&lt;code&gt;     o          =     2    3    4   5   6   7   8   9&#xA;     120 - 11*o =    98   87   76  65  54  43  32  21&#xA;     -9*x       =    90   81   72  63  54  36  27  18&#xA;       w        =     8    6    4   2   0   7   5   3&lt;/pre&gt;&lt;/code&gt;&#xA; So we see that when m=8 we have only two possibilities: o=5 and o=6.&lt;br&gt;&#xA;&lt;br&gt;&#xA; When m=8, o=5 we have &lt;code&gt;c = a + 7&lt;/code&gt; and &lt;code&gt;w = 2&lt;/code&gt;. The above table for &lt;code&gt;w&lt;/code&gt; shows that w=2 only when (n=9, h=7). This leaves no possibilities for c, as &lt;code&gt;a&lt;/code&gt; is minimum 2 and thus &lt;code&gt;c&lt;/code&gt; is minimum &lt;code&gt;9&lt;/code&gt;. Dead end.&lt;br&gt;&#xA;&lt;br&gt;&#xA; When m=8, o=6 we have &lt;code&gt;c = a + 6&lt;/code&gt; and &lt;code&gt;w = 0&lt;/code&gt;, which permits (n=2, h=6), forbidden as o=h, or (n=3, h=9), forbidden because again &lt;code&gt;a&lt;/code&gt; is minimum 2 and thus &lt;code&gt;c&lt;/code&gt; is minimum 8, but both 8 and 9 are occupied. Also a dead end.&lt;br&gt;&#xA;&lt;br&gt;&#xA; So we conclude &lt;strong&gt;m = 9&lt;/strong&gt;. Our equation is now &lt;code&gt;11*o - 9*x + w = 10.&lt;/code&gt; The restriction on &lt;code&gt;w&lt;/code&gt; is still as onerous as ever:&#xA; &lt;pre&gt;&lt;code&gt; &#xA;     o          =     2    3    4    5    6    7    8&#xA;     10 - 11*o  =   -12  -23  -34  -45  -56  -67  -78&#xA;     -9*x       =   -18  -27  -36  -45  -63  -72  -81&#xA;     w          =     6    4    2    0    7    5    3&lt;/pre&gt;&lt;/code&gt;&#xA; (When o=9 it conflicts with m; also you get a = c + 9 which is similarly impossible.) So we have only two possibilities here:&lt;br&gt;&#xA; &#xA; When (o=4, w=2) we have (n=8, h=4), but again o = h, BZZT NO.&lt;br&gt;&#xA;&lt;br&gt;&#xA; When (o=5, w=0) we have (n=2, h=6), a = c + 5. This is only possible with c = 3, a = 8, because n has 2 and m has 9. DING DING DING.&lt;br&gt;&#xA;&lt;br&gt;&#xA; This is the above solution and it is unique.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7552" LastEditorUserId="7552" LastEditDate="2016-02-09T17:07:55.787" LastActivityDate="2016-02-09T17:07:55.787" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26280" PostTypeId="1" AcceptedAnswerId="26281" CreationDate="2016-02-09T16:36:06.623" Score="7" ViewCount="249" Body="&lt;p&gt;Sotheby's New York. Article 522 - An Authentic Ornamental Sword Presented By King George V&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Article 522 reads: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;This sword's fascinating history has excited collectors from around the globe. It's find was recorded as one of the most unusual in history. The finder, a Mr. PooPong McPlop, was hurrying to water his pigs and stopped to pick a stray blade of grass that was growing in a nearby meadow. As he stooped down he slipped a disc in his spine, fell into a wheelbarrow which proceeded to deposit him in a nearby well. The rest is history and we now have a sword which we know is authentic due to its very original inscription. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It reads as follows: &quot;&lt;em&gt;On this day of 19th February 1928 this sword is being presented to Captain Sakhryn-Brown on his remarkable bravery in the Battle of the Somme during WW1. He single-handedly rescued 32 french soldiers and 19 British ones as they were cornered in an apple juice factory. The dear Captain receives this sword posthumously, from His Majesty King George V, after he was taken ill and unfortunately died from cyanide and a sharp writing instrument in 1926&lt;/em&gt;.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sword is extremely rare due to its sharpness and weight which when one divided by the other the number is exactly pi! It's several rubies are themselves worth slightly more than a crate of Kellog's Cornflakes.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To cut a long story short during the bidding, just as a heavyset gentlemen offered his bid of $34, 971 (thinking of how nice and red it would look, being run through the midriff of his business partner), someone from the back of the room stands up and screams &quot;It's a fake!!!&quot;. He, incidentally, is right. How did he know? &lt;/p&gt;&#xA;" OwnerUserId="19086" LastActivityDate="2016-02-09T16:48:12.073" Title="The Case of the Forged Sword" Tags="|riddle|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26281" PostTypeId="2" ParentId="26280" CreationDate="2016-02-09T16:48:12.073" Score="10" Body="&lt;p&gt;The crucial term is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; WW1.  In the year 1928, WW2 was still to come, and there was only a single world war known.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2016-02-09T16:48:12.073" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26282" PostTypeId="2" ParentId="26265" CreationDate="2016-02-09T17:00:21.297" Score="3" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; R=1 E=0 M=9 O=5 A=8 C=3 N=2 H=6&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which yields:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 902 + 382 = 10836 - 9552&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since MOON&amp;lt;9999, MAN&amp;lt;999, CAN&amp;lt;999, REACH &amp;lt;12000, hence R=1. Also from the above, E&amp;lt;2, but E=1 is impossible because R=1 already, hence E=0.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If M&amp;lt;=7, since MAN&amp;lt;999, CAN&amp;lt;999, MOON&amp;lt;7999, MAN+CAN+MOON &amp;lt; 10,000. Hence M&gt;7. We assume M=9.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We can simplify the equation given by writing, insteand of CAN, for instance, CAN -&gt; 100C + 10A +N, which allows the following simplification:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  3N-H = 10 (10 +9(A-C)-11 O)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since the rightmost term, 10 +9 (A-C)- 11 O, is an integer, also (3N-H)/10 must be an integer. It is easy that we can only have X=(3N-H)/10=0,1,2. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rewriting the above,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 10 = 11 O - 9(A-C) +X&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which can be easily solved by trial and error. It turns out that &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the only solution is O=5, A-C=5, X=0. In order to have X=3N-H=0, we can only have N=2,H=6 because the other possible solution, N=3,H=9 is prohibited by the fact that 9 is already allocated to M.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On the other hand,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A-C=5 has the unique solution A=8,C=3, given that the other possible solutions, 0,5 - 1,6 - 2-7, 4-9, would all use digits which are allocated.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A similar treatment&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; shows that, for M=8, there are no other possible  solutions. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19102" LastActivityDate="2016-02-09T17:00:21.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26283" PostTypeId="1" CreationDate="2016-02-09T17:19:57.043" Score="-3" ViewCount="161" Body="&lt;p&gt;Given:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;RCK = THE&lt;br&gt; TKRRKLG = LETTERS&lt;br&gt; SO = IN&lt;br&gt; HJHI = BABY&lt;br&gt; OZQHKL&#xA;  = NUMBER&lt;br&gt; RDB'G = TWO'S&lt;br&gt; OJQK = NAME&lt;br&gt; JLK = ARE&lt;br&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;YKXRKICRQAKLGLJFKTSK = _E_TEYHTM_ERSRA_ELIE ( DESCHAMP is BABY'S LAST NAME)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Find the baby's first name and middle name.&lt;/p&gt;&#xA;" OwnerUserId="19120" LastEditorUserId="19120" LastEditDate="2016-02-09T20:26:07.980" LastActivityDate="2016-02-09T20:26:07.980" Title="figuring out my grandsons name!" Tags="|cipher|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26284" PostTypeId="1" CreationDate="2016-02-09T17:31:08.483" Score="3" ViewCount="180" Body="&lt;p&gt;You have a bucket of 12lbs of a mixture of dry chemicals including both powders and coarse agglomerates.  You actually need a 4.5lb sample.  The rest will be desposed of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You pour the sample into your other bucket and notice a serious issue: when such samples are poured from one container to another, they segrate by particle size to some random extent.  In this case, the largest chunks poured first so the 4.5lb sample is far coarser than that remaining in the original bucket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You find a &lt;a href=&quot;https://en.wikipedia.org/wiki/Riffle_splitter&quot; rel=&quot;nofollow&quot;&gt;Riffle-type splitter&lt;/a&gt; which is designed to split any material poured into it into two identical size samples with the same particle size distribution.  This splitter, however, only can deposit the material into two tins which can only hold 1 lb each.  If you let the sample pour onto the floor (no tin) the sample will be contaminated.  Note: this means that you can only pour 2lbs at a time into the splitter but both 1lb tins it fills will have the sample distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the minimum number of tins you need in order to get your sample?  What is the minimum number of splits to get your sample with this many tins (to reduce splitting error)?&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="17250" LastEditDate="2016-02-10T15:34:04.127" LastActivityDate="2016-02-10T15:50:13.347" Title="Manage Sample Segregation" Tags="|mathematics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26285" PostTypeId="2" ParentId="26251" CreationDate="2016-02-09T18:22:13.917" Score="5" Body="&lt;p&gt;Without looking at any of the answers the strategy is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All the people who see an even number of blue noses go to section A.&lt;br&gt;&#xA; All the people who see an odd number of blue noses go to section B.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This works because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are &lt;em&gt;b&lt;/em&gt; blue noses.&lt;br&gt;&#xA; Those with a blue nose see &lt;em&gt;b-1&lt;/em&gt; blue noses.&lt;br&gt;&#xA; Those with a red nose see &lt;em&gt;b&lt;/em&gt; blue noses.&lt;br&gt;&#xA; So one set sees an even number the other an odd number.&lt;br&gt;&#xA; By separating on that criterion they group the red noses together and likewise the blue noses.&lt;br&gt;&#xA; Of course they could choose red noses, odd to section A, &lt;em&gt;etc&lt;/em&gt; and it'll still work.&lt;br&gt;&#xA; The main objective is they all agree on a single strategy.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastActivityDate="2016-02-09T18:22:13.917" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26286" PostTypeId="2" ParentId="26284" CreationDate="2016-02-09T18:23:00.613" Score="2" Body="&lt;p&gt;Answer (number of tins):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You need three 1 lb tins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assumptions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I've assumed that the splitter starts right away and you can't pour stuff into the splitter and then use the same Tin you've just poured from to catch stuff.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Terminology:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To make things simpler, I'll use the following terms: &lt;br&gt;tin 1/2/3= a (numbered) 1 pound tin&lt;br&gt;Bucket = final, 4.5 pound bucket&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Big picture:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Essentially, you need 3/8 out of every pound, while preserving the overall distribution. (3/8 * 12 = 4.5). So out of every two pounds we need 3/4 of a pound.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First pour:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Use the splitter to split the first two pounds between the Tin 1 and Tin 2. Empty Tin 2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second pour:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Use the splitter to split the pound in Tin 1 between Tin 2 and Tin 3. Pour Tin 2 (1/2 pound) into the bucket. You now have half a pound in the bucket and half a pound in Tin 3, both have the same distribution as the original 2 pounds poured out of the 12 pound bag.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Third pour:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Use the splitter to split the material in Tin 3 between Tin 1 and Tin 2. Pour one out, pour the other into the bucket. The bucket now has 3/4 pound of material that has the same distribution as the first two pounds of material.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To finalize:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Repeat with the other 10 pounds of raw material, two pounds at a time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the splitter could pour into the bucket:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Then you'd only need two tins. Each of pours two and three could split into the bucket and the empty tin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I haven't confirmed this is the minimum number of splits, but this algorithm does it in:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 18 Splits, 3 for each 2 pounds of the material.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastEditorUserId="1620" LastEditDate="2016-02-09T19:10:00.543" LastActivityDate="2016-02-09T19:10:00.543" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26287" PostTypeId="2" ParentId="26283" CreationDate="2016-02-09T18:28:20.983" Score="5" Body="&lt;p&gt;I substituted letters in Deschamp that are missing from the substitutions already performed in your question. That left one letter unknown. Removed the letters in Deschamp and ran it through an anagram solver to get &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pieter Teyler&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18493" LastActivityDate="2016-02-09T18:28:20.983" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26288" PostTypeId="2" ParentId="25208" CreationDate="2016-02-09T18:37:13.773" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Maid number 2. If no windows or doors - no dust gets in - no need to dust...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19086" LastEditorUserId="19086" LastEditDate="2016-02-10T12:27:13.287" LastActivityDate="2016-02-10T12:27:13.287" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26289" PostTypeId="1" AcceptedAnswerId="26294" CreationDate="2016-02-09T18:55:26.823" Score="16" ViewCount="1181" Body="&lt;p&gt;There is a small town with a strange custom. Residents each own a single pebble, painted blue or red, which they always carry in their pockets. Those with blue pebbles always tell the truth, and those with red pebbles always lie. Each resident also knows the color of every other resident's pebble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The town has an eccentric mayor, who is the sole resident not abiding by the custom. The mayor also carries a pebble (blue or red), but can say both true statements and false statements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One day the town's bank gets robbed. You are a detective, sent to investigate the crime. You know for sure the getaway car was either blue or red, but you would like to know which. The witnesses at the scene make the following statements:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Alex: I didn't get a good look at the getaway car.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Beth: The getaway car was the same color as Alex's pebble.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Carl: Everyone whose name starts with a vowel has the same color pebble.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Doug: Either Beth or Grace (or possibly both) have a pebble that is a different color than my own.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Emily: Irina's pebble is red.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Floyd: Between me, Grace, and Alex, we have an odd number of red pebbles.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Grace: The mayor is not Alex, Doug, Floyd, or me.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Hal: My pebble is blue and so is Carl's.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Irina: At least one of Emily, Hal, or Carl has a pebble the same color as mine.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, you do not know the mayor's name, nor can you see the color of any witness' pebble. Can you determine the color of the getaway car?&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2016-02-09T19:55:02.073" LastActivityDate="2016-02-09T21:40:12.670" Title="Town of liars and truth-tellers with eccentric mayor" Tags="|logical-deduction|liars|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26290" PostTypeId="2" ParentId="26245" CreationDate="2016-02-09T19:17:59.527" Score="3" Body="&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hit the pale ghost, kill it. Wait until another one turns pale, hit it, kill it, wait until another turns pale, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason it works:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If &quot;due to evolution and natural selection&quot; there are 99 camo ghosts and 1 pale ghost, it stands to reason (somewhat) that there should always be one pale ghost. If there wasn't a reason for it, then all the ghosts would be camo. Therefore, if all the ghosts are camo and you wait a while, one should become pale through some path of evolution/natural selection.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2016-02-09T19:17:59.527" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26291" PostTypeId="2" ParentId="26245" CreationDate="2016-02-09T19:37:02.957" Score="2" Body="&lt;p&gt;It says you only have to kill them with your baseball bat.  So,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Punch them all until they are all pale, then finish them off with your bat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2016-02-09T19:37:02.957" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26294" PostTypeId="2" ParentId="26289" CreationDate="2016-02-09T20:01:51.937" Score="14" Body="&lt;p&gt;The car is...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Blue&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning (using A through I to indicate the people):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let's look first at the cycle of C-E-H-I and start by assuming none of them are the mayor.&lt;br&gt;&lt;br&gt;Then, if E's pebble is blue, she's telling the truth and I's is clearly red. But if E's pebble is red, then she's lying and I's is clearly blue.&lt;br&gt;&lt;br&gt;Given that, C must be lying since E and I have two different pebbles. So C's pebble is red.&lt;br&gt;&lt;br&gt; Given that, H's pebble must also be red. And I can't have a red pebble because then the statement would be true, so we've got I with a blue pebble. So that gives us C-Red, E-Red, H-Red, I-Blue. But I's statement is not actually true, which is an impossible situation. That means one of C, E, H or I is the mayor. (I'm not sure it's possible to figure out which one, but it doesn't matter).&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Musical interlude (for organization).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Given that, We now know that G is telling the truth and has a blue pebble. That means that D can't have a red pebble or their statement would be true, so D has a blue pebble. Since D's statement must be true and B and G have blue pebbles, then B must have red and be lying. Therefore the getaway car is a different color than A's pebble.&lt;br&gt;&lt;br&gt;F says that A+F+G yields an odd number of red pebbles. Since G is blue, we either have F blue and A red or F red and A red. Either way, A is red.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Stunning conclusion:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Therefore, since A is red and the car is different than A, the getaway car must be blue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nice puzzle!&lt;/p&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2016-02-09T20:01:51.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26295" PostTypeId="2" ParentId="14864" CreationDate="2016-02-09T20:17:49.297" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's a trick question, neither is correct. Egg yolk is yellow.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17712" LastEditorUserId="13101" LastEditDate="2016-02-09T20:20:50.603" LastActivityDate="2016-02-09T20:20:50.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26296" PostTypeId="2" ParentId="26180" CreationDate="2016-02-09T20:22:24.997" Score="3" Body="&lt;p&gt;&lt;sup&gt;I doubted posting this, because it isn't quite an answer. But I spent some time pondering this puzzle and I wanted to share my insights, such as they are.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Computing an average... won't work&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;My initial thought was to have each mathematician compute the &quot;average light state&quot; in their cell; they proclaim &quot;bright mode&quot; if this average is higher than ½, and &quot;dark mode&quot; if it is lower.  I believe computations can get a little sketchy with infinities; it might be defined as $$A_i = \lim_{n \rightarrow \infty} \frac1n \sum_{k=1}^n L(i,k)$$ where $L(i,k) \in \{0,1\}$ denotes &quot;in mathematician $i$'s cell on day $k$, the light is switched on/off&quot;.  I wasn't sure whether this limit would converge; we might have to resolve this by defining a &lt;a href=&quot;https://en.wikipedia.org/wiki/Banach_limit&quot; rel=&quot;nofollow noreferrer&quot;&gt;Banach limit&lt;/a&gt; as suggested in &lt;a href=&quot;https://puzzling.stackexchange.com/users/19036/ron&quot;&gt;Ron&lt;/a&gt;'s answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This average would be completely determined by the infinite number of days spent in either bright or dark mode; neutral mode would not influence it at all.  We would have $\sum_i A_i \in \{1,2\}$, depending on the mode chosen by the warden.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Big question, however, is: if $A_i = ½$, what should mathematician $i$ do?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I wondered this myself, and it was also pointed out by &lt;a href=&quot;https://puzzling.stackexchange.com/users/2627/julian-rosen&quot;&gt;Julian Rosen&lt;/a&gt; in comments.  It is easy to show that the warden can give any two mathematicians an average of ½ regardless of the mode he picks; the third one then gets $A_i \in \{0,1\}$.  They won't be able to agree on a rule that works in all cases.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The ultrafilter&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I was only recently introduced to ultrafilters, by Julian Rosen's answer.  At first I thought there was something fishy going on; what if the warden follows the ½-½-0/1 tactic?  The answer is that the sets of days where the light is on for both half-mathematicians must be unequal (in fact, on &lt;a href=&quot;https://en.wikipedia.org/wiki/Almost&quot; rel=&quot;nofollow noreferrer&quot;&gt;almost&lt;/a&gt; all days their states must be different).  The ultrafilter leverages this.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;To make this more precise:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Let $L_0$ be the set of days where both half-mathematicians have equal lighting states.  Because the third mathematician's average light state must be either 0 or 1, $L_0$ is finite.  Let $L_1$ be the set of days where the first half-mathematician's light is on and let $L_2$ consist of the days where the second half-mathematician's light is on.  Both these sets are infinite, and together with $L_0$ they partition the set of all days spent in prison.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By properties 2 and 3 of a non-principal ultrafilter, either $L_1$ or $L_2$ is &quot;large&quot;; so by property 1, one and only one of the half-mathematicians spends a large number of days in a lit cell.  Crisis averted.&lt;/p&gt;&#xA;" OwnerUserId="13484" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2016-02-11T19:13:33.960" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="26297" PostTypeId="1" AcceptedAnswerId="26307" CreationDate="2016-02-09T20:30:14.310" Score="22" ViewCount="1498" Body="&lt;p&gt;This puzzle belongs to the puzzle series: &lt;a href=&quot;https://puzzling.stackexchange.com/search?q=hyper-modern%20is%3Aquestion&quot;&gt;hyper-modern art&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;After cracking the door code of the last room, the two friends in the gallery of hyper-modern art are moving on to the next room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Oh, look at this! This image is lifting my mood quite a bit after this monochrome depression lately.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yes, but I don't quite know why it's making me thirsty just to look at it.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Hmm, I can see what you mean. It also looks somehow familiar... What is the picture called, anyway?&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;code&gt;Sweet Dream&lt;/code&gt; - quite fitting, I'd say.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;I agree. But now you'll soon tell me that there is more to that than meets the eye on first sight, right?&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Of course! Would it be hyper-modern art otherwise? As with a lot of modern art, the picture hides a message to the public.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Really? A political statement hidden in art? Maybe I can see through it...&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Not so sure if it is political, but it is clearly a statement. A wish. A dream, if you will.&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LFOgr.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z8gQO.jpg&quot; alt=&quot;Sweet &amp;amp; Marvellous!&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/LFOgr.jpg&quot; rel=&quot;noreferrer&quot;&gt;Full resolution image, 1.3Mb, 4000 x 3000 pixels&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The goal of this puzzle is to derive the message coded into it. This message consists of 3 words and is in English. It is hidden in the picture and not in the digital content. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You could print the picture and solve the puzzle from the printed picture. The link beneath the image leads to a full-resolution version, which might be recommended for solving this puzzle.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Additionally to the actual message, there are also various hints 'hidden' in the image. Bonus for finding them all!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5840" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2016-02-09T23:30:47.317" Title="Sweet! Dream! Another hyper-modern art piece" Tags="|cipher|visual|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26298" PostTypeId="2" ParentId="26289" CreationDate="2016-02-09T20:42:38.387" Score="4" Body="&lt;p&gt;The car is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; blue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C has a blue pebble, but is the mayor and is lying. Suppose C is telling the truth: E cannot be blue, else I would be blue and that contradicts E's statement. If E is red, then I is blue, and that contradicts C's statement. Therefore, C is lying.&lt;br/&gt;&lt;br/&gt;Now, suppose C is not the mayor. Then C and H have red pebbles. If I has a blue pebble, then E has a blue pebble, but that contradicts E's statement. If I has a red pebble, then it has the same pebble as C and H, which is also a contradiction. Ergo, C is the mayor.&lt;br/&gt;&lt;br/&gt;EDIT: As pointed out in the comments, this doesn't show that the mayor is not H, I or E.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now that we know that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; G is telling the truth. If D is red, then D's statement is true, so D's pebble must be blue. This means that B != D, so B is red, which means the car is not the color of A.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So what color is A?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We know G is blue. If F is blue, then A must be red to satisfy the number of odd pebbles. If F is red, the number of red pebbles is even, so A must also be red. Thus, A is red, and the car is blue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17446" LastEditorUserId="17446" LastEditDate="2016-02-09T21:40:12.670" LastActivityDate="2016-02-09T21:40:12.670" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26299" PostTypeId="2" ParentId="26245" CreationDate="2016-02-09T20:46:11.283" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are ghosts, so they are already dead. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Done.&lt;/p&gt;&#xA;" OwnerUserId="19128" LastActivityDate="2016-02-09T20:46:11.283" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26300" PostTypeId="2" ParentId="26284" CreationDate="2016-02-09T20:59:09.647" Score="1" Body="&lt;p&gt;Assumptions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I have 2 buckets (items capable of holding more than 1 lb of material). Based on the OP wording &quot;your other bucket&quot;&lt;/li&gt;&#xA;&lt;li&gt;I can't measure exactly. I can either measure 2 pounds or &quot;everything left&quot;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I know that I need to measure out everything in my bucket to ensure that I have it split evenly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 bucket with 12 pounds of material&lt;br&gt;&#xA; 1 empty bucket&#xA; &lt;p&gt;&#xA; First set of 6 pours (all 12 pounds from original bucket):&lt;br&gt;&#xA; All poured using 1 tin and losing half my material. Tin is emptied into my other bucket.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This gives:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 bucket with 6 pounds of material&lt;br&gt;&#xA; 1 empty bucket&lt;br&gt;&#xA; 6 pours&lt;br&gt;&#xA; 1 tin used&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Step two:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Next set of 3 pours. All material needs to be saved.&#xA; &lt;p&gt;&#xA; Pour 7 uses 2 tins. One is emptied into the empty bucket, one is set aside.&lt;br&gt;&#xA; Pour 8 uses 2 tins (3 total tins now). One is emptied into the bucket, one is set aside.&lt;br&gt;&#xA; Pour 9 uses 2 tins (4 total tins now). One is emptied into the bucket. Now the 3 filled tins are emptied into the newly empty bucket.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This gives: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 bucket with 3 pounds of material&lt;br&gt;&#xA; 1 bucket also with 3 pounds of material&lt;br&gt;&#xA; 9 pours&lt;br&gt;&#xA; 4 tins&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Step three:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pick either bucket to start emptying. Other bucket will be poured into. Only need to save half the material, so only 1 tin used.&#xA; &lt;p&gt;&#xA; Pour 10 gets you 1 pound of material. Pour into bucket (now has 4 pounds in it).&lt;br&gt;&#xA; Pour 11 gets you 1/2 pound of material. Pour into bucket.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This gives:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 bucket with 4 1/2 pounds of material&lt;br&gt;&#xA; 11 pours&lt;br&gt;&#xA; 4 tins used&lt;br&gt;&#xA; A bit of a messy floor&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;TL/DR:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4 tins, 11 pours.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="375" LastEditorUserId="375" LastEditDate="2016-02-10T14:19:09.660" LastActivityDate="2016-02-10T14:19:09.660" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26301" PostTypeId="2" ParentId="26245" CreationDate="2016-02-09T21:01:45.750" Score="1" Body="&lt;p&gt;Ok.  This is a much better answer than my previous.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Recoil does not exist, so I can hit the ghosts as hard as I want, knocking one into another into another .... .&#xA; I line the ghosts up in a line and whack the first one hard into the second, the second hits the third, and so on...  After 5 hits, they are all dead except the one on the end.  I finish him with the 6th hit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19128" LastActivityDate="2016-02-09T21:01:45.750" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26302" PostTypeId="1" AcceptedAnswerId="26303" CreationDate="2016-02-09T21:16:06.310" Score="5" ViewCount="374" Body="&lt;p&gt;Wiggle me when you want to find me &lt;br&gt;&#xA;Pay attention to what's behind me &lt;br&gt;&#xA;I'm an extension of your will &lt;br&gt;&#xA;I can act, but I cannot feel&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who am I?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user17947" LastActivityDate="2016-02-10T03:11:36.433" Title="Wiggle me when you want to find me" Tags="|riddle|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26303" PostTypeId="2" ParentId="26302" CreationDate="2016-02-09T21:34:29.890" Score="10" Body="&lt;p&gt;How about this: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A mouse (pointer).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. &quot;Wiggle me when you want to find me&quot;  -  When you want to find your mouse pointer on the screen, you wiggle the mouse in your hand.&lt;br&gt;&#xA;&lt;br&gt;&#xA; 2. &quot;Pay attention to what's behind me&quot; - Whatever is behind the pointer is what you activate when you click.&lt;br&gt;&#xA;&lt;br&gt;&#xA; 3. &quot;I'm an extension of your will&quot; - A mouse is an input device, to exercise your will on the computer.&lt;br&gt;&#xA;&lt;br&gt;&#xA; 4. &quot;I can act, but I cannot feel&quot; - The mouse pointer 'acts' by clicking and moving, but has no 'feelings' as such.  The mouse in your hand only senses movement.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19128" LastEditorUserId="19128" LastEditDate="2016-02-09T22:01:42.100" LastActivityDate="2016-02-09T22:01:42.100" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26304" PostTypeId="2" ParentId="26302" CreationDate="2016-02-09T21:41:20.090" Score="5" Body="&lt;p&gt;Is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A mouse pointer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wiggle me when you want to find me &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To find it, most people wiggle it so its movement is caught and the pointer is located &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pay attention to what's behind me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You need to pay attention to the web page or at least what's on the computer screen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm an extension of your will &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People can use this external device willingly and how they want to willingly and force the pointer to do what the person does willingly&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can act, but I cannot feel&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are an object that can highlight, click, select and do lots of stuff but you can't express emotion&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9042" LastActivityDate="2016-02-09T21:41:20.090" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26305" PostTypeId="1" AcceptedAnswerId="26316" CreationDate="2016-02-09T22:54:56.840" Score="3" ViewCount="393" Body="&lt;p&gt;Here comes your second test from the CIA. It is for graduation, but you still have to do it to prove yourself worthy of being an agent. Agents from around the world have struggled with this mysterious code, which no one can possible understand. Once deciphered, translate the word to the appropriate message.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;derfcxsaqwszy67uhgt = ?&lt;/p&gt;&#xA;&lt;p&gt;o90plkifrtgvcd = ?&lt;/p&gt;&#xA;&lt;p&gt;htyujnbge34rdswaqwsxzr45tfdet56ygfr = ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The only hint from the instructor is:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;05NaN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;What does the code mean and what is the message?&lt;/p&gt;&#xA;" OwnerUserId="9042" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-02-10T15:25:24.647" Title="What's the message this time?" Tags="|cipher|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26306" PostTypeId="1" AcceptedAnswerId="26312" CreationDate="2016-02-09T23:05:55.133" Score="6" ViewCount="746" Body="&lt;p&gt;Oh so very long,&lt;br&gt;&#xA;Often very hard.&lt;br&gt;&#xA;Endurance is the Key.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Faster, Faster,&lt;br&gt;&#xA;Push harder, harder.&lt;br&gt;&#xA;Euphoria at the end.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some say that it hurts,&lt;br&gt;&#xA;But not if done right.&lt;br&gt;&#xA;Rush to the end, or take your time.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep your socks on,&lt;br&gt;&#xA;Stay hydrated.&lt;br&gt;&#xA;Strong to the end.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheering and screaming,&lt;br&gt;&#xA;right at the end.&lt;br&gt;&#xA;You must be tired.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe a game before,&lt;br&gt;&#xA;Thanks, that was fun.&lt;br&gt;&#xA;Now we are done.  &lt;/p&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="9951" LastEditDate="2016-02-10T01:56:16.733" LastActivityDate="2016-02-10T03:02:00.220" Title="That's what she said" Tags="|riddle|word|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26307" PostTypeId="2" ParentId="26297" CreationDate="2016-02-09T23:12:30.893" Score="18" Body="&lt;p&gt;The pattern of candy and blank spaces&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;matches the word lengths of Martin Luther King's &quot;I Have A Dream&quot; speech, starting at the line &quot;I have a dream that one day this nation will rise up....&quot; This is clued by the image at the bottom of a dark-colored chocolate square delivering a speech to a mixed group of light and dark squares. The orator is crowned as a &quot;king&quot;, and the candy images come from the game Candy Crush, which was developed by the company King.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At the bottom&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the light and dark squares encode the word &lt;code&gt;ORIGINAL&lt;/code&gt; using a Bacon code, while their shadows encode &lt;code&gt;ELEMENTS&lt;/code&gt; (noticed by @Will).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This suggests that we find&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the candies with shape/color combinations that match the original ones in the game Candy Crush. If we superimpose MLK's speech on the candies, the letters in those spots spell out &lt;code&gt;CHOCOLATE FOR EVERYONE&lt;/code&gt;, which is the message. The whole image was too large to upload, so here's a section of four lines near the middle: &lt;img src=&quot;https://i.stack.imgur.com/7OQ8R.jpg&quot; alt=&quot;1]&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2016-02-09T23:30:47.317" LastActivityDate="2016-02-09T23:30:47.317" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26309" PostTypeId="2" ParentId="26306" CreationDate="2016-02-10T00:21:32.127" Score="3" Body="&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is the tiebreaker game in tennis.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The idea:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Maybe a game before (when we were still tied), thanks that was fun; but now I lost, and there is cheering, on the other side. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19102" LastActivityDate="2016-02-10T00:21:32.127" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26310" PostTypeId="2" ParentId="26306" CreationDate="2016-02-10T00:39:10.187" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Is it an education?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Oh so very long,&lt;br&gt;&#xA;Often very hard.&lt;br&gt;&#xA;Endurance is the Key.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An education seems to take forever&lt;br&gt;&#xA; and is fraught with challenges,&lt;br&gt;&#xA; but you have to just stick with it to finish.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Faster, Faster,&lt;br&gt;&#xA;Push harder, harder.&lt;br&gt;&#xA;Euphoria at the end.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The further you get, the faster the pace&lt;br&gt;&#xA; and the greater the difficulty,&lt;br&gt;&#xA; but when it's over: &quot;Freeeeedom!&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some say that it hurts,&lt;br&gt;&#xA;But not if done right.&lt;br&gt;&#xA;Rush to the end, or take your time.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Forced learning is painful,&lt;br&gt;&#xA; but not for a keen mind.&lt;br&gt;&#xA; Graduate early or throw in a few gap years; your call.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cheering and screaming,&lt;br&gt;&#xA;right at the end.&lt;br&gt;&#xA;You must be tired.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hooting and hollering aplenty&lt;br&gt;&#xA; at graduation.&lt;br&gt;&#xA; Now you can sleep in until noon every day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Maybe a game before,&lt;br&gt;&#xA;Thanks, that was fun.&lt;br&gt;&#xA;Now we are done.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I don't know about you, but I played a lot of games in gym class,&lt;br&gt;&#xA; which was a nice way to follow art class. Thanks for the crayons, Mrs. D.&lt;br&gt;&#xA; I need to find a job :( &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19134" LastActivityDate="2016-02-10T00:39:10.187" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26311" PostTypeId="2" ParentId="25287" CreationDate="2016-02-10T01:21:14.933" Score="1" Body="&lt;h3&gt;I can switch the black knights with the white knights in just &lt;strong&gt;1&lt;/strong&gt; move:&lt;/h3&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; rotate the board 180 degrees!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19137" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-02-10T01:21:14.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26312" PostTypeId="2" ParentId="26306" CreationDate="2016-02-10T03:02:00.220" Score="5" Body="&lt;p&gt;Is this a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Marathon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Oh so very long,&lt;br&gt;&#xA;Often very hard.&lt;br&gt;&#xA;Endurance is the Key.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; marathon is long distance running race and very hard, and yes you need endurance to finish it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Faster, Faster,&lt;br&gt;&#xA;Push harder, harder.&lt;br&gt;&#xA;Euphoria at the end.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Faster in running and winning one or finishing one can give you a great feeling of success&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some say that it hurts,&lt;br&gt;&#xA;But not if done right.&lt;br&gt;&#xA;Rush to the end, or take your time.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; running will injure you if you don't stretch or warm up. Rush or slow pace at first then rush when close to finishing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Keep your socks on,&lt;br&gt;&#xA;Stay hydrated.&lt;br&gt;&#xA;Strong to the end.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; water to keep you going and socks of course are needed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cheering and screaming,&lt;br&gt;&#xA;right at the end.&lt;br&gt;&#xA;You must be tired.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; audience screams at the sidelines at also at the finish line. And you are very tired after the marathon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Maybe a game before,&lt;br&gt;&#xA;Thanks, that was fun.&lt;br&gt;&#xA;Now we are done. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; running is considered a game when we are still young or for a child &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16036" LastActivityDate="2016-02-10T03:02:00.220" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26313" PostTypeId="1" AcceptedAnswerId="26315" CreationDate="2016-02-10T03:05:25.470" Score="7" ViewCount="443" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A troop of soldiers: in formation bravely they stand &lt;br&gt;&#xA;  An ominous rumble is heard across the land &lt;br&gt;&#xA;  Then, boom! A blast, they are struck down - fallen, slain &lt;br&gt;&#xA;  But rise from the grave to face danger again&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explain what this is describing.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user17947" LastEditorUserId="8874" LastEditDate="2016-02-10T09:49:31.903" LastActivityDate="2016-02-10T09:49:31.903" Title="The riddle of the 300" Tags="|riddle|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26314" PostTypeId="2" ParentId="26302" CreationDate="2016-02-10T03:11:36.433" Score="3" Body="&lt;p&gt;Is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A loose tooth.  Its got the wiggling down, and you know it's there when it does.  When it wiggles the nerves behind it in your gums will get your attention.  Its an outgrowth of your head, the seat of your will. The nerve in the tooth itself are disconnected, but you could still chew, gently, with the tooth. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11921" LastActivityDate="2016-02-10T03:11:36.433" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26315" PostTypeId="2" ParentId="26313" CreationDate="2016-02-10T03:22:33.167" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bowling&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The soldiers are pins, in a set formation. The ball comes rumbling towards them and then knocks them all down at once (which is known as a strike). But they are then picked up (usually by a machine) and set up to be bowled at again. The title refers to the maximum possible score in a full game.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9523" LastEditorUserId="9523" LastEditDate="2016-02-10T03:29:46.143" LastActivityDate="2016-02-10T03:29:46.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26316" PostTypeId="2" ParentId="26305" CreationDate="2016-02-10T04:48:05.593" Score="7" Body="&lt;p&gt;When the strings are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; traced out on a keyboard, they look like &lt;a href=&quot;https://i.stack.imgur.com/5tkt0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;.&lt;br&gt;&#xA;&lt;em&gt;(In the image, the top keyboard is the first string, the second keyboard is the second string, and the bottom two are the last string. Red comes first, then blue, then yellow, then green, and finally purple. (Only the last one has more than two segments.)&lt;/em&gt;&lt;br&gt;&#xA;There's a clear hexagon-ish pattern that repeats. Sometimes it's cut off because the keys aren't on the keyboard, but it's always there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we replace each occurrence of the pattern with the starting letter, we get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - &lt;code&gt;day&lt;/code&gt;&lt;br&gt;&#xA; - &lt;code&gt;of&lt;/code&gt;&lt;br&gt;&#xA; - &lt;code&gt;h(?)ea(?)rt&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The anomalies represented by the question marks are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;an extra &lt;code&gt;t&lt;/code&gt; in the &lt;code&gt;h&lt;/code&gt; pattern, and an extra &lt;code&gt;x&lt;/code&gt; in the &lt;code&gt;a&lt;/code&gt; pattern.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The clue &lt;em&gt;may&lt;/em&gt; be referring to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; either the World Heart Day (29 September) or Valentine's Day (February 14th).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="11876" LastEditDate="2016-02-10T05:18:22.563" LastActivityDate="2016-02-10T05:18:22.563" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="26317" PostTypeId="2" ParentId="26172" CreationDate="2016-02-10T05:04:15.760" Score="1" Body="&lt;p&gt;The perfect gift is a  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Wedding Ring&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They've been talking about getting one for years&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Lots of talk goes around about getting engaged&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and now's your chance to give them what you know they want but will never get themselves&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;No one ever buys themselves a wedding ring&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A great gift for the procrastinator in your life&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You always want to make sure that this is the right person to marry&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;just in time for Valentine's Day!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A romantic time to get one&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Inquire about our wide selection of available styles*.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Lots of ring types are available&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;*square and triangle models currently unavailable&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;No such thing as a square of triangular ring&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19140" LastActivityDate="2016-02-10T05:04:15.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26318" PostTypeId="2" ParentId="26167" CreationDate="2016-02-10T05:36:47.483" Score="10" Body="&lt;p&gt;So, my wife has been thinking about this and proposed the following (partial) answer. You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Horse&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sound that horses make is &quot;neigh&quot;, which sounds like &quot;nay&quot; (i.e., &quot;no&quot; or &quot;you're wrong&quot;.) A horse would reply &quot;neigh&quot; to anything anyone would call it. But there are also some apparent connections to the individual clues:&#xA; &lt;br&gt;&lt;br&gt;&#xA; &lt;em&gt;Some call me royalty, but I say they're wrong.&lt;/em&gt;&lt;br&gt;&#xA; It could be that some called the horse &quot;filly&quot; (a young female horse), which sounds like &quot;Philip&quot; (Prince Philip, the Duke of Edinburgh). Alternatively, maybe some called the horse &quot;American Pharoah&quot; (the subject of a &lt;a href=&quot;https://puzzling.stackexchange.com/q/26055/17947&quot;&gt;recent riddle&lt;/a&gt; on Puzzling), or the names of any number of famous race horses whose names are related to royalty (&lt;a href=&quot;https://en.wikipedia.org/wiki/King_(horse)&quot; rel=&quot;nofollow noreferrer&quot;&gt;here's one example&lt;/a&gt;).&#xA;&lt;br&gt;&lt;br&gt;&#xA; &lt;em&gt;Some call me crazy, but I say they're wrong.&lt;/em&gt; &lt;br&gt;&#xA; &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Crazy_Horse&quot; rel=&quot;nofollow noreferrer&quot;&gt;Crazy Horse&lt;/a&gt;&quot; was the name of a famous native American tribal leader.&#xA;&lt;br&gt;&lt;br&gt;&#xA; &lt;em&gt;Some call me a ghost, but I say they're wrong.&lt;/em&gt;&lt;br&gt;&#xA; There's gotta be a race horse out there somewhere that's called &quot;ghost&quot;.&#xA;&lt;br&gt;&lt;br&gt;&#xA; &lt;em&gt;Some call me racist, but I say they're wrong.&lt;/em&gt;&lt;br&gt;&#xA; If a horse is a &lt;em&gt;race&lt;/em&gt; horse, does that make it racist?&#xA; &lt;br&gt;&lt;br&gt;&#xA; &lt;em&gt;Some call me a hat, but I say they're wrong.&lt;/em&gt;&lt;br&gt;&#xA; &lt;a href=&quot;https://en.wikipedia.org/wiki/Bowler_hat&quot; rel=&quot;nofollow noreferrer&quot;&gt;Derby hats&lt;/a&gt;, anyone?&#xA;&lt;br&gt;&lt;br&gt;&#xA; Hint 1: &lt;em&gt;I am a word&lt;/em&gt;&lt;br&gt;&#xA; &quot;Horse&quot; is a word&#xA;&lt;br&gt;&lt;br&gt;&#xA; Hint 2: &lt;em&gt;The answer involves rhymes&lt;/em&gt;&#xA;&lt;br&gt;&#xA; I have absolutely no idea what this refers to.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user17947" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2016-02-10T05:36:47.483" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26319" PostTypeId="1" AcceptedAnswerId="26373" CreationDate="2016-02-10T07:34:14.263" Score="2" ViewCount="218" Body="&lt;p&gt;I have won 3 at the same time&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did this 3 times, and actually won 4 the last time&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There 19 similar to me in the world with this achievement, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;but after 2015 I am above them all&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am from the Oceania&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 without Ramon, other 2 thanks to Ramon&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; from Oceania is real. Not a trick.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Related with sports, but not a person&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18685" LastEditorUserId="18685" LastEditDate="2016-02-11T07:39:43.447" LastActivityDate="2016-02-11T23:16:03.567" Title="What am I by SC #4" Tags="|riddle|word|" AnswerCount="2" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26320" PostTypeId="2" ParentId="26251" CreationDate="2016-02-10T08:06:23.033" Score="1" Body="&lt;p&gt;The fountain is the key.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assumptions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. There is chocolate fountain in the the middle of the room. &quot;We will open the door to the chocolate-fountain room&quot;&#xA; &lt;br/&gt;&lt;br/&gt;&#xA; 2. You can move as much as you want, till you place yourself on on of the sides of the room &quot;as soon as you take your place - we will mace you&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. First prisoner gets in and stand on one side.&#xA; &lt;br/&gt;&lt;br/&gt;&#xA; 2. Second prisoner gets in and goes around fountain not stopping.&#xA; &lt;br/&gt;&lt;br/&gt;&#xA; 3. Next prisoner gets in and:&#xA; &lt;br/&gt;&lt;br/&gt;&#xA; if prisoner moving around fountain is &quot;blue&quot;, he goes the &quot;same&quot; direction (clockwise or counter-clockwise)&#xA; &lt;br/&gt;&lt;br/&gt;&#xA; if prisoner moving around fountain is &quot;red&quot;, he goes the &quot;opposite&quot; direction.&#xA; &lt;br/&gt;&lt;br/&gt;&#xA; 4. Now one of the prisoners traveling around fountain knows his color. &#xA; &lt;br/&gt;&lt;br/&gt;&#xA; If there are more prisoners in queue, the one who knows goes to his group. New prisoner will start at step 3.&#xA; &lt;br/&gt;&lt;br/&gt;&#xA; If there are no more prisoners, the one who knows will check color of the other prisoner:&#xA; &lt;br/&gt;&lt;br/&gt;&#xA; If the other one is blue, the one who knows go to his group.&#xA; If the other one is red, the one who knows will circle around fountain, till the other one will leave.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19146" LastActivityDate="2016-02-10T08:06:23.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26321" PostTypeId="2" ParentId="26319" CreationDate="2016-02-10T11:30:13.620" Score="3" Body="&lt;p&gt;The answer should be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tiger Woods&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have won 3 at the same time, and actually won 4 the last time  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Woods victory at the 2013 Players Championship also marked a win in&lt;br&gt;&#xA; his 300th PGA Tour start. He also won golf tournaments in his 100th&lt;br&gt;&#xA; (in 2000) and 200th (in 2006) tour starts.&lt;br&gt;&#xA;&lt;br&gt;&#xA; He is one of five players (along with Gene Sarazen, Ben Hogan, Gary&lt;br&gt;&#xA; Player, and Jack Nicklaus) to have won all four professional major&lt;br&gt;&#xA; championships in his career, known as the Career Grand Slam, and was&lt;br&gt;&#xA; the youngest to do so  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There 19 similar to me in the world with this achievement,&#xA;but after 2015 I am above them all&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Woods is the only player to have won all four professional major&#xA; championships in a row&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am from the Ocenia&lt;br&gt;&#xA;Perhaps related to the following one..&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; “I am really good at playing video games,” Woods said. “That’s&#xA; basically how I pass my time.”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1 without Ramon, Another 2 thanks to Ramon&lt;br&gt;&#xA;Perhaps not sure though&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Woods visits with students from Hopeworks n’ Camden; From L-R – Sean&#xA; Dougherty, Ramon Alverado, Jonathan Lugo, Tiger Woods, and Luis Solis&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19141" LastEditorUserId="17250" LastEditDate="2016-02-10T11:39:00.000" LastActivityDate="2016-02-10T11:39:00.000" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26322" PostTypeId="2" ParentId="16333" CreationDate="2016-02-10T12:34:55.557" Score="5" Body="&lt;p&gt;This is a slightly lateral-thinking answer, and is definitely against the spirit of the question, if not its letter.  But there's a lot to recommend this approach, enough that I felt it was worth providing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I need to begin by admitting that this approach is only convenient if you happen to live in Westeros, from &lt;a href=&quot;https://en.wikipedia.org/wiki/Game_of_Thrones&quot; rel=&quot;nofollow noreferrer&quot;&gt;Game of Thrones&lt;/a&gt;.  (And if you do, my condolences).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, to make up for that requirement, it's worth pointing out that this approach requires no maths or probability calculations at all.  Additionally, it only ever requires a single roll.  No rerolls, no use of base-6 numbers, etc.  All of which is probably a bit of a relief to you if you do happen to live in Westeros, as statistically you've probably had no mathematical schooling at all.  (again, my condolences)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But those of us living in the real world can use this method too;  it just requires us to first purchase a very particular (and slightly expensive) licensed coin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the mythos of the show, one common coin is the &lt;a href=&quot;http://www.joelscoins.com/westeros.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Copper Star of Robert Baratheon&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/ylhTe.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ylhTe.jpg&quot; alt=&quot;Image of the coin&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Notably, the &quot;tails&quot; side of the coin depicts a seven-pointed star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So all one needs to do in order to generate an evenly distributed 1/7 chance with a die and a coin is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Number the points of that star from 1-7.&lt;/li&gt;&#xA;&lt;li&gt;Roll the die and the coin together into the rolling area.  &lt;/li&gt;&#xA;&lt;li&gt;If the coin lands face-up, flip it over to be face-down (without looking!).  &lt;/li&gt;&#xA;&lt;li&gt;Imagine a line drawn from the center of the coin to the center of the die. See which of the seven points of the star on the coin is closest to that line, and that's your result.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;See?  Probability can be easy, if you're not wedded to solving it with maths!&lt;/p&gt;&#xA;" OwnerUserId="9286" LastActivityDate="2016-02-10T12:34:55.557" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="26325" PostTypeId="1" AcceptedAnswerId="26329" CreationDate="2016-02-10T13:26:01.093" Score="0" ViewCount="309" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The following puzzle has a spelling mistake: DOUCE should be DOUZE.&lt;br&gt;&#xA;  The correct version is: &#xA;  &lt;a href=&quot;https://puzzling.stackexchange.com/questions/26360/french-alphametic-corrected&quot;&gt;French alphametic (corrected)&lt;/a&gt;&lt;br&gt;&#xA;  (Surprisingly the faulty version allows feasible solutions.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Every letter stands for a digit in base-10 representation, &#xA;different letters stand for different digits:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;         UN &#xA;         UN&#xA;       DEUX&#xA;   +  DOUCE&#xA;  ------------&#xA;      SEIZE&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which digit does each letter represent?&lt;br&gt;&#xA;(Please present the full analysis how these digits can be determined.)&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2016-03-03T10:19:33.680" Title="French alphametic (misspelled)" Tags="|no-computers|alphametic|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="26326" PostTypeId="1" AcceptedAnswerId="26336" CreationDate="2016-02-10T13:36:02.860" Score="1" ViewCount="2446" Body="&lt;p&gt;Have you ever noticed how, on a highway in conditions of heavy traffic, whenever you have to brake for any reason, the braking turns out to be sudden and hard? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same does not seem to happen on slow and/or less busy roads. Can you explain the difference?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S: strictly speaking, this is not a physics problem but in the theory of queues, but the queue tag does not exist. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Try to imagine under which (physical) conditions a smooth section of traffic is brought to a sudden stop rather than a mild slow-down.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19102" LastActivityDate="2016-02-10T18:25:45.507" Title="Sudden braking on the highway" Tags="|physics|" AnswerCount="4" CommentCount="5" ClosedDate="2016-02-12T07:00:42.213" ContentLicense="CC BY-SA 3.0" />
  <row Id="26327" PostTypeId="2" ParentId="26326" CreationDate="2016-02-10T13:58:53.740" Score="10" Body="&lt;p&gt;I have no illustration, but this happens due to the &quot;concertina effect&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Let's say we have 5 cars driving in a straight line, one behind the other, with driver A at front and driver B behind him, and so on with driver E at the back.&lt;/li&gt;&#xA;&lt;li&gt;Let's assume they are traveling at a moderate pace, all at the same speed, spaced two car-lengths from each other. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If driver A were to start braking normally (uniformly), driver B would take some time (some milliseconds) to respond to the brake lights of driver A, depending on his reaction speed. This already forces driver B to brake harder and less uniformly than driver A, as driver B has less distance than driver A to brake in. Now, this transpires with every following vehicle, until driver E experiences a sudden brake, much harsher than driver A would have braked, and therefor compensates by applying an even harder brake (not to mention his reaction time as well). This whole scenario of course neglects other factors which are apparent due to human error (driver not concentrating, driver riding very close to driver in front of him, incorrect handling of brakes, etc.). Therefor, due to the concertina effect, this happens. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why the concertina effect is not as apparent in slower traffic is because there is more time to react. It therefor becomes negligent, relative to driver and brake-mechanics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An interesting thought: if all cars at a red light would start to drive at the same time as the light turns green, the wait time would be the sum of red-light and braking time, with no added queuing (traffic) time. This might some day be the norm with interconnected vehicle communication.  &lt;/p&gt;&#xA;" OwnerUserId="10327" LastActivityDate="2016-02-10T13:58:53.740" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="26328" PostTypeId="2" ParentId="26326" CreationDate="2016-02-10T14:51:14.373" Score="6" Body="&lt;p&gt;Some students at Caltech did a summer research project on this about 30 years ago.  Unfortunately, I cant' find any links to it; I'm not sure the results were ever formally published.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They spent the summer analyzing traffic on the Santa Monica Freeway, particularly during rush hour.  What they found is that traffic will tend to run smoothly, but sometimes there will be a sudden stop for no apparent reason.  All the cars slow to a stop, then continue.  This creates a ripple of slowed cars that moves down the highway (against the flow of traffic).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further study found that this effect is most pronounced when traffic density is highest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The model they came up with is that when one driver slows briefly, the driver behind them will react, also slowing.  The amount of reaction by the following driver depends on the distance between the cars (and the velocity).  If the car in front is far enough away, then the car behind can ignore the glitch and the effect will not propagate.  If the cars are closer together, then the following driver will react by slowing more than the car in front.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this leads to a specific threshold density (which is a nearly-linear function of velocity) where if the density is below the threshold, then glitches in driving behavior will tend to fade out, while above the density, glitches will grow.  If the density is high enough, a simple glitch can grow into a full traffic snarl where everybody stops for no reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Highways are really the only places where this sort of phenomenon can build up.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2016-02-10T14:51:14.373" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26329" PostTypeId="2" ParentId="26325" CreationDate="2016-02-10T14:51:41.817" Score="7" Body="&lt;p&gt;sorry for the format I don't know how to put spoiler to multiple lines of formula. I am just putting spoiler to the answer below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;remaining numbers : 0-1-2-3-4-5-6-7-8-9&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;             UN&#xA;             UN&#xA;           DEUX&#xA;        + DOUCE&#xA;        -------&#xA;          SEIZE&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Giving 0 to E , and since we know S = D + 1 I random them as 8 - 9. I will try giving 2 to O . This is what we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E=0 , D=8 , S=9, O=2, remaining numbers : 1-3-4-5-6-7&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;              UN&#xA;              UN&#xA;            80UX&#xA;        +  82UC0&#xA;         -------&#xA;           90IZ0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now we know that :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N+N+X = 10 or 20&lt;/p&gt;&#xA;&#xA;&lt;p&gt;U+U+U+C + B = Z where B = 1 or 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;U + 0 = I but we know they can't be the same so there is a remainder which means:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;U + W = I where W = 1 or 2&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;        N N X possibilities:&#xA;        a1: 7+7+6&#xA;        a2: 3+3+4&#xA;        a3: 2+2+6&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;To determine whether if W or B can be 1 or 2 I pick a2 to give big numbers to net stage and I get;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E=0 , D=8 , S=9, O=2, N=3, X=4, remaining numbers : 1-5-6-7&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;             U3&#xA;             U3&#xA;           80U4&#xA;        + 82UC0&#xA;        -------&#xA;          90IZ0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;U+U+U+C + 1 = Z&lt;/p&gt;&#xA;&#xA;&lt;p&gt;U + W = I where W = 1 or 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;U can be 5 or 6, and I can 6 or 7 because of remaining numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UUUC possibilities:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;6661 +1 = 20 x Z can't be 0&#xA;6665 +1 = 24 x Z can't be 4&#xA;5557 +1 = 23 x Z can't be 3&#xA;5551 +1 = 17 so let's give our values and rewrite:&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ANSWER:&lt;/strong&gt;           &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;E=0 , D=8 , S=9, O=2, N=3, X=4, U=5, C=1, Z=7, I=6&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;             53&#xA;             53&#xA;           8054&#xA;        + 82510&#xA;        -------&#xA;          90670&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Is my solution but I am sure there are more possibilities and this is not a unique solution because of many possibilities.&lt;/p&gt;&#xA;" OwnerUserId="18685" LastActivityDate="2016-02-10T14:51:41.817" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26330" PostTypeId="1" AcceptedAnswerId="26331" CreationDate="2016-02-10T15:04:09.547" Score="3" ViewCount="529" Body="&lt;p&gt;I am 3/8 of PAKISTAN&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3/8 of ISTANBUL&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2/10 of ANTARCTICA&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;" OwnerUserId="11522" LastEditorUserId="11876" LastEditDate="2016-02-10T16:17:55.840" LastActivityDate="2016-02-10T16:17:55.840" Title="Can anyone tell me, Who am I?" Tags="|wordplay|letter-sequence|" AnswerCount="4" CommentCount="1" ClosedDate="2016-02-10T16:26:27.833" ContentLicense="CC BY-SA 3.0" />
  <row Id="26331" PostTypeId="2" ParentId="26330" CreationDate="2016-02-10T15:06:43.920" Score="7" Body="&lt;p&gt;Well you &lt;em&gt;could&lt;/em&gt; be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 100% of PAKISTAN (&lt;strong&gt;PAK&lt;/strong&gt;ISTAN + &lt;strong&gt;IST&lt;/strong&gt;ANBUL + &lt;strong&gt;AN&lt;/strong&gt;TARCTICA)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or possibly&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the Quebecois profanity &lt;em&gt;TABARNAC!&lt;/em&gt; (also works with spelling TABARNAK)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2016-02-10T15:28:32.490" LastActivityDate="2016-02-10T15:28:32.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26332" PostTypeId="2" ParentId="26330" CreationDate="2016-02-10T15:17:18.087" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; since you just give ratio and no info is given that they must be together, taking 3 letters from Pakistan and Istanbul, 2 letters from anttarctica:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PANTSUIT (&lt;strong&gt;PA&lt;/strong&gt;KISTA&lt;strong&gt;N&lt;/strong&gt; + I&lt;strong&gt;ST&lt;/strong&gt;ANB&lt;strong&gt;U&lt;/strong&gt;L + ANTARC&lt;strong&gt;TI&lt;/strong&gt;CA)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18685" LastActivityDate="2016-02-10T15:17:18.087" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26333" PostTypeId="2" ParentId="26330" CreationDate="2016-02-10T15:24:57.990" Score="4" Body="&lt;p&gt;Perhaps&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a PASTABAR? (&lt;strong&gt;PA&lt;/strong&gt;KI&lt;strong&gt;S&lt;/strong&gt;TAN + IS&lt;strong&gt;TA&lt;/strong&gt;N&lt;strong&gt;B&lt;/strong&gt;UL + ANT&lt;strong&gt;AR&lt;/strong&gt;CTICA)&lt;br&gt;&#xA;&lt;br&gt;&#xA; see for instance &lt;a href=&quot;https://www.facebook.com/mypastabar/&quot;&gt;https://www.facebook.com/mypastabar/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2016-02-10T15:24:57.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26334" PostTypeId="2" ParentId="26330" CreationDate="2016-02-10T15:41:49.790" Score="6" Body="&lt;p&gt;I'd vote for...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SATANICA, the album by Behemoth. Aren't all hidden messages about Satan? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastActivityDate="2016-02-10T15:41:49.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26335" PostTypeId="2" ParentId="26325" CreationDate="2016-02-10T16:55:25.127" Score="3" Body="&lt;p&gt;There are 13 solutions;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;C=3 D=1 E=0 I=5 N=6 O=9 S=2 U=4 X=8 Z=7&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;C=6 D=1 E=0 I=7 N=8 O=9 S=2 U=5 X=4 Z=3 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;C=7 D=3 E=0 I=2 N=1 O=6 S=4 U=9 X=8 Z=5 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;C=1 D=3 E=0 I=6 N=9 O=7 S=4 U=5 X=2 Z=8&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;C=7 D=5 E=0 I=1 N=9 O=4 S=6 U=8 X=2 Z=3 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;C=3 D=5 E=4 I=7 N=1 O=9 S=6 U=2 X=8 Z=0 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;7.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;C=3 D=7 E=1 I=9 N=5 O=4 S=8 U=6 X=0 Z=2 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;8.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;C=0 D=7 E=3 I=4 N=9 O=6 S=8 U=1 X=2 Z=5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;9.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;C=3 D=7 E=4 I=1 N=9 O=6 S=8 U=5 X=2 Z=0 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;10.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;C=1 D=8 E=0 I=6 N=3 O=2 S=9 U=5 X=4 Z=7 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;11.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;C=0 D=8 E=1 I=7 N=4 O=3 S=9 U=5 X=2 Z=6 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;12.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;C=4 D=8 E=2 I=1 N=5 O=3 S=9 U=7 X=0 Z=6 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;13.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;C=0 D=8 E=2 I=3 N=7 O=4 S=9 U=1 X=6 Z=5 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;found it by using a simple alphametic solver.&lt;/p&gt;&#xA;" OwnerUserId="18514" LastActivityDate="2016-02-10T16:55:25.127" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26336" PostTypeId="2" ParentId="26326" CreationDate="2016-02-10T17:12:31.050" Score="0" Body="&lt;p&gt;The system is prone, under certain conditions, to an instability. Let the first car in a long line of cars, pairwise separated by a distance &lt;code&gt;d&lt;/code&gt;, see a very remote obstacle, at distance &lt;code&gt;D&lt;/code&gt;. The driver starts to break immediately, as he tries to avoid brusque brakings. His braking space X_1 is then &lt;code&gt;D&lt;/code&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X_1 = D&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The braking space of the next car is not &lt;code&gt;D&lt;/code&gt;, it is instead&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X_2 = D + d - v T&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where &lt;code&gt;d&lt;/code&gt; is the (mean) distance between cars, &lt;code&gt;v&lt;/code&gt; is the car speed. and &lt;code&gt;T&lt;/code&gt; is an average reaction time. Under certain circumstances, &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X_2 &amp;lt; X_1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;or, &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;d -v T &amp;lt; 0.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The braking space available to the n-th car is,by induction, &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X_n = D + (n-1)(d - v T).&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If &lt;code&gt;d-vT &amp;lt; 0&lt;/code&gt;, it can occur that &lt;code&gt;X_n &amp;lt;= 0&lt;/code&gt;; since &lt;code&gt;d-vT&lt;/code&gt; does not depend on &lt;code&gt;n&lt;/code&gt;, this will occur for a &lt;strong&gt;finite&lt;/strong&gt; &lt;code&gt;n&lt;/code&gt;. This n-th driver will be obliged to brake brusquely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The condition &lt;code&gt;d- v T &amp;lt; 0&lt;/code&gt; is most easily satisfied when &lt;code&gt;d&lt;/code&gt; is small (heavy traffic), or &lt;code&gt;v&lt;/code&gt; is large (highways, as opposed to slower roads). &lt;/p&gt;&#xA;" OwnerUserId="19102" LastEditorUserId="19102" LastEditDate="2016-02-10T17:18:21.893" LastActivityDate="2016-02-10T17:18:21.893" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26337" PostTypeId="2" ParentId="26326" CreationDate="2016-02-10T18:11:49.037" Score="2" Body="&lt;p&gt;Not sure what you're going for, but here's my guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When driving on the highway, you're driving fast and expect to be driving fast with no interruptions. When something does happen to cause you to have to brake, it's a surprise, and because of your high speed, you react harder and brake faster. When driving on a slow neighborhood road, you're driving slowly and expect to be stopping/slowing more often, so you're ready.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Driving fast takes a harder deceleration to slow in a reasonable amount of time than while driving slowly&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13101" LastEditorUserId="13101" LastEditDate="2016-02-10T18:25:45.507" LastActivityDate="2016-02-10T18:25:45.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26338" PostTypeId="2" ParentId="26168" CreationDate="2016-02-10T18:43:42.993" Score="3" Body="&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Canyon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am spread out like a leaf.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Canyons spread out like a leaf where the branches of the river that form the canyon look like the veins that carry the water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That what forms me cuts me, but builds me up.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; River forms the canyon and it cuts through the ground. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That what cuts me seeks always to go lower, but lifts its bed higher.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because of &lt;a href=&quot;https://en.wikipedia.org/wiki/Downcutting&quot; rel=&quot;nofollow&quot;&gt;downcutting&lt;/a&gt;, the canyon gets deeper and the stream bed stays higher.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am commonly found where what builds me is less common.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Canyons are more common in arid than in wet areas.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19157" LastActivityDate="2016-02-10T18:43:42.993" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="26340" PostTypeId="1" CreationDate="2016-02-10T18:58:43.253" Score="10" ViewCount="551" Body="&lt;p&gt;Guess a 6 digits number in the form of $\ abcdef$ which holds Properties #1, #2 and #3. The real challenge is to find 4 other interesting properties of this number:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Property #1&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$\ a+d=b+e=c+f=9 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ ab+cd+ef=99 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ abc+def=999 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ abcd+efab+cdef=9999 $&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Property #2&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$\ abcdef^2= ghijklmnopq$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ ghijk+lmnopq=abcdef$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Property #3&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$\ def^2 - abc^2=fabcde $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ efa^2 - bcd^2=abcdef $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ fab^2 - cde^2=bcdefa$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Property #4 (found!)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ 1\times abcdef=abcdef$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\ 2\times abcdef=cdefab$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\ 3\times abcdef=bcdefa$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\ 4\times abcdef=efabcd$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\ 5\times abcdef=fabcde$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\ 6\times abcdef=defabc$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Property #5 &#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;hint: Improve the equation below to gain cyclic number similar to previous properties (p is a prime number):&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ \frac1p =0.\overline{abcdef}$&#xA;  ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Property #6&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;hint: following formula can be starting point for reproducing cyclic number( r and s are two constant numbers that you need to find):&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\ r^1\mod\ s=c $&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Property #7&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;hint: following formula can be starting point for reproducing cyclic number(p is a prime number):&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\ 10=3+(p \times a) $&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note: What I mean by $ abc $ is concatenation of numbers or mathematicaly speaking, $ abc $ is $ (100*a)+(10*b)+c $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spoiler:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Prime number which produce this cyclic number is p=7&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18675" LastEditorUserId="18675" LastEditDate="2016-02-13T09:58:02.123" LastActivityDate="2016-02-13T09:58:02.123" Title="Find the missing properties of cyclic number" Tags="|mathematics|number-sequence|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="26341" PostTypeId="2" ParentId="26107" CreationDate="2016-02-10T19:05:14.850" Score="3" Body="&lt;p&gt;My guess is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Physics&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To this riddle one shall add &quot;meta&quot;&lt;br&gt;&#xA;Like a program and an SE site&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you prepend &quot;meta&quot; to &quot;physics&quot;, you get &quot;metaphysics&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the answer you have some data&lt;br&gt;&#xA;Yet not always you will get it right&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In &quot;metaphysics&quot;, you have some data but it is not always facts and you often come up with scientific theories. Thus, you are not always right.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19157" LastActivityDate="2016-02-10T19:05:14.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26342" PostTypeId="2" ParentId="26248" CreationDate="2016-02-10T19:25:46.520" Score="0" Body="&lt;p&gt;It's a little convoluted but...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Before they are painted, each dwarf agrees to be numbered sequentially (starting with 1) - let's suppose there are M dwarves in all, and every dwarf knows every other dwarf's number.  On the first day after being painted, the odd numbered dwarves goes to the dining hall at a specific time.  On the second day, the even numbered dwarves go to the dining hall if and only if the dwarf with the number 1 lower than their number was present the day before and had a blue dot.  The odd numbered dwarves look into the hall and can then determine who among them has a blue or a red dot.  The process can then be repeated with the odd dwarves acting as the signal on the third and fourth days.  In the case where there is an odd number of dwarves - there will be one left out, but could be dealt with by assigning them an odd partner that will be present/absent on the second day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19159" LastActivityDate="2016-02-10T19:25:46.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26343" PostTypeId="2" ParentId="25960" CreationDate="2016-02-10T19:56:56.483" Score="6" Body="&lt;p&gt;Denote &lt;span class=&quot;math-container&quot;&gt;$C_i$&lt;/span&gt; the carry over from the sum of the digits in column &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$S_i$&lt;/span&gt; be the sum of the digits in column &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; plus &lt;span class=&quot;math-container&quot;&gt;$C_{i-1}$&lt;/span&gt;.  &lt;span class=&quot;math-container&quot;&gt;$C_i$&lt;/span&gt; is simply &lt;span class=&quot;math-container&quot;&gt;$S_i \div 10$&lt;/span&gt; rounded down.  Lastly, denot &lt;span class=&quot;math-container&quot;&gt;$D_i$&lt;/span&gt; to be the ones digit of &lt;span class=&quot;math-container&quot;&gt;$S_i$&lt;/span&gt;. In other words, &lt;span class=&quot;math-container&quot;&gt;$D_i = S_i \mod 10$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Since the sum of all the numbers is still a 5 digit number, we know that &lt;span class=&quot;math-container&quot;&gt;$H \in \{1,2,3\}$&lt;/span&gt;.  If &lt;span class=&quot;math-container&quot;&gt;$H=3$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$A=9$&lt;/span&gt;.  But then in the 4th column you have 3 &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;s, which would mean &lt;span class=&quot;math-container&quot;&gt;$C_4 \ge 2$&lt;/span&gt;, making a six digit number.  Therefore, &lt;span class=&quot;math-container&quot;&gt;$H \in \{1,2\}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;This means, &lt;span class=&quot;math-container&quot;&gt;$A \ge 3$&lt;/span&gt;.  Since &lt;span class=&quot;math-container&quot;&gt;$D \ne 0$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$A=3 \implies C_4=1$&lt;/span&gt;, which results in &lt;span class=&quot;math-container&quot;&gt;$A=4$&lt;/span&gt;.  Thus, &lt;span class=&quot;math-container&quot;&gt;$A \ne 3$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$A=4$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$C_4=1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$H=1$&lt;/span&gt;.  Then &lt;span class=&quot;math-container&quot;&gt;$S_4=12+D+C_3$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D_4=H=1$&lt;/span&gt;.  This can only happen if &lt;span class=&quot;math-container&quot;&gt;$D+C_3=9 \implies C_4=2$&lt;/span&gt;.  Thus, &lt;span class=&quot;math-container&quot;&gt;$A \ne 4$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$A=6$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$C_4=0$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$H=2$&lt;/span&gt;, or &lt;span class=&quot;math-container&quot;&gt;$C_4=3$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$H=1$&lt;/span&gt;.  &lt;span class=&quot;math-container&quot;&gt;$S_4=18+D+C_3=H$&lt;/span&gt;.  If &lt;span class=&quot;math-container&quot;&gt;$H=1$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$D+C_3=13$&lt;/span&gt;, which only works if &lt;span class=&quot;math-container&quot;&gt;$C_3$&lt;/span&gt; could be 4, but this is only possible when you are summing five or more numbers.  Thus, &lt;span class=&quot;math-container&quot;&gt;$H=2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_4=0$&lt;/span&gt;.  But &lt;span class=&quot;math-container&quot;&gt;$S_4 \gt 3 \times A = 18 \implies C_4 \gt 0$&lt;/span&gt;.  Thus &lt;span class=&quot;math-container&quot;&gt;$A \ne 6$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$A=7$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$C_4=1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$H=2$&lt;/span&gt;, or &lt;span class=&quot;math-container&quot;&gt;$C_4=4$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$H=1$&lt;/span&gt;.  Since &lt;span class=&quot;math-container&quot;&gt;$S_4=3\times 7+D+C_3=21+D+C_3$&lt;/span&gt;, we know that &lt;span class=&quot;math-container&quot;&gt;$C_4$&lt;/span&gt; can only be 4 when &lt;span class=&quot;math-container&quot;&gt;$D+C_3 \ge 19$&lt;/span&gt; (which can't happen), and is at least &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt;.  Thus, &lt;span class=&quot;math-container&quot;&gt;$C_4=1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$H=2$&lt;/span&gt;.  But &lt;span class=&quot;math-container&quot;&gt;$S_4\ge3\times7=21 \implies C_4\ge2$&lt;/span&gt;.  Thus &lt;span class=&quot;math-container&quot;&gt;$A \ne 7$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$A=8$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$C_4=2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$H=2$&lt;/span&gt;.  &lt;span class=&quot;math-container&quot;&gt;$S_4=3 \times 8 + D + C_3=24 + D +C_3$&lt;/span&gt;.  We know that &lt;span class=&quot;math-container&quot;&gt;$D_4=H=2$&lt;/span&gt; only works when &lt;span class=&quot;math-container&quot;&gt;$D+C_3=8 \implies C_4=3$&lt;/span&gt;.  Thus, &lt;span class=&quot;math-container&quot;&gt;$A \ne 8$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;After all this, we know that &lt;span class=&quot;math-container&quot;&gt;$A \in \{5, 9\}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$H \in \{1, 2\}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;h3&gt;Assume &lt;span class=&quot;math-container&quot;&gt;$A=5$&lt;/span&gt;, so &lt;span class=&quot;math-container&quot;&gt;$H=1$&lt;/span&gt;&lt;/h3&gt;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$A=5$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$S_4=15+D+C_3$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D_4=H=1$&lt;/span&gt;.  Thus, &lt;span class=&quot;math-container&quot;&gt;$D+C_3=6$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_4=2$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$C_3=0$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$P=1$&lt;/span&gt;, which is already taken by &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt;. If &lt;span class=&quot;math-container&quot;&gt;$C_3 = 1$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$S_4=3 \times 5 + D + C_3 = 15 + D + 1 = 16 + D$&lt;/span&gt;.  We know that &lt;span class=&quot;math-container&quot;&gt;$D_4=H=1$&lt;/span&gt; only works if &lt;span class=&quot;math-container&quot;&gt;$D=5$&lt;/span&gt;, which is already taken by &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;.   We know that &lt;span class=&quot;math-container&quot;&gt;$C_3 \ne 4$&lt;/span&gt; because the highest &lt;span class=&quot;math-container&quot;&gt;$S_3$&lt;/span&gt; can be is when &lt;span class=&quot;math-container&quot;&gt;$P=9$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_1=3$&lt;/span&gt;, which makes &lt;span class=&quot;math-container&quot;&gt;$S_3=35$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Thus &lt;span class=&quot;math-container&quot;&gt;$C_3 \in \{2,3\}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D \in \{4,3\}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Lets look at &lt;span class=&quot;math-container&quot;&gt;$S_3=3 \times P + 5 + C_2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$S_2=3 \times P + Y + C_1$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$P \in \{2,3,4\}$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$C_3=1$&lt;/span&gt;.  Thus, &lt;span class=&quot;math-container&quot;&gt;$P \ge 6$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$P=6$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$S_2=18+Y+C_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D_2=5$&lt;/span&gt;.  So &lt;span class=&quot;math-container&quot;&gt;$Y+C_1=7 \implies C_2=2$&lt;/span&gt;. Thus, &lt;span class=&quot;math-container&quot;&gt;$S_3=12+5+2=19 \implies E=9$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_3=1$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$P=7$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$S_2=21+Y+C_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D_2=5$&lt;/span&gt;. So &lt;span class=&quot;math-container&quot;&gt;$Y+C_1=4$&lt;/span&gt;.  Thus, &lt;span class=&quot;math-container&quot;&gt;$Y \in \{2,3\}$&lt;/span&gt;.  If &lt;span class=&quot;math-container&quot;&gt;$Y=2$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$S_1=6+S$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D_1=D$&lt;/span&gt;.  If &lt;span class=&quot;math-container&quot;&gt;$D=4$&lt;/span&gt; then &lt;span class=&quot;math-container&quot;&gt;$S=7$&lt;/span&gt; which is a contradiction.  Thus, &lt;span class=&quot;math-container&quot;&gt;$S=6$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D=3$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_3=3$&lt;/span&gt;. &lt;span class=&quot;math-container&quot;&gt;$S_3=21+5+C_2=26+C_2 \implies C_2=4$&lt;/span&gt;. This is impossible when &lt;span class=&quot;math-container&quot;&gt;$P=7$&lt;/span&gt;, so &lt;span class=&quot;math-container&quot;&gt;$Y=3$&lt;/span&gt;.  Then &lt;span class=&quot;math-container&quot;&gt;$C_1=1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$S_1=9+S$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D_1=D \in \{3,4\}$&lt;/span&gt;.  If &lt;span class=&quot;math-container&quot;&gt;$D=4$&lt;/span&gt; then &lt;span class=&quot;math-container&quot;&gt;$S=3$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$P=8$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$S_3=24+5+C_2=29+C_2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_3=3$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D=3$&lt;/span&gt;.  From &lt;span class=&quot;math-container&quot;&gt;$S_2=24+Y+C_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D_2=5$&lt;/span&gt;, we know that &lt;span class=&quot;math-container&quot;&gt;$Y+C_1 \in \{1,11\}$&lt;/span&gt;.  If &lt;span class=&quot;math-container&quot;&gt;$Y=0$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$C_1=0$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$Y \ne 1$&lt;/span&gt; since that is already taken by &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt;.  Thus, &lt;span class=&quot;math-container&quot;&gt;$Y+C_1=11 \implies Y=9$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_1=2$&lt;/span&gt;.  But &lt;span class=&quot;math-container&quot;&gt;$S_1=27+S$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D_1=D=3 \implies S=6$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_1=3$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$P=9$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$S_3=27+5+C_2=32+C_2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_3=3$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D=3$&lt;/span&gt;.  From &lt;span class=&quot;math-container&quot;&gt;$S_2=27+Y+C_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D_2=5$&lt;/span&gt;, we know that &lt;span class=&quot;math-container&quot;&gt;$Y+C_1=8$&lt;/span&gt;. If &lt;span class=&quot;math-container&quot;&gt;$C_1=1$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$Y=7$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_1=2$&lt;/span&gt;.  Thus, &lt;span class=&quot;math-container&quot;&gt;$C_1=2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$Y=6$&lt;/span&gt;. This makes &lt;span class=&quot;math-container&quot;&gt;$S=5$&lt;/span&gt; which is already taken by &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;h2&gt;Thus &lt;span class=&quot;math-container&quot;&gt;$A=9$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$H=2$&lt;/span&gt;&lt;/h2&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$S_4=27+D+C_3 \implies D+C_3=5$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_4=3$&lt;/span&gt;.  Since &lt;span class=&quot;math-container&quot;&gt;$D \ne 2$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$C_3 \ne 3$&lt;/span&gt;.  Any value of &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; will result in &lt;span class=&quot;math-container&quot;&gt;$C_3 \ge 1$&lt;/span&gt;, so  &lt;span class=&quot;math-container&quot;&gt;$C_3 \in \{1,2\}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D \in \{4,3\}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$P=8$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$S_3=32+C_2 \implies C_3=3$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$P=7$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$C_2 \ge 2 \implies S_3=29+C_2\ge31 \implies C_3=3$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$P=1$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$S_3=12+C_2 \implies C_3=1 \implies D=4$&lt;/span&gt;.  &lt;span class=&quot;math-container&quot;&gt;$S_2=3+Y+C_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D_2=A=9 \implies Y+C_1=6$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_2=0$&lt;/span&gt;.  Thus &lt;span class=&quot;math-container&quot;&gt;$S_3=12$&lt;/span&gt; so &lt;span class=&quot;math-container&quot;&gt;$D_3=E=2$&lt;/span&gt; which is already taken.&lt;/p&gt;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$P=3$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$S_3=18+C_2$&lt;/span&gt;.  If &lt;span class=&quot;math-container&quot;&gt;$C_2\ge2$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$C_3=2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D=3$&lt;/span&gt;, which is already taken by &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt;.  If &lt;span class=&quot;math-container&quot;&gt;$C_2=1$&lt;/span&gt; then &lt;span class=&quot;math-container&quot;&gt;$D_3=E=9$&lt;/span&gt; which is already taken. If &lt;span class=&quot;math-container&quot;&gt;$C_2=0$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$S_2=9+Y+C1 \implies Y+C_1=0$&lt;/span&gt;.  Thus, &lt;span class=&quot;math-container&quot;&gt;$Y=0$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$S=D$&lt;/span&gt;, which is not allowed.&lt;/p&gt;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$P=4$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$S_3=21+C_2 \implies C_3=2 \implies D=3$&lt;/span&gt;. &lt;span class=&quot;math-container&quot;&gt;$S_2=12+Y+C_1 \implies Y+C_1=7$&lt;/span&gt; so that &lt;span class=&quot;math-container&quot;&gt;$D_2=A=9$&lt;/span&gt;.  But then &lt;span class=&quot;math-container&quot;&gt;$C_2=1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D_3=2$&lt;/span&gt;, which is already taken by &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$P=5$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$S_3=24+C_2 \implies C_3=2 \implies D=3$&lt;/span&gt;. &lt;span class=&quot;math-container&quot;&gt;$S_2=15+Y+C_1 \implies Y+C_1=4$&lt;/span&gt; so that &lt;span class=&quot;math-container&quot;&gt;$D_2=A=9$&lt;/span&gt;.  But then &lt;span class=&quot;math-container&quot;&gt;$C_2=1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D_3=5$&lt;/span&gt; which is already taken by &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;h2&gt;Therefore &lt;span class=&quot;math-container&quot;&gt;$P=6$&lt;/span&gt;&lt;/h2&gt;&#xA;&lt;p&gt;From &lt;span class=&quot;math-container&quot;&gt;$S_2=18+Y+C_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D_2=A=9$&lt;/span&gt;, we know that &lt;span class=&quot;math-container&quot;&gt;$Y+C_1 \in \{1,11\}$&lt;/span&gt;.  If &lt;span class=&quot;math-container&quot;&gt;$Y+C_1=11$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$Y=8$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_1=3$&lt;/span&gt;.  But then &lt;span class=&quot;math-container&quot;&gt;$S_1=24+S$&lt;/span&gt;.  No value of &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; can make &lt;span class=&quot;math-container&quot;&gt;$C_1=3$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D\in \{4,3\}$&lt;/span&gt;.  Therefore, &lt;span class=&quot;math-container&quot;&gt;$Y+C_1=1$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$S_2=19$&lt;/span&gt;, so &lt;span class=&quot;math-container&quot;&gt;$C_2=1$&lt;/span&gt;. If &lt;span class=&quot;math-container&quot;&gt;$Y=0$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$C_1=0$&lt;/span&gt; as well, which is a contradiction, so &lt;span class=&quot;math-container&quot;&gt;$Y=1$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;This means that &lt;span class=&quot;math-container&quot;&gt;$S_3=18+9+C_2=28$&lt;/span&gt;, so &lt;span class=&quot;math-container&quot;&gt;$D_3=E=8$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_3=2$&lt;/span&gt;.  Thus, &lt;span class=&quot;math-container&quot;&gt;$D=3$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Lastly, &lt;span class=&quot;math-container&quot;&gt;$S_1=3+S=D=3 \implies S=0$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Full solution:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$$H=2$$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$$A=9$$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$$P=6$$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$$Y=1$$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$$E=8$$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$$D=3$$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$$S=0$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-02-11T18:36:54.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26344" PostTypeId="2" ParentId="26340" CreationDate="2016-02-10T20:09:39.500" Score="15" Body="&lt;p&gt;Given that $abc+def=999$, we must have $a+d=b+e=c+f=9$ because there's no way to produce a carry anywhere. From property 3, we must have $d&amp;gt;a$, $e&amp;gt;b$, $f&amp;gt;c$, otherwise the numbers on the right would be negative. $$fabcde=def^2-abc^2=(def+abc)(def-abc)=999(def-abc)=999(def+def-999)=999(200d+20e+2f-999)$$ But $$fabcde=1000(fab)+cde=1000(fab)+(999-fab)=999(fab+1)=999(100f+10a+b+1)=999(100f-10d-e+100)$$ So $200d+20e+2f-999=100f-10d-e+100$, which we can rearrange to get $30d+3e=157+14f$. $157+14f$ must be divisible by 3, while $f$ must be at least 5 and at most 9. The only possibility is that $f=7$. Then $abcdef=142857$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some other properties of the number $142857$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$2\times142857=285714$, which is the same number but shifted over. The same thing happens for 3, 4, 5, or 6 times.&lt;/li&gt;&#xA;&lt;li&gt;$7\times142857=999999$.&lt;/li&gt;&#xA;&lt;li&gt;If you multiply $142857$ by &lt;em&gt;any positive integer&lt;/em&gt;, add the last six digits to the rest, and continue until you have only six digits, the result will be $999999$ or a cyclic shift of $142857$. This results from a combination of the previous two properties.&lt;/li&gt;&#xA;&lt;li&gt;$0.\overline{142857}=\frac{1}{7}$, which is part of the explanation for why all of these properties hold.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2016-02-10T20:09:39.500" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26345" PostTypeId="1" AcceptedAnswerId="26398" CreationDate="2016-02-10T21:51:06.043" Score="13" ViewCount="605" Body="&lt;p&gt;Comes from the past but it won't last&lt;br&gt;&#xA;It might shine first but it will rust&lt;br&gt;&#xA;Ideally, there are three friends he can trust&lt;br&gt;&#xA;They keep him busy, so it won't gather dust&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is it?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Hint: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Title&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second stanza as the second hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It both stays and moves &lt;br&gt;&#xA; If it obeys the rules &lt;br&gt;&#xA; A genius baby proves &lt;br&gt;&#xA; When he gently pulls&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19157" LastEditorUserId="19157" LastEditDate="2016-02-12T15:29:43.037" LastActivityDate="2016-02-12T17:10:17.183" Title="Got the balls to answer?" Tags="|riddle|rhyme|poetry|" AnswerCount="4" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="26347" PostTypeId="1" AcceptedAnswerId="26369" CreationDate="2016-02-10T22:15:12.343" Score="4" ViewCount="424" Body="&lt;p&gt;I am the reason you are here&lt;br&gt;&#xA;The salvation of the digital world&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am quite simple &lt;br&gt;&#xA;as I am quite hyper at first&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet I can sit &lt;br&gt;&#xA;And create speeches and essays for everyone&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can fly like the wind &lt;br&gt;&#xA;No restrictions from gravity &lt;br&gt;&#xA;And a good mark in my criminal record&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You speak of me &lt;br&gt;&#xA;By mouth or text, who knows what's next?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Who am I?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9042" LastActivityDate="2016-02-14T21:15:07.357" Title="Stack Overflow Relationship... What am I?" Tags="|riddle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26348" PostTypeId="2" ParentId="26251" CreationDate="2016-02-11T00:05:45.387" Score="0" Body="&lt;p&gt;I'm only giving this answer because I saw a lateral-thinking tag.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the prisoners bloody the tips of their noses.  This gives them a red mark regardless of the previous color.  All the prisoners then group on the same wall.&lt;/p&gt;&#xA;" OwnerUserId="457" LastActivityDate="2016-02-11T00:05:45.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26350" PostTypeId="2" ParentId="26347" CreationDate="2016-02-11T09:40:09.663" Score="5" Body="&lt;p&gt;I think you are :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reputation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the reason you are here&#xA;The salvation of the digital world&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We are here for sharing our knowledge and gain reputation. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am quite simple&#xA;as I am quite hyper at first&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; At first, it is hard to gain reputation but after a while it becomes easier&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet I can sit&#xA;And create speeches and essays for everyone&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reputation details give all the information about us, badges and achievements are the reference for speeches and essays.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can fly like the wind&#xA;No restrictions from gravity&#xA;And a good mark in my criminal record&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reputation has no limits, you can go higher and earn everyday. And while gaining it, your record becomes better as well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You speak of me&#xA;By mouth or text, who knows what's next?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This I can't directly relate but, may be unlocking more privilages by gaining reputation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18685" LastActivityDate="2016-02-11T09:40:09.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26351" PostTypeId="1" AcceptedAnswerId="26355" CreationDate="2016-02-11T09:57:03.703" Score="8" ViewCount="1014" Body="&lt;p&gt;Alice and Bob play the following game with two heaps of 10,000 and 20,000 marbles.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Alice and Bob move alternatingly. Alice makes the first move.&lt;/li&gt;&#xA;&lt;li&gt;In every move, the active player may&lt;br /&gt;&#xA;(i) either remove  ≥ 1 marbles from the first heap,&lt;br /&gt;&#xA;(ii) or remove  ≥ 1 marbles from the second heap,&lt;br /&gt;&#xA;(iii) or remove  ≥ 1 marbles from the first heap and  ≥ 1 marbles from the second heap, where  +  is divisible by 2016.&lt;/li&gt;&#xA;&lt;li&gt;The player who takes the &lt;em&gt;last&lt;/em&gt; marble wins the game.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Which player is going to win this game?&lt;br /&gt;&#xA;(As usual, we assume that Alice and Bob both use optimal strategies.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="69582" LastEditDate="2020-12-27T16:04:07.637" LastActivityDate="2020-12-27T16:04:07.637" Title="Heaps of marbles" Tags="|mathematics|strategy|game|nim|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26353" PostTypeId="1" AcceptedAnswerId="26396" CreationDate="2016-02-11T11:38:07.390" Score="13" ViewCount="3415" Body="&lt;p&gt;My altered name is lurking slyly here&lt;br&gt;&#xA;As my self-referential words make clear.&lt;br&gt;&#xA;You've guessed my easy riddle now, I fear.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The title could have been: &quot;My 25% Odds Acrostic Riddle&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://www.google.ch/webhp?sourceid=chrome-instant&amp;amp;ion=1&amp;amp;espv=2&amp;amp;ie=UTF-8#q=acrostic&quot; rel=&quot;noreferrer&quot;&gt;https://www.google.ch/webhp?sourceid=chrome-instant&amp;amp;ion=1&amp;amp;espv=2&amp;amp;ie=UTF-8#q=acrostic&lt;/a&gt;&#xA; In this case, I used another marker&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hinting again, and summarizing clues from comments&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One piece of punctuation was a misdirection&lt;br&gt;&#xA; All parts of the puzzle (everything above &quot;A hint:&quot;) are significant&lt;br&gt;&#xA; This is not a usual acrostic. The title is a key or map to the solution  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For instance&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &quot;My&quot; in the title is a clue that the word &quot;my&quot; is the acrostic marker&lt;br&gt;&#xA; 3 2 1 is not an ordering and the digits may mean different things  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is always a trick&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The word &quot;self-referential&quot; applies in more than one way here&lt;br&gt;&#xA; There is a certain tricky sense in which this puzzle is indeed easy  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Want something specific?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are six letters in the solution.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastEditorUserId="17250" LastEditDate="2016-02-12T15:51:19.960" LastActivityDate="2016-02-12T15:51:19.960" Title="My 12:57 Acrostic Riddle" Tags="|riddle|word|" AnswerCount="10" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="26354" PostTypeId="2" ParentId="26353" CreationDate="2016-02-11T11:41:21.800" Score="10" Body="&lt;p&gt;Your name could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My alter&lt;strong&gt;ed na&lt;/strong&gt;me is lur&lt;strong&gt;king sly&lt;/strong&gt;ly here As my self-referential words make clear. You've guessed my easy riddle now, I fear.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But it probably isn't&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's probably &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MAY, based on the work some others have done (12:57 $\rightarrow$ &quot;three to one&quot; $\rightarrow$ 3 2 1 $\rightarrow$ Y A M + &quot;altered&quot; $\rightarrow$ MAY), as well as the &quot;acrostic&quot; part of the hint.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And yet! It could &lt;em&gt;also&lt;/em&gt; be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;em&gt;ESA&lt;/em&gt;&lt;/strong&gt; (this sounds like a stretch, but the last couple of attempts from myself and Kate Gregory were rejected) based on &quot;acrostic&quot; and OP's hint that his &lt;em&gt;marker&lt;/em&gt; had changed. Self-referential words could refer to the recurring &quot;my&quot;, and the first letter after each &quot;my&quot; are A, S, and E, in that order. Giving them the 3-2-1 treatment mentioned by Kate Gregory, we get Esa, which the internet &lt;em&gt;assures&lt;/em&gt; me is a name.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;sub&gt;If we're &lt;em&gt;also&lt;/em&gt; taking &quot;altered&quot; into account, the name could be &quot;SEA&quot;. Yes, I literally know a person named Sea.&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2016-02-11T17:36:46.730" LastActivityDate="2016-02-11T17:36:46.730" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="26355" PostTypeId="2" ParentId="26351" CreationDate="2016-02-11T11:57:15.973" Score="18" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alice wins&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is how:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alice first takes $9.112$ marbles from the first heap and $19.112$ from the second heap ($9.112 + 19.112 = 28.224 = 2.016 * 14$), leaving 888 marbles on each heap. From that on she just mirrors Bob's move keeping the heaps equal sized, making it impossible for Bob to win.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2016-02-11T11:57:15.973" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26356" PostTypeId="2" ParentId="26353" CreationDate="2016-02-11T12:00:54.993" Score="15" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Me &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;1.&lt;/strong&gt; it's a different ('altered') version of one's name. &lt;br/&gt;&lt;strong&gt;2.&lt;/strong&gt; It's self-referential. &lt;br/&gt; &lt;strong&gt;3.&lt;/strong&gt; It's the 12th and 57th letter of the question respectively (see the title - 12:57).&lt;br/&gt; &lt;strong&gt;4.&lt;/strong&gt; It's answered by the time you read the third line which says 'You've guessed my easy riddle now' i.e. before the third line, the answer is apparent.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19086" LastEditorUserId="19086" LastEditDate="2016-02-11T19:05:11.087" LastActivityDate="2016-02-11T19:05:11.087" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="26357" PostTypeId="2" ParentId="26353" CreationDate="2016-02-11T12:04:04.773" Score="7" Body="&lt;p&gt;I suspect your name is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;M&lt;/strong&gt;y altered name is lurking slyly here&lt;br&gt;&#xA;&lt;strong&gt;A&lt;/strong&gt;s my self-referential words make clear.&lt;br&gt;&#xA;&lt;strong&gt;Y&lt;/strong&gt;ou've guessed my easy riddle now, I fear.&lt;br&gt;&#xA;&lt;strong&gt;A&lt;/strong&gt;nswer?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How this fits (in my opinion): &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Maya is a common female name. The clues 'self-referential' and 'lurking slyly' make me think I'm onto something by picking up the initials of each line. Also, Maya sounds like the last name &lt;a href=&quot;https://puzzling.stackexchange.com/users/18924/hugh-meyers&quot;&gt;Meyers&lt;/a&gt; of the OP, which is where 'altered name' fits well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2016-02-11T12:04:04.773" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26358" PostTypeId="2" ParentId="26353" CreationDate="2016-02-11T12:41:29.437" Score="20" Body="&lt;p&gt;The time 12:57 is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Three minutes to 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And 25% odds are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Three to one&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the pattern of lines is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3, then 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the &quot;name&quot; is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 321&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since we have 3 lines, and the word acrostic in the title, we can use the first letter of each line, in this order:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Y A M&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But then it says &quot;altered&quot; meaning we can use whatever order we like after all.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Of the permutations AYM, AMY, MAY, MYA, YAM, and YMA, May is possibly a name,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I think&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Amy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is more clearly and obviously a name.&lt;/p&gt;&#xA;" OwnerUserId="1509" LastEditorUserId="1509" LastEditDate="2016-02-11T15:40:20.593" LastActivityDate="2016-02-11T15:40:20.593" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="26359" PostTypeId="2" ParentId="26353" CreationDate="2016-02-11T13:12:58.043" Score="7" Body="&lt;p&gt;Very similar to @KateGregory 's answer, my guess would be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; YAM&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3 to one ~ three two one ~ line three, line two, line one&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;thats not a good answer, I fear&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but something something here&lt;/p&gt;&#xA;&#xA;&lt;p&gt;rhyming is difficult, my dear&lt;/p&gt;&#xA;&#xA;&lt;p&gt;second guess &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; man (3 2 1 of name, as in &lt;em&gt;altered name&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18380" LastActivityDate="2016-02-11T13:12:58.043" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26360" PostTypeId="1" AcceptedAnswerId="26366" CreationDate="2016-02-11T13:19:09.673" Score="5" ViewCount="279" Body="&lt;p&gt;This is the corrected version of puzzle &#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/26325/french-alphametic&quot;&gt;French alphametic (misspelled)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every letter stands for a digit in base-10 representation, &#xA;different letters stand for different digits, &#xA;and &lt;strong&gt;the four summands and the sum are even&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;         UN &#xA;         UN&#xA;       DEUX&#xA;   +  DOUZE&#xA;  ------------&#xA;      SEIZE&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which digit does each letter represent?&lt;br&gt;&#xA;(Please present the full analysis how these digits can be determined.)&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2016-08-30T01:51:01.983" Title="French alphametic (corrected)" Tags="|no-computers|alphametic|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26362" PostTypeId="2" ParentId="26360" CreationDate="2016-02-11T14:01:44.863" Score="1" Body="&lt;p&gt;in order to be U+U+U+Z = Z , U+U+U needs to be 10 or 20, considering it will get a remainder of 1 or 2 from first line. so only alternatives are 3 or 6. I try giving 6.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;        N + N + X = 10 or 20&#xA;        118&#xA;        226 x&#xA;        334&#xA;        442&#xA;        550&#xA;        668 x&#xA;        776 x&#xA;        884&#xA;        992 - will try this&#xA;&#xA;              69&#xA;              69&#xA;            DE62&#xA;         + DO6ZE&#xA;         -------&#xA;           SEIZE&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I give 3 to E for making min +2 remainder&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;             69&#xA;             69&#xA;           D362&#xA;        + DO6Z3&#xA;        -------&#xA;          S3IZ3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here is what I've got now: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;N=9, X=2 , U=6 , E=3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;remaining: 1,4,5,7,8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so I has to 1. By brute force trying 7 to Z, I get:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;             69&#xA;             69&#xA;           D362&#xA;        + DO673&#xA;        -------&#xA;          S3173&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;N=9, X=2 , U=6 , E=3 , I=1 , Z=7&lt;/p&gt;&#xA;&#xA;&lt;p&gt;remaining: 4,5,8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have no alternative leaving to get the solution of:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;             69&#xA;             69&#xA;           4362&#xA;        + 48673&#xA;        -------&#xA;          53173&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;because D must be S-1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;N=9, X=2 , U=6 , E=3 , I=1 , Z=7 , D=4 , S=5 , O=8&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18685" LastActivityDate="2016-02-11T14:01:44.863" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26363" PostTypeId="2" ParentId="26353" CreationDate="2016-02-11T14:13:52.120" Score="2" Body="&lt;p&gt;the word is &quot;easy&quot;!! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My&lt;/strong&gt; altered name is &lt;strong&gt;lurking&lt;/strong&gt; slyly here&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As &lt;strong&gt;my&lt;/strong&gt; self-referential &lt;strong&gt;words&lt;/strong&gt; make clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You've guessed &lt;strong&gt;my&lt;/strong&gt; &lt;strong&gt;easy&lt;/strong&gt; riddle now, I fear.&lt;/p&gt;&#xA;" OwnerUserId="19172" LastActivityDate="2016-02-11T14:13:52.120" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26364" PostTypeId="1" AcceptedAnswerId="27510" CreationDate="2016-02-11T14:23:46.507" Score="7" ViewCount="506" Body="&lt;p&gt;You'll toss me on the ground&lt;br&gt;&#xA;I'll jump at you just in case&lt;br&gt;&#xA;You'll help me off a horse&lt;br&gt;&#xA;I'll lift you up instead&lt;br&gt;&#xA;You'll mount me on your boat&lt;br&gt;&#xA;I'll shout the time for you&lt;br&gt;&#xA;You'll wager your house with me&lt;br&gt;&#xA;Then take things back illegally&lt;br&gt;&#xA;&lt;br&gt;&#xA;Who/what am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is a single word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can wager other things as well, not just your house. Usually it'll just be money.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The ground will usually be grass, though I've seen people toss it on gravel many times.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I won't literally shout the time out loud, just make loudly sure you know it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The wager will take place in a specific family of casino/gambling games.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 6:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'll lift you up a short distance off the ground. You won't fly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 7:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You'll usually mount me on warships, rather than lowly boats, and you'll take me off while under way.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 8:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This item is tossed on the ground as part of a game.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 9:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is a four-letter word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18064" LastEditorUserId="18064" LastEditDate="2016-02-16T12:00:33.033" LastActivityDate="2016-06-18T04:55:27.100" Title="A word puzzle I made" Tags="|word|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26366" PostTypeId="2" ParentId="26360" CreationDate="2016-02-11T15:24:39.210" Score="4" Body="&lt;p&gt;Denote by $c_1$ the carry-over from the rightmost column, by $c_2$ the carry-over from the next column, by $c_3$ the carry-over from the middle &#xA;column, and by $c_4$ the carry-over from the fourth column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since all four summands and the sum are even, the equation $2N+X+E=10c_1+E$ must be solved with $N,X,E$ even. By distinguishing five cases on $N$, we get the following five possibilities:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;       N  X  c1 &#xA;     ------------&#xA;       0  0  0     (Case 0)&#xA;       2  6  1     (Case 1)&#xA;       4  2  1     (Case 2)&#xA;       6  8  2     (Case 3)&#xA;       8  4  2     (Case 4)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Case 0 has $N=X$, and therefore is illegal.&lt;br&gt;&#xA;The next column in the summation yields $3U+Z+c_1=10c_2+Z$, and hence &#xA;$U=(10c_2-c_1)/3$. Since $0\le c_2\le3$, this yields the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;       N  X  c1 U  c2&#xA;     -----------------&#xA;       2  6  1  3  1  (Case 1)&#xA;       4  2  1  3  1  (Case 2)&#xA;       6  8  2  6  2  (Case 3)&#xA;       8  4  2  6  2  (Case 4)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now Case 3 has $N=U$, and is illegal.&lt;br&gt;&#xA;The middle column in the summation yields $E+U+c_2=10c_3+I$.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;In Cases 1 and 2, this equation becomes $E+4=10c_3+I$. &#xA;Hence $I$ is also even. The five subcases $E=0,2,4,6,8$ yield $I=4,6,8,0,2$.&#xA;Since $E$ and $I$ must be different from $N$ and $X$, this leaves only the following subcases alive:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   N  X  c1 U  c2 E  I  c3&#xA; --------------------------&#xA;   2  6  1  3  1  0  4  0   (Case 1a)&#xA;   2  6  1  3  1  4  8  1   (Case 1b) &#xA;   4  2  1  3  1  6  0  1   (Case 2 )&#xA;&lt;/code&gt;&lt;/pre&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In Case 4, the equation becomes $E+8=10c_3+I$. &#xA;Hence $I$ is also even. The five subcases $E=0,2,4,6,8$ yield $I=8,0,2,4,6$.&#xA;Since $E$ and $I$ must be different from $N,X,U$, this leaves only the following subcase alive:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   N  X  c1 U  c2 E  I  c3&#xA; --------------------------&#xA;   8  4  2  6  2  2  0  1   (Case 4 )&#xA;&lt;/code&gt;&lt;/pre&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now let us turn to the fourth column, which yields $D+O+c_3=10c_4+E$.&#xA;Since $E$ is even, this means that $D+O+c_3$ is even.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In Case 4, the five even digits 0,2,4,6,8 ahave been assigned to $I,E,X,U,N$, respectively. Hence $D$ and $O$ both must be odd, which together with $c_3=1$ yields a contradiction.&lt;/li&gt;&#xA;&lt;li&gt;Similarly in Case 2, the only unassigned even digit is $8$. &#xA;In this case the equation $D+O+c_3=10c_4+E$ becomes $D+O=10c_4+5$, which forces $c_4=1$ and $\{D,O\}=\{7,8\}$.&lt;/li&gt;&#xA;&lt;li&gt;Similarly in Case 1a, the only unassigned even digit is $8$.&#xA;In this case the equation $D+O+c_3=10c_4+E$ becomes $D+O=10c_4$, which forces $c_4=1$ and $D+O=10$ with $O$ and $D$ odd and distinct.&lt;/li&gt;&#xA;&lt;li&gt;Finally in Case 1b, the only unassigned even digit is $0$.&#xA;In this case the equation $D+O+c_3=10c_4+E$ becomes $D+O=10c_4+3$, which forces $c_4=1$ and $\{D,O\}=\{0,3\}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In all surviving cases, we have $U=3$ and $c_4=1$.&#xA;In particular, this implies $S=D+1$, so that one of $S$ and $D$ is even.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In Case 1a, we conclude $S=8$, $D=7$ and $O=3$; this subcase has $U=O$ and yields a contradiction.&lt;/li&gt;&#xA;&lt;li&gt;In Case 1b, we conclude $D=0$ and $S=1$, which has $D=0$ as starting digit of a number and hence violates one of the basic principles of alphametic puzzles.&lt;/li&gt;&#xA;&lt;li&gt;In Case 2, we finally conclude $D=8$ and $S=9$, and $O=7$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Let us summarize:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;       N  X  c1 U  c2 E  I  c3 D  S  O&#xA;     -----------------------------------&#xA;       4  2  1  3  1  6  0  1  8  9  7   (Case 2 )&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Hence $I=0$, $X=2$, $U=3$, $N=4$, $E=6$, $O=7$, $D=8$ and $S=9$ have been assigned, and only the digits $1$ and $5$ remain open, and both can be legally assigned to $Z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This yields the following two solutions (that only differ in the value of $Z$):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;     34               34&#xA;     34               34&#xA;   8632     and     8632 &#xA;+ 87316          + 87356&#xA; ------           ------&#xA;  96016            96056&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2016-02-11T15:24:39.210" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26367" PostTypeId="2" ParentId="26360" CreationDate="2016-02-11T15:26:48.183" Score="4" Body="&lt;p&gt;Denote the following:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$S_i$&lt;/span&gt; to be the sum of the digits in column &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$D_i$&lt;/span&gt; to be the digit in column &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; of the sum.  Thus &lt;span class=&quot;math-container&quot;&gt;$D_i = S_i \mod 10$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$C_i$&lt;/span&gt; to be the carry over of &lt;span class=&quot;math-container&quot;&gt;$S_i$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Given that all the numbers are even, we know that &lt;span class=&quot;math-container&quot;&gt;$N,X,E \in \{0,2,4,6,8\}$&lt;/span&gt;.  Looking at &lt;span class=&quot;math-container&quot;&gt;$S_1=N+N+X+E \le 8+8+6+4=26 \implies C_1 \le 2$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Lets look at &lt;span class=&quot;math-container&quot;&gt;$S_2=U+U+U+Z+C_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D_2=Z$&lt;/span&gt;.  We know that &lt;span class=&quot;math-container&quot;&gt;$3\times U + C_1 \mod 10 = 0$&lt;/span&gt; in order to make this work.  If &lt;span class=&quot;math-container&quot;&gt;$C_1=0 \implies U=0$&lt;/span&gt;, which can't happen.  &lt;span class=&quot;math-container&quot;&gt;$C_1=1 \implies U=3$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_1=2 \implies U=6$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;We can also see that &lt;span class=&quot;math-container&quot;&gt;$C_4 \le 1$&lt;/span&gt; and since &lt;span class=&quot;math-container&quot;&gt;$S \ne D$&lt;/span&gt;, we have &lt;span class=&quot;math-container&quot;&gt;$C_4=1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$S=D+1$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Now look at &lt;span class=&quot;math-container&quot;&gt;$S_1=N+N+X+E$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D_1=E$&lt;/span&gt;.  This implies that &lt;span class=&quot;math-container&quot;&gt;$N+N+X \mod 10 = 0$&lt;/span&gt;.  If &lt;span class=&quot;math-container&quot;&gt;$N=0$&lt;/span&gt; then &lt;span class=&quot;math-container&quot;&gt;$X=0$&lt;/span&gt;, which can't happen.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$N=2 \implies X=6$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_1=1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$U=3$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_2=1$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$N=4 \implies X=2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_1=1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$U=3$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_2=1$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$N=6 \implies X=8$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_1=2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$U=6$&lt;/span&gt; which can't happen&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$N=8 \implies X=4$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_1=2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$U=6$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_2=2$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;We also know that &lt;span class=&quot;math-container&quot;&gt;$C_3 \le 1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_4=1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$S=D+1$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;h3&gt;Let assume &lt;span class=&quot;math-container&quot;&gt;$U=6$&lt;/span&gt;.&lt;/h3&gt;&#xA;&lt;p&gt;Thus, &lt;span class=&quot;math-container&quot;&gt;$N=8$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$X=4$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$C_1=C_2=2$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Then &lt;span class=&quot;math-container&quot;&gt;$S_3=E+6+2$&lt;/span&gt;.  The valid choices for &lt;span class=&quot;math-container&quot;&gt;$E \in \{2,0\}$&lt;/span&gt;  If &lt;span class=&quot;math-container&quot;&gt;$E=0$&lt;/span&gt; then &lt;span class=&quot;math-container&quot;&gt;$D_3=I=8$&lt;/span&gt; which is already used.  Thus &lt;span class=&quot;math-container&quot;&gt;$E=2$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$I=0$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_3=1$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;In &lt;span class=&quot;math-container&quot;&gt;$S_4=D+O+C_3=D+O+1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D_4=E=2$&lt;/span&gt;.  Thus, &lt;span class=&quot;math-container&quot;&gt;$D+O=11$&lt;/span&gt;.  But all the even numbers are already taken, so there is no way to make this equality work.&lt;/p&gt;&#xA;&lt;p&gt;Thus &lt;span class=&quot;math-container&quot;&gt;$U \ne 6$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;h3&gt;Thus &lt;span class=&quot;math-container&quot;&gt;$U=3$&lt;/span&gt;&lt;/h3&gt;&#xA;&lt;p&gt;So we know that &lt;span class=&quot;math-container&quot;&gt;$N \in \{2,4\}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$X \in \{6,2\}$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$C_1=C_2=1$&lt;/span&gt;.  In either case, &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt;, so no other letter can be &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt;. So &lt;span class=&quot;math-container&quot;&gt;$E \in \{0,4,6,8\}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$E=8$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$C_3=1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$S_4=D+O+C_3=D+O+1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D_4=E$&lt;/span&gt;.  This would require one of &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$O$&lt;/span&gt; to be 9, an the other 8.&lt;/p&gt;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$E=4$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$N=2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$X=6$&lt;/span&gt;.  In &lt;span class=&quot;math-container&quot;&gt;$S_3=E+U+C_2=4+3+1=8$&lt;/span&gt;, so &lt;span class=&quot;math-container&quot;&gt;$D_3=I=8$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_3=0$&lt;/span&gt;.  From &lt;span class=&quot;math-container&quot;&gt;$S_4=D+O+C_3=D+O$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D_4=E=4$&lt;/span&gt;.  The only combination that works are &lt;span class=&quot;math-container&quot;&gt;$D+O=9+5$&lt;/span&gt;.  But either solution for &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; results in an invalid &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; since &lt;span class=&quot;math-container&quot;&gt;$S=D+1$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$E=0$&lt;/span&gt;, then from &lt;span class=&quot;math-container&quot;&gt;$S_3=E+U+C_2=0+3+1=4$&lt;/span&gt; means that &lt;span class=&quot;math-container&quot;&gt;$D_3=I=4$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_3=0$&lt;/span&gt;.  Thus, &lt;span class=&quot;math-container&quot;&gt;$N=2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$X=6$&lt;/span&gt;.  From &lt;span class=&quot;math-container&quot;&gt;$S_4=D+O+C_3=D+O$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D_4=E=0$&lt;/span&gt;.  The only combinations of remaining numbers that work for &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$O$&lt;/span&gt; are &lt;span class=&quot;math-container&quot;&gt;$9$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;.  But neither result in a valid &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; since &lt;span class=&quot;math-container&quot;&gt;$S=D+1$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Thus &lt;span class=&quot;math-container&quot;&gt;$E=6$&lt;/span&gt;, then means that one of &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$O$&lt;/span&gt; is 8, and the other is 7.  Since &lt;span class=&quot;math-container&quot;&gt;$S=D+1$&lt;/span&gt;, if &lt;span class=&quot;math-container&quot;&gt;$D=7$&lt;/span&gt; then both &lt;span class=&quot;math-container&quot;&gt;$N=O=8$&lt;/span&gt;.  Thus, &lt;span class=&quot;math-container&quot;&gt;$D=8$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$O=7$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$S=9$&lt;/span&gt;. Also, &lt;span class=&quot;math-container&quot;&gt;$X=2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$N=4$&lt;/span&gt;. Also, from &lt;span class=&quot;math-container&quot;&gt;$S_3=E+U+C_2=6+3+1=10$&lt;/span&gt;, so &lt;span class=&quot;math-container&quot;&gt;$D_3=I=0$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The final solution is:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$$U=3, E=6, D=8, O=7, S=9, X=2, N=4, I=0, Z\in\{1,5\}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-02-11T15:26:48.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26368" PostTypeId="1" AcceptedAnswerId="26380" CreationDate="2016-02-11T16:50:01.797" Score="30" ViewCount="1705" Body="&lt;p&gt;One hundred tiles are arranged in a $10 \times 10$ square. Each tile is black on one side and white on the other side. Two types of move are allowed:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Flip over all four tiles in any $2 \times 2$ square&lt;/li&gt;&#xA;&lt;li&gt;Flip over all nine tiles in any $3 \times 3$ square.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is it always possible, given any starting configuration and using only these actions, to turn every tile white side up?&#xA;If so, how many moves are necessary? If not, how close can we be guaranteed to get?&lt;/p&gt;&#xA;" OwnerUserId="18508" LastEditorUserId="8874" LastEditDate="2016-02-11T17:34:03.793" LastActivityDate="2016-08-04T09:45:34.883" Title="One hundred tiles" Tags="|mathematics|combinatorics|reachability|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26369" PostTypeId="2" ParentId="26347" CreationDate="2016-02-11T16:55:59.603" Score="6" Body="&lt;p&gt;I think you are :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;HTML&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the reason you are here The salvation of the digital world&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;HTML is the way the website works, and quite literally sustains the digital world i.e. the internet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am quite simple as I am quite hyper at first&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;HTML is a relatively simple language, since it's hyper(text markup).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet I can sit And create speeches and essays for everyone&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Text&quot; is what written speeches and essays are written in.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can fly like the wind No restrictions from gravity And a good mark in my criminal record&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Mark- up!&quot;HTML can move as quickly as the speed of your internet connection.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You speak of me By mouth or text&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Language&quot; is used by mouth and text - orally and written. We speak and write about HTML constantly on StackOverflow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who knows what's next?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The future of computing is unknown. Who knows what the next creative use of HTML will be?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19179" LastEditorUserId="19179" LastEditDate="2016-02-14T21:15:07.357" LastActivityDate="2016-02-14T21:15:07.357" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26370" PostTypeId="2" ParentId="16333" CreationDate="2016-02-11T17:08:41.063" Score="1" Body="&lt;p&gt;I'm posting this solution since it is really different from what has been posted so far. Not because it is efficient. It isn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are working in rounds. Initially each day is in round 0.&#xA;Now we toss a coin for each day. If the coin lands on heads the day moves on to the next round. We repeat this for every new round, until we end up with a round having only one day left. This day is the answer.&#xA;If accidentally we end up with a round having no days left at all, we repeat the previous round.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does this yield a uniform distribution? We have treated each day the same way.&lt;/p&gt;&#xA;" OwnerUserId="12679" LastActivityDate="2016-02-11T17:08:41.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26371" PostTypeId="2" ParentId="26368" CreationDate="2016-02-11T17:32:24.270" Score="28" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No, this is not always possible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Argument:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Consider the $70$ cells in the seven rows $1,2$ and $4,5$ and $7,8$ &#xA; and $10$.&lt;br&gt;&#xA; Let us call these cells &lt;em&gt;special&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;============&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Every $2\times2$-move flips an even number ($2$ or $4$) of these &#xA;special cells.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;============&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Every $3\times3$-move flips an even number (always $6$) of these &#xA;special cells.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;============&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Suppose that in the initial coloring, an odd number of special &#xA; cells is white. As the flips do not change the parity of the number of white &#xA; special cells, it is impossible to reach the all white coloring.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2016-02-11T17:32:24.270" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26372" PostTypeId="2" ParentId="26353" CreationDate="2016-02-11T19:06:26.060" Score="1" Body="&lt;p&gt;Another answer (based off the authors comment to @got it):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ire. (It's a name) &lt;br/&gt; The first letter of the 3rd word after the 1st 'my' is 'I'. The first letter of the 2nd word after the 2nd 'my' is 'r'. And the first letter of the 1st word after the 3rd 'my' is 'e'.&lt;br/&gt; The 321 refers to the 'my' which is '123', as an anchor for the 3rd 2nd 1st words after them. &lt;br/&gt; Long shot.... &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19086" LastEditorUserId="19086" LastEditDate="2016-02-11T19:20:46.830" LastActivityDate="2016-02-11T19:20:46.830" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26373" PostTypeId="2" ParentId="26319" CreationDate="2016-02-11T23:16:03.567" Score="2" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Auckland City Football Club&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have won 3 at the same time&lt;br&gt;&#xA;I did this 3 times, and actually won 4 the last time&lt;br&gt;&#xA;There 19 similar to me in the world with this achievement,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Refers to winning the Treble (&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Treble_(association_football)#Annual_and_seasonal_trebles&quot; rel=&quot;nofollow&quot;&gt;https://en.m.wikipedia.org/wiki/Treble_(association_football)#Annual_and_seasonal_trebles&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but after 2015 I am above them all&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I suppose because they are the only team to do this 3 times.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am from the Oceania&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The club is in the Oceania Football Confederation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1 without Ramon, other 2 thanks to Ramon&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ramon is the club's current coach. If I had time right now to fact check I suppose I'd find he was only the coach two of the seasons that they did this&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This was very straightforward to find, especially with the hints. I picked New Zealand randomly based on the term &quot;Oceania&quot; and a search of &quot;New Zealand Ramon sports&quot; did the rest. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18493" LastActivityDate="2016-02-11T23:16:03.567" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26375" PostTypeId="1" AcceptedAnswerId="26376" CreationDate="2016-02-12T00:19:01.087" Score="1" ViewCount="342" Body="&lt;p&gt;The CIA fired you because you were too smart. Now you being recruited by the MI6 for as a spy to infiltrate the Martian space camp 123 miles away from Earth. The mission number is 456 and the 678 contestants who failed had to solve this Martian message in order to gain entry:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;t5thnht5ne3dc = ? &lt;br&gt;&#xA;    p0;/aq1zsw2xsw2xw2sxo9l.r4fvde3c = ? &lt;br&gt;&#xA;    i8k,sw2x = ?  &lt;br&gt;&#xA;    aq1zr4fve3dcsw2x = ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the secret message and what is its relationship to Mars?&lt;/p&gt;&#xA;" OwnerUserId="9042" LastEditorUserId="17250" LastEditDate="2016-03-04T12:53:05.990" LastActivityDate="2016-03-04T12:53:05.990" Title="A final test... again" Tags="|cipher|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26376" PostTypeId="2" ParentId="26375" CreationDate="2016-02-12T00:31:03.713" Score="14" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Every fourth letter (ignoring an extra 't' in the first line) spells out, &quot;The password is Ares.&quot; Ares is the Greek counterpart of Mars, the Roman god of war.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2016-02-12T00:31:03.713" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26377" PostTypeId="2" ParentId="26245" CreationDate="2016-02-12T01:05:47.860" Score="4" Body="&lt;p&gt;Here's my guess&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hit the pale ghost as hard as you can for a spectacular death.  The other ghosts will pale at seeing this, allowing you to kill them with one hit each.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9353" LastActivityDate="2016-02-12T01:05:47.860" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26378" PostTypeId="1" CreationDate="2016-02-12T02:34:29.950" Score="0" ViewCount="229" Body="&lt;p&gt;You put a metal rod halfway through a portal and open another portal so that the end of the rod that went through the first portal touches the other end of the rod. If you weld the two ends of the rod together, how will you remove the rod from the portals? Assume that the portals cannot close if there is something in the way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you can't put the rod through the first portal before opening the second one, both portals are already opened but spaced far apart and on movable surfaces so that once the rod is halfway through the first portal, the surface with the second portal is moved closer to bring the ends of the rod together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A friend asked me this question and I've been thinking about this for a long time now. I apologize for not giving more details before.&lt;/p&gt;&#xA;" OwnerUserId="19184" LastEditorUserId="19184" LastEditDate="2016-02-12T23:06:43.673" LastActivityDate="2016-02-12T23:06:43.673" Title="Connecting ends of a single object through portals" Tags="|lateral-thinking|enigmatic-puzzle|" AnswerCount="2" CommentCount="3" ClosedDate="2016-02-12T07:00:54.970" ContentLicense="CC BY-SA 3.0" />
  <row Id="26379" PostTypeId="2" ParentId="26378" CreationDate="2016-02-12T02:49:43.873" Score="3" Body="&lt;p&gt;Well, since portals (like the ones in the Portal video game and other science-fiction entertainment) don't exist, it is difficult to answer the question definitively.  It all depends on whatever fake physics are involved in making the portal in the first place.  Here is one simple answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Shut off the portal(s).  In most science fiction, this will sever the objects that are midway through the portal in the plane of the portal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19128" LastActivityDate="2016-02-12T02:49:43.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26380" PostTypeId="2" ParentId="26368" CreationDate="2016-02-12T03:16:12.520" Score="15" Body="&lt;p&gt;Label the tiles like chessboard notation, so that the bottom left is a1 and the top right is j10. Moves will be denoted by referring to a pair of opposite corners in the square being flipped.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If any square not on the j file or the 1st rank is black, we can make it white by flipping the 2x2 square with it at the top left corner (e.g. if a9 is black, flip a9-b8). This can be repeated until all squares except the j file and the 1st rank are white.&lt;br&gt;&#xA;&lt;br&gt;&#xA;Now if a1 is black, we can make it white with this sequence of moves:&lt;br&gt;&#xA;&lt;li&gt;a1-c3  &lt;/li&gt;&#xA;&lt;li&gt;b1-d3  &lt;/li&gt;&#xA;&lt;li&gt;a2-b3  &lt;/li&gt;&#xA;&lt;li&gt;b2-c3  &lt;/li&gt;&#xA;&lt;li&gt;c2-d3  &lt;/li&gt;&#xA;&lt;br&gt;&#xA;This flips a1 and d1 and leaves everything else the same. We can eliminate b1, c1, ... g1 and j10, j9, ... j4 by making a similar series of moves, so we only have to worry about five squares: h1, i1, j1, j2, and j3.&lt;br&gt;&#xA;&lt;br&gt;&#xA;If more than two of these squares are black, we can flip all five with h1-j3 and h2-i3 to reduce the number of black squares.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This guarantees that it is always possible to get down to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;two&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;black squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove that it is not always possible to reduce the number of black squares further:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Consider these three sets of cells:&lt;br&gt;&#xA;&lt;li&gt;Ranks 1, 2, 4, 5, 7, 8, and 10  &lt;/li&gt;&#xA;&lt;li&gt;Ranks 1, 3, 4, 6, 7, 9, and 10  &lt;/li&gt;&#xA;&lt;li&gt;Files b, c, e, f, h, and i  &lt;/li&gt;&#xA;&lt;br&gt;&#xA;As in @Gamow's answer, each one of these sets has the property that any move flips an even number of tiles in that set, so the parity of the number of black cells in each set cannot change.&lt;br&gt;&#xA;&lt;br&gt;&#xA;Suppose the initial position has two black squares, one on i1 and one on j1. Then the first two sets will always have an even number of black tiles, and the third will always have an odd number. A position with zero black tiles will have an even number in the third set, while a position with one black tile will have an odd number in at least one of the first two sets. Neither of these is possible, so there will always be at least two black tiles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2016-02-12T03:24:16.873" LastActivityDate="2016-02-12T03:24:16.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26381" PostTypeId="2" ParentId="26378" CreationDate="2016-02-12T05:16:19.187" Score="2" Body="&lt;ol&gt;&#xA;&lt;li&gt;Duplicate the situation with&#xA;a second rod and two more portals (C &amp;amp; D), ...&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ... completely apart form the original rod&#xA; and the first two portals (A &amp;amp; B).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Open yet another entry portal (E) ...&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ... that loops through portals A &amp;amp; B,&#xA; parallel to the original rod along its entire length.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;Open exit portal F for portal E so that ...&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ... portal F loops through portals C &amp;amp; D,&#xA; parallel to the second rod along its entire length.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;The original rod's local space,&#xA;which portals A &amp;amp; B had turned into a 3-dimensional&#xA;&amp;ldquo;surface&amp;rdquo; of a 4-dimensional cylinder, is now ...&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ... part of a 3-dimensional &amp;ldquo;surface&amp;rdquo; of a&#xA; &lt;a href=&quot;http://dr-mikes-maths.com/4d-torus.html#pictures&quot; rel=&quot;nofollow&quot;&gt;4-dimensional torus&lt;/a&gt;,&#xA; with both rods on the same surface.&#xA; You can now slide the original rod completely through portals E &amp;amp; F&#xA; to be next to the new rod, without touching portals A &amp;amp; B.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;After doing that, close portals E &amp;amp; F so that the original rod&#xA;is completely independent of portals A &amp;amp; B, even though it is now ...&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ... still linked to itself,&#xA; but through portals C &amp;amp; D just like the second rod&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(The second rod wasn't actually necessary, just the portals.)&lt;/p&gt;&#xA;" OwnerUserId="18129" LastEditorUserId="18129" LastEditDate="2016-02-12T05:27:24.933" LastActivityDate="2016-02-12T05:27:24.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26382" PostTypeId="1" AcceptedAnswerId="26390" CreationDate="2016-02-12T09:19:20.340" Score="9" ViewCount="1177" Body="&lt;p&gt;Ten thousand tiles are arranged in a $100\times100$ square. &#xA;Each tile is colored red, blue, yellow, or green, so that each $2\times2$ (contiguous) sub-square of the square contains all four colors.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; &#xA;  What are the possible color combinations for the four tiles in the corners of the square?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2016-02-12T15:48:06.107" Title="Ten thousand tiles" Tags="|mathematics|combinatorics|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26383" PostTypeId="1" AcceptedAnswerId="26385" CreationDate="2016-02-12T09:45:15.263" Score="-6" ViewCount="445" Body="&lt;p&gt;Alice: &quot;Bob is lying, so is Charlie!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bob: &quot;Charlie is lying, so is Alice!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Charlie: &quot;Alice is lying, so is Bob!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who is/are lying?&lt;/p&gt;&#xA;" OwnerUserId="19189" LastEditorUserId="19196" LastEditDate="2016-02-12T20:48:16.507" LastActivityDate="2016-02-14T13:28:01.830" Title="Catch the liar(s)" Tags="|logical-deduction|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="26384" PostTypeId="2" ParentId="26383" CreationDate="2016-02-12T10:00:02.937" Score="5" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Exactly one of them is telling the truth but we can't tell which.&#xA; For example, if Alice is truthful, then Bob is a liar because&#xA; he is wrong about her and right about Charlie. Charlie is right about&#xA; Bob but wrong about Alice. This works no matter who you start with.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The other cases:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No two can be telling the truth because each is calls the other two liars.&#xA; They cannot all be liars because then they would all be telling the truth. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastEditorUserId="18924" LastEditDate="2016-02-12T10:08:32.087" LastActivityDate="2016-02-12T10:08:32.087" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26385" PostTypeId="2" ParentId="26383" CreationDate="2016-02-12T10:03:25.397" Score="4" Body="&lt;p&gt;There are only eight cases so it's easy to check by hand which cases are true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;case 1: They are all lying&lt;/strong&gt;&lt;br&gt;&#xA;This is impossible because that would make their statements true&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;case 2: They are all telling the truth&lt;/strong&gt;&lt;br&gt;&#xA;this is also impossible because it contradicts all statements&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;case 3,4,5: two people are telling the truth&lt;/strong&gt;&lt;br&gt;&#xA;This is not possible because person 1 says person 2 lies and vice versa&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;case 6,7,8: two people are lying&lt;/strong&gt;&lt;br&gt;&#xA;This is possible! The two person that lie, lie about the fact that the truth teller is telling the truth, while the truth teller correctly points to the other two as liars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there is one truth teller and two liars, and the truth teller could be anyone&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2016-02-12T10:03:25.397" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26386" PostTypeId="2" ParentId="26382" CreationDate="2016-02-12T10:27:33.343" Score="10" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; One of each colour&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/XhfXa.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XhfXa.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Given the even number of rows and columns, this means that:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Whichever rows OR columns you swap either:&lt;br /&gt;&#xA; The first column begins and ends with colours A and D and every even column begins and ends with colours B and E (either way round)&lt;br /&gt;&#xA; &lt;strong&gt;OR&lt;/strong&gt; the first row begins and ends with colours A and B every even row begins and ends with colours D and E&#xA; Leaving every corner with one of the ABDE colours we started with.&#xA; &lt;a href=&quot;https://i.stack.imgur.com/PpT8A.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PpT8A.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA; Resulting in 24 possible permutations (4!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-02-12T10:27:33.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26387" PostTypeId="2" ParentId="26383" CreationDate="2016-02-12T11:05:39.043" Score="3" Body="&lt;p&gt;The statement of the puzzle has been changed several times.&#xA;My answer below fits with the following version of the puzzle.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Alice: &quot;If Bob is lying, Charlie is lying!&quot;&lt;br&gt;&#xA;  Bob: &quot;If Charlie is lying, Alice is lying!&quot;&lt;br&gt;&#xA;  Charlie: &quot;If Alice is lying, Bob is lying!&quot;  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Who is/are lying?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;We use the words &quot;predecessor&quot; and &quot;successor&quot; with respect to the cycle Alice -&gt; Bob -&gt; Charlie -&gt; Alice.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If person $X$ is lying, then the successor of $X$ is saying the truth (as his/her statement ends with &quot;then $X$ is lying&quot;).&lt;/li&gt;&#xA;&lt;li&gt;If two (or more) persons were lying, then one of them would be the successor of the other one; contradiction to the above statement.&lt;/li&gt;&#xA;&lt;li&gt;If exactly one of them is lying (say Alice) and the other two (say Bob and Charlie) are saying the truth, we get the contradiction from Charlie: &quot;If Alice is lying, Bob is lying!&quot; (Charlie says the truth; Alice lies; hence Bob must lie).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The only remaining possibility is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;All three are saying the truth (which indeed is fine).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-02-14T13:28:01.830" LastActivityDate="2016-02-14T13:28:01.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26388" PostTypeId="2" ParentId="26167" CreationDate="2016-02-12T12:22:31.190" Score="6" Body="&lt;p&gt;I would say you are a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;CAP&lt;/strong&gt;! More precisely, a &lt;strong&gt;DENTAL CAP&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some call me royalty, but I say they're wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Its called a DENTAL CROWN too! Even if it is made from Gold, calling it royalty? Wrong.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some call me crazy, but I say they're wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A dental cap mostly made of steel can make a person's appearance crazy but it would be wrong to say so.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some call me a ghost, but I say they're wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ceramic dental crowns blend so well, they do not appear do exist. Ghosts eh?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some call me racist, but I say they're wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dental cap are used to hide discoloring teeth and make the teeth appear white. Does that make it racist? Wrong.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some call me a hat, but I say they're wrong&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CAP is a kind of hat. Does that make a dental cap a hat? Wrong.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19191" LastActivityDate="2016-02-12T12:22:31.190" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26389" PostTypeId="2" ParentId="26353" CreationDate="2016-02-12T12:25:35.123" Score="1" Body="&lt;p&gt;Building up on the rest of solutions. The hints relative to the text itself I think have been covered already. Yet nobody seems to pay much attention to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the &quot;acrostic&quot; part of the title, meaning that the first letter of each line gives the answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore it should be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MAY&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19192" LastActivityDate="2016-02-12T12:25:35.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26390" PostTypeId="2" ParentId="26382" CreationDate="2016-02-12T12:48:35.723" Score="15" Body="&lt;p&gt;The entire square can be divided into 2500 2x2 squares, so it contains 2500 of each color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we ignore the first and last columns, we have a 98x100 rectangle which can be divided into 2450 2x2 squares. It must contain 2450 of each color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, if we ignore the first and last rows, we have 2450 of each color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, if we ignore all four edges, we have a 98x98 square, which has 2401 of each color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we start with the original square, add a copy of the 98x98 square, and remove both of the rectangles, we are left with just the corners. So the corners must be 1 of each color.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2016-02-12T12:48:35.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26391" PostTypeId="2" ParentId="26345" CreationDate="2016-02-12T13:33:27.503" Score="2" Body="&lt;p&gt;Sounds to me like&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The balls outside a pawnbroker's shop. They always come in 3's, they have a&#xA; long tradition, they do start shiny and will rust since they hang outside.&#xA; I suppose it might be busy inside a pawnbroker's but this is one of the&#xA; weak parts of my argument.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastActivityDate="2016-02-12T13:33:27.503" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26392" PostTypeId="2" ParentId="26353" CreationDate="2016-02-12T13:39:29.253" Score="5" Body="&lt;p&gt;I agree with manshu's answer (unfortunately, now deleted) but have a different reasoning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hugh Meyers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's why:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The 'acrostic' refers to the first &lt;em&gt;word&lt;/em&gt; of each line: My As You (technically &quot;You've&quot; for the last). Motivated by Kate Gregory's 3-2-1, the 12:57 is 3-to-1, which leaves 2 last. Taking this as the word order, we get You My As, which is a 'slyly lurking' and 'altered' version that sounds like &quot;Hugh Meyers&quot;, the OP's name. It's a self-referential riddle, referencing the OP's self; and 'easy' because the actual name is in full view.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The hints:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The '25% Odds' is 1:3, as Kate Gregory points out in comments below her answer. If we read this a &quot;one to three&quot;, it &lt;em&gt;could&lt;/em&gt; be considered a hint to swap the first and third words, reducing to the above line of reasoning. I'm not sure what to make of the second hint, except that it highlights the acrostic as a large part of the puzzle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9422" LastActivityDate="2016-02-12T13:39:29.253" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26393" PostTypeId="2" ParentId="26278" CreationDate="2016-02-12T14:15:01.403" Score="10" Body="&lt;p&gt;For the first body, I chose the left and back right pockets. For the second, clue 2, and for the third, pocket and clue.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Based on the clue for the third (brown and black hairs), and the fact that she's in pain from lifting heavy objects (like a 6'1&quot; human), I suspect Cindy Silver of being the murderer. As an employee, presumably she also would have known of the second victim's carrot allergy, too. No idea as to motives, though  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also of note:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The victims are slowly making up parts of a snowman. There was a top hat, a carrot nose, and the tree branch arms. Also the names - Winter, S. N. Bhal (snowball) and Andy K. Cane (Candy cane)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Attempt 2:&lt;br&gt;&#xA;The murderer is  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Namon, the janitor. Changes to the hotel's staff benefits policy since the owner's father died meant that he couldn't afford the medical/funeral costs for his mother, so he holds Vlad responsible. Now he's trying to drive the resort out of business by killing staff and guests.&lt;br&gt;&#xA;The black and brown hairs could be a combination of his and his second victim's, as an employee, he'd still know about the carrot allergy, and his name backwards is almost 'snowman' (I suspect his surname starts with 'S', fitting the general theme of the murders &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11717" LastEditorUserId="11717" LastEditDate="2016-02-16T13:05:59.220" LastActivityDate="2016-02-16T13:05:59.220" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26394" PostTypeId="2" ParentId="26168" CreationDate="2016-02-12T14:55:09.213" Score="1" Body="&lt;p&gt;Looks to me like&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A book &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Spread out like a leaf&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; These would be the leaves (pages) of the book&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cuts me up but builds me up&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The press trims the leaves and binds them together&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Seeks to go lower but lifts its bed higher&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The bed of the press during the printing and binding process&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Found where what builds me is less common&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Few printing presses in forests.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ta-da!&lt;/p&gt;&#xA;" OwnerUserId="18924" LastActivityDate="2016-02-12T14:55:09.213" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26395" PostTypeId="2" ParentId="26382" CreationDate="2016-02-12T15:48:06.107" Score="12" Body="&lt;p&gt;Not really any better than the others but a different way to look at it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider two squares in the top row colored A and B.   Then the two squares below that are C and D in some order, and below that A and B in some order, then below that C and D, etc.   Eventually, in the bottom row, you have a C and D.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this means is if you remove the whole bottom row and place it on top, the square will still satisfy the key property (of each $2 \times 2$ subsquare having four colors).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By symmetry, you can then remove the whole right side column and move it all the way to the left, and it still satisfies the key property.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the squares that use to be in the four corners are located in the first $2 \times 2$ subsquare in the upper left, and thus it is obvious that they are four different colors.&lt;/p&gt;&#xA;" OwnerUserId="2722" LastActivityDate="2016-02-12T15:48:06.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26396" PostTypeId="2" ParentId="26353" CreationDate="2016-02-12T15:50:11.577" Score="14" Body="&lt;p&gt;Answer?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; yes, it is.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  The title indeed gives away the key, &quot;My&quot;. It also tells how to use it with 12:57 which can also be stated as 3 to 1, homophone for 321.  For the &lt;strong&gt;3&lt;/strong&gt; &quot;my&quot;s you take the next &lt;strong&gt;2&lt;/strong&gt; words and use their &lt;strong&gt;1&lt;/strong&gt;st letter.  My &lt;strong&gt;a&lt;/strong&gt;ltered &lt;strong&gt;n&lt;/strong&gt;ame, my &lt;strong&gt;s&lt;/strong&gt;elf-referential &lt;strong&gt;w&lt;/strong&gt;ords, my &lt;strong&gt;e&lt;/strong&gt;asy &lt;strong&gt;r&lt;/strong&gt;iddle.  It is, in fact, easy because you give it away with the last line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9353" LastActivityDate="2016-02-12T15:50:11.577" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="26397" PostTypeId="2" ParentId="26345" CreationDate="2016-02-12T16:00:44.343" Score="2" Body="&lt;p&gt;This doesn't seem to make too much sense (as it doesn't fit the second hint), but I will give it a shot anyways...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Home plate in baseball&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It comes from the past but won't last because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The home plate is the place where you started at first (past), but also where you could potentially end up.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It will shine at first it will rust because &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Obviously it will get &quot;rust&quot; eventually&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And finally his three friends are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The other three bases, which helps the home plate to prevent the other team from scoring, or getting to the home plate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18995" LastActivityDate="2016-02-12T16:00:44.343" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26398" PostTypeId="2" ParentId="26345" CreationDate="2016-02-12T16:11:06.120" Score="9" Body="&lt;p&gt;How about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Newton's Cradle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Got the balls to answer?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's made of a series of balls suspended from a frame...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Comes from the past but it won't last&lt;br /&gt;&#xA;It might shine first but it will rust&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Early versions were made in the XVII Century. They're often made of some sort of metal&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ideally, there are three friends he can trust&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Newton's three laws of motion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They keep him busy, so it won't gather dust&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After starting one, they go one for a while&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It both stays and moves&lt;br /&gt;&#xA;If it obeys the rules &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The end balls fly out while all the others are stationary, while obeying the rules of momentum&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A genius baby proves&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Genius=Newton, Baby-&gt;Cradle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When he gently pulls&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; How you start the cradle, by pulling back and releasing one of the balls&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17446" LastEditorUserId="17446" LastEditDate="2016-02-12T16:19:17.977" LastActivityDate="2016-02-12T16:19:17.977" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="26399" PostTypeId="2" ParentId="26167" CreationDate="2016-02-12T16:27:28.757" Score="2" Body="&lt;p&gt;I'm not sure at all... So this is just a guess &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Word &quot;Belong&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some call me royalty, but I say they're wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Some say royalty belong, but I say they're wrong&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some call me crazy, but I say they're wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Some say crazy people belong, but I say they're wrong&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some call me a ghost, but I say they're wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Some say ghosts belong, but I say they're wrong&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some call me racist, but I say they're wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Some say racism belongs, but I say they're wrong&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some call me a hat, but I say they're wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Some say hats belong, but I say they're wrong&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19196" LastActivityDate="2016-02-12T16:27:28.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26400" PostTypeId="2" ParentId="26168" CreationDate="2016-02-12T16:49:27.583" Score="1" Body="&lt;p&gt;I'll Guess&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fossil&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I am spread out like a leaf.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Just a fossil formed from a plant&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That what forms me cuts me, but builds me up.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water Erodes the Shape into the rock&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That what cuts me seeks always to go lower, but lifts its bed higher.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Water Builds up over time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am commonly found where what builds me is less common.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fossils often found in deserts where water is less common&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19196" LastActivityDate="2016-02-12T16:49:27.583" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26401" PostTypeId="2" ParentId="26345" CreationDate="2016-02-12T17:10:17.183" Score="2" Body="&lt;p&gt;I know it's already been answered, but I wanted to include my answer that I think works moderately well before including the second stanza.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A clock&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Comes from the past but it won't last&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wall clocks have been around for centuries. Nicer clocks often become antique items and/or are considered family heirlooms and passed down for many years but will eventually wear out and cease functioning.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It might shine first but it will rust&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Without proper maintenance, the mechanisms in a clock can gather moisture, rust, and seize up.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ideally, there are three friends he can trust&lt;br&gt;&#xA;They keep him busy, so it won't gather dust&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Also related to the title -- wall clocks are often weight-driven, the larger ones with three weights.  While traditionally cylindrical, ball weights also exist. The slow descent of the weights can keep a clock running for several days at a time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It still (kind of almost) works with the second stanza, with the exception of a pretty specific line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It both stays and moves&lt;br&gt;&#xA;If it obeys the rules  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The pendulum and weights keep the gears moving, but the clock stays in place.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A genius baby proves&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yeah, I got nothing for this one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When he gently pulls&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To wind the clock, you pull the chain for each weight.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7586" LastActivityDate="2016-02-12T17:10:17.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26402" PostTypeId="2" ParentId="26278" CreationDate="2016-02-12T17:37:51.030" Score="7" Body="&lt;p&gt;I think the murderer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sally N Bhal&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reasoning stems from the observations made by @LogicianWithAHat but I think there is more to the picture. To prove my case, I'll give a story of my own:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Background 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Terry W. Terr and Cindy Silver are boyfriend and girlfriend, who are madly in love. Every year for the past three years, they have organised a skiing vacation at the same resort. Cindy enjoys cold weather vacations while Terry does not. However, they have reached a compromise where Terry always takes her skiing while she treats him to a gift involving something warmer (e.g, a sauna). Since 2013, they have driven up to the resort together in Terry's car.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Background 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sally N Bhal has been an employee at the ski resort for several years. Working there used to be great until Mr. Engelbel's father died. Then everything went downhill, employee benefits were cut and there were other cuts too which made it difficult to finance a mortgage, and look after a herd of dogs, especially being a part-time cleaner. Sally was desperate, she wanted out of this miserable life badly but didn't know where to go. Then, one day in 2013, she was struck dumb when the most handsome man she had ever seen checked into the resort, Terry W Terr. Sally became obsessed with Terry, she even bought a smartphone just to take pictures of him skiing and snowboarding whenever he visited. During the rest of the year she would flick through the albums she had made in the previous winter. As she did this, she knew in her heart that her future was with Terry and that was to be her way out of the resort. But, alas there was a sting in the tail as Terry already had a girlfriend, Cindy Silver. She was determined to be with him, so she tried to make herself look like Cindy to the point of dying and styling her hair in the same way this winter. This was a change even the normally unobservant Vlad Englebel noticed with his &quot;nice hair&quot; remark. Despite having an eidetic memory for useless information, the aloof owner was usually terrible when it came to faces and appearances so this really was remarkable. So caught up in the scheme was Sally that she was determined to take Cindy's place and be with Terry.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Background 3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Andy K. Cane also likes to ski at the resort although he came alone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Background 4&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Jack Jones has travelled to the ski resort with his friends, including the young Shane Shady. They plan on camping illegally in the adjacent forest. Shane has brought with him a large swiss army knife which he will use to start a campfire. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Setup 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; On one day of their trip, Cindy and Terry decide they will go into the forest to build a snowman. They flout the &quot;Do Not Enter&quot; sign as they feel it is not applicable to them, since they know the area so well. They build the giant snowballs for the snowman's head and body but realise they do not have any of the supplementary supplies. Cindy volunteers her scarf for the snowman as long as Terry promises to get a new one. Terry writes a reminder on a piece of paper, but is shivering due to being out in the cold so long, and hands it to Cindy because she is better at holding onto things. Terry notices the contrast in Cindy's red neck and white face due to the absence of the scarf. They then formulate a plan to get the rest of the items. Cindy volunteers to go to the lodge to try and find a carrot, a top hat and some coals while Terry will go further into the woods to find a tree branch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Murder 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cindy easily retrieves a carrot which she gets from the kitchen on a plate. While in the lodge, she 'conveniently' bumps into Sally and explains to her what she's doing and what she's trying to find. Sally says she knows where to find a top hat and tells Cindy to go down to the laundry room where she'll find a fireplace and possibly some coal Sally also promises to bring Cindy down some food s Cindy has expressed she is hungry. Cindy goes down to the laundry room but cannot find any loose coals and the fireplace is burning. She decides to wait for Sally, which turns out to be longer than expected. While waiting, she absentmindedly chews on the carrot to saitsfy her hunger. Meanwhile, Sally steals a top hat from Mr Hail's room which she saw earlier while cleaning. Sally takes the opportunity to murder Cindy as she will never have a better one, her weapon of choice is poison. When she returns downstairs she offers Cindy the top hat as well as some snacks she &quot;found&quot;. Unfortunately for Cindy the snacks are laiden with poison and she dies quickly. Sally decides that she will try to impersonate Cindy, convinced of the likeness that they now share, so she swaps clothes and identification with Cindy. She takes the top hat with her but decides the carrot is too far gone (Sally cannot touch carrots anyway because of her allergy) and goes outside to confront her love, Terry.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Setup 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; While out looking for a tree branch, Terry runs into Shane Shady who offers assistance with his knife to cut down two branches. Terrys thanks Shane and returns to the snowman and puts the branches in as arms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Murder 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When Sally returns to Terry she makes an attempt to deceive Terry but is quickly found out. Terry is angry and demands to know where Cindy is. Sally pleads with Terry that she is his one true love, even showing her the pictures of him on her smartphone. Terry pockets the phone, promising to show it to the police. In a panic, Sally drops the top hat and grabs one of the tree branches from the snowman. She uses it to hit Terry over the head killing him instantly. She runs off once she realises what she has done.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Murder 3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Andy K Cane has seen these events from a distance and runs over to confront Sally who is now in a severe panic. There is a struggle which ends in Sally goring Andy with the tree branch. She is covered in blood and decides it best to retreat to her home nearby.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Coverup 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When Shane Shady comes across the body, he is shocked when he realises he met the first victim just hours earlier. Worried about being implicated in a murder, he goes to Jack Jones who offers to hide his swiss-army knife for him.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Coverup 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As luck would have it, a freak avalanche happens which covers up Terry's body before it can be properly investigated. Unfortunately, Andy's body is not covered up and is discovered later by Sally who initially drags it back to her home. Later, she decides to bring it to the boiler room, where she can release the gas and hide the body in an explosion. To do this, especially to access various parts of the facility, she decides to don her worker's uniform. But, before she can fully enact the plan, the body is discovered. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Further explanations&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because Vlad is not very good at recognising faces and because he doesn't see his employees often enough, he misidentifies Sally due to her hairstyle and uniform. Sally decides to finish the stay at the resort as Cindy so as not to arouse suspicion. When she is first called into the detective's office she is still wearing her uniform which she has forgotten to take off while trying to dispose of Andy's body. Luckily for her, the detective doesn't know that Cindy is a guest and identifies her as a cleaner. As @LogicianWithAHat noted, her injury is from carrying around Andy's body. Also, thanks to @LogicianWithAHat for the remark about the black and brown hairs. Mr. Hail's fury with the resort is largely due to his missing top hat which he realises could implicate him in a murder so he doesn't mention it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18422" LastActivityDate="2016-02-12T17:37:51.030" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26403" PostTypeId="1" AcceptedAnswerId="41537" CreationDate="2016-02-12T21:01:28.553" Score="6" ViewCount="288" Body="&lt;p&gt;The king of the Gerbils has recently been killed by the Gerbil's greatest enemy - the snake. To become the next king our Gerbil, called Beastly (ahem), must find and slay the snake. So Beastly must set out on a journey to find him. Unfortunately he has no idea where the snake's abode is. To help, you will need to know the back story.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the snake slithered sneakily inside the gerbil's cave, he bit the king injecting him with poisonous venom. But this snake has plundered many gerbil caves and stolen a lot of gold. He modified his fangs to give a clue to any hunters - trying to give the gerbils a false hope as he is confident that no gerbil will solve the clue. Before embarking on his journey, Beastly checked the king's stomach (gross I know) where he was bitten. This is what he saw:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/E6Sjl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/E6Sjl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You realise that the each symbol forms a word. The words form a phrase. Each new colour is a different word. The symbols in black form the important words. The important words can be merged together to form one word which is the snake's abode.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;What is the phrase? Where does Beastly need to go to find the snake?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;" OwnerUserId="18250" LastEditorUserId="18250" LastEditDate="2016-09-23T20:40:43.050" LastActivityDate="2016-09-23T20:40:43.050" Title="The Gerbil's quest" Tags="|story|rebus|symbols|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26404" PostTypeId="2" ParentId="26403" CreationDate="2016-02-12T21:22:42.130" Score="6" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Look to right below us&quot;&lt;br&gt;&#xA; Eye symbol = look, Two dots = to, Right arrow = right, Down arrow = under/below, United States Flag = U.S. = us  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19128" LastActivityDate="2016-02-12T21:22:42.130" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26405" PostTypeId="2" ParentId="26167" CreationDate="2016-02-12T22:06:33.370" Score="2" Body="&lt;p&gt;Alright so this is pretty far fetched but pretty much fits it all.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Buster&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some call me royalty, but I say they're wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A man called Buster Howes was a royal marine, also Lionel (Buster) Crabb was in the royal navy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some call me crazy, but I say they're wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Busta Rhymes has a song titled &lt;a href=&quot;http://www.azlyrics.com/lyrics/bustarhymes/crazy.html&quot; rel=&quot;nofollow&quot;&gt;Crazy&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some call me a ghost, but I say they're wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Ghost Busters are a group of people who hunt ghosts&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some call me racist, but I say they're wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Buster Keaton was a silent film actor in the 30s, at which point they had very different ethics regarding other races when compared to today&#xA;Also Dave and Buster posted a tweet once that was considered racially insensitive and had to apologise.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some call me a hat, but I say they're wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Buster caps are a type of hat, also coincidentally RuneScape has a hat called the Skullbuster hat!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a word&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yep, definitely a word&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer involves rhymes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;b&gt;Busta Rhymes&lt;/b&gt; is a hip-hop artist&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18940" LastActivityDate="2016-02-12T22:06:33.370" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26406" PostTypeId="1" CreationDate="2016-02-12T23:04:34.180" Score="12" ViewCount="4825" Body="&lt;p&gt;Insert multiplication and division operators and parentheses into the following sequence to make the equation true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that you can combine digits to form numbers, such as 1 and 2 to make 12, or 4 and 5 to make 45, but &lt;strong&gt;you can't alter the order of the sequence.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1 2 3 4 5 6 7 8 9 = 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19200" LastEditorUserId="13484" LastEditDate="2016-03-07T09:33:04.947" LastActivityDate="2021-09-29T08:00:59.610" Title="123456789=1 problem" Tags="|mathematics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26409" PostTypeId="2" ParentId="26406" CreationDate="2016-02-12T23:18:49.833" Score="19" Body="&lt;p&gt;$$12 \times (3/4)/56 \times 7 \times 8/9 = 1$$&lt;/p&gt;&#xA;" OwnerUserId="11476" LastActivityDate="2016-02-12T23:18:49.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26410" PostTypeId="1" AcceptedAnswerId="26423" CreationDate="2016-02-13T02:06:46.643" Score="0" ViewCount="1008" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am known to hack the network &lt;br&gt;&#xA;  but I never do &lt;br&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Yet I can be number one &lt;br&gt;&#xA;  or number two if bored&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;My grade is always a C plus &lt;br&gt;&#xA;  never above that dang C&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I'm more famous here &lt;br&gt;&#xA;  than the whole wide world&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I rhyme with cheat &lt;br&gt;&#xA;  though I never do&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I got so much gold &lt;br&gt;&#xA;  I got more than all your medals combined &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I can see sharper than all of you &lt;br&gt;&#xA;  and as famous as Chuck Norris&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Think locally... what is here? And what really familar network am I talking about?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9042" LastEditorUserId="9042" LastEditDate="2016-02-13T14:37:27.583" LastActivityDate="2016-02-13T15:14:25.303" Title="Who am I, Hacking you everyday?" Tags="|riddle|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26411" PostTypeId="2" ParentId="26167" CreationDate="2016-02-13T06:30:01.103" Score="0" Body="&lt;p&gt;I'm not certain this is right, but it's the best I've been able to come up with, and it does fit the clues.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am a writer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some call me royalty, but I say they're wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stephen King is a famous writer, and King is a royal title.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some call me crazy, but I say they're wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many famous writers such as Ernest Hemingway and Edgar Allen Poe were somewhat crazy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some call me a ghost, but I say they're wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A ghost writer is someone who writes a book that is credit to another.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some call me racist, but I say they're wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mark Twain's Tom Sawyer and Huckleberry Finn have been called racist and put on banned books lists. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some call me a hat, but I say they're wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A writer's hat is an app to help with idea generation. Or a writer may wear different hats, depending on the type of writing being done.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; writer is a word, and writers produce words.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; poets are a type of writer, and some types of poetry are rhyming.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17297" LastActivityDate="2016-02-13T06:30:01.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26412" PostTypeId="1" AcceptedAnswerId="27473" CreationDate="2016-02-13T07:05:19.350" Score="10" ViewCount="669" Body="&lt;p&gt;[&lt;em&gt;Update: added a second stanza with additional clues&lt;/em&gt;]&lt;br&gt;&#xA;[&lt;em&gt;Update 2: changed title, added some clues, added a third stanza, and started a bounty&lt;/em&gt;]&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I am the question that no unerring answer knows &lt;br&gt; &#xA;  The thought of which does ofttimes give my asker pause &lt;br&gt; &#xA;  When resolution wavers, he'll reflect &lt;br&gt; &#xA;  On which solution heretofore considered is correct &lt;br&gt; &#xA;  Of heartache, of fortune, of death and dreams &lt;br&gt; &#xA;  In this oration, life's greatest riddles are the themes &lt;br&gt;&#xA;  &lt;br&gt;&#xA;  You hear me spoken time and time again &lt;br&gt;&#xA;  For centuries my lasting magic still remains &lt;br&gt;&#xA;  Fit for a prince, uttered in earnest, sometimes in jest &lt;br&gt;&#xA;  Of all my kind I surely am one of the best &lt;br&gt;&#xA;  &lt;br&gt;&#xA;  Now there's a bounty on my head &lt;br&gt;&#xA;  A bit more work, and I'll be dead &lt;br&gt;&#xA;  Come on Puzzling, you can win &lt;br&gt;&#xA;  If you remember all your sins &lt;br&gt;&#xA;  I puzzle the will, confound the sense &lt;br&gt;&#xA;  'Tis time to end all the suspense&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What am I?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Remarks:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;The bounty and coveted green check mark will be given to the first correct answer that explains most or all of the clues in the riddle.&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;There is only one correct answer, which I'm pretty sure will appear fairly obvious and straightforward once it is pointed out. If you have an idea for a solution but are unsure if it's correct, it probably isn't.&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerDisplayName="user17947" LastEditorDisplayName="user17947" LastEditDate="2016-02-15T07:23:30.403" LastActivityDate="2016-02-15T22:53:52.547" Title="A riddle fit for a prince (extended edition)" Tags="|riddle|poetry|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26413" PostTypeId="1" AcceptedAnswerId="27460" CreationDate="2016-02-13T07:33:30.920" Score="11" ViewCount="307" Body="&lt;p&gt;Two players take turns writing an X (first player) or an O (second player) in a 9 by 9 grid of cells. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A player wins if they manage to occupy four cells which form the vertices of a square of any size with horizontal and vertical sides. If this doesn't happen before the board is filled, then O wins.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Whom does this game favor?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2016-02-14T20:29:00.230" Title="Squic Squac Squo" Tags="|game|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26414" PostTypeId="2" ParentId="26410" CreationDate="2016-02-13T09:35:56.223" Score="1" Body="&lt;p&gt;I think the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; heat&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am known to hack the network &lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I never do &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sun can melt the whole earth if it comes a bit near.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet I can be number one &lt;/p&gt;&#xA;&#xA;&lt;p&gt;or number two if bored&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In night it will be number two because number one is moon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My grade is always a C plus &lt;/p&gt;&#xA;&#xA;&lt;p&gt;never above that dang C&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Heat movie has C+ grade in &lt;a href=&quot;http://www.avclub.com/review/the-heat-99486&quot; rel=&quot;nofollow&quot;&gt;avclub website&lt;/a&gt;!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I rhyme with cheat &lt;/p&gt;&#xA;&#xA;&lt;p&gt;though I never do&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Heat rhyme with cheat but it never cheats (Sun never leave us without heat)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I got so much gold &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got more than all your medals combined&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Who else can bring more heat (which is source of any sort of energy) to us? Also it can refer to golden color of sun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can see sharper than all of you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; maybe refers to heat vison AKA Infrared thermography which allow to see at night &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and as famous as Chuck Norris&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Maybe refer to the heat movie which is as famous as Chuck norris.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18675" LastEditorUserId="18675" LastEditDate="2016-02-13T09:47:07.470" LastActivityDate="2016-02-13T09:47:07.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26415" PostTypeId="2" ParentId="24240" CreationDate="2016-02-13T09:51:38.833" Score="0" Body="&lt;p&gt;I can beat Orp, and arguably beat NeedAName depending on how Infect scores.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Play &lt;strong&gt;1B&lt;/strong&gt; as land. Mana: B (20 Life)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Play &lt;strong&gt;B&lt;/strong&gt; as &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=397401&quot; rel=&quot;nofollow&quot;&gt;Dark Ritual&lt;/a&gt;. Mana: BBB&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Play &lt;strong&gt;1&lt;/strong&gt; as &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=46112&quot; rel=&quot;nofollow&quot;&gt;Myr Moonvessel&lt;/a&gt;. Mana: BB&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Play &lt;strong&gt;B&lt;/strong&gt; as &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=6085&quot; rel=&quot;nofollow&quot;&gt;Culling the Weak&lt;/a&gt;, targeting Myr Moonvessel. Mana: 1BBBBB&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Play &lt;strong&gt;BBB&lt;/strong&gt; as &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=12957&quot; rel=&quot;nofollow&quot;&gt;Infernal Contract&lt;/a&gt;, drawing &lt;strong&gt;B&lt;/strong&gt;, &lt;strong&gt;1G&lt;/strong&gt;, &lt;strong&gt;1WW&lt;/strong&gt;, and &lt;strong&gt;R&lt;/strong&gt;. Mana: 1BB (10 Life)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Cycle &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=243442&quot; rel=&quot;nofollow&quot;&gt;Edge of Autumn&lt;/a&gt; (&lt;strong&gt;1G&lt;/strong&gt;), sacrificing your land, drawing &lt;strong&gt;U&lt;/strong&gt;.  Mana: 1BB&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Play &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=233056&quot; rel=&quot;nofollow&quot;&gt;Gitaxian Probe&lt;/a&gt;``` (&lt;strong&gt;U&lt;/strong&gt;) for 2 life, drawing &lt;strong&gt;BB&lt;/strong&gt;.  Mana: 1BB (8 Life)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Play &lt;strong&gt;1B&lt;/strong&gt; as &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=382877&quot; rel=&quot;nofollow&quot;&gt;Cabal Ritual&lt;/a&gt;.  Mana: BBBBBB (Graveyard was &lt;strong&gt;1B&lt;/strong&gt;, &lt;strong&gt;B&lt;/strong&gt;, &lt;strong&gt;1&lt;/strong&gt;, &lt;strong&gt;B&lt;/strong&gt;, &lt;strong&gt;BBB&lt;/strong&gt;, &lt;strong&gt;1G&lt;/strong&gt;, &lt;strong&gt;U&lt;/strong&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Play &lt;strong&gt;B&lt;/strong&gt; as &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=44206&quot; rel=&quot;nofollow&quot;&gt;Blood Celebrant&lt;/a&gt;.  Mana: BBBBB&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Play &lt;strong&gt;B&lt;/strong&gt; as &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=198372&quot; rel=&quot;nofollow&quot;&gt;Death's Shadow&lt;/a&gt;.  Mana: BBBB&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Activate Blood Celebrant for R, G, B, B, B, B, B.  Mana: BBRG (1 Life)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Play &lt;strong&gt;R&lt;/strong&gt; as &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=146060&quot; rel=&quot;nofollow&quot;&gt;Crimson Wisps&lt;/a&gt; on Death's Shadow, giving it haste and drawing &lt;strong&gt;G&lt;/strong&gt;. Mana: BBG&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Play &lt;strong&gt;BB&lt;/strong&gt; as &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=214346&quot; rel=&quot;nofollow&quot;&gt;Glistening Oil&lt;/a&gt;, enchanting Death's Shadow and giving it Infect.  Mana: G&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Go to combat and swing, then play &lt;strong&gt;G&lt;/strong&gt; as &lt;a href=&quot;http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=382865&quot; rel=&quot;nofollow&quot;&gt;Berserk&lt;/a&gt; on Death’s Shadow, giving +12/+0 and trample&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Hit for 24 infect (0 cards in deck, &lt;strong&gt;1WW&lt;/strong&gt; in hand)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;```Presuming we're playing by Mental Magic's rule set, this is legal per &lt;a href=&quot;http://mentalmagic.neocities.org/&quot; rel=&quot;nofollow&quot;&gt;MM.3i&lt;/a&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="17389" LastEditDate="2016-02-13T10:06:14.363" LastActivityDate="2016-02-13T10:06:14.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26416" PostTypeId="1" AcceptedAnswerId="40297" CreationDate="2016-02-13T11:19:38.057" Score="5" ViewCount="562" Body="&lt;p&gt;A factory that manufactures clothing cuts multiple garments at once.&lt;/p&gt;&#xA;&lt;p&gt;Sometimes a number of markers are used for a job. Each marker may have a single size or multiple sizes depending on what is needed. So the length of each is different.&lt;/p&gt;&#xA;&lt;p&gt;To get the required number of garments a number of lays are laid out over each other to be cut at once. The number of lays can be different.&lt;/p&gt;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Marker1: 100 lays 10.5m each&lt;/li&gt;&#xA;&lt;li&gt;Marker2: 80 lays  11m each&lt;/li&gt;&#xA;&lt;li&gt;Marker3: 20 lays  5.2m each&lt;/li&gt;&#xA;&lt;li&gt;Marker4: 20 lays  6.1m each&lt;/li&gt;&#xA;&lt;li&gt;Marker5: 6 lays   2m each&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;So for Marker1 1050m of fabric is needed, for Marker2 another 880m of fabric is needed, and so on.&lt;/p&gt;&#xA;&lt;p&gt;Some markers can be combined, such as Marker3 and Marker4, where the number of lays is the same, to form a marker of 11.3m.&lt;/p&gt;&#xA;&lt;p&gt;Given the markers above, the total metres of fabric required would be 2168m.&lt;/p&gt;&#xA;&lt;p&gt;However the fabric does not come in one big roll. And this is what causes the problem. Perhaps 11 rolls of 100m will be supplied. This causes wastage of fabric, one roll can be laid 9 times on Marker1, leaving 5.5m which can be used on the other markers.&lt;/p&gt;&#xA;&lt;p&gt;The puzzle then is, how do you find the best way of using the following rolls of fabric on the markers above?&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;150m x 2&lt;/li&gt;&#xA;&lt;li&gt;120m x 4&lt;/li&gt;&#xA;&lt;li&gt;100m x 12&lt;/li&gt;&#xA;&lt;li&gt;50m x 5&lt;/li&gt;&#xA;&lt;li&gt;20m x 3&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;I work in a factory where they manufacture clothing. I wonder if there was some strategy to optimize cloth usage in general.&lt;/p&gt;&#xA;" OwnerUserId="19204" LastEditorUserId="69582" LastEditDate="2021-07-12T03:57:43.120" LastActivityDate="2021-07-12T03:57:43.120" Title="Optimize cloth usage" Tags="|mathematics|calculation-puzzle|optimization|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26420" PostTypeId="1" CreationDate="2016-02-13T13:05:55.187" Score="7" ViewCount="1878" Body="&lt;p&gt;I have found this puzzle and I wanted to share it with you to see if anyone can solve it, because I can't figure it out.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;**Can you put this all together?**&lt;br&gt;&#xA;  From: Agent J&lt;br&gt;&#xA;  To: adib&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Agent,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;See if you can piece all these intercepted pieces of intel together, our team is really struggling to do so. We've got an encrypted string, and this weird calendar image with a date circled.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Anyway, here's the encrypted string: &lt;code&gt;NXIS24CuEq@uEq@uEq@u&lt;/code&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;See if you can figure it out.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;J&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The calendar image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hQybX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="19205" LastEditorUserId="30633" LastEditDate="2016-12-19T19:54:39.950" LastActivityDate="2017-08-05T07:42:12.750" Title="Junk decryption-- a puzzle I found on a wargame site" Tags="|cipher|computer-puzzle|" AnswerCount="1" CommentCount="5" ClosedDate="2017-08-05T08:28:39.730" ContentLicense="CC BY-SA 3.0" />
  <row Id="26423" PostTypeId="2" ParentId="26410" CreationDate="2016-02-13T15:14:25.303" Score="5" Body="&lt;p&gt;You are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jon Skeet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am known to hack the network&#xA;but I never do&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He is a member of 15 communities in the Stack Exchange network.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet I can be number one&#xA;or number two if bored&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He is the top ranked user.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My grade is always a C plus&#xA;never above that dang C&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm not sure of this one, unless it has to do with his programming. He doesn't use B?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm more famous here&#xA;than the whole wide world&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He's a well known user of Stack Exchange&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I rhyme with cheat&#xA;though I never do&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Skeet rhymes with cheat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I got so much gold&#xA;I got more than all your medals combined&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He has 430 gold medals, which is more gold medals than many users have total.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can see sharper than all of you&#xA;and as famous as Chuck Norris&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He is an author of a C# book.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17297" LastActivityDate="2016-02-13T15:14:25.303" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26424" PostTypeId="2" ParentId="26364" CreationDate="2016-02-13T16:54:40.620" Score="2" Body="&lt;p&gt;answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a traitor of a criminal gang&lt;br&gt;&#xA;&lt;br&gt;&#xA; a criminal will attack a traitor thus toss it on the ground&lt;br&gt;&#xA; a traitor will jump back at you when the betrayal is revealed&lt;br&gt;&#xA; the traitor will pretend to receive help off a horse whilst capturing you instead.&lt;br&gt;&#xA; a traitor is someone you used to  trust as a companion hence &quot;mount on the boat&quot;&lt;br&gt;&#xA; The traitor is thought to help organise plans for the other  but will make it obvious for the person's enemies.&lt;br&gt;&#xA;the next line either refers to betting in bridge card game (thus &quot;with me&quot; as a partner or sharing houses literally using mortgage contracts.&lt;br&gt;&#xA; Finally the last line refers to the other person stealing from the traitor's house which is against the contract that the reader has made.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19209" LastEditorUserId="17250" LastEditDate="2016-02-15T14:23:43.780" LastActivityDate="2016-02-15T14:23:43.780" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="26425" PostTypeId="1" AcceptedAnswerId="26441" CreationDate="2016-02-13T17:22:49.977" Score="9" ViewCount="1486" Body="&lt;p&gt;The year is 2065. For solving this puzzle, you may assume that no significant changes have been made in political and geographical situations in Europe and Africa except for those that are described in the question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One country X in Europe started expanding its area. Occupying the neighbouring country was hard, but that country bordered one that was easy to occupy for a part because it could not finance an own army and other countries refused to help despite some treaties. However, the country was never occupied in full. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One country Y in Africa also started expanding its area and occupied an neighbouring country. Now countries X and Y both border the Mediterranean Sea. They start a war between each other. Now finally an international organisation steps in. They decide that there will be a war on sea between two large ships, and that the country that wins, will win the war. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The flags of countries are of course present on the ship. During this battle, country X manages to let the boat of country Y sink by a technique that was present in one of the occupied territories for over a thousand years. &#xA;The territory of country Y and is given to country X. Many artists made their impression of the war. However, later people thought that the war that was fought was actually a civil war in country X. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should be enough information to figure out what countries X and Y are [remember, all countries have still the same name as they have now excepted for the occupied territories], and why people later thought that the war was a civil war in country X. &lt;/p&gt;&#xA;" OwnerUserId="16322" LastEditorUserId="16322" LastEditDate="2016-02-14T11:55:45.533" LastActivityDate="2016-02-14T12:06:54.393" Title="A war between two countries, but it is not a civil war" Tags="|riddle|geography|" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26426" PostTypeId="1" AcceptedAnswerId="27545" CreationDate="2016-02-13T18:53:54.247" Score="3" ViewCount="321" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;3&#xA;  4&#xA;  9&#xA;  8&#xA;  9&#xA;  8&#xA;  13&#xA;  14&#xA;  13&#xA;  ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So what is the next number?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Hint&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the actual difference between numbers is just 1 and the sequence actually starts with 1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Hint&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Spelling is the key!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="18514" LastEditorUserId="18514" LastEditDate="2016-02-15T14:18:55.800" LastActivityDate="2016-02-17T14:53:28.587" Title="What is the next in this ordinary sequence?" Tags="|number-sequence|" AnswerCount="5" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="26427" PostTypeId="1" CreationDate="2016-02-13T19:39:56.963" Score="2" ViewCount="46" Body="&lt;p&gt;Here are the numbers.... What'cha gonna do? &#xA;(try and solve it...duh)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                                   1&#xA;                               1       1&#xA;                               2       1&#xA;                           1   2       1    1&#xA;                        1    1   1   2    2    1&#xA;                      ????????????????????????????           &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="19086" LastActivityDate="2016-02-13T19:39:56.963" Title="Pyramid of hopelessness" Tags="|number-sequence|" AnswerCount="0" CommentCount="0" ClosedDate="2016-02-13T19:43:10.483" ContentLicense="CC BY-SA 3.0" />
  <row Id="26429" PostTypeId="1" AcceptedAnswerId="26453" CreationDate="2016-02-13T20:06:00.187" Score="10" ViewCount="715" Body="&lt;p&gt;After her adventures in Wonderland and The Looking Glass Alice led a varied and interesting life. One day, when she had grown to be an attractive young woman, she found herself in the town of Cryptic. The people may have been well-meaning but they were full of odd tics and mannerisms and they spoke in such a confusing way that it gave her a headache trying to make sense of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result of several misunderstanding, evening found her alone and hungry in Cryptic with no dinner plans, nowhere to stay, and no means of transportation to somewhere more plainspoken. She was weighing her options when she heard someone call her name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Alice!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;She turned to see a man walking toward her. He was tall, blond, and muscular. He was too far away for her to see his eyes but she felt instinctively that they were deep and dreamy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Hello,&quot; she replied. &quot;I'm sorry... I don't... who are you?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He smiled and held out his hand. &quot;You can call me Will.&quot; His eyes &lt;em&gt;were&lt;/em&gt;&#xA;deep and dreamy. They were also of a shade of blue that she found particularly interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Happy to meet you, Will,&quot; she said with several layers of meaning. &quot;You wouldn't know of a place near here where I could buy a meal, would you?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will laughed. &quot;Join me for dinner. I have a great cook and after 6 o'clock this place is absolutely dead.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;But we've just met. Wait. How did you know my name?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;People love to gossip about pretty strangers. I'll tell the maid to set another place for you. You must come. I'll see you soon!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that, he dashed off. It was a tempting offer but what if he were to make a trial of her virtue? Or, worse, what if he didn't? She was just telling herself that there was only one way to find out when she realised that he hadn't told her where he lived. She ran in the direction he had gone and almost knocked over the mayor of Cryptic himself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Mr. Mayor,&quot; she said. &quot;A rich, handsome man named Will has invited me for dinner. Where does he live? ... And would it be proper to accept,&quot; she added as an afterthought.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Asked how?&quot; Said the mayor, for the moment too shaken to speak cryptically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alice repeated the conversation verbatim as she had an excellent memory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mayor looked shocked. &quot;If following this road to a dwelling of considerable dimension was your intention I'd lie to deny it would serve your intention. For the rest it were best just to say his home isn't here, the Hollow, my dear is where Will has chosen to stay.&quot; The mayor laid his finger aside of his nose then slunk off, winking and tugging his earlobe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question: should Alice follow the road or not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What term best describes the invitation?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note:&#xA;I am told that some have assumed that the puzzle is whether or not the mayor said that the road leads to Will's house. Ah, Gentle Reader, if only life were that simple. Look deeper. There is a mystery within the enigma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two hints to be used only if you really need them:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The bonus is a hint.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The people of Cryptic speak cryptically but Will is not from Cryptic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastEditorUserId="18924" LastEditDate="2016-02-14T15:24:50.037" LastActivityDate="2016-02-14T16:56:40.123" Title="A Question Concerning Alice In Cryptic" Tags="|riddle|word-problem|story|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26430" PostTypeId="2" ParentId="26426" CreationDate="2016-02-13T22:06:04.470" Score="0" Body="&lt;p&gt;Consider the following polynomial:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P(x) = 898-k-\frac{507793}{210}\cdot x-\frac{5}{192}\cdot kx^6+\frac{3013}{17280}\cdot kx^5-\frac{95}{128}\cdot kx^4+\frac{4523}{2268}\cdot kx^3-\frac{6515}{2016}\cdot kx^2+\frac{7129}{2520}\cdot kx+\frac{1092977}{420}\cdot x^2+\frac{1}{362880}\cdot kx^9-\frac{1}{8064}\cdot kx^8+\frac{29}{12096}\cdot kx^7-\frac{5}{4032}\cdot x^9-\frac{7552313}{5040}\cdot x^3+\frac{41377}{80}\cdot x^4-\frac{107677}{960}\cdot x^5+\frac{77}{5}\cdot x^6-\frac{4363}{3360}\cdot x^7+\frac{103}{1680}\cdot x^8$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a straightforward calculation to check that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$P(1) = 3$&lt;/li&gt;&#xA;&lt;li&gt;$P(2) = 4$&lt;/li&gt;&#xA;&lt;li&gt;$P(3) = 9$&lt;/li&gt;&#xA;&lt;li&gt;$P(4) = 8$&lt;/li&gt;&#xA;&lt;li&gt;$P(5) = 9$&lt;/li&gt;&#xA;&lt;li&gt;$P(6) = 8$&lt;/li&gt;&#xA;&lt;li&gt;$P(7) = 13$&lt;/li&gt;&#xA;&lt;li&gt;$P(8) = 14$&lt;/li&gt;&#xA;&lt;li&gt;$P(9) = 13$&lt;/li&gt;&#xA;&lt;li&gt;$P(10) = k$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So it might be that there are multiple solutions?&lt;/p&gt;&#xA;" OwnerUserId="16478" LastActivityDate="2016-02-13T22:06:04.470" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26431" PostTypeId="2" ParentId="26412" CreationDate="2016-02-14T00:30:50.103" Score="1" Body="&lt;p&gt;First thought on an answer was literally&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;A mistake&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, since I need to find the question, maybe it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What's wrong?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the question that no unerring answer knows&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer to this question can always be something else, and usually it's not just one thing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The thought of which does ofttimes give my asker pause&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When trying to find out what's wrong, the person thinking the question has to take a pause.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When resolution wavers, he'll reflect&#xA;On which solution heretofore considered is correct&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When unsure about a solution, the question &quot;what's wrong with my current solutions&quot; could be asked.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of heartache, of torment, suffering and dreams &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; heartache, torment and suffering - for all of these the question &quot;what's wrong&quot; might be asked (or &quot;what was wrong&quot;). If we fail at achieving dreams, we could ask &quot;what's wrong with me&quot; (lets not get into discussing if we should ;) )&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18972" LastEditorUserId="18972" LastEditDate="2016-02-14T01:13:43.283" LastActivityDate="2016-02-14T01:13:43.283" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26432" PostTypeId="1" AcceptedAnswerId="26434" CreationDate="2016-02-14T01:55:53.230" Score="8" ViewCount="504" Body="&lt;p&gt;While a one-dimensional sequence is represented by a list, a two-dimensional sequence (a sequence of sequences) can be represented by a matrix. Find the pattern behind the following sequence, where a section of the matrix with unknown coordinates is shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{bmatrix}&#xA;    &amp;amp;\vdots&amp;amp;\vdots&amp;amp;\vdots&amp;amp;\vdots&amp;amp;\vdots&amp;amp;\vdots&amp;amp;\vdots&amp;amp;\vdots&amp;amp;\vdots\\&#xA;    \dots&amp;amp; 1 &amp;amp; 5 &amp;amp; 3 &amp;amp; 7 &amp;amp; 1 &amp;amp; 9 &amp;amp; 5 &amp;amp; 13 &amp;amp; 3 &amp;amp;\dots\\&#xA;    \dots&amp;amp; 4 &amp;amp; 7 &amp;amp; 2 &amp;amp; 5 &amp;amp; 8 &amp;amp; 1 &amp;amp; 10 &amp;amp; 19 &amp;amp; 4 &amp;amp;\dots\\&#xA;    \dots&amp;amp; 1 &amp;amp; 5 &amp;amp; 9 &amp;amp; 13 &amp;amp; 2 &amp;amp; 6 &amp;amp; 10 &amp;amp; 14 &amp;amp; 3 &amp;amp;\dots\\&#xA;    \dots&amp;amp; 4 &amp;amp; 1 &amp;amp; 6 &amp;amp; 11 &amp;amp; 16 &amp;amp; 21 &amp;amp; 2 &amp;amp; 7 &amp;amp; 12 &amp;amp;\dots\\&#xA;    &amp;amp;\vdots&amp;amp;\vdots&amp;amp;\vdots&amp;amp;\vdots&amp;amp;\vdots&amp;amp;\vdots&amp;amp;\vdots&amp;amp;\vdots&amp;amp;\vdots\\&#xA;\end{bmatrix}$&lt;/p&gt;&#xA;" OwnerUserId="16100" LastEditorUserId="16100" LastEditDate="2016-02-16T22:21:52.997" LastActivityDate="2016-02-16T22:21:52.997" Title="Matrix sequence" Tags="|mathematics|pattern|calculation-puzzle|number-sequence|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26434" PostTypeId="2" ParentId="26432" CreationDate="2016-02-14T02:53:00.693" Score="6" Body="&lt;p&gt;The pattern is that the number in the $m$th row and $n$th column is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the number $n$, written in base $m$, reversed, and converted back into base 10. The section shown is the 2nd through 5th lines, 4th through 12th columns.&lt;br&gt;&#xA;&lt;br&gt;&#xA;For example, the $4$th row, $11$th column would be calculated like this: $11_{10}=23_4\rightarrow32_4=14_{10}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2016-02-14T02:53:00.693" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26436" PostTypeId="2" ParentId="15126" CreationDate="2016-02-14T06:18:44.190" Score="1" Body="&lt;h1&gt;3 is simple assuming an L cut is one cut.&lt;/h1&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2Jfca.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2Jfca.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5854" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-02-14T06:18:44.190" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26437" PostTypeId="1" AcceptedAnswerId="26452" CreationDate="2016-02-14T06:26:24.330" Score="5" ViewCount="237" Body="&lt;p&gt;Here's my first attempt at a puzzle.  The answer is a word.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A tawny pelt stretched toward the sky&lt;br&gt;&#xA;  It reached ground long after it died&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;My zestful hearts were next to fall&lt;br&gt;&#xA;  Our blood spread thin amidst it all  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Our culture was the last to go&lt;br&gt;&#xA;  Torn apart before they zapped the whole&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Don't ask who, instead ask why&lt;br&gt;&#xA;  And find the answer simple as pie&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19128" LastEditorUserId="19128" LastEditDate="2016-02-14T15:36:02.107" LastActivityDate="2016-02-23T22:28:15.180" Title="Riddle: What the Rock Glimpsed" Tags="|riddle|word|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26438" PostTypeId="2" ParentId="26412" CreationDate="2016-02-14T07:16:20.223" Score="2" Body="&lt;p&gt;Could it be . . .&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A paradox?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I am the question that no unerring answer knows &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A paradox presents, from the title, a question with no answer, so to answer it would be to err.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The thought of which does ofttimes give my asker pause&lt;br&gt;&#xA;When resolution wavers, he'll reflect &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is often instinctual to try to answer a paradox, before realizing the inherent, and multi-directional contradiction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On which solution heretofore considered is correct &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In order to realize the nature of contradiction of a paradox, both sides have first to be considered as the solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of heartache, of torment, suffering and dreams&lt;br&gt;&#xA;In this oration, life's greatest riddles are the themes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That enumeration could be considered some of life's greatest subjective riddles, but surely paradoxes are objectively so — they're unanswerable!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19213" LastActivityDate="2016-02-14T07:16:20.223" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26440" PostTypeId="2" ParentId="26426" CreationDate="2016-02-14T10:49:18.230" Score="2" Body="&lt;p&gt;Going for the obvious here.... (because someone has to..)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 14. &lt;br/&gt; &lt;br/&gt;The whole thing is very arbitrary. Goes from 3 -&gt; 4 then jumps up 6 and sequences 1 down 9 -&gt; 8 -&gt; 9 -&gt; 8 and then jumps up 6 again (from 8 to 14: just wrote this for the sake of some sort of pattern) and sequences up 13-&gt;14-&gt;13-&gt;14. &#xA; &lt;br/&gt;&lt;br/&gt;&#xA; Very disappointed if this is the answer :-{ &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19086" LastActivityDate="2016-02-14T10:49:18.230" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26441" PostTypeId="2" ParentId="26425" CreationDate="2016-02-14T11:49:07.070" Score="7" Body="&lt;p&gt;Here is my guess.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I assume that people believed in a civil war based on the similarities of the flags of both countries. One example of such a similarity is &lt;a href=&quot;https://en.wikipedia.org/wiki/Republic_of_Ireland&quot;&gt;Ireland&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Ivory_Coast&quot;&gt;Ivory Coast&lt;/a&gt; but they don't match the other hints, so let's look at them.&lt;br&gt;&lt;br&gt;There are 5 countries in &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/0/0e/Europe_countries_map_en_2.png&quot;&gt;Europe&lt;/a&gt; without a standing army: Andorra, Liechtenstein, Monaco, Vatican City and San Marino. The country which was attacked by X must share a border with one of these countries and have access to the Mediterranean Sea as well. There are 3 possibilities: Spain, France and Italy. Country X must share a border with one of these 3 countries and not have access to the Mediterranean Sea, which leves the following possibilities for X: Portugal, Belgium, Luxembourg, Germany, Switzerland and Austria.&lt;br&gt;&lt;br&gt;Similarly we can determine countries in &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/b/b7/African_continent-en.svg&quot;&gt;Africa&lt;/a&gt; which can gain access to the Mediterranean Sea by occupying another country: Western Sahara, Mauritania, Mali, Niger, Chad and Sudan.&lt;br&gt;&lt;br&gt;Among these countries there are 2 with very similar flags: &lt;a href=&quot;https://en.wikipedia.org/wiki/Belgium&quot;&gt;Belgium&lt;/a&gt; (=X) and &lt;a href=&quot;https://en.wikipedia.org/wiki/Chad&quot;&gt;Chad&lt;/a&gt; (=Y).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second guess based on &lt;strong&gt;wythagoras'&lt;/strong&gt; comment.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Based on the information that the country without an army currently has one, the best guess is Greece which has financial problems. That would leave &lt;a href=&quot;https://en.wikipedia.org/wiki/Romania&quot;&gt;Romania&lt;/a&gt; (=X) which actually has the same flag as &lt;a href=&quot;https://en.wikipedia.org/wiki/Chad&quot;&gt;Chad&lt;/a&gt; (=Y). Romania would have to occupy Bulgaria and then a part of Greece. I'm not sure how good &lt;a href=&quot;https://en.wikipedia.org/wiki/Greek_fire&quot;&gt;Greek fire&lt;/a&gt; would work as a weapon in 2065.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2016-02-14T12:06:54.393" LastActivityDate="2016-02-14T12:06:54.393" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26442" PostTypeId="1" AcceptedAnswerId="26445" CreationDate="2016-02-14T13:18:08.840" Score="7" ViewCount="701" Body="&lt;p&gt;On the table, there is a balance with two pans.&#xA;The display on the balance tells the difference between the weight in the left ban and the weight in the right pan (measured in gram).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also $2016$ coins on the table.&#xA;Cosmo tells Fredo: &lt;em&gt;&quot;There are exactly $99$ fake coins among these $2016$ coins. All genuine coins have the same weight. &#xA;Some fake coins weigh one gram less than the genuine coins, and&#xA;the other fake coins all weigh one gram more than the genuine coins.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cosmo points at the leftmost coin $x$ and asks Fredo to determine whether $x$ is fake.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Can Fredo decide whether coin $x$ is fake by using the balance at most twice?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2016-02-14T15:12:39.650" Title="2016 coins and a balance" Tags="|mathematics|strategy|weighing|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26443" PostTypeId="2" ParentId="22784" CreationDate="2016-02-14T13:57:36.760" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Vote 1 if you've had steak but didn't get it tonight.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rationale:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Whoever has steak first may have it once or many days.  Once they lose it the  second person to get steak has steak once or many days, and the first person is  voting 1.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Should the first person then get steak again, second steak-eater is voting 1 but first one stops.  The steak may flip between these two any number of times and only one will be voting for freedom.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Once the third prisoner to get steak has steak, both other prisoners are voting 1.  So they're freed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19220" LastEditorUserId="12408" LastEditDate="2016-02-14T15:12:06.507" LastActivityDate="2016-02-14T15:12:06.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26444" PostTypeId="1" AcceptedAnswerId="26449" CreationDate="2016-02-14T14:36:40.027" Score="12" ViewCount="1061" Body="&lt;p&gt;Every morning a couple has breakfast at one of three cafes: &#xA;Amy's, Bob's, or Cindy's.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;After eating at Amy's or Bob's, they flip a coin to see which &#xA;of the other two cafes they'll go to the next morning.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;However after Cindy's, they always go to Amy's.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Question: Over time, in what ratio do they attend the three cafes?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is an original problem.  I was working on odds for Monopoly boards and first wanted a small degenerate version of the game movement to make sure my methodology was sound.&lt;/p&gt;&#xA;" OwnerUserId="19220" LastEditorUserId="8874" LastEditDate="2016-02-14T19:59:36.380" LastActivityDate="2016-02-14T19:59:36.380" Title="Three Cafes Puzzle" Tags="|mathematics|probability|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26445" PostTypeId="2" ParentId="26442" CreationDate="2016-02-14T15:09:36.017" Score="14" Body="&lt;p&gt;Fredo should weigh the single coin against nothing, then all 2016 coins together against nothing. If the coin is real, then 2016 times its weight will be within 99 grams of the total weight. Otherwise, it will be at least 1917 grams off.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2016-02-14T15:09:36.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26446" PostTypeId="2" ParentId="26442" CreationDate="2016-02-14T15:12:39.650" Score="10" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes, he can. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Determined as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fredo puts all 2016 coins on the left pan and notes the &quot;difference&quot;.&#xA; This is the total weight of all coins, real and fake.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Next:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He puts coin x and the left pan and notes the difference. &#xA; If this number times 2016 is within 99 grams of the original&#xA; number then x is genuine. Otherwise it is a fake. A fake coin's&#xA; weight times 2016 will be off by almost 2,000.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastActivityDate="2016-02-14T15:12:39.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26447" PostTypeId="1" AcceptedAnswerId="26451" CreationDate="2016-02-14T15:17:54.723" Score="3" ViewCount="491" Body="&lt;p&gt;This continues the puzzle &quot;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/26442/2016-coins-and-a-balance/26445&quot;&gt;2016 coins and a balance&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, there is a balance with two pans on the table.&#xA;Again, the display on the balance tells the difference between the weight in the left ban and the weight in the right pan, measured in gram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, there are $2016$ coins on the table, of which exactly $99$ are fake.&#xA;All genuine coins have the same weight. &#xA;Some fake coins weigh one gram less than the genuine coins, and&#xA;the remaining fake coins all weigh one gram more than the genuine coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cosmo points at the rightmost coin $y$ and asks Fredo to determine whether $y$ is fake.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Can Fredo decide whether coin $y$ is fake by using the balance only &lt;strong&gt;once&lt;/strong&gt;?  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Part a:&lt;/strong&gt; If the  coins have rational weights.&lt;br&gt;&#xA;  &lt;strong&gt;Part b:&lt;/strong&gt; If the  coins have integer weights.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2016-02-14T17:57:30.160" Title="Another 2016 coins and a balance" Tags="|mathematics|strategy|weighing|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26448" PostTypeId="2" ParentId="26437" CreationDate="2016-02-14T15:22:37.870" Score="1" Body="&lt;p&gt;This makes me think of Canadian (or American) &lt;a href=&quot;https://en.m.wikipedia.org/wiki/First_Nations&quot; rel=&quot;nofollow&quot;&gt;First Nations&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A tawny pelt stretched toward the the sky&lt;br&gt;&#xA;It reached ground long after it died&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This could refer to a teepee, which is made of animal hide, which requires time to process before it can be &quot;stretched toward the sky&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My zestful heart was next to fall&lt;br&gt;&#xA;Our  spread thin amidst it all&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This could refer to animal blood used to decorate the teepee. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Our culture was the last to go&lt;br&gt;&#xA;Torn apart before they zapped the whole&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This could refer to the way European colonists suppressed and actively destroyed the First Nations culture, including the &lt;a href=&quot;http://www.thecanadianencyclopedia.ca/en/article/residential-schools/&quot; rel=&quot;nofollow&quot;&gt;residential schools&lt;/a&gt;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Don't ask who, instead ask why&lt;br&gt;&#xA;And find the answer simple as pie&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not too sure what this means. Is it referring to the reasons for the European cultural oppression?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2016-02-14T15:22:37.870" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26449" PostTypeId="2" ParentId="26444" CreationDate="2016-02-14T15:25:29.397" Score="12" Body="&lt;p&gt;Essentially, this puzzle asks for the steady state behavior of a &lt;a href=&quot;https://en.wikipedia.org/wiki/Markov_chain&quot;&gt;Markov Chain&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Amy:Bob:Cindy = 4:2:3&lt;br&gt;&#xA; In other words Prob[Amy]=4/9 and Prob[Bob]=2/9 and Prob[Cindy]=1/3.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Formally, let us denote the long term probabilities for ending up in Amy, Bob, Cindy respectively by $a,b,c$. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Then $a=b/2+c$ (as Amy is visited either with probability 1/2 after Bob, or with probability 1 after Cindy).&lt;/li&gt;&#xA;&lt;li&gt;Furthermore $b=a/2$ (as Bob is visited with probability 1/2 after Amy).&lt;/li&gt;&#xA;&lt;li&gt;Furthermore $c=a/2+b/2$&lt;/li&gt;&#xA;&lt;li&gt;And of course $a+b+c=1$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Solving the resulting equation system yields the above probabilities.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2016-02-14T15:25:29.397" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="26450" PostTypeId="2" ParentId="26444" CreationDate="2016-02-14T15:31:14.367" Score="5" Body="&lt;p&gt;In a steady state, they go to Amy's, Bob's, and Cindy's with probability $a$, $b$, $c$ respectively. For this to be true, the probability of each cafe must be $a$, $b$, $c$ the next day as well. Therefore $$a=0.5b+c\\b=0.5a\\c=0.5a+0.5b$$These equations (along with the fact that $a+b+c=1$) can be solved to find that $a=\frac{4}{9}$, $b=\frac{2}{9}$, $c=\frac{1}{3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more formal way to do this (find the steady state of a Markov process) would be to make a transition matrix and then find the eigenvector with eigenvalue 1, which represents the steady state. People have done this for Monopoly, for example &lt;a href=&quot;http://www.math.uiuc.edu/~bishop/monopoly.pdf&quot;&gt;here&lt;/a&gt;, &lt;a href=&quot;https://my.vanderbilt.edu/stacyfonstad/files/2011/10/monopoly.pdf&quot;&gt;here&lt;/a&gt;, and &lt;a href=&quot;http://web.williams.edu/Mathematics/sjmiller/public_html/hudson/Li_Markov%20Chains%20in%20the%20Game%20of%20Monopoly.pdf&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2016-02-14T15:31:14.367" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26451" PostTypeId="2" ParentId="26447" CreationDate="2016-02-14T15:54:21.470" Score="7" Body="&lt;p&gt;&lt;strong&gt;Part A:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Fredo doesn't weigh coin $y$ and also doesn't weigh some other coin, then one could be real and one could be fake, and it would be impossible to tell which was which.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Fredo doesn't weigh coin $y$ and does weigh all 2015 other coins, then one side will have $x&amp;gt;0$ coins more than the other (because 2015 is an odd number; if there was an even number here, weighing half against half and checking the parity of the result would work). The weight shown can vary freely by changing the weight of every coin, whether or not $y$ is fake. For example, increasing the weight of all coins by $\frac{1}{x}$ increases the reading by 1 gram. Because the reading is consistent with $y$ being real or fake, Fredo cannot determine whether it is fake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Fredo does weigh coin $y$ and puts at least one coin in the same pan, one could be real and one could be fake, and again he couldn't tell the difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Fredo weighs coin $y$ alone against exactly one other coin, they could both be real or both be heavy (or both light), and the scale would show zero in both cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Fredo weighs coin $y$ alone against nothing or against more than one coin, there is a difference in the number of coins, and the reading can vary freely like in the second case, so he can't determine whether the coin is genuine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are all the possible weighings (assuming the coins cannot be cut), and none of them allows Fredo to determine whether coin $y$ is real or fake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If coins can be cut, then Fredo should weigh coin $y$ against $\frac{1}{2015}$ of every other coin. The result will be at most $\frac{99}{2015}$ if $y$ is genuine, and at least $\frac{1917}{2015}$ if it is fake.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part B:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fredo should set aside coin $y$ and weigh all 2015 other coins together. The result will necessarily be within 99 grams of a multiple of 2015, which represents 2015 times the weight of a genuine coin. If the reading differs from that by an even number, then an even number of fake coins were weighed, so $y$ must be fake. On the other hand, if the reading differs by an odd number, then an odd number of fake coins were weighed, and $y$ must be genuine.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2016-02-14T17:57:30.160" LastActivityDate="2016-02-14T17:57:30.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26452" PostTypeId="2" ParentId="26437" CreationDate="2016-02-14T16:13:48.860" Score="7" Body="&lt;p&gt;This is a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; pizza&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A tawny pelt stretched toward the sky&#xA;It reached ground long after it died&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Traditionally pizza crusts (&quot;tawny pelt&quot;) are tossed in the air to stretch them&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;My zestful hearts were next to fall&#xA;Our blood spread thin amidst it all&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tomato sauce is definitely the blood.  The zestful hearts could be tomato pieces or maybe pepperoni slices. (The more I look at this, the &quot;hearts&quot; become unclear to me.  Artichoke hearts are also possible pizza toppings.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Our culture was the last to go&#xA;Torn apart before they zapped the whole&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Shredded mozzarella (&quot;culture&quot; refers to cheese cultures, and it is &quot;torn apart&quot;) is added before the pizza is baked quickly at high temperatures (&quot;zapped&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Don't ask who, instead ask why&#xA;And find the answer simple as pie&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pizza is often called a &quot;pie&quot; or &quot;pizza pie&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;edited to add: I forgot to mention the title.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;What the rock glimpsed&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pizzas are often baked on a pizza stone&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12092" LastActivityDate="2016-02-14T16:13:48.860" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26453" PostTypeId="2" ParentId="26429" CreationDate="2016-02-14T16:51:38.647" Score="5" Body="&lt;p&gt;My answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes, she should follow the road.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pulling apart the Mayor's first sentence implies that Alice will find a mansion (or other large dwelling) at the end of the road.  His second sentence implies that Will is from a place called The Hollow.&lt;br&gt;&#xA;&lt;br&gt;&#xA;If people from Cryptic speak cryptically, it stands to reason that people from The Hollow speak in a hollow manner somehow.  Perhaps we should hollow out each one of Will's statements, taking the first and last word from each.  If we do this we find:  &lt;em&gt;You will join dead people soon.&lt;/em&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; So that's super-creepy.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The questions remain: Should Alice accept the invitation or not?  Should Alice follow the road or not?&lt;br&gt;&#xA;&lt;br&gt;&#xA; If Will is implying that Cryptic is to be massacred, then Alice should leave.  If Will is implying that he will kill Alice at his home, then Alice should leave.  Either way, I think she needs to get out of town.  I'm kind of stuck here so I'm going to rely heavily on the Bonus.&lt;br&gt;&#xA;&lt;br&gt;&#xA;&quot;What word describe's Will's invitation?  Answer: &lt;em&gt;Forward&lt;/em&gt;.  So by that piece of information alone, I'm going to say that Alice should keep going on the road she is on.&lt;br&gt;&#xA;&lt;br&gt;&#xA;Run!  Run, Alice, run!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19128" LastEditorUserId="19128" LastEditDate="2016-02-14T16:56:40.123" LastActivityDate="2016-02-14T16:56:40.123" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26454" PostTypeId="1" AcceptedAnswerId="26455" CreationDate="2016-02-14T17:25:58.817" Score="17" ViewCount="2287" Body="&lt;p&gt;Determine the smallest integer $n \geq 0$ for which &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the decimal digit sum of n is a multiple of 17&lt;/li&gt;&#xA;&lt;li&gt;the decimal digit sum of $n+1$ is a multiple of 17.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;No computers! The puzzle has a nice direct solution.&lt;/p&gt;&#xA;" OwnerUserId="9125" LastActivityDate="2016-02-15T16:12:16.073" Title="Divisible by seventeen" Tags="|mathematics|number-theory|no-computers|" AnswerCount="5" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26455" PostTypeId="2" ParentId="26454" CreationDate="2016-02-14T17:54:41.900" Score="21" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We know that the digit sum of $n$ is a multiple of $17$, let us write that as ($d_1$ being the least significant digit): $$d_m + d_{m-1} + ... + d_2 + d_1 = x * 17$$ If $d_1$ would be smaller than $9$ then the digit sum of $n+1$ would be $$x * 17 + 1$$ which is obviously not divisible by $17$, so $d_1$ must be $9$. If $d_2$ would be smaller than $9$, then the digit sum of $n + 1$ wold be $$x * 17 - 9 + 1 = x * 17 - 8$$ which is not divisible by $17$ again, so $d_2$ is must be $9$ well. If $d_3$ would be smaller than $9$, then the digit sum of $n + 1$ would be $$x * 17 - 9 - 9 + 1 = x*17 - 17$$ which is divisible by $17$. This means we look for the smallest number with digit sum divisible by $17$ with the last 2 digits equal $9$ and the third last digit lower than $9$. This is obviously: $$8899$$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2016-02-14T18:03:07.017" LastActivityDate="2016-02-14T18:03:07.017" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26457" PostTypeId="2" ParentId="25439" CreationDate="2016-02-14T18:01:04.627" Score="13" Body="&lt;p&gt;Here's a stronger statement, with a quicker proof.  Given any graph (not necessarily complete or even connected) and any ordering of its edges, there's an increasing walk of length at least the average degree of the graph's vertices.  Proof:  Put a pedestrian on every vertex; at time k, let the pedestrians at either end of the kth edge change places.  When everyone's done walking, each pedestrian will have taken an increasing walk; if there are n vertices and m edges, there will have been n walks of total length 2m, thus average length 2m/n = average degree.&lt;/p&gt;&#xA;" OwnerUserId="19224" LastActivityDate="2016-02-14T18:01:04.627" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26458" PostTypeId="2" ParentId="26454" CreationDate="2016-02-14T18:39:45.087" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The integer in question must end with a bunch of 9s to leave the &quot;ones to carry&quot; to the previous digit. Hence it can be written as $abc..A9...9$ with $n$ 9s at the end and $A$ as the last digit less than 9. When we add 1 to this, it equals $abc..[A+1]0...0$, making the difference between both numbers' digit sums $9n-1$. 2 is the smallest value of $n$ allowing that expression to be divided by 17, so $n$ can be everything written as $17k+2$. Before bothering with over-17-digit numbers or something, we can simply settle for $n=2$ and fill in the few blanks we have. 9*2=18=17+1, so we at least need a bunch of digits with the sum 16, which can be the sum of 8+8 (7+9 is not allowed).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the smallest solution is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 8899.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastActivityDate="2016-02-14T18:39:45.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26459" PostTypeId="2" ParentId="26444" CreationDate="2016-02-14T19:10:47.927" Score="2" Body="&lt;p&gt;Let's call them A, B and C respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After A: With 1/2 probability B, 1/2 C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After B: With 1/2 probability C, 1/2 A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After C: The probability of going to A is 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Letting the probabilities of going to A, B and C be $P_A,P_B,P_C$ we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_A=P_B/2 + P_C$&lt;br&gt;&#xA;$P_B=P_A/2$&lt;br&gt;&#xA;$P_C=P_A/2 + P_B/2$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_A=P_A/4 + P_A/2 + P_A/4$&lt;br&gt;&#xA;$P_B=P_A/2$&lt;br&gt;&#xA;$P_C=P_A/2 + P_A/4$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1=9P_A/4$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_A=4/9$ (A to be visited 4/9 of the time)&lt;br&gt;&#xA;$P_B=2/9$ (B to be visited 2/9 of the time)&lt;br&gt;&#xA;$P_C=3/9$ (C to be visited 3/9 of the time) &lt;/p&gt;&#xA;" OwnerUserId="14016" LastActivityDate="2016-02-14T19:10:47.927" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27459" PostTypeId="1" AcceptedAnswerId="28300" CreationDate="2016-02-14T19:53:17.213" Score="8" ViewCount="376" Body="&lt;p&gt;In each of the following &lt;strong&gt;three&lt;/strong&gt; alphametic puzzles, every letter stands for a digit in base-10 representation, and different letters stand for different digits. Leading digits are always non-zero.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   FLOCK * 6 = GEESE&#xA;&#xA;   FLOCK * 7 = GEESE&#xA;&#xA;   FLOCK * 8 = GEESE&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which digit does each letter represent?&#xA;(Please present the full analysis how these digits can be determined.)&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="9404" LastEditDate="2016-03-04T11:07:07.727" LastActivityDate="2016-03-04T11:07:07.727" Title="Flock of Geese alphametic" Tags="|no-computers|alphametic|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27460" PostTypeId="2" ParentId="26413" CreationDate="2016-02-14T20:05:17.893" Score="9" Body="&lt;p&gt;My answer is that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; X can force a win on his 8th move.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof outline:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; By his 3rd move, X can put an L-shape on the board (three quarters of a square).  From this point on, O is no longer free to move but forced to try and block X completing a square.  Assuming O is trying to block X from winning, X's 4th move extends the L to a T-shape, which on move 5 he then either expands to a +-shape (to win on his 6th move) or to a new L-shape of double size (based on the T-shape's crossbar).  If he doubles the scale, moves 6 and 7 again respectively make up a T-shape and a +-shape, after which he wins.&#xA;&lt;p&gt;So, only O's first two moves are free for him to determine.  The first move determines the initial scale of the L/T/+.  The second move determines the orientation of the L- and T-shape or whether X has to double the scale on move 5.  X moves in such a way that O never gets an L-shape of his own, except on his last move, just before X wins.&#xA;&lt;p&gt;Depending on O's first move, X will try to either build a 2-by-2 or a 3-by-3 square.  Doubling scale thus results in either a 3-by-3 or a 5-by-5 square, potentially making use of the full size of the board.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I find it hard to make this really rigorous without constructing a full decision tree, but here are some worked examples that I believe to cover all possibilities:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; X starts in the centre (marked as &lt;code&gt;1&lt;/code&gt; below).  His next move (&lt;code&gt;2&lt;/code&gt;) depends on whether O puts his first move (&lt;code&gt;a&lt;/code&gt;) inside the 3-by-3 centre square (depicted on the left) or anywhere else on the board (depicted on the right).  The situation after X's 2nd move is then one of the following (up to mirroring and rotation):&#xA;&lt;pre&gt;&#xA;             .........              |              aaaaa....&#xA;             .........              |              .aaaa....&#xA;             .........              |              ..aaa....&#xA;             ...aa....              |              .........&#xA;             ....1....              |              ....1....&#xA;             .........              |              ....2....&#xA;             ....2....              |              .........&#xA;             .........              |              .........&#xA;             .........              |              .........&#xA;&lt;/pre&gt;&#xA; Player O can now follow one of three strategies, roughly:&lt;br&gt;&#xA; 1. Try to block the +-shape that X is trying to form: X doubles scale on his 4th move and wins on his 8th move.&lt;br&gt;&#xA; 2. Prevent X from making cell &lt;code&gt;2&lt;/code&gt; the centre of his +-shape: X instead centres it around &lt;code&gt;1&lt;/code&gt; and wins on his 6th move.&lt;br&gt;&#xA; 3. Anything else: X builds a +-shape centred around cell &lt;code&gt;2&lt;/code&gt; and wins on his 6th move.&lt;p&gt;&#xA; &lt;p&gt;I have tried to depict what each of these looks like in the six figures below (extended from the two figures above).&#xA; Note that I have indicated some of O's possible moves with &lt;code&gt;b&lt;/code&gt; but symmetric possibilities exist.  For example, in the leftmost figure, O can pick the spot marked &lt;code&gt;3&lt;/code&gt; and then X will simply pick the cell marked &lt;code&gt;b&lt;/code&gt; on the same row.&#xA;&#xA;&lt;pre&gt;&#xA; f...6...g   .........   .........  |  aaaaa....   aaaaa....   aaaaa....&#xA; .........   .........   .........  |  .aaaa....   .aaaa....   .aaaa....&#xA; ..b...b..   ..e.5.6..   ....b....  |  ..aaa....   ..aaa....   ..aaa....&#xA; ...aa....   ...aa....   ...aa....  |  ..8.4de..   ...e4d...   ....b....&#xA; 5...1.c.7   ..4.1.3..   ..6.1.c..  |  ...b13...   ...513...   ...61c...&#xA; .........   .........   .........  |  ..7b2c5..   ...62c...   ...523...&#xA; ..b.2.3..   ..d.2.c..   ..5.2.3..  |  .....d...   ....b....   ...e4d...&#xA; .........   .........   .........  |  ..g.6.f..   .........   .........&#xA; e...4.d.8   ....b....   ..e.4.d..  |  .........   .........   .........&#xA;&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The figures feel a little clumsy and not very elegant.  If I come up with a way to make the argument more insightful, I will edit.&lt;/p&gt;&#xA;" OwnerUserId="13484" LastEditorUserId="13484" LastEditDate="2016-02-14T20:29:00.230" LastActivityDate="2016-02-14T20:29:00.230" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27461" PostTypeId="2" ParentId="27459" CreationDate="2016-02-14T20:48:19.787" Score="3" Body="&lt;h1&gt;Partial answer (only first puzzle)&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;My analysis:&lt;br&gt;&#xA;&lt;strong&gt;F is 1&lt;/strong&gt; (assuming it can't be 0) because otherwise the result would be 6 digits, in all cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first case:&lt;br&gt;&#xA;GEESE is divisible by 6 which means that it's even, so &lt;strong&gt;E must be an even number&lt;/strong&gt;.&lt;br&gt;&#xA;GEESE is divisible by 6 which means it's also divisble by 3 which means that the digit sum must be divisible by three which means that &lt;strong&gt;G+S is divisble by three&lt;/strong&gt;.&lt;br&gt;&#xA;Because G is at least 6 times F then &lt;strong&gt;G is 6 or higher&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now by these conclusions the possibilities for GEESE is a list that I thought was small enough to check by hand:&lt;br&gt;&#xA;60030&#xA;60090&#xA;62232&#xA;62292&#xA;64434&#xA;64494&#xA;68838&#xA;68898&#xA;70020&#xA;70050&#xA;70080&#xA;72252&#xA;72282&#xA;74424&#xA;74454&#xA;74484&#xA;76626&#xA;76656&#xA;76686&#xA;78828&#xA;78858&#xA;80040&#xA;80070&#xA;82242&#xA;82272&#xA;84474&#xA;86646&#xA;86676&#xA;90030&#xA;90060&#xA;92202&#xA;92232&#xA;92262&#xA;94404&#xA;94434&#xA;94464&#xA;96606&#xA;96636&#xA;98808&#xA;98838&#xA;98868&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I checked every one of them what they were divided by 6 and if they would fit the equation. My result:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The solution is $15367 * 6 = 92202$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1748" LastEditorUserId="1748" LastEditDate="2016-02-15T06:04:27.270" LastActivityDate="2016-02-15T06:04:27.270" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27462" PostTypeId="1" AcceptedAnswerId="27726" CreationDate="2016-02-15T00:10:03.487" Score="5" ViewCount="1880" Body="&lt;p&gt;Underground&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is where I'm found&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Markets rise and fall with me&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As do actions of countries&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You will find me on the driver's lips, attached to the car, and part of  the road in cold parts of the US.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These hints are very specific and literally true without any logical stretches. Looking for correct reasoning.&lt;/p&gt;&#xA;" OwnerUserId="19179" LastEditorUserId="19179" LastEditDate="2016-02-22T19:14:51.067" LastActivityDate="2016-02-23T08:53:06.377" Title="Something Powerful" Tags="|riddle|" AnswerCount="6" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27463" PostTypeId="1" AcceptedAnswerId="27489" CreationDate="2016-02-15T00:33:33.737" Score="21" ViewCount="1605" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Excerpt of “Manipulating the Fabric of Space”&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;“Miscasting translocation&#xA;  spells can have many nasty consequences. The caster’s body could be&#xA;  warped or torn apart, his soul could be separated from his body -&#xA;  turning into an insubstantial wisp, a clone with part of the caster’s&#xA;  personality could appear, the caster could be transported into the&#xA;  endless and chaotic abyss or - &lt;strong&gt;worst of all...&lt;/strong&gt; the caster could&#xA;  accidentally summon a few &lt;em&gt;daemonia molestus&lt;/em&gt;, more commonly known as&#xA;  common imps.”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/amkfr.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/amkfr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After hours of practice Edward was finally feeling sure enough to attempt a simple dislocation spell, targeting a single cup on an empty table. It only took him a moment to speak the words and cast the spell - and the cup actually disappeared!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edward had the biggest grin on his face. “Hahaha, see? Told you that translocations are no problem for  someone with my talen--”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suddenly, he felt two weights land on top of his head before an annoying voice yelled into his right ear. “Hey!”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edward almost fell from his chair. He tried to grab whatever was sitting on his head and pull it down, only for them to dig their claws into both of his ears. “W-what… Who the heck are you?! Get off my head!”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two small creatures chuckled, before the one holding Edward’s right ear jumped down on his right shoulder, while the other remained in place, still clutching Edward's left ear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The small creature on Edward’s right shoulder then started explaining. “We’re imps! You have summoned us - and we came to play a little game with you. I’m Acredo and that’s my brother Bugia. Every sentence I say is the truth, while every sentence my brother says is a lie. As soon as you want to start the game you just tell us, and we’ll switch positions at random. You can ask us as many yes or no questions as you want, to detect who of us is my always lying brother, Bugia. But I should warn you… you only get one chance to guess who of us it is before the game is over. Every time you ask one of us a question that can’t be answered with just yes or no, like you just did, another one of our brothers appears after the imp you  asked finished answering your question. I will explain you the further rules once you accept to play our game”, he said. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As just mentioned, Edward could feel another weight land ontop of his head, quickly grabbing his free left ear with it’s sharp claws - before an equally annoying voice yelled “Hey!” into it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edward winced. “Stop yelling, I’m not deaf!”, he yelled. He crossed his arms and angrily stared down on his empty table. “And why should I bother to play your game?”, he asked the imp on his shoulder  in an annoyed tone. “I could always just get rid of you the ‘old fashioned’ way...”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All three imps chuckled and the imp on Edward’s right shoulder answered again. “Because we’ll stick around until your last days if you lose or don’t play! And if you kill an imp, two more show up! By the way, all of my brothers appearing because of your illegal questions(1) are completly unreliable. Each time you ask them a question, they'll decide between behaving like an imp who always tells the truth or an imp who always lies, randomly, and then answer the question”, he said.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;( (1) Illegal Questions: All questions that cannot be answered with yes or no.)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edward could feel a fourth weight landing on his left shoulder. “Hey brothers!”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sorcerer makes another futile attempt to get the imps off his shoulders and head but isn’t able to remove even one of them, only hurting himself in the process as they keep a firm hold on him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Willy, Edward’s familiar, who was sitting just next to the imp on Edwards right shoulder wiggles his antennae. “How many brothers do you have?”, he asked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Oh, we are a big family. Well, at least compared to humanoid families. I have 257 brothers, currently. And at this rate, your master will be the host of quite a few of them! Hehehe… well, he already is by now.”, the imp on Edward’s right shoulder answered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another imp appeared, this time grabbing the back of Edward’s head. “Hey! Nice to meet you ~”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edward’s face stated to turn red from anger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Nice to meet you too!”, Willy answered joyfully, before chatting with the imp on Edwards right shoulder again. “You know, I have many brothers too! And we all look quite similar as well. How do you imps always know who is who?”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edward suddenly slammed his fist against the table in front of him. “WILLY, JUST SHUT UP!!!”, he screamed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Oh, we don’t know who is who, actually!”, the imp on Edwards right shoulder answered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edward could feel a leathery tail wrapping around his throat, another imp appearing - hanging upside down against his chest. “Heya!”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In resignation, Edward just rests his head on his palms.”Ugh… FINE!! I’ll play! Explain the rules and let’s get this over with.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The imp on Edwards right shoulder nodded. “Very well! The rules are simple. As I mentioned, you have to find my always lying brother, Bugia, by asking us as many yes or no questions as you want. We’ll switch positions randomly before the game starts - but not physically. We can simply switch our consciousness between bodies at will, so you won’t be able to notice it. Oh, and I almost forgot to mention it, but &lt;strong&gt;every&lt;/strong&gt; time you ask one of us a question, that imp randomly chooses another imp who behaves differently and switches consciousness with it after answering the question. So, an imp who always lies won’t switch with another imp who always lies and instead choose an imp with a different behaviour, like for example one that always tells the truth. But as you already know, there is only one imp who always lies between us - my brother Bugia. Also, just to clarify: The switch happens instantly after the question was answered. Keep that in mind when you want to point out who Bugia is, since you’re always going to do that after the switch happened. And yes, you can win this game - without asking dozens of questions, too! Alright, lets begin! We’ll all shuffle our consciousness now - good luck!”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After thinking about it for a while, Edward grit his teeth. “This is impossible! The way the rules are laid out, it’s not possible to ever guess who Bugia is with 100% certainty!”, he mumbled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Willy started to wiggle with his antennae again. “Oh… Oh! Master! I think I know! I figured it out! Those imps really are tricky! I can solve the riddle with at most two questi--”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Shut up, Willy! You already caused enough problems - it’s your fault there are so many of them in the first place!”, Edward interrupted. “Uggh… Why does this kind of stuff always happen to me?”, Edward sighed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“It’s probably bad karma!”, one imp answered - before Edward could feel another imp land ontop of his head. “Hey there!”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“THAT WAS A RHETORIC QUESTION!!”&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edward’s Notes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There are now 7 Imps. One of them always tells the truth, one of them always lies and 5 of them give random answers (I'll refer to them as unreliable).&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Each time I ask one of the unreliable imps a question, they'll decide between behaving like an imp who always tells the truth or an imp who always lies, randomly, and answer the question.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;I can ask any imp a yes or no question. If I ask an “illegal” question that can’t be answered with yes or no a new imp will spawn that gives random answers.&lt;/li&gt;&#xA;&lt;li&gt;There was no mention about a limit of how many times I’m allowed to ask yes or no questions.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Apparently&lt;/em&gt;, they cannot tell each other apart. So they probably don’t know who’s a liar and who isn’t (except for themselves, obviously).&lt;/li&gt;&#xA;&lt;li&gt;After answering my question, the asked imp will instantly switch position with another random imp who has a different behaviour. Examples:&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;An imp who always lies, will switch with an imp who always tells the truth or an unreliable imp.&lt;/li&gt;&#xA;&lt;li&gt;An imp who always tells the truth will switch with an imp who always lies or an unreliable imp.&lt;/li&gt;&#xA;&lt;li&gt;A unreliable imp (who answers randomly) will switch with an imp who always lies  or one who always tells the truth.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;They never clarified if an imp knows with who he’s going to switch. Since they said it’s random, the imps probably don’t know their future position themselves.&lt;/li&gt;&#xA;&lt;li&gt;The imps switch consciousness between bodies, so it’s impossible for me to know who switched with who just by looking at them.&lt;/li&gt;&#xA;&lt;li&gt;My goal is to identify Bugia, the imp who always lies.&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Pointing out who Bugia is will always happen &lt;strong&gt;after&lt;/strong&gt; the imps switched consciousness. This means I’ll have to make sure in some way, that I know who is going to switch in.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;At least two informations seem to contradict themselves… and how am I supposed to find Bugia if I can only guess &lt;strong&gt;after&lt;/strong&gt; they switch consciousness? This makes no sense! Maybe I missunderstood something?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What questions could Willy have in mind, to solve this seemingly impossible game with at most 2 questions? And would it still work, now that there are seven imps - one more than when Willy mentioned having a solution?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Hint 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There seems to be a very subtle contradiction in Edward's notes. Maybe that's a good point to start?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the puzzle seems impossible to you as well, maybe you should analyze what the imps are saying a bit closer. It could be that they try to confuse you with &quot;context&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Trying to figure out what questions to ask the seven imps is the &lt;strong&gt;last&lt;/strong&gt; step! You have some work ahead of you before doing that.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14076" LastEditorUserId="14076" LastEditDate="2016-02-15T07:42:33.353" LastActivityDate="2016-02-19T21:07:11.600" Title="Edward and the seven imps' Game" Tags="|logical-deduction|story|liars|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="27464" PostTypeId="2" ParentId="26426" CreationDate="2016-02-15T00:42:05.157" Score="0" Body="&lt;p&gt;It's not 100% convincing, but worth a try:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3 4 9 8 9 8 13 14 13 ?&#xA; &lt;br/&gt;&#xA; Numbers 3, 6, 9: 9, 8, 13 (no idea. Typo? :)&#xA; &lt;br/&gt;&#xA; Numbers 2, 5, 8: 4, 9, 14 (+5)&#xA; &lt;br/&gt;&#xA; Numbers 1, 4, 7: 3, 8, 13 (+5)&#xA; &lt;br/&gt;&#xA; Number 10: 13 + 5 = &lt;strong&gt;18&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10628" LastActivityDate="2016-02-15T00:42:05.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27465" PostTypeId="2" ParentId="27462" CreationDate="2016-02-15T01:16:28.083" Score="10" Body="&lt;p&gt;This is just a Guess...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Salt&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Underground Is where I'm found&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Salt is found underground.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Markets rise and fall with me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Salt is a much needed material for markets/stores.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As do actions of countries&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Many countries have taken the initiative to try to reduce salt consumption in their specific country. Also there are many countries very dependent on salt as it used for so many things.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19196" LastEditorUserId="19196" LastEditDate="2016-02-15T01:43:11.390" LastActivityDate="2016-02-15T01:43:11.390" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="27466" PostTypeId="2" ParentId="27462" CreationDate="2016-02-15T01:22:13.917" Score="4" Body="&lt;p&gt;My guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tar.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Underground:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Can be produced from coal, wood, petroleum, or peat&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Markets:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Used at least pure form of heroine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Countries&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.oxforddictionaries.com/definition/english/don&amp;#39;t-spoil-the-ship-for-a-ha&amp;#39;p&amp;#39;orth-of-tar&quot; rel=&quot;nofollow&quot;&gt;&quot;don't spoil the ship for a ha'p'orth of tar &quot;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Attached to car&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; slang for wheels&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Streets&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; obviously on streets and it melts on too hot streets  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19227" LastEditorUserId="19227" LastEditDate="2016-02-15T01:27:43.180" LastActivityDate="2016-02-15T01:27:43.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27467" PostTypeId="2" ParentId="27462" CreationDate="2016-02-15T01:28:24.860" Score="13" Body="&lt;p&gt;Is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oil&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Underground&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is where I'm found&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To get to oil, you have to dig underground for this resource&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Markets rise and fall with me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Depending on OPEC (world's biggest distributor of oil) prices its oil, the markets can fall (when the price is really low like now) or rise (when the price gets high)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As do actions of countries&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Countries can react in many different ways depending on the oil industry&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to the hints...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oil might be seen when splashed onto the driver's face when fixing the engine, in the fuel tank of the car, and sometimes spilled on the road. The road might be the ocean as it serves as a road for ships and there are unfortunately, lots of oil spills&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9042" LastActivityDate="2016-02-15T01:28:24.860" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="27468" PostTypeId="2" ParentId="26454" CreationDate="2016-02-15T01:50:13.340" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 889,999,999,999,999,999,999&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;... is one solution. I can't find a smaller one, is this proof enough? ;)&lt;/p&gt;&#xA;" OwnerUserId="10628" LastActivityDate="2016-02-15T01:50:13.340" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27469" PostTypeId="2" ParentId="27459" CreationDate="2016-02-15T02:23:53.983" Score="3" Body="&lt;h2&gt;Part 3&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Assuming &lt;code&gt;F&lt;/code&gt; is at least 1, &lt;code&gt;FL&lt;/code&gt; must be either &lt;code&gt;12&lt;/code&gt; or &lt;code&gt;10&lt;/code&gt; for the product to be less than 10000.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &lt;code&gt;FL&lt;/code&gt; is &lt;code&gt;12&lt;/code&gt;, then &lt;code&gt;GEESE&lt;/code&gt; is at least 96000. &lt;code&gt;E&lt;/code&gt; must be even, so it can only be &lt;code&gt;6&lt;/code&gt; or &lt;code&gt;8&lt;/code&gt;. If &lt;code&gt;E&lt;/code&gt; is &lt;code&gt;6&lt;/code&gt;, then &lt;code&gt;FLOCK*8&lt;/code&gt; being between 96606 and 96686 requires that &lt;code&gt;O&lt;/code&gt; is &lt;code&gt;0&lt;/code&gt; and &lt;code&gt;C&lt;/code&gt; is &lt;code&gt;8&lt;/code&gt;. &lt;code&gt;K&lt;/code&gt; can't be &lt;code&gt;2&lt;/code&gt;, so it must be &lt;code&gt;7&lt;/code&gt; to make the units digit correct. But &lt;code&gt;12087*8=96696&lt;/code&gt; duplicates the digit 9, so &lt;code&gt;E&lt;/code&gt; being &lt;code&gt;6&lt;/code&gt; doesn't work. &lt;code&gt;E&lt;/code&gt; must be &lt;code&gt;8&lt;/code&gt;, making &lt;code&gt;O&lt;/code&gt; be &lt;code&gt;3&lt;/code&gt; and &lt;code&gt;C&lt;/code&gt; be &lt;code&gt;5&lt;/code&gt; or &lt;code&gt;6&lt;/code&gt;. &lt;code&gt;K&lt;/code&gt; can't be &lt;code&gt;1&lt;/code&gt;, so it is &lt;code&gt;6&lt;/code&gt; and &lt;code&gt;C&lt;/code&gt; is &lt;code&gt;5&lt;/code&gt;. The result &lt;code&gt;12356*8=98848&lt;/code&gt; works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &lt;code&gt;FL&lt;/code&gt; is &lt;code&gt;10&lt;/code&gt;, then &lt;code&gt;GEESE&lt;/code&gt; is at least 80000 and at most 87999. Again, &lt;code&gt;E&lt;/code&gt; must be even, so it has to be &lt;code&gt;2&lt;/code&gt;, &lt;code&gt;4&lt;/code&gt;, or &lt;code&gt;6&lt;/code&gt;. If &lt;code&gt;E&lt;/code&gt; is &lt;code&gt;2&lt;/code&gt;, then &lt;code&gt;O&lt;/code&gt; must also be &lt;code&gt;2&lt;/code&gt;, which is impossible. If &lt;code&gt;E&lt;/code&gt; is &lt;code&gt;4&lt;/code&gt;, then &lt;code&gt;O&lt;/code&gt; must be &lt;code&gt;5&lt;/code&gt;, and &lt;code&gt;C&lt;/code&gt; must be &lt;code&gt;6&lt;/code&gt; to avoid a repeat. Then &lt;code&gt;GEESE&lt;/code&gt; is either &lt;code&gt;84484&lt;/code&gt; or &lt;code&gt;84494&lt;/code&gt;, but neither of these numbers is divisible by 8. Finally, if &lt;code&gt;E&lt;/code&gt; is &lt;code&gt;6&lt;/code&gt;, then &lt;code&gt;O&lt;/code&gt; must be &lt;code&gt;8&lt;/code&gt;, which is already used as &lt;code&gt;G&lt;/code&gt;. So none of these cases works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only working answer is &lt;code&gt;12356*8=98848&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2016-02-15T02:23:53.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27470" PostTypeId="1" AcceptedAnswerId="27471" CreationDate="2016-02-15T03:06:52.933" Score="19" ViewCount="2080" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RKSBp.png&quot; alt=&quot;1]&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this cross shape, all twelve sides are the same length and all angles are right angles. How many cuts does it take to divide the cross into pieces that can be rearranged to form each of the following sets of squares?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a. Two squares, where the larger one has a side length two times the side length of the smaller.&lt;/li&gt;&#xA;&lt;li&gt;b. Two squares of the same size.&lt;/li&gt;&#xA;&lt;li&gt;c. Two squares, where the larger one has a side length three times the side length of the smaller.&lt;/li&gt;&#xA;&lt;li&gt;d. One square.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Each set of squares (a, b, c, d) is a separate problem. You start with an uncut cross each time and only have to produce one set of squares.&lt;/li&gt;&#xA;&lt;li&gt;All cuts must be straight lines.&lt;/li&gt;&#xA;&lt;li&gt;Cuts can go across multiple pieces, but pieces cannot be moved until all cuts are done.&lt;/li&gt;&#xA;&lt;li&gt;Pieces may be flipped or rotated, but cannot be resized or removed.&lt;/li&gt;&#xA;&lt;li&gt;Pieces cannot overlap or cover each other.&lt;/li&gt;&#xA;&lt;li&gt;After the pieces are moved, edges that join two pieces are ignored, as if the pieces fused together. The rest of the edges must form the set of squares exactly; squares must be formed entirely of piece edges, and extra edges or pieces are not allowed.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2016-02-27T01:14:41.017" LastActivityDate="2016-02-27T01:14:41.017" Title="Squaring a cross" Tags="|geometry|dissection|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27471" PostTypeId="2" ParentId="27470" CreationDate="2016-02-15T04:41:40.330" Score="26" Body="&lt;p&gt;Solutions are as in the following diagram:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/DZ9MY.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DZ9MY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addition: My original solution for E has 5 cuts and 9 separate pieces - a simpler solution is shown below, with only four cuts (to the cross) and five pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2nXdM.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2nXdM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dimensions are as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; a) side lengths = 2 and 1 —[2 cuts, 3 pieces]&#xA; b) side lengths = sqrt(5/2) —[2 cuts, 4 pieces]&#xA; c) side lengths = 3/sqrt(2) and 1/sqrt(2) —[2 cuts, 6 pieces]&#xA; d) side length = sqrt(5) —[2 cuts, 4 pieces]&#xA; e) side lengths = 3*sqrt(5/13) and 2*sqrt(5/13) —[4 cuts, 5 pieces]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note that these are not the only solutions &lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="1631" LastEditDate="2016-02-21T20:38:15.690" LastActivityDate="2016-02-21T20:38:15.690" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27473" PostTypeId="2" ParentId="26412" CreationDate="2016-02-15T09:10:51.223" Score="6" Body="&lt;p&gt;Your question is &quot;What am I?&quot; I'm guessing the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hamlet's question:&lt;br&gt;&#xA; &quot;to be or not to be, that is the question&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the question that no unerring answer knows&lt;br&gt;&#xA;The thought of which does ofttimes give my asker pause&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Both choices are open; both have pros and cons. The question is reflective, hence &quot;give my asker pause&quot;.&lt;br&gt;&#xA; &quot;... Must give us pause.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When resolution wavers, he'll reflect&lt;br&gt;&#xA;On which solution heretofore considered is correct&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Regardless of which is chosen (to be / not to be), the decision is questioned when the resolution to go through with it wavers.&lt;br&gt;&#xA; &quot;... Whether 'tis Nobler in the mind to suffer The Slings and Arrows of outrageous Fortune, Or to take Arms against a Sea of troubles, And by opposing end them: ...&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of heartache, of fortune, of death and dreams&lt;br&gt;&#xA;In this oration, life's greatest riddles are the themes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In Shakespeare's original, Hamlet's question is framed against his father's murder (heartache). It's been a long time since I studied Hamlet, so a lot of this is rusty, but fortune may be a reference to his uncle assuming the throne. Death and dreams are part of the line &quot;to sleep, perchance to dream&quot;.&lt;br&gt;&#xA; On fortune: &quot;... Slings and Arrows of outrageous Fortune, ...&quot;.&lt;br&gt;&#xA; On heartache, death and dreams: &quot;... The Heart-ache, and the thousand Natural shocks That Flesh is heir to? 'Tis a consummation Devoutly to be wished. To die, to sleep, To sleep, perchance to Dream; ...&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You hear me spoken time and time again&lt;br&gt;&#xA;For centuries my lasting magic still remains&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An allusion to the line &quot;to be or not to be&quot; being used in other contexts, or perhaps being quoted repeatedly due to the play being popular.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fit for a prince, uttered in earnest, sometimes in jest&lt;br&gt;&#xA;Of all my kind I surely am one of the best&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The character voicing the line was &lt;em&gt;Prince&lt;/em&gt; Hamlet. Earnest vs jest may be an allusion to how the line is used by other people in other contexts. Also Shakespeare (Hamlet's playwright) is &quot;widely regarded as the greatest writer in the English language and the world's pre-eminent dramatist&quot;, and among his plays, the line &quot;to be or not to be&quot; is one of the best known.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now there's a bounty on my head&lt;br&gt;&#xA;A bit more work, and I'll be dead&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;For in that sleep of death, what dreams may come, When we have shuffled off this mortal coil,&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Come on Puzzling, you can win&lt;br&gt;&#xA;If you remember all your sins&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;... The fair Ophelia? Nymph, in thy Orisons Be all my sins remembered.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I puzzle the will, confound the sense&lt;br&gt;&#xA;'Tis time to end all the suspense&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;... The undiscovered Country, from whose bourn No Traveller returns, Puzzles the will, And makes us rather bear those ills we have, ...&quot;&lt;br&gt;&#xA; The First Quarto version uses &quot;... puzzles the brain, and doth confound the sense, ...&quot;.&lt;br&gt;&#xA; About &lt;em&gt;'Tis&lt;/em&gt;: &quot;... Whether 'tis Nobler in the mind ...&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;References taken from&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/To_be,_or_not_to_be&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;, First Folio text except for one First Quarto quote.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9422" LastEditorUserId="9422" LastEditDate="2016-02-15T22:53:52.547" LastActivityDate="2016-02-15T22:53:52.547" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="27474" PostTypeId="1" AcceptedAnswerId="27505" CreationDate="2016-02-15T09:12:04.293" Score="2" ViewCount="610" Body="&lt;p&gt;This is variation of this question:&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/23126/what-is-the-value-of-this-coin-flipping-game&quot;&gt;What is the value of this coin flipping game?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine that you can play the game in a casino - you are flipping a fair coin. If you get tails, the game continues and the money you can win is doubled. If you get heads, the game ends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Money you can win at the start: $\$1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The casino has $\$1,000,000$ to pay you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can play this game only $1000$ times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If you get heads in the first round, you get $\$1$ and the game ends. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you get tails twice and then heads, you get $\$4$ ($1\times 2 \times 2$) and the game ends.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much you would pay for this game?&lt;/p&gt;&#xA;" OwnerUserId="18956" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2016-07-25T02:30:58.187" Title="What is the value of this other coin flipping game?" Tags="|mathematics|probability|coin-tossing|" AnswerCount="3" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="27475" PostTypeId="1" AcceptedAnswerId="27477" CreationDate="2016-02-15T10:21:36.067" Score="8" ViewCount="1245" Body="&lt;p&gt;Determine the smallest integer $n &amp;gt; 0$ for which &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the sum of the decimal digits of $n$ is divisible by $20$,&lt;/li&gt;&#xA;&lt;li&gt;$n$ is divisible by $20$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Question inspired by &lt;a href=&quot;https://puzzling.stackexchange.com/questions/26454/divisible-by-seventeen&quot;&gt;Divisible by seventeen&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="18514" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2016-02-15T14:37:40.560" Title="Divisible by 20" Tags="|mathematics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27476" PostTypeId="2" ParentId="26412" CreationDate="2016-02-15T10:22:46.890" Score="1" Body="&lt;p&gt;What about:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This question itself?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The tradition of this site and its users is said in the first stanza, of what the asker of the question and the answerers' experience is when the answer is still unknown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second stanza is about any question on Puzzling.SE.&#xA;Sometimes it's a puzzle, other times it's a riddle(a joke question). Any unworthy question is either tagged as duplicate, or downvoted to hell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third stanza is about this specific question.&#xA;What is going to happen soon, is described in poetic terms.&#xA;If you connect together the wrong(but still half-correct) answers, you will arrive at the correct answer.&lt;/p&gt;&#xA;" OwnerUserId="18117" LastActivityDate="2016-02-15T10:22:46.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27477" PostTypeId="2" ParentId="27475" CreationDate="2016-02-15T10:32:05.300" Score="6" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;3980&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The number must end in 0 and the other digits must add up to 20 (at least). Since $9+9&amp;lt;20$, the number must have at least four digits. It cannot begin with 1 since the maximum sum of the digits is $1+9+9 &amp;lt; 20$ and it cannot begin with 2 since $2+9+9=20$ but 2990 is not divisible by 20. Therefore $n &amp;gt; 3000$. Since $3+8+9=20$ is the only possible sum under these constraints (i.e, $n$ begins with 3 ends in 0 and the digit sum is 20), the first number that works is 3980.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18422" LastEditorUserId="18422" LastEditDate="2016-02-15T10:55:54.657" LastActivityDate="2016-02-15T10:55:54.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27478" PostTypeId="2" ParentId="27475" CreationDate="2016-02-15T11:23:03.177" Score="6" Body="&lt;p&gt;It's:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3980&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We must have 00, 20, 40, 60 or 80 as its last two digits to be divisible by 20.&lt;br&gt;&#xA; For the smallest number we'll want 80 to kick off the digit sum as large as possible.&lt;br&gt;&#xA; We now need smallest number of digits that sum with 8 to to give 20.&lt;br&gt;&#xA; That's 2 digits: 6, 6 or 7, 5 or 8, 4 or 9, 3.&lt;br&gt;&#xA; Of those pairs want the one with the lowest integer to take the thousand's place.&lt;br&gt;&#xA; That's 9 and 3 so we have 3980.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastActivityDate="2016-02-15T11:23:03.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27479" PostTypeId="2" ParentId="27474" CreationDate="2016-02-15T11:53:25.863" Score="2" Body="&lt;p&gt;The expected value of this game goes to infinity as below;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E(x)=0.5(1)+0.5(0.5\times 2+0.5(0.5\times 4+0.5(...))))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E(x)=0.5(2^x)+0.5(E(x+1))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E(x)=\infty $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you can play 1000 times at most;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;$E(x)=0.5x = 500$&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if there were $ 2^{1000}$$ but they have a million. so you can play the game getting tails all the time in a row 21 times at most.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you need to pay;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{n=0}^{20} (0.5^x(2^x-y))=0$ (for n=20, $2^n$ is assumed to be a million)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y=10.48 \$$&lt;/p&gt;&#xA;" OwnerUserId="18514" LastEditorUserId="18514" LastEditDate="2016-02-15T12:28:54.697" LastActivityDate="2016-02-15T12:28:54.697" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27480" PostTypeId="2" ParentId="27463" CreationDate="2016-02-15T12:11:29.057" Score="4" Body="&lt;p&gt;Pick any imp body to direct the questions to. The one clinging to Edward's chest is convenient. Pick a different imp body (e.g. the one on Edward's right shoulder) and call it &lt;em&gt;the target&lt;/em&gt;. Assume that the imp explaining the rules was telling the truth about the rules. (If the rules can't be trusted, this solution doesn't work.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is part of the rules:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;every time you ask one of us a question, that imp randomly chooses another imp who behaves differently and switches consciousness with it after answering the question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I will take that to mean that the choice is random and occurs while the question is asked (the choice is made before answering), and that switching occurs after answering. That is, the imp knows and is able to tell you (given an appropriate question) which position it will switch to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any given question, any imp will consistently tell the truth or say its opposite. Questions of the sort &quot;Would you say X is true?&quot; always produce truthful answers from an imp that is telling the truth. For a lying imp, &quot;Is X true?&quot; would give the untruthful answer, but then &quot;Would you say&quot; would invert the truth value, giving the truthful answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The imps always switch with a random imp of a different &quot;behaviour&quot;, so after every non-Bugia answerer, you have a 2/3 probability of getting Bugia, and each time Bugia answers, you are certain to get a non-Bugia imp. So given any &lt;em&gt;pair&lt;/em&gt; of successive answerers, there is a non-zero probability that one of them is Bugia. Since the switches are random, Bugia will eventually turn up as answerer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, consider just the times Bugia turns up as answerer. Bugia's &quot;behaviour&quot; is unique, so can switch with &lt;em&gt;any&lt;/em&gt; other imp. Since the switches are random, Bugia will eventually choose &lt;em&gt;the target&lt;/em&gt; to switch to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting this all together, eventually Bugia will eventually appear as the answerer and will switch to &lt;em&gt;the target&lt;/em&gt;. So keep asking the following question until you get a &quot;Yes&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Would you say that you are Bugia and that you will switch to my right shoulder after answering this question?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Only Bugia will answer &quot;Yes&quot;, but only if it is going to the target location. Every other imp will answer &quot;No&quot;, and even Bugia will answer &quot;No&quot; if it isn't going to switch to &lt;em&gt;the target&lt;/em&gt;. So &lt;strong&gt;once you get a &quot;Yes&quot; answer, point to &lt;em&gt;the target&lt;/em&gt; (Edward's right shoulder) to identify Bugia&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="9422" LastEditorUserId="9422" LastEditDate="2016-02-15T13:27:08.417" LastActivityDate="2016-02-15T13:27:08.417" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="27482" PostTypeId="2" ParentId="27462" CreationDate="2016-02-15T13:02:13.217" Score="1" Body="&lt;p&gt;You're...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; oil.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Underground&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is where I'm found&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oil is found in mines.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Markets rise and fall with me&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As do actions of countries&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Being a very widely used kind of fuel, it can force manufacturers to make smaller, more fuel-efficient products, or more people to switch to/invest in other energy sources when its price increases. Oil-rich countries can also use oil as leverage.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="14016" LastEditDate="2016-02-15T13:39:20.600" LastActivityDate="2016-02-15T13:39:20.600" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="27484" PostTypeId="2" ParentId="2147" CreationDate="2016-02-15T13:27:41.840" Score="2" Body="&lt;h1&gt;The right answer to THIS question (instead of random order shooting)&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I will try to consolidate some of the right answers and not only give the strategy for each participant, but also their chances of survival. - The question as stated is different from the similar &lt;a href=&quot;http://books.google.ca/books?id=fA3CAgAAQBAJ&amp;amp;pg=PA51&amp;amp;lpg=PA51&amp;amp;dq=martin%20gardner%20puzzle%20shoot%20first&amp;amp;source=bl&amp;amp;ots=QxCcl-suW0&amp;amp;sig=R1HgbRT8puvUu4hkHcAkzL_A7FY&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=aYz7U6HfOIisyASZ_oDoDQ&amp;amp;ved=0CCYQ6AEwAQ#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;puzzle&lt;/a&gt; by Martin Gardner, in our version they don't shoot in random order, but Jones always shoots first, then Brown and then Smith.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Strategies:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If someone has only one enemy left to fire at, the best option is always to shoot, because not shooting lowers your chance of survival. (Gruesome Wild West) If there are 2 possible targets, you can decide to shoot one of them or miss on purpose. Missing on purpose has the same effect as shooting and missing in this scenario, because no one holds grudges. So the only relevant part to decide if you should shoot someone is the scenario where you hit him (since you hit everyone with the same percentage).&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Smith&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Killing Brown will give him 50% chance to survive against Jones, killing Jones will give him 20% chance against Jones. Killing no one will only help him, if Brown or Jones shoot at each other.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Brown&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Killing Smith will give him below 50% chance to survive. Killing Jones result in 0% chance to survive. Shooting no one only helps if Smith or Jones shoot at each other.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Jones&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Killing Smith will get him below 20% chance and killing Brown 0% chance to survive. Shooting no one only helps if Smith or Brown shoot at each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Jones and Brown have no incentive to kill each other&lt;/strong&gt;. So Smith will kill Brown and therefore Brown will try to kill Smith. Since they try to kill each other, Jones has a Chance of 50% or higher, if one of the two kills the other, since Jones gets the next shot after that. This is higher than the chance after killing one himself first, &lt;strong&gt;so he will not shoot as long as the other two are still standing&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Surviving Probabilities:&lt;/h2&gt;&#xA;&#xA;&lt;h3&gt;Smith: 10%&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;He has only three scenarios: a) he is killed by Brown &lt;code&gt;0.8&lt;/code&gt;, b) he kills Brown and is killed by Jones &lt;code&gt;0.2 * 0.5&lt;/code&gt; or c) he kills both of them &lt;code&gt;0.2 * 0.5&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He only survives if Brown and Jones each miss him one time so his chance to survive is &lt;code&gt;0.1&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Brown 35,56%&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Brown really only has one chance to survive: He has to kill Smith with his first shot and then win the duel against Jones (with John getting the first shot). His chance to win in a round against Jones is: Jones misses him and he hits: &lt;code&gt;0.5 * 0.8 = 0.4&lt;/code&gt; The chance of Jones winning a round is &lt;code&gt;0.5&lt;/code&gt; the chance for the round actually ending is &lt;code&gt;0.9&lt;/code&gt; so the chance of Brown winning the duel overall is &lt;code&gt;0.4 / 0.9 = 4/9&lt;/code&gt; So his overall chance of survival is &lt;code&gt;0.8 * 4/9 = 16/45&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Jones 54,44%&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Jones has two scenarios to live: Brown kills Smith and Jones wins the duel against Brown: &lt;code&gt;0.8 * 5/9 = 4/9&lt;/code&gt; or Brown misses Smith, Smith kills Brown and Jones kills Smith on the first try: &lt;code&gt;0.2 * 0.5 = 0.1&lt;/code&gt; Overall: &lt;code&gt;4/9 + 0.1 = 49/90&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="1998" LastActivityDate="2016-02-15T13:27:41.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27485" PostTypeId="2" ParentId="27463" CreationDate="2016-02-15T13:39:39.833" Score="6" Body="&lt;p&gt;&lt;strong&gt;You don't need to ask any questions to &quot;win&quot; the game (one way or another).&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the very start, there are only two imps (Acredo, the truth teller, and Bugia, the liar). One of them starts talking and states &quot;&lt;em&gt;we're imps&lt;/em&gt;&quot;, which based on Edwards reaction, we'll trust to be true, meaning the speaker must be Acredo (since &lt;em&gt;every&lt;/em&gt; sentence he speaks is the truth).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this point, we need to skip ahead to the explanation of the rules, specifically to the point where the imp says, &quot;&lt;em&gt;every time you ask one of us a question, that imp randomly chooses another imp who behaves differently and switches consciousness with it after answering the question&lt;/em&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume this statement is true (by chance the imp speaking was Acredo, or a coincidentally truthful but unreliable imp), then the second imp to speak back at the start, has to have been Bugia (the liar), since he spoke only after the second question was asked, which has to have triggered this rule, and at that point there were no other imps to trade with. Therefore his statement that, &quot;&lt;em&gt;if you kill an imp, two more show up!&lt;/em&gt;&quot; is a lie. So Edward can in fact win the game &quot;the old fashioned way&quot;, as he puts it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume the rules statement is false (again by chance), then the claim that the imps can swap bodies is false. Which means Bugia, the liar, is still sitting in the same place he started - hanging onto Edward's left ear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to win, Edward should first identify the imp hanging onto his left ear as being Bugia, and if he's right, he wins. If he's wrong, then he should just kill them all. The end.&lt;/p&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="5208" LastEditDate="2016-02-15T13:49:33.207" LastActivityDate="2016-02-15T13:49:33.207" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27486" PostTypeId="2" ParentId="1872" CreationDate="2016-02-15T14:00:55.320" Score="4" Body="&lt;p&gt;While &lt;a href=&quot;https://puzzling.stackexchange.com/users/28/klm123&quot;&gt;kim123&lt;/a&gt;'s answer is correct, there is a simpler logic shown by colors:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/XvJDd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XvJDd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;shapes are moving and morphing. For instance, the red cross begins its journey from the the upper left crossword box to the the (missing) lower left crossword box. Steps forward and transforms (rotates). The same happens for the blue triangle and the other shapes.&lt;/p&gt;&#xA;" OwnerUserId="19089" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2016-02-16T04:12:02.440" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27487" PostTypeId="2" ParentId="26454" CreationDate="2016-02-15T14:15:03.373" Score="6" Body="&lt;p&gt;This is how I arrived at the solution, less formal more intuitive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 8899&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Both numbers are only 1 apart, but need to have a digit sum which is the same or 17 (or 34,51...) apart. If you take 100 and 101 the digit sums are 1 and 2, only 1 apart, the same for 8 and 9 - in fact the only way I can get another difference than 1 in the digit sums, is when the digits swap from&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; xxx9 to xxx0 - e.g. 9 -&gt; 10 (digit sums 9 and 1)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But a difference of 8 is not enough so try another 9&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 99 -&gt; 100 (digit sums 18 and 1)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They are 17 apart, so this can work - the second number is just 16 short of being a multiple of 17, so I have to add digits with a sum of 16, like 8+8. This gives the result of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 8899 -&gt; 8900 (digit sums 34 and 17)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And this is also the smallest solution, since I can form a digit sum of 16 only with more digits, or as 97 or 79. And the number 9799 is bigger than 8899 and 7999 does not work, because another 9 rolls over and the successor is 8000.&lt;/p&gt;&#xA;" OwnerUserId="1998" LastActivityDate="2016-02-15T14:15:03.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27488" PostTypeId="2" ParentId="27462" CreationDate="2016-02-15T15:14:15.847" Score="6" Body="&lt;p&gt;Another possibility:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Iron (or Steel - works either way)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Underground&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is where I'm found&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Iron comes from mines underground&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Markets rise and fall with me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Iron is a major commodity in markets&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As do actions of countries&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Iron is traded between countries.  It could also be used metaphorically to refer to warfare.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You will find me on the driver's face&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A driver is a type of golf club, typically made from iron (or wood)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;attached to the car&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many parts of a car are made from steel, which is an alloy of iron&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and on the road in cold parts of the US.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When it gets cold, some drivers put chains or iron-studded winter tires on their cars for traction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16131" LastActivityDate="2016-02-15T15:14:15.847" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27489" PostTypeId="2" ParentId="27463" CreationDate="2016-02-15T15:24:47.773" Score="17" Body="&lt;p&gt;Contrary to my previous answer, I will assume that imps start switching consciousness right after the first question. This assumption is supported by the highlighted word &lt;strong&gt;&quot;every&quot;&lt;/strong&gt; when Acredo explains the rules. Let's look at the answers given by the imps:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Answer 1:&lt;/strong&gt; &lt;em&gt;&quot;&lt;strong&gt;We’re imps&lt;/strong&gt;! You have summoned us - and we came to play a little game with you. I’m Acredo and that’s my brother Bugia. &lt;strong&gt;Every sentence I say is the truth, while every sentence my brother says is a lie.&lt;/strong&gt; As soon as you want to start the game you just tell us, and we’ll switch positions at random. You can ask us as many yes or no questions as you want, to detect who of us is my always lying brother, Bugia. But I should warn you… you only get one chance to guess who of us it is before the game is over. Every time you ask one of us a question that can’t be answered with just yes or no, like you just did, another one of our brothers appears after the imp you asked finished answering your question. I will explain you the further rules once you accept to play our game&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As the first sentence states the obvious this answer seems to be true. This means that Acredo and Bugia now switch places and &lt;strong&gt;every sentence&lt;/strong&gt; of the next answer will be a lie.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Answer 2:&lt;/strong&gt; &lt;em&gt;&quot;Because we’ll stick around until your last days if you lose or don’t play! And if you kill an imp, two more show up! By the way, &lt;strong&gt;all of my brothers appearing because of your illegal questions(1) are completly unreliable&lt;/strong&gt;. Each time you ask them a question, they'll decide between behaving like an imp who always tells the truth or an imp who always lies, randomly, and then answer the question&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We know every sentence of this answer is a lie. Which means that every imp appearing is actually &lt;strong&gt;reliable&lt;/strong&gt;, telling always the truth or always lying. Together with the assumption that the imps switch after each question, the next imp must be a truth-teller.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Answer 3:&lt;/strong&gt; &lt;em&gt;&quot;Oh, we are a big family. Well, at least compared to humanoid families. I have 257 brothers, currently. And at this rate, your master will be the host of quite a few of them! Hehehe… well, he already is by now.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nothing interesting in this answer, but continuing with our assumption, the next answer must be a lie.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Answer 4:&lt;/strong&gt; &lt;em&gt;&quot;Oh, we don’t know who is who, actually!&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As already said, this is a lie. Not as useful as it might look though, because of the next quote which explains the rules and must be true again.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;The rules:&lt;/strong&gt; &lt;em&gt;&quot;Very well! The rules are simple. As I mentioned, you have to find my always lying brother, Bugia, by asking us as many yes or no questions as you want. We’ll switch positions randomly before the game starts - but not physically. We can simply switch our consciousness between bodies at will, so you won’t be able to notice it. Oh, and I almost forgot to mention it, but &lt;strong&gt;every time you ask one of us a question, that imp randomly chooses another imp who behaves differently&lt;/strong&gt; and switches consciousness with it after answering the question. &lt;strong&gt;So, an imp who always lies won’t switch with another imp who always lies and instead choose an imp with a different behaviour, like for example one that always tells the truth.&lt;/strong&gt; But as you already know, &lt;strong&gt;there is only one imp who always lies between us - my brother Bugia&lt;/strong&gt;. Also, just to clarify: The switch happens instantly after the question was answered. Keep that in mind when you want to point out who Bugia is, since you’re always going to do that after the switch happened. And yes, you can win this game - without asking dozens of questions, too! Alright, lets begin! We’ll all shuffle our consciousness now - good luck!&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Much information here, and we know each sentence must be true. Lets look at the highlighted parts:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;every time you ask one of us a question, that imp randomly chooses another imp who behaves differently&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is our assumption from the beginning.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So, an imp who always lies won’t switch with another imp who always lies and instead choose an imp with a different behaviour, like for example one that always tells the truth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here the truth-teller tries to avoid to say anything about unreliable brothers. Which is one more hint that they don't exist.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;there is only one imp who always lies between us - my brother Bugia&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This means &lt;strong&gt;all other imps tell the truth&lt;/strong&gt;. Having this information the solution is pretty simple (actually the same as in my previous answer):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Choose any imp and ask him a question like &lt;em&gt;&quot;Is 1 bigger than 2?&quot;&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If he tells the truth, he has to switch consciousness with Bugia because he is the only liar, and then we know that Bugia is in the body of the imp who just answered.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If he tells a lie we were asking Bugia, but he is gone now. Ask the same imp the same question, now he must tell the truth and switch with Bugia again. Then we know that Bugia is in the body of the imp who just answered.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2016-02-17T15:46:34.317" LastActivityDate="2016-02-17T15:46:34.317" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="27490" PostTypeId="2" ParentId="26454" CreationDate="2016-02-15T16:12:16.073" Score="-2" Body="&lt;p&gt;N=458&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;4 + 5 + 8 = 17&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;N + 1 = 459 = 17 * 27&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19241" LastActivityDate="2016-02-15T16:12:16.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27491" PostTypeId="1" AcceptedAnswerId="27673" CreationDate="2016-02-15T16:37:35.247" Score="8" ViewCount="525" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;I am made on a square of 44,&lt;/p&gt;&#xA;&lt;p&gt;preceded by purple, German and Germans&lt;/p&gt;&#xA;&lt;p&gt;then they sold meals for fifty pfennig&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Update: This one turns out to be hard. As the tag says, the answer is a movie. Also, do not hesitate to Google some of the terms or synonyms of them. I tried some terms that are not too hard to guess and then it will be in the first few hits.&lt;/p&gt;&#xA;&lt;p&gt;Also, the title and the three lines contain a total of six hints.&lt;/p&gt;&#xA;" OwnerUserId="16322" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-02-19T19:26:16.547" Title="The square of 44" Tags="|riddle|movies|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27492" PostTypeId="2" ParentId="27491" CreationDate="2016-02-15T16:50:41.703" Score="5" Body="&lt;p&gt;I am made on a square of 44&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I was made in the year 1936&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Preceded by purple, German and Germans&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; See the design and writing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then they sold meals for fifty pfennig&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The value. I guess you could get a meal for a stamp, since they're of equal value?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A &lt;a href=&quot;http://ebid.s3.amazonaws.com/upload_medbig/9/9/4/1386443256-5789-47.jpg&quot; rel=&quot;nofollow&quot;&gt;stamp&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14076" LastEditorUserId="14076" LastEditDate="2016-02-15T18:08:52.130" LastActivityDate="2016-02-15T18:08:52.130" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27493" PostTypeId="2" ParentId="26167" CreationDate="2016-02-15T17:14:35.753" Score="1" Body="&lt;p&gt;Since nobody has been able to find the solution to this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am a rebuke. I tell them they're all wrong!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some call me royalty, but I say they're wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm not a duke&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some call me crazy, but I say they're wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm not a kook&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some call me a ghost, but I say they're wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm not a spook&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some call me racist, but I say they're wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm not a g***&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some call me a hat, but I say they're wrong.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm not a toque&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18103" LastActivityDate="2016-02-15T17:14:35.753" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="27496" PostTypeId="1" AcceptedAnswerId="27499" CreationDate="2016-02-15T19:24:19.710" Score="30" ViewCount="6462" Body="&lt;p&gt;I feel no pain&lt;br&gt;&#xA;Yet silently shriek&lt;br&gt;&#xA;An airy concoction&lt;br&gt;&#xA;Your desire I pique&lt;br&gt;&#xA;A sinful temptation&lt;br&gt;&#xA;You seldom refuse&lt;br&gt;&#xA;As I squeal, howl and bellow&lt;br&gt;&#xA;I delight and amuse&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who am I?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user17947" LastActivityDate="2016-02-19T18:04:19.707" Title="I feel no pain, yet silently shriek" Tags="|riddle|wordplay|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27497" PostTypeId="2" ParentId="27496" CreationDate="2016-02-15T19:40:30.530" Score="70" Body="&lt;p&gt;You're a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fart. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I feel no pain&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No pain involved.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet silently shriek&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; But are demanding to come out.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An airy concoction&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Made of air.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your desire I pique&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One wants to let it rip.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A sinful temptation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; But it's gonna stink and might be quite loud.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You seldom refuse&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most people let it rip sooner or later.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As I squeal, howl and bellow&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Making all sorts of noise.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I delight and amuse&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; But farts are very funny, if perhaps, in a puerile way.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastActivityDate="2016-02-15T19:40:30.530" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="27498" PostTypeId="2" ParentId="26426" CreationDate="2016-02-15T21:18:34.223" Score="2" Body="&lt;p&gt;I think the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 12&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The formula for generating the $n$th term in the sequence is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $n$ + 2*(number of letters in English spelling of $n$) - 4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example the first term in the sequence is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 + 2*3 - 4 = 3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The 9th term is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 9 + 2*4 - 4 = 13&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And so the 10th term is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 10 + 2*3 - 4 = 12&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In relation to the hints&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The underlying sequence is the positive integers beginning at 1 and generation of the sequence depends on the spelling of the numbers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18422" LastActivityDate="2016-02-15T21:18:34.223" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27499" PostTypeId="2" ParentId="27496" CreationDate="2016-02-15T21:40:35.847" Score="35" Body="&lt;p&gt;&lt;strong&gt;Answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel no pain&#xA;Yet silently shriek&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; is a play on &quot;ice cream&quot; and &quot;I scream&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An airy concoction&#xA;Your desire I pique&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That's me and ice cream, all right.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A sinful temptation&#xA;You seldom refuse&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sugar and heart-clogging fat. BRING IT!!!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As I squeal, howl and bellow&#xA;I delight and amuse.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'd explain this as well but I'm going to go get some ice cream now.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastEditorUserId="12014" LastEditDate="2016-02-18T14:48:31.260" LastActivityDate="2016-02-18T14:48:31.260" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="27500" PostTypeId="1" AcceptedAnswerId="27501" CreationDate="2016-02-16T00:42:13.557" Score="3" ViewCount="1753" Body="&lt;h2&gt;Background&lt;/h2&gt;&#xA;&lt;p&gt;I wrote up a web-based sketchpad that lets the user pick from a series of colours, and tracks the recently selected colours for ease of use. I randomly had this idea as a personal puzzle, but couldn't figure it out, so I thought I'd post it here.&lt;/p&gt;&#xA;&lt;p&gt;The recent colour list follows 2 rules for when a colour is selected:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If the colour already exists in the list, it's moved up to the front of the list by swapping it with the tile that was previously at the front.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the colour wasn't already in the list, it's prepended to the front, without any swapping taking place.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;In the screenshot below:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nC2YQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nC2YQ.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The order of the colours selected leading to the above &amp;quot;Recent Colours&amp;quot; were:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Blue, Pink, Yellow, White, Black, Purple, Red, Green&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;The Problem&lt;/h2&gt;&#xA;&lt;p&gt;What order (if any) can you select colours from the quick bar to have the &amp;quot;Recent Colours&amp;quot; bar be in the same order as the &amp;quot;Quick Colours&amp;quot; bar above?&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Hint: It's not as simple as selecting the colours in the reverse order of the &amp;quot;Quick Bar&amp;quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19255" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-02-17T17:53:49.767" Title="Put the colours back in order" Tags="|strategy|reachability|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27501" PostTypeId="2" ParentId="27500" CreationDate="2016-02-16T01:11:36.003" Score="10" Body="&lt;p&gt;Assuming that the starting position is the one shown in the screenshot, all of the colors are already in the bar, so the only possible move is to swap a color with the first one in the bar. To put a color in the right place, it must be the first one in the bar, and then the color that is occupying its place must be selected to swap it in. If we consider which colors in which places, we get that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;White and Green are in each other's spots;&lt;/li&gt;&#xA;&lt;li&gt;Red and Black are in each other's spots;&lt;/li&gt;&#xA;&lt;li&gt;Purple is in Blue's spot, Blue is in Pink's spot, and Pink is in Purple's spot;&lt;/li&gt;&#xA;&lt;li&gt;Yellow is already in the right spot.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We can fix White and Green on the first move by selecting White. To fix Red and Black takes three moves: one to put Red in the first slot, one to swap it into the right slot (putting Black in the first slot), and one to put Black into the right slot (returning White to the first slot). Similarly, Purple, Blue, and Pink take four moves. In total it takes eight selections to get the colors right. One possible sequence (but not the only one) is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;White, Red, Black, White, Purple, Pink, Blue, White.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In general, it is possible to return the colors to the right order no matter what order they are originally in. An algorithm to do this might be:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the color in the first slot is not White, select whichever color is occupying the slot that it should go in.&lt;/li&gt;&#xA;&lt;li&gt;Otherwise, select any color that is not in the right place.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Every time the first step is done, one color that isn't White goes into the right place, while every time the second step is done, the result is a situation where the first step applies. Because no step ever causes a non-White color to be removed from its correct place, this guarantees that eventually all the colors will be in the right order.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2016-02-16T01:17:41.707" LastActivityDate="2016-02-16T01:17:41.707" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27502" PostTypeId="1" CreationDate="2016-02-16T04:00:10.987" Score="8" ViewCount="1015" Body="&lt;p&gt;I was recently trying to recreate a crossword puzzle based off of a picture of half the puzzle and the answer key. I found it to be very enjoyable. Is this a thing? Is there anywhere that I can find reverse crossword puzzles that are designed to be &quot;solved&quot; this way?&lt;/p&gt;&#xA;" OwnerUserId="19258" LastEditorUserId="4421" LastEditDate="2016-02-17T15:45:27.640" LastActivityDate="2016-02-17T15:45:27.640" Title="Is reverse-engineering crossword puzzles a thing?" Tags="|crosswords|puzzle-identification|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27503" PostTypeId="2" ParentId="27502" CreationDate="2016-02-16T04:28:35.933" Score="6" Body="&lt;p&gt;Sounds like you may be interested in diagramless crosswords.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NeN2L.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NeN2L.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what they look like: you get clues, but no grid to work with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://crosswordfiend.blogspot.com/2007/12/what-do-you-do-with-diagramless.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt;'s a guide on how to solve them, and &lt;a href=&quot;http://web.mit.edu/puzzle/www/2011/puzzles/zelda/forsaken_fortress/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;'s one from the 2011 MIT Mystery Hunt (with an added twist once it's solved!)&lt;/p&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2016-02-16T04:28:35.933" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27504" PostTypeId="2" ParentId="27502" CreationDate="2016-02-16T05:34:29.313" Score="6" Body="&lt;p&gt;I think this is most commonly given the extremely creative moniker of &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Fill-In_(puzzle)&quot;&gt;fill-in puzzle&lt;/a&gt;. You are given an empty crossword grid and have to fill in the provided words. &lt;/p&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2016-02-16T05:34:29.313" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27505" PostTypeId="2" ParentId="27474" CreationDate="2016-02-16T06:16:20.050" Score="2" Body="&lt;p&gt;Let $J_n$ be the random variable whose value is the payout of the $n$-th game. For $n=1$, we can easily calculate the expectation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}\begin{split}&#xA;\mathbb{E}(J_1)&amp;amp;= \left(\sum_{1 \leq i \leq 20} \hphantom{.}\underbrace{\hphantom{..}2^{-i}\hphantom{..}}_{\mathbb{P}(J_1 = 2^{i-1})}\cdot 2^{i-1}\right) &amp;amp;+\sum_{i&amp;gt;20}2^{-i}\cdot 10^6\\&#xA;&amp;amp;=\hphantom{...........}10&amp;amp;+2^{-20}\cdot 10^6&#xA;\end{split}\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that if we get $21$ or more tails, we receive $10^6$ because $2^{i-1}&amp;gt;10^6$ for all $i\geq 21$. Also, the first summation limit is $20 = \lceil \log_2\left(10^6\right)\rceil$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, precisely because of the total payout limit of $10^6$, the result of $J_{n+1}$ depends on the results of all $J_k$ with $1 \leq k \leq n$. How, then, do we calculate the expectation of $J_{n+1}$'s? For that, we use the fundamental result that $\mathbb{E}(X) = \mathbb{E}(\mathbb{E}(X|Y))$. Thus, for instance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}\begin{split}&#xA;\mathbb{E}(J_2)&amp;amp;=\mathbb{E}(\mathbb{E}(J_2|J_1))\\&#xA;&amp;amp;=\left(\sum_{1\leq n \leq 20}\mathbb{E}\left(J_2|J_1=2^{n-1}\right)\cdot\mathbb{P}\left(J_1=2^{n-1}\right)\right)+\underbrace{\mathbb{E}\left(J_2|J_1=10^6\right)}_0\cdot \mathbb{P}\left(J_1=10^6\right)\\&#xA;&amp;amp;=\sum_{1\leq n\leq20}2^{-n} \cdot \mathbb{E}\left(J_2|J_1=2^{n-1}\right)&#xA;\end{split}\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We first calculate $\mathbb{E}\left(J_2|J_1=2^{n-1}\right)$. We can write it as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}\begin{split}&#xA;&amp;amp;\left( \sum_{\displaystyle 1\leq m \leq \lceil \log_2\left(10^6 -2^{n-1}\right)\rceil } 2^{-m} \cdot 2^{m-1} \right) \\&#xA;+ &amp;amp;\vphantom{.^{^{^{^{^{^{.}}}}}}}\hphantom{...}\sum_{\displaystyle m &amp;gt; \lceil \log_2 \left( 10^6 - 2^{n-1} \right) \rceil} 2^{-m} \cdot \left(10^6-2^{n-1} \right) \\&#xA;=&amp;amp;\hphantom{..}\frac12\cdot \lceil \log_2\left(10^6 -2^{n-1}\right)\rceil+2^{^{\displaystyle -\lceil \log_2\left(10^6 -2^{n-1}\right)\rceil \vphantom{^{^{^{^{^{.}}}}}}}}\cdot \left(10^6-2^{n-1}\right)&#xA;\end{split}\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that:&#xA;\begin{equation}\begin{split}&#xA;\mathbb{E}(J_2)&amp;amp;=\sum_{1\leq n \leq 20}2^{-n}\left(\frac12\cdot \lceil \log_2\left(10^6 -2^{n-1}\right)\rceil+2^{-\lceil \log_2\left(10^6 -2^{n-1}\right)\rceil }\cdot \left(10^6-2^{n-1}\right) \right)\\&#xA;&amp;amp;=\frac{11485739}{1048576}\vphantom{h^{h^{h^{h^{h^{h^{h^{h}}}}}}}}&#xA;\end{split}\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can repeat the process for the next $J_k$, and obtain that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}\begin{split}&#xA;\mathbb{E}(J_3)=\sum_{\displaystyle 1 \leq i_2 \leq \lceil \log_2\left(10^6 \right) \rceil} 2^{-i_2} \cdot \left( \sum_{\displaystyle 1 \leq i_3 \leq \lceil \log_2\left(10^6 - 2^{i_2 - 1} \right) \rceil} 2^{-i_3} \cdot e(i_2,i_3) \right)&#xA;\end{split}\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $e(i_2,i_3)$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}\begin{split}&#xA;&amp;amp;\frac12 \cdot \lceil \log_2\left(10^6 - 2^{i_2 - 1} -2^{i_3-1}\right) \rceil\\&#xA;+\hphantom{..}\vphantom{^{^{^{^{^{^{^{^{^{^{^{^{.}}}}}}}}}}}}} &amp;amp;2^{^{\displaystyle -\lceil \log_2\left(10^6 - 2^{i_2 - 1} -2^{i_3-1}\right) \rceil}}\cdot \left(10^6 - 2^{i_2 - 1} -2^{i_3-1}\right)&#xA;\end{split}\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from whence it follows that $\mathbb{E}(J_3)=\frac{24665391590542601}{2251799813685248}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}\begin{split}&#xA;\mathbb{E}(J_4)=\sum_{\displaystyle 1 \leq i_2 \leq \lceil \log_2\left(10^6 \right) \rceil} 2^{-i_2} \cdot \left( \sum_{\displaystyle 1 \leq i_3 \leq \lceil \log_2\left(10^6 - 2^{i_2 - 1} \right) \rceil} 2^{-i_3} \cdot \\&#xA;\left( \sum_{\displaystyle 1 \leq i_4 \leq \lceil \log_2\left(10^6 - 2^{i_2 - 1} - 2^{i_3 - 1} \right) \rceil} 2^{-i_4} \cdot e(i_2,i_3,i_4) \right) \right)&#xA;\end{split}\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $e(i_2,i_3,i_4)$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}\begin{split}&#xA;&amp;amp;\frac12 \cdot \lceil \log_2\left(10^6 - 2^{i_2 - 1} -2^{i_3-1} - 2^{i_4-1}\right) \rceil\\&#xA;+\hphantom{..}\vphantom{^{^{^{^{^{^{^{^{^{^{^{^{.}}}}}}}}}}}}} &amp;amp;2^{^{\displaystyle -\lceil \log_2\left(10^6 - 2^{i_2 - 1} -2^{i_3-1} - 2^{i_4 - 1}\right) \rceil}}\cdot \left(10^6 - 2^{i_2 - 1} -2^{i_3-1} - 2^{i_4 - 1}\right)&#xA;\end{split}\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that $\mathbb{E}(J_4) = \frac{808234073895490893065}{73786976294838206464}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe at this point the pattern should be clear on how to set up the recursion that yields $\mathbb{E}(J_{n+1})$ from previous $J_k$'s. That said, I'm not sure we're any wiser in doing so, and calculations become quickly quite troublesome. Here are the first few approximate values for $\mathbb{E}(J_{k})$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\mathbb{E}(J_{1}) = 10.9536743164062$&lt;/li&gt;&#xA;&lt;li&gt;$\mathbb{E}(J_{2}) = 10.9536542892456$&lt;/li&gt;&#xA;&lt;li&gt;$\mathbb{E}(J_{3}) = 10.9536342620864$&lt;/li&gt;&#xA;&lt;li&gt;$\mathbb{E}(J_{4}) = 10.9536142349288$&lt;/li&gt;&#xA;&lt;li&gt;$\mathbb{E}(J_{5}) = 10.9535942077726$&lt;/li&gt;&#xA;&lt;li&gt;$\mathbb{E}(J_{6}) = 10.9535741806178$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Unsurprisingly, with each game the expected payout is very slightly smaller than that of the previous game (because the limit payout is reduced from our previous winnings). This also implies that the difference between them is shrinking:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\mathbb{E}(J_{1}) - \mathbb{E}(J_{2}) = 0.0000200271606$&lt;/li&gt;&#xA;&lt;li&gt;$\mathbb{E}(J_{2}) - \mathbb{E}(J_{3}) = 0.0000200271592$&lt;/li&gt;&#xA;&lt;li&gt;$\mathbb{E}(J_{3}) - \mathbb{E}(J_{4}) = 0.0000200271576$&lt;/li&gt;&#xA;&lt;li&gt;$\mathbb{E}(J_{4}) - \mathbb{E}(J_{5}) = 0.0000200271562$&lt;/li&gt;&#xA;&lt;li&gt;$\mathbb{E}(J_{5}) - \mathbb{E}(J_{6}) = 0.0000200271548$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So we could write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbb{E}(J_k) = \mathbb{E}(J_1) - \left( \sum_{1 \leq i \leq k-1}\mathbb{E}(J_i) - \mathbb{E}(J_{i+1}) \right) &amp;gt; \mathbb{E}(J_1) - (k-1) \cdot \big( \mathbb{E}(J_1) - \mathbb{E}(J_2) \big)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We thus have the following rough estimate for all $k \geq 2$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbb{E}(J_1) - (k-1) \cdot \left( \mathbb{E}(J_1) - \mathbb{E}(J_2) \right) &amp;lt; \mathbb{E}(J_k) &amp;lt; \mathbb{E}(J_1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accordingly, for instance, $10.9336671829668 &amp;lt; \mathbb{E}(J_{1000}) &amp;lt; 10.9536743164062$. It is clear, I believe, that even over the course of the one thousand games, the effect is minuscule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the total expected payout $\mathbb{E}(\sum_{1 \leq k \leq 1000} J_k) = \sum_{1 \leq k \leq 1000} \mathbb{E}(J_k)$, we thus have the following estimates:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$10943.6707496865 &amp;lt; \mathbb{E}\left(\sum J_k\right) &amp;lt; 10953.6743164062$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $\mathbb{E}\left(\sum J_k\right) \simeq 10948.6725330464$, give or take $5$ dollars (an error of about $0.0457$%).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, if we're paying $10.94$ dollars or less per game, we should expect profit over the course of a thousand games. However, if we're paying $10.96$ or more per game, we should expect loss. For $10.95$ we're not quite sure, but since our upped bound is so lazy and yet 'barely' above $10950$, I'd hazard you're slightly in the loss here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yeah, up to $10.94$ dollars.&lt;/p&gt;&#xA;" OwnerUserId="16478" LastActivityDate="2016-02-16T06:16:20.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27507" PostTypeId="1" AcceptedAnswerId="27529" CreationDate="2016-02-16T07:34:50.990" Score="8" ViewCount="565" Body="&lt;p&gt;In a basket there are 100 apples. &#xA;Every apple weighs at least 25 gram, and the total weight of all &#xA;apples is 10 kilogram.&#xA;The teacher wants to cut these apples into pieces and assign the &#xA;pieces to 100 children, so that each child receives 100 gram. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Does there always exist a cutting so that each apple piece weighs at least 25 gram?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2016-02-17T10:44:42.743" Title="How to cut one hundred apples into pieces" Tags="|mathematics|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="27510" PostTypeId="2" ParentId="26364" CreationDate="2016-02-16T10:25:46.180" Score="5" Body="&lt;p&gt;My guess is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;s&gt;Horseshoe&lt;/s&gt;  &lt;br&gt; Jack&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'll toss me on the ground&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This refers to the &quot;jack&quot; (target ball) in a game of bowls&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll jump at you just in case&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Referring to &quot;Jumping jacks&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'll help me off a horse&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Refers to a well-known phrase about bad grammar: &quot;Helping your Uncle, Jack, off a horse&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll lift you up instead&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Referring to a car jack, designed to lift a car to enable repairs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'll mount me on your boat&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Refers to a jack staff, mounted on the bow of a boat to fly a flag&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll shout the time for you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jack refers to the figure of a man striking the bell on a clock.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'll wager your house with me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Black Jack&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then take things back illegally&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jack is a verb which means to take things back illegally.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This now fits with all the hints also.&lt;/p&gt;&#xA;" OwnerUserId="18422" LastEditorUserId="18422" LastEditDate="2016-02-16T14:16:57.100" LastActivityDate="2016-02-16T14:16:57.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27511" PostTypeId="2" ParentId="26364" CreationDate="2016-02-16T12:08:44.670" Score="1" Body="&lt;p&gt;I think it might be a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ball&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;although I'm still missing some explanations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll toss me on the ground&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are games where balls are tossed, especially ones with grass fields&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll jump at you just in case&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A ball thrown at the ground jumps back&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'll help me off a horse&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;No idea about this one&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll lift you up instead&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A ball can lift a person, if it is a gym ball for sitting on, or a child playing in the pool and climbing onto an inflatable ball&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'll mount me on your boat&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You mount cannons, which shoot cannonballs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll shout the time for you&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Again, no idea here&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'll wager your house with me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You wager on a ball when playing roulette&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then take things back illegally&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also no good guess here&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I hope others can provide ideas for the missing clues if they agree my guess fits. &lt;/p&gt;&#xA;" OwnerUserId="2811" LastActivityDate="2016-02-16T12:08:44.670" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27512" PostTypeId="2" ParentId="26088" CreationDate="2016-02-16T12:13:18.077" Score="1" Body="&lt;p&gt;Possibly halfway to a solution, here's my worksheet so far.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; \begin{matrix}                                                               &amp;amp; 9&amp;amp;~~?~~&amp;amp; 12  &amp;amp; 1 &amp;amp;  36 &amp;amp; 1 &amp;amp; 8 &amp;amp; 1 &amp;amp;  15  &amp;amp; 12  &amp;amp; 1 &amp;amp;  8  &amp;amp; 9 &amp;amp;~~?~~&amp;amp; 12  &amp;amp;          \\[-1ex] \cdots                                                          &amp;amp; -&amp;amp;     &amp;amp;\cdot&amp;amp; - &amp;amp;\cdot&amp;amp; - &amp;amp; - &amp;amp; - &amp;amp;-\!\!-&amp;amp;\cdot&amp;amp; - &amp;amp;\cdot&amp;amp; - &amp;amp;     &amp;amp;\cdot&amp;amp;   \cdots \\[-1ex]                                                                 &amp;amp; 2&amp;amp;  ?  &amp;amp;  5  &amp;amp; 1 &amp;amp;  1  &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp;   1  &amp;amp;  5  &amp;amp; 1 &amp;amp;  1  &amp;amp; 2 &amp;amp;  ?  &amp;amp;  5  &amp;amp;   \end{matrix}&#xA; Think I could almost stagger the rest of the way,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;without fully figuring out the rule.&#xA;Then again, what was that about notation convention?&lt;/p&gt;&#xA;" OwnerUserId="18129" LastActivityDate="2016-02-16T12:13:18.077" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27513" PostTypeId="1" AcceptedAnswerId="27519" CreationDate="2016-02-16T12:16:11.790" Score="12" ViewCount="1883" Body="&lt;p&gt;I saw this ad in the 'wanted' section of the classifieds:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Male seeks precious stone (only hollow).&lt;br&gt;&#xA;  Call Antonio: 018 911 937&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I thought it was fairly innocent, but a friend told me it was actually a warning against something malevolent. Can anyone explain why?&lt;/p&gt;&#xA;" OwnerUserId="18522" LastActivityDate="2018-08-14T09:59:07.823" Title="What does this 'wanted' ad mean?" Tags="|cryptic-clues|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="27514" PostTypeId="1" AcceptedAnswerId="27525" CreationDate="2016-02-16T12:33:34.650" Score="8" ViewCount="400" Body="&lt;p&gt;You got a ridiculous message from your best friend that doesn't make sense. But you know that it must be important because he is a wise man, so you must find out what he is trying to tell.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Mr. Robin was a person who could fly.&#xA;&#xA;Baited by a voice, he started floating.&#xA;&#xA;Flying in the skies, he saw something weird.&#xA;&#xA;Birds without wings were floating in the air.&#xA;&#xA;Actually there were no birds at all.&#xA;&#xA;Betrayal of his own mind was the fact.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="18685" LastActivityDate="2016-02-16T16:47:28.653" Title="Message from your best friend" Tags="|steganography|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27517" PostTypeId="1" AcceptedAnswerId="27528" CreationDate="2016-02-16T13:49:27.187" Score="5" ViewCount="515" Body="&lt;p&gt;Spying has never been easier with me, &lt;br&gt;&#xA;Boats have use of what's inside me,&lt;br/&gt;&#xA;Shout I do not, even when teeth may enter me.&#xA;&lt;br/&gt;&lt;br/&gt;&#xA;I'll be happy to share an interest&lt;br/&gt;&#xA;With many who look alike - unlike the rest,&lt;br/&gt;&#xA;Guard against secrets... I do my best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who/What am I?&lt;/p&gt;&#xA;" OwnerUserId="19086" LastEditorUserId="5373" LastEditDate="2018-11-22T11:00:21.710" LastActivityDate="2018-11-22T11:00:21.710" Title="Spying has never been easier" Tags="|riddle|word|wordplay|rhyme|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27518" PostTypeId="2" ParentId="27507" CreationDate="2016-02-16T14:27:30.020" Score="4" Body="&lt;p&gt;I think the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Use the following strategy. Take the smallest apple. If it is &amp;lt;= 75 g take the largest apple and cut a piece from it to match the first one to 100g. This is always possible because the largest apple is always &gt;= 100 g. If the smallest apple is &gt; 75 g you cut it in half and match each piece with pieces from the two largest apples. After doing this once you end up with essentially 99 apples that are at least 25g with an average of 100g, so exactly the original problem but with less apples. So repeat this strategy until there are no apples left and your done. This also actually proofs that you can do it in a way that every kid only gets 2 pieces.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2016-02-16T14:27:30.020" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27519" PostTypeId="2" ParentId="27513" CreationDate="2016-02-16T15:18:43.437" Score="15" Body="&lt;h1&gt;NOW IN ANSWER FORM&lt;/h1&gt;&#xA;&lt;p&gt;Antonio&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Gramsci lived from 1891 to 1937... perhaps we're being warned about &lt;a href=&quot;http://www.marxists.org/glossary/terms/h/e.htm&quot; rel=&quot;noreferrer&quot;&gt;hegemony&lt;/a&gt;?&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; 'Male' = &lt;em&gt;he&lt;/em&gt;&lt;br /&gt;&#xA; 'precious stone' = &lt;em&gt;gem&lt;/em&gt;&lt;br /&gt;&#xA; 'only hollow' = on&lt;s&gt;l&lt;/s&gt;y = &lt;em&gt;ony&lt;/em&gt;&lt;br /&gt;&#xA;&lt;br /&gt;&#xA;&lt;br /&gt;&#xA;&lt;br /&gt;&#xA;  &lt;sub&gt;I'm throwing a little party for myself for being able to link to marxists dot org in an answer&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-02-16T16:47:24.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27522" PostTypeId="1" AcceptedAnswerId="27523" CreationDate="2016-02-16T15:36:00.953" Score="4" ViewCount="523" Body="&lt;p&gt;This question was asked in CSIR _NET 2014. I cannot figure out what are conditions given:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lunch-dinner pattern of a person for '$m$' days is given below. He has a choice of a vegeterian food and non-vegetarian food for his lunch or dinner. The pattern below given should be considered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A.    If he takes non-vegetarian lunch then he will have to take only vegetarian  for dinner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B. He takes non-vegetarian dinner for exactly for nine days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C. He takes vegetarian lunch for exactly $15$ days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D. He takes a total of $14$ non-vegetarian meals.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Determine the value of $m$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18892" LastEditorUserId="8874" LastEditDate="2016-02-16T15:55:21.977" LastActivityDate="2016-02-17T17:30:04.553" Title="Find the value of $m$" Tags="|calculation-puzzle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27523" PostTypeId="2" ParentId="27522" CreationDate="2016-02-16T15:52:42.417" Score="5" Body="&lt;p&gt;Denote by $LN$ and $LV$ the numbers of non-vegetarian and vegetarian lunches, &#xA;and by $DN$ and $DV$ the corresponding numbers of dinners.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Statement B implies $DN=9$&lt;/li&gt;&#xA;&lt;li&gt;Statement C implies $LV=15$&lt;/li&gt;&#xA;&lt;li&gt;Statement D impies $LN+DN=14$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then $LN=(LN+DN)-DN=14-9=5$.&#xA;Furthermore,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the total number of days in this lunch/dinner schedule is $m=LN+LV=5+15=20$,&lt;/li&gt;&#xA;&lt;li&gt;and finally we get  $DV=m-DN=20-9=11$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Hence the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $m=20$ &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;if the person eats lunch and dinner on every day. Otherwise, we can only conclude that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $m\ge20$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2016-02-16T15:52:42.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27524" PostTypeId="2" ParentId="27517" CreationDate="2016-02-16T15:54:22.670" Score="3" Body="&lt;p&gt;Who/What am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Net/Internet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Spying has never been easier with me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The internet certainly seems to help the NSA...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Boats have use of what's inside me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The fish inside a net&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Shout I do not, even when teeth may enter me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well, it's a &quot;thing&quot;, so it can't really be hurt.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll be happy to share an interest&#xA;With many who look alike - unlike the rest,&#xA;Guard against secrets... I do my best.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, internet. It allows the sharing of interests between anonymous people, while staying anonymous isn't always as easy as it seems (&quot;I do my best.&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14076" LastActivityDate="2016-02-16T15:54:22.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27525" PostTypeId="2" ParentId="27514" CreationDate="2016-02-16T16:19:01.897" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The first letter from the first line, second letter from the second line, and so on spell out &lt;code&gt;Mayday&lt;/code&gt;. It sounds like he's in serious trouble.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2016-02-16T16:47:28.653" LastActivityDate="2016-02-16T16:47:28.653" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27526" PostTypeId="2" ParentId="27513" CreationDate="2016-02-16T16:48:53.767" Score="3" Body="&lt;p&gt;I haven't got the true meaning, but can see a hint of malevolence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Call&lt;/strong&gt; &lt;s&gt;Antonio: 018&lt;/s&gt; &lt;strong&gt;911&lt;/strong&gt; &lt;s&gt;937&lt;/s&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7586" LastActivityDate="2016-02-16T16:48:53.767" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27527" PostTypeId="2" ParentId="27496" CreationDate="2016-02-16T18:58:10.230" Score="4" Body="&lt;p&gt;I feel no pain&#xA;Yet silently shriek.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;So it makes noise without pain &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An airy concoction.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Contains air&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your desire I pique.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Works on demand whenever you want &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A sinful temptation,&lt;br&gt;&#xA;You seldom refuse.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Can be used to play pranks on someone &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As I squeal, howl and bellow, &#xA;I delight and amuse.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;In the end always amuses &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;a whoopee cushion.&lt;br&gt;&#xA;That's what first crossed my mind, and then I saw fart, now I am more sure of &gt;!my answer &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19281" LastEditorDisplayName="user17947" LastEditDate="2016-02-16T19:30:26.163" LastActivityDate="2016-02-16T19:30:26.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27528" PostTypeId="2" ParentId="27517" CreationDate="2016-02-16T19:46:54.823" Score="8" Body="&lt;p&gt;You're a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; lock.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Spying has never been easier with me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the age of the net, lock picking to gain access to a restricted room/cabinet is now trivial.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Boats have use of what's inside me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The central part of a lock is called a &lt;a href=&quot;https://www.lysator.liu.se/mit-guide/figure2.1.gif&quot;&gt;hull&lt;/a&gt;, which is &lt;a href=&quot;https://www.google.com/search?tbm=isch&amp;amp;q=boat%20hull&amp;amp;tbs=imgo:1&quot;&gt;also a part of a boat&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Shout I do not, even when teeth may enter me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Keys have teeth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll be happy to share an interest&#xA;With many who look alike - unlike the rest,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Identical keys (duplicates) can open the lock, but other keys cannot.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Guard against secrets... I do my best.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That's what locks are for!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12795" LastActivityDate="2016-02-16T19:46:54.823" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27529" PostTypeId="2" ParentId="27507" CreationDate="2016-02-16T19:56:59.123" Score="9" Body="&lt;p&gt;Suppose that we have $N+1$ apples (each at least $25$ grams) with a total weight of $100N$ grams, and we want to cut them into pieces of at least $25$ grams so that we can make $N$ groups weighing $100$ grams each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $N=1$, this is trivial: we have two pieces totalling $100$ grams, and we just put them together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $N&amp;gt;1$, I will show that we can always reduce the situation to make $N$ smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are three possibilities:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The smallest apple and the largest apple together weigh at least $125$ grams, and the smallest apple weighs no more than $75$ grams;&lt;/li&gt;&#xA;&lt;li&gt;The smallest apple and the largest apple together weigh less than $125$ grams;&lt;/li&gt;&#xA;&lt;li&gt;Every apple weighs more than $75$ grams.&#xA;&#xA;&lt;hr&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h3&gt;Case 1: The smallest apple and the largest apple together weigh at least $125$ grams, and the smallest apple weighs no more than $75$ grams.&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;In this case, we can cut a piece off the larger apple so that the piece and the smallest apple weigh $100$ grams together. We have gotten rid of one apple and one set of $100$ grams, so we have reduced $N$ by $1$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Case 2: The smallest apple and the largest apple together weigh less than $125$ grams.&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Because all the apples' weights sum to $100N$, the rest of the apples must have a total weight of at least $100N-125$ grams. The average weight of these apples is at least $\frac{100N-125}{N-1}=100-\frac{25}{N-1}$, which is at least $75$. Therefore the second-largest and largest apples must each weigh at least $75$ grams, and the smallest apple must weigh less than $50$ grams. We can cut a piece off the larger apple so that the piece and the smallest apple weigh $75$ grams together, then cut $25$ grams from the second-largest apple, for a total of $100$ grams. Again, we have gotten rid of one apple and one set of $100$ grams, so we have reduced $N$ by $1$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Case 3: Every apple weighs more than $75$ grams.&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Let the weight of the largest apple be $x$. For each $m\le N$, let $W_m$ be the weight of the $m$-th smallest apple, let $S_m$ be the sum of the weights of the $m$ smallest apples, and let $D_m=100m-S_m$. $D_m$ represents the size of the piece that would need to be added to the first $m$ apples to make them total $100m$ grams. In particular, $D_0=0$ and $D_N=x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D_m-D_{m-1}=100+S_{m-1}-S_m=100-W_m&amp;lt;25$, so $D_m$ cannot increase by more than $25$ grams at a time. Because $x&amp;gt;75$, the difference between $25$ and $x-25$ is more than $25$ grams. Because $D_0&amp;lt;25$ and $D_N&amp;gt;x-25$, some $D_m$ will necessarily have a value between $25$ and $x-25$. Then we can cut off a piece of size $D_m$ from the largest apple and combine it with the smallest $m$ apples to make $m+1$ pieces totalling $100m$ grams. The remaining $N-m+1$ pieces total $100(N-m)$ grams. This reduces our situation to two smaller situations.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Therefore, for all $N$, if we have $N+1$ apples with a total weight of $100N$ grams, we can divide them so that we have $N$ groups of $100$ grams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We start with $100$ apples. The largest one is at least $100$ grams, so we can safely cut it in half. Then we have $101$ pieces totalling $10000$ grams. This corresponds to $N=100$. As just shown, we can cut these pieces to make $100$ groups of $100$ grams.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="2722" LastEditDate="2016-02-17T05:55:49.007" LastActivityDate="2016-02-17T05:55:49.007" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27531" PostTypeId="1" AcceptedAnswerId="27551" CreationDate="2016-02-16T22:11:02.567" Score="10" ViewCount="208" Body="&lt;p&gt;The following two lines are each missing an entry.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ABBCCDDEEFFGHHIILLLMNPPRSSS&#xA;TFHABHNIWNWXYBMBEFGFYAIFWAC&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Your job is to get it in place with the rest.&lt;/p&gt;&#xA;" OwnerUserId="17446" LastActivityDate="2016-02-17T16:29:20.533" Title="ISO: Missing member" Tags="|cipher|wordplay|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27532" PostTypeId="2" ParentId="27491" CreationDate="2016-02-16T22:14:17.393" Score="3" Body="&lt;p&gt;I think this is the movie&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Race&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jesse Owens won four gold medals in the 1936 Olympics, which were held in Berlin that year.  I am still trying to tie in the meals and purple motif, but at least this may send someone down the right path.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14233" LastActivityDate="2016-02-16T22:14:17.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27533" PostTypeId="2" ParentId="27491" CreationDate="2016-02-16T23:37:50.293" Score="6" Body="&lt;p&gt;Well, since someone else went ahead and posted a partial answer, the fifty pfennig meals likely refer to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eintopfsonntag AKA Stew Sundays, a practice that was pushed as part of Winterhilfswerk (WHW), a drive to help support the Nationalsozialistische Volkswohlfahrt (National Socialist People’s Welfare Organization). Restaurants were required and families were encouraged to dull down their Sunday supper, donating the savings to WHW.  I found 50 pfennig referenced as the maximum amount to be spent on dinner.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7586" LastActivityDate="2016-02-16T23:37:50.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27534" PostTypeId="1" AcceptedAnswerId="27536" CreationDate="2016-02-17T00:13:47.147" Score="9" ViewCount="1268" Body="&lt;p&gt;Run up and down&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Run down and up&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By ancient Israelites &#xA;and classical Romans&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But not Africans or Middle Ages Europe&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Till today in every building&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electronic alternative is in use. Yet this remains in every building.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A pun: Don't &lt;em&gt;look&lt;/em&gt; to closely at the spelling.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19179" LastEditorUserId="19179" LastEditDate="2018-05-04T17:14:57.917" LastActivityDate="2018-05-04T17:14:57.917" Title="This way up! Who am I?" Tags="|word|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27535" PostTypeId="2" ParentId="27534" CreationDate="2016-02-17T00:49:39.880" Score="7" Body="&lt;p&gt;Guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A tablet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Run up and down...Run down and up&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When writing and reading a tablet, your eyes/hand go down and up on it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By ancient Israelites and classical Romans&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Both of these cultures used tablets as a primary method of writing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But not Africans or Middle Ages Europe&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Other forms of writing were used, such as paper.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Till today in every building...The electronic alternative is in use. Yet this remains in every building.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Electronic tablets such as the iPad are common today, and can be found in almost every business or house.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19212" LastActivityDate="2016-02-17T00:49:39.880" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27536" PostTypeId="2" ParentId="27534" CreationDate="2016-02-17T01:10:21.450" Score="11" Body="&lt;p&gt;How about &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stairs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Run up and down / run down and up&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can run up and down stairs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Israelites/Romans/Africans/Middle Ages Europe clue&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure. I can't (easily) find anything to corroborate the regional history of stairs across the world, but since this is the only clue that I can't totally fit, I'm still sure &quot;stairs&quot; is correct.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The electronic alternative is in use. Yet this remains in every building.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are elevators and escalators now, but there are also still stairs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;look&lt;/em&gt; could be another word for the homophone &lt;em&gt;stare&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;PLUS&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Up&quot; in the title, one of the two most common directions for stairs to go&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2016-02-17T16:07:49.787" LastActivityDate="2016-02-17T16:07:49.787" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="27537" PostTypeId="2" ParentId="27517" CreationDate="2016-02-17T02:39:13.383" Score="1" Body="&lt;p&gt;Is the answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A safe&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Spying has never been easier with me, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Safes contain sensitive information and can be helpful to spies&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Boats have use of what's inside me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Safes have locks, which is like the hull of the ship&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Shout I do not, even when teeth may enter me. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The lock(s)/key have teeth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll be happy to share an interest&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Find the right person, and all the stuff in a safe can be released&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With many who look alike - unlike the rest,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is in your best interest to show what is inside or how to open the safe with people you trust. In war, with people on your side&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Guard against secrets... I do my best&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Safes protect sensitive/valuable material from getting into the worng hands&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9042" LastActivityDate="2016-02-17T02:39:13.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27538" PostTypeId="1" AcceptedAnswerId="27541" CreationDate="2016-02-17T09:50:07.213" Score="2" ViewCount="273" Body="&lt;p&gt;I'm creating a new puzzle: two &lt;a href=&quot;https://en.wikipedia.org/wiki/15_puzzle&quot; rel=&quot;nofollow noreferrer&quot;&gt;sliding blocks 8-puzzle&lt;/a&gt;, one on top of the other. Several pieces are covered and cannot be moved until the covering piece moves away. The objective is to pair colors, no matter the final positions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YBv9U.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YBv9U.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;The Puzzle (by now it is not random, I'm testing yet)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CeMAw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CeMAw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;Objective is to pair colors&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/N5is2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/N5is2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;It consists of two layers (thanks GIMP...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.puzzlopia.com/puzzles/overlapped-8/play&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.puzzlopia.com/puzzles/overlapped-8/play&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact is that the puzzle can always be solved moving only the upper layer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The question is&lt;/strong&gt;: if the initial position is random on both layers, the shortest solution involves moving pieces from bottom layer or it always consists of moves only from upper layer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If optimal consists always only of upper pieces, then this puzzle reduces to an 8-puzzle and then I have to redesign it!&lt;/p&gt;&#xA;" OwnerUserId="17558" LastActivityDate="2016-02-17T13:12:25.123" Title="Is this two-sliding-block overlapped puzzle equivalent to the 8-puzzle?" Tags="|optimization|puzzle-creation|puzzle-identification|sliding-blocks|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="27539" PostTypeId="2" ParentId="27507" CreationDate="2016-02-17T10:44:42.743" Score="4" Body="&lt;p&gt;This is basically Ivo Beckers' solution. I'm not sure if the etiquette is to edit&#xA;his answer or to post mine and credit him. I'm choosing the latter. Ivo, the glory goes to you!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Instead of counting apples, I will count steps. At each step we&#xA; will give away 100g of apple. Call the current step N, begin with&#xA; N=100.&#xA;&#xA; As Ivo pointed out, at any point there are basically three cases.&#xA;&#xA; Easy case: All apples are 100g. Done.&#xA;&#xA; Next Easiest: The smallest apple is &amp;lt;= 75g&#xA; Apply Ivo's solution. Take the smallest apple and cut a piece to match&#xA; it from the largest apple which is guaranteed to be &gt; 100g since the&#xA; average weight of an apple is 100g. Now you have one fewer apples with an&#xA; average weight of 100g thus reducing the problem set by one. &#xA;&#xA; Hardest case: Apples are different weights and the smallest is &gt; 75g.&#xA; Cut a 25g slice from the smallest apple and put it aside. Next, take a&#xA; slice from the largest apple to combine with the (reduced) smallest apple&#xA; to make 100g. We now have one 25g slice and N-1 larger pieces with a total&#xA; weight of (N-1)*100 grams. Decrement N and move to the next step.&#xA; If we ever have 4 25g slices, we combine them, decrement N and continue.&#xA; At every step we have 100N grams of apple remaining consisting of&#xA; 0 - 3 25g slices totaling at most 75g and N larger pieces. The largest&#xA; piece is guaranteed to be at least 100 - 75/N grams so we can continue&#xA; until N = 2. At this point, we have two pieces totaling at least 125g&#xA; and at most 3 25g slices. Cut 25g slices from whichever is the largest piece&#xA; until we have 4. We now have 4 25g slices and two larger pieces which&#xA; also total 100g. Done.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastActivityDate="2016-02-17T10:44:42.743" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27540" PostTypeId="2" ParentId="26178" CreationDate="2016-02-17T11:15:02.700" Score="7" Body="&lt;p&gt;You are a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Heart&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At MyCorp U.S., Inc. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Corpus is latin for body - this is the corporation spoken of. I.e. My Corp U.S. = my Body&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm the pusher - one of the main keyboards on the org chart, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I push the blood round the body.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the one you want to follow. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Usually good to follow your heart..&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without me the corporation would be a cruel, harsh place, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Need to have a good heart and not be cruel. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and would quickly collapse. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Without me working - the person will collapse.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm generous and kind, but also have a sinister side. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Although usually people have a &lt;em&gt;generous&lt;/em&gt; and &lt;em&gt;kind&lt;/em&gt; hearts, they can have cruel hearts too.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One time I was arrested for some back-room deal, quite a serious infraction actually. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The heart was arrested once (i.e. heart attack). The heart &lt;a href=&quot;https://en.wikipedia.org/wiki/Ejection_fraction&quot;&gt;Ejection &lt;strong&gt;Fraction&lt;/strong&gt; Rate&lt;/a&gt; measures the pumping of the heart. &lt;em&gt;(Not sure about the back-room deal thing).&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Everyone was shocked!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Naturally, after a heart attack an electric shock was administered with a defibrillator.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19086" LastEditorUserId="19086" LastEditDate="2016-02-17T12:57:21.667" LastActivityDate="2016-02-17T12:57:21.667" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27541" PostTypeId="2" ParentId="27538" CreationDate="2016-02-17T13:12:25.123" Score="3" Body="&lt;p&gt;The optimal solution will involve only one layer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets assume that you have an optimal solution involving both layers.  The initial state of the top layer is $X$ and the initial state of the bottom layer is $Y$.  The optimal solution is a sequence of moves for the top layer $S_t=\{f_0, f_1, ... , f_n\}$ and a sequence of moves for the bottom layer $S_b=\{g_0, g_1, ... , g_m\}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S(A)=B$ be the application of the sequence of moves $S$ on the state $A$ and the resulting state is $B$.  Thus, $S_t(X)=W$ and $S_b(Y)=W$ since this is a solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $m^r$ be the reverse move of $m$.  If you reverse a move, you can undo the move.  For example, if applying move $m$ on state $A$ results in state $B$, then applying move $m^r$ on state $B$ results in state $A$.  Furthermore, if $S=\{a_0, ..., a_k\}$, then we will define a reversed sequence $S^r=\{a_k^r, ..., a_0^r\}$. If $S(A)=B$, then we can easily see that $S^r(B)=A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, we know that $S_b^r(W)=Y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if you have two sequences of moves where $S_0(A)=B$ and $S_1(B)=C$, then $S_1(S_0(A))=C$ and if $S=S_0 + S_1$, then $S(A)=C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let us construct a new sequence of moves $S_{t1}=S_t + S_b^r = \{f_0, ..., f_n, g_m^r, ..., g_0^r\}$.  We know that $S_t(X)=W$, and $S_b^r(W)=Y$.    Thus, $S_{t1}(X)=S_b^r(S_t(X))=S_b^r(W)=Y$.  Thus, $S_{t1}$ is also a solution, but only in the top layer.  It is also optimal since it has the same number of total moves as both $S_b$ and $S_t$.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2016-02-17T13:12:25.123" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27542" PostTypeId="1" AcceptedAnswerId="27543" CreationDate="2016-02-17T13:46:04.367" Score="1" ViewCount="342" Body="&lt;p&gt;Here are two alphabetic sequences:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A E F H I K L M N&#xA;B C D G J O P Q R&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;To which one would each of the remaining letters of the alphabet belong?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;S T U V W X Y Z&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And - why?&lt;/p&gt;&#xA;" OwnerUserId="17334" LastActivityDate="2016-02-17T13:52:40.803" Title="Two alphabetic sequences" Tags="|pattern|letter-sequence|" AnswerCount="1" CommentCount="1" ClosedDate="2016-02-17T15:35:44.400" ContentLicense="CC BY-SA 3.0" />
  <row Id="27543" PostTypeId="2" ParentId="27542" CreationDate="2016-02-17T13:52:40.803" Score="8" Body="&lt;p&gt;Set 1 &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;T V W X Y Z&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Set 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;S U&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Set one contains letters that are straight, set 2 contains curved letters&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18940" LastActivityDate="2016-02-17T13:52:40.803" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27544" PostTypeId="1" AcceptedAnswerId="30969" CreationDate="2016-02-17T14:20:12.987" Score="8" ViewCount="1428" Body="&lt;p&gt;Santa is exhausted, so he drops off his reindeer at the marked location below.&lt;br&gt;&#xA;He tells them he will be strolling the large divided meadow if they need him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where is Santa, and why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please reference this map:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rMy58.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rMy58.png&quot; alt=&quot;Where is Santa?&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hints:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Neither Santa nor his reindeer are in Brooklyn.&lt;br&gt;&#xA; Think outside the box but inside the correct -zoid, take everything into account.&lt;br&gt;&#xA; The reindeer are ensnared and cannot leave. They need your help!&lt;br&gt;&#xA; Santa thinks that its the best place in the area for reindeer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12014" LastEditorUserId="12014" LastEditDate="2016-02-19T13:05:59.460" LastActivityDate="2016-04-14T01:47:50.130" Title="Where have you gone Señor Santa Claus?" Tags="|lateral-thinking|knowledge|language|geography|cryptic-clues|" AnswerCount="8" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="27545" PostTypeId="2" ParentId="26426" CreationDate="2016-02-17T14:53:28.587" Score="0" Body="&lt;p&gt;I thought the question would be solved after giving Hint 1 and Hint 2 but I could not get the right one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Methodology&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since the sequence starts with 1 and increases by 1 (Hint 1) and spelling was important (Hint 2), You needed to check the spelling of 1, which is one. There are two vowels and one consonant and the generalization is found from here;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Equation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Number + (# of Vowels)x(# of Consonants)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For 10, ten has two consonants and one vowel so 10+2x1=&lt;strong&gt;12&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18514" LastActivityDate="2016-02-17T14:53:28.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27546" PostTypeId="2" ParentId="27534" CreationDate="2016-02-17T15:28:07.073" Score="3" Body="&lt;p&gt;My guess is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Line&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Run up and down / Run down and up&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;That's how to draw a line with a pencil&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By ancient Israelites and classical Romans / But not Africans or Middle Ages Europe&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Possibly &lt;em&gt;numeral systems&lt;/em&gt;? Both Roman And Israeli numeral systems contain vertical lines as numbers (I, II, III etc. for Roman, Hebrew number 6 is a straight vertical line), whereas African and European numeral systems do not.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Till today in every building / The electronic alternative is in use.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Electrical lines are in every building (not sure if wires or telephone lines are meant by this)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet this remains in every building.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Building lines are regulation to make sure buildings do not cross into the street or onto someone else's property.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A pun:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;A Line&quot; sounds like &quot;Align&quot;, which means to be the correct orientation (or &quot;This Way Up!&quot; from the question title).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11385" LastEditorUserId="11385" LastEditDate="2016-02-17T15:38:15.240" LastActivityDate="2016-02-17T15:38:15.240" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27547" PostTypeId="2" ParentId="27544" CreationDate="2016-02-17T16:14:28.490" Score="-3" Body="&lt;p&gt;given that he is exhausted, he has not moved from the marked location.&lt;/p&gt;&#xA;" OwnerUserId="19300" LastActivityDate="2016-02-17T16:14:28.490" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27548" PostTypeId="2" ParentId="27544" CreationDate="2016-02-17T16:18:49.367" Score="-2" Body="&lt;p&gt;Going for the obvious:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He's strolling in the large divided meadow because he's exhausted.&#xA;&lt;br/&gt;&lt;br/&gt; Ugh! An anti-puzzle??&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19086" LastActivityDate="2016-02-17T16:18:49.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27549" PostTypeId="1" AcceptedAnswerId="27550" CreationDate="2016-02-17T16:24:04.047" Score="4" ViewCount="341" Body="&lt;p&gt;Nothing's shocking through the looking glass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shade with teeth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your face manslaughter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weird revolution turned on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Commercial entertainment product 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="14233" LastEditorUserId="19051" LastEditDate="2016-02-17T16:44:41.320" LastActivityDate="2016-02-17T18:26:33.153" Title="Granted, these are obscure" Tags="|riddle|knowledge|music|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27550" PostTypeId="2" ParentId="27549" CreationDate="2016-02-17T16:26:07.560" Score="6" Body="&lt;h1&gt;EXTREMELY HELLA DONE EDITION&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;They are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Album titles&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;1A&lt;/strong&gt; - &lt;em&gt;Nothing's Shocking&lt;/em&gt; - Jane's Addiction (1988)&lt;br&gt;&#xA; &lt;strong&gt;1B&lt;/strong&gt; - &lt;em&gt;Through the Looking Glass&lt;/em&gt; - Siouxsie and the Banshees (1987)&lt;br&gt;&#xA; &lt;strong&gt;2A&lt;/strong&gt; - &lt;em&gt;Shade&lt;/em&gt; - Living Colour (2016)&lt;br&gt;&#xA; &lt;strong&gt;2B&lt;/strong&gt; - &lt;em&gt;With Teeth&lt;/em&gt; - Nine Inch Nails (2005)&lt;br&gt;&#xA; &lt;strong&gt;3A&lt;/strong&gt; - &lt;em&gt;In Your Face&lt;/em&gt; - Fishbone (1986)&lt;br&gt;&#xA; &lt;strong&gt;3B&lt;/strong&gt; - &lt;em&gt;Manslaughter&lt;/em&gt; - Body Count (2014)&lt;br&gt;&#xA; &lt;strong&gt;4A&lt;/strong&gt; - &lt;em&gt;Weird Revolution&lt;/em&gt; - Butthole Surfers (2001)&lt;br&gt;&#xA; &lt;strong&gt;4B&lt;/strong&gt; - &lt;em&gt;Turned On&lt;/em&gt; - Rollins Band (1990)&lt;br&gt;&#xA; &lt;strong&gt;5A&lt;/strong&gt; - &lt;em&gt;Commercial Entertainment Product&lt;/em&gt; - Emergency Broadcast Network (1992)&lt;br&gt;&#xA; &lt;strong&gt;5B&lt;/strong&gt; - &lt;em&gt;3&lt;/em&gt; - Violent Femmes (1989)&lt;br&gt;&#xA;&lt;br&gt;&#xA;  As for what &lt;em&gt;you&lt;/em&gt; are? Until I delve deeper, I'm just going to say you're a Gen Xer. You probably owned some black lace-up boots.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Connections:&lt;/strong&gt;&lt;br&gt;&#xA; - Some of these albums have featured people who were, at one point or another, members of Pigface&lt;br&gt;&#xA;&#xA;&lt;br&gt;&#xA; &lt;h1&gt;ACTUAL CONNECTION:&lt;/h1&gt;&#xA; All of these bands were &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_Lollapalooza_lineups_by_year#1991&quot; rel=&quot;nofollow&quot;&gt;at the first Lollapalooza&lt;/a&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; The &quot;grant&quot; in the title refers to Grant Park in Chicago, the current location of the festival&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2016-02-17T18:26:33.153" LastActivityDate="2016-02-17T18:26:33.153" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27551" PostTypeId="2" ParentId="27531" CreationDate="2016-02-17T16:29:20.533" Score="11" Body="&lt;p&gt;I think the missing entries are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; G in the top line and H in the bottom line (both as the 12th entry)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which fit into place to give the completed lists&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ABBCCDDEEFF&lt;b&gt;G&lt;/b&gt;GHHIILLLMNPPRSSS &lt;br&gt; TFHABHNIWNW&lt;b&gt;H&lt;/b&gt;XYBMBEFGFYAIFWAC&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To show this,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First, in the original list, write out each two letter pairing from the top line and bottom line in order: &lt;br&gt; &lt;br&gt; AT, BF, BH, CA, CB, DH, DN, EI, EW, FN, FW, GX, HY, HB, IM, IB, LE, LF, LG, MF, NY, PA, PI, RF, SW, SA, SC &lt;br&gt;&lt;br&gt; Now, for each of these bigrams, think of the second letter being encrypted by a Caesar cipher with the corresponding shift determined by the first letter. Specifically, the rule for decryption is &lt;br&gt;&lt;br&gt; &lt;b&gt;Decryption Rule: &lt;/b&gt; If the first letter is the $n$th letter of the alphabet, decrypt by shifting the second letter $n-1$ places back in the alphabet. &lt;br&gt;&lt;br&gt; So, for example, the first bigram AT remains the same, the next two bigrams BF and BH have their second letters shifted back &lt;b&gt;one&lt;/b&gt; place to become BE and BG, respectively. The next bigrams CA and CB have the second letter shifted back &lt;b&gt;two&lt;/b&gt; places (in which case the alphabet loops) to give CY and CZ, and so on. After decryption, the full sequence becomes &lt;br&gt;&lt;br&gt; AT, BE, BG, CY, CZ, DE, DK, EE, ES, FI, FR, GR, HR, HU, IE, IT, LT, LU, LV, MT, NL, PL, PT, RO, SE, SI, SK &lt;br&gt;&lt;br&gt; These are ISO 3166-1 alpha-2 country codes (hence the hint in the title) for members of the European Union, namely &lt;br&gt;&lt;br&gt; Austria, Belgium, Bulgaria, Cyprus, Czech Republic, Germany, Denmark, Estonia, Spain, Finland, France, Greece, Croatia, Hungary, Ireland, Italy, Lithuania, Luxembourg, Latvia, Malta, Netherlands, Poland, Portugal, Romania, Sweden, Slovenia and Slovakia &lt;br&gt;&lt;br&gt; There is one EU member missing from this list, the United Kingdom of Great Britain and Northern Ireland which has country code GB. But of course we must encrypt it again by shifting the B forward &lt;b&gt;six&lt;/b&gt; places to give GH. The list is sorted alphabetically, first by top row, then by bottom row, whence the idea for the placement.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18422" LastActivityDate="2016-02-17T16:29:20.533" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27552" PostTypeId="2" ParentId="27500" CreationDate="2016-02-17T16:43:57.043" Score="2" Body="&lt;p&gt;Not as comprehensive as the first one, but still...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To push a color into the last place, bring it to the front (by selecting it), then select the last one on the current list. Since the last place will never come into play again, we can select the color that's supposed to be at the 2nd-to-last place and the color coming 2nd-to-last then rinse and repeat. When only the first and 2nd colors are left untouched, we can either let it be or just select the 2nd on the current list. Starting from the position in the recent colors section in the image, it can be done by selecting the colors in this order:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Pink, Blue, Purple, Green, White, Red, Black, White.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="14016" LastEditDate="2016-02-17T17:53:49.767" LastActivityDate="2016-02-17T17:53:49.767" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27553" PostTypeId="1" AcceptedAnswerId="27556" CreationDate="2016-02-17T17:24:22.847" Score="6" ViewCount="514" Body="&lt;p&gt;Up and down. Not right or left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Top and bottom. Neither side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remaining ones are strange. Would think 'twas charmed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are we?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The title is a pun on the correct title (but would be too easy if I used the hidden title.) Solve it for a clue. Some science vocabulary required.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19179" LastActivityDate="2016-02-18T05:02:03.627" Title="Chocolate, Vanilla, Strawberry, Mint, Rocky Road, Bubblegum" Tags="|word|science|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27554" PostTypeId="2" ParentId="27522" CreationDate="2016-02-17T17:30:04.553" Score="2" Body="&lt;p&gt;Let the number of days in which the person had non-veg lunch and veg dinner be $x$, veg lunch and non-veg dinner $y$ and two veg meals $z$. Then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y$ = 9&lt;br&gt;&#xA;$y+z$ = 15&lt;br&gt;&#xA;$x+y$ = 14&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x+y+z = m$ = 20&lt;/p&gt;&#xA;" OwnerUserId="14016" LastActivityDate="2016-02-17T17:30:04.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27555" PostTypeId="2" ParentId="27553" CreationDate="2016-02-17T17:30:39.067" Score="9" Body="&lt;p&gt;I think the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quarks (or flavours of quark)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The title refers to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The term 'flavours' is used for different types of quark which are up, down, top, bottom, strange and charm as referenced in the text. In the case of the title, the flavours are flavours of ice cream.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18422" LastEditorUserId="18422" LastEditDate="2016-02-17T17:55:37.270" LastActivityDate="2016-02-17T17:55:37.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27556" PostTypeId="2" ParentId="27553" CreationDate="2016-02-17T17:47:36.877" Score="3" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Quarks&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Context:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There are 6 types (AKA flavours as hinted in the title) of quarks: up, down, top, bottom (not left, right or something), strange and charm.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="14016" LastEditDate="2016-02-17T18:02:16.933" LastActivityDate="2016-02-17T18:02:16.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27557" PostTypeId="2" ParentId="27544" CreationDate="2016-02-17T17:49:09.013" Score="6" Body="&lt;p&gt;He is strolling in &lt;a href=&quot;https://goo.gl/maps/Rwt87wMNa2D2&quot;&gt;Long Meadow&lt;/a&gt; in Prospect Park, just over &lt;a href=&quot;https://goo.gl/maps/fj3TdQfhxD42&quot;&gt;2 miles west&lt;/a&gt; of the reindeer landing spot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the name implies, it is a &lt;em&gt;large meadow&lt;/em&gt;, and it has been &lt;em&gt;divided&lt;/em&gt; into ballfields in the south and open meadow in the north.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for why Santa is there, &lt;a href=&quot;http://www.nycgo.com/venues/prospect-park-long-meadow&quot;&gt;this site&lt;/a&gt; claims that this is a popular location for children's birthday parties.  Santa is obviously doing some reconnaissance to populate his naughty/nice list.&lt;/p&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2016-02-17T17:49:09.013" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27558" PostTypeId="1" AcceptedAnswerId="27564" CreationDate="2016-02-17T18:04:56.240" Score="23" ViewCount="2527" Body="&lt;p&gt;Myself and two others had a debate earlier over what separates a logic puzzle and a riddle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Friend A was arguing that a logic puzzle was a puzzle that required any kind of logical and lateral thinking, although he didn't define what logical thinking is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Friend B was arguing that a logic puzzle is one that can be solved with an explicit set of instructions, thus resulting in a computer programme being able to solve a large number of logic puzzles using a single programme. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How the debate started&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of them gave a puzzle for us to solve:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You are in a room with 3 switches. One of the switches powers a light bulb in another room. The only way to check which switch powers the light is by walking into the room to check. You can only walk into the room once, and can not walk back into the room with the switches. How can you find out which switch powers the lightbulb? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is to switch one on, wait a while, switch it off and then switch a different one on. Walk into the room. If the light is warm, you know it was the switch that you turned on and off. If the light is on, you know it is the switch you turned on and did not turn back off. If the lightbulb is neither warm nor on then it is the other switch. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Friend B argued that this is a logic puzzle based on the fact that you had to use his idea of 'logical thinking' and lateral thinking. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Friend A argued that it is not a logic puzzle because of the fact that you have to think qualitatively about the fact that the light bulb will warm up, and it is in fact a riddle of some description. A ridiculously long code would need to be written in order for a computer to solve this puzzle, because the computer would have to have so much information stored in the code in order to think of the fact that a light bulb warms up - i.e. only humans can do that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who is correct? (Person B's argument convinced me, whereas person A's argument wasn't really an argument with any basis.)&lt;/p&gt;&#xA;" OwnerUserId="19307" LastEditorUserId="19307" LastEditDate="2016-02-18T20:32:21.360" LastActivityDate="2016-02-18T20:32:21.360" Title="How can a logic puzzle be defined?" Tags="|puzzle-identification|" AnswerCount="4" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="27559" PostTypeId="2" ParentId="27544" CreationDate="2016-02-17T18:45:31.503" Score="-1" Body="&lt;p&gt;He is in the North Pole, because today is not Christmas.&lt;/p&gt;&#xA;" OwnerUserId="11907" LastActivityDate="2016-02-17T18:45:31.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27560" PostTypeId="2" ParentId="27544" CreationDate="2016-02-17T19:28:45.110" Score="-1" Body="&lt;p&gt;Going a bit off the wall here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The marked location appears to be a BBQ place. I'll say that Santa is exhausted from dealing with all their workplace drama and dropped them off to be made into tasty meals. Santa is at the North Pole now, because he got his order to go. And the reindeer aren't much of anywhere, unless you count &lt;em&gt;INSIDE SANTA&lt;/em&gt; Muahahahahahaha!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16874" LastActivityDate="2016-02-17T19:28:45.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27561" PostTypeId="2" ParentId="27558" CreationDate="2016-02-17T21:02:55.033" Score="22" Body="&lt;p&gt;I think the most-common definition of a logic puzzle is one that requires the solver to make deductions using formal logic, like the famous &lt;a href=&quot;https://en.wikipedia.org/wiki/Zebra_Puzzle&quot;&gt;Einstein's puzzle&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this type of puzzle, you use only formal logic to come to conclusions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g.&lt;br&gt;&#xA;Given:  If &lt;strong&gt;A&lt;/strong&gt;, then &lt;strong&gt;B&lt;/strong&gt;&lt;br&gt;&#xA;Given:  If &lt;strong&gt;B&lt;/strong&gt;, then &lt;strong&gt;C&lt;/strong&gt;&lt;br&gt;&#xA;Deduction:  If &lt;strong&gt;A&lt;/strong&gt;, then &lt;strong&gt;C&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm no expert in formal logic, so the example is pretty simplistic, but it should serve to illustrate my point.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any puzzle that requires any kind of knowledge &lt;em&gt;not explicitly provided in the puzzle&lt;/em&gt; (aside from the rules of formal logic) would not generally be considered a logic puzzle.&lt;/p&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2016-02-17T21:02:55.033" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27562" PostTypeId="2" ParentId="27558" CreationDate="2016-02-17T21:17:53.830" Score="15" Body="&lt;p&gt;Persons A and B are trying to draw a semantic distinction over language that is intrinsically ambiguous. As a result, neither of them are wrong. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to have an argument of whether a puzzle is one of logic or not, you first need to clearly define what counts as &quot;logic,&quot; and subsequently, what counts as a &quot;logic puzzle.&quot; Neither of these things happened, which is the source of your argument. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I agree with GentlePurpleRain in principle - a logic puzzle is one where the answer is logically deducible from the information provided in the question. However, I think this still leaves semantic ambiguity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It could easily be considered not a logic problem. If we're restricting &quot;logic&quot; to mathematical or formulable steps, then it's certainly not - the crux of the problem is in realizing the light bulb will be hot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It could, however, easily be considered a logic problem, as well. You can logically deduce that flipping one switch does not give you enough information, and so you need to find another way to obtain the results. That, even, is mathematically expressible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, you &lt;em&gt;still&lt;/em&gt; need to build a way to actually do that, but the problem has logically expressible steps. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, this leaves the difference as nothing but semantic. &lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastActivityDate="2016-02-17T21:17:53.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27563" PostTypeId="2" ParentId="27544" CreationDate="2016-02-17T22:25:25.613" Score="8" Body="&lt;p&gt;I can't quite get to the right answer because...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; my Spanish isn't good enough.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I'll try to help the effort along.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Senor Santa implies Spanish language puns, so...&lt;br&gt;&#xA; Meadow = Nava&lt;br&gt;&#xA; Divided = Dividido&lt;br&gt;&#xA; Merry Christmas = &lt;em&gt;Feliz Navidad&lt;/em&gt;  (Nava dividido)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think that's the heart of the thing.  Here's some extras:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Feliz = Happy (There is also a shop named Feliz near this spot in Brooklyn, less than a mile to the northwest).&lt;br&gt;&#xA; That shop is next to Fort Greene Park; a divided meadow itself.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19128" LastActivityDate="2016-02-17T22:25:25.613" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27564" PostTypeId="2" ParentId="27558" CreationDate="2016-02-17T23:07:43.697" Score="16" Body="&lt;p&gt;Borrowing language from &lt;a href=&quot;https://en.wikipedia.org/wiki/Fuzzy_set&quot; rel=&quot;nofollow noreferrer&quot;&gt;fuzzy set theory&lt;/a&gt;, what you have is a crisp, binary decision imposed on a set of more than 2 values, at least one of which is fuzzy. Let me explain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your friends implicitly assume that puzzles can be crisply partitioned into logic and non-logic puzzles. This is not the case. There are at least 3 types: puzzles fully answerable by logical manipulation of the given information, puzzles requiring (substantial) logical manipulation but which also contain some lateral thinking, and puzzles that are fully or predominantly lateral thinking with very little formal logic, if any.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some examples:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;pure logic: &lt;a href=&quot;https://en.wikipedia.org/wiki/Zebra_Puzzle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Einstein's puzzle&lt;/a&gt;, as &lt;a href=&quot;https://puzzling.stackexchange.com/a/27561/9422&quot;&gt;GentlePurpleRain&lt;/a&gt; mentions;&lt;/li&gt;&#xA;&lt;li&gt;some logic, some lateral thinking: your light bulb puzzle; and&lt;/li&gt;&#xA;&lt;li&gt;pure lateral thinking, no formal logic: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16460/how-did-the-police-know&quot;&gt;How did the Police Know?&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note that the term &lt;em&gt;logic puzzle&lt;/em&gt; has been adopted by a certain class of &lt;a href=&quot;http://www.logic-puzzles.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;truth-table puzzles&lt;/a&gt;, so there is some justification to your Friend A's claim. On the other hand, semantically, it is just as justifiable to include type 2 in the term &lt;em&gt;logic puzzles&lt;/em&gt; as it is to consider smart phones to be &lt;em&gt;phones&lt;/em&gt; even though they are also &lt;em&gt;cameras&lt;/em&gt; and &lt;em&gt;music players&lt;/em&gt;, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Type 2 is obviously fuzzy, but there is a degree of fuzziness in the other types as well. Type 1 requires language comprehension, especially wordy forms like Einstein's Puzzle. Type 3 often requires &lt;em&gt;some&lt;/em&gt; logical thinking, even if it isn't mentioned in the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of trying to measure the amount of logic in the question, I suggest looking at &lt;strong&gt;how many answers can be considered correct&lt;/strong&gt;. Naively, logic puzzles have at most one correct answer, while non-logic puzzles have more than one correct answer. For example, the Police puzzle has many answers posted which are all consistent with the given information. Some of the answers are even contradictory. This is a hallmark of type 3 puzzles. However, for type 2 puzzles, it might only &lt;em&gt;seem&lt;/em&gt; to have one answer because others haven't yet been suggested. For example, instead of relying on heat-dissipation, one could rely on non-instantaneous fade-out of the bulb's filament, at least in principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;To sum up&lt;/strong&gt;, &lt;em&gt;logic puzzles&lt;/em&gt; constrain the set of correct answers by the information explicitly presented, while &lt;em&gt;non-logic puzzles&lt;/em&gt; (for want of a better term) allow the set of correct answers to be constrained (so to speak) by information sourced elsewhere. Under this definition, your light bulb puzzle is a non-logic puzzle.&lt;/p&gt;&#xA;" OwnerUserId="9422" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2016-02-18T03:06:25.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27565" PostTypeId="2" ParentId="27544" CreationDate="2016-02-17T23:47:57.933" Score="5" Body="&lt;p&gt;I think&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Santa is in the Greek underworld&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He is exhausted&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He is tired and about to die.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Looking at the map,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; there are lots of graveyards nearby.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He will be strolling the large divided meadow if they need him.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Greek underworld is divided into several sections: the Fields of Punishment, the Fields of Asphodel, and Elysium. Santa probably expects to be in Elysium. It's kind of hard to &quot;stroll&quot; the Fields of Punishment, and he probably won't be able to remember or help the reindeer in the Fields of Asphodel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Theory:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; His reindeer is bringing him to his funeral.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Dunno why he's called &quot;Señor&quot; though.&lt;/p&gt;&#xA;" OwnerUserId="12109" LastActivityDate="2016-02-17T23:47:57.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27567" PostTypeId="2" ParentId="27474" CreationDate="2016-02-18T00:11:39.107" Score="1" Body="&lt;p&gt;Oray's answer is close but there are some inconsistencies.  It starts out by saying the expected value of the game is infinity (which it clearly is not) but then calculates an expected value that is finite (which is close to correct but not quite).  The payout is as follows for the sequence of tails(T)/heads(H):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;H; pay = 1; p = 1/2&lt;/li&gt;&#xA;&lt;li&gt;TH; pay = 2; p = 1/4         &lt;/li&gt;&#xA;&lt;li&gt;TTH; pay = 4; p = 1/8&lt;/li&gt;&#xA;&lt;li&gt;TTTH; pay = 8; p = 1/16&lt;/li&gt;&#xA;&lt;li&gt;etc&lt;/li&gt;&#xA;&lt;li&gt;TT(19 tails)H; pay = 2^19; p = 2^(-20)&lt;/li&gt;&#xA;&lt;li&gt;TTT(20 tails); pay = 10^6; p = 2^(-20)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note that the probability of the last (20 tails or more) is 2^(-20) and not 2^(-21).  So the expected value is the sum of the payouts times the corresponding probabilities Sum[2^k * 2^-(k+1),{k,0,19}] + 10^6 * 2^(-20)&#xA;and gives approximately E[x] = $10.95&#xA;This is the expected amount you would win over a large number of plays.  How much you would be willing to pay to play, however, has a subjective component that is different for everyone.  I would want to make a profit so I certainly wouldn't pay more than 10 dollars!&lt;/p&gt;&#xA;" OwnerUserId="19251" LastActivityDate="2016-02-18T00:11:39.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27568" PostTypeId="1" AcceptedAnswerId="27569" CreationDate="2016-02-18T00:26:02.050" Score="3" ViewCount="122" Body="&lt;p&gt;One of us has the voice&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of us has the Katy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first went from the World to the Country&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We ran with McDaniels&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we've been heroes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some say we're crazy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For playing with the toys in the attic&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there's not a sweeter emotion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;than lovin' in an elevator&lt;/p&gt;&#xA;" OwnerUserId="18816" LastActivityDate="2016-02-18T00:38:15.703" Title="We are 5, former 8 from the 70s." Tags="|riddle|music|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27569" PostTypeId="2" ParentId="27568" CreationDate="2016-02-18T00:38:15.703" Score="5" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Aerosmith&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are 5, former 8 from the 70s.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There's 5 current members, with three additional past members (totalling 8) and they're from the 70s.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One of us has the voice&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Steven Tyler is the singer of the group and also a judge on American Idol (which is not quite The Voice).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One of us has the Katy&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Joe Perry shares a last name with Katy Perry (since she changed her name from Katy Hudson), though they are not related.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first went from the World to the Country&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They are a world-famous rock band and now Steven Tyler has released some country songs (though not with Aerosmith).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We ran with McDaniels&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Run DMC (with Darryl McDaniels as a member) rerecorded Walk This Way with Aerosmith.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And we've been heroes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They have a Guitar Hero dedicated to them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some say we're crazy&lt;br&gt;&#xA;For playing with the toys in the attic&lt;br&gt;&#xA;But there's not a sweeter emotion&lt;br&gt;&#xA;than lovin' in an elevator  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Crazy, Toys in the Attic, Sweet Emotion, and Love in an Elevator are all Aerosmith songs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2016-02-18T00:38:15.703" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27570" PostTypeId="2" ParentId="17681" CreationDate="2016-02-18T00:51:12.010" Score="1" Body="&lt;p&gt;The central Pentagon as A,B,C,D,E contains 540 DEGREES&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sum the 5 PAIRS supplementary angles ie. &#xA;2(180-A)+2(180-B)+2(180-C)+2(180-D)+2(180-E)= 1800 2(540) = 720&#xA;This 720 degrees represents the 'base' angles of the 5 triangles Which totals 5*180=900&#xA;900-720 = 180 (the 5 angles being sought.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The five Triangles at the points sum to 5*180 = 900  &lt;/p&gt;&#xA;" OwnerUserId="19318" LastActivityDate="2016-02-18T00:51:12.010" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27573" PostTypeId="1" AcceptedAnswerId="27622" CreationDate="2016-02-18T06:07:39.463" Score="14" ViewCount="345" Body="&lt;p&gt;A certain USA celebrity tweeted a cryptarithm,&#xA;which includes a cipher with an encrypted clue,&#xA;for friends in a nearby country.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{r} \qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad~~&#xA;\tt      M \, O \, R \, E  ~\\[-1ex]&#xA;\tt      M \, O \, R \, E  ~\\[-1ex]&#xA;\tt      M \, O \, R \, E  ~\\[-1ex]&#xA;\tt      M \, O \, R \, E  ~\\[-1ex]&#xA;\tt      E \, V \, E \, R  ~\\[-1ex]&#xA;\tt  +\, M \, O \, R \, E  ~\\[.5ex]&#xA;\hline \end{array}&#xA;\\&#xA;13.15 - \frac{18}{19.5} + 0.0 + \cdots + \frac3{15} - \frac45 + \cdots&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What did he or she want and how much?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Post-solution note&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The core encryption idea here is from @question_asker,&#xA;who let me run with it and promised not to solve this.&#xA;Anyone seen this system used before?&lt;/p&gt;&#xA;" OwnerUserId="18129" LastEditorUserId="9404" LastEditDate="2016-03-05T17:34:58.607" LastActivityDate="2016-03-05T17:34:58.607" Title="Ever more is never enough" Tags="|cipher|alphametic|" AnswerCount="3" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="27576" PostTypeId="2" ParentId="27573" CreationDate="2016-02-18T08:52:11.050" Score="3" Body="&lt;p&gt;I think she wants &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Forever more&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are 4 mores, making &quot;for&quot; as it sounds like &quot;four&quot;. Then just finish the statement. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12100" LastActivityDate="2016-02-18T08:52:11.050" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27577" PostTypeId="2" ParentId="27517" CreationDate="2016-02-18T10:36:47.273" Score="2" Body="&lt;p&gt;I am a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Keyhole!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Spying has never been easier with me, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Looking through the keyhole - great way to spy!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Boats have use of what's inside me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Locks are a key feature on canals for barges&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Shout I do not, even when teeth may enter me. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Keys have teeth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll be happy to share an interest&#xA;With many who look alike - unlike the rest,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Only keys that are the same can unlock me&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Guard against secrets... I do my best.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I enable the door to be locked!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19086" LastActivityDate="2016-02-18T10:36:47.273" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27580" PostTypeId="1" AcceptedAnswerId="27582" CreationDate="2016-02-18T12:48:23.980" Score="37" ViewCount="8849" Body="&lt;p&gt;Ten girls are sitting around a table.&#xA;Each of them picks a real number and whispers it to the two neighbors immediately to the left and to the right. (Hence: each girl communicates one number, and receives two numbers.)&#xA;Each girl then loudly announces the average of the two numbers she received.&#xA;The announced numbers in order around the circle are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1, 2, 3, 4, 5, 6, 7, 8, 9, 10&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What was the number picked by the girl who announced the average number 6?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2024-01-19T04:17:49.193" Title="Ten girls around a table" Tags="|mathematics|no-computers|" AnswerCount="13" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27581" PostTypeId="2" ParentId="27580" CreationDate="2016-02-18T13:05:30.957" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 6. For her to announce an average of 6, the two numbers she receives must add up to 12. This happens to be the average of the two averages announced to either side of her. This pattern holds around the circle, which makes sense if the announced average is also the number picked.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16874" LastActivityDate="2016-02-18T13:05:30.957" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27582" PostTypeId="2" ParentId="27580" CreationDate="2016-02-18T13:41:53.930" Score="28" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Consider 10 real numbers $a$ through $j$ and have each girl's whispered number be $2a, 2b, 2c$ etc (these do not have to be even numbers; $a$ through $j$ are not necessarily integers.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\dfrac{2a + 2c}2 = 1 \quad\text{or}\quad a + c = 1\\[10pt]\dfrac{2b + 2d}2 = 2\quad\text{or}\quad b + d = 2$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;etc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The girl who says 6 is sitting at g. Rearranging the &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$e + g = 5$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;equation into&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$g = 5-e$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and then substituting around the system of equations, we get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\begin{align}g &amp;amp;= 5-e \\&amp;amp;= 5 - (3-c) \\&amp;amp;= 2 + c \\&amp;amp;= 2 + (1-a) \\&amp;amp;= 3 - a \\&amp;amp;= 3 - (9-i) \\&amp;amp;= -6 + i \\&amp;amp;= -6 + (7-g) \\&amp;amp;= 1 - g\end{align}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;therefore&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$2g = 1$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is what she says to her neighbours.&lt;/p&gt;&#xA;" OwnerUserId="1509" LastEditorUserId="1509" LastEditDate="2018-04-28T23:13:37.763" LastActivityDate="2018-04-28T23:13:37.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27583" PostTypeId="1" AcceptedAnswerId="27587" CreationDate="2016-02-18T13:56:33.020" Score="35" ViewCount="7567" Body="&lt;p&gt;You see me all day&lt;br&gt;&#xA;I'm sharp&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I show you your way&lt;br&gt;&#xA;But you show me what to show you&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I exist, but not in your dimension&lt;br&gt;&#xA;I'm made up of down&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My older language is by rotation&lt;br&gt;&#xA;But now through light&lt;br&gt;&#xA;(though in the end it comes down to the same shocking tale) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I usually look a certain way&lt;br&gt;&#xA;But I change at certain times&lt;br&gt;&#xA;Into myself, my thin sibling, into time,&lt;br&gt;&#xA;and into ambiguity&lt;br&gt;&#xA;or something with which you grasp&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="10327" LastActivityDate="2016-02-20T12:34:15.653" Title="I'm made up of down, what am I?" Tags="|riddle|" AnswerCount="5" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="27585" PostTypeId="2" ParentId="5007" CreationDate="2016-02-18T14:09:25.367" Score="-1" Body="&lt;p&gt;Idk why, but I think she's a&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a chip&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Killed by:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;an iron&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;ALT:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Killed by a Iron, It was Mais con yelo (corn with ice and milk)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19333" LastActivityDate="2016-02-18T14:09:25.367" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27586" PostTypeId="2" ParentId="10101" CreationDate="2016-02-18T14:13:23.583" Score="-1" Body="&lt;p&gt;Answer?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Carbon Dioxide&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The thing you would love?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oxygen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19333" LastEditorUserId="17250" LastEditDate="2016-02-18T14:52:01.577" LastActivityDate="2016-02-18T14:52:01.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27587" PostTypeId="2" ParentId="27583" CreationDate="2016-02-18T14:24:36.990" Score="53" Body="&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A mouse pointer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You see me all day&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We use computers a lot!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm sharp&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's a pointer!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I show you your way&#xA;But you show me what to show you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Indicates where your cursor is, but you control it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I exist, but not in your dimension&#xA;I'm made up of down&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's made of light, but also the absence of light... only exists virtually and is visible only in the context of the light from surrounding pixels.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My older language is by rotation&#xA;But now through light&#xA;(though in the end it comes down to the same shocking tale)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Trackerballs used rotation, old mouse balls were also rotation based, but optical mice are now more common, although trackpads which also control pointers don't really fit this. But the rotation similarities don't stop there... Electron guns in CRTs 'cycled' through the scan lines. LCDs used polarization (a kind of rotation). Now it's usually LEDs. All these methods use electricity (shocking).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I usually look a certain way&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pointer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I change at certain times&#xA;Into myself, my thin sibling, into time,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A pointer, the text-select 'I' icon, the waiting/clock/hourglass/spinner symbol&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and into ambiguity&#xA;or something with which you grasp&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A question mark, the grabber hand.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mousy cursor pointer type thing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18522" LastEditorUserId="18522" LastEditDate="2016-02-18T16:33:51.177" LastActivityDate="2016-02-18T16:33:51.177" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="27588" PostTypeId="1" AcceptedAnswerId="30464" CreationDate="2016-02-18T14:39:11.757" Score="7" ViewCount="496" Body="&lt;p&gt;IIIIIIIIIIIIII&lt;br&gt;&#xA;TTTTTT&lt;br&gt;&#xA;LLLLLLL&lt;br&gt;&#xA;UUUUU&lt;br&gt;&#xA;JJJJJJJ&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the next logical sequence?&lt;/p&gt;&#xA;" OwnerUserId="19179" LastActivityDate="2016-04-07T03:13:11.343" Title="Letter Sequence Shape Shifter" Tags="|letter-sequence|" AnswerCount="6" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27589" PostTypeId="2" ParentId="27588" CreationDate="2016-02-18T14:43:33.133" Score="11" Body="&lt;p&gt;Sorry if that it is not the intendend solution but I've been trapped with this once : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the next sequence might be : &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&#xA;I&lt;br&gt;&#xA;T&lt;br&gt;&#xA;L&lt;br&gt;&#xA;U&lt;br&gt;&#xA;J&lt;br&gt;&#xA;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17888" LastActivityDate="2016-02-18T14:43:33.133" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27591" PostTypeId="2" ParentId="25608" CreationDate="2016-02-18T14:48:07.727" Score="8" Body="&lt;p&gt;The message says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Money is numbers and numbers never end.&lt;br&gt;&#xA; If it takes money to be happy,&lt;br&gt;&#xA; your search for happiness will never end.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;(which is a quote from Bob Marley)&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The encryption process works as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, produce an alphabet based on the &lt;a href=&quot;https://en.wikipedia.org/wiki/NATO_phonetic_alphabet&quot; rel=&quot;nofollow&quot;&gt;NATO phonetic alphabet&lt;/a&gt;, but by &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; grouping the alphabet by the last letter of the corresponding radio code (alpha, delta, india, ... bravo, echo, ...). Assign each group a number and each word a letter based on the order they appear (credit to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/25608/the-three-retirees-cipher/#comment81664_27591&quot;&gt;Zandar&lt;/a&gt; for working out this step).&lt;br&gt;&#xA; So &lt;strong&gt;A&lt;/strong&gt; = &lt;em&gt;alpha&lt;/em&gt; = &lt;code&gt;1A&lt;/code&gt; (first group, first word), &lt;strong&gt;B&lt;/strong&gt; = &lt;em&gt;beta&lt;/em&gt; = &lt;code&gt;2A&lt;/code&gt; (second group, first word), etc.&lt;br&gt;&#xA; Resulting in the following dictionary (code determined by column, then row):&lt;br&gt;&#xA; &lt;code&gt;-  1 2 3 4 5 6 7 8 9 0&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;A: A B C F G H N Q U W&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;B: D E M J     O     X&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;C: I K Y       V&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;D: L R&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;E: P T&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;F: S&lt;/code&gt;&lt;br&gt;&#xA; &lt;sub&gt;(not sure what they'd do if they needed a &lt;strong&gt;Z&lt;/strong&gt;, since that would require an 11th group...)&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From here, you can encrypt your message easily, however:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Once you have converted each letter into it's corresponding number/letter pair, if consecutive pairs use the same leading number, you drop the number in subsequent pairs (so &lt;code&gt;2A2B&lt;/code&gt; becomes &lt;code&gt;2AB&lt;/code&gt;).  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="5208" LastEditDate="2016-02-18T16:13:23.133" LastActivityDate="2016-02-18T16:13:23.133" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="27592" PostTypeId="2" ParentId="27588" CreationDate="2016-02-18T14:55:28.170" Score="4" Body="&lt;p&gt;IIIIIIIIIIIIII&lt;br&gt;&#xA;TTTTTT&lt;br&gt;&#xA;LLLLLLL&lt;br&gt;&#xA;UUUUU&lt;br&gt;&#xA;JJJJJJJ  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of these rows are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; basically a vertical line with a different configuration of &quot;extra stuff&quot;. The first row has the &quot;extra stuff&quot; as a disjoint second vertical line (noticing there are roughly twice as many characters in the first row as any other row).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The next row could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; CCCCCCC  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2016-02-18T14:55:28.170" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27594" PostTypeId="2" ParentId="27580" CreationDate="2016-02-18T15:27:46.713" Score="14" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;EDIT:&lt;/em&gt;&lt;/strong&gt; After the fact, realize this is the same proof as Kate's.  I will leave it here because I find it might be a bit clearer, but hers was first and also complete, so upvote that one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;~~~&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets the number that the girls say be $S_i$ and the number the girls think be $T_i$.  We then know $S_i=i$ and:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_1=\frac{T_{10}+T_2}{2}=1 \implies T_{10}+T_2=2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can repeat this for the rest and get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_2=T_1+T_3=4, S_3=T_2+T_4=6, S_4=T_3+T_5=8$$&#xA;$$S_5=T_4+T_6=10,S_6=T_5+T_7=12, S_7=T_6+T_8=14$$&#xA;$$S_8=T_7+T_9=16, S_9=T_8+T_{10}=18,S_{10}=T_9+T_1=20$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have 10 unknowns and 10 equations.  In fact, we have 2 sets of 5 equations with 5 unknowns since the even subscripts are never in an equation with odd subscripts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets solve the even subscript set first since $S_6=6$ and we are trying to solve $T_6$.&#xA;$$T_{10}+T_2=2 \implies T_2=2-T_{10}$$&#xA;$$T_2+T_4=6 \implies 2-T_{10}+T_4=6 \implies T_4=4+T_{10}$$&#xA;$$T_4+T_6=10 \implies 4+T_{10}+T_6=10 \implies T_6=6-T_{10}$$&#xA;$$T_6+T_8=14 \implies 6-T_{10}+T_8=14 \implies T_8=8+T_{10}$$&#xA;$$T_8+T_{10}=18 \implies 8+T_{10}+T_{10}=18 \implies T_{10}=5$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, $T_6=6-T_{10}=6-5=1$ and $S_6=6$, so the girl who said 6 thought 1.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="1592" LastEditDate="2016-02-19T12:38:17.360" LastActivityDate="2016-02-19T12:38:17.360" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27595" PostTypeId="2" ParentId="27583" CreationDate="2016-02-18T15:52:43.010" Score="11" Body="&lt;p&gt;&lt;strong&gt;You are:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An Up (or Down) Quark.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You see me all day:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The little buggers are everywhere, in all matter. &lt;sup&gt;[citation needed]&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm sharp:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quarks are a tiny, like the edge of an absurdly sharp blade. Some people would consider Quarks to be smart, as they are explained and though about by Physicists (ie: sharp people).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I show you your way&lt;br&gt;&#xA;But you show me what to show you:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; According to certain interpretations of Quantum Mechanics, the state of a quantum particle (ie: a Quark) is decided by whether it is observed or not. If it is unobserved, then it is in a state of Superposition, once it is observed its State Collapses.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I exist, but not in your dimension:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quarks are subatomic matter, they are &lt;em&gt;tiny&lt;/em&gt;. Relatively speaking, humans are &lt;strong&gt;huge&lt;/strong&gt;. Dimension here refers to size.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm made up of down:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hence Up or Down Quark (probably Up).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My older language is by rotation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Quark#Spin&quot;&gt;Quarks are imagined to 'spin' but this notion is misguided.&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But now through light:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I can't explain this. It might have something to do with Quarks &lt;a href=&quot;https://en.wikipedia.org/wiki/Quark#Strong_interaction_and_color_charge&quot;&gt;color charge&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(though in the end it comes down to the same shocking tale):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quantum Mechanics is sometimes considered bizarre and shocking. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I usually look a certain way&#xA;But I change at certain times&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again with Quantum State.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Into myself, my thin sibling, into time,&#xA;and into ambiguity&#xA;or something with which you grasp&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Changing states upon observation. 'Grasp' as in strong interaction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12656" LastActivityDate="2016-02-18T15:52:43.010" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27596" PostTypeId="2" ParentId="27580" CreationDate="2016-02-18T15:57:28.957" Score="9" Body="&lt;p&gt;Let's write down the sums:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_1 +a_3 = 2$&lt;br&gt;&#xA;$a_2 +a_4 = 4$&lt;br&gt;&#xA;...&lt;br&gt;&#xA;$a_9 +a_1 = 18$&lt;br&gt;&#xA;$a_{10} +a_2 = 20$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we can obtain two arithmetic series where the terms increase by 4:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_3,a_7,a_1,a_5,a_9$&lt;br&gt;&#xA;$a_4,a_8,a_2,a_6,a_{10}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The one who said 6 must have picked $a_7$, so we can find it using this and $a_7+a_9=14$ (or $a_7+a_5=10$), which means $a_7=1$.&lt;/p&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="14016" LastEditDate="2016-02-18T18:17:57.560" LastActivityDate="2016-02-18T18:17:57.560" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27598" PostTypeId="1" CreationDate="2016-02-18T16:20:37.140" Score="8" ViewCount="495" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A drink I do take,&lt;br&gt;&#xA;  to calm all my shakes,&lt;br&gt;&#xA;  and heave a sigh of relief.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The moon dims and fades,&lt;br&gt;&#xA;  then once again shines,&lt;br&gt;&#xA;  glimmering on that which fells trees.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A little flutter,&lt;br&gt;&#xA;  and now I do stutter,&lt;br&gt;&#xA;  my tongue swollen red, green and blue.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Good for the skin,&lt;br&gt;&#xA;  making it pearly and new,&lt;br&gt;&#xA;  as long as you pay heed&lt;br&gt;&#xA;  to whispers in the breeze,&lt;br&gt;&#xA;  that humming sound that drives you mad,&lt;br&gt;&#xA;  and the reason I am here alone.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;-From, a lawn chair in the sun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;to what is this letter referring? Why do you think the lawn chair is alone?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes:&#xA;Well, Easy Tiger figure out one part of the riddle, at least: The lawn chair is complaining that it's lonely. However, he is not quite right on WHY the owner has...shall we say...disappeared.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, keep in mind that every Stanza refers to a separate object that, when combined together, creates a story explaining what the lawn chair is complaining of. Listed below is the hints for each, as well as how the particular part of the story occurred, by way of an event that ties in to the particular object in question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stanza 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A drink I am sharing, with buscuits a latte. Tho others might pour a bit of &lt;strong&gt;ME&lt;/strong&gt; in their___(rhymes if pronounced with the right accent).  EVENT: SPILL/POUR&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Stanza 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the answer is hidden yet there in plain sight, (these words repeated X times 'fore the fight). EVENT: VERB IN THE STANZA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Stanza 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If I had cleared a Tin, I would be able to see; but, until then, my nose stays stuffy. EVENT: &quot;TRIGGERED&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Stanza 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First two lines help to explain why this event occured. The rest is a warning to not repeat the same mistakes. EVENT: WHY?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13697" LastEditorUserId="13697" LastEditDate="2016-07-27T16:42:04.067" LastActivityDate="2017-08-11T21:06:01.757" Title="What did the lawn chair complain of?" Tags="|riddle|wordplay|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27599" PostTypeId="2" ParentId="13324" CreationDate="2016-02-18T16:42:06.660" Score="1" Body="&lt;p&gt;There are two types of triangles; those with the point up, and those with the point down.  Since everything is symmetric, so all we need to do is count triangles pointing up (or down) and double it.  We will count the ones pointing up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, see that the six pointed star is just the superposition of one large triangle pointing up over one large triangle pointing down.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we can start counting the up triangles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Individual single-unit triangles pointing up are found in the large triangle pointing up, plus $3$ left over around the edge.  The large triangle pointing up has a base of $6$.  So, the number of individual triangles is $T_1=6+5+4+3+2+1+3=21+3=24$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next size of triangles are ones with a base of 2.  These triangles contain 4 triangles, with an upside down triangle at the centre.  Looking at the figure, we therefore only need to count the interior upside down triangles.  This time, there are no triangles around the edge, and less of them can fit in the large triangle.  We get $T_2=5+4+3+2+1=15$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeat with the rest, and you get $T_3=4+3+2+1=10$, $T_4=3+2+1=6$, $T_5=2+1=3$ and $T_6=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum of these is $T=24+15+10+6+3+1=59$.  Double this to get the total and you have $118$.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="1592" LastEditDate="2016-02-18T16:53:33.047" LastActivityDate="2016-02-18T16:53:33.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27600" PostTypeId="1" CreationDate="2016-02-18T16:44:03.293" Score="8" ViewCount="715" Body="&lt;p&gt;rested my Head on me leg and heaved a sigh of Relief.&lt;br&gt;&#xA;Asked the guy a Question I Didn't Understand.&lt;br&gt;&#xA;climbed A Hill and saw a Battered Cowboy.&lt;br&gt;&#xA;Then, decided to take a dip in the Pool.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I may not like cats very much, but I can't say this is any better: ___________&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To what am I referring?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It looks like some people are getting closer to a solution, so let me drop another hint: The most polar of opposites is a hypocrisy. And, in a way, the solutions are a form of hypocrisy, if you compare them to the title of this puzzle. ;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the Last Solution is a Cluster Of Words (see comments for details)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13697" LastEditorUserId="13697" LastEditDate="2016-07-27T16:52:20.640" LastActivityDate="2016-07-27T16:52:20.640" Title="Opposites can attract" Tags="|riddle|wordplay|" AnswerCount="1" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="27601" PostTypeId="2" ParentId="27558" CreationDate="2016-02-18T17:05:37.017" Score="1" Body="&lt;p&gt;This question is biased because the very 'riddle' (or 'logic puzzle') is biased: one piece of information is missing, i.e. the enunciation ought to include something like '[a lamp] whose one or more additional parameters other than visible light can be assessed'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whereas a lamp admittedly emits visible light, being able to measure or sense its temperature is not ascertained in all cases (the lamp might be out of reach or, if fitted with LEDs, its temperature increase might not be sensed straightforwardly). Without the additional temperature parameter, no computer could ever solve the riddle, for it could not suspect the very existence of the variable. If the 'can-the-temperature-be-sensed' parameter is registered as a boolean value, calculation will succeed only if it is true; otherwise it will lead to undecidability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tricky 'riddles' of the sort are not uncommon. If 'logic' is meant as a computable set of statements, then this one is no logic, whatever you may call it. Now if one thinks that 'logic' allows a recourse to undeclared variables, he may call it that —while he could not expect a computer to ever exploit loopholes.&lt;/p&gt;&#xA;" OwnerUserId="19340" LastActivityDate="2016-02-18T17:05:37.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27602" PostTypeId="2" ParentId="27573" CreationDate="2016-02-18T17:45:35.570" Score="2" Body="&lt;p&gt;This is just a partial answer, of course, but I think it's actually  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;something that's forever.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you look at the words, you can see that you get&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;four more&quot; and &quot;ever more&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which can be combined to form&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;former&quot; and &quot;evermore&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, if we take a look at the last 2 numbers in the sequence, we see that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One has a &lt;strong&gt;3&lt;/strong&gt; divided by &lt;strong&gt;(5*3)&lt;/strong&gt;, and the other has &lt;strong&gt;4/5&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since we have two words, and &quot;more is common enough&quot;, lets&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; substitute &quot;more&quot; for 5, and then apply the fractions to the words from earlier. If you subtract the &lt;strong&gt;last&lt;/strong&gt; &lt;strong&gt;three&lt;/strong&gt; letters from &quot;former&quot;, you get &lt;strong&gt;&quot;for&quot;&lt;/strong&gt;. IF you do the same with the &lt;strong&gt;last&lt;/strong&gt; &lt;strong&gt;four&lt;/strong&gt; letters from &quot;evermore&quot;, you get &quot;ever.&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, in most games like battleship, two letters joined with a dash reference a row and column, and can be combined together. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Thus, we end up with the word &lt;strong&gt;&quot;forever&quot;&lt;/strong&gt;, and the fact that &quot;MORE&quot; and 5 are somehow related.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13697" LastActivityDate="2016-02-18T17:45:35.570" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27603" PostTypeId="2" ParentId="26088" CreationDate="2016-02-18T17:51:55.793" Score="3" Body="&lt;p&gt;Your old friend question_asker here. I've got the answer, and it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $3.\overline{3}$ or $3.333...$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why, you ask?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Who knows!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, then how did I get the answer?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Brute force! Kind of!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(record scratch)&lt;/em&gt; lol what&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yeah, look, I went to college for like 1.5 semesters before dropping out, and during the semester I liked, I took Calculus I, and while I enjoyed it, it was the last of my formal math education.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Anyway, the kind (non-, as per their profile) doctor had &lt;em&gt;at one point&lt;/em&gt; left a hint on this question: &lt;em&gt;&quot;This sequence rolled out in reverse order as reciprocals for a probability problem &lt;a href=&quot;https://puzzling.stackexchange.com/questions/tagged/probability?sort=newest&quot;&gt;here&lt;/a&gt;.&quot;&lt;/em&gt; In a brief chat with Pangloss, they mentioned that the sequence was only of importance with regard to &lt;em&gt;games&lt;/em&gt; where &lt;em&gt;randomness&lt;/em&gt; was necessary (emphasis mine). From this I could narrow down that the problem I was looking for probably involved dice or a coin flip. Skimming those problems, I found that, lo and behold, there was &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16333/how-to-get-a-uniform-1-out-of-7-chance-using-only-a-coin-and-a-6-sided-die&quot;&gt;a problem that involved both&lt;/a&gt;, and specifically &lt;a href=&quot;https://puzzling.stackexchange.com/questions/16333/how-to-get-a-uniform-1-out-of-7-chance-using-only-a-coin-and-a-6-sided-die/26061#26061&quot;&gt;an answer that had the exact thing mentioned in the hint&lt;/a&gt;. I &quot;checked&quot; the &quot;math&quot; briefly and realized that it was, in fact, the reciprocals of the elements of the sequence in reverse. From this I could conclude that the missing value was $10/3$, and since it was mentioned that &lt;em&gt;the missing element required a notation convention not already seen in the puzzle&lt;/em&gt;, I realized it must be $3.\overline{3}$ or $3.333...$&lt;br&gt;&#xA;&lt;br&gt;&#xA; As such, with Pangloss's blessing, I am posting this answer and awaiting my parade.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;" OwnerUserId="17250" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2016-02-18T17:51:55.793" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27605" PostTypeId="5" CreationDate="2016-02-18T18:28:38.617" Score="0" Body="&lt;p&gt;A type of word problem where 2 or more words are combined or mashed together to form a new one, normally involving a set of minor alterations to get the correct result.&lt;/p&gt;&#xA;" OwnerUserId="13697" LastEditorUserId="13697" LastEditDate="2016-02-18T20:31:35.100" LastActivityDate="2016-02-18T20:31:35.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27606" PostTypeId="4" CreationDate="2016-02-18T18:28:38.617" Score="0" Body="This is for a type of word problem where multiple words are combined together to form new ones, resulting in a hidden phrase. Normally, slight alterations must be made to the words in order to correctly mash them together." OwnerUserId="13697" LastEditorUserId="13697" LastEditDate="2016-02-18T20:31:29.897" LastActivityDate="2016-02-18T20:31:29.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27607" PostTypeId="1" CreationDate="2016-02-18T18:52:02.613" Score="-5" ViewCount="472" Body="&lt;p&gt;&lt;em&gt;&lt;strong&gt;What bird can't fly or run very fast, and is different shades of brown, all at the same time?&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;If you're a man, stand on top of a tree, look down, put the two things you see together, and you might just think maybe the answer has to do with a part of you.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h3&gt;Hints (from the comments below):&lt;/h3&gt;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;  &lt;strong&gt;American&lt;/strong&gt;.  It's just a word but it's a part of the answer.  I think that part might be hard to guess because the riddle doesn't have any clues to that part.  The rest of the answer has a clue in the riddle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Multiple shades of brown all at once could mean a variety of different kinds of brown feathers on one bird.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18735" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-02-19T09:50:15.603" Title="The bird with no name (For mature people only)" Tags="|riddle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27608" PostTypeId="2" ParentId="27607" CreationDate="2016-02-18T19:35:43.743" Score="3" Body="&lt;p&gt;My answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Woody Woodpecker&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A man looking down would see the second word...  The first half is the content of the tree and second is a common slang for a male part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is not know for running or flying fast.  The content as mentioned has varying shades of brown.&lt;/p&gt;&#xA;" OwnerUserId="16187" LastEditorUserId="16187" LastEditDate="2016-02-18T19:40:38.833" LastActivityDate="2016-02-18T19:40:38.833" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="27610" PostTypeId="1" AcceptedAnswerId="27611" CreationDate="2016-02-18T19:38:48.633" Score="6" ViewCount="1270" Body="&lt;p&gt;Every day, people press me for advice, which I offer honestly. I never lie.&lt;br&gt;&#xA;Often, they don't listen.&lt;br&gt;&#xA;Sometimes they become scared of me and stop asking my opinion.&lt;br&gt;&#xA;But usually they come back in the end. After all, I'm the only reliable source of information on this subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who am I?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user17947" LastActivityDate="2016-02-19T13:52:05.527" Title="A scary oracle - who am I?" Tags="|riddle|" AnswerCount="7" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27611" PostTypeId="2" ParentId="27610" CreationDate="2016-02-18T19:43:23.510" Score="8" Body="&lt;p&gt;Are you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A scale (as in, the type that measures weight)?&lt;br&gt;&#xA;&lt;br&gt;&#xA; It's not unusual for people to get discouraged when the scale (which, provided it's calibrated correctly, never lies) tells them they're too heavy or too light. There's no other (reasonable, efficient, accurate, etc.) way to find out how much you weigh than by getting onto a scale.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2016-02-18T20:53:46.120" LastActivityDate="2016-02-18T20:53:46.120" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27612" PostTypeId="2" ParentId="27610" CreationDate="2016-02-18T19:50:41.127" Score="1" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Google&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People press the buttons and search for your advice. You show search results, well accurate(honest) as per the search keywords and page-rank algorithm. You don't lie. And yes, often people don't listen as they think that these search results are irrelevant.&lt;br&gt;&#xA;&lt;br&gt;&#xA; People are scared of you, as you contain their emails, search history, location and what not... Sometimes, people stop asking your opinion and use other search engines like Bing :-P&lt;br&gt;&#xA;&lt;br&gt;&#xA; But they come back, as you are reliable source of information on (almost) anything.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1766" LastEditorUserId="4421" LastEditDate="2016-02-18T22:08:37.417" LastActivityDate="2016-02-18T22:08:37.417" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27613" PostTypeId="2" ParentId="27598" CreationDate="2016-02-18T20:09:59.293" Score="6" Body="&lt;p&gt;The answer could be - &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Lawn Chair is alone because the trees have been cut down to make paper.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is what the poem means.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  - The first stanza is referring to a printer with RGB ink.&lt;br&gt;&#xA;  - The second stanza refers to refilling ink in the printer.&lt;br&gt;&#xA;  - The third stanza describes the action of scanning in a photocopier machine (made by cutting down trees = &quot;which fells trees&quot;). Both the printer and photocopier use paper (and hence trees).&lt;br&gt;&#xA;  - The fourth stanza refers to a tissue (&quot;making the skin pearly and new&quot;). The sound of tree-leaves in wind = &quot;whispers in the breeze&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19349" LastEditorUserId="4421" LastEditDate="2016-02-18T22:28:14.947" LastActivityDate="2016-02-18T22:28:14.947" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="27614" PostTypeId="2" ParentId="17211" CreationDate="2016-02-18T20:18:06.147" Score="4" Body="&lt;p&gt;The answer (Puzzle 3):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; parsley a.k.a &lt;em&gt;Petroselinum crispum&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I can't explain everything yet but I can give some information about hints. From &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Parsley&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;&lt;/em&gt;: &lt;br&gt;&#xA; - cultivated as a herb, a &lt;strong&gt;spice&lt;/strong&gt;, and a vegetable.&lt;br&gt;&#xA; - it grows as a biennial, in the first year, it forms a &lt;strong&gt;rosette&lt;/strong&gt; of &lt;strong&gt;tripinnate leaves&lt;/strong&gt;&lt;br&gt;&#xA; - Curly leaf parsley is often used as a &lt;strong&gt;garnish&lt;/strong&gt; &lt;br&gt;&#xA; Explanation for wordplays and portmanteaus:&lt;br&gt;&#xA; - Piece of Pie could be &quot;P&quot; and books can tell &quot;Lies&quot; and an obnoxious person is an &quot;arse&quot;. &lt;br&gt;&#xA; - a neat freak would need lots of soap. soap is made of &quot;lye&quot;&lt;br&gt;&#xA; - If a crook was to be caught, you might say he's fallen on his &quot;ass&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19157" LastEditorUserId="19157" LastEditDate="2016-02-18T20:36:14.867" LastActivityDate="2016-02-18T20:36:14.867" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="27615" PostTypeId="2" ParentId="27583" CreationDate="2016-02-18T20:23:12.790" Score="3" Body="&lt;p&gt;You see me all day&lt;br&gt;&#xA;I'm sharp &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; mobile phone&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I show you your way&lt;br&gt;&#xA;But you show me what to show you &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; GPS feature on mobile&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I exist, but not in your dimension&lt;br&gt;&#xA;I'm made up of down &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Electronic&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My older language is by rotation&lt;br&gt;&#xA;But now through light &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Older landline (rotations), newer mobile phones (LED touch screen)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(though in the end it comes down to the same shocking tale)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I usually look a certain way&lt;br&gt;&#xA;But I change at certain times&lt;br&gt;&#xA;Into myself, my thin sibling, into time,&lt;br&gt;&#xA;and into ambiguity&lt;br&gt;&#xA;or something with which you grasp  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; newer thin mobiles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A MOBILE PHONE!!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19351" LastEditorUserId="7586" LastEditDate="2016-02-18T22:42:36.057" LastActivityDate="2016-02-18T22:42:36.057" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27616" PostTypeId="2" ParentId="27610" CreationDate="2016-02-18T20:25:58.377" Score="2" Body="&lt;p&gt;Are you:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Nutrition facts&quot; information on food bought in supermarkets. Or any food nutrition labels in McDonalds etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19351" LastEditorUserId="17250" LastEditDate="2016-02-18T20:32:58.553" LastActivityDate="2016-02-18T20:32:58.553" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27617" PostTypeId="1" AcceptedAnswerId="27887" CreationDate="2016-02-18T20:36:53.930" Score="17" ViewCount="1172" Body="&lt;p&gt;I saw two neighbors between my eyes.&#xA;Then I lost my right eye, but I was still the same.&#xA;Then one neighbour left, but I was still no different.&#xA;Finally, I lost my left eye and now I no longer belong to any culture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who was I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Changed the wording a bit to make some points clearer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I come in many forms, where one form is hard to find, while a more common one might hurt you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; \ ( o . o ) /&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19350" LastEditorUserId="19350" LastEditDate="2016-02-24T09:09:43.093" LastActivityDate="2016-02-24T14:46:46.347" Title="Lost my eyes - A riddle" Tags="|riddle|" AnswerCount="5" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="27618" PostTypeId="2" ParentId="27610" CreationDate="2016-02-18T20:44:01.060" Score="9" Body="&lt;p&gt;This may be a bit abstract, but I'd like to think you are AN (oh no, that's practically a spoiler!)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alarm clock.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Every day, people press me for advice, which I offer honestly. I never lie. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many people have to get up at specific times every day, and they press their alarm clocks to tell them when to get up.  The alarm clock can only respond with the same level of accuracy that it was given.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Often, they don't listen.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people sleep through their alarms, when the volume is too low or if they are too deep in sleep.  Others simply ignore the alarm, wanting to sleep in a little later and snooze.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes they become scared of me and stop asking my opinion.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many people dread having to wake up in the morning to the ringing of their alarm, counting in their heads the hours of precious sleep they still might capture.  Some end up just trusting their bodies to wake up at the appropriate time, while others just throw their cares by the wayside.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But usually they come back in the end. After all, I'm the only reliable source of information on this subject.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Even if one goes without the use of an alarm clock, say to sleep in on vacation or on the weekend, they will return to using it when they have to, because it is a reliable tool in telling them how many hours have passed and one that tells them it's time to get up.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11452" LastActivityDate="2016-02-18T20:44:01.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27619" PostTypeId="1" AcceptedAnswerId="27623" CreationDate="2016-02-18T21:11:57.757" Score="17" ViewCount="3336" Body="&lt;p&gt;Nobody likes me.&lt;br&gt;&#xA;Nobody ever comes to see me just for fun.&lt;br&gt;&#xA;I'm nobody's hobby.&lt;br&gt;&#xA;People are always trying to avoid me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet every day people come to me.&lt;br&gt;&#xA;They spend precious time with me&lt;br&gt;&#xA;Though they could usually avoid me.&lt;br&gt;&#xA;After all, you usually know where I am.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more who come to me&lt;br&gt;&#xA;The worse I get.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But don't think I'm evil.&lt;br&gt;&#xA;I'm legal in all 50 states of the union&lt;br&gt;&#xA;(though I only exist with any frequency in some of them).&lt;br&gt;&#xA;I'm not addictive, not in the least.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you must know...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm slow as anything&lt;br&gt;&#xA;  Yet&lt;br&gt;&#xA;Often the fastest way to get anywhere is through me.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;" OwnerUserId="19179" LastActivityDate="2016-02-19T19:34:53.110" Title="Why Would One Willingly While Life Away?" Tags="|riddle|word|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27620" PostTypeId="2" ParentId="25818" CreationDate="2016-02-18T21:33:06.873" Score="1" Body="&lt;p&gt;It's kind of a partial match (it goes better before taking hints into account) and I'm posting after there's already an accepted answer, but I sort of like what went through my head upon reading this, so thought I'd share.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is (not really):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Harmony, which works in both the peaceful and the musical sense.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am loved by many,&lt;br&gt;&#xA;even if sometimes people are angry to me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most people dream of world peace, but some thrive on dissonance and chaos.  Everyone loves music, but there is the classic &quot;Dang kids and their music&quot; where every generation hates the next generation's taste to some extent.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can find me in different places,&lt;br&gt;&#xA;yet you won't always find me stable.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Even in peaceful nations, there are times of unrest. Harmony plays an integral part in all forms of music, sometimes including some measure of dissonance.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes you must use your talent,&lt;br&gt;&#xA;sometimes you need wisdom to be my best friend.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It takes a certain amount of talent to be successful at diplomacy, and some of the most world-famous wisemen are also regarded as the most peaceful (e.g. Buddha, Dalai Lama, Gandhi). Not everyone has the talent necessary to perform music. The best performers have extensive knowledge of music theory whether through study or practical application.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was born before NATO's first SACEUR is appointed,&lt;br&gt;&#xA;yet I am not old fashioned.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is where it gets shaky since SACEUR is so specific.  However, NATO does have to do with peace, so that part is alright-ish.  Harmony never goes out of style. :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can tear me apart,&lt;br&gt;&#xA;but then you will find radius and patella.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; H-ARM-O-NY -&gt;  &quot;H&quot; + &quot;ARM&quot; + &quot;OH&quot; + &quot;NEE (knee)&quot; -&gt; Radius is one of the two bones of your lower arm and Patella is your kneecap.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7586" LastEditorUserId="7586" LastEditDate="2016-02-18T21:41:45.797" LastActivityDate="2016-02-18T21:41:45.797" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27621" PostTypeId="2" ParentId="27619" CreationDate="2016-02-18T21:36:55.173" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An interstate highway. No one gets on the highway just for fun, and most people actively try to find ways not to get on the highway. People spend non-trivial portions of their lives in traffic on them. The more people on the highway, the worse traffic is, and the more time you spend! They exist in all 50 states, but states with higher populations and larger cities will have more around them. Despite all that, due to the higher speed limits on them, they are often the fastest route to a given destination.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16874" LastActivityDate="2016-02-18T21:36:55.173" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27622" PostTypeId="2" ParentId="27573" CreationDate="2016-02-18T21:37:39.757" Score="10" Body="&lt;p&gt;She wants&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;17,909 (Mexican) pesos&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you consider each of the numbers below the horizontal line as a letter (A=1, B=2, etc.), you get&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $M.O - \frac{R}{S.E} + 0.0 + \cdots + \frac{C}{O} - \frac{D}{E} + \cdots$&lt;br&gt;&#xA; (&lt;strong&gt;MORSE CODE&lt;/strong&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Taking that as a clue, and looking at the dots and dashes on that line (decimal points, ellipses, minus signs, fraction dividers), you get&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\cdot - - \cdot \quad \cdot \quad \cdots \quad - - - \quad \cdots$&lt;br&gt;&#xA; $\quad P \quad \quad E \quad S \quad \quad \;\; O \quad \quad \; \; S$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This gives you the alphametic&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\begin{array}{r} \qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad~~\tt      M \, O \, R \, E  ~\\[-1ex]\tt      M \, O \, R \, E  ~\\[-1ex]\tt      M \, O \, R \, E  ~\\[-1ex]\tt      M \, O \, R \, E  ~\\[-1ex]\tt      E \, V \, E \, R  ~\\[-1ex]\tt  +\, M \, O \, R \, E  ~\\[.5ex]\hline\tt    P\, E \, S \, O \, S  ~\\[-1ex] \end{array}\\$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which can be solved as&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\begin{array}{r} \qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad~~\tt      2 \, 0 \, 4 \, 7  ~\\[-1ex]\tt      2 \, 0 \, 4 \, 7  ~\\[-1ex]\tt      2 \, 0 \, 4 \, 7  ~\\[-1ex]\tt      2 \, 0 \, 4 \, 7  ~\\[-1ex]\tt      7 \, 6 \, 7 \, 4  ~\\[-1ex]\tt  +\, 2 \, 0 \, 4 \, 7  ~\\[.5ex]\hline\tt    1\, 7 \, 9 \, 0 \, 9  ~\\[-1ex] \end{array}\\$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2016-02-18T21:37:39.757" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="27623" PostTypeId="2" ParentId="27619" CreationDate="2016-02-18T21:38:58.617" Score="15" Body="&lt;p&gt;You are most definitely the bane of my morning and evening, you are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; traffic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nobody likes me.&lt;br&gt;&#xA;Nobody ever comes to see me just for fun.&lt;br&gt;&#xA;I'm nobody's hobby.&lt;br&gt;&#xA;People are always trying to avoid me.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; While the great (enter your nationality here) road can be enjoyed when it is wide and open, when it is full of cars, it is not fun.  You can take a leisurely drive at night, a fun road trip, etc, but no one goes to the roads eager to sit at lights or behind other cars.  In fact, people often turn away from backups if they have the chance, and new routing systems can actively subvert heavy traffic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet every day people come to me.&lt;br&gt;&#xA;They spend precious time with me&lt;br&gt;&#xA;Though they could usually avoid me.&lt;br&gt;&#xA;After all, you usually know where I am.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every day, people (like me!) commute to work on the road, where they spend far too much time to get where they are going.  Other forms of transit could help you avoid road traffic, in fact, you could avoid it almost anywhere-anywhere that isn't a road.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But don't think I'm evil.&lt;br&gt;&#xA;I'm legal in all 50 states of the union&lt;br&gt;&#xA;(though I only exist with any frequency in some of them).&lt;br&gt;&#xA;I'm not addictive, not in the least.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Traffic occurs on roads, because roads are in every state, as are the citizens who need to get to work.  In Montana, not so much, in LA, New York, Chicago- ouch.  You don't get hooked on traffic, you want to get away from it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The hint makes it look like Irishpanda's answer is closer.  It seems we were at work on our keyboards at roughly the same time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whatever the answer really is, please&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Don't make me drive home tonight...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11452" LastActivityDate="2016-02-18T21:38:58.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27624" PostTypeId="2" ParentId="27619" CreationDate="2016-02-18T21:39:51.337" Score="17" Body="&lt;p&gt;You are ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a traffic jam.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nobody likes me.&lt;br&gt;&#xA;Nobody ever comes to see me just for fun.&lt;br&gt;&#xA;I'm nobody's hobby.&lt;br&gt;&#xA;People are always trying to avoid me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nobody likes to be in a traffic jam, it's not fun or a hobby, everyone tries to avoid it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet every day people come to me.&lt;br&gt;&#xA;They spend precious time with me&lt;br&gt;&#xA;Though they could usually avoid me.&lt;br&gt;&#xA;After all, you usually know where I am.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people are in a traffic jam everyday, on their way to work. They know where it is, but usually can't avoid it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The more who come to me&lt;br&gt;&#xA;The worse I get.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The more people are there the worse is the jam.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But don't think I'm evil.&lt;br&gt;&#xA;I'm legal in all 50 states of the union&lt;br&gt;&#xA;(though I only exist with any frequency in some of them).&lt;br&gt;&#xA;I'm not addictive, not in the least.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's obviously neither evil nor illegal, and definitely not addictive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you must know...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As suggested by &lt;em&gt;Nyk 232&lt;/em&gt;, even though traffic jams are slow, using the highway is often faster despite the jam, than trying to go around the jam.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2016-02-18T22:02:37.300" LastActivityDate="2016-02-18T22:02:37.300" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27625" PostTypeId="2" ParentId="27610" CreationDate="2016-02-18T21:55:52.920" Score="7" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A mirror&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Every day, people press me for advice,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Does my hair look decent?  Do I need a shave?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which I offer honestly. I never lie.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The mirror shows your reflection as you are.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Often, they don't listen.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I'm not really losing my hair, it's just my imagination.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes they become scared of me and stop asking my opinion.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Screw you, I'm not gaining weight, it's just water retention.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But usually they come back in the end.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can't avoid your reflection forever.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After all, I'm the only reliable source of information on this subject.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I mean, you could take a photograph, but photos can be doctored, go out of date, etc.  Your reflection in the mirror is always accurate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16131" LastActivityDate="2016-02-18T21:55:52.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27626" PostTypeId="2" ParentId="27588" CreationDate="2016-02-18T21:58:35.537" Score="-1" Body="&lt;p&gt;The answer is BBBBBBBBBBBBBB because I appeared 14 times and the sequence was 11, 8, 9 , then 11 again thats why B was used&lt;/p&gt;&#xA;" OwnerUserId="19358" LastActivityDate="2016-02-18T21:58:35.537" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27627" PostTypeId="2" ParentId="27583" CreationDate="2016-02-18T22:00:22.680" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Sun. The Sun is always visible during the day. The Sun warps spacetime, and we are always falling towards it, falling down. It shows the way by its position in the sky, by the shadows it casts on a sundial (ok, reaching with that one). The sun looks the same until it sets, rises, or is eclipsed by the moon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19356" LastActivityDate="2016-02-18T22:00:22.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27628" PostTypeId="2" ParentId="27580" CreationDate="2016-02-18T23:56:43.410" Score="5" Body="&lt;p&gt;Let Girl&amp;lt;#&amp;gt; be the girl who announced #.  For example, Girl6 is the girl who announced 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, let Value&amp;lt;#&amp;gt; be the number chosen by, and whispered by, Girl&amp;lt;#&amp;gt;.  For example, Girl6 chose Value6, and whispered Value6 to Girl5 and Girl7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let x = Value6.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As shown below, x = Value6 = 1.  The girl who announced 6 chose 1.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Value4 = 10 - x.  This allows (Value4 + Value6)/2 = ((10 - x) + x)/2 = 10/2 = 5.&lt;br&gt;&#xA; Value2 = x - 4.  This allows (Value2 + Value4)/2 = ((x - 4) + (10 - x))/2 = 6/2 = 3.&lt;br&gt;&#xA; Value8 = 14 - x.  This allows (Value6 + Value8)/2 = (x + (14 - x))/2 = 14/2 = 7.&lt;br&gt;&#xA; Value10 = x + 4.  This allows (Value8 + Value10)/2 = ((14 - x) + (x + 4))/2 = 18/2 = 9.&lt;br&gt;&#xA; (Value2 + Value10)/2 = 1.  Thus, ((x - 4) + (x + 4))/2 = 1, or 2x/2 = 1, or x = 1.&lt;br&gt;&#xA;&lt;br&gt;&#xA; x = Value6 = 1.  The girl who announced 6 chose 1.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Check by substitution.&lt;/strong&gt;&lt;br&gt;&#xA; Girl..   Choice  Announced&lt;br&gt;&#xA; Girl1... __......  1&lt;br&gt;&#xA; Girl2....  -3......  2&lt;br&gt;&#xA; Girl3...  __......  3&lt;br&gt;&#xA; Girl4.....   9......  4&lt;br&gt;&#xA; Girl5...  __......  5&lt;br&gt;&#xA; &lt;strong&gt;Girl6&lt;/strong&gt;....   &lt;strong&gt;1&lt;/strong&gt;......  &lt;strong&gt;6&lt;/strong&gt;&lt;br&gt;&#xA; Girl7...  __......  7&lt;br&gt;&#xA; Girl8...  13......  8&lt;br&gt;&#xA; Girl9...  __......  9&lt;br&gt;&#xA; Girl10...   5....  10  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10046" LastEditorUserId="10046" LastEditDate="2016-02-19T14:46:31.627" LastActivityDate="2016-02-19T14:46:31.627" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="27629" PostTypeId="2" ParentId="27619" CreationDate="2016-02-19T00:04:39.097" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are &lt;strong&gt;bureaucracy&lt;/strong&gt; Nobody likes or enjoys bureaucracy and everyone tries to avoid, yet we have to spend a lot of time to go through bureaucracy. You could easily avoid by efficient employees. It is slow as hell but if you need anything you have to go through it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19362" LastActivityDate="2016-02-19T00:04:39.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27630" PostTypeId="2" ParentId="27580" CreationDate="2016-02-19T00:15:05.007" Score="13" Body="&lt;p&gt;Less rigorous and mathematical, but perhaps more intuitive approach: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) note that for each girl, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; there is no relationship at all between the number whispered and the number said; they are parts of different cycles. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(b) &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; in the group of girls we care about, the thought-of numbers averaged to $[1, 3, 5, 7, 9]$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(c)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Note that this is symmetrical, so the number that was a component of both 1 and 9 must be 5. From there, the rest follows: &#xA;&lt;br&gt;&#xA; $5 + a = 9\times2 \therefore a = 13$&lt;br&gt;&#xA; $13 + b = 7\times2 \therefore b = 1$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(d) to check the work, go the other way around the circle: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $5 + d = 1\times2 \therefore d = -3$&lt;br&gt;&#xA; $-3 + c = 3\times2 \therefore c = 9$&lt;br&gt;&#xA; $9 + b = 5\times2 \therefore b = 1.$&#xA;&lt;br&gt;&#xA; (The other 5 girls each thought of the number 1 higher than the girl sitting to one side of her).  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10355" LastEditorUserId="13494" LastEditDate="2016-02-20T16:02:54.473" LastActivityDate="2016-02-20T16:02:54.473" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27631" PostTypeId="2" ParentId="27610" CreationDate="2016-02-19T00:31:11.953" Score="1" Body="&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A person's conscience.  You are the only place to find out whether a person feels something is right or wrong. People are often afraid of finding out if they feel what they're doing is wrong, because doing what they feel is right could cause them adverse effects.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alternately you could be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A person's heart. In terms of old school metaphors, the heart tells you what you truly desire.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19364" LastEditorUserId="19364" LastEditDate="2016-02-19T01:35:44.053" LastActivityDate="2016-02-19T01:35:44.053" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27632" PostTypeId="2" ParentId="27583" CreationDate="2016-02-19T00:50:15.073" Score="4" Body="&lt;p&gt;You see me all day&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We look for direction constantly&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm sharp&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Made up of points&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I show you your way&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Uses magnetism to provide a constant bearing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But you show me what to show you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; By aligning the dial and the viewport&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I exist, but not in your dimension&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Getting fuzzy here&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm made up of down&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Technically magnets point the opposite direction. The end painted red is magnetized the opposite of north, which is why it is attracted to north.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My older language is by rotation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Compasses once rotated&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But now through light&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; on your phone!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(though in the end it comes down to the same shocking tale)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; electromagnetism&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I usually look a certain way&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The way we're used to&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I change at certain times&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Magnetic north moves, and sometimes the magnetic field flips 180º&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Into myself, my thin sibling, into time,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; North to South, South to North&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and into ambiguity&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What is it anyway?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or something with which you grasp&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Grasping for guidance&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Magnetic Compass &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19365" LastActivityDate="2016-02-19T00:50:15.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27634" PostTypeId="2" ParentId="26437" CreationDate="2016-02-19T01:49:41.437" Score="2" Body="&lt;p&gt;Since I already gave away the checkmark, I'll reveal the other clues here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First Clue: As Daphne B said, the first three stanzas describe a pizza, with the first of the couplet referring to the ingredient and the second the preparation:&lt;br&gt;&#xA; -Wheat Fields/Ground into flour once the grain is dried&lt;br&gt;&#xA; -Tomatoes/Spread as sauce in the middle&lt;br&gt;&#xA; -Cheese/Shredded on top and cooked&lt;br&gt;&#xA;&lt;br&gt;&#xA; Second Clue: The 7th line tells you that it is not a people, even though it sounds that way.  It also tells you to ask &quot;why.&quot;  If you take every letter appearing after a &quot;Y&quot; in the poem, you will find the letters spell &lt;em&gt;p-i-z-z-a&lt;/em&gt;.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Third Clue: 8th line.  As Daphne observed, 'pie' = 'pizza pie.'  Pretty easy.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Last Clue: The title has 2 clues, one tangential the other very obscure.  As Daphne said, it could refer to a pizza stone.  Also, the name &quot;Peter&quot; means Rock.  So what &quot;The Rock glimpsed&quot; is what &quot;Pete saw.&quot;  Pete-Saw.  Pizza.  Get it?  :P (Ok.  That last one is a real groaner, but I don't consider clues in the title to be necessary.  They are just for a little extra fun.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thank you to Daphne B for the excellent answer, GentlePurple Rain for biting on the mislead, f'' for the correction, Manshu for the edit, and for all the viewers who took the time to think about it.  Happy Puzzling.&lt;/p&gt;&#xA;" OwnerUserId="19128" LastActivityDate="2016-02-19T01:49:41.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27636" PostTypeId="2" ParentId="27607" CreationDate="2016-02-19T02:41:28.693" Score="4" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An &lt;a href=&quot;https://en.wikipedia.org/wiki/American_woodcock&quot; rel=&quot;nofollow&quot;&gt;American Woodcock&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's brown and has the slowest recorded flying speed of any bird.&lt;br&gt;&#xA; A tree is made of wood, and wood is a slang term for an erection, which ties in to the 2nd half of the name of the bird.&lt;br&gt;&#xA; A cock is slang for a penis, which a (naked) man would see when looking down from the top of a tree.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17297" LastEditorUserId="1766" LastEditDate="2016-02-19T09:50:15.603" LastActivityDate="2016-02-19T09:50:15.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27637" PostTypeId="1" AcceptedAnswerId="27913" CreationDate="2016-02-19T03:40:43.913" Score="5" ViewCount="1905" Body="&lt;h1&gt;A Ninja&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I speak through shadows,&lt;br&gt;&#xA;in cryptic tongues,&lt;br&gt;&#xA;and cryptic tongues only,&lt;br&gt;&#xA;to anyone who wants to know —&lt;br&gt;&#xA;but only if I have been told to say.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think not myself;&lt;br&gt;&#xA;offer no discretion of my own,&lt;br&gt;&#xA;but never trust the word of one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I was old,&lt;br&gt;&#xA;I would be far too clumsy, far too wrinkled,&lt;br&gt;&#xA;to hide among others without sight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I am young so, if you wished,&lt;br&gt;&#xA;I could hide in the smallest crevice&lt;br&gt;&#xA;or a fold in the world, hidden from sight&lt;br&gt;&#xA;among the unsavory sort.&lt;br&gt;&#xA;It is that I am young that I am so much easier to find.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My greatest opponent is he who imposes nothing,&lt;br&gt;&#xA;who requires nothing,&lt;br&gt;&#xA;enforces nothing,&lt;br&gt;&#xA;and is far more stealthy than me.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Than I, who weighs upon the world&lt;br&gt;&#xA;only barely more than a shadow&lt;br&gt;&#xA;as I roam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do know myself,&lt;br&gt;&#xA;know more of myself than you do,&lt;br&gt;&#xA;know more of the world than you do,&lt;br&gt;&#xA;and it is not my style to ask questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, I shall make one exception today:&lt;br&gt;&#xA;What am I?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Clues:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's not a ninja.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first line is really hard to guess, though it is almost a literal truth. Don't get bogged down by it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I fought my opponent once, or one of its major incarnations. It took a day off my life, but I — no, &lt;em&gt;we&lt;/em&gt; — won.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Time for more clues. One per paragraph.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cryptic tongues only, but only as of last year.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; [N]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If I was old, I would probably be having problems with my spine.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You could wish it so, but you probably don't.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; FIGHT THE MAN!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I can do what only old fat men with long beards were thought capable of.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I also know more about the Northern Plains Railroad than you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This tells you what kind of riddle it is.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Ok, some more clues. Getting close.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/questions/27637/riddle-of-a-ninja#comment82161_27713&quot;&gt;This comment&lt;/a&gt; should get you started. Think big and comprehensive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; NN(YN)YYNY(Y?)N.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer has one obvious, canonical name that is nine letters long, seven unique.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9960" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2020-04-28T17:18:42.870" Title="Riddle of a Ninja" Tags="|riddle|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27638" PostTypeId="2" ParentId="27637" CreationDate="2016-02-19T04:06:13.493" Score="3" Body="&lt;p&gt;&lt;em&gt;&lt;h1&gt;This answer is incorrect (my achievement)...confirmed by OP.&lt;/h1&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think Ninja is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Visible Light&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I speak through shadows,&lt;br&gt;&#xA;in cryptic tongues,&lt;br&gt;&#xA;and cryptic tongues only,&lt;br&gt;&#xA;to anyone who wants to know —&lt;br&gt;&#xA;but only if I have been told to say.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Shadow confirms the light. Cryptic tongue means that we can talk in code language with the help of it using torch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think not myself;&lt;br&gt;&#xA;offer no discretion of my own,&lt;br&gt;&#xA;but never trust the word of one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yup... the frequency defines the energy of light. And it is the property of the object that produces the light, not a property of light itself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I was old,&lt;br&gt;&#xA;I would be far too clumsy, far too wrinkled,&lt;br&gt;&#xA;to hide among others without sight.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.space.com/25732-redshift-blueshift.html&quot; rel=&quot;nofollow&quot;&gt;Redshift and Blue shift&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I am young so, if you wished,&lt;br&gt;&#xA;I could hide in the smallest crevice&lt;br&gt;&#xA;or a fold in the world, hidden from sight&lt;br&gt;&#xA;among the unsavory sort.&lt;br&gt;&#xA;It is that I am young that I am so much easier to find.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It says that even laser light (or any small thing) can produce light. And the whole world is covered with it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My greatest opponent is he who imposes nothing,&lt;br&gt;&#xA;who requires nothing,&lt;br&gt;&#xA;enforces nothing,&lt;br&gt;&#xA;and is far more stealthy than me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dark. Stealthy might mean the dark matter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Than I, who weighs upon the world&lt;br&gt;&#xA;only barely more than a shadow&lt;br&gt;&#xA;as I roam.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The pressure applied by light might make you think that light has weight (a lil' bit more than a shadow, almost negligible pressure and weight). &lt;a href=&quot;https://www.youtube.com/watch?v=Do1lm9IevYE&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is something to convince you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I do know myself,&lt;br&gt;&#xA;know more of myself than you do,&lt;br&gt;&#xA;know more of the world than you do,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yup.. we can only see the observable universe and light has seen more than that...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-02-19T05:04:16.970" LastActivityDate="2016-02-19T05:04:16.970" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="27640" PostTypeId="1" AcceptedAnswerId="27649" CreationDate="2016-02-19T08:00:48.853" Score="12" ViewCount="1400" Body="&lt;p&gt;How do I reconstruct the following exact long division in which the digits that have been replaced by letters except for the middle digit - the '8' - in the quotient. I've never seen one of these before and I am trying to solve gps coordinates using this problem. GPS coordinates are going to be: &#xA;N 38 4A.BC(D+E) W 121 0F.GH(C+G) ...so need to solve the problem so I can fill in A,B,C,D,E,F,G,H. Can someone please help. I've been banging my head against the wall for 2 days trying to figure where to start. Hopefully the JPG division problem posts below from an image upload.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/CvuJP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CvuJP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19382" LastEditorDisplayName="user88" LastEditDate="2016-07-31T16:37:14.837" LastActivityDate="2016-07-31T16:37:14.837" Title="Reconstructing long division problems" Tags="|geocaching|missing-digits|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27641" PostTypeId="2" ParentId="10559" CreationDate="2016-02-19T08:37:37.587" Score="3" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Charon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes we play catch;&#xA;Though you'd hope it more like fetch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd be the second; an honorable mention.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;We&quot; refers to the pair Charon and Pluto. From earth, they sometimes appear merged as a single blob, and sometimes separated by a small distance. Charon is the smaller among the two, and was discovered later, hence &quot;second&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Indeed it'd be, as if I've spent my life in detention.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Charon was unknown until 1978&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'd know where, if you found the tether;&#xA;Around about which, he and I dance together.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;he&quot; refers to the Sun. Charon orbits the Sun, as if tied to him with a tether&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cliches away; your usual meeting need not be so verbose.&#xA;PROPERLY told, I don't even have to come close.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Recently New Horizons came very close to Charon, and this created a lot of press coverage for Charon. However &quot;very close&quot; was 27,000 Kms!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Intrinsic you'd have a gut full of fear;&#xA;Ladies and gentlemen, boys and girls... Dying time's here!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The name Charon refers to the ferryman of the dead in Greek mythology&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Given me a name;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Charon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then please explain:&#xA;Where, when and what is,&#xA;That which I'll do,&#xA;Should you ever find out:&#xA;Me, I am true.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If I find out who you are, you will start appearing on search results of this site, I guess :P&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;About the hint :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Charon did not (before posting my answer) appear in any question or answer on Puzzling.SE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Extra Credits :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;For a while my imaginary sister, she had stolen my thunder. But now by the wise, she too is considered yet another blunder. What's her name? What can the wise see that we cannot?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sister is Pluto. It would seem that Pluto is more widely-known than Charon. However Pluto was recently demoted from the status of a  planet. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19380" LastEditorUserId="19380" LastEditDate="2016-02-19T08:51:24.123" LastActivityDate="2016-02-19T08:51:24.123" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27642" PostTypeId="2" ParentId="27588" CreationDate="2016-02-19T09:05:03.390" Score="2" Body="&lt;p&gt;The next sequence is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TTTTTT  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letters start as 'I' then become 'T' (I am not sure why) after that 'L' which could be seen as the beginning of a Turn counter clockwise, then comes 'U' which is mid turn and then 'J' which is the end of the turn. Next would be 'T' since it lapsed back again.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bonus:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Given that logic the sequence after that should be 'I' again&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I highly doubt this is the correct answer (since I am still missing something), but I'll still try since I am really interested in the correct solution&lt;/p&gt;&#xA;" OwnerUserId="19299" LastActivityDate="2016-02-19T09:05:03.390" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27643" PostTypeId="1" AcceptedAnswerId="27858" CreationDate="2016-02-19T10:53:18.490" Score="14" ViewCount="538" Body="&lt;p&gt;Solve this problem by solving it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Follow back, forwards.&lt;br&gt;&#xA;  Words.&lt;br&gt;&#xA;  Hare and the tortoise jumping skyward repeatedly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jump, happy word! Juggle by finger length.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastEditorUserId="4050" LastEditDate="2016-02-23T20:13:12.530" LastActivityDate="2016-02-24T00:33:56.920" Title="Find the Hidden Message: Follow back, forwards" Tags="|steganography|" AnswerCount="2" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="27644" PostTypeId="2" ParentId="27640" CreationDate="2016-02-19T11:08:14.350" Score="4" Body="&lt;p&gt;Solution:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A = 1, B = 0, C = 4, D = 2, E = 6, F = 8, G = 1, H = 2, J = 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Dividing the puzzle up into sections:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/CaNyo.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some reasoning (more to come when I'm back home with a scanner):  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;We know from the fact that the division is done in 3 steps rather than 5 that the solution is of the form F080X.&lt;br&gt;&#xA;8 * AXC is a 3 digit number. This means that &lt;strong&gt;A = 1&lt;/strong&gt;&lt;br&gt;&#xA;Sum 2, GXXX - XXH has a 2-digit solution. G has a maximum value of 1&lt;br&gt;&#xA;Similarly, Sum 1, XXBD - XXJ has a 2-digit solution; assuming no leading zero, this makes the first X equal to 1 (and the second to 0). This makes XXJ into 9XJ, making F either 8 or 9   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assuming no leading zeroes:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;G=1&lt;/strong&gt; This makes sum 2 into 10XX - 9XH = XX. &#xA;As sum 2 starts with '10', the solution to sum 1 is also 10. This means D and J must be equal. It also means that &lt;strong&gt;B = 0&lt;/strong&gt; and tells us the value of all the 'X' symbols. Sum 1 is now 100D - 99D = 10. If F is 9, D and J would have to be either 9 or 0, making C either 1 or 0. This would force G to be 0 for sum 2 to hold (as XXH would be less than 900), so &lt;strong&gt;F=8&lt;/strong&gt;. This also means that H is equal to J.&lt;br&gt;&#xA;Brute-forcing it slightly, the only number between 990 and 999 that has 8 as a factor is 992, giving us &lt;strong&gt;D=2&lt;/strong&gt;, &lt;strong&gt;J=2&lt;/strong&gt;, and &lt;strong&gt;C=4&lt;/strong&gt; (also the divisor AXC = 124).  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The rest is trivial-ish:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;We know that AXC is less than 1000 when multiplied by 8. This means the last digit of the solution to the division must be 9, giving us 80809. Since I got lazy when it came to determining the value of AXC, the rest is easy. (though it could be done long-hand)&lt;br&gt;&#xA;124*80809 gives you 10020316, and simply carrying out the long division gives you the remaining letters.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11717" LastEditorUserId="11717" LastEditDate="2016-02-19T13:48:25.110" LastActivityDate="2016-02-19T13:48:25.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27647" PostTypeId="2" ParentId="27598" CreationDate="2016-02-19T12:04:17.463" Score="6" Body="&lt;p&gt;My guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Lawn Chair is alone because its previous occupant was squashed by a tree.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My thinking:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A drink I do take,&#xA;to calm all my shakes,&#xA;and heave a sigh of relief.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'd like to think of this as a rainstorm filling and settling a lawn chair buffeted by the wind.  Alternatively, 'the shakes' could gruesomely refer to moment the owner violently had a tree dropped on them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The moon dims and fades,&#xA;then once again shines,&#xA;glimmering on that which fells trees.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The chair can still see the chainsaw, presumably dropped after the accident occurred.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A little flutter,&#xA;and now I do stutter,&#xA;my tongue swollen red, green and blue.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The best I can do here is the tongue is the fabric part of the chair.  The red = the blood stains, the green = grass/leaf stains, and the blue = maybe it was a blue chair originally?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Good for the skin,&#xA;making it pearly and new,&#xA;as long as you pay heed&#xA;to whispers in the breeze,&#xA;that humming sound that drives you mad,&#xA;and the reason I am here alone.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sitting in the shade of a huge tree is a great way to avoid sunburn, provided you make sure that the annoying humming noise isn't actually your neighbour cutting down the tree with a chainsaw.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19390" LastActivityDate="2016-02-19T12:04:17.463" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27648" PostTypeId="2" ParentId="27588" CreationDate="2016-02-19T12:12:59.633" Score="6" Body="&lt;p&gt;This is not really an answer but this might be the direction the OP is looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say each line out loud like&lt;br&gt;&#xA;IIIIIIIII = Is = ice&lt;br&gt;&#xA;TTTTTT = Ts = teas&lt;br&gt;&#xA;LLLLLLL = Ls = Els&lt;br&gt;&#xA;UUUUU = Us = Yous&lt;br&gt;&#xA;JJJJJJJ = Js = Jays  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we need to finish a phrase that goes something like &quot;Ice tea sells you sjay...&quot; but I can't find a solution.&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2016-02-19T12:12:59.633" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27649" PostTypeId="2" ParentId="27640" CreationDate="2016-02-19T12:31:50.627" Score="2" Body="&lt;p&gt;I ran a computer program and I was wondering why it didn't find a solution. It's because some numbers presented as 4-digit number are in fact 3-digit numbers.&lt;br&gt;&#xA;So I ran it again and it found 2350 solutions!&lt;br&gt;&#xA;20 solutions have XXBDXXXX mod AXC = 0. (The problem didn't say it had to be 0)&lt;br&gt;&#xA;Here's all 20 solutions with mod 0:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;124 * 80800 = 10019200 (A,B,C,D,E,F,G,H = 1,0,4,1,0,8,0,2)&#xA;124 * 80801 = 10019324 (A,B,C,D,E,F,G,H = 1,0,4,1,4,8,0,2)&#xA;124 * 80802 = 10019448 (A,B,C,D,E,F,G,H = 1,0,4,1,8,8,0,2)&#xA;124 * 80803 = 10019572 (A,B,C,D,E,F,G,H = 1,0,4,1,2,8,0,2)&#xA;124 * 80804 = 10019696 (A,B,C,D,E,F,G,H = 1,0,4,1,6,8,0,2)&#xA;124 * 80805 = 10019820 (A,B,C,D,E,F,G,H = 1,0,4,1,0,8,0,2)&#xA;124 * 80806 = 10019944 (A,B,C,D,E,F,G,H = 1,0,4,1,4,8,0,2)&#xA;124 * 80807 = 10020068 (A,B,C,D,E,F,G,H = 1,0,4,2,8,8,1,2)&#xA;124 * 80808 = 10020192 (A,B,C,D,E,F,G,H = 1,0,4,2,2,8,1,2)&#xA;124 * 80809 = 10020316 (A,B,C,D,E,F,G,H = 1,0,4,2,6,8,1,2)&#xA;111 * 90800 = 10078800 (A,B,C,D,E,F,G,H = 1,0,1,7,0,9,0,8)&#xA;111 * 90801 = 10078911 (A,B,C,D,E,F,G,H = 1,0,1,7,1,9,0,8)&#xA;111 * 90802 = 10079022 (A,B,C,D,E,F,G,H = 1,0,1,7,2,9,0,8)&#xA;111 * 90803 = 10079133 (A,B,C,D,E,F,G,H = 1,0,1,7,3,9,0,8)&#xA;111 * 90804 = 10079244 (A,B,C,D,E,F,G,H = 1,0,1,7,4,9,0,8)&#xA;111 * 90805 = 10079355 (A,B,C,D,E,F,G,H = 1,0,1,7,5,9,0,8)&#xA;111 * 90806 = 10079466 (A,B,C,D,E,F,G,H = 1,0,1,7,6,9,0,8)&#xA;111 * 90807 = 10079577 (A,B,C,D,E,F,G,H = 1,0,1,7,7,9,0,8)&#xA;111 * 90808 = 10079688 (A,B,C,D,E,F,G,H = 1,0,1,7,8,9,0,8)&#xA;111 * 90809 = 10079799 (A,B,C,D,E,F,G,H = 1,0,1,7,9,9,0,8)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Edit: WAIT! HOLD ON!&#xA;I made a small error. I double checked and there are 124 solutions that correctly result in a long division without requiring there be 3-digit numbers represented by a 4-digit number.&lt;br&gt;&#xA;Only 1 of those solutions has XXBDXXXX mod AXC = 0:&lt;br&gt;&#xA;124 * 80809 = 10020316 (A,B,C,D,E,F,G,H = 1,0,4,2,6,8,1,2)&lt;br&gt;&#xA;In all 124 solutions, the A..H are the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 2: I iterated over the big number (XXBDXXXX), which was stupid. If I iterate over AXC and FX8XX, I find only the above solution (124 * 80809) &lt;a href=&quot;http://ideone.com/iMXJCk&quot; rel=&quot;nofollow&quot;&gt;run it&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1598" LastEditorUserId="1598" LastEditDate="2016-02-19T14:45:29.330" LastActivityDate="2016-02-19T14:45:29.330" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="27650" PostTypeId="2" ParentId="5788" CreationDate="2016-02-19T13:30:56.963" Score="2" Body="&lt;p&gt;I know this is wrong, but it's oh-so-right.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is searching for letters exactly?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;What&quot; is fo[u]r letters exactly.&lt;br&gt;&#xA; And &lt;em&gt;for&lt;/em&gt; letters. (Erm, what? Exactly.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What do we get when using letters literally?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What we get when using letters literally.&lt;br&gt;&#xA; Do we get &quot;what&quot; when using letters literally?&lt;br&gt;&#xA; We do get &quot;what&quot; when using letters literally.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What does wordplay finally result in?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Evidently someone goes &quot;WHAT?&quot; because your wordplay confounds them.&lt;br&gt;&#xA; This wordplay also ends in &quot;What?&quot;, since that's the answer.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And what is a metaphorical series?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;What&quot; occurred four times in a series of metaphors. When? Just now.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a riddle&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;This&quot; is certainly a riddle, since &quot;What?&quot; is the answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9960" LastActivityDate="2016-02-19T13:30:56.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27651" PostTypeId="2" ParentId="27610" CreationDate="2016-02-19T13:52:05.527" Score="0" Body="&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Brain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Every day, people press me for advice, which I offer honestly. I never lie.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You think objectively, and you will just get meaningful and realistic ideas yourself. In your depth, you always know the truth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Often, they don't listen.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; But you still refuse the listen! You are trying to avoid your thoughts and try to follow your heart (strong reference to relationships,etc)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes they become scared of me and stop asking my opinion.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Brain is scumbag, so sometimes it may lead you doing stupid mistakes including CRIME or SUICIDE! This is the time you should stop thinking.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But usually they come back in the end. After all, I'm the only reliable source of information on this subject.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ih the end, brain is all you've got. It is your sanity, it is you. You can only trust yourself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18685" LastActivityDate="2016-02-19T13:52:05.527" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27652" PostTypeId="2" ParentId="27640" CreationDate="2016-02-19T13:52:58.783" Score="5" Body="&lt;h1&gt;Answer&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is a unique solution for the relevant part of the division:&lt;pre&gt;              8 0 8 0 9&lt;br&gt;      +----------------&lt;br&gt;1 2 4 | 1 0 0 2 0 X X X&lt;br&gt;          9 9 2&lt;br&gt;        -------&lt;br&gt;            1 0 0 X&lt;br&gt;              9 9 2&lt;br&gt;            -------&lt;br&gt;                1 X X X&lt;br&gt;                1 1 1 6&lt;br&gt;                -------&lt;/pre&gt;The remaining numbers cannot be found, because the remainder of the division is not specified. However, they are not required to find the values for interesting letters:&lt;pre&gt;A=1, B=0, C=4, D=2, E=6, F=8, G=1, H=2, J=2&lt;/pre&gt;The final coordinates point to a house in Folsom, CA.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;Explanation&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;First I will replace the $X$'s with different lowercase letters, so its easier to refer to them.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;              F a 8 b c&#xA;      +----------------&#xA;A d C | e f B D g h i j&#xA;          k m J&#xA;        -------&#xA;            G n g h&#xA;              o p H&#xA;            -------&#xA;                q r i j&#xA;                s t u E&#xA;                -------&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The first thing we can notice is, that there are no subtraction steps for $a$ and $b$. This means they are both $0$.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;              F 0 8 0 c&#xA;      +----------------&#xA;A d C | e f B D g h i j&#xA;          k m J&#xA;        -------&#xA;            G n g h&#xA;              o p H&#xA;            -------&#xA;                q r i j&#xA;                s t u E&#xA;                -------&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Next look at the first subtraction step: $efBD - kmJ = Gn$. The only possible value for $k$ is $9$. If it was lower, it wouldn't be possible to get a 4 digit number by adding $kmJ$ and $Gn$. If $k$ is $9$, then $ef$ must be $10$. The same applies for the second subtraction step: $Gngh - opH = qr$.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;              F 0 8 0 c&#xA;      +----------------&#xA;A d C | 1 0 B D g h i j&#xA;          9 m J&#xA;        -------&#xA;            1 0 g h&#xA;              9 p H&#xA;            -------&#xA;                q r i j&#xA;                s t u E&#xA;                -------&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now look at the product: $AdC * 8 = 9pH$. It's obvious that $A$ must be $1$. Otherwise the product would be too big to fit in 3 digits. Knowing this we can also deduce that in the third product ($1dC * c = stuE$), the letter $s$ is $1$ as well.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;              F 0 8 0 c&#xA;      +----------------&#xA;1 d C | 1 0 B D g h i j&#xA;          9 m J&#xA;        -------&#xA;            1 0 g h&#xA;              9 p H&#xA;            -------&#xA;                q r i j&#xA;                1 t u E&#xA;                -------&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Look again at the product $1dC * 8= 9pH$. We can see that $1dC$ must be in the range $113..124$ to get a 3 digit product starting with $9$. Knowing that range and looking at the second product $1dC * F = 9ph$ we can see that $F$ must be $8$ (because $113*9=1017$ and $124*7=868$). This also means that $9mJ$ and $9pH$ are equal.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;              8 0 8 0 c&#xA;      +----------------&#xA;1 d C | 1 0 B D g h i j&#xA;          9 m J&#xA;        -------&#xA;            1 0 g h&#xA;              9 m J&#xA;            -------&#xA;                q r i j&#xA;                1 t u E&#xA;                -------&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Knowing the range for $1dC$ and looking at the product $1dC * c = 1tuE$ we can also deduce, that $c$ must be $9$ (because $124*8=992$). As $123 * 80809 = 9939507$ we know that $1dC$ must be bigger than $123$, and there is only $124$ left in the possible range.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;              8 0 8 0 9&#xA;      +----------------&#xA;1 2 4 | 1 0 B D g h i j&#xA;          9 m J&#xA;        -------&#xA;            1 0 g h&#xA;              9 m J&#xA;            -------&#xA;                q r i j&#xA;                1 t u E&#xA;                -------&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now we can determine the result for the third product $124 * 9 = 1116$. We also know the values for $BDg$ because the dividend must be in the range $10020316$ (assuming remainder $0$) to $10020439$ (assuming remainder 123).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;              8 0 8 0 9&#xA;      +----------------&#xA;1 2 4 | 1 0 0 2 0 h i j&#xA;          9 m J&#xA;        -------&#xA;            1 0 g h&#xA;              9 m J&#xA;            -------&#xA;                q r i j&#xA;                1 1 1 6&#xA;                -------&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Using $1002 - 9mJ = 10$ we can determine that $mJ$ is $92$.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;              8 0 8 0 9&#xA;      +----------------&#xA;1 2 4 | 1 0 0 2 0 h i j&#xA;          9 9 2&#xA;        -------&#xA;            1 0 0 h&#xA;              9 9 2&#xA;            -------&#xA;                q r i j&#xA;                1 1 1 6&#xA;                -------&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We know that $h$ is either $3$ or $4$, so $q$ must be $1$ giving the solution at the top of the answer. This is the last digit which I could find without assuming a reminder.&lt;/p&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2016-02-19T16:04:43.837" LastActivityDate="2016-02-19T16:04:43.837" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27653" PostTypeId="1" CreationDate="2016-02-19T14:14:18.170" Score="-2" ViewCount="221" Body="&lt;p&gt;When I'm needed, I wish I wasn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I'm not needed I won't have what to do&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do I do, what am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clarificaton&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Life still occurs, birth will ultimately result in death&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt; (Due to being put on hold for being 'too broad' I decided to ease this with the following)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Think of someone who only has a job as long as people die unnatural deaths&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm the one that has to think about what the coroner tells me &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18816" LastEditorUserId="18816" LastEditDate="2016-02-20T11:50:46.373" LastActivityDate="2016-02-25T17:23:14.077" Title="I just wish I wasn't needed. What am I?" Tags="|riddle|" AnswerCount="3" CommentCount="22" ContentLicense="CC BY-SA 3.0" />
  <row Id="27654" PostTypeId="2" ParentId="27653" CreationDate="2016-02-19T14:34:40.773" Score="2" Body="&lt;p&gt;Could you be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A medical examiner/coroner? Since someone has to die for his services to be needed (performing autopsies), one would hope he would not want to be needed. Also, if no one dies, he has nothing to do.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16874" LastActivityDate="2016-02-19T14:34:40.773" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27655" PostTypeId="2" ParentId="27653" CreationDate="2016-02-19T14:35:02.077" Score="1" Body="&lt;p&gt;What do I do, what am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are an &quot;undertaker&quot;, who takes care of dead people&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I'm needed, I wish I wasn't.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No one wants to take care of dead people because seeing them is sad or there can be other self reasons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I'm not needed I won't have what to do&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When there are no bodies, you are not needed and won't have any job to do&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18685" LastActivityDate="2016-02-19T14:35:02.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27656" PostTypeId="1" AcceptedAnswerId="27657" CreationDate="2016-02-19T14:43:49.577" Score="3" ViewCount="156" Body="&lt;p&gt;A secret word that Hera found&lt;br&gt;&#xA;Pan scattered, just for fun.&lt;br&gt;&#xA;Your quest: restore the severed parts&lt;br&gt;&#xA;Then celebrate. You're done.&lt;br&gt;&#xA;But should you seek the letters here&lt;br&gt;&#xA;By Zeus, you'll not get one!  &lt;/p&gt;&#xA;" OwnerUserId="18924" LastEditorUserId="4421" LastEditDate="2016-02-19T20:32:34.330" LastActivityDate="2016-02-19T20:32:34.330" Title="The Gods Are Word-Crazy: Condensed and Corrected" Tags="|wordplay|poetry|anagram|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="27657" PostTypeId="2" ParentId="27656" CreationDate="2016-02-19T14:52:44.093" Score="6" Body="&lt;p&gt;This version made the hint a little more obvious. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are three letters that don't appear at all in the poem:  &lt;strong&gt;I, M, X&lt;/strong&gt;.&lt;br&gt;&#xA;&lt;br&gt;&#xA;  Anagramming these, we get &lt;strong&gt;MIX&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="4421" LastEditDate="2016-02-19T15:48:56.383" LastActivityDate="2016-02-19T15:48:56.383" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27658" PostTypeId="2" ParentId="27653" CreationDate="2016-02-19T15:03:21.403" Score="5" Body="&lt;p&gt;When I'm needed, I wish I wasn't.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not happy about killing/beheading/hanging people   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I'm not needed I won't have what to do&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; But, If i won't have to execute anyone, I'll be jobless.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What do I do, what am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Executioner&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19324" LastEditorUserId="16874" LastEditDate="2016-02-19T15:04:03.110" LastActivityDate="2016-02-19T15:04:03.110" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27659" PostTypeId="2" ParentId="27617" CreationDate="2016-02-19T15:15:09.607" Score="16" Body="&lt;p&gt;Are you the word &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; djinni?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It has two i's with two neighbors (letters) between then.  If you remove the right i you have&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the singular, djinn,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which makes sense since one &quot;neighbor&quot; (different than above use) left.  If you remove the left i you have&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; djnni,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which isn't a word at all.&lt;/p&gt;&#xA;" OwnerUserId="12795" LastActivityDate="2016-02-19T15:15:09.607" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27660" PostTypeId="2" ParentId="27580" CreationDate="2016-02-19T15:30:51.110" Score="4" Body="&lt;p&gt;We are given that the number $b_i$ announced by girl $i$ is the average of the numbers $x_{i^+}$ and $x_{i^-}$ chosen by the neighboring girls $i^+$ and $i^-$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2 b_i = x_{i^+} + x_{i^-},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{rcl}&#xA;i^+ &amp;amp; = &amp;amp; i + 1 - \left\lfloor \frac{i}{N} \right\rfloor N, \\&#xA;i^- &amp;amp; = &amp;amp; i - 1 + \left\lfloor \frac{N + 1 - i}{N} \right\rfloor N&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $N$ is the number of girls (that is, $i^+$ and $i^-$ loop around the interval $[1, N]$). This forms a linear equation system, which can be expressed on matrix form as $A\mathbf{x} = \mathbf{b}$ where the number at row $i$, column $j$ of the matrix $A$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_{i,j} = \begin{cases}&#xA;\frac{1}{2}, \ i-j \equiv_N 1,\\&#xA;\frac{1}{2}, \ j-i \equiv_N 1,\\&#xA;0  \ \ \mathrm{otherwise.}&#xA;\end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $N = 5$, $2A$ looks like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{ccccc}&#xA;    &amp;amp;0  &amp;amp;1  &amp;amp;0  &amp;amp;0  &amp;amp;1 \\&#xA;    &amp;amp;1  &amp;amp;0  &amp;amp;1  &amp;amp;0  &amp;amp;0 \\&#xA;    &amp;amp;0  &amp;amp;1  &amp;amp;0  &amp;amp;1  &amp;amp;0 \\&#xA;    &amp;amp;0  &amp;amp;0  &amp;amp;1  &amp;amp;0  &amp;amp;1 \\&#xA;    &amp;amp;1  &amp;amp;0  &amp;amp;0  &amp;amp;1  &amp;amp;0&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving this equation system for $\mathbf{x}$ with $N = 10$ and $\mathbf{b} = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]$ gives&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $x_6 = 1$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19401" LastEditorUserId="19401" LastEditDate="2016-02-22T10:15:40.647" LastActivityDate="2016-02-22T10:15:40.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27661" PostTypeId="1" AcceptedAnswerId="27671" CreationDate="2016-02-19T16:03:24.420" Score="13" ViewCount="750" Body="&lt;p&gt;Both $A$ and $B$ have numbered hats on their heads. $A$ and $B$ both cannot see his/her own hat, but they can see other one's hat. $A$ sees number on $B$'s hat as $5$ and $B$ sees $A$'s number as $4$. &#xA;Both $A$ and $B$ have been informed that $A$'s number is &lt;strong&gt;multiplication of two positive integers&lt;/strong&gt; and $B$'s number is &lt;strong&gt;addition of those two integers&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First $B$ is asked that whether he/she knows the two numbers on their hats? If the answer is &quot;No&quot;, Then the same question is asked from $A$. Likewise this question is repeatedly asked from both $A$ and $B$ until one says &quot;I know&quot;. Suppose that both $A$ and $B$ are ideal super genius. &#xA;Who will first say &quot;I know&quot; and At which question? &lt;/p&gt;&#xA;" OwnerUserId="19293" LastEditorUserId="10615" LastEditDate="2016-02-24T17:29:16.553" LastActivityDate="2016-02-24T17:29:16.553" Title="Who will find the number on their own hat first?" Tags="|logical-deduction|formation-of-numbers|meta-knowledge|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27663" PostTypeId="1" CreationDate="2016-02-19T16:42:23.350" Score="4" ViewCount="3625" Body="&lt;p&gt;A guard has to persuade 5 prisoners to stay inside a circle drawn on the ground. To discourage them, he can use his one-bullet gun. To always watch them, the guard stays outside the circle. What public threat can he announce to prevent any of the prisonners from escaping? &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The prisoners are brave (they try to escape if they think they have at least one chance in a billion).&lt;/li&gt;&#xA;&lt;li&gt;The prisoners are not suicidal (they stay if they are 100% sure that an escape attempt will kill them).&lt;/li&gt;&#xA;&lt;li&gt;The prisoners know that the gun has only one bullet.&lt;/li&gt;&#xA;&lt;li&gt;The prisoners know each other and can communicate privately.&lt;/li&gt;&#xA;&lt;li&gt;The prisoners can act simultaneously and collaborate on a common strategy if:&#xA;each of the collaborators staying thinks at least one prisoner will espcape, and each of the colloborators trying to escape thinks he/she has a chance.&lt;/li&gt;&#xA;&lt;li&gt;Most importantly, the prisoners are all rational and trust that the guard will behave in the way he announces.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A similar question has been closed as too broad and has been given an answer which is not completely satisfying. I have reformulated the question with more details about the hypothesis. Be as logical as the prisoners.&lt;/p&gt;&#xA;" OwnerUserId="19397" LastEditorUserId="19397" LastEditDate="2016-02-20T04:57:35.450" LastActivityDate="2016-02-20T05:16:05.237" Title="Discourage 5 prisoners from escaping with only 1 bullet" Tags="|logical-deduction|" AnswerCount="10" CommentCount="15" ClosedDate="2016-02-19T23:18:54.180" ContentLicense="CC BY-SA 3.0" />
  <row Id="27664" PostTypeId="2" ParentId="27663" CreationDate="2016-02-19T16:47:28.690" Score="9" Body="&lt;p&gt;Stated briefly:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The guard says &quot;I will shoot the first person who tries to escape.&quot;   No prisoner will move unless another prisoner has already started to move.  So they will all stay in place.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19128" LastActivityDate="2016-02-19T16:47:28.690" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27665" PostTypeId="2" ParentId="27663" CreationDate="2016-02-19T16:54:51.567" Score="4" Body="&lt;p&gt;Well, the obvious has already been stated, but let's suppose that the prisoners would, in that case, try to get someone else killed and then run (such as pushing them out of line or similar).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that case, they solution lies in a threat that will invalidate everyone's escape if triggered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here we go with a possibly slightly lateral option:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The guard handcuffs himself to each prisoner independently (there are 5, so maybe he could do 1 per arm / leg and one at the guard's neck). Then he threatens to kill himself if someone tries to escape. If the guard is dead, the prisoners (unless they are VERY strong) are constrained by a) moving together and b) carrying around the weight of a dead adult at least until finding tools to get rid of the cuffs, which likely means not being able to escape in time and also being accused of the guard's death, thus resulting in a worse sentence, likely death sentence, so escape is more likely to get them killed than succeding&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A slightly less violent option:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Same rationale, but chain them all together. The guard threatens to shoot the key away (by putting it into the chamber before shooting), rather than itself. Less effective (bc no body to carry around), but the guard lives.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14435" LastEditorUserId="14435" LastEditDate="2016-02-19T17:29:11.043" LastActivityDate="2016-02-19T17:29:11.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27666" PostTypeId="2" ParentId="27663" CreationDate="2016-02-19T17:16:21.310" Score="4" Body="&lt;p&gt;My attempt:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; State that &quot;I will shoot the first person who fails to get hold on the person in front of him/her.&quot;  Assume they formed a circle with hands placing on the shoulder on the person in front of them.  Each person will try their best to grab / hold the person in front of them thus no one will be able to escape.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="10037" LastEditDate="2016-02-19T17:27:20.520" LastActivityDate="2016-02-19T17:27:20.520" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="27667" PostTypeId="2" ParentId="27661" CreationDate="2016-02-19T17:23:56.040" Score="2" Body="&lt;p&gt;Presuming the stipulations I mentioned in my comment above (the two integers are distinct and we cannot have the same number on both hats):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; B gets it at the second question. B knows that his number is an addition of two of the factors of 4. This means the only possibilities are 4 and 5. Since we are assuming we can't have the same number on both hats, the only possibility is 5.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without restrictions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A gets on on round 3. A knows his possible numbers are 4 and 6. He passes on question #1. B knows his possibilities are 4 and 5. He does not have enough info to eliminate one, so he passes as well. Now, if A's number was 6, B would realize on round 2. In that case, B's possibilities would have been 5 and 7, and had it been 7, A would have got it on round 1 (7 is prime and only has 1 and 7 as factors. A's number could only be 8 in that case), since A did NOT, then the only choice left for B was 5. Since this was not the case, A knows his number is not 6, and therefore can only be 4.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16874" LastEditorUserId="16874" LastEditDate="2016-02-19T17:54:20.390" LastActivityDate="2016-02-19T17:54:20.390" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27668" PostTypeId="2" ParentId="27663" CreationDate="2016-02-19T17:30:34.297" Score="4" Body="&lt;h1&gt;First solution&lt;/h1&gt;&#xA;&lt;p&gt;I assume I have a bit of time before they coordinate anything.  So the first thing I do is:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &amp;quot;First person to talk or try to run gets shot!&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;That should keep them quiet and close by for a bit until they realize&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; if they all talk/run at the same time, I can't shoot anyone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So, while they are pondering that, I will then pick a prisoner and:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &amp;quot;Close your eyes and lie on your stomach or I will shoot you in two seconds if you don't comply!&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Repeat until all prisoners are&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;on their stomachs with their eyes closed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If I am quick enough and threatening enough, they will comply quickly without communicating to each other (via sign language?)  any type of coordinated attack/movement.&lt;/p&gt;&#xA;&lt;p&gt;Even if&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Some of them try to run while I am dealing with the others, I can call out to one of them running &amp;quot;You in the blue!  Down now or I shoot!&amp;quot;.  And repeat with all the ones running.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So long as they don't leave my effective hand gun range, I should be ok.&lt;/p&gt;&#xA;&lt;p&gt;Then, I will yell and berate them indefinitely, constantly reminding them that I can kill them at any time so they better not try anything funny.&lt;/p&gt;&#xA;&lt;h1&gt;Another solution:&lt;/h1&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &amp;quot;If I think any of you are trying to escape, I will close my eyes and spin around.  When I open my eyes, I will shoot the prisoner closest to my line of sight.&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;None of them will want to try to escape because they may be the one targeted.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-02-19T17:48:10.020" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27669" PostTypeId="2" ParentId="27663" CreationDate="2016-02-19T17:31:13.603" Score="18" Body="&lt;p&gt;Simple rule similar to that of Solocutor, with a patch in case of simultaneous escapes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The guard publicly assigns a number to each prisoner and says, &quot;If any prisoners try to escape, I will kill the prisoner among them with the smallest number.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning is by induction:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The prisoner with the smallest number, call him/her 1, surely won't try to escape. The prisoner with the second-smallest number, call her/him 2, realizes that 1 cannot attempt anything. Trying something is thus surely suicidal for 2 also. And so on for all prisoners. Defining a total order is the only solution: if all try to escape the rule announced in the threat must designate a precise prisoner so that he/she is sure to die, call him/her 1. Reasoning similarly if all but 1 try to escape, the rule must designate a prisoner among the four others, and so on. Making sure that all prisoners know this ordering is then sufficient.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19397" LastEditorUserId="19397" LastEditDate="2016-02-20T05:16:05.237" LastActivityDate="2016-02-20T05:16:05.237" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27670" PostTypeId="2" ParentId="27661" CreationDate="2016-02-19T17:51:11.063" Score="4" Body="&lt;p&gt;Answer:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;B finds his number first.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;A&lt;/strong&gt; knows the two numbers are either &lt;em&gt;1 &amp;amp; 4&lt;/em&gt; or &lt;em&gt;2 &amp;amp; 3&lt;/em&gt;, because those are the only positive integers that add to 5.  So &lt;strong&gt;A&lt;/strong&gt; knows his number is either &lt;em&gt;4 or 6&lt;/em&gt;.&lt;br&gt;&#xA; &lt;strong&gt;B&lt;/strong&gt; knows the two numbers are either &lt;em&gt;1 &amp;amp; 4&lt;/em&gt; or &lt;em&gt;2 &amp;amp; 2&lt;/em&gt;, because those are the only positive integers whose product is 4.  So &lt;strong&gt;B&lt;/strong&gt; knows his number is either &lt;em&gt;5 or 4&lt;/em&gt;.&lt;br&gt;&#xA; If &lt;strong&gt;B&lt;/strong&gt; saw a 6, he would know that his number was either 7 (6 + 1) or 5 (2 +3).&lt;br&gt;&#xA; If &lt;strong&gt;A&lt;/strong&gt; saw 4, he would know that his number was 4 or 3.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the questioning begins:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1.  &lt;strong&gt;B&lt;/strong&gt; is asked if he knows his number.  He doesn't, so he says &quot;no.&quot;&lt;br&gt;&#xA; 2.  &lt;strong&gt;A&lt;/strong&gt; is asked.  &lt;strong&gt;A&lt;/strong&gt; knows that &lt;strong&gt;B&lt;/strong&gt; sees either 6, 4.  Neither of those alone gives &lt;strong&gt;B&lt;/strong&gt; enough information to know his number, so &lt;strong&gt;A&lt;/strong&gt; says &quot;no.&quot;&lt;br&gt;&#xA; 3.  &lt;strong&gt;B&lt;/strong&gt; is asked.  &lt;strong&gt;B&lt;/strong&gt; knows that if &lt;strong&gt;A&lt;/strong&gt; saw a 4, &lt;strong&gt;A&lt;/strong&gt; would know his own number was 3 or 4.  But if &lt;strong&gt;A&lt;/strong&gt;'s number had been 3, then &lt;strong&gt;B&lt;/strong&gt; could only be 4, and &lt;strong&gt;B&lt;/strong&gt; would have answered right away in the first round.  So &lt;strong&gt;B&lt;/strong&gt; knows that &lt;strong&gt;A&lt;/strong&gt; knows that &lt;strong&gt;A&lt;/strong&gt; is not a 3.  So &lt;strong&gt;B&lt;/strong&gt; he knows his number is not 4, so it must be 5.  &quot;Yes! It is 5!&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19128" LastEditorUserId="19128" LastEditDate="2016-02-19T22:35:36.193" LastActivityDate="2016-02-19T22:35:36.193" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27671" PostTypeId="2" ParentId="27661" CreationDate="2016-02-19T18:02:32.550" Score="15" Body="&lt;p&gt;Let's start with some ASCII art as a reference:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;          +-----+                    +-----+&#xA;          |  4  |                    |  5  |&#xA;          | mul |                    | add |&#xA;        --+-----+--                --+-----+--&#xA;          | ° ° |                    | ° ° |&#xA;          |  A  |                    |  B  |&#xA;           \___/                      \___/&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h1&gt;First question to B&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;$B$ sees a $4$ on $A$'s hat which is the product of 2 positive integers. There are 2 possibilities:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$1*4$ which means the sum on $B$'s hat would be $1+4=5$&lt;/li&gt;&#xA;&lt;li&gt;$2*2$ which means the sum on $B$'s hat would be $2+2=4$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There is no unique solution therefore $B$ must answer &quot;No&quot;.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;First question to A&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;$A$ sees a $5$ on $B$'s hat which is the sum of 2 positive integers. There are 2 possibilities:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$1+4$ which means the product on $A$'s hat would be $1*4=4$&lt;/li&gt;&#xA;&lt;li&gt;$2+3$ which means the product on $A$'s hat would be $2*3=6$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Again no unique solution, but $A$ is not done yet. He has to check the thoughts of $B$ during his first question for each of the possibilities.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;First question to A - Assuming A's number is a 4, what were B's thoughts?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;We saw the possibilities for that already, and know there was no unique solution. Therefore $A$ knows that $B$ would answer &quot;No&quot; and must assume that a $4$ on his hat is possible.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;First question to A - Assuming A's number is a 6, what were B's thoughts?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Assuming $B$ sees a $6$ on $A$'s hat which is the product of 2 positive integers, there are 2 possibilities again:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$1*6$ which means the sum on $B$'s hat would be $1+6=7$&lt;/li&gt;&#xA;&lt;li&gt;$2*3$ which means the sum on $B$'s hat would be $2+3=5$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Again no unique solution, so $B$ would have to answer &quot;No&quot; and $A$ must assume that a $6$ on his hat is possible as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that from $A$'s point of view his hat can be either $4$ or $6$. No unique solution, so he must answer &quot;No&quot;.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Second question to B&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The possibilities for the number on $B$'s hat from $B$'s point of view are still the same: $5$ or $4$. But now he also knows $A$'s answer for his first question, so he has to analyze his thoughts as well.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Second question to B - Assuming B's number is 5, what were A's thoughts?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Again we know that already, and $B$ will come to the same conclusion. That a $5$ on his hat is possible.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Second question to B - Assuming B's number is 4, what were A's thoughts?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Assuming $A$ sees a $4$ on $B$'s hat which is the sum of 2 positive integers, there are 2 possibilities again:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$1+3$ which means the product on $A$'s hat would be $1*3=3$&lt;/li&gt;&#xA;&lt;li&gt;$2+2$ which means the product on $A$'s hat would be $2*2=4$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;2 possibilities, but $B$ must also analyze $A$'s thoughts based on $B$'s first answer. If $A$ would assume a $3$ on his hat, there would be only one possibility for a product ($1*3$). Therefore $A$ would know that $B$'s answer would be &quot;Yes&quot; during the first question. As this was not the case $A$ knows there can be no $3$ on his hat. Therefore $A$ would know there is a $4$ on his hat, which is a contradiction, because $A$ answered &quot;No&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $B$ knows that there is no $4$ on his hat. There is only one possibility left ($5$) and he can answer &quot;Yes&quot;.&lt;/p&gt;&#xA;" OwnerUserId="15632" LastActivityDate="2016-02-19T18:02:32.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27672" PostTypeId="1" AcceptedAnswerId="27678" CreationDate="2016-02-19T18:52:54.737" Score="7" ViewCount="11992" Body="&lt;p&gt;I'm a new fan of cubing. I've recently bought a speed cube and for now I have only mastered the beginner's method and solve it in barely a minute and a half (which is way over what a speedcuber is supposed to do). &lt;br&gt;&#xA;So I'm wondering, is there any recommended method to learn after the beginner's method in order to improve my time ?&lt;/p&gt;&#xA;" OwnerUserId="16904" LastEditorUserId="7586" LastEditDate="2016-05-20T01:47:24.563" LastActivityDate="2023-11-02T17:56:43.170" Title="Best cubing algorithm to learn after beginner's" Tags="|rubiks-cube|speedsolving|" AnswerCount="3" CommentCount="4" ClosedDate="2023-11-02T21:41:41.237" ContentLicense="CC BY-SA 3.0" />
  <row Id="27673" PostTypeId="2" ParentId="27491" CreationDate="2016-02-19T18:55:39.627" Score="4" Body="&lt;p&gt;I'm going to guess it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Das Veilchen vom Potsdamer Platz (The Violet of Potsdam Square)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Admittedly this answer is the result of following your Google hint. However, there's merit to it:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;made on a square of 44&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It was released in Germany in 1936&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Preceded by purple&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The title begins with &quot;The Violet&quot; (aside: an interesting contradiction arises here -- it was released in English in 1939, but the German title uses &quot;Veilchen&quot; meaning the flower as opposed to &quot;Violett&quot; for the color)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;then they sold meals for fifty pfennig&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The partial &lt;a href=&quot;https://puzzling.stackexchange.com/a/27533&quot;&gt;answer by feelinferrety&lt;/a&gt; seems sufficient here&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17147" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2016-02-19T19:26:16.547" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27674" PostTypeId="2" ParentId="27619" CreationDate="2016-02-19T19:08:59.353" Score="2" Body="&lt;p&gt;I think he is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a Doctor, we know where to find him but are reluctant to go.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19411" LastEditorUserId="17250" LastEditDate="2016-02-19T19:34:53.110" LastActivityDate="2016-02-19T19:34:53.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27675" PostTypeId="2" ParentId="27663" CreationDate="2016-02-19T19:14:13.910" Score="2" Body="&lt;p&gt;Simple:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The guard tells them that there are 20 trained gunmen watching. If they hear a gunshot, they will execute everyone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2016-02-19T19:14:13.910" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="27676" PostTypeId="1" CreationDate="2016-02-19T19:42:53.230" Score="3" ViewCount="351" Body="&lt;p&gt;So, I have a cipher challenge, maybe, depending on how difficult I can make this for you guys. Hope you like it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need a word, it may even be a sentence; I'm not sure. All I know is that it's somewhere, so I'm giving this to you. I can only provide you with this: &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&lt;img src=&quot;https://i.stack.imgur.com/7plvT.png&quot; alt=&quot;8 KB on disk (6,1113 bytes)&quot;&gt;&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;EnCt2400eb758cb0d3079825471a346850f1b376a25bb400eb758cb0d3079825471a3KjyVUREhcAE&#xA;  xXt1vx1Y542SX1MMrzAybclRTmj1UgMgWJ76QPhJfGf+Qynb0dbgZAJaSIr2wucW+76/UVMJ0KginZ9k&#xA;  j9lT7kc+gkxwb8TQq5O53tEsEIwEmS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you find this word, it is very important to me, so you must tell me. Thank you.&lt;/p&gt;&#xA;" OwnerUserId="13949" LastEditorUserId="4421" LastEditDate="2016-02-19T20:29:04.927" LastActivityDate="2023-06-19T08:28:52.407" Title="This man needs help" Tags="|cipher|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27677" PostTypeId="2" ParentId="27580" CreationDate="2016-02-19T20:19:06.833" Score="4" Body="&lt;p&gt;Let $g_n$ be the number the $n$th girl picked, so that $(g_{n-1}+g_{n+1})/2=n$ for $1\le n\le 10$ (where indices wrap around the circle, so $g_0=g_{10})$. Then&#xA;$$&#xA;\begin{align}&#xA;g_6 &#xA;&amp;amp;= (g_6-g_4)/2 + (g_6+g_4)/2\\&#xA;&amp;amp;= (g_6+(g_8-g_8)+(-g_{10}+g_{10})+(g_2-g_2)-g_4)/2 + (g_6+g_4)/2\\&#xA;&amp;amp;= (g_6+g_8)/2 -(g_8+g_{10})/2+(g_{10}+g_2)/2-(g_2+g_4)/2+(g_6+g_4)/2\\&#xA;&amp;amp;= 7-9+1-3+5=\boxed{1}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2016-02-19T22:38:26.550" LastActivityDate="2016-02-19T22:38:26.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27678" PostTypeId="2" ParentId="27672" CreationDate="2016-02-19T20:39:18.090" Score="6" Body="&lt;p&gt;I agree with @mchoy25 that you can bring your time down much more. Things like single finger turning, properly lubricating the cube, and knowing what the back of the cube looks like without rotating the cube will improve your time much more than switching algorithms. That said, if you want to switch to more complicated methods then I'd look up F2L. Basically, from beginner to master you'll just be reducing the number of steps taken while massively increasing the number of possible algorithms to perform/memorize for any given step&lt;/p&gt;&#xA;" OwnerUserId="18103" LastActivityDate="2016-02-19T20:39:18.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27679" PostTypeId="2" ParentId="27663" CreationDate="2016-02-19T20:49:47.890" Score="-1" Body="&lt;p&gt;Simple:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The guard should say: &quot;If anyone tries to escape i will shoot a random person&quot;&#xA; No one will try to escape because of rule #2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19416" LastEditorUserId="17250" LastEditDate="2016-02-19T22:50:43.103" LastActivityDate="2016-02-19T22:50:43.103" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27680" PostTypeId="2" ParentId="27580" CreationDate="2016-02-19T21:12:45.757" Score="11" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have used simple matrix multiplication while solving the problem;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let say who said 1 is A, 2 is B, 3 is C, 4 is B ... J said 10; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{J+B}{2}=1 \ for \ A$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;an so on;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{bmatrix}&#xA; &amp;amp;  &amp;amp;  1&amp;amp;  2&amp;amp;  &amp;amp;  &amp;amp;   \\ &#xA;  &amp;amp;  &amp;amp;  A&amp;amp;  B&amp;amp;  &amp;amp;  &amp;amp;   \\ &#xA; 10&amp;amp;  J&amp;amp;  &amp;amp;  &amp;amp;  C&amp;amp;  3&amp;amp;   \\ &#xA;9&amp;amp;  I&amp;amp;  &amp;amp;  &amp;amp;  D&amp;amp;  4&amp;amp;  \\ &#xA;  8&amp;amp;  H&amp;amp;  &amp;amp;  &amp;amp;  E&amp;amp;  5&amp;amp;  \\ &#xA; &amp;amp;  &amp;amp;  G&amp;amp;  F&amp;amp;  &amp;amp;  &amp;amp;   \\ &#xA;  &amp;amp;  &amp;amp;  7&amp;amp;  6&amp;amp;  &amp;amp;  &amp;amp;  &#xA;\end{bmatrix} &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So If we put every equation as a matrix form;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{bmatrix}&#xA; 1&amp;amp;  0&amp;amp;  1&amp;amp;  0&amp;amp;  0&amp;amp;  0&amp;amp;  0&amp;amp;  0&amp;amp;  0&amp;amp;   0\\ &#xA; 0&amp;amp;  1&amp;amp;  0&amp;amp;  1&amp;amp;  0&amp;amp;  0&amp;amp;  0&amp;amp;  0&amp;amp;  0&amp;amp;   0\\ &#xA; 0&amp;amp;  0&amp;amp;  1&amp;amp;  0&amp;amp;  1&amp;amp;  0&amp;amp;  0&amp;amp;  0&amp;amp;  0&amp;amp;   0\\ &#xA; 0&amp;amp;  0&amp;amp;  0&amp;amp;  1&amp;amp;  0&amp;amp;  1&amp;amp;  0&amp;amp;  0&amp;amp;  0&amp;amp;   0\\ &#xA; 0&amp;amp;  0&amp;amp;  0&amp;amp;  0&amp;amp;  1&amp;amp;  0&amp;amp;  1&amp;amp;  0&amp;amp;  0&amp;amp;   0\\ &#xA; 0&amp;amp;  0&amp;amp;  0&amp;amp;  0&amp;amp;  0&amp;amp;  1&amp;amp;  0&amp;amp;  1&amp;amp;  0&amp;amp;   0\\ &#xA; 0&amp;amp;  0&amp;amp;  0&amp;amp;  0&amp;amp;  0&amp;amp;  0&amp;amp;  1&amp;amp;  0&amp;amp;  1&amp;amp;   0\\ &#xA; 0&amp;amp;  0&amp;amp;  0&amp;amp;  0&amp;amp;  0&amp;amp;  0&amp;amp;  0&amp;amp;  1&amp;amp;  0&amp;amp;   1\\ &#xA; 1&amp;amp;  0&amp;amp;  0&amp;amp;  0&amp;amp;  0&amp;amp;  0&amp;amp;  0&amp;amp;  0&amp;amp;  1&amp;amp;   0\\ &#xA; 0&amp;amp;  1&amp;amp;  0&amp;amp;  0&amp;amp;  0&amp;amp;  0&amp;amp;  0&amp;amp;  0&amp;amp;  0&amp;amp;   1&#xA;\end{bmatrix}\cdot \begin{bmatrix}&#xA;J\\ &#xA;A\\ &#xA;B\\&#xA;C\\ &#xA;D\\ &#xA;E\\ &#xA;F\\ &#xA;G\\ &#xA;H\\ &#xA;I&#xA;\end{bmatrix}=\begin{bmatrix}&#xA;2\\ &#xA;4\\ &#xA;6\\ &#xA;8\\ &#xA;10\\ &#xA;12\\ &#xA;14\\ &#xA;16\\ &#xA;18\\ &#xA;20&#xA;\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we need to take inverse of the first matrix and multiply with the right hand side;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{bmatrix}&#xA;J (10)\\&#xA;A (1)\\ &#xA;B (2)\\ &#xA;C (3)\\ &#xA;D (4)\\ &#xA;E (5)\\ &#xA;F (6)\\ &#xA;G (7)\\ &#xA;H (8)\\ &#xA;I (9)&#xA;\end{bmatrix}=\begin{bmatrix}&#xA;5\\ &#xA;6\\ &#xA;-3\\ &#xA;-2\\ &#xA;9\\ &#xA;10\\ &#xA;1\\ &#xA;2\\ &#xA;13\\ &#xA;14&#xA;\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The one who announced 6 was F then the answer is 1. You may find the other numbers as above.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18514" LastActivityDate="2016-02-19T21:12:45.757" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27681" PostTypeId="2" ParentId="27663" CreationDate="2016-02-19T21:46:52.700" Score="0" Body="&lt;p&gt;My answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The guard tells the prisoners &quot;If any of you tries to escape, I will kill the man to his right.&quot; This way the prisoners are regulating themselves in their private communications. I think this works with rules 5 and 6.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19418" LastActivityDate="2016-02-19T21:46:52.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27683" PostTypeId="2" ParentId="27663" CreationDate="2016-02-19T23:13:33.580" Score="4" Body="&lt;p&gt;Tell each prisoner privately this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“If any prisoner escapes, I will kill YOU”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then once they are all together they won’t communicate this to each other, as that would be suicide because they don’t know what you told all the others. And all of them will try to keep each other for attempting to escape.&lt;/p&gt;&#xA;" OwnerUserId="19423" LastActivityDate="2016-02-19T23:13:33.580" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27684" PostTypeId="2" ParentId="27663" CreationDate="2016-02-19T23:20:15.493" Score="2" Body="&lt;p&gt;Handcuff them all together hand to foot in a circle. Tell them that the first person to try to move from their stop gets shot and you will kill them rest of them using the gun.&lt;/p&gt;&#xA;" OwnerUserId="19424" LastActivityDate="2016-02-19T23:20:15.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27686" PostTypeId="1" AcceptedAnswerId="27715" CreationDate="2016-02-19T23:36:35.117" Score="-5" ViewCount="1936" Body="&lt;p&gt;How do you make 82 with the numbers&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$4, 5, 6, 7, 8, 9, 0$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and 8 dots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Get creative! You can use any mathematical operation, and the dots can be used in any mathematical way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No ratios.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You must use every number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint #1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The dots are to be used in an unexpected way.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint #2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; How many ways can we use dots?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17970" LastEditorUserId="17970" LastEditDate="2016-02-20T04:04:28.063" LastActivityDate="2016-06-23T07:28:31.737" Title="Make 82 With Numbers and Dots" Tags="|mathematics|formation-of-numbers|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27687" PostTypeId="2" ParentId="27686" CreationDate="2016-02-19T23:59:10.907" Score="6" Body="&lt;p&gt;That's easy, if we remember our SATs...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 60 : 984 :: 5 : ?&lt;br&gt;&#xA;&lt;br&gt;&#xA; Thanks, analogies!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17389" LastActivityDate="2016-02-19T23:59:10.907" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="27688" PostTypeId="2" ParentId="27686" CreationDate="2016-02-20T00:15:41.870" Score="17" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$.7^0\cdot98.4\cdot.5\colon.6$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$0.7$ to the zeroth power, multiplied by $98.4$ and $0.5$, is $49.2$. The ratio of this number to $0.6$ is $82$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2016-02-20T00:15:41.870" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="27689" PostTypeId="1" AcceptedAnswerId="27690" CreationDate="2016-02-20T05:11:34.750" Score="7" ViewCount="2041" Body="&lt;p&gt;Title is self-explanatory. I know that this question is bit too dumb. But I don't know how to say for sure if a puzzle is enigma or cryptogram. Both of them has a message to decipher. So, finally, &lt;strong&gt;What is the difference between enigma puzzle and cryptograms?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-02-20T05:26:07.770" LastActivityDate="2016-02-20T05:58:17.617" Title="What is the difference between enigma puzzle and cryptograms?" Tags="|puzzle-identification|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27690" PostTypeId="2" ParentId="27689" CreationDate="2016-02-20T05:48:36.053" Score="7" Body="&lt;p&gt;A cryptogram is a puzzle where letters are replaced by some other symbols in a one-to-one cipher. The most common type involves letters being replaced with other letters, but numbers and symbols are also acceptable. So, for instance, A could be replaced with G; B could be replaced with I; C could be replaced with Q; and so on, with random letters chosen and none repeated. The goal of a cryptogram is often to decode the message.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Here is an example of a cryptogram unsolved:&lt;a href=&quot;https://i.stack.imgur.com/f3laI.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/f3laI.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;And here is the solution:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/PaMQb.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PaMQb.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You can see that each letter on the bottom corresponds to one letter on the top, and vice versa.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;An enigmatic puzzle is one where the solving method is not given. Some great examples of enigmatic puzzles are BmyGuest's &amp;quot;hyper-modern art&amp;quot; series, like &lt;a href=&quot;https://puzzling.stackexchange.com/questions/26297/sweet-dream-another-hyper-modern-art-piece&quot;&gt;this puzzle&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/questions/9915/what-is-the-title-of-this-modern-art-painting-if-not-stars-and-stripes&quot;&gt;this one&lt;/a&gt;. They are also common in puzzle hunts such as the MIT Mystery Hunt, ΣUMS, and P&amp;amp;A Magazine.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;Enigmatic puzzles&amp;quot; frequently involve a leap of intuition hinted at through the puzzle's framing; their solutions are usually a common English word or phrase.&lt;/p&gt;&#xA;&lt;p&gt;Here is an enigmatic puzzle &lt;a href=&quot;http://web.mit.edu/puzzle/www/2015/puzzle/even_the_blossoming_flowers/&quot; rel=&quot;noreferrer&quot;&gt;from the 2015 MIT Mystery Hunt&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Even The Blossoming Flowers&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/62ZLl.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/62ZLl.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Its solution:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The title is the same as the title of a Japanese poem called &amp;quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Iroha&quot; rel=&quot;noreferrer&quot;&gt;Iroha&lt;/a&gt;&amp;quot;; it is an alternate ordering for their syllabary. The letters in the grid are laid out in the same way as the syllabary; in this layout, each row represents a consonant (or possibly none) and each column represents the vowel that follows it. If you read off the letters in the order their corresponding syllables appear in the poem, you get the message &amp;quot;PLURAL OF WORD THAT FOLLOWS SHRINK GIFT SANDWICH OR BODY&amp;quot;. All of those words are commonly followed by &amp;quot;wrap&amp;quot;: &amp;quot;shrink wrap&amp;quot;, &amp;quot;gift wrap&amp;quot;, &amp;quot;sandwich wrap&amp;quot;, or &amp;quot;body wrap&amp;quot;. So the answer to the puzzle is &amp;quot;&lt;strong&gt;WRAPS&lt;/strong&gt;&amp;quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-02-20T05:58:17.617" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="27691" PostTypeId="2" ParentId="27580" CreationDate="2016-02-20T12:28:27.717" Score="4" Body="&lt;pre&gt;&#xA;&#xA;  &quot;2&quot; whispers v to .&#xA;                      ` .&#xA;                       . &quot;3&quot;  =  (v+w)/2&#xA;                     .'   .        .     `.&#xA;  &quot;4&quot; whispers w to '.    .        .       :&#xA;                      `.                    :&#xA;                       . &quot;5&quot;  =  (w+X)/2     :&#xA;                     .'   .        .          :&#xA;  &quot;6&quot; whispers X to '.    .        .           + --&gt;  3+9  =  (v+w+y+z)/2  =  12&#xA;                      `.                      :                            .&#xA;                       . &quot;7&quot;  =  (X+y)/2     :                             .&#xA;                     .'   .        .        :                             .&#xA;  &quot;8&quot; whispers y to '.    .        .       :                             .&#xA;                      `.                 .'                             .&#xA;                       . &quot;9&quot;  =  (y+z)/2                               .&#xA;                     .'   .        .                                 .&#xA; &quot;10&quot; whispers z to '.    .        .                                .&#xA;                      `.                                          .&#xA;                       . &quot;1&quot;  =  (z+v)/2                        .&#xA;                     .'   .        .                          .&#xA;  &quot;2&quot; whispers v to '     .        .                        .&#xA;                          .        .                     .&#xA;                   3+5+7+9+1  =  v+w+X+y+z  =  25     .&#xA;                                            :      .&#xA;                                  v+w+y+z   =  24&#xA;                                            :&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\:$   &lt;code&gt;:&lt;/code&gt; &lt;br&gt;&#xA; $\qquad\qquad\qquad\qquad\quad~~\!\;$ &lt;code&gt;&quot;6&quot; picked   X   =   25 - 24   =   1&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18129" LastEditorUserId="18129" LastEditDate="2016-02-20T12:58:23.740" LastActivityDate="2016-02-20T12:58:23.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27692" PostTypeId="1" AcceptedAnswerId="27695" CreationDate="2016-02-20T12:30:59.243" Score="15" ViewCount="3701" Body="&lt;p&gt;You call me as something physical, something you can touch.&lt;br&gt;&#xA;But to be technically correct, I don't weight very much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, my weight is zero, I'm not of matter at all.&lt;br&gt;&#xA;Yet, I can cause damage, inflicting material fall. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My creation involved once upon a time a feather.&lt;br&gt;&#xA;But I'm said to be a spawn of curiosity and weather.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm opposed to an animal but I can spread like a disease.&lt;br&gt;&#xA;If I'm plenty you'll nag, if I'm scarce you'll feel an unease.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some people haven't yet seen me, others see me all day.&lt;br&gt;&#xA;Sometimes I'm kept securely, sometimes thrown away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;br&gt;&#xA;&lt;em&gt;Apparently, the question caused a lot of headache for the community (as it should), so there's the natural suspicion that it might be poorly formulated. I understand that and I promise that I've taken a lot of consideration to make sure it isn't. Please view the other question of mine for verification.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Nevertheless, I'll collect the pointers and comments below to improve the solving experience. I hope it helps the community to reopen it.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feather...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; refers to the old style writing equipment.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Curiosity and weather...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; refers to Bejamin Franklin's interest for electricity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Feather vs. curiosity and weather...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; need to be viewed together as a polarity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Animal...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; is an actual, real species existing today, commonly spread over most areas of the world.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Opposed to an animal...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; refers to the term used to describe the opposite of our subject. If we would be looking for &lt;em&gt;recession&lt;/em&gt; (i.e. bear market), the opposite animal would be &lt;em&gt;bull&lt;/em&gt;. Remember, the subject is described &lt;strong&gt;without&lt;/strong&gt; the animal epithet - its original version is, however.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Spreading like a disease...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; refers to the rapid rate of spread but also the way, or rather nature, of the spread.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nagging vs being uneasy...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; refers to different qualities of the subjects, as the few appreciated ones are welcome, while the most of them aren't. It's possible to be displeased of the excessive number of such and simultaneously happy about the few that are wanted.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;People who hasn't seen it...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; are real people living today, although it's more and more unlikely to find such individuals. In fact, in the contemporary world, only remote tribes, extremely secluded communities and special cases as gravely handicapped, mentally challenged, infants and some of the elders may not be aware of the subject.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Keeping secure...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; refers both to the nature of the spread and to the period of storage.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Throwing away...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; refers to disposing of the unwanted or already consumed subjects.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; On average, I'd say that it's a very common concept used by you on daily basing, spoken of frequently and relied on since many years. It can be described by a single word or a few (all in all less than five, without disclosing the exact number).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4319" LastEditorUserId="40008" LastEditDate="2019-01-18T13:08:32.987" LastActivityDate="2019-01-18T13:08:32.987" Title="You speak of me as a thing but I'm not physical - or am I?" Tags="|riddle|rhyme|" AnswerCount="11" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="27694" PostTypeId="2" ParentId="27692" CreationDate="2016-02-20T13:33:56.043" Score="5" Body="&lt;h1&gt;This is wrong answer(my achievement)...confirmed by OP&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;My first thought is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Computer language&lt;/strong&gt; or &lt;strong&gt;code&lt;/strong&gt; or something like that&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You call me as something physical, something you can touch.&#xA;But to be technically correct, I don't weight very much.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Used in making machines that you can touch. They can have small size.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In fact, my weight is zero, I'm not of matter at all.&#xA;Yet, I can cause damage, inflicting material fall.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Don't have weight. You can hack other computer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My creation involved once upon a time a feather.&#xA;But I'm said to be a spawn of curiosity and weather.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I dont have any idea about feather and weather. But I am sure that it must have come in life because of curiosity (just like any other invention).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm opposed to an animal but I can spread like a disease.&#xA;If I'm plenty you'll nag, if I'm scarce you'll feel an unease.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can spread a virus by making its file. And viruses (biological) are spread by animals sometimes. If there are too many lines of code then it's hard to catch the mistake. If less number of lines then it can do lower number of functions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some people haven't yet seen me, others see me all day.&#xA;Sometimes I'm kept securely, sometimes thrown away.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people don't know anything about it (like me) while it's everyday work for other people. Sometimes top secret codes are kept securely. Sometimes if there is big mistake in the code then it is thrown away.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-02-20T13:55:22.007" LastActivityDate="2016-02-20T13:55:22.007" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="27695" PostTypeId="2" ParentId="27692" CreationDate="2016-02-20T13:54:43.900" Score="13" Body="&lt;h1&gt;New Answer&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;You are ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an &lt;strong&gt;email&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You call me as something physical, something you can touch.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mail is something physical.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But to be technically correct, I don't weight very much.&lt;br&gt;&#xA;In fact, my weight is zero, I'm not of matter at all.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Email doesn't weight anything.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet, I can cause damage, inflicting material fall.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Emails can cause damage in many ways. Either because the contained information has effects on the real world, or it can carry some form of malware.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My creation involved once upon a time a feather.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Letters were written with a quill.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I'm said to be a spawn of curiosity and weather.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Email needs electricity, which was discovered by Benjamin Franklin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm opposed to an animal but I can spread like a disease.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Physical mail is often referred to as snail mail, because it's slower than email.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I'm plenty you'll nag, if I'm scarce you'll feel an unease.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everyone who gets mail gets too much of it, usually in form of spam. Not hearing from a friend for a long time may cause unease.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some people haven't yet seen me, others see me all day.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people still don't use computers at all. Others get many emails every day &lt;em&gt;(sigh)&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I'm kept securely, sometimes thrown away.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some emails contain important information, is transferred using encryption, and stored securely. Spam is thrown away.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;Old Answer&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Maybe you are ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a &lt;strong&gt;word&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You call me as something physical, something you can touch.&lt;br&gt;&#xA;But to be technically correct, I don't weight very much.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A word doesn't weight anything.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In fact, my weight is zero, I'm not of matter at all.&lt;br&gt;&#xA;Yet, I can cause damage, inflicting material fall.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A wrong word can cause much damage.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My creation involved once upon a time a feather.&lt;br&gt;&#xA;But I'm said to be a spawn of curiosity and weather.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Words were earlier written using a quill.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm opposed to an animal but I can spread like a disease.&lt;br&gt;&#xA;If I'm plenty you'll nag, if I'm scarce you'll feel an unease.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A word can spread like a disease. Too much chatter can be annoying. Not having anyone to talk to when in need isn't good either.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some people haven't yet seen me, others see me all day.&lt;br&gt;&#xA;Sometimes I'm kept securely, sometimes thrown away.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are people which can't read, while it's common for others. Precious old books are kept securely, while promotional brochures are thrown away unread.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2016-02-22T20:50:49.193" LastActivityDate="2016-02-22T20:50:49.193" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="27696" PostTypeId="2" ParentId="27692" CreationDate="2016-02-20T13:58:51.447" Score="10" Body="&lt;p&gt;I think you are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Language&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You call me as something physical, something you can touch.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We speak of language as something that can touch us emotionally, it can affect our understanding of the world and stimulate us intellectually. In fact some written works have had a huge impact on our society.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But to be technically correct, I don't weight very much.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Written words could be said to weigh a small amount for the ink they use...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In fact, my weight is zero, I'm not of matter at all.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ..but in fact language itself is a concept, realised purely in our common understanding of it, and use of it for communications.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet, I can cause damage, inflicting material fall.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As well as all the huge benefits, language and communications can be used to coordinate all manner of damage: Destroying buildings; costing companies millions (while not 'material' itself, the consequences are); causing injury and bodily harm; causing hurt feelings and emotional harm, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My creation involved once upon a time a feather.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The written word used to be laid down with a quill, made from a feather.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I'm said to be a spawn of curiosity and weather.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When humans first started using language, the important day-to-day matters would have been amongst the first sounds/words uttered. Curiosity about something not understood would certainly have caused an early human to utter more of these sounds, while trying to explain, or elicit an explanation about it from another. Weather could also have been a common thing to talk about... we still love talking about the weather to this day!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm opposed to an animal but I can spread like a disease.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Language can be learned and taught to others, it spreads as humans interact with each other. But it is not a virus or any kind of parasite or infection.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I'm plenty you'll nag, if I'm scarce you'll feel an unease.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Too much talking and you'll ask for some quiet. If it's too quiet, this can bee uncomfortable or creepy in some situations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some people haven't yet seen me, others see me all day.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In some remote tribes language is only spoken and not written, but most people see written words for most of the day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I'm kept securely, sometimes thrown away.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some languages are dying out and groups exist purely to teach, learn, document, and preserve these languages so that they may live on. Other languages have died out and been forgotten.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Language.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18522" LastActivityDate="2016-02-20T13:58:51.447" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27697" PostTypeId="2" ParentId="27692" CreationDate="2016-02-20T14:00:05.503" Score="3" Body="&lt;h1&gt;I proudly present my another incorrect (but useful) attempt (confirmed by OP)&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;My Another answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; light&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You call me as something physical, something you can touch.&#xA;But to be technically correct, I don't weight very much.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Light don't weight apart from the pressure applied by it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In fact, my weight is zero, I'm not of matter at all.&#xA;Yet, I can cause damage, inflicting material fall.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It has negligible weight. It's not matter. Light have energy and can damage things.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My creation involved once upon a time a feather.&#xA;But I'm said to be a spawn of curiosity and weather.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Remember &quot;As &lt;em&gt;light&lt;/em&gt; as feather&quot;?...We found out about lightning (that it's made up of charge) by Benjamin Franklin's Experiment where he flew the kite in rainy weather.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm opposed to an animal but I can spread like a disease.&#xA;If I'm plenty you'll nag, if I'm scarce you'll feel an unease.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many animals are heavy. Light can spread. We don't feel alright if there is too much light and feel afraid if there is less.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some people haven't yet seen me, others see me all day.&#xA;Sometimes I'm kept securely, sometimes thrown away.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Blind people. In torch. By Sun.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-02-20T18:07:34.180" LastActivityDate="2016-02-20T18:07:34.180" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27698" PostTypeId="2" ParentId="27692" CreationDate="2016-02-20T14:38:47.750" Score="4" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an idea/thought&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You call me as something physical, something you can touch.&#xA;But to be technically correct, I don't weight very much.&#xA;In fact, my weight is zero, I'm not of matter at all.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One can touch upon an idea/thought. Yet it is not made up of matter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet, I can cause damage, inflicting material fall.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An idea can be quite destructive&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My creation involved once upon a time a feather.&#xA;But I'm said to be a spawn of curiosity and weather.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In old times thoughts and ideas were written down using quills. Curiosity leads to ideas. (I'm not sure about the weather connection)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm opposed to an animal but I can spread like a disease.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Do you mean to say coming up with ideas is a human ability (as opposed to other animals) ? I don't quite agree with this assumption though. Ideas can spread very fast.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I'm plenty you'll nag, if I'm scarce you'll feel an unease.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Too many thoughts in the head lead to confusion. A scarcity of thoughts or ideas can make one feel uneasy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some people haven't yet seen me, others see me all day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This one doesn't fit&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes I'm kept securely, sometimes thrown away.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some ideas are kept secret, while others are discarded or freely shared&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19380" LastActivityDate="2016-02-20T14:38:47.750" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27699" PostTypeId="2" ParentId="15362" CreationDate="2016-02-20T14:48:48.130" Score="0" Body="&lt;p&gt;Are you:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Nuclear Weapon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I go up and down, and am all around, but not many seem to know.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The weapons are carried all around (hopefully) by government by planes(up) and ships(down) and on ground transport (down). People are not aware of their presence around them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But if I crash and burn, to the ground, the world may ceases to flow.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Catastrophic consequences of detonating the weapon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For my existence has become a medial role, that people may try to ignore.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Medial role= central/critical role. People try to ignore the importance of such devastating weapons in our lives.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But those that dare play with me, their lives can improve or turn sore. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Those who have those weapons are protected from other countries (and thus improve their lives), or the lives turn sore on those the weapons are used.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I go up and down words are thrown around, as people think they might know.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Winging it here: Some people support the possession, some oppose having them around (words thrown around).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But when the time goes by and my job is done, the words of the 'wise' start to show.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As time goes by, job is done: Two cases here, one) Weapons are detonated, and two) No detonation. --&gt; In case one, now they are used (detonated), job is done. The 'wise' always said to be wary of MAD (Mutually Assured Destruction). In Case two, the weapons are not used. The job is done by assuring safety for a long time (the 'wise' always told you to have them in the first place!!).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19351" LastActivityDate="2016-02-20T14:48:48.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27700" PostTypeId="1" AcceptedAnswerId="27705" CreationDate="2016-02-20T15:30:07.473" Score="6" ViewCount="276" Body="&lt;p&gt;Every letter stands for a digit in base-10 representation, &#xA;different letters stand for different digits,&#xA;and leading digits are always non-zero.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\sqrt{CAREER} \,\, = \,\,  RUT $&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which digit does each letter represent? (Please present the full analysis how these digits can be determined.)&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="9404" LastEditDate="2016-03-03T10:18:17.223" LastActivityDate="2016-05-28T05:52:28.667" Title="Square-root of CAREER alphametic" Tags="|no-computers|alphametic|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27701" PostTypeId="1" AcceptedAnswerId="27702" CreationDate="2016-02-20T15:36:00.520" Score="19" ViewCount="1415" Body="&lt;p&gt;My Indian friend Jamal recently participated in the well-known TV quiz&#xA;show &lt;strong&gt;&lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Who_Wants_to_Be_a_Millionaire&quot;&gt;Who wants to be a millionaire?&lt;/a&gt;&lt;/em&gt;&lt;/strong&gt;. You might have heard or read about Jamal's spectacular success; the media reported on it extensively.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what Jamal told us about the 500.000 Euro question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For the 500.000 Euro question, I excluded answers A and B right away: A is &#xA;  the smallest of the eight, and does not have any of its own. Furthermore&#xA;  I recalled that B has two in fear and terror. Hence I had to decide myself &#xA;  between C and D. I was really stumped, so that meant I was going to have &#xA;  to use my phone-a-friend lifeline.  Fortunately, my friend Albert knows &#xA;  the weirdest things in the strangest sub-areas of science, and so this seemed&#xA;  to be the right question for him.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I read off the question to him, and only told him answers C and D.  There&#xA;  were a few seconds of silence, and I was afraid that the phone connection&#xA;  had broken down.  Then suddenly Albert's voice came back and told me that&#xA;  D is no good since those are all water deities, whereas the one in the&#xA;  question is &quot;as you like it&quot; and might well belong to C.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;How sure are you?&quot; I probed.  &quot;About nine...&quot; and the phone hung up.&#xA;  I was pretty sure he was going to say ninety-something, which was good&#xA;  enough for me. I happily smiled at Prem and said: &quot;In the 250.000 Euro&#xA;  question I chose for answer C, and also this time I will go for C!&quot;. &#xA;  Prem gave his usual &quot;Are you sure?&quot;, and I moved on to the final question. &#xA;  Good to have Albert as a friend.  Gosh, that's much better than good!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;What was the 500.000 Euro question?&lt;br&gt;&#xA;What were the four possible answers A, B, C, D?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2016-02-20T16:22:19.877" Title="Slumdog millionaire (14)" Tags="|knowledge|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="27702" PostTypeId="2" ParentId="27701" CreationDate="2016-02-20T16:19:30.060" Score="26" Body="&lt;p&gt;The question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Which planet has a moon named Rosalind?&lt;br&gt;&#xA;&lt;br&gt;&#xA;A. Mercury&lt;br&gt;&#xA;B. Mars&lt;br&gt;&#xA;C. Uranus&lt;br&gt;&#xA;D. Neptune&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Mercury is the smallest of the eight planets and doesn't have any moons. Mars has two moons, named Phobos (&quot;Fear&quot;) and Deimos (&quot;Terror&quot;). The moons of Uranus are named after Shakespeare characters, and Neptune's moons are named after water deities. Rosalind is a character from Shakespeare's play, &quot;As You Like It&quot;, so it is a moon of Uranus.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2016-02-20T16:19:30.060" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27703" PostTypeId="2" ParentId="27701" CreationDate="2016-02-20T16:22:19.877" Score="11" Body="&lt;p&gt;The question is something like:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Which planet has several moons named aftercharacters in Shakespeare's plays including 'As You Like It' &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer A:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mercury -- the smallest planet. Has no moons  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer B:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mars -- Two moons - Phobos(terror/fright) and Deimos (fear)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer C:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Uranus -- The moon Rosalind's name was in the 'As You Like it' and Portia , Juliet, Desdemona...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer D:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Neptune -- All its moons named after gods are water gods  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://www.fief.org/kathleen/Moons/Moons.html&quot;&gt;http://www.fief.org/kathleen/Moons/Moons.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7715" LastActivityDate="2016-02-20T16:22:19.877" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27704" PostTypeId="2" ParentId="27700" CreationDate="2016-02-20T16:28:40.680" Score="5" Body="&lt;p&gt;First of all, &lt;span class=&quot;math-container&quot;&gt;$R\geqslant 3$&lt;/span&gt; since CAREER has 6 digit numbers. In addition, we know that CAREER is simply square of RUT, so R cannot be 3 since the end of CAREER digit is R and it is &lt;span class=&quot;math-container&quot;&gt;$RUT^2$&lt;/span&gt;, so R can be 4,5,9,6 (last digits of any square of a number cannot be other value of 0,1,4,5,6,9).&lt;/p&gt;&#xA;&lt;p&gt;Moreover the R cannot be 5 since the last digit will be 5 and that will make T value is 5 which has to be different than R. So R can be 4,6,9.&lt;/p&gt;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$R=6$&lt;/span&gt;, T has to be 4.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;So CAREER becomes CA6EE6 and RUT becomes 6U4. The first digit of CAREER can only be 3 or 4. Since we have chosen T as 4. C can only be 3. So that makes U value as 1 or 2 because if we take &lt;span class=&quot;math-container&quot;&gt;$U&amp;gt;2$&lt;/span&gt; that makes C as 4. As a result, so only 614 and 624 are possible option for &lt;span class=&quot;math-container&quot;&gt;$R=6$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$R=9$&lt;/span&gt;, T has to be 3 or 7 so;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If R=9 and T=3; our CAREER becomes CA9EE9 and RUT becomes 9U3. C can only be 8 since our R value has to be different than C. That narrows down the value of U as 1,2,3,4. and since T=3, U value can only be 1,2,4.&lt;/p&gt;&#xA;&lt;p&gt;If R=9 and T=8; our CAREER becomes CA9EE9 and RUT becomes 9U7. C can only be 8 again. That narrows down the value of U as 1,2,3,4. U value can only be 1,2,3,4.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$R=4$&lt;/span&gt;; T has to be 2 or 8.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If R=4 and T=2; our CAREER becomes CA4EE4 and RUT becomes 4U2. C can only be 1 or 2 and since T =2 C can only be 1. That narrows U values down to 1,2,3,4. And C is 1 and T is 2, R  is 4, U can only be 3.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Since it is narrowed down quite a bit, we can try and find;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; CAREER = 376996 and RUT=614&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18514" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-02-20T16:55:29.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27705" PostTypeId="2" ParentId="27700" CreationDate="2016-02-20T16:45:04.723" Score="7" Body="&lt;p&gt;Since squares can only be $0, 1, 4, 5, 6, 9 \mod 10$, $R$ is one of those numbers. Further, since a three-digit number has a six-digit square, we have $R \geq 3$. So $R$ is $4, 5, 6$ or $9$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, since $T \neq R$, $R$ can't be $5$. This leaves $4$, $6$ or $9$ for $R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 1: $R=4$. Then $C=1$. Also, since the square of $T$ ends in a 4, $T$ is $2$ or $8$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt; &#xA;Case 1.1. $R=4$, $C=1$, $T=2$: $$104004+10000A+110E=161604+8040U+100U^2$$&#xA;$$10000A+110E=57600+8040U+100U^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $U\geq4$ would cause the RHS to be larger than 100000. So $U\leq3$, and since 1 and 2 are used we get $U=3$. Then $10000A+110E=82620$, which gives no solution for $A$ and $E$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt; &#xA;&#xA;&lt;p&gt;Case 1.2. $R=4$, $C=1$, $T=8$: $$104004+10000A+110E=166464+8160U+100U^2$$&#xA;$$10000A+110E=62460+8160U+100U^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $U\geq4$ would cause the RHS to be larger than 100000. So $U\leq3$, and since 1 is used we get $U=2$ or $U=3$. Then $10000A+110E=95940$ or $10000A+110E=82870$, neither of which gives no solution for $A$ and $E$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt; &#xA;&#xA;&lt;p&gt;Case 2. $R=6$. Since the square of $T$ ends in a 6, $T$ is $4$ or $6$, but 6 is taken. So $R=6$ and $T=4$. We look to the fourth digit of $(6U4)^2$, which is $12U+\left\lfloor \frac{4800+100U^2+80U}{1000} \right\rfloor$, but also $R=6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2U+\left\lfloor \frac{4800+100U^2+80U}{1000} \right\rfloor \equiv 6 \mod 10$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$U=1$ is possible. This gives $614^2=376996$, which is in fact an acceptable solution.  $U=2$ isn't possible, since this gives $9 \not\equiv 6 \mod 10$. $U=3$ isn't possible, since this gives $12 \not\equiv 6 \mod 10$. 4 and 6 are taken. $U=7$  isn't possible, since this gives $24 \not\equiv 6 \mod 10$. $U=8$  isn't possible, since this gives $27 \not\equiv 6 \mod 10$. $U=9$  isn't possible, since this gives $31 \not\equiv 6 \mod 10$. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Case 3. $R=9$. Since the square of $T$ ends in a 9, $T$ is $3$ or $7$. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Case 3.1. $R=9$, $T=3$. We look to the fourth digit of $(9U3)^2$, which is $18U+\left\lfloor \frac{5400+100U^2+60U}{1000} \right\rfloor$, but also $R=9$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $$18U+\left\lfloor \frac{5400+100U^2+60U}{1000} \right\rfloor \equiv 9 \mod 10$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$8U+\left\lfloor \frac{5400+100U^2+60U}{1000} \right\rfloor \equiv 9 \mod 10$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$U=1$ isn't possible, since this gives $13 \not\equiv 9 \mod 10$. $U=2$ isn't possible, since this gives $21 \not\equiv 9 \mod 10$. $3$ is already taken. $U=4$ is possible. However $943^2=889249$, which doesn't match the given form.  $U=5$ isn't possible, since this gives $48 \not\equiv 9 \mod 10$. $U=6$ isn't possible, since this gives $57 \not\equiv 9 \mod 10$. $U=7$ isn't possible, since this gives $66 \not\equiv 9 \mod 10$. $U=8$ isn't possible, since this gives $76 \not\equiv 9 \mod 10$. 9 is already taken. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Case 3.2. $R=9$, $T=7$. We look to the fourth digit of $(9U7)^2$, which is $18U+\left\lfloor \frac{5400+100U^2+60U}{1000} \right\rfloor$, but also $R=9$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $$18U+\left\lfloor \frac{5400+100U^2+60U}{1000} \right\rfloor \equiv 9 \mod 10$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$8U+\left\lfloor \frac{12600+100U^2+140U}{1000} \right\rfloor \equiv 9 \mod 10$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$8U+\left\lfloor \frac{2600+100U^2+140U}{1000} \right\rfloor \equiv 8 \mod 10$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$U=1$ isn't possible, since this gives $10 \not\equiv 8 \mod 10$. $U=2$ isn't possible, since this gives $19 \not\equiv 8 \mod 10$.  $U=3$ isn't possible, since this gives $27 \not\equiv 8 \mod 10$.  $U=4$ isn't possible, since this gives $36 \not\equiv 8 \mod 10$.  $U=5$ isn't possible, since this gives $45 \not\equiv 8 \mod 10$.   $U=6$ isn't possible, since this gives $54 \not\equiv 8 \mod 10$. $U=8$ isn't possible, since this gives $74 \not\equiv 8 \mod 10$.  7 and 9 are already taken.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Hence the only solution is $C=3$, $A=7$, $E=9$, $R=6$, $U=1$ and $T=4$. &lt;/p&gt;&#xA;" OwnerUserId="16322" LastEditorUserId="16322" LastEditDate="2016-05-28T05:52:28.667" LastActivityDate="2016-05-28T05:52:28.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27706" PostTypeId="1" CreationDate="2016-02-20T16:58:55.723" Score="12" ViewCount="603" Body="&lt;p&gt;Professor Halfbrain has spent his entire weekend by analyzing stacks of &lt;span class=&quot;math-container&quot;&gt;$52$&lt;/span&gt; cards that are numbered by &lt;span class=&quot;math-container&quot;&gt;$1,2,\ldots,52$&lt;/span&gt;.&#xA;Halfbrain always started with a stack having the cards face-up and in increasing order. A feasible move consists in picking a contiguous group of cards somewhere in the stack, taking out this group from the stack, and putting it back into the stack at some place (without changing the order within the group, and without turning over cards).&lt;/p&gt;&#xA;&lt;p&gt;Example: A typical move would be to start with &lt;span class=&quot;math-container&quot;&gt;$1,2,\ldots,52$&lt;/span&gt;, to take out the group &lt;span class=&quot;math-container&quot;&gt;$4,5,6,7,8,9$&lt;/span&gt;, and to put it back between &lt;span class=&quot;math-container&quot;&gt;$16$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$17$&lt;/span&gt;, so that the new stack becomes &lt;span class=&quot;math-container&quot;&gt;$1,2,3,10,11,12,13,14,15,16,\underline{4,5,6,7,8,9},17,18,\ldots,52$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The goal is to bring the cards in the stack from increasing order into decreasing order, and to do this by making as few moves as possible.&#xA;Professor Halfbrain has proved two extremely deep theorems on such rearrangement of stacks.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Professor Halfbrain's first theorem&lt;/strong&gt;: It is possible to bring the cards in the stack from increasing order into decreasing order by making at most &lt;span class=&quot;math-container&quot;&gt;$52$&lt;/span&gt; moves.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Professor Halfbrain's second theorem&lt;/strong&gt;: For bringing the cards in the stack from increasing order into decreasing order, one must make at least two moves.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This puzzle asks you to improve the two theorems of professor Halfbrain and to make them even deeper.&#xA;Find an integer &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;, so that &amp;quot;&lt;span class=&quot;math-container&quot;&gt;$52$&lt;/span&gt; moves&amp;quot; in the first theorem may be replaced by &amp;quot;&lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; moves&amp;quot;, and so that &amp;quot;at least two moves&amp;quot; in the second theorem may be replaced by &amp;quot;at least &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; moves&amp;quot; (again yielding true statements, of course).&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-11-19T13:41:29.930" Title="Professor Halfbrain and the 52 cards" Tags="|mathematics|combinatorics|reachability|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="27707" PostTypeId="2" ParentId="27692" CreationDate="2016-02-20T17:23:41.113" Score="14" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Electricity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You call me as something physical, something you can touch.&#xA;But to be technically correct, I don't weight very much.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Electricity doesn't weigh anything.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In fact, my weight is zero, I'm not of matter at all.&#xA;Yet, I can cause damage, inflicting material fall.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Electricity can certainly cause damage&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My creation involved once upon a time a feather.&#xA;But I'm said to be a spawn of curiosity and weather.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Static electricity experiments attracted feathers to amber rubbed with furs and the Benjamin Franklin experiment with lightning.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm opposed to an animal but I can spread like a disease.&#xA;If I'm plenty you'll nag, if I'm scarce you'll feel an unease.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Polar bear/ polarity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some people haven't yet seen me, others see me all day.&#xA;Sometimes I'm kept securely, sometimes thrown away.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Amazonian tribes exist that are presumably unfamiliar with electricity. Electricity can be stored in batteries and lots is wasted.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="392" LastActivityDate="2016-02-20T17:23:41.113" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="27708" PostTypeId="1" AcceptedAnswerId="27710" CreationDate="2016-02-20T17:53:24.663" Score="19" ViewCount="842" Body="&lt;p&gt;Uncle Mort was dead. Samuelson, the family lawyer, read the will. It stated that since he had no children of his own, his whole estate was to be divided between his two nieces, Roseanne and Chelsea and his nephew Sean if they &quot;could demonstrate a semblance of co-operation and intelligence&quot; by opening the combination safe in his office. They would have one chance. If they failed, the entire estate would go to charity. If they all agreed not to try, 90% of the estate would go to charity and they could split the remainder for &quot;at least being smart enough to realise they are idiots.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sean said: &quot;10% is good. Let's take it.&quot; Roseanne agreed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chelsea had a question: &quot;Can I, like, ask my boyfriend? He's so smart it's like weird or something.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Samuelson examined the will. &quot;There is no provision for that. I will stretch a point and allow it. Go ahead, young man.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The safe had buttons on the front numbered 1 to 100. To open it, three buttons had to be pressed in the correct order. It turned out that Uncle Mort had given his relatives some pretty broad clues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sean: I remember he gave me one of those sudoku puzzles. He circled a bunch of numbers. They were all in a line on the middle row. I can't remember how many there were and I lost the paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roseanne: Me too! And he said the numbers spelled a word in an &quot;asking&quot; code or something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chelsea: I forgot all about that until now. He did the same with me and then he wrote out the word. Then he told me: &quot;now there's no way you can forget this&quot; which just shows how dumb he is because I forgot it practically right away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The others agreed that Uncle Mort had said the same thing to each of them. After a great deal of thought, Sean remembered that the word was three letters long and Roseanne remembered that it was written oddly: &quot;Like, not all big or small letters?&quot; And Chelsea remembered that it was either: &quot;A capital letter, a small letter and a capital letter or small capital small.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were Chelsea's boyfriend, what combination would you suggest?&lt;/p&gt;&#xA;" OwnerUserId="18924" LastActivityDate="2016-02-21T20:18:03.687" Title="Uncle Mort's Idiots" Tags="|word-problem|story|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27709" PostTypeId="2" ParentId="27580" CreationDate="2016-02-20T17:58:22.817" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let $g_n$ be the number picked by the girl who said $n$. The average of $g_6$ and $g_4$ is two more than the average of $g_2$ and $g_4$, so $g_6$ must be four more than $g_2$. Symmetrically $g_6$ four less than $g_{10}$. $g_6$ is therefore the average of $g_2$ and $g_{10}$, but we already have already been told what this is.&lt;/p&gt;&#xA;" OwnerUserId="12593" LastActivityDate="2016-02-20T17:58:22.817" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27710" PostTypeId="2" ParentId="27708" CreationDate="2016-02-20T19:01:03.433" Score="23" Body="&lt;p&gt;I would suggest the combination&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 73 45 61.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Firstly, the &quot;asking&quot; code combined with the row of numbers suggests that they spelled out a word in &lt;a href=&quot;http://www.asciitable.com/&quot;&gt;ASCII code&lt;/a&gt;. The numbers on the front of the safe only go to 100, so Mort was probably using the hexadecimal notation for ASCII, since the decimal notations for standard letters go up to 122.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that all the numbers were on the same row in a Sudoku puzzle means that there were no repeated numbers and no zeroes. Combined with the restrictions on capital and small letters, this narrows the possible letters substantially, but ultimately there are still several possible words that could be made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual key is that Mort said to all three children that &quot;there's no way you can forget this.&quot; This indicates that there's some connection between the word and each child. As it happens, there's a three-letter word that they share right in their names - &quot;sea&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The capitalisation rules mean that the original word could have been &quot;SeA&quot; or &quot;sEa&quot;. Converting both to ASCII, the first becomes &quot;53 65 41&quot;, which has two 5s and is therefore impossible due to the Sudoku. &quot;sEa&quot; becomes &quot;73 45 61&quot;, and that is the most likely combination for the safe.&lt;/p&gt;&#xA;" OwnerUserId="12511" LastEditorUserId="12511" LastEditDate="2016-02-21T20:18:03.687" LastActivityDate="2016-02-21T20:18:03.687" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27711" PostTypeId="2" ParentId="27692" CreationDate="2016-02-20T19:21:00.273" Score="4" Body="&lt;p&gt;You are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the gravitational field&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You call me as something physical, something you can touch.&lt;br&gt;&#xA;But to be technically correct, I don't weight very much.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a field in its colloquial meaning is a physical thing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In fact, my weight is zero, I'm not of matter at all.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The gravitational field, like the electromagnetic field, is not considered a form of ordinary matter (technically, in quantum theory the two fields are mediated by massless particles called photons and gravitons).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet, I can cause damage, inflicting material fall.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gravity causes matter to fall, often causing damage.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My creation involved once upon a time a feather.&lt;br&gt;&#xA;But I'm said to be a spawn of curiosity and weather.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Newton wrote and published his groundbreaking theories about gravity, presumably using a quill pen, made of a feather. Like all great scientists, Newton was driven by curiosity, but I'm not sure what the reference to weather is about.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm opposed to an animal but I can spread like a disease.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This may be a reference to a bull plowing a field. The field opposes the animal, which is why force is required to overcome the opposition. The gravitational field spreads outward in waves (cf. the discovery of gravitational waves recently announced on the news), a bit like a disease spreading out from the epicenter of an epidemic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I'm plenty you'll nag, if I'm scarce you'll feel an unease.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When gravity is scarce (a zero gravity environment as experienced by astronauts in orbit around the Earth, or on board the &lt;a href=&quot;https://en.wikipedia.org/w/index.php?title=Vomit_comet&quot; rel=&quot;nofollow&quot;&gt;vomit comet&lt;/a&gt;), people often become nauseous.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some people haven't yet seen me, others see me all day.&lt;br&gt;&#xA;Sometimes I'm kept securely, sometimes thrown away.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure what this is referring to.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user17947" LastActivityDate="2016-02-20T19:21:00.273" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="27713" PostTypeId="2" ParentId="27637" CreationDate="2016-02-20T21:32:16.460" Score="1" Body="&lt;p&gt;This may not be a perfect match to everything, but my errant guess could lead someone down the right path.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that said, I think you are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an e-book, possibly with narration.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I speak through shadows,&lt;br&gt;&#xA;in cryptic tongues,&lt;br&gt;&#xA;and cryptic tongues only,&lt;br&gt;&#xA;to anyone who wants to know —&lt;br&gt;&#xA;but only if I have been told to say.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; E-books are read on lit screens, so their text shines through the darkness and illuminates the reader.  The e-books that come with narration capabilities will only narrate if you want it to.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think not myself;&lt;br&gt;&#xA;offer no discretion of my own,&lt;br&gt;&#xA;but never trust the word of one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No book, electronic or otherwise, thinks for itself; the book simply reports the thoughts, opinions, or ideas of the writer, at the time it was written.  The book, even if it is electronic, cannot choose to omit what may be false or harmful information, and that is why it is useful to read things from many sources.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I was old,&lt;br&gt;&#xA;I would be far too clumsy, far too wrinkled,&lt;br&gt;&#xA;to hide among others without sight.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Old&quot; books are made of paper and binding, which wrinkles with time.  Some people find that carrying around physical books is clumsy nowadays, especially with hardcovers.  The hint regarding the &lt;em&gt;spine&lt;/em&gt; touches on the fact that old books will have weak spines, whether the glue has simply worn out, or the spine has been weakened from the book being opened so much.  Physical books can be seen on a shelf, e-books cannot.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I am young so, if you wished,&lt;br&gt;&#xA;I could hide in the smallest crevice&lt;br&gt;&#xA;or a fold in the world, hidden from sight&lt;br&gt;&#xA;among the unsavory sort.&lt;br&gt;&#xA;It is that I am young that I am so much easier to find.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; E-books, being electronic, are able to be stored on electronic devices, where they don't occupy any physical space.  They could be stored in hidden directories that contain other questionable content.  But because they are in an electronic system, looking for them is much easier, as they can never be misplaced and you won't have to hunt through a library for them.  A simple search will return the item.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My greatest opponent is he who imposes nothing,&lt;br&gt;&#xA;who requires nothing,&lt;br&gt;&#xA;enforces nothing,&lt;br&gt;&#xA;and is far more stealthy than me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Radio and probably more specifically, podcasts, are other forms of electronic media.  However, both radio and podcasts are free, not requiring payment or a specific device to utilize them.  Radio waves are everywhere, and many radio shows are podcasted, so they in turn, are also everywhere, though we cannot see them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Than I, who weighs upon the world&lt;br&gt;&#xA;only barely more than a shadow&lt;br&gt;&#xA;as I roam.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, e-books are electronic, so have no physical weight, but their influence blankets the world more and more as they are instantly transferable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I do know myself,&lt;br&gt;&#xA;know more of myself than you do,&lt;br&gt;&#xA;know more of the world than you do,&lt;br&gt;&#xA;and it is not my style to ask questions.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;An e-book will be proficient in its own content, more than a person would be able to recognize.  Most books look to give answers and information, but this is a question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But, I shall make one exception today:&lt;br&gt;&#xA;What am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, an e-book.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Admittedly, this only works really well with a few of the stanzas, but getting this thought out here may allow room for another possibility to crawl into my head.  &lt;/p&gt;&#xA;" OwnerUserId="11452" LastActivityDate="2016-02-20T21:32:16.460" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27714" PostTypeId="1" AcceptedAnswerId="27724" CreationDate="2016-02-20T21:42:12.543" Score="20" ViewCount="1281" Body="&lt;p&gt;The &lt;em&gt;Prowler&lt;/em&gt; is a powerful piece in an alternative version of chess. In the standard board depicted below, we have an initial setup of white pieces and just one black prowler. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dwZET.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dwZET.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The special rules are the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The prowler can move to any square of the board not occupied by a piece of the same color. It can capture any piece, including the king;&lt;/li&gt;&#xA;&lt;li&gt;The prowler cannot put the opposite king in check;&lt;/li&gt;&#xA;&lt;li&gt;The game does not end when the king is captured: it ends when all pieces of one color are captured or by traditional draw rules (stalemate, repetition, etc). &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The question:&lt;/strong&gt; White starts the game. What is the minimum number &lt;strong&gt;N&lt;/strong&gt; of black prowlers you would need to force a win?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is not possible to force a win with N = 1. Easy way to demonstrate it: you can't capture the opposite queen AND the king, as long as they stay next to each other. So, if you capture one of them with the prowler, the other will capture you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19446" LastEditorUserId="15632" LastEditDate="2016-02-21T08:19:13.880" LastActivityDate="2016-02-21T21:08:43.647" Title="The Black Prowler (chess)" Tags="|chess|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="27715" PostTypeId="2" ParentId="27686" CreationDate="2016-02-20T22:31:53.490" Score="-4" Body="&lt;p&gt;Ok, I'll write my desired answer.&#xA;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$80 + .\dot{9}\dot{7} + .\dot{5} + .\dot{4}\dot{6}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which is basically...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$80 + 0.97979797... + 0.55555555... + 0.46464646...$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;that adds up to...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$80 + 198/99$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which finally equals to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$82$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17970" LastEditorUserId="12114" LastEditDate="2016-06-23T07:28:31.737" LastActivityDate="2016-06-23T07:28:31.737" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27716" PostTypeId="2" ParentId="27706" CreationDate="2016-02-20T23:44:57.650" Score="10" Body="&lt;p&gt;This is not a solution, but it is a demonstration that the answer is &lt;em&gt;not&lt;/em&gt; 51.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;5 cards can be reversed in 3 steps as follows:&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;code&gt;12345&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;34125 (move 34 to the left)&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;32541 (move 41 to the right)&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;54321 (move 54 to the left)&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;25 cards can be reversed in 18 steps by doing that on five groups of five cards, then doing the same thing to rearrange the groups. So 52 cards can be reversed by reversing the first 25 (18 steps), then the second 25 (18 steps), then swapping the last two cards (1 step), and reordering the three blocks (2 steps), for a total of at most 39 steps.'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The answer must be at least&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;26.&lt;br&gt;&#xA;&lt;br&gt;&#xA;Say a pair of consecutive cards is &lt;em&gt;good&lt;/em&gt; if the first card is greater than the second. A generic move (moving &lt;code&gt;x...y&lt;/code&gt;) looks like this:&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;code&gt;...ax...yb...cd... -&amp;gt; ...ab...cx...yd...&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;There are three places where new pairs are formed (&lt;code&gt;ab&lt;/code&gt;,&lt;code&gt;cx&lt;/code&gt;,&lt;code&gt;yd&lt;/code&gt;) and three places where pairs are lost (&lt;code&gt;ax&lt;/code&gt;,&lt;code&gt;yb&lt;/code&gt;,&lt;code&gt;cd&lt;/code&gt;). The total gain is the number of created good pairs minus the number of destroyed good pairs. For this gain to be greater than 2, all three created pairs must be good, and all of the destroyed pairs must not be good. This requires that &lt;code&gt;a&amp;gt;b&amp;gt;y&amp;gt;d&amp;gt;c&amp;gt;x&amp;gt;a&lt;/code&gt;, which is impossible. Therefore each step can only increase the number of good pairs by at most 2.&lt;br&gt;&#xA;&lt;br&gt;&#xA;When the deck is in order, there are no good pairs. When it is reversed, there are 51 good pairs. So it takes at least 26 steps to reverse the deck.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2016-02-21T01:29:57.267" LastActivityDate="2016-02-21T01:29:57.267" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="27717" PostTypeId="1" AcceptedAnswerId="27736" CreationDate="2016-02-21T00:55:02.050" Score="1" ViewCount="174" Body="&lt;p&gt;I am always on someone's cam,&lt;br&gt;&#xA;He controlls the way I am.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am tall,&lt;br&gt;&#xA;And I listen to mall,  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am very fat,&lt;br&gt;&#xA;Unlike a bat.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can light a fuse,&lt;br&gt;&#xA;Or abuse,&lt;br&gt;&#xA;I can gather food,&lt;br&gt;&#xA;Or chop down wood,&lt;br&gt;&#xA;I can mine,&lt;br&gt;&#xA;Or cut through slime,&lt;br&gt;&#xA;I can build,&lt;br&gt;&#xA;Or sell to a guild. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can change my look for fame,&lt;br&gt;&#xA;But my sizes will remain the same.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who am I ?&lt;/p&gt;&#xA;" OwnerUserId="19451" LastEditorUserId="11876" LastEditDate="2016-02-21T04:47:29.813" LastActivityDate="2016-02-21T18:29:19.923" Title="Who am I if I am controlled by someone?" Tags="|riddle|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27718" PostTypeId="1" CreationDate="2016-02-21T01:11:04.830" Score="4" ViewCount="975" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PvgPv.png&quot; alt=&quot;Colored Squares&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would appreciate any help that I can get on solving this riddle. I've been staring at this for hours and no patterns are occurring to me.&lt;/p&gt;&#xA;" OwnerUserId="19455" LastEditorUserId="32543" LastEditDate="2017-05-09T09:46:50.713" LastActivityDate="2017-05-09T09:46:50.713" Title="Can anyone help me solve this? I can't see a pattern anywhere" Tags="|pattern|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27719" PostTypeId="2" ParentId="27718" CreationDate="2016-02-21T04:00:37.800" Score="0" Body="&lt;p&gt;So, in trying to look for patterns, I noticed the white squares within each block of 9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0 2 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 3 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 1 ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure here. My gut says go with a box that has a single white square to complete it. But that doesn't exist... So I go with (A).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A) = 4 white squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0 2 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 3 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 1 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reading each column from top to bottom, left to right:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0 1 2;  2 3 1;  2 3 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm seeing some kind of pattern in the numbers, but not sure what to call it.&lt;/p&gt;&#xA;" OwnerUserId="19459" LastActivityDate="2016-02-21T04:00:37.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27720" PostTypeId="1" AcceptedAnswerId="27723" CreationDate="2016-02-21T04:14:00.303" Score="8" ViewCount="585" Body="&lt;p&gt;There are three words that sound the same&lt;br&gt;&#xA;But each has a different name&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the earliest words spoken by a baby, usually repeated twice.&lt;br&gt;&#xA;A part of speech hard to classify, sometimes used to show position&lt;br&gt;&#xA;And the foundation of the modern economy, as a verb&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you find the homophones?&lt;/p&gt;&#xA;" OwnerUserId="19179" LastEditorUserId="17482" LastEditDate="2016-02-21T10:23:52.287" LastActivityDate="2016-02-21T10:23:52.287" Title="Homophone Challenge" Tags="|word|english|language|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27721" PostTypeId="2" ParentId="27718" CreationDate="2016-02-21T04:36:52.877" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think it is C.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Reason&quot;: Take only the &lt;strong&gt;bottom left&lt;/strong&gt; and the &lt;strong&gt;bottom middle&lt;/strong&gt; field of each 3x3 square.&#xA;Now search for it in the big field &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For A: White + Purple -&gt; it's also on the bottom middle 3x3 field&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For B: White + Blue -&gt; it's also top middle&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For C: Red + White -&gt; doesn't exist&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For D: Blue + White -&gt; middle, left&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For E: Red + Blue -&gt; middle, middle&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For F: Purple + Red -&gt; middle, right&lt;/p&gt;&#xA;" OwnerUserId="4342" LastActivityDate="2016-02-21T04:36:52.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27723" PostTypeId="2" ParentId="27720" CreationDate="2016-02-21T05:03:25.253" Score="22" Body="&lt;p&gt;The words are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the earliest words spoken by a baby, usually repeated twice.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;bye&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A part of speech hard to classify, sometimes used to show position&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;by&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the foundation of the modern economy, as a verb&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;buy&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2016-02-21T05:03:25.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27724" PostTypeId="2" ParentId="27714" CreationDate="2016-02-21T05:28:11.047" Score="8" Body="&lt;p&gt;Black needs&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Three prowlers to force a win.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Assume there are only two.  White's first move is &lt;strong&gt;Nc3&lt;/strong&gt;, since this forces black to capture &lt;strong&gt;XxRa1&lt;/strong&gt; (or Q/K) - otherwise Rb1 would mean that Nc3/Rb1/Pb2 was a closed protected cycle and Qd1/Ke1 would be as well.  White's second move can then be &lt;strong&gt;Nf3&lt;/strong&gt;, which forces &lt;strong&gt;XxRh1&lt;/strong&gt; (or Q/K) by the same logic.  At this point, Qd1/Ke1 is a closed protected cycle.  But whichever is taken, Pd2/Pe2/Nc3/Nf3/QKd1 or QKe1 is still a protected closed cycle, and white can stall indefinitely with d1&gt;e1&gt;d1&gt;e1, etc.  Taking a different piece (Nc3 for example) leads to Pd2xXc3, while Qd1/Ke1 is still closed and protected.  If black chooses to capture Q or K earlier instead of Ra1 or Rh1, white just captures Qxe1/Kxd1, and now only needs to construct a single protected cycle - this means that black must capture both Rooks before white can move them, which gives white time to move both knights and set up the original Pd2/Pe2/Nc3/Nf3/QKd1 or QKe1.  Note that we do also need to prove that a tie can be forced if Black captures Q&gt;K, at which point neither piece can be our method of stalling.  Assume without loss of generality that Ra1 was the rook that survived to enable this play.  In this case, we have &lt;em&gt;at least&lt;/em&gt; Nc3/Ra1/Pb2/Pd2/Pe2/Pg2/Nf3/Bc1.  To force a tie without leaving any piece unprotected, just go Ra1b2 &gt; Pd2d3 &gt; Nf3d2 &gt; Nd2e4 &gt; Bc1g5, at which point Nc3/Ne4/Bg5-f6 is a tie.  Thus, &lt;strong&gt;white can always force a tie against two Prowlers.&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Aside: A counter play to Daedric's argument of zugzwang after capturing both rooks.&lt;/em&gt;&lt;br&gt;&#xA; &lt;strong&gt;Nc3 &gt; XxRa1 &gt; Nf3 &gt; XxRh1 &gt; Pe3 &gt; XxBc1&lt;/strong&gt; leaves Ke2-1 as a double protected stalling state.  The pieces involved in this are Nc3/Pd2/Qd1/Ke2-1/Nf3/Bf1/Pg2.  Each of these pieces is protected by some other piece for either King move.  Capturing Nc3 gets recaptured by Pd2.  Capturing Pd2 gets recaptured by Nf3.  Capturing Nf3 gets recaptured by Pg2.  Capturing Bf1 gets recaptured by Ke2-1, who is still protected by Q, who is protected by Nc3 - next turn, king moves back next to queen.  Capturing Pg2 gets recaptured by Bf1.  Each of these leaves QK forever adjacent, with Ke2-1 a draw.  The other options are capture Ke2-1, which gets recaptured by Qd1, which then becomes Qd1-e1, alternating being protected by each knight.  Or capture Qd1, recapture with Nc3 (which leaves everything still protected) then Nd1c3 &gt; Kd1-e1 is a draw.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; However, to prove that three is sufficient, you'd need to show that Nc3/Ne4/Bg5-f6, Pd2/Pe2/Bd3/Be3/Nf3-d4, or an equivalent setup can never be forced by white while capturing two prowlers during the loss of K,Q,R,R.  However, given that just a knight/bishop/pawns can't stall indefinitely, we can at least show that four prowlers is good enough.  &lt;em&gt;XxRa1 &gt; ??? &gt; XxRh1 &gt; ??? &gt; XxQd1 &gt; [captured] &gt; XxKd1 &gt; [captured]&lt;/em&gt; loosing only two prowlers can be followed by two more captures (either knight and either bishop) at the cost of a third prowler (there's no way to shield both knights and both bishops in time, so at least one of those captures is free) which prevents a stall.  Thus, we can at least say that &lt;strong&gt;four prowlers can always force a win.&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;While not a full proof, I can offer a (hopefully) convincing argument that three is sufficient.&lt;/strong&gt;  If white moves any pawn as their first move, black can capture whichever rook is on the same side of the board as that pawn.  If white's follow up is anything other than a knight move, black can respond by capturing queen or king (whichever is less guarded) forcing white to capture back, at which point black captures the other unguarded one, then can capture the other rook and one of the knights with only one lost prowler.  There are a lot of ways this can go down, but I feel confident in saying that pawn-first, non-knight second is not a winning (tying) line for white.  So, without loss of generality, let's say white starts on &lt;strong&gt;Nc3&lt;/strong&gt;.  Black can then go &lt;strong&gt;XxKe1&lt;/strong&gt;, forcing capture from the only protector, &lt;strong&gt;QxXe1&lt;/strong&gt;.  But now &lt;strong&gt;XxQe1&lt;/strong&gt; is a safe move, and black's only lost one prowler.  White needs to set up a protected cycle immediately, so &lt;strong&gt;Rb1&lt;/strong&gt; is the only sane play - anything else lets black snipe both rooks for no cost.  Black's response is &lt;strong&gt;XxRb1&lt;/strong&gt;, forcing the capture of &lt;strong&gt;NxXb1&lt;/strong&gt; - which is unprotected now, so &lt;strong&gt;XxNb1&lt;/strong&gt; snags a free knight.  At this point, there's no single move that can protect the other rook, so white makes any move &lt;strong&gt;???&lt;/strong&gt;, and black responds with &lt;strong&gt;XxRh1&lt;/strong&gt; (safely).  No two moves can protect both bishops, so black also gets to capture one of them for free.  From this point, white has one bishop, one knight, and a bunch of pawns.  There's no way to stall infinitely with those pieces, so black wins (eventually).  If white starts on a pawn move, followed by a knight move, black should just eat the rook on the same side as the pawn that moved and proceed as above from there.  Again, this isn't a solid proof, but it is a persuasive argument to the effect that black should be able to capture Q,K,R,R,N,B with only two losses - thus allowing them to win with three prowlers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="17389" LastEditDate="2016-02-21T21:08:43.647" LastActivityDate="2016-02-21T21:08:43.647" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27725" PostTypeId="2" ParentId="27717" CreationDate="2016-02-21T05:41:57.130" Score="1" Body="&lt;p&gt;Here's a not-entirely-finished answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An RPG character that's a member of a race like the Pandaren (WoW) or Baraka (TERA)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am always on someone's cam,&#xA;He controlls the way I am.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;cam&quot; meaning computer monitor. The player controls the character using this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am tall,&lt;br&gt;&#xA;And I listen to mall,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am very fat,&lt;br&gt;&#xA;Unlike a bat.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; These are tall and fat races. There's either a bat-like character or enemy that's thinner. For reference, Googling for bat with the RPG name gives &lt;a href=&quot;http://www.wow-petopia.com/images/skin_pix/batskinbrown01_v2.jpg&quot; rel=&quot;nofollow&quot;&gt;a comparatively thin-looking giant bat for WoW&lt;/a&gt;.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &quot;listen to mall&quot; probably refers to some NPC or god named &quot;Mall&quot; or similar that the character follows. Nothing comes up in searches so far, though.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can light a fuse,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://wowwiki.wikia.com/wiki/Quest:A_Short_Fuse&quot; rel=&quot;nofollow&quot;&gt;Quest item&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or abuse,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; PKing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can gather food,&#xA;Or chop down wood,&#xA;I can mine,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Standard MMORPG actions&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or cut through slime,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Basic MMORPG encounter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can build,&#xA;Or sell to a guild.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; More standard MMORPG actions&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can change my look for fame,&#xA;But my sizes will remain the same.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Equipment gives bonuses, is vital to progressing in these games and is the main customization point of a character. However, the underlying character shape is pretty much invariant.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Disclaimer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I haven't played an MMORPG in ages, and I've never played the ones mentioned here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9960" LastActivityDate="2016-02-21T05:41:57.130" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27726" PostTypeId="2" ParentId="27462" CreationDate="2016-02-21T09:34:28.323" Score="3" Body="&lt;p&gt;My guess is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Petrol&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Underground&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is where I'm found&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Found in fossils which are found beneath the surface of earth. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Markets rise and fall with me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Increase or decrease in price of petrol affects the stock market (or atleast the price of normal things). It is also used in many day-to-day things. &lt;a href=&quot;http://www.ranken-energy.com/products%20from%20petroleum.htm&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is the list.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As do actions of countries&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every country need it for transportation. Another explanation could be that many wars have been fought for it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You will find me on the driver's lips, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lipstick and lip balm both contain petroleum jelly. (Thanks to @f&quot; for suggestion)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;attached to the car, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Used in cars. (assuming the car is not using CNG or LPG)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and part of the road in cold parts of the US.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's hard for diesel engine to start in cold weather. So petrol is the choice. Also, the petrol may spill out of the petrol tank on the road. And it is also used in asphalt.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Something Powerful&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Powerful enough to blow. Powerful enough to be the reason of war.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-02-23T08:53:06.377" LastActivityDate="2016-02-23T08:53:06.377" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27728" PostTypeId="1" AcceptedAnswerId="27730" CreationDate="2016-02-21T10:49:18.587" Score="11" ViewCount="811" Body="&lt;p&gt;The teacher tells Summo and Prodo: &lt;em&gt;&quot;I have picked three positive integers $x\le y\le z$. I have whispered the sum $S=x+y+z$ into Summo's ear, and I have whispered the product $P=xyz$ into Prodo's ear.&quot;&lt;/em&gt; Now the following conversation takes place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summo: &lt;em&gt;&quot;I do not know $x,y,z$. But if I knew that your number $P$ is greater than my number $S$, then I would be able to determine $x,y,z$.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prodo: &lt;em&gt;&quot;Aha! Actually my number $P$ is less than your number $S$. And I am able to determine $x,y,z$.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What are these numbers $x,y,z$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="10615" LastEditDate="2016-02-24T04:01:36.943" LastActivityDate="2016-02-25T21:52:12.093" Title="Another sum and product puzzle" Tags="|mathematics|meta-knowledge|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27729" PostTypeId="1" AcceptedAnswerId="27800" CreationDate="2016-02-21T10:53:41.800" Score="18" ViewCount="812" Body="&lt;p&gt;I gave my friends a lateral thinking puzzle &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There are six eggs in the basket. Six people each take one of the eggs. How can it be that one egg is left in the basket?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My intended answer was&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One of them put the egg again inside the basket.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But one of my friend answered &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The last person took the basket with the last egg still inside.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clearly my friend's answer is more creative then my own (as lateral thinking puzzle demands). Now I am confused if I should take his answer for granted (as I should because I found his answer more creative) or should I say that his answer is wrong just because it was not my intended answer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition given by &lt;a href=&quot;https://en.wikipedia.org/wiki/Puzzle&quot;&gt;wikipedia on puzzle&lt;/a&gt; is: A puzzle is a game, problem, or toy that tests a person's ingenuity or knowledge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It got to me thinking if puzzles are just to find the intended answer of the question asker or is it to find the most appropriate and reasonable answer. Which one of them would be better? Or is there something I am missing?&lt;/p&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="12408" LastEditDate="2016-02-21T21:47:49.157" LastActivityDate="2017-05-19T15:19:22.923" Title="How to choose the best answer in puzzles?" Tags="|puzzle-creation|puzzle-theory|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="27730" PostTypeId="2" ParentId="27728" CreationDate="2016-02-21T11:23:05.473" Score="14" Body="&lt;p&gt;The fact $3z\geq x+y+z&amp;gt;xyz$ implies that $3&amp;gt;xy$, so $x=1$ and $y=1$ or $x=1$ and $y=2$. In the former one, we get $2+z&amp;gt;z$. Otherwise we get $3+z&amp;gt;2z$, so $z&amp;lt;3$ and then in fact $z=2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the possible tuples are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(1,1,z)$ and $(1,2,2)$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, Summo can only deduce what the tuple is if he knows that $P&amp;gt;S$. It follows that for his sum, there is exactly one possible tuple with $P&amp;gt;S$ and at least one with $P\leq S$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $S\geq 8$, we can split it in two triples $(2,3, S-5)$ and $(2,2,S-4)$. For both we have that $P&amp;gt;S$. Hence $S&amp;lt;8$. When $S\leq 5$, there are no tuples with $P&amp;gt;S$. We either have $(1,2,2)$ for a sum of 5 and a product of 4, or $(1,1,k)$ for a sum of $k+2$ and a product of $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $S=7$, then both $(2,2,3)$ and $(1,2,4)$ were possible. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hence $S=6$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now we can finish it. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The fact that $P&amp;lt;S$ now implies that $x=y=1$ and $z=4$, from the first part of my answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16322" LastEditorUserId="16322" LastEditDate="2016-02-25T20:35:31.737" LastActivityDate="2016-02-25T20:35:31.737" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27731" PostTypeId="1" AcceptedAnswerId="27734" CreationDate="2016-02-21T11:49:07.843" Score="10" ViewCount="490" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Using one of me is common in science and other areas,&lt;br&gt;&#xA;  but in other areas they sometimes use me the wrong way&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Using two of me is taboo in science, unless one of me is larger,&lt;br&gt; but in other areas they abuse me again by using two small variants of me&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Using three of me is taboo everywhere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="16322" LastActivityDate="2016-02-21T16:55:36.603" Title="One is good, two can be good, three is taboo" Tags="|riddle|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27732" PostTypeId="1" AcceptedAnswerId="27733" CreationDate="2016-02-21T15:53:34.057" Score="10" ViewCount="655" Body="&lt;p&gt;Yesterday a mysterious man came to my house. I didn't open the door, because I didn't know him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later I found a message under the door as show below:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1+1-1·0*1 n 1*9/9 y·w 2^2 n sqrt(49) sqrt(9) d·140/2·126/2·y 0*0 ur·fr 26/2 nd!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What does the message say?&lt;/p&gt;&#xA;" OwnerDisplayName="user18031" LastEditorUserId="13484" LastEditDate="2016-02-21T22:17:21.833" LastActivityDate="2016-02-21T22:17:21.833" Title="Encoding Extreme #2" Tags="|riddle|mathematics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27733" PostTypeId="2" ParentId="27732" CreationDate="2016-02-21T16:11:29.733" Score="7" Body="&lt;p&gt;&lt;strong&gt;Answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;I only wanted to be your friend!&quot;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reason:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Each equation represents &lt;a href=&quot;https://simple.wikipedia.org/wiki/Leet&quot; rel=&quot;nofollow&quot;&gt;Leet&lt;/a&gt; letters.&lt;/strong&gt;&lt;br&gt;&lt;br&gt;&#xA; 1+1-1 = 1 or I&lt;br&gt;&#xA; 0*1 = 0 or O&lt;br&gt;&#xA; 1*9/9 = 1 or L&lt;br&gt;&#xA; 2^2 = 4 or A&lt;br&gt;&#xA; sqrt(49) = 7 or T&lt;br&gt;&#xA; sqrt(9) = 3 or E&lt;br&gt;&#xA; 140/2 = 70 or TO&lt;br&gt;&#xA; 126/2 = 63 or BE (bE)&lt;br&gt;&#xA; 0*0 = 0 or O&lt;br&gt;&#xA; 26/2 = 13 or IE&lt;br&gt;&lt;br&gt;&#xA; &lt;strong&gt;Which combined with the letters included between equations creates the below text in Leet:&lt;/strong&gt;&lt;br&gt;&#xA; &lt;em&gt;&quot;1 0n1y w4n73d 70 63 y0ur fr13nd!&quot;&lt;/em&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12014" LastEditorUserId="12014" LastEditDate="2016-02-21T20:29:10.417" LastActivityDate="2016-02-21T20:29:10.417" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27734" PostTypeId="2" ParentId="27731" CreationDate="2016-02-21T16:55:36.603" Score="10" Body="&lt;p&gt;This is a long shot, but is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter 'k'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using one of me is common in science and other areas,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; k could be many things in science, such as a variable or a prefix to a unit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but in other areas they sometimes use me the wrong way&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'k' is sometimes used as an informal abbreviation for 'okay'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using two of me is taboo in science, unless one of me is larger,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If 'k' is a variable, it would not be good to name two variables k as you couldn't tell them apart, unless one was lowercase and the other capital. If 'k' is a unit prefix, there is no unit 'kk', but there is the unit 'kK' (kilokelvin).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but in other areas they abuse me again by using two small variants of me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The word 'kk' is also used as an informal way of saying 'okay'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using three of me is taboo everywhere.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Three k's would represent the KKK, a racist organization.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19212" LastActivityDate="2016-02-21T16:55:36.603" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27735" PostTypeId="2" ParentId="5900" CreationDate="2016-02-21T17:48:20.503" Score="-4" Body="&lt;p&gt;It's pretty easy to figure out I have worked out what it says: &quot;Sherlock I have found you&quot; it's an easy message code. You use a grid that shows the symbols in patterns. &lt;/p&gt;&#xA;" OwnerUserId="19469" LastActivityDate="2016-02-21T17:48:20.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27736" PostTypeId="2" ParentId="27717" CreationDate="2016-02-21T17:59:50.817" Score="5" Body="&lt;p&gt;You're&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the PC from Minecraft -- I think his name is Steve&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am always on someone's cam,&lt;br&gt;&#xA;He controlls the way I am.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Veedrac explained this -- on a computer screen, the player controls you&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am tall,&lt;br&gt;&#xA;And I listen to mall,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are taller than some of the other creatures (source: Minecraft Dimensions, &lt;a href=&quot;http://ata4.info/mcdim/&quot; rel=&quot;nofollow&quot;&gt;http://ata4.info/mcdim/&lt;/a&gt;).  &quot;mall&quot; is one of the music discs in Minecraft (source: Minecraft Wiki, &lt;a href=&quot;http://minecraft.gamepedia.com/Minecraft_Wiki&quot; rel=&quot;nofollow&quot;&gt;http://minecraft.gamepedia.com/Minecraft_Wiki&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am very fat,&lt;br&gt;&#xA;Unlike a bat.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again according to Minecraft Dimensions, bats are basically half a meter in all dimensions, but the PC is a meter wide&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can light a fuse,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In Minecraft TNT has a fuse&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or abuse,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This could mean semi-cheat stuff (&quot;abuse&quot; the mechanisms), using abusive language in chat, or just the fact that you're beating up/killing creatures&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can gather food,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Minecraft contains food&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or chop down wood,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Minecraft contains wood&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can mine,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yep, you can&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or cut through slime,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Minecraft contains slime&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can build,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yep, you can&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or sell to a guild.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think only mods have guilds, but a lot of people play Minecraft with a mod, so.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can change my look for fame,&lt;br&gt;&#xA;But my sizes will remain the same.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Different player skins are available but they don't change the PC's abilities/size&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who am I ?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; See the top of this answer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12092" LastEditorUserId="12092" LastEditDate="2016-02-21T18:29:19.923" LastActivityDate="2016-02-21T18:29:19.923" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27737" PostTypeId="1" AcceptedAnswerId="27749" CreationDate="2016-02-21T18:10:28.027" Score="10" ViewCount="251" Body="&lt;p&gt;Shafiqa chose a positive integer number $X\le100$, and Habibi is trying to guess the number.  Habibi can select two natural numbers $M$ and $N$ less than $100$ and ask about the greatest common divisor $\gcd(X+M,N)$ of $x+M$ and $N$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can Habibi always determine Shafiqa's number with at most seven questions?&lt;/p&gt;&#xA;" OwnerUserId="19470" LastEditorUserId="15632" LastEditDate="2016-02-21T18:24:15.080" LastActivityDate="2016-02-22T09:46:50.717" Title="The game of Shafiqa and Habibi" Tags="|strategy|" AnswerCount="3" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27738" PostTypeId="1" AcceptedAnswerId="27740" CreationDate="2016-02-21T18:27:55.353" Score="12" ViewCount="1462" Body="&lt;p&gt;Mahmoud and Zomhan play a game where they alternatively write numbers on a board. In each turn, a player takes the old number $N$ written on the board and replaces it either by the number $T$ of divisors of $n$ or by the difference $N-T$.  For example if $N=22$ is written on the blackboard then $T=4$ with the divisors 1, 2, 11, 22 of $N=22$, and the next player may write $T=4$ or $N-T=18$ on the blackboard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mahmoud is the first player to play, and whichever player is the first player to write the number $0$ is the winner of the game. Given that the initial number on the board is $1036$, determine which player has a winning strategy.&lt;/p&gt;&#xA;" OwnerUserId="19470" LastEditorUserId="10615" LastEditDate="2016-02-21T19:07:36.670" LastActivityDate="2016-02-21T19:07:36.670" Title="The game of 1036" Tags="|mathematics|game|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27739" PostTypeId="2" ParentId="27692" CreationDate="2016-02-21T18:39:04.910" Score="5" Body="&lt;p&gt;I think the answer is fire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can touch fire but it is insubstantial and does not weigh much, if anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can cause damage. &quot;Material fall&quot; = &quot;burning down&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It used to be created using a featherstick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a Maori myth about fire being discovered as a result of curiosity.&#xA;And lightning can start a fire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the &quot;opposing animal&quot; I am  thinking phoenix.&#xA;Fire can certainly spread like a disease.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Too much fire causes problems (e.g. forest fires) or simply being too hot, but lack of it (light or heat) means danger from predators or fear of the dark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some people in modern societies sometimes never see fire due to central heating.&#xA;But others use it all the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes fire is highly controlled and contained to do its work (e.g. pilot light).&#xA;&quot;Thrown away&quot; = &quot;put out&quot; ? As in &quot;put out the trash&quot;.&lt;/p&gt;&#xA;" OwnerUserId="19471" LastActivityDate="2016-02-21T18:39:04.910" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="27740" PostTypeId="2" ParentId="27738" CreationDate="2016-02-21T19:05:27.520" Score="17" Body="&lt;p&gt;A player has a winning strategy if and only if&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;They can write $0$, or&lt;/li&gt;&#xA;&lt;li&gt;They can write a number where their opponent does not have a winning strategy.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If $N$ is $1$ or $2$, the player can write $0$, so they win.&lt;br&gt;&#xA;If $N$ is $3$, the player must write $1$ or $2$, and then the opponent wins.&lt;br&gt;&#xA;If $N$ is $4$ or $9$ (3 divisors for each), the player can write $3$ and win.&lt;br&gt;&#xA;If $N$ is $6$ (4 divisors), the player must write $2$ or $4$, but the opponent wins from both.&lt;br&gt;&#xA;If $N$ is $11$ (2 divisors), the player must write $2$ or $9$, but the opponent wins from both.&lt;br&gt;&#xA;If $N$ is $12$ (6 divisors), the player writes $6$ and wins.&lt;br&gt;&#xA;If $N$ is $1024$ (11 divisors), the player can write $11$ and win.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If $N$ is $1036$ (12 divisors),&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;the only options are to write $12$ or $1024$. Both of these let the opponent win. So Mahmoud does not have a winning strategy, and Zomhan does.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2016-02-21T19:05:27.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27741" PostTypeId="1" AcceptedAnswerId="27743" CreationDate="2016-02-21T20:48:50.467" Score="2" ViewCount="383" Body="&lt;p&gt;Can one buy Prestige?&lt;br&gt;&#xA;Are Advantages sold?&lt;br&gt;&#xA;For a yearly fee&lt;br&gt;&#xA;Can one get Gold?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which company offers all of these?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Answer is the name of a company that offers the above.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19179" LastEditorUserId="12408" LastEditDate="2016-02-21T21:31:48.653" LastActivityDate="2016-02-21T21:31:48.653" Title="Who sells these?" Tags="|riddle|word|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27743" PostTypeId="2" ParentId="27741" CreationDate="2016-02-21T20:57:37.017" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; VISA &lt;a href=&quot;https://cafcu.org/?Cabinet=Main&amp;amp;Drawer=Loans+and+Visa+Cards&amp;amp;Folder=Visa+Credit+Cards&amp;amp;SubFolder=Visa+Advantage&quot;&gt;advantage&lt;/a&gt; &lt;a href=&quot;https://www.desjardins.com/ca/personal/loans-credit/credit-cards/visa-prestige-platinum/&quot;&gt;prestige&lt;/a&gt;, and &lt;a href=&quot;http://www.visa.ca/en/personal/cardsandproducts/visagold.jsp&quot;&gt;gold&lt;/a&gt; cards all exist and presumably have an annual fee.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="392" LastActivityDate="2016-02-21T20:57:37.017" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="27744" PostTypeId="5" CreationDate="2016-02-21T21:07:03.070" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2016-02-21T21:07:03.070" LastActivityDate="2016-02-21T21:07:03.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27745" PostTypeId="4" CreationDate="2016-02-21T21:07:03.070" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2016-02-21T21:07:03.070" LastActivityDate="2016-02-21T21:07:03.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27746" PostTypeId="1" CreationDate="2016-02-21T21:22:20.910" Score="5" ViewCount="1397" Body="&lt;p&gt;From the answers that I got over &lt;a href=&quot;https://puzzling.stackexchange.com/questions/27502/is-reverse-engineering-crossword-puzzles-a-thing&quot;&gt;here&lt;/a&gt;, I found out that the type of puzzle that I was looking for would be referred to as &lt;em&gt;diagramless fill-in puzzle&lt;/em&gt;. (A cross of a diagramless crossword and a fill-in puzzle.)&lt;/p&gt;&#xA;&lt;h3&gt;Where can I find such puzzles?&lt;/h3&gt;&#xA;" OwnerUserId="19258" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2017-12-29T17:20:57.260" Title="Where can I find diagramless fill-in puzzles?" Tags="|crosswords|reference-request|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="27747" PostTypeId="2" ParentId="27729" CreationDate="2016-02-21T21:51:57.747" Score="5" Body="&lt;p&gt;Suppose it was the &lt;em&gt;opposite&lt;/em&gt;: you had the most creative answer, and your friend gave the less creative answer. Then you could:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clarify your question by saying that six people each took one of the eggs and nobody put it again in the basket.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This way you narrowed the answers to the right one, which is also the most creative one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is a matter of opinion, but I'm in favor of the &lt;em&gt;most creative answer&lt;/em&gt; no matter what, specially for lateral thinking puzzles. So if your question admits more than one answer, you should accept the more creative one. &lt;/p&gt;&#xA;" OwnerUserId="19446" LastActivityDate="2016-02-21T21:51:57.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27748" PostTypeId="2" ParentId="27692" CreationDate="2016-02-21T23:02:06.093" Score="6" Body="&lt;p&gt;First time answering one of these.  I can't work out one of them, and the relation is a little shaky in another, but for the most part I think the answer fits well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; God&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You call me as something physical, something you can touch.&#xA;But to be technically correct, I don't weight very much.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; God is referred to as a being or entity, but to be technically correct it is an idea/belief&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In fact, my weight is zero, I'm not of matter at all.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://farm9.staticflickr.com/8813/18358215466_4e20ed7b62_o.jpg&quot; rel=&quot;noreferrer&quot;&gt;God doesn't weigh anything&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet, I can cause damage, inflicting material fall.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Earthquakes and other natural disasters are referred to as &quot;Acts of God&quot;.  Not too sure about &quot;inflicting&quot; material fall.  At a stretch, God created gravity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My creation involved once upon a time a feather. (&lt;em&gt;Haven't worked this one out&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Unknown (although I like the word &quot;creation&quot;).  I'm not a religious man myself and a quick Google didn't find any clear links to God or creation with feathers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I'm said to be a spawn of curiosity and weather.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People always blame weather and chance on God.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm opposed to an animal but I can spread like a disease.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; God reversed is Dog.  Religion spreads like disease.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I'm plenty you'll nag, if I'm scarce you'll feel an unease.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Godfellas&quot; rel=&quot;noreferrer&quot;&gt;Futurama&lt;/a&gt; tackled this, saying if God intervened and helped out then everyone would rely on divine intervention (constant prayer = nag).  On the contrary if there is no signs of divine intervention, people feel isolated and lost (uneasy).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some people haven't yet seen me, others see me all day.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people have never found religion, others see signs of God everywhere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I'm kept securely, sometimes thrown away.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Kept securely in ones heart/soul, or religion/God can be lost (thrown away).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; God&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19476" LastEditorUserId="19476" LastEditDate="2016-02-22T00:32:11.367" LastActivityDate="2016-02-22T00:32:11.367" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27749" PostTypeId="2" ParentId="27737" CreationDate="2016-02-21T23:07:20.363" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Yes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;He first asks the values from $gcd(X+0,30)$ to $gcd(X+3,30)$. Two of $X,X+1,X+2,X+3$ are bound to be divisible by 2, and at least one is a multiple of 3. If none of them are divisible by 5, $X+4$ is, but whether it's the case or not he can then determine the $mod 2,3$ and $5$ values of $X$. Thus, there are 3 or 4 possible $X$ solutions in the form of $n, n+30, n+60$, with $n+90$ in the mix if possible. If he asks the GCDs of $X$ and these values starting from the last, he can either find the solution directly (the first answer equal to the number he picked) or by eliminating the rest after 3 more questions at most.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An alternative solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;He first asks the values from $gcd(X+0,70)$ to $gcd(X+5,70)$, which allows him to determine the $mod 2,5$ and $7$ values of $X$. Thus, there are 1 or 2 possible $X$ solutions in the form of $n$ and $n+70$ if possible. If $n&amp;gt;29$, there's only one solution. Otherwise he asks the GCD of $X$ and $n+70$, after which he can either find the solution directly or by eliminating the rest.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="14016" LastEditDate="2016-02-22T02:44:38.863" LastActivityDate="2016-02-22T02:44:38.863" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27750" PostTypeId="1" AcceptedAnswerId="27756" CreationDate="2016-02-22T00:11:20.623" Score="2" ViewCount="486" Body="&lt;p&gt;One day, a person was in love. He wanted to find the person's phone number.&lt;/p&gt;&#xA;&lt;p&gt;So they start up a conversation.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Person: What's your phone number, sweetheart?&lt;/p&gt;&#xA;&lt;p&gt;Person's Love: I'm not going to give it that easily. If you replace the first digit of your phone number with the lowest odd digit, you'll get my number.&lt;/p&gt;&#xA;&lt;p&gt;Person: And the 4-digit code before it?&lt;/p&gt;&#xA;&lt;p&gt;Person's Love: You know, the product of the four digits in the code is actually the square root of my phone number.&lt;/p&gt;&#xA;&lt;p&gt;Person: Hey, that's insufficient. I'm a plumber, not God.&lt;/p&gt;&#xA;&lt;p&gt;Person's Love: But if I tell you the sum of digits of the code, I'd have told you too much.&lt;/p&gt;&#xA;&lt;p&gt;Person: I love you, dear.&lt;/p&gt;&#xA;&lt;p&gt;Person's Love: I love you too.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So what is the 4-digit code?&lt;/p&gt;&#xA;" OwnerUserId="17970" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-02-22T02:04:55.993" Title="Phone numbers and lovers" Tags="|logical-deduction|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27751" PostTypeId="1" AcceptedAnswerId="27779" CreationDate="2016-02-22T00:16:14.983" Score="3" ViewCount="146" Body="&lt;p&gt;The answer is a year, find it by working out this clue:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The cows first subtract one, almost too rushed brew gins?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18522" LastEditorUserId="18522" LastEditDate="2016-02-22T01:44:35.907" LastActivityDate="2016-02-22T14:35:45.417" Title="When did these cows brew gins in a hurry?" Tags="|cryptic-clues|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="27752" PostTypeId="1" AcceptedAnswerId="27754" CreationDate="2016-02-22T00:28:11.827" Score="16" ViewCount="2588" Body="&lt;p&gt;$12345$ players take part in a knockout tournament. In each round players are paired up; each pair plays a game with the winning player advancing to the next round (no ties). If there are an odd number of players at the start of a round, one is randomly selected to automatically progress. This continues until one player is delared champion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many games are played in total?&lt;/p&gt;&#xA;" OwnerUserId="18508" LastActivityDate="2016-04-19T14:58:21.667" Title="A knockout tournament" Tags="|mathematics|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27753" PostTypeId="2" ParentId="27752" CreationDate="2016-02-22T00:48:05.977" Score="14" Body="&lt;p&gt;Let $a_n$ be the number of players at the beginning of the $n$-th round, so that $a_1 = 12345$. Then $a_{n+1} = \lceil \frac{a_n}{2} \rceil$, and the number of games played in round $n$ is $g_n = \lfloor \frac{a_n}{2}\rfloor$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We thus have the following sequence for $a_n$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$12345, 6173, 3087, 1544, 772, 386, 193, 97, 49, 25, 13, 7, 4, 2, 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the following sequence for $g_n$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$6172, 3086, 1543, 772, 386, 193, 96, 48, 24, 12, 6, 3, 2, 1, 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a total of $12344$ games.&lt;/p&gt;&#xA;" OwnerUserId="16478" LastActivityDate="2016-02-22T00:48:05.977" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27754" PostTypeId="2" ParentId="27752" CreationDate="2016-02-22T01:06:39.480" Score="38" Body="&lt;p&gt;In every game one person loses. There is only one winner, so there must be $12344$ losers. Therefore there are $12344$ games.&lt;/p&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2016-02-22T01:06:39.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27755" PostTypeId="1" AcceptedAnswerId="27910" CreationDate="2016-02-22T01:37:02.133" Score="11" ViewCount="1124" Body="&lt;p&gt;A vicious villain am I&lt;br&gt;&#xA;The greatest of my kind&lt;br&gt;&#xA;Covering the face of the Earth&lt;br&gt;&#xA;A fiend of all mankind&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although I may appear weak and small&lt;br&gt;&#xA;Great power I am assigned&lt;br&gt;&#xA;Those who fail to heed my warning&lt;br&gt;&#xA;Condemnation they will find&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two natures do I have&lt;br&gt;&#xA;A son of God and of the Devil&lt;br&gt;&#xA;Though temples have been built to me&lt;br&gt;&#xA;Thieves are exalted to a higher level&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A loud cry sounds from the town&lt;br&gt;&#xA;Gloriously is my appearance made&lt;br&gt;&#xA;Considered me not your enemy&lt;br&gt;&#xA;And you will have nothing of which to be afraid&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer this puzzle and you will see&lt;br&gt;&#xA;All my secrets be made clear&lt;br&gt;&#xA;Consider, though, as you proceed&lt;br&gt;&#xA;All riddles are not as they appear&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every line is useful to determine the correct answer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another Hint (please only view after attempting):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is something that is tangible, and not a concept or property.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19212" LastEditorUserId="19212" LastEditDate="2016-02-22T14:54:27.973" LastActivityDate="2016-02-25T14:21:14.870" Title="A vicious villain" Tags="|riddle|lateral-thinking|enigmatic-puzzle|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27756" PostTypeId="2" ParentId="27750" CreationDate="2016-02-22T01:37:12.523" Score="4" Body="&lt;p&gt;The information given is that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The product of the digits is $X$, where $X$ is a number that the person knows, but that we do not know. $X^2$ starts with an odd digit.&lt;/li&gt;&#xA;&lt;li&gt;The person does not know the number given only the product, but if the sum of the digits was given, then the person would know the number.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The first thing to notice is that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If the digits of the code are rearranged, the product and sum stay the same. So all four digits &lt;em&gt;must be the same&lt;/em&gt;. Otherwise there would be multiple possible codes and the second bullet point would be false.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Given that, the possible values of $X$ are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The fourth powers of each digit: $0$,$1$,$16$,$81$,$256$,$625$,$1296$,$2401$,$4096$, or $6561$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can rule out the values whose squares start with an even digit, leaving&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$1$,$625$,$1296$, and $4096$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But if $X$ was&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$1$, $625$, or $4096$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;then there would only be one possible code that produced that product, and the person have known already. So we can eliminate these possibilities. This means $X$ must be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$1296$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the code must be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$6666$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2016-02-22T01:37:12.523" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27757" PostTypeId="2" ParentId="27617" CreationDate="2016-02-22T02:33:55.657" Score="6" Body="&lt;p&gt;Are you:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Jack?&#xA; &lt;a href=&quot;https://i.stack.imgur.com/viXTG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/viXTG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA; The jack has two eyes with two neighboring bodies in between. If you remove one of the eyes, there is still a complete body on the other side of the card. Ditto for removing the body that lost it's eye. However, when you remove the second eye, then it is no longer a complete body.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how this relates to the word culture though.&lt;/p&gt;&#xA;" OwnerUserId="18821" LastActivityDate="2016-02-22T02:33:55.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27758" PostTypeId="2" ParentId="27755" CreationDate="2016-02-22T02:42:18.933" Score="1" Body="&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Spelling errors&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A vicious villain am I&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Absolutely heinous indeed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The greatest of my kind&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There's nothing worse&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Covering the face of the Earth&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The planet is literally infested with people who can't spell.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A fiend of all mankind&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Even intelligent people are sometimes affected.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Although I may appear weak and small&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many people think it doesn't matter...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Great power I am assigned&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...but those people are wrong, it does matter. People don't get jobs, they get ridiculed, they miss out on business opportunities, they are ignored, etc. all because they spell something wrong and are branded an idiot by the reader, who subsequently treats the entire message with disdain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Those who fail to heed my warning&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some people don't even notice them...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Condemnation they will find&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oh yes - they will be condemned all right!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Two natures do I have&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well there's the one where people simply don't realise that &quot;their&quot; and &quot;there&quot;, and &quot;your&quot; and &quot;you're&quot; are actually different words, then there's the one where they just spell something wrong, for example &quot;definately&quot;. There's also the kind where British people use &quot;realise&quot; as I did above, and American people use &quot;realize&quot;, but technically these are alternate spellings rather than errors, although they may be used by mistake. There's also the one where people accidentally type a different word than they meant to, like &quot;thought&quot; instead of &quot;though&quot;, but that's a typing error, or 'typo'. So that's just two natures, since the last two don't count.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A son of God and of the Devil&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jesus couldn't spell for toffee, and the devil's in the details!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though temples have been built to me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We have plenty of temples to educate against spelling errors, and to mock them: web pages; info-graphics; books; comedy sketches; YouTube videos; etc. (oh, and schools, but I'm not sure they do as good a job as any of the others).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thieves are exalted to a higher level&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That's because it's spelt correctly, it's not &quot;Thiefs&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A loud cry sounds from the town&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every time someone has to suffer the eye-stabbing pain that results from spelling errors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Gloriously is my appearance made&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We revel in pointing them out.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Considered me not your enemy&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That's true - it's not your fault, it's the guy writing it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And you will have nothing of which to be afraid&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Those who embrace spelling errors by simply not knowing or caring have nothing to fear. They are oblivious to the horrible path of destruction they leave in their wake.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer this puzzle and you will see&lt;br&gt;&#xA;All my secrets be made clear&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Please tell me this means if I get this puzzle right I'll never have to suffer spelling errors again, and I'll never make any myself and have that awful feeling of disgust for myself when I realise my error.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Consider, though, as you proceede&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ahh, a sneaky clue... I see what you did there!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All riddles are not as they appear&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That's right, this entire riddle was just to point out that single spelling error on the previous line... devious... very devious indeed my friend!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now if that's not lateral thinking, I don't know what is :P&lt;/p&gt;&#xA;" OwnerUserId="18522" LastActivityDate="2016-02-22T02:42:18.933" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27759" PostTypeId="2" ParentId="27737" CreationDate="2016-02-22T03:00:35.413" Score="2" Body="&lt;p&gt;Even I like the answer of Nautilus, I have a different approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the very first round M isn't important, because you can swap M and X.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I set M to 1 and want to get the most possible information out of the GCD-Answer. Like Nautilus I came to N = 30 (because of 2*3*5) to &quot;filter&quot; these prime numbers. But this is improvable. I can get out more information if I choose a number beyond 50. (50 because 1+50*z is at maximum 2 times a number from 1 to 100)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I choose 30 I get 8 possible results for GCD(1+x, 30). For example N = 60 (2*2*3*5) or 72 (2*2*2*3*3) or 96 (2*2*2*2*2*3) would give me 12 different values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means: If I can build a binary search for the problem, I will find an answer within 6 rounds. This is a little bit complicated to explain (for me), but there are only 32 different possibilities where the GCD-answer is 1. And Sqrt(32) = 5.6568. This all counts only for the first round. Because if you can eliminate so much branches here, you can maybe get more information in the next rounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So first I think about an AI for this, but I don't want to overengineer and just bruteforced it with an algorithm to find a fingerprint in 6 rounds for all 100 X.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 6 rounds: (M=1,N=40), (M=2,N=30), (M=3,N=40), (M=4,N=80), (M=5,N=27), (M=6,N=48) gives a fingerprint&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is not all. Because you can eliminate more numbers per round than only 50%, you can build a shorter fingerprint.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 5 rounds: (M=1,N=40), (M=2,N=30), (M=3,N=40), (M=4,N=80), (M=6,N=48)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4342" LastEditorUserId="4342" LastEditDate="2016-02-22T03:17:10.883" LastActivityDate="2016-02-22T03:17:10.883" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27760" PostTypeId="2" ParentId="27692" CreationDate="2016-02-22T03:20:15.667" Score="4" Body="&lt;p&gt;My guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a hole&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You call me as something physical, something you can touch. But to be technically correct, I don't weight very much.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People refer to a hole as a physical object. &quot;There's a hole in my yard.&quot; Also, &quot;touching the void&quot; - void is another word for hole.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In fact, my weight is zero, I'm not of matter at all.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A hole is literally the absence of matter, and therefore weightless by definition.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet, I can cause damage, inflicting material fall.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If material falls in a hole, it can be damaged.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My creation involved once upon a time a feather.&#xA;But I'm said to be a spawn of curiosity and weather.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Electron_hole&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Electron_hole&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm opposed to an animal &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To seal something is to remove holes from it. And a seal is an animal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I can spread like a disease.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A hole in your clothes can spread.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I'm plenty you'll nag&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People complain about holes in the road.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;if I'm scarce you'll feel an unease.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Put someone in a box with no holes and they'll feel unease.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some people haven't yet seen me, others see me all day.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Could apply to just about anything, including holes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I'm kept securely&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Like the hole that lets you into a building, a door, which would normally have a lock on it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;sometimes thrown away.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Like a hole in a shoe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7684" LastEditorUserId="7684" LastEditDate="2016-02-22T03:26:00.773" LastActivityDate="2016-02-22T03:26:00.773" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27761" PostTypeId="2" ParentId="27755" CreationDate="2016-02-22T05:46:30.487" Score="8" Body="&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That villainous thief, &lt;a href=&quot;https://en.wikipedia.org/wiki/Carmen_Sandiego_%28character%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Carmen Sandiego&lt;/a&gt;.&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/3PIFP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3PIFP.png&quot; alt=&quot;Carmen Sandiego&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'd go through each line explaining this deduction, but that would be a waste of time because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The lines actually aren't important.  What's important is the first letter of each line.&#xA; &#xA; Each line begins with one of the following four letters:  &lt;code&gt;ATCG&lt;/code&gt;.  These are the letters used to describe the different nucleotides in a strand of DNA.  So rewritten to only include these first letters, the riddle becomes:&lt;P&gt;&#xA; &lt;code&gt;ATCA AGTC TATT AGCA AACA&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This, of course, doesn't get us a usable answer.  However, if we consult Wikipedia, we learn that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Nucleic_acid_notation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Groups of possible nucleotides in a particular position can be represented by other letters&lt;/a&gt;.  After exhaustive groupings and re-groupings, I found this intriguing one:&lt;P&gt;&#xA; &lt;code&gt;original:  ATCA AGTC TATT AGCA AACA&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;grouped:   H--A D--Y -ATT R-M- AM-A&lt;/code&gt;&lt;P&gt;&#xA;&#xA; And of course, &lt;code&gt;HADYATTRMAMA&lt;/code&gt; is an obvious anagram of &lt;code&gt;Tardy Hat Mama&lt;/code&gt;.&lt;P&gt;  Which, of course, could only be referring to that international thief who's never where you expected her to be when you expected her to be there (and who is absolutely rocking that fedora), Carmen Sandiego.&lt;P&gt;&#xA;&lt;strong&gt;QED.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9286" LastEditorUserId="9286" LastEditDate="2016-02-22T05:52:07.750" LastActivityDate="2016-02-22T05:52:07.750" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27762" PostTypeId="5" CreationDate="2016-02-22T05:52:43.693" Score="0" Body="&lt;p&gt;From Google, we have:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the mathematical study of the coding of information in the form of sequences of symbols, impulses, etc. and of how rapidly such information can be transmitted, for example through computer circuits or telecommunications channels.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11497" LastEditorUserId="25957" LastEditDate="2016-08-05T12:08:27.887" LastActivityDate="2016-08-05T12:08:27.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27763" PostTypeId="4" CreationDate="2016-02-22T05:52:43.693" Score="0" Body="For puzzles that use an aspect of information theory - the processing and transmission of information." OwnerUserId="25957" LastEditorUserId="25957" LastEditDate="2016-08-05T12:08:20.983" LastActivityDate="2016-08-05T12:08:20.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27764" PostTypeId="2" ParentId="27755" CreationDate="2016-02-22T06:27:38.667" Score="3" Body="&lt;p&gt;Answer is :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;SUN&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A vicious villain am I,&#xA;The greatest of my kind &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For all the damage it has done to the planets that are not in goldilock region(eg mercury,mars),it is a vicious villian. No body doubts its superiority in our solar system,all planets orbit around it and thus it is the greatest star of our solar system.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Covering the face of the Earth,A fiend of all mankind&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When it covers one face of earth, its day time on one side ,dark on the other,So it does cover the face of earth. The heat provided by sun was trapped by our atmosphere and our cold planet earth was transformed in to a warm planet that made life sustainable. So Sun had a huge part to play in Earths evolution and thus can be considered as a friend.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Although I may appear weak and small,&#xA;Great power I am assigned,&#xA;Those who fail to heed my warning,&#xA;Condemnation they will find&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It appears tiny in the sky, but in reality it is massive .It is so large that about 1,300,000 planet Earths can fit inside of it. And very powerful indeed because of all the hydrogen combustion and radiation. Its better if you stay away from it, anything that comes near it burns because of extreme tempratures.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Two natures do I have,&#xA;A son of God and of the Devil,&#xA;Though temples have been built to me,&#xA;Thieves are exalted to a higher level&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think there is a very famous temple called SUN temple in India and usually a lot of Hindus worship SUN in India.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A loud cry sounds from the town,&#xA;Gloriously is my appearance made,&#xA;Considered me not your enemy,&#xA;And you will have nothing of which to be afraid&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - Ohhh!! the sunrise is glorious indeed and darkness is compared to loud cry from the towns. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Title - A vicious villian and its devil nature :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There will be a time when the sun will run out of its fuel(hydrogen) and its core will collapse. That moment will give rise to a massive nuclear reaction and the sun will start expanding eventually engulfing earth. That will be our end. A vicious villian as a title is justified.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19151" LastEditorUserId="19151" LastEditDate="2016-02-22T09:12:43.157" LastActivityDate="2016-02-22T09:12:43.157" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="27765" PostTypeId="2" ParentId="27755" CreationDate="2016-02-22T06:40:24.470" Score="2" Body="&lt;h3&gt;This is a wrong answer(my achievement)...confirmed by OP&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I suppose the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Electric Charges&lt;/strong&gt; or &lt;strong&gt;Electricity&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A vicious villain am I&#xA;The greatest of my kind&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In case you get electric shock. (A person can die. What's worse than that?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Covering the face of the Earth&#xA;A fiend of all mankind&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everything is made up of atoms. Atoms have protons and electrons and they have electric charges.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Although I may appear weak and small&#xA;Great power I am assigned&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Protons and Electrons are small and their charges are weak. But the electric force between them is enough to make them bounded.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Those who fail to heed my warning&#xA;Condemnation they will find&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Corona Discharge. You know...the lightning and how it can destroy things.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Two natures do I have&#xA;A son of God and of the Devil&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Positive and Negative charge. They both are opposite (as God and Devil). Son may refer that they are part of atom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though temples have been built to me&#xA;Thieves are exalted to a higher level&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Temples refer to Labs. It is said that Edison stole the invention of light bulb from Nikola Tesla.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A loud cry sounds from the town&#xA;Gloriously is my appearance made&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sparks looks good.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Considered me not your enemy&#xA;And you will have nothing of which to be afraid&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you know how electrical appliances work then you should not be afraid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; manshu&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-02-22T06:50:40.270" LastActivityDate="2016-02-22T06:50:40.270" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27766" PostTypeId="1" AcceptedAnswerId="27771" CreationDate="2016-02-22T09:13:52.520" Score="19" ViewCount="1791" Body="&lt;p&gt;Define a number &lt;strong&gt;self-composable&lt;/strong&gt; if it may be computed using just the digits of the number itself (used just once) and the following operations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;basic operations ( $+, -, \times, \div$)&lt;/li&gt;&#xA;&lt;li&gt;less than basic operations ($x^y, \;\sqrt x, \;x!, \;x.y$ (decimal point))&lt;/li&gt;&#xA;&lt;li&gt;extended operations ($.x, \; .\overline{x}$)&lt;/li&gt;&#xA;&lt;li&gt;parentheses at will&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If the digits of the mathematical operation are in the same order as in the number itself, the number is said &lt;strong&gt;orderly self-composable&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, 25 is self-composable ($5^2 = 25$) and 343 is orderly self-composable, since $(3+4)^3 = 343$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2016 is self-composable too: find how.&lt;/p&gt;&#xA;" OwnerUserId="70" LastEditorUserId="4421" LastEditDate="2016-02-22T17:14:34.493" LastActivityDate="2016-02-22T18:03:41.393" Title="prove that 2016 is a self-composable number" Tags="|mathematics|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27767" PostTypeId="2" ParentId="27737" CreationDate="2016-02-22T09:23:59.623" Score="2" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes &lt;strong&gt;New starting point solves all within 5 guesses&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can achieve this by using a simple algorithm as follows&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;We try m=1,n=60, we eliminate all candidates that would've given&#xA;a different gcd, or those that are not dividable by the gcd (after&#xA;we add m).&lt;/li&gt;&#xA;&lt;li&gt;We try m=2,n=60, once again we eliminating all candidates that&#xA;don't fit.&lt;/li&gt;&#xA;&lt;li&gt;we try m=3,n=60, once again eliminating all candidates that don't&#xA;fit.&lt;/li&gt;&#xA;&lt;li&gt;-5 We try the largest number left, each time eliminating all&#xA;candidates that don't fit.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This works as proven by the following &lt;a href=&quot;https://jsfiddle.net/kjmta86k/11/&quot; rel=&quot;nofollow&quot;&gt;fiddle&lt;/a&gt;, which uses exactly this algorithm on all possibly starting numbers. (&lt;a href=&quot;https://jsfiddle.net/kjmta86k/12/&quot; rel=&quot;nofollow&quot;&gt;this fiddle&lt;/a&gt; is a bit more verbose, and prints all guesses for each number)&lt;/p&gt;&#xA;" OwnerUserId="16846" LastEditorUserId="16846" LastEditDate="2016-02-22T09:46:50.717" LastActivityDate="2016-02-22T09:46:50.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27768" PostTypeId="1" AcceptedAnswerId="27770" CreationDate="2016-02-22T09:39:01.127" Score="5" ViewCount="246" Body="&lt;p&gt;Can you discover the meaning of this image?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xPgMZ.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xPgMZ.png&quot; alt=&quot;puzzle&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19212" LastEditorUserId="12014" LastEditDate="2016-02-22T09:52:36.457" LastActivityDate="2016-02-22T10:05:46.143" Title="Magical color squares" Tags="|enigmatic-puzzle|visual|steganography|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27769" PostTypeId="2" ParentId="27766" CreationDate="2016-02-22T09:44:47.253" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; find divisors of 2016 that contains its own numbers: $2016/2 \rightarrow 1008/2 \rightarrow 504/2 \rightarrow 252/2 \rightarrow 126/126 \rightarrow 1$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;therefore;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; we can't use $2^4$ since $4$ is not in $2016$ but we can do $2^2$ for $2$:  $2^2 \times 2^2 \times 126 \times 1^0$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18685" LastEditorUserId="4421" LastEditDate="2016-02-22T17:21:21.413" LastActivityDate="2016-02-22T17:21:21.413" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27770" PostTypeId="2" ParentId="27768" CreationDate="2016-02-22T09:51:48.607" Score="6" Body="&lt;p&gt;I don't know if this is what you wanted to be found or there is more to it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; get the rgb colors and translate those into their ascii correspondents: &lt;br&gt;&lt;br&gt;&#xA;[105 - 110 - 32] - [97 - 114 - 99] - [117 - 109 - 32] - [118 - 101 - 114] - [105 - 116 - 97] - [115 -  32 - 101] - [115 - 116 - 33]&lt;br&gt;&#xA; which translates into : &lt;strong&gt;In arcum veritas est!&lt;/strong&gt; which further translates in &lt;strong&gt;The truth is in the bow/rainbow!&lt;/strong&gt; (I mean, I think it translates to this)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10692" LastEditorUserId="10692" LastEditDate="2016-02-22T10:05:46.143" LastActivityDate="2016-02-22T10:05:46.143" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27771" PostTypeId="2" ParentId="27766" CreationDate="2016-02-22T12:01:43.907" Score="19" Body="&lt;p&gt;I think I have it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $ 2016 = ({.2} / {.\overline1} + 0!) \times 6!$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also as pointed out by Matt in the comments, we can swap things around so that 2016 is orderly self-composable &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $ 2016 = ({.2} / .\overline{0!} + 1) \times 6!$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18422" LastEditorUserId="18422" LastEditDate="2016-02-22T18:03:41.393" LastActivityDate="2016-02-22T18:03:41.393" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="27772" PostTypeId="1" AcceptedAnswerId="27776" CreationDate="2016-02-22T12:22:50.980" Score="16" ViewCount="1734" Body="&lt;p&gt;The SET game works with a deck of $81$ cards. Each card contains a set of symbols with four attributes: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;color (red, green, purple), &lt;/li&gt;&#xA;&lt;li&gt;shading (empty, striped, or solid), &lt;/li&gt;&#xA;&lt;li&gt;shape (oval, squiggle, or rhombus), &lt;/li&gt;&#xA;&lt;li&gt;number of symbols (one, two, or three). &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All the cards are unique. A SET occurs when three cards satisfy the following property for each of the four attributes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The attribute on each card is the same, &lt;/li&gt;&#xA;&lt;li&gt;or the attribute on each card is different &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As an example for &quot;the same&quot;, all three cards may have striped shading.&#xA;As an example for &quot;different&quot;, the three cards may have  red, green, and purple color.&#xA;In order for three cards to form a SET, they must satisfy this property for all four attributes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the maximum number of SETs that can be present in six distinct cards.&lt;/p&gt;&#xA;" OwnerUserId="19470" LastEditorUserId="8874" LastEditDate="2016-02-22T21:15:53.433" LastActivityDate="2019-07-09T13:35:29.033" Title="The maximum number of SETs with six cards" Tags="|mathematics|combinatorics|game|set-game|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="27773" PostTypeId="1" AcceptedAnswerId="27778" CreationDate="2016-02-22T12:28:18.370" Score="13" ViewCount="1331" Body="&lt;p&gt;Aabzaari and Basharat play a game, alternating turns with Aabzaari going first. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the start of the game, there are 20 cookies on a red plate and 14 cookies on a blue plate. A legal move consists in &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;eating two cookies taken from the blue plate, or &lt;/li&gt;&#xA;&lt;li&gt;eating two cookies taken from the red plate, or&lt;/li&gt;&#xA;&lt;li&gt;moving one cookie from the red plate to the blue plate &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(but moving cookies from the blue plate to the red plate is not allowed). &#xA;The last player to make a legal move wins the game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which player can guarantee that he wins no matter what strategy the opponent chooses? &lt;/p&gt;&#xA;" OwnerUserId="19470" LastEditorUserId="82879" LastEditDate="2024-01-22T03:37:16.683" LastActivityDate="2024-01-22T03:37:16.683" Title="The game with cookies on a red and a blue plate" Tags="|mathematics|strategy|game|nim|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27774" PostTypeId="2" ParentId="27773" CreationDate="2016-02-22T12:50:12.173" Score="13" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Aabzaari always can win&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Note that, if the third move never was performed you have 17 times that two cookies are removed. Aabzaari has all odd-numbered turn so the 17th turn is Aabzaari's. Now all he has to do is take 2 cookies from either plate unless when Basharat performs the third move, then he also does it. This makes sure that both plates have multiples of two cookies on them all the time and that the game ends after Aabzaari takes the last two cookies. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2016-02-22T12:50:12.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27776" PostTypeId="2" ParentId="27772" CreationDate="2016-02-22T13:08:15.083" Score="22" Body="&lt;p&gt;For any two distinct cards, there exists aunique third card that completes the pair into a SET. &#xA;If I pick three distinct cards $A,B,C$, then I can add three cards $X,Y,Z$ such that ABX and ACY and BCZ form SETs.&#xA;Hence there exists a solution with three SETs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that three SETs are best possible, we argue with linear algebra.&#xA;We represent a card by a 4-dimensional vector $(w,x,y,z)$ over $\mathbb{Z}/3$.&#xA;The card deck then consists of 81 vectors corresponding to the 81 cards in the SET game. &#xA;Three vectors form a SET, if and only if they add up to the zero-vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider six vectors, and assume for the sake of contradiction that these six vectors contain four SETS. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Two such sets can overlap in at most one vector. (If they agree in two elements, then they must also agree in the third one, as the third one is uniquely defined.)&lt;/li&gt;&#xA;&lt;li&gt;Hence each of the six vectors is contained in at most two SETs. (Three SETs would need the considered vector plus three disjoint pairs, that is, at least seven vectors).&lt;/li&gt;&#xA;&lt;li&gt;Hence each vector is contained in exactly two SETs. (Four SETs need altogether $4\cdot3=12$ vectors, and each of the two vectors can only be used twice.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now let us wrap things up.&#xA;Consider some fixed vector $A$ so that ABC and ADE are SETs.&#xA;Then $B$ must be in two sets; the options AB? and CB? can only be completed as ABC. The option BDE collides with ADE. Hence only BDF (or the symmetric case BEF) remains.&#xA;Finally, $C$ must be in one more SET beyond ABC; the options AC? and BC? can only be completed by ABC itself; the option CDE collides with ADE, and the option CDF collides with BDF. Hence CD? is impossible, and only CEF remains.&#xA;We summarize:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The four sets are ABC and ADE and BDF and CEF.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the vectors, this means $A+B+C=0$ and $A+D+E=0$ and $B+D+F=0$ and $C+E+F=0$.&#xA;Adding up these four equations and dividing by 2 yields $A+B+C+D+E+F=0$.&#xA;Subtracting $A+B+C=0$ from this yields $D+E+F=0$, so that DEF is a feasible SET. But then DEF and ADE are both SETs, which implies $A=F$ so that the six cards/vectors are not pairwise distinct. Contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The maximum number of SETs that can be present in six distinct cards is 3.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-02-22T13:33:43.987" LastActivityDate="2016-02-22T13:33:43.987" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27777" PostTypeId="1" CreationDate="2016-02-16T23:34:42.953" Score="3" ViewCount="264" Body="&lt;p&gt;If you take the phonetic word form /fl* (where * denotes any vowel sound), you can create (at least) 6 valid English words:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;flay, flee, fly, flow, flue, flaw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which single-vowel phonetic word form can make the most valid English words by interchanging vowel sounds?&lt;/p&gt;&#xA;" OwnerDisplayName="Michael" LastEditorUserId="14435" LastEditDate="2016-02-22T14:00:41.200" LastActivityDate="2016-03-02T17:20:01.960" Title="What phonetic word form can generate the largest number of valid English words when adding single vowel sounds?" Tags="|wordplay|english|language|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="27778" PostTypeId="2" ParentId="27773" CreationDate="2016-02-22T14:28:47.610" Score="13" Body="&lt;p&gt;It seems to me that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Aabzaari always wins, regardless of the moves performed.&#xA; There are cookies for 17 turns so Aabzaari gets to eat the last two. The third move doesn't consume any cookies and &lt;em&gt;always&lt;/em&gt; allows a paired move, so it doesn't affect the parity of the turns.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18975" LastActivityDate="2016-02-22T14:28:47.610" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27779" PostTypeId="2" ParentId="27751" CreationDate="2016-02-22T14:35:45.417" Score="11" Body="&lt;p&gt;&lt;em&gt;The cows first subtract one,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cows are cattle, and if we move the first letter back one we get &lt;strong&gt;battle&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;almost too rushed brew gins?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Being too rushed can lead to being hasty. If we remove the last letter and add an anagram of &quot;gins&quot;, we get &lt;strong&gt;Hastings&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Putting them together, the clue indicates&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the Battle of Hastings, which took place in &lt;strong&gt;1066&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12511" LastActivityDate="2016-02-22T14:35:45.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27780" PostTypeId="2" ParentId="27773" CreationDate="2016-02-22T15:52:25.300" Score="4" Body="&lt;p&gt;Info 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If an $a,b$ combination loses Aabzari the game then $a+2,b; a,b+2$ and $a+1,b-1$ win, because starting from one of the latter three, Aabzaari can leave Basharat an $a,b$ game.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Info 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If $a-2,b ; a,b-2$ and $a-1,b+1$ all lose, $a,b$ wins. If at least one wins, it loses. The reason is, $a,b$ can turn into one of the three with Basharat as the player one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;We can find out that if $a,b$ wins, then $a-1, b+1$ loses and $a+1, b+1$ wins. Also $a,b-2$ loses, $a+2,b-2$ wins and $a+1,b-1$ loses. $0,(20+14)$ wins, because Aabzaari can make the first and last of the 17 moves until the end. Since $a,b$ wins, $a+1,b-1$ loses, $a+2,b-2$ wins and so on, $(0+20),(34-20)$ wins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="14016" LastEditDate="2016-02-22T16:05:57.810" LastActivityDate="2016-02-22T16:05:57.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27781" PostTypeId="1" CreationDate="2016-02-22T15:59:44.027" Score="6" ViewCount="2180" Body="&lt;p&gt;Can you fill in the missing numbers of these number sequences?&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;_, 3, _, 9, _, 14, _, _, 21, 27, _, 35, _&lt;/p&gt;&#xA;&lt;p&gt;_, 7, _, 15, _, 22, _, 30&lt;/p&gt;&#xA;&lt;p&gt;_, _, 5, 7, _, _, 12&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The three sequences may or may not be linked together in some way. If you can come up with the missing numbers, please provide an explanation.&lt;/p&gt;&#xA;&lt;p&gt;I have tried to solve this for about a week now and I've gotten nowhere, hopefully someone here comes up with something.&lt;/p&gt;&#xA;" OwnerUserId="16477" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-02-24T00:14:12.337" Title="Fill in the blanks of this number sequence" Tags="|number-sequence|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27782" PostTypeId="2" ParentId="27781" CreationDate="2016-02-22T16:14:17.817" Score="10" Body="&lt;p&gt;As a starting point, I think the bottom sequence is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1, 3, 5, 7, 8, 10, 12&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the calendar month numbers of months having 31 days.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5874" LastActivityDate="2016-02-22T16:14:17.817" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27783" PostTypeId="1" CreationDate="2016-02-22T17:13:41.050" Score="1" ViewCount="151" Body="&lt;p&gt;I've been thinking about this question for a while, but I have no idea how to approach it. I don't know what the answer is. The question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt; What is the shortest finite set of moves that, when repeated, will traverse every possible Rubik's cube permutation? &lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no restriction on repetition of states, so long as it can be shown that for each permutation, if this set of moves is repeated long enough, it will eventually reach that permutation. Obviously not including any permutation not attainable through legal rubik's cube moves.&lt;/p&gt;&#xA;" OwnerUserId="18103" LastActivityDate="2016-02-22T17:13:41.050" Title="Rubik's Cube Master Algorithm" Tags="|mathematics|rubiks-cube|algorithm|" AnswerCount="0" CommentCount="8" ClosedDate="2016-02-22T20:12:02.973" ContentLicense="CC BY-SA 3.0" />
  <row Id="27787" PostTypeId="1" AcceptedAnswerId="27789" CreationDate="2016-02-22T18:27:53.643" Score="-3" ViewCount="244" Body="&lt;ol&gt;&#xA;&lt;li&gt;Consider the &quot;equation&quot; $6145 - 1 = 6143$. Can you move two digits so as to create a valid equality?&lt;/li&gt;&#xA;&lt;li&gt;The equation $-127=-127$ is, of course, true. Suppose you were told again to move two digits as to leave a valid equation. Please note that you have to move two digits on the same side of this equation.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Moving means that you can change the position of any two digits independently, so that they still are a number/constant!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The moving doesn't require swapping of digits in both.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is a part of mathematical puzzles asked in the puzzling contest at my college(coordinated by me).&lt;/p&gt;&#xA;" OwnerUserId="1766" LastEditorUserId="1766" LastEditDate="2016-02-23T03:02:44.430" LastActivityDate="2016-02-23T03:02:44.430" Title="Equality and Fraternity" Tags="|mathematics|lateral-thinking|" AnswerCount="3" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27788" PostTypeId="2" ParentId="27787" CreationDate="2016-02-22T18:44:42.240" Score="1" Body="&lt;p&gt;The second one is fairly easy (makes me think I might not quite understand where you can &quot;move&quot; a digit):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $2-17 = 2-17$ OR $1-27 = 1-27$ OR $-217 = -217$ or pretty much anything else like that.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's a solution to the first one that takes advantage of the vagueness of the instructions. I'm not sure if it's &quot;legal&quot; though.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Remove the $1$ from the $6145$ and put it sideways, turning it into a subtraction sign. Then move the $1$ from $6143$ over to the other side. You'll end up getting $645-1-1=643$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2016-02-22T18:44:42.240" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27789" PostTypeId="2" ParentId="27787" CreationDate="2016-02-22T19:00:17.043" Score="4" Body="&lt;p&gt;First one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$614-1^{54}=613$ or $615-1^{43}=614$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$1-2^7=-127$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2016-02-22T19:00:17.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27790" PostTypeId="2" ParentId="27787" CreationDate="2016-02-22T19:39:22.187" Score="3" Body="&lt;p&gt;Lateral thinking says I can move these two digits in this way:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; By using the $1$ as a line instead of a number,&lt;br&gt;&#xA; 1. $614 − 5 \ne 6143$&lt;br&gt;&#xA; 2. $-127 \ne 72$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2016-02-22T19:39:22.187" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27791" PostTypeId="1" AcceptedAnswerId="27793" CreationDate="2016-02-22T19:58:40.290" Score="34" ViewCount="2794" Body="&lt;p&gt;What is the one-word solution to the following riddle?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Backward in your leg, forward in red ink.&lt;br&gt;&#xA;  Stung, piece and digit. What is the link?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastEditorUserId="4421" LastEditDate="2016-02-22T21:11:41.507" LastActivityDate="2016-02-22T21:11:41.507" Title="A Boring Riddle" Tags="|riddle|word|wordplay|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27792" PostTypeId="2" ParentId="27692" CreationDate="2016-02-22T20:04:50.983" Score="2" Body="&lt;p&gt;My (probably wrong) guess is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; information/data&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9353" LastActivityDate="2016-02-22T20:04:50.983" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27793" PostTypeId="2" ParentId="27791" CreationDate="2016-02-22T20:19:30.983" Score="39" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;BIT&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Backward in your leg&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The word appears in reverse in &lt;strong&gt;TIB&lt;/strong&gt;IA, which is a bone in your leg.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Forward in red ink&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In accounting, red ink indicates a DE&lt;strong&gt;BIT&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Stung&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you are stung by an insect, you might also say you are &lt;strong&gt;BIT&lt;/strong&gt;TEN.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Piece&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A small piece of something is a &lt;strong&gt;BIT&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Digit&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In binary, a digit (1 or 0) is called a &lt;strong&gt;BIT&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;TITLE: &lt;em&gt;A Boring Riddle&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For boring, or drilling, you use a &lt;strong&gt;BIT&lt;/strong&gt; on the end of the drill.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4421" LastActivityDate="2016-02-22T20:19:30.983" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="27794" PostTypeId="1" AcceptedAnswerId="27795" CreationDate="2016-02-22T20:37:02.487" Score="3" ViewCount="679" Body="&lt;p&gt;I'm like a giant ball on a cosmic trampoline.&lt;br&gt;&#xA;All bend to me and circle closer&lt;br&gt;&#xA;You could think of me as evil if you wish&lt;br&gt;&#xA;Since I destroy all those who come to me&lt;/p&gt;&#xA;" OwnerUserId="19179" LastEditorUserId="9042" LastEditDate="2016-02-22T22:23:08.860" LastActivityDate="2016-10-24T17:09:20.270" Title="A wrinkle in time" Tags="|riddle|word|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27795" PostTypeId="2" ParentId="27794" CreationDate="2016-02-22T20:42:12.340" Score="12" Body="&lt;p&gt;Are you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a &lt;a href=&quot;https://en.wikipedia.org/wiki/Black_hole&quot; rel=&quot;nofollow&quot;&gt;black hole&lt;/a&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm like a giant ball on a cosmic trampoline.&lt;sup&gt;1&lt;/sup&gt;&lt;br&gt;&#xA;All bend to me and circle closer&lt;sup&gt;2&lt;/sup&gt;&lt;br&gt;&#xA;You could think of me as evil if you wish&lt;sup&gt;3&lt;/sup&gt;&lt;br&gt;&#xA;Since I destroy all those who come to me&lt;sup&gt;4&lt;/sup&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Per the linked article, &lt;em&gt;&quot;A black hole is a region of spacetime exhibiting such strong gravitational effects that nothing—including particles and electromagnetic radiation such as light—can escape from inside it.&quot;&lt;/em&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;  &lt;em&gt;&quot;As predicted by general relativity, the presence of a mass deforms spacetime in such a way that the paths taken by particles bend towards the mass. At the event horizon of a black hole, this deformation becomes so strong that there are no paths that lead away from the black hole.&quot;&lt;/em&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; I don't know how to describe this concisely (nor do I understand how to recognize the scientific language in the linked article that says this, at a glance), but my understanding of a black hole is that it's &lt;strong&gt;[1]&lt;/strong&gt; a mass so dense that everything curves around it, including gravity—like a bowling ball on a trampoline. Anything coming near enough to it is &lt;strong&gt;[4]&lt;/strong&gt; sucked in—&lt;em&gt;&quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Spaghettification&quot; rel=&quot;nofollow&quot;&gt;spaghettified&lt;/a&gt;&quot;&lt;/em&gt;—by extreme tidal forces, rending it such that its constituent atoms no longer coalesce to anything recognizable.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;[2]&lt;/strong&gt; OK so I couldn't think of a clever way to explain this one, but, again, from my understanding, when things are pulled by gravitational forces toward a body in space, they sort of &quot;spiral&quot; toward it; in a lot of cases, they end up a distance away from the body, propelled eternally around it, but in the case of a black hole, they are sucked in rather than falling into an &quot;orbit&quot;.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;[3]&lt;/strong&gt; Uhh I don't know if I'm expected to explain this or not, but a) 'black' and 'darkness' are &lt;a href=&quot;https://en.wikipedia.org/wiki/Black#Darkness_and_evil&quot; rel=&quot;nofollow&quot;&gt;cultural signifiers of &quot;bad&quot; or &quot;evil&quot; things&lt;/a&gt;, b) black holes destroy things (destruction is generally considered bad, dunno if I need a citation here or if we can all agree that it's not the most popular noun at Word Thanksgiving), and c) black holes specifically &lt;a href=&quot;http://www.dailymail.co.uk/sciencetech/article-2079923/Nasa-finds-evil-eye-watching-Earth-deep-space.html&quot; rel=&quot;nofollow&quot;&gt;find themselves&lt;/a&gt; &lt;a href=&quot;https://www.quora.com/Are-black-holes-evil&quot; rel=&quot;nofollow&quot;&gt;mentioned near&lt;/a&gt; &lt;a href=&quot;https://answers.yahoo.com/question/index?qid=20110324105144AATdfJE&quot; rel=&quot;nofollow&quot;&gt;the concept of &quot;evil&quot;&lt;/a&gt; fairly often&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;sub&gt;Please forgive my non-sciencey explanation words.&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2016-02-26T15:55:41.607" LastActivityDate="2016-02-26T15:55:41.607" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27796" PostTypeId="1" AcceptedAnswerId="27797" CreationDate="2016-02-22T20:50:27.130" Score="0" ViewCount="141" Body="&lt;p&gt;All you do is knock me down&lt;br&gt;&#xA;Then stand me up again&lt;br&gt;&#xA;Then again you knock me down&lt;br&gt;&#xA;As for a hint - we're ten.&lt;/p&gt;&#xA;" OwnerUserId="19179" LastActivityDate="2016-02-22T20:53:33.987" Title="Knock 'em down, Knock 'em down" Tags="|riddle|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27797" PostTypeId="2" ParentId="27796" CreationDate="2016-02-22T20:53:33.987" Score="4" Body="&lt;p&gt;You sound like &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bowling pins&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All you do is knock me down&lt;br&gt;&#xA;Then stand me up again&lt;br&gt;&#xA;Then again you knock me down&lt;br&gt;&#xA;As for a hint - we're ten.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The game of bowling is played by rolling a ball down a lane to hit the ten pins at the end. The pins are knocked down (or not, in the case of a gutter ball) and then set up again (assuming you're at a bowling alley, this is done by mechanized process, though it could easily be done by hand if you were holding some kind of shady, back-alley bowling league).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastActivityDate="2016-02-22T20:53:33.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27798" PostTypeId="1" AcceptedAnswerId="28114" CreationDate="2016-02-22T20:57:27.287" Score="10" ViewCount="439" Body="&lt;p&gt;This is what happened to Jack and Dave.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KDlZb.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KDlZb.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;If you were Jack and Dave, what would you do?&lt;/p&gt;&#xA;" OwnerUserId="19179" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-03-02T16:57:40.577" Title="Can You Note the Notes?" Tags="|cipher|music|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="27799" PostTypeId="2" ParentId="27798" CreationDate="2016-02-22T21:20:41.497" Score="5" Body="&lt;p&gt;If we look at the notes, they are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; abca bcab&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We note that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The second bar has 5 beats, contradicting the common time (4 4)&#xA; in the first bar. I therefore assume that the bar separates words&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This leads me to think that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is a cryptogram for &quot;seas ease&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And if I were Jack and/or Dave I might&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Up anchor and sail away. Or something nautical like that.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastActivityDate="2016-02-22T21:20:41.497" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27800" PostTypeId="2" ParentId="27729" CreationDate="2016-02-22T22:51:58.707" Score="7" Body="&lt;p&gt;&lt;sup&gt;&#xA;(Not a guess at any specific answer that the question's poser had in mind.)&#xA;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Premise&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The more the merrier.&#xA;&amp;hairsp;&#xA;Different solutions have different virtues.&#xA;Eventually every solution will be appreciated&#xA;by some puzzle lover.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;&lt;sup&gt;&lt;sup&gt;&lt;/sup&gt;&lt;/sup&gt; Solutions have a variety of virtues&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Far from being exhaustive or mutually exclusive,&#xA;these qualities are&#xA;listed alphabetically&#xA;within two broad categories.&#xA;Edits welcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More for sport:&#xA;&lt;br&gt;&amp;ensp;&amp;nbsp;&amp;bull;&amp;nbsp; approval&#xA;&lt;br&gt;&amp;ensp;&amp;nbsp;&amp;bull;&amp;nbsp; brevity&#xA;&lt;br&gt;&amp;ensp;&amp;nbsp;&amp;bull;&amp;nbsp; creativity / inventiveness / originality&#xA;&lt;br&gt;&amp;ensp;&amp;nbsp;&amp;bull;&amp;nbsp; esoterica / sophistication&#xA;&lt;br&gt;&amp;ensp;&amp;nbsp;&amp;bull;&amp;nbsp; firstness&#xA;&lt;br&gt;&amp;ensp;&amp;nbsp;&amp;bull;&amp;nbsp;   &lt;del&gt;&amp;thinsp;humor&amp;hairsp;&lt;/del&gt;&#xA;                             &lt;del&gt;&amp;thinsp;&lt;strong&gt;humour&lt;/strong&gt;&amp;hairsp;&lt;/del&gt;&#xA;                                          &lt;strong&gt;mirth&lt;/strong&gt;&#xA;&lt;br&gt;&amp;ensp;&amp;nbsp;&amp;bull;&amp;nbsp; obviousness&#xA;&lt;br&gt;&amp;ensp;&amp;nbsp;&amp;bull;&amp;nbsp; simplicity / ease of calculation&#xA;&lt;br&gt;&amp;ensp;&amp;nbsp;&amp;bull;&amp;nbsp; subtlety&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &amp;bull;&amp;nbsp; surprise&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&lt;sup&gt;&lt;/sup&gt;&lt;/sup&gt;&#xA;More for the long term:&#xA;&lt;br&gt;&amp;ensp;&amp;nbsp;&amp;bull;&amp;nbsp; clarity&#xA;&lt;br&gt;&amp;ensp;&amp;nbsp;&amp;bull;&amp;nbsp; completeness&#xA;&lt;br&gt;&amp;ensp;&amp;nbsp;&amp;bull;&amp;nbsp; education / cross reference&#xA;&lt;br&gt;&amp;ensp;&amp;nbsp;&amp;bull;&amp;nbsp; generality&#xA;&lt;br&gt;&amp;ensp;&amp;nbsp;&amp;bull;&amp;nbsp; resourcefulness&#xA;&lt;br&gt;&amp;ensp;&amp;nbsp;&amp;bull;&amp;nbsp; variety, the spice of life&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;&lt;sup&gt;&lt;sup&gt;&lt;/sup&gt;&lt;/sup&gt; Brief case study of the kind of puzzle in question&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/q/25832/18129&quot;&gt;Relabeling two 20-sided dice without changing their&#xA; total&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This puzzle received a wonderful solution that was accepted by the poser,&#xA;who then proceeded to present the intended solution&#xA;so that it wouldn't be overlooked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three solutions have been posted to date, each with its virtues.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/a/25834/18129&quot;&gt;A complete solution that transforms the puzzle into&#xA;algebra&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This solution is so complete, educational, general and sophisticated,&#xA;who could ask for more?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;sup&gt;&lt;sup&gt;&lt;/sup&gt;&lt;/sup&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/a/25835/18129&quot;&gt;An old-fashioned solution&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This detective-like solution does not require the mathematics&#xA;or computer employed by the complete solution.&#xA;Then again, only one of several possible solutions is found&#xA;and a general solution is only alluded to.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;sup&gt;&lt;sup&gt;&lt;/sup&gt;&lt;/sup&gt;&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/a/25889/18129&quot;&gt;The poser's inventive&#xA;solution&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quite possibly the poser all along had subconsciously hoped that&#xA;nobody else would actually think of this gem of an approach,&#xA;one that even adds to the understanding of the other solutions.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h3&gt;&lt;sup&gt;&lt;sup&gt;&lt;/sup&gt;&lt;/sup&gt; Conclusion&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The more the merrier.&#xA;&amp;hairsp;&#xA;As in the case study:&#xA;a puzzle's creator can acknowledge an unintended excellent solution,&#xA;someone else can provide an incomplete solution for a different audience,&#xA;and the poser can present their original solution for posterity&#xA;and still blow our minds.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Corollary&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Some unexpected solutions simply deserve their own puzzles.&#xA;&amp;hairsp;&#xA;The original puzzle&amp;nbsp;(P)&#xA;may be restated as two different puzzles&#xA;tailored differently for the originally intended solution&amp;nbsp;(S)&#xA;and for the unexpected one&amp;nbsp;(S2).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;P &amp;#129106; P1.&lt;/strong&gt; &amp;hairsp;&#xA;   There are six eggs &lt;del&gt;&amp;thinsp;in the basket&amp;thinsp;&lt;/del&gt;&#xA;   &lt;strong&gt;on a shelf&lt;/strong&gt;.&#xA;   Six people each take one of the eggs.&#xA;   How can it be that one egg is left &lt;del&gt;&amp;thinsp;in the basket&amp;thinsp;&lt;/del&gt;&#xA;   &lt;strong&gt;on the shelf&lt;/strong&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;S &amp;#129106; S1.&lt;/strong&gt; &amp;hairsp;&#xA; One of them put the egg again &lt;del&gt;&amp;thinsp;inside the basket&amp;thinsp;&lt;/del&gt;&#xA; &lt;strong&gt;on the shelf&lt;/strong&gt;.&#xA; &lt;br&gt;&#xA; &amp;emsp;&amp;emsp;&amp;emsp;&amp;ensp;&amp;thinsp;&amp;thinsp; &lt;sup&gt;&lt;sup&gt;&lt;/sup&gt;&lt;/sup&gt;&#xA; (Presumably a person cannot take the shelf,&#xA;  or occupy it while holding their egg.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/a/27747/18129&quot;&gt;Chaotic&amp;rsquo;s answer&lt;/a&gt;&#xA;provides a way to restate part of the puzzle&#xA;so as to exclude the intended solution.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;P &amp;#129106; P2.&lt;/strong&gt; &amp;hairsp;&#xA;   There are six eggs in the basket.&#xA;   Six people each take one of the eggs&#xA;   &lt;strong&gt;and nobody put it again in the basket.&lt;/strong&gt;&#xA;   How can it be that one egg is left in the basket?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &amp;emsp;&amp;nbsp;&amp;nbsp;&amp;thinsp;&#xA; &lt;strong&gt;S2.&lt;/strong&gt; &amp;hairsp;&#xA; The last person took the basket with the last egg still inside.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;&lt;strong&gt;Examples of solutions that earned their own puzzles&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/a/49157/18129&quot;&gt;Gareth McCaughan&amp;rsquo;s&#xA;solution&lt;/a&gt;&#xA;(S2) received its own puzzle,&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/49151/unreflected-infinitely-simple-polygon-reflexivity&quot;&gt;&lt;em&gt;Unreflected infinitely simple polygon reflexivity&lt;/em&gt;&lt;/a&gt;&amp;nbsp;(P2),&#xA;because it was brilliantly different from,&#xA;even if not technically better than,&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/a/50724/18129&quot;&gt;the intended&#xA;solution&lt;/a&gt;&amp;nbsp;(S/S1)&#xA;to the puzzle restated as&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/49207/infinitely-simple-polygon-solipsism&quot;&gt;&lt;em&gt;Infinitely simple polygon solipsism&lt;/em&gt;&lt;/a&gt;&amp;nbsp;(P1).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/a/51783/18129&quot;&gt;boboquack&amp;rsquo;s&#xA;solution&lt;/a&gt;&#xA;(S2) at&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/51782/new-mathematics-forever&quot;&gt;&lt;em&gt;New Mathematics forever&lt;/em&gt;&lt;/a&gt;&amp;nbsp;(P2)&#xA;was an unforeseen superior solution to&#xA;the original statement of the puzzle reincarnated as&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/51932/if-6-was-9-or-100-was-64-or-m-was-n&quot;&gt;&lt;em&gt;If 6 was 9, or 100 was 64, or M&amp;nbsp;was&amp;nbsp;N&lt;/em&gt;&lt;/a&gt;&amp;nbsp;(P1).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="18129" LastEditorUserId="18129" LastEditDate="2017-05-19T15:19:22.923" LastActivityDate="2017-05-19T15:19:22.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27801" PostTypeId="2" ParentId="27772" CreationDate="2016-02-23T00:33:52.583" Score="12" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There can be at most three sets.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A more intuitive proof that does not rely on any higher math:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key is that &lt;strong&gt;for any given two cards, there is exactly one third card that completes the set&lt;/strong&gt;. Beginning with the simple construction of three sets as @Gamow did, we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A, B, C as unique cards, and find the unique cards X, Y, Z such that ABX, ACY, and BCZ are each sets.  We now have three sets in six cards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, imagine that there exists a fourth set.  We know that it cannot contain any two cards that are both in one of our existing sets, since then it would have to also contain the same third card per the bolded statement above.  We can therefore combine X with C, Y, Z; Y with B, X, Z; and Z with A, X, Y.  The only three-card set that could follow this rule is X, Y, Z.  Notably, A, B, C is NOT a set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since A, B, C is not a set, pick an attribute that causes them to not be a set.  In this attribute, two of these cards are the same and one is different - assume A and B are the same.  In this case, AC and BC contain the same two options in this attribute, which means that since they both form sets with Y and Z respectively, then Y and Z must be the same in this attribute as well.  However, AB is different and forms a set with X, so X must be different from Y and Z in this attribute.  But then X, Y, Z contains two cards which are the same and one which is different in an attribute - therefore, X, Y, Z cannot be a set.  Since that was the only possible option, there can be no other sets.&lt;/p&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="13484" LastEditDate="2016-02-23T11:31:57.903" LastActivityDate="2016-02-23T11:31:57.903" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27803" PostTypeId="2" ParentId="27794" CreationDate="2016-02-23T06:54:44.910" Score="8" Body="&lt;p&gt;Answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SUN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm like a giant ball on a cosmic trampoline.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Sun is indeed a giant ball and a star large enough that about 1,300,000 planet Earths can fit inside of it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All bend to me and circle closer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When the star is formed, it bends the curavature of space and time(the fabric of the cosmos). Because of this curvature, we feel the force of what we know today as GRAVITY. All the planets orbit around curvature formed by the SUN. Thank you Mr.Einstien :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You could think of me as evil if you wish&#xA;Since I destroy all those who come to me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For all the damage it has done to the planets that are not in goldilock region(eg mercury,mars), Life has literally ceased and the planets near to the sun are nothing more than hot rocky bodies .Its better if you stay away from it, anything that comes near it burns because of extreme tempratures.(5,778 K - surface temperature).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A wrinkle in time&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The title can also be justified. According to the people who study space, there will be a time when the sun will run out of its fuel and its core will collapse. The Sun is about 4.5 billion years old. It has used up about half of its nuclear fuel (hydrogen). In about 5 billion years from now, the sun will begin to die. However, for our massive massive universe which has billions and billions of stars, this is just another star that will die. So its just a wrinkle in time. Not to mention, if sun dies,we also die :) .But I think we still got plenty of time(4.5bln yrs) and who knows if humans could colonise some other planet in future.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19151" LastEditorUserId="19151" LastEditDate="2016-02-23T08:36:32.870" LastActivityDate="2016-02-23T08:36:32.870" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27804" PostTypeId="1" AcceptedAnswerId="27809" CreationDate="2016-02-23T07:10:39.497" Score="5" ViewCount="699" Body="&lt;p&gt;Freddy removes four &lt;strong&gt;consecutive even&lt;/strong&gt; numbers from the set $\{1,2,3,\ldots,n\}$. Freddy then computes $51.5625$ as the average of the remaining $n-4$ numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which four even numbers did Freddy remove?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-20T17:04:59.277" LastActivityDate="2024-01-22T07:15:01.720" Title="And the average is 51.5625" Tags="|mathematics|algebra|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27805" PostTypeId="2" ParentId="27804" CreationDate="2016-02-23T08:09:22.720" Score="6" Body="&lt;p&gt;sum of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; {1,2,3,4,...,&lt;strong&gt;100&lt;/strong&gt;} 1 to 100 = 5050&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;we need to know avg of n-4 &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 51.5625 x 96 = 4950&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so sum of 4 numbers must be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;5050 - 4950 = 100 , n+n+2+n+4+n+6 = 100 which is &lt;strong&gt;n = 22&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;numbers to be removed are : &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;22,24,26,28&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18685" LastActivityDate="2016-02-23T08:09:22.720" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27806" PostTypeId="1" AcceptedAnswerId="27808" CreationDate="2016-02-23T08:12:37.407" Score="2" ViewCount="299" Body="&lt;p&gt;Where the ice meets the flesh&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the time comes to an Halt&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lost in the heart of men&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tarnished by my creation&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Don't pay attention to the present&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's just a show&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same happens with my child&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Never to be the same&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ending the Riddle I am&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might need another view&lt;/p&gt;&#xA;" OwnerUserId="18816" LastActivityDate="2016-02-23T08:41:55.903" Title="Maybe you just need another perspective, Who am I posing as?" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27807" PostTypeId="2" ParentId="27804" CreationDate="2016-02-23T08:26:11.953" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 22 24 26 28&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Calculations:&lt;br&gt;&#xA;We know the average to be 51.5625, of 825/16.&lt;br&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To try and end up with that fraction, we need to see where the sum of the first n integers is high enough to create a possible solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For n = 20, sum is only 210, so no chance of approaching 825/16&lt;br&gt;&#xA;For n = 36, sum is 666, so no chance of approaching 1650/32.&lt;br&gt;&#xA;For n = 52, sum is 1378, so no chance of approaching 2475/48&lt;br&gt;&#xA;For n = 68, sum is 2346, so no chance of approaching 3300/64&lt;br&gt;&#xA;For n = 84, sum is 3570. so no chance of approaching 4125/80&lt;br&gt;&#xA;For n = 100, sum is 5050. We are looking for 4950/96, so this could have a possible solution.&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given we have a total sum of 5050 over de first 100 integers, and we only need 4950 from 96 numbers, we need to find which four consecutive integers add up to 100. Knowing the average is 25, it's fairly trivial to find &lt;strong&gt;22 24 26 28&lt;/strong&gt; as the series to be removed.&lt;/p&gt;&#xA;" OwnerUserId="2156" LastActivityDate="2016-02-23T08:26:11.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27808" PostTypeId="2" ParentId="27806" CreationDate="2016-02-23T08:36:08.357" Score="4" Body="&lt;p&gt;Answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Walter Elias &quot;Walt&quot; Disney &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;As the title suggest, see the first letters of each line.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Where the ice meets the flesh&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the time comes to an Halt&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lost in the heart of men&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tarnished by my creation&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Don't pay attention to the present&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's just a show&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Walt Disney is a name&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Same happens with my child&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Never to be the same&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; first child/pregnancy of her wife ended in miscarriage&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ending the Riddle I am&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might need another view&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; see first letter of each line and read it downwards&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16036" LastEditorUserId="16036" LastEditDate="2016-02-23T08:41:55.903" LastActivityDate="2016-02-23T08:41:55.903" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27809" PostTypeId="2" ParentId="27804" CreationDate="2016-02-23T09:13:11.140" Score="10" Body="&lt;p&gt;Note: 51.5625 = 825/16&lt;br /&gt;&#xA;To calculate the sum for a given &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; we can do &lt;span class=&quot;math-container&quot;&gt;$\frac{n \cdot (n+1)}{2}$&lt;/span&gt;.&lt;br /&gt;&#xA;Then we need to have that &lt;span class=&quot;math-container&quot;&gt;$\frac{n \cdot (n+1)}{2} - \frac{825 \cdot (n - 4)}{16}$&lt;/span&gt; is a number that can be expressed as four even consecutive numbers.&lt;/p&gt;&#xA;&lt;p&gt;Let &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; be the lowest of the four consecutive numbers.&lt;/p&gt;&#xA;&lt;p&gt;The four even consecutive numbers are of the form  &lt;span class=&quot;math-container&quot;&gt;$x+x+2+x+4+x+6= 4x + 12$&lt;/span&gt;.&lt;br /&gt;&#xA;This means that&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{\frac{n \cdot (n+1)}{2} - \frac{825 \cdot (n - 4)}{16}-12}{4}$$&lt;/span&gt; must be an positive even integer and is equal to &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;We can find then that &lt;span class=&quot;math-container&quot;&gt;$n=100$&lt;/span&gt; is the first solution giving the result 22. But we also find out that there are an unlimited number of solutions which are all of the form &lt;span class=&quot;math-container&quot;&gt;$n = 100 + 128k$&lt;/span&gt; for any positive integer &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt;. The first couple of solutions are for example:&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;n&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;x&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;100&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;22&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;228&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;3636&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;356&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;11356&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;484&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;23152&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;&#xA;&lt;p&gt;EDIT: As Tim Couwelier pointed out, the four consecutive numbers of course cannot exceed &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;, therefore the solution with &lt;span class=&quot;math-container&quot;&gt;$n=100$&lt;/span&gt; is the only solution&lt;/p&gt;&#xA;" OwnerUserId="1748" LastEditorUserId="1748" LastEditDate="2024-01-22T07:15:01.720" LastActivityDate="2024-01-22T07:15:01.720" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27810" PostTypeId="1" AcceptedAnswerId="27814" CreationDate="2016-02-23T09:15:05.367" Score="21" ViewCount="1635" Body="&lt;p&gt;What's the maximal number of dominos (2x1 tiles) that can be placed on a checkerboard (8x8 square) so that every domino covers exactly 2 squares of the checkerboard and no two dominos form a 2x2 square? &lt;/p&gt;&#xA;" OwnerUserId="19470" LastActivityDate="2016-02-24T12:33:18.790" Title="Dominos on a checkerboard" Tags="|mathematics|combinatorics|checkerboard|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27811" PostTypeId="1" AcceptedAnswerId="27822" CreationDate="2016-02-23T09:18:22.040" Score="1" ViewCount="185" Body="&lt;p&gt;1000 white stones and 1 black stone are arranged in a row. A move consists  in selecting one black stone and changing the color of the 2 neighbouring stones (or changing the color of 1 neighboring stone if the selected stone is at the end of the row).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find all possible initial positions of the black stone, so that all stones can be made black by moves.&lt;/p&gt;&#xA;" OwnerUserId="19470" LastActivityDate="2016-02-23T13:38:44.920" Title="Making all stones black" Tags="|mathematics|game|reachability|" AnswerCount="1" CommentCount="0" ClosedDate="2016-02-23T14:10:25.037" ContentLicense="CC BY-SA 3.0" />
  <row Id="27812" PostTypeId="2" ParentId="27804" CreationDate="2016-02-23T10:07:10.120" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; N=100 and terms are: 22,24,26,28.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As we know that $ \sum_{i=1}^n i $ $=$  $ \frac{n*(n+1)}{2} $&lt;br&gt;&#xA;Thus, average $\mu_1$ = $ \frac{(n+1)}{2} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given, four consecutive even numbers are removed. So, let the first number of the series be $x$, thus the next three are: $ x+2, x+4, x+6$.&lt;br&gt;&#xA;Their sum is: $4(x+3)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we can see that, the new &lt;strong&gt;SUM&lt;/strong&gt; of $n-4$ terms is :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\frac{n*(n+1)}{2}-4(x+3)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\frac{n*(n+1)-8(x+3)}{2}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, the new mean/average of remaining terms is $51.5625$ or $\frac{825}{16}$.&#xA;And the $\mu_2$ of $n-4$ terms :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\frac{n*(n+1)-8(x+3)}{2(n+4)}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus, :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\frac{n*(n+1)-8(x+3)}{2(n+4)} = \frac{825}{16}$ &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $ 8*( n(n+1) - 8(x+3) ) = 825(n-4)$ &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An equation of two variable, which must have integral solution for $n \gt 4 $ and $x \gt 2 $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I solved it using &lt;a href=&quot;http://ideone.com/49Owfx&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; (inefficient w.r.t Time complexity) code.&lt;/p&gt;&#xA;" OwnerUserId="1766" LastEditorUserId="1766" LastEditDate="2016-02-23T10:13:38.483" LastActivityDate="2016-02-23T10:13:38.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27813" PostTypeId="2" ParentId="27772" CreationDate="2016-02-23T10:45:24.247" Score="4" Body="&lt;p&gt;&lt;strong&gt;TL;DR:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt; is possible and easy to attain; &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt; or more is not possible, so &lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt; is the maximum.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;With &lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt; unique cards, it's possible to make &lt;span class=&quot;math-container&quot;&gt;${6\choose3} = 20$&lt;/span&gt; different combinations of &lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt; cards. But not all of those are valid Sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With cards &lt;span class=&quot;math-container&quot;&gt;$A, B, C, D, E, F$&lt;/span&gt; we can make the following combinations of three cards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{array}{llll} &#xA;ABC &amp;amp; ACE &amp;amp; BCD &amp;amp; BEF \\&#xA;ABD &amp;amp; ACF &amp;amp; BCE &amp;amp; CDE \\&#xA;ABE &amp;amp; ADE &amp;amp; BCF &amp;amp; CDF \\&#xA;ABF &amp;amp; ADF &amp;amp; BDE &amp;amp; CEF \\&#xA;ACD &amp;amp; AEF &amp;amp; BDF &amp;amp; DEF \\&#xA;\end{array}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since any two cards define a single, unique third card to complete a valid Set, once we pick a combination as a valid Set, all other combinations that contain any two of the cards in that Set can't be a valid Set itself.&lt;br&gt;&#xA;So if &lt;span class=&quot;math-container&quot;&gt;$ABC$&lt;/span&gt; is a valid Set, we can cross off all combinations containing either &lt;span class=&quot;math-container&quot;&gt;$AB$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$BC$&lt;/span&gt;, or &lt;span class=&quot;math-container&quot;&gt;$CA$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\require{cancel}\begin{array}{llll} &#xA;  \fbox{ABC} &amp;amp; \cancel{ACE} &amp;amp; \cancel{BCD} &amp;amp; BEF \\&#xA;\cancel{ABD} &amp;amp; \cancel{ACF} &amp;amp; \cancel{BCE} &amp;amp; CDE \\&#xA;\cancel{ABE} &amp;amp;         ADE  &amp;amp; \cancel{BCF} &amp;amp; CDF \\&#xA;\cancel{ABF} &amp;amp;         ADF  &amp;amp;         BDE  &amp;amp; CEF \\&#xA;\cancel{ACD} &amp;amp;         AEF  &amp;amp;         BDF  &amp;amp; DEF \\&#xA;\end{array}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if &lt;span class=&quot;math-container&quot;&gt;$DEF$&lt;/span&gt; was a valid Set, those two would be the only valid Sets among those combinations, since we would need to cross off all combinations containing &lt;span class=&quot;math-container&quot;&gt;$DE$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$EF$&lt;/span&gt;, or &lt;span class=&quot;math-container&quot;&gt;$FD$&lt;/span&gt; as invalid. But we're looking for more Sets, so &lt;span class=&quot;math-container&quot;&gt;$DEF$&lt;/span&gt; should not be a valid Set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Picking Sets and crossing off, we can come up with a maximum of four combinations that &lt;em&gt;might&lt;/em&gt; be valid Sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\require{cancel}\begin{array}{llll} &#xA;  \fbox{ABC} &amp;amp; \cancel{ACE} &amp;amp; \cancel{BCD} &amp;amp; \cancel{BEF} \\&#xA;\cancel{ABD} &amp;amp; \cancel{ACF} &amp;amp; \cancel{BCE} &amp;amp; \cancel{CDE} \\&#xA;\cancel{ABE} &amp;amp;   \fbox{ADE} &amp;amp; \cancel{BCF} &amp;amp; \cancel{CDF} \\&#xA;\cancel{ABF} &amp;amp; \cancel{ADF} &amp;amp; \cancel{BDE} &amp;amp;   \fbox{CEF} \\&#xA;\cancel{ACD} &amp;amp; \cancel{AEF} &amp;amp;   \fbox{BDF} &amp;amp; \cancel{DEF} \\&#xA;\end{array}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our candidates are &lt;span class=&quot;math-container&quot;&gt;$ABC, ADE, BDF, CEF$&lt;/span&gt;. All other possible combinations of four candidates are variants of this.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now that we have our Set candidates, let's check if they can actually be valid Sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we pick any of the four properties, it can have three possible values: &lt;span class=&quot;math-container&quot;&gt;$x, y, z$&lt;/span&gt;. It can either have the same value for all three cards, or all different values. Let's explore what that means for the other candidate Sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\require{cancel}&#xA;\begin{array}{llll}&#xA;ABC &amp;amp; ADE &amp;amp; BDF &amp;amp; CEF \\&#xA;\hline&#xA;xxx &amp;amp; xxx &amp;amp; xxx &amp;amp; xxx \\&#xA;xxx &amp;amp; xyz &amp;amp; xyz &amp;amp; \cancel{xzz} \\&#xA;xyz &amp;amp; xyz &amp;amp; yyy &amp;amp; \cancel{zzy} \\&#xA;xyz &amp;amp; xzy &amp;amp; yzx &amp;amp; zyx \\&#xA;\end{array}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this we see that the only valid Sets can occur if and only if a property has the same value on cards &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt;, on cards &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt;, and on cards &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt;. Since this has to hold for &lt;em&gt;all&lt;/em&gt; properties, this means the cards in these pairs are the same card, so &lt;span class=&quot;math-container&quot;&gt;$A = F$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$B = E$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$C = D$&lt;/span&gt;. But we specified we wanted unique cards, so this leaves us with an impossible solution.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Since it's relatively easy to find a solution with three valid Sets &amp;mdash; pick three cards &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; that are &lt;em&gt;not&lt;/em&gt; a Set, then for all combinations of two of those cards, add the card that makes it a Set, so you have Sets &lt;span class=&quot;math-container&quot;&gt;$ABD$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$BCE$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$ACF$&lt;/span&gt; &amp;mdash; &lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt; is the maximum of valid Sets for three unique cards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{array}{c|cccc|ccc} \text{card} &amp;amp; \text{number} &amp;amp; \text{shape}  &amp;amp; \text{colour}  &amp;amp; \text{fill} &amp;amp; \text{Set #1}  &amp;amp; \text{Set #2}  &amp;amp; \text{Set #3} \\&#xA; \hline &#xA; A &amp;amp; 1 &amp;amp; \text{oval} &amp;amp; \text{red}    &amp;amp; \text{solid}   &amp;amp; \checkmark &amp;amp;            &amp;amp; \checkmark \\&#xA; B &amp;amp; 1 &amp;amp; \text{oval} &amp;amp; \text{green}  &amp;amp; \text{striped} &amp;amp; \checkmark &amp;amp; \checkmark &amp;amp;            \\&#xA; C &amp;amp; 2 &amp;amp; \text{oval} &amp;amp; \text{purple} &amp;amp; \text{solid}   &amp;amp;            &amp;amp; \checkmark &amp;amp; \checkmark \\&#xA; D &amp;amp; 1 &amp;amp; \text{oval} &amp;amp; \text{purple} &amp;amp; \text{open}    &amp;amp; \checkmark &amp;amp;            &amp;amp;            \\&#xA; E &amp;amp; 3 &amp;amp; \text{oval} &amp;amp; \text{red}    &amp;amp; \text{open}    &amp;amp;            &amp;amp; \checkmark &amp;amp;            \\&#xA; F &amp;amp; 3 &amp;amp; \text{oval} &amp;amp; \text{green}  &amp;amp; \text{solid}   &amp;amp;            &amp;amp;            &amp;amp; \checkmark \\&#xA;\end{array}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditDate="2019-07-09T13:35:29.033" LastActivityDate="2019-07-09T13:35:29.033" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27814" PostTypeId="2" ParentId="27810" CreationDate="2016-02-23T10:55:45.837" Score="36" Body="&lt;p&gt;Looks like:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &amp;nbsp; 31 dominos can be placed &amp;emsp;&#xA; &lt;sub&gt;&lt;sub&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/IXQAz.png&quot; width=&quot;258&quot;&gt;&#xA; &lt;/sub&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;Thanks to @Gamow's comment, this number's maximality can be proved&#xA;by self-contradiction of the assumption that it is not maximal.&#xA;Any more dominos would cover all 64 squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Assumption to be disproved:&lt;/strong&gt; All squares can be covered with dominos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;ensp;&amp;hairsp;&#xA;A. As the top left corner must be covered, start with a horizontal domino&#xA;   there.  (Every other possible corner domino is equivalent to this&#xA;   by rotation and reflection.)  From here on:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Successive placements of a diagonal series of dominos&#xA; are forced to form a descending herringbone staircase in order to prevent&#xA; a 2x2 square from being formed in combination with each previous domino.&#xA; &lt;br&gt;&lt;sup&gt;&lt;sup&gt;&lt;sup&gt;&lt;/sup&gt;&lt;/sup&gt;&lt;/sup&gt;&#xA; B. The domino that neighbors A along the left edge must be vertical.&#xA; &lt;br&gt;&lt;sup&gt;&lt;sup&gt;&lt;sup&gt;&lt;/sup&gt;&lt;/sup&gt;&lt;/sup&gt;&#xA; C. The inside corner formed by A and B must be covered by&#xA;    a horizontal domino.&#xA; &lt;br&gt;&lt;sup&gt;&lt;sup&gt;&lt;sup&gt;&lt;/sup&gt;&lt;/sup&gt;&lt;/sup&gt;&#xA; D&amp;hairsp;&amp;minus;&amp;hairsp;&amp;hairsp;M. Likewise,&#xA;    until a horizontal domino-shaped hole at the bottom right corner,&#xA;    if filled, would form a 2x2 square&#xA;    with M.&lt;sub&gt;&lt;sub&gt;&lt;sub&gt;&lt;sub&gt;&lt;sub&gt;&lt;/sub&gt;&lt;/sub&gt;&lt;/sub&gt;&lt;/sub&gt;&lt;/sub&gt;&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/XSQyM.png&quot; width=&quot;200&quot;&gt;&amp;emsp;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/OmzR0.png&quot; width=&quot;200&quot;&gt;&amp;emsp;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/jPfJj.png&quot; width=&quot;200&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore the top left corner cannot be covered,&#xA;which negates the assumption&#xA;and proves that a 32-domino solution is impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;&lt;strong&gt;Further reading,&lt;/strong&gt; courtesy of @Fimpellizieri's comment:&#xA;&lt;br&gt;&lt;sup&gt;&lt;/sup&gt;&#xA;&lt;a href=&quot;http://www.cimat.mx/ciencia_para_jovenes/pensamiento_matematico/thurston.pdf&quot;&gt;Conway's Tiling Groups &lt;sub&gt;PDF&lt;/sub&gt;&lt;/a&gt;&#xA;[and height functions]&#xA;&amp;ndash;&amp;nbsp;W.&amp;hairsp;P.&amp;nbsp;Thurston&#xA;&lt;br&gt;&#xA;&lt;a href=&quot;http://olympiads.mccme.ru/lktg/2009/4/articles/conway.pdf&quot;&gt;Tiling with Polyominoes and Combinatorial Group Theory &lt;sub&gt;PDF&lt;/sub&gt;&lt;/a&gt;&#xA;&amp;ndash;&amp;nbsp;J.&amp;hairsp;H.&amp;nbsp;Conway&amp;nbsp;&amp;amp; J.&amp;hairsp;C.&amp;nbsp;Lagarias&#xA;&lt;br&gt;&#xA;&lt;a href=&quot;http://arxiv.org/pdf/1601.06354v1&quot;&gt;Domino Tilings of the Torus &lt;sub&gt;PDF&lt;/sub&gt;&lt;/a&gt;&#xA;[and the plane]&#xA;&lt;a href=&quot;http://arxiv.org/abs/1601.06354&quot;&gt;Abstract&lt;/a&gt;&#xA;&amp;ndash;&amp;nbsp;F.&amp;nbsp;de&amp;nbsp;Souza Lima Impellizieri&lt;/p&gt;&#xA;" OwnerUserId="18129" LastEditorUserId="18129" LastEditDate="2016-02-24T12:33:18.790" LastActivityDate="2016-02-24T12:33:18.790" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="27815" PostTypeId="1" AcceptedAnswerId="27816" CreationDate="2016-02-23T11:20:22.977" Score="3" ViewCount="178" Body="&lt;p&gt;My batty and quite senile great-grandmother (on my wife's side) left me a fish and chip shop in her final will and testament after she was beheaded for treason (...long story). &lt;br/&gt;&lt;br/&gt;&#xA;The executioner of the will (not of her head) informed us that she stipulated that there was one thing that may never be changed - that the shop may only ever sell their fish and chips (best haddock this side of the Atlantic btw) in portions of 6, 9 and 20.&lt;br/&gt;&lt;br/&gt;&#xA;In a community-wide initiative to advertise their shop they asked the following question: What would be the largest order of fish and chips that one cannot order from us?&lt;br/&gt; (a free albatross flavor ice-cube was given to those who provided a formula for any X, Y, Z combination...)&lt;/p&gt;&#xA;" OwnerUserId="19086" LastActivityDate="2016-02-24T00:23:20.993" Title="Pigging out in proportion" Tags="|mathematics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27816" PostTypeId="2" ParentId="27815" CreationDate="2016-02-23T11:45:16.690" Score="2" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $43$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Any integer greater than $1$ can be expressed as a sum of the form $2m + 3n$ where $m$ and $n$ are non-negative integers. This is easy to show as all even integers are already in the form $2m$ and odd integers $&amp;gt;1$ can be expressed in the form $2m + 3$. &lt;br&gt;&lt;br&gt; Hence, any integer, greater than $3$ and divisble by $3$ can be expressed in the form $6m + 9n = 3 \times (2m + 3n)$ where $m$ and $n$ are non-negative integers. &lt;br&gt;&lt;br&gt; Similarly, any number of the form $3k+2$ (for integer $k$) greater than $23$ can be expressed in the form $20 + 6m + 9n$ and any number of the form $3k+1$ (for integer $k$) greater than $43$ can be expressed in the form $(2\times 20) + 6m + 9n$ for non-negative integers $m$ and $n$. Hence, all orders of size greater than $43$ can be satisfied. &lt;br&gt;&lt;br&gt; Clearly, a order of size $43$ is unobtainable since any order with only portions of size $6$ and $9$ is divisible by $3$, any portion with a single $20$ and portions of $6$ and $9$ is of the form $3k+2$ for integer $k$ and having two portions of $20$ already brings us to an order of size $40$ whereby any extra portion provides us with an overshoot. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18422" LastEditorUserId="18422" LastEditDate="2016-02-23T11:58:33.303" LastActivityDate="2016-02-23T11:58:33.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27817" PostTypeId="1" CreationDate="2016-02-23T12:42:26.797" Score="-1" ViewCount="127" Body="&lt;p&gt;A shopkeeper keeps 25% of the amount given as commission and returns the remaining back, for example if you give him 1$ he will return back 3 quarters. What is the maximum number of quarters you can get back if you have ten dollars.&lt;/p&gt;&#xA;" OwnerUserId="19506" LastActivityDate="2016-02-23T12:45:42.293" Title="maximum number of quarters using 10 dollars" Tags="|logical-deduction|calculation-puzzle|" AnswerCount="1" CommentCount="2" ClosedDate="2016-02-23T15:21:54.873" ContentLicense="CC BY-SA 3.0" />
  <row Id="27818" PostTypeId="2" ParentId="27817" CreationDate="2016-02-23T12:45:42.293" Score="3" Body="&lt;p&gt;I must be missing something...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 30&lt;br/&gt;&#xA; Ask him for 7.50 change in quarters. That is 30 quarters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19086" LastActivityDate="2016-02-23T12:45:42.293" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="27820" PostTypeId="1" AcceptedAnswerId="36178" CreationDate="2016-02-23T13:04:57.623" Score="13" ViewCount="3158" Body="&lt;p&gt;You get a call in the middle of the night which wakes you up violently—a call which you were not expecting. Your cell phone rings loudly because you have forgotten to turn the ringer off from the day before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You pick up the phone. “Hello?” you mutter in a very drowsy voice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Sorry to call you this late during the night but we have an emergency. Please come to 678 Blazing Drive as soon as possible,” a familiar voice says to you as he hangs up the phone.  You hit the “End Call” button on your phone and rub your eyes. After a couple of seconds you remember the familiar voice, it is Kevin, a co-worker working under you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You quickly get dressed, hop in your car and begin to drive to the destination. As you reach the destination, you notice that there are a lot of police cars, a fire truck, and a forensics team all over the house. You drive up the driveway and get out of your car.  You see a lot of policemen and policewomen around, trying to calm the neighbors down and telling them to go back inside their homes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see a co-worker running towards you. “I’m glad you made it here so quickly,” your co-worker says.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“What seems to be the problem,” you ask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Well, earlier this morning there was a 911 call about a house on fire, and the firefighters rushed here to put out the fire. There were two people in the house, a mother and a father. The father did not survive and was declared dead at the scene; the mother is critically injured and is being rushed to the hospital. After the firefighters put out the fire, they did some investigating to see what started the fire. What they found was that the fire did not start naturally; there was gasoline involved. Someone set the fire and it has become an investigation now.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You thank your co-worker and entered the burned-down house. As you enter, you hear one of the firefighters say to the other, “Yeah, the fire started here in the living room. Gasoline was used to accelerate the fire, it seems.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(PLEASE CHOOSE 6 ROOMS)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Living Room&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You decide to enter the living room area. The love seat, couch, and recliner have been burnt severely; you can barely tell that it’s a couch anymore. Half of the coffee table has been disintegrated, as well as the TV. Some of the pictures seem to have been able to stay on the wall. You decide to take a look at the pictures.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The first picture is a family photo. You can barely make out the picture, but it is the mother, the father, and their four children. The mother is blonde, has long hair, and is in a red dress. The four children are blond, brunette, red-haired, and blond in that order, all males. You can’t make out what the father looks like because the fire has burned him and the left top side of the picture.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The second picture is of the blond son as a young kid playing with a toy plane outside.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The third picture is of the family eating dinner.  The father is on the very left, next to him is the mother, then the brown-haired son, red-haired son, and then on the very right is the blonde son with no one next to him on either side.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The fourth picture is a collage of the red-haired son playing baseball, soccer, swimming, and tennis.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The fifth picture is of the blond and brown-haired sons reading a book on the grass and the center of attention of all his friends respectively.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Kitchen/Dining Room&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You decide to enter the kitchen and dining room. The kitchen has been badly burned. You see that on the dining table there are five plates set up: a set on the very right, three more sets beside it, and the last set on the very right. The set up isn’t very neat; the dishes and utensils are kind of everywhere, not neatly placed.&lt;br&gt;&#xA;&lt;br&gt;&#xA; You head over to the kitchen, and everything has been damaged badly. There is nothing left in the area for you to check, so you decide to leave this area to explore more of the house.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Washroom&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You decide to check out the washroom. As you turn the knob on the door, the entire door just falls off. You quickly jump back as the door collapses in front of you. As you enter the washroom, you noticed that the washroom had not been burned at all; it is still very intact. “I guess that closed door saved the washroom,” you say to yourself. On the counter you see five toothbrushes and cups: a regular plain white cup, a cup with a lot of cartoon characters dancing, a sports cup, a cup with the periodic table on it, and a cup with famous gaming characters on it. In each cup there is a matching toothbrush, except for the white cup. The white cup has a plain blue and pink toothbrush in it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Master bedroom&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You decide to investigate the master bedroom. The fire destroyed everything along its path, but luckily some stuff survived the blaze. There is a picture on the wall of the brown-haired and blond-haired sons.  One of them is playing with friends while the other one is carrying what looks like Monopoly money around. On the bed you notice that the family's cordless home phone is lying in the middle of the bed. The phone is damaged but still seems to work. When you pick up the phone you notice that it is still on, as though they were trying to call someone. You check to see who they were trying to call but all you see is “#3” dialed. “I guess they might have accidentally hit some buttons while sleeping, since the phone was on the bed,” you think to yourself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bedroom 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You decide to enter the first bedroom upstairs. Looking around the room, you notice a half burnt poster, a picture frame on the bookshelf, and a mini statue next to the picture frame. You take a closer look at the poster.  Not much of it is left, but you can make out that a slim male is posing and at the bottom it says “Fi     09”. Some words on the poster have been burnt away and most of the picture of the male has been burnt as well so you can’t quite make out who it is. You then walk over to the picture frame. The picture frame is red, and has a picture of the mother and the father sitting together in a field. The mini statue beside the picture is that of a bull ready to charge. “Isn’t there a basketball team with a bull as their mascot?” you mutter to yourself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bedroom 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You decided to enter the second bedroom upstairs. Looking around the room, you notice a bookshelf which isn’t too badly damaged, a picture frame on the desk, and a poster on the wall. Taking a closer look at the bookshelf, you notice that the shelves are stacked with books. There are so many books that the bookshelf looks like it’s about to break. Just as you take a step back, the vibrations from the footsteps cause the badly damaged bookshelf to collapse and break apart. After the dust has cleared, you decided to take a look at the picture frame. The picture frame is white and has a picture of the mother and the father at a science fair. The poster on the wall is a picture of a ram’s head as a symbol in black, it is similar to a logo or a symbol for a team of some sort.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bedroom 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You decided to enter the third bedroom upstairs. Looking around the room, you notice a very nice wooden calendar on the wall. Besides the calendar, there is nothing else out of the ordinary in this room, only a bed, a chair and a desk. “This person doesn’t have much; maybe this person isn’t loved as much as the others,” you joke to yourself. Looking at the calendar, you notice that it has not been changed for a long time. Almost every day is filled with a task, but the dust and soot make it hard to read. Pinned to the calendar is a picture of the mother and the father at a festival with two other people you do not know. On the side of the calendar, there is a small area for people to write/draw whatever they want. A fish splashing out of the water is drawn in that area. “I guess this person likes fishing,” you say to yourself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bedroom 4&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You decided to enter the fourth bedroom upstairs. As you enter, you notice that this room is in bad condition. As you looked around for anything that survived, you noticed that this room has a lot of bookshelves but no books. The bookshelves contain a lot of DVD &amp;amp; CD cases in various colors like blue, black, green, white, grey, and see-through. The paper in between the cases’ sleeves has been burned, and the DVDs/CDs inside have been severely damaged. On one of the walls is a TV bracket that has survived the fire since it is made of steel, but the broken TV on the ground wasn’t as fortunate. On the desk there is a medium- sized statue of an odd-looking lion of some sort. On the statue is a small plaque that says, &quot;Collector’s edition.&quot; “This is a weird looking lion,” you say to yourself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As you are about to enter the next room, the ground starts shaking. “AHHH!” a woman screams. “Everybody run! This place is about to collapse!” another person yells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You look to your left and you see the entire house is coming down slowly. You start running as fast as you can towards the exit while trying to dodge all the objects falling around you. As you get closer to the entrance you jump towards the bright light coming through the exit. As your body hits the grass outside, the entire house collapses. Dust fills the air. You turn around, and all you see is a huge mess. “Guess that’s the end of my investigation” you say to yourself. Since the other victim is still in the hospital in critical condition, you decide that the next move should be to contact the four sons and ask them what they know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a couple of days, your team has found out where all four sons live and has contacted the police to bring all four of them in for interrogation. As you enter the room, you notice that there are four males sitting beside each other: one redhead, one blond, one brown-haired, and one bald, all looking around the same age.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You sit down and offer them a smoke or any drinks they would like. The brown-haired son accepts a cigarette, and the redhead kindly refuses both.  “Not what I usually do but sure I’ll take some,” says the bald son as he takes two cigarettes. The blond son asks if you have any rum but you tell him you do not; he does not ask for any other drink.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You take a quick glance at what each of them are wearing, before asking any questions. The bald son has on a full suit. “Looks like he’s very important,” you think to yourself. Both the brown-haired and blond-haired sons are dressed business casual, while the red-haired son is dressed in a dirty, greasy jumpsuit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You ask all four of them the same question: “What were you doing and where were you on the night of the fire?”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The blond one says, “Well, I was working from 8:00 a.m. to 4:00 p.m. and after that I went over to my parents' place because they asked me to come over for dinner.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“I worked and went over to my parents' place for dinner as well, but none of my other brothers were there. Mom did tell me that I was the third to arrive,” says the redhead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bald one answers, “I worked from 8:00 a.m. to 6:00 p.m. and then went to my parents' place for dinner after work. Mom told me that I was the second to arrive.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last son replies, “I worked the office time, and after that I also went to my parents place for dinner.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right when you are about to ask the next question, a fellow officer opens the door. “I have some terrible news, your mother did not make it. I’m sorry,” the officer says. The four sons get up immediately and rush towards the hospital, only a block away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are about to chase after them but something doesn’t sit right with you. You started thinking back on all the info you have gathered here and at the house. After a little while, it hits you. You know who started the fire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;WHO IS THE MURDERER?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus:&lt;/strong&gt; If someone is able to fill in the chart below, I will give them an upvote&#xA;&lt;a href=&quot;https://i.stack.imgur.com/6Y16z.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6Y16z.png&quot; alt=&quot;Chart 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&#xA;&lt;strong&gt;UPDATE&lt;/strong&gt;&#xA;&lt;br/&gt;updated chart that will help&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Cdf4Z.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Cdf4Z.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4645" LastEditorUserId="4645" LastEditDate="2016-05-25T15:57:20.503" LastActivityDate="2016-06-17T01:07:07.437" Title="Mysterious Murder Mystery - The Four Brothers" Tags="|riddle|word-problem|story|mystery|situation|" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="27821" PostTypeId="1" AcceptedAnswerId="27823" CreationDate="2016-02-23T13:25:25.207" Score="16" ViewCount="1192" Body="&lt;p&gt;There are two piles of balls. One contains 5 balls and the other contains 18 balls.&lt;/p&gt;&#xA;&lt;p&gt;A and B are playing the following game:&#xA;In each round, a player has to move balls from the larger pile to the smaller pile such that the number moved was a non-zero multiple of how many were previously in the smaller pile. A starts, and A and B alternate until a pile of balls is empty. The person who emptied it is the winner.&lt;/p&gt;&#xA;&lt;p&gt;Who will win this game?&lt;/p&gt;&#xA;" OwnerUserId="19293" LastEditorUserId="56715" LastEditDate="2020-12-15T16:28:58.697" LastActivityDate="2020-12-15T16:28:58.697" Title="Empty one pile of balls and win" Tags="|mathematics|strategy|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27822" PostTypeId="2" ParentId="27811" CreationDate="2016-02-23T13:38:44.920" Score="1" Body="&lt;p&gt;I suspect &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the only solution is when there is a single black stone with 500 white stones on each side.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You win by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; start in the middle.  Then selecting black stone(s) farthest from the middle until the ends are black.  Then start in the middle again and repeat until all the stones are black.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2016-02-23T13:38:44.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27823" PostTypeId="2" ParentId="27821" CreationDate="2016-02-23T13:46:07.017" Score="17" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A wins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A moves 5 balls over, which leaves piles of 10-13.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only legal move for B is moving 10, leaving 20-3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A responds by moving 9, leaving 11-12.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only legal move for B is moving 11, leaving 22-1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A moves all 22 balls, leaving 0-23.&lt;/p&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="17389" LastEditDate="2016-02-23T13:56:31.427" LastActivityDate="2016-02-23T13:56:31.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27824" PostTypeId="1" AcceptedAnswerId="27833" CreationDate="2016-02-23T13:52:48.913" Score="14" ViewCount="1981" Body="&lt;p&gt;Your answer should explain all the parts of this riddle. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Try&lt;br&gt;&#xA;  To try,&lt;br&gt;&#xA;  Vi&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer (as implied by the second tag) is a single short English word.&lt;/p&gt;&#xA;" OwnerUserId="18924" LastEditorUserId="18924" LastEditDate="2016-02-23T14:51:09.870" LastActivityDate="2016-02-23T19:06:57.457" Title="For Vivienne. A Very Short Riddle" Tags="|riddle|word|" AnswerCount="6" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27825" PostTypeId="1" CreationDate="2016-02-14T02:32:00.260" Score="-1" ViewCount="368" Body="&lt;p&gt;Without any restriction, can a &lt;strong&gt;&lt;em&gt;single sentence&lt;/em&gt;&lt;/strong&gt; be written or spoken  using the words &quot;to&quot;, &quot;too&quot;, and &quot;two&quot; in the same sentence consecutively used (one after the other) and be proper? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please explain and write your &lt;strong&gt;&lt;em&gt;example sentence&lt;/em&gt;&lt;/strong&gt; if this is possible. &lt;/p&gt;&#xA;" OwnerUserId="19360" OwnerDisplayName="1Fish_2Fish_RedFish_BlueFish" LastEditorUserId="4421" LastEditDate="2016-09-14T15:12:07.747" LastActivityDate="2016-09-14T15:12:07.747" Title="Can a single sentence be made using words to, too, and two consecutively (one after the other) in the same sentence properly?" Tags="|english|open-ended|" AnswerCount="2" CommentCount="2" FavoriteCount="0" ClosedDate="2016-02-23T17:29:40.070" ContentLicense="CC BY-SA 3.0" />
  <row Id="27830" PostTypeId="2" ParentId="27825" CreationDate="2016-02-23T14:06:31.590" Score="9" Body="&lt;p&gt;Peter: I won the game! I got to 2 points first!&lt;br/&gt;&#xA;Paul: But I got to two too!&lt;/p&gt;&#xA;" OwnerUserId="19086" LastActivityDate="2016-02-23T14:06:31.590" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27831" PostTypeId="2" ParentId="27824" CreationDate="2016-02-23T14:06:43.713" Score="6" Body="&lt;p&gt;Try&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Three (as in tri)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To try&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; two threes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Vi&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 6 (VI is 6 in Roman Numerals)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So Vivienne really likes&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mazdas? Specifically the Mazda 323 and Mazda 6&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9353" LastEditorUserId="9353" LastEditDate="2016-02-23T16:48:24.897" LastActivityDate="2016-02-23T16:48:24.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27832" PostTypeId="2" ParentId="27824" CreationDate="2016-02-23T14:11:55.713" Score="2" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BMW model# 525 or 526&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How did I get there?&lt;br/&gt;&lt;br/&gt;&#xA;TRY&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; = 5. a try in rugby is worth &lt;strong&gt;5&lt;/strong&gt; points&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;TO TRY&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; = 25&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;comma&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; = or&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Vi&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; = 6&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19086" LastActivityDate="2016-02-23T14:11:55.713" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27833" PostTypeId="2" ParentId="27824" CreationDate="2016-02-23T14:28:16.183" Score="16" Body="&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Based on the comment from the question poser:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You are within a league of the finish line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I could possibly see the solution being:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; league&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Try&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In rugby, a &quot;try&quot; is worth 5 points.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To try,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let's go ahead and make two of them now 5 5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Vi&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; And then interpret this as a Roman Numeral 6&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then we see&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A league is equal to 5556m.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;old answer left for posterity&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The number of days in a leap year.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Try&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a homonym of the prefix &quot;tri-&quot;, which indicates 3.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To Try,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a homonym of &quot;two&quot; and then &quot;tri-&quot; again, which indicates 6.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Vi&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; interpreted as Roman numerals, we get 6.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Composing this gives us&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 366: the number of days in a leap year (such as 2016).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's not yet clear to me if the title is related.&lt;/p&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2016-02-23T19:06:57.457" LastActivityDate="2016-02-23T19:06:57.457" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27834" PostTypeId="1" AcceptedAnswerId="27835" CreationDate="2016-02-23T14:58:46.377" Score="33" ViewCount="3420" Body="&lt;p&gt;Explain &quot;Two has three, and four has three, but three has only one&quot;.&lt;/p&gt;&#xA;" OwnerUserId="19510" LastActivityDate="2016-11-30T07:03:03.567" Title="A riddle I made up for my kids a while ago" Tags="|riddle|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27835" PostTypeId="2" ParentId="27834" CreationDate="2016-02-23T15:00:57.030" Score="36" Body="&lt;p&gt;I believe you are talking about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Homophones&lt;br&gt;&#xA;&lt;br&gt;&#xA; &quot;Two&quot; also has &quot;to&quot; and &quot;too&quot;, &quot;Four&quot; also has &quot;for&quot; and &quot;fore&quot;, but &quot;three&quot; has no homophones.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastActivityDate="2016-02-23T15:00:57.030" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="27836" PostTypeId="1" AcceptedAnswerId="27841" CreationDate="2016-02-23T15:20:33.593" Score="4" ViewCount="209" Body="&lt;p&gt;My uncle Todd is a little off his rocker. He's getting on in years, and sometimes I find him outside at night scratching numbers in the dirt with a stick. Last night I found him sitting quietly in front of his latest sets of numbers, gazing off at the city skyline that never seemed to rest. There were obvious marks where some numbers had once been but had been rendered incomprehensible by the breeze rolling in from the river.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most people just call him crazy, but I think there's a method to his scribbling. Can you uncover his pattern and fill in the missing numbers?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1, 2, 3, __, 5, __, __, 8, 9, 9, __, 11, __&lt;br&gt;&#xA;  __, 8, 8, 7, __, 4, 4, __, 1, 30, __, __, 29&lt;br&gt;&#xA;  1, __, __, 14, 18, 22, 27, __, 35, 39, __, 48, 52&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16275" LastActivityDate="2016-02-23T16:07:54.117" Title="A Lunatic's Sequences" Tags="|number-sequence|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27837" PostTypeId="1" AcceptedAnswerId="27839" CreationDate="2016-02-23T15:27:24.283" Score="13" ViewCount="699" Body="&lt;p&gt;At a recently held election one party fielded 3 candidates &lt;em&gt;A,B&lt;/em&gt; and &lt;em&gt;C&lt;/em&gt;. In this election people can vote for any number of candidates in a desired party. The voting pattern of 100 people as follows: The 3 candidates; &lt;em&gt;A,B&lt;/em&gt; and &lt;em&gt;C&lt;/em&gt; received $90, 60$ and $57$ votes respectively while 5 people did not vote for anyone. What is the most number of people who could have voted for all 3?&lt;/p&gt;&#xA;" OwnerUserId="19293" LastActivityDate="2016-02-23T15:33:21.843" Title="Vote for the election" Tags="|logical-deduction|combinatorics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27839" PostTypeId="2" ParentId="27837" CreationDate="2016-02-23T15:33:21.843" Score="15" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 56&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 57 is obviously a hard maximum, since that is the number of votes received by candidate C. But if 57 people voted for all 3, then the maximum number of total voters is 93: 57 for all 3, 3 for B, and 33 for A.&lt;br&gt;&lt;br&gt;&#xA; Next consider the case of 56 voting for all 3. This results in 1 for C, 4 for B, and 34 for A, for a total of 95 voters. (5 people abstained.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16275" LastActivityDate="2016-02-23T15:33:21.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27841" PostTypeId="2" ParentId="27836" CreationDate="2016-02-23T15:54:34.913" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It appears to be a lunar calendar, the first two line up pretty well with the month and day of new moons for 2016, the third with full Julian week.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, here we are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1, 2, 3, &lt;strong&gt;4&lt;/strong&gt;, 5, &lt;strong&gt;6&lt;/strong&gt;, &lt;strong&gt;7&lt;/strong&gt;, 8, 9, 9, &lt;strong&gt;10&lt;/strong&gt;, 11, &lt;strong&gt;12&lt;/strong&gt;&lt;br /&gt;&lt;strong&gt;9&lt;/strong&gt;, 8, 8, 7, &lt;strong&gt;6&lt;/strong&gt;, 4, 4, &lt;strong&gt;2&lt;/strong&gt;, 1, 30, &lt;strong&gt;30&lt;/strong&gt;, &lt;strong&gt;29&lt;/strong&gt;, 29&lt;br /&gt;1, &lt;strong&gt;6&lt;/strong&gt;, &lt;strong&gt;10&lt;/strong&gt;, 14, 18, 22, 27, &lt;strong&gt;31&lt;/strong&gt;, 35, 39, &lt;strong&gt;43&lt;/strong&gt;, 48, 52&lt;br /&gt;Courtesy: &lt;a href=&quot;http://www.calendar-12.com/moon_phases/2016&quot; rel=&quot;nofollow&quot;&gt;http://www.calendar-12.com/moon_phases/2016&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17446" LastEditorUserId="17446" LastEditDate="2016-02-23T16:07:54.117" LastActivityDate="2016-02-23T16:07:54.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27842" PostTypeId="2" ParentId="27824" CreationDate="2016-02-23T16:10:16.360" Score="2" Body="&lt;p&gt;Another answer: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Beef&lt;br/&gt;&lt;br/&gt;&#xA; The second (TO[2]), fifth (TRY[5]), fifth (TRY[5]) and sixth (Vi) letters of the alphabet.&lt;br/&gt;&lt;br/&gt; TRY = 5 because a try scores 5 points in rugby. Based off my first answer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19086" LastEditorUserId="19086" LastEditDate="2016-02-23T16:17:37.710" LastActivityDate="2016-02-23T16:17:37.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27843" PostTypeId="2" ParentId="27825" CreationDate="2016-02-23T16:23:04.503" Score="9" Body="&lt;p&gt;&quot;I think the number six horse ate an apple before the race.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I saw them give one to two too.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2016-02-23T16:23:04.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27844" PostTypeId="2" ParentId="27824" CreationDate="2016-02-23T16:51:08.330" Score="1" Body="&lt;p&gt;I believe the word is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Plan / Scheme&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Try to Try&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If we read this as &quot;Try to Try&quot;, we get to something extremely different than if we were to pause after the first try. Essentially, this hints that the narrator is telling someone to try, however, not to do x. He's telling someone to try to even attempt at doing x. Incentivating/Motivating of sorts. Before trying to do something, a plan(Trying to Try) must be formed, and only then it's executed (the 2nd try).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Vi&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm somewhat inclined to believe that this is misderection and just a calling. Usually on letters the receipient is identified by &quot;,receipientName&quot;, and as Vi can be short for Vivienne, that's my reasoning.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18816" LastActivityDate="2016-02-23T16:51:08.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27845" PostTypeId="2" ParentId="27643" CreationDate="2016-02-23T18:00:16.277" Score="1" Body="&lt;p&gt;If I &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Take every sixth word (&lt;em&gt;juggle by &quot;finger&quot; length?&lt;/em&gt;) starting with &quot;Follow&quot;, I &lt;em&gt;almost&lt;/em&gt; get &quot;Follow the jumping words.&quot;, but not quite. It involves moving &lt;em&gt;back&lt;/em&gt; through the words once I hit the end, as well as (maybe?) skipping the word &quot;skyward&quot; (i.e., &lt;em&gt;jumping&lt;/em&gt; it)&lt;br&gt;&#xA;&lt;br&gt;&#xA; I also almost get this if I follow &quot;back, fo[u]r[ ]w[o]rds&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2016-02-23T18:05:22.777" LastActivityDate="2016-02-23T18:05:22.777" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27846" PostTypeId="2" ParentId="27824" CreationDate="2016-02-23T19:06:06.843" Score="4" Body="&lt;p&gt;Perhaps this is an &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Exception&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Two &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Try blocks&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;between a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; to&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;in&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Vi (or vim)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;will result in an&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Exception&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4381" LastActivityDate="2016-02-23T19:06:06.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27847" PostTypeId="1" AcceptedAnswerId="27850" CreationDate="2016-02-23T20:05:13.373" Score="25" ViewCount="2313" Body="&lt;p&gt;On a 15x15 chessboard there are 15 rooks that do not attack each other (via ordinary rook moves). Then each of the rooks makes one move like that of a knight. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible that after all this is done, the 15 rooks still do not attack each other (via ordinary rook moves)?&lt;/p&gt;&#xA;" OwnerUserId="19470" LastEditorUserId="17482" LastEditDate="2016-02-24T07:54:48.550" LastActivityDate="2016-02-25T17:49:46.637" Title="Rooks on a 15x15 chessboard" Tags="|mathematics|combinatorics|chess|checkerboard|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="27848" PostTypeId="1" AcceptedAnswerId="27853" CreationDate="2016-02-23T20:17:30.173" Score="11" ViewCount="345" Body="&lt;p&gt;12 players took part in a chess tournament and any two of them played exactly one match against each other (1 point for victory, 1/2 point for draw, 0 point for loss). After the tournament the top-ranked three players together gained three times more points than the bottom-ranked five players together. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did the match between the seventh and the eighth player end?&lt;/p&gt;&#xA;" OwnerUserId="19470" LastEditorUserId="57212" LastEditDate="2022-10-22T15:12:57.603" LastActivityDate="2022-10-22T15:12:57.603" Title="The match between the seventh and the eighth player" Tags="|mathematics|combinatorics|game|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27849" PostTypeId="1" AcceptedAnswerId="27852" CreationDate="2016-02-23T20:41:59.447" Score="7" ViewCount="3006" Body="&lt;p&gt;&lt;strong&gt;Infomation/Story&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your prince is bewitched and asleep.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to feed him the good apples (from the truth-teller dwarves) to save him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But be careful. Only the truth-teller dwarves, who only tell the truth, have good apples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The liar dwarves, who only lie, have poisoned apples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can find all three good apples, your prince will be released.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Let's start then...&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Red Dwarf said:    Black has a good apple.&lt;/li&gt;&#xA;&lt;li&gt;Orange Dwarf said: I dug up a lump of gold!&lt;/li&gt;&#xA;&lt;li&gt;Yellow Dwarf said: Orange is in my group.&lt;/li&gt;&#xA;&lt;li&gt;Green Dwarf said:  The weather on the mountain is so whimsical. Yesterday, we&#xA;had a huge rainstorm.&lt;/li&gt;&#xA;&lt;li&gt;Blue Dwarf said:   White isn't in my group.&lt;/li&gt;&#xA;&lt;li&gt;Black Dwarf said:  Green isn't in my group.&lt;/li&gt;&#xA;&lt;li&gt;White Dwarf said:  Red's apple is a good one indeed.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, which ones are the good apples?&lt;/p&gt;&#xA;" OwnerUserId="19520" LastEditorUserId="37225" LastEditDate="2017-12-05T04:08:01.527" LastActivityDate="2017-12-05T04:08:01.527" Title="Who are the truth-teller dwarves?" Tags="|logical-deduction|liars|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27850" PostTypeId="2" ParentId="27847" CreationDate="2016-02-23T20:47:29.597" Score="39" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No&lt;br&gt;&#xA;&lt;br&gt;&#xA; Since there is (initially and finally) a rook in each row and column, we can say the sum of all rooks' X and Y positions must be equal to 2 $\times$ (15 + 14 + ... + 1) = 120.&lt;br&gt;&#xA;&lt;br&gt;&#xA; A knight's move will increment the rook's (X + Y) by +3, +1, -1, or -3, all of which are odd numbers.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The sum of 15 odd numbers (15 knight moves) is an odd number.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Since we are performing 15 moves, we are adding an odd number to the sum of the rooks' X and Y positions.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The final sum of the rooks' X and Y positions cannot be equal to the initial sum since performing 15 knight moves adds a non-zero number to that sum.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Since the initial sum is 120, the final sum cannot be 120 and therefore can't be a legal (non-attacking) position.&lt;br&gt;&#xA;&lt;br&gt;&#xA; This proof can be extended to any $N\times N$ board where $N$ is odd.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18103" LastEditorUserId="4421" LastEditDate="2016-02-25T17:49:46.637" LastActivityDate="2016-02-25T17:49:46.637" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27851" PostTypeId="2" ParentId="27848" CreationDate="2016-02-23T20:56:56.063" Score="2" Body="&lt;p&gt;Let say 8th to 12th players are the last 5 and they lost their all matches except against each other. So 8th has to win or draw &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(does not matter since total point would not change and I will go for win) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;against 9,10,11 and 12th, 9th would win against 10,11,12th etc and lost against 8th. So The total point of the last 5 players would be 8,6,4,2 and 0 at worst. That makes total 20. So the first three, 1st, 2nd and 3rd has to have 60 point at least. This is only possible if 1st, 2nd and 3rd would win all matches against other players except the matches between them. They can draw or win against each other. The total point would be 60. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4th, 5th, 6th, &lt;strong&gt;7th&lt;/strong&gt; games has to be lost against 1st, 2nd and 3rd. and win against 8th, 9th, 10th, 11th, 12th. In between 4th to 7th cannot be known and does not change the fact of the given in the question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;8th&lt;/strong&gt; has to lose against 1st to 7th as mentioned before. and win against the rest to become 8th.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18514" LastActivityDate="2016-02-23T20:56:56.063" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27852" PostTypeId="2" ParentId="27849" CreationDate="2016-02-23T21:02:10.813" Score="11" Body="&lt;p&gt;The truth telling dwarves are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Orange, Blue, and Yellow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let's break down one sentence in particular:&lt;br&gt;&#xA; &quot;X isn't in my group&quot; is said by two dwarves.&lt;br&gt;&#xA; Either:&lt;br&gt;&#xA; 1. The dwarf speaking is telling the truth, which means that X is a liar, or&lt;br&gt;&#xA; 2. The dwarf speaking is a liar, which means that X is also a liar&lt;br&gt;&#xA; Either way, X is a liar.&lt;br&gt;&#xA;&lt;br&gt;&#xA; From this, we can deduce that White and Green are liars because Black and Blue said so.&lt;br&gt;&#xA; Knowing about white means that red has a poison apple, and so is a liar.&lt;br&gt;&#xA; knowing about red means that black has a poison apple, and so is a liar.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Now we know that White, Green, Red, and Black are Liars. There are three remaining dwarves, and three remaining good apples. By elimination, Orange, Yellow, and Blue have the good apples!  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19491" LastEditorUserId="12109" LastEditDate="2016-02-23T22:22:34.400" LastActivityDate="2016-02-23T22:22:34.400" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27853" PostTypeId="2" ParentId="27848" CreationDate="2016-02-23T21:02:23.047" Score="7" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;7th beat 8th.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are 66 matches ($\frac{12 \times 11}2$).  Each match results in a total of 1 point gained, so there are 66 points to be gained total.  (This isn't necessary information, but is nice for checking our work.)&lt;br&gt;&#xA;&lt;br&gt;&#xA;Only one person can score a zero, and only one person can score an 11. (You can't have more than one person lose every match or more than one person win every match.)&lt;br&gt;&#xA;&lt;br&gt;&#xA;The highest possible score for the top 3 is 30. (Examples: [11, 10, and 9] or [11, 9.5, 9.5 ] or [10.5, 10.5, 9])  The important thing is that they each beat all the other players (9 $\times$ 3 = 27) and give up 3 points amongst themselves (27 + 3 = 30).&lt;br&gt;&#xA;&lt;br&gt;&#xA;That means the &lt;strong&gt;maximum&lt;/strong&gt; cumulative score the bottom 5 can achieve is 10 ($\frac{30}3$).&lt;br&gt;&#xA;&lt;br&gt;&#xA;Note that this is also the &lt;strong&gt;minimum&lt;/strong&gt; score that the bottom 5 can achieve. (They play 10 games against each other ($\frac{5 \times 4}2$), and someone has to win each match. So  there must be at least 10 points between the five of them.)  So these must be the &lt;strong&gt;exact&lt;/strong&gt; cumulative scores for the bottom 5 and the top 3 (&lt;strong&gt;&lt;em&gt;10 points and 30 points&lt;/em&gt;&lt;/strong&gt;).&lt;br&gt;&#xA;&lt;br&gt;&#xA;This leaves 26 points (30 + 10 + 26 = 66) for the middle 4, which is consistent with [5, 6, 7, 8].  So this is a valid scoring distribution given the premises.  This distribution is possible if everyone beat all the people below them in ranking, which means that 7th would have beat 8th.&lt;br&gt;&#xA;&lt;br&gt;&#xA;It doesn't really matter what 7th scored anyway, since &lt;strong&gt;&lt;em&gt;if 8th would have lost or tied to anyone outside the bottom 5, the cumulative score of the bottom 5 would be above 10, which would put the cumulative score of the top 3 above 30, which is impossible&lt;/em&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;TLDR:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The 'top-three-equals-three-times-the-bottom-five' restriction and the round-robin nature of the tournament requires that the top section be total winners and the bottom section be total losers (with respect to the other brackets).  8th had to lose to everyone above him.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19128" LastEditorUserId="4421" LastEditDate="2016-02-25T17:48:23.803" LastActivityDate="2016-02-25T17:48:23.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27854" PostTypeId="2" ParentId="26437" CreationDate="2016-02-23T22:07:35.850" Score="-1" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;An apple or some sort of fruit.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19524" LastEditorUserId="19128" LastEditDate="2016-02-23T22:28:15.180" LastActivityDate="2016-02-23T22:28:15.180" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27855" PostTypeId="2" ParentId="5006" CreationDate="2016-02-23T22:22:21.300" Score="0" Body="&lt;p&gt;To all you saying the answer to #3 is 2^20, let's simplify it to two trials.&#xA;You can only do 3 floors with two trials. There are four options. Breaking at the first, second, third floor, or not at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is 2^20-1&lt;/p&gt;&#xA;" OwnerUserId="19527" LastActivityDate="2016-02-23T22:22:21.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27858" PostTypeId="2" ParentId="27643" CreationDate="2016-02-24T00:12:45.030" Score="9" Body="&lt;p&gt;&lt;em&gt;Well, pineapple isn't tricky it you've too purple horse once understood.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Follow the words. Jumping back and forwards. Repeatedly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Starting at the start, write down each word, and then:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; jump forward by the number of characters in the word. Write down the next word. Jump back by the length of that word. Then forward, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The hint decodes to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jump by word length.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="5208" LastEditDate="2016-02-24T00:33:56.920" LastActivityDate="2016-02-24T00:33:56.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27859" PostTypeId="2" ParentId="27781" CreationDate="2016-02-24T00:14:12.337" Score="1" Body="&lt;p&gt;After I checked at least 1000 calendar systems (perceived), I have still no clue about the first two lines. But now I have good knowledge in things I never need again... i.e. 8-day-weeks :-/&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something productive now: The same sequence by Hellion also exists if you imagine a piano keyboard and start with the F-key as 1. The white keys are the numbers then. Maybe the others are something with guitar chords or so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the 52 as the scale for the first line mentioned by user265554 is very exceptional. 21, 27, ..., 35 should end in 40 or maybe 41 / 42.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now I suspend myself till google is able to solve this in 20 years.&lt;/p&gt;&#xA;" OwnerUserId="4342" LastActivityDate="2016-02-24T00:14:12.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27860" PostTypeId="2" ParentId="27815" CreationDate="2016-02-24T00:23:20.993" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All the orderable numbers can be written as $6x+9y+20z$, which can be incremented by 1 with $6(x+2)+9(y+1)+20(z-1)$, or if $z=0$, $6(x-2)+9(y-3)+20(z+2)$. A $6x+9y$ number can also be written as $6(x-3)+9(y+2)$, resulting in $6+9n$ or $6*2+9n$. Therefore $9k+3$ numbers can be legally incremented by 1 if $k&amp;gt;3$, $9k+6$ numbers if $k&amp;gt;4$ and $9k$ numbers if $k&amp;gt;4$. In addition, if an incrementable multiple of 3 has 1 added, 20*2 can be found in the new sum and even the next 2 numbers can be reached, though it doesn't necessarily make the newest number obtained adding the original one 3 incrementable. One thing is certain: any number starting from 9*5+6=51 is incrementable. $3k+1$ numbers smaller than 46 can't even be reached. All numbers from 44 to 51 are reachable, so the answer is 43.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastActivityDate="2016-02-24T00:23:20.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27861" PostTypeId="2" ParentId="27617" CreationDate="2016-02-24T01:03:24.187" Score="2" Body="&lt;p&gt;My only thought that seems to kind of fit so far is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Glasses/monocle. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fitting the riddle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Without one lens (eye), glasses become a monocle, which is still a vision aid.  Remove the other lens, and you don't have a monocle, so you aren't &lt;em&gt;culture&lt;/em&gt;d anymore.  As far as neighbors, I'm not sure.  The nose pads?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As for the hints:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are many types of glasses.  Monocles are hard to find.  Magnifying glasses can be dangerous... ?  &amp;nbsp; &amp;nbsp; :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19516" LastActivityDate="2016-02-24T01:03:24.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27862" PostTypeId="2" ParentId="27821" CreationDate="2016-02-24T02:06:36.597" Score="4" Body="&lt;p&gt;The total number of balls always remains 23, a prime number, meaning:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Being left with only 1 ball in the smaller pile is the only way to immediately win.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Scenario 1 can only be forced on the loser right away if he/she's left with 11 balls in the smaller one and 12 in the bigger one.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;2's condition can be forced on the loser right away only if (s)he leaves 2, 3, 4 or 6 (factors of 12 save 1 and itself) balls in the small pile, leading to a 12-ball pile.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The loser can be forced to leave a 3-ball pile if left with a 10-ball one, which can be done in the very first move by A.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="14016" LastActivityDate="2016-02-24T02:06:36.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27863" PostTypeId="5" CreationDate="2016-02-24T05:04:49.690" Score="0" Body="&lt;p&gt;The setup of meta-knowledge puzzles consist of several perfect logicians, each informed about different aspects of some secret piece of information. They then either have a conversation, or perform a series of actions, based on what they know. The goal is for the reader is either to determine what the secret is, or to figure out how these perfect logicians will behave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A common theme in these puzzles is knowledge about knowledge. To determine how the parties in the puzzle act, you have to think about what the parties know, and their knowledge about what everyone else knows, and their knowledge about their knowledge about what they know, and so on. A useful concept is the idea of &quot;common knowledge,&quot; which consists of what is known by everyone, and known to be known by everyone, and so on to all higher orders of knowledge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some famous examples are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;/questions/236/in-the-100-blue-eyes-problem-why-is-the-oracle-necessary&quot;&gt;Blue Eyed Islanders&lt;/a&gt;:  There are several people wearing hats who know everyone's hat but their own. They must leave the island on the first day that they figure out their eye color. An &quot;oracle&quot; visits the island, and tells the islanders something about the eye colors she sees. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;/questions/251/i-dont-know-the-two-numbers-but-now-i-do&quot;&gt;Sum and Product&lt;/a&gt;: There are two unknown numbers. One person is told their sum, and the other is told their product. They then have a conversation, and the reader must figure out the tow numbers from this conversation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;/questions/12229/explaining-cheryls-birthday&quot;&gt;Cheryl's Birthday&lt;/a&gt;: Cheryl tells Alice and Bob that her birthday is in a certain list of dates. She tells Alice the month, and Bob the day. After a conversation, you must figure out the birthday.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="10615" LastEditorUserId="11110" LastEditDate="2017-06-21T07:59:44.110" LastActivityDate="2017-06-21T07:59:44.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27864" PostTypeId="4" CreationDate="2016-02-24T05:04:49.690" Score="0" Body="Puzzles where perfect logicians are told different pieces of information, and the reader must figure out something based on their words and actions. " OwnerUserId="10615" LastEditorUserId="10615" LastEditDate="2016-02-24T05:04:49.690" LastActivityDate="2016-02-24T05:04:49.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27865" PostTypeId="1" AcceptedAnswerId="27868" CreationDate="2016-02-24T07:32:13.447" Score="13" ViewCount="1417" Body="&lt;p&gt;Professor Halfbrain has spent has spent the last few days with placing pawns on a &lt;span class=&quot;math-container&quot;&gt;$9\times9$&lt;/span&gt; chessboard; each of the &lt;span class=&quot;math-container&quot;&gt;$81$&lt;/span&gt; squares on the chessboard had side length &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;.&#xA;Halfbrain always started with an empty board, and then one by one placed (point-sized) pawns onto it. Every pawn was placed precisely in the middle of one of the little chessboard squares. Whenever Halfbrain placed a new pawn, its distance to each of the pawns placed before was at least &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Professor Halfbrain has proved two extremely deep theorems on such placements of pawns.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Professor Halfbrain's first theorem&lt;/strong&gt;: It is possible to place three pawns according to the above rules.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Professor Halfbrain's second theorem&lt;/strong&gt;: It is not possible to place &lt;span class=&quot;math-container&quot;&gt;$81$&lt;/span&gt; pawns according to the above rules.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This puzzle asks you to improve the two theorems of professor Halfbrain and to make them even deeper.&#xA;Find an integer &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;, so that &amp;quot;three pawns&amp;quot; in the first theorem may be replaced by &amp;quot;&lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; pawns&amp;quot;, and so that &amp;quot;&lt;span class=&quot;math-container&quot;&gt;$81$&lt;/span&gt; pawns&amp;quot; in the second theorem may be replaced by &amp;quot;&lt;span class=&quot;math-container&quot;&gt;$x+1$&lt;/span&gt; pawns&amp;quot; (again yielding true statements, of course).&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2021-12-03T14:40:50.653" Title="Professor Halfbrain and the 9x9 chessboard (Part 1)" Tags="|mathematics|combinatorics|checkerboard|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27866" PostTypeId="2" ParentId="27865" CreationDate="2016-02-24T08:04:30.817" Score="9" Body="&lt;p&gt;Here's what I got&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/DKsCP.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DKsCP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In this&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every dot represents a pawn. If I counted it correct then it is $9 \times 9$ board. Distance between two pawns is 2 units. And there are total $25$ pawns. To make this, I first placed the pawn on first row and first column and kept filling the grid according to the rules.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If I place the pawn on first row and second column then &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll get&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/GfqQF.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GfqQF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In this &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Total number of pawns are $22$ which is lower then the first case. Same can be done if I start placing the pawns from Row 2 and Column 2. In that case total number of pawns placed will be $18$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem 1 (corrected):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is possible to place 25 pawns according to the rules.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Theorem 2 (corrected):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is not possible to place 26 pawns according to the rules.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-02-24T08:29:38.927" LastActivityDate="2016-02-24T08:29:38.927" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27867" PostTypeId="1" AcceptedAnswerId="27870" CreationDate="2016-02-24T08:12:00.010" Score="10" ViewCount="951" Body="&lt;p&gt;Remove the message to reveal the message:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Tonbo etoto rono thart dofo rybo eu&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastActivityDate="2016-02-24T09:17:06.833" Title="No Quotes, Please. Prove me right" Tags="|steganography|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27868" PostTypeId="2" ParentId="27865" CreationDate="2016-02-24T08:36:27.603" Score="16" Body="&lt;p&gt;There is a simple way to see that manshu's answer &lt;span class=&quot;math-container&quot;&gt;$x=25$&lt;/span&gt; is correct.&lt;/p&gt;&#xA;&lt;h3&gt;Translate the problem into an easier problem:&lt;/h3&gt;&#xA;&lt;p&gt;Rather than consider a &lt;span class=&quot;math-container&quot;&gt;$9\times 9$&lt;/span&gt; board and pawns, consider a &lt;span class=&quot;math-container&quot;&gt;$10 \times 10$&lt;/span&gt; board and &lt;span class=&quot;math-container&quot;&gt;$2 \times 2$&lt;/span&gt; square pieces. We can associate to each pawn configuration on the &lt;span class=&quot;math-container&quot;&gt;$9 \times 9$&lt;/span&gt; board a unique configuration of squares in the &lt;span class=&quot;math-container&quot;&gt;$10 \times 10$&lt;/span&gt; board; consider the image below.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EUwUN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EUwUN.png&quot; alt=&quot;From pawns to 2x2 squares&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The blue squares are the squares we added to our board to make it &lt;span class=&quot;math-container&quot;&gt;$10 \times 10$&lt;/span&gt;. The &lt;span class=&quot;math-container&quot;&gt;$2 \times 2$&lt;/span&gt; square on the side is a piece for our new board, and the red square marks where the pawn lies for our previous board. It's easy to see this relation is a bijection.&lt;/p&gt;&#xA;&lt;p&gt;For example in the image below, to the left we have the &lt;span class=&quot;math-container&quot;&gt;$9 \times 9$&lt;/span&gt; board in which red squares mark pawn positions. To the right is the corresponding &lt;span class=&quot;math-container&quot;&gt;$10 \times 10$&lt;/span&gt; board with &lt;span class=&quot;math-container&quot;&gt;$2 \times 2$&lt;/span&gt; square pieces; notice the position of red squares is preserved and no pieces overlap.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/m7Xzp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/m7Xzp.png&quot; alt=&quot;An example of the bijection&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;h3&gt;Solution of the puzzle&lt;/h3&gt;&#xA;&lt;p&gt;Solving Professor Halfbrain's problem now becomes the same as solving the translated problem: How many &lt;span class=&quot;math-container&quot;&gt;$2 \times 2$&lt;/span&gt; square pieces can we place on a &lt;span class=&quot;math-container&quot;&gt;$10 \times 10$&lt;/span&gt; board without overlappings?&lt;br /&gt;&#xA;Now, our new board has a total of &lt;span class=&quot;math-container&quot;&gt;$100$&lt;/span&gt; squares, and each piece uses up &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt; squares. It's thus clear that no configuration can go beyond &lt;span class=&quot;math-container&quot;&gt;$25$&lt;/span&gt; pieces.&lt;br /&gt;&#xA;On the other hand, it is possible to place &lt;span class=&quot;math-container&quot;&gt;$25$&lt;/span&gt; pieces on the chessboard. (Just fill the board up until no space is left.)&lt;/p&gt;&#xA;" OwnerUserId="16478" LastEditorUserId="77411" LastEditDate="2021-12-03T14:40:50.653" LastActivityDate="2021-12-03T14:40:50.653" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27870" PostTypeId="2" ParentId="27867" CreationDate="2016-02-24T09:17:06.833" Score="15" Body="&lt;p&gt;We have the famous&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To be or not to be&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and we highlight&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;&lt;strong&gt;To&lt;/strong&gt;n&lt;strong&gt;b&lt;/strong&gt;o &lt;strong&gt;e&lt;/strong&gt;t&lt;strong&gt;o&lt;/strong&gt;to &lt;strong&gt;r&lt;/strong&gt;o&lt;strong&gt;no t&lt;/strong&gt;har&lt;strong&gt;t&lt;/strong&gt; d&lt;strong&gt;o&lt;/strong&gt;fo ry&lt;strong&gt;b&lt;/strong&gt;o &lt;strong&gt;e&lt;/strong&gt;u&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;to get&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not too hard for you&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19350" LastActivityDate="2016-02-24T09:17:06.833" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="27871" PostTypeId="1" AcceptedAnswerId="27950" CreationDate="2016-02-24T09:22:08.780" Score="18" ViewCount="2006" Body="&lt;p&gt;Two prisoners are imprisoned at opposite ends of the prison. They cannot meet, nor exchange any kind of information. However, one day they are brought together by the ward who tells them:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I shall give you a chance for getting free. In one hour you'll be separated again, and one of you will go back to his cell, while the other one will stay here. I shall show the former &lt;strong&gt;six&lt;/strong&gt; different numbers randomly chosen among the positive integers $ 1, ...., 245 $. He may think about the numbers and then will have to reject one of the six numbers and tell me the other five. &#xA;I shall write these five numbers down on a piece of paper, in one line and in the same order as I hear them. Then this guy will have to leave the room. The other man will be called in. I show him my piece of paper with the five numbers, and he must say a number. If he says the rejected number, then the two of you will be free. &quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can these two guys get free?&lt;/p&gt;&#xA;" OwnerUserId="19470" LastEditorUserId="8874" LastEditDate="2016-02-25T14:09:52.620" LastActivityDate="2016-02-25T21:30:41.257" Title="Two prisoners and the prison ward" Tags="|mathematics|strategy|game|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27872" PostTypeId="2" ParentId="27617" CreationDate="2016-02-24T09:59:15.200" Score="7" Body="&lt;p&gt;I saw two neighbors between my eyes. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; b&lt;strong&gt;illi&lt;/strong&gt;ard , English word. &quot;2 I&quot; are eyes, &quot;2 L&quot; are neighbours&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then I lost my right eye, but I was still the same. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; billard  - French for billiard&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then one neighbour left, but I was still no different. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; bilard - Polish for billiard&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally, I lost my left eye and now I no longer belong to any culture.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; blard - lost the meaning and no meaining at all in any language&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18685" LastEditorUserId="18685" LastEditDate="2016-02-24T10:15:16.207" LastActivityDate="2016-02-24T10:15:16.207" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27874" PostTypeId="2" ParentId="27871" CreationDate="2016-02-24T10:37:01.283" Score="14" Body="&lt;p&gt;This doesn't work. I will leave it in case someone can use it to create a real solution. The problem is this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Consider the sequence 121, 122, 123, 124, 125. Whatever scheme I came up with, it was possible to add a number to that sequence that would defeat the strategy. Try adding 120, 126, 1 or 245 as the sixth number and by my reckoning one of those should be unreachable by strategies like the one below.  The idea of using permutations to indicate a number between 1 and 120 is valid and Paul Sinclair has given an elegant explanation in his post. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The prisoner who is to leave announces that his strategy will be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To reject either the largest or the smallest number given. Of the remaining&#xA; five numbers, if there are fewer numbers larger than the largest &#xA; remaining numbers, then the largest was rejected. Likewise for smaller.&#xA; In case of a tie, the rejected number is smaller.&#xA; Thus no more than (245 - 5) / 2 = 120 numbers remain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The prisoners then agree on the following numbering system:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The base ordering of the five numbers is largest to smallest. This&#xA; represents the number 1. If the final two numbers are permuted,&#xA; this represents the number 2. If the third and fifth numbers are permuted, &#xA; this represents 3 and so on. The number of permutations is 5!, exactly&#xA; enough to solve the worst case.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastEditorUserId="18924" LastEditDate="2016-02-25T07:34:03.017" LastActivityDate="2016-02-25T07:34:03.017" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="27875" PostTypeId="1" AcceptedAnswerId="28096" CreationDate="2016-02-24T10:57:15.863" Score="10" ViewCount="1354" Body="&lt;p&gt;Your friend will pick a 4-letter word and you will make guesses in order to find it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-A word can contain only the letters A, B, C, and D, and they can be used more than once. (AAAA-DDDD). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-In your guess if at least three letters are in their correct places you will win a prize. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the minimum number of guesses in order to guarantee to win the prize?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: The question is compared to other question by many: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/546/clever-ways-to-solve-mastermind&quot;&gt;Mastermind&lt;/a&gt; &#xA;But I assure you that they are not the same! Mine is harder. Only feedback you will have is whether your guess is true(at least 3 letters in the right places)&#xA;or not. Have fun!&lt;/p&gt;&#xA;" OwnerUserId="19542" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2016-03-01T04:15:35.410" Title="Find the minimum number of guesses" Tags="|strategy|combinatorics|game|" AnswerCount="4" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="27876" PostTypeId="1" AcceptedAnswerId="28081" CreationDate="2016-02-24T11:56:46.717" Score="9" ViewCount="1336" Body="&lt;p&gt;This puzzle is the continuation of&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/27865/professor-halfbrain-and-the-9x9-chessboard&quot;&gt;Professor Halfbrain and the 9x9 chessboard (Part 1)&lt;/a&gt;.&#xA;The difference is that &amp;quot;distance at least &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt;&amp;quot; has now become &amp;quot;distance more than &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt;&amp;quot;.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Professor Halfbrain has spent the last few days with placing pawns on a &lt;span class=&quot;math-container&quot;&gt;$9\times9$&lt;/span&gt; chessboard; each of the &lt;span class=&quot;math-container&quot;&gt;$81$&lt;/span&gt; squares on the chessboard had side length &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;.&#xA;Halfbrain always started with an empty board, and then one by one placed (point-sized) pawns onto it. Every pawn was placed precisely in the middle of one of the little chessboard squares. Whenever Halfbrain placed a new pawn, its distance to each of the pawns placed before was &lt;strong&gt;more than &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Professor Halfbrain has proved two extremely deep theorems on such placements of pawns.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Professor Halfbrain's first theorem&lt;/strong&gt;: It is possible to place three pawns according to the above rules.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Professor Halfbrain's second theorem&lt;/strong&gt;: It is not possible to place &lt;span class=&quot;math-container&quot;&gt;$81$&lt;/span&gt; pawns according to the above rules.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This puzzle asks you to improve the two theorems of professor Halfbrain and to make them even deeper.&#xA;Find an integer &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;, so that &amp;quot;three pawns&amp;quot; in the first theorem may be replaced by &amp;quot;&lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; pawns&amp;quot;, and so that &amp;quot;&lt;span class=&quot;math-container&quot;&gt;$81$&lt;/span&gt; pawns&amp;quot; in the second theorem may be replaced by &amp;quot;&lt;span class=&quot;math-container&quot;&gt;$x+1$&lt;/span&gt; pawns&amp;quot; (again yielding true statements, of course).&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="63414" LastEditDate="2020-12-09T08:43:58.977" LastActivityDate="2020-12-27T03:58:07.563" Title="Professor Halfbrain and the 9x9 chessboard (Part 2)" Tags="|mathematics|combinatorics|checkerboard|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27877" PostTypeId="1" AcceptedAnswerId="27900" CreationDate="2016-02-24T11:57:07.830" Score="2" ViewCount="1746" Body="&lt;p&gt;&quot;Hey! If I were to add 6 to 13 I would get 7!&quot;, said Patrick slowly as he blew a stream of suspiciously maroon smoke out through his left nostril while calmly stabbing a funny-looking cigarette into an upturned potty.&lt;br/&gt;&#xA;&quot;Dude.... That is awesome!&quot; giggled Frank.&lt;br/&gt;&#xA;Provided that Patrick is not tripping, how does his statement make sense?&lt;/p&gt;&#xA;" OwnerUserId="19086" LastEditorUserId="19086" LastEditDate="2016-02-24T12:04:08.357" LastActivityDate="2016-02-24T21:15:03.480" Title="Adding makes no sense" Tags="|calculation-puzzle|" AnswerCount="6" CommentCount="0" ClosedDate="2016-02-25T02:54:34.927" ContentLicense="CC BY-SA 3.0" />
  <row Id="27878" PostTypeId="2" ParentId="27847" CreationDate="2016-02-24T12:05:57.300" Score="0" Body="&lt;p&gt;Yes, in at least three cases that I can think of:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The board wraps itself - i.e. from the top row, move 1 up, and you are now on the bottom row. In that case, all of the rooks make an identical move, and they are still safe.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You said all the rooks make one move. You didn't say, that they make ONLY one move. So, the rooks simple move out, then back to their original position.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The rooks move like knights. Presumably they also &lt;em&gt;take&lt;/em&gt; like knights. In that case, I think the solution is obvious ;).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="19545" LastActivityDate="2016-02-24T12:05:57.300" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27879" PostTypeId="2" ParentId="27877" CreationDate="2016-02-24T12:11:10.810" Score="14" Body="&lt;p&gt;This is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; base 12 clock arithmetic. If you start with 6 o'clock and add 13 hours you get 7 o'clock.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19546" LastEditorUserId="17250" LastEditDate="2016-02-24T12:37:27.683" LastActivityDate="2016-02-24T12:37:27.683" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27880" PostTypeId="2" ParentId="27876" CreationDate="2016-02-24T13:31:19.750" Score="3" Body="&lt;p&gt;If I use the same logic as was used &lt;a href=&quot;https://puzzling.stackexchange.com/a/27866/17250&quot;&gt;here&lt;/a&gt;, then I can&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Stagger the placement of the pawns to that, vertically and horizontally, they are no closer than 3 units away from each other, and diagonally, they are &lt;span class=&quot;math-container&quot;&gt;$\sqrt{5}$&lt;/span&gt; units away from each other. This gives us:&#xA;&lt;br /&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/MqFw9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MqFw9.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&#xA;&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; &lt;s&gt;Further staggering only results in a shifted/rotated version of the same configuration. Note that &lt;em&gt;no squares&lt;/em&gt; in the last column (or last row, if you attempted the same pattern rotated 90 degrees) can be used, as 2 is the &lt;em&gt;maximum&lt;/em&gt; distance from a &amp;quot;used&amp;quot; square.&lt;/s&gt; OK, it turns out I was extremely wrong about this—something told me that if it wasn't symmetrical in some way, or if there was inconsistent spacing, it was not an &amp;quot;optimized&amp;quot; configuration. I was wrong about that.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I believe I can say&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;strong&gt;Professor Halfbrain's first theorem:&lt;/strong&gt; It is possible to place 15 pawns according to the above rules.&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; &lt;strong&gt;Professor Halfbrain's second theorem:&lt;/strong&gt; It is not possible to place 16 pawns according to the above rules.&lt;br /&gt;&#xA;&#xA;&#xA;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-02-24T14:45:19.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27881" PostTypeId="2" ParentId="27876" CreationDate="2016-02-24T13:32:05.960" Score="4" Body="&lt;p&gt;Note: This solution is &lt;strong&gt;DISPROVED&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what I think:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/GDQaz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GDQaz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA; I have shown three configurations A, B, C. The main thing I kept in mind while making these grids is that the board should be closely packed with pawns. Now you may ask how is B different than C and A that it have more pawns.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think the reason is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/fV5TB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fV5TB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA; their slopes. The magnitude of slope for the line joining the pawns in slant position is more for B than A and it is more in such a way that the line contains most number of pawns by the given rules. In situation C, it makes both kind of angles, therefore having mixture of both kind of slopes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Professor Halfbrain's corrected first theorem: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is possible to place 16 pawns according to the rules.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Professor Halfbrain's corrected second theorem: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is not possible to place 17 pawns according to the rules.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-02-24T16:09:27.723" LastActivityDate="2016-02-24T16:09:27.723" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27882" PostTypeId="2" ParentId="27875" CreationDate="2016-02-24T13:35:08.910" Score="6" Body="&lt;p&gt;&lt;b&gt;Edit:&lt;/b&gt; I think I can back up Oray's solution with a proof and confirm that the answer is indeed &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $24$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;His algorithm suffices to give the upper bound so it only remains to show that it is also a lower bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lower bound&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Call a pair of words 'neighbours' if they differ by one letter e.g, $ABCD$ and $ABAD$ are neighbours. &lt;br&gt;&lt;br&gt; Let us consider the map $A \rightarrow 0, B \rightarrow 1, C \rightarrow 2, D \rightarrow 3$ so that each word maps to a point in 4-dimensional space, e.g, $BACD \rightarrow (1, 0, 2, 3)$. Then the set of words is a finite hypercube lattice (of side length 3) and the set of neighbours of a given word, $W$, is the set of points which lie on some axis-parallel line passing through the point corresponding to $W$. &lt;br&gt;&lt;br&gt; The set of words beginning with a given letter is a finite cubic lattice in this space. &lt;br&gt; &lt;b&gt;Key Observation:&lt;/b&gt; Each guess eliminates $10$ possible answers from the cube containing the guess and $1$ possible answer from each of the other cubes. &lt;br&gt;&lt;br&gt; Now, suppose that the number of guesses we make is less than $24$. Then there is at least one cube which contains at most $5$ guesses. &lt;br&gt;&lt;br&gt; We may ask the following question: what is the maximum number of words we can eliminate from a single cube with $5$ guesses? &lt;br&gt;&lt;br&gt; I propose that the answer is $45$. &lt;br&gt;To show this, notice that each cube further subdivides into four planes corresponding to fixing the second letter of the word. With $5$ guesses, there is at least one of these planes which contains at least two guesses. This pair of guesses will share at least two neighbours (call them intersections of the guesses). Alternatively, we can subdivide the cube lattice into planes in the other two directions (defined by fixing the third or fourth letter) and use the same logic to show that there exists a plane, in each of these directions, which contains at least two guesses with at least a pair of intersections. On face value this seems to constitute six intersections but really there are three possible scenarios to consider. &lt;br&gt;&lt;br&gt; (i) These intersections may be distinct, in which case there are a total of &lt;b&gt;six&lt;/b&gt; intersections. &lt;br&gt; (ii) One or more of these intersections may appear twice, in which case any double intersection counts twice (as it prevents two guesses from eliminating an additional possible answer). So again there are essentially a total of &lt;b&gt;six&lt;/b&gt; intersections. &lt;br&gt; (iii) One intersection may appear three times. This is a special case as the triple intersection may only count twice (it prevents two guesses from eliminating a possible answer but if these two guesses also intersect here it is to neither one's detriment). Hence, in this case there are essentially &lt;b&gt;five&lt;/b&gt; intersections. &lt;br&gt;&lt;br&gt; The point is that the maximum number of words in a given cube that can be eliminated as possible answers by $5$ guesses in that cube is $5 \times 10 - 5 = 45$. The maximum number of words in a given cube that can be covered by $18$ guesses in other cubes is $18$. Hence, if we are limited to at most $23$ guesses, then there is some cube with at most $45 + 18 = 63$ out of $64$ words eliminated as possible answers. Hence, the number of guesses we must make is at least $24$. &lt;br&gt;&lt;br&gt; &lt;b&gt;Note 1:&lt;/b&gt; If the above logic does not convince you, it does not take very much computing power to show that, in a given cube of words with $5$ guesses, we can eliminate at most $45$ possible answers since  ${{64}\choose{5}} = 7,624,512$ which is not terribly big. &lt;br&gt;&lt;br&gt; &lt;b&gt;Note 2:&lt;/b&gt; It should be fairly clear that it's not worth having less than five guesses in a given cube if we are limited to $23$ overall since we make up only one guess by guessing in a different cube and lose at least that by not guessing in the given cube. &lt;br&gt;&lt;br&gt; &lt;b&gt;Note 3:&lt;/b&gt; One thing that falls out from this analysis is that it may be possible, with $23$ guesses, to guarantee that either you win the prize or you definitely know your friend's word. I don't know if this is achievable but would be very interesting to see if so.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18422" LastEditorUserId="18422" LastEditDate="2016-02-29T11:07:23.657" LastActivityDate="2016-02-29T11:07:23.657" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27883" PostTypeId="2" ParentId="27876" CreationDate="2016-02-24T14:04:40.643" Score="8" Body="&lt;p&gt;It looked like the other answers had some room left to work with, so starting with them and messing around a bit, I managed:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/l2api.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/l2api.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I also found a symmetrical way to do this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/nZNKO.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nZNKO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Professor Halfbrain's first theorem:&lt;/strong&gt; It is possible to place 17 pawns according to the above rules.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Professor Halfbrain's second theorem:&lt;/strong&gt; It is not possible to place 18 pawns according to the above rules.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="375" LastEditorUserId="375" LastEditDate="2016-02-24T14:58:27.730" LastActivityDate="2016-02-24T14:58:27.730" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27884" PostTypeId="2" ParentId="27877" CreationDate="2016-02-24T14:08:32.453" Score="8" Body="&lt;p&gt;It's a bit of a stretch but it might be that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; they are filling baggies with carefully measured amounts of... oregano.&#xA; Perhaps because they and many of their friends are chefs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Patrick is announcing that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; as the thirteenth baggie has only a single ounce of oregano...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If [he] were to add six to thirteen, [he] would get seven&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly he is not tripping. His mastery of the subjunctive tense shows that he is of clear mind.&lt;/p&gt;&#xA;" OwnerUserId="18924" LastActivityDate="2016-02-24T14:08:32.453" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27885" PostTypeId="2" ParentId="27871" CreationDate="2016-02-24T14:20:46.837" Score="6" Body="&lt;p&gt;Not an answer, but too long to add as a comment to Hugh Meyer's answer. I don't completely follow his scheme for converting a permutation to a number either, but here is one that works:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a_1, a_2, a_3, a_4, a_5$ be the five numbers in the order given. Then  for $k = 1, 2, 3, 4$ define $d_k$ to be the count of numbers $a_1, ..., a_k$ which are less than $a_{k+1}$. So for example, $$\begin{align}d_1 &amp;amp;= \begin{cases} 0 &amp;amp; \text{if }a_2 &amp;lt; a_1\\1&amp;amp;\text{if }a_2 &amp;gt; a_1\end{cases}\\d_2 &amp;amp;= \begin{cases} 0 &amp;amp; \text{if }a_3 &amp;lt; a_1, a_2\\1&amp;amp;\text{if }a_1 &amp;lt; a_3 &amp;lt; a_2\text{ or }a_2 &amp;lt; a_3 &amp;lt; a_1\\2&amp;amp; \text{if }a_1, a_2 &amp;lt; a_3\end{cases}\end{align}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, $d_3$ is between $0$ and $3$, and $d_4$ is between $0$ and $4$.&lt;br&gt;&#xA;Then the number wanted is $1 + d_1 + 2\times d_2 + 6\times d_3 + 24 \times d_4$&lt;/p&gt;&#xA;" OwnerUserId="17915" LastEditorUserId="17915" LastEditDate="2016-02-25T21:30:41.257" LastActivityDate="2016-02-25T21:30:41.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27886" PostTypeId="1" AcceptedAnswerId="27895" CreationDate="2016-02-24T14:42:12.797" Score="4" ViewCount="315" Body="&lt;pre&gt;&lt;code&gt;I was so gifted since I was very little&#xA;But I was also full of pride and arrogance&#xA;Compassion and overconfidence were my downfall&#xA;Compassion and self regret were my redemption&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Who/What am I?&lt;/p&gt;&#xA;" OwnerUserId="18685" LastActivityDate="2016-02-26T07:25:14.347" Title="What/who am I series by SC #5" Tags="|riddle|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27887" PostTypeId="2" ParentId="27617" CreationDate="2016-02-24T14:46:46.347" Score="10" Body="&lt;p&gt;You are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; yoghourt&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and your 'eyes' &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; are letters 'o' rather than 'i', which commenters (including myself) tended to think&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you lose your right eye you become:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; yoghurt, the American spelling&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you lose a neighbour, you become:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; yogurt, a different spelling&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, losing your other eye makes you gibberish.&lt;/p&gt;&#xA;" OwnerUserId="18919" LastActivityDate="2016-02-24T14:46:46.347" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27888" PostTypeId="2" ParentId="27820" CreationDate="2016-02-24T14:47:37.630" Score="0" Body="&lt;p&gt;I choose Living Room.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is brown-haired. Because, Red-head was third to arrive and bald guy&#xA; was second. Bald guy works between 8 am and 6 pm so blond must be the&#xA; first to arrive(8 am 4 pm). Red-haired guy said none of his brother&#xA; were there and he was 3rd. That leaves the brown-haired guy, almost at&#xA; midnight, the last visitor. But I also suspected the little blond&#xA; brother. Asking for rom? Maybe brown was in this together with&#xA; blondie. I do not know it is something to think about.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sBn0V.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sBn0V.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19542" LastEditorUserId="19542" LastEditDate="2016-02-24T15:29:36.080" LastActivityDate="2016-02-24T15:29:36.080" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27889" PostTypeId="2" ParentId="27886" CreationDate="2016-02-24T14:50:48.487" Score="2" Body="&lt;p&gt;Might be a bit too philosophical but : &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The human race&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was so gifted since I was very little &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Humans have been gifted by intelligence, a bigger brain, which allowed to develop languages, technologies etc...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I was also full of pride and arrogance  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Humans always tried to rule on everything else on earth/nature&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Compassion and overconfidence were my downfall  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&#xA; humans are slowly destroying everything they need to keep beeing a specy, &gt;!cause we thought ressources were infinite (overconfidence). We know that and &gt;!regret past errors (compassion) yet do nothing.&#xA;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Compassion and self regret were my redemption &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The only redemption possible is that everyone starts making sacrifices (self &gt;!regret) to undo all the mistakes humanity made !&#xA; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17888" LastActivityDate="2016-02-24T14:50:48.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27890" PostTypeId="2" ParentId="27886" CreationDate="2016-02-24T15:01:41.903" Score="3" Body="&lt;h3&gt;This is a wrong answer(my achievement)...confirmed by OP&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I think the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Ravana&quot; rel=&quot;nofollow&quot;&gt;Ravan&lt;/a&gt; (Hindu Mythology)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was so gifted since I was very little&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ravana mastered the Vedas, the holy books, and also the arts and ways of Kshatriyas (warriors) when he was a child.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I was also full of pride and arrogance&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes, he kidnapped Sita for the pride of his family.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Compassion and overconfidence were my downfall&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He was overconfident that he would win the fight against Ram. And he was compassionate for his sister. That's why he kidnapped Sita.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Compassion and self regret were my redemption&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He did regret in the end while dying while still having compassion for his sister.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-02-26T07:25:14.347" LastActivityDate="2016-02-26T07:25:14.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27891" PostTypeId="2" ParentId="27877" CreationDate="2016-02-24T15:28:26.067" Score="33" Body="&lt;p&gt;Okay, well how about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Base 5031&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we work with this, then&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $6_{5031} + 13_{5031} = 19_{5031} = 5040_{10} = 7!$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2016-02-24T15:28:26.067" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27892" PostTypeId="2" ParentId="27820" CreationDate="2016-02-24T15:33:52.090" Score="3" Body="&lt;p&gt;I do not have a definitive answer, but here are my ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I chose living room, washroom, + the 4 non-master bedrooms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My suspect is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the brown son: Excluding delayed fires (which is possible, with a science fan son), it must have been caused by the last son arriving, and with no witnesses.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The claims are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Blonde: arrive 4pm&#xA; Bald: arrive 2nd at 6pm&#xA; Red: arrive 3rd and no one else there&#xA; Brown: arrive after &quot;office time&quot; (which I assume means 9-5, so 5pm)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assuming there is only 1 liar, the first 3 stories match well together, while the latter is refuted by both 2nd line and 3rd line at the same time. So&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Brown must have arrived after Red left, and no other witnesses were there. Additionally, he is a smoker, so he is more likely to have a fire-starting tool (no hard evidence, but still...)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But why? Well:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He is adopted, and felt like he was loved less than the other 3. His room is number 3. The two extra people in the picture are his original parents. The fish out of water is how he feels. His room feels like the one of an hard-worker (all days are full of tasks) that left the home abruptly, like running from home. While he's better off in life than Red, he lacks in terms of purpose, as he wasted most of his time on juvenile glory (he's admired by friends) and now he has nothing. Red is still working in the field he loves (sports), Bald is rich and affirmed, and Blond seems to be doing also OK.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now for the spreadsheet (age is birth order):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; brown: age=3 (fish is before bull as a zodiac sign over the year, assuming same year of birth), income=medium, personality=described already in answer, other=smoke, room=3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;-&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; blonde: age=1 (the missing child is taking the dinner picture, and it's usually the oldest to do so, but we don't know if him or bald, but lion is before aries in the same year. Also, the youngest are on parents' side so they can be spoon-fed when babies, and then they stay there), income=medium, personality=games?, room=4, other=video-games nerd (is the lion Griever?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;-&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; red: age=4 (bull is after fish), income=low (probably phys.ed. teacher), personality=sporty, room=1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;-&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; bald: age=1, income=high, personality=likes science, motors, he maybe a plane pilot, other=smokes cigar/pipes, room=2 (aries comes before lion)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14435" LastEditorUserId="14435" LastEditDate="2016-02-24T22:39:38.460" LastActivityDate="2016-02-24T22:39:38.460" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="27893" PostTypeId="2" ParentId="27877" CreationDate="2016-02-24T15:34:19.183" Score="-5" Body="&lt;p&gt;I'm stealing Tom's answer =) :  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it's 13:00 pm, adding 6 hrs will result in 19:00, which is 7 o'clock&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2016-02-24T15:34:19.183" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="27895" PostTypeId="2" ParentId="27886" CreationDate="2016-02-24T16:04:55.467" Score="5" Body="&lt;p&gt;I sense that you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Darth Vader/Anakin Skywalker&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was so gifted since I was very little&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Anakin was noted as a talented pilot, first seen through his podracing, then his participation in the Battle of Naboo, and subsequent air/space battles as an adult.  He was also gifted due to his high midichlorian count, which made him a strong user of the Force.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I was also full of pride and arrogance&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After being taken on as padawan to Obi-Wan, he went from annoying to arrogant.  Examples of his compassion &lt;em&gt;and&lt;/em&gt; arrogance include him driving at a forbidden relationship with Padme, and the emotion-filled lightsaber duels in which fought with a less than polished style.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Compassion and overconfidence were my downfall&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; His emotions, especially his desire to save Padme, were among what led Anakin to the dark side of the Force.  His confidence in himself against Obi-Wan in Episode III is what destroyed his former self, leading to him becoming a fully-fledged sith as Darth Vader.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Compassion and self regret were my redemption&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After Luke and Vader fought on the Second Death Star, Palpatine attacked Luke with Force Lightning, causing Luke to call out &quot;Father, please.&quot;  At this moment, compassion for a son he had never known blossomed and Vader attacked Palpatine to save Luke.  His love and compassion for Luke was enough to drag him out of the hate that was the dark side (this, along with the realization that Palpatine was merely using Vader at this point, to get to Luke).  Re-embracing the light side, Anakin/Vader was redeemed, albeit just before his death.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; At the time of writing, the answer is in your profile image, Sir SC.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11452" LastActivityDate="2016-02-24T16:04:55.467" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27896" PostTypeId="2" ParentId="27876" CreationDate="2016-02-24T17:13:53.873" Score="8" Body="&lt;p&gt;&lt;strong&gt;UPDATE:&lt;/strong&gt;  Based on the suggestion by f'' in the comments, my original statements can be greatly improved:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; To each pawn, we can associate its own square and the four squares surrounding it, making a plus-shaped tile with an area of 5 square units.  Any two such tiles that abut will automatically have a distance of at least &lt;span class=&quot;math-container&quot;&gt;$\sqrt{5}$&lt;/span&gt; between their centers (i.e., the pawn locations).  Moreover, the total area not on the original chessboard covered by these tiles will be at most 12 square tiles;  we can fit at most three such tiles along each edge, meaning that at most three &amp;quot;squares&amp;quot; from our plus-shaped tiles can stick out over the edge.  The total area covered by tiles is thus at most &lt;span class=&quot;math-container&quot;&gt;$9 \times 9 + 3\times 4 = 93$&lt;/span&gt; square units, and so &lt;span class=&quot;math-container&quot;&gt;$N &amp;lt; 93/5 = 18.6$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Thus, a significantly better version of Professor Half-Brain's Second Theorem is&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;strong&gt;It is not possible to place 19 pawns according to the above rules.&lt;/strong&gt;  Note that the bound is substantially better both because we have increased the &amp;quot;size&amp;quot; of the tiles and reduced the maximum amount of area they can cover.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Below this line are some weaker bounds from my original answer.&lt;/p&gt;&#xA;&lt;p&gt;A slightly different approach:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The closest any two pawns can be to each other, given the above rules, is &lt;span class=&quot;math-container&quot;&gt;$\sqrt{5}$&lt;/span&gt; (when the pawns are a &amp;quot;knight's move&amp;quot; away from each other.)  This means that we should be able to draw a circle of radius &lt;span class=&quot;math-container&quot;&gt;$\sqrt{5}/2$&lt;/span&gt; around each pawn, and that none of the circles will intersect.  Moreover, all of the circles will be contained in a square of side &lt;span class=&quot;math-container&quot;&gt;$8 + \sqrt{5}$&lt;/span&gt;, centered on the original chessboard.  (Note that the circle corresponding to a pawn on the edge will go &amp;quot;off the edge&amp;quot; of the original chessboard somewhat.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; We are thus presented with the problem of how many circles of radius &lt;span class=&quot;math-container&quot;&gt;$\sqrt{5}/2$&lt;/span&gt; can be packed into a square of side &lt;span class=&quot;math-container&quot;&gt;$8 + \sqrt{5}$&lt;/span&gt;, such that the center of each circle is on a grid point of the original chessboard.  Right off the bat, we can place a bound on the number of circles that can be fit.  Since none of the circles overlap, the area of all the circles can be no greater than the area of our new square.  The area of each circle is &lt;span class=&quot;math-container&quot;&gt;$5\pi/4$&lt;/span&gt;; thus, &lt;span class=&quot;math-container&quot;&gt;$N &amp;lt; (8 + \sqrt{5})^2/(5\pi/4) \approx 26.7.$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Thus, we can refine Professor Half-Brain's Second Theorem to read:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;strong&gt;It is not possible to place 27 pawns according to the above rules.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This can be refined slightly by noting that&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; if we remove the restriction that the pawns be on grid points, the best we can do is to pack the circles into a hexagonal grid. Thus, around each pawn we can draw a hexagon such that the radius of the circle (our original circles) inscribed in the hexagon is &lt;span class=&quot;math-container&quot;&gt;$\sqrt {5}/2 $&lt;/span&gt;.  Note that the hexagons can now reach out a little further from our original chessboard than the circles could;  the distance from the center to a vertex of one of these hexagons will be &lt;span class=&quot;math-container&quot;&gt;$\sqrt{5/3}$&lt;/span&gt;, and so all of the hexagons must now be contained inside a square of side &lt;span class=&quot;math-container&quot;&gt;$8 + 2 \sqrt{5/3}$&lt;/span&gt;.  The area of our hexagons is &lt;span class=&quot;math-container&quot;&gt;$5\sqrt {3}/ 2 $&lt;/span&gt;, and by the above logic, we must have &lt;span class=&quot;math-container&quot;&gt;$N &amp;lt; \frac {(8 + 2 \sqrt {5/3})^2} {5\sqrt {3}/2}\approx 25.9.$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Thus, a slightly refined version of Professor Half-Brain's Second Theorem is&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;strong&gt;It is not possible to place 26 pawns according to the above rules.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;In fact, if we view the problem in this way, we can actually look at the literature for guidance:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The smallest square into which one can pack &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; circles of radius 1 into a square has been solved for &lt;span class=&quot;math-container&quot;&gt;$n \leq 20$&lt;/span&gt;;  the solutions can be found &lt;a href=&quot;https://erich-friedman.github.io/packing/cirinsqu/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here.&lt;/a&gt;  In our case, we can rescale our circles so that they have radius 1;  the square in which they must all fit is therefore of side &lt;span class=&quot;math-container&quot;&gt;$(8 + \sqrt{5})/(\sqrt{5}/2) \approx 9.15542.$&lt;/span&gt;  Unfortunately, it is not known whether 21 circles can fit into a square of this size;  the best known packing is into a square of side 9.358...  So &lt;strong&gt;it seems likely that you cannot place 21 pawns according to the above rules&lt;/strong&gt;, but this is not a proof.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Note that none of these are strict upper bounds;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; the fact that we must place the pawns on square grid points will mean that we cannot optimally pack our circles/hexagons, and so some of the square's area will be &amp;quot;wasted&amp;quot;.  Basically, for the first two theorems, we have proven an upper bound among all possible pawn placements on the chessboard;  any placement for which the pawns are on grid points will necessarily be no greater than this bound.  Most likely, the smallest number of pawns that cannot be placed is less than 25 (or 21, if you believe the &amp;quot;non-proof&amp;quot; in the previous paragraph.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17035" LastEditorUserId="4254" LastEditDate="2020-12-07T03:28:15.903" LastActivityDate="2020-12-07T03:28:15.903" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27897" PostTypeId="1" AcceptedAnswerId="27909" CreationDate="2016-02-24T18:04:06.113" Score="6" ViewCount="513" Body="&lt;p&gt;A cross consists of six unit squares and looks as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;         O&#xA;       OOOO&#xA;         O&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(this may also be rotated by several right turns, or horizontally reflected or vertically reflected).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the maximum number of crosses than can be fit on a $ 10\times 11 $ chessboard without overlapping? (Every cross must cover exactly six of the little squares on the chessboard.)&lt;/p&gt;&#xA;" OwnerUserId="19470" LastActivityDate="2016-02-25T13:32:39.710" Title="Crosses on a 10x11 chessboard" Tags="|mathematics|combinatorics|checkerboard|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27898" PostTypeId="2" ParentId="27897" CreationDate="2016-02-24T18:19:26.160" Score="11" Body="&lt;h1&gt;Highly non-mathematical answer ahead&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I managed to get&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $14$&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/cb6hD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cb6hD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Seeing what seemed to be wasted space, I tried rotating a few (e.g., the leftmost bluish cross), but couldn't get any more in.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Not a proof, but some reasoning:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Corners are out. Since the shape has... let's call them &lt;em&gt;&quot;jutty-out parts&quot;&lt;/em&gt;... none can be packed flat against a side, so there will necessarily be unused squares on every single side.  Packing them as closely as possible (as I have at the bottom), each long side will have the previously-mentioned $2$ unused corner squares as well as $4$ unused squares which alternate with the end bits of the crosses. Each short side will have the previously-mentioned $2$ unused corner squares and &lt;em&gt;at least&lt;/em&gt; $4$ other unused squares.  This alone (unusable side squares) gives us a maximum $90$ squares that &lt;em&gt;could&lt;/em&gt; contain parts of crosses. If we could rearrange that remaining space, and if, instead of crosses, we were trying to pack $3\times2$ rectangles in that remaining space, then we could fit &lt;em&gt;exactly&lt;/em&gt; $15$ of them in there.&lt;br&gt;&#xA;&lt;br&gt;&#xA; This, in combination with the reasoning given in @M.M's post, shows that $14$ is the largest number of crosses that can be contained entirely within a $10\times11$ grid.&lt;br&gt;&#xA;&lt;br&gt;&#xA;  &lt;sub&gt;&lt;em&gt;extremely arnold schwarzenegger voice&lt;/em&gt; remember when I said I would never do another math puzzle again. i lied.&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2016-02-25T13:32:39.710" LastActivityDate="2016-02-25T13:32:39.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27899" PostTypeId="2" ParentId="27897" CreationDate="2016-02-24T18:23:41.253" Score="4" Body="&lt;p&gt;Manually:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 14&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/ccSbZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ccSbZ.png&quot; alt=&quot;My Solution&quot;&gt;&lt;/a&gt;&#xA;&lt;br&gt;&#xA; I cannot manage a better arrangement.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19516" LastActivityDate="2016-02-24T18:23:41.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27900" PostTypeId="2" ParentId="27877" CreationDate="2016-02-24T18:26:44.297" Score="32" Body="&lt;p&gt;Patrick has:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An extremely alkali solution (something bleach-like) of pH 13&lt;br&gt;&#xA; If he adds the correct amount of a moderately acidic solution (say, coffee or urine) of pH 6    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He'll get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A neutral solution of pH 7&lt;br&gt;&#xA; And Frank saying, &quot;You really do have a plan! Yeah science!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastEditorUserId="10118" LastEditDate="2016-02-24T21:15:03.480" LastActivityDate="2016-02-24T21:15:03.480" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27901" PostTypeId="1" AcceptedAnswerId="27903" CreationDate="2016-02-24T19:15:07.683" Score="10" ViewCount="4789" Body="&lt;p&gt;It is easy to divide an equilateral triangle into three equal, though not equilateral, triangles.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zvC9E.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zvC9E.png&quot; alt=&quot;Triangle divided&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;It is even simpler to divide a square into four equal squares.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/BC9PO.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BC9PO.png&quot; alt=&quot;Square divided&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;The difficult part is, &lt;strong&gt;whether you can divide a regular pentagon into five equal pentagons&lt;/strong&gt;?&lt;/p&gt;&#xA;&lt;p&gt;Note:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;code&gt;Equal&lt;/code&gt; means, equal in area.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1766" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-06-09T14:38:10.480" Title="Dividing the pentagon" Tags="|mathematics|geometry|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27902" PostTypeId="2" ParentId="27877" CreationDate="2016-02-24T19:21:20.000" Score="8" Body="&lt;p&gt;Well, they both have the...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; munchies, and they are looking at a take-away menu from their local Chinese.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Patrick noticed that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No. 6 on the menu is Shrimp, No. 13 on the menu is Fried Rice, and No. 7 is Shrimp Fried Rice. So he's quite right, if you add 6 to 13 you get 7.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18522" LastActivityDate="2016-02-24T19:21:20.000" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27903" PostTypeId="2" ParentId="27901" CreationDate="2016-02-24T19:32:37.303" Score="24" Body="&lt;p&gt;Easy, the only definition of a pentagon is that is must have 5 sides :)&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/l5N3C.jpg&quot; alt=&quot;splitted pentagon&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You could replicate it for orders 7,9, 11 and so on.&#xA;(Just figured it won't work for even numbers)&lt;/p&gt;&#xA;" OwnerUserId="19417" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-02-26T17:27:19.063" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="27904" PostTypeId="1" AcceptedAnswerId="27906" CreationDate="2016-02-24T19:38:06.367" Score="11" ViewCount="306" Body="&lt;p&gt;A little arithmetic to rock your brain.  Each pair of equations is separate from the others. Within each pair, the variable to be solved for will represent the same thing, however other variables may or (most likely) may not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;K1+K2+D=N&lt;br&gt;&#xA;  D+P+N+T+C=F&lt;br&gt;&#xA;  Solve for D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;K+E+C+P+J=Y&lt;br&gt;&#xA;  E+J+G=C&lt;br&gt;&#xA;  Solve for E&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;G+J+R+P=B&lt;br&gt;&#xA;  P+L+D+J=W&lt;br&gt;&#xA;  Solve for P&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is my first time posting so feedback is welcome.  Also, not sure how to tag it.&lt;/p&gt;&#xA;" OwnerUserId="9353" LastEditorUserId="9353" LastEditDate="2016-02-24T20:12:19.650" LastActivityDate="2016-02-24T20:12:19.650" Title="Group Arithmetic" Tags="|lateral-thinking|letter-sequence|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27905" PostTypeId="2" ParentId="27865" CreationDate="2016-02-24T19:41:24.450" Score="2" Body="&lt;p&gt;Not a math person, so this is more of an observation than a proof:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Professor Halfbrain's improved first theorem: For a square board of &lt;span class=&quot;math-container&quot;&gt;$S^2$&lt;/span&gt; spaces, it is possible to place &lt;span class=&quot;math-container&quot;&gt;$\left \lceil\frac{S}2 \right \rceil^2$&lt;/span&gt; pawns according to the above rules.&lt;/p&gt;&#xA;&lt;p&gt;Professor Halfbrain's improved second theorem: For a square board of &lt;span class=&quot;math-container&quot;&gt;$S^2$&lt;/span&gt; spaces, it is not possible to place more than &lt;span class=&quot;math-container&quot;&gt;$\left \lceil\frac{S}2 \right \rceil^2 + 1$&lt;/span&gt; pawns according to the above rules.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6ayNO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6ayNO.png&quot; alt=&quot;7x7 grid&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;For a 7x7 grid, 4 rows of 4 pawns can be placed while following Professor Halfbrain's rules. Increasing the grid to an 8x8 grid doesn't increase the max number of pawns that can be placed. So for a square grid of width 7 or 8, the max number of pawns is 4, or &lt;span class=&quot;math-container&quot;&gt;$\left \lceil\frac{\text{width}}2 \right \rceil$&lt;/span&gt;. For a 9x9 grid, this gives 25 pawns, for a 1x1 grid, this gives a maximum of 1 pawn.&lt;/p&gt;&#xA;" OwnerUserId="19560" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-02-25T17:45:10.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27906" PostTypeId="2" ParentId="27904" CreationDate="2016-02-24T19:46:35.290" Score="12" Body="&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Oops, I guess I technically didn't &quot;solve for&quot; the correct things. Changes made below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Kris + Kurt + Dave = &lt;a href=&quot;https://en.wikipedia.org/wiki/Nirvana_(band)&quot;&gt;Nirvana&lt;/a&gt;&lt;br&gt;&#xA; Dave + Pat + Nate + Taylor + Chris = &lt;a href=&quot;https://en.wikipedia.org/wiki/Foo_Fighters&quot;&gt;Foo Fighters&lt;/a&gt;&lt;br&gt;&#xA; $D = Dave\ Grohl$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Keith + Eric + Chris + Paul + Jimmy (or Jeff) = &lt;a href=&quot;https://en.wikipedia.org/wiki/The_Yardbirds&quot;&gt;Yardbirds&lt;/a&gt;&lt;br&gt;&#xA; Eric + Jack + Ginger = &lt;a href=&quot;https://en.wikipedia.org/wiki/Cream_(band)&quot;&gt;Cream&lt;/a&gt;&lt;br&gt;&#xA; $E = Eric\ Clapton$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; George + John + Ringo + Paul = &lt;a href=&quot;https://en.wikipedia.org/wiki/The_Beatles&quot;&gt;Beatles&lt;/a&gt;&lt;br&gt;&#xA; Paul + Linda + Denny + Jimmy = &lt;a href=&quot;https://en.wikipedia.org/wiki/Paul_McCartney_and_Wings&quot;&gt;Wings&lt;/a&gt;&lt;br&gt;&#xA; $P = Paul\ McCartney$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Extra Stuff:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Group&quot; in the title, &quot;rock&quot; in the opening sentence of the description... &lt;em&gt;ta-daaaaa&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2016-02-24T20:09:10.183" LastActivityDate="2016-02-24T20:09:10.183" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27907" PostTypeId="1" AcceptedAnswerId="27908" CreationDate="2016-02-24T20:23:14.063" Score="-1" ViewCount="218" Body="&lt;p&gt;You're a member of a new government with exotic rule-sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the rules governs when and how hair shall be cut.  The law states that on the second Thursday of every month hair shall be trimmed, and it shall be trimmed in the following manner:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the citizens shall be lined up alphabetically (last name, first name).  Going from the front of the line to the back, the following shaving rules shall be applied fairly to all citizens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If everyone in front of you is bald, you shall be shaved bald.  If this is not true one of two things can happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are bald you will be given a wig to hide your shame.  If you have hair it shall be permed to appear more wig like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are fourth in line and have immaculate hair.  If the hair states of the other contestants are randomized, what is the most common hair style you will be left with?  What percent chance is that?&lt;/p&gt;&#xA;" OwnerUserId="457" LastEditorUserId="457" LastEditDate="2016-02-24T20:39:12.990" LastActivityDate="2016-02-24T20:39:12.990" Title="What will your hair be?" Tags="|logical-deduction|lateral-thinking|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="27908" PostTypeId="2" ParentId="27907" CreationDate="2016-02-24T20:29:41.127" Score="4" Body="&lt;p&gt;Assuming the law of the land accounts for vacuous truth:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You will be shaved 100% of the time. For the first person in line, &quot;Everyone in front of you is bald&quot; is vacuously true, and so they will be shorn. Then the second has only a bald individual in front of them, and so on all the way down.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5169" LastActivityDate="2016-02-24T20:29:41.127" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27909" PostTypeId="2" ParentId="27897" CreationDate="2016-02-24T21:00:32.937" Score="8" Body="&lt;p&gt;To prove that Jeff B's answer is maximal:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As suggested by Slepz in comments, it is clear that no corner cell may be covered, and no two adjacent edge cells may be covered.  It follows that at most &lt;code&gt;5&lt;/code&gt; of the &lt;code&gt;9&lt;/code&gt; non-corner cells on each long edge may be covered, and at most &lt;code&gt;4&lt;/code&gt; of the &lt;code&gt;8&lt;/code&gt; non-corner cells on each short edge may be covered. This gives an upper bound of &lt;code&gt;5+5+4+4=18&lt;/code&gt; for the number of covered edge cells.&#xA; &lt;BR&gt;&lt;BR&gt;&#xA; &lt;em&gt;Lemma:&lt;/em&gt; It is not possible to achieve 18 covered edge cells&#xA; &lt;BR&gt;&lt;BR&gt;&#xA; &lt;em&gt;Proof:&lt;/em&gt; Following the same logic as the previous paragraph, 5+5+4+4 would be the only possible way to achieve 18.   By inspection we can see that the only way to cover 5 squares on the long edge is the green-yellow pattern in Jeff B's answer (with two possible orientations).&lt;br&gt;&#xA; &lt;BR&gt;&#xA; However, note that each yellow cross actually eliminates 2 non-corner cells on each short edge.  Considering both yellow crosses on one short edge, this means only the two central cells of the short edge remain. Since they are adjacent only at most one can be covered.&lt;br&gt;&#xA; &lt;BR&gt;&#xA; So we have shown that if 5 cells are covered on both long edges, then at most 3 cells may be covered on either short edge. This proves the lemma and reduces our upper bound for edge square coverage to &lt;code&gt;16&lt;/code&gt;.&#xA; &lt;BR&gt;&lt;BR&gt;&#xA; Finally, since there are &lt;code&gt;8 x 9 = 72&lt;/code&gt; non-edge squares, &lt;code&gt;72 + 16 = 88&lt;/code&gt; is now the upper bound on coverage for all squares; therefore placing 15 crosses (which would cover 90 squares) is not possible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19562" LastActivityDate="2016-02-24T21:00:32.937" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27910" PostTypeId="2" ParentId="27755" CreationDate="2016-02-24T21:01:14.723" Score="11" Body="&lt;h3&gt;Answer&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;You are: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;The red fox (&lt;em&gt;Vulpes vulpes&lt;/em&gt;)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;Explanation&lt;/h3&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first letter of each line gives us ATCAAGTCTATTAGCAAACA, which can be understood as a nucleotide sequence. Running a nucleotide search through &lt;a href=&quot;http://blast.ncbi.nlm.nih.gov/Blast.cgi&quot; rel=&quot;nofollow&quot;&gt;Basic Local Alignment Search Tool&lt;/a&gt; returns &lt;code&gt;Vulpes vulpes mitochondrion, complete genome&lt;/code&gt; (&lt;em&gt;Vulpes vulpes&lt;/em&gt;=red fox) and &lt;code&gt;Vulpes zerda mitochondrion, complete genome&lt;/code&gt; (&lt;em&gt;Vulpes zerda&lt;/em&gt;=fennec fox) as the the best matches for the sequence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A vicious villain am I&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Vicious villian = &lt;em&gt;Vulpes vulpes&lt;/em&gt; (alliteration!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The greatest of my kind&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; red foxes are the largest of the true foxes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Covering the face of the Earth&lt;br&gt;&#xA;A fiend of all mankind&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Red foxes are the most abundant wild member of Carnivora, being present across the entire Northern Hemisphere from the Arctic Circle to North Africa, North America and Eurasia.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Although I may appear weak and small&lt;br&gt;&#xA;Great power I am assigned&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; On average, adult red foxes measure 35–50 cm (14–20 in) high at the shoulder and 45 to 90 cm (18 to 35 in) in body length, but are relatively dangerous predators&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Those who fail to heed my warning&lt;br&gt;&#xA;Condemnation they will find&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Red foxes have distinctive vocalizations&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Two natures do I have&lt;br&gt;&#xA;A son of God and of the Devil&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In Christian theology, God made all the animals. Red is often associated with the color of the devil. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though temples have been built to me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Foxes are a common subject of Japanese folklore. There are temples called Inari shrines in honor of Inari Ōkami, the Japanese &lt;em&gt;kami&lt;/em&gt; (or spirit) of foxes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thieves are exalted to a higher level&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Foxes are popularly associated with thievery. There is a fox character from the popular TV show Dora the Explorer called Swiper who is a thief. Also, Robin Hood from Disney's Robin Hood adaptation is a fox.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A loud cry sounds from the town&lt;br&gt;&#xA;Gloriously is my appearance made&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Red foxes in urban areas would probably be a big disturbance&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Considered me not your enemy&lt;br&gt;&#xA;And you will have nothing of which to be afraid&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; red foxes tend not to be very aggressive toward humans unless provoked&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17825" LastEditorUserId="17825" LastEditDate="2016-02-25T14:21:14.870" LastActivityDate="2016-02-25T14:21:14.870" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27912" PostTypeId="1" AcceptedAnswerId="27923" CreationDate="2016-02-24T22:47:56.703" Score="2" ViewCount="266" Body="&lt;p&gt;&lt;strong&gt;[Inspired by the movie Bloodsport]&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Kumite is an illegal underground fighting tournament organized in the typical single elimination bracket style.  Think March Madness. All fights are fought one at a time on a central stage because the fighters want to watch other fighters to try to figure out their style in order to have a better chance at beating them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The champion of the tournament, Frank Dux, did so with with 56 consecutive knockouts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When did the tournament start?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;This is only my second puzzle, and I think I kind of screwed up the presentation.  Could have been a lot better.  Lesson learned about taking time to think out my presentation.&lt;/p&gt;&#xA;" OwnerUserId="9353" LastEditorUserId="17250" LastEditDate="2016-03-04T12:52:10.093" LastActivityDate="2016-03-04T12:52:10.093" Title="When did the tournament start?" Tags="|calculation-puzzle|" AnswerCount="2" CommentCount="7" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27913" PostTypeId="2" ParentId="27637" CreationDate="2016-02-24T23:00:51.270" Score="3" Body="&lt;p&gt;I think it might be:&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wikipedia&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&#xA;I speak through shadows,&lt;br/&gt;&#xA;in cryptic tongues,&lt;br/&gt;&#xA;and cryptic tongues only,&#xA;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wikipedia is available only over HTTPS, since 2015.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&#xA;to anyone who wants to know—&lt;br/&gt;&#xA;but only if I have been told to say.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Articles are added by voluntary contributors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&#xA;I think not myself;&lt;br/&gt;&#xA;offer no discretion of my own,&lt;br/&gt;&#xA;but never trust the word of one.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The content in Wikipedia comes from many different contributors.  Each article can have any number of authors. The content is expected to be established knowledge, with citations.  The hint [N] is a reference to the citation format. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&#xA;If I was old,&lt;br/&gt;&#xA;I would be far too clumsy, far too wrinkled,&lt;br/&gt;&#xA;to hide among others without sight.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All the information contained in Wikipedia would fill a lot of books.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&#xA;But I am young so, if you wished,&lt;br/&gt;&#xA;I could hide in the smallest crevice&lt;br/&gt;&#xA;or a fold in the world, hidden from sight&lt;br/&gt;&#xA;among the unsavory sort.&lt;br/&gt;&#xA;It is that I am young that I am so much easier to find.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Doesn't take up space, and is much easier to find information than searching through books in a library.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&#xA;My greatest opponent is he who imposes nothing,&lt;br/&gt;&#xA;who requires nothing,&lt;br/&gt;&#xA;enforces nothing,&lt;br/&gt;&#xA;and is far more stealthy than me.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Haven't figured this out...but one of the hints (&quot;I fought my opponent once, or one of its major incarnations. It took a day off my life, but I — no, we — won.&quot;) could refer to the SOPA protest blackout?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&#xA;Than I, who weighs upon the world&lt;br/&gt;&#xA;only barely more than a shadow&lt;br/&gt;&#xA;as I roam.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure how this fits either.  Maybe a reference to Wikipedia's small server infrastructure compared to some internet companies?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&#xA;I do know myself,&lt;br/&gt;&#xA;know more of myself than you do,&lt;br/&gt;&#xA;know more of the world than you do,&lt;br/&gt;&#xA;and it is not my style to ask questions.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wikipedia knows more about everything than I do.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&#xA;But, I shall make one exception today:&lt;br/&gt;&#xA;What am I?&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wikipedia!  And it fits the final hint: The answer has one obvious, canonical name that is nine letters long, seven unique.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&#xA;There are a few hints that I haven't been able to connect the dots on, but this seems to fit.  &lt;/p&gt;&#xA;" OwnerUserId="19348" LastEditorUserId="19348" LastEditDate="2016-02-25T22:53:28.980" LastActivityDate="2016-02-25T22:53:28.980" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27914" PostTypeId="1" AcceptedAnswerId="28136" CreationDate="2016-02-24T23:05:03.037" Score="15" ViewCount="2563" Body="&lt;p&gt;There's a game called the finger game. It's played by two players who each start with one finger on each hand up. The players take turns using one of their hands to tap one of their opponent's hands.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;When a player A taps player B's hand (or vice versa), B adds the number of fingers on A's hand to his own hand. Ex. A has one finger up, B has two. A taps B. A still has one finger up, B has three.  &lt;/li&gt;&#xA;&lt;li&gt;When either player has five fingers up on a hand, that hand is dead and out of the game. Dead hands have no fingers up.&lt;/li&gt;&#xA;&lt;li&gt;When a player would have more than five fingers up on a hand, they subtract five from the number of fingers they should have up. Ex. A has three fingers up, B has four. A taps B. B now has two fingers up.&lt;/li&gt;&#xA;&lt;li&gt;When a player only has one hand alive and their live hand has an even number of fingers up, they may skip their turn and pass half their fingers from the live hand to the dead hand, reviving it. &lt;/li&gt;&#xA;&lt;li&gt;If either player has two dead hands, they lose.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;With perfect play, who wins? The first player or the second?&lt;/p&gt;&#xA;" OwnerUserId="18103" LastEditorUserId="8874" LastEditDate="2016-03-08T17:47:29.027" LastActivityDate="2016-03-08T17:47:29.027" Title="The Finger Game" Tags="|logical-deduction|strategy|game|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="27915" PostTypeId="2" ParentId="27798" CreationDate="2016-02-24T23:06:09.300" Score="15" Body="&lt;p&gt;Just to show what can happen when important ciphers get into the wrong hands,&#xA;a partial decryption suggests that&#xA;if Jack and Dave were me, what they would do is look for a lawyer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/Ycxx9.png&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What were Jack and Dave thinking!, entering dance bars with A minor.&#xA;Time to face the music.&#xA;Apologies to anyone unfamiliar, or too familiar, with music theory,&#xA;and to the original intent of a yet-undeciphered interesting puzzle.&lt;/p&gt;&#xA;" OwnerUserId="18129" LastActivityDate="2016-02-24T23:06:09.300" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="27916" PostTypeId="1" AcceptedAnswerId="27935" CreationDate="2016-02-24T23:19:23.213" Score="75" ViewCount="6966" Body="&lt;p&gt;One day, you enter your math classroom. On the board, there are the words POP QUIZ.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dang it! Was it about fractions? you knew you were bad with fractions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You look down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{array}{|cc|} \hline \text{You can only use your brain, and paper. &#xA;} \\ \text{The number of digits in this box that are not 1 is} &amp;amp; \text{_______} \\ \text{The number of digits in this box that are not 2 is} &amp;amp; \text{_______} \\ \text{The number of digits in this box that are not 3 is} &amp;amp; \text{_______} \\ \text{The number of digits in this box that are not 4 is} &amp;amp; \text{_______} \\ \text{The number of digits in this box that are not 5 is} &amp;amp; \text{_______} \\ \text{The number of digits in this box that are not 6 is} &amp;amp; \text{_______} \\ \text{The number of digits in this box that are not 7 is} &amp;amp; \text{_______} \\ \text{The number of digits in this box that are not 8 is} &amp;amp; \text{_______} \\ \text{The number of digits in this box that are not 9 is} &amp;amp; \text{_______} \\ \hline \end{array}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All of the blanks are not mostly the same number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17970" LastEditorUserId="17970" LastEditDate="2016-06-28T21:30:22.393" LastActivityDate="2017-10-22T20:31:55.473" Title="Oh dang, I have to change my whole answer!" Tags="|logical-deduction|no-computers|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="27919" PostTypeId="1" AcceptedAnswerId="27986" CreationDate="2016-02-24T23:54:58.453" Score="12" ViewCount="398" Body="&lt;p&gt;A gardener has a sack of indistinguishable bulbs and a flat allotment. In the spring, each bulb will produce either a red, green or blue flower. She wishes to plant some of the bulbs so that when spring comes, some two flowers of the same colour will be exactly one metre apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many bulbs must she plant to achieve her goal?&lt;/p&gt;&#xA;" OwnerUserId="18508" LastActivityDate="2016-02-28T14:52:33.433" Title="Planting flowers of unknown colours" Tags="|mathematics|" AnswerCount="4" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="27922" PostTypeId="2" ParentId="27919" CreationDate="2016-02-25T00:19:54.833" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Seven bulbs suffice.  Plant them so that ABCD and AEFG are rhombuses with a 1-meter side and a sixty-degree angle at A, and such that CF is 1 meter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19559" LastActivityDate="2016-02-25T00:19:54.833" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27923" PostTypeId="2" ParentId="27912" CreationDate="2016-02-25T00:27:26.943" Score="5" Body="&lt;p&gt;The fight is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;(functionally) eternal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If it's a single elimination bracket tournament, and he's one 56 consecutive matches, there must've been $2^{56}$ combatants... which is a lot.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Given each match occurs on a single stage, they're happening in serial. So unless each fight took &lt;a href=&quot;https://www.wolframalpha.com/input/?i=age+of+universe+%2F+2%5E56&quot;&gt;less than 6 seconds&lt;/a&gt;, they would have had to have started before the birth of the universe.&lt;br&gt;&#xA;&lt;br&gt;&#xA; That being said, let's assume you could find 72 quadrillion willing combatants, who also happened to be immortal (how else do they live long enough to survive the billions of years of waiting between fights)... If I, as the organiser, charge each of them a \$10 entry fee, that's 720 quadrillion dollars.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Now assuming I was paid this at the start of the universe (I'm a smart business man and only accept up front payments), and invest the money in a term deposit at a measly 6% PA compounding... Now that the fight is over and I can pay Frank Dux his (very generous) \$1,000,000(!) prize money, I now have \$&lt;a href=&quot;http://www.wolframalpha.com/input/?i=continuous+interest+PV%3D720000000000000000,+rate%3D6%25,+periods%3D14000000000&quot;&gt;&lt;code&gt;[computation time exceeded... lets assume it's lots...]&lt;/code&gt;&lt;/a&gt; at my disposal.&lt;br&gt;&#xA;&lt;br&gt;&#xA; ...and finally it's time to live out my dream and swim like Scrooge McDuck in a sea of gold coins that exceeds the size of the universe by many orders of magnitude.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="5208" LastEditDate="2016-02-25T00:47:32.287" LastActivityDate="2016-02-25T00:47:32.287" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27925" PostTypeId="5" CreationDate="2016-02-25T02:59:01.143" Score="0" Body="&lt;h1&gt;What is a cipher?&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;A cipher is an encoded message, usually by substituting letters for other letters, numbers, or symbols.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;What makes a good cipher puzzle?&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;A cipher puzzle has to be solvable. There should be an intuitive and enjoyable solution path, and preferably one or more 'aha' moments. In case of multilayer ciphers, it's a good idea to provide the solver some confirmation after each step. See &lt;a href=&quot;https://puzzling.meta.stackexchange.com/questions/1717/code-puzzles-what-not-to-do&quot;&gt;Code Puzzles: What (Not) To Do?&lt;/a&gt; for more on this.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;How should I try to solve a cipher puzzle?&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Try to see what kind of cipher it is - whether you need to use rot13, or Morse code, or unscramble something. Then, if it has more than one layer of encoding, you can continue with the same for the next level.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;How should I answer the question once I have the solution to the cipher?&lt;/h1&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Make sure your solution fully fits the original message.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Answer the cipher fully, and provide a detailed explanation. Provide a line-by-line analysis if necessary to justify your solution.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Spoilertag content that directly relates to the answer, but do not write an answer entirely in a spoiler. Spoilertags hide part of a post by placing &lt;code&gt;&amp;gt;!&lt;/code&gt; at the beginning of a line.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h1&gt;A Guide&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;See this &lt;a href=&quot;https://puzzling.stackexchange.com/questions/52118/guide-to-codes-and-ciphers&quot;&gt;Guide to Codes and Ciphers&lt;/a&gt; for more info.&lt;/p&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="25412" LastEditDate="2017-05-27T14:02:24.277" LastActivityDate="2017-05-27T14:02:24.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27926" PostTypeId="4" CreationDate="2016-02-25T02:59:01.143" Score="0" Body="A cipher is a simple method of modifying messages to make them unreadable except to readers who know the key. Common ciphers are Caesar, Vigenere, Atbash, and several types of substitution cipher. Other ciphers include polygraphic ciphers such as Playfair and Foursquare and numerical ciphers such as Polybius and Nihilist. Cipher puzzles sometimes make use of data encodings like Morse, Binary, or Base64." OwnerUserId="11876" LastEditorUserId="83349" LastEditDate="2023-02-24T08:30:36.537" LastActivityDate="2023-02-24T08:30:36.537" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27927" PostTypeId="1" AcceptedAnswerId="27944" CreationDate="2016-02-25T04:07:17.050" Score="3" ViewCount="41807" Body="&lt;p&gt;Though I'm not alive, I had 4 legs when I was born.&lt;br&gt;&#xA;I lost a leg when I got too stressed and worn.&lt;br&gt;&#xA;I go up the hill with just 3 legs, and come down the hill with 4.&lt;br&gt;&#xA;If my owner doesn't lose some weight, I'll be going up the hill some more!&lt;/p&gt;&#xA;" OwnerUserId="19165" LastEditorUserId="19165" LastEditDate="2016-02-25T05:23:37.437" LastActivityDate="2017-02-28T12:18:00.073" Title="Up the hill with 3 legs and back down with 4" Tags="|riddle|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="27928" PostTypeId="1" AcceptedAnswerId="27941" CreationDate="2016-02-25T04:33:27.103" Score="7" ViewCount="1286" Body="&lt;p&gt;You have 10 coins: five heads, then five tails, all in a row.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;HHHHHTTTTT&lt;/code&gt; (H is for heads and T is for tails.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In each move, you can take two adjacent coins and move them somewhere else. You cannot switch the coins as you move them, and you may not move other coins. (This includes spreading them apart to make room!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If all ten coins are in a row, then you may move a group of two to either end. Otherwise, your next move must fill the gap left by the previous move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can you get them to alternate (either &lt;code&gt;HTHTHTHTHT&lt;/code&gt; or &lt;code&gt;THTHTHTHTH&lt;/code&gt;) in just &lt;strong&gt;5 moves&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example game that does not succeed:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Start: H H H H H T T T T T&#xA;    1: H H H H - - T T T T H T&#xA;    2: H - - H H H T T T T H T&#xA;    3: H T H H H H T T T - - T&#xA;    4: H T T H H H T T H T&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Any move from the last position would take coins out of the middle, so the last move cannot succeed in putting the coins adjacent to each other.&lt;/p&gt;&#xA;" OwnerUserId="19571" LastEditorUserId="11876" LastEditDate="2016-02-25T05:01:42.513" LastActivityDate="2016-02-25T10:02:39.173" Title="Sliding 10 coins to make them alternate" Tags="|strategy|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="27929" PostTypeId="2" ParentId="27928" CreationDate="2016-02-25T05:36:41.367" Score="1" Body="&lt;h1&gt;PARTIAL ANSWER&lt;/h1&gt;&#xA;&lt;p&gt;Some of this logic &lt;em&gt;may&lt;/em&gt; be wrong. Feel free to point out mistakes.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;The first move must be dedicated to taking a pair out of the row. That means we only can shuffle around 8 coins overall.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Assume we will end with &lt;code&gt;HTHTHTHTHT&lt;/code&gt;.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Since we only have enough moves to shuffle eight coins, and we need to get four alternating pairs in place, every move but one has to pick up two alternating coins.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Assume the first one will be the move that does not pick up two alternating coins.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Our only options for groups to move on turn 1 are:&lt;/p&gt;&#xA;&lt;p&gt;2 and 3, 4 and 5, 6 and 7, 8 and 9&lt;/p&gt;&#xA;&lt;p&gt;If you move either of the two middle groups, then you cannot replace them with an alternating pair on the next turn. WLOG, take group [2,3].&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;0 HHHHHTTTTT&#xA;1 H  HHTTTTTHH&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;This next move is forced since we must pick up an alternating pair going in the proper order.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;2  HTHHHTTTT  H&#xA;  (HTHTHTHTHT  )&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;The parentheses are what we need. Observe the differences between what we need and have.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;2  HTHHHTTTT  H&#xA;  (HTHTHTHTHT  )&#xA;   ...!..!.!!.!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;We must make at least three moves to switch around four groups, but one of the groups is a single coin completely isolated; it will require at least two moves to fix. Therefore, this is not going to work.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;If we end with &lt;code&gt;HTHTHTHTHT&lt;/code&gt;, then the first move will pick up alternating coins.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The only alternating coins to pick up are the middle two. Therefore this move is forced:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;0 HHHHHTTTTT&#xA;1 HHHH  TTTTHT&#xA; (HTHTHTHTHT)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;There is not a movable alternating pair in the right order.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Therefore we will not end with &lt;code&gt;HTHTHTHTHT&lt;/code&gt;.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Therefore we will end with &lt;code&gt;THTHTHTHTH&lt;/code&gt;.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-02-25T05:36:41.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27930" PostTypeId="2" ParentId="27928" CreationDate="2016-02-25T05:39:03.850" Score="4" Body="&lt;p&gt;I was able to do it in 6.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1. H H - - H T T T T T H H&#xA;2. H H T T H T - - T T H H&#xA;3. H - - T H T H T T T H H&#xA;4. H T H T H T H T T - - H&#xA;5. H T H T H T H - - T T H&#xA;6. H T H T H T H T H T&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="18821" LastEditorUserId="11876" LastEditDate="2016-02-25T05:39:36.257" LastActivityDate="2016-02-25T05:39:36.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27932" PostTypeId="2" ParentId="27912" CreationDate="2016-02-25T07:13:25.327" Score="4" Body="&lt;p&gt;3 minutes ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He knocked out all of the other fighters before the match started, and therefore won by default when no opponent could be found before the round timed out. Go Frank!&lt;/p&gt;&#xA;" OwnerUserId="19545" LastActivityDate="2016-02-25T07:13:25.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27933" PostTypeId="2" ParentId="27919" CreationDate="2016-02-25T07:36:48.500" Score="6" Body="&lt;p&gt;Suppose you were to plant just six bulbs.   Could we fail to have two flowers the same color a meter apart?  I'll show the answer is yes.  (This might be overly complicated, but I couldn't do it in less than four cases)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Call two bulbs adjacent if they are planted a meter apart.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;First note that if you plant a bulb $X$ adjacent to only two others, then we've already failed.  $X$ may come up the third color, and hence in the worst case scenario it is useless to our quest for two flowers the same color.&lt;/li&gt;&#xA;&lt;li&gt;Suppose we choose four points $ABCD$ on a unit rhombus with a unit diagonal.  Then these four points correspond to points on a triangular lattice.  The other two bulbs $E$ and $F$ must be adjacent to at least two of the points $ABCD$, but this is only possible if $E$ and $F$ are both on the triangular lattice as well.  But then $E$ and $F$ are forced into one of the following four locations marked x:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  .   .   .   .&#xA;.   .   x   .   .&#xA;  .   A   B   x  &#xA;.   x   C   D   .&#xA;  .   .   x   .&#xA;.   .   .   .&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both $E$ and $F$ are then adjacent to only two other flowers, putting us back in case 1.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;Suppose you have an equailateral triangle $ABC$ .  Any bulb adjacent to two of these puts us back in case 2, so that forces the other bulbs to be adjacent to at most one of the set $ABC$.  To avoid case 1, we need $D,E,F$ to be adjacent to $A,B,C$ respectively, and we also need $DEF$ to form an equilateral triangle.  However, if $ABC$ comes up red, green, blue and $DEF$ comes up green, blue, red, we're in trouble.&lt;/li&gt;&#xA;&lt;li&gt;Finally, consider bulb $A$.  To avoid case 1 we must have at least three bulbs adjacent to $A$.  Suppose $AB$, $AC$, and $AD$ are all adjacent.  To avoid a triangle (case 3), we know $BC$ are not adjacent, nor are $CD$, nor $BD$.  That means $B, C, D$ are all adjacent to the last two bulbs $E$ and $F$ (to avoid case 1), and $E$ and $F$ are not adjacent to each other or $A$ (to avoid case 3).  But if $A,E,F$ all come up blue and $B, C, D$ all come up green we're in trouble.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2722" LastActivityDate="2016-02-25T07:36:48.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27935" PostTypeId="2" ParentId="27916" CreationDate="2016-02-25T08:12:57.150" Score="52" Body="&lt;p&gt;One answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;1: 25&lt;/code&gt;&lt;br /&gt;&lt;code&gt;2: 18&lt;/code&gt;&lt;br /&gt;&lt;code&gt;3: 25&lt;/code&gt;&lt;br /&gt;&lt;code&gt;4: 24&lt;/code&gt;&lt;br /&gt;&lt;code&gt;5: 23&lt;/code&gt;&lt;br /&gt;&lt;code&gt;6: 24&lt;/code&gt;&lt;br /&gt;&lt;code&gt;7: 26&lt;/code&gt;&lt;br /&gt;&lt;code&gt;8: 25&lt;/code&gt;&lt;br /&gt;&lt;code&gt;9: 26&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To find this, I first noted that each of the blanks must be at least 8, since there are already 8 digits different from the given one. None of the blanks can be 8, though, since that would mean all of the digits in the blanks are the given digit, which would have to be 8, and there can't be 88 digits in the box - at most 27. If a blank were 9, all of the digits would have to be the given digit except, presumably, that 9. But then there would be 24 digits different from 9, which contains two different digits, '2' and '4,' so 9 is ruled out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, we know every blank is a 2-digit number, so there are indeed 27 digits in total. Thus the average number of digits that aren't a given digit is (9*27 - 27)/9 = 24.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I started by writing 24 in every blank. I then counted the number of each digit, and wrote what should be in the blanks in that case. Since only '2' and '4' appear in the blanks, the number of non-1 digits is 26, same as 3, 5, 6, 7, 8, and 9, while there are 17 non-2's and 17 non-4's. So I changed the blanks to 26, 17, 26, 17, 26, 26, 26, 26, 26. Then I counted again and corrected. After 11 iterations, I found that I hadn't changed any of the numbers, so I arrived at an answer, which I've written above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't prove that it's unique though. Also, my method may have ended in a cycle, so I was lucky to have found a fixed point.&lt;/p&gt;&#xA;" OwnerUserId="12006" LastActivityDate="2016-02-25T08:12:57.150" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27936" PostTypeId="1" AcceptedAnswerId="27937" CreationDate="2016-02-25T08:55:30.017" Score="4" ViewCount="116" Body="&lt;p&gt;Split me into 3, and you will feel like iron,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you see I am much stronger than it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Split me into 6, and you will feel like aluminium,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you see I can't be softened as easily as it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I stand next to a precious golden treasure,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I am still much more valuable than it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What/Who am I?&lt;/p&gt;&#xA;" OwnerUserId="18685" LastEditorUserId="4421" LastEditDate="2016-02-25T21:22:18.020" LastActivityDate="2016-02-25T21:22:18.020" Title="Who/What am I by SC #6" Tags="|riddle|word|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27937" PostTypeId="2" ParentId="27936" CreationDate="2016-02-25T08:56:50.677" Score="5" Body="&lt;p&gt;I think it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Platinum&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One thing to keep in mind is that Atomic number of platinum is 78&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Split me into 3, and you will feel like iron,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 78/3 = 26. And 26 is the atomic number of Iron.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But you see I am much more stronger than it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes. Bulk Modulus of platinum is 230 GPa while of the iron is 170 GPa&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Split me into 6, and you will feel like alluminium,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 78/6 = 13. And atomic number of aluminium is 13&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But you see I can't be softened as easy as it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Aluminium is more malleable than platinum&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I stand next to a precious golden treasure,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In &lt;a href=&quot;http://www.ptable.com/&quot; rel=&quot;nofollow&quot;&gt;periodic table&lt;/a&gt;, Platinum is right next to Gold which have atomic number 79&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I am still much more valuable than it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes, I think we all know that.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-02-25T09:20:15.977" LastActivityDate="2016-02-25T09:20:15.977" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27938" PostTypeId="1" AcceptedAnswerId="27946" CreationDate="2016-02-25T09:29:15.407" Score="15" ViewCount="2338" Body="&lt;p&gt;A weird pocket calculator has a numerical display (for digits 0-9) and only two buttons, inscripted $\boxed{D+}$ and $\boxed{D-}$. The first button doubles the displayed number and then adds 1. The second button doubles the displayed number and then subtracts 1. For example, if the display shows 6, then pressing $\boxed{D+}$ produces 13; if the display shows 6 and we press $\boxed{D-}$, we get 11. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that the initially displayed number is 1, and that we then perform exactly 22 button presses. How many different numbers that can possibly result from this? &lt;/p&gt;&#xA;" OwnerUserId="19470" LastActivityDate="2016-02-28T02:55:01.910" Title="A weird pocket calculator" Tags="|mathematics|combinatorics|no-computers|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27939" PostTypeId="1" AcceptedAnswerId="27943" CreationDate="2016-02-25T09:31:25.853" Score="18" ViewCount="2528" Body="&lt;p&gt;A rectangle is tiled by nine squares with side lengths $2,5,7,9,16,25,28,33$ and $36$ (without overlapping and without gaps). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the side lengths of the rectangle? What does the tiling look like?&lt;/p&gt;&#xA;" OwnerUserId="19470" LastActivityDate="2016-02-26T01:24:50.220" Title="Tiling a rectangle with nine squares" Tags="|mathematics|geometry|combinatorics|dissection|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27940" PostTypeId="2" ParentId="27938" CreationDate="2016-02-25T09:40:12.457" Score="8" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4194304&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Uhm... As someone pointed out this question has a no computer tag that I did not originally see. Please ignore my answer, I'll just leave it here to avoid the same mistake to others&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solved by simulation (the tree has 2^23 nodes, can do):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Set&amp;lt;Integer&amp;gt; values(int depth, Set&amp;lt;Integer&amp;gt; state) {&#xA;    if (depth &amp;gt;= 22) return state;&#xA;    else {&#xA;        Set&amp;lt;Integer&amp;gt; dPlus = state.stream().map(i -&amp;gt; i * 2 + 1).collect(Collectors.toSet());&#xA;        Set&amp;lt;Integer&amp;gt; dMinus = state.stream().map(i -&amp;gt; i * 2 - 1).collect(Collectors.toSet());&#xA;        Set&amp;lt;Integer&amp;gt; nextPlus = values(depth+1, dPlus);&#xA;        Set&amp;lt;Integer&amp;gt; nextMinus = values(depth+1, dMinus);&#xA;        nextPlus.addAll(nextMinus);&#xA;        return nextPlus;&#xA;    }&#xA;}&#xA;&#xA;void run(String[] args) {&#xA;    System.out.println(values(0, new HashSet&amp;lt;&amp;gt;(Arrays.asList(new Integer[]{1}))).size());&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="14435" LastEditorUserId="14435" LastEditDate="2016-02-25T09:42:21.177" LastActivityDate="2016-02-25T09:42:21.177" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27941" PostTypeId="2" ParentId="27928" CreationDate="2016-02-25T09:50:11.983" Score="11" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0) &lt;code&gt;H H H H H T T T T T - -&lt;/code&gt;&lt;br&gt;&#xA; 1) &lt;code&gt;H - - H H T T T T T H H&lt;/code&gt;&lt;br&gt;&#xA; 2) &lt;code&gt;H T T H H T T - - T H H&lt;/code&gt;&lt;br&gt;&#xA; 3) &lt;code&gt;H T T H - - T H T T H H&lt;/code&gt;&lt;br&gt;&#xA; 4) &lt;code&gt;H T T H T H T H T - - H&lt;/code&gt;&lt;br&gt;&#xA; 5) &lt;code&gt;- - T H T H T H T H T H&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The following ideas helped me to find the solution (sorted by obviousness):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First, you have to make room to move coins, ie, move two coins to the end. There is no point in closing the row before you are done.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As pointed out by Deusovi's partial answer, you really have to place a final pair with every move except the first.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; By beginning with [2,3], I can create two alternated pairs in the next move.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It all hangs on the final move that closes the gap. If I could just get rid of that...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I can by taking the coins from the other side.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1750" LastEditorUserId="1750" LastEditDate="2016-02-25T10:02:39.173" LastActivityDate="2016-02-25T10:02:39.173" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27943" PostTypeId="2" ParentId="27939" CreationDate="2016-02-25T10:04:27.707" Score="28" Body="&lt;p&gt;The dimensions are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 69 x 61&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the tiling looks like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/sQtJI.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sQtJI.png&quot; alt=&quot;solution&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Working out the dimensions of the rectangle is quite easy. We know its total area is $4209$ (i.e., $2^2 + 5^2 + 7^2 + 9^2 + 16^2 + 25^2 + 28^2 + 33^2 + 36^2$). This factorizes as $3 \times 23 \times 61$, and in order to fit in a square with a side length of 36, the rectangle must be $3 \times 23$ units long on one side, and $61$ units on the other. Fitting the squares into this rectangle only takes a few minutes.&lt;/p&gt;&#xA;" OwnerUserId="40" LastEditorUserId="40" LastEditDate="2016-02-25T10:11:10.817" LastActivityDate="2016-02-25T10:11:10.817" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27944" PostTypeId="2" ParentId="27927" CreationDate="2016-02-25T10:09:52.633" Score="6" Body="&lt;p&gt;Maybe&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A chair.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though I'm not alive, I had 4 legs when I was born.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4 legs and not alive. Sounds like a chair to me.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I lost a leg when I got too stressed and worn.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stressed and worn - a chair leg broke because of a large weight. (applying pressure causes stress and can wear down wood)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I go up the hill with just 3 legs, and come down the hill with 4.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ? I suppose that someone who fixes chairs has his office at the top of a hill for some reason.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If my owner doesn't lose some weight, I'll be going up the hill some more!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well, if the owner is heavy, he's bound to break that chair leg again sooner or later.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18972" LastActivityDate="2016-02-25T10:09:52.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27945" PostTypeId="2" ParentId="27939" CreationDate="2016-02-25T10:12:51.723" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. Size of rectangle = 61 x 69&lt;br&gt;&#xA; 2. Tiling: &lt;a href=&quot;https://i.stack.imgur.com/wYhFh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wYhFh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18724" LastEditorUserId="13148" LastEditDate="2016-02-25T17:49:25.153" LastActivityDate="2016-02-25T17:49:25.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27946" PostTypeId="2" ParentId="27938" CreationDate="2016-02-25T10:20:05.063" Score="21" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 4,194,304 (2&lt;sup&gt;22&lt;/sup&gt;) possible results&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why such a straightforward number?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each press doubles the number of possible sequences&#xA; and no two sequences can produce the same result.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why not?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each press redoubles the additive/subtractive&#xA; effect of all previous presses, which have already&#xA; been redoubled by any intervening presses.&#xA; (The first press's effect has been doubled the most often.)&#xA; After that, the new +1 or -1 will only make half the difference&#xA; of the immediately preceding press.&#xA; This is equivalent to a 22-digit binary number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The exact final result, $V\!$, equals a 23-bit binary number built from the&#xA;intermediate results $A$,$B$,...,$U$ of button presses $a$,$b$,...,$u$,&#xA;followed by one last button press, $v$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{lrrllll}&#xA;\rm\llap{1st} ~press&amp;amp; a \;{=}\;{\pm}\!\: 1 &amp;amp; ~A ~~\rlap{=} ~&amp;amp;  2(1) + a &amp;amp; ~~\llap{=}~~~ \phantom{ 2(   2( } 2{+}a        &amp;amp; ~~\llap{=}~~~ 2 + ~ a           \\&#xA;\rm\llap{2st} ~press&amp;amp; b \;{=}\;{\pm}\!\: 1 &amp;amp; ~B ~~\rlap{=} ~&amp;amp;~ 2A ~ + b &amp;amp; ~~\llap{=}~~~ \phantom{ 2( } 2(   2{+}a )+b    &amp;amp; ~~\llap{=}~~~ 4 + 2 a + ~ b     \\&#xA;\rm\llap{3st} ~press&amp;amp; c \;{=}\;{\pm}\!\: 1 &amp;amp; ~C ~~\rlap{=} ~&amp;amp;~ 2B ~ + c &amp;amp; ~~\llap{=}~~~           2(   2(   2{+}a )+b)+c &amp;amp; ~~\llap{=}~~~ 8 + 4 a + 2 b + c \\[-1ex]&#xA;\quad\vdots                                                                                                                                                \\[-1ex]&#xA;\rm\llap{22nd}~press&amp;amp; v \;{=}\;{\pm}\!\: 1 &amp;amp; ~V ~~\rlap{=} ~&amp;amp;~ 2U ~ + v &amp;amp; ~~\llap{=}~~~ \rlap{ 2 ( \cdots 2 ( 2{+}a ) + \cdots ) + v                     } \\[1ex]&#xA;                    &amp;amp;                      &amp;amp;      \rlap{=} ~&amp;amp;~ \rlap{      2^{22} + 2^{21} a + 2^{20} b + \cdots + 2^0 v                                 } \\[1ex]&#xA;                    &amp;amp;                      &amp;amp;      \rlap{=} ~&amp;amp;~ \rlap{      2^{22} \bigl( \frac12 {+} \frac a2 \bigr)&#xA;                                                                      + \, 2^{21} \bigl( \frac12 {+} \frac b2 \bigr) + \cdots&#xA;                                                                      + \, 2^1    \bigl( \frac12 {+} \frac v2 \bigr)&#xA;                                                                      + \, 2^0         ( 1 )                                                             } \\[1ex]&#xA;                    &amp;amp;                      &amp;amp;      \rlap{=} ~&amp;amp;~ \rlap{ 23 \;\!\textrm{-bit binary number} ~\mathit{abcdefghijklmnopqrstuv} \!\: 1_{\!\:2} } \\[-.5ex]&#xA;                    &amp;amp;                      &amp;amp;                &amp;amp;~ \rlap{ \textrm{where the}~{-}\!\textrm{1s of}~ a,b,\dots,v ~\textrm{are replaced by 0}\!\:\textrm{s} }&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each of the 4,194,304 ($2^{22}$) possible sequences of 22 button presses&#xA; has a unique combination of values for $a$,$b$,$c$,...,$v$,&#xA; which produces a unique value for $V\!$&#xA; and happens to cover all of the 4,194,304 odd numbers&#xA; from 1 through 8,388,607 ($2^{23}{-}1$).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sure enough, if every press were $\boxed{D-}$&#xA;then              $V \! = 000\textrm{...}001_2 = 1$,&#xA;the same as doubling the original 1 and diminishing that&#xA;by&amp;nbsp;1&amp;hairsp;&amp;mdash;&amp;hairsp;producing no net&#xA;change&amp;hairsp;&amp;mdash;&amp;hairsp;22 times.&#xA;And if every press were $\boxed{D+}$&#xA;then              $V \! = 11111111111111111111111_2$,&#xA;which amounts to $2{\times}1111111111111111111111_2 + 1$,&#xA;the same calculation as the 22nd and last repeated step of&#xA;doubling an intermediate result and adding&amp;nbsp;1 before the next step.&lt;/p&gt;&#xA;" OwnerUserId="18129" LastEditorUserId="18129" LastEditDate="2016-02-26T00:16:18.863" LastActivityDate="2016-02-26T00:16:18.863" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27950" PostTypeId="2" ParentId="27871" CreationDate="2016-02-25T12:41:07.713" Score="9" Body="&lt;p&gt;(1) We observe that $245=5\times49$. &#xA;Instead of working with the integers $1,\ldots,245$, we will use the integers $1,\ldots,49$ in the five colors red, blue, yellow, green and orange.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Among the six numbers randomly chosen by the ward, there will be at least two of the same color, say integers $x$ and $y$ both in color red. &#xA;Modulo $49$, exactly one of $x-y$ and $y-x$ is in the range $1,\ldots,24$;&#xA;assume that $y-x$ is in the range $1,\ldots,24$.&#xA;Then we reject $y$, and put $x$ as the first number on the list.&#xA;With this, the second prisoner knows that the rejected number has the &#xA;same color as the first number $x$ on the list and is one of $x+1,x+2,\ldots,x+24$ (taken modulo $49$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) There are $4!=24$ permutations of the other four integers.&#xA;This is enough to communicate the value of $y-x$ to the other prisoner.&#xA;(The two prisoners just have to agree on a bijection between permutations&#xA;of four objects and the numbers $1,2,\ldots,24$.)&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="18924" LastEditDate="2016-02-25T13:54:23.087" LastActivityDate="2016-02-25T13:54:23.087" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27951" PostTypeId="2" ParentId="1808" CreationDate="2016-02-25T12:59:56.213" Score="-1" Body="&lt;p&gt;@Alizerah Fallah. You already made a case for 5 weighings, but it's quite horrible. You're splitting the 132(12*11) possible cases into two cases of 36 and one of 60, and give up when the 60-case is hard to solve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I instead propose a first weighing of 5 vs 5 balls. This splits it into two 45-possibilities cases and one 42-possibilties case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weighing 3v3 does so as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thinking about writing a program to bruteforce this, it's quite dishearthening if the asker does not know the answer. I thought OP confirmed that it's doable in 5 in a comment on Alizerah's answer, but he did that himself.&lt;/p&gt;&#xA;" OwnerUserId="19527" LastActivityDate="2016-02-25T12:59:56.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27952" PostTypeId="2" ParentId="27871" CreationDate="2016-02-25T14:06:24.777" Score="7" Body="&lt;p&gt;Arrange the numbers from 1 to 245 clockwise around a circle, and mark the locations of the six numbers on it. Consider the gaps left between the marked numbers. There are six gaps, with a total length of 245. Each gap is at least 1, so the largest possible sum of two gaps is 241, and the second-largest gap is at most 120.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the largest gap (or possibly one that is tied for largest). The number that will be removed is the one on the clockwise edge of this gap. The next gap clockwise (before the number is removed) is at most 120, so its length can be encoded by the order of the five numbers to be written.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other prisoner receives the five numbers and similarly plots them on a circle. They will necessarily find that one of the five gaps is larger than any of the others. The removed number comes from this gap, and its value can be found by converting the order of the five numbers back into a value from 1 to 120, and counting that many places counterclockwise from the clockwise edge of the gap.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2016-02-25T14:06:24.777" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27953" PostTypeId="2" ParentId="26118" CreationDate="2016-02-25T14:30:59.223" Score="4" Body="&lt;p&gt;Welp, you need to check almost all rows and columns.&#xA;Proof by counterexample:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let X be a correct solution. Switch the topleft square with its right neighbour. All rows and boxes still check out but two columns do not. Same for rows if you switch two vertically aligned squares. So you can skip at most one row/column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;The only thing you can skip, after having checked all 9 of one item(say rows), is that you only need to check 8 of the other 2 items(columns and boxes), as it's already clear that there are 9 of each number. Alternatively, after fully checking rows and columns, you can skip boxes 1, 5 and 9(or 3 other boxes so that they do not align)(because there are 3 of the number in the 3 rows/columns it uses, and the other 2 boxes in those rows/columns are checked).&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: That's a mighty useful link, mathoverflow has thought this through a lot. Paraphrasing the pages of text found there:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you have checked the columns and rows, you can skip some boxes. Having checked the top 3 rows and boxes 1 and 2, you don't need to check box 3. Likewise you don't need to check box 6 if you have 4 and 5, and the bottom boxes are proven correct by having checked the columns and the first six boxes.&#xA;The last row is proven correct by the bottom 3 boxes and the two rows above it. So you can skip 1 row and 5 boxes for 21 checks total.&lt;/p&gt;&#xA;" OwnerUserId="19527" LastEditorUserId="19527" LastEditDate="2017-05-23T11:55:40.103" LastActivityDate="2017-05-23T11:55:40.103" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27954" PostTypeId="1" AcceptedAnswerId="27955" CreationDate="2016-02-25T14:38:08.650" Score="7" ViewCount="985" Body="&lt;p&gt;I hope you enjoy puzzling out the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Free three months&lt;br&gt;&#xA;  Jump the flower.&lt;br&gt;&#xA;  Originate the coil.&lt;br&gt;&#xA;  Reverse the addendum and finish with a bell sound.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastActivityDate="2016-02-25T15:03:05.793" Title="What Am I Talking About?" Tags="|riddle|word|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27955" PostTypeId="2" ParentId="27954" CreationDate="2016-02-25T14:42:33.667" Score="9" Body="&lt;p&gt;You're talking about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;SPRING.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Free three months&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The are three months in the season of spring; &quot;spring&quot; can also mean to free&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Jump the flower.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Jumping is &quot;springing&quot;, and flowers appear in spring; also, a spring, the body of water, is a &quot;flow-er&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Originate the coil.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Both &quot;originate&quot; and &quot;coil&quot; are synonyms of &quot;spring&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reverse the addendum and finish with a bell sound.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; P.S. is an addendum; a bell sound is a RING. Reverse and concatenate to get SPRING.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="11876" LastEditDate="2016-02-25T15:03:05.793" LastActivityDate="2016-02-25T15:03:05.793" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27959" PostTypeId="2" ParentId="27875" CreationDate="2016-02-25T17:10:17.940" Score="4" Body="&lt;p&gt;I have a set of size&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 30&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The set:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;pre&gt;BACB, CABC, ABDA, ACDA, BBCA, CADD, CBBA, DAAA, &lt;br&gt;CCDA, ABAD, ADAB, BDBD, CDAB, ADCC, CCCD, BDBA, &lt;br&gt;ABCB, DCBB, DDDC, AACB, BBDB, CCCA, BADD, DADD, &lt;br&gt;BCAC, ACDD, DBAD, CBBC, AABC, DDCC&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I found this using the &lt;a href=&quot;https://en.wikipedia.org/wiki/A*_search_algorithm#Bounded_relaxation&quot; rel=&quot;nofollow&quot;&gt;weighted A* algorithm&lt;/a&gt; with an epsilon of 5...meaning it is not guaranteed to be finding the optimal solution.  I'm trying to balance runtime demands with optimality; I'll gradually reduce the epsilon to 0 and see if anything beats the above solution.&lt;/p&gt;&#xA;" OwnerUserId="7681" LastActivityDate="2016-02-25T17:10:17.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27960" PostTypeId="2" ParentId="27939" CreationDate="2016-02-25T17:55:58.843" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;tricky is where to place the 2x2 square. It can't go on the corner or the sides because it will create a gap. Place it in the center. Experimentally add squares to it's sides. 9-2=7. 7-2=5. Continue fitting all squares until a rectangle is formed. Check widths and heights are the same. Answer:69*61&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19591" LastActivityDate="2016-02-25T17:55:58.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27962" PostTypeId="2" ParentId="27914" CreationDate="2016-02-25T18:36:50.207" Score="-3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The second player to move should win the game.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The second player can start out mirroring the moves of the first player, using the two hands not used by the first player. If this continues, either neither player will win or the first player will win as they are always one move ahead of the second. However, the second player can abandon this strategy when it benefits him, thereby gaining an advantage over player one. This will allow player two to win.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19517" LastActivityDate="2016-02-25T18:36:50.207" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27964" PostTypeId="1" AcceptedAnswerId="27966" CreationDate="2016-02-25T19:16:26.230" Score="7" ViewCount="503" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;If the aliens come to ask for this number, &lt;br&gt; we can give them the answer&lt;/p&gt;&#xA;&lt;p&gt;I am a pyramid, but not the kind Egyptians made&lt;/p&gt;&#xA;&lt;p&gt;My rule is relevant to chemistry&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;What number am I?&lt;/p&gt;&#xA;" OwnerUserId="16322" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-02-25T20:04:13.367" Title="What number can we give to the aliens?" Tags="|riddle|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27966" PostTypeId="2" ParentId="27964" CreationDate="2016-02-25T19:37:28.250" Score="9" Body="&lt;p&gt;You are the number &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 18.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If the aliens come to ask for this number,&lt;br&gt;&#xA;we can give them the answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Ramsey number $R(5,5)$ lies between 43 and 49; its exact value is unknown. Erdős asks us to imagine an alien force, vastly more powerful than us, landing on Earth and demanding the value of $R(5, 5)$ or they will destroy our planet. In that case, he claims, we should marshal all our computers and all our mathematicians and attempt to find the value. But suppose, instead, that they ask for $R(6, 6)$. In that case, he believes, we should attempt to destroy the aliens. See also: &lt;a href=&quot;https://en.wikipedia.org/wiki/Ramsey&amp;#39;s_theorem&quot;&gt;https://en.wikipedia.org/wiki/Ramsey's_theorem&lt;/a&gt;&lt;br&gt;&lt;br&gt;The Ramsey number $\bf{R(4,4)=18}$ is well-known, and can easily be given to the aliens.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a pyramid, but not the kind Egyptians made&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 18 is a pentagonal pyramidal number; see &lt;a href=&quot;https://en.wikipedia.org/wiki/Pentagonal_pyramidal_number&quot;&gt;https://en.wikipedia.org/wiki/Pentagonal_pyramidal_number&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My rule is relevant to chemistry&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The 18-electron rule is a rule in chemistry used primarily for predicting and rationalizing formulae for stable metal complexes, especially organometallic compounds; see also &lt;a href=&quot;https://en.wikipedia.org/wiki/18-Electron_rule&quot;&gt;https://en.wikipedia.org/wiki/18-Electron_rule&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="16322" LastEditDate="2016-02-25T20:04:13.367" LastActivityDate="2016-02-25T20:04:13.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27967" PostTypeId="1" CreationDate="2016-02-25T20:08:38.187" Score="0" ViewCount="3874" Body="&lt;p&gt;Is it possible to make the number 17 using only 2, 3, and 4?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the numbers only once and any operation, including factorials and other more advanced math. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And it would have to be exactly 17. &lt;/p&gt;&#xA;" OwnerUserId="19599" LastEditorUserId="12408" LastEditDate="2016-03-01T00:59:06.650" LastActivityDate="2016-03-03T17:36:10.387" Title="Make the number 17 using only 2, 3, and 4" Tags="|mathematics|formation-of-numbers|" AnswerCount="11" CommentCount="1" FavoriteCount="0" ClosedDate="2016-03-02T20:46:33.520" ContentLicense="CC BY-SA 3.0" />
  <row Id="27968" PostTypeId="2" ParentId="27967" CreationDate="2016-02-25T20:18:31.030" Score="3" Body="&lt;p&gt;Easily: $$4 + 3 + 2 + 4 + 4 = 17$$&lt;/p&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2016-02-25T20:18:31.030" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27969" PostTypeId="2" ParentId="27967" CreationDate="2016-02-25T20:26:01.970" Score="29" Body="&lt;p&gt;How about the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 34/2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9664" LastEditorUserId="16322" LastEditDate="2016-02-25T20:32:55.517" LastActivityDate="2016-02-25T20:32:55.517" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27970" PostTypeId="2" ParentId="27967" CreationDate="2016-02-25T20:31:55.993" Score="16" Body="&lt;p&gt;Without concatenating and/or reusing digits:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $2^4 + \lfloor \sqrt{3} \rfloor$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17147" LastActivityDate="2016-02-25T20:31:55.993" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27971" PostTypeId="2" ParentId="27967" CreationDate="2016-02-25T20:39:38.673" Score="23" Body="&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: This answer doesn't fit within the updated requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here you go:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;pre&gt;&#xA;      22     33333333&#xA;    2222           44&#xA;       2          44&#xA;       2          4&#xA;       2         44&#xA;       2         4 &#xA;       2         4 &#xA;   222222222     4 &lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2016-02-26T15:16:39.560" LastActivityDate="2016-02-26T15:16:39.560" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27973" PostTypeId="2" ParentId="27875" CreationDate="2016-02-25T20:56:29.293" Score="4" Body="&lt;p&gt;I have found the minimum set of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;24&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;pre&gt;AAAA AAAB ABBC ACCD&lt;br&gt;ACDD ADBC BABD BBCA&lt;br&gt;BBDB BCAC BDCB BDDA&lt;br&gt;CACC CADC CBAD CCBA&lt;br&gt;CCBB CDAD DABD DBCB&lt;br&gt;DBDA DCAC DDCA DDDB&lt;br&gt;&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the optimal solution (actually there are 4 solutions but they are mirrors so I share with AAAA version with you: BBBB CCCC and DDDD versions are possible too), found by using computer programming with my own algorithm that I had mentioned on my deleted post: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Take a word (from the table all possible 256 words), find the maximum number of matches with 13 possible words by changing one letter from A to D from each digit.&lt;/li&gt;&#xA;&lt;li&gt;Try to find matches in the Word Pool and count how many matches there are with all words,&lt;/li&gt;&#xA;&lt;li&gt;Choose a word randomly which has the highest amount of matches,&lt;/li&gt;&#xA;&lt;li&gt;Remove those matched words from the word pool with the word you chose randomly,&lt;/li&gt;&#xA;&lt;li&gt;Go back to the beginning and go on until you do not have any word left in your word pool and count your tries.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I wish I wanted to share the code, but it is too messy, but the algorithm is something like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result, I got&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;24&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is found after 8 hours of running my coding... I wanted to explain the result but this is purely found by computer.&lt;/p&gt;&#xA;" OwnerUserId="18514" LastEditorUserId="18514" LastEditDate="2016-02-27T09:26:41.053" LastActivityDate="2016-02-27T09:26:41.053" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27974" PostTypeId="2" ParentId="27967" CreationDate="2016-02-25T20:57:13.147" Score="6" Body="&lt;p&gt;Using each number exactly twice&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $2^4 + 2^{4*(3-3)}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19602" LastActivityDate="2016-02-25T20:57:13.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27975" PostTypeId="2" ParentId="27728" CreationDate="2016-02-25T21:52:12.093" Score="3" Body="&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(x,y,z) = (1,1,4)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's my reasoning (following the reasoning of Summo and Prodo).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Half of Summo's statement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Summo can't tell the numbers based on the sum, so that means they can't sum to either $3$ or $4$. Both of those would have unique digits ($1,1,1$ and $1,1,2$ respectively). That means $S \geq 5$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rest of Summo's statement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The rest of Summo's statement tells us that based on his sum, $P$ could be greater than $S$ for exactly one combo of $x,y,z$, while there may be multiple ways for $P$ to be less than $S$. This rules out $S=5$, since for both solutions ($1,2,2$ and $1,1,3$) we get $P$ smaller than $S$. $S$ could be $6$, since we have possible solutions $(1,1,4), (1,2,3) \text{ and } (2,2,2)$, where $1$ of those has product greater than $S$.&lt;br&gt;&lt;br&gt; I believe this means that $S=6$, since for any larger number there would be multiple products larger that $S$. Take $S=7$, for example. The possible solutions are $(1,1,5),(1,2,4),(1,3,3),(2,2,3)$. A number of those have products larger than $7$, and that will only get worse as $S$ gets higher.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Getting to Prodo:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Prodo now knows that $S=6$. She also says that her product is less than $S$, so $P\leq5$. For $S=6$, the only solution where $P\leq5$ is $(1,1,4)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2016-02-25T21:52:12.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27976" PostTypeId="1" AcceptedAnswerId="27978" CreationDate="2016-02-25T22:45:46.010" Score="7" ViewCount="1320" Body="&lt;p&gt;You're a cosmonaut dedicated to exploring and understanding strange new worlds. The latest world you've arrived on has a species with a very strange mating dance. After careful observation, you determine the rules for this dance are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;All aliens on the planet are ordered by height and given a number. Number 1 for the shortest, number 2 for the next shortest, etc. There are no ties.&lt;/li&gt;&#xA;&lt;li&gt;The aliens get into a line in a randomized order.&lt;/li&gt;&#xA;&lt;li&gt;The alien at the front of the line checks its number, finds the alien at that place in line, and trades place with them. Ex: if the front alien has number 5, it trades places with the fifth alien in line (itself being the first). They repeat this step indefinitely.&lt;/li&gt;&#xA;&lt;li&gt;If the shortest alien is at the front of the line, the dance ends and the orgy begins. These aliens will not stop dancing until this happens.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You start to wonder: how can this species exist? Surely given enough time, the random arrangement would result in a never-ending dance and, unable to mate, the species would die off. After pondering and observing a while, inspiration strikes. Now satisfied that this is impossible, you leave the planet in search of more interesting puzzles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did you know this species can't die off?&lt;/p&gt;&#xA;" OwnerUserId="18103" LastActivityDate="2016-02-26T09:14:29.467" Title="A Strange Planet with a Strange Dance" Tags="|mathematics|logical-deduction|computer-science|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27977" PostTypeId="2" ParentId="27976" CreationDate="2016-02-25T23:02:23.763" Score="2" Body="&lt;p&gt;The number of moves it takes in order for the shortest alien to be at the front of the line is O(n). Therefore, they keep dancing until enough aliens die off that the dance converges in a reasonable time. &lt;/p&gt;&#xA;" OwnerUserId="18821" LastActivityDate="2016-02-25T23:02:23.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27978" PostTypeId="2" ParentId="27976" CreationDate="2016-02-25T23:08:43.263" Score="28" Body="&lt;p&gt;Note that once an alien is in the spot matching its number, it will never swap with another alien again. Every step, one alien moves into its spot, so the number of aliens in the right spot increases by 1 every step. This number can't exceed the number of aliens, which is finite, so there must be a finite number of steps.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2016-02-25T23:08:43.263" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27982" PostTypeId="1" AcceptedAnswerId="27983" CreationDate="2016-02-26T03:23:18.197" Score="31" ViewCount="5440" Body="&lt;p&gt;Starting at 1 (which is 4 - $\sqrt4$ - 4/4), how many consecutive integers can you make using exactly four instances of the digit '4'?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basic rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Any operator symbol is &quot;free&quot;.&lt;/li&gt;&#xA;&lt;li&gt;Any printed '4' counts toward your four '4's.  &lt;/li&gt;&#xA;&lt;li&gt;Other digits are NOT allowed.&lt;/li&gt;&#xA;&lt;li&gt;Other characters like letters or miscellaneous punctuation are NOT allowed, unless you can provide some citation of an accepted mathematical definition.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Allowed operators (non-exhaustive):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;+ Addition&lt;/li&gt;&#xA;&lt;li&gt;- Subtraction or Negation&lt;/li&gt;&#xA;&lt;li&gt;* Multiplication&lt;/li&gt;&#xA;&lt;li&gt;/ Division&lt;/li&gt;&#xA;&lt;li&gt;$\sqrt4$ Square root (ignore the implicit '2' there)&lt;/li&gt;&#xA;&lt;li&gt;$\sqrt[4]4$ Radical (in this case you've used two '4's)&lt;/li&gt;&#xA;&lt;li&gt;^ Exponentiation&lt;/li&gt;&#xA;&lt;li&gt;! Factorial&lt;/li&gt;&#xA;&lt;li&gt;? Terminal function (4? = 4 + 3 + 2 + 1)&lt;/li&gt;&#xA;&lt;li&gt;44 Concatenation (which in this case consumes two '4's)&lt;/li&gt;&#xA;&lt;li&gt;|4| Absolute value&lt;/li&gt;&#xA;&lt;li&gt;. Decimal point&lt;/li&gt;&#xA;&lt;li&gt;If you can find a way to use calc, trig, matrices, whatever, by all means please do&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="19608" LastEditorUserId="69582" LastEditDate="2021-09-01T19:42:30.483" LastActivityDate="2021-09-01T19:42:30.483" Title="How many consecutive positive integers can you make using exactly four instances of the digit '4'?" Tags="|mathematics|calculation-puzzle|formation-of-numbers|" AnswerCount="9" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="27983" PostTypeId="2" ParentId="27982" CreationDate="2016-02-26T03:34:40.230" Score="78" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;All of them!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;For every positive integer $n$,$$\underbrace{\sec\arctan}_{n^2-1\text{ times}}\,\frac{44}{44}=n$$otherwise written as$$\sec\arctan(\sec\arctan(...\sec\arctan(\frac{44}{44})))$$so all positive integers can be made with four fours. (Idea from &lt;a href=&quot;https://puzzling.stackexchange.com/a/18749/12408&quot;&gt;this answer&lt;/a&gt;.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2016-02-26T16:45:51.213" CommentCount="24" ContentLicense="CC BY-SA 3.0" />
  <row Id="27984" PostTypeId="2" ParentId="6321" CreationDate="2016-02-26T04:19:57.930" Score="2" Body="&lt;p&gt;&lt;strong&gt;15 Letters&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note: the question doesn't specify explicitly whether the calculator needs to be inverted (although I think it intended it to be).  Also, I'm going to assume my calculator is set to decimal, as it often isn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on writing the alphabet, here are the letters I think I can reasonably use rightsideup:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a=0, b=6/8, d=0, e=3, g=9, i=1, l=1, o=0, q=9, s=5, z=2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note I allow a backwards &quot;E&quot; because of the cultural entrenchment of leetspeak.  Upsidedown:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a=0, b=8/9, d=0, e=3, g=6, h=4, i=1, l=1/7, o=0, q=6, s=5, t=7, z=2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since it seems that the set of upsidedown letters is a strict superset of the rightsideup ones, the ambiguity is irrelevant.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I have a script I wrote for bruteforce dictionary searches with regexes.  It uses a dictionary stored on my HDD, so there was no Googling, as required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The longest words &lt;code&gt;^[abdegiloqsz]{11,99}$&lt;/code&gt; returns are &lt;code&gt;babesiosises&lt;/code&gt;, &lt;code&gt;bibliologies&lt;/code&gt;, and &lt;code&gt;glossolalias&lt;/code&gt;, all 12 letters.  For the upsidedown words, &lt;code&gt;^[abdeghiloqstz]{15,99}$&lt;/code&gt; gives up &lt;code&gt;digestibilities&lt;/code&gt;, which is 15 letters.  Therefore, I claim:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;digestibilities = 53171719175361.0&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;N.B. the &lt;code&gt;i&lt;/code&gt;, &lt;code&gt;l&lt;/code&gt;, and &lt;code&gt;t&lt;/code&gt; are mapped by my scheme into &lt;code&gt;1&lt;/code&gt; and &lt;code&gt;7&lt;/code&gt;.  By the pigeonhole principle, there's a conflict.  However, if the current answer uses a &lt;code&gt;9&lt;/code&gt; for an &lt;code&gt;a&lt;/code&gt; (backwards character), then think probably I should be allowed a &lt;code&gt;4&lt;/code&gt; for a &lt;code&gt;t&lt;/code&gt; (one extra stroke).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N.B. if you don't like &lt;code&gt;digestibilities&lt;/code&gt;, here are some 14-letter words: &lt;code&gt;diabetologists&lt;/code&gt;, &lt;code&gt;disestablished&lt;/code&gt;, &lt;code&gt;disestablishes&lt;/code&gt;, (and more dubiously: &lt;code&gt;habitabilities&lt;/code&gt;, &lt;code&gt;illegibilities&lt;/code&gt;, &lt;code&gt;dilatabilities&lt;/code&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N.B. If anyone sees how more letters could be added (I'll admit I wasn't very thorough) or wants to give me a more comprehensive dictionary (ASCII; one word per line), I'd be happy to run the search again.&lt;/p&gt;&#xA;" OwnerUserId="4827" LastActivityDate="2016-02-26T04:19:57.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27986" PostTypeId="2" ParentId="27919" CreationDate="2016-02-26T06:32:45.390" Score="10" Body="&lt;p&gt;The problem in &lt;a href=&quot;https://en.wikipedia.org/wiki/Graph_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;graph theoretic&lt;/a&gt; terminology is to find the minimal number of &lt;a href=&quot;https://en.wikipedia.org/wiki/Vertex_%28graph_theory%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;vertices&lt;/a&gt; required for a &lt;a href=&quot;https://en.wikipedia.org/wiki/Unit_distance_graph&quot; rel=&quot;nofollow noreferrer&quot;&gt;unit distance graph&lt;/a&gt; with &lt;a href=&quot;https://en.wikipedia.org/wiki/Graph_coloring#Definition_and_terminology&quot; rel=&quot;nofollow noreferrer&quot;&gt;chromatic number&lt;/a&gt; of $4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Brooks&amp;#39;_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brokes' theorem&lt;/a&gt; states that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Any &lt;a href=&quot;https://en.wikipedia.org/wiki/Connectivity_%28graph_theory%29#Connected_graph&quot; rel=&quot;nofollow noreferrer&quot;&gt;connected&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Graph_%28discrete_mathematics%29#Undirected_graph&quot; rel=&quot;nofollow noreferrer&quot;&gt;undirected graph&lt;/a&gt; of &lt;a href=&quot;https://en.wikipedia.org/wiki/Degree_%28graph_theory%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;degree&lt;/a&gt; $d$ has chromatic number $k&amp;lt;=d$ unless the graph is either &lt;a href=&quot;https://en.wikipedia.org/wiki/Complete_graph&quot; rel=&quot;nofollow noreferrer&quot;&gt;complete&lt;/a&gt; or an &lt;a href=&quot;https://en.wikipedia.org/wiki/Cycle_graph&quot; rel=&quot;nofollow noreferrer&quot;&gt;odd cycle&lt;/a&gt; when $k=d+1$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unit distance graphs are already undirected. Furthermore any minimal solution will be connected since:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;unconnected vertices may be any colour and thus do not increase the chromatic number; and&lt;/li&gt;&#xA;&lt;li&gt;the chromatic number of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Graph_operations#Binary_operations&quot; rel=&quot;nofollow noreferrer&quot;&gt;disjoint union&lt;/a&gt; of two connected graphs is equal to the maximum chromatic number of those two graphs.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All odd cycle graphs are unit distance graphs (odd-sided, regular polygons of side $1$). However odd cycles always have degree $2$ and hence, by Brokes' theorem, a chromatic number of $3$ so no odd cycle is a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All even cycle graphs are unit distance graphs (even-sided, regular polygons of side $1$). However all even cycles have degree $2$ and hence a chromatic number less than or equal to $3$. In fact they have chromatic number $2$ which is easy to see: colour any vertex $colour 1$ and follow an edge to the next vetex and colour it $colour 2$, and continue following edges around the polygon alternating colours as you go until all vertices are coloured; the last vertex will be coloured $colour 2$ and hence will not conflict via the ultimate edge leading back to the first coloured vertex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can construct the complete graph containing $3$ vertices, $K_3$, as a unit distance graph - it is an equilateral triangle of side $1$, this also has degree $2$ and a chromatic number of $3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To construct a complete unit distance graph with more than $3$ vertices we would need to first find a point on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Plane_%28geometry%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Euclidean plane&lt;/a&gt; that lies at a distance of $1$ from all three of the vertices of an equilateral triangle of side $1$, but there is no such point - so there are no complete unit distance graphs with more than $3$ vertices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore any solution must have degree $&amp;gt;= 4$  (at least one vertex must have at least four neighbours).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any neighbours of a vertex $P$ lie on a &lt;a href=&quot;https://en.wikipedia.org/wiki/Unit_circle&quot; rel=&quot;nofollow noreferrer&quot;&gt;unit circle&lt;/a&gt; centred at $P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now not only know that the solution must be at least $5$, but that any valid instance will contain a vertex $P$ at the centre of a unit circle upon which a further four vertices $Q, R, S, T$ exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way $5$ could be the solution is if such an instance, with no additional vertices, exists with a chromatic number of $4$. We can check the few cases as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If no pair of $Q, R, S, T$ are $1$ apart the chromatic number would be $2$&lt;br&gt;- $P$ being $colour 1$ and $Q, R, S, T$ being $colour 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If only one pair of $Q, R, S, T$ are $1$ apart, say $Q$ and $R$, then the graph would be an &lt;a href=&quot;https://en.wikipedia.org/wiki/Equilateral_triangle&quot; rel=&quot;nofollow noreferrer&quot;&gt;equilateral triangle&lt;/a&gt; of side $1$ $PQR$ and two spokes $PS$ and $PT$, which has a chromatic number of $3$&lt;br&gt;- $P$ being $colour 1$, $Q$ being $colour 2$, $R$ being $colour 3$, and $S$ and $T$ being free to be either $colour 2$ or $colour 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If exactly two pairs of $Q, R, S, T$ are $1$ apart then there are two cases (ignoring label order isomorphisms):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$Q$ and $R$ are one pair and $S$ and $T$ are the other; or&lt;/li&gt;&#xA;&lt;li&gt;$Q$ and $R$ are one pair and $R$ and $S$ are the other.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Case 1 is two vertex-adjoined equilateral triangles $PQR$ and $PST$, which has a chromatic number of $3$&lt;br&gt;- $P$ being $colour 1$, $Q$ and $S$ (or $Q$ and $T$) being $colour 2$, and $R$ and $T$ (or $R$ and $S$) being $colour 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 2 is a unit &lt;a href=&quot;https://en.wikipedia.org/wiki/Polyiamond&quot; rel=&quot;nofollow noreferrer&quot;&gt;diamond&lt;/a&gt; $PQRS$ and a spoke $PT$ from one of the corners on the diagonal of length $1$, which has a chromatic number of $3$&lt;br&gt;- $P$ being $colour 1$, $Q$ and $S$ being $colour 2$, $R$ being $colour 3$, and $T$ being free to be either $colour 2$ or $colour 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If exactly three pairs of $Q, R, S, T$ are $1$ apart the only formation is the unit &lt;a href=&quot;https://en.wikipedia.org/wiki/Polyiamond&quot; rel=&quot;nofollow noreferrer&quot;&gt;triamond&lt;/a&gt; $QRST$ with $P$ at the only vertex of the central equilateral triangle that is not a corner of the triamond, which has a chromatic number of $3$&lt;br&gt;- $P$ being $colour 1$, $Q$ and $S$ being $colour 2$, and $R$ and $T$ being $colour 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order for $6$ to be the solution we would need to extend any of these instances by one vertex $U$ to yield a graph of chromatic number $4$ by forcing $Q, R, S, T$ to require three colours. This can only be done by placing $U$ at an unoccupied point $1$ away from two vertices already forced to be of equal colour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If no pair or one pair of $Q, R, S, T$ are $1$ apart no two are already forced to be of the same colour so it will not work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If two pairs of $Q, R, S, T$ are $1$ apart in case 1 we may swap the colour groups as indicated by the parenthesised options, hence no two are already actually forced to be of the same colour so it also wont work; in case 2 $Q$ and $S$ are forced to be the same colour but the only points one unit away from both are already occupied ($P$ and $R$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If exactly three pairs of $Q, R, S, T$ are $1$ apart there are two pairs which are already forced to be of the same colour:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$Q$ and $S$, but the only points one unit away from both are already occupied ($P$ and $R$); and&lt;/li&gt;&#xA;&lt;li&gt;$R$ and $T$, but the only points one unit away from both are already occupied ($P$ and $S$).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thus the solution is $&amp;gt; 6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since an example of a solution with $7$ vertices is provided in &lt;a href=&quot;https://puzzling.stackexchange.com/a/27922/16717&quot;&gt;ralphmerridew's answer&lt;/a&gt; and is illustrated below we can now say that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The solution is $7$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;in this example $P$ would be $A$ and $Q, R, S, T$ would be $B, E, D, G$ (in any order) this unit distance graph is known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Moser_spindle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Moser spindle&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/a/27922/16717&quot;&gt;ralphmerridew's solution&lt;/a&gt; in visual form - the three colours may be rotated, and additionally either or both of (BD) &amp;amp; (EG) may permute to give the same &quot;conflict&quot; on (CF):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/5/50/3_Colour_Flowers_Problem.svg&quot; alt=&quot;diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we label $A$ as the origin, with the unit vectors pointing right and down for convenience, and use $\theta$ to name the angle of rotation to transform the unit diamond* $ABCD$ to $AEFG$ then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C = (\sqrt{3}\cos(30), \sqrt{3}\sin(30)) = (\frac32, \sqrt{\frac34})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F = (\sqrt{3}\cos(30+\theta), \sqrt{3}\sin(30 + \theta))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for $C$ and $F$ to be one unit apart:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt{(\frac32-\sqrt{3}\cos(30+\theta))^2+(\sqrt{3}\sin(30+\theta)-\sqrt{\frac34})^2}=1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow(\frac32-\sqrt{3}\cos(30+\theta))^2+(\sqrt{3}\sin(30+\theta)-\sqrt{\frac34})^2=1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow\frac94-\sqrt{27}\cos(30+\theta)+3\cos^2(30+\theta)+3\sin^2(30+\theta)-3\sin(30+\theta)+\frac34 = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow\frac94-\sqrt{27}\cos(30+\theta)+3-3\sin(30+\theta)+\frac34 = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow\sqrt{27}\cos(30+\theta)+3\sin(30+\theta) = 5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow\sqrt{27}(\cos(30)\cos(\theta)-\sin(30)\sin(\theta))+3(\sin(30)\cos(\theta)+\cos(30)\sin(\theta)) = 5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow\sqrt{27}(\sqrt{\frac34}\cos(\theta)-\frac12\sin(\theta))+3(\frac12\cos(\theta)+\sqrt{\frac34}\sin(\theta)) = 5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow\frac92\cos(\theta)-\sqrt{\frac{27}4}\sin(\theta)+\frac32\cos(\theta)+\sqrt{\frac{27}4}\sin(\theta) = 5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow\cos(\theta) = \frac56$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$AEFG$ is a rotation about $A$ of $ABCD$ by $\theta=33.557$ degrees $= 0.58569$ radians (5sf)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Thus angle $BAG = 93.557$ degrees (5sf)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;*two unit equilateral triangles adjoined by one side forming a rhombus with 60 and 120 degree angles&lt;/p&gt;&#xA;" OwnerUserId="16717" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2016-02-28T14:52:33.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27987" PostTypeId="2" ParentId="27982" CreationDate="2016-02-26T07:57:38.680" Score="6" Body="&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/a/27983/18129&quot;&gt;@f''&amp;hairsp;&amp;lsquo;s solution&lt;/a&gt;&#xA;can be generalized for any positive even number of 4s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about odd numbers of 4s?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$ \begin{matrix} \underbrace{ \sec\arctan }_{ n^2-1 ~ \text{times} } \, \biggl( \cdots \root\of{\root\of{ \surd 4 }\,} \, \biggr) \end{matrix} = n $$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Three 4s to get 4 (or whatever number of nested &amp;radic;&amp;hairsp;s):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$ { \ln \ln 4 - \ln \ln \root\of{\root\of{\root\of{ \surd 4 }\,}\,} \over \ln \root\of 4 } = 4 $$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18129" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2016-02-26T18:28:10.553" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27988" PostTypeId="2" ParentId="27976" CreationDate="2016-02-26T09:09:23.457" Score="1" Body="&lt;p&gt;Obviously, the aliens can cheat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another solution is that the population is very small, or gets very small during the mating dance (aliens die during the dance and thus get removed from the line). It also makes sense to assume that the aliens are hermaphrodites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The line also does not need to be straight, which can reduce the travel time from one position to another.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: this assumes that &quot;aliens&quot; mean the inhabitants of the planet in question, not aliens from the inhabitants' point of view (that would be you, and the orgy would start when the shortest visiting cosmonaut arrived at the first position - not necessarily participating in the orgy).&lt;/p&gt;&#xA;" OwnerUserId="18915" LastEditorUserId="18915" LastEditDate="2016-02-26T09:14:29.467" LastActivityDate="2016-02-26T09:14:29.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27989" PostTypeId="1" AcceptedAnswerId="28004" CreationDate="2016-02-26T09:53:49.893" Score="36" ViewCount="1498" Body="&lt;p&gt;Xavier and Oliver were playing tic-tac-toe on the Titanic using a wooden board and metal X's and O's. As the stern of the ship sank lower and lower &#xA;they found, to their annoyance, that the pieces began to slide from left to right. After some thought they decided to turn the situation to their &#xA;advantage. They took turns holding the board level and added the following rules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the start of any player's turn he may allow the board to tilt causing all of the pieces to slide one square to the right. Any pieces leaving the &#xA;3x3 grid are removed. The player may then play a normal tic-tac-toe move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A player may not play two successive moves on the same square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A player may not allow the board to tilt if doing so will cause one or more of his opponent's pieces to be removed unless one or more of his own pieces &#xA;will also be removed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify, a player may tilt the board if a) one or more of his own pieces is in the right hand column or b) if there are no pieces in the right hand column.&#xA;Tilting only happens in the one direction: left to right. Tilting always takes place during a player's turn before he has placed his piece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming best play, can either player force a win?&lt;/p&gt;&#xA;" OwnerUserId="18924" LastEditorUserId="17482" LastEditDate="2016-02-26T14:30:12.877" LastActivityDate="2016-03-01T15:17:55.577" Title="Titanic Tic-Tac-Toe" Tags="|strategy|game|tic-tac-toe|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27991" PostTypeId="2" ParentId="27982" CreationDate="2016-02-26T10:32:06.067" Score="8" Body="&lt;p&gt;Solution for any odd number of fours different from one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\underbrace{\sec\arctan}_{(n+4)^2-1\text{ times}}\,\frac{44 \cdots}{44 \cdots} - 4=n$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19617" LastActivityDate="2016-02-26T10:32:06.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27992" PostTypeId="2" ParentId="15839" CreationDate="2016-02-26T11:11:19.760" Score="11" Body="&lt;h1&gt;n = 34&lt;/h1&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://legionmammal978.github.io/post/?dictWord=extemporizing&amp;amp;https://scrabble.hasbro.com/en-us/tools&quot; rel=&quot;noreferrer&quot;&gt;Extemporizing&lt;/a&gt; &lt;a href=&quot;http://legionmammal978.github.io/post/?dictWord=demythologizers&amp;amp;https://scrabble.hasbro.com/en-us/tools&quot; rel=&quot;noreferrer&quot;&gt;demythologizers&lt;/a&gt; &lt;a href=&quot;http://legionmammal978.github.io/post/?dictWord=demythologized&amp;amp;https://scrabble.hasbro.com/en-us/tools&quot; rel=&quot;noreferrer&quot;&gt;demythologized&lt;/a&gt; &lt;a href=&quot;http://legionmammal978.github.io/post/?dictWord=overhomogenized&amp;amp;https://scrabble.hasbro.com/en-us/tools&quot; rel=&quot;noreferrer&quot;&gt;overhomogenized&lt;/a&gt;, &lt;a href=&quot;http://legionmammal978.github.io/post/?dictWord=jackhammering&amp;amp;https://scrabble.hasbro.com/en-us/tools&quot; rel=&quot;noreferrer&quot;&gt;jackhammering&lt;/a&gt; &lt;a href=&quot;http://legionmammal978.github.io/post/?dictWord=azidothymidines&amp;amp;https://scrabble.hasbro.com/en-us/tools&quot; rel=&quot;noreferrer&quot;&gt;azidothymidines&lt;/a&gt;; &lt;a href=&quot;http://legionmammal978.github.io/post/?dictWord=puzzlingly&amp;amp;https://scrabble.hasbro.com/en-us/tools&quot; rel=&quot;noreferrer&quot;&gt;puzzlingly&lt;/a&gt; &lt;a href=&quot;http://legionmammal978.github.io/post/?dictWord=schizophrenic&amp;amp;https://scrabble.hasbro.com/en-us/tools&quot; rel=&quot;noreferrer&quot;&gt;schizophrenic&lt;/a&gt; &lt;a href=&quot;http://legionmammal978.github.io/post/?dictWord=jazzmen&amp;amp;https://scrabble.hasbro.com/en-us/tools&quot; rel=&quot;noreferrer&quot;&gt;jazzmen&lt;/a&gt; &lt;a href=&quot;http://legionmammal978.github.io/post/?dictWord=diphthongizing&amp;amp;https://scrabble.hasbro.com/en-us/tools&quot; rel=&quot;noreferrer&quot;&gt;diphthongizing&lt;/a&gt; &lt;a href=&quot;http://legionmammal978.github.io/post/?dictWord=hyperexcitement&amp;amp;https://scrabble.hasbro.com/en-us/tools&quot; rel=&quot;noreferrer&quot;&gt;hyperexcitement&lt;/a&gt; &lt;a href=&quot;http://legionmammal978.github.io/post/?dictWord=drizzlingly&amp;amp;https://scrabble.hasbro.com/en-us/tools&quot; rel=&quot;noreferrer&quot;&gt;drizzlingly&lt;/a&gt; &lt;a href=&quot;http://legionmammal978.github.io/post/?dictWord=overemphasized&amp;amp;https://scrabble.hasbro.com/en-us/tools&quot; rel=&quot;noreferrer&quot;&gt;overemphasized&lt;/a&gt; &lt;a href=&quot;http://legionmammal978.github.io/post/?dictWord=jabberwocky&amp;amp;https://scrabble.hasbro.com/en-us/tools&quot; rel=&quot;noreferrer&quot;&gt;jabberwocky&lt;/a&gt; &lt;a href=&quot;http://legionmammal978.github.io/post/?dictWord=jazziness&amp;amp;https://scrabble.hasbro.com/en-us/tools&quot; rel=&quot;noreferrer&quot;&gt;jazziness&lt;/a&gt;, &lt;a href=&quot;http://legionmammal978.github.io/post/?dictWord=hypothesizing&amp;amp;https://scrabble.hasbro.com/en-us/tools&quot; rel=&quot;noreferrer&quot;&gt;hypothesizing&lt;/a&gt; &lt;a href=&quot;http://legionmammal978.github.io/post/?dictWord=philosophizing&amp;amp;https://scrabble.hasbro.com/en-us/tools&quot; rel=&quot;noreferrer&quot;&gt;philosophizing&lt;/a&gt; &lt;a href=&quot;http://legionmammal978.github.io/post/?dictWord=jazzbos&amp;amp;https://scrabble.hasbro.com/en-us/tools&quot; rel=&quot;noreferrer&quot;&gt;jazzbos&lt;/a&gt;; &lt;a href=&quot;http://legionmammal978.github.io/post/?dictWord=quizzer&amp;amp;https://scrabble.hasbro.com/en-us/tools&quot; rel=&quot;noreferrer&quot;&gt;quizzer&lt;/a&gt; &lt;a href=&quot;http://legionmammal978.github.io/post/?dictWord=psychochemicals&amp;amp;https://scrabble.hasbro.com/en-us/tools&quot; rel=&quot;noreferrer&quot;&gt;psychochemicals&lt;/a&gt; &lt;a href=&quot;http://legionmammal978.github.io/post/?dictWord=hyperpolarized&amp;amp;https://scrabble.hasbro.com/en-us/tools&quot; rel=&quot;noreferrer&quot;&gt;hyperpolarized&lt;/a&gt;, &lt;a href=&quot;http://legionmammal978.github.io/post/?dictWord=contextualizing&amp;amp;https://scrabble.hasbro.com/en-us/tools&quot; rel=&quot;noreferrer&quot;&gt;contextualizing&lt;/a&gt; &lt;a href=&quot;http://legionmammal978.github.io/post/?dictWord=physicochemical&amp;amp;https://scrabble.hasbro.com/en-us/tools&quot; rel=&quot;noreferrer&quot;&gt;physicochemical&lt;/a&gt; &lt;a href=&quot;http://legionmammal978.github.io/post/?dictWord=jazzman&amp;amp;https://scrabble.hasbro.com/en-us/tools&quot; rel=&quot;noreferrer&quot;&gt;jazzman&lt;/a&gt; &lt;a href=&quot;http://legionmammal978.github.io/post/?dictWord=phosphatization&amp;amp;https://scrabble.hasbro.com/en-us/tools&quot; rel=&quot;noreferrer&quot;&gt;phosphatization&lt;/a&gt;; &lt;a href=&quot;http://legionmammal978.github.io/post/?dictWord=remythologizing&amp;amp;https://scrabble.hasbro.com/en-us/tools&quot; rel=&quot;noreferrer&quot;&gt;remythologizing&lt;/a&gt; &lt;a href=&quot;http://legionmammal978.github.io/post/?dictWord=chazzanim&amp;amp;https://scrabble.hasbro.com/en-us/tools&quot; rel=&quot;noreferrer&quot;&gt;chazzanim&lt;/a&gt; &lt;a href=&quot;http://legionmammal978.github.io/post/?dictWord=blackjacking&amp;amp;https://scrabble.hasbro.com/en-us/tools&quot; rel=&quot;noreferrer&quot;&gt;blackjacking&lt;/a&gt; &lt;a href=&quot;http://legionmammal978.github.io/post/?dictWord=huzzahing&amp;amp;https://scrabble.hasbro.com/en-us/tools&quot; rel=&quot;noreferrer&quot;&gt;huzzahing&lt;/a&gt; &lt;a href=&quot;http://legionmammal978.github.io/post/?dictWord=quizzes&amp;amp;https://scrabble.hasbro.com/en-us/tools&quot; rel=&quot;noreferrer&quot;&gt;quizzes&lt;/a&gt; &lt;a href=&quot;http://legionmammal978.github.io/post/?dictWord=psychologized&amp;amp;https://scrabble.hasbro.com/en-us/tools&quot; rel=&quot;noreferrer&quot;&gt;psychologized&lt;/a&gt; &lt;a href=&quot;http://legionmammal978.github.io/post/?dictWord=nephrectomizing&amp;amp;https://scrabble.hasbro.com/en-us/tools&quot; rel=&quot;noreferrer&quot;&gt;nephrectomizing&lt;/a&gt;, &lt;a href=&quot;http://legionmammal978.github.io/post/?dictWord=zizzed&amp;amp;https://scrabble.hasbro.com/en-us/tools&quot; rel=&quot;noreferrer&quot;&gt;zizzed&lt;/a&gt; &lt;a href=&quot;http://legionmammal978.github.io/post/?dictWord=pzazz&amp;amp;https://scrabble.hasbro.com/en-us/tools&quot; rel=&quot;noreferrer&quot;&gt;pzazz&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I finally did this. The sentence is divided into four parts, and is quite... musical.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Ones who make ideas less mythical (and improvise along the way) did so to drugs used to treat AIDS that are overly uniform and operate jackhammers.&lt;/li&gt;&#xA;&lt;li&gt;Jazz musicians confusingly diagnosed with schizophrenia who turn extreme excitement into a single syllable overemphasized (with light rain) characteristics of jazz with meaningless lyrics. Doing this, they expected to make jazz devotees act more serious.&lt;/li&gt;&#xA;&lt;li&gt;Now, chemicals affecting behavior belonging to ones who write tests increased in potential difference. This added context to the physical and chemical treatment of jazz musicians using phosphates.&lt;/li&gt;&#xA;&lt;li&gt;Leaders in prayer during Jewish services who create myths again and strike cheering quizzes with blackjacks psychologically interpreted the quality of excitement or attractiveness that surgically removes kidneys and makes buzzing sounds.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I used SOWPODS as my wordlist, with all words cross-checked against Hasbro's site. This is the longest possible sentence I could find.&lt;/p&gt;&#xA;" OwnerUserId="11098" LastEditorUserId="11098" LastEditDate="2018-06-09T02:03:56.507" LastActivityDate="2018-06-09T02:03:56.507" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27994" PostTypeId="2" ParentId="27938" CreationDate="2016-02-26T12:34:44.720" Score="6" Body="&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/a/27946/16717&quot;&gt;hmmn's answer&lt;/a&gt; is certainly correct, here is, I believe, a more formal approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A number, $x_0=1$, is displayed.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;we may press $D-$ and produce $2x_0-1=1$, or&lt;/li&gt;&#xA;&lt;li&gt;we may press $D+$ and produce $2x_0+1=3$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These are both odd integers, and odd integers may always be expressed as $2a+1$ where $a$ is an integer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fairly obviously, as soon as we press $D+$ we escape from $1$ and may never not increase the number displayed.&lt;br&gt;&lt;br&gt;Thus the only way to achieve a specific number with a specific quantity of button presses is to reach it with the final button press (e.g. $1$ only results from a sequence entirely of $D-$ and $3$ from a sequence of all $D-$ and a final $D+$).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If $x_n$ is $2a+1$ then $x_{n+1}$ is either&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$2(2a+1)-1$, or&lt;/li&gt;&#xA;&lt;li&gt;$2(2a+1)+1$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Two different odd integers from which we may branch again, $D+$ producing the bigger of the two.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The biggest number we may achieve with $m$ presses would be &lt;br&gt;&lt;br&gt;$$1 +\sum_{i=1}^{m}{2^i}=2^{m+1}-1$$ by only pressing $D+$.&lt;br&gt;&lt;br&gt;Thus with $m$ presses we may yield $2^m$ odd integers between $1$ and $2^{m+1}-1$ inclusive.&lt;br&gt;&lt;br&gt;There are exactly $\frac{2^{m+1}-1+1}{2}=\frac{2^{m+1}}{2}=2^m$ odd integers in this range, thus all such integers may be yielded.&lt;br&gt;&lt;br&gt;Here $m = 22$ so we may produce any of the odd integers from $1$ through $2^{22+1}-1=8,388,607$&lt;br&gt;&lt;br&gt;There are $2^{22}=4,194,304$ such numbers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16717" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2016-02-28T02:55:01.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27995" PostTypeId="1" AcceptedAnswerId="27997" CreationDate="2016-02-26T13:37:16.933" Score="9" ViewCount="604" Body="&lt;p&gt;Can you find the link in them? The answer should explain the link in all three of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/o3XuH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/o3XuH.png&quot; alt=&quot;Someone is dead&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-02-26T17:35:36.160" LastActivityDate="2016-02-26T17:35:36.160" Title="Linked pictures" Tags="|visual|connections-puzzle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27996" PostTypeId="2" ParentId="27995" CreationDate="2016-02-26T13:52:14.230" Score="4" Body="&lt;p&gt;The link is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Halloween. The painting is called Snap-Apple Night and depicts Irish folks celebrating on Halloween. The photo is of Indira Ghandi, who was assassinated on Halloween Day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16874" LastActivityDate="2016-02-26T13:52:14.230" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27997" PostTypeId="2" ParentId="27995" CreationDate="2016-02-26T13:52:49.007" Score="12" Body="&lt;p&gt;Well, on the left, there's a picture of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; apple bobbing, a common event on Halloween,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;on the right, there's a picture of &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Indira Gandhi, the only female Prime Minister of India, who was assassinated on Halloween, and&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;on the bottom, there's a string of characters that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; can be placed at the end of &lt;code&gt;https://www.youtube.com/watch?v=&lt;/code&gt; in order to find &lt;a href=&quot;https://www.youtube.com/watch?v=UMqZdYXvih4&quot;&gt;a cricket match between India and Sri Lanka&lt;/a&gt; which happened &lt;a href=&quot;http://www.espncricinfo.com/indvsl/engine/current/match/223634.html&quot;&gt;on Halloween, 2005&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2016-02-26T13:56:31.773" LastActivityDate="2016-02-26T13:56:31.773" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27998" PostTypeId="2" ParentId="27982" CreationDate="2016-02-26T14:10:54.227" Score="20" Body="&lt;p&gt;Well, for someone who's not a mathematics afficionado, the last bullet point (and its applications) seems like alien talk. So here's a list I've compiled which went until 40.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 = 4-sqrt(4)-4/4  &#xA;2 = 4-4+4-sqrt(4)&#xA;3 = (4+4+4)/4  &#xA;4 = 4-sqrt(4)-sqrt(4)+4  &#xA;5 = 4+sqrt(4)-4/4  &#xA;6 = 4+sqrt(4)-4+4  &#xA;7 = 4+sqrt(4)+4/4  &#xA;8 = 4+4+4-4  &#xA;9 = 4+4+4/4  &#xA;10 = 4+4+4-sqrt(4)  &#xA;11 = 4/.4+4/4  &#xA;12 = 4+4+sqrt(4)+sqrt(4)  &#xA;13 = 4?+4-4/4  &#xA;14 = 4?+4+4-4  &#xA;15 = 4?+4+4/4  &#xA;16 = 4*4-4+4  &#xA;17 = 4*4-4/4  &#xA;18 = 4*4-4/sqrt(4)  &#xA;19 = 4?+4?-4/4  &#xA;20 = 4?+4?-4+4  &#xA;21 = 4?+4?+4/4  &#xA;22 = 4?+4?+4/sqrt(4)  &#xA;23 = 4!-sqrt(4)+4/4  &#xA;24 = 4*4+4+4  &#xA;25 = 4!+sqrt(4)-4/4  &#xA;26 = 4!+sqrt(4)-4+4  &#xA;27 = 4!+sqrt(4)+4/4  &#xA;28 = 4!+4-4+4  &#xA;29 = 4!+4-4/4  &#xA;30 = 4!+4+4/sqrt(4)  &#xA;31 = 4??-4!+4-4  &#xA;32 = 4!+4+sqrt(4)+sqrt(4)  &#xA;33 = 4??-4!+4-sqrt(4)  &#xA;34 = 4*4*sqrt(4)+sqrt(4)  &#xA;35 = 4??-4!+sqrt(4)-sqrt(4)  &#xA;36 = 4!+4+4+4  &#xA;37 = 4??-4!+4+sqrt(4)  &#xA;38 = 4!+4*4-sqrt(4)  &#xA;39 = 4??-4?-4-sqrt(4)  &#xA;40 = 4!+4*sqrt(sqrt(4^4))  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'm not taking away from the brilliance of the top voted answer, just thought someone would enjoy going about it in this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;&lt;em&gt;I'm not certain if 41 can (or cannot) be done excluding the functions that fall under the last bullet point. I will turn this into community wiki if people can contribute to extend this list.&lt;/em&gt;&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out &lt;a href=&quot;https://puzzling.stackexchange.com/a/27999/12700&quot;&gt;MariusSiuram's answer&lt;/a&gt; (and its edit history) for a longer answer and an approach to extend this list. I've decided to take an early retirement at 40. ;)&lt;/p&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2016-02-29T08:24:57.080" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27999" PostTypeId="2" ParentId="27982" CreationDate="2016-02-26T14:29:04.857" Score="21" Body="&lt;h2&gt;Prefix edit&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Why not allow the &lt;a href=&quot;https://en.wikipedia.org/wiki/Double_factorial&quot; rel=&quot;nofollow&quot;&gt;double factorial&lt;/a&gt;? well, let's use it. For the record:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;0!! &amp;amp;= 1 \\&#xA;5!! &amp;amp;= 3 \cdot 5 = 15 \\&#xA;6!! &amp;amp;= 2 \cdot 4 \cdot 6 = 48 \\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the &lt;a href=&quot;https://en.wikipedia.org/wiki/Binomial_coefficient&quot; rel=&quot;nofollow&quot;&gt;choose operator&lt;/a&gt; (also known as binomial coefficient).&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Modular approach&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I had some fun doing it the long way, but then I decided to jump into a more modular / exploitable strategy. I will build it for a little bit, so bear with me for now :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, let's make the following list with one four:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;2 &amp;amp;= \sqrt 4 \\&#xA;3 &amp;amp;= \left(\sqrt 4\right)? \\&#xA;4 &amp;amp;= 4 \\&#xA;6 &amp;amp;= \left(\sqrt 4\right)?! \\&#xA;8 &amp;amp;= 4!! \\&#xA;10 &amp;amp;= 4? \\&#xA;21 &amp;amp;= \left(\sqrt 4\right)?!? \\&#xA;24 &amp;amp;= 4! \\&#xA;36 &amp;amp;= \left(4!!\right)? \\&#xA;48 &amp;amp;= \left(\left(\sqrt 4\right)?!\right)!! = 6!!\\&#xA;55 &amp;amp;= 4??&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can also consider $ 4 = \sqrt 4 + \sqrt 4 $, so there is no need to consider the &quot;extra fours&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From now on, I will use the fancy &quot;one-four&quot; substitution, and maybe some results will use less than four fours. But the translation from a compact equation to a &quot;four-fours&quot; equation is immediate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider the following list of two-four numbers (I purposely omit the ones that can be obtained with a signel four):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;0 &amp;amp;= 4 - 4 \\&#xA;1 &amp;amp;= \frac{4}{4} \\&#xA;5 &amp;amp;= 2 + 3 \\&#xA;7 &amp;amp;= 4 + 3 \\&#xA;9 &amp;amp;= 3 + 6 \\&#xA;11 &amp;amp;= 21 - 10 \\&#xA;12 &amp;amp;= 10 + 2 \\&#xA;13 &amp;amp;= 10 + 3 \\&#xA;14 &amp;amp;= 10 + 4 \\&#xA;15 &amp;amp;= 21 - 6 \\&#xA;16 &amp;amp;= 4 \cdot 4 \\&#xA;17 &amp;amp;= 21 - 4 \\&#xA;18 &amp;amp;= 21 - 3 \\&#xA;19 &amp;amp;= 21 - 2 \\&#xA;20 &amp;amp;= 24 - 4 \\&#xA;22 &amp;amp;= 24 - 2 \\&#xA;23 &amp;amp;= 21 + 2 \\&#xA;25 &amp;amp;= 21 + 4 \\&#xA;26 &amp;amp;= 24 + 2 \\&#xA;27 &amp;amp;= 24 + 3 \\&#xA;28 &amp;amp;= 24 + 4&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Up to this point we can get any integer below 28 with only two fours. My strategy will be obtaining numbers by combining the &quot;high part&quot; and &quot;low part&quot;. So, given any two-four &quot;high number&quot; $n$ we can generate all integers between $n - 28$ and $n + 28$. And the resulting formula uses up to four fours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can trivially consider the following property:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\forall n &amp;lt; 55: \quad (n+1)? - n? &amp;lt; 56 &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can have a &quot;dense-enough&quot; set of high numbers by simply using the $?$ operator to the list of &quot;two-four numbers&quot;. The maximum number at the moment is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 434 = 28? + 28 = (4! + 4)? + 4! + 4 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To continue, we should pick integers with a maximum distance of 56 between them. The next integer should be at most 463, because $ 462 - 28 = 434 $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next hand-picked &quot;high numbers&quot; (credit to @f'' for most of them!):&#xA;\begin{align}&#xA;441 &amp;amp;= 21^2 \\&#xA;465 &amp;amp;= (24 + 6)? \\&#xA;504 &amp;amp;= 21*24 \\&#xA;550 &amp;amp;= 55*10 \\&#xA;600 &amp;amp;= 24?+24? \\&#xA;630 &amp;amp;= \binom{36}{2} \\&#xA;665 &amp;amp;= 6! - 55 \\&#xA;720 &amp;amp;= 6! \\&#xA;775 &amp;amp;= 6! + 55&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My original post contained a exhaustive list up to 132 and some odd holes up to 148, holes that user @f fixed in the comments, so credit for him for that. But now that I present the alternative strategy, the original post seems overweighted and slow to load :(&lt;/p&gt;&#xA;" OwnerUserId="13065" LastEditorUserId="13065" LastEditDate="2016-06-08T13:08:01.393" LastActivityDate="2016-06-08T13:08:01.393" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28000" PostTypeId="2" ParentId="27919" CreationDate="2016-02-26T14:40:24.853" Score="1" Body="&lt;p&gt;Perhaps I am not understanding the question correctly but what about this, which appears to do it with 4 bulbs:  plant one bulb at the centre of a circle of 1 metre radius and then three other bulbs on the circumference of the circle.  There are only three colours so one of those on the circumference will be the same colour as the one in the centre.&lt;/p&gt;&#xA;" OwnerUserId="19621" LastActivityDate="2016-02-26T14:40:24.853" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28001" PostTypeId="1" AcceptedAnswerId="28075" CreationDate="2016-02-26T15:55:00.087" Score="11" ViewCount="345" Body="&lt;p&gt;A bottle washed ashore this morning and I found a note inside.  This seems like it may just be a sea shanty, but I feel there is meaning in the words, and I'd like to know what inspired it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A crew of seven, we set out, the driving wind a chill&lt;br&gt;&#xA;  Losing our captain, no one pouts, but all have fallen ill&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Another gone, we look to the cook- organize us, set us straight&lt;br&gt;&#xA;  And yet another, cries first mate, but alas, his loss will fill our plate&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Three remaining- &lt;em&gt;Is it raining?&lt;/em&gt;- We drown ourselves with a drink&lt;br&gt;&#xA;  We leave outside, we two, we hide, though the craft begins to sink&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Afloat atop the wreckage, their story lies with only one&lt;br&gt;&#xA;  If those clouds should block the sun, the story may be told by none&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can you determine the inspiration for each line, or what the lines are truly describing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For clarification:&lt;/strong&gt; There is something unique to take away from each line, one line possibly helping you find what to take away from the next.  The accepted answer should have each line represented and explained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Another bottle may wash up on shore, if the winds aren't blowing in the right direction.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="11452" LastEditorUserId="11452" LastEditDate="2016-02-26T16:23:36.477" LastActivityDate="2016-02-28T23:52:38.383" Title="A Story in the Sea Shanty?" Tags="|riddle|word|word-problem|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28002" PostTypeId="2" ParentId="27982" CreationDate="2016-02-26T16:41:10.287" Score="8" Body="&lt;p&gt;This formula will make any positive integer $ n $ from four fours:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ -\sqrt4\frac{\ln\left[\left(\ln\underbrace{\sqrt{\sqrt{\cdots\sqrt4}}}_{n}\right) / \ln4\right]}{\ln{4}} $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we allow the number $ 44 $ to count as two fours, then we can also have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\underbrace{\sec\arctan\cdots\sec\arctan}_{n^2-1}\,\frac{44}{44}$.&lt;/p&gt;&#xA;" OwnerUserId="10168" LastActivityDate="2016-02-26T16:41:10.287" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28003" PostTypeId="1" AcceptedAnswerId="28005" CreationDate="2016-02-26T18:07:57.557" Score="5" ViewCount="189" Body="&lt;p&gt;Hold on just a bit&lt;br&gt;&#xA;I need to talk to Lee&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;About a test for my Iq&lt;br&gt;&#xA;  That suited me to a T&lt;br&gt;&#xA;  Somehow I got a 6&lt;br&gt;&#xA;  When I should have got zt  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, if your brain's too tiny&lt;br&gt;&#xA;I'll give the riddle a curl&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When I was just 2&lt;br&gt;&#xA;  I played for an fc&lt;br&gt;&#xA;  And won 20 games&lt;br&gt;&#xA;  By the time I had turned 3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Warning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;When you figure this out, you'll wish you'd heeded my warning&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9353" LastActivityDate="2016-02-26T18:44:40.000" Title="Don't end up falling for the nonsense" Tags="|riddle|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28004" PostTypeId="2" ParentId="27989" CreationDate="2016-02-26T18:19:36.743" Score="31" Body="&lt;p&gt;The first player can always force a win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wrote &lt;a href=&quot;http://pastebin.com/GNFqy012&quot;&gt;this Python script&lt;/a&gt; to analyze the tree of possible game states reachable from an empty starting board. No matter what the second player does, the first player can make a move that eventually leads to a win in ten moves or less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I took the tree and &lt;a href=&quot;http://pastebin.com/UDQLddjh&quot;&gt;trimmed it down&lt;/a&gt; to only the optimal moves for the first player, plus all possible counter-moves for the second player. Excerpt:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[&quot;001&quot;, {&quot;120&quot;: [&quot;000&quot;, {&quot;122&quot;: [&quot;011&quot;, {&quot;012&quot;: [&quot;001&quot;, {}], &quot;002&quot;: [&quot;001&quot;, {}], &quot;001&quot;: [&quot;012&quot;, {}], &quot;000&quot;: [&quot;001&quot;, {}], &quot;111&quot;: [&quot;022&quot;, {}],&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Each string represents a possible move one of the players could make. The first character of the string is &quot;1&quot; if the player shifts the board, and &quot;0&quot; otherwise. The second character is the X coordinate of the piece placed, and the third character is the Y coordinate. Each list contains two items: first, the first player's optimal move; second, a dictionary of the second player's possible counter-moves. Each dictionary is keyed by the possible move the second player could take, and each value is the two-item list representing the first player's optimal counter-counter-move to that counter-move. If the dictionary is empty, that means the game is over and the first player won.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I took this data and wrote a simple &lt;a href=&quot;https://jsfiddle.net/6k65pmnj/&quot;&gt;interactive game&lt;/a&gt; that you can play against the unbeatable computer.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;edit: &lt;a href=&quot;https://jsfiddle.net/r50y76f5/1/&quot;&gt;here&lt;/a&gt; is an interactive version with a better interface.&lt;/p&gt;&#xA;" OwnerUserId="1497" LastEditorUserId="1497" LastEditDate="2016-03-01T15:17:55.577" LastActivityDate="2016-03-01T15:17:55.577" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="28005" PostTypeId="2" ParentId="28003" CreationDate="2016-02-26T18:29:43.673" Score="6" Body="&lt;p&gt;If I go to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; bit.ly (&quot;bit&quot; and &quot;Lee&quot; at the end of the first two lines) and add IqT6zt, I get &lt;a href=&quot;https://www.youtube.com/watch?v=dQw4w9WgXcQ&quot; rel=&quot;nofollow&quot;&gt;rickrolled&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if I go to tiny url (&quot;tiny&quot; and &quot;curl&quot; at the end of the next two lines) and add 2fcgames3, I get rickrolled again!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2016-02-26T18:44:40.000" LastActivityDate="2016-02-26T18:44:40.000" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28006" PostTypeId="1" AcceptedAnswerId="28007" CreationDate="2016-02-26T19:00:58.017" Score="3" ViewCount="505" Body="&lt;p&gt;You can't see me&lt;br&gt;&#xA;You can't hear me&lt;br&gt;&#xA;You can only feel me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet people always say&#xA;&quot;I see the &amp;lt;    &gt;.&quot;&lt;br&gt;&#xA;&quot;I hear the &amp;lt;     &gt;.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And no one thinks twice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I misspelled the title again. Oh well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19179" LastEditorUserId="19179" LastEditDate="2016-02-29T15:33:52.777" LastActivityDate="2016-02-29T16:09:33.527" Title="The Mysterious Effects of Blue" Tags="|word|" AnswerCount="4" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28007" PostTypeId="2" ParentId="28006" CreationDate="2016-02-26T19:25:29.893" Score="7" Body="&lt;p&gt;Are you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The wind?&lt;br&gt;&#xA;&lt;br&gt;&#xA; Technically, you can neither see nor hear the wind, you can only see or hear the effects of the wind on other things. But you can feel the wind, hoo boy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Perhaps &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the misspelled title was meant to use the word &quot;blew&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastActivityDate="2016-02-26T19:25:29.893" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28008" PostTypeId="2" ParentId="28006" CreationDate="2016-02-26T19:59:04.317" Score="2" Body="&lt;p&gt;I think you are talking about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;God&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can't see me  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I never saw any&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can't hear me  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I never heard any.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can only feel me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I can feel Him all around me.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet people always say &quot;I see the &amp;lt; &gt;.&quot;&#xA;&quot;I hear the &amp;lt; &gt;.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A person who is fully devoted to Him can be heard saying these sentences.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I misspelled the title again. Oh well.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;His 'affect' created us all.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-02-29T16:09:33.527" LastActivityDate="2016-02-29T16:09:33.527" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28009" PostTypeId="1" AcceptedAnswerId="28012" CreationDate="2016-02-26T21:32:54.537" Score="5" ViewCount="269" Body="&lt;p&gt;You have red, blue and green unit cubes. You will create a 2x2x2 cube by using 8 of these unit cubes. What is the maximum number of different cubes you can create? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: If a cube can be obtained by rotating another one, these two cubes are not considered to be different. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the problem was asked for only red and blue unit cubes, the answer would be 23.&lt;/p&gt;&#xA;" OwnerUserId="18514" LastEditorUserId="10615" LastEditDate="2016-02-26T22:46:28.587" LastActivityDate="2016-02-26T23:07:04.137" Title="Colored Cubes (Red, Blue and Green)" Tags="|logical-deduction|combinatorics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28010" PostTypeId="2" ParentId="28006" CreationDate="2016-02-26T21:42:59.320" Score="1" Body="&lt;p&gt;Are you  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Sky?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can't see me  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Though we claim to see the sky, it is air and is just invisible.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can't hear me  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sky itself makes no noise.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can only feel me.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We feel the air around us, and the sky is just air.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet people always say &quot;I see the &amp;lt; &gt;.&quot;&lt;br&gt;&#xA;&quot;I hear the &amp;lt; &gt;.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People say &quot;Look at the sky&quot; or point out noises from objects in the sky.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hints: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;I misspelled the title again&lt;/em&gt;: The sky affects many people different ways.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="9951" LastEditDate="2016-02-29T16:08:20.610" LastActivityDate="2016-02-29T16:08:20.610" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28011" PostTypeId="2" ParentId="28006" CreationDate="2016-02-26T23:01:37.237" Score="3" Body="&lt;p&gt;I was thinking&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; vacuum?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Literally speaking, it cannot be heard or seen.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Also, life cannot exist in a vacuum.&lt;br&gt;&#xA; However, many people refer to a vacuum cleaner as a &quot;vacuum&quot;.  It (the cleaner) can definitely be seen and heard.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The problem with my answer is that I don't have any connection to the title.&lt;/p&gt;&#xA;" OwnerUserId="12092" LastActivityDate="2016-02-26T23:01:37.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28012" PostTypeId="2" ParentId="28009" CreationDate="2016-02-26T23:07:04.137" Score="4" Body="&lt;p&gt;There are 24 orientations of the cube. From any orientation, the other 23 can be generated by rotations, which can be categorized as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;3 axes passing through opposite faces; for each axis, 90 degrees clockwise, 90 degrees counterclockwise, or 180 degrees. (9 total)&lt;/li&gt;&#xA;&lt;li&gt;6 axes passing through opposite edges; for each axis, 180 degrees. (6 total)&lt;/li&gt;&#xA;&lt;li&gt;4 axes passing through opposite vertices; for each axis, 120 degrees clockwise or 120 degrees counterclockwise. (8 total)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We start by building all possible cubes that use all three colors. The number of these is $3^8-3\cdot2^8+3=5796$. Then, if any two cubes can be converted to each other by a rotation, we put them in a set together. We want to find the number of sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If no rotation brings a cube to an identical one, then it is in a set of 24. However, if a cube is the same under some rotation, it is in a set of fewer than 24. We need to determine how many cubes are in smaller sets, and what the size of those sets is. Thankfully, no cube that has multiple rotational symmetries around different axes uses all three colors, so we can ignore the possibility of multiple rotational symmetries. Also, no cube that remains the same under a 90-degree rotation uses all three colors. We only need to consider the 180-degree rotations and the 120-degree rotations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 9 possible 180-degree rotations. For each one, a cube that remains the same under that rotation is defined by the colors of four subcubes, so there are $3^4-3\cdot2^4+3=36$ for each of the nine rotations. This is a total of $324$ cubes, which are in sets of 12.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 8 possible 120-degree rotations, but they come in clockwise-counterclockwise pairs. For each pair, a cube that remains the same under those rotations is also defined by the colors of four subcubes, so there are $36$ for each of the four pairs. This makes $144$ cubes, which are in sets of 8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remaining $5796-324-144=5328$ cubes are in sets of 24. The total number of sets is $\frac{324}{12}+\frac{144}{8}+\frac{5328}{24}=267$. Then we add the $23\cdot3-3=66$ cubes using at most two colors for a total of $\boxed{333}$.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2016-02-26T23:07:04.137" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28013" PostTypeId="1" AcceptedAnswerId="28019" CreationDate="2016-02-27T01:26:26.423" Score="7" ViewCount="379" Body="&lt;p&gt;&lt;em&gt;This is the lateral-thinking fifth part to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/27470/squaring-a-cross&quot;&gt;this question&lt;/a&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RKSBp.png&quot; alt=&quot;1]&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this cross shape, all twelve sides are the same length and all angles are right angles. What is the smallest number of cuts that can divide the cross into pieces that can be rearranged to form two squares such that the side length of the larger square is one-and-a-half times the side length of the other?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Each cut is a straight line, which can (but isn't required to) go across multiple pieces.&lt;/li&gt;&#xA;&lt;li&gt;Pieces cannot be altered except by cutting, until all cutting is done.&lt;/li&gt;&#xA;&lt;li&gt;After all cuts are made, pieces may be reflected, rotated, or moved, but cannot be resized, folded, removed, added, or altered in any other way.&lt;/li&gt;&#xA;&lt;li&gt;Pieces cannot overlap or cover themselves or each other.&lt;/li&gt;&#xA;&lt;li&gt;After the pieces are moved, edges that join two pieces are ignored, as if the pieces fused together. The rest of the edges must form the set of squares exactly; squares must be formed entirely of piece edges, and extra edges or pieces are not allowed.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.437" LastActivityDate="2016-02-27T05:33:25.173" Title="Squaring a cross one more time" Tags="|geometry|lateral-thinking|dissection|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28014" PostTypeId="1" AcceptedAnswerId="28017" CreationDate="2016-02-27T04:20:46.883" Score="5" ViewCount="781" Body="&lt;p&gt;I was snoozing downtown. A noise woke me up and I looked at my LED clock. I noted the product of the hour value and the minute value. I fell asleep. Another noise woke me up 55 minutes later. The product of the hour and the minute was the same as before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What time did I wake up?&lt;/p&gt;&#xA;" OwnerUserId="11497" LastEditorUserId="12408" LastEditDate="2016-02-27T16:12:30.207" LastActivityDate="2016-02-27T16:12:30.207" Title="What time did I wake up?" Tags="|mathematics|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28015" PostTypeId="2" ParentId="28014" CreationDate="2016-02-27T04:37:43.790" Score="2" Body="&lt;p&gt;The first time you woke up it was &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 12:55 AM.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second time you woke up it was &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1:50 AM. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There was a power surge/outage after you woke up the first time and the clock reset to 12:00 AM.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17297" LastActivityDate="2016-02-27T04:37:43.790" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28017" PostTypeId="2" ParentId="28014" CreationDate="2016-02-27T04:47:10.317" Score="13" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The minute value can't be the same after 55 minutes, so both the hour and the minute must have changed. The hour increased by 1 and the minute decreased by 5 (The hour couldn't have gone from 12 to 1 because the new product would be lower). If the first time was $h$ hours and $m$ minutes, then $(h+1)(m-5)=hm$. Expanding and rearranging, $m=5(h+1)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This has 10 solutions (possible values for the first wake-up time), which are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;code&gt;1:10&lt;/code&gt;, &lt;code&gt;2:15&lt;/code&gt;, &lt;code&gt;3:20&lt;/code&gt;, &lt;code&gt;4:25&lt;/code&gt;, &lt;code&gt;5:30&lt;/code&gt;, &lt;code&gt;6:35&lt;/code&gt;, &lt;code&gt;7:40&lt;/code&gt;, &lt;code&gt;8:45&lt;/code&gt;, &lt;code&gt;9:50&lt;/code&gt;, and &lt;code&gt;10:55&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2016-02-27T04:47:10.317" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28018" PostTypeId="2" ParentId="28014" CreationDate="2016-02-27T05:07:14.607" Score="8" Body="&lt;p&gt;6 additional times for a 24-hour display that goes from 00:00 to 23:59:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 00:00 , 00:01 , 00:02 , 00:03 , 00:04 , 00:05  : first wakening &lt;br&gt;&#xA; 00:55 , 00:56 , 00:57 , 00:58 , 00:59 , 01:00  : second wakening&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All other possibilities have been&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/a/28017/18129&quot;&gt;covered by @f''&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18129" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2016-02-27T05:07:14.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28019" PostTypeId="2" ParentId="28013" CreationDate="2016-02-27T05:33:25.173" Score="14" Body="&lt;p&gt;I can do it in two cuts:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/OcY3K.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OcY3K.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt; &lt;br&gt;&#xA; The outer boundary is 1.5 times the size of the inner.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-02-27T05:33:25.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28020" PostTypeId="1" AcceptedAnswerId="28259" CreationDate="2016-02-27T11:05:53.527" Score="8" ViewCount="956" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Now with many&lt;br&gt;&#xA;  Now with having lovers&lt;br&gt;&#xA;  Now with first swing&lt;br&gt;&#xA;  Now with some covers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The words above constitute a riddle whose answer is a single English word. Find the word and explain the title.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each of the four lines is a separate clue. It is also Socratic confusion &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I added the wordplay tag because there is a pun involved.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Final hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's strange but I thought revealing that this is an acrostic would&#xA; make it too easy. Well, as hexomino figured out from the first hint,&#xA; an acrostic is what it is. Go with that. Acrost it. The other clues&#xA; will follow more easily unless I have again misjudged.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I will post a solution tomorrow. You have been warned. :)&lt;/p&gt;&#xA;" OwnerUserId="18924" LastEditorUserId="18924" LastEditDate="2016-03-02T13:34:45.817" LastActivityDate="2016-03-04T10:26:59.513" Title="Five Clues In Four lines" Tags="|riddle|word|wordplay|poetry|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28021" PostTypeId="1" AcceptedAnswerId="28025" CreationDate="2016-02-27T11:30:44.157" Score="2" ViewCount="415" Body="&lt;p&gt;Two people randomly select two different 1-digit numbers (including 0 of course) and calculate their absolute difference. Then They share what they get, What is the probability of having the same absolute difference for these two?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, Player 1 take &quot;0&quot;, then take &quot;3&quot; (he cannot choose 0 again), the absolute difference is 3. Players 2 takes &quot;9&quot; then takes &quot;6&quot;, the absolute difference becomes 3. and they share their result of the difference. since both has the same result they yell &quot;yay&quot;... :)&lt;/p&gt;&#xA;" OwnerUserId="18514" LastEditorUserId="18514" LastEditDate="2016-02-27T14:31:26.423" LastActivityDate="2016-04-03T17:58:21.003" Title="A game with numbers" Tags="|logical-deduction|probability|counting|" AnswerCount="3" CommentCount="10" ClosedDate="2016-03-05T12:12:12.190" ContentLicense="CC BY-SA 3.0" />
  <row Id="28022" PostTypeId="2" ParentId="28021" CreationDate="2016-02-27T12:35:47.593" Score="2" Body="&lt;p&gt;The absolute difference $|a_1-a_2|$ for a player has the following distribution of probability:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$P(|a_1-a_2|=0)=\frac{10}{100}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$$P(|a_1-a_2|=k)=2\frac{10-k}{100}$$, for $k=1, \cdots ,9$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $P(|a_1-a_2|=|b_1-b_2|)=\frac 1 {100}+\sum_{k=1}^9 \left( 2\frac{10-k}{100} \right)^2=12.4\%$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17865" LastEditorUserId="17865" LastEditDate="2016-04-03T12:06:28.463" LastActivityDate="2016-04-03T12:06:28.463" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28023" PostTypeId="1" AcceptedAnswerId="28120" CreationDate="2016-02-27T13:03:42.427" Score="6" ViewCount="1039" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Once upon a time there was a woman named Alice. She had two husbands, Mr. Cantor and Alex.&lt;/p&gt;&#xA;&lt;p&gt;From Mr. Cantor, she had a son. He was very mischievous. All people in the house were bothered by him. But his fun-time ended with time.&lt;/p&gt;&#xA;&lt;p&gt;After some years, she had another son from Alex. He was even more mischievous than his brother. So of course, people in the home were even more bothered by him. Once he called his friends, Phat and Stuart at home. They destroyed all the thing in a whole room. Younger son and the whole family felt bad for his destructed room. Hearing so much sound, their landlord Yun Ong came to their home and told the whole family to not call such a devilry kids at home.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;What does this story reminds you of? Answer should have all characters explained.&lt;/p&gt;&#xA;&lt;p&gt;Note: If you think that characters in the story were real human beings then I would like to say 'LOL NO'&lt;/p&gt;&#xA;&lt;p&gt;Hint 1 :&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Added a tag and changed the name from Ally to Alice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Findings of our peers:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 1. &amp;quot;(Phat) Fat Boy&amp;quot; and &amp;quot;(Stuart) Little Boy&amp;quot; nukes were dropped on Japan.&amp;quot; (found by Doug.McFarlane)&lt;br /&gt;&#xA;2. Yun Ong: United Nations. (found by karobar)&lt;br /&gt;&#xA; I think that the way in which United Nations is named Yun Ong should give you some hint how other names are assigned.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-03-01T18:35:14.057" Title="The real story of two mischievous brothers" Tags="|riddle|story|history|" AnswerCount="1" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="28024" PostTypeId="2" ParentId="28014" CreationDate="2016-02-27T13:07:21.467" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The hour must have changed, otherwise the products wouldn't have been equal, so $hm=(h+1)(m-5) -&amp;gt; m=5(h+1)$. $h$ can range from 0 to 10. If both products are zero and we aren't necessarily following this formula, the first time can range from 00:00 to 00:05. 00:05 appears twice, thus there are 11+6-1=16 solutions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastActivityDate="2016-02-27T13:07:21.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28025" PostTypeId="2" ParentId="28021" CreationDate="2016-02-27T14:50:08.167" Score="5" Body="&lt;p&gt;Each person picks 2 numbers and finds their absolute difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 9 ways to get an absolute difference of 1 (1-0, 2-1, ..., 9-8).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 8 ways to get an absolute difference of 2 (2-0, 3-1, ..., 9-7).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are $w(d) = 10-d$ ways to get an absolute difference of $d$, with $1 \leq d \leq 9$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total number of digit combinations is ${10 \choose 2} = 45$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also the total number of 'ways': $\sum_{d=1}^{9} (10-d) = 90 - \frac{9 \cdot 10}{2} = 45$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We require both to pick the same absolute difference. The probability of this is the sum of the probabilities of both picking the same absolute difference, which works out to be $\frac{19}{135}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\sum_{d=1}^{9} \left( \frac{w(d)}{45} \cdot \frac{w(d)}{45} \right)&#xA;&amp;amp;= \sum_{d=1}^{9} \frac{(10 - d)^2}{45^2} \\&#xA;&amp;amp;= \frac{900 - 20(45) + \frac{(9)(9+1)(2 \cdot 9 + 1)}{6}}{45^2} \\&#xA;&amp;amp;= \frac{19}{135}&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="9422" LastActivityDate="2016-02-27T14:50:08.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28026" PostTypeId="2" ParentId="28021" CreationDate="2016-02-27T14:58:02.447" Score="6" Body="&lt;p&gt;The probability either player has a absolute difference of $k$ is $\dfrac{10-k}{\binom{10}{2}}$. For a match, both need to be the same, and so we consider:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_\limits{k=1}^9 \Big[\dfrac{10-k}{\binom{10}{2}}\Big]^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum of the first $n$ squares is $\dfrac{n(n+1)(2n+1)}{6}$, and so we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{4\cdot9\cdot10\cdot19}{6\cdot9^2\cdot10^2}=\dfrac{4\cdot19}{6\cdot9\cdot10}=\dfrac{19}{135}$$&lt;/p&gt;&#xA;" OwnerUserId="11497" LastEditorUserId="10615" LastEditDate="2016-04-03T17:58:21.003" LastActivityDate="2016-04-03T17:58:21.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28027" PostTypeId="1" AcceptedAnswerId="28034" CreationDate="2016-02-27T16:18:57.200" Score="8" ViewCount="324" Body="&lt;p&gt;At the beginning the blackboard contains a single integer $N$. If the blackboard contains some number $x$, then we may additionally write the two numbers $\displaystyle\frac x {x+2} $ and $ 2x + 1 $ on the blackboard. After some time you look at the blackboard and see the number 2016 written on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the possible starting values $x$ for this story to happen?&lt;/p&gt;&#xA;" OwnerUserId="19470" LastEditorUserId="8874" LastEditDate="2016-03-03T15:09:05.577" LastActivityDate="2016-03-03T15:09:05.577" Title="Blackboard problem with 2016" Tags="|mathematics|combinatorics|reachability|blackboard|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28028" PostTypeId="1" AcceptedAnswerId="28039" CreationDate="2016-02-27T16:24:40.653" Score="8" ViewCount="180" Body="&lt;p&gt;There are cubes with side lengths $ 1,2,3,...,N $ made from the same metal material, so that their weights are the the cube integers&#xA;$$ 1^3, 2^3, 3^3, .... ~ ...., (N-1)^3, N^3.$$&#xA;The cubes can be divided into two groups so that the heavier group weights exactly $20162016$ more than the lighter group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there exists an integer $ N \geq 1$ for which this story is possible?&lt;/p&gt;&#xA;" OwnerUserId="19470" LastActivityDate="2016-02-27T22:18:31.640" Title="Cube problem with twice 2016" Tags="|mathematics|combinatorics|number-theory|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28029" PostTypeId="2" ParentId="28027" CreationDate="2016-02-27T17:12:35.110" Score="6" Body="&lt;p&gt;Obviously we have $x=2016$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x=2$, then we can write $\dfrac24$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $x=4$ we can write $\dfrac46$, etc...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So any even number $\lt2016$.&lt;/p&gt;&#xA;" OwnerUserId="11497" LastActivityDate="2016-02-27T17:12:35.110" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="28030" PostTypeId="1" AcceptedAnswerId="28076" CreationDate="2016-02-27T17:29:29.337" Score="4" ViewCount="247" Body="&lt;p&gt;Originally posted on &lt;a href=&quot;https://gaming.stackexchange.com/questions/256907/stuck-on-a-think-different-questions&quot;&gt;Arqade&lt;/a&gt;. This question was posted by another user on Arqade due to being a game, which is correct. However due to the lack of attention this question has garnered so far I decided to post it at the home of the puzzlers.&lt;/p&gt;&#xA;&lt;p&gt;The issue is that the question doesn't seem to be solved mathematically, as the original poster claims the answer is &lt;strong&gt;not 16&lt;/strong&gt;, when mathematically it should be 16.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;3x = 24 &amp;lt;=&amp;gt; x = 8&lt;/p&gt;&#xA;&lt;p&gt;x + y = 25 &amp;lt;=&amp;gt; y = 17&lt;/p&gt;&#xA;&lt;p&gt;y + z = 8 &amp;lt;=&amp;gt; z = -9&lt;/p&gt;&#xA;&lt;p&gt;x+y+z = ? = 16&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gm25D.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gm25D.jpg&quot; alt=&quot;Problem&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18816" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-02-29T04:05:18.573" Title="Triple Equation, result not mathematical" Tags="|mathematics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28031" PostTypeId="1" AcceptedAnswerId="28035" CreationDate="2016-02-27T18:10:22.943" Score="10" ViewCount="351" Body="&lt;p&gt;Is there a way of filling a 4×4 table with 16 distinct integers from 1, 2, ..., 100 such that the products of the numbers in every row and in every column are all equal to each other?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="69582" LastEditDate="2021-03-13T21:57:00.963" LastActivityDate="2021-03-14T13:07:16.877" Title="Four-by-four table with equal row and column products" Tags="|mathematics|combinatorics|calculation-puzzle|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28032" PostTypeId="2" ParentId="28027" CreationDate="2016-02-27T18:20:14.073" Score="5" Body="&lt;p&gt;I tried tackling the problem numerically. I found that up to 13 layers of depth(the furthest I could go), there was no solution. First I inverted &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$  f(x) = 2x+1 -&amp;gt; f^{-1}(x) = \frac{x-1}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$  g(x) = \frac{x}{x+2} -&amp;gt; g^{-1}(x) = \frac{2x}{1-x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, I want to consider all possible inversions of 2016 at a depth n. The question we are trying to answer: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ h_1 ( h_2 ( h_3 ( h_4 ( h_5 ... h_n (2016 ))))))) \in \Z $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $h_i$ is either $f^{-1}(x)$ or $g^{-1}(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I did was considered all possible permutations of $f^{-1}$ and $g^{-1}$, and I find that up to n=13 there is no solution. That is, there is no integer which application of f(x) and g(x) reaches the value 2016.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for instance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f^{-1}(2016) = 1007.5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g^{-1}(2016) = -2.00099$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f^{-1}(f^{-1}(2016))=503.25$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g^{-1}(f^{-1}(2016))=-2.00199$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f^{-1}(g^{-1}(2016))=-1.5005$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g^{-1}(g^{-1}(2016))=-1.33355$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#include &amp;lt;stdio.h&amp;gt;&#xA;&#xA;&#xA;double finverse(double x)&#xA;{&#xA;return (x-1)/2;&#xA;}&#xA;&#xA;double ginverse(double x)&#xA;{&#xA;&#xA;return 2 * x/(1-x);&#xA;}&#xA;&#xA;void printBits(size_t const size, void const * const ptr)&#xA;{&#xA;unsigned char *b = (unsigned char*) ptr;&#xA;unsigned char byte;&#xA;int i, j;&#xA;&#xA;for (i=size-1;i&amp;gt;=0;i--)&#xA;{&#xA;    for (j=7;j&amp;gt;=0;j--)&#xA;    {&#xA;        byte = b[i] &amp;amp; (1&amp;lt;&amp;lt;j);&#xA;        byte &amp;gt;&amp;gt;= j;&#xA;        printf(&quot;%u&quot;, byte);&#xA;    }&#xA;}&#xA;puts(&quot;&quot;);&#xA;}&#xA;&#xA;int main(int argc, char **argv)&#xA;{&#xA;int num = 1;&#xA;int index;&#xA;double *previous = malloc(sizeof(double));&#xA;*previous = 2016;&#xA;double *current;&#xA;for(num=1;num&amp;lt;10;num++)&#xA;{&#xA;    printf(&quot;Depth is %d\n&quot;,num);&#xA;&#xA;current = malloc((int)pow(2,num)*sizeof(int));&#xA;for(index=0;index&amp;lt;pow(2,num);index++)&#xA;{&#xA;    //printf(&quot;Index is %d\n&quot;,index);&#xA;    if(index %2 == 0)&#xA;    *(current + index) = finverse(*(previous+index/2));&#xA;    else&#xA;    *(current + index) = ginverse(*(previous+index/2));&#xA;&#xA;    printf(&quot;Value is %g\n&quot;,*(current+index));&#xA;    if(*(current+index) == (int)(*(current+index)))&#xA;    {&#xA;        printf(&quot;\n\n\n\SOLUTION HAS BEEN FOUND\n\n\n&quot;);&#xA;        printf(&quot;%g &quot;,*(current+index));&#xA;        printBits(sizeof(int),&amp;amp;index);&#xA;        printf(&quot;\n&quot;);&#xA;    }&#xA;&#xA;}&#xA;previous = current;&#xA;}&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="18821" LastActivityDate="2016-02-27T18:20:14.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28034" PostTypeId="2" ParentId="28027" CreationDate="2016-02-27T18:47:08.730" Score="12" Body="&lt;p&gt;Obviously we're never going to get an irrational number. So any number we get can be expressed as $\frac{p}{q}$, where $p$ and $q$ are relatively prime integers ($q$ can be $1$, if the number is an integer). When we have $\frac{p}{q}$, we can write $\frac{2p+q}{q}$ and $\frac{p}{p+2q}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $p$ and $q$ are both odd, $2p+q$ and $p+2q$ are also odd, so any fraction we get from there will have an odd numerator and denominator. Since we want $\frac{2016}{1}$, which has an even numerator, a fraction with $p$ and $q$ odd is useless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $p$ is odd and $q$ is even, then $2p+q$ is even and $p+2q$ is odd. The fraction $\frac{p}{p+2q}$ will have an odd numerator and denominator, so it is useless. Similarly, if $p$ is even and $q$ is odd, $\frac{2p+q}{q}$ is useless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that we never get more than one useful fraction at a time. If $p$ is odd and $q$ even, we get $\frac{2p+q}{q}$, while if $q$ is odd and $p$ even, we get $\frac{p}{p+2q}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the case with $p$ odd and $q=2k$ (the other case is symmetrical). Then $\frac{2p+q}{q}=\frac{2p+2k}{2k}=\frac{p+k}{k}$ (which can't be reduced further because $p$ and $k=\frac{q}{2}$ are relatively prime). Note that the sum of the numerator and denominator is $p+k+k=p+q$, so it didn't change. &lt;em&gt;Useful moves don't change the sum of the numerator and denominator&lt;/em&gt;. We need to get $\frac{2016}{1}$, so the sum of the numerator and the denominator is $2017$. The only integer possibilities for $N$ are $\frac{2016}{1}=2016$ and $\frac{2018}{-1}=-2018$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2016$ obviously works. Running it with a computer shows that $-2018$ also works after 168 steps.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2016-02-27T18:47:08.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28035" PostTypeId="2" ParentId="28031" CreationDate="2016-02-27T19:05:29.360" Score="17" Body="&lt;p&gt;Notice that each row and each column contains the numbers 1, 1, 2, 3, 4, 5, 6, 7&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \begin{array}{ a b c d }&#xA;  4*7 &amp;amp; 1*6 &amp;amp; 3*1 &amp;amp; 2*5 &amp;amp; \\&#xA;  3*5 &amp;amp; 2*1 &amp;amp; 4*6 &amp;amp; 1*7 &amp;amp;\\&#xA;  1*1 &amp;amp; 4*5 &amp;amp; 2*7 &amp;amp; 3*6 &amp;amp;\\&#xA;  2*6 &amp;amp; 3*7 &amp;amp; 1*5 &amp;amp; 4*1 &amp;amp;\\&#xA;\end{array} $$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="18821" LastEditorUserId="8874" LastEditDate="2021-03-14T13:07:16.877" LastActivityDate="2021-03-14T13:07:16.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28036" PostTypeId="1" AcceptedAnswerId="28037" CreationDate="2016-02-27T19:14:23.280" Score="5" ViewCount="130" Body="&lt;p&gt;In this puzzle, I am offering you the paraphases of English palindromic phrases. That is, I take a phrase:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Egad, a base tone denotes a bad age!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;replace each fragment of the phrase with synonymous expressions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My god, a low voice is the hallmark of elderness!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Recreate&lt;/strong&gt; the initial palindromic phrases from the paraphrases below. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Return to such a peaceful position and meditate.&lt;/li&gt;&#xA;&lt;li&gt;Medic, rescue the antique amphoras and escape!&lt;/li&gt;&#xA;&lt;li&gt;The god of lightning ascended and noticed the strait of Egypt.&lt;/li&gt;&#xA;&lt;li&gt;Stand up and express your choice, noble man.&lt;/li&gt;&#xA;&lt;li&gt;If you meet mice, do not be afraid and do not signal approval.&lt;/li&gt;&#xA;&lt;li&gt;Is it time for the sweethearts to rebel?&lt;/li&gt;&#xA;&lt;li&gt;Wrong, a shiny ingot won't be a good building block for a fortress.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The line of reasoning for the example one may be as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Elderness&lt;/em&gt; may have to do with &lt;em&gt;age&lt;/em&gt;, so we should have &lt;em&gt;... age&lt;/em&gt; at the end of the phrase. It should start with &lt;em&gt;Ega...&lt;/em&gt; then, and we're looking for a swear expression similar to &lt;em&gt;My god&lt;/em&gt;. &#xA; Next, we have &lt;em&gt;voice&lt;/em&gt;. Looking for synonyms for &lt;em&gt;voice&lt;/em&gt;, we have &lt;em&gt;sound&lt;/em&gt;, &lt;em&gt;pitch&lt;/em&gt; or &lt;em&gt;tone&lt;/em&gt;. Reverting the letter order, it's &lt;em&gt;dnuos&lt;/em&gt;, &lt;em&gt;hctip&lt;/em&gt; or &lt;em&gt;enot&lt;/em&gt;. Only the last one makes sense as a part of a word, so we'll stick to that.&#xA; Now, we look for words and expressions that have &lt;em&gt;enot&lt;/em&gt; and match the meaning of &lt;em&gt;hallmark&lt;/em&gt;. After a while, we come up with &lt;em&gt;denote&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;EDIT: Whoa, cracked in an hour. Should I do more?&lt;/p&gt;&#xA;" OwnerUserId="19447" LastEditorUserId="19447" LastEditDate="2016-02-27T20:34:47.053" LastActivityDate="2016-02-28T03:12:22.703" Title="Recreating palindromes" Tags="|palindrome|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28037" PostTypeId="2" ParentId="28036" CreationDate="2016-02-27T19:51:51.493" Score="2" Body="&lt;p&gt;1) &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Resume so pacific a pose, muser.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2) &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nurse, save rare vases, run!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Zeus was deified, saw Suez.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rise to vote, Sir.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;5)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Do not start at rats to nod.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;6)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; INCORRECT Are we not drawn onwards, we Jews, drawn onward to new era? CORRECT Won't lovers revolt now?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;7)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No, it's a bar of gold, a bad log for a bastion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://www.cs.arizona.edu/icon/oddsends/palinsen.htm&quot; rel=&quot;nofollow&quot;&gt;https://www.cs.arizona.edu/icon/oddsends/palinsen.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11497" LastEditorUserId="11497" LastEditDate="2016-02-28T03:12:22.703" LastActivityDate="2016-02-28T03:12:22.703" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28038" PostTypeId="1" AcceptedAnswerId="28057" CreationDate="2016-02-27T21:46:32.020" Score="8" ViewCount="291" Body="&lt;p&gt;What is the minimum number of squares to be used to draw a 8x8 board which is divided into 64 unit squares.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The squares you are going to use can be of any size you want (see example below).&lt;/li&gt;&#xA;&lt;li&gt;There will be no extra lines or draws outside the table when you are done!&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example for 3x3:&lt;/strong&gt; First is drawn with 5 square, second is drawn with 4 squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nZOm7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nZOm7.jpg&quot; alt=&quot;Here is an example for 3x3&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18514" LastEditorUserId="8874" LastEditDate="2016-03-22T09:57:27.700" LastActivityDate="2016-03-22T09:57:27.700" Title="Bigger board with the least squares" Tags="|mathematics|geometry|combinatorics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28039" PostTypeId="2" ParentId="28028" CreationDate="2016-02-27T22:07:25.127" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Yes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;let $N=95$. The lighter group consists of $2^3$, $6^3$ through $12^3$, and $21^3$ through $34^3$, and has a total weight of $315792$. The heavier group contains all the other cubes, and has a weight of $20477808$. The difference is $20162016$, as desired.&lt;br&gt;&#xA;&lt;br&gt;&#xA;Alternatively, notice that for any $n$,&lt;br&gt;&#xA;$$(n+7)^3-(n+6)^3-(n+5)^3+(n+4)^3-(n+3)^3+(n+2)^3+(n+1)^3-n^3=48$$&lt;br&gt;&#xA;so out of any eight consecutive cubes, they can be divided into two groups where the heavier group weighs $48$ more than the lighter one. $20162016$ is divisible by $48$, so we can just keep adding eight cubes in this pattern until we reach it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2016-02-27T22:18:31.640" LastActivityDate="2016-02-27T22:18:31.640" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28040" PostTypeId="1" AcceptedAnswerId="28043" CreationDate="2016-02-27T22:12:34.147" Score="9" ViewCount="758" Body="&lt;p&gt;I am working on a series of book-trivia puzzles. Given an excerpt from the book directly mentioning the book title, players are asked to fill in the title.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Example&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Given the excerpt&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...now we were running down for it with a bright lookout day and night. It was about the last day of our outward voyage by the largest computation; some time that night, or at latest before noon of the morrow, we should sight the &lt;strong&gt;[book title]&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;recover the book title:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Robert Lewis Stevenson, &lt;em&gt;Treasure Island&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is an example of a book suitable for this puzzle, because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Words &lt;em&gt;outward voyage&lt;/em&gt; hint at heroes travelling by the sea, and &lt;em&gt;with a bright lookout&lt;/em&gt; hints that they are searching for a feature in the sea, i. e., a reef, a ship, or an island. This limits the search for the answer to the known works of fiction with &lt;em&gt;island&lt;/em&gt; in the name.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;More examples&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I will give more examples that I deem good below, both for the sake of clarity and for your entertainment.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The cub, who had thus received a name in the world, lay and watched. For a time the man-animals continued to make their mouth-noises. Then Grey Beaver took a knife from a sheath that hung around his neck, and went into the thicket and cut a stick. &lt;strong&gt;[Book title]&lt;/strong&gt; watched him. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;The cub&lt;/em&gt; and &lt;em&gt;man-animals&lt;/em&gt; hint at an animal story, and specifically, one told from the animal perspective. &lt;em&gt;Grey Beaver&lt;/em&gt; sounds like a Native American name / nickname. Combine the two to get &lt;em&gt;White Fang&lt;/em&gt; by Jack London.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;We have devoted several centuries to inventing and elaborating &lt;strong&gt;[book title]&lt;/strong&gt; as a universal language and method for expressing all intellectual concepts and all artistic values and reducing them to a common denominator.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The title of the book &lt;em&gt;is&lt;/em&gt; a universal language system and &lt;em&gt;is&lt;/em&gt; a method of expressing both science and art. Few authors have dared to combine science and art in one action, and even fewer named their books after such actions. The answer is &lt;em&gt;The Glass Bead Game&lt;/em&gt; by Herman Hesse.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;The human beings did not hate &lt;strong&gt;[book title]&lt;/strong&gt; any less now that it was prospering; indeed, they hated it more than ever. Every human being held it as an article of faith that the &lt;strong&gt;[book title]&lt;/strong&gt; would go bankrupt sooner or later, and, above all, that the windmill would be a failure. They would meet in the public-houses and prove to one another by means of diagrams that the windmill was bound to fall down, or that if it did stand up, then that it would never work. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, &lt;em&gt;human beings&lt;/em&gt; are contrasted to the main characters of the work of fiction, so they might be animals. If [book title] can go bankrupt, it should have a kind of &lt;em&gt;company&lt;/em&gt;, &lt;em&gt;enterprize&lt;/em&gt;, &lt;em&gt;factory&lt;/em&gt; etc in the title. Combine to get &lt;em&gt;Animal Farm&lt;/em&gt; by George Orwell.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;I have to catch everybody if they start to go over the cliff—I mean if they’re running and they don’t look where they’re going I have to come out from somewhere and catch them. That’s all I’d do all day. I’d just be &lt;strong&gt;[book title]&lt;/strong&gt; and all. I know it’s crazy, but that’s the only thing I’d really like to be. I know it’s crazy.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A monologue declares that the character wants to be &lt;em&gt;someone&lt;/em&gt;, and his main responsibility will be to catch people. Naturally, he is a &lt;em&gt;catcher&lt;/em&gt;, specifically, &lt;em&gt;The Catcher in the Rye&lt;/em&gt; by J. D. Salinger.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;Can't we,&quot; he pleaded to the world, to himself, to the officials, to the Machine, &quot;can't we take it back, can't we make it alive again? Can't we start over? Can't we-&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;He did not move. Eyes shut, he waited, shivering. He heard Travis breathe loud in the room; he heard Travis shift his rifle, click the safety catch, and raise the weapon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was &lt;strong&gt;[book title]&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Making alive again&lt;/em&gt; and &lt;em&gt;starting over&lt;/em&gt; hints at time travel stories. Then, in the second paragraph, &lt;em&gt;sounds&lt;/em&gt; are thouroughly listed: &lt;em&gt;loud breathing&lt;/em&gt;, &lt;em&gt;shifting a rifle&lt;/em&gt;, &lt;em&gt;clicking the catch&lt;/em&gt;; we are safe to assume that the book title will have a sound. Specifically, &lt;em&gt;A Sound of Thunder&lt;/em&gt; by Ray Bradbury.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;6&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;“You are a silly little boy,” said the &lt;strong&gt;[book title]&lt;/strong&gt;, “just an ignorant, silly little boy.” &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simon moved his swollen tongue but said nothing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Don’t you agree?” said the &lt;strong&gt;[book title]&lt;/strong&gt;. “Aren’t you just a silly little boy?” &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simon answered him in the same silent voice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Well then,” said the &lt;strong&gt;[book title]&lt;/strong&gt;, “you’d better run off and play with the others. They think you’re batty. You don’t want Ralph to think you’re batty, do you? You like Ralph a lot, don’t you? And Piggy, and Jack?”&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A work of art clearly depicting little boys and their wrongdoings. Too much conflict to be &lt;em&gt;Peter Pan&lt;/em&gt;. It's &lt;em&gt;Lord of the Flies&lt;/em&gt; by William Golding.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;7&quot;&gt;&#xA;&lt;li&gt;I’d rather you shot at tin cans in the back yard, but I know you’ll go after birds. Shoot all the bluejays you want, if you can hit ‘em, but remember it’s a sin &lt;strong&gt;[book title]&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; More of a construction puzzle, this one. The book title clearly has a verb first and a bird to contrast &lt;em&gt;bluejays&lt;/em&gt; second. It's &lt;em&gt;To Kill a Mockingbird&lt;/em&gt; by Harper Lee.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;8&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;Goodby Miss Kinnian and dr Strauss and evrybody…&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS. please tel prof Nemur not to be such a grouch when pepul laff at him and he woud have more frends. Its easy to have frends if you let pepul laff at you. Im going to have lots of frends where I go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PPS. please if you get a chanse put some &lt;strong&gt;[book tilte]&lt;/strong&gt; grave in the bak yard.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If a characteristic spelling, intentionally left out even in the placeholder, does not give this one away, think of things that are typically put at the graves. Food? Books? Anyway, flowers come to my mind first. Seeking for a work of fiction with &lt;em&gt;Flowers&lt;/em&gt; in the title, it's &lt;em&gt;Flowers for Algernon&lt;/em&gt; by Daniel Keyes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;9&quot;&gt;&#xA;&lt;li&gt;“The common people pray for rain, healthy children, and a summer that never ends,” Ser Jorah told her. “It is no matter to them if the high lords play their &lt;strong&gt;[book title]&lt;/strong&gt;, so long as they are left in peace.”&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First, there's &lt;em&gt;something&lt;/em&gt; that high lords &lt;em&gt;play&lt;/em&gt;, so we have a good chance of guessing right by looking for titles containing &lt;em&gt;Game&lt;/em&gt;. A medieval fantasy series with lots of high lords? &lt;em&gt;Game of Thrones&lt;/em&gt; by George R. R. Martin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;10&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;One of the recurring philosophical questions is: “Does a falling tree in the forest make a sound when there is no one to hear?”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which says something about the nature of philosophers, because there is always someone in a forest. It may only be a badger, wondering what that cracking noise was, or a squirrel a bit puzzled by all the scenery going upwards, but someone. At the very least, if it was deep enough in the forest, millions of &lt;strong&gt;[book title]&lt;/strong&gt; would have heard it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Might be the hardest ones. Who hears the sound in the forest, but for the people and animals? Insects? Possibly. Supernatural beings? More fitting for the title of a book raising philosophical questions. Millions of supernatural beings? &lt;em&gt;Small Gods&lt;/em&gt; by Terry Pratchett.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Request&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Try finding the answers without looking up in the Internet. I am fully aware that search engines trivialize this puzzle to the point where it can be done programmatically. However, I believe my examples have all the clues needed to answer the questions on your own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do provide some feedback: was it hard or not? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can, please suggest more books that can be plausibly used in the puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT. Most of the book titles have been correctly answered at this point. I'm updating the post with hints and answers. I'll still be happy with feedback.&lt;/p&gt;&#xA;" OwnerUserId="19447" LastEditorUserId="19447" LastEditDate="2016-03-01T22:37:13.293" LastActivityDate="2016-03-03T16:39:58.940" Title="Recover book titles and suggest more books" Tags="|knowledge|no-computers|" AnswerCount="7" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="28041" PostTypeId="2" ParentId="28040" CreationDate="2016-02-27T22:43:58.910" Score="8" Body="&lt;p&gt;6 is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lord of the Flies  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And 7 must be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To Kill a Mockingbird  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastActivityDate="2016-02-27T22:43:58.910" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28042" PostTypeId="2" ParentId="28040" CreationDate="2016-02-27T22:47:36.907" Score="8" Body="&lt;p&gt;I think 3 is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Animal Farm.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As stated already, 6 and 7 are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lord of the Flies and To Kill a Mockingbird.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;8 is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Flowers for Algernon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17297" LastActivityDate="2016-02-27T22:47:36.907" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28043" PostTypeId="2" ParentId="28040" CreationDate="2016-02-27T23:02:25.957" Score="7" Body="&lt;p&gt;I've got one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;“and in the eyes of the people there is the failure; and in the eyes of the hungry there is a growing &lt;strong&gt;[part of the book title]&lt;/strong&gt;. In the souls of the people &lt;strong&gt;[book title]&lt;/strong&gt; are filling and growing heavy, growing heavy for the vintage.” &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastActivityDate="2016-02-27T23:02:25.957" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28044" PostTypeId="2" ParentId="28038" CreationDate="2016-02-27T23:10:35.773" Score="5" Body="&lt;p&gt;Edit: I kept going back and forth between whether it should be 15 or 16 squares. I used color to convince myself that 15 does actually work.&#xA;1 is in the bottom left corner of a 8x8 square.&#xA;All of the other numbers are the bottom left corners of a 4x4 square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dCHP9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dCHP9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First consider that is is necessary for each of the number around the perimeter, 1,2,3,4,5,6,8,10,12 to have a square which has its lower left corner there. There are 18 lines in the image and each lower left corner can produce at most 2 of the lines. Additionally, only the number 1 spot can have an 8x8 square, all the rest must have smaller squares. This means that each square along the lower or left perimeter leaves partially complete lines and a second square must be added to complete the lines.&lt;/p&gt;&#xA;" OwnerUserId="18821" LastEditorUserId="18821" LastEditDate="2016-02-28T04:05:27.633" LastActivityDate="2016-02-28T04:05:27.633" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28045" PostTypeId="1" AcceptedAnswerId="28046" CreationDate="2016-02-27T23:19:30.257" Score="11" ViewCount="1638" Body="&lt;p&gt;There are 33 students in a class. Every pupil writes the number of other pupils in this class with the same first name as him- or herself onto the blackboard. This procedure is repeated, now with the number of students with the same surname.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end, of the 66 numbers on the blackboard, each of the numbers $0,1,2\ldots,9,10$ appears at least once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that at least two students in this class bear the same fore- and surname.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Remark: Every student in this class has exactly one first name and one last name (hence no middle name).&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; I've thought about Tony's answer a bit, especially about the part &quot;the only way this works is if there are 1 students with name A, 2 students with name B, 3 students with name C, etc up to 11 students with name K&quot;. I noticed that $1+2+\ldots+10+11=66=2\cdot33$ which means that these 1 to 11 students have to be equally distributed among first and last names (if you know what I mean). Is this just a coincidence or is this fact in any way relevant for solving the problem?&lt;/p&gt;&#xA;" OwnerUserId="19635" LastEditorUserId="8874" LastEditDate="2016-03-08T17:47:43.670" LastActivityDate="2016-03-08T17:47:43.670" Title="Students in a class with the same name" Tags="|mathematics|logical-deduction|combinatorics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28046" PostTypeId="2" ParentId="28045" CreationDate="2016-02-27T23:47:58.607" Score="11" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Each of the numbers&#xA;  0,1,2…,9,10 appears at least once.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In fact, the only way this works is if there are 1 students with name A, 2 students with name B, 3 students with name C, etc up to 11 students with name K&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The names are distributed as 7 first names and 4 last names, or 5 first names and 6 last names or vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look at the group that has 11 people with same same first(last) name. Then they have at most 7 different last(first) names, and by the pigeon hole principle at least 2 of them are in the same hole, have the same last name and first name.&lt;/p&gt;&#xA;" OwnerUserId="18821" LastActivityDate="2016-02-27T23:47:58.607" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="28048" PostTypeId="2" ParentId="28040" CreationDate="2016-02-28T03:35:46.677" Score="4" Body="&lt;p&gt;I think 1 is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; White Fang&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And 5 is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Sound of Thunder&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but that's technically a short story.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way, it was kind of hard, but mostly becuase if you haven't happened to read that particular book you haven't a chance. &lt;/p&gt;&#xA;" OwnerUserId="19637" LastEditorUserId="19637" LastEditDate="2016-02-28T03:42:10.617" LastActivityDate="2016-02-28T03:42:10.617" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28049" PostTypeId="2" ParentId="28040" CreationDate="2016-02-28T04:05:34.290" Score="5" Body="&lt;p&gt;4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Catcher in the Rye&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;9:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;(A) Game of Thrones&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2016-02-28T04:05:34.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28050" PostTypeId="2" ParentId="28040" CreationDate="2016-02-28T05:41:29.723" Score="2" Body="&lt;p&gt;Note: whoops, I didn't read closely enough to see that this answer was already here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8 is  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Flowers For Algernon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but not exactly because that doesn't fit the sentence structure; it must be  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Flowers on Algernon's&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or something like that.&lt;/p&gt;&#xA;" OwnerUserId="12092" LastEditorUserId="12092" LastEditDate="2016-02-28T05:48:34.503" LastActivityDate="2016-02-28T05:48:34.503" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28051" PostTypeId="1" AcceptedAnswerId="28052" CreationDate="2016-02-28T07:09:51.267" Score="12" ViewCount="2116" Body="&lt;p&gt;Is it possible to make 1 with &lt;strong&gt;any&lt;/strong&gt; 3 numbers, all the same, such as using 3, 3 and 3 or 6, 6 and 6?&#xA;Can you think of a proof which will work no matter which number it is?&#xA;Trig functions allowed.&lt;/p&gt;&#xA;" OwnerUserId="19640" LastEditorUserId="16322" LastEditDate="2016-02-28T14:18:08.900" LastActivityDate="2016-07-22T20:32:52.777" Title="Make 1 with any 3 of the same numbers" Tags="|formation-of-numbers|" AnswerCount="8" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28052" PostTypeId="2" ParentId="28051" CreationDate="2016-02-28T07:20:00.410" Score="24" Body="&lt;p&gt;It is possible.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $x^{x-x} = 1$ for all $x \in \mathbb{C}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See it in action &lt;a href=&quot;http://www.wolframalpha.com/input/?i=x%5E(x-x)&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="14470" LastEditDate="2016-02-29T19:34:55.293" LastActivityDate="2016-02-29T19:34:55.293" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="28053" PostTypeId="2" ParentId="28040" CreationDate="2016-02-28T07:32:32.153" Score="8" Body="&lt;p&gt;10:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Small Gods&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Paul Evans's answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Grapes of Wrath&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1 and 9 I thought were easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2, 5 and 8 I don't know, and I'm pretty sure I haven't read.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 and Paul Evans's I didn't know right off, but the clues were enough to guess. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should think programmatically creating these would be of the same order of difficulty as solving them, and vastly more entertaining.&lt;/p&gt;&#xA;" OwnerDisplayName="user19641" LastEditorUserId="4421" LastEditDate="2016-03-03T16:39:58.940" LastActivityDate="2016-03-03T16:39:58.940" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28054" PostTypeId="2" ParentId="28051" CreationDate="2016-02-28T09:01:23.613" Score="32" Body="&lt;p&gt;There are certainly many ways. Here are a few ones.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\cos\left(x\left(x-x\right)\right) = \cos\left(0\right) = 1$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\sec\left(x\left(x-x\right)\right) = \sec\left(0\right) = 1$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\left(x\left(x-x\right)\right)! = 0! = 1$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These all rely on the fact that $x\left(x-x\right)$ is 0 for every $x\in\mathbb C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally (and this could be done with only one copy of any complex number $x$),&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\lfloor\left|\sin\left(\left|x+x-x\right|\right)\right|\rfloor!=\lfloor\left|\sin\left(\left|x\right|\right)\right|\rfloor!=1$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where $\left|x\right|$ is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Absolute_value#Complex_numbers&quot;&gt;modulus&lt;/a&gt; of $x$ and $\lfloor r\rfloor$ is the value of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Floor_and_ceiling_functions&quot;&gt;floor function&lt;/a&gt; at $r$, where $r\in\left[0,1\right]$ for any $x$. The same works if $\sin$ is replaced with $\cos$ or $\lfloor r\rfloor$ with $\lceil r\rceil$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, if we are allowed to cheat and use set theoretical constructs, then&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\left|\{x\}\cup\{x\}\cup\{x\}\right|=\left|\{x\}\cap\{x\}\cap\{x\}\right|=\left|\{x\}\right|=1$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;will work for whatever (number) $x$: the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cardinality&quot;&gt;cardinality&lt;/a&gt; of a set containing only one element is 1. This is true even if $x$ is not a complex quantity.&lt;/p&gt;&#xA;" OwnerUserId="4394" LastEditorUserId="4394" LastEditDate="2016-02-28T14:46:38.417" LastActivityDate="2016-02-28T14:46:38.417" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28055" PostTypeId="2" ParentId="28051" CreationDate="2016-02-28T09:11:27.063" Score="1" Body="&lt;p&gt;This works, too, but it's in the category of trick question not mathematics&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1 - 1 + 1 = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;also, for $x \ne 0$ this works:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^0 - x^0 + x^0 = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;working the problem almost suggests a possible quadratic form, if $a$ and $b$ are each $1$, or as above if each evaluate to $1$, and if $x$ is also $1$, as long as $c$ is negative:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ax^2 + bx - 1 = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this works even for $x = 0$, but I still want the answer to be funny somehow not functional:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1^x - 1^x + 1^x = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here's a humorous answer, where $x = 6$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;why is $6$ afraid of $7$? because $((x + 1) - (x + 2)) ^ {(x + 3)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($6$ is afraid of $7$ because $7, 8, 9$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;evaluated: because $(7 - 8) ^ 9$.&#xA;or: because $(-1) ^ 9 = -1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-1$ to the power of any number is still $1$, positive if even exponent and negative if odd -- is $-1$ to the power of $0$ a positive $1$? such obscure theoretical math, is there even a right answer or did the first person ever to write a book about math just decide which sign she wanted to attribute to the unusual $(-1) ^ 0$ case?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to phrase the punchline to the joke could be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;because $-1$ is the loneliest number.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since $(-1) ^ 9 = -1$ you get the &quot;because $7, 8, 9$&quot; answer as the lazy student's &quot;obvious&quot; funny answer that appears to satisfy the problem, but the &quot;right&quot; funny answer expected by the puzzle could be to finish working the problem, arriving at the number $-1$, which would be most clearly suggested as the intended solution like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Why is $x$ afraid of $7$? Because $((x + 1) - (x + 2)) ^ {(x + 3)}$ is the loneliest number.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;solve for $x$. what is the loneliest number?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;most students will know to find an $x$ that results in &quot;$1$ is the loneliest number&quot; so when they see $x = 6$ it will cause confusion and humor on the way to the intended &quot;right&quot; funny answer that says being negative AND being alone is why $6$ is afraid of $7$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and as a bonus, the math teacher gets to joke about how NEGATIVE $1$ is surely a lonelier number than POSITIVE $1$, just ask any lonely math teacher!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is fun stuff for puzzles. Thanks for posing the question!&lt;/p&gt;&#xA;" OwnerUserId="19642" LastEditorUserId="4421" LastEditDate="2016-03-01T17:46:16.017" LastActivityDate="2016-03-01T17:46:16.017" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28056" PostTypeId="2" ParentId="28051" CreationDate="2016-02-28T10:43:30.713" Score="8" Body="&lt;p&gt;For $x \neq 0$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\left(\dfrac{x}{x}\right)^x=1^x=1$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alternatively, the following works for all $x$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\left(\dfrac{\cos(\sin(x))}{\cos(\sin(x))}\right)^x=1^x=1$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16322" LastEditorUserId="16322" LastEditDate="2016-07-22T20:32:52.777" LastActivityDate="2016-07-22T20:32:52.777" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28057" PostTypeId="2" ParentId="28038" CreationDate="2016-02-28T11:54:28.240" Score="7" Body="&lt;p&gt;The minimum number is 14 squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A solution with 14 squares it to place four 7x7 squares, four 6x6 squares, and four 5x5 squares touching the four corners of the boards, plus two 4x4 squares touching two opposite corners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that 14 is optimal, consider the 28 segments that touch the outer edge of the board but don't lie on it. Any square covers a maximum of two of these segments: no square can reach two segments adjoining two opposite sides, if it touches two segments adjoining the same side, it can't reach any other segments. So, 14 squares are needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also implies that in any 14-square solution, every square must touch the edge of the board, and the 8x8 square cannot be used.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="4551" LastEditDate="2016-02-28T12:15:47.300" LastActivityDate="2016-02-28T12:15:47.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28058" PostTypeId="1" AcceptedAnswerId="28071" CreationDate="2016-02-28T13:29:23.947" Score="6" ViewCount="277" Body="&lt;p&gt;Each line refers to one word - find them!&lt;br/&gt;&lt;br&gt;&#xA;26. Who needs a best friend?  &lt;br/&gt;&#xA;26. all powerful.&lt;br/&gt;&#xA;50. Doctors lose business.&lt;br/&gt;&#xA;63. Judging books now, are we?&lt;br/&gt;&#xA;73. I point and squeak!&lt;br/&gt;&#xA;94. Clay-making magic!&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1. The puzzle is not coming from a crossword.&lt;br&gt;&#xA;2. The words are not connected to each other. I.e. There are 6 separate puzzles...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19086" LastEditorUserId="19086" LastEditDate="2018-03-14T19:46:09.510" LastActivityDate="2018-03-14T19:46:09.510" Title="Try and find the words" Tags="|enigmatic-puzzle|wordplay|word-problem|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28059" PostTypeId="1" CreationDate="2016-02-28T13:55:00.800" Score="14" ViewCount="1096" Body="&lt;p&gt;I came across this puzzle that involves a treasure hunt in a poem. The puzzle is as follows:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A legend of nine,&lt;br&gt;&#xA;  In the Andaman,&lt;br&gt;&#xA;  A treasure you may find,&lt;br&gt;&#xA;  But, certainly, the blood of men.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Staring at the Sun,&lt;br&gt;&#xA;  At 30 to 6,&lt;br&gt;&#xA;  Al asked the captain,&lt;br&gt;&#xA;  “What is the fix?”&lt;br&gt;&#xA;  “The Sun” says the cap,&lt;br&gt;&#xA;  “And blood shall determine our fate”&lt;br&gt;&#xA;  And so he screams “Die, Al”&lt;br&gt;&#xA;  And so remain 8.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Down the direction, so gracefully learnt, Atop a lonely tree, they find a nest.&lt;br&gt;&#xA;  They hear the ill omens of the crows, And another life goes to waste. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Expectedly,&lt;br&gt;&#xA;  Another turn of events,&lt;br&gt;&#xA;  Left the first mate forever lost,&lt;br&gt;&#xA;  Along with the tents.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;At 45 to 4, onward they went,&lt;br&gt;&#xA;  Towards the oasis in sight.&lt;br&gt;&#xA;  The most eager of all shattered their illusion,&lt;br&gt;&#xA;  And died due to the snake bite.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Soon, fell upon them, the night,&lt;br&gt;&#xA;  The predators were approaching fast,&lt;br&gt;&#xA;  And with backs turned to the Polaris,&lt;br&gt;&#xA;  They headed towards the past. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;As the Sun lifted into the sky,&lt;br&gt;&#xA;  So did the spirits of the grey.&lt;br&gt;&#xA;  They were ready again to face the odds as before,&lt;br&gt;&#xA;  But with only mirrors leading the way. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;“R.I.P. comrade,” said the captain,&lt;br&gt;&#xA;  And engulfed himself in loneliness,&lt;br&gt;&#xA;  Only to find that he was back,&lt;br&gt;&#xA;  At that inn with his mistress;  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;“The inn is the centre of everything,”&lt;br&gt;&#xA;  He admitted, amazed,&lt;br&gt;&#xA;  And it is there where I should have realized,&lt;br&gt;&#xA;  The curse of the maze.” &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The above poem is famous in the islands of Andaman and Nicobar. The treasure is rumoured to be still there. Find the place where the treasure is buried.&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hints:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Use &quot;breadcrumbs&quot;&lt;br&gt;&#xA;The Sun and time are great direction specifying tools&lt;br&gt;&#xA; Read it aloud&lt;br&gt;&#xA; The inn has many names.    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;More Hints:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (Para 1) At every turn there is a death&lt;br&gt;&#xA; (Para 2) It points towards the north&lt;br&gt;&#xA; (Para 3) The key is in the missing rhyme&lt;br&gt;&#xA; (Para 7) Everything they did yesterday, with one minor change about the needle of the compass.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Even more Hints:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (Para 1) Every vertex is signified by a death.&lt;br&gt;&#xA; (Para 2) The sundial can be used to signify an angle, given the latitude and the time of a place.&lt;br&gt;&#xA; (Para 3) The direction is in the missing rhyme.&lt;br&gt;&#xA; (Para 5) The Sun still guides the way.&lt;br&gt;&#xA; (Para 6) Go back to the point where you started from.&lt;br&gt;&#xA; (Para 7) Everything they did yesterday but with lateral inversion about the needle of a compass.&lt;br&gt;&#xA; (Para 9) Properties of an equilateral triangle.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19644" LastEditorUserId="19644" LastEditDate="2016-03-15T09:36:01.287" LastActivityDate="2016-03-15T17:20:41.160" Title="Find the treasure location with Al and the Captain" Tags="|riddle|rhyme|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="28060" PostTypeId="1" CreationDate="2016-02-28T13:32:12.500" Score="5" ViewCount="169" Body="&lt;p&gt;I found a nice problem. Here is what it says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Kevin and Peter are given a piece of paper each where they write a number.&lt;/p&gt;&#xA;&lt;p&gt;Their friend John comes up and takes the two pieces of paper and opens them up to read the numbers on them (he reads them to himself and does not say them out loud). He then goes up to a blackboard nearby and writes down two numbers, one the &lt;em&gt;&lt;strong&gt;sum of the numbers&lt;/strong&gt;&lt;/em&gt; written by Kevin and Peter and the other, an &lt;em&gt;&lt;strong&gt;arbitrary number&lt;/strong&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;p&gt;He then asks Kevin if he knows Peter's number.  After some thought, Kevin replies, &amp;quot;No.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;Next John asks Peter if he knows Kevin's number, to which Peter, after some thought, says, &amp;quot;No.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;John  keeps asking them the same question alternatively until, at some point, one of them says that he knows the other person's number and at the same moment, the other person jumps up and says that he knows the latter's number.&lt;/p&gt;&#xA;&lt;p&gt;How is this possible presuming that there is absolutely no cheating and we assume that Kevin and Peter are very good at math and are deep thinkers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="Gautham" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-03-01T17:53:25.817" Title="Finding the numbers" Tags="|probability|" AnswerCount="0" CommentCount="2" ClosedDate="2016-02-28T16:53:09.067" ContentLicense="CC BY-SA 3.0" />
  <row Id="28061" PostTypeId="2" ParentId="28058" CreationDate="2016-02-28T14:37:28.587" Score="3" Body="&lt;p&gt;Not sure about the rest, but the fifth line is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; mouse, referring to the compute mouse and the animal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The others may be similar.&lt;/p&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2016-02-28T14:37:28.587" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28062" PostTypeId="2" ParentId="28051" CreationDate="2016-02-28T16:04:24.690" Score="7" Body="&lt;p&gt;Here's one that works for 0:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $x!^{x - x} \forall x \in \{0, 1, 2, 3, ...\}$ (adapted from Zerris' answer)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastActivityDate="2016-02-28T16:04:24.690" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28063" PostTypeId="2" ParentId="28058" CreationDate="2016-02-28T17:08:24.677" Score="2" Body="&lt;p&gt;A wild shot in the dark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are the 3rd and 4th:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Under - as in: no more bills after the patient goes &lt;em&gt;under&lt;/em&gt;.&lt;br&gt;&#xA; Cover - as in: don't judge a book by it's &lt;em&gt;cover&lt;/em&gt;.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In which case the first could be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cat - since we have &lt;em&gt;undercover mouse&lt;/em&gt;: a cat's toy.&lt;br&gt;&#xA; We also have &lt;em&gt;find it&lt;/em&gt; in the title...   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can't fit the rest so probably barking up the wrong tree.&lt;/p&gt;&#xA;" OwnerUserId="10118" LastActivityDate="2016-02-28T17:08:24.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28064" PostTypeId="1" AcceptedAnswerId="28069" CreationDate="2016-02-28T17:33:50.727" Score="5" ViewCount="178" Body="&lt;p&gt;How many different ways are there to make a bangle by selecting $6$ beads from $6$ green beads and $6$ blue beads?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The bangle is a closed loop, and rotations of the same bangle are to be counted only once. The upside-down version of a bangle is not considered to be the same bangle.)&lt;/p&gt;&#xA;" OwnerUserId="19293" LastEditorUserId="8874" LastEditDate="2016-03-01T14:34:08.690" LastActivityDate="2016-03-01T15:43:35.527" Title="Making a bangle from $6$ green and $6$ blue beads" Tags="|mathematics|combinatorics|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28065" PostTypeId="1" AcceptedAnswerId="28067" CreationDate="2016-02-28T17:36:41.400" Score="5" ViewCount="3220" Body="&lt;p&gt;A gang of thieves has discovered that the 5 digit combination to a safe is even and has exactly one odd&#xA;digit and exactly two digits equal to each other. How many possible combinations are there for them to&#xA;try?&lt;/p&gt;&#xA;" OwnerUserId="19293" LastEditorUserId="8874" LastEditDate="2016-03-22T09:07:54.293" LastActivityDate="2016-03-22T09:07:54.293" Title="Cracking a 5 digit password" Tags="|mathematics|combinatorics|password|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28066" PostTypeId="2" ParentId="28064" CreationDate="2016-02-28T17:50:31.933" Score="4" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 14&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Argument:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If we cut the bangle up, we get a green-blue string with six symbols; altogether, there are $2\cdot2\cdot2\cdot2\cdot2\cdot2=64$ such strings.&#xA;However, now we are over-counting, as many rotations yield rotated versions of the same bangle.&lt;/li&gt;&#xA;&lt;li&gt;First, there are the two strings $bbbbbb$ and $gggggg$ of period length 1; they both are counted only once. (This case yields two strings and two bangles.)&lt;/li&gt;&#xA;&lt;li&gt;There is one string $bgbgbg$ of period length 2; this string yields the same bangle as $gbgbgb$. (This case yields two strings and one bangle.)&lt;/li&gt;&#xA;&lt;li&gt;There are two strings of period 3: $bbgbbg$ and $ggbggb$.&#xA;These strings yield the same bangles as $gbbgbb$ and $bgbbgb$, respectively&#xA;as $bggbgg$ and $gbggbg$. (This case yields six strings and two bangles.)&lt;/li&gt;&#xA;&lt;li&gt;The remaining $54$ strings correspond to nine bangles, that each is counted in six rotated versions. (This case yields 54 strings and nine bangles.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Altogether, this yields $2+1+2+9=14$ bangles.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2016-02-28T17:50:31.933" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28067" PostTypeId="2" ParentId="28065" CreationDate="2016-02-28T18:08:04.237" Score="8" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $7200$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Argument:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There are four possible parity patterns for the combination (o=odd, e=even):&#xA;$oeeee$, $eoeee$, $eeoee$, and $eeeoe$.&lt;/li&gt;&#xA;&lt;li&gt;Every parity pattern contains four even digits, of which two must be equal to each other. This gives ${4\choose2}=6$ possibilities for picking the two equal positions.&lt;/li&gt;&#xA;&lt;li&gt;Once we have fixed the parity pattern and the two positions with equal digits, there remain five possibilities for the odd digit ($1,3,5,7,9$) and $5\cdot4\cdot3=60$ possibilities for fixing the three distinct even digits ($5$ possibilities for the leftmost even digit, $4$ for the next, and $3$ for the last.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Altogether, this yields $4\cdot 6\cdot 5\cdot 60=7200$ possible combinations (out of $100,000$ combinations overall with five digits).&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="1766" LastEditDate="2016-02-28T18:57:48.767" LastActivityDate="2016-02-28T18:57:48.767" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28068" PostTypeId="2" ParentId="28064" CreationDate="2016-02-28T18:14:28.013" Score="3" Body="&lt;p&gt;We can consider $4$ cases:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All green:&#xA;$gggggg$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One blue:&#xA;$gggggb$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two blues:&#xA;$ggggbb$&#xA;$gggbgb$&#xA;$ggbggb$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three blues:&#xA;$gggbbb$&#xA;$ggbgbb$&#xA;$gbgbgb$&#xA;$gbggbb$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By symmetry on green/blue, this gives a total of $14$ bangles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $ggbgbb$ and $gbggbb$ are different because of their orientations.&lt;/p&gt;&#xA;" OwnerUserId="11497" LastActivityDate="2016-02-28T18:14:28.013" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28069" PostTypeId="2" ParentId="28064" CreationDate="2016-02-28T18:34:39.537" Score="5" Body="&lt;p&gt;There are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 13 different bangles.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Imagine a 6 digit binary number where the 1s are green and the 0s are blue.&lt;br&gt;&#xA; This number can rotated to give the same bangle.&lt;br&gt;&#xA; Doubling for swaping blue with green when there's an unequal number gives:&lt;br&gt;&#xA; 2 for 111111 -  2 numbers.&lt;br&gt;&#xA; 1 for 101010 -  2 numbers.&lt;br&gt;&#xA; 2 for 110110 -  6 numbers.&lt;br&gt;&#xA; 1 for 111000 -  6 numbers.&lt;br&gt;&#xA; 2 for 111100 - 12 numbers.&lt;br&gt;&#xA; 2 for 111110 - 12 numbers.&lt;br&gt;&#xA; 1 for 110010 - 12 numbers.&lt;br&gt;&#xA; 2 for 111010 - 12 numbers.&lt;br&gt;&#xA; ---------------------------------------------&lt;br&gt;&#xA; 13 bangles for all 64 numbers.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastEditorUserId="10118" LastEditDate="2016-02-28T19:20:23.860" LastActivityDate="2016-02-28T19:20:23.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28070" PostTypeId="2" ParentId="28001" CreationDate="2016-02-28T19:28:57.317" Score="6" Body="&lt;p&gt;I think it is a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; seven letter word from which letters are removed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The last four I think are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; dine, din, in, I&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Need to add more till I get back to the beginning. Then I will connect to clues (I know them, just don't want to type them if this is all wrong)&lt;/p&gt;&#xA;" OwnerUserId="1509" LastActivityDate="2016-02-28T19:28:57.317" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28071" PostTypeId="2" ParentId="28058" CreationDate="2016-02-28T19:45:35.497" Score="5" Body="&lt;p&gt;&amp;#8203;26. Who needs a best friend? &lt;br&gt;&#xA;&amp;#8203;26. all powerful.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &amp;thinsp; &lt;code&gt;dog&lt;/code&gt;  (or its reverse)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Almost there:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;pre&gt;&#xA;             D    O   G&#xA;             4 + 15 + 7  =  26. Who needs a best friend?.&#xA; &#xA;             G    O   D&#xA;             7 + 15 + 4  =  26. all powerful.&#xA; &#xA;   A    P    P    L   E&#xA;   1 + 16 + 16 + 12 + 5  =  50. Doctors lose business.&#xA; &#xA;   C    O    V   E    R&#xA;   3 + 15 + 22 + 5 + 18  =  63. Judging books now, are we?&#xA; &#xA;   M    O    U    S   E&#xA;  13 + 15 + 21 + 19 + 5  =  73. I point and squeak!&#xA; &lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So close:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;pre&gt;&#xA;  P    O    R   C   E    L   A    I    N&#xA; 16 + 15 + 18 + 3 + 5 + 12 + 1 +  9 + 14  =  93. Clay-making magic!&#xA;&#xA;           C   A    S    S   I    U    S&#xA;           3 + 1 + 19 + 19 + 9 + 21 + 19  =  91. Clay-making magic!&#xA;&#xA;           R    O    W    L   I    N   G&#xA;          18 + 15 + 23 + 12 + 9 + 14 + 7  =  98. Clay-making magic!&#xA; &lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thank you, @IvanBarreto and @manshu!:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;pre&gt;&#xA;    P    O    T    T   E    R&#xA;   16 + 15 + 20 + 20 + 5 + 18   =  94. Clay-making magic!&#xA; &lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18129" LastEditorUserId="18129" LastEditDate="2016-02-28T21:28:10.813" LastActivityDate="2016-02-28T21:28:10.813" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28072" PostTypeId="1" AcceptedAnswerId="28074" CreationDate="2016-02-28T22:53:04.320" Score="0" ViewCount="216" Body="&lt;p&gt;This sounds simple. What is the easiest solution to the Monty Hall problem you have which does not use listing out every probability and also &lt;strong&gt;requires you use algorithms?&lt;/strong&gt; Also, no research or computers.&lt;/p&gt;&#xA;" OwnerUserId="19640" LastActivityDate="2016-03-01T19:24:40.007" Title="Simplist proof you have for the Monty Hall Problem" Tags="|no-computers|monty-hall|" AnswerCount="2" CommentCount="2" ClosedDate="2016-02-29T08:25:18.100" ContentLicense="CC BY-SA 3.0" />
  <row Id="28073" PostTypeId="2" ParentId="28072" CreationDate="2016-02-28T23:02:33.957" Score="6" Body="&lt;p&gt;You have a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\frac13$ chance of picking the correct door, Monty'll show you one of the other doors, you switch and lose.&lt;br&gt;&#xA; $\frac23$ chance of picking a wrong door, Monty'll show you the &lt;em&gt;other&lt;/em&gt; wrong door, you switch and win.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;EDIT: Now I have better understanding of what you've after.&lt;br&gt;&#xA;You have a robot:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Preprogrammed to &lt;em&gt;always&lt;/em&gt; swap.&lt;br&gt;&#xA; First move: Get rid of a door at random.  Say that as pick to Monty.&lt;br&gt;&#xA; Second move: Get rid of door Monty shows.&lt;br&gt;&#xA; Last move: Pick remaining door.&lt;br&gt;&#xA; Now Monty will never show a winning door.&lt;br&gt;&#xA; So the second move never gets rid of a win.&lt;br&gt;&#xA; So the only way to lose is if the robot randomly picks the winning door in the first move with probably $\frac13$.&lt;br&gt;&#xA; And therefore the probability of a win is $1 - \frac13 = \frac23$   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastEditorUserId="10118" LastEditDate="2016-02-29T00:14:59.717" LastActivityDate="2016-02-29T00:14:59.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28074" PostTypeId="2" ParentId="28072" CreationDate="2016-02-28T23:07:14.197" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you switch, if you had the car selected before, you now get a goat.  If you had a goat selected, the other is revealed, so you now get the car.  Thus, &lt;em&gt;the probabilities after switching are exactly equal to the reverse of your original probabilities&lt;/em&gt;.&lt;br&gt;&#xA;&lt;br&gt;&#xA; You originally had a $\frac13$ chance of getting the car, so switching gives you a $\frac23$ chance.  Not switching leaves your original odds unchanged, at $\frac13$.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The generalization of this would be for a Monty Hall problem with $x$ doors, of which one is the &quot;winning&quot; door.  After your initial guess, all but one other door is revealed.  In this case, your initial guess has $\frac1x$ odds of being right, while switching has $1-\frac1x$ odds.  As a result, switching improves your odds by $1-\frac2x$.  Note that, for all $x &amp;gt; 2$, this is a positive change.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="4421" LastEditDate="2016-03-01T19:24:40.007" LastActivityDate="2016-03-01T19:24:40.007" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28075" PostTypeId="2" ParentId="28001" CreationDate="2016-02-28T23:46:42.603" Score="11" Body="&lt;p&gt;I believe @Kate is on the right track, and that the answer is indeed:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a seven letter word, which has letters removed each line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Specifically:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SAILING - &quot;set out [in a] driving wind&quot;&lt;br&gt;&#xA; AILING - &quot;all have fallen ill&quot;&lt;br&gt;&#xA; ALIGN - &quot;set us straight&quot;&lt;br&gt;&#xA; GAIN - &quot;his loss [is our gain]&quot;&lt;br&gt;&#xA; GIN - &quot;drown ourselves with a drink&quot;&lt;br&gt;&#xA; IN - &quot;leave outside, ... we hide&quot;&lt;br&gt;&#xA; I - &quot;the story lies with only one&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of course that leaves the last line &quot;unexplained&quot;, but presumably that just means:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; none of the crew survived, since following the pattern would necessarily make the last line nothing at all... But if that's true, who sent the message in the bottle?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="12408" LastEditDate="2016-02-28T23:52:38.383" LastActivityDate="2016-02-28T23:52:38.383" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28076" PostTypeId="2" ParentId="28030" CreationDate="2016-02-29T04:05:18.573" Score="3" Body="&lt;p&gt;The answer of 16, suggested as incorrect in the question, is the solution - I entered it into the app and it passed.&lt;/p&gt;&#xA;" OwnerUserId="16717" LastActivityDate="2016-02-29T04:05:18.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28078" PostTypeId="1" AcceptedAnswerId="28080" CreationDate="2016-02-29T07:41:25.163" Score="15" ViewCount="4179" Body="&lt;p&gt;I was asked this question about finding a number and three clues were given which are as follows.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If this number is multiple of 5, it would lie between 1 to 19.&lt;/li&gt;&#xA;&lt;li&gt;If this number is not multiple of 8, it would lie between 20 to 29.&lt;/li&gt;&#xA;&lt;li&gt;If this number is not multiple of 10, it would lie between 30 to 39.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I have tried a lot, but couldn't find any solution. Hope someone can help me. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="19658" LastEditorUserId="19658" LastEditDate="2016-06-20T13:21:47.877" LastActivityDate="2016-06-20T13:21:47.877" Title="Find out the number using three clues" Tags="|mathematics|logical-deduction|calculation-puzzle|" AnswerCount="7" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28080" PostTypeId="2" ParentId="28078" CreationDate="2016-02-29T07:48:46.843" Score="26" Body="&lt;p&gt;The correct answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $32$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This answer meets the conditions, because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is not a multiple of $5$ and not in $[1, 19]$, it is a multiple of $8$, so it is not in $[20, 29]$, and it is not a multiple of $10$, so it is in $[30, 39]$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So how can we get the number?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Any multiple of $5$, must be in $[1, 19]$. However, all multiples of $5$ in that range are not multiples of $8$, so they must be in $[20, 29]$, so obviously that is not the answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But what about?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A non-multiple of $8$ in that range will also be a non-multiple of $10$, except for $20$, but $20$ is a multiple of $5$, so we have no valid answers here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Finally, we are looking for a non-multiple of $5$, which is also a multiple of $8$ and a non-multiple of $10$, and is in the range $[30, 39]$. The only one such in that range is $32$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus there is no other number that satisfies all condition.&lt;/p&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="5610" LastEditDate="2016-02-29T12:26:57.093" LastActivityDate="2016-02-29T12:26:57.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28081" PostTypeId="2" ParentId="27876" CreationDate="2016-02-29T08:20:55.273" Score="10" Body="&lt;p&gt;I've been thinking on this problem for a long time, and I think I nailed it. I will expand on the ideas I used for the solution of part 1 of this puzzle, and also on the reasoning of Michael Seifert's answer.&lt;/p&gt;&#xA;&lt;p&gt;We may convert this problem into an analagous problem, in which we have a &lt;span class=&quot;math-container&quot;&gt;$9 \times 9$&lt;/span&gt; board and cross pieces:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iOw9a.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iOw9a.png&quot; alt=&quot;A cross piece: the center square represents the pawn&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The central red square represents the pawn. We want to find the maximum number of cross pieces we can place on the board, subject to:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;No two pieces overlap;&lt;/li&gt;&#xA;&lt;li&gt;The central red square of a piece always lies inside the board.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;It is allowed for orange squares to not lie on the board. Because of this, a placement of crosses on the board may feature up to &lt;span class=&quot;math-container&quot;&gt;$12$&lt;/span&gt; 'extra' squares, which do not lie on the board itself. One example of such placement is given below.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LuJfl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LuJfl.png&quot; alt=&quot;Example extra squares on the boundary&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;This means that, in theory, up to &lt;span class=&quot;math-container&quot;&gt;$81 + 12 = 93$&lt;/span&gt; squares are available for use. Since each piece covers &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt; squares, we may therefore not place &lt;span class=&quot;math-container&quot;&gt;$19$&lt;/span&gt; pieces on the board, (&lt;span class=&quot;math-container&quot;&gt;$19 \times 5 = 95 &amp;gt; 93$&lt;/span&gt;). We will improve this estimate by analyzing the placement of crosses along the boundary (those are the ones that generate the additional squares).&lt;/p&gt;&#xA;&lt;p&gt;Let &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; be the big square (with side length &lt;span class=&quot;math-container&quot;&gt;$9$&lt;/span&gt;) that is the boundary of the board. We will consider the possible configurations of crosses along &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; and show that, for each of its sides, the net gain of extra squares will never reach &lt;span class=&quot;math-container&quot;&gt;$9$&lt;/span&gt; or more, so that one may not place &lt;span class=&quot;math-container&quot;&gt;$18$&lt;/span&gt; crosses on the board (because there will never be &lt;span class=&quot;math-container&quot;&gt;$81 + 9 = 90$&lt;/span&gt; or more squares available for use).&lt;/p&gt;&#xA;&lt;p&gt;Now, for each side of &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt;, the &lt;em&gt;outer squares&lt;/em&gt; of that side are the squares that are &lt;strong&gt;not&lt;/strong&gt; on the board &lt;strong&gt;and&lt;/strong&gt; contain an edge entirely on that side of &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt;. Notice that each side of &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; contains at most &lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt; outer squares. For each configuration of crosses along a side of &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt;, the net gain of extra squares (along that side) will be the number of outer squares minus the number of squares that must remain cross-free as a consequence of the configuration itself. We divide it in cases:&lt;/p&gt;&#xA;&lt;h2&gt;Case &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;&lt;/h2&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lRyG1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lRyG1.png&quot; alt=&quot;Case 1&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Each X indicates that one of the two squares must be free of cross pieces, so this case gives us a net gain of one additional square.&lt;/p&gt;&#xA;&lt;h2&gt;Case &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt;&lt;/h2&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/brY6W.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/brY6W.png&quot; alt=&quot;Case2&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The X indicates that one of the two squares must be free of cross pieces, so this case gives us a net gain of two additional squares.&lt;/p&gt;&#xA;&lt;h2&gt;Case &lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt;&lt;/h2&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Rw5Hi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Rw5Hi.png&quot; alt=&quot;Case 3&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Each X indicates that one of the two squares must be free of cross pieces, so this case gives us a net gain of one additional square.&lt;/p&gt;&#xA;&lt;h2&gt;Case &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt;&lt;/h2&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0u9Ai.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0u9Ai.png&quot; alt=&quot;Case 4&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Case &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt; is separated in two other cases.&lt;/p&gt;&#xA;&lt;p&gt;If a square touching the boundary is not covered by a cross piece, then at least two other squares touching it will also not be covered. This is shown in the picture to the bottom left. Hence, in this case, no additional squares will be gained.&lt;/p&gt;&#xA;&lt;p&gt;On the other hand, if both squares touching the boundary are covered, then two of the squares with the big X over them (in the bottom right picture) must be free of cross pieces. Therefore, this case results in a single additional square.&lt;/p&gt;&#xA;&lt;h2&gt;Case &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt;&lt;/h2&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/olJfv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/olJfv.png&quot; alt=&quot;Case 5&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Each X indicates that one of the two squares must be free of cross pieces, so this case gives us a net gain of one additional square.&lt;/p&gt;&#xA;&lt;p&gt;It’s easy to see that these cover all possible configurations of three crosses on a side of &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt;, up to symmetry (reflection). Additionally, if a side of &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; has two or fewer crosses, the net gain will necessarily be two or less. All of this implies that the net gain of each side is never above two, so the total net gain is never above &lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt;, and in particular never above &lt;span class=&quot;math-container&quot;&gt;$9$&lt;/span&gt;, as claimed.&lt;/p&gt;&#xA;&lt;p&gt;Now, an attentive reader might point out that cases &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt; allow for overlap of cross-free areas. In other words, it might happen that we count a cross-free square twice, so the ‘actual’ net gain is higher. Visually, it should look like this:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5tClF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5tClF.png&quot; alt=&quot;Miscounting cross-free squares&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;https://i.gyazo.com/4051635f03fcbf14eb113099462e6dff.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;gyazo.com&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;The first observation is that this ‘confluence’ may occur at most two times, on opposite corners of the board. The second observation is that if one of the sides of the confluence falls into case &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; or case &lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt;, the total net gain for both sides of the confluence will be at most &lt;span class=&quot;math-container&quot;&gt;$6 – 2 = 4$&lt;/span&gt;, so we’re still good. Thus, we need only consider confluences in which both sides fall into case &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;To that end, we refine our analysis of case &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; itself. If the lone square touching the boundary is cross-free, the total net gain of the confluence will once again be at most &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt; and we’re good. Suppose then that the square is covered; the situation looks like this:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/96HU0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/96HU0.png&quot; alt=&quot;Further analysis of case 2&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;https://i.gyazo.com/f5696f9c1d3df3c56de80736c87cb26a.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;gyazo.com&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;It’s easy to see that at least one of the purple squares must be cross-free. The following picture shows the whole confluence. At least one of the purple squares and at least one of the pink squares must be cross-free. Hence, the total net gain for both sides of the confluence is &lt;span class=&quot;math-container&quot;&gt;$6 – 3 =3$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FVoSR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FVoSR.png&quot; alt=&quot;A confluence in which both sides fall into case 2&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;https://i.gyazo.com/42e5082f88b6b84b523e36a4ec8b140d.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;gyazo.com&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;Finally, observe that a confluence like the one above may happen at most once; in other words, there can be no additional overlap with the purple or pink squares. With this, I believe all cases are covered, so the highest desired integer is &lt;span class=&quot;math-container&quot;&gt;$x = 17$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Sorry for the lengthy answer, but it ended up being the simplest approach I could find.&lt;/p&gt;&#xA;" OwnerUserId="16478" LastEditorUserId="11110" LastEditDate="2020-12-27T03:58:07.563" LastActivityDate="2020-12-27T03:58:07.563" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28082" PostTypeId="2" ParentId="28078" CreationDate="2016-02-29T08:31:36.757" Score="10" Body="&lt;p&gt;The answer is :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 32&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we have to choose a number from natural numbers, then :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Any number we choose say $x \gt 39$; if $x$ is a multiple of $10$, then $x$ also a multiple of $5$. So as per rule no. $1$, $x$ should be in the range $[1,19]$. So $x$ cannot be greater than $39$. &lt;/li&gt;&#xA;&lt;li&gt;But then, the only multiple of $10$ (and $5$) in the range as per rule $1$, $x$ should be 10. Then as per rule $2$, since $10$ is not a multiple of $8$, $x$ has to be in the range $[20,29]$. Thus, it causes a contradiction!&lt;/li&gt;&#xA;&lt;li&gt;And if we choose a number (this time $y$) which is not a multiple of $10$, then it has to be in the range $[30,39]$. (holding onto this branch for further possibilities...)&lt;/li&gt;&#xA;&lt;li&gt;Also, a point to note that if we do the same for rule $2$, i.e. choose a number $x$, which is not a multiple of $8$, then we need to make sure that $x$ is a multiple of $10$ (by rule $3$). But by rule $1$, it's another contradiction!!&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thus, &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We need a number $y$, which is in range $[30,39]$, a multiple of 8 and not a multiple of 5 or 10. So, $\{30,31,32,33,34,35,36,37,38,39\}$ we cancel : &lt;strike&gt; 30 , 35 &lt;/strike&gt;, and the only acceptable number(multiple of $8$) left is $32$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1766" LastEditorUserId="1766" LastEditDate="2016-02-29T08:39:48.403" LastActivityDate="2016-02-29T08:39:48.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28083" PostTypeId="1" AcceptedAnswerId="28104" CreationDate="2016-02-29T10:00:50.693" Score="15" ViewCount="1517" Body="&lt;p&gt;The answer to this riddle is a single English word.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Within, only holes. A void all about&lt;br&gt;&#xA;  Talking back like a grave, flip the first for a shout.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastEditorUserId="17482" LastEditDate="2016-02-29T11:28:00.703" LastActivityDate="2016-03-01T12:28:09.313" Title="Discuss Among Yourselves" Tags="|riddle|word|wordplay|" AnswerCount="5" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="28084" PostTypeId="2" ParentId="28083" CreationDate="2016-02-29T10:24:50.627" Score="2" Body="&lt;h3&gt;This is not the correct answer(my achievement)...confirmed by OP&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I think the riddle fits for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Trap&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Within, only holes. A void all about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Holes are used as traps for centuries. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Talking back like a grave, flip the first for a shout.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Traps are to kill someone (sometimes). Take mouse trap for example, it flips and kills the mouse with a shriek.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-02-29T11:41:17.793" LastActivityDate="2016-02-29T11:41:17.793" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28085" PostTypeId="2" ParentId="28083" CreationDate="2016-02-29T10:59:16.090" Score="3" Body="&lt;p&gt;I think the answer is...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DAY&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't have a complete explanation for all parts of the riddle, but for some...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The title is &lt;strong&gt;D&lt;/strong&gt;iscuss &lt;strong&gt;A&lt;/strong&gt;mong &lt;strong&gt;Y&lt;/strong&gt;ourselves.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Within, only holes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (Unsure) - possibly means the middle letter has a hole.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A void all about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Night is either side of day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Talking back like a grave,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Talking back -&gt; Backwards. 'Day' backwards is 'yad' which means 'to remember' in Hindi (Credit to manshu in comments), which is what graves are for. Also, graves are sad, which sounds like 'yad'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;flip the first for a shout.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Flipping the first letter of 'day' gives 'bay' - a kind of shout.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18522" LastEditorUserId="18522" LastEditDate="2016-02-29T11:35:47.163" LastActivityDate="2016-02-29T11:35:47.163" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28086" PostTypeId="2" ParentId="17074" CreationDate="2016-02-29T11:35:23.390" Score="4" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All the time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The clock only has 12 hours, and there are 24 hours in a day. Every orientation has at least 2 possible times.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19103" LastEditorUserId="22529" LastEditDate="2016-05-29T05:51:26.143" LastActivityDate="2016-05-29T05:51:26.143" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28088" PostTypeId="2" ParentId="28083" CreationDate="2016-02-29T11:53:00.017" Score="5" Body="&lt;p&gt;I think the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;hell&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Within, only holes. A void all about:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; hell = the abyss = void = bottomless pit / hole&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Talking back like a grave, flip the first for a shout.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I can't seem to relate it to &quot;Talking back from a grave&quot; but flipping the letter &quot;h&quot; would make the word appear like &lt;strong&gt;yell&lt;/strong&gt; which means shout. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19191" LastActivityDate="2016-02-29T11:53:00.017" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28089" PostTypeId="1" AcceptedAnswerId="28091" CreationDate="2016-02-29T11:55:40.893" Score="5" ViewCount="1825" Body="&lt;p&gt;I have been started to learn how to solve rubiks cube with Philip Marshall's ultimate solution. With this algorithm, I can easily solve cage and 6/8 corners, but however two corners comes up in an awkward position. Is there a special move for this ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tVBlZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tVBlZ.jpg&quot; alt=&quot;picture&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19661" LastEditorUserId="30887" LastEditDate="2017-03-08T12:05:06.083" LastActivityDate="2017-10-24T15:40:32.167" Title="A simple solution for two awkwardly formed corners, is needed" Tags="|rubiks-cube|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28090" PostTypeId="2" ParentId="28051" CreationDate="2016-02-29T12:07:09.253" Score="3" Body="&lt;p&gt;From a more computer science angle (depending on how your language handles boolean to integer conversions):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (int)(A == (A &amp;amp; A))&lt;br&gt;&#xA; Means: ConvertToInteger(Is A equal to (A bitwise-anded with A)?)&lt;br&gt;&#xA; Because: Bitwise-anding A with A always gives A, and that A always equals A, so the result is True, which typically evaluates to 1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Probably only works for integers, though technically it would work for floats if you had a language that allowed it.&lt;/p&gt;&#xA;" OwnerUserId="16474" LastActivityDate="2016-02-29T12:07:09.253" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28091" PostTypeId="2" ParentId="28089" CreationDate="2016-02-29T12:12:50.993" Score="5" Body="&lt;p&gt;What I always do in this situation is keep the blue face in this case on the left side. Notice that one of the corners needs to be rotated counter clock wise and the other clock wise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make it so the one that needs to be rotated clock wise is in the top left back position then do this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;U R -U -R U R -U -R&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then rotate the left face (just that one, DONT rotate the entire cube!) so the other is in the top back left corner and to the opposite:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R U -R -U R U -R -U&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finnally just rotate the left face&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Additional information:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;U R -U -R is a basic move that is used in a lot of other formulas. It's interesting to see what actually happens when doing this move:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;top-back-left and top-back-right are switched&lt;/li&gt;&#xA;&lt;li&gt;top-front-right and bottom-front-right are switched&lt;/li&gt;&#xA;&lt;li&gt;middle-front-right, top-middle-right and top-back-middle are switched&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;With some interesting other properties:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Doing it twice puts the corner pieces involved back on their original position but rotated&lt;/li&gt;&#xA;&lt;li&gt;Doing it three times puts the edge-pieces involved back in their original position AND rotation.&lt;/li&gt;&#xA;&lt;li&gt;Doing it 6 times puts the cube in it's original position.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why it works here&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the left-face only top-back-left is touched with this formula. Because we know that this move only rotates the top-back-left piece when done twice and we then rotate the left face and reverse the move twice we have essential rotated to different corners in the left face while leaving everything else intact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus&lt;/strong&gt;&lt;br&gt;&#xA;In the case you actually have 3 corners that are rotated, it's actually always the case that all need to be rotated clockwise or all need to be rotated counter clockwise. You could of course perform this formula twice to solve it but there is a faster way. In the case of 3 corners that need to be rotated clockwise:&#xA;Do the same as before by putting the corner you want rotated in the top-left-back and do U R -U -R U R -U -R. then rotate the left face and just repeat it two times for the other two corners. This works because we know that after 6 time U R -U -R the cube is in it's starting position while the top-left-back is rotated after every 2 moves.&lt;/p&gt;&#xA;" OwnerUserId="1748" LastEditorUserId="1748" LastEditDate="2016-02-29T12:43:09.737" LastActivityDate="2016-02-29T12:43:09.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28092" PostTypeId="2" ParentId="28020" CreationDate="2016-02-29T12:48:59.570" Score="13" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cast&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now with many&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Multicast, denoting a set of data sent across a computer network to many users at the same time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now with having lovers&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Opencast, denoting a method of mining and open corresponds to a non-monogamous relationship status &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now with first swing&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Forecast, denoting a prediction or estimate. 'Fore' is the call accompanying a golfer's swing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now with some covers&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Overcast, denoting cloud covering a large part of the sky. 'Over' is also some of the word 'covers'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fifth clue&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We can also read each clue without the 'now' to spell out a set of answers &lt;br&gt;&lt;br&gt;  Many = &lt;b&gt;C&lt;/b&gt; (100 in Roman numerals) &lt;br&gt; Having lovers = &lt;b&gt;A&lt;/b&gt;dulterous &lt;br&gt; First swing = &lt;b&gt;S&lt;/b&gt; (first letter of swing) &lt;br&gt; Some covers = &lt;b&gt;T&lt;/b&gt;ucked in &lt;br&gt;&lt;br&gt; These answers form an acrostic with the first letter of each answer combining to spell out the word &lt;b&gt;CAST&lt;/b&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'Socratic' is an anagram of acrostic (hence the confusion) which is the method used to spell out the answer in the last part.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18422" LastEditorUserId="18422" LastEditDate="2016-02-29T12:55:02.690" LastActivityDate="2016-02-29T12:55:02.690" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28094" PostTypeId="1" CreationDate="2016-02-29T13:14:55.753" Score="4" ViewCount="906" Body="&lt;p&gt;3 2 1 Go&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the key has already been viewed, be fooled you will, taken from the left to the right it shall be revealed!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;ily all sha llh aou&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;{3} to 13th&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18214" LastEditorUserId="18214" LastEditDate="2016-02-29T13:59:27.617" LastActivityDate="2016-03-01T17:07:43.690" Title="Can you do this?" Tags="|cipher|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28096" PostTypeId="2" ParentId="27875" CreationDate="2016-02-29T13:45:21.037" Score="4" Body="&lt;p&gt;I have my own algorithm. It generates immediately an answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;24&lt;/strong&gt;    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A minimum set&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;AAAA AABB ABDC ACBA ACAB ADCD&lt;br&gt;&#xA;BAAB BABA BBDC BCAA BCBB BDCD&lt;br&gt;&#xA;CACC CBAD CBBD CCCC CDDA CDDB&lt;br&gt;&#xA;DADD DBCA DBCB DCDD DDAC DDBC  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2-letter word&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If the game is pick 2-letter word. The rules like  &lt;a href=&quot;https://www.chess.com/blog/kurtgodden/8-lonely-rooks&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.chess.com/blog/kurtgodden/8-lonely-rooks&lt;/a&gt;&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/YOzqP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YOzqP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA; Select 1 piece will match 7 positions.&lt;br&gt;&#xA;&lt;br&gt;&#xA;  &lt;a href=&quot;https://i.stack.imgur.com/CWNSn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CWNSn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;br&gt;&#xA; 2-letter word. &lt;strong&gt;4 steps&lt;/strong&gt;:&lt;br&gt;&#xA; Step 1: Guess any.&lt;br&gt;&#xA; Step 2,3,4: Guess any but not the same row or column as in picture.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The 3-letter(&lt;strong&gt;8 steps&lt;/strong&gt;) and 4-letter(&lt;strong&gt;24 steps&lt;/strong&gt;) like Lonely Rooks puzzle in 3D and 4D.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Continue: 3-letter&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; From 4x4 square above we create 4 squares 4x4 by add A,B,C,D at the end. So now we have a cube 4x4x4 like this&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/JbfzC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JbfzC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/7hcpp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7hcpp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;br&gt;&#xA; Select 1 pieces will match 10 position.&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/CnM8p.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CnM8p.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA; 3-letter word. &lt;strong&gt;8 steps&lt;/strong&gt;:&lt;br&gt;&#xA; Step 1,2: Choose any 2 pieces (but not same row or column) from A,B,C,D =&gt; I random choose from &lt;strong&gt;A&lt;/strong&gt; then select &lt;strong&gt;AAA&lt;/strong&gt; and &lt;strong&gt;CCA&lt;/strong&gt;.&lt;br&gt;&#xA; Step 3,4: Choose any 2 pieces from 3 other boards (B,C,D left) =&gt; random choose &lt;strong&gt;B&lt;/strong&gt; then select &lt;strong&gt;CAB&lt;/strong&gt; and &lt;strong&gt;ACB&lt;/strong&gt; (not same row or column, but same row &amp;amp; column with previous board, board A).&lt;br&gt;&#xA; Step 5,6: Choose any from other 2 boards (C,D left) =&gt; choose from &lt;strong&gt;C&lt;/strong&gt; 2 pieces &lt;strong&gt;BBC&lt;/strong&gt; &amp;amp; &lt;strong&gt;DDC&lt;/strong&gt; (not same row or column)&lt;br&gt;&#xA; Step 7,8: Choose 2 other pieces left from &lt;strong&gt;D&lt;/strong&gt; =&gt; &lt;strong&gt;BDD&lt;/strong&gt; &amp;amp; &lt;strong&gt;DBD&lt;/strong&gt;&lt;br&gt;&#xA;&#xA; 2 other sets: &#xA; &lt;a href=&quot;https://i.stack.imgur.com/q5gML.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/q5gML.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/9OeNV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9OeNV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;br&gt;&#xA;  4-letter game have very same method like this to get a minimum set (&lt;strong&gt;24 steps&lt;/strong&gt;) A 4-D Cube 4x4x4x4 : D&lt;br&gt;&#xA; &lt;a href=&quot;https://www.youtube.com/watch?v=Aonf34s0Bqg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=Aonf34s0Bqg&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;https://www.youtube.com/watch?v=VOUw4V1bKQQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=VOUw4V1bKQQ&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;4-letter word&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; From 4 squares 4x4 above we create 16 squares 4x4 by add A,B,C,D at the end. So now we have a cube 4x4x4 like this&#xA; &lt;a href=&quot;https://i.stack.imgur.com/1S9IV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1S9IV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA; Select 1 piece will match 13 positions.&#xA; &lt;a href=&quot;https://i.stack.imgur.com/uqOKE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uqOKE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA; Use same method as 3-letter will get result:&lt;br&gt;&#xA; AAAA BBAA ABBA BABA DCCA CDDA&lt;br&gt;&#xA; BAAB ABAB AABB BBBB DCCB CDDB&lt;br&gt;&#xA; CCAC CCBC ADCC BDCC DADC DBDC&lt;br&gt;&#xA; DDAD DDBD CACD CBCD ACDD BCDD  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19647" LastEditorUserId="19647" LastEditDate="2016-03-01T04:15:35.410" LastActivityDate="2016-03-01T04:15:35.410" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="28097" PostTypeId="1" AcceptedAnswerId="30492" CreationDate="2016-02-29T13:47:37.263" Score="13" ViewCount="826" Body="&lt;p&gt;A nonsensical string of letters to decipher:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;b z z e a z d j j i z a i j a z z d d z z a d z e j z a d z j g a t e z a t z e d z z h a z z d&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Don't be satisfied with a nonsensical string of words,&#xA;which will be just halfway to the answer.&lt;/p&gt;&#xA;" OwnerUserId="18129" LastEditorUserId="8874" LastEditDate="2016-03-05T13:59:36.607" LastActivityDate="2016-09-22T21:37:01.570" Title="Some string of letters" Tags="|cipher|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28098" PostTypeId="2" ParentId="28094" CreationDate="2016-02-29T13:52:48.723" Score="5" Body="&lt;p&gt;It's:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You shall all hail&lt;br&gt;&#xA; Because I can read it scrambled, but don't see how &quot;3 2 1 Go&quot; makes it so.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;EDIT: Or is it, thanks to @manshu and @ABcDexter&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Shall hail all you.&lt;br&gt;&#xA; Because that starts in the 3rd group grabbing &quot;sha llh a&quot; which needs &quot;il&quot;.&lt;br&gt;&#xA; The gets &quot;all&quot; from the second group and finally &quot;you&quot; from the first.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastEditorUserId="10118" LastEditDate="2016-02-29T14:12:21.177" LastActivityDate="2016-02-29T14:12:21.177" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="28099" PostTypeId="2" ParentId="28078" CreationDate="2016-02-29T14:08:32.987" Score="5" Body="&lt;p&gt;Let's find all solutions using formal logic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a &lt;em&gt;logic&lt;/em&gt; statement of the form &quot;if $A$ then $B$&quot;, we can express it symbolically as $A \implies B$ or its equivalent, $\bar A \lor B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Numbering the given statements 1,2,3 respectively, we have&lt;br&gt;&#xA;$A_1$ = the truth value of &lt;em&gt;this number is a multiple of 5&lt;/em&gt;,&lt;br&gt;&#xA;$B_1$ = the truth value of &lt;em&gt;this number would lie between 1 to 19&lt;/em&gt;,&lt;br&gt;&#xA;and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $E$ be (the truth value of) the conjunction of the 3 statements. Then:&#xA;$$\begin{align}&#xA;E = &amp;amp; (\bar A_1 \lor B_1) (\bar A_2 \lor B_2) (\bar A_3 \lor B_3) \\&#xA;= &amp;amp; \bar A_1 \bar A_2 \bar A_3 \lor \bar A_1 \bar A_2  B_3&#xA;    \lor \bar A_1 B_2 \bar A_3 \lor \bar A_1 B_2  B_3 \\&#xA;  &amp;amp; \lor B_1 \bar A_2 \bar A_3 \lor B_1 \bar A_2  B_3&#xA;    \lor B_1 B_2 \bar A_3 \lor B_1 B_2  B_3&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since every multiple of 5 is also a multiple of 10, the conjunction $A_1 A_3$ must be false because it asserts that the number is a multiple of 5 and not a multiple of 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, $B_i B_j$ is false if $i \neq j$ because the intervals being disjoint means that the number cannot be simultaneously a member of two intervals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can therefore simplify to get $E = \bar A_1 \bar A_2  B_3 \lor B_1 \bar A_2 \bar A_3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first term requires a multiple of 8 from the interval 30 to 39 that is not a multiple of 5, so 32 is a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second term requires a multiple of 8 and 10 from the interval 1 to 19. There aren't any.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hence 32 is the only solution.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9422" LastEditorUserId="9422" LastEditDate="2016-02-29T14:26:53.870" LastActivityDate="2016-02-29T14:26:53.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28100" PostTypeId="1" AcceptedAnswerId="28101" CreationDate="2016-02-29T14:09:23.210" Score="6" ViewCount="444" Body="&lt;p&gt;Ten years before Alice's age is 4 times John's age. Ten years later, Alice's age is 2 times John's age. How old is John?&lt;/p&gt;&#xA;" OwnerUserId="19662" LastEditorUserId="12109" LastEditDate="2016-03-19T19:28:28.017" LastActivityDate="2016-03-20T14:16:52.880" Title="Find age of John" Tags="|mathematics|calculation-puzzle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28101" PostTypeId="2" ParentId="28100" CreationDate="2016-02-29T14:11:06.903" Score="9" Body="&lt;p&gt;John is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 20. Ten years earlier, John was 10 and Alice was 40. Ten years from now, Alice will be 60 and John will be 30.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastActivityDate="2016-02-29T14:11:06.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28102" PostTypeId="2" ParentId="28094" CreationDate="2016-02-29T14:11:54.223" Score="3" Body="&lt;p&gt;&lt;strong&gt;Explanation&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Can you do this?&quot; has &lt;strong&gt;13&lt;/strong&gt; visible characters &lt;br/&gt;3 2 1 go =&gt; the number &lt;strong&gt;3&lt;/strong&gt; is right above &quot;Here&quot;&lt;br/&gt;Here...&lt;br/&gt;&lt;br/&gt;The sentence was divided into 5 pieces made of three letters each&lt;br/&gt;&lt;br/&gt;ily all sha llh aou&lt;br/&gt;&lt;br/&gt;Until this point a little bit of guessing is needed&lt;br/&gt;&lt;br/&gt;Hint:&lt;br/&gt;&lt;br/&gt;{3} to 13th&lt;br/&gt;{ily} all sha llh....  a...  ou&lt;br/&gt;.........................13th&lt;br/&gt;&lt;br/&gt;&lt;strong&gt;{3} to 13th&lt;/strong&gt; means a group &lt;code&gt;{}&lt;/code&gt; containing &lt;code&gt;3&lt;/code&gt; letters to the &lt;code&gt;13th&lt;/code&gt; thirteenth position&lt;br/&gt;&lt;br/&gt;all sha llh  a{ily}    ou&lt;br/&gt;&lt;br/&gt;Thus:&lt;br/&gt;&lt;br/&gt;&lt;strong&gt;&lt;em&gt;all shall hail you&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18214" LastEditorUserId="18214" LastEditDate="2016-03-01T17:07:43.690" LastActivityDate="2016-03-01T17:07:43.690" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="28103" PostTypeId="1" CreationDate="2016-02-29T14:17:33.157" Score="0" ViewCount="91" Body="&lt;p&gt;Assuming that Average rain rained for 7 days is 7 cm.&#xA;First 6 days's average rain rained is 6 cm, then how much is rain rained on the 7th day of the week?&lt;/p&gt;&#xA;" OwnerUserId="19662" LastActivityDate="2016-02-29T14:24:54.313" Title="Average rain for 7 days" Tags="|calculation-puzzle|" AnswerCount="1" CommentCount="2" ClosedDate="2016-02-29T14:53:04.310" ContentLicense="CC BY-SA 3.0" />
  <row Id="28104" PostTypeId="2" ParentId="28083" CreationDate="2016-02-29T14:22:48.307" Score="20" Body="&lt;p&gt;Ooh, what about...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Moot?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Within, only holes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter O is the only letter in the middle of the word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A void all about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;M&quot; and &quot;T&quot; spoken out loud is &quot;empty.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Talking back like a grave,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reading moot backwards: toom sounds like &quot;tomb&quot; a grave&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;flip the first for a shout&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Woot!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19663" LastEditorUserId="1998" LastEditDate="2016-03-01T12:28:09.313" LastActivityDate="2016-03-01T12:28:09.313" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28105" PostTypeId="2" ParentId="28100" CreationDate="2016-02-29T14:23:20.220" Score="11" Body="&lt;p&gt;John is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 20&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Equations&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let age of Alice now be &lt;em&gt;a&lt;/em&gt; and age of John now be &lt;em&gt;j&lt;/em&gt;. Therefore 10 years ago, &lt;br&gt; $$a-10=4(j-10)$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After 10 years &lt;br&gt; $$a+10=2(j+10)$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solving two equations we get&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; j = 20 and a = 50. So today age of John is 20 years.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This proves that this is a $6^{th}$ grade math problem.&lt;/p&gt;&#xA;" OwnerUserId="17482" LastActivityDate="2016-02-29T14:23:20.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28106" PostTypeId="2" ParentId="28051" CreationDate="2016-02-29T14:23:44.983" Score="0" Body="&lt;p&gt;It is possible and simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\frac11 )\times1 =1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1\times1\times1 = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^{(x-x)} =1$  (except $x$ not $0$)&lt;/p&gt;&#xA;" OwnerUserId="19662" LastEditorUserId="4421" LastEditDate="2016-03-01T17:47:28.133" LastActivityDate="2016-03-01T17:47:28.133" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28107" PostTypeId="2" ParentId="28103" CreationDate="2016-02-29T14:24:54.313" Score="2" Body="&lt;p&gt;It's:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 13cm.&lt;br&gt;&#xA; First 6 days averaged 6cm so 36 cm of rain fell.&lt;br&gt;&#xA; Need 49cm for 7 day average of 7cm.&lt;br&gt;&#xA; $49 - 36 = 13$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastActivityDate="2016-02-29T14:24:54.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28108" PostTypeId="2" ParentId="28083" CreationDate="2016-02-29T14:34:10.537" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MOOT&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Within, only holes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter O's can be considered holes &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A void all about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A moot point is often void of importance&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Talking back like a grave&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A moot point is an open/unresolved question, you don't often get answers when talking to a grave&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;flip the first for a shout&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; WOOT WOOT!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2016-02-29T14:34:10.537" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28110" PostTypeId="2" ParentId="28051" CreationDate="2016-02-29T15:00:39.520" Score="1" Body="&lt;p&gt;This should work too:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $x^{x-x}=x^0=1$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19664" LastEditorUserId="17482" LastEditDate="2016-02-29T15:18:25.263" LastActivityDate="2016-02-29T15:18:25.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28111" PostTypeId="1" AcceptedAnswerId="28112" CreationDate="2016-02-29T15:37:19.093" Score="9" ViewCount="1002" Body="&lt;p&gt;They call me play, but I am no fun&lt;br&gt;&#xA;Under my civilized veneer&lt;br&gt;&#xA;I'm full of Death and destruction&lt;br&gt;&#xA;The ultimate in Betrayal, and bad wives too&lt;br&gt;&#xA;I wonder why anyone ever comes to&lt;br&gt;&#xA;Me knowth that all will end bad&lt;br&gt;&#xA;It always does. Yet again and again&lt;br&gt;&#xA;As if you don't know how it will end.&lt;br&gt;&#xA;March is coming. One ought to know better.&lt;/p&gt;&#xA;" OwnerUserId="19179" LastEditorUserId="19179" LastEditDate="2018-04-13T18:19:00.687" LastActivityDate="2018-04-13T18:19:00.687" Title="No Fun At All, Again" Tags="|riddle|wordplay|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28112" PostTypeId="2" ParentId="28111" CreationDate="2016-02-29T15:52:48.550" Score="12" Body="&lt;p&gt;I think that this is about:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The play Julius Caesar&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They call me play, but I am no fun&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This play is a tragedy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Under my civilized veneer&#xA;I'm full of Death and destruction&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is lots of killing and bloodshed in this play, but plays (especially Shakespearian plays) have formal, civilized connotations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The ultimate in Betrayal,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Et tu brute! &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and bad wives too&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Portia commits suicide, Calpurnia has premonitions about Caesar's betrayal. (the wives of Brutus and Caesar, respectively)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I wonder why anyone ever comes to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Me knowth that all will end bad&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It always does. Yet again and again&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As if you don't know how it will end.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is all about how we go to see plays over and over, but they always end the same way. We know it is a tragedy going in.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;March is coming. One aught to know better.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'The ides of march' is a famous phrase from the play, which is when Caesar is betrayed and killed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19491" LastActivityDate="2016-02-29T15:52:48.550" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28113" PostTypeId="2" ParentId="28078" CreationDate="2016-02-29T17:23:11.167" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's impossible for it to be divisible by 5 without being divisible by 8 or 10, so it can only be outside of the range if it's really a multiple of 5. Otherwise it can't be a multiple of 10, so it falls into the 3rd category, and this category only, thus it has to be div. by 8. The only suitable number would be 32.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastActivityDate="2016-02-29T17:23:11.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28114" PostTypeId="2" ParentId="27798" CreationDate="2016-02-29T19:16:20.967" Score="5" Body="&lt;p&gt;I would&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;run the other way for a(a) bee(b), they see(c) a(a) bee(b) see(c) a(a) beeeeeeeeee(b)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16471" LastActivityDate="2016-02-29T19:16:20.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28115" PostTypeId="2" ParentId="28078" CreationDate="2016-02-29T19:52:46.390" Score="6" Body="&lt;p&gt;Do you enjoy programming?  Try this in Java:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;public void findTheDangNumber() {&#xA;    for (int i = 1; true; i++){&#xA;        if (i % 5  == 0 &amp;amp;&amp;amp; (i &amp;lt;= 1  || i &amp;gt;= 19)) continue;&#xA;        if (i % 8  != 0 &amp;amp;&amp;amp; (i &amp;lt;= 20 || i &amp;gt;= 29)) continue;&#xA;        if (i % 10 != 0 &amp;amp;&amp;amp; (i &amp;lt;= 30 || i &amp;gt;= 39)) continue;&#xA;        System.out.println(&quot;answer: &quot; + i);&#xA;        break;&#xA;    }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What's happening is we are running until all of the conditions are met.  Each &quot;if&quot; line basically says, &quot;if it's divisible by the given number and is outside of the range, then skip to the next number.&quot;  If each of the &quot;if&quot; lines is satisfied, then we've found our answer and we print it out! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;: Thanks to @Ethan for help with the logic.&lt;/p&gt;&#xA;" OwnerUserId="18987" LastEditorUserId="18987" LastEditDate="2016-02-29T21:39:50.827" LastActivityDate="2016-02-29T21:39:50.827" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="28117" PostTypeId="1" AcceptedAnswerId="28121" CreationDate="2016-02-29T20:55:01.193" Score="9" ViewCount="573" Body="&lt;p&gt;Hangman game backward #8 - guess what was the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;fill in &quot;d&quot; - It does something horrible to the sand and stones, but it did the opposite to us.&lt;br&gt;&#xA;fill in &quot;f&quot; - To some it's rapidly dangerous, others it's steady dangerous.&lt;br&gt;&#xA;fill in &quot;p&quot; - It's from a drop of water, it's also from the American Beauty. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, your answer can be in the form of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The question was: _ p p l e&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First answer that fully explained each line will be accepted, enjoy!&lt;/p&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2016-03-01T13:59:31.833" Title="The hangman game backward #8" Tags="|riddle|knowledge|english|language|letters|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28118" PostTypeId="2" ParentId="28078" CreationDate="2016-02-29T21:05:00.423" Score="4" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $32$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here are my deductions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. The number is not a multiple of $8$ and $10$. If it were, then it must necessarily be a multiple of $5$ and be in the range $[1, 19]$. None of the numbers in that range fit the criteria.&lt;br /&gt;&lt;br /&gt;&#xA; 2. The number is a multiple of either $8$ or $10$. If it weren't, then it would have to be in both $[20, 29]$ and $[30, 39]$ &amp;mdash; a contradiction.&lt;br /&gt;&lt;br /&gt;&#xA; 3. From the second deduction we can narrow things down to the following numbers: $20, 32$.&lt;br /&gt;&lt;br /&gt;&#xA; 4. The answer must be $32$ since $20$ doesn't meet the first condition.&lt;br /&gt;&lt;br /&gt;&#xA; In hindsight, the first deduction is unneeded. I've kept it since that's how I initially processed the last two conditions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11500" LastActivityDate="2016-02-29T21:05:00.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28119" PostTypeId="1" AcceptedAnswerId="28122" CreationDate="2016-02-29T21:22:51.127" Score="42" ViewCount="3152" Body="&lt;p&gt;A chess king has been injured in battle against an evil wizard, and can no longer move northeast or southwest. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/p78fws.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/p78fws.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This king is on the North rim of a canyon, and must flee to safety on the South rim. The only way to cross is a bridge, which is a shaped like a standard chessboard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as the king is about to cross, the evil wizard casts a fire spell on all 64 squares of the bridge. Each square is destroyed with a probability of 50%, independently of the others. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the probability that the king can cross the canyon via the undestroyed squares?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To be clear, the king is allowed to start on any undestroyed square on the top row of the chessboard, and succeeds as long as he reaches any undestroyed square on the bottom row.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YeI0j.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YeI0j.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10615" LastActivityDate="2016-02-29T22:24:39.103" Title="Crippled King Crossing a Canyon" Tags="|probability|chess|checkerboard|graph-theory|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28120" PostTypeId="2" ParentId="28023" CreationDate="2016-02-29T21:28:02.200" Score="5" Body="&lt;p&gt;Alice&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Allied Powers/Allies of World War I and World War II&lt;/strong&gt; This link was just based on phonetics and context of the rest of the puzzle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mr. Cantor&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Central Powers:&lt;/strong&gt; &lt;em&gt;[husband of Alice]&lt;/em&gt; Opponent of the Allies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alice and Cantor's son &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;World War I:&lt;/strong&gt; &lt;em&gt;From Mr. Cantor, she had a son. He was very mischievous. All people in the house were bothered by him. But his fun-time ended with time.&lt;/em&gt; A war between the Central Powers and the Allies, where there was a great loss of lives and resources. The war involved the entire world, but the war eventually ended.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alex &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Axis Powers:&lt;/strong&gt; &lt;em&gt;[husband of Alice]&lt;/em&gt; Opponent of the Allies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alice and Alex's son&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;World War II:&lt;/strong&gt; &lt;em&gt;After some years, she had another son from Alex. He was even more mischievous than his brother. So of course, people in the home were even more bothered by him.&lt;/em&gt;  World War II had a greater toll on lives and economies than WWI, which set an uneasiness in the global zeitgeist.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Phat &amp;amp; Stuart&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Fat Man &amp;amp; Little Boy:&lt;/strong&gt; &lt;em&gt;Once he called his friends, Phat and Stuart at home.&lt;/em&gt; Like the atomic bombings of Hiroshima and Nagasaki. Stuart Little links to Little Boy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yun Ong&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;United Nations:&lt;/strong&gt; &lt;em&gt;Hearing so much sound, their landlord Yun Ong came to their home and told the whole family to not call such a devilry kids at home.&lt;/em&gt; This could be analogous to the signing of the Treaty on the Non-Proliferation of Nuclear Weapons by the UN in 1968&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17825" LastEditorUserId="17825" LastEditDate="2016-03-01T18:35:14.057" LastActivityDate="2016-03-01T18:35:14.057" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28121" PostTypeId="2" ParentId="28117" CreationDate="2016-02-29T21:42:42.233" Score="9" Body="&lt;p&gt;I think it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ri_ _le&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fill in &quot;d&quot; - It does something horrible to the sand and stones, but it did the opposite to us.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Riddle means 'to pierce with many holes' and we know what holes do. But here riddles are amazing. OP meant that riddle separates stones and sand, but bind us here in SE.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fill in &quot;f&quot; - To some it's rapidly dangerous, others it's steady dangerous.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Riffle means 'a rocky or shallow part of a stream or river where the water flows brokenly'. A fast flowing water is dangerous. It's even more dangerous when it's rocky and cold.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;fill in &quot;p&quot; - It's from a drop of water, it's also from the American Beauty.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ripple of water. American Beauty (album) by 'Grateful Dead' had a song named &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Ripple_(song)&quot; rel=&quot;nofollow&quot;&gt;Ripple&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All credit goes to my dictionary. :)&lt;/p&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-03-01T13:59:31.833" LastActivityDate="2016-03-01T13:59:31.833" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28122" PostTypeId="2" ParentId="28119" CreationDate="2016-02-29T22:24:39.103" Score="57" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Exactly 50%.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Consider another injured king (also can't move NE or SW) who is trying to get from the left side to the right side, but only travels on destroyed squares. By the &lt;a href=&quot;http://www.cut-the-knot.org/Curriculum/Games/HexTheory.shtml&quot;&gt;Hex Theorem&lt;/a&gt;, the second king can cross if and only if first king can't. Because the situation is symmetrical (every square has a 50% chance of being destroyed), each king can cross exactly half of the time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2016-02-29T22:24:39.103" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="28123" PostTypeId="2" ParentId="28045" CreationDate="2016-02-29T22:27:42.070" Score="3" Body="&lt;p&gt;Let the number of same first names for each written number be $n_0,n_1,...,n_{32}$ and that of the same surnames $m_0,m_1,...,m_{32}$. $n_0$ or $m_0$ denotes the number of unique names, $n_x$ or $m_x$ denotes the number of names common to $x+1$ people, and so on. Therefore, $n_x$ and $m_x$ must be divisible by $x+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum $n_0+m_0...+n_{10}+m_{10}$ can be 1+2+...+11=66 at a minimum, which is the total number of the written numbers. For there to be no full name overlap, the 11-strong group with the same first name/surname must all have different surnames/first names (if the first name is the same, the surnames must be different and vice versa), but adding up 11 existing same name/surname groups will give 66, which is greater than 33. In short, there IS at least one overlap.&lt;/p&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="14016" LastEditDate="2016-03-01T01:33:02.167" LastActivityDate="2016-03-01T01:33:02.167" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28124" PostTypeId="1" AcceptedAnswerId="28126" CreationDate="2016-02-29T23:54:46.137" Score="7" ViewCount="406" Body="&lt;p&gt;Fuzzy the fuzzball was very sad.&lt;br&gt;&#xA;He had lost his fuzz.&lt;br&gt;&#xA;Everyone knew if you lost your Fuzz you could no longer be a fuzzball.&lt;br&gt;&#xA;Fuzzy went to this friend Frizzy and shared his dilemma.&lt;br&gt;&#xA;&quot;Don't worry, Fuzzy,&quot; Frizzy said. &quot;You can buy new fuzz in the FuzzyFuzz store.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fuzzy went to the FuzzyFuzz store and asked to buy Fuzz.&lt;br&gt;&#xA;&quot;How much Fuzz do you need?&quot; the clerk asked.&lt;br&gt;&#xA;&quot;Enough to make me fuzzy,&quot; Fuzzy said. &quot;I'm a perfect sphere, 2.3456&quot; in diameter.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;We sell three types of Fuzz,&quot; said the clerk. &quot;Spartan fuzz is \$2.34 per cm². Standard fuzz is \$3.45  per cm². Super Fuzz is \$4.56 per cm².&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Ok,&quot; said Fuzzy. &quot;I have \$543.21. I will spend up to 67% of my money to buy the best Fuzz I can afford. I sure hope I can afford to become Super Fuzzy!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How fuzzy is Fuzzy the fuzzball after his purchase?&lt;/p&gt;&#xA;" OwnerUserId="19179" LastEditorUserId="19179" LastEditDate="2016-03-01T00:55:01.127" LastActivityDate="2018-11-22T20:27:25.477" Title="Fuzzy the Fuzzball" Tags="|calculation-puzzle|word-problem|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28125" PostTypeId="1" AcceptedAnswerId="28166" CreationDate="2016-03-01T00:10:02.940" Score="9" ViewCount="685" Body="&lt;p&gt;I'm all over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over you. Under you. Through you.&lt;br&gt;&#xA;&amp;lt; Insert Evil Laugh &gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I love black holes and dying cosmos.&lt;br&gt;&#xA;I come from evil politicians and ruthless dictators.&lt;br&gt;&#xA;I come from silly children and irresponsible criminals.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can cause horrible diseases.&lt;br&gt;&#xA;The government overregulates me.&lt;br&gt;&#xA;But that doesn't stop me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mwah Ha Ha.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nothing can stop me.&lt;br&gt;&#xA;Not walls. Not floors. Not Time or Space or Distance.&lt;br&gt;&#xA;I OWN THE WORLD!!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not that it does me much good.&lt;br&gt;&#xA;Sometimes I think of my existence as a giant treadmill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&#xA;Focus on these lines: (Though the correct answer will satisfy ALL clues.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Nothing can stop me. Not time or...distance.&#xA;The government regulates me. &#xA;It can (but often doesn't) cause horrible diseases.&lt;br&gt;&#xA;And disregard the evil laughter. It's just there for comic effect and to make the case more stark. This is not actually evil at all. Nor does it literally &quot;own the world.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19179" LastEditorUserId="19179" LastEditDate="2016-03-11T16:20:13.723" LastActivityDate="2016-03-11T16:20:13.723" Title="Preposition Hayday" Tags="|riddle|" AnswerCount="9" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28126" PostTypeId="2" ParentId="28124" CreationDate="2016-03-01T00:42:13.877" Score="8" Body="&lt;p&gt;So Fuzzy's total surface area is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$17.28$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; we multiply that by &lt;span class=&quot;math-container&quot;&gt;$2.54^2$&lt;/span&gt; to get cm. &lt;br/&gt;&lt;span class=&quot;math-container&quot;&gt;$111.483648$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We now find 67% of 543.21&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;363.9507&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now we divide the money into parts. How many &lt;span class=&quot;math-container&quot;&gt;$cm^2$&lt;/span&gt; total for each?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Super&lt;/strong&gt;: 79.91375 cm&lt;span class=&quot;math-container&quot;&gt;$^2$&lt;/span&gt; for \$363.9507&lt;br/&gt;&lt;strong&gt;Regular&lt;/strong&gt;: 105.49&lt;br/&gt;&lt;strong&gt;Spartan&lt;/strong&gt;: 155.53&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; since only one of them is obviously greater, we can say he can be completely covered in Spartan Fuzz.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17970" LastEditorUserId="52483" LastEditDate="2018-11-22T20:27:25.477" LastActivityDate="2018-11-22T20:27:25.477" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28128" PostTypeId="2" ParentId="28124" CreationDate="2016-03-01T01:03:21.977" Score="6" Body="&lt;p&gt;Fuzzy's:&lt;br&gt;&#xA;EDIT: Thanks to @bleh&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  2.3456 inches in diameter&lt;br&gt;&#xA;  Which is $\approx 2.9789cm$ in radius&lt;br&gt;&#xA;  For an total surface area of $4 \pi 2.9789^2 \approx 111.51cm^2$&lt;br&gt;&#xA;  He has $67\%$ of $\$543.21$ to spend so can afford $\frac{67}{100} *\frac{\$543.21}{111.51cm^2}\approx \$3.26/cm^2$&lt;br&gt;&#xA;  Spartan fuzz is $ \$2.34/cm^2$.&lt;br&gt;&#xA;  Standard fuzz is $\$3.45/cm^2 $&lt;br&gt;&#xA;  Super Fuzz costs $ \$4.56/cm^2$&lt;br&gt;&#xA;  So, Can we mix a bit of Spartan with Standard?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastEditorUserId="10118" LastEditDate="2016-03-01T01:20:59.050" LastActivityDate="2016-03-01T01:20:59.050" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28129" PostTypeId="2" ParentId="28125" CreationDate="2016-03-01T01:53:02.967" Score="5" Body="&lt;p&gt;You're:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Entropy &lt;em&gt;aka&lt;/em&gt; disorder&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You're everywhere - first and second law of thermodynamics.&lt;br&gt;&#xA; You create black holes and cause cosmos to die.&lt;br&gt;&#xA; Evil politicians, silly children, criminals,  &lt;em&gt;etc&lt;/em&gt; create &lt;em&gt;disorder&lt;/em&gt;.&lt;br&gt;&#xA; Disorder of the human body causes &lt;em&gt;all&lt;/em&gt; diseases.&lt;br&gt;&#xA; Governments try to establish &lt;em&gt;order&lt;/em&gt; but that doesn't stop the laws of thermodynamics.&lt;br&gt;&#xA;  Which state that &lt;strong&gt;Entropy&lt;/strong&gt; is always &lt;em&gt;increasing&lt;/em&gt;.&lt;br&gt;&#xA; It does you no good because &lt;strong&gt;Entropy&lt;/strong&gt; doesn't control anything - quite the opposite.&lt;br&gt;&#xA; And a treadmill has to create &lt;strong&gt;Entropy&lt;/strong&gt; too.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastEditorUserId="10118" LastEditDate="2016-03-01T02:21:32.200" LastActivityDate="2016-03-01T02:21:32.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28130" PostTypeId="1" AcceptedAnswerId="28132" CreationDate="2016-03-01T03:25:17.817" Score="8" ViewCount="289" Body="&lt;p&gt;&lt;strong&gt;Hint #1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An Earthquake + 200&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Birth of Mr. Green&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm MC Roman ML25&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #4&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Water in the US&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #5&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Birth of Miss Pretty&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #6&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Birth of &quot;Leonard&quot;!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #7&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can you bend it like Hint 7?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #8&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The very first Viking&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #9&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Do you recall the Iron Maiden Torture device?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint #10&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Well, back then the UK wanted to be in EU&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18816" LastActivityDate="2016-03-01T16:44:57.240" Title="What year connects all these 10 hints?" Tags="|word|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28131" PostTypeId="2" ParentId="3280" CreationDate="2016-03-01T03:41:54.757" Score="0" Body="&lt;p&gt;Open the door with the magnifying glass a jar, then open the door with the dragon (assuming it is opposite the door with the light and you have room to not be cooked alive), then completely open the magnifying door to expose the dragon who is probably interested in eating you to the light shining though the opposite door. Once the dragon is dead close the magnifying door and walk out the dragons door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least the dragon will be blinded by the intense light and you could use that to your advantage.&lt;/p&gt;&#xA;" OwnerUserId="12014" LastActivityDate="2016-03-01T03:41:54.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28132" PostTypeId="2" ParentId="28130" CreationDate="2016-03-01T03:57:48.680" Score="12" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1975&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Birth of Mr. Green&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;CeeLo Green was born in 1975.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;I'm MC Roman ML25&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;MCML is 1950 in Roman numerals, add 25 to get 1975.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;Water in the US&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Watergate scandal happened in 1975.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;Birth of Miss Pretty&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Angelina Jolie (&quot;pretty&quot; in French) was born in 1975. (from @dmg)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;6&quot;&gt;&#xA;&lt;li&gt;Birth of &quot;Leonard&quot;!&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Johnny Galecki, who plays &quot;Leonard&quot; in &lt;em&gt;The Big Bang Theory&lt;/em&gt;, was born in 1975. (from @IanMcDonald)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;7&quot;&gt;&#xA;&lt;li&gt;Can you bend it like Hint 7?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Can you bend it like Beckham?&quot; referring to David Beckham, who was born in 1975. (from @IanMcDonald)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;8&quot;&gt;&#xA;&lt;li&gt;The very first Viking&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The Viking 1 probe was launched in 1975.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;9&quot;&gt;&#xA;&lt;li&gt;Do you recall the Iron Maiden Torture device?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The band Iron Maiden was founded in 1975.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;10&quot;&gt;&#xA;&lt;li&gt;Well, back then the UK wanted to be in EU&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;In 1975, the United Kingdom voted to stay in the European Economic Community.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2016-03-01T16:44:57.240" LastActivityDate="2016-03-01T16:44:57.240" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="28133" PostTypeId="5" CreationDate="2016-03-01T05:52:28.357" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2016-03-01T05:52:28.357" LastActivityDate="2016-03-01T05:52:28.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28134" PostTypeId="4" CreationDate="2016-03-01T05:52:28.357" Score="0" Body="Puzzles that are asked about finding a number based on certain rules or clues." OwnerUserId="19658" LastEditorUserId="19658" LastEditDate="2016-03-01T09:46:09.070" LastActivityDate="2016-03-01T09:46:09.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28135" PostTypeId="2" ParentId="28078" CreationDate="2016-03-01T06:21:13.227" Score="4" Body="&lt;p&gt;Let's call the answer $N$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First suppose $N$ is a multiple of $5$.  Then $N$ must be in the range $1-19$, so $N$ is either $5, 10$, or $15$.  None of these is a multiple of $8$, and therefore $N$ must be in the range $20-29$.  That's a contradiction, so we know $N$ isn't a multiple of $5$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; we also know $N$ isn't a multiple of $10$, so $N$ is definitely in the range $30-39$.  That means $N$ is not in the range $20-29$, so $N$ must be a multiple of $8$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $N$ is a multiple of $8$ in the range $30-39$, so $N=32$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8930" LastActivityDate="2016-03-01T06:21:13.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28136" PostTypeId="2" ParentId="27914" CreationDate="2016-03-01T07:08:22.843" Score="3" Body="&lt;p&gt;I will assume that the active player does not have the option of tapping an opponent's dead hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With optimal play by both players the solution is that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The game is indefinite, with no winner.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This may be shown by:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Backward induction (&lt;a href=&quot;http://oyc.yale.edu/economics/econ-159&quot; rel=&quot;nofollow&quot;&gt;Ben Polack&lt;/a&gt; is right, this is almost always the answer!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The current state of a Finger Game may be represented as the numbers of fingers showing on each of the four hands in some fixed order relative to the current turn, say:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$($activePlayer'sLeftHand, activePlayer'sRightHand, opponent'sLeftHand, opponent'sRightHand$)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus if I am showing $(L,R) = (0,2)$ and you are showing $(L,R)=(3,1)$, then&lt;br&gt;if it is your turn to act the state would be $(3,1,0,2)$; whereas&lt;br&gt;if it is my turn to act the state would be $(0,2,3,1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Optimal play by both players results from each player choosing an optimal action at each state reachable given optimal previous actions*. Thus, in the previous example with you to act, you would certainly choose  $LR$ - to tap your left hand onto my right hand - and win.&lt;br&gt;But if were my turn to act what should I do?&lt;br&gt;I have three options: $split$, $RL$, or $RR$.&lt;br&gt;$RR$ would certainly be a blunder as it leads to $(3, 3, 0, 2)$ where you could play either $LR$ or $RR$ and win.&lt;br&gt;But what about $split$ and $RL$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are $5^{2^2}$ naive states the game could be in, but any of the form: $$(a,b,0,0)\lor(0,0,a,b)\ \forall\ a,b\in [0,4]$$ have already finished at some earlier point, leaving $(5^2-1)^2$ states&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now note that if the opponent has only one live hand showing $n\in[1,4]$ fingers then the current player to act cannot have two hands both showing $n$ fingers, since it would mean that the opponent previously tapped a dead hand.&lt;br&gt;This removes $2*4=8$ states:&lt;br&gt;$(1,1,0,1),(1,1,1,0),(2,2,0,2),(2,2,2,0),(3,3,0,3),(3,3,3,0),(4,4,0,4),(4,4,4,0)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also if the opponent has two live hands both showing $n\in[3,4]$ they could not have $split$ to get there and cannot have left the current player with two live hands both showing $n$ since it too means the opponent previously bought a dead hand back to life.&lt;br&gt;This removes another $2$ states:&lt;br&gt;$(3,3,3,3),(4,4,4,4)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore if the opponent has two live hands both showing $n\in[3,4]$ the current player cannot have two live hands showing $n$ and $n-2$ since the only ways to reach such a state would be either the opponent having tapped a dead hand, or from one of the two states removed in the previous paragraph.&lt;br&gt;This removes a final $2*2=4$ states:&lt;br&gt;$(1,3,3,3),(3,1,3,3),(2,4,4,4),(4,2,4,4)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the total number of valid states is  $(5^2-1)^2-8-2-4=562$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can now work backwards to see if $(1,1,1,1)$ is a winning state, a losing state or neither as follows.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Start with three empty sets: $states$, $winningStates$, and $losingStates$;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Place all $562$ states into $states$;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Move states from $states$ into $winningStates$ if &lt;em&gt;any&lt;/em&gt; action results in a win;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Repeatedly analyse $states$ twice:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Move states from $states$ into $losingStates$ if &lt;em&gt;all&lt;/em&gt; actions lead to states in $winningStates$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;move states from $states$ into $winningStates$ if &lt;em&gt;any&lt;/em&gt; action leads to a state in $losingStates$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Stop when no state is moved in either pass.&lt;br&gt;- we could solve the problem at hand by also stopping if $(1,1,1,1)$ is moved.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If we do this we see that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(1,1,1,1)$ is not moved - it is not a winning or losing state given optimal play by both players.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We may also see things such as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the example with me to act neither $split$ nor $RL$ lead to a state in $winningStates$&lt;br&gt;- I could play either of them without putting you in a winning position.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;br&gt;* A full strategy for optimal play would provide the action (or set of actions) for any possible state.&lt;/p&gt;&#xA;" OwnerUserId="16717" LastActivityDate="2016-03-01T07:08:22.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28137" PostTypeId="2" ParentId="27967" CreationDate="2016-03-01T09:09:58.350" Score="0" Body="&lt;p&gt;I guess this works:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $4^2 + 3 - 2$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16036" LastActivityDate="2016-03-01T09:09:58.350" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28138" PostTypeId="1" AcceptedAnswerId="28140" CreationDate="2016-03-01T09:31:16.103" Score="-1" ViewCount="166" Body="&lt;p&gt;For every $n, m \in \mathbb{N}; n,m &amp;gt; 1$, construct a method to produce $\lfloor\frac{n}{2}\rfloor$ using &lt;em&gt;n m&lt;/em&gt;s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use:&#xA;$$x+y$$&#xA;$$x-y$$&#xA;$$x*y$$&#xA;$$\frac{x}{y}$$&#xA;$$x^y$$&#xA;$$x\bmod y$$&#xA;$$\sqrt[k]{x} : k \in \mathbb{N}$$&#xA;$$x!$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't think of anything else now off the top of my head, but you get the general gist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NO TRIG FUNCTIONS.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, I mean to write something which tells you how to write an equation to form n/2 with n m's, it doesn't matter what n is or m is, it MUST work!&lt;/p&gt;&#xA;" OwnerUserId="19640" LastEditorUserId="10399" LastEditDate="2016-03-01T14:36:32.533" LastActivityDate="2016-03-01T14:36:32.533" Title="Form n/2 with n amount of ANY number" Tags="|formation-of-numbers|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28139" PostTypeId="2" ParentId="28138" CreationDate="2016-03-01T09:50:45.363" Score="3" Body="&lt;p&gt;This is easy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For even $n$ do $\frac m m + \frac m m + ...$ until you have $n m$s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT odd n is still incorrect, will fix it when i know a solution&lt;/strong&gt;&lt;br&gt;&#xA;For odd $n$ do $\frac {m+m} m + \frac m m + \frac m m + \frac m m ...$&lt;/p&gt;&#xA;" OwnerUserId="1748" LastEditorUserId="1748" LastEditDate="2016-03-01T09:58:24.067" LastActivityDate="2016-03-01T09:58:24.067" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28140" PostTypeId="2" ParentId="28138" CreationDate="2016-03-01T10:01:53.270" Score="6" Body="&lt;p&gt;With the new rules, my m/m &lt;a href=&quot;https://puzzling.stackexchange.com/questions/28138/form-n-2-with-n-amount-of-any-number/28140#comment83409_28138&quot;&gt;idea&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/a/28139/9422&quot;&gt;Ivo Beckers' continuation&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For even $n$: $\frac{n}{2} = \frac{m}{m} + \frac{m}{m} + ...$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For odd $n$: change the first term to $\frac{\sqrt{m}\sqrt{m}}{m}$.&lt;/p&gt;&#xA;" OwnerUserId="9422" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2016-03-01T10:03:47.187" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28141" PostTypeId="1" AcceptedAnswerId="28142" CreationDate="2016-03-01T10:30:33.037" Score="13" ViewCount="2143" Body="&lt;p&gt;Professor Halfbrain has spent the last weekend with filling the squares of a &lt;span class=&quot;math-container&quot;&gt;$99\times99$&lt;/span&gt; chessboard with real numbers from the interval &lt;span class=&quot;math-container&quot;&gt;$[-1,+1]$&lt;/span&gt;.&#xA;Whenever four squares form the corners of a rectangle (with sides parallel to the sides of the chessboard), then the four numbers in these squares had to add up to zero.&lt;/p&gt;&#xA;&lt;p&gt;Professor Halfbrain has proved two extremely deep theorems on this.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Professor Halfbrain's first theorem&lt;/strong&gt;: It is possible to fill the chessboard subject to the above rules, so that the sum of all numbers in all the squares is &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Professor Halfbrain's second theorem&lt;/strong&gt;: If we fill the chessboard subject to the above rules, then the sum of all numbers in all the squares is at most &lt;span class=&quot;math-container&quot;&gt;$9801$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This puzzle asks you to improve the two theorems of professor Halfbrain and to make them even deeper.&#xA;Find an integer &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;, so that &amp;quot;the sum in all the squares is &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt;&amp;quot; in the first theorem may be replaced by &amp;quot;the sum in all the squares is &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;&amp;quot;, and so that &amp;quot;the sum is at most &lt;span class=&quot;math-container&quot;&gt;$9801$&lt;/span&gt;&amp;quot; in the second theorem may be replaced by &amp;quot;the sum is at most &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;&amp;quot; (again yielding true statements, of course).&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-03-02T14:47:17.980" Title="Professor Halfbrain and the 99x99 chessboard (Part 1)" Tags="|mathematics|combinatorics|checkerboard|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28142" PostTypeId="2" ParentId="28141" CreationDate="2016-03-01T11:43:30.893" Score="12" Body="&lt;p&gt;This is a sketchy proof of what Ivo suspected in the comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take an arbitrary selection as below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/05PfW.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/05PfW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A+B+E+F = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A+B+I+K = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=&amp;gt; E+F=I+K$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E+F+I+K = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=&amp;gt;E +F = -I-K$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E = -F$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I = -K$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarily, we can obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F+K+E+I = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G+L+F+K = G+L + E+ I = 0 =&amp;gt; F+K-E-I = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F = -K$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E = -I$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F = I$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E = K$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for an arbitrary selection, we know that the diagonal elements are equal. Taking an arbitrary selection with odd sides (e.g. $A+C+I+L = 0$), we know that $A,C,I,L$ must be equal and hence 0. Since this can be done for all squares, they must all be 0.&lt;/p&gt;&#xA;" OwnerUserId="12901" LastActivityDate="2016-03-01T11:43:30.893" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28143" PostTypeId="2" ParentId="27982" CreationDate="2016-03-01T12:25:01.060" Score="6" Body="&lt;p&gt;All of them. Using S(n) the &lt;a href=&quot;https://en.wikipedia.org/wiki/Successor_function&quot; rel=&quot;noreferrer&quot;&gt;successor function&lt;/a&gt; used in the Peano axioms to define all natural numbers. And is equivalent to S(n) = n+1&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 = 44/44&#xA;2 = S(44/44)&#xA;3 = S( S( 44/44 ) )&#xA;&#xA;n+1 = S( n )&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1998" LastActivityDate="2016-03-01T12:25:01.060" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28144" PostTypeId="1" AcceptedAnswerId="28146" CreationDate="2016-03-01T14:06:35.490" Score="7" ViewCount="618" Body="&lt;p&gt;Which two mathematics related words are anagrams of &lt;strong&gt;ALERTING&lt;/strong&gt; and &lt;strong&gt;ALTERING&lt;/strong&gt; and &lt;strong&gt;RELATING&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="5208" LastEditDate="2016-03-08T23:40:47.413" LastActivityDate="2016-03-08T23:40:47.413" Title="Mathematics related anagrams" Tags="|word|knowledge|anagram|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28145" PostTypeId="1" AcceptedAnswerId="28155" CreationDate="2016-03-01T14:11:26.763" Score="7" ViewCount="271" Body="&lt;ul&gt;&#xA;&lt;li&gt;Every letter and every question mark  stands for a digit in base-10 representation&lt;/li&gt;&#xA;&lt;li&gt;Different letters stand for different digits&lt;/li&gt;&#xA;&lt;li&gt;Question marks are placeholders and stand for arbitrary digits (that also may occur for letters)&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Leading digits are always non-zero. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;        THIS&#xA;       *  IS&#xA;        ----&#xA;       ??TOO&#xA;      HARD?&#xA;      ------&#xA;      ??????&#xA;&lt;/code&gt;&lt;/pre&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Which digit does each letter represent? &#xA;(Please present the full analysis how these digits can be determined.)&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="8874" LastEditDate="2016-03-06T15:34:30.373" LastActivityDate="2016-03-06T15:34:30.373" Title="Multiplicative alphametic: This is too hard" Tags="|no-computers|alphametic|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28146" PostTypeId="2" ParentId="28144" CreationDate="2016-03-01T14:11:39.717" Score="16" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Integral and Triangle: A long sum, and a common shape.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19491" LastActivityDate="2016-03-01T14:11:39.717" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28147" PostTypeId="1" AcceptedAnswerId="28149" CreationDate="2016-03-01T15:07:07.687" Score="4" ViewCount="237" Body="&lt;p&gt;A friend of mine had an important message for me. Can you also read this important message below?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Fr s tH pTh t tH dRk sD Fr lDs t ngr Ngr ldS t hT Ht lDs t sffRng&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="18685" LastActivityDate="2016-03-01T15:11:10.970" Title="What is the message given?" Tags="|word|wordplay|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28148" PostTypeId="1" AcceptedAnswerId="28151" CreationDate="2016-03-01T15:09:55.107" Score="12" ViewCount="1350" Body="&lt;p&gt;This puzzle is the continuation of &#xA;&quot;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/28141/professor-halfbrain-and-the-99x99-chessboard-part-1&quot;&gt;Professor Halfbrain and the 99x99 chessboard (Part 1)&lt;/a&gt;&quot;.&#xA;The difference is that &quot;four corners of a rectangle&quot; has now become &quot;$2\times2$ subsquare&quot;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Professor Halfbrain has spent the last weekend with filling the squares of a $99\times99$ chessboard with real numbers from the interval $[-1,+1]$. &#xA;Whenever four squares form a &lt;strong&gt;$2\times2$ subsquare&lt;/strong&gt; of the chessboard, then the four numbers in these squares had to add up to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Professor Halfbrain has proved two extremely deep theorems on this.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Professor Halfbrain's first theorem&lt;/strong&gt;: It is possible to fill the chessboard subject to the above rules, so that the sum of all numbers in all the squares is $0$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Professor Halfbrain's second theorem&lt;/strong&gt;: If we fill the chessboard subject to the above rules, then the sum of all numbers in all the squares is at most $9801$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This puzzle asks you to improve the two theorems of professor Halfbrain and to make them even deeper.&#xA;Find an integer $y$, so that &quot;the sum in all the squares is $0$&quot; in the first theorem may be replaced by &quot;the sum in all the squares is $y$&quot;, and so that &quot;the sum is at most $9801$&quot; in the second theorem may be replaced by &quot;the sum is at most $y$&quot; (again yielding true statements, of course).&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.950" LastActivityDate="2016-03-01T21:15:46.347" Title="Professor Halfbrain and the 99x99 chessboard (Part 2)" Tags="|mathematics|combinatorics|checkerboard|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28149" PostTypeId="2" ParentId="28147" CreationDate="2016-03-01T15:11:10.970" Score="5" Body="&lt;p&gt;It's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;“Fear is the path to the dark side. Fear leads to anger. Anger leads to hate. Hate leads to suffering.” a famous quote from Yoda of Star Wars&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Just skimming the letters I could see it&lt;/p&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2016-03-01T15:11:10.970" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28150" PostTypeId="2" ParentId="28148" CreationDate="2016-03-01T15:23:52.030" Score="5" Body="&lt;p&gt;Another way of getting 99:&#xA;If we give each square an $x$ and $y$ coordinate, starting with (1,1) in the top left corner, we give each square with even $x$ and even $y$ a value of $-1$, each square with odd $x$ and $y$ a value of $1$, and each other square a value of $0$, then we have $50^2 \times 1 + 49^2 \times -1 = 99$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Old Answer:&#xA;Not sure if it's optimal, but I can get &lt;strike&gt;866&lt;/strike&gt; 65.6:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we give each square an $x$ and $y$ coordinate, starting with (1,1) in the top left corner, we give each square with even $x$ and even $y$ a value of $-1$, and each other square a value of $\frac13$. Each $2\times2$ square will contain exactly one square with value $-1$, and $3$ with value $\frac13$, for a total sum of $0$. We have $49^2$ -1's, and $99^2 - 49^2$ with value $\frac13$, for a total value of $65.6$.&lt;/p&gt;&#xA;" OwnerUserId="5169" LastEditorUserId="4421" LastEditDate="2016-03-01T20:08:21.157" LastActivityDate="2016-03-01T20:08:21.157" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28151" PostTypeId="2" ParentId="28148" CreationDate="2016-03-01T16:08:45.817" Score="14" Body="&lt;p&gt;&lt;strong&gt;Part 1&lt;/strong&gt;&lt;br&gt;&#xA;Since we know that all $2\times2$ squares will have a sum of all values equal to $0$, then we can say that the sum of the chessboard is equal to the sum of elements in the last row and column, because we can group all the other squares into $2\times2$ squares of sum $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can achieve a sum of $99$ by filling the board with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/T46kr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/T46kr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum of the last column and last row $= 99 \times 1 + 1-1 + 1-1 +... +1-1 = 99$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proving the upper, lower bound&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've shown a method to get the value $99$. We can get $-99$ by multiplying each square by $-1$.&lt;br&gt;I can further show that we definitely can't achieve more than $99$; this will be proof for the upper bound (also for lower bound).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will fill the chessboard with $2\times2$ tiles, where each tile covers a $2\times2$ square of which the sum is equal to $0$. Thanks to &lt;a href=&quot;https://puzzling.stackexchange.com/users/17915/paul-sinclair&quot;&gt;Paul Sinclair&lt;/a&gt; for noticing that the below tiling shows that you can place $49^2$ tiles and that we don't need to prove you can't place more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A valid tiling would be to leave a zig-zag across a diagonal, as shown in smaller scale: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SHRHL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SHRHL.png&quot; alt=&quot;tiling&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to now state that for every red L shape, the sum of all 3 squares is between $[-1, +1]$. This is obvious because the red L shape, along with the neighboring green square must have a sum of $0$, and the value of the green square is between $[-1, +1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this property we can now say that there are 50 blue squares on our board, and 49 red L shapes, and the sum of all these squares is the sum of our board. However the blue squares are each between $[-1, +1]$, and each red L shape is between $[-1, +1]$, so our board sum must be between $[-99, +99]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part 2&lt;/strong&gt;&lt;br&gt;&#xA;My first tiling allows us to get &lt;strong&gt;the sum of all numbers in all the squares is $= x$&lt;/strong&gt; as long as $-99 \le x \le 99$.&lt;br&gt;&#xA;Say $x = 99y$. Now lets change all my $1$'s to $y$'s, and all my $-1$'s to $-y$'s.&lt;br&gt;&#xA;Similarly, adding the squares gives the sum equal to $99y = x$, again $-1 \le y &amp;lt; 1$, so for this solution, $-99 \le x \le 99$.&lt;/p&gt;&#xA;" OwnerUserId="18972" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2016-03-01T21:15:46.347" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28152" PostTypeId="2" ParentId="28145" CreationDate="2016-03-01T16:19:28.837" Score="2" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;  4379&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt; *  79&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;------&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt; 39411&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;30653 &lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;------&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;345941&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the letters/digits are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;0 = A&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;1 = O&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;3 = H&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;4 = T&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;5 = D&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;6 = R&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;7 = I&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;9 = S&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When approaching this, I first decided to answer this question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any integer i between 1 and 9 inclusive, what s^2 + i*10 gives us a number where the ones and tens places hold the same digit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0 Would fit the bill, but ??TOO is != 0, so S can't be 0.&lt;br&gt;&#xA;1 no&lt;br&gt;&#xA;2 -- Maybe if I = 7  (O = 4)&lt;br&gt;&#xA;3 no&lt;br&gt;&#xA;4 Tens place of 4x4 is odd, so no value of I would work.&lt;br&gt;&#xA;S Tens place of 5 squared is not 0 or 5&lt;br&gt;&#xA;6 Tens place of 6x6 is odd, so no value of I would make this work.&lt;br&gt;&#xA;7 -- Maybe if I = 5  (O = 9)&lt;br&gt;&#xA;8 -- Maybe if I = 6  (O = 4)(or S = 1, but I can't be 1 because THIS is not  HARD?)&lt;br&gt;&#xA;9 -- Maybe if I = 7  (O = 1)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, I tried the possible combinations that I found.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Starting the highest possible values for S and I made the most sense to me, which led to the answer.  When testing out values of H when I = 7 and S = 9, 3 is the only one that fits.  We already know some of the other letters in this test case, so we don't have to test everything.  Finally, with only a few values to try, we find that T = 4 fits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was hoping that he last row would spell something if we backfilled some of the digits with letters.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; But all we get is htdsto.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17020" LastEditorUserId="17020" LastEditDate="2016-03-01T16:40:21.500" LastActivityDate="2016-03-01T16:40:21.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28153" PostTypeId="1" AcceptedAnswerId="28179" CreationDate="2016-03-01T16:22:09.693" Score="26" ViewCount="1127" Body="&lt;p&gt;&lt;em&gt;There once was a man called James&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;who liked to use different names&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;If you're slick and astute&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;by removing the fruit&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;your favorite flava remains&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What was his middle name?&lt;/p&gt;&#xA;" OwnerUserId="18522" LastEditorUserId="22922" LastEditDate="2016-09-01T14:29:10.820" LastActivityDate="2016-09-01T14:29:10.820" Title="What is James's middle name?" Tags="|riddle|wordplay|knowledge|limerick|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28154" PostTypeId="2" ParentId="28145" CreationDate="2016-03-01T16:27:09.243" Score="5" Body="&lt;p&gt;Let &lt;span class=&quot;math-container&quot;&gt;$P_i^j$&lt;/span&gt; be the product of the &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt;th digit in the first number and the &lt;span class=&quot;math-container&quot;&gt;$j$&lt;/span&gt;th digit in the second number.  Let &lt;span class=&quot;math-container&quot;&gt;$C_i^j$&lt;/span&gt; be the carry over of &lt;span class=&quot;math-container&quot;&gt;$P_i^j$&lt;/span&gt;, and let &lt;span class=&quot;math-container&quot;&gt;$D_i^j$&lt;/span&gt; be the digit in the result.&lt;/p&gt;&#xA;&lt;p&gt;For example, &lt;span class=&quot;math-container&quot;&gt;$P_3^2$&lt;/span&gt; is the product of H X I.&lt;/p&gt;&#xA;&lt;p&gt;First off, we can see that &lt;span class=&quot;math-container&quot;&gt;$S \notin \{0,1,5,6\}$&lt;/span&gt; since &lt;span class=&quot;math-container&quot;&gt;$S \ne O$&lt;/span&gt;. Also, if &lt;span class=&quot;math-container&quot;&gt;$I \ne 1$&lt;/span&gt; since the second product line is not identical to THIS.  So &lt;span class=&quot;math-container&quot;&gt;$I \notin \{0,1\}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;If S=3, then &lt;span class=&quot;math-container&quot;&gt;$P_1^1=D_1^1=O=9$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_1^1=0$&lt;/span&gt;.  We know that &lt;span class=&quot;math-container&quot;&gt;$D_2^1=O=9$&lt;/span&gt; as well, but no value for &lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt; results in &lt;span class=&quot;math-container&quot;&gt;$D_2^1=9$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;If S=4, then &lt;span class=&quot;math-container&quot;&gt;$P_1^1=16 \implies D_1^1=O=6$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_1^1=1$&lt;/span&gt;. But then we need &lt;span class=&quot;math-container&quot;&gt;$P_2^1=I \times S+C_1^1$&lt;/span&gt; to end in 6, which means &lt;span class=&quot;math-container&quot;&gt;$I \times S$&lt;/span&gt; ends in 5, which is impossible with an even S.&lt;/p&gt;&#xA;&lt;p&gt;Thus, &lt;span class=&quot;math-container&quot;&gt;$S \in \{2,7,8,9\}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;h2&gt;Assume S=2&lt;/h2&gt;&#xA;&lt;p&gt;Thus &lt;span class=&quot;math-container&quot;&gt;$P_1^1=D_1^1=O=4$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_1^1=0$&lt;/span&gt;. We know that &lt;span class=&quot;math-container&quot;&gt;$D_2^1=O=4$&lt;/span&gt; as well, so we know that &lt;span class=&quot;math-container&quot;&gt;$I=7$&lt;/span&gt;. The means that &lt;span class=&quot;math-container&quot;&gt;$D=0$&lt;/span&gt;.  Since &lt;span class=&quot;math-container&quot;&gt;$THIS \times 2$&lt;/span&gt; is a five digit number, we know &lt;span class=&quot;math-container&quot;&gt;$T \ge 5$&lt;/span&gt;.  Also, we know that &lt;span class=&quot;math-container&quot;&gt;$C_2^1=1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D_3^1=T$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$P_3^1=2H+1$&lt;/span&gt; so T is odd.&lt;/p&gt;&#xA;&lt;p&gt;If T=5, then &lt;span class=&quot;math-container&quot;&gt;$2H=4$&lt;/span&gt; only works with &lt;span class=&quot;math-container&quot;&gt;$H \in \{2,7\}$&lt;/span&gt;, both of which are taken.&lt;/p&gt;&#xA;&lt;p&gt;Thus, T=9 and &lt;span class=&quot;math-container&quot;&gt;$2H=8 \implies H=4$&lt;/span&gt;.  But then &lt;span class=&quot;math-container&quot;&gt;$THIS \times I=9472 \times 7 = 66304$&lt;/span&gt; which means H=A. Contradiction.&lt;/p&gt;&#xA;&lt;h2&gt;Assume S=7&lt;/h2&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$P_1^1=49 \implies D_1^1=O=9$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_1^1=4$&lt;/span&gt;.  To get &lt;span class=&quot;math-container&quot;&gt;$D_2^1=9$&lt;/span&gt;, we need &lt;span class=&quot;math-container&quot;&gt;$I \times S$&lt;/span&gt; to end in 5.  This only works when &lt;span class=&quot;math-container&quot;&gt;$I=5$&lt;/span&gt;.  So &lt;span class=&quot;math-container&quot;&gt;$P_2^1=I\ times S+C_1^1=5 \times 7+4+39 \implies C_2^1=3$&lt;/span&gt; which means &lt;span class=&quot;math-container&quot;&gt;$T=7H+3 \mod 10$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$H=1 \implies T=0$&lt;/span&gt;, but T can't be 0.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$H=2 \implies T=7$&lt;/span&gt;, but 7 is already taken by S&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$H=3 \implies T=4$&lt;/span&gt;, but &lt;span class=&quot;math-container&quot;&gt;$THIS \times I=4357\times 5=21785\implies A=7$&lt;/span&gt; is already taken by S.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$H=4 \implies T=1$&lt;/span&gt;, but &lt;span class=&quot;math-container&quot;&gt;$THIS \times I=1457\times 5=7285 \implies H=0$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$H=6 \implies T=5$&lt;/span&gt;, but 5 is already taken by I&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$H=8 \implies T=9$&lt;/span&gt;, but 9 is already taken by O&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Contradiction.&lt;/p&gt;&#xA;&lt;h2&gt;Assume S=8&lt;/h2&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$P_1^1=64 \implies D_1^1=O=4$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_1^1=6$&lt;/span&gt;.  To get &lt;span class=&quot;math-container&quot;&gt;$D_2^1=4$&lt;/span&gt;, we need &lt;span class=&quot;math-container&quot;&gt;$I \times S$&lt;/span&gt; to end in 8.  This only works when I=6.&lt;/p&gt;&#xA;&lt;p&gt;Now, &lt;span class=&quot;math-container&quot;&gt;$C_2^1=5$&lt;/span&gt;, so &lt;span class=&quot;math-container&quot;&gt;$T=8H+5 \mod 10$&lt;/span&gt;.  This means that T is odd and &lt;span class=&quot;math-container&quot;&gt;$H\ne5$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$H=1 \implies T=3$&lt;/span&gt;, but &lt;span class=&quot;math-container&quot;&gt;$THIS \times I = 3168 \times 6=19008 \implies R=A$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$H=2 \implies T=1$&lt;/span&gt;, but &lt;span class=&quot;math-container&quot;&gt;$THIS \times I = 1268 \times 6=4608 \implies H=0$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$H=3 \implies T=9$&lt;/span&gt;, but &lt;span class=&quot;math-container&quot;&gt;$THIS \times I = 9368 \times 6=56208 \implies H=5$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$H=7 \implies T=1$&lt;/span&gt;, but &lt;span class=&quot;math-container&quot;&gt;$THIS \times I = 1768 \times 6=11208 \implies H=1$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$H=9 \implies T=7$&lt;/span&gt;, but &lt;span class=&quot;math-container&quot;&gt;$THIS \times I = 7968 \times 6=47808 \implies H=4$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Contradiction&lt;/p&gt;&#xA;&lt;h2&gt;Therefore S=9&lt;/h2&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$P_1^1=81 \implies D_1^1=O=1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_1^1=8$&lt;/span&gt;.  To get &lt;span class=&quot;math-container&quot;&gt;$D_2^1=1$&lt;/span&gt;, we need &lt;span class=&quot;math-container&quot;&gt;$I \times S$&lt;/span&gt; to end in 3.  This only works when &lt;span class=&quot;math-container&quot;&gt;$I=7$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Now, &lt;span class=&quot;math-container&quot;&gt;$C_2^1=7$&lt;/span&gt;, so &lt;span class=&quot;math-container&quot;&gt;$T=9H+7 \mod 10$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$H=2 \implies T=5$&lt;/span&gt;, but &lt;span class=&quot;math-container&quot;&gt;$THIS \times I = 5279 \times 7=36953 \implies R=9$&lt;/span&gt; which is already taken by S&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$H=4 \implies T=3$&lt;/span&gt;, but &lt;span class=&quot;math-container&quot;&gt;$THIS \times I = 3479 \times 7=24353 \implies R=3$&lt;/span&gt; which is already taken by T&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$H=5 \implies T=2$&lt;/span&gt;, but &lt;span class=&quot;math-container&quot;&gt;$THIS \times I = 2579 \times 7=18053 \implies H=1$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$H=6 \implies T=1$&lt;/span&gt;, but 1 is already taken by O&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$H=8 \implies T=9$&lt;/span&gt;, but 9 is already taken by S&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Thus,  &lt;span class=&quot;math-container&quot;&gt;$H=3 \implies T=4$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;    4379&#xA;   *  79&#xA;    ----&#xA;   39411&#xA;  30653&#xA;  ------&#xA;  345941&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-03-01T16:27:09.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28155" PostTypeId="2" ParentId="28145" CreationDate="2016-03-01T16:45:24.363" Score="7" Body="&lt;p&gt;From the OO in TOO, we have S&lt;em&gt;S mod 10 = I&lt;/em&gt;S + (SS div 10) mod 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know S$\neq$0 (otherwise S=O=0). We also have I$\neq$1 (otherwise &quot;HARD?&quot; only has 4 digits).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Discarding solutions where more than one of I,S,O has the same digit, the above equation produces (I,S,O) = (5,7,9), (6,8,4), and (7,9,1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 1: (I,S,O) = (5,7,9)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In TOO, T = 7H+3 mod 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With H$\neq$0 in &quot;HARD?&quot; and T$\neq$0 in THIS, and discarding solutions where more than 1 letter has the same digit, we have (T,H) = (4,3) (R=7, discard) or (1,4) (&quot;HARD?&quot; will only have 5 digits, discard).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 2: (I,S,O) = (6,8,4)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In TOO, T = 8H+5 mod 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From TOO, T=8H+5 mod 10.&#xA;From &quot;HARD?&quot;, D=0 and R=6H+4 mod 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the equations for T and R, T$\neq$0, H$\neq$0, R$\neq$0 and eliminating duplicate digits, we have (T,H,R) = (9,3,2) (A=6, discard) or (1,7,3) (R=6, discard).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 3: (I,S,O) = (7,9,1)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following the same procedure as before, T=9H+7 mod 10, and R=7H+5 mod 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eliminating possibilities as before, we're left with (T,H,R) = (4,3,6) or (2,5,0) (H=5, discard).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only solution is THIS=4379, IS=79, ??TOO=39411, HARD?=30653, ??????=345941.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;            4379&#xA;           *  79&#xA;          ------&#xA;           39411&#xA;          30653&#xA;          ------&#xA;          345941&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Substituting our letters, we get the (totally meaningless) statements:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;            THIS&#xA;           *  IS&#xA;          ------&#xA;           HSTOO&#xA;          HARDH&#xA;          ------&#xA;          HTDSTO&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="9422" LastEditorUserId="10399" LastEditDate="2016-03-01T17:01:54.873" LastActivityDate="2016-03-01T17:01:54.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28156" PostTypeId="1" AcceptedAnswerId="28157" CreationDate="2016-03-01T17:10:07.843" Score="21" ViewCount="847" Body="&lt;p&gt;This is a riddle with diminishing returns.&lt;br&gt;&#xA;It's acrostic again. Can't kick that habit.&lt;br&gt;&#xA;No special knowledge is needed.&lt;br&gt;&#xA;When you find the answer, you will be in no doubt.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Faker! They cried and cut off his tail.&lt;br&gt;&#xA;  Call for the axe, put his head in a pail&lt;br&gt;&#xA;  Sharpen the axe and take off his top&lt;br&gt;&#xA;  Single him out, let him endlessly drop&lt;br&gt;&#xA;  Atop the trash heap. Take his last right away.&lt;br&gt;&#xA;  Nothing left. Now we're done. Goodbye and good day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastActivityDate="2016-03-01T17:19:55.770" Title="Reductio Ad Absurdum" Tags="|riddle|word|poetry|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28157" PostTypeId="2" ParentId="28156" CreationDate="2016-03-01T17:19:55.770" Score="16" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; PHONEY &lt;span class=&quot;math-container&quot;&gt;$~\to~$&lt;/span&gt; PHONE &lt;span class=&quot;math-container&quot;&gt;$~\to~$&lt;/span&gt; HONE &lt;span class=&quot;math-container&quot;&gt;$~\to~$&lt;/span&gt; ONE &lt;span class=&quot;math-container&quot;&gt;$~\to~$&lt;/span&gt; ON &lt;span class=&quot;math-container&quot;&gt;$~\to~$&lt;/span&gt; O&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;br&gt;&#xA;&lt;h2&gt;Argument:&lt;/h2&gt;&#xA;&lt;br&gt;&#xA;&lt;p&gt;Faker! They cried and cut off his tail.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; A faker is PHONEY; cutting of the tail Y yields PHONE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Call for the axe, put his head in a pail&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; You do calls with your PHONE; cutting of the head P yields HONE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Sharpen the axe and take off his top&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; To HONE is another word for to sharpen; taking of the top H yields ONE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Single him out, let him endlessly drop&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The ONE is singled out; dropping the end E yields ON&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Atop the trash heap. Take his last right away.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Atop is ON; taking the right N away leaves O&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Nothing left. Now we're done. Goodbye and good day.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; O or 0 or zero stands for nothing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-03-01T17:19:55.770" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28158" PostTypeId="1" AcceptedAnswerId="28159" CreationDate="2016-03-01T17:33:31.250" Score="0" ViewCount="234" Body="&lt;p&gt;What is the longest palindrome created?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For those who don't know&lt;/strong&gt;&#xA;A palindrome is a word or series of words that read the same forwards as backwards. For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;hannah&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A man a plan a canal Panama&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Some Rules&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The Palindrome must be in English. &lt;/li&gt;&#xA;&lt;li&gt;The Palindrome must make sense. By this I mean it cannot be simply a series of words that do not make a coherent sentence/string of sentences.&lt;/li&gt;&#xA;&lt;li&gt;Punctuation/capitals can be ignored.&lt;/li&gt;&#xA;&lt;li&gt;Odd phrasing/grammar is allowed as long as it still makes sense.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="19697" LastEditorUserId="17482" LastEditDate="2016-03-01T17:34:21.493" LastActivityDate="2016-03-01T18:04:03.480" Title="Longest Palindrome" Tags="|english|palindrome|" AnswerCount="1" CommentCount="7" ClosedDate="2016-03-01T21:34:21.453" ContentLicense="CC BY-SA 3.0" />
  <row Id="28159" PostTypeId="2" ParentId="28158" CreationDate="2016-03-01T17:37:38.900" Score="5" Body="&lt;p&gt;Here is a Palindromic Poem&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Mood's mode!&lt;br&gt;&#xA;Pallas, I won!&lt;br&gt;&#xA;(Diaper pane, sold entire.)&lt;br&gt;&#xA;Melt till ever sere, hide it.&lt;br&gt;&#xA;Drown a more vile note;&lt;br&gt;&#xA;(Tar of rennet.)&lt;br&gt;&#xA;Ah, trowel, baton, eras ago.&lt;br&gt;&#xA;The reward? A &quot;nisi.&quot; Two nag.&#xA;Otary tastes putrid, yam was green.&lt;br&gt;&#xA;Odes up and on; stare we.&lt;br&gt;&#xA;Rats nod. Nap used one-erg saw.&lt;br&gt;&#xA;(May dirt upset satyr?)&lt;br&gt;&#xA;A toga now; 'tis in a drawer, eh?&lt;br&gt;&#xA;Togas are notable.&lt;br&gt;&#xA;(Worth a tenner for Ate.)&lt;br&gt;&#xA;Tone liver. O Man, word-tied I.&lt;br&gt;&#xA;Here's revel!&lt;br&gt;&#xA;Little merit, Ned? Lose, Nap?&lt;br&gt;&#xA;Repaid now is all apedom's doom.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you are looking for a single sentence then...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lid of fade, metallic soot, emit Garret-simple, help mister ragtime to oscillate me, daffodil.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It have 75 letters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if nonsensical palindromes in one sentence are allowed then&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A man, a plan, a caret, a ban, a myriad, a sum, a lac, a liar, a hoop, a pint, a catalpa, a gas, an oil, a bird, a yell, a vat, a caw, a pax, a wag, a tax, a nay, a ram, a cap, a yam, a gay, a tsar, a wall, a car, a luger, a ward, a bin, a woman, a vassal, a wolf, a tuna, a nit, a pall, a fret, a watt, a bay, a daub, a tan, a cab, a datum, a gall, a hat, a fag, a zap, a say, a jaw, a lay, a wet, a gallop, a tug, a trot, a trap, a tram, a torr, a caper, a top, a tonk, a toll, a ball, a fair, a sax, a minim, a tenor, a bass, a passer, a capital, a rut, an amen, a ted, a cabal, a tang, a sun, an ass, a maw, a sag, a jam, a dam, a sub, a salt, an axon, a sail, an ad, a wadi, a radian, a room, a rood, a rip, a tad, a pariah, a revel, a reel, a reed, a pool, a plug, a pin, a peek, a parabola, a dog, a pat, a cud, a nu, a fan, a pal, a rum, a nod, an eta, a lag, an eel, a batik, a mug, a mot, a nap, a maxim, a mood, a leek, a grub, a gob, a gel, a drab, a citadel, a total, a cedar, a tap, a gag, a rat, a manor, a bar, a gal, a cola, a pap, a yaw, a tab, a raj, a gab, a nag, a pagan, a bag, a jar, a bat, a way, a papa, a local, a gar, a baron, a mat, a rag, a gap, a tar, a decal, a tot, a led, a tic, a bard, a leg, a bog, a burg, a keel, a doom, a mix, a map, an atom, a gum, a kit, a baleen, a gala, a ten, a don, a mural, a pan, a faun, a ducat, a pagoda, a lob, a rap, a keep, a nip, a gulp, a loop, a deer, a leer, a lever, a hair, a pad, a tapir, a door, a moor, an aid, a raid, a wad, an alias, an ox, an atlas, a bus, a madam, a jag, a saw, a mass, an anus, a gnat, a lab, a cadet, an em, a natural, a tip, a caress, a pass, a baronet, a minimax, a sari, a fall, a ballot, a knot, a pot, a rep, a carrot, a mart, a part, a tort, a gut, a poll, a gateway, a law, a jay, a sap, a zag, a fat, a hall, a gamut, a dab, a can, a tabu, a day, a batt, a waterfall, a patina, a nut, a flow, a lass, a van, a mow, a nib, a draw, a regular, a call, a war, a stay, a gam, a yap, a cam, a ray, an ax, a tag, a wax, a paw, a cat, a valley, a drib, a lion, a saga, a plat, a catnip, a pooh, a rail, a calamus, a dairyman, a bater, a canal - Panama!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-03-01T18:04:03.480" LastActivityDate="2016-03-01T18:04:03.480" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28162" PostTypeId="1" AcceptedAnswerId="28163" CreationDate="2016-03-01T19:16:25.217" Score="2" ViewCount="408" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am something that you eat&lt;br&gt;&#xA;  But I am also something which is called disease&lt;br&gt;&#xA;  I am something that is grown in the ground  but I am not veggie&lt;br&gt;&#xA;  But I am also something that can grow mainly on your feet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for such an awful rhyming...&lt;/p&gt;&#xA;" OwnerUserId="17482" LastActivityDate="2016-04-05T12:41:28.297" Title="Disease that you eat" Tags="|riddle|word|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28163" PostTypeId="2" ParentId="28162" CreationDate="2016-03-01T19:18:42.177" Score="2" Body="&lt;p&gt;You are a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fungus&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;I am something that you eat&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Edible_mushroom&quot; rel=&quot;nofollow&quot;&gt;Mushrooms can be a delicious fungus&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;But I am also something which is called disease&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One can have a fungal infection, like &lt;a href=&quot;https://en.wikipedia.org/wiki/Candidiasis&quot; rel=&quot;nofollow&quot;&gt;Thrush&lt;/a&gt; in the mouth, or Athlete's Foot* on the feet. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;I am something that is grown in the ground but I am not veggie&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fungus usually grows on or in the ground, though it can grow in a lot of places&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;But I am also something that can grow mainly on your feet.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; See above, re: Athlete's Foot.&lt;br&gt;&#xA;&lt;br&gt;&#xA; * &lt;em&gt;I am not linking to this one, because there's a gross-ass picture there&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastActivityDate="2016-03-01T19:18:42.177" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28164" PostTypeId="2" ParentId="28141" CreationDate="2016-03-01T19:29:12.230" Score="8" Body="&lt;p&gt;A clarification on Carl Löndahl's proof: He has shown that for any two adjacent squares, the numbers must add up to $0$. Hence, it must form an alternating pattern for some value $a$:&#xA;$$\begin{matrix} +a &amp;amp; -a &amp;amp; +a &amp;amp; -a &amp;amp; \ldots\\ -a &amp;amp; +a &amp;amp; -a &amp;amp; +a &amp;amp; \ldots\\ +a &amp;amp; -a &amp;amp; +a &amp;amp; -a &amp;amp; \ldots\\ -a &amp;amp; +a &amp;amp; -a &amp;amp; +a &amp;amp; \ldots\\\vdots &amp;amp; \vdots  &amp;amp; \vdots  &amp;amp; \vdots &amp;amp; \ddots\end{matrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now, consider a 3x3 square with $+a$ in one (and therefore all) corners. This gives $4a = 0$, so $a = 0$. Every square on the board must be $0$.&lt;/p&gt;&#xA;" OwnerUserId="17915" LastActivityDate="2016-03-01T19:29:12.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28165" PostTypeId="1" CreationDate="2016-03-01T20:36:31.727" Score="3" ViewCount="370" Body="&lt;p&gt;Disclaimer: I dont know the answer but it is definitely solvable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A legend of nine, In the Andaman, A treasure you may find, But, certainly, the blood of men.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Staring at the Sun, At 30 to 6, Al asked the captain, “What is the fix?” “The Sun” says the cap, “And blood shall determine our fate” And so he screams “Die, Al” And so remain 8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Down the direction, so gracefully learnt, Atop a lonely tree, they find a nest. They hear the ill omens of the crows, And another life goes to waste.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expectedly, Another turn of events, Left the first mate forever lost, Along with the tents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At 45 to 4, onward they went, Towards the oasis in sight. The most eager of all shattered their illusion, And died due to the snake bite. Soon, fell upon them, the night, The predators were approaching fast, And with backs turned to the Polaris, They headed towards the past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the Sun lifted into the sky, So did the spirits of the grey. They were ready again to face the odds as before, But with only mirrors leading the way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“R.I.P. comrade,” said the captain, And engulfed himself in loneliness, Only to find that he was back, At that inn with his mistress;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“The inn is the centre of everything,” He admitted, amazed, And it is there where I should have realized, The curse of the maze.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above poem is famous in the islands of Andaman and Nicobar. The treasure is rumoured to be still there. Find the place where the treasure is buried.&lt;/p&gt;&#xA;" OwnerUserId="19701" LastEditorUserId="19701" LastEditDate="2016-03-02T04:54:10.560" LastActivityDate="2016-03-02T04:54:10.560" Title="Legend of Nine Treasure Hunt in Andaman and Nicobar Islands" Tags="|riddle|enigmatic-puzzle|treasure-hunt|" AnswerCount="0" CommentCount="8" ClosedDate="2016-03-15T02:23:01.883" ContentLicense="CC BY-SA 3.0" />
  <row Id="28166" PostTypeId="2" ParentId="28125" CreationDate="2016-03-01T20:55:58.550" Score="7" Body="&lt;p&gt;I think you're&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; radiation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm all over.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Radiation is &lt;a href=&quot;http://www.nrc.gov/about-nrc/radiation/around-us/doses-daily-lives.html&quot; rel=&quot;nofollow&quot;&gt;everywhere&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Over you. Under you. Through you.&lt;br&gt;&#xA;&amp;lt; Insert Evil Laugh &gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; See above&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I love black holes and dying cosmos.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;&quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Hawking_radiation&quot; rel=&quot;nofollow&quot;&gt;Hawking radiation&lt;/a&gt; is black-body radiation that is predicted to be released by black holes, due to quantum effects near the event horizon.&quot;&lt;/em&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come from evil politicians and ruthless dictators.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; States worldwide have used or threatened to use nuclear weapons&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come from silly children and irresponsible criminals.&lt;br&gt;&#xA;I can cause horrible diseases.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Exposure to certain types of radiation can cause cancer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The government overregulates me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I don't know enough to claim &lt;em&gt;over&lt;/em&gt;-regulation, but many governments have some form of &lt;a href=&quot;http://www.nrc.gov/about-nrc/radiation/protects-you/reg-matls.html&quot; rel=&quot;nofollow&quot;&gt;nuclear regulation&lt;/a&gt; in effect&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But that doesn't stop me.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mwah Ha Ha.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nothing can stop me.&lt;br&gt;&#xA;Not walls. Not floors. Not Time or Space or Distance.&lt;br&gt;&#xA;I OWN THE WORLD!!!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Non-ionizing_radiation&quot; rel=&quot;nofollow&quot;&gt;Certain types of radiation&lt;/a&gt; can pass through walls, floors, and over great distances of space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not that it does me much good.&#xA;Sometimes I think of my existence as a giant treadmill.&lt;/p&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2016-03-02T12:44:39.620" LastActivityDate="2016-03-02T12:44:39.620" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28168" PostTypeId="2" ParentId="28125" CreationDate="2016-03-01T22:14:30.300" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Death&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm all over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over you. Under you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Death is everywhere in relation to someone&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Through you.&#xA;&amp;lt; Insert Evil Laugh &gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Whether people want to admit it or not, they cause death.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I love black holes and dying cosmos.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Causing the death of any life nearby&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come from evil politicians and ruthless dictators.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; War&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come from silly children&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Children love to kill bugs, whether stomping or otherwise; additionally, just playing will cause things like plants and small creatures to die (often by accident, and to the parents' chagrin)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and irresponsible criminals.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Murder is indeed a crime&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can cause horrible diseases.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many diseases are caused by the decay of a body&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The government overregulates me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Much debate occurs around when someone or a pet is allowed to die, from executions to abortion to suicide to war to assisted suicide...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But that doesn't stop me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People still die&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mwah Ha Ha.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mwah Ha Ha!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nothing can stop me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Nothing is certain except for death and taxes.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not walls. Not floors. Not Time or Space or Distance.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Walls can't keep out death, whether from disease or natural causes.&#xA; It doesn't matter how high up you are in the world (floors), you will still die.&#xA; Indeed, the longer you live and the more you experience, your chance of experiencing death in the near future increases.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I OWN THE WORLD!!!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Most of the Earth we are familiar with (i.e., rocks, soil) has at one point been part of a living being. The world as we know it is very much made of dead things.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not that it does me much good.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What does something that is dead care?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I think of my existence as a giant treadmill.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Circle of life/death&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19703" LastActivityDate="2016-03-01T22:14:30.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28169" PostTypeId="1" AcceptedAnswerId="28170" CreationDate="2016-03-01T22:14:30.973" Score="14" ViewCount="849" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Think sunburnt capital and Charles the great.&lt;br&gt;&#xA;  Think part of a train with a rather large weight.&lt;br&gt;&#xA;  Think of  a wagon but with what added on.&lt;br&gt;&#xA;  If I amused as I puzzled, that's the art in the con. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastActivityDate="2016-03-02T07:54:20.833" Title="Thinking Outside the Box" Tags="|riddle|word|poetry|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28170" PostTypeId="2" ParentId="28169" CreationDate="2016-03-01T22:51:50.760" Score="9" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; carton&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Think sunburnt capital and Charles the great.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; not entirely sure, possibly referring to Corton / Corton-Charlemagne / Burgundy, though this seems like a stretch&lt;br&gt;&#xA;&lt;br&gt;&#xA; Other possible stretch: Since &quot;great&quot; is not capitalized, you could be referring to Charles Dickens, and this could be referring to Sydney Carton  - Sydney being the capital of NSW which is known for bushfires.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Think part of a train with a rather large weight.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; car + ton = carton&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Think of a wagon but with what added on.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; cart + on = carton&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I amused as I puzzled, that's the art in the con.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;c&lt;/strong&gt; &lt;em&gt;art&lt;/em&gt; &lt;strong&gt;on&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17147" LastEditorUserId="17147" LastEditDate="2016-03-01T23:49:37.510" LastActivityDate="2016-03-01T23:49:37.510" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28171" PostTypeId="1" CreationDate="2016-03-01T23:05:31.213" Score="21" ViewCount="1132" Body="&lt;p&gt;This is something of a follow-on to Hugh Meyer's enjoyable &quot;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/27989/titanic-tic-tac-toe&quot;&gt;Titanic Tic-Tac-Toe&lt;/a&gt;.&quot; Fortunately, our intrepid heroes Xavier and Oliver have wrested away a bit of wreckage that some fellow named Jack was holding on to and, as they sit in the icy North Atlantic, they pass the time by playing their favourite game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since they are now subject to the random undulations of the waves, they've had to change up their play again. Their rules are now as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Each player may, before placing a marker, shift the board in any direction, not just to the right.&lt;/li&gt;&#xA;&lt;li&gt;As before, the player may only shift a row or column provided that it either (a) removes one of his tokens or (b) does not remove any of his opponent's.&lt;/li&gt;&#xA;&lt;li&gt;As before, a player may not play on the same square in consecutive turns. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;After some discussion and gentlemanly concession, Xavier and Oliver have added the following stipulation:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The player may not make a play so as to return the board to state that it has previously been after his turn. If that is the only such move they can make, they lose.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To provide some mathematical clarity, if $S_i$ is the state of the board after the $i$ turns, then for $i \ne j$, $S_i = S_j$ only if $i \not\equiv j \mod 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their hope is that this will keep them from getting caught in a vicious cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, will Oliver win a game before the &lt;em&gt;Carpathia&lt;/em&gt; arrives to save them? (Will Xavier?)&lt;/p&gt;&#xA;" OwnerUserId="17446" LastEditorUserId="20521" LastEditDate="2016-05-08T07:38:22.000" LastActivityDate="2017-06-24T12:34:09.963" Title="Titanicker Tic-Tac-Toe" Tags="|strategy|game|tic-tac-toe|" AnswerCount="2" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="28173" PostTypeId="1" AcceptedAnswerId="28186" CreationDate="2016-03-02T02:49:18.613" Score="6" ViewCount="826" Body="&lt;p&gt;What word am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;First composed of &quot;Badap ba daa ba da daa ba daa da deda daa Badap ba daa ba da daa ba daa da deda daa&quot;.&lt;br&gt;&#xA;  A benevolent dictator for life implicitly writes.&lt;br&gt;&#xA;  The beginning of people saying, &quot;There's people waiting to be crucified out here.&quot;&lt;br&gt;&#xA;  And the back end of completely naked men!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There's a clue in the title.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastEditorUserId="10118" LastEditDate="2016-03-02T10:08:35.037" LastActivityDate="2016-03-02T10:41:45.083" Title="My not very bad one" Tags="|riddle|word|" AnswerCount="4" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28174" PostTypeId="2" ParentId="28173" CreationDate="2016-03-02T03:10:10.393" Score="1" Body="&lt;p&gt;Are you &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Judgement?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First composed of &quot;Badap ba daa ba da daa ba daa da deda daa Badap ba daa ba da daa ba daa da deda daa&quot;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No clue. Maybe a song&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A benevolent dictator for life implicitly writes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A benevolent dictator must be a fair judge and therefore casts fair judgement in their writings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The beginning of people saying, &quot;There's people waiting to be crucified out here.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;They must be judged before being executed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the back end of completely naked men!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; men! looks like the last 4 letters of judgement&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18821" LastActivityDate="2016-03-02T03:10:10.393" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28175" PostTypeId="2" ParentId="28173" CreationDate="2016-03-02T05:03:18.730" Score="2" Body="&lt;p&gt;The Word is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Doctor&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First composed of &quot;Badap ba daa ba da daa ba daa da deda daa Badap ba daa ba da daa ba daa da deda daa&quot;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Possibly the song Seems James Bond Announcement. The First Composed of James Bond Series is Dr.No&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A benevolent dictator for life implicitly writes.The beginning of people saying, &quot;There's people waiting to be crucified out here.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Crucified sort of like in the Operation theater?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the back end of completely naked men!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everybody goes Naked in front of Doctor.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18742" LastActivityDate="2016-03-02T05:03:18.730" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28176" PostTypeId="2" ParentId="8481" CreationDate="2016-03-02T07:20:11.173" Score="-1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;25 * 2 = 50&lt;br&gt;&#xA;(50 /2) -1 =24&lt;br&gt;&#xA;24/2 = 12&lt;br&gt;&#xA;12/2=6&lt;br&gt;&#xA;(6/2) -1 = 2&lt;br&gt;&#xA;2*2 = 4&lt;br&gt;&#xA;4*2 = 8&lt;br&gt;&#xA;8*2 =16  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19712" LastEditorUserId="17482" LastEditDate="2016-03-02T07:48:14.493" LastActivityDate="2016-03-02T07:48:14.493" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28177" PostTypeId="2" ParentId="28169" CreationDate="2016-03-02T07:54:20.833" Score="5" Body="&lt;p&gt;Just to add a little detail. The first line refers to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sydney, Australia via the poem &lt;a href=&quot;https://en.m.wikipedia.org/wiki/My_Country&quot;&gt;My Country &lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This links to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sydney Carton, hero of A Tale Of Two Cities by the &quot;great&quot; Charles Dickens. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I wanted to lead with something tricky since the rest of the puzzle was so straightforward. The rest follows as in the accepted answer. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A car is part of a train and a ton is a large weight&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The third line is a trick of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Punctuation. A cart is a wagon. With what added? &quot;On.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The last line hammers it home. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; By placing &quot;art&quot; in &quot;con&quot; c(art)on&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The title &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Is the only direct reference to the meaning of the word itself. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The whole thing popped into my head as I was falling asleep so I posted as is. &lt;/p&gt;&#xA;" OwnerUserId="18924" LastActivityDate="2016-03-02T07:54:20.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28178" PostTypeId="1" AcceptedAnswerId="28180" CreationDate="2016-03-02T08:28:13.257" Score="20" ViewCount="1494" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lbEmJ.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lbEmJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find X, Y and Z from the above rebus. &lt;/p&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="11876" LastEditDate="2016-03-02T11:37:19.160" LastActivityDate="2016-03-02T12:05:52.197" Title="Rebus: Romance, Cats and a Pen" Tags="|visual|rebus|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28179" PostTypeId="2" ParentId="28153" CreationDate="2016-03-02T08:49:18.500" Score="29" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;His middle name is Ambrose.&lt;br/&gt;&#xA;The riddle is a limerick. Removing the fruit, &quot;Lime&quot;, leaves &quot;Rick&quot;. This points towards the musician Rick James, who used quite a few different names, one of them &quot;Slick Rick&quot;. His real name was James Ambrose Johnson.&#xA;Also, &quot;Favorite Flava&quot; is a song from his album &quot;Urban Rhapsody&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7545" LastActivityDate="2016-03-02T08:49:18.500" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28180" PostTypeId="2" ParentId="28178" CreationDate="2016-03-02T09:22:41.680" Score="16" Body="&lt;p&gt;I know one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Y is Sylvester Stallone. (Sylvester the cat, and he's on stall).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Update. Thanks to &lt;a href=&quot;https://puzzling.stackexchange.com/users/17482/manshu&quot;&gt;manshu&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/users/17147/will&quot;&gt;Will&lt;/a&gt; :-)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; X is James Cameron (Cameron is an anagram of &lt;code&gt;romance&lt;/code&gt;.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If X and Y are correct, then Z is Rambo : First Blood II. (X and Y wrote it together.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1766" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2016-03-02T12:05:52.197" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="28181" PostTypeId="2" ParentId="28178" CreationDate="2016-03-02T09:29:39.527" Score="9" Body="&lt;p&gt;X =&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; James Cameron&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 007 = James Bond. Therefore half of 007 = James. &lt;em&gt;Cameron&lt;/em&gt; is anagram of &lt;em&gt;Romance&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastActivityDate="2016-03-02T09:29:39.527" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28182" PostTypeId="2" ParentId="28178" CreationDate="2016-03-02T09:31:31.863" Score="12" Body="&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/a/28181/&quot;&gt;Manshu&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/a/28180/&quot;&gt;ABcDexter&lt;/a&gt; have done almost all the work&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their answers lead to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Z = Rambo: First Blood Part II&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which has a screenplay written by X and Y&lt;/p&gt;&#xA;" OwnerUserId="17147" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2016-03-02T09:31:31.863" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28183" PostTypeId="2" ParentId="27967" CreationDate="2016-03-02T09:33:22.653" Score="3" Body="&lt;p&gt;I try to make it simple !!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  4/.2  -  3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And Another solution&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (ln(e)/.2)  + (4*3)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13616" LastEditorUserId="13616" LastEditDate="2016-03-02T10:01:22.403" LastActivityDate="2016-03-02T10:01:22.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28184" PostTypeId="1" CreationDate="2016-03-02T10:13:18.557" Score="0" ViewCount="6253" Body="&lt;p&gt;How can we determine the number of magic squares with magic constant 0?&lt;/p&gt;&#xA;" OwnerUserId="19713" LastEditorUserId="17482" LastEditDate="2016-03-02T12:32:01.497" LastActivityDate="2016-03-03T02:11:55.947" Title="Number of magic squares with magic constant 0?" Tags="|mathematics|magic-square|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28185" PostTypeId="2" ParentId="28173" CreationDate="2016-03-02T10:26:38.553" Score="3" Body="&lt;p&gt;Sounds like every clue points to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Monty Python's Life of Brian&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First composed of &quot;Badap ba daa ba da daa ba daa da deda daa Badap ba daa ba da daa ba daa da deda daa&quot;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Always look on the bright side of life&quot; is sung by the crucified people&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A benevolent dictator for life implicitly writes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; no clue yet, the emperor in the movie is Pontius Pilatus&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The beginning of people saying, &quot;There's people waiting to be crucified out here.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Quote from the crucification scene&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the back end of completely naked men!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; also no clue&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19714" LastActivityDate="2016-03-02T10:26:38.553" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28186" PostTypeId="2" ParentId="28173" CreationDate="2016-03-02T10:41:45.083" Score="13" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Monty&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First composed of &quot;Badap ba daa ba da daa ba daa da deda daa Badap ba daa ba da daa ba daa da deda daa&quot;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The composer of the James Bond theme tune is &lt;b&gt;Monty&lt;/b&gt; Norman&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A benevolent dictator for life implicitly writes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Refers to BDFL Guido van Rossum who wrote most of the Python examples using phrases from &lt;b&gt;Monty&lt;/b&gt; Python.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The beginning of people saying, &quot;There's people waiting to be crucified out here.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;b&gt;Monty&lt;/b&gt; Python's Life of Brian&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the back end of completely naked men!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Full &lt;b&gt;Monty&lt;/b&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Title &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Monty is an anagram of 'My not'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18422" LastActivityDate="2016-03-02T10:41:45.083" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28187" PostTypeId="2" ParentId="27967" CreationDate="2016-03-02T11:22:03.677" Score="5" Body="&lt;p&gt;Exactly 17:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  $\lceil 4^2 - \cos(3) \rceil$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or slightly less (with an error of less than 0.06%):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  $ 4^2 - \cos (3) \approx 16.98999 $&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19714" LastEditorUserId="8874" LastEditDate="2016-03-03T17:36:10.387" LastActivityDate="2016-03-03T17:36:10.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28188" PostTypeId="2" ParentId="28184" CreationDate="2016-03-02T12:20:30.747" Score="13" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There are an infinite number of such squares&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Take the example square below:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;-3  2  1&#xA; 4  0 -4&#xA;-1 -2  3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;To generate a new square, simply multiply each element of this square by any positive integer. As there are an infinite number of positive integers, there are an infinite number of possible squares.&lt;/p&gt;&#xA;" OwnerUserId="3987" LastEditorUserId="3987" LastEditDate="2016-03-02T12:59:09.187" LastActivityDate="2016-03-02T12:59:09.187" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28189" PostTypeId="2" ParentId="27967" CreationDate="2016-03-02T12:35:09.110" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $2^{4}+\prod_{n=3}^{\infty}\frac{n}{n}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1598" LastActivityDate="2016-03-02T12:35:09.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28190" PostTypeId="2" ParentId="27967" CreationDate="2016-03-02T13:39:17.927" Score="2" Body="&lt;p&gt;Without using concatenation or rounding of any kind:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$(2 \times 3)? - 4=17$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; '?' is the &quot;termial function&quot; which is like factorial, but with addition instead of multiplication.  Thus, $(2 \times 3)? = 6? = 1+2+3+4+5+6=21$.  Take away 4 to get 17.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2016-03-02T13:39:17.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28191" PostTypeId="2" ParentId="28020" CreationDate="2016-03-02T13:45:50.610" Score="2" Body="&lt;p&gt;Now with many&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A synonym for &quot;many&quot; is &lt;strong&gt;C&lt;/strong&gt;OPIOUS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now with having lovers&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Somebody who had a lover is &lt;strong&gt;L&lt;/strong&gt;ADY CHATTERLEY&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now with first swing&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;U&lt;/strong&gt; is shaped like a &lt;a href=&quot;http://thebrianmountjoy.typepad.com/.a/6a00e3981b770d8833017ee97ed232970d-800wi&quot; rel=&quot;nofollow&quot;&gt;swing&lt;/a&gt;. Yes u is.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now with some covers&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Part of a cover is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cover_(philately)&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;E&lt;/strong&gt;NVELOPE&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So that fifth clue?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's the acrosticized version of all the answers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2016-03-02T18:13:55.603" LastActivityDate="2016-03-02T18:13:55.603" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28192" PostTypeId="2" ParentId="28020" CreationDate="2016-03-02T13:56:19.550" Score="3" Body="&lt;h3&gt;This is not a correct answer(my achievement)...confirmed by OP&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I think it is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Mark_Lowe&quot; rel=&quot;nofollow&quot;&gt;Mark Lowe&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now with many&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He previously played for the Seattle Mariners, Texas Rangers, Los Angeles Angels of Anaheim, Cleveland Indians, and Toronto Blue Jays.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now with having lovers&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is the pun on his last name I think as stated in hint 2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now with first swing&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It tells us that he is related to baseball&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now with some covers&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He is a relief pitcher. So we can say that he cover pitching for his team. (But I think it still needs better explanation)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As already stated in the hint 3 that it's acrostic. So  &lt;strong&gt;M&lt;/strong&gt;ANY H&lt;strong&gt;A&lt;/strong&gt;VING FI&lt;strong&gt;R&lt;/strong&gt;ST &lt;strong&gt;C&lt;/strong&gt;OVERS... It gives us MARC, pronounced as MARK (as much as I know).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-03-03T08:07:24.947" LastActivityDate="2016-03-03T08:07:24.947" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28193" PostTypeId="2" ParentId="28141" CreationDate="2016-03-02T14:47:17.980" Score="5" Body="&lt;p&gt;Pick any four points that form a rectangle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{matrix} &#xA;a_1 &amp;amp; \ldots &amp;amp; a_2\\ &#xA;\vdots &amp;amp; \ldots &amp;amp; \vdots\\&#xA;a_3 &amp;amp; \ldots &amp;amp; a_4&#xA;\end{matrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that the sum of these four points is zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_1+a_2+a_3+a_4=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the two additional points $b_1$ and $b_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{matrix} &#xA;a_1 &amp;amp; \ldots &amp;amp; a_2 &amp;amp; \ldots &amp;amp; b_1\\ &#xA;\vdots &amp;amp; \ldots &amp;amp; \vdots\\&#xA;a_3 &amp;amp; \ldots &amp;amp; a_4 &amp;amp; \ldots &amp;amp; b_2&#xA;\end{matrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These points form two additional rectangles (one large, and one smaller on the the right), so we get two more equations;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_1+a_3+b_1+b_2=0$$&#xA;$$a_2+a_4+b_1+b_2=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the difference of these equations shows us that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_1+a_3=a_2+a_4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting in our first equation, we get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_1+a_3+a_1+a_3=0 \implies a_1=-a_3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since our choice of points was arbitrary, we know that any two points that share a side of a rectangle must be the opposite sign.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_1=-a_2, a_1=-b_1, a_2=-b_1 \implies a_1=-a_1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way to satisfy these equations is if they are all zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus (again since our choice of points was arbitrary) all the points must be zero.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2016-03-02T14:47:17.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28194" PostTypeId="1" CreationDate="2016-03-02T15:08:05.147" Score="20" ViewCount="3886" Body="&lt;p&gt;The Blue Eyes puzzle is possibly the most popular induction-based puzzle known. Is it really known who was the first to come up with it (in any form).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The link at &lt;a href=&quot;http://www.xkcd.com/blue_eyes.html&quot;&gt;xkcd&lt;/a&gt; says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I didn't come up with the idea of this puzzle, but I've written and rewritten it over the the years to try to make a definitive version. The guy who told it to me originally was some dude on the street in Boston named Joel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, it doesn't claim that this stranger was the first to create the puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia provides another &lt;a href=&quot;https://en.wikipedia.org/wiki/Common_knowledge_(logic)#endnote_Sevitan&quot;&gt;link&lt;/a&gt; for the same. [I know, I just linked to another link ;)]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So can we safely conclude that the original author is not known? And is there any reward for satisfactorily proving that you are the original author of the puzzle?&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2016-09-05T02:28:05.367" Title="The origin of the blue eyes puzzle" Tags="|logical-deduction|puzzle-creation|puzzle-identification|blue-eyes|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28195" PostTypeId="2" ParentId="28194" CreationDate="2016-03-02T15:51:43.773" Score="10" Body="&lt;p&gt;The earliest occurrence of the puzzle that I am aware of is from 1958.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;George Gamow and Marvin Stern: &quot;Puzzle Math&quot;, Viking Press (February 7, 1958)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is the Amazon page for this book:&lt;br&gt;&#xA;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0670583359&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Puzzle-math-George-Gamow/dp/0670583359&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Chapter 1 of the book contains several mathematical puzzle stories on &#xA;the great Sultan Ibn-al-Kuz of Quasiababia. The third of these stories is&#xA;called &quot;FORTY UNFAITHFUL WIVES&quot;. and deals with the blue eyes puzzle.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The great Sultan Ibn-al-Kuz was very much worried about the large number &#xA;  of unfaithful wives among the population of his capital city. There were &#xA;  forty women who were openly deceiving their husbands, but, as often &#xA;  happens although all these cases were a matter of common knowledge, the &#xA;  husbands in question were ignorant of their wives’ behavior. In order to &#xA;  punish the wretched women, the sultan issued a proclamation which permitted &#xA;  the husbands of unfaithful wives to kill them, provided, however, that &#xA;  they were quite sure of the infidelity. The proclamation did not mention &#xA;  either the number or the names of the wives known to be unfaithful; it &#xA;  merely stated that such cases were known in the city and suggested that &#xA;  the husbands do something about it. However, to the great surprise of the &#xA;  entire legislative body and the city police, no wife killings were reported &#xA;  on the day of the proclamation, or on the days that followed. In fact, an &#xA;  entire month passed without any result, and it seemed the deceived husbands &#xA;  just did not care to save their honor.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;“O Great Sultan,” said the vizier to Ibn-al-Kuz, “shouldn’t we announce the &#xA;  names of the forty unfaithful wives, if the husbands are too lazy to pursue &#xA;  the cases themselves?”&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;“No,” said the sultan. “Let us wait. My people may be lazy, but they are &#xA;  certainly very intelligent and wise. I am sure action will be taken very &#xA;  soon.”&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;And, indeed, on the fortieth day after the proclamation, action suddenly &#xA;  broke out. That single night forty women were killed, and a quick check &#xA;  revealed that they were the forty who were known to have been deceiving &#xA;  their hus-bands.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;“I do not understand it,” exclaimed the vizier. “Why did these forty &#xA;  wronged husbands wait such a long time to take action, and why did they &#xA;  all finally take it on the same day?”&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;“Very simple, my dear Watson.” The sultan chuckled. “As a matter of fact &#xA;  I expected this good news exactly on that day. My people, as I suggested &#xA;  before, may be too lazy to organize the shadowing of their wives for the &#xA;  purpose of establishing their faithfulness or unfaithfulness, but they &#xA;  have certainly shown themselves intelligent enough to resolve the case by &#xA;  purely logical analysis.”&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;“I do not understand you, Great Sultan,” said the vizier.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;“Well, assume that there were not forty unfaithful wives, but only one. &#xA;  In this case, everybody with the exception of her husband knew the fact. &#xA;  Her husband, however, believing in the faithfulness of his wife, and &#xA;  knowing no other case of unfaithfulness (about which he would undoubtedly &#xA;  have heard) was under the impression that all wives in the city, including &#xA;  his own, were faithful. If he read the proclamation which stated that&#xA;  there are unfaithful wives in the city, he would realize it could mean &#xA;  only his own wife. Thus he would kill her the very first night. &#xA;  Do you follow me?”&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;“I do,” said the vizier.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;“Now let us assume,” continued the sultan, “that there were two deceived &#xA;  husbands, let us call them Abdula and Hadjibaba. Abdula knew all the time &#xA;  that Hadjibaba’s wife was deceiving himt and Hadjibaba knew the same about &#xA;  Abdula’s wife. But each thought his own wife was faithful.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;“On the day that the proclamation was published, Abdula said to himself, &#xA;  ‘Aha, tonight Hadjibaba will kill his wife.’ On the other hand, Hadjibaba &#xA;  thought the same about Abdula. However, the fact that next morning both &#xA;  wives were still alive proved to both Abdula and Hadjibaba that they were &#xA;  wrong in believing in the faithfulness of their wives. Thus during the &#xA;  second night two daggers would have found their target, and two women &#xA;  would have been dead.”&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;“I follow you so far” said the vizier, “but how about the case of three &#xA;  or more unfaithful wives?”&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;“Well, from now on we have what is called mathematical induction. I have &#xA;  just proved to you that, if there were only two unfaithful wives in the &#xA;  city, the husbands would have killed them on the second night, by force &#xA;  of purely logical deduction. Now suppose that there were three wives, &#xA;  Abdula’s Hadjibaba’s, and Faruk’s, who were unfaithful. Faruk knows, of &#xA;  course, that Abdula’s and Hadjibaba’s wives are deceiving them, and so he &#xA;  expects that these two characters will murder their wives on the second &#xA;  night. But they don’t. Why? Of course because his, Faruk’s, wife is &#xA;  unfaithful, tool and so in goes the dagger, or the three daggers, as a &#xA;  matter of fact.”&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;“O Great Sultan,” exclaimed the vizier, “you have certainly opened my &#xA;  eyes on that problem. Of course, if there were four unfaithful wives, &#xA;  each of the four wronged husbands would reduce the case to that of three &#xA;  and not kill his wife until the fourth day. And so on, and so on, up to &#xA;  forty wives.”&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;“I am glad,” said the sultan, “that you finally understand the situation. &#xA;  It is nice to have a vizier whose intelligence is so much inferior to &#xA;  that of the average citizen. But what if I tell you that the reported &#xA;  number of unfaithful wives was actually forty-one?”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Gamow and Stern state in a footnote that this puzzle was communicated &#xA;to them by Dr. Victor Ambarzuminian.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="4421" LastEditDate="2016-09-05T02:28:05.367" LastActivityDate="2016-09-05T02:28:05.367" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28197" PostTypeId="2" ParentId="28194" CreationDate="2016-03-02T16:01:06.493" Score="8" Body="&lt;p&gt;A version of this puzzle, involving three ladies with dirty faces laughing at one another, appears in J. E. Littlewood's &quot;A Mathematician's Miscellany,&quot; published in 1953. Littlewood writes that this is &quot;a well-known puzzle that swept Europe a good many years ago and in one form or another has appeared in a number of books.&quot; The puzzle involves just three ladies, but on the next page Littlewood says &quot;But further, what has not got into the books so far as I know, there is an extension, in principle, to $n$ ladies, all dirty and laughing.&quot;&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2016-03-02T16:17:11.727" LastActivityDate="2016-03-02T16:17:11.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28198" PostTypeId="2" ParentId="28194" CreationDate="2016-03-02T16:19:38.130" Score="15" Body="&lt;p&gt;For the muddy children variant of this problem, there are several &#xA;earlier sources.&#xA;For instance, A.A. Bennett (Problem No. 3734, American Mathematical &#xA;Monthly 42, 1935, page 256) formulated the following version back in 1935:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A car with $n&amp;gt;2$ passengers of different speeds of mental reaction &#xA;  passes through a tunnel, and each passenger acquires unconsciously a &#xA;  smudge of soot upon his forehead. Suppose that each passenger&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;laughs and continues to laugh as soon as and only so long as he sees a&#xA;  smudge upon the forehead of a fellow passenger;&lt;/li&gt;&#xA;  &lt;li&gt;can see the foreheads of all his fellows;&lt;/li&gt;&#xA;  &lt;li&gt;reasons correctly;&lt;/li&gt;&#xA;  &lt;li&gt;will clean his own forehead when and only when his reasoning forces&#xA;  him to conclude that he has a smudge;&lt;/li&gt;&#xA;  &lt;li&gt;knows that (1), (2), (3) and (4) hold for each of his fellows.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;Show that each passenger will eventually wipe his own forehead.&#xA;  (For the case of $n=3$, this has been proposed in conversation by &#xA;  Dr. Church of Princeton.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2016-03-02T16:19:38.130" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28199" PostTypeId="2" ParentId="27777" CreationDate="2016-03-02T16:54:04.853" Score="4" Body="&lt;p&gt;The question doesn't make this clear and I'm not good with phonetics, so I'll make these assumptions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;you're looking for 1 syllable words only&lt;/li&gt;&#xA;&lt;li&gt;each word must end in a 'vowel sound', which excludes hard consonant endings like 'ed'&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;With the above, I wrote a list of vowel-sound endings (I might have missed some):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;suffixes = ['aa', 'ough', 'ae', 'ai', 'ah', 'ir', 'ea', 'ar', 'au', 'o', 'aw', 'ay', 'ei', 'ew', 'ie', 'er', 'ear', 'ier', 'owe', 'ire', 'igh', 'oo', 'augh', 'y', 'aigh', 'ee', 'ye', 'ooh', 'ow', 'oor', 'ore', 'a', 'e', 'err', 'i', 'oe', 'oa', 'ur', 'eigh', 'u', 'urr', 'ue', 'ey', 'ou', 'or']&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And ran a python script which joins all possible prefixes ('thr','sp','tr','fl' etc) to all possible vowel-sound endings and checks if the result is a word. The winners are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;18 words:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;['pea', 'par', 'po', 'paw', 'pay', 'pew', 'pie', 'per', 'pear', 'pier', 'pee', 'pooh', 'pow', 'poor', 'pore', 'pa', 'pi', 'purr']&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;19 words:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;['tough', 'tai', 'tea', 'tar', 'tau', 'to', 'taw', 'tie', 'tear', 'tier', 'tire', 'too', 'tee', 'tow', 'tore', 'terr', 'ti', 'toe', 'tor']&#xA; ['baa', 'bough', 'bah', 'bar', 'bay', 'bear', 'bier', 'boo', 'by', 'bee', 'bye', 'bow', 'boor', 'bore', 'be', 'boa', 'bur', 'burr', 'bey']&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;23 words:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;['hah', 'ho', 'haw', 'hay', 'hew', 'hie', 'her', 'hear', 'howe', 'hire', 'high', 'haugh', 'hee', 'how', 'ha', 'he', 'herr', 'hi', 'hoe', 'heigh', 'hue', 'hey', 'hor']&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This was using &lt;a href=&quot;http://www-01.sil.org/linguistics/wordlists/english/wordlist/wordsEn.txt&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; dictionary, with others your mileage may vary.&lt;/p&gt;&#xA;" OwnerUserId="19698" LastEditorUserId="19698" LastEditDate="2016-03-02T17:20:01.960" LastActivityDate="2016-03-02T17:20:01.960" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28200" PostTypeId="2" ParentId="2907" CreationDate="2016-03-02T17:36:12.200" Score="1" Body="&lt;p&gt;I'm a bit late to the party, but here's a solution that costs 26:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UHf0a.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UHf0a.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19698" LastActivityDate="2016-03-02T17:36:12.200" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28201" PostTypeId="1" AcceptedAnswerId="28206" CreationDate="2016-03-02T17:43:46.507" Score="71" ViewCount="16778" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;As an alphabet aficionado, I can make you laugh long into the night.&lt;br&gt;&#xA;  As a make of Japanese vehicle, I can keep your surroundings clean and beautiful.&lt;br&gt;&#xA;  As an execution apparatus, I make you laugh despite not maturing.&lt;br&gt;&#xA;  As a mixed-up romance, I lead tens of millions.&lt;br&gt;&#xA;  As a gardening tool, I provide weekend evening entertainment and provide guidelines for proposing.&lt;br&gt;&#xA;  As a reddish-brown meadow, I dazzle you with tricks.&lt;br&gt;&#xA;  As a fighting knife, I make popular music and navigate a maze.&lt;br&gt;&#xA;  As a speech-impaired aquatic athlete, I am your friend.&lt;br&gt;&#xA;  As a renter, I am your doctor.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Who am I?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="4421" LastEditDate="2016-03-02T17:55:17.360" LastActivityDate="2016-03-03T07:24:03.220" Title="You've Probably Seen Me Naked" Tags="|riddle|word|wordplay|knowledge|" AnswerCount="5" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28202" PostTypeId="2" ParentId="28201" CreationDate="2016-03-02T19:00:13.660" Score="20" Body="&lt;p&gt;As a mixed-up romance, I lead tens of millions.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Possibly David &lt;strong&gt;Cameron&lt;/strong&gt; (anagram of &lt;em&gt;romance&lt;/em&gt;) who lead tens of millions as the Prime Minister of the UK?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a gardening tool, I provide weekend evening entertainment and provide guidelines for proposals.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; David Spade comedian and star of Rules of Engagement&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastEditorUserId="10118" LastEditDate="2016-03-02T19:46:51.760" LastActivityDate="2016-03-02T19:46:51.760" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28203" PostTypeId="2" ParentId="28201" CreationDate="2016-03-02T19:10:18.317" Score="18" Body="&lt;p&gt;As a fighting knife, I make popular music and navigate a maze.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is a &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Bowie_knife&quot;&gt;Bowie knife&lt;/a&gt;, and David Bowie is a singer and was in the movie &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Labyrinth_(film)&quot;&gt;Labyrinth&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastActivityDate="2016-03-02T19:10:18.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28204" PostTypeId="2" ParentId="28201" CreationDate="2016-03-02T19:15:23.027" Score="17" Body="&lt;p&gt;A reddish-brown meadow who dazzles with tricks could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; David Copperfield &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The alphabet aficionado who makes you laugh into the night is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; David Letterman&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Japanese vehicle/beautiful surroundings is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; David Suzuki&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2016-03-02T19:22:35.240" LastActivityDate="2016-03-02T19:22:35.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28205" PostTypeId="2" ParentId="28201" CreationDate="2016-03-02T19:25:14.983" Score="19" Body="&lt;p&gt;To supplement others' answers, many have seen&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The statue of David&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;naked&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; David Schwimmer played Ross on Friends&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; David Tennant plays the Doctor&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19703" LastActivityDate="2016-03-02T19:25:14.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28206" PostTypeId="2" ParentId="28201" CreationDate="2016-03-02T19:28:42.493" Score="68" Body="&lt;p&gt;This is a compilation of many previous answers and adding of my own. Most of the credit to &lt;a href=&quot;https://puzzling.stackexchange.com/a/28202/4421&quot;&gt;Paul Evans&lt;/a&gt;, &lt;a href=&quot;https://puzzling.stackexchange.com/a/28203/4421&quot;&gt;Sleafar&lt;/a&gt;, &lt;a href=&quot;https://puzzling.stackexchange.com/a/28205/4421&quot;&gt;charfellow&lt;/a&gt;, and &lt;a href=&quot;https://puzzling.stackexchange.com/a/28204/4421&quot;&gt;question_asker&lt;/a&gt; (in no particular order). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;David&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You've probably seen me naked.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/David_(Michelangelo)&quot; rel=&quot;noreferrer&quot;&gt;Statue of David&lt;/a&gt;&lt;/strong&gt; is a famous statue of King David in the nude.  (from charfellow)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As an alphabet aficionado, I can make you laugh long into the night. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/David_Letterman&quot; rel=&quot;noreferrer&quot;&gt;David Letterman&lt;/a&gt;&lt;/strong&gt;, comedian and former late-show talk host.  A &lt;em&gt;letter man&lt;/em&gt; is an alphabet aficionado.  (from question_asker)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a make of Japanese vehicle, I can keep your surroundings clean and beautiful.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/David_Suzuki&quot; rel=&quot;noreferrer&quot;&gt;David Suzuki&lt;/a&gt;&lt;/strong&gt;, environmental activist.  &lt;a href=&quot;https://en.wikipedia.org/wiki/Suzuki&quot; rel=&quot;noreferrer&quot;&gt;Suzuki&lt;/a&gt; is a make of vehicle from Japan.  (from question_asker)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As an execution apparatus, I make you laugh despite not maturing.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/David_Cross&quot; rel=&quot;noreferrer&quot;&gt;David Cross&lt;/a&gt;&lt;/strong&gt;, comedian and actor on &lt;a href=&quot;https://en.wikipedia.org/wiki/Arrested_Development_(TV_series)&quot; rel=&quot;noreferrer&quot;&gt;Arrested Development&lt;/a&gt;.  A &lt;em&gt;cross&lt;/em&gt; is an ancient method of execution, and &lt;em&gt;not maturing&lt;/em&gt; = &lt;em&gt;arrested development&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a mixed-up romance, I lead tens of millions.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/David_Cameron&quot; rel=&quot;noreferrer&quot;&gt;David Cameron&lt;/a&gt;&lt;/strong&gt;, prime minister of the UK.  &lt;em&gt;Cameron&lt;/em&gt; is an anagram of &lt;em&gt;romance&lt;/em&gt;.  (from Paul Evans)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a gardening tool, I provide weekend evening entertainment and provide guidelines for proposing.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/David_Spade&quot; rel=&quot;noreferrer&quot;&gt;David Spade&lt;/a&gt;&lt;/strong&gt;, a comedian on &lt;a href=&quot;https://en.wikipedia.org/wiki/Saturday_Night_Live&quot; rel=&quot;noreferrer&quot;&gt;Saturday Night Live&lt;/a&gt; and an actor in &lt;a href=&quot;https://en.wikipedia.org/wiki/Rules_of_Engagement_(TV_series)&quot; rel=&quot;noreferrer&quot;&gt;Rules of Engagement&lt;/a&gt;.&lt;br&gt;&#xA; &lt;em&gt;weekend evening entertainment&lt;/em&gt; = &lt;em&gt;Saturday Night Live&lt;/em&gt;&lt;br&gt;&#xA; &lt;em&gt;guidelines for proposing&lt;/em&gt; = &lt;em&gt;rules of engagement&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a reddish-brown meadow, I dazzle you with tricks.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/David_Copperfield_(illusionist)&quot; rel=&quot;noreferrer&quot;&gt;David Copperfield&lt;/a&gt;&lt;/strong&gt;, illusionist.  A &lt;em&gt;reddish-brown meadow&lt;/em&gt; is a &lt;em&gt;copper field&lt;/em&gt;.  (from question_asker)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a fighting knife, I make popular music and navigate a maze.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/David_Bowie&quot; rel=&quot;noreferrer&quot;&gt;David Bowie&lt;/a&gt;&lt;/strong&gt;, acclaimed singer, and actor in the movie &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Labyrinth_(film)&quot; rel=&quot;noreferrer&quot;&gt;Labyrinth&lt;/a&gt;&lt;/em&gt;. A &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Bowie_knife&quot; rel=&quot;noreferrer&quot;&gt;Bowie knife&lt;/a&gt;&lt;/em&gt; is a popular knife originally made for fighting, and a &lt;em&gt;maze&lt;/em&gt; is a &lt;em&gt;labyrinth&lt;/em&gt;. (from  Sleafar) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a speech-impaired aquatic athlete, I am your friend.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/David_Schwimmer&quot; rel=&quot;noreferrer&quot;&gt;David Schwimmer&lt;/a&gt;&lt;/strong&gt;, actor on the TV show &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Friends&quot; rel=&quot;noreferrer&quot;&gt;Friends&lt;/a&gt;&lt;/em&gt;. An &lt;em&gt;aquatic athlete&lt;/em&gt; is a &lt;em&gt;swimmer&lt;/em&gt;, which might be pronounced, &quot;&lt;em&gt;schwimmer&lt;/em&gt;&quot; by someone with a speech impairment. (from charfellow)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a renter, I am your doctor.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/David_Tennant&quot; rel=&quot;noreferrer&quot;&gt;David Tennant&lt;/a&gt;&lt;/strong&gt;, actor on the TV show &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Doctor_Who&quot; rel=&quot;noreferrer&quot;&gt;Doctor Who&lt;/a&gt;&lt;/em&gt;.  A &lt;em&gt;renter&lt;/em&gt; is a &lt;em&gt;tenant&lt;/em&gt;, and he plays the role of the &lt;em&gt;doctor&lt;/em&gt;. (from charfellow)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2016-03-02T20:28:15.553" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="28207" PostTypeId="2" ParentId="9000" CreationDate="2016-03-02T19:47:43.930" Score="1" Body="&lt;p&gt;Another possibility for #4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Calling it a day&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19726" LastActivityDate="2016-03-02T19:47:43.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28208" PostTypeId="2" ParentId="27967" CreationDate="2016-03-02T20:22:33.570" Score="4" Body="&lt;p&gt;$$&#xA;\pi\left(\Gamma\left(\frac{\mathrm{archav}(S(\theta(\mathrm{sinc}(4))))}{\sqrt{\zeta(2)}}+\Im\left\{\log(\mathrm{erf}(-3))\right\}\right)\right)=17&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\pi$ is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Prime-counting_function&quot; rel=&quot;nofollow&quot;&gt;prime-counting function&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;$\Gamma$ is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Gamma_function&quot; rel=&quot;nofollow&quot;&gt;gamma function&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;$\mathrm{archav}$ is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Versine&quot; rel=&quot;nofollow&quot;&gt;inverse haversine&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;$S$ is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Successor_function&quot; rel=&quot;nofollow&quot;&gt;successor function&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;$\theta$ is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Heaviside_step_function&quot; rel=&quot;nofollow&quot;&gt;Heaviside step function&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;$\mathrm{sinc}$ is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Sinc_function&quot; rel=&quot;nofollow&quot;&gt;sinc function&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;$\zeta$ is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Riemann_zeta_function&quot; rel=&quot;nofollow&quot;&gt;Riemann zeta function&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;$\Im$ is the imaginary part of a &lt;a href=&quot;https://en.wikipedia.org/wiki/Complex_number#Definition&quot; rel=&quot;nofollow&quot;&gt;complex number&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;$\mathrm{erf}$ is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Error_function&quot; rel=&quot;nofollow&quot;&gt;error function&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;How it Works&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;When I saw:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;any&lt;/em&gt; operation, including factorials and other &lt;em&gt;more advanced math&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I knew I had to do something crazy using a few of my favorite functions.  I started off with the prime-counting function.  Since it always outputs a whole number, even for noninteger inputs, it gives me a bit of leeway.  We just need to get in between $p_{17}=59$ and $p_{18}=61$, which conveniently bracket $60$, which seems like an easy target to reach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I played around with $5!/2=\Gamma(6)/2=60$ for a while.  Although it was easy to get the $5$ or $6$ I needed, I couldn't reach them using all the functions I wanted to include.  Since I needed a divisor of $2$ (or $\sqrt{4}$) this left me with at most two terms, which was not enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I discovered that $\Gamma(\sqrt{6}+\pi)\approx 60.6657\ldots$.  This is good for two reasons: first, it frees up the divisor outside the gamma function.  Second, $\zeta(2)=\pi^2/6$, which gives me a chance to include the zeta function.  At this point the expression in my head looks something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\pi\left(\Gamma\left(\frac{\pi}{\sqrt{\zeta(2)}}+\pi\right)\right)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that $\pi\approx 3.1415\ldots$ is not the same as $\pi(\cdot)$ the prime-counting function.)  I now just needed two ways to get $\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My idea for the first $\pi$ was to use obscure trigonometric functions.  Fortunately I can use their inverses to get the angle $\pi$ without using the notation $f^{-1}$ due to the $\mathrm{arc}\cdots$ naming convention.  I decided on the haversine, since $\mathrm{hav}(\pi)=1$ and $1$ is an easy number to reach.  I wanted to get to the $1$ using the successor function, like $S(0)=1$.  Now how do I get a zero?  I used the unit step function with a negative argument, in this case $\mathrm{sinc}(-4)\approx-0.1892\ldots$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second $\pi$ I wanted to play around with the logarithm and complex numbers; since $\log(z)=\log(|z|)+i \arg(z)$, any negative number gives an imaginary part of $\pi i$.  For bonus points I wanted to approximate $\log(-1)=\pi i$.  To get close to $-1$ I used the error function, which very quickly approaches $\pm 1$ for large inputs.  In fact, $\mathrm{erf}(-3)\approx -0.999978\ldots$, and $\log(\mathrm{erf}(-3))\approx-0.000022\ldots+\pi i$.  Of course, the real part doesn't matter since it gets chopped off anyway.&lt;/p&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="9718" LastEditDate="2016-03-02T20:59:41.297" LastActivityDate="2016-03-02T20:59:41.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28209" PostTypeId="1" AcceptedAnswerId="28242" CreationDate="2016-03-02T21:44:19.080" Score="8" ViewCount="1233" Body="&lt;p&gt;Opened or closed,&#xA;I continue to be.&lt;br&gt;&#xA;A mechanism of choice,&#xA;But I can be beastly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I come in all shapes and sizes,&#xA;Each with a common task.&lt;br&gt;&#xA;I cling to life by a thread,&#xA;What am I, I ask?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I can be made out of anything from A to Z,&#xA; and sometimes have eyes with which I can't see.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I hate to give another hint but there have been a ton of views and guesses, but here it is:&lt;br&gt;&#xA;&#xA; I have a few close friends, just like a pea,&#xA; and I jump through hoops for everyone around me.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is part 1 in a series by the way. The final part will use clues from each. &lt;/p&gt;&#xA;" OwnerUserId="19731" LastEditorUserId="19731" LastEditDate="2016-03-03T17:54:05.100" LastActivityDate="2016-03-03T17:54:05.100" Title="Large and in charge" Tags="|riddle|word|rhyme|english|poetry|" AnswerCount="11" CommentCount="1" ClosedDate="2016-03-03T17:15:21.107" ContentLicense="CC BY-SA 3.0" />
  <row Id="28210" PostTypeId="1" AcceptedAnswerId="28211" CreationDate="2016-03-02T21:48:03.483" Score="10" ViewCount="560" Body="&lt;p&gt;Which name does not belong on this list?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Charles Darwin&lt;br&gt;&#xA;  Anthony Burgess&lt;br&gt;&#xA;  Edward Albee&lt;br&gt;&#xA;  Alice Munro&lt;br&gt;&#xA;  Ignatius of Loyola&lt;br&gt;&#xA;  Phyllis Schlafly  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First hint:&lt;br&gt;&#xA;You actually don't need any hints&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second hint:&lt;br&gt;&#xA;Six&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third hint:&lt;br&gt;&#xA;If you have any doubts or debates concerning your answer, it is almost certainly wrong. &lt;/p&gt;&#xA;" OwnerUserId="18924" LastEditorUserId="4421" LastEditDate="2016-03-02T22:04:52.347" LastActivityDate="2016-03-03T10:30:02.413" Title="Read and Answer the Question" Tags="|pattern|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28211" PostTypeId="2" ParentId="28210" CreationDate="2016-03-02T21:55:46.897" Score="14" Body="&lt;p&gt;It's either&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ignatius of Loyola, because there are three words there instead of two,  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Phyllis Schlafly, because it's misspelled.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;OK, I know that's not the real answer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Charles Darwin, because all the others have a bug in the name:&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;Ant&lt;/strong&gt;hony Burgess&lt;br&gt;&#xA; Edward Al&lt;strong&gt;bee&lt;/strong&gt;&lt;br&gt;&#xA; A&lt;strong&gt;lice&lt;/strong&gt; Munro&lt;br&gt;&#xA; I&lt;strong&gt;gnat&lt;/strong&gt;ius of Loyola&lt;br&gt;&#xA; Phyllis Schla&lt;strong&gt;fly&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="4421" LastEditDate="2016-03-02T22:04:19.233" LastActivityDate="2016-03-02T22:04:19.233" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28212" PostTypeId="2" ParentId="28209" CreationDate="2016-03-02T21:56:02.317" Score="4" Body="&lt;p&gt;You could be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A trap&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Open or closed I continue to be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Whether open or closed, a trap is still a trap &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A mechanism of choice &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many traps operate through trickery. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be beastly &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A number of animals (beasts) make traps&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come in all shapes and sizes with one common purpose &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pits, snares, cages... Lots of types of traps all designed to catch the unwary. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cling to life by a thread &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I 'm thinking of spider webs here&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastEditorUserId="18924" LastEditDate="2016-03-02T21:57:28.610" LastActivityDate="2016-03-02T21:57:28.610" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28213" PostTypeId="2" ParentId="28210" CreationDate="2016-03-02T22:08:51.113" Score="6" Body="&lt;h3&gt;This is not the intended answer of OP&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I (don't) hope it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Anthony Burgess&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Only his name doesn't have letter &lt;strong&gt;&lt;em&gt;L&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&#xA; CHAR&lt;strong&gt;L&lt;/strong&gt;ES DARWIN&lt;br&gt;&#xA; Anthony Burgess&lt;br&gt;&#xA; EDWARD A&lt;strong&gt;L&lt;/strong&gt;BEE&lt;br&gt;&#xA; A&lt;strong&gt;L&lt;/strong&gt;ICE MUNRO&#xA; IGNATIUS of &lt;strong&gt;L&lt;/strong&gt;OYO&lt;strong&gt;L&lt;/strong&gt;A&lt;br&gt;&#xA; PHY&lt;strong&gt;LL&lt;/strong&gt;IS SCH&lt;strong&gt;L&lt;/strong&gt;AF&lt;strong&gt;L&lt;/strong&gt;Y&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-03-03T10:30:02.413" LastActivityDate="2016-03-03T10:30:02.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28214" PostTypeId="2" ParentId="28209" CreationDate="2016-03-02T22:09:51.880" Score="3" Body="&lt;p&gt;I think you are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a mouth &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Large and in charge &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rushing in to take charge, asked or unasked, is something people with large mouths do &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Opened or closed, I continue to be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A mouth can be opened or closed &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A mechanism of choice, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you inform others of your choice, you say it with your mouth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I can be beastly.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The thing we fear most about beasts are their mouths&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come in all shapes and sizes, Each with a common task.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; human mouths, beaks, snouts - there are many actual mouths in the animal kingdom. Their different tasks are important in evolutionary theory, e.g. the species of bird which differ by bill shape because one is for eating bugs and the other for nuts. Beyond that, there are mouths of jars etc. with different shapes and tasks &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cling to life by a thread, What am I, I ask?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The vocal &quot;cords&quot; are attached to the mouth, &quot;cord&quot; is synonymous of &quot;thread&quot; and people can die from a spasm in the pharynx area which shuts off air supply, possibly also in the vocal cords. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2811" LastEditorUserId="2811" LastEditDate="2016-03-02T22:15:52.960" LastActivityDate="2016-03-02T22:15:52.960" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28215" PostTypeId="2" ParentId="2907" CreationDate="2016-03-02T22:21:50.947" Score="5" Body="&lt;p&gt;I have another solution at 28 points - can't see an easy way to get lower, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZMUvx.png&quot; alt=&quot;zerris&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="17389" LastActivityDate="2016-03-02T22:21:50.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28216" PostTypeId="2" ParentId="28209" CreationDate="2016-03-02T22:24:43.100" Score="3" Body="&lt;p&gt;You could be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A mind or brain &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Open or closed I continue to be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People can be open or closed minded &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An instrument of choice &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We choose with our minds&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I can be beastly &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All the fiercest beasts are controlled by their minds (hence large and in charge)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come in all shapes and sizes with one common task &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; From elephants to fleas, they are all keeping their hosts alive &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cling to life by a thread &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The spinal cord is the brain's connection to the body&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastEditorUserId="18924" LastEditDate="2016-03-02T22:30:54.843" LastActivityDate="2016-03-02T22:30:54.843" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28218" PostTypeId="2" ParentId="28209" CreationDate="2016-03-02T23:20:49.723" Score="3" Body="&lt;p&gt;A new guess. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An eye - I have no idea why only my anatomical associations work for this riddle. I swear I tried others. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Large &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Large eyes are common&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and in charge&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; sight is the dominant sense of most people &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Opened or closed, I continue to be.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; eyes can be opened or closed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A mechanism of choice, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We choose between things by their looks &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I can be beastly.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The eye of the beast&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come in all shapes and sizes, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are different shapes in humans, like almond eyes, Asian eyes, round eyes, etc. And in animals, there are many sizes of eyes. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Each with a common task.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All eyes are for seeing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cling to life by a thread, What am I, I ask?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In cartoons, an eye popped out of the socket is held in place by the optical nerve&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2811" LastActivityDate="2016-03-02T23:20:49.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28219" PostTypeId="2" ParentId="28184" CreationDate="2016-03-03T00:11:51.560" Score="3" Body="&lt;p&gt;If you add a number $k$ to every element of a magic square, you get another magic square whose magic constant is the original constant plus $3k$. In particular, if you set $k$ to $-\frac 13$ of the original constant, you get a magic square whose constant is $0$. Thus every magic square induces one whose constant is $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for a square with constant $0$, if we name the following elements,&#xA;$$\begin{matrix}x &amp;amp; y &amp;amp; \cdot\\\cdot &amp;amp; z &amp;amp; \cdot\\\cdot &amp;amp; \cdot &amp;amp; \cdot\end{matrix}$$&#xA;we can start filling in the other elements from the requirement all columns, rows and diagonals sum to $0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{matrix}x &amp;amp; y &amp;amp; -x-y\\\cdot &amp;amp; z &amp;amp; \cdot\\\color{red}{w} &amp;amp; -y-z &amp;amp; -x-z\end{matrix}$$&#xA;With $w$, we run into an impasse: to make the diagonal $0$, we must have $w = x + y - z$, and to make the bottom row $0$, we must have $w = x + y + 2z$. These can both be realized only if $z = 0$. With that change, we can complete the pattern:&#xA;$$\begin{matrix}x &amp;amp; y &amp;amp; -x-y\\-2x -y &amp;amp; 0 &amp;amp; 2x+y\\x + y &amp;amp; -y &amp;amp; -x\end{matrix}$$&#xA;By the definition quoted by Ian MacDonald, any $x$ and $y$ would work. But I've always understood that the numbers in a magic square should also be distinct (&lt;a href=&quot;https://en.wikipedia.org/wiki/Magic_square&quot; rel=&quot;nofollow&quot;&gt;Wikipedia agrees&lt;/a&gt;). If we add that condition then we have a list of restrictions on the values of $x$ and $y$:&#xA;$$x \ne 0, y \ne 0, y \ne kx\text{ for }k \in \{1, -\frac 12, -1, -\frac 32, -2, -3\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we do not consider rotations/reflections as separate squares, then these restrictions allow only one magic square that consists of consecutive integers, the one that frodoskywalker has already produced: $$\begin{matrix}1 &amp;amp; 2 &amp;amp; -3\\-4 &amp;amp; 0 &amp;amp; 4\\3 &amp;amp; -2 &amp;amp; -1\end{matrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do consider rotations and reflections as separate squares, then there are 8 possible squares, depending on which of the 4 corners the $1$ is in, and whether the $2$ is clockwise or counter-clockwise from it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you allow non-consecutive integers, or non-integers, then there are infinitely many solutions, even if you don't allow multiples like frodoskywalker mentions. Just choose a non-zero $x$, then choose any value for $y$ other than the 7 that are restricted (and also relatively prime to $x$, if you want to avoid multiples of other squares).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum:&lt;/strong&gt; why there is only one magic square with consecutive integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any pattern with $x &amp;lt; 0$ is a rotation of $180^\circ$ from one with positive $x$, so we can assume $x &amp;gt; 0$. Further, if $|y| &amp;lt; x$, then $-x &amp;lt; y$, so $x &amp;lt; 2x + y = |-2x - y|$. So by flipping the square if needed, we can also assume $|y| &amp;gt; x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the sum of all the numbers in a magic square is the sum of the values of all three rows, and therefore is 3 times the magic number, which in this case is $0$. The only 9 consecutive integers that sum up to $0$ are those between $-4$ and $4$. So all the elements in our square need to be in this range. In particular $|2x + y| \le 4$. Consider the cases:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$x = 1$: $y \ne -2, -3$ by the restrictions, and $2x + y &amp;lt; 4$ gives $y \le 2$. So $y = 2$ and $y = -4$ are the only choices, and both give the same square.&lt;/li&gt;&#xA;&lt;li&gt;$x = 2$: $y \ne -1, -2, -3, -4$ by the restrictions, and $2x + y &amp;lt; 4$ gives $y &amp;lt; 0$, so there is no solution.&lt;/li&gt;&#xA;&lt;li&gt;$x = 3$: Since $|y| &amp;gt; x$, the only possibilities are $-4$ and $4$, but $2\cdot 1 + 4 &amp;gt; 4$, so it must be that $y = -4$, which gives the same pattern again.&lt;/li&gt;&#xA;&lt;li&gt;$x = 4$: $|y| &amp;gt; x$ cannot be satisfied.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Which exhausts all possibilities.&lt;/p&gt;&#xA;" OwnerUserId="17915" LastEditorUserId="17915" LastEditDate="2016-03-03T02:11:55.947" LastActivityDate="2016-03-03T02:11:55.947" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28220" PostTypeId="1" AcceptedAnswerId="28729" CreationDate="2016-03-03T02:35:17.210" Score="18" ViewCount="794" Body="&lt;p&gt;I recently discovered some old love letters, I estimate that some of them date back as much as 40 years. Here are some snippets from a few of them$\ldots$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We share so many &lt;em&gt;little&lt;/em&gt; memories that never fade, &lt;em&gt;persisting over long periods of time without changing&lt;/em&gt;. My love for you is how &lt;em&gt;they all combine&lt;/em&gt; into a beautiful whole that is so much more than any of them individually. I treasure these, holding on to them as if trying to &lt;em&gt;prevent anyone else from taking them away&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When I'm not with you I sometimes think about undressing you, &lt;em&gt;dismantling your clothing one thread at a time&lt;/em&gt; so that it lasts much longer before I have to return to my day. Your &lt;em&gt;natural beauty&lt;/em&gt; is so &lt;em&gt;bright and colourful&lt;/em&gt; I feel overwhelmed. It's as if you &lt;em&gt;shower&lt;/em&gt; me with &lt;em&gt;arrows&lt;/em&gt; and play my heart-&lt;em&gt;strings&lt;/em&gt; like a &lt;em&gt;violin&lt;/em&gt;, &lt;em&gt;tying&lt;/em&gt; me up in &lt;em&gt;knots&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It's like I have a &lt;em&gt;hunger&lt;/em&gt; for you, to know more about you, to fathom how you can possibly be so &lt;em&gt;incredible&lt;/em&gt;. The more I know you, the more I am amazed by the intricacies of your beauty and character. This &lt;em&gt;admiration&lt;/em&gt; only &lt;em&gt;whets&lt;/em&gt; my &lt;em&gt;desire&lt;/em&gt; further.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Who wrote these?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; Over the weekend, I went through some more papers from the same box. Most were boring but there also was another love letter. I've included a snippet from it below. Further, I noticed that some of the words were underlined in a different pen which had almost faded away, and I've added emphasis to those words in the snippets above and below.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Without you, I &lt;em&gt;cannot see&lt;/em&gt;. It's as if I've &lt;em&gt;never seen&lt;/em&gt;. I continue about my business, wishing you were with me, but really I'm just &lt;em&gt;marking time&lt;/em&gt;. &lt;em&gt;Creating&lt;/em&gt; new ways to &lt;em&gt;count the seconds&lt;/em&gt; until I will see you once again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I found another one$\ldots$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Before we met, my life felt like I was trudging up an infinite &lt;em&gt;mountain&lt;/em&gt;. Step after step; &lt;em&gt;hand-hold after foot-hold&lt;/em&gt;; toward a summit I wasn't sure was even there. But on the day I met you, I knew I had indeed reached the summit. I thought about it when I got home and realised how &lt;em&gt;unlikely&lt;/em&gt; our meeting was, how if any one of a million little things was different then it never would have happened, and that's how I knew.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18522" LastEditorUserId="8874" LastEditDate="2016-03-10T10:23:00.337" LastActivityDate="2016-03-10T10:23:00.337" Title="Old love letters - but who wrote them?" Tags="|wordplay|story|literature|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28221" PostTypeId="2" ParentId="28209" CreationDate="2016-03-03T03:04:57.173" Score="3" Body="&lt;p&gt;I think you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A word&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Opened or closed, I continue to be.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Both &quot;Opened&quot; and &quot;closed&quot; are words.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A mechanism of choice, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You say &quot;yes&quot; or &quot;no&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I can be beastly.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Words can be insulting, painful, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come in all shapes and sizes,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Words can be long/short and have different letter shapes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Each with a common task.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each word has a particular meaning&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cling to life by a thread, What am I, I ask?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure about this one...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Added hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can be made out of anything from A to Z, and sometimes have eyes with which I can't see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Words are composed of letters, and can include &quot;i&quot;s (which are not eyes)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19738" LastActivityDate="2016-03-03T03:04:57.173" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28222" PostTypeId="2" ParentId="28209" CreationDate="2016-03-03T03:05:02.147" Score="2" Body="&lt;p&gt;I think you are...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Money.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Opened or closed, I continue to be.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Accounts used to track money are opened and closed, but money is not contained or stored &lt;em&gt;in&lt;/em&gt; such accounts, they only track/measure it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A mechanism of choice, But I can be beastly.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We choose to use it as a mechanism for exchange. But it causes us so much grief, we're constantly trying to reform, change, regulate, etc. in order to mitigate the problems it causes us.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come in all shapes and sizes, Each with a common task.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cash, and other tokens representing money have many forms. Coins and notes of many shapes and sizes, plastic chips, shells, beads, stuffed toys, cheques, IOUs, etc. The common task is providing a trusted exchange mechanism.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cling to life by a thread, What am I, I ask?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is only the fragile trust of the people that keep any currency going. The abstract and virtual nature of money as a resource means that this thin thread can easily break, as we have seen in recent events, and as we continue to see more and more often.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be made out of anything from A to Z, and sometimes have eyes with which I can't see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Anything can be used as a money token, and over the years so many items have been. Some money has people's faces printed on it, the eyes that cannot see, for money is blind as to how it was earned - a drug dealer or a charity worker can take their notes and spend them without anyone knowing how they earned it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18522" LastActivityDate="2016-03-03T03:05:02.147" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28224" PostTypeId="1" AcceptedAnswerId="28228" CreationDate="2016-03-03T04:05:45.577" Score="11" ViewCount="1090" Body="&lt;p&gt;(I created this variant of a classic puzzle a long time ago, before the movie HPCOS was released.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Enough is enough. You've decided it's time to take a stand. Every year since that boy started attending Hogwarts, Dumbledore has given the House cup to Gryffindor by a &quot;last minute awarding&quot; of large numbers of points to him for ever more dubious reasons. The Gryffindors have even stopped trying, knowing that they'll get the cup no matter what happens. This year, Dumbledore has just announced an award of 400 points to the boy &quot;for parting his hair in a most interesting way&quot;, taking the cup away from your own house of Hufflepuff. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you've confronted Dumbledore, and not only your own house, but also all the Ravenclaws and Slytherins have joined you. Even the Gryffindors look embarassed. &quot;So, young Hufflepuff. I've been partial, have I? Well, then, let's put this nonsense to rest. We'll have a contest, between you and our fine Gryffindor champion. Mr. Finch! Bring the Urns of Peano from my office!&quot; Finch comes back carrying 3 large urns, and Dumbledore calls the two of you forward. You see that two of the urns are empty while the third is full of balls with numbers printed in a strange magical script. Indeed, you pull one out of the top and try to read the number, but it keeps going on around the ball repeatedly, yet somehow without ever overlapping itself. Finally you give up and put it back. Dumbledore explains &quot;In this urn is a ball for every natural number. Yes, infinitely many - a wonderful magic. As you can see, each of the urns has a dispenser track on the bottom where the lowest ten values in the urn are displayed. Go ahead, take the first ball.&quot; You pick it up and find the number 0. When you drop it back in the top of the urn, suddenly the balls on the track back up a space, while the 0 ball rolls through them to its original position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The contest&quot;, Dumbledore explains, &quot;is this: Each of your houses are assigned one of the empty urns. The two of you will act in turns. In each round, you, young Hufflepuff,&quot; (you suspect he doesn't even know your name) &quot;will take two balls from the track and place them in the Hufflepuff urn. Then our fine Gryffindor here will take a single ball from the Hufflepuff urn, and move it to the Gryffindor urn. This will continue until one of you can no longer take your turn.&quot; &quot;But, won't that take forever?&quot; asks your opponent. &quot;Indeed, it would&quot;, Dumbledore replies, &quot;But I shall cast 'Zeno's Accelerator' upon you. Fix a strategy in your mind, and when I cast the spell, your hands will follow it automatically. Each round will take but half the time of the round before, until an infinite number of rounds have been completed. In this way, the entirety of the urn may be emptied in only twice as much time as the first round takes.&quot; Your opponent's egghead friend can no longer contain herself. &quot;That's ridiculous! The energy it would take is infinite! The heat produced would kill us all!&quot;. &quot;Honestly, young lady,&quot;, Dumbledore replies, &quot;you should have learned by now that muggle physics takes a back seat to magic!&quot; The Gryffindor champion still appears confused. &quot;But, sir, if in every round, two balls are added to Hufflepuff, and I take one of them to put in Gryffindor, how will either house pull ahead? We'll just be tied in the end!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Oh, no, Gryffindor. You misapprehend! If even one ball remains in the Hufflepuff urn when the contest ends, Hufflepuff will be declared the winner. Let them accuse me of favoritism now!&quot; Your opponent looks very nervous, but suddenly appears to listen to something and calm down. A glance shows his egghead friend whispering something, apparently using a teleloquela charm. Obviously the fix is in. But fortunately, it appears that you've paid more attention to your arithmancy classes than Dumbledore did, for you spot a flaw in his plan. You smile as you fix your strategy in mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does Dumbledore expect Gryffindor to win? And what strategy can you use to defeat him?&lt;/p&gt;&#xA;" OwnerUserId="17915" LastEditorUserId="12408" LastEditDate="2016-03-04T20:20:25.733" LastActivityDate="2016-03-04T20:20:25.733" Title="For the honor of Hufflepuff" Tags="|mathematics|" AnswerCount="5" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="28225" PostTypeId="2" ParentId="28224" CreationDate="2016-03-03T04:17:41.997" Score="3" Body="&lt;p&gt;Gryffindor's strategy will be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; to take the lowest number in the Hufflepuff urn every time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and it will work because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; every marble will be guaranteed to be removed at most $n$ turns after you put it in, where $n$ is the number on the marble. If a marble is not put in, it will not be in at the end; if a marble is put in, then it will be removed and it will not be in. Therefore, no marbles will be in Hufflepuff at the end.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="11876" LastEditDate="2016-03-03T04:34:12.293" LastActivityDate="2016-03-03T04:34:12.293" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28227" PostTypeId="1" CreationDate="2016-03-03T05:18:41.323" Score="14" ViewCount="1418" Body="&lt;pre&gt;&lt;code&gt;10101010101001 01 010100010 010 10 01010 010101.. 1001010&#xA;10 0100 00 0 100 01001010100101010 0101 01010 01 010 100 0&#xA;1010010 010010 0 011  101  10 1 01 10 0 01 01 01 01 010 01 010&#xA;10010 01 01 01 01001000 10 01 0010110111111100 0 0 001 10 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is the exact source of what it's supposed to be, spaces and all. Apparently it's not binary (or at least, not ASCII text converted to binary).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that's all I know really. It's supposed to be simple and it's a &quot;code&quot; you have to &quot;crack&quot;, and the person who gave me this puzzle is really good at mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: &quot;...there is a phrase in mind and a way to decode it but the way to get to the phrase does extend outside the realm of just decoding the numbers themselves.&quot;&lt;/p&gt;&#xA;" OwnerUserId="19740" LastEditorUserId="5373" LastEditDate="2016-12-13T11:49:04.623" LastActivityDate="2018-07-07T20:39:07.823" Title="How would this binary-ish code be translated?" Tags="|mathematics|cipher|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28228" PostTypeId="2" ParentId="28224" CreationDate="2016-03-03T05:25:57.020" Score="5" Body="&lt;p&gt;Your strategy should be to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Start by picking balls 2 and 3 and leaving 0 and 1 on the track. Next take balls 4 and 5, etc. This way, when all infinite turns have occurred besides the last, you will pick up the 0 and 1 and put them in your urn. Then when Gryffindor takes the 0 or 1 the first urn will be empty and you can no longer take your turn. There will still be a ball left in Hufflepuff's urn.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit: My comments&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Even though this is the answer, it is not something that feels comfortable. The main reason is because it is hard for us to grasp what happens in a finite number of turns after an infinite number of turns. I think some of the confusion is due to two conflicting proofs: 1. In the end, all of the balls will be inside of Gryffindor's urn. Others have already stated this better, but you can come to this conclusion because there are an infinite number of balls in the initial urn, and every ball that is placed in the Hufflepuff urn is eventually removed and placed in Gryffindor urn because if the ball is placed in Hufflepuff's urn at turn n, it will be removed in turn 2n. The result which seems to contradict this statement is: 2 After Hufflepuff and Gryffindor have both taken their turns, there will be an equal number of balls in each of their urns. This is because every turn 2 balls are transferred into Hufflepuffs urn from the main urn, and 1 from Hufflepuff's into Gryffindor's. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first result seems to imply that the end result of the competition will be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Gryffindor: $\infty$ Hufflepuff: 0 Original Urn: 0 or $\infty$?. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whereas the second result seems to imply that the very end will be: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gryffindor: $\infty$/2 Hufflepuff: $\infty$/2 Original Urn: 0 or $\infty$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is definitely some confusion here about what happens with the initial urn. After all how can we deplete it?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;In the first case, we can ignore the initial urn, because we plainly see that Hufflepuff's urn is empty, therefore the competition is over. Whereas in the second case, we cannot be sure that anyone's urn is empty, and therefore the competition is still going on. Part of the logic block here is that we cannot invoke the second case due to lacking a proof that one of the three urns is empty. Therefore we cannot conclude that when Dumbledore magically makes an infinite number of turns occur, that there will be equal number of balls in Gryffindor and Hufflepuff's urns. Why is it different for the last turn to be Hufflepuff removing balls $\infty -1$ and $\infty$ vs 0 and 1? Well, if there are an infinite number of balls, how can there be a last ball? The trick lies in getting rid of infinity before the proof starts. So the reason this proof works, is because Dumbledore has swept away infinity and left us with something we can grasp. On the last turn the balls 0 and 1 will be in the original urn, and they will be the only balls in the urn. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18821" LastEditorUserId="7586" LastEditDate="2016-03-04T18:52:40.960" LastActivityDate="2016-03-04T18:52:40.960" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28230" PostTypeId="1" AcceptedAnswerId="28250" CreationDate="2016-03-03T06:09:41.083" Score="2" ViewCount="2996" Body="&lt;p&gt;This is based on a word game that as far as I know was invented by my friend's Dad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal is to make as long a word as you can, starting from any three-letter word, using the following rule: You must always change one of the letters in the word, and you must always add a new letter to the word, and the result must always be a word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for example, you could start from &lt;code&gt;cat&lt;/code&gt; to get&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;cat&#xA;fate             [change the c to an f, add an e]&#xA;grate            [change the f to an r, add a g]&#xA;graced           [change the t to a c, add a d]&#xA;cracked          [change the g to a c, add a k]&#xA;crackers         [change the d to an r, add an s]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For 8 letters. Note that you can change any letter, and the letter you add can be at the beginning, the end or somewhere else. However, you have the following restrictions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;you always have to add one letter and change one letter&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;you can't add a letter and then change it in the same move&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To be absolutely clear on that last rule: a move isn't allowed if it is possible to get from the longer word to the shorter one just by removing one letter. (Thanks to astralfenix for suggesting this clarification.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typical scrabble rules apply, e.g. no proper nouns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of disputes regarding words of length 8 or less we'll use &lt;a href=&quot;http://www.puzzlers.org/pub/wordlists/ospd.txt&quot; rel=&quot;nofollow&quot;&gt;this word list&lt;/a&gt;, which I think is the one used for scrabble. For words longer than 8 characters, any reasonable dictionary will do. Whoever gets the longest word first will win. I don't know the longest possible word.&lt;/p&gt;&#xA;" OwnerUserId="4017" LastEditorUserId="4017" LastEditDate="2016-03-03T13:32:20.650" LastActivityDate="2016-03-03T16:37:56.107" Title="Change a letter, add a letter" Tags="|english|open-ended|letters|word-game|" AnswerCount="7" CommentCount="7" ClosedDate="2021-05-10T17:51:46.010" ContentLicense="CC BY-SA 3.0" />
  <row Id="28232" PostTypeId="2" ParentId="28230" CreationDate="2016-03-03T07:35:19.687" Score="0" Body="&lt;pre&gt;&lt;code&gt;gat    &#xA;batt    &#xA;bates    &#xA;barter    &#xA;battery    &#xA;battered    &#xA;battersea&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="11522" LastActivityDate="2016-03-03T07:35:19.687" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28233" PostTypeId="2" ParentId="236" CreationDate="2016-03-03T07:51:24.673" Score="9" Body="&lt;h2&gt;Does the Guru's statement bring any new information?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The misleading thing here is that you might get tricked into the belief that the statement of the Guru just tells the people on the island that there is someone with blue eyes. But that is nothing new! The people already knew that by looking around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statement of the Guru says something deeper. It not only makes the people know that there is someone with blue eyes, it also makes them know that everybody else knows that there is someone with blue eyes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even deeper, it makes them know that everybody else knows that everybody else knows that everybody else knows (ad infinitum) that there is someone with blue eyes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Now that is a strong statement, because the people themselves only knew this up to a certain point!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;A small example&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;For instance suppose that we have 3 blue eyed people, &lt;code&gt;A&lt;/code&gt;, &lt;code&gt;B&lt;/code&gt; and &lt;code&gt;C&lt;/code&gt;, and no Guru. &lt;code&gt;A&lt;/code&gt; knows that there is someone with blue eyes. &lt;code&gt;A&lt;/code&gt; knows that &lt;code&gt;B&lt;/code&gt; knows that there is someone with blue eyes. But &lt;code&gt;A&lt;/code&gt; does &lt;em&gt;not&lt;/em&gt; know that &lt;code&gt;B&lt;/code&gt; knows that &lt;code&gt;C&lt;/code&gt; knows that there is someone with blue eyes, because &lt;code&gt;A&lt;/code&gt; doesn't know his own eye color. For this to know, &lt;code&gt;A&lt;/code&gt; needs the statement of the Guru.&lt;/p&gt;&#xA;" OwnerUserId="12679" LastActivityDate="2016-03-03T07:51:24.673" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="28234" PostTypeId="2" ParentId="28230" CreationDate="2016-03-03T08:19:05.940" Score="3" Body="&lt;pre&gt;&lt;code&gt;flu&#xA;plus      [change the f to a p, add an s]&#xA;blush     [change the p to a b, add an h]&#xA;slushy    [change the b to an s, add a y]&#xA;plushly   [change the s to a p, add an l]&#xA;slushily  [change the p to an s, add an i]&#xA;squashily [change the l to a q, add an a]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;reference to legitimate the word (adv):&#xA;&lt;a href=&quot;http://www.thefreedictionary.com/squashily&quot; rel=&quot;nofollow&quot;&gt;squashily&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18685" LastActivityDate="2016-03-03T08:19:05.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28235" PostTypeId="2" ParentId="28230" CreationDate="2016-03-03T09:08:51.467" Score="3" Body="&lt;p&gt;Here is a sequence that reaches 10 letters. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;lap         &#xA;lace        p-&amp;gt;c +e&#xA;laser       c-&amp;gt;s +r&#xA;basher      l-&amp;gt;b +h&#xA;bushier     a-&amp;gt;u +i&#xA;slushier    b-&amp;gt;l +s&#xA;slushiest   r-&amp;gt;s +t&#xA;squashiest  l-&amp;gt;q +a&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;All words are valid on &lt;a href=&quot;http://dictionary.reference.com/&quot; rel=&quot;nofollow&quot;&gt;Dictionary.com&lt;/a&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="12700" LastEditDate="2016-03-03T12:59:38.323" LastActivityDate="2016-03-03T12:59:38.323" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28236" PostTypeId="2" ParentId="28224" CreationDate="2016-03-03T09:54:11.910" Score="5" Body="&lt;p&gt;To follow on from the other answers, what Dumbledore has missed is that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; he never stated that the entire contest should take place under the Zeno-accelerated time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In fact,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you can just make your strategy to take all the balls from 2 upwards in ascending order. When the Zeno spell ends, there will still be 0 and 1 in the original urn, with all others in your opponent's urn. At this point, the concept of whose turn it was last is vague at best, but if you manage to argue that it's yours (after all, you don't have any balls left so you can't have just taken two) then you're free to take 0 and 1. Your opponent can only take one of these, leaving you with one at the end of the contest.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A countably infinite number of points to Hufflepuff!&lt;/p&gt;&#xA;" OwnerUserId="12511" LastActivityDate="2016-03-03T09:54:11.910" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28237" PostTypeId="1" AcceptedAnswerId="28270" CreationDate="2016-03-03T10:13:37.730" Score="6" ViewCount="221" Body="&lt;p&gt;Every letter and every question mark stands for a digit in &lt;strong&gt;base-9&lt;/strong&gt; representation. Different letters stand for different digits. Leading digits are always non-zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alphametic for the lovers of &lt;strong&gt;Mad&lt;/strong&gt; magazine:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;ALFRED $ \, \, \times \, \, $   E = NEUMAN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which digit does each letter represent? &#xA;(Please present the full analysis how these digits can be determined.)&lt;/p&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2016-03-03T20:26:33.890" Title="Alfred E. Neuman alphametic" Tags="|no-computers|alphametic|" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28238" PostTypeId="5" CreationDate="2016-03-03T10:16:06.337" Score="0" Body="&lt;p&gt;Alphametic puzzles involve mathematical equations where the digits have been replaced with letters or symbols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The puzzle is then to reverse this process, and to determine which letters represent which digits by using the information given via the puzzle to narrow down the options. The solution should ideally be unique.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example $A+B=AC$ has the unique solution $A=1, B=9$ and $C=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should use this tag for questions concerning verbal arithmetic creation, solutions, and strategies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Standard assumptions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Different letters represent different digits&lt;/li&gt;&#xA;&lt;li&gt;Leading digits are always non-zero&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Verbal_arithmetic&quot; rel=&quot;nofollow&quot;&gt;Verbal arithmetic&lt;/a&gt; at Wikipedia.&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="11497" LastEditDate="2016-08-19T15:21:26.800" LastActivityDate="2016-08-19T15:21:26.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28239" PostTypeId="4" CreationDate="2016-03-03T10:16:06.337" Score="0" Body="Alphametic puzzles involve equations where digits have been replaced with letters, where the goal is usually to reconstruct the original equation, matching letters to numbers." OwnerUserId="9404" LastEditorUserId="69582" LastEditDate="2021-02-06T18:30:56.750" LastActivityDate="2021-02-06T18:30:56.750" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28240" PostTypeId="1" AcceptedAnswerId="29734" CreationDate="2016-03-03T10:29:08.523" Score="9" ViewCount="380" Body="&lt;p&gt;The answer to this riddle is a single word.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Murray Abraham as Greek.&lt;br&gt;&#xA;  The sun seen from the dark side.&lt;br&gt;&#xA;  And someone who is always late.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The whole?&lt;br&gt;&#xA;  I just love Meryl Streep!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2016-03-27T16:50:01.910" Title="Loving Meryl Streep" Tags="|riddle|word|wordplay|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28241" PostTypeId="2" ParentId="28209" CreationDate="2016-03-03T11:15:50.123" Score="2" Body="&lt;p&gt;I would say you are a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Chain&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Large and in Charge&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chains secure objects, they are in charge! &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Opened or closed, I continue to be.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Whether open or closed, a chain is a chain. (-_-)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A mechanism of choice, But I can be beastly.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chains can be chosen for various purposes but the can be beastly. Example: large chains tied to anchors of boats.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come in all shapes and sizes, Each with a common task.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They come in all shapes and sizes but have a common task. To secure the object. Tie 'em up.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cling to life by a thread, What am I, I ask?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A chain is a kind of string / thread. If it breaks then technically the chain is dead. (X.X)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hints:&#xA;I can't relate it to first hint, but for the second: &#xA;I have a few close friends, just like a pea, and I jump through hoops for everyone around me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A chain has numerous links connected to it in succession &lt;strong&gt;~~~~~~&lt;/strong&gt; kind of like peas. The chains do go through hoops to seal the objects.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19191" LastActivityDate="2016-03-03T11:15:50.123" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28242" PostTypeId="2" ParentId="28209" CreationDate="2016-03-03T11:33:44.083" Score="8" Body="&lt;p&gt;You might be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a button&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Opened or closed, I continue to be.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you are buttoned up or not, you are still button.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A mechanism of choice, But I can be beastly.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Buttons are very often used, but some of them can be very tricky to unbutton.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come in all shapes and sizes, Each with a common task.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Buttons come in all shapes and sizes, but they are all used to connect two pieces of clothing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cling to life by a thread, What am I, I ask?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Buttons are sown to clothes by a thread.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I can be made out of anything from A to Z, and sometimes have eyes with which I can't see.&#xA; Buttons can be made from many materials and they can have the holes for thread to be sown.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I have a few close friends, just like a pea, and I jump through hoops for everyone around me.&#xA; Buttons usually come in rows, like on a shirt. When you button up the shirt, you pass each button through an opening/hoop.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19749" LastActivityDate="2016-03-03T11:33:44.083" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28243" PostTypeId="2" ParentId="28209" CreationDate="2016-03-03T11:37:26.137" Score="1" Body="&lt;p&gt;I think you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A switch!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Opened or closed, I continue to be.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A switch can be either opened or closed, but it is there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A mechanism of choice, But I can be beastly.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Switch is essentially a mechanism of choice. Yet sometimes it may not work, thus can be beastly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come in all shapes and sizes, Each with a common task.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;We can find switches in all shapes and sizes but the task is same.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cling to life by a thread, What am I, I ask?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Switch needs a wire (thread) to close/open the circuit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19750" LastActivityDate="2016-03-03T11:37:26.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28244" PostTypeId="2" ParentId="28209" CreationDate="2016-03-03T11:43:24.460" Score="1" Body="&lt;h3&gt;This is not the intended answer(my achievement)&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I would like to join in with the word&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Electricity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Opened or closed, I continue to be.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Depends on circuit&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A mechanism of choice, But I can be beastly.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It can kill you&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come in all shapes and sizes, Each with a common task.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Depends on the material in which it is flowing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cling to life by a thread, What am I, I ask?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We use wire for our household electricity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 1 : I can be made out of anything from A to Z, and sometimes have eyes with which I can't see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Electricity&lt;/strong&gt; has letter &lt;strong&gt;I&lt;/strong&gt; in it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2 : I have a few close friends, just like a pea, and I jump through hoops for everyone around me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Electrons jump from one atom to another while conducting electricity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Large and in charge&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Voltages can be large. And flow of electricity surely is caused by flow of charge.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-03-03T16:43:24.717" LastActivityDate="2016-03-03T16:43:24.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28245" PostTypeId="1" AcceptedAnswerId="28246" CreationDate="2016-03-03T12:36:24.580" Score="6" ViewCount="1066" Body="&lt;p&gt;You are a top secret agent trying to get into a enemy hide out. When you arrive at the enemy base, you see two guards at the door. You could take them out but that would make a ruckus and the enemies will know you are here, thus you wait in a bush. You see a man walk up towards the guards. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;1!&quot; yells the guard in a thundering voice, so loud that you can almost hear the exclamation mark. The man replies with &quot;1&quot; and is allowed him in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the man walks in and the door closes behind him, another person walks up to the guards. &quot;2!&quot; yells the guard, again with a resounding exclamation mark. The person then replied with &quot;2&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;These aren't the smartest bad guys I've ever went against&quot; you say to yourself as you start walking towards the guards. The guard yells &quot;3!&quot;, and you reply with confidence &quot;3&quot;. The guards look at each other and then shoot you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you lie there bleeding, you see another person walk up to the guards. The guard yells out &quot;0!&quot; and the person replies with &quot;1&quot;. The answer hits you as you slowly lose consciousness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the answer to their password system?&lt;/p&gt;&#xA;" OwnerUserId="4645" LastEditorUserId="8874" LastEditDate="2016-03-03T16:06:26.483" LastActivityDate="2016-03-03T17:30:27.247" Title="The Secret Agent Who Got Shot" Tags="|mathematics|word-problem|password|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28246" PostTypeId="2" ParentId="28245" CreationDate="2016-03-03T12:47:01.477" Score="17" Body="&lt;p&gt;This one is very literal.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1! = 1, 2! = 2, 3! = 6.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I would love to know how the guards vocalized the exclamation point.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12092" LastActivityDate="2016-03-03T12:47:01.477" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28247" PostTypeId="2" ParentId="28245" CreationDate="2016-03-03T12:48:36.167" Score="2" Body="&lt;h3&gt;This is not the intended answer(my achievement)&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;It can be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The rule is &lt;strong&gt;$3 - $(number of vowels in the english alphabet of the number)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One has two vowels (&lt;strong&gt;O&lt;/strong&gt; and &lt;strong&gt;E&lt;/strong&gt;). So 3-2=1. Two has one vowel (&lt;strong&gt;O&lt;/strong&gt;). So 3-1=2. Now three has two vowels (two &lt;strong&gt;E&lt;/strong&gt;'s). So 3-2=1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-03-03T16:29:02.640" LastActivityDate="2016-03-03T16:29:02.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28248" PostTypeId="2" ParentId="28230" CreationDate="2016-03-03T13:04:53.083" Score="1" Body="&lt;p&gt;Another sequence. You could go farther with hyphenated words. I believe the jury is out on whether re-spraining is hyphenated or not.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;man          +b, m-&amp;gt;r  &#xA;bran         +r, m-&amp;gt;b  &#xA;bring        +g, a-&amp;gt;i  &#xA;airing       +a, b-&amp;gt;i  &#xA;railing      +r, r-&amp;gt;l  &#xA;draining     +d, l-&amp;gt;n  &#xA;detaining    +e, r-&amp;gt;t  &#xA;retraining   +r, d-&amp;gt;r  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="18924" LastEditorUserId="4017" LastEditDate="2016-03-03T13:43:19.573" LastActivityDate="2016-03-03T13:43:19.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28249" PostTypeId="2" ParentId="28209" CreationDate="2016-03-03T13:07:22.257" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A bullet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Opened or closed, I continue to be.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fired or unfired.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A mechanism of choice, But I can be beastly.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Violent but likely the most convenient for its cause.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come in all shapes and sizes, Each with a common task.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Different calibers and shapes but all made to harm or kill.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cling to life by a thread, What am I, I ask?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The rifling that serves to extend the flight?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be made out of anything from A to Z, and sometimes have eyes with which I can't see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Buckshot?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have a few close friends, just like a pea, and I jump through hoops for everyone around me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Like peas bullets are next to eachother in a magazine. Jumping through hoops means getting fired maybe?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user19751" LastActivityDate="2016-03-03T13:07:22.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28250" PostTypeId="2" ParentId="28230" CreationDate="2016-03-03T13:09:32.717" Score="9" Body="&lt;p&gt;A sequence reaching 15 letters:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;fun&#xA;lung&#xA;suing&#xA;cluing&#xA;cloying&#xA;cooeying&#xA;conveying&#xA;conventing&#xA;conventions&#xA;conventional&#xA;conventionary&#xA;conventionally&#xA;conventionality&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="19698" LastEditorUserId="19698" LastEditDate="2016-03-03T13:28:17.247" LastActivityDate="2016-03-03T13:28:17.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28253" PostTypeId="2" ParentId="28240" CreationDate="2016-03-03T14:35:41.933" Score="2" Body="&lt;p&gt;I think the word is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oscar&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This part is incomplete:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Murray Abraham as Greek - refers to F Murray Abraham's role in Mighty Aphrodite&lt;br&gt;&#xA; The sun seen from the dark side - Dark Side of the Sun, Brad Pitt's first leading role.&lt;br&gt;&#xA; And someone who is always late - could refer to how Oscar bait movies are usually released very late in the year, so as to be fresh in Oscar voters' minds.&lt;br&gt;&#xA; I just love Meryl Steep! - Holds records for the most nominations (including 3 wins)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16874" LastEditorUserId="16874" LastEditDate="2016-03-03T14:51:44.213" LastActivityDate="2016-03-03T14:51:44.213" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28254" PostTypeId="1" CreationDate="2016-03-03T14:35:52.393" Score="22" ViewCount="2881" Body="&lt;p&gt;Alice and Bob play the following game with two (identical) standard decks of $52$ cards.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First Alice secretly arranges one deck of $52$ cards in a long row on the table. All cards are face-down, and Bob has no knowledge of the positions of any of the cards.&lt;/li&gt;&#xA;&lt;li&gt;Then the game goes through several rounds.&#xA;In every round, Bob first pays $1$ Euro to Alice.&#xA;Then Bob arranges the second deck of cards in a long row on the table, parallel to Alice's row and with all the cards face-up.&#xA;Then Alice tells Bob all the face-up cards in his row that agree with the corresponding face-down cards in her row.&#xA;Then the next round starts.&lt;/li&gt;&#xA;&lt;li&gt;The game ends, as soon as Bob knows the positions of all the face-down cards in Alice's row. &#xA;Bob then receives $x$ Euros from Alice.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What is the smallest value of $x$, for which Bob can still avoid (with absolute certainty) to lose any money?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-03T14:47:10.923" LastActivityDate="2019-04-05T07:41:07.920" Title="A game with 52 cards" Tags="|mathematics|strategy|combinatorics|game|" AnswerCount="6" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="28255" PostTypeId="2" ParentId="28254" CreationDate="2016-03-03T15:02:58.360" Score="19" Body="&lt;p&gt;An upper bound&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 51 Euro.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bob picks a random permutation of the deck and then rotates the subset of his deck that does not match Alice's. The worst case is when the whole deck is a rotation of his initial permutation, in which case he has to rotate 51 times.&lt;/p&gt;&#xA;" OwnerUserId="12901" LastActivityDate="2016-03-03T15:02:58.360" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28256" PostTypeId="2" ParentId="28254" CreationDate="2016-03-03T15:03:27.827" Score="6" Body="&lt;p&gt;I think the worst case scenario is 52 tries. That scenario is where each guess (until the last) is completely wrong. Your first guess is shifted by 1, and subsequent guesses are just shifting the cards, but in the wrong direction. On the 52nd guess, all the cards will line up. Since you can disregard the last guess (you know the guess), Bob has to pay 51 Euro.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's an unlikely scenario, since you're likely to make some right guesses during that time; you just skip over those cards when shifting. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2016-03-03T15:03:27.827" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28257" PostTypeId="2" ParentId="28230" CreationDate="2016-03-03T16:14:12.163" Score="1" Body="&lt;p&gt;same 8 words attempts:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;act             &#xA;race             +r, t -&amp;gt; e  &#xA;raven            +n, c -&amp;gt; v  &#xA;raving           +g, e -&amp;gt; i  &#xA;ranging          +g, v -&amp;gt; n  &#xA;reneging         +e, a -&amp;gt; e  &#xA;rendering        +d, g -&amp;gt; r  &#xA;meandering       +a, r -&amp;gt; m  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2016-03-03T16:14:12.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28258" PostTypeId="2" ParentId="28230" CreationDate="2016-03-03T16:37:56.107" Score="5" Body="&lt;p&gt;There are a few 15-letter words which can be reached. This sequence mainly avoids obscure words:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;UTE&#xA;BATE&#xA;CRATE&#xA;BERATE&#xA;ACERATE&#xA;ACCURATE&#xA;ACICULATE&#xA;PANICULATE&#xA;PARTICULATE&#xA;INARTICULATE&#xA;DISARTICULATE&#xA;DISARTICULATOR&#xA;DISARTICULATION&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I had a hunch that -TION would be a good way to get long words, and that gives this sequence:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;BON&#xA;CION&#xA;ANION&#xA;ACTION&#xA;DICTION&#xA;EVICTION&#xA;EMICATION&#xA;DEDICATION&#xA;DEIFICATION&#xA;AERIFICATION&#xA;METRIFICATION&#xA;GENTRIFICATION&#xA;DENITRIFICATION&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Cion&lt;/em&gt; is an obscure spelling of &lt;em&gt;scion&lt;/em&gt;).&lt;/p&gt;&#xA;" OwnerUserId="228" LastActivityDate="2016-03-03T16:37:56.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28259" PostTypeId="2" ParentId="28020" CreationDate="2016-03-03T17:11:07.983" Score="4" Body="&lt;p&gt;In hindsight I should have lead off with the acrostic designation. I completely misjudged the difficulty of this riddle. Let me show you how I thought the reasoning was going to go. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now with many &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hmm. It goes with &quot;many&quot; lots of synonyms there. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now with having lovers &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cheating? Infidelities? Liaisons? Affairs? Unfaithful?&lt;br&gt;&#xA; A forest of possibilities again. But not as big. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now with first swing &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Here is where I hoped the reader could get his feet on the ground.&lt;br&gt;&#xA; People seemed to get that this was sports-related. The sport where the first swing is a big deal is tennis. Or any racquet sport really. What I was going for was &quot;serve&quot; or &quot;service&quot;. I omitted the article to tip it towards &quot;service&quot;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now with some covers &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some? What type of covers?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At this point I figured someone would put it together. Forgive me. Some of the answers here seemed superhuman to this newcomer. But I stuck in an acrostic because it appealed to my warped sense of humor. To solve the acrostic &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Take the first letter of each line &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But... But... That doesn't...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; N N N N = 4N = foreign&lt;br&gt;&#xA; Hence&lt;br&gt;&#xA; Foreign Legion&lt;br&gt;&#xA; Foreign Affairs&lt;br&gt;&#xA; Foreign Service&lt;br&gt;&#xA; Foreign Films&lt;br&gt;&#xA;&lt;br&gt;&#xA; My comments often contain hints. Aced and let for tennis. Affair... Well, my heart was in the right place. Happy puzzling. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastEditorUserId="8874" LastEditDate="2016-03-04T10:26:59.513" LastActivityDate="2016-03-04T10:26:59.513" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28260" PostTypeId="2" ParentId="28245" CreationDate="2016-03-03T17:30:27.247" Score="11" Body="&lt;p&gt;You've already accepted, but this should work as well:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As the system could have been&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the digits required to represent the number in binary (1 -&gt; 1, requiring one digit, 2 -&gt; 10, requiring 2 digits, 0 -&gt; 0, requiring one digit). Maybe they were really concerned about memory management.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19703" LastActivityDate="2016-03-03T17:30:27.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28261" PostTypeId="2" ParentId="28254" CreationDate="2016-03-03T17:55:08.543" Score="3" Body="&lt;p&gt;The optimal worst case it requires&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 51 tries&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To show that, Try the game with 2 cards, let's call them A and B. In the worst case;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A B (all face down)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B A&lt;/p&gt;&#xA;&#xA;&lt;p&gt;would be the first try. So Alice would not tell anything since no match up is available. In the game rule it is stated that;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The &lt;strong&gt;game ends, as soon as Bob knows the positions&lt;/strong&gt; of all the face-down cards in Alice's row. Bob then receives xx Euros from Alice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So before the next game begins, Bob knows the positions and he gets the money, which should be 1€ at least! So with two cards, 1 try is enough to find at worst optimal case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 3 cards;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A B C &lt;/p&gt;&#xA;&#xA;&lt;p&gt;B C A&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then no response from Alice; Next try will be (unluckily)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A B C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C A B&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then Bob knows all face down cards automatically, so 2 tries would be enough. Then 2€ for 3 cards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 4 cards; it makes 3 tries and 3€ not to lose any money at least!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For 52 cards it has to require 51 tries and 51 Euros at the worst optimal case! At worst case, Bob will strangely never find any card until the last try, which is kinda very low chance!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18514" LastEditorUserId="18514" LastEditDate="2016-03-03T18:15:13.513" LastActivityDate="2016-03-03T18:15:13.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28262" PostTypeId="1" AcceptedAnswerId="28266" CreationDate="2016-03-03T18:14:33.490" Score="2" ViewCount="346" Body="&lt;p&gt;At the same time I can be low or high,&lt;br&gt;&#xA;I can be found where the ground meets the sky. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I may have a bald spot on top of my head,&lt;br&gt;&#xA;but pick a fight with me, and you may be dead. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Slow moving, slow growing, I eventually change,&lt;br&gt;&#xA;but I'll always have friends out on the range. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 2 of the series.&lt;/p&gt;&#xA;" OwnerUserId="19731" LastEditorUserId="19731" LastEditDate="2016-03-05T06:44:15.300" LastActivityDate="2016-03-05T06:44:15.300" Title="Small and resilient" Tags="|riddle|word|rhyme|poetry|literature|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28263" PostTypeId="1" CreationDate="2016-03-03T18:25:09.657" Score="-1" ViewCount="191" Body="&lt;p&gt;This is the Follow up Question to &lt;a href=&quot;https://puzzling.stackexchange.com/questions/28254/a-game-with-52-cards&quot;&gt;A game with 52 cards&lt;/a&gt;. So it requires to understand previous question! I believe it is a bit harder to find!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what is the chance to know the positions of all the face-down cards in Alice's row in the worst optimal case?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please indicate your answer as fraction.&lt;/p&gt;&#xA;" OwnerUserId="18514" LastActivityDate="2016-03-03T18:53:36.597" Title="A chance to know 52 cards" Tags="|mathematics|combinatorics|probability|" AnswerCount="2" CommentCount="2" ClosedDate="2016-03-04T11:05:15.293" ContentLicense="CC BY-SA 3.0" />
  <row Id="28264" PostTypeId="2" ParentId="28263" CreationDate="2016-03-03T18:26:26.617" Score="2" Body="&lt;p&gt;1 / 52! If you try all possible guesses. &lt;/p&gt;&#xA;" OwnerUserId="19731" LastActivityDate="2016-03-03T18:26:26.617" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="28265" PostTypeId="2" ParentId="28262" CreationDate="2016-03-03T18:29:19.987" Score="2" Body="&lt;p&gt;You could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the sun.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At the same time I can be low or high, I can be found where the ground meets the sky.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; During the day, the sun will be at different places.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I may have a bald spot on top of my head, But pick a fight with me and you may be dead.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Do fly to close to the sun. I have trouble getting the bald spot to fit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Slow moving, slow growing, I eventually change, But I'll always have friends out on the range.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sun will eventually turn into another stellar body when the hydrogen runs out (is transformed). There are several sun-like bodies in the universe, for instance Alpha Centauri.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12901" LastActivityDate="2016-03-03T18:29:19.987" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28266" PostTypeId="2" ParentId="28262" CreationDate="2016-03-03T18:29:29.273" Score="3" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A mountain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At the same time I can be low or high, I can be found where the ground meets the sky.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mountains can be seen on the horizon; some of them are low (usually older mountains, which have eroded), some high (usually younger ones)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I may have a bald spot on top of my head, But pick a fight with me and you may be dead.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Snow-capped peaks are something you see on mountains high enough - these resemble a &quot;bald spot&quot; on the top of someone's head. But people who climb mountains—seen as a challenge or fight—could &quot;lose the fight&quot; often die.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Slow moving, slow growing, I eventually change, But I'll always have friends out on the range.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mountains change very slowly, from parts of the earth moving as well as erosion; mountains are found in &quot;ranges&quot;, that is, a bunch of them all together.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2016-03-03T18:33:26.697" LastActivityDate="2016-03-03T18:33:26.697" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28267" PostTypeId="2" ParentId="28263" CreationDate="2016-03-03T18:53:36.597" Score="3" Body="&lt;p&gt;The chance of getting all of the cards right on your first guess is 1/52!, but if you get all but two, then you actually know the position of the last two (it's the opposite of the way you had them!). Three doesn't work, because you only know one of the ways that the three aren't arranged, but there are more than one other ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is the chance that you know all but two? If it was only the first two, then it would be a chance of 1/52! as well, because there is only one permutation that has them switched. Because it can be any two, and there are 52*51 possible positions for the two to be placed in the list, it is 51*52/52! = 1/49!.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we need to remember the chance that we got them all right to begin with, so the final answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (52*51 + 1)/52! = 2653/52!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19491" LastActivityDate="2016-03-03T18:53:36.597" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28268" PostTypeId="1" AcceptedAnswerId="28269" CreationDate="2016-03-03T19:23:49.457" Score="6" ViewCount="1709" Body="&lt;p&gt;$7*7~$ is $~25$&lt;br&gt;&#xA;$6*6~$ is $~18$&lt;br&gt;&#xA;$9*9~$ is $~41$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When is this true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Think inside the box.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19179" LastEditorUserId="8874" LastEditDate="2016-03-03T19:43:46.033" LastActivityDate="2016-03-08T06:25:25.203" Title="Multiplication Homework" Tags="|calculation-puzzle|lateral-thinking|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28269" PostTypeId="2" ParentId="28268" CreationDate="2016-03-03T19:35:19.953" Score="17" Body="&lt;p&gt;These are true, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if you define $~x*y~$ as the number of black squares on a chessboard with $x$ rows and $y$ columns, where the lower left corner square is black.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Equivalently,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; define $~x*y=\lceil xy/2 \rceil$, that is, as the product of $x$ and $y$, divided by $2$, and then rounded up to the next integer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With this,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $7*7 ~=~ \lceil 49/2\rceil ~=~  \lceil 24.5 \rceil ~=~ 25$&lt;br&gt;&#xA; $6*6 ~=~ \lceil 36/2\rceil ~=~  \lceil 18~~~\rceil ~=~ 18$&lt;br&gt;&#xA; $9*9 ~=~ \lceil 81/2\rceil ~=~  \lceil 40.5 \rceil ~=~ 41$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-04T10:10:48.763" LastActivityDate="2016-03-04T10:10:48.763" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28270" PostTypeId="2" ParentId="28237" CreationDate="2016-03-03T19:57:47.463" Score="10" Body="&lt;p&gt;Written vertically:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ALFRED&#xA;x    E&#xA;------&#xA;NEUMAN&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Trivially, we can see that $E \gt 1$ and $A \in \{1,2,3,4\}$ since other values would either create a leading zero or too many digits.  Also, $N$ is greater than both $A$ and $E$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A=4$, then $E=2$ and $N=8$.  But no value for $D$ can result in $D \times E$ yielding 8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A=3$, then $E=2$.  But $D \times E$ can have at most a carry over of 1, and $E\times E =E^2=4 \implies A\in\{4,5\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, $A\in\{2,3\}$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Assume $A=2$&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Then $E\in\{3,4\}$. If $E=4$, then $E^2=17$, which means $D\times E$ must carry over 4, which is impossible.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $E=3$, and $E^2=10$ which means $D \times E$ must carry over 2, which means $D=\in\{7,8\}$ and $N\in\{3,6\}$. Only the second pair of these doesn't result in a conflict, so $D=8$ and $N=6$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that $L\times E$ must result in no carry over.  This is only possible if $L\in\{0,1\}$ of the remaining valid values.  If $L=0$ then we need the carry over of $F\times E$ to be $E$ which is impossible.  Thus, $L=1$ and the carry over of $F\times E$ is 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, $F=7$. Since $F\times E=7\times 3=23$, and the only remaining values for $U$ are $\{0,4,5\}$, we know that the carry over must be 1 and $R=5$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then we have $217538 \times 3=645726 \implies U=7$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Thus $A=1$&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The ones column is $D \times E$ which yields a carry over less than both $D$ and $E$. Lets look at the second column where adding this carry over to $E\times E$ yields $A$ &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;if $E=2$, then $E^2=4$, requiring carry over of 6.&lt;/li&gt;&#xA;&lt;li&gt;if $E=3$, then $E^2=10$, requiring carry over of 1. Thus, $D\in\{4,5\}$ and $N\in\{3,6\}$.  Since $3$ is already taken, we know $D=5$ and $N=6$. But this requires a carry over of 3 in the $L\times E$ spot.&lt;/li&gt;&#xA;&lt;li&gt;if $E=4$, then $E^2=17$, requiring carry over of 3. &lt;/li&gt;&#xA;&lt;li&gt;if $E=5$, then $E^2=27$, requiring carry over of 3. &lt;/li&gt;&#xA;&lt;li&gt;if $E=6$, then $E^2=40$, requiring carry over of 1. Thus, $D=2 \implies N=3$.&lt;/li&gt;&#xA;&lt;li&gt;if $E=7$, then $E^2=54$, requiring carry over of 6. Thus $D=8 \implies N=2$.&lt;/li&gt;&#xA;&lt;li&gt;if $E=8$, then $E^2=71$, requiring carry over of 0. Thus $D=0 \implies N=0$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, the only ones that don't lead to immediate contradiction are $E\in\{4,5\}$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Assume $E=4$&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;$D=7 \implies N=1$, so $D=8$ and $N=5$ is the only way to get our carry over of 3.  We then need a carry over of 1 in the $L\times E$ spot.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we need to have $L\in\{2,3\}$. If $L=2$, then $L\times E=8$, so we need a carry over of 5 to get $E$.  This is impossible, so $L=3$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, $L\times E=13$ so in order to yield $E$ again, we need another carry over of 1 from the $F \times E$ spot.  Thus, $F=2$ and we need a carry over from the $R \times E$ spot. If this carry over is greater than 1, then $U$ will conflict, so it needs to be 1 and $U=0$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now we've run out of numbers that will result in a carry over of 1!&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Thus, $E=5$&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;We need a carry over of 3, so $D\in\{6,7\}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $D=6$, then $D\times E=6 \times 5=33$.  This means $N=3$ which is too small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $D=7$ and $D\times E=7\times 5= 38 \implies N=8$.  This means we also need a carry over of 3 from $L\times E$, so $L=6$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L\times E=6\times 5=33$ and we need this to yield $E=5$, so we need a carry over of 2 from the $F \times E$ spot.  This means we need $F\in\{3,4\} \implies F\times E\in\{16,22\}$.  If $F=3$, then we need $R\times E$ to yield a carry over of 4.  But 8 and 7 are already taken, and $R=6$ is insufficient.  Thus, $F=4$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remaining values are $\{0,2,3\}$. If $R=2$, then $M=4$.  If $R=3$, then $M=0$ and $U=4$.  Thus, $R=0$, then $M=3$ and $U=2$.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;164057&#xA;x    5&#xA;------&#xA;852318&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="1592" LastEditDate="2016-03-03T20:26:33.890" LastActivityDate="2016-03-03T20:26:33.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28271" PostTypeId="2" ParentId="28268" CreationDate="2016-03-03T20:05:52.507" Score="20" Body="&lt;p&gt;It is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The maximum number of knights on a board of that size (i.e. width is the first number and height is the second number) where none of them can attack another. (Assuming it's free-for-all chess for some reason)&lt;br&gt;&#xA; And it's inside the &quot;box&quot; because the box is a chess board.&#xA; &lt;br/&gt;&lt;br&gt;&#xA; It's also just how many black squares would be on a chess/checkers board, as @Sleafer pointed out. It turns out I over complicated it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You would continue the sequence&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$1*1=1\\2*2=4\text{ (or }2\text{ if counting squares)}\\3*3=5\\4*4=8\\5*5=13\\6*6=18\\7*7=25\\8*8=32\\9*9=41\\10*10=50$$&#xA; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18289" LastEditorUserId="18289" LastEditDate="2016-03-04T19:05:55.913" LastActivityDate="2016-03-04T19:05:55.913" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="28272" PostTypeId="1" CreationDate="2016-03-03T20:28:41.790" Score="-1" ViewCount="1666" Body="&lt;p&gt;I have been trying to get to solution for some time now, and it starts irritating me. Anyone?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/kNS3R.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kNS3R.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19761" LastEditorUserId="11876" LastEditDate="2016-03-04T01:23:46.660" LastActivityDate="2016-03-04T01:23:46.660" Title="Logic puzzle 4x4 grid" Tags="|mathematics|pattern|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28273" PostTypeId="2" ParentId="28272" CreationDate="2016-03-03T20:32:54.657" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;C&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the columns all add up to 14.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19128" LastEditorUserId="19128" LastEditDate="2016-03-03T20:33:51.267" LastActivityDate="2016-03-03T20:33:51.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28276" PostTypeId="1" CreationDate="2016-03-03T20:44:07.160" Score="5" ViewCount="1025" Body="&lt;p&gt;What is the number in the last star? Use logic to fin the missing number&#xA;&lt;a href=&quot;https://i.stack.imgur.com/CfhDj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CfhDj.png&quot; alt=&quot;picture&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19762" LastEditorUserId="11876" LastEditDate="2016-03-04T01:23:31.503" LastActivityDate="2016-03-04T01:23:31.503" Title="Tricky star question" Tags="|pattern|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28277" PostTypeId="2" ParentId="28276" CreationDate="2016-03-03T20:52:26.747" Score="16" Body="&lt;p&gt;It's:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You add the top 3 numbers and subtract the bottom 2 numbers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastActivityDate="2016-03-03T20:52:26.747" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28279" PostTypeId="2" ParentId="28268" CreationDate="2016-03-03T21:19:14.597" Score="21" Body="&lt;p&gt;Because $*$ means:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$\left\lceil\frac{n\times m}{2}\right\rceil$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="9718" LastEditDate="2016-03-08T06:25:25.203" LastActivityDate="2016-03-08T06:25:25.203" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28280" PostTypeId="2" ParentId="8594" CreationDate="2016-03-03T22:10:39.093" Score="0" Body="&lt;p&gt;Another solution is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;PRIME = 81957&lt;/strong&gt;, NUMBER = 245871  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4664" LastActivityDate="2016-03-03T22:10:39.093" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28281" PostTypeId="1" AcceptedAnswerId="28291" CreationDate="2016-03-03T22:45:25.237" Score="34" ViewCount="3035" Body="&lt;p&gt;A group of ten bandits stand in a flat desert, with no pair the same distance apart. Tensions grow, and at the crack of dawn each bandit fires a single bullet at the bandit closest to him. All have perfect aim and all those that are hit are killed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Must one of them live to tell the tale? How many of them could possibly survive?&lt;/p&gt;&#xA;" OwnerUserId="18508" LastEditorUserId="18508" LastEditDate="2016-03-04T17:35:15.830" LastActivityDate="2016-03-05T00:08:51.003" Title="Gunfire at dawn" Tags="|mathematics|geometry|" AnswerCount="6" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="28282" PostTypeId="2" ParentId="28281" CreationDate="2016-03-03T22:53:53.653" Score="-2" Body="&lt;p&gt;&quot;Nearest&quot; involves two points, so every pair of bandits would shoot each other. None would survive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Since I was marked down, I'll explain further. The important part is &quot;no pair the same distance apart&quot;, which means there would be no overlap on shooting. Folks could only survive if that condition wasn't present, and one or more bandits got two bullets. &lt;/p&gt;&#xA;" OwnerUserId="12771" LastEditorUserId="12771" LastEditDate="2016-03-03T23:48:19.650" LastActivityDate="2016-03-03T23:48:19.650" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28283" PostTypeId="2" ParentId="28281" CreationDate="2016-03-03T22:56:09.280" Score="5" Body="&lt;p&gt;I believe the max that can survive is 7. Arrange the bandits in approximately two half hexagons. For one hexagon, it's 5 bandits, one at each point and one in the middle. More than a side length away is the other hexagon, parallel, with one exterior point empty. Between the two is the last bandit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/H3XGP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/H3XGP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To satisfy the distance clause, just have the outer bandits be closer to the center than each other, and with very small differences the overall situation stands. &lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2016-03-03T23:57:30.130" LastActivityDate="2016-03-03T23:57:30.130" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28286" PostTypeId="2" ParentId="28281" CreationDate="2016-03-04T00:26:56.367" Score="2" Body="&lt;p&gt;6 survivors is the most. 2 sets of 5. 4 corners and 1 in the middle. All different distances apart from middle, but the 4 outsiders shoot the middle guy and the middle guy shoots the closest corner. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A pentagon and triangle work the same way and if you go with a hexagon you run into issues with the outsiders shooting each other. &lt;/p&gt;&#xA;" OwnerUserId="19731" LastActivityDate="2016-03-04T00:26:56.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28287" PostTypeId="2" ParentId="28281" CreationDate="2016-03-04T00:28:44.687" Score="2" Body="&lt;p&gt;I can guarantee at least three will survive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take one gunman, called Gunman 1. He is separated from Gunman 2 by a distance $X_1$. Gunman 3 is on the other side of Gunman 2, separated by a distance $X_2$, where $X_2&amp;lt;C_1$. Continue doing this until you reach Gunman 7. Each $X_n&amp;lt;X_{n-1}$. Therefore, Gunman $M$ will shoot Gunman $M+1$, with the exception of Gunman 7, who will shoot Gunman $6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take the remaining three gunmen and place them 90 degrees apart around Gunman 1, forming a sort of cross. Gunman 8 is a distance $A$ from Gunman 1, Gunman 9 is a distance $B$ from Gunman 1, and Gunman 10 is a distance $C$ from Gunman 1. Here, $C\neq A\neq B\neq C$, all of which are greater than $X_1$. They are separated so that the distance between any of these three gunmen is greater than the distance between each gunman and Gunman 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gunman 8, Gunman 9, and Gunman 10 will all shoot Gunman 1 and survive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AcVAf.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that you could fit more than these three guys around Gunman 1 - and I don't know just how many - but I don't know for sure.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/a/28291/5300&quot;&gt;Zerris has found a better related configuration.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5300" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2016-03-04T01:22:23.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28288" PostTypeId="1" CreationDate="2016-03-04T00:31:39.610" Score="1" ViewCount="140" Body="&lt;p&gt;Jack is a law-abiding US citizen who was filing his first taxes at age 21. He put his information into a free online e-filer and got the shock of his life. Jack only had \$3000 in income - but the return said that he owed \$2800 in taxes! How was this possible?&lt;/p&gt;&#xA;" OwnerUserId="19179" LastEditorDisplayName="user88" LastEditDate="2016-03-04T02:33:26.323" LastActivityDate="2016-03-04T02:33:26.323" Title="Too much taxes?" Tags="|lateral-thinking|" AnswerCount="1" CommentCount="3" ClosedDate="2016-03-04T03:13:56.997" ContentLicense="CC BY-SA 3.0" />
  <row Id="28289" PostTypeId="2" ParentId="28288" CreationDate="2016-03-04T00:38:36.460" Score="4" Body="&lt;p&gt;The first answer that comes to my mind is that Jack is a storekeeper who owes sales taxes. Even though his net income after deductions is only \$3,000, his gross sales could be around \$40,000, which had \$2,800 (7%) of unremitted sales taxes collected on it, and out of which \$37,000 was deducted as cost of goods sold and other business expenses.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastActivityDate="2016-03-04T00:38:36.460" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28290" PostTypeId="1" AcceptedAnswerId="28294" CreationDate="2016-03-04T00:38:52.877" Score="1" ViewCount="573" Body="&lt;p&gt;Sarah Lee was late for work. Again. This time it was no joke. She quickly got into her car, and saw that it was nearly out of gas. But there was no time to fill up. &quot;Fastest route will be 24 miles and 35 minutes,&quot; the GPS told her. Sarah saw that she only had 14 miles left on her car before she would be stuck. She presses the button on the GPS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recalculating...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can Sarah get to work without running out of gas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those who wanted clarification:&lt;br&gt;&#xA;Sarah Lee works in a physical building located some distance from her house.&lt;br&gt;&#xA;Sarah Lee needs to drive her car from her house to her place of work.&lt;br&gt;&#xA;Sarah Lee needs to get to work without running out of gas.&lt;br&gt;&#xA;Sarah Lee has no time to get more gas for her car before work.&lt;br&gt;&#xA;Sarah Lee can driver her car for 14 miles before running out of gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; How do GPSs calculate routes?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19179" LastEditorUserId="19179" LastEditDate="2016-03-08T15:37:07.360" LastActivityDate="2022-01-22T16:47:02.813" Title="Time Crunch Puzzle" Tags="|enigmatic-puzzle|" AnswerCount="5" CommentCount="10" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28291" PostTypeId="2" ParentId="28281" CreationDate="2016-03-04T00:51:29.957" Score="19" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Up to 7 can survive (but it is possible that none do).&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For maximum survival, set up a pentagon and a square such that each gunman is closer to the center of their shape than the other corners.  Then, have the same person be closest to each center.  All the corners shoot the center, and the two centers shoot the same corner. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All seven other corners survive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rough diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TGiAD.png&quot; alt=&quot;Living&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For minimum survival, just pair them all off.  Everybody shoots their partner, and they all die.  (These pairs are not actually parallel, since the distances along each side can't be the same, but that's easy enough to make happen)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/E3ZVx.png&quot; alt=&quot;Dying&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theoretical maximum would be 8 - since the two closest gunmen must shoot each other.  However, for this to work, we need everybody else to be closest to one of those two gunmen.  There's no way to have two people be closer to the center than each other if their angle around the center is less than or equal to 60*, so hexagons are right out.  Thus, we would need two pentagons such that one's corner is another's center.  Unfortunately, this causes some of the corners to be nearer to the corners of the other pentagon than their own center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example of failure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ENBaH.png&quot; alt=&quot;Failure&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="17389" LastEditDate="2016-03-04T06:05:08.953" LastActivityDate="2016-03-04T06:05:08.953" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28292" PostTypeId="2" ParentId="28290" CreationDate="2016-03-04T00:51:52.133" Score="5" Body="&lt;p&gt;Maybe her work was down a hill that was ten miles long that was 14 mile away and she could just coast down the hill for the last 10 miles.&lt;/p&gt;&#xA;" OwnerUserId="10856" LastActivityDate="2016-03-04T00:51:52.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28293" PostTypeId="1" AcceptedAnswerId="28297" CreationDate="2016-03-04T00:53:54.237" Score="28" ViewCount="15405" Body="&lt;p&gt;A police investigator walks into a room. He sees two dead bodies on the floor, those of the unfortunate duo Alice and Bob, the players in many physics thought experiments. They had been locked in the room together. The investigator notices several things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The door was sealed, and remained sealed since Alice and Bob were locked in one day before.&lt;/li&gt;&#xA;&lt;li&gt;There is a broken window high up on the wall, with shattered pieces of glass all over the floor. The investigator finds that it is impossible to reach. The window was not broken when Alice and Bob were locked in. There are no other exits besides this window and the door.&lt;/li&gt;&#xA;&lt;li&gt;There is a table by the window approximately four feet in height. Even an unlucky fall wouldn't have killed a person.&lt;/li&gt;&#xA;&lt;li&gt;Security camera footage shows that nobody - living or dead - entered or exited the room. No inanimate objects entered or exited the room, either (obviously, things like light and air would have passed through the broken window). The footage is accurate.&lt;/li&gt;&#xA;&lt;li&gt;There is a glass on the table, which has also been broken. It appears that it was a glass of water, because water was spilled all over the floor.&lt;/li&gt;&#xA;&lt;li&gt;There is nothing else in the room - including no weapons.&lt;/li&gt;&#xA;&lt;li&gt;Alice and Bob's bodies appear unharmed. There are no traces of poison.&lt;/li&gt;&#xA;&lt;li&gt;Prior to being locked in the room, Alice and Bob had everything they needed to survive - food, water, oxygen, etc.&lt;/li&gt;&#xA;&lt;li&gt;Alice and Bob died before the investigator opened the door.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The investigator relates all this information to his supervisor. The supervisor cannot solve the puzzle, yet the investigator, smiling, says that he knows the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did Alice and Bob die?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Note: This isn't a riddle of my own devising. A friend of mine got it from somewhere, and related it to me.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="5300" LastEditorUserId="5300" LastEditDate="2016-03-04T01:29:35.297" LastActivityDate="2016-03-07T09:33:16.297" Title="How did Alice and Bob die?" Tags="|lateral-thinking|" AnswerCount="16" CommentCount="24" ContentLicense="CC BY-SA 3.0" />
  <row Id="28294" PostTypeId="2" ParentId="28290" CreationDate="2016-03-04T00:58:28.743" Score="6" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The fastest route is 24 miles, but the shortest route is 14 or less.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13054" LastEditorUserId="2136" LastEditDate="2021-12-08T20:56:04.820" LastActivityDate="2021-12-08T20:56:04.820" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28296" PostTypeId="2" ParentId="28293" CreationDate="2016-03-04T01:42:30.197" Score="18" Body="&lt;p&gt;They were killed by:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lightning!!!!   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They both had their hands on the glass of water (probably due some complicated competition of a mathematical nature) and at the exact moment water poured out of the glass and onto the floor in a continuous steam forming a perfect circuit to ground - a bolt from Thor (who may of been a bit jealous) crashed through the window! Ever so lightly zapping them into another dimension.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The investigator knows because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He &lt;strong&gt;is&lt;/strong&gt; Thor (working in his day job)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastActivityDate="2016-03-04T01:42:30.197" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28297" PostTypeId="2" ParentId="28293" CreationDate="2016-03-04T01:46:02.767" Score="36" Body="&lt;p&gt;Alice and Bob...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...are actually fish! They were in a glass fishbowl on the table that had shattered, perhaps due to an unfortunate resonant frequency*. This frequency also broke the glass window. This explains the broken glass and water on the floor, and the two dead (but otherwise uninjured) fish :(  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit*&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For completeness sake I'm gonna say that a passing ambulance was blasting a siren at said resonant frequency.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14095" LastEditorUserId="14095" LastEditDate="2016-03-04T06:38:52.640" LastActivityDate="2016-03-04T06:38:52.640" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="28298" PostTypeId="2" ParentId="28293" CreationDate="2016-03-04T02:28:59.263" Score="14" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Unbeknownst to anyone there was a large block of dry ice in the room. It sublimates so fast that it filled the room with CO2 suffocating Alice and Bob. One of them had a glass of water which slipped from their hands to the floor. The extra pressure caused the window to break open. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15942" LastActivityDate="2016-03-04T02:28:59.263" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="28299" PostTypeId="2" ParentId="28293" CreationDate="2016-03-04T03:02:18.810" Score="6" Body="&lt;p&gt;Alice and Bob were trying to Science the Shit out of some hydrazine to create water for growing potatoes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately they forgot about the Oxygen they were breathing out and blew themselves up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take that Neil Armstrong!&lt;/p&gt;&#xA;" OwnerUserId="4055" LastActivityDate="2016-03-04T03:02:18.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28300" PostTypeId="2" ParentId="27459" CreationDate="2016-03-04T03:04:24.317" Score="4" Body="&lt;h1&gt;&lt;strong&gt;PART 2&lt;/strong&gt;&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;For FLOCK * 7 = GEESE, the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; F=1, L=0, O=9, C=4, K=8, G=7, E=6, S=3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F can only be 1 to maintain the number of digits, therefore, G=7/8/9 and L=0/2/3/4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since 7 is odd, we can produce numbers with patterns 700#0, 711#1, 722#2, etc. until 988#8. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started with 700#0. We can establish 70000 and 70070 are divisible by 7. If we keep adding either 70 or 1071 (which is 7 * 53) as needed, we can establish the pattern:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;70000, 70070, 71141, 72212, 72282, 73353, 74424, 74494, 75565, 76636, 77707, 77777, 78848, 79919, 79989 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finding the next number divisible by 7 with the format 800#0 and continuing the pattern, we get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;80010, 80080, 81151, 82222, 82292, 83363, 84434, 85505, 85575, 86646, 87717, 87787, 88858, 89929, 89999&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With 900#0, we get&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;90020, 90090, 91161, 92232, 93303, 93373, 94444, 95515, 95585, 96656, 97727, 97797, 98868, 99939&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Removing the ones with:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1s on any digit,&lt;/li&gt;&#xA;&lt;li&gt;repeated tens digit,&lt;/li&gt;&#xA;&lt;li&gt;repeated ten-thousands digit,&lt;/li&gt;&#xA;&lt;li&gt;0s formatted 7####,&lt;/li&gt;&#xA;&lt;li&gt;2s formatted 8####, and&lt;/li&gt;&#xA;&lt;li&gt;3s or 4s formatted 9####&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;72282, 73353, 74424, 74494, 75565, 76636, 78848, 79989, 80080, 82292, 83363, 84434, 85505, 85575, 86646, 89929, 90020, 90090, 95585, 96656, 97727, 98868&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I tried dividing it all by 7 and the only working solution is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 10948 * 7 = 76636&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19538" LastActivityDate="2016-03-04T03:04:24.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28301" PostTypeId="2" ParentId="28293" CreationDate="2016-03-04T03:11:46.860" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alice and Bob both suffered from a chronic condition that required them to take pills twice a day. They had taken with them into the room a glass of water for that purpose. At the beginning of their stay inside the locked room, Alice dozed off and Bob decided to scare her and wake her up. When Alice woke up, she was so startled that she screamed at just the right pitch to shatter both the window and the glass of water. Having lost the water they needed for their pills, they both died within hours.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19770" LastActivityDate="2016-03-04T03:11:46.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28302" PostTypeId="2" ParentId="28293" CreationDate="2016-03-04T03:28:06.450" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alice and Bob are / were fish and were in the glass of water.  A bird saw them or for other reasons flew into the window and broke it.  A piece of glass dropped and knocked the glass off the table and on to the floor thus dislodging the fish from the water and hence asphyxiating them. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19771" LastActivityDate="2016-03-04T03:28:06.450" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28303" PostTypeId="2" ParentId="28293" CreationDate="2016-03-04T03:54:21.207" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Perhaps there was a downpour of rain, and the two drowned?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19773" LastActivityDate="2016-03-04T03:54:21.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28304" PostTypeId="2" ParentId="28293" CreationDate="2016-03-04T06:55:19.613" Score="69" Body="&lt;p&gt;We are asking the wrong question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; *  The security tapes show that nobody else entered the room, and we are told that the inspector is the one to have unsealed the door.&lt;br&gt;&#xA; *  We are told everything which was in the room, and there was no method of communication that could have been used by Alice or Bob to have summoned the police themselves.&lt;br&gt;&#xA; *  So the most important question is:  since nobody found the bodies, and since Alice and Bob could not have summoned the police themselves, why was the investigator even there, unsealing the door?&lt;P&gt;&#xA;&#xA; The only rational answer to this critical conundrum (&quot;why was the inspector there?&quot;) is that the inspector must have already &lt;em&gt;known&lt;/em&gt; the bodies were there to be found.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And therefore..&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The investigator must have killed Alice and Bob some number of days in the past, and then sealed their bodies into the room in order to conceal his crime, and then returned to &quot;discover&quot; their bodies the next day.&lt;P&gt;&#xA; Critically, note that the puzzle never actually specifies that Alice and Bob were alive at the time of being sealed into the room.  Since we've established that they died at some point before the inspector unsealed the door, and that nobody killed them while they were sealed inside the room, the only logical conclusion is that they had died at some point before even being sealed into the room in the first place.&lt;P&gt;&#xA; As a result, the investigator gets away with his shocking crime, and gets to smile enigmatically as he informs his supervisor that he knows how it happened.  And then invite him out to dinner, and maybe back to his place for afters.  Maybe his supervisor would finally &lt;em&gt;notice&lt;/em&gt; him now!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9286" LastActivityDate="2016-03-04T06:55:19.613" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28305" PostTypeId="2" ParentId="28293" CreationDate="2016-03-04T07:13:17.120" Score="15" Body="&lt;p&gt;Alice and Bob&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;suffocated.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;there was no oxygen in the room to begin with. &quot;&lt;strong&gt;Prior to being locked in the room&lt;/strong&gt;, Alice and Bob had everything they needed to survive - food, water, &lt;strong&gt;oxygen&lt;/strong&gt;, etc.&quot; Presumably the air was vacuumed out after the room was sealed. The investigator opening the door caused air pressure to rapidly equalize, breaking the glass and the window.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19788" LastActivityDate="2016-03-04T07:13:17.120" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="28306" PostTypeId="1" AcceptedAnswerId="28588" CreationDate="2016-03-04T07:29:26.027" Score="10" ViewCount="847" Body="&lt;p&gt;I have a long history from east to west,&lt;br&gt;&#xA;to some I'm the worst and others the best.&lt;br&gt;&#xA;With a core made of resin I dance and I flutter,&lt;br&gt;&#xA;no one judges me when I speak with a stutter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I may be an object of legend or myth,&lt;br&gt;&#xA;at first you won't find me but maybe in fifth. &lt;br&gt;&#xA;For at any point, you summon power of three, &lt;br&gt;&#xA;And look to the wind, you may just find me. &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My presence strengthens each link in the chain, &lt;br&gt;&#xA;certain times that you use me, it may be in vain. &lt;br&gt;&#xA;Storms when you're older and peace as a child, &lt;br&gt;&#xA;with us two together, adventures are wild. &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A limited spectrum between ceiling and floor, &lt;br&gt;&#xA;Shot full of holes from twelve until four. &lt;br&gt;&#xA;In solving this riddle you're just out Of Time,&lt;br&gt;&#xA;you should hear my voice as it is sublime. &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be clear by now that I'm not a Paid Genius Poet,&lt;br&gt;&#xA;&quot;What am I, I ask?,&quot; by now you should know it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Edit) Rearranged stanzas and hints and got rid of spoilers &lt;code&gt;&amp;gt;!&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 3&lt;/p&gt;&#xA;" OwnerUserId="19731" LastEditorUserId="19731" LastEditDate="2016-03-08T07:20:02.973" LastActivityDate="2016-08-10T20:26:52.593" Title="Immovable object" Tags="|riddle|word|rhyme|poetry|literature|" AnswerCount="9" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28307" PostTypeId="2" ParentId="28293" CreationDate="2016-03-04T08:01:41.077" Score="9" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The room was pumped with carbon monoxide prior to Alice and Bob being locked in. They don't notice at first, but start to feel light headed and decide to throw the glass of water at the window to get some air. It's unfortunately too little too late and there isn't enough time for the carbon monoxide to fully disperse before Alice and Bob are killed by carbon monoxide poisoning.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19790" LastActivityDate="2016-03-04T08:01:41.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28308" PostTypeId="1" AcceptedAnswerId="29272" CreationDate="2016-03-04T08:17:29.107" Score="18" ViewCount="396" Body="&lt;p&gt;I found this sketch in the notes old Alembic took while exploring the ruins of the ancient city of Mictal. The man must have copied some writing he encountered among the ruins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YbxHI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If only I could figure out what these markings mean...&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE - hints&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is a set of instructions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(I'll give it another week before I post the next.)&lt;/p&gt;&#xA;" OwnerUserId="18558" LastEditorUserId="18558" LastEditDate="2016-03-19T19:05:52.687" LastActivityDate="2016-03-19T20:27:12.940" Title="The markings of Mictal" Tags="|enigmatic-puzzle|visual|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28310" PostTypeId="1" AcceptedAnswerId="28438" CreationDate="2016-03-04T09:35:25.877" Score="12" ViewCount="479" Body="&lt;p&gt;The answer to this riddle is a single English word.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Likely you'll find me when lending support&lt;br&gt;&#xA;  Or shedding some knots by a slip of some sort&lt;br&gt;&#xA;  Or adding the silver before fitting the frame&lt;br&gt;&#xA;  Keep calm and breathe is the key to this game.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you have doubts about your answer, it is not correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; question_asker was absolutely right. The the second line is boaty. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is a very common English word. The clues refer to less common meanings of this word. Nothing really obscure, but definitely not the most common meanings. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Trust the last line. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Trust the title. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Trust me to be a pain in the posterior. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastEditorUserId="18924" LastEditDate="2016-03-05T21:54:45.813" LastActivityDate="2016-03-06T14:13:58.960" Title="An Acrostiholic Riddle" Tags="|riddle|word|rhyme|" AnswerCount="8" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28312" PostTypeId="1" AcceptedAnswerId="28317" CreationDate="2016-03-04T10:15:06.213" Score="33" ViewCount="1013" Body="&lt;p&gt;This is a short riddle which has a short answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I look like I have no memory,&lt;br&gt;&#xA;but I am fine,&lt;br&gt;&#xA;but turn me around&lt;br&gt;&#xA;and I may knock you down.&lt;/p&gt;&#xA;" OwnerUserId="19547" LastEditorUserId="19165" LastEditDate="2016-03-05T06:10:07.733" LastActivityDate="2016-03-05T06:10:07.733" Title="A little abbreviated response" Tags="|riddle|wordplay|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28313" PostTypeId="2" ParentId="28293" CreationDate="2016-03-04T10:17:35.983" Score="58" Body="&lt;p&gt;The investigator knew, as the supervisor did not that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The room was not heated!  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There was a sudden, catastrophic cold snap the night before.&lt;br&gt;&#xA; The extreme cold caused the metal window frame to contract, breaking the glass.&lt;br&gt;&#xA; The water in a sealed glass container froze, breaking it.&lt;br&gt;&#xA; When temperatures returned to normal in the morning, the ice melted.&lt;br&gt;&#xA; Therefore, Alice and Bob died of hypothermia. &#xA; I &lt;strong&gt;know&lt;/strong&gt; this is the right answer because OP cleverly hid the word &quot;ice&quot; in&#xA; Al&lt;strong&gt;ice&lt;/strong&gt;'s name.  The remaining letters are an anagram of BLABO, an oil tank&#xA; cleaning system described &lt;a href=&quot;https://www.google.ch/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=1&amp;amp;cad=rja&amp;amp;uact=8&amp;amp;ved=0ahUKEwjR5NiU9KbLAhXJiRoKHWO8AesQFggdMAA&amp;amp;url=http%3A%2F%2Fwww.oreco.com%2FSolutions%2FCrude-oil-tank-cleaing.aspx&amp;amp;usg=AFQjCNG26BbwN_MTPtJJ2JY9br-G6tJ4FA&amp;amp;sig2=EzkolSK9k7Iywl7Oy_W_3w&quot;&gt;here&lt;/a&gt;.&lt;br&gt;&#xA; This is the last piece of the puzzle. Clearly, the reason the room wasn't &#xA; heated was because the oil tank for the furnace was being cleaned with BLABO.&lt;br&gt;&#xA; And the investigator smiled at this tragedy. That sick, sadistic so and so!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastEditorUserId="18924" LastEditDate="2016-03-05T12:52:36.040" LastActivityDate="2016-03-05T12:52:36.040" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28314" PostTypeId="1" AcceptedAnswerId="28316" CreationDate="2016-03-04T10:19:36.923" Score="2" ViewCount="500" Body="&lt;p&gt;There's a boat traveling on a river. When it's going with the river flow, it takes 10 minutes to cover 1km, but when going against the flow, it takes it 15 minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How long does it take to cover 1km when traveling in still water?&lt;/p&gt;&#xA;" OwnerUserId="19803" LastEditorUserId="17250" LastEditDate="2016-03-04T12:50:56.493" LastActivityDate="2016-03-04T13:36:34.530" Title="Boat traveling on a river" Tags="|calculation-puzzle|" AnswerCount="4" CommentCount="5" ClosedDate="2016-03-04T13:53:15.877" ContentLicense="CC BY-SA 3.0" />
  <row Id="28315" PostTypeId="2" ParentId="28314" CreationDate="2016-03-04T10:28:46.050" Score="5" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It will never travel 1km in still water&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A river will never have still water, as a river by definition is a stream of water that &lt;strong&gt;flows&lt;/strong&gt; into a larger body (a lake, or sea etc). Therefore it will never be still&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17034" LastActivityDate="2016-03-04T10:28:46.050" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28316" PostTypeId="2" ParentId="28314" CreationDate="2016-03-04T10:34:15.483" Score="6" Body="&lt;p&gt;The boat takes 12 min to cover 1km in still water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going downstream, the boat is moving at 6km/hr. Going upstream, it moves at 4km/hr. The difference is twice the speed of the river, which is flowing at 1km/hr.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the boat itself moves at 5km/hr in still water.&lt;/p&gt;&#xA;" OwnerUserId="12730" LastActivityDate="2016-03-04T10:34:15.483" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28317" PostTypeId="2" ParentId="28312" CreationDate="2016-03-04T10:37:25.157" Score="44" Body="&lt;p&gt;I think the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; OK&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I look like I have no memory&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; OK looks like 0K (zero kilobytes)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I am fine,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; OK is synonymous with fine.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but turn me around and I may knock you down.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; OK backwards is KO (an acronym for knock out)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18422" LastActivityDate="2016-03-04T10:37:25.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28318" PostTypeId="2" ParentId="28314" CreationDate="2016-03-04T10:40:30.377" Score="5" Body="&lt;p&gt;Let &lt;strong&gt;x&lt;/strong&gt; be the speed of boat and &lt;strong&gt;y&lt;/strong&gt; be the speed of river flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(x+y)*10 = 1&#xA;&#xA;(x-y)*15 = 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;That is&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(x+y)10 = (x-y)*15&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which gives &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;y = x/5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;No consider &lt;strong&gt;t&lt;/strong&gt; be the time it will take to cover 1Km in still water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(x+y)*10 = x*t  &#xA;&#xA;(x+x/5)*10 = x*t&#xA;&#xA;60x/5x = t&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So we get &lt;code&gt;t = 12&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The boat will take &lt;strong&gt;12&lt;/strong&gt; minutes to cover &lt;strong&gt;1Km&lt;/strong&gt; in still water.&lt;/p&gt;&#xA;" OwnerUserId="2730" LastActivityDate="2016-03-04T10:40:30.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28319" PostTypeId="2" ParentId="28310" CreationDate="2016-03-04T10:55:43.727" Score="3" Body="&lt;p&gt;Answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Luck&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Title suggests that it is an acrostic, which would lead to the answer being LOOK. The 'rhyme' tag suggests that the alternate spelling LUCK would be more correct, and it also fits the riddle more&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17034" LastActivityDate="2016-03-04T10:55:43.727" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28320" PostTypeId="1" AcceptedAnswerId="28324" CreationDate="2016-03-04T11:12:14.067" Score="6" ViewCount="187" Body="&lt;p&gt;A man walks up a hill at 2mph, and back down the same way at 4mph. Upon returning to where he started, what is his average speed?&lt;/p&gt;&#xA;" OwnerUserId="12730" LastActivityDate="2016-03-04T11:31:23.893" Title="A man on a hill" Tags="|mathematics|calculation-puzzle|" AnswerCount="1" CommentCount="5" ClosedDate="2016-03-04T12:14:24.137" ContentLicense="CC BY-SA 3.0" />
  <row Id="28322" PostTypeId="1" AcceptedAnswerId="28329" CreationDate="2016-03-04T11:16:50.480" Score="7" ViewCount="1004" Body="&lt;p&gt;First comes the bad and boring riddle (the answer is to be written in all capital letters):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Magnesium plus Gallium yields a joke.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Next comes the easy alphametic:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Solve the answer to the bad and boring riddle as an alphametic.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Every letter stands for a digit in base-10 representation, different letters stand for different digits, and leading digits are always non-zero.)&#xA;Which digit does each letter represent? Please present the full analysis how these digits can be determined.&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="8874" LastEditDate="2016-03-04T15:52:36.447" LastActivityDate="2016-03-04T18:36:35.253" Title="A bad and boring riddle, and an easy alphametic" Tags="|riddle|no-computers|alphametic|" AnswerCount="3" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28323" PostTypeId="1" AcceptedAnswerId="28328" CreationDate="2016-03-04T11:20:51.703" Score="8" ViewCount="186" Body="&lt;p&gt;Yesterday I posed an Alfred E. Neuman alphametic with multiplication, and &#xA;today I pose an Alfred E. Neuman alphametic with division for the lovers of Mad magazine: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every letter and every question mark stands for a digit in &lt;strong&gt;base-9&lt;/strong&gt; representation. Different letters stand for different digits. &#xA;Leading digits are always non-zero. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;ALFRED $\div$ E = NEUMAN&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which digit does each letter represent? &#xA;(Please present the full analysis how these digits can be determined.)&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="4421" LastEditDate="2016-03-04T17:25:17.347" LastActivityDate="2016-03-04T17:25:17.347" Title="A very different Alfred E. Neuman alphametic" Tags="|no-computers|alphametic|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28324" PostTypeId="2" ParentId="28320" CreationDate="2016-03-04T11:21:40.880" Score="7" Body="&lt;p&gt;The answer is :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$\frac{8}{3}$ mph&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's take the distance up (and down) the hill as $d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Time taken in reaching top of the hill is : $\frac{d}{2}$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Time taken in coming down the hill is : $\frac{d}{4}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus, average speed is: $\frac{total  \space distance}{total \space time \space taken}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\frac{d+d}{\frac{d}{2}+\frac{d}{4}}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On simplifying,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it gives $\frac{8}{3}$ or $2.67$ mph (approx).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1766" LastEditorUserId="17482" LastEditDate="2016-03-04T11:31:23.893" LastActivityDate="2016-03-04T11:31:23.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28325" PostTypeId="2" ParentId="28293" CreationDate="2016-03-04T11:54:49.937" Score="4" Body="&lt;p&gt;Alice and Bob were performing physics experiments. &#xA;There is a chance of fire in the room and must have produced lots of smoke and they would have started suffocating can't use oxygen because it will ignite fire more tried to break the window using the glass but it was too late for them, enough carbon was inside their lungs to suffocate them and they died.&lt;/p&gt;&#xA;" OwnerUserId="19808" LastEditorUserId="8874" LastEditDate="2016-03-04T12:05:01.540" LastActivityDate="2016-03-04T12:05:01.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28326" PostTypeId="2" ParentId="28281" CreationDate="2016-03-04T12:03:40.413" Score="6" Body="&lt;h2&gt;Maximization&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The maximum value of dead bandits clearly is $10$ (so that all are dead).&#xA;For instance, put five bandits &#xA;at the points $(100,0)$, $(200,0)$, $(400,0)$, $(800,0)$, $(1600,0)$ and &#xA;five bandits at the points $(100,1)$, $(200,2)$, $(400,4)$, $(800,8)$, $(1600,16)$. Then no pair is the same distance apart, and the five pairs &#xA;with equal $x$-coordinates kill each other.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Minimization&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The minimum value of dead bandits is more difficult to determine.&#xA;We reformulate the minimization problem as: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given a set $S$ of $10$ points in the plane, such that the distances &#xA;  between them are all distinct. For each point $p\in S$ we mark the point &#xA;  $q\in S-\{p\}$ that is nearest to $p$. Find the least possible number of &#xA;  marked points.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(1) We observe that each point $x\in S$ is the nearest to at most five &#xA;other points in $S$. &#xA;Indeed, for any six points $p_1,\ldots,p_6$ one of the angles $\angle p_ixp_j$ is at most $60^{\circ}$, in which case the distance $p_ip_j$ is smaller than one of the distances $xp_i$ and $xp_j$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) It follows that at least two points are marked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) Now suppose that exactly two points, say $x$ and $y$, are marked. &#xA;Then $x$ is the closest point to $y$, and $y$ is the closests point to $x$.&#xA;So by the observation (1) the remaining eight points in $S$ split into &#xA;two groups $S_x$ and $S_y$ of four points each, so that their closest points&#xA;are $x$ and $y$ respectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Denote $S_x=\{a_1,a_2,a_3,a_4\}$ and $S_y=\{b_1,b_2,b_3,b_4\}$, so that &#xA;the angles $\angle a_ixa_{i+1}$ are successively adjacent as well as the angles $\angle b_iyb_{i+1}$, and so that $a_1$ and $b_1$ lie on one side of the line through $xy$ while $a_4$ and $b_4$ lie on the other side of this line. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since all the angles $\angle a_ixa_{i+1}$ and $\angle b_iyb_{i+1}$ are greater than $60^{\circ}$, it follows that&#xA;$$ \angle a_1xy + \angle yxa_4 + \angle b_1yx + \angle xyb_4 ~&amp;lt;~ 360^{\circ}.$$&#xA;Therefore $\angle a_1xy + \angle b_1yx &amp;lt; 180^{\circ}$ or &#xA;$\angle yxa_4 + \angle xyb_4 &amp;lt; 180^{\circ}$.&#xA;Without loss of generality, let us assume the first inequality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) Next, note that the quadrilateral $xyb_1a_1$ is convex because &#xA;$a_1$ and $b_1$ are on different sides of the perpendicular bisector of $xy$. &#xA;From $a_1b_1 &amp;gt; a_1x$ and $yb_1&amp;gt;xy$ we get the inequalities &#xA;$\angle a_1xb_1 &amp;gt; \angle a_1b_1x$ and&#xA;$\angle yxb_1 &amp;gt; \angle xb_1y$.&#xA;Adding these two inequalities yields $\angle a_1xy &amp;gt; \angle a_1b_1y$.&#xA;A symmetric argument yields $\angle b_1yx &amp;gt; \angle b_1a_1x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These last two inequalities imply&#xA;$$180^{\circ} &amp;gt; \angle a_1xy +\angle b_1yx &amp;gt; \angle a_1b_1y + \angle b_1a_1x.$$&#xA;Hence the sum of the angles of the quadrilateral $xyb_1a_1$ is less &#xA;than $360^{\circ}$, which is a contradiction. &#xA;Consequently at least three points are marked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5) Finally, the configuration with ten points&#xA;$(-100,0)$, $(0,0)$, $(101,0)$, $(213,0)$, $(0,110)$, $(0,-111)$, $(-120,110)$, $(-120,-111)$, $(121,110)$, $(121,-111)$ shows that the minimum with three marked points indeed can be attained.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Summary&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The maximum number of dead bandits is $10$, and the minimum number of dead bandits is $3$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="17389" LastEditDate="2016-03-05T00:08:51.003" LastActivityDate="2016-03-05T00:08:51.003" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28327" PostTypeId="2" ParentId="28322" CreationDate="2016-03-04T12:25:57.040" Score="3" Body="&lt;p&gt;It is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MG+GA=LOL&lt;br&gt;&#xA;&#xA;Here Mg is Magnesium and Ga is Gallium and I interpreted joke as LOL.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First of all, it has many solutions. It is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 72+29=101, 63+68=101, 54+47=101, 45+56=101&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How I got it?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Now $L=1$ because well if two 'n digit numbers' are added then if they yield a $n+1$ digit number then the $(n+1)^{th}$ digit is always $1$. That being said, we now have MG+GA=1O1. All alphabets have different values, so M, G, A, O cannot be equal to $1$.&lt;br&gt;&#xA; Now as G and A cannot be 1 so they can have possible values such that their addition always gives 1 at the one's place So they can have possible values of (2 and 9), (3 and 8), (4 and 7) etc. And they will surely leave a carry of 1 at tens place. So,&lt;br&gt;&#xA;&lt;strong&gt;CASE 1 $\Rightarrow$ if G and A are 2 and 9 respectively.&lt;/strong&gt; In this case we have M2+29=1O1. Here M have to be greater than or equals to 7. It can't be 9 because A=9. It can't be 8 because then O will be 1 which should not be. So M=7. Therefore O=zero. And therefore 72+29=101&lt;br&gt;&#xA;&lt;strong&gt;CASE 2 $\Rightarrow$ if G and A are 3 and 8 respectively.&lt;/strong&gt; In this case we have M3+38=1O1. Here M have to be greater than or equals to 6. If M=9 then O is 3 which should not be. M cannot be 8 because 8 is reserved by A in this case. If M=7 then O=1, which cannot be. So if M=6 then O=zero.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I hope you have got the point by now.&lt;/p&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-03-04T13:00:24.080" LastActivityDate="2016-03-04T13:00:24.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28328" PostTypeId="2" ParentId="28323" CreationDate="2016-03-04T12:46:09.623" Score="7" Body="&lt;p&gt;I hate base-9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets re-write it as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;NEUMAN&#xA;x    E&#xA;------&#xA;ALFRED&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;OK, so first off, we know $E\notin\{0,1\}$.  Since the result is a 6 digit number, we know that $N\in\{1,2,3,4\}$. Since $E \times N$ shows up twice, once with a $D$ result and once with an $A$ result, we know that the fifth column ($E \times E$) must have a carry over.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, we will use the fact that the carry over for multiplying any two numbers must be less than the two numbers.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $N=4$, then $E=2$ and $A=8$.  But then $D=8$ too, which is impossible.&lt;/li&gt;&#xA;&lt;li&gt;If $N=3$, then $E=2$ and $D=6 \implies A \in \{7,8\}$.  Since $E^2=4$, we need a carry over of at least 4 from the $U \times E$ column, which is impossible.&lt;/li&gt;&#xA;&lt;li&gt;If $N=1$, then $E \times N=D \implies E=D$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Thus $N=2$&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;So $E=3$ and $D=6 \implies A \in \{7,8\}$.  The only value that works for the second column where $A \times E$ yields $E$ is 7 since $7\times 3=23$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the fifth column, $E \times E$ yields L.  Since $E^2=10$, we know that $L$ is simply the carry over from the previous column, which cannot be greater than 2.  Thus, $L\in\{0,1\}$ since 2 is already taken.  &lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Assume, $L=1$&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;So $U \times E$ has a carry over of 1, $U\in \{4,5\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remaining values are $\{0,4,5,8\}$.  And since $A=7$ the carry over from the second column  into the third is 2.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $M=0$, then $M \times 3 +2 =2 \implies R=2$, which is already taken.&lt;/li&gt;&#xA;&lt;li&gt;If $M=4$, then $M \times 3 +2 =13+2 \implies R=5$, leaving nothing for $U$.&lt;/li&gt;&#xA;&lt;li&gt;If $M=8$, then $M \times 3 +2 =26+2 \implies R=8$ too.&lt;/li&gt;&#xA;&lt;li&gt;If $M=5$, then $M \times 3 +2 =16+2 \implies R=8$. And $U=4$.  This leave $F=0$, but doing the math gets $234572 \times 3=714836 \implies F=4$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Thus $L=0$&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;$U \times E$ has a carry over of 0, so $U=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remaining values are $\{4,5,8\}$.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$M=8 \implies 231872 \times 3 = 705836$, so $R=8$&lt;/li&gt;&#xA;&lt;li&gt;$M=4 \implies 231472 \times 3 = 704536$, so $F=4$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thus, $M=5 \implies 231572 \times 3 = 704836$ so $R=8$ and $F=4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;231572&#xA;x    3&#xA;------&#xA;704836&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="1592" LastEditDate="2016-03-04T14:35:26.797" LastActivityDate="2016-03-04T14:35:26.797" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28329" PostTypeId="2" ParentId="28322" CreationDate="2016-03-04T12:59:11.607" Score="13" Body="&lt;p&gt;The solution to the bad and boring riddle is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mg + Ga = Gag, respectively in all capitals: $~~MG+GA=GAG$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The resulting easy alphametic has only a single solution in the decimal system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Since $10M+G\le99$ and $10G+A\le99$, their sum satisfies $100G+10A+G\le198$. This implies $G=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) The unit digits in the addition satisfy $G+A=G$ or $G+A=10+G$. This implies $A=0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) Finally, $10M+1=101-10=91$ implies $M=9$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The unique solution to the easy alphametic is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $~~MG+GA=GAG$ $~~~~~\Longleftrightarrow~~~~~$ $91+10=101$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2016-03-04T12:59:11.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28330" PostTypeId="1" AcceptedAnswerId="28340" CreationDate="2016-03-04T13:05:57.113" Score="3" ViewCount="153" Body="&lt;p&gt;Darth Sidious is too powerful. Anakin and Obi-Wan discusses and convinces Count Dooku (Darth Tyrannus) to fight against his own master. It is the only way they can be a true match against the Dark Lord of the Sith. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want you to solve this epic fight in &lt;strong&gt;Base10 alphametic, without the usage of computer&lt;/strong&gt;, with every details! There is only 1 solution I did(if I am not mistaken). Good luck, have fun!&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; ANAKIN&#xA; OBIWAN&#xA;+ DOOKU&#xA;-------&#xA;SIDIOUS&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="18685" LastEditorUserId="8874" LastEditDate="2016-03-04T15:45:42.743" LastActivityDate="2016-03-04T15:45:42.743" Title="Anakin, Obi-Wan and Dooku are a match for Sidious!" Tags="|no-computers|alphametic|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28331" PostTypeId="2" ParentId="28310" CreationDate="2016-03-04T13:10:24.567" Score="1" Body="&lt;p&gt;The answer is not:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mirror, Reflection, or Glass (or Gemination). The reasoning is available in the edit history if you want to see it. I'm leaving this open on the off chance that I figure out the real answer at some point.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2016-03-04T18:24:30.350" LastActivityDate="2016-03-04T18:24:30.350" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="28332" PostTypeId="2" ParentId="28314" CreationDate="2016-03-04T13:36:34.530" Score="2" Body="&lt;p&gt;let's calculate step by step&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Going downstream: 1/(10/60) Km/hr =&gt; 6Km/hr&lt;/li&gt;&#xA;&lt;li&gt;Going upstream 1/(15/60) km/hr =&gt; 4 Km/hr&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Downstream equation: speed of boat + speed of current = x + Y = 6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;upstream equation: speed of boat - speed of current = x - Y = 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On solving these two equations: speed of boat = 5Km/hr&#xA;And thus it will take (1/5)hr or 12 mins to cover 1Km in still water.&lt;/p&gt;&#xA;" OwnerUserId="19812" LastActivityDate="2016-03-04T13:36:34.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28333" PostTypeId="1" AcceptedAnswerId="28337" CreationDate="2016-03-04T13:40:11.513" Score="15" ViewCount="1634" Body="&lt;p&gt;In the late 1400's, a Count, who is hosting a dinner party, receives a message from his steward. &#xA;The note lists that the following relatives of the Count will be attending the party:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;The father's brother-in-law&lt;/li&gt;&#xA;  &lt;li&gt;The brother's father-in-law&lt;/li&gt;&#xA;  &lt;li&gt;The father-in-law's brother&lt;/li&gt;&#xA;  &lt;li&gt;The brother-in-law's father&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Count immediately assumes that there will be four guests. However, the steward's message makes him seem less sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;What is the least number of guests that the Count can expect?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="17744" LastEditorUserId="7586" LastEditDate="2017-08-07T17:00:34.207" LastActivityDate="2017-08-07T17:00:34.207" Title="Attendance at a dinner party" Tags="|lateral-thinking|family|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28334" PostTypeId="2" ParentId="28224" CreationDate="2016-03-04T13:40:56.473" Score="2" Body="&lt;p&gt;The concept of infinity involves that there is no last turn. That means for Gryffindor:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They will win with any strategy that guarantees a marble is taken in each turn. This may be &quot;Take the one with the lowest number.&quot; or &quot;Take a random one.&quot; When the game runs until infinity, then Hufflepuff's urn will be empty.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, the flaw in Dumbledore's plan is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;This will continue until one of you can no longer take your turn.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus Hufflepuff's strategy will be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Make sure that we can't take our turn anywhere between 1 and infinity, &lt;strong&gt;exclusive&lt;/strong&gt;. It can be as simple as &quot;Take the marbles numbered 0 and 1.&quot; This will work fine in the first turn, leaving 1 marble in Hufflepuff's and 1 marble in Gryffindor's urn. In the second turn, Hufflepuff's strategy will fail, marbles 0 and 1 are no longer available, and the game stops with 1 marble in Hufflepuff's urn.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All points for Hufflepuff!&lt;/p&gt;&#xA;" OwnerUserId="19811" LastActivityDate="2016-03-04T13:40:56.473" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28335" PostTypeId="2" ParentId="28333" CreationDate="2016-03-04T13:58:41.897" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Zero other than himself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The count is married to his aunt, thus his father's brother-in-law is himself.&#xA;The count is also married to his mum (he's a mormon or something, allowing multiple marriages). Thus his brother's father in law is himself. &#xA;The count's mum is also married to his brother, thus the father-in-law's brother is himself.&#xA;The count's sister is married to the count's son, thus the brother in law's father is himself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;O_o&lt;/p&gt;&#xA;" OwnerUserId="19698" LastEditorUserId="17482" LastEditDate="2016-03-04T13:59:46.433" LastActivityDate="2016-03-04T13:59:46.433" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28336" PostTypeId="2" ParentId="28333" CreationDate="2016-03-04T14:04:44.430" Score="5" Body="&lt;p&gt;If we do not consider any incest or multiple marriages then answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Three&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because his brother's father-in-law and his father-in-law's brother can be the same guy given that his brother marries a cousin of his wife.&lt;/p&gt;&#xA;" OwnerUserId="18377" LastActivityDate="2016-03-04T14:04:44.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28337" PostTypeId="2" ParentId="28333" CreationDate="2016-03-04T14:10:04.907" Score="16" Body="&lt;p&gt;It's:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They could all be the same person in a totally respectable manner.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The father's brother-in-law isn't necessary his uncle.  The father could have remarried.&lt;br&gt;&#xA; The brother's father-in-law could be that person if the father's new wife is his brother's wife's mother.&lt;br&gt;&#xA; The father-in-law's brother is also the same person if he's his wife's father's brother.&lt;br&gt;&#xA; The brother-in-law's father can still be the same person if his wife's her brother's mother remarried her ex's brother.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastActivityDate="2016-03-04T14:10:04.907" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28339" PostTypeId="2" ParentId="28310" CreationDate="2016-03-04T14:51:45.077" Score="0" Body="&lt;p&gt;This is some sort of joke answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it is an hairdresser.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Likely you'll find me when lending support&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If bartenders are the most famous supporters for men, hairdressers are for women :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or shedding some knots by a slip of some sort&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/kgfaZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kgfaZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or adding the silver before fitting the frame&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/xMLn2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xMLn2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Keep calm and breathe is the key to this game.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/zBXdP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zBXdP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hope to have stolen you a little smile ;)&lt;/p&gt;&#xA;" OwnerUserId="4652" LastActivityDate="2016-03-04T14:51:45.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28340" PostTypeId="2" ParentId="28330" CreationDate="2016-03-04T15:07:51.510" Score="3" Body="&lt;p&gt;Let &lt;span class=&quot;math-container&quot;&gt;$S_i$&lt;/span&gt; be the sum of the digits in the &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt;th column.  Let &lt;span class=&quot;math-container&quot;&gt;$C_i$&lt;/span&gt; be the carry over from &lt;span class=&quot;math-container&quot;&gt;$S_i + C_{i-1}$&lt;/span&gt; and let &lt;span class=&quot;math-container&quot;&gt;$D_i$&lt;/span&gt; be the resulting digit from &lt;span class=&quot;math-container&quot;&gt;$S_i$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Since the result is one more digit than the others, we know that &lt;span class=&quot;math-container&quot;&gt;$S_6 \ge 10$&lt;/span&gt;.  At most, &lt;span class=&quot;math-container&quot;&gt;$C_5=2$&lt;/span&gt;, so at most &lt;span class=&quot;math-container&quot;&gt;$A+O+2=19$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;h2&gt;Thus, we know that &lt;span class=&quot;math-container&quot;&gt;$S=1$&lt;/span&gt;&lt;/h2&gt;&#xA;&lt;p&gt;From &lt;span class=&quot;math-container&quot;&gt;$S_1=2N+U$&lt;/span&gt;, we know that &lt;span class=&quot;math-container&quot;&gt;$U$&lt;/span&gt; must be odd.  Also, &lt;span class=&quot;math-container&quot;&gt;$N\notin\{0,5\}$&lt;/span&gt; since either result in &lt;span class=&quot;math-container&quot;&gt;$U=S$&lt;/span&gt;.  Also, since &lt;span class=&quot;math-container&quot;&gt;$S=1$&lt;/span&gt; there must be a carry over so &lt;span class=&quot;math-container&quot;&gt;$C_1 \in \{1,2\}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;We can see &lt;span class=&quot;math-container&quot;&gt;$S_5=N+B+D+C_4$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D_5=D$&lt;/span&gt;.  Since &lt;span class=&quot;math-container&quot;&gt;$C_5=1$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$N+B+C_4=10$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Similarly, &lt;span class=&quot;math-container&quot;&gt;$S_4=A+I+O+C_3$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D_4=I$&lt;/span&gt;, so &lt;span class=&quot;math-container&quot;&gt;$A+O+C_3=10$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_4=1$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Lastly, &lt;span class=&quot;math-container&quot;&gt;$S_3=K+W+O+C_2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D_3=O$&lt;/span&gt;, so &lt;span class=&quot;math-container&quot;&gt;$K+W+C_2=10$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_3=1$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Since &lt;span class=&quot;math-container&quot;&gt;$C_3=1$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$S_4=A+O+C_3=10 \implies A+O=9$&lt;/span&gt;. Since &lt;span class=&quot;math-container&quot;&gt;$C_5=1$&lt;/span&gt; we know;&lt;/p&gt;&#xA;&lt;h2&gt;Therefore &lt;span class=&quot;math-container&quot;&gt;$I=0$&lt;/span&gt;&lt;/h2&gt;&#xA;&lt;p&gt;Also, &lt;span class=&quot;math-container&quot;&gt;$N+B+C_4=10 \implies N+B=9$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Since &lt;span class=&quot;math-container&quot;&gt;$I=0$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$S_2=I+A+K+C_1=A+K+C_1$&lt;/span&gt;.  Thus, &lt;span class=&quot;math-container&quot;&gt;$C_2 \in \{0,1\}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Let go with the remaining odd values of &lt;span class=&quot;math-container&quot;&gt;$U$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If &lt;span class=&quot;math-container&quot;&gt;$U=3$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$N\in\{4,9\}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$B\in\{5,0\}$&lt;/span&gt;. 0 is already taken, so &lt;span class=&quot;math-container&quot;&gt;$N=4$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$B=5$&lt;/span&gt;.  Thus, &lt;span class=&quot;math-container&quot;&gt;$C_1=1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$S_2=A+K+C_1=A+K+1$&lt;/span&gt; yields 3.  Thus, &lt;span class=&quot;math-container&quot;&gt;$A+K=12$&lt;/span&gt;.  But no remaining values allow for unique &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;If &lt;span class=&quot;math-container&quot;&gt;$U=7$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$N=2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$B=7$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;So, &lt;span class=&quot;math-container&quot;&gt;$U\in\{5,9\}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;h3&gt;Assume &lt;span class=&quot;math-container&quot;&gt;$U=5$&lt;/span&gt;&lt;/h3&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$N\in\{2,7\}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$B\in\{7,2\}$&lt;/span&gt;. If &lt;span class=&quot;math-container&quot;&gt;$N=2$&lt;/span&gt; than &lt;span class=&quot;math-container&quot;&gt;$C_1=1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$A+K+C_1=A+K+1$&lt;/span&gt; yields 5.  Thus, &lt;span class=&quot;math-container&quot;&gt;$A+K\in\{4,14\}$&lt;/span&gt;.  4 doesn't work, but 14 needs &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; to be &lt;span class=&quot;math-container&quot;&gt;$\{6,8\}$&lt;/span&gt;. Also, &lt;span class=&quot;math-container&quot;&gt;$C_2=1$&lt;/span&gt;, so &lt;span class=&quot;math-container&quot;&gt;$K+W+1=10 \implies K+W=9$&lt;/span&gt;.  Thus, &lt;span class=&quot;math-container&quot;&gt;$W\in\{3,1\}$&lt;/span&gt;, but 1 is already taken, so &lt;span class=&quot;math-container&quot;&gt;$W=3$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$K=6$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$A=8$&lt;/span&gt;.  But &lt;span class=&quot;math-container&quot;&gt;$A=8 \implies O=2$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;h2&gt;Thus &lt;span class=&quot;math-container&quot;&gt;$U=9$&lt;/span&gt;&lt;/h2&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$N\in\{1,6\}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$B\in\{8,3\}$&lt;/span&gt;.  But &lt;span class=&quot;math-container&quot;&gt;$S=1$&lt;/span&gt;, so &lt;span class=&quot;math-container&quot;&gt;$N=6$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$B=3$&lt;/span&gt;.  Thus, &lt;span class=&quot;math-container&quot;&gt;$C_1=2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$S_2=A+K+C_1=A+K+2$&lt;/span&gt; yields 9 and &lt;span class=&quot;math-container&quot;&gt;$C_2=0$&lt;/span&gt;.  We can't get 19 with the remaining values, so it must be &lt;span class=&quot;math-container&quot;&gt;$A+K+2=9 \implies A+K=7$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The only values for &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; are &lt;span class=&quot;math-container&quot;&gt;$\{2,5\}$&lt;/span&gt;.  Since &lt;span class=&quot;math-container&quot;&gt;$K+W+C_2=10$&lt;/span&gt;, we know that &lt;span class=&quot;math-container&quot;&gt;$K+W=10$&lt;/span&gt;, so &lt;span class=&quot;math-container&quot;&gt;$K\ne 5$&lt;/span&gt;. Thus, &lt;span class=&quot;math-container&quot;&gt;$A=5$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$O=4$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$K=2$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$W=8$&lt;/span&gt;. Leaving only &lt;span class=&quot;math-container&quot;&gt;$D=7$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt; 565206&#xA; 430856&#xA;+ 74429&#xA;-------&#xA;1070491&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-03-04T15:07:51.510" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28341" PostTypeId="2" ParentId="28293" CreationDate="2016-03-04T15:13:11.853" Score="7" Body="&lt;p&gt;I'd think this one would be obvious.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Outside the window was another small enclosed space, containing a fan. The fan was blowing SO HARD that it broke the window and knocked the glass of water over. Alice and Bob were unfortunate victims of &lt;a href=&quot;https://en.wikipedia.org/wiki/Fan_death&quot; rel=&quot;nofollow&quot;&gt;fan death&lt;/a&gt;. When in doubt... Fan death.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alternative answer to the question &quot;How did Alice and Bob die?&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mysteriously.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19818" LastEditorUserId="19818" LastEditDate="2016-03-04T18:35:07.627" LastActivityDate="2016-03-04T18:35:07.627" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28342" PostTypeId="1" AcceptedAnswerId="28347" CreationDate="2016-03-04T15:18:49.527" Score="13" ViewCount="3274" Body="&lt;p&gt;Sir Isaac Newton's book &quot;Arithmetica Universalis&quot; contains the following famous puzzle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In $4$ weeks, $12$ cows graze bare $3\frac13$ acres of pasture land, and &#xA;  in $9$ weeks, $21$ cows graze bare $10$       acres of pasture land.&#xA;  Accounting for the uniform growth rate of grass and assuming equal &#xA;  quantities of grass per acre when the pastures are put into use, how &#xA;  many cows will it take to graze bare $24$ acres of pasture land in a&#xA;  period of $18$ weeks?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nowadays, some routine knowledge of highschool algebra suffices to find &#xA;the answer to Sir Isaac's puzzle: It will take $36$ cows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This puzzle asks you to develop the general solution for general grazing cow puzzles of this type:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Sir Isaac Newton's generalized puzzle:&lt;/strong&gt; Assume that&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;it takes $c_1$ cows to graze bare $a_1$ acres of pasture in $w_1$ weeks; &lt;/li&gt;&#xA;  &lt;li&gt;it takes $c_2$ cows to graze bare $a_2$ acres of pasture in $w_2$ weeks; &lt;/li&gt;&#xA;  &lt;li&gt;it takes $c_3$ cows to graze bare $a_3$ acres of pasture in $w_3$ weeks.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;What simple algebraic relation does there exist between the nine &#xA;  quantities $c_1,c_2,c_3$ and $a_1,a_2,a_3$ and $w_1,w_2,w_3$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer can either be given as an ugly equation (by plodding and &#xA;laborious work), or in terms of a simple and elegant determinant &#xA;(by mathematical insight).&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-18T15:56:39.840" LastActivityDate="2018-05-10T09:07:11.910" Title="The grazing cows of Sir Isaac Newton" Tags="|mathematics|calculation-puzzle|algebra|puzzle-history|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28343" PostTypeId="1" CreationDate="2016-03-04T15:46:40.440" Score="2" ViewCount="674" Body="&lt;p&gt;I read through the wiki on the &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Unexpected_hanging_paradox&quot; rel=&quot;nofollow noreferrer&quot;&gt; unexpected hanging paradox &lt;/a&gt;.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A judge tells a condemned prisoner that he will be hanged at noon on one weekday in the following week but that the execution will be a surprise to the prisoner. He will not know the day of the hanging until the executioner knocks on his cell door at noon that day.&lt;/p&gt;&#xA;&lt;p&gt;Having reflected on his sentence, the prisoner draws the conclusion that he will escape from the hanging. His reasoning is in several parts. He begins by concluding that the &amp;quot;surprise hanging&amp;quot; can't be on Friday, as if he hasn't been hanged by Thursday, there is only one day left - and so it won't be a surprise if he's hanged on Friday. Since the judge's sentence stipulated that the hanging would be a surprise to him, he concludes it cannot occur on Friday.&lt;/p&gt;&#xA;&lt;p&gt;He then reasons that the surprise hanging cannot be on Thursday either, because Friday has already been eliminated and if he hasn't been hanged by Wednesday night, the hanging must occur on Thursday, making a Thursday hanging not a surprise either. By similar reasoning he concludes that the hanging can also not occur on Wednesday, Tuesday or Monday. Joyfully he retires to his cell confident that the hanging will not occur at all.&lt;/p&gt;&#xA;&lt;p&gt;The next week, the executioner knocks on the prisoner's door at noon on Wednesday — which, despite all the above, was an utter surprise to him. Everything the judge said came true.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I found the most of the arguments rely on the following facts.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The prisoner is a perfect logician.&lt;/li&gt;&#xA;&lt;li&gt;The judge is a perfect logician.&lt;/li&gt;&#xA;&lt;li&gt;The judge is truthful.&lt;/li&gt;&#xA;&lt;li&gt;Statements 1, 2 and 3 are common knowledge.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Is there any reason these are not included as part of the problem statement?&lt;/p&gt;&#xA;" OwnerUserId="4269" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2017-08-17T00:40:01.197" Title="Clarifications on the unexpected hanging paradox" Tags="|logical-deduction|puzzle-creation|explanation|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28344" PostTypeId="2" ParentId="28293" CreationDate="2016-03-04T15:50:50.540" Score="-3" Body="&lt;p&gt;Alice and Bob's last work assignment was participating at a physics experiment at CERN, and unbeknownst to them,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a tiny black hole had gotten stuck in Bob's throat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It slowly grew, until Bob sneezed. The extra velocity and the glass of water it hit caused it to grow rapidly, sucking out all the air in the room. Unfortunately Alice and Bob suffocated before the window burst in.&lt;/p&gt;&#xA;" OwnerUserId="12604" LastActivityDate="2016-03-04T15:50:50.540" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28345" PostTypeId="2" ParentId="28310" CreationDate="2016-03-04T16:34:10.027" Score="0" Body="&lt;p&gt;Is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Line&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To get support you may wait in a line or call on a phone line. &#xA;&#xA; Knots in a line (rope). &#xA;&#xA; Adding AG (silver) to line you get AliGne (align) which is something you'd do to a picture or painting when putting it in a frame. &#xA;&#xA; If you're waiting in a line you must keep patient. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm starting to think it may have something to do with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A ring or marriage or divorce or something. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19731" LastEditorUserId="19731" LastEditDate="2016-03-04T17:49:53.473" LastActivityDate="2016-03-04T17:49:53.473" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28346" PostTypeId="1" AcceptedAnswerId="28354" CreationDate="2016-03-04T17:07:27.670" Score="0" ViewCount="539" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Every friday, Bob is given a class test. The teacher doesn't check the students' answers by himself because there are 96 students in the class and he has other classes to manage as well. Instead, the teacher distributes those answer copies to other students in the same class such that no student gets his own copy. The teacher has a list according to which he distributes the copies. For example, Phineas always gets the copy of Ferb. And Ferb always gets the copy of Isabella. None of the students know who gets whose copy.  That is, neither Phineas knows that he is checking the copy of Ferb, nor does Ferb know that his copy is being checked by Phineas. However, it seems like someone is having fun with Bob because Bob gets deducted marks even if his answers are correct.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How can Bob find out about the classmate who checks his copy every time, without letting the teacher know about it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some things to remember:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I took 96 arbitrarily.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I am not asking who. I am asking &lt;strong&gt;how&lt;/strong&gt; can he find out about the classmate?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Isabella, Phineas and Ferb are just for examples. You can ignore them.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Just in case you think that I am Bob then I would say nope. This question just came to my mind accidentally.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17250" LastEditDate="2016-03-04T18:17:43.033" LastActivityDate="2016-03-21T19:56:40.420" Title="Who is having fun with Bob" Tags="|lateral-thinking|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28347" PostTypeId="2" ParentId="28342" CreationDate="2016-03-04T17:15:32.840" Score="8" Body="&lt;p&gt;First things first: we will declare some &lt;strong&gt;constants&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\text{COW_GRASS_PER_WEEK}=$ quantity of grass eaten by a cow in one week&lt;/li&gt;&#xA;&lt;li&gt;$\text{ACRE_GRASS_PER_WEEK}=$ quantity of grass that grows in one acre in one week&lt;/li&gt;&#xA;&lt;li&gt;$\text{INITIAL_GRASS_PER_ACRE}=$ initial quantity of grass per acre&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We will call them respectively $CGW$, $AGW$ and $IGA$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;variables&lt;/strong&gt; are going to be the same as in the question:&#xA;$c_1,c_2,c_3$ for the number of cow, &#xA;$a_1,a_2,a_3$ for the number of acres and &#xA;$w_1,w_2,w_3$ for the number of weeks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To resolve the example problem we have: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$c_1=12$, $a_1=\frac{10}{3}$,$ w_1=4$  &lt;/li&gt;&#xA;&lt;li&gt;$c_2=21$, $a_2=10$, $w_2=9$  &lt;/li&gt;&#xA;&lt;li&gt;$c_3=X $, $a_3=24$, $w_3=18$ &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We want the amount of grass eaten by the cows to be the same as &#xA;the amount of grass that grew in a fixed time period plus the initial &#xA;amount, so that we have the generalized equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c\cdot w\cdot CGW=a\cdot w\cdot AGW + a\cdot IGA$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our case:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$12\cdot4\cdot CGW=\frac{10}{3}\cdot4\cdot AGW + \frac{10}{3}\cdot IGA$,   &lt;/li&gt;&#xA;&lt;li&gt;$21\cdot9\cdot CGW=10\cdot9\cdot AGW + 10\cdot IGA$ and  &lt;/li&gt;&#xA;&lt;li&gt;$X\cdot18\cdot CGW=24\cdot18\cdot AGW + 24\cdot IGA$  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;which leads us to $CGW=\frac{10}{9}AGW$ and $IGA=12\cdot AGW$.&#xA;After substituting we get $20\cdot X \cdot AGW = 720 \cdot AGW$, &#xA;which finally implies the answer $X=36$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now how can we generalize this? &#xA;Here is where &lt;strong&gt;matrices&lt;/strong&gt; come into play!&#xA;We have 3 equations and 3 variables of the form:&#xA;$$c_i\cdot w_i\cdot CGW=a_i\cdot w_i\cdot AGW + a_i\cdot IGA$$ &#xA;respectively &#xA;$$a_i\cdot w_i\cdot AGW + a_i\cdot IGA - c_i\cdot w_i\cdot CGW = 0.$$&#xA;The resulting linear equation system has the following coefficient matrix:&#xA;$$\begin{bmatrix}a_1\cdot w_1 &amp;amp; a_1 &amp;amp; -c_1\cdot w_1\\a_2\cdot w_2 &amp;amp; a_2 &amp;amp; -c_2\cdot w_2\\a_3\cdot w_3 &amp;amp; a_3 &amp;amp; -c_3\cdot w_3\end{bmatrix}.$$&lt;br&gt;&#xA;If the determinant of this matrix was nonzero, then the matrix would &#xA;be invertible and we would have a unique solution for $CGW$, $AGW$, &#xA;and $IGA$. But the solution isn't unique, as we may scale all three &#xA;values by the same factor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently, the determinant must be zero, and we get the condition&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;\begin{eqnarray}&#xA;D &amp;amp;~=~&amp;amp; &#xA;a_2\cdot a_3\cdot w_1\cdot w_3\cdot c_1 +&#xA;a_1\cdot a_2\cdot w_2\cdot w_3\cdot c_3 +&#xA;a_1\cdot a_3\cdot w_1\cdot w_2\cdot c_2  &#xA;\\ &amp;amp;&amp;amp; {} -&#xA;a_1\cdot a_2\cdot w_1\cdot w_3\cdot c_3 -&#xA;a_1\cdot a_3\cdot w_2\cdot w_3\cdot c_2 -&#xA;a_2\cdot a_3\cdot w_1\cdot w_2\cdot c_1 ~=~ 0&#xA;\end{eqnarray}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As this equation is linear in each of the nine values &#xA;$c_1,c_2,c_3$ and $a_1,a_2,a_3$ and $w_1,w_2,w_3$, &#xA;it easily allows to compute any of these nine values from the other eight values.&lt;/p&gt;&#xA;" OwnerUserId="4652" LastEditorUserId="8874" LastEditDate="2016-03-09T10:07:08.173" LastActivityDate="2016-03-09T10:07:08.173" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28348" PostTypeId="2" ParentId="28346" CreationDate="2016-03-04T17:16:16.673" Score="2" Body="&lt;p&gt;Possible answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He could put his name in an answer and threaten the person. Then wait for the day after grading and see who's avoiding him and keeping an eye out for him. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He could say give me an A and I'll pay you $100 and see who comes to claim it. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19731" LastActivityDate="2016-03-04T17:16:16.673" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28351" PostTypeId="1" CreationDate="2016-03-04T17:34:16.980" Score="4" ViewCount="506" Body="&lt;p&gt;Here is the puzzle that has been torturing me for past few hours... I still have not come up with logical solution.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/5oY3h.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5oY3h.jpg&quot; alt=&quot;here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19831" LastEditorUserId="8874" LastEditDate="2016-03-08T17:47:18.820" LastActivityDate="2016-03-08T17:47:18.820" Title="Words logic puzzle" Tags="|logical-deduction|pattern|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28352" PostTypeId="2" ParentId="28351" CreationDate="2016-03-04T17:43:16.123" Score="9" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  82617&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first thing you need to realize is that the numbers on the right are not lined up with the items on the left.  So MERT does not equal 3174.  However, if you look closely at the words, you can see a couple of patterns...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The 2nd letter is limited to E and A, and the 3rd letter is limited to R and S.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By looking at the characters and numbers in those positions, you can see...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That there are 2 A's and 3 E's in the second position of the text, and 2 1's and 3 2's I the second position of the numbers, so there appears to be a direct correlation.  1 = A, 2 = E, and doing the same for the 3rd position, we get 6 = R and 7 = S.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At this point, you don't even need to continue to find the letters, because there is only one of the multiple choice items that fits the pattern once you convert TERAS to the number scheme.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ?2617, which is B.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, if you do continue to expand out, you will finally find that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  5 = M, 3 = F, 4 = K, and 8 = T, just by process of elimination. 9 has no corollary in this puzzle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14233" LastEditorUserId="14233" LastEditDate="2016-03-04T17:49:48.330" LastActivityDate="2016-03-04T17:49:48.330" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28353" PostTypeId="2" ParentId="28224" CreationDate="2016-03-04T17:48:02.400" Score="4" Body="&lt;p&gt;&lt;strong&gt;Sorry guys - had to write this.  It isn't any different from the accepted answer, but was fun.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...You smile as you fix your strategy in mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suddenly, Gryffindor's egghead friend pipes up, a worried look on her face.  &quot;Professor Dumbledore, sir.  It is not that I want to question your magic, but I am still concerned about the muggle physics and would not wish to see us all destroyed in the energy release.  In the interests of safety, can we avoid the spell and simply argue why our strategy will work?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dumbledore shrugs his shoulders, clearly not interested since his young protégé's friend has obviously figured out the problem and the result is no longer in doubt.  &quot;If young Hufflepuff agrees...&quot; he says reluctantly putting away his wand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have no wish to see your hands move that fast anyway, and wondering if you'd actually live an infinite lifetime during that crazy spell, forced to move balls forever, you agree to her proposal. &quot;Fine&quot;, you say.  &quot;Since I get to go first, my strategy would be...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But egghead quickly interrupts.  Now that she has the answer, she is quite excited to reveal it and will not wait for your dimwitted pandering.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;&lt;strong&gt;&lt;em&gt;Our&lt;/em&gt;&lt;/strong&gt; strategy is to simply take the lowest ball in Hufflepuffs's track.&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;She seems to also be unaware of your name. &lt;strong&gt;&lt;em&gt;How come no one knows who I am?&lt;/em&gt;&lt;/strong&gt; you wonder to yourself.  &lt;strong&gt;&lt;em&gt;Is a simply memory charm beyond these people?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Egghead continues without pause:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;&lt;strong&gt;&lt;em&gt;I&lt;/em&gt;&lt;/strong&gt; can prove that every ball in the his urn must eventually end up in our urn.  Thus, our urn has all the balls and Hufflepuff's must therefore be empty.&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While most the crowd cannot make sense of the answer, a few Ravenclaws are seen to be nodding sagely.  Dumbledore smiles; &quot;Well done!  I award 100 points to Gryffindor for outstanding use of logic!&quot;  There is a smattering of applause from the Gryffindor crowd, but the rest of the gathered students and teachers hang their heads in collective defeat.  &quot;This year's house cup,&quot; he continues, &quot;is once again is awarded to...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Excuse me Professor Dumbledore&quot;, you interject.  He pauses his delivery, clearly forgetting you were even there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yes, young Hufflepuff?  Do you have something to add?&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Actually, sir, I haven't had a chance to reveal my strategy sir.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;But haven't you heard young egghe.. I mean Gryffindor's strategy?  Can't you see that for any numbered ball you believe to be left in your cup, it will eventually end up in Gryffindor's cup, so at the end of the spell, your urn must be empty?&quot;  Egghead, smirks, clearly thinking along the same lines as Dumbledore, that you are confounded by infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Well, yes sir, I do see that, but if I may...&quot; you start. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Do you need me to cast my spell so that you may witness the result?&quot;  He asks pulling our his wand from beneath his robes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This time, both you and egghead jump in to stop him and both say &quot;That will not be necessary!&quot; over each other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Al-right&quot;, he says with a sigh, again reluctantly putting away his wand.  &quot;Now, if I may continue with the presentation of the house cup?&quot;  He asks, politely waiting for you to surrender the contest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I just would like to say my strategy if you don't mind first professor.&quot; you say demurely. &quot;I think you will find it interesting.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Go on if you must&quot;, he says, patiently waiting for you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;As I was going to say professor, my strategy would be to simply leave the first two balls.  I will take the lowest of the balls numbered 2 or higher.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You pause and look to egghead to see if she has figured it out.  Her puzzled look soon melts into understanding as she realizes your strategy.   Dumbledore, however, is still in doubt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yes, yes, and all those balls would end up in Gryffindor's urn, would they not?&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yes professor.  But the 0 and 1 would be the only balls remaining in the original urn.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;But every ball you put in your urn will eventually be taken and put into the Gryffindor urn correct?&quot; he says, getting impatient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Yes, except for the last.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;But there is no last!  We are talking about &lt;strong&gt;&lt;em&gt;all&lt;/em&gt;&lt;/strong&gt; the numbers!  Blast! This is not getting us anywhere.&quot; he says taking out his wand again.  &quot;I see I have no choice but to cast the spell and prove you are incorrect young Hufflepuff.&quot;  He begins the spell and you look over to egghead.  She is clearly distraught, but cannot stop the crazy old wizard from destroying the universe.  You have just enough time to concentrate on your strategy when he completes the spell and you feel a tightness in your chest as light washes all over you.  As it dissipates, the contest begins of its own accord.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You watch like a spectator as your hands calmly take the 2 and 3 out of the main urn and put them in your own.  You see scarhead take the 2 and put it in his urn.  the whole process takes less than 10 seconds.  Then your hands move of their own accord and take the 4 and 5, much quicker this time.  Four-eyes takes the 3, equally as fast.  By the time your hands are reaching for the 12 and 13, you can no longer keep up and close your eyes.  Even through your closed lids, you can see that your hands are moving so fast they are starting to glow.  You begin to wonder if it will ever end, and the entire school will explo... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;...and it is over.  You look down and your hands are smoking, but not in too much pain.  The original urn is also smoking, but when you look in the top, it is empty.  There are only 2 balls remaining in the track.   The 0 and the 1.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your urn is also empty.  You glance at Gryffindor's urn and see him also looking at his smoking hands in wonder.  He too peers in own urn and smiles.  The track contains 2-11 and is full.  Egghead is not so pleased; being quicker than her friend, she has seen where you are headed.  Unfortunately, Dumbledore has not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;As you can see, your urn is empty, and Gryffindor's is full.  While I appreciate the opportunity to practice my magic, I assume you will now admit defeat so that we can present the cup?&quot;  His is very proud of his skills with the Zeno's Accelerator spell and was happy to show it off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;But sir,&quot; you say quietly, &quot;the game is not over&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Whatever do you mean?&quot; he asks. &quot;Your urn is empty and his is full.&quot;  He is confused by your stubborness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The main urn is not empty professor.  I still have a move available to me.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He now sees for the first time that there is a 0 and 1 available.  &quot;Go on then, make your last move.&quot;  As wonderous a wizard he is, he has not seen the eventual outcome.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Gryffindor crowd, however has.  You take the final two balls and place them in your urn.  Egghead is hanging her head while Dumbledore's favourite takes the 0 from your urn and places it in his own.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I can no longer make any moves professor.&quot; you announce.  Finally, Dumbledore sees what has happened.  Your urn contains a single ball labelled 1.  The Gryffindor urn contains every other possible ball, but not the 1.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Well, young Hufflepuff,&quot; Dumbledore finally looks at you with more than passing interest.  &quot;I see you have outwitted Gryffindor in this test.  Congratulations!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crowd erupts!  Even some of the Gryffindor students are openly cheering for you!  As you get pats on the back and handshakes from many kids you never knew, Dumbledore approaches you.  &quot;What is your name again young man?&quot; he asks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Cedric, sir&quot; you say.  &quot;Cedric Diggory.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Well, congratulations Cedric.  You have made your house proud today, and I am certain you will do so again in the near future.&quot;  His eyes twinkle knowingly and you wonder what he has in store for you...&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2016-03-04T17:48:02.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28354" PostTypeId="2" ParentId="28346" CreationDate="2016-03-04T17:51:08.743" Score="13" Body="&lt;p&gt;He can &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Write his answers upside-down, and watch to see who turns the paper in order to read. Whoever does will be invited to have a &quot;strident discussion of proper grading procedure&quot; with Bob after school. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16874" LastActivityDate="2016-03-04T17:51:08.743" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28355" PostTypeId="2" ParentId="28342" CreationDate="2016-03-04T18:00:57.893" Score="4" Body="&lt;p&gt;We have $a_i$ acres of land, $c_i$ cows and $w_i$ weeks. Since the grass grows uniformly over the field, the $a_i$ also measures the amount of grass in, lets say, acre-equivalent mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now lets denote with $f$ the cow feed rate in acre-equivalent/week and with $g$ the grass growth rate in acre-equivalent/week.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want the cows to eat all the grass just at the end of our time frame. So we can write:&#xA;$$ a_i + a_i w_i g - c_i w_i f = 0$$&#xA;The first term is the amount of grass available at the beginning, the second term describes the growth of grass and the third term the amount of grass the cows eat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As per the problem, we have three sets of quantities and so three equations:&#xA;$$ a_1 + a_1 w_1 g - c_1 w_1 f = 0 $$&#xA;$$ a_2 + a_2 w_2 g - c_2 w_2 f = 0 $$&#xA;$$ a_3 + a_3 w_3 g - c_3 w_3 f = 0 $$&#xA;Lets assume that the first two sets are given, like in the example above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the first equation we can now express the grass growth rate:&#xA;$$ g = \frac{c_1 w_1 f - a_1}{a_1 w_1} $$&#xA;This can now be inserted into the second equation, from where we can get the cow feed rate:&#xA;$$ f = \frac{a_2 - a_1 \frac{a_2 w_2}{a_1 w_1}}{c_2 w_2 - c_1 w_1 \frac{a_2 w_2}{a_1 w_1}} $$&#xA;With this, both of the parameters are fixed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the quantity missing from the third set, we have to use one of the below equations:&#xA;$$ a_3 = \frac{c_3 w_3 f}{1 + w_3 g} $$&#xA;$$ c_3 = \frac{a_3 + a_3 w_3 g}{w_3 f} $$&#xA;$$ w_3 = \frac{a_3}{c_3 f - a_3 g} $$&lt;/p&gt;&#xA;" OwnerUserId="19749" LastActivityDate="2016-03-04T18:00:57.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28356" PostTypeId="2" ParentId="28346" CreationDate="2016-03-04T18:10:33.333" Score="4" Body="&lt;p&gt;There might be many legit ways to find out who marks Bob's paper but here's 2:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1) Paint the paper top red (or shed a corner, or any altering to the paper to make it outstanding to Bob but unnoticeable to the teacher)&lt;br&gt;&#xA; 2) Submit an empty test;  Person who marks an empty test will instantly marks 0 and hand in the paper quickly&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I have to say, Bob = Manshu.  &lt;/p&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2016-03-04T18:10:33.333" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28359" PostTypeId="2" ParentId="28322" CreationDate="2016-03-04T18:36:35.253" Score="5" Body="&lt;p&gt;I tried, I really did, but I can't make myself do alphametics. I did think of another solution to the riddle part which I am willing to share with anyone interested. It is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mg + Ga yields GaMg or Gamgee, a name from the Lord Of the Rings books. Gaffer and Sam are, as far as I recall, the only named members of the Gamgee clan. Gaffer is clearly too long so I would propose G A + M G = S A M as the equation. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am using the mobile app and can't figure out how to make this a wiki post.&lt;/p&gt;&#xA;" OwnerUserId="18924" LastActivityDate="2016-03-04T18:36:35.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28360" PostTypeId="1" AcceptedAnswerId="28763" CreationDate="2016-03-04T19:27:57.997" Score="9" ViewCount="1355" Body="&lt;p&gt;I'll start to count while you solve this problem,&lt;br&gt;&#xA;and just relax here at the top of my column. &lt;br&gt;&#xA;Everywhere I go I let out a good whistle, &lt;br&gt;&#xA;but don't mess with me cause I carry a pistol. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically speaking I reach senses of three, &lt;br&gt;&#xA;when you count those senses, you may find me. &lt;br&gt;&#xA;For if you hold me, I'm likely dead, &lt;br&gt;&#xA;why did you take me away from my bed? &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Out of the rainbow I cannot see blue, &lt;br&gt;&#xA;and my name can be split into two. &lt;br&gt;&#xA;The first part of me is a computer's worst dream, &lt;br&gt;&#xA;while the second one comes apart at the seam.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back in the 30's I caused a great craze, &lt;br&gt;&#xA;not one single chap did I cease to amaze. &lt;br&gt;&#xA;Everyone loved me and wanted a kiss, &lt;br&gt;&#xA;but then in the 40's screamed &quot;What did I miss?&quot; &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I came with a flash and out like a light, &lt;br&gt;&#xA;but I don't really mind because life is so bright. &lt;br&gt;&#xA;I've been explosive at one point in history, &lt;br&gt;&#xA;what I truly am is not much of a mystery. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 4. Good Luck!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint... Every line contains at least one hint (except for the last) so if you find something that doesn't fit a line, it may not be right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional Hint... a lot of the hints in the poem reference a part of the word.&lt;/p&gt;&#xA;" OwnerUserId="19731" LastEditorUserId="19731" LastEditDate="2016-03-10T20:33:53.243" LastActivityDate="2023-05-16T09:34:37.423" Title="Unstoppable force" Tags="|riddle|word|wordplay|rhyme|poetry|" AnswerCount="12" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="28361" PostTypeId="2" ParentId="28310" CreationDate="2016-03-04T20:30:27.733" Score="1" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Pane&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likely you'll find me when lending support&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;pane- door&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or shedding some knots by a slip of some sort&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;pane- strip of cloth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or adding the silver before fitting the frame&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;pane- mirror&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Keep calm and breathe is the key to this game.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;pane- piece- peace&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;LOOK &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;window pane&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19834" LastActivityDate="2016-03-04T20:30:27.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28362" PostTypeId="1" AcceptedAnswerId="41980" CreationDate="2016-03-04T21:16:29.270" Score="6" ViewCount="464" Body="&lt;p&gt;Inspired by &lt;a href=&quot;https://puzzling.stackexchange.com/questions/28254/a-game-with-52-cards&quot;&gt;A game with 52 cards&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Alice and Bob play the following game with two (identical) standard decks of $52$ cards.&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;First Alice secretly arranges one deck of $52$ cards in a long row on the table. All cards are face-down, and Bob has no knowledge of the positions of any of the cards.&lt;/li&gt;&#xA;  &lt;li&gt;Then the game goes through several rounds.&#xA;  In every round, Bob first pays $1$ Euro to Alice.&#xA;  Then Bob arranges the second deck of cards in a long row on the table, parallel to Alice's row and with all the cards face-up.&#xA;  Then Alice tells Bob all the face-up cards in his row that agree with the corresponding face-down cards in her row.&#xA;  Then the next round starts.&lt;/li&gt;&#xA;  &lt;li&gt;The game ends, as soon as Bob knows the positions of all the face-down cards in Alice's row. &#xA;  Bob then receives $x$ Euros from Alice.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The original problem asked for the minimum risk-free wager. Instead: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the minimum payout $x$ for Bob to break even on average?&lt;/p&gt;&#xA;" OwnerUserId="19843" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2016-09-04T03:29:23.710" Title="Corollary to &quot;A game with 52 cards&quot;" Tags="|mathematics|probability|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28363" PostTypeId="1" AcceptedAnswerId="28366" CreationDate="2016-03-04T21:34:26.190" Score="10" ViewCount="535" Body="&lt;p&gt;Is there an algorithm that you can apply &lt;em&gt;twice&lt;/em&gt; to an input string of $0$s and $1$s in order to reverse it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, is there a function $f : \mathrm B \to \mathrm B$ on the set of binary strings $\mathrm B = \{\varepsilon, 0, 1, 00, 01, 10, 11, 000, ...\}$ so that $f(f(b)) = \mathrm{reverse}(b)$ for all $b \in \mathrm B$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, describe such a function/algorithm. If not, prove why it doesn’t exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;strong&gt;BONUS&lt;/strong&gt;: for which other positive integers $n$ is there an algorithm that reverses a string when applied $n$ consecutive times?)&lt;/p&gt;&#xA;" OwnerUserId="11476" LastActivityDate="2016-03-04T22:44:56.740" Title="Half-reverse operation on binary strings" Tags="|mathematics|algorithm|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28364" PostTypeId="2" ParentId="28268" CreationDate="2016-03-04T21:43:45.353" Score="2" Body="&lt;p&gt;My take on it is you would change the counting base from decimal (base 10), so 7 * 7 = 49 in decimal, which is 25 in base 22 ((22 * 2) + 5).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus 6 * 6 = 36 in b10, which is 18 in b28, and 9 * 9 = 81 in b10, which is 41 in b20.&lt;/p&gt;&#xA;" OwnerUserId="19847" LastActivityDate="2016-03-04T21:43:45.353" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28365" PostTypeId="1" AcceptedAnswerId="28398" CreationDate="2016-03-04T21:46:36.380" Score="11" ViewCount="207" Body="&lt;pre&gt;&lt;code&gt;Older than history, as young as today, &#xA;I was here before you, and I'm here to stay.&#xA;&#xA;My fathers were high, my mother had might&#xA;I'm beloved of nobles, but never by right. &#xA;&#xA;Conceived in the garden, I walk in all weather&#xA;then disappear all black in a flutter of feather. &#xA;&#xA;I dally with dahlias, I rib with the rose,&#xA;I enter the forest surrounded by does. &#xA;&#xA;I'm awarded degrees, yet not educated, &#xA;they judge me as worst among the related. &#xA;&#xA;You always see me but hope we don't meet&#xA;I never respect you, but if you scream, I will bleed. &#xA;&#xA;I'll arrive in a glass, or I'll arrive in a wave&#xA;and then my sad ballads will emerge from the cave. &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;" OwnerUserId="2811" LastEditorUserId="2811" LastEditDate="2016-03-04T22:02:41.003" LastActivityDate="2016-03-05T13:20:29.840" Title="I only exist for a reason" Tags="|riddle|rhyme|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28366" PostTypeId="2" ParentId="28363" CreationDate="2016-03-04T21:58:09.987" Score="12" Body="&lt;p&gt;Consider the set of binary strings that are lexicographically earlier than their reverses. For example, &lt;code&gt;011&lt;/code&gt;, &lt;code&gt;1011&lt;/code&gt;, and &lt;code&gt;11010111&lt;/code&gt; are all members of this set, because &lt;code&gt;011&lt;/code&gt; is earlier than &lt;code&gt;110&lt;/code&gt;, &lt;code&gt;1011&lt;/code&gt; is earlier than &lt;code&gt;1101&lt;/code&gt;, and &lt;code&gt;11010111&lt;/code&gt; is earlier than &lt;code&gt;11101011&lt;/code&gt;. We can order this set by length, followed by lexicographical order. Let the first member (&lt;code&gt;01&lt;/code&gt;) be $c_1$, the second member (&lt;code&gt;001&lt;/code&gt;) be $c_2$, the third member (&lt;code&gt;011&lt;/code&gt;) be $c_3$, and so on. Let $d_i=\text{reverse}(c_i)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an integer $n$, define $f_n$ as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For every integer $i$ not a multiple of $n$, $f_n(c_i)=c_{i+1}$ and $f_n(d_i)=d_{i+1}$.&lt;/li&gt;&#xA;&lt;li&gt;For every integer $i$ that is a multiple of $n$, $f_n(c_i)=d_{i+1-n}$ and $f_n(d_i)=c_{i+1-n}$.&lt;/li&gt;&#xA;&lt;li&gt;Otherwise, $f_n(b)=b$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When $f_n$ is applied $n$ times, the result is the reverse of the input:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$f_n^{(n)}(c_i)=d_i$&lt;/li&gt;&#xA;&lt;li&gt;$f_n^{(n)}(d_i)=c_i$&lt;/li&gt;&#xA;&lt;li&gt;$f_n^{(n)}(b)=b$ for all $b$ that are not a $c_i$ or a $d_i$; the only such strings are the ones that are neither earlier than nor later than their reverses, so they are the same as their reverses.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2016-03-04T22:44:56.740" LastActivityDate="2016-03-04T22:44:56.740" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="28368" PostTypeId="1" CreationDate="2016-03-04T22:26:00.017" Score="0" ViewCount="350" Body="&lt;p&gt;You are an inmate in a concentration camp/prison. A sadistic guard is about to hang your son who tried to escape and wants you to pull the chair from underneath him. He says that if you don’t he will not only kill your son but some other innocent inmate as well. You don’t have any doubt that he means what he says. What will you do?&lt;/p&gt;&#xA;" OwnerUserId="18105" LastEditorUserId="17250" LastEditDate="2016-03-04T22:39:20.310" LastActivityDate="2016-03-04T22:49:14.420" Title="Two Sons, One Father" Tags="|riddle|lateral-thinking|" AnswerCount="1" CommentCount="1" ClosedDate="2016-03-05T02:20:47.063" ContentLicense="CC BY-SA 3.0" />
  <row Id="28369" PostTypeId="2" ParentId="28368" CreationDate="2016-03-04T22:49:14.420" Score="4" Body="&lt;p&gt;I think the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you will pull the chair out from underneath him, because it's under the guard, not the son.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19302" LastActivityDate="2016-03-04T22:49:14.420" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28370" PostTypeId="1" AcceptedAnswerId="28371" CreationDate="2016-03-04T22:49:54.363" Score="1" ViewCount="1312" Body="&lt;p&gt;3 couples on one side of a river. Aa Bb Cc (caps representing husbands, lowercase representing wives)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One boat in which 2 people can travel at a time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least one person must row the boat wherever it needs to be taken. The boat cannot travel to any side by water current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People can travel with someone of the other gender only if they are their spouse. No man or woman can travel with someone they are not married with. (This is what makes this puzzle different from &lt;a href=&quot;https://puzzling.stackexchange.com/questions/11450/n-couples-crossing-a-river&quot;&gt;$n$ couples crossing a river&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No woman can be left on a side with another man unless her husband is also present there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the method for all of them crossing the river?&lt;/p&gt;&#xA;" OwnerUserId="16342" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2016-12-06T01:12:11.000" Title="Crossing The River (Humans And Monsters Puzzle With A Twist)" Tags="|river-crossing|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28371" PostTypeId="2" ParentId="28370" CreationDate="2016-03-04T23:00:25.140" Score="2" Body="&lt;p&gt;Here goes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Aa cross over. A crosses back. &lt;br&gt;Departure side: A, Bb, Cc &lt;br&gt; Destination side: a&lt;br&gt;&lt;br&gt;Next, bc row over, c rows back alone. &lt;br&gt;Departure side: A, B, Cc &lt;br&gt; Destination side: ab&lt;br&gt;&lt;br&gt;AB row over together, Bb row back. &lt;br&gt;Departure side: Bb, Cc &lt;br&gt; Destination side: Aa&lt;br&gt;&lt;br&gt;Now the two men head over. BC to destination. Send a back. &lt;br&gt;Departure side: abc &lt;br&gt; Destination side: ABC&lt;br&gt;&lt;br&gt; Now it's simple. ab over, a back, then ac over and done!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2016-03-04T23:00:25.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28372" PostTypeId="2" ParentId="28254" CreationDate="2016-03-04T23:19:49.327" Score="4" Body="&lt;p&gt;If we can agree that not knowing a card is always less informative than knowing a card, then I guess we can show the lower bound too. I'm not exactly sure how obvious this assumption is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carl's suggested strategy maximizes information in the sense that in each turn there is no dead guess for any card. Even when guesses are wrong, on each turn and for each card at least one possibility is excluded. The only way to do better is by guessing some other card right (thus excluding more possibilities). Therefore, in the worst case scenario, on each turn and for each card we exclude exactly one possibility, so that after $51$ turns we are done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to think of this is as follows. &lt;strong&gt;Rather&lt;/strong&gt; than consider the worst case scenario for finding out the entire sequence of cards, consider the worst case scenario for finding out &lt;strong&gt;a single given card&lt;/strong&gt;, say the first. On each turn, we can exclude at least one possibility for the first card, so the worst case scenario cannot be worse than $51$ turns. On the other hand, the only way we can do better is if, before then, we guess some other card right &lt;strong&gt;and&lt;/strong&gt; that guess had not yet been made for the first card.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the worst case scenario, these correct guesses don't happen (here is where I've used the assumption). In other words, the worst case for guessing correctly the first card is $51$ turns. Surely, the worst case for guessing correctly the entire sequence can't be better, so it too must be $51$ turns.&lt;/p&gt;&#xA;" OwnerUserId="16478" LastActivityDate="2016-03-04T23:19:49.327" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28373" PostTypeId="2" ParentId="28310" CreationDate="2016-03-04T23:42:15.677" Score="6" Body="&lt;pre&gt;&lt;code&gt;Answer: &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eyes?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Likely you'll find me when lending support&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A friendly gaze&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or shedding some knots by a slip of some sort&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Running thread through the EYE of a needle, securing via SLIP-KNOT&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or adding the silver before fitting the frame&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You make a mirror, you're going to see your eyes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;LOOK&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; yup. With your eyes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19728" LastActivityDate="2016-03-04T23:42:15.677" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28378" PostTypeId="1" AcceptedAnswerId="28379" CreationDate="2016-03-05T02:16:16.403" Score="9" ViewCount="312" Body="&lt;p&gt;&lt;sub&gt;This was inspired by the octospiders in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Rendezvous_with_Rama&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rama series&lt;/a&gt; by Arthur C. Clarke. Reading the series is not a requirement for solving the puzzle and actually wouldn't help at all besides being able to say, &quot;I recognize that.&quot;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Nicole des Jardins comes across a subspecies from the octospiders' home planet. Although it uses color bands to communicate just like the octospiders, the language is completely different. As she's puzzling over the creature and trying to understand, Dr. Blue approaches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Do you understand?&quot; he asks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Not yet,&quot; replies Nicole, &quot;but I'm not one to give up easily.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;If I may offer some guidance...&quot; the doctor begins, waiting for Nicole to respond in the affirmative before continuing, &quot;This is a species we engineered using communication systems recovered from the communication equipment humans left on &lt;em&gt;Rama&lt;/em&gt;. Although it uses color bands so we can understand, it speaks in a language best understood by your own computers and bears no resemblance to our own language or grammar. With a little translation help of any of your computers, a human should have no trouble reading it.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I thought about that but all I got was gibberish - mostly just parentheses, actually.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;How many legs does an octospider have, Nicole?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Oh, right. Hang on, then.&quot; Nicole makes a few alterations on her pocket computer's program. When the result is displayed, she looks up at Dr. Blue. &quot;I see you've been watching ancient Earth movies again.&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The series of color bands that the little creature was displaying are shown below in the order they appeared. Can you figure out what it was saying to Nicole?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iwRpR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iwRpR.png&quot; alt=&quot;Color Seak&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9000" LastEditorUserId="12408" LastEditDate="2016-03-05T04:30:30.763" LastActivityDate="2021-04-02T23:06:24.097" Title="Octospider Cousins" Tags="|visual|computer-puzzle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28379" PostTypeId="2" ParentId="28378" CreationDate="2016-03-05T03:17:21.030" Score="8" Body="&lt;p&gt;Thanks to Alconja for the suggestion.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;caN yoU PaiNT wITH ALL ThE COlorS Of tHE wind&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;My first thought upon seeing this problem is, I wonder what the RGB and HSV values of those bands are.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The solution is to interpret each RGB value as a character in order. However you have to interpret them as octal values not decimal values.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The program I used gives R, G, B in the range 0-255&lt;/p&gt;&#xA;&lt;div class=&quot;s-table-container&quot;&gt;&#xA;&lt;table class=&quot;s-table&quot;&gt;&#xA;&lt;thead&gt;&#xA;&lt;tr&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;r&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;g&lt;/th&gt;&#xA;&lt;th style=&quot;text-align: center;&quot;&gt;b&lt;/th&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/thead&gt;&#xA;&lt;tbody&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;103&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;141&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;116&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;40&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;171&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;117&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;125&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;40&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;120&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;141&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;151&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;116&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;124&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;40&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;167&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;111&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;124&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;110&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;40&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;101&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;114&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;114&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;40&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;124&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;150&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;105&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;40&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;103&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;117&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;154&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;157&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;162&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;123&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;40&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;117&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;146&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;40&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;164&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;110&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;105&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;40&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;167&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;tr&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;151&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;156&lt;/td&gt;&#xA;&lt;td style=&quot;text-align: center;&quot;&gt;144&lt;/td&gt;&#xA;&lt;/tr&gt;&#xA;&lt;/tbody&gt;&#xA;&lt;/table&gt;&#xA;&lt;/div&gt;" OwnerUserId="18821" LastEditorUserId="69582" LastEditDate="2021-04-02T23:06:24.097" LastActivityDate="2021-04-02T23:06:24.097" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28381" PostTypeId="2" ParentId="28306" CreationDate="2016-03-05T04:07:44.480" Score="4" Body="&lt;p&gt;This is a bit of a shot in the dark.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Are you a dragon?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You're historically found in cultures throughout the world.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In China you're considered lucky, but in the Dark Ages you were to be feared in tales of knights' heroics.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Resin is a flammable substance,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Which ties into a dragon's flaming breath, as does the idea of no judgment when you spew red-hot flames at your enemies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Obviously you're mythical, and the first/fifth…&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; May be a reference to &lt;em&gt;The Elder Scrolls&lt;/em&gt;, where dragons first appeared in the 5th game, &lt;em&gt;Skyrim&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19853" LastActivityDate="2016-03-05T04:07:44.480" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28383" PostTypeId="1" AcceptedAnswerId="28384" CreationDate="2016-03-05T04:31:05.977" Score="7" ViewCount="239" Body="&lt;p&gt;On a $3\times3$ grid we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mFwqR.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mFwqR.png&quot; alt=&quot;33grid&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $8$ moves needed to swap the red and blue knights.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the minimum numbers of moves to swap the knights on a $4\times4$ grid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aWGQK.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aWGQK.png&quot; alt=&quot;44grid&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11497" LastEditorUserId="11497" LastEditDate="2016-03-05T04:51:06.173" LastActivityDate="2016-03-05T08:05:06.473" Title="Swapping knights" Tags="|mathematics|checkerboard|knight-moves|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28384" PostTypeId="2" ParentId="28383" CreationDate="2016-03-05T04:57:06.677" Score="4" Body="&lt;p&gt;It will take:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;20 moves to swap the knights&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This can be done as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/Xr4qJ.png&quot; alt=&quot;Zerris&quot;&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All credit goes to @klm123 for demonstrating a good way to visualize this in &lt;a href=&quot;https://puzzling.stackexchange.com/questions/25287/switch-the-knights/25313#25313&quot;&gt;a similar puzzle&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2016-03-05T08:05:06.473" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28385" PostTypeId="2" ParentId="28383" CreationDate="2016-03-05T05:17:51.843" Score="0" Body="&lt;p&gt;I believe it is, in chess terms,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 20 half moves, or 10 move pairs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In algebraic notation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;pre&gt;&#xA; 1.  N3b2 N2c3&#xA; 2.  Na1  Nc1 &#xA; 3.  Nb3  Nbd3 &#xA; 4.  Ncb4 Nc2 &#xA; 5.  N1d2 Nd4 &#xA; 6.  Nb1  Nd2 &#xA; 7.  Na3  Ncb1 &#xA; 8.  Nbc4 Nac3 &#xA; 9.  Ndb2 Nc2 &#xA; 10. Na4  Nd1 &#xA; &lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17297" LastEditorUserId="17297" LastEditDate="2016-03-05T05:23:43.937" LastActivityDate="2016-03-05T05:23:43.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28387" PostTypeId="1" AcceptedAnswerId="28764" CreationDate="2016-03-05T05:57:40.530" Score="5" ViewCount="456" Body="&lt;p&gt;This is the final part of a 5 part series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Links to the first 4:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 1: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/28209/large-and-in-charge&quot;&gt;Large and in charge&lt;/a&gt;  [solved]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 2: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/28262/small-and-resilient&quot;&gt;Small and resilient&lt;/a&gt;  [solved]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 3: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/28306/immovable-object&quot;&gt;Immovable object&lt;/a&gt;  [solved]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 4: &lt;a href=&quot;https://puzzling.stackexchange.com/questions/28360/unstoppable-force&quot;&gt;Unstoppable force&lt;/a&gt;  [solved]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(parts 1-4 can be solved out of order)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Now for Part 5:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Part 1---&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good problem solver peers into the past,&lt;br&gt;&#xA;once you find me, it's quiet at last. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Completing this riddle is the ultimate test,&lt;br&gt;&#xA;for after its done, you will need your rest. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 2---&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other day I screamed &quot;JSTYX&quot; out loud all confused,&lt;br&gt;&#xA;To figure out what it means, I'd be greatly amused. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a great riddler, I've not spoken this word,&lt;br&gt;&#xA;for if I had, that'd just be absurd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some twisting and thrashing, I'll have solved this problem, &lt;br&gt;&#xA;the four other parts, I'm sure that I've solved them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problems the key, I know it, I know it,&lt;br&gt;&#xA;But instead of sharing I might as well throw it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As with the past, a similar method involved,&lt;br&gt;&#xA;This time it's much harder, and has greatly evolved. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conclusion---&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With all of these pieces, you look back in time,&lt;br&gt;&#xA;This is the end of the puzzle, but not my last rhyme. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What/who am I?&lt;/p&gt;&#xA;" OwnerUserId="19731" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2016-03-10T23:06:17.760" Title="Breathing New Life [Part 5]" Tags="|riddle|lateral-thinking|wordplay|rhyme|poetry|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="28388" PostTypeId="1" AcceptedAnswerId="28391" CreationDate="2016-03-05T06:31:15.563" Score="6" ViewCount="627" Body="&lt;p&gt;At the time of this posting, I don't exist.  You can help fix that, though.  In fact, the author posits that if you're reading this, you've probably done something similar enough already.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That doesn't make your task any less delicate, however -- if you don't help me in just the right way you'll create something that &lt;em&gt;looks&lt;/em&gt; like me, but actually isn't.  Whether or not you actually manage to help me, you'll pick not-quite-me apart to try to convince yourself that you did.  Don't let that discourage you, though; you still get props for trying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's (hopefully) only one way that you can bring me into existence.  But the author and I promise, if you do then your success will stand high above the rest, and be made obvious for all to see.&lt;/p&gt;&#xA;" OwnerUserId="19856" LastActivityDate="2016-03-05T06:49:14.723" Title="I don't exist yet. Help me fix that!" Tags="|riddle|lateral-thinking|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28389" PostTypeId="2" ParentId="28388" CreationDate="2016-03-05T06:37:44.847" Score="1" Body="&lt;h3&gt;This is not the intended answer(my achievement)...confirmed by OP&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I think it is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Question&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;At the time of this posting, I don't exist.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is no question in the whole riddle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can help fix that, though.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As the riddle's answer is 'question' so it is fixed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In fact, the author posits that if you're reading this, you've probably done something similar enough already.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If we read a puzzle, we assume that there must be a question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That doesn't make your task any less delicate, however -- if you don't help me in just the right way you'll create something that looks like me, but actually isn't.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If we are wrong then we'll create an answer which will not be a question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whether or not you actually manage to help me, you'll pick me apart to try to convince yourself that you did. Don't let that discourage you, though; you still get props for trying.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We try to pick what is the question in the riddle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's (hopefully) only one way that you can bring me into existence. But I promise, if you do then your success will stand high above the rest, and be made obvious for all to see.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As the answer is 'question', so now it exists.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-03-05T06:49:14.723" LastActivityDate="2016-03-05T06:49:14.723" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28390" PostTypeId="2" ParentId="28365" CreationDate="2016-03-05T06:40:26.177" Score="2" Body="&lt;p&gt;I feel like the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water, rain, storm, weather related... they all seem to kinda fit&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Older than history, as young as today, &#xA;I was here before you, and I'm here to stay.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The atoms have been here for a long time, and so has the water in one form or another.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My fathers were high, my mother had might&#xA;I'm beloved of nobles, but never by right.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Father as in God or maybe Zeus if its about a storm.  Could also be a cloud.  Mother is Mother Earth.&#xA; Jesus turned water into wine? Not sure about right...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Conceived in the garden, I walk in all weather&#xA;then disappear all black in a flutter of feather. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Moisture in &quot;gardens&quot; (or forests) often condenses and brings rain or storms.&#xA; Storm clouds are dark.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I enter the forest surrounded by does. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the forest again with the rain idea.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm awarded degrees, yet not educated, &#xA;they judge me as worst among the related. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Degrees -- &gt; Weather  and judging as worst among the related, a storm cloud is &quot;worse&quot; than regular fluffy clouds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll arrive in a glass, or I'll arrive in a wave&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Water&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another guess I've got to fit into a few parts is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; money&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19731" LastActivityDate="2016-03-05T06:40:26.177" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28391" PostTypeId="2" ParentId="28388" CreationDate="2016-03-05T06:44:14.100" Score="7" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This answer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At the time of this posting, I don't exist&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It wasn't here when you posted the question. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can help fix that, though&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I just fixed that&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;you don't help me in just the right way you'll create something that looks like me... you'll pick not-quite-me apart to try to convince yourself that you did&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I might be wrong, making arguments that are lopsided. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's (hopefully) only one way that you can bring me into existence&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some questions may have more than one answer, but here we should be posting riddles, which have one right answer. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;if you do then your success will stand high above the rest, and be made obvious for all to see&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This answer will be marked as the correct one and will pop up against the rest,if there will be any others. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18064" LastActivityDate="2016-03-05T06:44:14.100" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28392" PostTypeId="2" ParentId="28306" CreationDate="2016-03-05T07:05:18.987" Score="2" Body="&lt;h3&gt;This is not the intended answer (my achievement)...confirmed by OP&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I think it is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fire&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I have a long history from east to west,&#xA;to some I'm the worst and others the best.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every culture had different thoughts on fire.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With a core made of resin I dance and I flutter,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Resin ties into a dragon's flaming breath. (borrowed from @brilliantlyInsane's answer)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;no one judges me when I speak with a stutter.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You know, the popping sound.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I may be an object of legend or myth,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many mythical stories use it as power&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;at first you won't find me but maybe in fifth.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is one of the five elements of nature.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-03-05T07:11:25.207" LastActivityDate="2016-03-05T07:11:25.207" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28393" PostTypeId="1" AcceptedAnswerId="28827" CreationDate="2016-03-05T07:17:18.837" Score="8" ViewCount="945" Body="&lt;p&gt;I was, for the most part, created to accompany something else.  It would &lt;em&gt;seem&lt;/em&gt; like I'm just playing second fiddle to a jive-talking glory hog, but that's not actually the case; we both wound up being really famous and respected.  In fact, I'm arguably more famous than my master, because I get around more--I'm more versatile and less likely to get in the way of people, and I don't need as much gas to get to the same place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now it's all right, it's okay; we'll live to see another day.  As of fairly recently (well after Stack Overflow was founded), both myself and my master are currently being preserved by a very powerful organization for all of time.  We'll both be stayin' alive for a very long time, which is more than can be said of the three men to whom I most owe my existence.  It's a shame that two of them are dead, as we really had a great thing going on; if I can't have them, I don't want nobody else.  The oldest guy is still around, at least, but it just doesn't feel the same without his brothers, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently, the three guys who made me originally worked with three others for a little while, though this was before my time.  They also had a younger brother who did his own separate thing, but I never met him (and he's dead too, apparently).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, who's taller, my master or me?  Well, that's actually an interesting question.  It depends on whether or not he's allowed to show you his penis.  Yes, really.  Though I suppose it doesn't really matter anyway, because we're both about the same height regardless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, not that it matters much to me, but everyone generally likes to pretend my master's little brother doesn't exist.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Hints&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's a very specific thing.  A famous thing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's older than I am, but I don't know if it's older than you (StackExchange attracts a pretty wide audience these days).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you tried to sell one of these without getting permission from a &lt;em&gt;different&lt;/em&gt; large organization, they would get very mad and might sue you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19856" LastEditorUserId="19856" LastEditDate="2016-03-07T12:01:26.620" LastActivityDate="2016-03-11T23:11:13.333" Title="More than a riddle" Tags="|riddle|knowledge|" AnswerCount="5" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28394" PostTypeId="2" ParentId="28393" CreationDate="2016-03-05T08:14:11.180" Score="0" Body="&lt;p&gt;Are you:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Subconscious &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If so it's 3am. I'll add explanation after I get a few hours of shuteye. &lt;/p&gt;&#xA;" OwnerUserId="19731" LastActivityDate="2016-03-05T08:14:11.180" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28395" PostTypeId="1" AcceptedAnswerId="28397" CreationDate="2016-03-05T08:32:56.440" Score="8" ViewCount="369" Body="&lt;p&gt;My friend ran into trouble when in New Zealand. He has been captured and is being moved around the country to a series of undisclosed locations. This is the last message I received from him.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;My friend, I met a Kiwi, Kat&#xA;We hit it off, and that was that &#xA;&#xA;But thinking I was quite the wag&#xA;I mocked her motherland’s new flag&#xA;&#xA;In hobbit prison now I rot&#xA;I need your help, it’s all I’ve got!&#xA;&#xA;I hope this puzzle’s not a dud &#xA;And fine ideas from you will flood…&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Help me decipher the message, so I can retrace his steps. Where should I look for my friend?&lt;/p&gt;&#xA;" OwnerUserId="58764" LastActivityDate="2016-03-05T10:26:05.023" Title="My friend is being held somewhere in New Zealand! Please help!" Tags="|riddle|lateral-thinking|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28396" PostTypeId="2" ParentId="28343" CreationDate="2016-03-05T09:18:28.040" Score="12" Body="&lt;p&gt;As you have posed the question, the four conditions clearly don't work. The prisoner can't conclude that he won't be hanged because the judge said he would and the judge is truthful. As a perfect logician he would recognize that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be quite tricky to come up with a problem statement that works well. The paradox rests, I believe, on a rather neat concealed Gödelization. If I can prove that an event would not be surprising, then it becomes surprising.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The circularity becomes clearer if we remove the part about the intervening days. Imagine if, at the conclusion of the trial, the judge pulled out a gun, pointed it at the prisoner and said: &quot;I will shoot you in 10 seconds and it will be a surprise.&quot; The prisoner reasons that he is perfectly safe because there is nothing less surprising than being shot by a man who is holding a gun and has just said he will shoot you. Having reasoned this way, he will, of course, be surprised when he is shot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gödel's theorem (loosely rendered) states that any logical system powerful enough to express normal integer arithmetic is powerful enough to form the statement: &quot;This statement is true and unprovable.&quot; Hence all such systems are either incomplete (there is at least one true statement which cannot be proved) or inconsistent (you can prove both a statement and its negation). For the latter, imagine that you could prove the Gödel statement. By proving it, you have made it false. If you can prove one false statement, you can prove anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Returning to the judge and the prisoner, if we replace the word &quot;surprising&quot; with the word &quot;unprovable&quot; then the parallel is clear. What does this mean for the set of four axioms? That is a bit murky. What do we really mean by &quot;perfect logician&quot;? Are all of the actions and emotions of a &quot;perfect logician&quot; defined by logical reasoning? Does this mean that &quot;surprising&quot; and &quot;unprovable&quot; are exactly the same for a perfect logician? If we accept this, then I think it implies that the judge must also have perfect knowledge of the prisoner's world-view or he could not say with certainty whether the prisoner would be surprised or not. In this case the axioms could be consistent in themselves but not consistent with the prisoner's reasoning. A perfect logician would accept that the judge had made a true yet unprovable statement about his future. Since the statement is unprovable, the prisoner would accept it as true but be unable to use it to form any expectations about his execution. Hence, he would still be surprised on his execution day but for a different reason.&lt;/p&gt;&#xA;" OwnerUserId="18924" LastEditorUserId="18924" LastEditDate="2016-03-05T10:46:47.790" LastActivityDate="2016-03-05T10:46:47.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28397" PostTypeId="2" ParentId="28395" CreationDate="2016-03-05T10:20:41.200" Score="7" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The last word of the first line of each paragraph refers to an IATA airport code in New Zealand:&lt;br&gt;&lt;br&gt;&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Kaitaia_Airport&quot;&gt;KAT - Kaitaia Airport&lt;/a&gt;&lt;br&gt;&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Wanganui_Airport&quot;&gt;WAG - Wanganui Airport&lt;/a&gt;&lt;br&gt;&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Rotorua_Regional_Airport&quot;&gt;ROT - Rotorua Regional Airport&lt;/a&gt;&lt;br&gt;&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Dunedin_Airport&quot;&gt;DUD - Dunedin Airport&lt;/a&gt;&lt;br&gt;&lt;br&gt;If you are fast enough, he might still be at the last listed airport.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2016-03-05T10:26:05.023" LastActivityDate="2016-03-05T10:26:05.023" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28398" PostTypeId="2" ParentId="28365" CreationDate="2016-03-05T10:25:37.143" Score="8" Body="&lt;p&gt;A little grim, but I believe the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Murder&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Older than history, as young as today, / I was here before you, and I'm here to stay.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People have been killing each other forever, and will likely continue doing so for a while yet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My fathers were high, my mother had might&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Murders are committed by the high and mighty.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I'm beloved of nobles, but never by right.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nothing nobility likes more than a good old fashioned bit of murder (but that doesn't make it right).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Conceived in the garden, I walk in all weather / then disappear all black in a flutter of feather.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A murder of crows.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I dally with dahlias, I rib with the rose,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; References to &lt;a href=&quot;https://en.wikipedia.org/wiki/Black_Dahlia&quot;&gt;The Black Dahlia&lt;/a&gt; murder, and to serial killer &lt;a href=&quot;https://en.wikipedia.org/wiki/Rosemary_West&quot;&gt;Rose West&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I enter the forest surrounded by does.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Murderers often dump their victims (John/Jane Does) in forests.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I'm awarded degrees, yet not educated, / they judge me as worst among the related.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Generally considered the worst crime, and is graded in degrees (first/second degree murder, etc).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You always see me but hope we don't meet&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Seen on the news/movies/etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I never respect you, but if you scream, I will bleed.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; From the idiom, to scream bloody murder.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I'll arrive in a glass, or I'll arrive in a wave / and then my sad ballads will emerge from the cave.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reference to musician &lt;a href=&quot;https://en.wikipedia.org/wiki/Nick_Cave_and_the_Bad_Seeds&quot;&gt;Nick Cave&lt;/a&gt;, who has an album called Murder Ballads.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="5208" LastEditDate="2016-03-05T13:20:29.840" LastActivityDate="2016-03-05T13:20:29.840" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28399" PostTypeId="1" AcceptedAnswerId="28423" CreationDate="2016-03-05T11:11:08.597" Score="14" ViewCount="1421" Body="&lt;p&gt;The colors of the rainbow were not enough, I needed 8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You probably heard about me before you were 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To learn my name you had to count to 25.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you needed already know me to learn to count to 24.&lt;/p&gt;&#xA;" OwnerUserId="19350" LastActivityDate="2016-03-07T21:18:50.443" Title="A counting riddle - Who am I?" Tags="|riddle|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28400" PostTypeId="2" ParentId="28310" CreationDate="2016-03-05T12:01:05.903" Score="1" Body="&lt;p&gt;A little tenuous, but I wanna say it's:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Bridge&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Likely you'll find me when lending support&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A bridge supports roads, railways, etc&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Or shedding some knots by a slip of some sort&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You'd need to be on the bride of your ship if you're slowing down to land at a slipway.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Or adding the silver before fitting the frame&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bugs me that I can't explain the silver, but you fit the frame of your glasses on the bridge of your nose...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Keep calm and breathe is the key to this game.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Referring to the card game.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, an acrostic:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bridges lines of a poem together into another word.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastActivityDate="2016-03-05T12:01:05.903" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28401" PostTypeId="2" ParentId="28343" CreationDate="2016-03-05T12:16:21.630" Score="2" Body="&lt;p&gt;The question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there any reason these are not included as part of the problem statement?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The problem with the suggested statements is that statement 1 and the portions of statement 4 related to the well-known status of statements 2 and 3 cannot both be true at the same time, in the posited scenario.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that the prisoner reaches a conclusion that an execution will not happen, when the judge has clearly stated that it &lt;em&gt;will&lt;/em&gt; happen during the coming week.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way for the judge's statement regarding the execution to be false would be if the judge is either mistaken about future events (i.e.: not a perfect logician) or else lying (i.e.: not truthful).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neither of these is the case as per statements 2 and 3, respectfully.  Additionally, statement 4 tells us that the prisoner &lt;em&gt;knows&lt;/em&gt; that neither of these are the case.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So taking statement 1 (prisoner is a perfect logician) and statement 4 together (The prisoner knows that the judge is a perfect logician and truthful), the prisoner cannot accept a conclusion which would require that the judge be either wrong or lying about the execution date.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, either statement 1 or statement 4 must be false.  My money is on statement 1 being the false one, personally.  That makes sense both logically and thematically, as the whole story is really a fairly simple morality play about a criminal who thinks they're more clever than they actually are, and gets their comeuppance.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you can also make the story's logic work if you prefer to state that the prisoner didn't know the judge was infallible and truthful, and assumed that they were wrong and/or lying.  It's not much of a bedtime story at that point, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, neither statement 2 nor statement 3 is actually important to the key problem in the scenario;  what's important is that we're told the prisoner believes that the judge is truthful and a perfect logician (statement 4), and that the prisoner is a perfect logician (statement 1), and yet somehow still comes to a conclusion which would require that the judge either has made a logical mistake or is lying.  That's a pretty huge logical mistake for our &quot;perfect logician&quot; prisoner to make.&lt;/p&gt;&#xA;" OwnerUserId="9286" LastActivityDate="2016-03-05T12:16:21.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28402" PostTypeId="1" AcceptedAnswerId="28403" CreationDate="2016-03-05T13:50:12.640" Score="11" ViewCount="386" Body="&lt;p&gt;Professor Halfbrain has spent the last few days (and sleepless nights)&#xA;with analyzing integer numbers of the form $~N_x(n):=n^x+x~$. &#xA;The professor computed and analyzed thousands and thousands of these &#xA;numbers, and made a number of fascinating observations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For $x=1$, the number $N_1(6)=6^1+1=~7$ is prime.&lt;/li&gt;&#xA;&lt;li&gt;For $x=2$, the number $N_2(3)=3^2+2=11$ is prime.&lt;/li&gt;&#xA;&lt;li&gt;For $x=3$, the number $N_3(4)=4^3+3=67$ is prime.&lt;/li&gt;&#xA;&lt;li&gt;For $x=4$, the number $N_4(1)=1^4+4=~5$ is prime.&lt;/li&gt;&#xA;&lt;li&gt;For $x=5$, the number $N_5(2)=2^5+5=37$ is prime.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And so on. And so on.&#xA;Finally, the professor detected the following extremely deep theorem.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Professor Halfbrain's theorem&lt;/strong&gt;:&lt;br&gt;&#xA;  For every positive integer $x$, there exists a positive integer $n$ &#xA;  so that $N_x(n)$ is prime.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This puzzle asks you to decide the correctness of this theorem.&#xA;If it is correct, then provide a convincing argument for it.&#xA;If it is incorrect, then state the smallest positive integer $x$ &#xA;for which the statement is violated.&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2016-03-05T16:44:45.767" Title="Professor Halfbrain and the prime numbers" Tags="|mathematics|number-theory|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28403" PostTypeId="2" ParentId="28402" CreationDate="2016-03-05T14:48:30.267" Score="10" Body="&lt;p&gt;Professor Halfbrain's theorem is incorrect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $x=27$ and any $n$, there is a factorization $$N_{27}(n)=(n^9+3)(n^{18}-3n^9+9)$$ as a product of two integers bigger than $1$, so $N_{27}(n)$ can never be prime. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, whenever $k\geq 3$ is odd, there is a factorization $$N_{k^k}(n)=\left(n^{k^{k-1}}+k\right)\left(\sum_{i=0}^{k-1} (-k)^{k-1-i}n^{i k^{k-1}}\right), $$&#xA;so $N_{k^k}(n)$ is never prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A computer search determined that $N_x$ does take a prime value for $x=6,7,\ldots,26$, so $x=27$ is the smallest counterexample. The first prime values are:&#xA;\begin{align*}&#xA;N_{6}(1)&amp;amp;=7\\N_{7}(16)&amp;amp;=268435463\\N_{8}(3)&amp;amp;=6569\\N_{9}(2)&amp;amp;=521\\N_{\&#xA;10}(1)&amp;amp;=11\\N_{11}(32)&amp;amp;=36028797018963979\\N_{12}(1)&amp;amp;=13\\N_{13}(118)&amp;amp;\&#xA;=859935929762876868984659981\\N_{14}(417)&amp;amp;=\&#xA;4807339234680508004200143948920808143\\N_{15}(2)&amp;amp;=32783\\N_{16}(1)&amp;amp;=\&#xA;17\\N_{17}(14)&amp;amp;=30491346729331195921\\N_{18}(1)&amp;amp;=19\\N_{19}(22)&amp;amp;=\&#xA;32064977213018365645815827\\N_{20}(81)&amp;amp;=\&#xA;147808829414345923316083210206383297621\\N_{21}(76)&amp;amp;=\&#xA;3141126580731587340586460636236874776597\\N_{22}(1)&amp;amp;=23\\N_{23}(12)&amp;amp;=\&#xA;6624737266949237011120151\\N_{24}(55)&amp;amp;=\&#xA;587089817274070447368135511875152587890649\\N_{25}(28)&amp;amp;=\&#xA;1509909033949224437981629384719597593\\N_{26}(15)&amp;amp;=\&#xA;3787675244106352329254150390651\\&#xA;\end{align*}&lt;/p&gt;&#xA;" OwnerUserId="2627" LastEditorUserId="2627" LastEditDate="2016-03-05T16:44:45.767" LastActivityDate="2016-03-05T16:44:45.767" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28404" PostTypeId="2" ParentId="28393" CreationDate="2016-03-05T15:18:29.257" Score="1" Body="&lt;p&gt;Are you:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pants or underwear &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because well:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I was, for the most part, created to accompany something else. It would seem like I'm just playing second fiddle to a jive-talking glory hog, but that's not actually the case; we both wound up being really famous and respected. In fact, I'm arguably more famous than my master, because I get around more.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pants &amp;amp; underwear are meant to be worn arround your low-back and &quot;playing second fiddle to a jive-talking glory hog&quot; seems like an allusion to a penis.&lt;br&gt;&#xA;&lt;br&gt;&#xA; The next line may refer to well-known companies that make jeans and/or underwear.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;But none of that matters any more, because both myself and my master are currently being watched after by very powerful people.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A reference to said companies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;We'll both stay alive for a very long time, which is more than can be said of (two out of three of) the men to whom I most owe my existence.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Styles of underwear &amp;amp; jeans don't vary too much over the years. There is some variation but not much. Today's jeans are quite similar to those from even the 60s. In that case, many in the industry are no longer living. Most likely, only 1 of the 3 that make today's models are from back then. Even the jeans themselves can be inherited, if well preserved.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Also, who's taller, my master or me? Well, that's actually an interesting question. It depends on whether or not he's allowed to show you his penis. Yes, really. Though I suppose it doesn't really matter anyway, because we're both about the same height regardless.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When the penis is brought outside the pants, for example when urinating, the &quot;master&quot; of the pants is bigger than the pants (or underwear) itself, because he is breaking outside of the manner in which they are to be worn. &quot;Though I suppose it doesn't really matter anyway, because we're both about the same height regardless&quot; - the pants &amp;amp; underwear surround the body parts of their &quot;master&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Finally, not that it matters much to me, but everyone generally likes to pretend my master's little brother doesn't exist.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because pants (and underwear) are meant to hide one's genitals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19865" LastEditorUserId="19865" LastEditDate="2016-03-07T18:10:33.583" LastActivityDate="2016-03-07T18:10:33.583" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28405" PostTypeId="2" ParentId="28360" CreationDate="2016-03-05T15:41:33.317" Score="2" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dynamite (or bombs)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;I'll start to count while you solve this problem,&#xA;and just relax here at the top of my column.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dynamite is basically a tube with a wick and some powder-fire reactive component inside, usually (but not limited to) Trinitrotoluene and/or gunpowder&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Everywhere I go I let out a good whistle,&#xA;but don't mess with me cause I carry a pistol.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you know there is explosive material in an area, you are going to react negatively or at least exercise some grade of caution, especially if you are the one controlling it because it is very dangerous.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;I've been explosive at one point in history,&#xA;what I truly am is not much of a mystery.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dynamite or other generic bombs are highly explosive!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19865" LastEditorUserId="7586" LastEditDate="2016-03-07T17:59:30.870" LastActivityDate="2016-03-07T17:59:30.870" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28406" PostTypeId="1" AcceptedAnswerId="28614" CreationDate="2016-03-05T15:56:51.270" Score="4" ViewCount="401" Body="&lt;p&gt;I am dead and lifeless&lt;br&gt;&#xA;I am a monster, cold&lt;br&gt;&#xA;Look at me a closer look&lt;br&gt;&#xA;Gasp and behold!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Older than your pyramids&lt;br&gt;&#xA;And far more than your Rome&lt;br&gt;&#xA;I am older than it all&lt;br&gt;&#xA;As old as your home  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it were not me, you know&lt;br&gt;&#xA;They'd stone you soon to death&lt;br&gt;&#xA;Giant boulders would crush you&lt;br&gt;&#xA;And take your last breath  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I came alone for you&lt;br&gt;&#xA;And you come, you four&lt;br&gt;&#xA;I'd be left standing and&lt;br&gt;&#xA;You all would be no more  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am father of them all&lt;br&gt;&#xA;Who wander in the sky&lt;br&gt;&#xA;And whoever wanders close&lt;br&gt;&#xA;Must invariably die  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINTS&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hints are right there in all the stanzas. But still, let me put them simply in non-poetic form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stanza1  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am a giant &lt;em&gt;dead&lt;/em&gt; monster. If you look at me in detail, you would be horrified.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Stanza2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am far older than Rome or the Egyptian pyramids. I am almost as old as your home is.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Stanza3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If it was not for my presence, &lt;em&gt;they&lt;/em&gt; would stone you to your death. Giant rocks would fall on you and leave no-one alive!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Stanza4&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you gather your 3 partners and try to come for a battle with me, it would be me who would be left while all of you would be obliterated.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Stanza5&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am the father of the vagabonds. If anyone comes too close to me, death is certain for him!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16342" LastEditorUserId="16342" LastEditDate="2016-03-06T06:15:46.667" LastActivityDate="2016-03-08T18:57:51.353" Title="A Monster Who Is Also A Savior" Tags="|riddle|poetry|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="28407" PostTypeId="1" CreationDate="2016-03-05T16:05:35.857" Score="5" ViewCount="2434" Body="&lt;p&gt;Been staring at this question for 20 minutes and can't come to a reasonable answer, any help would be really appreciated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2A7U2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2A7U2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19869" LastEditorUserId="11876" LastEditDate="2016-03-05T21:29:44.237" LastActivityDate="2018-09-25T12:33:57.647" Title="Logical Pattern" Tags="|pattern|visual|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28408" PostTypeId="2" ParentId="28406" CreationDate="2016-03-05T16:14:00.300" Score="1" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the moon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;I am dead and lifeless&#xA;I am a monster, cold&#xA;Look at me a closer look&#xA;Gasp and behold!&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It doesn't have life in it, or life itself, and is cold and big.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Older than your pyramids&#xA;And far more than your Rome&#xA;I am older than it all&#xA;As old as your home&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It was created when the earth was created, which is &quot;our home&quot; and far older than any of our civilizations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;If it were not me, you know&#xA;They'd stone you soon to death&#xA;Giant boulders would crush you&#xA;And take your last breath&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All the Craters of the moon exist there by a reason, and is because it doesn't have any atmosfere, but also has been a shield for earth from many generations and Eons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;If I came alone for you&#xA;And you come, you four&#xA;I'd be left standing and&#xA;You all would be no more&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Like astronauts?, without atmosphere, there is no chance to survive, plus the effect on the sea would affect the life on earth considerably, so we'll &quot;die&quot; if we lose it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;I am father of them all&#xA;Who wander in the sky&#xA;And whoever wanders close&#xA;Must invariably die&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No Atmosphere point all over again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19865" LastEditorUserId="19865" LastEditDate="2016-03-05T16:20:15.697" LastActivityDate="2016-03-05T16:20:15.697" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28409" PostTypeId="1" AcceptedAnswerId="28414" CreationDate="2016-03-05T16:48:53.387" Score="11" ViewCount="746" Body="&lt;p&gt;Find the smallest possible value of $ab+c$, where $a,b,c$ are positive integers with $a+bc=2016$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(No computers! The puzzle has a nice direct solution.)&lt;/p&gt;&#xA;" OwnerUserId="9125" LastActivityDate="2016-03-07T00:25:43.843" Title="Three positive integers" Tags="|mathematics|no-computers|number-theory|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28410" PostTypeId="2" ParentId="28407" CreationDate="2016-03-05T16:56:08.110" Score="0" Body="&lt;p&gt;I think that the correct answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;&lt;strong&gt;&gt;&amp;lt;&lt;/strong&gt;&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sequence is that:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first and the second columns both have a two-characters field, so the third should also have one (under the &quot;?&quot; sign). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As seen there, the middle field borrows characters from neighbouring fields.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &quot;?&quot; field borrows the 3rd and 4th characters from the top field of the third column.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This is why I think that the correct answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;&gt;&amp;lt;&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Don't get angry if it's not right. If possible, you could try again if available.)   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S: If I wasn't right, you could try the &quot;Try All&quot; method and try every possible answer, starting with the 2-character ones.&lt;/p&gt;&#xA;" OwnerUserId="19451" LastEditorUserId="42452" LastEditDate="2018-09-25T12:33:57.647" LastActivityDate="2018-09-25T12:33:57.647" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28411" PostTypeId="2" ParentId="28407" CreationDate="2016-03-05T16:58:39.620" Score="1" Body="&lt;p&gt;Ok so here it goes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot; &quot; &amp;lt; &gt; &quot; &quot; is the answer &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The right one has each option in the rows. The three options are &quot; &quot; &quot;&gt;&quot; and &quot;&amp;lt;&quot;&#xA; So you know it's _ _ _ &quot; &quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Part 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The left one has each option in the columns. &#xA; So you know it's &quot; &quot; _ _ &quot; &quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Part 3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In each of the columns the middle 2 share 2 alike and since top and bottom are the same you can rule out &quot; &quot; &gt;&gt; &quot; &quot;&#xA; For one of the 2 middle columns on each column you share 3 like symbols so at least one of the 2 in the middle are &quot;&gt;&quot;. So you have either &gt;&amp;lt; or &amp;lt;&gt;&#xA; In column one the shared column is column 3, in column 2 the shared is 2 so it must go back to 3 in the 3rd column so you have &quot; &quot; _ &gt; &quot; &quot;&#xA; The second spot must be &amp;lt; because it can't be &quot;&gt;&quot; but there's a second reason. In the column that doesn't share like symbols. 2 are the same and the other is different since both point right this one must point left. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot; &quot; &amp;lt; &gt; &quot; &quot; is your answer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A second answer fits though:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot; &quot; &gt; &amp;lt; &quot; &quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the first column the 2 character item is in row 1 so it takes chars 3 and 4 from row 3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the second column the 2 character item is in row 3 so it takes chars 3 and 4 from row 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the third column it's in row 2 and takes from rows 1 and 3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19731" LastEditorUserId="19731" LastEditDate="2016-03-05T17:05:14.300" LastActivityDate="2016-03-05T17:05:14.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28412" PostTypeId="2" ParentId="3215" CreationDate="2016-03-05T17:16:54.270" Score="8" Body="&lt;p&gt;When you knock on the door, listen to the guard's response and use his language.&lt;/p&gt;&#xA;" OwnerUserId="19856" LastActivityDate="2016-03-05T17:16:54.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28413" PostTypeId="2" ParentId="28409" CreationDate="2016-03-05T17:46:34.170" Score="9" Body="&lt;p&gt;The smallest possible value is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;AB+C = 96&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The smallest value will occur when  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A = 1 and B = C&lt;br&gt;&#xA;&lt;br&gt;&#xA; Unfortunately, 2016-1=2015 isn't a square, so do some factoring on paper to get the closest values of B and C as 31 * 65.  A = 1, B = 31, C = 65 gives A+BC = 2016 and AB+C = 96.  The theoretical minimum at A = 1, B = C would put B and C at around 45, for a total of just under 90.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="17389" LastEditDate="2016-03-05T19:14:10.753" LastActivityDate="2016-03-05T19:14:10.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28414" PostTypeId="2" ParentId="28409" CreationDate="2016-03-05T17:56:29.717" Score="8" Body="&lt;p&gt;Proving that Zerris's answer is correct:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ab+c\ge2\sqrt{abc}=2\sqrt{a(2016-a)}=2\sqrt{1008^2-(1008-a)^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because $2\sqrt{1008^2-1006^2}&amp;gt;96$, if $ab+c&amp;lt;96$, $\left|1008-a\right|$ must be greater than $1006$. The only possible values are $a=1$ and $a=2015$. Obviously $a=2015$ won't work, so $a=1$ and $bc=2015$. $2015=5\cdot13\cdot31$ only has four pairs of factors, and none of them sums to less than $96$.&lt;/p&gt;&#xA;" OwnerUserId="12408" LastEditorUserId="12408" LastEditDate="2016-03-05T17:58:36.030" LastActivityDate="2016-03-05T17:58:36.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28417" PostTypeId="2" ParentId="28399" CreationDate="2016-03-05T18:49:57.390" Score="11" Body="&lt;p&gt;Maybe it's a too-far-away Association, but here is my answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Crayola Crayons&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The colors of the rainbow were not enough, I needed 8&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 8 Colors were in the original color box 1903.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Source: &lt;a href=&quot;http://www.crayola.com/faq/your-history/what-were-the-original-eight-8-colors-in-the-1903-box-of-crayola-crayons/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.crayola.com/faq/your-history/what-were-the-original-eight-8-colors-in-the-1903-box-of-crayola-crayons/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You probably heard about me before you were 7&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes, when I was a child, I probably paint something with them&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To learn my name you had to count to 25.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes, because the name include an Y, the 25 letter in alphabet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But you needed already know me to learn to count to 24.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The actual box has 24 colors &lt;a href=&quot;https://i.stack.imgur.com/iSb2q.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iSb2q.jpg&quot; alt=&quot;crayola colors&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4342" LastEditorUserId="4342" LastEditDate="2016-03-05T18:55:46.873" LastActivityDate="2016-03-05T18:55:46.873" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28418" PostTypeId="2" ParentId="28407" CreationDate="2016-03-05T19:34:16.800" Score="4" Body="&lt;p&gt;This fits:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &gt;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each row and column contain a 2-word and 2 4-words, of which one has two of each symbol and one has three of one symbol and one of the other. Also in each row the 2-word is NOT the first letter of the first 4-word, and NOT the first letter of the second 4-word (NOT is used in the logical sense). In each column, the 2-word appears in the middle of each 4-word with the 3:1 ratio.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11497" LastActivityDate="2016-03-05T19:34:16.800" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28419" PostTypeId="2" ParentId="28399" CreationDate="2016-03-05T19:54:49.053" Score="2" Body="&lt;p&gt;A &lt;em&gt;bit&lt;/em&gt; far-stretched but I think it is about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Colors that a computer stores.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The colors of the rainbow were not enough, I needed 8.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/8-bit_color&quot; rel=&quot;nofollow&quot;&gt;8-bit color graphics&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You probably heard about me before you were 7.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the link given above, you can see in the table that numbers 0 and 1 are assigned to Blue color. 2,3,4 to Green color. 5,6,7 to Red. Clearly, the numbers are assigned till 7.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To learn my name you had to count to 25.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'Count' here means 'to include'. So it is about that the maximum number of colors that can be displayed at any one time, that is 256. And you can clearly see 25 here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But you needed already know me to learn to count to 24. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 24-bit color (also called 'true color') has 16,777,216 color variations and the human eye can discriminate up to ten million colors. So, we get all colors we can discriminate in it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-03-05T20:01:47.947" LastActivityDate="2016-03-05T20:01:47.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28420" PostTypeId="1" CreationDate="2016-03-05T21:02:40.340" Score="2" ViewCount="465" Body="&lt;p&gt;So here's how it works: Take the digits from 1991. You can use the digits in any order, only once (you can't make a 19 or 91), with any operation sign to get answers between 1-100. For example: $4=(1+1)^2-9+9$. You cannot use the zero power, but any other powers are good. Powers don't count as using one of the digits. Again, you may not form two digit numbers using your numbers. Logs may be used three times.&lt;/p&gt;&#xA;" OwnerUserId="19876" LastEditorUserId="12408" LastEditDate="2016-03-05T21:13:43.383" LastActivityDate="2016-03-09T15:28:08.623" Title="90s Number Puzzle" Tags="|mathematics|formation-of-numbers|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28421" PostTypeId="2" ParentId="28399" CreationDate="2016-03-05T21:15:49.440" Score="2" Body="&lt;p&gt;It's the &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; alphabet &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The colors of the rainbow were not enough, I needed 8.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the word &quot;alphabet&quot; is 8 letters long.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You probably heard about me before you were 7.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; most children learn the alphabet before age 7&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To learn my name you had to count to 25. &#xA;But you needed already know me to learn to count to 24.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; there are 26 letters in the alphabet. Also, children are often encouraged to wash their hands for 24-25 seconds, but since many learn their ABCs before learning to count as higher than 10, many children simply wash their hands while saying/singing the alphabet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19165" LastActivityDate="2016-03-05T21:15:49.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28423" PostTypeId="2" ParentId="28399" CreationDate="2016-03-05T21:36:23.803" Score="10" Body="&lt;p&gt;Are you a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Xylophone&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The colors of the rainbow were not enough, I needed 8.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  The xylophone needs at least 8 thingies, hence 8 colours for musical reasons. Also hidden in the text we can see &quot;were &lt;strong&gt;not e&lt;/strong&gt;nough&quot;, perhaps accidental, but perhaps hinting it is related to music. &lt;a href=&quot;https://i.stack.imgur.com/TKvtw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TKvtw.jpg&quot; alt=&quot;Picture of a children&amp;#39;s xylophone with coloured notes&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You probably heard about me before you were 7.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The image above shows a xylophone one would encounter at a young age, not sure what else 7 could refer to.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To learn my name you had to count to 25.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Here I guess that counting to 25, is as other suggested, actually reciting the alphabet up until the 25th letter, which is &lt;strong&gt;Y&lt;/strong&gt;, and you need it to spell the word x&lt;strong&gt;Y&lt;/strong&gt;lophone&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But you needed already know me to learn to count to 24.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When kids learn the alphabet, it is very common to learn by associating pictures with the letters. A common thing associated with the 24th letter, i.e. &lt;strong&gt;X&lt;/strong&gt;, is of course, a Xylophone. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19877" LastEditorUserId="10781" LastEditDate="2016-03-07T21:18:50.443" LastActivityDate="2016-03-07T21:18:50.443" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28424" PostTypeId="1" AcceptedAnswerId="28426" CreationDate="2016-03-05T22:00:05.293" Score="-4" ViewCount="116" Body="&lt;p&gt;Find solution, how work algoritem for last, control number in 8 places barcode.&#xA;Here is some results:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Here we got last number 8&lt;a href=&quot;https://i.stack.imgur.com/bVvS4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bVvS4.png&quot; alt=&quot;Example 1&quot; /&gt;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Here we got last number 7: &lt;a href=&quot;https://i.stack.imgur.com/yBCvL.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yBCvL.gif&quot; alt=&quot;Example 2&quot; /&gt;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Here we got last number 5 &lt;a href=&quot;https://i.stack.imgur.com/DVnCl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DVnCl.png&quot; alt=&quot;Example3 &quot; /&gt;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;What is last number of this 2 Barcode?&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://i.stack.imgur.com/4oaXZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4oaXZ.png&quot; alt=&quot;Barcode 1&quot; /&gt;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;and this&lt;/p&gt;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://i.stack.imgur.com/6y3Gn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6y3Gn.png&quot; alt=&quot;Barcode 2&quot; /&gt;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Try multiply some numbers with 3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19803" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-03-05T22:44:10.373" Title="Last number in barcode" Tags="|algorithm|" AnswerCount="2" CommentCount="3" ClosedDate="2016-03-06T03:41:53.913" ContentLicense="CC BY-SA 3.0" />
  <row Id="28425" PostTypeId="2" ParentId="28424" CreationDate="2016-03-05T22:36:05.370" Score="0" Body="&lt;p&gt;Pretty sure its:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1.  5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2.  0&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Just by looking at previous numbers. &lt;/p&gt;&#xA;" OwnerUserId="19731" LastActivityDate="2016-03-05T22:36:05.370" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28426" PostTypeId="2" ParentId="28424" CreationDate="2016-03-05T22:38:43.980" Score="1" Body="&lt;p&gt;The barcode checksum is derived through the following process:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Add the digits at odd positions, then multiply by 3.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Add the digits at even positions to your answer.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;10 minus your result, modulo 10, is the check digit.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So for the first one: $ 10 -( ((5+1+4+8)\times3 + 1 + 8 + 3) \mod 10)=7$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for the second: $10- (((5+0+2+4)\times 3 + 7 + 1 + 3&#xA;)\mod 10) = 6$&lt;/p&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="11876" LastEditDate="2016-03-05T22:44:10.373" LastActivityDate="2016-03-05T22:44:10.373" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28427" PostTypeId="1" AcceptedAnswerId="28436" CreationDate="2016-03-06T00:11:05.540" Score="5" ViewCount="231" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;To marry me forever is the exception&lt;br&gt;&#xA;  To divorce me is the rule&lt;br&gt;&#xA;  To love me is to show respect&lt;br&gt;&#xA;  To hate me is to show a lack&lt;br&gt;&#xA;  To have me you must make a promise&lt;br&gt;&#xA;  To keep me you must break it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18340" LastEditorUserId="8874" LastEditDate="2016-03-06T14:02:36.233" LastActivityDate="2016-03-07T17:54:27.797" Title="Rain &amp; Back pain" Tags="|riddle|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28428" PostTypeId="2" ParentId="28409" CreationDate="2016-03-06T03:55:24.113" Score="5" Body="&lt;p&gt;If $a=1$, then $b=31$ or $65$, $c=65$ or $31$ and $ab+c=96$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $96&amp;gt;a&amp;gt;1$, $ab+c$ is at the minimum and $a$ is &quot;known&quot;, then $bc$ is also known and...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(ab+c)^2$ (min) $= a^2b^2 + c^2 + 2abc$ (known), $(ab+c)^2 - 4abc$ (known) $= (ab-c)^2$. so $|ab-c|$ must also be as low as possible. Even if $ab=c$, then $b=[sqrt((2016-a)/a)]$, making $ab+c=2*[sqrt((2016-a)*a)]$. The product inside is at least 4028, easily preventing us from reaching a better result than 96.&lt;/p&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="14016" LastEditDate="2016-03-07T00:25:43.843" LastActivityDate="2016-03-07T00:25:43.843" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28429" PostTypeId="1" AcceptedAnswerId="28430" CreationDate="2016-03-06T06:46:56.967" Score="10" ViewCount="676" Body="&lt;p&gt;In our university somebody posted this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;BTHGIR ENIKLN   CGEKAR   ODEPWU   MONOTT   ESLEFT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The background of the whole thing has a grid-like pattern. Can somebody decipher this? It's somehow math related cause we will be celebrating math day this March.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/un2jW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/un2jW.jpg&quot; alt=&quot;That&amp;#39;s the pic for your reference&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18377" LastEditorUserId="17482" LastEditDate="2016-03-06T07:24:46.017" LastActivityDate="2016-03-06T07:24:46.017" Title="A math related Puzzle(?!)" Tags="|enigmatic-puzzle|cipher|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28430" PostTypeId="2" ParentId="28429" CreationDate="2016-03-06T06:59:35.310" Score="20" Body="&lt;p&gt;The message is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; KEEP WALKING - DO NOT TURN. RIGHT BECOMES LEFT.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we arrange the letters in&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a 6x6 grid, we can read the letters spiraling outward&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/ayHhy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ayHhy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17147" LastActivityDate="2016-03-06T06:59:35.310" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28431" PostTypeId="1" AcceptedAnswerId="28479" CreationDate="2016-03-06T07:46:52.633" Score="8" ViewCount="572" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I was a dreamer who dreamed to the end&lt;br&gt;&#xA;  A maker of song so that you had a friend&lt;br&gt;&#xA;  I went on singing, standing by soon and late&lt;br&gt;&#xA;  In the icy cold waters of the northernmost state.&lt;br&gt;&#xA;  I am a man  from a popular game.&lt;br&gt;&#xA;  I am a beast that a beauty could tame.&lt;br&gt;&#xA;  I was nicknamed a bone at the top of my fame.&lt;br&gt;&#xA;  A hillock in Georgia reveals my name.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Feel free to have a shot at it, particularly if you're a 60s and 70s buff. &lt;/p&gt;&#xA;" OwnerUserId="18924" LastEditorUserId="8874" LastEditDate="2016-03-07T15:55:21.617" LastActivityDate="2016-03-07T15:55:21.617" Title="I'm Asking An Easy One" Tags="|riddle|word|pattern|rhyme|" AnswerCount="3" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28432" PostTypeId="2" ParentId="28306" CreationDate="2016-03-06T10:18:46.250" Score="3" Body="&lt;p&gt;Are you &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;a violin?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have a long history from east to west&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The violin has roots in the East and has been used in Western music since antiquity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;to some I'm the worst and others the best.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;In capable hands, a violin can produce beautiful music. But the sounds produced by a beginning violinist can be pretty horrible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With a core made of resin I dance and I flutter,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Rosin is a &quot;core&quot; of resin that can be used to lubricate a violin bow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I may be an object of legend or myth&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Violins are often the subject of legends, like Nero's, which was supposedly played as Rome burned&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;at first you won't find me but maybe in fifth. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Beethoven's &lt;em&gt;Fifth&lt;/em&gt; heavily features violins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;no one judges me when I speak with a stutter. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; While violins can be played fluidly, they can also be played with a staccato.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For at any point, you summon power of three,&#xA;And look to the wind, you may just find me. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Violins are often played in &lt;em&gt;trios&lt;/em&gt;. There are &lt;em&gt;three&lt;/em&gt; string instruments in Bach's &lt;em&gt;Air&lt;/em&gt; on The G String, and Air on the G String is in the &lt;em&gt;Third&lt;/em&gt; Orchestral Suite&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19890" LastActivityDate="2016-03-06T10:18:46.250" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28433" PostTypeId="2" ParentId="28431" CreationDate="2016-03-06T11:27:16.887" Score="4" Body="&lt;p&gt;Answer is : &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jarvis Jerrell Jones. Who is a Linebacker for Pittsburgh Steelers of the National Football League&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a man from a popular game.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Football.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a beast that a beauty could tame.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Position:    Linebacker. A beast in this defense.Jack Lambert Award winner in 2012. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was nicknamed a bone at the top of my fame.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Jones is sometimes referred to as &quot;Dawg Bones&quot; Jones.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A hillock in Georgia reveals my name.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  an American football linebacker for the Pittsburgh Steelers of the National Football League (NFL).Jones was born in Columbus, Georgia and grew up in Lumpkin and Richland, Georgia. He attended Carver High School in Columbus, Georgia, where he led his high school football team to a state title in 2007, and was rated as a four-star recruit by Rivals.com.In addition to playing football, he was also an all-state basketball player.He was ranked the 59th best high school prospect by ESPN.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19141" LastEditorUserId="17482" LastEditDate="2016-03-06T11:30:25.637" LastActivityDate="2016-03-06T11:30:25.637" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28434" PostTypeId="2" ParentId="28427" CreationDate="2016-03-06T12:13:56.980" Score="2" Body="&lt;p&gt;It is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bonding in atoms&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To marry me forever is the exception&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An atom bonded to another atom forever is an exception.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To divorce me is the rule&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bonds are made and broken continuously in a mixture.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To love me is to show respect&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Show respect means to share electrons&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To hate me is to show a lack&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lack of electrons causes an atom to not to make a bond.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To have me you must make a promise&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Promise of sharing of electrons&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To keep me you must break it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To make a new bond, the atom have to break a bond.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastActivityDate="2016-03-06T12:13:56.980" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28435" PostTypeId="2" ParentId="28097" CreationDate="2016-03-06T12:23:01.650" Score="2" Body="&lt;p&gt;&lt;strong&gt;Wrap-up: the making of&#xA;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/28097&quot;&gt;&lt;em&gt;Some string of letters&lt;/em&gt;&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This is not a solution to the puzzle but provides notes from its poser.&#xA; This type of answer has been&#xA; &lt;a href=&quot;https://puzzling.meta.stackexchange.com/questions/5420/&quot;&gt;approved by the community&lt;/a&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Caution:&lt;/strong&gt; This post contains information about the solution.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&#xA;(This post has been adapted from its original form,&#xA; which consisted of just the first chart, presented as a hint.)&#xA;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;b z z e a z d j j i z a i j a z z d d z z a d z e j z a d z j g a t e z a t z e d z z h a z z d&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The mystery encryption scheme was devised to require trivial effort&#xA;to encode and decode,&#xA;with a twist intended to defy most automated deciphering techniques.&#xA;The twist is that...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...each original message letter can be encoded as 26 different&#xA; encryption-letter pairs,&#xA; so that a message with 200 letters can&#xA; generally be encoded without repeated pairs.&#xA; A longer message can deliberately repeat pairs for&#xA; less common letters, such as consonants early in the alphabet,&#xA; to serve as distractions from inevitably repeated pairs&#xA; that will represent frequent letters, such as E and&amp;nbsp;T.&#xA; &lt;br&gt;&lt;sup&gt;&lt;sup&gt;&lt;sup&gt;&lt;/sup&gt;&lt;/sup&gt;&lt;/sup&gt;&#xA; The pairs used in the puzzle&amp;rsquo;s cipher&#xA; were intentionally selected to provide clues.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As the accepted answer does not mention&#xA;what led to solution, here is how clues were meant to work.&#xA;Among other considerations, solving any letter-sequence cipher&#xA;includes two initial steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 1.&lt;/strong&gt; Convert each letter into its alphabetic position number.&#xA;&lt;br&gt;&lt;sup&gt;&lt;sup&gt;&lt;/sup&gt;&lt;/sup&gt;&#xA;&lt;strong&gt;Step 2.&lt;/strong&gt; Count how many times each letter/number appears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A chart that combines these two steps&#xA;for the cipher in question&#xA;reveals an odd distribution.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;                                                                 z&#xA;                                                                 z&#xA;                                                                 z&#xA;                                                                 z&#xA;                                                                 z&#xA;                                                                 z&#xA;                                                                 z&#xA;                                                                 z&#xA;                                                                 z&#xA;               a                                                 z&#xA;               a     d                                           z&#xA;               a     d                                           z&#xA;               a     d           j                               z&#xA;               a     d e         j                               z&#xA;               a     d e         j                               z&#xA;               a     d e       i j                   t           z&#xA;letter         a b   d e   g h i j                   t           z&#xA;&#xA;alphabetic #   1 2 3 4 5 6 7 8 9 10  .   .   .   .  20          26&#xA;&#xA;count          8 1   7 4   1 1 2 5                   2          17&#xA;               :.................:                   :...........:&#xA;              Almost all letters are               The only outliers&#xA;              early in the alphabet.               are 20th and last.&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The noticeable clustering was meant as a compelling clue&#xA;while a subtler clue is that the total number&#xA;of letters is 48, which reeks of doubling,&#xA;and that the letter counts divide into neat halves.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;letter         a b   d e   g h i j                   t           z&#xA;&#xA;count          8 1   7 4   1 1 2 5                   2          17&#xA;               :...............: :...............................:&#xA;               The left portion        The right portion also&#xA;             includes 24 letters.       includes 24 letters.&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Compare all that to how this distribution could have&#xA;easily been made more cryptic.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;                       e f g h i j k l m n o p q r s t u v w x y z&#xA;letter         a b c d e f g h i j k l m n o p q r s t u v w x y z&#xA;&#xA;alphabetic #   1 2 3 4 5 6 7 8 9 10  .   .   .   .   .   .   .  26&#xA;&#xA;count          1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The actual distribution was meant&#xA;to draw attention...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...to the letters j, t and z,&#xA; along with their suggestive alphabet positions&#xA;  of 10th, 20th and last.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The next two steps were meant to follow naturally.&#xA;&lt;br&gt;&lt;sup&gt;&lt;sup&gt;&lt;/sup&gt;&lt;/sup&gt;&#xA;&lt;strong&gt;Step 3.&lt;/strong&gt; Highlight the interesting letters.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;pre&gt;&#xA;                  zz  z jj z  j zz  zz  z jz  zj  t z tz  zz  zz&#xA;                 b  ea d  i ai a  dd  ad e  ad  ga e a  ed  ha  d  &lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A very even dispersion emerges, meant to strongly suggest...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...letter pairing, which further reveals that&#xA; each pair includes exactly one j, t, or z.&#xA; &lt;pre&gt;&#xA;      | z|z | z| j|j |z | j| z|z | z|z | z| j|z | z|j | t| z| t|z | z|z | z|z |&#xA;      |b | e|a |d | i| a|i |a | d|d | a|d |e | a|d | g|a |e |a | e|d | h|a | d| &lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 4.&lt;/strong&gt; Tally...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...the letter pairs.&#xA; &lt;pre&gt;&#xA;                                              z a             z d&#xA;                                              z a             z d&#xA;                                              z a             z d&#xA;                                              z a             z d     z e&#xA;                              j i     t a     z a             z d     z e&#xA;      j d     j e     j g     j i     t a     z a     z b     z d     z e     z h&#xA;&#xA;     10 4    10 5    10 7    10 9    20 1    26 1    26 2    26 4    26 5    26 8 &lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This was meant to be the &amp;ldquo;Aha!&amp;rdquo; moment,&#xA;where the distribution can be recognized&#xA;as having characteristics of a typical&#xA;letter-count distribution, especially if...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...z&amp;thinsp;=&amp;thinsp;26 is taken as z&amp;thinsp;=&amp;thinsp;0 (&lt;strong&gt;z&lt;/strong&gt;ero)&#xA; &lt;pre&gt;&#xA;       z a           z d&#xA;       z a           z d&#xA;       z a           z d&#xA;       z a           z d    z e&#xA;       z a           z d    z e                                   j i     t a&#xA;       z a    z b    z d    z e    z h     j d    j e     j g     j i     t a&#xA;&#xA;       0 1    0 2    0 4    0 5    0 8    10 4   10 5    10 7    10 9    20 1&#xA;         1      2      4      5      8      14     15      17      19      21&#xA;         a      b      d      e      h       n      o       q       s       u &lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This may indeed represent a 24-letter message&#xA;as it has 10 unique letters, including 4 vowels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 5.&lt;/strong&gt; Rewrite the cipher...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;. ...in terms of these pairs.&#xA; &lt;pre&gt;&#xA;       bz ze az dj ji za ij az zd dz za dz ej za dz jg at ez at ze dz zh az zd&#xA;&#xA;       b  e  a  n  s  a  s  a  d  d  a  d  o  a  d  q  u  e  u  e  d  h  a  d&#xA;&#xA;           beans       as     add    a   do    ad        queued         had &lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Terse as the puzzle statement is, it does say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Don&amp;rsquo;t be satisfied with a nonsensical string of words,&#xA;  which will be just halfway to the answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This validates the progress so far&#xA;and leads to completion by repeating steps 4 and 5.&#xA;Unlike the initial letter distribution with deliberate clues,&#xA;the chart at this stage would draw attention to nothing useful.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;               a     d&#xA;               a     d&#xA;               a     d&#xA;               a     d e&#xA;               a     d e                           s   u&#xA;letter         a b   d e     h           n o   q   s   u&#xA;&#xA;alphabetic #   1 2   4 5     8          14 15  17  19  21&#xA;&#xA;count          6 1   6 3     1           1 1   1   2   2&#xA;&lt;/pre&gt;&#xA;" OwnerUserId="18129" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2016-09-22T21:37:01.570" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28436" PostTypeId="2" ParentId="28427" CreationDate="2016-03-06T13:11:04.117" Score="10" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A library book (or similar lending scheme)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To marry me forever is the exception&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You don't normally have library books forever&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To divorce me is the rule&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You generally return them&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To love me is to show respect&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you love books, you show then respect by treating them well (regardless of how good a specific book is)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To hate me is to show a lack&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hating books demonstrates a lack of character (or something)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To have me you must make a promise&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You promise to bring the book back when you sign up at the library&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To keep me you must break it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; See above&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15942" LastActivityDate="2016-03-06T13:11:04.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28437" PostTypeId="1" AcceptedAnswerId="28586" CreationDate="2016-03-06T14:05:42.743" Score="8" ViewCount="1736" Body="&lt;p&gt;A series of numbers is given in the question and it could be anything. Have tried  things related to temperature values etc. but none worked. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Guessing the next number in this series looks highly unlikely.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;16 18 20 62&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Hint: Answer is a 7 letter word.&lt;/p&gt;&#xA;" OwnerUserId="19898" LastEditorUserId="8874" LastEditDate="2016-03-08T14:59:59.003" LastActivityDate="2016-08-28T23:58:21.980" Title="Treasure hunt question" Tags="|riddle|word|treasure-hunt|" AnswerCount="1" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28438" PostTypeId="2" ParentId="28310" CreationDate="2016-03-06T14:13:58.960" Score="9" Body="&lt;p&gt;I may be off base because I'm not very good at these but I looked at the title and &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Took the first letter of each line LOOK &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I was staring at that and a thought hit me. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An acrosticolic won't stop at one, right?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm keep staring. Right before it says it's  the key?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Keep Calm And Breathe is BACK backwards in acronym plus it meets with the K in LOOK&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think this fits. To support something can be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; To BACK it &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The hint says the next line is boats. I googled the word and &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BACKING is a way to slow a sail boat. &#xA; Knots is the speed of a boat so if you're slowing down you are shedding knots in a way and a slip is the place in a marina where you anchor &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And also to &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Put something on the back of to decorate or protect and the example in the Google def was a MIRROR&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So it can't be a coincidence. The answer has to be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BACK&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19899" LastActivityDate="2016-03-06T14:13:58.960" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28439" PostTypeId="1" AcceptedAnswerId="28440" CreationDate="2016-03-06T14:27:42.843" Score="17" ViewCount="11067" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Find a mathematical expression that yields the value &lt;span class=&quot;math-container&quot;&gt;$2016$&lt;/span&gt; while obeying the following rules:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Each of the digits &lt;span class=&quot;math-container&quot;&gt;$1,2,3,4,5,6,7,8,9$&lt;/span&gt; is used &lt;strong&gt;exactly once&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;Decimal points are allowed&lt;/li&gt;&#xA;&lt;li&gt;You may use brackets &amp;quot;(&amp;quot; and &amp;quot;)&amp;quot; to structure your expression, and to make it well-defined&lt;/li&gt;&#xA;&lt;li&gt;The only allowed mathematical operations are addition (+), subtraction (-), multiplication (*) and division (÷)&lt;/li&gt;&#xA;&lt;li&gt;The only allowed mathematical functions are square-roots and logarithms.&#xA;Logarithms must be written in the form &lt;span class=&quot;math-container&quot;&gt;$\log[b](x)$&lt;/span&gt; to denote the base-&lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; logarithm of number &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;br&gt;&#xA;Note that in particular the following is **not allowed**:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Juxtaposition of digits (as juxtaposing 1 and 3 to get &amp;quot;31&amp;quot;)&lt;/li&gt;&#xA;&lt;li&gt;other mathematical operations and functions (cube-roots, exponentiation, factorials, absolute values, trigonometric functions, etc)&lt;/li&gt;&#xA;&lt;li&gt;matrices and determinants&lt;/li&gt;&#xA;&lt;li&gt;integration, differentiation, limits&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2020-12-09T08:14:30.897" LastActivityDate="2023-06-20T12:46:40.433" Title="A truly amazing way of making the number 2016" Tags="|mathematics|calculation-puzzle|formation-of-numbers|" AnswerCount="18" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="28440" PostTypeId="2" ParentId="28439" CreationDate="2016-03-06T14:38:50.557" Score="48" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; ((9 * 8 * 7 * 6 * 2) / 3) + 5 - 4 - 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;And three more &lt;em&gt;à la Perry&lt;/em&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 1 * (2 - 3 + 4 - 5 + 6) * 7 * 8 * 9&lt;br /&gt;&#xA; (1 + 2 + 3 + 4 + 5 + 6 + 7) * 8 * 9&lt;br /&gt;&#xA; 1 * (2 + 3 + 4 + 5) * 6 * (7 + 8 + 9)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17609" LastEditorUserId="69582" LastEditDate="2020-12-08T23:39:49.773" LastActivityDate="2020-12-08T23:39:49.773" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28441" PostTypeId="2" ParentId="28439" CreationDate="2016-03-06T14:41:16.040" Score="11" Body="&lt;p&gt;How about:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(9 \cdot 8 \cdot 7)(6 - 2)\frac{5 - 3}{\sqrt4}\times1$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastEditorUserId="10118" LastEditDate="2016-03-06T14:58:27.733" LastActivityDate="2016-03-06T14:58:27.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28442" PostTypeId="2" ParentId="28439" CreationDate="2016-03-06T14:46:48.210" Score="49" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; (((1 - 2) x (3 + 4)) + 5 + 6) x 7 x 8 x 9&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11497" LastEditorUserId="69582" LastEditDate="2020-12-08T23:37:35.007" LastActivityDate="2020-12-08T23:37:35.007" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28443" PostTypeId="2" ParentId="28439" CreationDate="2016-03-06T14:49:32.997" Score="9" Body="&lt;p&gt;A simple answer:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 1·2·3·6·7·8 + 0·4·5·9 = 2016; based off of the fact that 2016=2&lt;sup&gt;5&lt;/sup&gt;·3&lt;sup&gt;2&lt;/sup&gt;·7.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If that feels too cheaty, then try this:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 1·2·3·6·7·8 + 4 + 5 - 9 = 2016&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11098" LastEditorUserId="11098" LastEditDate="2020-12-09T01:38:07.570" LastActivityDate="2020-12-09T01:38:07.570" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28444" PostTypeId="1" AcceptedAnswerId="28447" CreationDate="2016-03-06T14:52:03.953" Score="10" ViewCount="2557" Body="&lt;p&gt;This is a follow-up to &quot;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/28439/a-truly-amazing-way-of-making-the-number-2016&quot;&gt;A truly amazing way of making the number 2016&lt;/a&gt;&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For every positive integer $n$, find a mathematical expression that &#xA;  yields the value $n$ while obeying the following rules:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Each of the digits $1,2,3,4,5,6,7,8,9$ is used &lt;strong&gt;exactly once&lt;/strong&gt;&lt;/li&gt;&#xA;  &lt;li&gt;Decimal points are allowed&lt;/li&gt;&#xA;  &lt;li&gt;You may use brackets &quot;(&quot; and &quot;)&quot; to structure your expression, and &#xA;  to make it well-defined&lt;/li&gt;&#xA;  &lt;li&gt;The only allowed mathematical operations are addition (+), &#xA;  subtraction (-), multiplication (*), division (/); the minus sign may also be used as the sign of a negative number. &lt;/li&gt;&#xA;  &lt;li&gt;The only allowed mathematical functions are square-roots and logarithms.&#xA;  Logarithms must be written in the form $\log[b](x)$ to denote the base-$b$ logarithm of number $x$&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;Note that in particular the following is &lt;strong&gt;not allowed&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Juxtaposition of digits (as juxtaposing 1 and 3 to get &quot;31&quot;)&lt;/li&gt;&#xA;&lt;li&gt;using the digit 0, or using non-decimal digits&lt;/li&gt;&#xA;&lt;li&gt;other mathematical operations and functions (cube-roots, exponentiation, factorials, absolute values, trigonometric functions, etc)&lt;/li&gt;&#xA;&lt;li&gt;integration, differentiation, limits, matrices, and determinants&lt;/li&gt;&#xA;&lt;li&gt;rounding up, rounding down, rounding to the nearest integer&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2017-03-21T02:45:59.163" Title="A truly amazing way of making every possible positive integer" Tags="|mathematics|calculation-puzzle|formation-of-numbers|" AnswerCount="4" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28445" PostTypeId="2" ParentId="28439" CreationDate="2016-03-06T15:17:05.257" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; ((8 + 6 + 3 + (2 * 7)) * (9 + 4) * 5) + 1 = 2016&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="69582" LastEditDate="2020-12-08T23:25:52.427" LastActivityDate="2020-12-08T23:25:52.427" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28446" PostTypeId="2" ParentId="28444" CreationDate="2016-03-06T15:30:19.213" Score="6" Body="&lt;p&gt;Partial answer:$$1/\log[\sqrt4](\underbrace{\surd\surd\cdots\surd}_\text{$n$ times}2)+3-5-6+7-8+9=2^n$$Adjust the added values to get nearby integers. Also, changing the base of the logarithm and the radicand can produce powers of other numbers.&lt;/p&gt;&#xA;" OwnerUserId="11098" LastEditorUserId="11098" LastEditDate="2016-04-17T20:02:56.147" LastActivityDate="2016-04-17T20:02:56.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28447" PostTypeId="2" ParentId="28444" CreationDate="2016-03-06T15:40:45.287" Score="20" Body="&lt;p&gt;We can use:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\log[2]\left(\frac{1}{\log[3]\left(\underbrace{\sqrt{\sqrt{ \ldots \sqrt{5-\sqrt{4}}}}}_\text{square root repeated $n$ times}\right)}\right) + 6 - 7 - 8 + 9$$   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This works because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $a^{\overbrace{b \cdot b \ldots b}^\text{$n$ times}} = a^{b^n}$, and since $\sqrt{a} = a^{\frac12}$ we have that $\sqrt{\sqrt{ \ldots \sqrt{5 - \sqrt{4}}}} = 3^{\left(\frac12\right)^n} = 3^{\frac1{2^n}}$&lt;br&gt;&#xA; Now, $\text{log}[a](a^c) = c$, so $\text{log}[3]\left(3^{\frac1{2^n}}\right) = \frac1{2^n}$&lt;br&gt;&#xA; Finally, by the same token:&#xA;\begin{equation}\text{log}[2]\left(\frac1{\frac1{2^n}}\right) = \text{log}[2]\left(2^n\right) = n\end{equation}&#xA; The rest is simply adding the unneeded numbers in a zero sum.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you want a version with the numbers &lt;strong&gt;in order&lt;/strong&gt; for 1 to 9, we can do a small manipulation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$-1\cdot\log[2]\left(\log[3]\left(\underbrace{\sqrt{\sqrt{ \ldots \sqrt{-\sqrt{4}+5}}}}_\text{square root repeated $n$ times}\right)\right) + 6 - 7 - 8 + 9$$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastEditorUserId="30903" LastEditDate="2017-03-21T02:45:59.163" LastActivityDate="2017-03-21T02:45:59.163" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28448" PostTypeId="2" ParentId="28439" CreationDate="2016-03-06T15:58:16.033" Score="12" Body="&lt;p&gt;A variation on Jon Mark Perry's answer:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; (1 + 2 - 3 - 4) x (5 - 6) x 7 x 8 x 9&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I wonder how many such 'monotonic' solutions there are?&lt;/p&gt;&#xA;" OwnerUserId="19901" LastEditorUserId="69582" LastEditDate="2020-12-08T23:34:58.473" LastActivityDate="2020-12-08T23:34:58.473" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28449" PostTypeId="1" AcceptedAnswerId="28452" CreationDate="2016-03-06T16:24:20.550" Score="7" ViewCount="174" Body="&lt;p&gt;There is a list with $K$ distinct positive integers so that&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;exactly 1 number in the list is divisible by $K$,&lt;/li&gt;&#xA;&lt;li&gt;exactly 2 numbers in the list are divisible by $K-1$,&lt;/li&gt;&#xA;&lt;li&gt;exactly 3 numbers in the list are divisible by $K-2$,&lt;/li&gt;&#xA;&lt;li&gt;... ... ... ... ...&lt;/li&gt;&#xA;&lt;li&gt;exactly $K$ numbers in the list are divisible by $1$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What's the largest possible value of $K$ under these circumstances?&lt;/p&gt;&#xA;" OwnerUserId="19470" LastActivityDate="2016-03-06T20:36:56.513" Title="A list with many divisibilities" Tags="|mathematics|combinatorics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28450" PostTypeId="2" ParentId="28293" CreationDate="2016-03-06T16:28:40.023" Score="-1" Body="&lt;p&gt;There was a sound so horrible it killed Bob, Alice, the window, and glass of water.&lt;/p&gt;&#xA;" OwnerUserId="9993" LastActivityDate="2016-03-06T16:28:40.023" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28451" PostTypeId="1" AcceptedAnswerId="28455" CreationDate="2016-03-06T16:28:50.650" Score="12" ViewCount="400" Body="&lt;p&gt;True or false: The integers $ 1 , 2 , ... , 999 $ can be arranged in a row in such a way that no sum of any number of consecutive elements is an integer multiple of 1000.&lt;/p&gt;&#xA;" OwnerUserId="19470" LastActivityDate="2016-03-06T18:22:18.677" Title="Arranging the numbers from 1 to 999 in a row" Tags="|mathematics|combinatorics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28452" PostTypeId="2" ParentId="28449" CreationDate="2016-03-06T16:43:36.337" Score="6" Body="&lt;p&gt;The largest possible value is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;K = 5&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This can be shown by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Observing that K = 5 is possible (1, 2, 6, 12, 60 works)&lt;br&gt;&#xA;&lt;br&gt;&#xA; Observing that for K = 6, we need five numbers divisible by 2 and four numbers divisible by 3.  This means at minimum, three of our numbers must overlap and be divisible by 6, which is a failure because exactly one should be.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;f&quot; did a more eloquent job than me of extending this proof to all larger values of K:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;&quot;For K&gt;5, only one number isn't divisible by 2 and two numbers aren't divisible by 3. That means at most three numbers aren't divisible by 6, but there need to be five that aren't.&quot;&lt;/em&gt; - f&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="17389" LastEditDate="2016-03-06T19:53:08.070" LastActivityDate="2016-03-06T19:53:08.070" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28453" PostTypeId="2" ParentId="28444" CreationDate="2016-03-06T16:47:07.387" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\dfrac{(9-8)\times(7-6)-(5-4)}{3-2-1}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11497" LastEditorUserId="11497" LastEditDate="2016-03-06T17:38:49.050" LastActivityDate="2016-03-06T17:38:49.050" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="28454" PostTypeId="2" ParentId="24270" CreationDate="2016-03-06T16:58:50.180" Score="0" Body="&lt;p&gt;General &quot;Countdown&quot; arithmetic puzzle solution !!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some Unix Shells have a property called 'brace expansion'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usage varies between &lt;strong&gt;csh&lt;/strong&gt; and &lt;strong&gt;bash&lt;/strong&gt; but is something like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;echo {a,i}{s,t}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Outputs:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;as at is it&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This can be used to solve the &quot;Countdown&quot; letters puzzle, simply have 9 consecutive braces with all 9 letters in each.  If you don't know Countdown, you can use this to find Scrabble™ words, if you add a final brace at the end like &lt;strong&gt;{'\n'}&lt;/strong&gt; you even get one result per line so you can pipe output through &lt;strong&gt;sort&lt;/strong&gt; or &lt;strong&gt;usr/dict&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how do we do the Arithmetic puzzle ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we are given the integers 2,4,5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternate braces containing integers and operators:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;set array {2,4,5}{+,-,\*,/}{2,4,5}…{'\n'} # * escaped to avoid wildcard&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This gives:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2+2-4…&#xA;2-3*4…&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Every permutation of integer and operator.  Finally,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;foreach s in (array) {echo $s ; eval $s} | sort&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Evaluates all the expressions and even sorts them so you can find the &lt;strong&gt;integer you are looking for&lt;/strong&gt; and the expression which created it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Countdown&quot;™ Solved !!!&lt;/p&gt;&#xA;" OwnerUserId="19677" LastActivityDate="2016-03-06T16:58:50.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28455" PostTypeId="2" ParentId="28451" CreationDate="2016-03-06T17:43:00.607" Score="13" Body="&lt;p&gt;It is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;true&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Arrange them in this order:$$1,998,3,996,5,994,\ldots,995,4,997,2,999$$For $0\le n\le999$ the sum of the first $n$ terms, mod 1000, is $1000-\frac{n}{2}$ if $n$ is even, and $\frac{n+1}{2}$ if $n$ is odd. Note that these are all different from each other, so the difference between any two is never zero mod 1000. The sum of the terms from the $i$th through the $j$th is the sum of the first $j$ terms, minus the sum of the first $i-1$ terms. Because this is never zero mod 1000, no sum of consecutive elements is a multiple of 1000.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2016-03-06T17:43:00.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28457" PostTypeId="2" ParentId="28439" CreationDate="2016-03-06T19:52:44.947" Score="8" Body="&lt;p&gt;My favorite so far:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 9 * 8 * 7 * 6 * (5 - 4) / 3 * 2 * 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Had factorials been allowed, I'd have preferred&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 9 * 8 * 7 * 6! / 5! * 4 / 3 / 2 * 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19914" LastEditorUserId="69582" LastEditDate="2020-12-08T23:31:43.407" LastActivityDate="2020-12-08T23:31:43.407" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28458" PostTypeId="2" ParentId="28439" CreationDate="2016-03-06T19:54:40.190" Score="19" Body="&lt;p&gt;Since 2016 is a multiple of 504 = 7 x 8 x 9, any expression using 1, 2, 3, 4, 5, 6 that evaluates to 4 can just multiply those three afterwards and evaluate to 2016.&lt;/p&gt;&#xA;&lt;p&gt;So all the following are correct:&lt;/p&gt;&#xA;&lt;p&gt;9 x 8 x 7 x (6 - 5) x 4 x (3 - 2) x 1 = 2016&lt;br /&gt;&#xA;9 x 8 x 7 x (6 - 5) x 4 + 3 - 2 - 1 = 2016&lt;br /&gt;&#xA;9 x 8 x 7 x 6 / (5 + 4) x (3 + 2 + 1) = 2016&lt;/p&gt;&#xA;&lt;p&gt;There are probably at least fifty more variations that I didn't bother to work out.&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="69582" LastEditDate="2020-12-08T23:36:58.193" LastActivityDate="2020-12-08T23:36:58.193" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28459" PostTypeId="1" AcceptedAnswerId="28462" CreationDate="2016-03-06T20:20:06.490" Score="7" ViewCount="468" Body="&lt;p&gt;While I was writing on a file on my Linux portable computer, the thing I was writing suddenly started to change!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at the file:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;¶€µ€µ”€¶ ŧø ðø ænøŧħ€¶ @€n¢øð→nŋ €»ŧ¶€µ€@ đø¶ ŧøðæ←|&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What happened to my computer or what did I do to &quot;encode&quot; the text?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, what's written on the file?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I didn't found any help on Google.)&lt;/p&gt;&#xA;" OwnerDisplayName="user18031" LastEditorUserId="69582" LastEditDate="2023-11-02T16:04:39.117" LastActivityDate="2023-11-02T16:04:39.117" Title="Encoding Extreme #3" Tags="|cipher|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28460" PostTypeId="2" ParentId="28459" CreationDate="2016-03-06T20:28:40.263" Score="6" Body="&lt;p&gt;It's a simple substitution cipher where (most of) the symbols resemble the corresponding letter&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Examples:&lt;br&gt;&#xA; € = E, ¢ = C, ŧ = T, ¶ = R&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It appears to say&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; REMEMBER TO DO ANOTHER &quot;ENCODING EXTREME&quot; FOR TODAY.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17147" LastActivityDate="2016-03-06T20:28:40.263" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28461" PostTypeId="2" ParentId="28449" CreationDate="2016-03-06T20:36:56.513" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $K-1$ of them are div. by 2 and $K-2$ are div. by 3, so at least $K-3$ must be multiples of both (div. by 6), which contradicts the info given in the question. That can only be avoided if $K&amp;lt;6$. A legit list is possible if $K=5$ - the numbers can be 1, 6, 12, 14 and 60 for example.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastActivityDate="2016-03-06T20:36:56.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28462" PostTypeId="2" ParentId="28459" CreationDate="2016-03-06T21:27:09.267" Score="6" Body="&lt;p&gt;You were holding one of the “alternate graphics” keys on your Canadian Multilingual Standard keyboard:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/blQbL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/blQbL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note how the dark green glyph on the ‘R’ key is ¶, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only ‘E’ is different; I guess that one is vendor-specific or something.&lt;/p&gt;&#xA;" OwnerUserId="11476" LastActivityDate="2016-03-06T21:27:09.267" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28463" PostTypeId="1" AcceptedAnswerId="28486" CreationDate="2016-03-07T00:06:47.953" Score="11" ViewCount="289" Body="&lt;p&gt;Oh guys... I'm in front of an ATM and want to withdraw some money, but I forgot my 4 digit PIN. Can somebody help me out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a backup card with 6x6 fields, each field has exactly one character!&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    - A B C D -&#xA;    A - 1 1 3 G&#xA;    B 9 - 1 2 C&#xA;    C 9 1 - 1 D&#xA;    D 9 3 6 - -&#xA;    - L C M - -&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;How do I have to use it?&lt;/p&gt;&#xA;" OwnerUserId="4342" LastEditorUserId="30903" LastEditDate="2016-12-20T21:34:05.653" LastActivityDate="2016-12-20T21:34:05.653" Title="I forgot my PuzzlingCard PIN" Tags="|mathematics|calculation-puzzle|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28464" PostTypeId="1" AcceptedAnswerId="28467" CreationDate="2016-03-07T00:55:54.133" Score="5" ViewCount="341" Body="&lt;p&gt;A rich man is on the verge of dying. He has unlimited amount of money and intends to distribute some of it among his $n$ relatives with the following conditions :-&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Total money to be distributed must be a positive multiple of $j$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;No relative gets more than $mj-1$ dollars. (A relative may not get money)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Find total number of ways in which the rich man can distribute the money.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take $n=6, j=5,$ and $m=3$&lt;/p&gt;&#xA;" OwnerUserId="17970" LastEditorUserId="17970" LastEditDate="2016-03-07T01:15:06.987" LastActivityDate="2016-03-08T13:49:55.463" Title="Where's my money?" Tags="|combinatorics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28465" PostTypeId="2" ParentId="28439" CreationDate="2016-03-07T01:05:06.340" Score="37" Body="&lt;p&gt;For those interested, &lt;a href=&quot;http://pastebin.com/Up2U2VFh&quot;&gt;here&lt;/a&gt; is a brute-forced list of all &quot;monotonic&quot; expressions that use addition, subtraction, multiplication, and division. There's 835 of them in total.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; For extra fun, &lt;a href=&quot;http://pastebin.com/Xs7efEy5&quot;&gt;here&lt;/a&gt; is a list of the 1,151 expressions that use 9 through 1 in order!&lt;/p&gt;&#xA;" OwnerUserId="11098" LastEditorUserId="11098" LastEditDate="2016-03-07T17:08:42.207" LastActivityDate="2016-03-07T17:08:42.207" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28466" PostTypeId="1" CreationDate="2016-03-07T02:34:24.673" Score="7" ViewCount="1866" Body="&lt;p&gt;I think today I have faced a bug on rubik's cube. As it should be, I followed the steps Cross and then F2L, instead of OLL algorithms, I used R'FRF' and three corner swapping moves to complete last layer. After a long period, I saw that all corners and 2/4 edges solved correctly. There is not an algorithm to swap two edges. Is it possible ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.hizliresim.com/bDJ1Lb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.hizliresim.com/bDJ1Lb.png&quot; alt=&quot;picture&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.hizliresim.com/d3kr7X.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.hizliresim.com/d3kr7X.png&quot; alt=&quot;picture&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 11 days we have passed I got an original cube. I have been trying to reach same position for 11 days so I did it today. This time I took photos and if it is needed I can upload more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.hizliresim.com/YkNM3Z.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.hizliresim.com/YkNM3Z.jpg&quot; alt=&quot;picture&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19661" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:48.853" LastActivityDate="2022-04-05T13:05:35.130" Title="Can a Rubik's cube be put in a position not solvable by Fridrich's algorithm?" Tags="|rubiks-cube|reachability|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="28467" PostTypeId="2" ParentId="28464" CreationDate="2016-03-07T02:35:36.347" Score="8" Body="&lt;p&gt;Assuming that only whole dollars can be given,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;There are $mj$ possibilities for the first $n-1$ relatives, and $m$ for the last one to make the total divisible by $j$. The answer is $m^nj^{n-1}$. For the given values, this is $2278125$ ways.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12408" LastActivityDate="2016-03-07T02:35:36.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28468" PostTypeId="2" ParentId="28466" CreationDate="2016-03-07T03:05:35.103" Score="10" Body="&lt;p&gt;There is no route from this position to a solved cube, using any algorithm. The reason is fairly simple: two edges cannot be swapped on a Rubik's cube. You can only swap edges in groups of three, called &quot;cycles.&quot; (You may have heard the term 3-cycle before, and that's what it refers to.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Along the same vein, this position can't be &lt;em&gt;reached&lt;/em&gt; via any scramble. (If we could get there, we could get back.) As a result, you'll never see a position like this come up, if you start solved. &lt;strong&gt;Some positions are not possible on a Rubik's cube.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result, I suppose, &lt;em&gt;technically&lt;/em&gt; there are positions which can't be solved by Fridrich's algorithm, but you can't reach them if you start from a solved cube.&lt;/p&gt;&#xA;" OwnerDisplayName="user20" LastEditorDisplayName="user20" LastEditDate="2016-03-07T03:58:45.990" LastActivityDate="2016-03-07T03:58:45.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28469" PostTypeId="2" ParentId="28439" CreationDate="2016-03-07T04:48:32.150" Score="6" Body="&lt;p&gt;In the spirit of the question, I felt it was time for a solution that includes all allowed operations, including a decimal point, a logarithm and a square root (using &lt;a href=&quot;https://puzzling.stackexchange.com/a/28458/3941&quot;&gt;the hint from Joe Z&lt;/a&gt; about the general structure of a solution):&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 9. * 8 * 7 * (-6 + 5 + log[2](4) * 3 / √1)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;or, with a small tweak from the comment by &lt;a href=&quot;https://puzzling.stackexchange.com/users/12408/f&quot;&gt;@f''&lt;/a&gt;, this:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 9 * 7 * 4 * (2 - 3 / log[5](√((1 + .6) / 8)))&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;which not only has +, -, *, /, decimal, sqrt, and log, but makes every operation &amp;quot;do something&amp;quot;&lt;/p&gt;&#xA;" OwnerUserId="3941" LastEditorUserId="69582" LastEditDate="2020-12-08T23:29:53.453" LastActivityDate="2020-12-08T23:29:53.453" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28471" PostTypeId="2" ParentId="28466" CreationDate="2016-03-07T06:27:59.087" Score="5" Body="&lt;p&gt;Short answer is - No not by face turns alone from a solvable cube, but yes you may &quot;put&quot; the edges in a position that can not be solved by &lt;em&gt;any&lt;/em&gt; algorithm that uses face turns alone (which would include Fridrich's algorithm).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were assembling a Rubik's cube you would have:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A frame of three axes with the six centre faces on the ends of each axis;&lt;/li&gt;&#xA;&lt;li&gt;Eight corner pieces; and&lt;/li&gt;&#xA;&lt;li&gt;Twelve edge pieces.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You could then choose where to put each piece in the frame and in which orientation to put each piece. The first corner may go in one of eight locations, the second in one of seven, and so on until there is only one possible location for the last corner piece and each can be placed in one of three orientations. Similarly the first edge may go in one of twelve locations, the second in one of eleven, and so on and each can be placed in one of two orientations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are, therefore, $8!*3^8*12!*2^{12}=519,024,039,293,878,272,000$ ways to put the cube together, however:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$1/2$ of the ways you could place all the edges will be such that one edge is flipped relative to the others,&lt;/li&gt;&#xA;&lt;li&gt;$2/3$ of the ways you could place all the corners will be such that one corner is twisted relative to the others.&lt;/li&gt;&#xA;&lt;li&gt;$1/2$ of the ways you could place all the edges and corners together will have two corners swapped or, equivalently, two edges swapped (as in your case).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So only $1/2/3/2=1/12$ of the ways you could put the cube together will yield a solvable cube.&lt;br&gt;(Hence there are $8!*3^7*12!*2^{10}=43,252,003,274,489,856,000$ valid states)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for your case is that the parity of the permutations generated by face turns is even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to transform your posted cube from having two edges swapped to having two corners swapped would be to perform the following 11 turn transformation, given  that &lt;strong&gt;U&lt;/strong&gt; is white and &lt;strong&gt;F&lt;/strong&gt; is blue:&lt;br&gt;&lt;strong&gt;R' U2 L' B' L U2 R' F R' F' R'&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="16717" LastEditorUserId="16717" LastEditDate="2016-03-07T06:45:06.397" LastActivityDate="2016-03-07T06:45:06.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28472" PostTypeId="2" ParentId="28431" CreationDate="2016-03-07T06:43:32.710" Score="1" Body="&lt;p&gt;Is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; John Lennon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dreamer and song maker are obvious. &lt;br&gt;&#xA; &quot;with a little help from my friends&quot;&lt;br&gt;&#xA; Spent a lot of time in Alaska. &lt;br&gt;&#xA; Popular game can reference music industry. &lt;br&gt;&#xA; Beast references Beatles and the beauty is yoko. &lt;br&gt;&#xA; Georgia guidestones?&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19731" LastActivityDate="2016-03-07T06:43:32.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28473" PostTypeId="1" AcceptedAnswerId="28474" CreationDate="2016-03-07T07:09:46.780" Score="4" ViewCount="506" Body="&lt;p&gt;Many people have fond memories of us.  A little over two years ago, said memories resurfaced when one of us died.  It's a shame, because that means we'll have to deal with him again at some point, and having to look him in the face while doing so is &lt;em&gt;not&lt;/em&gt; gonna be fun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this summer, your happy memories of us will &lt;em&gt;probably&lt;/em&gt; be ruined.  In fact, people are already making fun of our successors on social media, and they haven't even had a chance to really prove themselves yet.  Their grand entrance to the public eye was...underwhelming, to say the least.&lt;/p&gt;&#xA;" OwnerUserId="19856" LastActivityDate="2016-03-07T07:23:06.800" Title="The gang's all here?" Tags="|riddle|lateral-thinking|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28474" PostTypeId="2" ParentId="28473" CreationDate="2016-03-07T07:23:06.800" Score="5" Body="&lt;p&gt;Are you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ghostbusters&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Many people have fond memories of us. A little over two years ago, said memories resurfaced when one of us died.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Harold Ramis died 24 February 2014&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's a shame, because that means we'll have to deal with him again at some point, and having to look him in the face while doing so is not gonna be fun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this summer, your happy memories of us will probably be ruined.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The new movie's coming out in July&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In fact, people are already making fun of our successors on social media, and they haven't even had a chance to really prove themselves yet.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;em&gt;trailer&lt;/em&gt; has many thumbs down on YouTube&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Their grand entrance to the public eye was...underwhelming, to say the least.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The original announcement was... not memorable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17147" LastActivityDate="2016-03-07T07:23:06.800" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28475" PostTypeId="1" CreationDate="2016-03-07T08:34:35.197" Score="0" ViewCount="1293" Body="&lt;p&gt;Suppose we get a rope that's about 40,000 kilometers long, and tie it snug around the Earth's equator (imagine a perfectly spherical and solid earth). Now let's add just 10 meters of slack to the rope and distribute it around the planet so that the rope is equidistant from the surface at all points (vertically).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much space is between the rope and the Earth's surface? Before solving numerically, take a ballpark guess - can you fit a mountain under the rope? Walk underneath it? Can you even slide a sheet of paper between the rope and the Earth's surface?&lt;/p&gt;&#xA;" OwnerUserId="12730" LastActivityDate="2016-05-24T19:25:58.337" Title="A rope around the world" Tags="|mathematics|calculation-puzzle|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28476" PostTypeId="2" ParentId="28475" CreationDate="2016-03-07T09:14:13.313" Score="4" Body="&lt;p&gt;Circumference is ~ 3 x Diameter. 10 meters in circumference means that the diameter is ~3 meters larger. So the rope should be roughly 1.5 meters above the ground.&lt;/p&gt;&#xA;" OwnerUserId="19792" LastActivityDate="2016-03-07T09:14:13.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28477" PostTypeId="2" ParentId="28427" CreationDate="2016-03-07T09:22:11.660" Score="1" Body="&lt;p&gt;My answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Master/Apprentice relationship in the Star Wars Universe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To marry me forever is the exception&lt;br&gt;&#xA;To divorce me is the rule&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eventually, the apprenticeship ends. In the Jedi Order, they are trained until they become Jedi Knights; in the Sith Order, generally one betrays the other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To love me is to show respect&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Generally the master chooses his/her apprentice, and in order to continue to work together harmoniously, the master and apprentice must share a platonic love and show respect and to one other. Examples can be seen in the movies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To hate me is to show a lack&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Introducing hate into the Master/Apprentice relationship means being disrespectful or unfaithful to each other and, to quote, &lt;code&gt;&quot;I find this lack of faith disturbing&quot;&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To have me you must make a promise&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Members of each order take an oath to follow their respective Code. Those in the Jedi Order follow the Jedi Code; those in the Sith Order follow the Sith Code.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To keep me you must break it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Sith Code contains the line, &lt;code&gt;&quot;...my chains are broken&quot;&lt;/code&gt;. From a certain point of view, it can be interpreted as being free, but the apprentice is not free until his chains are broken. In theory, s/he can keep the Master/Apprentice relationship forever by breaking the oath.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18685" LastEditorUserId="7586" LastEditDate="2016-03-07T17:54:27.797" LastActivityDate="2016-03-07T17:54:27.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28478" PostTypeId="2" ParentId="28293" CreationDate="2016-03-07T09:33:16.297" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &quot;room&quot; is a submarine.&lt;p&gt;&#xA; &#xA; When the exterior pressure got too high during a descent, the high window shattered inward and the submarine flooded, causing both Alice and Bob to drown (and incidentally breaking an otherwise unrelated Crimson Flounder™-brand drinking glass that happened to be present). The flooding also accounts for the water on the floor.  (In fact, there was rather a &lt;em&gt;lot&lt;/em&gt; of water on the floor.  And almost everywhere else.)&lt;P&gt;&#xA; &#xA; The room being a submarine also explains why the room would contain a supply of food and water, why there are no weapons to hand, and why the room would even be sealed in the first place.&lt;P&gt;&#xA;&#xA; Presumably the inspector accidentally discovered the remains of the room (and thereby the bodies of Alice and Bob) while engaging in some scuba training, in an effort to gain some impressive new skills which would cause his dreamy (but frustratingly tsundere) supervisor to &lt;em&gt;finally notice him&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9286" LastActivityDate="2016-03-07T09:33:16.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28479" PostTypeId="2" ParentId="28431" CreationDate="2016-03-07T10:26:44.477" Score="8" Body="&lt;p&gt;The answer might be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; King&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Explanation:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I was a dreamer who dreamed to the end  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Martin Luther &lt;strong&gt;King&lt;/strong&gt; with &quot;I have a dream&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A maker of song so that you had a friend  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Carole &lt;strong&gt;King&lt;/strong&gt;: &quot;You've Got A Friend&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I went on singing, standing by soon and late  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Song &quot;Stand by me&quot; by singer-songwriter Ben E. &lt;strong&gt;King&lt;/strong&gt;&lt;br&gt;&#xA; or Movie &quot;Stand by me&quot; based on novel by Stephen &lt;strong&gt;King&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the icy cold waters of the northernmost state.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alaskan &lt;strong&gt;King&lt;/strong&gt; crab&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a man from a popular game.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;strong&gt;king&lt;/strong&gt; in the game of chess&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am a beast that a beauty could tame.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;King&lt;/strong&gt; Kong with closing quote &quot;It was beauty killed the beast&quot; (courtesy of Hugh Meyers)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was nicknamed a bone at the top of my fame.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Elvis &quot;the pelvis&quot; Presley, the &lt;strong&gt;King&lt;/strong&gt; of Rock and Roll, (courtesy of Hugh Meyers)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A hillock in Georgia reveals my name.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; hilloc &lt;strong&gt;K in G&lt;/strong&gt; eorgia (found by @Deusovi)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-07T15:54:34.177" LastActivityDate="2016-03-07T15:54:34.177" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28480" PostTypeId="2" ParentId="28360" CreationDate="2016-03-07T10:38:30.967" Score="5" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Fuse&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll start to count while you solve this problem,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A fuse starts the countdown to an explosion&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and just relax here at the top of my column.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A fuse is at the top of a cannon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Everywhere I go I let out a good whistle, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A fuse makes a whistling sound&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but don't mess with me cause I carry a pistol.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Im connected to a cannon, kinda like a pistol&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've been explosive at one point in history, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They used to use blackpowder, which is explosive&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;what I truly am is not much of a mystery.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everybody knows what a fuse is&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19928" LastActivityDate="2016-03-07T10:38:30.967" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28481" PostTypeId="1" AcceptedAnswerId="28484" CreationDate="2016-03-07T10:50:43.247" Score="5" ViewCount="1547" Body="&lt;p&gt;You &lt;em&gt;might&lt;/em&gt; be holding me in your hands, but the author can't be sure of that because he's not standing right besides you.  What the author can be reasonably sure about is that your day job involves looking at me (or things similar to me) a lot.&lt;/p&gt;&#xA;" OwnerUserId="19856" LastEditorUserId="11876" LastEditDate="2016-03-07T11:30:14.517" LastActivityDate="2016-03-07T11:30:14.517" Title="You're looking at me right now" Tags="|riddle|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28483" PostTypeId="2" ParentId="28481" CreationDate="2016-03-07T10:56:47.697" Score="10" Body="&lt;p&gt;The answer seems to be this.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a laptop or a mobile phone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reason being-&#xA;You're looking at me right now.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes Sir, I am.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You might be holding me in your hands, but the author can't be sure of that because he's not standing right besides you. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He can't be sure about the device on which you are viewing this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What the author can be reasonably sure about is that your day job involves looking at me (or things similar to me) a lot.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This, because of the wide assumption that most of us on StackExchange belong to the group whose day job comprises staring at computer screens.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19656" LastEditorUserId="19656" LastEditDate="2016-03-07T11:09:03.010" LastActivityDate="2016-03-07T11:09:03.010" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28484" PostTypeId="2" ParentId="28481" CreationDate="2016-03-07T11:21:15.000" Score="12" Body="&lt;p&gt;You are a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Screen&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Author is pretty sure at our Day job we do keep coming back to Puzzling SE and we are looking at it right now&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;through a &lt;strong&gt;screen&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19191" LastActivityDate="2016-03-07T11:21:15.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28485" PostTypeId="2" ParentId="28393" CreationDate="2016-03-07T11:38:05.783" Score="1" Body="&lt;p&gt;I don't know who the &quot;master&quot; is you keep referring to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Elvis?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I'm pretty sure the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Disco&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The three you refer to who played a big part where two of them are dead are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Bee Gees of which some of their lyrics and songs are in the riddle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The large organisation preserving them for all time I'd say&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I-Tunes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19931" LastActivityDate="2016-03-07T11:38:05.783" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28486" PostTypeId="2" ParentId="28463" CreationDate="2016-03-07T12:11:10.593" Score="6" Body="&lt;p&gt;Your pin is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 9476 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Argument:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - The table lists the four digits (called $A,B,C,D$).&lt;br&gt;&#xA; - The upper right triangle lists the digit sums of the greatest common divisors; for instance, the greatest common divisor of $B=4$ and $D=6$ equals $2$.&lt;br&gt;&#xA; - The lower left triangle lists the least common multiples modulo $9$; for instance, the least common multiple of $B=4$ and $D=6$ is $12$, which yields $3$ when taken modulo $9$; similarly, the least common multiple of $C=7$ and $D=6$ modulo $9$ yields $6$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-07T12:30:02.523" LastActivityDate="2016-03-07T12:30:02.523" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28487" PostTypeId="5" CreationDate="2016-03-07T12:43:54.493" Score="0" Body="&lt;p&gt;The Frobenius problem asks for the largest amount of money that cannot be obtained by using only coins of specified denominations. Instead of coins, some definitions use stamps or Chicken McNuggets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formally, $F(a_1,\ldots,a_n)$ denotes the largest integer that cannot be represented by using only coins of specified denominations $a_1,\ldots,a_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The largest amount that cannot be represented by coins of denominations $2$ and $3$ is the amount $5$; in other words $F(2,3)=1$. All even numbers can be written as $2+2+\cdots+2$, and all odd numbers (except $1$) can be written as $3+(2+\cdots+2)$.&lt;/li&gt;&#xA;&lt;li&gt;The Chicken McNugget problem asks for $F(6,9,20)$, as McDonald's Chicken McNuggets boxes come with $6$, $9$, and $20$ nuggets. One can show that $F(6,9,20)=43$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-07T12:43:54.493" LastActivityDate="2016-03-07T12:43:54.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28488" PostTypeId="4" CreationDate="2016-03-07T12:43:54.493" Score="0" Body="The Frobenius problem asks for the largest amount that cannot be obtained by using only coins of specified denominations." OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-07T12:43:54.493" LastActivityDate="2016-03-07T12:43:54.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28489" PostTypeId="1" AcceptedAnswerId="28491" CreationDate="2016-03-07T12:49:22.417" Score="9" ViewCount="671" Body="&lt;p&gt;Find the appropriate fillers for the blanks...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;jafe, maap, maju, ____, ____, node&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The title says it loud and clear! ;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19656" LastEditorUserId="8874" LastEditDate="2016-03-07T15:53:16.860" LastActivityDate="2016-03-07T15:53:16.860" Title="This keeps coming back" Tags="|pattern|lateral-thinking|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28490" PostTypeId="1" AcceptedAnswerId="28524" CreationDate="2016-03-07T12:50:01.570" Score="51" ViewCount="6691" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/James_Joseph_Sylvester&quot;&gt;James Joseph Sylvester&lt;/a&gt; was one the greatest British mathematicians of &#xA;the 19th century, who made many fundamental contributions to number theory, &#xA;combinatorics, and invariant theory.&#xA;In 1884, he posed the following mathematical puzzle in the &lt;em&gt;Educational Times&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A man possesses a large quantity of stamps of only two denominations: &#xA;  $5$-cent stamps and $17$-cent stamps. &#xA;  What is the largest (finite) amount of cents which the man cannot make &#xA;  up with a combination of these stamps? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This was apparently the first occurrence of the Frobenius problem as a mathematical puzzle. What's the solution?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-18T15:56:12.323" LastActivityDate="2016-03-18T15:56:12.323" Title="A man possesses a large quantity of stamps" Tags="|mathematics|combinatorics|frobenius|puzzle-history|" AnswerCount="5" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28491" PostTypeId="2" ParentId="28489" CreationDate="2016-03-07T12:52:19.397" Score="10" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; juau, seoc&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Explanation:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first two letters of the twelve months. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2016-03-07T12:52:19.397" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28493" PostTypeId="1" AcceptedAnswerId="28495" CreationDate="2016-03-07T13:03:39.227" Score="15" ViewCount="1995" Body="&lt;p&gt;Here is a series of movies that have something in common other than being thoroughly entertaining.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VaY6z.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VaY6z.jpg&quot; alt=&quot;Find the next movie in this series&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the next movie in this series.&lt;/p&gt;&#xA;" OwnerUserId="19656" LastEditorUserId="8874" LastEditDate="2016-03-08T17:45:21.520" LastActivityDate="2016-03-08T17:45:21.520" Title="The next movie in this series is..?" Tags="|pattern|visual|rebus|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28494" PostTypeId="1" AcceptedAnswerId="28553" CreationDate="2016-03-07T13:08:59.830" Score="444" ViewCount="42559" Body="&lt;p&gt;Look, you don't know me. We've never met. But, like I said, I need you to listen...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had the most disturbing dream the other night... Woke up in a cold sweat and couldn't get back to sleep. It wasn't a nightmare as such, just ... eerie, I guess? I don't even remember the details, but it left me with a real sense of foreboding. Weird thing about dreams, so clear when you have them, but the details fade so quickly... Vivid, yet fuzzy... And the strange logic that applies... Inconsistent flow of time... Disjointed cause and effect...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm rambling... The thing is, when I woke up, I found an odd magazine clipping on my bedside table. It feels so familiar, like I've had it for quite a while, but for some reason, I can't remember how or when I acquired it. Wherever it came from, all I know is that it I gives me the same creepy feeling my dream did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, the strangest part about all this is... And perhaps you won't believe me... But, this piece of paper, I think...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...I think it's meant for &lt;em&gt;you&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here's a photo I took showing the whole page (click to enlarge):&#xA;&lt;a href=&quot;https://i.stack.imgur.com/tcgNT.jpg&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JCtyr.jpg&quot; alt=&quot;mysterious page&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also took a &lt;a href=&quot;https://i.stack.imgur.com/B7g38.png&quot;&gt;scan of the main picture&lt;/a&gt;, but my intuition says it's the least important part. I thought the text of the main &quot;article&quot; might be more useful to you, so I've transcribed it here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;RXZMHTHXCUTANEMZNVNQDTH&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;strong&gt;MWDEJULDYHPHVERHBPGPOMT&lt;/strong&gt;  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;pltdlgvqncltrr&lt;br&gt;&#xA;  umtmrqgba&lt;br&gt;&#xA;  mcdmwuygplpsdw&lt;br&gt;&#xA;  vaitzgifamqciw&lt;br&gt;&#xA;  wcogsplseplhbz&lt;br&gt;&#xA;  kgfibhsjyiebvt&lt;br&gt;&#xA;  jsurvclhzegko  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;nipzsbwaetoivh&lt;br&gt;&#xA;  onbvrvhtslls&lt;br&gt;&#xA;  lusihvlncaawlttfv&lt;br&gt;&#xA;  bngamiivvc&lt;br&gt;&#xA;  dyktpklecfjbijm&lt;br&gt;&#xA;  lpuhhvqtneqtrw&lt;br&gt;&#xA;  wfsgtfvqkyshgk&lt;br&gt;&#xA;  ngtdwiivvlitga&lt;br&gt;&#xA;  nrhqmkcmrleflmgl&lt;br&gt;&#xA;  uwlmfwgvv&lt;br&gt;&#xA;  unayficwvhoickso&lt;br&gt;&#xA;  wfxjbklwfmsnf  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;uwilt&lt;br&gt;&#xA;  RGMRI R VQKQ&lt;br&gt;&#xA;  LKH EOP A AGU A&lt;br&gt;&#xA;  kmuwzs&lt;br&gt;&#xA;  KNPW AGMR IIQ LBQP&lt;br&gt;&#xA;  KHAOP WFLZA WAK EZ  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;qikbfvmaovsncb&lt;br&gt;&#xA;  qaoelayjrklvum&lt;br&gt;&#xA;  owgarzmmndm&lt;br&gt;&#xA;  brghwcovoyiflark&lt;br&gt;&#xA;  apsblpqiflmhjbty&lt;br&gt;&#xA;  fifmayvzpij  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Hopefully this all means more to you than it does to me...&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Edit&lt;/em&gt;&lt;/strong&gt;: As per &lt;a href=&quot;http://meta.puzzling.stackexchange.com/questions/4530/etiquette-for-teaming-up-to-solve-a-puzzle&quot;&gt;this meta post&lt;/a&gt;, I've created a &lt;a href=&quot;https://chat.stackexchange.com/rooms/36672/this-is-important-i-need-you-to-listen&quot;&gt;chat room&lt;/a&gt; if people wish to collaborate and discuss theories or post their partial solutions/notes somewhere.&lt;/p&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="5208" LastEditDate="2016-03-08T00:16:47.100" LastActivityDate="2020-10-20T02:36:56.177" Title="This is important. I need you to listen…" Tags="|riddle|enigmatic-puzzle|visual|story|cipher|" AnswerCount="4" CommentCount="25" ContentLicense="CC BY-SA 3.0" />
  <row Id="28495" PostTypeId="2" ParentId="28493" CreationDate="2016-03-07T13:39:56.193" Score="22" Body="&lt;p&gt;Next in sequence is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Ace of aces&quot; or &quot;Ace Ventura&quot; (Ace)       &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sequence can also be extended in the other direction:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.imdb.com/title/tt0147800/&quot;&gt;Ten Things I Hate About You&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;http://www.imdb.com/title/tt0113986/&quot;&gt;Nine Months&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;http://www.imdb.com/title/tt3460252/&quot;&gt;The Hateful Eight&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;http://www.imdb.com/title/tt0054047/&quot;&gt;The Magnificent Seven&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;http://www.imdb.com/title/tt4180298/&quot;&gt;The Six Billion Dollar Man&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="16400" LastEditDate="2016-03-07T17:32:06.427" LastActivityDate="2016-03-07T17:32:06.427" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28496" PostTypeId="2" ParentId="28490" CreationDate="2016-03-07T13:44:20.957" Score="16" Body="&lt;p&gt;The largest solution you cannot obtain is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;63&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;This is because numbers that end in either 9 or 4 can be obtained by just taking two multiples of 17 and then using combinations of 5. As you multiple 17 more and more, you get different endings to which adding 5 will give you one more number ending. Every number from 63 and above is achievable using a combination of 17 and 5 from this. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Additionally:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I will be adding a graph/plot soon to confirm my answer. However, currently, where I am typing this answer from (my school), much of the frontend/backend required by websites to plot graphs gets blocked and so it isn't possible from here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19051" LastActivityDate="2016-03-07T13:44:20.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28497" PostTypeId="2" ParentId="28490" CreationDate="2016-03-07T13:45:34.980" Score="5" Body="&lt;p&gt;with 5-cent stamps and 17-cent stamps you can find any number that ends with 4 and 9 with the equation below after &lt;strong&gt;39&lt;/strong&gt; (considering you have at least one 5-cent and one 17-cent stamp);&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5n+34\ where\ n&amp;gt;0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, you can find any number that ends with 2,7,1,6,3,8,0,5 after the numbers stated in the equations below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5n+17\ where\ n\geqslant 0$ after &lt;strong&gt;17&lt;/strong&gt; ending with 2,7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5n+51\ where\ n\geqslant 0$ after &lt;strong&gt;51&lt;/strong&gt; ending with 1,6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5n+68\ where\ n\geqslant 0$ after &lt;strong&gt;68&lt;/strong&gt; ending with 8,3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5n+85\ where\ n\geqslant 0$ after &lt;strong&gt;85&lt;/strong&gt; ending with 0,5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5n\ where\ n\geqslant 0$ after &lt;strong&gt;0&lt;/strong&gt; ending with 0,5 since you do not have to use 17-cent stamp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result After 68, you can find any number you want by using 5 and 17-cents. So the next number below 68 by using the equation $5n+68$ will be the upper limit where $n=-1$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;63.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18514" LastEditorUserId="18514" LastEditDate="2016-03-07T14:40:53.057" LastActivityDate="2016-03-07T14:40:53.057" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28498" PostTypeId="1" AcceptedAnswerId="28506" CreationDate="2016-03-07T13:48:03.123" Score="11" ViewCount="1379" Body="&lt;p&gt;Suppose you have 240 bottles of juice, one of which has been poisoned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also have 5 blue parrots, all of them are immune to the poison, but will turn red 24 hours later and stays red forever upon tasting a sip of the poisoned juice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it is currently 9 am and you're hosting a huge party 50 hours from now, you'd like to serve out as many bottles as possible while knowing, with certainty, that they are non-poisonous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Assume that a parrot can take sips from a bottle as many times you want and you cannot pour the content of a bottle to another to make it as another bottle&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*EDIT 1: The parrots can only take the sips at 10 am for them to turn red for weird reasons involving their natural circadian rhythms.&lt;/p&gt;&#xA;" OwnerUserId="8882" LastEditorUserId="8882" LastEditDate="2016-03-07T15:34:07.810" LastActivityDate="2016-03-08T21:07:42.090" Title="How many bottles can you serve?" Tags="|mathematics|" AnswerCount="9" CommentCount="11" ClosedDate="2016-03-07T20:26:41.167" ContentLicense="CC BY-SA 3.0" />
  <row Id="28499" PostTypeId="2" ParentId="28493" CreationDate="2016-03-07T13:59:39.733" Score="7" Body="&lt;p&gt;Next movie is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ace Ventura&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because in every movie:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Movie names are J,Q,K &amp;amp; A and in each movie character name/title is in movie title : &lt;strong&gt;Jack&lt;/strong&gt; Reacher , Snow &lt;strong&gt;Queen&lt;/strong&gt; , &lt;strong&gt;King&lt;/strong&gt; Kong , &lt;strong&gt;Ace&lt;/strong&gt; Ventura&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18685" LastEditorUserId="18685" LastEditDate="2016-03-07T14:33:04.077" LastActivityDate="2016-03-07T14:33:04.077" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28500" PostTypeId="2" ParentId="28498" CreationDate="2016-03-07T14:13:51.083" Score="3" Body="&lt;p&gt;Naively, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The upper bound for wasted juice bottles is 24. You have 50 hours, which is greater than 24 x 2, which means you have ten opportunities to check bottles in groups of 24 (i.e., Divide bottles into 10 groups of 24; split those 10 groups into two groups; from the first group, assign each group of 24 to one parrot, and have that parrot take a sip from each of the bottles. If no parrot is red within 24 hours, repeat the process with the &lt;em&gt;other&lt;/em&gt; group-of-groups of bottles. Whichever parrot is red 24 hours later, remove that parrot's group of 24 bottles from service). You are now left with 216 Definitely Not Poisoned bottles of juice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2016-03-07T14:48:32.033" LastActivityDate="2016-03-07T14:48:32.033" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28501" PostTypeId="2" ParentId="28498" CreationDate="2016-03-07T14:13:56.297" Score="4" Body="&lt;p&gt;&lt;strong&gt;This answer is DISPROVED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;We can have maximum &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 235 bottles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Something that can be done is that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First let every parrot have 60 bottles such that one parrot always have 20 bottles in common with each of them. To explain this...let us say that 4 parrots drank 60 sips of bottles. So every bottle is covered. Now let the fifth parrot have 15 bottles of each of the other parrots. By doing this we can be sure that which bottle can have the poison. Now we have two cases based on the number of parrots who changed their colors.&lt;br&gt;&#xA;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/Zst5i.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zst5i.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&#xA;In this picture each of the four colored squares represents one of the four parrots. The square in the middle is the parrot  who takes 15 sips from each bottle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;CASE 1 - If only one parrot turns red&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Then we know that there are 45 bottles in which there can be the poison. This time we have only four parrot because one has turned red. Now we can repeat that process again. Giving 3 parrots 15 bottles each and giving the last parrot 15 bottles, 5 from each of the other parrot's bottles. In the end we have to throw away minimum 5 bottles...&lt;strong&gt;So we have 235 bottles for guests.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;CASE 2 - If two parrots turn red.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Then we know that 15 bottles are poisonous. This time we have only 3 parrots remaining. So this time we will give each parrot 5 bottles. One of them will turn red. &lt;strong&gt;So we will have 235 bottles for the guests&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-03-08T21:07:42.090" LastActivityDate="2016-03-08T21:07:42.090" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28502" PostTypeId="2" ParentId="28498" CreationDate="2016-03-07T14:21:07.477" Score="10" Body="&lt;p&gt;You can serve&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 239 bottles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using the following method:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Give each parrot 48 bottles and note which parrot got which.&lt;br&gt;&#xA;Then with 15 minutes apart divide the bottles uniquely between the parrots and let them drink.&lt;br&gt;&#xA;&lt;br&gt;&#xA;A: 48 + 12 of B (after 15m) + 12 of C (after 30m) + 12 of D (after 45m) + 12 of E (after 1 hour)&lt;br&gt;&#xA;B: 48 + same but with ACDE&lt;br&gt;&#xA;C: 48 + same but with ABDE&lt;br&gt;&#xA;D: 48 + same but with ABCE&lt;br&gt;&#xA;E: 48 + same but with ABCD&lt;br&gt;&#xA;&lt;br&gt;&#xA;case: A turns red after 1 hour and B after 1h + 15m&lt;br&gt;&#xA;12 bottles can contain the poison&lt;br&gt;&#xA;&lt;br&gt;&#xA;Divide the 12 bottles between the parots and then divide uniquely the others again (just as above)&lt;br&gt;&#xA;C: 4 + 1 of D (after 15m) + 1 of D (after 15m) + 1 of E (after 15m) + 1 of E (after 15m)&lt;br&gt;&#xA;D: 4 same but with CCEE&lt;br&gt;&#xA;E: 4 same but with CCDD&lt;br&gt;&#xA;&lt;br&gt;&#xA;case: C turns red after 1 hour and D after 1h + 15m&lt;br&gt;&#xA;&lt;br&gt;&#xA;Conclusion: you found the poison bottle and you can serve the other 239 bottles.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19938" LastEditorUserId="8874" LastEditDate="2016-03-07T14:23:40.613" LastActivityDate="2016-03-07T14:23:40.613" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28503" PostTypeId="2" ParentId="28498" CreationDate="2016-03-07T14:24:21.943" Score="1" Body="&lt;p&gt;I assume you ment the turn red after taking a sip from the poisonous jouce.&#xA;After 48 hours and some feeding time I end up with&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 230 bottles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assume you place all 240 bottles in one line and give them numbers from 1 to 240.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let parrot Alfred drink the first half (1 - 120) and parrot Bjorn the second and third quarter (61 - 180).&#xA; One day later you maybe will find one, two or none of them beeing red.&#xA; Here comes the clue: when overlapping the bottle-intervals of the red parrots  you will end up with one of the four quarters.&#xA; Let's assume the poison is in bottle 1 to 60.&#xA; Now do the same trick again, but use three parrots and assign them a diffrent array: parrot Charles drinks 1 to 30, parrot Duffy drinks 11 to 40 and parrot Edguy drinks 21 to 50.&#xA; Again: One day later overlap their drinking intervals and you will end up throwing 10 bottles away and having a random mixed blue-green quintet of parrots. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Aiye, Pirate!&lt;/p&gt;&#xA;" OwnerUserId="19936" LastEditorUserId="19936" LastEditDate="2016-03-07T14:48:59.040" LastActivityDate="2016-03-07T14:48:59.040" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28504" PostTypeId="2" ParentId="28498" CreationDate="2016-03-07T14:51:13.213" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pour drops from 120 bottles into the five birds' food bowls using a &lt;a href=&quot;https://en.wikipedia.org/wiki/Gray_code#Position_encoders&quot; rel=&quot;nofollow&quot;&gt;gray code&lt;/a&gt; pattern such that no two birds get juice from the same 120 bottles, thus dividing the 240 bottles into 32 ($2^5$) distinct groups. The algorithm for deciding which bird gets which bottle is relatively simple. Assign each bottle a number from zero to 239. Convert that number to binary. If the $n^{th}$ bit from the right of the bottle's number is 1, then the $n^{th}$ bird gets a sip from that bottle. Notice that this means that one in every 32 bottles will not be drank from by &lt;em&gt;any&lt;/em&gt; of the birds.&#xA;&#xA; Since 240 / 32 = 7.5, in the worst case we will have to discard 8 bottles, because we can't evenly divide the groups.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In other words,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; With 0 birds, we must discard all 240 bottles. With 1 bird, we must only discard 120. 2 birds --&gt; 60, 3 birds --&gt; 30, 4 birds --&gt; 15, 5 birds --&gt; $\lceil{}7.5\rceil{}$ = 8. So, you can serve at least 232 bottles, and at most 233.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of course, that's only in the first 25 hours. In the next 24 hours...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can repeat the process from before. In fact, you only need three parrots this time, because you only need a 3-bit gray code since there are only 8 options. So you can serve a total of 239 drinks. That is assuming that at least three birds weren't used in the first round, which cannot be guaranteed. The algorithm is the same from before, except with only those 8 bottles which may be poisoned and only those birds which have not changed color. Again, due to the nature of this test, we cannot guarantee we will still have any birds to use at this point, so in the worst case we only have 232 usable bottles, 1 out of every 32 times we'll lose 8 bottles, and the rest of the time we'll lose fewer than four bottles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How good is this solution?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's optimal most of the time. We have at worst 5 bits of resolution, and at best 10 bits of resolution. Since we only have 240 bottles, the identity of the poisoned one is contained in slightly under 8 bits of information. Since this solution depends on the probability distribution of the number of birds we'll have left over, it's possible to give the exact expectation of remaining bottles along with the variance. However, since that distribution is skewed by the fact that we have 240 bottles, and not 256, the math is mildly annoying, so I'm not going to do that calculation. Suffice it to say that about 31 out of every 32 times we'll lose $\le$ 4 bottles. Most of the time we'll lose only the poisoned bottle. E(usable bottles) will be nearly 239.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19818" LastEditorUserId="19818" LastEditDate="2016-03-07T18:33:57.680" LastActivityDate="2016-03-07T18:33:57.680" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28505" PostTypeId="2" ParentId="28498" CreationDate="2016-03-07T15:02:10.600" Score="1" Body="&lt;p&gt;I am getting &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 236 served, 4 lost&#xA; You have 240 bottles, divided by the 5 birds on 10AM Day 1 (1 hour)&#xA; &#xA; Each bird has 48, so one turns red leaving 4 birds&#xA; &#xA; Divide the 48 by 4, that gives us 12 with 3 birds left so on 10AM day (25 hour)&#xA; 2&#xA; &#xA; Divide the remaining 12 by 3, on 10 am day 3 (49 hour), giving you 4 bottles you need to drop&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19940" LastActivityDate="2016-03-07T15:02:10.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28506" PostTypeId="2" ParentId="28498" CreationDate="2016-03-07T15:04:45.357" Score="9" Body="&lt;p&gt;My reading of the question means we need to identify the poisoned bottle in at most 2 rounds of testing. Under these circumstances it's possible to safely serve  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 239 bottles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The procedure to achieve this is as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; on the first day:&lt;br&gt;&#xA; Give each parrot 16 bottles of their own.&lt;br&gt;&#xA; Give each pair of parrots a sip each from 8 bottles&lt;br&gt;&#xA; Give each set of 3 parrots a sip each from 4 bottles&lt;br&gt;&#xA; Give each set of 4 parrots a sip each from 2 bottles&lt;br&gt;&#xA; Give all five parrots a sip from a single bottle&lt;br&gt;&#xA; 29 bottles are left over&lt;br&gt;&#xA;&lt;br&gt;&#xA; On the second day we have n red parrots and 5-n blue parrots left&lt;br&gt;&#xA; We also have no more than 2^(5-n) bottles so we can uniquely identify the poisoned one with a simple binary test.&lt;br&gt;&#xA; This method works for up to 243 bottles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19942" LastEditorDisplayName="user20" LastEditDate="2016-03-07T19:54:36.627" LastActivityDate="2016-03-07T19:54:36.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28507" PostTypeId="2" ParentId="28498" CreationDate="2016-03-07T15:10:27.047" Score="2" Body="&lt;p&gt;Since it is 9 am now, in 50 hours, 10 am comes twice, so we will have two trials, assuming we have any birds left.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first trial idea comes from @manshu. We will split the bottles up into 4 groups of 60 and label them and number them.  The first group will be $A=\{a_1, ... a_{60}\}$ for example.&lt;br&gt;&#xA; We will then create a fifth group $E=\{a_1, ... a_{15}, b_1, ... b_{15}, c_1, ... c_{15}, d_1, ... d_{15}\}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now label all the parrots $A, B, C, D, E$ and give them a sip from all the bottles in their same named group at 10 am of the first day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 24 hours, you will have one of 2 cases:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1 parrot turns red.  You can exclude all bottles in $E$ and the other three groups.  This means that the poisoned bottle is one of 45 bottles in that group, and you have 4 birds remaining.&lt;br&gt;&#xA;2 parrots turn red.  This means that the poisoned bottle is one of 15 bottles in that group, and you have 3 birds remaining.&lt;br&gt;&#xA; Now we can number the bottles in binary and take the lower bits to decide which bird to feed it to.  For example, if the bottle number is 35, in binary it is 100011.  If you have 4 bird, then use the last 4 bits 0011 and feed it to the first and second bird, since the first and second bit are set in that number.  Repeat with all the bottles.&lt;br&gt;&#xA; If you started with 4 birds and 45 bottles, then you can see which birds turn and re-create the binary number.  If birds 3 and 4 change, then the binary number is 1100.  So you know that birds 12 (001100), and 28 (011100), and 44 (101100) might be poisonous.&lt;br&gt;&#xA; If you started with 3 birds and 15 bottles, then you can do a bit better and only end up with two possibilities.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus, at the end, you can guarantee &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;237 bottles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;are safe.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2016-03-07T15:10:27.047" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28508" PostTypeId="1" AcceptedAnswerId="28739" CreationDate="2016-03-07T15:21:15.850" Score="8" ViewCount="376" Body="&lt;p&gt;You find yourself trapped on a strange planet, whose residents use an English alphabet made of English symbols. You are given a piece of paper with the following sentences in it.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;./4)-5/ t- -+2 8-5/&#xA;9| y-+ =2/ 8/2/ t- )-61+/2 +3 t+26 8-5/ 6-.&#xA;t893 3-)9/ty 93 89784y =,0=6)/,&#xA;t8/ t=(4/ -| /4/5/6t3 )-6t=963 =)t969+5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What does the message say, and where does the planet's alphabet come from?&lt;/p&gt;&#xA;" OwnerUserId="15603" LastActivityDate="2016-03-10T12:25:55.433" Title="What alphabet is this?" Tags="|word|language|cipher|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28509" PostTypeId="2" ParentId="28508" CreationDate="2016-03-07T16:07:48.153" Score="5" Body="&lt;p&gt;I think the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Uranus&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I looked for commonalities between the words to get a substitute started.  It looks like:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;t&lt;/code&gt; and &lt;code&gt;y&lt;/code&gt; mean themselves.  They have not been changed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This gives a starting point.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The letter o is a common letter and I see it at the beginning of the phrase.  Used in &lt;code&gt;to&lt;/code&gt;.  This also helps with &lt;code&gt;u&lt;/code&gt;, giving the words &lt;code&gt;our&lt;/code&gt; and &lt;code&gt;you&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Judging by the number of letters, in the first sentence, I've deemed it to state:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Welcome to our home&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Working, from this assumption, I have a partial cypher.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; welcome to our home&lt;br&gt;&#xA; 9| you =re here to co61uer u3 tur6 home 6ow&lt;br&gt;&#xA; th93 3oc9ety 93 h97hly =,0=6ce,&lt;br&gt;&#xA; the t=(le o| eleme6t3 co6t=963 =ct969um  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, I know a few more letters.  I can assume:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;=&lt;/code&gt; is &lt;code&gt;a&lt;/code&gt;.  And the third sentence says: &quot;This society is highly advanced&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This gives me more to work with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I run it again.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; welcome to our home&lt;br&gt;&#xA; i| you are here to con1uer us turn home now&lt;br&gt;&#xA; this society is highly advanced&lt;br&gt;&#xA;  the ta(le o| elements contains actinium  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clearly, we're near the end.  Add the last few letters and:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; welcome to our home&lt;br&gt;&#xA; if you are here to conquer us turn home now&lt;br&gt;&#xA; this society is highly advanced&lt;br&gt;&#xA; the table of elements contains actinium  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now....&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Where can we find actinium?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is where it gets a little shaky, and I'm not certain.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; if you are here to conquer us turn home now&lt;br&gt;&#xA; this society is highly advanced&lt;br&gt;&#xA;&lt;br&gt;&#xA; A society with nuclear bombs could be considered highly advanced.&lt;br&gt;&#xA; actinium is found as a byproduct of uranium.  The catch, though, is that I'm basing the answer purely off the relationship between the words Uranium, and Uranus.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I used the following to decrypt, making it as I went:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;pre&gt;&lt;code&gt;void Main()&#xA; {&#xA;   string phrase = @&quot;./4)-5/ t- -+2 8-5/&#xA;   9| y-+ =2/ 8/2/ t- )-61+/2 +3 t+26 8-5/ 6-.&#xA;   t893 3-)9/ty 93 89784y =,0=6)/,&#xA;   t8/ t=(4/ -| /4/5/6t3 )-6t=963 =)t969+5&quot;;&#xA;&lt;br&gt;&#xA;   Dictionary Cypher = new Dictionary()&#xA;   {&#xA;     { '.', 'w' },&#xA;     { '/', 'e' },&#xA;     { '4', 'l' },&#xA;     { ')', 'c' },&#xA;     { '-', 'o' },&#xA;     { '5', 'm' },&#xA;     { 't', 't' },&#xA;     { '+', 'u' },&#xA;     { '2', 'r' },&#xA;     { '8', 'h' },&#xA;     { 'y', 'y' },&#xA;     { '=' , 'a' },&#xA;     { '9' , 'i' },&#xA;     { '3' , 's' },&#xA;     { '7' , 'g' },&#xA;     { ',' , 'd' },&#xA;     { '6' , 'n' },&#xA;     { '0' , 'v' },&#xA;     { '|' , 'f' },&#xA;     { '(' , 'b' },&#xA;     { '1' , 'q' },&#xA;   }; &#xA;&lt;br&gt;&#xA;   GetDecoded(phrase, Cypher).Dump();&#xA; }&#xA; &#xA; public string GetDecoded(string phrase, Dictionary cypher)&#xA; {&#xA;   StringBuilder result = new StringBuilder();&#xA;   for (int i = 0; i &amp;lt; phrase.Length; i++)&#xA;   {&#xA;     try&#xA;     {&#xA;       result.Append(cypher.Single(a =&gt; a.Key == phrase[i]).Value);&#xA;     }&#xA;     catch&#xA;     {&#xA;       result.Append(phrase[i]);&#xA;     }&#xA;   }&#xA;   return result.ToString();&#xA; } &lt;/code&gt;&lt;/pre&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19944" LastEditorUserId="19944" LastEditDate="2016-03-07T16:24:47.670" LastActivityDate="2016-03-07T16:24:47.670" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="28510" PostTypeId="2" ParentId="28439" CreationDate="2016-03-07T16:13:20.537" Score="12" Body="&lt;p&gt;Using all operations and functions&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $9 \cdot 8 \cdot 7 \cdot \sqrt{4} \cdot \log_2(1+\frac{6}{5-3})$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="356" LastEditorUserId="356" LastEditDate="2016-03-07T16:21:21.577" LastActivityDate="2016-03-07T16:21:21.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28512" PostTypeId="2" ParentId="28490" CreationDate="2016-03-07T17:09:22.003" Score="10" Body="&lt;p&gt;The general picture here is as follows: if you have two positive integers $m,n$ with no common factor then every integer bigger than $mn-m-n$ can be written as $am+bn$ with $a,b$ non-negative integers, but $mn-m-n$ itself can't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, suppose $mn-m-n = am+bn$. Then $(b+1)n$ is a multiple of $m$, and therefore (since $m,n$ have no common factor) so is $b+1$; so $b \geq m-1$. Similarly, $a \geq n-1$. So $am+bn \geq 2mn-m-n$, contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $mn-m-n$ isn't $am+bn$ with $a,b$ non-negative integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[In this particular case, this says: suppose $63=5a+17b$. Then $80=5a+17(b+1)$, so $17(b+1)$ is a multiple of 5, so $b+1$ is too, so $b\geq4$. And $68=5(a+1)+17b$, so $5(a+1)$ is a multiple of 17, so $a+1$ is too, so $a\geq16$. But then $5a+17b\geq80+68&amp;gt;63$, contradiction.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to show that any integer $k\geq mn-m-n+1=(m-1)(n-1)$ will do, we'll first do it allowing negative numbers of stamps, and then fix up the negativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that if you allow a negative number of stamps, you can do this for &lt;em&gt;any&lt;/em&gt; $k$. Let's do it for $k=1$; that is, find $a,b$ &lt;em&gt;not necessarily non-negative&lt;/em&gt; with $am+bn=1$. One way to do it is to run Euclid's algorithm on $m,n$ and keep track of how to write every number involved in the form $am+bn$. At the last step, one of those numbers is 1 and you're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[In this particular case, we'll find e.g. that $7\times5-2\times17=1$.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now we have $k=(ka)m+(kb)n$, but of course our coefficients may be negative. The remaining job will be to fix it up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[In this particular case, let's take $k=64$; we get $64=448\times5-128\times17$.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how to fix it up? If $am+bn=k$ then for any integer $t$ we have $(a+tn)m+(b-tm)n=k$. We'll try to choose $t$ to make both $a+tn$ and $b-tm$ non-negative. That means making $t\geq -a/n$ and $t\leq b/m$. If we can't do this then there's no integer between $-a/n$ and $b/m$, which means $b/m-(-a/n)&amp;lt;1$; that is, $b/m+a/n&amp;lt;1$; that is, $am+bn&amp;lt;mn$. The LHS here is our $k$, but (alas!) this isn't quite the inequality we're looking for; we'll have to be more precise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trick here is that $a+tn$ and $b-tm$ are always integers. So making them non-negative is the same as making them $&amp;gt;-1$; what we need, then, is $a+tn&amp;gt;-1$ and $b-tm&amp;gt;-1$; equivalently, $t&amp;gt;(-a-1)/n$ and $t&amp;lt;(b+1)/m$. If there is no integer (strictly) between $(-a-1)/n$ and $(b+1)/m$ then that interval must be no longer than 1 unit, so $(b+1)/m+(a+1)/n\leq1$. Multiplying through by $mn$ and using $am+bn=k$ gives us $k&amp;lt;=mn-m-n$, and we're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[In this particular case with k=64 as above, what we need is $448+17t\geq0$ and $-128-5t\geq0$, which we tweak to $448+17t&amp;gt;-1$ and $-128-5t&amp;gt;-1$, which are equivalent to $t&amp;gt;-449/17$ and $t&amp;lt;-129/5$. Those bounds are approximately -26.4 and -25.8 respectively, and as you can see -25 lies between them.]&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;What if the two positive integers $m, n$ &lt;em&gt;do&lt;/em&gt; have a factor in common? In the case of e.g. 15 and 21, both denominations are divisible by 3. This in turn means that any combination will be divisible by 3, and thus there are infinitely many values that you &lt;em&gt;cannot&lt;/em&gt; make up by a combination. In general, if the greatest common divisor of $m,n$ is $d$, then you can only make multiples of $d$; the largest that you can't will be $d((m/d)(n/d)-(m/d)-(n/d))=mn/d-m-n$.&lt;/p&gt;&#xA;" OwnerUserId="19114" LastEditorUserId="19114" LastEditDate="2016-03-10T13:12:55.800" LastActivityDate="2016-03-10T13:12:55.800" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28513" PostTypeId="2" ParentId="28498" CreationDate="2016-03-07T17:10:09.870" Score="3" Body="&lt;p&gt;Another optimal solution that I believe is equivalent to @user19942's, but more intuitive to me (as a programmer):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Number and label the the bottles with 5 digit serial numbers in base 3 (i.e. 00000, 00001, ..., 22212, 22220).  Each parrot will give you one digit of the poisoned bottle's serial number:&#xA; On the first day each parrot should sip from bottles having a '1' in their digit position, and on the second day they'll sip from bottles having a '2' in their position.  Each digit can then be decoded as a '1' or '2' if its corresponding parrot turns red tomorrow or the day after respectively, or a '0' if the relevant parrot is still blue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This solution also has the advantage of letting you plan all of the trials in advance - as @Sleafar has implied in a comment, the accepted answer requires you to instantaneously prepare the Parrots' breakfast beverages at 10AM on the second day after noting which ones have changed colour.&lt;/p&gt;&#xA;" OwnerUserId="2461" LastEditorUserId="2461" LastEditDate="2016-03-07T18:01:30.873" LastActivityDate="2016-03-07T18:01:30.873" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28514" PostTypeId="1" CreationDate="2016-03-07T17:20:00.210" Score="0" ViewCount="66" Body="&lt;p&gt;You are coming to the climax of a very mischievous game show, the curtain opens and you are presented with three doors, behind each door there is a prize of varying qualities. Behind one door, there is a brand new sports car valued at over a million dollars, behind the second door, there stands a  goat and behind the third door, there is another goat. The score stands at: one door has a sports car and the other two have goats. You randomly pick a door. The host then announces that he will now show you where ONE of the goats is.. he does so leaving the third door closed, he then gives you the option of switching your unopened door with the closed third one or just opening the door you chose first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the best choice to make and why? &lt;/p&gt;&#xA;" OwnerUserId="19947" LastActivityDate="2016-03-07T17:20:00.210" Title="One out of three?" Tags="|mathematics|logical-deduction|" AnswerCount="0" CommentCount="4" ClosedDate="2016-03-07T17:27:11.803" ContentLicense="CC BY-SA 3.0" />
  <row Id="28515" PostTypeId="2" ParentId="28439" CreationDate="2016-03-07T17:24:08.557" Score="11" Body="&lt;p&gt;I had code lying around that solves this exact problem, and the first thing it spat out was:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; (1 + 2 + 3 + 4 + 5 + 6 + 7) * 8 * 9&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19949" LastEditorUserId="69582" LastEditDate="2020-12-08T23:34:13.130" LastActivityDate="2020-12-08T23:34:13.130" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28516" PostTypeId="1" AcceptedAnswerId="28517" CreationDate="2016-03-07T17:58:15.077" Score="23" ViewCount="2227" Body="&lt;p&gt;You woke up this morning locked in a room.  The only apparent exit is a door with a terminal.  On the screen you see the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;gt; einn&#xA;Access granted. Door opened.&#xA;&amp;gt; três&#xA;Access granted. Door opened.&#xA;&amp;gt; dhá&#xA;Access granted. Door opened.&#xA;&amp;gt; seis&#xA;Access granted. Door opened.&#xA;&amp;gt; seven&#xA;Access granted. Door opened.&#xA;&amp;gt; cinq&#xA;Access granted. Door opened.&#xA;&amp;gt; vier&#xA;Access granted. Door opened.&#xA;&amp;gt; &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The last line seems to be waiting for some input from you...&lt;/p&gt;&#xA;" OwnerUserId="2071" LastActivityDate="2017-01-20T19:23:22.317" Title="Don't clear your terminal history" Tags="|word|pattern|story|geography|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28517" PostTypeId="2" ParentId="28516" CreationDate="2016-03-07T18:10:23.977" Score="22" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;twaalf&lt;/strong&gt;, the Dutch word for the number twelve&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The languages &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; go from West to East: Iceland $\rightarrow$ Portugal $\rightarrow$ Ireland $\rightarrow$ Spain $\rightarrow$ UK $\rightarrow$ France $\rightarrow$ Belgium.&lt;br&gt;&#xA; Hence the next language should be &lt;strong&gt;Dutch&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The numerical values&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; follow the 4-bit Gray code (see &lt;a href=&quot;https://en.wikipedia.org/wiki/Gray_code&quot;&gt;here&lt;/a&gt;):&#xA; 0000=0, 0001=1, 0011=3, 0010=2, 0110=6, 0111=7, 0101=5, 0100=4, 1100=12, &#xA; 1101=13, 1111=15, 1110=14, 1010=10, 1011=11, 1001=9, 1000=8.&lt;br&gt;&#xA; Hence the next number should be &lt;strong&gt;12&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="18250" LastEditDate="2016-11-15T21:26:13.270" LastActivityDate="2016-11-15T21:26:13.270" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28518" PostTypeId="2" ParentId="28516" CreationDate="2016-03-07T18:13:16.757" Score="3" Body="&lt;p&gt;To provide step one (mostly) but to add a guess: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inputs are numbers, following this pattern:    &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\begin{array}{l | c} \text{Language} &amp;amp; \text{Number} \\  \hline \text{Icelandic} &amp;amp; 1 \\  \text{Portuguese} &amp;amp; 3 \\  \text{Irish} &amp;amp; 2 \\ \text{Spanish} &amp;amp; 6 \\ \text{English} &amp;amp; 7 \\ \text{French} &amp;amp; 5 \\ \text{German} &amp;amp; 4 \\ \end{array}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, the answer might be  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;osiem&lt;/em&gt;  - Total guess as 8 hasn't been listed yet, and going eastward following the countries the languages are spoken in, so headed to Poland next.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="9951" LastEditDate="2016-03-07T18:51:05.990" LastActivityDate="2016-03-07T18:51:05.990" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28519" PostTypeId="1" AcceptedAnswerId="28520" CreationDate="2016-03-07T18:55:01.430" Score="12" ViewCount="2797" Body="&lt;p&gt;This was on the entrance exam at our local university last year.&lt;br&gt;&#xA;I still can't figure out the answer...&#xA;&lt;a href=&quot;https://i.stack.imgur.com/0MTwp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0MTwp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19953" LastEditorUserId="19470" LastEditDate="2016-03-19T14:56:46.647" LastActivityDate="2021-05-15T08:18:06.203" Title="Star and Numbers logic puzzle" Tags="|mathematics|pattern|" AnswerCount="3" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28520" PostTypeId="2" ParentId="28519" CreationDate="2016-03-07T18:58:41.677" Score="21" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 61&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each number on the point is the sum of the squares of the numbers connecting to it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19491" LastActivityDate="2016-03-07T18:58:41.677" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28521" PostTypeId="2" ParentId="28519" CreationDate="2016-03-07T18:59:57.083" Score="11" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $5^2 + 6^2 = 25+36$= &lt;strong&gt;$61$&lt;/strong&gt;&#xA;&lt;br&gt;&#xA; Its just square and add&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastActivityDate="2016-03-07T18:59:57.083" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28522" PostTypeId="2" ParentId="28360" CreationDate="2016-03-07T19:06:44.047" Score="11" Body="&lt;p&gt;I am&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Horatio Nelson&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll start to count while you solve this problem,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lord Horatio Nelson was the &lt;strong&gt;first&lt;/strong&gt; Vis&lt;strong&gt;count&lt;/strong&gt; Nelson.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and just relax here at the top of my column. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; His figure is on top of &lt;a href=&quot;https://en.wikipedia.org/wiki/Nelson&amp;#39;s_Column&quot;&gt;Nelson's Column&lt;/a&gt; in pretty relaxed pose.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Everywhere I go I let out a good whistle, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are &lt;a href=&quot;https://en.wikipedia.org/wiki/HMS_Nelson&quot;&gt;several ships and even ship classes&lt;/a&gt; named after Nelson, and ship's signal is commonly called &lt;em&gt;whistle&lt;/em&gt;. Alternatively, he was a naval officer and it is expected that ships where he was on signalled with a whistle as any other ship.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but don't mess with me cause I carry a pistol. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Only weapon you can expect naval officer to &lt;em&gt;always&lt;/em&gt; have on person, is his sword. But Nelson's one had a &lt;a href=&quot;http://www.bbc.com/news/uk-england-stoke-staffordshire-20612757&quot;&gt;pistol built in it&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've been explosive at one point in history, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Nelson&amp;#39;s_Pillar&quot;&gt;Nelson's Pillar&lt;/a&gt; in Dublin was destroyed by explosion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;what I truly am is not much of a mystery. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Horatio Nelson is a well-known historical figure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19952" LastActivityDate="2016-03-07T19:06:44.047" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28523" PostTypeId="2" ParentId="28439" CreationDate="2016-03-07T19:29:28.610" Score="5" Body="&lt;p&gt;Here's mine:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2016 = (9 * 8 * 7 * 4) + 5 * (1 + 2 + 3 - 6)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first expression is just the factorization of 2016, the rest is just to spend the other digits in a zero-sum.&lt;/p&gt;&#xA;" OwnerUserId="17562" LastActivityDate="2016-03-07T19:29:28.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28524" PostTypeId="2" ParentId="28490" CreationDate="2016-03-07T19:39:29.193" Score="128" Body="&lt;p&gt;For the more visually inclined, arrange all positive integers in a 5-wide chart, as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 1   2   3   4   5&#xA; 6   7   8   9  10&#xA;11  12  13  14  15&#xA;16  17  18  19  20&#xA;21  22  23  24  25&#xA;26  27  28  29  30&#xA;31  32  33  34  35&#xA;36  37  38  39  40&#xA;41  42  43  44  45&#xA;46  47  48  49  50&#xA;51  52  53  54  55&#xA;56  57  58  59  60&#xA;61  62  63  64  65&#xA;66  67  68  69  70&#xA;..  ..  ..  ..  ..&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;To start, all the values in the rightmost column are divisible by 5, and thus can be made up simply of a number of 5-cent stamps. They can all be eliminated.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 1   2   3   4  --&#xA; 6   7   8   9  --&#xA;11  12  13  14  --&#xA;16  17  18  19  --&#xA;21  22  23  24  --&#xA;26  27  28  29  --&#xA;31  32  33  34  --&#xA;36  37  38  39  --&#xA;41  42  43  44  --&#xA;46  47  48  49  --&#xA;51  52  53  54  --&#xA;56  57  58  59  --&#xA;61  62  63  64  --&#xA;66  67  68  69  --&#xA;..  ..  ..  ..  --&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note that if you can generate a solution for any particular value on the chart, you can also generate a solution for the value directly below it by adding one 5-cent stamp. Likewise, you can find a solution for the &lt;em&gt;next&lt;/em&gt; value down by adding &lt;em&gt;another&lt;/em&gt; 5-cent stamp, and this cascades down into infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, since a solution for the value 17 is simply a single 17-cent stamp, we can eliminate that and every value below it in the second column.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 1   2   3   4  --&#xA; 6   7   8   9  --&#xA;11  12  13  14  --&#xA;16  --  18  19  --&#xA;21  --  23  24  --&#xA;26  --  28  29  --&#xA;31  --  33  34  --&#xA;36  --  38  39  --&#xA;41  --  43  44  --&#xA;46  --  48  49  --&#xA;51  --  53  54  --&#xA;56  --  58  59  --&#xA;61  --  63  64  --&#xA;66  --  68  69  --&#xA;..  --  ..  ..  --&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Similarly, since a solution for the value 34 is two 17-cent stamps, we can eliminate that and every value below it in the fourth column.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 1   2   3   4  --&#xA; 6   7   8   9  --&#xA;11  12  13  14  --&#xA;16  --  18  19  --&#xA;21  --  23  24  --&#xA;26  --  28  29  --&#xA;31  --  33  --  --&#xA;36  --  38  --  --&#xA;41  --  43  --  --&#xA;46  --  48  --  --&#xA;51  --  53  --  --&#xA;56  --  58  --  --&#xA;61  --  63  --  --&#xA;66  --  68  --  --&#xA;..  --  ..  --  --&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Since a solution for the value 51 is three 17-cent stamps, we can eliminate that and every value below it in the first column.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 1   2   3   4  --&#xA; 6   7   8   9  --&#xA;11  12  13  14  --&#xA;16  --  18  19  --&#xA;21  --  23  24  --&#xA;26  --  28  29  --&#xA;31  --  33  --  --&#xA;36  --  38  --  --&#xA;41  --  43  --  --&#xA;46  --  48  --  --&#xA;--  --  53  --  --&#xA;--  --  58  --  --&#xA;--  --  63  --  --&#xA;--  --  68  --  --&#xA;--  --  ..  --  --&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Finally, since a solution for the value 68 is four 17-cent stamps, we can eliminate that and every value below it in the third column.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 1   2   3   4  --&#xA; 6   7   8   9  --&#xA;11  12  13  14  --&#xA;16  --  18  19  --&#xA;21  --  23  24  --&#xA;26  --  28  29  --&#xA;31  --  33  --  --&#xA;36  --  38  --  --&#xA;41  --  43  --  --&#xA;46  --  48  --  --&#xA;--  --  53  --  --&#xA;--  --  58  --  --&#xA;--  --  63  --  --&#xA;--  --  --  --  --&#xA;--  --  --  --  --&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Thus we can see that the highest possible value we can not make is 63 cents.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;EDIT: I sacrificed some rigor in order to make the above a more accessible read. What follows attempts to somewhat bridge the gap, but if you're satisfied with the above, you can stop reading here.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What I didn't actually prove above was that the remaining numbers in the last chart cannot be made with only 5- and 17-cent stamps. For that, note that to have a valid solution for a given value, you must have 0 or more 17-cent stamps. The first elimination above accounted for all the cases where there are 0 17-cent stamps (except for the trivial case of no stamps at all). This is apparent when you consider that stepping downward in the chart is equivalent to incrementing the count of 5-cent stamps for each step, while keeping the 17-cent stamp count constant. Likewise, the remaining 4 eliminations account for all possible values for 1, 2, 3, and 4 17-cent stamps. When we get to 5 or more 17-cent stamps, the minimum value is 5*17=85. However, since we've already determined that all values above 63 have valid solutions, no more eliminations can be made. Therefore, the remaining numbers must all have no solution.&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="5456" LastEditDate="2016-03-10T00:34:13.967" LastActivityDate="2016-03-10T00:34:13.967" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="28525" PostTypeId="1" AcceptedAnswerId="28543" CreationDate="2016-03-07T20:15:29.857" Score="7" ViewCount="306" Body="&lt;p&gt;Your goal here is to make me look dumb. Take heart. My nearest and dearest assure me that this task  is within the reach of a not particularly bright coral polyp. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reading the original is not necessary for this problem but if you are interested, it is &lt;a href=&quot;https://puzzling.stackexchange.com/questions/27708/uncle-morts-idiots&quot;&gt;here&lt;/a&gt;. In the original the object was to guess a sequence of three natural numbers between 1 and 100 based on the following information: the numbers were found in contiguous cells in a row of a correctly completed sudoku, and they spelled a three letter word in ASCII in which each letter was of a different case from the one before. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you work the above out, you will find multiple solutions. The solution set is surprisingly small but not small enough. I reduced the solution set to one by a further restriction that I have never been happy with. I gave some thought to what the final restriction &lt;em&gt;should&lt;/em&gt; have been but I was discouraged because it would not be possible to ask the question here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe it is within the scope of this community to ask the following: what is the smallest set of restrictions you can think of that, added to the conditions above will yield a unique solution. In keeping with the spirit of the original, your restrictions (ideally a single restriction) should not mention a specific number and should be something that a mathematical idiot might plausibly remember. An example of what I mean here is that a mathematical idiot might remember &quot;none of the numbers had any of those nasty sharp corners&quot; but definitely not &quot;the numbers were pairwise prime&quot;. Find the ideal final condition that I couldn't. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most elegant solution at the end of a week wins. I read the guidelines and believe that this fits. Doubtless I will be informed if I am mistaken. My hope is that answering this question will help us to learn to be better puzzlers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addendum:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The final restriction in the original puzzle was that &quot;there was no way&quot; the heirs could forget the word and I placed the word in each of their names. That left me thinking why I bothered with the sudoku thing in the first place. When it says that Uncle Mort circled part of a row in a sudoku, it sounds as if we're hardly getting any information at all. But if you look at the ascii table (&lt;a href=&quot;http://www.asciitable.com/mobile/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; for example) the restriction of no zeros or repeated digits and alternating case rules out anything in octal or decimal representation. We are left with hex and j-p and J-P are eliminated right off the bat. No letters with repeated digits like D=44 or f=66. If the first letter is in the A-I range the last must be in the Q-Z range. We are close, but not close enough. What reasonable restriction added to the above gives us a unique solution?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;br&gt;&#xA;A good restriction would be one that appears to give almost no information like: &quot;I remember Uncle Mort pointed to the box where the first three numbers were and told me the other numbers in the box were the day and year some movie guy died. No idea who.'&quot; (Thanks to Manshu for the idea.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't know the who died, you don't know the day, you don't know anything... or do you? If the person who died was a movie star he couldn't have died before 1900. He couldn't have died later than 1987 because there can't be repeated digits or a 0 in the year (sudoku restriction). Therefore, we know that none of the first three digits can be a 1 or a 9.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also know that the day can't have a 0, 1 or 9 in it. The wording suggests that the day and year are expressed by the six other digits, four for the year and two for the day. The day must then be in the range from 23 to 28, so none of the first 3 digits in the solution can be a 2.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dang! Wish I'd thought of this at the time!&lt;/p&gt;&#xA;" OwnerUserId="18924" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2016-03-08T17:01:15.827" Title="Uncle Mort Meta" Tags="|password|puzzle-creation|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28526" PostTypeId="1" AcceptedAnswerId="28536" CreationDate="2016-03-07T20:30:25.373" Score="12" ViewCount="401" Body="&lt;p&gt;This might, perhaps, be too easy, but this is my first riddle to post here.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Get Copernicus and cut him in half for the start&lt;br&gt;&#xA;  Divide the start by two and you've got one extra.&lt;br&gt;&#xA;  Divide it by eight, instead, and you've got the last.&lt;br&gt;&#xA;  Put all three together for a yummy snack.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19944" LastEditorUserId="19944" LastEditDate="2016-03-07T21:00:00.140" LastActivityDate="2016-03-07T21:41:38.163" Title="Gone Fission - Food for thought" Tags="|riddle|word|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28527" PostTypeId="2" ParentId="28525" CreationDate="2016-03-07T20:40:49.360" Score="1" Body="&lt;p&gt;It could be something like&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Chelsea: I remember once I was watching a movie with him named &quot;The Seven Year Itch&quot; and when the lead actress died, he said &quot;now there's no way you can forget this&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The interpretation is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Main actress of &quot;The Seven Year Itch&quot; is Marilyn Monroe. She died on August 5, 1962. So the code is 851962&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can also add the name of the country from which the Uncle Mort belongs so that you can clear the confusion between 58**** and 85****&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: I haven't seen the movie. So I am not sure if there is even a death scene or not but I took this just for the sake of the example.&lt;/p&gt;&#xA;" OwnerUserId="17482" LastActivityDate="2016-03-07T20:40:49.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28528" PostTypeId="2" ParentId="28519" CreationDate="2016-03-07T20:46:28.787" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Every number at the points is equal to the sum of the squares of the closest two numbers inside, so it's 25+36=61.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastActivityDate="2016-03-07T20:46:28.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28529" PostTypeId="2" ParentId="28290" CreationDate="2016-03-07T20:48:29.253" Score="7" Body="&lt;p&gt;Sarah Lee...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; has an amphibious vehicle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Her GPS is incorrect because...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's not taking bodies of water into account.  She can easily avoid the long drive around by simply driving straight into the water and boating across.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7586" LastActivityDate="2016-03-07T20:48:29.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28531" PostTypeId="2" ParentId="28360" CreationDate="2016-03-07T20:54:25.997" Score="3" Body="&lt;p&gt;My answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hydrogen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;I'll start to count while you solve this problem,&#xA;and just relax here at the top of my column.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'Start to count' -&gt; Hydrogen is element #1, also located at the top of column #1 of the periodic table.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Everywhere I go I let out a good whistle&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Feels far-fetched... Hydrogen is present in our atmosphere. Could refer to wind or people whistling.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;but don't mess with me cause I carry a pistol.&quot;&#xA;&quot;I've been explosive at one point in history,&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hydrogen is explosive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;what I truly am is not much of a mystery.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hydrogen is pretty common knowledge&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19417" LastActivityDate="2016-03-07T20:54:25.997" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28532" PostTypeId="1" AcceptedAnswerId="28540" CreationDate="2016-03-07T20:57:18.540" Score="3" ViewCount="332" Body="&lt;p&gt;These five things describe me:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Quintessential to freight&lt;/li&gt;&#xA;&lt;li&gt;Up there&lt;/li&gt;&#xA;&lt;li&gt;Eighteen rubber circles&lt;/li&gt;&#xA;&lt;li&gt;Everyone knows me&lt;/li&gt;&#xA;&lt;li&gt;Need considerable distance to stop&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am a vehicle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &quot;QUEEN&quot; part is another hint and is significant to the answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If I had &quot;asked&quot; this question in another particular SE community, they would know the answer right away.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19824" LastEditorUserId="19824" LastEditDate="2016-03-07T21:12:19.893" LastActivityDate="2016-03-07T21:29:23.800" Title="Original What Am I" Tags="|riddle|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28533" PostTypeId="2" ParentId="28290" CreationDate="2016-03-07T20:59:00.623" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; She drives across the field that she would normally drive around. Because $14^2 + 14^2 &amp;lt; 24^2$, it's possible there is some non-road surface separating her from work.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19703" LastActivityDate="2016-03-07T20:59:00.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28534" PostTypeId="2" ParentId="28532" CreationDate="2016-03-07T21:00:44.290" Score="0" Body="&lt;p&gt;2 answers seem possible, either:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; crown or royal 18 wheelers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;another possibility is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; bicycle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; people deliver things on bikes... Queen song... another SE is on bicycles... etc... but 18 circles doesnt quite fit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19731" LastEditorUserId="19731" LastEditDate="2016-03-07T21:09:49.883" LastActivityDate="2016-03-07T21:09:49.883" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28536" PostTypeId="2" ParentId="28526" CreationDate="2016-03-07T21:01:06.687" Score="12" Body="&lt;p&gt;The yummy snack is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bacon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Copernicus is the Chemical element. The atomic number is 112.&lt;br&gt;&#xA; 112 / 2 = 56 = Barium = &lt;strong&gt;Ba&lt;/strong&gt;&lt;br&gt;&#xA; 56 / 2 - 1 = 27 = Cobalt = &lt;strong&gt;Co&lt;/strong&gt;&lt;br&gt;&#xA; 56 / 8 = 7 = Nitro­gen = &lt;strong&gt;N&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4342" LastEditorUserId="4342" LastEditDate="2016-03-07T21:41:38.163" LastActivityDate="2016-03-07T21:41:38.163" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="28537" PostTypeId="2" ParentId="28290" CreationDate="2016-03-07T21:07:51.923" Score="9" Body="&lt;p&gt;Sarah Lee...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; should tailgate the **** out of a truck.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Tests proved that driving 10 feet behind a truck could result in a 45% gas usage reduction &lt;a href=&quot;http://www.treehugger.com/cars/drafting-behind-trucks-does-it-work.html&quot;&gt;(source)&lt;/a&gt;.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Which means that the 14 miles she could usually drive are divided by 0.55 times her usual gas mileage =&gt; 14 / 0.55 = 25.45 miles.&lt;br&gt;&#xA;&lt;br&gt;&#xA; That distance being longer than the 24 miles she needs to drive on the fastest route, she'll hopefully make it alive and on time!!! :D  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19417" LastEditorUserId="19417" LastEditDate="2016-03-07T21:22:50.907" LastActivityDate="2016-03-07T21:22:50.907" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28538" PostTypeId="1" AcceptedAnswerId="28539" CreationDate="2016-03-07T21:09:47.343" Score="0" ViewCount="2761" Body="&lt;p&gt;In my last school-class I did a logic-test with 48 exercises about finding the missing dominoes. I'm sure about 47, but this one broke my head, and I don't know yet the answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lw6RY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lw6RY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19503" LastEditorUserId="11876" LastEditDate="2016-03-07T21:24:35.550" LastActivityDate="2018-04-07T01:48:13.067" Title="Find the missing domino" Tags="|pattern|dominoes|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28539" PostTypeId="2" ParentId="28538" CreationDate="2016-03-07T21:20:22.393" Score="7" Body="&lt;p&gt;It is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3,6 since the sum of diagonally opposite squares sum up to 6.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the picture is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/HycWK.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HycWK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19350" LastEditorUserId="17482" LastEditDate="2016-03-07T21:24:27.410" LastActivityDate="2016-03-07T21:24:27.410" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="28540" PostTypeId="2" ParentId="28532" CreationDate="2016-03-07T21:21:23.823" Score="6" Body="&lt;p&gt;You're probably&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Boeing 747&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;the OP appears to be an active user of&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://aviation.stackexchange.com/&quot;&gt;Aviation&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;according to Wikipedia you're sometimes referred to as &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Queen of the Skies&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;you have 18&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; wheels&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6134" LastEditorUserId="-1" LastEditDate="2017-04-13T13:00:01.880" LastActivityDate="2016-03-07T21:29:23.800" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28541" PostTypeId="2" ParentId="28532" CreationDate="2016-03-07T21:22:00.783" Score="3" Body="&lt;p&gt;You are a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Boeing_747&quot; rel=&quot;nofollow noreferrer&quot;&gt;Boeing 747 jetliner&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. &lt;strong&gt;Quintessential to freight&lt;/strong&gt; - FedEx has their own fleet!&lt;br&gt;&#xA; 2. &lt;strong&gt;Up there&lt;/strong&gt; - Up in the air&lt;br&gt;&#xA; 3. &lt;strong&gt;Eighteen rubber circles&lt;/strong&gt; - Landing gear has 18 wheels&lt;br&gt;&#xA; 4. &lt;strong&gt;Everyone knows me&lt;/strong&gt; - Ask an amateur about a plane model, they will likely say &quot;747&quot;&lt;br&gt;&#xA; 5. &lt;strong&gt;Need considerable distance to stop&lt;/strong&gt; - Like, a whole airstrip worth of space!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bonus: Also known as...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;The Queen of the Skies&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bonus 2: There is...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An &lt;a href=&quot;https://aviation.stackexchange.com/&quot;&gt;Aviation SE&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7586" LastEditorUserId="-1" LastEditDate="2017-04-13T13:00:01.880" LastActivityDate="2016-03-07T21:22:00.783" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28542" PostTypeId="2" ParentId="28538" CreationDate="2016-03-07T21:28:33.223" Score="2" Body="&lt;p&gt;It could be &lt;em&gt;anything&lt;/em&gt; but let's guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You have a total of 15 on the inside circle and 12 on the outside circle.&lt;br&gt;&#xA; So say they have to add up to the same number for some obscure reason.&lt;br&gt;&#xA; Remember it could be &lt;em&gt;anything&lt;/em&gt; - so let's take a wild pot shot at:&lt;br&gt;&#xA; We have 2 dominoes that total 8 and two that total 4 and &lt;em&gt;one&lt;/em&gt; that totals 3.&lt;br&gt;&#xA; So for symmetry, lets say the missing domino has to total 3.&lt;br&gt;&#xA; So it's blank on the inside circle and 3 on the outside circle.&lt;br&gt;&#xA; For a total of 15 on both the inside and outside circle and two matching sums all around.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastActivityDate="2016-03-07T21:28:33.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28543" PostTypeId="2" ParentId="28525" CreationDate="2016-03-07T21:57:52.920" Score="2" Body="&lt;p&gt;So i limited it to two restrictions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Restrictions: &lt;strong&gt;Absolute value between hex ascii values is 3&lt;/strong&gt;, &lt;strong&gt;Letters are in alphabetical order&lt;/strong&gt;. For example the number 41 or 47 would be acceptable because abs(4-1)=3 and abs(1-4)=3. There are only 7 letters that meet this condition.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A:41&lt;/li&gt;&#xA;&lt;li&gt;G:47&lt;/li&gt;&#xA;&lt;li&gt;R:52&lt;/li&gt;&#xA;&lt;li&gt;X:58&lt;/li&gt;&#xA;&lt;li&gt;c:63&lt;/li&gt;&#xA;&lt;li&gt;i:69&lt;/li&gt;&#xA;&lt;li&gt;t:74&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;With these letters the only word that meets the original restrictions and the new restrictions is &lt;strong&gt;AiR&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10856" LastActivityDate="2016-03-07T21:57:52.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28544" PostTypeId="2" ParentId="28494" CreationDate="2016-03-07T22:42:38.147" Score="12" Body="&lt;p&gt;Putting this in an answer so I can properly spoiler it out without doing rot13 in a comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use this for what it is... ideas I've looked into on this... doesn't mean I'm going in the right direction.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I believe there's a word hidden in the picture and with what ive looked at on it, steganography comes up, so there may be something hidden in the file itself.... also solved the crossword and cant quite gather anything meaningful from the hex like digits below the pic even using letters from those respective words... words 2, 3, and 20 are missing from the page, but the rest of them likely go with the words in the crossword. Vitamins A &amp;amp; D only stands out with the letters used in the hexlike chars tho...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;more observations:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; confirmed pic isnt useful in the beginning of the puzzle... i bet the article decodes to instructions for manipulating the picture.&#xA; 2F, 19A, 4E, 11D, 4A, 25C, 11A, 24A, 1B, 3F, 20C --&gt; CF, TA, IE. TD, IA, YC, TA, KA, EB, _F, _C ???? focus on A and D. an anagram basically spits out &quot;at a data kit&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;unless someone wants to start a chat on this one and I'll delete this.&lt;/p&gt;&#xA;" OwnerUserId="19731" LastEditorUserId="19731" LastEditDate="2016-03-07T23:17:21.803" LastActivityDate="2016-03-07T23:17:21.803" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28545" PostTypeId="1" AcceptedAnswerId="28563" CreationDate="2016-03-07T22:46:40.443" Score="3" ViewCount="281" Body="&lt;p&gt;The following adjectives apply to a certain noun.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Complex&lt;br&gt;&#xA;  Alternative&lt;br&gt;&#xA;  Exempt&lt;br&gt;&#xA;  Deductive&lt;br&gt;&#xA;  Dependent&lt;br&gt;&#xA;  Fuzzy&lt;br&gt;&#xA;  Critical  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And most of all, &lt;code&gt;Certain&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the noun?&lt;/p&gt;&#xA;" OwnerUserId="19179" LastEditorUserId="17744" LastEditDate="2016-03-28T14:55:54.443" LastActivityDate="2016-03-28T14:55:54.443" Title="Certainly Certain" Tags="|riddle|word|" AnswerCount="5" CommentCount="0" FavoriteCount="0" ClosedDate="2016-03-28T19:59:30.847" ContentLicense="CC BY-SA 3.0" />
  <row Id="28546" PostTypeId="2" ParentId="28545" CreationDate="2016-03-07T23:17:17.497" Score="7" Body="&lt;p&gt;Could it be:&lt;br&gt;&#xA;EDIT Thanks to @feelinferrety&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reasoning&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because then we have:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Complex Reasoning&lt;br&gt;&#xA; Alternative Reasoning&lt;br&gt;&#xA; Exempt Reasoning&lt;br&gt;&#xA; Deductive Reasoning&lt;br&gt;&#xA; Dependent Reasoning&lt;br&gt;&#xA; Fuzzy Reasoning&lt;br&gt;&#xA; Critical Reasoning&lt;br&gt;&#xA; Certain Reasoning  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastEditorUserId="10118" LastEditDate="2016-03-08T01:01:00.910" LastActivityDate="2016-03-08T01:01:00.910" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28547" PostTypeId="2" ParentId="28494" CreationDate="2016-03-07T23:21:06.027" Score="21" Body="&lt;p&gt;I'm going to follow another user's example and post an answer to make notes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solving the crossword:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to Z. Dailey for the missing words. (1D, 4A, 11D)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/bPmc8.jpg&quot; alt=&quot;crossword&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7586" LastEditorUserId="7586" LastEditDate="2016-03-08T00:05:22.747" LastActivityDate="2016-03-08T00:05:22.747" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28549" PostTypeId="2" ParentId="28494" CreationDate="2016-03-07T23:57:41.343" Score="29" Body="&lt;p&gt;Here's what I found so far:&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Title&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The word &lt;strong&gt;listen&lt;/strong&gt; in the title looked pretty important from the beginning. It is also repeated in the first line of the question and there is a hint in the crossword (see below).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;Crossword&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The visible hints for horizontal entries give the words ISOLATED, THEOLOGY, PITY, THEORY and KEY. The vertical word EYES was given in the linked image from the question. Filling the remaining letters reveals 2 additional vertical words: TITLE and EIGHTH. The eighth word in the title is &lt;strong&gt;listen&lt;/strong&gt;.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/1DuXt.jpg&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1DuXt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;Continuation&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm stuck for now. The word &lt;strong&gt;listen&lt;/strong&gt; suggests that the rest of the puzzle might have to do something with music. The letters A-F could stand for notes (they are obviously not hex numbers as stated in the advertisement). Replacing the numbers below the image with letters from the crossword didn't work for me either (even when using A &amp;amp; D vitamins only).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2016-03-08T00:10:39.263" LastActivityDate="2016-03-08T00:10:39.263" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28550" PostTypeId="2" ParentId="28439" CreationDate="2016-03-08T00:04:22.397" Score="15" Body="&lt;p&gt;I have found all 3,020 expressions satisfying the given restrictions along with four others:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The digits 1-9 must be in order.&lt;/li&gt;&#xA;&lt;li&gt;You are not allowed to take the square root of &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt;.  Without this restriction we could have arbitrarily complex expressions.&lt;/li&gt;&#xA;&lt;li&gt;Every subexpression must have a rational value.  This excludes expressions that involve terms like &lt;span class=&quot;math-container&quot;&gt;$\log[2]\sqrt{2}=\frac{1}{2}$&lt;/span&gt;.  Without this restriction we could have arbitrarily complex expressions.&lt;/li&gt;&#xA;&lt;li&gt;Decimal points are not allowed, since it is not clear how we are allowed to use them.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;My algorithm works as follows.  I compute a table of values &lt;span class=&quot;math-container&quot;&gt;$\mathtt{terms}$&lt;/span&gt;, where each element &lt;span class=&quot;math-container&quot;&gt;$\mathtt{terms}[i,j]$&lt;/span&gt; represents the possible values of a subexpression starting at the &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt;-th number and spanning &lt;span class=&quot;math-container&quot;&gt;$j$&lt;/span&gt; numbers.  As an example, take the expression:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$((1 - (2 \times (3 - 4) + 5)) + 6) \times 7 \times 8 \times 9$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The term &lt;span class=&quot;math-container&quot;&gt;$3-4$&lt;/span&gt; is associated with &lt;span class=&quot;math-container&quot;&gt;$\mathtt{terms}[3,2]$&lt;/span&gt;, and the term &lt;span class=&quot;math-container&quot;&gt;$7 \times 8 \times 9$&lt;/span&gt; is associated with &lt;span class=&quot;math-container&quot;&gt;$\mathtt{terms}[7,3]$&lt;/span&gt;.  The entire expression is associated with &lt;span class=&quot;math-container&quot;&gt;$\mathtt{terms}[1,9]$&lt;/span&gt;, and the individual numbers are associated with &lt;span class=&quot;math-container&quot;&gt;$\mathtt{terms}[*,1]$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Each &lt;span class=&quot;math-container&quot;&gt;$\mathtt{terms}[i,j]$&lt;/span&gt; is a list of key-value pairs: the keys are the values of that subexpression, and the values are the ways of reaching that value.  For example:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{align}&#xA;\mathtt{terms}[1, 2] = \{\quad&#xA;3 &amp;amp;\to \{1+2\ (1)\},\\&#xA;-1 &amp;amp;\to \{1-2\ (1)\}, \\&#xA; 2 &amp;amp;\to \{1 \times 2\ (1)\}, \\&#xA;\tfrac{1}{2} &amp;amp;\to \{1 \div 2\ (1)\}&#xA;\quad\}&#xA;\end{align}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;A quick note: for binary operators (all of them except the square root) I store the operator, the left and right-hand operands, &lt;em&gt;and&lt;/em&gt; the size of the left operand.  The last one is necessary in the second part of the algorithm, and above and below it is shown in parenthesis.  Another example:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{align}&#xA;\mathtt{terms}[1, 3] = \{\quad&#xA;6 &amp;amp;\to \{1+5\ (1),\ 1\times 6\ (1),\ 3+3\ (2),\ 2\times 3\ (2)\},\\&#xA;0 &amp;amp;\to \{1+(-1)\ (1),\ 3-3\ (2),\ \sqrt{0}\},\\&#xA;1 &amp;amp;\to \{3\div 3\ (2),\ \log[3](3)\ (2),\ \sqrt{1}\}, \\&#xA;\vdots&#xA;\end{align}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;A few notes:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For all operators, only the values of the operands are stored instead of the entire subexpressions.  This is what makes this method efficient.&lt;/li&gt;&#xA;&lt;li&gt;Square roots do not need a length stored with them, since they apply to a single subexpression.&lt;/li&gt;&#xA;&lt;li&gt;The restriction on square roots of &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; is not applied at this stage (they add at most a constant overhead).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;The algorithm that computes &lt;span class=&quot;math-container&quot;&gt;$\mathtt{terms}$&lt;/span&gt; is as follows:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Loop over &lt;span class=&quot;math-container&quot;&gt;$\ell=1\ldots 9$&lt;/span&gt;:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Loop over &lt;span class=&quot;math-container&quot;&gt;$i=1\ldots 10 - \ell$&lt;/span&gt;:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If &lt;span class=&quot;math-container&quot;&gt;$\ell = 1$&lt;/span&gt;, initialize &lt;span class=&quot;math-container&quot;&gt;$\mathtt{vals}=\{i \to \{i\}\}$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Otherwise, initialize &lt;span class=&quot;math-container&quot;&gt;$\mathtt{vals}=\{\}$&lt;/span&gt; and loop through &lt;span class=&quot;math-container&quot;&gt;$k=1\ldots \ell-1$&lt;/span&gt;:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Get the possible values of the left-hand and right-hand operands.  These are: &lt;span class=&quot;math-container&quot;&gt;$$\begin{align}\mathtt{LHS}&amp;amp;=\mathtt{terms}[i,k] \\ \mathtt{RHS}&amp;amp;=\mathtt{terms}[i+k,\ell-k]\end{align}$$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;Construct all possible terms: &lt;span class=&quot;math-container&quot;&gt;$$a\circ b\ (k);\ a\in\mathtt{LHS},\ b\in\mathtt{RHS},\ \circ\in\{+,-,\times,\div,\log\}$$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;Add to &lt;span class=&quot;math-container&quot;&gt;$\mathtt{vals}$&lt;/span&gt; all terms that result in a rational value.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;Repeatedly take a the square root of each term in &lt;span class=&quot;math-container&quot;&gt;$\mathtt{vals}$&lt;/span&gt;.  If the result is a rational value, add it to &lt;span class=&quot;math-container&quot;&gt;$\mathtt{vals}$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Set &lt;span class=&quot;math-container&quot;&gt;$\mathtt{term}[i,\ell]=\mathtt{vals}$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Next we traverse &lt;span class=&quot;math-container&quot;&gt;$\mathtt{terms}$&lt;/span&gt; in reverse order.  Starting with &lt;span class=&quot;math-container&quot;&gt;$\mathtt{terms}[1,9]$&lt;/span&gt; we get all the ways to make &lt;span class=&quot;math-container&quot;&gt;$2016$&lt;/span&gt;, then recursively get all the ways to make each subexpression, ignoring &lt;span class=&quot;math-container&quot;&gt;$\sqrt{1}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\sqrt{0}$&lt;/span&gt;.  This results in a list of 32,282 expressions.  I then normalize the expressions using the following transformations:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{align}&#xA;\cdots+(x_1+\cdots+x_n)+\cdots &amp;amp;\to \cdots+x_1+\cdots+x_n+\cdots \\&#xA;\cdots\times(x_1\times\cdots\times x_n)\times\cdots &amp;amp;\to \cdots\times x_1\times\cdots\times x_n\times\cdots \\&#xA;\cdots+(x-y)+\cdots &amp;amp;\to ((\cdots+x)-y)+\cdots \\&#xA;\cdots\times(x\div y)\times\cdots &amp;amp;\to ((\cdots\times x)\div y)\times\cdots&#xA;\end{align}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;This prevent us from counting expressions like &lt;span class=&quot;math-container&quot;&gt;$(1+2)+3$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$1+(2+3)$&lt;/span&gt; as different.  After this step, there are only 3,020 distinct expressions.  The full list is available &lt;a href=&quot;https://gist.github.com/robertcampion/85decaebb49993757ab1&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. Note that this list uses &lt;em&gt;Mathematica&lt;/em&gt; notation: &lt;span class=&quot;math-container&quot;&gt;$\times\to\mathtt{*},$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\div\to\mathtt{/},$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\sqrt{x}\to\mathtt{Sqrt[}x\mathtt{]},$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\log[b](x)\to\mathtt{Log[}b\mathtt{,}x\mathtt{]}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Of particular note are the expressions:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;(1+Log[2,Log[3,4+5]])*6*7*8*Sqrt[9]&#xA;(1+Log[2,Log[3,4+5]+6])*7*8*9&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;(i.e:)&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;(1+\log[2](\log[3](4+5)))\times 6\times 7\times 8\times\sqrt{9} \\&#xA;(1+\log[2](\log[3](4+5)+6))\times 7\times 8\times 9&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Each of which use &lt;span class=&quot;math-container&quot;&gt;$\log$&lt;/span&gt; twice, with two different bases!&lt;/p&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="9718" LastEditDate="2023-06-20T12:46:40.433" LastActivityDate="2023-06-20T12:46:40.433" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28551" PostTypeId="2" ParentId="28439" CreationDate="2016-03-08T00:26:15.067" Score="4" Body="&lt;p&gt;This is my answer:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 1 * 2 * 3 * 4 * 6 * 7 * 8 / (9-5)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19967" LastEditorUserId="69582" LastEditDate="2020-12-08T23:25:18.523" LastActivityDate="2020-12-08T23:25:18.523" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28552" PostTypeId="1" AcceptedAnswerId="28554" CreationDate="2016-03-08T01:03:21.957" Score="9" ViewCount="196" Body="&lt;p&gt;Did we do something wrong? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are we to blame for modern song?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beliebers rule, but the Beatles I long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\\$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For teenagers, I fear&lt;/p&gt;&#xA;&#xA;&lt;p&gt;#yolo is what they hear&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good music is nowhere near.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\\$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do we need to go way way back?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Get music pure and not so black?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aha I know exactly which track!&lt;/p&gt;&#xA;" OwnerUserId="19350" LastEditorUserId="17147" LastEditDate="2016-03-08T18:02:23.947" LastActivityDate="2016-03-08T18:02:23.947" Title="Music was better closer to the 80's - Riddle" Tags="|riddle|music|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28553" PostTypeId="2" ParentId="28494" CreationDate="2016-03-08T01:13:23.463" Score="247" Body="&lt;p&gt;(&lt;em&gt;Thanks to f'' for the help in chat! He solved the majority of this puzzle - I just came in at the end.&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Step 1: Crossword&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The crossword was solved by feelinferrety, Z. Dailey, and Sleafar:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/1DuXt.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1DuXt.jpg&quot; alt=&quot;crosswordsolved&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Step 2: NOTHEX&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;In the caption, if you look at A and D only (as it says in the NOTHEX ad), and interpret them as crossword clues (only taking the letters not torn off) you get the message &amp;quot;THE TITLE IS THE KEY&amp;quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Step 3: Using The Key&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; When you use the newspaper title as a Vignere key for the first paragraph, you get the message&lt;br /&gt;&#xA; &amp;quot;you're not listening read it again and realise the truth&amp;quot;&lt;br /&gt;&#xA;&lt;br /&gt;&#xA; If you use the &lt;em&gt;question&lt;/em&gt; title instead,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;you get a message for all four paragraphs:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;well done you are&lt;br /&gt;&#xA;beginning&lt;br /&gt;&#xA;to see but we have&lt;br /&gt;&#xA;not finished yet&lt;br /&gt;&#xA;you need all your&lt;br /&gt;&#xA;synapses firing&lt;br /&gt;&#xA;for this to work&lt;br /&gt;&#xA;&lt;br /&gt;&#xA;a pirate's search&lt;br /&gt;&#xA;but in reverse&lt;br /&gt;&#xA;a map disguised here&lt;br /&gt;&#xA;in the verse&lt;br /&gt;&#xA;perfectly square&lt;br /&gt;&#xA;this text aligns&lt;br /&gt;&#xA;the mark is found&lt;br /&gt;&#xA;follow the signs&lt;br /&gt;&#xA;and now it dawns it's&lt;br /&gt;&#xA;not enough&lt;br /&gt;&#xA;dual maps required&lt;br /&gt;&#xA;one fore one aft&lt;br /&gt;&#xA;&lt;br /&gt;&#xA;first&lt;br /&gt;&#xA;ENEEE N SSWW&lt;br /&gt;&#xA;SWW WWW W NNN S&lt;br /&gt;&#xA;second&lt;br /&gt;&#xA;WWWW NNEE EEN NNWW&lt;br /&gt;&#xA;WWSWW SSSSS ESS WN&lt;br /&gt;&#xA;&lt;br /&gt;&#xA;but if it's broken&lt;br /&gt;&#xA;what difference&lt;br /&gt;&#xA;would it make&lt;br /&gt;&#xA;once you can answer&lt;br /&gt;&#xA;that the truth will&lt;br /&gt;&#xA;become clear&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Step 4: Following Directions&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;&amp;quot;In the verse / perfectly square&amp;quot; hints at the poems being more than just instructions. In fact, all of the text in the &amp;quot;article&amp;quot; is exactly 400 letters, which can be arranged into a 20×20 grid. &amp;quot;A pirate's search but in reverse&amp;quot; hints that we're starting at an X instead of ending at one. So, if you take the text and arrange it into a square grid, you get this:&lt;br /&gt;&#xA;&lt;code&gt;pltdlgvqncltrrumtmrq&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;gbamcdmwuygplpsdwvai&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;tzgifamqciwwcogsplse&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;plhbzkgfibhsjyiebvtj&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;survclhzegkonipzsbwa&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;etoivhonbvrvhtsllslu&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;sihvlncaawlttfvbngam&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;iivvcdyktpklecfjbijm&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;lpuhhvqtneqtrwwfsgtf&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;vqkyshgkngtdwiivvlit&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;ganrhqmkcmrleflmgluw&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;lmfwgvvunayficwvhoic&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;ksowfXjbklwfmsnfuwil&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;trgmrirvqkqlkheopaag&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;uakmuwzsknpwagmriiql&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;bqpkhaopwflzawakezqi&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;kbfvmaovsncbqaoelayj&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;rklvumowgarzmmndmbrg&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;hwcovoyiflarkapsblpq&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;iflmhjbtyfifmayvzpij&lt;/code&gt;&lt;br /&gt;&#xA;&lt;strong&gt;and for the plaintext:&lt;/strong&gt;&lt;br /&gt;&#xA;&lt;code&gt;welldoneyouarebeginn&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;ingtoseebutwehavenot&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;finishedyetyouneedal&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;lyoursynapsesfiringf&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;orthistoworkapirates&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;searchbutinreverseam&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;apdisguisedhereinthe&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;verseperfectlysquare&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;thisteXtalignsthemar&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;kisfoundfollowthesig&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;nsandnowitdawnsitsno&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;tenoughdualmapsrequi&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;redoneforeoneaftfirs&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;teneeensswwswwwwwwnn&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;nssecondwwwwnneeeenn&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;nwwwwswwsssssesswnbu&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;tifitsbrokenwhatdiff&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;erencewoulditmakeonc&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;eyoucananswerthatthe&lt;/code&gt;&lt;br /&gt;&#xA;&lt;code&gt;truthwillbecomeclear&lt;/code&gt;&lt;br /&gt;&#xA; Follow the directions in the third paragraph on the first grid (taking letters after every space), and you get the message &amp;quot;AMBIGRAM&amp;quot;. If you repeat the process with the second grid and the second set of directions (&amp;quot;dual maps required, one fore one aft&amp;quot;) you get &amp;quot;ISBROKEN&amp;quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Step 5: The Image&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;An ambigram is the same thing upside-down and right-side-up. The image is &lt;em&gt;almost&lt;/em&gt; an ambigram, but some pixels are changed. If you XOR the image with itself rotated 180°, you get the following: &lt;a href=&quot;https://i.stack.imgur.com/0phTN.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0phTN.png&quot; alt=&quot;pleasewakeup&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Step 6: Interpreting Everything&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt;Well, we appear to be in a coma. The messages on the borders of the &amp;quot;please wake up!&amp;quot; image are:&lt;br /&gt;&#xA; - We've been trying to get a message to you...&lt;br /&gt;&#xA; - You're in a coma...&lt;br /&gt;&#xA; - There was an accident...&lt;br /&gt;&#xA; - Open your eyes...&lt;br /&gt;&#xA; - We miss you...&lt;br /&gt;&#xA;&lt;br /&gt;&#xA;Even the crossword clues hint that you're in a coma: specifically, the unused clues for 7, 16, 21, and 26 across. (The solutions to those are &amp;quot;accident&amp;quot;, &amp;quot;pity&amp;quot;, &amp;quot;coma&amp;quot;, and &amp;quot;dream&amp;quot;.)&#xA;&lt;br /&gt;&#xA; So we have two options: waking up and having to talk to people or staying in weird coma-land where everything is puzzles. I know which one &lt;em&gt;I'd&lt;/em&gt; choose.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11876" LastEditorUserId="58672" LastEditDate="2020-10-20T02:36:56.177" LastActivityDate="2020-10-20T02:36:56.177" CommentCount="25" ContentLicense="CC BY-SA 4.0" />
  <row Id="28554" PostTypeId="2" ParentId="28552" CreationDate="2016-03-08T01:17:30.617" Score="14" Body="&lt;p&gt;The answer appears to be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Pachelbel&amp;#39;s_Canon&quot;&gt;Pachelbel's Canon&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Taking an acrostic of the riddle provides the first eight notes of the song.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastActivityDate="2016-03-08T01:17:30.617" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28555" PostTypeId="1" AcceptedAnswerId="28565" CreationDate="2016-03-08T01:25:07.053" Score="15" ViewCount="1079" Body="&lt;p&gt;Addition Hangman is a game for two players. The rules are&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The first player (the Chooser) chooses an addition problem $x+y=z$, where $x$, $y$, and $z$ are positive integers. He writes the addition $x+y=z$ in base $10$, replaces each digit by a blank space ($\underline{\hspace{8mm}}$), and shows the result to the second player.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The second player (the Guesser) starts off with some number of tokens. She may risk a token in order to guess a digit. If the guessed digit appears anywhere in the addition, the Chooser fills in the appropriate blank space(s), and the Guesser keeps her token. If the guessed digit does not appear, the Guesser loses her token.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Guesser wins if all of the blank spaces are filled in. The Chooser wins if the Guesser runs out of tokens.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: The Guesser receives the addition problem&#xA;  $$&#xA;\begin{array}{cc}&#xA;&amp;amp;\underline{\hspace{8mm}}\\&#xA;+&amp;amp;\underline{\hspace{8mm}}\\\hline&#xA;&amp;amp;\underline{\hspace{8mm}}&#xA;\end{array}&#xA;$$&#xA;  Each blank represents a (not necessarily distinct) digit between $1$ and $9$.&#xA;  What is the smallest number of tokens the Guesser needs in order to be certain she can win?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$\newcommand{\bl}{\underline{\hspace{8mm}}}$&lt;/p&gt;&#xA;" OwnerUserId="2627" LastActivityDate="2016-03-10T04:03:15.577" Title="Addition hangman" Tags="|mathematics|game|" AnswerCount="5" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28556" PostTypeId="1" AcceptedAnswerId="28637" CreationDate="2016-03-08T02:24:22.013" Score="12" ViewCount="609" Body="&lt;p&gt;I was hanging around at home last week, a little lost for something to do, when I got a phone call from Ernie. He was in a bit of a panic because a guest was going to drop in later in the day. He had inadvertently dismantled his oven for a certain experiment he was working on so had no way to bake a cake for the guest. &quot;Is there any way...&quot;, he pleaded, &quot;...that you would be able to bake something for me in your oven?&quot;. I was in a generous mood so told him I would be delighted to help and asked what time he needed the baking brought over. We settled on a time that afternoon and after checking that the guest didn't have any allergies I hung up and proceeded to the kitchen where I began to mix the ingredients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was just about to put the dough into a nice round tin when the phone rang again. It was Ernie. &quot;One little thing I neglected to mention.&quot; he said. &quot;The guest is Great Aunt Titania. Now as you know, Auntie Titania loves symmetry above all else, so you might want to make the cake in the shape of a perfect regular polygon.&quot; I winced a little - my last meeting with Titania had been a bit stressful - so I poured myself a wee brandy to calm my nerves. After a moments thought I could see a simple solution. I put the round cake tin back on the shelf, selected a more appropriately shaped one, poured in the dough, and popped the cake into the oven.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some time later I was just turning the cake out onto a rack to cool when I got another call from Ernie. &quot;Small change in plans&quot; he said. &quot; Unknown to me, Auntie Titania invited her sister Great Aunt Lusitania to drop in too. Now as you know Auntie Lusitania loves equality above all else, so to keep her happy you might want to have two cakes, both the same shape and size of course&quot;. I felt a sharp pain in my left temple, it didn't pay to leave Lusitania unsatisfied, but I still needed to keep Titania happy. I poured another brandy&#xA;to dull the pain. &quot;But I don't have enough ingredients (or time) to make a second cake&quot; I lamented. &quot;Oh that wont be a problem&quot;, Ernie responded, &quot;I am sure the aunts won't mind if you cut the original cake up and reassemble the pieces into different shapes&quot;.  After a little thought I could see a fairly simple solution so I made a series of straight, vertical cuts in the original cake and reassembled the pieces into two smaller identical cakes. I was quite proud that there was no waste left over when I had finished, so I poured myself a celebratory brandy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was just about to put the two (identical) cakes into a box to take them over to Ernie, when the phone rang again. I could tell from the ring-tone that it was Ernie, so I had a small brandy to calm my nerves before answering. &quot;Small change in plans&quot; he said. &quot;Unknown to me, Auntie Lusitania invite her sister Great Aunt Bismarkia to drop in. So I am afraid we will need three cakes - and as you know, Auntie Bismarkia loves efficiency above all else. You might want to use the smallest total length of cuts possible when re-dividing the cakes &quot;. I swigged a large drink from the bottle - my head was pounding. &quot;But I wasn't thinking about efficiency when I made the first cuts&quot;, I lamented, &quot;How can I possibly satisfy Bismarkia?&quot;. &quot;Oh I am sure she will compromise just a little.&quot;, Ernie responded, &quot;Instead of the smallest total length of cuts, why not just ensure you end up with the smallest total number of pieces instead&quot;. It took me quite a while to find a solution, then I made another series of straight, vertical cuts in the existing cake pieces and reassembled them into three still smaller identical cakes.  I was quite proud that there was no waste left over when I had finished, so I poured the last of the brandy into a glass to celebrate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I decided that driving wouldn't be the most responsible way to get the cakes to Ernie, so walked (slightly unsteadily) around to drop them off at the door. &quot;Perfect&quot;, Ernie said as soon as he clapped his eyes on the cakes, &lt;strong&gt;&quot;Regular polygons - that will make Auntie Titania happy, all three cakes are the same size and shape - that will satisfy Auntie Lusitania, and I see you have used the smallest possible total number of  pieces to achieve the goal - that should get Auntie Bismarkia's approval. You made a lucky initial choice of cake-tin as no other starting polygon would use less pieces.&quot;&lt;/strong&gt; I was offered a place at the table, but felt that the aunts might not quite approve of my somewhat slurred speech, so I headed back home.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/r3pSt.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/r3pSt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, possibly as a result of the slight over-consumption of brandy, my mind is now a complete blank regarding the solution I found. And to be honest, the kitchen was such a mess when I woke up this morning that I couldn't even tell which polygonal cake tin I had used to bake the original cake! Can you help? The best regular polygon to start with, the minimum number of pieces required and a simple diagram should help jog my memory.&lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="1631" LastEditDate="2016-03-08T20:19:29.837" LastActivityDate="2016-03-08T23:18:45.370" Title="Ernie and the Revisitation of the Aunts" Tags="|mathematics|geometry|word-problem|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28557" PostTypeId="2" ParentId="28555" CreationDate="2016-03-08T02:24:53.663" Score="5" Body="&lt;p&gt;I can do it with five tokens.&lt;br&gt;&#xA;Guess $4$.  If you find $4+\_=\_$, guess $5,6,7$ until you find one.&lt;br&gt;&#xA;Guess $6$.  If you find one, guess $3,2,1$.&lt;br&gt;&#xA;Guess $2$.  If you find $\_+2=\_$ guess $5,1$&lt;br&gt;&#xA;Guess $8$ and you are guaranteed to find one, so do not lose a token.  If you find $\_+8=\_$ it is $1+8=9$.  Otherwise, it is $1+7=8$ or $3+5=8$  One more guess&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditDate="2016-03-08T04:02:03.153" LastActivityDate="2016-03-08T04:02:03.153" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28558" PostTypeId="2" ParentId="28555" CreationDate="2016-03-08T02:42:18.373" Score="0" Body="&lt;p&gt;From what I can tell there is a worst possible outcome of 6 coins.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Guess 2, 4, 6, and 9 and you will either get it or get nothing... if you get nothing you know it is 1 + 7 = 8 or 3 + 5 = 8, if you get 2, you can obviously figure it out.  If you get 1, you know it is either 1 + 2 = 3, 1 + 3 = 4, 1 + 4 = 5, 1 + 8 = 9, 2 + 3 = 5, 2 + 5 = 7, 3 + 4 = 7.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;therefore&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it depends on what you picked up, but either way you guess the one that occurs the most for your respective situation you've landed in and you can't possibly get more than 2 more wrong. Therefore you can lose 5 coins and stay in the game.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19731" LastActivityDate="2016-03-08T02:42:18.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28559" PostTypeId="2" ParentId="28555" CreationDate="2016-03-08T03:01:04.550" Score="3" Body="&lt;p&gt;I think it can be done with &lt;strike&gt; 4 &lt;/strike&gt; 5. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Guess 1 first. If you hit ($1+1$) you're done, if you hit ($1+\_$) the remaining options for $(z,y)$ are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(9,8)$ $(8,7)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(7,6)$ $(6,5)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(5,4)$ $(4,3)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(3,2)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, guess $8,6,4$ and you will know the answer with certainty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $1$ yields $(\_+\_)$ guess $2$, again $(2+2)$ and you're done, $(2+\_)$ means&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(9,7)$ $(7,5)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(8,6)$ $(6,4)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(5,3)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then guess $7,6$ and you're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If 2 yields $(\_+\_)$ then guess $4$. If you get $(4+4)$ you're done. If you get $(4+\_)$ then the options are &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(7,3)$ $(9,5)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Guess $7$ and you're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $4$ yields $(\_+\_)$ the answer must be $(3+\_)$ with options&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(9,6)$ $(8,5)$ $(6,3)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Guess 6 and you will know the answer with certainty.&lt;/p&gt;&#xA;" OwnerUserId="19843" LastEditorUserId="19843" LastEditDate="2016-03-08T04:04:27.763" LastActivityDate="2016-03-08T04:04:27.763" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="28560" PostTypeId="2" ParentId="28556" CreationDate="2016-03-08T03:41:16.920" Score="4" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Two cuts total&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;like so&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the shape is irrelevant, pick any regular polygon you want.  For the first, slice horizontally in half.  For the second, set the two pieces next to each other and slice horizontally again, this time, a third of the way up.  Take the bottom third from each and stack together.  Voila, three cakes, all the same shape and size and are regular polygons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9353" LastActivityDate="2016-03-08T03:41:16.920" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="28561" PostTypeId="2" ParentId="28556" CreationDate="2016-03-08T03:53:44.750" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/jWB8C.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jWB8C.png&quot; alt=&quot;the cake is a lie&quot;&gt;&lt;/a&gt;&#xA; hexagon, 3 cuts, squares, ..., profit?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but if I'm wrong:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'll be like his aunts and go down with my ship.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19731" LastActivityDate="2016-03-08T03:53:44.750" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28562" PostTypeId="2" ParentId="28306" CreationDate="2016-03-08T04:59:42.013" Score="0" Body="&lt;p&gt;It feels a little like fitting the clues to the solution, rather than the other way around, but it could be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Resistance&lt;/strong&gt; (which fits the title, &quot;immovable object&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I have a long history from east to west, / to some I'm the worst and others the best.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Resistances (as in, an uprisings against incumbent governments) are common throughout history, and are seen as the best by supporters, and as a terrorist act by others.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;With a core made of resin I dance and I flutter, / no one judges me when I speak with a stutter.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Word play on RESIn i dANCE ==&gt; RESIstANCE ??&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I may be an object of legend or myth,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Plenty of legendary uprisings in history...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;at first you won't find me but maybe in fifth.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A reference to the US fifth amendment (right to remain silent), which could be seen as a form of resistance.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;It should be clear by now that I'm not a Paid Genius Poet, / &quot;What am I, I ask?,&quot; by now you should know it.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Paid Genius Poet's capitalisation, PGP, could reference &lt;a href=&quot;https://en.wikipedia.org/wiki/Pretty_Good_Privacy&quot; rel=&quot;nofollow&quot;&gt;Pretty Good Privacy&lt;/a&gt; (encryption tool that could be used by members of a resistance) or &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/P-glycoprotein&quot; rel=&quot;nofollow&quot;&gt;multidrug resistance protein 1&lt;/a&gt;&quot;...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;For at any point, you summon power of three, / And look to the wind, you may just find me.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Electrical resistors can be made by winding wire around another material (&quot;look to the wind&quot;), and on a resistor, there are coloured bands, where the third indicates the decimal multiplier (&quot;three&quot;, &quot;point&quot;, &quot;power&quot;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My presence strengthens each link in the chain,&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Resist against weakness?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;certain times that you use me, it may be in vain.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Resistance is futile&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Storms when you're older and peace as a child, / with us two together, adventures are wild.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Adults weather the storm, while kids resist peace &amp;amp; quiet?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5208" LastActivityDate="2016-03-08T04:59:42.013" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28563" PostTypeId="2" ParentId="28545" CreationDate="2016-03-08T06:30:56.717" Score="5" Body="&lt;p&gt;How about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; tax?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Complex&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; complex tax: many rules.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alternative&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; alternative tax: where the tax rules allow choice&lt;br&gt;&#xA; ADDED: perhaps a shortened form of &lt;a href=&quot;https://en.wikipedia.org/wiki/Alternative_minimum_tax&quot; rel=&quot;nofollow&quot;&gt;Alternative Minimum Tax&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Exempt&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; tax exempt: no tax payable&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Deductive&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; tax deductive: reduces the amount of tax payable (a tax deduction is a reduction in the income subject to tax)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Dependent&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; dependent tax: the tax is conditional on (some) criteria. (A dependent is also the label given to &lt;a href=&quot;https://apps.irs.gov/app/understandingTaxes/hows/tax_tutorials/mod04/tt_mod04_01.jsp&quot; rel=&quot;nofollow&quot;&gt;persons who entitle a taxpayer to claim a dependency exemption&lt;/a&gt;.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fuzzy&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; fuzzy tax: the amount of tax raised (or the applicability of criteria) is imprecise&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Critical&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; critical tax: the country depends on it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And most of all, Certain.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Our new Constitution is now established, and has an appearance that promises permanency; but in this world nothing can be said to be certain, except death and taxes.&quot;&#xA; &lt;a href=&quot;https://en.wikipedia.org/wiki/Death_and_taxes_%28idiom%29&quot; rel=&quot;nofollow&quot;&gt;- Benjamin Franklin, quoted in wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9422" LastEditorUserId="9422" LastEditDate="2016-03-10T01:07:41.437" LastActivityDate="2016-03-10T01:07:41.437" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28564" PostTypeId="2" ParentId="28306" CreationDate="2016-03-08T06:54:44.247" Score="1" Body="&lt;p&gt;Are you &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Lamp&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have a long history from east to west,&#xA;to some I'm the worst and others the best.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Long History, Yes. Worst for goons and criminals, best for the rest.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With a core made of resin I dance and I flutter,&#xA;no one judges me when I speak with a stutter.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The fire inside the lamp dances and flutters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I may be an object of legend or myth,&#xA;at first you won't find me but maybe in fifth.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Legends like Arabian Nights had magical lamps.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For at any point, you summon power of three, &#xA;And look to the wind, you may just find me. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The three wishes of the genie in the lamp.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Storms when you're older and peace as a child, &#xA;with us two together, adventures are wild. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Maybe, another reference to the Genie in the Lamp.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19656" LastActivityDate="2016-03-08T06:54:44.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28565" PostTypeId="2" ParentId="28555" CreationDate="2016-03-08T07:24:16.947" Score="17" Body="&lt;p&gt;The Guesser can guarantee a win with &lt;strong&gt;4&lt;/strong&gt; tokens.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Guess $1$. If it's a match, but the solution is still ambiguous, the problem must contain a $3$, $5$, $7$, or $9$ so those can be guessed in order, losing at most three tokens.&lt;/li&gt;&#xA;&lt;li&gt;If the first guess was a miss, guess $3$. If it's a match, but the solution is still ambiguous, the problem must contain a $5$, $7$, or $9$, so those can be guessed in order, losing at most two additional tokens (for three lost total).&lt;/li&gt;&#xA;&lt;li&gt;If the first two guesses were misses, guess $2$. If it's a match, but the solution is still ambiguous, the problem must contain $6$ or $7$, so those can be guessed in order, losing at most one additional token (for three lost total).&lt;/li&gt;&#xA;&lt;li&gt;If the first three guesses were all misses, we're down to our last token. But the problem must contain $4$, so we can guess that, which will reveal the solution (either $4 + 4 = 8$ or $4 + 5 = 9$).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There are some variations possible (for example, $3$ can be guessed before $1$, or the follow-up guesses after a $3$ match could be $4$, $5$, $6$). There's no way to guarantee a win with 3 tokens, because there's no set of three digits such that every possible solution contains at least one digit from the set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a flowchart representation of the guessing strategy, with all possible solutions to the $\_ + \_ = \_$ problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7ML8I.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7ML8I.png&quot; alt=&quot;Flowchart representation of guesses and all possible solutions&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Italicized numbers represent blanks that have only one possible way to fill them. Green rightward lines lead to possible outcomes for matching guesses, and red downward lines are taken when there's no match. Note that every solution is reachable with at most three incorrect guesses.&lt;/p&gt;&#xA;" OwnerUserId="9044" LastEditorUserId="9044" LastEditDate="2016-03-10T04:03:15.577" LastActivityDate="2016-03-10T04:03:15.577" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28566" PostTypeId="1" AcceptedAnswerId="28569" CreationDate="2016-03-08T07:44:01.320" Score="6" ViewCount="1029" Body="&lt;p&gt;A particularly easy Rebus. What is it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wspDD.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wspDD.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19656" LastEditorUserId="8874" LastEditDate="2016-03-08T10:00:02.410" LastActivityDate="2016-05-29T14:12:56.440" Title="The Rebus Challenge" Tags="|visual|rebus|" AnswerCount="4" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28567" PostTypeId="2" ParentId="28566" CreationDate="2016-03-08T07:47:25.837" Score="5" Body="&lt;p&gt;It could be either&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; damage control&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; out of control / no control&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17147" LastActivityDate="2016-03-08T07:47:25.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28568" PostTypeId="1" AcceptedAnswerId="28575" CreationDate="2016-03-08T07:49:57.383" Score="0" ViewCount="239" Body="&lt;p&gt;Using the numbers 1,2,3,4,5,6,7,8,9 only once in the complete set, Give me a set of perfect squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have found the set, you have the liberty to use all these numbers without a comma and give me a nine digit perfect square featuring all of the nine digits...JUST ONCE!&lt;/p&gt;&#xA;" OwnerUserId="19656" LastActivityDate="2016-03-08T08:53:32.877" Title="The Perfect Square Syndrome" Tags="|mathematics|calculation-puzzle|" AnswerCount="1" CommentCount="0" ClosedDate="2016-03-08T11:22:44.393" ContentLicense="CC BY-SA 3.0" />
  <row Id="28569" PostTypeId="2" ParentId="28566" CreationDate="2016-03-08T07:50:29.537" Score="11" Body="&lt;p&gt;Depending on how your keyboard is laid out:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Losing control&quot; or &quot;no alternative&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19731" LastActivityDate="2016-03-08T07:50:29.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28570" PostTypeId="2" ParentId="28306" CreationDate="2016-03-08T07:54:58.870" Score="1" Body="&lt;p&gt;This riddle strongly reminds me of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;TARDIS&lt;/strong&gt;(Time and Relative Dimension In Space) from the series &quot;Doctor Who&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have a long history from east to west,&#xA;to some I'm the worst and others the best.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The name says it all! It has a long history (It's a time machine!) and it has that history from east to west (It's also a transport device. It also travels through dimensions!). Its good for the universe and bad especially for the DALEKS.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With a core made of resin I dance and I flutter,&#xA;no one judges me when I speak with a stutter.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I do not remember about the TARDIS core but the dance and flutter reference may be to how it operates in the show. No one judges the TARDIS!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I may be an object of legend or myth,&#xA;at first you won't find me but maybe in fifth. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The TARDIS and Doctor Who is a kind of legendary series and both of them are known as legends or myth in the series as well. Not sure about the first and fifth reference.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For at any point, you summon power of three, &#xA;And look to the wind, you may just find me. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Power of three is an episode of Doctor Who and one can find TARDIS in the wind if it is really needed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My presence strengthens each link in the chain, &#xA;certain times that you use me, it may be in vain. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the show it is very creatively related to real life events hence the chain reference. Certain times it has gone in vain to use the TARDIS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Storms when you're older and peace as a child, &#xA;with us two together, adventures are wild. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There have been many wild adventures with the TARDIS, I'll provide the storm and peace explanation when I get it(Probably as the time passes.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A limited spectrum between ceiling and floor, &#xA;Shot full of holes from twelve until four. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The TARDIS is small on the outside and bigger on the inside, hence the limited spectrum reference.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In solving this riddle you're just out Of Time,&#xA;you should hear my voice as it is sublime.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &quot;out Of Time&quot; phrase just makes me think about the TARDIS again. The sound it makes when it arrives is indeed sublime and provides hope to the people who hear it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It should be clear by now that I'm not a Paid Genius Poet,&#xA;&quot;What am I, I ask?,&quot; by now you should know it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are not a paid genius poet and the biggest question on the series is &quot;Doctor Who?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19191" LastActivityDate="2016-03-08T07:54:58.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28572" PostTypeId="2" ParentId="28306" CreationDate="2016-03-08T08:01:07.790" Score="1" Body="&lt;p&gt;The answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fireworks&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have a long history from east to west,&#xA;to some I'm the worst and others the best.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fireworks comes from the east (asia) and has also been used in the west for many years. Some love it and some hate it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With a core made of resin I dance and I flutter,&#xA;no one judges me when I speak with a stutter.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; fireworks may have had a core of resin, not sure on that. Fireworks makes a stuttering sound when launched, but we're nice people so we dont judge it :D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I may be an object of legend or myth,&#xA;at first you won't find me but maybe in fifth. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fireworks was used in China because of the myth that there are ghosts(or something supernatural like that). At the fifth of november there is a fireworks in london to celebrate guy fawkey night. (Remember remember the fifth of november ;))&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For at any point, you summon power of three, &#xA;And look to the wind, you may just find me. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are 3 classes of firework: daytime, nighttime and novelty fireworks. Some fireworks is shot in the sky, thats why you can find it there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My presence strengthens each link in the chain, &#xA;certain times that you use me, it may be in vain. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fireworks is used at newyear, and you might say it makes the link between the old and new year better/stronger. Sometimes fireworks doesnt go off, so its in vain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Storms when you're older and peace as a child, &#xA;with us two together, adventures are wild. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When you're youngh you might only use sparklers, but when you're older you'll use fireworks that make alot of noise. Playing with fireworks can get wild.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A limited spectrum between ceiling and floor, &#xA;Shot full of holes from twelve until four. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can only shoot fireworks between 12 and 4 at night. This is a limited timespan. (may also mean that it only goes a certain height up in the air)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In solving this riddle you're just out Of Time,&#xA;you should hear my voice as it is sublime. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I toke too long in writing this stuff down, now the fireworks went off and i hear the bang.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It should be clear by now that I'm not a Paid Genius Poet,&#xA;&quot;What am I, I ask?,&quot; by now you should know it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are fireworks!!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19928" LastActivityDate="2016-03-08T08:01:07.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28574" PostTypeId="1" AcceptedAnswerId="28576" CreationDate="2016-03-08T08:49:11.970" Score="-2" ViewCount="137" Body="&lt;p&gt;Give the name of the person who said these amazing words...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;pbzzbafraf vf abg jung lbh frr vasebag bs lbhe rlrf vgf n tvsg bs frrvat jungf vasebag bs bguref rlr&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This person is a fictional ghost.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19656" LastEditorUserId="17147" LastEditDate="2016-03-08T08:50:50.607" LastActivityDate="2016-03-08T09:11:27.693" Title="What kind of Quote is this?" Tags="|cipher|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28575" PostTypeId="2" ParentId="28568" CreationDate="2016-03-08T08:53:32.877" Score="3" Body="&lt;p&gt;We have, for instance, the following set of squares:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;\begin{array}{c}25 = 5^2\\784 = 28^2\\1369 = 37^2\end{array}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As for the nine digit perfect square, there are several. Below, they are listed in 'reverse' increasing order.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;\begin{array}{l}549386721 = 23439^2\\735982641 = 27129^2\\385297641 = 19629^2\\529874361 = 23019^2\\245893761 = 15681^2\\382945761 = 19569^2\\697435281 = 26409^2\\597362481 = 24441^2\\627953481 = 25059^2\\672935481 = 25941^2\\653927184 = 25572^2\\326597184 = 18072^2\\537219684 = 23178^2\\847159236 = 29106^2\\842973156 = 29034^2\\923187456 = 30384^2\\412739856 = 20316^2\\589324176 = 24276^2\\139854276 = 11826^2\\215384976 = 14676^2\\375468129 = 19377^2\\361874529 = 19023^2\\743816529 = 27273^2\\615387249 = 24807^2\\157326849 = 12543^2\\587432169 = 24237^2\\254817369 = 15963^2\\152843769 = 12363^2\\523814769 = 22887^2\\714653289 = 26733^2\end{array}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16478" LastActivityDate="2016-03-08T08:53:32.877" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28576" PostTypeId="2" ParentId="28574" CreationDate="2016-03-08T08:53:45.113" Score="3" Body="&lt;p&gt;By ROT13 I get&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; commonsens is not what you see infront of your eyes its a gift of seeing whats infront of others eye&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Common sense is not what you see in front of your eyes. Its a gift of seeing what's in front of others eye&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This quote is by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jason Bourne (courtesy of @Prashant Kumar)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-03-08T09:11:27.693" LastActivityDate="2016-03-08T09:11:27.693" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28577" PostTypeId="1" AcceptedAnswerId="28578" CreationDate="2016-03-08T11:47:57.147" Score="10" ViewCount="1069" Body="&lt;p&gt;Given below is a picture.&#xA;Find the word(s), and use it/them to find the name of a famous personality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/09KRS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/09KRS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The personality is Indian i.e. from India, and perhaps one of the most famous Indian around the world right now.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; That in the picture, is not Chinese ;P&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Good Luck!&lt;/p&gt;&#xA;" OwnerUserId="19656" LastEditorUserId="17147" LastEditDate="2016-03-08T11:50:27.740" LastActivityDate="2016-03-08T11:58:10.437" Title="ANAGRAM in a REBUS" Tags="|visual|rebus|anagram|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28578" PostTypeId="2" ParentId="28577" CreationDate="2016-03-08T11:58:10.437" Score="12" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Narendra Modi, prime minister of India&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The image is two parts:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; MANDARIN and REDO, which together comprise the letters in NARENDRA MODI&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17147" LastActivityDate="2016-03-08T11:58:10.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28579" PostTypeId="2" ParentId="28566" CreationDate="2016-03-08T12:25:57.473" Score="4" Body="&lt;p&gt;Based on my keyboard layout it is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Lost control&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19984" LastEditorUserId="16225" LastEditDate="2016-03-08T15:00:33.060" LastActivityDate="2016-03-08T15:00:33.060" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28580" PostTypeId="1" CreationDate="2016-03-08T12:38:08.960" Score="7" ViewCount="1450" Body="&lt;p&gt;Four numbers are available: $1$, $3$, $4$ and $6$. Every number must be used once and only once with (some of) the operations $+$, $-$, $\times$, $\div$ to form the number $24$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's from the book &quot;Art of Exploitation&quot;, 2nd edition. Give it a try!&lt;/p&gt;&#xA;" OwnerDisplayName="user19985" LastEditorUserId="4421" LastEditDate="2016-03-08T18:07:48.227" LastActivityDate="2016-03-08T20:57:32.000" Title="Simple but interesting math puzzle" Tags="|mathematics|calculation-puzzle|formation-of-numbers|" AnswerCount="5" CommentCount="4" FavoriteCount="0" ClosedDate="2016-03-08T21:00:14.953" ContentLicense="CC BY-SA 3.0" />
  <row Id="28581" PostTypeId="2" ParentId="28580" CreationDate="2016-03-08T12:45:28.820" Score="20" Body="&lt;p&gt;An answer to this problem is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $6 \div (1 - \frac34)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19986" LastEditorUserId="4421" LastEditDate="2016-03-08T18:09:32.253" LastActivityDate="2016-03-08T18:09:32.253" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28582" PostTypeId="2" ParentId="28580" CreationDate="2016-03-08T12:50:49.377" Score="-4" Body="&lt;p&gt;Answer could also b:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $6 \times 4 \times 1^3$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19086" LastEditorUserId="4421" LastEditDate="2016-03-08T18:10:22.023" LastActivityDate="2016-03-08T18:10:22.023" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="28583" PostTypeId="2" ParentId="28580" CreationDate="2016-03-08T12:57:49.237" Score="-3" Body="&lt;p&gt;Bit of an off the wall answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (6-4)(3+1) &lt;strong&gt;forms&lt;/strong&gt; 24 if you calculate the values within the brackets, and leave the results as they are&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not mathematically correct at all I know, it's an answer that loosely fits the wording of the puzzle&lt;/p&gt;&#xA;" OwnerUserId="17034" LastActivityDate="2016-03-08T12:57:49.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28584" PostTypeId="1" AcceptedAnswerId="28605" CreationDate="2016-03-08T13:35:28.837" Score="16" ViewCount="749" Body="&lt;p&gt;Can you place seven (point-sized) pawns on a $7\times7$ checkerboard, so that&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;every pawn is placed precisely in the middle of one of the little checkerboard squares, and&lt;/li&gt;&#xA;&lt;li&gt;all distances between pairs are different?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2016-03-11T10:57:19.147" Title="Placing seven point-sized pawns" Tags="|mathematics|combinatorics|checkerboard|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28585" PostTypeId="1" AcceptedAnswerId="28597" CreationDate="2016-03-08T13:40:30.683" Score="9" ViewCount="250" Body="&lt;p&gt;Every letter stands for a digit in base-9 representation, &#xA;different letters stand for different digits,&#xA;and leading digits are always non-zero.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;       ALLS&#xA;    +  WELL&#xA;    +  THAT&#xA;    +   END&#xA;   ---------&#xA;      SWELL&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which digit does each letter represent? (Please present the full analysis how these digits can be determined.)&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="9404" LastEditDate="2016-03-08T14:13:14.063" LastActivityDate="2016-03-08T16:27:38.073" Title="An almost Shakespearian alphametic" Tags="|no-computers|alphametic|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="28586" PostTypeId="2" ParentId="28437" CreationDate="2016-03-08T13:55:36.270" Score="15" Body="&lt;p&gt;I think it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Sarcasm&lt;/strong&gt;&lt;br&gt;&#xA; These are the atomic numbers of the elements with symbols  S = 16 Ar = 18 Ca = 20 Sm = 62&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19547" LastEditorUserId="8874" LastEditDate="2016-03-08T13:58:55.977" LastActivityDate="2016-03-08T13:58:55.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28588" PostTypeId="2" ParentId="28306" CreationDate="2016-03-08T14:21:43.317" Score="10" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An ocarina&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have a long history from east to west,&lt;br&gt;&#xA;to some I'm the worst and others the best.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Ocarina&quot;&gt;&lt;em&gt;&quot;The ocarina belongs to a very old family of instruments, believed to date back over 12,000 years.&quot;&lt;/em&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With a core made of resin I dance and I flutter,&lt;br&gt;&#xA;no one judges me when I speak with a stutter.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They can be played legato or staccato. The ocarina has a &lt;em&gt;resonating&lt;/em&gt; chamber (on which &lt;em&gt;resin&lt;/em&gt; is a pun); being a wind instrument, it's a relative of the &lt;em&gt;flute&lt;/em&gt; (hence: &lt;em&gt;flutter&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I may be an object of legend or myth,&lt;br&gt;&#xA;at first you won't find me but maybe in fifth.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's present in the fifth &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/The_Legend_of_Zelda&quot;&gt;Legend of Zelda&lt;/a&gt;&lt;/em&gt; game released, &lt;em&gt;The Ocarina of Time&lt;/em&gt;, in which it was part of the game's lore.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For at any point, you summon power of three,&lt;br&gt;&#xA;And look to the wind, you may just find me.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; OK, we're into video game territory for sure here. The power of three could be the Triforce. It also could have to do with the three stones you have to gather from the Forest, Fire, and Water temples before you get the Ocarina of Time. &lt;em&gt;Wind&lt;/em&gt; has to do with the type of instrument it is.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My presence strengthens each link in the chain,&lt;br&gt;&#xA;certain times that you use me, it may be in vain.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sometimes when you use the ocarina in the game, nothing happens. And &quot;link&quot; refers to, well, &lt;strong&gt;Link&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Storms when you're older and peace as a child,&lt;br&gt;&#xA;with us two together, adventures are wild.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Adult Link saves areas from evil - prior to this, the areas are ravaged with fire and storms; afterward, they are calm, as they were when he was a child.  Also, the Song of Storms is a song that is taught only to Adult Link.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A limited spectrum between ceiling and floor,&lt;br&gt;&#xA;Shot full of holes from twelve until four.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; OK, this is back to regular ocarinae: The ocarina doesn't allow for a very wide range of notes, and also: &lt;em&gt;&quot;[v]ariations exist, but a typical ocarina is an enclosed space with &lt;strong&gt;four to twelve finger holes&lt;/strong&gt; and a mouthpiece that projects from the body.&quot;&lt;/em&gt; - &lt;em&gt;ibid.&lt;/em&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In solving this riddle you're just out Of Time,&lt;br&gt;&#xA;you should hear my voice as it is sublime. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Ocarina &lt;strong&gt;of Time&lt;/strong&gt;! Hooray! And &quot;sublime voice&quot; refers to the beautiful sound it makes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It should be clear by now that I'm not a Paid Genius Poet,&lt;br&gt;&#xA;&quot;What am I, I ask?,&quot; by now you should know it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Paid Genius Poet&quot; is an anagram of &lt;a href=&quot;https://en.wikipedia.org/wiki/Giuseppe_Donati&quot;&gt;Giuseppe Donati, inventor of the ocarina&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2016-03-08T14:46:19.687" LastActivityDate="2016-03-08T14:46:19.687" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="28589" PostTypeId="1" AcceptedAnswerId="28591" CreationDate="2016-03-08T14:22:10.607" Score="4" ViewCount="113" Body="&lt;p&gt;Every digit (in base-10 representation) stands for a letter,&#xA;different digits stand for different letters.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    233 + 8519 + 954 = 9706&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The whole is a famous Shakespeare quote.&#xA;Which letter does each digit represent? &lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="9404" LastEditDate="2016-03-08T14:29:36.860" LastActivityDate="2016-03-08T14:46:00.830" Title="A Shakesperian inverse alphametic" Tags="|knowledge|alphametic|literature|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28590" PostTypeId="2" ParentId="28306" CreationDate="2016-03-08T14:23:16.050" Score="1" Body="&lt;p&gt;I do have some hole left, maybe someone can improve it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you are the Levante , wich is a wind&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have a long history from east to west&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Blow from east to west&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;to some I'm the worst and others the best&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The winds are moist before the Strait of Gibraltar but dry after, carrying bad and good weather respectively&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With a core made of resin I dance and I flutter&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;don't know about the resin, but the &quot;dance and flutter&quot; stick with the idea&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;no one judges me when I speak with a stutter.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it can produce winds of 20-40 kt (10-20 m/s), which, if i remeber corectly, is not that much for a wind&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I may be an object of legend or myth,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I have done some search and it's actually linked to some legend&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;at first you won't find me but maybe in fifth. &#xA;For at any point, you summon power of three, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; i have no clue for this &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And look to the wind, you may just find me. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; quite literally!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My presence strengthens each link in the chain, &#xA;certain times that you use me, it may be in vain. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; maybe linked to sailing ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Storms when you're older and peace as a child, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; yet again terms linked to winds&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;with us two together, adventures are wild. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; again linked to sailing and adventuring&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A limited spectrum between ceiling and floor, &#xA;Shot full of holes from twelve until four. &#xA;In solving this riddle you're just out Of Time,&#xA;you should hear my voice as it is sublime. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; once again i have no clue here&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19980" LastEditorUserId="19980" LastEditDate="2016-03-08T14:34:35.293" LastActivityDate="2016-03-08T14:34:35.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28591" PostTypeId="2" ParentId="28589" CreationDate="2016-03-08T14:40:03.740" Score="6" Body="&lt;p&gt;The quote is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;Off with his head!&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Digit-to-letter mappings:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;2&lt;/strong&gt; &lt;em&gt;-&gt;&lt;/em&gt; &lt;em&gt;O&lt;/em&gt;, &lt;strong&gt;3&lt;/strong&gt; &lt;em&gt;-&gt;&lt;/em&gt; &lt;em&gt;f&lt;/em&gt;, &lt;strong&gt;8&lt;/strong&gt; &lt;em&gt;-&gt;&lt;/em&gt; &lt;em&gt;w&lt;/em&gt;, &lt;strong&gt;5&lt;/strong&gt; &lt;em&gt;-&gt;&lt;/em&gt; &lt;em&gt;i&lt;/em&gt;, &lt;strong&gt;1&lt;/strong&gt; &lt;em&gt;-&gt;&lt;/em&gt; &lt;em&gt;t&lt;/em&gt;, &lt;strong&gt;9&lt;/strong&gt; &lt;em&gt;-&gt;&lt;/em&gt; &lt;em&gt;h&lt;/em&gt;, &lt;strong&gt;4&lt;/strong&gt; &lt;em&gt;-&gt;&lt;/em&gt; &lt;em&gt;s&lt;/em&gt;, &lt;strong&gt;7&lt;/strong&gt; &lt;em&gt;-&gt;&lt;/em&gt; &lt;em&gt;e&lt;/em&gt;, &lt;strong&gt;0&lt;/strong&gt; &lt;em&gt;-&gt;&lt;/em&gt; &lt;em&gt;a&lt;/em&gt;, &lt;strong&gt;6&lt;/strong&gt; &lt;em&gt;-&gt;&lt;/em&gt; &lt;em&gt;d&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6134" LastEditorUserId="6134" LastEditDate="2016-03-08T14:46:00.830" LastActivityDate="2016-03-08T14:46:00.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28592" PostTypeId="2" ParentId="28584" CreationDate="2016-03-08T15:20:46.267" Score="10" Body="&lt;p&gt;Not really an answer but I'd like to add my thought process to this&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First I checked: How many possible distances are there really on a 7x7 checkerboard? To do this you can draw lines from (1,1) to (1,7) till (7,7), from (1,1) to (1,6) till (6,6) and so on. This mean $7+6+5+4+3+2 = 27$, However because 3,4,5 is a pythagorean triple it is one less because (1,1) to (4,5) = (1,1) to (6,1). There are 21 connections between 7 points. My intuition therefore says it is very unlikely that there are 21 connections that are all different because there are only 26 possibilities to begin with.&lt;br&gt;&lt;br&gt;This does tell us that if there is a solution that at least 3 points are in a square along the border. Why? There are 7 distances that go from border to border and only a maximum of 5 are not used of these. The two distances could share a point, therefore at least 3 points are on the border squares&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1748" LastEditorUserId="8874" LastEditDate="2016-03-08T16:09:19.423" LastActivityDate="2016-03-08T16:09:19.423" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28593" PostTypeId="1" AcceptedAnswerId="28606" CreationDate="2016-03-08T15:25:56.467" Score="11" ViewCount="469" Body="&lt;p&gt;Ms. Shipra is a renowned scientist in TIFR, Mumbai. She recently discovered a new way of gaining huge energy from the water-current at a much lower cost. Considering its future application, many companies were after her research paper. She finally decided to file the patent. On February 4th, 2016 she visits the Indian Patent Office in Mumbai. After 3 minutes of submitting the paper, one officer in the office found that the main paper is missing. Everybody, including who were present in the office and who entered and left the office during that period, was body-searched but the paper was not found.&#xA;Here is the link of the CCTV footage of 3 minutes during which the paper was stolen. (In this video, the first girl entering is Ms. Shipra.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://drive.google.com/folderview?id=0B1j9Jep9l0z5b2Zjc21WUjZsMFE&amp;amp;usp=drive_web&quot; rel=&quot;nofollow&quot;&gt;https://drive.google.com/folderview?id=0B1j9Jep9l0z5b2Zjc21WUjZsMFE&amp;amp;usp=drive_web&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you determine who stole the paper, and why was he/she not caught during searching?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My attempt:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The old man with the walking stick came with the stick in the left hand and after all the mess, he walks out with the stick in his right hand. The question still remains, where did he hide the documents? If we watch carefully, it looks like the guy in the black and white checkered shirt exchanged something with the old man. Is he also involved?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This question is taken from the event CSI (Crime Scene Investigation), Inquivesta.&#xA;Website: &lt;a href=&quot;http://www.inquivesta.iiserkol.ac.in&quot; rel=&quot;nofollow&quot;&gt;www.inquivesta.iiserkol.ac.in&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. This website plays a really annoying music in the background. Also, you cannot see the questions without registering to the fest.&lt;/p&gt;&#xA;" OwnerUserId="19644" LastEditorUserId="19644" LastEditDate="2016-03-11T05:17:46.340" LastActivityDate="2016-03-11T05:17:46.340" Title="Watch the video and figure out who stole the documents?" Tags="|visual|no-computers|mystery|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="28594" PostTypeId="2" ParentId="28360" CreationDate="2016-03-08T15:35:19.727" Score="4" Body="&lt;p&gt;you are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the Big Ben &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll start to count while you solve this problem,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it's a clock&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and just relax here at the top of my column.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it's on top of a tower&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Everywhere I go I let out a good whistle, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I bet it have many working reproduction&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but don't mess with me cause I carry a pistol.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;ok, here come the stretch, i did my research and in a videogame (s.t.a.l.k.e.r.) there is a pistol called Big Ben, i wouldn' t bet on it, but since Z.Dailey already referenced videogame...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've been explosive at one point in history, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;exploded&quot; in 1976&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;what I truly am is not much of a mystery.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; everybody know about the Big Ben&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19980" LastActivityDate="2016-03-08T15:35:19.727" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28595" PostTypeId="1" CreationDate="2016-03-08T15:44:20.833" Score="8" ViewCount="1637" Body="&lt;p&gt;I was the freest man around.&lt;br&gt;&#xA;Yet I never left my grounds&lt;br&gt;&#xA;(for years at a stretch)&lt;br&gt;&#xA;I could walk, fight, or run.&lt;br&gt;&#xA;I'm as able-bodied as anyone!&lt;br&gt;&#xA;And my riches I can't count!&lt;br&gt;&#xA;But I rarely ventured out.&lt;br&gt;&#xA;I'm stable mentally and emotionally&lt;br&gt;&#xA;And many dream of being me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/questions/28818/&quot;&gt;A Tail of One Lady&lt;/a&gt; (but not the same person.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19179" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.047" LastActivityDate="2016-03-15T18:11:12.400" Title="The Happy Prisoner" Tags="|riddle|rhyme|" AnswerCount="8" CommentCount="10" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28597" PostTypeId="2" ParentId="28585" CreationDate="2016-03-08T16:01:09.057" Score="5" Body="&lt;p&gt;First, we see that $S\in\{1,2\}$.  From the first column, we see that $S+L+T+D$ yields $L$.  Thus, $S+T+D\in \{10,20\}$.  Since S is so small, we can conclude that $S+T+D=10$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the second column, we see $L+L+A+N$ yields $L$. Since there is a carry over of 1, $L+A+N\in\{8,18\}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the third column, we see $L+E+H+E$ yields $E$.  Since the carry over is 1 or 2, we know $L+E+H\in\{7,8,17,18\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, $A+W+T$ yields $W$.  Again, carry over is 1 or 2, so $A+T\in\{7,8,17,18\}$.  But only $\{7,8\}$ are really possible, so we know that &lt;strong&gt;&lt;em&gt;S=1&lt;/em&gt;&lt;/strong&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, we know that $T+D=8 \implies A+T=7$, so we know that $A=D-1$.  Also, we know the carry over from the 3rd column must be 2, so $L+E+H\in\{17,18\}$.  The possible values are then:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$A \in \{2,4,5\}$&lt;/li&gt;&#xA;&lt;li&gt;$T \in \{5,3,2\}$&lt;/li&gt;&#xA;&lt;li&gt;$D \in \{3,5,6\}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thus, 5 is not an option for any other letter since it must be used by one of these three.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets look at the letter $L\in\{0,2,3,4,6,7,8\}$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $L=0$, then $L+E+H$ cannot be made 17 or 18.&lt;/li&gt;&#xA;&lt;li&gt;If $L=2$, then $A+N=6 \implies N=T-1$ which means either $N$ or $A$ must be 2.&lt;/li&gt;&#xA;&lt;li&gt;If $L=3$, then $A+N=5 \implies N=T-2 \implies N=0, T=2, A=5, D=6$.  But then $E+H=15$ cannot be made since 6 in taken.&lt;/li&gt;&#xA;&lt;li&gt;If $L=4$, then $A+N=4 \implies A=N=2$.  Thus, $A+N=14 \implies A=5, T=2, D=6, N=8$. Again, this makes $E+H=13$ impossible with the remaining digits. &lt;/li&gt;&#xA;&lt;li&gt;If $L=8$, then $A+N=10 \implies N=T+2$. Thus, $A=5, T=2, D=6, N=4$. Again, $E+H=8$ is impossible with the remaining digits. &lt;/li&gt;&#xA;&lt;li&gt;If $L=7$, then $A+N=11 \implies N=T+3$. Thus, either $A=2,T=5,D=3,N=8$ or $A=4,T=3,D=5,N=6$.  In both cases, $E+H=10$ are impossible.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thus $L=6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $A+N=12$ or $A+N=2$.  If $A+N=12$, then $N=T+4$.  Thus, $A=4, T=3, D=5, N=7$. Also, the carry over is 2 so that $L+E+H=17$.  Thus, $E+H=11$ so $E,H \in \{8,2\}$.  This leaves $W=0$, but we cannot have a 3 digit number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $A+N=2$.  Thus, $A=2, N=0, T=5, D=3$.  $L+E+H=18$ so $E+H+12$.  This requires $E,H\in\{4,6\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the valid solutions are:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 2661    &#xA; 8466&#xA; 5725&#xA;+ 403&#xA;-----&#xA;18466&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 2661    &#xA; 8766&#xA; 5425&#xA;+ 703&#xA;-----&#xA;18766&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A=2, D=3, L=6, N=0, S=1, T=5, W=8, E,H\in\{4,7\}$$&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2016-03-08T16:01:09.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28599" PostTypeId="2" ParentId="28585" CreationDate="2016-03-08T16:12:40.040" Score="2" Body="&lt;p&gt;Solution(s):  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A = 2&lt;br&gt;&#xA; L = 6&lt;br&gt;&#xA; S = 1&lt;br&gt;&#xA; W = 8&lt;br&gt;&#xA; E = 4/7&lt;br&gt;&#xA; T = 5&lt;br&gt;&#xA; H = 7/4&lt;br&gt;&#xA; N = 0&lt;br&gt;&#xA; D = 3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation&#xA;In order to make it easier my explanation are going to be made in base 10.&#xA;So if I say A+W = 10 this means in base 10, In base 9 it would be A+W = 11 (which is 10 in base 9). Headache already? :)&#xA;I will only specify base 9 where is needed&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; SWELL - WELL = S0000 (in base 9).   this means that&lt;br&gt;&#xA; &#xA;  ALLS +&lt;br&gt;&#xA;  THAT +&lt;br&gt;&#xA;   END =&lt;br&gt;&#xA; S0000 (base 9).&lt;br&gt;&#xA; &#xA; S has to be 1. In order for it to be 2, A+T+ carriage from the hundreds&#xA; should be above 18 (20 in base 9). Since they can be max 8+7 it means&#xA; the carriage from the tens should be 3 at least. And L+H+E can be max&#xA; 6+5+4. You cannot add up to 27 (30 in base 9) by adding the carriage&#xA; from the units position.&lt;br&gt;&#xA; &#xA;  ALL1 +&lt;br&gt;&#xA;  THAT +&lt;br&gt;&#xA;   END =&lt;br&gt;&#xA; 10000 (base 9).&lt;br&gt;&#xA;&#xA;   This means T+D = 8. &#xA; Also T+A+carriage from hundreds = 9 (10 in base 9). This means T+A = 7&#xA; or 8. It cannot be 8 because this will result in A = D. So T+A = 7.&lt;br&gt;&#xA; This results in D = A+1 so D &gt;= 1. But if D = 2 then A = 1 = S so D &gt;=&#xA; 3.   Let's start with D = 3. So A = 2 and T = 5&lt;br&gt;&#xA; &#xA;  2LL1 +&lt;br&gt;&#xA;  5H25 +&lt;br&gt;&#xA;   EN3 =&lt;br&gt;&#xA; 10000 (base 9)&lt;br&gt;&#xA;&#xA; this means L + 2 + N +  1 (carriage) = 9 or 18 (base 10). So L + N = 6/15 (base 10).&lt;br&gt;&#xA; &#xA; L + N = 6 can be obtained from the remaining digits only if one is 0&#xA; and one is 6.&lt;br&gt;&#xA; Let's say L = 6 and N = 0.&lt;br&gt;&#xA; &#xA;  2661 +&lt;br&gt;&#xA;  5H25 +&lt;br&gt;&#xA;   E03 =&lt;br&gt;&#xA; 10000 (base 9)&#xA; &#xA; this means H + 6 + E + 1 (carriage) = 18 (base 10). H + E  = 11 (base&#xA; 10). Only available values for a sum of 12 (base 10) are 7 and 4.&lt;br&gt;&#xA; Works in both combinations.   but let's say H = 7 and E = 4   Now to&#xA; the original sum&#xA; &#xA;  2661 +&lt;br&gt;&#xA;  W466 +&lt;br&gt;&#xA;  5725 +&lt;br&gt;&#xA;   403 =&lt;br&gt;&#xA; 1W466&lt;br&gt;&#xA; &#xA; W can be anything but the only available value is 8. &#xA; &#xA; So.   Solution 1&#xA; &#xA;  2661 +&lt;br&gt;&#xA;  8466 +&lt;br&gt;&#xA;  5725 +&lt;br&gt;&#xA;   403 =&lt;br&gt;&#xA; 18466  (base 9)&lt;br&gt;&#xA; &#xA; A = 2   L = 6   S = 1   W = 8   E = 4   T = 5   H = 7   N = 0   D = 3 &#xA; &#xA; &#xA; Solution 2 reversing H and E&#xA; &#xA; &#xA;  2661 +&lt;br&gt;&#xA;  8466 +&lt;br&gt;&#xA;  5725 +&lt;br&gt;&#xA;   403 =&lt;br&gt;&#xA; 18466  (base 9)&lt;br&gt;&#xA; &#xA; A = 2   L = 6   S = 1   W = 8   E = 7   T = 5   H = 4   N = 0   D = 3 &#xA; &#xA; &#xA; When we started this we assumed D = 3.   Let's increment. D = 4. From&#xA; T + D = 8 we get a &quot;No go&quot; since T = D = 4.   For D = 5 We get T = 3&#xA; and A = 4.&lt;br&gt;&#xA; &#xA;  4LL1 +&lt;br&gt;&#xA;  3H43 +&lt;br&gt;&#xA;   EH5 =&lt;br&gt;&#xA; 10000 (base 9)&#xA; &#xA; So L + H + 4 + 1 = 9 or 18 (base 10). L + H = 4 or 13. &#xA; &#xA; So L and H can be 6 and 7.   L = 6 results in&#xA; &#xA;  4661 +&lt;br&gt;&#xA;  3743 +&lt;br&gt;&#xA;   EN5 =&lt;br&gt;&#xA; 10000 (base 9).&lt;br&gt;&#xA; &#xA; This results in N = 7 which cannot be since H = 7.   same goes for L =&#xA; 7 and H = 6.&lt;br&gt;&#xA; &#xA; For D = 6 or more we get similar results following the same path.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19989" LastEditorUserId="19989" LastEditDate="2016-03-08T16:27:38.073" LastActivityDate="2016-03-08T16:27:38.073" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28600" PostTypeId="2" ParentId="28584" CreationDate="2016-03-08T16:22:00.207" Score="0" Body="&lt;p&gt;I would say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No this isn't possible. You should only have 26 different options of distance between pawns and the total number of relationships to take into account is 5040. Still thinking on it, but if I'm right feel free to tell me. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is unless:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The board itself isn't square ;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19731" LastEditorUserId="19731" LastEditDate="2016-03-08T16:29:30.800" LastActivityDate="2016-03-08T16:29:30.800" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28601" PostTypeId="2" ParentId="28595" CreationDate="2016-03-08T16:26:11.267" Score="2" Body="&lt;p&gt;&lt;strong&gt;Answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Howard_Hughes&quot; rel=&quot;nofollow&quot;&gt;Howard Hughes&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He was a famous, wealthy recluse.  As an aeronautics tycoon, billionaire, and pilot; he could fly anywhere in the world that he wanted; but due to mental issues, he stayed locked up in his home.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Here's a blow-by-blow if you need it:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was the freest man around.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;As a rich aviator, he could go anywhere he wanted...  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet I never left my grounds&#xA;(for years at a stretch)  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; but became notoriously reclusive in his later years.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I could walk, fight, or run.&lt;br&gt;&#xA;I'm as able-bodied as anyone!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;He was not physically disabled, but had psychological (and other) issues that caused him to isolate himself.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And my riches I can't count!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He was one of the richest men of his time.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I rarely ventured out.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;He was notoriously reclusive.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19128" LastEditorUserId="19128" LastEditDate="2016-03-08T18:35:30.490" LastActivityDate="2016-03-08T18:35:30.490" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28602" PostTypeId="2" ParentId="28595" CreationDate="2016-03-08T16:27:36.610" Score="3" Body="&lt;p&gt;It can be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Characters of GTA series&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I was the freest man around.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yup, free enough to kill people.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet I never left my grounds&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Can't leave the city&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I could walk, fight, or run.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A lot.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm as able-bodied as anyone!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yeah, in every single game.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And my riches I can't count!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Cheat codes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I rarely ventured out.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Can't go out of the city.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm stable mentally and emotionally&lt;br&gt;&#xA;And many dream of being me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Depends on the person who is playing the game&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-03-08T21:35:36.370" LastActivityDate="2016-03-08T21:35:36.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28604" PostTypeId="1" AcceptedAnswerId="28607" CreationDate="2016-03-08T16:55:32.527" Score="0" ViewCount="129" Body="&lt;p&gt;Here is your challenge; Formulate a yes or no question that cannot be answered without lying.... I have written one such question but there may be more..Good luck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember, the answer to the question must be restricted to yes or no.&lt;/p&gt;&#xA;" OwnerUserId="19947" LastActivityDate="2016-03-08T17:18:41.810" Title="No truthful answer?" Tags="|logical-deduction|wordplay|" AnswerCount="1" CommentCount="7" ClosedDate="2016-03-08T17:43:37.267" ContentLicense="CC BY-SA 3.0" />
  <row Id="28605" PostTypeId="2" ParentId="28584" CreationDate="2016-03-08T17:06:36.217" Score="10" Body="&lt;p&gt;Aight, second try. This time was computer generated, so let's see if my code is dummy or what:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pos{x=0, y=0}, Pos{x=6, y=6}, Pos{x=5, y=5}, Pos{x=1, y=2}, Pos{x=0, y=2}, Pos{x=2, y=6}, Pos{x=3, y=0}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and these are the used distances:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2.0, 5.830951894845301, 7.211102550927978, 7.0710678118654755, 5.385164807134504, 6.324555320336759, 3.0, 2.23606797749979, 6.4031242374328485, 2.8284271247461903, 8.48528137423857, 4.0, 1.0, 5.0, 3.605551275463989, 6.708203932499369, 4.123105625617661, 3.1622776601683795, 4.47213595499958, 1.4142135623730951, 6.082762530298219&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the curious, here is the code (minus syntactic sugar for brevity):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;class Pos {&#xA;    final int x;&#xA;    final int y;&#xA;}&#xA;&#xA;class State {&#xA;    final Set&amp;lt;Double&amp;gt; usedDistances;&#xA;    final Set&amp;lt;Pos&amp;gt; usedCells;&#xA;}&#xA;&#xA;void run(String[] args) {&#xA;    State initial = new State(new HashSet&amp;lt;&amp;gt;(), new HashSet&amp;lt;&amp;gt;());&#xA;    Set&amp;lt;State&amp;gt; visited = new HashSet&amp;lt;&amp;gt;();&#xA;    Deque&amp;lt;State&amp;gt; queue = new LinkedList&amp;lt;&amp;gt;();&#xA;    queue.add(initial);&#xA;    while (!queue.isEmpty()) {&#xA;        State state = queue.remove();&#xA;        if (!visited.contains(state)) {&#xA;            visited.add(state);&#xA;            if (isFinal(state)) {&#xA;                System.out.println(state);&#xA;                return;&#xA;            }&#xA;            else {&#xA;                Set&amp;lt;State&amp;gt; next = nextOf(state);&#xA;                queue.addAll(next);&#xA;            }&#xA;        }&#xA;    }&#xA;    System.out.println(&quot;nothing&quot;);&#xA;}&#xA;&#xA;private Set&amp;lt;State&amp;gt; nextOf(State state) {&#xA;    Set&amp;lt;State&amp;gt; res = new HashSet&amp;lt;&amp;gt;();&#xA;    for (int i = 0; i&amp;lt;7; i++) {&#xA;        for (int j = 0; j&amp;lt;7; j++) {&#xA;            Pos pos = new Pos(i, j);&#xA;            if (!state.usedCells.contains(pos)) {&#xA;                Set&amp;lt;Double&amp;gt; nextDistances = new HashSet&amp;lt;&amp;gt;(state.usedDistances);&#xA;                Set&amp;lt;Pos&amp;gt; nextPos = new HashSet&amp;lt;&amp;gt;(state.usedCells);&#xA;                boolean valid = true;&#xA;                for (Pos used: state.usedCells) {&#xA;                    double distance = computeDistance(pos, used);&#xA;                    if (nextDistances.contains(distance)) {&#xA;                        valid = false;&#xA;                        break;&#xA;                    }&#xA;                    else {&#xA;                        nextDistances.add(distance);&#xA;                    }&#xA;                }&#xA;                if (valid) {&#xA;                    nextPos.add(pos);&#xA;                    res.add(new State(nextDistances, nextPos));&#xA;                }&#xA;            }&#xA;        }&#xA;    }&#xA;    return res;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Finally, here is a picture of the solution:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;       -----------------------------&#xA;       | X |   |   | X |   |   |   |&#xA;       -----------------------------&#xA;       |   |   |   |   |   |   |   |&#xA;       -----------------------------&#xA;       | X | X |   |   |   |   |   |&#xA;       -----------------------------&#xA;       |   |   |   |   |   |   |   |&#xA;       -----------------------------&#xA;       |   |   |   |   |   |   |   |&#xA;       -----------------------------&#xA;       |   |   |   |   |   | X |   |&#xA;       -----------------------------&#xA;       |   |   | X |   |   |   | X |&#xA;       -----------------------------&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="14435" LastEditorUserId="8874" LastEditDate="2016-03-11T10:57:19.147" LastActivityDate="2016-03-11T10:57:19.147" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28606" PostTypeId="2" ParentId="28593" CreationDate="2016-03-08T17:10:11.870" Score="7" Body="&lt;p&gt;Going off of the OP's comments, I have determined that the culprit is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; indeed the man with the cane.  I'll call him Caneman&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This suspect first appears on camera at&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1:55, with a cane in his left hand.  This aids him as it appears he has a bad limp in his right leg, a hard and heavy limp.  The cane causes his left shoulder to be elevated.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After seeing this suspect continue in the way we first observed, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; at 2:08, Caneman loses the limp on his way to the table, in accordance with the distraction of the collision.  Once the testy man knocks down the man who is entering (on his phone?), Caneman quickly and lightly steps to the table.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At this point something subtle occurs.  While the testy man tries to help the fallen man up, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; at 2:13 Caneman turns his cane so that the handle is parallel to his body, instead of perpendicular, as it was when he walked with it.  It is held upright, slightly away from his body, and his right arm can be seen moving across his midsection.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then, those with swift eyes would see that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; by 2:18 the cane has switched hands, from left to right.  This obviously makes the right shoulder higher than the left, but likely wouldn't affect the lower body.  However, the hard, sharp limp from the right leg is now a slow, dragging limp in the left.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Further evidence of a fake is seen&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; at 2:32, when Caneman- with all his weight on his freshly lame left leg- spins all the way around on that leg, doing his best Michael Jackson impression.  Maybe that's because he really is a smooth criminal...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If everyone in and out of the office was body-searched for the missing paper, then it becomes obvious that the only place it could be is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; inside the Caneman's cane.  Canes like his are hollow pipes, and we see what appears to be him shoving it inside the handle during the scuffle.  The testy man and the man in the checkered shirt were likely involved, maybe even the man who discovers that the paper is missing.  But the Caneman is the one who swipes it and gets it out.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11452" LastActivityDate="2016-03-08T17:10:11.870" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28607" PostTypeId="2" ParentId="28604" CreationDate="2016-03-08T17:11:41.903" Score="2" Body="&lt;p&gt;A short list of possible answers:   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Will you answer this question &quot;No&quot;?&lt;br&gt;&#xA;&#xA; Is the correct answer to this question &quot;No&quot;? (from &lt;a href=&quot;https://puzzling.stackexchange.com/users/8874/gamow&quot;&gt;Gamow&lt;/a&gt;)&lt;br&gt;&#xA;&#xA; Will you answer this with &quot;No&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2016-03-08T17:18:41.810" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28608" PostTypeId="2" ParentId="28595" CreationDate="2016-03-08T17:18:16.313" Score="9" Body="&lt;p&gt;Is it  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bruce Wayne/Batman  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was the freest man around.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bruce Wayne could go wherever, whenever he wanted with his vast wealth. No one accused him of anything.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet I never left my grounds&lt;br&gt;&#xA;(for years at a stretch)    &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After the incidents of &lt;em&gt;Dark Knight&lt;/em&gt;, he became a recluse.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I could walk, fight, or run.&lt;br&gt;&#xA;I'm as able-bodied as anyone!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Batman was physically fit, and could do just about anything.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And my riches I can't count!   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Vastly wealthy as Bruce Wayne &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I rarely ventured out.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, never left the manor for several years due to shame/guilt/etc.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2016-03-08T17:18:16.313" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28609" PostTypeId="2" ParentId="28580" CreationDate="2016-03-08T17:20:33.660" Score="-2" Body="&lt;p&gt;Another possible answer could be...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(14-6) \times 3 = 24$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19997" LastEditorUserId="4421" LastEditDate="2016-03-08T18:10:52.963" LastActivityDate="2016-03-08T18:10:52.963" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28610" PostTypeId="1" AcceptedAnswerId="28613" CreationDate="2016-03-08T17:26:21.700" Score="5" ViewCount="218" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I could visit the darkness,&lt;br&gt;&#xA;  if the sky would but let me,&lt;br&gt;&#xA;  But I venture as far,&lt;br&gt;&#xA;  as explosions will get me  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I leave markings behind me,&lt;br&gt;&#xA;  when I breathe out just right,&lt;br&gt;&#xA;  but I don't really care,&lt;br&gt;&#xA;  if it's morning or night.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I rely on the touch,&lt;br&gt;&#xA;  of a woman or man&lt;br&gt;&#xA;  To give up my thoughts,&lt;br&gt;&#xA;  in the air, or on land.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If none of the people&lt;br&gt;&#xA;  I need can be found,&lt;br&gt;&#xA;  My record of deeds,&lt;br&gt;&#xA;  may be found on the ground.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who am I, and how do you know?&lt;/p&gt;&#xA;" OwnerUserId="19944" LastActivityDate="2016-03-08T18:15:11.923" Title="The Traveller's Lament" Tags="|riddle|word|rhyme|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28611" PostTypeId="2" ParentId="5117" CreationDate="2016-03-08T17:34:31.017" Score="2" Body="&lt;p&gt;I believe the answer to be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $267.187$ seconds = $4$ minutes $27.187$ seconds&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The strategy to get this time is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Overcharge production of workers every $20$ seconds until $60$ seconds is reached. Build the number of workers up to $11$ and let the $11$ workers mine until $800$ units of gold is reached. At that point, two of the workers should place markers on the ground to request additional bases. There is no benefit from providing mainframe support extension. There is no benefit from requesting airdrop of one base before another.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are several things which are a little unclear but to get to my answer I have made the following assumptions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (i) The initial number of workers is $5$ and the general has $50$ units of gold originally as in the first Optimal Building Strategy Puzzle. &lt;br&gt; (ii) Each worker is able to mine $40$ units of gold per minute (as in the original problem) and production of a new worker costs $50$ units of gold &lt;br&gt; (iii) Initially, producing a new worker takes $17$ seconds but after overcharging, the speed of this production increases by $50\%$ (this equates to dividing the production time for a single worker by $1.5$ after each overcharge). &lt;br&gt; (iv) The electricity of the mining base is at $100\%$ initially and this is true for any subsequent new mining base. &lt;br&gt; (v) &lt;b&gt;Important:&lt;/b&gt; If the overcharge finishes as a worker is being created then the new production rate applies to the completion of this worker, not the old production rate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we first consider the problem without parts 2 and 3 above, it's clear that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The solution to the problem is the same as the original where base creation is $100$ seconds instead of $120$. Hence, without the additional advances in technology, we know that it takes at most $275.5$ seconds for the general to have three active mining bases.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now let us consider the problem where we just take points 1 and 2 into account.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In this case, we assume that each mining base provides support for a maximum of $11$ workers. Since it takes six workers $100$ seconds to accumulate $400$ units of gold, it is inevitable with any reasonable strategy that the general will have enough gold to request the second additional base to be dropped before the first additional base arrives. Hence, we may simplify the problem by assuming that the general builds up his gold supply to $800$ units and requests both additional bases to be dropped simultaneously. Since there is no monetary demand on electricity, the general may as well adopt a strategy to overcharge the base every $20$ seconds, when possible until he has $11$ workers. With this strategy, we obtain the following table: &lt;br&gt;&lt;br&gt; Time &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; &amp;nbsp;&amp;nbsp;&amp;nbsp;| Action  &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; | Workers |  &amp;nbsp;&amp;nbsp;&amp;nbsp;  Gold &amp;nbsp;&amp;nbsp;&amp;nbsp; &amp;nbsp;  | Electricity (%) | Time to 800 | &lt;br&gt;---------------------------------------------------------------------------------------------------------------------&lt;br&gt;  &amp;nbsp;0.000000 &amp;nbsp;| Create Worker&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;5 &amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp; 0.000000&amp;nbsp; | &amp;nbsp;100.000000&amp;nbsp; | &amp;nbsp;240.000000&amp;nbsp; | &lt;br&gt; &amp;nbsp;0.000000 &amp;nbsp;| Begin Overcharge&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;5 &amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp; 0.000000&amp;nbsp; |  &amp;nbsp; 75.000000&amp;nbsp; | &amp;nbsp;240.000000&amp;nbsp; | &lt;br&gt;17.000000 &amp;nbsp;| Worker Created&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;6 &amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp;56.666667&amp;nbsp; | &amp;nbsp; 84.562500&amp;nbsp; | &amp;nbsp;185.833333&amp;nbsp; |&lt;br&gt;17.000000&amp;nbsp; | Create Worker&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;6 &amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp; 6.666667&amp;nbsp; | &amp;nbsp; 84.562500&amp;nbsp; | &amp;nbsp;198.333333&amp;nbsp; |&lt;br&gt;20.000000&amp;nbsp; | Overcharge Done&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;6 &amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp;18.666667&amp;nbsp; |  &amp;nbsp; 86.250000&amp;nbsp; | &amp;nbsp;195.333333&amp;nbsp; |&lt;br&gt;20.000000&amp;nbsp; | Begin Overcharge&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;6 &amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp;18.666667&amp;nbsp; |  &amp;nbsp; 61.250000&amp;nbsp; | &amp;nbsp;195.333333&amp;nbsp; |&lt;br&gt;29.333333&amp;nbsp;| Worker Created&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;7 &amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp;56.000000&amp;nbsp; | &amp;nbsp; 66.500000&amp;nbsp; | &amp;nbsp;159.428571&amp;nbsp; |&lt;br&gt;29.333333&amp;nbsp; | Create Worker&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;7 &amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp; 6.000000&amp;nbsp; | &amp;nbsp; 66.500000&amp;nbsp; | &amp;nbsp;170.142857&amp;nbsp; |&lt;br&gt;40.000000&amp;nbsp; | Overcharge Done&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;7 &amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp;49.777778&amp;nbsp; |  &amp;nbsp; 72.500000&amp;nbsp; | &amp;nbsp;160.761905&amp;nbsp; |&lt;br&gt;40.000000&amp;nbsp; | Begin Overcharge&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;7 &amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp;49.777778&amp;nbsp; |  &amp;nbsp; 47.500000&amp;nbsp; | &amp;nbsp;160.761905&amp;nbsp; |&lt;br&gt;40.444444&amp;nbsp;| Worker Created&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;8 &amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp;51.851851&amp;nbsp; | &amp;nbsp; 47.750000&amp;nbsp; | &amp;nbsp;140.277778&amp;nbsp; |&lt;br&gt;40.444444&amp;nbsp; | Create Worker&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;8 &amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp; 1.851851&amp;nbsp; | &amp;nbsp; 47.750000&amp;nbsp; | &amp;nbsp;149.652778&amp;nbsp; |&lt;br&gt;48.000000&amp;nbsp;| Worker Created&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;9 &amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp;42.148147&amp;nbsp; | &amp;nbsp; 52.000000&amp;nbsp; | &amp;nbsp;126.308642&amp;nbsp; |&lt;br&gt;49.308642&amp;nbsp; | Create Worker&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;9 &amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp; 0.000000&amp;nbsp; | &amp;nbsp; 52.736111&amp;nbsp; | &amp;nbsp;133.333333&amp;nbsp; |&lt;br&gt;56.864198&amp;nbsp;| Worker Created&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;10&amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp;45.333333&amp;nbsp; | &amp;nbsp; 56.986111&amp;nbsp; | &amp;nbsp;113.200000&amp;nbsp; |&lt;br&gt;57.564198&amp;nbsp; | Create Worker&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;10&amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp; 0.000000&amp;nbsp; | &amp;nbsp; 57.379861&amp;nbsp; | &amp;nbsp;120.000000&amp;nbsp; |&lt;br&gt;60.000000&amp;nbsp; | Overcharge Done&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;10&amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp;16.238680&amp;nbsp; |  &amp;nbsp; 58.750000&amp;nbsp; | &amp;nbsp;117.564198&amp;nbsp; |&lt;br&gt;60.000000&amp;nbsp; | Begin Overcharge&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;10&amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp;16.238680&amp;nbsp; |  &amp;nbsp; 33.750000&amp;nbsp; | &amp;nbsp;117.564198&amp;nbsp; |&lt;br&gt;63.413169&amp;nbsp;| Worker Created&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;11&amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp;38.993140&amp;nbsp; | &amp;nbsp; 35.669908&amp;nbsp; | &amp;nbsp;103.773663&amp;nbsp; |&lt;br&gt;167.186832| Base Requested&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;11&amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp;400.000000&amp;nbsp;| &amp;nbsp; 94.042953&amp;nbsp; | &amp;nbsp;  0.000000&amp;nbsp; |&lt;br&gt;167.186832| Base Requested&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;11&amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp; 0.000000&amp;nbsp; | &amp;nbsp; 94.042953&amp;nbsp; | &amp;nbsp;  0.000000&amp;nbsp; |&lt;br&gt;267.186832| Base Completed&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;11&amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp;733.333333&amp;nbsp;| &amp;nbsp;100.000000&amp;nbsp; | &amp;nbsp;  0.000000&amp;nbsp; |&lt;br&gt;267.186832| Base Completed&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;11&amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp;733.333333&amp;nbsp;| &amp;nbsp;100.000000&amp;nbsp; | &amp;nbsp;  0.000000&amp;nbsp; | &lt;br&gt;&lt;br&gt;There are some points here which need explaining. The first relates to the speed at which workers are created. Take for example the second worker that the general creates. At the outset, creation takes $17$ seconds but, $3$ seconds into production, the first overcharge is completed so the remaining completion time for the second worker is cut to $\frac{14}{1.5} = 9.333333$ seconds. The same logic is used throughout. &lt;br&gt;&lt;br&gt; The second point relates to the 'Time to 800' column. This represents the time it would take the general reach the goal of $800$ units of gold with the current stash of gold and number of workers after the 'Action' in the same row is executed. The most important times are when the general should opt to Create Worker or not. From this analysis, it is clear that he should build up the workers to the maximum allowed and let them work to mine $800$ units of gold. This process takes $267.186832$ seconds which is the best that can be achieved under these conditions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The final thing we need to take into account is point 3&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Is it worth providing additional processing overhead to the base at any point? &lt;br&gt;&lt;br&gt;To answer this, first note that it is only worth providing additional support if the general is to have at least $12$ workers at the base. To build a support extension, he must take a break between consecutive worker production at some point along the way. We assume that overcharging proceeds as before. In the following table let 'Support Workers' be the number of workers after which the base saves enough gold to provide the mainframe support extension, let $T_{all}$ be the total time taken to provide $11$ workers and the support extension under these conditions, let 'Gold' be the amount of gold accumulated at time $T_{all}$ and $T_{12}$ the time taken to begin creating the 12th worker. We can look at the following table to determine the smallest value of $T_{12}$ &lt;br&gt; Support Worker | &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; $T_{all}$ &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; | &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; Gold &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; | &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; $T_{12}$ &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; |&lt;br&gt; &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; 5 &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; | 86.525133 | 22.386831 | 90.288565 |&lt;br&gt; &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; 6 &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; | 82.920116 | 23.846529 | 86.486498 |&lt;br&gt; &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; 7 &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; | 81.325867 | 29.160689 | 84.167591 |&lt;br&gt; &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; 8 &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; | 80.549108 | 31.749887 | 83.087760 |&lt;br&gt; &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; 9 &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; | 82.641976 | 46.997531 | 83.051404 |&lt;br&gt; &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;10 &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; | 90.064198 |124.975309 | 90.064198 |&lt;br&gt; &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;11 &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; | 96.732286 |183.333333 | 96.732286 |&lt;br&gt;&lt;br&gt;From the table, it seems as though the best way to proceed is to build up to $9$, $10$ or $11$ workers and then call in for a support extension. The case with $9$ workers is quicker but it is not immediately clear whether the stash of extra gold will be a benefit in the $10$ or $11$ worker case. In the support worker $9$ case, the general will have enough gold to start creating the $12$th worker at $83.051404$ seconds. Going with this case, we can analyse what happens when he tries to create additional workers in the same way as the first table above. For this purpose, let us assume that at each $20$ second mark an overcharge on production has been requested. &lt;br&gt;&lt;br&gt; Time &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; &amp;nbsp;&amp;nbsp;&amp;nbsp;| Action  &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; | Workers |  &amp;nbsp;&amp;nbsp;&amp;nbsp;  Gold &amp;nbsp;&amp;nbsp;&amp;nbsp; &amp;nbsp;  | Electricity (%) | Time to 800 | &lt;br&gt;---------------------------------------------------------------------------------------------------------------------&lt;br&gt;  &amp;nbsp;83.051414&amp;nbsp;| Create Worker&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;11&amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp; 0.000000&amp;nbsp; | &amp;nbsp; 21.716420&amp;nbsp; | &amp;nbsp;109.090909&amp;nbsp; |&lt;br&gt;86.409439&amp;nbsp;| Worker Created&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;12&amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp;24.625514&amp;nbsp; | &amp;nbsp; 23.605309&amp;nbsp; | &amp;nbsp;96.921811&amp;nbsp; |&lt;br&gt;  &amp;nbsp;89.58125&amp;nbsp;| Create Worker&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;12&amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp; 0.000000&amp;nbsp; | &amp;nbsp; 25.389453&amp;nbsp; | &amp;nbsp;100.000000&amp;nbsp; |&lt;br&gt;92.939275&amp;nbsp;| Worker Created&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;13&amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp;26.864198&amp;nbsp; | &amp;nbsp; 27.278342&amp;nbsp; | &amp;nbsp;89.207977&amp;nbsp; |&lt;br&gt;  &amp;nbsp;95.608791&amp;nbsp;| Create Worker&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;13&amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp; 0.000000&amp;nbsp; | &amp;nbsp; 28.779945&amp;nbsp; | &amp;nbsp; 92.307692&amp;nbsp; |&lt;br&gt;98.966816&amp;nbsp;| Worker Created&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;14&amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp;29.102881&amp;nbsp; | &amp;nbsp; 30.668834&amp;nbsp; | &amp;nbsp;82.59612&amp;nbsp; |&lt;br&gt;100.000000&amp;nbsp;| Overcharge Done&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;14&amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp;38.745932&amp;nbsp; |  &amp;nbsp; 31.250000&amp;nbsp; | &amp;nbsp; 81.562936&amp;nbsp; |&lt;br&gt;100.000000&amp;nbsp;| Begin Overcharge&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;14&amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp;38.745932&amp;nbsp; |  &amp;nbsp;  6.250000&amp;nbsp; | &amp;nbsp; 81.562936&amp;nbsp; |&lt;br&gt;&amp;nbsp;101.205793&amp;nbsp;| Create Worker&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;14&amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp; 0.000000&amp;nbsp; | &amp;nbsp;  6.928259&amp;nbsp; | &amp;nbsp; 85.714286&amp;nbsp; |&lt;br&gt;103.444476&amp;nbsp;| Worker Created&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;15&amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp;20.894375&amp;nbsp; | &amp;nbsp;  8.187518&amp;nbsp; | &amp;nbsp;77.910563&amp;nbsp; |&lt;br&gt;&amp;nbsp;106.355039&amp;nbsp;| Create Worker&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;15&amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp; 0.000000&amp;nbsp; | &amp;nbsp;  9.824709&amp;nbsp; | &amp;nbsp; 80.000000&amp;nbsp; |&lt;br&gt;108.593722&amp;nbsp;| Worker Created&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;16&amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp;22.386831&amp;nbsp; | &amp;nbsp; 11.083968&amp;nbsp; | &amp;nbsp;72.901235&amp;nbsp; |&lt;br&gt;&amp;nbsp;111.182457&amp;nbsp;| Create Worker&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;16&amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp; 0.000000&amp;nbsp; | &amp;nbsp; 12.540131&amp;nbsp; | &amp;nbsp; 75.000000&amp;nbsp; |&lt;br&gt;113.421140&amp;nbsp;| Worker Created&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;17&amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp;23.879287&amp;nbsp; | &amp;nbsp; 13.79939&amp;nbsp; | &amp;nbsp;68.481239&amp;nbsp; |&lt;br&gt;&amp;nbsp;115.725909&amp;nbsp;| Create Worker&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;17&amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp; 0.000000&amp;nbsp; | &amp;nbsp; 15.095822&amp;nbsp; | &amp;nbsp; 70.588235&amp;nbsp; |&lt;br&gt;117.964592&amp;nbsp;| Worker Created&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;18&amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp;25.371742&amp;nbsp; | &amp;nbsp; 16.392255&amp;nbsp; | &amp;nbsp;64.552355&amp;nbsp; |&lt;br&gt;120.000000&amp;nbsp;| Overcharge Done&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;18&amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp;49.796638&amp;nbsp; |  &amp;nbsp; 17.500000&amp;nbsp; | &amp;nbsp; 62.516947&amp;nbsp; |&lt;br&gt;&amp;nbsp;120.016947&amp;nbsp;| Create Worker&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;18&amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp; 0.000000&amp;nbsp; | &amp;nbsp; 17.509533&amp;nbsp; | &amp;nbsp; 62.500000&amp;nbsp; |&lt;br&gt;121.509402&amp;nbsp;| Worker Created&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;|  &amp;nbsp; &amp;nbsp; &amp;nbsp;19&amp;nbsp; &amp;nbsp; &amp;nbsp;&amp;nbsp; | &amp;nbsp;17.909465&amp;nbsp; | &amp;nbsp; 18.349039&amp;nbsp; | &amp;nbsp;61.74399&amp;nbsp; |&lt;br&gt;&lt;br&gt;At this point we need to pause as the general has reached the quota that our support extension will allow. However, we notice that this is as far as we need to go because the mininum time to achieve a gold supply of $800$ units in this phase is attained after the creation of the $15$th worker where, if the general keeps the number of workers fixed, the bases will begin building at $181.355039 $seconds. After that point, even if the next support extension is given for free he will not be able to build the base quicker with extra workers. This may seem counterintuitive since the speed at which the base may produce workers increases. However, after a few overcharges, the bottleneck is not the rate at which new workers are produced but the time it takes to reach $50$ units of gold. If we let $W$ be the number of workers then the point at which it becomes impractical to create additional workers is when the time taken to mine 50 units of gold ($\frac{75}{W}$) is greater than the time saved getting to 800 units with the additional worker ($\frac{1200}{W(W+1)}$) i.e, when $W &amp;gt; 15$. &lt;br&gt; We can also run the same analysis for the cases when the support extension is built after the $10$th or $11$th worker. The advantage here is that for the next few worker creations the general doesn't need to wait to build up to $50$ units of gold. I won't write the full analysis here but, as before the optimal solution is to build to $15$ workers and then mine until $800$ units of gold. In the case that the support worker is $10$ our quickest time to get $800$ units is $179.724582$ seconds. In the case that the support worker is $11$ our quickest time to get $800$ units is $180.032626$ seconds. Both are better than the case above, but not by enough to beat the time given when just taking points 1 and 2 into account.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;P.S.&lt;/b&gt; Sorry about the wobbly tables. I don't really know how to format them.&lt;/p&gt;&#xA;" OwnerUserId="18422" LastEditorUserId="18422" LastEditDate="2016-03-10T11:20:35.273" LastActivityDate="2016-03-10T11:20:35.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28612" PostTypeId="2" ParentId="28545" CreationDate="2016-03-08T17:38:13.797" Score="10" Body="&lt;p&gt;You have to be talking about &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; My daughter &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Complex&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Way too complex for me to understand &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alternative&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Music, hairstyle: absolutely &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Exempt&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Here you must be referring to her attitude to household chores. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Deductive&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes. It's like there's a hole in my wallet. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Dependent&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Come income tax time, that's what goes on the form&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fuzzy&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Particularly under her bed. Don't get me started. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Critical&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You really do know her, don't you?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And most of all, Certain.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Certain about every topic on earth. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What do I win?&lt;/p&gt;&#xA;" OwnerUserId="18924" LastActivityDate="2016-03-08T17:38:13.797" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28613" PostTypeId="2" ParentId="28610" CreationDate="2016-03-08T18:09:03.407" Score="5" Body="&lt;p&gt;Is it ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ... an Airplane?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I could visit the darkness,&#xA;if the sky would but let me,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An Airplane could go to space, if air density allowed it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I venture as far,&#xA;as explosions will get me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Turbine fire could be seen as a set of continuous explosions&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I leave markings behind me,&#xA;when I breathe out just right,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Condensate stripes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I don't really care,&#xA;if it's morning or night.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Can fly any time of day&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I rely on the touch,&#xA;of a woman or man&#xA;To give up my thoughts,&#xA;in the air, or on land.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Needs a pilot to steer and read and interpret instruments&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If none of the people&#xA;I need can be found,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oh my god they're all dead!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My record of deeds,&#xA;may be found on the ground.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Black Box after crash&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19998" LastEditorUserId="19998" LastEditDate="2016-03-08T18:15:11.923" LastActivityDate="2016-03-08T18:15:11.923" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28614" PostTypeId="2" ParentId="28406" CreationDate="2016-03-08T18:45:29.147" Score="2" Body="&lt;p&gt;Is it ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ... Jupiter?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am dead and lifeless&#xA;I am a monster, cold&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well nothing lives on Jupiter, and it is -108°C&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Look at me a closer look&#xA;Gasp and behold!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Feel the menace of the EYE OF JUPITER!!!!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Older than your pyramids&#xA;And far more than your Rome&#xA;I am older than it all&#xA;As old as your home&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (very) roughly as old as all planets in the solar system, like Earth, our home.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If it were not me, you know&#xA;They'd stone you soon to death&#xA;Giant boulders would crush you&#xA;And take your last breath&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Vacuums up asteroids before they can hit any of the inner planets&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I came alone for you&#xA;And you come, you four&#xA;I'd be left standing and&#xA;You all would be no more&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A lot heavier than all the four inner planets (and maybe the outer planets, dunno) combined. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am father of them all&#xA;Who wander in the sky&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Romans shamelessly stole and renamed the population of the Greek pantheon, and thereby their names for the planets (Greek for wanderers). Zeus the father of gods, became Jupiter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And whoever wanders close&#xA;Must invariably die&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ... or wait until Elon Musk sells Eye-Diving Tickets to billionaires and opens a wellness spa on the ground below...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19998" LastEditorUserId="19998" LastEditDate="2016-03-08T18:57:51.353" LastActivityDate="2016-03-08T18:57:51.353" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28615" PostTypeId="2" ParentId="28580" CreationDate="2016-03-08T19:08:19.337" Score="1" Body="&lt;p&gt;A possible interpretation of the rules:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $1+3+4+6=24_{5}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="20001" LastActivityDate="2016-03-08T19:08:19.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28616" PostTypeId="1" AcceptedAnswerId="29578" CreationDate="2016-03-08T19:13:09.963" Score="23" ViewCount="1156" Body="&lt;h1&gt;What is the next song in the list?&lt;/h1&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Pedal To The Medal&lt;/li&gt;&#xA;&lt;li&gt;Not Inside -- Outside&lt;/li&gt;&#xA;&lt;li&gt;The Final Countdown&lt;/li&gt;&#xA;&lt;li&gt;Every Rose Has Its Thorn&lt;/li&gt;&#xA;&lt;li&gt;.........&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;The Puzzle&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;Within these dots, you shall find a magical, mystical, hidden message.  A (wo)man more focused on the music than these dots would have a jump start.  To determine what they mean, you must pay attention to whats already been played, for those songs hold the key.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IDzmO.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IDzmO.png&quot; alt=&quot;???decimal???&quot;&gt;&lt;/a&gt;&#xA;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of these things are not like the others, &lt;br&gt;&#xA;watch what you do, or you might have your druthers. &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;A Final Clue&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;too lag cof fee&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;Hints?&lt;/h1&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  - The first four songs reveal what type of puzzle this is.&lt;br&gt;&#xA;  - Once you know what type of puzzle it is, the hyphenated endings will begin to make sense as there is a unique attribute to this puzzle that may not be immediately obvious.&lt;br&gt;&#xA;  - A small hint may be steganographically hidden for after you have halfway decoded the puzzle.&lt;br&gt;&#xA;  - Pay attention to the features of what comes before the hyphens ;)&lt;br&gt;&#xA;  - Yahtzee!!!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19731" LastEditorUserId="19731" LastEditDate="2016-03-22T17:18:04.600" LastActivityDate="2016-03-25T12:39:28.060" Title="Musical in Nature" Tags="|mathematics|lateral-thinking|wordplay|enigmatic-puzzle|music|" AnswerCount="3" CommentCount="19" ContentLicense="CC BY-SA 3.0" />
  <row Id="28617" PostTypeId="1" AcceptedAnswerId="28618" CreationDate="2016-03-08T19:19:10.310" Score="5" ViewCount="272" Body="&lt;p&gt;I have split the alphabet into two sets using a rule:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;code&gt;A E F H I L M N O R S X&lt;/code&gt;&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;code&gt;B C D G J K P Q T U V W Y Z&lt;/code&gt;&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What is the rule?&lt;/p&gt;&#xA;" OwnerUserId="18567" LastActivityDate="2016-03-08T19:28:14.037" Title="What rule did I use to split the alphabet" Tags="|pattern|" AnswerCount="1" CommentCount="5" ClosedDate="2016-03-08T21:27:45.627" ContentLicense="CC BY-SA 3.0" />
  <row Id="28618" PostTypeId="2" ParentId="28617" CreationDate="2016-03-08T19:26:32.930" Score="7" Body="&lt;p&gt;The rule is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The top set when spelled would start with a vowel and the bottom set would start with a consonant. For example x would be spelled ex(vowel) and z would be spelled zee(consonant)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10856" LastEditorUserId="17482" LastEditDate="2016-03-08T19:28:14.037" LastActivityDate="2016-03-08T19:28:14.037" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28619" PostTypeId="2" ParentId="28595" CreationDate="2016-03-08T19:47:40.830" Score="2" Body="&lt;p&gt;Unlikely, but worth a shot:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Brian Wilson of the Beach Boys&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was the freest man around.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Free in his mind anyhow, also free from preconceived notions of what music should be, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet I never left my grounds&lt;br&gt;&#xA;(for years at a stretch)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He famously spent 3 straight years in bed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I could walk, fight, or run.&lt;br&gt;&#xA;I'm as able-bodied as anyone!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He never had physical problems, only mental and emotional&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And my riches I can't count!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He's made millions from album sales&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I rarely ventured out.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He was (and still is to a large degree) very reclusive&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16131" LastEditorUserId="16131" LastEditDate="2016-03-13T16:28:09.577" LastActivityDate="2016-03-13T16:28:09.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28620" PostTypeId="2" ParentId="28545" CreationDate="2016-03-08T20:07:00.080" Score="11" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Logic&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Complex&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Complex Logic: mental processes that facilitate the understanding of complicated problems using logic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alternative&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alternative Logic: the name given to formal systems that differ in a significant way from standard logical systems such as propositional and predicate logic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Exempt&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Exempt Logic: 'A priori' knowledge or justification as independent of experience.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Deductive&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Deductive Logic: the process of reasoning from one or more statements to reach a logically certain conclusion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Dependent&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dependent Logic: a logical formalism which adds dependence atoms to the language of first-order logic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fuzzy&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fuzzy Logic: an approach based on degrees of truth rather than true or false.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Critical&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Critical Logic: clear, rational thinking involving critique.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Certain&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Certain Logic: there's a certain logic to this..&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19834" LastActivityDate="2016-03-08T20:07:00.080" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28622" PostTypeId="1" AcceptedAnswerId="28624" CreationDate="2016-03-08T21:40:31.767" Score="14" ViewCount="11207" Body="&lt;p&gt;I don't know if anybody has played the game Tribond, but here's how it works: three words are given, and the player has to figure out what they have in common. Example: &quot;What do Earth, Pie, and Bread all have in common?&quot;  Answer: &quot;They all have crusts&quot;. Answers ARE permitted to rely on homophones, an example (not a very good one), being &quot;What do Apple, Cherry, and 3.14159...&quot; have in common? Answer: &quot;They're all pie/pi&quot; (Apple Pie, Cherry Pie, and pi)! So, here are two Tribond puzzles I've come up with (hopefully they take more than five seconds to solve):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do these three have in common:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Discussions, Sewing machines, screws?  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Printers, Airplanes, New York?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17712" LastEditorUserId="15942" LastEditDate="2018-12-22T14:40:25.233" LastActivityDate="2018-12-22T14:40:25.233" Title="Tribond: what do these three have in common?" Tags="|wordplay|tribond|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28624" PostTypeId="2" ParentId="28622" CreationDate="2016-03-08T21:43:53.380" Score="19" Body="&lt;p&gt;This would be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All have threads&lt;br&gt;&#xA; Screws are threaded, discussions in forums have threads and sewing machines use thread to sew. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All have jets.&lt;br&gt;&#xA; Printers have ink jets, airplanes have jet engines and The New York Jets are a sports franchise. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastEditorUserId="18924" LastEditDate="2016-03-08T21:48:39.180" LastActivityDate="2016-03-08T21:48:39.180" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28625" PostTypeId="5" CreationDate="2016-03-08T21:46:53.927" Score="0" Body="&lt;p&gt;For puzzles of the form &amp;quot;What do these three words have in common?&amp;quot; Three words are provided.&lt;/p&gt;&#xA;&lt;p&gt;For example, if the three words are:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Helium, Neon, Radon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;...then the connection is &lt;strong&gt;noble gases&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11497" LastEditorUserId="60644" LastEditDate="2020-08-30T12:06:13.157" LastActivityDate="2020-08-30T12:06:13.157" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28626" PostTypeId="4" CreationDate="2016-03-08T21:46:53.927" Score="0" Body="For puzzles of the form &quot;What do these three words have in common?&quot;" OwnerUserId="17712" LastEditorUserId="60644" LastEditDate="2020-08-30T12:06:13.157" LastActivityDate="2020-08-30T12:06:13.157" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28627" PostTypeId="1" AcceptedAnswerId="28628" CreationDate="2016-03-08T21:50:58.373" Score="5" ViewCount="146" Body="&lt;p&gt;I have two forms one bigger than the other.&lt;br&gt;&#xA;One is always hungry the other is always stuffed.&lt;br&gt;&#xA;One is very cuddly. The other is strong and rough. Especially if it is female.&lt;br&gt;&#xA;What am I? &lt;/p&gt;&#xA;" OwnerUserId="20014" LastEditorUserId="22922" LastEditDate="2016-09-01T10:34:06.883" LastActivityDate="2016-09-01T10:34:06.883" Title="The Same But Different Animal" Tags="|riddle|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28628" PostTypeId="2" ParentId="28627" CreationDate="2016-03-08T21:54:08.550" Score="8" Body="&lt;p&gt;You are a  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Bear (Teddy Bear and Real Bear)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have two forms one bigger than the other.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A stuffed bear (teddy bear) and a real bear.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One is always hungry the other is always stuffed.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Bears are constantly eating, while teddy bears are always stuffed (with fluff).  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One is very cuddly. The other is strong and rough. Especially if it is female.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Snuggle with your teddy, but don't fight a real bear, especially &quot;mamma bear&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2016-03-08T21:54:08.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28630" PostTypeId="1" AcceptedAnswerId="28631" CreationDate="2016-03-08T22:08:42.063" Score="-2" ViewCount="340" Body="&lt;p&gt;I am a game character.&lt;br&gt;&#xA;I like money.&lt;br&gt;&#xA;I am on lots of cards.&lt;br&gt;&#xA;In the game I am in, you pick either a ship, dog, wheelbarrow, hat, car, or thimble.&lt;br&gt;&#xA;Who am I?&lt;/p&gt;&#xA;" OwnerUserId="20014" LastEditorUserId="7586" LastEditDate="2016-03-08T22:59:10.307" LastActivityDate="2016-03-08T22:59:10.307" Title="Let's Play a Game" Tags="|riddle|" AnswerCount="1" CommentCount="1" ClosedDate="2016-03-09T01:00:19.827" ContentLicense="CC BY-SA 3.0" />
  <row Id="28631" PostTypeId="2" ParentId="28630" CreationDate="2016-03-08T22:10:00.063" Score="4" Body="&lt;p&gt;You are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Uncle Pennybags&lt;/strong&gt; from Monopoly. He is also known as &lt;strong&gt;Mr. Monopoly&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-03-08T22:14:35.697" LastActivityDate="2016-03-08T22:14:35.697" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="28632" PostTypeId="1" AcceptedAnswerId="28633" CreationDate="2016-03-08T22:25:59.937" Score="-5" ViewCount="344" Body="&lt;p&gt;I have a brother who is taller than me. I have been a good guy and a bad guy. I was once known as Jumpman. I was in lots of video games.&lt;/p&gt;&#xA;" OwnerUserId="20014" LastEditorUserId="5373" LastEditDate="2016-08-31T14:09:41.267" LastActivityDate="2016-08-31T14:09:41.267" Title="Mr. Who Am I? Please tell me" Tags="|riddle|video-games|" AnswerCount="1" CommentCount="7" ClosedDate="2016-03-09T01:00:07.887" ContentLicense="CC BY-SA 3.0" />
  <row Id="28633" PostTypeId="2" ParentId="28632" CreationDate="2016-03-08T22:30:40.697" Score="3" Body="&lt;p&gt;You are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mario&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And your brother is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Luigi&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastActivityDate="2016-03-08T22:30:40.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28635" PostTypeId="2" ParentId="15097" CreationDate="2016-03-08T22:33:42.820" Score="0" Body="&lt;p&gt;Alternate Universe Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If we ignore the 5th 'instinctive knowledge' statement about Duke Froggington  (taking him at his word, rather than reading his intentions) we get a happier solution.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Setup:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Duke Froggington always lies and is a perfect logician.  He knows that if he declares himself the guest of honor, he will partake of the strawberry pastry.  (By mmKings correct logic, Duke Froggington will not be dethroned if he declares himself the Guest of Honor right away.)  So when asked &quot;Do you wish to declare yourself the guest of honor?&quot; Duke Froggington &lt;strong&gt;must&lt;/strong&gt; say &quot;no,&quot; because the answer is &quot;yes&quot; and he always lies.  The phrasing of the question means that he must always work against his own interests.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So here's what happens:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Duke Froggington responds &quot;no.&quot;  The other animals all know that Duke Froggington will dethrone the next guest of honor if and only if it results in he himself being dethroned at some future point.  If one of the other animals declares itself the Guest of Honor, Duke Froggington will replace it, secure in his knowledge that the next animal will bump him off in turn (due to the odd/even math presented in mmKing's solution).  Therefore &lt;strong&gt;no one&lt;/strong&gt; will declare themselves the Guest of Honor first.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Everyone enjoys the tea party, and Mr. Bun remains the guest of honor, thanks to Duke Froggington's perverse nature keeping the other animals at bay. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19128" LastEditorUserId="19128" LastEditDate="2016-03-08T23:20:21.587" LastActivityDate="2016-03-08T23:20:21.587" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28636" PostTypeId="2" ParentId="28125" CreationDate="2016-03-08T22:35:53.207" Score="-1" Body="&lt;p&gt;It is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A Time MAchine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Because a time machine could cause a new disease or cancer. If it could teleport anywhere and to anytime it could change things or cause wars and the government wouldn't want that. Black Holes are pretty much what make time machines teleport.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="20014" LastEditorUserId="19179" LastEditDate="2016-03-09T03:52:08.283" LastActivityDate="2016-03-09T03:52:08.283" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28637" PostTypeId="2" ParentId="28556" CreationDate="2016-03-08T23:18:45.370" Score="8" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'll start with a star-shaped cake which is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Regular_polygon&quot; rel=&quot;noreferrer&quot;&gt;regular polygon&lt;/a&gt;. After 3 cuts the 6 parts can be arranged into 2 hexagons.&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/dhmm2.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dhmm2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; In the second step again 3 cuts are performed and the parts are rearranged to 3 triangles. If aligning of the parts before cutting is allowed the 3 cuts can be performed as a single cut as well.&lt;br&gt;&#xA;&lt;br&gt;&#xA;  &lt;a href=&quot;https://i.stack.imgur.com/2uGPS.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2uGPS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastActivityDate="2016-03-08T23:18:45.370" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28638" PostTypeId="2" ParentId="28125" CreationDate="2016-03-08T23:25:16.277" Score="3" Body="&lt;p&gt;Is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Greed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm all over. Over you. Under you. Through you. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Greed is everywhere&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I love black holes and dying cosmos. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Black holes suck up everything for themselves and dying cosmos can make black holes. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come from evil politicians and ruthless dictators. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Evil politicians and dictators (greedy beyond belief)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come from silly children and irresponsible criminals. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Children always want more &quot;But I want it!!!&quot;, and criminals steal&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can cause horrible disease. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Greed can cause famine by uneven distribution of wealth. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The government over regulates me. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They break up monopolies, but they are also greedy. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nothing can stop me. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Greed has been around a long time and very likely isn't going anywhere any time soon. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I OWN THE WORLD!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Greedy much?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not that it does me much good. I think of my existence as a giant treadmill. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Treadmill --&gt; people who are greedy always want more no matter how much they have. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19731" LastEditorUserId="19731" LastEditDate="2016-03-09T17:17:23.807" LastActivityDate="2016-03-09T17:17:23.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28640" PostTypeId="1" AcceptedAnswerId="28643" CreationDate="2016-03-09T00:08:06.690" Score="3" ViewCount="173" Body="&lt;p&gt;Two men sat at a square table, each with a very large number of stakes. They had entered a contest of wits, or so they called it. The winner of the contest would be the last person to place a stake onto the table without disturbing another.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;They didn't seem to mind that the stakes weren't all identical, but they reasoned that all of them were close to being four inches long, with all sides flat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second man sat smug in his chair, a clever strategy worked out in his head. However, when the first man had made his move, the second man's smile vanished.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;What is the best strategy to win this game?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;What did the first man do to disrupt the second's strategy?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="17744" LastEditorUserId="17744" LastEditDate="2016-03-09T00:39:02.310" LastActivityDate="2016-03-09T00:39:02.310" Title="Stakes on a table" Tags="|strategy|game|board-games|" AnswerCount="2" CommentCount="3" ClosedDate="2016-03-09T14:58:00.973" ContentLicense="CC BY-SA 3.0" />
  <row Id="28642" PostTypeId="2" ParentId="28640" CreationDate="2016-03-09T00:31:46.730" Score="2" Body="&lt;p&gt;It seems obvious:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first man placed his stake balanced precariously on one edge (or on the point of the stake), so that pretty much any touching of the table or placing of another stake would cause it to fall.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &quot;best strategy&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Go first and place a stake balanced very carefully so that your opponent is most likely to upset it no matter what they do.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2016-03-09T00:31:46.730" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28643" PostTypeId="2" ParentId="28640" CreationDate="2016-03-09T00:32:17.133" Score="7" Body="&lt;p&gt;The first man:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Placed a stake (maybe a very large one - but doesn't matter) in the very center of the table.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why the second man's smile vanished:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because they both had a &quot;very large number of stakes&quot; so no matter where the second man puts whatever stake - the first man could simply place the &lt;em&gt;same&lt;/em&gt; stake on the diagonal of the middle to that second (or 4th or 6th) &lt;em&gt;mirroring&lt;/em&gt; the second - he &lt;em&gt;always&lt;/em&gt; has a safe place to put a stake.  Until the second falters - then the first man wins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastActivityDate="2016-03-09T00:32:17.133" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28644" PostTypeId="1" AcceptedAnswerId="28647" CreationDate="2016-03-09T03:37:24.213" Score="7" ViewCount="263" Body="&lt;p&gt;In each pair of clues, the first clue will give you a word. The second clue will give you a different word comprising the first word, with one letter added, anywhere in the word. For example, clue one gives you &quot;bag&quot; and clue two gives you &quot;b&lt;strong&gt;R&lt;/strong&gt;ag&quot;. At the end, take all the letters you have added to the second clues. They will give you the answer to &quot;overall clue 1&quot;. Then solve &quot;overall clue 2&quot; by adding a letter to the answer to overall clue one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 Pound&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 Baby animal&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 Tunnel&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 Desire&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 Wok&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 Hurt&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 Ready&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 Complain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall 1 Ship&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall 2 Fetch&lt;/p&gt;&#xA;" OwnerUserId="17712" LastEditorUserId="17712" LastEditDate="2016-03-10T00:39:39.430" LastActivityDate="2016-03-10T02:32:13.880" Title="Add Letters to the Solutions" Tags="|wordplay|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28647" PostTypeId="2" ParentId="28644" CreationDate="2016-03-09T06:24:09.133" Score="4" Body="&lt;p&gt;1 Pound&lt;br&gt;2 Baby animal&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; lam&lt;br&gt;lam&lt;strong&gt;b&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1 Tunnel&lt;br&gt;2 Desire&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; cave&lt;br&gt;c&lt;strong&gt;r&lt;/strong&gt;ave&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1 Wok&lt;br&gt;2 Hurt&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; pan&lt;br&gt;pa&lt;strong&gt;i&lt;/strong&gt;n&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1 Ready&lt;br&gt;2 Complain&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ripe&lt;br&gt;&lt;strong&gt;g&lt;/strong&gt;ripe&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Overall:&lt;br&gt;&#xA;1 Ship&lt;br&gt;2 Fetch&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; brig&lt;br&gt;&#xA; bri&lt;strong&gt;n&lt;/strong&gt;g&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19843" LastEditorUserId="19843" LastEditDate="2016-03-10T02:32:13.880" LastActivityDate="2016-03-10T02:32:13.880" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28648" PostTypeId="2" ParentId="28360" CreationDate="2016-03-09T06:53:18.970" Score="2" Body="&lt;p&gt;You are ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a &lt;strong&gt;lightning&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll start to count while you solve this problem,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People start to count when they see a lightning, to find out how far away it is (waiting for the thunder).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and just relax here at the top of my column.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This might refer to a &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Tesla_coil&quot; rel=&quot;nofollow&quot;&gt;Tesla coil&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Everywhere I go I let out a good whistle,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lightnings cause &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Whistler_(radio)&quot; rel=&quot;nofollow&quot;&gt;Whistlers&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but don't mess with me cause I carry a pistol.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is for example the &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Colt_M1877&quot; rel=&quot;nofollow&quot;&gt;Colt M1877 &quot;Lightning&quot;&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I came with a flash and out like a light,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;A flash of lightning.&quot; (from dictionary.com)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I don't really mind because life is so bright.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A lightning is definitely bright.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've been explosive at one point in history,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lightnings can &lt;a href=&quot;https://youtu.be/9higGDEwLIg&quot; rel=&quot;nofollow&quot;&gt;cause explosions&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;what I truly am is not much of a mystery.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; See above. Additional hints are the titles, which as I assume are mixed up. &quot;Large and in charge&quot; would fit a lightning, &quot;Small and resilient&quot; a button, &quot;Immovable object&quot; a mountain and maybe &quot;Breathing new life&quot; for an ocarina. This would leave &quot;Unstoppable force&quot; for the final riddle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15632" LastActivityDate="2016-03-09T06:53:18.970" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28649" PostTypeId="2" ParentId="28125" CreationDate="2016-03-09T07:54:58.390" Score="2" Body="&lt;p&gt;You are an&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Idea&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm all over. Over you. Under you. Through you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ideas are everywhere&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I love black holes and dying cosmos. I come from evil politicians and ruthless dictators. I come from silly children and irresponsible criminals.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A lot of thought, Ideas have been out there about black holes, cosmoses, what happens inside a black hole and they fascinate mankind very much. Evil politicians, dictators, silly children and criminal have only one thing in common: an idea (although the type of idea may differ).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can cause horrible diseases. The government over regulates me. But that doesn't stop me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It only takes an idea to start a cigarette or mix something in the water which can led to horrible diseases. You try to implement a really crazy idea but there are lot of rules and regulations that will not stop you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nothing can stop me. Not walls. Not floors. Not Time or Space or Distance.&#xA;I OWN THE WORLD!!!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An idea is unstoppable. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not that it does me much good. Sometimes I think of my existence as a giant treadmill.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You &lt;strong&gt;think&lt;/strong&gt; of your existence as a giant treadmill. You can think anything. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19191" LastActivityDate="2016-03-09T07:54:58.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28650" PostTypeId="1" AcceptedAnswerId="28651" CreationDate="2016-03-09T08:19:33.400" Score="9" ViewCount="1248" Body="&lt;p&gt;Find that name which hides within this image.&#xA;Hint&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is a city. And somewhere on the lines of &lt;a href=&quot;https://puzzling.stackexchange.com/questions/28577/anagram-in-a-rebus&quot;&gt;my last picture puzzle&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RV1TS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RV1TS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good Luck!&lt;/p&gt;&#xA;" OwnerUserId="19656" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2016-03-09T13:57:51.850" Title="Mathematics is Prohibited" Tags="|visual|rebus|anagram|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28651" PostTypeId="2" ParentId="28650" CreationDate="2016-03-09T08:27:25.940" Score="10" Body="&lt;p&gt;The word is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BANGALORE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As before, there are two keywords&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;ALGEBRA&lt;/strong&gt; and &lt;strong&gt;NO&lt;/strong&gt;, which contain the letters in &lt;strong&gt;BANGALORE&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17147" LastActivityDate="2016-03-09T08:27:25.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28652" PostTypeId="1" AcceptedAnswerId="28653" CreationDate="2016-03-09T09:30:01.023" Score="12" ViewCount="1184" Body="&lt;p&gt;Every evening, Peter meets with some of his many friends in the pub.&#xA;One day he observes that over the last $n$ days:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;every evening he has met with exactly three of his friends;&lt;/li&gt;&#xA;&lt;li&gt;on no two of these evenings he has met with exactly the same three friends;&lt;/li&gt;&#xA;&lt;li&gt;for any two of these evenings, there is exactly one friend that was present at both evenings;&lt;/li&gt;&#xA;&lt;li&gt;none of his friends was present at all $n$ evenings.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What's the largest value that $n$ can possibly take?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2016-03-19T14:46:22.723" Title="Drinking beer in the pub" Tags="|mathematics|combinatorics|story|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28653" PostTypeId="2" ParentId="28652" CreationDate="2016-03-09T09:58:47.133" Score="10" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 7&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With one possible combination being&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ABC ADE BDF AFG BEG CDG CEF&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each friend cannot appear on more than 3 days. There needs to be a day without this friend, but still share another friend from each day. Since there only 3 friends each day, 3 is the maximum. Each new day we have to include a friend from each of the previous days. Since we can only reuse a person which has occurred less than 3 times, each of the persons on the 7th day can match up with 2 of the previous days. we cannot do this for day 8. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19103" LastActivityDate="2016-03-09T09:58:47.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28654" PostTypeId="2" ParentId="28420" CreationDate="2016-03-09T10:25:59.180" Score="9" Body="&lt;p&gt;Nobody is going to check this anyway, but here it goes. &#xA;I hope all the math is right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1 = 9 / 9 + 1 - 1$&lt;br&gt;&#xA;$2 = 9 / 9 * 1 + 1$&lt;br&gt;&#xA;$3 = 9 / 9 + 1 + 1$&lt;br&gt;&#xA;$4 = (1+1)^2 - 9 + 9$&lt;br&gt;&#xA;$5 = (1+1)^2 + 9 /9$&lt;br&gt;&#xA;$6 = (9*9)^{1/4} * (1 + 1) $&lt;br&gt;&#xA;$7 = 9^{1/2} + 9^{1/2} + 1 * 1$&lt;br&gt;&#xA;$8 = (1+1*9/9)^3$&lt;br&gt;&#xA;$9 = (9 / 9 + 1 + 1)^2$&lt;br&gt;&#xA;$10 = (9*9)^{1/2} + 1*1$&lt;br&gt;&#xA;$11 = (9*9)^{1/2} + 1+1$&lt;br&gt;&#xA;$12 = (9*1*1)^{1/2} + 9$&lt;br&gt;&#xA;$13 = (9*1)^{1/2} + 9 + 1$&lt;br&gt;&#xA;$14 = (9)^{1/2} + 9+1+1$&lt;br&gt;&#xA;$15 = 9*(1+1) - 9^{1/2}$&lt;br&gt;&#xA;$16 = (1+1*9/9)^4$&lt;br&gt;&#xA;$17 = 9*1 + 9 - 1$&lt;br&gt;&#xA;$18 = 9 + 9 +1 -1$&lt;br&gt;&#xA;$19 = 9 + 9 +1 *1$&lt;br&gt;&#xA;$20 = 9 + 9 +1 +1$&lt;br&gt;&#xA;$21 = (9 -1 -1) * 9^(1/2)$&lt;br&gt;&#xA;$22 = 9^{3/2} - (9^{1/2} +1+1)$&lt;br&gt;&#xA;$23 = 1 * 9^{3/2} - (9^{1/2} +1)$&lt;br&gt;&#xA;$24 = 1 * 1 * 9^{3/2} - 9^{1/2})$&lt;br&gt;&#xA;$25 = ((1+1)^2 + 9 /9)^2$&lt;br&gt;&#xA;$26 = (9*9)^{3/4} -1*1$&lt;br&gt;&#xA;$27 = (9 / 9 + 1 + 1)^3$&lt;br&gt;&#xA;$28 = (9 * 9)^{3/4} + 1 * 1$&lt;br&gt;&#xA;$29 = (9 * 9)^{3/4} + 1 + 1$&lt;br&gt;&#xA;$30 = 9^{1/2} * 1 * (9+1)$&lt;br&gt;&#xA;$31 = 9^{1/2} * (9+1) + 1$&lt;br&gt;&#xA;$32 = (1+1*9/9)^5$&lt;br&gt;&#xA;$33 = 9^{1/2} * (9 + 1 + 1)$&lt;br&gt;&#xA;$34 = 9^{3/2} + 9 - 1 - 1$&lt;br&gt;&#xA;$35 = 9^{3/2} * 1 + 9 - 1$&lt;br&gt;&#xA;$36 = ((9*9)^{1/4} * (1 + 1))^2$&lt;br&gt;&#xA;$37 = 9^{3/2} + 9 + 1 * 1$&lt;br&gt;&#xA;$38 = 9^{3/2} + 9 + 1 + 1$&lt;br&gt;&#xA;$39 = 9^{3/2} + 9^{1/2} * (1+1)^2$&lt;br&gt;&#xA;$40 = (9^{1/2} + 1) * (9 + 1)$&lt;br&gt;&#xA;$41 = (1+1)^5 + (9*9)^{1/2}$&lt;br&gt;&#xA;$42 = 9^{3/2} - 1 + (9^{1/2} + 1)^2$&lt;br&gt;&#xA;$43 = (1+1)^4 + (9/(9^{1/2}))^3$&lt;br&gt;&#xA;$44 = (9^{1/2} + 9^{1/2})^2 + (1+1)^3$&lt;br&gt;&#xA;$45 = (9^{1/2} + 1 + 1) * 9$&lt;br&gt;&#xA;$46 = 9^2 - 9^{3/2} - (1+1)^3$&lt;br&gt;&#xA;$47 = (9^{1/2} + 1)^2 * 9^{1/2} - 1$&lt;br&gt;&#xA;$48 = (9^{1/2} + 9^{1/2} + 1)^2 - 1$&lt;br&gt;&#xA;$49 = (9^{1/2} + 9^{1/2} + 1 * 1)^2$&lt;br&gt;&#xA;$50 = 9^2 - 9^{3/2} - (1+1)^2$&lt;br&gt;&#xA;$51 = 9^{3/2} + 9^{1/2} * (1+1)^3$&lt;br&gt;&#xA;$52 = 9^2 - 9^{3/2} - 1 - 1$&lt;br&gt;&#xA;$53 = 9^2 - 9^{3/2} - 1 * 1$&lt;br&gt;&#xA;$54 = 9^{1/2} * (1 + 1) * 9$&lt;br&gt;&#xA;$55 = (1+1)^6 - (9*9)^{1/2}$&lt;br&gt;&#xA;$56 = (9-1)^2 - 9 + 1$&lt;br&gt;&#xA;$57 = (9^{3/2} - (1 + 1)^3) * 9^{1/2}$&lt;br&gt;&#xA;$58 = (1+1)^6 - 9^{1/2} - 9^{1/2}$&lt;br&gt;&#xA;$59 = (9^{1/2} - 1)^5 + 9^{3/2} * 1$&lt;br&gt;&#xA;$60 = (9^{1/2} + 1)^3 - 9^{1/2} - 1$&lt;br&gt;&#xA;$61 = (9^{1/2} + 1)^3 - 9^{1/2} * 1$&lt;br&gt;&#xA;$62 = (9^{1/2} + 1)^3 - 9^{1/2} + 1$&lt;br&gt;&#xA;$63 = 9* (9-1-1)$&lt;br&gt;&#xA;$64 = (1+1*9/9)^6$&lt;br&gt;&#xA;$65 = (1+9/9)^6 + 1$&lt;br&gt;&#xA;$66 = (9^{1/2} + 1)^3 + 9^{1/2} - 1$&lt;br&gt;&#xA;$67 = (9^{1/2} + 1)^3 + 9^{1/2} * 1$&lt;br&gt;&#xA;$68 = (9^{1/2} + 1)^3 + 9^{1/2} + 1$&lt;br&gt;&#xA;$69 = (9^{3/2} - (1 + 1)^2) * 9^{1/2}$&lt;br&gt;&#xA;$70 = 9^2 - 9 - 1 - 1$&lt;br&gt;&#xA;$71 = 9* (9-1) - 1$&lt;br&gt;&#xA;$72 = 9* 1* (9-1)$&lt;br&gt;&#xA;$73 = 9* (9-1) + 1$&lt;br&gt;&#xA;$74 = 9^2 - 9 + 1 + 1$&lt;br&gt;&#xA;$75 = (9^{3/2} - 1 - 1) * 9^{1/2}$&lt;br&gt;&#xA;$76 = 9^2 - 9^{1/2} - 1 - 1$&lt;br&gt;&#xA;$77 = (9^{3/2} - 1) * 9^{1/2} - 1$&lt;br&gt;&#xA;$78 = (9^{3/2} - 1) * 9^{1/2} * 1$&lt;br&gt;&#xA;$79 = 9 * 9 - 1 - 1$&lt;br&gt;&#xA;$80 = 9 * 9 * 1 - 1$&lt;br&gt;&#xA;$81 = 9 * 9 * 1 * 1$&lt;br&gt;&#xA;$82 = 9 * 9 * 1 + 1$&lt;br&gt;&#xA;$83 = 9 * 9 + 1 + 1$&lt;br&gt;&#xA;$84 = (9^{3/2} + 1) * 9^{1/2} * 1$&lt;br&gt;&#xA;$85 = (9^{3/2} + 1) * 9^{1/2} + 1$&lt;br&gt;&#xA;$86 = 9^2 + 9^{1/2} + 1 + 1$&lt;br&gt;&#xA;$87 = ((1+1)^5 - 9^{1/2}) * 9^{1/2}$&lt;br&gt;&#xA;$88 = 9^2 + 9 - 1 - 1$&lt;br&gt;&#xA;$89 = 9* (9+1) - 1$&lt;br&gt;&#xA;$90 = 9* 1* (9+1)$&lt;br&gt;&#xA;$91 = 9* (9+1) + 1$&lt;br&gt;&#xA;$92 = 9^2 + 9 + 1 + 1$&lt;br&gt;&#xA;$93 = (1+1)^5 * 9^{1/2} - 9^{1/2}$&lt;br&gt;&#xA;$94 = (1+1)^6 + 9^{3/2} + 9^{1/2}$&lt;br&gt;&#xA;$95 = 9^2 + (\log_39)^4 - 1 -1$&lt;br&gt;&#xA;$96 = (1+1)^5 * (9*9)^{1/4} $&lt;br&gt;&#xA;$97 = 9*9 + (1+1)^4$&lt;br&gt;&#xA;$98 = 9^2 + (1+9^{1/2})^2 + 1$&lt;br&gt;&#xA;$99 = 9* (9+1+1)$&lt;br&gt;&#xA;$100 = (9+1)*(9+1)$&lt;/p&gt;&#xA;" OwnerUserId="19989" LastEditorUserId="19989" LastEditDate="2016-03-09T15:28:08.623" LastActivityDate="2016-03-09T15:28:08.623" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28656" PostTypeId="1" AcceptedAnswerId="28657" CreationDate="2016-03-09T10:44:01.450" Score="15" ViewCount="1379" Body="&lt;p&gt;A real life puzzle, yesterday on &lt;strong&gt;The Price is Right&lt;/strong&gt; a contestant played the following game:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;She could choose 5 cards from a pile where 2 cards were &quot;Car&quot;, 11 were &quot;C&quot;, 11 were &quot;A&quot; and 11 were &quot;R&quot;. &lt;/li&gt;&#xA;&lt;li&gt;If she chose a &quot;Car&quot; card (at least 1) or the letters &quot;C&quot;, &quot;A&quot;, and &quot;R&quot; (at least 1 of each), she would win a new car. According to the game, the letters &quot;C&quot;, &quot;A&quot;, and &quot;R&quot; in the second case did not need to be in order. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What were the odds she was going to win?&lt;/p&gt;&#xA;" OwnerUserId="9125" LastActivityDate="2016-03-10T00:49:35.313" Title="The Price is Right" Tags="|mathematics|probability|" AnswerCount="5" CommentCount="5" ClosedDate="2016-05-25T23:24:23.010" ContentLicense="CC BY-SA 3.0" />
  <row Id="28657" PostTypeId="2" ParentId="28656" CreationDate="2016-03-09T11:02:07.600" Score="15" Body="&lt;p&gt;If I'm not mistaken&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(\text{to win car}) = 1 - P(\text{only draw Cs}) - P(\text{only draw As}) - P(\text{only draw Rs}) - P(\text{only draw Cs or As}) - P(\text{only draw Cs or Rs}) - P(\text{only draw Rs or As})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(\text{only draw Cs}) = P(\text{only draw As}) = P(\text{only draw Rs}) = \frac{\binom{11}{5}}{\binom{35}{5}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(\text{only draw Cs or As}) = P(\text{only draw Cs or Rs}) = P(\text{only draw Rs or As}) = \frac{\binom{22}{5}}{\binom{35}{5}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(\text{to win car}) = 1 - \frac{3\binom{11}{5}}{\binom{35}{5}} - \frac{3\binom{22}{5}}{\binom{35}{5}} = \frac{244244}{324632} \approx 0.75$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;br&gt;&#xA;Thinking about it I believe that $P(\text{only draw Cs or As})$ also includes the chances of only Cs and only As. and they get counted twice so I have to actually add them again to make it right. So I believe it might actually be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(\text{to win car}) = 1 + \frac{3\binom{11}{5}}{\binom{35}{5}} - \frac{3\binom{22}{5}}{\binom{35}{5}} = \frac{247016}{324632} \approx 0.76091$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone confirm?&lt;/p&gt;&#xA;" OwnerUserId="1748" LastEditorUserId="1748" LastEditDate="2016-03-09T15:04:41.067" LastActivityDate="2016-03-09T15:04:41.067" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28658" PostTypeId="1" AcceptedAnswerId="28668" CreationDate="2016-03-09T11:24:28.037" Score="11" ViewCount="1162" Body="&lt;p&gt;I went away. :(&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I came back. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I went away. :(&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I came back. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I went away. :(&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I came back. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I went away. :(&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I came back. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I went away. :(&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I came back. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's a Rebus.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;EDIT-&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This ISN'T a lateral thinking problem or a riddle. Also the Smile emoticons hold significance. It's a phrase, that's why tagged under rebus.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19656" LastEditorUserId="19656" LastEditDate="2016-03-10T06:32:22.750" LastActivityDate="2016-03-13T10:39:03.817" Title="Don't Go.. Please" Tags="|wordplay|rebus|" AnswerCount="9" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28659" PostTypeId="2" ParentId="28360" CreationDate="2016-03-09T11:26:11.010" Score="0" Body="&lt;p&gt;My answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll start to count while you solve this problem,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Time counts up&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and just relax here at the top of my column.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Well this is the only weird sentence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Everywhere I go I let out a good whistle,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you let time pass, something good or bad will happen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but don't mess with me cause I carry a pistol.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Time can heal wounds, but also cause wounds&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I came with a flash and out like a light,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Big Bang&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I don't really mind because life is so bright.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Time doesn't care about anything&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've been explosive at one point in history,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes, the Big Bang&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;what I truly am is not much of a mystery.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Time is not a mystery.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="20035" LastActivityDate="2016-03-09T11:26:11.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28660" PostTypeId="2" ParentId="28652" CreationDate="2016-03-09T12:31:02.037" Score="7" Body="&lt;p&gt;Kruga's answer with an alternate proof.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 7 is indeed the maximum&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I use a constructive proof. Let the friends be represented by letters as in Kruga's answer. The letters do not imply any sort of order. Therefore, we have:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Day 1: ABC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the second day.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One friend from the first day must be present. The letters are arbitrary, so we pick A. Also, since all of the conditions are reflexive and commutative, the numbering of the days is arbitrary. By condition 3 (we note that condition 2 is a special case of condition 3) no other friend from the first day can be present; therefore, we have two new friends D and E.  &lt;br&gt;&lt;br&gt;&#xA; Day 2: ADE&lt;br&gt;&#xA;&lt;br&gt;&#xA; By the 4th condition, there must exist a day when A is not present. I will call this Day 3, but, again, the number is arbitrary. This day must have exactly 1 friend from Day 1. We can pick B since the names are arbitrary. There must also be 1 friend from Day 2, but not A. We can pick D. There cannot be a second friend from either day so the third friend must be one never seen before. Call this friend F. We have:  &lt;br&gt;&lt;br&gt;&#xA; Day 3: BDF&lt;br&gt;&#xA;&lt;br&gt;&#xA; Now we note that it is only possible to construct one more day with either A or B. Consider A. A cannot be paired with B or C due to Day 1 or with D or E due to Day 2 yet there must be one common friend with Day 3. The only possibility is F. In other words any new day containing A must contain F. By the third condition, there can be only one such day or A and F would be both be present on two occasions. The same logic applies to B.&lt;br&gt;&#xA;&lt;br&gt;&#xA; By condition 3, any day outside of the three we have listed must share exactly one friend with Day 1. We can generate a maximum of one more day containing A, a maximum of one more day containing B. If we want to generate more than two more days, the only remaining possibility is to use C. Call this new day Day 4. Day 4 must Contain C. It must have a friend in common with Day 2, but not A or Day 4 and Day 1 would have both C and A in common. Whichever you pick, the set of Day 1, Day 2 and Day 4 is isometric with Day 1, Day 2 and Day 3. Therefore, by the same reasoning only one more day can be generated including C after Day 4. Thus there is a maximum of 4 + 1 + 1 +1 = 7 combinations. Kruga has given one such combination. QED.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastEditorUserId="19470" LastEditDate="2016-03-19T14:46:22.723" LastActivityDate="2016-03-19T14:46:22.723" CommentCount="0" CommunityOwnedDate="2016-03-09T14:20:01.473" ContentLicense="CC BY-SA 3.0" />
  <row Id="28661" PostTypeId="2" ParentId="28656" CreationDate="2016-03-09T12:42:03.437" Score="3" Body="&lt;p&gt;Hmmm... Let's see... We have total $35$ cards and we need to choose only $5$ from it. So there must be $\binom{35}{5}$ (= $324632$) ways to do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what if I get only one &quot;Car&quot; card...there are $2$ ways to get it (because $\binom{2}{1}$)...and I need to choose any four cards of C, A and R out of those $33$ cards. So the probability when only one &quot;CAR&quot; card is chosen = $\dfrac{\binom{2}{1}\binom{33}{4}}{\binom{35}{5}} = \dfrac{81840}{324632}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what if I get two &quot;Car&quot; cards...there is only $1$ way to get it (because $\binom{2}{2}$)...and I need to choose any $3$ cards of C, A and R out of those $33$ cards. So the probability when two &quot;CAR&quot; cards is chosen = $\dfrac{\binom{2}{2}\binom{33}{3}}{\binom{35}{5}} = \dfrac{5456}{324632}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what if I don't get any &quot;Car&quot; card...then to win I must have C, A and R...all three of them. And if I have all three of them then two of them can be chosen twice. If it happens then one of them must be chosen once. If C is chosen once then the number of ways it can be chosen is $\binom{11}{1}$. So the probability when no CAR card is chosen and two cards are chosen twice is $3\dfrac{\binom{2}{0}\binom{11}{1}\binom{11}{2}\binom{11}{2}}{\binom{35}{5}} = \dfrac{99825}{324632}$. By multiplying by $3$ I covered the probabilities that even A and R can be chosen once. But it is also possible that one of them is chosen thrice and the other two are chosen once. So the probability when no CAR card is chosen and one card is chosen thrice is $3\dfrac{\binom{2}{0}\binom{11}{1}\binom{11}{1}\binom{11}{3}}{\binom{35}{5}} = \dfrac{59895}{324632}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These were the four probabilities...by adding them I get $\dfrac{247016}{324632}$...which simplifies to $0.7609$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Thanks to &lt;a href=&quot;https://puzzling.stackexchange.com/users/1748/ivo-beckers&quot;&gt;IvoBeckers&lt;/a&gt; for help.&lt;/p&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.600" LastActivityDate="2016-03-09T16:20:52.513" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="28662" PostTypeId="1" AcceptedAnswerId="28663" CreationDate="2016-03-09T12:42:53.843" Score="9" ViewCount="523" Body="&lt;p&gt;Here I come, with another Rebus.&lt;br&gt;&#xA;This is rather easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/w4WuM.jpg&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/w4WuM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Almost Everywhere, but We Don't Care.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19656" LastActivityDate="2016-03-09T12:58:11.550" Title="Musical Chemistry" Tags="|visual|rebus|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="28663" PostTypeId="2" ParentId="28662" CreationDate="2016-03-09T12:47:40.153" Score="9" Body="&lt;p&gt;It looks like it should be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;HTTP://&lt;/strong&gt; (H Ti Ti Pi Wall-Socket-What-Looks-Like-A-Colon Slash Slash)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2016-03-09T12:58:11.550" LastActivityDate="2016-03-09T12:58:11.550" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="28664" PostTypeId="1" AcceptedAnswerId="28666" CreationDate="2016-03-09T13:00:29.563" Score="11" ViewCount="1951" Body="&lt;p&gt;In 1989, someone going by the name [BLANK] made a maze game on the computer. It is said that the maze constantly changes its layout from time to time so that no one can make it to the end to claim the prize. Outside the maze on the wall, there is a cryptic message and some clues on how to beat the maze, but no one has cracked the code.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;GPTESTF&#xA;  &lt;br/&gt;FKBNG&#xA;  &lt;br/&gt;FKBNG&#xA;  &lt;br/&gt;KWDR&#xA;  &lt;br/&gt;NOW&#xA;  &lt;br/&gt;4303Q5&#xA;  &lt;br/&gt;020103112301180419&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;br/&gt;Solving this puzzle isn’t hard&#xA;  &lt;br/&gt;If you do, you will be rewarded with a crystal shard&#xA;  &lt;br/&gt;Look down and you can see&#xA;  &lt;br/&gt;The answer is in front of thee&#xA;  &lt;br/&gt;The first step is never right&#xA;  &lt;br/&gt;The second and third step is in the same direction that gives mankind light&#xA;  &lt;br/&gt;The next line is the opposite if you believe&#xA;  &lt;br/&gt;The forth step is the past tense of leave&#xA;  &lt;br/&gt;Once you have solved all these clues&#xA;  &lt;br/&gt;You are now halfway, don’t look so blue&#xA;  &lt;br/&gt;Now listen carefully as this is tricky&#xA;  &lt;br/&gt;Cause this maze is very picky&#xA;  &lt;br/&gt;Listen to what the last three clues say&#xA;  &lt;br/&gt;And you will be on your way&#xA;  &lt;br/&gt;The fifth step is as you see&#xA;  &lt;br/&gt;Next step is in the same direction as a falling tree&#xA;  &lt;br/&gt;The last step must be split in pairs&#xA;  &lt;br/&gt;Now there are two things you must compare&#xA;  &lt;br/&gt;Once you solve this, you are done&#xA;  &lt;br/&gt;Congratulations, you have won!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can you solve the puzzle and get the crystal shard from the maze?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Please show how each step was deciphered, as well as, the direction to go to complete the maze.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4645" LastEditorUserId="4645" LastEditDate="2016-03-09T13:51:24.380" LastActivityDate="2016-03-09T14:46:32.750" Title="The Unsolved Maze With The Annoying Rhyme" Tags="|riddle|pattern|rhyme|cipher|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28665" PostTypeId="2" ParentId="28664" CreationDate="2016-03-09T13:14:42.977" Score="3" Body="&lt;p&gt;Just eyeballing those encoded lines, I figure&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; GPTESTF = &lt;strong&gt;FORWARD&lt;/strong&gt; (first step is never right)&lt;br&gt;&#xA; FKBNG = &lt;strong&gt;RIGHT&lt;/strong&gt;&lt;br&gt;&#xA; FKBNG = &lt;strong&gt;RIGHT&lt;/strong&gt;&lt;br&gt;&#xA; KWDR = &lt;strong&gt;LEFT&lt;/strong&gt; (this fits with &quot;past tense of leave&quot; but is contradicted by the preceding line in the poem)&lt;br&gt;&#xA; NOW = (probably) just the word &lt;strong&gt;&quot;now&quot;&lt;/strong&gt; (it &quot;is as you see&quot;)&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;strong&gt;OK, this part of the solution I got from the one other answer:&lt;/strong&gt; &lt;em&gt;Apparently&lt;/em&gt;, trees fall diagonally down to the right. On a keyboard, diagonally down to the right from the keys 4303Q5 is &lt;strong&gt;REPEAT&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; The last one (020103112301180419) &quot;split in pairs&quot;, yields the A = 1, B = 2, etc. number codes for &quot;&lt;strong&gt;BACKWARDS&lt;/strong&gt;&quot;&lt;br&gt;&#xA;   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Put these all together and&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; FORWARD RIGHT RIGHT LEFT (and then REPEAT BACKWARDS, which gives us) LEFT RIGHT RIGHT FORWARD&lt;br&gt;&#xA;&lt;br&gt;&#xA; Though I don't know what the rule is that changes the encoding between the first and second and between the third and fourth lines. Below is the regular alphabet, and then the letters each letter is encoded as in lines 1, 2 &lt;strong&gt;and&lt;/strong&gt; 3, and 4, respectively.&lt;br&gt;&#xA;&lt;br&gt;&#xA;  &lt;code&gt;ABCDEFGHIJKLMNOPQRSTUVWXYZ&lt;/code&gt;&lt;br&gt;&#xA;  &lt;code&gt;S__F_G________P__T____E___&lt;/code&gt;&lt;br&gt;&#xA;  &lt;code&gt;______BNK________F_G______&lt;/code&gt;&lt;br&gt;&#xA;  &lt;code&gt;____WD_____K_______R______&lt;/code&gt;&lt;br&gt;&#xA;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2016-03-09T13:51:12.350" LastActivityDate="2016-03-09T13:51:12.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28666" PostTypeId="2" ParentId="28664" CreationDate="2016-03-09T13:29:07.153" Score="13" Body="&lt;p&gt;The first step is never right&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; FORWARD (GPTESTF, taking keys to the left on keyboard)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second and third step is in the same direction that gives mankind light&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; RIGHT, RIGHT (FKBNG, taking keys left+up on keyboard)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The next line is the opposite if you believe&#xA;The forth step is the past tense of leave&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; LEFT (KWDR, taking keys to the right on keyboard) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The fifth step is as you see&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; NOW (NOW, taking as you see and as it is)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Next step is in the same direction as a falling tree&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; REPEAT (4303Q5, taking keys right+down on keyboard)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The last step must be split in pairs &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; BACKWARDS (substitution cipher, A=1, B=2, ...)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Summary of solution:&lt;/h2&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; FORWARD, RIGHT, RIGHT, LEFT, LEFT, RIGHT, RIGHT, FORWARD&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="9125" LastEditDate="2016-03-09T14:46:32.750" LastActivityDate="2016-03-09T14:46:32.750" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28667" PostTypeId="2" ParentId="28650" CreationDate="2016-03-09T13:57:51.850" Score="3" Body="&lt;p&gt;I found the name.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;AL&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It wasn't all that hidden though.&lt;/p&gt;&#xA;" OwnerUserId="20038" LastActivityDate="2016-03-09T13:57:51.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28668" PostTypeId="2" ParentId="28658" CreationDate="2016-03-09T14:24:44.463" Score="11" Body="&lt;p&gt;Since it keeps coming back with smiles, here's another guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Many_happy_returns_%28greeting%29&quot;&gt;Many happy returns&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Previous attempt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the emoti(c)ons and movements, how about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Come and go as you please.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9422" LastEditorUserId="9422" LastEditDate="2016-03-10T06:41:42.920" LastActivityDate="2016-03-10T06:41:42.920" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28669" PostTypeId="1" AcceptedAnswerId="28688" CreationDate="2016-03-09T14:25:36.300" Score="4" ViewCount="216" Body="&lt;p&gt;Every letter stands for a digit in base-10 representation, &#xA;different letters stand for different digits,&#xA;and leading digits are always non-zero.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;     SQUARE&#xA;    + DANCE&#xA;  ----------&#xA;     DANCER&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which digit does each letter represent? (Please present the full analysis how these digits can be determined.)&lt;/p&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2016-03-10T18:48:13.147" Title="Square dance alphametic" Tags="|no-computers|alphametic|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28670" PostTypeId="2" ParentId="28658" CreationDate="2016-03-09T14:26:28.693" Score="6" Body="&lt;p&gt;In a parallel universe it could be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A &quot;tennis&quot; game.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I see:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;10 eyes&quot; going back and forth.&#xA; 10 Is on the left and 10 (pairs of) eyes on the right.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18129" LastEditorUserId="18129" LastEditDate="2016-03-09T14:32:38.307" LastActivityDate="2016-03-09T14:32:38.307" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28671" PostTypeId="2" ParentId="28656" CreationDate="2016-03-09T14:31:17.550" Score="5" Body="&lt;p&gt;I believe there is a :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 75,664% chance&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Based on the other answers, and the general consensus it's easier to calculate the 'no win' scenario, and deduct them from 1 to find the win odds:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Calculate what chances yield 'no CAR and no C'&lt;br&gt;&#xA;2) Calculate what chances yield 'no CAR and no A'&lt;br&gt;&#xA;3) Calculate what chances yield 'no CAR and no R'&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They will (obviously) all equal the same, so here's the math for option 1:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Card 1: 22/35 cards are OK, 11 C's and 2 CAR's aren not &lt;br&gt;&lt;br&gt;&#xA;  Card 2: 21/34 cards are OK&lt;br&gt;&lt;br&gt;&#xA;  Card 3: 20/33 cards are OK&lt;br&gt;&lt;br&gt;&#xA;  Card 4: 19/32 cards are OK&lt;br&gt;&lt;br&gt;&#xA;  Card 5: 18/31 cards are OK&lt;br&gt;&lt;br&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which makes the odds of '&lt;em&gt;no car and no C's&lt;/em&gt;' the product of the above five fractions:3.160.080 / 38.955.840 = 8.112%&#xA;&lt;br&gt;&lt;br&gt;&#xA;P(winning the car) = 1 - P(not winning the car) = 1 - P(no car and no C) - P(no car and no A) - P(no car and no R) = 1 - 8.112% - 8.112% - 8.112% = 75,664% odds of winning the car.&lt;/p&gt;&#xA;" OwnerUserId="2156" LastActivityDate="2016-03-09T14:31:17.550" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28672" PostTypeId="2" ParentId="28420" CreationDate="2016-03-09T14:33:39.877" Score="4" Body="&lt;p&gt;Using only addition and subtraction as the operators, only integer powers, and only employing $\log_39=2$ and $\log_99=1$ (avoiding the temptation to use $\log1=0$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;(1+1)^3+\log_39-9&amp;amp;=1\\&#xA;9+1+1-9&amp;amp;=2\\&#xA;9-(1+1)^2-\log_39&amp;amp;=3\\&#xA;9+(1+1)^2-9&amp;amp;=4\\&#xA;9-\log_39-1-1&amp;amp;=5\\&#xA;\log_39+\log_39+1+1&amp;amp;=6\\&#xA;9+1-\log_39-1&amp;amp;=7\\&#xA;9+(1+1)^3-9&amp;amp;=8\\&#xA;9+1+1-\log_39&amp;amp;=9\\&#xA;9+9-(1+1)^3&amp;amp;=10\\&#xA;9+\log_3{9}+1-1&amp;amp;=11\\&#xA;(1+1)^3+\log_3{9}+\log_3{9}&amp;amp;=12\\&#xA;9+\log_3{9}+1+1&amp;amp;=13 \\&#xA;9+9-(1+1)^2&amp;amp;=14\\&#xA;9+\log_39+(1+1)^2&amp;amp;=15\\&#xA;9+9-1-1&amp;amp;=16\\&#xA;(\log_39+1)^2+9-1&amp;amp;=17\\&#xA;9+9+1-1&amp;amp;=18\\&#xA;9+(1+1)^3+\log_39&amp;amp;=19\\&#xA;9+9+1+1&amp;amp;=20\\&#xA;(1+1)^5-9-\log_39&amp;amp;=21\\&#xA;9+9+(1+1)^2&amp;amp;=22\\&#xA;(1+1)^4+9-\log_39&amp;amp;=23\\&#xA;(\log_39+1)^3-\log_39-1&amp;amp;=24\\&#xA;(1+1)^5+\log_39-9&amp;amp;=25\\&#xA;9+9+(1+1)^3&amp;amp;=26\\&#xA;(1+1)^4+9+\log_39&amp;amp;=27\\&#xA;(\log_39+\log_39-1)^3+1&amp;amp;=28\\&#xA;(1+1)^5-\log_39-\log_99&amp;amp;=29\\&#xA;(\log_39+1)^3+\log_39+1&amp;amp;=30\\&#xA;(1+1)^5-\log_39+\log_99&amp;amp;=31\\&#xA;(1+1)^5+9-9&amp;amp;=32\\&#xA;(\log_39)^5+\log_99+1-1&amp;amp;=33\\&#xA;(\log_39)^5+\log_39+1-1&amp;amp;=34\\&#xA;(\log_39+1)^3+9-1&amp;amp;=35\\&#xA;(9+1)^2-(9-1)^2&amp;amp;=36\\&#xA;(\log_39+1)^3+9+1&amp;amp;=37\\&#xA;(\log_39)^6+1-(\log_39+1)^3&amp;amp;=38\\&#xA;(1+1)^5+9-\log_39&amp;amp;=39\\&#xA;(9+1+1)^2-9^2&amp;amp;=40\\&#xA;(\log_39)^5+9+1-1&amp;amp;=41\\&#xA;(1+1)^5+9+\log_99&amp;amp;=42\\&#xA;(\log_39)^5+9+1+1&amp;amp;=43\\&#xA;(\log_39)^5+(1+1)^3+(\log_39)^2&amp;amp;=44\\&#xA;(\log_39)^5+9+(1+1)^2&amp;amp;=45\\&#xA;(1+1)^6-9-9&amp;amp;=46\\&#xA;(9-1-1)^2-\log_39&amp;amp;=47\\&#xA;(\log_39)^5+(\log_39)^5-(1+1)^4&amp;amp;=48\\&#xA;(9-\log_39)^2+1-1&amp;amp;=49\\&#xA;(1+1)^5+9+9&amp;amp;=50\\&#xA;(9-1-1)^2+\log_39&amp;amp;=51\\&#xA;(1+1)^5+(\log_39)^4+(\log_39)^2&amp;amp;=52\\&#xA;(\log_39)^6-9-1-1&amp;amp;=53\\&#xA;(9-1)^2-9-1&amp;amp;=54\\&#xA;(\log_39)^6+1-9-1&amp;amp;=55\\&#xA;(9-1)^2+1-9&amp;amp;=56\\&#xA;(\log_39)^6+1+1-9&amp;amp;=57\\&#xA;(9-1-1)^2+9&amp;amp;=58\\&#xA;(9-1)^2-(\log_39)^2-1&amp;amp;=59\\&#xA;(\log_39)^5+(\log_39+1)^3+1&amp;amp;=60\\&#xA;(9-1)^2-\log_39-1&amp;amp;=61\\&#xA;(9-1)^2-log_99-1&amp;amp;=62\\&#xA;(9-1)^2+1-\log_39&amp;amp;=63\\&#xA;9^2-9-(1+1)^3&amp;amp;=64\\&#xA;9^2-(\log_39)^3-(1+1)^3&amp;amp;=65\\&#xA;(9-1)^2+\log_99+1&amp;amp;=66\\&#xA;(9-1)^2+\log_39+1&amp;amp;=67\\&#xA;9^2-9-(1+1)^2&amp;amp;=68\\&#xA;9^2-(1+1)^3-(\log_39)^2&amp;amp;=69\\&#xA;9^2-9-1-1&amp;amp;=70\\&#xA;(\log_39)^6+9-1-1&amp;amp;=71\\&#xA;9^2+1-9-1&amp;amp;=72\\&#xA;(\log_39)^6+9+1-1&amp;amp;=73\\&#xA;9^2+1+1-9&amp;amp;=74\\&#xA;(\log_39)^6+9+1+1&amp;amp;=75\\&#xA;9^2+(1+1)^2-9&amp;amp;=76\\&#xA;9^2-\log_39-1-1&amp;amp;=77\\&#xA;9^2-\log_99-1-1&amp;amp;=78\\&#xA;9^2+1-\log_39-1&amp;amp;=79\\&#xA;9^2+1-\log_99-1&amp;amp;=80\\&#xA;9^2+1+1-\log_39&amp;amp;=81\\&#xA;9^2+9-(1+1)^3&amp;amp;=82\\&#xA;9^2+\log_39+1-1&amp;amp;=83\\&#xA;9^2+\log_99+1+1&amp;amp;=84\\&#xA;9^2+\log_39+1+1&amp;amp;=85\\&#xA;9^2+9-(1+1)^2&amp;amp;=86\\&#xA;9^2+(1+1)^2+\log_39&amp;amp;=87\\&#xA;9^2+9-1-1&amp;amp;=88\\&#xA;9^2+(\log_39)^2+(1+1)^2&amp;amp;=89\\&#xA;(9+1)^2-9-1&amp;amp;=90\\&#xA;9^2+(1+1)^3+\log_39&amp;amp;=91\\&#xA;9^2+9+1+1&amp;amp;=92\\&#xA;9^2+(\log_39)^3+(1+1)^2&amp;amp;=93\\&#xA;9^2+9+(1+1)^2&amp;amp;=94\\&#xA;(9+1)^2-(\log_39)^2-1&amp;amp;=95\\&#xA;9^2+(1+1)^4-\log_99&amp;amp;=96\\&#xA;(9+1)^2-\log_39-1&amp;amp;=97\\&#xA;9^2+9^2-(1+1)^6&amp;amp;=98\\&#xA;(9+1)^2+1-\log_39&amp;amp;=99\\&#xA;(9+1)^2+\log_99-1&amp;amp;=100\\&#xA;\end{align}&lt;/p&gt;&#xA;" OwnerUserId="16717" LastActivityDate="2016-03-09T14:33:39.877" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28673" PostTypeId="1" AcceptedAnswerId="28679" CreationDate="2016-03-09T14:36:35.667" Score="10" ViewCount="378" Body="&lt;p&gt;There is a story that a poor college student sent a telegram with the words &quot;SEND MORE MONEY&quot; to his parents, asking for more money, and asking to fill in each letter with a different digit in order to figure out how much the student was asking for. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In another story an impoverished college girl received a telegram from her parents with the words &quot;ALAS LASS NO MORE CASH&quot;, asking her to fill in each letter with a different digit in order to figure out how much money she would receive.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;       ALAS&#xA;     + LASS&#xA;     +   NO&#xA;     + MORE&#xA;    ---------&#xA;       CASH&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(Every letter stands for a digit in base-10 representation, &#xA;different letters stand for different digits,&#xA;and leading digits are always non-zero.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the smallest amount of money the girl would receive?&lt;br&gt;&#xA;What is the largest amount of money the girl would receive?  &lt;/p&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2016-03-10T12:43:25.903" Title="Not the &quot;SEND MORE MONEY&quot; alphametic" Tags="|no-computers|alphametic|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28674" PostTypeId="2" ParentId="28652" CreationDate="2016-03-09T14:42:49.333" Score="3" Body="&lt;p&gt;No-go:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If at least 4 groups with the same friend can be found, there'll be no legit group without him, so there can be 3 such groups at most rather than 4.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Let the friends in the first group be $a, b$ and $c$. One of them (not always the same) must be found in all groups, but there must be only one of them in every group other than the first. If 3 of them have each of the three, $b$ and $c$ will be counted twice each, hence we can reach 9-2=7.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$a-b-c / a-d-e / a-f-g / b-d-f / b-e-g / c-d-g / c-e-f$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastActivityDate="2016-03-09T14:42:49.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28675" PostTypeId="2" ParentId="295" CreationDate="2016-03-09T14:55:42.293" Score="0" Body="&lt;p&gt;There is a mistake in the selected answer! The Sum is used for the situation when both Jorge (&lt;span class=&quot;math-container&quot;&gt;$70\%$&lt;/span&gt; chance to miss) and José (&lt;span class=&quot;math-container&quot;&gt;$30\%$&lt;/span&gt; chance to miss) miss, so it will be&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\sum_{n=1}^\infty (0.3\cdot 0.7)^n=\frac{1}{1-0.3\cdot 0.7} =0.2658&#xA;$$&lt;/span&gt;&#xA;by using the geometric series.&lt;/p&gt;&#xA;&lt;p&gt;Other calculations are simply adding and multiplying, so the final results will be:&lt;/p&gt;&#xA;&lt;p&gt;If Jorge will try to shoot Juan:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Juan &lt;span class=&quot;math-container&quot;&gt;$14.7\%$&lt;/span&gt;,&lt;/li&gt;&#xA;&lt;li&gt;José &lt;span class=&quot;math-container&quot;&gt;$57.0\%$&lt;/span&gt; (rounded),&lt;/li&gt;&#xA;&lt;li&gt;Jorge &lt;span class=&quot;math-container&quot;&gt;$28.3\%$&lt;/span&gt; (rounded).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Better result for Jorge if Jorge will miss intentionally (until José or Juan dies):&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Juan &lt;span class=&quot;math-container&quot;&gt;$21\%$&lt;/span&gt;,&lt;/li&gt;&#xA;&lt;li&gt;José &lt;span class=&quot;math-container&quot;&gt;$43.4\%$&lt;/span&gt; (rounded),&lt;/li&gt;&#xA;&lt;li&gt;Jorge &lt;span class=&quot;math-container&quot;&gt;$35.6\%$&lt;/span&gt; (rounded).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="20037" LastEditorUserId="66258" LastEditDate="2023-08-22T12:57:55.057" LastActivityDate="2023-08-22T12:57:55.057" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28679" PostTypeId="2" ParentId="28673" CreationDate="2016-03-09T16:10:55.350" Score="8" Body="&lt;p&gt;First off, we know that $C\ge 6$ because it is the sum of three distinct numbers.  Looking at the third column we see $L+A+O$ yields $A$, so there must be a carry over of at least one into the last column.  Thus, $C \ge 7$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, we know that if any of $A,L,M$ are greater than 5, then we would have an overflow into 5 digits, so $1 \le A,L,M \le 5$ and $7 \le C \le 9$.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Lower Bound&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;So for the lower bound, lets try for $C=7$.  Also, to make the sum the lowest, lets make $A,L,M \in \{1,2,3\}$.  For the lowest possible answer, set $A=1$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now lets try setting $S$ as low as possible.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Assume $S=0$&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The first column is $S+S+O+E=O+E$ yields $H$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lowest remaining values for $O,E$ are $\{4,5\}$ which makes $H=9$ and no carry over.  The second column therefore $A+S+N+R=1+N+R$ yielding $S=0$.  Thus, $N+R=9$ which can't be done with the remaining numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, $O+E \ge 14$ in order to get a valid value for $H$. This makes a carry over of 1.  Thus, $A+S+N+R+1=1+0+N+R+1=2+N+R$ yields $S=0$, so $N+R=8$.  Again, this is impossible with the remaining digits.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Assume $S=4$&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Since 1,2,3 are taken, the lowest value remaining for $S$ is $4$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also want to minimize $H$, but the lowest values remaining for $O$ and $E$ are 5 and 6 which makes $H=9$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The carry over of the first column comes from $S+S+O+E=4+4+5+6=19$, so it is 1.  Thus, $A+S+N+R+1=1+4+N+R+1=6+N+R$ yielding $S=4$.  With the remaining values, we can make this work with $N+R=8$, so $N=8$ and $R=0$.  The full column then gives $14$, so again a carry over of 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third column has $L+A+O+1=L+1+O+1=L+O+2$ yielding $A=1$. Thus, $L=3$ and $O=6$.  This leaves $M=2$ and $E=5$ for a final solution of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C=7, A=1, S=4, H=9, N=8, R=0, L=3, O=6, M=2, E=5$$&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 1314    &#xA; 3144&#xA;   86&#xA;+2605&#xA;-----&#xA; 7149&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h1&gt;Upper Bound&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;This time, lets maximize $C=9$.  We know that there is a carry over into the last column, so $A+L+M=8$.  This means that $A,L,M \in \{1,2,5\} \text{ or } \{1,3,4\}$. Lets try with $A=5$.  Thus $L,M \in \{1,2\}$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Assume $S=8$&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The highest remaining value for $S$.  The first column then has $S+S+O+E$ yielding $H$.  This is $8+8+O+E=16+O+E$.  If $O+E \le 3$ there is a carry over of 1 to the next column.  Since 1 and 2 are taken, only $O+E=3$ gives a carry over of 1.  But if this is the case, then $H=C=9$, so we know $O+E \gt 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the carry over will be 2 into the next column.  Thus, $A+S+N+R+2=5+8+N+R+2=15+N+R$ yields $S=8$.  So $N+R=3$ or $N+R=13$.  The first is impossible, so $N+R=13$, which means $N,R \in \{6,7\}$ are the only possible values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes that column sum to $28$, again giving a carry over of 2.  Thus, the third column is $L+A+O+2=L+O+2$ yielding $A=5$.  So, $L+O=3$ or $L+O=13$.  13 is impossible since $L \in \{1,2\}$.  But $L+O=3 \implies O=M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $S \ne 8$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Assume $S=7$&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The first column has $S+S+O+E=7+7+O+E=14+O+E$ yielding $H$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To maximize the result, the highest remaining value for $H$ is 8.  This means $O+E\in \{4,14\}$.  $14$ is impossible with the remaining digits, so $O,E \in \{0,4\}$ and the carry over to the second column is 1. This column then adds to $A+S+N+R+1=5+7+N+R+1=13+N+R$ yielding $S=7$.  Thus, $N+R \in \{4,14\}$ and now all the values are taken.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next highest value for $H$ is 6.  This, $O+E \in \{2,12\}$.  2 is not possible, so $O+E=12$.  With the remaining values, $O,E \in \{4,8\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next column gets a carry over of 2, so the next column sums to $14+N+R$ which yields $S=7$.  Thus, $N+R \in \{3,13\}$.  Since we can't make 13 with any remaining values, we know that $N,R \in \{0,3\} \implies N=3 \text{ and } R=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives a carry over of 1 to the third column, so $L+A+O+1=L+5+O+1=L+O+6$ yields $A=5$.  So, $L+O=9$.  This only works when $L=1$ and $O=8$ given $L\in\{1,2\}$.  Thus, $M=2$ and $E=4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution is then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C=9, A=5, S=7, H=6, N=3, R=0, L=1, O=8, M=2, E=4$$&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 5157&#xA; 1577&#xA;   38&#xA;+2804&#xA;-----&#xA; 9576&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="1592" LastEditDate="2016-03-10T12:43:25.903" LastActivityDate="2016-03-10T12:43:25.903" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28682" PostTypeId="2" ParentId="28555" CreationDate="2016-03-09T16:28:59.770" Score="-2" Body="&lt;p&gt;My answer is 8. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question states that the least number of tries by the Guesser but doesn't not factor in the intelligence of the said Guesser. So take for example if the solution was &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1+1=2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the Guesser guessed   &lt;strong&gt;9,8,7,6,5,4,3&lt;/strong&gt;   he would need at least 8 tokens to be guaranteed a win. This I believe is the worst case scenario.&lt;/p&gt;&#xA;" OwnerUserId="19870" LastActivityDate="2016-03-09T16:28:59.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28687" PostTypeId="2" ParentId="28656" CreationDate="2016-03-09T16:37:48.960" Score="6" Body="&lt;p&gt;Brute force method using this code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;string[] cards = &#xA;{ &quot;CAR&quot;, &quot;CAR&quot;, &#xA;  &quot;C&quot;, &quot;C&quot;, &quot;C&quot;, &quot;C&quot;, &quot;C&quot;, &quot;C&quot;, &quot;C&quot;, &quot;C&quot;, &quot;C&quot;, &quot;C&quot;, &quot;C&quot;, &#xA;  &quot;A&quot;, &quot;A&quot;, &quot;A&quot;, &quot;A&quot;, &quot;A&quot;, &quot;A&quot;, &quot;A&quot;, &quot;A&quot;, &quot;A&quot;, &quot;A&quot;, &quot;A&quot;, &#xA;  &quot;R&quot;, &quot;R&quot;, &quot;R&quot;, &quot;R&quot;, &quot;R&quot;, &quot;R&quot;, &quot;R&quot;, &quot;R&quot;, &quot;R&quot;, &quot;R&quot;, &quot;R&quot; };&#xA;&#xA;&#xA;long count=0;&#xA;long hits = 0;&#xA;for (int i = 0; i &amp;lt; 35; i++)&#xA;{&#xA;    for (int j = 0; j &amp;lt; 35; j++)&#xA;    {&#xA;        if (j == i) { continue; }&#xA;&#xA;        for (int k = 0; k &amp;lt; 35; k++)&#xA;        {&#xA;            if (k == i || k==j) { continue; }&#xA;&#xA;            for (int l = 0; l &amp;lt; 35; l++)&#xA;            {&#xA;                if (l == i || l == j || l==k) { continue; }&#xA;&#xA;                for (int m = 0; m &amp;lt; 35; m++)&#xA;                {&#xA;                    if (m == i || m == j || m == k || m==l) &#xA;                    {&#xA;                        continue;&#xA;                    }&#xA;&#xA;                    count++;&#xA;                    string temp = cards[i] + cards[j] + cards[k] + cards[l] + cards[m];&#xA;                    if (temp.Contains(&quot;C&quot;) &amp;amp;&amp;amp; temp.Contains(&quot;A&quot;) &amp;amp;&amp;amp; temp.Contains(&quot;R&quot;))&#xA;                    {&#xA;                        hits++;&#xA;                    }&#xA;                }&#xA;            }&#xA;        }&#xA;    }&#xA;}&#xA;&#xA;Console.WriteLine(hits + &quot;/&quot; + count);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Gives the following output:&lt;br&gt;&#xA;29641920/38955840&lt;br&gt;&#xA;which comes out to &#xA;76.0911% chance of winning&lt;/p&gt;&#xA;" OwnerUserId="9353" LastEditorUserId="4369" LastEditDate="2016-03-09T19:44:49.483" LastActivityDate="2016-03-09T19:44:49.483" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28688" PostTypeId="2" ParentId="28669" CreationDate="2016-03-09T16:54:57.627" Score="7" Body="&lt;p&gt;Without having checked all possible cases, I've found a solution.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;E=3, R=6, C=7, A=1, N=5, U=4, Q=2, D=9, S=8&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A useful observation is that we must have S=D-1 and neither can be 0. &lt;br&gt;Then it seems as though when you pick E, your hand is forced on the other letters. &lt;br&gt;&lt;br&gt;For example, if we say E=3 then we immediately have to have R=6 and C=7. &lt;br&gt; Then A+N+1 ends in a 7, but with the digits available this means than the combination A+N must be 1+5, 5+1, 4+2, 2+4.&lt;br&gt; The next step, U+A=N forces U to be 4, 6, 8 or 2 respectively. &lt;br&gt;6 is already taken as would be 2 and if we pick the 8 option then we have left no consecutive pair of non-zero digits for S and D. Therefore, we can only have A=1, N=5, U=4. &lt;br&gt;&lt;br&gt; This means we must have S=8, D=9 as they are the only pair of non-zero consecutive digits left and to make this work we are forced to make Q=2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Observations&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There may be an easier way to solve this as we notice &lt;br&gt;&lt;br&gt; SQUARE + DANCE = 10*DANCE + R &lt;br&gt;&lt;br&gt; and so &lt;br&gt;&lt;br&gt; SQUARE = 9*DANCE + R &lt;br&gt;&lt;br&gt; which should lead to some nice properties.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18422" LastActivityDate="2016-03-09T16:54:57.627" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28689" PostTypeId="1" AcceptedAnswerId="28694" CreationDate="2016-03-09T16:59:43.170" Score="5" ViewCount="961" Body="&lt;p&gt;Adventuring through an old house, I happened upon a journal.  On the front of the notepad, embossed, was a large number 3.  Inside the journal was just pages, and pages, of numbers.  It seemed simple to me, at first, and then cryptic again, but after spending a few hours trying to make some sense of it, it finally dawned on me what the journal was talking about.  It was really quite simple, in the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you figure it out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The journal reads:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;68 122 68 124 35 113 114 122 35 118&lt;br&gt;&#xA;  119 120 115 104 113 103 114 120 118 35&lt;br&gt;&#xA;  118 120 101 120 117 101 100 113 35 111&lt;br&gt;&#xA;  100 103 35 100 117 112 124 35 100 121&lt;br&gt;&#xA;  104 113 106 104 35 103 114 122 113 117&lt;br&gt;&#xA;  108 106 107 119 35 119 107 104 112 35&lt;br&gt;&#xA;  108 113 121 114 110 104 35 108 117 114&lt;br&gt;&#xA;  113 124 35 48 35 101 100 101 124 118&lt;br&gt;&#xA;  108 119 35 106 108 106 35 104 119 107&lt;br&gt;&#xA;  108 102 100 111 35 108 112 115 117 114&lt;br&gt;&#xA;  121 35 48 35 72 107 35 113 108 113&lt;br&gt;&#xA;  109 100 35 114 101 104 124 35 120 118&lt;br&gt;&#xA;  35 48 35 71 114 122 113 122 100 117&lt;br&gt;&#xA;  103 111 124 35 110 108 35 108 117 114&lt;br&gt;&#xA;  113 118 35 106 108 47 35 112 114 117&lt;br&gt;&#xA;  100 119 114 117 108 120 112 35 113 104&lt;br&gt;&#xA;  104 103 118 47 35 111 100 122 118 120&lt;br&gt;&#xA;  108 119 35 119 104 119 107 104 117 35&lt;br&gt;&#xA;  106 100 112 101 111 104 35 48 35 101&lt;br&gt;&#xA;  104 104 107 108 121 104 35 104 123 119&lt;br&gt;&#xA;  117 100 35 124 114 35 122 108 119 107&lt;br&gt;&#xA;  104 117 104 103 35 117 114 118 104 118&lt;br&gt;&#xA;  47 35 48 35 87 100 101 114 114 35&lt;br&gt;&#xA;  120 113 118 102 100 119 107 104 103 35&lt;br&gt;&#xA;  48 35 112 100 110 104 120 115 35 101&lt;br&gt;&#xA;  117 100 121 104 111 124 35 111 114 122&lt;br&gt;&#xA;  35 104 112 114 119 108 121 104 35 104&lt;br&gt;&#xA;  113 120 117 104 47 35 113 104 122 108&lt;br&gt;&#xA;  118 107 35 100 108 117 115 111 100 113&lt;br&gt;&#xA;  104 118 47 35 102 117 124 115 119 114&lt;br&gt;&#xA;  35 117 120 112 47 35 100 113 106 104&lt;br&gt;&#xA;  111 108 102 35 121 108 111 104 35 48&lt;br&gt;&#xA;  35 100 103 112 108 117 104 35 100 35&lt;br&gt;&#xA;  104 123 100 102 119 35 103 108 118 107&lt;br&gt;&#xA;  35 116 120 104 120 104 35 104 119 107&lt;br&gt;&#xA;  104 117 108 125 104 35 104 107 47 35&lt;br&gt;&#xA;  73 120 77 108 35 120 117 113 118 47&lt;br&gt;&#xA;  35 119 107 104 35 102 100 112 104 117&lt;br&gt;&#xA;  100 35 118 110 100 119 104 118 47 35&lt;br&gt;&#xA;  70 100 118 118 108 120 118 35 117 100&lt;br&gt;&#xA;  102 104 118 47 35 115 118 124 102 107&lt;br&gt;&#xA;  114 35 118 120 113 47 35 108 113 110&lt;br&gt;&#xA;  111 108 113 106 35 114 107 35 122 107&lt;br&gt;&#xA;  124 35 48 35 122 108 119 107 103 117&lt;br&gt;&#xA;  100 122 35 119 108 112 35 100 119 119&lt;br&gt;&#xA;  100 101 114 124 47 35 112 108 103 113&lt;br&gt;&#xA;  114 114 113 47 35 103 114 122 113 117&lt;br&gt;&#xA;  108 106 107 119 35 100 107 100 47 35&lt;br&gt;&#xA;  119 114 112 101 114 124 118 47 35 112&lt;br&gt;&#xA;  100 110 104 117 104 100 103 124 35 104&lt;br&gt;&#xA;  118 104 47 35 68 113 87 108 119 107&lt;br&gt;&#xA;  104 118 108 118 35 118 108 119 35 120&lt;br&gt;&#xA;  113 122 114 112 100 113 111 124 35 48&lt;br&gt;&#xA;  35 100 117 100 102 107 113 114 108 103&lt;br&gt;&#xA;  35 100 35 120 111 119 117 100 107 120&lt;br&gt;&#xA;  112 100 113 35 119 100 123 35 100 111&lt;br&gt;&#xA;  76 101 108 35 112 108 102 104 35 107&lt;br&gt;&#xA;  104 123 100 106 114 113 35 100 106 117&lt;br&gt;&#xA;  104 104 118 47 35 77 104 122 108 118&lt;br&gt;&#xA;  107 35 73 100 120 118 119 35 48 35&lt;br&gt;&#xA;  68 118 108 100 113 35 100 120 103 108&lt;br&gt;&#xA;  119 114 117 47 35 68 103 80 108 117&lt;br&gt;&#xA;  100 101 111 124 35 114 101 118 102 104&lt;br&gt;&#xA;  113 104 35 103 114 109 114 35 120 117&lt;br&gt;&#xA;  113 118 35 108 119 102 107 35 103 104&lt;br&gt;&#xA;  118 119 117 114 124 35 112 104 119 104&lt;br&gt;&#xA;  117 104 103 35 122 100 118 115 124 35&lt;br&gt;&#xA;  41 35 101 117 108 118 119 111 124 35&lt;br&gt;&#xA;  111 100 106 35 108 113 103 108 106 114&lt;br&gt;&#xA;  35 48 35 102 100 118 108 113 114 35&lt;br&gt;&#xA;  48 35 118 114 105 119 102 114 121 104&lt;br&gt;&#xA;  117 35 48 35 115 107 124 111 111 114&lt;br&gt;&#xA;  103 108 120 112&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19944" LastEditorUserId="19944" LastEditDate="2016-03-09T17:23:22.020" LastActivityDate="2016-03-09T19:21:46.553" Title="Cubic Compression Conundrum" Tags="|riddle|word|story|cipher|" AnswerCount="3" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28690" PostTypeId="2" ParentId="28689" CreationDate="2016-03-09T17:24:12.117" Score="4" Body="&lt;p&gt;(Partial answer) The numbers translate to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; AwAy now stupendous suburban lad army avenge downright them invoke irony - babysit gig ethical improv - Eh ninja obey us - Downwardly ki irons gi, moratorium needs, lawsuit tether gamble - beehive extra yo withered roses, - Taboo unscathed - makeup bravely low emotive enure, newish airplanes, crypto rum, angelic vile - admire a exact dish queue etherize eh, FuJi urns, the camera skates, Cassius races, psycho sun, inkling oh why - withdraw tim attaboy, midnoon, downright aha, tomboys, makeready ese, AnTithesis sit unwomanly - arachnoid a ultrahuman tax alIbi mice hexagon agrees, Jewish Faust - Asian auditor, AdMirably obscene dojo urns itch destroy metered waspy &amp;amp; bristly lag indigo - casino - softcover - phyllodium&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12901" LastActivityDate="2016-03-09T17:24:12.117" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28691" PostTypeId="1" AcceptedAnswerId="28708" CreationDate="2016-03-09T18:08:46.663" Score="4" ViewCount="546" Body="&lt;p&gt;I put pieces together but i'm not a strategist.&lt;br&gt;&#xA;I give advice but I'm not a life coach.&lt;br&gt;&#xA;I come along with a message but I'm not a Prophet.&lt;br&gt;&#xA;I am not religious but I am twisted.&lt;br&gt;&#xA;No, I'm not your teacher, parent, nor a doomsayer.&lt;br&gt;&#xA;I am not &lt;em&gt;who&lt;/em&gt; you think I am.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am...?  &lt;/p&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2016-03-09T23:08:12.120" Title="I am not who you think I am" Tags="|riddle|wordplay|" AnswerCount="3" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28692" PostTypeId="2" ParentId="28691" CreationDate="2016-03-09T18:20:29.000" Score="4" Body="&lt;p&gt;I am...?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://twistedmatrix.com/trac/&quot; rel=&quot;nofollow&quot;&gt;Twisted Matrix&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I put pieces together but i'm not a strategist.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Data transmission happens in packets. These packets are reconstituted into a whole when received.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I give advice but I'm not a life coach.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This framework provides some advice for how to use it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come along with a message but I'm not a Prophet.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Messages are sent via the protocol.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am not religious but I am twisted.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Twisted.. this gave you away, my friend.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, I'm not your teacher, parent, nor a doomsayer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is true. You are none of these things.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am not who you think I am.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The name suggests &quot;matrix&quot;, but it is actually a framework.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2016-03-09T18:20:29.000" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28693" PostTypeId="2" ParentId="28689" CreationDate="2016-03-09T18:42:39.020" Score="3" Body="&lt;p&gt;Every third character gives the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An unusual man with very big hair&lt;br&gt;&#xA; Enjoys walking round with me everywhere&lt;br&gt;&#xA; Touch me all over while you give me a squeeze&lt;br&gt;&#xA; Just make sure you know what to do with my keys&lt;br&gt;&#xA; This way and that I change what i do&lt;br&gt;&#xA; Maybe just someday ill do so for you&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="20047" LastEditorUserId="17250" LastEditDate="2016-03-09T18:50:23.710" LastActivityDate="2016-03-09T18:50:23.710" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28694" PostTypeId="2" ParentId="28689" CreationDate="2016-03-09T18:46:51.163" Score="3" Body="&lt;p&gt;Based on the answers provided by Carl Löndahl and lythamlatic,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It seems to be a poem written by &lt;a href=&quot;http://wpmedia.ottawacitizen.com/2015/01/prv0819-yankovic-01-vancouver-bc-see-tom-harrison-story-au.jpg?quality=55&amp;amp;strip=all&quot; rel=&quot;nofollow&quot;&gt;Weird Al's accordion&lt;/a&gt;, giving way to another meaning of &lt;strong&gt;compression&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The process is as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Carl Löndahl started by &lt;strong&gt;&lt;em&gt;subtracting 3&lt;/em&gt;&lt;/strong&gt; from each of the numbers and then taking the ASCII values, giving us this:&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;em&gt;AwAy now stupendous suburban lad army avenge downright them invoke irony - babysit gig ethical improv - Eh ninja obey us - Downwardly ki irons gi, moratorium needs, lawsuit tether gamble - beehive extra yo withered roses, - Taboo unscathed - makeup bravely low emotive enure, newish airplanes, crypto rum, angelic vile - admire a exact dish queue etherize eh, FuJi urns, the camera skates, Cassius races, psycho sun, inkling oh why - withdraw tim attaboy, midnoon, downright aha, tomboys, makeready ese, AnTithesis sit unwomanly - arachnoid a ultrahuman tax alIbi mice hexagon agrees, Jewish Faust - Asian auditor, AdMirably obscene dojo urns itch destroy metered waspy &amp;amp; bristly lag indigo - casino - softcover - phyllodium&lt;/em&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Then lythamlatic took each letter that is at a position (1-indexed) congruent to 0 (mod 3), giving us this:&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;em&gt;An unusual man with very big hair&lt;br&gt;&#xA; Enjoys walking round with me everywhere&lt;br&gt;&#xA; Touch me all over while you give me a squeeze&lt;br&gt;&#xA; Just make sure you know what to do with my keys&lt;br&gt;&#xA; This way and that I change what i do&lt;br&gt;&#xA; Maybe just someday ill do so for you&lt;/em&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; From there, it was just a matter of interpreting the poem (Weird Al is a strange man with big hair, he's known for having an accordion, which is a squeezable instrument with keys, that changes what notes it outputs based on which keys and buttons are being pressed, as well as whether it's being squeezed or stretched)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2016-03-09T19:21:46.553" LastActivityDate="2016-03-09T19:21:46.553" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28695" PostTypeId="2" ParentId="28360" CreationDate="2016-03-09T19:39:18.627" Score="4" Body="&lt;p&gt;Sounds to me a bit like...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A camera flash (bulb).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll start to count while you solve this problem,&lt;br&gt;&#xA;and just relax here at the top of my column.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some cameras may flash at you to display the countdown when using the delay feature.  You can mount a camera on a tripod, and the flash is mounted on top of the camera.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Everywhere I go I let out a good whistle,&lt;br&gt;&#xA;but don't mess with me cause I carry a pistol.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everyone knows that characteristic whine of a recharging flash (from movies or disposable cameras). Be careful, because cameras can and will shoot you!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Typically speaking I reach senses of three,&lt;br&gt;&#xA;when you count those senses, you may find me. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The flash (sight), pop (sound), and chemical smell of a burnt-out bulb.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For if you hold me, I'm likely dead,&lt;br&gt;&#xA;why did you take me away from my bed? &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are not likely to be holding the actual bulb unless it has burnt out and needs replacing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I came with a flash and out like a light,&lt;br&gt;&#xA;but I don't really mind because life is so bright. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This one is pretty self-explanatory. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've been explosive at one point in history,&lt;br&gt;&#xA;what I truly am is not much of a mystery.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Flash was once achieved through the ignition of flash powder, creating a small explosion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7586" LastActivityDate="2016-03-09T19:39:18.627" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28696" PostTypeId="1" AcceptedAnswerId="28699" CreationDate="2016-03-09T20:00:07.250" Score="7" ViewCount="598" Body="&lt;p&gt;You are just walking over a bridge. You have 3/8 of the bridge behind you, 5/8 ahead. Suddenly you hear a train approaching from behind (the 3/8 end), but it´s not yet on the bridge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know that you can turn around and run the shorter distance towards the train, or keep direction and run the longer distance. Either way, you would make it off the bridge just in time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You run at 10 kph (or whatever velocity unit you prefer). How fast is the &#xA;train?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;If you need a calculator, you are overthinking this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="20048" LastEditorUserId="17482" LastEditDate="2016-03-09T20:31:24.010" LastActivityDate="2016-03-09T21:36:56.167" Title="Being on a bridge, what´s the fastest train you can just escape?" Tags="|calculation-puzzle|" AnswerCount="3" CommentCount="3" ClosedDate="2016-03-09T22:46:36.520" ContentLicense="CC BY-SA 3.0" />
  <row Id="28697" PostTypeId="2" ParentId="28696" CreationDate="2016-03-09T20:09:52.447" Score="5" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the train is travelling at 40kph&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Since the train travels the entire bridge in the same amount of time you travel a quarter of it, it is travelling at four times your speed. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18103" LastEditorUserId="18103" LastEditDate="2016-03-09T20:12:50.633" LastActivityDate="2016-03-09T20:12:50.633" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28699" PostTypeId="2" ParentId="28696" CreationDate="2016-03-09T20:32:18.763" Score="12" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; At the start you do not know the distance of the train to the bridge, but if you run towards the short end, you will cover 3/8 of the length of the bridge in the time it takes the train to get to the start of the bridge. If you ran the other way, you would also cover 3/8 of the length bridge when the train gets to the start of the bridge. This means you are 3/8+3/8 along on the bridge when the train gets to the start. You then can cover the remaining 1/4 of the bridge in the time it takes the train to cover the entire bridge.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The train is moving 4x as fast as you (i.e., 40 kph)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8984" LastActivityDate="2016-03-09T20:32:18.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28700" PostTypeId="1" AcceptedAnswerId="28719" CreationDate="2016-03-09T21:32:23.420" Score="9" ViewCount="411" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;It's Rodent the first in an oil bloc found&lt;br&gt;&#xA;  A twitch that tags on to a story in sound&lt;br&gt;&#xA;  Like Wotan it turns an irate cop around.&lt;br&gt;&#xA;  It's a long span of time, in a visual way.&lt;br&gt;&#xA;  &lt;em&gt;That&lt;/em&gt; is the word that I'm trying to say.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is a single English word. There are five clues, all found in the first four lines. This is a cryptic crossword style puzzle. The clues are four ways to construct the word and a definition. &lt;/p&gt;&#xA;" OwnerUserId="18924" LastEditorUserId="17147" LastEditDate="2016-03-10T06:12:04.830" LastActivityDate="2016-03-10T06:35:30.720" Title="Can You Tell What the Answer Is?" Tags="|riddle|wordplay|cryptic-clues|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28701" PostTypeId="2" ParentId="28696" CreationDate="2016-03-09T21:35:03.643" Score="2" Body="&lt;p&gt;You run at 10 kph (or whatever velocity unit you prefer).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I choose $10 c$ where $c$ is the velocity of light!!!&lt;br&gt;&#xA; I've now entered another dimension and piddly sluggish trains don't bother me in the slightest.&lt;br&gt;&#xA; I can get to places &lt;em&gt;before&lt;/em&gt; any current event occurs in time-space.&lt;br&gt;&#xA; I simply go back in time to when this approaching train was built and redesign the &quot;cow plow&quot; to gently scoop me up into a waiting (my new design) comfy sofa.&lt;br&gt;&#xA; Then zap back and await a ready transport across the bridge.&lt;br&gt;&#xA; No fuss at all.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastEditorUserId="10118" LastEditDate="2016-03-09T21:36:56.167" LastActivityDate="2016-03-09T21:36:56.167" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28702" PostTypeId="2" ParentId="28658" CreationDate="2016-03-09T21:39:37.363" Score="3" Body="&lt;p&gt;To me it suggests &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Yoyo. It keeps going away and coming back on opposite sides of the string. Of course the string isn't usually curved. Maybe he's doing some kind of trick. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastActivityDate="2016-03-09T21:39:37.363" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28703" PostTypeId="2" ParentId="28658" CreationDate="2016-03-09T22:06:33.133" Score="3" Body="&lt;p&gt;To my eyes...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The smiley-faces [now turn your head sideways...] could be depicting 2 people (the 2 dots) jumping on a trampoline (the smile). &lt;br/&gt;&lt;br/&gt;&#xA;Sad = the 2 jumping &lt;strong&gt;up and away&lt;/strong&gt; from the trampoline,&lt;br/&gt;&#xA;Smiley = 2 landing (&lt;strong&gt;coming back down&lt;/strong&gt;) to the trampoline.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19086" LastEditorUserId="19086" LastEditDate="2016-03-10T17:56:04.160" LastActivityDate="2016-03-10T17:56:04.160" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28704" PostTypeId="2" ParentId="28658" CreationDate="2016-03-09T22:11:52.853" Score="2" Body="&lt;h3&gt;This is not the intended answer...confirmed by OP&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;It can be something about &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Buses or trains that leaves the station and comes back. Maybe that's its a re&lt;strong&gt;bus&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-03-13T10:39:03.817" LastActivityDate="2016-03-13T10:39:03.817" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28705" PostTypeId="2" ParentId="28658" CreationDate="2016-03-09T22:15:03.040" Score="5" Body="&lt;p&gt;Or could be a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; boomerang&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;maybe?&lt;/p&gt;&#xA;" OwnerUserId="10876" LastActivityDate="2016-03-09T22:15:03.040" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28706" PostTypeId="2" ParentId="28658" CreationDate="2016-03-09T22:45:59.840" Score="7" Body="&lt;p&gt;Based on the brackets that make up the smiley / frowny faces:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A sine wave:&#xA; &lt;a href=&quot;https://i.stack.imgur.com/7l4eu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7l4eu.png&quot; alt=&quot;Sine wave diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12014" LastEditorUserId="12014" LastEditDate="2016-03-09T22:52:39.860" LastActivityDate="2016-03-09T22:52:39.860" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28707" PostTypeId="2" ParentId="28691" CreationDate="2016-03-09T22:59:47.693" Score="2" Body="&lt;p&gt;My guess is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hint&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hints helps putting things together.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hint is definitely an advice, how to solve a problem&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; hints always gave us a message :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it is twisted, because didnt see on stackexchange, just if you twist it with your mouse :D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; they gave us a lot of hints :D&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10876" LastActivityDate="2016-03-09T22:59:47.693" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28708" PostTypeId="2" ParentId="28691" CreationDate="2016-03-09T23:08:12.120" Score="6" Body="&lt;p&gt;I guess you are...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clippy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I put pieces together but i'm not a strategist.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You hold together pieces of paper&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I give advice but I'm not a life coach.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You make recommendations on how to improve the document I am composing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come along with a message but I'm not a Prophet.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You pop up with that little text bubble. OR - You show up when I am writing a letter and say, &quot;Looks like you're writing a letter!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am not religious but I am twisted.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Paperclips are, by their very nature, twisted.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, I'm not your teacher, parent, nor a doomsayer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your suggestions are no replacement for the advice of a teacher or parent. Despite how most people may feel, your appearance does not denote the end of the world. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am not who you think I am.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clippy's official name is Clippit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am...?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One obnoxious office assistant.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7586" LastActivityDate="2016-03-09T23:08:12.120" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28711" PostTypeId="2" ParentId="28658" CreationDate="2016-03-10T00:18:02.017" Score="2" Body="&lt;p&gt;Is it a:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Blinking SIgN?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It goes away and comes back and the smiley faces and frownie faces make a SIN wave&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If so:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it may be a lateral thinking problem as well. Or dare I say vertical... ;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19731" LastEditorUserId="19731" LastEditDate="2016-03-10T00:59:57.237" LastActivityDate="2016-03-10T00:59:57.237" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28712" PostTypeId="2" ParentId="28360" CreationDate="2016-03-10T00:29:35.857" Score="2" Body="&lt;p&gt;You are a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Flower.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll start to count while you solve this problem,&#xA;and just relax here at the top of my column. &#xA;Everywhere I go I let out a good whistle, &#xA;but don't mess with me cause I carry a pistol.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Flower's have pistils...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Typically speaking I reach senses of three, &#xA;when you count those senses, you may find me. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The sense of smell is normally the 3rd of 5 senses listed, and the one most associated with flowers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For if you hold me, I'm likely dead, &#xA;why did you take me away from my bed? &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Flowers grow in beds, and when you pick them, they die.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I came with a flash and out like a light, &#xA;but I don't really mind because life is so bright. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; They appear from the dark earth suddenly and grow toward the light.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've been explosive at one point in history, &#xA;what I truly am is not much of a mystery.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Perhaps this is referring to the Pre-Cambrian Explosion?  When plants proliferated around the world?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5683" LastActivityDate="2016-03-10T00:29:35.857" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28713" PostTypeId="2" ParentId="28644" CreationDate="2016-03-10T00:49:28.500" Score="5" Body="&lt;p&gt;1 Pound&lt;BR&gt;2 Baby animal&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; lam&lt;BR&gt;lamb  -&gt; B&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1 Tunnel&lt;BR&gt;2 Desire&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; cave&lt;BR&gt;crave  -&gt; R&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1 Wok&lt;BR&gt;2 Hurt&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; pan&lt;BR&gt;pain  -&gt; I&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1 Ready&lt;BR&gt;2 Complain&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ripe&lt;BR&gt;gripe  -&gt; G&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Overall 1 Ship&lt;BR&gt;Overall 2 Fetch&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; brig&lt;BR&gt;bring&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All those years of playing Sid Meier's Pirates! has finally paid off. :)&lt;/p&gt;&#xA;" OwnerUserId="15516" LastActivityDate="2016-03-10T00:49:28.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28714" PostTypeId="2" ParentId="28656" CreationDate="2016-03-10T00:49:35.313" Score="2" Body="&lt;p&gt;Inspired by @Kevin's answer (and not just because we share a name), I decided to see if I could get a good estimate by simply running many trials of the game and recording my wins and losses. This method gives relatively good estimates very quickly. Running a million trials (which only takes my computer a couple of seconds), gets me within a tenth of a percent of the actual answer (76.09% most runs. Not as accurate as Kevin's Brute Force method, but still fun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the code in case anyone wants to try it out for themselves (written in C#):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;private static void Main()&#xA;{&#xA;    Console.WriteLine(&quot;Enter number of trials&quot;);&#xA;    int numberOfTrials;&#xA;    int.TryParse(Console.ReadLine(), out numberOfTrials);&#xA;&#xA;    Random random = new Random();&#xA;&#xA;    int count = 0;&#xA;    int wins = 0;&#xA;    for (int i = 0; i &amp;lt; numberOfTrials; i++)&#xA;    {&#xA;        List&amp;lt;string&amp;gt; cards = new List&amp;lt;string&amp;gt;&#xA;        { &quot;CAR&quot;, &quot;CAR&quot;,&#xA;          &quot;C&quot;, &quot;C&quot;, &quot;C&quot;, &quot;C&quot;, &quot;C&quot;, &quot;C&quot;, &quot;C&quot;, &quot;C&quot;, &quot;C&quot;, &quot;C&quot;, &quot;C&quot;,&#xA;          &quot;A&quot;, &quot;A&quot;, &quot;A&quot;, &quot;A&quot;, &quot;A&quot;, &quot;A&quot;, &quot;A&quot;, &quot;A&quot;, &quot;A&quot;, &quot;A&quot;, &quot;A&quot;,&#xA;          &quot;R&quot;, &quot;R&quot;, &quot;R&quot;, &quot;R&quot;, &quot;R&quot;, &quot;R&quot;, &quot;R&quot;, &quot;R&quot;, &quot;R&quot;, &quot;R&quot;, &quot;R&quot; };&#xA;        count++;&#xA;        if (EvaluateTrial(cards, random)) wins++;&#xA;    }&#xA;&#xA;    Console.WriteLine($&quot;{wins} / {count} = {wins / (double)count * 100}%&quot;);&#xA;}&#xA;&#xA;private static bool EvaluateTrial(List&amp;lt;string&amp;gt; cards, Random random)&#xA;{&#xA;    string hand = &quot;&quot;;&#xA;&#xA;    for (int i = 0; i &amp;lt; 5; i++)&#xA;    {&#xA;        string nextCard = cards[random.Next(0, cards.Count)];&#xA;        hand += nextCard;&#xA;        cards.Remove(nextCard);&#xA;        if (hand.Contains(&quot;C&quot;) &amp;amp;&amp;amp; hand.Contains(&quot;A&quot;) &amp;amp;&amp;amp; hand.Contains(&quot;R&quot;)) return true;&#xA;    }&#xA;&#xA;    return false;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="18040" LastActivityDate="2016-03-10T00:49:35.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28715" PostTypeId="2" ParentId="2523" CreationDate="2016-03-10T02:20:40.687" Score="1" Body="&lt;p&gt;I'm not sure if this should count as a quick counterstrategy for the opponent but.. if there exists a true random number generator that can produce numbers in $[0,1]$ to use for the strategy Penguino posted, can't the opponent just use that same random number generator to produce two real numbers for this game to make it a 50% chance of winning?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to think that using the defined function that maps all real numbers to a range where you can use a random number generator as defined in Penguino's answer is good, but knowing that the player must stick to their mapping function, the opponent can create two sets, one that maps to a win and one to a loss and flip a coin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the given $M(x)$, the opponent can flip a coin and pick $0$ and $1$ if heads and $0$ and $-1$ if tails.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I couldn't post this as a response to the previous answers because I'm still very new to puzzling.stackexchange, I don't have enough points...)&lt;/p&gt;&#xA;" OwnerUserId="13349" LastActivityDate="2016-03-10T02:20:40.687" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28716" PostTypeId="1" AcceptedAnswerId="28718" CreationDate="2016-03-10T02:54:04.487" Score="8" ViewCount="771" Body="&lt;p&gt;You are the ruler of a medieval kingdom. This time, you've decided to throw a royal party with &lt;em&gt;10,000&lt;/em&gt; bottles of wine, rather than just 1,000.&lt;/p&gt;&#xA;&lt;p&gt;You remember the problems the courtier gave you with the wine &lt;a href=&quot;https://puzzling.stackexchange.com/questions/12770/find-the-composite-poison-in-the-wine-bottles&quot;&gt;last time&lt;/a&gt; — you search the kingdom high and low for a wine bottler who can produce 10,000 bottles in time for your party, but alas, the courtier is your only choice.&lt;/p&gt;&#xA;&lt;p&gt;You plead with the courtier not to poison the wine this time around, under the threat of refusing to pay for the wine received — alas, on the day the courtier delivers the wine, it comes with a note containing exactly what you'd dreaded.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Your Majesty,&lt;/p&gt;&#xA;&lt;p&gt;I must apologize deeply; for you see, a demon has besotted me. I simply cannot help myself. I have once again poisoned one of the wine bottles.&lt;/p&gt;&#xA;&lt;p&gt;Whoever should drink the poison shall die at the stroke of midnight on the day that it touches their lips, but only if they drink it in the morning, before noon. Should they drink it in the evening (or indeed any time after noon), they will die at the stroke of noon the next day.&lt;/p&gt;&#xA;&lt;p&gt;As you had so threatened me, so I shall not request payment for this wine.&lt;/p&gt;&#xA;&lt;p&gt;— Your Humble Courtier&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You grumble under your breath. You're going to have to kill some prisoners in testing again — you're running out of death row inmates in your dungeon to use for this!&lt;/p&gt;&#xA;&lt;p&gt;It's currently morning, and your party is set to start at the stroke of noon two days hence. What's the fewest number of prisoners you can use to test to find the poisoned bottle, and possibly force the courtier to drink it himself?&lt;/p&gt;&#xA;" OwnerDisplayName="user88" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-03-11T03:56:25.213" Title="Bottles of wine, poisoned with time" Tags="|logical-deduction|strategy|word-problem|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28717" PostTypeId="2" ParentId="28220" CreationDate="2016-03-10T03:24:08.263" Score="2" Body="&lt;p&gt;Here is just some of the damage possible&#xA;when personal correspondence falls into the wrong hands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the spirit of partial answers,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; each love letter might be a different part of the answer,&#xA; perhaps even an actual alphabetical letter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If so, the second love letter might refer to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &amp;ldquo;bow&amp;rdquo; or the writer's beau, Bo.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;When I'm not with you I sometimes think about undressing you,&#xA;&lt;em&gt;dismantling your clothing one thread at a time&lt;/em&gt;&#xA;so that it lasts much longer before I have to return to my day.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; hair bow? ribbon bows instead of buttons for clothing?&#xA; &amp;ensp; [not off to a great start here]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Your &lt;em&gt;natural beauty&lt;/em&gt; is so &lt;em&gt;bright and colourful&lt;/em&gt; I feel overwhelmed.&#xA;It's as if you &lt;em&gt;shower&lt;/em&gt; me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; rainbow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;with &lt;em&gt;arrows&lt;/em&gt; and play my heart-&lt;em&gt;strings&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; archery bow, bowstring&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;like a &lt;em&gt;violin&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; violin bow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;tying&lt;/em&gt; me up&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; bow tie&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;in &lt;em&gt;knots&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; bowknot&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The fourth love letter might refer to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; C or &amp;ldquo;sea.&amp;rdquo;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Without you, I &lt;em&gt;cannot see&lt;/em&gt;. It's as if I've &lt;em&gt;never seen&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; see why it might be C?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I continue about my business, wishing you were with me,&#xA;but really I'm just &lt;em&gt;marking time&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; in music, the marking for &lt;em&gt;common time&lt;/em&gt; is&amp;thinsp;&#xA; &lt;sub&gt;&lt;sub&gt;&lt;sub&gt;&lt;sub&gt;&lt;img src=&quot;https://i.stack.imgur.com/z1xts.png&quot; height=&quot;30&quot;&gt;&lt;/sub&gt;&lt;/sub&gt;&lt;/sub&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Creating&lt;/em&gt; new ways to &lt;em&gt;count the seconds&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;sup&gt;&lt;sup&gt;14&lt;/sup&gt;&lt;/sup&gt;C (carbon-14) dating&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;until I will see you once again.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; double CC ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="18129" LastActivityDate="2016-03-10T03:24:08.263" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28718" PostTypeId="2" ParentId="28716" CreationDate="2016-03-10T06:06:38.497" Score="6" Body="&lt;p&gt;I think the same generalized strategy described in &lt;a href=&quot;https://puzzling.stackexchange.com/a/12964/9044&quot;&gt;this answer&lt;/a&gt; applies. Though the time of death occurring after the end of the subsequent window makes the bookkeeping a little trickier, you can &quot;pipeline&quot; the wine (drinking the next batch while waiting to see if the first will kill you), but I don't see a way to use that to further reduce the needed number of prisoners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can identify the poisoned bottle with &lt;strong&gt;six&lt;/strong&gt; prisoners:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assign each prisoner a number 1–6, and each bottle a 6-digit label in base 5 (000000, 000001, ... 304444).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are four poison-able time windows before the party: today morning ($t=0$; die at midnight tonight), today evening ($t=1$; die tomorrow at noon), tomorrow morning ($t=2$; die at midnight at the end of tomorrow), and tomorrow evening ($t=3$; die at the commencement of the party—cutting it close!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During time window $t$, prisoner $n$ should drink from every bottle where the $n^{th}$ digit equals $t$. Record when each prisoner dies—if a prisoner doesn't die when the party starts, consider it a $4$. The times of death will indicate the poisoned bottle, which, if at least one prisoner survives, you may have to pluck from the hands of an eager guest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If you're uncomfortable cutting it that close, you can skip the the final time window, which only requires using one additional prisoner with 7-digit base-4 labels on the wine, up to 2130033).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, say that the poisoned bottle is $134401$. The following would take place:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;At $t=0$ (6:00am on the first day), prisoner 2 would drink from this bottle.  Prisoner 2 would also drink from many other bottles, and other prisoners would drink from other bottles, but no other prisoner would drink from this bottle at this time because the only 0 (current time slot) is in the second prisoner's position.&lt;/li&gt;&#xA;&lt;li&gt;At $t=1$ (6:00pm on the first day), prisoners 1 and 6 would drink from this bottle&lt;/li&gt;&#xA;&lt;li&gt;At midnight on the first day, prisoner 2 dies.  This is the first data point.  You then know that prisoner drank the poison during $t=0$.  So start to recreate the bottle number by putting a 0 in that prisoner's position: $XXXX0X$.&lt;/li&gt;&#xA;&lt;li&gt;At $t=2$ (6:00am on the second day), no one drinks.&lt;/li&gt;&#xA;&lt;li&gt;At noon on the second day, prisoners 1 and 6 die.  Record the poisoned bottle number as $1XXX01$.&lt;/li&gt;&#xA;&lt;li&gt;At $t=3$ (6:00pm on the second day), prisoner 5 drinks.&lt;/li&gt;&#xA;&lt;li&gt;At midnight on the second day, no one dies.  &lt;/li&gt;&#xA;&lt;li&gt;At noon on the third day (start of the party), prisoner 5 dies.  Record the poisoned bottle number as $13XX01$.&lt;/li&gt;&#xA;&lt;li&gt;No one else dies at noon, so you know that the last 2 prisoners (3 and 4) didn't drink from this bottle at all.  Record the number as $134401$.  Go find that bottle and remove it from the party, hopefully before anyone has had a chance to drink from it.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9044" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2016-03-11T03:56:25.213" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28719" PostTypeId="2" ParentId="28700" CreationDate="2016-03-10T06:16:29.843" Score="6" Body="&lt;p&gt;The word is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Operatic&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's Rodent the first in an oil bloc found&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; OPEC, with &quot;rat I&quot; (&quot;rat the first&quot;) inside&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A twitch that tags on to a story in sound&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Twitch = tic, story in sound = opera&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Like Wotan&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wotan is a character in Wagner's &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Der_Ring_des_Nibelungen&quot; rel=&quot;nofollow&quot;&gt;Ring Cycle&lt;/a&gt;&lt;/em&gt; operas&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;it turns an irate cop around.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Operatic is an anagram of &quot;irate cop&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's a long span of time, in a visual way.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;era&quot; inside &quot;optic&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can You Tell What the Answer Is?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The title contains a reference to &lt;a href=&quot;https://en.wikipedia.org/wiki/William_Tell_(opera)&quot; rel=&quot;nofollow&quot;&gt;William Tell&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17147" LastEditorUserId="17147" LastEditDate="2016-03-10T06:35:30.720" LastActivityDate="2016-03-10T06:35:30.720" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28721" PostTypeId="2" ParentId="28658" CreationDate="2016-03-10T06:54:34.073" Score="5" Body="&lt;p&gt;I think the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;FOR EVERY ACTION, THERE IS AN EQUAL AND OPPOSITE REACTION&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reason&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;I went away and I came back are actions. And for all those action there is a reaction (expression can also be considered as reaction) in the form of a smilie which are equal and opposite (sad and happy)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastActivityDate="2016-03-10T06:54:34.073" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28722" PostTypeId="1" AcceptedAnswerId="28742" CreationDate="2016-03-10T07:20:10.527" Score="5" ViewCount="332" Body="&lt;p&gt;Checkout these notes from a random page of my diary and tell what it indirectly refers to.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;p&gt;I have to join the Karate class next week.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;I bumped my head on the wall yesterday.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;I had my break fast early today.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;I met my childhood friend Thomas yesterday.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;We partied so hard yesterday night.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;I attended the job interview last wednesday.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;I lost my mobile last week.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;I will buy a new one tomorrow.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Check the tags.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The first and last notes are different from the rest.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="2730" LastEditDate="2016-11-19T12:19:50.807" LastActivityDate="2016-11-19T12:19:50.807" Title="Rebus: My Random Diary Notes" Tags="|wordplay|rebus|movies|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28723" PostTypeId="2" ParentId="28722" CreationDate="2016-03-10T08:29:57.143" Score="0" Body="&lt;p&gt;The easiest and least probable answer would be  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Notebook  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But, since that seems improbable, it may be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Three Weeks  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All of these actions mentioned in the question fall in a time span of three weeks, the previous, the current, and the next.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19656" LastActivityDate="2016-03-10T08:29:57.143" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28724" PostTypeId="1" AcceptedAnswerId="28727" CreationDate="2016-03-10T08:51:28.580" Score="3" ViewCount="591" Body="&lt;p&gt;The country Dalgonia issues stamps of only ten different denominations: $134$, $135$, $136$, $137$, $138$, $139$, $140$, $141$, $142$, and $143$ cents. &#xA;What is the largest amount of cents which cannot be made up with a combination of these stamps? &lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-10T09:29:25.933" LastActivityDate="2016-03-10T09:44:25.997" Title="Ten types of stamps" Tags="|mathematics|combinatorics|no-computers|frobenius|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28725" PostTypeId="2" ParentId="28724" CreationDate="2016-03-10T09:27:38.533" Score="2" Body="&lt;p&gt;What your basically asking is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What is the Frobenius Number of these numbers: 134,135,136,137,138,139,140,141,142,143&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thankfully, Wolfram Alpha has a built-in function for that&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://www.wolframalpha.com/input/?i=FrobeniusNumber%5B%7B134,135,136,137,138,139,140,141,142,143%7D%5D&quot; rel=&quot;nofollow&quot;&gt;https://www.wolframalpha.com/input/?i=FrobeniusNumber%5B%7B134,135,136,137,138,139,140,141,142,143%7D%5D&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 2009&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2016-03-10T09:27:38.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28726" PostTypeId="2" ParentId="28724" CreationDate="2016-03-10T09:34:52.730" Score="1" Body="&lt;p&gt;I'm not sure if this even qualifies as a puzzle; it's more of a math problem. There's a simple formula given on &lt;a href=&quot;https://en.wikipedia.org/wiki/Coin_problem#Arithmetic_sequences&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; for finding the Frobenius number for a given arithmetic sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Gy5uM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gy5uM.png&quot; alt=&quot;Frobenius number for AP&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can now substitute the values $a=134$, $d=1$ and $s=9$. Using it, we get the answer 2009 as already given by @IvoBeckers&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2016-03-10T09:34:52.730" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28727" PostTypeId="2" ParentId="28724" CreationDate="2016-03-10T09:44:25.997" Score="10" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Answer = 2009&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;None of the numbers less than 134 can be obtained with this set of stamps.&lt;br&gt;&#xA;All of the numbers in the range 134-143 can be obtained with a single stamp.&lt;br&gt;&#xA;None of the numbers between 143 and 2 * 134 can be obtained with a combination of these stamps.&lt;br&gt;&#xA;More generally, numbers that lie in the range between (n-1)&lt;em&gt;143 and n&lt;/em&gt;(134) cannot be obtained.&lt;br&gt;&#xA;Because the stamps are in increments of 1, every value between 134n and 143n can be obtained.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; So we need to discover where 134 * n is less than or equal to 143 * (n-1)&lt;br&gt;&#xA; Solving 134n = 143n - 143&lt;br&gt;&#xA; Gives 143 = 9n&lt;br&gt;&#xA; n=15.89&lt;br&gt;&#xA; So , for who number values of n greater than 15, there is no number that exists between the highest number obtainable with a combination of n-1 stamps and the lowest number that can be obtained with n stamps. All values above this can be achieved.&#xA; We need the highest number that exists in the gap below this, i.e. 15 * 134 -1&lt;br&gt;&#xA; Answer = 2009&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2016-03-10T09:44:25.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28728" PostTypeId="2" ParentId="28722" CreationDate="2016-03-10T09:48:30.203" Score="1" Body="&lt;h3&gt;This is not the intended answer...confirmed by OP&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I think&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; These time spans gives us the idea of when the movies were released. So 'today' might mean that it is on the theaters right now, 'Yesterday' means that it has been 2 or 3 years and so on...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I have to join the Karate class next week.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Kung-fu Panda 3&lt;/strong&gt; releasing this April.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I bumped my head on the wall yesterday.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Strike Back&lt;/strong&gt; (released in 2010)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I had my break fast early today.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;No idea&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I met my childhood friend Thomas yesterday.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Thomas and Friends : Curious Cargo&lt;/strong&gt; released in 2012&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We partied so hard yesterday night.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Search Party&lt;/strong&gt; released in 2014&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I attended the job interview last Wednesday.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Steve Jobs&lt;/strong&gt; released in 2015 or&lt;br&gt;&#xA; &lt;strong&gt;Pirates of Silicon Valley&lt;/strong&gt; released in 1999&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I lost my mobile last week.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Rob the Mob&lt;/strong&gt; released in 2014&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I will buy a new one tomorrow.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Raees&lt;/strong&gt; releasing this July... (Raees = Rich) :p&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-03-13T10:39:20.680" LastActivityDate="2016-03-13T10:39:20.680" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28729" PostTypeId="2" ParentId="28220" CreationDate="2016-03-10T10:20:48.823" Score="9" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Richard Dawkins (&lt;a href=&quot;https://en.wikipedia.org/wiki/Richard_Dawkins&quot;&gt;https://en.wikipedia.org/wiki/Richard_Dawkins&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Explanation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;We share so many &lt;em&gt;little&lt;/em&gt; memories that never fade, &lt;em&gt;persisting over long periods of time without changing&lt;/em&gt;. My love for you is how &lt;em&gt;they all combine&lt;/em&gt; into a beautiful whole that is so much more than any of them individually. I treasure these, holding on to them as if trying to &lt;em&gt;prevent anyone else from taking them away&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;little&lt;/em&gt;, &lt;em&gt;they all combine&lt;/em&gt;, &lt;em&gt;prevent anyone else from taking them away&lt;/em&gt; allude to:&lt;br&gt;&#xA; &quot;The Selfish Gene&quot;, &lt;a href=&quot;https://en.wikipedia.org/wiki/The_Selfish_Gene&quot;&gt;https://en.wikipedia.org/wiki/The_Selfish_Gene&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;When I'm not with you I sometimes think about undressing you, &lt;em&gt;dismantling your clothing one thread at a time&lt;/em&gt; so that it lasts much longer before I have to return to my day. Your &lt;em&gt;natural beauty&lt;/em&gt; is so &lt;em&gt;bright and colourful&lt;/em&gt; I feel overwhelmed. It's as if you &lt;em&gt;shower&lt;/em&gt; me with &lt;em&gt;arrows&lt;/em&gt; and play my heart-&lt;em&gt;strings&lt;/em&gt; like a &lt;em&gt;violin&lt;/em&gt;, &lt;em&gt;tying&lt;/em&gt; me up in &lt;em&gt;knots&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;dismantling your clothing one thread at a time&lt;/em&gt;, &lt;em&gt;natural beauty&lt;/em&gt;, &lt;em&gt;bright and colourful&lt;/em&gt; allude to:&lt;br&gt;&#xA; &quot;Unweaving the Rainbow&quot;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Unweaving_the_Rainbow&quot;&gt;https://en.wikipedia.org/wiki/Unweaving_the_Rainbow&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It's like I have a &lt;em&gt;hunger&lt;/em&gt; for you, to know more about you, to fathom how you can possibly be so &lt;em&gt;incredible&lt;/em&gt;. The more I know you, the more I am amazed by the intricacies of your beauty and character. This &lt;em&gt;admiration&lt;/em&gt; only &lt;em&gt;whets&lt;/em&gt; my &lt;em&gt;desire&lt;/em&gt; further.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;hunger&lt;/em&gt;, &lt;em&gt;incredible&lt;/em&gt;, &lt;em&gt;admiration&lt;/em&gt;, &lt;em&gt;desire&lt;/em&gt; allude to:&lt;br&gt;&#xA; &quot;An Appetite for Wonder&quot;, &lt;a href=&quot;https://en.wikipedia.org/wiki/An_Appetite_for_Wonder&quot;&gt;https://en.wikipedia.org/wiki/An_Appetite_for_Wonder&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Without you, I &lt;em&gt;cannot see&lt;/em&gt;. It's as if I've &lt;em&gt;never seen&lt;/em&gt;. I continue about my business, wishing you were with me, but really I'm just &lt;em&gt;marking time&lt;/em&gt;. &lt;em&gt;Creating&lt;/em&gt; new ways to &lt;em&gt;count the seconds&lt;/em&gt; until I will see you once again.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;cannot see&lt;/em&gt;, &lt;em&gt;never seen&lt;/em&gt;, &lt;em&gt;marking time&lt;/em&gt;, &lt;em&gt;Creating&lt;/em&gt; allude to:&lt;br&gt;&#xA; &quot;The Blind Watchmaker&quot;, &lt;a href=&quot;https://en.wikipedia.org/wiki/The_Blind_Watchmaker&quot;&gt;https://en.wikipedia.org/wiki/The_Blind_Watchmaker&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Before we met, my life felt like I was trudging up an infinite &lt;em&gt;mountain&lt;/em&gt;. Step after step; &lt;em&gt;hand-hold after foot-hold&lt;/em&gt;; toward a summit I wasn't sure was even there. But on the day I met you, I knew I had indeed reached the summit. I thought about it when I got home and realised how &lt;em&gt;unlikely&lt;/em&gt; our meeting was, how if any one of a million little things was different then it never would have happened, and that's how I knew.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;em&gt;mountain&lt;/em&gt;, &lt;em&gt;hand-hold after foot-hold&lt;/em&gt;, &lt;em&gt;unlikely&lt;/em&gt;  allude to:&lt;br&gt;&#xA; &quot;Climbing Mount Improbable&quot;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Climbing_Mount_Improbable&quot;&gt;https://en.wikipedia.org/wiki/Climbing_Mount_Improbable&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2016-03-10T10:20:48.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28730" PostTypeId="1" AcceptedAnswerId="28733" CreationDate="2016-03-10T10:24:00.793" Score="6" ViewCount="3524" Body="&lt;p&gt;You have a standard chessboard, consisting of $64$ squares ($32$ white and $32$ black) each of dimensions $4\times4$ cm, and you have a standard coin of $2$ cm diameter with tail and head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You flip the coin and let it drop on the board, and you somehow manage to guarantee that no part of the coin will stay out of the board. (This means that the coin cannot go off the board or stay in any corner of the board with one half on the board and its other half outside the board.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the chances that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;if you get tail, the coin would be inside a random white square as no part of the coin touches a black square (as defined before),&lt;/li&gt;&#xA;&lt;li&gt;if you get head, the coin would be inside a random black square as no part of the coin touches a white square?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="18514" LastEditorUserId="8874" LastEditDate="2016-03-12T18:22:27.123" LastActivityDate="2018-10-07T07:56:44.220" Title="A chess board and a coin!" Tags="|mathematics|geometry|probability|checkerboard|" AnswerCount="5" CommentCount="5" ClosedDate="2016-03-10T21:13:45.927" ContentLicense="CC BY-SA 3.0" />
  <row Id="28732" PostTypeId="2" ParentId="28730" CreationDate="2016-03-10T10:44:24.573" Score="9" Body="&lt;p&gt;I'm not 100% percent sure but I think it's the following&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Being heads or tails is independent of landing on a single square. So my guess is we only have to determine what the chance of  landing on a single square is and divide it by two.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If it were an &lt;span class=&quot;math-container&quot;&gt;$\infty\times\infty$&lt;/span&gt; board this would be a lot easier. The chance of landing only on a single square is when the center of the coin is on the &lt;span class=&quot;math-container&quot;&gt;$2\times2$&lt;/span&gt; cm square area in the center of the square. This is &lt;span class=&quot;math-container&quot;&gt;$1/4$&lt;/span&gt; of the area of the square so the final chance is &lt;span class=&quot;math-container&quot;&gt;$0.25\times0.5 = 0.125$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; But the border makes it a somewhat harder. Basically there is a 1cm width area along the border of the board where the center of the coin can't be. so the total area where the coin can fall is &lt;span class=&quot;math-container&quot;&gt;$(4\times 8-2)\times(4\times8-2) = 900$&lt;/span&gt; cm&lt;span class=&quot;math-container&quot;&gt;${}^2$&lt;/span&gt;. And there are &lt;span class=&quot;math-container&quot;&gt;$64 \times 4$&lt;/span&gt; cm&lt;span class=&quot;math-container&quot;&gt;${}^2 = 256$&lt;/span&gt; cm&lt;span class=&quot;math-container&quot;&gt;${}^2$&lt;/span&gt; where the coin is allowed to fall so I think the final answer is &lt;span class=&quot;math-container&quot;&gt;$256/900\times0.5 = 0.14222$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1748" LastEditorUserId="42452" LastEditDate="2018-10-07T07:55:01.000" LastActivityDate="2018-10-07T07:55:01.000" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28733" PostTypeId="2" ParentId="28730" CreationDate="2016-03-10T10:44:42.653" Score="16" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$128~/~30^2 ~~\approx~~ 0.14222$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Explanation:&lt;/h2&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The side length of the chessboard is &lt;span class=&quot;math-container&quot;&gt;$4\times8=32$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let us consider the possibilities, if no part of the coin stays out of the board. Then the center of the coin must land at least &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;cm away from the boundary, and hence must fall into a &lt;span class=&quot;math-container&quot;&gt;$30\times30$&lt;/span&gt; square with area &lt;span class=&quot;math-container&quot;&gt;$30^2$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let us consider the possibilities where the coin lands inside one of the white squares. Then the center of the coin must be at least 1cm away from the boundary of such a white square, and hence fall into a small &lt;span class=&quot;math-container&quot;&gt;$2\times2$&lt;/span&gt; square right in the middle of that white square. As there are &lt;span class=&quot;math-container&quot;&gt;$32$&lt;/span&gt; white squares, the total area of these small &lt;span class=&quot;math-container&quot;&gt;$2\times2$&lt;/span&gt; squares is &lt;span class=&quot;math-container&quot;&gt;$4\times32=128$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We conclude that the probability that the coin lands inside one of the white squares is &lt;span class=&quot;math-container&quot;&gt;$p=128/30^2$&lt;/span&gt;. A symmetric argument shows that the probability that the coin lands inside one of the black squares is also &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Finally the probability that the coin lands heads is &lt;span class=&quot;math-container&quot;&gt;$1/2$&lt;/span&gt;, and the probability that the coin lands tails is &lt;span class=&quot;math-container&quot;&gt;$1/2$&lt;/span&gt;. Hence the probability that the coin lands (heads AND in black square) OR (tails AND in white square) amounts to &lt;span class=&quot;math-container&quot;&gt;$(1/2)p+(1/2)p=p$&lt;/span&gt;. Done.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="42452" LastEditDate="2018-10-07T07:50:11.080" LastActivityDate="2018-10-07T07:50:11.080" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28734" PostTypeId="2" ParentId="28730" CreationDate="2016-03-10T11:08:08.797" Score="5" Body="&lt;p&gt;I think the answer should be :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\frac{128}{900} $ &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As given, the coin cannot fall out of the board, nor can it hang. So the total area which will count as the sample space is $30*30=900$.&lt;br&gt;&#xA;And, inside each white(or black) square, the favourable area is one-fourth of $  4*4$ i.e. $4 \space cm^2$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, there are 32 such favourable small squares for each black and white.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the probability:&#xA;$\frac{total \space favourable \space area}{total\space possible\space area}$&lt;br&gt;&#xA;Using independence of the two events (head/tails is independent of falling at random place on chessboard)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\frac{1}{2} * \frac{32*4+32*4}{900}$ &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which, on simplifying gives.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\frac{128}{900} $ or  $0.14\overline{2} $&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vl7be.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vl7be.jpg&quot; alt=&quot;Just a scribble&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1766" LastEditorUserId="1766" LastEditDate="2016-03-10T11:13:12.670" LastActivityDate="2016-03-10T11:13:12.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28735" PostTypeId="2" ParentId="28722" CreationDate="2016-03-10T11:18:21.030" Score="0" Body="&lt;p&gt;Do soap operas count as movies?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Days of Our Lives (&lt;a href=&quot;http://www.nbc.com/days-of-our-lives&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9422" LastActivityDate="2016-03-10T11:18:21.030" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28736" PostTypeId="1" AcceptedAnswerId="28774" CreationDate="2016-03-10T11:32:16.650" Score="8" ViewCount="370" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;What is this:&lt;/strong&gt;&lt;br&gt;&#xA;  Whenever things go awry, you'll have to look-out for him/her/it, and he/she/it may help you.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/bEE65.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bEE65.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;&lt;strong&gt;HINT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first picture shows Dhanush, the singer of the viral youtube video song  &quot;Kolaveri Di&quot;. Dig deeper and you'll find the answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19656" LastEditorUserId="8874" LastEditDate="2016-03-11T17:17:18.963" LastActivityDate="2016-03-11T17:17:18.963" Title="Managing problems since the 80s" Tags="|riddle|visual|rebus|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28739" PostTypeId="2" ParentId="28508" CreationDate="2016-03-10T12:25:55.433" Score="5" Body="&lt;p&gt;I think that the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the alphabet belongs to the Texas instruments graphing calculator TI-89&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Khale_Kitha has decoded the message as  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; welcome to our home&lt;br&gt;&#xA; if you are here to conquer us turn home now&lt;br&gt;&#xA; this society is highly advanced&lt;br&gt;&#xA; the table of elements contains actinium  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The atomic number of actinium is 89, and all letters in the encoded message are from keys on the Texas instruments graphing calculator TI-89.&lt;/p&gt;&#xA;" OwnerUserId="9125" LastActivityDate="2016-03-10T12:25:55.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28740" PostTypeId="1" AcceptedAnswerId="28741" CreationDate="2016-03-10T12:40:02.520" Score="11" ViewCount="1361" Body="&lt;p&gt;The game show host puts seven cards with the numbers 1, 2, 4, 8, 16, 32, 64 into a bowler hat. The candidate picks cards up randomly from the hat, one card at a time, and without putting cards back, until the sum of the numbers on cards in his hand exceeds 124. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the most probable sum the candidate can get in the end? On which sum should he bet?&lt;/p&gt;&#xA;" OwnerUserId="9125" LastActivityDate="2016-03-11T11:51:20.477" Title="Seven cards in a bowler hat" Tags="|mathematics|probability|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28741" PostTypeId="2" ParentId="28740" CreationDate="2016-03-10T12:52:57.267" Score="17" Body="&lt;p&gt;The following sets of cards exceed $124$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$64 + 32 + 16 + 8 + 4 + 2 + 1 = 127$&lt;/li&gt;&#xA;&lt;li&gt;$64 + 32 + 16 + 8 + 4 + 2 = 126$&lt;/li&gt;&#xA;&lt;li&gt;$64 + 32 + 16 + 8 + 4 + 1 = 125$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thus, you must always pick at least $6$ cards to exceed $124$. The unpicked card in this case must be a $1$ or a $2$, otherwise you will need to pick all $7$ cards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that there is a $\frac{2}{7}$ chance that you will have a sum less than $127$. The most likely value, therefore, is $127$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expected value $E$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}E&#xA;&amp;amp; = 127 \times \frac{5}{7} + 126 \times \frac{1}{7} + 125 \times \frac{1}{7} \\&#xA;&amp;amp; = \frac{127 \times 5 + 126 + 125}{7} \\&#xA;&amp;amp; = \frac{886}{7} \\&#xA;&amp;amp; = 126 \frac{4}{7}&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="68" LastEditDate="2016-03-11T11:49:25.107" LastActivityDate="2016-03-11T11:49:25.107" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28742" PostTypeId="2" ParentId="28722" CreationDate="2016-03-10T13:11:10.507" Score="8" Body="&lt;p&gt;How about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Back to the Future&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;since&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first and last notes refer to future events, but the middle is all recounting past events.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastActivityDate="2016-03-10T13:11:10.507" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28743" PostTypeId="2" ParentId="28730" CreationDate="2016-03-10T13:19:59.273" Score="0" Body="&lt;p&gt;should be around&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 7.11%&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;let me try to explain&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let's say we go for the tail on pure white landing. &lt;P&gt;We have &lt;strong&gt;50%&lt;/strong&gt; chance to get tail.&lt;P&gt; Another &lt;strong&gt;50%&lt;/strong&gt; to land on a white square &lt;P&gt;Then we should figure out the chance of the coin landing not on the near square. Here we have 3 sub-set of possibility: &lt;P&gt;1) &quot;center&quot; which is the 6x6 grid on the inside; &lt;P&gt;2) the 4 &quot;corner&quot; &lt;P&gt;3) and the remaining &quot;side&quot;. &lt;P&gt;This happen because you said that the coin &lt;strong&gt;can't&lt;/strong&gt;  go outside the board, making the landing in corners and sides easyer for our purpose.For calculation purpose i say that we see where the center of the coin rest after the toss. &lt;p&gt; on a &quot;Center&quot; square we are surrounded by black square, this means that if our coin rest anywhere near 1cm from the border it will be touching the next square. Said that we can say that 4cm&lt;sup&gt;2&lt;/sup&gt; of the 16cm&lt;sup&gt;2&lt;/sup&gt; are good for us. &lt;p&gt; As I said on &quot;sides&quot; is easyer. we are surrounded only by 3 black square and the center of the coin can't rest too close the external border of the board cutting down the possible space to 4cm&lt;sup&gt;2&lt;/sup&gt; over 12cm&lt;sup&gt;2&lt;/sup&gt; max. &lt;p&gt; similar to the &quot;sides&quot;, the &quot;corner&quot; squares have 4cm&lt;sup&gt;2&lt;/sup&gt; over 9cm&lt;sup&gt;2&lt;/sup&gt; making this the easyest toss.&lt;p&gt;&lt;p&gt; putting all of this toghether we have: 256 cm&lt;sup&gt;2&lt;/sup&gt; over a total of 900cm&lt;sup&gt;2&lt;/sup&gt; of good surface where to land our coin, half of it is white and half of that is the time we get tail on our coin. do your math and 7.11%&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;i' m really sorry if i messed up something (more probably a lot) with my english but it isn't my first lenguage :/&lt;/p&gt;&#xA;" OwnerUserId="19980" LastActivityDate="2016-03-10T13:19:59.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28744" PostTypeId="2" ParentId="28730" CreationDate="2016-03-10T13:50:52.727" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If we know the probability &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; that the coin will fall inside a single square, then the answer to this question will be &lt;span class=&quot;math-container&quot;&gt;$\frac12p$&lt;/span&gt;, since given this event there is still a &lt;span class=&quot;math-container&quot;&gt;$50\%$&lt;/span&gt; chance that the colour condition will fail (this is so whether the coin is fair on not!). To know whether the coin is inside a single square, it suffices to know the coordinates of its centre: if both the &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; coordinates lie at least at distance &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; from the coordinate of dividing line between squares, that is of a number divisible by &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt;, then we are OK, and otherwise we are not.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Since the coin falls entirely inside the board, the centre has both coordinates in the range &lt;span class=&quot;math-container&quot;&gt;$[1,31]$&lt;/span&gt;. The &quot;good&quot; coordinates are in one of the ranges &lt;span class=&quot;math-container&quot;&gt;$[1,3],[5,7],\ldots,[29,31]$&lt;/span&gt;, which accumulate a length Of &lt;span class=&quot;math-container&quot;&gt;$8\times2=16$&lt;/span&gt;. The the probability for one coordinate to be good is &lt;span class=&quot;math-container&quot;&gt;$16/30=8/15$&lt;/span&gt;. Then &lt;span class=&quot;math-container&quot;&gt;$p=(8/15)^2$&lt;/span&gt;, and the outcome of the puzzle is &lt;span class=&quot;math-container&quot;&gt;$\frac12p=\frac{32}{225}\approx0.14222$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1521" LastEditorUserId="42452" LastEditDate="2018-10-07T07:56:44.220" LastActivityDate="2018-10-07T07:56:44.220" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28745" PostTypeId="2" ParentId="28740" CreationDate="2016-03-10T13:51:15.873" Score="10" Body="&lt;p&gt;The player has to pick at least $6$ cards whatever number on the card is at the beginning since the sum of biggest numbers does not exceed $124$ until $6$ cards are picked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the player picks $4$ or  $8$ or $16$ or $32$ or $64$ as a the last card, he has to pick all cards since the total would not exceed $124$ until the last card ispicked!&lt;br&gt;&#xA;The chance of the last card being any of these numbers is $\frac{5}{7}$ and the sum would be $127$ for these cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\frac{5}{7} &amp;gt; 50 \%$, you do not have to calculate the rest and the most probably sum would be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $127$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18514" LastEditorUserId="68" LastEditDate="2016-03-11T11:51:20.477" LastActivityDate="2016-03-11T11:51:20.477" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28746" PostTypeId="2" ParentId="26137" CreationDate="2016-03-10T14:59:35.763" Score="4" Body="&lt;p&gt;In each column, the result digit other than N occurs in the sum. Taking the common corresponding digits out of both parts doesn't change the equation or the mapping of letters to digits, but it leaves a convenient result: 1000000000. Let's work with this variant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number the columns 0 to 9 from left to right. E.g. SUN contains an N in column 9. Let c$i$ be the carry into column $i$. For example, if column 9 summed to 20, we carry 2 into column 8, so c$8$=2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Working our way from columns 1 to 9 (c0=N from column 0):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] N + I + c1 = 10N&#xA;[2] 2N + O + c2 = 10c1&lt;br&gt;&#xA;[3] E + I + N + S + c3 = 10c2&lt;br&gt;&#xA;[4] B + L + N + E + O + c4 = 10c3&lt;br&gt;&#xA;[5] 2L + U + I + E + N + c5 = 10c4&lt;br&gt;&#xA;[6] 3T + L + N + I + U + c6 = 10c5&lt;br&gt;&#xA;[7] 2T + 2E + S + O + I + B + c7 = 10c6&lt;br&gt;&#xA;[8] 2L + U + S + I + O + E + N + c8 = 10c7&lt;br&gt;&#xA;[9] 4E + 3N + T = 10c8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider [9]: LHS min, max respectively occur when (E,N,T) = (0,1,2) and (9,8,7), i.e. 5 $\leq$ LHS $\leq$ 67. But RHS is a multiple of 10, making LHS 10, 20, 30, 40, 50 or 60, so c8 $\in$ [1,6]. Working through each equation independently but carrying the c$i$ iteratively from [9] to [2], we have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;     c8 c7 c6 c5 c4 c3 c2 c1 c0=N&#xA;min: 1  3  2  2  2  2  1  1  1&#xA;max: 6  5  6  5  4  3  3  2  1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;S,L,U,B,E,N are all nonzero because they appear at the start of a word, so one of O,I,T,Y is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have &lt;strong&gt;N=1&lt;/strong&gt; from the 'carry' table above.&lt;br&gt;&#xA;From [1], I + c1 = 9, so I=7 or I=8, i.e. I is nonzero.&lt;br&gt;&#xA;From [9], T must be odd and hence nonzero.&lt;br&gt;&#xA;From [2], O must be at least 5 and hence nonzero.&lt;br&gt;&#xA;So by elimination, &lt;strong&gt;Y=0&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Revising c$i$, we get:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;     c8 c7 c6 c5 c4 c3 c2 c1 c0=N&#xA;min: 2  4  4  3  2  2  2  1  1&#xA;max: 4  5  6  5  4  3  2  1  1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note that c1 and c2 are now determined, so from [1] &lt;strong&gt;I=8&lt;/strong&gt;, and from [2] &lt;strong&gt;O=6&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set E to 2,3,4,5,7,9 in turn to find simultaneously valid but different S in [3], T in [9] and B in [7]. (E,S,T,B) can only be (2,7,9,3), (5,3,7,4) or (5,4,7,3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting each (E,S,T,B) in turn into [4] to get L and [5] to get U, we get (E,S,T,B,L,U) = (2,7,9,3,5,4) or (5,3,7,4,2,9). (The case with B=4 is rejected because c4=2 in [4] and c4=3 in [5].)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting each into [6], the case with U=9 requires c6=9 (reject - see 'carry' table). The other case has c6=5, which is legal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We therefore end up with &lt;strong&gt;NEBULOSITY = 1234567890&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="9422" LastActivityDate="2016-03-10T14:59:35.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28747" PostTypeId="1" AcceptedAnswerId="28748" CreationDate="2016-03-10T15:28:43.343" Score="7" ViewCount="399" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;As an animal I was&lt;br&gt;&#xA;  once named by a roe&lt;br&gt;&#xA;  On other occasions&lt;br&gt;&#xA;  I shift to and fro  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I live with my brethren,&lt;br&gt;&#xA;  together we stand,&lt;br&gt;&#xA;  If given free reign&lt;br&gt;&#xA;  We'll take o'er the land  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I'm plucky, I'm friendly&lt;br&gt;&#xA;  I have many names&lt;br&gt;&#xA;  I'm thorny, I'm driven&lt;br&gt;&#xA;  and girls use me for games  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;My homophonous friend&lt;br&gt;&#xA;  can create quite a treat&lt;br&gt;&#xA;  Just give it some friends&lt;br&gt;&#xA;  and add lots of heat.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?  Please explain.&lt;/p&gt;&#xA;" OwnerUserId="19944" LastActivityDate="2016-03-10T15:45:28.103" Title="The land-owners of yore" Tags="|riddle|word|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28748" PostTypeId="2" ParentId="28747" CreationDate="2016-03-10T15:41:42.873" Score="9" Body="&lt;p&gt;You are a  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Flower  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Paragraph 1:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The skunk from Bambi was named Flower, by Bambi the roe deer  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Paragraph 2:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Flowers grow in bunches, and spread almost indefinitely from generation to generation if not controlled  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Paragraph 3:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You pluck flowers, give them to friends (typically those of romantic interest), there are many species of flowering plant&lt;br&gt;&#xA;Some are thorny, not sure on the 'driven' part&lt;br&gt;&#xA;Games such as 'He loves me, he loves me not' require a flower  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Paragraph 4:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Flour can be used to make a cake!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11717" LastEditorUserId="11717" LastEditDate="2016-03-10T15:45:28.103" LastActivityDate="2016-03-10T15:45:28.103" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="28749" PostTypeId="2" ParentId="28740" CreationDate="2016-03-10T15:44:19.013" Score="4" Body="&lt;p&gt;You should bet on &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;127&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it is five times better than the either of the other choices.  All the matters is what the last card is.  If the last card is the 1, you will not draw it and your sum will be 126.  If your last card is the 2, you will not draw it and your sum will be 125.  If your last card is any other card, you will draw all 7 and your sum will be 127.  So, you have a  $\frac{1}{7}$ chance of getting 125, a  $\frac{1}{7}$ chance of getting 126 and a  $\frac{5}{7}$ chance of getting 127.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9353" LastActivityDate="2016-03-10T15:44:19.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28750" PostTypeId="2" ParentId="28736" CreationDate="2016-03-10T16:42:35.573" Score="3" Body="&lt;p&gt;Partial answer here as I can't figure out any more of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First image might refer to... &lt;strong&gt;CONFIRMED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Three&lt;/strong&gt;, from 3, the movie that &quot;Why This Kolaveri Di&quot; features in.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second image could be...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fries, McDonalds, M, french fries&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Third may be...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hitler, heil, leader, Third Reich, Fuhrer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fourth may be...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Piano, keys, ivory, keyboard&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7586" LastEditorUserId="7586" LastEditDate="2016-03-11T15:19:50.530" LastActivityDate="2016-03-11T15:19:50.530" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28751" PostTypeId="2" ParentId="28490" CreationDate="2016-03-10T17:21:25.570" Score="2" Body="&lt;p&gt;I solved this by considering modulus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once we can make a number x from the stamps, we can then make any x + 5y by adding y more 5-cent stamps. So, once we can make a number x for each modulus of 5, we can make all the remaining numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we can make a chart of n, where n is the number of 17-cent stamps, show its value in cents (n*17), and show which modulus of 5 it corresponds to. Once we have found all the modulus 5 values, we can make any number thereafter.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;n   n*17    n*17 modulus 5&#xA;0   0       0&#xA;1   17      2&#xA;2   34      4&#xA;3   51      1&#xA;4   68      3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Once we reach 68, we can create all modulus 5 values thereafter. Since 68 is the first number we can create with modulus 5 = 3, then the previous modulus 5 = 3 value (68-5=63) must have been the last number we couldn’t create.&lt;/p&gt;&#xA;" OwnerUserId="6137" LastActivityDate="2016-03-10T17:21:25.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28752" PostTypeId="1" AcceptedAnswerId="28755" CreationDate="2016-03-10T17:24:05.717" Score="16" ViewCount="1095" Body="&lt;p&gt;&quot;You won't believe this!&quot; my colleague said, shaking with excitement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He dragged me into his office.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Before I show you this discovery,&quot; he said, &quot;let me fill you in on Saharan mythology.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh, boy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dr. Henry had recently discovered the remains of a 10,000 year old civilization in the Sahara, and is naturally the leading expert in the field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;According to Saharan mythology,&quot; he started lecturing, &quot;Knowledge was stolen from the gods and brought to mankind by Daa, god of journeys.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nodding politely seemed the right thing to do at this point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The other gods where not happy about this, of course, so they banished Daa and made him mortal.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Reminds of Prometheus,&quot; I noted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;And of Adam and Eve,&quot; Dr. Henry said. &quot;But don't forget that the Saharans were around long before any other civilization. They might even be the ones who have brought civilization to the rest of mankind!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He was getting that look. I had to stop him before he got carried away. &quot;There was somerthing you wanted to show me?&quot; I prompted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He blinked at me. &quot;Right. After Daa was banished to the mortal world, he gave the Saharans three stone tablets, and made them swear an oath to protect the tablets with their lives.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Now he sounds like Moses.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;These tablets were said to hold the Knowledge of the gods, and the key to civilization.&quot; At this point, his voice became frantic. He licked his lips. &quot;And I have found them.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I barked a laugh. &quot;You found mythological stones made by a god?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;See for yourself,&quot; he said, leading me to a wooden crate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gEhg6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My smile faded a little. &quot;There's just two,&quot; I said.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;And they are in horrible condition,&quot; he added. &quot;But I can assure you, these are the very tablets the Saharans guarded for generations.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I leaned closer for a better look. &quot;What are all these lines?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;A later, unfortunate addition,&quot; Dr. Henry said. &quot;But you can still see some of the original markings.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;And these are suppose to hold the key to all civilization?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;According to Saharan mythology, yes.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I studied the stones. It just couldn't be... could it?&lt;/p&gt;&#xA;" OwnerUserId="18558" LastActivityDate="2016-03-12T22:00:15.720" Title="Origin of Civilization" Tags="|enigmatic-puzzle|visual|story|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28753" PostTypeId="2" ParentId="28669" CreationDate="2016-03-10T18:48:13.147" Score="3" Body="&lt;p&gt;To do this exhaustively was painful.  Perhaps there is a better way to eliminate some possibilities earlier?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will note that $C_i$ is the carry over of column $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have the following equations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$S+C_5 = D \implies C_5=1, D=S+1$ since otherwise, $S=D$ would be a contradiction&lt;/li&gt;&#xA;&lt;li&gt;$Q+D+C_4 =10+A$ since it must generate carry over $C_5=1$&lt;/li&gt;&#xA;&lt;li&gt;$U+A+C_3 \in \{N, 10+N\}$&lt;/li&gt;&#xA;&lt;li&gt;$A+N+C_2 \in \{C, 10+C\}$&lt;/li&gt;&#xA;&lt;li&gt;$R+C+C_1 \in \{E, 10+E\}$&lt;/li&gt;&#xA;&lt;li&gt;$E+E \in \{R, 10+R\}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Lets work with values of $E$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $E=1, R=2, C_1=0$, then &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$R+C+C_1 = 2+C = 10+E = 11 \implies C=9,C_2=1$.  &lt;/li&gt;&#xA;&lt;li&gt;$A+N+C_2=A+N+1=9 \implies A+N=8, C_3=0$.&lt;/li&gt;&#xA;&lt;li&gt;So $A \in \{0,3,5,8\}, N\in\{8,5,3,0\}$.&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $A=0, N=8$, then $U+A+C_3=U=N$ &lt;/li&gt;&#xA;&lt;li&gt;If $A=3, N=5$, then $U+A+C_3=U+3=5 \implies U=2$&lt;/li&gt;&#xA;&lt;li&gt;If $A=8, N=0$, then $U+A+C_3=U+8=10 \implies U=2$&lt;/li&gt;&#xA;&lt;li&gt;If $A=5, N=3$, then $U+A+C_3=U+5=13 \implies U=8 and C_4=1$&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Thus $Q+D+C_4=Q+9+1=10+Q \implies Q=A$&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If $E=2, R=4, C_1=0$, then  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$R+C+C_1=4+C=10+E=12 \implies C=8,C_2=1$.  &lt;/li&gt;&#xA;&lt;li&gt;$A+N+C_2=A+N+1 \in \{8,18\} \implies A+N \in \{7, 17\}$. 8 is taken so $A+N=7, C_3=0$.  &lt;/li&gt;&#xA;&lt;li&gt;If $A,N \in\{1,6\}$, then there are no options for $D=S+1$, so $A,N \in \{0,7\}$ and $S=5, D=6$.&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $A=0, N=7$, then $U+A+C_3=U=N$&lt;/li&gt;&#xA;&lt;li&gt;If $A=7, N=0$, then $U+A+C_3=U+7=10+N=10 \implies U=3, C_4=1$.&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$Q+D+C_4=Q+6+1=Q+7=7 \implies Q=0$&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If $E=4, R=8, C_1=0$, then&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$R+C+C_1=8+C=10+E=14 \implies C=6,C_2=1$.  &lt;/li&gt;&#xA;&lt;li&gt;$A+N+C_2=A+N+1 \in \{6,16\} \implies A+N \in \{5, 15\}$.  Thus, $A+N=5$. &lt;/li&gt;&#xA;&lt;li&gt;If $A,N \in \{2,3\}$, then there are no options for $D=S+1$, so $A,N \in \{0,5\}, C_3=0$.&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $A=0, N=5$, then $U+A+C_3=U=N$&lt;/li&gt;&#xA;&lt;li&gt;If $A=5, N=0$, then $U+A+C_3=U+5=10+N=10 \implies U=A$&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If $E=5, R=0, C_1=1$, then  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$R+C+C_1=0+C+1=E=5 \implies C=4,C_2=0$.  &lt;/li&gt;&#xA;&lt;li&gt;$A+N+C_2=A+N \in \{4,14\}$.&lt;/li&gt;&#xA;&lt;li&gt;So $A \in \{1,3,6,7\}, B \in \{3,1,7,6\}$&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $A=1, N=3, C_3=0$, so $U+A+C_3=U+1=N=3 \implies U=2, C_4=0$. But  $Q+D+C_4=Q+D=11$ leaves no possible values for $Q,D$.&lt;/li&gt;&#xA;&lt;li&gt;If $A=3, N=1, C_3=0$, so $U+A+C_3=U+3=10+N=11 \implies U=8, C_4=1$. But  $Q+D+C_4=Q+D+1=10+A=13$ leaves no solution for $Q+D=12$&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If $E=6, R=2, C_1=1$, then&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$R+C+C_1=2+C+1=E=6 \implies C=3,C_2=0$.  &lt;/li&gt;&#xA;&lt;li&gt;$A+N+C_2=A+N \in \{3,13\} \implies A+N=13, C_3=1$. &lt;/li&gt;&#xA;&lt;li&gt;If $A,N \in \{5,8\}$, then there are no options for $D=S+1$, so $A,N \in \{4,9\}$ and $D=8, S=7$.&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $A=4, N=9$, then $U+A+C_3=U+4=N=9 \implies U=5$&lt;/li&gt;&#xA;&lt;li&gt;If $A=9, N=4$, then $U+A+C_3=U+9=10+N=14 \implies U=5$ &lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;$Q+D+C_4 \in \{4, 14\}$ has no solution with the remaining digits $\{0,1,5\}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If $E=7, R=4, C_1=1$, then&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$R+C+C_1=4+C+1=E=7 \implies C=2,C_2=0$.  &lt;/li&gt;&#xA;&lt;li&gt;$A+N+C_2=A+N \in \{2,12\} \implies A+N=12, C_3=1$&lt;/li&gt;&#xA;&lt;li&gt;$A,C\in\{3,9\}$ and $S=5, D=6$.&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $A=3, N=9$, then $U+A+C_3=U+3+1=U+4=9 \implies U=5$&lt;/li&gt;&#xA;&lt;li&gt;If $A=9, N=3$, then $U+A+C_3=U+9+1=10+N \implies U=N$&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If $E=8, R=6, C_1=1$, then&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$R+C+C_1=6+C+1=E=8 \implies C=1,C_2=0$.&lt;/li&gt;&#xA;&lt;li&gt;$A+N+C_2=A+N \in \{1,11\} \implies A+N=11, C_3=1$.&lt;/li&gt;&#xA;&lt;li&gt;Thus, $A\in \{2,9,4,7\}, N\in\{9,2,7,4\}$.&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $A=2, N=9$, then $U+A+C_3=U+2+1=N=9 \implies U=6$&lt;/li&gt;&#xA;&lt;li&gt;If $A=9, N=2$, then $U+A+C_3=U+9+1=10+N \implies U=N$&lt;/li&gt;&#xA;&lt;li&gt;If $A=4, N=7$, then $U+A+C_3=U+4+1=N=7 \implies U=2$, which leaves no solution for $D=S+1$.&lt;/li&gt;&#xA;&lt;li&gt;If $A=7, N=4$, then $U+A+C_3=U+7+1=20+N=24 \implies U=6$&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If $E=9, R=8, C_1=1$, then&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$R+C+C_1=8+C+1=E=9 \implies C=0,C_2=0$.&lt;/li&gt;&#xA;&lt;li&gt;$A+N+C_2=A+N=10, C_3=1$.  &lt;/li&gt;&#xA;&lt;li&gt;So $A\in\{3,7,4,6\}, N\in\{7,3,6,4\}$.&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $A=3, N=7$, then $U+A+C_3=U+3+1=U+4=7 \implies U=3$.&lt;/li&gt;&#xA;&lt;li&gt;If $A=7, N=3$, then $U+A+C_3=U+7+1=U+8=13 \implies U=5, C_4=1$.  This leaves $S=1, D=2$.  &#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$Q+D+C_4=Q+2+1=Q+3=10+A=17 \implies Q=14$&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;If $A=4, N=6$, then $U+A+C_3=U+4+1=U+5=6 \implies U=1, C_4=0$. This leave $S=2, D=3$.&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$Q+D+C_4=Q+3=10+A=14 \implies Q=11$&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;If $A=3, N=7$, then $U+A+C_3=U+3+1=U+4=7 \implies U=3$.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thus, $E=3, R=6, C_1=0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also know that $R+C+C_1=6+C=10+E=13 \implies C=7, C_2=1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, $A+N+C_2=A+N+1 \in \{7,17\}$.  Thus, $A+N \in \{6,16\}$.  But 16 is not possible with 7 already taken, so $A+N=6, C_3=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that $A\in \{1,5,2,4\}, N\in\{5,1,4,2\}$.  Either way, the only solution for $D=S+1$ is $S=8, D=9$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $A=5, N=1$, then $U+A+C_3=U+5=10+N=11 \implies U=6$.&lt;/li&gt;&#xA;&lt;li&gt;If $A=2, N=4$, then $U+A+C_3=U+2=N=4 \implies U=2$&lt;/li&gt;&#xA;&lt;li&gt;If $A=4, N=2$, then $U+A+C_3=U+4=10+N=12 \implies U=8$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thus, $A=1, N=5$.  Therefore $U+A+C_3=U+1=N=5 \implies U=4, C_4=0$.  Then $Q+D+C_4=Q+9=10+A \implies Q=2$.  Thus, the solution is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E=3, R=6, C=7, S=8, D=9, A=1, N=5, U=4, Q=2$$&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;824163&#xA;+91573&#xA;------&#xA;915736&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2016-03-10T18:48:13.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28754" PostTypeId="1" AcceptedAnswerId="28759" CreationDate="2016-03-10T19:20:29.180" Score="6" ViewCount="542" Body="&lt;p&gt;The following clues lead to a famous quote.  Can you tell me what the quote is, and how you got there?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Emily Dickinson has decided that something of ours is swiss.  &lt;/li&gt;&#xA;  &lt;li&gt;Laura Regan's 2002 one-liner  &lt;/li&gt;&#xA;  &lt;li&gt;Moby Dick's 7th  &lt;/li&gt;&#xA;  &lt;li&gt;Who takes George Foreman's tears?  &lt;/li&gt;&#xA;  &lt;li&gt;Oscar Wao's Brief first word  &lt;/li&gt;&#xA;  &lt;li&gt;A conflicting conjunction  &lt;/li&gt;&#xA;  &lt;li&gt;A-ha's juicy first word  &lt;/li&gt;&#xA;  &lt;li&gt;Who's father art in heaven?  &lt;/li&gt;&#xA;  &lt;li&gt;Yogi Berra's advice, regarding what to do with forks.  &lt;/li&gt;&#xA;  &lt;li&gt;1984's slavery  &lt;/li&gt;&#xA;  &lt;li&gt;Peter's mother's sister, and his daughter  &lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hints: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - All clues lead to one word (or a contraction).&lt;br&gt;&#xA; - The clues are not in order.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19944" LastActivityDate="2016-03-10T20:30:09.050" Title="A wordsmith's bravado" Tags="|riddle|knowledge|literature|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28755" PostTypeId="2" ParentId="28752" CreationDate="2016-03-10T19:21:34.193" Score="20" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;I'm from the future&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;Time machine broke&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is written on the stones, but &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;you must rotate the picture or turn your head sideways.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Image:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &#xA; &lt;img src=&quot;https://i.stack.imgur.com/jvEOJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="20075" LastEditorUserId="5840" LastEditDate="2016-03-12T22:00:15.720" LastActivityDate="2016-03-12T22:00:15.720" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="28756" PostTypeId="2" ParentId="28754" CreationDate="2016-03-10T19:28:35.480" Score="9" Body="&lt;p&gt;Partial answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1984's slavery is &lt;strong&gt;freedom&lt;/strong&gt;&lt;br&gt;&#xA; Yogi Berra suggests you &lt;strong&gt;take&lt;/strong&gt; any forks you come to in the road&lt;br&gt;&#xA; &lt;strong&gt;our&lt;/strong&gt; father, who art in heaven&lt;br&gt;&#xA; &lt;strong&gt;They&lt;/strong&gt; (2002), starring Laura Regan&lt;br&gt;&#xA; &lt;strong&gt;&lt;em&gt;Take&lt;/strong&gt; On Me&lt;/em&gt;, a song once known as 'The Juicy Fruit Song'&lt;br&gt;&#xA; &lt;strong&gt;but&lt;/strong&gt;, the conflicting conjunction!&lt;br&gt;&#xA; &quot;&lt;strong&gt;They'll&lt;/strong&gt; take everything, even your tears.&quot; —George Foreman&lt;br&gt;&#xA; &lt;s&gt;I don't know if the first word of the &lt;em&gt;book&lt;/em&gt; is &quot;life&quot;, but the &lt;em&gt;title&lt;/em&gt; of the book is &lt;em&gt;The Brief Wondrous &lt;strong&gt;Life&lt;/strong&gt; of Oscar Wao&lt;/em&gt;&lt;/s&gt;&lt;br&gt;&#xA; &lt;em&gt;Our &lt;strong&gt;lives&lt;/strong&gt; are Swiss&lt;/em&gt; is a poem by Emily Dickinson&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;em&gt;I think I see where this is going&lt;/em&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &quot;They may take our lives, but they'll never take our freedom&quot; —&lt;strong&gt;Braveheart&lt;/strong&gt; (get it? Bravado??)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2016-03-10T19:55:31.430" LastActivityDate="2016-03-10T19:55:31.430" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28757" PostTypeId="1" AcceptedAnswerId="28765" CreationDate="2016-03-10T19:30:26.190" Score="11" ViewCount="289" Body="&lt;p&gt;Over the time people collect tons of characters in movies, songs and games.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure you don't need to know all of them neither for the resulting &lt;strong&gt;prime number&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RW8ZF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RW8ZF.jpg&quot; alt=&quot;the description is really unimportant&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are my first 2 hints. Yes, two, because the first is already part of my comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Halle berry&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hannibal&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4342" LastEditorUserId="4342" LastEditDate="2016-03-10T23:33:50.813" LastActivityDate="2016-03-10T23:56:36.607" Title="a prime number behind a picture" Tags="|wordplay|rebus|knowledge|" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="28758" PostTypeId="1" AcceptedAnswerId="28760" CreationDate="2016-03-10T19:32:59.623" Score="10" ViewCount="378" Body="&lt;p&gt;Allan, Brian, and Charlie had a chess set and an afternoon with nothing to do. A chess tournament was suggested but to three towering intellects such as these, regular chess is a frivolous, juvenile pursuit. Accordingly, they set themselves to play some high-stakes horsie-push. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those unfamiliar with the sport, the rules are as follows. The squares on a chessboard are numbered in sequence from one to sixty-four. The players each place a potato chip in a dish beside the board. These are the stakes: winner takes all. The first player (Allan, because it is his board) takes a knight and places it on any one of the first five squares. The next player, Brian, then must advance the knight at least one but no more than five squares in a straight line. If Allan's first move was to place the knight on square three, then Brian must move it to one of 4, 5, 6, 7, or 8. Charlie, who has funny looking ears and a yellow tooth, goes last following the same one-to-five squares rule as Brian. Play proceeds until a player is able to move the knight to the final square where he wins the game and three disks of salty deep fried starch. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But on this day Charlie had an objection. &quot;It's not fair. You guys are just going to gang up on me. &quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Don't be a wimp,&quot; countered Allan. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;If you don't play fair, I'm going to take my bag of chips and go home.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was a forceful argument. The source from which all the wagers were drawn was Charlie's bag of chips. Compromise became the rule of the day and it was agreed that Charlie could move the knight from one up to nine squares when it was his turn. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming perfect co-operation between Allan and Brian and the unlikely event of best play on both sides, who will win the game?&lt;/p&gt;&#xA;" OwnerUserId="18924" LastEditorUserId="18924" LastEditDate="2016-03-10T19:43:48.563" LastActivityDate="2016-03-10T19:57:23.200" Title="Who Wins At Horsie-Push?" Tags="|board-games|game-theory|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28759" PostTypeId="2" ParentId="28754" CreationDate="2016-03-10T19:42:17.977" Score="9" Body="&lt;p&gt;Since we've taken to partial answers for this, here are some question_asker does not yet have (at time of original writing):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Emily Dickinson has decided that something of ours is swiss.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; our &lt;strong&gt;lives&lt;/strong&gt; are Swiss&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Moby Dick's 7th&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Call me Ishmael. Some years ago - &lt;strong&gt;never&lt;/strong&gt; mind how long precisely&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Oscar Wao's Brief first word&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Our&lt;/strong&gt; hero was not one of those Dominican cats &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A conflicting conjunction&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;but&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A-ha's juicy first word&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;The Juicy Fruit Song&quot; was the original title of &quot;&lt;strong&gt;Take&lt;/strong&gt; On Me&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Peter's mother's sister, and his daughter&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Spider-Man's Aunt &lt;strong&gt;May&lt;/strong&gt;, after whom he named a daughter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The remaining words, courtesy of question_asker:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1984's slavery is &lt;strong&gt;freedom&lt;/strong&gt;&lt;br&gt;&#xA; Yogi Berra suggests you &lt;strong&gt;take&lt;/strong&gt; any forks you come to in the road&lt;br&gt;&#xA; &lt;strong&gt;our&lt;/strong&gt; father, who art in heaven&lt;br&gt;&#xA; &lt;strong&gt;They&lt;/strong&gt; (2002), starring Laura Regan&lt;br&gt;&#xA; &quot;&lt;strong&gt;They'll&lt;/strong&gt; take everything, even your tears.&quot; —George Foreman  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This gives the words needed for the quote (again, courtesy of question_asker):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;They may take our lives, but they'll never take our freedom&quot; —Braveheart (get it? Bravado??)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17147" LastEditorUserId="17147" LastEditDate="2016-03-10T20:30:09.050" LastActivityDate="2016-03-10T20:30:09.050" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="28760" PostTypeId="2" ParentId="28758" CreationDate="2016-03-10T19:57:23.200" Score="8" Body="&lt;p&gt;Here's the situation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Charlie can win if he gets a move with the knight at spot 55 or higher. Therefore, Charlie will win if he can get the knight to spot 53. Allan and Brian can only move the knight 10 spots between them (up to 63), but must move the knight at least 2 spots (up to 55). If Charlie has to put the knight at spot 54 or higher, he loses and Brian wins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Working backwards...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Moving back another 11 spots, that means Charlie can win if he gets the knight to spot 42. Similarly to the last statement, Allan and Brian have to move it at least 2 spots (up to 44) which would allow Charlie to get it to 53. But they can't move it more than 10 spots (up to 52), so Charlie will definitely be able to get it to 53.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Keep going...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We can now see that Charlie will win if he can get the knight to 31, 20 or 9, by the same logic as the previous steps. So Allan would need to place the knight in a spot where Charlie couldn't get it there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Allan places the knight on spot 5. Brian moves it to spot 9. Charlie must move it off of 9, but can only move it as far as 18. Thus it will be between 10 and 18. Allan and Brian can combine to move the knight to 20, thus after Charlie's next move it will be between 21 and 29. Allan and Brian keep cooperating to place the knight on the spots Charlie would need to get it to in order to win - 31, 42 then 53. On each of Charlie's moves, he has to move it to within reach of the next one (but not all the way there).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So finally...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Allan or Brian will win, depending on where Charlie pushes the knight once it gets to spot 53. If he pushes it to somewhere in the 54-58 range, Brian will win. If he pushes it somewhere in the 59-62 range, Allan will win.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1620" LastActivityDate="2016-03-10T19:57:23.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28761" PostTypeId="2" ParentId="28360" CreationDate="2016-03-10T20:58:03.013" Score="0" Body="&lt;p&gt;OK, I've been going back and forth over whether or not to post an answer, and @Z.Dailey suggested I do so, so I think you could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; an orchid&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or maybe&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a pansy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll start to count while you solve this problem,&lt;br&gt;&#xA;and just relax here at the top of my column.&lt;br&gt;&#xA;Everywhere I go I let out a good whistle,&lt;br&gt;&#xA;but don't mess with me cause I carry a pistol.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A &lt;a href=&quot;https://en.wikipedia.org/wiki/Column_(botany)&quot; rel=&quot;nofollow&quot;&gt;column&lt;/a&gt; is a part of some flowers, specifically orchids (but not pansies), which includes the &lt;em&gt;pistil&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Typically speaking I reach senses of three,&lt;br&gt;&#xA;when you count those senses, you may find me.&lt;br&gt;&#xA;For if you hold me, I'm likely dead,&lt;br&gt;&#xA;why did you take me away from my bed?   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Flowers appeal to sight and smell, and can be touched (or alternately, I suppose, tasted by bees?). If you hold a flower, it has very likely been pulled or cut from the ground (a &quot;flowerbed&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Out of the rainbow I cannot see blue,&lt;br&gt;&#xA;and my name can be split into two.&lt;br&gt;&#xA;The first part of me is a computer's worst dream,&lt;br&gt;&#xA;while the second one comes apart at the seam.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back in the 30's I caused a great craze,&lt;br&gt;&#xA;not one single chap did I cease to amaze.&lt;br&gt;&#xA;Everyone loved me and wanted a kiss,&lt;br&gt;&#xA;but then in the 40's screamed &quot;What did I miss?&quot;   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;a href=&quot;https://en.wikipedia.org/wiki/Pansy_Craze&quot; rel=&quot;nofollow&quot;&gt;Pansy Craze&lt;/a&gt; was a movement involving increased visibility of the gay community in Manhattan, pre-Hayes Code.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I came with a flash and out like a light,&lt;br&gt;&#xA;but I don't really mind because life is so bright.&lt;br&gt;&#xA;I've been explosive at one point in history,&lt;br&gt;&#xA;what I truly am is not much of a mystery.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Some orchids, like the &lt;a href=&quot;https://en.wikipedia.org/wiki/Dendrophylax_lindenii&quot; rel=&quot;nofollow&quot;&gt;Ghost Orchid&lt;/a&gt;, are very rare, and poached for profit. I feel like that sort of, but doesn't exactly fit the &quot;explosive at one point in history&quot; clue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastActivityDate="2016-03-10T20:58:03.013" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28762" PostTypeId="2" ParentId="28360" CreationDate="2016-03-10T21:31:37.790" Score="0" Body="&lt;p&gt;I've been thinking about this answer, for several days, but it seems to fit, and not fit, at the same time.  I might as well try...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you a&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; clover leaf or shamrock?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll start to count while you solve this problem,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; clovers are famous for being frequently counted, in the hopes of finding a 4-leaved one&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and just relax here at the top of my column.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; clover leaves sit at the top of the clover stalk&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Everywhere I go I let out a good whistle,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a clover leaf whistle (weak)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but don't mess with me cause I carry a pistol.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; or a pistil, rather&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Typically speaking I reach senses of three,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; typically they are found with 3 leaves, rather than 4, or more&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;when you count those senses, you may find me.&#xA;For if you hold me, I'm likely dead,&#xA;why did you take me away from my bed?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a picked flower quickly begins to die&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Out of the rainbow I cannot see blue,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; clovers come in many colors - red, green, white - but never blue&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and my name can be split into two.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; clover leaf is two words - shamrock &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first part of me is a computer's worst dream,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clover yellow mosaic virus - a virus is a computer's worst dream&lt;br&gt;&#xA; shamrock doesn't fit here&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;while the second one comes apart at the seam.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; leaves fall apart at the seams, when you mess with them&#xA; again, shamrock doesn't fit&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Back in the 30's I caused a great craze,&#xA;not one single chap did I cease to amaze.&#xA;Everyone loved me and wanted a kiss,&#xA;but then in the 40's screamed &quot;What did I miss?&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Dan Sullivan's Shamrock Band?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I came with a flash and out like a light,&#xA;but I don't really mind because life is so bright.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clovers/Shamrocks have a short lifespan, but they grow in the sunlight. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've been explosive at one point in history,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://physicsworld.com/cws/article/news/2008/aug/27/cold-atoms-explode-like-cloverleafs&quot; rel=&quot;nofollow&quot;&gt;clover-leaf shaped explosions?&lt;/a&gt; (weak)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;what I truly am is not much of a mystery. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Clearly a lie.  Very much a mystery!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19944" LastActivityDate="2016-03-10T21:31:37.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28763" PostTypeId="2" ParentId="28360" CreationDate="2016-03-10T21:46:21.320" Score="7" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tulip&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll start to count while you solve this problem,&#xA;and just relax here at the top of my column. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tulip flower is on top of the &quot;column&quot; — stem.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Everywhere I go I let out a good whistle, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Reference to &lt;em&gt;tiptoe through the tulips&lt;/em&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but don't mess with me cause I carry a pistol. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As several other answers already mentioned, &lt;em&gt;Pistil&lt;/em&gt; is a part of flower.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Typically speaking I reach senses of three,&#xA;when you count those senses, you may find me. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Visual, smell and touch senses. Two groups of 3 petals. It is common to &quot;count&quot; petals on a flower.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For if you hold me, I'm likely dead,&#xA;why did you take me away from my bed? &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You picked a flower from flowerbed and killed it. You monster.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Out of the rainbow I cannot see blue, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are no natural blue tulips. All &quot;blue&quot; ones are actually more or less violet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and my name can be split into two.&#xA;The first part of me is a computer's worst dream,&#xA;while the second one comes apart at the seam.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I cannot make head or tails of &quot;Tu&quot;, but &quot;lip&quot;s are coming apart at a seam&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Back in the 30's I caused a great craze,&#xA;not one single chap did I cease to amaze.&#xA;Everyone loved me and wanted a kiss,&#xA;but then in the 40's screamed &quot;What did I miss?&quot; &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Tulip_mania&quot;&gt;Tulip mania&lt;/a&gt; of 1630's&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I came with a flash and out like a light,&#xA;but I don't really mind because life is so bright. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Possible short flowering cycle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've been explosive at one point in history, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;USS Tulip exploded in 1864&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19952" LastActivityDate="2016-03-10T21:46:21.320" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28764" PostTypeId="2" ParentId="28387" CreationDate="2016-03-10T22:46:21.273" Score="4" Body="&lt;p&gt;The answer is  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tombstone&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A good problem solver peers into the past,&lt;br&gt;&#xA;once you find me, it's quiet at last.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;T&lt;/strong&gt;ulip&lt;br&gt;&#xA; &lt;strong&gt;O&lt;/strong&gt;carina&lt;br&gt;&#xA; &lt;strong&gt;M&lt;/strong&gt;ountain&lt;br&gt;&#xA; &lt;strong&gt;B&lt;/strong&gt;utton&#xA; I used the first letters of each of the previous puzzle's solutions to find the word &lt;code&gt;Tomb&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Completing this riddle is the ultimate test,&lt;br&gt;&#xA;for after its done, you will need your rest.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your final resting place is generally in a tomb&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The other day I screamed &quot;JSTYX&quot; out loud all confused,&lt;br&gt;&#xA;To figure out what it means, I'd be greatly amused.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Subtract 5 from each of the letters of &quot;JSTYX&quot; to get &quot;ENOTS&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The problems the key, I know it, I know it,&lt;br&gt;&#xA;But instead of sharing I might as well throw it.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;As with the past, a similar method involved,&lt;br&gt;&#xA;This time it's much harder, and has greatly evolved.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is STONE backwards&lt;br&gt;&#xA; A stone is often thrown, rather than shared.&lt;br&gt;&#xA; Again, we're looking into the past, but this time it's to reverse the word.&#xA; A stone is hard and is weathered from larger stones, over time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The only thing I can think of, on the titles, is that they're backwards:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Large and in charge&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The word tulip comes from the Persian word, delband, which means Turban - generally worn by a Persian ruler.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Small and resilient&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An ocarina is a small, handheld, instrument - often made of metal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Immovable object&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A mountain....well it's not going anywhere anytime soon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Unstoppable Force&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The goal of a button is to hold something in place - to keep it from moving&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Breathing New Life&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This one breaks the pattern, but tombs do feed carrion and other bugs that inhabit graveyards.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19944" LastEditorUserId="19944" LastEditDate="2016-03-10T23:06:17.760" LastActivityDate="2016-03-10T23:06:17.760" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28765" PostTypeId="2" ParentId="28757" CreationDate="2016-03-10T23:50:50.990" Score="10" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 397&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The pictures spell out a clue&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;b&gt;F&lt;/b&gt;arin Urlaub &lt;br&gt;&lt;b&gt;O&lt;/b&gt;nix &lt;br&gt;&lt;b&gt;U&lt;/b&gt;mberto Eco &lt;br&gt;&lt;b&gt;R&lt;/b&gt;emy &lt;br&gt;&lt;br&gt;&lt;b&gt;H&lt;/b&gt;alle Berry &lt;br&gt;&lt;b&gt;U&lt;/b&gt;rgot &lt;br&gt;&lt;b&gt;N&lt;/b&gt;ilsson &lt;br&gt;&lt;b&gt;D&lt;/b&gt;onkey Kong &lt;br&gt;&lt;b&gt;R&lt;/b&gt;udolph &lt;br&gt;&lt;b&gt;E&lt;/b&gt;lrond &lt;br&gt;&lt;b&gt;D&lt;/b&gt;arth Vader &lt;br&gt;&lt;br&gt;&lt;b&gt;M&lt;/b&gt;acguyver &lt;br&gt;&lt;b&gt;I&lt;/b&gt;ndiana Jones &lt;br&gt;&lt;b&gt;N&lt;/b&gt;ed Flanders &lt;br&gt;&lt;b&gt;U&lt;/b&gt;sain Bolt &lt;br&gt;&lt;b&gt;S&lt;/b&gt;cotty &lt;br&gt;&lt;br&gt;&lt;b&gt;T&lt;/b&gt;iger Woods &lt;br&gt;&lt;b&gt;H&lt;/b&gt;annibal Smith &lt;br&gt;&lt;b&gt;R&lt;/b&gt;exxar &lt;br&gt;&lt;b&gt;E&lt;/b&gt;lvis Presley &lt;br&gt;&lt;b&gt;E&lt;/b&gt;rnie &lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 397 is a cuban prime which may explain the background&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18422" LastEditorUserId="18422" LastEditDate="2016-03-10T23:56:36.607" LastActivityDate="2016-03-10T23:56:36.607" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28767" PostTypeId="2" ParentId="28616" CreationDate="2016-03-11T03:13:42.840" Score="14" Body="&lt;p&gt;Not a solution, but here's a conversion of the image to text, to make this puzzle accessible to the blind. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Braille:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;⠼⠁⠋⠉⠃⠉⠑⠋⠁⠤⠚ ⠼⠃⠃⠋⠙⠋⠁⠃⠃⠤⠚&#xA;⠼⠉⠉⠃⠙⠁⠋⠉⠉⠤⠁⠀⠼⠁⠃⠉⠋⠉⠉⠃⠁⠤⠚&#xA;⠼⠋⠙⠁⠉⠑⠃⠙⠋⠤⠁⠀⠼⠁⠙⠙⠃⠉⠋⠙⠁⠤⠚&#xA;⠼⠋⠑⠃⠃⠁⠙⠑⠋⠤⠚⠀⠼⠙⠁⠑⠉⠑⠋⠁⠙⠤⠁&#xA;&#xA;⠼⠁⠋⠉⠑⠃⠙⠋⠁⠤⠚⠀⠼⠋⠃⠙⠉⠉⠉⠃⠋⠤⠚&#xA;⠼⠋⠁⠉⠑⠙⠃⠁⠋⠤⠁⠀⠼⠁⠙⠋⠙⠋⠉⠙⠁⠤⠁&#xA;⠼⠃⠙⠉⠉⠙⠉⠙⠃⠤⠁⠀⠼⠙⠙⠙⠃⠉⠙⠙⠙⠤⠚&#xA;⠼⠁⠁⠉⠉⠉⠉⠁⠁⠤⠚⠀⠼⠙⠃⠋⠑⠁⠃⠃⠙⠤⠚&#xA;&#xA;⠼⠉⠙⠁⠃⠙⠉⠙⠉⠤⠁⠀⠼⠁⠁⠋⠙⠃⠉⠁⠁⠤⠚&#xA;⠼⠋⠋⠉⠃⠙⠑⠋⠋⠤⠚⠀⠼⠁⠉⠋⠙⠃⠉⠉⠁⠤⠚&#xA;⠼⠙⠃⠉⠉⠙⠉⠃⠙⠤⠚⠀⠼⠋⠁⠙⠉⠑⠃⠁⠋⠤⠚&#xA;⠼⠋⠑⠁⠃⠙⠃⠑⠋⠤⠚⠀⠼⠁⠉⠑⠃⠙⠉⠉⠁⠤⠁&#xA;&#xA;⠼⠑⠁⠃⠋⠙⠋⠁⠑⠤⠚⠀⠼⠃⠃⠃⠉⠙⠉⠃⠃⠤⠁&#xA;⠼⠁⠋⠉⠙⠉⠃⠋⠁⠤⠚⠀⠼⠋⠃⠑⠉⠁⠑⠃⠋⠤⠁&#xA;⠼⠁⠉⠃⠑⠃⠋⠉⠁⠤⠚⠀⠼⠁⠃⠉⠙⠙⠉⠃⠁⠤⠚&#xA;⠼⠙⠋⠑⠃⠙⠉⠋⠙⠤⠁⠀⠼⠁⠃⠙⠋⠉⠙⠃⠁⠤⠁&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Possible translation:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;16323561-0 22646122-0&#xA;33241633-1 12363321-0&#xA;64135246-1 14423641-0&#xA;65221456-0 41535614-1&#xA;&#xA;16352461-0 62433326-0&#xA;61354216-1 14646341-1&#xA;24334342-1 44423444-0&#xA;11333311-0 42651224-0&#xA;&#xA;34124343-1 11642311-0&#xA;66324566-0 13642331-0&#xA;42334324-0 61435216-0&#xA;65124256-0 13524331-1&#xA;&#xA;51264615-0 22234322-1&#xA;16343261-0 62531526-1&#xA;13252631-0 12344321-0&#xA;46524364-1 12463421-1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="9044" LastEditorUserId="20151" LastEditDate="2016-03-23T14:16:39.330" LastActivityDate="2016-03-23T14:16:39.330" CommentCount="11" CommunityOwnedDate="2016-03-11T03:13:42.840" ContentLicense="CC BY-SA 3.0" />
  <row Id="28768" PostTypeId="1" AcceptedAnswerId="28769" CreationDate="2016-03-11T05:16:35.087" Score="13" ViewCount="1270" Body="&lt;p&gt;A famous painting of Julius Caesar has been stolen from a museum. The thief, being a cryptic lover, has left these clues behind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“I have left the painting somewhere within the city.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PSkJ5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PSkJ5.png&quot; alt=&quot;image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The letter found alongside was which is supposed to give the location of the painting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ZQMDFTQEAGFTNDUPSQ&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you find out the painting?&lt;/p&gt;&#xA;" OwnerUserId="19644" LastEditorUserId="18144" LastEditDate="2016-03-11T08:59:52.883" LastActivityDate="2016-03-11T08:59:52.883" Title="Find the painting" Tags="|enigmatic-puzzle|mystery|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28769" PostTypeId="2" ParentId="28768" CreationDate="2016-03-11T05:23:27.317" Score="28" Body="&lt;p&gt;To read the... weird thing which seems to part of the letter (?)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Flip it horizontally:&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/xEJlr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xEJlr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This alludes to this kind of decoder&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/kWOBS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kWOBS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Move the inner circle so A lines up with M, and we get&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; NEARTHESOUTHBRIDGE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So we look there&lt;/p&gt;&#xA;" OwnerUserId="17147" LastActivityDate="2016-03-11T05:23:27.317" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28770" PostTypeId="1" AcceptedAnswerId="28772" CreationDate="2016-03-11T07:35:52.393" Score="0" ViewCount="281" Body="&lt;p&gt;$()+()+()+()+()=30$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How can you fill in the parentheses with the numbers $1, 3, 5, 7, 9, 11, 13$ to make the above equation true?&lt;/strong&gt; Repeats are allowed, but you must fill in all of the parentheses! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus if you already solved this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Can you do it without extra symbols? (no +, -, ÷, ｘ, &amp;lt;&amp;lt;, !, etc.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(&lt;strong&gt;EDIT:&lt;/strong&gt; You don't have to use all of the numbers.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT2&lt;/strong&gt;: Okay, the answer is not that there is no answer.  I'm not looking for a mathematical proof here.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT3&lt;/strong&gt;: The above information is all you need to solve the problem. Be creative!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possible solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $(5+1)+(5+1)+(5+1)+(5+1)+(5+1)=30$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My favorite solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ${{5}\choose{3}}$+(5)+(5)+(5)+(5)=30&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18144" LastEditorUserId="18144" LastEditDate="2016-03-11T19:22:20.543" LastActivityDate="2016-03-11T19:22:20.543" Title="Fill in the parentheses" Tags="|calculation-puzzle|formation-of-numbers|" AnswerCount="3" CommentCount="5" ClosedDate="2016-03-11T16:58:29.497" ContentLicense="CC BY-SA 3.0" />
  <row Id="28772" PostTypeId="2" ParentId="28770" CreationDate="2016-03-11T08:06:38.920" Score="4" Body="&lt;p&gt;Answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (3!) + (9) + (7) + (3) + (5) = 30&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For Bonus:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; sum of 5 odd numbers is ALWAYS an odd number so it is not possible unless something is done. What I am doing is using the number 9 as BASE. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (13) + (11) + (5) + (3) + (1) = 33 , base9 (&lt;strong&gt;33&lt;/strong&gt;) = base10 (&lt;strong&gt;30&lt;/strong&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or a more direct way with that logic;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (7+5+5+5+5)base9 = (30)base9&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18685" LastEditorUserId="18685" LastEditDate="2016-03-11T08:35:15.620" LastActivityDate="2016-03-11T08:35:15.620" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28773" PostTypeId="2" ParentId="28770" CreationDate="2016-03-11T08:43:58.050" Score="1" Body="&lt;p&gt;One answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Using base 11: 1 + 3 + 3 + 11 + 13 = 30&lt;br&gt;&#xA; Equivalent in base 10&lt;br&gt;&#xA; 1 + 3 + 3 + 12 + 14 = 33 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Using base 13: 1 + 3 + 5 + 11 + 13 = 30&lt;br&gt;&#xA; Equivalent in base 10&lt;br&gt;&#xA; 1 + 3 + 5 + 14 + 16 = 39  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And a third one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because you cannot add 5 odd numbers to get an even number in base 10 you have to use tricks.&lt;br&gt;&#xA; 3 + 5 + 3! + 7 + 9 = 30&lt;br&gt;&#xA; equivalent to&lt;br&gt;&#xA; 3 + 5 + 6 + 7 + 9 = 30&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[EDIT]&lt;/strong&gt;&lt;br&gt;&#xA;Bonus one or two:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Using base 9: 1 + 1 + 3 + 11 + 13 = 30&lt;br&gt;&#xA; Equivalent in base 10:&lt;br&gt;&#xA; 1 + 1 + 3 + 10 + 12 = 27&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Using base 7: 1 + 1 + 1 + 11 + 13 = 30&lt;br&gt;&#xA; Equivalent in base 10:&lt;br&gt;&#xA; 1 + 1 + 1 + 8 + 10 = 21&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19989" LastEditorUserId="19989" LastEditDate="2016-03-11T09:00:26.503" LastActivityDate="2016-03-11T09:00:26.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28774" PostTypeId="2" ParentId="28736" CreationDate="2016-03-11T09:03:43.140" Score="10" Body="&lt;p&gt;The answer might be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Control-Alt-Delete&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Explanation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Image 1: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;THREE&lt;/strong&gt; (The song &quot;Kolaveri Di&quot; is from a Tamil psychological thriller film called &quot;3&quot;.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Image 2: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;FINGER&lt;/strong&gt; (potato fingers)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Image 3: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;SALUTE&lt;/strong&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Image 4: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;KEYS&lt;/strong&gt; (piano keys)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Conclusion:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Three-finger_salute&quot;&gt;https://en.wikipedia.org/wiki/Three-finger_salute&lt;/a&gt;, &quot;Three-finger salute&quot; is a jocular term for the three-key command Control-Alt-Delete.&lt;br&gt;&#xA; 2. Whenever things go awry with my PC, I reboot it by Control-Alt-Delete.&lt;br&gt;&#xA; 3. Control-Alt-Delete came up in the 1980s with the MS-DOS operating system.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2016-03-11T09:03:43.140" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28776" PostTypeId="1" AcceptedAnswerId="28779" CreationDate="2016-03-11T09:41:07.323" Score="12" ViewCount="2923" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Niccol%C3%B2_Fontana_Tartaglia&quot;&gt;Niccolò Tartaglia&lt;/a&gt; was an Italian mathematician, engineer and bookkeeper, living in the 16th century in the Republic of Venice.&#xA;Tartaglia published many books, including his &quot;General Trattato di Numeri&quot;, an acclaimed compilation of the mathematics known at his time.&#xA;Among many other problems, the Trattato contained the following mathematical puzzle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A dishonest servant takes out three pints of wine from a barrel and replaces them with the same amount of water. He repeats this theft another two times, removing altogether nine pints and replacing them with water. As a result of this swindle, the diluted wine remaining in the barrel lost half of its former strength.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;How much wine was in the barrel before the thievery began?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Note: Tartaglia is also famous for his solution of the cubic equation. The cubic equation that is implicit in this puzzle, however, is of a simple form that could be solved already before Tartaglia's work.)&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-18T15:53:03.137" LastActivityDate="2016-03-18T15:53:03.137" Title="Turning wine into water" Tags="|mathematics|algebra|puzzle-history|" AnswerCount="5" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28777" PostTypeId="5" CreationDate="2016-03-11T09:50:06.280" Score="0" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Algebra&quot; rel=&quot;nofollow noreferrer&quot;&gt;Algebra&lt;/a&gt; is the study of mathematical symbols and the rules for manipulating these symbols. Puzzles involving equations, variables, polynomials, vectors and/or matrices should include this tag. General algebra questions are off-topic but can be asked on our sister site &lt;a href=&quot;https://math.stackexchange.com/&quot;&gt;Mathematics Stack Exchange&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Example&lt;/strong&gt;: Some birds are sitting in two trees. The birds in the first tree say to the others: &amp;quot;If one of you comes to our tree, our number will be twice your number.&amp;quot; The birds in the second tree said: &amp;quot;If one of you comes here, our number will be equal to yours.&amp;quot; How many birds are in each tree? (Answer: &lt;span class=&quot;math-container&quot;&gt;$7$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="60644" LastEditDate="2020-11-24T23:10:40.937" LastActivityDate="2020-11-24T23:10:40.937" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28778" PostTypeId="4" CreationDate="2016-03-11T09:50:06.280" Score="0" Body="A mathematical puzzle about the manipulation and solution of equations. Use with [mathematics]" OwnerUserId="8874" LastEditorUserId="69582" LastEditDate="2021-02-18T08:14:14.160" LastActivityDate="2021-02-18T08:14:14.160" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28779" PostTypeId="2" ParentId="28776" CreationDate="2016-03-11T09:55:28.437" Score="13" Body="&lt;p&gt;(I'm assuming throughout that the concentration of wine at the end needs to be 50%.)&lt;br&gt;&#xA;Unless I'm missing something:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are $n$ pints of wine to begin with. After the first 3 pints are taken, you have an $(n-3)/n$ fraction of wine after the water is put back. After the second 3 pints are taken and replaced, you just multiply by the same fraction again, so you have $(n-3)^2/n^2$, and after the last 3 pints you end up with an $(n-3)^3/n^3$ fraction. This leads to $(n-3)^3/n^3=0.5$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation can be rewritten into the linear equation &#xA;$(n-3)/n=1/\sqrt[3]{2}$, which then implies &#xA;$n=3\sqrt[3]{2}/(\sqrt[3]{2}-1) \approx 14.54$ pints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way is to rewrite the equation into the cubic equation $~~n^3/2 -9n^2 +27n -27=0$, which can be solved by numerical methods to give $n\approx14.54$ pints.&lt;/p&gt;&#xA;" OwnerUserId="19698" LastEditorUserId="8874" LastEditDate="2016-03-16T14:06:39.683" LastActivityDate="2016-03-16T14:06:39.683" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28780" PostTypeId="2" ParentId="28776" CreationDate="2016-03-11T10:06:16.817" Score="7" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$14.542$ pints if he replaces wine with water after taking 3 pints of wine.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let say you have 300 units of wine and you take off 3 pints of wine and add 3 pints of wine 3 times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I. Theft&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After you take off 3 pints of wine (3p) and add 3 pints of wine and add water the new concentration of wine would be;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{300-3p}{300}$ or $\frac{100-p}{100}$ or $(100-p) \%$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result, you would have taken off;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3p$ of wine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;II. Theft&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know that the concentration of wine is $\frac{100-p}{100}$. So If you take off 3 pints of wine (3p) again, you would actually take off $(100-p)\%$ wine and $p\%$ water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result, you would take off &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3p \frac{(100-p)}{100}=3p-\frac{3p^2}{100}$ of wine from the barrel as a wine! So previously it was $3p$ of wine and now the difference between I. Theft and II. Theft is $\frac{3p^2}{100}$ less than first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{p}{100}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result, you would have taken off;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3p-\frac{3p^2}{100}$ of wine&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;III. Theft&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know that the new concentration of wine is $\frac{100-p-(p-\frac{p^2}{100})}{100}$. So If you take off 3 pints of wine (3p) again, you would actually take off $(100-p-(p-\frac{p^2}{100}))\%$ wine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result you would taken off;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3p \frac{(100-p-(p-\frac{p^2}{100})}{100}=3p-\frac{6p^2}{100}+\frac{3p^3}{100^2}$ of wine!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;RESULT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the sum of wine taken from the barrel would be equal to 50% of wine, or 150 units;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$50\% \ of \ barrel=3p+(3p-\frac{3p^2}{100})+(3p-\frac{6p^2}{100}+\frac{3p^3}{100^2})=150$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p(3-\frac{3p}{100}+\frac{p^2}{100^2})=50$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so $p=20.630$ and since total amount of barrel is assumed to be 300;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The capacity of barrel in terms of pints is $\frac{300}{20.630}=14.542$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18514" LastEditorUserId="18514" LastEditDate="2016-03-11T14:00:12.717" LastActivityDate="2016-03-11T14:00:12.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28781" PostTypeId="2" ParentId="28776" CreationDate="2016-03-11T10:13:28.317" Score="8" Body="&lt;p&gt;Let $x$ be the starting amount of pints of wine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now just focus on the amount of water in the barrel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the first step there is 3 pints of water in the barrel. Because the content of the barrel is still $x$ pints the amount of water per pint is therefore $3/x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the second step we take $3$ times $3/x$ water so we take away $9/x$ water and we add $3$ pints of water so we end up with $6 - 9/x$ water in the barrel. This means there is $\frac{6 - 9/x}{x}$ water per pint in the barrel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the final step we take $3$ times $\frac{6 - 9/x}{x}$ water so we take away $3 \cdot \frac{6 - 9/x}{x}$ water and we add $3$ pints of water so we end up with $9 - 9/x - 3 \cdot \frac{6 - 9/x}{x}$ water in the barrel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know this is half the amount of the total barrel, so half of $x$, so we just need to solve the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$9 - 9/x - 3 \cdot \frac{6 - 9/x}{x} = \frac x 2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving this gives $x = 3 (2+ \sqrt[3]{2} +\sqrt[3]{4})$&lt;/p&gt;&#xA;" OwnerUserId="1748" LastEditorUserId="1748" LastEditDate="2016-03-11T10:42:40.173" LastActivityDate="2016-03-11T10:42:40.173" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28782" PostTypeId="5" CreationDate="2016-03-11T10:45:30.880" Score="0" Body="&lt;p&gt;Set is a card game designed by Marsha Falco. The deck consists of $81$ cards varying in four features:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;color (red, green, purple), &lt;/li&gt;&#xA;&lt;li&gt;shading (empty, striped, or solid), &lt;/li&gt;&#xA;&lt;li&gt;shape (oval, squiggle, or rhombus), &lt;/li&gt;&#xA;&lt;li&gt;number of symbols (one, two, or three). &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All the cards are unique. A SET occurs when three cards satisfy the following property for each of the four attributes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The attribute on each card is the same, &lt;/li&gt;&#xA;&lt;li&gt;or the attribute on each card is different &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As an example for &quot;the same&quot;, all three cards may have striped shading.&#xA;As an example for &quot;different&quot;, the three cards may have  red, green, and purple color.&#xA;In order for three cards to form a SET, they must satisfy this property for all four attributes. &lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-11T10:45:30.880" LastActivityDate="2016-03-11T10:45:30.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28783" PostTypeId="4" CreationDate="2016-03-11T10:45:30.880" Score="0" Body="A puzzle about the card game SET, played with a deck of 81 cards." OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-11T10:45:30.880" LastActivityDate="2016-03-11T10:45:30.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28784" PostTypeId="5" CreationDate="2016-03-11T10:47:53.023" Score="0" Body="" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-11T10:47:53.023" LastActivityDate="2016-03-11T10:47:53.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28785" PostTypeId="4" CreationDate="2016-03-11T10:47:53.023" Score="0" Body="A brain teaser is a puzzle that requires thought to solve, and often can be solved by a single insight." OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-11T10:47:53.023" LastActivityDate="2016-03-11T10:47:53.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28786" PostTypeId="1" AcceptedAnswerId="28787" CreationDate="2016-03-11T11:05:32.693" Score="4" ViewCount="1335" Body="&lt;p&gt;I have this puzzle that none of my colleagues (including me) can figure out and I can't sleep because of it :). Can someone please help:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/lChNF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lChNF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="20085" LastEditorUserId="8874" LastEditDate="2016-03-11T17:17:04.857" LastActivityDate="2016-03-11T17:17:04.857" Title="Fill the missing number puzzle" Tags="|pattern|number-sequence|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28787" PostTypeId="2" ParentId="28786" CreationDate="2016-03-11T11:09:43.013" Score="3" Body="&lt;p&gt;My guess would be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the missing number is $3$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Explanation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If we denote the four numbers by $a,b,c,d$, in clockwise order, starting with the lower left position (so that in the left picture $a=2$, $b=8$, $c=14$ and $d=9$),&#xA;then these number satisfy the relation&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $d*a-(b/a) = c~~~~$  Indeed, $9*2-(8/2)=14$ in the left picture.&lt;br&gt;&#xA;&lt;br&gt;&#xA; In the right picture, the unknown $x=?$ then should satisfy $x*3-(15/3)=4$, which yields the answer $x=3$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2016-03-11T11:09:43.013" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28788" PostTypeId="2" ParentId="28125" CreationDate="2016-03-11T11:52:20.517" Score="2" Body="&lt;p&gt;I think you are : &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gravity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm all over.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No denying it, just because we don't feel it's effects everywhere, it is omnipresent. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Over you. Under you. Through you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; See above. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I love black holes and dying cosmos.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Strongest in these areas. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come from evil politicians and ruthless dictators.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People often gravitate towards the history and characters of notorious politicians and dictators etc. Just look at how much we know about Hitler.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come from silly children and irresponsible criminals.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I guess there is (or can be) an endearment towards these characters although I will admit, it's a stretch. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can cause horrible diseases.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Brittle bone syndrome (Osteogenesis imperfecta) will be massively affected by gravity or lack thereof.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The government overregulates me.&#xA;But that doesn't stop me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Flights and aeronautic regulation after ever crash or accident. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nothing can stop me.&#xA;Not walls. Not floors. Not Time or Space or Distance.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Self explanatory. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19078" LastActivityDate="2016-03-11T11:52:20.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28789" PostTypeId="1" AcceptedAnswerId="28793" CreationDate="2016-03-11T12:17:08.203" Score="2" ViewCount="1533" Body="&lt;p&gt;I saw this riddle ealier today on 9GAG (and it's found on multiple news sites), Im not quite sure about the answer, but i'm wondering if someone can solve it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;100 prisoners are lined up by an executioner, who places a red or blue hat upon each of their heads.&#xA;The prisoners can see the hats of the people lined up in from of them, but they cannot look at the hats behind them, or at their own.&#xA;Starting at the back of the line, the executioner asks the last prisoner to state the colour of his hat.&#xA;In order to live, the prisoner must answer correctly. If he doesn't, he is killed 'instantly and silently.'&#xA;This means that the other prisoners will hear the answer, but will not know whether or not it was correct. &#xA;The night before the line-up, the prisoners can discuss a strategy to help them survive. What should they do?&lt;/p&gt;&#xA;" OwnerUserId="18683" LastEditorUserId="7586" LastEditDate="2016-03-11T16:59:03.043" LastActivityDate="2016-03-11T16:59:03.043" Title="The 100 Hat Riddle" Tags="|logical-deduction|hat-guessing|" AnswerCount="1" CommentCount="1" ClosedDate="2016-03-11T14:22:16.577" ContentLicense="CC BY-SA 3.0" />
  <row Id="28790" PostTypeId="1" CreationDate="2016-03-11T12:22:18.453" Score="7" ViewCount="336" Body="&lt;p&gt;I saw this riddle ealier today on 9GAG (and it's found on multiple news sites), I have found 2 answers with my collegues, but i'm wondering if someone else can solve it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;'One hundred prisoners have been newly ushered into prison. &#xA;The warden tells them that starting tomorrow, each of them will be placed in an isolated cell, unable to communicate amongst each other. &#xA;Each day, the warden will choose one of the prisoners uniformly at random with replacement, and place him in a central interrogation room containing only a light bulb with a toggle switch. &#xA;The prisoner will be able to observe the current state of the light bulb. If he wishes, he can toggle the light bulb. He also has the option of announcing that he believes all prisoners have visited the interrogation room at some point in time. &#xA;If this announcement is true, then all prisoners are set free, but if it is false, all prisoners are executed. The warden leaves and the prisoners huddle together to discuss their fate. &#xA;Can they agree on a protocol that will guarantee their freedom?' &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If time wouldn't matter, this would be a flawless solution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Together will all prisoners, choose 1 leader. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Step 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Whenever a prisoner enters the interrogation room he will turn on the light IF the light is &lt;code&gt;OFF&lt;/code&gt; AND he hasn't turned on the light before.&#xA; IF the light is &lt;code&gt;ON&lt;/code&gt;, he ignores it (and will wait for his next turn)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Step 3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; every time the Leader steps in the room he makes sure he turns the light &lt;code&gt;OFF&lt;/code&gt; and counts&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Step 4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; When the leader has entered the room 99 times while the light was on, it means everyone has been in the room. And he will call the guard.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18683" LastEditorUserId="18683" LastEditDate="2016-03-11T12:53:35.847" LastActivityDate="2016-03-11T13:02:44.247" Title="The Switch Riddle" Tags="|logical-deduction|" AnswerCount="0" CommentCount="4" ClosedDate="2016-03-11T14:22:37.810" ContentLicense="CC BY-SA 3.0" />
  <row Id="28792" PostTypeId="2" ParentId="28125" CreationDate="2016-03-11T12:29:07.877" Score="2" Body="&lt;p&gt;I think the best fit is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Misinformation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm all over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over you. Under you. Through you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are myths, lies and inaccuracies everywhere. I often find that things I believed have been proven untrue. (Are eggs good for you? The answer seems to change every ten years.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I love black holes and dying cosmos.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I take this figuratively. Mistaken ideas can be a huge drain of time and energy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come from evil politicians and ruthless dictators.&#xA;I come from silly children and irresponsible criminals.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; These are all examples of people who spread misinformation deliberately or through ignorance.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can cause horrible diseases.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Mistaken beliefs about sanitation or sexual contact can cause epidemics.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The government overregulates me.&#xA;But that doesn't stop me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are many laws regarding libel, slander and truth in advertising but misinformation is still rampant.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nothing can stop me.&#xA;Not walls. Not floors. Not Time or Space or Distance.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is very true of misinformation. It is spread by every medium. You can even argue that distance increases the effect as information can be further garbled as it is passed from person to person.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I OWN THE WORLD!!!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This may in itself be a piece of misinformation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not that it does me much good.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is little ultimate benefit to misinformation, even if it is well-intended like the &quot;instructive fairy tales&quot; people sometimes tell children.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I think of my existence as a giant treadmill.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It does seem like a huge, meaningless waste of effort: all the time that people spend generating and spreading misinformation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastActivityDate="2016-03-11T12:29:07.877" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28793" PostTypeId="2" ParentId="28789" CreationDate="2016-03-11T12:32:39.173" Score="3" Body="&lt;p&gt;The first prisoner has a 50% chance of surviving. On the night before, they can choose a hat color to begin with and colors to be odd or even. So say they say odd = blue and red = even and they begin with blue. If the first prisoner sees a odd number of blue hats, he/she says &quot;blue&quot; and vice versa. Then if the second prisoner also sees an odd number of hats, then he/she will know that his/her hat is red. Then if the third prisoner sees an even number of blue hats, they'll know that his/her hat is blue. And so on. But good luck to the first prisoner! This answer is partly from &lt;a href=&quot;http://ed.ted.com/lessons/can-you-solve-the-prisoner-hat-riddle-alex-gendler&quot; rel=&quot;nofollow&quot;&gt;http://ed.ted.com/lessons/can-you-solve-the-prisoner-hat-riddle-alex-gendler&lt;/a&gt; .&lt;/p&gt;&#xA;" OwnerUserId="20087" LastActivityDate="2016-03-11T12:32:39.173" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28794" PostTypeId="1" AcceptedAnswerId="28828" CreationDate="2016-03-11T12:40:25.960" Score="13" ViewCount="910" Body="&lt;p&gt;Seven cowboys play a coin tossing game with a fair coin where each player has a chance of 1/2 of winning. Each of them plays once against all other six. What is the probability that each cowboy wins exactly three games?&lt;/p&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="12408" LastEditDate="2016-07-25T02:30:23.970" LastActivityDate="2016-07-25T02:30:23.970" Title="Seven cowboys play a coin tossing game" Tags="|mathematics|combinatorics|probability|coin-tossing|" AnswerCount="5" CommentCount="5" ClosedDate="2016-03-16T13:23:50.930" ContentLicense="CC BY-SA 3.0" />
  <row Id="28796" PostTypeId="1" AcceptedAnswerId="28801" CreationDate="2016-03-11T12:55:23.250" Score="8" ViewCount="217" Body="&lt;p&gt;Roohullah wrote his favorite positive integer $m$ down on a piece of paper. Then Roohullah computed the value $m+m^2+m^3+.....+m^{2m-3}-4$. Roohullah noticed that this value was a prime number. Now I ask you: What are the possible values for the unit digit (last digit) of $m$?&lt;/p&gt;&#xA;" OwnerUserId="19470" LastEditorUserId="12408" LastEditDate="2016-07-25T02:18:38.390" LastActivityDate="2016-07-25T02:18:38.390" Title="Roohullah's prime" Tags="|mathematics|number-theory|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28797" PostTypeId="1" AcceptedAnswerId="28889" CreationDate="2016-03-11T13:02:26.647" Score="8" ViewCount="264" Body="&lt;p&gt;Cheherazad, the carpet salesman, has bought a rectangular piece of carpet. Unfortunately, Cheherazad's has lost his tape measure and he has no other measuring instruments. However, he finds that if he lays the carpet flat on the floor of his smaller rectangular (38 feet by 55 feet) storeroom, then each corner of the carpet touches a different wall of that room. Also if Cheherazad lays the carpet flat on the floor of his larger rectangular (50 feet by 55 feet) storeroom, each corner of the carpet touches a different wall of that room. What are the carpet dimensions?&lt;/p&gt;&#xA;" OwnerUserId="19470" LastEditorUserId="8874" LastEditDate="2016-03-13T15:40:22.387" LastActivityDate="2016-03-19T14:20:09.643" Title="Cheherazad, the carpet salesman" Tags="|mathematics|geometry|algebra|" AnswerCount="2" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28799" PostTypeId="1" AcceptedAnswerId="28809" CreationDate="2016-03-11T13:11:13.107" Score="5" ViewCount="1600" Body="&lt;p&gt;I own a normal domino game with 28 tiles. Each tile carries two of the numbers [0, 1, 2, 3, 4, 5, 6] (where 0 represents blank).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During cleanup five pieces fell out of the box. Two of them showed their backside. &lt;strong&gt;How big is the chance that you can arrange these five pieces into a valid domino line?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One valid example of such a line: &lt;strong&gt;&lt;code&gt;[2,5]&lt;/code&gt;&lt;/strong&gt; &lt;strong&gt;&lt;code&gt;[5,0]&lt;/code&gt;&lt;/strong&gt; &lt;code&gt;[0,3]&lt;/code&gt; &lt;code&gt;[3,4]&lt;/code&gt; &lt;strong&gt;&lt;code&gt;[4,4]&lt;/code&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CzOBI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CzOBI.jpg&quot; alt=&quot;domino tiles&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4342" LastEditorUserId="8874" LastEditDate="2016-03-11T14:17:13.567" LastActivityDate="2016-05-17T16:04:49.920" Title="Domino tile possibilities" Tags="|mathematics|combinatorics|probability|dominoes|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28800" PostTypeId="1" AcceptedAnswerId="28804" CreationDate="2016-03-11T13:23:54.003" Score="10" ViewCount="2055" Body="&lt;p&gt;You are a archeologist and have taken 8 of your students on a expedition trip to a ancient cave to try and find ancient treasure. After you and your team entered the cave, you travel as a group exploring the entire cave. There is only one tunnel left in the cave to explore. As you reach the end of the tunnel one of your students accidentally leans on a panel that opens up a secret entrance before you. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you enter you see ancient writing on the wall and a giant gem sitting on a pedestal. After sometime you and your students deciphered the ancient writing, it says &quot;Beware of the curse! The darts will control whether you lie or tell the truth and the green gas will turn you into a hideous beast. The only way to return back to normal is to escape alive&quot;. Some of your students laugh and call it a bluff designed to scare and prevent theft of the treasure hidden in the cave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You decided to take a chance and take the gem off of the pedestal. As you do, two darts fly out of nowhere, barely missing you, and hit two of your students. You jumped away from the pedestal just in time as green gas sprays all 8 of the students. After the gas clears, everyone but you, look exactly the same. You can't tell them apart. One of the students points behind you and says &quot;What's that?!&quot;. You turn around to see purple gas starting to fill the room, naturally you and your students start running back to the entrance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After escaping the tunnel you came from, you have to make a choice between 4 other tunnels. From what you recall it took around 4 minutes to travel from one end of the tunnel to the other. The gas will reach where you are in about 10 minutes. You decide to split up into groups, check each tunnel, and report back here to see which tunnel is the correct way out. Which means there is only have 2 minutes to discuss and make a decision before the gas reaches the groups location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7BJZ2.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7BJZ2.png&quot; alt=&quot;Ancient Cave&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember, two of the students can't control if they lie or not, and they all look the same so you can't tell them apart. How many groups should there be when you all split up? How many people per group? and how can you tell which is the correct tunnel to escape through? &lt;/p&gt;&#xA;" OwnerUserId="4645" LastEditorUserId="12014" LastEditDate="2016-03-11T14:00:41.873" LastActivityDate="2016-03-11T17:09:43.420" Title="The Expedition Into The Ancient Cave" Tags="|logical-deduction|strategy|liars|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="28801" PostTypeId="2" ParentId="28796" CreationDate="2016-03-11T14:09:08.903" Score="9" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The only possible unit digit is $5$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reasoning&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We'll assume $m&amp;gt;2$ so the result is non-negative. If $m$ is even then Roohullah's result is even and not prime; hence we'll only consider odd $m$. &lt;br&gt;&lt;br&gt;If $m$ ends in $1$ then Roohullah's result modulo $10$ is &lt;br&gt; $2m-3-4 \equiv 5 \bmod  10$. &lt;br&gt;&lt;br&gt; If $m$ ends in $3$, Roohullah's result modulo $10$ is $(3+9+7)+(1+3+9+7)\left(\frac{m-3}{2}\right)-4 \equiv 5 \bmod 10$&lt;br&gt;&lt;br&gt; If $m$ ends in $5$, Roohullah's result modulo $10$ is &lt;br&gt; $10m-15-4 \equiv 1 \bmod 10$ &lt;br&gt;&lt;br&gt; If $m$ ends in $7$, Roohullah's result modulo $10$ is $(7+9+3)+(1+7+9+3)\left(\frac{m-3}{2}\right)-4\equiv 5 \bmod 10$ &lt;br&gt;&lt;br&gt; If $m$ ends in $9$, Roohullah's result modulo $10$ is &lt;br&gt;$9+(1+9)\left(m-1 \right) -4\equiv 5 \bmod 10$ &lt;br&gt;&lt;br&gt; Since a result ending in $5$ won't be a prime number, the only possibility for the last digit of $m$ is $5$. To show that the prime criteria can be achieved, note that &lt;br&gt; $5 + 5^2 + 5^3 + \ldots + 5^7 - 4 = 97651$ which is prime. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18422" LastEditorUserId="19470" LastEditDate="2016-03-19T14:21:09.913" LastActivityDate="2016-03-19T14:21:09.913" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28803" PostTypeId="2" ParentId="28800" CreationDate="2016-03-11T14:28:52.380" Score="-1" Body="&lt;p&gt;Assuming that the darts &lt;strong&gt;would&lt;/strong&gt; control who lies, so the two students hit by darts will lie. Split the students and us into three groups of three. Each group goes down one tunnel, which will cover three of the four possibilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the students hit with darts don't know that they're lying, and we can only ask them one question after returning, then when everyone returns we can ask everyone from every group if they found the exit. If any group has all three people say &quot;yes&quot;, then that's the exit, otherwise there are two possibilities;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: two groups have mixed answers, so both groups have a liar. If either group had two &quot;yes&quot; then that's the exit. If both had two &quot;no&quot;, neither of them have the exit, and the exit is through the last, unchecked tunnel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2: one group has mixed answers, so it has both liars, so if it has two &quot;no&quot;, then those are the two lairs and that's the exit. If it has two &quot;yes&quot; then those are the two liars, and it's not the exit, and we go through the last, unchecked tunnel.&lt;/p&gt;&#xA;" OwnerUserId="18972" LastActivityDate="2016-03-11T14:28:52.380" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28804" PostTypeId="2" ParentId="28800" CreationDate="2016-03-11T14:33:17.377" Score="6" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;EDIT: Handled the unpredictable nature of the affected students this time.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thinking quickly, you decide to send &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 3 students down one tunnel, 3 down another, 2 down a third, and you check the forth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In 8 minutes all return back and you make your decision.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You know for certain the status of your own tunnel, so if it is the exit gather them up and leave.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Otherwise, that was not the exit, so it must be in the other tunnels.  Listen to what the students have to say.  The possibilities are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; * Both 3 person groups are unanimous in their report&lt;br&gt;&#xA; * One 3 person group is in disagreement and the 2 person group is unanimous&lt;br&gt;&#xA; * One 3 person group is in disagreement and the 2 person group is in disagreement&lt;br&gt;&#xA; * Both 3 person groups are in disagreement  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But these are all easily resolvable:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; * For the first case, both 3 person groups have a truth teller and thus are not lying.  If neither indicate the exit, the 2 person tunnel will be the exit, regardless of how they answered.&lt;br&gt;&#xA; * For the second case, we don't know if there are two liars in this group, or if there are two truth tellers.  Regardless, we know all the other groups are telling the truth since they have truth tellers and are in agreement.  Thus, infer the truth value of this group based on the other two groups.  For example, if the other two groups didn't find the exit, then we know this group did find the exit regardless of how they answer.&lt;br&gt;&#xA; * For the third case, we know that there are 2 truth tellers in the 3 person group and only one liar since the other liar must be in the 2 person group.  So use the majority decision of both 3 person groups to infer the status of the two person tunnel.&lt;br&gt;&#xA; * For the last case, if both 3 person groups are in disagreement, then they both have a single affected person.  But that means the other two people in the group are unaffected, so simply use the majority decision of all three groups.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus, you can easily infer the status of the tunnels and can safely make your escape!&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="1592" LastEditDate="2016-03-11T15:55:12.870" LastActivityDate="2016-03-11T15:55:12.870" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28805" PostTypeId="1" AcceptedAnswerId="28808" CreationDate="2016-03-11T15:07:43.487" Score="9" ViewCount="1172" Body="&lt;p&gt;Which computer-made song is this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0113H&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(And what computer made it?)&lt;/p&gt;&#xA;" OwnerUserId="19350" LastEditorUserId="17482" LastEditDate="2016-03-11T20:07:59.097" LastActivityDate="2016-03-11T20:07:59.097" Title="Which computer-made song is this?" Tags="|riddle|music|humor|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28806" PostTypeId="2" ParentId="28799" CreationDate="2016-03-11T15:16:56.637" Score="3" Body="&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $10%$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Henceforth the notation &lt;kbd&gt;2 $|$ 5&lt;/kbd&gt;-&lt;kbd&gt;5 $|$ &amp;nbsp;&lt;/kbd&gt; refers to the event that piece &lt;kbd&gt;2 $|$ 5&lt;/kbd&gt; is joined to &lt;kbd&gt;5 $|$ &amp;nbsp;&lt;/kbd&gt;. the notation &lt;kbd&gt;2 $|$ &amp;nbsp;&lt;/kbd&gt;' refers to a &quot;reduced&quot; piece, for example &lt;kbd&gt;2 $|$ 5&lt;/kbd&gt;-&lt;kbd&gt;5 $|$ &amp;nbsp;&lt;/kbd&gt; effectively can be considered as a single &lt;kbd&gt;2 $|$ &amp;nbsp;&lt;/kbd&gt;' piece, as long as we bear in mind that the actual &lt;kbd&gt;2 $|$ &amp;nbsp;&lt;/kbd&gt; piece hasn't been used. The notation &lt;kbd&gt;2 $|$ *&lt;/kbd&gt; refers to anything containing a 2. square brackets enclose a set of pieces that need to be joined.&lt;br&gt;&#xA;   Since there's a &lt;kbd&gt;4 $|$ 4&lt;/kbd&gt; piece, at least one of the face down pieces needs to have a 4. Call this piece $D$. Consider each possibility for $D$:&lt;/p&gt;&#xA;  &#xA;  &lt;hr&gt;&#xA;  &#xA;  &lt;p&gt;$D=$&lt;kbd&gt;4 $|$ &amp;nbsp;&lt;/kbd&gt;.&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;p&gt;If &lt;kbd&gt;2 $|$ 5&lt;/kbd&gt;-&lt;kbd&gt;5 $|$ &amp;nbsp;&lt;/kbd&gt;, then we have $[$&lt;kbd&gt;4 $|$ &amp;nbsp;&lt;/kbd&gt;', &lt;kbd&gt;2 $|$ &amp;nbsp;&lt;/kbd&gt;',$E]$.&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;If &lt;kbd&gt;4 $|$ &amp;nbsp;&lt;/kbd&gt;'-&lt;kbd&gt;2 $|$ &amp;nbsp;&lt;/kbd&gt;', then $E$ can be &lt;kbd&gt;2 $|$ *&lt;/kbd&gt; or &lt;kbd&gt;4 $|$ *&lt;/kbd&gt;, barring already used pieces.&lt;/li&gt;&#xA;  &lt;/ul&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;If not &lt;kbd&gt;2 $|$ 5&lt;/kbd&gt;-&lt;kbd&gt;5 $|$ &amp;nbsp;&lt;/kbd&gt;, then it must be that &lt;kbd&gt;4 $|$ &amp;nbsp;&lt;/kbd&gt;'-&lt;kbd&gt;&amp;nbsp; $|$ 5&lt;/kbd&gt;, therefore &lt;kbd&gt;4 $|$ 5&lt;/kbd&gt;' and &lt;kbd&gt;5 $|$ 2&lt;/kbd&gt; need to be joined but not on the fives. This can be done with &lt;kbd&gt;4 $|$ 5&lt;/kbd&gt;,&lt;kbd&gt;4 $|$ 2&lt;/kbd&gt;,&lt;kbd&gt;5 $|$ 5&lt;/kbd&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;Joining the 2 sets we get:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;&lt;kbd&gt;2 $|$ &amp;nbsp;&lt;/kbd&gt;,&lt;kbd&gt;2 $|$ 1&lt;/kbd&gt;,&lt;kbd&gt;2 $|$ 2&lt;/kbd&gt;,&lt;kbd&gt;2 $|$ 3&lt;/kbd&gt;,&lt;kbd&gt;2 $|$ 4&lt;/kbd&gt;,&lt;kbd&gt;2 $|$ 6&lt;/kbd&gt;,&lt;kbd&gt;4 $|$ 1&lt;/kbd&gt;,&lt;kbd&gt;4 $|$ 2&lt;/kbd&gt;,&lt;kbd&gt;4 $|$ 3&lt;/kbd&gt;,&lt;kbd&gt;4 $|$ 5&lt;/kbd&gt;,&lt;kbd&gt;4 $|$ 6&lt;/kbd&gt;,&lt;kbd&gt;&amp;nbsp; $|$ &amp;nbsp;&lt;/kbd&gt;,&lt;kbd&gt;5 $|$ 5&lt;/kbd&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;hr&gt;&#xA;  &#xA;  &lt;p&gt;$D=$&lt;kbd&gt;4 $|$ 1&lt;/kbd&gt;.&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;p&gt;If &lt;kbd&gt;2 $|$ 5&lt;/kbd&gt;-&lt;kbd&gt;5 $|$ &amp;nbsp;&lt;/kbd&gt;, then $[$&lt;kbd&gt;4 $|$ 1&lt;/kbd&gt;',&lt;kbd&gt;2 $|$ &amp;nbsp;&lt;/kbd&gt;',$E]$. The first 2 of these have no numbers in common, therefore new solutions are &lt;strong&gt;&lt;kbd&gt;4 $|$ 2&lt;/kbd&gt;,&lt;kbd&gt;1 $|$ 2&lt;/kbd&gt;,&lt;kbd&gt;1 $|$ &amp;nbsp;&lt;/kbd&gt;&lt;/strong&gt;.&lt;br&gt;&#xA;  Note that I don't include &lt;kbd&gt;4 $|$ &amp;nbsp;&lt;/kbd&gt;, as that would be a repeat of a solution already counted in $D=$&lt;kbd&gt;4 $|$ &amp;nbsp;&lt;/kbd&gt;, therefore is not new. &lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;If not &lt;kbd&gt;2 $|$ 5&lt;/kbd&gt;-&lt;kbd&gt;5 $|$ &amp;nbsp;&lt;/kbd&gt;, there are no solutions. &lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;hr&gt;&#xA;  &#xA;  &lt;p&gt;$D=$&lt;kbd&gt;4 $|$ 2&lt;/kbd&gt;.&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;p&gt;If &lt;kbd&gt;2 $|$ 5&lt;/kbd&gt;-&lt;kbd&gt;5 $|$ &amp;nbsp;&lt;/kbd&gt;, then we have $[$&lt;kbd&gt;4 $|$ 2&lt;/kbd&gt;',&lt;kbd&gt;2 $|$ &amp;nbsp;&lt;/kbd&gt;',$E]$.&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;p&gt;If &lt;kbd&gt;4 $|$ 2&lt;/kbd&gt;'-&lt;kbd&gt;2 $|$ &amp;nbsp;&lt;/kbd&gt;', then any new, not used &lt;kbd&gt;4 $|$ *&lt;/kbd&gt; or &lt;kbd&gt;&amp;nbsp; $|$ *&lt;/kbd&gt; will suffice. These are &lt;kbd&gt;4 $|$ 3&lt;/kbd&gt;,&lt;kbd&gt;4 $|$ 5&lt;/kbd&gt;,&lt;kbd&gt;4 $|$ 6&lt;/kbd&gt;,&lt;kbd&gt;&amp;nbsp; $|$ 1&lt;/kbd&gt;,&lt;kbd&gt;&amp;nbsp; $|$ 2&lt;/kbd&gt;,&lt;kbd&gt;&amp;nbsp; $|$ 3&lt;/kbd&gt;,&lt;kbd&gt;&amp;nbsp; $|$ 6&lt;/kbd&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;If not &lt;kbd&gt;4 $|$ 2&lt;/kbd&gt;'-&lt;kbd&gt;2 $|$ &amp;nbsp;&lt;/kbd&gt;' then solutions are &lt;kbd&gt;2 $|$ 2&lt;/kbd&gt;,&lt;kbd&gt;2 $|$ &amp;nbsp;&lt;/kbd&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;If not &lt;kbd&gt;2 $|$ 5&lt;/kbd&gt;-&lt;kbd&gt;5 $|$ &amp;nbsp;&lt;/kbd&gt;, then solutions are &lt;kbd&gt;4 $|$ 5&lt;/kbd&gt;,&lt;kbd&gt;5 $|$ 5&lt;/kbd&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;Joining the 2 sets we get:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;&lt;kbd&gt;4 $|$ 3&lt;/kbd&gt;,&lt;kbd&gt;4 $|$ 5&lt;/kbd&gt;,&lt;kbd&gt;4 $|$ 6&lt;/kbd&gt;,&lt;kbd&gt;&amp;nbsp; $|$ 1&lt;/kbd&gt;,&lt;kbd&gt;&amp;nbsp; $|$ 2&lt;/kbd&gt;,&lt;kbd&gt;&amp;nbsp; $|$ 3&lt;/kbd&gt;,&lt;kbd&gt;&amp;nbsp; $|$ 6&lt;/kbd&gt;,&lt;kbd&gt;5 $|$ 5&lt;/kbd&gt;,&lt;kbd&gt;2 $|$ 2&lt;/kbd&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;hr&gt;&#xA;  &#xA;  &lt;p&gt;$D=$&lt;kbd&gt;4 $|$ 3&lt;/kbd&gt;.&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;p&gt;If &lt;kbd&gt;2 $|$ 5&lt;/kbd&gt;-&lt;kbd&gt;5 $|$ &amp;nbsp;&lt;/kbd&gt;, then we have $[$&lt;kbd&gt;4 $|$ 3&lt;/kbd&gt;',&lt;kbd&gt;2 $|$ &amp;nbsp;&lt;/kbd&gt;',$E]$. Solutions are &lt;strong&gt;&lt;kbd&gt;3 $|$ 2&lt;/kbd&gt;,&lt;kbd&gt;3 $|$ &amp;nbsp;&lt;/kbd&gt;&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;If not &lt;kbd&gt;2 $|$ 5&lt;/kbd&gt;-&lt;kbd&gt;5 $|$ &amp;nbsp;&lt;/kbd&gt;, then $[$&lt;kbd&gt;4 $|$ 3&lt;/kbd&gt;',&lt;kbd&gt;5 $|$ 2&lt;/kbd&gt;,&lt;kbd&gt;5 $|$ &amp;nbsp;&lt;/kbd&gt;,$E]$ has no new solutions. &lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;hr&gt;&#xA;  &#xA;  &lt;p&gt;$D=$&lt;kbd&gt;4 $|$ 5&lt;/kbd&gt;.&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;p&gt;If &lt;kbd&gt;2 $|$ 5&lt;/kbd&gt;-&lt;kbd&gt;5 $|$ &amp;nbsp;&lt;/kbd&gt;, then $[$&lt;kbd&gt;4 $|$ 5&lt;/kbd&gt;',&lt;kbd&gt;2 $|$ &amp;nbsp;&lt;/kbd&gt;',$E]$. No new solutions.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;If not &lt;kbd&gt;2 $|$ 5&lt;/kbd&gt;-&lt;kbd&gt;5 $|$ &amp;nbsp;&lt;/kbd&gt;, then $[$&lt;kbd&gt;4 $|$ 5&lt;/kbd&gt;',&lt;kbd&gt;5 $|$ 2&lt;/kbd&gt;,&lt;kbd&gt;5 $|$ &amp;nbsp;&lt;/kbd&gt;,$E]$.&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;p&gt;If &lt;kbd&gt;5 $|$ &amp;nbsp;&lt;/kbd&gt;-&lt;kbd&gt;5 $|$ 4&lt;/kbd&gt;', then the only new solution to $[$&lt;kbd&gt;&amp;nbsp; $|$ 4&lt;/kbd&gt;',&lt;kbd&gt;5 $|$ 2&lt;/kbd&gt;,$E]$ is &lt;strong&gt;&lt;kbd&gt;&amp;nbsp; $|$ 2&lt;/kbd&gt;&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;If &lt;kbd&gt;4 $|$ 5&lt;/kbd&gt;'-&lt;kbd&gt;5 $|$ 2&lt;/kbd&gt;, no new sols. &lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;hr&gt;&#xA;  &#xA;  &lt;p&gt;$D=$&lt;kbd&gt;4 $|$ 6&lt;/kbd&gt;.&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;p&gt;If &lt;kbd&gt;2 $|$ 5&lt;/kbd&gt;-&lt;kbd&gt;5 $|$ &amp;nbsp;&lt;/kbd&gt;, then $[$&lt;kbd&gt;2 $|$ &amp;nbsp;&lt;/kbd&gt;',&lt;kbd&gt;4 $|$ 6&lt;/kbd&gt;',$E]$ has new solutions &lt;strong&gt;&lt;kbd&gt;2 $|$ 6&lt;/kbd&gt;,&lt;kbd&gt;&amp;nbsp; $|$ 6&lt;/kbd&gt;&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Otherwise, no new solutions.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Counting the &lt;strong&gt;bold&lt;/strong&gt; dominoes I get $30$. ${25\choose2} = 300$, therefore the probability is $\frac{30}{300} = 0.1$. Unless I've messed up somewhere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19698" LastEditorUserId="4421" LastEditDate="2016-03-11T18:28:08.653" LastActivityDate="2016-03-11T18:28:08.653" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28807" PostTypeId="2" ParentId="28797" CreationDate="2016-03-11T15:20:20.140" Score="6" Body="&lt;p&gt;Lets assume the carpet is $X \times Y$.  There are two angles $\theta$ and $\phi$, so that when the carpet is rotated by them, it can be inscribed inside another rectangle as stated in the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When rotated by $\theta$, the width changes to $X  \cos{\theta} + Y  \sin{\theta}$ and the height changes to $X  \sin{\theta} + Y  \cos{\theta}$.  Therefore, we know that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X  \cos{\theta} + Y  \sin{\theta} = 38$$&#xA;$$X  \sin{\theta} + Y  \cos{\theta} = 55$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly for $\phi$ except we don't know which is the length or height, so we know one of the following must be correct:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X  \cos{\phi} + Y \sin{\phi} = 50$$&#xA;$$X  \sin{\phi} + Y  \cos{\phi} = 55$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X  \cos{\phi} + Y  \sin{\phi} = 55$$&#xA;$$X  \sin{\phi} + Y  \cos{\phi} = 50$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using &lt;a href=&quot;https://www.wolframalpha.com/input/?i=X*cos(a)%2BY*sin(a)%3D38,+X*sin(a)%2BY*cos(a)%3D55,+X*cos(b)%2BY*sin(b)%3D50,X*sin(b)%2BY*cos(b)%3D55&quot; rel=&quot;nofollow&quot;&gt;wolframalpha&lt;/a&gt;, we get that $X=25$, $Y=50$ are the dimensions of the carpet.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="19470" LastEditDate="2016-03-19T14:20:09.643" LastActivityDate="2016-03-19T14:20:09.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28808" PostTypeId="2" ParentId="28805" CreationDate="2016-03-11T15:28:21.610" Score="19" Body="&lt;p&gt;The song is clearly supposed to be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hello&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which is a well known song by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Adele&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With some imagination it is pronounced the same as&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Dell, a well known computer brand&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2016-03-11T15:28:21.610" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28809" PostTypeId="2" ParentId="28799" CreationDate="2016-03-11T15:34:51.410" Score="3" Body="&lt;p&gt;This is a rather inelegant, brute-force way of solving it, but I think it works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ANSWER:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;30/300 = 10%.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Consider a graph with its vertices labelled 0 to 6.  A given domino tile connects two vertices in this graph (not necessarily distinct vertices, either;  the double-4 domino connects vertex #4 to itself.)  A line of dominoes will correspond to a path in this graph.  Finally, we can only have one line connecting any pair of vertices (again, not necessarily distinct vertices), since a standard set of dominoes only contains one of each tile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the dominoes that are exposed when they fell out of the box are connecting the following vertices:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tH7oz.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tH7oz.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We wish to find how many ways there are to select two more edges in this graph such that the resulting subgraph forms a path in the graph.  Note that this implies that the resulting subgraph can have at most two vertices with odd degree (shout-out to &lt;a href=&quot;https://en.wikipedia.org/wiki/Seven_Bridges_of_K%C3%B6nigsberg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Euler.&lt;/a&gt;)  We can break this down into sub-cases:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The resulting subgraph contains only the vertices (0, 2, 4, 5).  The ways that we can do this are as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) Vertex 4 is connected to Vertex 0 only.  Then Vertices 2 and 4 will have odd degree, and Vertices 0 and 5 have even degree.  The remaining edge must either connect a vertex with odd degree to a vertex with even degree, or connect a vertex to itself.  (In principle we could also connect two vertices of odd degree, but we already said that there was only one edge to Vertex 4.)  For the first case, we can only connect (2,0), since (2,5) is already taken;  for the second case, we can connect (0,0), (2,2), or (5,5).  So there are &lt;strong&gt;4 options&lt;/strong&gt; in this case.  (Note that my first answer for this case was wrong, and I didn't realize it until I closely examined @Marius's table of results.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Vertex 4 is connected to Vertex 2 only.  The above logic still holds, with the roles of Vertex 0 and Vertex 2 swapped.  So there are &lt;strong&gt;4 options&lt;/strong&gt; in this case as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c) Vertex 4 is attached to Vertex 5 only.  Then all four vertices have odd degree, and the remaining edge must connect two distinct vertices.  The only way left to do this is to attach Vertex 0 to Vertex 2 (by assumption, Vertex 4 is not attached to any of the others, and the only edge left to choose is the edge between Vertices 0 and 2.)  So there is &lt;strong&gt;1 option&lt;/strong&gt; in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(d)  Vertex 4 is attached to both Vertices 0 and 2.  This is &lt;strong&gt;one more valid option&lt;/strong&gt;, and there are no other choices to make.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(e)  Vertex 4 is attached to both Vertices 0 and 5.  This is &lt;strong&gt;one more valid option&lt;/strong&gt;, and there are no other choices to make.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(f)  Vertex 4 is attached to both Vertices 2 and 5.  This is &lt;strong&gt;one more valid option&lt;/strong&gt;, and there are no other choices to make.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The resulting graph contains one of the remaining vertices (1, 3, 6).  It cannot contain more than one such vertex, since this would necessarily result in a disconnected graph.  Let us denote this new vertex by &lt;em&gt;x&lt;/em&gt;.  We can distinguish then between the following cases:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a)  Vertex &lt;em&gt;x&lt;/em&gt; has degree 2.  Then one of the two edges attached to &lt;em&gt;x&lt;/em&gt; must go to Vertex 4 (since otherwise it will be disconnected) and the other must go to one of Vertices 0 or 2.  (As pointed out by @Trenin in the comments, it cannot go to Vertex 5, since this would result in four vertices with odd degree). Thus, there are two options for each value of &lt;em&gt;x&lt;/em&gt;, or a total of &lt;strong&gt;six options.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b)  Vertex &lt;em&gt;x&lt;/em&gt; has degree 1.  We can then distinguish between two sub-sub-cases:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i)  Vertex &lt;em&gt;x&lt;/em&gt; is connected to Vertex 4.  Then Vertex 4 will have odd degree, and so must be connected to another vertex of odd degree, which must be Vertex 0 or Vertex 2.  So there are two options for each value of &lt;em&gt;x&lt;/em&gt;, or &lt;strong&gt;six more options.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii)  Vertex &lt;em&gt;x&lt;/em&gt; is connected to Vertex 0 or 2.  Then Vertex 4 must be connected to the other one of those two vertices.  (Otherwise, there will be four vertices with odd degree.)  Thus, we have two more options for each value of &lt;em&gt;x&lt;/em&gt;, or &lt;strong&gt;six more options.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(iii) Vertex &lt;em&gt;x&lt;/em&gt; is connected to Vertex 5.  At this point, we have four vertices with odd degree (0, 2, 5, &lt;em&gt;x&lt;/em&gt;), and since we must also connect Vertex 4 to the rest of the graph, we cannot reduce the number of odd-degree vertices with our remaining edge.  So this option doesn't work at all.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; So, summing up:  there are $ 4 + 4 + 1 + 1 + 1 + 1 + 6 + 6 + 6 = 30$ ways to choose these two dominoes such that they form a connected path in this graph.  Since there are 25 unspecified dominoes in the set, there are $25 \times 24/2 = 300$ ways to choose two of them.  Thus, the chance of being able to assemble these into a line is &lt;strong&gt;30/300 = 10%.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; For the record, the options for domino choices are:  (4,0) &amp;amp; (2,2);  (4,0) &amp;amp; (2,0); (4,0) &amp;amp; (0,0); (4,0) &amp;amp; (5,5); (4,2) &amp;amp; (0,0);  (4,2) &amp;amp; (2,0); (4,2) &amp;amp; (2,2); (4,2) &amp;amp; (5,5); (4,5) &amp;amp; (0,2);  (4,0) &amp;amp; (4,2);  (4,0) &amp;amp; (4,5);  (4,2) &amp;amp; (4,5); (4,&lt;em&gt;x&lt;/em&gt;) &amp;amp; (0,&lt;em&gt;x&lt;/em&gt;); (4,&lt;em&gt;x&lt;/em&gt;) &amp;amp; (2,&lt;em&gt;x&lt;/em&gt;); (4,&lt;em&gt;x&lt;/em&gt;) &amp;amp; (4,0); (4,&lt;em&gt;x&lt;/em&gt;) &amp;amp; (4,2); (0,&lt;em&gt;x&lt;/em&gt;) &amp;amp; (2,4); and (2,&lt;em&gt;x&lt;/em&gt;) &amp;amp; (0,4).  Throughout, &lt;em&gt;x&lt;/em&gt; stands for 1, 3, or 6.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17035" LastEditorUserId="17035" LastEditDate="2016-03-11T16:49:25.680" LastActivityDate="2016-03-11T16:49:25.680" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28810" PostTypeId="2" ParentId="28770" CreationDate="2016-03-11T15:48:05.003" Score="1" Body="&lt;p&gt;I remember solving similar question before:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; By putting number 9 upside down, i get (13)+(7)+(6)+(3)+(1)=30&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2016-03-11T15:48:05.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28811" PostTypeId="2" ParentId="28125" CreationDate="2016-03-11T15:55:38.493" Score="-1" Body="&lt;p&gt;It's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Energy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm all over.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Energy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Over you. Under you. Through you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Energy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I love black holes and dying cosmos.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;...both of which involve enormous amounts of energy. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come from evil politicians and ruthless dictators.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It takes a lot of energy to maintain a fascist totalitarianism.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come from silly children and irresponsible criminals.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Again, energy. Silly children have loads of it, and irresponsible criminals use a lot making rash decisions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can cause horrible diseases.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Energy causes &lt;strong&gt;everything&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The government overregulates me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;DEFINITELY energy. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But that doesn't stop me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Of course not. Energy never stops. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nothing can stop me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Energy. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not walls. Not floors. Not Time or Space or Distance.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Energy. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I OWN THE WORLD!!!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Yep. Everything is owned by energy. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not that it does me much good.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Well, when you ARE everything, nothing does you much good.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes I think of my existence as a giant treadmill.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Yep. Energy is constantly moving at a constant speed (the speed of light). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="20092" LastActivityDate="2016-03-11T15:55:38.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28812" PostTypeId="2" ParentId="28799" CreationDate="2016-03-11T15:58:41.747" Score="3" Body="&lt;p&gt;I started answering in the math way, but 2 answers already appeared.&lt;br&gt;&#xA;So for the fun of it here it is &quot;the long way around&quot;:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0 = chain not possible&lt;br&gt;&#xA; 1 = chain formed&lt;br&gt;&#xA; X = cannot have this combination  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;      [0,0]||[0,1]||[0,2]||[0,3]||[0,4]||[0,5]||[0,6]||[1,1]||[1,2]||[1,3]||[1,4]||[1,5]||[1,6]||[2,2]||[2,3]||[2,4]||[2,5]||[2,6]||[3,3]||[3,4]||[3,5]||[3,6]||[4,4]||[4,5]||[4,6]||[5,5]||[5,6]||[6,6]&#xA;[0,0]|  X  | &#xA;[0,1]|  0  ||  X  | &#xA;[0,2]|  0  ||  0  ||  X  | &#xA;[0,3]|  0  ||  0  ||  0  ||  X  | &#xA;[0,4]|  1  ||  0  ||  1  ||  0  ||  X  | &#xA;[0,5]|  X  ||  X  ||  X  ||  X  ||  X  ||  X  | &#xA;[0,6]|  0  ||  0  ||  0  ||  0  ||  0  ||  X  ||  X  | &#xA;[1,1]|  0  ||  0  ||  0  ||  0  ||  0  ||  X  ||  0  ||  X  | &#xA;[1,2]|  0  ||  0  ||  0  ||  0  ||  1  ||  X  ||  0  ||  0  ||  X  | &#xA;[1,3]|  0  ||  0  ||  0  ||  0  ||  0  ||  X  ||  0  ||  0  ||  0  ||  X  | &#xA;[1,4]|  0  ||  1  ||  0  ||  0  ||  1  ||  X  ||  0  ||  0  ||  1  ||  0  ||  X  | &#xA;[1,5]|  0  ||  0  ||  0  ||  0  ||  0  ||  X  ||  0  ||  0  ||  0  ||  0  ||  0  ||  X  | &#xA;[1,6]|  0  ||  0  ||  0  ||  0  ||  0  ||  X  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  X  | &#xA;[2,2]|  0  ||  0  ||  0  ||  0  ||  1  ||  X  ||  0  ||  X  ||  0  ||  0  ||  0  ||  0  ||  0  ||  X  | &#xA;[2,3]|  0  ||  0  ||  0  ||  0  ||  1  ||  X  ||  0  ||  0  ||  0  ||  0  ||  1  ||  1  ||  0  ||  0  ||  X  | &#xA;[2,4]|  1  ||  1  ||  1  ||  1  ||  1  ||  X  ||  1  ||  0  ||  0  ||  0  ||  1  ||  0  ||  0  ||  1  ||  0  ||  X  | &#xA;[2,5]|  X  ||  X  ||  X  ||  X  ||  X  ||  X  ||  X  ||  X  ||  X  ||  X  ||  X  ||  X  ||  X  ||  X  ||  X  ||  X  ||  X  | &#xA;[2,6]|  0  ||  0  ||  0  ||  0  ||  1  ||  X  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  X  ||  X  | &#xA;[3,3]|  0  ||  0  ||  0  ||  0  ||  0  ||  X  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  X  ||  0  ||  X  | &#xA;[3,4]|  0  ||  0  ||  0  ||  1  ||  1  ||  X  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  1  ||  1  ||  X  ||  0  ||  0  ||  X  | &#xA;[3,5]|  0  ||  0  ||  0  ||  0  ||  0  ||  X  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  X  ||  0  ||  0  ||  0  ||  X  | &#xA;[3,6]|  0  ||  0  ||  0  ||  0  ||  0  ||  X  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  X  ||  0  ||  0  ||  0  ||  0  ||  X  | &#xA;[4,4]|  X  ||  X  ||  X  ||  X  ||  X  ||  X  ||  X  ||  X  ||  X  ||  X  ||  X  ||  X  ||  X  ||  X  ||  X  ||  X  ||  X  ||  X  ||  X  ||  X  ||  X  ||  X  ||  X  | &#xA;[4,5]|  0  ||  0  ||  1  ||  0  ||  1  ||  X  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  X  ||  0  ||  0  ||  0  ||  0  ||  0  ||  X  ||  X  | &#xA;[4,6]|  0  ||  0  ||  0  ||  0  ||  1  ||  X  ||  1  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  1  ||  X  ||  0  ||  0  ||  0  ||  0  ||  0  ||  X  ||  0  ||  X  | &#xA;[5,5]|  0  ||  0  ||  0  ||  0  ||  1  ||  X  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  1  ||  X  ||  0  ||  0  ||  0  ||  0  ||  0  ||  X  || 0/1 ||  0  ||  X  | &#xA;[5,6]|  0  ||  0  ||  0  ||  0  ||  0  ||  X  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  X  ||  0  ||  0  ||  0  ||  0  ||  0  ||  X  ||  0  ||  0  ||  0  ||  X  | &#xA;[6,6]|  0  ||  0  ||  0  ||  0  ||  0  ||  X  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  0  ||  X  ||  0  ||  0  ||  0  ||  0  ||  0  ||  X  ||  0  ||  0  ||  0  ||  0  ||  X  |  &#xA;      [0,0]||[0,1]||[0,2]||[0,3]||[0,4]||[0,5]||[0,6]||[1,1]||[1,2]||[1,3]||[1,4]||[1,5]||[1,6]||[2,2]||[2,3]||[2,4]||[2,5]||[2,6]||[3,3]||[3,4]||[3,5]||[3,6]||[4,4]||[4,5]||[4,6]||[5,5]||[5,6]||[6,6]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are $24*25 / 2$ combinations in total&lt;br&gt;&#xA; Now just count the 1. I got: &#xA;&lt;br&gt;&#xA;  - 31 if you allow forks. In this case the chance is $31/300 = 0.1033$.  So 10.33%&lt;br&gt;&#xA;  - 30 if you don't allow forks. In this case the chance is $30/300 = 0.10$.  So 10% &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19989" LastEditorUserId="19989" LastEditDate="2016-05-17T16:04:49.920" LastActivityDate="2016-05-17T16:04:49.920" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28813" PostTypeId="1" AcceptedAnswerId="29047" CreationDate="2016-03-11T16:48:39.627" Score="11" ViewCount="1137" Body="&lt;p&gt;I went to visit a friend the other day, and was invited inside.  When I entered, I spied an arrangement of various dominos, upon the table.  They were not arranged, standing up, as if she had been meaning to do anything spectacular with them, but the arrangement did seem to be important to her.  I inquired, of her, as to their meaning, but she simply smiled, and kept about her business.  Eventually, I let it go -- but I keep finding myself drawn to the sequence, as if it means something.&lt;/p&gt;&#xA;&lt;p&gt;Do you think you could help me figure it out?&lt;/p&gt;&#xA;&lt;p&gt;It looked like this:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Wemuq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Wemuq.jpg&quot; alt=&quot;Domino Sequence&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;h1&gt;Hints (Added as needed):&lt;/h1&gt;&#xA;&lt;h1&gt;1&lt;/h1&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; One time, upon the sea&lt;br /&gt;&#xA; I ordered a toody fruity&lt;br /&gt;&#xA; meal or three-ee&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h1&gt;2&lt;/h1&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; I could add another tag to this question, but it would give it away.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h1&gt;3&lt;/h1&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Something I should note.  This riddle is not nearly as complex as it seems.  A small group of children could possibly solve it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h1&gt;4&lt;/h1&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The number 14 has very little to do with this puzzle despite this sentence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h1&gt;5&lt;/h1&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; My friend's first name is Ella.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h1&gt;WARNING - SOLUTION BELOW&lt;/h1&gt;&#xA;&lt;p&gt;For anyone still trying to figure out how everything connects together, look at this.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/a1YTZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/a1YTZ.png&quot; alt=&quot;Solution&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19944" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-03-16T17:40:11.117" Title="The First Lady's Dominos" Tags="|riddle|visual|dominoes|" AnswerCount="5" CommentCount="25" ContentLicense="CC BY-SA 3.0" />
  <row Id="28814" PostTypeId="2" ParentId="28800" CreationDate="2016-03-11T16:57:26.963" Score="2" Body="&lt;p&gt;&lt;strong&gt;Bonus points:&lt;/strong&gt;&#xA;I think this solution would work even if you, the professor, had also been affected by the green gas or if EVERYONE has been darted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My answer is based on the assumption that the darted students are still in control of their actions, they can't just decide whether to lie or tell the truth. It also assumes that the tunnels aren't able to prevent involuntary non-return (such as being booby-trapped), as pointed out in the comments to the question post by Daedric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution is simple:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Split the group in 2 / 2 / 2 / 3 (one with you, doesn't matter which) and send a party down each tunnel. Instruct the parties to leave through the exit if they find it within 4 minutes of walking, and to come back to the center if it's a dead-end or it goes deeper in the mountain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What will happen is that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Only one party won't return to the center, and those have found the exit. Have everyone else walk down the tunnel people did not come back from. Everybody lives&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14435" LastEditorUserId="14435" LastEditDate="2016-03-11T17:09:43.420" LastActivityDate="2016-03-11T17:09:43.420" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28815" PostTypeId="1" AcceptedAnswerId="28820" CreationDate="2016-03-11T17:07:51.250" Score="6" ViewCount="187" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I have dealt with the bull many times&lt;br&gt;&#xA;  The first was the first of the titans&lt;br&gt;&#xA;  And the second was the second of the sword.&lt;br&gt;&#xA;  The third was the first of the fallen&lt;br&gt;&#xA;  And the fourth was the second of that set.&lt;br&gt;&#xA;  The latest was the first of two&lt;br&gt;&#xA;  though I am not involved in the second.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;" OwnerUserId="14233" LastEditorUserId="19470" LastEditDate="2016-03-19T15:02:12.753" LastActivityDate="2016-03-19T15:02:12.753" Title="The Torero (Bullfighter)" Tags="|riddle|knowledge|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="28816" PostTypeId="2" ParentId="282" CreationDate="2016-03-11T17:30:49.407" Score="-2" Body="&lt;p&gt;I suppose if the humans were willing to stretch the rules a little bit, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each guess could contain the color of the hat in front of the person guessing. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So if they knew &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Their hat was purple (because of the previous human's guess) and the next in line was a orange, they would guess &quot;Not Orange&quot;. If the next in line was purple, they would guess &quot;Purple&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first guess would be a 50/50 guess, but would give the next human (and so forth) the necessary information to survive.&lt;/p&gt;&#xA;" OwnerUserId="20095" LastEditorUserId="46903" LastEditDate="2018-09-16T19:05:37.233" LastActivityDate="2018-09-16T19:05:37.233" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28817" PostTypeId="2" ParentId="28393" CreationDate="2016-03-11T18:05:16.770" Score="2" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A Beetle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;In fact, I'm arguably more famous than my master, because I get around more--I'm more versatile and less likely to get in the way of people, and I don't need as much gas to get to the same place.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I think the VW Beetle could be seen as more well known than Volkswagen, and Volkswagen has a recent bad rep for their recent emissions scandal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Both myself and my master are currently being preserved by a very powerful organization for all of time.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This might be a reference to the Beatles and preservation of their music?  Or a reference to the John Lennon Museum?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;It's a shame that two of them are dead, as we really had a great thing going on; if I can't have them, I don't want nobody else. The oldest guy is still around, at least, but it just doesn't feel the same without his brothers, too.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  Of the Beatles, John Lennon and George Harrison are dead, Paul McCartney and Ringo Starr are still alive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Apparently, the three guys who made me originally worked with three others for a little while, though this was before my time. They also had a younger brother who did his own separate thing, but I never met him (and he's dead too, apparently).&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'm not sure how to actually match this clue, the history of the formation of the Beatles doesn't seem to match well but I might be missing something.  Also not sure who the younger brother is.  Maybe this is a reference to Volkswagen, I'll have to do some research.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Also, who's taller, my master or me? Well, that's actually an interesting question. It depends on whether or not he's allowed to show you his penis. Yes, really. Though I suppose it doesn't really matter anyway, because we're both about the same height regardless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, not that it matters much to me, but everyone generally likes to pretend my master's little brother doesn't exist.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Also not sure if this matches well, but it could be talking about the emissions scandal and measuring emissions values?  And the &quot;master's little brother&quot; could be one of the companies that Volkswagen owns, but most of them are well known (Audi, Porsche, Bugatti, Lamborghini) so I'm not sure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll probably update this answer soon as I do more research...&lt;/p&gt;&#xA;" OwnerUserId="13349" LastActivityDate="2016-03-11T18:05:16.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28818" PostTypeId="1" CreationDate="2016-03-11T18:47:17.527" Score="9" ViewCount="596" Body="&lt;p&gt;I am the luckiest lady around!&lt;br&gt;&#xA;I am the most constricted of all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am the richest lady on earth!&lt;br&gt;&#xA;Yet I own nothing at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The world bends to my will!&lt;br&gt;&#xA;Yet I possess no power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every girl wishes to be in my place!&lt;br&gt;&#xA;But I chose not to come and only leave at death.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/28595&quot;&gt;The Happy Prisoner&lt;/a&gt;   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The title is a pun, but I doubt anyone will catch it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19179" LastEditorUserId="19179" LastEditDate="2018-05-04T17:20:03.637" LastActivityDate="2018-05-04T17:20:03.637" Title="A Tail of One Lady" Tags="|riddle|" AnswerCount="4" CommentCount="12" ContentLicense="CC BY-SA 4.0" />
  <row Id="28819" PostTypeId="2" ParentId="28794" CreationDate="2016-03-11T19:18:28.580" Score="6" Body="&lt;p&gt;Making the events independent, we get a probabilty that is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_{i,j} = \mathbb{P}(\text{player }i\text{ wins }j\text{ games}) = {6 \choose j}\cdot 2^{-6}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(note: if six players wins 3 games, then the last player will also win 3 games)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for all players, assuming independent events (which clearly is not true), we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\prod_i p_{i,3} = \left({6 \choose 3}\cdot 2^{-6} \right)^6 \approx 0.000931.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I run a simulation and got $\approx 0.0013$, so it is off by about 40 %.&lt;/p&gt;&#xA;" OwnerUserId="12901" LastActivityDate="2016-03-11T19:18:28.580" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28820" PostTypeId="2" ParentId="28815" CreationDate="2016-03-11T19:26:38.880" Score="6" Body="&lt;p&gt;You could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ron Perlman&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first of the Titans could be &lt;em&gt;&lt;a href=&quot;http://www.imdb.com/title/tt0104029/?ref_=nm_flmg_dr_13&quot;&gt;Cronos&lt;/a&gt;&lt;/em&gt;&lt;br&gt;&#xA; The second of the sword could be &lt;em&gt;&lt;a href=&quot;http://www.imdb.com/title/tt0187738/?ref_=nm_flmg_dr_10&quot;&gt;Blade II&lt;/a&gt;&lt;/em&gt;&lt;br&gt;&#xA; The first and second of the fallen could be &lt;em&gt;&lt;a href=&quot;http://www.imdb.com/title/tt0167190/?ref_=nm_flmg_dr_9&quot;&gt;Hellboy&lt;/a&gt;&lt;/em&gt; and &lt;em&gt;&lt;a href=&quot;http://www.imdb.com/title/tt0411477/?ref_=nm_flmg_dr_7&quot;&gt;Hellboy II&lt;/a&gt;&lt;/em&gt;&lt;br&gt;&#xA; The latest that's the first of two could be &lt;em&gt;&lt;a href=&quot;http://www.imdb.com/title/tt1663662/?ref_=nm_flmg_dr_6&quot;&gt;Pacific Rim&lt;/a&gt;&lt;/em&gt;, since you are only rumored to be in the second&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because the bull is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;http://www.imdb.com/name/nm0868219/&quot;&gt;Guillermo del Toro&lt;/a&gt;, who wrote or directed each of the above&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2016-03-11T19:32:04.900" LastActivityDate="2016-03-11T19:32:04.900" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="28821" PostTypeId="2" ParentId="28794" CreationDate="2016-03-11T19:42:29.373" Score="1" Body="&lt;p&gt;Each cowboy would have 64 possible outcomes, of which 20 are 3 wins. 5/16^7 = 1.86e-8 ... Not very good odds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said I'm fairly certain my math is off somehow... At minimum, pretty sure I'm double counting all of the games, still even doubled, those odds would be pretty abysmal.&lt;/p&gt;&#xA;" OwnerUserId="8746" LastActivityDate="2016-03-11T19:42:29.373" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28822" PostTypeId="2" ParentId="28794" CreationDate="2016-03-11T20:01:47.497" Score="2" Body="&lt;p&gt;I'm not positive this is right, but I hopefully it's close. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $1/(6!)\approx 0.00139$&lt;br&gt;&lt;br&gt;&#xA; Let's say cowboy $i$ wins against cowboys $(i+1 \mod 7)$, $(i+2 \mod 7$),  $(i+3 \mod 7$), and loses against all others. The number of cyclically invariant ways to order the $7$ cowboys is $6!$, only one assignment will match the above requirements.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19843" LastActivityDate="2016-03-11T20:01:47.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28823" PostTypeId="2" ParentId="28776" CreationDate="2016-03-11T20:14:05.840" Score="1" Body="&lt;p&gt;Here is an alternate line of reasoning based on &lt;em&gt;astralfenix&lt;/em&gt;'s very helpful answer to solving the puzzle:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;p = Capacity of barrel in pints&#xA;((p - 3) / p) ** 3 = 0.5&#xA;(p - 3) / p = 0.5 ** (1 / 3)&#xA;C = 0.5 ** (1 / 3)&#xA;(p - 3) / p = C&#xA;p - 3 = C * p&#xA;p - 3 - C * p = 0&#xA;p - C * p = 3&#xA;1 * p - C * p = 3&#xA;(1 - C) * p = 3&#xA;p = 3 / (1 - C)&#xA;p = 3 / (1 - 0.5 ** (1 / 3))&#xA;p = 14.541966305589222&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1661" LastActivityDate="2016-03-11T20:14:05.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28824" PostTypeId="2" ParentId="28794" CreationDate="2016-03-11T20:18:40.853" Score="1" Body="&lt;p&gt;There are $7$ cowboys, and a total of $21$ games played $(6 + 5 + 4 + 3 + 2 + 1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is only one possible configuration (and $6!$ different reordering of players that can generate the same outcome) where each player wins $3$ games.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure how to format this well, but one such configuration is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Player 1 wins against 2, 3, 4, loses to 5, 6, 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Player 2 wins against 3, 4, 5, loses to 6, 7, 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Player 3 wins against 4, 5, 6, loses to 7, 1, 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Player 4 wins against 5, 6, 7, loses to 1, 2, 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Player 5 wins against 6, 7, 1, loses to 2, 3, 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Player 6 wins against 7, 1, 2, loses to 3, 4, 5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Player 7 wins against 1, 2, 3, loses to 4, 5, 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this particular configuration to happen, all 21 games have to proceed with this configuration (or be one that is congruent (same outcome of wins/losses with player numbers swapped, basically a different reordering of players).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives us a probability of this, but it doesn't match simulation results:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\dfrac{6!}{2^{21}} = \dfrac{5040}{2097152} \approx 0.00034332275$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$6!$ for the number of different orderings of players that produce this one outcome, $2^{21}$ for the number of different possible outcomes for 21 games.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I also can't seem to edit the math to make it a spoiler...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Welp, I wrote a simulation in R, and I get a probability of around $0.0012$ as well...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Updated R simulation - can try $n = 3$ and $n = 5$:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;all3wins = 0&#xA;&#xA;n = 7&#xA;p = 1&#xA;q = p + 1&#xA;g = (n*(n-1))/2&#xA;neededwins = (n-1)/2&#xA;winsmatrix = array(0,dim = c(n,1))&#xA;gamematrix = array(0, dim = c(g,2))&#xA;threematrix = array(3,dim = c(n,1))&#xA;&#xA;## Filling in the game matrix with correct pairings of 1v1&#xA;for (j in 1:g){&#xA;  gamematrix[j,1] = p&#xA;  gamematrix[j,2] = q&#xA;  q = q + 1&#xA;  if (q &amp;gt; n){&#xA;    p = p + 1&#xA;    q = p + 1&#xA;  }&#xA;}&#xA;&#xA;## Simulation part&#xA;for (qweq in 1:1000000){&#xA;winsmatrix = array(0,dim = c(n,1))&#xA;  # Loops through 21 games&#xA;  # Picks one of two winners from participants in game&#xA;  # Adds one win to their row in winmatrix&#xA;  for (j in 1:g)&#xA;  {&#xA;    w = gamematrix[j,sample(1:2,1)]&#xA;    winsmatrix[w] = winsmatrix[w] + 1&#xA;  }&#xA;&#xA;  # Counts how many times everyone gets 3 wins&#xA;  isall3win = all(winsmatrix == neededwins)&#xA;  all3wins = all3wins + isall3win&#xA;}&#xA;probability = all3wins/qweq&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I tried $n = 3$ and $n = 5$ cases, and I got them to be this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 3 players, getting 1 win per player is $\dfrac{2!}{2^3} = \dfrac{1}{4} = 0.25$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 5 players, my simulation gets this... $0.0235 \approx \dfrac{4!}{2^{10}} = 0.0234375$ (the mathematical expression is just something I came up with that fits).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes me think that this should work for $n$ players:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ p = \dfrac{(n-1)!}{2^t}$ where $t = \dfrac{n(n-1)}{2} = $ number of games played, but this doesn't match simulation results for n = 7.&lt;/p&gt;&#xA;" OwnerUserId="13349" LastEditorUserId="13349" LastEditDate="2016-03-12T02:49:20.617" LastActivityDate="2016-03-12T02:49:20.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28825" PostTypeId="2" ParentId="28818" CreationDate="2016-03-11T21:05:13.807" Score="23" Body="&lt;p&gt;You're: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lady Luck.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the luckiest lady around!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Self explanatory.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the most constricted of all.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You're completely &lt;strong&gt;constricted&lt;/strong&gt; by the laws of probably.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the richest lady on earth!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All riches come from some sort of luck. Even if you're born into it, your birth alone had some luck involved.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet I own nothing at all.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Luck has no possessions.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The world bends to my will!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everything, right down to quantum mechanics has some factor (however large or small) of probability/luck involved.  Even the earth's orbit has a slight wobble to it's &lt;strong&gt;bend&lt;/strong&gt; around the sun do to probability.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet I possess no power.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; But luck itself doesn't have any power.  It disperses energy in a random manner.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Every girl wishes to be in my place!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All girls (and boys) want to be the lucky one.    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I chose not to come and only leave at death.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Luck doesn't &lt;strong&gt;chose&lt;/strong&gt; to do anything and when you're dead &quot;your &lt;em&gt;luck&lt;/em&gt; is up&quot;.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pun:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The title and hint refer to this being a story (&lt;strong&gt;tale&lt;/strong&gt;) of Lady Luck and the luck of heads or &lt;strong&gt;tail&lt;/strong&gt;s in a coin toss.  You might not &lt;strong&gt;catch it&lt;/strong&gt; (the coin, riddle or pun) if you're &lt;em&gt;unlucky&lt;/em&gt;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastEditorUserId="10118" LastEditDate="2016-03-11T23:47:30.207" LastActivityDate="2016-03-11T23:47:30.207" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28827" PostTypeId="2" ParentId="28393" CreationDate="2016-03-11T22:52:51.850" Score="6" Body="&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The song &lt;em&gt;Stayin Alive&lt;/em&gt; by the Bee Gees   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was, for the most part, created to accompany something else.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It was the theme for &lt;em&gt;Saturday Night Fever&lt;/em&gt;.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It would seem like I'm just playing second fiddle to a jive-talking glory hog, but that's not actually the case; we both wound up being really famous and respected. In fact, I'm arguably more famous than my master, because I get around more--I'm more versatile and less likely to get in the way of people, and I don't need as much gas to get to the same place.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &quot;master&quot; would be &lt;em&gt;Saturday Night Fever&lt;/em&gt;, the movie for which it was made. The movie was a critical success, and the song was a huge hit. The movie made John Travolta a household name, and catapulted the Bee Gees to the top of the charts. The song still gets airplay.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But now it's all right, it's okay; we'll live to see another day. As of fairly recently (well after Stack Overflow was founded), both myself and my master are currently being preserved by a very powerful organization for all of time. We'll both be stayin' alive for a very long time, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The movie (and by extension the song as well) was recently added to the Library of Congress' National Film Registry.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is more than can be said of the three men to whom I most owe my existence. It's a shame that two of them are dead, as we really had a great thing going on; if I can't have them, I don't want nobody else. The oldest guy is still around, at least, but it just doesn't feel the same without his brothers, too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Bee Gees. They were a group of three Gibb brothers, the younger two of which have since passed. Only Barry Gibb, the oldest, survives.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Apparently, the three guys who made me originally worked with three others for a little while, though this was before my time. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The band has had 3 other people as members, but by the time of the movie, it was only the brothers Gibb.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They also had a younger brother who did his own separate thing, but I never met him (and he's dead too, apparently).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Andy Gibb was the youngest of the brothers, but was never part of the band, instead becoming a solo artist.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, who's taller, my master or me? Well, that's actually an interesting question. It depends on whether or not he's allowed to show you his penis. Yes, really. Though I suppose it doesn't really matter anyway, because we're both about the same height regardless.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not entirely sure here. I imagine we're talking in terms of popularity, or in sales? Not sure. I am sure that this is referring to the movies R-rated theatrical version versus its PG-rated cut that was made to accomodate the younger crowd that made the song such a hit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally, not that it matters much to me, but everyone generally likes to pretend my master's little brother doesn't exist.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The movie had a sequel, also called &lt;em&gt;Stayin Alive&lt;/em&gt;, is widely considered to be a terrible movie, even winning some Razzies as a result.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16874" LastEditorUserId="16874" LastEditDate="2016-03-11T23:11:13.333" LastActivityDate="2016-03-11T23:11:13.333" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28828" PostTypeId="2" ParentId="28794" CreationDate="2016-03-11T23:00:50.140" Score="7" Body="&lt;p&gt;It's pretty easy to calculate the exact result using a computer. I haven't found an easy way to calculate it by hand (yet), but if someone wants to check his result before posting:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\frac{2640}{2097152} = \frac{165}{131072} = 0.00125885009765625$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; Optimized version, counting only matching combinations. It's now possible to verify, that the number of matches for all odd $n$ up to $11$ is listed in &lt;a href=&quot;https://oeis.org/A007079&quot; rel=&quot;nofollow&quot;&gt;A007079&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;public class Main {&#xA;    private static final int SIZE = 7;&#xA;    private static int[] rowSums = new int[SIZE + 1];&#xA;    private static long count = 0;&#xA;&#xA;    private static void rec(int row, int col, int sum) {&#xA;        if (row &amp;gt;= SIZE) {&#xA;            ++ count;&#xA;        } else {&#xA;            if (col &amp;gt;= SIZE) {&#xA;                if (sum == SIZE / 2) {&#xA;                    rec(row + 1, row + 2, rowSums[row + 1]);&#xA;                }&#xA;            } else {&#xA;                if (sum &amp;lt; SIZE / 2) {&#xA;                    rec(row, col + 1, sum  + 1);&#xA;                }&#xA;                if (rowSums[col] &amp;lt; SIZE / 2) {&#xA;                    ++ rowSums[col];&#xA;                    rec(row, col + 1, sum);&#xA;                    -- rowSums[col];&#xA;                }&#xA;            }&#xA;        }&#xA;    }&#xA;&#xA;    public static void main(String[] args) {&#xA;        rec(0, 1, 0);&#xA;        System.out.printf(&quot;%d / %d\n&quot;, count, 1L &amp;lt;&amp;lt; (SIZE * (SIZE - 1) / 2));&#xA;    }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="15632" LastEditDate="2016-03-12T12:44:40.380" LastActivityDate="2016-03-12T12:44:40.380" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28829" PostTypeId="2" ParentId="28776" CreationDate="2016-03-11T23:15:43.497" Score="3" Body="&lt;p&gt;Each time wine is replaced the wine remaining is $\frac{n-3}{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{(n-3)^3}{n^3} = \frac{1}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{2(n-3)^3}{n^3} = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2(n-3)^3 = n^3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt[3]{2(n-3)^3} = \sqrt[3]{n^3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(n-3)\sqrt[3]{2} = n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n\sqrt[3]{2} - n - 3\sqrt[3]{2} = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n\sqrt[3]{2} - n = 3\sqrt[3]{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n(\sqrt[3]{2} - 1) = 3\sqrt[3]{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n = \frac{3\sqrt[3]{2}}{\sqrt[3]{2} - 1}$&lt;/p&gt;&#xA;" OwnerUserId="9696" LastActivityDate="2016-03-11T23:15:43.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28830" PostTypeId="2" ParentId="28595" CreationDate="2016-03-11T23:23:46.430" Score="1" Body="&lt;p&gt;You are:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Morgan Freeman  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was the freest man around.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Free&quot;man  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet I never left my grounds&lt;br&gt;&#xA;(for years at a stretch)  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Can't figure this one out&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I could walk, fight, or run.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wanted&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm as able-bodied as anyone!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Invictus?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And my riches I can't count!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Million Dollar Baby  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I rarely ventured out.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I'll put the bucket list here&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm stable mentally and emotionally  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Teachers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And many dream of being me.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DreamCatcher&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also the title:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Shawshank Redemption (credit to manshu)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2016-03-11T23:23:46.430" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28832" PostTypeId="2" ParentId="28595" CreationDate="2016-03-11T23:39:33.473" Score="0" Body="&lt;p&gt;It better not be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Gordon Freeman&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Howard Hughes is a way better answer. ;P&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19128" LastActivityDate="2016-03-11T23:39:33.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28833" PostTypeId="2" ParentId="28818" CreationDate="2016-03-11T23:41:15.020" Score="4" Body="&lt;p&gt;It better not be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chell (from the Portal series).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the luckiest lady around!&#xA;I am the most constricted of all.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Only person around. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the richest lady on earth!&#xA;Yet I own nothing at all.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, only person around.  Unlike most video game characters, has zero inventory space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The world bends to my will!&#xA;Yet I possess no power.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;She has no poowers, but bends the world with her portal gun.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Every girl wishes to be in my place!&#xA;But I chose not to come and only leave at death.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;She chose not to come to the party with cake, she is in the facility against her will, and only leaves when either she or the boss dies. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A Tail of One Lady  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;She has a ponytail.&lt;br&gt;&#xA;Also, didn't she get mocked for being born with a tail?  I can't remember.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19128" LastEditorUserId="19128" LastEditDate="2016-03-11T23:46:53.783" LastActivityDate="2016-03-11T23:46:53.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28835" PostTypeId="1" AcceptedAnswerId="28837" CreationDate="2016-03-12T01:31:07.433" Score="7" ViewCount="401" Body="&lt;p&gt;Is it possible to checkmate an opponent with a king and a pawn without the pawn promoting to a queen or rook? I've heard that it is possible, but I don't know what the strategy is.&lt;/p&gt;&#xA;" OwnerUserId="20088" LastEditorUserId="11876" LastEditDate="2016-03-12T01:56:43.170" LastActivityDate="2016-03-12T06:30:23.897" Title="Checkmate with Pawn and King" Tags="|chess|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="28837" PostTypeId="2" ParentId="28835" CreationDate="2016-03-12T02:14:28.593" Score="7" Body="&lt;p&gt;The checkmated king is in one of his corners with one of his bishops on the same rank right beside him (blocking him). The checkmating pawn is in front of the bishop (checkmating the king) with the winning king two squares in front of the checkmated king on the same file protecting his pawn and blocking the checkmated king's move forward.  Of course there can be no piece available to capture the checkmating pawn. &lt;a href=&quot;https://i.stack.imgur.com/UE16n.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UE16n.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10118" LastEditorUserId="10118" LastEditDate="2016-03-12T06:30:23.897" LastActivityDate="2016-03-12T06:30:23.897" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="28841" PostTypeId="1" AcceptedAnswerId="28842" CreationDate="2016-03-12T02:37:48.257" Score="4" ViewCount="156" Body="&lt;p&gt;Is it possible to have a two knights + king vs king checkmate if the king is very smart? I know that if the king goes to the corner, it is possible to checkmate, but is it possible to checkmate in any other way?&lt;/p&gt;&#xA;" OwnerUserId="20088" LastActivityDate="2016-03-12T02:43:39.270" Title="2 Knight Checkmate" Tags="|chess|" AnswerCount="1" CommentCount="5" ClosedDate="2016-03-12T10:28:48.833" ContentLicense="CC BY-SA 3.0" />
  <row Id="28842" PostTypeId="2" ParentId="28841" CreationDate="2016-03-12T02:43:39.270" Score="3" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Two_knights_endgame#Two_knights_cannot_force_checkmate&quot; rel=&quot;nofollow&quot;&gt;No&lt;/a&gt;, two knights cannot force a checkmate.&lt;/p&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2016-03-12T02:43:39.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28843" PostTypeId="1" AcceptedAnswerId="28853" CreationDate="2016-03-12T07:58:26.263" Score="5" ViewCount="269" Body="&lt;p&gt;There are many who wear me&lt;br&gt;&#xA;I protect them from harm&lt;br&gt;&#xA;Others say I'm lucky&lt;br&gt;&#xA;And use me for charm&lt;br&gt;&#xA;I can cause an attraction&lt;br&gt;&#xA;Of a physical sort&lt;br&gt;&#xA;While some folks will throw me&lt;br&gt;&#xA;For amusement and sport&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who am I - can you guess?&lt;br&gt;&#xA;I'm so much like you&lt;br&gt;&#xA;You'll nail it, for sure&lt;br&gt;&#xA;If you think this one through&lt;/p&gt;&#xA;" OwnerDisplayName="user17947" LastActivityDate="2016-03-12T23:05:57.890" Title="A nebulous who am I riddle" Tags="|riddle|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28844" PostTypeId="2" ParentId="28652" CreationDate="2016-03-12T08:01:02.650" Score="6" Body="&lt;p&gt;Here's a &amp;ldquo;seating arrangement&amp;rdquo;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 7 friends A,B,C,G,P,R,Y meet on $n\!$ =&amp;hairsp;7 evenings &lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/4wiJ9.png&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Vertices ($\bullet$) = evenings.&amp;nbsp;&#xA;Large labeled triangles = friends.&amp;nbsp;&#xA;Small labeled triangles = outward-pointing corners of large triangles,&#xA;                          where they meet other triangles.&amp;nbsp;&#xA;(Center vertex =  first evening, where/when&#xA; the orange triangles A,B,C are the first three friends to meet.)&#xA;&lt;br&gt;&amp;bull;&lt;sup&gt;&lt;sup&gt;&amp;ensp;&lt;/sup&gt;&lt;/sup&gt;&#xA;    3 friends/triangles meet on/at each evening/vertex.&#xA;&lt;br&gt;&amp;bull;&lt;sup&gt;&lt;sup&gt;&amp;ensp;&lt;/sup&gt;&lt;/sup&gt;&#xA;    Each pair of evenings/vertices are connected by exactly one&#xA;    friend&amp;thinsp;/&amp;thinsp;triangle-side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/0FpvA.png&quot;&gt;&#xA; &lt;br&gt;&lt;strong&gt;Left:&lt;/strong&gt;&lt;sup&gt;&lt;sup&gt;&amp;ensp;&lt;/sup&gt;&lt;/sup&gt;&#xA; No friend F can be present on 4 or more evenings because that would require&#xA; at least 4 other friends to be present on some other single evening.&#xA; (Some evening E is required where F is not present;&#xA;  E is required to share a friend with each of F's 4 evenings,&#xA;  but all other friends on F's 4 evenings would be different&#xA;  from each other or else they'd be with F on two evenings.)&#xA; &lt;br&gt;&lt;strong&gt;Center:&lt;/strong&gt;&lt;sup&gt;&lt;sup&gt;&amp;ensp;&lt;/sup&gt;&lt;/sup&gt;&#xA; At most $n\!$ =&amp;hairsp;7 evenings remain in play.&#xA; (Each of the first 3 friends can be present on at most 2 other evenings.)&#xA; The light blue triangles show possible other friends, as long as their&#xA; extra corners can be connected back to the 6 perimeter vertices.&#xA; &lt;br&gt;&lt;strong&gt;Right:&lt;/strong&gt;&lt;sup&gt;&lt;sup&gt;&amp;ensp;&lt;/sup&gt;&lt;/sup&gt;&#xA; The solution unfolded with multiple images of each friend/triangle&#xA; so that all evenings/vertices can be seen.&#xA; (The nights/vertices are equivalent to a &lt;a href=&quot;https://en.wikipedia.org/wiki/Four_color_theorem#Generalizations&quot; rel=&quot;nofollow noreferrer&quot;&gt;7-colored torus&lt;/a&gt;.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18129" LastEditorUserId="18129" LastEditDate="2016-03-12T13:15:23.963" LastActivityDate="2016-03-12T13:15:23.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28845" PostTypeId="1" AcceptedAnswerId="28887" CreationDate="2016-03-12T09:24:55.680" Score="6" ViewCount="371" Body="&lt;p&gt;Let $f(x)$ be a function such that&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$f(1111)=4$&lt;/li&gt;&#xA;&lt;li&gt;$f(1234)=3$&lt;/li&gt;&#xA;&lt;li&gt;$f(4567)=2$&lt;/li&gt;&#xA;&lt;li&gt;$f(1357)=4$&lt;/li&gt;&#xA;&lt;li&gt;$f(6518)=4$&lt;/li&gt;&#xA;&lt;li&gt;$f(3817)=6$&lt;/li&gt;&#xA;&lt;li&gt;$f(8008)=6$&lt;/li&gt;&#xA;&lt;li&gt;$f(1000)=4$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Then, $f(2014)=\,\,?$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Hint: Each digit is valuable.)&lt;/p&gt;&#xA;" OwnerUserId="12690" LastEditorUserId="8874" LastEditDate="2016-03-13T14:15:59.247" LastActivityDate="2016-03-13T14:15:59.247" Title="Function(al) Puzzle" Tags="|mathematics|pattern|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28846" PostTypeId="1" AcceptedAnswerId="29251" CreationDate="2016-03-12T09:29:27.310" Score="4" ViewCount="1385" Body="&lt;p&gt;Say there are 100 perfectly intelligent prisoners (who don't have incredibly good memories) who are jailed for life (and also live eternally). The warden decides to play a game with them. There is a room in the prison which has one 3x3 Rubik's Cube and a robot which can solve it. Every day, the warden chooses a prisoner at random, regardless of whether he has entered the room before or not, and leaves him there until he comes out (of his own accord). While inside, he can do anything to the 3x3 Rubik's Cube in there and let the machine solve it but is not allowed to touch anything else or to take anything in / leave with something (That includes making a visible mark on ANYTHING, though you're allowed to turn the cube).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one prisoner comes up to the warden while in the room and correctly says, &quot;All 100 prisoners have entered this room since the game began,&quot; all of the prisioners will be set free. However, if the prisoner is incorrect, all of them will be immediately executed. The prisoners are all confined in their own cells with no way of communication with any other prisoner. However, all the prisoners have one tool each, a permanent marker, which &lt;em&gt;must not leave their cell&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The day before the game begins, the warden lets the prisioners go into the courtyard to formulate an action plan. If the prisoners all want to escape, what will their plan be?&lt;/p&gt;&#xA;" OwnerUserId="19640" LastEditorUserId="5373" LastEditDate="2016-10-24T00:30:16.500" LastActivityDate="2020-08-29T09:53:44.613" Title="One Hundred Prisoners and a Rubik's Cube" Tags="|logical-deduction|strategy|rubiks-cube|" AnswerCount="8" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28847" PostTypeId="1" AcceptedAnswerId="29221" CreationDate="2016-03-12T09:40:44.953" Score="3" ViewCount="360" Body="&lt;p&gt;Last night, you went to a party and someone knocked you out, then grabbed you and locked you in a small room. You're trapped!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After awaking and examining the room, you see that it's a storage locker. On one of the shelves inside the room, there's a small box with a tiny machine inside. It looks like a lock picker, but much more complex. You observe that you need a 3-digit password to use the machine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inside the box, there is also a note:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;496620796f752077616e7420746f2075736520746865206d616368696e652e2e2e205573652074686520636f6465 111-111-111--111-1--111-11&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the password? Solve it quickly before you &lt;strong&gt;die&lt;/strong&gt; (In 1 or 2 days)!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not that hard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This puzzle is &lt;strong&gt;not&lt;/strong&gt; related to math.&lt;/p&gt;&#xA;" OwnerDisplayName="user18031" LastEditorDisplayName="user18031" LastEditDate="2016-06-14T13:35:38.147" LastActivityDate="2016-06-14T13:35:38.147" Title="Encoding Extreme #5" Tags="|number-sequence|password|" AnswerCount="3" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28849" PostTypeId="2" ParentId="28846" CreationDate="2016-03-12T10:07:12.070" Score="15" Body="&lt;p&gt;They simply decide on 98 different valid Rubik's cube states at the beginning of the game and denote them 1 to 98. On the first day, the prisoner leaves the cube in any state apart from the pre-defined 98. We can call this state 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After this, everyone (including the 1st prisoner) follows the same strategy. Whenever a prisoner sees the cube for the first time, he takes it from the current state to the next one. So suppose it is currently in 23rd state, he just takes it to the 24th state. If a prisoner has already seen the cube he does nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a prisoner finds the cube in the 98th state the first time he sees it, then he knows that everyone else has already seen it. Hence he can declare it.&lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2016-03-12T10:07:12.070" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28850" PostTypeId="2" ParentId="28846" CreationDate="2016-03-12T10:19:59.523" Score="4" Body="&lt;p&gt;Before leaving their cell for the first time, the prisoner breaks the permanent marker and wets their finger with the ink. If they have ever left their cell before, they take no action while in the cell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are no cube faces marked, they smear the ink on one and leave the cube unsolved. If there are any marked faces:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;if the cube is solved, mark a face and scramble the cube&lt;/li&gt;&#xA;&lt;li&gt;if the cube is scrambled, solve it&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;After this, if 50 faces are marked and the cube is solved, declare victory.&lt;/p&gt;&#xA;" OwnerUserId="3987" LastActivityDate="2016-03-12T10:19:59.523" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28851" PostTypeId="2" ParentId="5626" CreationDate="2016-03-12T13:19:40.077" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The dragon can always survive by drinking from well 5 before the duel and flying to the 6th well and drinking from it after the duel. (Details are mentioned in other answers before)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The knight can ensure its survival by drinking from well 1 before the duel and drinking from well 1 again after the duel and drinking from well 2 after it.&lt;br&gt;&#xA; &lt;strong&gt;Option 1&lt;/strong&gt;&lt;br&gt;&#xA; If dragon gave him pond water or poison from well 1, it will have no effect and the knight will still be poisoned from level 1. After it, drinking from well 1 will also do nothing but drinking from well 2 after it will cure it.&lt;br/&gt;&#xA; &lt;strong&gt;Option 2&lt;/strong&gt;&lt;br&gt;&#xA; If dragon gave him any poison from level 2 to level 6, it will cure him. He will again get poisoned after drinking from well 1 and again get cured by drinking from well 2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="20103" LastEditorUserId="44915" LastEditDate="2018-05-28T11:09:16.930" LastActivityDate="2018-05-28T11:09:16.930" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28853" PostTypeId="2" ParentId="28843" CreationDate="2016-03-12T14:17:46.857" Score="6" Body="&lt;p&gt;You are a &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; horseshoe &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are many who wear me&#xA;I protect them from harm&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; they are horses&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Others say I'm lucky&#xA;And use me for charm&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it is a symbol of luck &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can cause an attraction&#xA;Of a physical sort&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A horseshoe shaped magnet &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While some folks will throw me&#xA;For amusement and sport&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Throwing horseshoes onto pegs, I don't know the exact name&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'll nail it, for sure&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A reference to the nails with which the horseshoe is attached to the hoof &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Update, two last clues (after comment)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm so much like you&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the letter U&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nebulous (I had to google that one, hope it counts)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Horseshoe Nebula, a celestial object&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2811" LastEditorUserId="2811" LastEditDate="2016-03-12T23:05:57.890" LastActivityDate="2016-03-12T23:05:57.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28854" PostTypeId="1" AcceptedAnswerId="28975" CreationDate="2016-03-12T14:20:44.120" Score="7" ViewCount="249" Body="&lt;p&gt;Every letter stands for a digit in base-10 representation, &#xA;different letters stand for different digits,&#xA;and leading digits are always non-zero.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;      SQUARE&#xA;    -  DANCE&#xA;  ------------&#xA;      DANCER&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This square dance alphametic has a unique solution under the addition and a unique solution under the subtraction. This makes the square dance alphametic special and interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which digit does each letter represent? (Please present the full analysis how these digits can be determined.)&lt;/p&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2016-03-15T15:10:36.423" Title="Subtraction square dance alphametic" Tags="|no-computers|alphametic|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28855" PostTypeId="1" AcceptedAnswerId="28856" CreationDate="2016-03-12T14:29:20.050" Score="7" ViewCount="531" Body="&lt;p&gt;There are six fruit baskets with apples, bananas and plums on the table.&#xA;The number of bananas in each basket is equal to the total number of apples in the other five baskets together.&#xA;The number of apples in each basket is equal to the total number of plums in the other five baskets together.&#xA;Altogether the six baskets contain at least 1000 pieces of fruit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the minimum number of pieces of fruit in the six baskets together?&lt;/p&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2016-03-14T17:10:20.327" Title="Six fruit baskets" Tags="|mathematics|algebra|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28856" PostTypeId="2" ParentId="28855" CreationDate="2016-03-12T14:53:00.113" Score="11" Body="&lt;p&gt;The minimum greater than 1000 is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1023&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; EDITED to include information from comments.&lt;br&gt;&#xA; Thanks to ryanp16 and manshu for asking for clarifications.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Let $b$ be the total number of bananas.&lt;br&gt;&#xA; And let $b_1, b_2 \ldots b_6$ be the number of bananas in baskets $1, 2, \ldots 6$.&lt;br&gt;&#xA; Also let $a$ be the total number of apples.&lt;br&gt;&#xA; And let $a_1, a_2 \ldots a_6$ be the number of apples in baskets $1, 2, \ldots 6$.&lt;br&gt;&#xA; Now:&lt;br&gt;&#xA; $$b = \sum_{i=1}^6{b_i} = \sum_{i=1}^6{a_i} - a_1 + \sum_{i=1}^6{a_i} - a_2 + \ldots + \sum_{i=1}^6{a_i} - a_6$$&lt;br&gt;&#xA; $$= 6\cdot\sum_{i=1}^6{a_i} - \sum_{i=1}^6{a_i} = 5\cdot\sum_{i=1}^6{a_i}$$&lt;br&gt;&#xA; $$ \therefore b = 5a$$&lt;br&gt;&#xA;  Likewise $a = 5p$ where $p$ is the total number of plums.&lt;br&gt;&#xA;&lt;br&gt;&#xA; So the total amount of fruit is: $$b + a + p = 5a + 5p + p = 25p + 6p = 31p$$&lt;br&gt;&#xA; And $1023 = 33 \cdot 31$ is the lowest multiple of $31$ greater than $1000$.&lt;br&gt;&#xA;&lt;br&gt;&#xA;  One way to arrange the fruit is to have $33$ plums and $165 = 5 \cdot 33$ bananas basket $1$.&lt;br&gt;&#xA;  And $33$ apples and $132 = 4 \cdot 33$ bananas in the other five baskets.&lt;br&gt;&#xA;&lt;br&gt;&#xA;  The total is then: $$33 \cdot (1 + 5) + 132 \cdot 5 + 165 = 1023$$   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastEditorUserId="10118" LastEditDate="2016-03-14T00:53:37.490" LastActivityDate="2016-03-14T00:53:37.490" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="28857" PostTypeId="2" ParentId="28855" CreationDate="2016-03-12T15:08:25.017" Score="2" Body="&lt;p&gt;We have 6 baskets. So banana, apple and plum are contained in these 6 basket. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the question&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b=5a$$ and $$a=5p$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now here $b$, $a$ and $p$ are only for one basket. Taking them for 6 baskets and adding them we get: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$$6b+6a+6p \geq 1000$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Putting values&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$$30(5p)+30p+6p \geq 1000$$ $$\Rightarrow 186p\geq 1000$$ As $p$ is an integer. Therefore $p=6$ and $a=30$ and $b=150$&lt;br&gt;&#xA;&#xA;Now we get $$6b+6a+6p=1116$$&lt;br&gt;&#xA; &#xA; Therefore my final answer is $1116$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastActivityDate="2016-03-12T15:08:25.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28858" PostTypeId="1" AcceptedAnswerId="28875" CreationDate="2016-03-12T15:10:25.007" Score="7" ViewCount="620" Body="&lt;p&gt;Alice and Bob play the following number guessing game.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First Alice picks an integer $n$ with exactly $250$ positive divisors. &#xA;These divisors include $1$ and $n$, and are denoted as &#xA;$1=d_1&amp;lt;d_2&amp;lt;d_3&amp;lt;\cdots&amp;lt;d_{249}&amp;lt;d_{250}=n$.&lt;/li&gt;&#xA;&lt;li&gt;Then Bob may query these divisors by paying a fee of $x$ Euro per query to Alice. Such a query simply consists of an index $i$ with $1\le i\le250$.&lt;/li&gt;&#xA;&lt;li&gt;If Bob already knows the value of the divisor $d_{251-i}$ (since he has asked it in an earlier query, or since he is able to deduce it from the knowledge gained so far), then Alice answers the query with the word &quot;BOOM&quot;.&lt;/li&gt;&#xA;&lt;li&gt;If Bob does not know the value of $d_{251-i}$, then Alice answers the query with the value of $d_i$.&lt;/li&gt;&#xA;&lt;li&gt;The game ends, as soon as Bob knows $n$ and announces it to Alice. Bob then receives $600$ Euros from Alice.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What is the largest fee $x$, for which Bob can still avoid (with absolute certainty) to lose any money?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-12T15:16:31.710" LastActivityDate="2019-07-02T23:30:52.877" Title="Guessing a number with 250 divisors" Tags="|mathematics|strategy|game|number-theory|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28859" PostTypeId="2" ParentId="26124" CreationDate="2016-03-12T15:22:01.063" Score="3" Body="&lt;p&gt;There will be no right answer for this question since it is subjective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The oft-used adjectives are &lt;a href=&quot;https://en.wikipedia.org/wiki/Knights_and_Knaves&quot; rel=&quot;nofollow noreferrer&quot;&gt;Knights and Knaves&lt;/a&gt; for truth-tellers and liars respectively (as &lt;a href=&quot;https://puzzling.stackexchange.com/users/4234/peter-lefanu-lumsdaine&quot;&gt;Peter LeFanu Lumsdaine&lt;/a&gt; and &lt;a href=&quot;https://puzzling.stackexchange.com/users/4692/alexwlchan&quot;&gt;alexwlchan&lt;/a&gt; pointed out). Often the normals are described as such - since they are usually not normal per se, but rather do something like lie or tell the truth according to a uniform random variable (are &quot;unpredictables&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I quite like these somewhat archaic English terms &lt;strong&gt;soothsayers&lt;/strong&gt;, &lt;strong&gt;mythomaniacs&lt;/strong&gt;, and &lt;strong&gt;quixotics&lt;/strong&gt; for truth-tellers, liars, and unpredicatables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could also try to give them individual names; some with somewhat relevant meanings one could use are as follows (I'm sure there are more).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the truth-teller:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Alethea (Greek: aletheia = truth)&lt;/li&gt;&#xA;&lt;li&gt;Amin (Arabic: amin = truthful)&lt;/li&gt;&#xA;&lt;li&gt;Sati (Sanskrit: sati = truthful)&lt;/li&gt;&#xA;&lt;li&gt;Sidika (Turkish: sidika = truth)&lt;/li&gt;&#xA;&lt;li&gt;Verity (English: verity = truth)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For the liar:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Bellona (Latin: bellare = fight)&lt;/li&gt;&#xA;&lt;li&gt;Drake (Old Norse: draki = dragon)&lt;/li&gt;&#xA;&lt;li&gt;Lileth (Akkadian: lilitu = of the night)&lt;/li&gt;&#xA;&lt;li&gt;Mark / Marcus / Martin (Greek: Mars = the god of war)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For the normal / unpredictable&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Alf / Alfred (Old Norse: Alfr = elf)&lt;/li&gt;&#xA;&lt;li&gt;Caprice (English capricious = unpredictable)&lt;/li&gt;&#xA;&lt;li&gt;Dougal (Gaelic dubh = dark&quot;, gall &quot;stranger&quot;)&lt;/li&gt;&#xA;&lt;li&gt;Nemo (Latin: nemo = nobody)&lt;/li&gt;&#xA;&lt;li&gt;Peter (Greek: petros = stone)&lt;/li&gt;&#xA;&lt;li&gt;Thomas (Aramaic: a'oma' = twin)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Or how about being a little tongue in cheek and going for something like: Frank (to be frank); Phoebe (fibber) or Lyle (liar); and Norman (normal).&lt;/p&gt;&#xA;" OwnerUserId="16717" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2016-03-12T22:13:47.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28860" PostTypeId="1" CreationDate="2016-03-12T16:59:12.533" Score="7" ViewCount="408" Body="&lt;p&gt;Find three popular phrases:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;raport&lt;/li&gt;&#xA;  &lt;li&gt;ga me&lt;/li&gt;&#xA;  &lt;li&gt;hbianrdd = btwusoh&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11497" LastEditorUserId="12408" LastEditDate="2016-04-15T15:10:40.223" LastActivityDate="2016-08-31T15:21:20.847" Title="English phrase rebuses" Tags="|rebus|english|" AnswerCount="3" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28861" PostTypeId="2" ParentId="28813" CreationDate="2016-03-12T17:00:07.910" Score="3" Body="&lt;h1&gt;This is not the solution!!!&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I staring hours to this and did heavy calculations and associations, but I got nothing. Here are some of my impressions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Positions&lt;/strong&gt;&lt;br&gt;&#xA;I'm not sure if the position of the bottom elements &lt;kbd&gt;3 | 2&lt;/kbd&gt;, &lt;kbd&gt;3 | 1&lt;/kbd&gt; is important. (If it is a sequence, why they are not left aligned?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Math and Tables&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Decimal: 55, 34, 53, 45, 53, 65, 33, 44, 23, 44, 22, 54, 32, 31&#xA;&#xA;HEX+DEC: E5, C4, E3, D5, E3, F5, C3, D4, B3, D4, B2, E4, C2, C1&#xA;  asDec: 229, 196, 227, 213, 227, 245, 195, 212, 179, 212, 178, 228, 194, 193&#xA;&#xA;DEC+HEX: 5E, 3D, 5C, 4E, 5C, 6E, 3C, 4D, 2C, 4D, 2B, 5D, 3B, 3A&#xA;  asDec: 94, 61, 92, 78, 92, 110, 60, 77, 44, 77, 43, 93, 59, 58&#xA;&#xA;HEX+HEX: EE, CD, EC, DE, EC, FE, CC, DD, BC, DD, BB, ED, CB, CA&#xA;  asDec: 238, 205, 236, 222, 236, 254, 204, 221, 188, 221, 187, 237, 203, 202&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;abs(d(x) - d(x+1)) &lt;code&gt;21, 19, 8, 8, 12, 32, 11, 21, 21, 22, 32, 22, 1&lt;/code&gt;&lt;br&gt;&#xA;If 32 is a space it could be a text&#xA;Other combinations I tried as well, but it ends up in a mess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pure Number association&lt;/strong&gt;&lt;br&gt;&#xA;I tried submarine numbers, president birthdays, game(i.e. soccer) scores, and so on&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hints&lt;/strong&gt;&lt;br&gt;&#xA;Both hints have 14 &quot;words&quot; like the number of tiles. Maybe it's only coincidence. But therefore it's to conspicuous to add &lt;code&gt;-ee&lt;/code&gt; to the first hint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;br&gt;&#xA;&quot;I could &lt;code&gt;add a not-her tag&lt;/code&gt; to this question, but it would &lt;code&gt;give it a way&lt;/code&gt;&quot;  Then it is not the First Lady but the president. (Obama?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;br&gt;&#xA;&quot;One time, up&quot;. - Is there something we have to set &quot;up&quot;-side?&lt;br&gt;&#xA;&quot;ordered&quot; like &quot;to sort&quot;?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Tags&lt;/strong&gt;&lt;br&gt;&#xA;A &quot;visual rebus&quot; means no or not directly math. Perhaps I miss combinatoric. I don't know.&lt;/p&gt;&#xA;" OwnerUserId="4342" LastActivityDate="2016-03-12T17:00:07.910" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28864" PostTypeId="2" ParentId="28860" CreationDate="2016-03-12T17:47:55.950" Score="5" Body="&lt;p&gt;&lt;strong&gt;Partial answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;ga me&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It’s a game of two halves&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or possibly (As @Hugh pointed out)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Get in the game&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-03-12T19:13:21.763" LastActivityDate="2016-03-12T19:13:21.763" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="28866" PostTypeId="2" ParentId="28860" CreationDate="2016-03-12T18:02:13.363" Score="10" Body="&lt;p&gt;raport&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is an anagram for both &lt;strong&gt;parrot&lt;/strong&gt; and &lt;strong&gt;raptor&lt;/strong&gt;, so I have three guesses:&lt;br&gt;&#xA;&lt;br&gt;&#xA; - Flipping the bird&lt;br&gt;&#xA; - Kill two birds with one stone&lt;br&gt;&#xA; - Birds of a feather flock together&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;ga me&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; total guess, but I'm going with &lt;em&gt;waiting game&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;hbianrdd = btwusoh&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;hbianrdd&lt;/strong&gt; is an anagram of &lt;strong&gt;bird hand&lt;/strong&gt;&lt;br&gt;&#xA; &lt;strong&gt;btwusoh&lt;/strong&gt; is an anagram of &lt;strong&gt;two bush&lt;/strong&gt;&lt;br&gt;&#xA; i.e,&lt;br&gt;&#xA; &lt;strong&gt;h&lt;/strong&gt;&lt;em&gt;bi&lt;/em&gt;&lt;strong&gt;an&lt;/strong&gt;&lt;em&gt;rd&lt;/em&gt;&lt;strong&gt;d&lt;/strong&gt; = &lt;strong&gt;b&lt;/strong&gt;&lt;em&gt;tw&lt;/em&gt;&lt;strong&gt;us&lt;/strong&gt;&lt;em&gt;o&lt;/em&gt;&lt;strong&gt;h&lt;/strong&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; So it's &lt;em&gt;A bird in the hand is worth two in the bush&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17147" LastEditorUserId="17147" LastEditDate="2016-03-12T23:47:10.373" LastActivityDate="2016-03-12T23:47:10.373" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28867" PostTypeId="1" AcceptedAnswerId="28868" CreationDate="2016-03-12T18:02:22.980" Score="13" ViewCount="746" Body="&lt;p&gt;Don't be fooled by the easy appearance of this riddle. The answer gets you to first base but to get a gold star, you must find and explain &lt;em&gt;all&lt;/em&gt; of the clues. It took a good deal of effort to compose this so I intend to make you work to get the answer. If you find twelve or fewer, try it again. My advice would be to take it in stages. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Trailing and getting a man at a bag&lt;br&gt;&#xA;  At the end of a tail it's an earmark or flag.&lt;br&gt;&#xA;  Giving a touch, give it a name&lt;br&gt;&#xA;  Greatest in code but it's kind of a game&lt;br&gt;&#xA;  A part of a lace maybe something you spray&lt;br&gt;&#xA;  Take a gathering back in advantage I say.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rate yourself:&lt;br&gt;&#xA;(Spoilers ahead!)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; How many clues did you find, excluding this section?&lt;br&gt;&#xA;    0 - 11    Low wattage performance&lt;br&gt;&#xA;    12 - 16   Staggering to the finish line&lt;br&gt;&#xA;    17 - 21     Join the montage of champions!&lt;br&gt;&#xA;    22 and up   The &quot;taggant agita greeting&quot; award  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastEditorUserId="18924" LastEditDate="2016-03-14T19:42:31.363" LastActivityDate="2016-03-14T19:42:31.363" Title="&quot;Try and Guess&quot; Is Not Enough" Tags="|riddle|word|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28868" PostTypeId="2" ParentId="28867" CreationDate="2016-03-12T18:14:57.310" Score="10" Body="&lt;p&gt;Is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; tag&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Trailing and getting a man at a bag&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Trailing And Getting (TAG) &lt;br&gt;&#xA; tagging someone out at a base &lt;br&gt;&#xA; aT A baG&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At the end of a tail it's an earmark or flag.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tagging of animals&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Giving a touch, give it a name&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; tagging someone&lt;br&gt;&#xA; Giving A Touch (TAG)&#xA; givinG A Touch&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Greatest in code but it's kind of a game&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Coding tags&lt;br&gt;&#xA; Tag is a game&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A part of a lace maybe something you spray&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; parts of lace are called tags &lt;br&gt;&#xA; Tag body spray&lt;br&gt;&#xA; Tagging via graffiti &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Take a gathering back in advantage I say. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Take A Gathering (TAG) &lt;br&gt;&#xA; Back referencing tag backwards in GAThering&lt;br&gt;&#xA; AdvanTAGe&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;also&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TAG appears as the first letters to series of words a few times in the pre-text to the poem/riddle. As well as in the title... &quot;Try And Guess&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But wait. There's more. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The first letter of each line reads TAGGAT&lt;br&gt;&#xA; &quot;try iT AGain&quot; and &quot;take it in sTAGes&quot; &lt;br&gt; &#xA; Man at a baG AT the &lt;br&gt;&#xA; The Answer Gets &lt;br&gt;&#xA; Took A Good&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wouldn't surprise me if there's more. That's far more than twelve though. ;)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Do I geT A Gold star? Or at least do I geT A Green checkmark?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19731" LastEditorUserId="19731" LastEditDate="2016-03-12T19:29:49.700" LastActivityDate="2016-03-12T19:29:49.700" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="28869" PostTypeId="2" ParentId="28858" CreationDate="2016-03-12T18:38:48.683" Score="4" Body="&lt;p&gt;You can solve for n in at most &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1 questions. Therefore x = 600&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is how this works:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First consider that n = $\prod_{j=1}^{k}\pi_j^{g_j}$ where $\pi_j$ is a prime number, $g_j$ is its multiplicity, and k is the total number of unique primes. The number of divisors is N = $\prod_{j=1}^{k}(g_j+1)$ Therefore $g_j$ = {1,4,4,4}; {1,4,24}; {1,124}; {4,4,9}; {4,49}; or {249} since N = 250. These are the possible multiplicities of the prime factors of n. I will query  $d_{248}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Furthermore:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let's consider the information gained by finding $d_{248}$ and factoring it. By comparing it to the multiplicity list above there are 2 possibilities, either the lowest prime number does not have multiplicity 1, and we will get multiplicities like =  {1,4,4,2} Then obviously $n = d_{248} * \pi_4^2$. If the smallest prime number does have multiplicity 1 then we will get multiplicities like {1,3,4,4} and n = $d_{248} * \pi_2$ Therefore we know all the prime numbers and their multiplicities of the factorization of n. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; n = $2^4*3^4*5^9$ Then $d_{248} = 2^4*3^3*5^9$ We see that this list goes with the {4,4,9} possibility. Therefore the term with a multiplicity of 3 is one too low, and we conclude that n = $d_{248}*3$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit, I realized this does not work for the case n = $3^4 * 5 * 11^4 * 13^4$ because&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $d_{248}= 3^4 * 11^4 * 13^4$ and you cannot determine what the missing factor is (5 or 7)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18821" LastEditorUserId="18821" LastEditDate="2016-03-12T19:18:46.637" LastActivityDate="2016-03-12T19:18:46.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28871" PostTypeId="2" ParentId="28867" CreationDate="2016-03-12T21:18:00.450" Score="6" Body="&lt;p&gt;Though a green tick was awarded, I was never one to let a good thing go to waste. Therefore, please accept the short addenda following: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trailing and getting a man at a bag&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Trailing is &quot;tagging along&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At the end of a tail it's an earmark or flag.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The colored bit at the end of an animal's tail is a tag. Earmarking and flagging are synonyms for tagging &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Greatest in code&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; HTML and other tags are in angle brackets which look like &quot;greater than&quot; and &quot;less than&quot; sign making the tag look greater than its surroundings. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Giving a touch, give it a name&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Tag can also be nickname&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Thanks and goodbye is the message I send.&lt;br&gt;&#xA; In this couplet that comes as a tag at the end.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastActivityDate="2016-03-12T21:18:00.450" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="28872" PostTypeId="1" AcceptedAnswerId="28876" CreationDate="2016-03-12T21:46:08.603" Score="12" ViewCount="531" Body="&lt;p&gt;Fill in the blanks with &lt;em&gt;only words&lt;/em&gt; - no digits allowed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is false to say that the letter 't' appears ________ times in this sentence. &lt;br/&gt;&#xA;  It is _____ to say that the letter 'f' appears _____ times in this paragraph. &lt;br/&gt;&#xA;  It is ______ to say that the letter 't' appears forty-six times in this paragraph. &lt;br/&gt;&#xA;  It is true to say that the letter 'e' appears ______ times in this sentence. &lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Note: All statements only refer to the four lines in the yellow box.)&lt;/p&gt;&#xA;" OwnerUserId="19086" LastEditorUserId="4421" LastEditDate="2016-03-16T16:26:14.910" LastActivityDate="2016-03-16T16:26:14.910" Title="Blankety-Blank-Blank" Tags="|counting|" AnswerCount="5" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28873" PostTypeId="1" CreationDate="2016-03-12T22:07:26.540" Score="6" ViewCount="4697" Body="&lt;p&gt;I have the following image &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bVKEN.jpg&quot; alt=&quot;colorful puzzle image&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm new to solving puzzles and figuring them out, and I have no idea how to approach decryption of this image or how to solve it. I've tried running it on some image decrypting software, but they didn't really find anything. All I have to go off is the fact it's a forensics problem and the hint: you'll know for certain when you've found the flag. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be very appreciated. Thank you very much! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: for clarification, this picture is found on the Internet. However, this image is bigger than the original image on the Internet. There is something hidden within this or a message. I just can't figure out what it is. THIS particular image has something unique in it that is different from the ones on the Internet. &lt;/p&gt;&#xA;" OwnerUserId="20111" LastEditorUserId="20110" LastEditDate="2016-03-13T12:30:43.547" LastActivityDate="2016-03-29T08:54:23.097" Title="Image steganography/ hidden message inside image" Tags="|visual|computer-puzzle|steganography|cipher|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28875" PostTypeId="2" ParentId="28858" CreationDate="2016-03-12T23:33:39.620" Score="7" Body="&lt;p&gt;Surely I am missing something. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I believe he needs two guesses. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first guess &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 249. We know that $d_{2}$ is the smallest prime factor. We look at the factors of $d_{249}$ including itself and 1. There are two cases. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First case:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If $d_{2}$ is not a factor of $d_{249}$ then $d_{2}$ times every factor $d_{249}$ will be a new factor of n. Therefore, if there are 125 factors of $d_{249}$ then all we know is that $d_{2}$ is a prime number less than the smallest prime factor of $d_{249}$. Let the smallest prime factor of n be q. $d_{248}$ is n divided by q. We ask for this number and solve for $d_{2}$, the smallest prime factor of $d_{248}$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second case:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If there are more than 125 factors of $d_{249}$ then it must be that $d_{2}$ is a factor of $d_{249}$. We don't need a second question in this case. $d_{2}$ is the smallest prime factor of $d_{249}$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can't do any better than asking for 249 because anything less than 250 misses at least one prime factor.  Call it p. If it is unique, that is p squared is not a factor of n, then in the worst case you can only know what range p is in but not p itself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can see this as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Consider the case where n has only two prime factors and n = $p_{1}$ * $p_{2}^{124}$. If $p_{1}$ &amp;lt; $p_{2}$, then only the even numbered divisors will contain $p_{1}$. If $p_{2}$ &amp;lt; $p_{1}$ &amp;lt; $p_{2}^{2}$ then only the odd numbered divisors will contain $p_{1}$. Therefore, whatever divisor we choose with our first guess we cannot guarantee that it will contain $p_{1}$. We can only know that it lies between successive powers of $p_{2}$. Since $p_{2}$ can be arbitrarily large, we cannot guarantee that only one prime lies in that range.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastEditorUserId="18924" LastEditDate="2016-03-16T13:34:46.640" LastActivityDate="2016-03-16T13:34:46.640" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28876" PostTypeId="2" ParentId="28872" CreationDate="2016-03-12T23:50:05.907" Score="11" Body="&lt;p&gt;One solution is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is false to say that the letter 't' appears &lt;strong&gt;two&lt;/strong&gt; times in this sentence.&lt;br&gt;&#xA; It is &lt;strong&gt;true&lt;/strong&gt; to say that the letter 'f' appears &lt;strong&gt;four&lt;/strong&gt; times in this paragraph.&lt;br&gt;&#xA; It is &lt;strong&gt;true&lt;/strong&gt; to say that the letter 't' appears forty-six times in this paragraph.&lt;br&gt;&#xA; It is true to say that the letter 'e' appears &lt;strong&gt;twelve&lt;/strong&gt; times in this sentence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of course:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &lt;strong&gt;two&lt;/strong&gt; in the first sentence can be replaced with any number that has one 't' (and no 'f') in it except &lt;strong&gt;twelve&lt;/strong&gt;.&lt;br&gt;&#xA; The last sentence can't change - it's the only solution.&lt;br&gt;&#xA; I'm sure there's waggle room in the two middle sentences which then means the first sentence might have to be revisited.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastEditorUserId="10118" LastEditDate="2016-03-14T14:25:16.000" LastActivityDate="2016-03-14T14:25:16.000" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28879" PostTypeId="1" AcceptedAnswerId="28880" CreationDate="2016-03-13T05:57:55.483" Score="8" ViewCount="1424" Body="&lt;p&gt;You have a 6 x 6 grid. You start at the top left corner and your &quot;finish square&quot; is the bottom right corner. You can only move to the right or move down. How many ways are there for you to get to the &quot;finish square&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0Zpkb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0Zpkb.png&quot; alt=&quot;6 x 6 grid&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20087" LastEditorUserId="65935" LastEditDate="2020-03-18T15:59:49.500" LastActivityDate="2020-03-18T16:30:34.290" Title="How many paths are there?" Tags="|mathematics|combinatorics|visual|" AnswerCount="1" CommentCount="2" ClosedDate="2020-03-18T16:46:14.177" ContentLicense="CC BY-SA 4.0" />
  <row Id="28880" PostTypeId="2" ParentId="28879" CreationDate="2016-03-13T06:05:53.883" Score="16" Body="&lt;p&gt;Assuming that each &amp;quot;movement&amp;quot; moves me exactly one grid square, there are&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; 252 possible paths.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; All paths will have 5 right moves and 5 down moves. Thus, this problem is equivalent to the permutations of RRRRRDDDDD, which is &lt;span class=&quot;math-container&quot;&gt;$\frac{10!}{5!5!}=252$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;Alternative way of solving (that arrives at the same solution).&lt;/h2&gt;&#xA;&lt;p&gt;Since this puzzle is tagged &amp;quot;Visual&amp;quot; and not &amp;quot;Math,&amp;quot; I'll provide a more visual solution as well. We will find how many paths there are from each square. The number of paths from the initial square is, of course, the solution.&lt;/p&gt;&#xA;&lt;p&gt;To begin with, there is exactly one path from the goal to the goal (namely: you're already there!). Additionally, if you are already in the final row or column, there is only one path to the goal (follow that row or column to the end).&lt;/p&gt;&#xA;&lt;p&gt;Our table so far looks like this:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;?   ?   ?   ?   ?   1&#xA;?   ?   ?   ?   ?   1&#xA;?   ?   ?   ?   ?   1&#xA;?   ?   ?   ?   ?   1&#xA;?   ?   ?   ?   ?   1&#xA;1   1   1   1   1   1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Then, observe that for each of those question marks, since we can choose to move to either the space below us or to the right, each space has a number of paths equal to the sum of its two reachable neighbors. We can start by filling in a &amp;quot;2&amp;quot; in the one space whose neighbors we both know, and then work row by row and column by column.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;252 126 56  21  6   1&#xA;126 70  35  15  5   1&#xA;56  35  20  10  4   1&#xA;21  15  10  6   3   1&#xA;6   5   4   3   2   1&#xA;1   1   1   1   1   1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="16990" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-03-16T16:45:34.210" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28881" PostTypeId="2" ParentId="28872" CreationDate="2016-03-13T06:44:40.020" Score="12" Body="&lt;p&gt;A somewhat less rational solution than the one provided by Paul Evans:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is false to say that the letter 't' appears &lt;strong&gt;pi&lt;/strong&gt; times in this sentence.&lt;br&gt;&#xA; It is &lt;strong&gt;false&lt;/strong&gt; to say that the letter 'f' appears &lt;strong&gt;pi&lt;/strong&gt; times in this paragraph.&lt;br&gt;&#xA; It is &lt;strong&gt;possible&lt;/strong&gt; to say that the letter 't' appears forty-six times in this paragraph.&lt;br&gt;&#xA; It is true to say that the letter 'e' appears &lt;strong&gt;at least two&lt;/strong&gt; times in this sentence.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As Paul Evans points out in his answer, there are many possible solutions. I just decided on an irrational one.&lt;/p&gt;&#xA;" OwnerUserId="16990" LastActivityDate="2016-03-13T06:44:40.020" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28882" PostTypeId="1" AcceptedAnswerId="28883" CreationDate="2016-03-13T07:44:07.913" Score="6" ViewCount="8187" Body="&lt;p&gt;A bear fell from a height of $10$ metres from ground. &lt;br&gt;&#xA;It reached ground in $\sqrt2$ seconds. But somehow, it didn’t get hurt. &#xA;&lt;br&gt;&#xA;What is the colour of the bear?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Note : There is no &lt;code&gt;lateral-thinking&lt;/code&gt; tag. &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I solved this question a year ago, felt like sharing it. It can have another tag : 'geography'.  &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1766" LastEditorUserId="1766" LastEditDate="2016-03-14T10:29:06.087" LastActivityDate="2018-08-29T13:53:09.127" Title="Color of the bear" Tags="|mathematics|physics|" AnswerCount="9" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="28883" PostTypeId="2" ParentId="28882" CreationDate="2016-03-13T08:06:16.613" Score="14" Body="&lt;p&gt;The bear is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; White&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I haven't a clue, but every question that cares about the color of a bear refers to polar bears.  Every single one.  Won't anybody think about the brown bears or the black bears?!  They want to be famous too!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or, more reasonably...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because for it to reach the ground in $\sqrt2$ seconds from free fall, it would need to be accelerating at 10 $m/s^2$ ($x = ½at^2$).  Interestingly, acceleration on earth is &lt;a href=&quot;https://www.newscientist.com/article/dn24068-gravity-map-reveals-earths-extremes/&quot;&gt;a bit less than that&lt;/a&gt;.  This is made worse by air resistance and many other unpleasant factors.  But the bear is certainly &lt;em&gt;closest&lt;/em&gt; to falling at that acceleration in the Arctic Ocean, which - conveniently enough - has polar bears.  Not getting hurt when falling from such a height probably involves hitting &quot;ground&quot; that can give way significantly, thus decreasing the impulse at the end.  I'd suggest a small iceberg.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="17389" LastEditDate="2016-03-13T08:21:30.970" LastActivityDate="2016-03-13T08:21:30.970" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="28885" PostTypeId="1" AcceptedAnswerId="29175" CreationDate="2016-03-13T09:55:02.807" Score="15" ViewCount="705" Body="&lt;p&gt;A rebus has gone missing. &amp;nbsp;  What rebus, and under which&#xA;$~\boxed{\sf~ \raise{.5ex}\underline\quad \, ? ~\raise{.1ex}\strut}$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \color{#274}{\boxed{ \quad \begin{matrix}                         \\[-1ex]&#xA;&amp;amp; \kern-1em \rm ADAD &amp;amp; &amp;amp; \raise{2.5ex}\strut&#xA;\kern-2em\raise{1.2ex}\color{black}{\boxed{\sf~ here? ~\raise{.1ex}\strut}} \\[-1ex]&#xA;&amp;amp; \kern-1em \rm ADAD &amp;amp; &amp;amp; &amp;amp; &amp;amp; \kern-1em\rm LLL                        \\[-1ex]&#xA;&amp;amp;                    &amp;amp; &amp;amp; &amp;amp; &amp;amp; \kern-1em\rm LLL &amp;amp; &amp;amp; \kern-1em x  \,  y \\[-1ex]&#xA;&amp;amp;                    &amp;amp; &amp;amp; &amp;amp; &amp;amp;                  &amp;amp; &amp;amp; \kern-1em z \:\: t \\[-1ex]&#xA;\rm XCXC &amp;amp;           &amp;amp; &amp;amp; &amp;amp; &amp;amp;                  &amp;amp; &amp;amp;                    \\[-1ex]&#xA;\rm XCXC &amp;amp;           &amp;amp; &amp;amp; &amp;amp; &amp;amp;                  &amp;amp; &amp;amp; &amp;amp; &amp;amp;&#xA;\kern-2em\raise{-2ex}\color{black}{\boxed{\sf~ perhaps? ~\raise{.1ex}\strut\raise{-.2ex}\strut}} \\[-1ex]&#xA;&amp;amp; &amp;amp; \kern-2em\raise{1.4ex}\color{black}{\boxed{\sf~ maybe? ~\raise{.1ex}\strut\raise{-.2ex}\strut}}&#xA;    &amp;amp; &amp;amp; \kern-1em\rm TTT &amp;amp; &amp;amp;                                         \\[-1ex]&#xA;&amp;amp; &amp;amp; &amp;amp; &amp;amp; \kern-1em\rm TTT &amp;amp; &amp;amp;&#xA;\kern-1em\raise{-1.2ex}\color{black}{\boxed{\sf~ or? ~\raise{.1ex}\strut}} \\[-1ex]&#xA;&amp;amp; &amp;amp; &amp;amp; &amp;amp; &amp;amp; &amp;amp; &amp;amp; &amp;amp;                           &amp;amp; &amp;amp; &amp;amp; \kern-1em\rm Sn      \\[-1ex]&#xA;&amp;amp; &amp;amp; &amp;amp; &amp;amp; &amp;amp; &amp;amp; &amp;amp; &amp;amp;\kern-1em ~ 2e/6           &amp;amp; &amp;amp; &amp;amp; \kern-1em\rm Sn      \\[0ex]&#xA;&amp;amp; &amp;amp; &amp;amp; &amp;amp; &amp;amp; &amp;amp; &amp;amp; &amp;amp;\kern-1em e/2 {-} \!\; e/6 &amp;amp; &amp;amp;                        \\[-1ex]&#xA;&amp;amp; &amp;amp; &amp;amp; &amp;amp; &amp;amp; &amp;amp; &amp;amp; &amp;amp;                           &amp;amp; &amp;amp; \kern-1em\rm TT        \\[-1ex]&#xA;&amp;amp; &amp;amp; &amp;amp; &amp;amp; &amp;amp; &amp;amp; &amp;amp; &amp;amp;                           &amp;amp; &amp;amp; &amp;amp; &amp;amp; \kern-1em\rm ZZZZZ \\[-1ex]&#xA;&amp;amp; &amp;amp; &amp;amp; &amp;amp; &amp;amp; &amp;amp; &amp;amp; &amp;amp;   \raise{-1.7ex}\strut    &amp;amp; &amp;amp; &amp;amp; &amp;amp; \kern-1em\rm ZZZZZ \\&#xA;\end{matrix} \quad }} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Naturally, the answer may be rebused more than one way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nothing is under three of the&#xA;$~\raise{-1.2ex}{^{^\boxed{\sf~ \raise{.5ex}\underline\quad \, ? ~\raise{.1ex}\strut}}}~\!$s.&#xA;&amp;nbsp;&#xA;Hint from a comment:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The clues&#xA; $\rm\raise{-.3ex}{\color{#274}{~ \atop TT}}$&#xA; and&#xA; $\rm\raise{2.3ex}{\color{#274}{\raise{-.5ex}{Sn} \atop Sn}}$&#xA; lead to stereotypical New Zealand pronunciations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18129" LastEditorUserId="18129" LastEditDate="2016-09-22T21:38:03.517" LastActivityDate="2016-09-22T21:38:40.297" Title="The Case of the Missing Rebus" Tags="|pattern|rebus|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28886" PostTypeId="2" ParentId="28873" CreationDate="2016-03-13T13:03:55.033" Score="8" Body="&lt;p&gt;&lt;strong&gt;The solution wouldn't have been possible without 2012rcampion's contribution.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most common form of steganography for digital images is embedding a binary message in the least significant bit (LSB) of the pixels. However, in order for the information to survive, you must save it to a lossless format The image in the question is png so that is a good starting point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good initial test is checking the LSB of each colour plane, but here I found the alpha channel was modified in a more boldly way. The alpha values for column 128 are locally and serially modified, which implies intend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Python code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import matplotlib.pyplot as plt&#xA;import numpy as np&#xA;from PIL import Image&#xA;&#xA;img = Image.open('nBXonA7.png')&#xA;data = np.array(img)&#xA;&#xA;plt.pcolor(data[...,3])&#xA;plt.colorbar()&#xA;plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And the result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dy387.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dy387.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Converting the values to ASCII we see the string resembles a base64 encoded stream.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;alpha = data[...,3]&#xA;secret = alpha[alpha != 255]&#xA;stream = ''.join(map(chr, secret))&#xA;print(stream)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which prints&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;AAAARooqvsPACc+zK/EjP5YzNh0LN+ETNx8yAo3AAY1preECIs4H&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Decoding the reverse of this string reveals a gzip compressed stream (header 1f 8b), which when decompressed reveals the flag. All the credits goes to 2012rcampion for this one.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import zlib&#xA;print(zlib.decompress(stream[::-1].decode('base64'), 16+zlib.MAX_WBITS))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And the result&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;image_peek_a_boo&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note that the flag's last character is a newline.&lt;/p&gt;&#xA;" OwnerUserId="20122" LastEditorUserId="20122" LastEditDate="2016-03-29T08:54:23.097" LastActivityDate="2016-03-29T08:54:23.097" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="28887" PostTypeId="2" ParentId="28845" CreationDate="2016-03-13T13:52:43.057" Score="6" Body="&lt;p&gt;Perhaps &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $~~~f(2014)=2$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Argument:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The digits $0,1,2,3,4,5,6,7,8$ respectively contribute $1,1,0,2,0,0,1,1,2$ to the function value; that is $f(0)=1$, $f(1)=1$, $f(2)=0$, $\ldots$, $f(8)=2$.&lt;br&gt;&#xA; With this, $f(2014)=f(2)+f(0)+f(1)+f(4)=0+1+1+0=2$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2016-03-13T13:52:43.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28888" PostTypeId="2" ParentId="28882" CreationDate="2016-03-13T14:32:22.817" Score="10" Body="&lt;p&gt;To fall at that rate it must either be on another planet, accelerated by something other than gravity, or be falling inside an evacuated chamber. The last of those seems the least unreasonable. But since it is unharmed by the vacuum, it must have been dead to begin with. And since it is unharmed by the landing, I think it must be a teddy bear, which could be any colour but is most typically brown.&lt;/p&gt;&#xA;" OwnerUserId="228" LastActivityDate="2016-03-13T14:32:22.817" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28889" PostTypeId="2" ParentId="28797" CreationDate="2016-03-13T14:46:37.387" Score="3" Body="&lt;p&gt;(1). We denote the width and length of the carpet by $x$ and $y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2). Assume that the carpet $EFGH$ lies in the rectangular storeroom $ABCD$, with corner $E$ on wall $AB$, corner $F$ on $BC$, corner $G$ on $CD$, and corner $H$ on $DA$. Then &#xA;$$\Delta AEH ~\equiv~ \Delta CGF ~\sim~ \Delta BFE ~\equiv~ \Delta DHG,$$&#xA;where $\equiv$ denotes congruence and $\sim$ denotes similarity.&#xA;Let $k=y/x$, $|AE|=a$ and $|AH|=b$. This implies $|BE|=kb$ and $|DH|=ka$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3). The data of the first storeroom now yields with (2) that &#xA;$a+kb=50$ and $ka+b=55$, which can be rewritten as $a=(55k−50)/(k^2−1)$ &#xA;and $b=(50k−55)/(k^2−1)$. This leads to&#xA;$x^2 = a^2+b^2 = (5525k^2−11000k+5525)/(k^2−1)^2$, and hence to&#xA;$$ x^2(k^2−1)2 ~=~ 5525k^2−11000k+5525.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4). Similarly, the data of the second storeroom yields with (2) that&#xA;$$ x^2(k^2−1)2 ~=~ 4469k^2−8360k+4469.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5). By combining the two equations in (3) and (4) we get&#xA;$5525k^2−11000k+5525 = 4469k^2−8360k+4469$, which implies $k=2$ or $k=1/2$. &#xA;Without loss of generality we have $y=2x$ and $a+2b=50$ and $2a+b=55$.&lt;br&gt;&#xA;Then $a=20$ and $b=15$, which implies $x=\sqrt{15^2+20^2}=25$ and $y=50$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summarizing, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the carpet is $25$ feet by $50$ feet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2016-03-13T14:46:37.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28890" PostTypeId="1" AcceptedAnswerId="28945" CreationDate="2016-03-13T15:02:50.480" Score="20" ViewCount="994" Body="&lt;p&gt;Yesterday I met professor Halfbrain at the coffee house. The professor told me that he had been spending his time with computing powers of $2016$ and combining them with powers of $32$.&#xA;The professor was happy to report that after many hours of computing and computing and computing, he had managed to prove the following theorem.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Professor Halfbrain's theorem:&lt;/strong&gt; For every integer $n\ge1$, the decimal representation of the number $2016^n$ has the same number of digits as the decimal representation of $2016^n+32^n$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is Halfbrain's theorem indeed true, or has the professor once again made one of his well-known mathematical blunders?&lt;/p&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2016-03-22T04:07:58.410" Title="Professor Halfbrain and the powers of 2016" Tags="|mathematics|number-theory|" AnswerCount="5" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28892" PostTypeId="1" AcceptedAnswerId="28894" CreationDate="2016-03-13T15:59:49.287" Score="4" ViewCount="1895" Body="&lt;p&gt;I came across this question in FB. I tried to solve but unable to get the answer for it. Feeling so demotivated. This is the question:&lt;/p&gt;&#xA;&lt;p&gt;I need to turn the whole board WHITE.&lt;br&gt; Rules of the game:&lt;br&gt;&#xA;1.When you select a square, the color of that square switches (from Black to White or vice-versa).&lt;br&gt;&#xA;2.In the same move, the colors of the squares at the TOP, LEFT, RIGHT, BOTTOM of the selected square, if available on the board, switch (from Black to White or vice-versa) as well.&#xA;Which squares should I select to win the game.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KrMz0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KrMz0.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20124" LastEditorUserId="-1" LastEditorDisplayName="user20" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-03-20T19:52:27.827" Title="Color Configuration 3x3 board" Tags="|board-games|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28893" PostTypeId="1" AcceptedAnswerId="28898" CreationDate="2016-03-13T16:02:35.877" Score="7" ViewCount="429" Body="&lt;p&gt;The SET game works with a deck of $81$ cards. Each card contains a set of symbols with four attributes: &#xA;color (red, green, purple), &#xA;shading (empty, striped, or solid), &#xA;shape (oval, squiggle, or rhombus), &#xA;number of symbols (one, two, or three). &#xA;All the cards are unique. A SET occurs when three cards satisfy the following property for each of the four attributes: &#xA;The attribute on each card is the same, &#xA;or the attribute on each card is different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two players play a game with the 81 SET game deck. The game starts when first player selects 12 cards from the deck and puts them on the table. Then there are rounds where first player removes a SET (3 cards) from the table, and second player replaces the 3 cards by cards from the deck. First player wins, if after 27 rounds he ends up with 27 SETs. Second player wins, if in some round the 12 cards on the table contain no SET.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who wins, first player or second player with the optimal strategy?&lt;/p&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2016-03-13T19:48:40.663" Title="Two players playing the SET game" Tags="|mathematics|strategy|game|set-game|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28894" PostTypeId="2" ParentId="28892" CreationDate="2016-03-13T16:11:03.653" Score="12" Body="&lt;p&gt;This kind of puzzle is commonly referred to as &quot;lights out&quot;, after a board game by that name.  I googled that phrase and came upon a solver at &lt;a href=&quot;http://www.ueda.info.waseda.ac.jp/~n-kato/lightsout/&quot;&gt;http://www.ueda.info.waseda.ac.jp/~n-kato/lightsout/&lt;/a&gt;.  So I can't take credit for the answer but it's pretty simple:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1, 3, 7, 9&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12092" LastActivityDate="2016-03-13T16:11:03.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28895" PostTypeId="2" ParentId="28882" CreationDate="2016-03-13T17:09:02.840" Score="7" Body="&lt;p&gt;The bear is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; White. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Assuming these number are precise measurements, he couldn't have fallen $10$m in $\sqrt2$ seconds since that would mean he was falling faster and further than the laws of gravity permit (see Zerris' answer for the equation).  So he must have fallen less than $10$m.  In fact, he fell &lt;em&gt;at most&lt;/em&gt; $9.807$m.  And he wouldn't even have fallen that far if he was slowed down by any air resistance, scraping any parts of the tree, braking any branches or any farting.  This could happen if there was a snow drift over the ground.  That would also explain why he wasn't hurt.  He hit the ground through a cushion of soft fluffy snow.  So he's a white polar bear as expected.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastEditorUserId="10118" LastEditDate="2016-03-15T10:51:21.437" LastActivityDate="2016-03-15T10:51:21.437" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28898" PostTypeId="2" ParentId="28893" CreationDate="2016-03-13T19:37:54.970" Score="10" Body="&lt;p&gt;The winner is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The second player&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using the following strategy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Note that for any two cards, there is exactly one third card that turns them into a SET.  Thus, each distinct pair of cards on the table makes one card unsafe to add without creating a new SET.  That card might not be possible to add (already there, taken with previous SET) but that turns out not to matter.&lt;br&gt;&#xA;&lt;br&gt;&#xA; No matter what twelve cards are originally selected, after taking a SET, there are nine cards left.  The number of distinct pairs is 36, so there are &lt;em&gt;at least&lt;/em&gt; 33 cards left in the deck that don't make any SETs with cards already on the table (take the deck of 81, subtract 9 on board, 3 taken, and at most 36 unsafe).  Select one of these safe cards.  There are then nine new pairs formed (and one card used) so &lt;em&gt;at least&lt;/em&gt; 23 cards in the deck are still safe.  Do this again forming ten new pairs and using a card, at least 12 cards are still safe to pick from.  This means we can select three cards so as to be assured of adding no new SETs to the table after the first set is taken.&lt;br&gt;&#xA;&lt;br&gt;&#xA; When the second SET is taken, do the exact same thing.  The only difference is that there are three cards fewer in the deck, so only 9 are certain to be safe for the third card replacement.  Note that since our last three cards added no new SETs, the second SET must have been entirely contained in the original twelve cards that the first player laid out.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Repeat again for the third and fourth SETs taken, with at least 6 and 3 cards safe for the last replacements respectively.  Each time, we have enough freedom that we can ensure we add no new SETs - and that means that once all twelve original cards have been taken (after four SETs) we can arrange to have no SETs on the board.  Thus, &lt;strong&gt;player two wins by the fifth round.&lt;/strong&gt; They might even win earlier if the original twelve cards do not contain four distinct sets.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Since this strategy at no point cares what twelve cards started on the table or how the SETs were taken, it will beat any strategy from the first player.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17389" LastEditorUserId="17389" LastEditDate="2016-03-13T19:48:40.663" LastActivityDate="2016-03-13T19:48:40.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28899" PostTypeId="2" ParentId="28882" CreationDate="2016-03-13T19:42:19.240" Score="4" Body="&lt;p&gt;Ok, I will throw my hat in the ring. The trick to the problem, I believe, is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You wouldn't expect this much acceleration because of the effect of atmospheric drag.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The drop took place where drag is the least, at the top of one of the highest Himalayan peaks. The reduced drag more than makes up for the increased distance from the center of the Earth.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The bear was unhurt because,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; it fell into the snow covering the mountain top.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the colour of the bear&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; is really undetermined since, if you're going to load a bear into an aircraft that is basically a rocket and fly it up over the Himalayan mountains, you can pick any bear you want. However, I will say it was black and white since panda bears are indigenous to the area and because I happen to like panda bears and, yes, I know they are not true bears and that there are brown bears indigenous to the same area but I don't care. Black and white is my answer and I'm sticking to it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastEditorUserId="18924" LastEditDate="2016-03-13T20:52:32.723" LastActivityDate="2016-03-13T20:52:32.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28900" PostTypeId="2" ParentId="1941" CreationDate="2016-03-13T22:40:30.147" Score="6" Body="&lt;p&gt;John enters the cage, noting the orientation North, stands in the center of the cage facing north and waits.  As soon as the light comes on, he starts hopping in place, up and down.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the cage is spinning so slowly, and the amount of time John's feet contact the floor would be minimal, John should be able to keep his north-facing orientation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the cage is opened, John knows which way is north, and drinks the glass in this direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember, John is a runner, likely in good shape, and shouldn't have any problem jumping in place for a few minutes at a time when his life is on the line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the best answer I can come up with (thanks to the person who suggested juggling).&lt;/p&gt;&#xA;" OwnerUserId="20132" LastEditorUserId="22419" LastEditDate="2016-06-08T08:15:37.713" LastActivityDate="2016-06-08T08:15:37.713" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28902" PostTypeId="2" ParentId="28882" CreationDate="2016-03-14T02:35:57.950" Score="3" Body="&lt;p&gt;The bear is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;BLACK&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It's falling into deep snow hence it's a polar bear. Polar bears are black with transparent hair which makes them appear to be white. Of course, it could also be a common brown or black bear being dropped into a deep snowdrift - the presence of snow implies it is Winter, hence the bear would likely be hibernating.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19780" LastActivityDate="2016-03-14T02:35:57.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28903" PostTypeId="1" AcceptedAnswerId="78715" CreationDate="2016-03-14T02:53:36.273" Score="5" ViewCount="470" Body="&lt;p&gt;I am the thing most needed in all the world, but so fast nothing can catch me.&lt;br&gt;&#xA;I myself can go the distance, even until the nearest star.&lt;br&gt;&#xA;Articles have been written about me, though not one in English.&lt;br&gt;&#xA;I give a soothing ending to an evening's rest.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But be warned, I stand in judgment of all things measurable,&lt;br&gt;&#xA;and this day, I could be at your throat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The title references a song whose sister knows me.  Ask her, and she will show you the way.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am not light, but light and her maker are some 28.6% of my being&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="1887" LastEditDate="2016-03-15T01:48:37.580" LastActivityDate="2019-01-22T17:46:27.637" Title="Giving life to the hills" Tags="|riddle|wordplay|" AnswerCount="7" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28904" PostTypeId="2" ParentId="1963" CreationDate="2016-03-14T03:40:37.830" Score="3" Body="&lt;p&gt;I am doing this for the eights only:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$8  \ -  \ \sqrt{\sqrt{8 + 8}} \ = \ 6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-\sqrt{\sqrt{8 + 8}} \ + \  8 \ = \ 6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\sqrt{8 + (8 - 8)!})! \ = \ 6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\sqrt{(8 - 8)! + 8})! \ = \ 6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$((\sqrt{8 + 8})!/8)! \ = \ 6$&lt;/p&gt;&#xA;" OwnerUserId="13203" LastEditorUserId="13203" LastEditDate="2016-03-14T04:47:43.767" LastActivityDate="2016-03-14T04:47:43.767" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28905" PostTypeId="2" ParentId="28882" CreationDate="2016-03-14T03:41:35.257" Score="9" Body="&lt;p&gt;The bear is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;transparent, white, red, orange, yellow, green, purple, or black&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;it's a &lt;a href=&quot;https://en.wikipedia.org/wiki/Tardigrade&quot;&gt;tardigrade&lt;/a&gt;, or &quot;water bear&quot;. Tardigrades are notable for their hardiness, and would be the most likely of all creatures called &quot;bears&quot; to survive whatever conditions caused the force of gravity to be so strong, as well as the fall itself.&lt;br&gt;&#xA;Note that the Wikipedia article doesn't mention the possible colours. I found them &lt;a href=&quot;http://www.tardigrada.net/newsletter/tardigrades.htm&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4473" LastActivityDate="2016-03-14T03:41:35.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28907" PostTypeId="2" ParentId="28882" CreationDate="2016-03-14T05:52:23.470" Score="3" Body="&lt;p&gt;The colour is black, blue and various tinges of yellow, being the colour of bruises, though of course one won't see the colour because of the fur. Specifically, S=ut+1/2.at^^2 where a is the acceleration, normally about 9.81 m.s^^-2 at the earth's surface. Assuming zero initial speed, this simplifies to a=10/9.81 metres.s^^-2. which may be an approximation of 1, or if it is precise then the acceleration is rather greater than normal, which is marginally (not precisely) higher at the poles, partly because of distance from the gravitational centre (F=m1.m2/R^^2), and partly because gravity is minutely offset at lower latitudes due to centripetal force. So if the north pole, then obviously a polar bear with white fur.&lt;/p&gt;&#xA;" OwnerUserId="20137" LastActivityDate="2016-03-14T05:52:23.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28909" PostTypeId="2" ParentId="28847" CreationDate="2016-03-14T09:41:23.700" Score="1" Body="&lt;p&gt;I'm not sure on what the answer is, but I think I know what the first bit is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It says 'If you want to use the machine... Use the code'. if you convert 496620796f752077616e7420746f2075736520746865206d616368696e652e2e2e205573652074686520636f6465 as hexadecimal into ASCII. I don't know if it is relevant but putting '111-111-111--111-1--111-11' into google's calculator gives '99'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="20141" LastActivityDate="2016-03-14T09:41:23.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28910" PostTypeId="2" ParentId="28882" CreationDate="2016-03-14T10:06:07.187" Score="1" Body="&lt;p&gt;Answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; White.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From 2nd law of motion &#xA;$ s= ut+ \frac{1}{2}a(t^2) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $s$ is distance, $u$ is initial velocity and $a$ is acceleration and &lt;code&gt;t&lt;/code&gt; is the time.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting the values in the equation.&#xA;initial velocity $u$ is 0.&lt;br&gt;&#xA;$10 = \frac{1}{2} * a * 2$  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; that will result in a=10.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here, for free falling acceleration $a$ is &lt;strong&gt;g&lt;/strong&gt; (acceleration due to gravity).&lt;br&gt;&#xA;so $g=10$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we know, acceleration due to gravity is slightly higher in poles as compared to equator(because Earth is oblate spheroid in shape).&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;So the location is poles and polar bears are white in color.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="20142" LastEditorUserId="11876" LastEditDate="2016-03-14T11:19:19.613" LastActivityDate="2016-03-14T11:19:19.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28911" PostTypeId="2" ParentId="28903" CreationDate="2016-03-14T10:14:19.850" Score="12" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Light&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the thing most needed in all the world, but so fast nothing can catch me.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Light, as we know it.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I myself can go the distance, even until the nearest star.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Light travels from stars billions of light years away too.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Articles have been written about me, though not one in English.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not one, Many.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I give a soothing ending to an evening's rest.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Sunset is soothing to the eyes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But be warned, I stand in judgment of all things measurable,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As cited by @manshu Speed of light is a universal constant(c) and is therefore used as reference in measurements, especially time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and this day, I could be at your throat.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In the form of a jewel, reflecting or refracting light.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19656" LastEditorUserId="19656" LastEditDate="2016-03-14T10:35:26.680" LastActivityDate="2016-03-14T10:35:26.680" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28912" PostTypeId="1" AcceptedAnswerId="28918" CreationDate="2016-03-14T11:38:26.820" Score="10" ViewCount="2251" Body="&lt;p&gt;Solve This Rebus, the correct answer is an English phrase. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;JACK THE RI_ _ER  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All the best :)&lt;/p&gt;&#xA;" OwnerUserId="19656" LastActivityDate="2016-03-14T23:38:25.353" Title="Criminal Jack is here to haunt" Tags="|wordplay|rebus|" AnswerCount="9" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28913" PostTypeId="2" ParentId="28912" CreationDate="2016-03-14T11:40:20.097" Score="2" Body="&lt;p&gt;Shooting in the dark here: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; JACK THE RISKER&lt;br&gt;&#xA; JACK THE RINGER&lt;br&gt;&#xA; JACK THE RIGGER&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19989" LastEditorUserId="19989" LastEditDate="2016-03-14T11:46:31.880" LastActivityDate="2016-03-14T11:46:31.880" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28914" PostTypeId="2" ParentId="28912" CreationDate="2016-03-14T11:47:09.190" Score="11" Body="&lt;h3&gt;This is not the intended answer(my achievement)...confirmed by OP&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;There had been a movie named &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Jack the Ripper&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the phrase can be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Pee in space&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or maybe &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Pee in space suit&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastEditorUserId="17482" LastEditDate="2016-03-14T12:47:01.437" LastActivityDate="2016-03-14T12:47:01.437" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28915" PostTypeId="2" ParentId="28912" CreationDate="2016-03-14T11:53:57.373" Score="4" Body="&lt;p&gt;Perhaps&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Peace in Space&lt;/strong&gt; (&lt;a href=&quot;http://www.peaceinspace.com/&quot;&gt;http://www.peaceinspace.com/&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Argument:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. The space in JACK THE RI__ER corresponds to two Ps, which sounds like PEACE&lt;br&gt;&#xA; 2. The outer space treaty (&lt;a href=&quot;https://en.wikipedia.org/wiki/Outer_Space_Treaty&quot;&gt;https://en.wikipedia.org/wiki/Outer_Space_Treaty&lt;/a&gt;) is sometimes called PEACE IN SPACE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2016-03-14T11:53:57.373" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28916" PostTypeId="2" ParentId="28912" CreationDate="2016-03-14T11:55:51.427" Score="9" Body="&lt;p&gt;It may be...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rest In Peace&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Only rest of the word is 'P's only, and Jack is the &lt;strong&gt;RIP&lt;/strong&gt;per..&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19191" LastEditorUserId="19191" LastEditDate="2016-03-14T11:59:39.340" LastActivityDate="2016-03-14T11:59:39.340" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28917" PostTypeId="2" ParentId="28912" CreationDate="2016-03-14T12:02:50.107" Score="4" Body="&lt;p&gt;I think it will be &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;The missing piece ( The missing Ps)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It should be JACK THE RIPPER with two Ps missing (JACK THE RI__PER).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;A Piece Of Apron&quot; is known as the missing piece in the Jack the Ripper case file.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="2730" LastEditDate="2016-03-14T12:09:33.437" LastActivityDate="2016-03-14T12:09:33.437" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28918" PostTypeId="2" ParentId="28912" CreationDate="2016-03-14T12:04:35.293" Score="17" Body="&lt;p&gt;How about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;No peace for the wicked&lt;/strong&gt; (no Ps, Jack the Ripper could be considered wicked, etc.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastActivityDate="2016-03-14T12:04:35.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28919" PostTypeId="2" ParentId="28912" CreationDate="2016-03-14T12:05:26.287" Score="5" Body="&lt;p&gt;I think it may be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No peace for the wicked&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19547" LastActivityDate="2016-03-14T12:05:26.287" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28920" PostTypeId="2" ParentId="28890" CreationDate="2016-03-14T12:56:39.820" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The answer is the theorem is true. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is my, quite possibly naive, proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rewriting the first expression for the number of digits $ d $, where $ a $ is a real number between 0.1 and 1, then:&#xA;$$ 2016^n = a \times 10^d \\&#xA;d = \log_{10}{\left({\frac{2016^n}{a}}\right)} \\&#xA;=\log_{10}{\left(2016^n\right)}-\log_{10}{a}&#xA;$$&#xA;Similarly for the second expression, this time where $ 0.1 &amp;lt;= b &amp;lt;= 1 $:&#xA;$$ 2016^n + 32^n = b \times 10^d \\&#xA;d = \log_{10}{\left({\frac{2016^n+32^n}{b}}\right)} \\&#xA;=\log_{10}{\left(2016^n+32^n\right)}-\log_{10}{b}&#xA;$$&#xA;Clearly $ -\log_{10}a $ and $ -\log_{10}b $ will both be real numbers between 0 and 1.  Therefore if the number of digits is the same then:&#xA;$$ \log_{10}{\left(2016^n\right)} + x = \log_{10}{\left(2016^n+32^n\right)} $$&#xA;where $ x = \log_{10}b $ - $ \log_{10}a $ and therefore $ -1 &amp;lt;= x &amp;lt;= 1 $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuing on this basis:&#xA;$$ x = \log_{10}{\left(2016^n+32^n\right)} - \log_{10}{\left(2016^n\right)} \\&#xA;= \log_{10}{\left(\frac{2016^n+32^n}{2016^n}\right)} \\&#xA;= \log_{10}{\left(1 + 63^{-n}\right)}&#xA;$$&#xA;For any integer $ n \geq  0 $ the second term will be insignificant and this relationship will always be true.  In fact for $ n = 0, x \approx 0.3 $ and at $ \lim_{n \rightarrow \infty} x\left(n\right) = 0 \quad\blacksquare $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  Alas, this is definitely not a proof.  If it were a proof then the last line would mean that you could replace 32 with any base for any $ n $ and that is clearly not correct. &lt;/p&gt;&#xA;" OwnerUserId="13335" LastEditorUserId="13335" LastEditDate="2016-03-15T12:13:07.443" LastActivityDate="2016-03-15T12:13:07.443" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28921" PostTypeId="2" ParentId="28903" CreationDate="2016-03-14T14:00:50.960" Score="12" Body="&lt;p&gt;Probably I am the only one who thinks this way but:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It could be the letter &quot;t&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;I am the thing most needed in all the world, but so fast nothing can catch me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;most&quot; needs the letter &quot;t&quot; or it is incomplete.&#xA; &quot;nothing&quot; caught &quot;t&quot;. There it is, caught right in the middle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I myself can go the distance, even until the nearest star.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is in &quot;distance&quot; and in &quot;star&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Articles have been written about me, though not one in English.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The word &quot;Articles&quot; surrounds &quot;t&quot; so it is about it in that sense. There is no &quot;t&quot; in English.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I give a soothing ending to an evening's rest.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The last letter, or ending, of &quot;rest&quot; is &quot;t&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But be warned, I stand in judgment of all things measurable,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This could refer to the expression &quot;to a t&quot; meaning that it exactly correct, thus it is the standard for exactness in measures of all sorts. And of course &quot;t&quot; stands at the end of the word &quot;judgment&quot;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and this day, I could be at your throat.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Indeed it is at both ends of the word &quot;throat&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastEditorUserId="18924" LastEditDate="2016-03-15T06:53:52.903" LastActivityDate="2016-03-15T06:53:52.903" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28922" PostTypeId="1" AcceptedAnswerId="28923" CreationDate="2016-03-14T14:55:43.347" Score="16" ViewCount="1549" Body="&lt;p&gt;First a bit of background. I'm a collector of twisty puzzles. I've been collecting them for almost 3.5 years now. &lt;a href=&quot;http://twistypuzzles.com/forum/viewtopic.php?f=14&amp;amp;t=26889&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here is my current collection of 209 puzzles.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you may be able to tell by the linked collection thread, I like to catalogue my puzzles with their respective brand/producer and year (if pre-2005). There are a few puzzles of which I don't know the brand or year. That doesn't matter too much, although if anyone has any information about &lt;a href=&quot;http://twistypuzzles.com/forum/viewtopic.php?f=14&amp;amp;t=30037&quot; rel=&quot;nofollow noreferrer&quot;&gt;the producer/year of those&lt;/a&gt; I would also appreciate it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For one puzzle however, I'm unable to find the name. I bought it from eBay with the title &quot;Interweaving Rings Puzzle - London, England&quot;, but I know that this is not the official name of the puzzle, and the seller also didn't knew the name. I haven't got a response on the TwistyPuzzles Forum yet, but after seeing &lt;a href=&quot;https://puzzling.stackexchange.com/questions/23796/help-with-the-name-of-this-puzzle&quot;&gt;Help with the name of this puzzle&lt;/a&gt; I thought that perhaps someone here knows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jtZzF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jtZzF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's an equator-type puzzle with sliding tiles, and (as correctly pointed out by the eBay seller) an 'interweaving rings' puzzle.&lt;/p&gt;&#xA;" OwnerUserId="13190" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.923" LastActivityDate="2016-11-12T08:45:40.973" Title="What is the name (and brand / year) of this puzzle?" Tags="|puzzle-identification|twisty-puzzles|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28923" PostTypeId="2" ParentId="28922" CreationDate="2016-03-14T15:04:20.720" Score="25" Body="&lt;p&gt;After searching around, going through Hungarian Rings, I finally stumbled on &lt;strong&gt;Radosza's Rings&lt;/strong&gt;, which seem to be educational toys. &lt;a href=&quot;http://www.radoszarings.com/eng/turista_eng.html&quot;&gt;Link&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2016-03-14T15:04:20.720" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28924" PostTypeId="2" ParentId="28872" CreationDate="2016-03-14T15:24:10.943" Score="4" Body="&lt;p&gt;I decided to put the extra challenge of fitting many words into the blanks.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is false to say that the letter 't' appears &lt;strong&gt;to be written more than thirty thousand&lt;/strong&gt; times in this sentence.&lt;br&gt;&#xA; It is &lt;strong&gt;incorrect but not impossible&lt;/strong&gt; to say that the letter 'f' appears &lt;strong&gt;to be formatted using a different font four hundred fifty five&lt;/strong&gt; times in this paragraph.&lt;br&gt;&#xA; It is &lt;strong&gt;inconsequential whether or not it is easy&lt;/strong&gt; to say that the letter 't' appears forty-six times in this paragraph.&lt;br&gt;&#xA; It is true to say that the letter 'e' appears &lt;strong&gt;less than twenty two times but greater than three&lt;/strong&gt; times in this sentence. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastActivityDate="2016-03-14T15:24:10.943" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28925" PostTypeId="2" ParentId="28813" CreationDate="2016-03-14T15:35:09.807" Score="2" Body="&lt;p&gt;Is the answer perhaps&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Fats Domino&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Argument:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The hints are pointing at Ella Fitzgerald (the &quot;First Lady of Song&quot;) and at Strange Fruit&lt;br&gt;&#xA; Ella Fitzgerald won 14 Grammy awards&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-14T16:01:36.343" LastActivityDate="2016-03-14T16:01:36.343" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28926" PostTypeId="1" AcceptedAnswerId="28927" CreationDate="2016-03-14T15:43:22.533" Score="4" ViewCount="624" Body="&lt;p&gt;What is the largest number of queens that can be placed on a regular $8\times8$ chessboard, if the following rules are met:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;A queen can be either black or white, and there can be unequal numbers of each type.&lt;/li&gt;&#xA;  &lt;li&gt;A queen must &lt;em&gt;not&lt;/em&gt; be threatened by other queens of the &lt;strong&gt;same&lt;/strong&gt; color.&lt;/li&gt;&#xA;  &lt;li&gt;Queens threaten all squares in the same row, column, or diagonal (as in chess). Also, threats are blocked by other queens.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bonus:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Would this number change if rule 1 was changed to enforce equal numbers of black and white queens?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11679" LastEditorUserId="11679" LastEditDate="2016-03-14T18:03:33.697" LastActivityDate="2016-03-16T20:37:24.837" Title="Black and white queens on an 8x8 chessboard" Tags="|mathematics|combinatorics|chess|checkerboard|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28927" PostTypeId="2" ParentId="28926" CreationDate="2016-03-14T15:46:54.080" Score="11" Body="&lt;p&gt;I'll guess&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;32 queens&lt;/strong&gt;.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Every other row is filled with alternating queens. Starting queens on each filled row alternate.&lt;br&gt;&#xA;&lt;br&gt;&#xA;  &lt;code&gt;W B W B W B W B&lt;/code&gt;&lt;br&gt;&#xA;  &lt;code&gt;- - - - - - - -&lt;/code&gt;&lt;br&gt;&#xA;  &lt;code&gt;B W B W B W B W&lt;/code&gt;&lt;br&gt;&#xA;  &lt;code&gt;- - - - - - - -&lt;/code&gt;&lt;br&gt;&#xA;  &lt;code&gt;W B W B W B W B&lt;/code&gt;&lt;br&gt;&#xA;  &lt;code&gt;- - - - - - - -&lt;/code&gt;&lt;br&gt;&#xA;  &lt;code&gt;B W B W B W B W&lt;/code&gt;&lt;br&gt;&#xA;   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think this is ideal, since for a single queen, cutting off all lines of sight would require:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;B - B - B&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;- - - - -&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;- B W B -&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;- - - - -&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;B - B - B&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Filling white queens into the spaces and repeating the pattern gives us the 32 solution. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="4421" LastEditDate="2016-03-16T20:37:24.837" LastActivityDate="2016-03-16T20:37:24.837" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="28928" PostTypeId="2" ParentId="28903" CreationDate="2016-03-14T16:25:57.583" Score="8" Body="&lt;p&gt;The title refers to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;The Sound of Music&lt;/strong&gt; (&lt;em&gt;&quot;The hills are alive / with [title]&quot;&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the thing most needed in all the world, but so fast nothing can catch me.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Music is not really something that can be &quot;caught&quot; in the physical sense&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I myself can go the distance, even until the nearest star.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We broadcast radio waves that carry music off into space   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Articles have been written about me, though not one in English.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Like the &quot;Light&quot; answer, not one, but &lt;em&gt;many&lt;/em&gt; articles have been written about music&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I give a soothing ending to an evening's rest.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It can be very nice to listen to soothing music at night.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But be warned, I stand in judgment of all things measurable,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Music has been written on all manner of topics. Music has been written to criticize, eulogize, and praise.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and this day, I could be at your throat.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your throat could produce music, through singing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastActivityDate="2016-03-14T16:25:57.583" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28929" PostTypeId="1" AcceptedAnswerId="28940" CreationDate="2016-03-14T16:47:54.250" Score="21" ViewCount="1386" Body="&lt;p&gt;A pond contains $24$ waterlilies that are arranged in a rectangular $2\times12$ grid (that is, two rows with twelve waterlilies). One evening $24$ frogs give a croaking concerto for the residents of the pond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the beginning of the concert, the $k$th frog is sitting on the $k$th lily pad (where $1\le k\le24$). There is a short break in the middle of the concert, and after this break none of the frogs returns to its old lily pad. It turns out that after the break each of the $24$ lily pads again carries one frog, and that every frog is now sitting on some new pad that is horizontally or vertically adjacent to its old pad from before the break.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; How many different seating arrangements are there after the break that match the above description. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(The answer to this question will be a square number. A good solution will clearly explain the reason why a square number shows up here.)&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-14T17:26:49.053" LastActivityDate="2016-03-16T23:11:37.627" Title="The frog concerto" Tags="|mathematics|combinatorics|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28930" PostTypeId="2" ParentId="28813" CreationDate="2016-03-14T16:57:24.940" Score="9" Body="&lt;h1&gt;Partial answer&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The first clue indicates&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a relationship between 1 and C (&quot;one... sea&quot;), 2 and D (&quot;toody&quot;), and 3 and E (&quot;three-ee&quot;). Given this, it seems likely that the numbers correspond to letters in the same order in the alphabet. A Caesar cipher doesn't produce a legible result, but there's another sequence of letters starting with C: a musical scale, with 1 being C and 6 being A (possibly 0 would be B, but it doesn't appear).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using this, we can convert the dominoes to&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a sequence of notes: GGEFGEFGGEAGEEFFDEFFDDGFEDEC. Given the &quot;first lady&quot; in the title and the later hint about Ella, this is likely part of one of the songs of Ella Fitzgerald, as suggested by &lt;a href=&quot;https://puzzling.stackexchange.com/a/28925/12511&quot;&gt;Gamow&lt;/a&gt; (possibly one of her collaborations with Fats Domino). Unfortunately, I don't recognise the tune when I hum it and I can't listen to her songs at the moment to check.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12511" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.573" LastActivityDate="2016-03-14T16:57:24.940" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28931" PostTypeId="2" ParentId="28855" CreationDate="2016-03-14T17:10:20.327" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The size of each group of bananas is equal to the sum of 5 different groups of apples (rotating), so the total number of bananas AND apples is (5*6+6)/6=6 times the total number of apples, which is equivalent to these 6 different groups of apple repeated 6 times each. Similarly, the size of each group of apples is equal to the sum of 5 different groups of plums (rotating), so the total number of fruits is (5*36+6)/6=31 times the total number of plums. Therefore the total must be divisible by 31, and the answer is 1023.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastActivityDate="2016-03-14T17:10:20.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28932" PostTypeId="2" ParentId="28903" CreationDate="2016-03-14T17:34:02.443" Score="2" Body="&lt;p&gt;I guess it is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Thought/Mind/Brain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I am the thing most needed in all the world, but so fast nothing can catch me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Brain is the most needed thing. One second you are thinking about batman and the next second about Emma Watson. Such is the speed of brain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I myself can go the distance, even until the nearest star.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many theories in cosmology are based on the &lt;strong&gt;thought&lt;/strong&gt; processes. That means we are not certain about many things but we only have some ideas. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Articles have been written about me, though not one in English.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; One reasoning to this sentence can be that articles on brain must be in the subject of Psychology (not in English). Other reasoning is (like other answers) that there are many articles, not only one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I give a soothing ending to an evening's rest.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yeah. It is self explanatory.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But be warned, I stand in judgment of all things measurable,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yup. We as human thinks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and this day, I could be at your throat.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Throat rhymes with thought.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nicki Minaj had song &lt;strong&gt;&lt;em&gt;Mind On My Money&lt;/em&gt;&lt;/strong&gt; and she also featured in The Weeknd's song &lt;em&gt;The Hills&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17482" LastActivityDate="2016-03-14T17:34:02.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28933" PostTypeId="2" ParentId="28872" CreationDate="2016-03-14T18:02:39.547" Score="4" Body="&lt;p&gt;As others have done, I decided to go for a roundabout answer.&#xA;Let's start of with whether or not we know an answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It is false to say that the letter 't' appears &lt;strong&gt;an indeterminate number of&lt;/strong&gt; times in this sentence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So it seems we know something about the answer. What else do we know about the answer?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It is &lt;strong&gt;quite silly&lt;/strong&gt; to say that the letter 'f' appears &lt;strong&gt;burrito&lt;/strong&gt; times in this paragraph.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, that rules out &lt;em&gt;that&lt;/em&gt; type of answer. How about a true (and logical) one?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It is &lt;strong&gt;entirely truthful, to say little of other matters, particularly so of the town’s terrible titans of typography, who typically rant tumultuously about the excess of the letter ‘t’ in run-on sentences such as the present example (for which I have to give them credit where it is quite due, for I myself seem to have violated the unspoken rule)&lt;/strong&gt;, to say the letter t appears forty-six times in this paragraph.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...Can you give a shorter answer, please? That was obnoxious.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;It is true to say that the letter 'e' appears &lt;strong&gt;at predetermined&lt;/strong&gt; times in this sentence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I guess I will give up on getting a good answer, then.&lt;/p&gt;&#xA;" OwnerUserId="19703" LastActivityDate="2016-03-14T18:02:39.547" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28936" PostTypeId="2" ParentId="28903" CreationDate="2016-03-14T18:59:06.993" Score="4" Body="&lt;p&gt;I have a guess stuck in my head and can't seem to get it out, so I'll run with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Love.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the thing most needed in all the world, but so fast nothing can catch me.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;All you need is Love&lt;/strong&gt;, and it's not something you can catch; it finds you and is sometimes fleeting.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I myself can go the distance, even until the nearest star.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; What could stop love from reaching &lt;strong&gt;across the universe&lt;/strong&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Articles have been written about me, though not one in English.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hmm... Love is a language of its own.  (&lt;strong&gt;P.S. I Love You&lt;/strong&gt;?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I give a soothing ending to an evening's rest.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After &lt;strong&gt;a hard day's night&lt;/strong&gt;, it's best to come to bed with the one you love.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But be warned, I stand in judgment of all things measurable,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Uh...without an applicable Beatles song in mind, I'll just opt for ABBA's &lt;strong&gt;The Winner Takes it All&lt;/strong&gt;, which sadly happens in the game of love.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and this day, I could be at your throat.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Yesterday&lt;/strong&gt;, love may have been an easy game to play.  That may not be the case &quot;this day&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The title references a song whose sister knows me. Ask her, and she will show you the way.--  The Fool on the Hill is on side one of the Beatles' &lt;em&gt;Magical Mystery Tour&lt;/em&gt;, and on side two (the sister side?) is All You Need is Love.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now you can see why my answer is stuck in my head, not that I'm complaining.&lt;/p&gt;&#xA;" OwnerUserId="11452" LastActivityDate="2016-03-14T18:59:06.993" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28937" PostTypeId="2" ParentId="28903" CreationDate="2016-03-14T19:19:07.093" Score="0" Body="&lt;p&gt;Since my previous answer has been marked wrong by OP, here is another shot at it.&lt;br&gt;&#xA;My answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Faith or Prayer or Religion &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the thing most needed in all the world, but so fast nothing can catch me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Faith is needed in this world of chaos, and it isn't material so can't be caught.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I myself can go the distance, even until the nearest star.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No Clue!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Articles have been written about me, though not one in English.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Perhaps the OP is referring to a religion, most probably Christianity. Bible was written in Hebrew, not English.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I give a soothing ending to an evening's rest.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Prayers and chants have mesmerizing effects.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But be warned, I stand in judgment of all things measurable,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Perhaps a reference to the Judgement Day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and this day, I could be at your throat.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In faith or fear, You may be praying.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19656" LastActivityDate="2016-03-14T19:19:07.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28938" PostTypeId="2" ParentId="28890" CreationDate="2016-03-14T19:35:48.597" Score="8" Body="&lt;p&gt;I haven't been able to produce a definite proof but nevertheless I'll post my reasoning here in case someone manages to take it further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have that the number of digits of $2016^n$ is $$1 +\left \lfloor \log_{10}(2016^n) \right \rfloor = 1 + \left \lfloor n \cdot \log_{10}(2016) \right \rfloor \approx 1 + \left \lfloor 3.3045n \right \rfloor$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the number of digits of $32^n$ is $$1 + \left \lfloor n \cdot \log_{10}(32) \right \rfloor \approx 1 + \left \lfloor 1.5051n \right \rfloor$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, for large $n$ the number of digits of $2016^n$ is approximately $2.2$ times the number of digits of $32^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, when adding up two numbers (in base $10$) using the schoolgrade algorithm, the largest carry-over for any given digit addition is $1$. This means that, if by adding up $32^n$ to $2016^n$ a new digit is obtained, then that digit must be $1$. More importantly, this means that &lt;strong&gt;all digits&lt;/strong&gt; of $2016^n$ which are beyond the leftmost digit of $32^n$ must be $9$’s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can check for low values of $n$ that the theorem does hold, and so with the previous observation this means that, if the theorem were untrue for some $n_0$, then over half of the digits of $2016^{n_0}$ must be $9$’s;  the leftmost digits of $2016^{n_0}$  would be a long sequence of $9$’s. This seems very, very unlikely. I have tried in many ways to show the impossibility of this, but have not managed thus far.&lt;/p&gt;&#xA;" OwnerUserId="16478" LastActivityDate="2016-03-14T19:35:48.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28939" PostTypeId="2" ParentId="25248" CreationDate="2016-03-14T20:01:40.613" Score="13" Body="&lt;p&gt;Is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Daisies&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I.e.'s are an example of something, used for clarification. &lt;br&gt;&#xA; Dais is a platform on which people speak. &lt;br&gt;&#xA; And I know this is a stretch but a lowercase a can look like a d without the top vertical line so the vowels are close to reading diie. &lt;br&gt;&#xA; Alternative pointed out by Hugh is aïe is French for &quot;dang.&quot; &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Worth a shot. &lt;/p&gt;&#xA;" OwnerUserId="19731" LastEditorUserId="19731" LastEditDate="2016-03-15T19:41:31.133" LastActivityDate="2016-03-15T19:41:31.133" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="28940" PostTypeId="2" ParentId="28929" CreationDate="2016-03-14T20:02:49.170" Score="27" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $54289$, or $233^2$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If we label the frogs 01-12 on the top and 24-13 on the bottom&lt;br&gt;&#xA; 01 02 03 04 05 06 07 08 09 10 11 12&lt;br&gt;&#xA; 24 23 22 21 20 19 18 17 16 15 14 13&lt;br&gt;&#xA;&lt;br&gt;&#xA; We see that every odd frog must end on an even position and every even frog must end on an odd position. So the positioning of the even and odd frogs are independent of each other. They are also symmetrical to each other, so the total number of even positions equals the total number of odd positions. Therefore, we can consider only the odd frogs and square the answer to get the total number of seating arrangements.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, to calculate the position for 24 frogs, I broke it up into 2 parts:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Part 1: Frog 01 moves to position 24. In this case, the number of positions is the same as calculating the positions for the remaining 11 odd frogs in a 2 x 11 rectangle.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Part 2: Frog 01 moves to position 02. This forces frog 23 to position 24. In this case, the number of positions is the same as calculating the positions for the remaining 10 odd frogs in a 2 x 10 rectangle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $F(x) = F(x-1) + F(x-2)$, so we have a Fibonacci sequence.&lt;br&gt;&#xA; $F(1) = 1$&lt;br&gt;&#xA; $F(2) = 2$&lt;br&gt;&#xA; $F(12) = 233$, so there are 233 positions for 12 odd frogs. Square this to give the total positions for 24 frogs, and we arrive at our answer, $54289$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="375" LastEditorUserId="375" LastEditDate="2016-03-16T19:16:22.477" LastActivityDate="2016-03-16T19:16:22.477" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="28941" PostTypeId="2" ParentId="28912" CreationDate="2016-03-14T21:37:10.953" Score="1" Body="&lt;p&gt;Will be two PP's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Peace in Space will fit&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="20152" LastActivityDate="2016-03-14T21:37:10.953" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28942" PostTypeId="2" ParentId="28912" CreationDate="2016-03-14T22:46:23.250" Score="5" Body="&lt;p&gt;So sorry officer, can't fully recollect his name 'cause...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; ...he scared the &lt;em&gt;P's&lt;/em&gt; outta me!!!   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastEditorUserId="10118" LastEditDate="2016-03-14T23:38:25.353" LastActivityDate="2016-03-14T23:38:25.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28943" PostTypeId="2" ParentId="28890" CreationDate="2016-03-15T02:38:40.060" Score="7" Body="&lt;p&gt;This isn't quite an answer, but it brings some heavy machinery to bear on the problem and shows that, at least theoretically, the question could be resolved by checking its truth for the first few values (i.e. the first $10^{15}$ values would suffice). Indeed, more strongly, we can say that Professor Halfbrain's theorem holds at least for all but finitely many $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that, basically, we are looking for a solution to the following equation:&#xA;$$2016^n &amp;lt; 10^k\leq 2016^n + 32^n$$&#xA;for non-negative integers $k$ and $n$. We can attack this by first taking logarithms:&#xA;$$\log(2016)\cdot n &amp;lt; \log(10)\cdot k \leq \log(2016^n+32^n)$$&#xA;Now, the last term is sort of nasty, but we can deal with it know that the logarithm is concave and has derivative $\frac{1}x$, giving that $\log(x+y)\leq \log(x)+\frac{y}x$. Noting that $\frac{32}{2016}=\frac{1}{63}$, we can weaken the above equation to&#xA;$$\log(2016)\cdot n &amp;lt; \log(10)\cdot k \leq \log(2016)\cdot n + \frac{1}{63^n}$$&#xA;which implies that:&#xA;$$\left|\log(2016)\cdot n - \log(10)\cdot k \right| &amp;lt; \frac{1}{63^n}.$$&#xA;And, if we let $\Lambda=\log(2016)\cdot n -\log(10)\cdot k$ and take logarithms of both sides, we get:&#xA;$$\log |\Lambda | &amp;lt; -\log(63)\cdot n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is both good news and bad news. The good news is that this is called a linear form in logarithms and is well studied in mathematics (the crucial theorem in the area being &lt;a href=&quot;https://en.wikipedia.org/wiki/Baker%27s_theorem&quot;&gt;Baker's Theorem&lt;/a&gt;). The better news is that we only need an exponential bound, whereas Baker's Theorem gives a stronger bound. The bad news is that I'm not well versed in the field, so it's hard to find explicit results that I can use. However in &lt;a href=&quot;http://gdz.sub.uni-goettingen.de/dms/load/img/?PID=GDZPPN002210908&amp;amp;physid=PHYS_0024&quot;&gt;Baker and Wüstholz (1993)&lt;/a&gt; they give a form into which it's rather easy to plug constants:&#xA;$$\log |\Lambda | \geq -(16nd)^{2(n+2)}\log(A_1)\log(A_2)\log(B)$$&#xA;where $d=1$ since the things we're taking logarithms of are rational and $n=2$ since there are two logarithms. The $A_1$ and $A_2$ happen to work out to $2016$ and $10$ respectively, since the minimal defining polynomial for an integers $n$ is $x-n$. Finally, $B$ is an upper bound for the coefficients in the linear form. Since it's rather obvious that $k\geq n$, but we're lazy, we'll just take $B$ to be $n$, which weakens the result. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting this all in gives:&#xA;$$\log |\Lambda | \geq -2.31956\cdot 10^{13} \cdot \log(n)$$&#xA;and the left hand quantity is greater than $-n\log(63)$ when $n$ gets big, implying that there may only be finitely many exceptions to this statement. In particular, no exception may have $n\geq 1.89\times 10^{14}$. This is infeasible to check with computer, but it's very possible that stronger bounds exist that I'm unaware of that might resolve it, since I used a bound applicable for all algebraic numbers and for all $n$ (and moreover, which is applicable in the non-homogenous case, whereas we are in the homogenous case. I'm not an expert in this field, so I'm using the coarse, but accessible results).&lt;/p&gt;&#xA;" OwnerUserId="4765" LastActivityDate="2016-03-15T02:38:40.060" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28944" PostTypeId="2" ParentId="25248" CreationDate="2016-03-15T04:18:47.307" Score="2" Body="&lt;p&gt;My random guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wingspan&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A wing of a building is a place to make a speech.&lt;br&gt; A pan can serve to clarify &lt;em&gt;butter&lt;/em&gt; by melting it. &lt;br&gt;Ai! as an exclamation isn't usually considered a curse but it's close.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I also considered &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Walleye&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I couldn't figure out how that would be a curse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this isn't it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It might have to do with &lt;em&gt;letters&lt;/em&gt; &lt;br&gt;oic: Oh, I see? &lt;br&gt;icy:I see why? &lt;br&gt;our:Oh, you are? &lt;br&gt;ucy: You see why?&lt;br&gt; I can't really figure out the other letters that might be with TV though.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="20154" LastEditorUserId="20154" LastEditDate="2016-03-15T04:23:59.480" LastActivityDate="2016-03-15T04:23:59.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28945" PostTypeId="2" ParentId="28890" CreationDate="2016-03-15T05:30:34.610" Score="20" Body="&lt;p&gt;This answer is an improvement on &lt;a href=&quot;https://puzzling.stackexchange.com/a/28943/9718&quot;&gt;Milo Brandt's answer&lt;/a&gt;, and is based on &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0022314X85711419&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Formes linéaires en deux logarithmes et déterminants d′interpolation&quot; (M. Laurent, M. Mignotte, and Y. Nesterenko)&lt;/a&gt; (which I got from &lt;a href=&quot;https://mathoverflow.net/questions/33257/strongest-known-version-of-bakers-theorem&quot;&gt;this MathOverflow answer&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will prove Professor Halfbrain's theorem by contradiction.  Suppose that there is some $n$ for which the theorem does not hold.  Then we must have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;2016^n + 32^n \ge 10^k &amp;gt; 2016^n&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some integer $k$.  That is, when adding $32^n$ we must have &quot;passed over&quot; a power of ten in order to increase the number of digits.  Like Milo, we start by taking logarithms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\log\left(2016^n + 32^n\right) &amp;gt; k\log 10 \ge n\log 2016&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This step is valid since all parts of the inequality are positive, and $\log x$ is strictly increasing for all $x&amp;gt;0$.  The logarithm is also strictly concave, which means that it is &lt;a href=&quot;https://en.wikipedia.org/wiki/Concave_function#Properties&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;bounded above by its first-order Taylor approximation&quot;&lt;/a&gt;; that is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\log(x) &amp;lt; \log(x_0) + \log'(x_0)(x-x_0) = \log(x_0) + \frac{x-x_0}{x_0},\ x \ne x_0 \\&#xA;\log(x_0+y) &amp;lt; \log(x_0) + \frac{y}{x_0},\ y \ne 0&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, we can write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\log\left(2016^n + 32^n\right) &amp;lt; n\log 2016 + \frac{32^n}{2016^n} = n \log 2016 + 63^{-n} \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can add this to our previous inequality:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;n \log 2016 + 63^{-n} &amp;gt; \log\left(2016^n + 32^n\right) &amp;gt; k\log 10 \ge n\log 2016&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that a counterexample to Professor Halfbrain's theorem, which must satisfy our original inequality, also satisfies the weaker inequality:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;n \log 2016 + 63^{-n} &amp;gt; k\log 10 \ge n\log 2016&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can weaken this inequality even further by subtracting $63^{-n}$ from the right-hand side:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;n \log 2016 + 63^{-n} &amp;gt; k\log 10 &amp;gt; n\log 2016 - 63^{-n} \\&#xA;63^{-n} &amp;gt; k\log 10 - n \log 2016 &amp;gt; - 63^{-n} \\&#xA;63^{-n} &amp;gt; \left| k\log 10 - n \log 2016\right|&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick aside: this is a &lt;em&gt;weaker&lt;/em&gt; inequality than our original one; since has a larger range it could be true for more values of $n$.  However, we will eventually use this to prove (by contradiction) a &lt;em&gt;stronger&lt;/em&gt; statement than Professor Halfbrain's theorem.  Essentially we will say that, not only do $2016^n$ and $2016^n+32^n$ have the same number of digits, but that no power of $2016$ is &quot;close&quot; to a power of ten.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The paper mentioned above discusses the so-called linear form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Lambda = b_2\log\alpha_2 - b_1\log\alpha_1&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\alpha_1$ and $\alpha_2$ are nonzero algebraic numbers, and $b_1$ and $b_2$ are positive integers.  In our case we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;\alpha_1 &amp;amp;= 2016 &amp;amp; b_1 &amp;amp;= n \\&#xA;\alpha_2 &amp;amp;= 10 &amp;amp; b_2 &amp;amp;= k&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a couple more quantities that we will need later: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;D = \frac{[\mathbb{Q}(\alpha_1,\alpha_2) : \mathbb{Q}]}{[\mathbb{R}(\alpha_1,\alpha_2) : \mathbb{R}]}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The notation $[E:F]$ is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Degree_of_a_field_extension&quot; rel=&quot;nofollow noreferrer&quot;&gt;degree of a field extension&lt;/a&gt;.  Since our $\alpha_1$ and $\alpha_2$ are integers, they don't actually extend $\mathbb{Q}$ or $\mathbb{R}$, and $D=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, we need $h(\alpha_i)$, where $h$ is the logarithmic height.  For an algebraic number of degree $d$ whose minimal polynomial over the integers is $a\prod_{i=1}^{d}\left(X-a^{(i)}\right)$ (where the $a^{(i)}$ are the roots of the polynomial) the height is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;h(\alpha) = \frac{1}{d}\left(\log |a|+\sum_{i=1}^{d}\log^+\left|a^{(i)}\right|\right)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\log^+(x)=\max\{\log x,0\}$.  Since both our $\alpha_i$ are integers, their minimal polynomials are simply $X-\alpha_i$ (and $a=d=1$).  Therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;h(\alpha_i) = \frac{1}{1}\left(\log 1+\log^+\left|\alpha_i\right|\right)=\log\alpha_i&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we need the quantity $b'$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;b'=\frac{b_1}{D\log A_2}+\frac{b_2}{D\log A_1}=\frac{b_1}{\log A_2}+\frac{b_2}{\log A_1}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the $A_i$ are real numbers satisfying $A_i&amp;gt;1$ and:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;A_i \ge \max\left\{h(\alpha_i),\frac{|\log \alpha_i|}{D},\frac{1}{D}\right\}=\max\left\{\log\alpha_i,\frac{|\log \alpha_i|}{1},\frac{1}{1}\right\}=\log \alpha_i&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In the last step we assume $\alpha_i\ge e$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Corollary 2 to Theorem 2 of the paper states that, if $\alpha_1$ and $\alpha_2$ are multiplicatively independent, positive real numbers (which they are), then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\log|\Lambda| \ge -C_2D^4\left(\max\left\{\log b'+0.14,\frac{h_2}{D},\frac{1}{2}\right\}\right)^2\log A_1\log A_2&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($\alpha_1$ and $\alpha_2$ are multiplicatively independent if &quot;no integral-exponent power product of them is equal to 1, unless all exponents are zero&quot; &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0022404900001560&quot; rel=&quot;nofollow noreferrer&quot;&gt;[source]&lt;/a&gt;; that is, the only solution to $\alpha_1^{x_1}\alpha_2^{x_2}=1,\ \{x_1,x_2\}\in \mathbb{Z}$ is $x_1=x_2=0$.)  Pairs of values for the constants $C_2$ and $h_2$ are given in a table later in the document.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can simplify Corollary 2 using the values of $D$, etc. that we determined earlier:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{multline}&#xA;-\log|\Lambda| \le C_2\left(\max\left\{\log\left(\frac{n}{\log\log 10}+\frac{k}{\log\log 2016}\right)+0.14,h_2\right\}\right)^2\cdot \\ (\log\log 2016)(\log\log 10)&#xA;\end{multline}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the signs of both sides have been reversed, in order to make the following argument easier to follow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that $10^k$ is the power of ten just above $2016^n$, that is, $k = \lfloor n\log_{10}2016 \rfloor + 1$.  This means $k &amp;lt; n\log_{10}2016 + 1$.  We can use the same &quot;weakening&quot; argument from before to write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\log\left(\frac{n}{\log\log 10}+\frac{k}{\log\log 2016}\right) \\&#xA;&amp;lt; \log\left(n\left[\frac{1}{\log\log 10}+\frac{\log_{10}2016}{\log\log 2016}\right] + \frac{1}{\log\log 2016}\right) \\&#xA;&amp;lt; \log n + \log\left(\frac{1}{\log\log 10}+\frac{\log_{10}2016}{\log\log 2016}\right) + \frac{1}{n\left(\frac{\log\log 2016}{\log\log 10}+\log_{10}2016\right)} \\&#xA;&amp;lt; \log(n) + 1.040 + \frac{0.175}{n}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;-\log|\Lambda| &amp;lt; C_2\left(\max\left\{\log n + 1.28 + \frac{0.175}{n},h_2\right\}\right)^2\cdot 1.693&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We reverse the signs again to simplify our reasoning:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\log|\Lambda| &amp;gt; -1.693C_2\left(\max\left\{\log n + 1.28 + \frac{0.175}{n},h_2\right\}\right)^2&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we bring back our original equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;63^{-n} &amp;gt; \left| k\log 10 - n \log 2016\right| = |\Lambda|&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any counterexample to Professor Halfbrain's theorem must satisfy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;-n\log 63 &amp;gt; \log|\Lambda| &amp;gt; -1.693C_2\left(\max\left\{\log n + 1.28 + \frac{0.175}{n},h_2\right\}\right)^2&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore we only need to check values of $n$ that satisfy the inequality:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;-n\log 63 &amp;gt; -1.693C_2\left(\max\left\{\log n + 1.28 + \frac{0.175}{n},h_2\right\}\right)^2 \\&#xA;n &amp;lt; 0.409C_2\left(\max\left\{\log n + 1.28 + \frac{0.175}{n},h_2\right\}\right)^2&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point we substitute values of $C_2$ and $h_2$ from the table:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;h_2 = 10,\quad C_2=32.31 \\&#xA;n &amp;lt; \max\left\{13.2\left(\log n + 1.28 + \frac{0.175}{n}\right)^2, 1321\right\} \\&#xA;n \le 1320&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, there are no large counterexamples to Professor Halfbrain's theorem, and we only need to check the cases $n=1$ to $1320$, which can be done with a short computer program; for example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#include &amp;lt;stdio.h&amp;gt;&#xA;#include &amp;lt;gmp.h&amp;gt;&#xA;#define NMAX 1320&#xA;&#xA;int main(int argc, char *argv[]) {&#xA;  int n;&#xA;  mpz_t pow_2016, pow_32, sum;&#xA;&#xA;  mpz_init(pow_2016);&#xA;  mpz_init(pow_32);&#xA;  mpz_init(sum);&#xA;&#xA;  for(n = 1; n &amp;lt;= NMAX; n++) {&#xA;    mpz_ui_pow_ui(pow_2016, 2016, n);&#xA;    mpz_ui_pow_ui(pow_32,   32,   n);&#xA;    mpz_add(sum, pow_2016, pow_32);&#xA;&#xA;    size_t len_2016 = mpz_sizeinbase(pow_2016, 10);&#xA;    size_t len_sum  = mpz_sizeinbase(sum, 10);&#xA;    if(len_2016 &amp;lt; len_sum) {&#xA;      printf(&quot;failure at n = %d\n&quot;, n);&#xA;      break;&#xA;    }&#xA;  }&#xA;&#xA;  if(n &amp;gt; NMAX) {&#xA;    printf(&quot;success\n&quot;);&#xA;  }&#xA;&#xA;  mpz_clear(pow_2016);&#xA;  mpz_clear(pow_32);&#xA;  mpz_clear(sum);&#xA;&#xA;  return 0;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The result:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;robert@unity:~/c$ gcc -o 2016 2016.c -lgmp -lm &amp;amp;&amp;amp; ./2016&#xA;success&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Therefore, Professor Halfbrain is correct (this time).&lt;/p&gt;&#xA;" OwnerUserId="9718" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:31.357" LastActivityDate="2016-03-15T21:52:03.890" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28946" PostTypeId="2" ParentId="25248" CreationDate="2016-03-15T06:20:56.143" Score="2" Body="&lt;h1&gt;Not an Answer - Just Some Thoughts&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Almost a curse: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It occurred to me that some people use &quot;a-hole&quot; as a curse. So &quot;U A O&quot; might be considered close to &quot;You a-hole&quot; depending on how you pronounce it. Possibly even &quot;U A&quot; might be considered almost a curse but that would not make me happy if it were the solution.&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Another possibility for cursing could be &quot;O I U&quot; as in &quot;Oi, you!&quot; This is a British expression meaning basically &quot;hey, you&quot; but it is rather aggressive and might be considered almost a curse. &quot;Oi&quot; or &quot;Oy&quot; is also a Yiddish exclamation as in &quot;Oy vey&quot; or &quot;oy gevalt&quot; but it's really more dismay than cursing.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clarifying: &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; google &quot;ion clarify&quot;. Lots of hits, so potentially anything ending in ion could qualify. Doesn't work at all with my &quot;almost a curse&quot; idea of course.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where famous people speak &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; might refer to politics. The Hill is a name for Washington DC. I couldn't go any further with this but are there other four letter names for places where politicians speak?&lt;br&gt;&#xA; A quad is a place, but not particularly one where famous people speak.&lt;br&gt;&#xA; The United Nations General Assembly UNGA is a place where famous people speak but &quot;ungainly&quot; doesn't fit with any other clues. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastEditorUserId="18924" LastEditDate="2016-03-19T21:45:36.563" LastActivityDate="2016-03-19T21:45:36.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28947" PostTypeId="2" ParentId="28882" CreationDate="2016-03-15T06:31:45.643" Score="3" Body="&lt;p&gt;The bear's color is obviously&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; somewhere between silvery grey and chocolate brown,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the bear is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Koala&quot; rel=&quot;nofollow&quot;&gt;koala bear&lt;/a&gt;, and they come in a range of colors from silvery grey to chocolate brown;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and I know this because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. The question mentions specifically that there is no lateral-thinking tag. However, it is rather odd to write such a thing. If indeed OP wanted us to solve the puzzle based on a straightforward use of logic, he/she would simply not have said anything about lateral thinking. Saying that there is no lateral-thinking tag can only be a trick meant to distract us from the reality that the puzzle should indeed be solved using lateral thinking.&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; 2. Tellingly, OP used the British spellings for &quot;colour&quot; and &quot;metre&quot;, as also used in Australia, home of the koala. The same spellings are also used in Canada, home to polar bears and other bear species, but that would be too obvious, and would not require any lateral thinking, which we are not supposed to use, and which we therefore (as I explained above) are in fact supposed to use.&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; 3. Australia has a lot of sand. The koala could have fallen off a tree (which makes sense since it actually spends much of its life on trees, unlike other bears which normally do not have any business being 10 metres above the ground) and into a big pile of sand. That is why it was not hurt.&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; 4. Other answers have mentioned all sorts of other bears. No one has mentioned koala bears yet. Clearly I'm on to something! My answer also involves the cutest kind of bear, possibly tied with Hugh Meyers' suggestion of a panda.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user17947" LastActivityDate="2016-03-15T06:31:45.643" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28952" PostTypeId="2" ParentId="25248" CreationDate="2016-03-15T07:01:41.287" Score="3" Body="&lt;p&gt;My attempt at this!&#xA;The word is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Bookshelf&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My last three letters serve to clarify.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;ELF&lt;/strong&gt;'s clarifying magic can be seen &lt;a href=&quot;http://www.elfcosmetics.com.au/products/e-l-f-essential-clarifying-pressed-powder#.VuevE_l97IU&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My first four letters are where famous people speak.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;BOOK&lt;/strong&gt; is where famous people speak out their auto-biographies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Dropping all consonants, I am almost a curse.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;OOE&lt;/strong&gt; or Out Of Endurance is almost a curse if you're playing Dark age of Camelot, because if you don't have an endurance potion, you will die. &lt;a href=&quot;http://www.urbandictionary.com/define.php?term=ooe&quot; rel=&quot;nofollow&quot;&gt;SEE HERE&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19656" LastEditorUserId="19656" LastEditDate="2016-03-15T07:26:38.070" LastActivityDate="2016-03-15T07:26:38.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28954" PostTypeId="1" AcceptedAnswerId="28955" CreationDate="2016-03-15T08:17:01.903" Score="10" ViewCount="1458" Body="&lt;p&gt;Honest Oscar was a cruel criminal, but he always spoke the truth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After being swindled by Lying Larry (another cruel criminal, and a much less honest one), Oscar confronted Larry on the roof of a tall building.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;You've lied to me,&quot; said Oscar, lifting his handgun. &quot;And I am going to kill you now. But first, I want you to talk.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was a dramatic pause.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;If what you say is true,&quot; said Oscar, &quot;I will shoot you. If you lie or remain silent, I will push you off the roof.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A sly grin spread across Larry's face. &quot;You are going to push me off the roof.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oscar considered this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then, true to his word, Honest Oscar killed Lying Larry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="18558" LastActivityDate="2017-03-28T09:54:38.613" Title="An Honest Death" Tags="|logical-deduction|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28955" PostTypeId="2" ParentId="28954" CreationDate="2016-03-15T08:26:06.173" Score="13" Body="&lt;p&gt;He shot him &lt;strong&gt;and&lt;/strong&gt; pushed him off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He never said he would push him &lt;strong&gt;only&lt;/strong&gt; if he lied or remain silent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, he said &lt;em&gt;I am going to kill you now.&lt;/em&gt;, so killing him, pushing him off the roof and shooting him was the only way to keep his promises.&lt;/p&gt;&#xA;" OwnerUserId="12165" LastEditorUserId="12165" LastEditDate="2017-03-28T09:54:38.613" LastActivityDate="2017-03-28T09:54:38.613" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28959" PostTypeId="2" ParentId="28892" CreationDate="2016-03-15T11:27:32.333" Score="1" Body="&lt;p&gt;The answer to that puzzle is the following moves:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;First select the square number 5 to turn all the board to black.&lt;/li&gt;&#xA;&lt;li&gt;Then select all the squares positionned in a corner which are 1, 3, 7 and 9. We obtain like that the inverse of the initial board.&lt;/li&gt;&#xA;&lt;li&gt;Then simply select the square number 5 and all the board is white.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="20112" LastActivityDate="2016-03-15T11:27:32.333" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28960" PostTypeId="1" AcceptedAnswerId="28962" CreationDate="2016-03-15T11:34:09.213" Score="6" ViewCount="202" Body="&lt;p&gt;Here is an easy Rebus to wake the group and save the day.&#xA;Happy Solving!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Robert Downey Jr. + CMnSiSNi&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19656" LastActivityDate="2016-03-15T11:56:05.607" Title="The One We All Love" Tags="|wordplay|rebus|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28962" PostTypeId="2" ParentId="28960" CreationDate="2016-03-15T11:43:46.800" Score="5" Body="&lt;p&gt;The answer might be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Man of Steel, &lt;a href=&quot;http://www.imdb.com/title/tt0770828/&quot; rel=&quot;nofollow&quot;&gt;http://www.imdb.com/title/tt0770828/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Argument:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. Robert Downey Jr plays in the movie Iron Man, &lt;a href=&quot;http://www.imdb.com/title/tt0371746/&quot; rel=&quot;nofollow&quot;&gt;http://www.imdb.com/title/tt0371746/&lt;/a&gt;&lt;br&gt;&#xA; 2. carbon, manganese, silicon, sulfur, nickel are standard ingredients for steel alloys, &lt;a href=&quot;https://en.wikipedia.org/wiki/Alloy_steel&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Alloy_steel&lt;/a&gt;&lt;br&gt;&#xA; 3. Iron + alloys = steel&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-15T11:56:05.607" LastActivityDate="2016-03-15T11:56:05.607" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="28964" PostTypeId="1" AcceptedAnswerId="30156" CreationDate="2016-03-15T11:47:18.167" Score="7" ViewCount="542" Body="&lt;p&gt;I'm implementing a Douglas Engel's puzzle, &lt;a href=&quot;http://www.puzzlopia.com/puzzles/color-wheels/play&quot; rel=&quot;nofollow noreferrer&quot;&gt;Color Wheels&lt;/a&gt; (reproduced with permission) and now I would like to check solutions of users.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those of you that don't know the puzzle, here you can see that it consists of two intersecting wheels with three common pieces. The wheels can rotate so there is an interchange of pieces. As you can imagine, after mixing the pieces, the objective is to rearrange to original position, pretty much the same as in Rubik's Cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ImT6T.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ImT6T.png&quot; alt=&quot;Picture from Douglas Engel page&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two problems:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I am mixing the pieces with 90 alternative random movements (45 each wheel). But, is there any state beyond that? I mean, is any valid state reachable with these 90 movements?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I want to give 3 stars to those who solve the puzzle with optimal solution. I've mixed pieces randomly so I don't know the optimal (and solving the puzzle with javascript would require an algorithm and an implementation of it). What I would like is to know if there is a &quot;God number&quot; for this puzzle, a minimum number of steps to solve any configuration (like in Rubik's Cube)*.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(Please note that my implementation takes into account the orientation of pieces, but the original puzzle &quot;Color Wheels&quot; doesn't. By now, I am interested in the version &lt;strong&gt;without orientations&lt;/strong&gt;, like in the image above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*This means that sometimes I will give 3 stars to solutions that are not truly optimal, but that's ok.&lt;/p&gt;&#xA;" OwnerUserId="17558" LastEditorUserId="11876" LastEditDate="2016-03-15T16:08:44.673" LastActivityDate="2016-04-02T02:09:02.080" Title="Engel's Enigma Optimals" Tags="|combinatorics|puzzle-creation|twisty-puzzles|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="28968" PostTypeId="2" ParentId="28595" CreationDate="2016-03-15T13:08:11.673" Score="3" Body="&lt;p&gt;Here's my &lt;strong&gt;shot&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I do think this answer is more related to sports especially Football rather than movies. Hence my answer is &lt;strong&gt;David Beckham&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was the freest man around.&#xA;Yet I never left my grounds&#xA;(for years at a stretch)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He played the &lt;strong&gt;midfielder&lt;/strong&gt; position in Football. Midfielders are the freest men around and travel the longest distance in a Football match. They never leave the ground, or it can be a reference to David's outfield appearance record. &lt;a href=&quot;https://en.wikipedia.org/wiki/David_Beckham&quot; rel=&quot;nofollow&quot; title=&quot;Beckham Wiki&quot;&gt;Here's the Wiki link&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I could walk, fight, or run.&#xA;I'm as able-bodied as anyone!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I guess its pretty easy to relate these with the answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And my riches I can't count!&#xA;But I rarely ventured out.&#xA;I'm stable mentally and emotionally&#xA;And many dream of being me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We can say David Beckham is kind of &quot;Richy Rich Rich&quot;. Rarely ventured out can again be a reference to his outfield appearance record, and of course he is mentally and emotionally stable (but his OCD keeps me bothering about this mental stability relation). May people do dream about being Beckham.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19191" LastActivityDate="2016-03-15T13:08:11.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28970" PostTypeId="2" ParentId="28890" CreationDate="2016-03-15T14:06:43.420" Score="1" Body="&lt;p&gt;I've already posted an answer which is clearly wrong but this is a different approach so I think it warrants another answer.  The answer is simpler than some of the others posted but may be incomplete.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is still that the theorem is true.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We start with the observation that for the theorem to be true $ 32^n $ must be smaller than the difference between the next higher power of 10 and $ 2016^n $.  The next higher power of 10 is simply $ 10^{\lfloor\log_{10}(2016^n)\rfloor+1} $, so the theorem is true iff (for all $ n $):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;10^{\lfloor\log_{10}(2016^n)\rfloor+1} - 2016^n &amp;gt; 32^n \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get rid of the pesky floor function we replace the $+1$ with $ +x $, such that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\lfloor\log_{10}(2016^n)\rfloor+1 = \log_{10}(2016^n)+x&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ x $ is equal to $ 1 $ less the fractional part of $ \log_{10}(2016^n) $, and is always in the range $ 0 &amp;lt; x \leq 1 $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The previous inequality can be reduced to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;10^{\log_{10}(2016^n) + x} - 2016^n &amp;gt; 32^n \\&#xA;2016^n10^x-2016^n - 32^n &amp;gt; 0 \\&#xA;32^n\left(-1 - 63^n + 63^n10^x\right) &amp;gt; 0 \\&#xA;63^n\left(10^x-1\right) -1 &amp;gt; 0 \\&#xA;10^x - 1 &amp;gt; 63^{-n}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $ x $ is in the range $ 0 &amp;lt; x \leq 1 $, the first term is in the range $ 1 &amp;lt; 10^x \le 10 $. A number of observations can be made:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;First, clearly as $ x \to 0 $ the inequality does not hold for any $ n $.  This is intuitively what we would expect, i.e. as $ 2016^n $ approaches a power of $ 10 $, the $ 32^n $ comes closer to rolling it over to the next power of $ 10 $.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Second, as $ n \to\infty $, the inequality is more likely to be true ($ x $ can be smaller and smaller while still producing a positive overall result).  This observation is dramatically demonstrated by changing to above inequality to an equality, and solving for the largest value of $ x $ that makes the inequality false:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ x = \log_{10}(1 + 63^{-n}) \approx \frac{63^{-n}}{\log 10} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is apparent that the critical value of $ x $ rapidly approaches $ 0 $, even for quite small values of $ n $.  For example, when $ n = 5 $ the inequality becomes $ x &amp;gt; 4.4 \times 10^{-10} $; again, $ x $ needs to be larger than this tiny value to provide a counterexample to the theorem.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So how small does $ x $ actually get?  Remember $ x $ is equal to $ 1 $ less the fractional part of $ \log_{10}(2016^n) = n\log_{10} 2016 $.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, $ x $ follows a cycle that hits its minimum where:&#xA;$$&#xA;n = \left\lfloor\frac{10^m}{\log_{10}(2016)}\right\rfloor&#xA;$$&#xA;where $ m $ is a non-zero positive integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first such case is $ n = 3 $, followed by $ n = 30 $, $ n = 302 $... The log function apparently cannot be expressed as a decimal fraction, so $ x $ will never be exactly zero but will get closer as the approximation is improved with each power of 10, but only by one digit with each power of 10.  Therefore although the minimum $ x $ approaches $ 0 $ it does so far more slowly than the required $ x $ to break the theorem - by many orders of magnitude - and the gap just keeps on widening.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At $ n = 3 $ the theorem holds and as the disparity between the minimum $ x $ and the required minimum $ x $ only widens with increasing $ n $, we can say conclusively that no value of $ n $ will break the theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be taken a step further, because knowing that $ n = 3 $ is the point of vulnerability, we can explore other bases other than $ 32 $.  By doing this we can demonstrate that any integer base $ b &amp;lt; 1218 $ will also not break the theorem for any value of $ n $ e.g.:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 2016^3 = 8.19 \times 10^9 \\&#xA;2016^3 + 1217^3 = 9.99 \times 10^9 \\&#xA;2016^3 + 1218^3 = 1.00 \times 10^{10}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The weakness of the proof is that $ \frac{1}{\log_{10}\left(2016\right)} $ is assumed to not be representable as a decimal fraction.  If it is, then at some point the minimum $ x $ &lt;em&gt;will&lt;/em&gt; reach zero but the required minimum will never reach zero so at that point the theorem fails.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Actually $ \frac{1}{\log_{10}\left(2016\right)} $ can be shown to be not just unrepresentable as a decimal fraction but, in fact, irrational:&#xA;$$&#xA;\log_{10}\left(2016\right) = \frac{a}{b} \\&#xA;2016^b = 10^a&#xA;$$&#xA;which is impossible since $ 6 \cdot 6 = 36 $ and therefore the least significant digit of all powers of $ 2016 $ will be  $ 6 $ and never $ 0 $.  The reciprocal is obviously also irrational.  The proof is therefore complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...and just in case anyone is still interested in this (although I  am probably just talking to myself), as a corollary of the above proof the original theorem can be generalized as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For any number with an integer base $ x \geq 1 $, raised to an integer power $ n \geq 1 $, the maximum integer base $ y \geq 1 $, raised to the same power, that can be added such that the result has the same number of digits when represented as a radix $ b $ number is:&#xA;  $$&#xA;y = \lfloor{x\left(b^{b-m\log_b{x}} - 1\right)^\frac{1}{m}}\rfloor&#xA;$$&#xA;  where $ m = \lfloor{b\log_x{b}}\rfloor $, $ m \neq 0 $ and $ b-m\log_b{x} \leq 1 $&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For $ m = 0 $ there is no solution and for $ b-m\log_b{x} &amp;gt; 1 $ the theorem breaks down.  This occurs adjacent to the discontinuities that exist in the this expression due to the floor function. It might be possible to modify the theorem to remove, or limit, the second condition.&lt;/p&gt;&#xA;" OwnerUserId="13335" LastEditorUserId="9718" LastEditDate="2016-03-22T04:07:58.410" LastActivityDate="2016-03-22T04:07:58.410" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28973" PostTypeId="2" ParentId="28847" CreationDate="2016-03-15T14:49:22.827" Score="1" Body="&lt;p&gt;Going to give it a shot here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Converting: 49 66 20 79 6f 75 20 77 61 6e 74 20 74 6f 20 75 73 65 20 74 68 65 20 6d 61 63 68 69 6e 65 2e 2e 2e 20 55 73 65 20 74 68 65 20 63 6f 64 65 from hex to ascii it reveals &quot;If you want to use the machine... Use the code&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The next step:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The numbers '111-111-111--111-1--111-11' at the end could signify 777--71--73 so I'm going to assume that means subtraction or addition as those are the only real ways to end up with a 3 digit number at the end. If it's subtraction it's 633, but if the double subtraction represents addition it would be 921. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since the puzzle doesn't state whether or not you die if you enter the code wrong at first I'm assuming you could try both numbers and make it out ok either way. &lt;/p&gt;&#xA;" OwnerUserId="19731" LastActivityDate="2016-03-15T14:49:22.827" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28974" PostTypeId="2" ParentId="1660" CreationDate="2016-03-15T14:54:38.127" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Select a random road and...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ask either one of the trolls:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Would the other troll point me on this road in order for me to live?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Expanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1. &lt;strong&gt;You chose the wrong road and asked the liar.&lt;/strong&gt;&lt;br&gt;&#xA;    He would answer Yes. Because the truth teller would tell you &quot;No&quot; when asked if this is the right road.&lt;br&gt;&#xA; 2. &lt;strong&gt;You chose the wrong road and asked the truth teller.&lt;/strong&gt;&lt;br&gt;&#xA;    He would answer Yes. Because the liar would say &quot;No&quot; when asked if this is the right road.&lt;br&gt;&#xA; 3. &lt;strong&gt;You chose the right road and asked the liar.&lt;/strong&gt;&lt;br&gt;&#xA;    He would answer No. Because the truth teller would tell you &quot;Yes&quot; when asked if this is the right road.&lt;br&gt;&#xA; 4. &lt;strong&gt;You chose the right road and asked the truth teller.&lt;/strong&gt;&lt;br&gt;&#xA;    He would answer No. Because the liar would say &quot;No&quot; when asked if this is the right road.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Conclusion:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The answer is Yes from either one of the trolls: take the other road.&lt;br&gt;&#xA; The answer is No from either one of the trolls: take the road you selected.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19989" LastActivityDate="2016-03-15T14:54:38.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28975" PostTypeId="2" ParentId="28854" CreationDate="2016-03-15T15:10:36.423" Score="4" Body="&lt;p&gt;There is a single valid solution which is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $R=0, E=8, C=2, A=7, N=4, S=6, D=5$ and $Q=3$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Clearly, from the units part of equation &lt;br&gt;$R= E-E =0$ &lt;br&gt;and from the $10$s &lt;br&gt; $C \equiv - E (\text{mod } 10)$. &lt;br&gt; Also, as before, we can take advantage of the fact that $D=S-1$ so that we must have a pair of consecutive digits at the end of calculation (provided we proceed from right to left as below).&lt;br&gt;&lt;br&gt; As with the previous alphametic, if we select $E$ the rest is determined to a large extent. This time, I'll go through each of the cases explicitly. &lt;br&gt;&lt;br&gt; $E=1 \rightarrow C=9$ and since $A-N-1 \equiv 9 (\text{mod } 10)$ we get $A \equiv N (\text{mod } 10)$ which is not allowed. &lt;br&gt;&lt;br&gt; $E=2 \rightarrow C=8$ which means $A-N-1 \equiv 8(\text{mod } 10)$  and $A = N-1$. Then, $U-A-1 \equiv N (\text{mod } 10) \rightarrow U=2N(\text{mod }10)$.&lt;br&gt; Since $0$, $2$ and $8$ are taken, this means that $U=4$ or $6$ but $U=6$ isn't feasible since then $N=3$ (since it can't be $8$) and $A=2$ (already taken). Hence, $U=4 \rightarrow N=7 \rightarrow A=6$. But this solution fails to preserve a pair of consecutive digits for $D$ and $S$. &lt;br&gt;&lt;br&gt; $E=3 \rightarrow C=7 \rightarrow A-N-1 \equiv 7 (\text{mod } 10)$. Now, we could have $(A,N) = (9,1)$ or $A=N-2$. If the former, then $U=2$ and $Q \equiv D (\text{mod }10)$ which is not allowed. &lt;br&gt; If the latter then $U-A-1 \equiv N (\text{mod } 10) \rightarrow U \equiv 2N-1(\text{mod }10)$ which mean that $U$ must be $1, 5$ or $9$. We divide this into subcases &lt;br&gt; (i) $U=1 \rightarrow N=6 \rightarrow A=4 \rightarrow S=9$ and $D=8$ (since they are the only remaining consecutive digits). But then $Q-9 \equiv 4 (\text{mod }10) \rightarrow Q=E=3$ which is not allowed.&lt;br&gt;(ii) $U=5 \rightarrow N=8 \rightarrow A=6 \rightarrow S=2, D=1$ but then $Q-2 = 6 \rightarrow Q=N=8$ also not allowed &lt;br&gt; (iii) $U=9 \rightarrow N=5 \rightarrow A=E=3$ which is not allowed.&lt;br&gt;&lt;br&gt;$E=4 \rightarrow C=6 \rightarrow A-N-1 \equiv 6 (\text{mod }10)$. So $(A,N) = (9,2), (8,1), (2,5)$ or $(5,8)$ &lt;br&gt;(i) $A=9 \rightarrow U=1 \rightarrow Q=D$ which is not allowed. &lt;br&gt;(ii) $A=8 \rightarrow U=9 \rightarrow S=3, D=2 \rightarrow Q-2 \equiv 8 (\text{mod } 10) \rightarrow Q=R=0$ not allowed. &lt;br&gt;(iii) $A=2 \rightarrow U=8$ which means there are no consecutive digits left for $S$ and $D$. &lt;br&gt; (iv) $A=5 \rightarrow U=4=E$, not allowed. &lt;br&gt;&lt;br&gt; $E=5 \rightarrow C=5$, not allowed. &lt;br&gt;&lt;br&gt; $E=6 \rightarrow C=4 \rightarrow (A,N) = (2,7), (3,8), (7,2)$ or $(8,3)$ then &lt;br&gt; (i) $A=2 \rightarrow U=0=R$, not allowed &lt;br&gt;(ii) $A=3 \rightarrow U=2$ but there are no remaining consecutive digits for $S$ and $D$. &lt;br&gt; (iii) $A=7 \rightarrow U=9$ but again no remaining consecutive digits for $S$ and $D$. &lt;br&gt;(iv) $A=8 \rightarrow U=1 \rightarrow S=3, D=2 \rightarrow Q=1=U$, not allowed. &lt;br&gt;&lt;br&gt; $E=7 \rightarrow C=3 \rightarrow (A,N) = (2,8), (5,1), (6,2), (8,4) or (9,5)$ then &lt;br&gt;(i) $A=2 \rightarrow U=1 \rightarrow D=4,5$ and $Q-D=3$ which means $Q=7$ or $8$ so $Q=E$ or $Q=N$, both of which are not allowed. &lt;br&gt; (ii) $A=5 \rightarrow U=6 \rightarrow S=9, D=8$ but then $Q=3=C$, not allowed. &lt;br&gt; (iii) $A=6 \rightarrow U=8 \rightarrow S=5, D=4$ but then $Q=0=R$, not allowed &lt;br&gt; (iv) $A=8 \rightarrow U=2 \rightarrow S=6, D=5$ but then $Q=4=N$, not allowed. &lt;br&gt; (v) $A=9 \rightarrow U=4 \rightarrow S=2, D=1 \rightarrow Q=1=D$, not allowed. &lt;br&gt;&lt;br&gt; $E=8 \rightarrow C=2 \rightarrow (A,N) = (2,9), (4,1), (6,3), (7,4)$ or $(9,6)$ then &lt;br&gt; (i)$A=2 \rightarrow U=2=A$, not allowed &lt;br&gt; (ii) $A=4 \rightarrow U=5 \rightarrow S=7, D=6 \rightarrow Q=0=R$, not allowed &lt;br&gt; (iii) $A=6 \rightarrow U=9 \rightarrow S=5, D=4 \rightarrow Q=0=R$, not allowed &lt;br&gt; (iv) $A=7 \rightarrow U=1 \rightarrow S=6, D=5 \rightarrow Q=3$ &lt;b&gt;which is a valid solution&lt;/b&gt; &lt;br&gt; (v) $A=9 \rightarrow U=5 \rightarrow S=4, D=3 \rightarrow Q=3=D$, not allowed. &lt;br&gt;&lt;br&gt; Finally $E=9 \rightarrow C=1 \rightarrow (A,N) = (4,2), (5,3), (6,4), (7,5)$ or $(8,6)$ then &lt;br&gt; (i) $A=4 \rightarrow U=6 \rightarrow S=8, D=7$ but then $Q=1=C$, not allowed. &lt;br&gt;(ii) $A=5 \rightarrow U=8 \rightarrow S=7, D=6 \rightarrow Q=1=C$, not allowed &lt;br&gt; (iii) $A=6 \rightarrow U=0=R$, not allowed. &lt;br&gt; (iv) $A=7 \rightarrow U=2 \rightarrow S=4, D=3 \rightarrow Q=0=R$, not allowed. &lt;br&gt; (v) $A=8 \rightarrow U=4 \rightarrow S=3, D=2 \rightarrow Q=1=C$, not allowed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18422" LastActivityDate="2016-03-15T15:10:36.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28976" PostTypeId="2" ParentId="28818" CreationDate="2016-03-15T15:20:59.633" Score="5" Body="&lt;p&gt;You could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Queen Elizabeth II&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the luckiest lady around!&#xA;I am the most constricted of all.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Many people consider coins good luck, her portrait is on British coinage. As a portrait on a coin she cannot go anywhere so she is very restricted. Also, she herself is lucky to be the queen, but she is restricted in the things she can do by her station.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the richest lady on earth!&#xA;Yet I own nothing at all.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Again, as a portrait on a money, she &quot;has&quot; lots of money yet doesn't own it. Also, as queen she is wealthy, yet her wealth ultimately belongs to the British nation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The world bends to my will!&#xA;Yet I possess no power.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The world's behavior is influenced by the money with her portrait on it. Yet as queen, she's a figurehead that doesn't have political power.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Every girl wishes to be in my place!&#xA;But I chose not to come and only leave at death.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Every girl dreams of being the Queen. However, the Queen was born into her position and didn't choose it, and will be Queen until her death.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The title is a pun, but I doubt anyone will catch it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;The &quot;tail&quot; in the title refers to the coins that bear her image. Also &quot;catch it&quot; might be referring to catching a coin after it is flipped in the air.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="20179" LastActivityDate="2016-03-15T15:20:59.633" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28977" PostTypeId="2" ParentId="1660" CreationDate="2016-03-15T15:21:27.583" Score="-3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Which road would the other troll tell me to go to for this destination?&quot;&lt;br&gt;&#xA; Then take the opposite road.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="20181" LastEditorUserId="12109" LastEditDate="2016-03-15T15:54:55.650" LastActivityDate="2016-03-15T15:54:55.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28979" PostTypeId="1" AcceptedAnswerId="29013" CreationDate="2016-03-15T16:03:41.437" Score="15" ViewCount="4740" Body="&lt;p&gt;Wendy Testaburger and Sally Turner play a game with two unfair coins. A coin flip with Wendy's coin shows head with probability $\frac {1} {100}$. A coin flip with Sally's coin shows head with probability $p$. Wendy does the first coin flip. Wendy and Sally flip their coins alternately until a coin shows heads. The girl with the coin showing head wins the game. What value of $p$ makes this game a fair game?&lt;/p&gt;&#xA;" OwnerUserId="9125" LastEditorUserId="12408" LastEditDate="2016-07-25T02:28:49.250" LastActivityDate="2016-07-25T02:28:49.250" Title="Unfair coins at South Park Elementary" Tags="|mathematics|probability|coin-tossing|" AnswerCount="8" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28981" PostTypeId="2" ParentId="25828" CreationDate="2016-03-15T16:12:35.140" Score="2" Body="&lt;p&gt;Looks like 5468 or 5408 to me... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qlWox.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qlWox.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20186" LastActivityDate="2016-03-15T16:12:35.140" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28982" PostTypeId="2" ParentId="28954" CreationDate="2016-03-15T16:14:43.137" Score="2" Body="&lt;p&gt;Are all false answers lies? If I tell you &quot;you're going to do X&quot; it's less a statement of fact than a prediction. So if upon hearing these words you proceed to do Y instead of X, even though my statement was false, it's not that I've &lt;em&gt;lied&lt;/em&gt; so much as that I was merely incorrect, especially since the chief factor in the outcome of my prediction is &lt;em&gt;your&lt;/em&gt; behavior and not my own. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oscar's threat only covers the conditions that Larry's statement is (a) true or (b) dishonest; he is silent on how he will kill Larry if his statement is false but sincere. In this case, Oscar is free to kill Larry any way he chooses (although if Oscar decides to push Larry off the roof, Larry's prediction becomes true, and Oscar must also shoot him to fulfill his promise).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we might also take that a prediction is an expression of personal belief, i.e., that when Larry says &quot;You are going to push me off the roof&quot;, it should be understood as &quot;&lt;em&gt;I believe that&lt;/em&gt; you are going to push me off the roof&quot;. In this case, Larry's statement is honest if he does believe Oscar will push him off the roof, and a lie if Larry believes that he won't. Taken this way, Oscar can now evaluate Larry's statement in both relevant dimensions (true/false and sincere/lie), giving us four possible outcomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;True and sincere prediction: Oscar must shoot Larry&#xA;False and sincere prediction: Oscar may kill Larry by any means he chooses&#xA;False and insincere prediction: Oscar must push Larry off the roof&#xA;True and insincere prediction: Oscar must shoot Larry and push him off the roof&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since Oscar can't know for sure whether Larry really believes in his stated prediction, his best course of action is to push him off the roof &lt;em&gt;and&lt;/em&gt; shoot him, which fulfills Oscar's promise in every eventuality.&lt;/p&gt;&#xA;" OwnerUserId="20189" LastActivityDate="2016-03-15T16:14:43.137" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28983" PostTypeId="2" ParentId="28979" CreationDate="2016-03-15T16:15:35.777" Score="3" Body="&lt;p&gt;&lt;strike&gt;First instinct, but I may be wrong on this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1/100.  Then both of them have the same chance of winning.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;/strike&gt;&#xA;&lt;strong&gt;[EDIT]&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second instinct :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1/99&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why my second instinct?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; let's limit to 1 toss each.&lt;br&gt;&#xA; Wendy's chance. $1/100$.&lt;br&gt;&#xA; Sally's chance $99 * p/100 $. For this to be equal chances $p = 1/99$.&lt;br&gt;&#xA;&lt;br&gt;&#xA; Note: will try my third instinct for 2 tosses each.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19989" LastEditorUserId="19989" LastEditDate="2016-03-15T16:35:31.657" LastActivityDate="2016-03-15T16:35:31.657" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="28984" PostTypeId="2" ParentId="28872" CreationDate="2016-03-15T16:23:06.030" Score="5" Body="&lt;p&gt;Wouldn't this be the &lt;em&gt;simplest&lt;/em&gt; solution?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It is false to say that the letter 't' appears &lt;strong&gt;zero&lt;/strong&gt; times in this sentence.&lt;br&gt;&#xA; It is &lt;strong&gt;false&lt;/strong&gt; to say that the letter 'f' appears &lt;strong&gt;zero&lt;/strong&gt; times in this paragraph.&lt;br&gt;&#xA; It is &lt;strong&gt;false&lt;/strong&gt; to say that the letter 't' appears forty-six times in this paragraph.&lt;br&gt;&#xA; It is true to say that the letter 'e' appears &lt;strong&gt;twelve&lt;/strong&gt; times in this sentence.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17024" LastActivityDate="2016-03-15T16:23:06.030" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28986" PostTypeId="2" ParentId="28979" CreationDate="2016-03-15T16:27:26.647" Score="12" Body="&lt;p&gt;The game is fair for&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; p = $\frac1{99}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If Wendy's coin has odds $\frac1n$, a fair game is achieved when Sally's coin has odds $\frac1{n-1}$. It's easy to see that if Wendy's coin has $\frac12$ odds, Sally needs a coin that always comes up heads. If Wendy's coin comes up heads $\frac13$ of the time, and Sally's comes up heads $\frac12$ of the time, there's a $\frac13$ chance Wendy wins on turn 1, and a $\frac12 \times \frac23=\frac13$ chance Sally wins on turn 2 ($\frac12$ as her odds of getting heads, and $\frac23$ as the odds of the game lasting long enough for there to be a turn 2). Then there's a $\frac13 \times \frac13 = \frac19$ chance Wendy wins on turn 3, a $\frac12 \times \frac29=\frac19$ Sally wins on turn 4... for any pair of turns, Wendy and Sally will have equal odds. We can see this trend holds for any $\frac1n$, leading us to the above answer of $\frac1{99}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7459" LastEditorUserId="4421" LastEditDate="2016-03-16T20:59:20.487" LastActivityDate="2016-03-16T20:59:20.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28987" PostTypeId="2" ParentId="28979" CreationDate="2016-03-15T16:39:04.500" Score="8" Body="&lt;p&gt;The answer is 0.0101010101...&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Proof&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Say that $w$ is the chance of Wendy’s coin coming up heads; $s$ is the chance for Sally’s coin. Wendy will win with probability:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p(W)= w + w's'w + w's'w's'w + w's'w's'w's'w ... = w ( 1 + (w's')^2 + (w's')^3 ... )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sally wins with probability:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p(S) = w's + w's'w's + w's'w's'w's ... = s ( w' + w'^2s'^1 + w'^3s'^2 +  ....)&#xA;= sw'(1 + (w's')^2 + (w's')^3 ... $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set them equal to each other:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$sw'(1 + (w's')^2 + (w's')^3 ...  =  w ( 1 + (w's')^2 + (w's')^3$$&#xA;$$sw' = w$$&#xA;$$w= 0.01, w'=0.99$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $s = 0.01 / 0.99 = 1/99 = 0.0101010101 ...$.&lt;/p&gt;&#xA;" OwnerUserId="19698" LastEditorUserId="359" LastEditDate="2016-03-16T12:17:23.563" LastActivityDate="2016-03-16T12:17:23.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28988" PostTypeId="2" ParentId="28979" CreationDate="2016-03-15T16:40:30.473" Score="12" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$p = \frac{1}{99}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Let $p(W,k)$ be the probability of Wendy winning on her $k$th turn and $p(S,k)$ be the probability of Sally winning on her $k$th turn. Then, &lt;br&gt;&lt;br&gt; $p(W,1) = \frac{1}{100}, p(W,2) = \frac{99}{100}(1-p)\frac{1}{100}$ and, by simple induction $p(W, k) = \left(\frac{99}{100}(1-p) \right)^{k-1} \left(\frac{1}{100} \right)$. &lt;br&gt;&lt;br&gt; Also, by a simple induction it can be shown that &lt;br&gt; $p(S, k) = \left(\frac{99}{100}(1-p) \right)^{k-1}\left(\frac{99p}{100} \right)$. &lt;br&gt;&lt;br&gt; We must then enforce &lt;br&gt;&lt;br&gt; $\sum_{k=0}^{\infty} p(W,k) = \sum_{k=0}^{\infty} p(S,k)$ &lt;br&gt; which implies $\frac{1}{100} = \frac{99p}{100}$ as required &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18422" LastActivityDate="2016-03-15T16:40:30.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28989" PostTypeId="1" AcceptedAnswerId="29002" CreationDate="2016-03-15T16:45:57.713" Score="10" ViewCount="3049" Body="&lt;p&gt;This is a follow-up question to my older puzzle &quot;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/28439/a-truly-amazing-way-of-making-the-number-2016&quot;&gt;A truly amazing way of making the number 2016&lt;/a&gt;&quot; and to my other older puzzle &quot;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/28444/a-truly-amazing-way-of-making-every-possible-positive-integer&quot;&gt;A truly amazing way of making every possible positive integer&lt;/a&gt;&quot;. My favorite solution to the first puzzle observed that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$~~~~2016 ~=~ (1+2+3+4+5+6+7)*8*9$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that in this solution the digits occur in a neat increasing order!&#xA;The solution to the second puzzle gives for every positive integer $n$ a mathematical expression that yields the value $n$ while obeying the following five rules:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Each of the digits $1,2,3,4,5,6,7,8,9$ is used &lt;strong&gt;exactly once&lt;/strong&gt; (while digit $0$ is not used).&lt;/li&gt;&#xA;  &lt;li&gt;Decimal points are allowed.&lt;/li&gt;&#xA;  &lt;li&gt;One may use brackets &quot;(&quot; and &quot;)&quot; to structure your expression, and &#xA;  to make it well-defined.&lt;/li&gt;&#xA;  &lt;li&gt;The only allowed mathematical operations are addition (+), &#xA;  subtraction (-), multiplication (*), division (/); the minus sign may also be used as the sign of a negative number. &lt;/li&gt;&#xA;  &lt;li&gt;The only allowed mathematical functions are square-roots and logarithms.&#xA;  Logarithms must be written in the form $\log[b](x)$ to denote the base-$b$ logarithm of number $x$.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Note that juxtaposition of digits, non-decimal digits, cube-roots, exponentiation, factorials, absolute values, trigonometric functions, rounding, limits, etc are strictly forbidden in this puzzle.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now in this breathtaking follow-up puzzle, I ask you to find a solution that combines the good properties of the solutions for the two older puzzles:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;New puzzle:&lt;/strong&gt; For every positive integer $n$, find a mathematical expression &lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;that yields the value $n$,&lt;/li&gt;&#xA;  &lt;li&gt;that obeys the five rules 1.-5.as listed above,&lt;/li&gt;&#xA;  &lt;li&gt;and that contains the digits in increasing order (when read from left to right).&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(The solution does not require any tricks. No lateral thinking. &#xA;The expression/string can be written in a single line, with the nine digits occurring in increasing order.)&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.923" LastActivityDate="2018-02-26T12:57:47.930" Title="A breathtaking way of making every possible positive integer" Tags="|mathematics|formation-of-numbers|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28990" PostTypeId="1" AcceptedAnswerId="28991" CreationDate="2016-03-15T17:07:59.903" Score="22" ViewCount="1995" Body="&lt;p&gt;I can be a hopeless optimist, or crookedly straight&lt;br&gt;&#xA;My three beginnings are half of an eight&lt;br&gt;&#xA;The rest of me works hard as a fool&lt;br&gt;&#xA;I may be pretty ugly, but also nerdishly cool&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who am I?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user17947" LastActivityDate="2016-03-15T18:46:37.103" Title="A hopeless optimist - who am I?" Tags="|riddle|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28991" PostTypeId="2" ParentId="28990" CreationDate="2016-03-15T17:16:59.390" Score="31" Body="&lt;p&gt;You are an&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oxymoron&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can be a hopeless optimist, or crookedly straight&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'Hopeless optimist' and 'crookedly straight' are oxymorons - phrases with terms that contradict each other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My three beginnings are half of an eight&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;oxy&lt;/strong&gt; is half of oxygen, the element with atomic number 8. Also, three is not half of eight, making the statement itself a sort of oxymoron&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The rest of me works hard as a fool  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;moron&lt;/strong&gt; is another word for fool; &quot;rest&quot; also has another meaning that is incongruous with &quot;hard work&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I may be pretty ugly, but also nerdishly cool&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 'pretty ugly' and 'nerdishly cool' are two more oxymorons (though, in this case, 'pretty' could be read as being used as an adverb, not an adjective, which would be a true oxymoron)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2016-03-15T18:46:37.103" LastActivityDate="2016-03-15T18:46:37.103" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="28992" PostTypeId="2" ParentId="28059" CreationDate="2016-03-15T17:20:41.160" Score="6" Body="&lt;h1&gt;Going to take a stab at it, but I'm not 100% by any means.&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - Second stanza infers getting direction from a sundial.&lt;br&gt;&#xA; - Starting the day at Andaman, at 30 to 6, a sundial would be pointing ~262.5 degrees or fairly close to WbS. &lt;br&gt;&#xA; - Looking at this as a local search for treasure there would be no way to find it as there is no starting point in Andaman to work with, so we must look at the large scale map. &lt;br&gt;&#xA; - On that bearing we would end up on the northern point of Sri Lanka. &lt;br&gt;&#xA; - Stanza 3 leaves us with a missing rhyme, and what rhymes with nest? guess we're headed west. &lt;br&gt;&#xA; - This takes us to the southern tip of India. &lt;br&gt;&#xA; - Stanza 4 lets us know were changing course again, and in stanza 5 we see that we need to travel on a bearing of 45 to 4. &lt;br&gt;&#xA; - Using our handy sundial again we end up going just off NE at ~48.75 degrees.&lt;br&gt;&#xA; - This leads us to land around Sittwe. &lt;br&gt;&#xA; - Stanza 6 lets us know were going South, with our backs to Polaris, this will obviously lead us right back to Andaman.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;This is where things start to change...&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - Now were going to be using mirrors to guide the way, using a lateral inversion of the day before's trip, we find ourselves heading just off of a bearing of EbS.&lt;br&gt;&#xA; - Using this, we come to shore on Letsok-aw, Burma. &lt;br&gt;&#xA; - Following our trip opposite of the day before, now we must head east. &lt;br&gt;&#xA; - This takes us to the northern end of Kanmaw. &lt;br&gt;&#xA; - Now for our trip, slightly off of bearing NW, we und up at Sabi. &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;Finale?&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - For our last part of the trip we must head south, and the island due south of Sabi is Money Island, Burma.  Money == Treasure&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I believe the original question asked something about how many people died or something.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Its been stated that one die at every turn, so there should only be the 1 left at the end. (but at least he got the treasure and the mistress.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19731" LastActivityDate="2016-03-15T17:20:41.160" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28997" PostTypeId="2" ParentId="28595" CreationDate="2016-03-15T18:11:12.400" Score="3" Body="&lt;p&gt;I will wager a guess that you are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Adam&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was the freest man around. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In his creation story, Adam was the first man on Earth, and was free to do anything except eat from one tree.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet I never left my grounds&lt;br&gt;&#xA;(for years at a stretch)  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Having everything he could want in the Garden of Eden, Adam had no need to go anywhere.  However, he was eventually forced to leave after the trick of the serpent.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I could walk, fight, or run.&lt;br&gt;&#xA;I'm as able-bodied as anyone!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again, he was the first man, and made in God's own image, so he was likely able-bodied, even if there were few comparisons available.  He &lt;em&gt;was&lt;/em&gt; able-bodied enough to have another body, that of Eve, created from his own, but that was &lt;em&gt;probably&lt;/em&gt; more of God's work than Adam's.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And my riches I can't count!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; God gave man dominion over all other creation, and in Eden, there would be no want, instead plentiful...everything.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I rarely ventured out.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He didn't leave the garden.  Though, when he left Eve's side, the serpent tricked her into eating the forbidden fruit and they were both expelled from the garden.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm stable mentally and emotionally  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; He wasn't crazy, he just became more aware after eating the fruit.  Like seeing he was naked and covering himself, for instance- not crazy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And many dream of being me.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I am sure many men would dream of being the only man on Earth, in the paradise of the Garden of Eden, made in God's own image, and with an eternal mate.  Could be worse.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, Related:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/a/28998/11452&quot;&gt;https://puzzling.stackexchange.com/a/28998/11452&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11452" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2016-03-15T18:11:12.400" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28998" PostTypeId="2" ParentId="28818" CreationDate="2016-03-15T18:12:33.670" Score="4" Body="&lt;p&gt;In line with my answer on the related question, I think you are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eve&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the luckiest lady around!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eve was blessed by God to enjoy the pleasures of life and Earth, but you could call His blessing luck.  Plus, being made from a rib is pretty lucky.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the most constricted of all.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;She was tempted by the serpent.  Some serpents are constrictors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am the richest lady on earth!&lt;br&gt;&#xA;Yet I own nothing at all.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eve was the first woman in her creation story, so she could claim anything she liked.  However, the world (and Eve) were really God's creation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The world bends to my will!&lt;br&gt;&#xA;Yet I possess no power.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; God gave man (and &lt;em&gt;wo&lt;/em&gt;man) dominion over all creatures that crawled the Earth, though neither Adam nor Eve possessed any especial powers themselves.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Every girl wishes to be in my place!&lt;br&gt;&#xA;But I chose not to come and only leave at death.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Eve was the only woman on Earth, and had all to herself an &lt;s&gt; Abel&lt;/s&gt; &lt;em&gt;ahem&lt;/em&gt;, able partner, as well as a garden of paradise to seemingly enjoy forever.  Her expulsion from the garden was the first fall from grace of mankind.  Outside the garden, Adam and Eve would be exposed to pains, trials, and death.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who am I? &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Eve&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Related:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://puzzling.stackexchange.com/a/28997/11452&quot;&gt;https://puzzling.stackexchange.com/a/28997/11452&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11452" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2016-03-15T18:12:33.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29002" PostTypeId="2" ParentId="28989" CreationDate="2016-03-15T18:37:30.670" Score="24" Body="&lt;p&gt;Simple&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $ \kern4em\llap{ 1 } = \log\, [ 1/2 \!~ ]  \Bigl( \log\, [ \!\: 3{-}4{+}5{+}6{+}7{-}8 \!~ ]  \bigl(       \sqrt 9  ~ \bigr)  \Bigr) \raise{-2ex}\strut $&#xA; &lt;br&gt;&#xA; $ \kern4em\llap{ 2 } = \log\, [ 1/2 \!~ ] \biggl( \log\, [ \!\: 3{-}4{+}5{+}6{+}7{-}8 \!~ ]  \Bigl( \sqrt{\surd 9} ~ \Bigr) \biggr) $&#xA; &lt;br&gt;&#xA; $\kern4.5em \vdots$&#xA; &lt;br&gt;&#xA; $ \kern4em\llap{ n } = \log\, [ 1/2 \!~ ] \Biggl( \log\, [ \!\: 3{-}4{+}5{+}6{+}7{-}8 \!~ ] \biggl( ~ \underbrace { \!\!\sqrt{\sqrt{\cdots\surd 9 }\,} \!\!\!\!\!\! }_n ~\quad \biggr) \Biggr) $&#xA; &lt;br&gt;&#xA; $ \kern4em     {   } = \log_{\frac12} \log_{\!\; 9} \!\: 9 \!\: \raise{1.3ex}( \!\!\: \strut^\frac12 \! \raise{1.3ex})^n \raise{-3ex}\strut $&#xA; &lt;br&gt;&#xA; $ \kern4em     {   } = \log_{\frac12} \bigl( \frac12 \bigr)^n $&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Complex&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $ \kern4em\llap{ n } = \log\, [ 1/2 \!~ ] \Bigl( \log\, [ \!\: 3{+}4{+}5{-}6{-}7 \!\: ] \biggl( \sqrt{\sqrt{\cdots\sqrt{ 8{-}9 \raise{1.8ex}{\,} } \strut \,}\,} ~ \biggr) \Bigr) $&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Side note&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $ \kern4em\llap{ \raise{.2ex}\biggl( n } = \log_{ \, \ln \!\!\: \surd e } \ln \sqrt{\sqrt{ \cdots \surd e \, }\,} \; \raise{.2ex}\biggr) $&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Original&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$ 1 = \log[1/2] \Bigl( \log[3] \Bigl( \!\, \log[4] \, \bigl( (-5)-6+7+8 \bigr) ~\cdot~                   \root\of{\surd 9 \,}        \, \Bigr)  \Bigr) $$&#xA; $$ 2 = \log[1/2] \left( \log[3] \left(      \log[4] \, \bigl( (-5)-6+7+8 \bigr) ~\cdot~          \root\of{\root\of{\surd 9 \,}\,}    ~\, \right) \right) $$&#xA; $$ 3 = \log[1/2] \left( \log[3] \left(      \log[4] \, \bigl( (-5)-6+7+8 \bigr) ~\cdot~ \root\of{\root\of{\root\of{\surd 9 \,}\,}\,} ~\, \right) \right) $$&#xA; $$\vdots \qquad\qquad$$&#xA; $$\begin{array}{rl} n \kern-1em &amp;amp; = \: \log_{\frac12} \log_3 \raise{.5ex}\Bigl( \bigl( \log_4 4 \bigr) \root\of {9 \,} ^{ \bigl( \frac12 \bigr)^n } \; \raise{.5ex}\Bigr) \\[1ex]             &amp;amp; = \: \log_{\frac12} \log_3 \raise{.5ex}\Bigl( 3^{ \bigl( \frac12 \bigr)^n } \; \raise{.5ex}\Bigr) \\[1ex]             &amp;amp; = \: \log_{\frac12} \left( \frac12 \right)^n \end{array}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18129" LastEditorUserId="18129" LastEditDate="2016-03-17T14:23:35.623" LastActivityDate="2016-03-17T14:23:35.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29003" PostTypeId="2" ParentId="28989" CreationDate="2016-03-15T18:50:18.433" Score="12" Body="&lt;p&gt;How about:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  $n = \log [1 - 2 + 3]\left(\left(-4 + 5\right)\text{/}\left({\log\left[\sqrt{-6 + 7 + 8}\right]\left(\sqrt {\sqrt {\ldots\sqrt { \sqrt9}}}\right)}\right)\right)$&lt;br&gt;&#xA; $\;\;\;= \log[2]\left(1/\left(\log[3]\left(\underbrace{\sqrt {\sqrt {\ldots\sqrt3}}}_{\text{$\sqrt{\phantom0}$repeated $n$ times}}\right)\right)\right)$&lt;br&gt;&#xA; $\;\;\;= \log_2\left(1/\left(\log_3\left(3^\frac1{2^n}\right)\right)\right)$&lt;br&gt;&#xA; $\;\;\;= \log_2\left(1/\left(\frac1{2^n}\right)\right)$&lt;br&gt;&#xA; $\;\;\;= \log_2\left(2^n\right)$&lt;br&gt;&#xA; $\;\;\;= n$&lt;br&gt;&#xA; Also, by using &lt;em&gt;no&lt;/em&gt; repetitions of $\sqrt{\phantom0}$ (as per the second line) we get $n = 0$&lt;br&gt;&#xA; Likewise, by substituting the $/$ with $\cdot$ we get, as a bonus, $-n$ for all $n$ repetitions of $\sqrt{\phantom0}$   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10118" LastEditorUserId="10118" LastEditDate="2016-03-16T08:45:57.857" LastActivityDate="2016-03-16T08:45:57.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29004" PostTypeId="1" AcceptedAnswerId="29006" CreationDate="2016-03-15T18:52:50.833" Score="11" ViewCount="3807" Body="&lt;pre&gt;&lt;code&gt;      XXIII  &#xA;     -------  =  II  &#xA;       VII&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;All you need to do is move one matchstick from the roman numerals to make the equation true.&lt;/p&gt;&#xA;" OwnerUserId="1766" LastEditorUserId="8874" LastEditDate="2016-03-16T09:05:54.647" LastActivityDate="2016-03-17T21:38:34.133" Title="Ummm what day was it, a matchstick puzzle" Tags="|matches|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29006" PostTypeId="2" ParentId="29004" CreationDate="2016-03-15T19:02:44.067" Score="19" Body="&lt;p&gt;You could move...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The last &lt;strong&gt;I&lt;/strong&gt; from the numerator onto the top of the result, like so:&lt;br&gt;&#xA;&lt;br&gt;&#xA; XXII&lt;br&gt;&#xA; ---- = $\pi$&lt;br&gt;&#xA; VII&lt;br&gt;&#xA; It's an approximation, but yesterday was Pi Day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="1766" LastEditDate="2016-03-15T19:12:32.103" LastActivityDate="2016-03-15T19:12:32.103" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="29007" PostTypeId="2" ParentId="29004" CreationDate="2016-03-15T19:09:08.477" Score="17" Body="&lt;p&gt;If in &amp;ldquo;make the equation true&amp;rdquo;&#xA;only the words &lt;em&gt;make&lt;/em&gt; and &lt;em&gt;true&lt;/em&gt; are taken literally:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  XXII&#xA; &lt;br&gt;&#xA; &amp;mdash;&amp;mdash; &amp;hairsp; &amp;ne; &amp;hairsp; II&#xA; &lt;br&gt;&#xA; &amp;hairsp; VII&#xA; &lt;br&gt;&lt;br&gt;&#xA; As &lt;a href=&quot;https://puzzling.stackexchange.com/questions/29004/ummm-what-day-was-it-a-matchstick-puzzle/29007?noredirect=1#comment86138_29007&quot;&gt;quintopia points out&lt;/a&gt;,&#xA; this is no longer an equation.&#xA; (A deserved custard $\pi$ in my face.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18129" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:51.923" LastActivityDate="2016-03-17T21:38:34.133" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="29008" PostTypeId="2" ParentId="29004" CreationDate="2016-03-15T19:12:09.070" Score="12" Body="&lt;p&gt;I think I'm allowed to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;X\III&lt;br&gt;&#xA;------ = II&lt;br&gt;&#xA;VII&lt;br&gt;&#xA; It's hard to tell, but I removed part of an 'X' to make a single piece (placing the spare piece in the division sign), showing an informal &quot;14&quot;. 14 divided by 7 equals 2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;shrug&lt;/em&gt;&#xA;&quot;proof&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&quot;XXIII&quot; is 23 and consists of 7 &quot;sticks&quot;, where each &quot;X&quot; is two sticks. The division sign is made of 5 sticks (possibly just a coincidence, but &quot;meh&quot;). The &quot;VII&quot; is 7 and consists of 4 &quot;sticks, where each &quot;V&quot; is represented by two sticks. &lt;br&gt; &lt;br&gt; Remove one stick from the second X in &quot;XXIII&quot; to create &quot;XIIII&quot; (7-1 sticks) and place that stick in the division sign (which now consists of 5+1 sticks).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="20149" LastEditorUserId="20149" LastEditDate="2016-03-15T19:31:25.333" LastActivityDate="2016-03-15T19:31:25.333" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="29009" PostTypeId="1" AcceptedAnswerId="29014" CreationDate="2016-03-15T20:30:18.347" Score="15" ViewCount="394" Body="&lt;p&gt;Sparkle and shimmer, you're losing a grand.&lt;br&gt;&#xA;Refuse and you'll find that steeped leaves I demand.&lt;br&gt;&#xA;Measure the liquid and then you are gone.&lt;br&gt;&#xA;Fatless and weightless the east end is done.&lt;br&gt;&#xA;Burning and brightened you call out: &quot;Left, left!&quot;&lt;br&gt;&#xA;It just isn't right, me alone and bereft.  &lt;/p&gt;&#xA;" OwnerUserId="18924" LastEditorUserId="18924" LastEditDate="2016-03-15T20:37:45.113" LastActivityDate="2016-03-16T00:53:16.490" Title="A Second Reduction" Tags="|riddle|word|wordplay|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29010" PostTypeId="2" ParentId="28979" CreationDate="2016-03-15T20:36:40.413" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let $f(n)$ be Wendy's chance of winning at her $n$th turn. $f(1)=1/100$ with $f(n+1) = f(n)*(1-p)*99/100$ in general. Her total chance of winning is the sum of the infinite series consisting of all $f(n)$ values. If we let it be $A$, we get $A*(1-p)*99/100 + 1/100 = A$. Since $A$ is supposed to equal 1/2, $p=1/99$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastActivityDate="2016-03-15T20:36:40.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29011" PostTypeId="1" AcceptedAnswerId="29131" CreationDate="2016-03-15T21:08:26.723" Score="4" ViewCount="876" Body="&lt;p&gt;Here's a good puzzle to end the day with.  See if you can figure out what the image sequence is telling you.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/37MCy.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/37MCy.jpg&quot; alt=&quot;Image 1&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vOu3q.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vOu3q.jpg&quot; alt=&quot;Image 2&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xqGsF.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xqGsF.jpg&quot; alt=&quot;Image 3&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;h1&gt;Hints:&lt;/h1&gt;&#xA;&lt;h1&gt;1&lt;/h1&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The first picture has to do with the 24th nickname, though it's not official, or anything.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h1&gt;2&lt;/h1&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; The third image refers to a place - but not a specific place.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19944" LastEditorUserId="-1" LastEditDate="2020-06-17T08:22:35.137" LastActivityDate="2016-03-17T18:17:00.423" Title="Rebuses are so cliché" Tags="|visual|rebus|" AnswerCount="7" CommentCount="17" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29012" PostTypeId="2" ParentId="29004" CreationDate="2016-03-15T21:16:04.083" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;XXIII&lt;br&gt;&#xA;  ----- = II&lt;br&gt;&#xA;  VII  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;can be changed to become...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;XXIII&lt;br&gt;&#xA;----- &gt; II&lt;br&gt;&#xA;VII  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;in which...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; I slightly move one of the sticks in the equals sign to become a greater-than sign.  Because I cannot visually represent it, the top line will be angled downward to meet its bottom compatriot.  Like so, but reversed: &amp;ang;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7586" LastActivityDate="2016-03-15T21:16:04.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29013" PostTypeId="2" ParentId="28979" CreationDate="2016-03-15T21:21:24.013" Score="26" Body="&lt;p&gt;Wendy wins on the first flip with probability $1/100$. Otherwise, the game keeps going and Sally has probability $p$ to win on the next flip, which has overall probability $99/100 \times p$. If not, the game returns to the start, which has no effect on fairness. So, to be fair, these two probabilities should be equal, which happens for $p=1/99$.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2016-03-15T21:21:24.013" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29014" PostTypeId="2" ParentId="29009" CreationDate="2016-03-15T21:52:46.270" Score="19" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;GLITTER&lt;/strong&gt; -&gt; Sparkle and shimmer&lt;br&gt;&#xA;&lt;strong&gt;LITTER&lt;/strong&gt; -&gt; Not refUse, but rather REFuse (garbage)&lt;br&gt;&#xA;&lt;strong&gt;LITER&lt;/strong&gt; -&gt; A measure of liquid&lt;br&gt;&#xA;&lt;strong&gt;LITE&lt;/strong&gt; -&gt; Fatless and weightless (diet foods often labeled &quot;lite&quot;)&lt;br&gt;&#xA;&lt;strong&gt;LIT&lt;/strong&gt; -&gt; Burning and brightened, like a fire&lt;br&gt;&#xA;&lt;strong&gt;IT&lt;/strong&gt; -&gt; &quot;It&quot; isn't right&lt;br&gt;&#xA;&lt;strong&gt;I&lt;/strong&gt; -&gt; Me, alone and bereft  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reduction is as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;GLITTER&lt;/strong&gt; -&gt; &lt;em&gt;You're losing a grand&lt;/em&gt; = &lt;strong&gt;G&lt;/strong&gt; as in &quot;I got 50 G&quot;&lt;br&gt;&#xA;&lt;strong&gt;LITTER&lt;/strong&gt; -&gt; &lt;em&gt;steeped leaves I demand&lt;/em&gt; = tea = &lt;strong&gt;T&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;LITER&lt;/strong&gt; -&gt; &lt;em&gt;you are gone&lt;/em&gt; = ...&quot;&lt;strong&gt;R&lt;/strong&gt;&quot; gone&lt;br&gt;&#xA;&lt;strong&gt;LITE&lt;/strong&gt; -&gt; &lt;em&gt;east end is done&lt;/em&gt; = rightmost letter = &lt;strong&gt;E&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;LIT&lt;/strong&gt; -&gt; &lt;em&gt;you call out: &quot;Left, left!&quot;&lt;/em&gt; = leftmost letter = &lt;strong&gt;L&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;IT&lt;/strong&gt; -&gt; &lt;em&gt;just isn't right&lt;/em&gt; = rightmost letter = &lt;strong&gt;T&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;I&lt;/strong&gt; -&gt; &lt;em&gt;*insert your favorite song about loneliness here*&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7586" LastEditorUserId="7586" LastEditDate="2016-03-16T00:53:16.490" LastActivityDate="2016-03-16T00:53:16.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29015" PostTypeId="2" ParentId="29011" CreationDate="2016-03-15T22:26:30.460" Score="4" Body="&lt;p&gt;The path underneath this beautiful blue arch is not accessible for women.&lt;/p&gt;&#xA;" OwnerUserId="11473" LastActivityDate="2016-03-15T22:26:30.460" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29016" PostTypeId="2" ParentId="28979" CreationDate="2016-03-15T23:30:13.277" Score="5" Body="&lt;p&gt;The key simplification is that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; since the game repeats from the beginning if no-one wins after they each have a turn, we only have to make sure that they chance of them each winning on their first toss is equal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore it must hold that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;1/100 = (1 - 1/100)p&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;p = 1/99&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5081" LastEditorUserId="5081" LastEditDate="2016-03-16T22:51:31.393" LastActivityDate="2016-03-16T22:51:31.393" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29017" PostTypeId="1" AcceptedAnswerId="29018" CreationDate="2016-03-16T00:56:31.440" Score="6" ViewCount="505" Body="&lt;p&gt;I lost my password to a critical website. All I have is a really obfuscating password hint. Can you please help me figure it out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My password hint: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Third letter is r, middle is a.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In my email I found the process I used to make the password:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Set the Table&lt;br&gt;&#xA;  That holds all things&lt;br&gt;&#xA;  Find the two most important places&lt;br&gt;&#xA;  To the right of the bottom one&lt;br&gt;&#xA;  Take the first three letters. Keep the capitalization intact.&#xA;  Now add the first two letters of her first name. Capitalization intact.&lt;br&gt;&#xA;  Take away 49. Add the symbol. All caps.&lt;br&gt;&#xA;  Add two and take the square root. Use the other number of it rounded to the nearest whole number.&#xA;  Now place the number between the doubled letters.&lt;br&gt;&#xA;  You're set!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19179" LastEditorUserId="8874" LastEditDate="2016-03-16T16:32:41.397" LastActivityDate="2016-03-16T16:36:08.757" Title="Help - I lost my password" Tags="|logical-deduction|password|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29018" PostTypeId="2" ParentId="29017" CreationDate="2016-03-16T01:28:17.147" Score="10" Body="&lt;p&gt;&lt;strong&gt;ANSWER:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;CurMa14AG&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Set the Table&lt;br&gt;&#xA;That holds all things&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Periodic Table of Elements&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Find the two most important places&lt;br&gt;&#xA;To the right of the bottom one&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wasn't really sure what this means specifically, but combined with the password hint and the logic of the next line, I made an assumption and went with it. It may have something to do with which version is being referenced.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Take the first three letters. Keep the capitalization intact. Now add the first two letters of her first name. Capitalization intact.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The only thing that makes sense here (and just so happens to fit the hint) is &lt;strong&gt;Curium&lt;/strong&gt; (&lt;em&gt;Cm&lt;/em&gt;), named after Pierre and &lt;strong&gt;Marie&lt;/strong&gt;.  So, we get &lt;strong&gt;CurMa&lt;/strong&gt;. &quot;r&quot; is the 3rd letter, and we have an &quot;a&quot;, so there are probably only 4 characters after this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Take away 49. Add the symbol. All caps.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chemical number of &lt;em&gt;Cm&lt;/em&gt; is 96. Take away 49, and you have 47, the chemical number for Silver (&lt;em&gt;Ag&lt;/em&gt;) -&gt; All caps -&gt; &lt;strong&gt;AG&lt;/strong&gt;.   We have &lt;strong&gt;CurMaAG&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Add two and take the square root. Use the other number of it rounded to the nearest whole number. Now place the number between the doubled letters.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Add two to 47 yields 49, and taking the square root yields 7. Nitrogen (&lt;em&gt;N&lt;/em&gt;) has atomic number 7. The &quot;other number for it&quot; might refer to the Standard atomic weight of Nitrogen, which is 14.007. Rounded to the nearest whole number yields 14. Place between the doubled letters, and we get &lt;strong&gt;CurMa14AG&lt;/strong&gt;. Voila!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7586" LastEditorUserId="7586" LastEditDate="2016-03-16T16:36:08.757" LastActivityDate="2016-03-16T16:36:08.757" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="29019" PostTypeId="2" ParentId="29011" CreationDate="2016-03-16T06:55:42.437" Score="5" Body="&lt;p&gt;My theory:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Overarching his view.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Breakdown:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Overarching: an arch over, his: not a woman, view: a picture of a spectacular vista.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="20154" LastActivityDate="2016-03-16T06:55:42.437" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29020" PostTypeId="2" ParentId="29011" CreationDate="2016-03-16T07:39:54.933" Score="1" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; City Women Rock (it's not rocket science)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11497" LastActivityDate="2016-03-16T07:39:54.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29021" PostTypeId="1" AcceptedAnswerId="29389" CreationDate="2016-03-16T07:51:16.803" Score="40" ViewCount="1107" Body="&lt;p&gt;There is a simple way to generate every one of the thirty-seven numbers below. They form a closed set; that is, there are no more numbers with this property. My apologies in advance if this doesn't go as planned. This is my first number pattern and my puzzles tend to be either impossible to solve or trivially easy. Good luck!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;  3   5  17  27  29  33  &#xA; 35  37  43  50  60  62  &#xA; 69  74  82  96 100 115&#xA;117 121 129 135 153 169  &#xA;181 193 210 235 242 258  &#xA;262 270 272 276 288 297  &#xA;312 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; These numbers represent a pattern of which you all should be aware. In one sense I can definitely say that this puzzle can be solved with ease. As stated in the comments, it does not depend on any national or cultural pre knowledge. The order and arrangement of the numbers is not important. No special equipment is needed to derive these numbers. Again, good luck!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastEditorUserId="18924" LastEditDate="2016-03-21T17:41:30.103" LastActivityDate="2016-03-22T14:34:01.407" Title="No, It's Not A Polynomial" Tags="|pattern|" AnswerCount="3" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="29022" PostTypeId="2" ParentId="29011" CreationDate="2016-03-16T10:31:53.723" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Archman&lt;/strong&gt; is a town in south-western Turkmenistan. &lt;strong&gt;Archman&lt;/strong&gt; lies in the Ahal Province on the northern rim of the Kopet Dag mountain range.&lt;br&gt;&#xA; The first picture shows an &lt;strong&gt;arch&lt;/strong&gt;. &lt;br&gt;&#xA; The second picture shows not a woman meaning &lt;strong&gt;man&lt;/strong&gt; &lt;br&gt;&#xA; The third picturen shows a picture of the suroundings of town &lt;strong&gt;Archman&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2016-03-16T10:31:53.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29023" PostTypeId="1" CreationDate="2016-03-16T10:55:12.667" Score="1" ViewCount="655" Body="&lt;p&gt;There lived a famous scientist named Professor Bob. The opposition companies were keen to steal his secret findings. So he created a robot named Jack to protect the secret data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the night, he kept Jack with the secret data in the middle of an empty, closed room with a locked door, no windows and a table with a glass of diesel if the robot wanted to recharge.(Not possible to demolish the house or break the door)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The robot would kill any unidentifiable person. It had an on/off button which was with Professor Bob. It could not be stolen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next morning, when Professor Bob entered the room, he found footprints on the ceiling, the diesel spilt on the robot's head and the secret data stolen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the same hour, Detective Chuck investigated and found out what had happened.How was the data stolen?&#xA; Think outside the box and answer the question.&lt;/p&gt;&#xA;" OwnerUserId="20191" LastEditorUserId="20191" LastEditDate="2016-03-16T11:15:29.500" LastActivityDate="2016-03-18T20:34:58.917" Title="How was the data stolen?" Tags="|lateral-thinking|" AnswerCount="7" CommentCount="3" ClosedDate="2016-03-30T04:31:11.550" ContentLicense="CC BY-SA 3.0" />
  <row Id="29026" PostTypeId="2" ParentId="29023" CreationDate="2016-03-16T12:04:32.677" Score="2" Body="&lt;p&gt;Just to list a few possible answers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - The robot was in it's 'off' state.&lt;br&gt;- The professor has an identical twin&lt;br&gt; - The robot was shown a mirror, killed itself by pouring the diesel over, thief stole items and make footprints on ceiling as distraction&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2156" LastActivityDate="2016-03-16T12:04:32.677" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="29028" PostTypeId="1" AcceptedAnswerId="29032" CreationDate="2016-03-16T12:07:19.500" Score="10" ViewCount="844" Body="&lt;p&gt;(This puzzle continues the puzzle &quot;&lt;a href=&quot;https://puzzling.stackexchange.com/questions/28929/the-frog-concerto&quot;&gt;The frog concerto&lt;/a&gt;&quot;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is another pond that contains $39$ waterlilies that are arranged in a rectangular $3\times13$ grid. One saturday evening $39$ frogs give a croaking concerto for the residents of the pond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the beginning of the concert, the $k$th frog is sitting on the $k$th lily pad (where $1\le k\le39$). There is a short break in the middle of the concert, and after this break none of the frogs returns to its old lily pad. It turns out that after the break each of the $39$ lily pads again carries one frog, and that every frog is now sitting on some new pad that is horizontally or vertically adjacent to its old pad from before the break.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; How many different seating arrangements are there after the break that match the above description. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(The answer to this question will be a square number. A good solution will clearly explain the reason why a square number shows up here.)&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.530" LastActivityDate="2016-03-18T09:08:23.503" Title="The second frog concerto" Tags="|mathematics|combinatorics|" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29029" PostTypeId="2" ParentId="29023" CreationDate="2016-03-16T12:25:06.353" Score="2" Body="&lt;p&gt;Here is my wild wild attempt - &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Insight one - &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Everything is metallic. The room, table, glass. Jack himself is made of metal. Obviously the diesel is not. -_-&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Just like Nolan actually physically rotated the set in one of the famous scenes of Inception (the second dream level, where they are in the hotel), the enemies came and rotated the room in circles using huge jacks and levers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As the room starts to turn, Jack starts walking to stay level with the ground. The footprints on the ceiling are his own.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The table and glass don't fall as they are stuck with the surface on which they were kept due to magnetism.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Diesel starts falling due to gravity. Now, the table is stuck to the ceiling due to magnetism, and Jack is now on the ceiling (which has turned to now become the floor).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That explains - &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Diesel falling on Jack's head. If the villains rotate the room at a specific speed, it can be ensured that by the time the diesel starts falling due to gravity, it falls on Jack's head.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a result - &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The villains can hear Jack walking. But they hear nothing else! They assume the room has nothing but Jack.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So - &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;They deduce that the data is not stored in physical files or sheets of paper. But is stored in Jack's hard disk memory.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Relieved, they hack into his system and get the data.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14035" LastActivityDate="2016-03-16T12:25:06.353" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="29030" PostTypeId="1" AcceptedAnswerId="29038" CreationDate="2016-03-16T12:41:56.450" Score="10" ViewCount="418" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Earth girls are easy&lt;br&gt;&#xA;  Lets dance upside down, freak out!&lt;br&gt;&#xA;  We are family&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I talking about?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Spoiler for second stanza&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Risque baseline hint&lt;br&gt;&#xA; Good times, chic cosmic thing!&lt;br&gt;&#xA; You might get lucky  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11921" LastEditorUserId="8874" LastEditDate="2016-03-16T13:26:03.903" LastActivityDate="2016-03-22T09:07:03.800" Title="A river runs through it. Do you copy?" Tags="|riddle|haiku|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29032" PostTypeId="2" ParentId="29028" CreationDate="2016-03-16T12:44:29.177" Score="28" Body="&lt;p&gt;Someone suggested I put this as an answer, so here goes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are &lt;strong&gt;no solutions&lt;/strong&gt; to the question as stated.&lt;br&gt;&#xA; Hence the answer is &lt;strong&gt;$0$&lt;/strong&gt; (which indeed is a square number).  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Argument:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you set up the $3\times13$ board in a checkerboard pattern, there are $20$ black squares and $19$ white squares.&lt;br&gt;&#xA; Every black square frog must move to a white square, and vice versa. But as there are $20$ black square frogs and only $19$ white squares, you'll end up with at least one white square with two or more frogs on it. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19698" LastEditorUserId="8874" LastEditDate="2016-03-18T09:08:23.503" LastActivityDate="2016-03-18T09:08:23.503" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29034" PostTypeId="1" AcceptedAnswerId="29035" CreationDate="2016-03-16T12:48:35.393" Score="14" ViewCount="5780" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3uQ9j.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3uQ9j.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know the answer, but I'm sure someone here can figure it out.&lt;br&gt;&#xA;Have fun!&lt;/p&gt;&#xA;" OwnerUserId="20213" LastEditorUserId="8874" LastEditDate="2016-03-16T14:11:13.027" LastActivityDate="2016-03-16T14:11:13.027" Title="What is the next set of boxes in this sequence?" Tags="|pattern|visual|" AnswerCount="2" CommentCount="0" ClosedDate="2023-05-25T17:11:03.510" ContentLicense="CC BY-SA 3.0" />
  <row Id="29035" PostTypeId="2" ParentId="29034" CreationDate="2016-03-16T12:54:41.250" Score="20" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (b)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Arguments:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ignore the squares, and just look at the diagonal lines inside the squares.&lt;br&gt;&#xA; The first four boxes spell out S,H,E,L.&lt;br&gt;&#xA; Box (b) contains an &quot;L&quot; and completes the word to SHELL.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2016-03-16T12:54:41.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29036" PostTypeId="2" ParentId="29034" CreationDate="2016-03-16T12:57:01.517" Score="7" Body="&lt;p&gt;I think the answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (b)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If you look at the shapes formed by the diagonal lines in each of the diagrams, they form capital letters - S, H, E, L. (b) also contains L while (d) seems to contain C but it makes more sense that it would be spelling the word SHELL rather than SHELC&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18422" LastActivityDate="2016-03-16T12:57:01.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29038" PostTypeId="2" ParentId="29030" CreationDate="2016-03-16T13:17:19.327" Score="10" Body="&lt;p&gt;The river seems to be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Nile&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Nile Rodgers&lt;/strong&gt; is all over this puzzle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Earth girls are easy&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Nile Rodgers&lt;/strong&gt; did the soundtrack for &quot;Earth girls are easy&quot;&lt;br&gt;&#xA; &lt;a href=&quot;https://en.wikipedia.org/wiki/Earth_Girls_Are_Easy&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Earth_Girls_Are_Easy&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lets dance upside down, freak out!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Nile Rodgers&lt;/strong&gt; produced David Bowie's &quot;Let's Dance&quot;&lt;br&gt;&#xA; &lt;strong&gt;Nile Rodgers&lt;/strong&gt; produced Diana Ross's &quot;Upside Down&quot;&lt;br&gt;&#xA; &lt;strong&gt;Nile Rodgers&lt;/strong&gt; was a member of Chic with &quot;Le Freak&quot;&lt;br&gt;&#xA;  &lt;a href=&quot;https://en.wikipedia.org/wiki/Let%27s_Dance_(David_Bowie_album)&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Let%27s_Dance_(David_Bowie_album)&lt;/a&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;https://en.wikipedia.org/wiki/Upside_Down&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Upside_Down&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;https://en.wikipedia.org/wiki/Le_Freak&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Le_Freak&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are family&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Nile Rodgers&lt;/strong&gt; composed &quot;We Are Family&quot;&lt;br&gt;&#xA; &lt;a href=&quot;https://en.wikipedia.org/wiki/We_Are_Family_%28song%29&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/We_Are_Family_%28song%29&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Risque baseline hint  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Risqué&quot; was the third studio of Chic&lt;br&gt;&#xA; Chic was famous for funky bass lines&lt;br&gt;&#xA; &lt;a href=&quot;https://en.wikipedia.org/wiki/Risqu%C3%A9_%28album%29&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Risqu%C3%A9_%28album%29&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Good times, chic cosmic thing!   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Good times&quot; was composed by Bernard Edwards and &lt;strong&gt;Nile Rodgers&lt;/strong&gt;&lt;br&gt;&#xA; &lt;strong&gt;Nile Rodgers&lt;/strong&gt; produced &quot;Cosmic Thing&quot; for &quot;The B-52's&quot;&lt;br&gt;&#xA; &lt;a href=&quot;https://en.wikipedia.org/wiki/Good_Times_%28Chic_song%29&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Good_Times_%28Chic_song%29&lt;/a&gt;&lt;br&gt;&#xA; &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmic_Thing&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Cosmic_Thing&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You might get lucky &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Nile Rodgers&lt;/strong&gt; wrote &quot;Get Lucky&quot; for Daft Punk and Pharrell Williams&lt;br&gt;&#xA; &lt;a href=&quot;https://en.wikipedia.org/wiki/Get_Lucky_%28Daft_Punk_song%29&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Get_Lucky_%28Daft_Punk_song%29&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-22T09:07:03.800" LastActivityDate="2016-03-22T09:07:03.800" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29039" PostTypeId="1" AcceptedAnswerId="29041" CreationDate="2016-03-16T14:09:15.853" Score="10" ViewCount="2067" Body="&lt;p&gt;Following &lt;a href=&quot;https://puzzling.stackexchange.com/questions/28979/unfair-coins-at-south-park-elementary&quot;&gt;Unfair coins at South Park Elementary&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wendy and Sally have another friend, called Timmy. Although Timmy &lt;strong&gt;has no coin&lt;/strong&gt;, he wants to play this game in a fair condition with them. Wendy does the first coin flip then Sally does. Is it possible to play this game in a fair way for Timmy, even without him having a coin? If so, what should be the values of $p_1$ (Wendy's coin shows head probability) and $p_2$ (Sally's Head probability) that make this game a fair game for all of them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; Timmy cannot use Sally's or Wendy's coin and one of them wins at the end!&lt;/p&gt;&#xA;" OwnerUserId="18514" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:47.373" LastActivityDate="2016-07-25T02:28:54.187" Title="Unfair coins at South Park Elementary v2" Tags="|mathematics|probability|coin-tossing|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29040" PostTypeId="2" ParentId="29039" CreationDate="2016-03-16T14:12:53.743" Score="6" Body="&lt;p&gt;Unless I'm missing something&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Because Tim has no coin, he can't possibly have an outcome of &quot;heads&quot;, therefore his winning chance is always 0.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This means&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wendy and Sally must also have a winning chance of 0 so $p_1 = p_2 = 0$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1748" LastActivityDate="2016-03-16T14:12:53.743" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29041" PostTypeId="2" ParentId="29039" CreationDate="2016-03-16T14:13:27.623" Score="26" Body="&lt;p&gt;As Timmy has a &quot;head&quot; on his shoulders with probability of 1, he always wins at his turn. Then &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Wendy = 1/3 and Sally = 1/2 makes the game fair.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="8874" LastEditDate="2016-03-16T14:16:33.573" LastActivityDate="2016-03-16T14:16:33.573" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29042" PostTypeId="2" ParentId="29039" CreationDate="2016-03-16T14:32:21.657" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are 3 kids so for this to be fair, each of them should have the&#xA; change 1/3.&lt;br&gt;&#xA; So Wendy goes first then p1 = 1/3.&lt;br&gt;&#xA; Sally goes second.&lt;br&gt;&#xA; so Her chance is 2/3 * p2 = 1/3. This results in p2 = 1/2.&lt;br&gt;&#xA; And &quot;Timmah&quot; wins if neither Wendy or Sally's coins turn out to be Head.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19989" LastActivityDate="2016-03-16T14:32:21.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29043" PostTypeId="2" ParentId="29039" CreationDate="2016-03-16T14:35:38.310" Score="6" Body="&lt;p&gt;Yes, we can have a game in which&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $p_1 = \frac{1}{2}$ and  $p_2 =\frac{1}{2}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We have two tosses, one each from Wendy and Sally.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; After two turns, the number of heads(with proper ordering) is counted by Timmy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 0 heads means Wendy wins, 1 heads (on Sally's coin only) means Sally wins and 2 heads means Timmy wins. Here, each case has probability $\frac{1}{4}.$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assuming that the results from two coins tosses are independent. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In case of $\{Head, Tail\} $, i.e. Head on Wendy's coin and then Tail on Sally's coin, then we redo the whole two-step process!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In simple words,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\{Tail, Tail\} =$ Wendy wins.&lt;br&gt;&#xA; $\{Tail, Head\} =$ Sally wins.&lt;br&gt;&#xA; $\{Head, Head\} =$ Timmy wins.&lt;br&gt;&#xA; $\{Head, Tail\} =$ Again two coins are tossed in the same order( Wendy then Sally)      &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus, there's no need of biased coins.&lt;/p&gt;&#xA;" OwnerUserId="1766" LastEditorUserId="1766" LastEditDate="2016-03-16T18:26:12.590" LastActivityDate="2016-03-16T18:26:12.590" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="29044" PostTypeId="2" ParentId="29023" CreationDate="2016-03-16T14:39:49.797" Score="0" Body="&lt;p&gt;As suggested, you might want to narrow down the condition.  For example I can come up with below answers and you might have to update your question to exclude these possibilities:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1) The thief hide in the room's ceiling until the robot's refill time and attacked the robot and steal the data&lt;br&gt;&#xA; 2) The thief is able to create a remote to control Jack, make him open the door and split diesel on his own head.  He then creates footprint with diesel on the ceiling to confuse investigator.&lt;br&gt;&#xA; 3) Bob is being Hypnotize and the rest go as 2).  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2016-03-16T14:39:49.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29045" PostTypeId="2" ParentId="25248" CreationDate="2016-03-16T15:18:52.900" Score="7" Body="&lt;p&gt;My guess is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Discourage&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My last three letters serve to clarify.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; With age usually comes clarification on most things due to gained experiences and knowledge.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My first four letters are where famous people speak.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Famous people speak in movies which are often held on a disc (DVD, Blue-ray, etc.).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Dropping all consonants, I am almost a curse.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Removing all consonants gives: i, o, u, a and e or another way to look at it: I O U A&amp;amp;E. This can be seen as 'I owe you A&amp;amp;E' (the British, Irish, New Zealand, and Hong Kong term for a hospital's emergency department) which is &lt;strong&gt;almost&lt;/strong&gt; an offensive word or phrase used to express anger or annoyance (a definition of curse). Personally, I would interpret this phrase as someone telling me they want to beat me up/send me to A&amp;amp;E.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="20200" LastActivityDate="2016-03-16T15:18:52.900" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29047" PostTypeId="2" ParentId="28813" CreationDate="2016-03-16T15:36:21.290" Score="7" Body="&lt;p&gt;So I think it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A-Tisket, A-Tasket&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some reasoning.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Thanks to the partial answer by Zander, I found that the tune was in fact &lt;em&gt;A-Tisket, A-Tasket&lt;/em&gt;, which was indeed covered by Ella Fitzgerald. It's also a nursery rhyme, thus the hint about a group of children. &lt;em&gt;Tisket&lt;/em&gt; was also her propellent to national fame, which is why she smiled at in while you were there. For too long, I got hung up on the Fats Domino idea, and the only song she covered by him was &lt;em&gt;I'm Walkin'&lt;/em&gt;, which sadly has no relationship with the tune outlined in the dominoes. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="20151" LastActivityDate="2016-03-16T15:36:21.290" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29050" PostTypeId="2" ParentId="28929" CreationDate="2016-03-16T15:50:46.660" Score="4" Body="&lt;p&gt;I think the answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$384199201$$ or $$19601^{2}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lets consider &lt;em&gt;nx2&lt;/em&gt; chessboard.&#xA;Frog is checker. Checkers move from black to white and vice-versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let &lt;em&gt;w(i)&lt;/em&gt; be the number of possible rearrangement of checkers from black to white on &lt;em&gt;ix2&lt;/em&gt; chessboard. Final answer will be &lt;em&gt;w(12)^2&lt;/em&gt; because case is symmetrical and independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now obviously:&#xA;$$w(0) = 1$$&#xA;$$w(1) = 1$$&#xA;$$w(2) = w(1) + 2*w(0)$$&#xA;And:&#xA;$$w(i) = w(i-1) + 2*w(i-2)+2*w(i-3)+...+2*2(0)$$&#xA;$$w(i) = w(i-1) + 2*[w(i-2) + w(i-3) + … + w(0)]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because either:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;checker on first row changes it place and checkers on other &lt;em&gt;(i-1)&lt;/em&gt; chessboard have all the possibilities in the world&lt;/li&gt;&#xA;&lt;li&gt;checker on first &lt;em&gt;m&lt;/em&gt; row jump places in cycle in either direction (2 possibilities) and checkers on &lt;em&gt;2x(i-m)&lt;/em&gt; chessboard have all the possibilities in the world&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now calculating all elements directly, answer to w(12) and w(12)^2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No case is duplicated because for every element summed there is further checker horizontal position change forbidden in further sum elements.&lt;/p&gt;&#xA;" OwnerUserId="20216" LastActivityDate="2016-03-16T15:50:46.660" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="29051" PostTypeId="2" ParentId="28813" CreationDate="2016-03-16T17:02:20.163" Score="2" Body="&lt;p&gt;I wonder if you mean &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Blue Monday. I don't have it handy to listen to.  Ella did a song called Blue Monday that I'm not familiar with. Fats Domino did a completely different song with the same title. I know the song by Fats and it's nothing like the notes in the puzzle but maybe it's the Ella song. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastActivityDate="2016-03-16T17:02:20.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29052" PostTypeId="2" ParentId="28254" CreationDate="2016-03-16T17:32:27.720" Score="3" Body="&lt;p&gt;The smallest value of x, for which Bob can still avoid (with absolute certainty) to lose any money is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; €51&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He starts off knowing Alice's order is one of 52! orders. Let us imagine Bob can keep a list of this vast amount of orders. When he pays €1 he receives some information: &quot;these positions are correct and the rest are not&quot;. He now deletes impossible orders and will have a smaller number of possible orders left - those forming the intersection of those orders that match at positions he has matched so far and those that do not match any of his attempts so far at positions he has not yet matched.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, in any round, Bob:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;places a matched card in an as yet unmatched position;&lt;/li&gt;&#xA;&lt;li&gt;places an unmatched card in a matched position; or&lt;/li&gt;&#xA;&lt;li&gt;places an unmatched card in a position he has already tried for said card&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;...then he will receive some information he already knew (he knew that card did not go there) and hence will, in expectation, reduce his list of possible orders by less than if he avoided doing so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, since he will know Alice's order if he matches all but two positions, he should&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;avoid possible orders that include any pairwise transpositions of two positions of previously unmatched cards&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;...as there is a chance he will get this information for free. Note, however, that this does not affect the worst case performance, but does affect the expected performance (see this &lt;a href=&quot;https://puzzling.stackexchange.com/questions/28362/corollary-to-a-game-with-52-cards&quot;&gt;follow up question&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way, and by far the easiest, to assure he avoids all four of these scenarios is the &lt;a href=&quot;https://puzzling.stackexchange.com/a/28255/16717&quot;&gt;cycling unmatched cards method given by Carl Löndahl&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since Bob is avoiding at least 1. 2. &amp;amp; 3 to maximise his information and the information gained from making a match is at least as much as the information gained from making a non-match, the worst case is necessarily not matching any cards all the way to round N-2 and then matching some number (possibly zero) of the cards in round N-1 which then reduces his possible orders to 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we label Bob's order as&lt;br/&gt;(0,1,2,3,...,N-2,N-1)&lt;br/&gt;then for a cycling of unmatched cards there are two worst case orders for Alice to have chosen, if we cycle rightwards then&lt;br/&gt;(1,2,3,...,N-2,N-1,0) has no matches in round N-1, and&lt;br/&gt;(2,3,...,N-2,N-1,0,1) matches all in round N-1;&lt;br/&gt;if instead we cycle leftwards they would be&lt;br/&gt;(N-1,0,1,2,..,N-2) and&lt;br/&gt;(N-2,N-1,0,1,2,...) respectively.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As an example of the fourth point one can compare Carl's method with choosing the first of the possible orders when they are listed in lexicographical order for all cases of a four card game. If we label Bob's starting choice as (0,1,2,3) and have him cycle leftwards, then his costs for each of the 24 orders Alice could choose are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KN3WV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KN3WV.png&quot; alt=&quot;Comparison of cycle unmatched with next lexicographically available order&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When using the first lexicographically listed order, when Bob does not match any in the first round using (0,1,2,3) then his next round choice will be (1,0,3,2), which, while being a possible order, happens to be two pairwise transpositions of his first round attempt. This helps him in one of the two worst cases he would experience from cycling unmatched cards (if he was cycling rightwards it would help him with the other instead), but hinders him in three cases. The worst case scenario is the same, but the expected cost rises from $\frac{43}{24}$ to $\frac{45}{24}$.&lt;/p&gt;&#xA;" OwnerUserId="16717" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:52.160" LastActivityDate="2016-03-18T03:24:11.807" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="29053" PostTypeId="1" AcceptedAnswerId="29194" CreationDate="2016-03-16T17:43:48.933" Score="9" ViewCount="585" Body="&lt;p&gt;I'm on a roll now. I just lost my whole keychain!&lt;br&gt;&#xA;Please help me figure out this password ASAP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Luckily, I saved the process I used to get the password in my email. This is what I wrote:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Add one to ten and get a pet&lt;br&gt;&#xA;  Switch to the alternative&lt;br&gt;&#xA;  Now make it a candy&lt;br&gt;&#xA;  Put its first year squared between the parts&lt;br&gt;&#xA;  Then flip it with 'S'tandard 'E'ncryption.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On the site I am trying to log-in to, I found my Password Hint.&lt;br&gt;&#xA;Password Hint: 2X&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please use the hint to check your answer before posting. If the process was done correctly, then the hint will confirm that the password is correct.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Don't bother trying to work with the password hint first. It only makes sense once you have the complete password. Follow the process to get the password, then double check it with the hint. I only gave the hint to prevent people from posting many wrong answers.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alright, I'll add a real hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Add one to ten&quot; will result in a number. But it's not 11. ;) Think outside the standard mathematical rules.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19179" LastEditorUserId="19179" LastEditDate="2018-05-04T16:54:48.357" LastActivityDate="2018-05-04T16:54:48.357" Title="Another Lost Password" Tags="|lateral-thinking|password|" AnswerCount="1" CommentCount="17" ContentLicense="CC BY-SA 4.0" />
  <row Id="29055" PostTypeId="2" ParentId="28929" CreationDate="2016-03-16T20:01:14.690" Score="3" Body="&lt;p&gt;Let's assume they were initially arranged like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;_1__2__3__4__5__6__7__8__9_10_11_12&lt;br&gt;&#xA;13_14_15_16_17_18_19_20_21_22_23_24  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x_n$be the number of arrangements for $n$ columns, and $x_{m,m+1}$ the one for two aligned rows with one being one unit longer. If we start from the upper left corner, then go down and rinse and repeat to place our numbers we'll get something like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$13-1......................................x_{11}$ (13 top left, 1 bottom left etc.)&lt;br&gt;&#xA;$2-14......................................x_{10}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$13-14....................................x_{10,11}$&lt;br&gt;&#xA;$13-14-1-2.........................x_{10}$&lt;br&gt;&#xA;$13-14-1-15-2.................x_{9,10}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2-1.........................................$(equivalent to 13-14, 90-degree rotation)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_0 = 1$&lt;br&gt;&#xA;$x_1 = 1$&lt;br&gt;&#xA;$x_2 = 1 + 1 + 2*1=4$&lt;br&gt;&#xA;$x_{0,1} = 1$&lt;br&gt;&#xA;$x_3 = 9$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Induction time:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_n = x_{n-1} + x_{n-2} + 2*x_{n-2,n-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_{m,m+1} = x_{m} + x_{m-1,m}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2*x_{n-2,n-1} = 2*x_{n-2} + 2*x_{n-3,n-2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2*x_{n-2,n-1} = 2*x_{n-2} + 2*x_{n-3} + ... + 2*x_1 + 2*x_{0,1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_n    = x_{n-1} + 3*x_{n-2} + 2*x_{n-3} + ...            + 2*x_1 + 2$&#xA;$x_{n-1} =            x_{n-2} + 3*x_{n-3} + 2*x_{n-4} + ... + 2*x_1 + 2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_n = 2*x_{n-1} + 2*x_{n-2} -   x_{n-3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_0 = 1$&lt;br&gt;&#xA;$x_4 = 25$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, $x_k = F(k)^2$. If every value before $x_n$ follows this rule, $x_{n-2} = a^2$ and $x_{n-3}=b^2$, then $x_{n-1} = (a+b)^2$. $x_n = 2a^2+2b^2+4ab + 2a^2 - b^2 = (2a+b)^2 = [a+(a+b)]^2$, so our formula is always true and $x_{12} = 54289$.&lt;/p&gt;&#xA;" OwnerUserId="14016" LastEditorUserId="14016" LastEditDate="2016-03-16T20:12:26.213" LastActivityDate="2016-03-16T20:12:26.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29056" PostTypeId="2" ParentId="29039" CreationDate="2016-03-16T20:19:53.770" Score="0" Body="&lt;p&gt;If Wendy wins with probability $\dfrac1{p_1}$, then Sally's coin must be $p_2=\dfrac1{p_1-1}$, as &lt;a href=&quot;https://puzzling.stackexchange.com/questions/28979/unfair-coins-at-south-park-elementary&quot;&gt;Xnor points out&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Any of these values will work, as they agree to split their winnings with Timmy in a $2:1$ ratio.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11497" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:48.113" LastActivityDate="2016-03-16T20:19:53.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29057" PostTypeId="1" AcceptedAnswerId="29062" CreationDate="2016-03-16T20:26:06.943" Score="2" ViewCount="296" Body="&lt;p&gt;I'm doing cryptanalysis and I suspect that the message is a substitution cipher with English plaintext. What words could fit the pattern &lt;code&gt;slls&lt;/code&gt;?&lt;/p&gt;&#xA;" OwnerUserId="20223" LastEditorUserId="11876" LastEditDate="2016-03-16T21:22:09.090" LastActivityDate="2016-03-16T22:47:59.220" Title="Possible list of 4 letter words for this part of a message" Tags="|cipher|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29058" PostTypeId="2" ParentId="29057" CreationDate="2016-03-16T20:32:39.203" Score="1" Body="&lt;p&gt;Seems to me that any palindromic, 4 letter word should suffice:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a &lt;a href=&quot;https://en.wiktionary.org/wiki/Appendix:English_palindromes#Four_letters&quot; rel=&quot;nofollow&quot;&gt;list&lt;/a&gt; that includes words like:&#xA;poop, peep, noon, naan, abba&lt;/p&gt;&#xA;" OwnerUserId="20151" LastEditorUserId="20151" LastEditDate="2016-03-16T20:38:33.357" LastActivityDate="2016-03-16T20:38:33.357" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="29059" PostTypeId="2" ParentId="29011" CreationDate="2016-03-16T20:38:50.477" Score="2" Body="&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Bob Marley song: No Woman, No Cry&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Missouri = Misery&lt;br&gt;&#xA; No Woman&lt;br&gt;&#xA; Heaven/Paradise&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9353" LastActivityDate="2016-03-16T20:38:50.477" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="29060" PostTypeId="2" ParentId="29057" CreationDate="2016-03-16T21:25:09.887" Score="1" Body="&lt;p&gt;Here's a list using &lt;a href=&quot;http://www.quinapalus.com/cgi-bin/qat?pat=1221&amp;amp;ent=Search&amp;amp;dict=0&quot; rel=&quot;nofollow&quot;&gt;Qat&lt;/a&gt;, a word pattern matcher tool:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;abba anna beeb boob deed ecce esse goog keek kook maam naan noon oppo otto peep poop sees toot&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="11876" LastActivityDate="2016-03-16T21:25:09.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29061" PostTypeId="2" ParentId="24283" CreationDate="2016-03-16T22:21:59.437" Score="2" Body="&lt;p&gt;1 is Conway's Life, 2 I don't know, but it sounds like MineSweeper, 3 I think is this, but I don't know it's name:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/11iRZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/11iRZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The colours tell you how many of each colour are in each row/column, and are in the order red/blue/green/yellow, so they only tell you how many with no indication of position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure if the solution is unique for this one, but here's one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/MpETq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MpETq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11497" LastEditorUserId="11497" LastEditDate="2016-03-16T22:30:19.573" LastActivityDate="2016-03-16T22:30:19.573" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="29062" PostTypeId="2" ParentId="29057" CreationDate="2016-03-16T22:47:59.220" Score="1" Body="&lt;p&gt;if you use 'union' dictionary from &lt;a href=&quot;http://www.quinapalus.com/cgi-bin/qat?pat=1221&amp;amp;ent=Search&amp;amp;dict=5&quot; rel=&quot;nofollow&quot;&gt;Qat&lt;/a&gt; you get&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;abba acca adda affa akka alla amma anna atta beeb boob cttc deed ecce elle esse ette goog immi issi keek kook lccl maam naan noon oppo otto peep poop sccs sees svvs teet toot tsst uttu&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so I suppose it depends what you are decoding.&lt;/p&gt;&#xA;" OwnerUserId="11497" LastActivityDate="2016-03-16T22:47:59.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29063" PostTypeId="2" ParentId="28929" CreationDate="2016-03-16T23:11:37.627" Score="2" Body="&lt;p&gt;This is Joel Rondeau's solution. I just thought I would try to make it easier to follow:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the frogs first arrive, they are pleased to find that their lillypads have been tastefully decorated in a black and white checkerboard pattern. As they leave for the break, each white frog anticipates returning to one of the 2 or 3 black pads adjacent to its white one, and similarly for the black frogs. But when they return, they are surprised to find that some scallawag has swapped the color scheme on every lillypad. The ones that were white before are now black, and vice versa. Each frog is somewhat disconcerted to see that their new pad is the same color as their old.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the surprises are not over, for after they move in place, they realize there is a now a mirror above them, angled so that from the audience, the top row appears on the bottom, and the bottom row on top. Indeed, in the mirror, the color scheme appears unchanged. But those frogs who had chosen to move down or up appear in the mirror as to have not moved at all. Those whose chose to move right or left appear in the mirror to have moved diagonally right or left instead, staying on their same color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this view, the white frogs rearranged themselves on the white pads, and the black frogs rearranged themselves independently on the black pads. The two colors are symmetrically treated, so the number of arrangements on each is the same. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now assuming that there are $n$ rows, let $F_n$ be the number of allowable arrangements of the $n$ white frogs. The leftmost frog has only two choices (as viewed by mirror): stay still, or move to the pad in column 2. If it stays in place, then the remaining $n-1$ frogs can arrange themselves in the remaining places in $F_{n-1}$ ways. If the frog moves, then the column 2 frog is the only one that can reach the column 1 position, which it must do if each frog is to sit on its own pad. So white frogs 1 and 2 swap places. The remaining $n-2$ frogs can arrange themselves in $F_{n-2}$ ways. Therefore&#xA;$$F_n = F_{n-1} + F_{n-2}$$.&#xA;Now if $n=1$, there is only one column, and so the frog has only one choice: stay put. $F_1 = 1$. If $n=2$, then the two frogs can either stay put or swap places. $F_2 = 2$. This is the pattern for the Fibonacci numbers. The recursion fomula can be used to calculate $F_{12} = 233$. There are 233 ways the white frogs can rearrange, and 233 ways the black frogs can rearrange, giving $233^2 = 54289$ total ways the frogs can rearrange.&lt;/p&gt;&#xA;" OwnerUserId="17915" LastActivityDate="2016-03-16T23:11:37.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29064" PostTypeId="1" AcceptedAnswerId="29752" CreationDate="2016-03-17T05:59:25.503" Score="33" ViewCount="1570" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LXwOF.jpg&quot; alt=&quot;pan up and to the left&quot;&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Agent&lt;/strong&gt;: OK, let's test this out... pan over to that art on the wall... zoom in... Enhance!&lt;br&gt;&#xA;  &lt;strong&gt;Agent&lt;/strong&gt;: &lt;em&gt;[under breath]&lt;/em&gt; I've always wanted to say that...&lt;br&gt;&#xA;  &lt;strong&gt;Agent&lt;/strong&gt;: Excellent, it's crystal clea-- Wait... What's that reflection?&lt;br&gt;&#xA;  &lt;strong&gt;Me&lt;/strong&gt;: It looks like a face.&lt;br&gt;&#xA;  &lt;strong&gt;Agent&lt;/strong&gt;: Ok, pan up and zoom in... Now, enhance!&lt;br&gt;&#xA;  &lt;strong&gt;Agent&lt;/strong&gt;: Wow! Incredible! There's a slight fringing, but the detail is unbelievable! How does it work?&lt;br&gt;&#xA;  &lt;strong&gt;Me&lt;/strong&gt;: Well it's based off a recent discovery I made about the fundamental recursive nature of the quantum membrane--&lt;br&gt;&#xA;  &lt;strong&gt;Agent&lt;/strong&gt;: Shut up. I don't care. Zoom in on that reflection in the eye... ENHANCE!  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;[...8 enhancements later...]&lt;/em&gt;  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Agent&lt;/strong&gt;: Ok, pivot by ninety degrees... There! What's that hanging on the wall? Zoom in...&lt;br&gt;&#xA;  &lt;strong&gt;Me&lt;/strong&gt;: &lt;em&gt;*sigh*&lt;/em&gt; It's still just the same cross-stitch art as before, only from yet another slightly different angle. Nothing interesting.&lt;br&gt;&#xA;  &lt;strong&gt;Agent&lt;/strong&gt;: What would you know about detective work? It's probably some sort of clue... Enhance!&lt;br&gt;&#xA;  &lt;strong&gt;Agent&lt;/strong&gt;: OK, crop that and print it out... Let me have a look.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DafF0.png&quot; alt=&quot;corrupted cross-stitch&quot;&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Agent&lt;/strong&gt;: What the-- That makes no sense. Clearly your software is rubbish. Get out! You're fired!&lt;br&gt;&#xA;  &lt;strong&gt;Me&lt;/strong&gt;: Hmmm... Looks like the enhancement algorithm just has an off by one error in the fractal decompression subroutine--&lt;br&gt;&#xA;  &lt;strong&gt;Agent&lt;/strong&gt;: What? Are you still here? Security!!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What does the cross-stitch say?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hints&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although you need to focus almost entirely on the cross-stitch, there's still a couple of clues to be found in the other parts:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Take another look at the first picture (though, as I mentioned in the comments, you can ignore the text at the bottom right)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;the &lt;em&gt;enhancement algorithm&lt;/em&gt; just has an &lt;em&gt;off by one&lt;/em&gt; error&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="5208" LastEditDate="2016-03-21T09:06:07.737" LastActivityDate="2016-03-29T05:35:26.903" Title="Pan. Zoom. ENHANCE! ...hmmm" Tags="|enigmatic-puzzle|visual|" AnswerCount="1" CommentCount="51" ContentLicense="CC BY-SA 3.0" />
  <row Id="29066" PostTypeId="2" ParentId="29023" CreationDate="2016-03-17T06:54:30.170" Score="1" Body="&lt;p&gt;I have a hunch that &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The thieves magnetized the ceiling. The robot, finding himself upside down, made an unreasonable assumption such as the room had been rotated or he was asleep and either tried to wake himself by dumping the oil on his head or drink it upside down. This incapacitated him in some way, allowing the thieves to operate unhindered. Maybe the robot assumed he was malfunctioning and shut himself down. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastActivityDate="2016-03-17T06:54:30.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29067" PostTypeId="1" AcceptedAnswerId="29082" CreationDate="2016-03-17T07:48:25.863" Score="12" ViewCount="899" Body="&lt;p&gt;Here is another &lt;strong&gt;elementary&lt;/strong&gt; REBUS for you to solve.&#xA;As the Caption states, it is Something Very Necessary.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;THIJKFELMNO &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Be Fast ;) &lt;/p&gt;&#xA;" OwnerUserId="19656" LastActivityDate="2016-03-17T10:15:00.740" Title="Something Very Necessary" Tags="|wordplay|rebus|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29068" PostTypeId="1" AcceptedAnswerId="29080" CreationDate="2016-03-17T08:30:17.333" Score="5" ViewCount="2237" Body="&lt;p&gt;This is a good one, a rebus as I like it. Plus, this is the Question #5000.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  1. Mon&#xA;  2. Tue&#xA;  3. Wed&#xA;  4. Thu&#xA;  5. Fri&#xA;  6. Sat&#xA;  7. Sun&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Happy Solving. :)&lt;/p&gt;&#xA;" OwnerUserId="19656" LastEditorUserId="8874" LastEditDate="2016-03-17T10:21:53.930" LastActivityDate="2016-03-17T18:53:24.440" Title="The 5000th Question" Tags="|wordplay|rebus|" AnswerCount="5" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="29069" PostTypeId="2" ParentId="29067" CreationDate="2016-03-17T08:33:23.823" Score="3" Body="&lt;p&gt;How about &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Tritium&quot; rel=&quot;nofollow&quot;&gt;Tritium&lt;/a&gt; (T) in heavy water - said to be a &quot;radiation hazard when ... ingested via ... water&quot;.&lt;br&gt;&#xA; The letters T is on the outside, with FE interspersed among HIJKLMNO, i.e. H to O (H2O = water). Iron (Fe) with water could be considered &lt;a href=&quot;https://en.wikipedia.org/wiki/Heavy_water&quot; rel=&quot;nofollow&quot;&gt;heavy water&lt;/a&gt;. Heavy water normally contains deuterium, but tritium is hydrogen-3 is arguably even heavier.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's elementary because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the answer relates to element symbols.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's &quot;very necessary&quot; because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; H2O (water) is necessary to ongoing life.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And &quot;be fast&quot; because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; T is toxic. On the outside, T is not considered dangerous because &quot;its beta particles are unable to penetrate the skin&quot;. Ingested via water, though, it's a radiation hazard.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9422" LastEditorUserId="9422" LastEditDate="2016-03-17T09:43:20.253" LastActivityDate="2016-03-17T09:43:20.253" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="29070" PostTypeId="2" ParentId="29068" CreationDate="2016-03-17T08:43:04.323" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; September 5000 or December 5000&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; According to &lt;a href=&quot;http://5000.calendardonkey.com/&quot; rel=&quot;nofollow&quot;&gt;Calendar Donkey&lt;/a&gt;, these are the listings for these months.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11497" LastActivityDate="2016-03-17T08:43:04.323" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29072" PostTypeId="2" ParentId="29068" CreationDate="2016-03-17T08:48:43.293" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Counting down the days.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As, one can notice...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Each day of the week is written with numeric bullets and the word &lt;code&gt;day&lt;/code&gt; is omitted from each one of them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This might be a slight hint towards the upcoming movie, &lt;a href=&quot;http://www.imdb.com/title/tt2975590/&quot; rel=&quot;nofollow&quot;&gt;Batman v Superman&lt;/a&gt;, almost a week to go!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1766" LastActivityDate="2016-03-17T08:48:43.293" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29073" PostTypeId="1" CreationDate="2016-03-17T08:59:31.110" Score="5" ViewCount="504" Body="&lt;p&gt;You awake in a small room to the sound of dozens of clocks ticking and chiming. Once you stir, you hear a familiar evil cackle over a speaker system somewhere in the room. It's the self-proclaimed &quot;evil&quot; Doctor Doozit, an old adversary of yours. You wonder what he has for you today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I see you are finally awake! Welcome to my hidden lair in the mountains. If you want to escape, you have to simply complete a couple of challenges for me. The first one is simple: tell me the time!&quot; The doctor breaks of for a quick cackle before resuming his spiel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I want you to tell me the time &lt;strong&gt;accurate to the second.&lt;/strong&gt; When you are ready, press the button on the wall and tell me what the time is as of pressing the button.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You look around the room. It is 4 meter x 4 meter with a small barred window on one wall. You are a bit of a camping enthusiast so you pull out your keys with a pocket knife and a compass and determine the window to be on the south wall. You look out of it and see the sun high in the sky, so roughly midday. There is a door on the north wall with the previously mentioned button on it. The east and west walls are covered in clocks with various different times set on them, obviously the doctor taunting you. Your eyes drift to a table in the middle of the room. On it is an old computer in a state of disrepair, your phone, an old chunky phone with an oversized aerial, an old alarm clock and a pad of paper with a pen. You try the old phone with no luck. As you reach for your phone, the doctor chimes in again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;If I hear any words out of you other than the time, I will purge the room.&quot; Just as well, your phone has no signal when you switch it on anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a brief bit of work, you speak the correct time and walk through the door, much to the annoyance of the doctor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How were you able to determine the correct time?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="20229" LastEditorUserId="9718" LastEditDate="2016-03-19T02:22:52.347" LastActivityDate="2016-03-19T02:22:52.347" Title="The evil Doctor Doozit with time" Tags="|enigmatic-puzzle|story|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="29074" PostTypeId="2" ParentId="29023" CreationDate="2016-03-17T09:12:26.530" Score="1" Body="&lt;p&gt;Combining Yuganka Sharan's answer with Tim Couwelier's,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Rotate the room as described by Yuganka Sharan to get the diesel on Jack. The footprints indicate the diesel also spilled on the floor. Jack investigates the diesel and sees its own reflection as Tim Couwelier suspected.&lt;br&gt;&#xA; Jack deduces that it is a reflection and that there is no one else in the room, so the reflection must be itself. However, its image processor could not identify the diesel-modified reflection, so Jack incapacitated itself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The room is now unguarded, leaving the data vulnerable to theft.&lt;/p&gt;&#xA;" OwnerUserId="9422" LastActivityDate="2016-03-17T09:12:26.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29075" PostTypeId="2" ParentId="29068" CreationDate="2016-03-17T09:18:22.367" Score="3" Body="&lt;p&gt;My guess is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Endless days, the days are endless or some variation of that since it is a list of the beginnings of the days only. I note that there is no period at the end of the letters as would be usual for abbreviations. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastActivityDate="2016-03-17T09:18:22.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29080" PostTypeId="2" ParentId="29068" CreationDate="2016-03-17T09:59:03.467" Score="17" Body="&lt;p&gt;My guess is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your days are numbered&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18422" LastActivityDate="2016-03-17T09:59:03.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29082" PostTypeId="2" ParentId="29067" CreationDate="2016-03-17T10:08:44.507" Score="19" Body="&lt;p&gt;The answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; TRUST. As @Lawrence has outlined, HIJKLMNO is H2O, also known as water, and FE is iron. These are chemical symbols, which fits the 'Elementary' hint.&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; When water interacts so intimately with iron, the iron corrodes to form RUST. Append that to the T in front and you get TRUST!&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Why is that necessary? Because humans are social creatures, and without it, our most intimate relationships will corrode, just as the water corrodes the iron....&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Apologies about the formatting! Am on my phone.&lt;/p&gt;&#xA;" OwnerUserId="14095" LastEditorUserId="14095" LastEditDate="2016-03-17T10:15:00.740" LastActivityDate="2016-03-17T10:15:00.740" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29086" PostTypeId="2" ParentId="29073" CreationDate="2016-03-17T11:18:14.820" Score="7" Body="&lt;p&gt;I'm guessing&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the hero takes the oversized aerial from the old phone and uses it to boost the reception of his own phone, thus getting a signal and being able to determine the time either by using the internal clock or going to a site like dateandtime.com or by calling a time service (which will speak the time without the hero having to say anything). If the aerial alone doesn't work, it can be extended by using cables from the non-functioning computer.&#xA;&lt;br&gt;&#xA; If the old phone is not a real dinosaur, it may be a satellite phone which would greatly improve the chances of getting a signal in a hidden lair in the mountains. Satellite phones are much chunkier than regular cell phones and have large aerials.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastEditorUserId="18924" LastEditDate="2016-03-17T11:37:50.423" LastActivityDate="2016-03-17T11:37:50.423" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="29087" PostTypeId="1" AcceptedAnswerId="29146" CreationDate="2016-03-17T11:35:00.993" Score="8" ViewCount="249" Body="&lt;p&gt;Can you work out what famous phrase is this?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Fourteen years of jail, Alice.&lt;br&gt;&#xA;  wards&lt;br&gt;&#xA;  You are guilty, Alice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19656" LastActivityDate="2016-03-17T20:26:33.693" Title="From the Land Of Wonders" Tags="|wordplay|rebus|" AnswerCount="2" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29089" PostTypeId="1" AcceptedAnswerId="29101" CreationDate="2016-03-17T11:52:47.263" Score="6" ViewCount="1668" Body="&lt;p&gt;&quot;Hello&quot; you say&#xA;&lt;br/&gt;and just and friendly I'd speak back.&#xA;&lt;br/&gt;Yet it's a lie, I only quote&#xA;&lt;br/&gt;that welcoming into my throat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surely, you think, they're just for show,&#xA;&lt;br/&gt;those fangs you see both high and low&#xA;&lt;br/&gt;and though my heart's as black as night&#xA;&lt;br/&gt;you needn't worry - I won't bite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather I'm much like a fairy tale troll&#xA;&lt;br/&gt;swallowing anything that comes to me whole.&#xA;&lt;br/&gt;My skin is thicker than any bruit,&#xA;&lt;br/&gt;yet I haven't life and alone I'm mute. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;" OwnerUserId="4645" LastActivityDate="2016-03-18T00:32:04.280" Title="Riddle De Do, Riddle De Dum" Tags="|riddle|word|rhyme|" AnswerCount="6" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="29090" PostTypeId="1" AcceptedAnswerId="29106" CreationDate="2016-03-17T12:03:08.830" Score="9" ViewCount="3183" Body="&lt;p&gt;Little Johnny Red is standing on the edge of a lake, and the place &#xA;where he stands is $1$ yard above the water level.&#xA;Johnny's toy boat is floating in the lake, and the boy is pulling &#xA;it ashore with a string (that is more than $2$ yards long). &#xA;When Johnny pulls in $1$ yard of string, will the boat advance &#xA;towards the shore by (i) more than one yard, (ii) exactly one yard, &#xA;or (iii) less than one yard?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&#xA;(Note: This puzzle is tagged as [math]. The solution does neither require lateral thinking nor physics.)&#xA;&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="8874" LastEditorUserId="8874" LastEditDate="2016-03-17T17:33:31.263" LastActivityDate="2016-03-18T17:42:33.630" Title="Toy boat on a string" Tags="|mathematics|" AnswerCount="12" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="29091" PostTypeId="2" ParentId="29090" CreationDate="2016-03-17T12:07:52.417" Score="1" Body="&lt;p&gt;I would say&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; More than one yard - even if the yard of string only accounted for less than one yard of distance traveled, surely momentum would carry the boat further forward. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(This answer was posted before multiple changes were made to the question)&lt;/p&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2016-03-17T19:02:33.377" LastActivityDate="2016-03-17T19:02:33.377" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="29092" PostTypeId="2" ParentId="29090" CreationDate="2016-03-17T12:18:17.433" Score="4" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (iii) Less than one yard.&#xA; The water surface, the string and the quay form a right triangle, with the string being the hypothenuse. Imagine an one-yard-long segment of the hypothenuse casting a &quot;shadow&quot; onto the cathetus, the shadow would be shorter than the original pieve (by the ratio of hypothenuse/cathetus).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is valid as long as&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the string is still longer than sqrt(2) yards; from there on, it is more than one yard (see @Hugh Meyers' answer).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16487" LastEditorUserId="16487" LastEditDate="2016-03-17T12:33:04.957" LastActivityDate="2016-03-17T12:33:04.957" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29093" PostTypeId="2" ParentId="2827" CreationDate="2016-03-17T12:18:22.813" Score="0" Body="&lt;p&gt;How about &#xA;cutting the cake in an even X making 4 slices. then cutting a concentric circle making them 8 pieces.&lt;/p&gt;&#xA;" OwnerUserId="20234" LastEditorUserId="20234" LastEditDate="2020-06-03T10:33:52.793" LastActivityDate="2020-06-03T10:33:52.793" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29094" PostTypeId="2" ParentId="29089" CreationDate="2016-03-17T12:19:13.483" Score="3" Body="&lt;p&gt;This looks to me like :&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A &lt;strong&gt;Telephone&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Hello&quot; you say &#xA;and just and friendly I'd speak back. &#xA;Yet it's a lie, I only quote &#xA;that welcoming into my throat.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We say &quot;Hello&quot; and the other end replies &quot;hello&quot; as well. It's a lie because the phone doesn't speak it only quotes the person who talks to its throat(Receiver.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Surely, you think, they're just for show, &#xA;those fangs you see both high and low &#xA;and though my heart's as black as night &#xA;you needn't worry - I won't bite.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The fangs are the buttons on the receiver end as well as the keypad. The heart is black might be a reference to its battery or the chip sets, and yes the phone doesn't bite.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rather I'm much like a fairy tale troll &#xA;swallowing anything that comes to me whole. &#xA;My skin is thicker than any bruit, &#xA;yet I haven't life and alone I'm mute.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Telephones can seem ugly trolls or they can be used for trolling. Bruit means spreading a rumor widely. Thicker skin can be a reference to &quot;phones can spread rumors very rapidly.&quot;&lt;br&gt;&#xA; Alone, its always mute. it only comes to life when someone calls.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; a Telephone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19191" LastEditorUserId="17250" LastEditDate="2016-03-17T12:35:46.267" LastActivityDate="2016-03-17T12:35:46.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29095" PostTypeId="2" ParentId="29090" CreationDate="2016-03-17T12:19:50.563" Score="9" Body="&lt;p&gt;Ignoring catenary curvature, and momentum of the boat, I would say&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; More than a yard.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Imagine Johnny's hand is directly above a point A on the shore. Boat to A and A to Johnny's hand and Johnny's hand to the boat forms a right-angled triangle with the string as hypotenuse. Imagine that Johnny is 0 yards tall and the string is 2 yards in length. The boat is $sqrt(2^2 - 1^2) = sqrt(3)$ from the shore. When Johnny reels in a yard of string, the boat is right at the bank. It has traveled $sqrt(3)$ yards, which is more than one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More generally,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If the distance from Johnny's hand to A is x and the length of the string is y then the square of the distance of the boat from the shore is $sqrt(y^2 - x^2)$. Because the bank is raised, we know both that x is constant and &gt; 0. Therefore the distance from the shore must decrease at a rate strictly greater than the decrease in the length of the string.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Just to note,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; We don't actually know the height of Johnny's hand above water level. The point is only that since the shore is a yard above water level and Johnny is standing, then the distance above water level is somewhere between 1 yard and however tall it is reasonable for Johnny to be. The point being that under any reasonable definition of the word &quot;standing&quot; and reasonable geometry of Johnny (e.g. his arms don't hang below his feet) we know that the length of the string is strictly greater than the distance of the boat to shore. So, one more try at clarity.&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA; Call the length of the string y and the distance from shore x. The height of Johnny's hand above water is a constant $n &amp;gt;= 1$. We have:&lt;br&gt;&#xA; $y^2 = x^2 + n^2$&lt;br&gt;&#xA; When y decreases by 1, the left hand side of the equation decreases by 2y - 1&lt;br&gt;&#xA; the right hand side must decrease by more than 2x -1 to maintain equality since y is strictly greater than x. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastEditorUserId="18924" LastEditDate="2016-03-17T13:53:39.003" LastActivityDate="2016-03-17T13:53:39.003" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="29096" PostTypeId="2" ParentId="29090" CreationDate="2016-03-17T12:29:48.113" Score="5" Body="&lt;p&gt;Answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; More than 1yard.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let's take a simple case. The rope is 6y long.   this means the boat&#xA; is $\sqrt{6^2 - 1^2} = \sqrt{35}$ that is aproximately $5.916$ yards&#xA; from the shore.&lt;br&gt;&#xA; Pull 1y of rope, the rest of the rope is 5y.   The&#xA; boat is $\sqrt{5^2 - 1^2} = \sqrt{24}$ that is aproximately $4.899$&#xA; yards from the shore.&lt;br&gt;&#xA; The difference is more than 1y.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19989" LastEditorUserId="19989" LastEditDate="2016-03-17T20:38:02.577" LastActivityDate="2016-03-17T20:38:02.577" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29098" PostTypeId="2" ParentId="29090" CreationDate="2016-03-17T12:42:09.283" Score="5" Body="&lt;p&gt;The boat is pulled&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; more than one yard.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Suppose the boat begins $10$ yards from shore. The string forms a hypotenuse of a right triangle with legs $10$ yards and $1$ yard, resulting in a length of $\sqrt{10^2+1^2}=\sqrt{101}$. When the string is pulled one yard, the hypotenuse is reduced by $1$ to $\sqrt{101}-1$. We now have a new right triangle with legs $x$ and $1$, yielding $\sqrt{101}-1=\sqrt{x^2+1}$. Rearranging for yields $x^2=101-2\sqrt{101}$, so $x\approx{}8.994$. Therefore,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the boat has been pulled ever so slightly more than one yard.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16275" LastActivityDate="2016-03-17T12:42:09.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29099" PostTypeId="2" ParentId="29090" CreationDate="2016-03-17T12:48:55.050" Score="13" Body="&lt;p&gt;It has to be (mathematically)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (i)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/P6PTl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/P6PTl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is our boat and geometry of it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all we know that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x-y=1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since the Little Johhny pulled 1 yard the boat and he is 1 meter above from the sea level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b^2+1=y^2$ and $ 1+a^2+2ab+b^2=x^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and these are the simple the Pythagorean theorem of an triangle, so by using both equations;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^2-y^2=(x-y)(x+y)=1+a^2+2ab+b^2-(b^2+1)=a^2+2ab$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x+y=a(a+2b)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is asking about $a$ whether if greater, less or equal to $1$; &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$y$ has to be greater than $1$ since $b$ should be greater than $0$.&lt;/li&gt;&#xA;&lt;li&gt;$x$ has to be greater than $2$ since $x-y=1$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Therefore, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a(a+2b)&amp;gt;3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the largest value of $a$ is possible only when $b\rightarrow0$, which makes $x$ and $y$ value as $2$ and $1$ sequently, that makes $a$ value as;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a^2=3$ or $a=\sqrt3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the smallest value of $a$ is possible only while $b\rightarrow  \infty$ that makes $a$ value;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a\geqslant1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the range of a becomes;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt3 &amp;gt; a\geqslant1$&lt;/p&gt;&#xA;" OwnerUserId="18514" LastEditorUserId="18514" LastEditDate="2016-03-18T17:42:33.630" LastActivityDate="2016-03-18T17:42:33.630" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="29100" PostTypeId="2" ParentId="5747" CreationDate="2016-03-17T12:53:27.657" Score="0" Body="&lt;p&gt;Nail won't be there as trees do grow they they also expand ,so then the trunk will expand the nail won't be there as they expand inside out.&lt;/p&gt;&#xA;" OwnerUserId="20236" LastActivityDate="2016-03-17T12:53:27.657" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29101" PostTypeId="2" ParentId="29089" CreationDate="2016-03-17T12:53:50.387" Score="15" Body="&lt;p&gt;&quot;Hello&quot; you say&#xA;and just and friendly I'd speak back.&#xA;Yet it's a lie, I only quote&#xA;that welcoming into my throat.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; An Echo&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Surely, you think, they're just for show,&#xA;those fangs you see both high and low&#xA;and though my heart's as black as night&#xA;you needn't worry - I won't bite.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Stalagmites/tites&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rather I'm much like a fairy tale troll&#xA;swallowing anything that comes to me whole.&#xA;My skin is thicker than any bruit,&#xA;yet I haven't life and alone I'm mute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the other hints into account:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are a cave/ a cave mouth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="20235" LastActivityDate="2016-03-17T12:53:50.387" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29103" PostTypeId="2" ParentId="29090" CreationDate="2016-03-17T13:34:57.453" Score="3" Body="&lt;p&gt;Define $D(r)$ as the distance from the boat to the shore when the rope is of length $r$. From the Pythagorean theorem, $D(r) = \sqrt{r^2-1}$. Now, if the derivative of this function is always greater than $1$, then $D(r)$ will change faster than $r$; that is, decreasing $r$ by $1$ will always decrease $D(r)$ by more than $1$. This would mean that pulling the rope $1$ yard would always bring the boat more than $1$ yard closer to the shore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we see that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{align} D'(r) &amp;amp; =\frac{d}{dr}(r^2-1)^\frac{1}{2} \\ &amp;amp; =\frac{1}{2}(r^2-1)^{-\frac{1}{2}} \frac{d}{dr}(r^2-1) \\ &amp;amp; = \frac{2r}{2\sqrt{r^2-1}} \\ &amp;amp; = \frac{r}{\sqrt{r^2-1}}\\ \end{align}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $r&amp;gt;\sqrt{r^2-1}$, this derivative is always greater than $1$. Thus when Johnny pulls in $1$ yard of string, the boat will always move more than $1$ yard towards shore.&lt;/p&gt;&#xA;" OwnerUserId="12511" LastActivityDate="2016-03-17T13:34:57.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29105" PostTypeId="2" ParentId="29089" CreationDate="2016-03-17T14:19:32.937" Score="1" Body="&lt;p&gt;What comes to mind is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A fun-house ride. The only problem is, it seems too literal. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Hello&quot; you say &#xA;and just and friendly I'd speak back. &#xA;Yet it's a lie, I only quote &#xA;that welcoming into my throat.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Lots of people when they go into these rides yell &quot;hello!&quot; to hear the echo, I suppose. I've never really understood this. The ride usually has some sort of greeting that plays over and over in a loop, so it's a quote rather than a real welcome.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Surely, you think, they're just for show, &#xA;those fangs you see both high and low &#xA;and though my heart's as black as night &#xA;you needn't worry - I won't bite.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The entrance is often a mouth. The ride is usually dark inside and, though it may look scary, is actually safe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rather I'm much like a fairy tale troll &#xA;swallowing anything that comes to me whole. &#xA;My skin is thicker than any bruit, &#xA;yet I haven't life and alone I'm mute.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The cars or boats of the ride go into the mouth hole. The sides are plywood and canvas - thicker than any &quot;brute&quot;'s hide. It is operated by the guy at the switches and it goes dead whenever the power is turned off.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18924" LastActivityDate="2016-03-17T14:19:32.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29106" PostTypeId="2" ParentId="29090" CreationDate="2016-03-17T14:47:41.857" Score="19" Body="&lt;p&gt;Proof by self-explanatory picture:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/sjhSh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sjhSh.png&quot; alt=&quot;Toy boat&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2722" LastActivityDate="2016-03-17T14:47:41.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29107" PostTypeId="2" ParentId="29090" CreationDate="2016-03-17T14:48:05.253" Score="15" Body="&lt;p&gt;The answer is on the pic below:    &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  &lt;a href=&quot;https://i.stack.imgur.com/Utc16.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Utc16.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14016" LastActivityDate="2016-03-17T14:48:05.253" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29110" PostTypeId="1" AcceptedAnswerId="29125" CreationDate="2016-03-17T14:54:40.443" Score="10" ViewCount="1080" Body="&lt;p&gt;The following rebus regards something commonly done, in certain places; it's debatable as to whether or not this is a fair practice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Zzvxo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zzvxo.png&quot; alt=&quot;Image 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19944" LastActivityDate="2016-03-17T16:35:21.973" Title="Is it fair to do this?" Tags="|visual|rebus|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29115" PostTypeId="2" ParentId="29110" CreationDate="2016-03-17T15:29:07.097" Score="2" Body="&lt;p&gt;My lame guess is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Ordering around&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;systematisation (creating a system / creating order) + Tintinnabulation (ringing sound),  which is then made round. So... Order ring around.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19698" LastActivityDate="2016-03-17T15:29:07.097" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29116" PostTypeId="2" ParentId="29090" CreationDate="2016-03-17T15:38:11.297" Score="10" Body="&lt;p&gt;More than 1 yard. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine if instead of pulling the boat in by 1 yard of rope, the rope broke 1 yard from the end. The 1 yard falls to the water (pivoting at the original boat location), and the remainder's tip falls to the water. As the shortest distance was the original rope, the two tips have to move away from each other.  Hence, more than 1 yard. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6fuWc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6fuWc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2806" LastEditorUserId="2806" LastEditDate="2016-03-17T15:44:16.763" LastActivityDate="2016-03-17T15:44:16.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29117" PostTypeId="1" AcceptedAnswerId="29174" CreationDate="2016-03-17T15:41:37.367" Score="3" ViewCount="175" Body="&lt;p&gt;While I was in school this morning, I found a way to encode text... &lt;em&gt;unusually&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 15 minutes, I drew this sketch:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 ~ * ¬ +*-*++*+* ¬&#xA;5 ~ *+*-----*++++++* ¬&#xA;4 ~ *++++*+*++*+*&#xA;1 ~ *+++++++*+++++*&#xA;#!&#xA;&#xA;1i/2l/3k/4e/5t/6h/7s/8n/9c/o+1/d+2/g+3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What does the sketch say? Also, explain how to decode it.&lt;/p&gt;&#xA;" OwnerDisplayName="user18031" LastEditorDisplayName="user18031" LastEditDate="2016-03-18T14:46:59.947" LastActivityDate="2016-03-18T15:13:38.790" Title="Encoding Extreme #6" Tags="|cipher|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29118" PostTypeId="2" ParentId="29110" CreationDate="2016-03-17T15:45:26.370" Score="4" Body="&lt;p&gt;Is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Over ordering&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;systematization&lt;/strong&gt; = making a system/&lt;strong&gt;order&lt;/strong&gt; and &#xA; &lt;strong&gt;tintinnabulation&lt;/strong&gt; = chime/&lt;strong&gt;ring&lt;/strong&gt;. There is a line separating them like a fraction with order over ring...over ordering.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Something commonly done, in certain places; it's debatable as to whether or not this is a fair practice.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; People often take other out to dinner and offer to pay for their tab. It is polite to not order too much as it will become expensive. However, some people will take advantage of the free meal and overorder. It's debatable if it's fair because the person &lt;em&gt;did&lt;/em&gt; offer to pay in the first place. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10965" LastActivityDate="2016-03-17T15:45:26.370" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29120" PostTypeId="1" AcceptedAnswerId="29129" CreationDate="2016-03-17T16:00:44.230" Score="6" ViewCount="663" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lH1IT.png&quot; alt=&quot;Triangle with sides x,y,z. Three equal length line segments parallel to the sides intersect at point W and are bounded by the triangle.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lengths of the triangle sides satisfy $x &amp;lt; y &amp;lt; z$. Lines that look parallel indeed are parallel. Triangles that look similar indeed are similar. The three red lines are concurrent (meet at a single point) and congruent (have the same length). What is their length?&lt;/p&gt;&#xA;" OwnerDisplayName="user20247" LastEditorUserId="9125" LastEditDate="2016-03-17T16:34:25.367" LastActivityDate="2016-03-21T17:22:15.050" Title="Geometry puzzle" Tags="|mathematics|geometry|" AnswerCount="2" CommentCount="3" ClosedDate="2016-03-21T16:31:15.680" ContentLicense="CC BY-SA 3.0" />
  <row Id="29121" PostTypeId="1" AcceptedAnswerId="29122" CreationDate="2016-03-17T16:00:50.483" Score="5" ViewCount="443" Body="&lt;p&gt;What does the following rebus represent?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Supenlla&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17712" LastEditorUserId="8874" LastEditDate="2016-03-17T16:21:09.553" LastActivityDate="2016-03-28T22:52:17.850" Title="What am I? A rebus" Tags="|rebus|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29122" PostTypeId="2" ParentId="29121" CreationDate="2016-03-17T16:07:10.727" Score="8" Body="&lt;p&gt;I'd imagine it's&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Su&lt;strong&gt;pen&lt;/strong&gt;lla&lt;br&gt;&#xA; &quot;pen&quot; in &quot;Sulla&quot;&lt;br&gt;&#xA; Peninsula (misspelled as peninsulla)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4353" LastEditorUserId="10399" LastEditDate="2016-03-17T16:09:48.033" LastActivityDate="2016-03-17T16:09:48.033" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29123" PostTypeId="2" ParentId="28979" CreationDate="2016-03-17T16:12:26.663" Score="-2" Body="&lt;p&gt;You can't make this a fair game because if you give 1/100 to one person and 1/99 to another, after like 200 chances the 1/99 person would be in advantage. If the chances are the same(1/100), the person starting first have the advantage.&lt;/p&gt;&#xA;" OwnerUserId="20214" LastActivityDate="2016-03-17T16:12:26.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29125" PostTypeId="2" ParentId="29110" CreationDate="2016-03-17T16:35:21.973" Score="14" Body="&lt;p&gt;My guess:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Grading on a bell curve&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It may not be fair because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It doesn't matter how well you do in the course, only how well you do compared to the other students in your class. If you're in a class with a bunch of underachievers, it would be easy to get a top score and thus a passing grade. Meanwhile, your friend might have class amongst the smarties and do just as well as you, but receive a lower grade.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7586" LastActivityDate="2016-03-17T16:35:21.973" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="29126" PostTypeId="2" ParentId="29090" CreationDate="2016-03-17T16:56:23.153" Score="2" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; More than a yard&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Imagine the scenario the other way around. The boat is $n$ yards from the shore, so the rope is $\sqrt{1+n^2}$ yards long. Johnny wants to pull the boat in $1$yd, so the rope will be $\sqrt{1+(n-1)^2}=\sqrt{1+n^2-2n+1}$ yards long. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Maths bit&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\sqrt{1+n^2}-\sqrt{1+n^2-2n+1}=x\to 1+n^2-2\sqrt{(1+n^2)(1+n^2-2n+1)}+1+n^2-2n+1=x^2$, which implies $x\lt1$ by the AM-GM inequality.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11497" LastActivityDate="2016-03-17T16:56:23.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29127" PostTypeId="2" ParentId="29011" CreationDate="2016-03-17T17:29:48.730" Score="2" Body="&lt;p&gt;Is it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The gateway to no-(wo)man's land&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The Arch is sometimes called the gateway to the west. &lt;br&gt;&#xA; The second picture was listed as gender unspecific, but was looking for a gender specific answer. &lt;br&gt;&#xA; The third picture was hinted at being a simple answer &lt;br&gt;&#xA; Possible coincidence, but there's a startrek GATEWAYs book called NO MANS LAND. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Worth a shot. &lt;/p&gt;&#xA;" OwnerUserId="19731" LastActivityDate="2016-03-17T17:29:48.730" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29128" PostTypeId="1" AcceptedAnswerId="29178" CreationDate="2016-03-17T17:37:19.147" Score="7" ViewCount="446" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;You can mix &lt;em&gt;me&lt;/em&gt; up, but that is not &lt;em&gt;my&lt;/em&gt; goal.&lt;br&gt;&#xA;  You can find &lt;em&gt;me&lt;/em&gt; in my recent past.&lt;br&gt;&#xA;  The fans disgust the most famous of them all.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is my first riddle, so no idea how easy or hard it will be (came up with it while cycling home). I also came up with five hints, so if the correct answer isn't posted by tomorrow, I will spoil the hints one by one every ± 12 hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There is a difference between the same regular font and &lt;em&gt;italic&lt;/em&gt; font words, they are not the same.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13190" LastEditorUserId="19470" LastEditDate="2016-03-19T14:39:37.980" LastActivityDate="2016-04-19T12:34:41.200" Title="There are just too many for me!" Tags="|riddle|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="29129" PostTypeId="2" ParentId="29120" CreationDate="2016-03-17T17:40:19.163" Score="4" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Let $a$ be the proportion of the dimensions of the triangle including the line from $y$ to $z$ to those of the full triangle. Define $b$ and $c$ similarly for the lines from $x$ to $z$, and from $x$ to $y$. Then, from similarity, $ax=by=cz$. &lt;br /&gt;&lt;br /&gt; Also, if we look at side $x$, we can see from similarity that the section from the bottom to the $xz$ line has length $bx$, the section from the top to the $xy$ line has length $cx$, and the overlap of those two sections (the section between the two red lines) has length $(1-a)x$. Thus &lt;br /&gt;&lt;br /&gt; $bx + cx - (1-a)x = x$ &lt;br /&gt;&lt;br /&gt; $\implies a+b+c=2$ &lt;br /&gt;&lt;br /&gt;Substituting $b=\frac{ax}{y}$ and $c=\frac{ax}{z}$, we have &lt;br /&gt;&lt;br /&gt;$a+\frac{ax}{y}+\frac{ax}{z}=2$ &lt;br /&gt;&lt;br /&gt;$\implies a(\frac{yz+xz+xy}{yz})=2$ &lt;br /&gt;&lt;br /&gt; $\implies ax=\frac{2xyz}{xy+yz+zx}$&lt;br&gt;&lt;br&gt;Since the three lines are congruent, this is the length of all of them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12511" LastEditorUserId="16322" LastEditDate="2016-03-21T17:22:15.050" LastActivityDate="2016-03-21T17:22:15.050" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29130" PostTypeId="2" ParentId="29120" CreationDate="2016-03-17T17:42:17.620" Score="2" Body="&lt;p&gt;Answer &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\frac{2*x*y*z}{x*y + x*z + y*z}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof:&#xA;Let's convene to the notations on the image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And let's make the length of the red lines $q$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/t5mNz.png&quot; alt=&quot;&quot;&gt;&#xA; &#xA; $AxH = AyA$&lt;br&gt;&#xA; $BxH = BzB$&lt;br&gt;&#xA; &#xA; This means $AB = AxBx + AyBz$ which translates to $AyBz = x-q$&lt;br&gt;&#xA; &#xA; In a similar logic $AxCz = y-q$&lt;br&gt;&#xA; &#xA; Triangle AyBzH is similar to ABC This results in&#xA; &#xA; $\frac{AyBz}{x} = \frac{BzH}{z}$&lt;br&gt;&#xA; &#xA; Triangle AxCzH is similar to ABC   This results in&#xA; &#xA; $\frac{AxCz}{y} = \frac{CzH}{z}$&lt;br&gt;&#xA; &#xA; Adding the 2 above and replacing $AxBz$ with $y-q$ and $AyBz$ with&#xA; $x-q$  we end up with&lt;br&gt;&#xA; &#xA; $\frac{x-q}{x} + \frac{y-q}{y} = \frac{BzH}{z} + \frac{CzH}{z}$&lt;br&gt;&#xA; &#xA; This is the same as&lt;br&gt;&#xA; &#xA; $1- \frac{q}{x} + 1 - \frac{q}{y} = \frac{BzH + CzH}{z}$&lt;br&gt;&#xA; But $BzH +CzH = q$&lt;br&gt;&#xA; So &#xA; $1- \frac{q}{x} + 1 - \frac{q}{y} = \frac{q}{z}$&lt;br&gt;&#xA; &#xA; Resolving the equation for unknown $q$ we get&lt;br&gt;&#xA; &#xA; $q = \frac{2*x*y*z}{x*y + x*z + y*z}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19989" LastEditorUserId="19989" LastEditDate="2016-03-21T16:14:47.707" LastActivityDate="2016-03-21T16:14:47.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29131" PostTypeId="2" ParentId="29011" CreationDate="2016-03-17T18:17:00.423" Score="5" Body="&lt;p&gt;The first picture is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The St. Louis Arch, in Missouri, the &lt;strong&gt;Show&lt;/strong&gt;-Me State&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second picture is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not her, but &lt;strong&gt;him&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The third picture is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The great outdoors, or simply &lt;strong&gt;out&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The rebus could be&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Show him out&lt;/strong&gt; (i.e., point a man toward the door, so that he leaves)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastActivityDate="2016-03-17T18:17:00.423" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29132" PostTypeId="2" ParentId="29128" CreationDate="2016-03-17T18:17:30.097" Score="10" Body="&lt;p&gt;A guess:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A Rubik's cube?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can mix me up but that is not my goal&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can mix a cube up, but the goal is to &quot;unmix&quot; it (order it correctly)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can find me in my recent past&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A previous arrangement of the cube can be found by undoing recent moves&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The fans disgust the most famous of them all&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Not sure. Perhaps it is disgusting how hard it is?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Additionally:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The title might refer to the 43,252,003,274,489,856,000 possible combinations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your profile pic has a Rubik's cube in it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17712" LastActivityDate="2016-03-17T18:17:30.097" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29133" PostTypeId="2" ParentId="29087" CreationDate="2016-03-17T18:18:16.813" Score="0" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alice Is Taught to Grow Upwards, Towards the Light&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="20154" LastActivityDate="2016-03-17T18:18:16.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29134" PostTypeId="1" AcceptedAnswerId="29135" CreationDate="2016-03-17T18:20:46.193" Score="0" ViewCount="184" Body="&lt;p&gt;Two vertical posts are in level ground at 150 feet apart.  A bendable cable,&#xA;but one that does not stretch from its original length, is 150 feet and &#xA;1 inch in length.  The cable is fastened in between the two vertical posts at ground level on each side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The middle point of the cable is lifted up until it is taut, and the &#xA;ground and the middle sections of the cable form an obtuse isosceles &#xA;triangular shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there room enough for you to crawl through in the space between &#xA;the ground and underneath&#xA;the middle of the cable? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;*  The length of the cable has been edited.  *&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="13203" LastEditorUserId="13203" LastEditDate="2016-03-18T02:06:55.310" LastActivityDate="2016-03-18T02:06:55.310" Title="Can you crawl between the ground and under the middle of the cable?" Tags="|geometry|" AnswerCount="2" CommentCount="6" ClosedDate="2016-03-17T19:32:51.140" ContentLicense="CC BY-SA 3.0" />
  <row Id="29135" PostTypeId="2" ParentId="29134" CreationDate="2016-03-17T18:32:07.063" Score="3" Body="&lt;p&gt;I think&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Probably? I mean, assuming you can pretty much just lie on the ground and scoot?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The isosceles triangle is also two right triangles with bases of $900$ inches and hypotenuses of $900.25$ inches, making the altitude about $21$ inches (actually $21.214676523$ish)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;WITH NEW CABLE LENGTH&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; still yeah. I mean it's even longer. of course.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="17250" LastEditDate="2016-03-17T23:43:11.063" LastActivityDate="2016-03-17T23:43:11.063" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29137" PostTypeId="1" AcceptedAnswerId="29139" CreationDate="2016-03-17T18:40:21.993" Score="2" ViewCount="4658" Body="&lt;p&gt;This is my first Puzzle posting on here. If I can figure out how to do a spoiler, I will add in a hint. Best of luck!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I heard this one a while back, and I don't know how well known it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;-75 = 19&lt;br&gt;&#xA;    19 = 8&lt;br&gt;&#xA;    8 = 5&lt;br&gt;&#xA;    5 = 4&lt;br&gt;&#xA;    4 = 4    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Example 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;252 = 18&lt;br&gt;&#xA;    18 = 8&lt;br&gt;&#xA;    8 = 5&lt;br&gt;&#xA;    5 = 4&lt;br&gt;&#xA;    4 = 4    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Figure out the process of how I am go form any number and get it to be 4=4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that you have figured it out how I am doing it by getting 74 all the way down to 4=4&lt;/p&gt;&#xA;" OwnerUserId="20242" LastEditorUserId="8874" LastEditDate="2016-03-17T18:57:48.820" LastActivityDate="2016-03-17T18:58:41.140" Title="Every number eventually equals 4" Tags="|logical-deduction|pattern|number-sequence|" AnswerCount="3" CommentCount="10" ClosedDate="2016-03-17T19:43:50.753" ContentLicense="CC BY-SA 3.0" />
  <row Id="29138" PostTypeId="2" ParentId="29134" CreationDate="2016-03-17T18:40:40.057" Score="0" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes. Because each side of the cable can be considered an isosceles triangle with a right angle. We know the ground is 75 feet or 900 inches on each side, and the taut cable above it would be 900.25 inches. &lt;br&gt;900² = 810,000 &lt;br&gt;900.25² = 810,450.0625&lt;br&gt; The positive square root of 450.0625 is about 21.2146765...&lt;br&gt; That should be enough room for most fit people, though it might be tight.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="20154" LastActivityDate="2016-03-17T18:40:40.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29139" PostTypeId="2" ParentId="29137" CreationDate="2016-03-17T18:44:37.377" Score="8" Body="&lt;p&gt;This puzzle is solved by&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Counting the letters in the English representation of each number. Every number eventually counts down to &quot;four&quot;, which has four letters and becomes the terminus.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I feel like the explanation of the algorithm supersedes the meaningfulness of providing a solution from a specific example, but if I am required to do so for the sake of a green check, here it is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 74 = seventy-four : 11 letters&lt;br&gt;&#xA; 11 = eleven : 6 letters&lt;br&gt;&#xA; 6 = six : 3 letters&lt;br&gt;&#xA; 3 = three : 5 letters&lt;br&gt;&#xA; 5 = five : 4 letters&lt;br&gt;&#xA; 4 = four : 4 letters &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10399" LastEditorUserId="10399" LastEditDate="2016-03-17T18:58:41.140" LastActivityDate="2016-03-17T18:58:41.140" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29140" PostTypeId="2" ParentId="29137" CreationDate="2016-03-17T18:44:38.617" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;74 = 11&lt;br&gt;&#xA;  11 = 6&lt;br&gt;&#xA;  6 = 3&lt;br&gt;&#xA;  3 = 5&lt;br&gt;&#xA;  5 = 4&lt;br&gt;&#xA;  4 = 4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8874" LastActivityDate="2016-03-17T18:44:38.617" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29141" PostTypeId="2" ParentId="29137" CreationDate="2016-03-17T18:45:22.090" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Seventy four = 11&lt;br&gt;eleven=6&lt;br&gt;six=3&lt;br&gt;three=5&lt;br&gt;five=4&lt;br&gt;four=4 :-)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="20154" LastActivityDate="2016-03-17T18:45:22.090" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29142" PostTypeId="2" ParentId="29068" CreationDate="2016-03-17T18:53:24.440" Score="2" Body="&lt;p&gt;How about&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Counting the days 'til the weekend.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="20154" LastActivityDate="2016-03-17T18:53:24.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29143" PostTypeId="1" AcceptedAnswerId="29161" CreationDate="2016-03-17T19:06:20.100" Score="10" ViewCount="1419" Body="&lt;p&gt;Recently a tablet from ancient rome was discovered containing a list of cryptic numbers. Most of the numbers are easy to read, but there are scratches over the last three making them difficult to decipher:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2&#xA;4&#xA;2&#xA;&#xA;2&#xA;1&#xA;1&#xA;5&#xA;1&#xA;&#xA;11&#xA;1&#xA;2&#xA;5&#xA;2&#xA;&#xA;2000&#xA;10&#xA;2000&#xA;2&#xA;220&#xA;&#xA;200&#xA;1&#xA;2&#xA;4&#xA;20&#xA;&#xA;1100&#xA;1&#xA;110&#xA;2&#xA;????&#xA;????&#xA;????&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What are the last three numbers?&lt;/p&gt;&#xA;" OwnerUserId="5896" LastActivityDate="2016-03-17T23:31:08.240" Title="A Roman Emergency" Tags="|riddle|mathematics|cipher|" AnswerCount="5" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29144" PostTypeId="2" ParentId="29143" CreationDate="2016-03-17T19:44:46.527" Score="7" Body="&lt;h2&gt;Not an answer&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I doubt that an ancient Roman tablet would have numbers in Hindu-Arabic, so I've translated the numbers to Roman Numerals, in case it helps someone else solve the puzzle.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;code&gt;II&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;IV&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;II&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;code&gt;II&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;I&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;I&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;V&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;I&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;code&gt;XI&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;I&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;II&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;V&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;II&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;code&gt;MM&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;X&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;MM&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;II&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;CCXX&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;code&gt;CC&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;I&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;II&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;IV&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;XX&lt;/code&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA; &lt;code&gt;MC&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;I&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;CX&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;II&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;????&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;????&lt;/code&gt;&lt;br&gt;&#xA; &lt;code&gt;????&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19944" LastEditorUserId="7586" LastEditDate="2016-03-17T19:53:03.723" LastActivityDate="2016-03-17T19:53:03.723" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="29145" PostTypeId="2" ParentId="29143" CreationDate="2016-03-17T20:13:16.120" Score="1" Body="&lt;h1&gt;Not an Answer: Alternate Formatting&lt;/h1&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Single line breaks -&gt; spaces, double line breaks -&gt; new row:&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;code&gt;II IV II&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;II I I V I&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;XI I II V II&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;MM X MM II CCXX&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;CC I II IV XX&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;MC I CX II ???? ???? ????&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;OR&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Only double line breaks are preserved:&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;code&gt;IIIVII&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;IIIIVI&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;XIIIIVII&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;MMXMMIICCXX&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;CCIIIIVXX&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;MCICXII????????????&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;OR  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Order of placement in each individual list is maintained but lists are otherwise combined (AKA rows are kept but column is reduced to 1):&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;code&gt;IIIIXIMMCCMC&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;IVIIIXII&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;IIIIMMIICX&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;VVIIIVII&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;IIICCXXXX????&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;????&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;????&lt;/code&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7586" LastEditorUserId="7586" LastEditDate="2016-03-17T22:25:34.693" LastActivityDate="2016-03-17T22:25:34.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29146" PostTypeId="2" ParentId="29087" CreationDate="2016-03-17T20:26:33.693" Score="13" Body="&lt;p&gt;It is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &quot;Sentence first -- verdict afterwards!&quot; - &lt;a href=&quot;https://ebooks.adelaide.edu.au/c/carroll/lewis/alice/chapter12.html&quot;&gt;the Queen&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11775" LastActivityDate="2016-03-17T20:26:33.693" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="29147" PostTypeId="2" ParentId="29089" CreationDate="2016-03-17T20:32:42.700" Score="3" Body="&lt;p&gt;You are...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;a cave.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Hello&quot; you say&lt;br&gt;&#xA;and just and friendly I'd speak back. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;You talk into the mouth and it echoes back.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Surely, you think, they're just for show,&lt;br&gt;&#xA;those fangs you see both high and low&lt;br&gt;&#xA;and though my heart's as black as night&lt;br&gt;&#xA;you needn't worry - I won't bite.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;Stalactites and stalagmites are the fangs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rather I'm much like a fairy tale troll&lt;br&gt;&#xA;swallowing anything that comes to me whole.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; As you walk in you are swallowed whole.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My skin is thicker than any bruit,&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The whole earth is the thick skin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yet I haven't life and alone I'm mute.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; No sound unless someone or something makes a sound inside.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="20260" LastEditorUserId="14016" LastEditDate="2016-03-17T21:00:24.797" LastActivityDate="2016-03-17T21:00:24.797" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29148" PostTypeId="1" AcceptedAnswerId="29151" CreationDate="2016-03-17T20:34:19.377" Score="3" ViewCount="642" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am an acid, found in all of your friends.&lt;br&gt;&#xA;  I tell you how to look, but I don't make amends.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;My man named Fred, he told you who I am.&lt;br&gt;&#xA;  But no one understood me until I had an exam!  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I've only got one vowel to my name, for sure.&lt;br&gt;&#xA;  But write me out the long way and you'll find a lot more.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Shout out to my people A, G, C, and T!&lt;br&gt;&#xA;  All of your bases are belong to me!  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who am I?&lt;/p&gt;&#xA;" OwnerUserId="19944" LastEditorUserId="19944" LastEditDate="2016-03-17T20:47:20.593" LastActivityDate="2016-03-17T20:47:20.593" Title="The Acid Chain Rap" Tags="|riddle|word|rhyme|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29149" PostTypeId="2" ParentId="29148" CreationDate="2016-03-17T20:39:47.487" Score="3" Body="&lt;p&gt;You are&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DNA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am an acid, found in all of your friends.&#xA;I tell you how to look, but I don't make amends. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DNA stands for Deoxyribonucleic acid and encodes near everything about us.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My man named Fred, he told you who I am.&#xA;But no one understood me until I had an exam!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First discovered by Frederick Griffith.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've got a couple of vowels to my name, for sure.&#xA;But write me out the long way and you'll find a lot more. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Again DNA stands for Deoxyribonucleic acid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Shout out to my people A, G, C, and T!&#xA;All of your bases are belong to me!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A, G, C and T are 4 base pairs, with which the information is encoded in the DNA.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19749" LastEditorUserId="19749" LastEditDate="2016-03-17T20:43:04.440" LastActivityDate="2016-03-17T20:43:04.440" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29150" PostTypeId="2" ParentId="29148" CreationDate="2016-03-17T20:39:54.877" Score="3" Body="&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You are DNA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am an acid, found in all of your friends.&lt;br&gt;&#xA;I tell you how to look, but I don't make amends. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DNA tells your body how to make you look&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My man named Fred, he told you who I am.&lt;br&gt;&#xA;But no one understood me until I had an exam! &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; First found by Friedrich Miescher&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've got a couple of vowels to my name, for sure.&lt;br&gt;&#xA;But write me out the long way and you'll find a lot more. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Deoxyribonucleic acid&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Shout out to my people A, G, C, and T!&lt;br&gt;&#xA;All of your bases are belong to me! &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The four components of DNA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2016-03-17T20:39:54.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29151" PostTypeId="2" ParentId="29148" CreationDate="2016-03-17T20:40:56.533" Score="10" Body="&lt;p&gt;You are &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; DNA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am an acid, found in all of your friends.&#xA; I tell you how to look, but I don't make amends. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The &quot;A&quot; in DNA stands for acid.  DNA is also found in everyone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My man named Fred, he told you who I am.&#xA; But no one understood me until I had an exam! &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Friedrich Meischer first discovered DNA, but it was not fully understood until Watson and Crick identified it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've got a couple of vowels to my name, for sure.&#xA; But write me out the long way and you'll find a lot more. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Deoxyribonucleic acid&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Shout out to my people A, G, C, and T!&#xA; All of your bases are belong to me! &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The four bases found in DNA are Adenine, Thymine, Guanine and Cytosine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14233" LastEditorUserId="14233" LastEditDate="2016-03-17T20:45:19.647" LastActivityDate="2016-03-17T20:45:19.647" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29152" PostTypeId="2" ParentId="29089" CreationDate="2016-03-17T21:09:58.577" Score="0" Body="&lt;p&gt;Definitely&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; A cave:&#xA; A dead, thick-skinned object with echoes, a black heart of darkness, non biting fangs of stalactites and stalagmites&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11757" LastEditorUserId="11757" LastEditDate="2016-03-18T00:32:04.280" LastActivityDate="2016-03-18T00:32:04.280" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29153" PostTypeId="2" ParentId="29089" CreationDate="2016-03-17T21:16:10.403" Score="0" Body="&lt;p&gt;I think you're:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Siri&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Hello&quot; you say&lt;br&gt;&#xA;and just and friendly I'd speak back.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Speak to Siri and she (or he) will reply  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet it's a lie, I only quote&lt;br&gt;&#xA;that welcoming into my throat.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; All the answers are pre-programmed  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Surely, you think, they're just for show,&lt;br&gt;&#xA;those fangs you see both high and low   &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Those sound wave when you speak that looks like fangs  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and though my heart's as black as night &#xA;you needn't worry - I won't bite.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The phone itself&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rather I'm much like a fairy tale troll&lt;br&gt;&#xA;swallowing anything that comes to me whole.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You can tell / ask Siri about anything&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My skin is thicker than any bruit, &#xA;yet I haven't life and alone I'm mute.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Phone case, and Siri is only a program and do not speak if not activated&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What am I?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Siri&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2016-03-17T21:16:10.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29154" PostTypeId="2" ParentId="29143" CreationDate="2016-03-17T21:32:54.430" Score="5" Body="&lt;p&gt;Based on @Khale_Kitha's partial answer...&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Alphametics?&lt;br&gt;&#xA; Let's turn things on their sides...&lt;br&gt;&#xA; $$II \ge II$$ &lt;br&gt;&#xA; $$II - I &amp;gt; I$$ &lt;br&gt;&#xA; $$XI - II &amp;gt; II$$&lt;br&gt;&#xA; $$MM  x  MM = CCXX$$ &lt;br&gt;&#xA; $$CC - II \ge XX$$&lt;br&gt;&#xA; $$MC - CX = ????$$&lt;br&gt;&#xA; $$????$$&lt;br&gt;&#xA; $$????$$&lt;br&gt;&#xA; So the last line seems like it &lt;em&gt;could&lt;/em&gt; be 1100, 1, and 110 (MC - CX = MC - CX?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solving time!&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The only valid value for $M$ is $8$; $C$ is $7$ and $X$ is $4$&lt;br&gt;&#xA; Because $$77 - 44 \ge II,$$&lt;br&gt;&#xA; I = $1$ or $2$ or $3$&lt;br&gt;&#xA; Because $XI - II &amp;gt; II$, $X &amp;gt; 2*I$, so $I = 1